Problems
1. Derive Green’s function for Poisson’s equation on infinite three dimensional space. What
is the condition at infinity required to ensure vanishing contribution from the boundary
integral?
2. Show that Green’s function (10.7.37) satisfies the boundary condition (10.7.35).
3. Use (10.7.39) to obtain the solution of Laplace’s equation on the upper half plane subject
to
u(x, 0) = h(x)
4. Use the method of eigenfunction expansion to determine G(
r;
r
0
) if
∇
2
G = δ(
r
− r
0
),
0 < x < 1,
0 < y < 1
subject to the following boundary conditions
G(0, y;
r
0
) = G
x
(1, y;
r
0
) = G
y
(x, 0;
r
0
) = G
y
(x, 1;
r
0
) = 0
5. Solve the above problem inside a unit cube with zero Dirichlet boundary condition on
all sides.
6. Derive Green’s function for Poisson’s equation on a circle by using the method of images.
7. Use the above Green’s function to show that Laplace’s equation inside a circle of radius
ρ with
u(r, θ) = h(θ)
for r = ρ
is given by Poisson’s formula
u(r, θ) =
1
2π
2π
0
h(θ
0
)
ρ
2
− r
2
r
2
+ ρ
2
− 2ρr cos(θ − θ
0
)
dθ
0
.
8. Determine Green’s function for the right half plane and use it to solve Poisson’s equation.
9. Determine Green’s function for the upper half plane subject to
∂G
∂y
= 0
on y = 0.
Use it to solve Poisson’s equation
∇
2
u = f
∂u
∂y
= h(x),
on y = 0.
Ignore the contributions at infinity.
10. Use the method of images to solve
∇
2
G = δ(
r
− r
0
)
in the first quadrant with G = 0 on the boundary.
243
10.8
Wave Equation on Infinite Domains
Consider the wave equation
∂
2
u
∂t
2
= c
2
∇
2
u + Q(x, t)
(10.8.1)
with initial conditions
u(x, 0) = f (x)
(10.8.2)
u
t
(x, 0) = g(x)
(10.8.3)
where x = (x, y, z). The spatial domain here is infinite, i.e. x
R
3
. The solution for semi-
infinite or finite domains can be obtained using the method of images from this most general
case.
If we consider a concentrated source at x = x
0
≡ (x
0
, y
0
, z
0
) and at t = t
0
. The Green’s
function G(x, t; x
0
, t
0
) is the solution to
∂
2
G
∂t
2
= c
2
∇
2
G + δ(x
− x
0
)δ(t
− t
0
).
(10.8.4)
Since the time variable increases in one direction, we require
G(x, t; x
0
t
0
) = 0
for
t < t
0
(causality principle)
(10.8.5)
We may also translate the time variable to the origin so that
G(x, t; x
0
t
0
) = G(x, t
− t
0
; x
0
, 0)
(10.8.6)
We will solve for the Green’s function G using Fourier transforms since the domain is infinite.
Hence, we need the following results about Fourier transforms:
f (x, y, z) =
F (w
1
, w
2
, w
3
)e
−i(w
1
, w
2
, w
3
)·(x, y, z)
dw
1
dw
2
dw
3
(10.8.7)
or
f (x) =
F (w)e
−iw·x
dw
and
244
F (w) =
1
(2π)
3
f (x)e
iw·x
dx
where
dx = dxdydz
(10.8.8)
For the delta function, the results are
F [δ(x − x
0
)] =
1
(2π)
3
δ(x
− x
0
)e
iw·x
dx =
e
iw·x
0
(2π)
3
(10.8.9)
and (formally)
δ(x
− x
0
) =
e
iw·x
0
(2π)
3
e
−iw·x
dw =
1
(2π)
3
e
−iw·(x−x
0
)
dw.
(10.8.10)
We take the Fourier transform of the Green’s function G(w, t; x
0
, t
0
) and solve for it from
the system
∂
2
G
∂t
2
− c
2
∇
2
G = δ(x
− x
0
)δ(t
− t
0
),
G(x, t, x
0
, t
0
) = 0
if
t < t
0
.
