Partial differential equations, graduate level problems and solutions by igor yanovsky

Partial Diﬀerential Equations: Graduate Level Problems
and Solutions
Igor Yanovsky
1

Partial Diﬀerential Equations Igor Yanovsky, 2005 2
Disclaimer: This handbook is intended to assist graduate students with qualifying
examination preparation. Please be aware, however, that the handbook might contain,
and almost certainly contains, typos as well as incorrect or inaccurate solutions. I can
not be made responsible for any inaccuracies contained in this handbook.

Contents
1 Trigonometric Identities 6
2 Simple Eigenvalue Problem 8
3 Separation of Variables:
Quick Guide 9
4 Eigenvalues of the Laplacian: Quick Guide 9
5 First-Order Equations 10
5.1 Quasilinear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
5.2 Weak Solutions for Quasilinear Equations . . . . . . . . . . . . . . . . . 12
5.2.1 Conservation Laws and Jump Conditions . . . . . . . . . . . . . 12
5.2.2 Fans and Rarefaction Waves . . . . . . . . . . . . . . . . . . . . . 12
5.3 General Nonlinear Equations . . . . . . . . . . . . . . . . . . . . . . . . 13
5.3.1 Two Spatial Dimensions . . . . . . . . . . . . . . . . . . . . . . . 13
5.3.2 Three Spatial Dimensions . . . . . . . . . . . . . . . . . . . . . . 13
6 Second-Order Equations 14
6.1 Classiﬁcation by Characteristics . . . . . . . . . . . . . . . . . . . . . . . 14
6.2 Canonical Forms and General Solutions . . . . . . . . . . . . . . . . . . 14
6.3 Well-Posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
7 Wave Equation 23
7.1 The Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 23
7.2 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7.3 Initial/Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . 24
7.4 Duhamel’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7.5 The Nonhomogeneous Equation . . . . . . . . . . . . . . . . . . . . . . . 24
7.6 Higher Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
7.6.1 Spherical Means . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
7.6.2 Application to the Cauchy Problem . . . . . . . . . . . . . . . . 26
7.6.3 Three-Dimensional Wave Equation . . . . . . . . . . . . . . . . . 27
7.6.4 Two-Dimensional Wave Equation . . . . . . . . . . . . . . . . . . 28
7.6.5 Huygen’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 28
7.7 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
7.8 Contraction Mapping Principle . . . . . . . . . . . . . . . . . . . . . . . 30
8 Laplace Equation 31
8.1 Green’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
8.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
8.3 Polar Laplacian in R2 for Radial Functions . . . . . . . . . . . . . . . . 32
8.4 Spherical Laplacian in R3
and Rn
for Radial Functions . . . . . . . . . . 32
8.5 Cylindrical Laplacian in R3 for Radial Functions . . . . . . . . . . . . . 33
8.6 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8.7 Maximum Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
8.8 The Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 34
8.9 Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
8.10 Green’s Function and the Poisson Kernel . . . . . . . . . . . . . . . . . . 42

8.11 Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . 44
8.12 Eigenvalues of the Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . 44
9 Heat Equation 45
9.1 The Pure Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . 45
9.1.1 Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 45
9.1.2 Multi-Index Notation . . . . . . . . . . . . . . . . . . . . . . . . 45
9.1.3 Solution of the Pure Initial Value Problem . . . . . . . . . . . . . 49
9.1.4 Nonhomogeneous Equation . . . . . . . . . . . . . . . . . . . . . 50
9.1.5 Nonhomogeneous Equation with Nonhomogeneous Initial Condi-
tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
9.1.6 The Fundamental Solution . . . . . . . . . . . . . . . . . . . . . 50
10 Schr¨odinger Equation 52
11 Problems: Quasilinear Equations 54
12 Problems: Shocks 75
13 Problems: General Nonlinear Equations 86
13.1 Two Spatial Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
13.2 Three Spatial Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 93
14 Problems: First-Order Systems 102
15 Problems: Gas Dynamics Systems 127
15.1 Perturbation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
15.2 Stationary Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
15.3 Periodic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
15.4 Energy Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
16 Problems: Wave Equation 139
16.1 The Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 139
16.2 Initial/Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . 141
16.3 Similarity Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
16.4 Traveling Wave Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 156
16.5 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
16.6 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
16.7 Wave Equation in 2D and 3D . . . . . . . . . . . . . . . . . . . . . . . . 187
17 Problems: Laplace Equation 196
17.1 Green’s Function and the Poisson Kernel . . . . . . . . . . . . . . . . . . 196
17.2 The Fundamental Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 205
17.3 Radial Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
17.4 Weak Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
17.5 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
17.6 Self-Adjoint Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
17.7 Spherical Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
17.8 Harmonic Extensions, Subharmonic Functions . . . . . . . . . . . . . . . 249

18 Problems: Heat Equation 255
18.1 Heat Equation with Lower Order Terms . . . . . . . . . . . . . . . . . . 263
18.1.1 Heat Equation Energy Estimates . . . . . . . . . . . . . . . . . . 264
19 Contraction Mapping and Uniqueness - Wave 271
20 Contraction Mapping and Uniqueness - Heat 273
21 Problems: Maximum Principle - Laplace and Heat 279
21.1 Heat Equation - Maximum Principle and Uniqueness . . . . . . . . . . . 279
21.2 Laplace Equation - Maximum Principle . . . . . . . . . . . . . . . . . . 281
22 Problems: Separation of Variables - Laplace Equation 282
23 Problems: Separation of Variables - Poisson Equation 302
24 Problems: Separation of Variables - Wave Equation 305
25 Problems: Separation of Variables - Heat Equation 309
26 Problems: Eigenvalues of the Laplacian - Laplace 323
27 Problems: Eigenvalues of the Laplacian - Poisson 333
28 Problems: Eigenvalues of the Laplacian - Wave 338
29 Problems: Eigenvalues of the Laplacian - Heat 346
29.1 Heat Equation with Periodic Boundary Conditions in 2D
(with extra terms) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
30 Problems: Fourier Transform 365
31 Laplace Transform 385
32 Linear Functional Analysis 393
32.1 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
32.2 Banach and Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 393
32.3 Cauchy-Schwarz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 393
32.4 H¨older Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
32.5 Minkowski Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
32.6 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

1 Trigonometric Identities
cos(a + b) = cos a cos b − sina sinb
cos(a − b) = cos a cos b + sina sinb
sin(a + b) = sin a cos b + cos a sinb
sin(a − b) = sin a cos b − cos a sinb
cos a cos b =
cos(a + b) + cos(a − b)
2
sin a cos b =
sin(a + b) + sin(a − b)
2
sin a sinb =
cos(a − b) − cos(a + b)
2
cos 2t = cos2
t − sin2
t
sin2t = 2 sint cos t
cos2 1
2
t =
1 + cos t
2
sin2 1
2
t =
1 − cos t
2
1 + tan2
t = sec2
t
cot2
t + 1 = csc2
t
cos x =
eix
+ e−ix
2
sinx =
eix
− e−ix
2i
coshx =
ex + e−x
2
sinhx =
ex − e−x
2
d
dx
cosh x = sinh(x)
d
dx
sinh x = cosh(x)
cosh2
x − sinh2
x = 1
du
a2 + u2
=
1
a
tan−1 u
a
+ C
du
√
a2 − u2
= sin−1 u
a
+ C
L
−L
cos
nπx
L
cos
mπx
L
dx =
0 n = m
L n = m
L
−L
sin
nπx
L
sin
mπx
L
dx =
0 n = m
L n = m
L
−L
sin
nπx
L
cos
mπx
L
dx = 0
L
0
cos
nπx
L
cos
mπx
L
dx =
0 n = m
L
2 n = m
L
0
sin
nπx
L
sin
mπx
L
dx =
0 n = m
L
2 n = m
L
0
einx
eimx
dx =
0 n = m
L n = m
L
0
einx
dx =
0 n = 0
L n = 0
sin2
x dx =
x
2
−
sin x cos x
2
cos2
x dx =
x
2
+
sin x cos x
2
tan2
x dx = tanx − x
sinx cos x dx = −
cos2
x
2
ln(xy) = ln(x) + ln(y)
ln
x
y
= ln(x) − ln(y)
ln xr
= r lnx
ln x dx = x ln x − x
x ln x dx =
x2
2
ln x −
x2
4
R
e−z2
dz =
√
π
R
e−z2
2 dz =
√
2π

A =
a b
c d
, A−1
=
1
det(A)
d −b
−c a

2 Simple Eigenvalue Problem
X + λX = 0
Boundary conditions Eigenvalues λn Eigenfunctions Xn
X(0) = X(L) = 0 nπ
L
2
sin nπ
L x n = 1, 2, . . .
X(0) = X (L) = 0
(n−1
2
)π
L
2
sin
(n−1
2
)π
L x n = 1, 2, . . .
X (0) = X(L) = 0
(n−1
2
)π
L
2
cos
(n−1
2
)π
L x n = 1, 2, . . .
X (0) = X (L) = 0 nπ
L
2
cos nπ
L x n = 0, 1, 2, . . .
X(0) = X(L), X (0) = X (L) 2nπ
L
2
sin 2nπ
L x n = 1, 2, . . .
cos 2nπ
L x n = 0, 1, 2, . . .
X(−L) = X(L), X (−L) = X (L) nπ
L
2
sin nπ
L x n = 1, 2, . . .
cos nπ
L x n = 0, 1, 2, . . .
X − λX = 0
Boundary conditions Eigenvalues λn Eigenfunctions Xn
X(0) = X(L) = 0, X (0) = X (L) = 0 nπ
L
4
sin nπ
L x n = 1, 2, . . .
X (0) = X (L) = 0, X (0) = X (L) = 0 nπ
L
4
cos nπ
L x n = 0, 1, 2, . . .

3 Separation of Variables:
Quick Guide
Laplace Equation: u = 0.
X (x)
X(x)
= −
Y (y)
Y (y)
= −λ.
X + λX = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
Y (θ) + λY (θ) = 0.
Wave Equation: utt − uxx = 0.
X (x)
X(x)
=
T (t)
T(t)
= −λ.
X + λX = 0.
utt + 3ut + u = uxx.
T
T
+ 3
T
T
+ 1 =
X
X
= −λ.
X + λX = 0.
utt − uxx + u = 0.
T
T
+ 1 =
X
X
= −λ.
X + λX = 0.
utt + μut = c2uxx + βuxxt, (β > 0)
X
X
= −λ,
1
c2
T
T
+
μ
c2
T
T
= 1 +
β
c2
T
T
X
X
.
4th Order: utt = −k uxxxx.
−
X
X
=
1
k
T
T
= −λ.
X − λX = 0.
Heat Equation: ut = kuxx.
T
T
= k
X
X
= −λ.
X +
λ
k
X = 0.
4th Order: ut = −uxxxx.
T
T
= −
X
X
= −λ.
X − λX = 0.
4 Eigenvalues of the Lapla-
cian: Quick Guide
Laplace Equation: uxx +uyy +λu = 0.
X
X
+
Y
Y
+ λ = 0. (λ = μ2
+ ν2
)
X + μ2
X = 0, Y + ν2
Y = 0.
uxx + uyy + k2
u = 0.
−
X
X
=
Y
Y
+ k2
= c2
.
X + c2
X = 0,
Y + (k2
− c2
)Y = 0.
uxx + uyy + k2u = 0.
−
Y
Y
=
X
X
+ k2
= c2
.
Y + c2
Y = 0,
X + (k2
− c2
)X = 0.

5 First-Order Equations
5.1 Quasilinear Equations
Consider the Cauchy problem for the quasilinear equation in two variables
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u),
with Γ parameterized by (f(s), g(s), h(s)). The characteristic equations are
dx
dt
= a(x, y, z),
dy
dt
= b(x, y, z),
dz
dt
= c(x, y, z),
with initial conditions
x(s, 0) = f(s), y(s, 0) = g(s), z(s, 0) = h(s).
In a quasilinear case, the characteristic equations for dx
dt and dy
dt need not decouple from
the dz
dt equation; this means that we must take the z values into account even to ﬁnd
the projected characteristic curves in the xy-plane. In particular, this allows for the
possibility that the projected characteristics may cross each other.
The condition for solving for s and t in terms of x and y requires that the Jacobian
matrix be nonsingular:
J ≡
xs ys
xt yt
= xsyt − ysxt = 0.
In particular, at t = 0 we obtain the condition
f (s) · b(f(s), g(s), h(s)) − g (s) · a(f(s), g(s), h(s)) = 0.
Burger’s Equation. Solve the Cauchy problem
ut + uux = 0,
u(x, 0) = h(x).
(5.1)
The characteristic equations are
dx
dt
= z,
dy
dt
= 1,
dz
dt
= 0,
and Γ may be parametrized by (s, 0, h(s)).
x = h(s)t + s, y = t, z = h(s).
u(x, y) = h(x − uy) (5.2)
The characteristic projection in the xt-plane1 passing through the point (s, 0) is the
line
x = h(s)t + s
along which u has the constant value u = h(s). Two characteristics x = h(s1)t + s1
and x = h(s2)t + s2 intersect at a point (x, t) with
t = −
s2 − s1
h(s2) − h(s1)
.
1
y and t are interchanged here

From (5.2), we have
ux = h (s)(1 − uxt) ⇒ ux =
h (s)
1 + h (s)t
Hence for h (s) < 0, ux becomes inﬁnite at the positive time
t =
−1
h (s)
.
The smallest t for which this happens corresponds to the value s = s0 at which h (s)
has a minimum (i.e.−h (s) has a maximum). At time T = −1/h (s0) the solution u
experiences a “gradient catastrophe”.

5.2 Weak Solutions for Quasilinear Equations
5.2.1 Conservation Laws and Jump Conditions
Consider shocks for an equation
ut + f(u)x = 0, (5.3)
where f is a smooth function of u. If we integrate (5.3) with respect to x for a ≤ x ≤ b,
we obtain
d
dt
b
a
u(x, t) dx + f(u(b, t)) − f(u(a, t)) = 0. (5.4)
This is an example of a conservation law. Notice that (5.4) implies (5.3) if u is C1, but
(5.4) makes sense for more general u.
Consider a solution of (5.4) that, for ﬁxed t, has a jump discontinuity at x = ξ(t).
We assume that u, ux, and ut are continuous up to ξ. Also, we assume that ξ(t) is C1
in t.
Taking a < ξ(t) < b in (5.4), we obtain
d
dt
ξ
a
u dx +
b
ξ
u dx + f(u(b, t)) − f(u(a, t))
= ξ (t)ul(ξ(t), t) − ξ (t)ur(ξ(t), t) +
ξ
a
ut(x, t) dx +
b
ξ
ut(x, t) dx
+ f(u(b, t)) − f(u(a, t)) = 0,
where ul and ur denote the limiting values of u from the left and right sides of the shock.
Letting a ↑ ξ(t) and b ↓ ξ(t), we get the Rankine-Hugoniot jump condition:
ξ (t)(ul − ur) + f(ur) − f(ul) = 0,
ξ (t) =
f(ur) − f(ul)
ur − ul
.
5.2.2 Fans and Rarefaction Waves
For Burgers’ equation
ut +
1
2
u2
x
= 0,
we have f (u) = u, f ˜u
x
t
=
x
t
⇒ ˜u
x
t
=
x
t
.
For a rarefaction fan emanating from (s, 0) on xt-plane, we have:
u(x, t) =
⎧
⎪⎨
⎪⎩
ul, x−s
t ≤ f (ul) = ul,
x−s
t , ul ≤ x−s
t ≤ ur,
ur, x−s
t ≥ f (ur) = ur.

5.3 General Nonlinear Equations
5.3.1 Two Spatial Dimensions
Write a general nonlinear equation F(x, y, u, ux, uy) = 0 as
F(x, y, z, p, q) = 0.
Γ is parameterized by
Γ : f(s)
x(s,0)
, g(s)
y(s,0)
, h(s)
z(s,0)
, φ(s)
p(s,0)
, ψ(s)
q(s,0)
We need to complete Γ to a strip. Find φ(s) and ψ(s), the initial conditions for p(s, t)
and q(s, t), respectively:
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0
• h (s) = φ(s)f (s) + ψ(s)g (s)
dx
dt
= Fp
dy
dt
= Fq
dz
dt
= pFp + qFq
dp
dt
= −Fx − Fzp
dq
dt
= −Fy − Fzq
We need to have the Jacobian condition. That is, in order to solve the Cauchy problem
in a neighborhood of Γ, the following condition must be satisﬁed:
f (s) · Fq[f, g, h, φ, ψ](s) − g (s) · Fp[f, g, h, φ, ψ](s) = 0.
5.3.2 Three Spatial Dimensions
Write a general nonlinear equation F(x1, x2, x3, u, ux1, ux2, ux3 ) = 0 as
F(x1, x2, x3, z, p1, p2, p3) = 0.
Γ : f1(s1, s2)
x1(s1,s2,0)
, f2(s1, s2)
x2(s1,s2,0)
, f3(s1, s2)
x3(s1,s2,0)
, h(s1, s2)
z(s1,s2,0)
, φ1(s1, s2)
p1(s1,s2,0)
, φ2(s1, s2)
p2(s1,s2,0)
, φ3(s1, s2)
p3(s1,s2,0)
We need to complete Γ to a strip. Find φ1(s1, s2), φ2(s1, s2), and φ3(s1, s2), the initial
conditions for p1(s1, s2, t), p2(s1, s2, t), and p3(s1, s2, t), respectively:
• F f1(s1, s2), f2(s1, s2), f3(s1, s2), h(s1, s2), φ1, φ2, φ3 = 0
•
∂h
∂s1
= φ1
∂f1
∂s1
+ φ2
∂f2
∂s1
+ φ3
∂f3
∂s1
•
∂h
∂s2
= φ1
∂f1
∂s2
+ φ2
∂f2
∂s2
+ φ3
∂f3
∂s2
dx1
dt
= Fp1
dx2
dt
= Fp2
dx3
dt
= Fp3
dz
dt
= p1Fp1 + p2Fp2 + p3Fp3
dp1
dt
= −Fx1 − p1Fz
dp2
dt
= −Fx2 − p2Fz
dp3
dt
= −Fx3 − p3Fz

6 Second-Order Equations
6.1 Classification by Characteristics
Consider the second-order equation in which the derivatives of second-order all occur
linearly, with coefficients only depending on the independent variables:
a(x, y)uxx + b(x, y)uxy + c(x, y)uyy = d(x, y, u, ux, uy). (6.1)
The characteristic equation is
dy
dx
=
b ±
√
b2 − 4ac
2a
.
• b2
− 4ac > 0 ⇒ two characteristics, and (6.1) is called hyperbolic;
• b2 − 4ac = 0 ⇒ one characteristic, and (6.1) is called parabolic;
• b2
− 4ac < 0 ⇒ no characteristics, and (6.1) is called elliptic.
These definitions are all taken at a point x0 ∈ R2; unless a, b, and c are all constant,
the type may change with the point x0.
6.2 Canonical Forms and General Solutions
➀ uxx − uyy = 0 is hyperbolic (one-dimensional wave equation).
➁ uxx − uy = 0 is parabolic (one-dimensional heat equation).
➂ uxx + uyy = 0 is elliptic (two-dimensional Laplace equation).
By the introduction of new coordinates μ and η in place of x and y, the equation
(6.1) may be transformed so that its principal part takes the form ➀, ➁, or ➂.
If (6.1) is hyperbolic, parabolic, or elliptic, there exists a change of variables μ(x, y) and
η(x, y) under which (6.1) becomes, respectively,
uμη = ˜d(μ, η, u, uμ, uη) ⇔ u¯x¯x − u¯y¯y = ¯d(¯x, ¯y, u, u¯x, u¯y),
uμμ = ˜d(μ, η, u, uμ, uη),
uμμ + uηη = ˜d(μ, η, u, uμ, uη).
Example 1. Reduce to canonical form and find the general solution:
uxx + 5uxy + 6uyy = 0. (6.2)
Proof. a = 1, b = 5, c = 6 ⇒ b2 − 4ac = 1 > 0 ⇒ hyperbolic ⇒ two
characteristics.
The characteristics are found by solving
dy
dx
=
5 ± 1
2
=
3
2
to find y = 3x + c1 and y = 2x + c2.

Let μ(x, y) = 3x − y, η(x, y) = 2x − y.
μx = 3, ηx = 2,
μy = −1, ηy = −1.
u = u(μ(x, y), η(x, y));
ux = uμμx + uηηx = 3uμ + 2uη,
uy = uμμy + uηηy = −uμ − uη,
uxx = (3uμ + 2uη)x = 3(uμμμx + uμηηx) + 2(uημμx + uηηηx) = 9uμμ + 12uμη + 4uηη,
uxy = (3uμ + 2uη)y = 3(uμμμy + uμηηy) + 2(uημμy + uηηηy) = −3uμμ − 5uμη − 2uηη,
uyy = −(uμ + uη)y = −(uμμμy + uμηηy + uημμy + uηηηy) = uμμ + 2uμη + uηη.
Inserting these expressions into (6.2) and simplifying, we obtain
uμη = 0, which is the Canonical form,
uμ = f(μ),
u = F(μ) + G(η),
u(x, y) = F(3x − y) + G(2x − y), General solution.
Example 2. Reduce to canonical form and ﬁnd the general solution:
y2
uxx − 2yuxy + uyy = ux + 6y. (6.3)
Proof. a = y2, b = −2y, c = 1 ⇒ b2 −4ac = 0 ⇒ parabolic ⇒ one characteristic.
The characteristics are found by solving
dy
dx
=
−2y
2y2
= −
1
y
to ﬁnd −
y2
2
+ c = x.
Let μ = y2
2 + x. We must choose a second constant function η(x, y) so that η is not
parallel to μ. Choose η(x, y) = y.
μx = 1, ηx = 0,
μy = y, ηy = 1.
u = u(μ(x, y), η(x, y));
ux = uμμx + uηηx = uμ,
uy = uμμy + uηηy = yuμ + uη,
uxx = (uμ)x = uμμμx + uμηηx = uμμ,
uxy = (uμ)y = uμμμy + uμηηy = yuμμ + uμη,
uyy = (yuμ + uη)y = uμ + y(uμμμy + uμηηy) + (uημμy + uηηηy)
= uμ + y2
uμμ + 2yuμη + uηη.

Inserting these expressions into (6.3) and simplifying, we obtain
uηη = 6y,
uηη = 6η, which is the Canonical form,
uη = 3η2
+ f(μ),
u = η3
+ ηf(μ) + g(μ),
u(x, y) = y3
+ y · f
y2
2
+ x + g
y2
2
+ x , General solution.

Problem (F’03, #4). Find the characteristics of the partial diﬀerential equation
xuxx + (x − y)uxy − yuyy = 0, x > 0, y > 0, (6.4)
and then show that it can be transformed into the canonical form
(ξ2
+ 4η)uξη + ξuη = 0
whence ξ and η are suitably chosen canonical coordinates. Use this to obtain the general
solution in the form
u(ξ, η) = f(ξ) +
η
g(η ) dη
(ξ2 + 4η )
1
2
where f and g are arbitrary functions of ξ and η.
Proof. a = x, b = x − y, c = −y ⇒ b2
− 4ac = (x − y)2
+ 4xy > 0 for x > 0,
y > 0 ⇒ hyperbolic ⇒ two characteristics.
➀ The characteristics are found by solving
dy
dx
=
b ±
√
b2 − 4ac
2a
=
x − y ± (x − y)2 + 4xy
2x
=
x − y ± (x + y)
2x
=
2x
2x = 1
−2y
2x = −y
x
⇒ y = x + c1,
dy
y
= −
dx
x
,
ln y = ln x−1
+ ˜c2,
y =
c2
x
.➁ Let μ = x − y and η = xy
μx = 1, ηx = y,
μy = −1, ηy = x.
u = u(μ(x, y), η(x, y));
ux = uμμx + uηηx = uμ + yuη,
uy = uμμy + uηηy = −uμ + xuη,
uxx = (uμ + yuη)x = uμμμx + uμηηx + y(uημμx + uηηηx) = uμμ + 2yuμη + y2
uηη,
uxy = (uμ + yuη)y = uμμμy + uμηηy + uη + y(uημμy + uηηηy) = −uμμ + xuμη + uη − yuημ + xyuηη,
uyy = (−uμ + xuη)y = −uμμμy − uμηηy + x(uημμy + uηηηy) = uμμ − 2xuμη + x2
uηη,
Inserting these expressions into (6.4), we obtain
x(uμμ + 2yuμη + y2
uηη) + (x − y)(−uμμ + xuμη + uη − yuημ + xyuηη) − y(uμμ − 2xuμη + x2
uηη) = 0,
(x2
+ 2xy + y2
)uμη + (x − y)uη = 0,
(x − y)2
+ 4xy uμη + (x − y)uη = 0,
(μ2
+ 4η)uμη + μuη = 0, which is the Canonical form.

➂ We need to integrate twice to get the general solution:
(μ2
+ 4η)(uη)μ + μuη = 0,
(uη)μ
uη
dμ = −
μ
μ2 + 4η
dμ,
ln uη = −
1
2
ln (μ2
+ 4η) + ˜g(η),
ln uη = ln (μ2
+ 4η)−1
2 + ˜g(η),
uη =
g(η)
(μ2 + 4η)
1
2
,
u(μ, η) = f(μ) +
g(η) dη
(μ2 + 4η)
1
2
, General solution.

6.3 Well-Posedness
Problem (S’99, #2). In R2 consider the unit square Ω deﬁned by 0 ≤ x, y ≤ 1.
Consider
a) ux + uyy = 0;
b) uxx + uyy = 0;
c) uxx − uyy = 0.
Prescribe data for each problem separately on the boundary of Ω so that each of these
problems is well-posed. Justify your answers.
Proof. • The initial / boundary value problem for the HEAT EQUATION is well-
posed:
⎧
⎪⎨
⎪⎩
ut = u x ∈ Ω, t > 0,
u(x, 0) = g(x) x ∈ Ω,
u(x, t) = 0 x ∈ ∂Ω, t > 0.
Existence - by eigenfunction expansion.
Uniqueness and continuous dependence on the data -
by maximum principle.
The method of eigenfunction expansion and maximum
principle give well-posedness for more general problems:
⎧
⎪⎨
⎪⎩
ut = u + f(x, t) x ∈ Ω, t > 0,
u(x, 0) = g(x) x ∈ Ω,
u(x, t) = h(x, t) x ∈ ∂Ω, t > 0.
It is also possible to replace the Dirichlet boundary condition u(x, t) = h(x, t) by a
Neumann or Robin condition, provided we replace λn, φn by the eigenvalues and eigen-
functions for the appropriate boundary value problem.
a) • Relabel the variables (x → t, y → x).
We have the BACKWARDS HEAT EQUATION:
ut + uxx = 0.
Need to deﬁne initial conditions u(x, 1) = g(x), and
either Dirichlet, Neumann, or Robin boundary conditions.
b) • The solution to the LAPLACE EQUATION
u = 0 in Ω,
u = g on ∂Ω
exists if g is continuous on ∂Ω, by Perron’s method. Maximum principle gives unique-
ness.
To show the continuous dependence on the data, assume
u1 = 0 in Ω,
u1 = g1 on ∂Ω;
u2 = 0 in Ω,
u2 = g2 on ∂Ω.

Then (u1 − u2) = 0 in Ω. Maximum principle gives
max
Ω
(u1 − u2) = max
∂Ω
(g1 − g2). Thus,
max
Ω
|u1 − u2| = max
∂Ω
|g1 − g2|.
Thus, |u1 − u2| is bounded by |g1 − g2|, i.e. continuous dependence on data.
• Perron’s method gives existence of the solution to the POISSON EQUATION
u = f in Ω,
∂u
∂n = h on ∂Ω
for f ∈ C∞
(Ω) and h ∈ C∞
(∂Ω), satisfying the compatibility condition ∂Ω h dS =
Ω f dx. It is unique up to an additive constant.
c) • Relabel the variables (y → t).
The solution to the WAVE EQUATION
utt − uxx = 0,
is of the form u(x, y) = F(x + t) + G(x − t).
The existence of the solution to the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − uxx = 0 0 < x < 1, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) 0 < x < 1
u(0, t) = α(t), u(1, t) = β(t) t ≥ 0.
is given by the method of separation of variables
(expansion in eigenfunctions)
and by the parallelogram rule.
Uniqueness is given by the energy method.
Need initial conditions u(x, 0), ut(x, 0).
Prescribe u or ux for each of the two boundaries.

Problem (F’95, #7). Let a, b be real numbers. The PDE
uy + auxx + buyy = 0
is to be solved in the box Ω = [0, 1]2.
Find data, given on an appropriate part of ∂Ω, that will make this a well-posed prob-
lem.
Cover all cases according to the possible values of a and b. Justify your statements.
Proof.
➀ ab < 0 ⇒ two sets of characteristics ⇒ hyperbolic.
Relabeling the variables (y → t), we have
utt +
a
b
uxx = −
1
b
ut.
The solution of the equation is of the form
u(x, t) = F(x + −a
b t) + G(x − −a
b t).
Existence of the solution to the initial/boundary
value problem is given by the method of separation
of variables (expansion in eigenfunctions)
and by the parallelogram rule.
Uniqueness is given by the energy method.
Need initial conditions u(x, 0), ut(x, 0).
Prescribe u or ux for each of the two boundaries.
➁ ab > 0 ⇒ no characteristics ⇒ elliptic.
The solution to the Laplace equation with boundary conditions u = g on ∂Ω exists
if g is continuous on ∂Ω, by Perron’s method.
To show uniqueness, we use maximum principle. Assume there are two solutions u1
and u2 with with u1 = g(x), u2 = g(x) on ∂Ω. By maximum principle
max
Ω
(u1 − u2) = max
∂Ω
(g(x) − g(x)) = 0. Thus, u1 = u2.
➂ ab = 0 ⇒ one set of characteristics ⇒ parabolic.
• a = b = 0. We have uy = 0, a first-order ODE.
u must be specified on y = 0, i.e. x -axis.
• a = 0, b = 0. We have uy + buyy = 0, a second-order ODE.
u and uy must be specified on y = 0, i.e. x -axis.
• a > 0, b = 0. We have a Backwards Heat Equation.
ut = −auxx.
Need to define initial conditions u(x, 1) = g(x), and
either Dirichlet, Neumann, or Robin boundary conditions.

• a < 0, b = 0. We have a Heat Equation.
ut = −auxx.
The initial / boundary value problem for the heat equation is well-posed:
⎧
⎪⎨
⎪⎩
ut = u x ∈ Ω, t > 0,
u(x, 0) = g(x) x ∈ Ω,
u(x, t) = 0 x ∈ ∂Ω, t > 0.
Existence - by eigenfunction expansion.
Uniqueness and continuous dependence on the data -
by maximum principle.

7 Wave Equation
The one-dimensional wave equation is
utt − c2
uxx = 0. (7.1)
The characteristic equation with a = −c2, b = 0, c = 1 would be
dt
dx
=
b ±
√
b2 − 4ac
2a
= ±
√
4c2
−2c2
= ±
1
c
,
and thus
t = −
1
c
x + c1 and t =
1
c
x + c2,
μ = x + ct η = x − ct,
which transforms (7.1) to
uμη = 0. (7.2)
The general solution of (7.2) is u(μ, η) = F(μ)+G(η), where F and G are C1
functions.
Returning to the variables x, t we find that
u(x, t) = F(x + ct) + G(x − ct) (7.3)
solves (7.1). Moreover, u is C2 provided that F and G are C2.
If F ≡ 0, then u has constant values along the lines x−ct = const, so may be described
as a wave moving in the positive x-direction with speed dx/dt = c; if G ≡ 0, then u is
a wave moving in the negative x-direction with speed c.
7.1 The Initial Value Problem
For an initial value problem, consider the Cauchy problem
utt − c2
uxx = 0,
u(x, 0) = g(x), ut(x, 0) = h(x).
(7.4)
Using (7.3) and (7.4), we find that F and G satisfy
F(x) + G(x) = g(x), cF (x) − cG (x) = h(x). (7.5)
If we integrate the second equation in (7.5), we get cF(x) − cG(x) =
x
0 h(ξ) dξ + C.
Combining this with the first equation in (7.5), we can solve for F and G to find
F(x) = 1
2 g(x) + 1
2c
x
0 h(ξ) dξ + C1
G(x) = 1
2 g(x) − 1
2c
x
0 h(ξ) dξ − C1,
Using these expressions in (7.3), we obtain d’Alembert’s Formula for the solution
of the initial value problem (7.4):
u(x, t) =
1
2
(g(x + ct) + g(x − ct)) +
1
2c
x+ct
x−ct
h(ξ) dξ.
If g ∈ C2 and h ∈ C1, then d’Alembert’s Formula defines a C2 solution of (7.4).

7.2 Weak Solutions
Equation (7.3) defines a weak solution of (7.1) when F and G are not C2 functions.
Consider the parallelogram with sides that are
segments of characteristics. Since
u(x, t) = F(x + ct) + G(x − ct), we have
u(A) + u(C) =
= F(k1) + G(k3) + F(k2) + G(k4)
= u(B) + u(D),
which is the parallelogram rule.
7.3 Initial/Boundary Value Problem
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0 0 < x < L, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) 0 < x < L
u(0, t) = α(t), u(L, t) = β(t) t ≥ 0.
(7.6)
Use separation of variables to obtain an expansion in eigenfunctions. Find u(x, t) in
the form
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos
nπx
L
+ bn(t) sin
nπx
L
.
7.4 Duhamel’s Principle
⎧
⎪⎨
⎪⎩
utt − c2
uxx = f(x, t)
u(x, 0) = 0
ut(x, 0) = 0.
⇒
⎧
⎪⎨
⎪⎩
Utt − c2
Uxx = 0
U(x, 0, s) = 0
Ut(x, 0, s) = f(x, s)
u(x, t) =
t
0
U(x, t−s, s) ds.
⎧
⎪⎨
⎪⎩
an + λnan = fn(t)
an(0) = 0
an(0) = 0
⇒
⎧
⎪⎨
⎪⎩
ãn + λnãn = 0
ãn(0, s) = 0
ãn(0, s) = fn(s)
an(t) =
t
0
ãn(t−s, s) ds.
7.5 The Nonhomogeneous Equation
Consider the nonhomogeneous wave equation with homogeneous initial conditions:
utt − c2
uxx = f(x, t),
u(x, 0) = 0, ut(x, 0) = 0.
(7.7)
Duhamel’s Principle provides the solution of (7.7):
u(x, t) =
1
2c
t
0
x+c(t−s)
x−c(t−s)
f(ξ, s) dξ ds.
If f(x, t) is C1
in x and C0
in t, then Duhamel’s Principle provides a C2
solution of
(7.7).

We can solve (7.7) with nonhomogeneous initial conditions,
utt − c2
uxx = f(x, t),
u(x, 0) = g(x), ut(x, 0) = h(x),
(7.8)
by adding together d’Alembert’s formula and Duhamel’s principle gives the solution:
u(x, t) =
1
2
(g(x + ct) + g(x − ct)) +
1
2c
x+ct
x−ct
h(ξ) dξ +
1
2c
t
0
x+c(t−s)
x−c(t−s)
f(ξ, s) dξ ds.

7.6 Higher Dimensions
7.6.1 Spherical Means
For a continuous function u(x) on Rn, its spherical mean or average on a sphere of
radius r and center x is
Mu(x, r) =
1
ωn |ξ|=1
u(x + rξ)dSξ,
where ωn is the area of the unit sphere Sn−1 = {ξ ∈ Rn : |ξ| = 1} and dSξ is surface
measure. Since u is continuous in x, Mu(x, r) is continuous in x and r, so
Mu(x, 0) = u(x).
Using the chain rule, we find
∂
∂r
Mu(x, r) =
1
ωn |ξ|=1
n
i=1
uxi (x + rξ) ξi dSξ =
To compute the RHS, we apply the divergence theorem in Ω = {ξ ∈ Rn
: |ξ| < 1},
which has boundary ∂Ω = Sn−1 and exterior unit normal n(ξ) = ξ. The integrand is
V · n where V (ξ) = r−1
∇ξu(x + rξ) = ∇xu(x + rξ). Computing the divergence of V ,
we obtain
div V (ξ) = r
n
i=1
uxixi (x + rξ) = r xu(x + rξ), so,
=
1
ωn |ξ|<1
r xu(x + rξ) dξ =
r
ωn
x
|ξ|<1
u(x + rξ) dξ (ξ = rξ)
=
r
ωn
1
rn x
|ξ |<r
u(x + ξ ) dξ (spherical coordinates)
=
1
ωnrn−1 x
r
0
ρn−1
|ξ|=1
u(x + ρξ) dSξ dρ
=
1
ωnrn−1
ωn x
r
0
ρn−1
Mu(x, ρ) dρ =
1
rn−1 x
r
0
ρn−1
Mu(x, ρ) dρ.
If we multiply by rn−1
, differentiate with respect to r, and then divide by rn−1
,
we obtain the Darboux equation:
∂2
∂r2
+
n − 1
r
∂
∂r
Mu(x, r) = xMu(x, r).
Note that for a radial function u = u(r), we have Mu = u, so the equation provides the
Laplacian of u in spherical coordinates.
7.6.2 Application to the Cauchy Problem
We want to solve the equation
utt = c2
u x ∈ Rn
, t > 0, (7.9)
u(x, 0) = g(x), ut(x, 0) = h(x) x ∈ Rn
.
We use Poisson’s method of spherical means to reduce this problem to a partial differ-
ential equation in the two variables r and t.

Suppose that u(x, t) solves (7.9). We can view t as a parameter and take the spherical
mean to obtain Mu(x, r, t), which satisfies
∂2
∂t2
Mu(x, r, t) =
1
ωn |ξ|=1
utt(x + rξ, t)dSξ =
1
ωn |ξ|=1
c2
u(x + rξ, t)dSξ = c2
Mu(x, r, t).
Invoking the Darboux equation, we obtain the Euler-Poisson-Darboux equation:
∂2
∂t2
Mu(x, r, t) = c2 ∂2
∂r2
+
n − 1
r
∂
∂r
Mu(x, r, t).
The initial conditions are obtained by taking the spherical means:
Mu(x, r, 0) = Mg(x, r),
∂Mu
∂t
(x, r, 0) = Mh(x, r).
If we find Mu(x, r, t), we can then recover u(x, t) by:
u(x, t) = lim
r→0
Mu(x, r, t).
7.6.3 Three-Dimensional Wave Equation
When n = 3, we can write the Euler-Poisson-Darboux equation as 2
∂2
∂t2
rMu(x, r, t) = c2 ∂2
∂r2
rMu(x, r, t) .
For each fixed x, consider V x
(r, t) = rMu(x, r, t) as a solution of the one-dimensional
wave equation in r, t > 0:
∂2
∂t2
V x
(r, t) = c2 ∂2
∂r2
V x
(r, t),
V x
(r, 0) = rMg(x, r) ≡ Gx
(r), (IC)
V x
t (r, 0) = rMh(x, r) ≡ Hx
(r), (IC)
V x
(0, t) = lim
r→0
rMu(x, r, t) = 0 · u(x, t) = 0. (BC)
Gx
(0) = Hx
(0) = 0.
We may extend Gx
and Hx
as odd functions of r and use d’Alembert’s formula for
V x(r, t):
V x
(r, t) =
1
2
Gx
(r + ct) + Gx
(r − ct) +
1
2c
r+ct
r−ct
Hx
(ρ) dρ.
Since Gx
and Hx
are odd functions, we have for r < ct:
Gx
(r − ct) = −Gx
(ct − r) and
r+ct
r−ct
Hx
(ρ) dρ =
ct+r
ct−r
Hx
(ρ) dρ.
After some more manipulations, we find that the solution of (7.9) is given by the
Kirchhoff’s formula:
u(x, t) =
1
4π
∂
∂t
t
|ξ|=1
g(x + ctξ)dSξ +
t
4π |ξ|=1
h(x + ctξ)dSξ.
If g ∈ C3(R3) and h ∈ C2(R3), then Kirchhoff’s formula defines a C2-solution of (7.9).
2
It is seen by expanding the equation below.

7.6.4 Two-Dimensional Wave Equation
This problem is solved by Hadamard’s method of descent, namely, view (7.9) as a special
case of a three-dimensional problem with initial conditions independent of x3.
We need to convert surface integrals in R3 to domain integrals in R2.
u(x1, x2, t) =
1
4π
∂
∂t
2t
ξ2
1+ξ2
2 <1
g(x1 + ctξ1, x2 + ctξ2)dξ1dξ2
1 − ξ2
1 − ξ2
2
+
t
4π
2
ξ2
1+ξ2
2 <1
h(x1 + ctξ1, x2 + ctξ2)dξ1dξ2
1 − ξ2
1 − ξ2
2
If g ∈ C3
(R2
) and h ∈ C2
(R2
), then this equation deﬁnes a C2
-solution of (7.9).
7.6.5 Huygen’s Principle
Notice that u(x, t) depends only on the Cauchy data g, h on the surface of the hyper-
sphere {x + ctξ : |ξ| = 1} in Rn
, n = 2k + 1; in other words we have sharp signals.
If we use the method of descent to obtain the solution for n = 2k, the hypersurface
integrals become domain integrals. This means that there are no sharp signals.
The fact that sharp signals exist only for odd dimensions n ≥ 3 is known as Huygen’s
principle.
3
3
For x ∈ Rn
:
∂
∂t |ξ|=1
f(x + tξ)dSξ =
1
tn−1
|y|≤t
f(x + y)dy
∂
∂t |y|≤t
f(x + y)dy = tn−1
|ξ|=1
f(x + tξ)dSξ

7.7 Energy Methods
Suppose u ∈ C2(Rn × (0, ∞)) solves
utt = c2 u x ∈ Rn, t > 0,
u(x, 0) = g(x), ut(x, 0) = h(x) x ∈ Rn,
(7.10)
where g and h have compact support.
Deﬁne energy for a function u(x, t) at time t by
E(t) =
1
2 Rn
(u2
t + c2
|∇u|2
) dx.
If we diﬀerentiate this energy function, we obtain
dE
dt
=
d
dt
1
2 Rn
u2
t + c2
n
i=1
u2
xi
dx =
Rn
ututt + c2
n
i=1
uxi uxit dx
=
Rn
ututt dx + c2
n
i=1
uxi ut
∂Rn
−
Rn
c2
n
i=1
uxixi ut dx
=
Rn
ut(utt − c2
u) dx = 0,
or
dE
dt
=
d
dt
1
2 Rn
u2
t + c2
n
i=1
u2
xi
dx =
Rn
ututt + c2
n
i=1
uxi uxit dx
=
Rn
ututt + c2
∇u · ∇ut dx
=
Rn
ututt dx + c2
∂Rn
ut
∂u
∂n
ds −
Rn
ut u dx
=
Rn
ut(utt − c2
u) dx = 0.
Hence, E(t) is constant, or E(t) ≡ E(0).
In particular, if u1 and u2 are two solutions of (7.10), then w = u1 −u2 has zero Cauchy
data and hence Ew(0) = 0. By discussion above, Ew(t) ≡ 0, which implies w(x, t) ≡
const. But w(x, 0) = 0 then implies w(x, t) ≡ 0, so the solution is unique.

7.8 Contraction Mapping Principle
Suppose X is a complete metric space with distance function represented by d(·, ·).
A mapping T : X → X is a strict contraction if there exists 0 < α < 1 such that
d(Tx, Ty) ≤ α d(x, y) ∀ x, y ∈ X.
An obvious example on X = Rn
is Tx = αx, which shrinks all of Rn
, leaving 0 fixed.
The Contraction Mapping Principle. If X is a complete metric space and T :
X → X is a strict contraction, then T has a unique fixed point.
The process of replacing a differential equation by an integral equation occurs in
time-evolution partial differential equations.
The Contraction Mapping Principle is used to establish the local existence and unique-
ness of solutions to various nonlinear equations.

8 Laplace Equation
Consider the Laplace equation
u = 0 in Ω ⊂ Rn
(8.1)
and the Poisson equation
u = f in Ω ⊂ Rn
. (8.2)
Solutions of (8.1) are called harmonic functions in Ω.
Cauchy problems for (8.1) and (8.2) are not well posed. We use separation of variables
for some special domains Ω to ﬁnd boundary conditions that are appropriate for (8.1),
(8.2).
Dirichlet problem: u(x) = g(x), x ∈ ∂Ω
Neumann problem:
∂u(x)
∂n
= h(x), x ∈ ∂Ω
Robin problem:
∂u
∂n
+ αu = β, x ∈ ∂Ω
8.1 Green’s Formulas
Ω
∇u · ∇v dx =
∂Ω
v
∂u
∂n
ds −
Ω
v u dx (8.3)
∂Ω
v
∂u
∂n
− u
∂v
∂n
ds =
Ω
(v u − u v) dx
∂Ω
∂u
∂n
ds =
Ω
u dx (v = 1 in (8.3))
Ω
|∇u|2
dx =
∂Ω
u
∂u
∂n
ds −
Ω
u u dx (u = v in (8.3))
Ω
uxvx dxdy =
∂Ω
vuxn1 ds −
Ω
vuxx dxdy n = (n1, n2) ∈ R2
Ω
uxk
v dx =
∂Ω
uvnk ds −
Ω
uvxk
dx n = (n1, . . ., nn) ∈ Rn
.
Ω
u 2
v dx =
∂Ω
u
∂ v
∂n
ds −
∂Ω
v
∂u
∂n
ds +
Ω
u v dx.
Ω
u 2
v − v 2
u dx =
∂Ω
u
∂ v
∂n
− v
∂ u
∂n
ds +
∂Ω
u
∂v
∂n
− v
∂u
∂n
ds.

8.2 Polar Coordinates
Polar Coordinates. Let f : Rn → R be continuous. Then
Rn
f dx =
∞
0 ∂Br(x0)
f dS dr
for each x0 ∈ Rn. In particular
d
dr Br(x0)
f dx =
∂Br(x0)
f dS
for each r > 0.
u = u(x(r, θ), y(r, θ))
x(r, θ) = r cos θ
y(r, θ) = r sin θ
ur = uxxr + uyyr = ux cos θ + uy sinθ,
uθ = uxxθ + uyyθ = −uxr sin θ + uyr cos θ,
urr = (ux cos θ + uy sinθ)r = (uxxxr + uxyyr) cosθ + (uyxxr + uyyyr) sinθ
= uxx cos2
θ + 2uxy cos θ sinθ + uyy sin2
θ,
uθθ = (−uxr sinθ + uyr cos θ)θ
= (−uxxxθ − uxyyθ)r sinθ − uxr cos θ + (uyxxθ + uyyyθ)r cos θ − uyr sin θ
= (uxxr sin θ − uxyr cos θ)r sinθ − uxr cos θ + (−uyxr sin θ + uyyr cos θ)r cos θ − uyr sin θ
= r2
(uxx sin2
θ − 2uxy cos θ sinθ + uyy cos2
θ) − r(ux cos θ + uy sinθ).
urr + 1
r2 uθθ
= uxx cos2 θ + 2uxy cos θ sinθ + uyy sin2
θ + uxx sin2
θ − 2uxy cos θ sinθ + uyy cos2 θ − 1
r (ux cos θ + uy sin θ)
= uxx + uyy − 1
r ur.
uxx + uyy = urr +
1
r
ur +
1
r2
uθθ.
∂2
∂x2
+
∂2
∂y2
=
∂2
∂r2
+
1
r
∂
∂r
+
1
r2
∂2
∂θ2
.
8.3 Polar Laplacian in R2
for Radial Functions
u =
1
r
rur r
=
∂2
∂r2
+
1
r
∂
∂r
u.
8.4 Spherical Laplacian in R3
and Rn
u =
∂2
∂r2
+
n − 1
r
∂
∂r
u.
In R3: 4
u =
1
r2
r2
ur r
=
1
r
ru rr
=
∂2
∂r2
+
2
r
∂
∂r
u.
4
These formulas are taken from S. Farlow, p. 411.

8.5 Cylindrical Laplacian in R3
u =
1
r
rur r
=
∂2
∂r2
+
1
r
∂
∂r
u.
8.6 Mean Value Theorem
Gauss Mean Value Theorem. If u ∈ C2
(Ω) is harmonic in Ω, let ξ ∈ Ω and pick
r > 0 so that Br(ξ) = {x : |x − ξ| ≤ r} ⊂ Ω. Then
u(ξ) = Mu(ξ, r) ≡
1
ωn |x|=1
u(ξ + rx) dSx,
where ωn is the measure of the (n − 1)-dimensional sphere in Rn
.
8.7 Maximum Principle
Maximum Principle. If u ∈ C2(Ω) satisﬁes u ≥ 0 in Ω, then either u is a constant,
or
u(ξ) < sup
x∈Ω
u(x)
for all ξ ∈ Ω.
Proof. We may assume A = supx∈Ω u(x) ≤ ∞, so by continuity of u we know that
{x ∈ Ω : u(x) = A} is relatively closed in Ω. But since
u(ξ) ≤
n
ωn |x|≤1
u(ξ + rx) dx,
if u(ξ) = A at an interior point ξ, then u(x) = A for all x in a ball about ξ, so
{x ∈ Ω : u(x) = A} is open. The connectedness of Ω implies u(ξ) < A or u(ξ) ≡ A for
all ξ ∈ Ω.
The maximum principle shows that u ∈ C2
(Ω) with u ≥ 0 can attain an interior
maximum only if u is constant. In particular, if Ω is compact, and u ∈ C2(Ω) ∩ C(Ω)
satisﬁes u ≥ 0 in Ω, we have the weak maximum principle:
max
x∈Ω
u(x) = max
x∈∂Ω
u(x).

8.8 The Fundamental Solution
A fundamental solution K(x) for the Laplace operator is a distribution satisfying
K(x) = δ(x) (8.4)
where δ is the delta distribution supported at x = 0. In order to solve (8.4), we should
ﬁrst observe that is symmetric in the variables x1, . . ., xn, and δ(x) is also radially
symmetric (i.e., its value only depends on r = |x|). Thus, we try to solve (8.4) with a
radially symmetric function K(x). Since δ(x) = 0 for x = 0, we see that (8.4) requires
K to be harmonic for r > 0. For the radially symmetric function K, Laplace equation
becomes (K = K(r)):
∂2K
∂r2
+
n − 1
r
∂K
∂r
= 0. (8.5)
The general solution to (8.5) is
K(r) =
c1 + c2 log r if n = 2
c1 + c2r2−n if n ≥ 3.
(8.6)
After we determine c2, we ﬁnd the fundamental solution for the Laplace operator:
K(x) =
1
2π log r if n = 2
1
(2−n)ωn
r2−n if n ≥ 3.
• We can derive, (8.6) for any given n. For intance, when n = 3, we have:
K +
2
r
K = 0.
Let
K =
1
r
w(r),
K =
1
r
w −
1
r2
w,
K =
1
r
w −
2
r2
w +
2
r3
w.
Plugging these into , we obtain:
1
r
w = 0, or
w = 0.
Thus,
w = c1r + c2,
K =
1
r
w(r) = c1 +
c2
r
.
See the similar problem, F’99, #2, where the fundamental solution for ( − I) is
found in the process.

Find the Fundamental Solution of the Laplace Operator for n = 3
We found that starting with the Laplacian in R3 for a radially symmetric function K,
K +
2
r
K = 0,
and letting K = 1
r w(r), we obtained the equation: w = c1r + c2, which implied:
K = c1 +
c2
r
.
We now ﬁnd the constant c2 that ensures that for v ∈ C∞
0 (R3
), we have
R3
K(|x|) v(x) dx = v(0).
Suppose v(x) ≡ 0 for |x| ≥ R and let Ω = BR(0); for small > 0 let
Ω = Ω − B (0).
K(|x|) is harmonic ( K(|x|) = 0) in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪
∂B (0)):
Ω
K(|x|) v dx =
∂Ω
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
=0, since v≡0 for x≥R
+
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS.
lim
→0 Ω
K(|x|) v dx =
Ω
K(|x|) v dx. Since K(r) = c1 +
c2
r
is integrable at x = 0.
On ∂B (0), K(|x|) = K( ). Thus, 5
∂B (0)
K(|x|)
∂v
∂n
dS = K( )
∂B (0)
∂v
∂n
dS ≤ c1 +
c2
4π 2
max ∇v → 0, as → 0.
∂B (0)
v(x)
∂K(|x|)
∂n
dS =
∂B (0)
c2
2
v(x) dS
=
∂B (0)
c2
2
v(0) dS +
∂B (0)
c2
2
[v(x) − v(0)] dS
=
c2
2
v(0) 4π 2
+ 4πc2 max
x∈∂B (0)
v(x) − v(0)
→0, (v is continuous)
= 4πc2 v(0) → −v(0).
Thus, taking 4πc2 = −1, i.e. c2 = − 1
4π , we obtain
Ω
K(|x|) v dx = lim
→0 Ω
K(|x|) v dx = v(0),
that is K(r) = − 1
4πr is the fundamental solution of .
5
In R3
, for |x| = ,
K(|x|) = K( ) = c1 +
c2
.
∂K(|x|)
∂n
= −
∂K( )
∂r
=
c2
2
, (since n points inwards.)
n points toward 0 on the sphere |x| = (i.e., n = −x/|x|).

Show that the Fundamental Solution of the Laplace Operator is given by.
K(x) =
1
2π log r if n = 2
1
(2−n)ωn
r2−n
if n ≥ 3.
(8.7)
Proof. For v ∈ C∞
0 (Rn), we want to show
Rn
K(|x|) v(x) dx = v(0).
Ω = Ω − B (0).
K(|x|) is harmonic ( K(|x|) = 0) in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪
∂B (0)):
Ω
K(|x|) v dx =
∂Ω
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
+
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS.
lim
→0 Ω
K(|x|) v dx =
Ω
K(|x|) v dx. Since K(r) is integrable at x = 0.
On ∂B (0), K(|x|) = K( ). Thus, 6
∂B (0)
K(|x|)
∂v
∂n
dS = K( )
∂B (0)
∂v
∂n
dS ≤ K( ) ωn
n−1
max ∇v → 0, as → 0.
∂B (0)
v(x)
∂K(|x|)
∂n
dS =
∂B (0)
−
1
ωn
n−1
v(x) dS
=
∂B (0)
−
1
ωn
n−1
v(0) dS +
∂B (0)
−
1
ωn
n−1
[v(x) − v(0)] dS
= −
1
ωn
n−1
v(0) ωn
n−1
− max
x∈∂B (0)
v(x) − v(0)
= −v(0).
Thus,
Ω
K(|x|) v dx = lim
→0 Ω
K(|x|) v dx = v(0).
6
Note that for |x| = ,
K(|x|) = K( ) =
1
2π
log if n = 2
1
(2−n)ωn
2−n
if n ≥ 3.
∂K(|x|)
∂n
= −
∂K( )
∂r
= −
1
2π
if n = 2
1
ωn
n−1 if n ≥ 3,
= −
1
ωn
n−1
n points toward 0 on the sphere |x| = (i.e., n = −x/|x|).

8.9 Representation Theorem
Representation Theorem, n = 3.
Let Ω be bounded domain in R3
and let n be the unit exterior normal to ∂Ω. Let
u ∈ C2(Ω). Then the value of u at any point x ∈ Ω is given by the formula
u(x) =
1
4π ∂Ω
1
|x − y|
∂u(y)
∂n
− u(y)
∂
∂n
1
|x − y|
dS −
1
4π Ω
u(y)
|x − y|
dy. (8.8)
Proof. Consider the Green’s identity:
Ω
(u w − w u) dy =
∂Ω
u
∂w
∂n
− w
∂u
∂n
dS,
where w is the harmonic function
w(y) =
1
|x − y|
,
which is singular at x ∈ Ω. In order to be able to apply Green’s identity, we consider
a new domain Ω :
Ω = Ω − B (x).
Since u, w ∈ C2(Ω ), Green’s identity can be applied. Since w is harmonic ( w = 0)
in Ω and since ∂Ω = ∂Ω ∪ ∂B (x), we have
−
Ω
u(y)
|x − y|
dy =
∂Ω
u(y)
∂
∂n
1
|x − y|
−
1
|x − y|
∂u(y)
∂n
dS (8.9)
+
∂B (x)
u(y)
∂
∂n
1
|x − y|
−
1
|x − y|
∂u(y)
∂n
dS. (8.10)
We will show that formula (8.8) is obtained by letting → 0.
lim
→0
−
Ω
u(y)
|x − y|
dy = −
Ω
u(y)
|x − y|
dy. Since
1
|x − y|
is integrable at x = y.
The ﬁrst integral on the right of (8.10) does not depend on . Hence, the limit as → 0
of the second integral on the right of (8.10) exists, and in order to obtain (8.8), need
lim
→0 ∂B (x)
u(y)
∂
∂n
1
|x − y|
−
1
|x − y|
∂u(y)
∂n
dS = 4πu(x).
∂B (x)
u(y)
∂
∂n
1
|x − y|
−
1
|x − y|
∂u(y)
∂n
dS =
∂B (x)
1
2
u(y) −
1 ∂u(y)
∂n
dS
=
∂B (x)
1
2
u(x) dS +
∂B (x)
1
2
[u(y) − u(x)] −
1 ∂u(y)
∂n
dS
= 4πu(x) +
∂B (x)
1
2
[u(y) − u(x)] −
1 ∂u(y)
∂n
dS.

7
The last integral tends to 0 as → 0:
∂B (x)
1
2
[u(y) − u(x)] −
1 ∂u(y)
∂n
dS ≤
1
2
∂B (x)
u(y) − u(x) +
1
∂B (x)
∂u(y)
∂n
dS
≤ 4π max
y∈∂B (x)
u(y) − u(x)
→0, (u continuous in Ω)
+ 4π max
y∈Ω
∇u(y)
→0, (|∇u| is finite)
.
7
Note that for points y on ∂B (x),
1
|x − y|
=
1
and
∂
∂n
1
|x − y|
=
1
2
.

Representation Theorem, n = 2.
Let Ω be bounded domain in R2 and let n be the unit exterior normal to ∂Ω. Let
u ∈ C2
(Ω). Then the value of u at any point x ∈ Ω is given by the formula
u(x) =
1
2π Ω
u(y) log|x − y| dy +
1
2π ∂Ω
u(y)
∂
∂n
log |x − y| − log |x − y|
∂u(y)
∂n
dS.(8.11)
Ω
(u w − w u) dy =
∂Ω
u
∂w
∂n
− w
∂u
∂n
dS,
w(y) = log |x − y|,
which is singular at x ∈ Ω. In order to be able to apply Green’s identity, we consider
a new domain Ω :
Ω = Ω − B (x).
Since u, w ∈ C2(Ω ), Green’s identity can be applied. Since w is harmonic ( w = 0)
in Ω and since ∂Ω = ∂Ω ∪ ∂B (x), we have
−
Ω
u(y) log |x − y| dy (8.12)
=
∂Ω
u(y)
∂
∂n
∂u(y)
∂n
dS
+
∂B (x)
u(y)
∂
∂n
∂u(y)
∂n
dS.
lim
→0
−
Ω
u(y) log|x − y| dy = −
Ω
u(y) log |x − y| dy. since log |x − y| is integrable at x = y.
lim
→0 ∂B (x)
u(y)
∂
∂n
∂u(y)
∂n
dS = 2πu(x).
∂B (x)
u(y)
∂
∂n
∂u(y)
∂n
dS =
∂B (x)
1
u(y) − log
∂u(y)
∂n
dS
=
∂B (x)
1
u(x) dS +
∂B (x)
1
[u(y) − u(x)] − log
∂u(y)
∂n
dS
= 2πu(x) +
∂B (x)
1
[u(y) − u(x)] − log
∂u(y)
∂n
dS.

8
The last integral tends to 0 as → 0:
∂B (x)
1
[u(y) − u(x)] − log
∂u(y)
∂n
dS ≤
1
∂B (x)
u(y) − u(x) + log
∂B (x)
∂u(y)
∂n
dS
≤ 2π max
y∈∂B (x)
u(y) − u(x)
+ 2π log max
y∈Ω
∇u(y)
.
8
log |x − y| = log and
∂
∂n
log |x − y| =
1
.

Representation Theorems, n > 3 can be obtained in the same way. We use the
Green’s identity with
w(y) =
1
|x − y|n−2
,
which is a harmonic function in Rn with a singularity at x.
The fundamental solution for the Laplace operator is (r = |x|):
K(x) =
1
2π log r if n = 2
1
(2−n)ωn
r2−n if n ≥ 3.
Representation Theorem. If Ω ∈ Rn
is bounded, u ∈ C2
(Ω), and x ∈ Ω, then
u(x) =
Ω
K(x − y) u(y) dy +
∂Ω
u(y)
∂K(x − y)
∂n
− K(x − y)
∂u(y)
∂n
dS.(8.13)
Ω
(u w − w u) dy =
∂Ω
u
∂w
∂n
− w
∂u
∂n
dS,
w(y) = K(x − y),
which is singular at y = x. In order to be able to apply Green’s identity, we consider a
new domain Ω :
Ω = Ω − B (x).
Since u, K(x − y) ∈ C2(Ω ), Green’s identity can be applied. Since K(x − y) is
harmonic ( K(x − y) = 0) in Ω and since ∂Ω = ∂Ω ∪ ∂B (x), we have
−
Ω
K(x − y) u(y) dy =
∂Ω
u(y)
∂K(x − y)
∂n
− K(x − y)
∂u(y)
∂n
dS (8.14)
+
∂B (x)
u(y)
∂K(x − y)
∂n
− K(x − y)
∂u(y)
∂n
dS.(8.15)
lim
→0
−
Ω
K(x − y) u(y) dy = −
Ω
K(x − y) u(y) dy. since K(x − y) is integrable at x = y.
lim
→0 ∂B (x)
u(y)
∂K(x − y)
∂n
− K(x − y)
∂u(y)
∂n
dS = −u(x).

∂B (x)
u(y)
∂K(x − y)
∂n
− K(x − y)
∂u(y)
∂n
dS =
∂B (x)
u(y)
∂K( )
∂n
− K( )
∂u(y)
∂n
dS
=
∂B (x)
u(x)
∂K( )
∂n
dS +
∂B (x)
∂K( )
∂n
[u(y) − u(x)] − K( )
∂u(y)
∂n
dS
= −
1
ωn
n−1
∂B (x)
u(x) dS −
1
ωn
n−1
∂B (x)
[u(y) − u(x)] dS −
∂B (x)
K( )
∂u(y)
∂n
dS
= −
1
ωn
n−1
u(x)ωn
n−1
−u(x)
−
1
ωn
n−1
∂B (x)
[u(y) − u(x)] dS −
∂B (x)
K( )
∂u(y)
∂n
dS.
9 The last two integrals tend to 0 as → 0:
−
1
ωn
n−1
∂B (x)
[u(y) − u(x)] dS −
∂B (x)
K( )
∂u(y)
∂n
dS
≤
1
ωn
n−1
max
y∈∂B (x)
u(y) − u(x) ωn
n−1
+ K( ) max
y∈Ω
∇u(y) ωn
n−1
.
8.10 Green’s Function and the Poisson Kernel
With a slight change in notation, the Representation Theorem has the following special
case.
Theorem. If Ω ∈ Rn is bounded, u ∈ C2(Ω) C1(Ω) is harmonic, and ξ ∈ Ω, then
u(ξ) =
∂Ω
u(x)
∂K(x − ξ)
∂n
− K(x − ξ)
∂u(x)
∂n
dS. (8.16)
Let ω(x) be any harmonic function in Ω, and for x, ξ ∈ Ω consider
G(x, ξ) = K(x − ξ) + ω(x).
If we use the Green’s identity (with u = 0 and ω = 0), we get:
0 =
∂Ω
u
∂ω
∂n
− ω
∂u
∂n
ds. (8.17)
Adding (8.16) and (8.17), we obtain:
u(ξ) =
∂Ω
u(x)
∂G(x, ξ)
∂n
− G(x, ξ)
∂u(x)
∂n
dS. (8.18)
Suppose that for each ξ ∈ Ω we can ﬁnd a function ωξ(x) that is harmonic in Ω and
satisﬁes ωξ(x) = −K(x − ξ) for all x ∈ ∂Ω. Then G(x, ξ) = K(x − ξ) + ωξ(x) is a
fundamental solution such that
G(x, ξ) = 0 x ∈ ∂Ω.
9
K(x − y) = K( ) =
1
2π
log if n = 2
1
(2−n)ωn
2−n
if n ≥ 3.
∂K(x − y)
∂n
= −
∂K( )
∂r
= −
1
2π
if n = 2
1
ωn
n−1 if n ≥ 3,
= −
1
ωn
n−1

G is called the Green’s function and is useful in satisfying Dirichlet boundary conditions.
The Green’s function is diﬃcult to construct for a general domain Ω since it requires
solving the Dirichlet problem ωξ = 0 in Ω, ωξ(x) = −K(x − ξ) for x ∈ ∂Ω, for each
ξ ∈ Ω.
From (8.18) we ﬁnd 10
u(ξ) =
∂Ω
u(x)
∂G(x, ξ)
∂n
dS.
Thus if we know that the Dirichlet problem has a solution u ∈ C2
(Ω), then we can
calculate u from the Poisson integral formula (provided of course that we can compute
G(x, ξ)).
10
If we did not assume u = 0 in our derivation, we would have (8.13) instead of (8.16), and an
extra term in (8.17), which would give us a more general expression:
u(ξ) =
Ω
G(x, ξ) u dx +
∂Ω
u(x)
∂G(x, ξ)
∂n
dS.

8.11 Properties of Harmonic Functions
Liouville’s Theorem. A bounded harmonic function defined on all of Rn must be a
constant.
8.12 Eigenvalues of the Laplacian
Consider the equation
u + λu = 0 in Ω
u = 0 on ∂Ω,
(8.19)
where Ω is a bounded domain and λ is a (complex) number. The values of λ for which
(8.19) admits a nontrivial solution u are called the eigenvalues of in Ω, and the
solution u is an eigenfunction associated to the eigenvalue λ. (The convention
u + λu = 0 is chosen so that all eigenvalues λ will be positive.)
Properties of the Eigenvalues and Eigenfunctions for (8.19):
1. The eigenvalues of (8.19) form a countable set {λn}∞
n=1 of positive numbers with
λn → ∞ as n → ∞.
2. For each eigenvalue λn there is a finite number (called the multiplicity of λn) of
linearly independent eigenfunctions un.
3. The first (or principal) eigenvalue, λ1, is simple and u1 does not change sign in Ω.
4. Eigenfunctions corresponding to distinct eigenvalues are orthogonal.
5. The eigenfunctions may be used to expand certain functions on Ω in an infinite
series.

9 Heat Equation
The heat equation is
ut = k u for x ∈ Ω, t > 0, (9.1)
with initial and boundary conditions.
9.1 The Pure Initial Value Problem
9.1.1 Fourier Transform
If u ∈ C∞
0 (Rn), define its Fourier transform u by
u(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ
u(x) dx for ξ ∈ Rn
.
We can differentiate û:
∂
∂ξj
u(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ
(−ixj)u(x) dx = (−ixj) u (ξ).
Iterating this computation, we obtain
∂
∂ξj
k
u(ξ) = (−ixj)k u (ξ). (9.2)
Similarly, integrating by parts shows
∂u
∂xj
(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ ∂u
∂xj
(x) dx = −
1
(2π)
n
2 Rn
∂
∂xj
(e−ix·ξ
)u(x) dx
=
1
(2π)
n
2 Rn
(iξj)e−ix·ξ
u(x) dx
= (iξj)u(ξ).
Iterating this computation, we obtain
∂ku
∂xk
j
(ξ) = (iξj)k
u(ξ). (9.3)
Formulas (9.2) and (9.3) express the fact that Fourier transform interchanges differen-
tiation and multiplication by the coordinate function.
9.1.2 Multi-Index Notation
A multi-index is a vector α = (α1, . . ., αn) where each αi is a nonnegative integer.
The order of the multi-index is |α| = α1 + . . . + αn. Given a multi-index α, define
Dα
u =
∂|α|
u
∂xα1
1 · · ·∂xαn
n
= ∂α1
x1
· · ·∂αn
xn
u.
We can generalize (9.3) in multi-index notation:
Dαu(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ
Dα
u(x) dx =
(−1)|α|
(2π)
n
2 Rn
Dα
x (e−ix·ξ
)u(x) dx
=
1
(2π)
n
2 Rn
(iξ)α
e−ix·ξ
u(x) dx
= (iξ)α
u(ξ).
(iξ)α
= (iξ1)α1
· · ·(iξn)αn
.

Parseval’s theorem (Plancherel’s theorem).
Assume u ∈ L1(Rn) ∩ L2(Rn). Then u, u∨ ∈ L2(Rn) and
||u||L2(Rn) = ||u∨
||L2(Rn) = ||u||L2(Rn), or
∞
−∞
|u(x)|2
dx =
∞
−∞
|u(ξ)|2
dξ.
Also,
∞
−∞
u(x) v(x)dx =
∞
−∞
u(ξ) v(ξ) dξ.
The properties (9.2) and (9.3) make it very natural to consider the fourier transform
on a subspace of L1
(Rn
) called the Schwartz class of functions, S, which consists of the
smooth functions whose derivatives of all orders decay faster than any polynomial, i.e.
S = {u ∈ C∞
(Rn
) : for every k ∈ N and α ∈ Nn
, |x|k
|Dα
u(x)| is bounded on Rn
}.
For u ∈ S, the Fourier transform u exists since u decays rapidly at ∞.
Lemma. (i) If u ∈ L1(Rn), then u is bounded. (ii) If u ∈ S, then u ∈ S.
Deﬁne the inverse Fourier transform for u ∈ L1(Rn):
u∨
(ξ) =
1
(2π)
n
2 Rn
eix·ξ
u(x) dx for ξ ∈ Rn
, or
u(x) =
1
(2π)
n
2 Rn
eix·ξ
u(ξ) dξ for x ∈ Rn
.
Fourier Inversion Theorem (McOwen). If u ∈ S, then (u)∨
= u; that is,
u(x) =
1
(2π)
n
2 Rn
eix·ξ
u(ξ) dξ =
1
(2π)n
R2n
ei(x−y)·ξ
u(y) dy dξ = (u)∨
(x).
Fourier Inversion Theorem (Evans). Assume u ∈ L2
(Rn
). Then, u = (u)∨
.

Shift: Let u(x − a
y
) = v(x), and determinte v(ξ):
u(x − a)(ξ) = v(ξ) =
1
√
2π R
e−ixξ
v(x) dx =
1
√
2π R
e−i(y+a)ξ
u(y) dy
=
1
√
2π R
e−iyξ
e−iaξ
u(y) dy = e−iaξ
u(ξ).
u(x − a)(ξ) = e−iaξ
u(ξ).
Delta function:
δ(x)(ξ) =
1
√
2π R
e−ixξ
δ(x) dx =
1
√
2π
, since u(x) =
R
δ(x − y) u(y) dy .
δ(x − a)(ξ) = e−iaξ
δ(ξ) =
1
√
2π
e−iaξ
. (using result from ‘Shift’)
Convolution:
(f ∗ g)(x) =
Rn
f(x − y)g(y) dy,
(f ∗ g)(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ
Rn
f(x − y) g(y) dydx =
1
(2π)
n
2 Rn Rn
e−ix·ξ
f(x − y) g(y) dydx
=
1
(2π)
n
2 Rn Rn
e−i(x−y)·ξ
f(x − y) dx e−iy·ξ
g(y) dy
=
1
(2π)
n
2 Rn
e−iz·ξ
f(z) dz ·
Rn
e−iy·ξ
g(y) dy = (2π)
n
2 f(ξ)g(ξ).
(f ∗ g)(ξ) = (2π)
n
2 f(ξ) g(ξ).
Gaussian: (completing the square)
e−x2
2 (ξ) =
1
√
2π R
e−ixξ
e−x2
2 dx =
1
√
2π R
e−x2 +2ixξ
2 dx =
1
√
2π R
e−x2+2ixξ−ξ2
2 dx e−ξ2
2
=
1
√
2π R
e−
(x+iξ)2
2 dx e−ξ2
2 =
1
√
2π R
e
−y2
2 dy e−ξ2
2 =
1
√
2π
√
2πe−ξ2
2 = e−ξ2
2 .
e−x2
2 (ξ) = e−ξ2
2 .
Multiplication by x:
−ixu(ξ) =
1
√
2π R
e−ixξ
− ixu(x) dx =
d
dξ
u(ξ).
xu(x)(ξ) = i
d
dξ
u(ξ).

