Symmetry and Blow-Up for 3D Modified CBF Equations in Sobolev–Gevrey Spaces

Abstract

We study the three-dimensional modified critical homogeneous convective Brinkman–Forchheimer equations. Local existence and uniqueness are obtained via the Fixed Point Theorem, and under the assumption of finite maximal existence time, we establish blow-up criteria in Sobolev–Gevrey spaces. The model preserves Euclidean invariances (translations and rotations) and, when $\alpha =0$, the critical scaling symmetry of the $3D$ Navier-Stokes system. The blow-up thresholds are shown to depend mainly on the Gevrey structure rather than on any loss of symmetry.

1. Introduction

The critical homogeneous convective Brinkman–Forchheimer system in three spatial dimensions is described by the following equations (refer to [1,2]):

$$\left\{\begin{array}{cc}{\partial }_{t}u-\mu \Delta u+u·\nabla u+\alpha u+\beta {\left|u\right|}^{2}u+\nabla p=0& \mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3\\ \mathrm{div}u=0,& \mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3\\ u(0,x)=u_0\left(x\right)& \mathrm{in}{\mathbb{R}}^3,\end{array}\right.$$

In this formulation, $u(t,x)$ denotes the velocity field at time $t>0$ and position $x\in {\mathbb{R}}^3$, while $p(t,x)$ represents the associated pressure field. The parameters $\mu ,\alpha ,\beta >0$ correspond to the effective viscosity (Brinkman term), the permeability of the porous medium (Darcy term), and the nonlinear drag (Forchheimer term), respectively. The model preserves the basic Euclidean invariances (translations and rotations), and when $\alpha =0$, it shares the critical scaling of the $3D$ Navier-Stokes system [1,3,4].

In recent years, many authors have proposed modifications to the classical form of the convective Brinkman–Forchheimer equations in three dimensions; see, for instance, [5,6].

The present study explores an alternative formulation by replacing the nonlinear term ${\left|u\right|}^{2}u$ with the component-wise cubic term ${\sum }_{k=1}^3{u}_k^3{e}_k$. For a more detailed discussion, the reader may consult [7,8]. The modified system we investigate is expressed as follows:

$$\left\{\begin{array}{cccc}& {\partial }_{t}u-\mu \Delta u+u·\nabla u+\alpha u+\beta {\displaystyle \sum _{k=1}^3}{u}_k^3{e}_k+\nabla p=0,& & \mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3,\\ & \mathrm{div}u=0,& & \mathrm{in}{\mathbb{R}}^{+}\times {\mathbb{R}}^3,\\ & u(0,x)=u_0\left(x\right)& & \mathrm{in}{\mathbb{R}}^3.\end{array}\right.$$

(1)

Here, $(e_1,e_2,e_3)$ denotes the canonical basis of ${\mathbb{R}}^3$. This change in the term is technically necessary, and it does not affect the scaling of the equation nor the different properties of such a solution. The component-wise cubic nonlinearity also respects coordinate-permutation and parity symmetries, so the geometric invariances of the flow are retained.

In this work, we study the blow-up criterion of solutions. In recent years, several authors studied this phenomenon of many fluid mechanics equations in various spaces. In the case of a three-dimensional magnetohydrodynamic system $\left(MHD\right)$, Benameur and Selmi in [9] established the blow-up result in the Lei–Lin–Gevrey spaces ${\mathcal{X}}_{a,\sigma }^{-1}\left({\mathbb{R}}^3\right)$. Also, in [10] the authors give a lower bound of solutions in Sobolev–Gevrey spaces. While, we refer to [11,12] and the references therein for the result related to Navier–Stokes equations $\left(NSE\right)$.

Although many results exist on the Navier–Stokes and classical Brinkman–Forchheimer equations, the influence of Gevrey regularity on symmetry-preserving blow-up remains unclear. This paper fills this gap by establishing new blow-up criteria in Sobolev–Gevrey spaces and showing how analytic regularity controls singularity formation.

The present system differs from the classical convective Brinkman–Forchheimer and Navier–Stokes equations by replacing the nonlinear term ${\left|u\right|}^{2}u$ with the component-wise cubic term ${\sum }_{k=1}^3{u}_k^3{e}_k$, which preserves symmetry but modifies the nonlinear structure. This change, together with the use of Sobolev–Gevrey spaces, allows a finer analysis of blow-up behavior and extends previous results obtained for the Navier–Stokes and $\left(MHD\right)$ systems.

The result that concerns us is that in [13] Benameur answered this question: Do the blow-up phenomena depend on the chosen space or on the nonlinear part $(u·\nabla u)$ in the case of Navier–Stokes equations $\left(NSE\right)$? To answer this question, the author gives an exponential type of blow-up profile in Sobolev–Gevrey spaces $H_{a,\sigma }^s\left({\mathbb{R}}^3\right)$ with $s>{\displaystyle \frac{3}{2}}$ and proves that the blow-up phenomena depends on the chosen space. Also, in [14] the authors improve this result with less regularity on the initial data $s=1$. To ameliorate these previous results, the author in [8] studied this type of explosion in the case of critical homogeneous convective Brinkman–Forchheimer equations and in the Sobolev–Gevrey spaces $H_{a,\sigma }^s\left({\mathbb{R}}^3\right)$ with $s>{\displaystyle \frac{3}{2}}$, where the nonlinear parts of this equation are given by $u·\nabla u+{\beta \left|u\right|}^{2}u$.

Inspired by these works [8,13,14] in this present work, we focus on the previous question. Precisely, we study the blow-up phenomena in the case of three-dimensional modified homogeneous convective Brinkman–Forchheimer equations and in the Sobolev–Gevrey spaces $H_{a,\sigma }^s\left({\mathbb{R}}^3\right)$ with Sobolev index $s=1$.

The convective Brinkman–Forchheimer model is widely used to describe viscous flows through porous media, such as petroleum extraction, groundwater filtration, and biological tissues. Understanding the blow-up and stability mechanisms in such models provides valuable insight into how viscosity, damping, and nonlinear convection interact in complex porous flows. The present study refines previous mathematical results by showing that singularity formation depends mainly on the Gevrey regularity rather than on symmetry breaking. This clarification contributes to the theoretical understanding of nonlinear fluid dynamics and highlights the importance of analytic regularity in predicting finite-time explosion in physical systems.

Before presenting our main findings, we begin by defining the non-homogeneous Sobolev–Gevrey spaces. Given parameters $a>0, \gamma \in (0,1)$, and $s\in \mathbb{R}$, we consider the space

$$H_{a,\gamma }^s\left({\mathbb{R}}^3\right):=\left\{u\in L^2\left({\mathbb{R}}^3\right)|{\left(1+{\left|\xi \right|}^2\right)}^{s/2}{e}^{{a\left|\xi \right|}^{\gamma }}\widehat{u}\left(\xi \right)\in L^2\left({\mathbb{R}}^3\right)\right\},$$

which is equipped with the norm

$${\parallel u\parallel }_{H_{a,\gamma }^s\left({\mathbb{R}}^3\right)}:={\left({\int }_{{\mathbb{R}}^3}{\left(1+{\left|\xi \right|}^2\right)}^s{e}^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}\left(\xi \right)\right|}^{2}d\xi \right)}^{1/2}.$$

The corresponding inner product on this space is defined by

$${\langle u,v\rangle }_{H_{a,\gamma }^s\left({\mathbb{R}}^3\right)}:={\langle e^{{a\left|D\right|}^{\gamma }}u,e^{{a\left|D\right|}^{\gamma }}v\rangle }_{H^s\left({\mathbb{R}}^3\right)},$$

where the operator $\left|D\right|={(-\Delta )}^{1/2}$ denotes the square root of the negative Laplacian, and it is defined by the Fourier transform

$$\widehat{{(-\Delta )}^{1/2}u}\left(\xi \right)=\left|\xi \right|\widehat{u}\left(\xi \right).$$

Since $e^{{a\left|\xi \right|}^{\gamma }}$ is a radial multiplier, the $H_{a,\gamma }^s$ norm is rotation-invariant; this symmetry-compatibility is used in our estimates.

With these definitions in place, we now turn to the statement of our primary results. The first of these addresses the issue of local well-posedness and is stated as follows:

Theorem 1.

For $a>0$ and $\gamma \in (0,1)$. Suppose the initial data $u_0\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ satisfies the divergence-free condition $\mathrm{div}{u}_0=0$. Then, there exists a positive time $T>0$ such that Equation (1) admits a unique solution $u\in C([0,T],H_{a,\gamma }^1\left({\mathbb{R}}^3\right))$.

Now, let us suppose that the maximal time of existence $T_{a,\gamma }^{*}$, as given by Theorem 1, is finite. Under this assumption, we establish blow-up criteria for the solution $u\in C([0,T_{a,\gamma }^{*}),H_{a,\gamma }^1\left({\mathbb{R}}^3\right))$, which corresponds to the maximal solution of Equation (1). Our main result in this regard is stated in the following theorem:

Theorem 2.

