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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

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$g$ is odd. You're integrating over a symmetric interval. Draw a picture if don't see why this implies that the integral is zero. –  mrf Apr 21 '12 at 12:17
    
Yeah, just realized what I was missing (sorry this is a really bad question!) I just split the integral $\int_{0}^{t}$ and $\int_{-t}^{0}$ and change variable sorry about that –  hmmmm Apr 21 '12 at 12:18
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up vote 2 down vote accepted

The reason $$\int_{-t}^{t} g(x') dx'=0$$ is that $g$ is odd (as mrf said). Substitution $u=-x'$ yields $$\int_{0}^{t} g(x') dx' = \int_{0}^{-t} g(-u) (-du) =- \int_{-t}^0 g(u) du $$ which added to $\int_{-t}^{0} g(x') dx'$ results in $0$.

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