Page 681 - Advanced_Engineering_Mathematics o'neil

18.8 The Neumann Problem 661

and
∂u
(a, y) = g(y) for 0 ≤ y ≤ b.
∂x
For the rectangle, the normal derivative is ∂u/∂x on the vertical sides and ∂u/∂y on the horizontal
sides. As a necessary (but not sufficient) condition for a solution to exist, we assume that
b
g(y)dy = 0.
0
It will be instructive to see how this assumption plays a role in this problem having a solution.
Let u(x, y) = X(x)Y(y), and substitute into Laplace’s equation and also into the boundary
conditions to obtain

X + λX = 0; X (0) = 0
and
Y − λY = 0;Y (0) = Y (b) = 0.

This Sturm-Liouville problem for Y has eigenvalues and eigenfunction
2
n π 2 nπy
λ n =− ,Y n (y) = cos
b 2 b
for n = 0,1,2,··· . Notice that Y(y) is constant for n = 0.
Now the problem for X is
2
n π 2

X − X = 0; X (0) = 0.
b 2
This problem for X has only a boundary condition at x = 0, so we must look at cases in solving
for X.
If n =0, the differential equation for X is just X =0, so X(x)=cx +d. Then X (0)=d =0,

so X(x) is constant in this case.
If n is a positive integer, then the differential equation for X has the general solution
X(x) = ce nπx/b + de −nπx/b .
Then
nπ nπ

X (0) = c − d = 0,
b b
so c = d. This means that X(x) must be have the form
nπ

X(x) = c cosh x .
b
We now have functions
u 0 (x, y) = constant

and, for each positive integer n,
nπ nπ

u n (x, y) = X n (x)Y n (y) = c n cosh x cos y .
b b
We have used the zero boundary conditions on the top, bottom, and left sides of the rectangle. To
satisfy the last boundary condition (on the right side) attempt a superposition
∞

u(x, y) = u n (x, y)
n=0
∞

nπ nπ
= c 0 + c n cosh x cos y .
b b
n=1

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October 14, 2010 15:27 THM/NEIL Page-661 27410_18_ch18_p641-666