Prove that all the eigenvalues of the following Sturm–Liouville problem are positive: $u'' + (\lambda -x^2)u = 0$ $\hspace{0.5cm}$ $0< x < \infty$ Prove that all the eigenvalues of the following Sturm–Liouville problem are
positive
$u'' + (\lambda -x^2)u = 0$ $\hspace{0.5cm}$ $0< x < \infty$
$u'(0) = \lim_{x \to \infty}{u(x)} =0$
I'm trying to solve this Sturm-Liouville problem. But I haven't gotten a solution yet.
I tried to solve it analogously to what I did in the following problem:
$v'' + \lambda v= 0$ $\hspace{0.5cm} 0<x<L$
$v(0) = v(\pi) = 0$
In this case you can show that if $\lambda = 0$, then $v(x) = 0$. Then $\lambda = 0$ is not an eigenvalue of the problem. If $\lambda \neq 0$, then the general solution to the problem is $v(x) = Ae^{i\sqrt{\lambda}x} + Be^{-i\sqrt{\lambda}x}$. This solution satisfies the conditions of the problem if and only if \begin{cases} A +B =0\\ Ae^{i\sqrt{\lambda}\pi} + Be^{-i\sqrt{\lambda}\pi}=0 \end{cases} This system has a non-trivial solution if and only if $e^{-i\sqrt{\lambda}\pi} - e^{i\sqrt{\lambda}\pi}=0$. From this it is obtained that the eigenvalues are $\lambda_n = n^2$ and since $B=-A$, the eigenfunctions are $v_n{(x)} = 2iA\sin(nx)$.
However, my problem here is different in the sense that the coefficient $\lambda -x^2$ is variable by $x$.
I think that if $\lambda<0$, let's say $\lambda = -k^2$, then $u(x) = C_{1} e^{(\sqrt{k^2 + x^2})x} + C_{2}e^{-(\sqrt{k^2 + x^2})x}$ 
If $\lambda = 0$, then $u'' -x^2 u = 0 \Rightarrow  u(x) =  C_{1} e^{(x)x} + C_{2}e^{-(x)x}$
If $\lambda> 0$, let's say $\lambda = k^2$, then $u(x) = C_{1} e^{(\sqrt{x^2 - k^2})x} + C_{2}e^{-(\sqrt{x^2 - k^2})x}$
Now I'm going to consider these cases and see which of them is possible, however I do not know if this form is correct.
How can I solve this problem?
I need some help or a way to solve this problem.
I would like to know a solution.
Any help is appreciated
 A: The solutions you give unfortunately do not work. Just plug them in the equation and use the chain and product rule and you will see.
The problem is asking to prove that the eigenvalues are all positive, not necessarily to solve the problem.
Suppose $\lambda=\lambda_0\leq 0$ is an eigenvalue with eigenvector $u=u_0$.
Then
$$u_0'' = (x^2-\lambda_0)u_0,\;\;u_0'(0)=0.\;\;\;(1)$$
Assume $u_0(0)>0$, then (1) tells us $u''_0(0)>0$ and $u_0'$ and $u_0$ are increasing functions on $(0,\infty)$ and $u_0$ cannot limit to zero as $x\rightarrow\infty$.
Assume $u_0(0)<0$, then (1) tells us $u''_0(0)<0$ and $u_0'$ and $u_0$ are decreasing functions on $(0,\infty)$ and $u_0$ cannot limit to zero as $x\rightarrow\infty$.
So $u_0$ cannot be an eigenvector.
A: The operator $Lf = -f''+x^2f$ is the Hermite operator, and it is associated with the Quantum Harmonic Oscillator. You can show that, if $f,g$ are twice continuously differentiable with $f, Lf, g, Lg \in L^2[0,\infty)$, then
$$
       \langle Lf,g\rangle = \langle f,Lg\rangle+(-f'\overline{g}+f\overline{g}')|_{x=0}^{\infty}
$$
where $\langle\cdot,\cdot\rangle$ is the $L^2[0,\infty)$ inner product given by
$$\langle h,k\rangle = \int_{0}^{\infty}h(x)\overline{k(x)}dx.$$
Furthermore, it is automatic that the evaluation terms at $\infty$ vanish.
The condition that every $f\in \mathcal{D}(L)$ also satisfies $f'(0)=0$ is then enough to force the symmetry condition $\langle Lf,g\rangle=\langle f,Lg\rangle$ for all $f,g\in\mathcal{D}(L)$. The eigenfunctions are the even order Hermite functions $h_n(x)$ where $n=0,2,4,6,\cdots,$ because of the condition $f'(0)=0$.
