# Weird eigensystem of ODE with regular singularity

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?

• 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. – LutzL Dec 25 '18 at 14:57
• @LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me. – xiaohuamao Dec 25 '18 at 16:40