I have to show that if f and g are odd the $u(0,t)=0$ where
$u(x,t)=\frac{1}{2}(f(x-t)+f(x+t))+\frac{1}{2}\int_{x-t}^{x+t} g(x') dx'$ subject to $u(x,0)=f(x)$ and $u_t(x,o)=g(x)$ so:
$u(0,t)=\frac{1}{2}(f(-t)+f(t))+\frac{1}{2}\int_{-t}^{t} g(x') dx'$ and as f is odd we get:
$u(0,t)=\frac{1}{2}\int_{-t}^{t} g(x') dx'$ but I am not sure why this is zero, I'm pretty sure this should be obvious but i am missing something?
Thanks for any help