Local smooth periodic Euler and Navier–Stokes

Let $S_N$ be the orthogonal Fourier cutoff to modes $\mathrm{\mid }k{\mathrm{\mid }}_{\mathrm{\infty }}\le N$, and let $X_N$ be its finite-dimensional real divergence-free range. On $X_N$ solve ${\mathrm{\partial }}_t{u}_N+S_{N}P((u_N\cdot \mathrm{\nabla })u_N)=\nu \mathrm{\Delta }{u}_N$, $u_N(0)=S_N{u}_0$. The coefficient vector field is polynomial, hence locally Lipschitz by the finite-monomial estimate in the proof of Continuation of bounded polynomial ODE solutions; [F6] gives a unique local real solution, and its right-hand side preserves $X_N$.

Put $\mathrm{\Lambda }$ for the Fourier multiplier $w_k$. Pair the Galerkin equation with ${\mathrm{\Lambda }}^{2m}{u}_N$; self-adjointness of $S_N,P$, their commutation with $\mathrm{\Lambda }$, and $P{u}_N=S_N{u}_N=u_N$ remove the projections. The principal transport term is ${\int }_{{\mathbb{T}}^d}({\mathrm{\Lambda }}^m{u}_N)\cdot (u_N\cdot \mathrm{\nabla }{\mathrm{\Lambda }}^m{u}_N)dx=\frac{1}{2}\int u_N\cdot \mathrm{\nabla }\mathrm{\mid }{\mathrm{\Lambda }}^m{u}_N{\mathrm{\mid }}^{2}dx=0$ by periodicity and $\mathrm{div}{u}_N=0$. The remaining commutator is bounded by [F5], while viscosity contributes $-\nu \mathrm{\parallel }\mathrm{\nabla }{u}_N{\mathrm{\parallel }}_{H^m}^2$, giving $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^m}^2+\nu \mathrm{\parallel }\mathrm{\nabla }{u}_N{\mathrm{\parallel }}_{H^m}^2\le C_m\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^s}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^m}^2$.

Define a canonical extraction $E$ for a bounded real sequence $(b_n{)}_{n\ge 1}$. Let $B\ge 1$ be the least integer with $\mathrm{\mid }{b}_n\mathrm{\mid }\le B$ for every $n$, and put $I_0=[-B,B]$. Having defined $I_{\mathrm{\ell }}$, bisect it into its two closed halves and define $I_{\mathrm{\ell }+1}$ to be the left half if that half contains $b_n$ for infinitely many indices $n$, and the right half otherwise. At least one half contains infinitely many such indices, so this rule is defined and unique. Set $e_0=0$ and let $e_{\mathrm{\ell }}$ be the least $n>e_{\mathrm{\ell }-1}$ with $b_n\in I_{\mathrm{\ell }}$. These minima exist because the corresponding index sets are infinite. Then $E(b)(\mathrm{\ell })=e_{\mathrm{\ell }}$ is strictly increasing and $(b_{e_{\mathrm{\ell }}})$ is Cauchy: for $p,q\ge \mathrm{\ell }$, both values lie in $I_{\mathrm{\ell }}$, whose length is $2B{2}^{-\mathrm{\ell }}$. It converges, since [F8] gives a convergent subsequence of this bounded Cauchy sequence and the Cauchy inequality forces the entire sequence to have the same limit. All intervals and indices are uniquely specified; this defines a function $E$ without any choice principle.

At $m=s$, set $Y_N=\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^s}$. When $Y_N>0$, the inequality in step 1.2 gives $Y_N^{\mathrm{\prime }}\le C_s{Y}_N^2$; if $Y_N(0)=0$, uniqueness of the finite-dimensional ODE makes $u_N\equiv 0$. Otherwise, integrating $(1\mathrm{/}{Y}_N{)}^{\mathrm{\prime }}\ge -C_s$ shows $Y_N(t)\le 2\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}$ for $0\le t\le T$, where $T=(2C_s\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}{)}^{-1}$. For zero datum take $T=1$. These bounds do not use $\nu$ or higher norms.

For every integer $m>s$, step 1.2 and step 2.1 give $(\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^m}^2{)}^{\mathrm{\prime }}\le 4C_m\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}\mathrm{\parallel }{u}_N{\mathrm{\parallel }}_{H^m}^2$. Multiplying by $e^{-4C_m\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}t}$ and differentiating proves $\mathrm{\parallel }{u}_N(t){\mathrm{\parallel }}_{H^m}\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^m}{e}^{2C_m\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}T}$ on the same interval. Parseval [F2] and the $H^s$ bound control the Euclidean coefficient norm of $u_N$; [F7] therefore extends each finite-dimensional solution through $[0,T]$, even though the constants $C_m$ may depend on $m$.

