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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 $$ \underbrace{\sup_{0 \leq t < T} \frac{1}{2} \int_{{\Bbb R}^3} |u(t,x)|^2\ dx}_{\mbox{kinetic energy}}+ \underbrace{\frac{1}{2}\int_0^T \int_{{\Bbb R}^3} |\nabla u(t,x)|^2\ dx dt}_{\mbox{cumulative energy dissipation}}\leq \underbrace{\|u_0\|_{L^2({\Bbb R}^3)}}_{E}$$ 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:

  1. Why global regularity for NS equations is hard, Terry Tao, what's new blog
    1. Robinson, J., Rodrigo, J., & Sadowski, W. (2016). The Three-Dimensional Navier–Stokes Equations: Classical Theory (Cambridge Studies in Advanced Mathematics). Cambridge: Cambridge University Press. doi:10.1017/CBO9781139095143

Thanks

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closed as off-topic by Michael Renardy, Pace Nielsen, David Handelman, Deane Yang, Ben McKay Apr 2 '18 at 9:38

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    $\begingroup$ I wonder whether the users who voted to close this question as not of research level have bothered reading it. $\endgroup$ – Alex M. Apr 2 '18 at 11:13
  • $\begingroup$ What do you mean by "exclusive dependency"? The fact that there is such a thing as a "critical space" depends on the scaling properties of the nonlinear term in Navier-Stokes, so I don't understand what you mean by that the critical LWP depends exclusively on the properties of linear heat equation. $\endgroup$ – Willie Wong Apr 2 '18 at 14:03
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    $\begingroup$ Ah, now we are getting somewhere. You are misreading the quote from Robinson. He points out that the time $T^*$ (defining the interval $[0,T*)$ on which the local solution exists) depends only on certain bounds of the linearized flow. This is interesting as, in general, one may expect that the rate at which the nonlinear solution deviates from the linear solution to be dependent on the fine properties of the initial data. What is shown is that when working with the critical space, the rate of deviation is somehow "uniform" across functions in the critical space. That this is in fact true... $\endgroup$ – Willie Wong Apr 2 '18 at 14:25
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    $\begingroup$ ... is based on the analysis of the nonlinear structure of the NS equations. It is certainly not the case that the proof of local existence in the critical space uses only linear estimates from the heat equation! $\endgroup$ – Willie Wong Apr 2 '18 at 14:27
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    $\begingroup$ You need to put a reference to the originating blog if you choose to copy/paste entire paragraphs from there ("Note that the control ..."). This is not an option. $\endgroup$ – Hannes Apr 4 '18 at 11:25