Local and Global Solutions of the 3D-NSE in Homogeneous Lei–Lin–Gevrey Spaces

Abstract

This paper investigates the existence and uniqueness of local and global solutions to the incompressible three-dimensional Navier–Stokes equations within the framework of homogeneous Lei–Lin–Gevrey spaces ${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$, where $\rho \in [-1,0),a>0$, and $\gamma \in (0,1)$. These function spaces combine the critical scaling structure of the Lei–Lin spaces with the exponential regularity of Gevrey classes, thereby enabling a refined treatment of analytic regularity and frequency localization. The main results are obtained under the assumption of small initial data in the critical Lei–Lin space ${\mathcal{X}}^{\rho }({\mathbb{R}}^3)$, extending previous works and improving regularity thresholds. In particular, we establish that for suitable initial data, the Navier–Stokes system admits unique solutions globally in time. The influence of the Gevrey parameter $\gamma$ on the high-frequency behavior of solutions is also discussed. This work contributes to a deeper understanding of regularity and decay properties in critical and supercritical regimes.

1. Introduction and Principal Results

The three-dimensional incompressible Navier–Stokes equations are a fundamental model in fluid dynamics:

$$\left\{\begin{array}{c}{\partial }_{t}u-\mu \mathrm{\Delta }u+u·\nabla u=-\nabla p\mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3\hfill \\ divu=0\mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3\hfill \\ u(0,x)=u^0(x)\mathrm{in}{\mathbb{R}}^3,\hfill \end{array}\right.$$

(1)

where u is the velocity field, p the pressure, and $\mu >0$ is the viscosity. Despite their classical formulation, the global existence and smoothness of solutions in three dimensions remain open for general initial data.

Several functional frameworks—such as Sobolev spaces ${\dot{H}}^s({\mathbb{R}}^3)$, Lebesgue spaces $L^3({\mathbb{R}}^3)$, and Gevrey classes—have been used to investigate existence and regularity. However, these spaces may not fully capture the interaction between analytic smoothing and frequency decay, which is key to understanding fine-scale structures in fluid flows.

To address this, Lei and Lin introduced the Fourier-based critical space ${\mathcal{X}}^{\rho }({\mathbb{R}}^3)$, defined by integrability of ${|\xi |}^{\rho }\widehat{u}(\xi )$ in $L^1$. This space preserves the scaling invariance of the Navier–Stokes equations and improves control over low frequencies compared to classical Sobolev spaces.

Building on this idea, the homogeneous Lei–Lin–Gevrey space ${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$, with exponential weight $e^{{a|\xi |}^{\gamma }}$, was proposed to incorporate analytic regularity. Here, $a>0$, $\gamma \in (0,1)$, and $\rho \in [-1,0)$. These spaces allow better tracking of high-frequency behavior and enable refined analysis of smoothing effects due to viscosity.

The parameter $\gamma$ quantifies the level of analyticity: values of $\gamma$ close to 1 correspond to stronger smoothing effects, while values near 0 represent milder analytic behavior. Compared to standard Gevrey spaces, the Lei–Lin–Gevrey framework combines critical scaling with exponential regularity and is particularly effective for analyzing nonlinear structures in the Fourier domain.

Significant advances have been made for small initial data in critical spaces. Fujita and Kato [1] proved global existence and uniqueness in ${\dot{H}}^{1/2}({\mathbb{R}}^3)$, later extended by Chemin [2] to ${\dot{H}}^s({\mathbb{R}}^3)$ for $s>1/2$. Kato also obtained similar results in $L^3({\mathbb{R}}^3)$ [3]. An in-depth overview is given in the monograph by Cannone [4].

Recent contributions by Schorlepp et al. [5], Cannone and Karch [6], and Sawada and Taniuchi [7] further investigate analytic regularity, Gevrey smoothing, and symmetry-breaking effects in incompressible flow models.

In [8], Lei and Lin introduced the space

$${\mathcal{X}}^{-1}({\mathbb{R}}^3):=\left\{u\in {\mathcal{S}}^{\prime }({\mathbb{R}}^3):{\int }_{{\mathbb{R}}^3}{|\xi |}^{-1}|\widehat{u}(\xi )|d\xi <\infty \right\},$$

and established the global solvability of the Navier–Stokes system for small initial data in this space.

This was generalized in [9] by introducing the non-homogeneous Lei–Lin space

$${\mathcal{X}}^{s,\rho }({\mathbb{R}}^3):=\left\{u\in {\mathcal{S}}^{\prime }({\mathbb{R}}^3):{\int }_{{\mathbb{R}}^3}{(|\xi |}^s+{|\xi |}^{\rho })|\widehat{u}(\xi )|d\xi <\infty \right\},$$

with existence and uniqueness results for $s=\{-1,0\}$ and $\rho \ge -1$.

More recently, in [10], the homogeneous Lei–Lin–Gevrey space

$${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3):=\left\{u\in {\mathcal{S}}^{\prime }({\mathbb{R}}^3):{\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}|\widehat{u}(\xi )|d\xi <\infty \right\}$$

was introduced. For $\rho =-1$, global existence and uniqueness were shown under small data assumptions.

This paper extends those results by proving the existence and uniqueness of solutions for $\rho \in [-1,0)$ in $X_{a,\gamma }^{\rho }({\mathbb{R}}^3)$, thereby bridging critical and subcritical regimes while maintaining analytic control.

The main results are summarized below.

Theorem 1. 

Let $\rho \in [-1,0)$ and $\gamma \in (0,1)$. Suppose the initial data $u^0\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$, with $u^0\ne 0$. Then there exists a time $T>0$ such that the system (1) admits a unique solution $u\in \mathcal{C}([0,T],{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))$ satisfying $u\in L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$.

And

Theorem 2. 

Let $\rho \in [-1,0),\gamma \in (0,1)$, and $a>0$ be fixed. Assume that the initial data $u^0$ belongs to the homogeneous Lei–Lin–Gevrey space ${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$, and satisfies the smallness condition:

$$\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)}<\mu$$

for some positive constant $\mu >0$ depending on the viscosity. Then, the Navier–Stokes system (1) admits a unique global-in-time solution

$$u\in \mathcal{C}({\mathbb{R}}^{+},{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L^1({\mathbb{R}}^{+},{\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3)).$$

Moreover, this solution depends continuously on the initial data.

To proceed, we first establish several technical lemmas that will be useful in proving the main results of this paper.

2. Technical Lemmas

Lemma 1. 

For $a,b\in \mathbb{R}$ and $s\in (0,1]$, we have

$${|a|}^s\le {max(|a-b|}^s{,|b|}^s{)+smin(|a-b|}^s{,|b|}^s).$$

(2)

Proof. 

Without the loss of generality, we assume that $a,b\in [0,+\infty )$ and $b\le a$.

If $a\ge 2b$, using the mean value theorem, we get $a^s={(a-b)}^s+sb{c}^{s-1}$, with $b\le a-b\le c\le a$.

Then $c^{s-1}\le b^{s-1}$ and we have (2).

