Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 10 pps

v(x, τ) =
1
2π

∞
−∞
f(ξ)π(δ(x − ξ + τ) + δ(x −ξ −τ)) dξ
+
1
c
1
2π

∞
−∞
g(ξ)π(H(x −ξ + τ) −H(x −ξ − τ)) dξ
v(x, τ) =
1
2
(f(x + τ) + f(x −τ)) +
1
2c

x+τ
x−τ
g(ξ) dξ
Finally we make the change of variables t = τ/c, u(x, t) = v(x, τ) to obtain D’Alembert’s solution of the wave equation,
u(x, t) =
1
2

(f(x −ct) + f(x + ct)) +
1
2c

x+ct
x−ct
g(ξ) dξ.
Solution 44.6
With the change of variables
τ = ct,
∂
∂τ
=
∂t
∂τ
∂
∂t
=
1
c
∂
∂t
, v(x, τ) = u(x, t),
the problem becomes
v
ττ
= v
xx
, v(x, 0) = f(x), v
τ

(x, 0) =
1
c
g(x).
We take the Laplace transform in τ of the equation, (we consider x to be a parameter),
s
2
V (x, s) −sv(x, 0) − v
τ
(x, 0) = V
xx
(x, s),
V
xx
(x, s) −s
2
V (x, s) = −sf(x) −
1
c
g(x),
Now we have an ordinary differential equation for V (x, s), (now we consider s to be a parameter). We impose the
boundary conditions that the solution is bounded at x = ±∞. Consider the Green’s function problem
g
xx
(x; ξ) − s
2
g(x; ξ) = δ(x − ξ), g(±∞; ξ) bounded.
1934
e
sx

is a homogeneous solution that is bounded at x = −∞.
e
−sx
is a homogeneous solution that is bounded at
x = +∞. The Wronskian of these solutions is
W (x) =




e
sx
e
−sx
s
e
sx
−s
e
−sx




= −2s.
Thus the Green’s function is
g(x; ξ) =

−
1

2s
e
sx
e
−sξ
for x < ξ,
−
1
2s
e
sξ
e
−sx
for x > ξ,
= −
1
2s
e
−s|x−ξ|
.
The solution for V (x, s) is
V (x, s) = −
1
2s

∞
−∞
e
−s|x−ξ|
(−sf(ξ) −

1
c
g(ξ)) dξ,
V (x, s) =
1
2

∞
−∞
e
−s|x−ξ|
f(ξ) dξ +
1
2cs

∞
−∞
e
−s|x−ξ|
g(ξ)) dξ,
V (x, s) =
1
2

∞
−∞
e
−s|ξ|
f(x −ξ) dξ +
1

2c

∞
−∞
e
−s|ξ|
s
g(x −ξ)) dξ.
Now we take the inverse Laplace transform and interchange the order of integration.
v(x, τ) =
1
2
L
−1


∞
−∞
e
−s|ξ|
f(x −ξ) dξ

+
1
2c
L
−1


∞

−∞
e
−s|ξ|
s
g(x −ξ)) dξ

v(x, τ) =
1
2

∞
−∞
L
−1

e
−s|ξ|

f(x −ξ) dξ +
1
2c

∞
−∞
L
−1

e
−s|ξ|
s


g(x −ξ)) dξ
v(x, τ) =
1
2

∞
−∞
δ(τ −|ξ|)f(x −ξ) dξ +
1
2c

∞
−∞
H(τ −|ξ|)g(x −ξ)) dξ
v(x, τ) =
1
2
(f(x −τ) + f(x + τ)) +
1
2c

τ
−τ
g(x −ξ) dξ
1935
v(x, τ) =
1
2
(f(x −τ) + f(x + τ)) +

1
2c

−x+τ
−x−τ
g(−ξ) dξ
v(x, τ) =
1
2
(f(x −τ) + f(x + τ)) +
1
2c

x+τ
x−τ
g(ξ) dξ
Now we write make the change of variables t = τ/c, u(x, t) = v(x, τ ) to obtain D’Alembert’s solution of the wave
equation,
u(x, t) =
1
2
(f(x −ct) + f(x + ct)) +
1
2c

x+ct
x−ct
g(ξ) dξ.
Solution 44.7
1. We take the Laplace transform of Equation 44.1.

