Solution of Laplace Equation for Spherical Coordinate

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Laplace Equation in Spherical Coordinates

Laplace Equation in spherical coordinates $(r, \theta, \phi)$ becomes

$$\begin{aligned} \nabla^2 V = \frac{1}{r^2}\frac{\partial}{\partial r}\Big( r^2 \frac{\partial V}{\partial r}\Big) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta} \Big( \sin\theta \frac{\partial V}{\partial \theta}\Big) + \frac{1}{r^2 \sin^2 \theta}\frac{\partial^2 V}{\partial \phi^2}=0\end{aligned}$$

$$\begin{aligned} \nabla^2 V = \frac{1}{r}\frac{\partial^2}{\partial r^2}\Big(r V\Big) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\Big(\sin\theta \frac{\partial V}{\partial \theta}\Big) + \frac{1}{r^2 \sin^2\theta}\frac{\partial V}{\partial \phi^2}=0\end{aligned}$$

Step 1: Separation of variables assumption

Assume a product form for the potential $V$.

$$\begin{aligned} \boxed{V = \frac{R(r)}{r}\Theta(\theta)\Phi(\theta)}\end{aligned}$$

Here, we put $\frac{R(r)}{r}$ for easy calculation. Then the result equation becomes

$$\begin{aligned} \frac{\Phi \Theta}{r}\frac{d^2}{d r^2}R(r)+\frac{R\Phi}{r^3 \sin\theta}\frac{d}{d \theta}\Big(\sin\theta \frac{d \Theta(\theta)}{d \theta}\Big)+\frac{R\Theta}{r^3\sin^2\theta}\frac{d \Phi(\phi)}{d \phi^2}=0 \label{eq:1} \end{aligned}$$

Step 2: Find the separated equations

If we multiply $\frac{1}{R \Theta\Phi}$ to $Eq.\eqref{eq:1}$.

$$\begin{aligned} \frac{1}{rR}\frac{d^2}{dr^2}R(r)+\frac{1}{r^3\sin{\theta}}\frac{1}{\Theta }\frac{d}{d\theta}\Big(\sin\theta \frac{d\Theta(\theta)}{d\theta}\Big)+\frac{1}{r^3\sin^2\theta}\frac{1}{\Phi}\frac{d\Phi(\phi)}{d\phi^2} =0\end{aligned}$$

Nothing separates until we multiply $r^3\sin^2\theta$.

$$\begin{aligned} \frac{r^2 \sin^2\theta}{R}\frac{d^2}{dr^2}R(r) + \frac{\sin\theta}{\Theta}\frac{d}{d\theta}\Big(\sin\theta \frac{d\Theta(\theta)}{d\theta}\Big) + \frac{1}{\Phi}\frac{d\Phi(\phi)}{d\phi^2} =0\end{aligned}$$

Now, $\Phi$ term has completely separated. If we take partial derivative with respect to $\phi$

$$\begin{aligned} \frac{\partial}{\partial \phi}\Big(\frac{1}{\Phi}\frac{d^2 \Phi}{d\phi^2}\Big) =0\end{aligned}$$

since the first two terms depend only on $r$ and $\theta$. Therefor

$$\begin{aligned} \frac{1}{\Phi}\frac{d^2 \Phi}{d\phi^2} = -m^2\end{aligned}$$

The equation for $\Phi$ is then,

$$\begin{aligned} \boxed{\frac{d^2\Phi}{d\phi^2}+m^2 \Phi=0} \label{eq:sv1}\end{aligned}$$

The remainder of the Laplace equation is

$$\begin{aligned} \frac{r^2 R^{\prime\prime}}{R}+\frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\Big(\sin{\theta}\frac{d\Theta}{d\theta}\Big) -\frac{m^2}{\sin^2\theta}=0\end{aligned}$$

Now, we can separate the functions between $r$ and $\theta$. Let us define each term as $\alpha$.

$$\begin{aligned} {\frac{r^2 R^{\prime\prime}}{R}=\alpha} \label{eq:sv_2}\\ {\frac{1}{\Theta \sin\theta}\frac{d}{d\theta}\Big(\sin{\theta}\frac{d\Theta}{d\theta}\Big)-\frac{m^2}{\sin^2\theta}=-\alpha } \label{eq:sv_3}\end{aligned}$$

Step 3:Solving the equations for each variables

Eq.[eq:sv1] for $\Phi$

$$\begin{aligned} \sin^2\theta \{\frac{r^2 R\prime\prime}{R} + \frac{1}{\Theta \sin\theta}\frac{d}{d\theta}(\sin\theta\frac{d\Theta}{d\theta}) \}=-\frac{\Phi\prime\prime}{\Phi}= m^2\end{aligned}$$

We let $m^2$ to be positive because we want periodic answer for $Q$. Hence

$$\begin{aligned} \Phi \propto e^{\pm im\phi}\end{aligned}$$

Eq.[eq:sv_2] for R

$$\begin{aligned} r^2 R^{\prime\prime}-\alpha R = 0\end{aligned}$$

Notice that re-scaling $r$ leaves this equation unchanged. This is a clue that powers of $r$ may work. Substituting a power-law solution, $\frac{R(r)}{r}=r^{l}$; (notice that in the beginning, we divide extra $r$ for convenience).

