Conditional monotone finite-volume parabolic limit identification

The strong convergence ${\bar{U}}_m\to u$ and the reconstruction estimate in [A2] imply $w_m\to u$ strongly in $L^p(0,T;H)$. The sequence is bounded in $L^p(0,T;V)$, so [F3] gives, after subsequence extraction, $w_m\rightharpoonup u$ in $L^p(0,T;V)$ and ${\xi }_m=\mathrm{\nabla }{w}_m\rightharpoonup \mathrm{\nabla }u$ in $L^p(Q;{\mathbb{R}}^d)$.

Fix $\phi \in C_c^{\mathrm{\infty }}([0,T)\times \mathrm{\Omega })$. Strong convergence of ${\bar{U}}_m$ gives ${\int }_Q{\bar{U}}_m{\mathrm{\partial }}_t\phi \to {\int }_{Q}u{\mathrm{\partial }}_t\phi$; weak convergence of $q_m$ gives ${\int }_Q{q}_m\cdot \mathrm{\nabla }\phi \to {\int }_{Q}q\cdot \mathrm{\nabla }\phi$. The source and initial terms are fixed, and [A2] gives $r_m(\phi )\to 0$. Passing to the limit in the discrete weak identity therefore yields $-{\int }_{Q}u{\mathrm{\partial }}_t\phi +{\int }_{Q}q\cdot \mathrm{\nabla }\phi ={\int }_{Q}f\phi +{\int }_{\mathrm{\Omega }}{u}_0\phi (\cdot ,0)\mathrm{.}$

Step 1.1 gives ${\xi }_m\rightharpoonup \mathrm{\nabla }u$ in $L^p$, the statement gives $q_m=a(x,{\xi }_m)\rightharpoonup q$ in $L^{p^{\mathrm{\prime }}}$, and the assumed energy limsup is exactly the remaining hypothesis of [F4]. Thus [F4] implies $q=a(x,\mathrm{\nabla }u)$ almost everywhere.

For $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\theta \in C_c^{\mathrm{\infty }}(0,T)$, use $\phi (t,x)=\theta (t)v(x)$ in step 2.1. The weak identity shows in distributions that $u^{\mathrm{\prime }}=g$, where $⟨g(t),v⟩={\int }_{\mathrm{\Omega }}f(t,x)v(x)dx-{\int }_{\mathrm{\Omega }}q(t,x)\cdot \mathrm{\nabla }v(x)dx$. By H"older and the Poincar'e norm equivalence in [F2], $\mathrm{\parallel }g(t){\mathrm{\parallel }}_{V^{\ast }}\le C_P\mathrm{\parallel }f(t){\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(\mathrm{\Omega })}+\mathrm{\parallel }q(t){\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(\mathrm{\Omega })}$ for almost every $t$. Thus $g\in L^{p^{\mathrm{\prime }}}(0,T;V^{\ast })$. The identity extends from smooth $v$ to every $v\in V$ by the defining density of $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ in $W_0^{1,p}(\mathrm{\Omega })$ and the displayed bound, so it is the distributional derivative from [F1]. Hence $u\in W_p(0,T;V)$.

By [A1], [A3], step 3.1, and the assumed $u\in L^p(0,T;V)$, the continuous-trace theorem [F5] gives a unique representative $u\in C([0,T];H)$.

Fix $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and $\theta \in C_c^{\mathrm{\infty }}([0,T))$. The scalar function $t\mapsto (u(t),v{)}_H$ is bounded on $(0,T)$ by [A3], and its distributional derivative $t\mapsto ⟨u^{\mathrm{\prime }}(t),v⟩$ lies in $L^1(0,T)$ by step 3.1 and finite $T$. Thus it belongs to $W^{1,1}(0,T)$, so [A1] and [F6] give an absolutely continuous representative and integration by parts on $[0,T]$. This representative agrees almost everywhere with the continuous scalar function from step 4.1; since both are continuous, they agree everywhere, including at $t=0$. Apply step 2.1 to $\phi =\theta v$ and use $⟨u^{\mathrm{\prime }},v⟩=\int fv-\int q\cdot \mathrm{\nabla }v$ from step 3.1. The bulk terms cancel and leave $(u(0)-u_0,v{)}_H\theta (0)=0$. Choose $\theta (0)=1$. This holds for every smooth compactly supported $v$, which is dense in $H$; therefore $u(0)=u_0$ in $H$.

By step 3.1 and the definition of the induced operator in [F2], $u^{\mathrm{\prime }}+A(u)=f$ in $L^{p^{\mathrm{\prime }}}(0,T;V^{\ast })$ after substituting $q=a(x,\mathrm{\nabla }u)$ from step 2.2. Step 4.1 gives the required continuous $H$ representative and step 5.1 gives its initial trace. These are exactly the weak evolution solution conditions of [F2].