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

Question

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?

Answer 1

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 $c_n\ne 0 \implies\lambda_n=\lambda$.

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