Multiplication of ux by x: (using the above result)
xux(x)(ξ) =
1
√
2π R
e−ixξ
xux(x) dx =
1
√
2π
e−ixξ
xu
∞
−∞
= 0
−
1
√
2π R
(−iξ)e−ixξ
x + e−ixξ
u dx
=
1
√
2π
iξ
R
e−ixξ
x u dx −
1
√
2π R
e−ixξ
u dx
= iξ xu(x)(ξ) − u(ξ) = iξ i
d
dξ
u(ξ) − u(ξ) = −ξ
d
dξ
u(ξ) − u(ξ).
xux(x)(ξ) = −ξ
d
dξ
u(ξ) − u(ξ).
Table of Fourier Transforms: 11
e−ax2
2 (ξ) =
1
√
a
e−ξ2
2a , (Gaussian)
eibxf(ax)(ξ) =
1
a
f
ξ − b
a
,
f(x) =
1, |x| ≤ L
0, |x| > L,
f(x)(ξ) =
1
√
2π
2 sin(ξL)
ξ
,
e−a|x|(ξ) =
1
√
2π
2a
a2 + ξ2
, (a > 0)
1
a2 + x2
(ξ) =
√
2π
2a
e−a|ξ|
, (a > 0)
H(a − |x|)(ξ) =
2
π
1
ξ
sinaξ,
H(x)(ξ) =
1
√
2π
πδ(ξ) +
1
iξ
,
H(x) − H(−x) (ξ) =
2
π
1
iξ
, (sign)
1(ξ) =
√
2πδ(ξ).
11
Results with marked with were taken from W. Strauss, where the deﬁnition of Fourier Transform
is diﬀerent. An extra multiple of 1√
2π
was added to each of these results.

9.1.3 Solution of the Pure Initial Value Problem
Consider the pure initial value problem
ut = u for t > 0, x ∈ Rn
u(x, 0) = g(x) for x ∈ Rn
.
(9.4)
We take the Fourier transform of the heat equation in the x-variables.
(ut)(ξ, t) =
1
(2π)
n
2 Rn
e−ix·ξ
ut(x, t) dx =
∂
∂t
u(ξ, t)
u(ξ, t) =
n
j=1
(iξj)2
u(ξ, t) = −|ξ|2
u(ξ, t).
The heat equation therefore becomes
∂
∂t
u(ξ, t) = −|ξ|2
u(ξ, t),
which is an ordinary differential equation in t, with the solution u(ξ, t) = Ce−|ξ|2t
.
The initial condition u(ξ, 0) = g(ξ) gives
u(ξ, t) = g(ξ) e−|ξ|2t
,
u(x, t) = g(ξ) e−|ξ|2t
∨
=
1
(2π)
n
2
g ∗ e−|ξ|2t ∨
=
1
(2π)
n
2
g ∗
1
(2π)
n
2 Rn
e−|ξ|2t
eix·ξ
dξ
=
1
(4π2)
n
2
g ∗
Rn
eix·ξ−|ξ|2t
dξ =
1
(4π2)
n
2
g ∗ e−|x|2
4t
π
t
n
2
=
1
(4πt)
n
2
g ∗ e−
|x|2
4t =
1
(4πt)
n
2 Rn
e−
|x−y|2
4t g(y) dy.
Thus, 12 solution of the initial value problem (9.4) is
u(x, t) =
Rn
K(x, y, t) g(y) dy =
1
(4πt)
n
2 Rn
e−
|x−y|2
4t g(y) dy.
Uniqueness of solutions for the pure initial value problem fails: there are nontrivial
solutions of (9.4) with g = 0. 13 Thus, the pure initial value problem for the heat
equation is not well-posed, as it was for the wave equation. However, the nontrivial
solutions are unbounded as functions of x when t > 0 is fixed; uniqueness can be
regained by adding a boundedness condition on the solution.
12
Identity (Evans, p. 187.) :
Rn
eix·ξ−|ξ|2
t
dξ = e−
|x|2
4t
π
t
n
2
.
13
The following function u satisfies ut = uxx for t > 0 with u(x, 0) = 0:
u(x, t) =
∞
k=0
1
(2k)!
x2k dk
dtk
e−1/t2
.

9.1.4 Nonhomogeneous Equation
Consider the pure initial value problem with homogeneous initial condition:
ut = u + f(x, t) for t > 0, x ∈ Rn
u(x, 0) = 0 for x ∈ Rn
.
(9.5)
Duhamel’s principle gives the solution:
u(x, t) =
t
0 Rn
˜K(x − y, t − s) f(y, s) dyds.
9.1.5 Nonhomogeneous Equation with Nonhomogeneous Initial Conditions
Combining two solutions above, we find that the solution of the initial value problem
ut = u + f(x, t) for t > 0, x ∈ Rn
u(x, 0) = g(x) for x ∈ Rn.
(9.6)
is given by
u(x, t) =
Rn
˜K(x − y, t) g(y) dy +
t
0 Rn
˜K(x − y, t − s) f(y, s) dyds.
9.1.6 The Fundamental Solution
Suppose we want to solve the Cauchy problem
ut = Lu x ∈ Rn, t > 0
u(x, 0) = g(x) x ∈ Rn
.
(9.7)
where L is a differential operator in Rn with constant coefficients. Suppose K(x, t) is
a distribution in Rn
for each value of t ≥ 0, K is C1
in t and satisfies
Kt − LK = 0,
K(x, 0) = δ(x).
(9.8)
We call K a fundamental solution for the initial value problem. The solution of
(9.7) is then given by convolution in the space variables:
u(x, t) =
Rn
K(x − y, t) g(y) dy.

For operators of the form ∂t −L, the fundamental solution of the initial value problem,
K(x, t) as defined in (9.8), coincides with the “free space” fundamental solution, which
satisfies
∂t − L K(x, t) = δ(x, t),
provided we extend K(x, t) by zero to t < 0. For the heat equation, consider
˜K(x, t) =
⎧
⎨
⎩
1
(4πt)n/2 e−
|x|2
4t t > 0
0 t ≤ 0.
(9.9)
Notice that ˜K is smooth for (x, t) = (0, 0).
˜K defined as in (9.9), is the fundamental solution of the “free space” heat
equation.
Proof. We need to show:
∂t − K(x, t) = δ(x, t). (9.10)
To verify (9.10) as distributions, we must show that for any v ∈ C∞
0 (Rn+1): 14
Rn+1
˜K(x, t) − ∂t − v dx dt =
Rn+1
δ(x, t) v(x, t) dx dt ≡ v(0, 0).
To do this, let us take > 0 and define
˜K (x, t) =
⎧
⎨
⎩
1
(4πt)n/2 e−
|x|2
4t t >
0 t ≤ .
Then ˜K → ˜K as distributions, so it suffices to show that (∂t − ) ˜K → δ as distribu-
tions. Now
˜K − ∂t − v dx dt =
∞
Rn
˜K(x, t) − ∂t − v(x, t) dx dt
= −
∞
Rn
˜K(x, t) ∂tv(x, t) dx dt −
∞
Rn
˜K(x, t) v(x, t) dx dt
= −
Rn
˜K(x, t) v(x, t) dx
t=∞
t=
+
∞
Rn
∂t
˜K(x, t) v(x, t) dx dt −
∞
Rn
˜K(x, t) v(x, t) dx dt
=
∞
Rn
∂t − ˜K(x, t) v(x, t) dx dt +
Rn
˜K(x, ) v(x, ) dx.
But for t > , (∂t − ) ˜K(x, t) = 0; moreover, since limt→0+ ˜K(x, t) = δ0(x) = δ(x),
we have ˜K(x, ) → δ0(x) as → 0, so the last integral tends to v(0, 0).
14
Note, for the operator L = ∂/∂t, the adjoint operator is L∗
= −∂/∂t.

10 Schrödinger Equation
Problem (F’96, #5). The Gauss kernel
G(t, x, y) =
1
(4πt)
1
2
e−
(x−y)2
4t
is the fundamental solution of the heat equation, solving
Gt = Gxx, G(0, x, y) = δ(x − y).
By analogy with the heat equation, find the fundamental solution H(t, x, y) of the
Schrödinger equation
Ht = iHxx, H(0, x, y) = δ(x − y).
Show that your expression H(x) is indeed the fundamental solution for the
Schrödinger equation. You may use the following special integral
∞
−∞
e
−ix2
4 dx =
√
−i4π.
Proof. • Remark: Consider the initial value problem for the Schrödinger equation
ut = i u x ∈ Rn, t > 0,
u(x, 0) = g(x) x ∈ Rn.
If we formally replace t by it in the heat kernel, we obtain the Fundamental
Solution of the Schrödinger Equation: 15
H(x, t) =
1
(4πit)
n
2
e−
|x|2
4it (x ∈ Rn
, t = 0)
u(x, t) =
1
(4πit)
n
2 Rn
e−
|x−y|2
4it g(y) dy.
In particular, the Schrödinger equation is reversible in time, whereas the heat equation
is not.
• Solution: We have already found the fundamental solution for the heat equation
using the Fourier transform. For the Schrödinger equation is one dimension, we have
∂
∂t
u(ξ, t) = −iξ2
u(ξ, t),
which is an ordinary differential equation in t, with the solution u(ξ, t) = Ce−iξ2t
.
The initial condition u(ξ, 0) = g(ξ) gives
u(ξ, t) = g(ξ) e−iξ2t
,
u(x, t) = g(ξ) e−iξ2t
∨
=
1
√
2π
g ∗ e−iξ2t ∨
=
1
√
2π
g ∗
1
√
2π R
e−iξ2t
eix·ξ
dξ
=
1
2π
g ∗
R
eix·ξ−iξ2t
dξ = (need some work) =
=
1
√
4πit
g ∗ e−
|x|2
4it =
1
√
4πit R
e−
|x−y|2
4it g(y) dy.
15
Evans, p. 188, Example 3.

• For the Schrödinger equation, consider
˜Ψ(x, t) =
⎧
⎨
⎩
1
(4πit)n/2 e−
|x|2
4it t > 0
0 t ≤ 0.
(10.1)
Notice that ˜Ψ is smooth for (x, t) = (0, 0).
˜Ψ defined as in (10.1), is the fundamental solution of the Schrödinger equa-
tion. We need to show:
∂t − i Ψ(x, t) = δ(x, t). (10.2)
To verify (10.2) as distributions, we must show that for any v ∈ C∞
0 (Rn+1): 16
Rn+1
˜Ψ(x, t) − ∂t − i v dx dt =
Rn+1
δ(x, t) v(x, t) dx dt ≡ v(0, 0).
To do this, let us take > 0 and define
˜Ψ (x, t) =
⎧
⎨
⎩
1
(4πit)n/2 e−
|x|2
4it t >
0 t ≤ .
Then ˜Ψ → ˜Ψ as distributions, so it suffices to show that (∂t − i )˜Ψ → δ as distribu-
tions. Now
˜Ψ − ∂t − i v dx dt =
∞
Rn
˜Ψ(x, t) − ∂t − i v(x, t) dx dt
=
∞
Rn
∂t − i ˜Ψ(x, t) v(x, t) dx dt +
Rn
˜Ψ(x, ) v(x, ) dx.
But for t > , (∂t − i )˜Ψ(x, t) = 0; moreover, since limt→0+ ˜Ψ(x, t) = δ0(x) = δ(x),
we have ˜Ψ(x, ) → δ0(x) as → 0, so the last integral tends to v(0, 0).
16
Note, for the operator L = ∂/∂t, the adjoint operator is L∗
= −∂/∂t.

11 Problems: Quasilinear Equations
Problem (F’90, #7). Use the method of characteristics to find the solution of the
first order partial differential equation
x2
ux + xyuy = u2
which passes through the curve u = 1, x = y2. Determine where this solution becomes
singular.
Proof. We have a condition u(x = y2) = 1. Γ is parametrized by Γ : (s2, s, 1).
dx
dt
= x2
⇒ x =
1
−t − c1(s)
⇒ x(0, s) =
1
−c1(s)
= s2
⇒ x =
1
−t + 1
s2
=
s2
1 − ts2
,
dy
dt
= xy ⇒
dy
dt
=
s2y
1 − ts2
⇒ y =
c2(s)
1 − ts2
⇒ y(s, 0) = c2(s) = s ⇒ y =
s
1 − ts2
,
dz
dt
= z2
⇒ z =
1
−t − c3(s)
⇒ z(0, s) =
1
−c3(s)
= 1 ⇒ z =
1
1 − t
.
Thus,
x
y
= s ⇒ y =
x
y
1 − tx2
y2
⇒ t =
y2
x2
−
1
x
.
⇒ u(x, y) =
1
1 − y2
x2 + 1
x
=
x2
x2 + x − y2
.
The solution becomes singular when y2
= x2
+ x.
It can be checked that the solution satisfies the PDE and u(x = y2) = y4
y4+y2 −y2 = 1.
Problem (S’91, #7). Solve the first order PDE
fx + x2
yfy + f = 0
f(x = 0, y) = y2
using the method of characteristics.
Proof. Rewrite the equation
ux + x2
yuy = −u,
u(0, y) = y2
.
Γ is parameterized by Γ : (0, s, s2).
dx
dt
= 1 ⇒ x = t,
dy
dt
= x2
y ⇒
dy
dt
= t2
y ⇒ y = se
t3
3 ,
dz
dt
= −z ⇒ z = s2
e−t
.
Thus, x = t and s = ye−t3
3 = ye−x3
3 , and
u(x, y) = (ye−x3
3 )2
e−x
= y2
e−2
3
x3−x
.
The solution satisfies both the PDE and initial conditions.

Problem (S’92, #1). Consider the Cauchy problem
ut = xux − u + 1 − ∞ < x < ∞, t ≥ 0
u(x, 0) = sinx − ∞ < x < ∞
and solve it by the method of characteristics. Discuss the properties of the solution; in
particular investigate the behavior of |ux(·, t)|∞ for t → ∞.
Proof. Γ is parametrized by Γ : (s, 0, sins). We have
dx
dt
= −x ⇒ x = se−t
,
dy
dt
= 1 ⇒ y = t,
dz
dt
= 1 − z ⇒ z = 1 −
1 − sins
et
.
Thus, t = y, s = xey, and
u(x, y) = 1 −
1
ey
+
sin(xey)
ey
.
It can be checked that the solution satisﬁes the PDE and the initial condition.
As t → ∞, u(x, t) → 1. Also,
|ux(x, y)|∞ = | cos(xey
)|∞ = 1.
ux(x, y) oscillate between −1 and 1. If x = 0, ux = 1.
Problem (W’02, #6). Solve the Cauchy problem
ut + u2
ux = 0, t > 0,
u(0, x) = 2 + x.
Proof. Solved

Problem (S’97, #1). Find the solution of the Burgers’ equation
ut + uux = −x, t ≥ 0
u(x, 0) = f(x), −∞ < x < ∞.
Proof. Γ is parameterized by Γ : (s, 0, f(s)).
dx
dt
= z,
dy
dt
= 1 ⇒ y = t,
dz
dt
= −x.
Note that we have a coupled system:
˙x = z,
˙z = −x,
which can be written as a second order ODE:
¨x + x = 0, x(s, 0) = s, ˙x(s, 0) = z(0) = f(s).
Solving the equation, we get
x(s, t) = s cos t + f(s) sint, and thus,
z(s, t) = ˙x(t) = −s sint + f(s) cos t.
x = s cos y + f(s) siny,
u = −s sin y + f(s) cos y.
⇒
x cos y = s cos2
y + f(s) siny cos y,
u siny = −s sin2
y + f(s) cos y siny.
⇒ x cos y − u siny = s(cos2
y + sin2
y) = s.
⇒ u(x, y) = f(x cosy − u siny) cosy − (x cos y − u siny) siny.
Problem (F’98, #2). Solve the partial diﬀerential equation
uy − u2
ux = 3u, u(x, 0) = f(x)
using method of characteristics. (Hint: ﬁnd a parametric representation of the solu-
tion.)
Proof. Γ is parameterized by Γ : (s, 0, f(s)).
dx
dt
= −z2
⇒
dx
dt
= −f2
(s)e6t
⇒ x = −
1
6
f2
(s)e6t
+
1
6
f2
(s) + s,
dy
dt
= 1 ⇒ y = t,
dz
dt
= 3z ⇒ z = f(s)e3t
.

Thus,
x = −1
6 f2(s)e6y + 1
6 f2(s) + s,
f(s) = z
e3y
⇒ x = −
1
6
z2
e6y
e6y
+
1
6
z2
e6y
+ s =
z2
6e6y
−
z2
6
+ s,
⇒ s = x −
z2
6e6y
+
z2
6
.
⇒ z = f x −
z2
6e6y
+
z2
6
e3y
.
⇒ u(x, y) = f x −
u2
6e6y
+
u2
6
e3y
.

Problem (S’99, #1) Modified Problem. a) Solve
ut +
u3
3 x
= 0 (11.1)
for t > 0, −∞ < x < ∞ with initial data
u(x, 0) = h(x) =
−a(1 − ex), x < 0
−a(1 − e−x), x > 0
where a > 0 is constant. Solve until the first appearance of discontinuous derivative
and determine that critical time.
b) Consider the equation
ut +
u3
3 x
= −cu. (11.2)
How large does the constant c > 0 has to be, so that a smooth solution (with no discon-
tinuities) exists for all t > 0? Explain.
Proof. a) Characteristic form: ut + u2ux = 0. Γ : (s, 0, h(s)).
dx
dt
= z2
,
dy
dt
= 1,
dz
dt
= 0.
x = h(s)2
t + s, y = t, z = h(s).
u(x, y) = h(x − u2
y) (11.3)
The characteristic projection in the xt-plane17 passing through the point (s, 0) is the
line
x = h(s)2
t + s
along which u has the constant value u = h(s).
The derivative of the initial data is discontinuous, and that leads to a
rarefaction-like behavior at t = 0. However, if the question meant to ask to
determine the first time when a shock forms, we proceed as follows.
Two characteristics x = h(s1)2t + s1 and x = h(s2)2t + s2 intersect at a point (x, t)
with
t = −
s2 − s1
h(s2)2 − h(s1)2
.
From (11.3), we have
ux = h (s)(1 − 2uuxt) ⇒ ux =
h (s)
1 + 2h(s)h (s)t
Hence for 2h(s)h (s) < 0, ux becomes infinite at the positive time
t =
−1
2h(s)h (s)
.
The smallest t for which this happens corresponds to the value s = s0 at which h(s)h (s)
has a minimum (i.e.−h(s)h (s) has a maximum). At time T = −1/(2h(s0)h (s0)) the
17

solution u experiences a “gradient catastrophe”.
Therefore, need to ﬁnd a minimum of
f(x) = 2h(x)h (x) =
−2a(1 − ex
) · aex
−2a(1 − e−x) · (−ae−x)
=
−2a2
ex
(1 − ex
), x < 0
2a2e−x(1 − e−x), x > 0
f (x) =
−2a2
ex
(1 − 2ex
), x < 0
−2a2e−x(1 − 2e−x), x > 0
= 0 ⇒
x = ln(1
2) = − ln(2), x < 0
x = ln(2), x > 0
⇒
f(ln(1
2)) = −2a2
eln( 1
2
)
(1 − eln( 1
2
)
) = −2a2
(1
2 )(1
2) = −a2
2 , x < 0
f(ln(2)) = 2a2(1
2 )(1 − 1
2 ) = a2
2 , x > 0
⇒ t = −
1
min{2h(s)h (s)}
=
2
a2
Proof. b) Characteristic form: ut + u2ux = −cu. Γ : (s, 0, h(s)).
dx
dt
= z2
= h(s)2
e−2ct
⇒ x = s +
1
2c
h(s)2
(1 − e−2ct
),
dy
dt
= 1 ⇒ y = t,
dz
dt
= −cz ⇒ z = h(s)e−ct
(⇒ h(s) = uecy
).
Solving for s and t in terms of x and y, we get:
t = y, s = x −
1
2c
h(s)2
(1 − e−2cy
).
Thus,
u(x, y) = h x −
1
2c
u2
e2cy
(1 − e−2cy
) · e−cy
.
ux = h (s)e−cy
· (1 −
1
c
uuxe2cy
(1 − e−2cy
)),
ux =
h (s)e−cy
1 + 1
c h (s)ecyu · (1 − e−2cy)
=
h (s)e−cy
1 + 1
c h (s)h(s)(1 − e−2cy)
.
Thus, c > 0 that would allow a smooth solution to exist for all t > 0 should satisfy
1 +
1
c
h (s)h(s)(1 − e−2cy
) = 0.
We can perform further calculations taking into account the result from part (a):
min{2h(s)h (s)} = −
a2
2
.

Problem (S’99, #1). Original Problem. a). Solve
ut +
u3
x
3
= 0 (11.4)
for t > 0, −∞ < x < ∞ with initial data
u(x, 0) = h(x) =
−a(1 − ex
), x < 0
−a(1 − e−x), x > 0
where a > 0 is constant.
Proof. Rewrite the equation as
F(x, y, u, ux, uy) =
u3
x
3
+ uy = 0,
F(x, y, z, p, q) =
p3
3
+ q = 0.
Γ is parameterized by Γ : (s, 0, h(s), φ(s), ψ(s)).
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
φ(s)3
3
+ ψ(s) = 0,
ψ(s) = −
φ(s)3
3
.
• h (s) = φ(s)f (s) + ψ(s)g (s)
aes = φ(s), x < 0
−ae−s
= φ(s), x > 0
⇒
ψ(s) = −a3e3s
3 , x < 0
ψ(s) = a3e−3s
3 , x > 0
Therefore, now Γ is parametrized by
Γ : (s, 0, −a(1 − es
), aes
, −a3e3s
3 ), x < 0
Γ : (s, 0, −a(1 − e−s
), −ae−s
, a3e−3s
3 ), x > 0
dx
dt
= Fp = p2
=
a2
e2s
a2
e−2s
⇒ x(s, t) =
a2
e2s
t + c4(s)
a2
e−2s
t + c5(s)
⇒ x =
a2
e2s
t + s
a2
e−2s
t + s
dy
dt
= Fq = 1 ⇒ y(s, t) = t + c1(s) ⇒ y = t
dz
dt
= pFp + qFq = p3
+ q =
a3
e3s
− a3e3s
3 = 2
3 a3
e3s
, x < 0
−a3e−3s + a3e−3s
3 = −2
3 a3e−3s, x > 0
⇒ z(s, t) =
2
3 a3
e3s
t + c6(s), x < 0
−2
3 a3e−3st + c7(s), x > 0
⇒ z =
2
3 a3
e3s
t − a(1 − es
), x < 0
−2
3 a3e−3st − a(1 − e−s), x > 0
dp
dt
= −Fx − Fzp = 0 ⇒ p(s, t) = c2(s) ⇒ p =
aes
, x < 0
−ae−s, x > 0

dq
dt
= −Fy − Fzq = 0 ⇒ q(s, t) = c3(s) ⇒ q =
−a3e3s
3 , x < 0
a3e−3s
3 , x > 0
Thus,
u(x, y) =
2
3a3e3sy − a(1 − es), x < 0
−2
3 a3
e−3s
y − a(1 − e−s
), x > 0
where s is deﬁned as
x =
a2e2sy + s, x < 0
a2
e−2s
y + s, x > 0.
b). Solve the equation
ut +
u3
x
3
= −cu. (11.5)
F(x, y, u, ux, uy) =
u3
x
3
+ uy + cu = 0,
F(x, y, z, p, q) =
p3
3
+ q + cz = 0.
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
φ(s)3
3
+ ψ(s) + ch(s) = 0,
ψ(s) = −
φ(s)3
3
− ch(s) =
−
φ(s)3
3 + ca(1 − ex
), x < 0
−φ(s)3
3 + ca(1 − e−x), x > 0
• h (s) = φ(s)f (s) + ψ(s)g (s)
aes
= φ(s), x < 0
−ae−s = φ(s), x > 0
⇒
ψ(s) = −a3e3s
3 + ca(1 − ex
), x < 0
ψ(s) = a3e−3s
3 + ca(1 − e−x), x > 0
Therefore, now Γ is parametrized by
Γ : (s, 0, −a(1 − es), aes, −a3e3s
3 + ca(1 − ex), x < 0
Γ : (s, 0, −a(1 − e−s), −ae−s, a3e−3s
3 + ca(1 − e−x), x > 0

dx
dt
= Fp = p2
dy
dt
= Fq = 1
dz
dt
= pFp + qFq = p3
+ q
dp
dt
= −Fx − Fzp = −cp
dq
dt
= −Fy − Fzq = −cq
We can proceed solving the characteristic equations with initial conditions above.

Problem (S’95, #7). a) Solve the following equation, using characteristics,
ut + u3
ux = 0,
u(x, 0) =
a(1 − ex
), for x < 0
−a(1 − e−x), for x > 0
where a > 0 is a constant. Determine the first time when a shock forms.
Proof. a) Γ is parameterized by Γ : (s, 0, h(s)).
dx
dt
= z3
,
dy
dt
= 1,
dz
dt
= 0.
x = h(s)3
t + s, y = t, z = h(s).
u(x, y) = h(x − u3
y) (11.6)
The characteristic projection in the xt-plane18
passing through the point (s, 0) is the line
x = h(s)3
t + s
along which u has a constant value u = h(s).
Characteristics x = h(s1)3t+s1 and x = h(s2)3t+s2 intersect at a point (x, t) with
t = −
s2 − s1
h(s2)3 − h(s1)3
.
From (11.6), we have
ux = h (s)(1 − 3u2
uxt) ⇒ ux =
h (s)
1 + 3h(s)2h (s)t
Hence for 3h(s)2h (s) < 0, ux becomes infinite at the positive time
t =
−1
3h(s)2h (s)
.
The smallest t for which this happens corresponds to the value s = s0 at which
h(s)2h (s) has a minimum (i.e.−h(s)2h (s) has a maximum). At time T = −1/(3h(s0)2h (s0))
the solution u experiences a “gradient catastrophe”.
Therefore, need to find a minimum of
f(x) = 3h(x)2
h (x) =
−3a2
(1 − ex
)2
aex
= −3a3
ex
(1 − ex
)2
, x < 0
−3a2
(1 − e−x
)2
ae−x
= −3a3
e−x
(1 − e−x
)2
, x > 0
f (x) =
−3a3
ex
(1 − ex
)2
− ex
2(1 − ex
)ex
= −3a3
ex
(1 − ex
)(1 − 3ex
), x < 0
−3a3
− e−x
(1 − e−x
)2
+ e−x
2(1 − e−x
)e−x
= −3a3
e−x
(1 − e−x
)(−1 + 3e−x
), x > 0
= 0
The zeros of f (x) are
x = 0, x = − ln 3, x < 0,
x = 0, x = ln 3, x > 0.
We check which ones give the minimum of f(x) :
⇒
f(0) = −3a3
, f(− ln3) = −3a3 1
3 (1 − 1
3)2
= −4a3
9 , x < 0
f(0) = −3a3, f(ln3) = −3a3 1
3 (1 − 1
3)2 = −4a3
9 , x > 0
18

⇒ t = −
1
min{3h(s)2h (s)}
= −
1
min f(s)
=
1
3a3
.

b) Now consider
ut + u3
ux + cu = 0
with the same initial data and a positive constant c. How large does c need to be in
order to prevent shock formation?
b) Characteristic form: ut + u3ux = −cu. Γ : (s, 0, h(s)).
dx
dt
= z3
= h(s)3
e−3ct
⇒ x = s +
1
3c
h(s)3
(1 − e−3ct
),
dy
dt
= 1 ⇒ y = t,
dz
dt
= −cz ⇒ z = h(s)e−ct
(⇒ h(s) = uecy
).
⇒ z(s, t) = h x −
1
3c
h(s)3
(1 − e−3ct
) e−ct
,
⇒ u(x, y) = h x −
1
3c
u3
e3cy
(1 − e−3cy
) e−cy
.
ux = h (s) · e−cy
· 1 −
1
c
u2
uxe3cy
(1 − e−3cy
) ,
ux =
h (s)e−cy
1 + 1
c h (s)u2e2cy(1 − e−3cy)
=
h (s)e−cy
1 + 1
c h (s)h(s)2(1 − e−3cy)
.
Thus, we need
1 +
1
c
h (s)h(s)2
(1 − e−3cy
) = 0.
We can perform further calculations taking into account the result from part (a):
min{3h(s)2
h (s)} = −3a3
.

Problem (F’99, #4). Consider the Cauchy problem
uy + a(x)ux = 0,
u(x, 0) = h(x).
Give an example of an (unbounded) smooth a(x) for which the solution of the Cauchy
problem is not unique.
Proof. Γ is parameterized by Γ : (s, 0, h(s)).
dx
dt
= a(x) ⇒ x(t) − x(0) =
t
0
a(x)dt ⇒ x =
t
0
a(x)dt + s,
dy
dt
= 1 ⇒ y(s, t) = t + c1(s) ⇒ y = t,
dz
dt
= 0 ⇒ z(s, t) = c2(s) ⇒ z = h(s).
Thus,
u(x, t) = h x −
y
0
a(x)dy
Problem (F’97, #7). a) Solve the Cauchy problem
ut − xuux = 0 − ∞ < x < ∞, t ≥ 0,
u(x, 0) = f(x) − ∞ < x < ∞.
b) Find a class of initial data such that this problem has a global solution for all t.
Compute the critical time for the existence of a smooth solution for initial data, f,
which is not in the above class.
Proof. a) Γ is parameterized by Γ : (s, 0, f(s)).
dx
dt
= −xz ⇒
dx
dt
= −xf(s) ⇒ x = se−f(s)t
,
dy
dt
= 1 ⇒ y = t,
dz
dt
= 0 ⇒ z = f(s).
⇒ z = f xef(s)t
,
⇒ u(x, y) = f xeuy
.
Check:
ux = f (s) · (euy
+ xeuy
uxy)
uy = f (s) · xeuy(uyy + u)
⇒
ux − f (s)xeuy
uxy = f (s)euy
uy − f (s)xeuyuyy = f (s)xeuyu
⇒
ux = f (s)euy
1−f (s)xyeuy
uy =
f (s)euyxu
1−f (s)xyeuy
⇒ uy − xuux =
f (s)euyxu
1 − f (s)xyeuy
− xu
f (s)euy
1 − f (s)xyeuy
= 0.
u(x, 0) = f(x).

b) The characteristics would intersect when 1 − f (s)xyeuy
= 0. Thus,
tc =
1
f (s)xeutc
.

Problem (F’96, #6). Find an implicit formula for the solution u of the initial-value
problem
ut = (2x − 1)tux + sin(πx) − t,
u(x, t = 0) = 0.
Evaluate u explicitly at the point (x = 0.5, t = 2).
uy + (1 − 2x)yux = sin(πx) − y.
Γ is parameterized by Γ : (s, 0, 0).
dx
dt
= (1 − 2x)y = (1 − 2x)t ⇒ x =
1
2
(2s − 1)e−t2
+
1
2
, ⇒ s = (x −
1
2
)et2
+
1
2
,
dy
dt
= 1 ⇒ y = t,
dz
dt
= sin(πx) − y = sin
π
2
(2s − 1)e−t2
+
π
2
− t.
⇒ z(s, t) =
t
0
sin
π
2
(2s − 1)e−t2
+
π
2
− t dt + z(s, 0),
z(s, t) =
t
0
sin
π
2
(2s − 1)e−t2
+
π
2
− t dt.
⇒ u(x, y) =
y
0
sin
π
2
(2s − 1)e−y2
+
π
2
− y dy
=
y
0
sin
π
2
(2x − 1)ey2
e−y2
+
π
2
− y dy
=
y
0
sin
π
2
(2x − 1) +
π
2
− y dy =
y
0
sin(πx) − y dy,
⇒ u(x, y) = y sin(πx) −
y2
2
.
Note: This solution does not satisfy the PDE.
Problem (S’90, #8). Consider the Cauchy problem
ut = xux − u, −∞ < x < ∞, t ≥ 0,
u(x, 0) = f(x), f(x) ∈ C∞
.
Assume that f ≡ 0 for |x| ≥ 1.
Solve the equation by the method of characteristics and discuss the behavior of the
solution.
uy − xux = −u,
Γ is parameterized by Γ : (s, 0, f(s)).
dx
dt
= −x ⇒ x = se−t
,
dy
dt
= 1 ⇒ y = t,
dz
dt
= −z ⇒ z = f(s)e−t
.
⇒ u(x, y) = f(xey
)e−y
.

The solution satisﬁes the PDE and initial conditions.
As y → +∞, u → 0. u = 0 for |xey| ≥ 1 ⇒ u = 0 for |x| ≥ 1
ey .

Problem (F’02, #4). Consider the nonlinear hyperbolic equation
uy + uux = 0 − ∞ < x < ∞.
a) Find a smooth solution to this equation for initial condition u(x, 0) = x.
b) Describe the breakdown of smoothness for the solution if u(x, 0) = −x.
Proof. a) Γ is parameterized by Γ : (s, 0, s).
dx
dt
= z = s ⇒ x = st + s ⇒ s =
x
t + 1
=
x
y + 1
.
dy
dt
= 1 ⇒ y = t,
dz
dt
= 0 ⇒ z = s.
⇒ u(x, y) =
x
y + 1
; solution is smooth for all positive time y.
b) Γ is parameterized by Γ : (s, 0, −s).
dx
dt
= z = −s ⇒ x = −st + s ⇒ s =
x
1 − t
=
x
1 − y
.
dy
dt
= 1 ⇒ y = t,
dz
dt
= 0 ⇒ z = −s.
⇒ u(x, y) =
x
y − 1
; solution blows up at time y = 1.

Problem (F’97, #4). Solve the initial-boundary value problem
ut + (x + 1)2
ux = x for x > 0, t > 0
u(x, 0) = f(x) 0 < x < +∞
u(0, t) = g(t) 0 < t < +∞.
uy + (x + 1)2
ux = x for x > 0, y > 0
u(x, 0) = f(x) 0 < x < +∞
u(0, y) = g(y) 0 < y < +∞.
• For region I, we solve the following characteristic equations with Γ is parameterized
19 by Γ : (s, 0, f(s)).
dx
dt
= (x + 1)2
⇒ x = −
s + 1
(s + 1)t − 1
− 1,
dy
dt
= 1 ⇒ y = t,
dz
dt
= x = −
s + 1
(s + 1)t − 1
− 1,
⇒ z = −ln|(s + 1)t − 1| − t + c1(s),
⇒ z = −ln|(s + 1)t − 1| − t + f(s).
In region I, characteristics are of the form
x = −
s + 1
(s + 1)y − 1
− 1.
Thus, region I is bounded above by the line
x = −
1
y − 1
− 1, or y =
x
x + 1
.
Since t = y, s = x−xy−y
xy+y+1 , we have
u(x, y) = −ln
x − xy − y
xy + y + 1
+ 1 y − 1 − y + f
x − xy − y
xy + y + 1
,
⇒ u(x, y) = −ln
−1
xy + y + 1
− y + f
x − xy − y
xy + y + 1
.
• For region II, Γ is parameterized by Γ : (0, s, g(s)).
dx
dt
= (x + 1)2
⇒ x = −
1
t − 1
− 1,
dy
dt
= 1 ⇒ y = t + s,
dz
dt
= x = −
1
t − 1
− 1,
⇒ z = −ln|t − 1| − t + c2(s),
⇒ z = −ln|t − 1| − t + g(s).
19
Variable t as a third coordinate of u and variable t used to parametrize characteristic equations
are two diﬀerent entities.

Since t = x
x+1 , s = y − x
x+1 , we have
u(x, y) = −ln
x
x + 1
− 1 −
x
x + 1
+ g y −
x
x + 1
.
Note that on y = x
x+1 , both solutions are equal if f(0) = g(0).

Problem (S’93, #3). Solve the following equation
ut + ux + yuy = sint
for 0 ≤ t, 0 ≤ x, −∞ < y < ∞ and with
u = x + y for t = 0, x ≥ 0 and
u = t2
+ y for x = 0, t ≥ 0.
Proof. Rewrite the equation as (x ↔ x1, y ↔ x2, t ↔ x3):
ux3 + ux1 + x2ux2 = sinx3 for 0 ≤ x3, 0 ≤ x1, −∞ < x2 < ∞,
u(x1, x2, 0) = x1 + x2,
u(0, x2, x3) = x2
3 + x2.
• For region I, we solve the following characteristic equations with Γ is parameterized
20
by Γ : (s1, s2, 0, s1 + s2).
dx1
dt
= 1 ⇒ x1 = t + s1,
dx2
dt
= x2 ⇒ x2 = s2et
,
dx3
dt
= 1 ⇒ x3 = t,
dz
dt
= sin x3 = sin t
⇒ z = − cos t + s1 + s2 + 1.
Since in region I, in x1x3-plane, characteristics are of the form x1 = x3 + s1, region
I is bounded above by the line x1 = x3. Since t = x3, s1 = x1 − x3, s2 = x2e−x3 , we
have
u(x1, x2, x3) = − cos x3 + x1 − x3 + x2e−x3
+ 1, or
u(x, y, t) = − cos t + x − t + ye−t
+ 1, x ≥ t.
• For region II, we solve the following characteristic equations with Γ is parameterized
by Γ : (0, s2, s3, s2 + s2
3).
dx1
dt
= 1 ⇒ x1 = t,
dx2
dt
= x2 ⇒ x2 = s2et
,
dx3
dt
= 1 ⇒ x3 = t + s3,
dz
dt
= sin x3 = sin(t + s3) ⇒ z = − cos(t + s3) + cos s3 + s2 + s2
3.
Since t = x1, s3 = x3 − x1, s2 = x2e−x3 , we have
u(x1, x2, x3) = − cos x3 + cos(x3 − x1) + x2e−x3
+ (x3 − x1)2
, or
u(x, y, t) = − cos t + cos(t − x) + ye−t
+ (t − x)2
, x ≤ t.
Note that on x = t, both solutions are u(x = t, y) = − cos x + ye−x
+ 1.
20

Problem (W’03, #5). Find a solution to
xux + (x + y)uy = 1
which satisﬁes u(1, y) = y for 0 ≤ y ≤ 1. Find a region in {x ≥ 0, y ≥ 0} where u is
uniquely determined by these conditions.
Proof. Γ is parameterized by Γ : (1, s, s).
dx
dt
= x ⇒ x = et
.
dy
dt
= x + y ⇒ y − y = et
.
dz
dt
= 1 ⇒ z = t + s.
The homogeneous solution for the second equation is yh(s, t) = c1(s)et
. Since the
right hand side and yh are linearly dependent, our guess for the particular solution is
yp(s, t) = c2(s)tet
. Plugging in yp into the diﬀerential equation, we get
c2(s)tet
+ c2(s)et
− c2(s)tet
= et
⇒ c2(s) = 1.
Thus, yp(s, t) = tet and
y(s, t) = yh + yp = c1(s)et
+ tet
.
Since y(s, 0) = s = c1(s), we get
y = set
+ tet
.
With and , we can solve for s and t in terms of x and y to get
t = ln x,
y = sx + x ln x ⇒ s =
y − x ln x
x
.
u(x, y) = t + s = ln x +
y − x ln x
x
.
u(x, y) =
y
x
.
We have found that the characteristics in the xy-plane are of the form
y = sx + x ln x,
where s is such that 0 ≤ s ≤ 1. Also, the characteristics originate from Γ.
Thus, u is uniquely determined in the region between the graphs:
y = x ln x,
y = x + x ln x.

12 Problems: Shocks
Example 1. Determine the exact solution to Burgers’ equation
ut +
1
2
u2
x
= 0, t > 0
with initial data
u(x, 0) = h(x) =
⎧
⎪⎨
⎪⎩
1 if x < −1,
0 if − 1 < x < 1,
−1 if x > 1.
Proof. Characteristic form: ut + uux = 0.
The characteristic projection in xt-plane passing through the point (s, 0) is the line
x = h(s)t + s.
• Rankine-Hugoniot shock condition at s = −1:
shock speed: ξ (t) =
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
0 − 1
2
0 − 1
=
1
2
.
The “1/slope” of the shock curve = 1/2. Thus,
x = ξ(t) =
1
2
t + s,
and since the jump occurs at (−1, 0), ξ(0) = −1 = s. Therefore,
x =
1
2
t − 1.
• Rankine-Hugoniot shock condition at s = 1:
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
1
2 − 0
−1 − 0
= −
1
2
.
The “1/slope” of the shock curve = −1/2. Thus,
x = ξ(t) = −
1
2
t + s,
and since the jump occurs at (1, 0), ξ(0) = 1 = s. Therefore,
x = −
1
2
t + 1.
• At t = 2, Rankine-Hugoniot shock condition at s = 0:
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
1
2 − 1
2
−1 − 1
= 0.
The “1/slope” of the shock curve = 0. Thus,
x = ξ(t) = s,
and since the jump occurs at (x, t) = (0, 2), ξ(2) = 0 = s. Therefore,
x = 0.

➡ For t < 2, u(x, t) =
⎧
⎪⎨
⎪⎩
1 if x < 1
2 t − 1,
0 if 1
2 t − 1 < x < −1
2 t + 1,
−1 if x > −1
2 t + 1.
➡ and for t > 2, u(x, t) =
1 if x < 0,
−1 if x > 0.

ut +
1
2
u2
x
= 0, t > 0
with initial data
u(x, 0) = h(x) =
⎧
⎪⎨
⎪⎩
−1 if x < −1,
0 if − 1 < x < 1,
1 if x > 1.
x = h(s)t + s.
For Burgers’ equation, for a rarefaction fan emanating from (s, 0) on xt-plane, we have:
u(x, t) =
⎧
⎪⎨
⎪⎩
ul, x−s
t ≤ ul,
x−s
t , ul ≤ x−s
t ≤ ur,
ur, x−s
t ≥ ur.
➡ u(x, t) =
⎧
⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎩
−1, x < −t − 1,
x+1
t , −t − 1 < x < −1, i.e. − 1 < x+1
t < 0
0, −1 < x < 1,
x−1
t , 1 < x < t + 1, i.e. 0 < x−1
t < 1
1, x > t + 1.

ut +
1
2
u2
x
= 0, t > 0
with initial data
u(x, 0) = h(x) =
2 if 0 < x < 1,
0 if otherwise.
x = h(s)t + s.
• Shock: Rankine-Hugoniot shock condition at s = 1:
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
0 − 2
0 − 2
= 1.
The “1/slope” of the shock curve = 1. Thus,
x = ξ(t) = t + s,
and since the jump occurs at (1, 0), ξ(0) = 1 = s. Therefore,
x = t + 1.
• Rarefaction: A rarefaction emanates from (0, 0) on xt-plane.
➡ For 0 < t < 1, u(x, t) =
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
0 if x < 0,
x
t if 0 < x < 2t,
2 if 2t < x < t + 1.
0 if x > t + 1.
Rarefaction catches up to shock at t = 1.
• Shock: At (x, t) = (2, 1), ul = x/t, ur = 0. Rankine-Hugoniot shock condition:
ξ (t) =
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
0 − 1
2 (x
t )2
0 − x
t
=
1
2
x
t
,
dxs
dt
=
x
2t
,
x = c
√
t,
and since the jump occurs at (x, t) = (2, 1), x(1) = 2 = c. Therefore, x = 2
√
t.
➡ For t > 1, u(x, t) =
⎧
⎪⎨
⎪⎩
0 if x < 0,
x
t if 0 < x < 2
√
t,
0 if x > 2
√
t.

ut +
1
2
u2
x
= 0, t > 0
with initial data
u(x, 0) = h(x) =
1 + x if x < 0,
0 if x > 0.
x = h(s)t + s.
➀ For s > 0, the characteristics are x = s.
➁ For s < 0, the characteristics are x = (1 + s)t + s.
• There are two ways to look for the solution on the left half-plane. One is to notice
that the characteristic at s = 0− is x = t and characteristic at s = −1 is x = −1 and
that characteristics between s = −∞ and s = 0−
are intersecting at (x, t) = (−1, −1).
Also, for a fixed t, u is a linear function of x, i.e. for t = 0, u = 1 + x, allowing
a continuous change of u with x. Thus, the solution may be viewed as an ‘implicit’
rarefaction, originating at (−1, −1), thus giving rise to the solution
u(x, t) =
x + 1
t + 1
.
Another way to find a solution on the left half-plane is to solve ➁ for s to find
s =
x − t
1 + t
. Thus, u(x, t) = h(s) = 1 + s = 1 +
x − t
1 + t
=
x + 1
t + 1
.
• Shock: At (x, t) = (0, 0), ul = x+1
t+1 , ur = 0. Rankine-Hugoniot shock condition:
ξ (t) =
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
0 − 1
2 (x+1
t+1 )2
0 − x+1
t+1
=
1
2
x + 1
t + 1
,
dxs
dt
=
1
2
x + 1
t + 1
,
x = c
√
t + 1 − 1,
and since the jump occurs at (x, t) = (0, 0), x(0) = 0 = c − 1, or c = 1. Therefore,
the shock curve is x =
√
t + 1 − 1.
➡ u(x, t) =
x+1
t+1 if x <
√
t + 1 − 1,
0 if x >
√
t + 1 − 1.

ut +
1
2
u2
x
= 0, t > 0
with initial data
u(x, 0) = h(x) =
⎧
⎪⎨
⎪⎩
u0 if x < 0,
u0 · (1 − x) if 0 < x < 1,
0 if x ≥ 1,
where u0 > 0.
x = h(s)t + s.
➀ For s > 1, the characteristics are x = s.
➁ For 0 < s < 1, the characteristics are x = u0(1 − s)t + s.
➂ For s < 0, the characteristics are x = u0t + s.
The characteristics emanating from (s, 0), 0 < s < 1 on xt-plane intersect at (1, 1
u0
).
Also, we can check that the characteristics do not intersect before t = 1
u0
for this
problem:
tc = min
−1
h (s)
=
1
u0
.
• To ﬁnd solution in a triangular domain between x = u0t and x = 1, we note that
characteristics there are x = u0 · (1 − s)t + s. Solving for s we get
s =
x − u0t
1 − u0t
. Thus, u(x, t) = h(s) = u0 · (1 − s) = u0 · 1 −
x − u0t
1 − u0t
=
u0 · (1 − x)
1 − u0t
.
We can also ﬁnd a solution in the triangular domain as follows. Note, that the charac-
teristics are the straight lines
dx
dt
= u = const.
Integrating the equation above, we obtain
x = ut + c
Since all characteristics in the triangular domain meet at (1, 1
u0
), we have c = 1 − u
u0
,
and
x = ut + 1 −
u
u0
or u =
u0 · (1 − x)
1 − u0t
.
➡ For 0 < t <
1
u0
, u(x, t) =
⎧
⎪⎨
⎪⎩
u0 if x < u0t,
u0·(1−x)
1−u0t if u0t < x < 1,
0 if x > 1.

• Shock: At (x, t) = (1, 1
u0
), Rankine-Hugoniot shock condition:
ξ (t) =
F(ur) − F(ul)
ur − ul
=
1
2 u2
r − 1
2 u2
l
ur − ul
=
0 − 1
2 u2
0
0 − u0
=
1
2
u0,
ξ(t) =
1
2
u0t + c,
and since the jump occurs at (x, t) = (1, 1
u0
), x 1
u0
= 1 = 1
2 +c, or c = 1
2. Therefore,
the shock curve is x = u0t+1
2 .

➡ For t >
1
u0
, u(x, t) =
u0 if x < u0t+1
2 ,
0 if x > u0t+1
2 .
Problem. Show that for u = f(x/t) to be a nonconstant solution of ut + a(u)ux = 0,
f must be the inverse of the function a.
Proof. If u = f(x/t),
ut = −f
x
t
·
x
t2
and ux = f
x
t
·
1
t
.
Hence, ut + a(u)ux = 0 implies that
−f
x
t
·
x
t2
+ a f
x
t
f
x
t
·
1
t
= 0
or, assuming f is not identically 0 to rule out the constant solution, that
a f
x
t
=
x
t
.
This shows the functions a and f to be inverses of each other.

13 Problems: General Nonlinear Equations
13.1 Two Spatial Dimensions
Problem (S’01, #3). Solve the initial value problem
1
2
u2
x − uy = −
x2
2
,
u(x, 0) = x.
You will ﬁnd that the solution blows up in ﬁnite time. Explain this in terms of the
characteristics for this equation.
F(x, y, z, p, q) =
p2
2
− q +
x2
2
= 0.
Γ is parameterized by Γ : (s, 0, s, φ(s), ψ(s)).
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
F(s, 0, s, φ(s), ψ(s)) = 0,
φ(s)2
2
− ψ(s) +
s2
2
= 0,
ψ(s) =
φ(s)2
+ s2
2
.
• h (s) = φ(s)f (s) + ψ(s)g (s),
1 = φ(s).
⇒ ψ(s) =
s2
+ 1
2
.
Therefore, now Γ is parametrized by Γ : (s, 0, s, 1, s2+1
2 ).
dx
dt
= Fp = p,
dy
dt
= Fq = −1 ⇒ y(s, t) = −t + c1(s) ⇒ y = −t,
dz
dt
= pFp + qFq = p2
− q,
dp
dt
= −Fx − Fzp = −x,
dq
dt
= −Fy − Fzq = 0 ⇒ q(s, t) = c2(s) ⇒ q =
s2
+ 1
2
.
Thus, we found y and q in terms of s and t. Note that we have a coupled system:
x = p,
p = −x,
which can be written as two second order ODEs:
x + x = 0, x(s, 0) = s, x (s, 0) = p(s, 0) = 1,
p + p = 0, p(s, 0) = 1, p (s, 0) = −x(s, 0) = −s.

Solving the two equations separately, we get
x(s, t) = s · cos t + sint,
p(s, t) = cos t − s · sint.
From this, we get
dz
dt
= p2
− q = cos t − s · sint
2
−
s2 + 1
2
= cos2
t − 2s cos t sint + s2
sin2
t −
s2 + 1
2
.
z(s, t) =
t
0
cos2
t − 2s cos t sint + s2
sin2
t −
s2
+ 1
2
dt + z(s, 0),
z(s, t) =
t
2
+
sint cos t
2
+ s cos2
t +
s2t
2
−
s2 sint cos t
2
−
t(s2 + 1)
2
t
0
+ s,
=
sin t cos t
2
+ s cos2
t −
s2 sint cos t
2
t
0
+ s,
=
sint cos t
2
+ s cos2
t −
s2 sin t cos t
2
− s + s =
=
sint cos t
2
+ s cos2
t −
s2
sin t cos t
2
.
Plugging in x and y found earlier for s and t, we get
u(x, y) =
sin(−y) cos(−y)
2
+
x − sin(−y)
cos(−y)
cos2
(−y) −
(x − sin(−y))2
cos2(−y)
·
sin(−y) cos(−y)
2
= −
sin y cos y
2
+
x + siny
cos y
cos2
y +
(x + sin y)2
cos2 y
·
siny cos y
2
= −
sin y cos y
2
+ (x + sin y) cosy +
(x + siny)2 sin y
2 cosy
= x cos y +
sin y cos y
2
+
(x + sin y)2 siny
2 cos y
.

Problem (S’98, #3). Find the solution of
ut +
u2
x
2
=
−x2
2
, t ≥ 0, −∞ < x < ∞
u(x, 0) = h(x), −∞ < x < ∞,
where h(x) is smooth function which vanishes for |x| large enough.
F(x, y, z, p, q) =
p2
2
+ q +
x2
2
= 0.
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
F(s, 0, h(s), φ(s), ψ(s)) = 0,
φ(s)2
2
+ ψ(s) +
s2
2
= 0,
ψ(s) = −
φ(s)2
+ s2
2
.
• h (s) = φ(s)f (s) + ψ(s)g (s),
h (s) = φ(s).
⇒ ψ(s) = −
h (s)2
+ s2
2
.
Therefore, now Γ is parametrized by Γ : (s, 0, s, h (s), −h (s)2+s2
2 ).
dx
dt
= Fp = p,
dy
dt
= Fq = 1 ⇒ y(s, t) = t + c1(s) ⇒ y = t,
dz
dt
= pFp + qFq = p2
+ q,
dp
dt
= −Fx − Fzp = −x,
dq
dt
= −Fy − Fzq = 0 ⇒ q(s, t) = c2(s) ⇒ q = −
h (s)2 + s2
2
.
Thus, we found y and q in terms of s and t. Note that we have a coupled system:
x = p,
p = −x,
which can be written as a second order ODE:
x + x = 0, x(s, 0) = s, x (s, 0) = p(s, 0) = h (s).
Solving the equation, we get
x(s, t) = s cos t + h (s) sint,
p(s, t) = x (s, t) = h (s) cos t − s sin t.

From this, we get
dz
dt
= p2
+ q = h (s) cost − s sint
2
−
h (s)2 + s2
2
= h (s)2
cos2
t − 2sh (s) cost sin t + s2
sin2
t −
h (s)2
+ s2
2
.
z(s, t) =
t
0
h (s)2
cos2
t − 2sh (s) cos t sint + s2
sin2
t −
h (s)2
+ s2
2
dt + z(s, 0)
=
t
0
h (s)2
cos2
t − 2sh (s) cos t sint + s2
sin2
t −
h (s)2
+ s2
2
dt + h(s).
We integrate the above expression similar to S 01#3 to get an expression for z(s, t).
Plugging in x and y found earlier for s and t, we get u(x, y).

Problem (S’97, #4).
Describe the method of the bicharacteristics for solving the initial value problem
∂
∂x
u(x, y)
2
+
∂
∂y
u(x, y)
2
= 2 + y,
u(x, 0) = u0(x) = x.
Assume that | ∂
∂xu0(x)| < 2 and consider the solution such that ∂u
∂y > 0.
Apply all general computations for the particular case u0(x) = x.
Proof. We have
u2
x + u2
y = 2 + y
u(x, 0) = u0(x) = x.
Rewrite the equation as
F(x, y, z, p, q) = p2
+ q2
− y − 2 = 0.
Γ is parameterized by Γ : (s, 0, s, φ(s), ψ(s)).
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
F(s, 0, s, φ(s), ψ(s)) = 0,
φ(s)2
+ ψ(s)2
− 2 = 0,
φ(s)2
+ ψ(s)2
= 2.
• h (s) = φ(s)f (s) + ψ(s)g (s),
1 = φ(s).
⇒ ψ(s) = ±1.
Since we have a condition that q(s, t) > 0, we choose q(s, 0) = ψ(s) = 1.
Therefore, now Γ is parametrized by Γ : (s, 0, s, 1, 1).
dx
dt
= Fp = 2p ⇒
dx
dt
= 2 ⇒ x = 2t + s,
dy
dt
= Fq = 2q ⇒
dy
dt
= 2t + 2 ⇒ y = t2
+ 2t,
dz
dt
= pFp + qFq = 2p2
+ 2q2
= 2y + 4 ⇒
dz
dt
= 2t2
+ 4t + 4,
⇒ z =
2
3
t3
+ 2t2
+ 4t + s =
2
3
t3
+ 2t2
+ 4t + x − 2t =
2
3
t3
+ 2t2
+ 2t + x,
dp
dt
= −Fx − Fzp = 0 ⇒ p = 1,
dq
dt
= −Fy − Fzq = 1 ⇒ q = t + 1.
We solve y = t2
+ 2t, a quadratic equation in t, t2
+ 2t − y = 0, for t in terms of y to
get:
t = −1 ± 1 + y.
⇒ u(x, y) =
2
3
(−1 ± 1 + y)3
+ 2(−1 ± 1 + y)2
+ 2(−1 ± 1 + y) + x.
Both u± satisfy the PDE. ux = 1, uy = ±
√
y + 1 ⇒ u2
x + u2
y = y + 2
u+ satisﬁes u+(x, 0) = x . However, u− does not satisfy IC, i.e. u−(x, 0) = x−4
3.

Problem (S’02, #6). Consider the equation
ux + uxuy = 1,
u(x, 0) = f(x).
Assuming that f is differentiable, what conditions on f insure that the problem is
noncharacteristic? If f satisfies those conditions, show that the solution is
u(x, y) = f(r) − y +
2y
f (r)
,
where r must satisfy y = (f (r))2
(x − r).
Finally, show that one can solve the equation for (x, y) in a sufficiently small neighbor-
hood of (x0, 0) with r(x0, 0) = x0.
Proof. Solved.
In order to solve the Cauchy problem in a neighborhood of Γ, need:
f (s) · Fq[f, g, h, φ, ψ](s) − g (s) · Fp[f, g, h, φ, ψ](s) = 0,
1 · h (s) − 0 · 1 +
1 − h (s)
h (s)
= 0,
h (s) = 0.
Thus, h (s) = 0 ensures that the problem is noncharacteristic.
To show that one can solve y = (f (s))2(x − s) for (x, y) in a sufficiently small
neighborhood of (x0, 0) with s(x0, 0) = x0, let
G(x, y, s) = (f (s))2
(x − s) − y = 0,
G(x0, 0, x0) = 0,
Gr(x0, 0, x0) = −(f (s))2
.
Hence, if f (s) = 0, ∀s, then Gs(x0, 0, x0) = 0 and we can use the implicit function
theorem in a neighborhood of (x0, 0, x0) to get
G(x, y, h(x, y)) = 0
and solve the equation in terms of x and y.

Problem (S’00, #1). Find the solutions of
(ux)2
+ (uy)2
= 1
in a neighborhood of the curve y = x2
2 satisfying the conditions
u x,
x2
2
= 0 and uy x,
x2
2
> 0.
Leave your answer in parametric form.
F(x, y, z, p, q) = p2
+ q2
− 1 = 0.
Γ is parameterized by Γ : (s, s2
2 , 0, φ(s), ψ(s)).
• F(f(s), g(s), h(s), φ(s), ψ(s)) = 0,
F s,
s2
2
, 0, φ(s), ψ(s) = 0,
φ(s)2
+ ψ(s)2
= 1.
• h (s) = φ(s)f (s) + ψ(s)g (s),
0 = φ(s) + sψ(s),
φ(s) = −sψ(s).
Thus, s2
ψ(s)2
+ ψ(s)2
= 1 ⇒ ψ(s)2
=
1
s2 + 1
.
Since, by assumption, ψ(s) > 0, we have ψ(s) = 1√
s2+1
.
Therefore, now Γ is parametrized by Γ : s, s2
2 , 0, −s√
s2+1
, 1√
s2+1
.
dx
dt
= Fp = 2p =
−2s
√
s2 + 1
⇒ x =
−2st
√
s2 + 1
+ s,
dy
dt
= Fq = 2q =
2
√
s2 + 1
⇒ y =
2t
√
s2 + 1
+
s2
2
,
dz
dt
= pFp + qFq = 2p2
+ 2q2
= 2 ⇒ z = 2t,
dp
dt
= −Fx − Fzp = 0 ⇒ p =
−s
√
s2 + 1
,
dq
dt
= −Fy − Fzq = 0 ⇒ q =
1
√
s2 + 1
.
Thus, in parametric form,
z(s, t) = 2t,
x(s, t) =
−2st
√
s2 + 1
+ s,
y(s, t) =
2t
√
s2 + 1
+
s2
2
.

13.2 Three Spatial Dimensions
Problem (S’96, #2). Solve the following Cauchy problem21:
ux + u2
y + u2
z = 1,
u(0, y, z) = y · z.
ux1 + u2
x2
+ u2
x3
= 1,
u(0, x2, x3) = x2 · x3.
Write a general nonlinear equation
F(x1, x2, x3, z, p1, p2, p3) = p1 + p2
2 + p2
3 − 1 = 0.
Γ : 0
x1(s1,s2,0)
, s1
x2(s1,s2,0)
, s2
x3(s1,s2,0)
, s1s2
z(s1,s2,0)
, φ1(s1, s2)
p1(s1,s2,0)
, φ2(s1, s2)
p2(s1,s2,0)
, φ3(s1, s2)
p3(s1,s2,0)
• F f1(s1, s2), f2(s1, s2), f3(s1, s2), h(s1, s2), φ1, φ2, φ3 = 0,
F 0, s1, s2, s1s2, φ1, φ2, φ3 = φ1 + φ2
2 + φ2
3 − 1 = 0,
⇒ φ1 + φ2
2 + φ2
3 = 1.
•
∂h
∂s1
= φ1
∂f1
∂s1
+ φ2
∂f2
∂s1
+ φ3
∂f3
∂s1
,
⇒ s2 = φ2.
•
∂h
∂s2
= φ1
∂f1
∂s2
+ φ2
∂f2
∂s2
+ φ3
∂f3
∂s2
,
⇒ s1 = φ3.
Thus, we have: φ2 = s2, φ3 = s1, φ1 = −s2
1 − s2
2 + 1.
Γ : 0
x1(s1,s2,0)
, s1
x2(s1,s2,0)
, s2
x3(s1,s2,0)
, s1s2
z(s1,s2,0)
, −s2
1 − s2
2 + 1
p1(s1,s2,0)
, s2
p2(s1,s2,0)
, s1
p3(s1,s2,0)
21
This problem is very similar to an already hand-written solved problem F’95 #2.

dx1
dt
= Fp1 = 1 ⇒ x1 = t,
dx2
dt
= Fp2 = 2p2 ⇒
dx2
dt
= 2s2 ⇒ x2 = 2s2t + s1,
dx3
dt
= Fp3 = 2p3 ⇒
dx3
dt
= 2s1 ⇒ x3 = 2s1t + s2,
dz
dt
= p1Fp1 + p2Fp2 + p3Fp3 = p1 + 2p2
2 + 2p2
3 = −s2
1 − s2
2 + 1 + 2s2
2 + 2s2
1
= s2
1 + s2
2 + 1 ⇒ z = (s2
1 + s2
2 + 1)t + s1s2,
dp1
dt
= −Fx1 − p1Fz = 0 ⇒ p1 = −s2
1 − s2
2 + 1,
dp2
dt
= −Fx2 − p2Fz = 0 ⇒ p2 = s2,
dp3
dt
= −Fx3 − p3Fz = 0 ⇒ p3 = s1.
Thus, we have
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
x1 = t
x2 = 2s2t + s1
x3 = 2s1t + s2
z = (s2
1 + s2
2 + 1)t + s1s2
⇒
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
t = x1
s1 = x2 − 2s2t
s2 = x3 − 2s1t
z = (s2
1 + s2
2 + 1)t + s1s2
⇒
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
t = x1
s1 = x2−2x1x3
1−4x2
1
s2 = x3−2x1x2
1−4x2
1
z = (s2
1 + s2
2 + 1)t + s1s2
⇒ u(x1, x2, x3) =
x2 − 2x1x3
1 − 4x2
1
2
+
x3 − 2x1x2
1 − 4x2
1
2
+ 1 x1 +
x2 − 2x1x3
1 − 4x2
1
x3 − 2x1x2
1 − 4x2
1
.
Problem (F’95, #2). Solve the following Cauchy problem
ux + uy + u3
z = x + y + z,
u(x, y, 0) = xy.
Proof. Solved

Problem (S’94, #1). Solve the following PDE for f(x, y, t):
ft + xfx + 3t2
fy = 0
f(x, y, 0) = x2
+ y2
.
Proof. Rewrite the equation as (x → x1, y → x2, t → x3, f → u):
x1ux1 + 3x2
3ux2 + ux3 = 0,
u(x1, x2, 0) = x2
1 + x2
2.
F(x1, x2, x3, z, p1, p2, p3) = x1p1 + 3x2
3p2 + p3 = 0.
Γ : s1
x1(s1,s2,0)
, s2
x2(s1,s2,0)
, 0
x3(s1,s2,0)
, s2
1 + s2
2
z(s1,s2,0)
, φ1(s1, s2)
p1(s1,s2,0)
, φ2(s1, s2)
p2(s1,s2,0)
, φ3(s1, s2)
p3(s1,s2,0)
F s1, s2, 0, s2
1 + s2
2, φ1, φ2, φ3 = s1φ1 + φ3 = 0,
⇒ φ3 = s1φ1.
•
∂h
∂s1
= φ1
∂f1
∂s1
+ φ2
∂f2
∂s1
+ φ3
∂f3
∂s1
,
⇒ 2s1 = φ1.
•
∂h
∂s2
= φ1
∂f1
∂s2
+ φ2
∂f2
∂s2
+ φ3
∂f3
∂s2
,
⇒ 2s2 = φ2.
Thus, we have: φ1 = 2s1, φ2 = 2s2, φ3 = 2s2
1.
Γ : s1
x1(s1,s2,0)
, s2
x2(s1,s2,0)
, 0
x3(s1,s2,0)
, s2
1 + s2
2
z(s1,s2,0)
, 2s1
p1(s1,s2,0)
, 2s2
p2(s1,s2,0)
, 2s2
1
p3(s1,s2,0)
dx1
dt
= Fp1 = x1 ⇒ x1 = s1et
,
dx2
dt
= Fp2 = 3x2
3 ⇒
dx2
dt
= 3t2
⇒ x2 = t3
+ s2,
dx3
dt
= Fp3 = 1 ⇒ x3 = t,
dz
dt
= p1Fp1 + p2Fp2 + p3Fp3 = p1x1 + p23x2
3 + p3 = 0 ⇒ z = s2
1 + s2
2,
dp1
dt
= −Fx1 − p1Fz = −p1 ⇒ p1 = 2s1e−t
,
dp2
dt
= −Fx2 − p2Fz = 0 ⇒ p2 = 2s2,
dp3
dt
= −Fx3 − p3Fz = −6x3p2 ⇒
dp3
dt
= −12ts2 ⇒ p3 = −6t2
s2 + 2s2
1.
With t = x3, s1 = x1e−x3
, s2 = x2 − x3
3, we have
u(x1, x2, x3) = x2
1e−2x3
+ (x2 − x3
3)2
. f(x, y, t) = x2
e−2t
+ (y − t3
)2
.
The solution satisﬁes the PDE and initial condition.

Problem (F’93, #3). Find the solution of the following equation
ft + xfx + (x + t)fy = t3
f(x, y, 0) = xy.
Proof. Rewrite the equation as (x → x1, y → x2, t → x3, f → u):
x1ux1 + (x1 + x3)ux2 + ux3 = x3
,
u(x1, x2, 0) = x1x2.
Method I: Treat the equation as a QUASILINEAR equation.
Γ is parameterized by Γ : (s1, s2, 0, s1s2).
dx1
dt
= x1 ⇒ x1 = s1et
,
dx2
dt
= x1 + x3 ⇒
dx2
dt
= s1et
+ t ⇒ x2 = s1et
+
t2
2
+ s2 − s1,
dx3
dt
= 1 ⇒ x3 = t,
dz
dt
= x3
3 ⇒
dz
dt
= t3
⇒ z =
t4
4
+ s1s2.
Since t = x3, s1 = x1e−x3 , s2 = x2 − s1et
− t2
2 + s1 = x2 − x1 −
x2
3
2 + x1e−x3 , we have
u(x1, x2, x3) =
x4
3
4
+ x1e−x3
(x2 − x1 −
x2
3
2
+ x1e−x3
), or
f(x, y, t) =
t4
4
+ xe−t
(y − x −
t2
2
+ xe−t
).
Method II: Treat the equation as a fully NONLINEAR equation.
F(x1, x2, x3, z, p1, p2, p3) = x1p1 + (x1 + x3)p2 + p3 − x3
3 = 0.
Γ : s1
x1(s1,s2,0)
, s2
x2(s1,s2,0)
, 0
x3(s1,s2,0)
, s1s2
z(s1,s2,0)
, φ1(s1, s2)
p1(s1,s2,0)
, φ2(s1, s2)
p2(s1,s2,0)
, φ3(s1, s2)
p3(s1,s2,0)
F s1, s2, 0, s1s2, φ1, φ2, φ3 = s1φ1 + s1φ2 + φ3 = 0,
⇒ φ3 = −s1(φ1 + φ2).
•
∂h
∂s1
= φ1
∂f1
∂s1
+ φ2
∂f2
∂s1
+ φ3
∂f3
∂s1
,
⇒ s2 = φ1.
•
∂h
∂s2
= φ1
∂f1
∂s2
+ φ2
∂f2
∂s2
+ φ3
∂f3
∂s2
,
⇒ s1 = φ2.
Thus, we have: φ1 = s2, φ2 = s1, φ3 = −s2
1 − s1s2.
Γ : s1
x1(s1,s2,0)
, s2
x2(s1,s2,0)
, 0
x3(s1,s2,0)
, s1s2
z(s1,s2,0)
, s2
p1(s1,s2,0)
, s1
p2(s1,s2,0)
, −s2
1 − s1s2
p3(s1,s2,0)

dx1
dt
= Fp1 = x1 ⇒ x1 = s1et
,
dx2
dt
= Fp2 = x1 + x3 ⇒
dx2
dt
= s1et
+ t ⇒ x2 = s1et
+
t2
2
+ s2 − s1,
dx3
dt
= Fp3 = 1 ⇒ x3 = t,
dz
dt
= p1Fp1 + p2Fp2 + p3Fp3 = p1x1 + p2(x1 + x3) + p3 = x3
3 = t3
⇒ z =
t4
4
+ s1s2,
dp1
dt
= −Fx1 − p1Fz = −p1 − p2 = −p1 − s1 ⇒ p1 = 2s1e−t
− s1,
dp2
dt
= −Fx2 − p2Fz = 0 ⇒ p2 = s1,
dp3
dt
= −Fx3 − p3Fz = 3x2
3 − p2 = 3t2
− s1 ⇒ p3 = t3
− s1t − s2
1 − s1s2.
With t = x3, s1 = x1e−x3
, s2 = x2 − s1et
− t2
2 + s1 = x2 − x1 −
x2
3
2 + x1e−x3
, we have
u(x1, x2, x3) =
x4
3
4
+ x1e−x3
(x2 − x1 −
x2
3
2
+ x1e−x3
), or
f(x, y, t) =
t4
4
+ xe−t
(y − x −
t2
2
+ xe−t
).
22 The solution satisﬁes the PDE and initial condition.
22
Variable t in the derivatives of characteristics equations and t in the solution f(x,y, t) are diﬀerent
entities.

Problem (F’92, #1). Solve the initial value problem
ut + αux + βuy + γu = 0 for t > 0
u(x, y, 0) = ϕ(x, y),
in which α, β and γ are real constants and ϕ is a smooth function.
Proof. Rewrite the equation as (x → x1, y → x2, t → x3)23:
αux1 + βux2 + ux3 = −γu,
u(x1, x2, 0) = ϕ(x1, x2).
Γ is parameterized by Γ : (s1, s2, 0, ϕ(s1, s2)).
dx1
dt
= α ⇒ x1 = αt + s1,
dx2
dt
= β ⇒ x2 = βt + s2,
dx3
dt
= 1 ⇒ x3 = t,
dz
dt
= −γz ⇒
dz
z
= −γdt ⇒ z = ϕ(s1, s2)e−γt
.
J ≡ det
∂(x1, x2, x3)
∂(s1, s2, t)
=
1 0 0
0 1 0
α β 1
= 1 = 0 ⇒ J is invertible.
Since t = x3, s1 = x1 − αx3, s2 = x2 − βx3, we have
u(x1, x2, x3) = ϕ(x1 − αx3, x2 − βx3)e−γx3
, or
u(x, y, t) = ϕ(x − αt, y − βt)e−γt
.
The solution satisﬁes the PDE and initial condition.24
23
24
Chain Rule: u(x1, x2, x3) = ϕ(f(x1, x2, x3), g(x1, x2, x3)), then ux1 = ∂ϕ
∂f
∂f
∂x1
+ ∂ϕ
∂g
∂g
∂x1
.

