# Solve Laplace's equation in spherical coordinates, $\nabla^2 u(r,\theta,\phi)=0$, in the general case.

To find the equilibrium temperature distribution of a sphere I am required to solve the following PDE BVP (Laplace's equation in spherical coordinates).

$$\nabla^2 u(r,\theta,\phi)=0, 0<r<a,0<\theta<2\pi,0<\phi<\pi$$ $$u(a,\theta,\phi)=f(\theta,\phi), 0<\phi<\pi$$

Note that I am required to solve this problem for the general case.

Work thus far

Note that in spherical coordinates $\nabla^2 u(r,\theta,\phi)=u_{rr}+\frac{2}{r}u_r+\frac{1}{r^2}[u_{\phi \phi}+\cot(\phi)u_{\phi}+\csc^2(\phi)u_{\theta \theta}]=0$

We use separation of variables letting $u(r,\theta,\phi)=w(r)v(\theta,\phi)$. Substituting in
\begin{align} 0&=u_{rr}+\frac{2}{r}u_r+\frac{1}{r^2}[u_{\phi \phi}+\cot(\phi)u_{\phi}+\csc^2(\phi)u_{\theta \theta}] \\ 0&=w''v+\frac{2}{r}w'v+\frac{1}{r^2}[wv_{\phi \phi}+\cot(\phi)wv_{\phi}+\csc^2(\phi)wv_{\theta \theta}] \\ \frac{r^2}{w}(-w''-\frac{2}{r}w')&=\frac{1}{v}[v_{\phi \phi}+\cot(\phi)v_{\phi}+\csc^2(\phi)v_{\theta \theta}]=-\lambda \\ \end{align} to obtain the following ODEs \begin{align} 0&=r^2w''+2rw'-\lambda w\\ 0&=v_{\phi \phi}+\cot(\phi)v_{\phi}+\csc^2(\phi)v_{\theta \theta}+\lambda v\\ \end{align} We now do another separation of variables $v(\theta,\phi)=g(\theta)h(\phi)$. We get the following separation $$\frac{h''+\cot(\phi)h'+\lambda h}{\csc^2(\phi)h}=\frac{-g''}{g}=\mu$$ and obtain the following ODEs \begin{align} 0&=g''+\mu g\\ 0&=h'' + \cot(\phi)h'+\lambda h-\mu \csc^2(\phi)h\\ \end{align} Solving the eigenvalue problem $g''+\mu g=0$ with periodic boundary conditions $g(0)=g(2\pi)$ and $g'(0)=g'(2\pi)$, we obtain the eigenvalues $\mu_n=n^2$ and the eigenvectors $g_n(\theta)=\cos(n\theta)$ or $\sin(n\theta)$ for $n=0,1,2,3,...$.

What order do I solve the eigenvalue problems to find the eigenvalues $\mu,\lambda$? How do I transform the ODE $0=h'' + \cot(\phi)h'+\lambda h-\mu \csc^2(\phi)h$ into a easier ODE to solve.

Note that following posts are not relevant as

• There is a formula for the solution of Laplace's equation in 3d-ball. Why not use it and then translate it into spherical coordinates? – Michał Miśkiewicz Mar 8 '17 at 18:53
• @MichałMiśkiewicz Could you provide a reference? So far all the cases I see solve the problem in spherical coordinates and assume some symmetry . – AzJ Mar 8 '17 at 19:07
• There is a solution to Laplaces equation in spherical coordinates here. Why did you want to write the Laplacian in cylindrical coordinates may I ask? – Rumplestillskin Mar 9 '17 at 3:57
• @Rumplestillskin That version of Laplace's equation I used was given as part of the question. The formula for cylindrical coordinates is different (is does not has sines and cosines)en.wikipedia.org/wiki/Laplace's_equation I also checked and the two definitions are equivalent. I will look over your solution but how can I adapt it to the boundary condition $u(a,\theta,\phi)=f(\theta,\phi), 0<\phi<\pi$. Since posting the initial problem I am close to a series solution to the answer if you know how to transform the ODE for $h$ into Lagrange's PDE that would be very helpful. – AzJ Mar 9 '17 at 4:21
• You can look up Gilbarg and Trudinger's Elliptic PDEs of second order, chapter 2. I guess any book should work; search for Green's function and Poisson integral. – Michał Miśkiewicz Mar 9 '17 at 21:22

