On Symmetry Breaking for the Navier–Stokes Equations

Abstract

Inspired by an open question by Chemin, Zhang and Zhang about the regularity of the 3D Navier–Stokes equations with one initially small component, we investigate symmetry breaking and symmetry preservation. Our results fall in three classes. First we prove strong symmetry breaking. Specifically, we demonstrate third component norm inflation (3rdNI) and Isotropic Norm Inflation (INI) starting from zero third component. Second we prove symmetry breaking for initially zero third component, even in the presence of a favorable initial pressure gradient. Third we study certain symmetry preserving solutions with a shear flow structure. Specifically, we give applications to the inviscid limit and exhibit explicit solutions that inviscidly damp to the Kolmogorov flow.

Appendix A. Heuristics for the Structure of 2.75D Shear Flows

In this appendix we give some heuristics about the derivation of 2.75D shear flows. As mentioned earlier, they are rotated versions of the parallel flows introduced by Wang [45]. Here we outline another derivation based on the analysis of the following reduced Navier–Stokes system

$$\begin{aligned} \left\{ \begin{aligned} \partial _t{ u^{\textrm{h}}}+u^{\textrm{h}}\cdot \nabla _{\textrm{h}} u^{\textrm{h}}+\nabla _{\textrm{h}} P&=\Delta u ^{\textrm{h}},\\ \partial _3 P&=0,\\ \textrm{div}_{\textrm{h}} u ^{\textrm{h}}&=0,\\ u^{\textrm{h}}|_{t=0}&=u^{\textrm{h}}_{\textrm{in}}. \end{aligned}\right. . \end{aligned}$$

(A.1)

We dub that system the ‘2.75D Navier–Stokes equations’.¹⁶ It is well-known that solutions to this system with $H^1$ data are smooth.¹⁷

Consider initial data $u^{\textrm{h}}_{\textrm{in}}$ of the form

$$\begin{aligned} u^{\textrm{h}}_{\textrm{in}}=\Bigl (\partial _2\phi (x), \,-\partial _1\phi (x)\Bigr ). \end{aligned}$$

(A.2)

Let us look for solutions of system (A.1) under the form

$$\begin{aligned} u^{\textrm{h}}=\Bigl (\partial _2\Phi (t, x), \,-\partial _1\Phi (t, x)\Bigr )=\nabla _{\textrm{h}}^{\perp } \Phi , \end{aligned}$$

(A.3)

where $\nabla _{\textrm{h}}^{\perp }:=(\partial _2, -\partial _1)$. Notice that the pressure is now given by

$$\begin{aligned} \left\{ \begin{aligned}&\Delta _{\textrm{h}} P =-\textrm{div}_{\textrm{h}}(u^{\textrm{h}}\cdot \nabla _h u^{\textrm{h}})=2 \det \bigl (\textrm{Hessian}_{\textrm{h}}~~\Phi \bigr )\\&\partial _3 P=0, \end{aligned}\right. \end{aligned}$$

where we used the vector identities

$$\begin{aligned} \textrm{div}_{\textrm{h}} \Bigl ((\nabla _{\textrm{h}}^{\perp } \Phi )\cdot \nabla _{\textrm{h}} (\nabla _{\textrm{h}}^\perp \Phi )\Bigr )&= \textrm{div}_{\textrm{h}}\textrm{div}_{\textrm{h}}\Bigl ((\nabla _{\textrm{h}}^{\perp } \Phi ) \otimes (\nabla _{\textrm{h}}^{\perp } \Phi ) \Bigr ) \\&= (\nabla _{\textrm{h}}\nabla _{\textrm{h}}^\perp \Phi ): (\nabla _{\textrm{h}}\nabla _{\textrm{h}}^\perp \Phi )^{\textrm{T}}=-2 \det \bigl (\textrm{Hessian}_{\textrm{h}}~~\Phi \bigr ), \end{aligned}$$

where

$$\begin{aligned} \textrm{Hessian}_{\textrm{h}}{:=}\left( \begin{array}{cc} \partial _{1}^2 &{} \partial _{1}\partial _{2}\\ \partial _{1}\partial _{2} &{} \partial _{2}^2 \end{array} \right) . \end{aligned}$$

In order to have $\partial _3 P=0$, one has to satisfy

$$\begin{aligned} \partial _3\det \bigl (\textrm{Hessian}_{\textrm{h}}~~\Phi \bigr )=0. \end{aligned}$$