Problem (F’94, #2). Find the solution of the Cauchy problem
ut(x, y, t) + aux(x, y, t) + buy(x, y, t) + c(x, y, t)u(x, y, t) = 0
u(x, y, 0) = u0(x, y),
where 0 < t < +∞, −∞ < x < +∞, −∞ < y < +∞,
a, b are constants, c(x, y, t) is a continuous function of (x, y, t), and u0(x, y) is a con-
tinuous function of (x, y).
Proof. Rewrite the equation as (x → x1, y → x2, t → x3):
aux1 + bux2 + ux3 = −c(x1, x2, x3)u,
u(x1, x2, 0) = u0(x1, x2).
Γ is parameterized by Γ : (s1, s2, 0, u0(s1, s2)).
dx1
dt
= a ⇒ x1 = at + s1,
dx2
dt
= b ⇒ x2 = bt + s2,
dx3
dt
= 1 ⇒ x3 = t,
dz
dt
= −c(x1, x2, x3)z ⇒
dz
dt
= −c(at + s1, bt + s2, t)z ⇒
dz
z
= −c(at + s1, bt + s2, t)dt
⇒ ln z = −
t
0
c(aξ + s1, bξ + s2, ξ)dξ + c1(s1, s2),
⇒ z(s1, s2, t) = c2(s1, s2)e−
t
0 c(aξ+s1,bξ+s2,ξ)dξ
⇒ z(s1, s2, 0) = c2(s1, s2) = u0(s2, s2),
⇒ z(s1, s2, t) = u0(s1, s2)e− t
0 c(aξ+s1,bξ+s2,ξ)dξ
.
J ≡ det
∂(x1, x2, x3)
∂(s1, s2, t)
=
1 0 0
0 1 0
a b 1
Since t = x3, s1 = x1 − ax3, s2 = x2 − bx3, we have
u(x1, x2, x3) = u0(x1 − ax3, x2 − bx3)e−
x3
0 c(aξ+x1−ax3,bξ+x2−bx3,ξ)dξ
= u0(x1 − ax3, x2 − bx3)e−
x3
0 c(x1+a(ξ−x3),x2+b(ξ−x3),ξ)dξ
, or
u(x, y, t) = u0(x − at, y − bt)e−
t
0 c(x+a(ξ−t),y+b(ξ−t),ξ)dξ
.

Problem (F’89, #4). Consider the ﬁrst order partial diﬀerential equation
ut + (α + βt)ux + γet
uy = 0 (13.1)
in which α, β and γ are constants.
a) For this equation, solve the initial value problem with initial data
u(x, y, t = 0) = sin(xy) (13.2)
for all x and y and for t ≥ 0.
b) Suppose that this initial data is prescribed only for x ≥ 0 (and all y) and consider
(13.1) in the region x ≥ 0, t ≥ 0 and all y. For which values of α, β and γ is it possible
to solve the initial-boundary value problem (13.1), (13.2) with u(x = 0, y, t) given for
t ≥ 0?
For non-permissible values of α, β and γ, where can boundary values be prescribed in
order to determine a solution of (13.1) in the region (x ≥ 0, t ≥ 0, all y).
Proof. a) Rewrite the equation as (x → x1, y → x2, t → x3):
(α + βx3)ux1 + γex3
ux2 + ux3 = 0,
u(x1, x2, 0) = sin(x1x2).
Γ is parameterized by Γ : (s1, s2, 0, sin(s1s2)).
dx1
dt
= α + βx3 ⇒
dx1
dt
= α + βt ⇒ x1 =
βt2
2
+ αt + s1,
dx2
dt
= γex3
⇒
dx2
dt
= γet
⇒ x2 = γet
− γ + s2,
dx3
dt
= 1 ⇒ x3 = t,
dz
dt
= 0 ⇒ z = sin(s1s2).
J ≡ det
∂(x1, x2, x3)
∂(s1, s2, t)
=
1 0 0
0 1 0
βt + α γet
1
Since t = x3, s1 = x1 −
βx2
3
2 − αx3, s2 = x2 − γex3 + γ, we have
u(x1, x2, x3) = sin((x1 −
βx2
3
2
− αx3)(x2 − γex3
+ γ)), or
u(x, y, t) = sin((x −
βt2
2
− αt)(y − γet
+ γ)).
b) We need a compatibility condition between the initial and boundary values to hold
on y-axis (x = 0, t = 0):
u(x = 0, y, 0) = u(0, y, t = 0),
0 = 0.

14 Problems: First-Order Systems
Problem (S’01, #2a). Find the solution u =
u1(x, t)
u2(x, t)
, (x, t) ∈ R × R,
to the (strictly) hyperbolic equation
ut −
1 0
5 3
ux = 0,
satisfying
u1(x, 0)
u2(x, 0)
=
eixa
0
, a ∈ R.
Ut +
−1 0
−5 −3
Ux = 0,
U(x, 0) =
u(1)(x, 0)
u(2)
(x, 0)
=
eixa
0
.
The eigenvalues of the matrix A are λ1 = −1, λ2 = −3 and the corresponding
eigenvectors are e1 =
2
−5
, e2 =
0
1
. Thus,
Λ =
−1 0
0 −3
, Γ =
2 0
−5 1
, Γ−1
=
1
det Γ
· Γ =
1
2 0
5
2 1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Thus, the transformed problem is
Vt +
−1 0
0 −3
Vx = 0,
V (x, 0) = Γ−1
U(x, 0) =
1
2 0
5
2 1
eixa
0
=
1
2
eixa 1
5
.
We have two initial value problems
v
(1)
t − v
(1)
x = 0,
v(1)
(x, 0) = 1
2 eixa
;
v
(2)
t − 3v
(2)
x = 0,
v(2)
(x, 0) = 5
2 eixa
,
which we solve by characteristics to get
v(1)
(x, t) =
1
2
eia(x+t)
, v(2)
(x, t) =
5
2
eia(x+3t)
.
We solve for U: U = ΓV = Γ
v(1)
v(2) =
2 0
−5 1
1
2eia(x+t)
5
2 eia(x+3t) .
Thus, U =
u(1)
(x, t)
u(2)
(x, t)
=
eia(x+t)
−5
2 eia(x+t)
+ 5
2 eia(x+3t) .
Can check that this is the correct solution by plugging it into the original equation.

Part (b) of the problem is solved in the Fourier Transform section.

Problem (S’96, #7). Solve the following initial-boundary value problem in the do-
main x > 0, t > 0, for the unknown vector U =
u(1)
u(2) :
Ut +
−2 3
0 1
Ux = 0. (14.1)
U(x, 0) =
sinx
0
and u(2)
(0, t) = t.
Proof. The eigenvalues of the matrix A are λ1 = −2, λ2 = 1 and the corresponding
1
0
, e2 =
1
1
. Thus,
Λ =
−2 0
0 1
, Γ =
1 1
0 1
, Γ−1
=
1
det Γ
· Γ =
1 −1
0 1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
−2 0
0 1
Vx = 0, (14.2)
V (x, 0) = Γ−1
U(x, 0) =
1 −1
0 1
sin x
0
=
sinx
0
. (14.3)
Equation (14.2) gives traveling wave solutions of the form
v(1)
(x, t) = F(x + 2t), v(2)
(x, t) = G(x − t).
We can write U in terms of V :
U = ΓV =
1 1
0 1
v(1)
v(2) =
1 1
0 1
F(x + 2t)
G(x − t)
=
F(x + 2t) + G(x − t)
G(x − t)
.
(14.4)

• For region I, (14.2) and (14.3) give two initial value problems (since any point in
region I can be traced back along both characteristics to initial conditions):
v
(1)
t − 2v
(1)
x = 0,
v(1)
(x, 0) = sinx;
v
(2)
t + v
(2)
x = 0,
v(2)
(x, 0) = 0.
which we solve by characteristics to get traveling wave solutions:
v(1)
(x, t) = sin(x + 2t), v(2)
(x, t) = 0.
➡ Thus, for region I,
U = ΓV =
1 1
0 1
sin(x + 2t)
0
=
sin(x + 2t)
0
.
• For region II, solutions of the form F(x+2t) can be traced back to initial conditions.
Thus, v(1) is the same as in region I. Solutions of the form G(x −t) can be traced back
to the boundary. Since from (14.4), u(2)
= v(2)
, we use boundary conditions to get
u(2)
(0, t) = t = G(−t).
Hence, G(x − t) = −(x − t).
➡ Thus, for region II,
U = ΓV =
1 1
0 1
sin(x + 2t)
−(x − t)
=
sin(x + 2t) − (x − t)
−(x − t)
.
Solutions for regions I and II satisfy (14.1).
Solution for region I satisﬁes both initial conditions.
Solution for region II satisﬁes given boundary condition.

Problem (S’02, #7). Consider the system
∂
∂t
u
v
=
−1 2
2 2
∂
∂x
u
v
. (14.5)
Find an explicit solution for the following mixed problem for the system (14.5):
u(x, 0)
v(x, 0)
=
f(x)
0
for x > 0,
u(0, t) = 0 for t > 0.
You may assume that the function f is smooth and vanishes on a neighborhood of x = 0.
Ut +
1 −2
−2 −2
Ux = 0,
U(x, 0) =
u(1)
(x, 0)
u(2)
(x, 0)
=
f(x)
0
.
The eigenvalues of the matrix A are λ1 = −3, λ2 = 2 and the corresponding eigen-
vectors are e1 =
1
2
, e2 =
−2
1
. Thus,
Λ =
−3 0
0 2
, Γ =
1 −2
2 1
, Γ−1
=
1
det Γ
· Γ =
1
5
1 2
−2 1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
−3 0
0 2
Vx = 0, (14.6)
V (x, 0) = Γ−1
U(x, 0) =
1
5
1 2
−2 1
f(x)
0
=
f(x)
5
1
−2
. (14.7)
Equation (14.6) gives traveling wave solutions of the form:
v(1)
(x, t) = F(x + 3t), v(2)
(x, t) = G(x − 2t). (14.8)
U = ΓV =
1 −2
2 1
v(1)
v(2) =
1 −2
2 1
F(x + 3t)
G(x − 2t)
=
F(x + 3t) − 2G(x − 2t)
2F(x + 3t) + G(x − 2t)
.
(14.9)

• For region I, (14.6) and (14.7) give two initial value problems (since value at any
point in region I can be traced back along both characteristics to initial conditions):
v
(1)
t − 3v
(1)
x = 0,
v(1)(x, 0) = 1
5 f(x);
v
(2)
t + 2v
(2)
x = 0,
v(2)(x, 0) = −2
5 f(x).
v(1)
(x, t) =
1
5
f(x + 3t), v(2)
(x, t) = −
2
5
f(x − 2t).
➡ Thus, for region I, U = ΓV =
1 −2
2 1
1
5 f(x + 3t)
−2
5 f(x − 2t)
=
1
5 f(x + 3t) + 4
5 f(x − 2t)
2
5 f(x + 3t) − 2
5 f(x − 2t)
.
Thus, v(1) is the same as in region I. Solutions of the form G(x−2t) can be traced back
to the boundary. Since from (14.9),
u(1)
= v(1)
− 2v(2)
, we have
u(1)
(x, t) = F(x + 3t) − 2G(x − 2t) =
1
5
f(x + 3t) − 2G(x − 2t).
The boundary condition gives
u(1)
(0, t) = 0 =
1
5
f(3t) − 2G(−2t),
2G(−2t) =
1
5
f(3t),
G(t) =
1
10
f −
3
2
t ,
G(x − 2t) =
1
10
f −
3
2
(x − 2t) .
➡ Thus, for region II, U = ΓV =
1 −2
2 1
1
5 f(x + 3t)
1
10 f(−3
2 (x − 2t))
=
1
5 f(x + 3t) − 1
5 f(−3
2 (x − 2t))
2
5 f(x + 3t) + 1
10f(−3
2 (x − 2t))
.

Problem (F’94, #1; S’97, #7). Solve the initial-boundary value problem
ut + 3vx = 0,
vt + ux + 2vx = 0
in the quarter plane 0 ≤ x, t < ∞, with initial conditions 25
u(x, 0) = ϕ1(x), v(x, 0) = ϕ2(x), 0 < x < +∞
and boundary condition
u(0, t) = ψ(t), t > 0.
Proof. Rewrite the equation as Ut + AUx = 0:
Ut +
0 3
1 2
Ux = 0, (14.10)
U(x, 0) =
u(1)(x, 0)
u(2)
(x, 0)
=
ϕ1(x)
ϕ2(x)
.
vectors are e1 =
−3
1
, e2 =
1
1
. Thus,
Λ =
−1 0
0 3
, Γ =
−3 1
1 1
, Γ−1
=
1
det Γ
· Γ =
1
4
−1 1
1 3
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
−1 0
0 3
Vx = 0, (14.11)
V (x, 0) = Γ−1
U(x, 0) =
1
4
−1 1
1 3
ϕ1(x)
ϕ2(x)
=
1
4
−ϕ1(x) + ϕ2(x)
ϕ1(x) + 3ϕ2(x)
. (14.12)
v(1)
(x, t) = F(x + t), v(2)
(x, t) = G(x − 3t). (14.13)
U = ΓV =
−3 1
1 1
v(1)
v(2) =
−3 1
1 1
F(x + t)
G(x − 3t)
=
−3F(x + t) + G(x − 3t)
F(x + t) + G(x − 3t)
.
(14.14)
25
In S’97, #7, the zero initial conditions are considered.

• For region I, (14.11) and (14.12) give two initial value problems (since value at
any point in region I can be traced back along characteristics to initial conditions):
v
(1)
t − v
(1)
x = 0,
v(1)(x, 0) = −1
4 ϕ1(x) + 1
4 ϕ2(x);
v
(2)
t + 3v
(2)
x = 0,
v(2)(x, 0) = 1
4ϕ1(x) + 3
4ϕ2(x),
v(1)
(x, t) = −
1
4
ϕ1(x + t) +
1
4
ϕ2(x + t), v(2)
(x, t) =
1
4
ϕ1(x − 3t) +
3
4
ϕ2(x − 3t).
U = ΓV =
−3 1
1 1
−1
4 ϕ1(x + t) + 1
4ϕ2(x + t)
1
4 ϕ1(x − 3t) + 3
4ϕ2(x − 3t)
=
1
4
3ϕ1(x + t) − 3ϕ2(x + t) + ϕ1(x − 3t) + 3ϕ2(x − 3t)
−ϕ1(x + t) + ϕ2(x + t) + ϕ1(x − 3t) + 3ϕ2(x − 3t)
.
• For region II, solutions of the form F(x + t) can be traced back to initial conditions.
Thus, v(1)
is the same as in region I. Solutions of the form G(x−3t) can be traced back
to the boundary. Since from (14.14),
u(1)
= −3v(1)
+ v(2)
, we have
u(1)
(x, t) =
3
4
ϕ1(x + t) −
3
4
ϕ2(x + t) + G(x − 3t).
The boundary condition gives
u(1)
(0, t) = ψ(t) =
3
4
ϕ1(t) −
3
4
ϕ2(t) + G(−3t),
G(−3t) = ψ(t) −
3
4
ϕ1(t) +
3
4
ϕ2(t),
G(t) = ψ −
t
3
−
3
4
ϕ1 −
t
3
+
3
4
ϕ2 −
t
3
,
G(x − 3t) = ψ −
x − 3t
3
−
3
4
ϕ1 −
x − 3t
3
+
3
4
ϕ2 −
x − 3t
3
.
U = ΓV =
−3 1
1 1
−1
4 ϕ1(x + t) + 1
4ϕ2(x + t)
ψ(−x−3t
3 ) − 3
4ϕ1(−x−3t
3 ) + 3
4ϕ2(−x−3t
3 )
=
3
4 ϕ1(x + t) − 3
4 ϕ2(x + t) + ψ(−x−3t
3 ) − 3
4 ϕ1(−x−3t
3 ) + 3
4 ϕ2(−x−3t
3 )
−1
4ϕ1(x + t) + 1
4ϕ2(x + t) + ψ(−x−3t
3 ) − 3
4ϕ1(−x−3t
3 ) + 3
4ϕ2(−x−3t
3 )
.

Problem (F’91, #1). Solve explicitly the following initial-boundary value problem for
linear 2×2 hyperbolic system
ut = ux + vx
vt = 3ux − vx,
where 0 < t < +∞, 0 < x < +∞ with initial conditions
u(x, 0) = u0(x), v(x, 0) = v0(x), 0 < x < +∞,
and the boundary condition
u(0, t) + bv(0, t) = ϕ(t), 0 < t < +∞,
where b = 1
3 is a constant.
What happens when b = 1
3 ?
Proof. Let us change the notation (u ↔ u(1)
, v ↔ u(2)
). Rewrite the equation as
Ut +
−1 −1
−3 1
Ux = 0, (14.15)
U(x, 0) =
u(1)
(x, 0)
u(2)(x, 0)
=
u
(1)
0 (x)
u
(2)
0 (x)
.
vectors are e1 =
1
1
, e2 =
1
−3
. Thus,
Λ =
−2 0
0 2
, Γ =
1 1
1 −3
, Γ−1
=
1
4
3 1
1 −1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
−2 0
0 2
Vx = 0, (14.16)
V (x, 0) = Γ−1
U(x, 0) =
1
4
3 1
1 −1
u(1)(x, 0)
u(2)
(x, 0)
=
1
4
3u
(1)
0 (x) + u
(2)
0 (x)
u
(1)
0 (x) − u
(2)
0 (x)
.
(14.17)
v(1)
(x, t) = F(x + 2t), v(2)
(x, t) = G(x − 2t). (14.18)

U = ΓV =
1 1
1 −3
v(1)
v(2) =
1 1
1 −3
F(x + 2t)
G(x − 2t)
=
F(x + 2t) + G(x − 2t)
F(x + 2t) − 3G(x − 2t)
.
(14.19)
• For region I, (14.16) and (14.17) give two initial value problems (since value at any
point in region I can be traced back along characteristics to initial conditions):
v
(1)
t − 2v
(1)
x = 0,
v(1)(x, 0) = 3
4 u
(1)
0 (x) + 1
4u
(2)
0 (x);
v
(2)
t + 2v
(2)
x = 0,
v(2)(x, 0) = 1
4u
(1)
0 (x) − 1
4 u
(2)
0 (x),
v(1)
(x, t) =
3
4
u
(1)
0 (x + 2t) +
1
4
u
(2)
0 (x + 2t); v(2)
(x, t) =
1
4
u
(1)
0 (x − 2t) −
1
4
u
(2)
0 (x − 2t).
U = ΓV =
1 1
1 −3
3
4u
(1)
0 (x + 2t) + 1
4 u
(2)
0 (x + 2t)
1
4u
(1)
0 (x − 2t) − 1
4 u
(2)
0 (x − 2t)
=
3
4u
(1)
0 (x + 2t) + 1
4 u
(2)
0 (x + 2t) + 1
4u
(1)
0 (x − 2t) − 1
4 u
(2)
0 (x − 2t)
3
4u
(1)
0 (x + 2t) + 1
4 u
(2)
0 (x + 2t) − 3
4u
(1)
0 (x − 2t) + 3
4 u
(2)
0 (x − 2t)
.
Thus, v(1)
is the same as in region I. Solutions of the form G(x−2t) can be traced back
to the boundary. The boundary condition gives
u(1)
(0, t) + bu(2)
(0, t) = ϕ(t).
Using (14.19),
v(1)
(0, t) + G(−2t) + bv(1)
(0, t) − 3bG(−2t) = ϕ(t),
(1 + b)v(1)
(0, t) + (1 − 3b)G(−2t) = ϕ(t),
(1 + b)
3
4
u
(1)
0 (2t) +
1
4
u
(2)
0 (2t) + (1 − 3b)G(−2t) = ϕ(t),
G(−2t) =
ϕ(t) − (1 + b) 3
4 u
(1)
0 (2t) + 1
4 u
(2)
0 (2t)
1 − 3b
,
G(t) =
ϕ(−t
2) − (1 + b) 3
4u
(1)
0 (−t) + 1
4 u
(2)
0 (−t)
1 − 3b
,
G(x − 2t) =
ϕ(−x−2t
2 ) − (1 + b) 3
4 u
(1)
0 (−(x − 2t)) + 1
4 u
(2)
0 (−(x − 2t))
1 − 3b
.
U = ΓV =
1 1
1 −3
⎛
⎝
3
4 u
(1)
0 (x + 2t) + 1
4u
(2)
0 (x + 2t)
ϕ(−x−2t
2
)−(1+b) 3
4
u
(1)
0 (−(x−2t))+1
4
u
(2)
0 (−(x−2t))
1−3b
⎞
⎠
=
⎛
⎝
3
4u
(1)
0 (x + 2t) + 1
4 u
(2)
0 (x + 2t) +
ϕ(−x−2t
2
)−(1+b) 3
4
u
(1)
0 (−(x−2t))+1
4
u
(2)
0 (−(x−2t))
1−3b
3
4 u
(1)
0 (x + 2t) + 1
4 u
(2)
0 (x + 2t) −
3ϕ(−x−2t
2
)−3(1+b) 3
4
u
(1)
0 (−(x−2t))+1
4
u
(2)
0 (−(x−2t))
1−3b
⎞
⎠ .
The following were performed, but are arithmetically complicated:

If b = 1
3 , u(1)
(0, t) + 1
3u(2)
(0, t) = F(2t) + G(−2t) + 1
3 F(2t) − G(−2t) = 4
3 F(2t) = ϕ(t).
Thus, the solutions of the form v(2) = G(x − 2t) are not deﬁned at x = 0, which leads
to ill-posedness.

Problem (F’96, #8). Consider the system
ut = 3ux + 2vx
vt = −vx − v
in the region x ≥ 0, t ≥ 0. Which of the following sets of initial and boundary data
make this a well-posed problem?
a) u(x, 0) = 0, x ≥ 0
v(x, 0) = x2
, x ≥ 0
v(0, t) = t2
, t ≥ 0.
b) u(x, 0) = 0, x ≥ 0
v(x, 0) = x2
, x ≥ 0
u(0, t) = t, t ≥ 0.
c) u(x, 0) = 0, x ≥ 0
v(x, 0) = x2
, x ≥ 0
u(0, t) = t, t ≥ 0
v(0, t) = t2
, t ≥ 0.
Proof. Rewrite the equation as Ut + AUx = BU. Initial conditions are same for
(a),(b),(c):
Ut +
−3 −2
0 1
Ux =
0 0
0 −1
U,
U(x, 0) =
u(1)(x, 0)
u(2)
(x, 0)
=
0
x2 .
The eigenvalues of the matrix A are λ1 = −3, λ2 = 1, and the corresponding eigen-
vectors are e1 =
1
0
, e2 =
1
−2
. Thus,
Λ =
−3 0
0 1
, Γ =
1 1
0 −2
, Γ−1
=
1
2
2 1
0 −1
.
Let U = ΓV . Then,
Ut + AUx = BU,
ΓVt + AΓVx = BΓV,
Vt + Γ−1
AΓVx = Γ−1
BΓV,
Vt + ΛVx = Γ−1
BΓV.
Vt +
−3 0
0 1
Vx =
0 1
0 −1
V, (14.20)
V (x, 0) = Γ−1
U(x, 0) =
1
2
2 1
0 −1
0
x2 =
x2
2
1
−1
. (14.21)
Equation (14.20) gives traveling wave solutions of the form
v(1)
(x, t) = F(x + 3t), v(2)
(x, t) = G(x − t). (14.22)

U = ΓV =
1 1
0 −2
v(1)
v(2) =
1 1
0 −2
F(x + 3t)
G(x − t)
=
F(x + 3t) + G(x − t)
−2G(x − t)
.
(14.23)
• For region I, (14.20) and (14.21) give two initial value problems (since a value at any
point in region I can be traced back along both characteristics to initial conditions):
v
(1)
t − 3v
(1)
x = v(2),
v(1)(x, 0) = x2
2 ;
v
(2)
t + v
(2)
x = −v(2),
v(2)(x, 0) = −x2
2 ,
which we do not solve here. Thus, initial conditions for v(1) and v(2) have to be defined.
Since (14.23) defines u(1) and u(2) in terms of v(1) and v(2), we need to define two initial
conditions for U.
Thus, v(1)
is the same as in region I. Solutions of the form G(x − t) are traced back to
the boundary at x = 0. Since from (14.23), u(2)(x, t) = −2v(2)(x, t) = −2G(x − t), i.e.
u(2)
is written in term of v(2)
only, u(2)
requires a boundary condition to be defined on
x = 0.
Thus,
a) u(2)
(0, t) = t2
, t ≥ 0. Well-posed.
b) u(1)(0, t) = t, t ≥ 0. Not well-posed.
c) u(1)
(0, t) = t, u(2)
(0, t) = t2
, t ≥ 0. Not well-posed.

Problem (F’02, #3). Consider the ﬁrst order system
ut + ux + vx = 0
vt + ux − vx = 0
on the domain 0 < t < ∞ and 0 < x < 1. Which of the following sets of initial-
boundary data are well posed for this system? Explain your answers.
a) u(x,0) = f(x), v(x,0) = g(x);
b) u(x,0) = f(x), v(x,0) = g(x), u(0,t) = h(x), v(0,t) = k(x);
c) u(x,0) = f(x), v(x,0) = g(x), u(0,t) = h(x), v(1,t) = k(x).
Proof. Rewrite the equation as Ut+AUx = 0. Initial conditions are same for (a),(b),(c):
Ut +
1 1
1 −1
Ux = 0,
U(x, 0) =
u(1)
(x, 0)
u(2)(x, 0)
=
f(x)
g(x)
.
The eigenvalues of the matrix A are λ1 =
√
2, λ2 = −
√
2 and the corresponding
1
−1 +
√
2
, e2 =
1
−1 −
√
2
. Thus,
Λ =
√
2 0
0 −
√
2
, Γ =
1 1
−1 +
√
2 −1 −
√
2
, Γ−1
=
1
2
√
2
1 +
√
2 1
−1 +
√
2 −1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
√
2 0
0 −
√
2
Vx = 0, (14.24)
V (x, 0) = Γ−1
U(x, 0) =
1
2
√
2
1 +
√
2 1
−1 +
√
2 −1
f(x)
g(x)
=
1
2
√
2
(1 +
√
2)f(x) + g(x)
(−1 +
√
2)f(x) − g(x)
.
(14.25)
v(1)
(x, t) = F(x −
√
2t), v(2)
(x, t) = G(x +
√
2t). (14.26)
However, we can continue and obtain the solutions. We have two initial value problems
v
(1)
t +
√
2v
(1)
x = 0,
v(1)(x, 0) = (1+
√
2)
2
√
2
f(x) + 1
2
√
2
g(x);
v
(2)
t −
√
2v
(2)
x = 0,
v(2)(x, 0) = (−1+
√
2)
2
√
2
f(x) − 1
2
√
2
g(x),
v(1)
(x, t) =
(1 +
√
2)
2
√
2
f(x −
√
2t) +
1
2
√
2
g(x −
√
2t),
v(2)
(x, t) =
(−1 +
√
2)
2
√
2
f(x +
√
2t) −
1
2
√
2
g(x +
√
2t).

We can obtain general solution U by writing U in terms of V :
U = ΓV = Γ
v(1)
v(2) =
1 1
−1 +
√
2 −1 −
√
2
1
2
√
2
(1 +
√
2)f(x −
√
2t) + g(x −
√
2t)
(−1 +
√
2)f(x +
√
2t) − g(x +
√
2t)
.
(14.27)
• In region I, the solution is obtained by solving two initial value problems(since a
value at any point in region I can be traced back along both characteristics to initial
conditions).
• In region II, the solutions of the form v(2) = G(x+
√
2t) can be traced back to initial
conditions and those of the form v(1)
= F(x −
√
2t), to left boundary. Since by (14.27),
u(1) and u(2) are written in terms of both v(1) and v(2), one initial condition and one
boundary condition at x = 0 need to be prescribed.
• In region III, the solutions of the form v(2) = G(x +
√
2t) can be traced back to
right boundary and those of the form v(1)
= F(x −
√
2t), to initial condition. Since by
(14.27), u(1)
and u(2)
are written in terms of both v(1)
and v(2)
, one initial condition
and one boundary condition at x = 1 need to be prescribed.
• To obtain the solution for region IV, two boundary conditions, one for each bound-
ary, should be given.
Thus,
a) No boundary conditions. Not well-posed.
b) u(1)(0, t) = h(x), u(2)(0, t) = k(x). Not well-posed.
c) u(1)
(0, t) = h(x), u(2)
(1, t) = k(x). Well-posed.

Problem (S’94, #3). Consider the system of equations
ft + gx = 0
gt + fx = 0
ht + 2hx = 0
on the set x ≥ 0, t ≥ 0, with the following initial-boundary values:
a) f, g, h prescribed on t = 0, x ≥ 0; f, h prescribed on x = 0, t ≥ 0.
b) f, g, h prescribed on t = 0, x ≥ 0; f − g, h prescribed on x = 0, t ≥ 0.
c) f + g, h prescribed on t = 0, x ≥ 0; f, g, h prescribed on x = 0, t ≥ 0.
For each of these 3 sets of data, determine whether or not the system is well-posed.
Justify your conclusions.
Proof. The third equation is decoupled from the ﬁrst two and can be considered sepa-
rately. Its solution can be written in the form
h(x, t) = H(x − 2t),
and therefore, h must be prescribed on t = 0 and on x = 0, since the characteristics
propagate from both the x and t axis.
We rewrite the ﬁrst two equations as (f ↔ u1, g ↔ u2):
Ut +
0 1
1 0
Ux = 0,
U(x, 0) =
u(1)
(x, 0)
u(2)(x, 0)
.
vectors are e1 =
−1
1
, e2 =
1
1
. Thus,
Λ =
−1 0
0 1
, Γ =
−1 1
1 1
, Γ−1
=
1
2
−1 1
1 1
.
Let U = ΓV . Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
−1 0
0 1
Vx = 0, (14.28)
V (x, 0) = Γ−1
U(x, 0) =
1
2
−1 1
1 1
u(1)
(x, 0)
u(1)
(x, 0)
. (14.29)
v(1)
(x, t) = F(x + t), v(2)
(x, t) = G(x − t). (14.30)

U = ΓV =
−1 1
1 1
v(1)
v(2) =
−1 1
1 1
F(x + t)
G(x − t)
=
−F(x + t) + G(x − t)
F(x + t) + G(x − t)
.
(14.31)
• For region I, (14.28) and (14.29) give two initial value problems (since a value at any
point in region I can be traced back along both characteristics to initial conditions).
Thus, initial conditions for v(1) and v(2) have to be defined. Since (14.31) defines u(1)
and u(2)
in terms of v(1)
and v(2)
, we need to define two initial conditions for U.
• For region II, solutions of the form F(x + t) can be traced back to initial conditions.
Thus, v(1)
is the same as in region I. Solutions of the form G(x − t) are traced back
to the boundary at x = 0. Since from (14.31), u(2)(x, t) = v(1)(x, t) + v(2)(x, t) =
F(x + t) + G(x − t), i.e. u(2)
is written in terms of v(2)
= G(x − t), u(2)
requires a
boundary condition to be defined on x = 0.
a) u(1)
, u(2)
prescribed on t = 0; u(1)
prescribed on x = 0.
Since u(1)
(x, t) = −F(x + t) + G(x − t),
u(2)(x, t) = F(x + t) + G(x − t), i.e. both
u(1)
and u(2)
are written in terms of F(x + t)
and G(x − t), we need to define two initial
conditions for U (on t = 0).
A boundary condition also needs to be prescribed
on x = 0 to be able to trace back v(2)
= G(x − t).
Well-posed.
b) u(1), u(2) prescribed on t = 0; u(1) − u(2) prescribed on x = 0.
As in part (a), we need to define two initial
conditions for U.
Since u(1)
− u(2)
= −2F(x + t), its definition
on x = 0 leads to ill-posedness. On the
contrary, u(1)
+ u(2)
= 2G(x − t) should be
defined on x = 0 in order to be able to trace
back the values through characteristics.
Ill-posed.
c) u(1) + u(2) prescribed on t = 0; u(1), u(2) prescribed on x = 0.
Since u(1) + u(2) = 2G(x − t), another initial
condition should be prescribed to be able to
trace back solutions of the form v(2) = F(x + t),
without which the problem is ill-posed.
Also, two boundary conditions for both u(1)
and u(2)
define solutions of both v(1)
= G(x − t)
and v(2) = F(x + t) on the boundary. The former
boundary condition leads to ill-posedness.
Ill-posed.

Problem (F’92, #8). Consider the system
ut + ux + avx = 0
vt + bux + vx = 0
for 0 < x < 1 with boundary and initial conditions
u = v = 0 for x = 0
u = u0, v = v0 for t = 0.
a) For which values of a and b is this a well-posed problem?
b) For this class of a, b, state conditions on u0 and v0 so that the solution u, v will be
continuous and continuously diﬀerentiable.
Proof. a) Let us change the notation (u ↔ u(1), v ↔ u(2)). Rewrite the equation as
Ut +
1 a
b 1
Ux = 0, (14.32)
U(x, 0) =
u(1)
(x, 0)
u(2)
(x, 0)
=
u
(1)
0 (x)
u
(2)
0 (x)
,
U(0, t) =
u(1)
(0, t)
u(2)(0, t)
= 0.
The eigenvalues of the matrix A are λ1 = 1 −
√
ab, λ2 = 1 +
√
ab.
Λ =
1 −
√
ab 0
0 1 +
√
ab
.
Let U = ΓV , where Γ is a matrix of eigenvectors. Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
1 −
√
ab 0
0 1 +
√
ab
Vx = 0, (14.33)
V (x, 0) = Γ−1
U(x, 0).
The equation (14.33) gives traveling wave solutions of the form:
v(1)
(x, t) = F(x − (1 −
√
ab)t), v(2)
(x, t) = G(x − (1 +
√
ab)t). (14.34)
We also have U = ΓV , i.e. both u(1)
and u(2)
(and their initial and boundary conditions)
are combinations of v(1) and v(2).
In order for this problem to be well-posed, both sets of characteristics should emanate
from the boundary at x = 0. Thus, the eigenvalues of the system are real (ab > 0) and
λ1,2 > 0 (ab < 1). Thus,
0 < ab < 1.
b) For U to be C1, we require the compatibility condition, u
(1)
0 (0) = 0, u
(2)
0 (0) =
0.

Problem (F’93, #2). Consider the initial-boundary value problem
ut + ux = 0
vt − (1 − cx2
)vx + ux = 0
on −1 ≤ x ≤ 1 and 0 ≤ t, with the following prescribed data:
u(x, 0), v(x, 0),
u(−1, t), v(1, t).
For which values of c is this a well-posed problem?
Proof. Let us change the notation (u ↔ u(1)
, v ↔ u(2)
).
The first equation can be solved with u(1)(x, 0) = F(x) to get a solution in the form
u(1)
(x, t) = F(x − t), which requires u(1)
(x, 0) and u(1)
(−1, t) to be defined.
With u(1) known, we can solve the second equation
u
(2)
t − (1 − cx2
)u(2)
x + F(x − t) = 0.
Solving the equation by characteristics, we obtain
the characteristics in the xt-plane are of the form
dx
dt
= cx2
− 1.
We need to determine c such that the prescribed
data u(2)(x, 0) and u(2)(1, t) makes the problem to
be well-posed. The boundary condition for u(2)
(1, t)
requires the characteristics to propagate to the
left with t increasing. Thus, x(t) is a decreasing
function, i.e.
dx
dt
< 0 ⇒ cx2
− 1 < 0 for − 1 < x < 1 ⇒ c < 1.
We could also do similar analysis we have done in other problems on first order sys-
tems involving finding eigenvalues/eigenvectors of the system and using the fact that
u(1)(x, t) is known at both boundaries (i.e. values of u(1)(1, t) can be traced back either
to initial conditions or to boundary conditions on x = −1).

Problem (S’91, #4). Consider the ﬁrst order system
ut + aux + bvx = 0
vt + cux + dvx = 0
for 0 < x < 1, with prescribed initial data:
u(x, 0) = u0(x)
v(x, 0) = v0(x).
a) Find conditions on a, b, c, d such that there is a full set of characteristics and, in
this case, ﬁnd the characteristic speeds.
b) For which values of a, b, c, d can boundary data be prescribed on x = 0 and for which
values can it be prescribed on x = 1? How many pieces of data can be prescribed on
each boundary?
Proof. a) Let us change the notation (u ↔ u(1), v ↔ u(2)). Rewrite the equation as
Ut +
a b
c d
Ux = 0, (14.35)
U(x, 0) =
u(1)(x, 0)
u(2)
(x, 0)
=
u
(1)
0 (x)
u
(2)
0 (x)
.
The system is hyperbolic if for each value of u(1) and u(2) the eigenvalues are real
and the matrix is diagonalizable, i.e. there is a complete set of linearly independent
eigenvectors. The eigenvalues of the matrix A are
λ1,2 =
a + d ± (a + d)2 − 4(ad − bc)
2
=
a + d ± (a − d)2 + 4bc
2
.
We need (a − d)2 + 4bc > 0. This also makes the problem to be diagonalizable.
Let U = ΓV , where Γ is a matrix of eigenvectors. Then,
Ut + AUx = 0,
ΓVt + AΓVx = 0,
Vt + Γ−1
AΓVx = 0,
Vt + ΛVx = 0.
Vt +
λ1 0
0 λ2
Vx = 0, (14.36)
v(1)
(x, t) = F(x − λ1t), v(2)
(x, t) = G(x − λ2t). (14.37)
The characteristic speeds are dx
dt = λ1, dx
dt = λ2.
b) We assume (a + d)2
− 4(ad − bc) > 0.
a + d > 0, ad − bc > 0 ⇒ λ1, λ2 > 0 ⇒ 2 B.C. on x = 0.
a + d > 0, ad − bc < 0 ⇒ λ1 < 0, λ2 > 0 ⇒ 1 B.C. on x = 0, 1 B.C. on x = 1.
a + d < 0, ad − bc > 0 ⇒ λ1, λ2 < 0 ⇒ 2 B.C. on x = 1.

a + d < 0, ad − bc < 0 ⇒ λ1 < 0, λ2 > 0 ⇒ 1 B.C. on x = 0, 1 B.C. on x = 1.
a + d > 0, ad − bc = 0 ⇒ λ1 = 0, λ2 > 0 ⇒ 1 B.C. on x = 0.
a + d < 0, ad − bc = 0 ⇒ λ1 = 0, λ2 < 0 ⇒ 1 B.C. on x = 1.
a + d = 0, ad − bc < 0 ⇒ λ1 < 0, λ2 > 0 ⇒ 1 B.C. on x = 0, 1 B.C. on
x = 1.

Problem (S’94, #2). Consider the differential operator
L
u
v
=
ut + 9vx − uxx
vt − ux − vxx
on 0 ≤ x ≤ 2π, t ≥ 0, in which the vector
u(x, t)
v(x, t)
consists of two functions that
are periodic in x.
a) Find the eigenfunctions and eigenvalues of the operator L.
b) Use the results of (a) to solve the initial value problem
L
u
v
= 0 for t ≥ 0,
u
v
=
eix
0
for t = 0.
Proof. a) We find the ”space” eigenvalues and eigenfunctions. We rewrite the system
as
Ut +
0 9
−1 0
Ux +
−1 0
0 −1
Uxx = 0,
and find eigenvalues
0 9
−1 0
Ux +
−1 0
0 −1
Uxx = λU. (14.38)
Set U =
u(x, t)
v(x, t)
=
n=∞
n=−∞ un(t)einx
n=∞
n=−∞ vn(t)einx . Plugging this into (14.38), we get
0 9
−1 0
inun(t)einx
invn(t)einx +
−1 0
0 −1
−n2
un(t)einx
−n2vn(t)einx = λ
un(t)einx
vn(t)einx ,
0 9
−1 0
inun(t)
invn(t)
+
−1 0
0 −1
−n2
un(t)
−n2vn(t)
= λ
un(t)
vn(t)
,
0 9in
−in 0
un(t)
vn(t)
+
n2 0
0 n2
un(t)
vn(t)
= λ
un(t)
vn(t)
,
n2
− λ 9in
−in n2 − λ
un(t)
vn(t)
= 0,
(n2
− λ)2
− 9n2
= 0,
which gives λ1 = n2+3n, λ2 = n2−3n, are eigenvalues, and v1 =
3i
1
, v2 =
3i
−1
,
are corresponding eigenvectors.

b) We want to solve
u
v t
+ L
u
v
= 0, L
u
v
=
9vx − uxx
−ux − vxx
. We have
u
v t
= −L
u
v
= −λ
u
v
, i.e.
u
v
= e−λt
. We can write the solution as
U(x, t) =
un(t)einx
vn(t)einx =
∞
n=−∞
ane−λ1t
v1einx
+ bne−λ2t
v2einx
=
∞
n=−∞
ane−(n2+3n)t 3i
1
einx
+ bne−(n2−3n)t 3i
−1
einx
.
U(x, 0) =
∞
n=−∞
an
3i
1
einx
+ bn
3i
−1
einx
=
eix
0
,
⇒ an = bn = 0, n = 1;
a1 + b1 =
1
3i
and a1 = b1 ⇒ a1 = b1 =
1
6i
.
⇒ U(x, t) =
1
6i
e−4t 3i
1
eix
+
1
6i
e2t 3i
−1
eix
=
1
2(e−4t
+ e2t
)
1
6i(e−4t − e2t)
eix
.
26 27
26
ChiuYen’s and Sung-Ha’s solutions give similar answers.
27
Questions about this problem:
1. Needed to ﬁnd eigenfunctions, not eigenvectors.
2. The notation of L was changed. The problem statement incorporates the derivatives wrt. t into L.
3. Why can we write the solution in this form above?

Problem (W’04, #6). Consider the ﬁrst order system
ut − ux = vt + vx = 0
in the diamond shaped region −1 < x + t < 1, −1 < x − t < 1. For each of
the following boundary value problems state whether this problem is well-posed. If it is
well-posed, ﬁnd the solution.
a) u(x + t) = u0(x + t) on x − t = −1, v(x − t) = v0(x − t) on x + t = −1.
b) v(x + t) = v0(x + t) on x − t = −1, u(x − t) = u0(x − t) on x + t = −1.
Proof. We have
ut − ux = 0,
vt + vx = 0.
• u is constant along the characteristics: x + t = c1(s).
Thus, its solution is u(x, t) = u0(x + t).
It the initial condition is prescribed at x − t = −1,
the solution can be determined in the entire region
by tracing back through the characteristics.
• v is constant along the characteristics: x − t = c2(s).
Thus, its solution is v(x, t) = v0(x − t).
It the initial condition is prescribed at x + t = −1,
the solution can be determined in the entire region
by tracing forward through the characteristics.

15 Problems: Gas Dynamics Systems
15.1 Perturbation
Problem (S’92, #3). 28 29
Consider the gas dynamic equations
ut + uux + (F(ρ))x = 0,
ρt + (uρ)x = 0.
Here F(ρ) is a given C∞-smooth function of ρ. At t = 0, 2π-periodic initial data
u(x, 0) = f(x), ρ(x, 0) = g(x).
a) Assume that
f(x) = U0 + εf1(x), g(x) = R0 + εg1(x)
where U0, R0 > 0 are constants and εf1(x), εg1(x) are “small” perturbations. Lin-
earize the equations and given conditions for F such that the linearized problem is
well-posed.
b) Assume that U0 > 0 and consider the above linearized equations for 0 ≤ x ≤ 1,
t ≥ 0. Construct boundary conditions such that the initial-boundary value problem is
well-posed.
Proof. a) We write the equations in characteristic form:
ut + uux + F (ρ)ρx = 0,
ρt + uxρ + uρx = 0.
Consider the special case of nearly constant initial data
u(x, 0) = u0 + εu1(x, 0),
ρ(x, 0) = ρ0 + ερ1(x, 0).
Then we can approximate nonlinear equations by linear equations. Assuming
u(x, t) = u0 + εu1(x, t),
ρ(x, t) = ρ0 + ερ1(x, t)
remain valid with u1 = O(1), ρ1 = O(1), we ﬁnd that
ut = εu1t, ρt = ερ1t,
ux = εu1x, ρx = ερ1x,
F (ρ) = F (ρ0 + ερ1(x, t)) = F (ρ0) + ερ1F (ρ0) + O(ε2
).
Plugging these into , gives
εu1t + (u0 + εu1)εu1x + F (ρ0) + ερ1F (ρ0) + O(ε2
) ερ1x = 0,
ερ1t + εu1x(ρ0 + ερ1) + (u0 + εu1)ερ1x = 0.
Dividing by ε gives
u1t + u0u1x + F (ρ0)ρ1x = −εu1u1x − ερ1ρ1xF (ρ0) + O(ε2
),
ρ1t + u1xρ0 + u0ρ1x = −εu1xρ1 − εu1ρ1x.
28
See LeVeque, Second Edition, Birkh¨auser Verlag, 1992, p. 44.
29
This problem has similar notation with S’92, #4.

For small ε, we have
u1t + u0u1x + F (ρ0)ρ1x = 0,
ρ1t + u1xρ0 + u0ρ1x = 0.
This can be written as
u1
ρ1 t
+
u0 F (ρ0)
ρ0 u0
u1
ρ1 x
=
0
0
.
u0 − λ F (ρ0)
ρ0 u0 − λ
= (u0 − λ)(u0 − λ) − ρ0F (ρ0) = 0,
λ2
− 2u0λ + u2
0 − ρ0F (ρ0) = 0,
λ1,2 = u0 ± ρ0F (ρ0), u0 > 0, ρ0 > 0.
For well-posedness, need λ1,2 ∈ R or F (ρ0) ≥ 0.
b) We have u0 > 0, and λ1 = u0 + ρ0F (ρ0), λ2 = u0 − ρ0F (ρ0).
• If u0 > ρ0F (ρ0) ⇒ λ1 > 0, λ2 > 0 ⇒ 2 BC at x = 0.
• If u0 = ρ0F (ρ0) ⇒ λ1 > 0, λ2 = 0 ⇒ 1 BC at x = 0.
• If 0 < u0 < ρ0F (ρ0) ⇒ λ1 > 0, λ2 < 0 ⇒ 1 BC at x = 0, 1 BC at x = 1.
15.2 Stationary Solutions
Problem (S’92, #4). 30 Consider
ut + uux + ρx = νuxx,
ρt + (uρ)x = 0
for t ≥ 0, −∞ < x < ∞.
Give conditions for the states U+, U−, R+, R−, such that the system has
stationary solutions (i.e. ut = ρt = 0) satisfying
lim
x→+∞
u
ρ
=
U+
R+
, lim
x→−∞
u
ρ
=
U−
R−
.
Proof. For stationary solutions, we need
ut = −
u2
2 x
− ρx + νuxx = 0,
ρt = −(uρ)x = 0.
Integrating the above equations, we obtain
−
u2
2
− ρ + νux = C1,
−uρ = C2.
30
This problem has similar notation with S’92, #3.

Conditions give ux = 0 at x = ±∞. Thus
U2
+
2
+ R+ =
U2
−
2
+ R−,
U+R+ = U−R−.

15.3 Periodic Solutions
Problem (F’94, #4). Let u(x, t) be a solution of the Cauchy problem
ut = −uxxxx − 2uxx, −∞ < x < +∞, 0 < t < +∞,
u(x, 0) = ϕ(x),
where u(x, t) and ϕ(x) are C∞
functions periodic in x with period 2π;
i.e. u(x + 2π, t) = u(x, t), ∀x, ∀t.
Prove that
||u(·, t)|| ≤ Ceat
||ϕ||
where ||u(·, t)|| =
2π
0 |u(x, t)|2 dx, ||ϕ|| =
2π
0 |ϕ(x)|2 dx, C, a are some constants.
Proof. METHOD I: Since u is 2π-periodic, let
u(x, t) =
∞
n=−∞
an(t)einx
.
Plugging this into the equation, we get
∞
n=−∞
an(t)einx
= −
∞
n=−∞
n4
an(t)einx
+ 2
∞
n=−∞
n2
an(t)einx
,
an(t) = (−n4
+ 2n2
)an(t),
an(t) = an(0)e(−n4+2n2)t
.
Also, initial condition gives
u(x, 0) =
∞
n=−∞
an(0)einx
= ϕ(x),
∞
n=−∞
an(0)einx
= |ϕ(x)|.
||u(x, t)||2
2 =
2π
0
u2
(x, t) dx =
2π
0
∞
n=−∞
an(t)einx
∞
m=−∞
an(t)eimx
dx
=
∞
n=−∞
a2
n(t)
2π
0
einx
e−inx
dx = 2π
∞
n=−∞
a2
n(t) = 2π
∞
n=−∞
a2
n(0)e2(−n4+2n2)t
≤ 2π
∞
n=−∞
a2
n(0)
∞
n=−∞
e2(−n4+2n2)t
= 2π
∞
n=−∞
a2
n(0)
||ϕ||2
e2t
∞
n=−∞
e−2(n2−1)2t
= C1, (convergent)
= C2e2t
||ϕ||2
.
⇒ ||u(x, t)|| ≤ Cet
||ϕ||.

METHOD II: Multiply this equation by u and integrate
uut = −uuxxxx − 2uuxx,
1
2
d
dt
(u2
) = −uuxxxx − 2uuxx,
1
2
d
dt
2π
0
u2
dx = −
2π
0
uuxxxx dx −
2π
0
2uuxx dx,
1
2
d
dt
||u||2
2 = −uuxxx
2π
0
=0
+ uxuxx
2π
0
=0
−
2π
0
u2
xx dx −
2π
0
2uuxx dx,
1
2
d
dt
||u||2
2 = −
2π
0
u2
xx dx −
2π
0
2uuxx dx (−2ab ≤ a2
+ b2
)
≤ −
2π
0
u2
xx dx +
2π
0
(u2
+ u2
xx) dx =
2π
0
u2
dx = ||u||2
,
⇒
d
dt
||u||2
≤ 2||u||2
,
||u||2
≤ ||u(0)||2
e2t
,
||u|| ≤ ||u(0)||et
.
METHOD III: Can use Fourier transform. See ChiuYen’s solutions, that have both
Method II and III.

Let f(x) ∈ C∞ be a 2π-periodic function, i.e., f(x) = f(x + 2π) and denote by
||f||2
=
2π
0
|f(x)|2
dx
the L2-norm of f.
a) Express ||dp
f/dxp
||2
in terms of the Fourier coeﬃcients of f.
b) Let q > p > 0 be integers. Prove that ∀ > 0, ∃K = N(ε, p, q), constant, such that
dp
f
dxp
2
≤ ε
dq
f
dxq
2
+ K||f||2
.
c) Discuss how K depends on ε.
Proof. a) Let 31
f(x) =
∞
−∞
fneinx
,
dp
f
dxp
=
∞
−∞
fn(in)p
einx
,
dp
f
dxp
2
=
2π
0
∞
−∞
fn(in)p
einx 2
dx =
2π
0
|i2
|p
∞
−∞
fnnp
einx 2
dx
=
2π
0
∞
−∞
fnnp
einx 2
dx = 2π
∞
n=0
f2
nn2p
.
b) We have
dpf
dxp
2
≤ ε
dqf
dxq
2
+ K||f||2
,
2π
∞
n=0
f2
nn2p
≤ ε 2π
∞
n=0
f2
nn2q
+ K 2π
∞
n=0
f2
n,
n2p
− εn2q
≤ K,
n2p
(1 − εnq
)
< 0, for n large
≤ K, some q > 0.
Thus, the above inequality is true for n large enough. The statement follows.
31
Note:
L
0
einx
eimx
dx =
0 n = m
L n = m

Problem (S’90, #5). 32
Consider the ﬂame front equation
ut + uux + uxx + uxxxx = 0
with 2π-periodic initial data
u(x, 0) = f(x), f(x) = f(x + 2π) ∈ C∞
.
a) Determine the solution, if f(x) ≡ f0 = const.
b) Assume that
f(x) = 1 + εg(x), 0 < ε 1, |g|∞ = 1, g(x) = g(x + 2π).
Linearize the equation. Is the Cauchy problem well-posed for the linearized equation,
i.e., do its solutions v satisfy an estimate
||v(·, t)|| ≤ Keα(t−t0)
||v(·, t0)||?
c) Determine the best possible constants K, α.
Proof. a) The solution to
ut + uux + uxx + uxxxx = 0,
u(x, 0) = f0 = const,
is u(x, t) = f0 = const.
b) We consider the special case of nearly constant initial data
u(x, 0) = 1 + εu1(x, 0).
Then we can approximate the nonlinear equation by a linear equation. Assuming
u(x, t) = 1 + εu1(x, t),
remain valid with u1 = O(1), from , we ﬁnd that
εu1t + (1 + εu1)εu1x + εu1xx + εu1xxxx = 0.
Dividing by ε gives
u1t + u1x + εu1u1x + u1xx + u1xxxx = 0.
For small ε, we have
u1t + u1x + u1xx + u1xxxx = 0.
Multiply this equation by u1 and integrate
u1u1t + u1u1x + u1u1xx + u1u1xxxx = 0,
d
dt
u2
1
2
+
u2
1
2 x
+ u1u1xx + u1u1xxxx = 0,
1
2
d
dt
2π
0
u2
1 dx +
u2
1
2
2π
0
=0
+
2π
0
u1u1xx dx +
2π
0
u1u1xxxx dx = 0,
1
2
d
dt
||u1||2
2 + u1u1x
2π
0
=0
−
2π
0
u2
1x dx + u1u1xxx
2π
0
=0
− u1xu1xx
2π
0
=0
+
2π
0
u2
1xx dx = 0,
1
2
d
dt
||u1||2
2 =
2π
0
u2
1x dx −
2π
0
u2
1xx dx.
32
S’90 #5, #6, #7 all have similar formulations.

Since u1 is 2π-periodic, let
u1 =
∞
n=−∞
an(t)einx
. Then,
u1x = i
∞
n=−∞
nan(t)einx
⇒ u2
1x = −
∞
n=−∞
nan(t)einx
2
,
u1xx = −
∞
n=−∞
n2
an(t)einx
⇒ u2
1xx =
∞
n=−∞
n2
an(t)einx
2
.
Thus,
1
2
d
dt
||u1||2
2 =
2π
0
u2
1x dx −
2π
0
u2
1xx dx
= −
2π
0
nan(t)einx
2
dx −
2π
0
n2
an(t)einx
2
dx
= −2π n2
an(t)2
− 2π n4
an(t)2
= −2π an(t)2
(n2
+ n4
) ≤ 0.
⇒ ||u1(·, t)||2 ≤ ||u1(·, 0)||2,
where K = 1, α = 0.
Problem (W’03, #4). Consider the PDE
ut = ux + u4
for t > 0
u = u0 for t = 0
for 0 < x < 2π. Define the set A = {u = u(x) : û(k) = 0 if k < 0}, in which
{û(k, t)}∞
−∞ is the Fourier series of u in x on [0, 2π].
a) If u0 ∈ A, show that u(t) ∈ A.
b) Find differential equations for û(0, t), û(1, t), and û(2, t).
Proof. a) Solving
ut = ux + u4
u(x, 0) = u0(x)
by the method of characteristics, we get
u(x, t) =
u0(x + t)
(1 − 3t(u0(x + t))3)
1
3
.
Since u0 ∈ A, u0k = 0 if k < 0. Thus,
u0(x) =
∞
k=0
u0k eikx
2 .
Since
uk =
1
2π
2π
0
u(x, t) e−ikx
2 dx,

we have
u(x, t) =
∞
k=0
uk eikx
2 ,
that is, u(t) ∈ A.

15.4 Energy Estimates
Problem (S’90, #6). Let U(x, t) ∈ C∞ be 2π-periodic in x. Consider the linear
equation
ut + Uux + uxx + uxxxx = 0,
u(x, 0) = f(x), f(x) = f(x + 2π) ∈ C∞
.
a) Derive an energy estimate for u.
b) Prove that one can estimate all derivatives ||∂p
u/∂xp
||.
c) Indicate how to prove existence of solutions. 33
Proof. a) Multiply the equation by u and integrate
uut + Uuux + uuxx + uuxxxx = 0,
1
2
d
dt
(u2
) +
1
2
U(u2
)x + uuxx + uuxxxx = 0,
1
2
d
dt
2π
0
u2
dx +
1
2
2π
0
U(u2
)x dx +
2π
0
uuxx dx +
2π
0
uuxxxx dx = 0,
1
2
d
dt
||u||2
+
1
2
Uu2
2π
0
=0
−
1
2
2π
0
Uxu2
dx + uux
2π
0
−
2π
0
u2
x dx
+uuxxx
2π
0
− uxuxx
2π
0
+
2π
0
u2
xx dx = 0,
1
2
d
dt
||u||2
−
1
2
2π
0
Uxu2
dx −
2π
0
u2
x dx +
2π
0
u2
xx dx = 0,
1
2
d
dt
||u||2
=
1
2
2π
0
Uxu2
dx +
2π
0
u2
x dx −
2π
0
u2
xx dx ≤ (from S’90, #5) ≤
≤
1
2
2π
0
Uxu2
dx ≤
1
2
max
x
Ux
2π
0
u2
dx.
⇒
d
dt
||u||2
≤ max
x
Ux||u||2
,
||u(x, t)||2
≤ ||u(x, 0)||2
e(maxx Ux)t
.
This can also been done using Fourier Transform. See ChiuYen’s solutions where the
above method and the Fourier Transform methods are used.
33

Problem (S’90, #7). 34
Consider the nonlinear equation
ut + uux + uxx + uxxxx = 0,
u(x, 0) = f(x), f(x) = f(x + 2π) ∈ C∞
.
a) Derive an energy estimate for u.
b) Show that there is an interval 0 ≤ t ≤ T, T depending on f,
such that also ||∂u(·, t)/∂x|| can be bounded.
Proof. a) Multiply the above equation by u and integrate
uut + u2
ux + uuxx + uuxxxx = 0,
1
2
d
dt
(u2
) +
1
3
(u3
)x + uuxx + uuxxxx = 0,
1
2
d
dt
2π
0
u2
dx +
1
3
2π
0
(u3
)x dx +
2π
0
uuxx dx +
2π
0
uuxxxx dx = 0,
1
2
d
dt
||u||2
+
1
3
u3
2π
0
=0
−
2π
0
u2
x dx +
2π
0
u2
xx dx = 0,
1
2
d
dt
||u||2
=
2π
0
u2
x dx −
2π
0
u2
xx dx ≤ 0, (from S’90, #5)
⇒ ||u(·, t)|| ≤ ||u(·, 0)||.
b) In order to ﬁnd a bound for ||ux(·, t)||, diﬀerentiate with respect to x:
utx + (uux)x + uxxx + uxxxxx = 0,
Multiply the above equation by ux and integrate:
uxutx + ux(uux)x + uxuxxx + uxuxxxxx = 0,
1
2
d
dt
2π
0
(ux)2
dx +
2π
0
ux(uux)x dx +
2π
0
uxuxxx dx +
2π
0
uxuxxxxx dx = 0.
We evaluate one of the integrals in the above expression using the periodicity:
2π
0
ux(uux)x dx = −
2π
0
uxxuux =
2π
0
ux(u2
x + uuxx) =
2π
0
u3
x +
2π
0
uuxuxx,
⇒
2π
0
uxxuux = −
1
2
2π
0
u3
x,
⇒
2π
0
ux(uux)x =
1
2
2π
0
u3
x.
We have
1
2
d
dt
||ux||2
+
2π
0
u3
x dx +
2π
0
uxuxxx dx +
2π
0
uxuxxxxx dx = 0.
34

Let w = ux, then
1
2
d
dt
||w||2
= −
2π
0
w3
dx −
2π
0
wwxx dx −
2π
0
wwxxxx dx
= −
2π
0
w3
dx +
2π
0
w2
x dx −
2π
0
w2
xx dx ≤ −
2π
0
w3
dx,
⇒
d
dt
||ux||2
= −
2π
0
u3
x dx.

16 Problems: Wave Equation
16.1 The Initial Value Problem
Example (McOwen 3.1 #1). Solve the initial value problem:
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0,
u(x, 0) = x3
g(x)
, ut(x, 0) = sinx
h(x)
.
Proof. D’Alembert’s formula gives the solution:
u(x, t) =
1
2
(g(x + ct) + g(x − ct)) +
1
2c
x+ct
x−ct
h(ξ) dξ
=
1
2
(x + ct)3
+
1
2
(x − ct)3
+
1
2c
x+ct
x−ct
sin ξ dξ
= x3
+ 2xc2
t2
−
1
2c
cos(x + ct) +
1
2c
cos(x − ct) =
= x3
+ 2xc2
t2
+
1
c
sin x sinct.
Problem (S’99, #6). Solve the Cauchy problem
utt = a2uxx + cos x,
u(x, 0) = sin x, ut(x, 0) = 1 + x.
(16.1)
Proof. We have a nonhomogeneous PDE with nonhomogeneous initial conditions:
⎧
⎪⎪⎪⎨
⎪⎪⎪⎩
utt − c2uxx = cos x
f(x,t)
,
u(x, 0) = sin x
g(x)
, ut(x, 0) = 1 + x
h(x)
.
The solution is given by d’Alembert’s formula and Duhamel’s principle.35
uA
(x, t) =
1
2
(g(x + ct) + g(x − ct)) +
1
2c
x+ct
x−ct
h(ξ) dξ
=
1
2
(sin(x + ct) + sin(x − ct)) +
1
2c
x+ct
x−ct
(1 + ξ) dξ
= sinx cos ct +
1
2c
ξ +
ξ2
2
ξ=x+ct
ξ=x−ct
= sinx cos ct + xt + t.
uD
(x, t) =
1
2c
t
0
x+c(t−s)
x−c(t−s)
f(ξ, s) dξ ds =
1
2c
t
0
x+c(t−s)
x−c(t−s)
cos ξ dξ ds
=
1
2c
t
0
sin[x + c(t − s)] − sin[x − c(t − s)] ds =
1
c2
(cos x − cos x cos ct).
u(x, t) = uA
(x, t) + uD
(x, t) = sinx cos ct + xt + t +
1
c2
(cos x − cos x cos ct).
35
Note the relationship: x ↔ ξ, t ↔ s.

We can check that the solution satisﬁes equation (16.1). Can also check that uA
, uD
satisfy
uA
tt − c2
uA
xx = 0,
uA(x, 0) = sinx, uA
t (x, 0) = 1 + x;
uD
tt − c2
uD
xx = cos x,
uD(x, 0) = 0, uD
t (x, 0) = 0.

16.2 Initial/Boundary Value Problem
Problem 1. Consider the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0 0 < x < L, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) 0 < x < L
u(0, t) = 0, u(L, t) = 0 t ≥ 0.
(16.2)
Proof. Find u(x, t) in the form
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos
nπx
L
+ bn(t) sin
nπx
L
.
• Functions an(t) and bn(t) are determined by the boundary conditions:
0 = u(0, t) =
a0(t)
2
+
∞
n=1
an(t) ⇒ an(t) = 0. Thus,
u(x, t) =
∞
n=1
bn(t) sin
nπx
L
. (16.3)
• If we substitute (16.3) into the equation utt − c2uxx = 0, we get
∞
n=1
bn(t) sin
nπx
L
+ c2
∞
n=1
nπ
L
2
bn(t) sin
nπx
L
= 0, or
bn(t) +
nπc
L
2
bn(t) = 0,
whose general solution is
bn(t) = cn sin
nπct
L
+ dn cos
nπct
L
. (16.4)
Also, bn(t) = cn(nπc
L ) cos nπct
L − dn(nπc
L ) sin nπct
L .
• The constants cn and dn are determined by the initial conditions:
g(x) = u(x, 0) =
∞
n=1
bn(0) sin
nπx
L
=
∞
n=1
dn sin
nπx
L
,
h(x) = ut(x, 0) =
∞
n=1
bn(0) sin
nπx
L
=
∞
n=1
cn
nπc
L
sin
nπx
L
.
By orthogonality, we may multiply by sin(mπx/L) and integrate:
L
0
g(x) sin
mπx
L
dx =
L
0
∞
n=1
dn sin
nπx
L
sin
mπx
L
dx = dm
L
2
,
L
0
h(x) sin
mπx
L
dx =
L
0
∞
n=1
cn
nπc
L
sin
nπx
L
sin
mπx
L
dx = cm
mπc
L
L
2
.
Thus,
dn =
2
L
L
0
g(x) sin
nπx
L
dx, cn =
2
nπc
L
0
h(x) sin
nπx
L
dx. (16.5)
The formulas (16.3), (16.4), and (16.5) deﬁne the solution.

Example (McOwen 3.1 #2). Consider the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − uxx = 0 0 < x < π, t > 0
u(x, 0) = 1, ut(x, 0) = 0 0 < x < π
u(0, t) = 0, u(π, t) = 0 t ≥ 0.
(16.6)
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos nx + bn(t) sinnx.
0 = u(0, t) =
a0(t)
2
+
∞
n=1
an(t) ⇒ an(t) = 0. Thus,
u(x, t) =
∞
n=1
bn(t) sinnx. (16.7)
• If we substitute this into utt − uxx = 0, we get
∞
n=1
bn(t) sinnx +
∞
n=1
bn(t)n2
sinnx = 0, or
bn(t) + n2
bn(t) = 0,
whose general solution is
bn(t) = cn sinnt + dn cos nt. (16.8)
Also, bn(t) = ncn cos nt − ndn sinnt.
1 = u(x, 0) =
∞
n=1
bn(0) sinnx =
∞
n=1
dn sin nx,
0 = ut(x, 0) =
∞
n=1
bn(0) sinnx =
∞
n=1
ncn sinnx.
By orthogonality, we may multiply both equations by sinmx and integrate:
π
0
sin mx dx = dm
π
2
,
π
0
0 dx = ncn
π
2
.
Thus,
dn =
2
nπ
(1 − cos nπ) =
4
nπ , n odd,
0, n even,
and cn = 0. (16.9)
Using this in (16.8) and (16.7), we get
bn(t) =
4
nπ cos nt, n odd,
0, n even,

u(x, t) =
4
π
∞
n=0
cos(2n + 1)t sin(2n + 1)x
(2n + 1)
.

We can sum the series in regions bouded by characteristics. We have
u(x, t) =
4
π
∞
n=0
(2n + 1)
, or
u(x, t) =
2
π
∞
n=0
sin[(2n + 1)(x + t)]
(2n + 1)
+
2
π
∞
n=0
sin[(2n + 1)(x − t)]
(2n + 1)
. (16.10)
The initial condition may be written as
1 = u(x, 0) =
4
π
∞
n=0
sin(2n + 1)x
(2n + 1)
for 0 < x < π. (16.11)
We can use (16.11) to sum the series in (16.10).
In R1, u(x, t) =
1
2
+
1
2
= 1.
Since sin[(2n + 1)(x − t)] = − sin[(2n + 1)(−(x − t))], and 0 < −(x − t) < π in R2,
in R2, u(x, t) =
1
2
−
1
2
= 0.
Since sin[(2n + 1)(x + t)] = sin[(2n + 1)(x + t − 2π)] = − sin[(2n + 1)(2π − (x + t))],
and 0 < 2π − (x + t) < π in R3,
in R3, u(x, t) = −
1
2
+
1
2
= 0.
Since 0 < −(x − t) < π and 0 < 2π − (x + t) < π in R4,
in R4, u(x, t) = −
1
2
−
1
2
= −1.

Problem 2. Consider the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0 0 < x < L, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) 0 < x < L
ux(0, t) = 0, ux(L, t) = 0 t ≥ 0.
(16.12)
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos
nπx
L
+ bn(t) sin
nπx
L
.
ux(x, t) =
∞
n=1
−an(t)
nπ
L
sin
nπx
L
+ bn(t)
nπ
L
cos
nπx
L
,
0 = ux(0, t) =
∞
n=1
bn(t)
nπ
L
⇒ bn(t) = 0. Thus,
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos
nπx
L
. (16.13)
• If we substitute (16.13) into the equation utt − c2
uxx = 0, we get
a0(t)
2
+
∞
n=1
an(t) cos
nπx
L
+ c2
∞
n=1
an(t)
nπ
L
2
cos
nπx
L
= 0,
a0(t) = 0 and an(t) +
nπc
L
2
an(t) = 0,
whose general solutions are
a0(t) = c0t + d0 and an(t) = cn sin
nπct
L
+ dn cos
nπct
L
. (16.14)
Also, a0(t) = c0 and an(t) = cn(nπc
L ) cos nπct
L − dn(nπc
L ) sin nπct
L .
g(x) = u(x, 0) =
a0(0)
2
+
∞
n=1
an(0) cos
nπx
L
=
d0
2
+
∞
n=1
dn cos
nπx
L
,
h(x) = ut(x, 0) =
a0(0)
2
+
∞
n=1
an(0) cos
nπx
L
=
c0
2
+
∞
n=1
cn
nπc
L
cos
nπx
L
.
By orthogonality, we may multiply both equations by cos(mπx/L), including m = 0,
and integrate:
L
0
g(x) dx = d0
L
2
,
L
0
g(x) cos
mπx
L
dx = dm
L
2
,
L
0
h(x) dx = c0
L
2
,
L
0
h(x) cos
mπx
L
dx = cm
mπc
L
L
2
.
Thus,
dn =
2
L
L
0
g(x) cos
nπx
L
dx, cn =
2
nπc
L
0
h(x) cos
nπx
L
dx, c0 =
2
L
L
0
h(x) dx.
(16.15)
The formulas (16.13), (16.14), and (16.15) deﬁne the solution.

Example (McOwen 3.1 #3). Consider the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − uxx = 0 0 < x < π, t > 0
u(x, 0) = x, ut(x, 0) = 0 0 < x < π
ux(0, t) = 0, ux(π, t) = 0 t ≥ 0.
(16.16)
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos nx + bn(t) sinnx.
ux(x, t) =
∞
n=1
−an(t)n sinnx + bn(t)n cos nx,
0 = ux(0, t) =
∞
n=1
bn(t)n ⇒ bn(t) = 0. Thus,
u(x, t) =
a0(t)
2
+
∞
n=1
an(t) cos nx. (16.17)
• If we substitute (16.17) into the equation utt − uxx = 0, we get
a0(t)
2
+
∞
n=1
an(t) cos nx +
∞
n=1
an(t)n2
cos nx = 0,
a0(t) = 0 and an(t) + n2
an(t) = 0,
whose general solutions are
a0(t) = c0t + d0 and an(t) = cn sinnt + dn cos nt. (16.18)
Also, a0(t) = c0 and an(t) = cnn cos nt − dnn sinnt.
x = u(x, 0) =
a0(0)
2
+
∞
n=1
an(0) cos nx =
d0
2
+
∞
n=1
dn cos nx,
0 = ut(x, 0) =
a0(0)
2
+
∞
n=1
an(0) cos nx =
c0
2
+
∞
n=1
cnn cos nx.
By orthogonality, we may multiply both equations by cos mx, including m = 0, and
integrate:
π
0
x dx = d0
π
2
,
π
0
x cos mx dx = dm
π
2
,
π
0
0 dx = c0
π
2
,
π
0
0 cos mx dx = cmm
π
2
.
Thus,
d0 = π, dn =
2
πn2
(cos nπ − 1), cn = 0. (16.19)
Using this in (16.18) and (16.17), we get
a0(t) = d0 = π, an(t) =
2
πn2
(cos nπ − 1) cosnt,

u(x, t) =
π
2
+
2
π
∞
n=1
(cos nπ − 1) cos nt cos nx
n2
.

We can sum the series in regions bouded by characteristics. We have
u(x, t) =
π
2
+
2
π
∞
n=1
(cos nπ − 1) cosnt cos nx
n2
, or
u(x, t) =
π
2
+
1
π
∞
n=1
(cos nπ − 1) cos[n(x − t)]
n2
+
1
π
∞
n=1
(cos nπ − 1) cos[n(x + t)]
n2
. (16.20)
The initial condition may be written as
u(x, 0) = x =
π
2
+
2
π
∞
n=1
(cos nπ − 1) cosnx
n2
for 0 < x < π,
which implies
x
2
−
π
4
=
1
π
∞
n=1
(cos nπ − 1) cosnx
n2
for 0 < x < π, (16.21)
We can use (16.21) to sum the series in (16.20).
In R1, u(x, t) =
π
2
+
x − t
2
−
π
4
+
x + t
2
−
π
4
= x.
Since cos[n(x − t)] = cos[n(−(x − t))], and 0 < −(x − t) < π in R2,
in R2, u(x, t) =
π
2
+
−(x − t)
2
−
π
4
+
x + t
2
−
π
4
= t.
Since cos[n(x+t)] = cos[n(x+t−2π)] = cos[n(2π−(x+t))], and 0 < 2π−(x+t) < π
in R3,
in R3, u(x, t) =
π
2
+
x − t
2
−
π
4
+
2π − (x + t)
2
−
π
4
= π − t.
Since 0 < −(x − t) < π and 0 < 2π − (x + t) < π in R4
in R4, u(x, t) =
π
2
+
−(x − t)
2
−
π
4
+
2π − (x + t)
2
−
π
4
= π − x.

