Laplace's equation on an annulus with Nuemann BCs

[nathan12343]

Homework Statement

Solve Laplace's equation inside a circular annulus $(a<r<b)$ subject to the boundary conditions $\frac{\partial{u}}{\partial{r}}(a,\theta) = f(\theta)\text{, }\frac{\partial{u}}{\partial{r}}(b,\theta) = g(\theta)$

Homework Equations

Assume solutions of the form $u(r,\theta) = G(r)\phi(\theta)$. This leads to an equation for phi and G, both of which can be solved by substitution:

$$\frac{d^2\phi}{{d\theta}^2} = -\lambda^2\phi$$
$$\phi = A\cos{\lambda\theta} + B\sin{\lambda\theta}$$

$$r^2\frac{d^2G}{{dr}^2} + r\frac{dG}{dr} - n^2G = 0$$
$$G = c_{1n}r^{-n} + c_{2n}r^n\text{ for } n\ne0$$
$$G = c_{10} + c_{20}\ln(r)\text{ for } n=0$$

The Attempt at a Solution

Periodic boundary conditions require that u and its derivative with respect to theta be continuous between -pi and pi. Then means $\lambda = n$ for all n greater than or equal to zero. We can write,

$$u(r,\theta) = &\, A_{01} + A_{02}\ln(r) + \sum_{n=1}^{\infty}(A_{n1}r^n + A_{n2}r^{-n})\cos{n\theta} + (B_{n1}r^n + B_{n2}r^{-n})\sin{n\theta}$$
$$\frac{\partial{u(r,\theta)}}{\partial{r}} = &\, A_{02}r^{-1} + \sum_{n=1}^{\infty}(nA_{n1}r^{n-1} + -nA_{n2}r^{-n-1})\cos{n\theta} + (nB_{n1}r^{n-1} - nB_{n2}r^{-n-1})\sin{n\theta}$$

I know that I can set everything in parentheses in front of the cosines and sines above equal to some constant when I set r=a,b to enforce the boundary conditions on the edge of the annulus. This leaves me with two equations in two unknowns for all the A's and B's with n greater than 1. My problem is how to set [itex]A_{02}[/tex] such that it will work at both boundaries. It seems like I really need two constants to match to the boundary for when $$n=0$$. Am I missing something?<br /> <br /> Thanks for your help![/itex]

 

[nathan12343]

Please disregard this - I meant to post in Calculus & Beyond.