For every integer $m\ge s+2$, the product estimate [F4] and the bounds in step 3.1 give a uniform bound on ${\mathrm{\partial }}_t{u}_N=-S_{N}P(u_N\cdot \mathrm{\nabla }{u}_N)+\nu \mathrm{\Delta }{u}_N$ in $C([0,T],H^{m-2})$. Thus each fixed Fourier coefficient is uniformly bounded and equicontinuous in time.

Enumerate the tuples $(k,j,\epsilon ,a,b)$ with $k\in {\mathbb{Z}}^d$, $1\le j\le d$, $\epsilon \in \{0,1\}$, integers $a\ge 0,b\ge 1$ with $\gcd (a,b)=1$ and $a\mathrm{/}b\le T$, by increasing height ${\sum }_r\mathrm{\mid }{k}_r\mathrm{\mid }+a+b$ and lexicographically within each finite height. Here $\epsilon =0,1$ specifies the real or imaginary part of component $j$ of ${\widehat{u}}_N(k)$ at time $a\mathrm{/}b$. Denote the resulting bounded real sequences in $N\ge 1$ by $b^1,b^2,\dots$. Starting with $h_0(n)=n$, define recursively $h_i=h_{i-1}\circ E(b^i\circ h_{i-1})$ and put $N_{\mathrm{\ell }}=h_{\mathrm{\ell }}(\mathrm{\ell })$. The successor in this recursion is a specified function of $h_{i-1}$, so this is ordinary recursion, not a succession of unspecified dependent selections. Each $h_i$ is strictly increasing and its range is contained in every preceding range. Moreover $N_{\mathrm{\ell }+1}=h_{\mathrm{\ell }}(E(b^{\mathrm{\ell }+1}\circ h_{\mathrm{\ell }})(\mathrm{\ell }+1))\ge h_{\mathrm{\ell }}(\mathrm{\ell }+1)>N_{\mathrm{\ell }}$. For each fixed $i$, the tail $(N_{\mathrm{\ell }}{)}_{\mathrm{\ell }\ge i}$ is an increasing subsequence of the range of $h_i$, on which $b^i$ converges. Thus this one subsequence converges at every enumerated coefficient-time coordinate. Countable Choice remains the standing analytic Fourier assumption; the diagonal construction itself does not require dependent choice.

The equicontinuity from step 4.1 upgrades convergence at rational times to uniform-in-time convergence of each fixed Fourier coefficient. Given $\epsilon >0$, choose a finite rational time grid so that every time is close enough to a grid point to make each of the two equicontinuity errors below $\epsilon \mathrm{/}3$; choose the subsequence indices large enough that the finitely many grid values differ by less than $\epsilon \mathrm{/}3$. The triangle inequality makes each coordinate uniformly Cauchy, and scalar completeness supplies its uniform limit.

For each integer $m\ge s$, Parseval [F2] and the uniform $H^{m+1}$ bound give $\mathrm{\parallel }(I-S_K)u_N(t){\mathrm{\parallel }}_{H^m}\le K^{-1}\mathrm{\parallel }{u}_N(t){\mathrm{\parallel }}_{H^{m+1}}$ for $K\ge 1$. The right side tends to zero uniformly in $N,t$ as $K\to \mathrm{\infty }$. Combining this tail estimate with the finite-mode uniform convergence in step 6.1 proves that the diagonal subsequence is Cauchy in $C([0,T],H^m)$ for every integer $m\ge s$. Completeness of the coefficient space [F1] yields a limit $u\in {\bigcap }_{m\ge s}C([0,T],H^m)$; finite sums show ${\sup }_t\mathrm{\parallel }u(t){\mathrm{\parallel }}_{H^m}\le {\sup }_N{\sup }_t\mathrm{\parallel }{u}_N(t){\mathrm{\parallel }}_{H^m}$, and coefficientwise limits preserve reality, divergence-free constraints, and $u(0)=u_0$.

Fix integer $m\ge s$. Product convergence follows from $u_N\to u$ in $C_t{H}^m$ and [F4]: $(u_N\cdot \mathrm{\nabla })u_N-(u\cdot \mathrm{\nabla })u=(u_N-u)\cdot \mathrm{\nabla }{u}_N+u\cdot \mathrm{\nabla }(u_N-u)\to 0$ in $C_t{H}^{m-1}$. Also $\mathrm{\Delta }{u}_N\to \mathrm{\Delta }u$ in $C_t{H}^{m-2}$.

