Neumann Problems on Rectangles

MATH 467 Partial Differential Equations

J. Robert Buchanan

Department of Mathematics

Fall 2017

Objectives

In this lesson we will learn:

▶ to solve Laplace’s equation on two-dimensional domains

with Neumann boundary conditions, ▶ to compare the solutions on domains with Dirichlet

boundary conditions to solution on domains with Neumann boundary conditions.

Boundary Value Problem

Consider Laplace’s equation on the rectangle Ω = {(x, y) | 0 < x < a, 0 < y < b} with Neumann boundary conditions:

△u = 0 for (x, y) ∈ Ω
u_{y}(x, 0) = u_{y}(x, b) = 0 for 0 < x < a
u_{x}(0, y) = 0 for 0 < y < b
u_{x}(a, y) = f (y) for 0 < y < b.

Physical Interpretation: the steady-state heat distribution in Ω when the rectangular region is insulated along its bottom, top, and left edges and there is a flow of heat on the right edge.

Product Solutions

Assume u(x, y) = X(x)Y (y) solves Laplace’s equation. Separation of variables induces the following ODEs for X(x) and Y (y).

X′′(x) − σX(x) = 0 with X′(0) = 0
Y′′(y) + σY (y) = 0 with Y′(0) = 0 = Y′(b),

where σ is a constant.

Taking the second BVP, the only nontrivial solutions are:

Y_{n}(y) = cos \frac{nπy}{b}
σ_{n} = \frac{n^{2}π^{2}}{b^{2}}

for n = 0, 1, 2, . . ..

This implies X_{n}(x) = cosh \frac{nπx}{b} for n = 0, 1, 2, . . ..

Series Solution

The product solutions:

u_{n}(x, y) = cosh \frac{nπx}{b} cos \frac{nπy}{b} for n = 0, 1, 2, …

solve Laplace’s equation and satisfy the three homogeneous boundary conditions.

By the Principle of Superposition,

u(x, y) = c_{0} + ∑_{n=1}^{∞}c_{n} cosh \frac{nπx}{b} cos \frac{nπy}{b}.

The coefficients c_{n} can be determined from the remaining nonhomogeneous boundary condition.

Determining the Coefficients

Differentiate the formal series solution and let x = a.

u_{x}(a, y) = ∑_{n=1}^{∞}c_{n} (\frac{nπ}{b}) sinh \frac{nπa}{b} cos \frac{nπy}{b} = f (y)

Remarks:

▶ The coefficient c_{0} was lost during the differentiation. ▶ The infinite series can be regarded as a cosine series for

f (y) if the integral of f over [0, b] vanishes, i.e., if

∫_{0}^{b} f (y) dy = 0.

Further Remarks

▶ If ∫_{0}^{b} f (y) dy ≠ 0 then a solution to the BVP does not exist.

▶ Consider the physics:

▶ If the definite integral vanishes then there is no net flux of heat across the boundary at x = a and hence a steady-state (time independent) heat distribution can evolve.

▶ If the definite integral does not vanish, then there is a net flux of heat in or out of Ω and no time independent temperature distribution can exist.

▶ Even if the definite integral vanishes, the solution can be determined only up to the addition of an arbitrary constant. Thus Laplace’s equation on a rectangle with Neumann boundary conditions on all four edges has no unique solution.

▶ This type of boundary value problem is ill-posed.

Assuming ∫_{0}^{b} f (y) dy = 0

c_{n} = \frac{2}{nπ sinh \frac{nπa}{b}} ∫_{0}^{b} f (y) cos \frac{nπy}{b} dy,

for n ∈ ℕ.

u(x, y) = c_{0} + ∑_{n=1}^{∞}c_{n} cosh \frac{nπx}{b} cos \frac{nπy}{b}

where c_{0} is an arbitrary constant.

Example

Find a solution to the Neumann boundary value problem on the unit square:

△u = 0 for 0 < x < 1 and 0 < y < 1
u_{y}(x, 0) = u_{y}(x, 1) = 0 for 0 < x < 1
u_{x}(0, y) = 0 for 0 < y < 1

u_{x}(1, y) = y − 1/2 for 0 < y < 1.

Solution (1 of 2)

Check: ∫_{0}^{1} (y − \frac{1}{2}) dy = [\frac{y^{2}}{2} − \frac{y}{2}]_{y=0}^{y=1} = 0.

Using the Euler-Fourier coefficient formula:

c_{n} = \frac{2}{nπ sinh(nπ)} ∫_{0}^{1} (y − \frac{1}{2})cos (nπy) dy
= \frac{−2}{n^{2}π^{2} sinh(nπ)} ∫_{0}^{1} sin(nπy) dy
= \frac{2((−1)^{n} − 1)}{n^{3}π^{3} sinh(nπ)}.

Solution (2 of 2)

u(x, y) = c_{0} − \frac{4}{π^{3}} ∑_{n=1}^{∞} \frac{cosh ((2n − 1)πx) cos ((2n − 1)πy)}{(2n − 1)^{3} sinh((2n − 1)π)}

where c_{0} is an arbitrary constant.

Homework

▶ Read Section x.y ▶ Exercises: