# Solution to Laplace’s Equation

This post deals with the familiar (to the physics student) Laplace’s equation. I am solving this equation in the context of physics, instead of a pure mathematical perspective. This problem is considered most extensively in the context of electrostatics. This equation is usually considered in the spherical polar coordinate system. A lot of finer details are also considered in a mathematical physics course where the topic of spherical harmonics is discussed. Assuming spherical-polar coordinates, Laplace’s equation is

$\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\bigg\{r^{2}\frac{\partial \psi}{\partial r}\bigg\}+\frac{1}{r^{2}\sin{\theta}}\frac{\partial}{\partial \theta}\bigg\{\sin{\theta}\frac{\partial \psi}{\partial \theta}\bigg\}+\frac{1}{r^{2}\sin{\theta}}\frac{\partial^{2}\psi}{\partial \phi^{2}}=0, (1)$

Suppose that the function $\psi\rightarrow \psi(r,\theta,\phi)$. Furthermore, by the method of separation of variables we suppose that the solution is a product of eigenfunctions of the form $R(r)Y(\theta,\phi)$. Hence, Laplace’s equation becomes

$\displaystyle \frac{Y}{r^{2}}\frac{d}{dr}\bigg\{r^{2}\frac{d^{2}R}{dr^{2}}\bigg\}+\frac{R}{r^{2}\sin{\theta}}\frac{\partial}{\partial \theta}\bigg\{\sin{\theta}\frac{\partial Y}{\partial \theta}\bigg\}+\frac{R}{r^{2}\sin^{\theta}}\bigg\{\frac{\partial^{2}Y}{\partial \phi^{2}}\bigg\}=0. (2)$

Furthermore, we can separate further the term $Y(\theta, \phi)$ into $\Theta(\theta)\Phi(\phi)$. Rewriting (2) and multiplying by $r^{2}\sin^{2}{\theta}R^{-1}Y^{-1}$, we get

$\displaystyle \frac{1}{R}\frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}+\frac{\sin{\theta}}{\Theta}\frac{d}{d\theta}\bigg\{\sin{\theta}\frac{d\Theta}{d\theta}\bigg\}+\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=0.(3)$

Bringing the radial and angular component to the other side of the equation and setting the azimuthal component equal to a separation constant $-m^{2}$, yielding

$\displaystyle -\frac{1}{R}\frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}-\frac{\sin{\theta}}{\Theta}\frac{d}{d\theta}\bigg\{\sin{\theta}\frac{d\Theta}{d\theta}\bigg\}=\frac{1}{\Phi}\frac{d^{2}\Phi}{d\phi^{2}}=-m^{2}.$

Solving the right-hand side of the equation we get

$\displaystyle \Phi(\phi)=A\exp({+im\phi}). (4)$

Now, we set the azimuthal component equal to $m^{2}$ and carry out the derivative in the angular component to get the Associated Legendre equation:

$\displaystyle \frac{d}{d\mu}\bigg\{(1-\mu^{2})\frac{d\Theta}{d\mu}\bigg\}+\bigg\{l(l+1)-\frac{m^{2}}{1-\mu^{2}}\bigg\}\Theta=0, (5)$

where I have let $\mu=\cos{\theta}$. The solutions to this equation are known as the associated Legendre functions. However, instead of solving this difficult equation by brute force methods (i.e. power series method), we consider the case for which $m^{2}=0$. In this case, Eq.(5) simplifies to Legendre’s differential equation discussed previously. Instead of quoting the explicit form of the Legendre polynomials, we equivalently state the Rodrigues formula for these polynomials

$\displaystyle P_{l}(\mu)=\frac{1}{2^{l}l!}\frac{d^{l}}{d\mu^{l}}(\mu^{2}-1)^{l}. (6)$

However, the solutions to Eq.(5) are the associated Legendre functions, not polynomials. Therefore, we use the following to determine the associated Legendre functions from the Legendre polynomials:

$\displaystyle {\Theta_{l}}^{m}(\mu) = (1-\mu^{2})^{\frac{|m|}{2}}\frac{d^{|m|}}{d\mu^{|m|}}P_{l}(\mu). (7)$

For real solutions $|m|\leq l$ and for complex solutions $-l \leq m \leq l$, thus we can write the solutions as series whose indices run from 0 to l for the real-values and from -l to l for the complex-valued solutions.

Now we turn to the radial part which upon substitution of $-l(l+1)$ in place of the angular component, we get

$\displaystyle \frac{d}{dr}\bigg\{r^{2}\frac{dR}{dr}\bigg\}-l(l+1)R=0,$

and if we evaluate the derivatives in the first term we get an Euler equation of the form

$\displaystyle r^{2}\frac{d^{2}R}{dr^{2}}+2r\frac{dR}{dr}-l(l+1)R=0. (8)$

Let us assume that the solution can be represented as

$\displaystyle R(r)=\sum_{j=0}^{\infty}a_{j}r^{j}.$

Taking the necessary derivatives and substituting into Eq.(8) gives

$\displaystyle r^{2}\sum_{j=0}^{\infty}j(j-1)a_{j}r^{j-2}+\sum_{j=0}^{\infty}2rja_{j}r^{j-1}-l(l+1)\sum_{j=0}^{\infty}a_{j}r^{j}=0. (9)$

Simplifying the powers of r we find that

$\displaystyle \sum_{j=0}^{\infty}[j(j-1)+2j-l(l+1)]a_{j}r^{j}=0.$

Now, since $\displaystyle \sum_{j=0}^{\infty}a_{j}r^{j}\neq 0$, this means that

$\displaystyle j(j-1)+2j-l(l+1)=0. (10)$

We can simplify this further to get

$\displaystyle j^{2}-l^{2}+j-l=0.$

Factoring out $(j-l)$, we arrive at

$\displaystyle (j-l)((j+l)+1)=0 \implies j=l ; j=-l-1.$

Since this must be true for all values of l, the solution therefore becomes

$\displaystyle R(r)= \sum_{l=0}^{\infty}Br^{l}+Cr^{-l-1}. (11)$

Thus, we can write the solution to Laplace’s equation as

$\displaystyle \psi(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=0}^{\infty}[Br^{l}+Cr^{-l-1}]\Theta_{l}^{m}(\mu)A\exp({im\phi}), (12)$

for real solutions, and

$\displaystyle \psi(r,\theta,\phi)=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}[Br^{l}+Cr^{-l-1}]\Theta_{l}^{m}(\mu)A\exp({im\phi}), (13)$

for complex solutions.

## 3 thoughts on “Solution to Laplace’s Equation”

1. Jason White says:

How did you factor j + l out of j^2 + l^2 + j + l = 0?

(j + l)(j + l + 1) = j^2 + l^2 + j + l + 2jl

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1. Hello, thanks for pointing that out. It looks like I made a few sign mistakes that carried through to the values of j. I have updated my solution with my correction.

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2. Jason White says:

for some reason my l’s look like larger 1’s…

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