Weak Solutions for Navier–Stokes Equations with Initial Data in Weighted $L^2$ Spaces

Abstract

We show the existence of global weak solutions to the three dimensional Navier–Stokes equations with initial velocity in the weighted spaces $L^2_{w_\gamma }$, where $w_\gamma (x)=(1+\vert x\vert )^{-\gamma }$ and $0<\gamma \leqq 2$, using new energy controls. As an application we give a new proof of the existence of global weak discretely self-similar solutions to the three dimensional Navier–Stokes equations for discretely self-similar initial velocities which are locally square integrable.

1 Introduction

Infinite-energy weak Leray solutions to the Navier–Stokes equations were introduced by Lemarié-Rieusset in 1999 [8] (they are presented more completely in [9] and [10]). This has allowed demonstration of the existence of local weak solutions for a uniformly locally square integrable initial data.

Other constructions of infinite-energy solutions for locally uniformly square integrable initial data were given in 2006 by Basson [1] and in 2007 by Kikuchi and Seregin [7]. These solutions allowed Jia and Sverak [6] to construct in 2014 the self-similar solutions for large (homogeneous of degree -1) smooth data. Their result has been extended in 2016 by Lemarié-Rieusset [10] to solutions for rough locally square integrable data. We remark that an homogeneous (of degree -1) and locally square integrable data is automatically uniformly locally $L^2$.

Recently, Bradshaw and Tsai [2] and Chae and Wolf [3] considered the case of solutions which are self-similar according to a discrete subgroup of dilations. Those solutions are related to an initial data which is self-similar only for a discrete group of dilations; in contrast to the case of self-similar solutions for all dilations, such initial data, when locally $L^2$, is not necessarily uniformly locally $L^2$, therefore their results are no consequence of constructions described by Lemarié-Rieusset in [10].

In this paper, we construct an alternative theory to obtain infinite-energy global weak solutions for large initial data, which include the discretely self-similar locally square integrable data. More specifically, we consider the weights

$$\begin{aligned} w_\gamma (x)=\frac{1}{(1+\vert x\vert )^\gamma } \end{aligned}$$

with $0<\gamma$, and the spaces

$$\begin{aligned} L^2_{w_\gamma }=L^2(w_\gamma \, \mathrm{d}x). \end{aligned}$$

Our main theorem is the following one:

Theorem 1

Let $0<\gamma \leqq 2$. If $\mathbf{u}_{0}$ is a divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and if ${\mathbb {F}}$ is a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,+\infty ), L^2_{w_\gamma })$, then the Navier–Stokes equations with initial value $\mathbf{u}_0$

$$\begin{aligned} (NS) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{u}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

have a global weak solution $\mathbf{u}$ such that:

• for every $0<T<+\infty$, $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$

• the pressure p is related to $\mathbf{u}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

  $$\begin{aligned} p =\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j-F_{i,j}) \end{aligned}$$

  where, for every $0<T<+\infty$, $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j)$ belongs to $L^{4}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j F_{i,j}$ belongs to $L^{2}((0,T),L^{2}_{w_\gamma })$

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

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0. \end{aligned}$$

• the solution $\mathbf{u}$ is suitable: there exists a non-negative locally finite measure $\mu$ on $(0,+\infty )\times {\mathbb {R}}^3$ such that

  $$\begin{aligned}&\partial _t\left( \frac{\vert \mathbf{u}\vert ^2}{2}\right) =\Delta \left( \frac{\vert \mathbf{u}\vert ^2}{2}\right) -\vert \mathbf{\nabla }\mathbf{u}\vert ^2\\&\quad - \mathbf{\nabla }\cdot \left( \Big (\frac{\vert \mathbf{u}\vert ^2}{2}+p\Big )\mathbf{u}\right) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}})-\mu . \end{aligned}$$

In particular, we have the energy controls

$$\begin{aligned} \begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 +2\int _0^t \Vert \mathbf{\nabla }\mathbf{u}(s,.)\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 -\int _0^t \int \mathbf{\nabla }\vert \mathbf{u}\vert ^2\cdot \mathbf{\nabla }w_\gamma \, \mathrm{d}x \, \mathrm{d}s +\int _0^t\int (\vert \mathbf{u}\vert ^2+2p) \mathbf{u}\cdot \mathbf{\nabla }(w_\gamma )\, \, \mathrm{d}x \, \mathrm{d}s \\&\qquad -2\sum _{i=1}^3\sum _{j=1}^3 \int _0^t\int F_{i,j} (\partial _i u_j) w_\gamma +F_{i,j} u_i \partial _j(w_\gamma ) \, \mathrm{d}x \, \mathrm{d}s \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2\leqq & {} \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm{d}s \\&+\, C_\gamma \int _0^t \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^2 + \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^6 \, \mathrm{d}s \end{aligned}$$

Remark

We use the following notations: the vector $\mathbf{u}$ is given by its coordinates $\mathbf{u}=(\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3)$. The operator $\mathbf{u}\cdot \nabla$ is the differential operator $\mathbf{u}_1\partial _1+\mathbf{u}_2\partial _2+\mathbf{u}_3\partial _3$. Thus, $\nabla \cdot (f\mathbf{u})=f \nabla \cdot \mathbf{u}+ \mathbf{u}\cdot \nabla f$.

For ${\mathbb {F}}=(F_{i,j})$, we write $\nabla \cdot {\mathbb {F}}$ for the vector $\displaystyle (\sum _{i=1}^3 \partial _i F_{i,1}, \sum _{i=1}^3 \partial _i F_{i,2}, \sum _{i=1}^3 \partial _i F_{i,3} )$.

For the vector fields $\mathbf{b}$ and $\mathbf{u}$, we define $\mathbf{b}\otimes \mathbf{u}$ as $(b_iu_j)_{1\leqq i\leqq 3, 1\leqq j\leqq 3}$. Thus, if $\mathbf{b}$ is divergence free (that is if $\mathbf{\nabla }\cdot \mathbf{b}=0$) we have $\mathbf{\nabla }\cdot (\mathbf{b}\otimes \mathbf{u})=(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}$.

A key tool for proving Theorem 1 and for applying it to the study of discretely self-similar solutions is given by the following a priori estimates for an advection-diffusion problem:

Theorem 2

Let $0<\gamma \leqq 2$. Let $0<T<+\infty$. Let $\mathbf{u}_{0}$ be a divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}$ be a time-dependent divergence free vector-field ($\mathbf{\nabla }\cdot \mathbf{b}=0$) such that $\mathbf{b}\in L^3((0,T),L^3_{w_{3\gamma /2}})$.

Let $\mathbf{u}$ be a solution of the following advection-diffusion problem:

$$\begin{aligned} (AD) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

such that

• $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

• the pressure p is related to $\mathbf{u}$, $\mathbf{b}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

  $$\begin{aligned} p=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j-F_{i,j}) \end{aligned}$$

  where $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j)$ belongs to $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j F_{i,j}$ belongs to $L^{2}((0,T),L^{2}_{w_\gamma })$;

• the map $t\in [0,T)\mapsto \mathbf{u}(t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$ :

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0; \end{aligned}$$

• there exists a non-negative locally finite measure $\mu$ on $(0,T)\times {\mathbb {R}}^3$ such that

  $$\begin{aligned} \partial _t\left( \frac{\vert \mathbf{u}\vert ^2}{2}\right) =\Delta \left( \frac{\vert \mathbf{u}\vert ^2}{2}\right) -\vert \mathbf{\nabla }\mathbf{u}\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}\vert ^2}{2}\mathbf{b}\right) -\mathbf{\nabla }\cdot (p\mathbf{u}) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}})-\mu . \end{aligned}$$

  (1)

Then, we have the energy controls

$$\begin{aligned}\begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 +2\int _0^t \Vert \mathbf{\nabla }\mathbf{u}(s,.)\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 -\int _0^t \int \mathbf{\nabla }\vert \mathbf{u}\vert ^2\cdot \mathbf{\nabla }w_\gamma \, \mathrm{d}x \, \mathrm{d}s +\int _0^t\int \vert \mathbf{u}\vert ^2 \mathbf{b}\cdot \mathbf{\nabla }(w_\gamma ) \, \mathrm{d}x \, \mathrm{d}s \\&\qquad + 2\int _0^t\int p\mathbf{u}\cdot \mathbf{\nabla }(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s -2\sum _{i=1}^3\sum _{j=1}^3 \int _0^t\int F_{i,j} (\partial _i u_j) w_\gamma \\&\qquad +F_{i,j} u_i \partial _j(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 + \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm{d}s \\&\qquad + C_\gamma \int _0^t (1+ \Vert \mathbf{b}(s,. )\Vert _{L^3_{w_{3\gamma /2}}}^2 ) \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^2 \, \mathrm{d}s, \end{aligned} \end{aligned}$$

where $C_\gamma$ depends only on $\gamma$ (and not on T, and not on $\mathbf{b}$, $\mathbf{u}$, $\mathbf{u}_0$ nor ${\mathbb {F}}$).

In particular, we shall prove the following stability result:

Theorem 3

Let $0<\gamma \leqq 2$. Let $0<T<+\infty$. Let $\mathbf{u}_{0,n}$ be divergence-free vector fields such that $\mathbf{u}_{0,n}\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}_n$ be tensors such that ${\mathbb {F}}_n\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}_n$ be time-dependent divergence free vector-fields such that $\mathbf{b}_n\in L^3((0,T),L^3_{w_{3\gamma /2}})$.

Let $\mathbf{u}_n$ be solutions of the advection-diffusion problems

$$\begin{aligned} (AD_n) \left\{ \begin{array}{l} \partial _t \mathbf{u}_n= \Delta \mathbf{u}_n -(\mathbf{b}_n\cdot \mathbf{\nabla })\mathbf{u}_n- \mathbf{\nabla }p_n +\mathbf{\nabla }\cdot {\mathbb {F}}_n \\ \\ \mathbf{\nabla }\cdot \mathbf{u}_n=0, \quad \quad \mathbf{u}_n(0,.)=\mathbf{u}_{0,n} \end{array}\right. \end{aligned}$$

such that

• $\mathbf{u}_n$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_n$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

• the pressure $p_n$ is related to $\mathbf{u}_n$, $\mathbf{b}_n$ and ${\mathbb {F}}_n$ by the formula

  $$\begin{aligned} p_n=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_{n,i}u_{n,j}-F_{n,i,j}); \end{aligned}$$

• the map $t\in [0,T)\mapsto \mathbf{u}_n(t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$:

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}_n(t,.)-\mathbf{u}_{0,n}\Vert _{L^2_{w_\gamma }}=0. \end{aligned}$$

• there exists a non-negative locally finite measure $\mu _n$ on $(0,T)\times {\mathbb {R}}^3$ such that

  $$\begin{aligned}&\partial _t\left( \frac{\vert \mathbf{u}_n\vert ^2}{2}\right) =\Delta \left( \frac{\vert \mathbf{u}_n\vert ^2}{2}\right) -\vert \mathbf{\nabla }\mathbf{u}_n\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}_n\vert ^2}{2}\mathbf{b}_n\right) \\&\quad -\mathbf{\nabla }\cdot (p_n\mathbf{u}_n) + \mathbf{u}_n\cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_n)-\mu _n; \end{aligned}$$

If $\mathbf{u}_{0,n}$ is strongly convergent to $\mathbf{u}_{0,\infty }$ in $L^2_{w_\gamma }$, if the sequence ${\mathbb {F}}_n$ is strongly convergent to ${\mathbb {F}}_\infty$ in $L^2((0,T), L^2_{w_\gamma })$, and if the sequence $\mathbf{b}_n$ is bounded in $L^3((0,T), L^3_{w_{3\gamma /2}})$, then there exists $p_\infty$, $\mathbf{u}_\infty$, $\mathbf{b}_\infty$ and an increasing sequence $(n_k)_{k\in {\mathbb {N}}}$ with values in ${\mathbb {N}}$ such that

• $\mathbf{u}_{n_k}$ converges *-weakly to $\mathbf{u}_\infty$ in $L^\infty ((0,T), L^2_{w_\gamma })$, $\mathbf{\nabla }\mathbf{u}_{n_k}$ converges weakly to $\mathbf{\nabla }\mathbf{u}_\infty$ in $L^2((0,T),L^2_{w_\gamma })$;

• $\mathbf{b}_{n_k}$ converges weakly to $\mathbf{b}_\infty$ in $L^3((0,T), L^3_{w_{3\gamma /2}})$, $p_{n_k}$ converges weakly to $p_\infty$ in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})+L^{2}((0,T),L^{2}_{w_\gamma })$;

• $\mathbf{u}_{n_k}$ converges strongly to $\mathbf{u}_\infty$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$ such that for every $T_0\in (0,T)$ and every $R>0$, we have

  $$\begin{aligned} \lim _{k\rightarrow +\infty } \int _0^{T_0} \int _{\vert y\vert <R} \vert \mathbf{u}_{n_k}(s,y)-\mathbf{u}_\infty (s,y)\vert ^2\, \mathrm{d}s\, \mathrm{d}y=0. \end{aligned}$$

Moreover, $\mathbf{u}_\infty$ is a solution of the advection-diffusion problem

$$\begin{aligned} (AD_\infty ) \left\{ \begin{array}{l} \partial _t \mathbf{u}_\infty = \Delta \mathbf{u}_\infty -(\mathbf{b}_\infty \cdot \mathbf{\nabla })\mathbf{u}_\infty - \mathbf{\nabla }p_\infty +\mathbf{\nabla }\cdot {\mathbb {F}}_\infty \\ \\ \mathbf{\nabla }\cdot \mathbf{u}_\infty =0, \quad \quad \mathbf{u}_\infty (0,.)=\mathbf{u}_{0,\infty } \end{array}\right. \end{aligned}$$

and is such that

• the map $t\in [0,T)\mapsto \mathbf{u}_\infty (t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$ :

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}_\infty (t,.)-\mathbf{u}_{0,\infty }\Vert _{L^2_{w_\gamma }}=0; \end{aligned}$$

• there exists a non-negative locally finite measure $\mu _\infty$ on $(0,T)\times {\mathbb {R}}^3$ such that

  $$\begin{aligned}&\partial _t\left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\right) =\Delta \left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\right) -\vert \mathbf{\nabla }\mathbf{u}_\infty \vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\mathbf{b}_\infty \right) \\&\quad -\mathbf{\nabla }\cdot (p_\infty \mathbf{u}_\infty ) + \mathbf{u}_\infty \cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_\infty )-\mu _\infty . \end{aligned}$$

2 Notations

Throughout the text, $C_\gamma$ is a positive constant whose value may change from line to line but which depends only on $\gamma$.

3 The Weights $\varvec{w}_{\varvec{\delta }}$

We consider the weights $w_\delta =\frac{1}{(1+\vert x\vert )^\delta }$ where $0<\delta$ and $x\in {\mathbb {R}}^3$. A very important feature of those weights is the control of their gradients:

$$\begin{aligned} \vert \mathbf{\nabla }w_\delta (x)\vert =\delta \frac{w_\delta (x)}{1+\vert x\vert } \end{aligned}$$

(2)

From this control, we can infer the following Sobolev embedding:

Lemma 1

(Sobolev embeddings) Let $\delta >0$. If $f\in L^2_{w_\delta }$ and $\mathbf{\nabla }f\in L^2_{w_\delta }$ then $f\in L^6_{w_{3\delta }}$ and

$$\begin{aligned} \Vert f\Vert _{L^6_{w_{3\delta }}}\leqq C_\delta (\Vert f\Vert _{L^2_{w_\delta }}+ \Vert \mathbf{\nabla }f\Vert _{L^2_{w_\delta }}). \end{aligned}$$

Proof

Since both f and $w_{\delta /2}$ are locally in $H^1$, we write

$$\begin{aligned} \partial _i(f w_{\delta /2})=w_{\delta /2} \partial _i f+ f \partial _i(w_{\delta /2})= w_{\delta /2} \partial _i f -\frac{\delta }{2} \frac{x_i}{\vert x\vert } \frac{1}{1+\vert x\vert }w_{\delta /2} f, \end{aligned}$$

and thus

$$\begin{aligned} \Vert w_{\delta /2}f\Vert _2^2+ \Vert \mathbf{\nabla }(w_{\delta /2}f)\Vert _2^2\leqq \left( 1+\frac{\delta ^2}{2}\right) \Vert w_{\delta /2}f\Vert _2^2 + 2 \Vert w_{\delta /2}\mathbf{\nabla }f\Vert _2^2. \end{aligned}$$

Thus, $w_{\delta /2} f$ belongs to $L^6$ (since $H^1\subset L^6$), or equivalently $f\in L^6_{w_{3\delta }}$. $\quad \square$

We shall mainly be interested in the case $\delta \leqq 2$. An important property for $0<\delta <3$ is

Lemma 2

(Muckenhoupt weights) If $0<\delta <3$ and $1<p<+\infty$, then $w_\delta$ belongs to the Muckenhoupt class ${\mathcal {A}}_p$.