If $b\le a\le 2b$, using also the mean value theorem, we get $a^s=b^s+s(a-b)c^{s-1}$, where $b\le c\le a$. As $b\le a$, we have $a-b\le b\le c\le a$ and $c^{s-1}\le b^{s-1}\le {(a-b)}^{s-1}$ and we get (2). □

We deduce the following result:

Corollary 1. 

For all $\xi ,\eta \in {\mathbb{R}}^3$ and for all $\gamma \in (0,1)$, we have the following:

$$\begin{array}{c}\hfill e^{{a|\xi |}^{\gamma }}\le e^{amax\left({|\xi -\eta |}^{\gamma },{|\eta |}^{\gamma }\right)}{e}^{a\gamma min\left({|\xi -\eta |}^{\gamma },{|\eta |}^{\gamma }\right)}.\end{array}$$

Lemma 2. 

Let $\rho \in [-1,0)$, $\gamma \in (0,1)$. If $u\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)\cap {\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3)$, then $u\in {\mathcal{X}}_{a,\gamma }^s({\mathbb{R}}^3)$ for all $s\in (\rho ,\rho +2)$ and we have

$${\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^s}\le {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2-s}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{s-\rho }{2}}}.$$

In particular:

$${\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^0}\le {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}$$

(3)

and

$${\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}.$$

(4)

Proof. 

We use the Hölder inequality for ${\displaystyle p=\frac{2}{\rho +2-s}}$ and ${\displaystyle q=\frac{2}{s-\rho }}$, we get

$${\int }_{{\mathbb{R}}^3}{|\xi |}^s{e}^{{a|\xi |}^{\gamma }}|\widehat{u}{(\xi )|d\xi \le \parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2-s}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{s-\rho }{2}}}.$$

□

Lemma 3. 

For $\rho \in [-1,0)$, $\gamma \in (0,1)$. If $u,v\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)\cap {\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3)$, then

$${\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}+{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}.$$

(5)

In particular, for $u=v$, we have

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le {2\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +3}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1-\rho }{2}}}.$$

Proof. 

Using the inequality ${(x+y)}^r\le x^r+y^r$, for $x,y\ge 0$ and $0<r<1$, we get for $\rho \in [-1,0)$ and $\gamma \in (0,1)$

$$\begin{array}{ccc}\hfill {\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}& \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{u}*\widehat{v}(\xi )|d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}{\int }_{{\mathbb{R}}^3}|\widehat{u}(\xi -y)||\widehat{v}(y)|dyd\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}\left({\int }_{{\mathbb{R}}^3}{|\xi -y|}^{\rho +1}{e}^{{a|\xi -y|}^{\gamma }}|\widehat{u}(\xi -y)|d\xi \right)e^{{a|y|}^{\gamma }}|\widehat{v}(y)|dy\hfill \\ & & +{\int }_{{\mathbb{R}}^3}\left({\int }_{{\mathbb{R}}^3}{e}^{{a|\xi -y|}^{\gamma }}|\widehat{u}(\xi -y)|d\xi \right){|y|}^{\rho +1}{e}^{{a|y|}^{\gamma }}|\widehat{v}(y)|dy\hfill \\ & \le & {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^0}+{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^0}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}.\hfill \end{array}$$

Using the inequalities (3)–(4), we get

$${\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le {\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}+{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}.$$

□

Lemma 4. 

Let $\rho \in (-1,0)$, $\gamma \in (0,1)$, and let $u,v\in L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })\cap L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})$. Then, for every $t\in [0,T]$, we have the following:

$$\begin{array}{c}{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}\hfill \\ \le T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ +T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}.\hfill \end{array}$$

In particular, for $u=v$, we have

$$\begin{array}{ccc}\hfill {∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}& \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +3}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1-\rho }{2}}}.\hfill \end{array}$$

Proof. 

Firstly, we want to prove the following inequality

$${∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}\le {\int }_0^t{\parallel u\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau .$$

The inequality follows from the triangle inequality applied to the integral, the boundedness of the Leray projector $\mathbb{P}$ on ${\mathcal{X}}_{a,\gamma }^{\rho }$, and the fact that

$${\parallel \nabla ·(u\otimes v)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\le {\parallel u\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}.$$

Additionally, since the exponential weight $e^{{a|\xi |}^{\gamma }}$ already captures analytic regularity, the heat semigroup $e^{\mu (t-\tau )\mathrm{\Delta }}$ acts as a contraction on ${\mathcal{X}}_{a,\gamma }^{\rho }$.

Secondly, using the inequality (5) and Hölder inequality, we get

$$\begin{array}{c}{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}\hfill \\ \le {\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}d\tau +{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}d\tau \hfill \\ \le {\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{-\rho }\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ +{\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{-\rho }\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \le T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ +T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}.\hfill \end{array}$$

□

Lemma 5. 

Let $\rho \in (-1,0)$, $\gamma \in (0,1)$, and $u,v\in L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })({\mathbb{R}}^3)\cap L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})({\mathbb{R}}^3)$. Then, the following time-integrated estimate holds:

$$\begin{array}{c}{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt\hfill \\ \le {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ +{\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}.\hfill \end{array}$$

In particular, for $u=v$, we have

$${\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes u))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt\le {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +3}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1-\rho }{2}}}.$$

Proof. 

$$\begin{array}{c}{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt\hfill \\ \le {\int }_0^T{\int }_0^t{∥e^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(\mathrm{div}(u\otimes v))∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau dt\hfill \\ \le {\int }_0^T{\int }_0^t{∥e^{\mu (t-\tau )\mathrm{\Delta }}\mathrm{div}(u\otimes v)∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau dt\hfill \\ \le {\int }_0^T{\int }_0^t{\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +3}{e}^{-{\mu (t-\tau )|\xi |}^2}{e}^{{a|\xi |}^{\gamma }}|\widehat{u\otimes v}|(\tau ,\xi )d\tau dtd\xi \hfill \\ \le {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +3}{\int }_0^T{\int }_0^t{e}^{-{\mu (t-\tau )|\xi |}^2}{e}^{{a|\xi |}^{\gamma }}|\widehat{u\otimes v}|(\tau ,\xi )d\tau dtd\xi .\hfill \end{array}$$

Using the Fubini Theorem, inequality (5), and the Cauchy–Shwarz inequality, we get

$$\begin{array}{c}{\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +3}{\int }_0^T{\int }_0^t{e}^{-{\mu (t-\tau )|\xi |}^2}{e}^{{a|\xi |}^{\gamma }}|\widehat{u\otimes v}|(\tau ,\xi )d\tau dtd\xi \hfill \\ ={\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +3}{e}^{{a|\xi |}^{\gamma }}{\int }_0^T|\widehat{u\otimes v}|(\tau ,\xi )\left({\int }_{\tau }^T{e}^{-{\mu (T-\tau )|\xi |}^2}dt\right)d\tau d\xi \hfill \\ ={\displaystyle \frac{1}{\mu }}{\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}{\int }_0^T|\widehat{u\otimes v}|(\tau ,\xi )\left(1-e^{-{\mu (T-\tau )|\xi |}^2}\right)d\tau d\xi \hfill \\ \le {\displaystyle \frac{1}{\mu }}{\int }_0^T{\parallel u\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau \hfill \\ \le {\displaystyle \frac{1}{\mu }}{\int }_0^T{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}d\tau +{\displaystyle \frac{1}{\mu }}{\int }_0^T{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}d\tau \hfill \\ \le {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ +{\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel u\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel u\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}.\hfill \end{array}$$

□

Lemma 6. 