(10.8.11)
We get the following O.D.E.:
∂
2
G
∂t
2
+ c
2
w
2
G =
e
iw·x
0
(2π)
3
δ(t
− t
0
)
where
w
2
= w
· w,
(10.8.12)
with
G(w, t; x
0
, t
0
) = 0
for
t < t
0
(10.8.13)
So for t > t
0
,
∂
2
G
∂t
2
+ c
2
w
2
G = 0,
(10.8.14)
Hence, the transform of the Green’s function is
G =
0
t < t
0
A cos cw(t
− t
0
) + B sin cw(t
− t
0
) t > t
0
(10.8.15)
245
Since
G is continuous at t = t
0
,
A = 0
To solve for B, we integrate the O.D.E.
t
+
0
t
−
0
∂
2
G
∂t
2
+ c
2
w
2
G =
e
iw·x
0
(2π)
3
δ(t
− t
0
)
dt
So
∂G
∂t
t
+
0
t
−
0
+ 0 =
e
iw·x
0
(2π)
3
,
but
∂G
∂t
t
−
0
= 0,
so
cwB cos cw(t
− t
0
)
t
+
0
=
e
iw·x
0
(2π)
3
⇒ B =
e
iw·x
0
cw(2π)
3
.
Using the inverse transform of G =
e
iw·x
0
cw(2π)
3
sin cw(t
− t
0
),
we get
G(x, t; x
0
, t
0
) =
0
t < t
0
1
(2π)
3
∞
−∞
∞
−∞
∞
−∞
e
−iw·(x−x
0
)
sin cw(t
− t
0
)
cw
dw t > t
0
(10.8.16)
where
w = (w
2
1
+ w
2
2
+ w
2
3
)
1/2
=
|w|
To evaluate the integral
1
(2π)
3
∞
−∞
∞
−∞
∞
−∞
e
−iw·(x−x
0
)
sin cw(t
− t
0
)
cw
dw,
we introduce spherical coordinates with the origin w = 0,
φ = 0 corresponding to the w
3
axis, and we integrate in the direction (x
− x
0
). This yields w
· (x − x
0
) =
|w| |x − x
0
| cos φ,
and letting ρ =
|x − x
0
| we obtain w · (x − x
0
) = wρ cos φ. With the angle θ measured from
the positive w
1
axis, the volume differential becomes
dw
≡ dw
1
dw
2
dw
3
= w
2
sin φ dφ dθ dw,
and the integration limits for infinite space become 0 < φ < π,
0 < θ < 2π,
0 < w <
∞.
Our integrand is independent of θ (based on our selection of coordinates) yielding
G(x, t; x
0
, t
0
) =
1
(2π)
2
∞
0
π
0
e
−iwρ cos φ
sin cw(t
− t
0
)
cw
w
2
sin φ dφ dw.
(10.8.17)
Integrating first with respect to φ
246
G(x, t; x
0
, t
0
) =
1
iρc(2π)
2
∞
0
sin cw(t
− t
0
)
π
0
e
−iwρ cos φ
(iwρ sin φ) dφ dw
=
1
iρc(2π)
2
∞
0
sin cw(t
− t
0
)
e
−iwρ cos φ
|
π
0
dw
=
1
iρc(2π)
2
∞
0
sin cw(t
− t
0
)
e
iwρ
− e
−iwρ
=2i sin wρ
dw
=
2
ρc(2π)
2
∞
0
sin(wρ) sin cw(t
− t
0
) dw
=
1
ρc(2π)
2
∞
0
(cos w [ρ
− c(t − t
0
)]
− cos w [ρ + c(t − t
0
)]) dw
Since
1
2π
∞
−∞
e
−iwz
dw = δ(z), using the real part of e
−iwz
and the evenness of the cosine
function we see
∞
0
cos wz dz = δ(z),
(10.8.18)
Hence
G(x, t; x
0
, t
0
) =
1
4π
2
ρc
[δ(ρ
− c(t − t
0
))
− δ(ρ + c(t − t
0
))]
for
t > t
0
(10.8.19)
Since
ρ + c(t
− t
0
) > 0,
we get
G(x, t; x
0
, t
0
) =
0
t < t
0
1
4π
2
ρc
δ(ρ
− c(t − t
0
)) t > t
0
(10.8.20)
where
ρ =
|x − x
0
|
To solve the wave equation
∂
2
u
∂t
2
= c
2
∇
2
u + Q(x, t),
u(x, 0) = f (x), u
t
(x, 0) = g(x),
(10.8.21)
using Green’s function, we proceed as follows.