s
ˆ
φ −φ(x, 0) = a
2
ˆ
φ
xx
ˆ
φ
xx
−
s
a
2
ˆ
φ = 0 (44.3)
We take the Laplace transform of the initial condition, φ(0, t) = f(t), and use that
ˆ
φ(x, s) vanishes as x → ∞
to obtain boundary conditions for
ˆ
φ(x, s).
ˆ
φ(0, s) =
ˆ
f(s),
ˆ
φ(∞, s) = 0
The solutions of Equation 44.3 are
exp


±
√
s
a
x

.
The solution that satisfies the boundary conditions is
ˆ
φ(x, s) =
ˆ
f(s) exp

−
√
s
a
x

.
1936
We write this as the product of two Laplace transforms.
ˆ
φ(x, s) =
ˆ
f(s)L

x
2a

√
πt
3/2
exp

−
x
2
4a
2
t

We invert using the convolution theorem.
φ(x, t) =
x
2a
√
π

t
0
f(t −τ)
1
τ
3/2
exp

−
x
2

4a
2
τ

dτ.
2. Consider the case f(t) = 1.
φ(x, t) =
x
2a
√
π

t
0
1
τ
3/2
exp

−
x
2
4a
2
τ

dτ
ξ =
x
2a

√
τ
, dξ = −
x
4aτ
3/2
φ(x, t) = −
2
√
π

x/(2a
√
t)
∞
e
−ξ
2
dξ
φ(x, t) = erfc

x
2a
√
t

Now consider the case in which f(t) = 1 for 0 < t < T , with f(t) = 0 for t > T . For t < T , φ is the same as
before.
φ(x, t) = erfc


x
2a
√
t

, for 0 < t < T
1937
Consider t > T .
φ(x, t) =
x
2a
√
π

t
t−T
1
τ
3/2
exp

−
x
2
4a
2
τ

dτ
φ(x, t) = −

2
√
π

x/(2a
√
t)
x/(2a
√
t−T )
e
−ξ
2
dξ
φ(x, t) = erf

x
2a
√
t −T

− erf

x
2a
√
t

Solution 44.8
u

t
= κu
xx
, x > 0, t > 0,
u
x
(0, t) −αu(0, t) = 0, u(x, 0) = f(x).
First we find the partial differential equation that v satisfies. We start with the partial differential equation for u,
u
t
= κu
xx
.
Differentiating this equation with respect to x yields,
u
tx
= κu
xxx
.
Subtracting α times the former equation from the latter yields,
u
tx
− αu
t
= κu
xxx
− ακu
xx
,
∂

∂t
(u
x
− αu) = κ
∂
2
∂x
2
(u
x
− αu) ,
v
t
= κv
xx
.
1938
Thus v satisfies the same partial differential equation as u. This is because the equation for u is linear and homogeneous
and v is a linear combination of u and its derivatives. The problem for v is,
v
t
= κv
xx
, x > 0, t > 0,
v(0, t) = 0, v(x, 0) = f

(x) −αf(x).
With this new boundary condition, we can solve the problem with the Fourier sine transform. We take the sine transform
of the partial differential equation and the initial condition.
ˆv

t
(ω, t) = κ

−ω
2
ˆv(ω, t) +
1
π
ωv(0, t)

,
ˆv(ω, 0) = F
s
[f

(x) −αf(x)]
ˆv
t
(ω, t) = −κω
2
ˆv(ω, t)
ˆv(ω, 0) = F
s
[f

(x) −αf(x)]
Now we have a first order, ordinary differential equation for ˆv. The general solution is,
ˆv(ω, t) = c
e
−κω

