Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$.
Let's follow the ubiquitous argument when solving eigensystem related to 1st kind & 2nd kind solutions to this equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained.

However, seen from this condition for $\alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that SL theory claims.

What is wrong here? Am I missing some solutions or else?

Consider the following Sturm–Liouville (SL) eigenvalue problem defined in $(-\infty,0]$ or $[0,\infty)$ or $(-\infty,+\infty)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading to Kummer's equation with two independent solutions (1st & 2nd kind) $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$.
Let's then follow the ubiquitous argument. 

Requiring nondivergence at $x=0$, the 2nd kind solution is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained. Seen from this condition for $\alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum. But I'm not sure whether it starts from the largest eigenvalue.

Question

Is it possible to know the limit circle/point classification of this ODE near $0,\pm\infty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.

Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained.

However, seen from this condition for $\alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that SL theory claims.

What is wrong here? Am I missing some solutions?

Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$.
Let's follow the ubiquitous argument when solving eigensystem related to 1st kind & 2nd kind solutions to this equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained.

However, seen from this condition for $\alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that SL theory claims.

What is wrong here? Am I missing some solutions or else?

Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'+qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained.

However, seen from this condition for $\alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that Sturm–Liouville theory claims.

What is wrong here? Am I missing some solutions?

Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'-qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=(x/2+a)^2+a$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained.

However, seen from this condition for $\alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that SL theory claims.

What is wrong here? Am I missing some solutions?