Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 2 potx

This solution demonstrates the domain of dependence of the solution. The first term is an integral over the triangle
domain {(ξ, τ) : 0 < τ < t, x − cτ < ξ < x + cτ}. The second term involves only the points (x ± ct, 0). The third
term is an integral on the line segment {(ξ, 0) : x −ct < ξ < x + ct}. In totallity, this is just the triangle domain. This
is shown graphically in Figure 45.4.
x-ct x+ct
Domain of
Dependence
x,t
Figure 45.4: Domain of dependence for the wave equation.
Solution 45.18
Single Sum Representation. First we find the eigenfunctions of the homogeneous problem ∆u − k
2
u = 0. We
substitute the separation of variables, u(x, y) = X(x)Y (y) into the partial differential equation.
X

Y + XY

− k
2
XY = 0
X

X
= k
2
−
Y

Y

= −λ
2
We have the regular Sturm-Liouville eigenvalue problem,
X

= −λ
2
X, X(0) = X(a) = 0,
2014
which has the solutions,
λ
n
=
nπ
a
, X
n
= sin

nπx
a

, n ∈ N.
We expand the solution u in a series of these eigenfunctions.
G(x, y; ξ, ψ) =
∞

n=1
c
n

(y) sin

nπx
a

We substitute this series into the partial differential equation to find equations for the c
n
(y).
∞

n=1

−

nπ
a

2
c
n
(y) + c

n
(y) − k
2
c
n
(y)

sin


nπx
a

= δ(x − ξ)δ(y −ψ)
The series expansion of the right side is,
δ(x − ξ)δ(y −ψ) =
∞

n=1
d
n
(y) sin

nπx
a

d
n
(y) =
2
a

a
0
δ(x − ξ)δ(y −ψ) sin

nπx
a


dx
d
n
(y) =
2
a
sin

nπξ
a

δ(y −ψ).
The the equations for the c
n
(y) are
c

n
(y) −

k
2
+

nπ
a

2

c

n
(y) =
2
a
sin

nπξ
a

δ(y −ψ), c
n
(0) = c
n
(b) = 0.
The homogeneous solutions are {cosh(σ
n
y), sinh(σ
n
y)}, where σ
n
=

k
2
(nπ/a)
2
. The solutions that satisfy the
boundary conditions at y = 0 and y = b are, sinh(σ
n
y) and sinh(σ

n
(y − b)), respectively. The Wronskian of these
2015
solutions is,
W (y) =




sinh(σ
n
y) sinh(σ
n
(y −b))
σ
n
cosh(σ
n
y) σ
n
cosh(σ
n
(y −b))




= σ
n
(sinh(σ

n
y) cosh(σ
n
(y −b)) − sinh(σ
n
(y −b)) cosh(σ
n
y))
= σ
n
sinh(σ
n
b).
The solution for c
n
(y) is
c
n
(y) =
2
a
sin

nπξ
a

sinh(σ
n
y
<

) sinh(σ
n
(y
>
− b))
σ
n
sinh(σ
n
b)
.
The Green function for the partial differential equation is
G(x, y; ξ, ψ) =
2
a
∞

n=1
sinh(σ
n
y
<
) sinh(σ
n
(y
>
− b))
σ
n
sinh(σ

n
b)
sin

nπx
a

sin

nπξ
a

.
Solution 45.19
We take the Fourier cosine transform in x of the partial differential equation and the boundary condition along y = 0.
G
xx
+ G
yy
− k
2
G = δ(x − ξ)δ(y −ψ)
−α
2
ˆ
G(α, y) −
1
π
ˆ
G

x
(0, y) +
ˆ
G
yy
(α, y) − k
2
ˆ
G(α, y) =
1
π
cos(αξ)δ(y −ψ)
ˆ
G
yy
(α, y) − (k
2
+ α
2
)
ˆ
G(α, y) ==
1
π
cos(αξ)δ(y −ψ),
ˆ
G(α, 0) = 0
Then we take the Fourier sine transform in y.
−β
2

