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660 CHAPTER 18 The Potential Equation

Use the lemma as follows. If g(x, y) = 1 and f = u (a harmonic function on D), then the
double integral in the lemma is zero because its integrand vanishes, and the line integral is just
the line integral of the normal derivative of u over the boundary C of D. For a Neumann problem,
this normal derivative is a given function g, so the lemma tells us that

∂u
ds = gds = 0.
C ∂n C
This means that vanishing of the integral of the given normal derivative over the boundary of the
region is a necessary condition for the Neumann problem to have a solution. Put another way, if
the integral of g over C is not zero, this Neumann problem has no solution.
This result can be extended to the case that D is not a bounded region and C is not a closed
curve. This occurs, for example, with the right quarter plane given by x ≥ 0, y ≥ 0. Here the
region is unbounded, and its boundary curve consists of the nonnegative x- and y-axes.

EXAMPLE 18.7
We will solve a Neumann problem for a square:
2
∇ u = 0for0 < x <,0 < y < 1,

∂u 0 on the left, top and lower sides,
=
∂n y 2 on the right side of the square.
This means that
∂u ∂u ∂u
(x,0) = (x,1) = (0, y) = 0
∂n ∂n ∂n
while
∂u
2
(1, y) = y for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
∂n
First take the line integral of ∂u/∂n about the boundary of D, which consists of four straight
line segments:
2
∂u 2 1
ds = y dy = = 0.
C ∂n o 3
Therefore this Neumann problem has no solution.

Dirichlet problems may also fail to have solutions, depending on the region and the given
function at values on the boundary. However, for “simple” regions such as rectangles and disks,
and “reasonable” data functions on the boundary, Dirichlet problems have solutions. The last
example shows that, even for a simple region (a square) and reasonably well-behaved normal
derivative on the boundary, a Neumann problem may be ill posed (no solution).
We will analyze Neumann problems for rectangles and disks.

18.8.1 A Neumann Problem for a Rectangle
We will consider the Neumann problem
2
∇ u(x, y) = 0for0 < x < a,0 < y < b,
∂u ∂u
(x,0) = (x,b) = 0for 0 ≤ x ≤ a,
∂y ∂y
∂u
(0, y) = 0for0 ≤ y ≤ b,
∂x

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October 14, 2010 15:27 THM/NEIL Page-660 27410_18_ch18_p641-666