Integrated pressure recovery and MINI pressure-primitive convergence

The weakly continuous representative of $u$ obeys $\mathrm{\parallel }u(t){\mathrm{\parallel }}_2\le U$ for every $t\in [0,T]$. Indeed, if $\mathrm{\parallel }u(t_0){\mathrm{\parallel }}_2>U$, take $w=u(t_0)\mathrm{/}\mathrm{\parallel }u(t_0){\mathrm{\parallel }}_2\in H_{\sigma }$. Weak continuity makes $(u(t),w)$ continuous and greater than $U$ on a relative neighborhood of $t_0$ with positive time measure, contradicting the essential bound $U$.

Define $g(t)\in V^{\ast }$ by $g(t)(v)=(f(t),v{)}_2-a_{\nu }(u(t),v)-c(u(t);u(t),v)$. The first two terms are strongly measurable bounded linear images of Bochner functions; the last is a continuous quadratic map $V\to V^{\ast }$ by the trilinear bound in [F3], hence is strongly measurable. Poincare and [F3] give $\mathrm{\parallel }g(t){\mathrm{\parallel }}_{V^{\ast }}\le C_P\mathrm{\parallel }f(t){\mathrm{\parallel }}_2+2\nu \mathrm{\parallel }u(t){\mathrm{\parallel }}_V+C_c\mathrm{\parallel }u(t){\mathrm{\parallel }}_2^{1\mathrm{/}2}\mathrm{\parallel }u(t){\mathrm{\parallel }}_V^{3\mathrm{/}2}\mathrm{.}$ Holder makes $g\in L^{4\mathrm{/}3}(0,T;V^{\ast })\subset L^1(0,T;V^{\ast })$ and yields $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^1{V}^{\ast }}\le C_P\sqrt{T}F+2\nu \sqrt{T}E+C_c{T}^{1\mathrm{/}4}{U}^{1\mathrm{/}2}{E}^{3\mathrm{/}2}$.

For a mixed MINI solution define ${\widehat{u}}_h$ by affine interpolation of the velocity nodes and let bars denote right-constant reconstructions. On each time cell, ${\mathrm{\partial }}_t{P}_h={\bar{p}}_h$ and ${\mathrm{\partial }}_t{\widehat{u}}_h=D_{\tau }{u}_h^n$. Integrating the discrete equation over complete or partial cells gives, for every $t$ and $v_h\in V_h$, $({\widehat{u}}_h(t)-u_h^0,v_h{)}_2+{\int }_0^t(a_{\nu }({\bar{u}}_h,v_h)+c({\bar{u}}_h;{\bar{u}}_h,v_h))ds+b(v_h,P_h(t))={\int }_0^t(f_h,v_h{)}_{2}ds\mathrm{.}$ Because $P_h(t)\in Q_h$, the uniform inf-sup estimate applies. Convexity gives $\mathrm{\parallel }{\widehat{u}}_h(t){\mathrm{\parallel }}_2\le U_h$; [F2], Holder, and [F3] bound the mass, force, viscosity and convection terms by $2C_P{U}_h$, $C_P\sqrt{T}F$, $2\nu \sqrt{T}{E}_h$, and $C_c{T}^{1\mathrm{/}4}{U}_h^{1\mathrm{/}2}{E}_h^{3\mathrm{/}2}$ times $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_V$. This proves the displayed uniform primitive bound.

Let $u_{h,j}^0=P_{Z_j}{u}_0$. For any $ϵ>0$, choose $v\in V_{\sigma }$ with $\mathrm{\parallel }{u}_0-v{\mathrm{\parallel }}_2<ϵ$; the Fortin map satisfies ${\mathrm{\Pi }}_{j}v\in Z_j$ and $\mathrm{\parallel }{\mathrm{\Pi }}_{j}v-v{\mathrm{\parallel }}_2\to 0$. Best approximation of the orthogonal projector gives $\mathrm{\parallel }{u}_{h,j}^0-u_0{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\mathrm{\Pi }}_{j}v-u_0{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\mathrm{\Pi }}_{j}v-v{\mathrm{\parallel }}_2+ϵ$. Letting $j\to \mathrm{\infty }$ and then $ϵ\downarrow 0$ proves $u_{h,j}^0\to u_0$ strongly.

Set $G(t)={\int }_0^{t}g(s)ds$. By [F5], $G$ is norm-continuous in $V^{\ast }$, $G(0)=0$, and ${\sup }_t\mathrm{\parallel }G(t){\mathrm{\parallel }}_{V^{\ast }}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^1{V}^{\ast }}$.