ˆ
ˆ
G(α, β) +
β
π
ˆ
ˆ
G(α, 0) − (k
2
+ α
2
)
ˆ
ˆ
G(α, β) =
1
π
2
cos(αξ) sin(βψ)
ˆ
ˆ
G = −
cos(αξ) sin(βψ)
π
2
(k
2
+ α
2
+ β

2
)
2016
We take two inverse transforms to find the solution. For one integral representation of the Green function we take the
inverse sine transform followed by the inverse cosine transform.
ˆ
ˆ
G = −cos(αξ)
sin(βψ)
π
1
π(k
2
+ α
2
+ β
2
)
ˆ
ˆ
G = −cos(αξ)F
s
[δ(y −ψ)]F
c

1
√
k
2
+ α

2
e
−
√
k
2
+α
2
y

ˆ
G(α, y) = −cos(αξ)
1
2π

∞
0
δ(z −ψ)
1
√
k
2
+ α
2

exp

−
√
k

2
+ α
2
|y −z|

− exp

−
√
k
2
+ α
2
(y + z)

dz
ˆ
G(α, y) = −
cos(αξ)
2π
√
k
2
+ α
2

exp

−
√

k
2
+ α
2
|y −ψ|

− exp

−
√
k
2
+ α
2
(y + ψ)

G(x, y; ξ, ψ) = −
1
π

∞
0
cos(αξ)
√
k
2
+ α
2

exp


−
√
k
2
+ α
2
|y −ψ|

− exp

−
√
k
2
+ α
2
(y + ψ)

dα
For another integral representation of the Green function, we take the inverse cosine transform followed by the inverse
sine transform.
ˆ
ˆ
G(α, β) = −sin(βψ)
cos(αξ)
π
1
π(k
2

+ α
2
+ β
2
)
ˆ
ˆ
G(α, β) = −sin(βψ)F
c
[δ(x − ξ)]F
c

1

k
2
+ β
2
e
−
√
k
2
+β
2
x

ˆ
G(x, β) = −sin(βψ)
1

2π

∞
0
δ(z −ξ)
1

k
2
+ β
2

e
−
√
k
2
+β
2
|x−z|
+
e
−
√
k
2
+β
2
(x+z)


dz
ˆ
G(x, β) = −sin(βψ)
1
2π
1

k
2
+ β
2

e
−
√
k
2
+β
2
|x−ξ|
+
e
−
√
k
2
+β
2
(x+ξ)


G(x, y; ξ, ψ) = −
1
π

∞
0
sin(βy) sin(βψ)

k
2
+ β
2

e
−
√
k
2
+β
2
|x−ξ|
+
e
−
√
k
2
+β
2
(x+ξ)


dβ
2017
Solution 45.20
The problem is:
G
rr
+
1
r
G
r
+
1
r
2
G
θθ
=
δ(r −ρ)δ(θ −ϑ)
r
, 0 < r < ∞, 0 < θ < α,
G(r, 0, ρ, ϑ) = G(r, α, ρ, ϑ) = 0,
G(0, θ, ρ, ϑ) = 0
G(r, θ, ρ, ϑ) → 0 as r → ∞.
Let w = r
e
iθ
and z = x + iy. We use the conformal mapping, z = w
π/α

to map the sector to the uppe r half z plane.
The problem is (x, y) space is
G
xx
+ G
yy
= δ(x − ξ)δ(y −ψ), −∞ < x < ∞, 0 < y < ∞,
G(x, 0, ξ, ψ) = 0,
G(x, y, ξ, ψ) → 0 as x, y → ∞.
We will solve this problem with the method of images. Note that the solution of,
G
xx
+ G
yy
= δ(x − ξ)δ(y −ψ) − δ(x − ξ)δ(y + ψ), −∞ < x < ∞, −∞ < y < ∞,
G(x, y, ξ, ψ) → 0 as x, y → ∞,
satisfies the condition, G(x, 0, ξ, ψ) = 0. Since the infinite space Green fun ction for the Laplacian in two di mens ions is
1
4π
ln

