LECTURE 18: SOLVING THE LAPLACE EQUATIONS III AND POISSON EQUATIONS
MINGFENG ZHAO
August 05, 2015
Solving the Laplace equation in a wedge with cut-outs
Example 1. Consider the following problem:

1
1


urr + ur + 2 uθθ = 0, a < r < b, 0 < θ < α,


r
r

(1)
BC
:
u
(r,
0)
=
u
(r,
α), a ≤ r ≤ b,
θ
θ





u(b, θ) = 0, u(a, θ) = f (θ), 0 < θ < α.
First, let’s look the following problem:

1
1


urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,


r
r

(2)
BC : uθ (r, 0) = uθ (r, α), a ≤ r ≤ b,





u(b, θ) = 0, 0 < θ < α.
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (2), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
That is, we have
R00 (r) + 1r R0 (r)
Θ00 (θ)
=−
= λ.
R(r)
Θ(θ)
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since uθ (r, 0) = uθ (r, α) = 0, then Θ0 (0) = Θ0 (α) = 0. So
Θ(θ) satisfies:
(3)


 Θ00 + λΘ = 0,

 Θ0 (0) = Θ0 (α) = 0.
For the eigenvalue problem (3), we know that the eigenvalues are λ =
nπ S
are Θ(θ) = C cos
θ , with n ∈ N {0}.
α
1
nπ 2
α
, and the corresponding eigenfunctions
2
MINGFENG ZHAO
nπ For each n ≥ 0, when λ =
α


 C1 + C2 ln r,
1
, since R00 (r) + R0 (r) − λR(r) = 0, then R(r) =

r



 C1 + C2 ln b = 0,
C1 r
nπ
α
if n = 0,
.
+ C2 r−
nπ
α
, if n ≥ 1.


if n = 0,
 C[− ln b + ln r],
Since u(b, θ) = 0, then
, that is, R(r) =
i
h


2nπ
nπ
 C b nπ
 C r nπ
− nπ
α − e α b r− α
, if n ≥ 1.
1 α + C2 b α = 0, if n ≥ 1.
In summary, for any n ≥ 0, we can find a non-zero solution to (2):


 − ln b + ln r,
if n = 0,
h nπ
i
nπ un (r, θ) =
2nπ

 r α − b α cos
θ , if n ≥ 1.
α
if n = 0,
Let
u(r, θ) = a0 [− ln b + ln r] +
∞
X
i
nπ h nπ
2nπ
nπ
θ
an r α − b α r− α cos
α
n=0
be the solution to (1), it’s easy to see that uθ (r, 0) = uθ (r, α) = u(b, θ) = 0. Since u(a, θ) = f (θ), then
∞
X
h nπ
i
nπ 2nπ
nπ
an a α − b α a− α cos
θ .
α
n=0
f (θ) = a0 [− ln b + ln a] +
Then we get
Z
1 2
a0 [− ln b + ln a] = ·
2 α
α
f (θ) dθ,
h
and an a
nπ
α
−b
2nπ
α
− nπ
α
a
i
0
Z
2
=
α
α
f (θ) cos
nπ α
0
dθ,
for all n ≥ 1.
That is,
a0 =
1
α[− ln b + ln a]
Z
α
f (θ) dθ,
and an =
α a
0
Z
2
h
nπ
α
−b
2nπ
α
a−
i
nπ
α
α
f (θ) cos
nπ 0
Solving the Laplace equation with Dirichlet problem in a disc
Consider the following problem:








 BC :
(4)








1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < 2π,
r
r
u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π), 0 < r ≤ a,
u(a, θ) = f (θ),
0 ≤ θ < 2π,
u(r, θ) is bounded as r → 0.
First, let’s look the following problem:

1
1


urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,


r
r

(5)
BC : u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π), 0 < r ≤ a,





u(r, θ) is bounded as r → 0.
α
dθ,
for all n ≥ 1.
LECTURE 18: SOLVING THE LAPLACE EQUATIONS III AND POISSON EQUATIONS
3
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (5), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
That is, we have
R00 (r) + 1r R0 (r)
Θ00 (θ)
=−
= λ.
R(r)
Θ(θ)
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since u(r, 0) = u(r, 2π) and uθ (r, 0) = uθ (r, 2π), then
Θ(0) = Θ(2π) and Θ0 (0) = Θ0 (2π). So Θ(θ) satisfies:


 Θ00 + λΘ = 0,
(6)

 Θ(0) = Θ(2π), Θ0 (0) = Θ0 (2π).
For the eigenvalue problem (6), we know that the eigenvalues are λ = n2 , and the corresponding eigenfunctions are
S
Θ(θ) = C1 cos(nx) + C2 sin(nx), with n ∈ N {0}.


