10.10
Green’s Function for the Wave Equation on a Cube
Solving the wave equation in
R
3
with Cartesian coordinates, we select a rectangular domain
D = {x = (x, y, z) : x [0, α] , y [0, β] , z [0, γ]}
So that the wave equation is
∂
2
u(x, t)
∂t
2
− c
2
∇
2
u(x, t) = Q(x, t)
x
D, t > 0
(10.10.1)
u(x, t) = f (x, t)
x∂
D
(10.10.2)
u(x, 0) = g(x),
x
D
(10.10.3)
u
t
(x, 0) = h(x)
x
D
(10.10.4)
Defining the wave operator
L =
∂
2
∂t
2
− c
2
∇
2
(10.10.5)
we seek a Green’s function G(x, t, x
0
, t
0
) such that
L [G(x, t; x
0
, t
0
)] = δ(x
− x
0
)δ(t
− t
0
)
(10.10.6)
Also,
G(x, t; x
0
, t
0
) = 0
if
t < t
0
(10.10.7)
G(x, t
+
0
; x
0
, t
0
) = 0,
(10.10.8)
and
G
t
(x, t
+
0
; x
0
, t
0
) = δ(x
− x
0
)
(10.10.9)
We require the translation property
256
G(x, t; x
0
, t
0
) = G(x, t
− t
0
; x
0
, 0)
(10.10.10)
and spatial symmetry
G(x, t
− t
0
; x
0
, 0) = G(x
0
, t
− t
0
; x, 0)
(10.10.11)
provided the difference t
− t
0
is the same in each case.
We have shown that the solution to the wave equation is
u(x, t) =
t
0
D
G(x, t; x
0
, t
0
)Q(x
0
, t
0
) dx
0
dt
0
+
D
[h(x
0
)G(x, t; x
0
, 0)
− g(x
0
)G
t
0
(x, t; x
0
, 0)] dx
0
(10.10.12)
−c
2
t
0
∂D
f (x
0
, t
0
)
∇
x
0
G(x, t; x
0
, t
0
)
− G(x, t; x
0
, t
0
)
∇
x
0
f (x
0
, t
0
))
· n ds
0
dt
0
We therefore must find the Green’s function to solve our problem. We begin by finding the
Green’s function for the Helmholtz operator
ˆ
Lu
≡ ∇
2
u + c
2
u = 0
(10.10.13)
Where u is now a spatial function of x on
D. The required Green’s function G
c
(x, x
0
)
satisfies
∇
2
G
c
+ c
2
G
c
= δ(x
− x
0
)
(10.10.14)
with homogeneous boundary conditions
G
c
(x) = 0
for
x∂
D
(10.10.15)
We will use eigenfunction expansion to find the Green’s function for the Helmholtz equation.
We let the eigenpairs
{φ
N
, λ
N
} be such that
∇
2
φ
N
+ λ
2
N
φ
N
= 0
(10.10.16)
257
Hence the eigenfunctions are
φ
mn
(x, y, z) = sin
πx
α
sin
mπy
β
sin
nπz
γ
(10.10.17)
and the eigenvlaues are
λ
mn
= π
2
α
2
+
m
β
2
+
m
γ
2
for
, m, n = 1, 2, 3, . . .
(10.10.18)
We know that these eigenfunctions form a complete, orthonormal set which satisfy the
homogeneous bounday conditions on ∂
D. Since G
c
also satisfies the homogeneous boundary
conditions, we expand G
c
in terms of the eigenfunctions φ
N
, where N represents the index
set
{, m, n}, so that
G
c
(x; x
0
) =
N
A
N
φ
N
(x)
(10.10.19)
Substituting into the PDE (10.10.14) we see
N
A
N
(c
2
− λ
2
N
)φ
N
(x) = δ(x
− x
0
)
(10.10.20)
If we multiply the above equation by φ
M
and integrate over
D, using the fact that
D
φ
N
φ
M
dx = δ
N M
(10.10.21)
A
N
=
φ
N
(x
0
)
c
2
− λ
2
N
(10.10.22)
So that
G
c
(x; x
0
) =
N
φ
N
(x
0
)φ
N
(x)
c
2
− λ
2
N
(10.10.23)
There are two apparent problems with this form. The first is, it appears to not be symmetric
in x and x
0
; However, if we note that the Helmholtz equation involved no complex numbers
explicitly, φ
N
and φ
N
are distinct eigenfunctions corresponding to the eigenvalue λ
N
and
258
that the above expansion contains both the terms
φ
N
(x
0
)φ
N
(x)
c
2
− λ
2
N
and
φ
N
(x
0
)φ
N
(x)
c
2
− λ
2
N
, so that
the Green’s function is, in fact, symmetric and real.
