📂Partial Differential Equations

General Solution to the Laplace Equation in Spherical Coordinates

Theorem

In spherical coordinates, the Laplace equation is as follows:

$$\nabla ^2 f = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial f}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial^2 \phi}=0$$

Explanation

Assuming that $f$ can be separated into variables as $f(r,\theta,\phi)=R(r)\Theta (\theta)\Phi (\phi)$, the general solution for the radial component can be derived by solving the Euler differential equation as follows:

$$R(r)=\sum \limits_{l=0}^{\infty}R_{l}(r)=\sum \limits_{l=0}^{\infty}\left( A_{l}r^{l}+\frac{ B_{l}}{r^{l+1}} \right)$$

The solutions for the polar angle $\theta$ and the azimuthal angle $\phi$ are specifically called spherical harmonics, as follows:

$$Y_{l}^{m}(\theta,\phi)=e^{im\phi}P_{l}^{m}(\cos \theta)$$

Here, $P_{l}^{m}$ is the associated Legendre polynomial. Where $l$ is a non-negative integer, and $m$ is an integer that satisfies $-l\le m \le l$. Therefore, the general solution for $\theta$ and $\phi$ components is as follows:

$$\Theta (\theta)\Phi (\phi)=\sum\limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}e^{im\phi}P_{l}^{m}(\cos\theta)$$

Combining the two results above, the general solution of the Laplace equation in spherical coordinates can be obtained as follows:

$$\begin{align*} f(r,\theta,\phi)&=R(r)\Theta (\theta)\Phi (\phi) \\ &=\sum \limits_{l=0}^{\infty}\sum\limits_{m=-l}^{l}\left( A_{l}r^{l}+\frac{ B_{l}}{r^{l+1}} \right)e^{im\phi}P_{l}^{m}(\cos\theta) \end{align*}$$

If the radial component has symmetry, the solution to the Laplace equation is a spherical harmonic. In the case of symmetry regarding the azimuthal angle $\phi$, the solution is as follows:

$$f(r,\theta) = \sum \limits_{l=0} ^\infty \left( A_{l} r^l + \dfrac{B_{l}}{r^{l+1} } \right) P_{l}(\cos \theta)$$

Here, $P_{l}(x)$ is the Legendre polynomial.