# [SOLVED]General soln to Laplace on an annulus

#### dwsmith

##### Well-known member
I am up to here but now I am stuck. The $u(b,\theta) = 0$ is giving me trouble with Fourier coefficients.
$$\frac{a_0}{2} + \frac{b_0\ln b}{2} = \frac{1}{\pi}\int_{-\pi}^{\pi}0d\theta = A_0$$
$$c_nb^n + d_nb^{-n} = \frac{1}{\pi}\int_{-\pi}^{\pi}0d\theta = A_n$$
$$e_nb^n + f_nb^{-n} = \frac{1}{\pi}\int_{-\pi}^{\pi}0d\theta = B_n$$
Consider Laplace's equation on a circular annulus of inner radius $a$ and outer radius $b$ subject to the boundary conditions
$$u(a,\theta) = f(\theta)\quad u(b,\theta) = 0.$$
Show that the solution is given by
$$A_0\frac{\ln r - \ln b}{\ln a - \ln b} + \sum_{n = 1}^{\infty}\left[A_n\cos n\theta + B_n\sin n\theta\right]\left(\frac{a}{r}\right)^n\frac{b^{2n} - r^{2n}}{b^{2n} - a^{2n}}$$
where $A_n$ and $B_n$ are the Fourier series coefficients of $f(\theta)$.
$$\nabla^2u = u_{rr} + \frac{1}{r}u_r + \frac{1}{r^2}u_{\theta\theta}$$
Suppose $u$ is of the form $u(r,\theta) = R(r)\Theta(\theta)$.
By the method of separation of variables,
$$\frac{r^2R'' + rR'}{R} = - \frac{\Theta''}{\Theta} = \lambda^2.$$
That is, $\Theta(\theta) = A\cos\lambda_n\theta + B\sin\lambda_n\theta$.
Let's look at the case when $\lambda = 0$.
$$\Theta'' = 0\iff \Theta(\theta) = \alpha\theta + \beta$$
Since we are dealing with an annular region, $\Theta(0) = \theta(2\pi)$, i.e. $\Theta$ must be periodic with period $2\pi$.
$$\Theta(0) = \beta = 2\alpha\pi + \beta = \Theta(2\pi)\Rightarrow 2\alpha\pi = 0\iff \alpha = 0$$
Therefore, $\Theta(\theta) = b$ where $b$ is constant when $\lambda = 0$.
For $\lambda\neq 0$, the boundary conditions must also be periodic with period $2\pi$.
$$\text{B.C.} = \begin{cases} \Theta(-\pi) = \Theta(\pi)\\ \Theta'(-\pi) = \Theta'(\pi) \end{cases}$$
Using the first boundary condition, we have
$$\Theta(-\pi) = A\cos\lambda\pi - B\sin\lambda\pi = A\cos\lambda\pi + B\sin\lambda\pi = \Theta(\pi)$$
$$2B\sin\lambda\pi = 0$$
Therefore, we could have $B = 0$ or $\lambda = n$ where $n\in\mathbb{Z}$.
$$\Theta'(-\pi) = A\lambda\sin\lambda\pi + B\lambda\cos\lambda\pi = -A\lambda\sin\lambda\pi + B\lambda\cos\lambda\pi = \Theta'(\pi)$$
$$2A\lambda\sin\lambda\pi = 0$$
This time we could have $A = 0$ or $\lambda = n$ where $n\in\mathbb{Z}$.
Since we aren't looking for the trivial solution, $\lambda_n = n$.
$$\Theta(\theta) = \begin{cases} b, & \text{if }\lambda = 0\\ A_n\cos n\theta + B_n\sin n\theta, & \text{if }\lambda\neq 0 \end{cases}$$
Let's look at the $\lambda = 0$ case first.
$$r^2R'' + rR' = 0\iff r^n[n(n - 1) + n] = 0\Rightarrow n^2 = 0$$
Since we have 0 with multiplicity 2, the solution is
$$R(r) = C + D\ln r.$$
The second ODE is of the Cauchy-Euler type.
$$r^m(m^2 - \lambda_n^2) = 0\iff m = \pm\lambda_n = \pm n$$
That is, $R(r) = Er^{|n|}$.
$$R(r) = \begin{cases} C, & \text{if }\lambda = 0\\ \ln r, & \text{if }\lambda = 0\\ r^n, & \text{if }\lambda \neq 0\\ \frac{1}{r^n}, & \text{if }\lambda \neq 0\\ \end{cases}$$
Therefore, the general solution is
$$u(r,\theta) = \frac{a_0}{2} + \frac{b_0\ln r}{2} + \sum_{n = 1}^{\infty}\left[(c_nr^n + d_nr^{-n})\cos n\theta + (e_nr^n + f_nr^{-n})\sin n\theta\right].$$
The boundary conditions for $u$ are
$$\text{B.C.} = \begin{cases} u(a,\theta) = f(\theta)\\ u(b,\theta) = 0 \end{cases}$$

Last edited: