1
$\begingroup$

Find the eigenvalues and eigenfunctions for the following kernel: $$K(x,t)=-2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right) \in \mathcal{L}_2([0,2 \pi ]^2).$$

What I have: $$y(x)=\int_0^{2 \pi} -2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right) y(t)dt = \\ -2\int_0^{2\pi} \left( \sum_{n=1}^\infty \frac{\sin(nx) \sin(nt)}{n}+\sum_{n=1}^\infty \frac{\cos(nx) \cos(nt)}{n} \right)y(t)dt$$ by Fourier series. It sort of seems that eigenfunctions should be in the form $y(x)=A\cos(nx)+B\sin(nx)$ but I don't have any arguments for that. I know that the eigenvalues should be $\lambda_n=- \frac{n}{2 \pi}$.

Also I've found a similar question (Eigenvalues of an integral operator with non-degenerate kernel) but its kernel is much simpler. Not sure if the same technique would work in my case. What corresponding differential equation could I get from this: $y''(x)+\lambda y(x)=0$, $y(0)=y(2 \pi)=0$? However I wouldn't get the same eigenvalues $\lambda_n=- \frac{n}{2 \pi}$. Help?

$\endgroup$

1 Answer 1

2
$\begingroup$

It is a convolution operator, convolution in $L^2(\Bbb{R/2\pi Z})$, ie. $f$ is $L^2_{loc}$ and $2\pi$-periodic and $$(Tf)(x)= \int_0^{2\pi} f(t) h(x-t)dt, \qquad h(x)=-2 \log \left( \sin \left( \frac{1}{2} \left( x \right) \right) \right)$$

The complex exponentials $e^{i n x},n\in \Bbb{Z}$ are an orthonormal basis of eigenfunctions with eigenvalues $\lambda_n=(T e^{-in x})(0)$, such that $$h = \frac1{2\pi}\sum_n \lambda_n e^{inx}$$ We find the $\lambda_n$ from $$h = \lim_{r\to 1^-}\Re(-2 \log \left(\frac{e^{ix/2}}{2i} (1-r e^{-ix})\right))$$ (convergence in $L^2(\Bbb{R/2\pi Z})$) and the Taylor series of $-\log(1-z)$

Decompose $$f = \sum_n c_n e^{inx}, \qquad Tf=\sum_n \lambda_n c_n e^{inx}$$ (convergence in $L^2(\Bbb{R/2\pi Z})$)

Then $Tf= \lambda f$ iff $c_n\ne 0 \implies\lambda_n=\lambda$.


Since we are keeping only the real part $h =\lim_{r\to 1^-}\Re(-2\log \left(1-r e^{-ix}\right))$ then from the Taylor series of $-\log(1-z)$ we have for $r\in (0,1)$ $$\Re(-2\log \left(1-r e^{-ix}\right))= \Re(2\sum_{n\ge 1} r^n e^{-inx}/n)=\sum_{n\ne 0} r^n e^{inx}/|n|$$ and $$h = \lim_{r\to 1^-} \sum_{n\ne 0} r^n e^{inx}/|n|= \sum_{n\ne 0} e^{inx}/|n|$$ which converges without problem in $L^2[0,2\pi]$. So the $\lambda_n$ are $0$ and $2\pi/|n|$ and the eigenfunctions are the linear combination of complex exponentials with the same eigenvalue ie. $ae^{inx}+be^{-inx}$.

$\endgroup$
7
  • $\begingroup$ Sorry but I don't understand how all of this could help me calculate the eigenvalues $\lambda_n$ and eigenfunctions $y_n$. $\endgroup$
    – Karagum
    Dec 26, 2020 at 16:26
  • $\begingroup$ The Fourier series of $h$ is found from the Taylor series of $-\log(1-z)$. I explained the necessary and sufficient condition for $f$ to be an eigenfunction. $\endgroup$
    – reuns
    Dec 26, 2020 at 16:28
  • $\begingroup$ It all looks like general theoretical ideas that I don't find clear. $\endgroup$
    – Karagum
    Dec 27, 2020 at 16:03
  • $\begingroup$ I'm still waiting for explanations. It is totally unclear what you did in your solution. Where you take the limit as $r \to 1^{-}$, the limit doesn't mean a lot. Anyway, I'd like to discuss on your solution since I've been trying to understand your solution and solve it by myself for over a week. :) $\endgroup$
    – Karagum
    Dec 29, 2020 at 16:47
  • $\begingroup$ Since we are keeping only the real part $h =\lim_{r\to 1^-}\Re(-2\log \left(1-r e^{-ix}\right))$ then from the Taylor series of $-\log(1-z)$ we have for $r\in (0,1)$ $$\Re(-2\log \left(1-r e^{-ix}\right))= \Re(2\sum_{n\ge 1} r^n e^{-inx}/n)=\sum_{n\ne 0} r^n e^{inx}/|n|$$ and $$h = \lim_{r\to 1^-} \sum_{n\ne 0} r^n e^{inx}/|n|= \sum_{n\ne 0} e^{inx}/|n|$$ which converges without problem in $L^2[0,2\pi]$. So the $\lambda_n$ are $0$ and $1/|n|$ and the eigenfunctions are the linear combination of complex exponentials with the same eigenvalue ie. $ae^{inx}+be^{-inx}$. @Karagum $\endgroup$
    – reuns
    Dec 29, 2020 at 16:52

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.