(x − ξ)
2
+ (y −ψ)
2

,
the solution of this problem is,
G(x, y, ξ, ψ) =
1

4π
ln

(x − ξ)
2
+ (y −ψ)
2

−
1
4π
ln

(x − ξ)
2
+ (y + ψ)
2

=
1
4π
ln

(x − ξ)
2
+ (y −ψ)
2
(x − ξ)
2
+ (y + ψ)

2

.
2018
Now we solve for x and y in the conformal mapping.
z = w
π/α
= (r
e
iθ
)
π/α
x + iy = r
π/α
(cos(θπ/α) + i sin(θπ/α))
x = r
π/α
cos(θπ/α), y = r
π/α
sin(θπ/α)
We substitute these expressions into G(x, y, ξ, ψ) to obtain G(r, θ, ρ, ϑ).
G(r, θ, ρ, ϑ) =
1
4π
ln

(r
π/α
cos(θπ/α) − ρ
π/α

cos(ϑπ/α))
2
+ (r
π/α
sin(θπ/α) − ρ
π/α
sin(ϑπ/α))
2
(r
π/α
cos(θπ/α) − ρ
π/α
cos(ϑπ/α))
2
+ (r
π/α
sin(θπ/α) + ρ
π/α
sin(ϑπ/α))
2

=
1
4π
ln

r
2π/α
+ ρ
2π/α

− 2r
π/α
ρ
π/α
cos(π(θ −ϑ)/α)
r
2π/α
+ ρ
2π/α
− 2r
π/α
ρ
π/α
cos(π(θ + ϑ)/α)

=
1
4π
ln

(r/ρ)
π/α
/2 + (ρ/r)
π/α
/2 − cos(π(θ −ϑ)/α)
(r/ρ)
π/α
/2 + (ρ/r)
π/α
/2 − cos(π(θ + ϑ)/α)


=
1
4π
ln

e
π ln(r/ρ)/α
/2 +
e
π ln(ρ/r)/α
/2 − cos(π(θ −ϑ)/α)
e
π ln(r/ρ)/α
/2 +
e
π ln(ρ/r)/α
/2 − cos(π(θ + ϑ)/α)

G(r, θ, ρ, ϑ) =
1
4π
ln


cosh

π/α
ln
r

ρ

− cos(π(θ −ϑ)/α)
cosh

π/α
ln
r
ρ

− cos(π(θ + ϑ)/α)


Now recall that the solution of
∆u = f(x),
subject to the boundary condition,
u(x) = g(x),
is
u(x) =
 
f(xi)G(x; xi) dA
ξ
+

g(xi)∇
ξ
G(x; xi) ·
ˆ
n ds
ξ

.
2019
The normal directions along the l ower and upper edges of the sector are −
ˆ
θ and
ˆ
θ, respectively. The gradient in polar
coordinates is
∇
ξ
= ˆρ
∂
∂ρ
+
ˆ
ϑ
ρ
∂
∂ϑ
.
We only need to compute the
ˆ
ϑ component of the gradient of G. This is
1
ρ
∂
∂ρ
G =
sin(π(θ −ϑ)/α)
4αρ


cosh

π
α
ln
r
ρ

− cos(π(θ −ϑ)/α)

+
sin(π(θ −ϑ)/α)
4αρ

cosh

π
α
ln
r
ρ

− cos(π(θ + ϑ)/α)

Along ϑ = 0, this is
1
ρ
G
ϑ

(r, θ, ρ, 0) =
sin(πθ/α)
2αρ

cosh

π
α
ln
r
ρ

− cos(πθ/α)

.
Along ϑ = α, this is
1
ρ
G
ϑ
(r, θ, ρ, α) = −
sin(πθ/α)
2αρ

cosh

π
α
ln
r

ρ

+ cos(πθ/α)

.
2020
The solution of our problem is
u(r, θ) =

c
∞
−
sin(πθ/α)
2αρ

cosh

π
α
ln
r
ρ

+ cos(πθ/α)

dρ +

∞
c
−

sin(πθ/α)
2αρ

cosh

π
α
ln
r
ρ

− cos(πθ/α)

dρ
u(r, θ) =

∞
c
−sin(πθ/α)
2αρ

cosh

π
α
ln
r
ρ

− cos(πθ/α)


