📂Odinary Differential Equations

Hermite Differential Equations and Series Solutions

Definition

The differential equation given below is referred to as the Hermite Differential Equation.

$$y^{\prime \prime}-2xy^{\prime}+2ny=0,\quad n=0,1,2,\cdots$$

The solution to the Hermite Differential Equation is called the Hermite Polynomial, and it is commonly denoted as $H_{n}(x)$.

$$\begin{align*} H_{0}(x) &= 1 \\ H_{1}(x) &= 2x \\ H_{2}(x) &= 4x^{2} - 2 \\ H_{3}(x) &= 8x^{3} - 12x \\ H_{4}(x) &= 16x^{4} - 48x^{2} + 12 \\ H_{5}(x) &= 32x^{5} - 160x^{3} + 120x \\ \vdots& \end{align*}$$

Explanation

Since Hermite was French, the correct pronunciation is [에르미트]. In English, it may be pronounced as [Her-mit] or [Her-mite].

The form mentioned above is specifically called the Physicist’s Hermite Function, and there is another expression known as the Probabilist’s Hermite Function.

It involves the independent variable $x$ in the coefficients, and if we assume that the solution takes the form of a power series, it can be solved. Solutions to the Chebyshev equation are known as Chebyshev polynomials and are commonly denoted as $T_{n}(x)$.

Solution

$$y^{\prime \prime}-2xy^{\prime}+2\lambda y=0$$

Let us assume the solution to the given Hermite Differential Equation is the following series.

$$y=\sum \limits _{n=0}^{\infty} a_{n}x^{n}$$

Although we start by assuming a series solution, towards the end of the solution process, we realize that the terms of $y$ are finite. To substitute into the differential equation, we find $y^{\prime}, y^{\prime \prime}$ as follows.

$$\begin{align*} y^{\prime} =&\ \sum \limits _{n=1}^{\infty} na_{n}x^{n-1} \\ y^{\prime \prime} =&\ \sum \limits _{n=2}^{\infty}n(n-1)a_{n}x^{n-2} \end{align*}$$

Upon substitution into the differential equation, the result is as follows.

$$\sum \limits _{n=2}^{\infty} n(n-1)a_{n}x^{n-2} -2\sum \limits _{n=1}^{\infty}na_{n}x^{n}+2\lambda \sum \limits _{n=0}^{\infty}a_{n}x^{n}=0$$

To match the order of $x$, changing the index of the first series gives

$$\sum \limits _{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n} -2\sum \limits _{n=1}^{\infty}na_{n}x^{n}+2\lambda \sum \limits _{n=0}^{\infty}a_{n}x^{n}=0$$

Taking out the term $n=0$ and combining the series into one gives the following.

$$2a_{2}+2\lambda a_{0}+\sum \limits _{n=1}^{\infty}\left[ (n+2)(n+1)a_{n+2} -2na_{n}+2\lambda a_{n}\right]x^{n}=0$$

For the above equation to hold, the coefficients of all terms must be $0$, thus giving the two conditions below.

$$\begin{align*} 2a_{2}+2\lambda a_{0} =&\ 0 \\ (n+2)(n+1)a_{n+2}-2na_{n} +2\lambda a_{n} =&\ 0 \end{align*}$$

However, by substituting $n=0$ into the equation below, we find it is essentially one condition. Sorting out $a_{n+2}$, we obtain the following recursion formula.

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

From this recursive formula, we understand that terms higher than $n=2$ are expressed as either $a_{0}$ or $a_{1}$. First, calculating for the even term $n$ gives the following.

$$\begin{equation} \begin{aligned} a_{2} =&\ \frac{2(-\lambda)}{2\cdot 1}a_{0} \\ a_{4} =&\ \frac{2(2-\lambda)}{4\cdot 3}a_{2}=\frac{2^{2}(2-\lambda)(-\lambda)}{4!}a_{0} \\ a_{6} =&\ \frac{2(4-\lambda)}{6\cdot 5}a_{4}=\frac{2^{3}(4-\lambda)(2-\lambda)(-\lambda)}{6!}a_{0} \\ \vdots & \\ a_{2n} =&\ \frac{2^{n}(2n-2-\lambda)(2n-4-\lambda)\cdots(2-\lambda)(-\lambda)}{(2n)!}a_{0} \end{aligned} \label{1} \end{equation}$$

Calculating for the odd term gives

$$\begin{align*} a_{3} =&\ \frac{2(1-\lambda)}{3\cdot 2}a_{1} \\ a_{5} =&\ \frac{2(3-\lambda)}{5\cdot 4}a_{3}=\frac{2^{2}(3-\lambda)(1-\lambda)}{5!}a_{1} \\ a_{7} =&\ \frac{2(5-\lambda)}{7\cdot 6}a_{5}=\frac{2^{3}(5-\lambda)(3-\lambda)(1-\lambda)}{7!}a_{1} \\ \vdots & \\ a_{2n+1} =&\ \frac{2^{n}(2n-1-\lambda)(2n-3-\lambda)\cdots(3-\lambda)(1-\lambda)}{(2n+1)!}a_{1} \end{align*}$$

Based on the above results, the series solution can be divided into two major parts.

$$\begin{align*} y =&\ \sum \limits _{n=0}^{\infty}a_{n}x^{n} \\ =&\ \left[ a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots\right]+\left[ a_{1}x + a_{3}x^{3}+a_{5}x^{5}+\cdots \right] \\ =&\ a_{0}\left[ 1-\lambda x^{2} + \frac{(2-\lambda)(-\lambda)}{3!}x^{4}+\cdots \right]+a_{1}\left[x + \frac{1-\lambda}{3}x^{3}+\frac{2^{2}(3-\lambda)(1-\lambda)}{5!}x^{5} + \cdots \right] \end{align*}$$

However, if $\lambda$ is a non-negative integer, one of the two parts remains finite. For example, if $\lambda=4$, as seen in $\eqref{1}$, it is $a_{6}=a_{8}=\cdots=a_{2n}=0$. Therefore, when the constant $\lambda$ of the differential equation is a non-negative integer, a non-divergent solution can be obtained. The solution for each lambda is denoted as $H_{\lambda}$.

$$\begin{align*} H_{0}(x) =&\ a_{0} \\ H_{1}(x) =&\ a_{1}x \\ H_{2}(x) =&\ a_{0}(1-2x^{2}) \\ H_{3}(x) =&\ a_{1}\left( x -\frac{2}{3}x^{3} \right) \\ H_{4}(x) =&\ a_{0} \left( 1-4x^{2}+\frac{4}{3}x^{4} \right) \\ H_{5}(x) =&\ a_{1}\left( x+-\frac{4}{3}x^{3}+\frac{4}{15}x^{5} \right) \\ \vdots & \end{align*}$$

Now, by substituting values into $a_{0}$, $a_{1}$ such that the coefficient of the highest term $x^{n}$ of $H_{n}(x)$ becomes $2^{n}$, we obtain the following Physicist’s Hermite Polynomial.

$$\begin{align*} H_{0}(x) =&\ 1 \\ H_{1}(x) =&\ 2x \\ H_{2}(x) =&\ 4x^{2}-2 \\ H_{3}(x) =&\ 8x^{3}-12x \\ H_{4}(x) =&\ 16x^{4}-48x^{2}+12 \\ H_{5}(x) =&\ 32x^{5}-160x^{3}+120x \\ \vdots & \end{align*}$$

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