Global smooth two-dimensional periodic Navier–Stokes

By [F1], local smooth solutions exist and are unique for every smooth restart datum. Uniqueness makes solutions with the same data agree on overlaps, so their union defines a unique maximal smooth solution $u$ on an interval $[0,T^{\ast })$, where $T^{\ast }$ is the supremum of its right endpoints.

For $0\le t<T^{\ast }$, [F2] and integration from $0$ to $t$ give $\mathrm{\parallel }\omega (t){\mathrm{\parallel }}_2^2+2\nu {\int }_0^t\mathrm{\parallel }\mathrm{\nabla }\omega (s){\mathrm{\parallel }}_2^{2}ds=\mathrm{\parallel }{\omega }_0{\mathrm{\parallel }}_2^2$. Therefore, with $B(t)=\mathrm{\parallel }\omega (t){\mathrm{\parallel }}_{H^1}$, ${\int }_0^{t}B(s{)}^{2}ds\le \mathrm{\parallel }{\omega }_0{\mathrm{\parallel }}_2^2(t+(2\nu {)}^{-1})$.

Fix an integer $m\ge 4$ and put $Y(t)=\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H^m}$ and $S(t)={\sum }_{k\mathrm{\ne }0}\mathrm{\mid }k\mathrm{\mid }\mathrm{\mid }\widehat{u}(t,k)\mathrm{\mid }$. Pair the full equation after applying ${\mathrm{\Lambda }}^m$ with ${\mathrm{\Lambda }}^{m}u$. The pressure term vanishes because $\mathrm{div}{\mathrm{\Lambda }}^{m}u=0$ and periodic integration by parts applies. Writing ${\mathrm{\Lambda }}^m(u\cdot \mathrm{\nabla }u)=u\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^{m}u+[{\mathrm{\Lambda }}^m,u\cdot \mathrm{\nabla }]u$, the first term has integral $\frac{1}{2}\int u\cdot \mathrm{\nabla }\mathrm{\mid }{\mathrm{\Lambda }}^{m}u{\mathrm{\mid }}^{2}d\mu =0$ by incompressibility and periodicity. The viscous term is $-\nu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{H^m}^2\le 0$, while [F4] and Cauchy–Schwarz bound the remaining commutator pairing by $C_{m}S{Y}^2$. Hence $\frac{1}{2}(Y^2{)}^{\mathrm{\prime }}\le C_{m}S{Y}^2$. For $Y_{ϵ}=(Y^2+{ϵ}^2{)}^{1\mathrm{/}2}$ with $ϵ>0$, differentiation gives $Y_{ϵ}^{\mathrm{\prime }}\le C_{m}S{Y}^2\mathrm{/}{Y}_{ϵ}\le C_{m}S{Y}_{ϵ}$.

Set $Z_{ϵ}=\log (e+Y_{ϵ})\ge 1$. By [F3], $S\le C_m(1+B\sqrt{\log (e+Y)})\le C_m(1+B\sqrt{Z_{ϵ}})$. Step 1.3 therefore gives $Z_{ϵ}^{\mathrm{\prime }}\le C_m(1+B\sqrt{Z_{ϵ}})$ and $(\sqrt{Z_{ϵ}}{)}^{\mathrm{\prime }}\le C_m(1+B)$, since $Z_{ϵ}\ge 1$.

Integrating step 2.1 and applying Cauchy–Schwarz in time with step 1.2 yields $\sqrt{Z_{ϵ}(t)}\le \sqrt{Z_{ϵ}(0)}+C_m[t+\mathrm{\parallel }{\omega }_0{\mathrm{\parallel }}_2\sqrt{t(t+\frac{1}{2\nu })}]\mathrm{.}$ Letting $ϵ\downarrow 0$ proves the displayed estimate in the Statement for each $m\ge 4$ and every $t<T^{\ast }$.

If $T^{\ast }<\mathrm{\infty }$, step 3.1 with $m=4$ and $t<T^{\ast }$ has a finite right-hand side bounded uniformly by replacing $t$ with $T^{\ast }$. Thus ${\sup }_{t<T^{\ast }}\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H^4}<\mathrm{\infty }$. By [F5], the solution extends smoothly past $T^{\ast }$, contradicting maximality. Hence $T^{\ast }=\mathrm{\infty }$.

The local theorem [F1] gives smoothness and uniqueness on every restart interval, so the maximal solution is smooth on each compact time interval. Fix any $T>0$ and let $\tau$ be the supremum of times through which two global smooth solutions agree on $[0,\tau ]$. They agree at $\tau$ by continuity. Starting [F1] from their common value at $\tau$ gives a positive local interval on which uniqueness extends their agreement; therefore $\tau$ cannot be less than $T$. As $T$ is arbitrary, they agree for all time. Since $T^{\ast }=\mathrm{\infty }$, step 3.1 supplies the asserted bound for every finite $t$ and every integer $m\ge 4$.