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I would like to compute the eigenvalues and eigenfunctions for the Fredholm integral operator

$$K(g) = \int_0^1 e^{xt} g(t) \,dt.$$

The sources I've checked* seem to say that the process is fairly involved. Has anything been published on this kernel? Or, if not, am I correct that it's going to be a hard thing to do?

* See, e.g., equations (12) and on here: https://www.encyclopediaofmath.org/index.php/Fredholm_equation

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  • $\begingroup$ $k$ is operating on which space? $\endgroup$ – Martin Argerami Feb 13 '12 at 5:16
  • $\begingroup$ @MartinArgerami: $L^2(\mathbb{R})$. $\endgroup$ – Antonio Vargas Feb 13 '12 at 5:33
  • $\begingroup$ In that case I would expect your case to be harder than those in the link you provided: because they require the kernel to be in $L^2$ of the product space, while yours isn't. $\endgroup$ – Martin Argerami Feb 13 '12 at 15:54
  • $\begingroup$ @MartinArgerami: Sorry, I should have been more clear, I meant $L^2([0,1])$. The kernel of the operator in question, $e^{xt}$, is certainly in $L^2([0,1]^2)$, and the operator acts on $L^2([0,1])$. The link definitely applies here. $\endgroup$ – Antonio Vargas Feb 13 '12 at 17:02
  • $\begingroup$ Just to clarify, is the eigenvalue problem defined as $$\int_0^1 K(x,t) g(t) dt= \lambda g(x)$$ where in your case $K(x,t)=e^{x t}$? $\endgroup$ – Yuriy S Oct 19 '18 at 13:10
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We need to solve the eigenproblem:

$$\int_0^1 K(x,t) g(t) dt= \lambda g(x)\tag{1}$$

While I doubt that there's an exact solution in terms of known function, we can find an approximate one by exploiting the method described here for a certain class of kernels:

$$K(x,t)=\sum_{j=0}^n \phi_j(x) \psi_j(t)$$

The approximation is achieved by Taylor expansion of the exponential kernel:

$$e^{xt} \approx \sum_{j=0}^n \frac{x^j t^j}{j!}$$

From symmetry considerations, let us set:

$$ \phi_j(x)=\frac{x^j}{\sqrt{j!}} \\ \psi_j(t)=\frac{t^j}{\sqrt{j!}}$$

We need to compute:

$$a_{jk}=\int_0^1 \psi_j(t) \phi_k(t) dt=\frac{1}{\sqrt{j!k!}} \int_0^1 t^{j+k} dt=\frac{1}{(j+k+1)\sqrt{j!k!}}$$

Then the (approximate) eigenvalues are defined from the equation:

$$\det \{ a_{jk} - \lambda \delta_{jk} \}=0$$

In other words, we simply need to diagonalize the matrix $\{a_{jk}\}$.

The eigenfunctions will be:

$$g(x)=\sum_{j=0}^n C_j \phi_j(x)=\sum_{j=0}^n C_j \frac{x^j}{\sqrt{j!}}$$

Where $\{C_j\}$ are eigenvectors of the matrix.

As far as I see, the eigenfunctions for $n \to \infty$ should converge to the exact eigenfunctions, and in case they are analytical, the expansion above defines their Taylor series.


Here's a numerical check in Mathematica. For $n$ as small as $10$ we obtain a perfectly looking solution (the blue curve is the left hand side of the equation (1) and the orange dashed curve is the right hand side):

Nm = 10;
A = Table[N[1/(j + k + 1)/Sqrt[j! k!], 100], {j, 0, Nm}, {k, 0, Nm}];
{ln, Cn} = N[Eigensystem[A], 20];
g[n_, x_] := Sum[Cn[[n]][[j + 1]]/Sqrt[j!] x^j, {j, 0, Nm}];
Plot[{NIntegrate[Exp[x t] g[1, t], {t, 0, 1}], ln[[1]] g[1, x]}, {x, 
  0, 1}, PlotStyle -> {Thick, Dashed}]

enter image description here

And two more eigenfunctions:

enter image description here

enter image description here

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