Consider the eigenvalue problem in with the DE is
and the boundary conditions are
Can Green's Formula be used to show all the eigenvalues are real?
Not sure if this is correct and what to do next.
Your "Green's Formula" looks like the typical inner product used to show that a self-adjoint operator has real eigenvalues. You might want to double-check that formula, though, as you seem to be assuming that the adjoint of is That's what you'd be wanting to prove! I would try to show that the operator , acting on any function satisfying your boundary conditions, is a self-adjoint operator. Then you know that the eigenvalues are real.
Alternatively, you could go the long route: find the eigenvalues, and get a condition on them that must be true in order not to have trivial solutions (since eigenvectors aren't allowed, by definition, to be zero), and show that that condition implies the eigenvalues are real.
Those are my 3/4 baked thoughts.
Here's how I would do it, actually: take a look at this link. You can see that for your BVP, the function vanishes at the endpoints, so this formal definition of the adjoint works. Make sure to look at the example (Stürm-Liouville operator) to get a feel for how to do it. Your problem is considerably simpler.
Well, you want to form the inner product
where are eigenvectors corresponding to different eigenvalues, and you want to show that that inner product is zero. I think the usual procedure here is to look at the expression
The expression on the left looks like your Green's formula, so I think you might be able to show that it's zero. And then, since the eigenvalues are different (and real), I think your result will follow. You might need the self-adjoint-ness of the operator, though.