The energy theorem [F8] gives ${\sum }_n\mathrm{\parallel }{u}_{h,j}^n-u_{h,j}^{n-1}{\mathrm{\parallel }}_2^2\le C$ uniformly. Therefore $\mathrm{\parallel }{\widehat{u}}_{h,j}-{\bar{u}}_{h,j}{\mathrm{\parallel }}_{L^2{L}^2}^2=\frac{1}{3}{\sum }_n{\tau }_n^j\mathrm{\parallel }{u}_{h,j}^n-u_{h,j}^{n-1}{\mathrm{\parallel }}_2^2\le C{\tau }_{\max ,j}⟶0.$ The pressure bound in step 1.3 is uniform in $j$, so $(P_{h,j})$ is bounded in the separable Hilbert space $L^2(0,T;Q)$. By [A1] and [F7], a further subsequence converges weakly to some $P_{\ast }\in L^2(0,T;Q)$. If $K$ is the uniform $L^{\mathrm{\infty }}Q$ bound and $A_{ϵ}=\{t:\mathrm{\parallel }{P}_{\ast }(t){\mathrm{\parallel }}_2>K+ϵ\}$ had positive measure, testing weak convergence against ${\mathbf{1}}_{A_{ϵ}}{P}_{\ast }$ gives ${\int }_{A_{ϵ}}\mathrm{\parallel }{P}_{\ast }(t){\mathrm{\parallel }}_2^{2}dt={\lim }_j{\int }_{A_{ϵ}}(P_{h,j}(t),P_{\ast }(t){)}_{2}dt\le K{\int }_{A_{ϵ}}\mathrm{\parallel }{P}_{\ast }(t){\mathrm{\parallel }}_{2}dt\mathrm{.}$ The left side is at least $(K+ϵ){\int }_{A_{ϵ}}\mathrm{\parallel }{P}_{\ast }(t){\mathrm{\parallel }}_{2}dt$, a contradiction. Hence every $A_{ϵ}$ is null and $P_{\ast }\in L^{\mathrm{\infty }}Q$ with bound $K$.

Conversely suppose the actual $u_t^{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$ is represented in $L^{4\mathrm{/}3}(0,T;V^{\ast })$ and set $r=g-u_t^{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. For each $z\in V_{\sigma }$, the scalar weak equation [F1] gives $r(t)(z)=0$ as a time distribution. The space $V$ is separable: with its equivalent full $H^1$ norm it embeds isometrically into a finite product of scalar $L^2(D)$ spaces by listing each velocity component and its first derivatives; restricting rational-box step functions makes that product separable. The subspace $V_{\sigma }=\ker \mathrm{div}$ is closed, hence separable (a countable base for the ambient metric space restricts to a countable base on $V_{\sigma }$, and Countable Choice selects one point from each nonempty base member). For a countable dense family $(z_k)$ in $V_{\sigma }$, each scalar $r(\cdot )(z_k)\in L^1(0,T)$ is zero almost everywhere by [F6]; outside the countable union of these null sets, continuity of $r(t)\in V^{\ast }$ makes it vanish on all of $V_{\sigma }$.

For every $t$, define the full-test residual $R(t)\in V^{\ast }$ by $R(t)(v)=G(t)(v)-(u(t)-u_0,v{)}_2$. This functional is bounded with $\mathrm{\parallel }R(t){\mathrm{\parallel }}_{V^{\ast }}\le C_P(U+\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_2)+\mathrm{\parallel }g{\mathrm{\parallel }}_{L^1{V}^{\ast }}$. For each fixed $v\in V$, the scalar map $t\mapsto R(t)(v)$ is continuous: orthogonally project $v$ onto the closed subspace $H_{\sigma }$; since $u(t)\in H_{\sigma }$, its pairing with $v$ equals its pairing with that projection and is continuous by [F1].