+
sin(πθ/α)
2αρ

cosh

π
α
ln
r
ρ

+ cos(πθ/α)

dρ
u(r, θ) = −
1
α
sin

πθ
α

cos

πθ
α



∞
c
1
ρ

cosh
2

π
α
ln
r
ρ

− cos
2

πθ
α


dρ
u(r, θ) = −
1
α
sin

πθ
α


cos

πθ
α


∞
ln(c/r)
1
cosh
2

πx
α

− cos
2

πθ
α

dx
u(r, θ) = −
2
α
sin

πθ
α


cos

πθ
α


∞
ln(c/r)
1
cosh

2πx
α

− cos

2πθ
α

dx
Solution 45.21
First consider the Green function for
u
t
− κu
xx
= 0, u(x, 0) = f(x).
The differential equation and initial condition is
G
t

= κG
xx
, G(x, 0; ξ) = δ(x − ξ).
The Green function is a solution of the homogeneous heat equation for the initial condition of a unit amount of heat
concentrated at the point x = ξ. You can verify that the Green function is a solution of the heat equation for t > 0
and that it has the property:

∞
−∞
G(x, t; ξ) dx = 1, for t > 0.
This property demonstrates that the total amount of heat is the constant 1. At time t = 0 the heat is concentrated at
the point x = ξ. As time increases, the heat diffuses out from this point.
2021
The solution for u(x, t) is the linear combination of the Green functions that satisfies the initial condition u(x, 0) =
f(x). This linear combination is
u(x, t) =

∞
−∞
G(x, t; ξ)f(ξ) dξ.
G(x, t; 1) and G(x, t; −1) are plotted in Figure 45.5 for the domain t ∈ [1/100 1/4], x ∈ [−2 2] and κ = 1.
0.1
0.2
-2
-1
0
1
2
0
1

2
0.1
0.2
Figure 45.5: G(x, t; 1) and G(x, t; −1)
Now we consider the problem
u
t
= κu
xx
, u(x, 0) = f(x) for x > 0, u(0, t) = 0.
2022
Note that the solution of
G
t
= κG
xx
, x > 0, t > 0,
G(x, 0; ξ) = δ(x − ξ) − δ(x + ξ),
satisfies the boundary condition G(0, t; ξ) = 0. We write the solution as the difference of infinite space Green functions.
G(x, t; ξ) =
1
√
4πκt
e
−(x−ξ)
2
/(4κt)
−
1
√

4πκt
e
−(x+ξ)
2
/(4κt)
=
1
√
4πκt

e
−(x−ξ)
2
/(4κt)
−
e
−(x+ξ)
2
/(4κt)

G(x, t; ξ) =
1
√
4πκt
e
−(x
2
+ξ
2
)/(4κt)

sinh

xξ
2κt

Next we consider the problem
u
t
= κu
xx
, u(x, 0) = f(x) for x > 0, u
x
(0, t) = 0.
Note that the solution of
G
t
= κG
xx
, x > 0, t > 0,
G(x, 0; ξ) = δ(x − ξ) + δ(x + ξ),
satisfies the boundary condition G
x
(0, t; ξ) = 0. We write the solution as the sum of infinite space Green functions.
G(x, t; ξ) =
1
√
4πκt
e
−(x−ξ)
2

/(4κt)
+
1
√
4πκt
e
−(x+ξ)
2
/(4κt)
G(x, t; ξ) =
1
√
4πκt
e
−(x
2
+ξ
2
)/(4κt)
cosh

xξ
2κt

The Green functions for the two boundary conditions are shown in Figure 45.6.
2023
0.05
0.1
0.15
0.2

0.25
0
0.2
0.4
0.6
0.8
1
0
1
2
0.05
0.1
0.15
0.2
0.25
0.05
0.1
0.15
0.2
0.25
0
0.2
0.4
0.6
0.8
1
0
1
2
0.05