2
t
.
The solution subject to the initial condition is,
ˆv(ω, t) = F
s
[f

(x) −αf(x)]
e
−κω
2
t
.
Now we take the inverse sine transform to find v. We utilize the Fourier cosine transform pair,
F
−1
c

e
−κω
2
t

=

π
κt
e
−x

2
/(4κt)
,
to write ˆv in a form that is suitable for the convolution theorem.
ˆv(ω, t) = F
s
[f

(x) −αf(x)] F
c


π
κt
e
−x
2
/(4κt)

1939
Recall that the Fourier sine convolu tion theorem is,
F
s

1
2π

∞
0
f(ξ) (g(|x − ξ|) − g(x + ξ)) dξ


= F
s
[f(x)]F
c
[g(x)].
Thus v(x, t) is
v(x, t) =
1
2
√
πκt

∞
0
(f

(ξ) − αf(ξ))

e
−|x−ξ|
2
/(4κt)
−
e
−(x+ξ)
2
/(4κt)

dξ.

With v determined, we have a first order, ordinary differential equation for u,
u
x
− αu = v.
We solve this equation by multiplying by the integrating factor and integrating.
∂
∂x

e
−αx
u

=
e
−αx
v
e
−αx
u =

x
e
−αξ
v(x, t) dξ + c(t)
u =

x
e
−α(ξ−x)
v(x, t) dξ +

e
αx
c(t)
The solution that vanishes as x → ∞ is
u(x, t) = −

∞
x
e
−α(ξ−x)
v(ξ, t) dξ.
1940
Solution 44.9

∞
0
ω
e
−cω
2
sin(ωx) dω = −
∂
∂x

∞
0
e
−cω
2
cos(ωx) dω

= −
1
2
∂
∂x

∞
−∞
e
−cω
2
cos(ωx) dω
= −
1
2
∂
∂x

∞
−∞
e
−cω
2
+ıωx
dω
= −
1
2
∂
∂x


∞
−∞
e
−c(ω+ıx/(2c))
2
e
−x
2
/(4c)
dω
= −
1
2
∂
∂x
e
−x
2
/(4c)

∞
−∞
e
−cω
2
dω
= −
1
2


π
c
∂
∂x
e
−x
2
/(4c)
=
x
√
π
4c
3/2
e
−x
2
/(4c)
u
t
= u
xx
, x > 0, t > 0,
u(0, t) = g(t), u(x, 0) = 0.
We take the Fourier sine transform of the partial differential equation and the initial condition.
ˆu
t
(ω, t) = −ω
2

ˆu(ω, t) +
ω
π
g(t), ˆu(ω, 0) = 0
Now we have a first order, ordinary differential equation for ˆu(ω, t).
∂
∂t

e
ω
2
t
ˆu
t
(ω, t)

=
ω
π
g(t)
e
ω
2
t
ˆu(ω, t) =
ω
π
e
−ω
2

t

t
0
g(τ)
e
ω
2
τ
dτ + c(ω)
e
−ω
2
t
1941
The initial condition is satisfied for c(ω) = 0.
ˆu(ω, t) =
ω
π

t
0
g(τ)
e
−ω
2
(t−τ)
dτ
We take the inverse sine transform to find u.
u(x, t) = F

−1
s

ω
π

t
0
g(τ)
e
−ω
2
(t−τ)
dτ

u(x, t) =

t
0
g(τ)F
−1
s

ω
π
e
−ω
2
(t−τ)


dτ
u(x, t) =

t
0
g(τ)
x
2
√
π(t −τ)
3/2
e
−x
2
/(4(t−τ))
dτ
u(x, t) =
x
2
√
π

t
0
g(τ)
e
−x
2
/(4(t−τ))
(t −τ)

