Consider the eigenvalue equation for the Hill operator $$L(y)= -y''+ v(x) y = \lambda y, \quad x\in \mathbb{R},$$ where $v(x)$ is any potential and $\lambda$ is the spectral parameter. If $v(x) \equiv 0$, the spectrum of $L$ subject to periodic boundary conditions (bc): $y(0)=y(\pi)$, $ y'(0)=y'(\pi)$ and anti-periodic bc: $ y(0)=-y(\pi)$, $ y'(0)=-y'(\pi)$ coincides with the spectrum subject to Dirichlet bc: $y(0)=y(\pi)=0$; that is $\lambda_n=n^2$, $n \in \mathbb{N}$, if $n$ is even or $n$ is odd, respectively.
My concern is about the inverse problem: If we know that the spectrum with respect to above boundary conditions coincides as above, what can we say about the potential $v(x)$? Does it have to be identically zero?
Thanks!
No. the potential $ v(x) $ can be recovered from the spectral data..
let us suppose we know the energies $ E_{n} $ or the Spectral staircase $ N(E) $
then $$ v^{-1}(x)= 2\sqrt{\pi} \frac{d^{1/2}}{dx^{1/2}}N(x) $$
if the boundary condition $ y(0)=0 $ is included
and $$ v^{-1}(x)= \sqrt{\pi} \frac{d^{1/2}}{dx^{1/2}}N(x)$$
if we asume the potential is even $ v(x)=v-x) $