Explain why no boundary conditions are needed The Question:
Consider the operator
$$My(x) \equiv \frac{d}{dx}\biggl(a(x) \frac{dy}{dx} \biggr) + b(x)y(x) \; \; \; \; \; \; \; \; , \; \; \; \; \; \; \; \; \alpha<x<\beta $$
Suppose that $a$ satisfies $a(\alpha)=a(\beta)=0$, and we want to solve
$$My(x)=\lambda y(x)$$
I have to explain why no boundary conditions need to be given.
I get how the operator is already in Sturm-Liouville form, so that it is self-adjoint, but I honestly don't get how you can solve an ODE with no boundary conditions.
 A: There are cases where boundary conditions are required, and there are cases where they are not. For example, the ordinary Legendre equation
$$
                          -((1-x^2)y')'=\lambda y
$$
does require conditions. For example, for $\lambda=0$, there are solutions
\begin{align}
                        (1-x^2)y' & = C \\
                       y' & = \frac{C}{1-x^2} = \frac{1}{2}\left( \frac{C}{1-x}+\frac{C}{1+x}\right) \\
               y & = \frac{C}{2}\ln\frac{1+x}{1-x}+D.
\end{align}
And both of these solutions are in $L^2[-1,1]$. Requiring boundedness of the solutions at $x=\pm 1$ is the classical condition leading to a well-posed selfadjoint problem. That leads to $C=0$. Without that, the problem is not selfadjoint.
On the other hand, the Associated Legendre equation given below does not require any endpoint conditions for $m=1,2,3,4,\cdots$:
$$
     -\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\frac{m^2}{1-x^2}y=\lambda y.
$$
In this case, the problem is selfadjoint with no enpoint conditions; all that is required is that the solutions be in $L^2[-1,1]$.
The distinction between these cases is that the first equation is in the limit circle case at both endpoints, and the second equation is in the limit point case at both endpoints. This classification determines whether or not some kind of condition will be required at either or both endpoints in order to achieve a selfadjoint problem in $L^2$.
