Month: January 2018

Solution to the Hermite Differential Equation

One typically finds the Hermite differential equation in the context of an infinite square well potential and the consequential solution of the Schrödinger equation. However, I will consider this equation is its “raw” mathematical form viz.

\displaystyle \frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+\lambda y(x) =0. (1)

First we will consider the more general case, leaving \lambda undefined. The second case will consider in a future post \lambda = 2n, n\in \mathbb{Z}^{+}, where \mathbb{Z}^{+}=\bigg\{x\in\mathbb{Z}|x > 0\bigg\}.

PART I: 

Let us assume the solution has the form

\displaystyle y(x)=\sum_{j=0}^{\infty}a_{j}x^{j}. (2)

Now we take the necessary derivatives

\displaystyle y^{\prime}(x)=\sum_{j=1}^{\infty}ja_{j}x^{j-1}, (3)

\displaystyle y^{\prime \prime}(x)=\sum_{j=2}^{\infty} j(j-1)a_{j}x^{j-2}, (4)

where upon substitution yields the following

\displaystyle \sum_{j=2}^{\infty}j(j-1)a_{j}x^{j-2}-\sum_{j=1}^{\infty}2ja_{j}x^{j}+\sum_{j=0}^{\infty}\lambda a_{j}x^{j}=0, (5)

Introducing the dummy variable m=j-2 and using this and its variants we arrive at

\displaystyle \sum_{j=0}^{\infty}(j+2)(j+1)a_{j+2}x^{j}-\sum_{j=0}^{\infty}2ja_{j}x^{j}+\sum_{j=0}^{\infty}\lambda a_{j}x^{j}=0. (6)

Bringing this under one summation sign…

\displaystyle \sum_{j=0}^{\infty}[(j+2)(j+1)a_{j+2}-2ja_{j}+\lambda a_{j}]x^{j}=0. (7)

Since \displaystyle \sum_{j=0}^{\infty}x^{j}\neq 0, we therefore require that

\displaystyle (j+2)(j+1)a_{j+2}=(2j - \lambda)a_{j}, (8)

or

\displaystyle a_{j+2}=\frac{(2j-\lambda)a_{j}}{(j+2)(j+1)}. (9)

This is our recurrence relation. If we let j=0,1,2,3,... we arrive at two linearly independent solutions (one even and one odd) in terms of the fundamental coefficients a_{0} and a_{1} which may be written as

\displaystyle y_{even}(x)= a_{0}\bigg\{1+\sum_{j=0}^{j/2}\frac{(-1)^{j}(\lambda -2j)!}{(j+2)!}x^{j}\bigg\}, (10)

and

\displaystyle y_{odd}(x)=a_{1}\bigg\{\sum_{j=0}^{(j-1)/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j}\bigg\}. (11)

Thus, our final solution is the following

\displaystyle y(x)=y_{even}(x)+y_{odd}(x), (12.1)

 

\displaystyle y(x)=a_{0}\bigg\{1+\sum_{j=0}^{j/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j+2}\bigg\}+a_{1}\bigg\{x+\sum_{j=1}^{(j-1)/2}\frac{(-1)^{j}(\lambda-2j)!}{(j+2)!}x^{j+2}\bigg\}. (12.2)

 

 

 

Legendre Polynomials

Some time ago, I wrote a post discussing the solution to Legendre’s ODE. In that post, I discussed what is an alternative definition of Legendre polynomials in which I stated Rodriguez’s formula:

\displaystyle \frac{1}{2^{p}p!}\frac{d^{p}}{dx^{p}}\bigg\{(x^{2}-1)^{p}\bigg\}, (0.1)

where

\displaystyle P_{p}(x)=\sum_{n=0}^{\alpha}\frac{(-1)^{n}(2p-2n)!}{2^{p}{n!}(p-n)!(p-2n)!} (0.2),

and

\displaystyle P_{p}(x)=\sum_{n=0}^{\beta}\frac{(-1)^{n}(2p-2n)!}{2^{p}{n!}(p-n)!(p-2n)!} (0.3)

in which I have let \displaystyle \alpha=p/2 and \displaystyle \beta=(p-1)/2 corresponding to the even and odd expressions for the Legendre polynomials.

However, in this post I shall be using the approach of the generating function. This will be from a purely mathematical perspective, so I am not applying this to any particular topic of physics.

Consider a triangle with sides \displaystyle X,Y.Z and angles \displaystyle \theta, \phi, \lambda. The law of cosines therefore maintains that

\displaystyle Z^{2}=X^{2}+Y^{2}-2XY\cos{(\lambda)}. (1)

We can factor out \displaystyle X^{2} from the left hand side of Eq.(1), take the square root and invert this yielding

\displaystyle \frac{1}{Z}=\frac{1}{X}\bigg\{1+\bigg\{\frac{Y}{X}\bigg\}^{2}-2\bigg\{\frac{Y}{X}\bigg\}\cos{(\lambda)}\bigg\}^{-1/2}. (2)

Now, we can expand this by means of the binomial expansion. Let \displaystyle \kappa \equiv \bigg\{\frac{Y}{X}\bigg\}^{2}-2\bigg\{\frac{Y}{X}\bigg\}\cos{(\lambda)}, therefore the binomial expansion is

\displaystyle \frac{1}{(1+\kappa)^{1/2}}=1-\frac{1}{2}\kappa+\frac{3}{8}\kappa^{2}-\frac{5}{16}\kappa^{3}+... (3)

Hence if we expand this in terms of the sides and angle(s) of the triangle and group by powers of \displaystyle (y/x) we get

\displaystyle \frac{1}{Z}=\frac{1}{X}\bigg\{1+\bigg\{\frac{Y}{X}\bigg\}\cos{(\lambda)}+\bigg\{\frac{Y}{X}\bigg\}^{2}\frac{1}{2}(3\cos^{2}{(\lambda)}-1)+\bigg\{\frac{Y}{X}\bigg\}^{3}\frac{1}{2}(5\cos^{3}{(\lambda)}-3\cos{(\lambda)}\bigg\}.(4)

Notice the coefficients, these are precisely the expressions for the Legendre polynomials. Therefore, we see that

\displaystyle \frac{1}{Z}=\frac{1}{X}\bigg\{\sum_{l=0}^{\infty}\bigg\{\frac{Y}{X}\bigg\}^{l}P_{l}(\cos{(\lambda)}\bigg\}, (5)

or

\displaystyle \frac{1}{Z}=\frac{1}{\sqrt[]{X^{2}+Y^{2}-2XY\cos{(\lambda)}}}=\sum_{l=0}^{\infty}\frac{Y^{l}}{X^{l+1}}P_{l}(\cos{(\lambda)}. (6)

Thus we see that the generating function \displaystyle 1/Z generates the Legendre polynomials. Two prominent uses of these polynomials includes gravity and its application to the theory of potentials of a spherical mass distributions, and the other is that of electrostatics. For example, suppose we have the potential equation

\displaystyle V(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\rho(R)\frac{\hat{\mathcal{R}}}{\mathcal{R_{0}}}d\tau. (7.1)

We may use the result of the generating function to get the following result for the electric potential due an arbitrary charge distribution

\displaystyle V(\mathcal{R})=\frac{1}{4\pi\epsilon_{0}}\sum_{l=0}^{\infty}\frac{\mathcal{R}^{l}}{\mathcal{R_{0}}^{l+1}}\int P_{l}(\cos{(\lambda)}). (7.2)

(For more details, see Chapter 3 of Griffith’s text: Introduction to Electrodynamics.)