Assume $$x=r\sin\theta\cos\phi,\;\;\; y=r\sin\theta\sin\phi,\;\;\; z=r\cos\theta, \\ 0 \le r < \infty, \;\;\; 0 \le \theta \le \phi,\;\;\; 0 \le \phi \le 2\pi.$$ Because the spherical coordinate system is an orthogonal coordinate system (meaning that the coordinate curves where only one variable varies are mutually orthogonal where they meet,) the Laplacian is determined by the metric scale factors $$m_r = 1,\;\; m_{\theta}=r,\;\; m_{\phi} = r\sin\theta.$$ The Laplacian is \begin{eqnarray*} & & \nabla^{2}F(r,\phi,\theta) \\ & = & \frac{1}{m_{r}m_{\phi}m_{\theta}}\left[ \frac{\partial}{\partial r}\left(\frac{m_{\phi}m_{\theta}}{m_{r}} \frac{\partial F}{\partial r}\right) +\frac{\partial}{\partial\phi}\left(\frac{m_{r}m_{\theta}}{m_{\phi}} \frac{\partial F}{\partial\phi}\right) +\frac{\partial}{\partial\theta}\left(\frac{m_{r}m_{\phi}}{m_{\theta}} \frac{\partial F}{\partial\theta}\right) \right] \\ % & = & \frac{1}{r^{2}\sin\theta}\left[ % \frac{\partial}{\partial r}\left(r^{2}\sin\theta % \frac{\partial F}{\partial r}\right) % +\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\theta} % \frac{\partial F}{\partial \phi}\right) % +\frac{\partial}{\partial\theta}\left(\sin\theta % \frac{\partial F}{\partial\theta}\right) % \right] % \\ & = & \frac{1}{r^{2}}\frac{\partial}{\partial r} \left(r^{2}\frac{\partial F}{\partial r}\right) + \frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}F}{\partial\phi^{2}} + \frac{1}{r^{2}\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial F}{\partial\theta}\right). \end{eqnarray*} Standard separation of variables equations are obtained by setting $F(r,\theta,\phi)=R(r)\Theta(\theta)\Phi(\phi)$. The separated equations involve a separation parameter $m=0,\pm 1,\pm 2,\cdots$ imposed by periodicity in $\phi$, and a separation parameter $l$ determined by the ODE in $\theta$. The separated equations are: \begin{eqnarray*} -\frac{d^2}{d\phi^2}\Phi & = & m^2\Phi, \\ -\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) +\frac{m^2}{\sin\theta}\Theta & = & l\sin\theta \Theta, \\ -\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+l R & = & 0. \end{eqnarray*} The equation in $\theta$ is the Associated Legendre equation, and this equation determines the possible values of $l$. The radial equation is determined by the requirement that $R$ be bounded near $r=0$. The equation in $\theta$ is transformed to standard Legendre form by setting $x=-\cos\theta$ so that $x$ varies between $-1$ and $1$ as $\theta$ varies between $0$ and $\pi$. Then $X$ defined by $X(-\cos\theta)=\Theta(\theta)$ satisfies the Associated Legendre Equation: $$-\frac{d}{dx}(1-x^2)\frac{dX}{dx}+\frac{m^2}{1-x^2}X=l X.$$ For a given $m$, the values $l$ for which there are bounded solutions $X$ are those given by $l=n(n+1)$ where $n=m,m+1,m+2,\cdots$. The resulting bounded solutions are scalar multiples of $$e^{im\phi}r^{n}\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^{n}.$$