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.
1 Answer
It appears that what you need is the tensor Bezoutian for operator polynomials. Its definition and relation to the counting of the common eigenvalues is briefly reviewed in the following article (Theorem 9), where also references are given for further details:
Lancaster, Peter, Common eigenvalues, divisors, and multiples of matrix polynomials: A review, Linear Algebra Appl. 84, 139160 (1986). ZBL0627.15004.
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).

$\begingroup$ Thank you @Igor Khavkine for this great answer. I've taken a look at this paper, I found that it deals with bounded operator case. I 'm not sure that this can be generalized to unbounded operator case. Thank you in advance to clarify more things. $\endgroup$– GustaveMar 13, 2020 at 18:33

1$\begingroup$ @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. $\endgroup$ Mar 13, 2020 at 22:41

$\begingroup$ Thank you sir for the answer. Best regards. $\endgroup$– GustaveMar 15, 2020 at 17:28