 C1 + C2 ln r,
if n = 0,
1
.
For each n ≥ 0, when λ = n2 , since R00 (r) + R0 (r) − λR(r) = 0, then R(r) =

r
nπ
 C r nπ
α + C r− α ,
if n ≥ 1.
1
2


 C,
if n = 0,
Since u(r, θ) is bounded as r → 0, then C2 = 0, that is, R(r) =

 Crn , if n ≥ 1.
In summary, for any n ≥ 0, we can find a non-zero solution to (5):

 1,
if n = 0,
un (r, θ) =
 rn [C cos(nθ) + C sin(nθ)], if n ≥ 1.
1
2
Let
u(r, θ) = a0 +
∞
X
[an rn cos(nθ) + bn rn sin(nθ)]
n=1
be the solution to (4), it’s easy to see that u(r, 0) = u(r, 2π) and uθ (r, 0) = uθ (r, 2π). Since u(a, θ) = f (θ), then
∞
a0 X
f (θ) =
+
[an an cos(nθ) + bn an sin(nθ)].
2
n=1
Then we get
Z
1 2π
f (θ) dθ,
a0 =
π 0
1
an a =
π
Z
a−n
an =
π
Z
n
2π
f (θ) cos(nθ) dθ,
0
2π
1
and bn a =
π
Z
a−n
and bn =
π
Z
n
f (θ) sin(nθ) dθ,
for all n ≥ 1.
f (θ) sin(nθ) dθ,
for all n ≥ 1.
0
That is,
1
a0 =
π
Z
2π
f (θ) dθ,
0
2π
f (θ) cos(nθ) dθ,
0
0
2π
4
MINGFENG ZHAO
The Poisson’s integral formula
Recall that the solution to the following problem:

1
1


urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < 2π,


r
r



 BC : u(r, 0) = u(r, 2π), u (r, 0) = u (r, 2π), 0 < r ≤ a,
θ
θ
(7)


u(a, θ) = f (θ), 0 ≤ θ < 2π,






u(r, θ) is bounded as r → 0.
is given by:
∞
a0 X
+
[an rn cos(nθ) + bn rn sin(nθ)],
2
n=1
u(r, θ) =
where
1
a0 =
π
Z
2π
f (θ) dθ,
0
a−n
an =
π
2π
Z
f (θ) cos(nθ) dθ,
0
a−n
and bn =
π
Z
2π
f (θ) sin(nθ) dθ,
for all n ≥ 1.
0
Then we have
u(r, θ)
=
=
=
=
Let z =
1
2π
Z
1
2π
Z
2π
f (τ ) dτ +
0
∞ −n Z
X
a
n=1
2π
f (τ ) dτ +
0
π
∞
X
a−n rn
π
n=1
2π
f (τ ) cos(nτ ) dτ
rn cos(nθ) +
0
a−n
π
2π
Z
f (τ ) sin(nτ ) dτ
0
2π
Z
f (τ )[cos(nτ ) cos(nθ) + sin(nτ ) sin(nθ)] dτ
0
Z
∞
X
a−n rn 2π
f (τ ) dτ +
f (τ ) cos[n(τ − θ)] dτ
π
0
0
n=1
#
"
Z
∞
1 2π
1 X r n
cos[n(θ − τ )] dτ.
+
f (τ )
π 0
2 n=1 a
1
2π
Z
2π
r n
r i(θ−τ )
e
, then
cos[n(θ − τ )] = Re z n for all n 6= 1, which implies that
a
a
∞ n
X
r
n=1
a
cos[n(θ − τ )]
∞
X
=
n=1
Re z n = Re
∞
X
zn
n=1
1
If |z| < 1, that is, r < a
1−z
ar cos(θ − τ )
.
a2 − 2ar cos(θ − τ ) + r2
=
Re
=
Therefore, we get
u(r, θ)
=
1
π
Z
0
2π
1
ar cos(θ − τ )
f (τ )
+
2 a2 − 2ar cos(θ − τ ) + r2
dτ
rn sin(nθ)
LECTURE 18: SOLVING THE LAPLACE EQUATIONS III AND POISSON EQUATIONS
=
a2 − r2
2π
Z
0
2π
5
f (τ )
dτ,
a2 − 2ar cos(θ − τ ) + r2
which is called the Poisson integral formula.
Solving the Laplace equation with Neumann problem in a disc
Consider the following problem:
(8)









