Numerical methods for integral eigenvalue equation

Question

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?

Answer 1

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 $\xi$.