$$\begin{aligned} r^2 (r^{(l+1) \prime\prime})-\alpha r^{l+1}&=0 \\ (l(l+1)-\alpha)r^{l+1} &= 0\\ \alpha &= l(l+1)\end{aligned}$$

with $a=l(l+1)$, we have a solution for every number $l$.Since it is a quadratic equation, there are two values of $l$.

$$\begin{aligned} l^2 + l - \alpha =0\end{aligned}$$

Let $l$ have some value $x$, so that $\alpha = x(x+1)$. Then, the value $l=-(x+1)$ gives the same $\alpha$.

$$\begin{aligned} \alpha = (-(x+1))(-(x+1)+1)=x(x+1)\end{aligned}$$

with $\alpha = l(l+1)$, we now have to solve for $\Theta$.

$$\begin{aligned} \frac{1}{\sin\theta}\frac{d}{d\theta}\Big( \sin\theta \frac{d\Theta}{d\theta}\Big) + \Big( l(l+1) - \frac{m^2}{\sin^2\theta}\Big)\Theta = 0 \label{eq:theta_1}\end{aligned}$$

• Legendre polynomials: The solution to the $\theta$ equation with $m=0$.
  

• associated Legendre polynomials: Solutions with general $m$.

Eq.[eq:sv_3] for $\Theta$

Legendre polynomials, $m=0$
Now $\sin\theta d\theta = d\cos\theta$. Hence, define $x=\cos\theta$. Then,

$$\begin{aligned} \frac{d}{d\theta}&=\frac{dx}{d\theta}\frac{d}{dx}\\ &=-\sin\theta \frac{d}{dx}\\ \frac{d}{dx}&=-\frac{1}{\sin\theta}\frac{d}{d\theta}\end{aligned}$$

When we substitute this to Eq.[eq:theta_1] with $m=0$,

$$\begin{aligned} \frac{d}{dx}\Big((1-x^2)\frac{d\Theta}{dx}\Big) + l(l+1)\Theta = 0\end{aligned}$$

Notice that $\sin^2\theta=1-\cos^2\theta=1-x^2$. The solutions are polynomials, $P_l(x)=P_l(\cos\theta)$

Legendre Polynomial

Let us consider azimuthally symmetric case, meaning $m=0$.

$$\begin{aligned} \frac{d}{dx}[(1-x^2)\frac{dP}{dx}]+[l(l+1)]P=0\end{aligned}$$

This equation is called Legendre Equation. The range of $x$ is $-1 \leq x \leq 1$ because the range of $\cos\theta$ is $0 \leq \cos\theta \leq \pi$

$$\begin{aligned} \sum_{j=0}^{\infty} & \{[(a+j)(a+j-1)]a_j x^{\alpha+j-2} \nonumber \\ & -[(a+j)(a+j+1)+2(a+j)-l(l+1)]a_jx^{\alpha+j}\}\end{aligned}$$

$$\begin{aligned} l=0 P_0(x)&=1 \nonumber \\ l=1 P_1(x)&=x \nonumber \\ l=2 P_2(x)&=\frac{1}{2}(3x^2-1) \nonumber \\ l=3 P_3(x)&=\frac{1}{2}(5x^3-3x) \nonumber \\ l=4 P_4(x)&=\frac{1}{8}(35x^4-30x^2+3) \end{aligned}$$

$\rightarrow$ $P_1$ is an even function for even $l$, and odd for odd $l$.

$\rightarrow$ Legendre polynomials can be represented by Rodrigues’ formula

$$\begin{aligned} P_l(x)=\frac{1}{2^{l}l!}\frac{d^l}{dx^l}(x^2-1)^l\end{aligned}$$

Example

$$\begin{aligned} P_2(x)&=\frac{1}{2^2 2!}\frac{d^2}{dx^2}(x^2-1)^2 \nonumber \\ &=\frac{1}{8} \frac{d^2}{dx^2}(x^4-2x^2+1) \nonumber \\ &=\frac{1}{8}(12x^2-4)=\frac{3x^2-1}{2} \nonumber\end{aligned}$$

$\rightarrow$ $P_l$ is orthogonal

$$\begin{aligned} \int_{-1}^{1} P_l(x) P_{l\prime}(x)dx =\frac{2}{2l+1}\delta_{ll\prime}\end{aligned}$$

This means that any function defined for $[-1,1]$ can be represented by expansion by $P_i$

$$\begin{aligned} f(x) = \sum_{l=0}^{\infty}A_l P_l(x), \quad \quad A_l=\frac{2l+1}{2}\int_{-1}^{1}f(x)P_l(x)dx\end{aligned}$$

Step 4:Putting all together.

For general $m$, the full solution is

$$\boxed{V(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} \Big( A_{lm}r^l + \frac{B_{lm}}{r^{l+1}} \Big) P_{l}^{m}( \cos \theta) e^{i m \phi}}$$