Rectangular no-slip Stokes velocity and distribution pressure

Apply [F3] to obtain the unique $u\in V$ and the stated variational identity and gradient bound. Define the vector functional $T=f+\mu \mathrm{\Delta }u$ on vector tests by $⟨T,\psi ⟩={\int }_{\mathrm{\Omega }}f\cdot \psi dx-\mu {\int }_{\mathrm{\Omega }}\mathrm{\nabla }u:\mathrm{\nabla }\psi dx\mathrm{.}$ It satisfies $\mathrm{\mid }⟨T,\psi ⟩\mathrm{\mid }\le (\mathrm{\parallel }f{\mathrm{\parallel }}_2+\mu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2)\mathrm{\parallel }\psi {\mathrm{\parallel }}_{H^1}$. For each fixed compact support $K$, $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{H^1}\le C_K{p}_{K,1}(\psi )$ componentwise, so $T$ is continuous on every ${\mathcal{D}}_K$ and hence belongs to $D^{\mathrm{\prime }}(\mathrm{\Omega };{\mathbb{R}}^d)$ with the stated inductive-limit topology. The variational identity shows that $T$ annihilates every compactly supported smooth divergence-free vector field, since each lies in $V$.

Each interval contains a central compact subinterval. Affinely rescale $b(s)=e^{-1\mathrm{/}(1-s^2)}$ on $\mathrm{\mid }s\mathrm{\mid }<1$, extended by zero. Every derivative is the exponential times a finite sum of rational terms with poles of finite order at $\mathrm{\mid }s\mathrm{\mid }=1$, and the exponential tends to zero faster than every such power; thus the extension is smooth. Normalize its positive finite integral to fix interval bumps ${\theta }_i$, independently of all input tests and support sets, and set $\theta (x)={\prod }_{i=1}^d{\theta }_i(x_i)$. Then $\theta \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and ${\int }_{\mathrm{\Omega }}\theta dx=1$.

The existence and test-space continuity in [F4] alone assert no Sobolev estimate. Choose a particular inverse explicitly, and verify its needed properties locally. In dimension one set $\mathcal{R}h(x)={\int }_{a_1}^{x}h(s)ds$ for $\int h=0$. In dimension $d>1$, put $x^{\mathrm{\prime }}=(x_2,\dots ,x_d)$, ${\mathrm{\Omega }}^{\mathrm{\prime }}={\prod }_{i=2}^d(a_i,b_i)$ and $H(x^{\mathrm{\prime }})={\int }_{a_1}^{b_1}h(s,x^{\mathrm{\prime }})ds$. Define recursively $(\mathcal{R}h{)}_1(x_1,x^{\mathrm{\prime }})={\int }_{a_1}^{x_1}(h(s,x^{\mathrm{\prime }})-{\theta }_1(s)H(x^{\mathrm{\prime }}))ds,(\mathcal{R}h{)}_j(x_1,x^{\mathrm{\prime }})={\theta }_1(x_1)({\mathcal{R}}^{\mathrm{\prime }}H{)}_{j-1}(x^{\mathrm{\prime }})(2\le j\le d),$ where ${\mathcal{R}}^{\mathrm{\prime }}$ is the same construction on ${\mathrm{\Omega }}^{\mathrm{\prime }}$ using ${\theta }_2,\dots ,{\theta }_d$. Smoothness and differentiation under these finite integrals follow from the smooth zero extension of the tests to a neighborhood of the closed box. Fubini gives ${\int }_{{\mathrm{\Omega }}^{\mathrm{\prime }}}H={\int }_{\mathrm{\Omega }}h=0$. The first-component derivative is $h-{\theta }_{1}H$; the sum of the other divergence terms is ${\theta }_{1}H$ by induction, proving $\mathrm{div}(\mathcal{R}h)=h$. If $h$ is supported in a fixed compact $K$, choose closed intervals inside each side containing its coordinate projection and $\mathrm{supp}{\theta }_i$. Their product contains every output support: the first primitive vanishes before this interval and after it because its full first-coordinate integral is zero, and the other components have this support by induction. For $p_m$ the supremum of derivatives of order at most $m$, $p_m(H)\le (b_1-a_1)p_m(h)$. Differentiating the displayed formula bounds each output derivative by a fixed bump-derivative constant times $p_m(h)$, a finite-integral length times that bound, or $p_m({\mathcal{R}}^{\mathrm{\prime }}H)$. The one-dimensional primitive has $p_m(\mathcal{R}h)\le \max (1,b_1-a_1)p_m(h)$; induction therefore gives $p_m(\mathcal{R}h)\le C_{\mathrm{\Omega },m}{p}_m(h)$. All operations are linear. This proves that this chosen map has the right-inverse and support-continuity properties, without extracting a quantitative Sobolev estimate from [F4].

For $\varphi \in \mathcal{D}(\mathrm{\Omega })$ put ${\varphi }_0=\varphi -({\int }_{\mathrm{\Omega }}\varphi dx)\theta$ and define $⟨p,\varphi ⟩=-⟨T,\mathcal{R}{\varphi }_0⟩$. This is linear. For tests supported in fixed $K$, ${\varphi }_0$ is supported in $K\cup \mathrm{supp}\theta$, and step 2.1 puts $\mathcal{R}{\varphi }_0$ in a common compact support with every output seminorm bounded by input seminorms. Continuity of $T$ on that test space proves that $p$ is continuous on every ${\mathcal{D}}_K$, hence a distribution. Its construction gives $⟨p,\theta ⟩=0$.

