Homework 3 Solutions

October 4, 2017

Exercise 2.19

We are asked to solve Laplace’s equation uxx+uyy = 0 in the semi infinite strip. We again attempt a separation of

variables solution u(x, y) = X(x)Y (y) so that− XX = YY = λ. As we showed in Problem 1.1, in order for u(x, y) = 0

at x = 0 and x = 1, we must have λ = nπ. Then the nth solutions are

Xn(x) = A sin(nπx)

Yn(y) = Cenπy +De−nπy.

Since y is unbounded in the strip, we must require that C = 0 for the solution to be finite. Then the general solution

will have the form

u(x, y) =

∞∑n=1

αne−nπy sin(nπx). (1)

If we assume that the boundary condition u(x, 0) = f(x) has a Fourier expansion f(x) =∑∞n=1 an sin(nπx), then

we clearly have

αn = an = 2

∫ 1

0

f(x) sin(nπx)dx.

Note that |an| < B for some constant B since f is bounded. Thus the sum in Eq. (1) converges uniformly. Hence,

u(x, y) =

∞∑n=1

(2

∫ 1

0

f(x) sin(nπx)dx

)e−nπy sin(nπx)

= 2

∫ 1

0

f(x)

( ∞∑n=1

sin2(nπx)e−nπy

)dx, (2)

where the interchange of summation and integration is permitted since the convergence in Eq. (1) is uniform.

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Exercise 2.20

We will solve Laplace’s equation,∂2u

∂r2+

1

r

∂u

∂r+

1

r2∂2u

∂θ2= 0,

in the annulus defined by {(r, θ) : ρ < r < 1}. We first seek a separable solution u(r, θ) = F (r)G(θ), which leads to

the equalityr2F (r) + rF (r)

F (r)= − G(θ)

G(θ)= λ.

The angular solution leads has the form G(θ) = Cei√λθ + De−i

√λθ. Demanding that G(θ) = G(θ + 2π) requires

that 1 = ei√λ2π , which implies that

√λ = m ∈ Z. Turning to the radial equation, first consider the case when m = 0.

Then we are left with an essentially first-order differential equation, rF (r) + F (r) = 0, which is easily seen to have

the solution

F0(r) = A0 +B0 log r.

For m 6= 0, direct substitution shows that

Fm = Amrm +Bmr

−m

solves the radial equation. We therefore suppose that the general solution has the form

u(r, θ) = A0 +B0 log r +∑m6=0

(Amr

m +Bmr−m) (Cmeimθ +Dme

−imθ)= A0 +B0 log r +

∑m6=0

(Amr

m + Bmr−m)eimθ. (3)

We now impose the boundary conditions u(1, θ) = f(θ) and u(ρ, θ) = g(θ). Assume that f(θ) and g(θ) have

Fourier series

f(θ) ∼∑n

aneinθ, g(θ) ∼

∑n

bneinθ. (4)

By comparing coefficients with Eq. (3) we must have that

a0 = A0, an = Am + Bm (n 6= 0)

b0 = A0 +B0 log ρ, bn = Amρm + Bmρ

−m (n 6= 0). (5)

These can be inverted to give

A0 = a0, Am =anρ−m − bm

ρ−m − ρm(n 6= 0)

B0 =b0 − a0log ρ

, Bm = −anρm − bm

ρ−m − ρm(n 6= 0). (6)

Substituting into Eq. (3) yields

u(r, θ) = a0 +b0 − a0log ρ

log r +∑m6=0

(am

(ρ/r)−m − (ρ/r)m

ρ−m − ρm+ bm

r−m − rm

ρ−m − ρm

)eimθ. (7)

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Observe that the since 1 < r < ρ, the coefficients of am and bm are always positive and less than one. Thus,

|u(r, θ)− f(θ)| =

∣∣∣∣∣∣b0 − a0log ρlog r +

∑m 6=0

(am

[(ρ/r)−m − (ρ/r)m

ρ−m − ρm− 1

]+ bm

r−m − rm

ρ−m − ρm

)eimθ

∣∣∣∣∣∣≤ |b0 − a0|

log ρlog r +

∑m 6=0

(|am|

[1− (ρ/r)−m − (ρ/r)m

ρ−m − ρm

]+ |bm|

r−m − rm

ρ−m − ρm

). (8)

We want to take the limit as r → 1, but care must be taken because of the infinite sum in m. Since both the sums∑m 6=0 |am| and

∑m 6=0 |bm| converge, for any ε > 0 there must exist some integer M such that

∑|m|>M |am| < ε

and∑|m|>M |bm| ≤ ε. Then using the facts that

1− (ρ/r)−m − (ρ/r)m

ρ−m − ρm≤ 1,

r−m − rm

ρ−m − ρm≤ 1,

we can continue the bound on |u(r, θ)− f(θ)| as

|u(r, θ)− f(θ)| ≤ |b0 − a0|log ρ

log r +∑|m|≤Mm6=0

(|am|

[1− (ρ/r)−m − (ρ/r)m

ρ−m − ρm

]+ |bm|

r−m − rm

ρ−m − ρm

)+ 2ε. (9)

With the sum being finite, we can now take a limit r → 1 on both sides to obtain limr→1 |u(r, θ) − f(θ)| ≤ 2ε. But

since ε is arbitrary and the bound on |u(r, θ)− f(θ)| holds for all θ, we have

u(r, θ)→ f(θ) uniformly in θ as r → 1. (10)

Now consider the convolution (Pr ∗ f)(θ) =∑∞m=−∞ r|m|ame

imθ. From Theorem 4.1 in the textbook, we have that

(Pr ∗ f)(θ)→ f(θ) uniformly in θ as r → 1 (11)

since f(θ) is continuous. From the triangle inequality

|u(r, θ)− (Pr ∗ f)(θ)| ≤ |u(r, θ)− f(θ)|+ |(Pr ∗ f)(θ)− f(θ)|,

we combine Eqns. (10) and (11) to conclude

u(r, θ)→ (Pr ∗ f)(θ) uniformly in θ as r → 1. (12)

An analogous argument gives u(r, θ)→ (Pρ/r ∗ g)(θ) uniformly in θ as r → ρ.

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