Let $\rho \in (-1,0),\gamma \in (0,1)$, and $u,v\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)\cap {\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3)$. Then there exists a constant $C_{a,\gamma }>0$ such that

$${\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +1}{C}_{a,\gamma }\left({\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\right).$$

In particular, for $u=v$, we have

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +2}{C}_{a,\gamma }{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}.$$

Proof. 

$$\begin{array}{ccc}\hfill {\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}& =& {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|uv(\xi )|d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{u}*\widehat{v}|(\xi )|d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}\left({\int }_{{\mathbb{R}}^3}|\widehat{u}|(\xi -\eta )|\widehat{v}|(\eta )d\eta \right)d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}\left({\int }_{|\eta |<|\xi -\eta |}|\widehat{u}|(\xi -\eta )|\widehat{v}|(\eta )d\eta \right)d\xi \hfill \\ & & +{\int }_{{\mathbb{R}}^3}{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}\left({\int }_{|\eta |>|\xi -\eta |}|\widehat{u}|(\xi -\eta )|\widehat{v}|(\eta )d\eta \right)d\xi .\hfill \end{array}$$

Using Corollary 1 and the well known inequality ${|\xi |}^{\rho +1}\le 2^{\rho +1}max\left({|\xi -\eta |}^{\rho +1},{|\eta |}^{\rho +1}\right)$ for all $\rho \in (-1,0)$, we get

$$\begin{array}{ccc}\hfill {\parallel uv\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}& \le & 2^{\rho +1}{\int }_{{\mathbb{R}}^3}\left({\int }_{|\eta |<|\xi -\eta |}{|\xi -\eta |}^{\rho +1}{e}^{{a|\xi -\eta |}^{\gamma }}|\widehat{u}|(\xi -\eta )e^{{a\gamma |\eta |}^{\gamma }}|\widehat{v}|(\eta )d\eta \right)d\xi \hfill \\ & & +2^{\rho +1}{\int }_{{\mathbb{R}}^3}\left({\int }_{|\eta |>|\xi -\eta |}{e}^{{a\gamma |\xi -\eta |}^{\gamma }}|\widehat{u}{|(\xi -\eta )|\eta |}^{\rho +1}{e}^{{a|\eta |}^{\gamma }}|\widehat{v}|(\eta )d\eta \right)d\xi \hfill \\ & \le & 2^{\rho +1}{∥\left({|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{u}|(\xi )\right)*\left(e^{{a\gamma |\xi |}^{\gamma }}|\widehat{v}|(\xi )\right)∥}_{L^1}\hfill \\ & & +2^{\rho +1}{∥\left(e^{{a\gamma |\xi |}^{\gamma }}|\widehat{u}|(\xi )\right)*\left({|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{v}|(\xi )\right)∥}_{L^1}\hfill \\ & \le & 2^{\rho +1}{∥{|\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{u}|(\xi ){\parallel }_{L^1}\parallel e^{{a\gamma |\xi |}^{\gamma }}|\widehat{v}|(\xi )∥}_{L^1}\hfill \\ & & +2^{\rho +1}{∥e^{{a\gamma |\xi |}^{\gamma }}|\widehat{u}|(\xi )|{\parallel }_{L^1}{\parallel |\xi |}^{\rho +1}{e}^{{a|\xi |}^{\gamma }}|\widehat{v}|(\xi )∥}_{L^1}\hfill \\ & \le & 2^{\rho +1}\left({\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}{\parallel v\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}+{\parallel u\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\right).\hfill \end{array}$$

(6)

Moreover, we have

$$\begin{array}{ccc}\hfill {\parallel u\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}& =& {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma |\xi |}^{\gamma }}|\widehat{u}|(\xi )d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{|\xi |}^{-\rho }{e}^{{a(\gamma -1)|\xi |}^{\gamma }}{|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}|\widehat{u}|(\xi )d\xi \hfill \\ & \le & C_{a,\gamma }{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }},\hfill \end{array}$$

where $C_{a,\gamma }={sup}_{\xi \in {\mathbb{R}}^3}{|\xi |}^{-\rho }{e}^{{a(\gamma -1)|\xi |}^{\gamma }},a>0,\gamma \in (0,1)$. Similarly, we have

$$\begin{array}{ccc}\hfill {\parallel v\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}& \le & C_{a,\gamma }{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }},\hfill \end{array}$$

We deduce that

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +1}{C}_{a,\gamma }\left({\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\right).$$

□

Lemma 7. 

For $\rho \in (-1,0)$, $\gamma \in (0,1)$ and $u\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)\cap {\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3)$, we have

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +2}{C}_{\gamma }{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}},$$

where $C_{\gamma }={sup}_{\xi \in {\mathbb{R}}^3}{|\xi |}^{-\rho }{e}^{a(\gamma -\sqrt{\gamma }){|\xi |}^{\gamma }},a>0,\gamma \in (0,1)$.

Proof. 

From the inequality (6) and for $u=v$, we have

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +2}{\parallel u\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}.$$

Using the inequality (4), we get

$$\parallel u^2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le 2^{\rho +2}{\parallel u\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}.$$

The Cauchy–Shwarz inequality yields

$$\begin{array}{ccc}\hfill {\parallel u\parallel }_{{\mathcal{X}}_{a\gamma ,\gamma }^0({\mathbb{R}}^3)}& =& {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma |\xi |}^{\gamma }}|\widehat{u}|(\xi )d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}\left({|\xi |}^{-\rho }{e}^{a(\gamma -\sqrt{\gamma }){|\xi |}^{\gamma }}\right)\left({|\xi |}^{\rho }{e}^{a\sqrt{\gamma }{|\xi |}^{\gamma }}|\widehat{u}|(\xi )\right)d\xi \hfill \\ & \le & C_{\gamma }{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }({\mathbb{R}}^3)}.\hfill \end{array}$$

Then, we can deduce the desired result. □

3. Proof of Theorem 1

We take the case where $\rho =-1$ was studied in [9]. Now, suppose that $\rho \in (-1,0)$ and $\gamma \in (0,1)$. Our first objective is to establish the existence of a local solution. To do so, we begin by decomposing the initial condition into components with higher and lower frequency terms. For the lower frequencies, we construct a regular solution to the corresponding linear system (1). For the higher frequencies, we analyze a partial differential equation closely related to (1), but with small initial data in the space ${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$. We then apply the Fixed Point Theorem to demonstrate the existence of a solution.