For the convergence of convection, use the separate Sobolev routes in [F3]: $H_0^1\hookrightarrow L^6$ by the $p=2$ Sobolev exponent in dimension three, and by the $p=3\mathrm{/}2$ inequality plus finite-measure Holder in dimension two. Interpolation gives $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^3}\le \mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^2}^{1\mathrm{/}2}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^6}^{1\mathrm{/}2}$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^4}\le \mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^2}^{1\mathrm{/}4}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2{L}^6}^{3\mathrm{/}4}$. The energy bounds give uniform $L^2{L}^6$ norms, while strong $L^2{L}^2$ convergence in [F9] therefore implies strong $L^2{L}^3$ and $L^2{L}^4$ convergence. For fixed $v\in V$, the gradient term in the skew form pairs weakly convergent gradients with ${\bar{u}}_{h,j}\otimes v$ converging strongly in $L^2{L}^2$; its other term pairs $\mathrm{\nabla }v$ with ${\bar{u}}_{h,j}\otimes {\bar{u}}_{h,j}$ converging strongly in $L^1{L}^2$, since $\mathrm{\parallel }{\bar{u}}_{h,j}\otimes {\bar{u}}_{h,j}-u\otimes u{\mathrm{\parallel }}_{L^1{L}^2}\le (\mathrm{\parallel }{\bar{u}}_{h,j}{\mathrm{\parallel }}_{L^2{L}^4}+\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2{L}^4})\mathrm{\parallel }{\bar{u}}_{h,j}-u{\mathrm{\parallel }}_{L^2{L}^4}\to 0.$ Replacing $v$ by ${\mathrm{\Pi }}_{j}v$ changes the integral by at most $C\mathrm{\parallel }{\mathrm{\Pi }}_{j}v-v{\mathrm{\parallel }}_V\mathrm{\parallel }{\bar{u}}_{h,j}{\mathrm{\parallel }}_{L^{2}V}^2$, which tends to zero. No divergence-free property of the test $v$ is used.

Fix $z\in V_{\sigma }$. The scalar weak equation [F1] says that $(u(t),z{)}_2$ has distributional derivative $g(t)(z)$ and initial trace $(u_0,z{)}_2$. Since $g(z)\in L^1$, [F6] gives an absolutely-continuous representative equal to $(u_0,z{)}_2+G(t)(z)$. Weak continuity in [F1] identifies that representative with $(u(t),z{)}_2$ for every $t$. Hence $R(t)(z)=0$ for every $z\in V_{\sigma }$ and every $t$, and $R(0)=0$.

For each $t$, the map $q\mapsto -R(t)(Bq)$ is a bounded real-linear functional on $Q$ of norm at most $C_{\mathrm{d}\mathrm{i}\mathrm{v}}\mathrm{\parallel }R(t){\mathrm{\parallel }}_{V^{\ast }}$. Riesz representation [F7] therefore defines a unique $P(t)\in Q$ by $(P(t),q{)}_2=-R(t)(Bq)$ for all $q\in Q$. The uniform bound from step 3.1 gives the stated $L^{\mathrm{\infty }}Q$ estimate. For fixed $q$, $t\mapsto (P(t),q{)}_2=-R(t)(Bq)$ is continuous, so $P\in C_w([0,T];Q)$ and $P(0)=0$. Choose a countable orthonormal basis of the separable Hilbert space $Q$. Each finite-coordinate projection of $P$ is strongly measurable because its coordinates are continuous; the projections converge in norm to $P(t)$ at every $t$. Thus $P$ is strongly measurable and belongs to $L^{\mathrm{\infty }}(0,T;Q)$.

For $v\in V$, let $q=\mathrm{div}v\in Q$ by [F2]. Then $v-Bq\in V_{\sigma }$, so step 4.1 gives $R(t)(v)=R(t)(Bq)=-(P(t),q{)}_2=b(v,P(t))$. Substitution of the definition of $R$ proves the integrated identity for every $t$ and $v$. If $P_1,P_2$ both satisfy it, set $v=Bq$ for arbitrary $q\in Q$ to obtain $(P_1(t)-P_2(t),q{)}_2=0$ for every $q$, hence $P_1(t)=P_2(t)$.

The formula $⟨p,\eta ⟩=-{\int }_0^{T}P(t){\eta }^{\mathrm{\prime }}(t)dt$ defines a continuous $Q$-valued distribution because its norm is bounded by $\mathrm{\parallel }P{\mathrm{\parallel }}_{L^{\mathrm{\infty }}Q}\mathrm{\parallel }{\eta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^1}$. Differentiating step 6.1 against scalar tests gives $u_t+a_{\nu }(u,\cdot )+c(u;u,\cdot )+b(\cdot ,p)=f$ in $V^{\ast }$-valued distributions. If another $Q$-valued distribution has this property, test the difference against $Bq$ for each $q\in Q$; since $\mathrm{div}Bq=q$, the $Q$ inner product of the difference with every $q$ vanishes, so the difference is zero.

