# $K(x,t)=-2 \log \left( \sin \left( \frac{1}{2} \left( x-t \right) \right) \right)$ eigenvalues and eigenfunctions

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?

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. $$ae^{inx}+be^{-inx}$$.

• Sorry but I don't understand how all of this could help me calculate the eigenvalues $\lambda_n$ and eigenfunctions $y_n$. Dec 26, 2020 at 16:26
• 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. Dec 26, 2020 at 16:28
• It all looks like general theoretical ideas that I don't find clear. Dec 27, 2020 at 16:03
• 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. :) Dec 29, 2020 at 16:47
• 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 Dec 29, 2020 at 16:52