# 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

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

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