Page 122 - 6099

The radial equation
0
2
r R” + rR − λR = 0
√
2
α
is an ODE with the solutions R(r) = r , where α − λ = 0 or α = ± λ = ±πn/β.
The negative exponent is rejected again because we are looking for a solution u(r, θ)
that is continuous in the wedge as well as its boundary: the function r −n/β is infinite
at the origin (which is a boundary point of the wedge). Thus we end up with the
series
∞
X πnθ
u(r, θ) = A n r πn/β sin . (10.56)
β
n=1
Finally, the inhomogeneous boundary condition requires that
∞
X πn πnθ
h(θ) = A n α −1+πn/β sin .
β β
n=1
This is just a Fourier sine series in the interval [0, β], so its coefficients are given by
the formula
Z β
2 πnθ
A n = a 1−πn/β h(θ) sin dθ. (10.57)
πn 0 β
The complete solution is given by (10.56) and (10.57).

Example 10.5 The Annulus
The Dirichlet problem for an annulus (see Figure 10.8) is

2
2
2
2
u xx + u yy = 0 in 0 < a < x + y < b ,
2
2
2
u = g(θ) for x + y = a ,
2
2
2
u = h(θ) for x + y = b .
u = h

u = g

Figure 10.8: The annulus

115