Lax-Wendroff finite volume scheme derivation 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:
$$u=\text{function of x,t}\\\hat{u}=\frac{1}{\Delta x}\int_{x_{i-1/2}}^{x_{i+1/2}}u\thinspace dx \text{ (the average flux through volume)}$$
$$\frac{\partial u}{\partial t}=-\frac{\partial f(u)}{\partial x}\rightarrow \\\hat{u}_j^{n+1}=\hat{u}_j^n - \frac{\Delta t}{\Delta x}(F_{j+1/2}^n-F_{j-1/2}^n)\\
F_{j+1/2}=\frac{1}{2}\thinspace (f_{j+1}+f_j)-\frac{1}{2}\thinspace a^2_{j+1/2}\frac{\Delta t}{\Delta x}\thinspace (\hat{u}_{j+1}-\hat{u}_j)\\
a_{j+1/2}=\begin{cases} 
      \frac{f_{j+1}-f_j}{\hat{u}_{j+1}-\hat{u}_j} & if \enspace \hat{u}_{j+1}\neq \hat{u}_j \\
      f'(u_j) & if \enspace \hat{u}_{j+1}=\hat{u}_j  
   \end{cases}$$
I know that in a finite difference Lax-Wendroff is derived 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?
 A: To derive the Lax-Wendroff method in a finite volume-type formulation first of we have to understand that we are looking for the solution $u(x,t)$ of the conservation law:
$$ \begin{equation} \int_{x_1}^{x_2} u (x,t) dx = \int_{x_1}^{x_2} u(x,0) dx - \int_0^t (f(u(x_2,t)) - f(u(x_1,t)) dt . \end{equation} \tag{1}\label{eq1}$$
Which, under some regularity conditions it reduces to the EDP problem
$$\partial_t u(x,t) + \partial_x f(u(x,t))=0$$
with a given initial data $u(x,0)$.
Now, the integral law \eqref{eq1} holds for all control volume $[x_1, x_2] \times [0, t)$ in the domain. So if we look at the Lax-Wendroff scheme for linear advection (which is derived from finite differences as you pointed), we can see that it can be understood as a conservative scheme:
$$ U_i^{n+1} = U_i^n - \frac{\Delta t}{\Delta x} ( F_{i+1/2} - F_{i-1/2}), \tag{2}\label{eq2}$$
with
$$F_{i+1/2} = f(U_{i+1/2}^{n+1/2})$$
and
$$U_{i+1/2}^{n+1/2} = \frac{1+\Delta t/\Delta x}{2} U^n_i - \frac{1-\Delta t/\Delta x}{2} U^n_{i+1}. \tag{3}\label{eq3}$$
At this point, remember that we are doing this for the linear advection equation, which means $f(u) = a u$ with $a$ the speed of the wave. I recommend you to substitute \eqref{eq3} in \eqref{eq2} with the linear advection to reproduce the classical Lax-Wendroff scheme.
All of this, helps us to generalize \eqref{eq3} to
$$U_{i+1/2}^{n+1/2} = \frac{1}{\Delta x} \int_{x_{i}}^{x_{i+1}}u_{i+1/2}(x,\Delta t/2) dx,$$
being $u_{i+1/2}(x,t)$ the exact solution of the Riemann problem with initial data $u_L = U_i^n$ and $u_R = U_{i+1}^n$. Again, if we use $f(u) =au$ we obtain the result we already know.
Now, we can use the conservation law \eqref{eq1} to evaluate this integral under the control volume $[-\Delta x/2, \Delta x/2]\times [0 , \Delta t/2]$:
$$\int_{-\Delta x /2} ^{\Delta x/2} u_{i+1/2} ( x , \frac{\Delta t}{2} ) dx =\int_{-\Delta x /2} ^{\Delta x/2} u_{i+1/2} ( x , 0 ) dx  + \int_0^{\Delta t/2} f(u_{i+1/2} (-\frac{\Delta x}{2} , t )) dt \\
 - \int_0^{\Delta t/2} f(u_{i+1/2} (-\frac{\Delta x}{2} , t )) dt.$$
We know, from the initial data
$$\int_{-\Delta x /2} ^{\Delta x/2} u_{i+1/2} ( x , 0 ) dx = \frac{\Delta x}{2} ( U_i^n + U_{i+1}^n).$$
Also under the CFL condition, the waves in $[0, \Delta t/2]$ coming from points $x_i$ and $x_{i+1}$ do not interact, so
$$\int_0^{\Delta t/2}  f(u_{i+1/2} (\frac{\Delta x}{2} , t )) dt  = f (U_{i+1}^n )\frac{\Delta t }{2}.$$
Finally, we have derived the Ritchmyer two-step Lax-Wendroff scheme (which is the most common way to see the L-W scheme), and it is a conservative method \eqref{eq2} with
$$F_{i+1/2} = f(U_{i+1/2}^{n+1/2})$$
and
$$U_{i+1/2}^{n+1/2} = \frac{1}{2}( U_i^n + U_{i+1}^n + \frac{\Delta t}{\Delta x} ( f( U_i^n) - f(U_{i+1}^n))$$
