# Solving a functional equation $2 f(2x)=f(x)(1+\cos(x))+f(x+\pi)(1-\cos(x))$

I am trying to solve the following functional equation, which appears in some of my physics calculations : $f(x)=\frac{1}{2}\left(f(\frac{x}{2})(1+\cos(\frac{x}{2}))+f(\frac{x}{2}+\pi)(1-\cos(\frac{x}{2}))\right)$,

for a function $f$ defined on $(0,2\pi)$.

I am interested in all functions satisfying this equation, but let us start by finding all functions which are defined on $[0,2\pi]$, which are continuous and differentiable.

Constant functions are solutions.

Let us note $f(0)=a$ and $f(2\pi)=b$. My numerics, starting from some function and iterating until I get to a fixed point seem to point towards a unique solution depending only on $a$ and $b$. If $a=b$, this would be the constant function.

If $a\neq b$, the solution seems to have the following properties (conjectures) :

• Monotonous
• $f(\pi)=(f(0)+f(2\pi))/2$
• $f(\pi-x)+f(\pi+x)=2*f(\pi)$, so $f(\pi)$ is a symmetric point of the curve
• First derivative is $0$ in $0$ and $2\pi$

The last point can be proven easily by deriving the equation, which gives $f'(x)=\frac{1}{4}\left(f'(\frac{x}{2})(1+\cos(\frac{x}{2}))+f'(\frac{x}{2}+\pi)(1-\cos(\frac{x}{2}))+\sin(\frac{x}{2})(f(\frac{x}{2}+\pi)-f(\frac{x}{2}+)\right)$ Evaluating at $0$ and $2\pi$ yields $0$.

Evaluating the original equation at $0$ or $2\pi$ yields consistent results, at $\pi$ gives $f(\pi)=(f(\pi/2)+f(3\pi/2))/2$, which is consistent with the second conjecture at this particular point.

So far I haven't managed to prove the other conjectures on the solution. Note that all these properties are shared by $Cos(x/2)+C$, but this is not a solution, so there are other things needed to characterize a solution. Here is a plot of an approximate numerical solution.

Any ideas how to solve this? Thanks!

• No one has an idea? I have rephrased the question and my attempts a bit. – cestyx Jun 2 '16 at 7:00
• Could you try a Fourier solution? Like assume $f(x)=\sum_n a_n e^{inx}$ and see what conditions the $a_n$ might have to satisfy. – πr8 Jun 2 '16 at 7:06
• Thanks for the suggestion. I actually found out that the function is not periodic if it is defined on a larger interval (I have updated the plot of the function) and proved that $f(k*2\pi)=f(2\pi)$ for $k>1$. – cestyx Jun 3 '16 at 9:35
• A series expansion leads to $a_{n/2}=a_n+\frac{1}{2}(a_{n-1}+a_{n+1})$ ($n$ even). However there is no solution which keeps only small $n$ terms, and starting with some initial conditions $a_0$, $a_1$, it is not possible to iteratively compute other values with such an expression. – cestyx Jun 3 '16 at 12:54
• The functional equation in the title is algebraically equivalent to the one in the body, but not obtained by "scaling the domain of $f$"; could you please confirm whether the second term in the body should contain the factor $f(\frac{x + \pi}{2})$ instead of $f(\frac{x}{2} + \pi)$? – Andrew D. Hwang Jun 3 '16 at 14:58