3/2
dτ
Solution 44.10
The problem is
u
xx
+ u
yy
= 0, 0 < x, 0 < y < 1,
u(x, 0) = u(x, 1) = 0, u(0, y) = f(y).
We take the Fourier sine transform of the partial differential equation and the boundary conditions.
−ω
2
ˆu(ω, y) +
k
π
u(0, y) + ˆu
yy
(ω, y) = 0
ˆu
yy
(ω, y) −ω
2
ˆu(ω, y) = −
k
π
f(y), ˆu(ω, 0) = ˆu(ω, 1) = 0
1942
This is an inhomogeneous, ordinary differential equation that we can solve with Green functions. The homogeneous
solutions are

{cosh(ωy), sinh(ωy)}.
The homogeneous solutions that satisfy the left and right boundary conditions are
y
1
= sinh(ωy), y
2
= sinh(ω(y −1)).
The Wronskian of these two solutions is,
W (x) =




sinh(ωy) sinh(ω(y −1))
ω cosh(ωy) ω cosh(ω(y −1))




= ω (sinh(ωy) cosh(ω(y −1)) −cosh(ωy) sinh(ω(y −1)))
= ω sinh(ω).
The Green function is
G(y|η) =
sinh(ωy
<
) sinh(ω(y
>
− 1))
ω sinh(ω)
.

The solution of the ordinary differential equation for ˆu(ω, y) is
ˆu(ω, y) = −
ω
π

1
0
f(η)G(y|η) dη
= −
1
π

y
0
f(η)
sinh(ωη) sinh(ω(y −1))
sinh(ω)
dη −
1
π

1
y
f(η)
sinh(ωy) sinh(ω(η −1))
sinh(ω)
dη.
With some uninteresting grunge, you can show that,
2


∞
0
sinh(ωη) sinh(ω(y −1))
sinh(ω)
sin(ωx) dω = −2
sin(πη) sin(πy)
(cosh(πx) −cos(π(y − η)))(cosh(πx) − cos(π(y + η)))
.
1943
Taking the inverse Fourier sine transform of ˆu(ω, y) and interchanging the order of integration yields,
u(x, y) =
2
π

y
0
f(η)
sin(πη) sin(πy)
(cosh(πx) −cos(π(y − η)))(cosh(πx) − cos(π(y + η)))
dη
+
2
π

1
y
f(η)
sin(πy) sin(πη)
(cosh(πx) −cos(π(η − y)))(cosh(πx) − cos(π(η + y)))
dη.

u(x, y) =
2
π

1
0
f(η)
sin(πη) sin(πy)
(cosh(πx) −cos(π(y − η)))(cosh(πx) − cos(π(y + η)))
dη
Solution 44.11
The problem for u(x, y) is,
u
xx
+ u
yy
= 0, −∞ < x < ∞, y > 0,
u(x, 0) = g(x).
We take the Fourier transform of the partial differential equ ation and the boundary condition.
−ω
2
ˆu(ω, y) + ˆu
yy
(ω, y) = 0, ˆu(ω, 0) = ˆg(ω).
This is an ordinary differential equation for ˆu(ω, y). So far we only have one boundary condition. In order that u
is bounded we impose the second boundary condition ˆu(ω, y) is bounded as y → ∞. The general solution of the
differential equation is
ˆu(ω, y) =

c

1
(ω)
e
ωy
+c
2
(ω)
e
−ωy
, for ω = 0,
c
1
(ω) + c
2
(ω)y, for ω = 0.
Note that
e
ωy
is the bounded solution for ω < 0, 1 is the bounded solution for ω = 0 and
e
−ωy
is the bounded solution
for ω > 0. Thus the bounded solution is
ˆu(ω, y) = c(ω)
e
−|ω|y
.
1944
The boundary condition at y = 0 determines the constant of integration.
ˆu(ω, y) = ˆg(ω)

e
−|ω|y
Now we take the inverse Fourier transform to obtain the solution for u(x, y). To do this we use the Fourier transform
pair,
F

