We recall that the weights
\(\{ w_{\mathrm {k}} \}_{\mathrm {k}\in {\mathcal {K}}}\) satisfy
\(\sum _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}=1\). We further denote the vector of weights by
\(\vec {w}:=(w_{1},\ldots ,w_{K})\), and let
\({\overline{\mu }}^{*}=({\overline{\mu }}^{*}_1,\ldots ,{\overline{\mu }}^{*}_K) \in \Lambda (\vec {w},{\bar{\mu }})\) correspond to an optimal transport plan between
\({\bar{\mu }}\) and
\(\hat{{\bar{\mu }}}^K\). Because
\(\mu _{t}\) is a dissipative statistical solution, there exists
\((\mu _{1,t}^*,\ldots ,\mu _{K,t}^*) \in \Lambda (\vec {w},\mu _{t})\), such that
$$\begin{aligned} \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}& \Bigg [\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \Big ( u\cdot \partial _{t} \phi _\mathrm {k}+ f(u) \cdot \partial _{x}\phi _\mathrm {k}\Big ) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _{{\mathcal {F}}} \int \limits _{D} {\bar{u}} \cdot \phi _\mathrm {k}(0,\cdot ) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ]=0, \end{aligned}$$
(12)
for every
\(\phi _\mathrm {k}\in C_c^\infty ([0,T)\times D;{\mathbb {R}}^m)\),
\(\mathrm {k}\in {\mathcal {K}}\). Recalling that each function
\(u^{st}_{\mathrm {k}}\) is a Lipschitz-continuous solution of the perturbed conservation law (
5), considering its weak formulation yields
$$\begin{aligned} \int \limits _0^T\int \limits _D\Big (u^{st}_{\mathrm {k}} \cdot \partial _{t}\phi _\mathrm {k}+ f(u^{st}_{\mathrm {k}}) \cdot \partial _{x} \phi _\mathrm {k}\Big )~\mathrm {d}x\mathrm {d}t+ \int \limits _D{\bar{u}}^{st}_{\mathrm {k}}\cdot \phi _\mathrm {k}(0,\cdot ) ~ \mathrm {d}x+ \int \limits _0^T\int \limits _D{\mathcal {R}}^{st}_{\mathrm {k}} \cdot \phi _\mathrm {k}~\mathrm {d}x\mathrm {d}t= 0, \end{aligned}$$
(13)
for every
\(\phi _\mathrm {k}\in C_c^\infty ([0,T)\times D;{\mathbb {R}}^m)\). As
\((\mu _{\mathrm {k},t}^*)_{\mathrm {k}\in {\mathcal {K}}}\) are probability measures on
\({\mathcal {F}}\), we obtain (after changing order of integration)
$$\begin{aligned} 0 =&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \Big (u^{st}_{\mathrm {k}}\cdot \partial _{t}\phi _\mathrm {k}+ f(u^{st}_{\mathrm {k}})\cdot \partial _{x} \phi _\mathrm {k}\Big ) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _{{\mathcal {F}}} \int \limits _{D} {\bar{u}}^{st}_{\mathrm {k}}\cdot \phi _\mathrm {k}(0,\cdot ) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}}) + \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} {\mathcal {R}}^{st}_{\mathrm {k}}\cdot \phi _\mathrm {k} ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t\Bigg ]. \end{aligned}$$
(14)
Subtracting (
14) from (
12) and using the Lipschitz-continuous test function
\(\phi _\mathrm {k}(t,x):= {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}(t,x)) \phi (t)\), where
\(\phi \in C_c^\infty ([0,T);{\mathbb {R}}_+)\) yields
$$\begin{aligned} 0 =&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \Big ((u- u^{st}_{\mathrm {k}}) \cdot \partial _{t} ({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) \phi ) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} (f(u)-f(u^{st}_{\mathrm {k}}))\cdot \partial _{x} ({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) \phi ) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _{{\mathcal {F}}} \int \limits _{D} ({\bar{u}}-{\bar{u}}^{st}_{\mathrm {k}}) \cdot {{\,\mathrm{D}\,}}\eta ({\bar{u}}^{st}_{\mathrm {k}}) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}}) - \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} {\mathcal {R}}^{st}_{\mathrm {k}} \cdot {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t\Bigg ]. \end{aligned}$$
(15)
We compute the partial derivatives of
\({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}(t,x))\phi (t)\) using product and chain rule
$$\begin{aligned}&\partial _{t} ({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}})\phi ) = \partial _{t}u^{st}_{\mathrm {k}} \cdot H_{} \eta (u^{st}_{\mathrm {k}}) \phi + \partial _{t} \phi {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}), \end{aligned}$$
(16)
$$\begin{aligned}&\partial _{x} ({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}})\phi ) = \partial _{x}u^{st}_{\mathrm {k}} \cdot H_{}\eta (u^{st}_{\mathrm {k}}) \phi . \end{aligned}$$
(17)
Next, we multiply (
5) by
\({{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}})\). Upon using the chain rule for Lipschitz-continuous functions and the relationship
\({{\,\mathrm{D}\,}}q( u^{st}_{\mathrm {k}})={{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) {{\,\mathrm{D}\,}}f(u^{st}_{\mathrm {k}})\) we derive the relation
$$\begin{aligned} {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}})\cdot {\mathcal {R}}^{st}_{\mathrm {k}} = \partial _{t} \eta (u^{st}_{\mathrm {k}}) + \partial _{x} q(u^{st}_{\mathrm {k}}). \end{aligned}$$
(18)
Let us consider the weak form of (
18) and integrate w.r.t.
