Proving that a Sturm-Liouville problem is in the limit-point/-circle case I would like to understand techniques anybody is able to detail to me on how one may actually prove that a particular Sturm-Liouville (S-L) problem, i.e., of the form
\begin{equation}
L[y](x):=-\frac{d}{dx}\big(p(x)\frac{dy}{dx}\big)+q(x)y(x)=\lambda w(x)y(x)
\end{equation}
on the interval $[0,\infty)$ and with boundary condition of the form $y(0)\cos(\alpha)-(py')(0)\sin(\alpha)=0$, is in the limit-point (l.p.) or limit-circle (l.c.) case at $\infty$. For example, if $p(x)=w(x)=1/(x+1)=1/q(x)$ and $\alpha=0$, how might one set about proving that $L$ is l.p. or l.c.?
I have tried considering the following: suppose $\{u,v\}$ is a fundamental system for $L[y]=\lambda y$ on $[0,\infty)$ satisfying
\begin{equation}
u(0)=(pv')(0)=\cos(\alpha), (pu')(0)=-v(0)=-\sin(\alpha),
\end{equation}
so that for any $b>0$, $0\leq\beta<2\pi$ and $\lambda\in\mathbb{C}\backslash\mathbb{R}$ there is unique $m=m_{b,\beta}(\lambda)\in\mathbb{C}$ such that $y=u+mv$ satisfies
\begin{equation}
y(b)\cos(\beta)-(py')(b)\sin(\beta)=0.
\end{equation}
Then $m$ is on a circle of radius $(2$Im$\lambda\int_0^bw|v|^2)^{-1}$; if in the limit $b\rightarrow\infty$ this radius becomes $0$ then we are in l.p., and if it has non-zero limit then we are in l.c.. So the relevant behaviour of the S-L problem is governed by whether or not this solution $v$ is globally in the weighted space $L^2(0,\infty;w)$. However I have no idea how to use the structure of such a particular S-L problem as that given above, to determine how this $v$ behaves. Does anybody have any ideas? Should I try to calculate exactly what the solution $v$ is, or can I simply use some trick just to determine its (non-)global square integrability? Or is there indeed a completely alternative way to proceed regarding this? Thanks in advance.
 A: You are looking at the following eigenvalue problem on $[0,\infty)$:
$$
        Lf= - \frac{d}{dx}\frac{1}{1+x}\frac{df}{dx}+(x+1)f=\lambda\frac{1}{x+1}f
$$
If you can explicitly solve for any particular $\lambda$, then you can determine if the equation is in the limit point or limit circle case at $\infty$. If you are in the limit circle case, then all eigenfunctions for all eigenvalues will be in $L^{2}_{1/(1+x)}[0,\infty)$. If some eigenfunction isn't, then you're in the limit point case. It's often convenient to look at $\lambda=0$. Consider the case where $\lambda = 0$ and make the substitution
$$
             f(x) = F(\frac{1}{2}(1+x)^{2}).
$$
Then you want to solve
$$
         -\frac{1}{1+x}\frac{d}{dx}\frac{1}{1+x}\frac{df}{dx}+f = 0, \\
                -F''((1+x)^{2}/2)+F((1+x)^{2}/2) = 0 \\
               F(s) = Ae^{s}+Be^{-s} \\
               f(x)=Ae^{(1+x)^{2}/2}+Be^{-(1+x)^{2}/2}
$$
You can check that this is the general solution for $\lambda=0$. Only the solution where $A=0$ is in $L^{2}_{1/(1+x)}[0,\infty)$ near $\infty$. Therefore $L$ is in the limit point case, because it cannot be in the limit circle case or all eigenfunctions would be in $L^{2}_{1/(1+x)}[0,\infty)$.
