legendre polynomial

Legendre Differential Equation(#4) : A friendly introduction

/ 153armstrong / Leave a comment

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) .

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Legendre Differential Equation(#2): A friendly introduction

/ 153armstrong / 1 Comment

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

or

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

/ 153armstrong / 2 Comments

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)

(i)+(ii)+(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 ?