Smooth Sobolev continuation criterion

The local theorems [F1] and [F2] give nonempty smooth solution intervals. Uniqueness on overlaps lets their union define a unique maximal solution on $[0,T^{\ast })$ in either branch. Assume the corresponding bounds in the Statement and $T^{\ast }<\mathrm{\infty }$. Write $z=u$ or $z=q$, respectively, and $M={\sup }_{t<T^{\ast }}\mathrm{\parallel }z(t){\mathrm{\parallel }}_{H^s}<\mathrm{\infty }$.

For any integer $m\ge s$, apply ${\mathrm{\Lambda }}^m$ to the incompressible projected equation and pair with ${\mathrm{\Lambda }}^{m}u$. The projection disappears by [F3]; periodicity and $\mathrm{div}u=0$ cancel the principal transport term, [F5] bounds the commutator pairing by $C_m\mathrm{\parallel }u{\mathrm{\parallel }}_{H^s}\mathrm{\parallel }u{\mathrm{\parallel }}_{H^m}^2$, and viscosity contributes $-\nu \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{H^m}^2\le 0$. For the symmetric system, symmetry gives $\int ({\mathrm{\Lambda }}^{m}q{)}^{\mathsf{T}}{A}_j(q){\mathrm{\partial }}_j{\mathrm{\Lambda }}^{m}q=-\frac{1}{2}\int ({\mathrm{\Lambda }}^{m}q{)}^{\mathsf{T}}({\mathrm{\partial }}_j{A}_j(q)){\mathrm{\Lambda }}^{m}q$. Linearity of $A_j$ and [F4] bound this by $C_m\mathrm{\parallel }q{\mathrm{\parallel }}_{H^s}\mathrm{\parallel }q{\mathrm{\parallel }}_{H^m}^2$, since $s>1+d\mathrm{/}2$. The remaining commutators satisfy the same bound by [F5], componentwise. Thus in both branches $\frac{1}{2}(\mathrm{\parallel }z{\mathrm{\parallel }}_{H^m}^2{)}^{\mathrm{\prime }}\le C_{m}M\mathrm{\parallel }z{\mathrm{\parallel }}_{H^m}^2$ on $[0,T^{\ast })$, with constants depending only on the fixed orders, dimension and, in the symmetric branch, the fixed matrices.

Multiplying the inequality in step 2.1 by $e^{-2C_{m}Mt}$ gives $\mathrm{\parallel }z(t){\mathrm{\parallel }}_{H^m}\le \mathrm{\parallel }z(0){\mathrm{\parallel }}_{H^m}{e}^{C_{m}M{T}^{\ast }}=:K_m<\mathrm{\infty }$ for every integer $m\ge s$ and $t<T^{\ast }$. The data are smooth, so all $K_m$ are finite.

For every integer $m\ge s$, [F3], [F4], and step 3.1 bound the right-hand side of the evolution equation uniformly in $H^m$: use $K_{m+1}$ for products with one spatial derivative, and $\nu K_{m+2}$ for the viscous term. Thus $\mathrm{\parallel }{z}_t(t){\mathrm{\parallel }}_{H^m}\le D_m<\mathrm{\infty }$. Integrating on compact subintervals gives $\mathrm{\parallel }z(t)-z(t^{\mathrm{\prime }}){\mathrm{\parallel }}_{H^m}\le D_m\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$. Completeness in [F3] supplies an endpoint limit $z_{\ast }\in H^m$ as $t\uparrow T^{\ast }$. The inclusions $H^{m+1}\subset H^m$ are continuous by the weights, so these unique limits are compatible for every $m$. Hence one $z_{\ast }$ lies in all these spaces, and [F4] makes it smooth with convergence in every $C^j$. In the incompressible branch, divergence-free constraints and reality pass to the limit. In the symmetric branch, uniform convergence and the assumed lower bound give $r_{\ast }(x)\ge {\inf }_{t<T^{\ast },x}r(t,x)>0$, so the inverse in [F2] gives a smooth strictly positive density at the endpoint.

Apply [F1] to $u_{\ast }$ or [F2] to the density and velocity recovered from $q_{\ast }$, with time translated by $T^{\ast }$. This supplies a smooth solution on $[T^{\ast },T^{\ast }+\delta ]$ for some $\delta >0$, and positivity in the symmetric branch. Join it to the old solution with value $z_{\ast }$ at $T^{\ast }$. The joined curve is continuous in every $H^m$ by step 4.1. Its evolution right-hand side is continuous there, because each spatial derivative is bounded between the indicated Sobolev orders and the products are continuous by [F4]. Passing the integral equation to $T^{\ast }$ on each side shows that the joined curve has the same first time derivative from both sides and satisfies the equation across the join.

Inductively differentiating the polynomial evolution right-hand sides in step 5.1 expresses every higher time derivative as a finite sum of products of spatial derivatives of earlier time derivatives, with the fixed bounded Fourier map $P$ in the incompressible branch. Since all spatial Sobolev orders are available, [F3] and [F4] make each such derivative continuous across the join. Embedding then gives all mixed derivatives continuously, hence a smooth extension. In the symmetric branch the smooth positive inverse density map preserves this regularity. This proves both asserted continuation conclusions without assuming a quantitative restart-time formula from either local theorem.