0.1
0.15
0.2
0.25
Figure 45.6: Green functions for the boundary conditions u(0, t) = 0 and u
x
(0, t) = 0.
Solution 45.22
a) The Green function problem is
G
tt
− c
2
G
xx
= δ(t − τ)δ(x − ξ), 0 < x < L, t > 0,
G(0, t; ξ, τ) = G
x
(L, t; ξ, τ) = 0,
G(x, t; ξ, τ) = 0 for t < τ.
The condition that G is zero for t < τ makes this a causal Green function. We solve this problem by expanding G in
a series of eigenfunctions of the x variable. The coefficients in the expansion will be functions of t. First we find the
eigenfunctions of x in the h omogeneous problem. We substitute the separation of variables u = X(x)T(t) into the
2024
homogeneous partial differential equation.
XT

= c
2
X


T
T

c
2
T
=
X

X
= −λ
2
The eigenvalue problem is
X

= −λ
2
X, X(0) = X

(L) = 0,
which has the solutions,
λ
n
=
(2n − 1)π
2L
, X
n
= sin


(2n − 1)πx
2L

, n ∈ N.
The series expansion of the Green function has the form,
G(x, t; ξ, τ) =
∞

n=1
g
n
(t) sin

(2n − 1)πx
2L

.
We determine the coefficients by subs tituting the expansion into the Green function differential equation.
G
tt
− c
2
G
xx
= δ(x − ξ)δ(t − τ)
∞

n=1


g

n
(t) +

(2n − 1)πc
2L

2
g
n
(t)

sin

(2n − 1)πx
2L

= δ(x − ξ)δ(t − τ)
We need to expand the right side of the equation in the sine series
δ(x − ξ)δ(t −τ) =
∞

n=1
d
n
(t) sin

(2n − 1)πx
2L


d
n
(t) =
2
L

L
0
δ(x − ξ)δ(t −τ) sin

(2n − 1)πx
2L

dx
d
n
(t) =
2
L
sin

(2n − 1)πξ
2L

δ(t − τ)
2025
By equating coefficients in the sine series, we obtain ordinary differential equation Green function problems for the g
n
’s.

g

n
(t; τ) +

(2n − 1)πc
2L

2
g
n
(t; τ) =
2
L
sin

(2n − 1)πξ
2L

δ(t − τ)
From the causality condition for G, we have the causality conditions for the g
n
’s,
g
n
(t; τ) = g

n
(t; τ) = 0 for t < τ.
The continuity and jump conditions for the g

n
are
g
n
(τ
+
; τ) = 0, g

n
(τ
+
; τ) =
2
L
sin

(2n − 1)πξ
2L

.
A set of homogeneous solutions of the ordinary differential equation are

cos

(2n − 1)πct
2L

, sin

(2n − 1)πct

2L

Since the continuity and jump c onditions are given at the point t = τ, a handy set of solutions to use for this problem
is the fundamental set of solutions at that point:

cos

(2n − 1)πc(t − τ)
2L

,
2L
(2n − 1)πc
sin

(2n − 1)πc(t − τ)
2L

The solution that satisfies the causality condition and the continuity and jump conditions is,
g
n
(t; τ) =
4
(2n − 1)πc
sin

(2n − 1)πξ
2L

sin


(2n − 1)πc(t − τ)
2L

H(t − τ).
Substituting this into the sum yields,
G(x, t; ξ, τ) =
4
πc
H(t − τ)
∞

n=1
1
2n − 1
sin

(2n − 1)πξ
2L

sin

(2n − 1)πc(t − τ)
2L

sin

(2n − 1)πx
2L


.
We use trigonometric identities to write this in terms of traveling waves.
2026
G(x, t; ξ, τ) =
1
πc
H(t − τ)
∞

n=1
1
2n − 1

sin

(2n − 1)π((x − ξ) − c(t − τ))
2L

+ sin

(2n − 1)π((x − ξ) + c(t − τ))
2L

−sin

(2n − 1)π((x + ξ) − c(t − τ))
2L

− sin


(2n − 1)π((x + ξ) + c(t − τ))
2L


b) Now we consider the Green function with the boundary conditions,
u
x
(0, t) = u
x
(L, t) = 0.
First we find the eigenfunctions in x of the homogeneous problem. The eigenvalue problem is
X

