# Solving the Laplacian in Spherical Coordinates (#1)

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

$x = rsin\theta cos\phi$
$y = rsin\theta cos\phi$
$z = rcos\theta$

where

$0 \leq r < \infty$
$0 \leq \theta \leq \pi$
$0 \leq \phi < 2\pi$

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

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

To solve it we use the method of separation of variables.

$f = R(r)\Theta(\theta)\Phi(\phi)$

Plugging in the value of $f$ into the Laplacian, we get that :

$\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0$

Dividing throughout by $R\Theta\Phi$ and multiplying throughout by $r^2$, further simplifies into:

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0$

It can be observed that the first expression in the differential equation is merely a function of $r$ and the remaining a function of $\theta$ and $\phi$ only. Therefore, we equate the first expression to be $\lambda = l(l+1)$ and the second to be $-\lambda = -l(l+1)$. The reason for choosing the peculiar value of $l(l+1)$ is explained in another post.

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0$ (1)

The first expression in (1) the Euler-Cauchy equation in $r$.

$\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R$

The general solution of this has been in discussed in a previous post and it can be written as:

$R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}$

The second expression in (1) takes the form as follows:

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0$

The following observation can be made similar to the previous analysis

$\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0$ (2)

The first expression in the above equation (2) is the Associated Legendre Differential equation.

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2$

$sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0$

The general solution to this differential equation can be given as:
$\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)$

The solution to the second term in the equation (2) is a trivial one:

$\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi$
$\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}$

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

$R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)$

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# Euler-Cauchy equation

While trying to solve for the Laplacian in polar coordinates, one encounters the famous Euler-Cauchy differential equation.

$x^{2}{\frac {d^{2}y}{dx^{2}}}+2x{\frac {dy}{dx}}-l(l+1)y=0$

or

${\frac {d^{2}y}{dx^{2}}}+ \frac{2}{x}{\frac {dy}{dx}}- \frac{l(l+1)}{x^2}y=0$

How does one find a solution to this differential equation ? Well, most places that I have read simply dictate that the solutions to this differential equation as $y = x^l, x^{-(l+1)}$ without explaining why only these two are the only solutions. The purpose of this post is to explain why!

We can clearly see that x = 0 is a singular point. Therefore a simple power series solution won’t work. Hence we use the frobenius method to get rid of the singularity i.e

$y= \sum\limits_{n=0}^{\infty} a_{n}x^{n + \lambda}$

Let’s now compute its derivatives wrt x.

$\frac{dy}{dx}= \sum\limits_{n=0}^{\infty} (n+\lambda)a_{n}x^{n +\lambda -1}$

$\frac{d^{2}y}{dx^{2}}= \sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n + \lambda -2}$

Putting the values for $y,\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ back into the differential equation, we get the following form.

$x^{2}{\sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n-2}}$

$+2x{\sum\limits_{n=0}^{\infty}(n+\lambda)a_{n}x^{n-1}}$

$- l(l+1){\sum\limits_{n=0}^{\infty} a_{n}x^{n}}=0$

Bringing the $x,x^{2}$ terms inside the summation, we obtain the following form:

$\sum\limits_{n=0}^{\infty}x^n \left({(n+\lambda)(n+\lambda -1)a_n + 2(n+\lambda)a_n -l(l+1)a_n}\right)=0$

$\sum\limits_{n=0}^{\infty}a_n x^n \left({(n+\lambda)(n+\lambda +1) -l(l+1)}\right) = 0$

Obviously, the coefficients of the summation cannot be zero and x=0 is a trivial solution. Therefore, we get the indicial equation.

$(n+\lambda)(n+\lambda+1) - l(l+1)= 0$ or

$(n+\lambda)(n+\lambda+1) = l(l+1)$

(i) When n = 0, the indicial equation becomes

$\lambda(\lambda+1) = l(l+1)$

The values of the $\lambda$ that solve for this equation are $l,-(l+1)$

$\lambda = l , -(l+1)$

(ii) When n = 1, the indicial equation becomes

$(\lambda + 1)(\lambda+2) = l(l+1)$

The values of the $\lambda$ that solve for this equation are $(l-1),-(l+2)$

$\lambda = (l-1) , -(l+2)$

$\vdots$

And in general:

$\lambda_n = (l-n), -(l+n+1)$

$\lambda_{n_1} = (l-n)$, $\lambda_{n_2} = -(l+n+1)$

Here is the crux of it all.
Lets now write down the solution for this differential equation in its glory:

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + \lambda_{n_1}} + b_n x^{n+ \lambda_{n_{2}}}\right)$

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + (l-n)} + b_n x^{n-(l+n+1)}\right)$

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{l} + b_n x^{-(l+1)}\right)$

$y = \left(a_0 + a_1 + a_2 + \hdots \right) x^{l} + \left(b_0 + b_1 + b_2 + \hdots \right) x^{-(l+1)}$

$y = A x^{l} + B x^{-(l+1)}$

where A and B are constants.

This is the general solution to the Euler-Cauchy differential equation. 😀