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.
The Laplacian in Spherical coordinates in its ultimate glory is written as follows:
To solve it we use the method of separation of variables.
Plugging in the value of into the Laplacian, we get that :
Dividing throughout by and multiplying throughout by , further simplifies into:
It can be observed that the first expression in the differential equation is merely a function of and the remaining a function of and only. Therefore, we equate the first expression to be and the second to be . The reason for choosing the peculiar value of is explained in another post.
The first expression in (1) the Euler-Cauchy equation in .
The general solution of this has been in discussed in a previous post and it can be written as:
The second expression in (1) takes the form as follows:
The following observation can be made similar to the previous analysis
The first expression in the above equation (2) is the Associated Legendre Differential equation.
The general solution to this differential equation can be given as:
The solution to the second term in the equation (2) is a trivial one:
Therefore the general solution to the Laplacian in Spherical coordinates is given by:
If f and g are n-times differentiable functions, then :
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:
You must have noticed a pattern in the above expressions. The coefficients seem are the one in the binomial expansion of
Therefore we can write the expression for the n-derivative of fg as the following expression:
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.
When you are working with Spherical harmonics, then the Legendre Differential Equation does not appear in its natural form i.e
Instead, it appears in this form:
It seems daunting but the above is the same as the LDE. We can arrive at it by taking and proceeding as follows:
Now, applying chain rule, we obtain that
Now simplifying the above expression, we obtain that:
Plugging in the values of and into the Legendre Differential Equation,
Now if we do some algebra and simplify the trigonometric identities, we will arrive at the following expression for the Legendre Differential Equation:
If we take the solution for the LDE as , then the solution to the LDE in the above form is merely .
This post is just a note on the notation that is used across internet sources and books while referring to the LDE.
If one takes , then it follows that . The differential equation can be rewritten as follows:
Now the first two terms must seem familiar to you from the chain rule. ( ). Ergo,
Now, putting back the value of p :
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.
Now there is something about the Legendre differential equation that drove me crazy. What is up with the l(l+1) !!!
To understand why let’s take this form of the LDE and arrive at the above:
If we do a power series expansion and following the same steps as the previous post, we end up with the following recursion relation.
Here’s the deal: We want a convergent solution for our differential solution. This means that as .
Hence we obtain that
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:
We will find solutions for this differential equation using the power series expansion i.e
We will plug in these expressions for the derivatives into the differential equation.
** Note: Begin
Let’s take .
As n -> . , -> .
As n -> , -> .
Again performing a change of variables from to n.
** Note: End
(iii) can now be written as follows.
x = 0 is a trivial solution and therefore we get the indicial equation:
We get the following recursion relation on the coefficients of the power series expansion.
Next post: What do these coefficients mean ?
While trying to solve for the Laplacian in polar coordinates, one encounters the famous Euler-Cauchy differential equation.
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 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
Let’s now compute its derivatives wrt x.
Putting the values for and back into the differential equation, we get the following form.
Bringing the terms inside the summation, we obtain the following form:
Obviously, the coefficients of the summation cannot be zero and x=0 is a trivial solution. Therefore, we get the indicial equation.
(i) When n = 0, the indicial equation becomes
The values of the that solve for this equation are
(ii) When n = 1, the indicial equation becomes
The values of the that solve for this equation are
And in general:
Here is the crux of it all.
Lets now write down the solution for this differential equation in its glory:
where A and B are constants.
This is the general solution to the Euler-Cauchy differential equation. 😀