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

  • 1
    $\begingroup$ Welcome to Math.SE. Here, we expect that people who posts questions like that can give us some details of what they have tried to do and what exactly is stopping them from solving the problem. $\endgroup$ – Ron Gordon May 8 '13 at 13:05
  • $\begingroup$ @John, I guess you have just attempted an edit of your question. If this is the case, please do not do it as anonymous, but rather after having logged in. $\endgroup$ – Andreas Caranti May 8 '13 at 13:27

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 $\hat{H}$.

  • $\begingroup$ Perhaps he meant all eigenvalues are positive? $\endgroup$ – TCL May 8 '13 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.