Example (McOwen 3.1 #4). Consider the initial boundary value problem
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0 for x > 0, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x > 0
u(0, t) = 0 for t ≥ 0,
(16.22)
where g(0) = 0 = h(0). If we extend g and h as odd functions on −∞ < x < ∞, show
that d’Alembert’s formula gives the solution.
Proof. Extend g and h as odd functions on −∞ < x < ∞:
˜g(x) =
g(x), x ≥ 0
−g(−x), x < 0
˜h(x) =
h(x), x ≥ 0
−h(−x), x < 0.
Then, we need to solve
˜utt − c2˜uxx = 0 for − ∞ < x < ∞, t > 0
˜u(x, 0) = ˜g(x), ˜ut(x, 0) = ˜h(x) for − ∞ < x < ∞.
(16.23)
To show that d’Alembert’s formula gives the solution to (16.23), we need to show that
the solution given by d’Alembert’s formula satisﬁes the boundary condition ˜u(0, t) = 0.
˜u(x, t) =
1
2
(˜g(x + ct) + ˜g(x − ct)) +
1
2c
x+ct
x−ct
˜h(ξ) dξ,
˜u(0, t) =
1
2
(˜g(ct) + ˜g(−ct)) +
1
2c
ct
−ct
˜h(ξ) dξ
=
1
2
(˜g(ct) − ˜g(ct)) +
1
2c
(H(ct) − H(−ct))
= 0 +
1
2c
(H(ct) − H(ct)) = 0,
where we used H(x) =
x
0
˜h(ξ) dξ; and since ˜h is odd, then H is even.
Example (McOwen 3.1 #5). Find in closed form (similar to d’Alembet’s formula)
the solution u(x, t) of
⎧
⎪⎨
⎪⎩
utt − c2
uxx = 0 for x, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x > 0
u(0, t) = α(t) for t ≥ 0,
(16.24)
where g, h, α ∈ C2 satisfy α(0) = g(0), α (0) = h(0), and α (0) = c2g (0). Verify that
u ∈ C2
, even on the characteristic x = ct.
Proof. As in (McOwen 3.1 #4), we can extend g and h to be odd functions. We want
to transform the problem to have zero boundary conditions.
Consider the function:
U(x, t) = u(x, t) − α(t). (16.25)

Then (16.24) transforms to:
⎧
⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎩
Utt − c2Uxx = −α (t)
fU (x,t)
U(x, 0) = g(x) − α(0)
gU (x)
, Ut(x, 0) = h(x) − α (0)
hU (x)
U(0, t) = 0
αu(t)
.
We use d’Alembert’s formula and Duhamel’s principle on U.
After getting U, we can get u from u(x, t) = U(x, t) + α(t).

Example (Zachmanoglou, Chapter 8, Example 7.2). Find the solution of
⎧
⎪⎨
⎪⎩
utt − c2uxx = 0 for x > 0, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x > 0
ux(0, t) = 0 for t > 0.
(16.26)
Proof. Extend g and h as even functions on −∞ < x < ∞:
˜g(x) =
g(x), x ≥ 0
g(−x), x < 0
˜h(x) =
h(x), x ≥ 0
h(−x), x < 0.
˜utt − c2˜uxx = 0 for − ∞ < x < ∞, t > 0
˜u(x, 0) = ˜g(x), ˜ut(x, 0) = ˜h(x) for − ∞ < x < ∞.
(16.27)
To show that d’Alembert’s formula gives the solution to (16.27), we need to show that
the solution given by d’Alembert’s formula satisﬁes the boundary condition ˜ux(0, t) = 0.
˜u(x, t) =
1
2
(˜g(x + ct) + ˜g(x − ct)) +
1
2c
x+ct
x−ct
˜h(ξ) dξ.
˜ux(x, t) =
1
2
(˜g (x + ct) + ˜g (x − ct)) +
1
2c
[˜h(x + ct) − ˜h(x − ct)],
˜ux(0, t) =
1
2
(˜g (ct) + ˜g (−ct)) +
1
2c
[˜h(ct) − ˜h(−ct)] = 0.
Since ˜g is even, then g is odd.
Problem (F’89, #3). 36
Let α = c, constant. Find the solution of
⎧
⎪⎨
⎪⎩
utt − c2
uxx = 0 for x > 0, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x > 0
ut(0, t) = αux(0, t) for t > 0,
(16.28)
where g, h ∈ C2 for x > 0 and vanish near x = 0.
Hint: Use the fact that a general solution of (16.28) can be written as the sum of two
traveling wave solutions.
Proof. D’Alembert’s formula is derived by plugging in the following into the above
equation and initial conditions:
u(x, t) = F(x + ct) + G(x − ct).
As in (Zachmanoglou 7.2), we can extend g and h to be even functions.
36
Similar to McOwen 3.1 #5. The notation in this problem is changed to be consistent with McOwen.

Example (McOwen 3.1 #6). Solve the initial/boundary value problem
⎧
⎪⎨
⎪⎩
utt − uxx = 1 for 0 < x < π and t > 0
u(x, 0) = 0, ut(x, 0) = 0 for 0 < x < π
u(0, t) = 0, u(π, t) = −π2/2 for t ≥ 0.
(16.29)
Proof. If we first find a particular solution of the nonhomogeneous equation, this re-
duces the problem to a boundary value problem for the homogeneous equation ( as in
(McOwen 3.1 #2) and (McOwen 3.1 #3) ).
Hint: You should use a particular solution depending on x!
❶ First, find a particular solution. This is similar to the method of separation of
variables. Assume
up(x, t) = X(x),
which gives
−X (x) = 1,
X (x) = −1.
The solution to the above ODE is
X(x) = −
x2
2
+ ax + b.
The boundary conditions give
up(0, t) = b = 0,
up(π, t) = −
π2
2
+ aπ + b = −
π2
2
, ⇒ a = b = 0.
Thus, the particular solution is
up(x, t) = −
x2
2
.
This solution satisfies the following:
⎧
⎪⎨
⎪⎩
uptt − upxx = 1
up(x, 0) = −x2
2 , upt(x, 0) = 0
up(0, t) = 0, up(π, t) = −π2
2 .
❷ Second, we find a solution to a boundary value problem for the homogeneous equa-
tion:
⎧
⎪⎨
⎪⎩
utt − uxx = 0
u(x, 0) = x2
2 , ut(x, 0) = 0
u(0, t) = 0, u(π, t) = 0.
This is solved by the method of Separation of Variables. See Separation of Variables
subsection of “Problems: Separation of Variables: Wave Equation” McOwen 3.1 #2.
The only difference there is that u(x, 0) = 1.
We would find uh(x, t). Then,
u(x, t) = uh(x, t) + up(x, t).

Problem (S’02, #2). a) Given a continuous function f on R which vanishes for
|x| > R, solve the initial value problem
utt − uxx = f(x) cos t,
u(x, 0) = 0, ut(x, 0) = 0, −∞ < x < ∞, 0 ≤ t < ∞
by first finding a particular solution by separation of variables and then adding the
appropriate solution of the homogeneous PDE.
b) Since the particular solution is not unique, it will not be obvious that the solution
to the initial value problem that you have found in part (a) is unique. Prove that it is
unique.
Proof. a) ❶ First, find a particular solution by separation of variables. Assume
up(x, t) = X(x) cost,
which gives
−X(x) cost − X (x) cost = f(x) cos t,
X + X = −f(x).
The solution to the above ODE is written as X = Xh +Xp. The homogeneous solution
is
Xh(x) = a cos x + b sinx.
To find a particular solution, note that since f is continuous, ∃G ∈ C2
(R), such that
G + G = −f(x).
Thus,
Xp(x) = G(x).
⇒ X(x) = Xh(x) + Xp(x) = a cos x + b sinx + G(x).
up(x, t) = a cos x + b sinx + G(x) cos t.
It can be verified that this solution satisfies the following:
uptt − upxx = f(x) cost,
up(x, 0) = a cos x + b sinx + G(x), upt(x, 0) = 0.
❷ Second, we find a solution of the homogeneous PDE:
⎧
⎪⎨
⎪⎩
utt − uxx = 0,
u(x, 0) = −a cos x − b sinx − G(x)
g(x)
, ut(x, 0) = 0
h(x)
.
The solution is given by d’Alembert’s formula (with c = 1):
uh(x, t) = uA
(x, t) =
1
2
(g(x + t) + g(x − t)) +
1
2
x+t
x−t
h(ξ) dξ
=
1
2
− a cos(x + t) − b sin(x + t) − G(x + t) + − a cos(x − t) − b sin(x − t) − G(x − t)
= −
1
2
a cos(x + t) + b sin(x + t) + G(x + t) −
1
2
a cos(x − t) + b sin(x − t) + G(x − t) .

It can be verified that the solution satisfies the above homogeneous PDE with the
boundary conditions. Thus, the complete solution is:
u(x, t) = uh(x, t) + up(x, t).
Alternatively, we could use Duhamel’s principle to find the solution: 37
u(x, t) =
1
2
t
0
x+(t−s)
x−(t−s)
f(ξ) cos s dξ ds.
However, this is not how it was suggested to do this problem.
b) The particular solution is not unique, since any constants a, b give the solution.
However, we show that the solution to the initial value problem is unique.
Suppose u1 and u2 are two solutions. Then w = u1 − u2 satisfies:
wtt − wxx = 0,
w(x, 0) = 0, wt(x, 0) = 0.
D’Alembert’s formula gives
w(x, t) =
1
2
(g(x + t) + g(x − t)) +
1
2
x+t
x−t
h(ξ) dξ = 0.
Thus, the solution to the initial value problem is unique.
37
Note the relationship: x ↔ ξ, t ↔ s.

16.3 Similarity Solutions
Problem (F’98, #7). Look for a similarity solution of the form
v(x, t) = tα
w(y = x/tβ
) for the differential equation
vt = vxx + (v2
)x. (16.30)
a) Find the parameters α and β.
b) Find a differential equation for w(y) and show that this ODE can be reduced to first
order.
c) Find a solution for the resulting first order ODE.
Proof. We can rewrite (16.30) as
vt = vxx + 2vvx. (16.31)
We look for a similarity solution of the form
v(x, t) = tα
w(y), y =
x
tβ
.
vt = αtα−1
w + tα
w yt = αtα−1
w + tα
−
βx
tβ+1
w = αtα−1
w − tα−1
βyw ,
vx = tα
w yx = tα
w t−β
= tα−β
w ,
vxx = (tα−β
w )x = tα−β
w yx = tα−β
w t−β
= tα−2β
w .
Plugging in the derivatives we calculated into (16.31), we obtain
αtα−1
w − tα−1
βyw = tα−2β
w + 2(tα
w)(tα−β
w ),
αw − βyw = t1−2β
w + 2tα−β+1
ww .
The parameters that would eliminate t from equation above are
β =
1
2
, α = −
1
2
.
With these parameters, we obtain the differential equation for w(y):
−
1
2
w −
1
2
yw = w + 2ww ,
w + 2ww +
1
2
yw +
1
2
w = 0.
We can write the ODE as
w + 2ww +
1
2
(yw) = 0.
Integrating it with respect to y, we obtain the first order ODE:
w + w2
+
1
2
yw = c.

16.4 Traveling Wave Solutions
Consider the Korteweg-de Vries (KdV) equation in the form 38
ut + 6uux + uxxx = 0, −∞ < x < ∞, t > 0. (16.32)
We look for a traveling wave solution
u(x, t) = f(x − ct). (16.33)
We get the ODE
−cf + 6ff + f = 0. (16.34)
We integrate (16.34) to get
−cf + 3f2
+ f = a, (16.35)
where a is a constant. Multiplying this equality by f , we obtain
−cff + 3f2
f + f f = af .
Integrating again, we get
−
c
2
f2
+ f3
+
(f )2
2
= af + b. (16.36)
We are looking for solutions f which satisfy f(x), f (x), f (x) → 0 as x → ±∞. (In
which case the function u having the form (16.33) is called a solitary wave.) Then
(16.35) and (16.36) imply a = b = 0, so that
−
c
2
f2
+ f3
+
(f )2
2
= 0, or f = ±f c − 2f.
The solution of this ODE is
f(x) =
c
2
sech2
[
√
c
2
(x − x0)],
where x0 is the constant of integration. A solution of this form is called a soliton.
38
Evans, p. 174; Strauss, p. 367.

Problem (S’93, #6). The generalized KdV equation is
∂u
∂t
=
1
2
(n + 1)(n + 2)un ∂u
∂x
−
∂3u
∂x3
,
where n is a positive integer. Solitary wave solutions are sought in which u = f(η),
where η = x − ct and
f, f , f → 0, as |η| → ∞;
c, the wave speed, is constant.
Show that
f 2
= fn+2
+ cf2
.
Hence show that solitary waves do not exist if n is even.
Show also that, when n = 1, all conditions of the problem are satisﬁed provided c > 0
and
u = −c sech2 1
2
√
c(x − ct) .
Proof. • We look for a traveling wave solution
u(x, t) = f(x − ct).
We get the ODE
−cf =
1
2
(n + 1)(n + 2)fn
f − f ,
Integrating this equation, we get
−cf =
1
2
(n + 2)fn+1
− f + a, (16.37)
where a is a constant. Multiplying this equality by f , we obtain
−cff =
1
2
(n + 2)fn+1
f − f f + af .
Integrating again, we get
−
cf2
2
=
1
2
fn+2
−
(f )2
2
+ af + b. (16.38)
We are looking for solutions f which satisfy f, f , f → 0 as x → ±∞. Then (16.37)
and (16.38) imply a = b = 0, so that
−
cf2
2
=
1
2
fn+2
−
(f )2
2
,
(f )2
= fn+2
+ cf2
.
• We show that solitary waves do not exist if n is even. We have
f = ± fn+2 + cf2 = ±|f| fn + c,
∞
−∞
f dη = ±
∞
−∞
|f| fn + c dη,
f
∞
−∞
= ±
∞
−∞
|f| fn + c dη,
0 = ±
∞
−∞
|f| fn + c dη.

Thus, either ➀ |f| ≡ 0 ⇒ f = 0, or
➁ fn + c = 0. Since f → 0 as x → ±∞, we have c = 0 ⇒ f = 0.
Thus, solitary waves do not exist if n is even.

• When n = 1, we have
(f )2
= f3
+ cf2
. (16.39)
We show that all conditions of the problem are satisﬁed provided c > 0, including
u = −c sech2 1
2
√
c(x − ct) , or
f = −c sech2 η
√
c
2
= −
c
cosh2
[η
√
c
2 ]
= −c cosh
η
√
c
2
−2
.
We have
f = 2c cosh
η
√
c
2
−3
· sinh
η
√
c
2
·
√
c
2
= c
√
c cosh
η
√
c
2
−3
· sinh
η
√
c
2
,
(f )2
=
c3
sinh2 η
√
c
2
cosh6 η
√
c
2
,
f3
= −
c3
cosh6 η
√
c
2
,
cf2
=
c3
cosh4 η
√
c
2
.
Plugging these into (16.39), we obtain: 39
c3
sinh2 η
√
c
2
cosh6 η
√
c
2
= −
c3
cosh6 η
√
c
2
+
c3
cosh4 η
√
c
2
,
c3
sinh2 η
√
c
2
cosh6 η
√
c
2
=
−c3
+ c3
cosh2 η
√
c
2
cosh6 η
√
c
2
,
c3
sinh2 η
√
c
2
cosh6 η
√
c
2
=
c3
sinh2 η
√
c
2
cosh6 η
√
c
2
.
Also, f, f , f → 0, as |η| → ∞, since
f(η) = −c sech2 η
√
c
2
= −
c
cosh2
[η
√
c
2 ]
= −c
2
e[ η
√
c
2
]
+ e−[ η
√
c
2
]
2
→ 0, as |η| → ∞.
Similarly, f , f → 0, as |η| → ∞.
39
cosh2
x − sinh2
x = 1.
cosh x =
ex
+ e−x
2
, sinh x =
ex
− e−x
2

Problem (S’00, #5). Look for a traveling wave solution of the PDE
utt + (u2
)xx = −uxxxx
of the form u(x, t) = v(x − ct). In particular, you should ﬁnd an ODE for v. Under
the assumption that v goes to a constant as |x| → ∞, describe the form of the solution.
Proof. Since (u2
)x = 2uux, and (u2
)xx = 2u2
x + 2uuxx, we have
utt + 2u2
x + 2uuxx = −uxxxx.
We look for a traveling wave solution
u(x, t) = v(x − ct).
We get the ODE
c2
v + 2(v )2
+ 2vv = −v ,
c2
v + 2((v )2
+ vv ) = −v ,
c2
v + 2(vv ) = −v , (exact diﬀerentials)
c2
v + 2vv = −v + a, s = x − ct
c2
v + v2
= −v + as + b,
v + c2
v + v2
= a(x − ct) + b.
Since v → C = const as |x| → ∞, we have v , v → 0, as |x| → ∞. Thus, implies
c2
v + v2
= as + b.
Since |x| → ∞, but v → C, we have a = 0:
v2
+ c2
v − b = 0.
v =
−c2 ±
√
c4 + 4b
2
.

Problem (S’95, #2). Consider the KdV-Burgers equation
ut + uux = uxx + δuxxx
in which > 0, δ > 0.
a) Find an ODE for traveling wave solutions of the form
u(x, t) = ϕ(x − st)
with s > 0 and
lim
y→−∞
ϕ(y) = 0
and analyze the stationary points from this ODE.
b) Find the possible (finite) values of
ϕ+ = lim
y→∞
ϕ(y).
Proof. a) We look for a traveling wave solution
u(x, t) = ϕ(x − st), y = x − st.
We get the ODE
−sϕ + ϕϕ = ϕ + δϕ ,
−sϕ +
1
2
ϕ2
= ϕ + δϕ + a.
Since ϕ → 0 as y → −∞, then ϕ , ϕ → 0 as y → −∞. Therefore, at y = −∞, a = 0.
We found the following ODE,
ϕ +
δ
ϕ +
s
δ
ϕ −
1
2δ
ϕ2
= 0.
In order to find and analyze the stationary points of an ODE above, we write it as a
first-order system.
φ1 = ϕ,
φ2 = ϕ .
φ1 = ϕ = φ2,
φ2 = ϕ = −
δ
ϕ −
s
δ
ϕ +
1
2δ
ϕ2
= −
δ
φ2 −
s
δ
φ1 +
1
2δ
φ2
1.
φ1 = φ2 = 0,
φ2 = −δ φ2 − s
δ φ1 + 1
2δ φ2
1 = 0;
⇒
φ1 = φ2 = 0,
φ2 = −s
δ φ1 + 1
2δ φ2
1 = 0;
⇒
φ1 = φ2 = 0,
φ2 = −1
δ φ1(s − 1
2φ1) = 0.
Stationary points: (0, 0), (2s, 0), s > 0.
φ1 = φ2 = f(φ1, φ2),
φ2 = −
δ
φ2 −
s
δ
φ1 +
1
2δ
φ2
1 = g(φ1, φ2).

In order to classify a stationary point, need to ﬁnd eigenvalues of a linearized system
at that point.
J(f(φ1, φ2), g(φ1, φ2)) =
∂f
∂φ1
∂f
∂φ2
∂g
∂φ1
∂g
∂φ2
=
0 1
−s
δ + 1
δ φ1 −δ
.
• For (φ1, φ2) = (0, 0) :
det(J|(0,0) − λI) =
−λ 1
−s
δ −δ − λ
= λ2
+ δ λ + s
δ = 0.
λ± = −2δ ±
2
4δ2 − s
δ .
If
2
4δ > s ⇒ λ± ∈ R, λ± < 0.
⇒ (0,0) is Stable Improper Node.
If
2
4δ < s ⇒ λ± ∈ C, Re(λ±) < 0.
⇒ (0,0) is Stable Spiral Point.
• For (φ1, φ2) = (2s, 0) :
det(J|(2s,0) − λI) =
−λ 1
s
δ −δ − λ
= λ2
+ δ λ − s
δ = 0.
λ± = −2δ ±
2
4δ2 + s
δ .
⇒ λ+ > 0, λ− < 0.
⇒ (2s,0) is Untable Saddle Point.
b) Since
lim
y→−∞
ϕ(y) = 0 = lim
t→∞
ϕ(x − st),
we may have
lim
y→+∞
ϕ(y) = lim
t→−∞
ϕ(x − st) = 2s.
That is, a particle may start oﬀ at an unstable node (2s, 0) and as t increases, approach
the stable node (0, 0).
A phase diagram with (0, 0) being a stable spiral point, is shown below.

Problem (F’95, #8). Consider the equation
ut + f(u)x = uxx
where f is smooth and > 0. We seek traveling wave solutions to this equation,
i.e., solutions of the form u = φ(x − st), under the boundary conditions
u → uL and ux → 0 as x → −∞,
u → uR and ux → 0 as x → +∞.
Find a necessary and suﬃcient condition on f, uL, uR and s for such traveling waves
to exist; in case this condition holds, write an equation which deﬁnes φ implicitly.
Proof. We look for traveling wave solutions
u(x, t) = φ(x − st), y = x − st.
The boundary conditions become
φ → uL and φ → 0 as x → −∞,
φ → uR and φ → 0 as x → +∞.
Since f(φ(x − st))x = f (φ)φ , we get the ODE
−sφ + f (φ)φ = φ ,
−sφ + (f(φ)) = φ ,
−sφ + f(φ) = φ + a,
φ =
−sφ + f(φ)
+ b.
We use boundary conditions to determine constant b:
At x = −∞, 0 = φ =
−suL + f(uL)
+ b ⇒ b =
suL − f(uL)
.
At x = +∞, 0 = φ =
−suR + f(uR)
+ b ⇒ b =
suR − f(uR)
.
s =
f(uL) − f(uR)
uL − uR
.
40
40
For the solution for the second part of the problem, refer to Chiu-Yen’s solutions.

Problem (S’02, #5; F’90, #2). Fisher’s Equation. Consider
ut = u(1 − u) + uxx, −∞ < x < ∞, t > 0.
The solutions of physical interest satisfy 0 ≤ u ≤ 1, and
lim
x→−∞
u(x, t) = 0, lim
x→+∞
u(x, t) = 1.
One class of solutions is the set of “wavefront” solutions. These have the form u(x, t) =
φ(x + ct), c ≥ 0.
Determine the ordinary diﬀerential equation and boundary conditions which φ must
satisfy (to be of physical interest). Carry out a phase plane analysis of this equation,
and show that physically interesting wavefront solutions are possible if c ≥ 2, but not if
0 ≤ c < 2.
Proof. We look for a traveling wave solution
u(x, t) = φ(x + ct), s = x + ct.
We get the ODE
cφ = φ(1 − φ) + φ ,
φ − cφ + φ − φ2
= 0,
◦ φ(s) → 0, as s → −∞,
◦ φ(s) → 1, as s → +∞,
◦ 0 ≤ φ ≤ 1.
y1 = φ,
y2 = φ .
y1 = φ = y2,
y2 = φ = cφ − φ + φ2
= cy2 − y1 + y2
1.
y1 = y2 = 0,
y2 = cy2 − y1 + y2
1 = 0;
⇒
y2 = 0,
y1(y1 − 1) = 0.
Stationary points: (0, 0), (1, 0).
y1 = y2 = f(y1, y2),
y2 = cy2 − y1 + y2
1 = g(y1, y2).
at that point.
J(f(y1, y2), g(y1, y2)) =
∂f
∂y1
∂f
∂y2
∂g
∂y1
∂g
∂y2
=
0 1
2y1 − 1 c
.

• For (y1, y2) = (0, 0) :
det(J|(0,0) − λI) =
−λ 1
−1 c − λ
= λ2
− cλ + 1 = 0.
λ± =
c ±
√
c2 − 4
2
.
If c ≥ 2 ⇒ λ± ∈ R, λ± > 0.
(0,0) is Unstable Improper (c > 2) / Proper (c = 2) Node.
If 0 ≤ c < 2 ⇒ λ± ∈ C, Re(λ±) ≥ 0.
(0,0) is Unstable Spiral Node.
• For (y1, y2) = (1, 0) :
det(J|(1,0) − λI) =
−λ 1
1 c − λ
= λ2
− cλ − 1 = 0.
λ± =
c ±
√
c2 + 4
2
.
If c ≥ 0 ⇒ λ+ > 0, λ− < 0.
(1,0) is Unstable Saddle Point.
By looking at the phase plot, a particle may start oﬀ at an unstable node (0, 0) and as
t increases, approach the unstable node (1, 0).

Problem (F’99, #6). For the system
∂tρ + ∂xu = 0
∂tu + ∂x(ρu) = ∂2
xu
look for traveling wave solutions of the form ρ(x, t) = ρ(y = x−st), u(x, t) = u(y =
x − st). In particular
a) Find a first order ODE for u.
b) Show that this equation has solutions of the form
u(y) = u0 + u1 tanh(αy + y0),
for some constants u0, u1, α, y0.
Proof. a) We rewrite the system:
ρt + ux = 0
ut + ρxu + ρux = uxx
We look for traveling wave solutions
ρ(x, t) = ρ(x − st), u(x, t) = u(x − st), y = x − st.
We get the system of ODEs
−sρ + u = 0,
−su + ρ u + ρu = u .
The first ODE gives
ρ =
1
s
u ,
ρ =
1
s
u + a,
where a is a constant, and integration was done with respect to y. The second ODE
gives
−su +
1
s
u u +
1
s
u + a u = u ,
−su +
2
s
uu + au = u . Integrating, we get
−su +
1
s
u2
+ au = u + b.
u =
1
s
u2
+ (a − s)u − b.
b) Note that the ODE above may be written in the following form:
u + Au2
+ Bu = C,
which is a nonlinear first order equation.

Problem (S’01, #7). Consider the following system of PDEs:
ft + fx = g2
− f2
gt − gx = f2
− g
a) Find a system of ODEs that describes traveling wave solutions of the PDE
system; i.e. for solutions of the form f(x, t) = f(x − st) and g(x, t) = g(x − st).
b) Analyze the stationary points and draw the phase plane for this ODE system in the
standing wave case s = 0.
Proof. a) We look for traveling wave solutions
f(x, t) = f(x − st), g(x, t) = g(x − st).
We get the system of ODEs
−sf + f = g2
− f2
,
−sg − g = f2
− g.
Thus,
f =
g2 − f2
1 − s
,
g =
f2
− g
−1 − s
.
b) If s = 0, the system becomes
f = g2
− f2
,
g = g − f2.
Relabel the variables f → y1, g → y2.
y1 = y2
2 − y2
1 = 0,
y2 = y2 − y2
1 = 0.
Stationary points: (0, 0), (−1, 1), (1, 1).
y1 = y2
2 − y2
1 = φ(y1, y2),
y2 = y2 − y2
1 = ψ(y1, y2).
at that point.
J(φ(y1, y2), ψ(y1, y2)) =
∂φ
∂y1
∂φ
∂y2
∂ψ
∂y1
∂ψ
∂y2
=
−2y1 2y2
−2y1 1
.
• For (y1, y2) = (0, 0) :
det(J|(0,0) − λI) =
−λ 0
0 1 − λ
= −λ(1 − λ) = 0.
λ1 = 0, λ2 = 1; eigenvectors: v1 =
1
0
, v2 =
0
1
.
(0,0) is Unstable Node.

• For (y1, y2) = (−1, 1) :
det(J|(−1,1) − λI) =
2 − λ 2
2 1 − λ
= λ2 − 3λ − 2 = 0.
λ± =
3
2
±
√
17
2
.
λ− < 0, λ+ > 0.
(-1,1) is Unstable Saddle Point.
• For (y1, y2) = (1, 1) :
det(J|(1,1) − λI) =
−2 − λ 2
−2 1 − λ
= λ2
+ λ + 2 = 0.
λ± = −
1
2
± i
√
7
2
.
Re(λ±) < 0.
(1,1) is Stable Spiral Point.

16.5 Dispersion
Problem (S’97, #8). Consider the following equation
ut = (f(ux))x − αuxxxx, f(v) = v2
− v, (16.40)
with constant α.
a) Linearize this equation around u = 0 and find the principal mode solution of the
form eωt+ikx
. For which values of α are there unstable modes, i.e., modes with ω = 0
for real k? For these values, find the maximally unstable mode, i.e., the value of k with
the largest positive value of ω.
b) Consider the steady solution of the (fully nonlinear) problem. Show that the resulting
equation can be written as a second order autonomous ODE for v = ux and draw the
corresponding phase plane.
Proof. a) We have
ut = (f(ux))x − αuxxxx,
ut = (u2
x − ux)x − αuxxxx,
ut = 2uxuxx − uxx − αuxxxx.
However, we need to linearize (16.40) around u = 0. To do this, we need to linearize f.
f(u) = f(0) + uf (0) +
u2
2
f (0) + · · · = 0 + u(0 − 1) + · · · = −u + · · · .
Thus, we have
ut = −uxx − αuxxxx.
Consider u(x, t) = eωt+ikx.
ωeωt+ikx
= (k2
− αk4
)eωt+ikx
,
ω = k2
− αk4
.
To find unstable nodes, we set ω = 0, to get
α =
1
k2
.
• To find the maximally unstable mode, i.e., the value of k with the largest positive
value of ω, consider
ω(k) = k2
− αk4
,
ω (k) = 2k − 4αk3
.
To find the extremas of ω, we set ω = 0. Thus,the extremas are at
k1 = 0, k2,3 = ±
1
2α
.
To find if the extremas are maximums or minimums, we set ω = 0:
ω (k) = 2 − 12αk2
= 0,
ω (0) = 2 > 0 ⇒ k = 0 is the minimum.
ω ±
1
2α
= −4 < 0 ⇒ k = ±
1
2α
is the maximum unstable mode.
ω ±
1
2α
=
1
4α
is the largest positive value of ω.

b) Integrating , we get
u2
x − ux − αuxxx = 0.
Let v = ux. Then,
v2
− v − αvxx = 0, or
v =
v2
− v
α
.
y1 = v,
y2 = v .
y1 = v = y2,
y2 = v =
v2
− v
α
=
y2
1 − y1
α
.
y1 = y2 = 0,
y2 =
y2
1 −y1
α = 0;
⇒
y2 = 0,
y1(y1 − 1) = 0.
Stationary points: (0, 0), (1, 0).
y1 = y2 = f(y1, y2),
y2 =
y2
1 − y1
α
= g(y1, y2).
at that point.
J(f(y1, y2), g(y1, y2)) =
∂f
∂y1
∂f
∂y2
∂g
∂y1
∂g
∂y2
=
0 1
2y1−1
α 0
.
• For (y1, y2) = (0, 0), λ± = ± −1
α .
If α < 0, λ± ∈ R, λ+ > 0, λ− < 0. ⇒ (0,0) is Unstable Saddle Point.
If α > 0, λ± = ±i 1
α ∈ C, Re(λ±) = 0. ⇒ (0,0) is Spiral Point.
• For (y1, y2) = (1, 0), λ± = ± 1
α .
If α < 0, λ± = ±i −1
α ∈ C, Re(λ±) = 0. ⇒ (1,0) is Spiral Point.
If α > 0, λ± ∈ R, λ+ > 0, λ− < 0. ⇒ (1,0) is Unstable Saddle Point.

16.6 Energy Methods
Problem (S’98, #9; S’96, #5). Consider the following initial-boundary value
problem for the multi-dimensional wave equation:
utt = u in Ω × (0, ∞),
u(x, 0) = f(x),
∂u
∂t
(x, 0) = g(x) for x ∈ Ω,
∂u
∂n
+ a(x)
∂u
∂t
= 0 on ∂Ω.
Here, Ω is a bounded domain in Rn and a(x) ≥ 0. Deﬁne the Energy integral for this
problem and use it in order to prove the uniqueness of the classical solution of the prob-
lem.
Proof.
d ˜E
dt
= 0 =
Ω
(utt − u)ut dx =
Ω
uttut dx −
∂Ω
∂u
∂n
ut ds +
Ω
∇u · ∇ut dx
=
Ω
1
2
∂
∂t
(u2
t ) dx +
Ω
1
2
∂
∂t
|∇u|2
dx +
∂Ω
a(x)u2
t ds.
Thus,
−
∂Ω
a(x)u2
t dx
≤0
=
1
2
∂
∂t Ω
u2
t + |∇u|2
dx.
Let Energy integral be
E(t) =
1
2 Ω
u2
t + |∇u|2
dx.
In order to prove that the given E(t) ≤ 0 from scratch, take its derivative with respect
to t:
dE
dt
(t) =
Ω
ututt + ∇u · ∇ut dx
=
Ω
ututt dx +
∂Ω
ut
∂u
∂n
ds −
Ω
ut u dx
=
Ω
ut(utt − u) dx
=0
−
∂Ω
a(x)u2
t dx ≤ 0.
Thus, E(t) ≤ E(0).
To prove the uniqueness of the classical solution, suppose u1 and u2 are two solutions
of the initial boundary value problem. Let w = u1 − u2. Then, w satisﬁes
wtt = w in Ω × (0, ∞),
w(x, 0) = 0, wt(x, 0) = 0 for x ∈ Ω,
∂w
∂n
+ a(x)
∂w
∂t
= 0 on ∂Ω.
We have
Ew(0) =
1
2 Ω
(wt(x, 0)2
+ |∇w(x, 0)|2
) dx = 0.

Ew(t) ≤ Ew(0) = 0 ⇒ Ew(t) = 0. Thus, wt = 0, wxi = 0 ⇒ w(x, t) = const = 0.
Hence, u1 = u2.
Problem (S’94, #7). Consider the wave equation
1
c2(x)
utt = u x ∈ Ω
∂u
∂t
− α(x)
∂u
∂n
= 0 on ∂Ω × R.
Assume that α(x) is of one sign for all x (i.e. α always positive or α always negative).
For the energy
E(t) =
1
2 Ω
1
c2(x)
u2
t + |∇u|2
dx,
show that the sign of dE
dt is determined by the sign of α.
Proof. We have
dE
dt
(t) =
Ω
1
c2(x)
ututt + ∇u · ∇ut dx
=
Ω
1
c2(x)
ututt dx +
∂Ω
ut
∂u
∂n
ds −
Ω
ut u dx
=
Ω
ut
1
c2(x)
utt − u dx
=0
+
∂Ω
1
α(x)
u2
t dx
=
∂Ω
1
α(x)
u2
t dx =
> 0, if α(x) > 0, ∀x ∈ Ω,
< 0, if α(x) < 0, ∀x ∈ Ω.

Problem (F’92, #2). Let Ω ∈ Rn
. Let u(x, t) be a smooth solution of the following
initial boundary value problem:
utt − u + u3
= 0 for (x, t) ∈ Ω × [0, T]
u(x, t) = 0 for (x, t) ∈ ∂Ω × [0, T].
a) Derive an energy equality for u. (Hint: Multiply by ut and integrate over Ω ×
[0, T].)
b) Show that if u|t=0 = ut|t=0 = 0 for x ∈ Ω, then u ≡ 0.
Proof. a) Multiply by ut and integrate:
0 =
Ω
(utt − u + u3
)ut dx =
Ω
uttut dx −
∂Ω
∂u
∂n
ut ds
=0
+
Ω
∇u · ∇ut dx +
Ω
u3
ut dx
=
Ω
1
2
∂
∂t
(u2
t ) dx +
Ω
1
2
∂
∂t
|∇u|2
dx +
Ω
1
4
∂
∂t
(u4
) dx =
1
2
d
dt Ω
u2
t + |∇u|2
+
1
2
u4
dx.
Thus, the Energy integral is
E(t) =
Ω
u2
t + |∇u|2
+
1
2
u4
dx = const = E(0).
b) Since u(x, 0) = 0, ut(x, 0) = 0, we have
E(0) =
Ω
ut(x, 0)2
+ |∇u(x, 0)|2
+
1
2
u(x, 0)4
dx = 0.
Since E(t) = E(0) = 0, we have
E(t) =
Ω
ut(x, t)2
+ |∇u(x, t)|2
+
1
2
u(x, t)4
dx = 0.
Thus, u ≡ 0.

Problem (F’04, #3). Consider a damped wave equation
utt − u + a(x)ut = 0, (x, t) ∈ R3 × R,
u|t=0 = u0, ut|t=0 = u1.
Here the damping coeﬃcient a ∈ C∞
0 (R3) is a non-negative function and u0, u1 ∈
C∞
0 (R3
). Show that the energy of the solution u(x, t) at time t,
E(t) =
1
2 R3
|∇xu|2
+ |ut|2
dx
is a decreasing function of t ≥ 0.
Proof. Take the derivative of E(t) with respect to t. Note that the boundary integral
is 0 by Huygen’s principle.
dE
dt
(t) =
R3
∇u · ∇ut + ututt dx
=
∂R3
ut
∂u
∂n
ds
=0
−
R3
ut u dx +
R3
ututt dx
=
R3
ut(− u + utt) dx =
R3
ut(−a(x)ut) dx =
R3
−a(x)u2
t dx ≤ 0.
Thus, dE
dt ≤ 0 ⇒ E(t) ≤ E(0), i.e. E(t) is a decreasing function of t.

Problem (W’03, #8). a) Consider the damped wave equation for high-speed waves
(0 < << 1) in a bounded region D
2
utt + ut = u
with the boundary condition u(x, t) = 0 on ∂D. Show that the energy functional
E(t) =
D
2
u2
t + |∇u|2
dx
is nonincreasing on solutions of the boundary value problem.
b) Consider the solution to the boundary value problem in part (a) with initial data
u (x, 0) = 0, ut(x, 0) = −αf(x), where f does not depend on and α < 1. Use part
(a) to show that
D
|∇u (x, t)|2
dx → 0
uniformly on 0 ≤ t ≤ T for any T as → 0.
c) Show that the result in part (b) does not hold for α = 1. To do this consider
the case where f is an eigenfunction of the Laplacian, i.e. f + λf = 0 in D and
f = 0 on ∂D, and solve for u explicitly.
Proof. a)
dE
dt
=
D
2 2
ututt dx +
D
2∇u · ∇ut dx
=
D
2 2
ututt dx +
∂D
2
∂u
∂n
ut ds
=0, (u=0 on ∂D)
−
D
2 uut dx
= 2
D
( 2
utt − u)ut dx = = −2
D
|ut|2
dx ≤ 0.
Thus, E(t) ≤ E(0), i.e. E(t) is nonincreasing.
b) From (a), we know dE
dt ≤ 0. We also have
E (0) =
D
2
(ut(x, 0))2
+ |∇u (x, 0)|2
dx
=
D
2
( −α
f(x))2
+ 0 dx =
D
2(1−α)
f(x)2
dx → 0 as → 0.
Since E (0) ≥ E (t) = D
2(ut)2 + |∇u |2 dx, then E (t) → 0 as → 0.
Thus, D |∇u |2
dx → 0 as → 0.
c) If α = 1,
E (0) =
D
2(1−α)
f(x)2
dx =
D
f(x)2
dx.
Since f is independent of , E (0) does not approach 0 as → 0. We can not conclude
that D |∇u (x, t)|2
dx → 0.

Problem (F’98, #6). Let f solve the nonlinear wave equation
ftt − fxx = −f(1 + f2
)−1
for x ∈ [0, 1], with f(x = 0, t) = f(x = 1, t) = 0 and with smooth initial data f(x, t) =
f0(x).
a) Find an energy integral E(t) which is constant in time.
b) Show that |f(x, t)| < c for all x and t, in which c is a constant.
Hint: Note that
f
1 + f2
=
1
2
d
df
log(1 + f2
).
Proof. a) Since f(0, t) = f(1, t) = 0, ∀t, we have ft(0, t) = ft(1, t) = 0. Let
dE
dt
= 0 =
1
0
ftt − fxx + f(1 + f2
)−1
ft dx
=
1
0
fttft dx −
1
0
fxxft dx +
1
0
fft
1 + f2
dx
=
1
0
fttft dx − [fx ft
=0
]1
0 +
1
0
fxftx dx +
1
0
fft
1 + f2
dx
=
1
0
1
2
∂
∂t
(f2
t ) dx +
1
0
1
2
∂
∂t
(f2
x) dx +
1
0
1
2
∂
∂t
(ln(1 + f2
)) dx
=
1
2
d
dt
1
0
f2
t + f2
x + ln(1 + f2
) dx.
Thus,
E(t) =
1
2
1
0
f2
t + f2
x + ln(1 + f2
) dx.
b) We want to show that f is bounded. For smooth f(x, 0) = f0(x), we have
E(0) =
1
2
1
0
ft(x, 0)2
+ fx(x, 0)2
+ ln(1 + f(x, 0)2
) dx < ∞.
Since E(t) is constant in time, E(t) = E(0) < ∞. Thus,
1
2
1
0
ln(1 + f2
) dx ≤
1
2
1
0
f2
t + f2
x + ln(1 + f2
) dx = E(t) < ∞.
Hence, f is bounded.

Problem (F’97, #1). Consider initial-boundary value problem
utt + a2
(x, t)ut − u(x, t) = 0 x ∈ Ω ⊂ Rn
, 0 < t < +∞
u(x) = 0 x ∈ ∂Ω
u(x, 0) = f(x), ut(x, 0) = g(x) x ∈ Ω.
Prove that L2-norm of the solution is bounded in t on (0, +∞).
Here Ω is a bounded domain, and a(x, t), f(x), g(x) are smooth functions.
Proof. Multiply the equation by ut and integrate over Ω:
ututt + a2
u2
t − ut u = 0,
Ω
ututt dx +
Ω
a2
u2
t dx −
Ω
ut u dx = 0,
1
2
d
dt Ω
u2
t dx +
Ω
a2
u2
t dx −
∂Ω
ut
∂u
∂n
ds
=0, (u=0, x∈∂Ω)
+
Ω
∇u · ∇ut dx = 0,
1
2
d
dt Ω
u2
t dx +
Ω
a2
u2
t dx +
1
2
d
dt Ω
|∇u|2
dx = 0,
1
2
d
dt Ω
u2
t + |∇u|2
dx = −
Ω
a2
u2
t dx ≤ 0.
Let Energy integral be
E(t) =
Ω
u2
t + |∇u|2
dx.
We have dE
dt ≤ 0, i.e. E(t) ≤ E(0).
E(t) ≤ E(0) =
Ω
ut(x, 0)2
+ |∇u(x, 0)|2
dx =
Ω
g(x)2
+ |∇f(x)|2
dx < ∞,
since f and g are smooth functions. Thus,
E(t) =
Ω
u2
t + |∇u|2
dx < ∞,
Ω
|∇u|2
dx < ∞,
Ω
u2
dx ≤ C
Ω
|∇u|2
dx < ∞, by Poincare inequality.
Thus, ||u||2 is bounded ∀t.

Problem (S’98, #4). a) Let u(x, y, z, t), −∞ < x, y, z < ∞ be a solution of the
equation
⎧
⎪⎨
⎪⎩
utt + ut = uxx + uyy + uzz
u(x, y, z, 0) = f(x, y, z),
ut(x, y, z, 0) = g(x, y, z).
(16.41)
Here f, g are smooth functions which vanish if x2 + y2 + z2 is large enough. Prove
that it is the unique solution for t ≥ 0.
b) Suppose we want to solve the same equation (16.41) in the region z ≥ 0, −∞ <
x, y < ∞, with the additional conditions
u(x, y, 0, t) = f(x, y, t)
uz(x, y, 0, t) = g(x, y, t)
with the same f, g as before in (16.41). What goes wrong?
Proof. a) Suppose u1 and u2 are two solutions. Let w = u1 − u2. Then,
⎧
⎪⎨
⎪⎩
wtt + wt = w,
w(x, y, z, 0) = 0,
wt(x, y, z, 0) = 0.
Multiply the equation by wt and integrate:
wtwtt + w2
t = wt w,
R3
wtwtt dx +
R3
w2
t dx =
R3
wt w dx,
1
2
d
dt R3
w2
t dx +
R3
w2
t dx =
∂R3
wt
∂w
∂n
dx
=0
−
R3
∇w · ∇wt dx,
1
2
d
dt R3
w2
t dx +
R3
w2
t dx = −
1
2
d
dt R3
|∇w|2
dx,
d
dt R3
w2
t + |∇w|2
dx
E(t)
= −2
R3
w2
t dx ≤ 0,
dE
dt
≤ 0,
E(t) ≤ E(0) =
R3
wt(x, 0)2
+ |∇w(x, 0)|2
dx = 0,
⇒ E(t) =
R3
w2
t + |∇w|2
dx = 0.
Thus, wt = 0, ∇w = 0, and w = constant. Since w(x, y, z, 0) = 0, we have w ≡ 0.
b)

Problem (F’94, #8). The one-dimensional, isothermal fluid equations with viscosity
and capillarity in Lagrangian variables are
vt − ux = 0
ut + p(v)x = εuxx − δvxxx
in which v(= 1/ρ) is specific volume, u is velocity, and p(v) is pressure. The coefficients
ε and δ are non-negative.
Find an energy integral which is non-increasing (as t increases) if ε > 0 and con-
served if ε = 0.
Hint: if δ = 0, E = u2/2 − P(v) dx where P (v) = p(v).
Proof. Multiply the second equation by u and integrate over R. We use ux = vt.
Note that the boundary integrals are 0 due to finite speed of propagation.
uut + up(v)x = εuuxx − δuvxxx,
R
uut dx +
R
up(v)x dx = ε
R
uuxx dx − δ
R
uvxxx dx,
1
2 R
∂
∂t
(u2
) dx +
∂R
up(v) ds
=0
+
R
uxp(v) dx
= ε
∂R
uux dx
=0
−ε
R
u2
x dx − δ
∂R
uvxx dx
=0
+δ
R
uxvxx dx,
1
2 R
∂
∂t
(u2
) dx +
R
vtp(v) dx = −ε
R
u2
x dx + δ
R
vtvxx dx,
1
2 R
∂
∂t
(u2
) dx +
R
∂
∂t
P(v) dx = −ε
R
u2
x dx + δ
∂R
vtvx dx
=0
−δ
R
vxtvx dx,
1
2 R
∂
∂t
(u2
) dx +
R
∂
∂t
P(v) dx +
δ
2 R
∂
∂t
(v2
x) dx = −ε
R
u2
x dx,
d
dt R
u2
2
+ P(v) +
δ
2
v2
x dx = −ε
R
u2
x dx ≤ 0.
E(t) =
R
u2
2
+ P(v) +
δ
2
v2
x dx
is nonincreasing if ε > 0, and conserved if ε = 0.

Problem (S’99, #5). Consider the equation
utt =
∂
∂x
σ(ux) (16.42)
with σ(z) a smooth function. This is to be solved for t > 0, 0 ≤ x ≤ 1, with
periodic boundary conditions and initial data u(x, 0) = u0(x) and ut(x, 0) = v0(x).
a) Multiply (16.42) by ut and get an expression of the form
d
dt
1
0
F(ut, ux) = 0
that is satisfied for an appropriate function F(y, z) with y = ut, z = ux,
where u is any smooth, periodic in space solution of (16.42).
b) Under what conditions on σ(z) is this function, F(y, z), convex in its variables?
c) What à priori inequality is satisfied for smooth solutions when F is convex?
d) Discuss the special case σ(z) = a2z3/3, with a > 0 and constant.
Proof. a) Multiply by ut and integrate:
ututt = utσ(ux)x,
1
0
ututt dx =
1
0
utσ(ux)x dx,
d
dt
1
0
u2
t
2
dx = utσ(ux)
1
0
=0, (2π-periodic)
−
1
0
utxσ(ux) dx =
Let Q (z) = σ(z), then d
dt Q(ux) = σ(ux)uxt. Thus,
= −
1
0
utxσ(ux) dx = −
d
dt
1
0
Q(ux) dx.
d
dt
1
0
u2
t
2
+ Q(ux) dx = 0.
b) We have
F(ut, ux) =
u2
t
2
+ Q(ux).
41 For F to be convex, the Hessian matrix of partial derivatives must be positive definite.
41
A function f is convex on a convex set S if it satisfies
f(αx + (1 − α)y) ≤ αf(x) + (1 − α)f(y)
for all 0 ≤ α ≤ 1 and for all x, y ∈ S.
If a one-dimensional function f has two continuous derivatives, then f is convex if and only if
f (x) ≥ 0.
In the multi-dimensional case the Hessian matrix of second derivatives must be positive semi-definite,
that is, at every point x ∈ S
yT
∇2
f(x) y ≥ 0, for all y.
The Hessian matrix is the matrix with entries
[∇2
f(x)]ij ≡
∂2
f(x)
∂xi∂xj
.
For functions with continuous second derivatives, it will always be symmetric matrix: fxixj = fxj xi .

The Hessian matrix is
∇2
F(ut, ux) =
Futut Futux
Fuxut Fuxux
=
1 0
0 σ (ux)
.
yT
∇2
F(x) y = y1 y2
1 0
0 σ (ux)
y1
y2
= y2
1 + σ (ux)y2
2 ≥
need
0.
Thus, for a Hessian matrix to be positive deﬁnite, need σ (ux) ≥ 0, so that the above
inequality holds for all y.
c) We have
d
dt
1
0
F(ut, ux) dx = 0,
1
0
F(ut, ux) dx = const,
1
0
F(ut, ux) dx =
1
0
F(ut(x, 0), ux(x, 0)) dx,
1
0
u2
t
2
+ Q(ux) dx =
1
0
v2
0
2
+ Q(u0x) dx.
d) If σ(z) = a2
z3
/3, we have
F(ut, ux) =
u2
t
2
+ Q(ux) =
u2
t
2
+
a2
u4
x
12
,
d
dt
1
0
u2
t
2
+
a2
u4
x
12
dx = 0,
1
0
u2
t
2
+
a2
u4
x
12
dx = const,
1
0
u2
t
2
+
a2
u4
x
12
dx =
1
0
v0
2
2
+
a2
u0
4
x
12
dx.

Problem (S’96, #8). 42
Let u(x, t) be the solution of the Korteweg-de Vries equation
ut + uux = uxxx, 0 ≤ x ≤ 2π,
with 2π-periodic boundary conditions and prescribed initial data
u(x, t = 0) = f(x).
a) Prove that the energy integral
I1(u) =
2π
0
u2
(x, t) dx
is independent of the time t.
b) Prove that the second “energy integral”,
I2(u) =
2π
0
1
2
u2
x(x, t) +
1
6
u3
(x, t) dx
is also independent of the time t.
c) Assume the initial data are such that I1(f) + I2(f) < ∞. Use (a) + (b) to prove
that the maximum norm of the solution, |u|∞ = supx |u(x, t)|, is bounded in time.
Hint: Use the following inequalities (here, |u|p is the Lp
-norm of u(x, t) at ﬁxed time
t):
• |u|2
∞ ≤
π
6
(|u|2
2 + |ux|2
2) (one of Sobolev’s inequalities);
• |u|3
3 ≤ |u|2
2 |u|∞ (straightforward).
Proof. a) Multiply the equation by u and integrate. Note that all boundary terms are
0 due to 2π-periodicity.
uut + u2
ux = uuxxx,
2π
0
uut dx +
2π
0
u2
ux dx =
2π
0
uuxxx dx,
1
2
d
dt
2π
0
u2
dx +
1
3
2π
0
(u3
)x dx = uuxx
2π
0
−
2π
0
uxuxx dx,
1
2
d
dt
2π
0
u2
dx +
1
3
u3 2π
0
= −
1
2
2π
0
(u2
x)x dx,
1
2
d
dt
2π
0
u2
dx = −
1
2
u2
x
2π
0
= 0.
I1(u) =
2π
0
u2
dx = C.
Thus, I1(u) =
2π
0 u2
(x, t) dx is independent of the time t.
Alternatively, we may diﬀerentiate I1(u):
dI1
dt
(u) =
d
dt
2π
0
u2
dx =
2π
0
2uut dx =
2π
0
2u(−uux + uxxx) dx
=
2π
0
−2u2
ux dx +
2π
0
2uuxxx dx =
2π
0
−
2
3
(u3
)x dx + 2uuxx
2π
0
−
2π
0
2uxuxx dx
= −
2
3
u3 2π
0
−
2π
0
(u2
x)x dx = −u2
x
2π
0
= 0.
42
Also, see S’92, #7.

b) Note that all boundary terms are 0 due to 2π-periodicity.
dI2
dt
(u) =
d
dt
2π
0
1
2
u2
x +
1
6
u3
dx =
2π
0
uxuxt +
1
2
u2
ut dx =
We diﬀerentiate the original equation with respect to x:
ut = −uux + uxxx
utx = −(uux)x + uxxxx.
=
2π
0
ux(−(uux)x + uxxxx) dx +
1
2
2π
0
u2
(−uux + uxxx) dx
=
2π
0
−ux(uux)x dx +
2π
0
uxuxxxx dx −
1
2
2π
0
u3
ux dx +
1
2
2π
0
u2
uxxx dx
= −uxuux
2π
0
+
2π
0
uxxuux dx + uxuxxx
2π
0
−
2π
0
uxxuxxx dx
−
1
2
2π
0
u4
4 x
dx +
1
2
u2
uxx
2π
0
−
1
2
2π
0
2uuxuxx dx
=
2π
0
uxxuux dx −
2π
0
uxxuxxx dx −
1
2
u4
4
2π
0
−
2π
0
uuxuxx dx
= −
2π
0
uxxuxxx dx = −u2
xx
2π
0
+
2π
0
uxxxuxx dx =
2π
0
uxxxuxx dx = 0,
since −
2π
0 uxxuxxx dx = +
2π
0 uxxuxxx dx. Thus,
I2(u) =
2π
0
1
2
u2
x(x, t) +
1
6
u3
(x, t) dx = C,
and I2(u) is independent of the time t.
c) From (a) and (b), we have
I1(u) =
2π
0
u2
dx = ||u||2
2,
I2(u) =
2π
0
1
2
u2
x +
1
6
u3
dx =
1
2
||ux||2
2 +
1
6
||u||3
3.
Using given inequalities, we have
||u||2
∞ ≤
π
6
(||u||2
2 + ||ux||2
2) ≤
π
6
I1(u) + 2I2(u) −
1
3
||u||3
3
≤
π
6
I1(u) +
π
3
I2(u) +
π
18
||u||2
2 ||u||∞ ≤
π
6
I1(u) +
π
3
I2(u) +
π
18
I1(u)||u||∞
= C + C1||u||∞.
⇒ ||u||2
∞ ≤ C + C1||u||∞,
⇒ ||u||∞ ≤ C2.
Thus, ||u||∞ is bounded in time.
Also see Energy Methods problems for higher order equations (3rd and
4th) in the section on Gas Dynamics.

16.7 Wave Equation in 2D and 3D
Problem (F’97, #8); (McOwen 3.2 #90). Solve
utt = uxx + uyy + uzz
with initial conditions
u(x, y, z, 0) = x2
+ y2
g(x)
, ut(x, y, z, 0) = 0
h(x)
.
Proof.
➀ We may use the Kirchhoﬀ’s formula:
u(x, t) =
1
4π
∂
∂t
t
|ξ|=1
g(x + ctξ) dSξ +
t
4π |ξ|=1
h(x + ctξ) dSξ
=
1
4π
∂
∂t
t
|ξ|=1
(x1 + ctξ1)2
+ (x2 + ctξ2)2
dSξ + 0 =
➁ We may solve the problem by Hadamard’s method of descent, since initial con-
ditions are independent of x3. We need to convert surface integrals in R3
to domain
integrals in R2. Speciﬁcally, we need to express the surface measure on the upper half
of the unit sphere S2
+ in terms of the two variables ξ1 and ξ2. To do this, consider
f(ξ1, ξ2) = 1 − ξ2
1 − ξ2
2 over the unit disk ξ2
1 + ξ2
2 < 1.
dSξ = 1 + (fξ1 )2 + (fξ2 )2 dξ1dξ2 =
dξ1dξ2
1 − ξ2
1 − ξ2
2
.

u(x1, x2, t) =
1
4π
∂
∂t
2t
ξ2
1+ξ2
2<1
g(x1 + ctξ1, x2 + ctξ2) dξ1dξ2
1 − ξ2
1 − ξ2
2
+
t
4π
2
ξ2
1+ξ2
2 <1
h(x1 + ctξ1, x2 + ctξ2) dξ1dξ2
1 − ξ2
1 − ξ2
2
=
1
4π
∂
∂t
2t
ξ2
1+ξ2
2<1
(x1 + tξ1)2 + (x2 + tξ2)2
1 − ξ2
1 − ξ2
2
dξ1dξ2 + 0,
=
1
2π
∂
∂t
t
ξ2
1+ξ2
2 <1
x2
1 + 2x1tξ1 + t2ξ2
1 + x2
2 + 2x2tξ2 + t2ξ2
2
1 − ξ2
1 − ξ2
2
dξ1dξ2
=
1
2π
∂
∂t ξ2
1+ξ2
2<1
tx2
1 + 2x1t2ξ1 + t3ξ2
1 + tx2
2 + 2x2t2ξ2 + t3ξ2
2
1 − ξ2
1 − ξ2
2
dξ1dξ2
=
1
2π ξ2
1+ξ2
2<1
x2
1 + 4x1tξ1 + 3t2ξ2
1 + x2
2 + 4x2tξ2 + 3t2ξ2
2
1 − ξ2
1 − ξ2
2
dξ1dξ2
=
1
2π ξ2
1+ξ2
2<1
(x2
1 + x2
2) + 4t(x1ξ1 + x2ξ2) + 3t2(ξ2
1 + ξ2
2)
1 − ξ2
1 − ξ2
2
dξ1dξ2
=
1
2π
(x2
1 + x2
2)
ξ2
1+ξ2
2 <1
dξ1dξ2
1 − ξ2
1 − ξ2
2
❶
+
4t
2π ξ2
1+ξ2
2 <1
x1ξ1 + x2ξ2
1 − ξ2
1 − ξ2
2
dξ1dξ2
❷
+
3t2
2π ξ2
1+ξ2
2<1
ξ2
1 + ξ2
2
1 − ξ2
1 − ξ2
2
dξ1dξ2
❸
=
❶ =
1
2π
(x2
1 + x2
2)
ξ2
1+ξ2
2 <1
dξ1dξ2
1 − ξ2
1 − ξ2
2
=
1
2π
(x2
1 + x2
2)
2π
0
1
0
r dr dθ
√
1 − r2
=
1
2π
(x2
1 + x2
2)
2π
0
−2
1
0
−1
2 du dθ
u
1
2
u = 1 − r2
, du = −2r dr
=
1
2π
(x2
1 + x2
2)
2π
0
1 dθ = x2
1 + x2
2.
❷ =
4t
2π ξ2
1+ξ2
2<1
x1ξ1 + x2ξ2
1 − ξ2
1 − ξ2
2
dξ1dξ2 =
4t
2π
1
−1
√
1−ξ2
2
−
√
1−ξ2
2
x1ξ1 + x2ξ2
1 − ξ2
1 − ξ2
2
dξ1dξ2
= 0.
❸ =
3t2
2π ξ2
1+ξ2
2 <1
ξ2
1 + ξ2
2
1 − ξ2
1 − ξ2
2
dξ1dξ2 =
3t2
2π
2π
0
1
0
(r cos θ)2
+ (r sinθ)2
√
1 − r2
r drdθ
=
3t2
2π
2π
0
1
0
r3
√
1 − r2
drdθ u = 1 − r2
, du = −2r dr
=
3t2
2π
2π
0
2
3
dθ =
t2
π
2π
0
dθ = 2t2
.
⇒ u(x1, x2, t) = ❶ + ❷ + ❸ = x2
1 + x2
2 + 2t2
.
➂ We may guess what the solution is:
u(x, y, z, t) =
1
2
(x + t)2
+ (y + t)2
+ (x − t)2
+ (y − t)2
= x2
+ y2
+ 2t2
.

Check:
u(x, y, z, 0) = x2
+ y2
.
ut(x, y, z, t) = (x + t) + (y + t) − (x − t) − (y − t),
ut(x, y, z, 0) = 0.
utt(x, y, z, t) = 4,
ux(x, y, z, t) = (x + t) + (x − t),
uxx(x, y, z, t) = 2,
uy(x, y, z, t) = (y + t) + (y − t),
uyy(x, y, z, t) = 2,
uzz(x, y, z, t) = 0,
utt = uxx + uyy + uzz.

Consider the two-dimensional wave equation wtt = a2 w, with initial data which van-
ish for x2
+y2
large enough. Prove that w(x, y, t) satisﬁes the decay |w(x, y, t)| ≤ C·t−1
.
(Note that the estimate is not uniform with respect to x, y since C may depend on x, y).
Proof. Suppose we have the following problem with initial data:
utt = a2
u x ∈ R2
, t > 0,
u(x, 0) = g(x), ut(x, 0) = h(x) x ∈ R2
.
The result is the consequence of the Huygens’ principle and may be proved by Hadamard’s
method of descent: 43
u(x, t) =
1
4π
∂
∂t
2t
ξ2
1+ξ2
2<1
g(x1 + ctξ1, x2 + ctξ2) dξ1dξ2
1 − ξ2
1 − ξ2
2
+
t
4π
2
ξ2
1+ξ2
2 <1
h(x1 + ctξ1, x2 + ctξ2) dξ1dξ2
1 − ξ2
1 − ξ2
2
=
1
2π |ξ|2<c2t2
th(x + ξ) + g(x + ξ)
1 −
|ξ|2
c2t2
dξ1dξ2
c2t2
+
t
2π |ξ|2<c2t2
∇g(x + ξ) · (ct, ct)
1 −
|ξ|2
c2t2
dξ1dξ2
c2t2
.
For a given x, let T(x) be so large that T > 1 and supp(h + g) ⊂ BT (x). Then for
t > 2T we have:
|u(x, t)| =
1
2π |ξ|2<c2T 2
tM + M + 2Mct
1 − c2T 2
c2T 24
dξ1dξ2
c2t2
=
πc2
T2
2π
M
3/4
1
c2t
+
M
3/4
1
c2Tt
+
2Mc
c2t
.
⇒ u(x, t) ≤ C1/t for t > 2T.
For t ≤ 2T:
|u(x, t)| =
1
2π |ξ|2<c2t2
2TM + M + 4McT
1 − |ξ|2
c2t2
dξ1dξ2
c2t2
=
1
2π
(2TM + M + 4Mct)2π
ct
0
r dr/c2
t2
1 − r2
c2t2
=
M(2T + 1 + 4cT)
2
1
0
−du
u1/2
=
M(2T + 1 + 4cT)
2
2 ≤
M(2T + 1 + 4cT)2T
t
.
Letting C = max(C1, M(2T + 1 + 4cT)2T), we have |u(x, t)| ≤ C(x)/t.
• For n = 3, suppose g, h ∈ C∞
0 (R3
). The solution is given by the Kircchoﬀ’s
formula. There is a constant C so that u(x, t) ≤ C/t for all x ∈ R3 and t > 0. As
McOwen suggensts in Hints for Exercises, to prove the result, we need to estimate the
43
Nick’s solution follows.

area of intersection of the sphere of radius ct with the support of the functions g and
h.

Problem (S’95, #6). Spherical waves in 3-d are waves symmetric about the origin;
i.e. u = u(r, t) where r is the distance from the origin. The wave equation
utt = c2
u
then reduces to
1
c2
utt = urr +
2
r
ur. (16.43)
a) Find the general solutions u(r, t) by solving (16.43). Include both the incoming waves
and outgoing waves in your solutions.
b) Consider only the outgoing waves and assume the finite out-flux condition
0 < lim
r→0
r2
ur < ∞
for all t. The wavefront is defined as r = ct. How is the amplitude of the wavefront
decaying in time?
Proof. a) We want to reduce (16.43) to the 1D wave equation. Let v = ru. Then
vtt = rutt,
vr = rur + u,
vrr = rurr + 2ur.
Thus, (16.43) becomes
1
c2
1
r
vtt =
1
r
vrr,
1
c2
vtt = vrr,
vtt = c2
vrr,
which has the solution
v(r, t) = f(r + ct) + g(r − ct).
Thus,
u(r, t) =
1
r
v(r, t) =
1
r
f(r + ct)
incoming, (c>0)
+
1
r
g(r − ct)
outgoing, (c>0)
.
b) We consider u(r, t) = 1
r g(r − ct):
0 < lim
r→0
r2
ur < ∞,
0 < lim
r→0
r2 1
r
g (r − ct) −
1
r2
g(r − ct) < ∞,
0 < lim
r→0
rg (r − ct) − g(r − ct) < ∞,
0 < −g(−ct) < ∞,
0 < −g(−ct) = G(t) < ∞,
g(t) = −G
t
−c
.

The wavefront is deﬁned as r = ct. We have
u(r, t) =
1
r
g(r − ct) = −
1
r
G
r − ct
−c
= −
1
ct
G(0).
|u(r, t)| =
1
t
−
1
c
G(0) .
The amplitude of the wavefront decays like 1
t .

Problem (S’00, #8). a) Show that for a smooth function F on the line, while
u(x, t) = F(ct + |x|)/|x| may look like a solution of the wave equation utt = c2 u
in R3
, it actually is not. Do this by showing that for any smooth function φ(x, t) with
compact support
R3×R
u(x, t)(φtt − φ) dxdt = 4π
R
φ(0, t)F(ct) dt.
Note that, setting r = |x|, for any function w which only depends on r one has
w = r−2
(r2
wr)r = wrr + 2
r wr.
b) If F(0) = F (0) = 0, what is the true solution to utt = u with initial conditions
u(x, 0) = F(|x|)/|x| and ut(x, 0) = F (|x|)/|x|?
c) (Ralston Hw) Suppose u(x, t) is a solution to the wave equation utt = c2
u in
R3 × R with u(x, t) = w(|x|, t) and u(x, 0) = 0. Show that
u(x, t) =
F(|x| + ct) − F(|x| − ct)
|x|
for a function F of one variable.
Proof. a) We have
R3 R
u (φtt − φ) dxdt = lim
→0 R
dt
|x|>
u (φtt − φ) dx
= lim
→0 R
dt
|x|>
φ (utt − u) dx +
|x|=
∂u
∂n
φ − u
∂φ
∂n
dS .
The ﬁnal equality is derived by integrating by parts twice in t, and using Green’s
theorem:
Ω
(v u − u v) dx =
∂Ω
v
∂u
∂n
− u
∂v
∂n
ds.
Since dS = 2
sinφ dφ dθ and ∂
∂n = − ∂
∂r , substituting u(x, t) = F(|x| + ct)/|x|
gives:
R3 R
u (φtt − φ) dxdt =
R
4πφF(ct) dt.
Thus, u is not a weak solution to the wave equation.
b)
c) We want to show that v(|x|, t) = |x|w(|x|, t) is a solution to the wave equation in
one space dimension and hence must have the from v = F(|x| +ct) +G(|x| −ct). Then
we can argue that w will be undeﬁned at x = 0 for some t unless F(ct) + G(−ct) = 0
for all t.
We work in spherical coordinates. Note that w and v are independent of φ and θ. We
have:
vtt(r, t) = c2
w = c2 1
r2
(r2
wr)r = c2 1
r2
(2rwr + r2
wrr),
⇒ rwtt = c2
rwrr + 2wr.
Thus we see that vtt = c2
vrr, and we can conclude that
v(r, t) = F(r + ct) + G(r − ct) and
w(r, t) =
F(r + ct) + G(r − ct)
r
.

limr→0 w(r, t) does not exist unless F(ct) + G(−ct) = 0 for all t. Hence
w(r, t) =
F(ct + r) + G(ct − r)
r
, and
u(x, t) =
F(ct + |x|) + G(ct − |x|)
|x|
.

17 Problems: Laplace Equation
A fundamental solution K(x) for the Laplace operator is a distribution satisfying 44
K(x) = δ(x)
The fundamental solution for the Laplace operator is
K(x) =
1
2π log |x| if n = 2
1
(2−n)ωn
|x|2−n if n ≥ 3.
17.1 Green’s Function and the Poisson Kernel
Green’s function is a special fundamental solution satisfying 45
G(x, ξ) = δ(x) for x ∈ Ω
G(x, ξ) = 0 for x ∈ ∂Ω,
(17.1)
To construct the Green’s function,
➀ consider wξ(x) with wξ(x) = 0 in Ω and wξ(x) = −K(x − ξ) on ∂Ω;
➁ consider G(x, ξ) = K(x − ξ) + wξ(x), which is a fundamental solution satisfying
(17.1).
Problem 1. Given a particular distribution solution to the set of Dirichlet problems
uξ(x) = δξ(x) for x ∈ Ω
uξ(x) = 0 for x ∈ ∂Ω,
how would you use this to solve
u = 0 for x ∈ Ω
u(x) = g(x) for x ∈ ∂Ω.
Proof. uξ(x) = G(x, ξ), a Green’s function. G is a fundamental solution to the Laplace
operator, G(x, ξ) = 0, x ∈ ∂Ω. In this problem, it is assumed that G(x, ξ) is known for
Ω. Then
u(ξ) =
Ω
G(x, ξ) u dx +
∂Ω
u(x)
∂G(x, ξ)
∂nx
dSx
for every u ∈ C2
(Ω). In particular, if u = 0 in Ω and u = g on ∂Ω, then we obtain
the Poisson integral formula
u(ξ) =
∂Ω
∂G(x, ξ)
∂nx
g(x) dSx,
44
We know that u(x) = Rn K(x−y)f(y)dy is a distribution solution of u = f when f is integrable
and has compact support. In particular, we have
u(x) =
Rn
K(x − y) u(y) dy whenever u ∈ C∞
0 (Rn
).
The above result is a consequence of:
u(x) =
Ω
δ(x − y)u(y) dy = ( K) ∗ u = K ∗ ( u) =
Ω
K(x − y) u(y) dy.
45
Green’s function is useful in satisfying Dirichlet boundary conditions.

where H(x, ξ) =
∂G(x,ξ)
∂nx
is the Poisson kernel.
Thus if we know that the Dirichlet problem has a solution u ∈ C2
(Ω), then we can
calculate u from the Poisson integral formula (provided of course that we can compute
G(x, ξ)).

Dirichlet Problem on a Half-Space. Solve the n-dimensional Laplace/Poisson
equation on the half-space with Dirichlet boundary conditions.
Proof. Use the method of reﬂection to construct Green’s function. Let Ω be an
upper half-space in Rn. If x = (x , xn), where x ∈ Rn−1, we can see
|x − ξ| = |x − ξ∗
|, and hence K(x − ξ) = K(x − ξ∗
). Thus
G(x, ξ) = K(x − ξ) − K(x − ξ∗
)
is the Green’s function on Ω. G(x, ξ) is harmonic in Ω,
and G(x, ξ) = 0 on ∂Ω.
To compute the Poisson kernel, we must diﬀerentiate G(x, ξ)
in the negative xn direction. For n ≥ 2,
∂
∂xn
K(x − ξ) =
xn − ξn
ωn
|x − ξ|−n
,
so that the Poisson kernel is given by
−
∂
∂xn
G(x, ξ) xn=0
=
2ξn
ωn
|x − ξ|−n
, for x ∈ Rn−1
.
Thus, the solution is
u(ξ) =
∂Ω
∂G(x, ξ)
∂nx
g(x) dSx =
2ξn
ωn Rn−1
g(x )
|x − ξ|n
dx .
If g(x ) is bounded and continuous for x ∈ Rn−1, then u(ξ) is C∞ and harmonic in Rn
+
and extends continuously to Rn
+ such that u(ξ ) = g(ξ ).

