Section 4.9 Hermite Differential Equation
The differential equation
\begin{equation}
\frac{d^{2}y}{dx^{2}}-2x\frac{\,dy}{\,dx}+2\lambda y=0\tag{4.9.1}
\end{equation}
is known as Hermite differential equation. Where \(\lambda\) is any constant. Such type of equation yields on solving the harmonic oscillator problems. It has no singularity at \(x=0\text{.}\) Let us assume the solution of eqn. (4.9.1) as
\begin{equation}
y=\sum\limits_{k=0}^{\infty}a_{k}x^{m+k}\tag{4.9.2}
\end{equation}
so that
\begin{equation*}
\frac{\,dy}{\,dx} =\sum a_{k}(m+k)x^{m+k-1}
\end{equation*}
and
\begin{equation*}
\frac{d^{2}y}{dx^{2}} = \sum a_{k}(m+k)(m+k-1)x^{m+k-2}
\end{equation*}
Substituting these values in equation (4.9.2), we have
\begin{equation*}
\sum a_{k}(m+k)(m+k-1)x^{m+k-2}-2x\sum a_{k}(m+k)x^{m+k-1}
\end{equation*}
\begin{equation*}
+2\lambda\sum a_{k}x^{m+k}=0
\end{equation*}
or,
\begin{equation*}
\sum a_{k}(m+k)(m+k-1)x^{m+k-2}
\end{equation*}
\begin{equation}
-2\sum a_{k}[m+k-\lambda]x^{m+k}=0\tag{4.9.3}
\end{equation}
Equating the coefficient of lowest power of \(x\) i.e., \(x^{m-2}\) to zero, we get -
\begin{equation*}
a_{o}m(m-1)=0
\end{equation*}
[for \(k=0\)] which is an indicial equation.
\begin{equation}
\because \quad a_{o}\neq 0, \quad m=0,1\tag{4.9.4}
\end{equation}
Equating the coefficient of \(x^{m-1}\) to zero, we get -
\begin{equation*}
a_{1}(m+1)m=0
\end{equation*}
[for \(k=1\)] \(\because m=0\text{,}\) \(a_{1}\) may or may not be zero. But \(m \neq -1\text{,}\) hence \(a_{1}\) must be zero. \([\because m = 1]\) Equating the coefficient of general term \(x^{m+k} \) to zero, we obtain the recurrence relations among the coefficients.
\begin{equation*}
a_{k+2} (m+k+2)(m+k+1)-2a_{k}(m+k-\lambda)=0
\end{equation*}
or,
\begin{equation}
a_{k+2} = \frac{2(m+k-\lambda)}{ (m+k+2)(m+k+1)}a_{k} \tag{4.9.5}
\end{equation}
- For \(m=0\text{,}\)\begin{equation} a_{k+2} = \frac{2(k-\lambda)}{(k+2)(k+1)}a_{k}\tag{4.9.6} \end{equation}Now,\begin{equation*} a_{2}=\frac{2(-\lambda)}{2\cdot1}a_{o} =-\frac{2\lambda}{2!}a_{o};\quad a_{3} \end{equation*}\begin{equation*} =\frac{2(1-\lambda)}{3\cdot2}a_{1}=-\frac{2(\lambda-1)}{3!}a_{1}; \end{equation*}\begin{equation*} a_{4}=\frac{2(2-\lambda)}{4\cdot3}a_{2}=-\frac{2^{2}(2-\lambda)\cdot\lambda}{4\cdot3\cdot2}a_{o} \end{equation*}\begin{equation*} =\frac{2^{2}\lambda(\lambda-2)}{4!}a_{o}; \end{equation*}\begin{equation*} a_{5}=\frac{2^{2}(\lambda-1)(\lambda-3)}{5!}a_{1}; \end{equation*}\begin{equation*} a_{6}=\frac{2^{2}\lambda(\lambda-2)(\lambda-4)}{6!}a_{o} \end{equation*}and so on. Thus\begin{equation*} a_{2r}= \frac{-2^{r}\lambda(\lambda-2)\cdots(\lambda-2r+2)}{(2r)!}a_{o} \end{equation*}and\begin{equation*} a_{2r+1}= \frac{(-2)^{r}\lambda(\lambda-1)(\lambda-3)\cdots(\lambda-2r+1)}{(2r+1)!}a_{1}. \end{equation*}Thus the series solution for \(m=0\) is\begin{equation*} y=a_{o}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \end{equation*}\begin{equation*} =a_{o}+a_{2}x^{2}+a_{4}x^{4}+\cdots+a_{1}x+a_{3}x^{3}+a_{5}x^{5}+\cdots \end{equation*}\begin{equation*} =a_{o}\left[1-\frac{2\lambda}{2!}x^{2}+\frac{2^{2}\cdot\lambda(\lambda-2)}{4!}x^{4}-\cdots\right. \end{equation*}\begin{equation*} \left.+\frac{-2^{r}\cdot\lambda(\lambda-2)\cdots(\lambda-2r+2)}{(2r)!}x^{2r}+\cdots\right] \end{equation*}\begin{equation*} +a_{1}x\left[1-\frac{2(\lambda-1)}{3!}x^{2}+\cdots\right. \end{equation*}\begin{equation} \left.+\frac{-2^{r}(\lambda-1)(\lambda-3)\cdots(\lambda-2r+1)}{(2r+1)!}x^{2r}+\cdots\right]\tag{4.9.7} \end{equation}
- For \(m=1\text{,}\)\begin{equation} a_{k+2} = \frac{2(1+k-\lambda)}{(k+3)(k+2)}a_{k}\tag{4.9.8} \end{equation}hence,\begin{equation*} a_{2}=\frac{2(1-\lambda)}{3\cdot2}a_{o}=-\frac{2(\lambda-1)}{3!}a_{o}; \end{equation*}\begin{equation*} a_{4}=\frac{2(3-\lambda)}{5\cdot3}a_{2}=\frac{2^{2}(\lambda-3)(\lambda-1)}{5!}a_{o} \end{equation*}and\begin{equation*} a_{1}=a_{3}=a_{5}=\cdots=0. \end{equation*}
\begin{equation*}
y=a_{o}x\left[1-\frac{2(\lambda-1)}{3!}x^{2}+\frac{2^{2}(\lambda-1)(\lambda-3)}{5!}x^{4}-\cdots\right.
\end{equation*}
\begin{equation}
\left.+\frac{(-2)^{r}(\lambda-1)(\lambda-3)\cdots(\lambda-2r+1)}{(2r+1)!}x^{2r}+\cdots\right] =y_{2} \hspace{2pt}(say)\tag{4.9.9}
\end{equation}
Equation (4.9.9) is a part of solution given in equation (4.9.7), which is not possible, because solutions (4.9.7) and (4.9.9) are two different solutions of the same equation. Hence \(a_{1}\) must be zero in solution (4.9.7).
\begin{equation*}
\therefore y=a_{o}x\left[1-\frac{2\lambda}{2!}x^{2}+\frac{2^{2}\lambda(\lambda-2)}{4!}x^{4}-\cdots\right.
\end{equation*}
\begin{equation}
\left.+\frac{(-2)^{r}\lambda(\lambda-2)\cdots(\lambda-2r+2)}{(2r)!}x^{2r}+\cdots\right]=y_{1}\hspace{2pt} \text{(say)}\tag{4.9.10}
\end{equation}
The general solution of equation (4.9.1) is then given by
\begin{equation}
y=Ay_{1}+By_{2}\tag{4.9.11}
\end{equation}
Where \(A\) and \(B\) are arbitrary constants.
