Finding a function as a solution to a system of integral equations

Question

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!

Answer 1

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

$\int^\pi_0 f(\theta)d\theta=0$

Answer 2

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.

Answer 3

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.