Proof

We recall that a weight w belongs to ${\mathcal {A}}_p({\mathbb {R}}^3)$ for $1<p<+\infty$ if and only if it satisfies the reverse Hölder inequality

$$\begin{aligned} \!\!\! \sup _{x\in {\mathbb {R}}^3, R>0} \!\! \left( \frac{1}{\vert B(x,R)\vert } \!\int _{B(x,R)}\! \!\!\!\!w(y)\, \mathrm{d}y\right) ^{\frac{1}{p}}\!\! \left( \frac{1}{\vert B(x,R)\vert }\! \int _{B(x,R)} \!\frac{\mathrm{d}y}{w(y)^{\frac{1}{p-1}}}\right) ^{1-\frac{1}{p}} \!\! \!\!<+\infty .\nonumber \\ \end{aligned}$$

(3)

For all $0<R\leqq 1$ the inequality $\vert x-y\vert <R$ implies $\frac{1}{2} (1+\vert x\vert ) \leqq 1+\vert y\vert \leqq 2 (1+\vert x\vert )$, thus we can control the left side in (3) for $w_\delta$ by $4^{\frac{\delta }{p}}$.

For all $R > 1$ and $\vert x\vert >10 R$, we have that the inequality $\vert x-y\vert <R$ implies $\frac{9}{10} (1+\vert x\vert ) \leqq 1+\vert y\vert \leqq \frac{11}{10} (1+\vert x\vert )$, thus we can control the left side in (3) for $w_\delta$ by $( \frac{11}{9})^{\frac{\delta }{p}}$.

Finally, for $R>1$ and $\vert x\vert \leqq 10 R$, we write

$$\begin{aligned} \begin{aligned}&\left( \frac{1}{\vert B(x,R)\vert } \!\int _{B(x,R)}\! \!\!\!\! w(y)\, \mathrm{d}y\right) ^{\frac{1}{p}} \left( \frac{1}{\vert B(x,R)\vert }\! \int _{B(0,R)} \!\frac{\mathrm{d}y}{w(y)^{\frac{1}{p-1}}}\right) ^{1-\frac{1}{p}} \\&\quad \leqq \left( \frac{1}{\vert B(0, R)\vert } \!\int _{B(x,11\, R)}\! \!\!\!\!w(y)\, \mathrm{d}y\right) ^{\frac{1}{p}}\!\! \left( \frac{1}{\vert B(0,R)\vert }\! \int _{B(0,11 \,R)} \!\frac{\mathrm{d}y}{w(y)^{\frac{1}{p-1}}}\right) ^{1-\frac{1}{p}} \\&\quad = \left( \frac{1}{R^3} \int _0^{11\, R} r^2 \frac{dr}{(1+r)^\delta }\right) ^{\frac{1}{p}} \left( \frac{1}{R^3} \int _0^{11\, R} r^2(1+r)^{\frac{\delta }{p-1}}\, dr\right) ^{1-\frac{1}{p}} \\&\quad \leqq c_{\delta , p} \left( \frac{1}{R^3} \int _0^{11\, R} r^2 \frac{dr}{ r^\delta }\right) ^{\frac{1}{p}} \left( \left( \frac{1}{R^3} \int _0^{11\, R} r^2 dr \right) ^{1-\frac{1}{p}} \right. \\&\qquad \left. + \left( \frac{1}{R^3} \int _0^{11\, R} r^{2+\frac{\delta }{p-1}}\, dr\right) ^{1-\frac{1}{p}} \right) \\&\quad = c_{\delta , p} \frac{11^3}{ (3-\delta )^{\frac{1}{p}} } \left( \frac{ (11R)^{ -\frac{\delta }{p} } }{ 3^{1-\frac{1}{p}}} + \frac{ 1 }{ (3+\frac{\delta }{p-1})^{1-\frac{1}{p}}} \right) . \end{aligned} \end{aligned}$$

The lemma is proved. $\quad \square$

Lemma 3

If $0<\delta <3$ and $1<p<+\infty$, then the Riesz transforms $R_i$ and the Hardy–Littlewood maximal function operator are bounded on $L^p_{w_\delta }=L^p(w_\delta (x)\, \mathrm{d}x)$:

$$\begin{aligned} \Vert R_jf\Vert _{L^p_{w_\delta }}\leqq C_{p,\delta } \Vert f\Vert _{L^p_{w_\delta }} \text { and } \Vert {\mathcal {M}}_f\Vert _{L^p_{w_\delta }}\leqq C_{p,\delta } \Vert f\Vert _{L^p_{w_\delta }}. \end{aligned}$$

Proof

The boundedness of the Riesz transforms or of the Hardy–Littlewwod maximal function on $L^p(w_\gamma \, \mathrm{d}x)$ are basic properties of the Muckenhoupt class ${\mathcal {A}}_p$ [5].

$\square$

We will use strategically the next corollary, which is specially useful to obtain discretely self-similar solutions.

Corollary 1

(Non-increasing kernels) Let $\theta \in L^1({\mathbb {R}}^3)$ be a non-negative radial function which is radially non-increasing. Then, if $0<\delta <3$ and $1<p<+\infty$, we have, for $f\in L^p_{w_\delta }$, the inequality

$$\begin{aligned} \Vert \theta *f\Vert _{L^p_{w_\delta }} \leqq C_{p,\delta } \Vert f\Vert _{L^p_{w_\delta }} \Vert \theta \Vert _1. \end{aligned}$$

Proof

We have the well-known inequality for radial non-increasing kernels [4]

$$\begin{aligned} \vert \theta *f(x)\vert \leqq \Vert \theta \Vert _1 {\mathcal {M}}_f(x) \end{aligned}$$

so that we may conclude with Lemma 3. $\quad \square$

We illustrate the utility of Lemma 3 with the following corollaries:

Corollary 2

Let $0<\gamma < \frac{5}{2}$ and $0<T<+\infty$. Let ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}$ be a time-dependent divergence free vector-field ($\mathbf{\nabla }\cdot \mathbf{b}=0$) such that $\mathbf{b}\in L^3((0,T),L^3_{w_{3\gamma /2}})$.

Let $\mathbf{u}$ be a solution of the following advection-diffusion problem:

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }q +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \end{array}\right. \end{aligned}$$

(4)

such that $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$, and the pressure q belongs to ${\mathcal {D}}'( (0,T) \times {\mathbb {R}} ^3 )$.

Then, the gradient of the pressure $\mathbf{\nabla }q$ is necessarily related to $\mathbf{u}$, $\mathbf{b}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

$$\begin{aligned} \mathbf{\nabla }q= \mathbf{\nabla }\left( \sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j-F_{i,j}) \right) \end{aligned}$$

and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j)$ belongs to $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j F_{i,j}$ belongs to $L^{2}((0,T),L^{2}_{w_\gamma })$.

Proof

We define

$$\begin{aligned} p = \left( \sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j-F_{i,j}) \right) . \end{aligned}$$

As $0< \gamma < \frac{5}{2}$ we can use Lemma 3 to obtain $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_i u_j)$ belongs to $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j F_{i,j}$ belongs to $L^{2}((0,T),L^{2}_{w_\gamma })$.

Taking the divergence in (4), we obtain $\Delta (q-p)=0$. We take a test function $\alpha \in {\mathcal {D}}({\mathbb {R}})$ such that $\alpha (t)= 0$ for all $|t| \geqq \varepsilon$, and a test function $\beta \in {\mathcal {D}}({\mathbb {R}}^3)$; then the distribution $\mathbf{\nabla }q *( \alpha \otimes \beta )$ is well defined on $(\varepsilon , T-\varepsilon ) \times {\mathbb {R}}^3$.

We fix $t \in (\varepsilon , T-\varepsilon )$ and define

$$\begin{aligned} A_{\alpha ,\beta ,t}=( \mathbf{\nabla }q * (\alpha \otimes \beta )- \mathbf{\nabla }p * (\alpha \otimes \beta )) (t,.). \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} A_{\alpha ,\beta ,t}=&( \mathbf{u}*(-\partial _t\alpha \otimes \beta +\alpha \otimes \Delta \beta ) +(-\mathbf{u}\otimes \mathbf{b}+{\mathbb {F}}) \cdot (\alpha \otimes \mathbf{\nabla }\beta ))(t,.)\\&- ( p * (\alpha \otimes \mathbf{\nabla }\beta )) (t,.) . \end{aligned} \end{aligned}$$

(5)

Convolution with a function in ${\mathcal {D}}({\mathbb {R}}^3)$ is a bounded operator on $L^2_{w_\gamma }$ and on $L^{6/5}_{w_{6\gamma /5}}$ (as, for $\varphi \in {\mathcal {D}}({\mathbb {R}}^3)$ we have $\vert f*\varphi \vert \leqq C_\varphi {\mathcal {M}}_f$). Thus, we may conclude from (5) that $A_{\alpha ,\beta ,t}\in L^2_{w_\gamma }+ L^{6/5}_{w_{6\gamma /5}}$. If $\max \{ \gamma , \frac{\gamma + 2 }{2} \}<\delta <5/2$ , we have $A_{\alpha ,\beta ,t}\in L^{6/5}_{w_{6\delta /5}}$.

In particular, $A_{\alpha ,\beta ,t}$ is a tempered distribution. As we have

$$\begin{aligned} \Delta A_{\alpha ,\beta ,t}=(\alpha \otimes \beta )*(\mathbf{\nabla }\Delta (q-p))(t,.)=0, \end{aligned}$$

we find that $A_{\alpha ,\beta ,t}$ is a polynomial. We remark that for all $1<r<+\infty$ and $0< \delta < 3$, $L^r_{w_\delta }$ does not contain non-trivial polynomials. Thus, $A_{\alpha ,\beta ,t}= 0$. We then use an approximation of identity $\frac{1}{\varepsilon ^4} \alpha (\frac{t}{\varepsilon })\beta (\frac{x}{\varepsilon })$ and conclude that $\mathbf{\nabla }(q-p)=0$.

$\square$

Actually, we can answer a question posed by Bradshaw and Tsai in [2] about the nature of the pressure for self-similar solutions of the Navier–Stokes equations. In effect, we have the next corollary.

Corollary 3

Let $1<\gamma < \frac{5}{2}$ and $0<T<+\infty$. Let ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$.

Let $\mathbf{u}$ be a solution of the following problem:

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{u}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }q +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \end{array}\right. \end{aligned}$$

such that $\mathbf{u}$ belongs to $L^\infty ([0,+\infty ), L^2)_{loc}$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2([0,+\infty ),L^2)_{loc}$, and the pressure q is in ${\mathcal {D}}'( (0,T) \times {\mathbb {R}} ^3 )$.

We suppose that there exists $\lambda >1$ such that $\lambda ^2 {\mathbb {F}}(\lambda ^2 t,\lambda x)={\mathbb {F}}(t,x)$ and $\lambda \mathbf{u}(\lambda ^2 t,\lambda x)=\mathbf{u}(t,x)$. Then, the gradient of the pressure $\mathbf{\nabla }q$ is necessarily related to $\mathbf{u}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

$$\begin{aligned} \mathbf{\nabla }q= \mathbf{\nabla }\left( \sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j-F_{i,j}) \right) \end{aligned}$$

and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j)$ belongs to $L^{4}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j F_{i,j}$ belongs to $L^{2}((0,T),L^{2}_{w_\gamma })$.

Proof

We shall use Corollary 2, and thus we need to show that $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma } \cap L^3((0,T), L^3_{3\gamma /2}))$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$. In fact,

$$\begin{aligned} \Vert u \Vert _{L^\infty ((0,T),L^2_{w_\gamma })}\leqq & {} \sup _{0 \leqq t \leqq T} \int _{|x|<1} |\mathbf{u}(t,x)|^2 \, \mathrm{d}x \\&\quad + c \sup _{0 \leqq t \leqq T} \sum _{k\in {\mathbb {N}}} \int _{\lambda ^{k-1}<|x|<\lambda ^k} \frac{|\mathbf{u}(t,x)|^2}{\lambda ^{\gamma k } } \, \mathrm{d}x \end{aligned}$$

and

$$\begin{aligned}&\sup _{0 \leqq t \leqq T} \sum _{k\geqq 1} \int _{\lambda ^{k-1}<|x|<\lambda ^k} \frac{|\mathbf{u}(t,x)|^2}{\lambda ^{\gamma k } } \, \mathrm{d}x \\&\quad \leqq \sup _{0 \leqq t \leqq T} \sum _{k\in {\mathbb {N}}} \lambda ^{ (1-\gamma ) k } \int _{\lambda ^{-1}<|x|< 1} |\mathbf{u}(\frac{t}{\lambda ^{2k}},x)|^2 \, \mathrm{d}x \\&\quad \leqq c \sup _{0 \leqq t \leqq T} \int _{\lambda ^{-1}<|x|< 1} |\mathbf{u}(t,x)|^2 \, \mathrm{d}x < +\infty . \end{aligned}$$

For $\mathbf{\nabla }\mathbf{u}$, we compute for $k\in {\mathbb {N}}$,

$$\begin{aligned} \int _0^T \int _{\lambda ^{k-1}<\vert x\vert<\lambda ^{k}} \vert \mathbf{\nabla }\mathbf{u}(t,x)\vert ^2\, \mathrm{d}t\, \mathrm{d}x=\lambda ^{k}\int _0^{\frac{T}{\lambda ^{2k}}}\int _{\frac{1}{\lambda }<\vert x\vert <1} \vert \mathbf{\nabla }\mathbf{u}(t,x)\vert ^2\, \mathrm{d}x\, \mathrm{d}t. \end{aligned}$$

We may conclude that $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$, since for $\gamma >1$ we have $\sum _{k\in {\mathbb {N}}} \lambda ^{ (1-\gamma ) k }<+\infty$.

Now, we use the Sobolev embedding described in Lemma 1 to get that $\mathbf{u}$ belongs to $L^2((0,T),L^6_{w_{3\gamma }})$, and thus (by interpolation with $L^\infty ((0,T),L^2_{w_\gamma }))$ to $L^4((0,T),L^3_{w_{3\gamma /2}})$.