Let $\epsilon >0$ and $r>0$. We choose $N\in \mathbb{N}$ such that

$${\int }_{|\xi |>N}{|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}|\widehat{u^0}(\xi )|d\xi <{\displaystyle \frac{r}{5}}.$$

For this choice of $N\in \mathbb{N}$, we define the low-frequency component by

$$v^0={\mathcal{F}}^{-1}({\chi }_{|\xi |<N}\widehat{u^0}(\xi )).$$

The unique solution of the linear system

$$\left\{\begin{array}{c}{\partial }_{t}v-\mu \mathrm{\Delta }v=0,\hfill \\ v(0,x)=v^0(x),\hfill \end{array}\right.$$

(7)

is given by $v=e^{\mu t\mathrm{\Delta }}{v}^0$, which satisfies

$${\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}={\int }_{|\xi |<N}{|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}{e}^{-{\mu t|\xi |}^2}|\widehat{u^0}(\xi )|d\xi \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }},\forall t\ge 0,$$

(8)

and

$${\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\le {\displaystyle \frac{1}{\mu }}{\int }_{|\xi |<N}{|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}\left(1-e^{-{\mu T|\xi |}^2}\right)|\widehat{u^0}(\xi )|d\xi .$$

Using the monotone convergence theorem, we deduce that

$$\underset{T\to 0^{+}}{lim}{\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}=0.$$

Then, there exists $T=T(\epsilon )$ such that

$${\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}<\epsilon .$$

(9)

Now, let u be a solution of the system (1) and v the solution of system (7). Then $w=u-v$ satisfies the following system:

$$\left\{\begin{array}{c}{\partial }_{t}w-\mu \mathrm{\Delta }w+(v+w)·\nabla (v+w)=-\nabla p,\hfill \\ w(0,x)=w^0(x),\hfill \end{array}\right.$$

(10)

where the high-frequency component of the initial data is given by

$$w^0={\mathcal{F}}^{-1}({\chi }_{|\xi |>N}\widehat{u^0}(\xi )),$$

which satisfies

$$\parallel w^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}<{\displaystyle \frac{r}{5}}.$$

(11)

The integral form of the solution w to this system is

$$w=e^{\mu t\mathrm{\Delta }}{w}^0-{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}((v+w)·\nabla (v+w))d\tau .$$

To establish the existence of w using the Fixed Point Theorem, we define the operator

$$\phi (w)=e^{\mu t\mathrm{\Delta }}{w}^0-{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}((v+w)·\nabla (v+w))d\tau .$$

We consider the space

$${\mathcal{X}}_T=\mathcal{C}([0,T],{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L^1([0,T],{\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$$

with the norm

$${\parallel f\parallel }_{{\mathcal{X}}_T}={\parallel f\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}+{\parallel f\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}.$$

For the chosen $r>0$, let $S_r$ be the subset of ${\mathcal{X}}_T$ defined by

$$S_r=\{w\in {\mathcal{X}}_T{:\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\le {r\mathrm{and}\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\le r\}.$$

Under certain conditions on the initial data $u^0$ and the time T, we aim to show that $\phi ({\mathcal{X}}_T)\subset {\mathcal{X}}_T$. To do this, let $w\in S_r$ and demonstrate that $\phi (w)\in S_r$.

We select $u^0$ and T, small enough to satisfy

$$\left\{\begin{array}{c}2\left(1+{\displaystyle \frac{1}{\mu }}\right)T^{{\displaystyle \frac{1+\rho }{2}}}{\epsilon }^{{\displaystyle \frac{1-\rho }{2}}}{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +3}{2}}}<{\displaystyle \frac{r}{5}},\hfill \\ 2\left(1+{\displaystyle \frac{1}{\mu }}\right)T^{{\displaystyle \frac{1+\rho }{2}}}r\le {\displaystyle \frac{1}{5}},\hfill \\ \left(1+{\displaystyle \frac{1}{\mu }}\right)T^{{\displaystyle \frac{1+\rho }{2}}}\left({\epsilon }^{{\displaystyle \frac{1}{2}}}\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}+{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\epsilon }^{-{\displaystyle \frac{\rho }{2}}}\right)<{\displaystyle \frac{1}{5}},\hfill \\ 2\left(1+{\displaystyle \frac{1}{\mu }}\right)T^{{\displaystyle \frac{1+\rho }{2}}}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{-{\displaystyle \frac{\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]<{\displaystyle \frac{1}{2}}.\hfill \end{array}\right.$$

(12)

We have

$$\begin{array}{ccc}\hfill {\parallel \phi (w)(t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}& \le & \underset{I_0}{\underbrace{\parallel e^{\mu t\mathrm{\Delta }}{w}^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}}}\hfill \\ & & +\underset{I_1}{\underbrace{{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(v·\nabla v)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}}}+\underset{I_2}{\underbrace{{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(w·\nabla w)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}}}\hfill \\ & & +\underset{I_3}{\underbrace{{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(v·\nabla w)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}}}+\underset{I_4}{\underbrace{{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(w·\nabla v)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}}}.\hfill \end{array}$$

From the inequalities (11), we obtain

$$I_0\le {\displaystyle \frac{r}{5}}.$$

Applying Lemma 4, the inequalities (8), (9), and (12), we further have the following:

$$\begin{array}{ccc}\hfill I_1& \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +3}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1-\rho }{2}}}\hfill \\ & \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}{\epsilon }^{{\displaystyle \frac{1-\rho }{2}}}{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +3}{2}}}<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Hence, using the fact that $u\in S_r$, Lemma 4, and the inequality (12) we get

$$\begin{array}{ccc}\hfill I_2& \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +3}{2}}}{\left({\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1-\rho }{2}}}\hfill \\ & \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}{r}^2\le {\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Also, using the fact that $u\in S_r$, Lemma 4, and the inequalities (8), (9), and (12), we obtain

$$\begin{array}{ccc}\hfill I_3& & \le T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ & & +T^{{\displaystyle \frac{1+\rho }{2}}}{\left({\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left({\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\right)}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ & & \le T^{{\displaystyle \frac{1+\rho }{2}}}r\left({\epsilon }^{{\displaystyle \frac{1}{2}}}\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}+{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\epsilon }^{{\displaystyle \frac{-\rho }{2}}}\right)<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Similarly, we have

$$\begin{array}{ccc}\hfill I_4& & \le T^{{\displaystyle \frac{1+\rho }{2}}}r\left({\epsilon }^{{\displaystyle \frac{1}{2}}}\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}+{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\epsilon }^{{\displaystyle \frac{-\rho }{2}}}\right)<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Thus,

$${\parallel \phi (w)(t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\le \sum _{k=0}^4{I}_k\le r.$$

On the other hand, we can show

$$\begin{array}{ccc}\hfill {\parallel \phi (w)(t)\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}& \le & \underset{J_0}{\underbrace{{\int }_0^T{\parallel e^{\mu t\mathrm{\Delta }}{w}^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt}}\hfill \\ & & +\underset{J_1}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(v·\nabla v)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt}}+\underset{J_2}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(w·\nabla w)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt}}\hfill \\ & & +\underset{J_3}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(v·\nabla w)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt}}+\underset{J_4}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}(w·\nabla v)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}dt}}.\hfill \end{array}$$

