# Laplace's Equation

In this article we discuss Laplaces equation in cylindrical and spherical coordinates. The general Laplace Equation can be written as \begin{align} \nabla ^2 u(x)=0 \end{align} without specifying the coordinate system or on how many variables $u(x)$ with $x \in \mathbb{R}^n$ depends.

## in cylindrical coordinates

The nabla operator in cylindrical coordinates reads \begin{align} \mathcal{L}=\nabla^2=\frac{\partial ^2 }{\partial r^2} + \frac{1}{r} \frac{\partial }{\partial r} + \frac{1}{r^2}\frac{\partial ^2 }{\partial \theta ^2} + \frac{\partial ^2 u}{\partial z ^2}=0 \end{align} To solve Lapalce's equation, we make the ansatz $u=R(r) \cdot \Theta (\theta) \cdot Z(z)$ \begin{align} R'' \Theta Z + \frac{1}{r} R' \Theta Z + \frac{1}{r^2} R \Theta '' Z+ R \theta Z''=0 \end{align} divide by $u$ and separate variables \begin{align} \frac{R''}{R} + \frac{1}{r} \frac{R'}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta}+ \frac{Z''}{Z}=0\\ \frac{R''}{R} + \frac{1}{r} \frac{R'}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta}=- \frac{Z''}{Z}=-\lambda^2 \end{align} The first equation we derive is a

harmonic differential equation \begin{align} Z''-\lambda^2 Z=0 \end{align} with the solution \begin{align} Z(z)=A \, \cos \lambda z + B \, \sin \lambda z \end{align}

we plug in $\lambda ^2$ and mulitply by $r^2$ \begin{align} \frac{R''}{R} + \frac{1}{r} \frac{R'}{R} + \frac{1}{r^2} \frac{\Theta''}{\Theta}+\lambda^2=0 \\ r^2 \frac{R''}{R} + r \frac{R'}{R} + \frac{\Theta''}{\Theta}+\lambda^2 r^2=0 \\ \end{align} now we are able to separate the remaining variables \begin{align} r^2 \frac{R''}{R} + r \frac{R'}{R} + \lambda^2 r^2=-\frac{\Theta''}{\Theta}=\mu^2 \\ \end{align} to get another

harmonic differential equation \begin{align} \Theta '' +\mu^2 \Theta=0 \\ \end{align} with the solution \begin{align} \Theta(\theta)=C \, \cos m \theta + D \, \sin m \theta \end{align}

and the ODE \begin{align} r^2 \frac{R''}{R} + r \frac{R'}{R} + \lambda^2 r^2-\mu^2 =0\\ r^2 R'' + r R' + (\lambda^2 r^2-\mu^2)R =0\\ \end{align} that can be transformed into Bessel's Differential Equation, like shown in this article by the substitution $x=\lambda r$.

Bessel's differential equation \begin{align} x^2 R'' + r R' + (x^2-\mu^2)R =0\\ \end{align} with the solution \begin{align} R(r)=E \, J_{+\mu}(\lambda r) + F \, J_{-\mu}(\lambda r) \end{align} a linear combination of the Bessel polynomials.

## in spherical coordinates

The nabla operater in spherical coordinates reads \begin{align} \mathcal{L}=\nabla^2 = \underbrace{{1 \over r^2}{\partial \over \partial r}\left(r^2 {\partial \over \partial r}\right) }_{r-term} + \underbrace{{1 \over r^2\sin\theta}{\partial \over \partial \theta}\left(\sin\theta {\partial \over \partial \theta}\right)}_{\theta-term} + \underbrace{{1 \over r^2\sin^2\theta}{\partial^2 \over \partial \varphi^2}}_{\varphi-term} = 0 \end{align} To remember the form of the operater notice that there is an overall factor of $1/r^2$ and that the operater contains terms, each having a derivative with respect to one of the variables $r, \theta$ or $\varphi$ only. To separate variables of Laplace's equation in spherical coordinates, we make the ansatz $u(r,\theta,\varphi)=R(r)\Theta(\theta)\Phi(\varphi)$ \begin{align} \nabla^2 u = \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial R(r)}{\partial r}\right) \Theta(\theta)\Phi(\varphi) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right) R(r)\Phi(\varphi) + \frac{1}{r^2\sin^2\theta} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} R(r)\Theta(\theta) = 0 \end{align} Now we multiply by $r^2$ and divide by $R(r)\Theta(\theta)\Phi(\varphi)$ \begin{align} \frac{1}{R(r)}\frac{\partial}{\partial r} \left(r^2 \frac{\partial R(r)}{\partial r}\right) + \frac{1}{\Theta(\theta) \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right) + \frac{1}{\Phi(\varphi) \sin^2\theta} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = 0 \end{align} We separate the the $r$ dependent part and set it equal to $\lambda$ \begin{align} \frac{1}{R(r)}\frac{\partial}{\partial r} \left(r^2 \frac{\partial R(r)}{\partial r}\right)= - \frac{1}{\Theta(\theta) \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right) - \frac{1}{\Phi(\varphi) \sin^2\theta} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = \lambda \end{align} After differentiation and rearranging equation \begin{align} \frac{1}{R(r)}\frac{\partial}{\partial r} \left(r^2 \frac{\partial R(r)}{\partial r}\right)= \lambda \end{align}

we arrive at a Euler-Cauchy differential equation \begin{align} r^2 R'' + 2r R' - \lambda R =0 \end{align} with the solution \begin{align} R(r)=A \, r^{\lambda_1} + B \, r^{\lambda_2} \end{align}

