Weighted Energy Estimates for the Incompressible Navier–Stokes Equations and Applications to Axisymmetric Solutions Without Swirl

Abstract

We consider a family of weights which permit to generalize the Leray procedure to obtain weak suitable solutions of the 3D incompressible Navier–Stokes equations with initial data in weighted $L^2$ spaces. Our principal result concerns the existence of regular global solutions when the initial velocity is an axisymmetric vector field without swirl such that both the initial velocity and its vorticity belong to $L^2 ( (1+ r^2)^{-\frac{\gamma }{2}} dx )$, with $r= \sqrt{x_1^2 + x_2^2}$ and $\gamma \in (0, 2)$.

1 Introduction

In 1934, Leray [20] proved global existence of weak solutions for the 3D incompressible Navier–Stokes equations

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

in the case of a fluid filling the whole space whose initial velocity $\mathbf{u }_0$ is in $L^2$. Leray’s strategy is to regularize the initial value and to mollify the non-linearity through convolution with a bump function: let $\theta _\epsilon (x)=\frac{1}{\epsilon ^3}\theta (\frac{x}{\epsilon })$, where $\theta \in {\mathcal {D}}({\mathbb {R}}^3)$, $\theta$ is non-negative and radially decreasing and $\int \theta \, dx=1$; the mollified equations are then

$$\begin{aligned} (NS_\epsilon ) \left\{ \begin{matrix} \partial _t \mathbf{u }_\epsilon = \Delta \mathbf{u }_\epsilon -( (\theta _\epsilon *\mathbf{u }_\epsilon ) \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon - {{\varvec{\nabla }}}p_\epsilon \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }_\epsilon =0, \qquad \qquad \mathbf{u }_\epsilon (0,.)=\theta _\epsilon *\mathbf{u }_{0}. \end{matrix}\right. \end{aligned}$$

Standard methods give existence of a smooth solution on an interval $[0,T_\epsilon ]$ where $T_\epsilon \approx \epsilon ^3 \Vert \theta _\epsilon *\mathbf{u }_0\Vert _2^{-2}$. Then, the energy equality

$$\begin{aligned} \Vert \mathbf{u }_\epsilon (t,.)\Vert _2^2+2\int _0^t \Vert {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _2^2\, ds=\Vert \theta _\epsilon *\mathbf{u }_0\Vert _2^2 \end{aligned}$$

allows one to extend the existence time and to get a global solution $\mathbf{u }_\epsilon$; moreover, the same energy equality allows one to use a compactness argument and to get a subsequence $\mathbf{u }_{\epsilon _k}$ that converges to a solution $\mathbf{u }$ of the Navier–Stokes equations (NS) which satisfies the energy inequality

$$\begin{aligned} \Vert \mathbf{u }(t,.)\Vert _2^2+2\int _0^t \Vert {{\varvec{\nabla }}}\otimes \mathbf{u }\Vert _2^2\, ds\le \Vert \mathbf{u }_0\Vert _2^2. \end{aligned}$$

Weak solutions of equations (NS) that satisfy this energy inequality are called Leray solutions.

There are many ways to extend Leray’s results to settings where $\mathbf{u }_0$ has infinite energy. A natural one is based on a splitting $\mathbf{u }=\mathbf{v }+\mathbf{w }$ where $\mathbf{v }$ satisfies an equation

$$\begin{aligned} \partial _t\mathbf{v }=\Delta \mathbf{v }+F(\mathbf{v }), \text { where }\nabla \cdot F(\mathbf{v })=0, \end{aligned}$$

that is easy to solve and $\mathbf{w }$ satisfies perturbed Navier–Stokes equations

$$\begin{aligned} \partial _t\mathbf{w }+\mathbf{w }\cdot \nabla \mathbf{w }= \Delta \mathbf{w }-\nabla q-\mathbf{v }\cdot \nabla \mathbf{w }-\mathbf{w }\cdot \nabla \mathbf{v }-\mathbf{v }\cdot \nabla \mathbf{v }-F(\mathbf{v }) \end{aligned}$$

for which Leray’s formalism still holds. For instance, if $\mathbf{u }_0\in L^p$ with $2\le p<3$, Calderón [5, 16] splits $\mathbf{u }_0$ into $\mathbf{v }_0+\mathbf{w }_0$ where $\mathbf{v }_0$ is a divergence-free vector field which is small in $L^3$ and $\mathbf{w }_0$ belongs to $L^2$. Then, $\mathbf{v }$ is the mild solution of the Navier–Stokes problem with $\mathbf{v }_0$ as initial value. Another recent example is the way Seregin and Šverák [21] deal with global weak solutions for large initial values in $L^3$, by splitting $\mathbf{u }$ into $\mathbf{v }+\mathbf{w }$, where $\mathbf{v }=e^{t\Delta }\mathbf{u }_0$ and $\mathbf{w }$ which has a finite energy.

Another way to extend Leray’s method is to consider weighted energy inequalities in $L^2(\Phi \, dx)$, or similarly energy inequalities for $\sqrt{\Phi }\mathbf{u }$. $\sqrt{\Phi }\mathbf{u }$ is solution of

$$\begin{aligned} \left\{ \begin{matrix} \partial _t (\sqrt{\Phi }\mathbf{u })= \sqrt{\Phi }\Delta \mathbf{u }-\sqrt{\Phi }(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- \sqrt{\Phi }{{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \sqrt{\Phi }\mathbf{u }(0,.)=\sqrt{\Phi }\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

The problem is that the non-linear part of the equation $-\sqrt{\Phi }(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- \sqrt{\Phi }{{\varvec{\nabla }}}p$ will then contribute to the energy balance, in contrast to the case of the Leray method. More precisely, we may write

$$\begin{aligned} -\sqrt{\Phi }(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }= -\mathbf{u }\cdot \nabla (\sqrt{\Phi }\mathbf{u })- \sqrt{\Phi }\mathbf{u }\cdot (\sqrt{\Phi }\mathbf{u }\cdot \nabla \frac{1}{\sqrt{\Phi }}). \end{aligned}$$

The advection term $\mathbf{u }\cdot \nabla (\sqrt{\Phi }\mathbf{u })$ corresponds to a transport by a divergence-free vector field and will not contribute to the energy; in order to control the impact of $\sqrt{\Phi }\mathbf{u }\cdot (\sqrt{\Phi }\mathbf{u }\cdot \nabla \frac{1}{\sqrt{\Phi }})$ on $\sqrt{\Phi }\mathbf{u }$, it is natural to assume that $\nabla \frac{1}{\sqrt{\Phi }}$ is bounded; assuming that $\Phi$ is positive, we find that $\frac{1}{\sqrt{\Phi (\frac{\lambda x}{|x|})}}\le \frac{1}{\sqrt{\Phi (0)}}+ C \lambda$, and thus that $\frac{1}{1+\vert x\vert ^2}\le C \Phi (x)$.

Recently, Bradshaw et al. [3] and Fernández-Dalgo and Lemarié-Rieusset [10] used Leray’s procedure to find a global weak solution to the equations (NS) when $\mathbf{u }_0$ is no longer assumed to have finite energy but only to satisfy the weaker assumption

$$\begin{aligned} \int \vert \mathbf{u }_0(x)\vert ^2\ \frac{dx}{1+\vert x\vert ^2}<+\infty . \end{aligned}$$

The solutions then satisfy, for every finite positive T,

$$\begin{aligned} \sup _{0\le t\le T} \int \vert \mathbf{u }(t,x)\vert ^2\ \frac{dx}{1+\vert x\vert ^2} +\int _0^T \int \vert {{\varvec{\nabla }}}\otimes \mathbf{u }(t,x)\vert ^2\ \frac{dx}{1+\vert x\vert ^2} <+\infty . \end{aligned}$$

For the proof, a precise description of the presssure is needed, as it interfers as well in the energy balance; this point has been discussed in [4, 11]. The scheme of proof of existence of such weak solutions can easily be generalized to equations which behave like the Navier–Stokes equations, for instance the magneto-hydrodynamic equations [8, 9].

An application of solutions in $L^2(\frac{1}{1+\vert x\vert ^2}dx)$ is given in [10]: a simple proof of existence of discretely self-similar solutions to the Navier–Stokes problem when the initial value is locally square integrable (and discretely homogeneous: $\lambda \mathbf{u }_0(\lambda x)=\mathbf{u }_0(x)$ for some $\lambda >1$). This existence was first proved by Chae and Wolf [6], and Bradshaw and Tsai [2], as a generalization of the result of Jia and Šverák [14] for a regular homogeneous initial value ($\lambda \mathbf{u }_0(\lambda x)=\mathbf{u }_0(x)$ for every $\lambda >1$), see [18] for the case of locally square-integrable homogeneous initial value. If $\mathbf{u }_0$ is homogeneous and locally square-integrable, then it belongs to $L^2_{\mathrm{uloc}}$; thus, the proof of Jia and Šverák relied on the control of weak solutions in the space $L^2_{\mathrm{uloc}}$ of uniformly locally square integrable vector fields, following the theory developed in [17]. If $\mathbf{u }_0$ is discretely homogeneous and locally square-integrable, then it may fail to belong to $L^2_{\mathrm{uloc}}$, but it belongs to $L^2(\frac{1}{1+\vert x\vert ^\gamma }\, dx)$ for $\gamma >1$.

Whereas the cases of finite energy and of infinite energy sound very similar, this similarity breaks down when we consider higher regularity.

When we consider solutions in function spaces with decaying weights, the growth of solutions can be amplified by the non-linearities. The authors in [3, 10] used the transport structure of the non-linearity $( \mathbf{u }\cdot {{\varvec{\nabla }}}) \mathbf{u }$ to get good controls for the velocity in some weighted spaces. When dealing with derivatives of the velocity, one loses the transport structure of non-linearities. The problem comes from the stretching term $\omega \cdot \nabla \mathbf{u }$ in the equations for the vorticity

$$\begin{aligned} \partial _t \omega = \Delta \omega + ( \omega \cdot {{\varvec{\nabla }}}) \mathbf{u }- (\mathbf{u }\cdot {{\varvec{\nabla }}}) \omega . \end{aligned}$$

In the case when $\mathbf{u }_0$ belongs to the classical Sobolev space $H^1$, for which local existence of a unique mild solution is known, this stretching term may potentially lead to blow-up in finite time, since it has a non-linear impact on the growth of $\Vert \omega \Vert _2$. There are two cases when this impact can be controlled: the case of 2D fluids (as the stretching term is equal to 0) and the case of axisymmetric vector fields with no swirl [15, 19]. In the case of weighted estimates, one cannot even get local control of the size of the vorticity in $L^2(\Phi \, dx)$ in general, but we shall show that we have global existence of a weak solution such that $\Vert \sqrt{\Phi }\omega (t,.)\Vert _2$ remains bounded on every bounded interval of times, when we work in 2D or when we consider axisymmetric vector fields with no swirl and weights that depend only on the distance to the symmetry axis.

2 Main Results

We shall first prove global existence in the weighted $L^2$ setting, in dimension d with $2\le d\le 4$ when the weight $\Phi$ satisfies some basic assumptions that allow the use of Leray’s projection operator and of energy estimates:

Definition 2.1

An adapted weight function $\Phi$ on ${\mathbb {R}}^d$ ($2\le d\le 4$) is a continuous Lipschitz function $\Phi$ such that:

• $(\mathbf{H }1)$ $0 < \Phi \le 1$.

• $(\mathbf{H }2)$ There exists $C_1>0$ such that $|{{\varvec{\nabla }}}\Phi | \le C_1 \Phi ^{\frac{3}{2}}$

• $(\mathbf{H }3)$ There exists $r \in (1, 2 ]$ such that $\Phi ^r \in {\mathcal {A}}_{r}$ (where ${\mathcal {A}}_r$ is the Muckenhoupt class of weights). In the case $d=4$, we require $r<2$ as well.

• $(\mathbf{H }4)$ There exists $C_2>0$ such that $\Phi (x)\le \Phi (\frac{x}{\lambda }) \le C_2 \lambda ^2 \Phi (x)$, for all $\lambda \ge 1$.

Examples of adapted weights can easily be given by radial slowly decaying functions:

• $d=2$, $\Phi (x)=\frac{1}{(1+\vert x\vert )^\gamma }$ where $0\le \gamma <2$

• $d=3$ or $d=4$, $\Phi (x)=\frac{1}{(1+\vert x\vert )^\gamma }$ where $0\le \gamma \le 2$

• $d=3$, $\Phi (x)=\frac{1}{(1+ r)^\gamma }$ where $r=\sqrt{x_1^2+x_2^2}$ and $0\le \gamma <2$.

The following result concerns the existence of weak solutions belonging to a weighted $L^2$ space, where the weight permits to consider initial data with a weak decay at infinity.