Problem (F’95, #3): Neumann Problem on a Half-Space.
a) Consider the Neumann problem in the upper half plane,
Ω = {x = (x1, x2) : −∞ < x1 < ∞, x2 > 0}:
u = ux1x1 + ux2x2 = 0 x ∈ Ω,
ux2 (x1, 0) = f(x1) − ∞ < x1 < ∞.
Find the corresponding Green’s function and conclude that
u(ξ) = u(ξ1, ξ2) =
1
2π
∞
−∞
ln [(x1 − ξ1)2
+ ξ2
2] · f(x1) dx1
is a solution of the problem.
b) Show that this solution is bounded in Ω if and only if
∞
−∞ f(x1) dx1 = 0.
Proof. a) Notation: x = (x, y), ξ = (x0, y0). Since K(x−ξ) = 1
2π log |x−ξ|, n = 2.
➀ First, we ﬁnd the Green’s function. We have
K(x − ξ) =
1
2π
log (x − x0)2 + (y − y0)2.
Let G(x, ξ) = K(x − ξ) + ω(x).
Since the problem is Neumann, we need:
G(x, ξ) = δ(x − ξ),
∂G
∂y ((x, 0), ξ) = 0.
G((x, y), ξ) =
1
2π
log (x − x0)2 + (y − y0)2 + ω((x, y), ξ),
∂G
∂y
((x, y), ξ) =
1
2π
y − y0
(x − x0)2 + (y − y0)2
+ ωy((x, y), ξ),
∂G
∂y
((x, 0), ξ) = −
1
2π
y0
(x − x0)2 + y2
0
+ ωy((x, 0), ξ) = 0.
Let
ω((x, y), ξ) =
a
2π
log (x − x0)2 + (y + y0)2. Then,
∂G
∂y
((x, 0), ξ) = −
1
2π
y0
(x − x0)2 + y2
0
+
a
2π
y0
(x − x0)2 + y2
0
= 0.
Thus, a = 1.
G((x, y), ξ) =
1
2π
log (x − x0)2 + (y − y0)2 +
1
2π
log (x − x0)2 + (y + y0)2.
46
➁ Consider Green’s identity (after cutting out B (ξ) and having → 0):
Ω
(u G − G u
=0
) dx =
∂Ω
u
∂G
∂n
=0
−G
∂u
∂n
dS
46
Note that for the Dirichlet problem, we would have gotten the “-” sign instead of “+” in front of
ω.

Since ∂u
∂n = ∂u
∂(−y) = −f(x), we have
Ω
u δ(x − ξ) dx =
∞
−∞
G((x, y), ξ) f(x) dx,
u(ξ) =
∞
−∞
G((x, y), ξ) f(x) dx.
For y = 0, we have
G((x, y), ξ) =
1
2π
log (x − x0)2 + y2
0 +
1
2π
log (x − x0)2 + y2
0
=
1
2π
2 log (x − x0)2 + y2
0
=
1
2π
log (x − x0)2
+ y2
0 .
Thus,
u(ξ) =
1
2π
∞
−∞
log (x − x0)2
+ y2
0 f(x) dx.
b) Show that this solution is bounded in Ω if and only if
∞
−∞ f(x1) dx1 = 0.
Consider the Green’s identity:
Ω
u dxdy =
∂Ω
∂u
∂n
dS = −
∞
−∞
∂u
∂y
dx =
∞
−∞
f(x) dx = 0.
Note that the Green’s identity applies to bounded domains Ω.
R
−R
f dx1 +
2π
0
∂u
∂r
R dθ = 0.
???

McOwen 4.2 # 6. For n = 2, use the method of reflections to find the Green’s
function for the first quadrant Ω = {(x, y) : x, y > 0}.
Proof. For x ∈ ∂Ω,
|x − ξ(0)
| · |x − ξ(2)
| = |x − ξ(1)
| · |x − ξ(3)
|,
|x − ξ(0)
| =
|x − ξ(1)
| · |x − ξ(3)
|
|x − ξ(2)|
.
But ξ(0) = ξ, so for n = 2,
G(x, ξ) =
1
2π
log |x − ξ| −
1
2π
log
|x − ξ(1)| · |x − ξ(3)|
|x − ξ(2)|
.
G(x, ξ) = 0, x ∈ ∂Ω.
Problem. Use the method of images to solve
G = δ(x − ξ)
in the first quadrant with G = 0 on the boundary.
Proof. To solve the problem in the first quadrant
we take a reflection to the fourth quadrant
and the two are reflected to the left half.
G = δ(x − ξ(0)
) − δ(x − ξ(1)
) − δ(x − ξ(2)
) + δ(x − ξ(3)
).
G =
1
2π
log
|x − ξ(0)| |x − ξ(3)|
|x − ξ(1)| |x − ξ(2)|
=
1
2π
log
(x − x0)2 + (y − y0)2 (x + x0)2 + (y + y0)2
(x − x0)2 + (y + y0)2 (x + x0)2 + (y − y0)2
.
Note that on the axes G = 0.

Problem (S’96, #3). Construct a Green’s function for the following mixed Dirichlet-
Neumann problem in Ω = {x = (x1, x2) ∈ R2 : x1 > 0, x2 > 0}:
u =
∂2
u
∂x2
1
+
∂2
u
∂x2
2
= f, x ∈ Ω,
ux2 (x1, 0) = 0, x1 > 0,
u(0, x2) = 0, x2 > 0.
Proof. Notation: x = (x, y), ξ = (x0, y0). Since K(x − ξ) = 1
2π log |x − ξ|, n = 2.
K(x − ξ) =
1
2π
log (x − x0)2 + (y − y0)2.
Let G(x, ξ) = K(x − ξ) + ω(x).
At (0, y), y > 0,
G (0, y), ξ =
1
2π
log x2
0 + (y − y0)2 + ω(0, y) = 0.
Also,
Gy (x, y), ξ =
1
2π
1
2 · 2(y − y0)
(x − x0)2 + (y − y0)2
+ wy(x, y)
=
1
2π
y − y0
(x − x0)2 + (y − y0)2
+ wy(x, y).
At (x, 0), x > 0,
Gy (x, 0), ξ = −
1
2π
y0
(x − x0)2 + y2
0
+ wy(x, 0) = 0.
We have
ω((x, y), ξ) =
a
2π
log (x + x0)2 + (y − y0)2
+
b
2π
log (x − x0)2 + (y + y0)2
+
c
2π
log (x + x0)2 + (y + y0)2.
Using boundary conditions, we have
0 = G((0, y), ξ) =
1
2π
log x2
0 + (y − y0)2 + ω(0, y)
=
1
2π
log x2
0 + (y − y0)2 +
a
2π
log x2
0 + (y − y0)2 +
b
2π
log x2
0 + (y + y0)2 +
c
2π
log x2
0 + (y + y0)2.
Thus, a = −1, c = −b. Also,
0 = Gy((x, 0), ξ) = −
1
2π
y0
(x − x0)2 + y2
0
+ wy(x, 0)
= −
1
2π
y0
(x − x0)2 + y2
0
−
(−1)
2π
y0
(x + x0)2 + y2
0
+
b
2π
y0
(x − x0)2 + y2
0
+
(−b)
2π
y0
(x + x0)2 + y2
0
.
Thus, b = 1, and
G((x, y), ξ) =
1
2π
log (x − x0)2 + (y − y0)2 + ω(x) =
1
2π
log (x − x0)2 + (y − y0)2

− log (x + x0)2 + (y − y0)2 + log (x − x0)2 + (y + y0)2 − log (x + x0)2 + (y + y0)2 .
It can be seen that G((x, y), ξ) = 0 on x = 0, for example.

Dirichlet Problem on a Ball. Solve the n-dimensional Laplace/Poisson equation on
the ball with Dirichlet boundary conditions.
Proof. Use the method of reflection to construct Green’s function.
Let Ω = {x ∈ Rn : |x| < a}. For ξ ∈ Ω, define ξ∗ = a2ξ
|ξ|2 as its reflection in ∂Ω; note
ξ∗
/∈ Ω.
|x − ξ∗
|
|x − ξ|
=
a
|ξ|
for |x| = a. ⇒ |x − ξ| =
|ξ|
a
|x − ξ∗
|. (17.2)
From (17.2) we conclude that for x ∈ ∂Ω (i.e. |x| = a),
K(x − ξ) =
⎧
⎨
⎩
1
2π log
|ξ|
a |x − ξ∗
| if n = 2
a
|ξ|
n−2
K(x − ξ∗
) if n ≥ 3.
(17.3)
Define for x, ξ ∈ Ω:
G(x, ξ) =
⎧
⎨
⎩
K(x − ξ) − 1
2π log |ξ|
a |x − ξ∗
| if n = 2
K(x − ξ) − a
|ξ|
n−2
K(x − ξ∗
) if n ≥ 3.
Since ξ∗ is not in Ω, the second terms on the RHS are harmonic
in x ∈ Ω. Moreover, by (17.3) we have G(x, ξ) = 0 if x ∈ ∂Ω.
Thus, G is the Green’s function for Ω.
u(ξ) =
∂Ω
∂G(x, ξ)
∂nx
g(x) dSx =
a2
− |ξ|2
aωn |x|=a
g(x)
|x − ξ|n
dSx.

17.2 The Fundamental Solution
Problem (F’99, #2). ➀ Given that Ka(x − y) and Kb(x − y) are the kernels for
the operators ( − aI)−1
and ( − bI)−1
on L2
(Rn
), where 0 < a < b, show that
( − aI)( − bI) has a fundamental solution of the form c1Ka + c2Kb.
➁ Use the preceding to ﬁnd a fundamental solution for 2
− , when n = 3.
Proof. METHOD ❶:
➀
( − aI)u = f ( − bI)u = f
u = Ka f u = Kb f
fundamental solution ⇔ kernel
⇒ u = Kaf u = Kbf if u ∈ L2
,
( − aI)u = (−|ξ|2
− a)u = f ( − bI)u = (−|ξ|2
− b)u = f
⇒ u = −
1
(ξ2 + a)
f(ξ) u = −
1
(ξ2 + b)
f(ξ)
⇒ Ka = −
1
ξ2 + a
Kb = −
1
ξ2 + b
( − aI)( − bI)u = f,
2
− (a + b) + abI u = f,
u =
1
(ξ2 + a)(ξ2 + b)
f(ξ) = Knewf(ξ),
Knew =
1
(ξ2 + a)(ξ2 + b)
=
1
b − a
−
1
ξ2 + b
+
1
ξ2 + a
=
1
b − a
(Kb − Ka),
Knew =
1
b − a
(Kb − Ka),
c1 =
1
b − a
, c2 = −
1
b − a
.
➁ n = 3 is not relevant (may be used to assume Ka, Kb ∈ L2
).
For 2 − , a = 0, b = 1 above, or more explicitly
( 2
− )u = f,
(ξ4
+ ξ2
)u = f,
u =
1
(ξ4 + ξ2)
f,
K =
1
(ξ4 + ξ2)
=
1
ξ2(ξ2 + 1)
= −
1
ξ2 + 1
+
1
ξ2
= K1 − K0.

METHOD ❷:
• For u ∈ C∞
0 (Rn) we have:
u(x) =
Rn
Ka(x − y) ( − aI) u(y) dy, ➀
u(x) =
Rn
Kb(x − y) ( − bI) u(y) dy. ➁
Let
u(x) = c1( − bI) φ(x), for ➀
u(x) = c2( − aI) φ(x), for ➁
for φ(x) ∈ C∞
0 (Rn
). Then,
c1( − bI)φ(x) =
Rn
Ka(x − y) ( − aI) c1( − bI)φ(y) dy,
c2( − aI)φ(x) =
Rn
Kb(x − y) ( − bI) c2( − aI)φ(y) dy.
We add two equations:
(c1 + c2) φ(x) − (c1b + c2a)φ(x) =
Rn
(c1Ka + c2Kb) ( − aI) ( − bI) φ(y) dy.
If c1 = −c2 and −(c1b + c2a) = 1, that is, c1 = 1
a−b, we have:
φ(x) =
Rn
1
a − b
(Ka − Kb) ( − aI) ( − bI) φ(y) dy,
which means that 1
a−b(Ka − Kb) is a fundamental solution of ( − aI)( − bI).
• 2 − = ( − 1) = ( − 0I)( − 1I).
( − 0I) has fundamental solution K0 = − 1
4πr in R3
.
To ﬁnd K, a fundamental solution for ( − 1I), we need to solve for a radially
symmetric solution of
( − 1I)K = δ.
In spherical coordinates, in R3, the above expression may be written as:
K +
2
r
K − K = 0.
Let
K =
1
r
w(r),
K =
1
r
w −
1
r2
w,
K =
1
r
w −
2
r2
w +
2
r3
w.
Plugging these into , we obtain:
1
r
w −
1
r
w = 0, or
w − w = 0.

Thus,
w = c1er
+ c2e−r
,
K =
1
r
w(r) = c1
er
r
+ c2
e−r
r
.
Ω = Ω − B (0).
Note: ( − I)K(|x|) = 0 in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪ ∂B (0)):
Ω
K(|x|) v − v K(|x|) dx =
∂Ω
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
+
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
We add − Ω K(|x|) v dx + Ω v K(|x|) dx to LHS to get:
Ω
K(|x|)( − I)v − v ( − I)K(|x|)
= 0, in Ω
dx =
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS.
lim
→0 Ω
K(|x|)( − I)v dx =
Ω
K(|x|)( − I)v dx. Since K(r) = c1
er
r
+ c2
e−r
r
On ∂B (0), K(|x|) = K( ). Thus, 47
∂B (0)
K(|x|)
∂v
∂n
dS = K( )
∂B (0)
∂v
∂n
dS ≤ c1
e
+ c2
e−
4π 2
max ∇v → 0, as → 0.
∂B (0)
v(x)
∂K(|x|)
∂n
dS =
∂B (0)
1
− c1e + c2e−
+
1
2
c1e + c2e−
v(x) dS
=
1
− c1e + c2e−
+
1
2
c1e + c2e−
∂B (0)
v(x) dS
=
1
− c1e + c2e−
+
1
2
c1e + c2e−
∂B (0)
v(0) dS
+
1
− c1e + c2e−
+
1
2
c1e + c2e−
∂B (0)
[v(x) − v(0)] dS
→
1
2
c1e + c2e−
v(0) 4π 2
→ 4π(c1 + c2)v(0) = −v(0).
Thus, taking c1 = c2, we have c1 = c2 = − 1
8π , which gives
Ω
K(|x|)( − I)v dx = lim
→0 Ω
K(|x|)( − I)v dx = v(0),
47
In R3
, for |x| = ,
K(|x|) = K( ) = c1
e
+ c2
e−
.
∂K(|x|)
∂n
= −
∂K( )
∂r
= −c1
e
−
e
2
− c2 −
e−
−
e−
2
=
1
− c1e + c2e−
+
1
2
c1e + c2e−
,
since n points inwards. n points toward 0 on the sphere |x| = (i.e., n = −x/|x|).

that is K(r) = − 1
8π
er
r + e−r
r = − 1
4πr cosh(r) is the fundamental solution of
( − I).
By part (a), 1
a−b(Ka − Kb) is a fundamental solution of ( − aI)( − bI).
Here, the fundamental solution of ( − 0I)( − 1I) is 1
−1 (K0 − K) = − − 1
4πr +
1
4πr cosh(r) = 1
4πr 1 − cosh(r) .

Problem (F’91, #3). Prove that
−
1
4π
cos k|x|
|x|
is a fundamental solution for ( + k2
) in R3
where |x| = x2
1 + x2
2 + x2
3,
i.e. prove that for any smooth function f(x) with compact support
u(x) = −
1
4π R3
cos k|x − y|
|x − y|
f(y) dy
is a solution to ( + k2
)u = f.
0 (Rn
), we want to show that for K(|x|) = − 1
4π
cos k|x|
|x| ,
we have ( + k2)K = δ, i.e.
Rn
K(|x|) ( + k2
)v(x) dx = v(0).
Ω = Ω − B (0).
( + k2
)K(|x|) = 0 in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪ ∂B (0)):
Ω
K(|x|) v − v K(|x|) dx =
∂Ω
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
+
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS
We add Ω k2 K(|x|) v dx − Ω v k2 K(|x|) dx to LHS to get:
Ω
K(|x|)( + k2
)v − v ( + k2
)K(|x|)
= 0, in Ω
dx =
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
dS.
lim
→0 Ω
K(|x|)( + k2
)v dx =
Ω
K(|x|)( + k2
)v dx. Since K(r) = −
cos kr
4πr
On ∂B (0), K(|x|) = K( ). Thus, 48
∂B (0)
K(|x|)
∂v
∂n
dS = K( )
∂B (0)
∂v
∂n
dS ≤ −
cos k
4π
4π 2
max ∇v → 0, as → 0.
48
In R3
, for |x| = ,
K(|x|) = K( ) = −
cos k
4π
.
∂K(|x|)
∂n
= −
∂K( )
∂r
=
1
4π
−
k sin k
−
cos k
2
= −
1
4π
k sin k +
cos k
,
since n points inwards. n points toward 0 on the sphere |x| = (i.e., n = −x/|x|).

∂B (0)
v(x)
∂K(|x|)
∂n
dS =
∂B (0)
−
1
4π
k sink +
cos k
v(x) dS
= −
1
4π
k sink +
cos k
∂B (0)
v(x) dS
= −
1
4π
k sink +
cos k
∂B (0)
v(0) dS −
1
4π
k sin k +
cos k
∂B (0)
[v(x) − v(0)] dS
= −
1
4π
k sink +
cos k
v(0) 4π 2
−
1
4π
k sin k +
cos k
[v(x) − v(0)] 4π 2
→ − cos k v(0) → −v(0).
Thus,
Ω
K(|x|)( + k2
)v dx = lim
→0 Ω
K(|x|)( + k2
)v dx = v(0),
that is, K(r) = − 1
4π
cos kr
r is the fundamental solution of + k2
.
Problem (F’97, #2). Let u(x) be a solution of the Helmholtz equation
u + k2
u = 0 x ∈ R3
satisfying the “radiation” conditions
u = O
1
r
,
∂u
∂r
− iku = O
1
r2
, |x| = r → ∞.
Prove that u ≡ 0.
Hint: A fundamental solution to the Helmholtz equation is 1
4πr eikr.
Use the Green formula.
Proof. Denote K(|x|) = 1
4πr eikr
, a fundamental solution. Thus, ( + k2
)K = δ.
Let x0 be any point and Ω = BR(x0); for small > 0 let
Ω = Ω − B (x0).
( + k2
)K(|x|) = 0 in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪ ∂B (x0)):
Ω
u ( + k2
)K − K( + k2
)u dx
= 0
=
∂Ω
u
∂K
∂n
− K
∂u
∂n
dS +
∂B (x0)
u
∂K
∂n
− K
∂u
∂n
dS
→u(x0), as →0
.
(It can be shown by the method previously used that the integral over B (x0) ap-
proaches u(x0) as → 0.) Taking the limit when → 0, we obtain
−u(x0) =
∂Ω
u
∂K
∂n
− K
∂u
∂n
dS =
∂Ω
u
∂
∂r
eik|x−x0|
4π|x − x0|
−
eik|x−x0|
4π|x − x0|
∂u
∂r
dS
=
∂Ω
u
∂
∂r
eik|x−x0|
4π|x − x0|
− ik
eik|x−x0|
4π|x − x0|
= O( 1
|x|2 ); (can be shown)
−
eik|x−x0|
4π|x − x0|
∂u
∂r
− iku dS
= O
1
R
· O
1
R2
· 4πR2
− O
1
R
· O
1
R2
· 4πR2
= 0.
Taking the limit when R → ∞, we get u(x0) = 0.

Problem (S’02, #1). a) Find a radially symmetric solution, u, to the equation in
R2,
u =
1
2π
log |x|,
and show that u is a fundamental solution for 2
, i.e. show
φ(0) =
R2
u 2
φ dx
for any smooth φ which vanishes for |x| large.
b) Explain how to construct the Green’s function for the following boundary value in
a bounded domain D ⊂ R2 with smooth boundary ∂D
w = 0 and
∂w
∂n
= 0 on ∂D,
2
w = f in D.
Proof. a) Rewriting the equation in polar coordinates, we have
u =
1
r
rur r
+
1
r2
uθθ =
1
2π
log r.
For a radially symmetric solution u(r), we have uθθ = 0. Thus,
1
r
rur r
=
1
2π
log r,
rur r
=
1
2π
r log r,
rur =
1
2π
r log r dr =
r2 log r
4π
−
r2
8π
,
ur =
r log r
4π
−
r
8π
,
u =
1
4π
r log r dr −
1
8π
r dr =
1
8π
r2
log r − 1 .
u(r) =
1
8π
r2
log r − 1 .
We want to show that u deﬁned above is a fundamental solution of 2
for n = 2. That
is
R2
u 2
v dx = v(0), v ∈ C∞
0 (Rn
).
See the next page that shows that u deﬁned as u(r) = 1
8π r2
log r is the
Fundamental Solution of 2
. (The − 1
8π r2
term does not play any role.)
In particular, the solution of
2
ω = f(x),
if given by
ω(x) =
R2
u(x − y) 2
ω(y) dy =
1
8π R2
|x − y|2
log |x − y| − 1 f(y) dy.

b) Let
K(x − ξ) =
1
8π
|x − ξ|2
log |x − ξ| − 1 .
We use the method of images to construct the Green’s function.
Let G(x, ξ) = K(x − ξ) + ω(x). We need G(x, ξ) = 0 and ∂G
∂n (x, ξ) = 0 for x ∈ ∂Ω.
Consider wξ(x) with 2wξ(x) = 0 in Ω, wξ(x) = −K(x − ξ) and
∂wξ
∂n (x) = −∂K
∂n (x − ξ)
on ∂Ω. Note, we can ﬁnd the Greens function for the upper-half plane, and then
make a conformal map onto the domain.

Problem (S’97, #6). Show that the fundamental solution of 2
in R2
is given by
V (x1, x2) =
1
8π
r2
ln(r), r = |x − ξ|,
and write the solution of
2
w = F(x1, x2).
Hint: In polar coordinates, = 1
r
∂
∂r (r ∂
∂r )+ 1
r2
∂2
∂θ2 ; for example, V = 1
2π (1+ln(r)).
Proof. Notation: x = (x1, x2). We have
V (x) =
1
8π
r2
log(r),
In polar coordinates: (here, Vθθ = 0)
V =
1
r
rVr r
=
1
r
r
1
8π
r2
log(r)
r r
=
1
8π
1
r
r 2r log(r) + r
r
=
1
8π
1
r
2r2
log(r) + r2
r
=
1
8π
1
r
4r + 4r log r
=
1
2π
(1 + log r).
The fundamental solution V (x) for 2 is the distribution satisfying: 2V (r) = δ(r).
2
V = ( V ) =
1
2π
(1 + log r) =
1
2π
(1 + log r) =
1
2π
1
r
r(1 + log r)r r
=
1
2π
1
r
r
1
r r
=
1
2π
1
r
(1)r = 0 for r = 0.
Thus, 2V (r) = δ(r) ⇒ V is the fundamental solution.
The approach above is not rigorous. See the next page that shows that
V deﬁned above is the Fundamental Solution of 2
.
The solution of
2
ω = F(x),
if given by
ω(x) =
R2
V (x − y) 2
ω(y) dy =
1
8π R2
|x − y|2
log |x − y| F(y) dy.

Show that the Fundamental Solution of 2
in R2
is given by:
K(x) =
1
8π
r2
ln(r), r = |x − ξ|, (17.4)
0 (Rn), we want to show
Rn
K(|x|) 2
v(x) dx = v(0).
Ω = Ω − B (0).
K(|x|) is biharmonic ( 2
K(|x|) = 0) in Ω . Consider Green’s identity (∂Ω = ∂Ω ∪
∂B (0)):
Ω
K(|x|) 2
v dx =
∂Ω
K(|x|)
∂ v
∂n
− v
∂ K(|x|)
∂n
ds +
∂Ω
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
ds
+
∂B (0)
K(|x|)
∂ v
∂n
− v
∂ K(|x|)
∂n
ds +
∂B (0)
K(|x|)
∂v
∂n
− v
∂K(|x|)
∂n
ds.
lim
→0 Ω
K(|x|) 2
v dx =
Ω
K(|x|) v2
dx. Since K(r) is integrable at x = 0.
On ∂B (0), K(|x|) = K( ). Thus, 49
∂B (0)
K(|x|)
∂ v
∂n
dS = K( )
∂B (0)
∂ v
∂n
dS ≤ K( ) ωn
1
max
x∈Ω
∇( v)
=
1
8π
2
log( ) ωn max
x∈Ω
∇( v) → 0, as → 0.
∂B (0)
v(x)
∂ K(|x|)
∂n
dS =
∂B (0)
−
1
2π
v(x) dS
=
∂B (0)
−
1
2π
v(0) dS +
∂B (0)
−
1
2π
[v(x) − v(0)] dS
= −
1
2π
v(0) 2π − max
x∈∂B (0)
v(x) − v(0)
= −v(0).
∂B (0)
K(|x|)
∂v
∂n
dS = K( )
∂B (0)
∂v
∂n
dS ≤
1
2π
(1 + log ) 2π max
x∈Ω
|∇v| → 0, as → 0.
∂B (0)
v
∂K(|x|)
∂n
dS =
∂B (0)
−
1
4π
log −
1
8π
v(x) dS
≤
4π
log +
1
2
· 2π max
x∈∂B (0)
| v| → 0, as → 0.
49
Note that for |x| = ,
K(|x|) = K( ) =
1
8π
2
log , K =
1
2π
(1 + log ),
∂K(|x|)
∂n
= −
∂K( )
∂r
= −
1
4π
log −
1
8π
,
∂ K
∂n
= −
∂ K
∂r
= −
1
2π
.

⇒
Ω
K(|x|) 2
v dx = lim
→0 Ω
K(|x|) 2
v dx = v(0).

17.3 Radial Variables
Problem (F’99, #8). Let u = u(x, t) solve the following PDE in two spatial dimen-
sions
− u = 1
for r < R(t), in which r = |x| is the radial variable, with boundary condition
u = 0
on r = R(t). In addition assume that R(t) satisﬁes
dR
dt
= −
∂u
∂r
(r = R)
with initial condition R(0) = R0.
a) Find the solution u(x, t).
b) Find an ODE for the outer radius R(t), and solve for R(t).
Proof. a) Rewrite the equation in polar coordinates:
− u = −
1
r
(rur)r +
1
r2
uθθ = 1.
For a radially symmetric solution u(r), we have uθθ = 0. Thus,
1
r
(rur)r = −1,
(rur)r = −r,
rur = −
r2
2
+ c1,
ur = −
r
2
+
c1
r
,
u(r, t) = −
r2
4
+ c1 log r + c2.
Since we want u to be deﬁned for r = 0, we have c1 = 0. Thus,
u(r, t) = −
r2
4
+ c2.
u(R(t), t) = −
R(t)2
4
+ c2 = 0 ⇒ c2 =
R(t)2
4
. Thus,
u(r, t) = −
r2
4
+
R(t)2
4
.
b) We have
u(r, t) = −
r2
4
+
R(t)2
4
,
∂u
∂r
= −
r
2
,
dR
dt
= −
∂u
∂r
(r = R) =
R
2
, (from )
dR
R
=
dt
2
,
log R =
t
2
,
R(t) = c1e
t
2 , R(0) = c1 = R0. Thus,

R(t) = R0e
t
2 .

Problem (F’01, #3). Let u = u(x, t) solve the following PDE in three spatial di-
mensions
u = 0
for R1 < r < R(t), in which r = |x| is the radial variable, with boundary conditions
u(r = R(t), t) = 0, and u(r = R1, t) = 1.
In addition assume that R(t) satisﬁes
dR
dt
= −
∂u
∂r
(r = R)
with initial condition R(0) = R0 in which R0 > R1.
a) Find the solution u(x, t).
b) Find an ODE for the outer radius R(t).
Proof. a) Rewrite the equation in spherical coordinates (n = 3, radial functions):
u =
∂2
∂r2
+
2
r
∂
∂r
u =
1
r2
(r2
ur)r = 0.
(r2
ur)r = 0,
r2
ur = c1,
ur =
c1
r2
,
u(r, t) = −
c1
r
+ c2.
u(R(t), t) = −
c1
R(t)
+ c2 = 0 ⇒ c2 =
c1
R(t)
,
u(R1, t) = −
c1
R1
+ c2 = 1.
This gives
c1 =
R1R(t)
R1 − R(t)
, c2 =
R1
R1 − R(t)
.
u(r, t) = −
R1R(t)
R1 − R(t)
·
1
r
+
R1
R1 − R(t)
.
b) We have
u(r, t) = −
R1R(t)
R1 − R(t)
·
1
r
+
R1
R1 − R(t)
,
∂u
∂r
=
R1R(t)
R1 − R(t)
·
1
r2
,
dR
dt
= −
∂u
∂r
(r = R) = −
R1R(t)
R1 − R(t)
·
1
R(t)2
= −
R1
(R1 − R(t)) R(t)
(from )
Thus, an ODE for the outer radius R(t) is
dR
dt = R1
(R(t)−R1) R(t),
R(0) = R0, R0 > R1.

Problem (S’02, #3). Steady viscous ﬂow in a cylindrical pipe is described by the
equation
(u · ∇)u +
1
ρ
∇p −
η
ρ
u = 0
on the domain −∞ < x1 < ∞, x2
2 +x2
3 ≤ R2
, where u = (u1, u2, u3) = (U(x2, x3), 0, 0)
is the velocity vector, p(x1, x2, x3) is the pressure, and η and ρ are constants.
a) Show that ∂p
∂x1
is a constant c, and that U = c/η.
b) Assuming further that U is radially symmetric and U = 0 on the surface of the pipe,
determine the mass Q of ﬂuid passing through a cross-section of pipe per unit time in
terms of c, ρ, η, and R. Note that
Q = ρ
{x2
2+x2
3≤R2}
U dx2dx3.
Proof. a) Since u = (u1, u2, u3) = (U(x2, x3), 0, 0), we have
(u · ∇)u = (u1, u2, u3) ·
∂u1
∂x1
,
∂u2
∂x2
,
∂u3
∂x3
= (U(x2, x3), 0, 0) · (0, 0, 0) = 0.
Thus,
1
ρ
∇p −
η
ρ
u = 0,
∇p = η u,
∂p
∂x1
,
∂p
∂x2
,
∂p
∂x3
= η( u1, u2, u3),
∂p
∂x1
,
∂p
∂x2
,
∂p
∂x3
= η(Ux2x2 + Ux3x3 , 0, 0).
We can make the following observations:
∂p
∂x1
= η (Ux2x2 + Ux3x3 )
indep. of x1
,
∂p
∂x2
= 0 ⇒ p = f(x1, x3),
∂p
∂x3
= 0 ⇒ p = g(x1, x2).
Thus, p = h(x1). But ∂p
∂x1
is independent of x1. Therefore, ∂p
∂x1
= c.
∂p
∂x1
= η U,
U =
1
η
∂p
∂x1
=
c
η
.

b) Cylindrical Laplacian in R3
for radial functions is
U =
1
r
rUr r
,
1
r
rUr r
=
c
η
,
rUr r
=
cr
η
,
rUr =
cr2
2η
+ c1,
Ur =
cr
2η
+
c1
r
.
For Ur to stay bounded for r = 0, we need c1 = 0. Thus,
Ur =
cr
2η
,
U =
cr2
4η
+ c2,
0 = U(R) =
cR2
4η
+ c2,
⇒ U =
cr2
4η
−
cR2
4η
=
c
4η
(r2
− R2
).
Q = ρ
{x2
2+x2
3≤R2}
U dx2dx3 =
cρ
4η
2π
0
R
0
(r2
− R2
) rdrdθ = −
cρ
4η
2π
0
R4
4
dθ
= −
cρR4π
8η
.
It is not clear why Q is negative?

17.4 Weak Solutions
A function u ∈ H2
0 (Ω) is a weak solution of the biharmonic equation
⎧
⎪⎨
⎪⎩
2u = f in Ω
u = 0 on ∂Ω
∂u
∂n = 0 on ∂Ω
provided
Ω
u v dx =
Ω
fv dx
for all test functions v ∈ H2
0 (Ω). Prove that for each f ∈ L2
(Ω), there exists a unique
weak solution for this problem. Here, H2
0 (Ω) is the closure of all smooth functions in
Ω which vanish on the boundary and with ﬁnite H2
norm: ||u||2
2 = Ω(u2
xx + u2
xy +
u2
yy) dxdy < ∞.
Hint: use Lax-Milgram lemma.
Proof. Multiply the equation by v ∈ H2
0 (Ω) and integrate over Ω:
2
u = f,
Ω
2
u v dx =
Ω
f v dx,
∂Ω
∂ u
∂n
v ds −
∂Ω
u
∂v
∂n
ds
= 0
+
Ω
u v dx =
Ω
f v dx,
Ω
u v dx
a(u,v)
=
Ω
f v dx
L(v)
.
Denote: V = H2
0 (Ω). Check the following conditions:
❶ a(·, ·) is continuous: ∃γ > 0, s.t. |a(u, v)| ≤ γ||u||V ||v||V , ∀u, v ∈ V ;
❷ a(·, ·) is V-elliptic: ∃α > 0, s.t. a(v, v) ≥ α||v||2
V , ∀v ∈ V ;
❸ L(·) is continuous: ∃Λ > 0, s.t. |L(v)| ≤ Λ||v||V , ∀v ∈ V.

We have 50
❶ |a(u, v)|2
=
Ω
u v dx
2
≤
Ω
( u)2
dx
Ω
( v)2
dx ≤ ||u||2
H2
0(Ω)||v||2
H2
0(Ω).
❷ a(v, v) =
Ω
( v)2
dx ≥ ||v||H2
0(Ω).
❸ |L(v)| =
Ω
f v dx ≤
Ω
|f| |v| dx ≤
Ω
f2
dx
1
2
Ω
v2
dx
1
2
= ||f||L2(Ω)||v||L2(Ω) ≤ ||f||L2(Ω)
Λ
||v||H2
0(Ω).
Thus, by Lax-Milgram theorem, there exists a weak solution u ∈ H2
0 (Ω).
Also, we can prove the stability result.
α||u||2
H2
0(Ω) ≤ a(u, u) = |L(u)| ≤ Λ||u||H2
0(Ω),
⇒ ||u||H2
0(Ω) ≤
Λ
α
.
Let u1, u2 be two solutions so that
a(u1, v) = L(v),
a(u2, v) = L(v)
for all v ∈ V . Subtracting these two equations, we see that:
a(u1 − u2, v) = 0 ∀v ∈ V.
Applying the stability estimate (with L ≡ 0, i.e. Λ = 0), we conclude that
||u1 − u2||H2
0 (Ω) = 0, i.e. u1 = u2.
50
Cauchy-Schwarz Inequality:
|(u, v)| ≤ ||u||||v|| in any norm, for example |uv|dx ≤ ( u2
dx)
1
2 ( v2
dx)
1
2 ;
|a(u, v)| ≤ a(u, u)
1
2 a(v, v)
1
2 ;
|v|dx = |v| · 1 dx = ( |v|2
dx)
1
2 ( 12
dx)
1
2 .
Poincare Inequality:
||v||H2(Ω) ≤ C
Ω
( v)2
dx.
Green’s formula:
Ω
( u)2
dx =
Ω
(u2
xx + u2
yy + 2uxxuyy) dxdy =
Ω
(u2
xx + u2
yy − 2uxxy uy) dxdy =
Ω
(u2
xx + u2
yy + 2|uxy|2
) dxdy ≥ ||u||2
H2
0(Ω).

17.5 Uniqueness
Problem. The solution of the Robin problem
∂u
∂n
+ αu = β, x ∈ ∂Ω
for the Laplace equation is unique when α > 0 is a constant.
Proof. Let u1 and u2 be two solutions of the Robin problem. Let w = u1 − u2. Then
w = 0 in Ω,
∂w
∂n
+ αw = 0 on ∂Ω.
Consider Green’s formula:
Ω
∇u · ∇v dx =
∂Ω
v
∂u
∂n
ds −
Ω
v u dx.
Setting w = u = v gives
Ω
|∇w|2
dx =
∂Ω
w
∂w
∂n
ds −
Ω
w w dx
=0
.
Boundary condition gives
Ω
|∇w|2
dx
≥0
= −
∂Ω
αw2
ds
≤0
.
Thus, w ≡ 0, and u1 ≡ u2. Hence, the solution to the Robin problem is unique.
Problem. Suppose q(x) ≥ 0 for x ∈ Ω and consider solutions u ∈ C2
(Ω) ∩ C1
(Ω) of
u − q(x)u = 0 in Ω.
Establish uniqueness theorems for
a) the Dirichlet problem: u(x) = g(x), x ∈ ∂Ω;
b) the Neumann problem: ∂u/∂n = h(x), x ∈ ∂Ω.
Proof. Let u1 and u2 be two solutions of the Dirichlet or Neumann problem.
Let w = u1 − u2. Then
w − q(x)w = 0 in Ω,
w = 0 or
∂w
∂n
= 0 on ∂Ω.
Ω
∇u · ∇v dx =
∂Ω
v
∂u
∂n
ds −
Ω
v u dx.
Ω
|∇w|2
dx =
∂Ω
w
∂w
∂n
ds
=0, Dirichlet or Neumann
−
Ω
w w dx.

Ω
|∇w|2
dx
≥0
= −
Ω
q(x)w2
dx
≤0
.
Thus, w ≡ 0, and u1 ≡ u2. Hence, the solution to the Dirichlet and Neumann problems
are unique.
Problem (F’02, #8; S’93, #5).
Let D be a bounded domain in R3. Show that a solution of the boundary value problem
2
u = f in D,
u = u = 0 on ∂D
is unique.
Proof. Method I: Maximum Principle. Let u1, u2 be two solutions of the boundary
value problem. Deﬁne w = u1 − u2. Then w satisﬁes
2
w = 0 in D,
w = w = 0 on ∂D.
So w is harmonic and thus achieves min and max on ∂D ⇒ w ≡ 0.
So w is harmonic, but w ≡ 0 on ∂D ⇒ w ≡ 0. Hence, u1 = u2.
Method II: Green’s Identities. Multiply the equation by w and integrate:
w 2
w = 0,
Ω
w 2
w dx = 0,
∂Ω
w
∂( w)
∂n
ds
=0
−
Ω
∇w∇( )w dx = 0,
−
∂Ω
∂w
∂n
w ds
=0
+
Ω
( w)2
dx = 0.
Thus, w ≡ 0. Now, multiply w = 0 by w. We get
Ω
|∇w|2
dx = 0.
Thus, ∇w = 0 and w is a constant. Since w = 0 on ∂Ω, we have w ≡ 0.
Problem (F’97, #6).
a) Let u(x) ≥ 0 be continuous in closed bounded domain Ω ⊂ Rn, u is continuous in
Ω,
u = u2
and u|∂Ω = 0.
Prove that u ≡ 0.
b) What can you say about u(x) when the condition u(x) ≥ 0 in Ω is dropped?

Proof. a) Multiply the equation by u and integrate:
u u = u3
,
Ω
u u dx =
Ω
u3
dx,
∂Ω
u
∂u
∂n
ds
=0
−
Ω
|∇u|2
dx =
Ω
u3
dx,
Ω
u3
+ |∇u|2
dx = 0.
Since u(x) ≥ 0, we have u ≡ 0.
b) If we don’t know that u(x) ≥ 0, then u can not be nonnegative on the entire
domain Ω. That is, u(x) < 0, on some (or all) parts of Ω. If u is nonnegative on Ω,
then u ≡ 0.

Problem (W’02, #5). Consider the boundary value problem
u +
n
k=1
αk
∂u
∂xk
− u3
= 0 in Ω,
u = 0 on ∂Ω,
where Ω is a bounded domain in Rn with smooth boundary. If the αk’s are constants,
and u(x) has continuous derivatives up to second order, prove that u must vanish
identically.
Proof. Multiply the equation by u and integrate:
u u +
n
k=1
αk
∂u
∂xk
u − u4
= 0,
Ω
u u dx +
Ω
n
k=1
αk
∂u
∂xk
u dx −
Ω
u4
dx = 0,
∂Ω
u
∂u
∂n
ds
= 0
−
Ω
|∇u|2
dx +
Ω
n
k=1
αk
∂u
∂xk
u dx
➀
−
Ω
u4
dx = 0.
We will show that ➀ = 0.
Ω
αk
∂u
∂xk
u dx =
∂Ω
αk u2
ds
= 0
−
Ω
αk u
∂u
∂xk
dx,
⇒ 2
Ω
αk
∂u
∂xk
u dx = 0,
⇒
Ω
n
k=1
αk
∂u
∂xk
u dx = 0.
Thus, we have
−
Ω
|∇u|2
dx −
Ω
u4
dx = 0,
Ω
|∇u|2
+
Ω
u4
dx = 0.
Hence, |∇u|2
= 0 and u4
= 0. Thus, u ≡ 0.
Note that
Ω
n
k=1
αk
∂u
∂xk
u dx =
Ω
α · ∇u u dx =
∂Ω
α · nu2
ds
= 0
−
Ω
α · ∇u u dx,
and thus,
Ω
α · ∇u u dx = 0.

Problem (W’02, #9). Let D = {x ∈ R2
: x1 ≥ 0, x2 ≥ 0}, and assume that f is
continuous on D and vanishes for |x| > R.
a) Show that the boundary value problem
u = f in D,
u(x1, 0) =
∂u
∂x1
(0, x2) = 0
can have only one bounded solution.
b) Find an explicit Green’s function for this boundary value problem.
Proof. a) Let u1, u2 be two solutions of the boundary value problem. Deﬁne w =
u1 − u2. Then w satisﬁes
w = 0 in D,
w(x1, 0) =
∂w
∂x1
(0, x2) = 0.
D
∇u · ∇v dx =
∂D
v
∂u
∂n
ds −
D
v u dx.
D
|∇w|2
dx =
∂D
w
∂w
∂n
ds −
D
w w dx,
D
|∇w|2
dx =
Rx1
w
∂w
∂n
ds +
Rx2
w
∂w
∂n
ds +
|x|>R
w
∂w
∂n
ds −
D
w w dx
=
Rx1
w(x1, 0)
=0
∂w
∂x2
ds +
Rx2
w(0, x2)
∂w
∂x1
=0
ds +
|x|>R
w
=0
∂w
∂n
ds −
D
w w
=0
dx,
D
|∇w|2
dx = 0 ⇒ |∇w|2
= 0 ⇒ w = const.
Since w(x1, 0) = 0 ⇒ w ≡ 0. Thus, u1 = u2.
b) The similar problem is solved in the appropriate section (S’96, #3).
Notice whenever you are on the boundary with variable x,
|x − ξ(0)
| =
|x − ξ(1)||x − ξ(3)|
|x − ξ(2)|
.
So, G(x, ξ) =
1
2π
log |x − ξ| − log
|x − ξ(1)||x − ξ(3)|
|x − ξ(2)|
is the Green’s function.

Problem (F’98, #4). In two dimensions x = (x, y), deﬁne the set Ωa as
Ωa = Ω+
∪ Ω−
in which
Ω+
= {|x − x0| ≤ a} ∩ {x ≥ 0}
Ω−
= {|x + x0| ≤ a} ∩ {x ≤ 0} = −Ω+
and x0 = (1, 0). Note that Ωa consists of two components when 0 < a < 1
and a single component when a > 1. Consider the Neumann problem
∇2
u = f, x ∈ Ωa
∂u/∂n = 0, x ∈ ∂Ωa
in which
Ω+
f(x) dx = 1
Ω−
f(x) dx = −1
a) Show that this problem has a solution for 1 < a, but not for 0 < a < 1.
(You do not need to construct the solution, only demonstrate solveability.)
b) Show that maxΩa |∇u| → ∞ as a → 1 from above. (Hint: Denote L to be
the line segment L = Ω+ ∩ Ω−, and note that its length |L| goes to 0 as a → 1.)
Proof. a) We use the Green’s identity. For 1 < a,
0 =
∂Ωa
∂u
∂n
ds =
Ωa
u dx =
Ωa
f(x) dx
=
Ω+
f(x) dx +
Ω−
f(x) dx = 1 − 1 = 0.
Thus, the problem has a solution for 1 < a.
For 0 < a < 1, Ω+
and Ω−
are disjoint. Consider Ω+
:
0 =
∂Ω+
∂u
∂n
ds =
Ω+
u dx =
Ω+
f(x) dx = 1,
0 =
∂Ω−
∂u
∂n
ds =
Ω−
u dx =
Ω−
f(x) dx = −1.
We get contradictions.
Thus, the solution does not exist for 0 < a < 1.

b) Using the Green’s identity, we have: (n+ is the unit normal to Ω+)
Ω+
u dx =
∂Ω+
∂u
∂n+
ds =
L
∂u
∂n+
ds,
Ω−
u dx =
∂Ω−
∂u
∂n−
ds =
L
∂u
∂n−
ds = −
L
∂u
∂n+
ds.
Ω+
u dx −
Ω−
u dx = 2
L
∂u
∂n+
ds,
Ω+
f(x) dx −
Ω−
f(x) dx = 2
L
∂u
∂n+
ds.
2 = 2
L
∂u
∂n+
ds,
1 =
L
∂u
∂n+
ds ≤
L
∂u
∂n+
ds ≤
L
∂u
∂n+
2
+
∂u
∂τ
2
≤ |L| max
L
|∇u| ≤ |L| max
Ωa
|∇u|.
Thus,
max
Ωa
|∇u| ≥
1
|L|
.
As a → 1 (L → 0) ⇒ maxΩa |∇u| → ∞.

Problem (F’00, #1). Consider the Dirichlet problem in a bounded domain D ⊂ Rn
with smooth boundary ∂D,
u + a(x)u = f(x) in D,
u = 0 on ∂D.
a) Assuming that |a(x)| is small enough, prove the uniqueness of the classical solution.
b) Prove the existence of the solution in the Sobolev space H1(D) assuming that f ∈
L2(D).
Note: Use Poincare inequality.
Proof. a) By Poincare Inequality, for any u ∈ C1
0 (D), we have ||u||2
2 ≤ C||∇u||2
2.
Consider two solutions of the Dirichlet problem above. Let w = u1 − u2. Then, w
satisﬁes
w + a(x)w = 0 in D,
w = 0 on ∂D.
w w + a(x)w2
= 0,
w w dx + a(x)w2
dx = 0,
− |∇w|2
dx + a(x)w2
dx = 0,
a(x)w2
dx = |∇w|2
dx ≥
1
C
w2
dx, (by Poincare inequality)
a(x)w2
dx −
1
C
w2
dx ≥ 0,
|a(x)| w2
dx −
1
C
w2
dx ≥ 0,
|a(x)| −
1
C
w2
dx ≥ 0.
If |a(x)| < 1
C ⇒ w ≡ 0.
b) Consider
F(v, u) = −
Ω
(v u + a(x)vu) dx = −
Ω
vf(x) dx = F(v).
F(v) is a bounded linear functional on v ∈ H1,2
(D), D = Ω.
|F(v)| ≤ ||f||2||v||2 ≤ ||f||2C||v||H1,2(D)
So by Riesz representation, there exists a solution u ∈ H1,2
0 (D) of
− < u, v >=
Ω
v u + a(x)vu dx =
Ω
vf(x) dx = F(v) ∀v ∈ H1,2
0 (D).

Problem (S’91, #8). Define the operator
Lu = uxx + uyy − 4(r2
+ 1)u
in which r2 = x2 + y2.
a) Show that ϕ = er2
satisfies Lϕ = 0.
b) Use this to show that the equation
Lu = f in Ω
∂u
∂n
= γ on ∂Ω
has a solution only if
Ω
ϕf dx =
∂Ω
ϕγ ds(x).
Proof. a) Expressing Laplacian in polar coordinates, we obtain:
Lu =
1
r
(rur)r − 4(r2
+ 1)u,
Lϕ =
1
r
(rϕr)r − 4(r2
+ 1)ϕ =
1
r
(2r2
er2
)r − 4(r2
+ 1)er2
=
1
r
(4rer2
+ 2r2
· 2rer2
) − 4r2
er2
− 4er2
= 0.
b) We have ϕ = er2
= ex2+y2
= ex2
ey2
. From part (a),
Lϕ = 0,
∂ϕ
∂n
= ∇ϕ · n = (ϕx, ϕy) · n = (2xex2
ey2
, 2yex2
ey2
) · n = 2er2
(x, y) · (−y, x) = 0.
51
Consider two equations:
Lu = u − 4(r2
+ 1)u,
Lϕ = ϕ − 4(r2
+ 1)ϕ.
Multiply the first equation by ϕ and the second by u and subtract the two equations:
ϕLu = ϕ u − 4(r2
+ 1)uϕ,
uLϕ = u ϕ − 4(r2
+ 1)uϕ,
ϕLu − uLϕ = ϕ u − u ϕ.
Then, we start from the LHS of the equality we need to prove and end up with RHS:
Ω
ϕf dx =
Ω
ϕLu dx =
Ω
(ϕLu − uLϕ) dx =
Ω
(ϕ u − u ϕ) dx
=
Ω
(ϕ
∂u
∂n
− u
∂ϕ
∂n
) ds =
Ω
ϕ
∂u
∂n
ds =
Ω
ϕγ ds.
51
The only shortcoming in the above proof is that we assume n = (−y, x), without giving an expla-
nation why it is so.

17.6 Self-Adjoint Operators
Consider an mth-order diﬀerential operator
Lu =
|α|≤m
aα(x)Dα
u.
The integration by parts formula
Ω
uxk
v dx =
∂Ω
uvnk ds −
Ω
uvxk
dx n = (n1, . . ., nn) ∈ Rn
,
with u or v vanishing near ∂Ω is:
Ω
uxk
v dx = −
Ω
uvxk
dx.
We can repeat the integration by parts with any combination of derivatives
Dα = (∂/∂x1)α1 · · ·(∂/∂xn)αn:
Ω
(Dα
u)v dx = (−1)m
Ω
uDα
v dx, (m = |α|).
We have
Ω
(Lu) v dx =
Ω |α|≤m
aα(x)Dα
u v dx =
|α|≤m Ω
aα(x) v Dα
u dx
=
|α|≤m
(−1)|α|
Ω
Dα
(aα(x) v) u dx =
Ω |α|≤m
(−1)|α|
Dα
(aα(x) v)
L∗(v)
u dx
=
Ω
L∗
(v) u dx,
for all u ∈ Cm
(Ω) and v ∈ C∞
0 .
The operator
L∗
(v) =
|α|≤m
(−1)|α|
Dα
(aα(x) v)
is called the adjoint of L.
The operator is self-adjoint if L∗ = L.
Also, L is self-adjoint if 52
Ω
vL(u) dx =
Ω
uL(v) dx.
52
L = L∗
⇔ (Lu|v) = (u|L∗
v) = (u|Lv).

Consider the Laplace operator in the wedge 0 ≤ x ≤ y with boundary conditions
∂f
∂x
= 0 on x = 0
∂f
∂x
− α
∂f
∂y
= 0 on x = y.
a) For which values of α is this operator self-adjoint?
b) For such a value of α, suppose that
f = e−r2/2
cos θ
with these boundary conditions. Evaluate
CR
∂
∂r
f ds
in which CR is the circular arc of radius R connecting the boundaries x = 0 and x = y.
Proof. a) We have
Lu = u = 0
∂u
∂x
= 0 on x = 0
∂u
∂x
− α
∂u
∂y
= 0 on x = y.
The operator L is self-adjoint if:
Ω
(u Lv − v Lu) dx = 0.
Ω
(u Lv − v Lu) dx =
Ω
(u v − v u) dx =
∂Ω
u
∂v
∂n
− v
∂u
∂n
ds
=
x=0
u
∂v
∂n
− v
∂u
∂n
ds +
x=y
u
∂v
∂n
− v
∂u
∂n
ds
=
x=0
u (∇v · n) − v (∇u · n) ds +
x=y
u (∇v · n) − v (∇u · n) ds
=
x=0
u (vx, vy) · (−1, 0) − v (ux, uy) · (−1, 0) ds
+
x=y
u (vx, vy) · (1/
√
2, −1/
√
2) − v (ux, uy) · (1/
√
2, −1/
√
2) ds
=
x=0
u (0, vy) · (−1, 0) − v (0, uy) · (−1, 0) ds
= 0
+
x=y
u (αvy, vy) · (1/
√
2, −1/
√
2) − v (αuy, uy) · (1/
√
2, −1/
√
2) ds
=
x=y
uvy
√
2
(α − 1) −
vuy
√
2
(α − 1) ds =
need
0.

Thus, we need α = 1 so that L is self-adjoint.
b) We have α = 1. Using Green’s identity and results from part (a), (∂f
∂n = 0 on
x = 0 and x = y):
Ω
f dx =
∂Ω
∂f
∂n
ds =
∂CR
∂f
∂n
ds +
x=0
∂f
∂n
=0
ds +
x=y
∂f
∂n
=0
ds =
∂CR
∂f
∂r
ds.
Thus,
∂CR
∂f
∂r
ds =
Ω
f dx =
R
0
π
2
π
4
e−r2/2
cos θ r drdθ
= 1 −
1
√
2
R
0
e−r2/2
r dr = 1 −
1
√
2
(1 − e−R2/2
).

Problem (F’99, #1). Suppose that u = 0 in the weak sense in Rn
and that there
is a constant C such that
{|x−y|<1}
|u(y)| dy < C, ∀x ∈ Rn
.
Show that u is constant.
Proof. Consider Green’s formula:
Ω
∇u · ∇v dx =
∂Ω
v
∂u
∂n
ds −
Ω
v u dx
For v = 1, we have
∂Ω
∂u
∂n
ds =
Ω
u dx.
Let Br(x0) be a ball in Rn
. We have
0 =
Br(x0)
u dx =
∂Br(x0)
∂u
∂n
ds = rn−1
|x|=1
∂u
∂r
(x0 + rx) ds
= rn−1
ωn
∂
∂r
1
ωn |x|=1
u(x0 + rx) ds.
Thus, 1
ωn |x|=1 u(x0 + rx) ds is independent of r. Hence, it is constant.
By continuity, as r → 0, we obtain the Mean Value property:
u(x0) =
1
ωn |x|=1
u(x0 + rx) ds.
If |x−y|<1 |u(y)| dy < C ∀x ∈ Rn, we have |u(x)| < C in Rn.
Since u is harmonic and bounded in Rn, u is constant by Liouville’s theorem. 53
53
Liouville’s Theorem: A bounded harmonic function deﬁned on Rn
is a constant.

Problem (S’01, #1). For bodies (bounded regions B in R3
) which are not perfectly
conducting one considers the boundary value problem
0 = ∇ · γ(x)∇u =
3
j=1
∂
∂xj
γ(x)
∂u
∂xj
u = f on ∂B.
The function γ(x) is the “local conductivity” of B and u is the voltage. We deﬁne
operator Λ(f) mapping the boundary data f to the current density at the boundary by
Λ(f) = γ(x)
∂u
∂n
,
and ∂/∂n is the inward normal derivative (this formula deﬁnes the current density).
a) Show that Λ is a symmetric operator, i.e. prove
∂B
gΛ(f) dS =
∂B
fΛ(g) dS.
b) Use the positivity of γ(x) > 0 to show that Λ is negative as an operator, i.e., prove
∂B
fΛ(f) dS ≤ 0.
Proof. a) Let
∇ · γ(x)∇u = 0 on Ω,
u = f on ∂Ω.
∇ · γ(x)∇v = 0 on Ω,
v = g on ∂Ω.
Λ(f) = γ(x)
∂u
∂n
, Λ(g) = γ(x)
∂v
∂n
.
Since ∂/∂n is inward normal derivative, Green’s formula is:
−
∂Ω
v
=g
γ(x)
∂u
∂n
dS −
Ω
∇v · γ(x)∇u dx =
Ω
v∇ · γ(x)∇u dx.
We have
∂Ω
gΛ(f) dS =
∂Ω
gγ(x)
∂u
∂n
dS = −
Ω
∇v · γ(x)∇u dx −
Ω
v ∇ · γ(x)∇u
=0
dx
=
∂Ω
uγ(x)
∂v
∂n
dS +
Ω
u ∇ · γ(x)∇v
=0
dx
=
∂Ω
fγ(x)
∂v
∂n
dS =
∂Ω
fΛ(g) dS.
b) We have γ(x) > 0.
∂Ω
fΛ(f) dS =
∂Ω
uγ(x)
∂u
∂n
dS = −
Ω
u ∇ · γ(x)∇u
=0
dx −
Ω
γ(x)∇u · ∇u dx
= −
Ω
γ(x)|∇u|2
≥0
≤ 0.

Problem (S’01, #4). The Poincare Inequality states that for any bounded domain
Ω in Rn there is a constant C such that
Ω
|u|2
dx ≤ C
Ω
|∇u|2
dx
for all smooth functions u which vanish on the boundary of Ω.
a) Find a formula for the “best” (smallest) constant for the domain Ω in terms of the
eigenvalues of the Laplacian on Ω, and
b) give the best constant for the rectangular domain in R2
Ω = {(x, y) : 0 ≤ x ≤ a, 0 ≤ y ≤ b}.
Proof. a) Consider Green’s formula:
Ω
∇u · ∇v dx =
∂Ω
v
∂u
∂n
ds −
Ω
v u dx.
Setting u = v and with u vanishing on ∂Ω, Green’s formula becomes:
Ω
|∇u|2
dx = −
Ω
u u dx.
Expanding u in the eigenfunctions of the Laplacian, u(x) = anφn(x), the formula
above gives
Ω
|∇u|2
dx = −
Ω
∞
n=1
anφn(x)
∞
m=1
−λmamφm(x) dx =
∞
m,n=1
λmanam
Ω
φnφm dx
=
∞
n=1
λn|an|2
.
Also,
Ω
|u|2
dx =
Ω
∞
n=1
anφn(x)
∞
m=1
amφm(x) =
∞
n=1
|an|2
.
Comparing and , and considering that λn increases as n → ∞, we obtain
λ1
Ω
|u|2
dx = λ1
∞
n=1
|an|2
≤
∞
n=1
λn|an|2
=
Ω
|∇u|2
dx.
Ω
|u|2
dx ≤
1
λ1 Ω
|∇u|2
dx,
with C = 1/λ1.
b) For the rectangular domain Ω = {(x, y) : 0 ≤ x ≤ a, 0 ≤ y ≤ b} ⊂ R2, the
eigenvalues of the Laplacian are
λmn = π2 m2
a2
+
n2
b2
, m, n = 1, 2, . . ..
λ1 = λ11 = π2 1
a2
+
1
b2
,
⇒ C =
1
λ11
=
1
π2
1
( 1
a2 + 1
b2 )
.

Problem (S’01, #6). a) Let B be a bounded region in R3
with smooth boundary ∂B.
The “conductor” potential for the body B is the solution of Laplace’s equation outside
B
V = 0 in R3
/B
subject to the boundary conditions, V = 1 on ∂B and V (x) tends to zero as |x| → ∞.
Assuming that the conductor potential exists, show that it is unique.
b) The “capacity” C(B) of B is defined to be the limit of |x|V (x) as |x| → ∞. Show
that
C(B) = −
1
4π ∂B
∂V
∂n
dS,
where ∂B is the boundary of B and n is the outer unit normal to it (i.e. the normal
pointing “toward infinity”).
c) Suppose that B ⊂ B. Show that C(B ) ≤ C(B).
Proof. a) Let V1, V2 be two solutions of the boundary value problem. Define W =
V1 − V2. Then W satisfies
⎧
⎪⎨
⎪⎩
W = 0 in R3
/B
W = 0 on ∂B
W → 0 as |x| → ∞.
B
∇u · ∇v dx =
∂B
v
∂u
∂n
ds −
B
v u dx.
Setting W = u = v gives
B
|∇W|2
dx =
∂B
W
=0
∂W
∂n
ds −
B
W W
=0
dx = 0.
Thus, |∇W|2 = 0 ⇒ W = const. Since W = 0 on ∂B, W ≡ 0, and V1 = V2.
b & c) For (b)&(c), see the solutions from Ralston’s homework (a few pages
down).

Problem (W’03, #2). Let L be the second order diﬀerential operator L = − a(x)
in which x = (x1, x2, x3) is in the three-dimensional cube C = {0 < xi < 1, i = 1, 2, 3}.
Suppose that a > 0 in C. Consider the eigenvalue problem
Lu = λu for x ∈ C
u = 0 for x ∈ ∂C.
a) Show that all eigenvalues are negative.
b) If u and v are eigenfunctions for distinct eigenvalues λ and μ, show that u and v
are orthogonal in the appropriate product.
c) If a(x) = a1(x1) + a2(x2) + a3(x3) ﬁnd an expression for the eigenvalues and eigen-
vectors of L in terms of the eigenvalues and eigenvectors of a set of one-dimensional
problems.
Proof. a) We have
u − a(x)u = λu.
Multiply the equation by u and integrate:
u u − a(x)u2
= λu2
,
Ω
u u dx −
Ω
a(x)u2
dx = λ
Ω
u2
dx,
∂Ω
u
∂u
∂n
ds
=0
−
Ω
|∇u|2
dx −
Ω
a(x)u2
dx = λ
Ω
u2
dx,
λ =
− Ω(|∇u|2 + a(x)u2) dx
Ω u2 dx
< 0.
b) Let λ, μ, be the eigenvalues and u, v be the corresponding eigenfunctions. We have
u − a(x)u = λu. (17.5)
v − a(x)v = μv. (17.6)
Multiply (17.5) by v and (17.6) by u and subtract equations from each other
v u − a(x)uv = λuv,
u v − a(x)uv = μuv.
v u − u v = (λ − μ)uv.
Integrating over Ω gives
Ω
v u − u v dx = (λ − μ)
Ω
uv dx,
∂Ω
v
∂u
∂n
− u
∂v
∂n
=0
dx = (λ − μ)
Ω
uv dx.
Since λ = μ, u and v are orthogonal on Ω.

c) The three one-dimensional eigenvalue problems are:
u1x1x1
(x1) − a(x1)u1(x1) = λ1u1(x1),
u2x2x2
(x2) − a(x2)u2(x2) = λ2u2(x2),
u3x3x3
(x3) − a(x3)u3(x3) = λ3u3(x3).
We need to derive how u1, u2, u3 and λ1, λ2, λ3 are related to u and λ.

17.7 Spherical Means
Problem (S’95, #4). Consider the biharmonic operator in R3
2
u ≡
∂2
∂x2
+
∂2
∂y2
+
∂2
∂z2
2
u.
a) Show that 2 is self-adjoint on |x| < 1 with the following boundary conditions on
|x| = 1:
u = 0,
u = 0.
Proof. a) We have
Lu = 2
u = 0
u = 0 on |x| = 1
u = 0 on |x| = 1.
The operator L is self-adjoint if:
Ω
(u Lv − v Lu) dx = 0.
Ω
(u Lv − v Lu) dx =
Ω
(u 2
v − v 2
u) dx
=
∂Ω
u
∂ v
∂n
ds
=0
−
Ω
∇u · ∇( v) dx −
∂Ω
v
∂ u
∂n
ds
=0
+
Ω
∇v · ∇( u) dx
= −
∂Ω
v
∂u
∂n
ds
=0
+
Ω
u v dx +
∂Ω
u
∂v
∂n
ds
=0
−
Ω
v u dx = 0.

b) Denote |x| = r and deﬁne the averages
S(r) = (4πr2
)−1
|x|=r
u(x) ds,
V (r) =
4
3
πr3
−1
|x|≤r
u(x) dx.
Show that
d
dr
S(r) =
r
3
V (r).
Hint: Rewrite S(r) as an integral over the unit sphere before diﬀerentiating; i.e.,
S(r) = (4π)−1
|x |=1
u(rx ) dx .
c) Use the result of (b) to show that if u is biharmonic, i.e. 2
u = 0, then
S(r) = u(0) +
r2
6
u(0).
Hint: Use the mean value theorem for u.
b) Let x = x/|x|. We have 54
S(r) =
1
4πr2
|x|=r
u(x) dSr =
1
4πr2
|x |=1
u(rx ) r2
dS1 =
1
4π |x |=1
u(rx ) dS1.
dS
dr
=
1
4π |x |=1
∂u
∂r
(rx ) dS1 =
1
4π |x |=1
∂u
∂n
(rx ) dS1 =
1
4πr2
|x|=r
∂u
∂n
(x) dSr
=
1
4πr2
|x|≤r
u dx.
where we have used Green’s identity in the last equality. Also
r
3
V (r) =
1
4πr2
|x|≤r
u dx.
c) Since u is biharmonic (i.e. u is harmonic), u has a mean value property. We
have
d
dr
S(r) =
r
3
V (r) =
r
3
4
3
πr3
−1
|x|≤r
u(x) dx =
r
3
u(0),
S(r) =
r2
6
u(0) + S(0) = u(0) +
r2
6
u(0).
54
Change of variables:
Surface integrals: x = rx in R3
:
|x|=r
u(x) dS =
|x |=1
u(rx ) r2
dS1.
Volume integrals: ξ = rξ in Rn
:
|ξ |<r
h(x + ξ ) dξ =
|ξ|<1
h(x + rξ) rn
dξ.

Problem (S’00, #7). Suppose that u = u(x) for x ∈ R3
is biharmonic;
i.e. that 2u ≡ ( u) = 0. Show that
(4πr2
)−1
|x|=r
u(x) ds(x) = u(0) + (r2
/6) u(0)
through the following steps:
a) Show that for any smooth f,
d
dr |x|≤r
f(x) dx =
|x|=r
f(x) ds(x).
b) Show that for any smooth f,
d
dr
(4πr2
)−1
|x|=r
f(x) ds(x) = (4πr2
)−1
|x|=r
n · ∇f(x, y) ds
in which n is the outward normal to the circle |x| = r.
c) Use step (b) to show that
d
dr
(4πr2
)−1
|x|=r
f(x) ds(x) = (4πr2
)−1
|x|≤r
f(x) dx.
d) Combine steps (a) and (c) to obtain the final result.
Proof. a) We can express the integral in Spherical Coordinates: 55
|x|≤R
f(x) dx =
R
0
2π
0
π
0
f(φ, θ, r) r2
sinφ dφ dθ dr.
d
dr |x|≤R
f(x) dx =
d
dr
R
0
2π
0
π
0
f(φ, θ, r) r2
sinφ dφ dθ dr = ???
=
2π
0
π
0
f(φ, θ, r) R2
sinφ dφ dθ
=
|x|=R
f(x) dS.
55
Differential Volume in spherical coordinates:
d3
ω = ω2
sin φ dφ dθ dω.
Differential Surface Area on sphere:
dS = ω2
sin φ dφ dθ.

b&c) We have
d
dr
1
4πr2
|x|=r
f(x) dS =
d
dr
1
4πr2
|x |=1
f(rx ) r2
dS1 =
1
4π
d
dr |x |=1
f(rx ) dS1
=
1
4π |x |=1
∂f
∂r
(rx ) dS1 =
1
4π |x |=1
∂f
∂n
(rx ) dS1
=
1
4πr2
|x|=r
∂f
∂n
(x) dS =
1
4πr2
|x|=r
∇f · n dS
=
1
4πr2
|x|≤r
f dx.
Green’s formula was used in the last equality.
Alternatively,
d
dr
1
4πr2
|x|=r
f(x) dS =
d
dr
1
4πr2
2π
0
π
0
f(φ, θ, r) r2
sinφ dφ dθ
=
d
dr
1
4π
2π
0
π
0
f(φ, θ, r) sinφ dφ dθ
=
1
4π
2π
0
π
0
∂f
∂r
(φ, θ, r) sin φ dφ dθ
=
1
4π
2π
0
π
0
∇f · n sin φ dφ dθ
=
1
4πr2
2π
0
π
0
∇f · n r2
sinφ dφ dθ
=
1
4πr2
|x|=r
∇f · n dS
=
1
4πr2
|x|=r
f dx.
d) Since f is biharmonic (i.e. f is harmonic), f has a mean value property. From
(c), we have 56
d
dr
1
4πr2
|x|=r
f(x) ds(x) =
1
4πr2
|x|≤r
f(x) dx =
r
3
1
4
3πr3
|x|≤r
f(x) dx
=
r
3
f(0).
1
4πr2
|x|=r
f(x) ds(x) =
r2
6
f(0) + f(0).
56
Note that part (a) was not used. We use exactly the same derivation as we did in S’95 #4.

Consider smooth solutions of u = k2u in dimension d = 2 with k > 0.
a) Show that u satisfies the following ‘mean value property’:
Mx (r) +
1
r
Mx(r) − k2
Mx(r) = 0,
in which Mx(r) is defined by
Mx(r) =
1
2π
2π
0
u(x + r cos θ, y + r sinθ) dθ
and the derivatives (denoted by ) are in r with x fixed.
b) For k = 1, this equation is the modified Bessel equation (of order 0)
f +
1
r
f − f = 0,
for which one solution (denoted as I0) is
I0(r) =
1
2π
2π
0
er sin θ
dθ.
Find an expression for Mx(r) in terms of I0.
Proof. a) Laplacian in polar coordinates written as:
u = urr +
1
r
ur +
1
r2
uθθ.
Thus, the equation may be written as
urr +
1
r
ur +
1
r2
uθθ = k2
u.
Mx(r) =
1
2π
2π
0
u dθ,
Mx(r) =
1
2π
2π
0
ur dθ,
Mx (r) =
1
2π
2π
0
urr dθ.
Mx (r) +
1
r
Mx(r) − k2
Mx(r) =
1
2π
2π
0
urr +
1
r
ur − k2
u dθ
= −
1
2πr2
2π
0
uθθ dθ = −
1
2πr2
uθ
2π
0
= 0.
b) Note that w = er sin θ
satisfies w = w, i.e.
w = wrr +
1
r
wr +
1
r2
wθθ
= sin2
θ er sin θ
+
1
r
sinθ er sin θ
+
1
r2
(−r sinθ er sin θ
+ r2
cos2
θ er sin θ
) = er sin θ
= w.
Thus,
Mx(r) = ey 1
2π
2π
0
er sin θ
dθ = ey
I0.

57
57
Check with someone about the last result.

17.8 Harmonic Extensions, Subharmonic Functions
Problem (S’94, #8). Suppose that Ω is a bounded region in R3 and that u = 1 on
∂Ω. If u = 0 in the exterior region R3
/Ω and u(x) → 0 as |x| → ∞, prove the
following:
a) u > 0 in R3/Ω;
b) if ρ(x) is a smooth function such that ρ(x) = 1 for |x| > R and ρ(x) = 0 near ∂Ω,
then for |x| > R,
u(x) = −
1
4π R3/Ω
( (ρu))(y)
|x − y|
dy.
c) lim|x|→∞ |x|u(x) exists and is non-negative.
Proof. a) Let Br(0) denote the closed ball {x : |x| ≥ r}.
Given ε > 0, we can ﬁnd r large enough that Ω ∈ BR1 (0) and maxx∈∂BR1
(0) |u(x)| < ε,
since |u(x)| → 0 as |x| → ∞.
Since u is harmonic in BR1 − Ω, it takes its maximum and minimum on the boundary.
Assume
min
x∈∂BR1
(0)
u(x) = −a < 0 (where |a| < ε).
We can ﬁnd an R2 such that maxx∈BR2
(0) |u(x)| < a
2 ; hence u takes a minimum inside
BR2 (0) − Ω, which is impossible; hence u ≥ 0.
Now let V = {x : u(x) = 0} and let α = minx∈V |x|. Since u cannot take a minimum
inside BR(0) (where R > α), it follows that u ≡ C and C = 0, but this contradicts
u = 1 on ∂Ω. Hence u > 0 in R3 − Ω.
b) For n = 3,
K(|x − y|) =
1
(2 − n)ωn
|x − y|2−n
= −
1
4π
1
|x − y|
.
Since ρ(x) = 1 for |x| > R, then for x /∈ BR, we have (ρu) = u = 0. Thus,
−
1
4π R3/Ω
( (ρu))(y)
|x − y|
dy
= −
1
4π BR/Ω
( (ρu))(y)
|x − y|
dy
=
1
4π BR/Ω
∇y
1
|x − y|
· ∇y(ρu) dy −
1
4π ∂(BR/Ω)
∂
∂n
ρu
1
|x − y|
dSy
= −
1
4π BR/Ω
1
|x − y|
ρu dy +
1
4π ∂(BR/Ω)
∂
∂n
1
|x − y|
ρu dSy −
1
4π ∂(BR/Ω)
∂
∂n
ρu
1
|x − y|
dSy
= ??? = u(x) −
1
4πR2
∂B
u dSy
→0, as R→∞
−
1
4πR ∂B
∂u
∂n
dSy
→0, as R→∞
= u(x).
c) See the next problem.

