Page 683 - Advanced_Engineering_Mathematics o'neil

18.8 The Neumann Problem 663

This is a Fourier expansion of f (θ) on [−π,π]. Notice that the constant term in this expansion
is zero. But this constant term is exactly
1 π
f (θ)dθ,
π −π
so we would have a contradiction if this integral did not vanish, as we have assumed. For the
other coefficients, we need
1 π
na n R n−1 = f (ξ)cos(nξ)dξ
π −π
and
1 π
b n = f (ξ)sin(nξ)dξ
π −π
for n = 1,2,···. Then
1 π
a n = f (ξ)cos(nξ)dξ
nπ R n−1 −π
and
1 π
b n = f (ξ)sin(nξ)dξ.
nπ R n−1 −π
Upon inserting these coefficients, the solution is
∞
r
1 R 1 n π
u(r,θ) = a 0 + [cos(nξ)cos(nθ) + sin(nξ)sin(nθ)] f (ξ)dξ.
2 π n R 0π
n=1
We can also write this solution as
∞ π
r
1 R 1 n
u(r,θ) = a 0 + cos(n(ξ − θ)) f (ξ)dξ.
2 π n R
n=1 −π
The term a 0 /2 is an arbitrary constant, written as a 0 /2 simply because of the context of a Fourier
series.

EXAMPLE 18.8
Solve the Neumann problem for the unit disk about the origin:
2
2
2
∇ u(x, y) = 0for x + y < 1
and
∂u 2 2 2
(x, y) = xy for x + y = 1.
∂n
Switch to polar coordinates, letting U(r,θ) = u(r cos(θ),r sin(θ)). Now the problem is
2
∇ U(r,θ) = 0for 0 ≤r < 1,−π ≤ θ ≤ π
and
∂U
2
(1,θ) = cos(θ)sin (θ).
∂r
First, compute
π
2
cos(θ)sin (θ)dθ = 0,
−π

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October 14, 2010 15:27 THM/NEIL Page-663 27410_18_ch18_p641-666