For the initial term,

$$J_0\le {\displaystyle \frac{r}{5}}.$$

From Lemma 4 and the inequalities (12), we have

$$\begin{array}{ccc}\hfill J_1& \le & {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}{\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +3}{2}}}{\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1-\rho }{2}}}\hfill \\ & \le & {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +3}{2}}}{\epsilon }^{{\displaystyle \frac{1-\rho }{2}}}<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Also, using Lemma 4, the fact that $w\in S_r$ and the inequalities (12) we obtain

$$\begin{array}{ccc}\hfill J_2& \le & {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}{\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +3}{2}}}{\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1-\rho }{2}}}\hfill \\ & \le & {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}{r}^2\le {\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Using Lemma 4, the fact that $w\in S_r$ and the inequalities (8), (9), and (12), we get

$$\begin{array}{ccc}\hfill J_3& \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}{\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ \hfill & & +{\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}{\parallel w\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel w\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ & \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}r\left({\epsilon }^{{\displaystyle \frac{1}{2}}}\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)}^{{\displaystyle \frac{1}{2}}}+{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\epsilon }^{{\displaystyle \frac{-\rho }{2}}}\right)<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Similarly, we have

$$\begin{array}{ccc}\hfill J_4& \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}r\left({\epsilon }^{{\displaystyle \frac{1}{2}}}\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)}^{{\displaystyle \frac{1}{2}}}+{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\epsilon }^{{\displaystyle \frac{-\rho }{2}}}\right)<{\displaystyle \frac{r}{5}}.\hfill \end{array}$$

Then

$${\parallel \phi (w)(t)|\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}\le \sum _{k=0}^4{J}_k\le r$$

and finally, we verify that $\phi (S_r)\subset S_r$.

Now, we show the following contraction to use the Fixed Point Theorem:

$$\parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_T}\le {\displaystyle \frac{1}{2}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.$$

For $w_1,w_2\in S_r$, we have

$$\begin{array}{ccc}& & \parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\hfill \\ & & ={∥-{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}\left((v+w_2)·\nabla (v+w_2)-(v+w_1)·\nabla (v+w_1)\right)d\tau ∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}\hfill \\ & & \le \underset{K_1}{\underbrace{{\int }_0^t{∥e^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}((v+w_2)·\nabla (w_2-w_1))∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau }}+\underset{K_2}{\underbrace{{\int }_0^t{∥e^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}((w_2-w_1)·\nabla (v+w_1))∥}_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau }}.\hfill \end{array}$$

Using Lemma 5, we have

$$\begin{array}{ccc}\hfill K_1& \le & T^{{\displaystyle \frac{1+\rho }{2}}}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\parallel w_2-w_1{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\parallel w_2-w_1\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}\hfill \\ & & +T^{{\displaystyle \frac{1+\rho }{2}}}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}\parallel w_2-w_1{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel w_2-w_1\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\hfill \end{array}$$

Hence, using the concavity of the logarithmic function

$$\begin{array}{c}\hfill x^t{y}^{1-t}\le x+y,\forall t\in [0,1].\end{array}$$

(13)

and the fact that $w\in S_r$ and the inequalities (8) and (9), we get

$$\begin{array}{ccc}\hfill K_1& \le & T^{{\displaystyle \frac{1+\rho }{2}}}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}\hfill \\ & & +T^{{\displaystyle \frac{1+\rho }{2}}}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}\hfill \\ & \le & T^{{\displaystyle \frac{1+\rho }{2}}}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.\hfill \end{array}$$

With similar estimations, we obtain

$$K_2\le T^{{\displaystyle \frac{1+\rho }{2}}}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.$$

Thus,

$$\begin{array}{cc}& \parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_{a,\sigma }^{\rho }}\hfill \\ \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.\hfill \end{array}$$

(14)

Additionally, we have

$$\begin{array}{ccc}\hfill \parallel \phi (w_2)-\phi (w_1){\parallel }_{L_T^1({\mathcal{X}}_{a,\sigma }^{\rho +2})}& \le & \underset{L_1}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}\left((v+w_2)·\nabla (w_2-w_1)\right)d\tau ∥}_{{\mathcal{X}}_{a,\sigma }^{\rho +2}}dt}}\hfill \\ & & +\underset{L_2}{\underbrace{{\int }_0^T{∥{\int }_0^t{e}^{\mu (t-\tau )\mathrm{\Delta }}\mathbb{P}\left((w_2-w_1)·\nabla (v+w_1)\right)d\tau ∥}_{{\mathcal{X}}_{a,\sigma }^{\rho +2}}dt}}.\hfill \end{array}$$

Using Lemma 5, the fact that $w\in S_r$ and inequalities (8), (9) and (13), we obtain

$$\begin{array}{ccc}\hfill L_1& \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\parallel w_2-w_1{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}{\parallel w_2-w_1\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}\hfill \\ & & +{\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}\parallel w_2-w_1{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel w_2-w_1\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ & \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{\rho +2}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{-\rho }{2}}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}\hfill \\ & & +{\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\parallel v+w_2{\parallel }_{L_T^{\infty }({\mathcal{X}}_{a,\gamma }^{\rho })}^{{\displaystyle \frac{1}{2}}}\parallel v+w_2{\parallel }_{L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2})}^{{\displaystyle \frac{1}{2}}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}\hfill \\ & \le & {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.\hfill \end{array}$$

With similar estimations, we obtain

$$L_2\le {\displaystyle \frac{T^{\frac{1+\rho }{2}}}{\mu }}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.$$

Thus,

$$\begin{array}{cc}& \parallel \phi (w_2)(t)-\phi (w_1)(t){\parallel }_{L_T^1({\mathcal{X}}_{a,\sigma }^{\rho +2})}\hfill \\ \le & {\displaystyle \frac{2T^{\frac{1+\rho }{2}}}{\mu }}\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.\hfill \end{array}$$

(15)

Combining estimates (14) and (15), we conclude that

$$\begin{array}{cc}& \parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_T}\hfill \\ \le & 2T^{{\displaystyle \frac{1+\rho }{2}}}\left(1+{\displaystyle \frac{1}{\mu }}\right)\left[{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{\rho +2}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{-\rho }{2}}}+{\left(\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+r\right)}^{{\displaystyle \frac{1}{2}}}{\left(\epsilon +r\right)}^{{\displaystyle \frac{1}{2}}}\right]{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.\hfill \end{array}$$

With the choice of r and $\epsilon$ in (12), we get

$$\parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_T}\le {\displaystyle \frac{1}{2}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}.$$

Since $\phi (S_r)\subset S_r$ and $\parallel \phi (w_2)-\phi (w_1){\parallel }_{{\mathcal{X}}_T}\le {\displaystyle \frac{1}{2}}{\parallel w_2-w_1\parallel }_{{\mathcal{X}}_T}$, it follows from the Banach Fixed Point Theorem that there exists a unique $w\in S_r$ such that $u=v+w$ is the solution of the Navier–Stokes system (1), with $u\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$.