Ralston Hw. a) Suppose that u is a smooth function on R3
and u = 0 for |x| > R.
If limx→∞ u(x) = 0, show that you can write u as a convolution of u with the − 1
4π|x|
and prove that limx→∞ |x|u(x) = 0 exists.
b) The “conductor potential” for Ω ⊂ R3 is the solution to the Dirichlet problem v =
0. The limit in part (a) is called the “capacity” of Ω. Show that if Ω1 ⊂ Ω2, then the
capacity of Ω2 is greater or equal the capacity of Ω1.
Proof. a) If we define
v(x) = −
1
4π R3
u(y)
|x − y|
dy,
then (u − v) = 0 in all R3
, and, since v(x) → 0 as |x| → ∞, we have lim|x|→∞(u(x)−
v(x)) = 0. Thus, u − v must be bounded, and Liouville’s theorem implies that it is
identically zero. Since we now have
|x|u(x) = −
1
4π R3
|x| u(y)
|x − y|
dy,
and |x|/|x − y| converges uniformly to 1 on {|y| ≤ R}, it follows that
lim
|x|→∞
|x|u(x) = −
1
4π R3
u(y) dy.
b) Note that part (a) implies that the limit lim|x|→∞ |x|v(x) exists, because we can
apply (a) to u(x) = φ(x)v(x), where φ is smooth and vanishes on Ω, but φ(x) = 1 for
|x| > R.
Let v1 be the conductor potential for Ω1 and v2 for Ω2. Since vi → ∞ as |x| → ∞ and
vi = 1 on ∂Ωi, the max principle says that 1 > vi(x) > 0 for x ∈ R3
− Ωi. Consider
v2 − v1. Since Ω1 ⊂ Ω2, this is defined in R3 − Ω2, positive on ∂Ω2, and has limit 0 as
|x| → ∞. Thus, it must be positive in R3
− Ω2. Thus, lim|x|→∞ |x|(v2 − v1) ≥ 0.
Problem (F’95, #4). 58
Let Ω be a simply connected open domain in R2
and u = u(x, y) be subharmonic there, i.e. u ≥ 0 in Ω. Prove that if
DR = {(x, y) : (x − x0)2
+ (y − y0)2
≤ R2
} ⊂ Ω
then
u(x0, y0) ≤
1
2π
2π
0
u(x0 + R cos θ, y0 + R sinθ) dθ.
Proof. Let
M(x0, R) =
1
2π
2π
0
u(x0 + R cos θ, y0 + R sin θ) dθ,
w(r, θ) = u(x0 + R cos θ, y0 + R sin θ).
Differentiate M(x0, R) with respect to R:
d
dr
M(x0, R) =
1
2πR
2π
0
wr(R, θ)Rdθ,
58
See McOwen, Sec.4.3, p.131, #1.

59
59
See ChiuYen’s solutions and Sung Ha’s solutions (in two places). Nick’s solutions, as started above,
have a very simplistic approach.

Ralston Hw (Maximum Principle).
Suppose that u ∈ C(Ω) satisfies the mean value property in the connected open set Ω.
a) Show that u satisfies the maximum principle in Ω, i.e.
either u is constant or u(x) < supΩ u for all x ∈ Ω.
b) Show that, if v is a continuous function on a closed ball Br(ξ) ⊂ Ω and has the
mean value property in Br(ξ), then u = v on ∂Br(ξ) implies u = v in Br(ξ). Does this
imply that u is harmonic in Ω?
Proof. a) If u(x) is not less than supΩ u for all x ∈ Ω, then the set
K = {x ∈ Ω : u(x) = sup
Ω
u}
is nonempty. This set is closed because u is continuous. We will show it is also open.
This implies that K = Ω because Ω is connected. Thus u is constant on Ω.
Let x0 ∈ K. Since Ω is open, ∃δ > 0, s.t. Bδ(x0) = {x ∈ Rn : |x − x0| ≤ δ} ⊂ Ω. Let
supΩ u = M. By the mean value property, for 0 ≤ r ≤ δ
M = u(x0) =
1
A(Sn−1) |ξ|=1
u(x0 + rξ)dSξ, and 0 =
1
A(Sn−1) |ξ|=1
(M − u(x0 + rξ))dSξ.
Sinse M −u(x0 +rξ) is a continuous nonnegative function on ξ, this implies M −u(x0 +
rξ) = 0 for all ξ ∈ Sn−1. Thus u = 0 on Bδ(x0).
b) Since u − v has the mean value property in the open interior of Br(ξ), by part
a) it satisfies the maximum principle. Since it is continuous on Br(ξ), its supremum
over the interior of Br(ξ) is its maximum on Br(ξ), and this maximum is assumed at a
point x0 in Br(ξ). If x0 in the interior of Br(ξ), then u −v is constant ant the constant
must be zero, since this is the value of u −v on the boundary. If x0 is on the boundary,
then u − v must be nonpositive in the interior of Br(ξ).
Applying the same argument to v − u, one finds that it is either identically zero or
nonpositive in the interior of Br(ξ). Thus, u − v ≡ 0 on Br(ξ).
Yes, it does follow that u harmonic in Ω. Take v in the preceding to be the harmonic
function in the interior of Br(ξ) which agrees with u on the boundary. Since u = v on
Br(ξ), u is harmonic in the interior of Br(ξ). Since Ω is open we can do this for every
ξ ∈ Ω. Thus u is harmonic in Ω.

Ralston Hw. Assume Ω is a bounded open set in Rn
and the Green’s function, G(x, y),
for Ω exists. Use the strong maximum principle, i.e. either u(x) < supΩ u for all x ∈ Ω,
or u is constant, to prove that G(x, y) < 0 for x, y ∈ Ω, x = y.
Proof. G(x, y) = K(x, y) + ω(x, y). For each x ∈ Ω, f(y) = ω(x, y) is continuous on Ω,
thus, bounded. So |ω(x, y)| ≤ Mx for all y ∈ Ω. K(x − y) → −∞ as y → x. Thus,
given Mx, there is δ > 0, such that K(x − y) < −Mx when |x − y| = r and 0 < r ≤ δ.
So for 0 < r ≤ δ the Green’s function with x fixed satisfies, G(x, y) is harmonic on
Ω − Br(x), and G(x, y) ≤ 0 on the boundary of Ω − Br(x). Since we can choose r as
small as we wish, we get G(x, y) < 0 for y ∈ Ω − {x}.
Problem (W’03, #6). Assume that u is a harmonic function in the half ball
D = {(x, y, z) : x2
+y2
+z2
< 1, z ≥ 0} which is continuously differentiable, and satis-
fies u(x, y, 0) = 0. Show that u can be extended to be a harmonic function in the whole
ball. If you propose and explicit extension for u, explain why the extension is harmonic.
Proof. We can extend u to all of n-space by defining
u(x , xn) = −u(x , −xn)
for xn < 0. Define
ω(x) =
1
aωn |y|=1
a2 − |x|2
|x − y|n
v(y)dSy
ω(x) is continuous on a closed ball B, harmonic in B.
Poisson kernel is symmetric in y at xn = 0. ⇒ ω(x) = 0, (xn = 0).
ω is harmonic for x ∈ B, xn ≥ 0,with the same boundary values ω = u.
ω is harmonic ⇒ u can be extended to a harmonic function on the interior of B.
Ralston Hw. Show that a bounded solution to the Dirichlet problem in a half
space is unique. (Note that one can show that a bounded solution exists for any
given bounded continuous Dirichlet data by using the Poisson kernel for the half space.)
Proof. We have to show that a function, u, which is harmonic in the half-space, con-
tinuous, equal to 0 when xn = 0, and bounded, must be identically 0. We can extend
u to all of n-space by defining
u(x , xn) = −u(x , −xn)
for xn < 0. This extends u to a bounded harmonic function on all of n-space (by the
problem above). Liouville’s theorem says u must be constant, and since u(x , 0) = 0,
the constant is 0. So the original u must be identically 0.
Ralston Hw. Suppose u is harmonic on the ball minus the origin, B0 = {x ∈ R3
:
0 < |x| < a}. Show that u(x) can be extended to a harmonic function on the ball
B = {|x| < a} iff lim|x|→0 |x|u(x) = 0.
Proof. The condition lim|x|→0 |x|u(x) = 0 is necessary, because harmonic functions are
continuous.
To prove the converse, let v be the function which is continuous on {|x| ≤ a/2},
harmonic on {|x| < a/2}, and equals u on {|x| = a/2}. One can construct v using the
Poisson kernel. Since v is continuous, it is bounded, and we can assume that |v| ≤ M.
Since lim|x|→0 |x|u(x) = 0, given > 0, we can choose δ, 0 < δ < a/2 such that
− < |x|u(x) < when |x| < δ. Note that u, v − 2 /|x|, and v + 2 /|x| are harmonic

on {0 < |x| < a/2}. Choose b, 0 < b < min( , a/2), so that /b > M. Then on both
{|x| = a/2} and {|x| = b} we have v − 2 /|x| < u(x) < v + 2 /|x|. Thus, by
max principle these inequalities hold on {b ≤ |x| ≤ a/2}. Pick x with 0 < |x| ≤ a/2.
u(x) = v(x). v is the extension of u on {|x| < a/2}, and u is extended on {|x| < a}.

18 Problems: Heat Equation
McOwen 5.2 #7(a). Consider
⎧
⎪⎨
⎪⎩
ut = uxx for x > 0, t > 0
u(x, 0) = g(x) for x > 0
u(0, t) = 0 for t > 0,
where g is continuous and bounded for x ≥ 0 and g(0) = 0.
Find a formula for the solution u(x, t).
Proof. Extend g to be an odd function on all of R:
˜g(x) =
g(x), x ≥ 0
−g(−x), x < 0.
˜ut = ˜uxx for x ∈ R, t > 0
˜u(x, 0) = ˜g(x) for x ∈ R.
The solution is given by: 60
˜u(x, t) =
R
K(x, y, t)g(y) dy =
1
√
4πt
∞
−∞
e−
(x−y)2
4t ˜g(y) dy
=
1
√
4πt
∞
0
e−(x−y)2
4t ˜g(y) dy +
0
−∞
e−(x−y)2
4t ˜g(y) dy
=
1
√
4πt
∞
0
e−
(x−y)2
4t g(y) dy −
∞
0
e−
(x+y)2
4t g(y) dy
=
1
√
4πt
∞
0
e
−x2+2xy−y2
4t − e
−x2−2xy−y2
4t g(y) dy
=
1
√
4πt
∞
0
e−
(x2+y2)
4t e
xy
2t − e−xy
2t g(y) dy.
u(x, t) =
1
√
4πt
∞
0
e−
(x2+y2)
4t 2 sinh
xy
2t
g(y) dy.
Since sinh(0) = 0, we can verify that u(0, t) = 0.
60
In calculations, we use:
0
−∞
ey
dy =
∞
0
e−y
dy, and g(−y) = −g(y).

McOwen 5.2 #7(b). Consider
⎧
⎪⎨
⎪⎩
ut = uxx for x > 0, t > 0
u(x, 0) = g(x) for x > 0
ux(0, t) = 0 for t > 0,
where g is continuous and bounded for x ≥ 0.
Find a formula for the solution u(x, t).
Proof. Extend g to be an even function 61
on all of R:
˜g(x) =
g(x), x ≥ 0
g(−x), x < 0.
˜ut = ˜uxx for x ∈ R, t > 0
˜u(x, 0) = ˜g(x) for x ∈ R.
The solution is given by: 62
˜u(x, t) =
R
K(x, y, t)g(y) dy =
1
√
4πt
∞
−∞
e−
(x−y)2
4t ˜g(y) dy
=
1
√
4πt
∞
0
e−(x−y)2
4t ˜g(y) dy +
0
−∞
e−(x−y)2
4t ˜g(y) dy
=
1
√
4πt
∞
0
e−
(x−y)2
4t g(y) dy +
∞
0
e−
(x+y)2
4t g(y) dy
=
1
√
4πt
∞
0
e
−x2+2xy−y2
4t + e
−x2−2xy−y2
4t g(y) dy
=
1
√
4πt
∞
0
e−
(x2+y2)
4t e
xy
2t + e−xy
2t g(y) dy.
u(x, t) =
1
√
4πt
∞
0
e−
(x2+y2)
4t 2 cosh
xy
2t
g(y) dy.
To check that the boundary condition holds, we perform the calculation:
ux(x, t) =
1
√
4πt
∞
0
d
dx
e−
(x2 +y2)
4t 2 cosh
xy
2t
g(y) dy
=
1
√
4πt
∞
0
−
2x
4t
e−(x2+y2)
4t 2 cosh
xy
2t
+ e−(x2+y2)
4t 2
y
2t
sinh
xy
2t
g(y) dy,
ux(0, t) =
1
√
4πt
∞
0
0 · e−y2
4t 2 cosh0 + e−y2
4t 2
y
2t
sinh 0 g(y) dy = 0.
61
Even extensions are always continuous. Not true for odd extensions. g odd is continuous if g(0) =
0.
62
In calculations, we use:
0
−∞
ey
dy =
∞
0
e−y
dy, and g(−y) = g(y).

The initial value problem for the heat equation on the whole real line is
ft = fxx t ≥ 0
f(t = 0, x) = f0(x)
with f0 smooth and bounded.
a) Write down the Green’s function G(x, y, t) for this initial value problem.
b) Write the solution f(x, t) as an integral involving G and f0.
c) Show that the maximum values of |f(x, t)| and |fx(x, t)| are non-increasing
as t increases, i.e.
sup
x
|f(x, t)| ≤ sup
x
|f0(x)| sup
x
|fx(x, t)| ≤ sup
x
|f0x(x)|.
When are these inequalities actually equalities?
Proof. a) The fundamental solution
K(x, y, t) =
1
√
4πt
e−
|x−y|2
4t .
The Green’s function is: 63
G(x, t; y, s) =
1
(2π)n
π
k(t − s)
n
2
e
−
(x−y)2
4k(t−s) .
b) The solution to the one-dimensional heat equation is
u(x, t) =
R
K(x, y, t) f0(y) dy =
1
√
4πt R
e−
|x−y|2
4t f0(y) dy.
c) We have
sup
x
|u(x, t)| =
1
√
4πt R
e−
(x−y)2
4t f0(y) dy ≤
1
√
4πt R
e−
(x−y)2
4t f0(y) dy
=
1
√
4πt R
e−y2
4t f0(x − y) dy
≤ sup
x
|f0(x)|
1
√
4πt R
e−y2
4t dy z =
y
√
4t
, dz =
dy
√
4t
≤ sup
x
|f0(x)|
1
√
4πt R
e−z2 √
4t dz
= sup
x
|f0(x)|
1
√
π R
e−z2
dz
=
√
π
= sup
x
|f0(x)|.
63
The Green’s function for the heat equation on an inﬁnite domain; derived in R. Haberman using
the Fourier transform.

ux(x, t) =
1
√
4πt R
−
2(x − y)
4t
e−
(x−y)2
4t f0(y) dy =
1
√
4πt R
−
d
dy
e−
(x−y)2
4t f0(y) dy
=
1
√
4πt
− e−
(x−y)2
4t f0(y)
∞
−∞
= 0
+
1
√
4πt R
e−
(x−y)2
4t f0y(y) dy,
sup
x
|u(x, t)| ≤
1
√
4πt
sup
x
|f0x(x)|
R
e−(x−y)2
4t dy =
1
√
4πt
sup
x
|f0x(x)|
R
e−z2 √
4t dz
= sup
x
|f0x(x)|.
These inequalities are equalities when f0(x) and f0x(x) are constants, respectively.

Problem (S’01, #5). a) Show that the solution of the heat equation
ut = uxx, −∞ < x < ∞
with square-integrable initial data u(x, 0) = f(x), decays in time, and there is a constant
α independent of f and t such that for all t > 0
max
x
|ux(x, t)| ≤ αt−3
4
x
|f(x)|2
dx
1
2
.
b) Consider the solution ρ of the transport equation ρt +uρx = 0 with square-integrable
initial data ρ(x, 0) = ρ0(x) and the velocity u from part (a). Show that ρ(x, t) remains
square-integrable for all ﬁnite time
R
|ρ(x, t)|2
dx ≤ eCt
1
4
R
|ρ0(x)|2
dx,
where C does not depend on ρ0.
Proof. a) The solution to the one-dimensional homogeneous heat equation is
u(x, t) =
1
√
4πt R
e−
(x−y)2
4t f(y) dy.
Take the derivative with respect to x, we get 64
ux(x, t) =
1
√
4πt R
−
2(x − y)
4t
e−
(x−y)2
4t f(y) dy = −
1
4t
3
2
√
π R
(x − y)e−
(x−y)2
4t f(y) dy.
|ux(x, t)| ≤
1
4t
3
2
√
π R
(x − y)e−(x−y)2
4t f(y) dy (Cauchy-Schwarz)
≤
1
4t
3
2
√
π R
(x − y)2
e−
(x−y)2
2t dy
1
2
||f||L2(R) z =
x − y
√
2t
, dz = −
dy
√
2t
=
1
4t
3
2
√
π R
− z2
(2t)
3
2 e−z2
dz
1
2
||f||L2(R)
=
(2t)
3
4
4t
3
2
√
π R
z2
e−z2
dz
M<∞
1
2
||f||L2(R)
= Ct−3
4 M
1
2 ||f||L2(R) = αt−3
4 ||f||L2(R).
b) Note:
max
x
|u| = max
x
1
√
4πt R
e−(x−y)2
4t f(y) dy ≤
1
√
4πt R
e−(x−y)2
2t dy
1
2
||f||L2(R)
≤
1
√
4πt R
− e−z2 √
2t dz
1
2
||f||L2(R) z =
x − y
√
2t
, dz = −
dy
√
2t
=
(2t)
1
4
2π
1
2 t
1
2 R
e−z2
dz
=
√
π
1
2
||f||L2(R) = Ct−1
4 ||f||L2(R).
65
64
Cauchy-Schwarz: |(u, v)| ≤ ||u||||v|| in any norm, for example |uv|dx ≤ ( u2
dx)
1
2 ( v2
dx)
1
2
65
See Yana’s and Alan’s solutions.

Let u(x, t) be a bounded solution to the Cauchy problem for the heat equation
ut = a2
uxx, t > 0, x ∈ R, a > 0,
u(x, 0) = ϕ(x).
Here ϕ(x) ∈ C(R) satisﬁes
lim
x→+∞
ϕ(x) = b, lim
x→−∞
ϕ(x) = c.
Compute the limit of u(x, t) as t → +∞, x ∈ R. Justify your argument carefully.
Proof. For a = 1, the solution to the one-dimensional homogeneous heat equation is
u(x, t) =
1
√
4πt R
e−(x−y)2
4t ϕ(y) dy.
We want to transform the equation to vt = vxx. Make a change of variables: x = ay.
u(x, t) = u(x(y), t) = u(ay, t) = v(y, t). Then,
vy = uxxy = aux,
vyy = auxxxy = a2
uxx,
v(y, 0) = u(ay, 0) = ϕ(ay).
Thus, the new problem is:
vt = vyy, t > 0, y ∈ R,
v(y, 0) = ϕ(ay).
v(y, t) =
1
√
4πt R
e−
(y−z)2
4t ϕ(az) dz.
Since ϕ is continuous, and limx→+∞ ϕ(x) = b, limx→−∞ ϕ(x) = c, we have
|ϕ(x)| < M, ∀x ∈ R. Thus,
|v(y, t)| ≤
M
√
4πt R
e−z2
4t dz s =
z
√
4t
, ds =
dz
√
4t
=
M
√
4πt R
e−s2 √
4t ds =
M
√
π R
e−s2
ds
√
π
= M.
Integral in converges uniformly ⇒ lim = lim. For ψ = ϕ(a·):
v(y, t) =
1
√
4πt
∞
−∞
e−
(y−z)2
4t ψ(z) dz =
1
√
4πt
∞
−∞
e−z2
4t ψ(y − z) dz
=
1
√
4πt
∞
−∞
e−s2
ψ(y − s
√
4t)
√
4t ds
=
1
√
π
∞
−∞
e−s2
ψ(y − s
√
4t) ds.

lim
t→+∞
v(y, t) =
1
√
π
∞
0
e−s2
lim
t→+∞
ψ(y − s
√
4t) ds +
1
√
π
0
−∞
e−s2
lim
t→+∞
ψ(y − s
√
4t) ds
=
1
√
π
∞
0
e−s2
c ds +
1
√
π
0
−∞
e−s2
b ds = c
1
√
π
√
π
2
+ b
1
√
π
√
π
2
=
c + b
2
.

Problem. Consider
ut = kuxx + Q, 0 < x < 1
u(0, t) = 0,
u(1, t) = 1.
What is the steady state temperature?
Proof. Set ut = 0, and integrate with respect to x twice:
kuxx + Q = 0,
uxx = −
Q
k
,
ux = −
Q
k
x + a,
u = −
Q
k
x2
2
+ ax + b.
Boundary conditions give
u(x) = −
Q
2k
x2
+ 1 +
Q
2k
x.

18.1 Heat Equation with Lower Order Terms
McOwen 5.2 #11. Find a formula for the solution of
ut = u − cu in Rn
× (0, ∞)
u(x, 0) = g(x) on Rn
.
(18.1)
Show that such solutions, with initial data g ∈ L2(Rn), are unique, even when c is
negative.
Proof. McOwen. Consider v(x, t) = ectu(x, t). The transformed problem is
vt = v in Rn × (0, ∞)
v(x, 0) = g(x) on Rn.
(18.2)
Since g is continuous and bounded in Rn, we have
v(x, t) =
Rn
K(x, y, t) g(y) dy =
1
(4πt)
n
2 Rn
e−
|x−y|2
4t g(y) dy,
u(x, t) = e−ct
v(x, t) =
1
(4πt)
n
2 Rn
e−
|x−y|2
4t
−ct
g(y) dy.
u(x, t) is a bounded solution since v(x, t) is.
To prove uniqueness, assume there is another solution v of (18.2). w = v − v satisﬁes
wt = w in Rn
× (0, ∞)
w(x, 0) = 0 on Rn.
(18.3)
Since bounded solutions of (18.3) are unique, and since w is a nontrivial solution, w is
unbounded. Thus, v is unbounded, and therefore, the bounded solution v is unique.

18.1.1 Heat Equation Energy Estimates
Problem (F’94, #3). Let u(x, y, t) be a twice continuously diﬀerential solution of
ut = u − u3
in Ω ⊂ R2
, t ≥ 0
u(x, y, 0) = 0 in Ω
u(x, y, t) = 0 in ∂Ω, t ≥ 0.
Prove that u(x, y, t) ≡ 0 in Ω × [0, T].
Proof. Multiply the equation by u and integrate:
uut = u u − u4
,
Ω
uut dx =
Ω
u u dx −
Ω
u4
dx,
1
2
d
dt Ω
u2
dx =
∂Ω
u
∂u
∂n
ds
=0
−
Ω
|∇u|2
dx −
Ω
u4
dx,
1
2
d
dt
||u||2
2 = −
Ω
|∇u|2
dx −
Ω
u4
dx ≤ 0.
Thus,
||u(x, y, t)||2 ≤ ||u(x, y, 0)||2 = 0.
Hence, ||u(x, y, t)||2 = 0, and u ≡ 0.

Problem (F’98, #5). Consider the heat equation
ut − u = 0
in a two dimensional region Ω. Define the mass M as
M(t) =
Ω
u(x, t) dx.
a) For a fixed domain Ω, show M is a constant in time if the boundary conditions are
∂u/∂n = 0.
b) Suppose that Ω = Ω(t) is evolving in time, with a boundary that moves at velocity
v, which may vary along the boundary. Find a modified boundary condition (in terms
of local quantities only) for u, so that M is constant.
Hint: You may use the fact that
d
dt Ω(t)
f(x, t) dx =
Ω(t)
ft(x, t) dx +
∂Ω(t)
n · v f(x, t) dl,
in which n is a unit normal vector to the boundary ∂Ω.
Proof. a) We have
ut − u = 0, on Ω
∂u
∂n = 0, on ∂Ω.
We want to show that d
dt M(t) = 0. We have 66
d
dt
M(t) =
d
dt Ω
u(x, t) dx =
Ω
ut dx =
Ω
u dx =
∂Ω
∂u
∂n
ds = 0.
b) We need d
dt M(t) = 0.
0 =
d
dt
M(t) =
d
dt Ω(t)
u(x, t) dx =
Ω(t)
ut dx +
∂Ω(t)
n · v u ds
=
Ω(t)
u dx +
∂Ω(t)
n · v u ds =
∂Ω(t)
∂u
∂n
ds +
∂Ω(t)
n · v u ds
=
∂Ω(t)
∇u · n ds +
∂Ω(t)
n · v u ds =
∂Ω(t)
n · (∇u + vu) ds.
Thus, we need:
n · (∇u + vu) ds = 0, on ∂Ω.
66
The last equality below is obtained from the Green’s formula:
Ω
u dx =
Ω
∂u
∂n
ds.

Problem (S’95, #3). Write down an explicit formula for a function u(x, t) solving
ut + b · ∇u + cu = u in Rn
× (0, ∞)
u(x, 0) = f(x) on Rn.
(18.4)
where b ∈ Rn
and c ∈ R are constants.
Hint: First transform this to the heat equation by a linear change of the dependent
and independent variables. Then solve the heat equation using the fundamental solution.
Proof. Consider
• u(x, t) = eα·x+βt
v(x, t).
ut = βeα·x+βt
v + eα·x+βt
vt = (vt + βv)eα·x+βt
,
∇u = αeα·x+βt
v + eα·x+βt
∇v = (αv + ∇v)eα·x+βt
,
∇ · (∇u) = ∇ · (αv + ∇v)eα·x+βt
= (α · ∇v + v)eα·x+βt
+ (|α|2
v + α · ∇v)eα·x+βt
= v + 2α · ∇v + |α|2
v)eα·x+βt
.
Plugging this into (18.4), we obtain
vt + βv + b · (αv + ∇v) + cv = v + 2α · ∇v + |α|2
v,
vt + b − 2α · ∇v + β + b · α + c − |α|2
v = v.
In order to get homogeneous heat equation, we set
α =
b
2
, β = −
|b|2
4
− c,
which gives
vt = v in Rn
× (0, ∞)
v(x, 0) = e− b
2
·x
f(x) on Rn.
The above PDE has the following solution:
v(x, t) =
1
(4πt)
n
2 Rn
e−
|x−y|2
4t e− b
2
·y
f(y) dy.
Thus,
u(x, t) = e
b
2
·x−( |b|2
4
+c)t
v(x, t) =
1
(4πt)
n
2
e
b
2
·x−( |b|2
4
+c)t
Rn
e−|x−y|2
4t e− b
2
·y
f(y) dy.

Problem (F’01, #7). Consider the parabolic problem
ut = uxx + c(x)u (18.5)
for −∞ < x < ∞, in which
c(x) = 0 for |x| > 1,
c(x) = 1 for |x| < 1.
Find solutions of the form u(x, t) = eλt
v(x) in which
∞
−∞ |u|2
dx < ∞.
Hint: Look for v to have the form
v(x) = ae−k|x|
for |x| > 1,
v(x) = b coslx for |x| < 1,
for some a, b, k, l.
Proof. Plug u(x, t) = eλtv(x) into (18.5) to get:
λeλt
v(x) = eλt
v (x) + ceλt
v(x),
λv(x) = v (x) + cv(x),
v (x) − λv(x) + cv(x) = 0.
• For |x| > 1, c = 0. We look for solutions of the form v(x) = ae−k|x|
.
v (x) − λv(x) = 0,
ak2
e−k|x|
− aλe−k|x|
= 0,
k2
− λ = 0,
k2
= λ,
k = ±
√
λ.
Thus, v(x) = c1e−
√
λx + c2e
√
λx. Since we want
∞
−∞ |u|2 dx < ∞:
u(x, t) = aeλt
e−
√
λx
.
• For |x| < 1, c = 1. We look for solutions of the form v(x) = b coslx.
v (x) − λv(x) + v(x) = 0,
−bl2
cos lx + (1 − λ)b coslx = 0,
−l2
+ (1 − λ) = 0,
l2
= 1 − λ,
l = ±
√
1 − λ.
Thus, (since cos(−x) = cos x)
u(x, t) = beλt
cos (1 − λ)x.
• We want v(x) to be continuous on R, and at x = ±1, in particular. Thus,
ae−
√
λ
= b cos (1 − λ),
a = be
√
λ
cos (1 − λ).
• Also, v(x) is symmetric:
∞
−∞
|u|2
dx = 2
∞
0
|u|2
dx = 2
1
0
|u|2
dx +
∞
1
|u|2
dx < ∞.

Problem (F’03, #3). ❶ The function
h(X, T) = (4πT)−1
2 e−X2
4T
satisfies (you do not have to show this)
hT = hXX.
Using this result, verify that for any smooth function U
u(x, t) = e
1
3
t3−xt
∞
−∞
U(ξ) h(x − t2
− ξ, t) dξ
satisfies
ut + xu = uxx.
❷ Given that U(x) is bounded and continuous everywhere on −∞ ≤ x ≤ ∞, establish
that
lim
t→0
∞
−∞
U(ξ) h(x − ξ, t) dξ = U(x)
❸ and show that u(x, t) → U(x) as t → 0. (You may use the fact that
∞
0 e−ξ2
dξ =
1
2
√
π.)
Proof. We change the notation: h → K, U → g, ξ → y. We have
K(X, T) =
1
√
4πT
e−X2
4T
❶ We want to verify that
u(x, t) = e
1
3
t3−xt
∞
−∞
K(x − y − t2
, t) g(y) dy.
satisfies
ut + xu = uxx.
We have
ut =
∞
−∞
d
dt
e
1
3
t3−xt
K(x − y − t2
, t) g(y) dy
=
∞
−∞
(t2
− x) e
1
3
t3−xt
K + e
1
3
t3−xt
KX · (−2t) + KT g(y) dy,
xu =
∞
−∞
x e
1
3
t3−xt
K(x − y − t2
, t) g(y) dy,
ux =
∞
−∞
d
dx
e
1
3
t3−xt
K(x − y − t2
, t) g(y) dy
=
∞
−∞
− t e
1
3
t3−xt
K + e
1
3
t3−xt
KX g(y) dy,
uxx =
∞
−∞
d
dx
− t e
1
3
t3−xt
K + e
1
3
t3−xt
KX g(y) dy
=
∞
−∞
t2
e
1
3
t3−xt
K − t e
1
3
t3−xt
KX − t e
1
3
t3−xt
KX + e
1
3
t3−xt
KXX g(y) dy.

Plugging these into , most of the terms cancel out. The remaining two terms cancel
because KT = KXX.
❷ Given that g(x) is bounded and continuous on −∞ ≤ x ≤ ∞, we establish that 67
lim
t→0
∞
−∞
K(x − y, t) g(y) dy = g(x).
Fix x0 ∈ Rn, ε > 0. Choose δ > 0 such that
|g(y) − g(x0)| < ε if |y − x0| < δ, y ∈ Rn
.
Then if |x − x0| < δ
2 , we have: ( R K(x, t) dx = 1)
R
K(x − y, t) g(y) dy − g(x0) ≤
R
K(x − y, t) [g(y) − g(x0)] dy
≤
Bδ(x0)
K(x − y, t) g(y) − g(x0) dy
≤ ε R K(x−y,t) dy = ε
+
R−Bδ(x0)
K(x − y, t) g(y) − g(x0) dy
Furthermore, if |x − x0| ≤ δ
2 and |y − x0| ≥ δ, then
|y − x0| ≤ |y − x| +
δ
2
≤ |y − x| +
1
2
|y − x0|.
Thus, |y − x| ≥ 1
2 |y − x0|. Consequently,
= ε + 2||g||L∞
R−Bδ(x0)
K(x − y, t) dy
≤ ε +
C
√
t R−Bδ(x0)
e−
|x−y|2
4t dy
≤ ε +
C
√
t R−Bδ(x0)
e−
|y−x0|2
16t dy
= ε +
C
√
t
∞
δ
e− r2
16t r dr → ε + 0 as t → 0+
.
Hence, if |x − x0| < δ
2 and t > 0 is small enough, |u(x, t) − g(x0)| < 2ε.
67
Evans, p. 47, Theorem 1 (c).

Problem (S’93, #4). The temperature T(x, t) in a stationary medium, x ≥ 0, is
governed by the heat conduction equation
∂T
∂t
=
∂2T
∂x2
. (18.6)
Making the change of variable (x, t) → (u, t), where u = x/2
√
t, show that
4t
∂T
∂t
=
∂2T
∂u2
+ 2u
∂T
∂u
. (18.7)
Solutions of (18.7) that depend on u alone are called similarity solutions. 68
Proof. We change notation: the change of variables is (x, t) → (u, τ), where t = τ.
After the change of variables, we have T = T(u(x, t), τ(t)).
u =
x
2
√
t
⇒ ut = −
x
4t
3
2
, ux =
1
2
√
t
, uxx = 0,
τ = t ⇒ τt = 1, τx = 0.
∂T
∂t
=
∂T
∂u
∂u
∂t
+
∂T
∂τ
,
∂T
∂x
=
∂T
∂u
∂u
∂x
,
∂2
T
∂x2
=
∂
∂x
∂T
∂x
=
∂
∂x
∂T
∂u
∂u
∂x
=
∂2
T
∂u2
∂u
∂x
∂u
∂x
+
∂T
∂u
∂2
u
∂x2
=0
=
∂2
T
∂u2
∂u
∂x
2
.
Thus, (18.6) gives:
∂T
∂u
∂u
∂t
+
∂T
∂τ
=
∂2T
∂u2
∂u
∂x
2
,
∂T
∂u
−
x
4t
3
2
+
∂T
∂τ
=
∂2
T
∂u2
1
2
√
t
2
,
∂T
∂τ
=
1
4t
∂2T
∂u2
+
x
4t
3
2
∂T
∂u
,
4t
∂T
∂τ
=
∂2T
∂u2
+
x
√
t
∂T
∂u
,
4t
∂T
∂τ
=
∂2
T
∂u2
+ 2u
∂T
∂u
.
68
This is only the part of the qual problem.

19 Contraction Mapping and Uniqueness - Wave
Recall that the solution to
utt − c2uxx = f(x, t),
u(x, 0) = g(x), ut(x, 0) = h(x),
(19.1)
is given by adding together d’Alembert’s formula and Duhamel’s principle:
u(x, t) =
1
2
(g(x + ct) + g(x − ct)) +
1
2c
x+ct
x−ct
h(ξ) dξ +
1
2c
t
0
x+c(t−s)
x−c(t−s)
f(ξ, s) dξ ds.
Problem (W’02, #8). a) Find an explicit solution of the following Cauchy problem
∂2u
∂t2 − ∂2u
∂x2 = f(t, x),
u(0, x) = 0, ∂u
∂x (0, x) = 0.
(19.2)
b) Use part (a) to prove the uniqueness of the solution of the Cauchy problem
∂2u
∂t2 − ∂2u
∂x2 + q(t, x)u = 0,
u(0, x) = 0, ∂u
∂x (0, x) = 0.
(19.3)
Here f(t, x) and q(t, x) are continuous functions.
Proof. a) It was probably meant to give the ut initially. We rewrite (19.2) as
utt − uxx = f(x, t),
u(x, 0) = 0, ut(x, 0) = 0.
(19.4)
Duhamel’s principle, with c = 1, gives the solution to (19.4):
u(x, t) =
1
2c
t
0
x+c(t−s)
x−c(t−s)
f(ξ, s) dξ ds =
1
2
t
0
x+(t−s)
x−(t−s)
f(ξ, s) dξ ds.
b) We use the Contraction Mapping Principle to prove uniqueness.
Define the operator
T(u) =
1
2
t
0
x+(t−s)
x−(t−s)
−q(ξ, s) u(ξ, s) dξ ds.
on the Banach space C2,2, || · ||∞.
We will show |Tun − Tun+1| < α||un − un+1|| where α < 1. Then {un}∞
n=1:
un+1 = T(un) converges to a unique fixed point which is the unique solution of PDE.
|Tun − Tun+1| =
1
2
t
0
x+(t−s)
x−(t−s)
−q(ξ, s) un(ξ, s) − un+1(ξ, s) dξ ds
≤
1
2
t
0
||q||∞||un − un+1||∞ 2(t − s) ds
≤ t2
||q||∞||un − un+1||∞ ≤ α||un − un+1||∞, for small t.
Thus, T is a contraction ⇒ ∃ a unique fixed point.
Since Tu = u, u is the solution to the PDE.

Problem (F’00, #3). Consider the Goursat problem:
Find the solution of the equation
∂2u
∂t2
−
∂2u
∂x2
+ a(x, t)u = 0
in the square D, satisfying the boundary conditions
u|γ1 = ϕ, u|γ2 = ψ,
where γ1, γ2 are two adjacent sides D. Here a(x, t), ϕ and ψ are continuous functions.
Prove the uniqueness of the solution of this Goursat problem.
Proof. The change of variable μ = x + t, η = x − t
transforms the equation to
˜uμη + ã(μ, η)˜u = 0.
We integrate the equation:
η
0
μ
0
˜uμη(u, v) du dv = −
η
0
μ
0
ã(μ, η) ˜udu dv,
η
0
˜uη(μ, v) − ˜uη(0, v) dv = −
η
0
μ
0
ã(μ, η) ˜udu dv,
˜u(μ, η) = ˜u(μ, 0) + ˜u(0, η) − u(0, 0) −
η
0
μ
0
ã(μ, η) ˜udu dv.
We change the notation. In the new notation:
f(x, y) = ϕ(x, y) −
x
0
y
0
a(u, v)f(u, v) du dv,
f = ϕ + Kf,
f = ϕ + K(ϕ + Kf),
· · ·
f = ϕ +
∞
n=1
Kn
ϕ,
f = Kf ⇒ f = 0,
max
0<x<δ
|f| ≤ δ max |a| max|f|.
For small enough δ, the operator K is a contraction. Thus, there exists a unique fixed
point of K, and f = Kf, where f is the unique solution.

20 Contraction Mapping and Uniqueness - Heat
The solution of the initial value problem
ut = u + f(x, t) for t > 0, x ∈ Rn
u(x, 0) = g(x) for x ∈ Rn
.
(20.1)
is given by
u(x, t) =
Rn
˜K(x − y, t) g(y) dy +
t
0 Rn
˜K(x − y, t − s) f(y, s) dyds
where
˜K(x, t) =
⎧
⎨
⎩
1
(4πt)
n
2
e−
|x|2
4t for t > 0,
0 for t ≤ 0.
Problem (F’00, #2). Consider the Cauchy problem
ut − u + u2
(x, t) = f(x, t), x ∈ RN
, 0 < t < T
u(x, 0) = 0.
Prove the uniqueness of the classical bounded solution assuming that T is small
enough.
Proof. Let {un} be a sequence of approximations to the solution, such that
S(un) = un+1 =
use Duhamel s principle
t
0 Rn
K(x − y, t − s) f(y, s) − u2
n(y, s) dy ds.
We will show that S has a fixed point |S(un) − S(un+1)| ≤ α|un − un+1|, α < 1
⇔ {un} converges to a uniques solution for small enough T.
Since un, un+1 ∈ C2
(Rn
) ∩ C1
(t) ⇒ |un+1 + un| ≤ M.
|S(un) − S(un+1)| ≤
t
0 Rn
K(x − y, t − s) u2
n+1 − u2
n dy ds
=
t
0 Rn
K(x − y, t − s) un+1 − un un+1 + un dy ds
≤ M
t
0 Rn
K(x − y, t − s) un+1 − un dy ds
≤ MM1
t
0
un+1(x, s) − un(x, s) ds
≤ MM1T ||un+1 − un||∞ < ||un+1 − un||∞ for small T.
Thus, S is a contraction ⇒ ∃ a unique fixed point u ∈ C2
(Rn
) ∩ C1
(t) such that
u = limn→∞ un. u is implicitly defined as
u(x, t) =
t
0 Rn
K(x − y, t − s) f(y, s) − u2
(y, s) dy ds.

Problem (S’97, #3). a) Let Q(x) ≥ 0 such that
∞
x=−∞ Q(x) dx = 1,
and define Q = 1
Q(x
). Show that (here ∗ denotes convolution)
||Q (x) ∗ w(x)||L∞ ≤ ||w(x)||L∞.
In particular, let Qt(x) denote the heat kernel (at time t), then
||Qt(x) ∗ w1(x) − Qt(x) ∗ w2(x)||L∞ ≤ ||w1(x) − w2(x)||L∞.
b) Consider the parabolic equation ut = uxx + u2
subject to initial conditions
u(x, 0) = f(x). Show that the solution of this equation satisfies
u(x, t) = Qt(x) ∗ f(x) +
t
0
Qt−s(x) ∗ u2
(x, s) ds. (20.2)
c) Fix t > 0. Let {un(x, t)}, n = 1, 2, . . . the fixed point iterations for the solution of
(20.2)
un+1(x, t) = Qt(x) ∗ f(x) +
t
0
Qt−s(x) ∗ u2
n(x, s) ds. (20.3)
Let Kn(t) = sup0≤m≤n ||um(x, t)||L∞. Using (a) and (b) show that
||un+1(x, t) − un(x, t)||L∞ ≤ 2 sup
0≤τ≤t
Kn(τ) ·
t
0
||un(x, s) − un−1(x, s)||L∞ ds.
Conclude that the fixed point iterations in (20.3) converge if t is sufficiently small.
Proof. a) We have
||Q (x) ∗ w(x)||L∞ =
∞
−∞
Q (x − y)w(y) dy ≤
∞
−∞
Q (x − y)w(y) dy
≤ ||w||∞
∞
−∞
Q (x − y) dy = ||w||∞
∞
−∞
1
Q
x − y
dy
= ||w||∞
∞
−∞
1
Q
y
dy z =
y
, dz =
dy
= ||w||∞
∞
−∞
Q(z) dz = ||w(x)||∞.

Qt(x) = 1√
4πt
e−x2
4t , the heat kernel. We have 69
||Qt(x) ∗ w1(x) − Qt(x) ∗ w2(x)||L∞ =
∞
−∞
Qt(x − y)w1(y) dy −
∞
−∞
Qt(x − y)w2(y) dy
∞
=
1
√
4πt
∞
−∞
e−
(x−y)2
4t w1(y) dy −
∞
−∞
e−
(x−y)2
4t w2(y) dy
∞
≤
1
√
4πt
∞
−∞
e−
(x−y)2
4t w1(y) − w2(y) dy
≤ w1(y) − w2(y) ∞
1
√
4πt
∞
−∞
e−
(x−y)2
4t dy
z =
x − y
√
4t
, dz =
−dy
√
4t
= w1(y) − w2(y) ∞
1
√
4πt
∞
−∞
e−z2 √
4t dz
= w1(y) − w2(y) ∞
1
√
π
∞
−∞
e−z2
dz
√
π
= w1(y) − w2(y) ∞
.
69
Note:
∞
−∞
Qt(x) dx =
1
√
4πt
∞
−∞
e−
(x−y)2
4t dy =
1
√
4πt
∞
−∞
e−z2 √
4t dz =
1
√
π
∞
−∞
e−z2
dz = 1.

b) Consider
ut = uxx + u2
,
u(x, 0) = f(x).
We will show that the solution of this equation satisﬁes
u(x, t) = Qt(x) ∗ f(x) +
t
0
Qt−s(x) ∗ u2
(x, s) ds.
t
0
Qt−s(x) ∗ u2
(x, s) ds =
t
0 R
Qt−s(x − y) u2
(y, s) dy ds
=
t
0 R
Qt−s(x − y) us(y, s) − uyy(y, s) dy ds
=
t
0 R
d
ds
Qt−s(x − y)u(y, s) −
d
ds
Qt−s(x − y) u(y, s) − Qt−s(x − y)uyy(y, s) dy ds
=
R
Q0(x − y)u(y, t) dy −
R
Qt(x − y)u(y, 0) dy
−
t
0 R
d
ds
Qt−s(x − y) u(y, s) +
d2
dy2
Qt−s(x − y)u(y, s)
= 0, since Qt satisfies heat equation
dy ds
= u(x, t) −
R
Qt(x − y)f(y) dy Note: lim
t→0+
Q(x, t) = δ0(x) = δ(x).
= u(x, t) − Qt(x) ∗ f(x). lim
t→0+ R Q(x − y, t)v(y) dy = v(0).
Note that we used: Dα(f ∗ g) = (Dαf) ∗ g = f ∗ (Dαg).
c) Let
un+1(x, t) = Qt(x) ∗ f(x) +
t
0
Qt−s(x) ∗ u2
n(x, s) ds.
||un+1(x, t) − un(x, t)||L∞ =
t
0
Qt−s(x) ∗ u2
n(x, s) − u2
n−1(x, s) ds
∞
≤
t
0
Qt−s(x) ∗ u2
n(x, s) − u2
n−1(x, s) ∞
ds
≤
(a)
t
0
u2
n(x, s) − u2
n−1(x, s) ∞
ds
≤
t
0
un(x, s) − un−1(x, s) ∞
un(x, s) + un−1(x, s) ∞
ds
≤ sup
0≤τ≤t
un(x, s) + un−1(x, s) ∞
t
0
un(x, s) − un−1(x, s) ∞
ds
≤ 2 sup
0≤τ≤t
Kn(τ) ·
t
0
||un(x, s) − un−1(x, s)||L∞ ds.
Also, ||un+1(x, t) − un(x, t)||L∞ ≤ 2t sup
0≤τ≤t
Kn(τ) · ||un(x, s) − un−1(x, s)||L∞.

For t small enough, 2t sup0≤τ≤t Kn(τ) ≤ α < 1. Thus, T deﬁned as
Tu = Qt(x) ∗ f(x) +
t
0
Qt−s(x) ∗ u2
(x, s) ds
is a contraction, and has a unique ﬁxed point u = Tu.

Problem (S’99, #3). Consider the system of equations
ut = uxx + f(u, v)
vt = 2vxx + g(u, v)
to be solved for t > 0, −∞ < x < ∞, and smooth initial data with compact support:
u(x, 0) = u0(x), v(x, 0) = v0(x).
If f and g are uniformly Lipschitz continuous, give a proof of existence and unique-
ness of the solution to this problem in the space of bounded continuous functions with
||u(·, t)|| = supx |u(x, t)|.
Proof. The space of continuous bounded functions forms a complete metric space so
the contraction mapping principle applies.
First, let v(x, t) = w x√
2
, t , then
ut = uxx + f(u, w)
wt = wxx + g(u, w).
These initial value problems have the following solutions (K is the heat kernel):
u(x, t) =
Rn
˜K(x − y, t) u0(y) dy +
t
0 Rn
˜K(x − y, t − s) f(u, w) dyds,
w(x, t) =
Rn
˜K(x − y, t) w0(y) dy +
t
0 Rn
˜K(x − y, t − s) g(u, w) dyds.
By the Lipshitz conditions,
|f(u, w)| ≤ M1||u||,
|g(u, w)| ≤ M2||w||.
Now we can show the mappings, as defined below, are contractions:
T1u =
Rn
˜K(x − y, t) u0(y) dy +
t
0 Rn
˜K(x − y, t − s) f(u, w) dyds,
T2w =
Rn
˜K(x − y, t) w0(y) dy +
t
0 Rn
˜K(x − y, t − s) g(u, w) dyds.
|T1(un) − T1(un+1)| ≤
t
0 Rn
˜K(x − y, t − s) f(un, w) − f(un+1, w) dy ds
≤ M1
t
0 Rn
˜K(x − y, t − s) un − un+1 dy ds
≤ M1
t
0
sup
x
un − un+1
Rn
˜K(x − y, t − s)dy ds
≤ M1
t
0
sup
x
un − un+1 ds ≤ M1t sup
x
un − un+1
< sup
x
un − un+1 for small t.
We used the Lipshitz condition and R
˜K(x − y, t − s) dy = 1.
Thus, for small t, T1 is a contraction, and has a unique fixed point. Thus, the solution
is defined as u = T1u.
Similarly, T2 is a contraction and has a unique fixed point. The solution is defined as
w = T2w.

21 Problems: Maximum Principle - Laplace and Heat
21.1 Heat Equation - Maximum Principle and Uniqueness
Let us introduce the “cylinder” U = UT = Ω × (0, T). We know that harmonic (and
subharmonic) functions achieve their maximum on the boundary of the domain. For
the heat equation, the result is improved in that the maximum is achieved on a certain
part of the boundary, parabolic boundary:
Γ = {(x, t) ∈ U : x ∈ ∂Ω or t = 0}.
Let us also denote by C2;1
(U) functions satisfying ut, uxixj ∈ C(U).
Weak Maximum Principle. Let u ∈ C2;1
(U) ∩ C(U) satisfy u ≥ ut in U.
Then u achieves its maximum on the parabolic boundary of U:
max
U
u(x, t) = max
Γ
u(x, t). (21.1)
Proof. • First, assume u > ut in U. For 0 < τ < T consider
Uτ = Ω × (0, τ), Γτ = {(x, t) ∈ Uτ : x ∈ ∂Ω or t = 0}.
If the maximum of u on Uτ occurs at x ∈ Ω and t = τ, then ut(x, τ) ≥ 0 and
u(x, τ) ≤ 0, violating our assumption; similarly, u cannot attain an interior maximum
on Uτ . Hence (21.1) holds for Uτ : maxUτ
u = maxΓτ u. But maxΓτ u ≤ maxΓ u
and by continuity of u, maxU u = limτ→T maxUτ
u. This establishes (21.1).
• Second, we consider the general case of u ≥ ut in U. Let u = v + εt for ε > 0.
Notice that v ≤ u on U and v − vt > 0 in U. Thus we may apply (21.1) to v:
max
U
u = max
U
(v + εt) ≤ max
U
v + εT = max
Γ
v + εT ≤ max
Γ
u + εT.
Letting ε → 0 establishes (21.1) for u.

Problem (S’98, #7). Prove that any smooth solution, u(x, y, t) in the unit box
Ω = {(x, y) | − 1 ≤ x, y ≤ 1}, of the following equation
ut = uux + uuy + u, t ≥ 0, (x, y) ∈ Ω
u(x, y, 0) = f(x, y), (x, y) ∈ Ω
satisﬁes the weak maximum principle,
max
Ω×[0,T ]
u(x, y, t) ≤ max{ max
0≤t≤T
u(±1, ±1, t), max
(x,y)∈Ω
f(x, y)}.
Proof. Suppose u satisﬁes given equation. Let u = v + εt for ε > 0. Then,
vt + ε = vvx + vvy + εt(vx + vy) + v.
Suppose v has a maximum at (x0, y0, t0) ∈ Ω × (0, T). Then
vx = vy = vt = 0 ⇒ ε = v ⇒ v > 0
⇒ v has a minimum at (x0, y0, t0), a contradiction.
Thus, the maximum of v is on the boundary of Ω × (0, T).
Suppose v has a maximum at (x0, y0, T), (x0, y0) ∈ Ω. Then
vx = vy = 0, vt ≥ 0 ⇒ ε ≤ v ⇒ v > 0
⇒ v has a minimum at (x0, y0, T), a contradiction. Thus,
max
Ω×[0,T ]
v ≤ max{ max
0≤t≤T
v(±1, ±1, t), max
(x,y)∈Ω
f(x, y)}.
Now
max
Ω×[0,T ]
u = max
Ω×[0,T ]
(v + εt) ≤ max
Ω×[0,T ]
v + εT ≤ max{ max
0≤t≤T
v(±1, ±1, t), max
(x,y)∈Ω
f(x, y)} + εT
≤ max{ max
0≤t≤T
u(±1, ±1, t), max
(x,y)∈Ω
f(x, y)} + εT.
Letting ε → 0 establishes the result.

21.2 Laplace Equation - Maximum Principle
Problem (S’91, #6). Suppose that u satisﬁes
Lu = auxx + buyy + cux + duy − eu = 0
with a > 0, b > 0, e > 0, for (x, y) ∈ Ω, with Ω a bounded open set in R2.
a) Show that u cannot have a positive maximum or a negative minimum in the in-
terior of Ω.
b) Use this to show that the only function u satisfying Lu = 0 in Ω, u = 0 on ∂Ω
and u continuous on Ω is u = 0.
Proof. a) For an interior (local) maximum or minimum at an interior point (x, y), we
have
ux = 0, uy = 0.
• Suppose u has a positive maximum in the interior of Ω. Then
u > 0, uxx ≤ 0, uyy ≤ 0.
With these values, we have
auxx
≤0
+ buyy
≤0
+ cux
=0
+ duy
=0
−eu
<0
= 0,
which leads to contradiction. Thus, u can not have a positive maximum in Ω.
• Suppose u has a negative minimum in the interior of Ω. Then
u < 0, uxx ≥ 0, uyy ≥ 0.
With these values, we have
auxx
≥0
+ buyy
≥0
+ cux
=0
+ duy
=0
−eu
>0
= 0,
which leads to contradiction. Thus, u can not have a negative minimum in Ω.
b) Since u can not have positive maximum in the interior of Ω, then maxu = 0 on Ω.
Since u can not have negative minimum in the interior of Ω, then min u = 0 on Ω.
Since u is continuous, u ≡ 0 on Ω.

22 Problems: Separation of Variables - Laplace Equation
Problem 1: The 2D LAPLACE Equation on a Square.
Let Ω = (0, π) × (0, π), and use separation of variables to solve the boundary value
problem
⎧
⎪⎨
⎪⎩
uxx + uyy = 0 0 < x, y < π
u(0, y) = 0 = u(π, y) 0 ≤ y ≤ π
u(x, 0) = 0, u(x, π) = g(x) 0 ≤ x ≤ π,
where g is a continuous function satisfying g(0) = 0 = g(π).
Proof. Assume u(x, y) = X(x)Y (y), then substitution in the PDE gives X Y +XY =
0.
X
X
= −
Y
Y
= −λ.
• From X + λX = 0, we get Xn(x) = an cos nx + bn sin nx. Boundary conditions
give
u(0, y) = X(0)Y (y) = 0
u(π, y) = X(π)Y (y) = 0
⇒ X(0) = 0 = X(π).
Thus, Xn(0) = an = 0, and
Xn(x) = bn sin nx, n = 1, 2, . . ..
−n2
bn sinnx + λbn sinnx = 0,
λn = n2
, n = 1, 2, . . ..
• With these values of λn we solve Y − n2Y = 0 to find Yn(y) = cn cosh ny +
dn sinhny.
u(x, 0) = X(x)Y (0) = 0 ⇒ Y (0) = 0 = cn.
Yn(x) = dn sinh ny.
• By superposition, we write
u(x, y) =
∞
n=1
ãn sin nx sinhny,
which satifies the equation and the three homogeneous boundary conditions. The
boundary condition at y = π gives
u(x, π) = g(x) =
∞
n=1
ãn sinnx sinh nπ,
π
0
g(x) sinmx dx =
∞
n=1
ãn sinhnπ
π
0
sin nx sinmx dx =
π
2
ãm sinhmπ.

˜an sinh nπ =
2
π
π
0
g(x) sinnx dx.

Problem 2: The 2D LAPLACE Equation on a Square. Let Ω = (0, π)×(0, π),
and use separation of variables to solve the mixed boundary value problem
⎧
⎪⎨
⎪⎩
u = 0 in Ω
ux(0, y) = 0 = ux(π, y) 0 < y < π
u(x, 0) = 0, u(x, π) = g(x) 0 < x < π.
0.
X
X
= −
Y
Y
= −λ.
• Consider X + λX = 0.
If λ = 0, X0(x) = a0x + b0.
If λ > 0, Xn(x) = an cos nx + bn sin nx.
ux(0, y) = X (0)Y (y) = 0
ux(π, y) = X (π)Y (y) = 0
⇒ X (0) = 0 = X (π).
Thus, X0(0) = a0 = 0, and Xn(0) = nbn = 0.
X0(x) = b0, Xn(x) = an cos nx, n = 1, 2, . . ..
−n2
an cos nx + λan cos nx = 0,
λn = n2
, n = 0, 1, 2, . . ..
• With these values of λn we solve Y − n2Y = 0.
If n = 0, Y0(y) = c0y + d0.
If n = 0, Yn(y) = cn coshny + dn sinhny.
u(x, 0) = X(x)Y (0) = 0 ⇒ Y (0) = 0.
Thus, Y0(0) = d0 = 0, and Yn(0) = cn = 0.
Y0(y) = c0y, Yn(y) = dn sinh ny, n = 1, 2, . . ..
• We have
u0(x, y) = X0(x)Y0(y) = b0c0y = ã0y,
un(x, y) = Xn(x)Yn(y) = (an cos nx)(dn sinh ny) = ãn cos nx sinhny.
By superposition, we write
u(x, y) = ã0y +
∞
n=1
ãn cos nx sinhny,
which satifies the equation and the three homogeneous boundary conditions. The fourth
boundary condition gives
u(x, π) = g(x) = ã0π +
∞
n=1
ãn cos nx sinh nπ,

π
0 g(x) dx =
π
0 ã0π + ∞
n=1 ãn cos nx sinh nπ dx = ã0π2,
π
0 g(x) cosmx dx = ∞
n=1 ãn sinhnπ
π
0 cos nx cos mx dx = π
2 ãm sinh mπ.
ã0 =
1
π2
π
0
g(x) dx,
ãn sinh nπ =
2
π
π
0
g(x) cosnx dx.

Problem (W’04, #5) The 2D LAPLACE Equation in an Upper-Half Plane.
Consider the Laplace equation
∂2u
∂x2
+
∂2u
∂y2
= 0, y > 0, −∞ < x < +∞
∂u(x, 0)
∂y
− u(x, 0) = f(x),
where f(x) ∈ C∞
0 (R1
).
Find a bounded solution u(x, y) and show that u(x, y) → 0 when |x| + y → ∞.
0.
X
X
= −
Y
Y
= −λ.
• Consider X + λX = 0.
If λ = 0, X0(x) = a0x + b0.
If λ > 0, Xn(x) = an cos
√
λnx + bn sin
√
λnx.
Since we look for bounded solutions as |x| → ∞, we have a0 = 0.
• Consider Y − λnY = 0.
If λn = 0, Y0(y) = c0y + d0.
If λn > 0, Yn(y) = cne−
√
λny
+ dne
√
λny
.
Since we look for bounded solutions as y → ∞, we have c0 = 0, dn = 0. Thus,
u(x, y) = ã0 +
∞
n=1
e−
√
λny
ãn cos λnx + ˜bn sin λnx .
Initial condition gives:
f(x) = uy(x, 0) − u(x, 0) = −ã0 −
∞
n=1
( λn + 1) ãn cos λnx + ˜bn sin λnx .
f(x) ∈ C∞
0 (R1
), i.e. has compact support [−L, L], for some L > 0. Thus the coefficients
ãn, ˜bn are given by
L
−L
f(x) cos λnx dx = −( λn + 1)ãnL.
L
−L
f(x) sin λnx dx = −( λn + 1)˜bnL.
Thus, u(x, y) → 0 when |x| + y → ∞. 70
70
Note that if we change the roles of X and Y in , the solution we get will be unbounded.

Problem 3: The 2D LAPLACE Equation on a Circle.
Let Ω be the unit disk in R2 and consider the problem
u = 0 in Ω
∂u
∂n = h on ∂Ω,
where h is a continuous function.
Proof. Use polar coordinates (r, θ)
urr + 1
r ur + 1
r2 uθθ = 0 for 0 ≤ r < 1, 0 ≤ θ < 2π
∂u
∂r (1, θ) = h(θ) for 0 ≤ θ < 2π.
r2
urr + rur + uθθ = 0.
Let r = e−t
, u(r(t), θ).
ut = urrt = −e−t
ur,
utt = (−e−t
ur)t = e−t
ur + e−2t
urr = rur + r2
urr.
Thus, we have
utt + uθθ = 0.
Let u(t, θ) = X(t)Y (θ), which gives X (t)Y (θ) + X(t)Y (θ) = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
• From Y (θ) + λY (θ) = 0, we get Yn(θ) = an cos nθ + bn sin nθ.
λn = n2
, n = 0, 1, 2, . . ..
• With these values of λn we solve X (t) − n2X(t) = 0.
If n = 0, X0(t) = c0t + d0. ⇒ X0(r) = −c0 log r + d0.
If n = 0, Xn(t) = cnent + dne−nt ⇒ Xn(r) = cnr−n + dnrn.
• We have
u0(r, θ) = X0(r)Y0(θ) = (−c0 log r + d0)a0,
un(r, θ) = Xn(r)Yn(θ) = (cnr−n
+ dnrn
)(an cos nθ + bn sinnθ).
But u must be finite at r = 0, so cn = 0, n = 0, 1, 2, . . ..
u0(r, θ) = d0a0,
un(r, θ) = dnrn
(an cos nθ + bn sinnθ).
u(r, θ) = ã0 +
∞
n=1
rn
(ãn cos nθ + ˜bn sinnθ).
ur(1, θ) =
∞
n=1
n(ãn cos nθ + ˜bn sinnθ) = h(θ).
The coefficients an, bn for n ≥ 1 are determined from the Fourier series for h(θ).
a0 is not determined by h(θ) and therefore may take an arbitrary value. Moreover,

the constant term in the Fourier series for h(θ) must be zero [i.e.,
2π
0 h(θ)dθ = 0].
Therefore, the problem is not solvable for an arbitrary function h(θ), and when it is
solvable, the solution is not unique.

Problem 4: The 2D LAPLACE Equation on a Circle.
Let Ω = {(x, y) ∈ R2 : x2 + y2 < 1} = {(r, θ) : 0 ≤ r < 1, 0 ≤ θ < 2π},
and use separation of variables (r, θ) to solve the Dirichlet problem
u = 0 in Ω
u(1, θ) = g(θ) for 0 ≤ θ < 2π.
urr + 1
r ur + 1
r2 uθθ = 0 for 0 ≤ r < 1, 0 ≤ θ < 2π
u(1, θ) = g(θ) for 0 ≤ θ < 2π.
r2
Let r = e−t, u(r(t), θ).
ut = urrt = −e−t
ur,
utt = (−e−t
ur)t = e−t
ur + e−2t
urr = rur + r2
urr.
Thus, we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
λn = n2, n = 0, 1, 2, . . ..
• With these values of λn we solve X (t) − n2
X(t) = 0.
If n = 0, Xn(t) = cnent
+ dne−nt
⇒ Xn(r) = cnr−n
+ dnrn
.
• We have
u0(r, θ) = X0(r)Y0(θ) = (−c0 log r + d0)a0,
+ dnrn
u0(r, θ) = d0a0,
un(r, θ) = dnrn
(an cos nθ + bn sinnθ).
u(r, θ) = ã0 +
∞
n=1
rn
u(1, θ) = ã0 +
∞
n=1
(ãn cos nθ + ˜bn sin nθ) = g(θ).

˜a0 =
1
π
π
0
g(θ) dθ,
˜an =
2
π
π
0
g(θ) cosnθ dθ,
˜bn =
2
π
π
0
g(θ) sinnθ dθ.

Problem (F’94, #6): The 2D LAPLACE Equation on a Circle.
Find all solutions of the homogeneous equation
uxx + uyy = 0, x2
+ y2
< 1,
∂u
∂n
− u = 0, x2
+ y2
= 1.
Hint: = 1
r
∂
∂r (r ∂
∂r ) + 1
r2
∂2
∂θ2 in polar coordinates.
Proof. Use polar coordinates (r, θ):
urr + 1
r ur + 1
r2 uθθ = 0 for 0 ≤ r < 1, 0 ≤ θ < 2π
∂u
∂r (1, θ) − u(1, θ) = 0 for 0 ≤ θ < 2π.
Since we solve the equation on a circle, we have periodic conditions:
u(r, 0) = u(r, 2π) ⇒ X(r)Y (0) = X(r)Y (2π) ⇒ Y (0) = Y (2π),
uθ(r, 0) = uθ(r, 2π) ⇒ X(r)Y (0) = X(r)Y (2π) ⇒ Y (0) = Y (2π).
Also, we want the solution to be bounded. In particular, u is bounded for r = 0.
r2
Let r = e−t, u(r(t), θ), we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
• From Y (θ) + λY (θ) = 0, we get Yn(θ) = an cos
√
λθ + bn sin
√
λθ.
Using periodic condition: Yn(0) = an,
Yn(2π) = an cos( λn 2π) + bn sin( λn 2π) = an ⇒ λn = n ⇒ λn = n2
.
Thus, Yn(θ) = an cos nθ + bn sinnθ.
X(t) = 0.
u must be finite at r = 0 ⇒ cn = 0, n = 0, 1, 2, . . ..
u(r, θ) = ã0 +
∞
n=1
rn
0 = ur(1, θ) − u(1, θ) = −ã0 +
∞
n=1
(n − 1)(ãn cos nθ + ˜bn sinnθ).
Calculating Fourier coefficients gives −2πã0 = 0 ⇒ ã0 = 0.
π(n − 1)an = 0 ⇒ ãn = 0, n = 2, 3, . . ..
a1, b1 are constants. Thus,
u(r, θ) = r(ã1 cos θ + ˜b1 sin θ).

a) Let (r, θ) be polar coordinates on the plane,
i.e. x1 + ix2 = reiθ
. Solve the boudary value problem
u = 0 in r < 1
∂u/∂r = f(θ) on r = 1,
beginning with the Fourier series for f (you may assume that f is continuously dif-
ferentiable). Give your answer as a power series in x1 + ix2 plus a power series in
x1 − ix2. There is a necessary condition on f for this boundary value problem to be
solvable that you will ﬁnd in the course of doing this.
b) Sum the series in part (a) to get a representation of u in the form
u(r, θ) =
2π
0
N(r, θ − θ )f(θ ) dθ .
Proof. a) Green’s identity gives the necessary compatibility condition on f:
2π
0
f(θ) dθ =
r=1
∂u
∂r
dθ =
∂Ω
∂u
∂n
ds =
Ω
u dx = 0.
Use polar coordinates (r, θ):
urr + 1
r ur + 1
r2 uθθ = 0 for 0 ≤ r < 1, 0 ≤ θ < 2π
∂u
∂r (1, θ) = f(θ) for 0 ≤ θ < 2π.
Since we solve the equation on a circle, we have periodic conditions:
uθ(r, 0) = uθ(r, 2π) ⇒ X(r)Y (0) = X(r)Y (2π) ⇒ Y (0) = Y (2π).
Also, we want the solution to be bounded. In particular, u is bounded for r = 0.
r2
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
√
λθ + bn sin
√
λθ.
.

If n = 0, Xn(t) = cnent
+ dne−nt
⇒ Xn(r) = cnr−n
+ dnrn
.
u must be finite at r = 0 ⇒ cn = 0, n = 0, 1, 2, . . ..
u(r, θ) = ã0 +
∞
n=1
rn
Since
ur(r, θ) =
∞
n=1
nrn−1
(ãn cos nθ + ˜bn sinnθ),
the boundary condition gives
ur(1, θ) =
∞
n=1
n (ãn cos nθ + ˜bn sinnθ) = f(θ).
ãn =
1
nπ
2π
0
f(θ) cos nθ dθ,
˜bn =
1
nπ
2π
0
f(θ) sin nθ dθ.
ã0 is not determined by f(θ) (since
2π
0 f(θ) dθ = 0). Therefore, it may take an
arbitrary value. Moreover, the constant term in the Fourier series for f(θ) must be zero
[i.e.,
2π
0 f(θ)dθ = 0]. Therefore, the problem is not solvable for an arbitrary function
f(θ), and when it is solvable, the solution is not unique.
b) In part (a), we obtained the solution and the Fourier coefficients:
ãn =
1
nπ
2π
0
f(θ ) cos nθ dθ ,
˜bn =
1
nπ
2π
0
f(θ ) sinnθ dθ .
u(r, θ) = ã0 +
∞
n=1
rn
(ãn cos nθ + ˜bn sinnθ)
= ã0 +
∞
n=1
rn 1
nπ
2π
0
f(θ ) cos nθ dθ cos nθ +
1
nπ
2π
0
f(θ ) sin nθ dθ sinnθ
= ã0 +
∞
n=1
rn
nπ
2π
0
f(θ ) cos nθ cos nθ + sin nθ sinnθ dθ
= ã0 +
∞
n=1
rn
nπ
2π
0
f(θ ) cos n(θ − θ) dθ
= ã0 +
2π
0
∞
n=1
rn
nπ
cos n(θ − θ )
N(r,θ−θ )
f(θ ) dθ .

Problem (S’92, #6). Consider the Laplace equation
uxx + uyy = 0
for x2 + y2 ≥ 1. Denoting by x = r cos θ, y = r sinθ polar coordinates, let f = f(θ) be
a given smooth function of θ. Construct a uniformly bounded solution which satisfies
boundary conditions
u = f for x2
+ y2
= 1.
What conditions has f to satisfy such that
lim
x2+y2→∞
(x2
+ y2
)u(x, y) = 0?
urr + 1
r ur + 1
r2 uθθ = 0 for r ≥ 1
u(1, θ) = f(θ) for 0 ≤ θ < 2π.
Since we solve the equation on outside of a circle, we have periodic conditions:
uθ(r, 0) = u(r, 2π) ⇒ X(r)Y (0) = X(r)Y (2π) ⇒ Y (0) = Y (2π).
Also, we want the solution to be bounded. In particular, u is bounded for r = ∞.
r2
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
√
λθ + bn sin
√
λθ.
.
u must be finite at r = ∞ ⇒ c0 = 0, dn = 0, n = 1, 2, . . ..
u(r, θ) = ã0 +
∞
n=1
r−n
(ãn cos nθ + ˜bn sin nθ).
f(θ) = u(1, θ) = ã0 +
∞
n=1

⎧
⎪⎨
⎪⎩
2πã0 =
2π
0 f(θ) dθ,
πãn =
2π
0 f(θ) cos nθ dθ,
π˜bn =
2π
0 f(θ) sinnθ dθ.
⇒
⎧
⎪⎨
⎪⎩
f0 = ã0 = 1
2π
2π
0 f(θ) dθ,
fn = ãn = 1
π
2π
0 f(θ) cos nθ dθ,
˜fn = ˜bn = 1
π
2π
0 f(θ) sinnθ dθ.
• We need to find conditions for f such that
lim
x2+y2→∞
(x2
+ y2
)u(x, y) = 0, or
lim
r→∞
r2
u(r, θ) =
need
0,
lim
r→∞
r2
f0 +
∞
n=1
r−n
(fn cos nθ + ˜fn sin nθ) =
need
0.
Since
lim
r→∞
∞
n>2
r2−n
(fn cos nθ + ˜fn sin nθ) = 0,
we need
lim
r→∞
r2
f0 +
2
n=1
r2−n
(fn cos nθ + ˜fn sinnθ) =
need
0.
Thus, the conditions are
fn, ˜fn = 0, n = 0, 1, 2.

Problem (F’96, #2): The 2D LAPLACE Equation on a Semi-Annulus.
Solve the Laplace equation in the semi-annulus
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
u = 0, 1 < r < 2, 0 < θ < π,
u(r, 0) = u(r, π) = 0, 1 < r < 2,
u(1, θ) = sinθ, 0 < θ < π,
u(2, θ) = 0, 0 < θ < π.
Hint: Use the formula = 1
r
∂
∂r (r ∂
∂r ) + 1
r2
∂2
∂θ2 for the Laplacian in polar coordinates.
urr +
1
r
ur +
1
r2
uθθ = 0 1 < r < 2, 0 < θ < π,
r2
With r = e−t, we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
√
λθ + bn sin
√
λθ.
un(r, 0) = 0 = Xn(r)Yn(0) = 0, ⇒ Yn(0) = 0,
un(r, π) = 0 = Xn(r)Yn(π) = 0, ⇒ Yn(π) = 0.
Thus, 0 = Yn(0) = an, and Yn(π) = bn sin
√
λπ = 0 ⇒
√
λ = n ⇒ λn = n2
.
Thus, Yn(θ) = bn sinnθ, n = 1, 2, . . ..
If n > 0, Xn(t) = cnent + dne−nt ⇒ Xn(r) = cnr−n + dnrn.
• We have,
u(r, θ) =
∞
n=1
Xn(r)Yn(θ) =
∞
n=1
(˜cnr−n
+ ˜dnrn
) sinnθ.
Using the other two boundary conditions, we obtain
sinθ = u(1, θ) =
∞
n=1
(˜cn + ˜dn) sinnθ ⇒
˜c1 + ˜d1 = 1,
˜cn + ˜dn = 0, n = 2, 3, . . ..
0 = u(2, θ) =
∞
n=1
(˜cn2−n
+ ˜dn2n
) sinnθ ⇒ ˜cn2−n
+ ˜dn2n
= 0, n = 1, 2, . . ..
Thus, the coeﬃcients are given by
c1 =
4
3
, d1 = −
1
3
;
cn = 0, dn = 0.

u(r, θ) =
4
3r
−
r
3
sinθ.

