0
$\begingroup$

I came across this exercise and I am stuck on how to solve it:

Consider a Fredholm Integral of the second kind: $u(x)=f(x)+\lambda\int_{0}^{L}K(x,t)u(t)dt$.

Define the 2nd iterate Kernel $K_2(x,y)=\int_{0}^{L}K(x,t)K(t,s)dt$. If $\lambda$ is an eigenvalue for $K_2$, then we must have that $\sqrt\lambda$ or $-\sqrt{\lambda}$ is an eigenvalue of the original kernel $K(x,t)$. I want to show this, and this seems like a relatively straightforward problem but I am struggling.

I am assuming the Kernel is defined on $L^2([0,L]^2)$.

$\endgroup$
  • 1
    $\begingroup$ Do you mean that if $\int_{0}^{L}K_2(x,t)u(t)dt=\lambda u(t)$, then $\int_{0}^{L}K(x,t)v(t)dt=\pm\sqrt{\lambda}v$ for some choice of $\pm$ and some $v$? $\endgroup$ – DisintegratingByParts Mar 4 '18 at 6:48
  • $\begingroup$ Yes, I suppose that is a better way to write it. $\endgroup$ – Kernel_Dirichlet Mar 5 '18 at 2:16
2
+50
$\begingroup$

For any linear operator $K$ on a complex vector space, if $$ K^2 v = \lambda v, \;\; v \ne 0, $$ one has $(K-\sqrt{\lambda}I)(K+\sqrt{\lambda}I)v=0$. So either $(a) Kv=-\sqrt{\lambda}v$ or (b) $w=(K+\sqrt{\lambda}I)v \ne 0$ and $Kw=\sqrt{\lambda}w$.

$\endgroup$
  • $\begingroup$ Thanks, it was indeed much simpler than I thought! $\endgroup$ – Kernel_Dirichlet Mar 6 '18 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.