How would one show that if $\omega$ is the vorticity associated to $\partial_t u+u\cdot \nabla u \nu \Delta u +\nabla p=0$ (with smooth, compactly supported initial data) and $$\omega\in L^\infty([0,T],H^1(\mathbb{R}^3))\cap L^2([0,T],H^2(\mathbb{R}^3))$$ then $u$ is a classical, smooth solution of the equation? Here $H^k$ is the Sobolev space of functions with $k$ weak derivatives in $L^2$.
1 Answer
For instance, you could prove that $u$ belong to the same space as $\omega$ in your assumption : in that case $u$ is a strong solution and becomes instantaneously smooth. To prove that $u$ have this regularity (at least locally, which should be sufficient to prove regularity properties), you can rely on the BiotSavart operator which maps $L^p(\mathbf{R}^3)$ to $L^q(\mathbf{R}^3)$, for $1<p<3$ and $1+\frac{1}{q}=\frac{2}{3}+\frac{1}{p}$. You can use this operator to understand the Sobolev regularity that your assumption implies on $u$.
EDIT, answering to the comment of the OP below : instantaneously smooth means $\mathscr{C}^\infty((0,T]\times\mathbf{R}^3)$. The $L^\infty(0,T;H^1(\mathbf{R}^3))\cap L^2(0,T;H^2(\mathbf{R}^3))$ regularity for $u$ is sufficient to trigger a bootstrap argument to achieve this regularity. Basically you differentiate the equation (in $x$) and iterate the same estimate to get $L^\infty(0,T;H^k(\mathbf{R}^3))\cap L^2(0,T;H^2(\mathbf{R}^{k+1}))$ for every integer $k\in\mathbf{N}$. This gives you the regularity in the space variable by Sobolev embedding, and then you do the same kind of computations in the time variable. This is for instance done in detail in Chapter 7 (Regularity of strong solutions) of the book The ThreeDimensional NavierStokes Equations: Classical Theory of RobinsonRodrigoSadowski.

$\begingroup$ OK, but how does the fact that $u$ belongs to the same space as $\omega$ imply that it is smooth and what do you mean by instantaneously smooth? Sorry, I don't have much experience with these things. $\endgroup$ Commented Nov 23, 2020 at 23:26

2$\begingroup$ I completed the answer above, with a reference. $\endgroup$ Commented Nov 24, 2020 at 8:43