Theorem 1

Let $d \in \{2, 3, 4 \}$. Consider a weight $\Phi$ satisfying $(\mathbf{H }1){-}(\mathbf{H }4)$. Let $\mathbf{u }_0$ be a divergence free vector field, such that $\mathbf{u }_0$ belongs to $L^2(\Phi \, dx, {\mathbb {R}}^d)$. Then, there exists a global solution $\mathbf{u }$ of the problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

such that

• $\mathbf{u }$ belongs to $L^\infty ((0,T), L^2 (\Phi dx) )$ and ${{\varvec{\nabla }}}\otimes \mathbf{u }$ belongs to $L^2((0,T),L^2 (\Phi dx) )$, for all $T>0$,

• $p = \sum _{1 \le i,j \le d} R_i R_j (u_i u_j)$,

• the map $t\in [0,+\infty )\mapsto \mathbf{u }(t,.)$ is weakly continuous from $[0,+\infty )$ to $L^2 (\Phi \, dx)$, and strongly continuous at $t=0$,

• For $d \in \{ 2, 3 \}$, $\mathbf{u }$ satisfies the local energy inequality: there exists a locally finite non-negative measure $\mu$ such that

  $$\begin{aligned} \partial _t\left( \frac{ | \mathbf{u }| ^2}{2}\right) = \Delta \left( \frac{ | \mathbf{u }| ^2}{2}\right) - | {{\varvec{\nabla }}}\otimes \mathbf{u }| ^2 - {{\varvec{\nabla }}}\cdot \left( \frac{ | \mathbf{u }| ^2}{2}\mathbf{u }\right) -{{\varvec{\nabla }}}\cdot (p\mathbf{u }) -\mu , \end{aligned}$$

  and we have $\mu =0$ when $d=2$.

We observe that we do not prove the local energy inequality for the solutions in dimension 4. We refer the papers [7, 22, 23] for more information on suitable solutions in dimension 4.

If we consider the problem of higher regularity, the case of dimension $d=2$ is easy, while, in the case $d=3$, one must restrict the study to the case of axisymmetric flows with no swirl (to circumvent the stretching effect in the evolution of the vorticity).

Theorem 2

(Case $d=2$). Let $\Phi$ be a weight satisfying $(\mathbf{H }1){-}(\mathbf{H }4)$. Let $\mathbf{u }_0$ be a divergence free vector field, such that $\mathbf{u }_0, {{\varvec{\nabla }}}\otimes \mathbf{u }_0$ belong to $L^2 (\Phi dx)$. Then there exists a global solution $\mathbf{u }$ of the problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

such that

• $\mathbf{u }$ and ${{\varvec{\nabla }}}\otimes \mathbf{u }$ belong to $L^\infty ((0,T), L^2 (\Phi \, dx) )$ and $\Delta \mathbf{u }$ belongs to $L^2((0,T),L^2 (\Phi \, dx) )$, for all $T>0$,

• the maps $t\in [0,+\infty )\mapsto \mathbf{u }(t,.)$ and $t\in [0,+\infty )\mapsto {{\varvec{\nabla }}}\otimes \mathbf{u }(t,.)$ are weakly continuous from $[0,+\infty )$ to $L^2 (\Phi dx)$, and are strongly continuous at $t=0$.

Theorem 3

(Case $d=3$). Let $\Phi$ be a weight satisfying $(\mathbf{H }1){-}(\mathbf{H }4)$. Let $\mathbf{u }_0$ be a divergence free axisymmetric vector field without swirl, such that $\mathbf{u }_0, {{\varvec{\nabla }}}\otimes \mathbf{u }_0$ belong to $L^2 (\Phi \, dx)$. Assume moreover that $\Phi$ depends only on $r=\sqrt{x_1^2+x_2^2}$. Then there exists a time $T>0$, and a local solution $\mathbf{u }$ on (0, T) of the problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

such that

• $\mathbf{u }$ is axisymmetric without swirl, $\mathbf{u }$ and ${{\varvec{\nabla }}}\otimes \mathbf{u }$ belong to $L^\infty ((0,T), L^2 (\Phi \, dx) )$ and $\Delta \mathbf{u }$ belongs to $L^2((0,T),L^2 (\Phi \, dx) )$,

• the maps $t\mapsto \mathbf{u }(t,.)$ and $t\mapsto {{\varvec{\nabla }}}\mathbf{u }(t,.)$ are weakly continuous from [0, T) to $L^2 (\Phi \, dx)$, and are strongly continuous at $t=0$.

An extra condition on the weight permits to obtain a global existence result. Moreover, if the vorticity is more integrable at time $t=0$, it will remain so in positive times. The next theorem precise these conditions on the weight.

Theorem 4

(Case $d=3$). Let $\Phi$ be a weight satisfying $(\mathbf{H }1){-}(\mathbf{H }4)$. Assume moreover that $\Phi$ depends only on $r=\sqrt{x_1^2+x_2^2}$. Let $\Psi$ be another continuous weight (that depends only on r) such that $\Phi \le \Psi \le 1$, $\Psi \in {\mathcal {A}}_2$ and there exists $C_1>0$ such that

$$\begin{aligned} \vert {{\varvec{\nabla }}}\Psi \vert \le C_1 \sqrt{\Phi }\Psi \text { and } \vert \Delta \Psi \vert \le C_1 \Phi \Psi . \end{aligned}$$

Let $\mathbf{u }_0$ be a divergence free axisymmetric vector field without swirl, such that $\mathbf{u }_0,$ belongs to $L^2 (\Phi dx)$ and ${{\varvec{\nabla }}}\otimes \mathbf{u }_0$ belongs to $L^2 (\Psi dx)$. Then there exists a global solution $\mathbf{u }$ of the problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

such that

• $\mathbf{u }$ is axisymmetric without swirl, $\mathbf{u }$ belongs to $L^\infty ((0,T), L^2 (\Phi \, dx) )$, ${{\varvec{\nabla }}}\otimes \mathbf{u }$ belong to $L^\infty ((0,T), L^2 (\Psi \, dx) )$ and $\Delta \mathbf{u }$ belongs to $L^2((0,T),L^2 (\Psi \, dx) )$, for all $T>0$,

• the maps $t\in [0,+\infty )\mapsto \mathbf{u }(t,.)$ and $t\in [0,+\infty )\mapsto {{\varvec{\nabla }}}\otimes \mathbf{u }(t,.)$ are weakly continuous from $[0,+\infty )$ to $L^2 (\Phi \, dx)$ and to $L^2(\Psi \, dx)$ respectively, and are strongly continuous at $t=0$.

Example: we can take $\Phi (x)=\frac{1}{(1+ r)^\gamma }$ and $\Psi (x)=\frac{1}{(1+ r^2)^{\delta /2}}$ with $0\le \delta \le \gamma <2$. Of course, $\Phi \approx \frac{1}{(1+ r^2)^{\gamma /2}}$. The case $\delta <\gamma$ means that, if $\omega$ has a better decay at initial time, it will keep this better decay at all times.

3 Some Lemmas on Weights

Let us first recall the definition of Muckenhoupt weights: for $1<q<+\infty$, a positive weight $\Phi$ belongs to ${\mathcal {A}}_q({\mathbb {R}}^d)$ if and only if

$$\begin{aligned} \sup _{x\in {\mathbb {R}}^d, \rho >0} \left( \frac{1}{\vert B(x,\rho )\vert }\int _{B(x,\rho )} \Phi \, dx\right) ^{\frac{1}{q}} \left( \frac{1}{\vert B(x,\rho )\vert }\int _{B(x,\rho )} \Phi ^{-\frac{1}{q-1}} \, dx \right) ^{1-\frac{1}{q}} <+\infty .\end{aligned}$$

(1)

We refer to the Chapter 9 in [13].

Due to the Hölder inequality, we have ${\mathcal {A}}_q({\mathbb {R}}^d) \subset {\mathcal {A}}_r({\mathbb {R}}^d)$ if $q\le r$.

One easily cheks that $w_\gamma =\frac{1}{(1+\vert x\vert )^\gamma }$ belongs to ${\mathcal {A}}_q({\mathbb {R}}^d)$ if and only if

$$\begin{aligned} -d(q-1)<\gamma <d. \end{aligned}$$

Thus, $\Phi =w_\gamma$ is an adapted weight if and only if $0\le \gamma \le 2$ and $\gamma <d$.

One may of course replace in inequality (1) the balls $B(x,\rho )$ by the cubes $Q(x,\rho )=]x_1-\rho ,x_1+\rho [\times \dots \times ]x_d-\rho ,x_d+\rho [$. Thus, we can see that, if $\Phi (x) = \Psi (x_1, x_2)$ and $1< q < + \infty$, then $\Phi \in {\mathcal {A}}_q({\mathbb {R}}^3)$ if and only if $\Psi \in {\mathcal {A}}_q({\mathbb {R}}^2)$. In particular, $\Phi (x)=\frac{1}{(1+ r)^\gamma }$ is an adapted weight on ${\mathbb {R}}^3$ if and only if $0\le \gamma <2$.

Lemma 3.1

Let $\Phi$ satisfy $(\mathbf{H }1)$ and $(\mathbf{H } 2)$ and let $1\le r<+\infty$. Then:

• (a) $\sqrt{\Phi }f\in H^1$ if and only if $f\in L^2(\Phi \, dx)$ and ${{\varvec{\nabla }}}f\in L^2(\Phi \, dx)$; moreover we have

  $$\begin{aligned} \Vert \sqrt{\Phi }f\Vert _{H^1}\approx \left( \int \Phi (\vert f\vert ^2+\vert {{\varvec{\nabla }}}f\vert ^2)\, dx\right) ^{1/2} \end{aligned}$$

• (b) $\Phi f\in W^{1,r}$ if and only if $f\in L^r(\Phi ^r\, dx)$ and ${{\varvec{\nabla }}}f\in L^r(\Phi ^r\, dx)$; moreover we have

  $$\begin{aligned} \Vert \Phi f\Vert _{W^{1,r}}\approx \left( \int \Phi ^r (\vert f\vert ^r+\vert {{\varvec{\nabla }}}f\vert ^r)\, dx\right) ^{1/r} \end{aligned}$$

Proof

This is obvious since $\vert {{\varvec{\nabla }}}\Phi \vert \le C_1 \Phi ^{3/2}\le C_1\Phi$ and $\vert {{\varvec{\nabla }}}(\sqrt{\Phi })\vert =\frac{1}{2} \frac{\vert {{\varvec{\nabla }}}\Phi \vert }{\Phi } \sqrt{\Phi }\le \frac{1}{2} C_1\sqrt{\Phi }$. $\square$

Lemma 3.2

If $\Phi \in {\mathcal {A}}_s$ then we have for all $\theta \in (0,1)$, $\Phi ^\theta \in {\mathcal {A}}_p$ with $\theta = \frac{p-1}{s-1}$. In particular, if a weight $\Phi$ satisfies $(\mathbf{H }3)$, we obtain $\Phi \in {\mathcal {A}}_p$ with $p = 1 + \frac{r-1}{r} =2 - \frac{1}{r} <2$, and so $\Phi \in {\mathcal {A}}_2$.

Proof

As $\frac{1}{p} = \frac{1}{s}+ \frac{s-p}{ps}$, we find by the Hölder inequality

$$\begin{aligned} \begin{aligned}&\left( \int _{Q} \Phi ^{\frac{p-1}{s-1}} \, dx\right) ^{\frac{1}{p}} \left( \int _Q \Phi ^{- (\frac{p-1}{s-1}) (\frac{1}{p-1} )} dx \right) ^{1-\frac{1}{p}} \\&\quad = \left( \int _{Q} \left( \Phi ^{\frac{1}{s} } \left( \Phi ^{-\frac{1}{s-1}}\right) ^{ \frac{s-p}{ps}} \right) ^p \, dx\right) ^{\frac{1}{p}} \left( \int _Q \Phi ^{- \left( \frac{p-1}{s-1}\right) \left( \frac{1}{p-1} \right) } dx \right) ^{1-\frac{1}{p}} \\&\quad \le \left( \int _Q \Phi \, dx\right) ^{\frac{1}{s}} \left( \int _{Q} \Phi ^{-\frac{1}{s-1}} \, dx \right) ^{\frac{1}{p}-\frac{1}{s} +1-\frac{1}{p}} \end{aligned} \end{aligned}$$

$\square$

Let us recall that for a weight $w\in {\mathcal {A}}_q$ ($1<q<+\infty$), the Riesz transforms and the Hardy–Littlewood maximal function are bounded on $L^q(w\, dx)$. We thus have the following inequalities:

Lemma 3.3

Let $\Phi$ satisfy $(\mathbf{H }1)$, $(\mathbf{H }2)$ and $(\mathbf{H }3)$. Then:

• (a) for $j=1,\dots , d$, the Riesz transforms $R_j$ satisfy that $\Vert \sqrt{\Phi }R_jf\Vert _2\le C \Vert \sqrt{\Phi }f\Vert _2$ and $\Vert \sqrt{\Phi }R_jf\Vert _{H^1}\le C \Vert \sqrt{\Phi }f\Vert _{H^1}$;