For the uniqueness of this solution, take $u_1$ and $u_2$, two solutions of (1) such that $u_1,u_2\in \mathcal{C}([0,T],{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L_T^1({\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$ and $u_1^0=u_2^0$.

Let $v=u_1-u_2$. We have

$${\partial }_{t}v-\mu \mathrm{\Delta }v+u_1·\nabla v+v·\nabla u_2=-\nabla (p_1-p_2).$$

Taking the Fourier transform and using $\mathrm{div}{p}_1=\mathrm{div}{p}_2=0$, we get

$${\partial }_t\widehat{v}+\mu {|\xi |}^2\widehat{v}+\widehat{u_1·\nabla v}+\widehat{v·\nabla u_2}=0.$$

The inner product with $\overline{\widehat{v}}$ yields

$${\displaystyle \frac{1}{2}}{\partial }_t|\widehat{v}{|}^2+{\mu |\xi |}^2{|\widehat{v}|}^2+\langle \widehat{u_1·\nabla v},\widehat{v}\rangle +\langle \widehat{v·\nabla u_2},\widehat{v}\rangle =0.$$

Taking the real part of this relation, we find

$${\partial }_t|\widehat{v}{|}^2+{2\mu |\xi |}^2{|\widehat{v}|}^2+2\mathrm{Re}\langle \widehat{u_1·\nabla v},\widehat{v}\rangle +2\mathrm{Re}\langle \widehat{v·\nabla u_2},\widehat{v}\rangle =0.$$

Hence

$${\partial }_t|\widehat{v}{|}^2+{2\mu |\xi |}^2|\widehat{v}{|}^2\le 2|\widehat{u_1·\nabla v}||\widehat{v}|+2|\widehat{v·\nabla u_2}||\widehat{v}|.$$

Let $\epsilon >0$ and $\alpha >0$, we have ${\displaystyle |\widehat{v}{|}^{2\alpha }{\partial }_t|\widehat{v}{|}^2=\frac{1}{1+\alpha }{\partial }_t{|\widehat{v}|}^{2+2\alpha }}$ and

$${\displaystyle \frac{{\partial }_t(|\widehat{v}{{|}^{1+\alpha }+\epsilon )}^2}{(1+\alpha )(|\widehat{v}{|}^{1+\alpha }+\epsilon )}=\frac{|\widehat{v}{|}^{2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }{\partial }_t{|\widehat{v}|}^2+\frac{2\epsilon }{1+\alpha }\frac{{\partial }_t{|\widehat{v}|}^{1+\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}$$

Hence

$${\displaystyle \frac{{\partial }_t(|\widehat{v}{{|}^{1+\alpha }+\epsilon )}^2}{(1+\alpha )(|\widehat{v}{|}^{1+\alpha }+\epsilon )}}-{\displaystyle \frac{2\epsilon }{1+\alpha }}{\displaystyle \frac{{\partial }_t{|\widehat{v}|}^{1+\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}+{2\mu |\xi |}^2{\displaystyle \frac{|\widehat{v}{|}^{2+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}\le 2|\widehat{u_1·\nabla v}|{\displaystyle \frac{|\widehat{v}{|}^{1+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}+2|\widehat{v·\nabla u_2}|{\displaystyle \frac{|\widehat{v}{|}^{1+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}.$$

By integrating with respect to t, we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{1+\alpha }}|\widehat{v}{|}^{1+\alpha }-{\displaystyle \frac{\epsilon }{1+\alpha }}ln\left({\displaystyle \frac{|\widehat{v}{|}^{1+\alpha }+\epsilon }{\epsilon }}\right)+\mu {|\xi |}^2{\int }_0^t{\displaystyle \frac{|\widehat{v}{|}^{2+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}d\tau & \le & {\int }_0^t|\widehat{u_1·\nabla v}|{\displaystyle \frac{|\widehat{v}{|}^{1+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}d\tau \hfill \\ & & +{\int }_0^t|\widehat{v·\nabla u_2}|{\displaystyle \frac{|\widehat{v}{|}^{1+2\alpha }}{|\widehat{v}{|}^{1+\alpha }+\epsilon }}d\tau .\hfill \end{array}$$

Taking the limits as $\epsilon ,\alpha \to 0^{+}$, we get

$$|\widehat{v}{|+\mu |\xi |}^2{\int }_0^t|\widehat{v}|d\tau \le {\int }_0^t|\widehat{u_1·\nabla v}|d\tau +{\int }_0^t|\widehat{v·\nabla u_2}|d\tau .$$

By multiplying by ${|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}$ and integrating with respect to $\xi$, we obtain

$$\begin{array}{cc}\hfill {\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau & \le {\int }_0^t\parallel u_1{·\nabla v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau +{\int }_0^t{\parallel v·\nabla u_2\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau \hfill \\ \hfill & \le {\int }_0^t\parallel u_1{\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau +{\int }_0^t{\parallel u_2\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau .\hfill \end{array}$$

(16)

On the other hand, using inequality (5) and Hölder’s inequality, we obtain

$$\begin{array}{ccc}\hfill \parallel u_1{\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}& \le & \parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}\hfill \\ & & +{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{\rho +2}{2}}}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{-\rho }{2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}\hfill \\ & \le & C_{\rho ,\mu }\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{\rho +2}}}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{\rho +2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\displaystyle \frac{\mu }{4}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}\hfill \\ & & +{\displaystyle \frac{1}{\mu }}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{\rho +2}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{-\rho }+{\displaystyle \frac{\mu }{4}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}.\hfill \end{array}$$

Similarly, we have

$$\parallel u_2{\otimes v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}\le C_{\rho ,\mu }\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{\rho +2}}}\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{\rho +2}}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\displaystyle \frac{\mu }{4}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}+{\displaystyle \frac{1}{\mu }}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{\rho +2}\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{-\rho }+{\displaystyle \frac{\mu }{4}}{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}.$$

Using these estimates in inequality (16), we get

$$\begin{array}{ccc}\hfill {\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}& \le & C_{\rho ,\mu }{\int }_0^t{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\left(\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{\rho +2}}}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{\rho +2}}}+{\displaystyle \frac{1}{\mu }}\parallel u_1{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{\rho +2}{\parallel u_1\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{-\rho }\right)d\tau .\hfill \\ & & +C_{\rho ,\mu }{\int }_0^t{\parallel v\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\left(\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{\rho +2}}}\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{\rho +2}}}+{\displaystyle \frac{1}{\mu }}\parallel u_2{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{\rho +2}{\parallel u_2\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{-\rho }\right)d\tau .\hfill \end{array}$$

Since $t\mapsto \parallel u_i{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{\parallel u_i\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}\in L^1([0,T])$ for $i=1,2$, applying Gronwall’s lemma allows us to conclude that $v=0$ on $[0,T]$ and hence we prove uniqueness.