= −λ
2
X, X

(0) = X

(L) = 0,
which has the solutions,
λ
0
= 0, X
0
= 1,
λ
n
=
nπ

L
, X
n
= cos

nπx
L

, n = 1, 2, . . . .
The series expansion of the Green function for t > τ has the form,
G(x, t; ξ, τ) =
1
2
g
0
(t) +
∞

n=1
g
n
(t) cos

nπx
L

.
(Note the factor of 1/2 in front of g
0
(t). With this, the integral formulas for all the coefficients are the same.) We

2027
determine the coefficients by substituting the expansion into the partial differential equation.
G
tt
− c
2
G
xx
= δ(x − ξ)δ(t − τ)
1
2
g

0
(t) +
∞

n=1

g

n
(t) +

nπc
L

2
g
n

(t)

cos

nπx
L

= δ(x − ξ)δ(t − τ)
We expand the right side of the equation in the cosine series.
δ(x − ξ)δ(t −τ) =
1
2
d
0
(t) +
∞

n=1
d
n
(t) cos

nπx
L

d
n
(t) =
2
L


L
0
δ(x − ξ)δ(t −τ) cos

nπx
L

dx
d
n
(t) =
2
L
cos

nπξ
L

δ(t − τ)
By equating coefficients in the cosine series, we obtain ordinary differential equations for the g
n
.
g

n
(t; τ) +

nπc
L


2
g
n
(t; τ) =
2
L
cos

nπξ
L

δ(t − τ), n = 0, 1, 2, . . .
From the causality condition for G, we have the causality condiions for the g
n
,
g
n
(t; τ) = g

n
(t; τ) = 0 for t < τ.
The continuity and jump conditions for the g
n
are
g
n
(τ
+
; τ) = 0, g


n
(τ
+
; τ) =
2
L
cos

nπξ
L

.
The homogeneous solutions of the ordinary differential equation for n = 0 and n > 0 are respectively,
{1, t},

cos

nπct
L

, sin

nπct
L

.
2028
Since the continuity and jump c onditions are given at the point t = τ, a handy set of solutions to use for this problem
is the fundamental set of solutions at that point:

{1, t − τ},

cos

nπc(t − τ)
L

,
L
nπc
sin

nπc(t − τ)
L

.
The solutions that satisfy the causality condition and the continuity and jump conditions are,
g
0
(t) =
2
L
(t − τ)H(t − τ),
g
n
(t) =
2
nπc
cos


nπξ
L

sin

nπc(t − τ)
L

H(t − τ).
Substituting this into the sum yields,
G(x, t; ξ, τ) = H(t − τ)

t − τ
L
+
2
πc
∞

n=1
1
n
cos

nπξ
L

sin

nπc(t − τ)

L

cos

nπx
L


.
We can write this as the sum of traveling waves.
G(x, t; ξ, τ) =
t − τ
L
H(t −τ ) +
1
2πc
H(t −τ )
∞

n=1
1
n

−sin

nπ((x − ξ) − c(t − τ))
2L

+ sin


nπ((x − ξ) + c(t − τ))
2L

− sin

nπ((x + ξ) − c(t − τ))
2L

+ sin

nπ((x + ξ) + c(t − τ))
2L


2029
Solution 45.23
First we derive Green’s id entity for this problem. We consider the integral of uL[v] − L[u]v on the domain 0 < x < 1,
0 < t < T .

T
0

1
0
(uL[v] − L[u]v) dx dt

T
0

1

0

u(v
tt
− c
2
v
xx
− (u
tt
− c
2
u
xx
)v

dx dt

T
0

1
0

∂
∂x
,
∂
∂t


·

−c
2
(uv
x
− u
x
v), uv
t
− u
t
v


dx dt
Now we can use the divergence theorem to write this as an integral along the boundary of the domain.