Write ${\mathrm{\ell }}_1=b_1-a_1$ and $Jg(x_1,x^{\mathrm{\prime }})={\int }_{a_1}^{x_1}g(s,x^{\mathrm{\prime }})ds$. Cauchy–Schwarz and Fubini give $\mathrm{\parallel }Jg{\mathrm{\parallel }}_2\le {\mathrm{\ell }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_2$, ${\mathrm{\partial }}_{1}Jg=g$ and ${\mathrm{\partial }}_{x_j}Jg=J({\mathrm{\partial }}_{x_j}g)$ for $j>1$, hence $\mathrm{\parallel }Jg{\mathrm{\parallel }}_{H^1}\le \sqrt{1+{\mathrm{\ell }}_1^2}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^1}$. They also give $\mathrm{\parallel }H{\mathrm{\parallel }}_{H^1({\mathrm{\Omega }}^{\mathrm{\prime }})}\le \sqrt{{\mathrm{\ell }}_1}\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$. With $B_1^2=\mathrm{\parallel }{\theta }_1{\mathrm{\parallel }}_{L^2(a_1,b_1)}^2+\mathrm{\parallel }{\theta }_1^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(a_1,b_1)}^2$, the product rule and Fubini give $\mathrm{\parallel }{\theta }_{1}G{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\le B_1\mathrm{\parallel }G{\mathrm{\parallel }}_{H^1({\mathrm{\Omega }}^{\mathrm{\prime }})}$, also for vector $G$. Thus the first component in step 2.1 is bounded in $H^1$ by $\sqrt{1+{\mathrm{\ell }}_1^2}(1+B_1\sqrt{{\mathrm{\ell }}_1})\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1}$; the other components together are bounded by $B_1\sqrt{{\mathrm{\ell }}_1}{C}_{{\mathrm{\Omega }}^{\mathrm{\prime }}}\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1}$ if the inductive inverse bound is $C_{{\mathrm{\Omega }}^{\mathrm{\prime }}}$. The one-dimensional primitive starts the induction, proving $\mathrm{\parallel }\mathcal{R}h{\mathrm{\parallel }}_{H^1}\le C_{\mathrm{\Omega }}\mathrm{\parallel }h{\mathrm{\parallel }}_{H^1}$ with a constant independent of the input support. Also $\mathrm{\parallel }{\varphi }_0{\mathrm{\parallel }}_{H^1}\le (1+\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }\theta {\mathrm{\parallel }}_{H^1})\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{H^1}$, since $\mathrm{\mid }{\int }_{\mathrm{\Omega }}\varphi \mathrm{\mid }\le \mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2$. Therefore $\mathrm{\mid }⟨p,\varphi ⟩\mathrm{\mid }\le C_{\mathrm{\Omega }}(\mathrm{\parallel }f{\mathrm{\parallel }}_2+\mu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2)\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{H^1}$. For any $H_0^1$ element choose smooth tests converging in $H^1$; this bound makes their pressure values Cauchy with the same limit for any such sequence. The resulting extension is a bounded linear functional on $H_0^1$, so $p\in H^{-1}$.

For compactly supported smooth vector $\psi$, ${\int }_{\mathrm{\Omega }}\mathrm{div}\psi dx=0$. The field $\psi -\mathcal{R}(\mathrm{div}\psi )$ is divergence-free by step 2.1, so step 1.1 implies $⟨T,\mathcal{R}(\mathrm{div}\psi )⟩=⟨T,\psi ⟩$. By the definition of distribution gradient, $⟨\mathrm{\nabla }p,\psi ⟩=-⟨p,\mathrm{div}\psi ⟩=⟨T,\psi ⟩$, hence $\mathrm{\nabla }p=T$ and $-\mu \mathrm{\Delta }u+\mathrm{\nabla }p=f$. If $p_1,p_2$ both work, then $\mathrm{\nabla }(p_1-p_2)=0$. Every $h\in D_0(\mathrm{\Omega })$ is a divergence by [F4], so the real pairing convention gives $⟨p_1-p_2,h⟩=0$. Decomposing any test as ${\varphi }_0+({\int }_{\mathrm{\Omega }}\varphi dx)\theta$ shows $p_1-p_2$ is constant. Uniqueness of the variational velocity in [F3] makes $f\mapsto u$ linear: any linear combination of two velocities solves the variational equation for the same combination of their forcings. With the bumps fixed, $T$ and the normalized $p$ are then linear in $f$ as well. By [F5] and [F3], $\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1}\le \sqrt{1+4R^2}2R\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{/}\mu$, while step 4.1 and [F3] give $\mathrm{\parallel }[p]\mathrm{\parallel }\le \mathrm{\parallel }p{\mathrm{\parallel }}_{H^{-1}}\le C_{\mathrm{\Omega }}(1+2R)\mathrm{\parallel }f{\mathrm{\parallel }}_2$. These prove the bounded linear map into the stated product.