Introduction to Laplace’s Equation

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Spring 2016

Objectives

In this lesson we will learn about

▶ the partial differential equations and boundary value

problems known as Laplace’s and Poisson’s equations, ▶ techniques for solving Laplace’s and Poisson’s equations,

and ▶ applications of Laplace’s and Poisson’s equations.

The Laplacian Operator

Let x = (x_{1}, x_{2}, …, x_{n}) ∈ ℝ^{n} and let function u : Ω ⊂ ℝ^{n} → ℝ, then the Laplacian of u is

Δu = ∇^{2}u = \frac{∂^{2}u}{∂x_{1}^{2}} + \frac{∂^{2}u}{∂x_{2}^{2}} + ⋯ + \frac{∂^{2}u}{∂x_{n}^{2}}.

If n = 2 we will often write x = (x, y) and

Δu = \frac{∂^{2}u}{∂x^{2}} + \frac{∂^{2}u}{∂y^{2}}.

If n = 3 we will often write x = (x, y, z) and

Δu = \frac{∂^{2}u}{∂x^{2}} + \frac{∂^{2}u}{∂y^{2}} + \frac{∂^{2}u}{∂z^{2}}.

Open Sets

A set Ω ⊂ ℝ^{n} is open if for every x ∈ Ω there exists δ > 0 such that

{y | ∥x − y∥ < δ} ⊂ Ω.

Connected Sets

A set Ω ⊂ ℝ^{n} is connected if for every x, y ∈ Ω there exists a path connecting x and y which lies completely within Ω.

Laplace’s Equation

If u : Ω ⊂ ℝ^{n} → ℝ where Ω is an open, connected set then u satisfies Laplace’s equation if

Δu = 0.

Function u is said to be a harmonic function.

If function f is defined on Ω then u satisfies Poisson’s equation if

Δu = f .

Example

Let u(x, y) = e^{x} cos y and show u is harmonic on Ω = ℝ^{2}.

Δu = e^{x} cos y + (e^{x}(− cos y)) = 0

for all (x, y) ∈ Ω.

Example

Let u(x, y) = e^{x} cos y and show u is harmonic on Ω = ℝ^{2}.

Δu = e^{x} cos y + (e^{x}(− cos y)) = 0

for all (x, y) ∈ Ω.

Boundary Conditions

▶ If Ω ⊂ ℝ^{n} is an open set then the set of boundary points of

Ω will be denoted ∂Ω. ▶ Laplace’s and Poisson’s equations must be supplemented

with boundary conditions.

Dirichlet BCs: u(x) = ϕ(x) for x ∈ ∂Ω.

Neumann BCs: \frac{∂u}{∂n}(x) = ϕ(x) for x ∈ ∂Ω where n is the unit outward normal vector to ∂Ω.

Mixed BCs: u(x) + α\frac{∂u}{∂n}(x) = ϕ(x) for x ∈ ∂Ω where α is constant and n is the unit outward normal vector to ∂Ω.

Boundary Conditions

▶ If Ω ⊂ ℝ^{n} is an open set then the set of boundary points of

Ω will be denoted ∂Ω. ▶ Laplace’s and Poisson’s equations must be supplemented

with boundary conditions.

Dirichlet BCs: u(x) = ϕ(x) for x ∈ ∂Ω.

Neumann BCs: \frac{∂u}{∂n}(x) = ϕ(x) for x ∈ ∂Ω where n is the unit outward normal vector to ∂Ω.

Mixed BCs: u(x) + α\frac{∂u}{∂n}(x) = ϕ(x) for x ∈ ∂Ω where α is constant and n is the unit outward normal vector to ∂Ω.

Dirichlet Problem on a Rectangle

Define

Ω = {(x, y) | 0 < x < a, 0 < y < b} = (0, a) × (0, b) where a and b are constants. In this case ∂Ω = {(x, y) | x = 0, a and 0 ≤ y ≤ b, or y = 0, b and 0 ≤ x ≤ a}. Consider the following example of Laplace’s equation.

Δu = 0 for (x, y) ∈ Ω
u(0, y) = 0
u(a, y) = f (y)
u(x, 0) = 0
u(x, b) = 0

This problem can be solved using separation of variables and Fourier series.

