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)

Legendre Differential Equation(#4) : A friendly introduction

When you are working with Spherical harmonics, then the Legendre Differential Equation does not appear in its natural form i.e

(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0

Instead, it appears in this form:

\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0

It seems daunting but the above is the same as the LDE. We can arrive at it by taking x = cos(\theta) and proceeding as follows:

\frac{dy}{dx} = \frac{dy}{d(cos\theta)} = \frac{-1}{sin\theta}\frac{dy}{d \theta}

\frac{d^2 y}{dx^2} = \frac{d}{d(cos\theta)}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right) = \frac{-1}{sin\theta}\frac{d}{d \theta}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right)

Now, applying chain rule, we obtain that

\frac{d^2 y}{dx^2} = \frac{-1}{sin\theta} \left( \frac{-1}{sin\theta} \frac{d^2 y}{d\theta^2} - \frac{cos\theta}{sin^2 \theta} \frac{dy}{d\theta}      \right)

Now simplifying the above expression, we obtain that:

\frac{d^2 y}{dx^2} = \frac{1}{sin^2\theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} \right)

Plugging in the values of \frac{dy}{dx} and \frac{d^2 y}{dx^2}  into the Legendre Differential Equation,

(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0

(1-cos^2 \theta)y^{''} -2cos\theta y^{'} + l(l+1)y = 0

\frac{1- cos^2 \theta}{sin^2 \theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta}\frac{dy}{d\theta} \right) +\frac{2cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0

Now if we do some algebra and simplify the trigonometric identities, we will arrive at the following expression for the Legendre Differential Equation:

\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0

If we take the solution for the LDE as f(x) , then the solution to the LDE in the above form is merely f(cos\theta) .

Legendre Differential Equation(#3): A friendly introduction

This post is just a note on the notation that is used across internet sources and books while referring to the LDE.

(1-x^2)y^{''} - 2xy^{'} + l(l+1) y = 0

If one takes p(x) = 1-x^2 , then it follows that p^{'}(x) = -2x . The differential equation can be rewritten as follows:

p(x)y^{''} + p^{'}(x)y^{'} + l(l+1) y = 0

Now the first two terms must seem familiar to you from the chain rule. ( (py)^{'} = py^{'} + yp^{'} ). Ergo,

(py^{'})^{'} + l(l+1)y = 0


\frac{d}{dx}(p \frac{dy}{dx}) + l(l+1)y = 0

Now, putting back the value of p :

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

And you will see this form of the LDE also in many places and I thought it was worth mentioning how one ended up in that form.

Legendre Differential Equation(#2): A friendly introduction

Now there is something about the Legendre differential equation that drove me crazy. What is up with the l(l+1) !!!

(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0

To understand why let’s take this form of the LDE and arrive at the above:

(1-x^2)y^{''} -2xy^{'} + \lambda y = 0

y = \sum\limits_{n=0}^{\infty} a_n x^n

If we do a power series expansion and following the same steps as the previous post, we end up with the following recursion relation.

(n+2)(n+1)a_{n+2} = (\lambda -n(n+1))a_n


a_{n+2} = a_n \frac{\lambda - n(n+1)}{(n+1)(n+2)}

Here’s the deal: We want a convergent solution for our differential solution. This means that as n \rightarrow l , a_{n+2} \rightarrow 0.

Hence we obtain that

\lambda = l(l+1)

Legendre Differential equation (#1) : A friendly introduction

In this series of posts about Legendre differential equation, I would like to de-construct the differential equation down to the very bones. The motivation for this series is to put all that I know about the LDE in one place and also maybe help someone as a result.

The Legendre differential equation is the following:

(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0

where y^{'} = \frac{dy}{dx} and y^{''} = \frac{d^{2}y}{dx}

We will find solutions for this differential equation using the power series expansion i.e
y = \sum\limits_{n=0}^{\infty} a_n x^n

y^{'} = \sum\limits_{n=0}^{\infty} na_n x^{n-1}

y^{''} = \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}

We will plug in these expressions for the derivatives into the differential equation.

l(l+1)y = l(l+1)\sum\limits_{n=0}^{\infty} a_n x^n – (i)

-2xy^{'} = -2\sum\limits_{n=0}^{\infty} na_n x^{n} – (ii)

(1-x^2)y^{''} = (1-x^2)\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}

= \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2} - \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n} – (iii)

** Note: Begin

\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}

Let’s take \lambda = n-2 .
As n -> 0. , \lambda -> -2.
As n -> \infty , \lambda -> \infty.

\sum\limits_{\lambda = -2}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}

= 0 + 0 + \sum\limits_{\lambda = 0}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}

Again performing a change of variables from \lambda to n.

= \sum\limits_{n= 0}^{\infty} (n+2)(n+1)a_n x^{n}

** Note: End

(iii) can now be written as follows.

\sum\limits_{n=0}^{\infty} x^n \left((n+1)(n+2)a_{n+2} - n(n-1)a_n \right)  – (iv)


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

x = 0 is a trivial solution and therefore we get the indicial equation:

(n+2)(n+1)a_{n+2} - (l(l+1)-n(n+1))a_n = 0

(n+2)(n+1)a_{n+2} = (l^2 - n^2 + l - n)a_n = 0

(n+2)(n+1)a_{n+2} = ((l-n)(l+n)+ l - n)a_n = 0

(n+2)(n+1)a_{n+2} = (l-n)(l+n+1)a_n = 0

We get the following recursion relation on the coefficients of the power series expansion.

a_{n+2} = a_n \frac{(l+n+1)(l-n)}{(n+1)(n+2)}

Next post: What do these coefficients mean ?

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


{\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}}


- 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)


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. 😀