For any conservative solution
\((u,F)\in [W^{1,\infty }([0,T]\times {\mathbb {R}})]^2\) with initial data
\((u_0,F_0)\), the characteristic equation
$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}x(t) = u(t,x(t)), \qquad x(0) = x_0, \end{aligned}$$
is uniquely solvable. Furthermore, the classical method of characteristics implies that the solution is unique and given by
$$\begin{aligned} u(t,x(t))&=u_0(x_0)+\frac{1}{2}(F_0(x_0)-\frac{1}{2}F_\infty )t,\\ F(t,x(t))&=F_0(x_0). \end{aligned}$$
Introduce
\(({\tilde{u}}^0, {\tilde{F}}^0)(t)=T_t{\mathbb {P}}_ {\varDelta x } (u_0,F_0)\) and recall that
\(x_j(t)\) denotes the characteristic starting at the grid point
\(x_j\), then
$$\begin{aligned} ({\tilde{u}}^0(t,x_j(t)),{\tilde{F}}^0(t,x_j(t)))=(u(t,x_j(t)), F(t,x_j(t)))\quad \text { for all } j\in {\mathbb {N}}. \end{aligned}$$
Moreover, for all
\(t\in [0,T]\),
$$\begin{aligned} \Vert {\tilde{u}}_x^0(t,\cdot )\Vert _\infty \le \Vert u_x(t,\cdot )\Vert _\infty \quad \text { and }\quad \Vert {\tilde{F}}_x^0(t,\cdot )\Vert _\infty \le \Vert F_x(t,\cdot )\Vert _\infty . \end{aligned}$$
For
\(n\ge 1\), define
\(({\tilde{u}}^n, {\tilde{F}}^n)\) by
$$\begin{aligned} ({\tilde{u}}^n,{\tilde{F}}^n)(t) = {\left\{ \begin{array}{ll} T_{(t-t^n)}{\mathbb {P}}_ {\varDelta x } ({\tilde{u}}^{n-1},{\tilde{F}}^{n-1})(t^n), &{}\quad t\ge t^n,\\ ({\tilde{u}}^{n-1},{\tilde{F}}^{n-1})(t), &{}\quad t<t^n. \end{array}\right. } \end{aligned}$$
Then, following the same lines, one has
$$\begin{aligned} \Vert {\tilde{u}}^n_x(t,\cdot )\Vert _\infty \le \Vert {\tilde{u}}^{n-1}_x(t,\cdot )\Vert _\infty \le \Vert u_x(t,\cdot )\Vert _\infty \end{aligned}$$
and
$$\begin{aligned} \Vert {\tilde{F}}^n_x(t,\cdot )\Vert _\infty \le \Vert {\tilde{F}}^{n-1}_x(t,\cdot )\Vert _\infty \le \Vert F_x(t,\cdot )\Vert _\infty . \end{aligned}$$
Since
\((u_{ {\varDelta x } }(t,\cdot ),F_{ {\varDelta x } }(t,\cdot ))={\mathbb {P}}_ {\varDelta x } ({\tilde{u}}^n (t,\cdot ), {\tilde{F}}^n(t,\cdot ))\) for all
\(t\le t^n\), we end up with
$$\begin{aligned} \sup _{t\in [0,T]}\big (\Vert u_{ {\varDelta x } ,x}(t,\cdot )\Vert _\infty +\Vert F_{ {\varDelta x } ,x}(t,\cdot )\Vert _\infty \big )\le \sup _{t\in [0,T]}\big (\Vert u_x(t,\cdot )\Vert _\infty +\Vert F_x(t,\cdot )\Vert _\infty \big ). \end{aligned}$$
It is left to show (
2.11). We start by comparing (
u,
F) with
\(({\tilde{u}}^0, {\tilde{F}}^0)\). To that end let
\((t,x)\in [0,T]\times {\mathbb {R}}\) such that
\(x_j(t)\le x\le x_{j+1}(t)\). Then there exists
\(x^0\) and
\({\tilde{x}}^0\) in
\([x_j,x_{j+1}]\) such that
$$\begin{aligned} u(t,x)&= u_0(x^0)+\frac{1}{2}\left( F_0(x^0)-\frac{1}{2}F_\infty \right) t,\\ F(t,x)&= F_0(x^0), \end{aligned}$$
and
