(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

[tex]

\frac{d^2\phi}{{d\theta}^2} = -\lambda^2\phi

[/tex]

[tex]

\phi = A\cos{\lambda\theta} + B\sin{\lambda\theta}

[/tex]

[tex]

r^2\frac{d^2G}{{dr}^2} + r\frac{dG}{dr} - n^2G = 0

[/tex]

[tex]

G = c_{1n}r^{-n} + c_{2n}r^n\text{ for } n\ne0

[/tex]

[tex]

G = c_{10} + c_{20}\ln(r)\text{ for } n=0

[/tex]

3. 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 [itex]\lambda = n[/itex] for all n greater than or equal to zero. We can write,

[tex]

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}

[/tex]

[tex]

\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}

[/tex]

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 [tex]n=0[/tex]. Am I missing something?

Thanks for your help!

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# Laplace's equation on an annulus with Nuemann BCs

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