Let $a>0$ and $\gamma \in (0,1)$. Consider $u\in C([0,T_{a,\gamma }^{*}),H_{a,\gamma }^1\left({\mathbb{R}}^3\right))$ as the maximal solution to Equation (1), as constructed in Theorem 1. If the maximal time $T_{a,\gamma }^{*}$ is finite, then the following assertions hold:

(i)

  $lim{\displaystyle \underset{t\nearrow T_{a,\gamma }^{*}}{sup}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=\infty .$

(ii)

  ${\int }_t^{T_{a,\gamma }^{*}}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1\left({\mathbb{R}}^3\right)}^{2}dz=\infty ,\forall t\in [0,T_{a,\gamma }^{*}).$

(iii)

  ${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2,\forall t\in [0,T_{a,\gamma }^{*}).$

(iv)

  ${\displaystyle \frac{C_3}{{(T_{a,\gamma }^{*}-t)}^{\frac{2({\gamma }_0+\gamma )+\gamma {\gamma }_0}{3{\gamma }_0}}}}exp\left({\displaystyle \frac{a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)\le {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}.$

Here, $C_2=C_2(\beta ,\mu ,\gamma ,u_0)>0$ and $C_3=C_3(\beta ,\mu ,a,\gamma ,{\gamma }_0,u_0)>0$. Moreover, ${\displaystyle \frac{2}{{\gamma }_0}}$ denotes the integer part of ${\displaystyle \frac{2}{\gamma }}$.

These blow-up criteria are formulated in a symmetry-aware manner and remain stable under translations and rotations.

The structure of the paper is as follows: In Section 2, we revisit some classical results and present new auxiliary lemmas. Section 3 is dedicated to establishing the local well-posedness of the system. In Section 4, we prove Theorem 2. The proof relies on embedding results for Sobolev spaces, Lemma (1), and an energy estimate from the literature (see [7,8]).

$${\parallel u\left(t\right)\parallel }_{L^2}^2+2\mu {\int }_0^t{\parallel \nabla u\parallel }_{L^2}^2+2\alpha {\int }_0^t{\parallel u\parallel }_{L^2}^2+{\displaystyle \frac{2}{3}}\beta {\int }_0^t{\parallel u\parallel }_{L^4}^4\le {\parallel u^0\parallel }_{L^2}^2,\forall t\ge 0.$$

(2)

Equation (2) is the standard energy estimate established in [7,8].

We emphasize that the thresholds we obtain are governed by the Gevrey regularity, rather than by any loss of isotropic symmetry.

2. Preliminary Results

Lemma 1

([13]). For $\sigma >{\displaystyle \frac{3}{2}}$, we have

$$\begin{array}{c}\hfill \parallel \widehat{u}{\parallel }_{L^1\left({\mathbb{R}}^3\right)}\le C_{\sigma }{\parallel u\parallel }_{L^2\left({\mathbb{R}}^3\right)}^{1-{\displaystyle \frac{3}{2\sigma }}}{\parallel u\parallel }_{{\dot{H}}^{\sigma }\left({\mathbb{R}}^3\right)}^{{\displaystyle \frac{3}{2\sigma }}},\end{array}$$

with $C_{\sigma }=2\sqrt{{\displaystyle \frac{\pi }{3}}}\left({({\displaystyle \frac{2\sigma }{3}}-1)}^{{\displaystyle \frac{3}{4\sigma }}}+{({\displaystyle \frac{2\sigma }{3}}-1)}^{-1+{\displaystyle \frac{3}{4\sigma }}}\right)$.

Moreover, for all ${\sigma }_0>{\displaystyle \frac{3}{2}}$, there is $B\left({\sigma }_0\right)>0$ such that

$$\begin{array}{c}\hfill C_{\sigma }\le B\left({\sigma }_0\right),\forall \sigma \ge {\sigma }_0.\end{array}$$

Before proceeding with the next lemma, we recall the definition of the homogeneous Sobolev–Gevrey space that will be frequently used in the following estimates:

Definition 1.

For $a>0$ and $\gamma \in (0,1)$, we denote by ${\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ the homogeneous Sobolev–Gevrey space equipped with the norm

$${\parallel u\parallel }_{{\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right)}={\left({\int }_{{\mathbb{R}}^3}{\left|\xi \right|}^2{e}^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}\left(\xi \right)\right|}^{2}d\xi \right)}^{1/2}.$$

Lemma 2.

Let $a>0,\gamma \in (0,1)$ and ${\gamma }^{\prime }={\displaystyle \frac{1+\gamma }{2}}$. Then, for any $u\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$, we have

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}& \le & C_{\gamma }{\parallel u\parallel }_{H_{a,\gamma }^1}\hfill \end{array}$$

(3)

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}& \le & C_{{\gamma }^{\prime }}{\parallel u\parallel }_{{\dot{H}}_{a{\gamma }^{\prime },\gamma }^1}.\hfill \end{array}$$

(4)

where $C_{\gamma }={\left[{\displaystyle \frac{4\pi }{\gamma {\left(2a(1-\gamma )\right)}^{\frac{1}{\gamma }}}}\Gamma \left({\displaystyle \frac{1}{\gamma }}\right)\right]}^{1/2}$ and $C_{{\gamma }^{\prime }}={\left[{\displaystyle \frac{4\pi }{\gamma {\left(2a({\gamma }^{\prime }-\gamma )\right)}^{\frac{1}{\gamma }}}}\Gamma \left({\displaystyle \frac{1}{\gamma }}\right)\right]}^{1/2}$.

Proof. 

From the Cauchy–Shwarz inequality, firstly, we have

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}& =& {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u}\left(\xi \right)\right|d\xi \hfill \\ & =& {\int }_{{\mathbb{R}}^3}\left({\displaystyle \frac{e^{{(a\gamma -a)\left|\xi \right|}^{\gamma }}}{\left|\xi \right|}}\right)\left(\left|\xi \right|e^{{a\left|\xi \right|}^{\gamma }}\left|\widehat{u}\left(\xi \right)\right|\right)d\xi \hfill \\ & \le & {\left({\int }_{{\mathbb{R}}^3}{\displaystyle \frac{e^{{2(a\gamma -a)\left|\xi \right|}^{\gamma }}}{{\left|\xi \right|}^2}}d\xi \right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^3}\left({\left|\xi \right|}^2{e}^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}\left(\xi \right)\right|}^2\right)d\xi \right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ & \le & C_{\gamma }{\parallel u\parallel }_{{\dot{H}}_{a,\gamma }^1}\hfill \\ & \le & C_{\gamma }{\parallel u\parallel }_{H_{a,\gamma }^1}.\hfill \end{array}$$

Secondly, we have

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}& =& {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u}\left(\xi \right)\right|d\xi \hfill \\ & =& {\int }_{{\mathbb{R}}^3}\left({\displaystyle \frac{e^{-a({\gamma }^{\prime }-\gamma ){\left|\xi \right|}^{\gamma }}}{\left|\xi \right|}}\right)\left(\left|\xi \right|e^{a{\gamma }^{\prime }{\left|\xi \right|}^{\gamma }}\left|\widehat{u}\left(\xi \right)\right|\right)d\xi \hfill \\ & \le & {\left({\int }_{{\mathbb{R}}^3}{\displaystyle \frac{e^{-2a({\gamma }^{\prime }-\gamma ){\left|\xi \right|}^{\gamma }}}{{\left|\xi \right|}^2}}d\xi \right)}^{1/2}{\left({\int }_{{\mathbb{R}}^3}{\left|\xi \right|}^2{e}^{2a{\gamma }^{\prime }{\left|\xi \right|}^{\gamma }}{\left|\widehat{u}\left(\xi \right)\right|}^{2}d\xi \right)}^{1/2}\hfill \\ & \le & C_{{\gamma }^{\prime }}{\parallel u\parallel }_{{\dot{H}}_{a{\gamma }^{\prime },\gamma }^1}.\hfill \end{array}$$

□

Lemma 3.

Let $a>0$ and $\gamma \in (0,1)$. Then, for any $u_i\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ with $i=1,2$, we have

$$\begin{array}{ccc}\hfill & & \prod _{i=1}^2{u}_i\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\hfill \end{array}$$

(5)

$$\begin{array}{c}\parallel \prod _{i=1}^2{u}_i{\parallel }_{H_{a,\gamma }^1}\le 2\sum _{k=1}^2\parallel u_k{\parallel }_{H_{a,\gamma }^1}\prod _{j\ne k}^2{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_j\right)\parallel }_{L^1}\hfill \end{array}$$

(6)

$$\begin{array}{cc}\parallel \prod _{i=1}^2{u}_i{\parallel }_{H_{a,\gamma }^1}\le \hfill & \hfill C_{\gamma }\prod _{i=1}^2{\parallel u_i\parallel }_{H_{a,\gamma }^1}.\end{array}$$

(7)

Proof. 