• (b) for $j=1,\dots , d$, the Riesz transforms $R_j$ satisfy that $\Vert \Phi R_jf\Vert _r\le C \Vert \Phi f\Vert _r$ and $\Vert \Phi R_jf\Vert _{W^{1,r}}\le C \Vert \Phi f\Vert _{W^{1,r}}$;

• (c) if ${\mathbb {P}}$ is the Leray projection operator on divergence-free vector fields, then for a vector field $\mathbf{u }$ we have $\Vert \sqrt{\Phi }{\mathbb {P}} \mathbf{u }\Vert _2\le C \Vert \sqrt{\Phi }\mathbf{u }\Vert _2$ and $\Vert \sqrt{\Phi }{\mathbb {P}} \mathbf{u }\Vert _{H^1}\le C \Vert \sqrt{\Phi }\mathbf{u }\Vert _{H^1}$;

• (d) if $d \in \{2,3,4 \}$, then for a vector field $\mathbf{u }$ we have

  $$\begin{aligned} \Vert \sqrt{\Phi }\, \mathbf{u }\Vert _{H^1} \approx \Vert \sqrt{\Phi }\, \mathbf{u }\Vert _2+\Vert \sqrt{\Phi }{{\varvec{\nabla }}}\cdot \mathbf{u }\Vert _2+\Vert \sqrt{\Phi }{{\varvec{\nabla }}}\wedge \mathbf{u }\Vert _2. \end{aligned}$$

• (e) Let $\theta _\epsilon (x)=\frac{1}{\epsilon ^d}\theta (\frac{x}{\epsilon })$, where $\theta \in {\mathcal {D}}({\mathbb {R}}^d)$, $\theta$ is non-negative and radially decreasing and $\int \theta \, dx=1$. Then we have $\Vert \sqrt{\Phi }\, (\theta _\epsilon *f)\Vert _2\le C\Vert \sqrt{\Phi }\, f\Vert _2$ and $\Vert \sqrt{\Phi }\, (\theta _\epsilon *f)\Vert _{H^1}\le C ( \Vert \sqrt{\Phi }\ f \Vert _{L^2} + \Vert \sqrt{\Phi }\ {{\varvec{\nabla }}}f \Vert _{L^2})$ (where the constant C does not depend on $\epsilon$ nor f).

Proof

(a) is a consequence of $\Phi \in {\mathcal {A}}_2$ and of Lemma 3.1 (since $\partial _k(R_jf)=R_j(\partial _kf)$). Similarly, (b) is a consequence of $\Phi ^r\in {\mathcal {A}}_r$ and of Lemma 3.1.

(c) is a consequence of (a): if $\mathbf{v }={\mathbb {P}}\mathbf{u }$, then $v_j=\sum _{k=1}^d R_jR_k(u_k)$.

(d) is a consequence of (a): if ${\mathcal {R}}=(R_1,\dots , R_d)$, we have the identity

$$\begin{aligned} -\Delta \mathbf{u }={{\varvec{\nabla }}}\wedge ({{\varvec{\nabla }}}\wedge \mathbf{u })-{{\varvec{\nabla }}}({{\varvec{\nabla }}}\cdot \mathbf{u }) \end{aligned}$$

so that

$$\begin{aligned} \partial _k\mathbf{u }=R_k{\mathcal {R}}\wedge ({{\varvec{\nabla }}}\wedge \mathbf{u })-R_k{\mathcal {R}}({{\varvec{\nabla }}}\cdot \mathbf{u }). \end{aligned}$$

(e) is a consequence of $\Phi \in {\mathcal {A}}_2$ and of Lemma 3.1: Theorem 2.1.10 in Chapter 2 of [12] states that we have $\vert \theta _\epsilon *f\vert \le {\mathcal {M}}_f$ (where ${\mathcal {M}}_f$ is the Hardy–Littlewood maximal function of f) and, similarly, $\vert \partial _k(\theta _\epsilon *f)\vert \le {\mathcal {M}}_{\partial _k f}$. $\square$

A final lemma states that $\Phi$ is slowly decaying at infinity:

Lemma 3.4

Let $\Phi$ satisfy $(\mathbf{H }1)$ and $(\mathbf{H }2)$. Then there exists a constant $C_3$ such that

$$\begin{aligned} \frac{1}{(1+\vert x\vert )^2}\le C_3 \Phi . \end{aligned}$$

If $d=3$ and $\Phi$ depends only on $r=\sqrt{x_1^2+x_2^2}$, then

$$\begin{aligned} \frac{1}{(1+\vert r\vert )^2}\le C_3 \Phi . \end{aligned}$$

Proof

We define $x_0=\frac{1}{\vert x\vert } x$ and $g(\lambda )=\Phi (\lambda x_0)$. We have

$$\begin{aligned} g'(\lambda ) =x_0\cdot {{\varvec{\nabla }}}\Phi (\lambda x_0)\ge - C_1 (\Phi (\lambda x_0))^{3/2}=-C_1 g(\lambda )^{3/2}. \end{aligned}$$

Thus

$$\begin{aligned} C_1 \lambda \ge -\int _0^\lambda g'(\mu )g(\mu )^{-3/2}\, d\mu = 2(g(\lambda )^{-1/2}-g(0)^{-1/2}) \end{aligned}$$

and we get

$$\begin{aligned} \Phi (x)^{-1/2} \le \Phi (0)+\frac{C_1}{2} \vert x\vert \le \sqrt{C_3}(1+\vert x\vert ). \end{aligned}$$

If $\Phi$ depends only on r, we find that

$$\begin{aligned} \frac{1}{(1+\vert r\vert )^2}\le C_3 \Phi (x_1,x_2,0)=C_3 \Phi (x). \end{aligned}$$

$\square$

4 Proof of Theorem 1 (The Case of $L^2(\Phi \, dx)$)

4.1 A Priori Controls

Let $\phi \in {\mathcal {D}}({\mathbb {R}}^d)$ be a real-valued test function which is equal to 1 in a neighborhood of 0 and let $\phi _\epsilon (x)= \phi (\epsilon x)$. Let

$$\begin{aligned} \mathbf{u }_{0, \epsilon } = {\mathbb {P}}( \phi _\epsilon \mathbf{u }_0 ) . \end{aligned}$$

Thus, $\mathbf{u }_{0, \epsilon }$ is divergence free and converges to $\mathbf{u }_{0}$ in $L^2(\Phi \, dx)$ since $\Phi \in {\mathcal {A}}_2$.

Let $\theta _\epsilon (x)=\frac{1}{\epsilon ^d}\theta (\frac{x}{\epsilon })$, where $\theta \in {\mathcal {D}}({\mathbb {R}}^d)$, $\theta$ is non-negative and radially decreasing and $\int \theta \, dx=1$. We denote $\mathbf{b }_\epsilon = \mathbf{u }_\epsilon *\theta _\epsilon$. Let $\mathbf{u }_\epsilon$ be the unique global solution of the problem

$$\begin{aligned} (NS_\epsilon ) \left\{ \begin{matrix} \partial _t \mathbf{u }_\epsilon = \Delta \mathbf{u }_\epsilon -( \mathbf{b }_\epsilon \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon - {{\varvec{\nabla }}}p_\epsilon \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }_\epsilon =0, \qquad \qquad \mathbf{u }_\epsilon (0,.)=\mathbf{u }_{0, \epsilon } \end{matrix}\right. \end{aligned}$$

which belongs to ${\mathcal {C}}([0,+\infty ), L^2({\mathbb {R}}^d))\cap L^2((0,+\infty ),\dot{H}^1({\mathbb {R}}^d))$.

We want to demonstrate that

$$\begin{aligned} \begin{aligned} \Vert \sqrt{\Phi } \mathbf{u }_\epsilon (t) \Vert _{L^2}^2&+ \int _0^t \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2}^2 \, ds \le \Vert \sqrt{\Phi } \mathbf{u }_{0,\epsilon } \Vert _{L^2}^2 + C_\Phi \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^2 + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^{2d} \, ds , \end{aligned} \end{aligned}$$

(2)

where $C_\Phi$ does not depend on $\epsilon$ nor on $\mathbf{u }_0$. (When $d=4$, the inequality will hold only if $\Vert \sqrt{\Phi } \mathbf{u }_\epsilon (t) \Vert _{L^2}$ remains small enough).

Since $\sqrt{\Phi },{{\varvec{\nabla }}}\sqrt{\Phi } \in L^\infty$, pointwise multiplication by $\sqrt{\Phi }$ maps boundedly $H^1$ to $H^1$ and $H^{-1}$ to $H^{-1}$. Thus, $\sqrt{\Phi } \mathbf{u }_\epsilon \in L^2 H^1$ and $\sqrt{\Phi } \partial _t \mathbf{u }_\epsilon \in L^2 H^{-1}$, we can calculate $\int \partial _t \mathbf{u }_\epsilon \cdot \mathbf{u }_\epsilon \Phi \,dx$ and obtain:

$$\begin{aligned} \begin{aligned}&\int \frac{ | \mathbf{u }_\epsilon (t,x) | ^2}{2} \Phi \, dx + \int _0^t \int | {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon | ^2\, \, \Phi dx\, ds \\&\quad = \int \frac{ | \mathbf{u }_{0, \epsilon } (x) | ^2}{2} \Phi \, dx - \int _0^t \int ({{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon ) \, \cdot ( {{\varvec{\nabla }}}\Phi \otimes \mathbf{u }_\epsilon ) \, dx\, ds \\&\qquad + \int _0^t \int \left( \frac{| \mathbf{u }_\epsilon |^2 }{2} \mathbf{b }_\epsilon + p_\epsilon \mathbf{u }_\epsilon \right) \, \cdot {{\varvec{\nabla }}}\Phi \, dx\, ds. \end{aligned} \end{aligned}$$

(3)

We use the fact that $|{{\varvec{\nabla }}}\Phi | \le C_0 \Phi ^{\frac{3}{2}} \le C_0 \Phi ,$ in order to control the following term

$$\begin{aligned} \left| - \int _0^t \int ( {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon ) \, \cdot ( {{\varvec{\nabla }}}\Phi \otimes \mathbf{u }_\epsilon ) dx\, ds \right| \le \frac{1}{8} \int _0^t \Vert \sqrt{\Phi }\, {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2}^2 + C \int _0^t \Vert \sqrt{\Phi }\, \mathbf{u }_\epsilon \Vert _{L^2}^2. \end{aligned}$$

Now, we analyze the integrals containing the pressure term. We distinguish two cases:

• Case 1: $d=2$ and $r \in (1, 2]$, or $d=3$ and $r \in [\frac{6}{5}, 2]$, or $d=4$ and $r\in [\frac{4}{3},2)$. For those values of d and r we have

  $$\begin{aligned} 0 \le \frac{d}{2}-\frac{d}{2r} \le 1 \text { and } \dot{H}^{\frac{d}{2}-\frac{d}{2r}}\subset L^{2r} \end{aligned}$$

  and

  $$\begin{aligned} 0 \le \frac{d}{r}-\frac{d}{2} \le 1\text { and } \dot{H}^{\frac{d}{r}-\frac{d}{2}}\subset L^{\frac{r}{r-1}}. \end{aligned}$$

  Using the continuity of the Riesz transforms on $L^r(\Phi ^r dx)$,

  $$\begin{aligned} \begin{aligned} \int _0^t \int ( \frac{| \mathbf{u }_\epsilon |^2 |\mathbf{b }_\epsilon | }{2} + |p_\epsilon | |\mathbf{u }_\epsilon | ) \, | {{\varvec{\nabla }}}\Phi | \, dx\, ds&\le \int _0^t \Vert \Phi ( | \mathbf{u }_\epsilon |\, | \mathbf{b }_\epsilon | + |p_\epsilon | ) \Vert _r \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{r}{r-1}} \\&\le C \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2r} \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2r} \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{r}{r-1}} ds \end{aligned} \end{aligned}$$

  Using the Sobolev embedding $\dot{H}^{\frac{d}{2}-\frac{d}{2r}}\subset L^{2r}$, the fact that $|{{\varvec{\nabla }}}\sqrt{\Phi }| \le C \sqrt{\Phi }$, and the continuity of the maximal function operator on $L^2 (\Phi dx)$, we have

  $$\begin{aligned} \begin{aligned}&\Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2r} \\&\quad \le C \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}^{1-(\frac{d}{2}- \frac{d}{2r})} \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Phi } \mathbf{b }_\epsilon ) \Vert _{2}^{\frac{d}{2}- \frac{d}{2r}} \\&\quad \le C' \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}^{1-(\frac{d}{2}- \frac{d}{2r})} ( \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{b }_\epsilon \Vert _{2})^{\frac{d}{2}- \frac{d}{2r}} \\&\quad \le C'' \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-(\frac{d}{2}- \frac{d}{2r})} ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{2})^{\frac{d}{2}- \frac{d}{2r}} , \end{aligned} \end{aligned}$$

  and

  $$\begin{aligned}&\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2r} \\&\quad \le C \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-(\frac{d}{2}- \frac{d}{2r})} ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{2})^{\frac{d}{2}- \frac{d}{2r}}. \end{aligned}$$