We have a linear differential operator
247
L =
∂
2
∂t
2
− c
2
∇
2
(10.8.22)
where
L = L
1
− c
2
L
2
(10.8.23)
with
L
1
=
∂
2
∂t
2
(10.8.24)
and
L
2
=
∇
2
(10.8.25)
We have the following Green’s formulae for L
1
and L
2
:
t
2
t
1
[uL
1
v
− vL
1
u] dt = uv
t
− vu
t
t
2
t
1
;
(10.8.26)
[uL
2
v
− vL
2
u] dx =
(u
∇v − v∇u) · n ds
(10.8.27)
Since
Lu = Q(x, t)
and
LG = δ(x
− x
0
)δ(t
− t
0
),
and
uLv
− vLu = uL
1
v
− vL
1
u
− c
2
(uL
2
v
− vL
2
u),
we see
t
2
t
1
[uLv
− vLu] dx dt =
[uv
t
− vu
t
]
t
2
t
1
dx
− c
2
t
2
t
1
(u
∇v − v∇u) · n ds dt
(10.8.28)
It can be shown that Maxwell’s reciprocity holds spatially for our Green’s function, provided
the elasped times between the points x and x
0
are the same. In fact, for the infinite domain.
G(x, t; x
0
, t
0
) =
0
t < t
0
1
4π
2
c
|x − x
0
|
δ [
|x − x
0
| − c(t − t
0
)] t > t
0
248
or
G(x, t; x
0
, t
0
) =
0
t < t
0
1
4π
2
c
|x
0
− x|
δ [
|x
0
− x| − c(t
0
− t)] t > t
0
which is
= G(x
0
, t; x, t
0
)
(10.8.29)
We now let u = u(x, t) be the solution to
Lu = Q(x, t)
subject to
u(x, 0) = f (x),
u
t
(x, 0) = g(x),
and
v = G(x, t
0
; x
0
, t) = G(x
0
, t
0
; x, t)
be the solution to
Lv = δ(x
− x
0
)δ(t
− t
0
)
subject to homogenous boundary conditions and the causality principle
G(x, t
0
; x
0
, t) = 0
for
t
0
< t
If we integrate in time from t
1
= 0 to t
2
= t
+
0
(a point just beyond the appearance of our
point source at (t = t
0
), we get
t
+
0
0
[u(x, t)δ(x
− x
0
)δ(t
− t
0
)
− G(x, t
0
; x
0
, t)Q(x, t)] dx dt =
[uG
t
− Gu
t
]
t
+
0
0
dx
− c
2
t
+
0
0
(u
∇G − G∇u) · n ds
dt
(10.8.30)
At t = t
+
0
, G
t
= G = 0, and using reciprocity, we see
u(x
0
, t
0
) =
t
+
0
0
G(x, t
0
; x, t)Q(x, t) dx dt
249
+
[u
t
(x, 0)G(x
0
, t
0
; x, 0)
− u(x, 0)G
t
(x
0
, t
0
; x, 0)] dx
−c
2
t
+
0
0
(u(x, t)
∇G(x
0
, t
0
; x, t)
− G(x
0
, t
0
; x, t)
∇u(x, t)) · n ds
dt
(10.8.31)
Taking the limit as t
+
0
→ t and interchanging (x
0
, t
0
) with (x, t) yields
u(x, t) =
t
0
G(x, t; x
0
, t
0
)Q(x
0
, t
0
) dx
0
dt
0
+
[g(x
0
)G(x, t; x
0
, 0)
− f(x
0
)G
t
0
(x, t; x
0
, 0)] dx
0
−c
2
t
0
(u(x
0
, t
0
)
∇
x
0
G(x, t; x
0
, t
0
)
− G(x, t; x
0
, t
0
)
∇
x
0
u(x
0
, t
0
))
· n ds
0
dt
0
,
(10.8.32)
where
∇
x
0
represents the gradient with respect to the source location x
0
.
