Solution $y(x)$ for $\sin(x) = \int_0^{2\pi} \max(y(t), y(x+t)) dt$

I've been banging my head against a wall for a few weeks to find a feasible solution for $$y(x)$$.

$$\sin(x) = \int_0^{2\pi} \max(y(t), y(x+t)) dt$$

I don't think there is an unique solution, but I will accept any solution. I've tried various numerical minimization methods, convolution, guess the solution, etc, but nothing seems to work.

Does anybody have any ideas?

• try to use the fact $max(y(t),y(x+t))=|\frac{y(t)+y(x+t)}{2}|+|\frac{y(t)-y(x+t)}{2}|$ – vidyarthi Dec 6 at 17:22

If $$x=0$$ then $$\begin{equation*} 0= \sin(0)=\int_{0}^{2\pi}\max(y(t),y(t))\, \mathrm{d}t = \int_{0}^{2\pi}y(t)\, \mathrm{d}t . \end{equation*}$$ If $$x = \frac{3\pi}{2}$$ then $$\begin{equation*} -1 = \sin\left(\frac{3\pi}{2}\right) = \int_{0}^{2\pi}\max\left(y(t),y\left(\frac{3\pi}{2}+t\right)\right)\, \mathrm{d}t \ge \int_{0}^{2\pi}y(t)\, \mathrm{d}t = 0. \end{equation*}$$ Consequently, it does not exist any solution $$y(x)$$.
• I used that $\max(a,b) \ge a$. – JanG Dec 8 at 14:11