Using the embedding $\dot{H}^{\frac{d}{r}-\frac{d}{2}}\subset L^{\frac{r}{r-1}}$, we also have

$$\begin{aligned} \begin{aligned} \Vert&\sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{r}{r-1}} \\&\le C \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-( \frac{d}{r}-\frac{d}{2} ) } \Vert {{\varvec{\nabla }}}\otimes ( \sqrt{\Phi } \mathbf{u }_\epsilon ) \Vert _{L^2}^{\frac{d}{r}-\frac{d}{2} } \\&\le C \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-( \frac{d}{r}-\frac{d}{2} ) } (\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2})^{ \frac{d}{r}-\frac{d}{2} } . \end{aligned} \end{aligned}$$

Hence, we find

$$\begin{aligned} \begin{aligned} \int _0^t \int&\left( \frac{| \mathbf {u }_\epsilon |^2 |\mathbf {b }_\epsilon | }{2} + |p_\epsilon | |\mathbf {u }_\epsilon | \right) \, | {{\mathbf {\nabla }}}\Phi | \, dx \, ds\\ \le&C \int _0^t \Vert \sqrt{\Phi } \mathbf {u }_\epsilon \Vert _{2}^{3- \frac{d}{2} } (\Vert \sqrt{\Phi } \mathbf {u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\mathbf {\nabla }}}\otimes \mathbf {u }_\epsilon \Vert _{L^2})^{ \frac{d}{2} }\, ds . \end{aligned} \end{aligned}$$

Using the Young inequality, we then find for $d=2$ or $d=3$

$$\begin{aligned} \begin{aligned} \int _0^t \int&\left( \frac{| \mathbf {u }_\epsilon |^2 |\mathbf {b }_\epsilon | }{2} + |p_\epsilon | |\mathbf {u }_\epsilon | \right) \, | {{\mathbf {\nabla }}}\Phi | \, dx \, ds\\ \le \frac{1}{8}&\int _0^t \Vert \sqrt{\Phi }{{\mathbf {\nabla }}}\otimes \mathbf {u }_\epsilon \Vert _{L^2}^2 \, ds + C_\Phi \int _0^t \Vert \sqrt{\Phi } \mathbf {u }_\epsilon \Vert _{L^2}^2 + \Vert \sqrt{\Phi } \mathbf {u }_\epsilon \Vert _{L^2}^{\frac{12-2d}{4-d}} \, ds, \end{aligned} \end{aligned}$$

where, as $d \in \{2,3\}$, we have $\frac{12-2d}{4-d}= 2d$.

When $d=4$, provided that $\Vert \sqrt{\Phi }\, \mathbf{u }_\epsilon \Vert _2<\epsilon _0$ with $C\epsilon _0<\frac{1}{8}$ we find

$$\begin{aligned} \begin{aligned} \int _0^t \int&\left( \frac{| \mathbf{u }_\epsilon |^2 |\mathbf{b }_\epsilon | }{2} + |p_\epsilon | |\mathbf{u }_\epsilon | \right) \, | {{\varvec{\nabla }}}\Phi | \, dx \, ds \le \frac{1}{8} \int _0^t \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2}^2 \, ds+ \frac{1}{8}\int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^2 \, ds, \end{aligned} \end{aligned}$$

• Case 2: $d=3$ and $r \in ( 1, \frac{6}{5} )$, or $d=4$ and $r\in (1,\frac{4}{3})$. Let $q = \frac{dr}{d-r}$; for those values of d, r and q, we have

  $$\begin{aligned} W^{1,r}\subset L^{q} 0 \le d-\frac{d}{r} \le 1\text { and } \dot{H}^{ d (1-\frac{1}{r})}\subset L^{ \frac{2r}{2-r}}. \end{aligned}$$

  and

  $$\begin{aligned} 0 \le \frac{d}{r}-\frac{d}{2}-1 \le 1\text { and } \dot{H}^{\frac{d}{r}-\frac{d}{2}-1}\subset L^{\frac{q}{q-1}}. \end{aligned}$$

Using the continuity of the Riesz transforms on $L^r(\Phi ^r dx)$, we have

$$\begin{aligned} \begin{aligned}&\int _0^t \int \left( \frac{| \mathbf{u }_\epsilon |^2 |\mathbf{b }_\epsilon | }{2} + |p_\epsilon | |\mathbf{u }_\epsilon | \right) \, | {{\varvec{\nabla }}}\Phi | \, dx\, ds \\&\quad \le \int _0^t \Vert \Phi | \mathbf{u }_\epsilon |^2 \Vert _{q} \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{\frac{q}{ q-1}} ds + \int _0^t \Vert \Phi p_\epsilon \Vert _q \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{q}{q-1}} ds \\&\quad \le C \int _0^t \Vert \Phi | \mathbf{u }_\epsilon |^2 \Vert _{W^{1,r}} \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{\frac{q}{q-1}} ds + \sum _{ij} \int _0^t \Vert \Phi b_{\epsilon , i} u_{\epsilon , j} \Vert _{W^{1,r}} \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{q}{q-1}} ds. \end{aligned} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned}&\Vert \Phi b_{\epsilon , i} u_{\epsilon , j} \Vert _{W^{1,r}} \\&\quad \le \Vert \Phi b_{\epsilon , i} u_{\epsilon , j} \Vert _r+\sum _{k} ( \Vert b_{\epsilon ,i} u_{\epsilon ,j} \, \partial _k \Phi \Vert _{L^r} + \Vert \Phi \, b_{\epsilon ,i} \, \partial _k u_{\epsilon ,j} \Vert _{L^r} + \Vert \Phi \, u_{\epsilon ,i} \, \partial _k b_{\epsilon ,j} \Vert _{L^r}) \\&\quad \le C ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{2r}{2-r}} \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}+ \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{\frac{2r}{2-r}} \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{\frac{2r}{2-r}} \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{b }_\epsilon \Vert _2), \\&\quad \le C' ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2} ) (\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{ \dot{H}^{ d (1-\frac{1}{r})}} + \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{\dot{H}^{ d (1-\frac{1}{r})}}). \end{aligned} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} \Vert&\sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{ \dot{H}^{ d (1-\frac{1}{r})}}\\&\le C \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}^{1-(d- \frac{d}{r})} \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Phi } \mathbf{b }_\epsilon ) \Vert _{2}^{d- \frac{d}{r}} \\&\le C' \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}^{1-(d- \frac{d}{r})} ( \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{b }_\epsilon \Vert _{2})^{d- \frac{d}{r}} \\&\le C'' \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-(d- \frac{d}{r})} ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2})^{d- \frac{d}{r}} , \end{aligned} \end{aligned}$$

and finally we get

$$\begin{aligned} \begin{aligned} \sum _{i,j} \Vert&\Phi b_{\epsilon , i} u_{\epsilon , j} \Vert _{W^{1,r}} + \Vert \Phi \vert u_{\epsilon }\vert ^2 \Vert _{W^{1,r}} \le C \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{1-(d- \frac{d}{r})} ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2})^{1+d- \frac{d}{r}} . \end{aligned} \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \begin{aligned} \Vert&\sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{\frac{q}{q-1}} \\&\le C \Vert \sqrt{\Phi } \mathbf{b }_\epsilon \Vert _{2}^{2-(\frac{d}{r}- \frac{d}{2})} \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Phi } \mathbf{b }_\epsilon ) \Vert _{2}^{\frac{d}{r}- \frac{d}{2}-1} \\&\le C' \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{2-(\frac{d}{r}- \frac{d}{2})} ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2})^{\frac{d}{r}- \frac{d}{2}-1}. \end{aligned} \end{aligned}$$

Hence, we find again

$$\begin{aligned} \begin{aligned}&\int _0^t \int \left( \frac{| \mathbf{u }_\epsilon |^2 |\mathbf{b }_\epsilon | }{2} + |p_\epsilon | |\mathbf{u }_\epsilon | \right) \, | {{\varvec{\nabla }}}\Phi | \, dx \, ds \le C \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}^{3- \frac{d}{2} } (\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} + \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2})^{ \frac{d}{2} }\, ds . \end{aligned} \end{aligned}$$

and we conclude in the same way as for the first case.

In the Case 1 and Case 2, we have found

$$\begin{aligned} \begin{aligned} \int _0^t \int \left( \frac{| \mathbf{u }_\epsilon |^2 |\mathbf{b }_\epsilon | }{2} + |p_\epsilon | |\mathbf{u }_\epsilon | \right) \, | {{\varvec{\nabla }}}\Phi | \, dx \, ds \le \frac{1}{8} \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^2 + C_\Phi \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^2 + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^{2d} \, ds. \end{aligned} \end{aligned}$$

From these controls, we get inequality (3), and thus inequality (2). Inequality (2) gives us a control on the size of $\Vert \sqrt{\Phi }\, \mathbf{u }_\epsilon \Vert _2$ on an interval of time that does not depend on $\epsilon$:

Lemma 4.1

If $\alpha$ is a continuous non-negative function on [0, T) which satisfies, for three constants $A,B \in (0, + \infty )$ and $b \in [1, \infty )$,

$$\begin{aligned} \alpha (t)\le A + B\int _0^t \alpha (s) + \alpha (s)^b\, ds. \end{aligned}$$

Let $0<T_1<T$ and $T_0=\min ( T_1, \frac{1}{3^b (A ^{b-1}+ (BT_1)^{b-1})})$. We have, for every $t\in [0,T_0]$, $\alpha (t)\le 3A$.

Proof

We try to estimate the first time $T^*<T_1$ (if it exists) for which we have

$$\begin{aligned} \alpha (T^*) =3A. \end{aligned}$$

We have

$$\begin{aligned} \alpha \le \frac{A}{BT_1} + (\frac{BT_1}{A})^{b-1} \alpha ^b. \end{aligned}$$

We thus find

$$\begin{aligned} \alpha (T^*)\le 2A + T^* (3A)^b (1+(\frac{BT_1}{A})^{b-1} ) \end{aligned}$$

and thus

$$\begin{aligned} T^* 3^b (A^{b-1}+ (BT_1)^{b-1})\ge 1. \end{aligned}$$

$\square$

By Lemma 4.1 and (2), we thus find that there exists a constant $C_\Phi \ge 1$ such that if $T_0$ satisfies

• if $d=2$, $C_\Phi \left( 1+\Vert \mathbf{u }_0\Vert _{L^2(\Phi dx)}^2 \right) \, T_0 \le 1$

• if $d=3$, $C_\Phi \left( 1+\Vert \mathbf{u }_0\Vert _{L^2(\Phi dx)}^2 \right) ^2 \, T_0 \le 1$

• if $d=4$ and $\Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}\le \frac{1}{C_\Phi }$, $C_\Phi \, T_0 \le 1$

then

$$\begin{aligned} \sup _{0\le t\le T_0} \Vert \ \mathbf{u }_\epsilon (t,.)\Vert _{L^2(\Phi dx)}^2 + \int _0^{T_0} \Vert {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2(\Phi \, dx)}^2\, ds \le C_\Phi (1 + \Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}^2 ). \end{aligned}$$

(4)

4.2 Passage to the Limit and Local Existence

We know that $\mathbf{u }_\epsilon$ is bounded in $L^\infty ((0,T_0 ), L^2(\Phi \, dx))$ and ${{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon$ is bounded in $L^2((0,T_0 ), L^2(\Phi \, dx))$. This will alow us to use a simple variant of the Aubin–Lions theorem:

Lemma 4.2

(Aubin–Lions theorem). Let $s >0$, $1<q$ and $\sigma <0$. Let $(f_n)$ be a sequence of functions on $(0,T)\times {\mathbb {R}}^d$ such that, for all $T_0\in (0,T)$ and all $\varphi \in {\mathcal {D}}({\mathbb {R}}^d)$,

• $\varphi f_n$ is bounded in $L^2((0,T_0), H^s)$

• $\varphi \partial _t f_n$ is bounded in $L^q((0,T_0), H^\sigma )$ .

Then, there exists a subsequence $(f_{n_k})$ such that $f_{n_k}$ is strongly convergent in $L^2_{\mathrm{loc}}([0,T)\times {\mathbb {R}}^d)$. More precisely: if we denote $f_\infty$ the limit, then for all $T_0\in (0,T)$ and all $R_0>0$,

$$\begin{aligned} \lim _{n_k\rightarrow +\infty } \int _0^{T_0} \int _{ | x | \le R_0} | f_{n_k}-f_\infty |^2 \, dx\, dt=0. \end{aligned}$$

For a proof of the Lemma, see [1, 18].

We want to verify that $\varphi \partial _t\mathbf{u }_\epsilon$ is bounded in $L^\alpha ((0,T_0 ), H^{-s})$ for some $s \in (-\infty ,0)$ and some $\alpha >1$.