∂Ω

−c
2
(uv
x
− u
x
v), uv
t
− u
t

v

· n ds
The domain and the outward normal vectors are shown in Figure 45.7.
Writing out the boundary integrals, Green’s identity for this problem is,

T
0

1
0

u(v
tt
− c
2
v
xx
) − (u
tt
− c
2
u
xx
)v

dx dt = −

1
0

(uv
t
− u
t
v)
t=0
dx
+

0
1
(uv
t
− u
t
v)
t=T
dx − c
2

T
0
(uv
x
− u
x
v)
x=1
dt + c
2


1
T
(uv
x
− u
x
v)
x=0
dt
The Green function problem is
G
tt
− c
2
G
xx
= δ(x − ξ)δ(t − τ), 0 < x, ξ < 1, t, τ > 0,
G
x
(0, t; ξ, τ) = G
x
(1, t; ξ, τ) = 0, t > 0, G(x, t; ξ, τ) = 0 for t < τ.
If we consider G as a function of (ξ, τ) with (x, t) as parameters, then it satisfies:
G
ττ
− c
2
G
ξξ

= δ(x − ξ)δ(t − τ),
G
ξ
(x, t; 0, τ) = G
ξ
(x, t; 1, τ) = 0, τ > 0, G(x, t; ξ, τ) = 0 for τ > t.
2030
x=0
x=1
t=0
t=T
n=(0,-1)
n=(1,0)
n=(0,1)
n=(-1,0)
Figure 45.7: Outward normal vectors of the domain.
Now we apply Green’s identity for u = u(ξ, τ), (the solution of the wave equation), and v = G(x, t; ξ, τ), (the Green
function), and integrate in the (ξ, τ) variables. The left side of Green’s identity becomes:

T
0

1
0

u(G
ττ
− c
2
G

ξξ
) − (u
ττ
− c
2
u
ξξ
)G

dξ dτ

T
0

1
0
(u(δ(x − ξ)δ(t −τ)) −(0)G) dξ dτ
u(x, t).
Since the normal derivative of u and G vanish on the sides of the domain, the i ntegrals along ξ = 0 and ξ = 1 in
Green’s identity vanish. If we take T > t, then G is zero for τ = T and the integral along τ = T vanishes. The one
remaining integral is
−

1
0
(u(ξ, 0)G
τ
(x, t; ξ, 0) −u
τ
(ξ, 0)G(x, t; ξ, 0) dξ.

2031
Thus Green’s identity allows us to write the solution of the inhomogeneous problem.
u(x, t) =

1
0
(u
τ
(ξ, 0)G(x, t; ξ, 0) − u(ξ, 0)G
τ
(x, t; ξ, 0)) dξ.
With the specified initial conditions this becomes
u(x, t) =

1
0
(G(x, t; ξ, 0) −ξ
2
(1 − ξ)
2
G
τ
(x, t; ξ, 0)) dξ.
Now we substitute in the Green function that we found in the previous exercise. The Green function and its derivative
are,
G(x, t; ξ, 0) = t +
∞

n=1
2

nπc
cos(nπξ) sin(nπct) cos(nπx),
G
τ
(x, t; ξ, 0) = −1 − 2
∞

n=1
cos(nπξ) cos(nπct) cos(nπx).
The integral of the first term is,

1
0

t +
∞

n=1
2
nπc
cos(nπξ) sin(nπct) cos(nπx)

dξ = t.
The integral of the second term is

1
0
ξ
2
(1 − ξ)

2

1 + 2
∞

n=1
cos(nπξ) cos(nπct) cos(nπx)

dξ =
1
30
− 3
∞

n=1
1
n
4
π
4
cos(2nπx) cos(2nπct).
Thus the solution is
u(x, t) =
1
30
+ t − 3
∞

n=1
1

n
4
π
4
cos(2nπx) cos(2nπct).
2032
For c = 1, the solution at x = 3/4, t = 7/2 is,
u(3/4, 7/2) =
1
30
+
7
2
− 3
∞

n=1
1
n
4
π
4
cos(3nπ/2) cos(7nπ).
Note that the summand is nonzero only for even terms.
u(3/4, 7/2) =
53
15
−
3
16π

