A shrinking clamped-stream correction on the square

The product formula shows that $R_{\mathrm{\ell }}=1$ whenever some $d_i<\mathrm{\ell }\mathrm{/}3$, and $R_{\mathrm{\ell }}=0$ whenever every $d_i\ge \mathrm{\ell }$. The supporting union has area at most $4\mathrm{\ell }$. Smoothness follows from the finite product. Because $b=0$ on each side, the fundamental theorem in its normal coordinate gives $\mathrm{\mid }b(x,y)\mathrm{\mid }\le C{d}_i(x,y)$ for every side $i$, and $\mathrm{\nabla }b$ and all fixed higher derivatives are bounded on the closed square.

Expand $D^k(b{R}_{\mathrm{\ell }})$ by the product rule. Every spatial derivative on a cutoff costs at most $C{\mathrm{\ell }}^{-1}$. If at least one cutoff is differentiated, choose that side $i$; on that term's support $d_i\le \mathrm{\ell }$, so an undifferentiated $b$ supplies a factor $C\mathrm{\ell }$. Thus the terms with no derivative on $b$ have magnitude at most $C{\mathrm{\ell }}^{1-k}$. If $j\ge 1$ derivatives hit $b$, the other factors cost at most $C{\mathrm{\ell }}^{-(k-j)}\le C{\mathrm{\ell }}^{1-k}$. Terms with no differentiated cutoff have the same bound for $k\ge 1$. All terms are supported in the union from step 1.1. Therefore $\mathrm{\parallel }{D}^k(b{R}_{\mathrm{\ell }}){\mathrm{\parallel }}_2\le C{\mathrm{\ell }}^{3\mathrm{/}2-k}$ for $k=1,2,3$; for $k=1$ the pointwise bound also gives $\mathrm{\parallel }\mathrm{\nabla }(b{R}_{\mathrm{\ell }}){\mathrm{\parallel }}_{\mathrm{\infty }}\le C$. Intersections of strips require no disjointness: the bound uses any differentiated side and the total union area.

With $\mathrm{\ell }=\sqrt{t}$, a time derivative of $\rho (d_i\mathrm{/}\mathrm{\ell })$ is bounded by $C{t}^{-1}$ and is supported where $d_i\le \mathrm{\ell }$. One extra spatial derivative costs at most $C{\mathrm{\ell }}^{-1}$, offset by $b=O(\mathrm{\ell })$ if it does not hit $b$. If it hits $b$, its derivative is bounded. Hence $\mathrm{\mid }{\mathrm{\partial }}_t\mathrm{\nabla }(b{R}_{\sqrt{t}})\mathrm{\mid }\le C{t}^{-1}$ on the same union of strips, and its $L^2$ norm is at most $C{t}^{-1}{\mathrm{\ell }}^{1\mathrm{/}2}$. Differentiating $r$ and applying step 2.1 with $k=1,2,3$ gives every asserted bound for $r$, $r_t$ and $\mathrm{\Delta }r$.

Near each boundary point $R_{\mathrm{\ell }}=1$, so the boundary value of $b{R}_{\mathrm{\ell }}$ is zero and its normal derivative equals that of $b$. Subtraction therefore cancels the stated stream trace and normal derivative. The tangential derivative of a zero boundary stream is zero, so its rotation has both zero normal and zero tangential velocity on each side. For $x\in J$ and sufficiently small $t$, the cutoffs of the vertical sides vanish identically and $R$ depends only on $y$. Since $b(x,0)=b(x,1)=0$, also $b_x,b_{xx}$ vanish on those edges and satisfy $\mathrm{\mid }{b}_x\mathrm{\mid }+\mathrm{\mid }{b}_{xx}\mathrm{\mid }\le C\min (y,1-y)$. Consequently ${\mathrm{\partial }}_{x}r=-m\mathrm{rot}(b_{x}R)$ is bounded by $C_J\mathrm{\mid }m\mathrm{\mid }$: the only factor $R_y=O({\mathrm{\ell }}^{-1})$ is multiplied by $b_x=O(\mathrm{\ell })$ in its supporting strip. This proves the last estimate and completes the proof.