The three terms represent respectivley the contributions due to the source, the initial con-
ditions, and the boudnary conditions. For our infinte domain, the last term goes away.
Hence, our complete solution for the infinite domain wave equation is given by
u(x, t) =
1
4π
2
c
t
0
1
|x − x
0
|
δ [
|x − x
0
| − c(t − t
0
)] Q(x
0
, t
0
) dx
0
dt
0
+
1
4π
2
c
g(x
0
)
|x − x
0
|
δ [
|x − x
0
| − c(t − t
0
)]
−
f (x
0
)
|x − x
0
|
∂
∂t
0
δ [
|x − x
0
| − c(t − t
0
)] dx
0
(10.8.33)
250
10.9
Heat Equation on Infinite Domains
Now consider the Heat Equation
∂u
∂t
= κ
∇
2
u + Q(x, t)
(10.9.1)
with initial condition
u(x, 0) = g(x)
(10.9.2)
where x = (x, y, z). The spatial domain is infinite, i.e. x
R
3
. If we consider a concentrated
source at x = x
0
≡ (x
0
, y
0
, z
0
) and at t = t
0
, the Green’s function is the solution to
∂G
∂t
= κ
∇
2
G + δ(x
− x
0
)δ(t
− t
0
)
(10.9.3)
From the causality principle
G(x, t; x
0
, t
0
) = 0
if
t < t
0
(10.9.4)
We may also translate the time variable to the origin so
G(x, t; x
0
t
0
) = G(x, t
− t
0
; x
0
, 0)
(10.9.5)
We will solve for the Green’s function G using Fourier transofrom because we lack boundary
conditions. For our infinite domain, we take the Fourier transforms of the Green’s function
G(w, t; x
0
, t
0
).
We get the following O.D.E.
∂G
∂t
+ κw
2
G =
e
iw·x
0
(2π)
3
δ(t
− t
0
)
(10.9.6)
where
w
2
= w
· w,
with
G(w, t; x
0
, t
0
) = 0
for
t < t
0
(10.9.7)
So, for t > t
0
,
∂G
∂t
+ κw
2
G = 0
(10.9.8)
251
Hence, the transform of the Green’s function is
G =
0
t < t
0
Ae
−κw
2
(t−t
0
)
t > t
0
(10.9.9)
By integrating the ODE form t
−
0
to t
+
0
we get
G(t
+
0
)
− G(t
−
0
) =
e
iw·x
0
(2π)
3
,
but
G(t
−
0
) = 0,
so
A = e
iw·x
0
/(2π)
3
.
(10.9.10)
Hence,
G(w, t; x
0
, t
0
) =
e
iw·x
0
(2π)
3
e
−κw
2
(t−t
0
)
(10.9.11)
Using the inverse Fourier transform, we get
G(x, t; x
0
, t
0
) =
0
t < t
0
∞
−∞
e
−κw
2
(t−t
0
)
(2π)
3
e
−iw·(x−x
0
)
dw t > t
0
(10.9.12)
Recognizing this Fourier transofrm of the Green’s fucntion as a Gaussian, we obtain
G(x, t; x
0
, t
0
) =
0
t < t
0
1
(2π)
3
π
κ(t
− t
0
)
3/2
e
−
|x−x0|2
4κ(t−t0)
t > t
0
(10.9.13)
To solve the heat equation
∂u
∂t
= κ
∇
2
u + Q(x, t)
(10.9.14)
using Green’s fucntion we proceed as follows. We have a linear differential operator
L =
∂
∂t
− κ∇
2
(10.9.15)
where
252
L = L
1
− κL
2
(10.9.16)
with
L
1
=
∂
∂t
(10.9.17)
and
L
2
=
∇
2
.