We also see a potential problem when λ
2
N
= c
2
. As a function of c, G
c
is analytic except
for simple poles at c =
±λ
N
, for which we have nontrivial solutions to the homogeneous
Helmholtz equation we must use modified Green’s functions as before when zero was an
eigenvalue.
We now use the Green’s function for the Helmholtz equation to find G(x, t; x
0
, t
0
), the
Green’s function for the wave equation. We notice that G
c
(x; x
0
)e
−ic
2
t
is a solution of the
wave equation for a point source located at x
0
, since
∂
2
∂t
2
G
c
e
−ic
2
t
− c
2
∇
2
x
G
c
e
−ic
2
t
=
−c
4
G
c
e
−ic
2
t
− c
2
−c
2
+ δ(x
− x
0
)
e
−ic
2
t
=
−c
2
δ(x
− x
0
)e
−ic
2
t
(10.10.24)
So that
∇
2
x
G
c
e
−ic
2
t
−
1
c
2
∂
2
∂t
2
G
c
e
−ic
2
t
= δ(x
− x
0
)e
−ic
2
t
(10.10.25)
Using the integral representation of the δ function
δ(t
− t
0
) =
1
2π
∞
−∞
e
−icw(t−t
0
)
dw
(10.10.26)
and using linearlity we obtain
G(x, t; x
0
, t
0
) =
1
2π
∞
−∞
G
c
(x; x
0
)e
−ic
2
(t−t
0
)
d(c
2
)
(10.10.27)
Although we have a form for the Green’s function, recalling the form for G
c
, we note we
cannot integrate along the real axis due to the poles at the eigenvalues
±λ
N
. If we write the
expansion for G
c
we get
G(x, t; x
0
, t
0
) =
1
2π
∞
−∞
N
φ
N
(x)φ
N
(x
0
)
c
2
− λ
2
N
e
−ic
2
(t−t
0
)
d(c
2
)
(10.10.28)
Changing variables with w
N
= cλ
N
, w = c
2
259
G(x, t; x
0
, t
0
) =
c
2
2π
N
φ
N
(x)φ
N
(x
0
)
∞
−∞
e
−iw(t−t
0
)
w
2
N
− w
2
dw,
(10.10.29)
We must integrate along a contour so that G(x, t; x
0
, t
0
) = 0 when t > t
0
, so we select
x + i x (
−∞, ∞) as the contour. For t > t
0
, we close the contour with an (infinite) semi-
circle in the lower half-plane without changing the value of the integral, and using Cauchy’s
formula we obtain
2π
w
N
sin[w
N
(t
− t
0
)]. If t < t
0
, we clsoe the contour with a semi-circle in
the upper half-plane in which there are no poles, and the integral equals zero.
Hence
G(x, t; x
0
, t
0
) = c
2
N
sin[w
N
(t
− t
0
)]
w
N
H(t
− t
0
)φ
N
(x
0
)φ
N
(x)
(10.10.30)
where H is the Heaviside function, and w
N
= cλ
N
.
260
SUMMARY
Let
Ly = −(py
)
+ qy.
• To get Green’s function for Ly = f,
0 < x < 1,
y(0)
− h
0
y
(0) = y(1)
− h
1
y
(1) = 0.
Step 1: Solve
Lu = 0,
u(0)
− h
0
u
(0) = 0,
and
Lv = 0,
v(1)
− h
1
v
(1) = 0.
Step 2:
G(x; s) =
u(s)v(x)
0
≤ s ≤ x ≤ 1
u(x)v(s)
0
≤ x ≤ s ≤ 1.
Step 3:
y =
1
0
G(x; s)f (s)ds,
where
LG = δ(x − s).
G satisfies homogeneous boundary conditions and a jump
∂G(s
+
; s)
∂x
−
∂G(s
−
; s)
∂x
=
−
1
p(s)
.
• To solve Ly − λry = f,
0 < x < 1,
y(x) = λ
1
0
G(x; s)r(s)y(s)ds + F (x),
where
F (x) =
1
0
G(x; s)f (s)ds.
