Consider the eigenvalue problem of the following 2nd-order ODE $$(x/2+a)^2y(x)-xy'(x)-x^2y''(x)=\lambda^2y(x),$$ in which $y\in(-\infty,+\infty)$ and parameter $a>0$. It has a regular singularity $x=0$ and an irregular singularity at $\infty$. This ODE is from some physical modelling and we hence basically hope for something like Dirichlet b.c. at $\pm\infty$.

It is solved by making the substitution $y(x)=e^{x/2}x^{\sqrt{a^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation of $u(x)$ that has two (1st kind & 2nd kind) independent solutions. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial and eigenvalue $\lambda$ is attained.

However, it's obvious that $y(x)$ diverges at $+\infty$ because of $e^{x/2}$, which doesn't fulfill the b.c. we wanted. But $y(0)=0$ is automatically satisfied, which seems that we can instead use $Y(x)=y(x)\theta(-x)$ as the eigenfunction.

My question is whether this $C^0$ continuous solution $Y(x)$ is a valid eigenfunction. Or I missed any other solution? And is it normal to have such an eigenfunction 'cut off' by a singularity?

  • $\begingroup$ So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=\sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues. $\endgroup$ – LutzL Dec 25 '18 at 14:57
  • $\begingroup$ @LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me. $\endgroup$ – xiaohuamao Dec 25 '18 at 16:40

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