solution is global. It also follows from that the sequence $\{(v_{n,m}; u_n; {\partial}_t u_n)\}$ contains a subsequence such that $$\begin{aligned}
&(v_{n,m}; u_n; {\partial}_t u_n) \rightharpoonup (v; u; {\partial}_t u) \quad \ast\mbox{-weakly in } L_\infty(0,T;{{\cal H}}),\label{uv-conv}
 \\
&v_{n,m} \rightharpoonup v \quad \mbox{weakly in } L_2(0,T;V).   \label{v_conv}\end{aligned}$$ Moreover, by the Aubin-Dubinsky theorem (see, e.g., [@sim Corollary 4]) we can assert that $$\begin{aligned}
&u_n \rightarrow u \quad \mbox{strongly in } C(0,T; H^{1-{\varepsilon}}_0({{\Omega}})\times H^{1-{\varepsilon}}_0({{\Omega}})\times H^{2-{\epsilon}}_0({{\Omega}}))
\label{u-strong}\end{aligned}$$ for every ${\varepsilon}>0$. Besides, the standard trace theorem yields $$\label{v_conv-b}
v_{n,m} \rightharpoonup v \quad \mbox{weakly in } L_2\left(0,T; \left[H^{1/2}(\partial{{\cal O}})\right]^3\right),$$ thus, we have $$\label{ut-conv}
{\partial}_t u_n \rightharpoonup {\partial}_t u \quad \mbox{weakly in } L_2(0,T; H^{1/2}_*({{\Omega}})).$$

One can see that $(v_{n,m}; u_n; {\partial}_t u_n)(t)$ satisfies (\[weak\_sol\_def\]) with the test function $\phi$ of the form $$\label{phi-pq}
\phi=\phi_{p,q}=\sum_{i=1}^p \gamma_i(t)\psi_i
+\sum_{j=1}^q\delta_j(t)\hat\phi_j
+\sum_{j=1}^q\eta_j(t)\tilde\phi_j,$$ where $p\le m$, $q\le n$ and $\gamma_i$, $\delta_j$, $\eta_j$ are scalar absolutely continuous functions on $[0,T]$ with time derivatives from $L_2(0,T)$, such that $\gamma_i(T)=\delta_j(T)=\eta_j(T)=0$. Using (\[uv-conv\]) and (\[v\_conv\]), we can easily pass to the limit in linear terms. Limit transition in the nonlinear terms can be done the same way as in [@Vor1957] (see also [@puel-AMO96] or [@KochLa_2002] for the same type argument). Thus, one can show that $(v; u; 