Firstly, we have

$$\begin{array}{ccc}\hfill \parallel \prod _{i=1}^2{u}_i{\parallel }_{H_{a,\gamma }^1}& =& {\int }_{{\mathbb{R}}^3}\left(1+{\left|\xi \right|}^2\right)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u_1{u}_2}\left(\xi \right)\right|}^{2}d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}\left(1+{\left|\xi \right|}^2\right)e^{{2a\left|\xi \right|}^{\gamma }}\left(\right|\widehat{u_1}|\ast |\widehat{u_2}{\left|\right)}^2\left(\xi \right)d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{\left({\int }_{\eta }{\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right|\widehat{u_2}\left(\eta \right)|d\eta \right)}^{2}d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}({\int }_{\left|\eta \right|<|\xi -\eta |}{\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right|\widehat{u_2}\left(\eta \right)|d\eta \hfill \\ & & +{\int }_{\left|\eta \right|>|\xi -\eta |}{\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right|\widehat{u_2}\left(\eta \right)|d\eta {)}^{2}d\xi \hfill \end{array}$$

Using the following inequalities [13]:

$$\begin{array}{ccc}& & {(1+|\xi |}^2{)}^{{\displaystyle \frac{1}{2}}}\le \sqrt{2}{\left[1+max{\left(|\xi -\eta |,\left|\eta \right|\right)}^2\right]}^{{\displaystyle \frac{1}{2}}}\hfill \\ & & e^{{a\left|\xi \right|}^{\gamma }}\le e^{amax\left(\right|\xi -{\eta |,|\eta \left|\right)}^{\gamma }}{e}^{a\gamma min\left(\right|\xi -{\eta |,|\eta \left|\right)}^{\gamma }},\hfill \end{array}$$

we obtain

$$\begin{array}{ccc}& & \parallel \prod _{i=1}^2{u}_i{\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le {\int }_{{\mathbb{R}}^3}(\sqrt{2}{\int }_{\left|\eta \right|<|\xi -\eta |}{(1+|\xi -\eta |}^2{)}^{{\displaystyle \frac{1}{2}}}{e}^{{a|\xi -\eta |}^{\gamma }}|\widehat{u_1}(\xi -\eta )|e^{{a\gamma \left|\eta \right|}^{\gamma }}\left|\widehat{u_2}\left(\eta \right)\right|d\eta \hfill \\ & & +\sqrt{2}{\int }_{\left|\eta \right|>|\xi -\eta |}{e}^{{a\gamma |\xi -\eta |}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right(1+{\left|\eta \right|}^2{)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\eta \right|}^{\gamma }}\left|\widehat{u_2}\left(\eta \right)\right|d\eta {)}^{2}d\xi \hfill \\ & & \le 2{\int }_{{\mathbb{R}}^3}{\left({\int }_{\left|\eta \right|<|\xi -\eta |}{(1+|\xi -\eta |}^2{)}^{{\displaystyle \frac{1}{2}}}{e}^{{a|\xi -\eta |}^{\gamma }}|\widehat{u_1}(\xi -\eta )|e^{{a\gamma \left|\eta \right|}^{\gamma }}\left|\widehat{u_2}\left(\eta \right)\right|d\eta \right)}^{2}d\xi \hfill \\ & & +2{\int }_{{\mathbb{R}}^3}{\left({\int }_{\left|\eta \right|>|\xi -\eta |}{e}^{{a\gamma |\xi -\eta |}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right(1+{\left|\eta \right|}^2{)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\eta \right|}^{\gamma }}\left|\widehat{u_2}\left(\eta \right)\right|d\eta \right)}^{2}d\xi \hfill \\ & & \le 2\parallel \left({\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}\left|\widehat{u_1}\left(\xi \right)\right|\right)\ast \left(e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_2}\left(\xi \right)\right|\right){\parallel }_{L^2}^2\hfill \\ & & +2\parallel \left({\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}\left|\widehat{u_2}\left(\xi \right)\right|\right)\ast \left(e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_1}\left(\xi \right)\right|\right){\parallel }_{L^2}^2\hfill \\ & & \le 2\parallel {\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}|\widehat{u_2}\left(\xi \right)|{\parallel }_{L^2}^2\parallel e^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u_2}\left(\xi \right)|{\parallel }_{L^1}^2+2\parallel {\left(1+{\left|\xi \right|}^2\right)}^{{\displaystyle \frac{1}{2}}}{e}^{{a\left|\xi \right|}^{\gamma }}|\widehat{u_2}\left(\xi \right)|{\parallel }_{L^2}^2\parallel e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_1}\left(\xi \right)\right|{\parallel }_{L^1}^2\hfill \\ & & \le 2\parallel u_2{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right){\parallel }_{L^1}+2\parallel u_1{\parallel }_{H_{a,\gamma }^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right)\parallel }_{L^1}.\hfill \end{array}$$

Then, Equations (5) and (6) are proved.

Secondly, combining Equations (6) and (3), we can deduce that Equation (7) and the proof of Lemma 3 is finished. □

Lemma 4.

Let $a>0$ and $\gamma \in (0,1)$. Then, for any $u_i\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ with $i=1,2$, we have

$$\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}\left(u_1{u}_2\right)\right){\parallel }_{L^1}\le \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right)\parallel }_{L^1}.$$

Proof. 

From the elementary inequality

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

(8)

we have

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{\gamma |.|}^{\gamma }}\left(u_1{u}_2\right)\right){\parallel }_{L^1}& =& {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_1{u}_2}\left(\xi \right)\right|d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left(|\widehat{u_1}|\ast |\widehat{u_2}|\left(\xi \right)\right)d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}\left({\int }_{{\mathbb{R}}^3}{e}^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u_1}(\xi -\eta )\left|\right|\widehat{u_2}\left(\eta \right)|d\eta \right)d\xi \hfill \\ & \le & \parallel \left(e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_1}\left(\xi \right)\right|\right)\ast \left(e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u_2}\left(\xi \right)\right|\right){\parallel }_{L^1}.\hfill \end{array}$$

Therefore, Young’s inequality implies

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{\gamma |.|}^{\gamma }}\left(u_1{u}_2\right)\right){\parallel }_{L^1}& \le & \parallel e^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u_1}{\left(\xi \right)|\parallel }_{L^1}\parallel e^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u_2}\left(\xi \right){|\parallel }_{L^1}\hfill \\ & \le & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right)\parallel }_{L^1}.\hfill \end{array}$$

□

Lemma 5.

Let $a>0$ and $\gamma \in (0,1)$. Then, for any $u_i\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ with $i=1,2,3$, we have

$$\begin{array}{ccc}\hfill & & \prod _{i=1}^3{u}_i\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill & & \parallel \prod _{i=1}^3{u}_i{\parallel }_{H_{a,\gamma }^1}\le 4\sum _{k=1}^3\parallel u_k{\parallel }_{H_{a,\gamma }^1}\prod _{j\ne k}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_j\right)\parallel }_{L^1}\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill & & \parallel \prod _{i=1}^3{u}_i{\parallel }_{H_{a,\gamma }^1}\le C_{\gamma }\prod _{i=1}^3{\parallel u_i\parallel }_{H_{a,\gamma }^1}.\hfill \end{array}$$

(11)

Proof. 

Firstly, using Equation (6) twice and Lemma 4 we get

$$\begin{array}{ccc}\hfill \parallel \prod _{i=1}^3{u}_i{\parallel }_{H_{a,\gamma }^1}& & =\parallel u_1\left(u_2{u}_3\right){\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le 2\parallel \left(u_2{u}_3\right){\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right){\parallel }_{L^1}+2\parallel u_1{\parallel }_{H_{a,\gamma }^1}{\parallel \mathcal{F}\left(e^{{\gamma |.|}^{\gamma }}\left(u_2{u}_3\right)\right)\parallel }_{L^1}\hfill \\ & & \le 2\left[2\parallel u_3{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right){\parallel }_{L^1}+2\parallel u_2{\parallel }_{H_{a,\gamma }^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_3\right)\parallel }_{L^1}\right]{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right)\parallel }_{L^1}\hfill \\ & & +2\parallel u_1{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_3\right)\parallel }_{L^1}\hfill \\ & & \le 4\parallel u_3{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right)\parallel }_{L^1}\hfill \\ & & +4\parallel u_2{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_3\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_1\right)\parallel }_{L^1}\hfill \\ & & +4\parallel u_1{\parallel }_{H_{a,\gamma }^1}\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_2\right){\parallel }_{L^1}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}{u}_3\right)\parallel }_{L^1}.\hfill \end{array}$$

Then, Equations (9) and (10) are proved.