Separation of Variables

Assume u(x, y) = X(x)Y (y), then

X′′(x)Y (y) + X(x)Y′′(y) = 0
\frac{X′′(x)Y (y)}{X(x)Y (y)}+ \frac{X(x)Y′′(y)}{X(x)Y (y)} = 0
−\frac{X′′(x)}{X(x)} = \frac{Y′′(y)}{Y (y)} = c

where c is a constant. The boundary conditions become

u(0, y) = X(0)Y (y) = 0 ⟹ X(0) = 0
u(x, 0) = X(x)Y (0) = 0 ⟹ Y (0) = 0
u(x, b) = X(x)Y (b) = 0 ⟹ Y (b) = 0.

Separation of Variables

Assume u(x, y) = X(x)Y (y), then

X′′(x)Y (y) + X(x)Y′′(y) = 0
\frac{X′′(x)Y (y)}{X(x)Y (y)}+ \frac{X(x)Y′′(y)}{X(x)Y (y)} = 0
−\frac{X′′(x)}{X(x)} = \frac{Y′′(y)}{Y (y)} = c

where c is a constant. The boundary conditions become

u(0, y) = X(0)Y (y) = 0 ⟹ X(0) = 0
u(x, 0) = X(x)Y (0) = 0 ⟹ Y (0) = 0
u(x, b) = X(x)Y (b) = 0 ⟹ Y (b) = 0.

Boundary Value Problem for Y (y)

Separation of variables and the Dirichlet boundary conditions imply the following BVP for Y (y):

Y (y) − c Y (y) = 0 for 0 < y < b
Y (0) = 0
Y (b) = 0.

Previously we have seen the only nontrivial solutions occur when

c = −λ_{n}^{2} = −\frac{n^{2}π^{2}}{b^{2}}

and Y_{n}(y) = sin \frac{nπy}{b} for n ∈ ℕ.

Boundary Value Problem for X(x)

Function X(x) must satisfy the following boundary value problem.

X′′(x) + c X(x) = X′′(x) − \frac{n^{2}π^{2}}{b^{2}}X(x) = 0
X(0) = 0

The general solution takes the form

X(x) = A cosh \frac{nπx}{b} + B sinh \frac{nπx}{b}

and A = 0 in order to satisfy the boundary condition. Thus X_{n}(x) = sinh \frac{nπx}{b}.

Product Solution

▶ Define the function

u_{n}(x, y) = sinh \frac{nπx}{b} sin \frac{nπy}{b}

for n ∈ ℕ. ▶ Each u_{n}(x, y) satisfies Laplace’s equation on the rectangle

and the boundary conditions for (x, y) = (x, 0) and (x, y) = (x, b) for 0 ≤ x ≤ a and for (x, y) = (0, y) for 0 ≤ y ≤ b.

▶ By the Principal of Superposition

u(x, y) = ∑_{n=1}^{N}c_{n} sinh \frac{nπx}{b} sin \frac{nπy}{b}

will also satisfy these equations for any N ≥ 1 and constants c_{n}.

Boundary (a, y) for 0 ≤ y ≤ b

▶ Extend f (y) as an odd, 2b-periodic function. ▶ For 0 ≤ y ≤ b

f (y) = u(a, y)
= ∑_{n=1}^{∞}c_{n} sinh \frac{nπa}{b} sin \frac{nπy}{b}.

▶ Multiply both sides of the equation by sin \frac{mπy}{b} and integrate over [0, b].

∫_{0}^{b} f (y) sin \frac{mπy}{b} dy = ∑_{n=1}^{∞}c_{n} sinh \frac{nπa}{b} ∫_{0}^{b} sin \frac{nπy}{b} sin \frac{mπy}{b} dy
= \frac{b}{2} c_{m} sinh \frac{mπa}{b}

Boundary (a, y) for 0 ≤ y ≤ b

▶ Extend f (y) as an odd, 2b-periodic function. ▶ For 0 ≤ y ≤ b

f (y) = u(a, y)
= ∑_{n=1}^{∞}c_{n} sinh \frac{nπa}{b} sin \frac{nπy}{b}.

▶ Multiply both sides of the equation by sin \frac{mπy}{b} and integrate over [0, b].