x,
t and
\(\mathrm {d}\mu _{\mathrm {k},t}^*\) for
\(\mathrm {k}\in {\mathcal {K}}\). Upon changing the order of integration we have
$$\begin{aligned} 0&= \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u^{st}_{\mathrm {k}}) \partial _{t} \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&\quad+ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} {\mathcal {R}}^{st}_{\mathrm {k}} \cdot {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t + \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta ({\bar{u}}^{st}_{\mathrm {k}}) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ] \end{aligned}$$
(19)
for any
\(\phi \in C_c^\infty ([0,T);{\mathbb {R}}_+)\). Since
\(\mu _{\mathrm {k},t}^*\) is a dissipative statistical solution it satisfies
$$\begin{aligned} 0 \leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Big [ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u) \partial _{t}\phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t + \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta ({\bar{u}}) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Big ]. \end{aligned}$$
(20)
Hence, subtracting (
19) from (
20) and using the definition of the relative entropy from Definition
4.1 yields
$$\begin{aligned}&0 \leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u|u^{st}_{\mathrm {k}}) \partial _{t}\phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&\quad +\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \Big ((u-u^{st}_{\mathrm {k}} )\cdot {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) \partial _{t}\phi \Big ) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&\quad - \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} {\mathcal {R}}^{st}_{\mathrm {k}} \cdot {{\,\mathrm{D}\,}}\eta (u^{st}_{\mathrm {k}}) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t + \int \limits _{{\mathcal {F}}} \int \limits _{D} \Big (\eta ({\bar{u}}) - \eta ({\bar{u}}^{st}_{\mathrm {k}})\Big ) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ]. \end{aligned}$$
(21)
After subtracting (
15) from (
21) and using (
16), (
17) we are led to
$$\begin{aligned} 0&\leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u|u^{st}_{\mathrm {k}}) \partial _{t}\phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&-\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} (u-u^{st}_{\mathrm {k}} ) \cdot H\eta (u^{st}_{\mathrm {k}}) \partial _{t}u^{st}_{\mathrm {k}} \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&-\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \Big ( f(u) -f(u^{st}_{\mathrm {k}}) \Big ) \cdot H \eta (u^{st}_{\mathrm {k}}) \partial _{x}u^{st}_{\mathrm {k}} \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta ({\bar{u}}|{\bar{u}}^{st}_{\mathrm {k}}) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ]. \end{aligned}$$
(22)
Rearranging (
5) yields
$$\begin{aligned} \partial _{t} u^{st}_{\mathrm {k}} = - {{\,\mathrm{D}\,}}f(u^{st}_{\mathrm {k}})\partial _{x} u^{st}_{\mathrm {k}} + {\mathcal {R}}^{st}_{\mathrm {k}}. \end{aligned}$$
(23)
Plugging (
23) into (
22) and after rearranging we have
$$\begin{aligned} 0 \leqslant&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u|u^{st}_{\mathrm {k}}) \partial _{t}\phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&-\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} \partial _{x}u^{st}_{\mathrm {k}} \cdot H \eta (u^{st}_{\mathrm {k}}) f(u|u^{st}_{\mathrm {k}}) \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&-\int \limits _0^T\int \limits _{{\mathcal {F}}} \int \limits _{D} (u-u^{st}_{\mathrm {k}} )\cdot H \eta (u^{st}_{\mathrm {k}}) {\mathcal {R}}^{st}_{\mathrm {k}} \phi ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&+ \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta ({\bar{u}}|{\bar{u}}^{st}_{\mathrm {k}}) \phi (0) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ], \end{aligned}$$
(24)
where we have used
\({{\,\mathrm{D}\,}}f \cdot H\eta = H\eta {{\,\mathrm{D}\,}}f.\) Up to now, the choice of
\(\phi (t)\) was arbitrary. We fix
\(s>0\) and
\(\epsilon >0\) and define
\(\phi\) as follows
$$\begin{aligned} \phi (\sigma ):= {\left\{ \begin{array}{ll} 1 \qquad &{}: \sigma<s,\\ 1- \frac{\sigma -s}{\epsilon } &{}: s<\sigma <s+\epsilon , \\ 0 &{}: \sigma > s+\epsilon . \end{array}\right. } \end{aligned}$$
According to Theorem
2.5 (a) we have that the mapping
$$\begin{aligned} t \mapsto \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u|u^{st}_{\mathrm {k}}(t,\cdot )) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \end{aligned}$$
(25)
is measurable for all
\(\mathrm {k}\in {\mathcal {K}}\). Moreover, due to the quadratic bound on the relative entropy, cf. (
8), Lebesgue’s differentiation theorem states that a.e.