Problem (S’98, #8): The 2D LAPLACE Equation on a Semi-Annulus.
Solve
⎧
⎪⎨
⎪⎩
u = 0, 1 < r < 2, 0 < θ < π,
u(r, 0) = u(r, π) = 0, 1 < r < 2,
u(1, θ) = u(2, θ) = 1, 0 < θ < π.
urr +
1
r
ur +
1
r2
uθθ = 0 for 1 < r < 2, 0 < θ < π,
r2
With r = e−t
, we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
un(r, 0) = 0 = Xn(r)Yn(0) = 0, ⇒ Yn(0) = 0,
un(r, π) = 0 = Xn(r)Yn(π) = 0, ⇒ Yn(π) = 0.
Thus, 0 = Yn(0) = an, and Yn(θ) = bn sin nθ.
λn = n2
, n = 1, 2, . . ..
If n > 0, Xn(t) = cnent + dne−nt ⇒ Xn(r) = cnr−n + dnrn.
• We have,
u(r, θ) =
∞
n=1
Xn(r)Yn(θ) =
∞
n=1
(˜cnr−n
+ ˜dnrn
) sinnθ.
u(1, θ) = 1 =
∞
n=1
(˜cn + ˜dn) sinnθ,
u(2, θ) = 1 =
∞
n=1
(˜cn2−n
+ ˜dn2n
) sinnθ,
which give the two equations for ˜cn and ˜dn:
π
0
sin nθ dθ =
π
2
(˜cn + ˜dn),
π
0
sin nθ dθ =
π
2
(˜cn2−n
+ ˜dn2n
),
that can be solved.

Problem (F’89, #1). Consider Laplace equation inside a 90◦
sector of a circular
annulus
u = 0 a < r < b, 0 < θ <
π
2
subject to the boundary conditions
∂u
∂θ
(r, 0) = 0,
∂u
∂θ
(r,
π
2
) = 0,
∂u
∂r
(a, θ) = f1(θ),
∂u
∂r
(b, θ) = f2(θ),
where f1(θ), f2(θ) are continuously diﬀerentiable.
a) Find the solution of this equation with the prescribed
boundary conditions using separation of variables.
Proof. a) Use polar coordinates (r, θ)
urr +
1
r
ur +
1
r2
uθθ = 0 for a < r < b, 0 < θ <
π
2
,
r2
With r = e−t, we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
√
λθ + bn sin
√
λθ.
unθ(r, 0) = Xn(r)Yn(0) = 0 ⇒ Yn(0) = 0,
unθ(r,
π
2
) = Xn(r)Yn(
π
2
) = 0 ⇒ Yn(
π
2
) = 0.
Yn(θ) = −an
√
λn sin
√
λnθ + bn
√
λn cos
√
λnθ. Thus, Yn(0) = bn
√
λn = 0 ⇒ bn = 0.
Yn(π
2 ) = −an
√
λn sin
√
λn
π
2 = 0 ⇒
√
λn
π
2 = nπ ⇒ λn = (2n)2.
Thus, Yn(θ) = an cos(2nθ), n = 0, 1, 2, . . ..
In particular, Y0(θ) = a0t + b0. Boundary conditions give Y0(θ) = b0.
• With these values of λn we solve X (t) − (2n)2
X(t) = 0.
If n > 0, Xn(t) = cne2nt + dne−2nt ⇒ Xn(r) = cnr−2n + dnr2n.
u(r, θ) = ˜c0 log r + ˜d0 +
∞
n=1
(˜cnr−2n
+ ˜dnr2n
) cos(2nθ).
ur(r, θ) =
˜c0
r
+
∞
n=1
(−2n˜cnr−2n−1
+ 2n ˜dnr2n−1
) cos(2nθ).

f1(θ) = ur(a, θ) =
˜c0
a
+ 2
∞
n=1
n(−˜cna−2n−1
+ ˜dna2n−1
) cos(2nθ),
f2(θ) = ur(b, θ) =
˜c0
b
+ 2
∞
n=1
n(−˜cnb−2n−1
+ ˜dnb2n−1
) cos(2nθ).
which give the two equations for ˜cn and ˜dn:
π
2
0
f1(θ) cos(2nθ) dθ =
π
2
n(−˜cna−2n−1
+ ˜dna2n−1
),
π
2
0
f2(θ) sin(2nθ) dθ =
π
2
n(−˜cnb−2n−1
+ ˜dnb2n−1
).
b) Show that the solution exists if and only if
a
π
2
0
f1(θ) dθ − b
π
2
0
f2(θ) dθ = 0.
Proof. Using Green’s identity, we obtain:
0 =
Ω
u dx =
∂Ω
∂u
∂n
=
π
2
0
∂u
∂r
(b, θ) dθ +
0
π
2
−
∂u
∂r
(a, θ) dθ +
b
a
−
∂u
∂θ
(r, 0) dr +
a
b
∂u
∂θ
r,
π
2
dr
=
π
2
0
f2(θ) dθ +
π
2
0
f1(θ) dθ + 0 + 0
=
π
2
0
f1(θ) dθ +
π
2
0
f2(θ) dθ.
c) Is the solution unique?
Proof. No, since the boundary conditions are Neumann. The solution is unique only
up to a constant.

Problem (S’99, #4). Let u(x, y) be harmonic inside the unit disc,
with boundary values along the unit circle
u(x, y) =
1, y > 0
0, y ≤ 0.
Compute u(0, 0) and u(0, y).
Proof. Since u is harmonic, u = 0. Use polar coordinates (r, θ)
⎧
⎪⎨
⎪⎩
urr + 1
r ur + 1
r2 uθθ = 0 0 ≤ r < 1, 0 ≤ θ < 2π
u(1, θ) =
1, 0 < θ < π
0, π ≤ θ ≤ 2π.
r2
With r = e−t
, we have
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
λn = n2, n = 1, 2, . . ..
X(t) = 0.
If n > 0, Xn(t) = cnent
+ dne−nt
⇒ Xn(r) = cnr−n
+ dnrn
.
• We have
u0(r, θ) = X0(r)Y0(θ) = (−c0 log r + d0)a0,
+ dnrn
u0(r, θ) = ã0,
un(r, θ) = rn
(ãn cos nθ + ˜bn sin nθ).
u(r, θ) = ã0 +
∞
n=1
rn
u(1, θ) = ã0 +
∞
n=1
(ãn cos nθ + ˜bn sin nθ) =
1, 0 < θ < π
0, π ≤ θ ≤ 2π,
and the coefficients ãn and ˜bn are determined from the above equation.
71
71
See Yana’s solutions, where Green’s function on a unit disk is constructed.

23 Problems: Separation of Variables - Poisson Equation
Problem (F’91, #2): The 2D POISSON Equation on a Quarter-Circle.
Solve explicitly the following boundary value problem
uxx + uyy = f(x, y)
in the domain Ω = {(x, y), x > 0, y > 0, x2 + y2 < 1}
with boundary conditions
u = 0 for y = 0, 0 < x < 1,
∂u
∂x
= 0 for x = 0, 0 < y < 1,
u = 0 for x > 0, y > 0, x2
+ y2
= 1.
Function f(x, y) is known and is assumed to be continuous.
⎧
⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
urr + 1
r ur + 1
r2 uθθ = f(r, θ) 0 ≤ r < 1, 0 ≤ θ < π
2
u(r, 0) = 0 0 ≤ r < 1,
uθ(r, π
2 ) = 0 0 ≤ r < 1,
u(1, θ) = 0 0 ≤ θ ≤ π
2 .
We solve
r2
utt + uθθ = 0.
X (t)
X(t)
= −
Y (θ)
Y (θ)
= λ.
√
λθ + bn sin
√
λθ. Boundary
conditions:
u(r, 0) = X(r)Y (0) = 0
uθ(r, π
2 ) = X(r)Y (π
2 ) = 0
⇒ Y (0) = Y
π
2
= 0.
Thus, Yn(0) = an = 0, and Yn(π
2 ) =
√
λnbn cos
√
λn
π
2 = 0
⇒
√
λn
π
2 = nπ − π
2 , n = 1, 2, . . . ⇒ λn = (2n − 1)2.
Thus, Yn(θ) = bn sin(2n − 1)θ, n = 1, 2, . . .. Thus, we have
u(r, θ) =
∞
n=1
Xn(r) sin[(2n − 1)θ].

We now plug this equation into with inhomogeneous term and obtain
∞
n=1
Xn(t) sin[(2n − 1)θ] − (2n − 1)2
Xn(t) sin[(2n − 1)θ] = f(t, θ),
∞
n=1
Xn(t) − (2n − 1)2
Xn(t) sin[(2n − 1)θ] = f(t, θ),
π
4
Xn(t) − (2n − 1)2
Xn(t) =
π
2
0
f(t, θ) sin[(2n − 1)θ] dθ,
Xn(t) − (2n − 1)2
Xn(t) =
4
π
π
2
0
f(t, θ) sin[(2n − 1)θ] dθ.
The solution to this equation is
Xn(t) = cne(2n−1)t
+ dne−(2n−1)t
+ Unp(t), or
Xn(r) = cnr−(2n−1)
+ dnr(2n−1)
+ unp(r),
where unp is the particular solution of inhomogeneous equation.
u must be finite at r = 0 ⇒ cn = 0, n = 1, 2, . . .. Thus,
u(r, θ) =
∞
n=1
dnr(2n−1)
+ unp(r) sin[(2n − 1)θ].
Using the last boundary condition, we have
0 = u(1, θ) =
∞
n=1
dn + unp(1) sin[(2n − 1)θ],
⇒ 0 =
π
4
(dn + unp(1)),
⇒ dn = −unp(1).
u(r, θ) =
∞
n=1
− unp(1)r(2n−1)
+ unp(r) sin[(2n − 1)θ].
The method used to solve this problem is similar to section
Problems: Eigenvalues of the Laplacian - Poisson Equation:
1) First, we find Yn(θ) eigenfunctions.
2) Then, we plug in our guess u(t, θ) = X(t)Y (θ) into the equation utt + uθθ = f(t, θ)
and solve an ODE in X(t).
Note the similar problem on 2D Poisson equation on a square domain. The prob-
lem is used by first finding the eigenvalues and eigenfunctions of the Laplacian, and
then expanding f(x, y) in eigenfunctions, and comparing coefficients of f with the gen-
eral solution u(x, y).
Here, however, this could not be done because of the circular geometry of the domain.
In particular, the boundary conditions do not give enough information to find explicit
representations for μm and νn. Also, the condition u = 0 for x > 0, y > 0, x2
+y2
= 1

can not be used.
72
72
ChiuYen’s solutions have attempts to solve this problem using Green’s function.

24 Problems: Separation of Variables - Wave Equation
Example (McOwen 3.1 #2). We considered the initial/boundary value problem and
solved it using Fourier Series. We now solve it using the Separation of Variables.
⎧
⎪⎨
⎪⎩
utt − uxx = 0 0 < x < π, t > 0
u(x, 0) = 1, ut(x, 0) = 0 0 < x < π
u(0, t) = 0, u(π, t) = 0 t ≥ 0.
(24.1)
Proof. Assume u(x, t) = X(x)T(t), then substitution in the PDE gives XT −X T = 0.
X
X
=
T
T
= −λ.
• From X + λX = 0, we get Xn(x) = an cos nx + bn sin nx. Boundary conditions
give
u(0, t) = X(0)T(t) = 0
u(π, t) = X(π)T(t) = 0
⇒ X(0) = X(π) = 0.
Thus, Xn(0) = an = 0, and Xn(x) = bn sinnx, λn = n2, n = 1, 2, . . ..
• With these values of λn, we solve T +n2
T = 0 to find Tn(t) = cn sinnt+dn cos nt.
Thus,
u(x, t) =
∞
n=1
˜cn sin nt + ˜dn cos nt sin nx,
ut(x, t) =
∞
n=1
n˜cn cos nt − n ˜dn sin nt sinnx.
• Initial conditions give
1 = u(x, 0) =
∞
n=1
˜dn sin nx,
0 = ut(x, 0) =
∞
n=1
n˜cn sinnx.
π
0
sin mx dx = ˜dm
π
2
,
π
0
0 dx = n˜cn
π
2
,
which gives the coefficients
˜dn =
2
nπ
(1 − cos nπ) =
4
nπ , n odd,
0, n even,
and ˜cn = 0.
Plugging the coefficients into a formula for u(x, t), we get
u(x, t) =
4
π
∞
n=0
(2n + 1)
.

Example. Use the method of separation of variables to ﬁnd the solution to:
⎧
⎪⎨
⎪⎩
utt + 3ut + u = uxx, 0 < x < 1
u(0, t) = 0, u(1, t) = 0,
u(x, 0) = 0, ut(x, 0) = x sin(2πx).
Proof. Assume u(x, t) = X(x)T(t), then substitution in the PDE gives
XT + 3XT + XT = X T,
T
T
+ 3
T
T
+ 1 =
X
X
= −λ.
• From X + λX = 0, Xn(x) = an cos
√
λnx + bn sin
√
λnx. Boundary conditions
give
u(0, t) = X(0)T(t) = 0
u(1, t) = X(1)T(t) = 0
⇒ X(0) = X(1) = 0.
Thus, Xn(0) = an = 0, and Xn(x) = bn sin
√
λnx.
Xn(1) = bn sin
√
λn = 0. Hence,
√
λn = nπ, or λn = (nπ)2
, n = 1, 2, . . ..
λn = (nπ)2
, Xn(x) = bn sinnπx.
• With these values of λn, we solve
T + 3T + T = −λnT,
T + 3T + T = −(nπ)2
T,
T + 3T + (1 + (nπ)2
)T = 0.
We can solve this 2nd-order ODE with the following guess, T(t) = cest to obtain
s = −3
2 ± 5
4 − (nπ)2. For n ≥ 1, 5
4 − (nπ)2 < 0. Thus, s = −3
2 ± i (nπ)2 − 5
4.
Tn(t) = e−3
2
t
cn cos (nπ)2 −
5
4
t + dn sin (nπ)2 −
5
4
t .
u(x, t) = X(x)T(t) =
∞
n=1
e−3
2
t
cn cos (nπ)2 −
5
4
t + dn sin (nπ)2 −
5
4
t sinnπx.
0 = u(x, 0) =
∞
n=1
cn sin nπx.
By orthogonality, we may multiply this equations by sin mπx and integrate:
1
0
0 dx =
1
2
cm ⇒ cm = 0.

Thus,
u(x, t) =
∞
n=1
dne−3
2
t
sin (nπ)2 −
5
4
t sinnπx.
ut(x, t) =
∞
n=1
−
3
2
dne−3
2
t
sin (nπ)2 −
5
4
t + dne−3
2
t
(nπ)2 −
5
4
cos (nπ)2 −
5
4
t sinnπx,
x sin(2πx) = ut(x, 0) =
∞
n=1
dn (nπ)2 −
5
4
sinnπx.
By orthogonality, we may multiply this equations by sin mπx and integrate:
1
0
x sin(2πx) sin(mπx) dx = dm
1
2
(mπ)2 −
5
4
,
dn =
2
(nπ)2 − 5
4
1
0
x sin(2πx) sin(nπx) dx.
u(x, t) = e−3
2
t
∞
n=1
dn sin (nπ)2 −
5
4
t sin nπx.
Problem (F’04, #1). Solve the following initial-boundary value problem for the wave
equation with a potential term,
⎧
⎪⎨
⎪⎩
utt − uxx + u = 0 0 < x < π, t < 0
u(0, t) = u(π, t) = 0 t > 0
u(x, 0) = f(x), ut(x, 0) = 0 0 < x < π,
where
f(x) =
x if x ∈ (0, π/2),
π − x if x ∈ (π/2, π).
The answer should be given in terms of an inﬁnite series of explicitly given functions.
Proof. Assume u(x, t) = X(x)T(t), then substitution in the PDE gives
XT − X T + XT = 0,
T
T
+ 1 =
X
X
= −λ.
• From X + λX = 0, Xn(x) = an cos
√
λnx + bn sin
√
λnx. Boundary conditions
give
u(0, t) = X(0)T(t) = 0
u(π, t) = X(π)T(t) = 0
⇒ X(0) = X(π) = 0.
Thus, Xn(0) = an = 0, and Xn(x) = bn sin
√
λnx.
Xn(π) = bn sin
√
λnπ = 0. Hence,
√
λn = n, or λn = n2
, n = 1, 2, . . ..
λn = n2
, Xn(x) = bn sinnx.

• With these values of λn, we solve
T + T = −λnT,
T + T = −n2
T,
Tn + (1 + n2
)Tn = 0.
The solution to this 2nd-order ODE is of the form:
Tn(t) = cn cos 1 + n2 t + dn sin 1 + n2 t.
u(x, t) = X(x)T(t) =
∞
n=1
cn cos 1 + n2 t + dn sin 1 + n2 t sinnx.
ut(x, t) =
∞
n=1
− cn( 1 + n2) sin 1 + n2 t + dn( 1 + n2) cos 1 + n2 t sin nx.
f(x) = u(x, 0) =
∞
n=1
cn sin nx.
0 = ut(x, 0) =
∞
n=1
dn( 1 + n2) sinnx.
π
0
f(x) sinmx dx = cm
π
2
,
π
0
0 dx = dm
π
2
1 + m2,
which gives the coeﬃcients
cn =
2
π
π
0
f(x) sinnx dx =
2
π
π
2
0
x sinnx dx +
2
π
π
π
2
(π − x) sinnx dx
=
2
π
− x
1
n
cos nx
π
2
0
+
1
n
π
2
0
cos nx dx +
2
π
−
π
n
cos nx
π
π
2
+ x
1
n
cos nx
π
π
2
−
1
n
π
π
2
cos nx dx
=
2
π
−
π
2n
cos
nπ
2
+
1
n2
sin
nπ
2
−
1
n2
sin0
+
2
π
−
π
n
cos nπ +
π
n
cos
nπ
2
+
π
n
cos nπ −
π
2n
cos
nπ
2
−
1
n2
sin nπ +
1
n2
sin
nπ
2
=
2
π
1
n2
sin
nπ
2
+
2
π
1
n2
sin
nπ
2
=
4
πn2
sin
nπ
2
=
⎧
⎪⎨
⎪⎩
0, n = 2k
4
πn2 , n = 4m + 1
− 4
πn2 , n = 4m + 3
=
0, n = 2k
(−1)
n−1
2
4
πn2 , n = 2k + 1.
dn = 0.
u(x, t) =
∞
n=1
cn cos 1 + n2 t sinnx.

25 Problems: Separation of Variables - Heat Equation
Solve the initial-boundary value problem
⎧
⎪⎨
⎪⎩
ut = uxx 0 < x < 2, t > 0
u(x, 0) = x2 − x + 1 0 ≤ x ≤ 2
u(0, t) = 1, u(2, t) = 3 t > 0.
Find limt→+∞ u(x, t).
Proof. ➀ First, we need to obtain function v that satisﬁes vt = vxx and takes 0
boundary conditions. Let
• v(x, t) = u(x, t) + (ax + b), (25.1)
where a and b are constants to be determined. Then,
vt = ut,
vxx = uxx.
Thus,
vt = vxx.
We need equation (25.1) to take 0 boundary conditions for v(0, t) and v(2, t):
v(0, t) = 0 = u(0, t) + b = 1 + b ⇒ b = −1,
v(2, t) = 0 = u(2, t) + 2a − 1 = 2a + 2 ⇒ a = −1.
v(x, t) = u(x, t) − x − 1. (25.2)
The new problem is
⎧
⎪⎨
⎪⎩
vt = vxx,
v(x, 0) = (x2
− x + 1) − x − 1 = x2
− 2x,
v(0, t) = v(2, t) = 0.
➁ We solve the problem for v using the method of separation of variables.
Let v(x, t) = X(x)T(t), which gives XT − X T = 0.
X
X
=
T
T
= −λ.
From X + λX = 0, we get Xn(x) = an cos
√
λx + bn sin
√
λx.
v(0, t) = X(0)T(t) = 0
v(2, t) = X(2)T(t) = 0
⇒ X(0) = X(2) = 0.
Hence, Xn(0) = an = 0, and Xn(x) = bn sin
√
λx.
Xn(2) = bn sin 2
√
λ = 0 ⇒ 2
√
λ = nπ ⇒ λn = (nπ
2 )2.
Xn(x) = bn sin
nπx
2
, λn =
nπ
2
2
.

With these values of λn, we solve T + nπ
2
2
T = 0 to ﬁnd
Tn(t) = cne−( nπ
2
)2t
.
v(x, t) =
∞
n=1
Xn(x)Tn(t) =
∞
n=1
˜cn e−( nπ
2
)2t
sin
nπx
2
.
Coeﬃcients ˜cn are obtained using the initial condition:
v(x, 0) =
∞
n=1
˜cn sin
nπx
2
= x2
− 2x.
˜cn =
2
0
(x2
− 2x) sin
nπx
2
dx =
0 n is even,
− 32
(nπ)3 n is odd.
⇒ v(x, t) =
∞
n=2k−1
−
32
(nπ)3
e−( nπ
2
)2t
sin
nπx
2
.
We now use equation (25.2) to convert back to function u:
u(x, t) = v(x, t) + x + 1.
u(x, t) =
∞
n=2k−1
−
32
(nπ)3
e−( nπ
2
)2t
sin
nπx
2
+ x + 1.
lim
t→+∞
u(x, t) = x + 1.

Let u(x, t) be the solution of the initial-boundary value problem for the heat equation
⎧
⎪⎨
⎪⎩
ut = uxx 0 < x < L, t > 0
u(x, 0) = f(x) 0 ≤ x ≤ L
ux(0, t) = ux(L, t) = A t > 0 (A = Const).
Find v(x) - the limit of u(x, t) when t → ∞. Show that v(x) is one of the iniﬁnitely
many solutions of the stationary problem
vxx = 0 0 < x < L
vx(0) = vx(L) = A.
• v(x, t) = u(x, t) + (ax + b), (25.3)
where a and b are constants to be determined. Then,
vt = ut,
vxx = uxx.
Thus,
vt = vxx.
We need equation (25.3) to take 0 boundary conditions for vx(0, t) and vx(L, t).
vx = ux + a.
vx(0, t) = 0 = ux(0, t) + a = A + a ⇒ a = −A,
vx(L, t) = 0 = ux(L, t) + a = A + a ⇒ a = −A.
We may set b = 0 (inﬁnitely many solutions are possible, one for each b).
v(x, t) = u(x, t) − Ax. (25.4)
The new problem is
⎧
⎪⎨
⎪⎩
vt = vxx,
v(x, 0) = f(x) − Ax,
vx(0, t) = vx(L, t) = 0.
X
X
=
T
T
= −λ.
√
λx + bn sin
√
λx.
vx(0, t) = X (0)T(t) = 0
vx(L, t) = X (L)T(t) = 0
⇒ X (0) = X (L) = 0.

Xn(x) = −an
√
λ sin
√
λx + bn
√
λ cos
√
λx.
Hence, Xn(0) = bn
√
λn = 0 ⇒ bn = 0; and Xn(x) = an cos
√
λx.
Xn(L) = −an
√
λ sinL
√
λ = 0 ⇒ L
√
λ = nπ ⇒ λn = (nπ
L )2
.
Xn(x) = an cos
nπx
L
, λn =
nπ
L
2
.
With these values of λn, we solve T + nπ
L
2
T = 0 to find
T0(t) = c0, Tn(t) = cne−( nπ
L
)2t
, n = 1, 2, . . ..
v(x, t) =
∞
n=1
Xn(x)Tn(t) = ˜c0 +
∞
n=1
˜cn e−( nπ
L
)2t
cos
nπx
L
.
Coefficients ˜cn are obtained using the initial condition:
v(x, 0) = ˜c0 +
∞
n=1
˜cn cos
nπx
L
= f(x) − Ax.
L˜c0 =
L
0
(f(x) − Ax) dx =
L
0
f(x) dx −
AL2
2
⇒ ˜c0 =
1
L
L
0
f(x) dx −
AL
2
,
L
2
˜cn =
L
0
(f(x) − Ax) cos
nπx
L
dx ⇒ ˜cn =
1
L
L
0
(f(x) − Ax) cos
nπx
L
dx.
⇒ v(x, t) =
1
L
L
0
f(x) dx −
AL
2
+
∞
n
˜cn e−( nπ
L
)2t
cos
nπx
L
.
u(x, t) = v(x, t) + Ax.
u(x, t) =
1
L
L
0
f(x) dx −
AL
2
+
∞
n
˜cn e−( nπ
L
)2t
cos
nπx
L
+ Ax.
lim
t→+∞
u(x, t) = Ax + b, b arbitrary.
To show that v(x) is one of the inifinitely many solutions of the stationary problem
vxx = 0 0 < x < L
vx(0) = vx(L) = A,
we can solve the boundary value problem to obtain v(x, t) = Ax+b, where b is arbitrary.

Heat Equation with Nonhomogeneous Time-Independent BC in N-dimensions.
The solution to this problem takes somewhat diﬀerent approach than in the last few prob-
lems, but is similar.
Consider the following initial-boundary value problem,
⎧
⎪⎨
⎪⎩
ut = u, x ∈ Ω, t ≥ 0
u(x, 0) = f(x), x ∈ Ω
u(x, t) = g(x), x ∈ ∂Ω, t > 0.
Proof. Let w(x) be the solution of the Dirichlet problem:
w = 0, x ∈ Ω
w(x) = g(x), x ∈ ∂Ω
and let v(x, t) be the solution of the IBVP for the heat equation with homogeneous
BC:
⎧
⎪⎨
⎪⎩
vt = v, x ∈ Ω, t ≥ 0
v(x, 0) = f(x) − w(x), x ∈ Ω
v(x, t) = 0, x ∈ ∂Ω, t > 0.
Then u(x, t) satisﬁes
u(x, t) = v(x, t) + w(x).
lim
t→∞
u(x, t) = w(x).

Nonhomogeneous Heat Equation with Nonhomogeneous Time-Independent
BC in N dimensions.
Describe the method of solution of the problem
⎧
⎪⎨
⎪⎩
ut = u + F(x, t), x ∈ Ω, t ≥ 0
u(x, 0) = f(x), x ∈ Ω
u(x, t) = g(x), x ∈ ∂Ω, t > 0.
Proof. ❶ We first find u1, the solution to the homogeneous heat equation (no F(x, t)).
Let w(x) be the solution of the Dirichlet problem:
w = 0, x ∈ Ω
w(x) = g(x), x ∈ ∂Ω
and let v(x, t) be the solution of the IBVP for the heat equation with homogeneous
BC:
⎧
⎪⎨
⎪⎩
vt = v, x ∈ Ω, t ≥ 0
v(x, 0) = f(x) − w(x), x ∈ Ω
v(x, t) = 0, x ∈ ∂Ω, t > 0.
Then u1(x, t) satisfies
u1(x, t) = v(x, t) + w(x).
lim
t→∞
u1(x, t) = w(x).
❷ The solution to the homogeneous equation with 0 boundary conditions is given by
Duhamel’s principle.
u2t = u2 + F(x, t) for t > 0, x ∈ Rn
u2(x, 0) = 0 for x ∈ Rn.
(25.5)
Duhamel’s principle gives the solution:
u2(x, t) =
t
0 Rn
˜K(x − y, t − s) F(y, s) dy ds
Note: u2(x, t) = 0 on ∂Ω may not be satisfied.
u(x, t) = v(x, t) + w(x) +
t
0 Rn
˜K(x − y, t − s) F(y, s) dy ds.

Problem (S’98, #5). Find the solution of
⎧
⎪⎨
⎪⎩
ut = uxx, t ≥ 0, 0 < x < 1,
u(x, 0) = 0, 0 < x < 1,
u(0, t) = 1 − e−t, ux(1, t) = e−t − 1, t > 0.
Prove that limt→∞ u(x, t) exists and ﬁnd it.
• v(x, t) = u(x, t) + (ax + b) + (c1 cos x + c2 sin x)e−t
, (25.6)
where a, b, c1, c2 are constants to be determined. Then,
vt = ut − (c1 cos x + c2 sinx)e−t
,
vxx = uxx + (−c1 cos x − c2 sin x)e−t
.
Thus,
vt = vxx.
We need equation (25.6) to take 0 boundary conditions for v(0, t) and vx(1, t):
v(0, t) = 0 = u(0, t) + b + c1e−t
= 1 − e−t
+ b + c1e−t
.
Thus, b = −1, c1 = 1, and (25.6) becomes
v(x, t) = u(x, t) + (ax − 1) + (cos x + c2 sin x)e−t
. (25.7)
vx(x, t) = ux(x, t) + a + (− sinx + c2 cos x)e−t
,
vx(1, t) = 0 = ux(1, t) + a + (− sin1 + c2 cos 1)e−t
= −1 + a + (1 − sin 1 + c2 cos 1)e−t
.
Thus, a = 1, c2 = sin 1−1
cos 1 , and equation (25.7) becomes
v(x, t) = u(x, t) + (x − 1) + (cos x +
sin 1 − 1
cos 1
sin x)e−t
. (25.8)
Initial condition tranforms to:
v(x, 0) = u(x, 0) + (x − 1) + (cos x +
sin 1 − 1
cos 1
sin x) = (x − 1) + (cos x +
sin1 − 1
cos 1
sinx).
The new problem is
⎧
⎪⎨
⎪⎩
vt = vxx,
v(x, 0) = (x − 1) + (cos x + sin 1−1
cos 1 sinx),
v(0, t) = 0, vx(1, t) = 0.
X
X
=
T
T
= −λ.

√
λx + bn sin
√
λx.
Using the first boundary condition, we have
v(0, t) = X(0)T(t) = 0 ⇒ X(0) = 0.
Hence, Xn(0) = an = 0, and Xn(x) = bn sin
√
λx. We also have
vx(1, t) = X (1)T(t) = 0 ⇒ X (1) = 0.
Xn(x) =
√
λbn cos
√
λx,
Xn(1) =
√
λbn cos
√
λ = 0,
cos
√
λ = 0,
√
λ = nπ +
π
2
.
Thus,
Xn(x) = bn sin nπ +
π
2
x, λn = nπ +
π
2
2
.
With these values of λn, we solve T + nπ + π
2
2
T = 0 to find
Tn(t) = cne−(nπ+π
2
)2t
.
v(x, t) =
∞
n=1
Xn(x)Tn(t) =
∞
n=1
˜bn sin nπ +
π
2
x e−(nπ+π
2
)2t
.
u(x, t) = v(x, t) − (x − 1) − (cos x +
sin 1 − 1
cos 1
sin x)e−t
.
u(x, t) =
∞
n=1
˜bn sin nπ +
π
2
x e−(nπ+π
2
)2t
− (x − 1) − (cos x +
sin 1 − 1
cos 1
sin x)e−t
.
Coefficients ˜bn are obtained using the initial condition:
u(x, 0) =
∞
n=1
˜bn sin nπ +
π
2
x − (x − 1) − (cos x +
sin1 − 1
cos 1
sinx).
➂ Finally, we can check that the differential equation and the boundary conditions are
satisfied:
u(0, t) = 1 − (1 + 0)e−t
= 1 − e−t
.
ux(x, t) =
∞
n=1
˜bn nπ +
π
2
cos nπ +
π
2
x e−(nπ+π
2
)2t
− 1 + (sinx −
sin1 − 1
cos 1
cos x)e−t
,
ux(1, t) = −1 + (sin1 −
sin 1 − 1
cos 1
cos 1)e−t
= −1 + e−t
.
ut =
∞
n=1
−˜bn nπ +
π
2
2
sin nπ +
π
2
x e−(nπ+π
2
)2t
+ (cos x +
sin1 − 1
cos 1
sin x)e−t
= uxx.

Problem (F’02, #6). The temperature of a rod insulated at the ends with an ex-
ponentially decreasing heat source in it is a solution of the following boundary value
problem:
⎧
⎪⎨
⎪⎩
ut = uxx + e−2t
g(x) for (x, t) ∈ [0, 1] × R+
ux(0, t) = ux(1, t) = 0
u(x, 0) = f(x).
Find the solution to this problem by writing u as a cosine series,
u(x, t) =
∞
n=0
an(t) cos nπx,
and determine limt→∞ u(x, t).
Proof. Let g accept an expansion in eigenfunctions
g(x) = b0 +
∞
n=1
bn cos nπx with bn = 2
1
0
g(x) cosnπx dx.
Plugging in the PDE gives:
a0(t) +
∞
n=1
an(t) cosnπx = −
∞
n=1
n2
π2
an(t) cos nπx + b0e−2t
+ e−2t
∞
n=1
bn cos nπx,
which gives
a0(t) = b0e−2t,
an(t) + n2π2an(t) = bne−2t, n = 1, 2, . . ..
Adding homogeneous and particular solutions of the above ODEs, we obtain the solu-
tions
a0(t) = c0 − b0
2 e−2t
,
an(t) = cne−n2π2t
− bn
2−n2π2 e−2t
, n = 1, 2, . . .,
for some constants cn, n = 0, 1, 2, . . .. Thus,
u(x, t) =
∞
n=0
cne−n2π2t
−
bn
2 − n2π2
e−2t
cos nπx.
Initial condition gives
u(x, 0) =
∞
n=0
cn −
bn
2 − n2π2
cos nπx = f(x),
As, t → ∞, the only mode that survives is n = 0:
u(x, t) → c0 +
b0
2
as t → ∞.

Problem (F’93, #4). a) Assume f, g ∈ C∞
. Give the compatibility conditions which
f and g must satisfy if the following problem is to possess a solution.
u = f(x) x ∈ Ω
∂u
∂n
(s) = g(s) s ∈ ∂Ω.
Show that your condition is necessary for a solution to exist.
b) Give an explicit solution to
⎧
⎪⎨
⎪⎩
ut = uxx + cos x x ∈ [0, 2π]
ux(0, t) = ux(2π, t) = 0 t > 0
u(x, 0) = cos x + cos 2x x ∈ [0, 2π].
c) Does there exist a steady state solution to the problem in (b) if
ux(0) = 1 ux(2π) = 0 ?
Explain your answer.
Proof. a) Integrating the equation and using Green’s identity gives:
Ω
f(x) dx =
Ω
u dx =
∂Ω
∂u
∂n
ds =
∂Ω
g(s) ds.
b) With
• v(x, t) = u(x, t) − cos x
the problem above transforms to
⎧
⎪⎨
⎪⎩
vt = vxx
vx(0, t) = vx(2π, t) = 0
v(x, 0) = cos 2x.
We solve this problem for v using the separation of variables. Let v(x, t) = X(x)T(t),
which gives XT = X T.
X
X
=
T
T
= −λ.
√
λx + bn sin
√
λx.
Xn(x) = −
√
λnan sin
√
λx +
√
λnbn cos
√
λx.
vx(0, t) = X (0)T(t) = 0
vx(2π, t) = X (2π)T(t) = 0
⇒ X (0) = X (2π) = 0.
Hence, Xn(0) =
√
λnbn = 0, and Xn(x) = an cos
√
λnx.
Xn(2π) = −
√
λnan sin
√
λn2π = 0 ⇒
√
λn = n
2 ⇒ λn = (n
2 )2. Thus,
Xn(x) = an cos
nx
2
, λn =
n
2
2

With these values of λn, we solve T + n
2
2
T = 0 to find
Tn(t) = cne−( n
2
)2t
.
v(x, t) =
∞
n=0
Xn(x)Tn(t) =
∞
n=0
ãn e−( n
2
)2t
cos
nx
2
.
Initial condition gives
v(x, 0) =
∞
n=0
ãn cos
nx
2
= cos 2x.
Thus, ã4 = 1, ãn = 0, n = 4. Hence,
v(x, t) = e−4t
cos 2x.
u(x, t) = v(x, t) + cos x = e−4t
cos 2x + cos x.
c) Does there exist a steady state solution to the problem in (b) if
ux(0) = 1 ux(2π) = 0 ?
Explain your answer.
c) Set ut = 0. We have
uxx + cos x = 0 x ∈ [0, 2π]
ux(0) = 1, ux(2π) = 0.
uxx = − cos x,
ux = − sin x + C,
u(x) = cos x + Cx + D.
Boundary conditions give:
1 = ux(0) = C,
0 = ux(2π) = C ⇒ contradiction
There exists no steady state solution.
We may use the result we obtained in part (a) with uxx = cos x = f(x). We
need
Ω
f(x) dx =
∂Ω
∂u
∂n
ds,
2π
0
cos x dx
=0
= ux(2π) − ux(0) = −1
given
.

Problem (F’96, #7). Solve the parabolic problem
u
v t
=
1 1
2
0 2
u
v xx
, 0 ≤ x ≤ π, t > 0
u(x, 0) = sinx, u(0, t) = u(π, t) = 0,
v(x, 0) = sin x, v(0, t) = v(π, t) = 0.
Prove the energy estimate (for general initial data)
π
x=0
[u2
(x, t) + v2
(x, t)] dx ≤ c
π
x=0
[u2
(x, 0) + v2
(x, 0)] dx
for come constant c.
Proof. We can solve the second equation for v and then use the value of v to solve the
first equation for u. 73
➀ We have
⎧
⎪⎨
⎪⎩
vt = 2vxx, 0 ≤ x ≤ π, t > 0
v(x, 0) = sinx,
v(0, t) = v(π, t) = 0.
Assume v(x, t) = X(x)T(t), then substitution in the PDE gives XT = 2X T.
T
T
= 2
X
X
= −λ.
From X + λ
2 X = 0, we get Xn(x) = an cos λ
2 x + bn sin λ
2 x.
v(0, t) = X(0)T(t) = 0
v(π, t) = X(π)T(t) = 0
⇒ X(0) = X(π) = 0.
Thus, Xn(0) = an = 0, and Xn(x) = bn sin λ
2 x.
Xn(π) = bn sin λ
2 π = 0. Hence λ
2 = n, or λ = 2n2
.
λ = 2n2
, Xn(x) = bn sinnx.
With these values of λn, we solve T + 2n2T = 0 to get Tn(t) = cne−2n2t.
Thus, the solution may be written in the form
v(x, t) =
∞
n=1
ãne−2n2t
sin nx.
From initial condition, we get
v(x, 0) =
∞
n=1
ãn sinnx = sinx.
Thus, ã1 = 1, ãn = 0, n = 2, 3, . . ..
v(x, t) = e−2t
sin x.
73
Note that if the matrix was fully inseparable, we would have to find eigenvalues and eigenvectors,
just as we did for the hyperbolic systems.

➁ We have
⎧
⎪⎨
⎪⎩
ut = uxx − 1
2 e−2t
sinx, 0 ≤ x ≤ π, t > 0
u(x, 0) = sin x,
u(0, t) = u(π, t) = 0.
Let u(x, t) = ∞
n=1 un(t) sinnx. Plugging this into the equation, we get
∞
n=1
un(t) sinnx +
∞
n=1
n2
un(t) sinnx = −
1
2
e−2t
sin x.
For n = 1:
u1(t) + u1(t) = −
1
2
e−2t
.
Combining homogeneous and particular solution of the above equation, we obtain:
u1(t) =
1
2
e−2t
+ c1e−t
.
For n = 2, 3, . . .:
un(t) + n2
un(t) = 0,
un(t) = cne−n2t
.
Thus,
u(x, t) =
1
2
e−2t
+ c1e−t
sinx +
∞
n=2
cne−n2t
sinnx =
1
2
e−2t
sinx +
∞
n=1
cne−n2t
sinnx.
From initial condition, we get
u(x, 0) =
1
2
sin x +
∞
n=1
cn sinnx = sin x.
Thus, c1 = 1
2 , cn = 0, n = 2, 3, . . ..
u(x, t) =
1
2
sinx (e−2t
+ e−t
).
To prove the energy estimate (for general initial data)
π
x=0
[u2
(x, t) + v2
(x, t)] dx ≤ c
π
x=0
[u2
(x, 0) + v2
(x, 0)] dx
for come constant c, we assume that
u(x, 0) =
∞
n=1
an sinnx, v(x, 0) =
∞
n=1
bn sinnx.

The general solutions are obtained by the same method as above
u(x, t) =
1
2
e−2t
sinx +
∞
n=1
cne−n2 t
sinnx,
v(x, t) =
∞
n=1
bne−2n2t
sinnx.
π
x=0
[u2
(x, t) + v2
(x, t)] dx =
π
x=0
1
2
e−2t
sinx +
∞
n=1
cne−n2t
sinnx
2
+
∞
n=1
bne−2n2t
sinnx
2
dx
≤
∞
n=1
(b2
n + a2
n)
π
x=0
sin2
nx dx ≤
π
x=0
[u2
(x, 0) + v2
(x, 0)] dx.

26 Problems: Eigenvalues of the Laplacian - Laplace
The 2D LAPLACE Equation (eigenvalues/eigenfuctions of the Laplacian).
Consider
⎧
⎪⎨
⎪⎩
uxx + uyy + λu = 0 in Ω
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
(26.1)
Proof. We can solve this problem by separation of variables.
Let u(x, y) = X(x)Y (y), then substitution in the PDE gives X Y + XY + λXY = 0.
X
X
+
Y
Y
+ λ = 0.
Letting λ = μ2 + ν2 and using boundary conditions, we ﬁnd the equations for X and
Y :
X + μ2
X = 0 Y + ν2
Y = 0
X(0) = X(a) = 0 Y (0) = Y (b) = 0.
The solutions of these one-dimensional eigenvalue problems are
μm =
mπ
a
νn =
nπ
b
Xm(x) = sin
mπx
a
Yn(y) = sin
nπy
b
,
where m, n = 1, 2, . . .. Thus we obtain solutions of (26.1) of the form
λmn = π2 m2
a2
+
n2
b2
umn(x, y) = sin
mπx
a
sin
nπy
b
,
where m, n = 1, 2, . . ..
Observe that the eigenvalues {λmn}∞
m,n=1 are positive. The smallest eigenvalue λ11
has only one eigenfunction u11(x, y) = sin(πx/a) sin(πy/b); notice that u11 is positive
in Ω. Other eigenvalues λ may correspond to more than one choice of m and n; for
example, in the case a = b we have λnm = λnm. For this λ, there are two linearly
independent eigenfunctions. However, for a particular value of λ there are at most
ﬁnitely many linearly independent eigenfunctions. Moreover,
b
0
a
0
umn(x, y) um n (x, y) dx dy =
b
0
a
0
sin
mπx
a
sin
nπy
b
sin
m πx
a
sin
n πy
b
dx dy
=
a
2
b
0 sin nπy
b sin n πy
b dy
0
=
ab
4 if m = m and n = n
0 if m = m or n = n .
In particular, the {umn} are pairwise orthogonal. We could normalize each umn by a
scalar multiple (i.e. multiply by 4/ab) so that ab/4 above becomes 1.
Let us change the notation somewhat so that each eigenvalue λn corresponds to a
particular eigenfunction φn(x). If we choose an orthonormal basis of eigenfunctions in
each eigenspace, we may arrange that {φn}∞
n=1 is pairwise orthonormal:
Ω
φn(x)φm(x) dx =
1 if m = n
0 if m = n.

In this notation, the eigenfunction expansion of f(x) deﬁned on Ω becomes
f(x) ∼
∞
n=1
anφn(x), where an =
Ω
f(x)φn(x) dx.

Problem (S’96, #4). Let D denote the rectangular
D = {(x, y) ∈ R2
: 0 < x < a, 0 < y < b}.
Find the eigenvalues of the following Dirichlet problem:
( + λ)u = 0 in D
u = 0 on ∂D.
Proof. The problem may be rewritten as
⎧
⎪⎨
⎪⎩
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
We may assume that the eigenvalues λ are positive, λ = μ2
+ ν2
. Then,
λmn = π2 m2
a2
+
n2
b2
umn(x, y) = sin
mπx
a
sin
nπy
b
, m, n = 1, 2, . . ..
Problem (W’04, #1). Consider the diﬀerential equation:
∂2u(x, y)
∂x2
+
∂2u(x, y)
∂y2
+ λu(x, y) = 0 (26.2)
in the strip {(x, y), 0 < y < π, −∞ < x < +∞} with boundary conditions
u(x, 0) = 0, u(x, π) = 0. (26.3)
Find all bounded solutions of the boundary value problem (26.4), (26.5) when
a) λ = 0, b) λ > 0, c) λ < 0.
Proof. a) λ = 0. We have
uxx + uyy = 0.
Assume u(x, y) = X(x)Y (y), then substitution in the PDE gives
X Y + XY = 0.
u(x, 0) = X(x)Y (0) = 0
u(x, π) = X(x)Y (π) = 0
⇒ Y (0) = Y (π) = 0.
Method I: We have
X
X
= −
Y
Y
= −c, c > 0.
From X + cX = 0, we have Xn(x) = an cos
√
cx + bn sin
√
cx.
From Y − cY = 0, we have Yn(y) = cne−
√
cy
+ dne
√
cy
.
Y (0) = cn + dn = 0 ⇒ cn = −dn.

Y (π) = cne−
√
cπ
− cne
√
cπ
= 0 ⇒ cn = 0 ⇒ Yn(y) = 0.
⇒ u(x, y) = X(x)Y (y) = 0.
Method II: We have
X
X
= −
Y
Y
= c, c > 0.
From X − cX = 0, we have Xn(x) = ane−
√
cx + bne
√
cx.
Since we look for bounded solutions for −∞ < x < ∞, an = bn = 0 ⇒ Xn(x) = 0.
From Y + cY = 0, we have Yn(y) = cn cos
√
cy + dn sin
√
cy.
Y (0) = cn = 0,
Y (π) = dn sin
√
cπ = 0 ⇒
√
c = n ⇒ c = n2.
⇒ Yn(y) = dn sin nx = 0.
⇒ u(x, y) = X(x)Y (y) = 0.
b) λ > 0. We have
X
X
+
Y
Y
+ λ = 0.
Letting λ = μ2 + ν2, and using boundary conditions for Y , we ﬁnd the equations:
X + μ2
X = 0 Y + ν2
Y = 0
Y (0) = Y (π) = 0.
Xm(x) = am cos μmx + bm sinμmx.
νn = n, Yn(y) = dn sin ny, where m, n = 1, 2, . . ..
u(x, y) =
∞
m,n=1
umn(x, y) =
∞
m,n=1
(am cos μmx + bm sinμmx) sinny.
c) λ < 0. We have
uxx + uyy + λu = 0,
u(x, 0) = 0, u(x, π) = 0.
u ≡ 0 is the solution to this equation. We will show that this solution is unique.
Let u1 and u2 be two solutions, and consider w = u1 − u2. Then,
w + λw = 0,
w(x, 0) = 0, w(x, π) = 0.
Multiply the equation by w and integrate:
w w + λw2
= 0,
Ω
w w dx + λ
Ω
w2
dx = 0,
∂Ω
w
∂w
∂n
ds
=0
−
Ω
|∇w|2
dx + λ
Ω
w2
dx = 0,
Ω
|∇w|2
dx
≥0
= λ
Ω
w2
dx
≤0
.

Thus, w ≡ 0 and the solution u(x, y) ≡ 0 is unique.

Problem (F’95, #5). Find all bounded solutions
for the following boundary value problem in the strip
0 < x < a, −∞ < y < ∞,
( + k2
)u = 0 (k = Const > 0),
u(0, y) = 0, ux(a, y) = 0.
In particular, show that when ak ≤ π,
the only bounded solution to this problem is u ≡ 0.
Proof. Let u(x, y) = X(x)Y (y), then we have X Y + XY + k2XY = 0.
X
X
+
Y
Y
+ k2
= 0.
Letting k2
= μ2
+ ν2
and using boundary conditions, we ﬁnd:
X + μ2
X = 0, Y + ν2
Y = 0.
X(0) = X (a) = 0.
μm =
(m − 1
2 )π
a
,
Xm(x) = sin
(m − 1
2 )πx
a
Yn(y) = cn cos νny + dn sinνny,
where m, n = 1, 2, . . .. Thus we obtain solutions of the form
k2
mn =
(m − 1
2 )π
a
2
+ν2
n, umn(x, y) = sin
(m − 1
2)πx
a
cn cos νny+dn sin νny ,
where m, n = 1, 2, . . ..
u(x, y) =
∞
m,n=1
umn(x, y) =
∞
m,n=1
sin
(m − 1
2 )πx
a
cn cos νny + dn sinνny .
• We can take an alternate approach and prove the second part of the question. We
have
X Y + XY + k2
XY = 0,
−
Y
Y
=
X
X
+ k2
= c2
.
We obtain Yn(y) = cn cos cy + dn sin cy. The second equation gives
X + k2
X = c2
X,
X + (k2
− c2
)X = 0,
Xm(x) = ame
√
c2−k2x
+ bme
√
c2−k2x
.
Thus, Xm(x) is bounded only if k2 − c2 > 0, (if k2 − c2 = 0, X = 0, and Xm(x) =
amx + bm, BC’s give Xm(x) = πx, unbounded), in which case
Xm(x) = am cos k2 − c2 x + bm sin k2 − c2 x.

Boundary conditions give Xm(0) = am = 0.
Xm(x) = bm k2 − c2 cos k2 − c2 x,
Xm(a) = bm k2 − c2 cos k2 − c2 a = 0,
k2 − c2 a = mπ −
π
2
, m = 1, 2, . . .,
k2
− c2
=
π
a
m −
1
2
2
,
k2
=
π
a
2
m −
1
2
2
+ c2
,
a2
k2
> π2
m −
1
2
2
,
ak > π m −
1
2
, m = 1, 2, . . ..
Thus, bounded solutions exist only when ak > π
2 .
Problem (S’90, #2). Show that the boundary value problem
∂2u(x, y)
∂x2
+
∂2u(x, y)
∂y2
+ k2
u(x, y) = 0, (26.4)
where −∞ < x < +∞, 0 < y < π, k > 0 is a constant,
u(x, 0) = 0, u(x, π) = 0 (26.5)
has a bounded solution if and only if k ≥ 1.
Proof. We have
uxx + uyy + k2
u = 0,
X Y + XY + k2
XY = 0,
−
X
X
=
Y
Y
+ k2
= c2
.
We obtain Xm(x) = am cos cx + bm sin cx. The second equation gives
Y + k2
Y = c2
Y,
Y + (k2
− c2
)Y = 0,
Yn(y) = cne
√
c2−k2y
+ dne
√
c2−k2y
.
Thus, Yn(y) is bounded only if k2
−c2
> 0, (if k2
−c2
= 0, Y = 0, and Yn(y) = cny+dn,
BC’s give Y ≡ 0), in which case
Yn(y) = cn cos k2 − c2 y + dn sin k2 − c2 y.
Boundary conditions give Yn(0) = cn = 0.
Yn(π) = dn sin
√
k2 − c2 π = 0 ⇒
√
k2 − c2 = n ⇒ k2 − c2 = n2 ⇒
k2
= n2
+ c2
, n = 1, 2, . . .. Hence, k > n, n = 1, 2, . . ..

Thus, bounded solutions exist if k ≥ 1.
Note: If k = 1, then c = 0, which gives trivial solutions for Yn(y).
u(x, y) =
∞
m,n=1
Xm(x)Yn(y) =
∞
m,n=1
sin ny Xm(x).

McOwen, 4.4 #7; 266B Ralston Hw. Show that the boundary value problem
−∇ · a(x)∇u + b(x)u = λu in Ω
u = 0 on ∂Ω
has only trivial solution with λ ≤ 0, when b(x) ≥ 0 and a(x) > 0 in Ω.
Proof. Multiplying the equation by u and integrating over Ω, we get
Ω
−u∇ · a∇u dx +
Ω
bu2
dx = λ
Ω
u2
dx.
Since ∇ · (ua∇u) = u∇ · a∇u + a|∇u|2
, we have
Ω
−∇ · (ua∇u) dx +
Ω
a|∇u|2
dx +
Ω
bu2
dx = λ
Ω
u2
dx. (26.6)
Using divergence theorem, we obtain
∂Ω
− u
=0
a
∂u
∂n
ds +
Ω
a|∇u|2
dx +
Ω
bu2
dx = λ
Ω
u2
dx,
Ω
a
>0
|∇u|2
dx +
Ω
b
≥0
u2
dx = λ
≤0
Ω
u2
dx,
Thus, ∇u = 0 in Ω, and u is constant. Since u = 0 on ∂Ω, u ≡ 0 on Ω.
Similar Problem I: Note that this argument also works with Neumann B.C.:
−∇ · a(x)∇u + b(x)u = λu in Ω
∂u/∂n = 0 on ∂Ω
Using divergence theorem, (26.6) becomes
∂Ω
−ua
∂u
∂n
=0
ds +
Ω
a|∇u|2
dx +
Ω
bu2
dx = λ
Ω
u2
dx,
Ω
a
>0
|∇u|2
dx +
Ω
b
≥0
u2
dx = λ
≤0
Ω
u2
dx.
Thus, ∇u = 0, and u = const on Ω. Hence, we now have
Ω
b
≥0
u2
dx = λ
≤0
Ω
u2
dx,
which implies λ = 0. This gives the useful information that for the eigenvalue problem74
−∇ · a(x)∇u + b(x)u = λu
∂u/∂n = 0,
λ = 0 is an eigenvalue, its eigenspace is the set of constants, and all other λ’s are
positive.
74
In Ralston’s Hw#7 solutions, there is no ‘-’ sign in front of ∇ · a(x)∇u below, which is probably a
typo.

Similar Problem II: If λ ≤ 0, we show that the only solution to the problem below
is the trivial solution.
u + λu = 0 in Ω
u = 0 on ∂Ω
Ω
u u dx + λ
Ω
u2
dx = 0,
∂Ω
u
=0
∂u
∂n
ds −
Ω
|∇u|2
dx + λ
≤0
Ω
u2
dx = 0.
Thus, ∇u = 0 in Ω, and u is constant. Since u = 0 on ∂Ω, u ≡ 0 on Ω.

27 Problems: Eigenvalues of the Laplacian - Poisson
The ND POISSON Equation (eigenvalues/eigenfunctions of the Laplacian).
Suppose we want to ﬁnd the eigenfunction expansion of the solution of
u = f in Ω
u = 0 on ∂Ω,
when f has the expansion in the orthonormal Dirichlet eigenfunctions φn:
f(x) ∼
∞
n=1
anφn(x), where an =
Ω
f(x)φn(x) dx.
Proof. Writing u = cnφn and inserting into −λu = f, we get
∞
n=1
−λncnφn =
∞
n=1
anφn(x).
Thus, cn = −an/λn, and
u(x) = −
∞
n=1
anφn(x)
λn
.
The 1D POISSON Equation (eigenvalues/eigenfunctions of the Laplacian).
For the boundary value problem
u = f(x)
u(0) = 0, u(L) = 0,
the related eigenvalue problem is
φ = −λφ
φ(0) = 0, φ(L) = 0.
The eigenvalues are λn = (nπ/L)2, and the corresponding eigenfunctions are sin(nπx/L),
n = 1, 2, . . ..
Writing u = cnφn = cn sin(nπx/L) and inserting into −λu = f, we get
∞
n=1
−cn
nπ
L
2
sin
nπx
L
= f(x),
L
0
∞
n=1
−cn
nπ
L
2
sin
nπx
L
sin
mπx
L
dx =
L
0
f(x) sin
mπx
L
dx,
−cn
nπ
L
2 L
2
=
L
0
f(x) sin
nπx
L
dx,
cn = −
2
L
L
0 f(x) sin(nπx/L) dx
(nπ/L)2
.

u(x) = cn sin(nπx/L) =
∞
n=1
−
2
L
L
0 f(ξ) sin(nπx/L) sin(nπξ/L) dξ
(nπ/L)2
,
u =
L
0
f(ξ) −
2
L
∞
n=1
sin(nπx/L) sin(nπξ/L)
(nπ/L)2
= G(x,ξ)
dξ.
See similar, but more complicated, problem in Sturm-Liouville Problems (S’92, #2(c)).

Example: Eigenfunction Expansion of the GREEN’s Function.
Suppose we ﬁx x and attempt to expand the Green’s function G(x, y) in the orthonormal
eigenfunctions φn(y):
G(x, y) ∼
∞
n=1
an(x)φn(y), where an(x) =
Ω
G(x, z)φn(z) dz.
Proof. We can rewrite u + λu = 0 in Ω, u = 0 on ∂Ω, as an integral equation 75
u(x) + λ
Ω
G(x, y)u(y) dy = 0.
Suppose, u(x) = cnφn(x). Plugging this into , we get
∞
m=1
cmφm(x) + λ
Ω
∞
n=1
an(x)φn(y)
∞
m=1
cmφm(y) dy = 0,
∞
m=1
cmφm(x) + λ
∞
n=1
an(x)
∞
m=1
cm
Ω
φn(y)φm(y) dy = 0,
∞
n=1
cnφn(x) +
∞
n=1
λan(x)cn = 0,
∞
n=1
cn φn(x) + λan(x) = 0,
an(x) = −
φn(x)
λn
.
Thus,
G(x, y) ∼
∞
n=1
−
φn(x)φn(y)
λn
.
75
See the section: ODE - Integral Equations.

The 2D POISSON Equation (eigenvalues/eigenfunctions of the Laplacian).
Solve the boundary value problem
⎧
⎪⎨
⎪⎩
uxx + uyy = f(x, y) for 0 < x < a, 0 < y < b
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
(27.1)
f(x, y) ∈ C2, f(x, y) = 0 if x = 0, x = a, y = 0, y = b,
f(x, y) =
2
√
ab
∞
m,n=1
cmn sin
mπx
a
sin
nπy
b
.
Proof. ➀ First, we ﬁnd eigenvalues/eigenfunctions of the Laplacian.
⎧
⎪⎨
⎪⎩
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
X
X
+
Y
Y
+ λ = 0.
Y :
X + μ2
X = 0 Y + ν2
Y = 0
X(0) = X(a) = 0 Y (0) = Y (b) = 0.
μm =
mπ
a
νn =
nπ
b
Xm(x) = sin
mπx
a
Yn(y) = sin
nπy
b
,
where m, n = 1, 2, . . .. Thus we obtain eigenvalues and normalized eigenfunctions of
the Laplacian:
λmn = π2 m2
a2
+
n2
b2
φmn(x, y) =
2
√
ab
sin
mπx
a
sin
nπy
b
,
where m, n = 1, 2, . . .. Note that
f(x, y) =
∞
m,n=1
cmnφmn.
➁ Second, writing u(x, y) = ˜cmnφmn and inserting into −λu = f, we get
−
∞
m,n=1
λmn˜cmnφmn(x, y) =
∞
m,n=1
cmnφmn(x, y).
Thus, ˜cmn = − cmn
λmn
.
u(x, y) = −
∞
n=1
cmn
λmn
φmn(x, y),

with λmn, φmn(x) given above, and cmn given by
b
0
a
0
f(x, y)φmn dx dy =
b
0
a
0
∞
m ,n =1
cm n φm n φmn dx dy = cmn.

28 Problems: Eigenvalues of the Laplacian - Wave
In the section on the wave equation, we considered an initial boundary value problem
for the one-dimensional wave equation on an interval, and we found that the solu-
tion could be obtained using Fourier series. If we replace the Fourier series by an
expansion in eigenfunctions, we can consider an initial/boundary value problem for the
n-dimensional wave equation.
The ND WAVE Equation (eigenvalues/eigenfunctions of the Laplacian).
Consider
⎧
⎪⎨
⎪⎩
utt = u for x ∈ Ω, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x ∈ Ω
u(x, t) = 0 for x ∈ ∂Ω, t > 0.
Proof. For g, h ∈ C2(Ω) with g = h = 0 on ∂Ω, we have eigenfunction expansions
g(x) =
∞
n=1
anφn(x) and h(x) =
∞
n=1
bnφn(x).
Assume the solution u(x, t) may be expanded in the eigenfunctions with coeﬃcients
depending on t: u(x, t) = ∞
n=1 un(t)φn(x). This implies
∞
n=1
un(t)φn(x) = −
∞
n=1
λnun(t)φn(x),
un(t) + λnun(t) = 0 for each n.
Since λn > 0, this ordinary diﬀerential equation has general solution
un(t) = An cos λnt + Bn sin λnt. Thus,
u(x, t) =
∞
n=1
An cos λnt + Bn sin λnt φn(x),
ut(x, t) =
∞
n=1
− λnAn sin λnt + λnBn cos λnt φn(x),
u(x, 0) =
∞
n=1
Anφn(x) = g(x),
ut(x, 0) =
∞
n=1
λnBnφn(x) = h(x).
Comparing with , we obtain
An = an, Bn =
bn
√
λn
.
Thus, the solution is given by
u(x, t) =
∞
n=1
an cos λnt +
bn
√
λn
sin λnt φn(x),

with
an =
Ω
g(x)φn(x) dx,
bn =
Ω
h(x)φn(x) dx.

The 2D WAVE Equation (eigenvalues/eigenfunctions of the Laplacian).
Let Ω = (0, a) × (0, b) and consider
⎧
⎪⎨
⎪⎩
utt = uxx + uyy for x ∈ Ω, t > 0
u(x, 0) = g(x), ut(x, 0) = h(x) for x ∈ Ω
u(x, t) = 0 for x ∈ ∂Ω, t > 0.
(28.1)
⎧
⎪⎨
⎪⎩
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
X
X
+
Y
Y
+ λ = 0.
Y :
X + μ2
X = 0 Y + ν2
Y = 0
X(0) = X(a) = 0 Y (0) = Y (b) = 0.
μm =
mπ
a
νn =
nπ
b
Xm(x) = sin
mπx
a
Yn(y) = sin
nπy
b
,
the Laplacian:
λmn = π2 m2
a2
+
n2
b2
φmn(x, y) =
2
√
ab
sin
mπx
a
sin
nπy
b
,
where m, n = 1, 2, . . ..
➁ Second, we solve the Wave Equation (28.1) using the “space” eigenfunctions.
For g, h ∈ C2(Ω) with g = h = 0 on ∂Ω, we have eigenfunction expansions 76
g(x) =
∞
n=1
anφn(x) and h(x) =
∞
n=1
bnφn(x).
Assume u(x, t) = ∞
un(t) + λnun(t) = 0 for each n.
76
In 2D, φn is really φmn, and x is (x, y).

Since λn > 0, this ordinary diﬀerential equation has general solution
un(t) = An cos λnt + Bn sin λnt. Thus,
u(x, t) =
∞
n=1
An cos λnt + Bn sin λnt φn(x),
ut(x, t) =
∞
n=1
− λnAn sin λnt + λnBn cos λnt φn(x),
u(x, 0) =
∞
n=1
Anφn(x) = g(x),
ut(x, 0) =
∞
n=1
λnBnφn(x) = h(x).
Comparing with , we obtain
An = an, Bn =
bn
√
λn
.
Thus, the solution is given by
u(x, t) =
∞
m,n=1
amn cos λmnt +
bmn
√
λmn
sin λmnt φmn(x),
with λmn, φmn(x) given above, and
amn =
Ω
g(x)φmn(x) dx,
bmn =
Ω
h(x)φmn(x) dx.

McOwen, 4.4 #3; 266B Ralston Hw. Consider the initial-boundary value problem
⎧
⎪⎨
⎪⎩
utt = u + f(x, t) for x ∈ Ω, t > 0
u(x, t) = 0 for x ∈ ∂Ω, t > 0
u(x, 0) = 0, ut(x, 0) = 0 for x ∈ Ω.
Use Duhamel’s principle and an expansion of f in eigenfunctions to obtain a (formal)
solution.
Proof. a) We expand u in terms of the Dirichlet eigenfunctions of Laplacian in
Ω.
φn + λnφn = 0 in Ω, φn = 0 on ∂Ω.
Assume
u(x, t) =
∞
n=1
an(t)φn(x), an(t) =
Ω
φn(x)u(x, t) dx.
f(x, t) =
∞
n=1
fn(t)φn(x), fn(t) =
Ω
φn(x)f(x, t) dx.
an(t) =
Ω
φn(x)utt dx =
Ω
φn( u + f) dx =
Ω
φn u dx +
Ω
φnf dx
=
Ω
φnu dx +
Ω
φnf dx = −λn
Ω
φnu dx +
Ω
φnf dx
fn
= −λnan(t) + fn(t).
an(0) =
Ω
φn(x)u(x, 0) dx = 0.
an(0) =
Ω
φn(x)ut(x, 0) dx = 0.
77 Thus, we have an ODE which is converted and solved by Duhamel’s principle:
⎧
⎪⎨
⎪⎩
an + λnan = fn(t)
an(0) = 0
an(0) = 0
⇒
⎧
⎪⎨
⎪⎩
ãn + λnãn = 0
ãn(0, s) = 0
ãn(0, s) = fn(s)
an(t) =
t
0
ãn(t−s, s) ds.
With the anzats ãn(t, s) = c1 cos
√
λnt + c2 sin
√
λnt, we get c1 = 0, c2 = fn(s)/
√
λn,
or
ãn(t, s) = fn(s)
sin
√
λnt
√
λn
.
Duhamel’s principle gives
an(t) =
t
0
ãn(t − s, s) ds =
t
0
fn(s)
sin(
√
λn(t − s))
√
λn
ds.
u(x, t) =
∞
n=1
φn(x)
√
λn
t
0
fn(s) sin( λn(t − s)) ds.
77
We used Green’s formula: ∂Ω
φn
∂u
∂n − u∂φn
∂n ds = Ω
(φn u − φnu) dx.
On ∂Ω, u = 0; φn = 0 since eigenfunctions are Dirichlet.

⎧
⎪⎨
⎪⎩
utt = a(t)uxx + f(x, t) 0 ≤ x ≤ π, t ≥ 0
u(0, t) = u(π, t) = 0 t ≥ 0
u(x, 0) = g(x), ut(x, 0) = h(x) 0 ≤ x ≤ π,
where the coeﬃcient a(t) = 0.
a) Express (formally) the solution of this problem by the method of eigenfunction ex-
pansions.
b) Show that this problem is not well-posed if a ≡ −1.
Hint: Take f = 0 and prove that the solution does not depend continuously on the
initial data g, h.
Proof. a) We expand u in terms of the Dirichlet eigenfunctions of Laplacian in
Ω.
φnxx + λnφn = 0 in Ω, φn(0) = φn(π) = 0.
That gives us the eigenvalues and eigenfunctions of the Laplacian: λn = n2, φn(x) =
sinnx.
Assume
u(x, t) =
∞
n=1
un(t)φn(x), un(t) =
Ω
φn(x)u(x, t) dx.
f(x, t) =
∞
n=1
fn(t)φn(x), fn(t) =
Ω
φn(x)f(x, t) dx.
g(x) =
∞
n=1
gnφn(x), gn =
Ω
φn(x)g(x) dx.
h(x) =
∞
n=1
hnφn(x), hn =
Ω
φn(x)h(x) dx.
un(t) =
Ω
φn(x)utt dx =
Ω
φn(a(t)uxx + f) dx = a(t)
Ω
φnuxx dx +
Ω
φnf dx
= a(t)
Ω
φnxxu dx +
Ω
φnf dx = −λna(t)
Ω
φnu dx +
Ω
φnf dx
fn
= −λna(t)un(t) + fn(t).
un(0) =
Ω
φn(x)u(x, 0) dx =
Ω
φn(x)g(x) dx = gn.
un(0) =
Ω
φn(x)ut(x, 0) dx =
Ω
φn(x)h(x) dx = hn.
Thus, we have an ODE which is converted and solved by Duhamel’s principle:
⎧
⎪⎨
⎪⎩
un + λna(t)un = fn(t)
un(0) = gn
un(0) = hn.

Note: The initial data is not 0; therefore, the Duhamel’s principle is not applicable.
Also, the ODE is not linear in t, and it’s solution is not obvious. Thus,
u(x, t) =
∞
n=1
un(t)φn(x),
where un(t) are solutions of .

b) Assume we have two solutions, u1 and u2, to the PDE:
⎧
⎪⎨
⎪⎩
u1tt + u1xx = 0,
u1(0, t) = u1(π, t) = 0,
u1(x, 0) = g1(x), u1t(x, 0) = h1(x);
⎧
⎪⎨
⎪⎩
u2tt + u2xx = 0,
u2(0, t) = u2(π, t) = 0,
u2(x, 0) = g2(x), u2t(x, 0) = h2(x).
Note that the equation is elliptic, and therefore, the maximum principle holds.
In order to prove that the solution does not depend continuously on the initial data
g, h, we need to show that one of the following conditions holds:
max
Ω
|u1 − u2| > max
∂Ω
|g1 − g2|,
max
Ω
|ut1 − ut2| > max
∂Ω
|h1 − h2|.
That is, the diﬀerence of the two solutions is not bounded by the diﬀerence of initial
data.
By the method of separation of variables, we may obtain
u(x, t) =
∞
n=1
(an cos nt + bn sinnt) sinnx,
u(x, 0) =
∞
n=1
an sinnx = g(x),
ut(x, 0) =
∞
n=1
nbn sinnx = h(x).
Not complete.
We also know that for elliptic equations, and for Laplace equation in particular, the
value of the function u has to be prescribed on the entire boundary, i.e. u = g on
∂Ω, which is not the case here, making the problem under-determined. Also, ut is
prescribed on one of the boundaries, making the problem overdetermined.

29 Problems: Eigenvalues of the Laplacian - Heat
The ND HEAT Equation (eigenvalues/eigenfunctions of the Laplacian).
Consider the initial value problem with homogeneous Dirichlet condition:
⎧
⎪⎨
⎪⎩
ut = u for x ∈ Ω, t > 0
u(x, 0) = g(x) for x ∈ Ω
u(x, t) = 0 for x ∈ ∂Ω, t > 0.
Proof. For g ∈ C2(Ω) with g = 0 on ∂Ω, we have eigenfunction expansion
g(x) =
∞
n=1
anφn(x)
Assume the solution u(x, t) may be expanded in the eigenfunctions with coeﬃcients
depending on t: u(x, t) = ∞
∞
n=1
un(t)φn(x) = −λn
∞
n=1
un(t)φn(x),
un(t) + λnun(t) = 0, which has the general solution
un(t) = Ane−λnt
. Thus,
u(x, t) =
∞
n=1
Ane−λnt
φn(x),
u(x, 0) =
∞
n=1
Anφn(x) = g(x).
Comparing with , we obtain An = an. Thus, the solution is given by
u(x, t) =
∞
n=1
ane−λnt
φn(x),
with an =
Ω
g(x)φn(x) dx.
Also
u(x, t) =
∞
n=1
ane−λnt
φn(x) =
∞
n=1 Ω
g(y)φn(y) dy e−λnt
φn(x)
=
Ω
∞
n=1
e−λnt
φn(x)φn(y)
K(x,y,t), heat kernel
g(y) dy

The 2D HEAT Equation (eigenvalues/eigenfunctions of the Laplacian).
Let Ω = (0, a) × (0, b) and consider
⎧
⎪⎨
⎪⎩
ut = uxx + uyy for x ∈ Ω, t > 0
u(x, 0) = g(x) for x ∈ Ω
u(x, t) = 0 for x ∈ ∂Ω, t > 0.
(29.1)
⎧
⎪⎨
⎪⎩
u(0, y) = 0 = u(a, y) for 0 ≤ y ≤ b,
u(x, 0) = 0 = u(x, b) for 0 ≤ x ≤ a.
X
X
+
Y
Y
+ λ = 0.
Y :
X + μ2
X = 0 Y + ν2
Y = 0
X(0) = X(a) = 0 Y (0) = Y (b) = 0.
μm =
mπ
a
νn =
nπ
b
Xm(x) = sin
mπx
a
Yn(y) = sin
nπy
b
,
the Laplacian:
λmn = π2 m2
a2
+
n2
b2
φmn(x, y) =
2
√
ab
sin
mπx
a
sin
nπy
b
,
where m, n = 1, 2, . . ..
➁ Second, we solve the Heat Equation (29.1) using the “space” eigenfunctions.
For g ∈ C2(Ω) with g = 0 on ∂Ω, we have eigenfunction expansion
g(x) =
∞
n=1
anφn(x).
un(t) = Ane−λnt
. Thus,
u(x, t) =
∞
n=1
Ane−λnt
φn(x),
u(x, 0) =
∞
n=1
Anφn(x) = g(x).

