Please me with the following problem.

$\displaystyle \frac{\partial u^2}{\partial t^2} = c^2 \frac{\partial u^2}{\partial x^2} \;\;,\;\; u(0,t)=u(L,t)=0 \;\;,\;\; u(x,0)=f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0)=g(x)$

$\displaystyle f(x) \;and\; g(x) \;are\; symmetric\; about\; x=\frac{L}{2} \Rightarrow f(L-x)=f(x) \;\;and\;\; g(L-x)=g(x)$

Show $\displaystyle u(x,t+\frac{L}{c})=-u(x,t)$

Relevant equations

$\displaystyle u(x,t)=\frac{1}{2}[f(x+ct)+f(x-ct)]+\frac{1}{2}[G(x+ct)-G(x-ct)]$

$\displaystyle G(x)=\frac{1}{c}[G(x+ct)-G(x-ct)]$

$\displaystyle u(-x,t)=-u(x,t) \;,\; u(x+2L,t)=u(x,t) \;\;,\;\; u(x-L,t)=u(x+L,t)$

The attempt at a solution

u(x,t) is periodic with T=2L.

$\displaystyle u(x, t+\frac{L}{c} ) =\frac{1}{2}[f(x+c (t+\frac{L}{c}) )+f(x-c(t+\frac{L}{c}) )]+\frac{1}{2}[G(x+c(t+\frac{L}{c}) )-G(x-c(t+\frac{L}{c}) )]$

$\displaystyle \Rightarrow u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x-L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x-L)-ct)]$

$\displaystyle u(x-L,t)=u(x+L,t)$

$\displaystyle \Rightarrow u(x, t+\frac{L}{c} ) =\frac{1}{2}[ f((x+L)+ct )+f((x+L)-ct)]+\frac{1}{2}[G((x+L)+ct)-G((x+L)-ct)]$

I can see odd and even function with symmetric at the middle of the period like sin(x) and cos(x) resp. That $\displaystyle sin(x+\pi)=-sin(x) \;and\; cos(x+\pi)=-cos(x)$

I just don't know how to express in mathematical terms. Can someone at least get me hints or answer?

Thanks
Alan