• Properties of Dirac delta function
• Fredholm alternative
261
• To solve Lu = f,
0 < x < 1,
u(0) = A,
u(1) = B,
and λ = 0 is not an eigenvalue:
Find G such that
LG = δ(x − s),
G(0; s) = A,
G(1; s) = B.
u(s) =
1
0
G(x; s)f (x)dx + BG
x
(1; s)
− AG
x
(0; s)
• To solve Lu = f,
0 < x < 1 subject to homogeneous boundary conditions
and λ = 0 is an eigenvalue:
Find ˆ
G such that
L ˆ
G = δ(x
− s) −
v
h
(x)v
h
(s)
(
L
0
v
2
h
(x)dx
,
where v
h
is the solution of the homogeneous
Lv
h
= 0,
The solution is
u(x) =
L
0
f (s) ˆ
G(x; s)ds.
• Solution of Poisson’s equation
∇
2
u = f (
r)
subject to homogeneous boundary conditions.
The Green’s function must be the solution of
∇
2
G(
r;
r
0
) = δ(x
− x
0
)δ(y
− y
0
)
where
r = (x, y)
subject to the same homogeneous boundary conditions.
The solution is then
u(
r) =
f (
r
0
)G(
r;
r
0
)d
r
0
.
262
• To solve Poisson’s equation with nonhomogeneous boundary conditions
∇
2
u = f (
r)
u = h(
r),
on the boundary,
Green’s function as before
∇
2
G = δ(x
− x
0
)δ(y
− y
0
),
with homogeneous boundary conditions
G(
r;
r
0
) = 0.
The solution is then
u(
r) =
f (
r
0
)G(
r;
r
0
)d
r
0
+
.
h(
r
0
)
∇G · nds.
• To solve
∇
2
u = f,
on infinite space with no boundary.
Green’s function should satisfy
∇
2
G = δ(x
− x
0
)δ(y
− y
0
),
on infinite space with no boundary.
For 2 dimensions
G(r) =
1
2π
ln r,
lim
r→∞
u
− r ln r
∂u
∂r
= 0.
u(
r) =
f (
r
0
)G(
r;
r
0
)d
r
0
.
For 3 dimensional space
G =
−
1
4π
r.
For upper half plane
G =
1
4π
ln
(x
− x
0
)
2
+ (y
− y
0
)
2
(x
− x
0
)
2
+ (y + y
0
)
2
,
method of images
u(
r) =
f (
r
0
)G(
r;
r
0
)d
r
0
+
∞
−∞
h(x
0
)
y
π
(x
− x
0
)
2
+ y
2
dx
0
.
263
• To solve the wave equation on infinite domains
∂
2
u
∂t
2
= c
2
∇
2
u + Q(x, t)
u(x, 0) = f (x)
u
t
(x, 0) = g(x)
G(x, t; x
0
, t
0
) =
0
t < t
0
1
4π
2
ρc
δ(ρ
− c(t − t
0
)) t > t
0
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
• To solve the heat equation on infinite domains
∂u
∂t
= κ
∇
2
u + Q(x, t)
u(x, 0) = g(x)
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
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
264
• To solve the Wave Equation on a Cube
∂
2
u(x, t)
∂t
2
− c
2
∇
2
u(x, t) = Q(x, t)
x
D, t > 0
u(x, t) = f (x, t)
x∂
D
u(x, 0) = g(x),
x
D
u
t
(x, 0) = h(x)
x
D
D = {x = (x, y, z) : x [0, α] , y [0, β] , z [0, γ]}
G(x, t; x
0
, t
0
) = c
2
N
sin[w
N
(t
− t
0
)]
w
N
H(t
− t
0
)φ
N
(x
0
)φ
N
(x)
where H is the Heaviside function, w
N
= cλ
N
, and λ
N
and φ
N
are the eigenvalues and
eigenfunctions of Helmholtz equation.
u(x, t) =
t
0
D
G(x, t; x
0
, t
0
)Q(x
0
, t
0
) dx
0
dt
0
+
D
[h(x
0
)G(x, t; x
0
, 0)
− g(x
0
)G
t
0
(x, t; x
0
, 0)] dx
0
−c
2
t
0
∂D
f (x
0
, t
0
)
∇
x
0
G(x, t; x
0
, t
0
)
− G(x, t; x
0
, t
0
)
∇
x
0
f (x
0
, t
0
))
· n ds
0
dt
0
265
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