(10.9.18)
We have the following Green’s formula for L
2
:
[uL
2
v
− vL
2
u] dx =
(u
∇v − v∇u) · nds.
(10.9.19)
However, for L
1
, since it is not self-adjoint, we have no such result. Nevertheless, integrating
by parts, we obtain
t
2
t
1
uL
1
v dt = uv
t
2
t
1
−
t
2
t
1
vL
1
u dt
(10.9.20)
So that if we introduce the adjoint opeator L
∗
1
=
−∂/∂t
We obtain
t
2
t
1
[uL
∗
1
v
− vL
1
u] =
−uv
t
2
t
1
(10.9.21)
Since
Lu = Q(x, t),
LG = δ(x
− x
0
)δ(t
− t
0
),
(10.9.22)
defining
L
∗
=
−
∂
∂t
− κ∇
2
,
(10.9.23)
we see
t
2
t
1
[uL
∗
v
− vL
∗
u] dx dt =
−
uv
t
2
t
1
dx+
κ
t
2
t
1
(v
∇u − u∇v) · n ds dt.
(10.9.24)
253
To get a representation for u(x, t) in terms of G, we consider the source-varying Green’s
function, which using translation, is
G(x, t
1
; x
1
, t) = G(x,
−t; x
1
,
−t
1
)
(10.9.25)
and by causality
G(x, t
1
; x
1
, t) = 0
if
t > t
1
.
(10.9.26)
Hence
∂
∂t
− κ∇
2
G(x, t
1
; x
1
, t) = δ(x
− x
1
)δ(t
− t
1
)
(10.9.27)
So
L
∗
[G(x, t
1
; x
1
, t)] = δ(x
− x
1
)δ(t
− t
1
)
(10.9.28)
where G(x, t
1
; x
1
, t) is called the adjoint Green’s fucntion. Furthermore,
G
∗
(x, t; x
1
, t
1
) = G(x, t; x
1
, t), and if
t > t
1
, G
∗
= G = 0.
(10.9.29)
We let u = u(x, t) be the solution to Lu = Q(x, t) subject to u(x, 0) = g(x),
and v = G(x, t
0
; x
0
, t) be the source-varying Green’s function satisfying
L
∗
v = δ(x
− x
0
)δ(t
− t
0
)
(10.9.30)
subject to homogenous boundary conditions and
G(x, t
0
; x
0
, t) = 0
for
t
0
> t.
(10.9.31)
Integrating from t
1
= 0 to t
2
= t
+
0
, our Green’s formula becomes
t
+
0
0
[uδ(x
− x
0
)δ(t
− t
0
)
− G(x, t
0
; x
0
, t)Q(x, t)] dx dt
=
u(x, 0)G(x, t
0
; x
0
, 0) dx
+κ
t
+
0
0
[G(x, t
0
; x
0
, t)
∇u − u∇G(x, t
0
; x
0
, t)]
· n ds dt.
(10.9.32)
Since G = 0 for t > t
0
, solving for u(x, t), replacing the upper limit of integration t
+
0
with
t
0
, and using reciprocity (interchanging x and x
0
, t
and
t
0
) yields
254
u(x, t) =
t
0
G(x, t; x
0
, t
0
)Q(x
0
, t
0
) dx
0
dt
0
+
G(x, t; x
0
, 0)g(x
0
) dx
0
+κ
t
0
[G(x, t; x
0
, t
0
)
∇
x
0
u(x
0
, t
0
)
− u(x
0
, t
0
)
∇
x
0
G(x, t; x
0
, t
0
)]
· n ds
0
dt
0
(10.9.33)
From our previous result, the solution for the infinite domain heat equation is given by
u(x, t) =
t
0
0
1
4πκ(t
− t
0
)
3/2
e
−
|x−x0|2
4κ(t−t0)
Q(x
0
, t
0
) dx
0
dt
0
+
f (x
0
)
1
4πκt
3/2
e
−
|x−x0|2
4κt
dx
0
(10.9.34)
255
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