Problem source

For $n=0,1,2,\ldots,$ the functions $y_{n}$ satisfy the differential equation 
\[
\frac{\mathrm{d}^{2}y_{n}}{\mathrm{d}x^{2}}-\omega^{2}x^{2}y_{n}=-(2n+1)\omega y_{n},
\]
where $\omega$ is a positive constant, and $y_{n}\rightarrow0$ and $\mathrm{d}y_{n}/\mathrm{d}x\rightarrow0$ as $x\rightarrow+\infty$
and as $x\rightarrow-\infty.$ Verify that these conditions are satisfied, for $n=0$ and $n=1,$ by 
\[
y_{0}(x)=\mathrm{e}^{-\lambda x^{2}}\qquad\mbox{ and }\qquad y_{1}(x)=x\mathrm{e}^{-\lambda x^{2}}
\]
for some constant $\lambda,$ to be determined. 
Show that 
\[
\frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right)=2(m-n)\omega y_{m}y_{n},
\]
and deduce that, if $m\neq n,$ 
\[
\int_{-\infty}^{\infty}y_{m}(x)y_{n}(x)\,\mathrm{d}x=0.
\]

Solution source

\begin{align*}
&& y_0(x) &= e^{-\lambda x^2} \\
&& \lim_{x \to \pm \infty} y_0(x) &= 0 \Leftrightarrow \lambda > 0 \\
&& \lim_{x \to \pm \infty} y'_0(x) &= \lim_{x \to \pm \infty} 2x\lambda e^{-\lambda x^2} \\
&&&= 0\Leftrightarrow \lambda > 0 \\
&& y''_0(x) &= 4x^2 \lambda^2 e^{-\lambda x^2} + 2\lambda e^{-\lambda x^2} \\
\\
&& y''_0 - \omega^2 x^2 y_0+(2\cdot 0 + 1) \omega y_0 &= e^{-\lambda x^2} \l 4x^2 \lambda^2 + 2 \lambda - \omega^2 x^2 + \omega\r \\
&&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} 
\end{align*}
Therefore $y_0$ satisfies if $\lambda = \frac{\omega}{2}$

Similarly for $y_1$,
\begin{align*}
&& y_1(x) &= xe^{-\lambda x^2} \\
&& \lim_{x \to \pm \infty} y_1(x) &= 0 \Leftrightarrow \lambda > 0 \\
&& \lim_{x \to \pm \infty} y'_1(x) &= \lim_{x \to \pm \infty} \l -2x^2 \lambda e^{-\lambda x^2} + e^{-\lambda x^2} \r \\
&&&= 0\Leftrightarrow \lambda > 0 \\
&& y''_0(x) &= e^{-\lambda x^2} \l 4x^3 \lambda^2-4x\lambda - 2x\lambda \r \\
&&&= e^{-\lambda x^2} \l 4x^3 \lambda^2-6x\lambda \r 
\\
&& y''_1 - \omega^2 x^2 y_1+(2\cdot 1 + 1) \omega y_1 &= e^{-\lambda x^2} \l 4x^3\lambda^2-6x\lambda-\omega^2x^3+3\omega x\r \\
&&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} 
\end{align*}
Therefore $y_1$ satisfies if $\lambda = \frac{\omega}{2}$

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) &= y'_my'_n+y_my''_n - y'_ny'_m-y_ny''_m \\
&= y_my''_n - y_ny''_m \\
&= y_m(\omega^2 x^2 y_n - (2n+1)\omega y_n) - y_n(\omega^2 x^2 y_m - (2m+1)\omega y_m) \\
&= y_my_n (2m-2n)\omega \\
&= 2(m-n) \omega y_my_n
\end{align*}

Therefore:

\begin{align*}
\int_{-\infty}^{\infty} y_m(x)y_n(x) \d x &=  \int_{-\infty}^{\infty} \frac{1}{2(m-n)} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) \d x \\
&= \frac{1}{2(m-n)} \left [ y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right]_{-\infty}^{\infty} \\
&\to 0
\end{align*}

This condition is known as Orthogonality. In fact this question is talking about a Sturm-Liouville orthogonality condition, in particular for the quantum harmonic oscillator, and the eigenfunctions are related to Hermite polynomials.