In particular, $\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j)$ belongs to $L^{4}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$, since we have

$$\begin{aligned} \Vert ( \mathbf{u}\otimes \mathbf{u}) w_\gamma \Vert _{ L^{6/5}} \leqq \Vert \sqrt{w_\gamma } \mathbf{u}\Vert _{ L^{2} } \Vert \sqrt{w_\gamma } \mathbf{u}\Vert _{ L^{3} } \leqq \Vert \sqrt{w_\gamma } \mathbf{u}\Vert ^{\frac{3}{2}}_{ L^{2} } \Vert \sqrt{w_\gamma } \mathbf{u}\Vert ^{\frac{1}{2}}_{ L^{6} .} \end{aligned}$$

$\square$

4 A Priori Estimates for the Advection-Diffusion Problem

4.1 Proof of Theorem 2

Let $0<t_0<t_1<T$. We take a function $\alpha \in {\mathcal {C}}^\infty ({\mathbb {R}})$ which is non-decreasing, with $\alpha (t)$ equal to 0 for $t<1/2$ and equal to 1 for $t>1$. For $0<\eta < \min (\frac{t_0}{2},T-t_1)$, we define

$$\begin{aligned} \alpha _{\eta ,t_0,t_1}(t)=\alpha \Big ( \frac{t-t_0}{\eta }\Big )-\alpha \Big (\frac{t-t_1}{\eta }\Big ) . \end{aligned}$$

We take as well a non-negative function $\phi \in {\mathcal {D}}({\mathbb {R}}^3)$ which is equal to 1 for $\vert x\vert \leqq 1$ and to 0 for $\vert x\vert \geqq 2$. For $R>0$, we define $\phi _R(x)=\phi (\frac{x}{R})$. Finally, we define, for $\varepsilon >0$, $w_{\gamma ,\varepsilon }= \left( 1+\sqrt{\varepsilon ^2+\vert x\vert ^2} \right) ^{-\gamma }$. We have $\alpha _{\eta ,t_0,t_1}(t)\phi _R(x) w_{\gamma ,\varepsilon }(x)\in {\mathcal {D}}((0,T)\times {\mathbb {R}}^3)$ and $\alpha _{\eta ,t_0,t_1}(t)\phi _R(x) w_{\gamma ,\varepsilon }(x) \geqq 0$. Thus, using the local energy balance (1) and the fact that $\mu \geqq 0$, we find

$$\begin{aligned}&-\iint \frac{\vert \mathbf{u}\vert ^2}{2} \partial _t\alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq -\sum _{i=1}^3 \iint \partial _i\mathbf{u}\cdot \mathbf{u}\, \alpha _{\eta ,t_0,t_1} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \iint \vert \mathbf{\nabla }\mathbf{u}\vert ^2\, \, \alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon } \mathrm{d}x\, \mathrm{d}s \\&\qquad + \sum _{i=1}^3 \iint \frac{\vert \mathbf{u}\vert ^2}{2} b_i \alpha _{\eta ,t_0,t_1} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad +\sum _{i=1}^3 \iint \alpha _{\eta ,t_0,t_1} pu_i (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3\sum _{j=1}^3 \iint F_{i,j} u_j \alpha _{\eta ,t_0,t_1} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3 \sum _{j=1}^3\iint F_{i,j}\partial _iu_j\ \alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s. \end{aligned}$$

We remark that, independently of $R>1$ and $\varepsilon >0$, we have (for $0<\gamma \leqq 2$)

$$\begin{aligned} \vert w_{\gamma ,\varepsilon }\partial _i \phi _R\vert +\vert \phi _R \partial _iw_{\gamma ,\varepsilon }\vert \leqq C_\gamma \frac{w_\gamma (x)}{1+\vert x\vert }\leqq C_\gamma w_{3\gamma /2}(x). \end{aligned}$$

Moreover, we know that $\mathbf{u}$ belongs to $L^\infty ((0,T),L^2_{w_\gamma })\cap L^2((0,T), L^6_{w_{3\gamma }})$ hence to $L^4((0,T),L^3_{w_{3\gamma /2}})$. Since $T<+\infty$, we have as well $\mathbf{u}\in L^3((0,T), L^3_{w_{3\gamma /2}})$. (This is the same type of integrability as required for $\mathbf{b}$). Moreover, we have $p u_i\in L^1_{w_{3\gamma /2}}$ since $w_\gamma p\in L^2 ((0,T), L^{6/5}+L^2)$ and $w_{\gamma /2} \mathbf{u}\in L^2((0,T), L^2\cap L^6)$. All those remarks will allow us to use dominated convergence.

We first let $\eta$ go to 0. We find that

$$\begin{aligned}&- \lim _{\eta \rightarrow 0} \iint \frac{\vert \mathbf{u}\vert ^2}{2} \partial _t\alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq -\sum _{i=1}^3 \int _{t_0}^{t_1}\int \partial _i\mathbf{u}\cdot \mathbf{u}\, (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \int _{t_0}^{t_1} \int \vert \mathbf{\nabla }\mathbf{u}\vert ^2\, \, \phi _R w_{\gamma ,\varepsilon } \mathrm{d}x\, \mathrm{d}s \\&\qquad + \sum _{i=1}^3 \int _{t_0}^{t_1}\int \frac{\vert \mathbf{u}\vert ^2}{2} b_i (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad +\sum _{i=1}^3 \int _{t_0}^{t_1} \int pu_i (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3\sum _{j=1}^3 \int _{t_0}^{t_1}\int F_{i,j} u_j (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3 \sum _{j=1}^3\int _{t_0}^{t_1} \int F_{i,j}\partial _iu_j\ \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s. \end{aligned}$$

Let us define

$$\begin{aligned} A_{R,\varepsilon }(t)=\int \vert \mathbf{u}(t,x)\vert ^2 \phi _R(x) w_{\gamma ,\varepsilon }(x)\, \mathrm{d}x. \end{aligned}$$

As we have

$$\begin{aligned} -\iint \frac{\vert \mathbf{u}\vert ^2}{2} \partial _t\alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon } \, \mathrm{d}x\, \mathrm{d}s=-\frac{1}{2}\int \partial _t\alpha _{\eta ,t_0,t_1} A_{R,\varepsilon }(s) \, \mathrm{d}s \end{aligned}$$

we find that, when $t_0$ and $t_1$ are Lebesgue points of the measurable function $A_{R,\varepsilon }$

$$\begin{aligned} \lim _{\eta \rightarrow 0} -\iint \frac{\vert \mathbf{u}\vert ^2}{2} \partial _t\alpha _{\eta ,t_0,t_1} \phi _R w_{\gamma ,\varepsilon } \, \mathrm{d}x\, \mathrm{d}s=\frac{1}{2} ( A_{R,\varepsilon }(t_1)- A_{R,\varepsilon }(t_0)) . \end{aligned}$$

Then, by continuity, we can let $t_0$ go to 0 and thus replace $t_0$ by 0 in the inequality. Moreover, if we let $t_1$ go to t, then by weak continuity, we find that $A_{R,\varepsilon }(t)\leqq \lim _{t_1\rightarrow t } A_{R,\varepsilon }(t_1)$, so that we may as well replace $t_1$ by $t\in (0,T)$. Thus we find that for every $t\in (0,T)$, we have

$$\begin{aligned}&\int \frac{\vert \mathbf{u}(t,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \nonumber \\&\quad \leqq \int \frac{\vert \mathbf{u}_0(x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \nonumber \\&\qquad -\sum _{i=1}^3 \int _0^t \int \partial _i\mathbf{u}\cdot \mathbf{u}\, (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \nonumber \\&\qquad - \int _0^t \int \vert \mathbf{\nabla }\mathbf{u}\vert ^2\, \, \phi _R w_{\gamma ,\varepsilon } \mathrm{d}x\, \mathrm{d}s \nonumber \\&\qquad + \sum _{i=1}^3 \int _0^t \int \frac{\vert \mathbf{u}\vert ^2}{2} b_i (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \nonumber \\&\qquad +\sum _{i=1}^3 \int _0^t \int pu_i (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \nonumber \\&\qquad - \sum _{i=1}^3\sum _{j=1}^3 \int _0^t \int F_{i,j} u_j (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \nonumber \\&\qquad - \sum _{i=1}^3 \sum _{j=1}^3\int _0^t \int F_{i,j}\partial _iu_j\ \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s. \end{aligned}$$

(6)

Thus, letting R go to $+\infty$ and then $\varepsilon$ go to 0, we find by dominated convergence that, for every $t\in (0,T)$, we have

$$\begin{aligned}\begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 +2\int _0^t \Vert \mathbf{\nabla }\mathbf{u}(s,.)\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 -\int _0^t \int \mathbf{\nabla }\vert \mathbf{u}\vert ^2\cdot \mathbf{\nabla }w_\gamma \, \mathrm{d}x\, \mathrm{d}s\\&\qquad +\int _0^t\int (\vert \mathbf{u}\vert ^2 \mathbf{b}+2p\mathbf{u}) \cdot \mathbf{\nabla }(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s\\&\qquad -2\sum _{i=1}^3\sum _{j=1}^3 \int _0^t\int F_{i,j} (\partial _i u_j) w_\gamma +F_{i,j} u_i \partial _j(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s. \end{aligned} \end{aligned}$$

Now we write

$$\begin{aligned} \left| \int _0^t\int \mathbf{\nabla }\vert \mathbf{u}\vert ^2\cdot \mathbf{\nabla }w_\gamma \, \mathrm{d}s\, \mathrm{d}s\right|\leqq & {} 2\gamma \int _0^t\int \vert \mathbf{u}\vert \vert \mathbf{\nabla }\mathbf{u}\vert \, w_\gamma \, \mathrm{d}x\, \mathrm{d}s \\\leqq & {} \frac{1}{4} \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s+4\gamma ^2 \int _0^t \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s . \end{aligned}$$

Writing

$$\begin{aligned} p_1=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j) \text { and } p_2=-\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(F_{i,j}), \end{aligned}$$

and using the fact that $w_{6\gamma /5}\in {\mathcal {A}}_{6/5}$ and $w_\gamma \in {\mathcal {A}}_2$, we get

$$\begin{aligned}\begin{aligned}&\left| \int _0^t\int (\vert \mathbf{u}\vert ^2 \mathbf{b}+2p_1\mathbf{u}) \cdot \mathbf{\nabla }(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s\right| \leqq \gamma \int _0^t\int (\vert \mathbf{u}\vert ^2 \vert \mathbf{b}\vert +2\vert p_1\vert \,\vert \mathbf{u}\vert ) \, w_\gamma ^{3/2}\, \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq \gamma \int _0^t \Vert w_\gamma ^{1/2} \mathbf{u}\Vert _6 (\Vert w_\gamma \vert \mathbf{b}\vert \vert \mathbf{u}\vert \Vert _{6/5}+\Vert w_\gamma p_1\Vert _{6/5}) \mathrm{d}s \\&\quad \leqq C_\gamma \int _0^t \Vert w_\gamma ^{1/2} \mathbf{u}\Vert _6 \Vert w_\gamma \vert \mathbf{b}\vert \vert \mathbf{u}\vert \Vert _{6/5} \, \mathrm{d}s \\&\quad \leqq C_\gamma \int _0^t \Vert w_\gamma ^{1/2} \mathbf{u}\Vert _6 \Vert w_\gamma ^{1/2} \mathbf{b}\Vert _3 \Vert w_\gamma ^{1/2}\mathbf{u}\Vert _2 \, \mathrm{d}s \\&\quad \leqq C_\gamma ' \int _0^t (\Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}+\Vert \mathbf{u}\Vert _{L^2_{w_\gamma }})\ \Vert \mathbf{b}\Vert _{L^3_{w_{3\gamma /2}}}\Vert \mathbf{u}\Vert _{L^2_{w_\gamma }} \, \mathrm{d}s \\&\quad \leqq \frac{1}{4} \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s + C''_\gamma \int _0^t \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }}^2 ( \Vert \mathbf{b}\Vert _{L^3_{w_{3\gamma /2}}} + \Vert \mathbf{b}\Vert _{L^3_{w_{3\gamma /2}}}^2)\, \mathrm{d}s \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\left| \int _0^t\int 2p_2\mathbf{u}\cdot \mathbf{\nabla }(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s\right| \\&\quad \leqq 2 \gamma \int _0^t\int \vert p_2\vert \, \vert \mathbf{u}\vert \, w_\gamma \, \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq \gamma \int _0^t \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }}^2+\Vert p_2\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq C_\gamma \int _0^t \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }}^2+\Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s. \end{aligned} \end{aligned}$$

Finally, we have

$$\begin{aligned} \begin{aligned}&\left| 2\sum _{i=1}^3\sum _{j=1}^3 \int _0^t\int F_{i,j} (\partial _i u_j) w_\gamma +F_{i,j} u_i \partial _j(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s\right| \\&\quad \leqq 2 \int _0^t\int \vert F\vert \, (\vert \mathbf{\nabla }\mathbf{u}\vert +\gamma \vert \mathbf{u}\vert ) \, w_\gamma \, \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq \frac{1}{4} \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s + C_\gamma \int _0^t \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }}^2+\Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s. \end{aligned} \end{aligned}$$

We have obtained

$$\begin{aligned} \begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 + \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm{d}s \\&\qquad + C_\gamma \int _0^t \Bigg (1+ \Vert \mathbf{b}(s,. )\Vert _{L^3_{w_{3\gamma /2}}}^2 \Bigg ) \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^2 \, \mathrm{d}s \end{aligned} \end{aligned}$$

(7)

and Theorem 2 is proven. $\quad \square$

4.2 Passive Transportation

From inequality (7), we have the following direct consequence:

Corollary 4

Under the assumptions of Theorem 2, we have

$$\begin{aligned} \sup _{0<t<T} \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }} \leqq (\Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}+ C_\gamma \Vert {\mathbb {F}}\Vert _{L^2((0,T), L^2_{w_\gamma })}) \ e^{C_\gamma (T+ T^{1/3} \Vert \mathbf{b}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}})}^2)} \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2((0,T),L^2_{w_\gamma )}} \leqq (\Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}\!+\! C_\gamma \Vert {\mathbb {F}}\Vert _{L^2((0,T), L^2_{w_\gamma })}) \ e^{C_\gamma (T+ T^{1/3} \Vert \mathbf{b}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}})}^2)}, \end{aligned}$$

where the constant $C_\gamma$ depends only on $\gamma$.

Another direct consequence is the following uniqueness result for the advection-diffusion problem with a (locally in time), bounded $\mathbf{b}$:

Corollary 5

Let $0<\gamma < 2$. Let $0<T<+\infty$. Let $\mathbf{u}_{0}$ be a divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}$ be a time-dependent divergence free vector-field ($\mathbf{\nabla }\cdot \mathbf{b}=0$) such that $\mathbf{b}\in L^3((0,T),L^3_{w_{3\gamma /2}})$. Assume moreover that $\mathbf{b}$ belongs to $L^2_t L^\infty _x(K)$ for every compact subset K of $(0,T)\times {\mathbb {R}}^3$.

Let $(\mathbf{u}_1, p_1)$ and $(\mathbf{u}_2,p_2)$ be two solutions of the following advection-diffusion problem:

$$\begin{aligned} (AD) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

such that, for $k=1$ and $k=2$,

• $\mathbf{u}_k$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_k$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

• the pressure $p_k$ is related to $\mathbf{u}_k$, $\mathbf{b}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

  $$\begin{aligned} p_k=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_{k,j}-F_{i,j}); \end{aligned}$$

• the map $t\in [0,T)\mapsto \mathbf{u}_k(t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$ :

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}_k(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0. \end{aligned}$$

Then $\mathbf{u}_1=\mathbf{u}_2$.

Proof

Let $\mathbf{v}=\mathbf{u}_1-\mathbf{u}_2$ and $q=p_1-p_2$. Then we have

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{v}= \Delta \mathbf{v}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{v}- \mathbf{\nabla }q \\ \\ \mathbf{\nabla }\cdot \mathbf{v}=0, \quad \quad \mathbf{v}(0,.)=0. \end{array}\right. \end{aligned}$$

Moreover on every compact subset K of $(0,T)\times {\mathbb {R}}^3$, $\mathbf{b}\otimes \mathbf{v}$ is in $L^2_t L^2_x$, while it belongs globally to $L^{3}_t L^{6/5}_{w_{6\gamma /5}}$. Writing, for $\varphi , \psi \in {\mathcal {D}}((0,T)\times {\mathbb {R}}^3)$ such that $\psi =1$ on the neigborhood of the support of $\varphi$,

$$\begin{aligned} \varphi q=q_1+q_2=\varphi \sum _{i=1}^3\sum _{j=1}^3 R_iR_j(\psi b_iv_j)+\varphi \sum _{i=1}^3\sum _{j=1}^3 R_iR_j((1-\psi ) b_i v_j), \end{aligned}$$

we find that $\Vert q_1\Vert _{L^2L^2}\leqq C_{\varphi ,\psi } \Vert \psi \mathbf{b}\otimes \mathbf{v}\Vert _{L^2 L^2}$ and

$$\begin{aligned} \Vert q_2\Vert _{L^3 L^\infty } \leqq C_{\varphi ,\psi } \Vert \mathbf{b}\otimes \mathbf{v}\Vert _{L^3 L^{6/5}_{w_{6\gamma /5}}} \end{aligned}$$

with

$$\begin{aligned} C_{\varphi ,\psi }\leqq C \Vert \varphi \Vert _\infty \Vert 1-\psi \Vert _\infty \sup _{x\in \mathrm{Supp}\, \varphi } \left( \int _{y\in \mathrm{Supp }\, (1-\psi )} \left( \frac{ (1+\vert y\vert )^\gamma }{\vert x-y\vert ^3}\right) ^6 \right) ^{1/6}<+\infty . \end{aligned}$$

Thus, we may take the scalar product of $\partial _t \mathbf{v}$ with $\mathbf{v}$ and find that

$$\begin{aligned} \partial _t(\frac{\vert \mathbf{v}\vert ^2}{2})=\Delta \Big (\frac{\vert \mathbf{v}\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{v}\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{v}\vert ^2}{2}\mathbf{b}\right) -\mathbf{\nabla }\cdot (q\mathbf{v}) . \end{aligned}$$

Thus we are under the assumptions of Theorem 2 and we may use Corollary 4 to find that $\mathbf{v}=0$. $\quad \square$

4.3 Active Transportation

We begin with the following lemma:

Lemma 4

Let $\alpha$ be a non-negative bounded measurable function on [0, T) such that, for two constants $A,B\geqq 0$, we have

$$\begin{aligned} \alpha (t)\leqq A + B\int _0^t \alpha (s)+\alpha (s)^3\, \mathrm{d}s. \end{aligned}$$

If $T_0>0$ and $T_1=\min (T,T_0, \frac{1}{8B (A+2BT_0)^2})$, we have, for every $t\in [0,T_1]$, $\alpha (t)\leqq \sqrt{ 2} (A+2BT_0)$.