4. Proof of Theorem 2

The proof of the Theorem is divided into two main steps.

• Step 1: Establishing the existence of a global solution for small initial data in homogeneous Lei–Lin–Gevrey space ${\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$.

We aim to prove the following result:

Proposition 1. 

Let $\rho \in (-1,0)$, $\gamma \in (0,1)$ and let the initial data $u^0\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$ satisfy $\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}({\mathbb{R}}^3)<{\displaystyle \frac{\mu }{C_{a,\gamma }{2}^{\rho +2}}}$. Then there exists a unique global solution $u\in \mathcal{C}({\mathbb{R}}^{+},{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L^1({\mathbb{R}}^{+},{\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$ to the Navier–Stokes system (1), such that

$${\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\displaystyle \frac{(\mu -C_{a,\gamma }{2}^{\rho +2}){\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}}{2}}{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le {\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}.$$

Proof. 

Let $u\in \mathcal{C}([0,T^{*}),{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L_{loc}^1([0,T^{*}),{\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$ be the maximal solution of the system (1), as established in Theorem 1 with initial data $u^0\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$.

Suppose that the initial norm satisfies $\parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}<{\displaystyle \frac{\mu }{C_{a,\gamma }{2}^{\rho +1}}}$. Applying the Fourier transform with respect to the spatial variable to the Navier–Stokes equations and multiplying the result by $\overline{\widehat{u}}$, we obtain the following:

$${\partial }_t\widehat{u}\overline{\widehat{u}}(t,\xi )+\mu {|\xi |}^2\widehat{u}\overline{\widehat{u}}(t,\xi )+\widehat{(u·\nabla u)}\overline{\widehat{u}}(t,\xi )=0.$$

Taking the real part yields the following:

$${\partial }_t|\widehat{u}{(t,\xi )|}^2+{2\mu |\xi |}^2{|\widehat{u}(t,\xi )|}^2+2Re\left(\widehat{(u·\nabla u)}\overline{\widehat{u}}(t,\xi )\right)=0,$$

and thus we obtain the following:

$${\partial }_t|\widehat{u}{(t,\xi )|}^2+{2\mu |\xi |}^2{|\widehat{u}(t,\xi )|}^2\le 2|\widehat{(u·\nabla u)}(t,\xi )||\widehat{u}(t,\xi )|.$$

For $\alpha >0$ and $\epsilon >0$, we have

$${\displaystyle |\widehat{u}{(t,\xi )|}^{2\epsilon }{\partial }_t|\widehat{u}{(t,\xi )|}^2=\frac{1}{1+\epsilon }{\partial }_t{|\widehat{u}(t,\xi )|}^{2+2\epsilon }}$$

and

$${\displaystyle \frac{{\partial }_t(|\widehat{u}(t,\xi ){{|}^{1+\epsilon }+\alpha )}^2}{(1+\epsilon )(|\widehat{u}(t,\xi ){|}^{1+\epsilon }+\alpha )}=\frac{|\widehat{u}{(t,\xi )|}^{2\epsilon }}{|\widehat{u}{(t,\xi )|}^{1+\epsilon }+\alpha }{\partial }_t{|\widehat{u}(t,\xi )|}^2+\frac{2\alpha }{1+\epsilon }\frac{{\partial }_t{|\widehat{u}(t,\xi )|}^{1+\epsilon }}{|\widehat{u}{(t,\xi )|}^{1+\epsilon }+\alpha }.}$$

We deduce after integration with respect to t that

$$\begin{array}{ccc}& & |\widehat{u}{(t,\xi )|}^{1+\epsilon }-\alpha ln\left({\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{1+\epsilon }+\alpha }{|\widehat{u}{(0,\xi )|}^{1+\epsilon }+\alpha }}\right)+\mu (1+\epsilon ){|\xi |}^2{\int }_0^t{\displaystyle \frac{|\widehat{u}{(\tau ,\xi )|}^{2+2\epsilon }}{|\widehat{u}{(\tau ,\xi )|}^{1+\epsilon }+\alpha }}d\tau \hfill \\ & & \le |\widehat{u}{(0,\xi )|}^{1+\epsilon }+(1+\epsilon ){\int }_0^t|\widehat{u·\nabla u}(\tau ,\xi )|{\displaystyle \frac{|\widehat{u}{(\tau ,\xi )|}^{1+2\epsilon }}{|\widehat{u}{(\tau ,\xi )|}^{1+\epsilon }+\epsilon }}d\tau .\hfill \end{array}$$

Taking the limits as $\epsilon ,\alpha \to 0^{+}$, we get

$$|\widehat{u}(t,\xi )|+\mu {\int }_0^t{|\xi |}^2|\widehat{u}(\tau ,\xi )|d\tau \le |\widehat{u}(0,\xi )|+{\int }_0^t|\widehat{u·\nabla u})(\tau ,\xi )|d\tau .$$

By multiplying by ${|\xi |}^{\rho }{e}^{{a|\xi |}^{\gamma }}$ and integrating with respect to variable space $\xi$, we obtain

$${\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\int }_0^t{\parallel div(u\otimes u)(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau .$$

Lemma 6 gives

$$\begin{array}{cc}\hfill {\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau & \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\int }_0^t{\parallel (u\otimes u)(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau \hfill \\ \hfill & \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+2^{\rho +2}{C}_{a,\gamma }{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau .\hfill \end{array}$$

Let $\alpha ={\displaystyle \frac{\mu +2^{\rho +2}{C}_{a,\gamma }{\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}}{2}}$ and $T_{*}=sup\{t\in [0,T^{*}):\parallel u(·,t){\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}<\alpha \}$. For $t\in [0,T_{*})$, we have

$${\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\alpha {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau .$$

Thus,

$${\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+(\mu -\alpha ){\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le {\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}<\alpha .$$

This implies that $T^{*}=T_{*}$.

If $T<T^{*}$, then

$${\parallel u(·,T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+(\mu -\alpha ){\int }_0^T{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le {\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}.$$

Hence, $T^{*}=\infty$. □

• Step 2: We aim to prove Theorem 2.