Secondly, combining Equations (10) and (3), we can deduce that Equation (11) and the proof of Lemma 5 is finished. □

Lemma 6.

Let $a>0$ and $\gamma \in (0,1)$. Then, for any $u_i\in L^{\infty }\left([0,T],H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\right)$ with $i=1,2,3$, we have

$$\begin{array}{ccc}\hfill \parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(u_1·\nabla u_2)dz{\parallel }_{H_{a,\gamma }^1}& \le & C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_1{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}{\parallel u_2\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\hfill \\ \hfill \parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}\left[u_1{u}_2{u}_3\right]dz{\parallel }_{H_{a,\gamma }^1}& \le & C_{\gamma }T\parallel u_1{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\parallel u_2{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}{\parallel u_3\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}.\hfill \end{array}$$

Proof. 

Firstly, using Equation (7), we get

$$\begin{array}{ccc}\hfill \parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(u_1·\nabla u_2)dz{\parallel }_{H_{a,\gamma }^1}& \le & {\int }_0^t\parallel e^{\mu (t-z)\Delta }\mathrm{div}(u_1\otimes u_2){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_0^t{\left({\int }_{{\mathbb{R}}^3}{\left|\xi \right|}^2{e}^{-{2\mu (t-z)\left|\xi \right|}^2}\left(1+{\left|\xi \right|}^2\right)e^{{2a\left|\xi \right|}^{\gamma }}|\widehat{(u_1\otimes u_2}{\left)(z,\xi )\right|}^{2}d\xi \right)}^{1/2}dz\hfill \\ & \le & {\int }_0^t{\displaystyle \frac{c}{\sqrt{\mu (t-z)}}}{\parallel u_1\otimes u_2\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_0^t{\displaystyle \frac{C_{\gamma }}{\sqrt{\mu (t-z)}}}\parallel u_1{\parallel }_{H_{a,\gamma }^1}{\parallel u_2\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_1{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}{\parallel u_2\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}.\hfill \end{array}$$

Secondly, using Equation (11), we obtain

$$\begin{array}{ccc}\hfill \parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}\left(u_1{u}_2{u}_3\right)dz{\parallel }_{H_{a,\gamma }^1}& \le & {\int }_0^t\parallel e^{\mu (t-z)\Delta }\left(u_1{u}_2{u}_3\right){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_0^t{\left({\int }_{{\mathbb{R}}^3}{e}^{-{2\mu (t-z)\left|\xi \right|}^2}\left(1+{\left|\xi \right|}^2\right)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\mathcal{F}\left(u_1{u}_2{u}_3\right)(z,\xi )\right|}^{2}d\xi \right)}^{1/2}dz\hfill \\ & \le & {\int }_0^t{\left({\int }_{{\mathbb{R}}^3}\left(1+{\left|\xi \right|}^2\right)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\mathcal{F}\left(u_1{u}_2{u}_3\right)(z,\xi )\right|}^{2}d\xi \right)}^{1/2}dz\hfill \\ & \le & {\int }_0^t{\parallel u_1{u}_2{u}_3\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & C_{\gamma }T\parallel u_1{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\parallel u_2{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}{\parallel u_3\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}.\hfill \end{array}$$

□

3. Proof of Theorem 1

The proof relies on the Fixed Point Theorem applied to a suitable operator in the Sobolev–Gevrey space $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$. We establish uniform estimates for the nonlinear terms and show that the operator is a contraction for sufficiently small $T>0$.

According to the Duhamel formula, the integral form of Equation (1) is given by

$$u\left(t\right)=e^{\mu t\Delta }{u}_0+\mathcal{A}\left(u\right)+\mathcal{B}(u,u)+\mathcal{C}\left(u\right)$$

where

$$\begin{array}{ccc}& & e^{\mu t\Delta }{u}_0={\mathcal{F}}^{-1}\left(\xi \to e^{\mu t\Delta }\widehat{u_0}\left(\xi \right)\right)\hfill \\ & & \mathcal{A}\left(u\right)=-\alpha {\int }_0^t{e}^{\mu (t-z)\Delta }udz\hfill \\ & & \mathcal{B}(u,u)=-{\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(u·\nabla u)dz\hfill \\ & & \mathcal{C}\left(u\right)=-\beta \sum _{j=1}^3{\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}{u}_j^3{e}_{j}dz.\hfill \end{array}$$

For a given $T>0$, we introduce the operator

$$\begin{array}{ccc}\hfill \mathcal{O}:& & {\mathcal{S}}_T⟶C_T\left(H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\right)\hfill \\ & & u⟼\mathcal{O}\left(u\right)=e^{\mu t\Delta }{u}_0+\mathcal{A}\left(u\right)+\mathcal{B}(u,u)+\mathcal{C}\left(u\right)\hfill \end{array}$$

where ${\mathcal{S}}_T$ is the closed subset of $C_T\left(H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\right)$ defined by

$${\mathcal{S}}_T=\left\{u\in C_T\left(H_{a,\gamma }^1\left({\mathbb{R}}^3\right)\right){;\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\le 2{\parallel u_0\parallel }_{H_{a,\gamma }^1}\right\}.$$

To apply the Fixed Point Theorem, let $T>0$ be sufficiently small such that the following condition is satisfied:

$$\alpha T+4C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_0{\parallel }_{H_{a,\gamma }^1}+12C_{\gamma }\beta T{\parallel u_0\parallel }_{H_{a,\gamma }^1}^2\le {\displaystyle \frac{1}{2}}.$$

(12)

Firstly, we wish to give the stability condition of the Fixed Point argument:

$$\mathcal{O}\left({\mathcal{S}}_T\right)\subset {\mathcal{S}}_T.$$

To do this, use the fact that $u\in {\mathcal{S}}_T$, Lemma 6 and Equation (12), we obtain

$$\begin{array}{ccc}\hfill {\parallel \mathcal{O}\left(u\right)\left(t\right)\parallel }_{H_{a,\gamma }^1}& \le & \parallel e^{\mu t\Delta }{u}_0{\parallel }_{H_{a,\gamma }^1}+{\parallel \mathcal{A}\left(u\right)\parallel }_{H_{a,\gamma }^1}+{\parallel B(u,u)\parallel }_{H_{a,\gamma }^1}+{\parallel \mathcal{C}\left(u\right)\parallel }_{H_{a,\gamma }^1}\hfill \\ & \le & \parallel u_0{\parallel }_{H_{a,\gamma }^1}+{\alpha T\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^2+C_{\gamma }\beta T{\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^3\hfill \\ & \le & \parallel u_0{\parallel }_{H_{a,\gamma }^1}+2\alpha T\parallel u_0{\parallel }_{H_{a,\sigma }^1}+4C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_0{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^2+8C_{\gamma }\beta T{\parallel u_0\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^3\hfill \\ & \le & \left(1+2\alpha T+4C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_0{\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+8C_{\gamma }\beta T{\parallel u_0\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^2\right){\parallel u_0\parallel }_{H_{a,\gamma }^1}\hfill \\ & \le & 2\parallel u_0{\parallel }_{H_{a,\gamma }^1}.\hfill \end{array}$$

Secondly, let $w=v-u$. We shall prove that $\mathcal{O}$ is a contraction mapping on ${\mathcal{S}}_T$. For this purpose, we choose $T>0$ sufficiently small so that Equation (12) holds. Using Lemma 6 and the above condition, we obtain, for all $u,v\in {\mathcal{S}}_T$,

$$\begin{array}{ccc}& & {\parallel \mathcal{O}\left(v\right)\left(t\right)-\mathcal{O}\left(u\right)\left(t\right)\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le \parallel \alpha {\int }_0^t{e}^{\mu (t-z)\Delta }wdz{\parallel }_{H_{a,\gamma }^1}+\parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(v·\nabla w)dz{\parallel }_{H_{a,\gamma }^1}+\parallel {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(w·\nabla u)dz{\parallel }_{H_{a,\gamma }^1}\hfill \\ & & +\sum _{j=1}^3\parallel \beta {\int }_0^t{e}^{\mu (t-z)\Delta }\mathbb{P}(v_j^3-u_j^3)e_{j}dz{\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le \alpha {\int }_0^t\parallel e^{\mu (t-z)\Delta }w{\parallel }_{H_{a,\gamma }^1}dz+{\int }_0^t\parallel e^{\mu (t-z)\Delta }\mathbb{P}(v·\nabla w){\parallel }_{H_{a,\gamma }^1}dz+{\int }_0^t\parallel e^{\mu (t-z)\Delta }\mathbb{P}(w·\nabla u){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & & +\beta \sum _{j=1}^3{\int }_0^t\parallel e^{\mu (t-z)\Delta }\mathbb{P}(v_j^2{w}_j+v_j{u}_j{w}_j+u_j^2{w}_j)e_j{\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & & \le {\alpha T\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}{\parallel v\parallel }_{H_{a,\gamma }^1}{\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}{\parallel u\parallel }_{H_{a,\gamma }^1}{\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\hfill \\ & & +C_{\gamma }\beta T\left({\parallel v\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^2+{\parallel v\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}{\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+{\parallel u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}^2\right){\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\hfill \\ & & \le {\alpha T\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+4C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_0{\parallel }_{H_{a,\gamma }^1}{\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}+12C_{\gamma }\beta T\parallel u_0{\parallel }_{H_{a,\gamma }^1}^2{\parallel w\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\hfill \\ & & \le \left(\alpha T+4C_{\gamma }{\mu }^{-{\displaystyle \frac{1}{2}}}{T}^{{\displaystyle \frac{1}{2}}}\parallel u_0{\parallel }_{H_{a,\gamma }^1}+12C_{\gamma }\beta T{\parallel u_0\parallel }_{H_{a,\gamma }^1}^2\right){\parallel v-u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}\hfill \\ & & \le {\displaystyle \frac{1}{2}}{\parallel v-u\parallel }_{L_T^{\infty }\left(H_{a,\gamma }^1\right)}.\hfill \end{array}$$

