minmod slope total variation diminishing This is exercise 6.5 of the book Finite Volume Methods for Hyperbolic Problems by R.J. LeVeque (2002).

Show that the minmod slope guarantees that $$
TV(q^n(·, t_n)) ≤ TV(Q^n) \tag{6.23}
$$ will be satisfied in general, and hence the minmod method is TVD.

with the minmod slope defined as
$$minmod(a,b) =\left\{\begin{array}{lll}a&\text{if } |a|<|b|&\text{and } ab>0\\b&\text{if } |b|<|a|&\text{and }ab>0\\0&\text{if } ab\leq 0&\end{array} \right.$$
$$\sigma_i^n= minmod\left(\frac{Q_i^n−Q_{i-1}^n}{\Delta x},\frac{Q_{i+1}^n−Q_i^n}{\Delta x}\right).$$
and the total variation diminution (TVD) property defined by $TV(Q_{n+1}) \leq TV(Q_n)$.
It is known that
$TV(Q)=\sum_i|Q_i-Q_{i-1}|$ (definition of total variation)
$TV(Q^{n+1})\leq TV(q^n\left(\cdot,t_{n+1})\right)$ (given in exercise 6.4)
$TV(q^n\left(\cdot,t_{n+1})\right)= TV(q^n\left(\cdot,t_{n})\right)$ (follows from REA algorithm 2nd step)
$TV(q^n\left(\cdot,t_{n})\right)=\sup\sum_i|q^n(x_i)-q^n(x_{i-1})|$ (defitinion)
My idea was
$$\begin{align}
TV(Q^{n+1})&\leq TV(q^n\left(\cdot,t_{n+1})\right)\\
&=TV(q^n\left(\cdot,t_{n})\right)\\
&=\sup\sum_i|q^n(x_i)-q^n(x_{i-1})|\\
&\leq \sup\sum_i|Q_i^n-Q_{i-1}^n|\\
&=TV(Q^n)
\end{align}$$
But nowhere I've used something from minmod. What am I missing?
Edit
The REA algorithm is the following:

*

*Reconstruct $q^n(\cdot,t_n)=Q_i^n+\sigma_i^n(x-x_i)$, $\forall x\in C_i$, where $q^n$ is a piecewise polynomial function, $Q_i^n$ the cell average and $\sigma_i^n$ the slope.


*Evolve the hyperbolic equation exactly with this initial data to obtain $q^n(x,t_{n+1})$ a time $\Delta t$ later.


*Average this function over each grid cell to obtain new cell averages $Q_i^{n+1}=\frac{1}{\Delta x}\int\limits_{C_{i}}q^n(\cdot,t_{n+1})dx$
 A: We consider the advection equation $q_t + cq_x = 0$ with $c>0$. For the REA slope-limiter method with minmod slope $\sigma_i^n$ specified in OP, the reconstructed piecewise linear data of the book's Eq. (6.11) reads \begin{aligned}
q^n(x, t_n) &= \sum_{i=-\infty}^\infty \left(Q_i^n + \sigma_i^n (x-x_i)\right) \Bbb I_i(x) \\
&= Q_{-\infty}^n + \sum_{i=-\infty}^\infty \left(Q_{i+1}^n - \sigma_{i+1}^n x_{i+1} - Q_i^n + \sigma_i^n x_i\right) H(x-x_{i+1/2})  \\ &\; + x \sum_{i=-\infty}^\infty (\sigma_{i+1}^n - \sigma_i^n)\, H(x-x_{i+1/2}) \, ,
\end{aligned}
where $\Bbb I_i(x)$ is the indicator function of the $i$th finite volume $[x_{i-1/2}, x_{i+1/2}[$, and $H$ is the Heaviside step function. In the case of zero slope $\sigma_i^n \equiv 0$, a straightforward computation of the total variation yields $$
TV(q^n(\cdot, t_n)) = \sum_{i=-\infty}^\infty |Q_{i+1}^n - Q_{i}^n| = TV(Q^n) \, ,
$$ cf. Eq. (6.21) of the book. The present exercise consists in proving that the above equality becomes an inequality of the form $TV(q^n(\cdot, t_n)) \leq TV(Q^n)$ in the case of minmod slope $\sigma_i^n \not\equiv 0$ defined in OP --- in fact, the evaluation of $TV(q^n(\cdot, t_n))$ in OP is incorrect. To understand how things work, one could start with the computation of $TV(q^n(\cdot, t_n))$ for very simple piecewise linear functions $q^n$, e.g. the case of one single nonzero state $Q_0^n \neq 0$, then the case of two successive nonzero states $Q_0^n, Q_1^n \neq 0$, etc.
This way, you'll be able to tackle the case of arbitrary data $Q^n$.
From the above inequality (Eq. (6.23) of the book) and Eqs. (6.24)-(6.25), one shows that the minmod slope-limiter method is TVD.
