I have an integral equation which is not exactly an eigenvalue type equation, but similar:


$$\int d^3 y\, K(x,y,\lambda) f(y) = f(x)$$


Here $\lambda$ can be thought of as an eigenvalue, so it is expected that only for a discrete set of $\lambda$ we have a solution. $K(x,y,\lambda)$ is some fairly simple expression involving ratios of polynomials and also $\lambda$ appears in a rational fashion.


What would be the typical numerical methods to find the lowest eigenvalue $\lambda$ and eigenfunction $f(x)$?


Actually, we can formulate the question generally: how does one find the eigenfunction corresponding to zero eigenvalue of a linear integral operator that depends on some $\lambda$? The task involves finding the particular $\lambda$ for which the operator has a zero eigenvalue to begin with.


$${\cal O}(\lambda) v = 0$$


Generally ${\cal O}(\lambda)$ doesn't have zero eigenvalues, but for a discrete set of $\lambda$ it does. How can I find it numerically? Probably an iterative procedure would be most cost effective. Is this true?



Numerical methods for integral eigenvalue equation
To solve $\int d^3 y\, K(x,y,\lambda) f(y) = f(x)$ I would discretize the coordinates $x\mapsto x\_n$, $y\mapsto y\_m$, $K(x,y,\lambda)\mapsto K(x\_n,y\_m,\lambda)\equiv K\_{nm}(\lambda)$ and solve the determinant equation
$$\text{det}\,[I-K\_{nm}(\lambda)]=0$$
for $\lambda$, when the indices $n,m$ vary over a finite range.


Things would simplify greatly if $K(x,y,\lambda)$ depends only on the difference $x-y$, because then you could Fourier transform and find a $\lambda$ such that $\hat{K}(\xi,\lambda)=1$ for some $\xi$.