2c
x
2
+ c
2

=
e
−c|ω|
,
and the convolution theorem,
F

1
2π

∞
−∞
f(ξ)g(x − ξ) dξ

=
ˆ
f(ω)ˆg(ω).
u(x, y) =

1
2π

∞
−∞
g(ξ)
2y
(x −ξ)
2
+ y
2
dξ.
Solution 44.12
Since the derivative of u is specified at x = 0, we take the cosine transform of the partial differential equation and the
initial condition.
ˆu
t
(ω, t) = κ

−ω
2
ˆu(ω, t) −
1
π
u
x
(0, t)

− a
2

ˆu(ω, t), ˆu(ω, 0) = 0
ˆu
t
+

κω
2
+ a
2

ˆu =
κ
π
f(t), ˆu(ω, 0) = 0
This first order, ordinary differential equation for ˆu(ω, t) has the solution,
ˆu(ω, t) =
κ
π

t
0
e
−(κω
2
+a
2
)(t−τ)
f(τ) dτ.
1945
We take the inverse Fourier cosine transform to find the solution u(x, t).

u(x, t) =
κ
π
F
−1
c


t
0
e
−(κω
2
+a
2
)(t−τ)
f(τ) dτ

u(x, t) =
κ
π

t
0
F
−1
c

e
−κω

2
(t−τ)

e
−a
2
(t−τ)
f(τ) dτ
u(x, t) =
κ
π

t
0

π
κ(t −τ)
e
−x
2
/(4κ(t−τ))
e
−a
2
(t−τ)
f(τ) dτ
u(x, t) =

κ
π


t
0
e
−x
2
/(4κ(t−τ))−a
2
(t−τ)
√
t −τ
f(τ) dτ
Solution 44.13
Mathematically stated we h ave
u
tt
= c
2
u
xx
, 0 < x < L, t > 0,
u(x, 0) = u
t
(x, 0) = 0,
u(0, t) = f(t), u(L, t) = 0.
We take the Laplace transform of the partial differential equation and the boundary conditions.
s
2
ˆu(x, s) −su(x, 0) −u
t

(x, 0) = c
2
ˆu
xx
(x, s)
ˆu
xx
=
s
2
c
2
ˆu, ˆu(0, s) =
ˆ
f(s), ˆu(L, s) = 0
Now we have an ordinary differential equation. A set of solutions is

cosh

sx
c

, sinh

sx
c

.
The solution that satisfies the right boundary condition is
ˆu = a sinh


s(L −x)
c

.
1946
The left boundary condition determines the multiplicative constant.
ˆu(x, s) =
ˆ
f(s)
sinh(s(L −x)/c)
sinh(sL/c)
If we can find the inverse Laplace transform of
ˆu(x, s) =
sinh(s(L −x)/c)
sinh(sL/c)
then we can use the convolution theorem to write u in terms of a single integral. We proceed by expanding this function
in a sum.
sinh(s(L −x)/c)
sinh(sL/c)
=
e
s(L−x)/c
−
e
−s(L−x)/c
e
sL/c
−
e

−sL/c
=
e
−sx/c
−
e
−s(2L−x)/c
1 −
e
−2sL/c
=

e
−sx/c
−
e
−s(2L−x)/c

∞

n=0
e
−2nsL/c
=
∞

n=0
e
−s(2nL+x)/c
−

∞

n=0
e
−s(2(n+1)L−x)/c
=
∞

n=0
e
−s(2nL+x)/c
−
∞

n=1
e
−s(2nL−x)/c
Now we use the Laplace transform pair:
L[δ(x −a)] =
e
−sa
.
L
−1

sinh(s(L −x)/c)
sinh(sL/c)