If there exist a function $\Psi : {{\mathbb {T}}}\times {{\mathbb {R}}}_+\rightarrow {{\mathbb {R}}}$ and a constant $\lambda \in {\mathbb {Z}}$ such that

$$\begin{aligned} {\mathcal {L}}_\lambda \Phi (t, x)=\Psi (t, x_3) \quad \textrm{with}\quad {\mathcal {L}}_\lambda := \partial _1-\lambda \partial _2, \end{aligned}$$

(A.4)

then we have

$$\begin{aligned} \partial _1 {\mathcal {L}}_\lambda \Phi = \partial _2 {\mathcal {L}}_\lambda \Phi =0 \quad \mathrm{i.e.}\quad \left( \begin{array}{c} \partial _{1}^2 \Phi \\ \partial _{1}\partial _{2} \Phi \end{array} \right) =\lambda \left( \begin{array}{c} \partial _{1}\partial _{2} \Phi \\ \partial _{2}^2 \Phi \end{array}\right) , \end{aligned}$$

and thus

$$\begin{aligned} \det \bigl (\textrm{Hessian}_{\textrm{h}}~~\Phi \bigr )=0. \end{aligned}$$

(A.5)

In the following, we will focus on the case (A.4) for the Cauchy problem (A.1). Concerning the initial data, we also need to look for a function $\psi : {{\mathbb {T}}}\rightarrow {{\mathbb {R}}}$ such that

$$\begin{aligned} {\mathcal {L}}_{\lambda }\phi (x) -\psi (x_3)=0. \end{aligned}$$

(A.6)

Recalling that the velocity $u^{\textrm{h}}=\nabla ^\perp _{\textrm{h}} \Phi$ and taking into consideration (A.4), one has

$$\begin{aligned} u^{\textrm{h}}\cdot \nabla _{\textrm{h}} u^{\textrm{h}}=(-1, \lambda )\,\Psi \partial _{2}^2\Phi \end{aligned}$$

and P is a constant. Finally, we are lead to considering the following system

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t{\partial _2 \Phi }-\Psi (t, x_3)\, \partial _{2}^2\Phi = \Delta \partial _2 \Phi \quad&\hbox {in}\ ~~{{\mathbb {R}}}_+\times {{\mathbb {T}}}^3,\\&\partial _t{(\lambda \partial _2 \Phi +\Psi (t, x_3))}-\lambda \Psi (t, x_3)\, \partial _{2}^2\Phi = \Delta (\lambda \partial _2 \Phi +\Psi (t, x_3)) \quad&\hbox {in}\ ~~{{\mathbb {R}}}_+\times {{\mathbb {T}}}^3,\\&(\Phi , \Psi )|_{t=0}= (\phi , \psi ) \quad \textrm{with}\quad {\mathcal {L}}_{\lambda }\phi -\psi (x_3)=0, \end{aligned}\right. \end{aligned}$$

which can be simplified as

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t{\partial _2 \Phi }-\Psi (t, x_3)\, \partial _{2}^2\Phi = \Delta \partial _2 \Phi \quad&\hbox {in}\ ~~{{\mathbb {R}}}_+\times {{\mathbb {T}}}^3,\\&\partial _t\Psi (t, x_3) = \partial _{3}^2\Psi (t, x_3)\quad&\hbox {in}\ ~~{{\mathbb {R}}}_+\times {{\mathbb {T}}},\\&(\Phi , \Psi )|_{t=0}= (\phi , \psi ) \quad \textrm{with}\quad {\mathcal {L}}_{\lambda }\phi -\psi (x_3)=0. \end{aligned}\right. \end{aligned}$$

(A.7)

In conclusion, $\Psi (t, x_3)= ({\mathcal {K}}\star \psi ) (t, x_3)$, where ${\mathcal {K}}$ is the one-dimensional heat kernel see (1.20), and $\partial _2 \Phi$ satisfies the linear transport-heat equation

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t v + V\cdot \nabla v = \Delta v\quad \textrm{in} ~~{{\mathbb {R}}}_+\times {{\mathbb {T}}}^3 \quad \textrm{with}\quad V=(0, -\Psi (t, x_3), 0),\\&v_{\textrm{in}}=\partial _2 \phi . \end{aligned}\right. \end{aligned}$$

(A.8)

Take $\psi (x_3)=g(x_3)$ and

$$\begin{aligned} \phi (x)=\phi (\lambda x_1+ x_2, x_3)=\int \limits _0^{\lambda x_1+x_2} f(y_1, x_3)\,d y_1+ x_1 \psi (x_3). \end{aligned}$$

Obviously, $\phi (x)$ and $\psi (x_3)$ satisfy (A.6), so the associated solution $\partial _2 \Phi (t, x)$ of (A.8) for which $u^{\textrm{h}}$ given by (A.3) solves problem (A.1).