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


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)

The generalized product rule ( Leibniz Formula )

If f and g are n-times differentiable functions, then :

(fg)^{'} = fg^{'} + gf^{'}

Now, we would like to find out a generalized expression for the n-th derivative of fg. In order to arrive at that formulation lets calculate a few derivatives to see whether we can find a pattern:

(fg)^{'} = fg^{'} + gf^{'}

(fg)^{''} = \left(fg^{'} + gf^{'}\right)^{'} = fg^{''} + 2 f^{'}g^{'} + gf^{''}

(fg)^{'''} = \left(fg^{''} + 2 f^{'}g^{'} + gf^{''} \right)^{'} = fg^{'''} + 3 f^{''}g^{'} + 3 f^{'}g^{''} + gf^{'''}

(fg)^{''''} = fg^{''''} + 4 f^{'''}g^{'} + 6f^{''}g^{''} + 4 f^{'}g^{'''} + gf^{''''}


You must have noticed a pattern in the above expressions. The coefficients seem are the one in the binomial expansion of (x+y)^n


Therefore we can write the expression for the n-derivative of fg as the following expression:

(fg)^n = \sum\limits_{i=0}^{n} \binom{n}{i} f^{(i)}g^{(n-i)}
where (i) means to differentiate i-times.

This is also known as Leibniz Formula.

** This plays an important role when we start discussing about the Associated Legendre Differential Equation.