If we rename $\lambda=\ell(\ell+1)$ we can write this solution in the form \begin{align} R(r)=A \, r^{\ell} + \frac{B}{r^{(\ell + 1)}} \end{align} Now we have to seperate the equation \begin{align} \frac{1}{\Theta(\theta) \sin\theta} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right) + \frac{1}{\Phi(\varphi) \sin^2\theta} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = -\lambda \end{align} we multiply by $\sin^2 \theta$ and rearrange \begin{align} \frac{\sin\theta}{\Theta(\theta)} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right)+\lambda \sin^2 \theta + \frac{1}{\Phi(\varphi)} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = 0 \end{align} Now we have separated the $\theta-term$, let's set it equal to $m^2$ \begin{align} \frac{\sin\theta}{\Theta(\theta)} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right)+\lambda \sin^2 \theta = \frac{1}{\Phi(\varphi)} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = m^2 \end{align} this gives a

harmonic differential equation \begin{align} \frac{\partial^2 \Phi(\varphi)}{\partial \varphi^2} = m^2 \Phi(\varphi) \end{align} with the solution \begin{align} \Phi(\varphi)=C \, \cos m\varphi + D \, \sin m \varphi \end{align}

the remaining eqauation \begin{align} \frac{\sin\theta}{\Theta(\theta)} \frac{\partial}{\partial \theta} \left(\sin\theta \frac{\partial \Theta(\theta)}{\partial \theta}\right)+\lambda \sin^2 \theta = m^2 \end{align} can be recognized as the associated Legendre's differential equation after the transformation $\xi = \cos \theta$ \begin{align} \frac{d \Theta}{d \theta}&=\frac{d \Theta}{d \xi}\frac{d \xi}{d \theta}=\frac{d \Theta}{d \xi} (- \sin \theta)=\frac{d \Theta}{d \xi} (-\sqrt{1-\xi^2})\\ \frac{d^2 \Theta}{d \theta^2}&=\frac{d^2 \Theta}{d \xi^2} \left( \frac{d \xi}{d \theta} \right)^2+\frac{d \Theta}{d \xi}\frac{d^2 \xi}{d \theta^2}=\frac{d^2 \Theta}{d \xi^2} \left( - \sin \theta \right)^2+\frac{d \Theta}{d \xi} (- \cos \theta) \end{align} differenting and dividing by $\sin^2 \theta$ yields \begin{align} \frac{1}{\Theta(\theta)} \frac{\cos \theta}{\sin\theta} \frac{\partial \Theta(\theta)}{\partial \theta}+ \frac{1}{\Theta(\theta)} \frac{\partial^2 \Theta(\theta)}{\partial \theta^2}+\lambda = \frac{m^2}{\sin^2 \theta} \end{align} and after transformation \begin{align} \frac{1}{\Theta(\theta)} \frac{\cos \theta}{\sin\theta} \frac{d \Theta}{d \xi} (- \sin \theta) + \frac{1}{\Theta(\theta)} \left( \frac{d^2 \Theta}{d \xi^2} \left( - \sin \theta \right)^2+\frac{d \Theta}{d \xi} (- \cos \theta) \right)+\lambda = \frac{m^2}{\sin^2 \theta} \end{align} Multiplying by $\Theta$ and rearranging we have \begin{align} \frac{d^2 \Theta}{d \xi^2} \left( - \sin \theta \right)^2-2 \cos \theta \frac{d \Theta}{d \xi} +\lambda \Theta(\theta) - \frac{m^2}{\sin^2 \theta} \Theta(\theta)=0 \end{align} using $\xi=\cos \theta$, $1-\xi^2=\sin \theta$ and $\lambda=\ell(\ell +1)$ results in the familiar form of

associated Legendre's differential equation \begin{align} (1-\xi^2) \frac{d^2 \Theta}{d \xi^2} -2 \xi \frac{d \Theta}{d \xi} + \left(\ell(\ell+1) - \frac{m^2}{1-\xi^2} \right) \Theta(\theta)=0 \end{align} with the solution \begin{align} \Theta(\xi)=E \, P_{\ell}^m(\xi)+ F \, Q_{\ell}^m(\xi) \end{align} the so called associated Legendre polynomials. We can transform back and write $\Theta$ as function of $\theta$ \begin{align} \Theta(\cos \theta)=E \, P_{\ell}^m(\cos \theta)+ F \, Q_{\ell}^m(\cos \theta) \end{align}