 BC :
1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < 2π,
r
r
u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π), 0 < r ≤ a,








 AC :
ur (a, θ) = f (θ), 0 ≤ θ < 2π, u(r, θ) is bounded as r → 0
Z 2π
f (θ) dθ = 0.
0
First, let’s look the following problem:
(9)






1
1
urr + ur + 2 uθθ = 0, 0 < r < a, 0 < θ < α,
r
r
BC : u(r, 0) = u(r, 2π), uθ (r, 0) = uθ (r, 2π), 0 < r ≤ a,





u(r, θ) is bounded as r → 0.
Let u(r, θ) = R(r)Θ(θ) be a non-zero separated solution to (9), then
1
1
R00 (r)Θ(θ) + R0 (r)Θ(θ) + 2 R(r)Θ00 (θ) = 0.
r
r
That is, we have
R00 (r) + 1r R0 (r)
Θ00 (θ)
=−
= λ.
R(r)
Θ(θ)
It’s easy to see that λr = λθ = 0, that is, λ is a constant. Since u(r, 0) = u(r, 2π) and uθ (r, 0) = uθ (r, 2π), then
Θ(0) = Θ(2π) and Θ0 (0) = Θ0 (2π). So Θ(θ) satisfies:
(10)


 Θ00 + λΘ = 0,

 Θ(0) = Θ(2π), Θ0 (0) = Θ0 (2π).
For the eigenvalue problem (10), we know that the eigenvalues are λ = n2 , and the corresponding eigenfunctions are
S
Θ(θ) = C1 cos(nx) + C2 sin(nx), with n ∈ N {0}.
6
MINGFENG ZHAO


 C1 + C2 ln r,
if n = 0,
1 0
.
R (r) − λR(r) = 0, then R(r) =

r
 C r nπ
− nπ
1 α + C2 r α , if n ≥ 1.


 C,
if n = 0,
Since u(r, θ) is bounded as r → 0, then C2 = 0, that is, R(r) =

 Crn , if n ≥ 1.
In summary, for any n ≥ 0, we can find a non-zero solution to (9):

 1,
if n = 0,
un (r, θ) =
 rn [C cos(nθ) + C sin(nθ)], if n ≥ 1.
For each n ≥ 0, when λ = n2 , since R00 (r) +
1
2
Let
u(r, θ) = a0 +
∞
X
[an rn cos(nθ) + bn rn sin(nθ)]
n=1
be the solution to (8), it’s easy to see that u(r, 0) = u(r, 2π) and uθ (r, 0) = uθ (r, 2π). Since ur (a, θ) = f (θ), then
f (θ) =
∞
X
[nan an−1 cos(nθ) + nbn an−1 sin(nθ)].
n=1
Then we have
Z 2π
f (θ) dθ = 0,
nan a
n−1
0
1
=
π
Z
2π
f (θ) cos(nθ) dθ,
n−1
and nbn a
0
1
=
π
Z
2π
f (θ) sin(nθ) dθ,
for all n ≥ 1.
0
That is,
a1−n
an =
nπ
Z
2π
f (θ) sin(nθ) dθ,
0
a1−n
and bn =
nπ
Z
2π
f (θ) sin(nθ) dθ,
for all n ≥ 1.
0
Solving the Poisson equation with Dirichlet boundary condition in a rectangle by using
eigenfunction expansions
For the Poisson equation in a rectangle D = [0, a] × [0, b], since ∂D consists of four line segments, then