In Case 1, we have that $\Phi \mathbf{b }_\epsilon \otimes \mathbf{u }_\epsilon$ and $\Phi p_\epsilon =\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_{\epsilon ,i}u_{\epsilon ,j})$ are bounded in $L^{\alpha _1}((0,T_0 ),L^{r})$, where $\alpha _1 = \frac{2r}{ dr-d}$, so that $\alpha _1 \in [2, \infty )$ if $d=2$, $\alpha _1 \in [\frac{4}{3}, 4]$ if $d=3$ and $\alpha _1 \in (1,2]$ if $d=4$.

In Case 2, we have that $\Phi \mathbf{b }_\epsilon \otimes \mathbf{u }_\epsilon$ and $\Phi p_\epsilon$ are bounded in $L^{\alpha _2}((0,T_0 ),W^{1,r})$, where $\alpha _2 = \frac{2r}{r + dr-d}$ and thus it is bounded in $L^{\alpha _2} L^{q}$, with $q = \frac{dr}{d-r}$. We have $\alpha _2 \in (\frac{4}{3}, 2)$ if $d=3$ and $\alpha _2 \in (1,2)$ if $d=4$.

Let $\varphi \in {\mathcal {D}}({\mathbb {R}}^d)$. We have that $\varphi \mathbf{u }_\epsilon$ is bounded in $L^2((0,T_0 ), H^1)$; moreover, writing

$$\begin{aligned} \begin{aligned} \partial _t \mathbf{u }_\epsilon&= \Delta \mathbf{u }_\epsilon - \left( \sum _{j=1}^3 \partial _j(b_{\epsilon ,j}\mathbf{u }_\epsilon ) +{{\varvec{\nabla }}}p_{\epsilon }\right) \end{aligned} \end{aligned}$$

and using the embeddings $L^r\subset \dot{H}^{\frac{d}{2} - \frac{d}{r}} \subset H^{-1}$ (in Case 1) or $L^{\frac{dr}{d-r}}\subset H^{-(\frac{d}{r}-\frac{d}{2}-1)}\ \subset H^{-1}$ (in Case 2) we see that $\varphi \partial _t \mathbf{u }_\epsilon$ is bounded in $L^{\alpha _i}((0,T_0), H^{-2})$.

Thus, by the Aubin–Lions theorem, there exist $\mathbf{u }$ and a sequence $(\epsilon _k)_{k\in {\mathbb {N}}}$ converging to 0 such that $\mathbf{u }_{\epsilon _k}$ converges strongly to $\mathbf{u }$ in $L^2_{\mathrm{loc}} ([0,T_0 )\times {\mathbb {R}}^3)$: for every ${\tilde{T}} \in (0,T_0 )$ and every $R>0$, we have

$$\begin{aligned} \lim _{k\rightarrow +\infty } \int _0^{{{\tilde{T}}}} \int _{ | y | <R} | \mathbf{u }_{\epsilon _k}-\mathbf{u }|^2\, dx\, ds =0 . \end{aligned}$$

Then, we have that $\mathbf{u }_{\epsilon _k}$ converge *-weakly to $\mathbf{u }$ in $L^\infty ((0,T_0 ), L^2(\Phi dx))$, ${{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon _k}$ converges weakly to ${{\varvec{\nabla }}}\otimes \mathbf{u }$ in $L^2((0,T_0 ),L^2(\Phi dx))$, and $\mathbf{u }_{\epsilon _k}$ converges weakly to $\mathbf{u }$ in $L^3((0,T_0 ), L^3(\Phi ^{\frac{3}{2}} dx))$. We deduce that $\mathbf{b }_{\epsilon _k}\otimes \mathbf{u }_{\epsilon _k}$ is weakly convergent in $(L^{6/5}L^{6/5})_{\mathrm{loc}}$ to $\mathbf{b }\otimes \mathbf{u }$ and thus in ${\mathcal {D}}'((0,T_0 )\times {\mathbb {R}}^d)$; as in Case 1, it is bounded in $L^{\alpha _1}((0,T_0 ),L^{r})$, and in Case 2 it is bounded in $L^{\alpha _2}((0,T_0 ),W^{1,r})$, it is weakly convergent in these spaces respectively (as ${\mathcal {D}}$ is dense in their dual spaces).

By the continuity of the Riesz transforms on $L^{r}(\Phi ^{r} dx)$ and on $W^{1,r}(\Phi ^{r} dx)$, we find that in the Case 1 and Case 2, $p_{\epsilon _k}$ is convergent to the distribution $p= \sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_{i}u_{j})$. We have obtained

$$\begin{aligned} \partial _t \mathbf{u }= \Delta \mathbf{u }+ ( \mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p. \end{aligned}$$

Moreover, we have seen that $\partial _t \mathbf{u }$ is locally in $L^1 H^{-2}$, and thus $\mathbf{u }$ has representative such that $t\mapsto \mathbf{u }(t,.)$ is continuous from $[0,T_0 )$ to ${\mathcal {D}}'({\mathbb {R}}^d)$ and coincides with $\mathbf{u }(0,.)+\int _0^t \partial _t \mathbf{u }\, ds$.

In the sense of distributions, we have

$$\begin{aligned} \mathbf{u }(0,.)+\int _0^t \partial _t \mathbf{u }\, ds=\mathbf{u }=\lim _{k\rightarrow +\infty } \mathbf{u }_{\epsilon _k}=\lim _{k\rightarrow +\infty } \mathbf{u }_{0,\epsilon _k}+ \int _0^t \partial _t \mathbf{u }_{n_k}\, ds=\mathbf{u }_{0}+\int _0^t \partial _t\mathbf{u }\, ds, \end{aligned}$$

hence, $\mathbf{u }(0,.)=\mathbf{u }_{0}$, and $\mathbf{u }$ is a solution of (NS).

Now, we want to prove the energy balance. In the case of dimension 2, we remark that, since $\sqrt{\Phi }\mathbf{u }\in L^\infty L^2 \cap L^2 H^1$, we have by interpolation that $\sqrt{\Phi }\mathbf{u }\in L^4 L^4$, and then we can define $((\mathbf{u }\cdot {{\varvec{\nabla }}}) \mathbf{u }) \cdot \mathbf{u }$. The equality

$$\begin{aligned} \partial _t\left( \frac{ | \mathbf{u }| ^2}{2}\right) =\Delta \left( \frac{ | \mathbf{u }| ^2}{2}\right) - | {{\varvec{\nabla }}}\mathbf{u }| ^2- {{\varvec{\nabla }}}\cdot \left( \frac{ | \mathbf{u }| ^2}{2}\mathbf{u }\right) -{{\varvec{\nabla }}}\cdot (p\mathbf{u }) \end{aligned}$$

is then easy to prove.

Let us consider the case $d = 3$. We define

$$\begin{aligned} A_\epsilon = - \partial _t\left( \frac{ | \mathbf{u }_\epsilon | ^2}{2}\right) +\Delta \left( \frac{ | \mathbf{u }_\epsilon | ^2}{2}\right) -{{\varvec{\nabla }}}\cdot \left( \frac{ | \mathbf{u }_\epsilon | ^2}{2}\mathbf{u }_\epsilon \right) -{{\varvec{\nabla }}}\cdot (p_\epsilon \mathbf{u }_\epsilon ) = | {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon | ^2. \end{aligned}$$

As $u_{\epsilon _k}$ is locally strongly convergent in $L^2 L^2$; and locally bounded in $L ^\infty L^2$, it is then locally strongly convergent in $L^{p'} L^2$, with $p' < \infty$. Then, as $\sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon$ is bounded in $L^2((0,T), L^2)$, by the Gagliardo-Nirenberg interpolation inequalities we obtain $\mathbf{u }_{\epsilon _k}$ is locally strongly convergent in $L^{p'} L^{q'}$ with $\frac{2}{p'}+ \frac{3}{q'} > \frac{d}{2}$.

In Case 1, we know that $p_{\epsilon _k}$ is locally weakly convergent in $L^{\alpha _1} L^r$ and by the remark above, $\mathbf{u }_{\epsilon _k}$ is locally strongly convergent in $L^{\frac{\alpha _1}{\alpha _1 -1}} L^{\frac{r}{r-1}}$, and hence $p_{\epsilon _k} \mathbf{u }_{\epsilon _k}$ converges in the sense of distributions.

In Case 2, we know that $p_{\epsilon _k}$ is locally weakly convergent in $L^{\alpha _2} L^{q}$ and and by the remark above, $\mathbf{u }_{\epsilon _k}$ is locally strongly convergent in $L^{\frac{\alpha _2}{\alpha _2-1}} L^{\frac{q}{q -1}}$, and hence $p_{\epsilon _k} \mathbf{u }_{\epsilon _k}$ converges in the sense of distributions.

Thus, $A_{\epsilon _k}$ is convergent in ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$ to

$$\begin{aligned} A= - \partial _t\left( \frac{ | \mathbf{u }| ^2}{2}\right) +\Delta \left( \frac{ | \mathbf{u }| ^2}{2}\right) -{{\varvec{\nabla }}}\cdot \left( \frac{ | \mathbf{u }| ^2}{2}\mathbf{u }\right) -{{\varvec{\nabla }}}\cdot (p \mathbf{u }), \end{aligned}$$

and $A =\lim _{k \rightarrow +\infty } | {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon _k} | ^2$. If $\theta \in {\mathcal {D}}((0,T)\times {\mathbb {R}}^d)$ is non-negative, we have that $\sqrt{\theta }{{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon _k}$ is weakly convergent in $L^2L^2$ to $\sqrt{\theta }{{\varvec{\nabla }}}\otimes \mathbf{u }$, so that

$$\begin{aligned} \iint A \theta \, dx\, ds=\lim _{\epsilon _k\rightarrow +\infty }\iint A_{\epsilon _k} \theta \, dx\, ds= \lim _{k\rightarrow +\infty }\iint | {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon _k} | ^2 \theta \, dx\, ds \ge \iint | {{\varvec{\nabla }}}\otimes \mathbf{u }| ^2 \theta \, dx\, ds . \end{aligned}$$

Hence, there exists a non-negative locally finite measure $\mu$ on $(0,T)\times {\mathbb {R}}^3$ such that $A= | {{\varvec{\nabla }}}\mathbf{u }| ^2 +\mu$, i.e. such that

$$\begin{aligned} \partial _t(\frac{ | \mathbf{u }| ^2}{2})=\Delta (\frac{ | \mathbf{u }| ^2}{2})- | {{\varvec{\nabla }}}\mathbf{u }| ^2- {{\varvec{\nabla }}}\cdot \left( \frac{ | \mathbf{u }| ^2}{2}\mathbf{u }\right) -{{\varvec{\nabla }}}\cdot (p\mathbf{u }) -\mu . \end{aligned}$$

4.3 Strong Convergence to the Initial Data

We use again inequalities (2) and (4). We know that on $(0,T_0)$ we have a control of $\Vert \mathbf{u }_\epsilon \Vert _{L^2(\Phi \, dx)}$ that holds uniformly in $\epsilon$ and t. Thus, inequality (2) gives us

$$\begin{aligned} \Vert \mathbf{u }_\epsilon (t,.)\Vert _{L^2(\Phi \, dx)}\le \Vert \mathbf{u }_{0,\epsilon }\Vert _{L^2(\Phi \, dx)} + C_\Phi t (1+\Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}^2+\Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}^{2d}). \end{aligned}$$

Since $\mathbf{u }_{\epsilon _k}= \mathbf{u }_{0, \epsilon _k}+ \int _0^t \partial _t \mathbf{u }_{\epsilon _k}\, ds$, we see that $\mathbf{u }_{\epsilon _k}(t,.)$ is convergent to $\mathbf{u }(t,.)$ in ${\mathcal {D}}'({\mathbb {R}}^d)$, hence is weakly convergent in $L^2(\Phi \, dx)$ (as it is bounded in $L^2(\Phi dx)$); on the other hand, $\mathbf{u }_{0,\epsilon _k}$ is strongly convergent to $\mathbf{u }_0$ in $L^2(\Phi \, dx)$. Thus, we have

$$\begin{aligned} \Vert \mathbf{u }(t,.)\Vert _{L^2(\Phi \, dx)}\le \Vert \mathbf{u }_{0}\Vert _{L^2(\Phi \, dx)} + C_\Phi t (1+\Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}^2+\Vert \mathbf{u }_0\Vert _{L^2(\Phi \, dx)}^{2d}). \end{aligned}$$

In particular,

$$\begin{aligned} \limsup _{t\rightarrow 0} \Vert \mathbf{u }(t,.)\Vert _{L^2(\Phi \, dx)}\le \Vert \mathbf{u }_{0}\Vert _{L^2(\Phi \, dx)} . \end{aligned}$$

Moreover, we have $\mathbf{u }= \mathbf{u }_{0}+ \int _0^t \partial _t \mathbf{u }\, ds$, so that $\mathbf{u }(t,.)$ is convergent to $\mathbf{u }_0$ in ${\mathcal {D}}'({\mathbb {R}}^d)$, hence is weakly convergent in $L^2(\Phi \, dx)$. Thus, we have

$$\begin{aligned} \Vert \mathbf{u }_{0}\Vert _{L^2(\Phi \, dx)}\le \liminf _{t\rightarrow 0} \Vert \mathbf{u }(t,.)\Vert _{L^2(\Phi \, dx)}. \end{aligned}$$

This gives $\Vert \mathbf{u }_{0} \Vert _{L^2(\Phi dx)}^2 = \lim _{t\rightarrow 0} \Vert \mathbf{u }(t,. )\Vert _{L^2(\Phi dx)}^2$, which allows to turn the weak convergence into a strong convergence. $\square$

4.4 Global Existence Using a Scaling Argument

Let $\lambda >0$, then $\mathbf{u }_{\epsilon }$ is a solution of the Cauchy initial value problem for the approximated Navier–Stokes equations $(NS_\epsilon )$ on (0, T) with initial value $\mathbf{u }_{0, \epsilon }$ if and only if $\mathbf{u }_{{\epsilon }, \lambda } (t,x)=\lambda \mathbf{u }_{\epsilon } (\lambda ^2 t,\lambda x)$ is a solution for the approximated Navier–Stokes equations $(NS_{\lambda \epsilon })$ on $(0,T/\lambda ^2)$ with initial value $\mathbf{u }_{0,{\epsilon },\lambda }(x)=\lambda \mathbf{u }_{0, {\epsilon }}(\lambda x)$. We shall write $\mathbf{u }_{0,\lambda }=\lambda \mathbf{u }_0(\lambda x)$.