Proof

We write $\alpha \leqq 1+\alpha ^3$. We define

$$\begin{aligned} \Phi (t)= A+2B T_0 + 2B \int _0^t \alpha ^3\, \mathrm{d}s \text { and } \Psi (t)=A+2BT_0+ 2B \int _0^t \Phi ^3(s)\, \mathrm{d}s. \end{aligned}$$

We have, for $t\in [0,T_1]$, $\alpha \leqq \Phi \leqq \Psi$. Since $\Psi$ is ${\mathcal {C}}^1$, we may write

$$\begin{aligned} \Psi '(t)= 2B \Phi (t)^3 \leqq 2B \Psi (t)^3 \end{aligned}$$

and thus

$$\begin{aligned} \frac{1}{\Psi (0)^2}-\frac{1}{\Psi (t)^2}\leqq 4Bt. \end{aligned}$$

We thus find

$$\begin{aligned} \Psi (t)^2 \leqq \frac{\Psi (0)^2}{1-4B \Psi (0)^2 t}\leqq 2 \Psi (0)^2. \end{aligned}$$

The lemma is proven. $\quad \square$

Corollary 6

Assume that $\mathbf{u}_0$, $\mathbf{u}$, p, ${\mathbb {F}}$ and $\mathbf{b}$ satisfy assumptions of Theorem 2. Assume moreover that $\mathbf{b}$ is the inequality in the next line expresses in which way $\mathbf {b}$ is controlled by $\mathbf{u}$: for every $t\in (0,T)$,

$$\begin{aligned} \Vert \mathbf{b}(t,.)\Vert _{L^3_{w_{3\gamma /2}}}\leqq C_0 \Vert \mathbf{u}(t,.)\Vert _{L^3_{w_{3\gamma /2}}}. \end{aligned}$$

Then there exists a constant $C_\gamma \geqq 1$ such that if $T_0<T$ is such that

$$\begin{aligned} C_\gamma (1+C_0^4) \left( 1+C_0^4+\Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2+\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, T_0\leqq 1 \end{aligned}$$

then

$$\begin{aligned} \sup _{0\leqq t\leqq T_0} \Vert \ \mathbf{u}(t,.)\Vert _{L^2_{w_\gamma }}^2 \leqq C_\gamma \Bigg (1+ C_0^4 + \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \Bigg ) \end{aligned}$$

and

$$\begin{aligned} { \int _0^{T_0} \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s }\leqq C_\gamma \Bigg (1+ C_0^4 + \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\Bigg ). \end{aligned}$$

Proof

We start from inequality (7):

$$\begin{aligned} \begin{aligned}&\Vert \mathbf{u}(t,. )\Vert _{L^2_{w_\gamma }}^2 + \int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm{d}s \\&\qquad + C_\gamma \int _0^t \Bigg (1+ \Vert \mathbf{b}(s,. )\Vert _{L^3_{w_{3\gamma /2}}}^2 \Bigg ) \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^2 \, \mathrm{d}s \end{aligned} \end{aligned}$$

We write

$$\begin{aligned} \Vert \mathbf{b}(s,. )\Vert _{L^3_{w_{3\gamma /2}}}^2 \leqq C_0^2 \Vert \mathbf{u}(s,. )\Vert _{L^3_{w_{3\gamma /2}}}^2 \leqq C_0^2 C_\gamma \Vert \mathbf{u}\Vert _{L^2_{w_\gamma }} (\Vert u\Vert _{L^2_{w_\gamma }}+\Vert \mathbf{\nabla }\mathbf{u}\Vert _{L^2_{w_\gamma }}). \end{aligned}$$

This gives

$$\begin{aligned} \begin{aligned} \begin{aligned}&\Vert \mathbf {u}(t,. ) \Vert _{L^2_{w_\gamma }}^2 + \frac{1}{2}\int _0^t \Vert \mathbf {\nabla }\mathbf {u}\Vert ^2_{L^2_{w_\gamma }} \, \mathrm {d}s \\ {}&\quad \leqq \Vert \mathbf {u}_0 \Vert _{L^2_{w_\gamma }}^2 + C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm {d}s \\ {}&\qquad + C_\gamma \int _0^t \Vert \mathbf {u}(s,. )\Vert _{L^2_{w_\gamma }}^2 + C_0^2 \Vert \mathbf {u}(s,. )\Vert _{L^2_{w_\gamma }}^4 + C_0^4 \Vert \mathbf {u}(s,. )\Vert _{L^2_{w_\gamma }}^6 \, \mathrm {d}s \\ {}&\quad \leqq \Vert \mathbf {u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^t \Vert {\mathbb {F}}(s,.)\Vert ^2_{L^2_{w_\gamma }}\, \mathrm {d}s \\ {}&\qquad +2 C_\gamma \int _0^t \Vert \mathbf {u}(s,. )\Vert _{L^2_{w_\gamma }}^2 + C_0^4 \Vert \mathbf {u}(s,. )\Vert _{L^2_{w_\gamma }}^6 \, \mathrm {d}s. \end{aligned}\end{aligned}\end{aligned}$$

For $t\leqq T_0$, we get

$$\begin{aligned}&\Vert \mathbf{u}(t,. ) \Vert _{L^2_{w_\gamma }}^2 + \frac{1}{2}\int _0^t \Vert \mathbf{\nabla }\mathbf{u}\Vert ^2_{L^2_{w_\gamma }} \, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_0 \Vert _{L^2_{w_\gamma }}^2 +C_\gamma \int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\qquad + C_\gamma (1+C_0^4) \int _0^t \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^2+ \Vert \mathbf{u}(s,. )\Vert _{L^2_{w_\gamma }}^6\, \mathrm{d}s \end{aligned}$$

and we may conclude with Lemma 4. $\quad \square$

5 Stability of Solutions for the Advection-Diffusion Problem

5.1 The Rellich Lemma

We recall the Rellich lemma:

Lemma 5

(Rellich) If $s>0$ and $(f_n)$ is a sequence of functions on ${\mathbb {R}}^d$ such that

• the family $(f_n)$ is bounded in $H^s({\mathbb {R}}^d)$,

• there is a compact subset of ${\mathbb {R}}^d$ such that the support of each $f_n$ is included in K,

then there exists a subsequence $(f_{n_k})$ such that $f_{n_k}$ is strongly convergent in $L^2({\mathbb {R}}^d)$.

We shall use a variant of this lemma (see [9]):

Lemma 6

(space-time Rellich) If $s>0$, $\sigma \in {\mathbb {R}}$ and $(f_n)$ is a sequence of functions on $(0,T)\times {\mathbb {R}}^d$ such that, for all $T_0\in (0,T)$ and all $\varphi \in {\mathcal {D}}({\mathbb {R}}^3)$,

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

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

then there exists a subsequence $(f_{n_k})$ such that $f_{n_k}$ is strongly convergent in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$ : if $f_\infty$ is the limit, we have for all $T_0\in (0,T)$ and all $R_0>0$

$$\begin{aligned} \lim _{n_k\rightarrow +\infty } \int _0^{T_0} \int _{\vert x\vert \leqq R} \vert f_{n_k}-f_\infty \vert ^2 \, \mathrm{d}x\, \mathrm{d}t=0. \end{aligned}$$

Proof

With no loss of generality, we may assume that $\sigma <\min (1,s)$. Define g by $g_n(t,x)=\alpha (t)\varphi (x) f_n(t,x)$ if $t>0$ and $g_n(t,x)=\alpha (t)\varphi (x) f_n(-t,x)$ if $t<0$, where $\alpha \in {\mathcal {C}}^\infty$ on (0, T), is equal to 1 on $[0,T_0]$ and equal to 0 for $t> \frac{T+T_0}{2}$, and $\varphi (x)=1$ on $B(0,R_0)$. Then the support of $g_n$ is contained in $[-\frac{T+T_0}{2},\frac{T+T_0}{2}]\times \mathrm{Supp} \,\varphi$. Moreover, $g_n$ is bounded in $L^2_t H^s$ and $\partial _t g_n$ is bounded in $L^2 H^\sigma$ so that $g_n$ is bounded in $H^\rho ({\mathbb {R}}\times {\mathbb {R}}^3)$ with $\rho = \frac{s}{s+1-\sigma }$ (just write $(1+\tau ^2+\xi ^2)^{\frac{s}{s+1-\sigma }} \leqq \left( (1+\tau ^2)(1+\xi ^2)^\sigma \right) ^{\frac{s}{s+1-\sigma }} \left( (1+\xi ^2)^s\right) ^{\frac{1-\sigma }{s+1-\sigma }}$). By the Rellich lemma, we know that there is a subsequence $g_{n_k}$ which is strongly convergent in $L^2({\mathbb {R}}\times {\mathbb {R}}^3)$, thus a subsequence $f_{n_k}$ which is strongly convergent in $L^2((0,T_0)\times B(0,R_0))$.

We then iterate this argument for an increasing sequence of times $T_0<T_1<\dots <T_N\rightarrow T$ and an increasing sequence of radii $R_0<R_1<\dots <R_N\rightarrow +\infty$ and finish the proof by the classical diagonal process of Cantor. $\quad \square$

5.2 Proof of Theorem 3

Assume that $\mathbf{u}_{0,n}$ is strongly convergent to $\mathbf{u}_{0,\infty }$ in $L^2_{w_\gamma }$ and that the sequence ${\mathbb {F}}_n$ is strongly convergent to ${\mathbb {F}}_\infty$ in $L^2((0,T), L^2_{w_\gamma })$, and assume that the sequence $\mathbf{b}_n$ is bounded in $L^3((0,T), L^3_{w_{3\gamma /2}})$. Then, by Theorem 2 and Corollary 4, we know that $\mathbf{u}_n$ is bounded in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_n$ is bounded in $L^2((0,T), L^2_{w_\gamma })$. In particular, writing $p_n=p_{n,1}+ p_{n,2}$ with

$$\begin{aligned} p_{n,1}=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_{n,i}u_{n,j}) \text { and } p_{n,2}=-\sum _{i=1}^3\sum _{j=1}^3 R_iR_j( F_{n,i,j}), \end{aligned}$$

we get that $p_{n,1}$ is bounded in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $p_{n,2}$ is bounded in $L^{2}((0,T),L^{2}_{w_\gamma })$.

If $\varphi \in {\mathcal {D}}({\mathbb {R}}^3)$, we find that $\varphi \mathbf{u}_n$ is bounded in $L^2((0,T), H^1)$ and, writing

$$\begin{aligned} \partial _t \mathbf{u}_n= \Delta \mathbf{u}_n - \left( \sum _{i=1}^3 \partial _i(b_{n,i}\mathbf{u}_n) +\mathbf{\nabla }p_{n,1}\right) + \left( \mathbf{\nabla }\cdot {\mathbb {F}}_n-\mathbf{\nabla }p_{n,2}\right) , \end{aligned}$$

$\varphi \partial _t\mathbf{u}_n$ is bounded in $L^2 L^2 + L^2 W^{-1,6/5}+ L^2 H^{-1}\subset L^2((0,T), H^{-2})$. Thus, by Lemma 6, there exist $\mathbf{u}_\infty$ and an increasing sequence $(n_k)_{k\in {\mathbb {N}}}$ with values in ${\mathbb {N}}$ such that $\mathbf{u}_{n_k}$ converges strongly to $\mathbf{u}_\infty$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$, and for every $T_0\in (0,T)$ and every $R>0$, we have

$$\begin{aligned} \lim _{k\rightarrow +\infty } \int _0^{T_0} \int _{\vert y\vert <R} \vert \mathbf{u}_{n_k}(s,y)-\mathbf{u}_\infty (s,y)\vert ^2\, \mathrm{d}y\, \mathrm{d}s=0. \end{aligned}$$

As $\mathbf{u}_n$ is bounded in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_n$ is bounded in $L^2((0,T), L^2_{w_\gamma })$, the convergence of $\mathbf{u}_{n_k}$ to $\mathbf{u}_\infty$ in ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$ implies that $\mathbf{u}_{n_k}$ converges *-weakly to $\mathbf{u}_\infty$ in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_{n_k}$ converges weakly to $\mathbf{\nabla }\mathbf{u}_\infty$ in $L^2((0,T),L^2_{w_\gamma })$.

By Banach–Alaoglu’s theorem, we may assume that there exists $\mathbf{b}_\infty$ such that $\mathbf{b}_{n_k}$ converges weakly to $\mathbf{b}_\infty$ in $L^3((0,T), L^3_{w_{3\gamma /2}})$. In particular $b_{n_k,i} u_{n_k,j}$ is weakly convergent in $(L^{6/5}L^{6/5})_\mathrm{loc}$ and thus in ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$; as it is bounded in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$, it is weakly convergent in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ to $b_{\infty ,i}u_{\infty ,j}$. Let

$$\begin{aligned} p_{\infty ,1}=\sum _{i=1}^3\sum _{j=1}^3 R_i R_j(b_{\infty ,i}u_{\infty ,j}) \text { and } p_{\infty ,2}=-\sum _{i=1}^3\sum _{j=1}^3 R_i R_j( F_{\infty ,i,j}). \end{aligned}$$

As the Riesz transforms are bounded on $L^{6/5}_{w_{\frac{6\gamma }{5}}}$ and on $L^2_{w_\gamma }$, we find that $p_{n_k,1}$ is weakly convergent in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ to $p_{\infty ,1}$ and that $p_{n_k,2}$ is strongly convergent in $L^2((0,T),L^2_{w_\gamma })$ to $p_{\infty ,2}$.

In particular, we find that in ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$,

$$\begin{aligned} \partial _t \mathbf{u}_\infty =\Delta \mathbf{u}_\infty -\sum _{i=1}^3 \partial _i (b_{\infty ,i}\mathbf{u}_\infty )-\mathbf{\nabla }(p_{\infty ,1}+p_{\infty ,2})+\mathbf{\nabla }\cdot {\mathbb {F}}_\infty . \end{aligned}$$

In particular, $\partial _t\mathbf{u}_\infty$ is locally in $L^2 H^{-2}$, and thus $\mathbf{u}_\infty$ has representative such that $t\mapsto \mathbf{u}_\infty (t,.)$ is continuous from [0, T) to ${\mathcal {D}}'({\mathbb {R}}^3)$ and coincides with $\mathbf{u}_\infty (0,.)+\int _0^t \partial _t \mathbf{u}_\infty \, \mathrm{d}s$. In ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$, we have that

$$\begin{aligned}&\mathbf{u}_\infty (0,.)+\int _0^t \partial _t \mathbf{u}_\infty \, \mathrm{d}s=\mathbf{u}_\infty =\lim _{n_k\rightarrow +\infty } \mathbf{u}_{n_k}\\ =&\lim _{n_k\rightarrow +\infty } \mathbf{u}_{0,n_k} + \int _0^t \partial _t \mathbf{u}_{n_k}\, \mathrm{d}s=\mathbf{u}_{0, \infty }+\int _0^t \partial _t\mathbf{u}_\infty \, \mathrm{d}s \end{aligned}$$

Thus, $\mathbf{u}_\infty (0,.)=\mathbf{u}_{0,\infty }$, and $\mathbf{u}_\infty$ is a solution of $(AD_\infty )$.