The Fourier cutoff converges uniformly on the compact time image of the continuous curve $t\mapsto P((u\cdot \mathrm{\nabla })u)(t)$ in $H^{m-2}$: choose a finite $\epsilon$-net of that image, use strong Fourier-tail convergence at each net point, and use $\mathrm{\parallel }{S}_N\mathrm{\parallel }\le 1$ to control every point by its net representative. Together with step 8.1 and $\mathrm{\parallel }P\mathrm{\parallel }\le 1$ from [F3], this gives $S_{N}P((u_N\cdot \mathrm{\nabla })u_N)\to P((u\cdot \mathrm{\nabla })u)$ uniformly in $H^{m-2}$.

Each finite-dimensional curve satisfies its integrated ODE. The right-hand sides converge uniformly in $H^{m-2}$ by the product-convergence and uniform cutoff-convergence arguments above, and $S_N{u}_0\to u_0$ by the Fourier tail of the smooth datum. A continuous curve into the complete coefficient Hilbert space has a Riemann integral: uniform continuity makes its Riemann sums Cauchy and completeness gives their limit; the triangle inequality on the sums gives $\mathrm{\parallel }{\int }_0^t{G}_N-{\int }_0^{t}G{\mathrm{\parallel }}_{H^{m-2}}\le T{\sup }_s\mathrm{\parallel }{G}_N(s)-G(s){\mathrm{\parallel }}_{H^{m-2}}$. Passing to the limit therefore yields $u(t)=u_0+{\int }_0^t[-P((u\cdot \mathrm{\nabla })u)+\nu \mathrm{\Delta }u](s)ds$.

The integrand in step 10.1 is continuous in $H^{m-2}$. For its integral $U(t)$, the difference quotient error satisfies $\mathrm{\parallel }[U(t+h)-U(t)]\mathrm{/}h-G(t)\mathrm{\parallel }\le {\sup }_{s\text{ between }t\text{ and }t+h}\mathrm{\parallel }G(s)-G(t)\mathrm{\parallel }\to 0$ by continuity. Hence $u_t=-P((u\cdot \mathrm{\nabla })u)+\nu \mathrm{\Delta }u$ in $C([0,T],H^{m-2})$. Since $m\ge s$ was arbitrary, the equation holds at every spatial Sobolev order covered by the Statement.

The bounds $u\in C_t{H}^m$ for every integer $m$ and the embeddings in [F4] give continuous spatial derivatives of every order. Since step 11.1 holds in $H^{m-2}$ for arbitrary $m$, $u_t$ is continuous in every $H^r$ by choosing $m\ge r+2$, so $u\in C_t^1{H}^r$ for every $r$. If $u$ has $j$ continuous time derivatives in every Sobolev order, differentiating the right side $-P((u\cdot \mathrm{\nabla })u)+\nu \mathrm{\Delta }u$ $j$ times produces a finite sum of products of spatial derivatives of those time derivatives; [F4] makes each product continuous in every lower Sobolev order. Induction on $j$ therefore gives all mixed derivatives continuously to the endpoints, hence $u\in C^{\mathrm{\infty }}([0,T]\times {\mathbb{T}}^d)$.

Since $d\le 3$ and $s\ge 4$, we have $s>d\mathrm{/}2+1$. The pressure recovery statement [F9] therefore applies to the smooth divergence-free field from step 12.1 and yields the unique mean-zero $\pi =Q(u\otimes u)$ with $P(u\cdot \mathrm{\nabla }u)=u\cdot \mathrm{\nabla }u+\mathrm{\nabla }\pi$. Substitution into the projected equation gives $u_t+(u\cdot \mathrm{\nabla })u=-\mathrm{\nabla }\pi +\nu \mathrm{\Delta }u$.

Let $u,v$ be any two smooth solutions with the same initial datum and viscosity. They satisfy the hypotheses of [F10], so $\mathrm{\parallel }u(t)-v(t){\mathrm{\parallel }}_2^2\le 0\cdot \exp (2{\int }_0^t\mathrm{\parallel }\mathrm{\nabla }v(s){\mathrm{\parallel }}_{\mathrm{\infty }}ds)=0$ for each $t\in [0,T]$. Thus $u=v$ throughout the interval. This proves uniqueness conditional on the actual smooth-solution difference-energy input.

The construction, regularity, pressure recovery, and uniqueness arguments above establish the claimed smooth solution and same-interval all-order bounds. The existence time was chosen in step 2.1 solely from $d$, $s$, and $\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{H^s}$ and was independent of $\nu$ and higher norms.