Steady State Temperature Distribution in Unbounded Region (Laplace's Equation in Spherical Coordinates)

I'm trying to find the steady state temperature distribution in the infinite region outside a sphere of unit radius centred on the origin, where the temperature takes the value $u = f(\theta$) on the spherical surface, $((r ,\theta , \phi)$ being spherical polar coordinates), and $u \to 0$ as $r \to \infty$.

My textbook tells me that the solution is

$$u(r, \theta) = \sum\limits_{n = 0}^\infty \dfrac{A_n}{r^{n + 1}}P_n(\cos(\theta)),$$

where $$A_n = \dfrac{(2n + 1)}{2} \int_0^\pi f(\theta)P_n(\cos(\theta))\sin(\theta) \ d \theta,$$

where $P_n$ is the $n^{th}$ Legendre Polynomial, but I don't understand how it came to this solution.

I would greatly appreciate it if people could please take the time to explain how this problem is solved, so that I understand how it is done for similar types of problems.

• Separate variables, construct separable solutions, project the boundary condition out onto the separable solutions. Everything except actually constructing the separable solutions is the same as it would be in easier problems (e.g. Laplace equation on a rectangle). – Ian Aug 23 '18 at 20:33
• @Ian Hi Ian. I've been conducting research on this problem, and I've realised I need to use Laplace's equation with spherical coordinates. But doesn't this problem only depend on $\theta$? So would I start by writing Laplace's equation in terms of just $\theta$? And how would I structure the boundary conditions? – The Pointer Aug 23 '18 at 20:35
• The BC depends only on $\theta$ but the solution depends also on $r$. It does not depend on $\phi$. There's a BC at $r=r_0$ (the radius of the sphere) and at $r \to \infty$. You may also need to enforce a periodicity requirement in $\theta$, not sure if that will work itself out or not. – Ian Aug 23 '18 at 20:37
• @Ian It should have been $u = f(\theta)$ in the problem statement. – The Pointer Aug 23 '18 at 20:40
• @Ian But how would I first write out Laplace's equation here? Would I just write it in terms of $r$ and $\theta$? – The Pointer Aug 23 '18 at 20:41

The proposed solution is a classical expansion for the solution of Laplace equation with axial boundary conditions. In this symmetry, as noted in the comments, the eqation to be solved can be written as \begin{equation} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{ \sin(\theta)} \dfrac{\partial}{\partial{\theta}}\left( \sin(\theta)\dfrac{\partial{u}}{\partial{\theta}} \right) = 0 \end{equation} Here is a brief description of the method. Solutions obtained by separation of the variables: $u(r,\theta)=R(r)\Theta(\theta)$ must verify \begin{equation} \frac{1}{R(r)}\dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{R(r)}}{\partial{r}} \right)=-\frac{1}{\Theta(\theta)}\dfrac{1}{ \sin(\theta)} \dfrac{\partial}{\partial{\theta}}\left( \sin(\theta)\dfrac{\partial{\Theta(\theta)}}{\partial{\theta}} \right)=A \end{equation} where $A$ is a constant. The $\Theta$-ODE solutions are Legendre functions of $\cos\theta$. The non-divergence condition for $0\le \theta\le 2\pi$ implies \begin{equation} \frac{1}{2}\sqrt{1+4A}-\frac{1}{2}=\ell \end{equation} where $\ell$ is a positive integer, thus $\Theta(\theta)=b_\ell P_\ell (\cos\theta)$ (where $b_\ell$ is a constant). Then, the solution for $R$ which does not diverge for $r\to\infty$ is $R(r)=b_\ell r^{-\frac{1}{2}\sqrt{1+4A}-\frac{1}{2}}=b_\ell r^{-\ell-1}$. A general solution for the Laplace equation can thus be written as a superposition \begin{equation} u(r,\theta)=\sum_{\ell=0}a_lr^{-\ell-1}P_\ell \left( \cos\theta \right) \end{equation} Finally the constants $a_\ell$ are determined by the boundary condition $u(1,\theta)=f(\theta)$. Using the orthogonality properties of the Legendre polynomials, we obtain the given result.