∫_{0}^{b} f (y) sin \frac{mπy}{b} dy = ∑_{n=1}^{∞}c_{n} sinh \frac{nπa}{b} ∫_{0}^{b} sin \frac{nπy}{b} sin \frac{mπy}{b} dy
= \frac{b}{2} c_{m} sinh \frac{mπa}{b}

General Solution

The solution to

Δu = 0 for (x, y) ∈ Ω
u(0, y) = 0
u(a, y) = f (y)
u(x, 0) = 0
u(x, b) = 0

is therefore u(x, y) = ∑_{n=1}^{∞}(\frac{2}{b sinh \frac{nπa}{b}} ∫_{0}^{b} f (y) sin \frac{nπy}{b} dy)sinh \frac{nπx}{b} sin \frac{nπy}{b}.

General Dirichlet Problem on a Rectangle

The general boundary value problem:

Δu = 0 for (x, y) ∈ Ω
u(0, y) = g_{1}(y)
u(a, y) = g_{2}(y)
u(x, 0) = f_{1}(x)
u(x, b) = f_{2}(x)

can be decomposed into four simpler boundary value problems.

Decomposition

For (x, y) ∈ Ω,

Δu_{1} = 0
u_{1}(0, y) = g_{1}(y)
u_{1}(a, y) = 0
u_{1}(x, 0) = 0
u_{1}(x, b) = 0

Δu_{2} = 0
u_{2}(0, y) = 0
u_{2}(a, y) = g_{2}(y)
u_{2}(x, 0) = 0
u_{2}(x, b) = 0

Δu_{3} = 0
u_{3}(0, y) = 0
u_{3}(a, y) = 0
u_{3}(x, 0) = f_{1}(x)
u_{3}(x, b) = 0

Δu_{4} = 0
u_{4}(0, y) = 0
u_{4}(a, y) = 0
u_{4}(x, 0) = 0
u_{4}(x, b) = f_{2}(x)

If u_{i}(x, y) solves the ith boundary value problem, then

u(x, y) = u_{1}(x, t) + u_{2}(x, y) + u_{3}(x, y) + u_{4}(x, y)

solves the general Dirichlet boundary value problem on Ω.

Solutions

u_{1}(x, y) = ∑_{n=1}^{∞}(\frac{2}{b sinh \frac{nπa}{b}} ∫_{0}^{b} g_{1}(y) sin \frac{nπy}{b} dy)sinh \frac{nπ(a − x)}{b} sin \frac{nπy}{b}

u_{2}(x, y) = ∑_{n=1}^{∞}(\frac{2}{b sinh \frac{nπa}{b}} ∫_{0}^{b} g_{2}(y) sin \frac{nπy}{b} dy)sinh \frac{nπx}{b} sin \frac{nπy}{b}

u_{3}(x, y) = ∑_{n=1}^{∞}(\frac{2}{a sinh \frac{nπb}{a}} ∫_{0}^{a} f_{1}(x) sin \frac{nπx}{a} dx)sinh \frac{nπ(b − y)}{a} sin \frac{nπx}{a}

u_{4}(x, y) = ∑_{n=1}^{∞}(\frac{2}{a sinh \frac{nπb}{a}} ∫_{0}^{a} f_{2}(x) sin \frac{nπx}{a} dx)sinh \frac{nπy}{a} sin \frac{nπx}{a}

Example

Let Ω = (0, 1) × (0, 1) and solve the Dirichlet boundary value problem for Laplace’s equation on Ω.

Δu = 0 for (x, y) ∈ Ω
u(0, y) = 0
u(1, y) = 1 − y
u(x, 0) = x
u(x, 1) = 0

Solution

Let a = b = 1, then using the notation prescribed earlier the solution can be expressed as

u(x, y) = u_{2}(x, y) + u_{3}(x, y)

where

u_{2}(x, y) = ∑_{n=1}^{∞}(\frac{2}{sinh(nπ)} ∫_{0}^{1}(1 − y) sin(nπy) dy)sinh(nπx) sin(nπy)
= ∑_{n=1}^{∞} \frac{2}{nπ sinh(nπ)} sinh(nπx) sin(nπy)
u_{3}(x, y) = ∑_{n=1}^{∞}(\frac{2}{sinh(nπ)} ∫_{0}^{1} x sin(nπx) dx)sinh(nπ(1 − y)) sin(nπx)
= − ∑_{n=1}^{∞} \frac{2(−1)^{n}}{nπ sinh(nπ)} sinh(nπ(1 − y)) sin(nπx)

Illustration

u(x, y) = ∑_{n=1}^{∞} \frac{2}{nπ sinh(nπ)} sinh(nπx) sin(nπy)

− ∑_{n=1}^{∞} \frac{2(−1)^{n}}{nπ sinh(nπ)} sinh(nπy) sin(nπx)

Laplace’s Equation on a Disk

Consider the Dirichlet boundary value problem:

Δu = 0 for x^{2} + y^{2} < a^{2}
u(x, y) = ϕ(x, y) for x^{2} + y^{2} = a^{2}.