4
∞

n=1
1
n
4
cos(3nπ) cos(14nπ)
=
53
15
−
3
16π
4
∞

n=1
(−1)
n
n
4
=
53
15
−
3
16π
4
−7π

4
720
u(3/4, 7/2) =
12727
3840
2033
Chapter 46
Conformal Mapping
2034
46.1 Exercises
Exercise 46.1
Use an appropriate conformal map to find a non-trivial solution to Laplace’s equation
u
xx
+ u
yy
= 0,
on the wedge bounded by the x-axis and the line y = x with boundary conditions:
1. u = 0 on both sides.
2.
du
dn
= 0 on both sides (where n is the inward normal to the boundary).
Exercise 46.2
Consider
u
xx
+ u
yy
= δ(x − ξ)δ(y −ψ),

on the quarter plane x, y > 0 with u(x, 0) = u(0, y) = 0 (and ξ, ψ > 0).
1. Use image sources to find u(x, y; ξ, ψ).
2. Compare this to the solution which would be obtained using conformal maps and the Green function for the upper
half plane.
3. Finally use this ide a and conformal mapping to discover how image sources are arrayed when the domain is now
the wedge bounded by the x-axis and the l ine y = x (with u = 0 on both sides).
Exercise 46.3
ζ = ξ + ıη is an analytic function of z, ζ = ζ(z). We assume that ζ

(z) is nonzero on the domain of interest. u(x, y)
is an arbitrary smooth fu nction of x and y. When expressed in terms of ξ and η, u(x, y) = υ(ξ, η). In Exercise 8.15
we showed that
∂
2
υ
∂ξ
2
+
∂
2
υ
∂η
2
=




dζ
dz





−2

∂
2
u
∂x
2
+
∂
2
u
∂y
2

.
2035
1. Show that if u satisfie s Laplace’s equation in the z-plane,
u
xx
+ u
yy
= 0,
then υ satisfies Laplace’s equation in the ζ-plane,
υ
ξξ
+ υ

ηη
= 0,
2. Show that if u satisfie s Helmholtz’s equation in the z-plane,
u
xx
+ u
yy
= λu,
then in the ζ-plane υ satisfies
υ
ξξ
+ υ
ηη
= λ




dz
dζ




2
υ.
3. Show that if u satisfie s Poisson’s eq uation in the z-plane,
u
xx
+ u

yy
= f(x, y),
then υ satisfies Poisson’s equation in the ζ-plane,
υ
ξξ
+ υ
ηη
=




dz
dζ




2
φ(ξ, η),
where φ(ξ, η) = f(x, y).
4. Show that if in the z-plane, u satisfies the Green function problem,
u
xx
+ u
yy
= δ(x − x
0
)δ(y −y
0

),
then in the ζ-plane, υ satisfies the Green function problem,
υ
ξξ
+ υ
ηη
= δ(ξ −ξ
0
)δ(η −η
0
).
2036
Exercise 46.4
A semi-ci rcular rod of infinite extent is maintained at temperature T = 0 on the flat side and at T = 1 on the curved
surface:
x
2
+ y
2
= 1, y > 0.
Use the conformal mapping
w = ξ + ıη =
1 + z
1 − z
, z = x + ıy,
to formulate the problem in terms of ξ and η. Solve the problem in terms of these variables. This problem is solved
with an eigenfunction expansion in Exercise 37.24. Verify that the two solutions agree.
Exercise 46.5
Consider Laplace’s equation on the domain −∞ < x < ∞, 0 < y < π, subject to the mixed boundary conditions,
u = 1 on y = 0, x > 0,

u = 0 on y = π, x > 0,
u
y
= 0 on y = 0, y = π, x < 0.
Because of the mixed boundary conditions, (u and u
y
are given on separate parts of the same boundary), this problem
cannot be solved with separation of variables. Verify that the conformal map,
ζ = cosh
−1
(
e
z
),
with z = x + ıy, ζ = ξ + ıη maps the infinite interval into the semi-infinite interval, ξ > 0, 0 < η < π. Solve Laplace’s
equation with the appropriate boundary conditions in the ζ plane by inspection. Write the solution u in terms of x and
y.
2037