I would like to solve next problem: Show that all the eigenfunctions of the Sturm-Liouville problem are positive:
$$u''+(\lambda-x^2)u=0$$ $$0<x<∞$$
$$u'(0)=\lim_{x \to \infty}u(x)=0$$
Any hint?
The problem is from the book An introduction to partial differential equations (Pinchover, Rubinstein) and it says eigenfunctions not eigenvalues. I know that Rayleigh quotient can be used to show that all the eigenvalues are positive. Thanks
The statement of the problem is incorrect as it stands. Any even eigenfunction of the quantum harmonic oscillator (with the hamiltonian $\hat{H}=-\frac{d^2}{dx^2}+x^2$) is an eigenfunction of the problem you describe.
More explicitly, we know that $$ \hat{H}\psi_n=\lambda_n\psi_n$$ where $\lambda_n=2n+1$, $\psi_n(x)=e^{-x^2/2}H_n(x)$ and $H_n(x)$ denotes the $n$th Hermite polynomial. If $n$ is even, then $\psi_n(x)$ automatically satisfies all conditions of your problem. But $H_{2k}(x)$ has exactly $k$ simple zeros on $(0,\infty)$, so we have a contradiction. The simplest counterexample is $u(x)=e^{-x^2/2}(2x^2-1)$.
Maybe you meant eigenvalues instead of eigenfunctions? Then the idea is to consider the quantity $$\int_0^{\infty}\psi(x)(\hat{H}\psi)(x)\,dx=\int_{0}^{\infty}\left(\psi'(x)^2+x^2\psi(x)^2\right)dx>0$$ and look at what happens with the left side when $\psi(x)$ is an eigenfunction of $\hat{H}$.