Let $T$ be an integral operator with kernel $K(x,y)=e^{|x-y|}$ on $L^2(-1,1)$. How can we find the eigenfunctions and eigenvalues of $T$?

Even though I am not sure whether the following arguments are correct, here is what I have done :

Say $Tf(x) = \displaystyle\int_{-1}^1 e^{|x-y|}f(y)dy.$ When $x-y \gt 0$ we have $Tf(x) = e^x \displaystyle\int_{-1}^1 e^{-y} f(y)dy. $ Suppose that we have $Tf(x) = \lambda f(x)$ for some scalar $ \lambda$.

Evaluating the integral, we have $Tf(x) = e^x (e-e^{-1}) = \lambda f(x)$.

Now, taking derivatives of both sides, $e^x(e-e^{-1}) = \lambda f'(x)$. Thus, we have $\lambda f(x) = \lambda f'(x) = \dots$. This seems like an ODE problem but we need some boundary coundtion and here is where I am stuck.

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    $\begingroup$ As you are not new on this site, you should know that we praise to see personal work mentionned... $\endgroup$ – Jean Marie Apr 19 at 6:05
  • $\begingroup$ You should distinguish within the integral whether $y>x$ or $y<x$. That is, split the integral. And why have you changed the kernel? Before it was $|y-x|$. $\endgroup$ – amsmath Apr 19 at 6:38
  • $\begingroup$ Thanks for providing this work : your problem is that you haven't well converted your kernel operator : you cannot say "when $x-y>0$" and leave aside the alternative case. You have to tackle both cases altogether. Besides, the idea to transform your issue into a differential equation is a very good track. The inital conditions are in fact "hidden" in conditions e.g., in the bounds of integrals. $\endgroup$ – Jean Marie Apr 19 at 6:50
  • $\begingroup$ How can I consider $2$ cases altogether? $\endgroup$ – Ninja Apr 19 at 8:43
  • $\begingroup$ I wasn't aware of your question because you hadn't prefixed it by arrowbase followed by my pseudo. I will write it in an "answer"... which is not an answer because it takes at once a certain volume. $\endgroup$ – Jean Marie Apr 19 at 9:12

This is not an answer : Just some calculations for the asker I couldn't place into a comment.

Clearly, considering the two cases

  • $x-y>0 \ \iff \ y<x$

  • $x-y<0 \ \iff \ y>x$

the integral can be split in the following way :

$$(Tf)(x)=e^{x} \underbrace{\int_{y=-1}^{y=x} e^{-y}f(y)dy}_{f_1(x)} \ + \ e^{-x} \underbrace{\int_{y=x}^{y=1} e^{y}f(y)dy}_{f_2(x)}\tag{1}$$

In particular :

$$(f_1)'(x))=e^{-x}f(x) \ \ \text{and} \ \ (f_2)'(x))=-e^{x}f(x).\tag{2}$$

We have

$$(Tf)'(x) \ = \ e^{x}(f_1(x)+f'_1(x))+e^{-x}(-f_2(x)+f'_2(x))\tag{3}$$

Taking (2) into account, one gets :


i.e., by cancellation :


Differentiating (5) gives :



$$(Tf)''(x)=e^{x}(f_1(x)+e^{-x}f(x) )-e^{-x}(-f_2(x)-e^{x}f(x))\tag{7}$$

$$(Tf)''(x)=\underbrace{e^{x}f_1(x)+e^{-x}f_2(x)}_{Tf(x)}+2 f(x)\tag{8}$$

Knowing that $(Tf)(x)=\lambda f(x)$ (eigenvalue/eigenfunction), can you take it from here (classical second order differential equation) ?

  • $\begingroup$ Thank you for your effort. How can we use the boundary counditions? $\endgroup$ – Ninja Apr 19 at 10:41
  • $\begingroup$ What boundary conditions are you talking about? Just plug your two-parameter solution of the differential equation into the original equation $Tf = \lambda f$ and see for which values of the parameters this is true. $\endgroup$ – amsmath Apr 22 at 15:16

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