uxx + uyy = f (x, y), (x, y) ∈ D,



(11)
BC : u(x, 0) = f1 (x), u(x, b) = f2 (x), 0 ≤ x ≤ a,





u(0, y) = g1 (y), u(a, y) = g2 (y), 0 ≤ y ≤ b.
To solve the problem (11):
Step I: Choose any “nice” function w(x, y) such that w(0, y) = g1 (y) and w(a, y) = g2 (y) for 0 ≤ y ≤ b. For example,
we can just choose
w(x, y) =
g2 (y) − g1 (y)
x + g1 (y).
a
LECTURE 18: SOLVING THE LAPLACE EQUATIONS III AND POISSON EQUATIONS
7
Step II: Let v(x, y) = u(x, y) − w(x, y), then
vxx + vyy
=
uxx + uyy − wxx − wyy
=
f (x, y) − wxx − wyy
=: f˜(x, y)
v(x, 0)
v(x, b)
v(0, y)
v(a, y)
(12)
=
u(x, 0) − w(x, 0)
=
f1 (x) − w(x, 0)
=
f˜1 (x)
=
v(x, b) − w(x, b)
=
f2 (x) − w(x, b)
=
f˜2 (x)
=
u(0, y) − w(0, y)
=
g1 (y) − g1 (y)
=
0
=
u(a, y) − w(a, y)
=
g2 (y) − g2 (y)
=
0.
Then v satisfies the following problem:



vxx + vyy = f˜(x, y), (x, y) ∈ D,



BC : v(x, 0) = f˜1 (x), v(x, b) = f˜2 (x),





v(0, y) = v(a, y) = 0, 0 ≤ y ≤ b.
0 ≤ x ≤ a,
To solve (12), we are going to use the method of eigenfunction expansions, which is a “generalization” of
“variation of parameters”:
Step II-1: First let’s look at the problem:



vxx + vyy = 0, (x, y) ∈ D,
(13)

 BC: v(0, y) = v(a, y) = 0, 0 ≤ y ≤ b.
8
MINGFENG ZHAO
By using the separation of variables, we know that the “ general ” solution to (13) is:
v(x, y) =
∞
nπ X
nπ nπ
an e a y + bn e− a y sin
x .
a
n=1
Step II-2: For each y > 0, we can think f˜(·, y) is a function of x, then we can write it as an “infinite” linear combination
n nπ o∞
x
of sin
, that is,
a
n=1
∞
nπ X
f˜(x, y) =
x ,
f˜n (t) sin
a
n=1
where
Z
a
nπ f˜(x, t) sin
x dx, ∀n ≥ 1.
a
0
n nπ o∞
Step II-3: Write f˜1 (x) and f˜2 (x) as “infinite” linear combinations of sin
x
, that is,
a
n=1
∞
∞
nπ nπ X
X
f˜1 (x) =
x , and f˜2 (x) =
x ,
f˜1,n sin
f˜2,n sin
a
a
n=1
n=1
2
f˜n (t) =
a
where
2
f˜1,n =
a
Z
0
a
nπ x dx,
f˜1 (x) sin
a
2
and f˜2,n =
a
Z
0
a
nπ x dx,
f˜2 (x) sin
a
∀n ≥ 1.
Step II-4: Let
v(x, y) =
∞
X
vn (y) sin
n=1
nπ x ,
a
be the solution to (12). It’s easy to see that v(0, y) = v(a, y) = 0 for all 0 ≤ y ≤ b. Moreover, “informally”
we have
∞
X
n=1
vn00 (y) sin
∞ nπ X
nπ nπ 2
x −
x
vn (y) sin
a
a
a
n=1
= vxx + vyy = f˜(x, y)
=
∞
X
vn (0) sin
n=1
nπ x
a
n=1
vn (b) sin
nπ x
a
nπ x
f˜n (y) sin
a
n=1
= v(x, 0) = f˜1 (x)
=
∞
X
∞
X
∞
X
nπ f˜1,n sin
x
a
n=1
= v(x, b) = f˜2 (x)
=
∞
X
nπ f˜2,n sin
x .
a
n=1
LECTURE 18: SOLVING THE LAPLACE EQUATIONS III AND POISSON EQUATIONS
(14)
9
Then for any n ≥ 1, vn (y) satisfies the following ODE with the initial value problem:

n2 π 2


 vn00 (y) − 2 vn (y) − f˜n (y) = 0,
a


 vn (0) = f˜1,n , vn (b) = f˜2,n .
So if for any n ≥ 1, vn (y) is the solution to (14), then v(x, y) =
∞
X
vn (y) sin
n=1
nπ x is the solution to (12).
a
Step III: The solution to (11) is given by:
u(x, y)
=
=
w(x, y) + v(x, y)
w(x, y) +
∞
X
n=1
vn (y) sin
nπ x .
a
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca