Scaling the NS equation: supercriticality and energy estimate

There is a ‘rescaling transformation’ that is particularly significant for the Navier–Stokes equations when they are posed on the whole space, but is also important in the local regularity theory. Suppose first that $u(x,t)$ is a solution of the linear heat equation $$\partial_t u-\Delta u=0,~~x\in \mathbb{R}^3$$ It is simple to check that for any $\lambda > 0$ and any $\alpha \in \mathbb{R}$ the function $$u_{\lambda, \alpha}=\lambda^\alpha u(\lambda x,\lambda^2 t)$$ is again a solution of the heat equation. When we consider the Navier–Stokes equations we need the nonlinear term $(u\cdot \nabla)u$ to transform in the same way as $\partial_t u$ and $\Delta u$; it is easy to check that this requires the choice $\alpha =1$. Therefore if $u(x, 0) = u_0(x)$ gives rise to a solution $u(x, t )$ of the Navier–Stokes equations on $\mathbb{R}^3$, with corresponding pressure $p(x, t )$, then the rescaled functions $$\lambda u(\lambda x,\lambda^2 t)~~and ~~\lambda^2p(\lambda x,\lambda^2 t)$$ still solve the equations, but with rescaled initial data $u_{0,\lambda}(x) = \lambda u_0(\lambda x)$. For $\lambda$ this corresponds to shrinking (spatial) distances by a factor of $\lambda^{-1}$ and increasing the speed by a factor of $\lambda$; it is clear that the time =(distance/speed) should therefore shrink by a factor of $\lambda^{−2}$. Note that if an initial condition $u_0$ gives rise to the solution $u(x, t )$ and the rescaled initial condition $u_{0,\lambda}$ gives rise to the solution $u_\lambda(x, t )$ then $$u_\lambda \mbox{ is regular on} [0, T_\lambda] \Leftrightarrow u \mbox{ is regular on} [0, \lambda^2T_{\lambda}].$$ Spaces of functions in which the norm is unchanged by the rescaling $$u(x) \rightarrow \lambda u(\lambda x)$$ are termed ‘critical spaces’. These are the natural spaces in which to try to prove ‘small data’ results, i.e. the global existence of smooth solutions when the norm of the initial condition is small, since the norm of the data is unaffected by the aformentioned rescaling transformation. We give two examples of such spaces: the Sobolev space $\dot{H}^{1/2}$ and the Lebesgue space $L^3$. Critical spaces are important since in some cases local existence in a critical space can be used to deduce global existence. ** Now, let us take the same scaling transformation in the context of **supercritical space such as $L^2$ and blow up this fine-scale behaviour by $\lambda$ to create a coarse-scale solution to Navier-Stokes. Given that the fine-scale solution could (in the worst-case scenario) be as bad as an arbitrary smooth vector field with kinetic energy and cumulative energy dissipation at most $E$, the rescaled unit-scale solution can be as bad as an arbitrary smooth vector field with kinetic energy and cumulative energy dissipation at most $E \lambda$, as a simple change-of-variables shows. Note that the control given by our two key quantities has worsened by a factor of $\lambda$; because of this worsening, we say that these quantities are supercritical – they become increasingly useless for controlling the solution as one moves to finer and finer scales. This should be contrasted with critical quantities (such as the energy for two-dimensional Navier-Stokes), which are invariant under scaling and thus control all scales equally well (or equally poorly), and subcritical quantities, control of which becomes increasingly powerful at fine scales (and increasingly useless at very coarse scales). Knowing that the energy estimate for the solution $u$ is given by $$\underbrace{\sup_{0 \leq t < T} \frac{1}{2} \int_{{\Bbb R}^3} |u(t,x)|^2\ dx}_{\mbox{kinetic energy}}+ \underbrace{\frac{1}{2}\int_0^T \int_{{\Bbb R}^3} |\nabla u(t,x)|^2\ dx dt}_{\mbox{cumulative energy dissipation}}\leq \underbrace{\|u_0\|_{L^2({\Bbb R}^3)}^2}_{E}$$ Would you please write the same energy estimate associated to $\lambda u$ and explain in mathematical terms how exactly in that case the energy estimate will fail to control the kinetic and cumulative dissapation energy in contrast with critical spaces case. References:

Thanks

Blow up the solution $u$ by $\lambda>1$ means applying the following transformation: $$u(x,t) \rightarrow u_\lambda(x)=\lambda u(\lambda x,\lambda^2 t)$$ taking now the norm of $u_\lambda$ in $L^2$ $$\|u_\lambda\|_{L^2}^2=\int_{\mathbb{R}^3}|u_\lambda(x)|^2dx=\int_{\mathbb{R}^3}|\lambda u(\lambda x,\lambda^2 t)|^2dx=\int_{\mathbb{R}^3}\lambda^2u^2(\lambda x,\lambda^2 t)dx=\int_{\mathbb{R}^3}\lambda^2\lambda^{-3}u^2(y,\lambda^2 t)dy=\lambda^{-1}\|u\|_{L^2}^2$$ Then $$\sup_{t\in [0,\lambda^{-2}T]}\frac{1}{2}\lambda^{-1}\|u(t)\|_{L^2}^2\leq E$$ repeating this scaling procedure $n$ times to obtain $$\sup_{t\in [0,\lambda^{-n2}T]}\frac{1}{2}\|u(t)\|_{L^2}^2\leq \lambda^{n}E \rightarrow \infty \mbox{ when n goes to } \infty$$ the aformentioned scaling is a kind of shrinking the space from fine to finer (magnifying glass: term used by Tao in his blog) since $$\lambda\times u(x)=\frac{\frac{x}{\lambda}}{\frac{t}{\lambda^2}}=u_\lambda(x).$$