Comparing with , we obtain An = an. Thus, the solution is given by
u(x, t) =
∞
m,n=1
amne−λmnt
φmn(x),
with λmn, φmn given above and amn = Ω g(x)φmn(x) dx.

Problem (S’91, #2). Consider the heat equation
ut = uxx + uyy
on the square Ω = {0 ≤ x ≤ 2π, 0 ≤ y ≤ 2π} with
periodic boundary conditions and with initial data
u(0, x, y) = f(x, y).
a) Find the solution using separation of variables.
⎧
⎪⎨
⎪⎩
u(0, y) = u(2π, y) for 0 ≤ y ≤ 2π,
u(x, 0) = u(x, 2π) for 0 ≤ x ≤ 2π.
X
X
+
Y
Y
+ λ = 0.
Letting λ = μ2 + ν2 and using periodic BC’s, we ﬁnd the equations for X and Y :
X + μ2
X = 0 Y + ν2
Y = 0
X(0) = X(2π) Y (0) = Y (2π).
μm = m νn = n
Xm(x) = eimx
Yn(y) = einy
,
where m, n = . . ., −2, −1, 0, 1, 2, . . .. Thus we obtain eigenvalues and normalized eigen-
functions of the Laplacian:
λmn = m2
+ n2
φmn(x, y) = eimx
einy
,
where m, n = . . ., −2, −1, 0, 1, 2, . . ..
➁ Second, we solve the Heat Equation using the “space” eigenfunctions.
Assume u(x, y, t) = ∞
m,n=−∞ umn(t)eimxeiny. This implies
umn(t) + (m2
+ n2
)umn(t) = 0, which has the general solution
un(t) = cmne−(m2+n2)t
. Thus,
u(x, y, t) =
∞
m,n=−∞
cmne−(m2+n2)t
eimx
einy
.

u(x, y, 0) =
∞
m,n=−∞
cmneimx
einy
= f(x, y),
2π
0
2π
0
f(x, y)eimx
einy
dxdy =
2π
0
2π
0
∞
m,n=−∞
cmneimx
einy
eim x
ein y
dxdy
= 2π
2π
0
∞
n=−∞
cmneiny
ein y
dy = 4π2
cmn.
cmn =
1
4π2
2π
0
2π
0
f(x, y)e−imx
e−iny
dxdy = fmn.

b) Show that the integral Ω u2(x, y, t) dxdy is decreasing in t, if f is not constant.
Proof. We have
ut = uxx + uyy
Multiply the equation by u and integrate:
uut = u u,
1
2
d
dt
u2
= u u,
1
2
d
dt Ω
u2
dxdy =
Ω
u u dxdy =
∂Ω
u
∂u
∂n
ds
=0, (periodic BC)
−
Ω
|∇u|2
dxdy
= −
Ω
|∇u|2
dxdy ≤ 0.
Equality is obtained only when ∇u = 0 ⇒ u = constant ⇒ f = constant.
If f is not constant, Ω u2 dxdy is decreasing in t.

Problem (F’98, #3). Consider the eigenvalue problem
d2φ
dx2
+ λφ = 0,
φ(0) −
dφ
dx
(0) = 0, φ(1) +
dφ
dx
(1) = 0.
a) Show that all eigenvalues are positive.
b) Show that there exist a sequence of eigenvalues λ = λn, each of which satisfies
tan
√
λ =
2
√
λ
λ − 1
.
c) Solve the following initial-boundary value problem on 0 < x < 1, t > 0
∂u
∂t
=
∂2u
∂x2
,
u(0, t) −
∂u
∂x
(0, t) = 0, u(1, t) +
∂u
∂x
(1, t) = 0,
u(x, 0) = f(x).
You may call the relevant eigenfunctions φn(x) and assume that they are known.
Proof. a) • If λ = 0, the ODE reduces to φ = 0. Try φ(x) = Ax + B.
From the first boundary condition,
φ(0) − φ (0) = 0 = B − A ⇒ B = A.
Thus, the solution takes the form φ(x) = Ax+A. The second boundary condition gives
φ(1) + φ (1) = 0 = 3A ⇒ A = B = 0.
Thus the only solution is φ ≡ 0, which is not an eigenfunction, and 0 not an eigenvalue.
• If λ < 0, try φ(x) = esx
, which gives s = ±
√
−λ = ±β ∈ R.
Hence, the family of solutions is φ(x) = Aeβx +Be−βx. Also, φ (x) = βAeβx −βBe−βx.
The boundary conditions give
φ(0) − φ (0) = 0 = A + B − βA + βB = A(1 − β) + B(1 + β), (29.2)
φ(1)+φ (1) = 0 = Aeβ
+Be−β
+βAeβ
−βBe−β
= Aeβ
(1+β)+Be−β
(1−β). (29.3)
From (29.2) and (29.3) we get
1 + β
1 − β
= −
A
B
and
1 + β
1 − β
= −
B
A
e−2β
, or
A
B
= e−β
.
From (29.2), β =
A + B
A − B
and thus,
A
B
= e
A+B
B−A , which has no solutions.
b) Since λ > 0, the anzats φ = esx gives s = ±i
√
λ and the family of solutions takes
the form
φ(x) = A sin(x
√
λ) + B cos(x
√
λ).
Then, φ (x) = A
√
λ cos(x
√
λ) − B
√
λ sin(x
√
λ). The first boundary condition gives
φ(0) − φ (0) = 0 = B − A
√
λ ⇒ B = A
√
λ.

Hence, φ(x) = A sin(x
√
λ) + A
√
λ cos(x
√
λ). The second boundary condition gives
φ(1) + φ (1) = 0 = A sin(
√
λ) + A
√
λ cos(
√
λ) + A
√
λ cos(
√
λ) − Aλ sin(
√
λ)
= A (1 − λ) sin(
√
λ) + 2
√
λ cos(
√
λ)
A = 0 (since A = 0 implies B = 0 and φ = 0, which is not an eigenfunction). Therefore,
−(1 − λ) sin(
√
λ) = 2
√
λ cos(
√
λ), and thus tan(
√
λ) = 2
√
λ
λ−1 .
c) We may assume that the eigenvalues/eigenfunctins of the Laplacian, λn and φn(x),
are known. We solve the Heat Equation using the “space” eigenfunctions.
⎧
⎪⎨
⎪⎩
ut = uxx,
u(0, t) − ux(0, t) = 0, u(1, t) + ux(1, t) = 0,
u(x, 0) = f(x).
For f, we have an eigenfunction expansion
f(x) =
∞
n=1
anφn(x).
un(t) = Ane−λnt
. Thus,
u(x, t) =
∞
n=1
Ane−λnt
φn(x),
u(x, 0) =
∞
n=1
Anφn(x) = f(x).
Comparing with , we have An = an. Thus, the solution is given by
u(x, t) =
∞
n=1
ane−λnt
φn(x),
with
an =
1
0
f(x)φn(x) dx.

Problem (W’03, #3); 266B Ralston Hw. Let Ω be a smooth domain in three
dimensions and consider the initial-boundary value problem for the heat equation
⎧
⎪⎨
⎪⎩
ut = u + f(x) for x ∈ Ω, t > 0
∂u/∂n = 0 for x ∈ ∂Ω, t > 0
u(x, 0) = g(x) for x ∈ Ω,
in which f and g are known smooth functions with
∂g/∂n = 0 for x ∈ ∂Ω.
a) Find an approximate formula for u as t → ∞.
Proof. We expand u in terms of the Neumann eigenfunctions of Laplacian in Ω.
φn + λnφn = 0 in Ω,
∂φn
∂n
= 0 on ∂Ω.
Note that here λ1 = 0 and φ1 is the constant V −1/2, where V is the volume of Ω.
Assume
u(x, t) =
∞
n=1
an(t)φn(x), an(t) =
Ω
φn(x)u(x, t) dx.
f(x) =
∞
n=1
fnφn(x), fn =
Ω
φn(x)f(x) dx.
g(x) =
∞
n=1
gnφn(x), gn =
Ω
φn(x)g(x) dx.
an(t) =
Ω
φn(x)ut dx =
Ω
φn( u + f) dx =
Ω
φn u dx +
Ω
φnf dx
=
Ω
φnu dx +
Ω
φnf dx = −λn
Ω
φnu dx +
Ω
φnf dx
fn
= −λnan + fn.
an(0) =
Ω
φn(x)u(x, 0) dx =
Ω
φng dx = gn.
78
Thus, we solve the ODE:
an + λnan = fn
an(0) = gn.
For n = 1, λ1 = 0, and we obtain a1(t) = f1t + g1.
For n ≥ 2, the homogeneous solution is anh
= ce−λnt. The anzats for a particular
solution is anp = c1t + c2, which gives c1 = 0 and c2 = fn/λn. Using the initial
condition, we obtain
an(t) = gn −
fn
λn
e−λnt
+
fn
λn
.
78
φn
∂u
∂n − u∂φn
∂n ds = Ω
On ∂Ω, ∂u
∂n
= 0; ∂φn
∂n
= 0 since eigenfunctions are Neumann.

u(x, t) = (f1t + g1)φ1(x) +
∞
n=2
gn −
fn
λn
e−λnt
+
fn
λn
φn(x).
If f1 = 0
Ω
f(x) dx = 0 , lim
t→∞
u(x, t) = g1φ1 +
∞
n=2
fnφn
λn
.
If f1 = 0
Ω
f(x) dx = 0 , lim
t→∞
u(x, t) ∼ f1φ1t.

b) If g ≥ 0 and f > 0, show that u > 0 for all t > 0.

Problem (S’97, #2). a) Consider the eigenvalue problem for the Laplace operator
in Ω ∈ R2 with zero Neumann boundary condition
∂u
∂n = 0 on ∂Ω.
Prove that λ0 = 0 is the lowest eigenvalue and that it is simple.
b) Assume that the eigenfunctions φn(x, y) of the problem in (a) form a complete
orthogonal system, and that f(x, y) has a uniformly convergent expansion
f(x, y) =
∞
n=0
fnφn(x, y).
Solve the initial value problem
ut = u + f(x, y)
subject to initial and boundary conditions
u(x, y, 0) = 0,
∂u
∂n
u|∂Ω = 0.
What is the behavior of u(x, y, t) as t → ∞?
c) Consider the problem with Neumann boundary conditions
vxx + vyy + f(x, y) = 0 in Ω
∂v
∂nv = 0 on ∂Ω.
When does a solution exist? Find this solution, and ﬁnd its relation with the behavior
of lim u(x, y, t) in (b) as t → ∞.
Proof. a) Suppose this eigenvalue problem did have a solution u with λ ≤ 0.
Multiplying u + λu = 0 by u and integrating over Ω, we get
Ω
u u dx + λ
Ω
u2
dx = 0,
∂Ω
u
∂u
∂n
=0
ds −
Ω
|∇u|2
dx + λ
Ω
u2
dx = 0,
Ω
|∇u|2
dx = λ
≤0
Ω
u2
dx,
Thus, ∇u = 0 in Ω, and u is constant in Ω. Hence, we now have
0 = λ
≤0
Ω
u2
dx.
For nontrivial u, we have λ = 0. For this eigenvalue problem, λ = 0 is an eigenvalue,
its eigenspace is the set of constants, and all other λ’s are positive.

b) We expand u in terms of the Neumann eigenfunctions of Laplacian in Ω. 79
φn + λnφn = 0 in Ω,
∂φn
∂n
= 0 on ∂Ω.
u(x, y, t) =
∞
n=1
an(t)φn(x, y), an(t) =
Ω
φn(x, y)u(x, y, t) dx.
an(t) =
Ω
φn(x, y)ut dx =
Ω
φn( u + f) dx =
Ω
φn u dx +
Ω
φnf dx
=
Ω
φnu dx +
Ω
φnf dx = −λn
Ω
φnu dx +
Ω
φnf dx
fn
= −λnan + fn.
an(0) =
Ω
φn(x, y)u(x, y, 0) dx = 0.
80 Thus, we solve the ODE:
an + λnan = fn
an(0) = 0.
For n = 1, λ1 = 0, and we obtain a1(t) = f1t.
For n ≥ 2, the homogeneous solution is anh
= ce−λnt
. The anzats for a particular
solution is anp = c1t + c2, which gives c1 = 0 and c2 = fn/λn. Using the initial
condition, we obtain
an(t) = −
fn
λn
e−λnt
+
fn
λn
.
u(x, t) = f1φ1t +
∞
n=2
−
fn
λn
e−λnt
+
fn
λn
φn(x).
If f1 = 0
Ω
f(x) dx = 0 , lim
t→∞
u(x, t) =
∞
n=2
fnφn
λn
.
If f1 = 0
Ω
f(x) dx = 0 , lim
t→∞
u(x, t) ∼ f1φ1t.
c) Integrate v + f(x, y) = 0 over Ω:
Ω
f dx = −
Ω
v dx = −
Ω
∇ · ∇v dx =1
−
∂Ω
∂v
∂n
ds =2
0,
where we used 1
divergence theorem and 2
Neumann boundary conditions. Thus, the
solution exists only if
Ω
f dx = 0.
79
We use dx dy → dx.
80
φn
∂u
∂n
− u∂φn
∂n
ds = Ω
On ∂Ω, ∂u
∂n
= 0; ∂φn
∂n
= 0 since eigenfunctions are Neumann.

Assume v(x, y) = ∞
n=0 anφn(x, y). Since we have f(x, y) = ∞
n=0 fnφn(x, y), we
obtain
−
∞
n=0
λnanφn +
∞
n=0
fnφn = 0,
−λnanφn + fnφn = 0,
an =
fn
λn
.
v(x, y) = ∞
n=0( fn
λn
)φn(x, y).

29.1 Heat Equation with Periodic Boundary Conditions in 2D
(with extra terms)
Problem (F’99, #5). In two spatial dimensions, consider the diﬀerential equation
ut = −ε u − 2
u
with periodic boundary conditions on the unit square [0, 2π]2
.
a) If ε = 2 ﬁnd a solution whose amplitude increases as t increases.
b) Find a value ε0, so that the solution of this PDE stays bounded as t → ∞, if ε < ε0.
Proof. a) Eigenfunctions of the Laplacian.
The periodic boundary conditions imply a Fourier Series solution of the form:
u(x, t) =
m,n
amn(t)ei(mx+ny)
.
ut =
m,n
amn(t)ei(mx+ny)
,
u = uxx + uyy = −
m,n
(m2
+ n2
) amn(t)ei(mx+ny)
,
2
u = uxxxx + 2uxxyy + uyyyy =
m,n
(m4
+ 2m2
n2
+ n4
) amn(t)ei(mx+ny)
=
m,n
(m2
+ n2
)2
amn(t)ei(mx+ny)
.
Plugging this into the PDE, we obtain
amn(t) = ε(m2
+ n2
)amn(t) − (m2
+ n2
)2
amn(t),
amn(t) − [ε(m2
+ n2
) − (m2
+ n2
)2
]amn(t) = 0,
amn(t) − (m2
+ n2
)[ε − (m2
+ n2
)]amn(t) = 0.
The solution to the ODE above is
amn(t) = αmn e(m2+n2)[ε−(m2+n2 )]t
.
u(x, t) =
m,n
αmn e(m2+n2)[ε−(m2+n2 )]t
ei(mx+ny)
oscillates
.
When ε = 2, we have
u(x, t) =
m,n
αmn e(m2+n2)[2−(m2+n2)]t
ei(mx+ny)
.
We need a solution whose amplitude increases as t increases. Thus, we need those
αmn > 0, with
(m2
+ n2
)[2 − (m2
+ n2
)] > 0,
2 − (m2
+ n2
) > 0,
2 > m2
+ n2
.
Hence, αmn > 0 for (m, n) = (0, 0), (m, n) = (1, 0), (m, n) = (0, 1).
Else, αmn = 0. Thus,
u(x, t) = α00 + α10et
eix
+ α01et
eiy
= 1 + et
eix
+ et
eiy
= 1 + et
(cos x + i sinx) + et
(cos y + i siny).

b) For ε ≤ ε0 = 1, the solution stays bounded as t → ∞.

Suppose that a and b are constants with a ≥ 0, and consider the equation
ut = uxx + uyy − au3
+ bu (29.4)
in which u(x, y, t) is 2π-periodic in x and y.
a) Let u be a solution of (29.4) with
||u(t = 0)|| =
2π
0
2π
0
|u(x, y, t = 0)|2
dxdy1/2
< .
Derive an explicit bound on ||u(t)|| and show that it stays finite for all t.
b) If a = 0, construct the normal modes for (29.4); i.e. find all solutions of the form
u(x, y, t) = eλt+ikx+ily
.
c) Use these normal modes to construct a solution of (29.4) with a = 0 for the initial
data
u(x, y, t = 0) =
1
1 − 1
2 eix
+
1
1 − 1
2e−ix
.
Proof. a) Multiply the equation by u and integrate:
ut = u − au3
+ bu,
uut = u u − au4
+ bu2
,
Ω
uut dx =
Ω
u u dx −
Ω
au4
dx +
Ω
bu2
dx,
1
2
d
dt Ω
u2
dx =
∂Ω
u
∂u
∂n
ds
=0, u periodic on [0,2π]2
−
Ω
|∇u|2
dx −
Ω
au4
dx
≤0
+
Ω
bu2
dx,
d
dt
||u||2
2 ≤ 2b ||u||2
2,
||u||2
2 ≤ ||u(x, 0)||2
2 e2bt
,
||u||2 ≤ ||u(x, 0)||2 ebt
≤ ε ebt
.
Thus, ||u|| stays finite for all t.
b) Since a = 0, plugging u = eλt+ikx+ily into the equation, we obtain:
ut = uxx + uyy + bu,
λ eλt+ikx+ily
= (−k2
− l2
+ b) eλt+ikx+ily
,
λ = −k2
− l2
+ b.
Thus,
ukl = e(−k2−l2+b)t+ikx+ily
,
u(x, y, t) =
k,l
akl e(−k2−l2+b)t+ikx+ily
.

c) Using the initial condition, we obtain:
u(x, y, 0) =
k,l
akl ei(kx+ly)
=
1
1 − 1
2eix
+
1
1 − 1
2 e−ix
=
∞
k=0
1
2
eix
k
+
∞
k=0
1
2
e−ix
k
=
∞
k=0
1
2k
eikx
+
∞
k=0
1
2k
e−ikx
,
= 2 +
∞
k=1
1
2k
eikx
+
−∞
k=−1
1
2−k
eikx
.
Thus, l = 0, and we have
∞
k=−∞
ak eikx
= 2 +
∞
k=1
1
2k
eikx
+
−∞
k=−1
1
2−k
eikx
,
⇒ a0 = 2; ak =
1
2k
, k > 0; ak =
1
2−k
, k < 0
⇒ a0 = 2; ak =
1
2|k|
, k = 0.
u(x, y, t) = 2ebt
+
+∞
k=−∞, k=0
1
2|k|
e(−k2+b)t+ikx
.
81
81
Note a similar question formulation in F’92 #3(b).

Problem (S’00, #3). Consider the initial-boundary value problem for u = u(x, y, t)
ut = u − u
for (x, y) ∈ [0, 2π]2, with periodic boundary conditions and with
u(x, y, 0) = u0(x, y)
in which u0 is periodic. Find an asymptotic expansion for u for t large with terms
tending to zero increasingly rapidly as t → ∞.
Proof. Since we have periodic boundary conditions, assume
u(x, y, t) =
m,n
umn(t) ei(mx+ny)
.
Plug this into the equation:
m,n
umn(t) ei(mx+ny)
=
m,n
(−m2
− n2
− 1) umn(t) ei(mx+ny)
,
umn(t) = (−m2
− n2
− 1) umn(t),
umn(t) = amn e(−m2−n2−1)t
,
u(x, y, t) =
m,n
amn e−(m2+n2+1)t
ei(mx+ny)
.
Since u0 is periodic,
u0(x, y) =
m,n
u0mn ei(mx+ny)
, u0mn =
1
4π2
2π
0
2π
0
u0(x, y) e−i(mx+ny)
dxdy.
Initial condition gives:
u(x, y, 0) =
m,n
amn ei(mx+ny)
= u0(x, y),
m,n
amn ei(mx+ny)
=
m,n
u0mn ei(mx+ny)
,
⇒ amn = u0mn.
u(x, y, t) =
m,n
u0mn e−(m2+n2+1)t
ei(mx+ny)
.
u0mn e−(m2+n2+1)t ei(mx+ny) → 0 as t → ∞, since e−(m2+n2+1)t → 0 as t → ∞.

30 Problems: Fourier Transform
Problem (S’01, #2b). Write the solution of initial value problem
Ut −
1 0
5 3
Ux = 0,
for general initial data
u(1)
(x, 0)
u(2)(x, 0)
=
f(x)
0
as an inverse Fourier transform.
You may assume that f is smooth and rapidly decreasing as |x| → ∞.
Proof. Consider the original system:
u
(1)
t − u(1)
x = 0,
u
(2)
t − 5u(1)
x − 3u(2)
x = 0.
Take the Fourier transform in x. The transformed initial value problems are:
u
(1)
t − iξu(1)
= 0, u(1)
(ξ, 0) = f(ξ),
u
(2)
t − 5iξu(1)
− 3iξu(2)
= 0, u(2)
(ξ, 0) = 0.
Solving the ﬁrst ODE for u(1) gives:
u(1)
(ξ, t) = f(ξ)eiξt
.
With this u(1)
, the second initial value problem becomes
u
(2)
t − 3iξu(2)
= 5iξf(ξ)eiξt
, u(2)
(ξ, 0) = 0.
The homogeneous solution of the above ODE is:
u
(2)
h (ξ, t) = c1e3iξt
.
With u
(2)
p = c2eiξt as anzats for a particular solution, we obtain:
iξc2eiξt
− 3iξc2eiξt
= 5iξf(ξ)eiξt
,
−2iξc2eiξt
= 5iξf(ξ)eiξt
,
c2 = −
5
2
f(ξ).
⇒ u(2)
p (ξ, t) = −
5
2
f(ξ)eiξt
.
u(2)
(ξ, t) = u
(2)
h (ξ, t) + u(2)
p (ξ, t) = c1e3iξt
−
5
2
f(ξ)eiξt
.
We ﬁnd c1 using initial conditions:
u(2)
(ξ, 0) = c1 −
5
2
f(ξ) = 0 ⇒ c1 =
5
2
f(ξ).
Thus,
u(2)
(ξ, t) =
5
2
f(ξ) e3iξt
− eiξt
.

u(1)
(x, t) and u(2)
(x, t) are be obtained by taking inverse Fourier transform:
u(1)
(x, t) = u(1)
(ξ, t)
∨
=
1
√
2π Rn
eixξ
f(ξ) eiξt
dξ,
u(2)
(x, t) = u(2)
(ξ, t)
∨
=
1
√
2π Rn
eixξ 5
2
f(ξ) e3iξt
− eiξt
dξ.

Problem (S’02, #4). Use the Fourier transform on L2
(R) to show that
du
dx
+ cu(x) + u(x − 1) = f (30.1)
has a unique solution u ∈ L2(R) for each f ∈ L2(R) when |c| > 1 - you may assume
that c is a real number.
Proof. u ∈ L2(R). Define its Fourier transform u by
u(ξ) =
1
√
2π R
e−ixξ
u(x) dx for ξ ∈ R.
du
dx
(ξ) = iξu(ξ).
We can find u(x − 1)(ξ) in two ways.
• Let u(x − 1
y
) = v(x), and determinte v(ξ):
u(x − 1)(ξ) = v(ξ) =
1
√
2π R
e−ixξ
v(x) dx =
1
√
2π R
e−i(y+1)ξ
u(y) dy
=
1
√
2π R
e−iyξ
e−iξ
u(y) dy = e−iξ
u(ξ).
• We can also write the definition for u(ξ) and substitute x − 1 later in calculations:
u(ξ) =
1
√
2π R
e−iyξ
u(y) dy =
1
√
2π R
e−i(x−1)ξ
u(x − 1) dx
=
1
√
2π R
e−ixξ
eiξ
u(x − 1) dx = eiξ
u(x − 1)(ξ),
⇒ u(x − 1)(ξ) = e−iξ
u(ξ).
Substituting into (30.1), we obtain
iξu(ξ) + cu(ξ) + e−iξ
u(ξ) = f(ξ),
u(ξ) =
f(ξ)
iξ + c + e−iξ
.
u(x) =
f(ξ)
iξ + c + e−iξ
∨
= f B
∨
=
1
√
2π
f ∗ B,
where B =
1
iξ + c + e−iξ
,
⇒ B =
1
iξ + c + e−iξ
∨
=
1
√
2π R
eixξ
iξ + c + e−iξ
dξ.
For |c| > 1, u(ξ) exists for all ξ ∈ R, so that u(x) = (u(ξ))∨
and this is unique by the
Fourier Inversion Theorem.
Note that in Rn
, becomes
u(x − 1)(ξ) = v(ξ) =
1
(2π)
n
2 Rn
e−ix·ξ
v(x) dx =
1
(2π)
n
2 Rn
e−i(y+1)·ξ
u(y) dy
=
1
(2π)
n
2 Rn
e−iy·ξ
e−i1·ξ
u(y) dy = e−i1·ξ
u(ξ) = e(−i j ξj)
u(ξ).

Problem (F’96, #3). Find the fundamental solution for the equation
ut = uxx − xux. (30.2)
Hint: The Fourier transform converts this problem into a PDE which can be solved
using the method of characteristics.
Proof. u ∈ L2
(R). Define its Fourier transform u by
u(ξ) =
1
√
2π R
e−ixξ
u(x) dx for ξ ∈ R.
ux(ξ) = iξ u(ξ),
uxx(ξ) = (iξ)2
u(ξ) = −ξ2
u(ξ).
We find xux(ξ) in two steps:
➀ Multiplication by x:
−ixu(ξ) =
1
√
2π R
e−ixξ
− ixu(x) dx =
d
dξ
u(ξ).
⇒ xu(x)(ξ) = i
d
dξ
u(ξ).
➁ Using the previous result, we find:
xux(x)(ξ) =
1
√
2π R
e−ixξ
xux(x) dx =
1
√
2π
e−ixξ
xu
∞
−∞
= 0
−
1
√
2π R
(−iξ)e−ixξ
x + e−ixξ
u dx
=
1
√
2π
iξ
R
e−ixξ
x u dx −
1
√
2π R
e−ixξ
u dx
= iξ xu(x)(ξ) − u(ξ) = iξ i
d
dξ
u(ξ) − u(ξ) = −ξ
d
dξ
u(ξ) − u(ξ).
⇒ xux(x)(ξ) = −ξ
d
dξ
u(ξ) − u(ξ).
Plugging these into (30.2), we get:
∂
∂t
u(ξ, t) = −ξ2
u(ξ, t) − − ξ
d
dξ
u(ξ, t) − u(ξ, t) ,
ut = −ξ2
u + ξuξ + u,
ut − ξuξ = −(ξ2
− 1)u.
We now solve the above equation by characteristics.
We change the notation: u → u, t → y, ξ → x. We have
uy − xux = −(x2
− 1)u.
dx
dt
= −x ⇒ x = c1e−t
, (c1 = xet
)
dy
dt
= 1 ⇒ y = t + c2,
dz
dt
= −(x2
− 1)z = −(c2
1e−2t
− 1)z ⇒
dz
z
= −(c2
1e−2t
− 1)dt
⇒ log z =
1
2
c2
1e−2t
+ t + c3 =
x2
2
+ t + c3 =
x2
2
+ y − c2 + c3 ⇒ z = ce
x2
2
+y
.

Changing the notation back, we have
u(ξ, t) = ce
ξ2
2
+t
.
Thus, we have
u(ξ, t) = ce
ξ2
2
+t
.
We use Inverse Fourier Tranform to get u(x, t): 82
u(x, t) =
1
√
2π R
eixξ
u(ξ, t) dξ =
1
√
2π R
eixξ
ce
ξ2
2
+t
dξ
=
c
√
2π
et
R
eixξ
e
ξ2
2 dξ =
c
√
2π
et
R
eixξ+ξ2
2 dξ
=
c
√
2π
et
R
e
2ixξ+ξ2
2 dξ =
c
√
2π
et
R
e
(ξ+ix)2
2 dξ e
x2
2
=
c
√
2π
et
e
x2
2
R
e
y2
2 dy =
c
√
2π
et
e
x2
2
√
2π = c et
e
x2
2 .
u(x, t) = c et
e
x2
2 .
Check:
ut = c et
e
x2
2 ,
ux = c et
xe
x2
2 ,
uxx = c et
e
x2
2 + x2
e
x2
2 .
Thus,
ut = uxx − xux,
c et
e
x2
2 = c et
e
x2
2 + x2
e
x2
2 − x c et
xe
x2
2 .
82
We complete the square for powers of exponentials.

Problem (W’02, #4). a) Solve the initial value problem
∂u
∂t
+
n
k=1
ak(t)
∂u
∂xk
+ a0(t)u = 0, x ∈ Rn
,
u(0, x) = f(x)
where ak(t), k = 1, . . ., n, and a0(t) are continuous functions, and f is a continuous
function. You may assume f has compact support.
b) Solve the initial value problem
∂u
∂t
+
n
k=1
ak(t)
∂u
∂xk
+ a0(t)u = f(x, t), x ∈ Rn
,
u(0, x) = 0
where f is continuous in x and t.
Proof. a) Use the Fourier transform to solve this problem.
u(ξ, t) =
1
(2π)
n
2 Rn
e−ix·ξ
u(x, t) dx for ξ ∈ R.
∂u
∂xk
= iξku.
Thus, the equation becomes:
ut + i n
k=1 ak(t)ξku + a0(t)u = 0,
u(ξ, 0) = f(ξ),
or
ut + i a(t) · ξ u + a0(t)u = 0,
ut = − i a(t) · ξ + a0(t) u.
This is an ODE in u with solution:
u(ξ, t) = ce−
t
0 (ia(s)·ξ+a0(s)) ds
, u(ξ, 0) = c = f(ξ). Thus,
u(ξ, t) = f(ξ) e− t
0
(i a(s)·ξ+a0(s)) ds
.
Use the Inverse Fourier transform to get u(x, t):
u(x, t) = u(ξ, t)∨
= f(ξ) e−
t
0 (i a(s)·ξ+a0(s)) ds
∨
=
(f ∗ g)(x)
(2π)
n
2
,
where g(ξ) = e− t
0
(i a(s)·ξ+a0(s)) ds
.
g(x) =
1
(2π)
n
2 Rn
eix·ξ
g(ξ) dξ =
1
(2π)
n
2 Rn
eix·ξ
e− t
0
(ia(s)·ξ+a0(s)) ds
dξ.
u(x, t) =
(f ∗ g)(x)
(2π)
n
2
=
1
(2π)n
Rn Rn
ei(x−y)·ξ
e−
t
dξ f(y) dy.
b) Use Duhamel’s Principle and the result from (a).
u(x, t) =
t
0
U(x, t − s, s) ds, where U(x, t, s) solves
∂U
∂t
+
n
k=1
ak(t)
∂U
∂xk
+ a0(t)U = 0,
U(x, 0, s) = f(x, s).

u(x, t) =
t
0
U(x, t − s, s) ds =
1
(2π)n
t
0 Rn Rn
ei(x−y)·ξ
e− t−s
dξ f(y, s) dy ds.

Problem (S’93, #2). a) Define the Fourier transform 83
f(ξ) =
∞
−∞
eixξ
f(x) dx.
State the inversion theorem. If
f(ξ) =
⎧
⎪⎨
⎪⎩
π, |ξ| < a,
1
2 π, |ξ| = a,
0, |ξ| > a,
where a is a real constant, what f(x) does the inversion theorem give?
b) Show that
f(x − b) = eiξb
f(x),
where b is a real constant. Hence, using part (a) and Parseval’s theorem, show that
1
π
∞
−∞
sin a(x + z)
x + z
sina(x + ξ)
x + ξ
dx =
sina(z − ξ)
z − ξ
,
where z and ξ are real constants.
Proof. a) • The inverse Fourier transform for f ∈ L1
(Rn
):
f∨
(ξ) =
1
2π
∞
−∞
e−ixξ
f(x) dx for ξ ∈ R.
Fourier Inversion Theorem: Assume f ∈ L2
(R). Then
f(x) =
1
2π
∞
−∞
e−ixξ
f(ξ) dξ =
1
2π
∞
−∞
∞
−∞
ei(y−x)ξ
f(y) dy dξ = (f)∨
(x).
• Parseval’s theorem (Plancherel’s theorem) (for this definition of the Fourier
transform). Assume f ∈ L1(Rn) ∩ L2(Rn). Then f, f∨ ∈ L2(Rn) and
1
2π
||f||L2(Rn) = ||f∨
||L2(Rn) = ||f||L2(Rn), or
∞
−∞
|f(x)|2
dx =
1
2π
∞
−∞
|f(ξ)|2
dξ.
Also,
∞
−∞
f(x) g(x)dx =
1
2π
∞
−∞
f(ξ) g(ξ) dξ.
• We can write
f(ξ) =
π, |ξ| < a,
0, |ξ| > a.
83
Note that the Fourier transform is defined incorrectly here. There should be ‘-’ sign in e−ixξ
.
Need to be careful, since the consequences of this definition propagate throughout the solution.

f(x) = (f(ξ))∨
=
1
2π
∞
−∞
e−ixξ
f(ξ) dξ =
1
2π
−a
−∞
0 dξ +
1
2π
a
−a
e−ixξ
π dξ +
1
2π
∞
a
0 dξ
=
1
2
a
−a
e−ixξ
dξ = −
1
2ix
e−ixξ
ξ=a
ξ=−a
= −
1
2ix
e−iax
− eiax
=
sinax
x
.
b) • Let f(x − b
y
) = g(x), and determinte g(ξ):
f(x − b)(ξ) = g(ξ) =
R
eixξ
g(x) dx =
R
ei(y+b)ξ
f(y) dy
=
R
eiyξ
eibξ
f(y) dy = eibξ
f(ξ).
• With f(x) = sin ax
x (from (a)), we have
1
π
∞
−∞
sina(x + z)
x + z
sin a(x + s)
x + s
dx =
1
π
∞
−∞
f(x + z)f(x + s) dx (x = x + s, dx = dx)
=
1
π
∞
−∞
f(x + z − s)f(x ) dx (Parseval’s)
=
1
π
1
2π
∞
−∞
f(x + z − s)f(x ) dξ part (b)
=
1
2π2
∞
−∞
f(ξ) e−i(z−s)ξ
f(ξ) dξ
=
1
2π2
a
−a
f(ξ)
2
e−i(z−s)ξ
dξ
=
1
2π2
a
−a
π2
e−i(z−s)ξ
dξ
=
1
2
a
−a
e−i(z−s)ξ
dξ
=
1
−2i(z − s)
e−i(z−s)ξ ξ=a
ξ=−a
=
ei(z−s)a − e−i(z−s)a
2i(z − s)
=
sin a(z − s)
z − s
.

Problem (F’03, #5). ❶ State Parseval’s relation for Fourier transforms.
❷ Find the Fourier transform ˆf(ξ) of
f(x) =
eiαx/2
√
πy, |x| ≤ y
0, |x| > y,
in which y and α are constants.
❸ Use this in Parseval’s relation to show that
∞
−∞
sin2
(α − ξ)y
(α − ξ)2
dξ = πy.
What does the transform ˆf(ξ) become in the limit y → ∞?
❹ Use Parseval’s relation to show that
sin(α − β)y
(α − β)
=
1
π
∞
−∞
sin(α − ξ)y
(α − ξ)
sin(β − ξ)y
(β − ξ)
dξ.
Proof. • f ∈ L2(R). Deﬁne its Fourier transform u by
f(ξ) =
1
√
2π R
e−ixξ
f(x) dx for ξ ∈ R.
❶ Parseval’s theorem (Plancherel’s theorem):
Assume f ∈ L1
(Rn
) ∩ L2
(Rn
). Then f, f∨
∈ L2
(Rn
) and
||f||L2(Rn) = ||f∨
||L2(Rn) = ||f||L2(Rn), or
∞
−∞
|f(x)|2
dx =
∞
−∞
|f(ξ)|2
dξ.
Also,
∞
−∞
f(x) g(x)dx =
∞
−∞
f(ξ) g(ξ) dξ.
❷ Find the Fourier transform of f:
f(ξ) =
1
√
2π R
e−ixξ
f(x) dx =
1
√
2π
y
−y
e−ixξ eiαx
2
√
πy
dx =
1
2π
√
2y
y
−y
ei(α−ξ)x
dx
=
1
2π
√
2y
1
i(α − ξ)
ei(α−ξ)x
x=y
x=−y
=
1
2iπ
√
2y(α − ξ)
ei(α−ξ)y
− e−i(α−ξ)y
=
siny(α − ξ)
π
√
2y(α − ξ)
.
❸ Parseval’s theorem gives:
∞
−∞
|f(ξ)|2
dξ =
∞
−∞
|f(x)|2
dx,
∞
−∞
sin2
y(α − ξ)
π22y(α − ξ)2
dξ =
y
−y
e2iαx
4πy
dx,
∞
−∞
sin2
y(α − ξ)
(α − ξ)2
dξ =
π
2
y
−y
dx,
∞
−∞
sin2
y(α − ξ)
(α − ξ)2
dξ = πy.

❹ We had
f(ξ) =
siny(α − ξ)
π
√
2y(α − ξ)
.
• We make change of variables: α − ξ = β − ξ . Then, ξ = ξ + α − β. We have
f(ξ) = f(ξ + α − β) =
siny(β − ξ )
(β − ξ )
, or
f(ξ + α − β) =
siny(β − ξ)
(β − ξ)
.
• We will also use the following result.
Let f(ξ + a
ξ
) = g(ξ), and determinte g(ξ)∨
:
f(ξ + a)∨
= g(ξ)∨
=
1
√
2π R
eixξ
g(ξ) dξ =
1
√
2π R
eix(ξ −a)
f(ξ ) dξ
= e−ixa
f(x).
• Using these results, we have
1
π
∞
−∞
sin(α − ξ)y
(α − ξ)
sin(β − ξ)y
(β − ξ)
dξ =
1
π
(π 2y)2
∞
−∞
f(ξ) f(ξ + α − β) dξ
= 2πy
∞
−∞
f(x) e−(α−β)ix
f(x) dx
= 2πy
∞
−∞
f(x)2
e−(α−β)ix
dx
= 2πy
y
−y
e2iαx
4πy
e−(α−β)ix
dx
=
1
2
y
−y
e−(α−β)ix
dx
=
1
−2i(α − β)
e−(α−β)ix x=y
x=−y
=
1
−2i(α − β)
e−(α−β)iy
− e(α−β)iy
=
sin(α − β)y
α − β
.

Problem (S’95, #5). For the Laplace equation
f ≡
∂2
∂x2
+
∂2
∂y2
f = 0 (30.3)
in the upper half plane y ≥ 0, consider
• the Dirichlet problem f(x, 0) = g(x);
• the Neumann problem ∂
∂y f(x, 0) = h(x).
Assume that f, g and h are 2π periodic in x and that f is bounded at inﬁnity.
Find the Fourier transform N of the Dirichlet-Neumann map. In other words,
ﬁnd an operator N taking the Fourier transform of g to the Fourier transform of h; i.e.
Ngk = hk.
Proof. We solve the problem by two methods.
❶ Fourier Series.
Since f is 2π-periodic in x, we can write
f(x, y) =
∞
n=−∞
an(y) einx
.
Plugging this into (30.3), we get the ODE:
∞
n=−∞
− n2
an(y)einx
+ an(y)einx
= 0,
an(y) − n2
an(y) = 0.
Initial conditions give: (g and h are 2π-periodic in x)
f(x, 0) =
∞
n=−∞
an(0)einx
= g(x) =
∞
n=−∞
gneinx
⇒ an(0) = gn.
fy(x, 0) =
∞
n=−∞
an(0)einx
= h(x) =
∞
n=−∞
hneinx
⇒ an(0) = hn.
Thus, the problems are:
an(y) − n2
an(y) = 0,
an(0) = gn, (Dirichlet)
an(0) = hn. (Neumann)
⇒ an(y) = bneny
+ cne−ny
, n = 1, 2, . . .; a0(y) = b0y + c0.
an(y) = nbneny
− ncne−ny
, n = 1, 2, . . .; a0(y) = b0.
Since f is bounded at y = ±∞, we have:
bn = 0 for n > 0,
cn = 0 for n < 0,
b0 = 0, c0 arbitrary.

• n > 0:
an(y) = cne−ny
,
an(0) = cn = gn, (Dirichlet)
an(0) = −ncn = hn. (Neumann)
⇒ −ngn = hn.
• n < 0:
an(y) = bneny
,
an(0) = bn = gn, (Dirichlet)
an(0) = nbn = hn. (Neumann)
⇒ ngn = hn.
−|n|gn = hn, n = 0.
• n = 0 : a0(y) = c0,
a0(0) = c0 = g0, (Dirichlet)
a0(0) = 0 = h0. (Neumann)
Note that solution f(x, y) may be written as
f(x, y) =
∞
n=−∞
an(y) einx
= a0(y) +
−1
n=−∞
an(y) einx
+
∞
n=1
an(y) einx
= c0 +
−1
n=−∞
bneny
einx
+
∞
n=1
cne−ny
einx
=
g0 + −1
n=−∞ gneny
einx
+ ∞
n=1 gne−ny
einx
, (Dirichlet)
c0 + −1
n=−∞
hn
n eny
einx
+ ∞
n=1 −hn
n e−ny
einx
. (Neumann)
❷ Fourier Transform. The Fourier transform of f(x, y) in x is:
f(ξ, y) =
1
√
2π
∞
−∞
e−ixξ
f(x, y) dx,
f(x, y) =
1
√
2π
∞
−∞
eixξ
f(ξ, y) dξ.
(iξ)2
f(ξ, y) + fyy(ξ, y) = 0,
fyy − ξ2
f = 0. The solution to this ODE is:
f(ξ, y) = c1eξy
+ c2e−ξy
.
For ξ > 0, c1 = 0; for ξ < 0, c2 = 0.
• ξ > 0 : f(ξ, y) = c2e−ξy
, fy(ξ, y) = −ξc2e−ξy
,
c2 = f(ξ, 0) =
1
√
2π
∞
−∞
e−ixξ
f(x, 0) dx =
1
√
2π
∞
−∞
e−ixξ
g(x) dx = g(ξ), (Dirichlet)
−ξc2 = fy(ξ, 0) =
1
√
2π
∞
−∞
e−ixξ
fy(x, 0) dx =
1
√
2π
∞
−∞
e−ixξ
h(x) dx = h(ξ). (Neumann)
⇒ −ξg(ξ) = h(ξ).

• ξ < 0 : f(ξ, y) = c1eξy
, fy(ξ, y) = ξc1eξy
,
c1 = f(ξ, 0) =
1
√
2π
∞
−∞
e−ixξ
f(x, 0) dx =
1
√
2π
∞
−∞
e−ixξ
g(x) dx = g(ξ), (Dirichlet)
ξc1 = fy(ξ, 0) =
1
√
2π
∞
−∞
e−ixξ
fy(x, 0) dx =
1
√
2π
∞
−∞
e−ixξ
h(x) dx = h(ξ). (Neumann)
⇒ ξg(ξ) = h(ξ).
−|ξ|g(ξ) = h(ξ).

Problem (F’97, #3). Consider the Dirichlet problem in the half-space xn > 0,
n ≥ 2:
u + a
∂u
∂xn
+ k2
u = 0, xn > 0
u(x , 0) = f(x ), x = (x1, . . ., xn−1).
Here a and k are constants.
Use the Fourier transform to show that for any f(x ) ∈ L2(Rn−1) there exists a
solution u(x , xn) of the Dirichlet problem such that
Rn
|u(x , xn)|2
dx ≤ C
for all 0 < xn < +∞.
Proof. 84
Denote ξ = (ξ , ξn). Transform in the ﬁrst n − 1 variables:
−|ξ |2
u(ξ , xn) +
∂2u
∂x2
n
(ξ , xn) + a
∂u
∂xn
(ξ , xn) + k2
u(ξ , xn) = 0.
Thus, the ODE and initial conditions of the transformed problem become:
uxnxn + auxn + (k2
− |ξ |2
)u = 0,
u(ξ , 0) = f(ξ ).
With the anzats u = cesxn , we obtain s2 + as + (k2 − |ξ |2) = 0, and
s1,2 =
−a ± a2 − 4(k2 − |ξ |2)
2
.
Choosing only the negative root, we obtain the solution: 85
u(ξ , xn) = c(ξ ) e
−a−
√
a2−4(k2−|ξ |2)
2
xn
. u(ξ , 0) = c = f(ξ ). Thus,
u(ξ , xn) = f(ξ ) e
−a−
√
a2−4(k2−|ξ |2)
2
xn
.
Parseval’s theorem gives:
||u||2
L2(Rn−1) = ||u||2
L2(Rn−1) =
Rn−1
|u(ξ , xn)|2
dξ
=
Rn−1
f(ξ ) e
−a−
√
a2−4(k2−|ξ |2)
2
xn 2
dξ ≤
Rn−1
f(ξ )
2
dξ
= ||f||2
L2(Rn−1) = ||f||2
L2(Rn−1) ≤ C,
since f(x ) ∈ L2
(Rn−1
). Thus, u(x , xn) ∈ L2
(Rn−1
).
84
Note that the last element of x = (x , xn) = (x1, . . . , xn−1, xn), i.e. xn, plays a role of time t.
As such, the PDE may be written as
u + utt + aut + k2
u = 0.
85
Note that a > 0 should have been provided by the statement of the problem.

Problem (F’89, #7). Find the following fundamental solutions
a)
∂G(x, y, t)
∂t
= a(t)
∂2G(x, y, t)
∂x2
+ b(t)
∂G(x, y, t)
∂x
+ c(t)G(x, y, t) for t > 0
G(x, y, 0) = δ(x − y),
where a(t), b(t), c(t) are continuous functions on [0, +∞], a(t) > 0 for t > 0.
b)
∂G
∂t
(x1, . . ., xn, y1, . . ., yn, t) =
n
k=1
ak(t)
∂G
∂xk
for t > 0,
G(x1, . . ., xn, y1, . . ., yn, 0) = δ(x1 − y1)δ(x2 − y2) . . .δ(xn − yn).
Proof. a) We use the Fourier transform to solve this problem.
Transform the equation in the ﬁrst variable only. That is,
G(ξ, y, t) =
1
√
2π R
e−ixξ
G(x, y, t) dx.
The equation is transformed to an ODE, that can be solved:
Gt(ξ, y, t) = −a(t) ξ2
G(ξ, y, t) + i b(t) ξ G(ξ, y, t) + c(t) G(ξ, y, t),
Gt(ξ, y, t) = − a(t) ξ2
+ i b(t) ξ + c(t) G(ξ, y, t),
G(ξ, y, t) = c e
t
0 [−a(s)ξ2+i b(s)ξ+c(s)] ds
.
We can also transform the initial condition:
G(ξ, y, 0) = δ(x − y)(ξ) = e−iyξ
δ(ξ) =
1
√
2π
e−iyξ
.
Thus, the solution of the transformed problem is:
G(ξ, y, t) =
1
√
2π
e−iyξ
e
t
.
The inverse Fourier transform gives the solution to the original problem:
G(x, y, t) = G(ξ, y, t)
∨
=
1
√
2π R
eixξ
G(ξ, y, t) dξ
=
1
√
2π R
eixξ 1
√
2π
e−iyξ
e
t
0
[−a(s)ξ2+i b(s)ξ+c(s)] ds
dξ
=
1
2π R
ei(x−y)ξ
e
t
dξ.
b) Denote x = (x1, . . ., xn), y = (y1, . . ., yn). Transform in x:
G(ξ, y, t) =
1
(2π)
n
2 Rn
e−ix·ξ
G(x, y, t) dx.
The equation is transformed to an ODE, that can be solved:
Gt(ξ, y, t) =
n
k=1
ak(t) iξk G(ξ, y, t),
G(ξ, y, t) = c ei
t
0 [ n
k=1 ak(s) ξk] ds
.

We can also transform the initial condition:
G(ξ, y, 0) = δ(x1 − y1)δ(x2 − y2) . . .δ(xn − yn) (ξ) = e−iy·ξ
δ(ξ) =
1
(2π)
n
2
e−iy·ξ
.
Thus, the solution of the transformed problem is:
G(ξ, y, t) =
1
(2π)
n
2
e−iy·ξ
ei t
0 [ n
k=1 ak(s) ξk] ds
.
The inverse Fourier transform gives the solution to the original problem:
G(x, y, t) = G(ξ, y, t)
∨
=
1
(2π)
n
2 Rn
eix·ξ
G(ξ, y, t) dξ
=
1
(2π)
n
2 Rn
eix·ξ 1
(2π)
n
2
e−iy·ξ
ei t
0 [ n
k=1 ak(s) ξk] ds
dξ
=
1
(2π)n
Rn
ei(x−y)·ξ
ei t
0 [ n
k=1 ak(s) ξk] ds
dξ.

Problem (W’02, #7). Consider the equation
∂2
∂x2
1
+ · · · +
∂2
∂x2
n
u = f in Rn
, (30.4)
where f is an integrable function (i.e. f ∈ L1(Rn)), satisfying f(x) = 0 for |x| ≥ R.
Solve (30.4) by Fourier transform, and prove the following results.
a) There is a solution of (30.4) belonging to L2(Rn) if n > 4.
b) If Rn f(x) dx = 0, there is a solution of (30.4) belonging to L2
(Rn
) if n > 2.
Proof.
u = f,
−|ξ|2
u(ξ) = f(ξ),
u(ξ) = −
1
|ξ|2
f(ξ), ξ ∈ Rn
,
u(x) = −
f(ξ)
|ξ|2
∨
.
a) Then
||u||L2(Rn) =
Rn
|f(ξ)|2
|ξ|4
dξ
1
2
≤
|ξ|<1
|f(ξ)|2
|ξ|4
dξ
A
+
|ξ|≥1
|f(ξ)|2
|ξ|4
dξ
B
1
2
.
Notice, ||f||2 = ||f||2 ≥ B, so B < ∞.
Use polar coordinates on A.
A =
|ξ|<1
|f(ξ)|2
|ξ|4
dξ =
1
0 Sn−1
|f|2
r4
rn−1
dSn−1 dr =
1
0 Sn−1
|f|2
rn−5
dSn−1 dr.
If n > 4,
A ≤
Sn−1
|f|2
dSn−1 = ||f||2
2 < ∞.
||u||L2(Rn) = ||u||L2(Rn) = (A + B)
1
2 < ∞.

b) We have
u(x, t) = −
f(ξ)
|ξ|2
∨
= −
1
(2π)
n
2
∞
−∞
eix·ξ f(ξ)
|ξ|2
dξ
= −
1
(2π)
n
2
∞
−∞
eix·ξ
|ξ|2
1
(2π)
n
2
∞
−∞
e−iy·ξ
f(y) dy dξ
= −
1
(2π)n
∞
−∞
f(y)
∞
−∞
ei(x−y)·ξ
|ξ|2
dξ dy
= −
1
(2π)n
∞
−∞
f(y)
1
0 Sn−1
ei(x−y)r
r2
rn−1
dSn−1 dr dy
= −
1
(2π)n
∞
−∞
f(y)
1
0 Sn−1
ei(x−y)r
rn−3
dSn−1 dr
≤ M < ∞, if n>2.
dy.
|u(x, t)| =
1
(2π)n
∞
−∞
M f(y) dy < ∞.

Problem (F’02, #7). For the right choice of the constant c, the function
F(x, y) = c(x + iy)−1 is a fundamental solution for the equation
∂u
∂x
+ i
∂u
∂y
= f in R2
.
Find the right choice of c, and use your answer to compute the Fourier transform
(in distribution sense) of (x + iy)−1
.
Proof. 86
=
∂
∂x
+ i
∂
∂y
∂
∂x
− i
∂
∂y
.
F1(x, y) = 1
2π log |z| is the fundamental solution of the Laplacian. z = x + iy.
F1(x, y) = δ,
∂
∂x
+ i
∂
∂y
∂
∂x
− i
∂
∂y
F(x, y) = δ.
hx + ihy = e−i(xξ1+yξ2)
.
Suppose h = h(xξ1 + yξ2) or h = ce−i(xξ1+yξ2).
⇒ c − iξ1 e−i(xξ1+yξ2)
− i2
ξ2 e−i(xξ1+yξ2)
= −ic(ξ1 − iξ2) e−i(xξ1+yξ2)
≡ e−i(xξ1+yξ2)
,
⇒ −ic(ξ1 − iξ2) = 1,
⇒ c = −
1
i(ξ1 − iξ2)
,
⇒ h(x, y) = −
1
i(ξ1 − iξ2)
e−i(xξ1+yξ2)
.
Integrate by parts:
1
x + iy
(ξ) =
R2
e−i(xξ1+yξ2) 1
i(ξ1 − iξ2)
∂
∂x
+ i
∂
∂y
1
(x + iy) − 0
dxdy
=
1
i(ξ1 − iξ2)
=
1
i(ξ2 + iξ1)
.
86
Alan solved in this problem in class.

31 Laplace Transform
If u ∈ L1(R+), we define its Laplace transform to be
L[u(t)] = u#
(s) =
∞
0
e−st
u(t) dt (s > 0).
In practice, for a PDE involving time, it may be useful to perform a Laplace transform
in t, holding the space variables x fixed.
The inversion formula for the Laplace transform is:
u(t) = L−1
[u#
(s)] =
1
2πi
c+i∞
c−i∞
est
u#
(s) ds.
Example: f(t) = 1.
L[1] =
∞
0
e−st
· 1 dt = −
1
s
e−st
t=∞
t=0
=
1
s
for s > 0.
Example: f(t) = eat.
L[eat
] =
∞
0
e−st
eat
dt =
∞
0
e(a−s)t
dt =
1
a − s
e(a−s)t
t=∞
t=0
=
1
s − a
for s > a.
Convolution: We want to find an inverse Laplace transform of 1
s · 1
s2+1
.
L−1 1
s
L[f]
·
1
s2 + 1
L[g]
= f ∗ g =
t
0
1 · sint dt = 1 − cos t.
Partial Derivatives: u = u(x, t)
L[ut] =
∞
0
e−st
ut dt = e−st
u(x, t)
t=∞
t=0
+ s
∞
0
e−st
u dt = sL[u] − u(x, 0),
L[utt] =
∞
0
e−st
utt dt = e−st
ut
t=∞
t=0
+ s
∞
0
e−st
ut dt = −ut(x, 0) + sL[ut]
= s2
L[u] − su(x, 0) − ut(x, 0),
L[ux] =
∞
0
e−st
ux dt =
∂
∂x
L[u],
L[uxx] =
∞
0
e−st
uxx dt =
∂2
∂x2
L[u].
Heat Equation: Consider
ut − u = 0 in U × (0, ∞)
u = f on U × {t = 0},
and perform a Laplace transform with respect to time:
L[ut] =
∞
0
e−st
ut dt = sL[u] − u(x, 0) = sL[u] − f(x),
L[ u] =
∞
0
e−st
u dt = L[u].

Thus, the transformed problem is: sL[u] − f(x) = L[u]. Writing v(x) = L[u], we
have
− v + sv = f in U.
Thus, the solution of this equation with RHS f is the Laplace transform of the solution
of the heat equation with initial data f.

Table of Laplace Transforms: L[f] = f#
(s)
L[sinat] =
a
s2 + a2
, s > 0
L[cos at] =
s
s2 + a2
, s > 0
L[sinhat] =
a
s2 − a2
, s > |a|
L[coshat] =
s
s2 − a2
, s > |a|
L[eat
sinbt] =
b
(s − a)2 + b2
, s > a
L[eat
cos bt] =
s − a
(s − a)2 + b2
, s > a
L[tn
] =
n!
sn+1
, s > 0
L[tn
eat
] =
n!
(s − a)n+1
, s > a
L[H(t − a)] =
e−as
s
, s > 0
L[H(t − a) f(t − a)] = e−as
L[f],
L[af(t) + bg(t)] = aL[f] + bL[g],
L[f(t) ∗ g(t)] = L[f] L[g],
L
t
0
g(t − t) f(t ) dt = L[f] L[g],
L
df
dt
= sL[f] − f(0),
L
d2
f
dt2
= s2
L[f] − sf(0) − f (0), f =
df
dt
L
dnf
dtn
= sn
L[f] − sn−1
f(0) − . . . − fn−1
(0),
L[f(at)] =
1
a
f# s
a
,
L[ebt
f(t)] = f#
(s − b),
L[tf(t)] = −
d
ds
L[f],
L
f(t)
t
=
∞
s
f#
(s ) ds ,
L
t
0
f(t ) dt =
1
s
L[f],
L[J0(at)] = (s2
+ a2
)−1
2 ,
L[δ(t − a)] = e−sa
.
Example: f(t) = sint. After integrating by parts twice, we obtain:
L[sint] =
∞
0
e−st
sint dt = 1 − s2
∞
0
e−st
sin t dt,
⇒
∞
0
e−st
sint dt =
1
1 + s2
.

Example: f(t) = tn
.
L[tn
] =
∞
0
e−st
tn
dt = −
tne−st
s
∞
0
+
n
s
∞
0
e−st
tn−1
dt =
n
s
L[tn−1
]
=
n
s
n − 1
s
L[tn−2
] = . . . =
n!
sn
L[1] =
n!
sn+1
.

ut − uxx + au = 0, t > 0, x > 0
u(x, 0) = 0, x > 0
u(0, t) = g(t), t > 0,
where g(t) is continuous function with a compact support, and a is constant.
Find the explicit solution of this problem.
Proof. We solve this problem using the Laplace transform.
L[u(x, t)] = u#
(x, s) =
∞
0
e−st
u(x, t) dt (s > 0).
L[ut] =
∞
0
e−st
ut dt = e−st
u(x, t)
t=∞
t=0
+ s
∞
0
e−st
u dt
= su#
(x, s) − u(x, 0) = su#
(x, s), (since u(x, 0) = 0)
L[uxx] =
∞
0
e−st
uxx dt =
∂2
∂x2
u#
(x, s),
L[u(0, t)] = u#
(0, s) =
∞
0
e−st
g(t) dt = g#
(s).
Plugging these into the equation, we obtain the ODE in u#
:
su#
(x, s) −
∂2
∂x2
u#
(x, s) + au#
(x, s) = 0.
(u#)xx − (s + a)u# = 0,
u#(0, s) = g#(s).
This initial value problem has a solution:
u#
(x, s) = c1e
√
s+a x
+ c2e−
√
s+a x
.
Since we want u to be bounded as x → ∞, we have c1 = 0, so
u#
(x, s) = c2e−
√
s+a x
. u#
(0, s) = c2 = g#
(s), thus,
u#
(x, s) = g#
(s)e−
√
s+a x
.
To obtain u(x, t), we take the inverse Laplace transform of u#(x, s):
u(x, t) = L−1
[u#
(x, s)] = L−1
g#
(s)
L[g]
e−
√
s+a x
L[f]
= g ∗ f
= g ∗ L−1
e−
√
s+a x
= g ∗
1
2πi
c+i∞
c−i∞
est
e−
√
s+a x
ds ,
u(x, t) =
t
0
g(t − t )
1
2πi
c+i∞
c−i∞
est
e−
√
s+a x
ds dt .

Problem (F’04, #8). The function y(x, t) satisfies the partial differential equation
x
∂y
∂x
+
∂2y
∂x∂t
+ 2y = 0,
and the boundary conditions
y(x, 0) = 1, y(0, t) = e−at
,
where a ≥ 0. Find the Laplace transform, y(x, s), of the solution, and hence derive
an expression for y(x, t) in the domain x ≥ 0, t ≥ 0.
Proof. We change the notation: y → u. We have
xux + uxt + 2u = 0,
u(x, 0) = 1, u(0, t) = e−at
.
The Laplace transform is defined as:
L[u(x, t)] = u#
(x, s) =
∞
0
e−st
u(x, t) dt (s > 0).
L[xux] =
∞
0
e−st
xux dt = x
∞
0
e−st
ux dt = x(u#
)x,
L[uxt] =
∞
0
e−st
uxt dt = e−st
ux(x, t)
t=∞
t=0
+ s
∞
0
e−st
ux dt
= s(u#
)x − ux(x, 0) = s(u#
)x, (since u(x, 0) = 0)
L[u(0, t)] = u#
(0, s) =
∞
0
e−st
e−at
dt =
∞
0
e−(s+a)t
dt = −
1
s + a
e−(s+a)t
t=∞
t=0
=
1
s + a
.
Plugging these into the equation, we obtain the ODE in u#:
(x + s)(u#
)x + 2u#
= 0,
u#
(0, s) = 1
s+a ,
which can be solved:
(u#
)x
u#
= −
2
x + s
⇒ log u#
= −2 log(x + s) + c1 ⇒ u#
= c2elog(x+s)−2
=
c2
(x + s)2
.
From the initial conditions:
u#
(0, s) =
c2
s2
=
1
s + a
⇒ c2 =
s2
s + a
.
u#
(x, s) =
s2
(s + a)(x + s)2
.
To obtain u(x, t), we take the inverse Laplace transform of u#(x, s):
u(x, t) = L−1
[u#
(x, s)] = L−1 s2
(s + a)(x + s)2
=
1
2πi
c+i∞
c−i∞
est s2
(s + a)(x + s)2
ds.
u(x, t) =
1
2πi
c+i∞
c−i∞
est s2
(s + a)(x + s)2
ds.

Problem (F’90, #1). Using the Laplace transform, or any other convenient method,
solve the Volterra integral equation
u(x) = sinx +
x
0
sin(x − y)u(y) dy.
Proof. Rewrite the equation:
u(t) = sint +
t
0
sin(t − t )u(t ) dt ,
u(t) = sint + (sint) ∗ u.
Taking the Laplace transform of each of the elements in :
L[u(t)] = u#
(s) =
∞
0
e−st
u(t) dt,
L[sint] =
1
1 + s2
,
L[(sint) ∗ u] = L[sint] ∗ L[u] =
u#
1 + s2
.
Plugging these into the equation:
u#
=
1
1 + s2
+
u#
1 + s2
=
u# + 1
1 + s2
.
u#
(s) =
1
s2
.
To obtain u(t), we take the inverse Laplace transform of u#(s):
u(t) = L−1
[u#
(s)] = L−1 1
s2
= t.
u(t) = t.

Problem (F’91, #5). In what follows, the Laplace transform of x(t) is denoted
either by x(s) or by Lx(t). ❶ Show that, for integral n ≥ 0,
L(tn
) =
n!
sn+1
.
❷ Hence show that
LJ0(2
√
ut) =
1
s
e−u/s
,
where
J0(z) =
∞
n=0
(−1)n
(1
2z)2n
n!n!
is a Bessel function. ❸ Hence show that
L
∞
0
J0(2
√
ut)x(u) du =
1
s
x
1
s
. (31.1)
❹ Assuming that
LJ0(at) =
1
√
a2 + s2
,
prove with the help of (31.1) that if t ≥ 0
∞
0
J0(au)J0(2
√
ut) du =
1
a
J0
t
a
.
Hint: For the last part, use the uniqueness of the Laplace transform.
Proof.
❶ L[tn
] =
∞
0
e−st
g
tn
f
dt = −
tn
e−st
s
∞
0
= 0
+
n
s
∞
0
e−st
tn−1
dt =
n
s
L[tn−1
]
=
n
s
n − 1
s
L[tn−2
] = . . . =
n!
sn
L[1] =
n!
sn+1
.
❷ LJ0(2
√
ut) = L
∞
n=0
(−1)n
un
tn
n!n!
=
∞
n=0
(−1)n
un
n!n!
L[tn
] =
∞
n=0
(−1)n
un
n!sn+1
=
1
s
∞
n=0
(−1)n
n!
u
s
n
=
1
s
e−u
s .
❸ L
∞
0
J0(2
√
ut) x(u) du =
∞
0
L[J0(2
√
ut)] x(u) du =
1
s
∞
0
e−u
s x(u) du
=
1
s
x# 1
s
,
where
x#
(s) =
∞
0
e−us
x(u) du.

32 Linear Functional Analysis
32.1 Norms
|| · || is a norm on a vector space X if
i) ||x|| = 0 iff x = 0.
ii) ||αx|| = |α| · ||x|| for all scalars α.
iii) ||x + y|| ≤ ||x|| + ||y|| (the triangle inequality).
The norm induces the distance function d(x, y) = ||x − y|| so that X is a metric space,
called a normed vector space.
32.2 Banach and Hilbert Spaces
A Banach space is a normed vector space that is complete in that norm’s metric. I.e.
a complete normed linear space is a Banach space.
A Hilbert space is an inner product space for which the corresponding normed space
is complete. I.e. a complete inner product space is a Hilbert space.
Examples: 1) Let K be a compact set of Rn and let C(K) denote the space of continuous
functions on K. Since every u ∈ C(K) achieves maximum and minimum values on K,
we may define
||u||∞ = max
x∈K
|u(x)|.
|| · ||∞ is indeed a norm on C(K) and since a uniform limit of continuous functions is
continuous, C(K) is a Banach space. However, this norm cannot be derived from an
inner product, so C(K) is not a Hilbert space.
2) C(K) is not a Banach space with || · ||2 norm. (Bell-shaped functions on [0, 1] may
converge to a discontinuous δ-function). In general, the space of continuous functions
on [0, 1], with the norm || · ||p, 1 ≤ p < ∞, is not a Banach space, since it is not
complete.
3) Rn and Cn are real and complex Banach spaces (with a Eucledian norm).
4) Lp
are Banach spaces (with || · ||p norm).
5) The space of bounded real-valued functions on a set S, with the sup norm || · ||S are
Banach spaces.
6) The space of bounded continuous real-valued functions on a metric space X is a
Banach space.
32.3 Cauchy-Schwarz Inequality
|(u, v)| ≤ ||u||||v|| in any norm, for example |uv|dx ≤ ( u2
dx)
1
2 ( v2
dx)
1
2
|a(u, v)| ≤ a(u, u)
1
2 a(v, v)
1
2
|v|dx = |v| · 1 dx = ( |v|2
dx)
1
2 ( 12
dx)
1
2
32.4 Hölder Inequality
Ω
|uv| dx ≤ ||u||p||v||q,
which holds for u ∈ Lp
(Ω) and v ∈ Lq
(Ω), where 1
p + 1
q = 1. In particular, this shows
uv ∈ L1
(Ω).

32.5 Minkowski Inequality
||u + v||p ≤ ||u||p + ||v||p,
which holds for u, v ∈ Lp(Ω). In particular, it shows u + v ∈ Lp(Ω).
Using the Minkowski Inequality, we find that || · ||p is a norm on Lp
(Ω).
The Riesz-Fischer theorem asserts that Lp(Ω) is complete in this norm, so Lp(Ω) is a
Banach space under the norm || · ||p.
If p = 2, then L2(Ω) is a Hilbert space with inner product
(u, v) =
Ω
uv dx.
Example: Ω ∈ Rn
bounded domain, C1
(¯Ω) denotes the functions that, along with
their first-order derivatives, extend continuously to the compact set ¯Ω. Then C1
(¯Ω) is
a Banach space under the norm
||u||1,∞ = max
x∈¯Ω
(|∇u(x)| + |u(x)|).
Note that C1(Ω) is not a Banach space since ||u||1,∞ need not be finite for u ∈ C1(Ω).
32.6 Sobolev Spaces
A Sobolev space is a space of functions whose distributional derivatives (up to some
fixed order) exist in an Lp
-space.
Let Ω be a domain in Rn, and let us introduce
< u, v >1=
Ω
(∇u · ∇v + uv) dx, (32.1)
||u||1,2 =
√
< u, u >1 =
Ω
(|∇u|2
+ |u|2
) dx
1
2
(32.2)
when these expressions are defined and finite. For example, (32.1) and (32.2) are defined
for functions in C1
0 (Ω). However, C1
0 (Ω) is not complete under the norm (32.2), and so
does not form a Hilbert space.
Divergence Theorem
∂Ω
A · n dS =
Ω
div A dx
Trace Theorem
u L2(∂Ω) ≤ C u H1(Ω) Ω smooth or square
Poincare Inequality
u p ≤ C ∇u p 1 ≤ p ≤ ∞
Ω
|u(x)|2
dx ≤ C
Ω
|∇u(x)|2
dx u ∈ C1
0 (Ω), H1,2
0 (Ω) i.e. p = 2
u − uΩ p ≤ ∇u p u ∈ H1,p
0 (Ω)

uΩ =
1
|Ω| Ω
u(x) dx (Average value of u over Ω), |Ω| is the volume of Ω
Notes
∂u
∂n
= ∇u · n = n1
∂u
∂x1
+ n2
∂u
∂x2
|∇u|2
= u2
x1
+ u2
x2
Ω
∇|u| dx =
Ω
|u|
u
∇u dx
√
ab ≤
a + b
2
⇒ ab ≤
a2 + b2
2
⇒ ||∇u||||u|| ≤
||∇u||2 + ||u||2
2
u∇u = ∇(
u2
2
)
Ω
(uxy)2
dx =
Ω
uxxuyy dx ∀u ∈ H2
0 (Ω) Ω square
Problem (F’04, #6). Let q ∈ C1
0 (R3
). Prove that the vector field
u(x) =
1
4π R3
q(y)(x − y)
|x − y|3
dy
enjoys the following properties: 87
a) u(x) is conservative;
b) div u(x) = q(x) for all x ∈ R3;
c) |u(x)| = O(|x|−2
) for large x.
Furthermore, prove that the proverties (1), (2), and (3) above determine the vector field
u(x) uniquely.
Proof. a) To show that u(x) is conservative, we need to show that curl u = 0.
The curl of V is another vector field defined by
curl V = ∇ × V = det
⎛
⎝
e1 e2 e3
∂1 ∂2 ∂3
V1 V2 V3
⎞
⎠ =
∂V3
∂x2
−
∂V2
∂x3
,
∂V1
∂x3
−
∂V3
∂x1
,
∂V2
∂x1
−
∂V1
∂x2
.
Consider
V (x) =
x
|x|3
=
(x1, x2, x3)
(x2
1 + x2
2 + x2
3)
3
2
.
Then,
u(x) =
1
4π R3
q(y) V (x − y) dy,
curl u(x) =
1
4π R3
q(y) curlx V (x − y) dy.
curl V (x) = curl
(x1, x2, x3)
(x2
1 + x2
2 + x2
3)
3
2
=
−3
2 · 2x2x3
(x2
1 + x2
2 + x2
3)
5
2
−
−3
2 · 2x3x2
(x2
1 + x2
2 + x2
3)
5
2
,
−3
2 · 2x3x1
(x2
1 + x2
2 + x2
3)
5
2
−
−3
2 · 2x1x3
(x2
1 + x2
2 + x2
3)
5
2
,
−3
2 · 2x1x2
(x2
1 + x2
2 + x2
3)
5
2
−
= (0, 0, 0).
87
McOwen, p. 138-140.

Thus, curl u = 1
4π R3 q(y) · 0 dy = 0, and u(x) is conservative.
b) Note that the Laplace kernel in R3 is − 1
4πr .
u(x) =
1
4π R3
q(y)(x − y)
|x − y|3
dy =
1
4π R3
q(r) r
r3
r dr =
R3
q(r)
4πr
dr = q.
c) Consider
F(x) = −
1
4π R3
q(y)
|x − y|
dy.
F(x) is O(|x|−1) as |x| → ∞.
Note that u = ∇F, which is clearly O(|x|−2
) as |x| → ∞.

Partial differential equations, graduate level problems and solutions by igor yanovsky

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