Basis Eigenfunctions

Suppose that the ${\mit\Phi}_r(\theta,\phi)$ are well-behaved solutions of the eigenvalue equation

$$D\,{\mit\Phi}_r +\lambda_r\,{\mit\Phi}_r=0,$$ (12.230)

subject to the boundary conditions

$$\frac{1}{\sin\theta}\,\frac{\partial {\mit\Phi}_r(\theta,\phi_\pm)}{\partial \phi} =0.$$ (12.231)

It immediately follows from Equations (12.234) and (12.235) that

$$\int_{\mit\Omega} ({\mit\Phi}_r^{\,\ast}\,D\,{\mit\Phi}_r - {\mit... ...a_r^{\,\ast})\int_{\mit\Omega} \vert{\mit\Phi}_r\vert^{\,2}\,d{\mit\Omega} = 0.$$ (12.232)

Hence, we deduce that the $\lambda_r$ are real. It follows that we can choose the ${\mit\Phi}_r$ to be real functions. Equations (12.234) and (12.235) also yield

$$\int_{\mit\Omega} {\mit\Phi}_r\,D\,{\mit\Phi}_r\,d{\mit\Omega} = ... ...heta}\,\frac{\partial {\mit\Phi}_r}{\partial\phi}\right)^2\right]d{\mit\Omega},$$ (12.233)

which implies that the $\lambda_r$ are positive. Integration of Equation (12.235), subject to the boundary condition (12.236), gives

$$\lambda_r\int_{\mit\Omega} {\mit\Phi}_r\,d{\mit\Omega} = 0.$$ (12.234)

Because $\lambda_r$ is positive, this implies that

$$\int_{\mit\Omega} {\mit\Phi}_r\,d{\mit\Omega} = 0.$$ (12.235)

Finally, Equations (12.234) and (12.235) yield

$$\int_{\mit\Omega} ({\mit\Phi}_s\,D\,{\mit\Phi}_r-{\mit\Phi}_r\,D\... ...da_s-\lambda_r)\int_{\mit\Omega} {\mit\Phi}_s\,{\mit\Phi}_r\,d{\mit\Omega} = 0.$$ (12.236)

It follows that

$$\int_{\mit\Omega} {\mit\Phi}_s\,{\mit\Phi}_r\,d{\mit\Omega} = 0$$ (12.237)

if $\lambda_s\neq \lambda_r$ . As is well known (Riley 1974), if ${\mit\Phi}_r$ and ${\mit\Phi}_{r'}$ are linearly independent solutions of (12.235) corresponding to the same eigenvalue, $\lambda_r$ , then it is always possible to choose linear combinations of them that satisfy

$$\int_{\mit\Omega} {\mit\Phi}_r\,{\mit\Phi}_{r'}\,d{\mit\Omega} = 0.$$ (12.238)

This argument can be extended to multiple linearly independent solutions corresponding to the same eigenvalue. Hence, we conclude that it is possible to choose the ${\mit\Phi}_r$ such that they satisfy the orthonormality condition

$$\lambda_r\int_{\mit\Omega} {\mit\Phi}_r\,{\mit\Phi}_s\,d{\mit\Ome... ...a_s\int_{\mit\Omega} {\mit\Phi}_r\,{\mit\Phi}_s\,d{\mit\Omega} = \delta_{r\,s}.$$ (12.239)

Let $F(\theta,\phi)$ be a well-behaved function. Suppose that

$$\frac{1}{\sin\theta}\,\frac{\partial F(\theta,\phi_\pm)}{\partial \phi} =0.$$ (12.240)

We can automatically satisfy the previous boundary condition by writing

$$F= \sum_{r=1,\infty} a_r\,{\mit\Phi}_r.$$ (12.241)

(Note that $F$ is undetermined to an arbitrary additive constant which is chosen so as to ensure that $\int_{\mit\Omega} F\,d{\mit\Omega}=0.$ ) Here, $\lambda_1$ is the smallest eigenvalue of Equation (12.235), $\lambda_2$ the next smallest eigenvalue, and so on. It follows from Equation (12.244) that

$$a_r = \lambda_r\int_{\mit\Omega} F\,{\mit\Phi}_r\,d{\mit\Omega}.$$ (12.242)

Suppose that the ${\mit\Psi}_r(\theta,\phi)$ are well-behaved solutions of the eigenvalue equation

$$D\,{\mit\Psi}_r +\mu_r\,{\mit\Psi}_r=0,$$ (12.243)

subject to the boundary conditions

$${\mit\Psi}_r(\theta,\phi_\pm)=0.$$ (12.244)

Using analogous arguments to those employed previously, we can show that the $\mu_r$ are real and positive, and that the ${\mit\Psi}_r$ can be chosen so as to satisfy the orthonormality constraint

$$\mu_r\int_{\mit\Omega} {\mit\Psi}_r\,{\mit\Psi}_s\,d{\mit\Omega}=\mu_s\int_{\mit\Omega} {\mit\Psi}_r\,{\mit\Psi}_s\,d{\mit\Omega} = \delta_{r\,s}.$$ (12.245)

Let $F(\theta,\phi)$ be a well-behaved function. Suppose that

$$F(\theta,\phi_\pm) =0.$$ (12.246)

We can automatically satisfy the previous boundary condition by writing

$$F= \sum_{r=1,\infty} a_r\,{\mit\Psi}_r.$$ (12.247)

It follows from Equation (12.250) that

$$a_r = \mu_r\int_{\mit\Omega} F\,{\mit\Psi}_r\,d{\mit\Omega}.$$ (12.248)