Consequently, Fixed Point Theorem guarantees that there is a unique solution $u\in C([0,T],H_{a,\gamma }^1\left({\mathbb{R}}^3\right))$. □

4. Proof of Theorem 2

Before presenting the proof, we note that the argument follows the same analytical techniques as in [8,14], relying on standard energy estimates in the Sobolev–Gevrey setting and on symmetry-preserving estimates to obtain the blow–up criteria.

For $a>0$ and $\gamma \in (0,1)$. Let $u\in C([0,T_{a,\gamma }^{*}),H_{a,\gamma }^1\left({\mathbb{R}}^3\right))$ be the maximal solution to Equation (1) given by Theorem 1. Suppose that the maximal time $T_{a,\gamma }^{*}$ of existence of the solution is finite.

4.1. Proof of Theorem 2 (i)

We want to prove the inequality $\left(i\right)$ by contradiction. For this, assume that the inequality $\left(i\right)$ is not valid. So that, if the function u is assumed to be bounded, then there exists $M>0$ such that

$$\begin{array}{c}\hfill {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}\le M,\forall t\in [0.T_{a,\gamma }^{*}).\end{array}$$

(13)

Firstly, taking the $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ inner product of the first equation of Equation (1) with u, we obtain

$$\begin{array}{ccc}\hfill & & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2\hfill \\ & & \le \beta \sum _{j=1}^3|<u_j^3{e}_j;u_j{e}_j{>}_{H_{a,\gamma }^1}|+|<u·\nabla u;u{>}_{H_{a,\gamma }^1}|.\hfill \end{array}$$

(14)

By applying Equations (11)–(7) and the fact that $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ be an algebra space, we infer

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2& \le & \beta \parallel u^3{\parallel }_{H_{a,\gamma }^1}{\parallel u\parallel }_{H_{a,\gamma }^1}{+\parallel u\otimes u\parallel }_{H_{a,\gamma }^1}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}\hfill \\ & \le & C_{\gamma }{\beta \parallel u\parallel }_{H_{a,\gamma }^1}^4+C_{\gamma }{\parallel u\parallel }_{H_{a,\gamma }^1}^2{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}.\hfill \end{array}$$

Hölder’s inequality gives

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\displaystyle \frac{\mu }{2}}{\parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2& \le & C_{\gamma }{\beta \parallel u\parallel }_{H_{a,\gamma }^1}^4+{\displaystyle \frac{C_{\gamma }^2}{2\mu }}{\parallel u\parallel }_{H_{a,\gamma }^1}^4.\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+2\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2& \le & 2C_{\gamma }{\beta \parallel u\parallel }_{H_{a,\gamma }^1}^4+C_{\gamma }^2{\mu }^{-1}{\parallel u\parallel }_{H_{a,\gamma }^1}^4\hfill \\ & \le & \left(2C_{\gamma }\beta +C_{\gamma }^2{\mu }^{-1}\right){\parallel u\parallel }_{H_{a,\gamma }^1}^4.\hfill \end{array}$$

Hence, integrating over $(0,t)$ and using Equation (13), one gets

$$\begin{array}{ccc}& & {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\mu {\int }_0^t{\parallel \nabla u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{2}dz+2\alpha {\int }_0^t{\parallel u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{2}dz\hfill \\ & & \le \parallel u_0{\parallel }_{H_{a,\gamma }^1}^2+\left(2C_{\gamma }\beta +C_{\gamma }^2{\mu }^{-1}\right){\int }_0^t{\parallel u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{4}dz\hfill \\ & & \le \parallel u_0{\parallel }_{H_{a,\gamma }^1}^2+\left(2C_{\gamma }\beta +C_{\gamma }^2{\mu }^{-1}\right)M^4.\hfill \end{array}$$

Then, for all $t\in [0,T_{a,\gamma }^{*})$ one deduces

$$\begin{array}{c}\hfill {\int }_0^{T_{a,\gamma }^{*}}{\parallel \nabla u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{2}dz\le M_1\end{array}$$

(15)

with ${\displaystyle M_1=\frac{\parallel u_0{\parallel }_{H_{a,\gamma }^1}^2+\left(2C_{\gamma }\beta +C_{\gamma }^2{\mu }^{-1}\right)M^4}{\mu }}$.

Secondly, suppose that $\left({\ell }_n\right)$ is a sequence in $(0,T_{a,\gamma }^{*})$ such that ${\ell }_n\to T_{a,\gamma }^{*}$ as $n\to \infty$. We want to prove that

$$\underset{n,m\to \infty }{lim\; sup}{\parallel u\left({\ell }_n\right)-u\left({\ell }_m\right)\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=0$$

where $0<{\ell }_n<{\ell }_m<T_{a,\gamma }^{*},\forall n,m\in \mathbb{N}$ with $n<m$.

According to the integral form of Equation (1), we have

$$u\left({\ell }_n\right)=e^{\mu {\ell }_n\Delta }{u}_0-\alpha {\int }_0^{{\ell }_n}{e}^{\mu ({\ell }_n-z)\Delta }udz-{\int }_0^{{\ell }_n}{e}^{\mu ({\ell }_n-z)\Delta }\mathbb{P}(u·\nabla u)dz-\beta {\int }_0^{{\ell }_n}{e}^{\mu ({\ell }_n-z)\Delta }\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)dz$$

and

$$u\left({\ell }_m\right)=e^{\mu {\ell }_m\Delta }{u}_0-\alpha {\int }_0^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }udz-{\int }_0^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }\mathbb{P}(u·\nabla u)dz-\beta {\int }_0^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)dz.$$

Then,

$$\begin{array}{ccc}\hfill u\left({\ell }_n\right)-u\left({\ell }_m\right)& =& \underset{I_{n,m}}{\underbrace{\left(e^{\mu {\ell }_n\Delta }-e^{\mu {\ell }_m\Delta }\right)u_0}}\hfill \\ & & +\underset{J_{n,m}}{\underbrace{\alpha {\int }_0^{{\ell }_n}\left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)udz}}+\underset{J_{m,n}}{\underbrace{\alpha {\int }_{{\ell }_n}^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }udz}}\hfill \\ & & +\underset{K_{n,m}}{\underbrace{{\int }_0^{{\ell }_n}\left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)\mathbb{P}(u·\nabla u)dz}}+\underset{K_{m,n}}{\underbrace{{\int }_{{\ell }_n}^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }\mathbb{P}(u·\nabla u)dz}}\hfill \\ & & +\underset{L_{n,m}}{\underbrace{\beta {\int }_0^{{\ell }_n}\left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)dz}}+\underset{L_{m,n}}{\underbrace{\beta {\int }_{{\ell }_n}^{{\ell }_m}{e}^{\mu ({\ell }_m-z)\Delta }\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)dz}}.\hfill \end{array}$$

Now, let us prove that $I_{n,m},J_{n,m},J_{m,n},K_{n,m},K_{m,n},L_{n,m}$ and $L_{m,n}$ tend to zero in $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$, as $n,m$ go to ∞.

