Entropy inequality under an assumed viscous limit

By [F2], the chain rule and compatibility give, pointwise for each $ϵ>0$, ${\mathrm{\partial }}_t\eta (u^{ϵ})+{\sum }_j{\mathrm{\partial }}_{x_j}{q}_j(u^{ϵ})=ϵ{\eta }^{\mathrm{\prime }}(u^{ϵ})\mathrm{\Delta }{u}^{ϵ}\mathrm{.}$ Applying the chain rule twice in each spatial coordinate and the product rule yields ${\eta }^{\mathrm{\prime }}(u^{ϵ})\mathrm{\Delta }{u}^{ϵ}=\mathrm{\Delta }\eta (u^{ϵ})-{\eta }^{\mathrm{\prime }\mathrm{\prime }}(u^{ϵ})\mathrm{\mid }\mathrm{\nabla }{u}^{ϵ}{\mathrm{\mid }}^2$. Since ${\eta }^{\mathrm{\prime }\mathrm{\prime }}\ge 0$, the entropy residual is at most $ϵ\mathrm{\Delta }\eta (u^{ϵ})$.

Multiply by $\psi \ge 0$ and integrate. By [F4], the integrals can be taken iteratively in time and each spatial coordinate. Applying [F3] to the smooth compactly supported products gives integration by parts twice in each spatial coordinate and once in time, with $\psi$ vanishing near $T$, and yields ${\int }_0^T\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{\mathrm{\Omega }}(\eta (u^{ϵ}){\psi }_t+{\sum }_j{q}_j(u^{ϵ}){\psi }_{x_j})dxdt+{\int }_{\mathrm{\Omega }}\eta (u_0^{ϵ})\psi (0,x)dx\ge -ϵ{\int }_0^T\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_{\mathrm{\Omega }}\eta (u^{ϵ})\mathrm{\Delta }\psi dxdt\mathrm{.}$ These integrations by parts are the one-dimensional FTC applied to the smooth compactly supported products in each coordinate and in time; no spatial boundary term remains.

Continuity of $u^{ϵ}$ up to $t=0$ and its almost-everywhere bound on $(0,T)\times \mathrm{\Omega }$ imply $\mathrm{\mid }{u}_0^{ϵ}\mathrm{\mid }\le M$ on $\mathrm{\Omega }$: a value larger than $M$ at any $x_0\in \mathrm{\Omega }$ would, by continuity, give a relative open neighborhood in $(0,T)\times \mathrm{\Omega }$ of positive measure where $\mathrm{\mid }{u}^{ϵ}\mathrm{\mid }>M$. For every compact $K\subseteq \mathrm{\Omega }$, this bound and the strong initial convergence give $0\le {\int }_K(\mathrm{\mid }{u}_0\mathrm{\mid }-M{)}_{+}dx\le {\int }_K\mathrm{\mid }{u}_0-u_0^{ϵ}\mathrm{\mid }dx\to 0$, so $(\mathrm{\mid }{u}_0\mathrm{\mid }-M{)}_{+}=0$ almost everywhere on each such $K$. The countable compact exhaustion $K_N=\{x\in \mathrm{\Omega }:\mathrm{\mid }x\mathrm{\mid }\le N,\mathrm{dist}(x,{\mathbb{R}}^d\setminus \mathrm{\Omega })\ge 1\mathrm{/}N\}$ (taking the distance to the empty set to be $+\mathrm{\infty }$) covers $\mathrm{\Omega }$, hence $\mathrm{\mid }{u}_0\mathrm{\mid }\le M$ almost everywhere on $\mathrm{\Omega }$. The functions $\eta$ and $q_j$ are Lipschitz and bounded on $[-M,M]$. Thus [A2] implies local $L^1$ convergence of $\eta (u^{ϵ})$ and $q_j(u^{ϵ})$ to $\eta (u)$ and $q_j(u)$, and of $\eta (u_0^{ϵ})$ to $\eta (u_0)$. For a given test, choose $0<\tau <T$ and compact $K\subseteq \mathrm{\Omega }$ with its support in $[0,\tau ]\times K$. The same bounded ranges imply absolute integrability of $\eta (u)$ and all $q_j(u)$ on this slab, as required by [F1]. For $0<\delta <\tau$, the spacetime terms converge on $[\delta ,\tau ]\times K$ by the interior $L^1$ convergence and these Lipschitz bounds. On $(0,\delta )\times K$, their difference is bounded in absolute integral by $2\delta {\lambda }_d(K)({\sup }_{[-M,M]}\mathrm{\mid }\eta \mathrm{\mid }\mathrm{\parallel }{\psi }_t{\mathrm{\parallel }}_{\mathrm{\infty }}+{\sum }_{j=1}^d{\sup }_{[-M,M]}\mathrm{\mid }{q}_j\mathrm{\mid }\mathrm{\parallel }{\psi }_{x_j}{\mathrm{\parallel }}_{\mathrm{\infty }}),$ uniformly in $ϵ$. Letting $ϵ\downarrow 0$ first and then $\delta \downarrow 0$ proves convergence for the complete initial-boundary test. Moreover, the absolute value of the right-hand side in step 2.1 is at most $ϵ{\sup }_{[-M,M]}\mathrm{\mid }\eta \mathrm{\mid }\int \mathrm{\mid }\mathrm{\Delta }\psi \mathrm{\mid }$, which tends to zero. Passing to the limit yields the stated entropy inequality.