Next, we define

$$\begin{aligned} A_n= & {} =\vert \mathbf{\nabla }\mathbf{u}_n\vert ^2 +\mu _n\\= & {} - \partial _t\Big (\frac{\vert \mathbf{u}_n\vert ^2}{2}\Big )+\Delta \Big (\frac{\vert \mathbf{u}_n\vert ^2}{2}\Big )-\mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}_n\vert ^2}{2}\mathbf{b}_n\right) -\mathbf{\nabla }\cdot (p_n\mathbf{u}_n) + \mathbf{u}_n\cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_n) . \end{aligned}$$

As $\mathbf{u}_n$ is bounded in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_n$ is bounded in $L^2((0,T), L^2_{w_\gamma })$, it is bounded in $L^2((0,T), L^6_{w_{3\gamma /2}})$ and by interpolation with $L^\infty ((0,T), L^2_{w_\gamma })$ it is bounded in $L^{10/3}((0,T), L^{10/3}_{w_{5\gamma /3}})$. Thus, $u_{n_k}$ is locally bounded in $L^{10/3}L^{10/3}$ and locally strongly convergent in $L^2 L^2$; it is then strongly convergent in $L^3 L^3$. Thus, $A_{n_k}$ is convergent in ${\mathcal {D}}'((0,T)\times {\mathbb {R}}^3)$ to

$$\begin{aligned} A_\infty = - \partial _t\left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\right) +\Delta \left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\right) -\mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\mathbf{b}_\infty \right) -\mathbf{\nabla }\cdot (p_\infty \mathbf{u}_\infty ) + \mathbf{u}_\infty \cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_\infty ) . \end{aligned}$$

In particular, $A_\infty =\lim _{n_k\rightarrow +\infty } \vert \mathbf{\nabla }\mathbf{u}_{n_k}\vert ^2 +\mu _{n_k}$. If $\Phi \in {\mathcal {D}}((0,T)\times {\mathbb {R}}^3)$ is non-negative, we have

$$\begin{aligned} \iint A_\infty \Phi \, \mathrm{d}x\, \mathrm{d}s= & {} \lim _{n_k\rightarrow +\infty }\iint A_{n_k} \Phi \, \mathrm{d}x\, \mathrm{d}s\\\geqq & {} \limsup _{n_k\rightarrow +\infty }\iint \vert \mathbf{\nabla }\mathbf{u}_{n_k}\vert ^2 \Phi \, \mathrm{d}x\, \mathrm{d}s \geqq \iint \vert \mathbf{\nabla }\mathbf{u}_{\infty }\vert ^2 \Phi \, \mathrm{d}x\, \mathrm{d}s \end{aligned}$$

(since $\sqrt{\Phi }\mathbf{\nabla }\mathbf{u}_{n_k}$ is weakly convergent to $\sqrt{\Phi }\mathbf{\nabla }\mathbf{u}_\infty$ in $L^2L^2$). Thus, there exists a non-negative locally finite measure $\mu _\infty$ on $(0,T)\times {\mathbb {R}}^3$ such that $A_\infty =\vert \mathbf{\nabla }\mathbf{u}_\infty \vert ^2 +\mu _\infty$, that is such that

$$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}_\infty \vert ^2}{2}\Big )= & {} \Delta \Big (\frac{\vert \mathbf{u}_\infty \vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}_\infty \vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}_\infty \vert ^2}{2}\mathbf{b}_\infty \right) \\&-\mathbf{\nabla }\cdot (p_\infty \mathbf{u}_\infty ) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_\infty )-\mu _\infty . \end{aligned}$$

Finally, we start from inequality (6):

$$\begin{aligned} \int \frac{\vert \mathbf{u}_n(t,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\leqq & {} \int \frac{\vert \mathbf{u}_{0,n}(x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \\&-\,\sum _{i=1}^3 \int _0^t \int \partial _i\mathbf{u}_n\cdot \mathbf{u}_n\, (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\quad - \int _0^t \int \vert \mathbf{\nabla }\mathbf{u}_n\vert ^2\, \, \phi _R w_{\gamma ,\varepsilon } \mathrm{d}x\, \mathrm{d}s \\&\quad + \sum _{i=1}^3 \int _0^t \int \frac{\vert \mathbf{u}_n\vert ^2}{2} b_{n,i} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\quad +\sum _{i=1}^3 \int _0^t \int p_nu_{n,i} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\quad - \sum _{i=1}^3\sum _{j=1}^3 \int _0^t \int F_{n,i,j} u_{n,j} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\quad - \sum _{i=1}^3 \sum _{j=1}^3\int _0^t \int F_{n,i,j}\partial _iu_{n,j}\ \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s. \end{aligned}$$

This gives

$$\begin{aligned} \begin{aligned}&\limsup _{n_k\rightarrow +\infty }\int \frac{\vert \mathbf{u}_{n_k}(t,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x +\int _0^t \int \vert \mathbf{\nabla }\mathbf{u}_{n_k}\vert ^2\, \, \phi _R w_{\gamma ,\varepsilon } \mathrm{d}x\, \mathrm{d}s \\&\quad \leqq \int \frac{\vert \mathbf{u}_{0,\infty }(x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \\&\qquad -\sum _{i=1}^3 \int _0^t \int \partial _i\mathbf{u}_\infty \cdot \mathbf{u}_\infty \, (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad + \sum _{i=1}^3 \int _0^t \int \frac{\vert \mathbf{u}_\infty \vert ^2}{2} b_{\infty ,i} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad +\sum _{i=1}^3 \int _0^t \int p_\infty u_{\infty ,i} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3\sum _{j=1}^3 \int _0^t \int F_{\infty ,i,j} u_{\infty ,j} (w_{\gamma ,\varepsilon }\partial _i \phi _R+\phi _R \partial _iw_{\gamma ,\varepsilon })\, \mathrm{d}x\, \mathrm{d}s \\&\qquad - \sum _{i=1}^3 \sum _{j=1}^3\int _0^t \int F_{\infty ,i,j}\partial _iu_{\infty ,j}\ \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x\, \mathrm{d}s. \end{aligned} \end{aligned}$$

As we have

$$\begin{aligned} \mathbf{u}_{n_k}= \mathbf{u}_{0, n_k}+ \int _0^t \partial _t \mathbf{u}_{n_k}\, \mathrm{d}s, \end{aligned}$$

we see that $\mathbf{u}_{n_k}(t,.)$ is convergent to $\mathbf{u}_\infty (t,.)$ in ${\mathcal {D}}'({\mathbb {R}}^3)$, hence is weakly convergent in $L^2_\mathrm{loc}$ (as it is bounded in $L^2_{w_\gamma }$), so that:

$$\begin{aligned} \int \frac{\vert \mathbf{u}_\infty (t,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \leqq \limsup _{n_k\rightarrow +\infty }\int \frac{\vert \mathbf{u}_{n_k}(t,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x . \end{aligned}$$

Similarly, as $\mathbf{\nabla }\mathbf{u}_{n_k}$ is weakly convergent in $L^2L^2_{w_\gamma }$, we have

$$\begin{aligned} \int _0^t \int \frac{\vert \mathbf{\nabla }\mathbf{u}_\infty (s,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \, \mathrm{d}s \leqq \limsup _{n_k\rightarrow +\infty } \int _0^t\int \frac{\vert \mathbf{\nabla }\mathbf{u}_{n_k}(s,x)\vert ^2}{2} \phi _R w_{\gamma ,\varepsilon }\, \mathrm{d}x \, \mathrm{d}s. \end{aligned}$$

Thus, letting R go to $+\infty$ and then $\varepsilon$ go to 0, we find by dominated convergence that, for every $t\in (0,T)$, we have

$$\begin{aligned} \begin{aligned}&\Vert \mathbf{u}_\infty (t,. )\Vert _{L^2_{w_\gamma }}^2 +2\int _0^t \Vert \mathbf{\nabla }\mathbf{u}_\infty (s,.)\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \\&\quad \leqq \Vert \mathbf{u}_{0, \infty } \Vert _{L^2_{w_\gamma }}^2 -\int _0^t \int \mathbf{\nabla }\vert \mathbf{u}_\infty \vert ^2\cdot \mathbf{\nabla }w_\gamma \, \mathrm{d}x\, \mathrm{d}s\\&\qquad +\int _0^t\int (\vert \mathbf{u}_\infty \vert ^2 \mathbf{b}_\infty +2p_\infty \mathbf{u}_\infty ) \cdot \mathbf{\nabla }(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s\\&\qquad -2\sum _{i=1}^3\sum _{j=1}^3 \int _0^t\int F_{\infty ,i,j} (\partial _i u_{\infty ,j}) w_\gamma +F_{\infty ,i,j} u_{\infty ,i} \partial _j(w_\gamma )\, \mathrm{d}x\, \mathrm{d}s. \end{aligned} \end{aligned}$$

Letting t go to 0, we find

$$\begin{aligned} \limsup _{t\rightarrow 0} \Vert \mathbf{u}_\infty (t,. )\Vert _{L^2_{w_\gamma }}^2\leqq \Vert \mathbf{u}_{0,\infty } \Vert _{L^2_{w_\gamma }}^2 . \end{aligned}$$

On the other hand, we know that $\mathbf{u}_\infty$ is weakly continuous in $L^2_{w_\gamma }$ and thus we have

$$\begin{aligned} \Vert \mathbf{u}_{0,\infty } \Vert _{L^2_{w_\gamma }}^2 \leqq \liminf _{t\rightarrow 0} \Vert \mathbf{u}_\infty (t,. )\Vert _{L^2_{w_\gamma }}^2 . \end{aligned}$$

This gives $\Vert \mathbf{u}_{0,\infty } \Vert _{L^2_{w_\gamma }}^2 = \lim _{t\rightarrow 0} \Vert \mathbf{u}_\infty (t,. )\Vert _{L^2_{w_\gamma }}^2$, which allows to turn the weak convergence into a strong convergence. Theorem 3 is proven. $\quad \square$

6 Solutions of the Navier–Stokes Problem with Initial Data in $\varvec{L}^{\varvec{2}}_{\varvec{w}_{\varvec{\gamma }}}$

We now prove Theorem 1. The idea is to approximate the problem by a Navier–Stokes problem in $L^2$, then use the a priori estimates (Theorem 2) and the stability theorem (Theorem 3) to find a solution to the Navier–Stokes problem with data in $L^2_{w_\gamma })$.

6.1 Approximation by Square Integrable Data

Lemma 7

(Leray’s projection operator) Let $0<\delta <3$ and $1<r<+\infty$. If $\mathbf{v}$ is a vector field on ${\mathbb {R}}^3$ such that $\mathbf{v}\in L^r_{w_{\delta }}$, then there exists a unique decompostion

$$\begin{aligned} \mathbf{v}=\mathbf{v}_\sigma +\mathbf{v}_\nabla \end{aligned}$$

such that

• $\mathbf{v}_\sigma \in L^r_{w_\delta }$ and $\mathbf{\nabla }\cdot \mathbf{v}_\sigma =0$,

• $\mathbf{v}_\nabla \in L^r_{w_\delta }$ and $\mathbf{\nabla }\wedge \mathbf{v}_\nabla =0$.

We shall write $\mathbf{v}_\sigma ={\mathbb {P}}\mathbf{v}$, where ${\mathbb {P}}$ is Leray’s projection operator.

Similarly, if $\mathbf{v}$ is a distribution vector field of the type $\mathbf{v}=\mathbf{\nabla }\cdot {\mathbb {G}}$ with ${\mathbb {G}}\in L^r_{w_\delta }$ then there exists a unique decompostion

$$\begin{aligned} \mathbf{v}=\mathbf{v}_\sigma +\mathbf{v}_\nabla \end{aligned}$$

such that

• there exists ${\mathbb {H}}\in L^r_{w_\delta }$ such that $\mathbf{v}_\sigma \ = \mathbf{\nabla }\cdot {\mathbb {H}}$ and $\mathbf{\nabla }\cdot \mathbf{v}_\sigma =0$,

• there exists $q\in L^r_{w_\delta }$ such that $\mathbf{v}_\nabla =\mathbf{\nabla }q$ (and thus $\mathbf{\nabla }\wedge \mathbf{v}_\nabla =0$).

We shall still write $\mathbf{v}_\sigma ={\mathbb {P}}\mathbf{v}$. Moreover, the function q is given by

$$\begin{aligned} q=- \sum _{i=1}^3\sum _{j=1}^3 R_i R_j (G_{i,j}). \end{aligned}$$

Proof

As $w_\delta \in {\mathcal {A}}_r$ the Riesz transforms are bounded on $L^r_{w_\delta }$. Using the identity

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

we find (if the decomposition exists) that

$$\begin{aligned} \Delta \mathbf{v}_\sigma = -\mathbf{\nabla }\wedge (\mathbf{\nabla }\wedge \mathbf{v}_\sigma ) =-\mathbf{\nabla }\wedge (\mathbf{\nabla }\wedge \mathbf{v}) \text { and } \Delta \mathbf{v}_\nabla = \mathbf{\nabla }(\mathbf{\nabla }\cdot \mathbf{v}_\nabla )=\mathbf{\nabla }(\mathbf{\nabla }\cdot \mathbf{v}). \end{aligned}$$

This proves the uniqueness. By linearity, we just have to prove that $\mathbf{v}=0\implies \mathbf{v}_\nabla =0$. We have $\Delta \mathbf{v}_\nabla =0$, and thus $\mathbf{v}_\nabla$ is harmonic; as it belongs to ${\mathcal {S}}'$, we find that it is a polynomial. But a polynomial which belongs to $L^r_{w_\delta }$ must be equal to 0. Similarly, if $\mathbf{v}_\nabla =\mathbf{\nabla }q$, then $\Delta q=\mathbf{\nabla }\cdot \mathbf{v}_\nabla =\mathbf{\nabla }\cdot \mathbf{v}=0$; thus q is harmonic and belongs to $L^r_{w_\delta }$, hence $q=0$.

For the existence, it is enough to check that $v_{\nabla ,i} =-\sum _{j=1}^3 R_iR_j v_j$ in the first case and $\mathbf{v}_\nabla =\mathbf{\nabla }q$ with $q= \sum _{i=1}^3\sum _{j=1}^3 R_i R_j (G_{i,j})$ in the second case fulfill the conclusions of the lemma. $\quad \square$

Lemma 8

Let $0<\gamma < 2$. Let $\mathbf{u}_{0}$ be a divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,+\infty ), L^2_{w_\gamma })$. Let $\phi \in {\mathcal {D}}({\mathbb {R}}^3)$ be a non-negative function which is equal to 1 for $\vert x\vert \leqq 1$ and to 0 for $\vert x\vert \geqq 2$. For $R>0$, we define $\phi _R(x)=\phi (\frac{x}{R})$, $\mathbf{u}_{0,R}={\mathbb {P}}(\phi _R \mathbf{u}_0)$ and ${\mathbb {F}}_R=\phi _R {\mathbb {F}}$. Then $\mathbf{u}_{0,R}$ is a divergence-free square integrable vector field and $\lim _{R\rightarrow +\infty } \Vert \mathbf{u}_{0,R}-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0$. Similarly, ${\mathbb {F}}_R$ belongs to $L^2 L^2$ and $\lim _{R\rightarrow +\infty } \Vert {\mathbb {F}}_R-{\mathbb {F}}\Vert _{L^2((0,+\infty ),L^2_{w_\gamma })}=0$.

Proof

By dominated convergence, we have $\lim _{R\rightarrow +\infty } \Vert \phi _R \mathbf{u}_0-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0$. We conclude by writing $\mathbf{u}_{0,R}-\mathbf{u}_0={\mathbb {P}}(\phi _R\mathbf{u}_0-\mathbf{u}_0)$. $\quad \square$

6.2 Leray’s Mollification

We want to solve the Navier–Stokes equations with initial value $\mathbf{u}_0$:

$$\begin{aligned} (NS) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{u}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

We begin with Leray’s method [11] for solving the problem in $L^2$:

$$\begin{aligned} (NS_R) \left\{ \begin{array}{l} \partial _t \mathbf{u}_R= \Delta \mathbf{u}_R -(\mathbf{u}_R\cdot \mathbf{\nabla })\mathbf{u}_R- \mathbf{\nabla }p_R +\mathbf{\nabla }\cdot \mathbb {F_R} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}_R=0, \quad \quad \mathbf{u}_R(0,.)=\mathbf{u}_{0,R} \end{array}\right. \end{aligned}$$

The idea of Leray is to mollify the non-linearity by replacing $\mathbf{u}_R\cdot \mathbf{\nabla }$ by $(\mathbf{u}_R*\theta _\varepsilon )\cdot \mathbf{\nabla }$, where $\theta (x)=\frac{1}{\varepsilon ^3}\theta (\frac{x}{\varepsilon })$, $\theta \in {\mathcal {D}}({\mathbb {R}}^3)$, $\theta$ is non-negative and radially decreasing and $\int \theta \, \mathrm{d}x=1$. We thus solve the problem

$$\begin{aligned} (NS_{R,\varepsilon }) \left\{ \begin{array}{l} \partial _t \mathbf{u}_{R,\varepsilon }= \Delta \mathbf{u}_{R,\varepsilon } -((\mathbf{u}_{R,\varepsilon }*\theta _\varepsilon )\cdot \mathbf{\nabla })\mathbf{u}_{R,\varepsilon }- \mathbf{\nabla }p_{R,\varepsilon } +\mathbf{\nabla }\cdot {\mathbb {F}}_{R} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}_{R,\varepsilon }=0, \quad \quad \mathbf{u}_{R,\varepsilon }(0,.)=\mathbf{u}_{0,R} \end{array}\right. \end{aligned}$$

The classical result of Leray states that the problem $(NS_{R,\varepsilon })$ is well-posed:

Lemma 9

Let $\mathbf{v}_0\in L^2$ be a divergence-free vector field. Let ${\mathbb {G}}\in L^2((0,+\infty ), L^2)$. Then the problem

$$\begin{aligned} (NS_{\varepsilon }) \left\{ \begin{array}{l} \partial _t \mathbf{v}_\varepsilon = \Delta \mathbf{v}_\varepsilon -((\mathbf{v}_\varepsilon *\theta _\varepsilon )\cdot \mathbf{\nabla })\mathbf{v}_\varepsilon - \mathbf{\nabla }q_\varepsilon +\mathbf{\nabla }\cdot {\mathbb {G}} \\ \\ \mathbf{\nabla }\cdot \mathbf{v}_\varepsilon =0, \quad \quad \mathbf{v}_\varepsilon (0,.)=\mathbf{v}_0 \end{array}\right. \end{aligned}$$

has a unique solution $\mathbf{v}_\varepsilon$ in $L^\infty ((0,+\infty ), L^2)\cap L^2((0,+\infty ),\dot{H}^1)$. Moreover, this solution belongs to ${\mathcal {C}}([0,+\infty ), L^2)$.