We have seen that

$$\begin{aligned} \begin{aligned} \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon , \lambda } (t) \Vert _{L^2}^2 + \int _0^t \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon , \lambda } \Vert _{L^2}^2 \le \Vert \sqrt{\Phi } \mathbf{u }_{0, \epsilon , \lambda } \Vert _{L^2}^2 + C_\Phi \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon , \lambda } \Vert _{L^2}^2 + \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon , \lambda } \Vert _{L^2}^{2d} \, ds \end{aligned} \end{aligned}$$

(under the extra condition, when $d=4$, that $\Vert \sqrt{\Phi } \mathbf{u }_{\epsilon , \lambda } (t) \Vert _{L^2}$ remains smaller than $\epsilon _0$).

By Lemma 4.1, we thus found that there exists a constant $C_\Phi \ge 1$ such that if $T_\lambda$ satisfies

• if $d=2$, $C_\Phi \left( 1+\Vert \mathbf{u }_{0,\lambda }\Vert _{L^2(\Phi dx)}^2 \right) \, T_\lambda = 1$

• if $d=3$, $C_\Phi \left( 1+\Vert \mathbf{u }_{0,\lambda }\Vert _{L^2(\Phi dx)}^2 \right) ^2 \, T_\lambda = 1$

• if $d=4$ and $\Vert u_{0,\lambda }\Vert _{L^2(\Phi \, dx)}\le \frac{1}{C_\Phi }$, $C_\Phi \, T_\lambda = 1$

then

$$\begin{aligned} \sup _{0\le t\le T_\lambda } \Vert \ \mathbf{u }_{\epsilon ,\lambda }(t,.)\Vert _{L^2(\Phi dx)}^2 + \int _0^{T_\lambda } \Vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon ,\lambda } \Vert _{L^2(\Phi \, dx)}^2\, ds \le C_\Phi (1 + \Vert \mathbf{u }_{0,\lambda }\Vert _{L^2(\Phi \, dx)}^2 ). \end{aligned}$$

(5)

It gives that the solutions $\mathbf{u }_\epsilon$ are controlled, uniformly in $\epsilon$, on $(0, \lambda ^2 T_\lambda )$ since for $t \in (0, T_\lambda )$,

$$\begin{aligned} \int \vert \mathbf{u }_{{\epsilon }, \lambda }(t,x)\vert ^2 \Phi (x)\, dx = \int \vert \mathbf{u }_\epsilon (\lambda ^2 t, y)\vert ^2 \Phi (\frac{y}{\lambda }) \lambda ^{2-d} \, dy\ge \lambda ^{2-d}\int \vert \mathbf{u }_{{\epsilon }}(\lambda ^2 t,x)\vert ^2 \Phi (x)\, dx \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _0^{T_\lambda } \int \vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon , \lambda }(t,x) \vert ^2 \Phi (x)\, dx\, dt=&\int _0^{\lambda ^2 T_\lambda } \int \vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon , \lambda }(s,y) \vert ^2 \Phi (\frac{y}{\lambda }) \lambda ^{2-d}\, dy\, ds \\ \ge&\lambda ^{2-d} \int _0^{\lambda ^2 T_\lambda } \int \vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon }(t,x) \vert ^2 \Phi ( x )\, dx\, dt\\ \int _0^{T_\lambda } \int \vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon , \lambda }(t,x) \vert ^2 \Phi (x) \, dx\, dt \ge&C_{\lambda } \int _0^{\lambda ^2 T_\lambda }\Vert {{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon } \Vert _{L^2(\Phi dx)}^2\, ds. \end{aligned} \end{aligned}$$

Moreover, we have $\lim _{ \lambda \rightarrow +\infty } \Vert \mathbf{u }_{0,\lambda }\Vert _{L^2(\Phi \, dx)}=0$ when $d=4$ and $\lim _{ \lambda \rightarrow +\infty } \lambda ^2 T_\lambda = + \infty$ when $2\le d\le 4$. Indeed, we have

$$\begin{aligned} \int \lambda ^2\vert \mathbf{u }_0(\lambda x)\vert ^2 \Phi (x)\, dx=\lambda ^{2-d} \int \vert \mathbf{u }_0( x)\vert ^2 \Phi (\frac{x}{\lambda })\, dx = \lambda ^{4-d} \int \vert \mathbf{u }_0( x)\vert ^2 \frac{\Phi (\frac{x}{\lambda })}{\lambda ^2\Phi (x)} \Phi (x)\, dx \end{aligned}$$

Since $\frac{\Phi (\frac{x}{\lambda })}{\lambda ^2\Phi (x)} \le \min \{ C_2,\frac{1}{\lambda ^2\Phi (x)} \}$ by hypothesis $(\mathbf{H }4)$, we find by dominated convergence that $\Vert \mathbf{u }_{0,\lambda }\Vert _{L^2(\Phi \, dx)}=o(\lambda ^{\frac{4-d}{2}})$ and thus $\lim _{ \lambda \rightarrow +\infty } \lambda ^2 T_\lambda = + \infty$ .

Thus, if we consider a finite time T and a sequence $\epsilon _k$, we may choose $\lambda$ such that $\lambda ^2 T_\lambda >T$ (and such that $\Vert \mathbf{u }_{0,\lambda } \Vert _{L^2(\Phi \, dx)}<\epsilon _0$ if $d=4$); we have a uniform control of $\mathbf{u }_{\epsilon ,\lambda }$ and of ${{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon ,\lambda }$ on $(0,T_\lambda )$, hence a uniform control of $\mathbf{u }_{\epsilon }$ and of ${{\varvec{\nabla }}}\otimes \mathbf{u }_{\epsilon }$ on (0, T). We may exhibit a solution on (0, T) using the Rellich–Lions theorem by extracting a subsequence $\epsilon _{k_n}$. A diagonal argument permits then to obtain a global solution.

Theorem 1 is proved. $\square$

5 Proof of Theorem 2 (The Case $d=2$)

In the case of dimension $d=2$, the Navier–Stokes equations are well-posed in $H^1$ and we don’t need to mollify the equations. Thus, we may approximate the Navier–Stokes equations with

$$\begin{aligned} (NS_\epsilon ) \left\{ \begin{matrix} \partial _t \mathbf{u }_\epsilon = \Delta \mathbf{u }_\epsilon -( \mathbf{u }_\epsilon \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon - {{\varvec{\nabla }}}p_\epsilon \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }_\epsilon =0, \qquad \qquad \mathbf{u }_\epsilon (0,.)=\mathbf{u }_{0, \epsilon } \end{matrix}\right. \end{aligned}$$

with

$$\begin{aligned} \mathbf{u }_{0, \epsilon } = {\mathbb {P}}( \phi _\epsilon \mathbf{u }_0 ) . \end{aligned}$$

Then the vorticity $\omega _\epsilon$ is solution of

$$\begin{aligned} \left\{ \begin{array}{ll} \partial _t \omega _\epsilon = \Delta \omega _\epsilon -( \mathbf{u }_\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon \\ \\ {{\varvec{\nabla }}}\cdot \omega _\epsilon =0, \qquad \qquad \omega _\epsilon (0,.)=\omega _{0, \epsilon } \end{array}\right. \end{aligned}$$

with

$$\begin{aligned} \omega _{0, \epsilon } = {{\varvec{\nabla }}}\wedge ( \phi _\epsilon \mathbf{u }_0 ) . \end{aligned}$$

$\mathbf{u }_{0,\epsilon }$ belongs to $H^1$, so we know that we have a global solution $\mathbf{u }_\epsilon$. We then just have to prove that, for every finite time $T_0$, we have a uniform control of the norms $\Vert \omega _\epsilon \Vert _{L^\infty ((0,T_0), L^2(\Phi \, dx))}$ and $\Vert {{\varvec{\nabla }}}\omega _\epsilon \Vert _{L^2 ((0,T_0),L^2(\Phi \, dx))}$.

We can calculate $\int \partial _t \omega _\epsilon \cdot \omega _\epsilon \Phi \,dx$ so that

$$\begin{aligned} \begin{aligned}&\int \frac{ | \omega _\epsilon (t,x) | ^2}{2} \Phi \, dx + \int _0^t \int | {{\varvec{\nabla }}}\omega _\epsilon | ^2\, \, \Phi dx\, ds \\&\quad = \int \frac{ | \omega _{0, \epsilon } (x) | ^2}{2} \Phi \, dx - \int _0^t \int {{\varvec{\nabla }}}(\frac{| \omega _\epsilon |^2}{2} )\, \, \cdot {{\varvec{\nabla }}}\Phi dx\, ds \\&\quad \quad + \int _0^t \int \frac{| \omega _\epsilon |^2}{2} \mathbf{u }_\epsilon \, \cdot {{\varvec{\nabla }}}\Phi \, dx\, ds. \end{aligned} \end{aligned}$$

As

$$\begin{aligned} \begin{aligned} \int _0^t \int \frac{| \omega _\epsilon |^2}{2} \mathbf{u }_\epsilon \, \cdot {{\varvec{\nabla }}}\Phi \, dx\, ds&\le \int _0^t \Vert \sqrt{\Phi } \omega _\epsilon \Vert _{L^{\frac{8}{3}} }^2 \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^4} \\&\le \int _0^t (\Vert \sqrt{\Phi } \omega _\epsilon \Vert _{L^2 }^{3/4} \Vert {{\varvec{\nabla }}}(\sqrt{\Phi } \omega _\epsilon ) \Vert _{L^2 }^{1/4} )^2 \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^4} \end{aligned} \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} \Vert \sqrt{\Phi } \omega _\epsilon (t) \Vert _{L^2}^2 + \int _0^t \Vert \sqrt{\Phi } {{\varvec{\nabla }}}\omega _\epsilon \Vert _{L^2}^2 \le \Vert \sqrt{\Phi } \omega _{0, \epsilon } \Vert _{L^2}^2 + C_\Phi \int _0^t \Vert \sqrt{\Phi } \omega _\epsilon \Vert _{L^2}^2 (1 + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^4}^\frac{4}{3}) \, ds \end{aligned} \end{aligned}$$

We can conclude that, for all $T> 0$ and for all $t \in (0,T)$,

$$\begin{aligned} \begin{aligned} \Vert \sqrt{\Phi } \omega _{\epsilon } (t) \Vert _{L^2}^2 + \int _0^t\Vert \sqrt{\Phi } {{\varvec{\nabla }}}\omega _{\epsilon } \Vert _{L^2}^2 \le \Vert \sqrt{\Phi } \omega _{0, \epsilon } \Vert _{L^2}^2 e^{ C_\Phi \sup _{\epsilon >0} \int _0^t (1+ \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon } \Vert _{L^4})^{\frac{4}{3}} \, ds } \end{aligned} \end{aligned}$$

Thus, we have uniform controls on (0, T). $\square$

6 Proof of Theorems 3 and 4 (The Axisymmetric Case)

6.1 Axisymmetry

In ${\mathbb {R}}^3$, we consider the usual coordinates $(x_1,x_2,x_3)$ and the cylindrical coordinates $(r,\theta ,z)$ given by the formulas $x_1 = r \cos \theta ,$ $x_2 = r \sin \theta$ and $x_3=z$.