=
∞


n=0
δ(t −(2nL + x)/c) −
∞

n=1
δ(t −(2nL −x)/c)
1947
We write ˆu in the form,
ˆu(x, s) = L[f(t)]L

∞

n=0
δ(t −(2nL + x)/c) −
∞

n=1
δ(t −(2nL −x)/c)

.
By the convolution theorem we have
u(x, t) =

t
0
f(τ)

∞


n=0
δ(t −τ − (2nL + x)/c) −
∞

n=1
δ(t −τ − (2nL − x)/c)

dτ.
We can simplify this a bit. First we determine which Dirac delta functions have their singularities in the range τ ∈ (0 t).
For the first sum, this condition is
0 < t −(2nL + x)/c < t.
The right inequality is always satisfied. The left inequality becomes
(2nL + x)/c < t,
n <
ct −x
2L
.
For the second sum, the condition is
0 < t −(2nL −x)/c < t.
Again the right inequality is always satisfied. The left ineq uality becomes
n <
ct + x
2L
.
We change the index range to reflect the nonzero contributions and do the integration.
u(x, t) =

t
0
f(τ)




ct−x
2L


n=0
δ(t −τ − (2nL + x)/c)

ct+x
2L


n=1
δ(t −τ − (2nL − x)/c)


dτ.
1948
u(x, t) =

ct−x
2L


n=0
f(t −(2nL + x)/c)

ct+x

2L


n=1
f(t −(2nL −x)/c)
Solution 44.14
We take the Fourier transform of the partial differential equ ation and the boundary conditions.
−ω
2
ˆ
φ +
ˆ
φ
yy
= 0,
ˆ
φ(ω, 0) =
1
2π
e
−ıωξ
,
ˆ
φ(ω, l) = 0
We solve this boundary value problem.
ˆ
φ(ω, y) = c
1
cosh(ω(l − y)) + c
2

sinh(ω(l − y))
ˆ
φ(ω, y) =
1
2π
e
−ıωξ
sinh(ω(l − y))
sinh(ωl)
We take the inverse Fourier transform to obtain an expression for the solution.
φ(x, y) =
1
2π

∞
−∞
e
ıω(x−ξ)
sinh(ω(l − y))
sinh(ωl)
dω
1949
Chapter 45
Green Functions
45.1 Inhomogeneous Equations and Homogeneous Boundary Con-
ditions
Consider a linear differential equation on the domain Ω subject to h omogeneous boundary conditions.
L[u(x)] = f(x) for x ∈ Ω, B[u(x)] = 0 for x ∈ ∂Ω (45.1)
For example, L[u] might be
L[u] = u

t
− κ∆u, or L[u] = u
tt
− c
2
∆u.
and B[u] might be u = 0, or ∇u · ˆn = 0.
If we find a Green function G(x; xi) that satisfies
L[G(x; xi)] = δ(x −xi), B[G(x; xi)] = 0
then the solution to Equation 45.1 is
u(x) =

Ω
G(x; xi)f(xi) dxi.
1950
We verify that this solution satisfies the equation and boundary condition.
L[u(x)] =

Ω
L[G(x; xi)]f(xi) dxi
=

Ω
δ(x −xi)f(xi) dxi
= f(x)
B[u(x)] =

Ω
B[G(x; xi)]f(xi) dxi
=


Ω
0 f(xi) dxi
= 0
45.2 Homogeneous Equations and Inhomogeneous Boundary Con-
ditions
Consider a homogeneous linear differential equation on the domain Ω subject to inhomogeneous boundary conditions,
L[u(x)] = 0 for x ∈ Ω, B[u(x)] = h(x) for x ∈ ∂Ω. (45.2)
If we find a Green function g(x; xi) that satisfies
L[g(x; xi)] = 0, B[g(x; xi)] = δ(x − xi)
then the solution to Equation 45.2 is
u(x) =

∂Ω
g(x; xi)h(xi) dxi.
1951
We verify that this solution satisfies the equation and boundary condition.
L[u(x)] =