Remark: since the boundary of Ω is not a rectangle, we cannot use separation of variables directly. Instead we must convert to polar coordinates before using separation of variables.

In polar coordinates (r, θ) the Laplacian operator can be expressed as

Δu = u_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^{2}}u_{θθ},

for r > 0.

Laplace’s Equation on a Disk

Consider the Dirichlet boundary value problem:

Δu = 0 for x^{2} + y^{2} < a^{2}
u(x, y) = ϕ(x, y) for x^{2} + y^{2} = a^{2}.

Remark: since the boundary of Ω is not a rectangle, we cannot use separation of variables directly. Instead we must convert to polar coordinates before using separation of variables.

In polar coordinates (r, θ) the Laplacian operator can be expressed as

Δu = u_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^{2}}u_{θθ},

for r > 0.

Dirichlet BVP in Polar Coordinates

u_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^{2}}u_{θθ} = 0 for 0 < r < a and −∞ < θ < ∞
u(a, θ) = f (θ) = ϕ(a cos θ, a sin θ)

Remarks: ▶ The boundary conditions are periodic in θ with period 2π. ▶ The solution u(r, θ) should be 2π-periodic in θ. ▶ The Laplacian is not defined for r = 0, but we wish for the

solution to remain finite as r → 0^{+}.

Separation of Variables in Polar Coordinates

Assume u(r, θ) = R(r)T (θ), then

R′′(r)T (θ) + \frac{1}{r}R′(r)T (θ) + \frac{1}{r^{2}}R(r)T′′(θ) = 0
\frac{R′′(r)T (θ)}{R(r)T (θ)}+ \frac{1}{r} \frac{R′(r)T (θ)}{R(r)T (θ)}+ \frac{1}{r^{2}} \frac{R(r)T′′(θ)}{R(r)T (θ)} = 0
\frac{r^{2}R′′(r)}{R(r)}+ \frac{r R′(r)}{R(r)} = −\frac{T′′(θ)}{T (θ)} = c

where c is a constant.

Implied Ordinary Differential Equations

T′′(θ) + c T (θ) = 0

case c = 0: the only nontrivial 2π-periodic solution is

T_{0}(θ) = A_{0}

case c = λ^{2} > 0: the only nontrivial 2π-periodic solution is

T_{n}(θ) = A_{n} cos (nθ) + B_{n} sin(nθ)

and c = λ_{n}^{2} = n^{2} for n ∈ ℕ.

Euler’s Equation

The implied ordinary differential equation for R(r) is

r^{2}R′′(r) + r R′(r) − c R(r) = 0,

which is known as Euler’s equation. case c = 0:

0 = r^{2}R′′(r) + r R′(r)
R_{0}(r) = C_{0} lnr + D_{0}

case c = n^{2} > 0:

0 = r^{2}R′′(r) + r R′(r) − n^{2} R(r)
R_{n}(r) = C_{n}r^{−n} + D_{n}r^{n}

In order for the product solution to be bounded as r → 0^{+} we must choose C_{0} = C_{1} = ⋯ = 0.

Define the function

u_{n}(r, θ) = R_{n}(r)T_{n}(θ) = r^{n} [A_{n} cos (nθ) + B_{n} sin(nθ)] for n = 0, 1, . . .. By the Principal of Superposition the function

u(r, θ) = A_{0} + ∑_{n=1}^{N}r^{n} [A_{n} cos (nθ) + B_{n} sin(nθ)]

satisfies the Laplacian on the disk for any choice of constants A_{0}, A_{n}, and B_{n}. Fourier series techniques can be used to satisfy the boundary condition.

Boundary Condition

Since u(a, θ) = f (θ) then

f (θ) = A_{0} + ∑_{n=1}^{∞}a^{n} [A_{n} cos (nθ) + B_{n} sin(nθ)]
A_{0} = \frac{1}{2π} ∫_{−π}^{π} f (θ) dθ = \frac{α_{0}}{2}
a^{n}A_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) cos (nθ) dθ = α_{n}
a^{n}B_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) sin(nθ) dθ = β_{n}

and

u(r, θ) = \frac{α_{0}}{2} + ∑_{n=1}^{∞}(\frac{r}{a})^{n} [α_{n} cos (nθ) + β_{n} sin(nθ)].