\(t \in (0,T)\) is a Lebesgue point of (
25). Thus, letting
\(\epsilon \rightarrow 0\) we obtain
$$\begin{aligned}&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\int \limits _{{\mathcal {F}}} \int \limits _{D} \eta (u|u^{st}_{\mathrm {k}}(s,\cdot )) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},s}^*(u) \nonumber \\&\quad \leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ -\int \limits _0^{s} \int \limits _{{\mathcal {F}}} \int \limits _{D} \partial _{x}u^{st}_{\mathrm {k}} \cdot H \eta (u^{st}_{\mathrm {k}}) f(u|u^{st}_{\mathrm {k}}) ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&\qquad -\int \limits _0^{s} \int \limits _{{\mathcal {F}}} \int \limits _{D} (u-u^{st}_{\mathrm {k}} )\cdot H \eta (u^{st}_{\mathrm {k}}) {\mathcal {R}}^{st}_{\mathrm {k}} ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t \nonumber \\&\qquad + \int \limits _{{\mathcal {F}}} \int \limits _{D} \eta ({\bar{u}}|{\bar{u}}^{st}_{\mathrm {k}}) ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}})\Bigg ] . \end{aligned}$$
(26)
The left hand side of (
26) is bounded from below using (
8). The first term on the right hand is estimated using the
\(L^\infty (D)\)-norm of the spatial derivative. We estimate the second term on the right hand side by Young’s inequality. Finally, we apply (
8) and then (
9) to both terms. The last term on the right hand side is estimated using (
8). We, thus, end up with
$$\begin{aligned}&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ A^{-1}\int \limits _{{\mathcal {F}}} \int \limits _{D} |u-u^{st}_{\mathrm {k}}(s,\cdot )|^2 ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},s}^*(u) \Bigg ]\\&\quad \leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^{s} \Big (A^2B \Vert \partial _{x}u^{st}_{\mathrm {k}}(t,\cdot ) \Vert _{L^\infty (D)} + A^2B\Big ) \int \limits _{{\mathcal {F}}} \int \limits _{D} |u-u^{st}_{\mathrm {k}}(t,\cdot )|^2 ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t\\&\qquad + \int \limits _0^{s} \int \limits _{{\mathcal {F}}} \int \limits _{D} |{\mathcal {R}}^{st}_{\mathrm {k}}|^2 ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t + A\int \limits _{{\mathcal {F}}} \int \limits _{D} |{\bar{u}}-{\bar{u}}^{st}_{\mathrm {k}}|^2 ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}}) \Bigg ]. \end{aligned}$$
Upon using Grönwall’s inequality we obtain
$$\begin{aligned}&\sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _{{\mathcal {F}}} \int \limits _{D} |u-u^{st}_{\mathrm {k}}(s,\cdot )|^2 ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},s}^*(u) \Bigg ]\\&\quad \leqslant \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ A \Big ( \int \limits _0^{s} \int \limits _{{\mathcal {F}}} \int \limits _{D} |{\mathcal {R}}^{st}_{\mathrm {k}}|^2 ~ \mathrm {d}x\mathrm {d}\mu _{\mathrm {k},t}^*(u) \mathrm {d}t + A \int \limits _{{\mathcal {F}}} \int \limits _{D} |{\bar{u}}-{\bar{u}}^{st}_{\mathrm {k}}|^2 ~ \mathrm {d}x\mathrm {d}{\bar{\mu }}_{\mathrm {k}}^*({\bar{u}}) \Big ) \\&\qquad \times \exp \Big ( \int \limits _0^s A \Big (A^2B \Vert \partial _{x}u^{st}_{\mathrm {k}}(t,\cdot ) \Vert _{L^\infty (D)} + A^2 B\Big ) \, {\text {dt}} \Bigg ]. \end{aligned}$$
Using
\(\max \limits _{\mathrm {k}\in {\mathcal {K}}} \Vert \partial _{x}u^{st}_{\mathrm {k}}\Vert _{ L^{\infty }((0,s)\times D)}=:L\) and recalling that
\(({\overline{\mu }}^{*}_\mathrm {k})_{\mathrm {k}\in {\mathcal {K}}}\) corresponds to an optimal transport plan and that
\((\mu _{\mathrm {k},s}^*)_{\mathrm {k}\in {\mathcal {K}}}\) corresponds to an admissible transport plan, we finally obtain
$$\begin{aligned} W_{2}(\mu _{s}, {\hat{\mu }}_{s}^K)^2 \leqslant&\,A \Bigg ( \sum \limits _{\mathrm {k}\in {\mathcal {K}}} w_{\mathrm {k}}\Bigg [ \int \limits _0^{s} \int \limits _{D} |{\mathcal {R}}^{st}_{\mathrm {k}}|^2 ~ \mathrm {d}x\mathrm {d}t\Bigg ] + A W_{2}({\bar{\mu }},\hat{{\bar{\mu }}}^K)^2 \Bigg )\\&\times \exp \Big ( \int \limits _0^s \Big (A^3 B L + A^3 B \Big ) \mathrm {d}t\Big ). \end{aligned}$$
\(\square\)