I'm looking for a solution to the following problem -

$\int_{-\infty}^{\infty} K(x-y) f(y) = \lambda f(x)$

Consider $K(x-y) = \left\{ \begin{array}{lr} e^{-(x-y)} & : x > y \\ 0 & : x < y \end{array} \right.$

In this case, we can actually find the eigenvalues and eigenvectors analytically, since $f(x) = e^{i \omega x}$ is an eigenvector:

$\int_{-\infty}^x e^{-(x-y)} e^{i \omega y} = \frac{e^{i \omega x}}{1 + i \omega}$

But now, if I try to do this numerically, i.e., if I discretize the integral: $\tilde{K}(x_i,y_j) = K(x_i - y_j)$, then I end up with a triangular matrix with 1's on the diagonal, and all the eigenvalues are thus 1.

I don't really see how to resolve this issue numerically, so any help would be much appreciated.

  • $\begingroup$ What quadrature rule are you using? $\endgroup$ Commented Jun 17, 2015 at 23:54
  • $\begingroup$ I just use the trapezoidal rule with a large cut-off value. But I don't see how that would make a difference since the matrix is triangular and so without even implementing a numerical routine, the eigenvalues would all seem to be 1 $\endgroup$
    – Aegon
    Commented Jun 17, 2015 at 23:59
  • $\begingroup$ If you are using a cutoff value, then you cannot choose discretization points near that cutoff value, because then the approximation will not be good. So your matrix should not be triangular in the traditional sense. $\endgroup$ Commented Jun 18, 2015 at 0:15
  • $\begingroup$ To be more explicit: There will be a band of 1s, but it should lie above the diagonal, not on it. $\endgroup$ Commented Jun 18, 2015 at 0:21
  • 1
    $\begingroup$ You could try discretizing with respect to a basis for which each integral can be computed exactly. You should also check if the operator is normal; if not then the eigenproblem is numerically unstable. $\endgroup$ Commented Jun 18, 2015 at 0:34


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