TITLE: A question related to Wave Equation
QUESTION [5 upvotes]: Let $L>0$. Suppose $f, g$ are $C^2$ functions on $\mathbb{R}$ such that 
$$f(t)+f(-t)+\int_{-t}^t g(s)\,ds=0$$ and 
$$f(L+t)+f(L-t)+\int_{L-t}^{L+t} g(s)\,ds=0$$
for all $t\in \mathbb{R}.$
Does it follow that $f, g$ are odd periodic functions of period $2L$?

REPLY [2 votes]: Lemma.  If $o(t)$ is odd and $e(t)$ is even and $o(t)=e(t)$ for all $t$, then $o(t)=e(t)=0$ for all $t$. 
To prove, simply substitute $-t$ for $t$ in the equation and get $-o(t)=e(t)$, so $-o(t)=o(t)$ and $o(t)=e(t)=0$ for all $t$.
Now in our first equation $f(t)+f(-t)$ is even and $-\int_{-t}^t g(s)\,ds$ is odd. So by Lemma we have $f(t)+f(-t)=0$ for all $t$ and $f$ is odd. 
In the second equation $f(L-t)+f(L+t)$ is even and $-\int_{L-t}^{L+t} g(s)\,ds$ is odd. So by Lemma, $f(L-t)=-f(L+t)$. Substituting $L+t$ for $t$, we get
$f(-t)=-f(2L+t)$. Since we have proved that $f(-t)=-f(t)$, we get $f(t)=f(2L+t)$.
Now the first equation becomes $\int_{-t}^t g(s)\,ds=0$. Differentiating we get $g(t)+g(-t)=0$,i.e. $g$ is odd. The second equation becomes $\int_{L-t}^{L+t} g(s)\,ds=0$. Differentiating we get $g(L+t)+g(L-t)=0$, so $g$ is of period $2L$, as we did for $f$.