Scaling symmetry of Navier–Stokes

Statement

If $(u,p,f)$ satisfies the incompressible Navier–Stokes equations with kinematic viscosity $\nu$ , then for $\lambda>0$ ,

u_\lambda(t,x)=\lambda u(\lambda^2t,\lambda x),\quad p_\lambda(t,x)=\lambda^2p(\lambda^2t,\lambda x),\quad f_\lambda(t,x)=\lambda^3f(\lambda^2t,\lambda x)

satisfies the same equations with the same $\nu$ on the correspondingly rescaled domain.

Verification

Each of $\partial_tu_\lambda$ , $(u_\lambda\cdot\nabla)u_\lambda$ , $\nabla p_\lambda$ , and $\nu\Delta u_\lambda$ equals $\lambda^3$ times the original term at $(\lambda^2t,\lambda x)$ . The divergence equals $\lambda^2(\nabla\cdot u)$ there and remains zero. Initial data transform as $u_{\lambda,0}(x)=\lambda u_0(\lambda x)$ .

Integral scaling

In dimension $d$ , kinetic energy scales as $E_\lambda(t)=\lambda^{2-d}E(\lambda^2t)$ . More generally, on $I_\lambda=\lambda^{-2}I$ ,

\|u_\lambda\|_{L^p_t(I_\lambda;L^q_x)} =\lambda^{1-2/p-d/q}\|u\|_{L^p_t(I;L^q_x)}.

The exponent is zero when $2/p+d/q=1$ ; such a norm is called scaling critical. These identities require the indicated norms to be defined.

References

Charles Fefferman, Existence and Smoothness of the Navier–Stokes Equation.