How to Solve the Laplace Equation with Azimuthal Symmetry in Spherical Coordinates Using Separation of Variables
Theorem
When there is azimuthal symmetry in spherical coordinates, the general solution of the Laplace equation is as follows.
$$ V(r,\theta) = \sum \limits_{l=0} ^\infty \left( A_{l} r^l + \dfrac{B_{l}}{r^{l+1} } \right) P_{l}(\cos \theta) $$
Proof
Step 0
When finding the potential, if the boundary condition is easily expressed in spherical coordinates, we must solve the Laplace equation in spherical coordinates. The Laplace equation in spherical coordinates is as follows. (Reference 1, Reference 2)
$$ \nabla ^2 V = \dfrac{1}{r} \dfrac{\partial}{\partial r} \left( r^2 \dfrac{\partial V}{\partial r} \right) +\dfrac{ 1}{r^2 \sin \theta } \dfrac{ \partial}{\partial \theta} \left( \sin \theta \dfrac{ \partial V}{\partial \theta }\right) + \dfrac{1}{r^2 \sin ^2 \theta} \dfrac{ \partial ^2 V}{\partial \phi^2 }=0 $$
Now suppose the potential $V$ is a function independent of $\phi$. In other words, this is the assumption that the value of $V$ does not change when only $\phi$ varies while the other values remain the same. Then the amount of change of $V$ with respect to $\phi$ is $0$, and since this means $\dfrac{\partial V}{\partial \phi}=0$, the third term vanishes.
$$ \dfrac{1}{r} \dfrac{\partial}{\partial r} \left( r^2 \dfrac{ \partial V}{\partial r} \right) +\dfrac{ 1}{r^2 \sin \theta } \dfrac{ \partial}{\partial \theta} \left( \sin \theta \dfrac{ \partial V}{\partial \theta }\right)=0 \tag{1} $$
Let us assume that the potential $V(r, \theta)$ is a separable function. This means we assume that $V$ consists of the product of a function $R(r)$ of $r$ only and a function $\Theta (\theta) $ of $\theta$ only. Substituting $V(r,\theta)=R(r) \Theta (\theta)$ into $(1)$ and dividing both sides by $V$ and rearranging, we obtain the following form.
$$ \dfrac{1}{R} \dfrac{d}{d r} \left( r^2 \dfrac{d R}{d r} \right) +\dfrac{ 1}{ \Theta \sin \theta } \dfrac{ d}{d \theta} \left( \sin \theta \dfrac{ d \Theta}{d \theta }\right) = 0 $$
Since each term depends only on $r$ and $\theta$ respectively, both terms are constants. Even if the value of $r$ changes, the values of the second term and the right-hand side do not change. Therefore the value of the first term must also always be the same, which means it is a constant term. The second term is also a constant term for the same reason.
$$ \dfrac{1}{R} \dfrac{d}{d r} \left( r^2 \dfrac{d R}{d r} \right) =l(l+1) $$
$$ \dfrac{ 1}{\Theta \sin \theta } \dfrac{d}{d \theta} \left( \sin \theta \dfrac{d \Theta }{d \theta }\right) = -l(l+1) $$
The complicated partial differential equation of $(1)$ has been transformed into two simple ordinary differential equations. Solving each differential equation to find $R(r)$ and $\Theta (\theta)$ and multiplying them gives us the $V(r,\theta)$ we want.
Step 1
$$ \dfrac{d}{dr} \left( r^2 \dfrac{dR}{dr} \right) = l(l+1) R $$ $$ \implies 2r \dfrac{dR}{dr} +r^2 \dfrac{d^2 R}{dr^2} =l(l+1)R $$
$$\implies r^2 \dfrac{d^2 R}{dr^2} + 2r\dfrac{dR}{dr} - l(l+1)R=0$$
This is in the form of an Euler equation, and its solution can be checked here, but in this article we will solve it more simply. Using the fact that the solution of the above differential equation comes out in the form $r^k$, we substitute $R=r^k$ into the first line. Then
$$ \dfrac{ d}{dr} \left( r^2 \dfrac{ d r^k}{dr} \right) = l(l+1)r^k$$ $$ \implies \dfrac{d}{dr} (k r^{k+1} ) = l(l+1) r^k$$ $$ \implies k(k+1)r^k=l(l+1)r^k $$ $$ \therefore k(k+1)=l(l+1) $$ $$ \implies k^2+ k -l(l+1)=0$$
Using the quadratic formula, $k=l$ or $k=-(l+1)$. Therefore $r^l$ and $\dfrac{1}{r^{l+1} }$ are solutions of the differential equation. Since the general solution is a linear combination of the two solutions, $R(r)=Ar^l + \dfrac{B}{r^{l+1} }$, where $A,B$ are constants.
Step 2
$$ \dfrac{d}{d\theta} \left( \sin \theta \dfrac{d \Theta }{d \theta} \right) =-l(l+1)\sin \theta \Theta $$
Since solving the differential equation with respect to $\theta$ is difficult, we introduce only the result. If you want to know the detailed solution, refer to here. The solution of the above differential equation is the Legendre polynomial in $\cos \theta $.
$$ \Theta (\theta) = P_{l}( \cos \theta) $$
Here $P_{l}(x)$ is as follows.
Rodrigues’ formula $$ P_{l}(x) := \dfrac{1}{2^l l!} \left( \dfrac{d}{dx} \right) ^l (x^2-1)^l $$
l$ is a positive integer, $P_{0}(x)=1$
The Legendre polynomials calculated according to the above formula are as follows.
$$P_{0}(x) =1$$ $$P_{1}(x)=x$$ $$P_2(x) = \dfrac{3x^2-1}{2}$$ $$P_{3}(x) = \dfrac{5x^3 -3x}{2}$$ $$\vdots$$
Since it is a second-order differential equation, we must find two solutions, but there is only one solution for each value of $l$. The remaining second solution is not a physically meaningful solution because it diverges at $\theta=0$ and $\theta=\pi$. Therefore we need only consider the first solution given by the Legendre polynomial.
Step 3.
Combining the results of Step 1 and Step 2, the potential is
$$ V(r,\theta) =\left( Ar^l ++ \dfrac{B}{r^{l+1}} \right) P_{l}( \cos \theta) $$
Since there is one solution for each value of $l$, the general solution is the sum of all of them.
$$ V(r,\theta) = \sum \limits_{l=0} ^\infty \left( A_{l} r^l + \dfrac{B_{l}}{r^{l+1} } \right) P_{l}(\cos \theta) $$
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