6.3 Proof of Theorem 1 (Local Existence)

We use Lemma 9 and find a solution $\mathbf{u}_{R,\varepsilon }$ to the problem $(NS_{R,\varepsilon })$. Then we check that $\mathbf{u}_{R,\varepsilon }$ fulfills the assumptions of Theorem 2 and of Corollary 6:

• $\mathbf{u}_{R,\varepsilon }$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_{R,\varepsilon }$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

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

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}_{R,\varepsilon } (t,.)-\mathbf{u}_{0, R}\Vert _{L^2_{w_\gamma }}=0, \end{aligned}$$

• on $(0,T)\times {\mathbb {R}}^3$, $\mathbf{u}_{R,\varepsilon }$ fulfills the energy equality

  $$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}_{R,\varepsilon }\vert ^2}{2}\Big )= & {} \Delta \Big (\frac{\vert \mathbf{u}_{R,\varepsilon }\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}_{R,\varepsilon }\vert ^2\\&- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}\vert ^2}{2}\mathbf{b}_{R,\varepsilon }\right) \\&-\mathbf{\nabla }\cdot (p_{R,\varepsilon }\mathbf{u}_{R,\varepsilon }) + \mathbf{u}_{R,\varepsilon }\cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_R). \end{aligned}$$

  with $\mathbf{b}_{R,\varepsilon }=\mathbf{u}_{R,\varepsilon }*\theta _\varepsilon$;

• $\mathbf{b}_{R,\varepsilon }$ is controlled by $\mathbf{u}_{R,\varepsilon }$ : for every $t\in (0,T)$,

  $$\begin{aligned} \Vert \mathbf{b}_{R,\varepsilon }(t,.)\Vert _{L^3_{w_{3\gamma /2}}}\leqq \Vert {\mathcal {M}}_{\mathbf{u}_{R,\varepsilon }(t,.)}\Vert _{L^3_{w_{3\gamma /2}}} \leqq C_0 \Vert \mathbf{u}_{R,\varepsilon }(t,.)\Vert _{L^3_{w_{3\gamma /2}}}. \end{aligned}$$

Thus, we know that, for every time $T_0$ such that

$$\begin{aligned} C_\gamma (1+C_0^4) \left( 1+C_0^4+\Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}^2+\int _0^{T_0} \Vert {\mathbb {F}}_R\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, T_0\leqq 1, \end{aligned}$$

we have

$$\begin{aligned} \sup _{0\leqq t\leqq T_0} \Vert \ \mathbf{u}_{R,\varepsilon }(t,.)\Vert _{L^2_{w_\gamma }}^2 \leqq C_\gamma (1+ C_0^4 + \Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}_R\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s ) \end{aligned}$$

and

$$\begin{aligned} { \int _0^{T_0} \Vert \mathbf{\nabla }\mathbf{u}_{R,\varepsilon }\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s }\leqq C_\gamma (1+ C_0^4 + \Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}_R\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s). \end{aligned}$$

Moreover, we have that

$$\begin{aligned} \Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}\leqq C_\gamma \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }} \text { and } \Vert {\mathbb {F}}_R\Vert _{L^2_{w_\gamma }} \leqq \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}, \end{aligned}$$

so that

$$\begin{aligned} \Vert \mathbf{b}_{R,\varepsilon }\Vert _{L^3((0,T_0), L^3_{w_{3\gamma /2}}}\leqq & {} C_\gamma \Vert \mathbf{u}_{R,\varepsilon }\Vert _{L^3((0,T_0), L^3_{w_{3\gamma /2}}} \\\leqq & {} C'_\gamma T_0^{ \frac{1}{12}} \left( (1+\sqrt{ T_0}) \Vert \mathbf{u}_{R,\varepsilon }\Vert _{L^\infty ((0,T_0), L^2_{w_\gamma })} \right. \\&\left. + \Vert \mathbf{\nabla }\mathbf{u}_{R,\varepsilon }\Vert _{L^2((0,T_0), L^2_{w_\gamma })}\right) \\\leqq & {} C''_\gamma \sqrt{1+ C_0^4 + \Vert \mathbf{u}_{0}\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s}. \end{aligned}$$

Let $R_n\rightarrow +\infty$ and $\varepsilon _n\rightarrow 0$. Let $\mathbf{u}_{0,n}=\mathbf{u}_{0,R_n}$, ${\mathbb {F}}_n={\mathbb {F}}_{R_n}$, $\mathbf{b}_n=\mathbf{b}_{R_n,\varepsilon _n}$ and $\mathbf{u}_n=\mathbf{u}_{R_n,\varepsilon _n}$. We may then apply Theorem 3, since $\mathbf{u}_{0,n}$ is strongly convergent to $\mathbf{u}_{0}$ in $L^2_{w_\gamma }$, ${\mathbb {F}}_n$ is strongly convergent to ${\mathbb {F}}$ in $L^2((0,T_0), L^2_{w_\gamma })$, and the sequence $\mathbf{b}_n$ is bounded in $L^3((0,T_0), L^3_{w_{3\gamma /2}})$. Thus there exists p, $\mathbf{u}$, $\mathbf{b}$ and an increasing sequence $(n_k)_{k\in {\mathbb {N}}}$ with values in ${\mathbb {N}}$ such that

• $\mathbf{u}_{n_k}$ converges *-weakly to $\mathbf{u}$ in $L^\infty ((0,T_0), L^2_{w_\gamma })$, $\mathbf{\nabla }\mathbf{u}_{n_k}$ converges weakly to $\mathbf{\nabla }\mathbf{u}$ in $L^2((0,T_0),L^2_{w_\gamma })$;

• $\mathbf{b}_{n_k}$ converges weakly to $\mathbf{b}$ in $L^3((0,T_0), L^3_{w_{3\gamma /2}})$, $p_{n_k}$ converges weakly to p in $L^{3}((0,T_0),L^{6/5}_{w_{\frac{6\gamma }{5}}})+L^{2}((0,T_0),L^{2}_{w_\gamma })$;

• $\mathbf{u}_{n_k}$ converges strongly to $\mathbf{u}$ in $L^2_\mathrm{loc}([0,T_0)\times {\mathbb {R}}^3)$.

Moreover, $\mathbf{u}$ is a solution of the advection-diffusion problem

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_{0}, \end{array}\right. \end{aligned}$$

and is such that

• the map $t\in [0,T_0)\mapsto \mathbf{u}(t,.)$ is weakly continuous from $[0,T_0)$ to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$ :

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}(t,.)-\mathbf{u}_{0}\Vert _{L^2_{w_\gamma }}=0; \end{aligned}$$

• there exists a non-negative locally finite measure $\mu$ on $(0,T_0)\times {\mathbb {R}}^3$ such that

  $$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )=\Delta \Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}\vert ^2}{2}\mathbf{b}\right) -\mathbf{\nabla }\cdot (p \mathbf{u}) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}})-\mu , \end{aligned}$$

Finally, as $\mathbf{b}_n=\theta _{\varepsilon _n}*(\mathbf{u}_n-\mathbf{u})+ \theta _{\varepsilon _n}*\mathbf{u}$, we see that $\mathbf{b}_{n_k}$ is strongly convergent to $\mathbf{u}$ in $L^3_\mathrm{loc}([0,T_0)\times {\mathbb {R}}^3)$, so that $\mathbf{b}=\mathbf{u}$ : thus, $\mathbf{u}$ is a solution of the Navier–Stokes problem on $(0,T_0)$. (It is easy to check that

$$\begin{aligned} p=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(u_iu_j-F_{i,j}) \end{aligned}$$

as $u_{i,n_k} u_{j,n_k}$ is weakly convergent to $u_i u_j$ in $L^{4}((0,T_0),L^{6/5}_{w_{\frac{6\gamma }{5}}})$ and $w_{\frac{6\gamma }{5}} \in {\mathcal {A}}_{6/5}$.)

6.4 Proof of Theorem 1 (Global Existence)

In order to finish the proof, we shall use the scaling properties of the Navier–Stokes equations : if $\lambda >0$, then $\mathbf{u}$ is a solution of the Cauchy initial value problem for the Navier–Stokes equations on (0, T) with initial value $\mathbf{u}_0$ and forcing tensor ${\mathbb {F}}$ if and only if $\mathbf{u}_\lambda (t,x)=\lambda \mathbf{u}(\lambda ^2 t,\lambda x)$ is a solution of the Navier–Stokes equations on $(0,T/\lambda ^2)$ with initial value $\mathbf{u}_{0,\lambda }(x)=\lambda \mathbf{u}_0(\lambda x)$ and forcing tensor ${\mathbb {F}}_\lambda (t,x)=\lambda ^2 {\mathbb {F}}(\lambda ^2 t,\lambda x)$.

We take $\lambda >1$ and for $n\in {\mathbb {N}}$ we consider the Navier–Stokes problem with initial value $\mathbf{v}_{0,n}=\lambda ^n \mathbf{u}_0(\lambda ^n \cdot )$ and forcing tensor ${\mathbb {F}}_n=\lambda ^{2n}{\mathbb {F}}(\lambda ^{2n} \cdot , \lambda ^n \cdot )$. Then we have seen that we can find a solution $\mathbf{v}_n$ on $(0,T_n)$, with

$$\begin{aligned} C_\gamma \left( 1+\Vert \mathbf{v}_{0,n}\Vert _{L^2_{w_\gamma }}^2+\int _0^{+\infty } \Vert {\mathbb {F}}_n\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, T_n = 1. \end{aligned}$$

Of course, we have $\mathbf{v}_n(t,x)=\lambda ^{n}\mathbf{u}_n(\lambda ^{2n}t,\lambda ^n x)$ where $\mathbf{u}_n$ is a solution of the Navier–Stokes equations on $(0,\lambda ^{2n}T_n)$ with initial value $\mathbf{u}_0$ and forcing tensor ${\mathbb {F}}$.

Lemma 10

$$\begin{aligned} \lim _{n\rightarrow +\infty } \frac{\lambda ^n}{ 1+\Vert \mathbf{v}_{0,n}\Vert _{L^2_{w_\gamma }}^2+\int _0^{+\infty } \Vert {\mathbb {F}}_n\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s}=+\infty . \end{aligned}$$

Proof

We have

$$\begin{aligned} \Vert \mathbf{v}_{0,n}\Vert _{L^2_{w_\gamma }}^2=\int \vert \mathbf{u}_0(x)\vert ^2 \lambda ^{n(\gamma -1)} \frac{(1+\vert x\vert )^\gamma }{(\lambda ^n+\vert x\vert )^\gamma } w_\gamma (x)\, \mathrm{d}x. \end{aligned}$$

We have

$$\begin{aligned} \lambda ^{n(\gamma -1)}\leqq \lambda ^n \end{aligned}$$

as $\gamma \leqq 2$ and we have, by dominated convergence,

$$\begin{aligned} \lim _{n\rightarrow +\infty } \int \vert \mathbf{u}_0(x)\vert ^2 \frac{(1+\vert x\vert )^\gamma }{(\lambda ^n+\vert x\vert )^\gamma } w_\gamma (x)\, \mathrm{d}x=0. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \int _0^{+\infty } \Vert {\mathbb {F}}_{n}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s =\int _0^{+\infty }\int \vert {\mathbb {F}}(s,x)\vert ^2 \lambda ^{n(\gamma -1)} \frac{(1+\vert x\vert )^\gamma }{(\lambda ^n+\vert x\vert )^\gamma } w_\gamma (x)\, \mathrm{d}x\, \mathrm{d}s=o(\lambda ^n). \end{aligned}$$

Thus, $\lim _{n\rightarrow +\infty } \lambda ^{2n} T_n=+\infty$.

Now, for a given $T>0$, if $\lambda ^{2n}T_n>T$ for $n\geqq n_T$, then $\mathbf{u}_n$ is a solution of the Navier-Stokes problem on (0, T). Let $\mathbf{w}_n(t,x)=\lambda ^{n_T} \mathbf{u}_n(\lambda ^{2n_T}t, \lambda ^{n_T}x)$. For $n\geqq n_T$, $\mathbf{w}_n$ is a solution of the Navier-Stokes problem on $(0,\lambda ^{-2n_T} T)$ with initial value $\mathbf{v}_{0,n_T}$ and forcing tensor ${\mathbb {F}}_{n_T}$. As $\lambda ^{-2n_T}T\leqq T_{n_T}$, we have

$$\begin{aligned} C_\gamma \left( 1+\Vert \mathbf{v}_{0,n_T}\Vert _{L^2_{w_\gamma }}^2+\int _0^{+\infty } \Vert {\mathbb {F}}_{n_T}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, \lambda ^{-2n_T} T \leqq 1. \end{aligned}$$

By Corollary 6, we have

$$\begin{aligned} \sup _{0\leqq t\leqq \lambda ^{-2n_T} T} \Vert \ \mathbf{w}_n(t,.)\Vert _{L^2_{w_\gamma }}^2 \leqq C_\gamma \left( 1+ \Vert \mathbf{v}_{0,n_T}\Vert _{L^2_{w_\gamma }}^2 +\int _0^{\lambda ^{-2n_T}T} \Vert {\mathbb {F}}_{n_T}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \right) \end{aligned}$$

and

$$\begin{aligned} { \int _0^{\lambda ^{-2n_T}T} \Vert \mathbf{\nabla }\mathbf{w}_n\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s }\leqq C_\gamma \left( 1+ \Vert \mathbf{v}_{0,n_T}\Vert _{L^2_{w_\gamma }}^2 +\int _0^{\lambda ^{-2n_T}T} \Vert {\mathbb {F}}_{n_T}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) . \end{aligned}$$

We have

$$\begin{aligned} \Vert \mathbf{w}_n\Vert _{L^2_{w_\gamma }}^2= & {} \int \vert \mathbf{u}_n(\lambda ^{2n_T}t,x)\vert ^2 \lambda ^{n_T(\gamma -1)} \frac{(1+\vert x\vert )^\gamma }{(\lambda ^{n_T}+\vert x\vert )^\gamma } w_\gamma (x)\, \mathrm{d}x\\\geqq & {} \lambda ^{-n_T \gamma } \Vert \mathbf{u}_n(\lambda ^{2n_T}t,.)\Vert _{L^2_{w_\gamma }}^2 \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _0^{\lambda ^{-2n_T}T} \Vert \mathbf{\nabla }\mathbf{w}_n\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s =&\int _0^T\int \vert \mathbf{\nabla }\mathbf{u}_n(s,x)\vert ^2 \lambda ^{n_T(\gamma -1)} \frac{(1+\vert x\vert )^\gamma }{(\lambda ^{n_T}+\vert x\vert )^\gamma } w_\gamma (x)\, \mathrm{d}x \, \mathrm{d}s \\ \geqq&\lambda ^{-n_T } \int _0^T\Vert \mathbf{\nabla }\mathbf{u}_n \Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s. \end{aligned} \end{aligned}$$

Thus, we have a uniform control of $\mathbf{u}_n$ and of $\mathbf{\nabla }\mathbf{u}_n$ on (0, T) for $n\geqq n_T$. We may then apply the Rellich lemma (Lemma 6) and Theorem 3 to find a subsequence $\mathbf{u}_{n_k}$ that converges to a global solution of the Navier–Stokes equations. Theorem 1 is proven. $\quad \square$

7 Solutions of the Advection-Diffusion Problem with Initial Data in $L^2_{w_\gamma }$

The proof of Theorem 1 on the Navier–Stokes problem can be easily adapted to the case of the advection-diffusion problem:

Theorem 4

Let $0<\gamma \leqq 2$. Let $0<T<+\infty$. Let $\mathbf{u}_{0}$ be a divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}$ be a tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}$ be a time-dependent divergence free vector-field ($\mathbf{\nabla }\cdot \mathbf{b}=0$) such that $\mathbf{b}\in L^3((0,T),L^3_{w_{3\gamma /2}})$.