A_{0} = \frac{1}{2π} ∫_{−π}^{π} f (θ) dθ = \frac{α_{0}}{2}
a^{n}A_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) cos (nθ) dθ = α_{n}
a^{n}B_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) sin(nθ) dθ = β_{n}

and

u(r, θ) = \frac{α_{0}}{2} + ∑_{n=1}^{∞}(\frac{r}{a})^{n} [α_{n} cos (nθ) + β_{n} sin(nθ)].

A_{0} = \frac{1}{2π} ∫_{−π}^{π} f (θ) dθ = \frac{α_{0}}{2}
a^{n}A_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) cos (nθ) dθ = α_{n}
a^{n}B_{n} = \frac{1}{π} ∫_{−π}^{π} f (θ) sin(nθ) dθ = β_{n}

and

u(r, θ) = \frac{α_{0}}{2} + ∑_{n=1}^{∞}(\frac{r}{a})^{n} [α_{n} cos (nθ) + β_{n} sin(nθ)].

Example

Solve Laplace’s equation on the unit disk with the following Dirichlet boundary condition.

Δu = 0 for x^{2} + y^{2} < 1
u(1, θ) = π − θ for −π < θ < π

Solution

The solution takes the form

u(r, θ) = \frac{α_{0}}{2} + ∑_{n=1}^{∞}r^{n} [α_{n} cos (nθ) + β_{n} sin(nθ)]

where

α_{0} = \frac{1}{π} ∫_{−π}^{π}(π − θ) dθ = 2π
α_{n} = \frac{1}{π} ∫_{−π}^{π}(π − θ) cos (nθ) dθ = 0
β_{n} = \frac{1}{π} ∫_{−π}^{π}(π − θ) sin(nθ) dθ = \frac{2(−1)^{n}}{n} .

Solution

The solution takes the form

u(r, θ) = \frac{α_{0}}{2} + ∑_{n=1}^{∞}r^{n} [α_{n} cos (nθ) + β_{n} sin(nθ)]

where

α_{0} = \frac{1}{π} ∫_{−π}^{π}(π − θ) dθ = 2π
α_{n} = \frac{1}{π} ∫_{−π}^{π}(π − θ) cos (nθ) dθ = 0
β_{n} = \frac{1}{π} ∫_{−π}^{π}(π − θ) sin(nθ) dθ = \frac{2(−1)^{n}}{n} .

Illustration

Complex Arithmetic (1 of 3)

∑_{n=0}^{∞}z^{n} = \frac{1}{1 − z} (if |z| < 1)

∑_{n=0}^{∞} \frac{z^{n+1}}{n + 1} = − ln(1 − z)

− ∑_{n=1}^{∞} \frac{z^{n}}{n} = ln(1 − z)

− ∑_{n=1}^{∞} \frac{z^{n}}{n} = ln(1 − z)

− ∑_{n=1}^{∞} \frac{1}{n}(−re^{iθ})^{n} = ln(1 + re^{iθ}) (if |r| < 1)

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n}e^{inθ} = ln(1 + re^{iθ})

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} Im(e^{inθ}) = Im(ln(1 + re^{iθ}))

− ∑_{n=1}^{∞} \frac{z^{n}}{n} = ln(1 − z)

− ∑_{n=1}^{∞} \frac{1}{n}(−re^{iθ})^{n} = ln(1 + re^{iθ}) (if |r| < 1)

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n}e^{inθ} = ln(1 + re^{iθ})

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} Im(e^{inθ}) = Im(ln(1 + re^{iθ}))

− ∑_{n=1}^{∞} \frac{z^{n}}{n} = ln(1 − z)

− ∑_{n=1}^{∞} \frac{1}{n}(−re^{iθ})^{n} = ln(1 + re^{iθ}) (if |r| < 1)

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n}e^{inθ} = ln(1 + re^{iθ})

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} Im(e^{inθ}) = Im(ln(1 + re^{iθ}))

Natural Logarithm of a Complex Number

Suppose z = a + ib ∈ ℂ and w = α + iβ ∈ ℂ such that

e^{w} = z

then w = lnz.

e^{α+iβ} = a + ib
e^{α}e^{iβ} = a + ib
e^{α}(cos β + i sin β) = a + ib

Thus a = e^{α} cos β and b = e^{α} sin β and

a^{2} + b^{2} = e^{2α} ⟹ α = \sqrt{a^{2} + b^{2}}
\frac{b}{a} = tanβ ⟹ β = tan^{−1} \frac{b}{a}.