$$\begin{aligned} {\tilde{u}}^0(t,x)&= {\tilde{u}}^0(0,{\tilde{x}}^0)+\frac{1}{2}\left( {\tilde{F}}^0(0,{\tilde{x}}^0)-\frac{1}{2}F_\infty \right) t\\ {\tilde{F}}^0(t,x)&= {\tilde{F}}^0(0,{\tilde{x}}^0). \end{aligned}$$
Using
\(({\tilde{u}}^0(0,x),{\tilde{F}}^0(0,x))= {\mathbb {P}}_ {\varDelta x } (u_0(x),F_0(x))\), we have,
$$\begin{aligned} |u(t,x)-{\tilde{u}}^0(t,x)|&\le |u_0(x^0)-{\tilde{u}}^0(0,x^0)| + |{\tilde{u}}^0(0,x^0)-{\tilde{u}}^0(0,{\tilde{x}}^0)|\\&\quad + \frac{1}{2}t\left( |F_0(x^0)-{\tilde{F}}^0(0,x^0)| + |{\tilde{F}}^0(0,x^0)-{\tilde{F}}^0(0,{\tilde{x}}^0)|\right) \\&\le 2\left( \Vert u_{0,x}\Vert _\infty + \frac{1}{2}t\Vert F_{0,x}\Vert _\infty \right) {\varDelta x } , \end{aligned}$$
and
$$\begin{aligned} |F(t,x)-{\tilde{F}}^0(t,x)|&\le |F_0(x^0)-F^0(0,x^0)| + |{\tilde{F}}^0(0,x^0)-{\tilde{F}}^0(0,{\tilde{x}}^0)|\\&\le 2\Vert F_{0,x}\Vert _\infty {\varDelta x } . \end{aligned}$$
Combining the last two inequalities, we have
$$\begin{aligned} \sup _{t\in [0,T]}\big (\Vert u(t,\cdot ) - {\tilde{u}}^0(t,\cdot )\Vert _\infty + \Vert F(t,\cdot )-{\tilde{F}}^0(t,\cdot )\Vert _\infty \big )\le 2L(1+\frac{1}{2}t) {\varDelta x } , \end{aligned}$$
where
\(L=\sup _{t\in [0,T]} \big (\Vert u_x(t,\cdot )\Vert _\infty +\Vert F_x(t,\cdot )\Vert _\infty \big )\). Moreover, we have by the same argument for
\(t\ge t^n\) that
$$\begin{aligned}&\Vert {\tilde{u}}^{n}(t,\cdot ) - {\tilde{u}}^{n-1}(t,\cdot )\Vert _\infty + \Vert {\tilde{F}}^{n}(t,\cdot )-{\tilde{F}}^{n-1}(t,\cdot )\Vert _\infty \\&\quad \le 2(\Vert {\tilde{u}}^{n-1}_x(t^n,\cdot )\Vert _\infty + \Vert {\tilde{F}}^{n-1}_x(t^n,\cdot )\Vert _\infty )(1+\frac{1}{2}(t-t^n)) {\varDelta x } \\&\quad \le 2L(1+\frac{1}{2} (t-t^n)) {\varDelta x } . \end{aligned}$$
Since
\((u_{ {\varDelta x } }(t,\cdot ),F_{ {\varDelta x } }(t,\cdot ))= {\mathbb {P}}_ {\varDelta x } ({\tilde{u}}^n(t,\cdot ), {\tilde{F}}^n(t,\cdot ))\) for all
\(t\le t^n\), we have for all
\(t^n\le t\le t^{n+1}\) that
$$\begin{aligned}&\Vert u(t,\cdot ) - u_ {\varDelta x } (t,\cdot )\Vert _\infty + \Vert F(t,\cdot )- F_ {\varDelta x } (t,\cdot )\Vert _\infty \\&\quad \le \Vert u(t,\cdot ) - {\tilde{u}}_0(t,\cdot )\Vert _\infty + \Vert F(t,\cdot )-{\tilde{F}}^0(t,\cdot )\Vert _\infty \\&\qquad + \sum _{l=1}^{n}\left( \Vert {\tilde{u}}^{l}(t,\cdot ) - {\tilde{u}}^{l-1}(t,\cdot )\Vert _\infty + \Vert {\tilde{F}}^{l}(t,\cdot )-{\tilde{F}}^{l-1}(t,\cdot )\Vert _\infty \right) \\&\qquad + \Vert {\tilde{u}}^n(t,\cdot )-u_ {\varDelta x } (t,\cdot )\Vert _\infty +\Vert {\tilde{F}}^n(t,\cdot )-F_ {\varDelta x } (t,\cdot )\Vert _\infty \\&\quad \le 2L(1+\frac{1}{2} t) {\varDelta x } +\sum _{l=1}^{n} 2L(1+\frac{1}{2} (t-t^{l})) {\varDelta x } +2L {\varDelta x } \\&\quad \le C(\sqrt{ {\varDelta x } }+ {\varDelta x } ), \end{aligned}$$
where we have used (
2.4).
\(\square \)