∂Ω
L[g(x; xi)]h(xi) dxi
=

∂Ω
0 h(xi) dxi
= 0
B[u(x)] =

∂Ω
B[g(x; xi)]h(xi) dxi

=

∂Ω
δ(x −xi)h(xi) dxi
= h(x)
Example 45.2.1 Consider the Cauchy problem for the homogeneous heat equation.
u
t
= κu
xx
, −∞ < x < ∞, t > 0
u(x, 0) = h(x), u(±∞, t) = 0
We find a Green function that satisfies
g
t
= κg
xx
, −∞ < x < ∞, t > 0
g(x, 0; ξ) = δ(x − ξ), g(±∞, t; ξ) = 0.
Then we write the solution
u(x, t) =

∞
−∞
g(x, t; ξ)h(ξ) dξ.
To find the Green function for this problem, we apply a Fourier transform to the equation and boundary condition
1952
for g.
ˆg
t

= −κω
2
ˆg, ˆg(ω, 0; ξ) = F[δ(x −ξ)]
ˆg(ω, t; ξ) = F[δ(x −ξ)]
e
−κω
2
t
ˆg(ω, t; ξ) = F[δ(x −ξ)]F


π
κt
exp

−
x
2
4κt

We invert using the convolution theorem.
g(x, t; ξ) =
1
2π

∞
−∞
δ(ψ −ξ)

π

κt
exp

−
(x −ψ)
2
4κt

dψ
=
1
√
4πκt
exp

−
(x −ξ)
2
4κt

The solution of the heat equation is
u(x, t) =
1
√
4πκt

∞
−∞
exp


−
(x −ξ)
2
4κt

h(ξ) dξ.
45.3 Eigenfunction Expansions for Elliptic Equations
Consider a Green function problem for an elliptic equation on a finite domain.
L[G] = δ(x −xi), x ∈ Ω (45.3)
B[G] = 0, x ∈ ∂Ω
Let the set of functions {φ
n
} be orthonormal and complete on Ω. (Here n is the multi-index n = n
1
, . . . , n
d
.)

Ω
φ
n
(x)φ
m
(x) dx = δ
nm
1953
In addition, let the φ
n
be eigenfunctions of L subject to the homogeneous boundary conditions.
L [φ

n
] = λ
n
φ
n
, B [φ
n
] = 0
We expand the Green function in the eigenfunctions.
G =

n
g
n
φ
n
(x)
Then we expand the Dirac Delta function.
δ(x −xi) =

n
d
n
φ
n
(x)
d
n
=


Ω
φ
n
(x)δ(x −xi) dx
d
n
= φ
n
(xi)
We substitute the series expansions for the Green function and the Dirac D elta function into Equation 45.3.

n
g
n
λ
n
φ
n
(x) =

n
φ
n
(xi)φ
n
(x)
We equate coefficients to solve for the g
n
and hence determine the Green function.
g

n
=
φ
n
(xi)
λ
n
G(x; xi) =

n
φ
n
(xi)φ
n
(x)
λ
n
Example 45.3.1 Consider the Green function for the reduced wave equation, ∆u − k
2
u in the rectangle, 0 ≤ x ≤ a,
0 ≤ y ≤ b, and vanishing on the sides.
1954
First we find the eigenfunctions of the operator L = ∆ −k
2
= 0. Note that φ = X(x)Y (y) is an eigenfunction of
L if X is an eigenfunction of
∂
2
∂x
2

and Y is an eigenfunction of
∂
2
∂y
2
. Thus we consider the two regular Sturm-Liouville
eigenvalue problems:
X