Fix $v\in V$ and $\eta \in C_c^{\mathrm{\infty }}(0,T)$. Test the integrated identity of step 1.3 with ${\mathrm{\Pi }}_{j}v$ and integrate in $t$ against $\eta$. Scalar Fubini rewrites each primitive integral using the bounded weight $s\mapsto {\int }_s^T\eta (t)dt$. Strong convergence of ${\widehat{u}}_{h,j}$ from [F9] and step 2.2, strong convergence of initial projections from step 1.4, weak $L^{2}V$ convergence from [F9], convergence of force averages from [F9], Fortin convergence ${\mathrm{\Pi }}_{j}v\to v$ in $V$, the convection convergence in step 3.2, and weak convergence $P_{h,j}\rightharpoonup P_{\ast }$ pass every term to the limit. In the pressure term, weak convergence of $P_{h,j}$ is paired with the strongly convergent tests $\eta \mathrm{div}{\mathrm{\Pi }}_{j}v\to \eta \mathrm{div}v$ in $L^2(0,T;Q)$. The result is the continuous integrated identity of step 6.1 with $P_{\ast }$ in place of $P$, as a scalar time distribution.

For fixed $v\in V$, step 6.1 says $(u(t),v{)}_2+b(v,P(t))=(u_0,v{)}_2+{\int }_0^{t}g(s)(v)ds$. The right side is absolutely continuous. Multiply this scalar identity by a $C^1$ time function vanishing near $T$ and integrate by parts; for a finite sum of tests $\phi (t)={\sum }_{k=1}^m{\eta }_k(t)v_k$ this gives the stated initial-trace formula. Such finite sums are dense in $C^1([0,T];V)$ for tests vanishing near $T$: since $\phi$ vanishes on a terminal interval, so does $\phi \mathrm{\prime }$. On successively finer partitions, piecewise-linear interpolation of this uniformly continuous derivative has finite-dimensional range, converges uniformly to $\phi \mathrm{\prime }$, and remains zero on a terminal interval. Write an interpolant as ${\sum }_{k=1}^m{\chi }_k(t)v_k$ with fixed $v_k\in V$ and continuous piecewise-linear scalar functions ${\chi }_k$. Then ${\phi }_m(t)=-{\sum }_{k=1}^m({\int }_t^T{\chi }_k(s)ds)v_k$ is a finite sum with $C^1$ scalar coefficients, vanishes near $T$, and converges to $\phi$ in $C^1$. Every term in the weak-form identity is continuous in that norm by [F1]–[F5], so the identity extends to every stated $\phi$.

Subtract step 6.1 from the limit identity in step 7.2 and set $v=Bq$ for each $q\in Q$. Then $(P_{\ast }-P,q{)}_2=0$ as a scalar time distribution. Since $P_{\ast }-P\in L^2(0,T;Q)$ and $Q$ has a countable dense subset, [F6] and the countable-union result make $P_{\ast }=P$ almost everywhere. For $\eta \in C_c^{\mathrm{\infty }}(0,T)$, the identity ${\mathrm{\partial }}_t{P}_{h,j}={\bar{p}}_{h,j}$ gives ${\int }_0^T({\bar{p}}_{h,j},q{)}_2\eta dt=-{\int }_0^T(P_{h,j},q{)}_2{\eta }^{\mathrm{\prime }}dt⟶-{\int }_0^T(P,q{)}_2{\eta }^{\mathrm{\prime }}dt=(⟨p,\eta ⟩,q{)}_2\mathrm{.}$ This is the claimed weak convergence as $Q$-valued time distributions.

To prove the forward direction of the instantaneous-pressure criterion, suppose $p\in L^{4\mathrm{/}3}(0,T;Q)$. The full equation in step 7.1 gives $u_t^{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}=g-b(\cdot ,p)$ as a $V^{\ast }$-valued distribution. Since $\mathrm{\mid }\mathrm{div}v\mathrm{\mid }\le \sqrt{d}\mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }$, the map $q\mapsto b(\cdot ,q)$ is bounded from $Q$ to $V^{\ast }$; [F5] therefore makes $b(\cdot ,p)$ Bochner $L^{4\mathrm{/}3}{V}^{\ast }$. As $g\in L^{4\mathrm{/}3}{V}^{\ast }$ by step 1.2, the actual full derivative has that same regularity.

The continuous residual-recovery lemma [F10] applies to $r\in L^{4\mathrm{/}3}(0,T;V^{\ast })$ and gives $p_s\in L^{4\mathrm{/}3}(0,T;Q)$ with $r(t)(v)=b(v,p_s(t))$ for all $v\in V$ almost everywhere. Thus $p_s$ is a full pressure distribution for $u$. Uniqueness in step 7.1 identifies $p_s$ with $p$, proving the reverse direction. Therefore the exact necessary-and-sufficient condition is the stated regularity of the actual full derivative.