We denote $(\mathbf{e }_1 , \mathbf{e }_2,\mathbf{e }_3)$ the usual canonical basis

$$\begin{aligned} \mathbf{e }_1=(1, 0, 0),\ \mathbf{e }_2=(0, 1, 0), \mathbf{e }_3=(0, 0, 1). \end{aligned}$$

We attach to the point x (with $r\ne 0$) another orthonormal basis

$$\begin{aligned} \mathbf{e }_ r = \frac{\partial x}{\partial r } = \cos \theta \, \mathbf{e }_1 + \sin \theta \, \mathbf{e }_2, \, \, \, \mathbf{e }_\theta = \frac{1}{ r } \frac{\partial x}{\partial \theta } = -\sin \theta \, \mathbf{e }_1 + \cos \theta \, \mathbf{e }_2, \, \, \, \mathbf{e }_z = \frac{\partial x}{\partial z} = \mathbf{e }_3. \end{aligned}$$

For a vector field $\mathbf{u }= (u_1, u_2, u_3 ) =u_1 \mathbf{e }_1 + u_2 \mathbf{e }_2 + u_3 \mathbf{e }_3$, we can see that

$$\begin{aligned} \mathbf{u }= (u_1 \cos \theta + u_2 \sin \theta ) \, \mathbf{e }_ r + (-u_1 \sin \theta + u_2 \cos \theta ) \, \mathbf{e }_\theta + u_3 \, \mathbf{e }_z . \end{aligned}$$

We will denote $(u_ r , u_\theta , u_z)_p$ the coordinates of $\mathbf{u }$ in the basis $(\mathbf{e }_ r , \mathbf{e }_\theta , \mathbf{e }_z)$. We will consider axially symmetric (axisymmetric) vector fields $\mathbf{u }$ without swirl and axisymmetric scalar functions a, which means that

$$\begin{aligned} \mathbf{u }= u_r(r,z) \, \mathbf{e }_r + u_z(r,z) \, \mathbf{e }_z \, \, \, \, \, \text {and} \, \, \, \, \, a = a(r,z). \end{aligned}$$

6.2 The $H^1$ Case

We will use the following well known results of Ladyzhenskaya [15, 18].

Proposition 6.1

Let $\mathbf{u }_0$ be a divergence free axisymmetric vector field without swirl, such that $\mathbf{u }_0$ belongs to $H^1$. Then, the following problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

has a unique solution $\mathbf{u }\in {\mathcal {C}}([0,+\infty ), H^1)$. This solution is axisymmetric without swirl. Moreover, $\mathbf{u }, {{\varvec{\nabla }}}\otimes \mathbf{u }$ belong to $L^\infty ((0,+\infty ), L^2)$, and ${{\varvec{\nabla }}}\otimes \mathbf{u }, \Delta \mathbf{u }$ belong to $L^2((0,+\infty ),L^2)$.

If $\mathbf{u }_0\in H^2$, we have the inequality

$$\begin{aligned} \int \frac{ | \omega (t) |^2 }{r^2} dx \le \int \frac{ | \omega _0 |^2 }{r^2} \le \Vert {{\varvec{\nabla }}}\otimes \omega _0 \Vert _2^2. \end{aligned}$$

6.3 A Priori Controls

Let $\phi \in {\mathcal {D}}({\mathbb {R}}^2)$ be a real-valued radial function which is equal to 1 in a neighborhood of 0 and let $\phi _\epsilon (x)= \phi (\epsilon (x_1,x_2))$. For $\epsilon \in (0,1]$, let

$$\begin{aligned} \mathbf{u }_{0, \epsilon } = {\mathbb {P}}( \phi _\epsilon \mathbf{u }_0 ) . \end{aligned}$$

Thus, $\mathbf{u }_{0, \epsilon }$ is a divergence free axisymmetric without swirl vector field which belongs to $H^1$. As we have

$$\begin{aligned} \omega _{0, \epsilon } = {{\varvec{\nabla }}}\wedge \mathbf{u }_{0, \epsilon } = {{\varvec{\nabla }}}\wedge (\phi _\epsilon \mathbf{u }_0) = \phi _\epsilon \omega _0 + \epsilon ({{\varvec{\nabla }}}\phi )(\epsilon x) \wedge \mathbf{u }_0, \end{aligned}$$

using $\Phi \in {\mathcal {A}}_2$ and $\vert \epsilon {{\varvec{\nabla }}}\phi (\epsilon x)\vert \le C \frac{1}{r} \mathbb {1}_{ r\ge \frac{1}{C\epsilon }}\le C' \mathbb {1}_{ r \ge \frac{1}{C\epsilon }}\sqrt{\Phi }$, we can see that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \Vert \mathbf{u }_0 - \mathbf{u }_{0, \epsilon } \Vert _{L^2 (\Phi \, dx)} + \Vert \omega _0 - \omega _{0, \epsilon } \Vert _{L^2 (\Psi \, dx)} =0. \end{aligned}$$

Let $\mathbf{u }_\epsilon$ be the global solution of the problem

$$\begin{aligned} (NS_\epsilon ) \left\{ \begin{matrix} \partial _t \mathbf{u }_\epsilon = \Delta \mathbf{u }_\epsilon -(\mathbf{u }_\epsilon \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon - {{\varvec{\nabla }}}p_\epsilon \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }_\epsilon =0, \qquad \qquad \mathbf{u }_\epsilon (0,.)=\mathbf{u }_{0, \epsilon } \end{matrix}\right. \end{aligned}$$

given by the Proposition 6.1. We denote $\omega _\epsilon = {{\varvec{\nabla }}}\wedge \, \mathbf{u }_\epsilon$, then

$$\begin{aligned} \partial _t \mathbf{u }_\epsilon = \Delta \mathbf{u }_\epsilon + ( \mathbf{u }_\epsilon \cdot {{\varvec{\nabla }}})\mathbf{u }_\epsilon - {{\varvec{\nabla }}}p_\epsilon \end{aligned}$$

(6)

and

$$\begin{aligned} \partial _t \omega _\epsilon = \Delta \omega _\epsilon + ( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon - (\mathbf{u }_\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon \end{aligned}$$

(7)

As $\sqrt{\Psi } \omega _\epsilon \in L^2 H^1$ (because $\sqrt{\Psi },{{\varvec{\nabla }}}\sqrt{\Psi } \in L^\infty$) and $\sqrt{\Psi } \partial _t \omega _\epsilon \in L^2 H^{-1}$, we can calculate $\int \partial _t \omega _\epsilon \cdot \omega _\epsilon \Psi \,dx$ using (7) so that

$$\begin{aligned} \begin{aligned}&\int \frac{ | \omega _\epsilon (t,x) | ^2}{2} \Psi \, dx + \int _0^t \int | {{\varvec{\nabla }}}\otimes \omega _\epsilon | ^2\, \, \Psi dx\, ds \\&\quad = \int \frac{ | \omega _{0, \epsilon } (x) | ^2}{2} \Psi \, dx - \int _0^t \int {{\varvec{\nabla }}}(\frac{| \omega _\epsilon |^2}{2} )\, \, \cdot {{\varvec{\nabla }}}\Psi dx\, ds \\&\qquad + \int _0^t \int \frac{| \omega _\epsilon |^2}{2} \mathbf{u }_\epsilon \, \cdot {{\varvec{\nabla }}}\Psi \, - (\omega _\epsilon \cdot \mathbf{u }_\epsilon )\omega _\epsilon \cdot {{\varvec{\nabla }}}\Psi \, dx\\&\qquad - \int _0^t \int (( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon ) \cdot \mathbf{u }_\epsilon \, \, \Psi \, dx\, ds\\&\quad \le \int \frac{ | \omega _{0, \epsilon } (x) | ^2}{2} \Psi \, dx +\frac{1}{8} \int _0^t \int | {{\varvec{\nabla }}}\otimes \omega _\epsilon | ^2\, \, \Psi dx\, ds + C \int _0^t \Vert \sqrt{\Psi }\, \omega _\epsilon \Vert _2^2\, ds \\&\qquad + C \int _0^t \Vert \sqrt{\Psi }\, \omega _\epsilon \Vert _2 \Vert \sqrt{\Psi }\, \omega _\epsilon \Vert _6 \Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _3\, ds \\&\qquad - \int _0^t \int (( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon ) \cdot \mathbf{u }_\epsilon \, \, \Psi \, dx\, ds \\&\quad \le \int \frac{ | \omega _{0, \epsilon } (x) | ^2}{2} \Psi \, dx +\frac{1}{4} \int _0^t \int | {{\varvec{\nabla }}}\otimes \omega _\epsilon | ^2\, \, \Psi dx\, ds + C \int _0^t \Vert \sqrt{\Psi }\, \omega _\epsilon \Vert _2^2\, ds \\&\qquad + C' \int _0^t \Vert \sqrt{\Psi }\, \omega _\epsilon \Vert _2^2 (\Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _3+ (\Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _3^{4/3}) \, ds \\&\qquad - \int _0^t \int (( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon ) \cdot \mathbf{u }_\epsilon \, \, \Psi \, dx\, ds \end{aligned} \end{aligned}$$

As $\omega _\epsilon = \omega _{\epsilon , \theta } \, \mathbf{e }_\theta$, we have

$$\begin{aligned} \omega _\epsilon \cdot {{\varvec{\nabla }}}\omega _\epsilon = -\frac{ \omega _{\epsilon , \theta } ^2 }{r} \, \mathbf{e }_r . \end{aligned}$$

In order to control $\mathbf{u }_\epsilon \cdot ( \omega _\epsilon \cdot {{\varvec{\nabla }}}\omega _\epsilon )$, we split the domain of integration in a domain where r is small and a domain where r is large. The support of $\phi _1$ is contained in $\{x\ /\ r<R\}$ for some $R>0\}$, and the support of $1-\phi _1$ is contained in $\{x\ /\ r>R_0\}$ for some $R_0>0\}$. We have

$$\begin{aligned} \inf _{r<R} \Phi (x)= \inf _{\sqrt{x_1^2+x_2^2}<R}\Phi (x_1,x_2,0)>0 \end{aligned}$$

and similarly

$$\begin{aligned} \inf _{r<R} \Psi (x)= \inf _{\sqrt{x_1^2+x_2^2}<R}\Psi (x_1,x_2,0)>0. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \inf _{r>R_0} r^2 \Phi (x)=\inf _{\sqrt{x_1^2+x_2^2}>R_0} (x_1^2+x_2^2) \Phi (x_1,x_2,0)\ge \inf _{\vert x\vert>R_0} \vert x\vert ^2 \Phi (x)>0. \end{aligned}$$

We then write:

$$\begin{aligned} \begin{aligned}&- \int _0^t \int (( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon ) \cdot \mathbf{u }_\epsilon \, \, \Psi \, dx\, ds \\&\quad = \int _0^t \int \phi _1( \, ( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \mathbf{u }_\epsilon ) \cdot \omega _\epsilon \, ) \, \Psi \, dx\, ds + \int _0^t \int ( \omega _\epsilon \cdot \mathbf{u }_\epsilon ) ( \omega _\epsilon \cdot {{\varvec{\nabla }}}\phi _1) \Psi \, dx\, ds \\&\qquad +\int _0^t \int \phi _1 ( \omega _\epsilon \cdot \mathbf{u }_\epsilon ) \omega _\epsilon \cdot {{\varvec{\nabla }}}\Psi dx \, ds \\&\qquad - \int _0^t \int (1-\phi _1) ( \mathbf{u }_\epsilon \cdot ( \omega _\epsilon \cdot {{\varvec{\nabla }}}\omega _\epsilon ) ) \Psi dx \, ds \\&\quad \le C \int _0^t \int \vert \omega _\epsilon \vert ^2 \vert {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \vert \, \Psi ^{3/2} \, dx\, ds + C \int _0^t \int \vert \omega _\epsilon \vert ^2 \vert \mathbf{u }_\epsilon \vert \, \sqrt{\Phi }\, \Psi \, dx\, ds . \end{aligned} \end{aligned}$$

As $\Psi \in {\mathcal {A}}_2$, we have $\Vert \sqrt{\Psi }{{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _2\approx \Vert \sqrt{\Psi }\omega _\epsilon \Vert _2$; moreover,

$$\begin{aligned} \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Phi }\mathbf{u }_\epsilon )\Vert _2\le C(\Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _2+\Vert \sqrt{\Psi }\omega _\epsilon \Vert _2) \end{aligned}$$

and

$$\begin{aligned} \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Psi }\omega _\epsilon )\Vert _2\le C(\Vert \sqrt{\Psi }\omega _\epsilon \Vert _2+\Vert \sqrt{\Psi }{{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _2), \end{aligned}$$

and thus we get

$$\begin{aligned} \begin{aligned}&- \int _0^t \int (( \omega _\epsilon \cdot {{\varvec{\nabla }}}) \omega _\epsilon ) \cdot \mathbf{u }_\epsilon \, \, \Psi \, dx\, ds \\&\quad \le C \int _0^t \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \mathbf{u }_\epsilon \Vert _{L^2} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^3} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^6} \, ds \\&\qquad + C \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^6} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^3} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2} \, ds\\&\quad \le C' \int _0^t \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2}^{\frac{3}{2}} (\Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2} + \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{L^2} )^{\frac{3}{2}} \, ds\\&\qquad + C' \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2}^{\frac{3}{2}} (\Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2} + \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{L^2} )^{\frac{1}{2}} \, ds\\&\quad \le C'' \int _0^t ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} +\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} ^{4/3}) \Vert \sqrt{\Psi }\omega _\epsilon \Vert _2^2+ \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{2}^{3} + \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{2}^6 \, ds\\&\qquad + \frac{1}{8} \int _0^t \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{2}^2\, ds \end{aligned} \end{aligned}$$