Let us estimate $\parallel I_{n,m}{\parallel }_{H_{a,\gamma }^1}$, we have

$$\begin{array}{ccc}\hfill \parallel I_{n,m}{\parallel }_{H_{a,\gamma }^1}^2& =& \parallel \left(e^{\mu {\ell }_n\Delta }-e^{\mu {\ell }_m\Delta }\right)u_0{\parallel }_{H_{a,\gamma }^1}^2\hfill \\ & =& {\int }_{{\mathbb{R}}^3}\left(e^{-\mu {\ell }_n{\left|\xi \right|}^2}-e^{-\mu {\ell }_m{\left|\xi \right|}^2}\right){(1+|\xi |}^2{)e^{{2a\left|\xi \right|}^{\gamma }}\widehat{u_0\left(\xi \right)}|}^{2}d\xi \hfill \\ & \le & {\int }_{{\mathbb{R}}^3}\left(e^{-\mu {\ell }_n{\left|\xi \right|}^2}-e^{-\mu T_{a,\gamma }^{*}{\left|\xi \right|}^2}\right){(1+|\xi |}^2{)e^{{2a\left|\xi \right|}^{\gamma }}\widehat{u_0\left(\xi \right)}|}^{2}d\xi .\hfill \end{array}$$

Put $F_n\left(\xi \right)=\left(e^{-\mu {\ell }_n{\left|\xi \right|}^2}-e^{-\mu T_{a,\gamma }^{*}{\left|\xi \right|}^2}\right){(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u_0\left(\xi \right)}\right|}^2$ and $F\left(\xi \right)={(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u_0\left(\xi \right)}\right|}^2$. We have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}0\le F_n\le F,\forall n\in \mathbb{N},\hfill \\ F\in L^1\left({\mathbb{R}}^3\right),\sin \mathrm{ce}{u}_0\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right),\hfill \\ F_n\to 0\mathrm{as}n\to \infty ,\mathrm{a}.\mathrm{e}.\hfill \end{array}\right.\end{array}$$

The Dominated Convergence Theorem gives

$$\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}{F}_n\left(\xi \right)d\xi =0$$

and

$$\underset{n,m\to \infty }{lim}{\parallel I_{n,m}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel J_{n,m}{\parallel }_{H_{a,\gamma }^1}$ using the Cauchy–Schwarz inequality, we have

$$\begin{array}{ccc}\hfill \parallel J_{n,m}{\parallel }_{H_{a,\gamma }^1}& \le & \alpha {\int }_0^{{\ell }_n}{\parallel \left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \alpha {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(e^{-\mu ({\ell }_m-z){\left|\xi \right|}^2}-e^{-\mu ({\ell }_n-z){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & \alpha {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{e}^{-2\mu ({\ell }_n-z){\left|\xi \right|}^2}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & \alpha {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & \alpha {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & \alpha {\int }_0^{T_{a,\gamma }^{*}}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & \alpha {\left(T_{a,\gamma }^{*}\right)}^{1/2}{\left({\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^{2}d\xi dz\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Put

$$G_n(\xi ,z)={\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^2$$

and

$$G(\xi ,z)={(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u}(\xi ,z)\right|}^2.$$

We have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}0\le G_n\le G,\forall n\in \mathbb{N},\hfill \\ G\in L^1\left({\mathbb{R}}^3\right)\mathrm{by}\mathrm{the}\mathrm{inequality}\left(13\right),\hfill \\ G_n\to 0\mathrm{as}n\to \infty ,\mathrm{a}.\mathrm{e}.\hfill \end{array}\right.\end{array}$$

By using the Dominated Convergence Theorem, one deduces

$$\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3}{G}_n\left(\xi \right)d\xi =0$$

and

$$\underset{n,m\to \infty }{lim}{\parallel J_{n,m}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel J_{m,n}{\parallel }_{H_{a,\gamma }^1}$ using Equation (13), we have

$$\begin{array}{ccc}\hfill \parallel J_{m,n}{\parallel }_{H_{a,\gamma }^1}& \le & \alpha {\int }_{{\ell }_n}^{{\ell }_m}{\parallel e^{\mu ({\ell }_m-z)\Delta }u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \alpha {\int }_{{\ell }_n}^{{\ell }_m}{\parallel u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \alpha {\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \alpha M(T_{a,\gamma }^{*}-{\ell }_n).\hfill \end{array}$$

As

$$\underset{n\to \infty }{lim}\alpha M(T_{a,\gamma }^{*}-{\ell }_n)=0,$$

then

$$\underset{n,m\to \infty }{lim\; sup}{\parallel J_{m,n}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel K_{n,m}{\parallel }_{H_{a,\gamma }^1}$ using the Cauchy–Schwarz inequality, we have

$$\begin{array}{ccc}\hfill \parallel K_{n,m}{\parallel }_{H_{a,\gamma }^1}& \le & {\int }_0^{{\ell }_n}{\parallel \left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)\mathbb{P}(u·\nabla u)\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(e^{-\mu ({\ell }_m-z){\left|\xi \right|}^2}-e^{-\mu ({\ell }_n-z){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{e}^{-2\mu ({\ell }_n-z){\left|\xi \right|}^2}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & {\int }_0^{T_{a,\gamma }^{*}}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & \le & {\left(T_{a,\gamma }^{*}\right)}^{1/2}{\left({\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}d\xi dz\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Put

$$H_n(\xi ,z)={\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^2$$

and

$$H(\xi ,z)={(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^2.$$

We have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}0\le H_n\le H,\forall n\in \mathbb{N},\hfill \\ H_n\to 0\mathrm{as}n\to \infty ,\mathrm{a}.\mathrm{e}.\hfill \end{array}\right.\end{array}$$

Hence, using the fact that the space $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ is algebra and Equations (13)–(15) we obtain

$$\begin{array}{ccc}\hfill {\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}H(\xi ,z)dzd\xi & =& {\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}{\left|\widehat{u·\nabla u}(\xi ,z)\right|}^{2}dzd\xi \hfill \\ & =& {\int }_0^{T_{a,\gamma }^{*}}{\parallel u·\nabla u\parallel }_{H_{a,\gamma }^1}^{2}dz\hfill \\ & \le & c{\int }_0^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{H_{a,\gamma }^1}^2{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}^{2}dz\hfill \\ & \le & cM{\int }_0^{T_{a,\gamma }^{*}}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}^{2}dz\hfill \\ & \le & cM{M}_1<\infty .\hfill \end{array}$$

The Dominated Convergence Theorem gives

$$\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{H}_n(\xi ,z)d\xi =0$$

and

$$\underset{n,m\to \infty }{lim}{\parallel K_{n,m}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel K_{m,n}{\parallel }_{H_{a,\gamma }^1}$ using he fact that the space $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ is algebra, the inequalities (13)–(15) and the Cauchy–Shwarz inequality, we have

$$\begin{array}{ccc}\hfill \parallel K_{m,n}{\parallel }_{H_{a,\gamma }^1}& \le & {\int }_{{\ell }_n}^{{\ell }_m}{\parallel e^{\mu ({\ell }_m-z)\Delta }\mathbb{P}(u·\nabla u)\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_{{\ell }_n}^{{\ell }_m}{\parallel e^{\mu ({\ell }_m-z)\Delta }(u·\nabla u)\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_{{\ell }_n}^{{\ell }_m}{\parallel u·\nabla u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & {\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel u·\nabla u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & c{\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{H_{a,\gamma }^1}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & cM{\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & cM{(T_{a,\gamma }^{*}-{\ell }_n)}^{1/2}{\left({\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}^{2}dz\right)}^{1/2}\hfill \\ & \le & cM{(T_{a,\gamma }^{*}-{\ell }_n)}^{1/2}{\left({\int }_0^{T_{a,\gamma }^{*}}{\parallel \nabla u\parallel }_{H_{a,\gamma }^1}^{2}dz\right)}^{1/2}\hfill \\ & \le & cM{\left(M_1\right)}^{1/2}{(T_{a,\gamma }^{*}-{\ell }_n)}^{1/2}.\hfill \end{array}$$

As

$$\underset{n\to \infty }{lim}cM{\left(M_1\right)}^{1/2}{(T_{a,\gamma }^{*}-{\ell }_n)}^{1/2}=0.$$

Then,

$$\underset{n,m\to \infty }{lim\; sup}{\parallel K_{m,n}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel L_{n,m}{\parallel }_{H_{a,\gamma }^1}$, we have

$$\begin{array}{ccc}& & \parallel L_{n,m}{\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le \beta {\int }_0^{{\ell }_n}\parallel \left(e^{\mu ({\ell }_m-z)\Delta }-e^{\mu ({\ell }_n-z)\Delta }\right)\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & & \le \beta {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(e^{-\mu ({\ell }_m-z){\left|\xi \right|}^2}-e^{-\mu ({\ell }_n-z){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi \right)}^{1/2}dz\hfill \\ & & \le \beta {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{e}^{-2\mu ({\ell }_n-z){\left|\xi \right|}^2}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & & \le \beta {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu ({\ell }_m-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & & \le \beta {\int }_0^{{\ell }_n}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & & \le \beta {\int }_0^{T_{a,\gamma }^{*}}{\left({\int }_{{\mathbb{R}}^3}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi \right)}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & & \le \beta {\left(T_{a,\gamma }^{*}\right)}^{1/2}{\left({\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2)e^{{2a\left|\xi \right|}^{\gamma }}|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}d\xi dz\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Put

$$P_n(\xi ,z)={\left(1-e^{-\mu (T_{a,\gamma }^{*}-{\ell }_n){\left|\xi \right|}^2}\right)}^2{(1+|\xi |}^2\left)e^{{2a\left|\xi \right|}^{\gamma }}\right|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^2$$

and

$$P(\xi ,z)={(1+|\xi |}^2\left)e^{{2a\left|\xi \right|}^{\gamma }}\right|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^2.$$

