Series Solutions: Hermite's Equation

Hermite's Equation of order k has the form


where k is usually a non-negative integer.

We know from the previous section that this equation will have series solutions which both converge and solve the differential equation everywhere.

Hermite's Equation is our first example of a differential equation, which has a polynomial solution.

As usual, the generic form of a power series is

\begin{displaymath}y(t)=\sum_{n=0}^\infty a_n t^n.\end{displaymath}

We have to determine the right choice for the coefficients (an).

As in other techniques for solving differential equations, once we have a "guess" for the solutions, we plug it into the differential equation. Recall that

\begin{displaymath}y'(t)=\sum_{n=1}^\infty n a_n t^{n-1}, \end{displaymath}


\begin{displaymath}y''(t)=\sum_{n=2}^\infty n(n-1) a_n t^{n-2}.\end{displaymath}

Plugging this information into the differential equation we obtain:

\begin{displaymath}\sum_{n=2}^\infty n(n-1)a_n t^{n-2}-2t \sum_{n=1}^\infty n a_n t^{n-1}+2k \sum_{n=0}^\infty a_n t^n=0,\end{displaymath}

or after rewriting slightly:

\begin{displaymath}\sum_{n=2}^\infty n(n-1)a_n t^{n-2}- \sum_{n=1}^\infty 2n a_n t^{n}+ \sum_{n=0}^\infty 2k a_n t^n=0.\end{displaymath}

Next we shift the first summation up by two units:

\begin{displaymath}\sum_{n=0}^\infty (n+2)(n+1)a_{n+2} t^{n}- \sum_{n=1}^\infty 2n a_n t^{n}+\sum_{n=0}^\infty 2k a_n t^n=0.\end{displaymath}

Before we can combine the terms into one sum, we have to overcome another slight obstacle: the second summation starts at n=1, while the other two start at n=0.

Evaluate the 0th term for the second sum: $2\cdot 0\cdot a_0\cdot t^0=0$. Consequently, we do not change the value of the second summation, if we start at n=0 instead of n=1:

\begin{displaymath}\sum_{n=1}^\infty 2n a_n t^{n}=\sum_{n=0}^\infty 2n a_n t^{n}.\end{displaymath}

Thus we can combine all three sums as follows:

\begin{displaymath}\sum_{n=0}^\infty \left(\phantom{\int}(n+2)(n+1)a_{n+2}- 2n a_n + 2k a_n\phantom{\int}\right) t^n=0.\end{displaymath}

Therefore our recurrence relations become:

\begin{displaymath}(n+2)(n+1)a_{n+2}- 2n a_n + 2k a_n=0\mbox{ for all } n=0,1,2,3,\ldots\end{displaymath}

After simplification, this becomes

\begin{displaymath}a_{n+2}= \frac{2(n-k)}{(n+2)(n+1)}a_n=0\mbox{ for all } n=0,1,2,3,\ldots\end{displaymath}

Let us look at the special case, where k=5, and the initial conditions are given as $y(0)=a_0=0,\ y'(0)=a_1=1$. In this case, all even coefficients will be equal to zero, since a0=0 and each coefficient is a multiple of its second predecessor.


What about the odd coefficients? a1=1, consequently

\begin{displaymath}a_3=\frac{2(1-5)}{2\cdot 3}=-\frac{4}{3},\end{displaymath}


\begin{displaymath}a_5=\frac{2(3-5)}{4\cdot 5} a_3=(-\frac{1}{5})(-\frac{4}{3})=\frac{4}{15}.\end{displaymath}

What about a7:

\begin{displaymath}a_7=\frac{2(5-5)}{6\cdot 7}a_5=0.\end{displaymath}

Since a7=0, all odd coefficients from now on will be equal to zero, since each coefficient is a multiple of its second predecessor.


Consequently, the solution has only 3 non-zero coefficients, and hence is a polynomial. This polynomial


(or a multiple of this polynomial) is called the Hermite Polynomial of order 5.

It turns out that the Hermite Equation of positive integer order k always has a polynomial solution of order k. We can even be more precise: If k is odd, the initial value problem $a_0=0,\ a_1=1$ will have a polynomial solution, while for k even, the initial value problem $a_0=1,\ a_1=0$ will have a polynomial solution.

Exercise 1:

Find the Hermite Polynomials of order 1 and 3.


Exercise 2:

Find the Hermite Polynomials of order 2, 4 and 6.


Exercise 3:

Consider the Hermite Equation of order 5:


Find the solution satisfying the initial conditions a0=1, a1=0.


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