TITLE: Why do we take the odd extension?
QUESTION [6 upvotes]: When we have the initial and boundary value problem $$u_{tt}(x,t)-c^2u_{xx}(x,t)=0, x>0, t>0 \\ u(0,t)=0 \\ u(x,0)=f(x), x \geq 0 \\ u_t(x,0)=g(x), x \geq 0$$ 
can we apply Green's theorem or does it have to stand that $x \in \mathbb{R}$ to use it?? 
Because in my notes they take the odd extension and I don't know why... 
Could you explain it to me?? 
$$$$ 
EDIT: 
In my notes they do it as followed: 
$$w_{tt}-c^2w_{xx}=0, x \in \mathbb{R}, t>0 \\ w(x,0)=f_{\text{odd}}(x), x \in \mathbb{R} \\ w_t(x,0)=g_{\text{odd}} (x), x \in \mathbb{R}$$ 
$$w(x,t)=\frac{1}{2}(f_{\text{odd}}(x-ct)+f_{\text{odd}}(x+ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}g_{\text{odd}}(s)ds \\ w(0,t)=\frac{1}{2}(f_{\text{odd}}(-ct)+f_{\text{odd}}(ct))+\frac{1}{2c}\int_{-ct}^{ct}g_{\text{odd}}(s)ds=0$$ 
So, for $x>0, t>0$ 
$$u(x,t)=w(x,t)=\frac{1}{2}(f_{\text{odd}}(x-ct)+f_{\text{odd}}(x+ct))+\frac{1}{2c}\int_{x-ct}^{x+ct}g_{\text{odd}}(s)ds$$ 
$$u(x,t)=\left\{\begin{matrix}
\frac{1}{2}{(f(x-ct)+f(x+ct))+\frac{1}{2c}{\int_{x-ct}^{x+ct}g(s)ds, \ \ x-ct \geq 0}}\\ \\
\frac{1}{2}(-f(ct-x)+f(x+ct))+\frac{1}{2c}\int_{ct-x}^{x+ct}g(s)ds \ \ \ \ \ \ \ \ \ \ \ \ \ 
\end{matrix}\right.$$ 
Why have we taken the odd extension although we solve $u$ and not $w$ ??

REPLY [3 votes]: Don't have time to write a long answer, but the short answer is because of boundary conditions. That is, in your first list of equations, you have the boundary condition $u(t,0)=0$. You don't have this boundary condition for the equations involving $w$. However, because you've chosen the initial conditions of $w$ carefully, the initial conditions you are interested are satisfied easily. Carefully check you final expression to see that it is a solution and satisfies the particular boundary condition.  The general sort of method here is called the Method of Images (http://en.wikipedia.org/wiki/Method_of_images).