sturm-liouville with negative weight function

Question

Suppose the following SL problem

$$ y''(t) + \lambda w(t)y(t)=0 \\ y(0)=0, \qquad y'(1)=0 $$ with $w\in C([0,1])$ a weight function.

If we take $w=f$, for a certain $f\in C([0,1]), f>0$, it is well-known that all the eigenvalues are positive and ordered.
If now we take $w=-f$, can we infer that

1. the eigenvalues are all negative;
2. the eigenvalues are those for the choice $w=f$, but with opposite sign?

Can you also provide references for this?

Thank you

Answer 1

Suppose that $$ \int_0^1 y''(t)y(t)dt+\lambda \int_0^{1}w(t)y(t)^2dt=0. $$ Then, $$ \left.y'(t)y(t)\right|_{0}^{1}-\int_0^1y'(t)^2dt+\lambda\int_0^1w(t)y(t)^2dt=0 \\ \lambda\int_0^1w(t)y(t)^2dt=\int_0^1y'(t)^2dt \\ \lambda=\frac{\int_0^1y'(t)^2dt}{\int_{0}^{1}w(t)y(t)^2dt}. $$ If $w < 0$, then $\lambda < 0$.

Try looking at $w\equiv 1$ and at $w\equiv -1$.