Then the advection-diffusion problem

$$\begin{aligned} (AD) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{b}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

has a solution $\mathbf{u}$ such that:

• $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

• the pressure p is related to $\mathbf{u}$, $\mathbf{b}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

  $$\begin{aligned} p=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j(b_iu_j-F_{i,j}); \end{aligned}$$

• the map $t\in [0,T)\mapsto \mathbf{u}(t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$:

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0; \end{aligned}$$

• there exists a non-negative locally finite measure $\mu$ on $(0,T)\times {\mathbb {R}}^3$ such that

  $$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )=\Delta \Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}\vert ^2}{2}\mathbf{b}\right) -\mathbf{\nabla }\cdot (p\mathbf{u}) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}})-\mu . \end{aligned}$$

Proof

Again, we define $\phi _R(x)=\phi (\frac{x}{R})$, $\mathbf{u}_{0,R}={\mathbb {P}}(\phi _R \mathbf{u}_0)$ and ${\mathbb {F}}_R=\phi _R {\mathbb {F}}$. Moreover, we define $\mathbf{b}_R={\mathbb {P}}(\phi _R \mathbf{b})$. We then solve the mollified problem

$$\begin{aligned} (AD_{R,\varepsilon }) \left\{ \begin{array}{l} \partial _t \mathbf{u}_{R,\varepsilon }= \Delta \mathbf{u}_{R,\varepsilon } -((\mathbf{b}_R*\theta _\varepsilon )\cdot \mathbf{\nabla })\mathbf{u}_{R,\varepsilon }- \mathbf{\nabla }p_{R,\varepsilon } +\mathbf{\nabla }\cdot {\mathbb {F}}_{R,\varepsilon } \\ \\ \mathbf{\nabla }\cdot \mathbf{u}_{R,\varepsilon }=0, \quad \quad \mathbf{u}_{R,\varepsilon }(0,.)=\mathbf{u}_{0,R}, \end{array}\right. \end{aligned}$$

for which we easily find a unique solution $\mathbf{u}_{R,\varepsilon }$ in $L^\infty ((0,T), L^2)\cap L^2((0,T),\dot{H}^1)$. Moreover, this solution belongs to ${\mathcal {C}}([0,T), L^2)$.

Again, $\mathbf{u}_{R,\varepsilon }$ fulfills the assumptions of Theorem 2:

• $\mathbf{u}_{R,\varepsilon }$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_{R,\varepsilon }$ belongs to $L^2((0,T),L^2_{w_\gamma })$

• the map $t\in [0,T)\mapsto \mathbf{u}_{R, \varepsilon }(t,.)$ is weakly continuous from [0, T) to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$:

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}_{R, \varepsilon }(t,.)-\mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}=0. \end{aligned}$$

• on $(0,T)\times {\mathbb {R}}^3$, $\mathbf{u}_{R,\varepsilon }$ fulfills the energy equality:

  $$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}_{R,\varepsilon }\vert ^2}{2}\Big )= & {} \Delta \Big (\frac{\vert \mathbf{u}_{R,\varepsilon }\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}_{R,\varepsilon }\vert ^2- \mathbf{\nabla }\cdot \left( \frac{\vert \mathbf{u}\vert ^2}{2}\mathbf{b}_{R,\varepsilon }\right) \\&-\,\mathbf{\nabla }\cdot (p_{R,\varepsilon }\mathbf{u}_{R,\varepsilon }) + \mathbf{u}_{R,\varepsilon }\cdot (\mathbf{\nabla }\cdot {\mathbb {F}}_R). \end{aligned}$$

  with $\mathbf{b}_{R,\varepsilon }=\mathbf{b}_R*\theta _\varepsilon$.

Thus, by Corollary 4 we know that,

$$\begin{aligned} \sup _{0<t<T} \Vert \mathbf{u}_{R,\varepsilon } \Vert _{L^2_{w_\gamma }} \leqq (\Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}+ C_\gamma \Vert {\mathbb {F}}_R\Vert _{L^2((0,T), L^2_{w_\gamma })}) \ e^{C_\gamma (T+ T^{1/3} \Vert \mathbf{b}_{R,\varepsilon } \Vert _{L^3((0,T), L^3_{w_{3\gamma / 2}})}^2)} \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{\nabla }\mathbf{u}_{R,\varepsilon } \Vert _{L^2((0,T),L^2_{w_\gamma } ) } \leqq (\Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma } }+ C_\gamma \Vert {\mathbb {F}}_R \Vert _{L^2((0,T), L^2_{w_\gamma })}) \ e^{C_\gamma (T+ T^{1/3} \Vert \mathbf{b}_{R,\varepsilon } \Vert _{L^3((0,T), L^3_{w_{3\gamma /2}})}^2)}, \end{aligned}$$

where the constant $C_\gamma$ depends only on $\gamma$.

Moreover, we have that

$$\begin{aligned} \Vert \mathbf{u}_{0,R}\Vert _{L^2_{w_\gamma }}\leqq C_\gamma \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}, \Vert {\mathbb {F}}_R\Vert _{L^2_{w_\gamma }} \leqq \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }} \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{b}_{R,\varepsilon }\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}} ) } \leqq \Vert {\mathcal {M}}_{\mathbf{b}_R}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}} ) }\leqq C_\gamma ' \Vert \mathbf{b}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}} ) }. \end{aligned}$$

Let $R_n\rightarrow +\infty$ and $\varepsilon _n\rightarrow 0$. Let $\mathbf{u}_{0,n}=\mathbf{u}_{0,R_n}$, ${\mathbb {F}}_n={\mathbb {F}}_{R_n}$, $\mathbf{b}_n=\mathbf{b}_{R_n,\varepsilon _n}$ and $\mathbf{u}_n=\mathbf{u}_{R_n,\varepsilon _n}$. We may then apply Theorem 3, since $\mathbf{u}_{0,n}$ is strongly convergent to $\mathbf{u}_{0}$ in $L^2_{w_\gamma }$, ${\mathbb {F}}_n$ is strongly convergent to ${\mathbb {F}}$ in $L^2((0,T), L^2_{w_\gamma })$, and the sequence $\mathbf{b}_n$ is strongly convergent to $\mathbf{b}$ in $L^3((0,T), L^3_{w_{3\gamma /2}})$. Thus there exists p, $\mathbf{u}$ and an increasing sequence $(n_k)_{k\in {\mathbb {N}}}$ with values in ${\mathbb {N}}$ such that

• $\mathbf{u}_{n_k}$ converges *-weakly to $\mathbf{u}$ in $L^\infty ((0,T), L^2_{w_\gamma })$, $\mathbf{\nabla }\mathbf{u}_{n_k}$ converges weakly to $\mathbf{\nabla }\mathbf{u}$ in $L^2((0,T),L^2_{w_\gamma })$;

• $p_{n_k}$ converges weakly to p in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})+L^{2}((0,T),L^{2}_{w_\gamma })$;

• $\mathbf{u}_{n_k}$ converges strongly to $\mathbf{u}$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$.

We then easily finish the proof. $\quad \square$

8 Application to the Study of $\lambda$-Discretely Self-similar Solutions

We may now apply our results to the study of $\lambda$-discretely self-similar solutions for the Navier–Stokes equations.

Definition 1

Let $\mathbf{u}_0\in L^2_\mathrm{loc}({\mathbb {R}}^3)$. We say that $\mathbf{u}_0$ is a $\lambda$-discretely self-similar function ($\lambda$-DSS) if there exists $\lambda >1$ such that $\lambda \mathbf{u}_0(\lambda x)= \mathbf{u}_0$.

A vector field $\mathbf{u}\in L^2_\mathrm{loc}([0,+\infty )\times {\mathbb {R}}^3)$ is $\lambda$-DSS if there exists $\lambda >1$ such that $\lambda \mathbf{u}(\lambda ^2 t,\lambda x)=\mathbf{u}(t,x)$.

A forcing tensor ${\mathbb {F}}\in L^2_\mathrm{loc}([0,+\infty )\times {\mathbb {R}}^3)$ is $\lambda$-DSS if there exists $\lambda >1$ such that $\lambda ^2 {\mathbb {F}}(\lambda ^2 t,\lambda x)={\mathbb {F}}(t,x)$.

We shall speak of self-similarity if $\mathbf{u}_0$, $\mathbf{u}$ or ${\mathbb {F}}$ are $\lambda$-DSS for every $\lambda >1$.

Examples

• Let $\gamma >1$ and $\lambda >1$. Then, for two positive constants $A_{\gamma ,\lambda }$ and $B_{\gamma ,\lambda }$, we have : if $\mathbf{u}_0\in L^2_\mathrm{loc}({\mathbb {R}}^3)$ is $\lambda$-DSS, then $\mathbf{u}_0\in L^2_{w_\gamma }$ and

  $$\begin{aligned} A_{\gamma ,\lambda } \int _{ 1<\vert x\vert \leqq \lambda } \vert \mathbf{u}_0(x)\vert ^2\, \mathrm{d}x \leqq \int \vert \mathbf{u}_0(x)\vert ^2 w_\gamma (x)\, \mathrm{d}x \leqq B_{\gamma ,\lambda } \int _{ 1<\vert x\vert \leqq \lambda } \vert \mathbf{u}_0(x)\vert ^2\, \mathrm{d}x. \end{aligned}$$

• $\mathbf{u}_0\in L^2_\mathrm{loc}$ is self-similar if and only if it is of the form $\mathbf{u}_0=\frac{\mathbf{w}_0(\frac{x}{\vert x\vert })}{\vert x\vert }$ with $\mathbf{w}_0\in L^2(S^2)$.

• ${\mathbb {F}}$ belongs to $L^2((0,+\infty ),L^2_{w_\gamma })$ with $\gamma >1$ and is self-similar if and only if it is of the form ${\mathbb {F}}(t,x)=\frac{1}{t} {\mathbb {F}}_0(\frac{x}{\sqrt{t}})$ with $\int \vert {\mathbb {F}}_0(x)\vert ^2 \frac{1}{\vert x\vert }\, \mathrm{d}x <+\infty$.

Proof

• If $\mathbf{u}_0$ is $\lambda$-DSS and if $k\in {\mathbb {Z}}$ we have

  $$\begin{aligned} \!\!\ \!\!\ \int _{\lambda ^k<\vert x\vert<\lambda ^{k+1}} \!\!\ \!\!\ \vert \mathbf{u}_0(x)\vert ^2 w_\gamma (x)\, \mathrm{d}x \leqq \frac{\lambda ^k}{(1+\lambda ^k)^\gamma } \int _{1<\vert x\vert <\lambda } \!\!\ \!\!\ \vert \mathbf{u}_0(x)\vert ^2 \, \mathrm{d}x \end{aligned}$$

  with $\sum _{k\in {\mathbb {Z}}} \frac{\lambda ^k}{(1+\lambda ^k)^\gamma }<+\infty$ for $\gamma >1$.

• If $\mathbf{u}_0$ is self-similar, we have $\mathbf{u}_0(x ) =\frac{1}{\vert x\vert } \mathbf{u}_0(\frac{x}{\vert x\vert })$. From this equality, we find that, for $\lambda >1$

  $$\begin{aligned} \int _{1<\vert x\vert <\lambda } \vert \mathbf{u}_0(x)\vert ^2\, \mathrm{d}x =(\lambda -1) \int _{S^2} \vert \mathbf{u}_0(\sigma ) \vert ^2 \, \mathrm{d}\sigma . \end{aligned}$$

• If ${\mathbb {F}}$ is self-similar, then it is of the form ${\mathbb {F}}(t,x)=\frac{1}{t} {\mathbb {F}}_0(\frac{x}{\sqrt{t}})$. Moreover, we have

  $$\begin{aligned} \int _0^{+\infty } \!\!\! \int \vert {\mathbb {F}}(t,x)\vert ^2\, w_\gamma (x) \, \mathrm{d}x\, \mathrm{d}s= & {} \int _0^{+\infty } \!\!\ \int \vert {\mathbb {F}}_0(x)\vert ^2 w_\gamma (\sqrt{t}\, x) \, \mathrm{d}x \, \frac{\mathrm{d}t}{\sqrt{t}} \\= & {} C_\gamma \int \vert {\mathbb {F}}_0(x)\vert ^2\, \frac{\mathrm{d}x}{\vert x\vert }. \end{aligned}$$

  with $C_\gamma =\int _0^{+\infty } \frac{1}{(1+ \sqrt{\theta })^\gamma } \frac{\mathrm{d}\theta }{\sqrt{\theta }}<+\infty$. $\quad \square$

In this section, we are going to give a new proof of the results of Chae and Wolf [3] and Bradshaw and Tsai [2] on the existence of $\lambda$-DSS solutions of the Navier–Stokes problem (and of Jia and Šverák [6] for self-similar solutions) :

Theorem 5

Let $4/3<\gamma < 2$ and $\lambda >1$. If $\mathbf{u}_{0}$ is a $\lambda$-DSS divergence-free vector field (such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$) and if ${\mathbb {F}}$ is a $\lambda$-DSS tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that ${\mathbb {F}}\in L^2_\mathrm{loc}([0,+\infty )\times {\mathbb {R}}^3)$, then the Navier–Stokes equations with initial value $\mathbf{u}_0$

$$\begin{aligned} (NS) \left\{ \begin{array}{l} \partial _t \mathbf{u}= \Delta \mathbf{u}-(\mathbf{u}\cdot \mathbf{\nabla })\mathbf{u}- \mathbf{\nabla }p +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

have a global weak solution $\mathbf{u}$ such that

• $\mathbf{u}$ is a $\lambda$-DSS vector field;

• for every $0<T<+\infty$, $\mathbf{u}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

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

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{u}(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0; \end{aligned}$$

• the solution $\mathbf{u}$ is suitable, and there exists a non-negative locally finite measure $\mu$ on $(0,+\infty )\times {\mathbb {R}}^3$ such that

  $$\begin{aligned} \partial _t\Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )=\Delta \Big (\frac{\vert \mathbf{u}\vert ^2}{2}\Big )-\vert \mathbf{\nabla }\mathbf{u}\vert ^2- \mathbf{\nabla }\cdot \left( (\frac{\vert \mathbf{u}\vert ^2}{2}+p)\mathbf{u}\right) + \mathbf{u}\cdot (\mathbf{\nabla }\cdot {\mathbb {F}})-\mu . \end{aligned}$$

8.1 The Linear Problem

Following Chae and Wolf, we consider an approximation of the problem that is consistent with the scaling properties of the equations: let $\theta$ be a non-negative and radially decreasing function in ${\mathcal {D}}({\mathbb {R}}^3)$ with $\int \theta \, \mathrm{d}x=1$. We define $\theta _{\varepsilon ,t}(x)=\frac{1}{(\varepsilon \sqrt{t})^3}\ \theta (\frac{x}{\varepsilon \sqrt{t}})$. We then will study the “mollified” problem

$$\begin{aligned} (NS_\varepsilon ) \left\{ \begin{array}{l} \partial _t \mathbf{u}_\varepsilon = \Delta \mathbf{u}_\varepsilon -((\mathbf{u}_\varepsilon *\theta _{\varepsilon ,t}) \cdot \mathbf{\nabla })\mathbf{u}_\varepsilon - \mathbf{\nabla }p_\varepsilon +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{u}=0, \quad \quad \mathbf{u}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

and begin with the linearized problem

$$\begin{aligned} (LNS_\varepsilon ) \left\{ \begin{array}{l} \partial _t \mathbf{v}= \Delta \mathbf{v}- ((\mathbf{b}*\theta _{\varepsilon ,t})\cdot \mathbf{\nabla })\mathbf{v}- \mathbf{\nabla }q +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{v}=0, \quad \quad \mathbf{v}(0,.)=\mathbf{u}_0. \end{array}\right. \end{aligned}$$

Lemma 11

Let $1<\gamma < 2$. Let $\lambda >1$ Let $\mathbf{u}_{0}$ be a $\lambda$-DSS divergence-free vector field such that $\mathbf{u}_0\in L^2_{w_\gamma }({\mathbb {R}}^3)$ and ${\mathbb {F}}$ be a $\lambda$-DSS tensor ${\mathbb {F}}(t,x)=\left( F_{i,j}(t,x)\right) _{1\leqq i,j\leqq 3}$ such that, for every $T>0$, ${\mathbb {F}}\in L^2((0,T), L^2_{w_\gamma })$. Let $\mathbf{b}$ be a $\lambda$-DSS time-dependent divergence free vector-field ($\mathbf{\nabla }\cdot \mathbf{b}=0$) such that, for every $T>0$, $\mathbf{b}\in L^3((0,T),L^3_{w_{3\gamma /2}})$.