We finally find that

$$\begin{aligned} \begin{aligned}&\Vert \sqrt{\Psi } \omega _\epsilon (t) \Vert _{L^2}^2 + \int _0^t \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{L^2}^2\, ds \\&\quad \le \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}^2 + C \int (1+\Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _3+ (\Vert \sqrt{\Phi }\mathbf{u }_\epsilon \Vert _3^{4/3}) \Vert \sqrt{\Psi }\omega _\epsilon \Vert _2^2 \, ds \\&\qquad + C \int _0^t ( \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}\ +\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} ^{4/3}) \Vert \sqrt{\Psi }\omega _\epsilon \Vert _2^2+ \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{2}^{3} + \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{2}^6 \, ds \\&\quad \le \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}^2 \\&\qquad + C' \int _0^t (1+ \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2}\ +\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{2} ^{4/3}) \Vert \sqrt{\Psi }\omega _\epsilon \Vert _2^2+ \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{2}^6 \, ds \end{aligned} \end{aligned}$$

(8)

We already know that $\Vert \sqrt{\Phi } \mathbf{u }_\epsilon (t) \Vert _{L^2}$ remains bounded (independently of $\epsilon$) on every bounded interval, so that we may again use Lemma 4.1 and control $\sup _{0\le t\le T_0} \Vert \ \omega _\epsilon (t,.)\Vert _{L^2(\Psi dx)}^2 + \int _0^{T_0} \Vert {{\varvec{\nabla }}}\omega _\epsilon \Vert _{L^2(\Psi dx)}^2\, ds$ for some $T_0$, where both $T_0$ and the control don’t depend on $\epsilon$. The control is then transferred to the limit $\omega$ since $\omega =\lim \omega _{\epsilon _k}=\lim {{\varvec{\nabla }}}\wedge \mathbf{u }_{\epsilon _k}$. This proves local existence of a regular solution and Theorem 3 is proved.

6.4 The Case of a Very Regular Initial Value

We present a result apparently more restrictive that our main Theorem (Theorem 4), but we will see that it implies almost directly our main Theorem.

Proposition 6.2

Let $\Phi$ be a weight satisfying $(\mathbf{H }1){-}(\mathbf{H }4)$. Assume moreover that $\Phi$ depends only on $r=\sqrt{x_1^2+x_2^2}$. Let $\Psi$ be another continuous weight (that depends only on r) such that $\Phi \le \Psi \le 1$, $\Psi \in {\mathcal {A}}_2$ and there exists $C_1>0$ such that

$$\begin{aligned} \vert {{\varvec{\nabla }}}\Psi \vert \le C_1 \sqrt{\Phi }\Psi \text { and } \vert \Delta \Psi \vert \le C_1 \Phi \Psi . \end{aligned}$$

Let $\mathbf{u }_0$ be a divergence free axisymmetric vector field without swirl, such that $\mathbf{u }_0,$ belongs to $L^2 (\Phi dx)$, ${{\varvec{\nabla }}}\otimes \mathbf{u }_0$ and $\Delta \mathbf{u }_0$ belong to $L^2 (\Psi dx)$. Then there exists a global solution $\mathbf{u }$ of the problem

$$\begin{aligned} (NS) \left\{ \begin{matrix} \partial _t \mathbf{u }= \Delta \mathbf{u }-(\mathbf{u }\cdot {{\varvec{\nabla }}})\mathbf{u }- {{\varvec{\nabla }}}p \\ \\ {{\varvec{\nabla }}}\cdot \mathbf{u }=0, \qquad \qquad \mathbf{u }(0,.)=\mathbf{u }_0 \end{matrix}\right. \end{aligned}$$

such that

• $\mathbf{u }$ is axisymmetric without swirl, $\mathbf{u }$ belongs to $L^\infty ((0,T), L^2 (\Phi \, dx) )$, ${{\varvec{\nabla }}}\otimes \mathbf{u }$ belong to $L^\infty ((0,T), L^2 (\Psi \, dx) )$ and $\Delta \mathbf{u }$ belongs to $L^2((0,T),L^2 (\Psi \, dx) )$, for all $T>0$,

• the maps $t\in [0,+\infty )\mapsto \mathbf{u }(t,.)$ and $t\in [0,+\infty )\mapsto {{\varvec{\nabla }}}\otimes \mathbf{u }(t,.)$ are weakly continuous from $[0,+\infty )$ to $L^2 (\Phi \, dx)$ and to $L^2(\Psi \, dx)$ respectively, and are strongly continuous at $t=0$.

Proof

Ladyzhenskaya’s inequality for axisymmetric fields with no swirl (Proposition 6.1) gives

$$\begin{aligned} \begin{aligned} \int \frac{| \omega _\epsilon (t)|^2}{r^2} dx \le \int \frac{|\omega _{0, \epsilon }|^2}{r^2} dx. \end{aligned} \end{aligned}$$

(9)

As we have

$$\begin{aligned} \partial _i \omega _{0, \epsilon } = \phi _\epsilon \partial _i \omega _0 + \epsilon \partial _i \phi (\epsilon x) \omega _0 + \epsilon ({{\varvec{\nabla }}}\phi )(\epsilon x) \wedge \partial _i \mathbf{u }_0 + \epsilon ^2 ({{\varvec{\nabla }}}\partial _i \phi )(\epsilon x) \wedge \mathbf{u }_0, \end{aligned}$$

we can see that

$$\begin{aligned} \lim _{\epsilon \rightarrow 0 } \Vert {{\varvec{\nabla }}}\otimes \omega _{0, \epsilon } - {{\varvec{\nabla }}}\otimes \omega _{0} \Vert _{L^2 (\Psi \, dx)} = 0. \end{aligned}$$

As

$$\begin{aligned} \int \frac{|\omega _{0, \epsilon } - \omega _{0} |^2}{r^2} dx \le C (\int _{ 0< r< 1 } |{{\varvec{\nabla }}}\otimes \omega _{0, \epsilon } - {{\varvec{\nabla }}}\otimes \omega _{0} |^2 \Psi \, dx + \int _{ 1< r < + \infty } |\omega _{0, \epsilon } - \omega _{0} |^2 \Psi \, dx), \end{aligned}$$

we also have

$$\begin{aligned} \lim _{\epsilon \rightarrow 0 } \int \frac{|\omega _{0, \epsilon } - \omega _{0} |^2}{r^2} dx =0. \end{aligned}$$

We know that

$$\begin{aligned} \begin{aligned}&\int \frac{ | \omega _\epsilon (t,x) | ^2}{2} \Psi \, dx + \int _0^t \int | {{\varvec{\nabla }}}\otimes \omega _\epsilon | ^2\, \, \Psi dx\, ds \\&\quad = \int \frac{ | \omega _{0, \epsilon } (x) | ^2}{2} \Psi \, dx - \int _0^t \int {{\varvec{\nabla }}}(\frac{| \omega _\epsilon |^2}{2} )\, \, \cdot {{\varvec{\nabla }}}\Psi dx\, ds \\&\qquad + \int _0^t \int \frac{| \omega _\epsilon |^2}{2} \mathbf{u }_\epsilon \, \cdot {{\varvec{\nabla }}}\Psi \, dx\, ds \\&\qquad - \int _0^t \int ( \omega _\epsilon \cdot \mathbf{u }_\epsilon ) \omega _\epsilon \, \cdot {{\varvec{\nabla }}}\Psi \, dx\, ds - \int _0^t \int \mathbf{u }_\epsilon ( \omega _\epsilon \cdot {{\varvec{\nabla }}}\omega _\epsilon ) \, \Psi \, dx\, ds \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned}&\Vert \sqrt{\Psi } \omega _\epsilon (t) \Vert _{L^2}^2 + 2 \int _0^t \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\omega _\epsilon \Vert _{L^2}^2 \\&\quad \le \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}^2 + 2 \int _0^t \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2} \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\omega _\epsilon \Vert _{L^2} \\&\qquad + \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^3} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^3}^2 \\&\qquad + \int _0^t \frac{1}{r} |\mathbf{u }_{r, \epsilon }| |\omega _\epsilon |^2 \Psi \, dx \, ds \end{aligned} \end{aligned}$$

Furthermore, we have

$$\begin{aligned} \int _0^t \int \frac{1-\phi _1(x)}{r} |\mathbf{u }_{r, \epsilon }| |\omega _\epsilon |^2 \Psi \, dx \, ds \le \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^3} \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^3}^2 \end{aligned}$$

and

$$\begin{aligned} \int _0^t \int \frac{ \phi _1(x)}{r} |\mathbf{u }_{\epsilon ,r}| |\omega _\epsilon |^2 dx \, ds \le C \int _0^t \Vert \frac{\omega _\epsilon }{r} \Vert _{L^2} \Vert \sqrt{\Psi } \mathbf{u }_\epsilon \Vert _{L^\infty } \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2}, \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \Vert \frac{\omega _\epsilon }{r} \Vert _{L^2} \le C \Vert \frac{\omega _{0, \epsilon } }{r} \Vert _{L^2}&\le C ( \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}+ \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _{0, \epsilon } \Vert _{L^2} ) \\&\le C' ( \Vert \sqrt{\Phi } \mathbf{u }_{0} \Vert _{L^2} + \Vert \sqrt{\Psi } \omega _{0} \Vert _{L^2}+\Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _{0} \Vert _{L^2} ) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert \sqrt{\Psi } \mathbf{u }_\epsilon \Vert _{L^\infty }^2&\le C \Vert {{\varvec{\nabla }}}\otimes (\sqrt{\Psi }\mathbf{u }_\epsilon ) \Vert _{2} \Vert \Delta (\sqrt{\Psi }\mathbf{u }_\epsilon ) \Vert _{2} \\&\le C' (\Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2} + \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2} + \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{L^2})^2. \end{aligned} \end{aligned}$$

Then, if we denote $A_0 = \Vert \sqrt{\Phi } \mathbf{u }_{0} \Vert _{L^2} + \Vert \sqrt{\Psi } \omega _0 \Vert _{L^2} + \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _0 \Vert _{L^2}$, we get

$$\begin{aligned} \begin{aligned}&\Vert \sqrt{\Psi } \omega _\epsilon (t) \Vert _{L^2}^2 + \int _0^t \Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _\epsilon \Vert _{L^2}^2 \\&\quad \le \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}^2 + C \int _0^t \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^2}^2 \\&\qquad + C_\Phi \int _0^t \Vert \sqrt{\Psi } \omega _\epsilon \Vert _{L^2}^2 (1 + A_0 + A_0^2 + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^3} + \Vert \sqrt{\Phi } \mathbf{u }_\epsilon \Vert _{L^3}^2) \, ds \end{aligned} \end{aligned}$$

So, we conclude that, for all $T> 0$ and for all $t \in (0,T)$,

$$\begin{aligned} \begin{aligned}&\Vert \sqrt{\Psi } \omega _{\epsilon } (t) \Vert _{L^2}^2 + \int _0^t\Vert \sqrt{\Psi } {{\varvec{\nabla }}}\otimes \omega _{\epsilon } \Vert _{L^2}^2 \\&\quad \le \left( \Vert \sqrt{\Psi } \omega _{0, \epsilon } \Vert _{L^2}^2 + C_\Phi \sup _{\epsilon>0} \int _0^T \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon } \Vert _{L^2}^2\right) e^{ C_\Phi \sup _{\epsilon >0} \int _0^t \left( 1 + A_0^2 + \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon } \Vert _{L^3} + \Vert \sqrt{\Phi } \mathbf{u }_{\epsilon } \Vert _{L^3}^2\right) \, ds } \end{aligned} \end{aligned}$$

Thus, we can obtain a solution on (0, T) using the Aubin–Lions Theorem and finish with a diagonal argument to get a global solution. $\square$

6.5 End of the Proof

We begin by consider a local solution $\mathbf{v }$ on $(0,T_0)$ with initial value $\mathbf{u }_0$ given by Theorem 3, which is continuous from $(0, T_0)$ to ${\mathcal {D}}'$. We take $T_1 \in (0,T_0)$ such that ${{\varvec{\nabla }}}\otimes ({{\varvec{\nabla }}}\wedge \mathbf{v })(T_1,.) \in L^2(\Phi dx)$. We consider a solution $\mathbf{w }$ on $(T_1, + \infty )$ and initial value $\mathbf{v }(T_1)$ given by Proposition  6.2. Our global solution is defined as $\mathbf{u }=\mathbf{v }$ on $(0,T_1)$ and $\mathbf{u }=\mathbf{w }$ on $(T_1, + \infty )$. $\square$