Let $u\in \mathcal{C}([0,T_{a,\gamma }^{*}),{\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3))\cap L_{loc}^1([0,T_{a,\gamma }^{*}),{\mathcal{X}}_{a,\gamma }^{\rho +2}({\mathbb{R}}^3))$ be the maximal solution to the system (1), as provided by Theorem 1, corresponding to the initial data $u^0\in {\mathcal{X}}_{a,\gamma }^{\rho }({\mathbb{R}}^3)$. Suppose that $\parallel u^0{\parallel }_{{\mathcal{X}}^{\rho }}<\mu$ for $\rho \in (-1,0)$. Then,

$$\begin{array}{cc}\hfill {\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau & \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\int }_0^t{\parallel \mathrm{div}(u\otimes u)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau \hfill \\ \hfill & \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\int }_0^t{\parallel u\otimes u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +1}}d\tau .\hfill \end{array}$$

Using Lemma 7, we obtain

$${\parallel u(·,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(·,\tau )\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+2^{\rho +2}{C}_{\gamma }{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}^{{\displaystyle \frac{1}{2}}}d\tau .$$

Using the classical Cauchy–Schwarz inequality, we obtain the following:

$${\parallel u(.,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_0^t{\parallel u(.,\tau )\parallel }_{{\mathcal{X}}_{a,\sigma }^{\rho +2}}d\tau \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+2^{\rho +3}{\mu }^{-1}{C}_{\gamma }{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }}^2{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau +{\displaystyle \frac{\mu }{2}}{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau ,$$

which simplifies to

$${\parallel u(.,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\displaystyle \frac{\mu }{2}}{\int }_0^t{\parallel u(.,\tau )\parallel }_{{\mathcal{X}}_{a,\sigma }^{\rho +2}}d\tau \le \parallel u^0{\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+2^{\rho +3}{\mu }^{-1}{C}_{\gamma }{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }}^2{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau .$$

Applying Gronwall’s lemma, we derive

$${\parallel u(.,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\le {\parallel u^0\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{e}^{2^{\rho +3}{\mu }^{-1}{C}_{\gamma }{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }}^{2}d\tau }.$$

(17)

Since $0<a\sqrt{\gamma }<a$, it follows from the embedding property of the homogeneous Lei–Lin–Gevrey spaces that

$${\mathcal{X}}_{a,\gamma }^{\rho }\hookrightarrow {\mathcal{X}}_{a\sqrt{\gamma },\gamma }^{\rho }.$$

Therefore, we conclude

$$T_{a,\gamma }^{*}=T_{a\sqrt{\gamma },\gamma }^{*}.$$

Inductively, we obtain for all $n\in \mathbb{N}$,

$$T_{a,\gamma }^{*}=T_{a\sqrt{\gamma },\gamma }^{*}=\dots =T_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{*},$$

and by the Dominated Convergence Theorem, we get

$$\underset{n\to \infty }{lim}\parallel u^0{\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{\rho }}={\parallel u^0\parallel }_{{\mathcal{X}}^{\rho }}<\mu .$$

Thus, there exists an integer $n_0\in \mathbb{N}$ such that for all $n\ge n_0$,

$$\parallel u^0{\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{\rho }}<\mu .$$

Applying Proposition 1 (Step 1), we have for all $n\ge n_0$:

$${\parallel u(t)\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{\rho }}+{\displaystyle \frac{(\mu -C_{a,\gamma }{2}^{\rho +1}){\parallel u^0\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{\frac{n}{2}},\gamma }^{\rho }}}{2}}{\int }_0^t{\parallel u\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{\rho +2}}d\tau \le {\parallel u^0\parallel }_{{\mathcal{X}}_{a{(\gamma )}^{{\displaystyle \frac{n}{2}}},\gamma }^{\rho }}.$$

(18)

Now, assuming $T_{a,\gamma }^{*}$ is finite, we aim to show that ${\int }_0^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau =\infty$.

Assume, by contradiction, that ${\int }_0^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau <\infty$. For any $\epsilon >0$, there exists a time $T\in (0,T_{a,\gamma }^{*})$ such that

$${\int }_T^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau <\epsilon .$$

(19)

For $T<t<T_{a,\gamma }^{*}$, we have

$${\parallel u(.,t)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\mu {\int }_T^t{\parallel u(.,\tau )\parallel }_{{\mathcal{X}}_{a,\sigma }^{\rho +2}}d\tau \le {\parallel u(T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\int }_T^t{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho +2}}d\tau .$$

Letting $M(t)={sup}_{T\le z\le t}{\parallel u(z)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}$, we obtain from (19) that

$$M(t)\le {\parallel u(T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+\epsilon {\int }_T^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}d\tau .$$

From inequalities (17) and (18), we find

$$M(t)\le {\parallel u(T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}+{\epsilon \parallel u(T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}{e}^{2^{\rho +3}\mu C_{\gamma }}\le {\parallel u(T)\parallel }_{{\mathcal{X}}_{a,\gamma }^{\rho }}\left(1+\epsilon e^{2^{\rho +3}\mu C_{\gamma }}\right)=M_T.$$

Thus, $M_t\le M_T$, which is a contradiction. Therefore, $T_{a,\gamma }^{*}=\infty$.

5. Discussion and Perspectives

The results obtained in this paper confirm the effectiveness of Lei–Lin–Gevrey spaces in establishing existence and uniqueness of global solutions to the three-dimensional incompressible Navier–Stokes equations under small initial data. These spaces allow one to combine the scaling properties of critical spaces with an exponential weight in frequency, enabling analytic smoothing effects to be tracked more precisely.

The Gevrey parameter $\gamma \in (0,1)$ plays a central role: it controls the degree of analyticity imposed on the initial data and governs the decay of high-frequency components in the solution. When $\gamma$ is close to 1, the exponential weight enforces stronger analyticity, whereas values closer to 0 retain properties closer to classical Lei–Lin spaces. This interpolation between low and high regularity regimes provides a flexible framework for analyzing the nonlinear structure of the Navier–Stokes equations in the Fourier domain.

One limitation of the present approach is that it relies heavily on the structure of ${\mathbb{R}}^3$ and the associated Fourier analysis. Extending this method to periodic settings such as ${\mathbb{T}}^3$, where the frequency spectrum is discrete, presents technical challenges. While similar analytic smoothing has been investigated in periodic domains (see, e.g., [11,12]), adapting the Lei–Lin–Gevrey framework to such settings would require a refined analysis of the interaction between the exponential weight and the discrete Laplacian.

Furthermore, the structure of the estimates derived here suggests that similar techniques could apply to other dissipative systems exhibiting parabolic behavior. Examples include magnetohydrodynamic (MHD) flows (see [13]), rotating fluids, or systems with anisotropic diffusion. In these contexts, the use of analytically weighted spaces could facilitate better control of nonlinear terms and provide insight into stability and long-time behavior.

These observations open promising directions for future work, particularly in the extension of the Gevrey approach to systems with additional physical complexity or constrained geometry.

6. Conclusions

We have established the existence and uniqueness of both local and global solutions for the three-dimensional incompressible Navier–Stokes equations within the homogeneous Lei–Lin–Gevrey framework ${\mathcal{X}}_{a,\gamma }^{\rho }$ under small initial data in the critical space ${\mathcal{X}}^{\rho }$.

Our analysis leverages the interplay between scaling invariance and analytic smoothing, offering sharper decay rates and a more transparent treatment of nonlinear interactions. These results improve upon prior approaches by introducing a parameter-dependent analytic structure that bridges Sobolev, Gevrey, and Lei–Lin theories.

The methods developed here pave the way for future studies in periodic domains or bounded geometries, where adaptation to compactness methods or spectral decompositions is necessary. Another promising direction involves applying this framework to magnetohydrodynamic systems, rotating fluids, or anisotropic models exhibiting similar scaling behavior.