Then the advection-diffusion problem

$$\begin{aligned} (LNS_\varepsilon ) \left\{ \begin{array}{l} \partial _t \mathbf{v}= \Delta \mathbf{v}- ((\mathbf{b}*\theta _{\varepsilon ,t})\cdot \mathbf{\nabla })\mathbf{v}- \mathbf{\nabla }q +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{v}=0, \quad \quad \mathbf{v}(0,.)=\mathbf{u}_0 \end{array}\right. \end{aligned}$$

has a unique solution $\mathbf{v}$ such that:

• for every positive T, $\mathbf{v}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{v}$ belongs to $L^2((0,T),L^2_{w_\gamma })$;

• the pressure p is related to $\mathbf{v}$, $\mathbf{b}$ and ${\mathbb {F}}$ through the Riesz transforms $R_i =\frac{\partial _i}{\sqrt{-\Delta }}$ by the formula

  $$\begin{aligned} p=\sum _{i=1}^3\sum _{j=1}^3 R_iR_j((b_i*\theta _{\varepsilon ,t})v_j-F_{i,j}); \end{aligned}$$

• the map $t\in [0,+\infty )\mapsto \mathbf{v}(t,.)$ is weakly continuous from $[0,+\infty )$ to $L^2_{w_\gamma }$, and is strongly continuous at $t=0$:

  $$\begin{aligned} \lim _{t\rightarrow 0} \Vert \mathbf{v}(t,.)-\mathbf{u}_0\Vert _{L^2_{w_\gamma }}=0. \end{aligned}$$

This solution $\mathbf{v}$ is a $\lambda$-DSS vector field.

Proof

As we have $\vert \mathbf{b}(t,.)*\theta _{\varepsilon ,t}\vert \leqq {\mathcal {M}}_{\mathbf{b}(t,.)}$ and thus

$$\begin{aligned} \Vert \mathbf{b}(t,.)*\theta _{\varepsilon ,t}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}})}\leqq C_\gamma \Vert \mathbf{b}\Vert _{L^3((0,T), L^3_{w_{3\gamma /2}})}, \end{aligned}$$

we see that we can use Theorem 4 to get a solution $\mathbf{v}$ on (0, T).

As clearly $\mathbf{b}*\theta _{\varepsilon ,t}$ belongs to $L^2_t L^\infty _x(K)$ for every compact subset K of $(0,T)\times {\mathbb {R}}^3$, we can use Corollary 5 to see that $\mathbf{v}$ is unique.

Let $\mathbf{w}(t,x)=\frac{1}{\lambda }\mathbf{v}(\frac{t}{\lambda ^2}, \frac{x}{\lambda })$. As $b*\theta _{\varepsilon ,t}$ is still $\lambda$-DSS, we see that $\mathbf{w}$ is solution of $(LNS_\varepsilon )$ on (0, T), so that $\mathbf{w}=\mathbf{v}$. This means that $\mathbf{v}$ is $\lambda$-DSS. $\quad \square$

8.2 The Mollified Navier–Stokes Equations

The solution $\mathbf{v}$ provided by Lemma 11 belongs to $L^3((0,T), L^3_{w_{3\gamma /2}})$ (as $\mathbf{v}$ belongs to $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{v}$ belongs to $L^2((0,T),L^2_{w_\gamma })$). Thus we have a mapping $L_\varepsilon : \mathbf{b}\mapsto \mathbf{v}$ which is defined from

$$\begin{aligned} X_{T,\gamma }=\{\mathbf{b}\in L^3((0,T), L^3_{w_{3\gamma /2}})\ /\ \mathbf{b}\text { is } \lambda -\text {DSS}\} \end{aligned}$$

to $X_{T,\gamma }$ by $L_\varepsilon (\mathbf{b})=\mathbf{v}$.

Lemma 12

For $4/3<\gamma$, $X_{T,\gamma }$ is a Banach space for the equivalent norms $\Vert \mathbf{b}\Vert _{L^3((0,T),L^3_{w_{3\gamma /2}})}$ and $\Vert \mathbf{b}\Vert _{L^3((0,T/{\lambda ^2}),\times B(0,\frac{1}{\lambda }))}$.

Proof

We have

$$\begin{aligned} \int _0^T \int _{B(0,1)} \vert \mathbf{b}(t,x)\vert ^3\, \mathrm{d}x\, \mathrm{d}t= \lambda ^2 \int _0^{\frac{T}{\lambda ^2}} \int _{B(0,\frac{1}{\lambda })} \vert \mathbf{b}(t,x)\vert ^3\, \mathrm{d}x\, \mathrm{d}t \end{aligned}$$

and , for $k\in {\mathbb {N}}$,

$$\begin{aligned} \int _0^T \int _{\lambda ^{k-1}<\vert x\vert<\lambda ^{k}} \vert \mathbf{b}(t,x)\vert ^3\, \mathrm{d}x\, \mathrm{d}t=\lambda ^{2k}\int _0^{\frac{T}{\lambda ^{2k}}}\int _{\frac{1}{\lambda }<\vert x\vert <1} \vert \mathbf{b}(t,x)\vert ^3\, \mathrm{d}x\, \mathrm{d}t. \end{aligned}$$

We may conclude, since for $\gamma >4/3$ we have $\sum _{k\in {\mathbb {N}}} \lambda ^{k (2-\frac{3\gamma }{2})}<+\infty$. $\quad \square$

Lemma 13

For $4/3<\gamma < 2$, the mapping $L_\varepsilon$ is continuous and compact on $X_{T,\gamma }$.

Proof

Let $\mathbf{b}_n$ be a bounded sequence in $X_{T,\gamma }$ and let $\mathbf{v}_n=L_\varepsilon (\mathbf{b}_n)$. We remark that the sequence $\mathbf{b}_n(t,.)*\theta _{\varepsilon ,t}$ is bounded in $X_{T,\gamma }$. Thus, by Theorem 2 and Corollary 4, the sequence $\mathbf{v}_n$ is bounded in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{v}_n$ is bounded in $L^2((0,T),L^2_{w_\gamma })$.

We now use Theorem 3 and get that then there exists $q_\infty$, $\mathbf{v}_\infty$, $\mathbf{B}_\infty$ and an increasing sequence $(n_k)_{k\in {\mathbb {N}}}$ with values in ${\mathbb {N}}$ such that

• $\mathbf{v}_{n_k}$ converges *-weakly to $\mathbf{v}_\infty$ in $L^\infty ((0,T), L^2_{w_\gamma })$, $\mathbf{\nabla }\mathbf{v}_{n_k}$ converges weakly to $\mathbf{\nabla }\mathbf{v}_\infty$ in $L^2((0,T),L^2_{w_\gamma })$;

• $\mathbf{b}_{n_k}*\theta _{\varepsilon ,t}$ converges weakly to $\mathbf{B}_\infty$ in $L^3((0,T), L^3_{w_{3\gamma /2}})$;

• the associated pressures $q_{n_k}$ converge weakly to $q_\infty$ in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})+L^{2}((0,T),L^{2}_{w_\gamma })$;

• $\mathbf{v}_{n_k}$ converges strongly to $\mathbf{v}_\infty$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$ : for every $T_0\in (0,T)$ and every $R>0$, we have

  $$\begin{aligned} \lim _{k\rightarrow +\infty } \int _0^{T_0} \int _{\vert y\vert <R} \vert \mathbf{v}_{n_k}(s,y)-\mathbf{v}_\infty (s,y)\vert ^2\, \mathrm{d}s\, \mathrm{d}y=0. \end{aligned}$$

As $\sqrt{w_\gamma }\mathbf{v}_n$ is bounded in $L^\infty ((0,T),L^2)$ and in $L^2((0,T), L^6)$, it is bounded in $L^{10/3}((0,T)\times {\mathbb {R}}^3)$. The strong convergence of $\mathbf{v}_{n_k}$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$ then implies the strong convergence of $\mathbf{v}_{n_k}$ in $L^3_\mathrm{loc}((0,T)\times {\mathbb {R}}^3)$.

Moreover, $\mathbf{v}_\infty$ is still $\lambda$-DSS (a property that is stable under weak limits).We find that $\mathbf{v}_\infty \in X_{T,\gamma }$ and that

$$\begin{aligned} \lim _{n_k\rightarrow +\infty } \int _0^{\frac{T}{\lambda ^2}} \int _{B(0,\frac{1}{\lambda })} \vert \mathbf{v}_{n_k}(s,y)-\mathbf{v}_\infty (s,y)\vert ^3\, \mathrm{d}s\, \mathrm{d}y=0. \end{aligned}$$

This proves that $L_\varepsilon$ is compact.

If we assume moreover that $\mathbf{b}_n$ is convergent to $\mathbf{b}_\infty$ in $X_{T,\gamma }$, then necessarily we have $\mathbf{B}_\infty =\mathbf{b}_\infty *\theta _{\varepsilon ,t}$, and $\mathbf{v}_\infty = L_\varepsilon (\mathbf{b}_\infty )$. Thus, the relatively compact sequence $\mathbf{v}_n$ can have only one limit point; thus it must be convergent. This proves that $L_\varepsilon$ is continuous. $\quad \square$

Lemma 14

Let $4/3<\gamma < 2$. If, for some $\mu \in [0,1]$, $\mathbf{v}$ is a solution of $\mathbf{v}=\mu L_\varepsilon (\mathbf{v})$ then

$$\begin{aligned} \Vert \mathbf{v}\Vert _{X_{T, \gamma }}\leqq C_{\mathbf{u}_0,{\mathbb {F}},\gamma , T}, \end{aligned}$$

where the constant $C_{\mathbf{u}_0,{\mathbb {F}},\gamma , T}$ depends only on $\mathbf{u}_0$, ${\mathbb {F}}$, $\gamma$ and T (but not on $\mu$ nor on $\varepsilon$).

Proof

We have $\mathbf{v}=\mu \mathbf{w}$; with

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{w}= \Delta \mathbf{w}-((\mathbf{v}*\theta _{\varepsilon ,t}) \cdot \mathbf{\nabla })\mathbf{w}- \mathbf{\nabla }q +\mathbf{\nabla }\cdot {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{w}=0, \quad \quad \mathbf{w}(0,.)=\mathbf{u}_0. \end{array}\right. \end{aligned}$$

Multiplying by $\mu$, we find that

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t \mathbf{v}= \Delta \mathbf{v}-((\mathbf{v}*\theta _{\varepsilon ,t}) \cdot \mathbf{\nabla })\mathbf{v}- \mathbf{\nabla }(\mu q) +\mathbf{\nabla }\cdot \mu {\mathbb {F}} \\ \\ \mathbf{\nabla }\cdot \mathbf{v}=0, \quad \quad \mathbf{v}(0,.) =\mu \mathbf{u}_0. \end{array}\right. \end{aligned}$$

We then use Corollary 6. We choose $T_0\in (0,T)$ such that

$$\begin{aligned} C_\gamma \left( 1+\Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2+\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, T_0\leqq 1. \end{aligned}$$

Then, as

$$\begin{aligned} C_\gamma \left( 1+\Vert \mu \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2+\int _0^{T_0} \ \Vert \mu {\mathbb {F}} \Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^2\, T_0\leqq 1. \end{aligned}$$

we know that

$$\begin{aligned} \sup _{0\leqq t\leqq T_0} \Vert \ \mathbf{v}(t,.)\Vert _{L^2_{w_\gamma }}^2 \leqq C_\gamma \left( 1 + \mu ^2 \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2 +\mu ^2\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s \right) \end{aligned}$$

and

$$\begin{aligned} { \int _0^{T_0} \Vert \mathbf{\nabla }\mathbf{v}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s }\leqq C_\gamma \left( 1 +\mu ^2 \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2 +\mu ^2\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) . \end{aligned}$$

In particular, we have

$$\begin{aligned} \int _0^{T_0} \Vert \mathbf{v}\Vert _{L^3_{w_{3\gamma /2}}}^3\, \mathrm{d}s \leqq C_\gamma T_0^{1/4} \left( 1 + \Vert \mathbf{u}_0\Vert _{L^2_{w_\gamma }}^2 +\int _0^{T_0} \Vert {\mathbb {F}}\Vert _{L^2_{w_\gamma }}^2\, \mathrm{d}s\right) ^{\frac{3}{2}}. \end{aligned}$$

As $\mathbf{v}$ is $\lambda$-DSS, we can go back from $T_0$ to T. $\quad \square$

Lemma 15

Let $4/3<\gamma \leqq 2$. There is at least one solution $\mathbf{u}_\varepsilon$ of the equation $\mathbf{u}_\varepsilon =L_\varepsilon (\mathbf{u}_\varepsilon )$.

Proof

Obvious due to the Leray–Schauder principle (and the Schaefer theorem), since $L_\varepsilon$ is continuous and compact and since we have uniform a priori estimates for the fixed points of $\mu L_\varepsilon$ for $0\leqq \mu \leqq 1$. $\quad \square$

8.3 Proof of Theorem 5

We may now finish the proof of Theorem 5. We consider the solutions $\mathbf{u}_\varepsilon$ of $\mathbf{u}_\varepsilon =L_\varepsilon (\mathbf{u}_\varepsilon )$.

By Lemma 14, $\mathbf{u}_\varepsilon$ is bounded in $L^3((0,T), L^3_{w_{3\gamma /2}})$, and so is $\mathbf{u}_\varepsilon *\theta _{\varepsilon ,t}$. We then know, by Theorem 2 and Corollary 4, that the familly $\mathbf{u}_\varepsilon$ is bounded in $L^\infty ((0,T), L^2_{w_\gamma })$ and $\mathbf{\nabla }\mathbf{u}_\varepsilon$ is bounded in $L^2((0,T),L^2_{w_\gamma })$.

We now use Theorem 3 and get that then there exists p, $\mathbf{u}$, $\mathbf{B}$ and a decreasing sequence $(\varepsilon _k)_{k\in {\mathbb {N}}}$ (converging to 0) with values in $(0,+\infty )$ such that

• $\mathbf{u}_{\varepsilon _k}$ converges *-weakly to $\mathbf{u}$ in $L^\infty ((0,T), L^2_{w_\gamma })$, $\mathbf{\nabla }\mathbf{u}_{\varepsilon _k}$ converges weakly to $\mathbf{\nabla }\mathbf{u}$ in $L^2((0,T),L^2_{w_\gamma })$;

• $\mathbf{u}_{\varepsilon _k}*\theta _{\varepsilon _k,t}$ converges weakly to $\mathbf{B}$ in $L^3((0,T), L^3_{w_{3\gamma /2}})$;

• the associated pressures $p_{\varepsilon _k}$ converge weakly to p in $L^{3}((0,T),L^{6/5}_{w_{\frac{6\gamma }{5}}})+L^{2}((0,T),L^{2}_{w_\gamma })$;

• $\mathbf{u}_{\varepsilon _k}$ converges strongly to $\mathbf{u}$ in $L^2_\mathrm{loc}([0,T)\times {\mathbb {R}}^3)$.

Moreover we easily see that $\mathbf{B}=\mathbf{u}$. Indeed, we have that $\mathbf{u}*\theta _{\varepsilon ,t}$ converges strongly in $L^2_\mathrm{loc}((0,T)\times {\mathbb {R}}^3)$ as $\varepsilon$ goes to 0 (since it is bounded by ${\mathcal {M}}_\mathbf{u}$ and converges, for each fixed t, strongly in $L^2_\mathrm{loc}({\mathbb {R}}^3)$); moreover, we have $\vert (\mathbf{u}-\mathbf{u}_\varepsilon )*\theta _{\varepsilon ,t}\vert \leqq {\mathcal {M}}_{\mathbf{u}-\mathbf{u}_\varepsilon }$, so that the strong convergence of $\mathbf{u}_{\varepsilon _k}$ to $\mathbf{u}$ is kept by convolution with $\theta _{\varepsilon ,t}$ as far as we work on compact subsets of $(0,T)\times {\mathbb {R}}^3$ (and thus don’t allow t to go to 0).

Thus, Theorem 5 is proven. $\quad \square$