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[SOLVED] General soln to Laplace on an annulus

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
\text{B.C.} = \begin{cases}
u(a,\theta) = f(\theta)\\
u(b,\theta) = 0
\end{cases}
$$
 
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