Bounded-domain residual error with explicit channel lift

Put $e=v-u$ and $\pi =q-p$. Subtracting the exact equation for $u$ from the residual equation for $v$ gives $e_t+(v\cdot \mathrm{\nabla })e+(e\cdot \mathrm{\nabla })u+{\rho }_0^{-1}\mathrm{\nabla }\pi -\nu \mathrm{\Delta }e=r$, because $(v\cdot \mathrm{\nabla })v-(u\cdot \mathrm{\nabla })u=(v\cdot \mathrm{\nabla })e+(e\cdot \mathrm{\nabla })u$.

In the channel case, $\mathrm{\ell }{\mathrm{\mid }}_{y=0}=a{e}_x$, $\mathrm{\ell }{\mathrm{\mid }}_{y=h}=b{e}_x$, $\mathrm{div}\mathrm{\ell }=0$, ${\mathrm{\ell }}_t=(a^{\mathrm{\prime }}(1-y\mathrm{/}h)+b^{\mathrm{\prime }}y\mathrm{/}h)e_x$, $\mathrm{\Delta }\mathrm{\ell }=0$, and $(\mathrm{\ell }\cdot \mathrm{\nabla })\mathrm{\ell }=0$, since $\mathrm{\ell }$ depends only on $y$ in its first component and that component is affine in $y$. Thus $v=w-\mathrm{\ell }$ is divergence-free and has the exact wall trace.

Set $Y(t)=\mathrm{\parallel }e(t){\mathrm{\parallel }}_2^2$ and $R(t)=\mathrm{\parallel }r(t){\mathrm{\parallel }}_2$. Pair the equation in step 1.1 with $e$ and integrate over $D$. The transport integral is $\frac{1}{2}{\int }_D\mathrm{div}(v\mathrm{\mid }e{\mathrm{\mid }}^2)=0$ because $\mathrm{div}v=0$ and $e=0$ on $\mathrm{\partial }D$; the pressure integral is ${\rho }_0^{-1}{\int }_D\mathrm{div}(\pi e)=0$ because $\mathrm{div}e=0$ and $e=0$ on $\mathrm{\partial }D$; and the diffusion term is $-\nu {\int }_{D}e\cdot \mathrm{\Delta }e=\nu {\int }_D\mathrm{\mid }\mathrm{\nabla }e{\mathrm{\mid }}^2$ because ${\int }_D\mathrm{div}({\sum }_i{e}_i\mathrm{\nabla }{e}_i)=0$. These three identities use [F2]. The bilinear gradient norm gives $\mathrm{\mid }(e\cdot \mathrm{\nabla })u\cdot e\mathrm{\mid }\le L(t)\mathrm{\mid }e{\mathrm{\mid }}^2$, and the pointwise Euclidean bound followed by scalar Cauchy–Schwarz in [F3] gives ${\int }_D\mathrm{\mid }e\cdot r\mathrm{\mid }\le {\int }_D\mathrm{\mid }e\mathrm{\mid }\mathrm{\mid }r\mathrm{\mid }\le \mathrm{\parallel }e{\mathrm{\parallel }}_2\mathrm{\parallel }r{\mathrm{\parallel }}_2=\sqrt{Y}R$. Pointwise in $x$, the difference quotients of $\mathrm{\mid }e(t,x){\mathrm{\mid }}^2$ converge uniformly on $\bar{D}$ to $2e\cdot e_t$ as the time increment tends to zero, because $e,e_t$ are continuous on the compact space-time closure; integrating preserves this uniform limit since $D$ has finite volume. Therefore $Y^{\mathrm{\prime }}=2{\int }_{D}e\cdot e_t$ and $\frac{1}{2}{Y}^{\mathrm{\prime }}(t)+\nu \mathrm{\parallel }\mathrm{\nabla }e(t){\mathrm{\parallel }}_2^2\le L(t)Y(t)+R(t)\sqrt{Y(t)}$.

Expanding the convective term in $\mathcal{R}(w-\mathrm{\ell },q)$ gives $\mathcal{R}(w-\mathrm{\ell },q)=\mathcal{R}(w,q)-{\mathrm{\ell }}_t-(\mathrm{\ell }\cdot \mathrm{\nabla })w-(w\cdot \mathrm{\nabla })\mathrm{\ell }+(\mathrm{\ell }\cdot \mathrm{\nabla })\mathrm{\ell }+\nu \mathrm{\Delta }\mathrm{\ell }$. For this explicit lift, $(\mathrm{\ell }\cdot \mathrm{\nabla })w={\mathrm{\ell }}_1{\mathrm{\partial }}_{x}w$ and $(w\cdot \mathrm{\nabla })\mathrm{\ell }=w_2(b-a)h^{-1}{e}_x$, while the last two lift-only terms vanish by step 1.2.

For $\delta >0$, let $y_{\delta }(t)=\sqrt{Y(t)+{\delta }^2}$. The radicand is positive; the identity $\mathrm{\mid }{y}_{\delta }(t+h)-y_{\delta }(t)\mathrm{\mid }=\mathrm{\mid }Y(t+h)-Y(t)\mathrm{\mid }\mathrm{/}(y_{\delta }(t+h)+y_{\delta }(t))$ first gives continuity of $y_{\delta }$, and the same difference-of-squares identity in the difference quotient gives $y_{\delta }^{\mathrm{\prime }}=Y^{\mathrm{\prime }}\mathrm{/}(2y_{\delta })$. Dropping the nonnegative diffusion term in step 2.1 and using $Y\mathrm{/}{y}_{\delta }\le y_{\delta }$ and $\sqrt{Y}\mathrm{/}{y}_{\delta }\le 1$ yields $y_{\delta }^{\mathrm{\prime }}\le L{y}_{\delta }+R$. The function $L$ is continuous because $\bar{D}$ is compact and $D_{x}u$ is continuous; for $A(t)={\int }_0^{t}L(s)ds$, the difference quotient is the average of $L$ on the interval between $t$ and $t+h$, so continuity gives $A^{\mathrm{\prime }}=L$. Thus [F4] and [F5] give $(e^{-A}{y}_{\delta }{)}^{\mathrm{\prime }}\le e^{-A}R$. Integrate on $[0,t]$ using [F6] and let $\delta \downarrow 0$. This proves the stated bound for $e=v-u$.

On the channel cell $[0,L_x]\times [0,h]\times [0,L_z]$, repeat the energy calculation in step 2.1 with $e=v-u$ and residual $r_{\mathrm{\ell }}$ from step 2.2. For each of its divergence terms, integrate each coordinate derivative by the fundamental theorem [F6]: the paired $x$ and $z$ face values agree by periodicity, while the $y=0,h$ contributions vanish because $e=0$ at both walls. Thus the transport and pressure terms vanish and the diffusion term is $\nu \mathrm{\parallel }\mathrm{\nabla }e{\mathrm{\parallel }}_2^2$, exactly as in step 2.1; no $C^2$ Euclidean boundary assertion about the rectangular cell is needed. The same norm regularization and integrating-factor argument in step 3.1 therefore gives the stated estimate for $v=w-\mathrm{\ell }$ with residual $r_{\mathrm{\ell }}$. Since $w-u=(v-u)+\mathrm{\ell }$, [F8] gives $\mathrm{\parallel }w-u{\mathrm{\parallel }}_2\le \mathrm{\parallel }v-u{\mathrm{\parallel }}_2+\mathrm{\parallel }\mathrm{\ell }{\mathrm{\parallel }}_2$, which is the asserted channel estimate.