We have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}0\le P_n\le P,\forall n\in \mathbb{N},\hfill \\ P_n\to 0\mathrm{as}n\to \infty ,\mathrm{a}.\mathrm{e}.\hfill \end{array}\right.\end{array}$$

Hence, using Equations (11)–(13) we obtain

$$\begin{array}{ccc}\hfill {\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}P(\xi ,z)dzd\xi & =& {\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{(1+|\xi |}^2\left)e^{{2a\left|\xi \right|}^{\gamma }}\right|\mathcal{F}(\left(\sum _{j=1}^3{u}_j^3{e}_j\right)(\xi ,z){|}^{2}dzd\xi \hfill \\ & \le & C_{\gamma }^2{\int }_0^{T_{a,\gamma }^{*}}{\parallel u\parallel }_{H_{a,\gamma }^1}^{6}dz\hfill \\ & \le & C_{\gamma }^2{M}^6<\infty .\hfill \end{array}$$

The Dominated Convergence Theorem gives

$$\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^3\times [0,T_{a,\gamma }^{*}[}{P}_n\left(\xi \right)d\xi =0$$

and

$$\underset{n,m\to \infty }{lim}{\parallel L_{n,m}\parallel }_{H_{a,\gamma }^1}^2=0.$$

For the estimation of $\parallel L_{m,n}{\parallel }_{H_{a,\gamma }^1}$ using Equations (11)–(13) we have

$$\begin{array}{ccc}\hfill \parallel L_{m,n}{\parallel }_{H_{a,\gamma }^1}& \le & \beta {\int }_{{\ell }_n}^{{\ell }_m}\parallel e^{\mu ({\ell }_m-z)\Delta }\mathbb{P}\left(\sum _{j=1}^3{u}_j^3{e}_j\right){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \beta {\int }_{{\ell }_n}^{{\ell }_m}\parallel \left(\sum _{j=1}^3{u}_j^3{e}_j\right){\parallel }_{H_{a,\gamma }^1}dz\hfill \\ & \le & \beta {\int }_{{\ell }_n}^{{\ell }_m}{\parallel u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{3}dz\hfill \\ & \le & C_{\gamma }\beta {\int }_{{\ell }_n}^{T_{a,\gamma }^{*}}{\parallel u\left(z\right)\parallel }_{H_{a,\gamma }^1}^{3}dz\hfill \\ & \le & C_{\gamma }{M}^3\beta (T_{a,\gamma }^{*}-{\ell }_n).\hfill \end{array}$$

As

$$\underset{n\to \infty }{lim}{C}_{\gamma }{M}^3\beta (T_{a,\gamma }^{*}-{\ell }_n)=0,$$

then,

$$\underset{n,m\to \infty }{lim\; sup}{\parallel L_{m,n}\parallel }_{H_{a,\gamma }^1}^2=0.$$

Finally, we can deduce that ${\left(u\left({\ell }_n\right)\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in the space $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$. The space $H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ is a Banach space. Then, there exists $u_0^{*}\in H_{a,\gamma }^1\left({\mathbb{R}}^3\right)$ such that

$$\underset{n\to \infty }{lim\; sup}{\parallel u\left({\ell }_n\right)-u_0^{*}\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=0.$$

Note that the limit $u_0^{*}$ does not depend on $\left({\ell }_n\right)$. Hence, we prove that the solution u can be extend to the interval $[0,T_{a,\gamma }^{*}+T_1^{*}]$. This is a contradiction, and so

$$\begin{array}{c}\hfill lim\underset{t\nearrow T_{a,\gamma }^{*}}{sup}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=\infty .\end{array}$$

(16)

This completes the proof of the Theorem 2 $\left(i\right)$.

Remark 1.

Taking into consideration the energy estimate Equation (2), we have

$${\parallel u\left(t\right)\parallel }_{L^2}\le {\parallel u_0\parallel }_{L^2},\forall t\in [0,T_{a,\gamma }^{*}).$$

Then,

$$\begin{array}{c}\hfill lim\underset{t\nearrow T_{a,\gamma }^{*}}{sup}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=\infty \iff lim\underset{t\nearrow T_{a,\gamma }^{*}}{sup}{\parallel u\left(t\right)\parallel }_{{\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right)}=\infty .\end{array}$$

4.2. Proof of Theorem 2 (ii)

From Equation (14), we have

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2\hfill \\ & & \le \beta \sum _{k=1}^3|<u_k^3{e}_k;u_k{e}_k{>}_{H_{a,\gamma }^1}|+|<u\left(t\right)·\nabla u\left(t\right);u\left(t\right){>}_{H_{a,\gamma }^1}|\hfill \end{array}$$

Using Equations (10)–(6), we obtain

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+\alpha {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2\hfill \\ & & \le \beta \sum _{k=1}^3|<e^{{a|.|}^{\gamma }}{u}_k^3{e}_k;e^{{a|.|}^{\gamma }}{u}_k{e}_k{>}_{H^1}|+|<e^{{a|.|}^{\gamma }}\mathrm{div}(\mathrm{u}\otimes \mathrm{u})\left(\mathrm{t}\right);{\mathrm{e}}^{{\mathrm{a}|.|}^{\gamma }}\mathrm{u}\left(\mathrm{t}\right){>}_{{\mathrm{H}}^1}|\hfill \\ & & \le \beta \sum _{k=1}^3|<e^{{a|.|}^{\gamma }}{u}_k^3{e}_k;e^{{a|.|}^{\gamma }}{u}_k{e}_k{>}_{H^1}|+|<e^{{a|.|}^{\gamma }}(u\otimes u)\left(t\right);\nabla e^{{a|.|}^{\gamma }}u\left(t\right){>}_{H^1}|\hfill \\ & & \le \beta \parallel \sum _{k=1}^3{u}_k^3{e}_k\left(t\right){\parallel }_{H_{a,\gamma }^1}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}+{\parallel (u\otimes u)\left(t\right)\parallel }_{H_{a,\gamma }^1}{\parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}\hfill \\ & & \le 4\beta \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}^2\parallel u\left(t\right){\parallel }_{H_{a,\gamma }^1}^2+4\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}\parallel u\left(t\right){\parallel }_{H_{a,\gamma }^1}{\parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}.\hfill \end{array}$$

Hölde’s inequality implies

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\displaystyle \frac{\mu }{2}}{\parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\alpha \parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2\le (4\beta +8{\mu }^{-1})\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}^2{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2.\end{array}$$

or equivalently

$$\begin{array}{c}\hfill {\displaystyle \frac{d}{dt}}{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{\mu \parallel \nabla u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2+{2\alpha \parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2\le (8\beta +8{\mu }^{-1})\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}^2{\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^2.\end{array}$$

Thereby, for any $0\le t\le T<T_{a,\gamma }^{*}$ by applying Gronwall Lemma, one deduces

$${\parallel u\left(T\right)\parallel }_{H_{a,\gamma }^1}^2\le {\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}^{2}exp\left((8\beta +8{\mu }^{-1}){\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{2}dz\right).$$

Applying Equation (16), we infer

$$\begin{array}{c}\hfill {\int }_t^{T_{a,\gamma }^{*}}{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1\left({\mathbb{R}}^3\right)}^{2}dz=\infty ,\forall 0\le t<T_{a,\gamma }^{*}.\end{array}$$

(17)

This completes the proof of Theorem 2 $\left(ii\right)$.