Thus

lnz = \sqrt{a^{2} + b^{2}} + i tan^{−1} \frac{b}{a}.

Natural Logarithm of a Complex Number

Suppose z = a + ib ∈ ℂ and w = α + iβ ∈ ℂ such that

e^{w} = z

then w = lnz.

e^{α+iβ} = a + ib
e^{α}e^{iβ} = a + ib
e^{α}(cos β + i sin β) = a + ib

Thus a = e^{α} cos β and b = e^{α} sin β and

a^{2} + b^{2} = e^{2α} ⟹ α = \sqrt{a^{2} + b^{2}}
\frac{b}{a} = tanβ ⟹ β = tan^{−1} \frac{b}{a}.

Thus

lnz = \sqrt{a^{2} + b^{2}} + i tan^{−1} \frac{b}{a}.

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} sin(nθ) = Im(ln(1 + r cos θ + ir sin θ))

− ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} sin(nθ) = tan^{−1} (\frac{r sin θ}{1 + r cos θ})

u(r, θ) = π + 2 ∑_{n=1}^{∞} \frac{(−1)^{n}}{n} r^{n} sin(nθ)

= π − 2 tan^{−1} (\frac{r sin θ}{1 + r cos θ})

Laplace’s Equation on a Sector of an Annulus

Find the solution to the following boundary value problem.

Δu = 0 for 1 < x^{2} + y^{2} < 4 with x > 0 and y > 0
u(x, 0) = 0
u(0, y) = 0
u(x, y) = 2xy for x^{2} + y^{2} = 1
u(x, y) = tan^{−1} \frac{y}{x} (\frac{π}{2} − tan^{−1} \frac{y}{x}) for x^{2} + y^{2} = 4

Polar Coordinates

u_{rr} + \frac{1}{r}u_{r} + \frac{1}{r^{2}}u_{θθ} = 0 for 1 < r < 2 with 0 < θ < π/2
u(r, 0) = 0
u(r, π/2) = 0
u(1, θ) = sin(2θ)
u(2, θ) = θ (\frac{π}{2} − θ)

We will again use separation of variables.

Separation of Variables

Assuming u(r, θ) = R(r)T (θ) then

R′′(r)T (θ) + \frac{1}{r}R′(r)T (θ) + \frac{1}{r^{2}}R(r)T′′(θ) = 0
\frac{r^{2}R′′(r)}{R(r)}+ \frac{rR′(r)}{R(r)} = −\frac{T′′(θ)}{T (θ)} = c.

This implies the following boundary value problem for the angular factor of the solution.

T′′(θ) + cT (θ) = 0
T (0) = 0
T (π/2) = 0

Eigenfunctions and Eigenvalues

The only nontrivial solutions to

T′′(θ) + cT (θ) = 0
T (0) = 0
T (π/2) = 0

are

T_{n}(θ) = sin(2nθ)

with c = λ_{n}^{2} = 4n^{2} for n ∈ ℕ. Using the eigenvalues in the radial factor of the product solution yields

r^{2}R′′(r) + r R′(r) − 4n^{2}R(r) = 0
R_{n}(r) = A_{n}r^{−2n} + B_{n}r^{2n}.

Eigenfunctions and Eigenvalues

The only nontrivial solutions to

T′′(θ) + cT (θ) = 0
T (0) = 0
T (π/2) = 0

are

T_{n}(θ) = sin(2nθ)

with c = λ_{n}^{2} = 4n^{2} for n ∈ ℕ. Using the eigenvalues in the radial factor of the product solution yields

r^{2}R′′(r) + r R′(r) − 4n^{2}R(r) = 0
R_{n}(r) = A_{n}r^{−2n} + B_{n}r^{2n}.

Define the product solution

u_{n}(r, θ) = R_{n}(r)T_{n}(θ) = (A_{n}r^{−2n} + B_{n}r^{2n})sin(2nθ).

