Finding a function as a solution to a system of integral equations In my research, I reached a point where I need to find a nonnegative function, that satisfies the following:
For any $\phi\in[0,\pi]$:
$[\int^\pi_\phi f(\theta)d\theta][ \int^{\pi-2\phi}_{-\phi} f(\theta)d\theta]+[\int^0_{\pi-\phi} f(\theta)d\theta ][\int^{2\phi}_{\pi+\phi} f(\theta)d\theta]-[\int^\phi_0 f(\theta)d\theta]^2-[\int^{\pi-\phi}_\pi f(\theta)d\theta]^2 = \pi^2\cos^2\frac{\phi}{2}$
I would be very glad if somebody can help me, and thank you in advance for your help!
 A: I think that condition can be satisfied only when $\sin^2\frac{\phi}{2}=\frac{1}{2}$.  However, I suggest a possible, slight modification of the constraint, so that a solution can be found, see later.
Let us start by writing $f(\theta)$ as a Fourier series involving only cosines (which are even functions):
\begin{equation}
f(\theta)=\sum_{n=0}^{+\infty}f_n\cos(n\theta)
\end{equation}
Let us impose the normalization:
\begin{equation}
\int_{0}^{\pi}d\theta \,\,f(\theta)=\sum_{n=0}^{+\infty}f_n\int_{0}^{\pi}\cos(n\theta)=\sum_{n=0}^{+\infty}f_n\frac{\sin{(\pi n)}}{n} \stackrel{!}{=}\frac{\pi}{2}
\end{equation}
this expression vanishes except when $n=0$, in that case
\begin{equation}
\frac{\sin{(\pi n)}}{n}=\pi
\end{equation}
or, in other words, $f_0=\frac{1}{2}$. Hence,
\begin{equation}
f(\theta)=\frac{1}{2}+\sum_{n=1}^{+\infty}f_n\cos(n\theta)
\end{equation}
Now we notice that, for $n\neq 0$,
\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos\left(n(\theta-\phi)\right)=\frac{\sin(n\phi )}{n}[1-(-1)^n],
\end{equation}
\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos(n\theta)=0
\end{equation}
Hence the first terms become
\begin{equation}
\int_{0}^{\pi}d\theta\,\,f(\theta)+\int_{0}^{\pi}d\theta\,\,f(\theta-\phi)=\frac{\pi}{2}+\frac{\pi}{2}+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]
\end{equation}
Now let us consider the other term.
\begin{equation}
\begin{split}
\int_0^{\pi}\int_0^{\pi}d\theta dt\,\,\left(\frac{1}{2}+\sum_{n= 1}^{+\infty}f_n\cos(n\theta)\right)\left(\frac{1}{2}+\sum_{m=1}^{+\infty}f_m\cos(m(t-\phi))\right) \\ \\
=\int_0^{\pi}\int_0^{\pi}d\theta dt\,\,\bigg\{\frac{1}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\cos(n(t-\phi))+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\cos(n\theta)\\ \\
+\sum_{n,m\neq0}f_nf_m\cos(n\theta)\cos(m(t-\phi))\bigg\}
\end{split}
\end{equation}

Let us integrate first over $\theta$. After the integration the previous integral becomes
\begin{equation}
\int_0^{\pi}dt\,\,\left[\frac{\pi}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\cos(n(t-\phi)) \right]=\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]
\end{equation}

Your condition then is
\begin{equation}
\begin{split}
\pi+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]-\frac{2}{\pi}\bigg\{\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]\bigg\}\\
\\
=\pi\sin^2\frac{\phi}{2}=\frac{\pi}{2}(1-\sin\phi)
\end{split}
\end{equation}
Now you see that the terms involving the sine cancel out, leaving
\begin{equation}
\sin^2\frac{\phi}{2}=\frac{1}{2} \hspace{3mm} \text{or} \hspace{3mm} \sin\phi=0
\end{equation}
However, if you could modify your condition as
\begin{equation}
\begin{split}
\int_0^{\pi}d\theta\,\,f(\theta-\phi)+\int_0^{\pi}d\theta\,\, f(\theta)-\frac{2}{\pi}\xi\int_0^{\pi}\int_0^{\pi}d\theta dt\,\, f(\theta)f(t-\phi)\\
=\pi\sin^2\frac{\phi}{2}+\frac{\pi}{2}(1-\xi)
\end{split}
\end{equation}
(Notice that you initial model has $\xi=1$), then you have
\begin{equation}
\begin{split}
\pi+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]-\frac{2}{\pi}\xi \bigg\{\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]\bigg\}\\
\\
=\pi\sin^2\frac{\phi}{2}+\frac{\pi}{2}(1-\xi)=\frac{\pi}{2}(1-\sin\phi)+\frac{\pi}{2}(1-\xi)
\end{split}
\end{equation}
which implies
\begin{equation}
(1-\xi)\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]=-\frac{\pi}{2}\sin\phi
\end{equation}
This condition is satisfied by setting $f_n=0$ for all $n>1$ and by choosing $f_1$ such that
\begin{equation}
2(1-\xi)f_1\sin\phi=-\frac{\pi}{2}\sin\phi
\end{equation}
from which
\begin{equation}
f_1=\frac{\pi}{4}\frac{1}{1-\xi}
\end{equation}
I hope the modified model could work for you.
A: Too long for a comment.
If  $\int_0^{\pi}f(\theta)d\theta = \frac{\pi}{2}$ then the integral equations reduce to
$$
\cases{
\frac{\pi}{2}\int_0^{\pi}f(t-\phi)dt = \frac{\pi^2}{2}\cos^2\left(\frac{\phi}{2}\right)\\
\frac{\pi}{2}+\int_0^{\pi}f(t-\phi)dt = \pi
}
$$
then $\int_0^{\pi}f(t-\phi)dt = \frac{\pi}{2}$. It seems that $f(\cdot)=C_0$. Contradictory.
A: Unfortunately, this is inconsistent:
If I put $\phi = 0$, I get
$\int^\pi_0 f(\theta)d\theta=\frac{\pi}{\sqrt[]{2}}$
while if I put $\phi=\pi$, I get
$\int^\pi_0 f(\theta)d\theta=0$