= λX, X(0) = X(a) = 0
Y

= λY, Y (0) = Y (b) = 0
This leads us to the eigenfunctions
φ
mn
= sin

mπx
a

sin

nπy
b

.
We use the orthogonality relation

2π

0
sin

mπx
a

sin

nπx
a

dx =
a
2
δ
mn
to make the eigenfunctions orthonormal.
φ
mn
=
2
√
ab
sin

mπx
a

sin


nπy
b

, m, n ∈ Z
+
The φ
mn
are eigenfunctions of L.
L [φ
mn
] = −


mπ
a

2
+

nπ
b

2
+ k
2

φ
mn
By expanding the Green function and the Dirac Delta function in the φ
mn

and substituting into the differential equation
we obtain the solution.
G =
∞

m,n=1
2
√
ab
sin

mπξ
a

sin

nπψ
b

2
√
ab
sin

mπx
a

sin

nπy

b

−


mπ
a

2
+

nπ
b

2
+ k
2

G(x, y; ξ, ψ) = −4ab
∞

m,n=1
sin

mπx
a

sin

mπξ

a

sin

nπy
b

sin

nπψ
b

(mπb)
2
+ (nπa)
2
+ (kab)
2
1955
Example 45.3.2 Consider the Green function for Laplace’s equation, ∆u = 0 in the disk, |r| < a, and vanishing at
r = a.
First we find the eigenfunctions of the operator
∆ =
∂
2
∂r
2
+
1
r

∂
∂r
+
1
r
2
∂
2
∂θ
2
.
We will look for eigenfunctions of the form φ = Θ(θ)R(r). We choose the Θ to be eigenfunctions of
d
2
dθ
2
subject to
the periodic boundary conditions in θ.
Θ

= λΘ, Θ(0) = Θ(2π), Θ

(0) = Θ

(2π)
Θ
n
=
e
inθ

, n ∈ Z
We determine R(r) by requiring that φ be an eigenfunction of ∆.
∆φ = λφ
(Θ
n
R)
rr
+
1
r
(Θ
n
R)
r
+
1
r
2
(Θ
n
R)
θθ
= λΘ
n
R
Θ
n
R

+

1
r
Θ
n
R

+
1
r
2
(−n
2
)Θ
n
R = λΘR
For notational convenience, we denote λ = −µ
2
.
R

+
1
r
R

+

µ
2
−

n
2
r
2

R = 0, R(0) bounded, R(a) = 0
The general solution for R is
R = c
1
J
n
(µr) + c
2
Y
n
(µr).
The left boundary condition demands that c
2
= 0. The right boundary condition determines the eigenvalues.
R
nm
= J
n

j
n,m
r
a

, µ

nm
=
j
n,m
a
1956
Here j
n,m
is the m
th
positive root of J
n
. This leads us to the eigenfunctions
φ
nm
=
e
inθ
J
n

j
n,m
r
a

We use the orthogonality relations

2π
0

e
−imθ
e
inθ
dθ = 2πδ
mn
,

1
0
rJ
ν
(j
ν,m
r)J
ν
(j
ν,n
r) dr =
1
2
(J

ν
(j
ν,n
))
2
δ
mn

to make the eigenfunctions orthonormal.
φ
nm
=
1
√
πa|J

n
(j
n,m
)|
e
inθ
J
n

j
n,m
r
a

, n ∈ Z, m ∈ Z
+
The φ
nm
are eigenfunctions of L.
∆φ
nm
= −


j
n,m
a

2
φ
nm
By expanding the Green function and the Dirac Delta function in the φ
nm
and substituting into the differential equation
we obtain the solution.
G =
∞

n=−∞
∞

m=1
1
√
πa|J

n
(j
n,m
)|
e
−inϑ
J

n

j
n,m
ρ
a

1
√
πa|J

n
(j
n,m
)|
e
inθ
J
n

j
n,m
r
a

−

j
n,m
a


2
G(r, θ; ρ, ϑ) = −
∞

n=−∞
∞

m=1
1
π(j
n,m
J

n
(j
n,m
))
2
e
in(θ−ϑ)
J
n

j
n,m
ρ
a

J

n

j
n,m
r
a

1957