# Common eigenvalues for two Sturm-Liouville problem

Does exist in literature any results concerning the common eigenvalues for the two eigenvalue problems of the form $$y''(x)=\lambda^2 y(x)+\lambda a(x)y(x), \ x\in(0,1),$$$$z''(x)=\lambda^2 z(x)-\lambda a(x)z(x), \ x\in(0,1),$$ with boundary conditions $$y(0)=y(1)=z(0)=z(1)=0.$$ Things are trivial if $$a$$ is constant function, but for $$a=a(x)$$ I couldnt find any way to handle this problem. Any suggestions?. Thank you.

• Do you mean the Dirichlet boundary condition (everything $=0$) ? Mar 12, 2020 at 13:11
• @DenisSerre Yes thank you sir. I have edited the post. Mar 12, 2020 at 13:47

Your operator polynomials (in $$\lambda^{-1}$$, to make the constant coefficient exactly equal to $$1$$, as in Theorem 9 above) are $$P_x^\pm(\lambda^{-1}) = 1 \pm \lambda^{-1} a(x) - \lambda^{-2}\partial_x^2$$. Their Bezoutian works out to be the oprator matrix $$B = [B_{ij}]$$, whose operator entries are defined by the identity $$\frac{P^+_{x_1}(\lambda^{-1}) P^-_{x_2}(\mu^{-1}) - P^+_{x_1}(\mu^{-1}) P^-_{x_2}(\lambda^{-1})}{\lambda^{-1} - \mu^{-1}} = \sum_{i=0}^1 \sum_{j=0}^1 \lambda^{-i} \mu^{-j} B_{ij} .$$ In this particular case, we get $$B = \begin{bmatrix} a(x_1) + a(x_2) & \partial_{x_1}^2 - \partial_{x_2}^2 \\ \partial_{x_1}^2 - \partial_{x_2}^2 & a(x_2) \partial_{x_1}^2 + a(x_1) \partial_{x_2}^2 \end{bmatrix} ,$$ acting on $$[\begin{smallmatrix} u(x_1,x_2) \\ v(x_1,x_2) \end{smallmatrix}]$$, with $$u$$ and $$v$$ satisfying Dirichlet boundary conditions on the square $$(x_1,x_2) \in [0,1]^2$$.
According to the theorem in Lancaster's review, the dimension of the kernel of $$B$$ counts the number of common eigenvalues of $$P^\pm(\lambda)$$ (excluding $$\lambda=0$$, I think, but that value is never an eigenvalue under Dirichlet boundary conditions).
• @Gustave, just a remark. If you multiply both of your operator polynomials by $(\partial_x^2)^{-1}$ (with the inverse defined by Dirichlet boundary conditions), then the operator coefficients of the powers of $\lambda$ become bounded. It's not hard to work out the new corresponding $B$ operator matrix. Mar 13, 2020 at 22:41