3
$\begingroup$

Let $u_0 \in \dot{H}^{1/2}(\mathbb{R}^3)$. The Fujita-Kato theorem gives rise to a local unique solution $(t,x) \mapsto u(t,x)$ to the Navier-Stokes equations $$\left\{ \begin{array}{ccc} \partial _t u + u\cdot \nabla u- \Delta u + \nabla p&=&0\\ div \;u&=&0 \\ u(t=0)&=&u_0. \end{array} \right.$$ The Fujita-Kato asserts that there exists a time $T^*=T(\|u_0\|_{\dot{H}^{1/2}})$ such that the solution $u$ is defined (and unique) on $[0,T^*)$. Itis also known that there is a constant $c_0$ such that if $\|u_0\|_{\dot H^{1/2}} < c_0$ then $u$ is globally defined.

Assume that $\|u_0\|_{\dot H^{1/2}} \gg c_0$. Is there a constant $K=K(\|u_0\|_{\dot H^{1/2}})$ such that if $T^* > K$ then $T^*= + \infty$? If yes, where can I read a proof of it?

$\endgroup$
1
  • $\begingroup$ Incidentally, I am not sure if you are quoting Fujita-Kato correctly. The time $T^*$ in the local existence paper of Fujita-Kato is not necessarily a function merely of $\|u_0\|_{\dot{H}^{1/2}}$. For any $s > 1/2$ you can choose $T^*$ as a function of $\|u_0\|_{H^{s}}$, but this doesn't hold in the critical case. (See Chapter 7 of Lemarie-Rieusset's "The Navier Stokes Problem in the 21st Century". Especially Theorems 7.1, 7.3, and 7.4.) $\endgroup$ Sep 6, 2018 at 17:37

1 Answer 1

6
$\begingroup$

$\dot{H}^{1/2}$ is critical with respect to scaling.

Let $\tilde{u}(t,x) = \lambda u(\lambda^2 t, \lambda x)$. Then $\tilde{u}$ solves the Navier-Stokes equation up to $\tilde{T}^* = T^* \lambda^{-2}$, with pressure $\tilde{p} = \lambda^2 p(\lambda^2 t, \lambda x)$.

One can check that the corresponding initial data satisfies $$ \|\tilde{u}\|_{\dot{H}^{1/2}(\mathbb{R}^3)} = \|u\|_{\dot{H}^{1/2}(\mathbb{R}^3)} $$

By letting $\lambda \searrow 0$ if you have a solution $u$ that exists up to time $1$, you will have a solution $\tilde{u}$ with the same $\dot{H}^{1/2}$ initial energy that exists for arbitrarily long time.

This means that the statement you asked about is equivalent to global existence of Navier-Stokes for all initial data with finite $\dot{H}^{1/2}$ norm. Which tells you that what you are asking about is certainly not a "known result".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.