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Aug 22, 2021 at 22:22 comment added nicoguaro Yes, that's a good approach.
Aug 22, 2021 at 22:12 comment added fibonatic @nicoguaro Can't one combine this with discretizing the problem? Discretizing does allow one to solve it using eigenvalues, but discretizing always introduces some errors (especially for the bigger magnitude eigenvalues). For example using a uniform grid of 50 points in $z$ for the same example as in my answer yields $-0.65120$, $-0.35716$, $0.13156$, $0.81296$, $1.6842$, $2.7418$, $3.9813$, $5.3977$, $6.9851$ and $8.7370$. But one could use these obtained values as initial guesses for the root finding, which I think would have better convergence compared to increasing grid size.
Aug 22, 2021 at 21:03 history edited fibonatic CC BY-SA 4.0
Added example numerical calculation results
Aug 22, 2021 at 4:36 comment added nicoguaro But, turning to a root solving problem is commonly a bad idea. Indeed, that problem is sometimes solved by eigenvalues.
Aug 22, 2021 at 3:56 comment added fibonatic @nicoguaro I know, that it why one should look for the roots of the norm of the LHS of $(3)$ (as a function of $\lambda$).
Aug 21, 2021 at 23:39 comment added nicoguaro But $\lambda$ is not known.
Aug 21, 2021 at 23:36 history answered fibonatic CC BY-SA 4.0