Find inverse of $A(x) = x(t) + \int_{-\pi}^\pi x(t-s) \sin^2(2s)\,ds$ in $L_{2}[0, 2\pi]$

Solution 1:

Any $x\in L^2(0, 2\pi)$ can be decomposed as a Fourier series: $$x(t)=\sum_{n\in \mathbb Z} c_ke^{ikt}$$ Let's look at the action of $A$ on the Fourier basis. Using, for $k\geq 1$, $$\int_{-\pi}^\pi e^{k(t-s)}\sin^2(2s)ds = \left \{ \begin{split} &\pi &\text{ if } k=0\\ &-\frac{\pi}2e^{ikt} &\text{ if } k=\pm 4\\ &0 &\text{ otherwise} \end{split} \right.\tag{1}$$ we obtain $$Ax(t)=x(t)+\pi c_0 -\frac{\pi} 2 \left( c_4e^{4it} +c_4e^{-4it}\right)$$ So if you visualize $A$ as being an infinite matrix (its action on the Fourier basis), you have a diagonal matrix. The eigenvectors are the Fourier basis, the eigenvalues are all $1$, except for $\{1, e^{4it}, e^{-4it}\}$, for which the eigenvalues are $\{1+\pi, 1-\frac \pi 2, 1-\frac \pi 2\}$.

Inverting a diagonal matrix is easy. In this case, for any $y\in L^2(0,2\pi)$ with the following Fourier decomposition $y(t)=\sum_{k\in\mathbb Z}d_ke^{ikt}$, $$\begin{split} A^{-1} y(t)&=\left(\sum_{k\neq \{0,4, -4\}}d_ke^{ikt}\right)+\frac 1 {1+\pi}d_0 +\frac 1 {1-\frac \pi 2}d_4e^{4it} +\frac 1 {1-\frac \pi 2}d_{-4}e^{-4it}\\ &= \left(\sum_{k\in\mathbb Z}d_ke^{ikt}\right)-\frac \pi {1+\pi}d_0 + \frac 1{1-\frac\pi 2}\frac{\pi} 2d_4e^{4it} +\frac 1{1-\frac\pi 2}\frac{\pi} 2d_{-4}e^{-4it}\\ &=y(t)-\frac \pi {1+\pi}d_0-\frac 1 {1-\frac \pi 2}\int_{-\pi}^{\pi}y(t-s)\sin^2(2s)ds+\frac{\pi}{1-\frac \pi 2}d_0 \end{split}$$ where we have conveniently used $(1)$ again to rewrite the $d_4$ and $d_{-4}$ terms a convolution with $y$. Note that this has generated an extra $d_0$ term, which we can now group with the one we already had.

Finally, noting that $d_0=\frac 1 {2\pi}\int_{-\pi}^{\pi}y(s)ds$, we conclude that $$A^{-1}y(t) = y(t)+\frac{6\pi^2}{(1+\pi)(2-\pi)}\int_{-\pi}^{\pi}y(s)ds -\frac 1{1-\frac \pi 2}\int_{-\pi}^{\pi}y(t-s)\sin^2(2s)ds$$ I'm not great at computing, so you might spot a few mistakes here and there. But you get the general idea.