By the Principal of Superposition a linear combination of product solutions will also solve Laplace’s equation and satisfy the boundary conditions at θ = 0 and θ = π/2.

u(r, θ) = ∑_{n=1}^{N}(A_{n}r^{−2n} + B_{n}r^{2n})sin(2nθ)

Now we must choose the coefficients A_{n} and B_{n} so that the boundary conditions at r = 1 and r = 2 are satisfied. Fourier series will be employed.

Define the product solution

u_{n}(r, θ) = R_{n}(r)T_{n}(θ) = (A_{n}r^{−2n} + B_{n}r^{2n})sin(2nθ).

By the Principal of Superposition a linear combination of product solutions will also solve Laplace’s equation and satisfy the boundary conditions at θ = 0 and θ = π/2.

u(r, θ) = ∑_{n=1}^{N}(A_{n}r^{−2n} + B_{n}r^{2n})sin(2nθ)

Now we must choose the coefficients A_{n} and B_{n} so that the boundary conditions at r = 1 and r = 2 are satisfied. Fourier series will be employed.

Boundary Condition at r = 1

u(1, θ) = ∑_{n=1}^{∞}(A_{n} + B_{n})sin(2nθ)

sin(2θ) = ∑_{n=1}^{∞}(A_{n} + B_{n})sin(2nθ)

The last equation implies the system of equations

A_{1} + B_{1} = 1
A_{n} + B_{n} = 0 for n ≥ 2.

Boundary Condition at r = 2

u(2, θ) = ∑_{n=1}^{∞}(A_{n}2^{−2n} + B_{n}2^{2n})sin(2nθ)

θ ( \frac{π}{2} − θ) = ∑_{n=1}^{∞}(A_{n}2^{−2n} + B_{n}2^{2n})sin(2nθ)

Multiply both sides of the last equation by sin(2mθ) and integrate over [0, π/2].

∫_{0}^{π/2} θ ( \frac{π}{2} − θ) sin(2mθ) dθ = ∑_{n=1}^{∞}(A_{n}2^{−2n} + B_{n}2^{2n})∫_{0}^{π/2} sin(2nθ) sin(2mθ) dθ

\frac{1 − (−1)^{m}}{4m^{3}} = \frac{π/2}{2} (A_{m}2^{−2m} + B_{m}2^{2m})
\frac{1 − (−1)^{m}}{πm^{3}} = A_{m}2^{−2m} + B_{m}2^{2m}

Systems of Equations (1 of 2)

A_{1} + B_{1} = 1
\frac{1}{4}A_{1} + 4B_{1} = \frac{2}{π}

which implies A_{1} = \frac{8(2π − 1)}{15π} and B_{2} = \frac{π − 8}{15π}. For n = (2k + 1) (n odd and greater than 1)

A_{2k+1} + B_{2k+1} = 0
2^{−4k−2}A_{2k+1} + 2^{4k+2}B_{2k+1} = \frac{−2}{π(2k + 1)^{3}}

which implies A_{2k+1} = \frac{2^{4k+3}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} and

B_{2k+1} = −\frac{2^{4k+3}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} .

Systems of Equations (2 of 2)

For n = 2k (n even)

A_{2k} + B_{2k} = 0
2^{−4k}A_{2k} + 2^{4k}B_{2k} = 0

which implies A_{2k} = B_{2k} = 0.

u(r, θ) = (\frac{8(2π − 1)r^{−2}}{15π} + \frac{(π − 8)r^{2}}{15π}) sin(2θ)

+ ∑_{n=1}^{∞} \frac{2^{4k+3}r^{−4k−2}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} sin((4n + 2)θ)

− \frac{2^{4k+3}r^{4k+2}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} sin((4n + 2)θ)

Systems of Equations (2 of 2)

For n = 2k (n even)

A_{2k} + B_{2k} = 0
2^{−4k}A_{2k} + 2^{4k}B_{2k} = 0

which implies A_{2k} = B_{2k} = 0.

u(r, θ) = (\frac{8(2π − 1)r^{−2}}{15π} + \frac{(π − 8)r^{2}}{15π}) sin(2θ)

+ ∑_{n=1}^{∞} \frac{2^{4k+3}r^{−4k−2}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} sin((4n + 2)θ)

− \frac{2^{4k+3}r^{4k+2}}{π(2k + 1)^{3}(2^{4(2k+1)} − 1)} sin((4n + 2)θ)

Illustration Homework

▶ Read Section 6.1–6.4 ▶ Exercises: 1–9