Solving an IVP with Green's Theorem: Wondering?

[mathmari]

Hey!

If we have the initial and boundary value problem $$u_{tt}(x,t)-c^2u_{xx}(x,t)=0, x>0, t>0 \\ u(0,t)=0 \\ u(x,0)=f(x), x\geq 0 \\ u_t(x,0)=g(x)$$ and we want to apply Green's theorem do we have to expand the problem to $x \in \mathbb{R}$ ?? (Wondering)

 

[Prove It]

Hey!

If we have the initial and boundary value problem $$u_{tt}(x,t)-c^2u_{xx}(x,t)=0, x>0, t>0 \\ u(0,t)=0 \\ u(x,0)=f(x), x\geq 0 \\ u_t(x,0)=g(x)$$ and we want to apply Green's theorem do we have to expand the problem to $x \in \mathbb{R}$ ?? (Wondering)

Couldn't you just separate the variables?

 

[mathmari]

Yes, but I want to use this method.

- - - Updated - - -

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 $w$ ?? (Wondering)