Derivation of Lax-Wendroff finite volume scheme

[fahraynk]

I'm trying to figure out how the finite volume version of Lax-Wendroff scheme is derived.

Here is the PDE and Lax-Wendfroff scheme, assume initial conditions are given:
$$u=\text{function of x,t}\\\hat{u}=\frac{1}{\Delta x}\int_{x_{i-1/2}}^{x_{i+1/2}}u\thinspace dx \text{ (the average flux through volume)}$$
$$\frac{\partial u}{\partial t}=-\frac{\partial f(u)}{\partial x}\rightarrow \\\hat{u}_j^{n+1}=\hat{u}_j^n - \frac{\Delta t}{\Delta x}(F_{j+1/2}^n-F_{j-1/2}^n)\\
F_{j+1/2}=\frac{1}{2}\thinspace (f_{j+1}+f_j)-\frac{1}{2}\thinspace a^2_{j+1/2}\frac{\Delta t}{\Delta x}\thinspace (\hat{u}_{j+1}-\hat{u}_j)\\
a_{j+1/2}=\begin{cases}
\frac{f_{j+1}-f_j}{\hat{u}_{j+1}-\hat{u}_j} & if \enspace \hat{u}_{j+1}\neq \hat{u}_j \\
f'(u_j) & if \enspace \hat{u}_{j+1}=\hat{u}_j
\end{cases}$$

I know that in a finite difference Lax-Wendroff is derived from the original PDE and a taylors expansion like this :

$$u_t=-cu_x \rightarrow u_{tt}=c^2u_{xx}\\
\text {taylors expansion :}\thinspace u(t+\Delta t, x)= u+\Delta t\thinspace u_t + \frac{\Delta t^2}{2}u_{tt} \rightarrow \\u^{m+1}_n=u^m_n-c\Delta t \thinspace u_x+\frac{c^2\Delta t^2}{2}u_{xx}$$

I know in finite volume we are measuring the average flux of $u$, so I attempt to get the equation into the right form using $\hat{u}$ by dividing by $\Delta x$ and taking an integral with respect to x.

$$\hat{u}_n^{m+1}=\hat{u}_n^m-\frac{c\Delta t}{\Delta x}(u(x_{i+1/2},t)-u(x_{i-1/2},t)+\frac{c^2\Delta t^2}{2\Delta x^2}u_x|_{x_{i-1/2}}^{x_{i+1/2}}$$ Not sure what to do though from here.

I also tried just expanding $f$ like this :
$$
f(u(x+\frac{\Delta x}{2}))=f(u(x))+\frac{\Delta x}{2}f_u u_x+(\frac{\Delta x}{2})^2(f_{uu}u_x^2+f_u u_{xx})
$$

I don't see where this is going either. I just can't find it derived after a ton of google searching. Can anyone show me or link me to a derivation?

 

[Chestermiller]

In Lax-Wendoff, isn't the spatial derivative approximated by $\frac{f_{j+1}-f_{j-1}}{2\Delta x}$?