4.3. Proof of Theorem 2 (iii)

Applying the Fourier transform with respect to the spatial variable in the first part of the Equation (1) and multiplying by $\overline{\widehat{u}}$, we obtain

$$\begin{array}{ccc}& & {\partial }_t\widehat{u}\overline{\widehat{u}}(t,\xi )+{\mu \left|\xi \right|}^2\widehat{u}\overline{\widehat{u}}(t,\xi )+\alpha \widehat{u}\overline{\widehat{u}}(t,\xi )|+\mathcal{F}(u·\nabla u)(t,\xi )·\overline{\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )}\hfill \\ & & +\beta \left[\mathcal{F}\left(\sum _{k=1}^3{u}_k^3{e}_k\right)(t,\xi )·\overline{\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )}\right]=0.\hfill \end{array}$$

Taking the real part yields

$$\begin{array}{ccc}& & {\partial }_t|\widehat{u}{(t,\xi )|}^2+{2\mu \left|\xi \right|}^2|\widehat{u}{(t,\xi )|}^2+2\alpha {\left|\widehat{u}(t,\xi )\right|}^2+2\mathcal{R}e\left[\mathcal{F}(u·\nabla u)(t,\xi )·\overline{\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )}\right]\hfill \\ & & +\beta [\mathcal{F}\left(\sum _{k=1}^3{u}_k^3{e}_k\right)(t,\xi )·\overline{\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )]}=0.\hfill \end{array}$$

This implies

$$\begin{array}{ccc}\hfill & & {\partial }_t|\widehat{u}{(t,\xi )|}^2+{2\mu \left|\xi \right|}^2|\widehat{u}{(t,\xi )|}^2+2\alpha {\left|\widehat{u}(t,\xi )\right|}^2\hfill \\ & & \le 2\left|\mathcal{F}(u·\nabla u)(t,\xi )\right|\left|\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )\right|\hfill \\ & & +2\beta \left|\mathcal{F}\left(\sum _{k=1}^3{u}_k^3{e}_k\right)(t,\xi )\right|\left|\mathcal{F}\left(\sum _{k=1}^3{u}_k{e}_k\right)(t,\xi )\right|.\hfill \end{array}$$

(18)

Now, for $\epsilon >0$ and $\delta >0$, we have

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

Then,

$$\begin{array}{c}\hfill {\displaystyle \frac{{\partial }_t\left(|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon \right)}{(1+\delta )\left(|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon \right)}}={\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta }{\partial }_t{\left|\widehat{u}(t,\xi )\right|}^2}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}+{\displaystyle \frac{2\epsilon }{(1+\delta )}}{\displaystyle \frac{{\partial }_t{\left|\widehat{u}(t,\xi )\right|}^{1+\delta }}{\left(\right|\widehat{u}(t,\xi ){|}^{1+\delta }+\epsilon )}}.\end{array}$$

(19)

Multiplying Equation (18) by ${\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta }}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}$ and using the Equation (19), we conclude

$$\begin{array}{ccc}& & {\displaystyle \frac{{\partial }_t{\left(|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon \right)}^2}{(1+\delta )\left(|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon \right)}}-{\displaystyle \frac{2\epsilon }{(1+\delta )}}{\displaystyle \frac{{\partial }_t{\left|\widehat{u}(t,\xi )\right|}^{1+\delta }}{\left(\right|\widehat{u}(t,\xi ){|}^{1+\delta }+\epsilon )}}+\mu {\left|\xi \right|}^2{\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta +2}}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}+\alpha {\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta +2}}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}\hfill \\ & & \le \left|\mathcal{F}(u·\nabla u)(t,\xi )\right|{\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta +1}}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}\hfill \\ & & +\beta \left|\mathcal{F}\left(\sum _{k=1}^3{u}_k^3{e}_k\right)(t,\xi )\right|{\displaystyle \frac{|\widehat{u}{(t,\xi )|}^{2\delta +1}}{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}.\hfill \end{array}$$

By integrating from $[t,T]\subset [0,T_{a,\gamma }^{*})$, one gets

$$\begin{array}{ccc}& & {\displaystyle \frac{|\widehat{u}{(T,\xi )|}^{1+\delta }}{1+\delta }}-{\displaystyle \frac{2\epsilon }{1+\delta }}ln\left({\displaystyle \frac{|\widehat{u}{(T,\xi )|}^{1+\delta }+\epsilon }{|\widehat{u}{(t,\xi )|}^{1+\delta }+\epsilon }}\right)+\mu {\left|\xi \right|}^2{\int }_t^T{\displaystyle \frac{|\widehat{u}{(z,\xi )|}^{2\delta +2}}{|\widehat{u}{(z,\xi )|}^{1+\delta }+\epsilon }}dz+\alpha {\int }_t^T{\displaystyle \frac{|\widehat{u}{(z,\xi )|}^{2\delta +2}}{|\widehat{u}{(z,\xi )|}^{1+\delta }+\epsilon }}dz\hfill \\ & & \le {\displaystyle \frac{2|\widehat{u}{(t,\xi )|}^{1+\delta }}{1+\delta }}\hfill \\ & & +{\int }_t^T\left|\mathcal{F}(u·\nabla u)(z,\xi )\right|{\displaystyle \frac{|\widehat{u}{(z,\xi )|}^{2\delta +1}}{|\widehat{u}{(z,\xi )|}^{1+\delta }+\epsilon }}dz\hfill \\ & & +\beta {\int }_t^T\left|\sum _{k=1}^3\mathcal{F}\left(u_k^3{e}_k\right)(z,\xi )\right|{\displaystyle \frac{|\widehat{u}{(z,\xi )|}^{2\delta +1}}{|\widehat{u}{(z,\xi )|}^{1+\delta }+\epsilon }}dz.\hfill \end{array}$$

Taking as $\epsilon ,\delta \to 0^{+}$, we obtain

$$\begin{array}{ccc}& & |\widehat{u}{(T,\xi )|+\mu |\xi |}^2{\int }_t^T|\widehat{u}(z,\xi )|dz+\alpha {\int }_t^T|\widehat{u}(z,\xi )|dz\hfill \\ & & \le |\widehat{u}(t,\xi )|+{\int }_t^T\left|\mathcal{F}(u·\nabla u)(z,\xi )\right|dz+\beta {\int }_t^T\left|\sum _{k=1}^3\mathcal{F}\left(u_k^3{e}_k\right)(z,\xi )\right|dz\hfill \\ & & \le |\widehat{u}(t,\xi )|+{\int }_t^T|\widehat{u}|{\ast }_{\xi }|\widehat{\nabla u}|(z,\xi )dz+\beta {\int }_t^T\left|\sum _{k=1}^3\mathcal{F}\left(u_k^3{e}_k\right)(z,\xi )\right|dz\hfill \\ & & \le |\widehat{u}(t,\xi )|+{\int }_t^T{\int }_{\eta }|\widehat{u}(z,\xi -\eta )\left|\right|\widehat{\nabla u}(z,\eta )|d\eta dz+\beta {\int }_t^T\left|\sum _{k=1}^3\mathcal{F}\left(u_k^3{e}_k\right)(z,\xi )\right|d\eta dz.\hfill \end{array}$$

Multiplying the above inequality by $e^{{a\gamma \left|\xi \right|}^{\gamma }}$ and using the Equation (8), we have

$$\begin{array}{ccc}& & e^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u}{(T,\xi )|+\mu |\xi |}^2{\int }_t^T{e}^{{a\gamma \left|\xi \right|}^{\gamma }}|\widehat{u}(z,\xi )|dz+\alpha {\int }_t^T{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u}(z,\xi )\right|dz\hfill \\ & & \le e^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\widehat{u}(t,\xi )\right|\hfill \\ & & +{\int }_t^T{\int }_{\eta }{e}^{{a\gamma |\xi -\eta |}^{\gamma }}|\widehat{u}(z,\xi -\eta )|e^{{a\gamma \left|\eta \right|}^{\gamma }}\left|\widehat{\nabla u}(z,\eta )\right|d\eta dz\hfill \\ & & +\beta {\int }_t^T{e}^{{a\gamma \left|\xi \right|}^{\gamma }}\left|\sum _{k=1}^3\mathcal{F}\left(u_k^3{e}_k\right)(z,\xi )\right|dz.\hfill \end{array}$$

Integrating with respect the space variable $\xi \in {\mathbb{R}}^3$ and using product Young inequality, we can deduce

$$\begin{array}{ccc}& & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(T\right){\parallel }_{L^1}+\mu {\int }_t^T\parallel \mathcal{F}(e^{{a\gamma |.|}^{\gamma }}\Delta u)\left(z\right){\parallel }_{L^1}dz+\alpha {\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}dz\hfill \\ & & \le \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}\hfill \\ & & +{\int }_t^T\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right){\parallel }_{L^1}{\parallel \mathcal{F}(e^{{a\gamma |.|}^{\gamma }}\nabla u)\left(z\right)\parallel }_{L^1}dz\hfill \\ & & +\beta {\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{3}dz.\hfill \end{array}$$

The Cauchy–Schwarz inequality yields

$$\begin{array}{ccc}& & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(T\right){\parallel }_{L^1}+\mu {\int }_t^T\parallel \mathcal{F}(e^{{a\gamma |.|}^{\gamma }}\Delta u)\left(z\right){\parallel }_{L^1}dz+\alpha {\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}dz\hfill \\ & & \le \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}\hfill \\ & & +{\int }_t^T\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right){\parallel }_{L^1}^{{\displaystyle \frac{3}{2}}}{\parallel \mathcal{F}(e^{{a\gamma |.|}^{\gamma }}\Delta u)\left(z\right)\parallel }_{L^1}^{{\displaystyle \frac{1}{2}}}dz\hfill \\ & & +\beta {\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{3}dz.\hfill \end{array}$$

Hölde’s inequality gives

$$\begin{array}{ccc}& & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(T\right){\parallel }_{L^1}+{\displaystyle \frac{\mu }{2}}{\int }_t^T\parallel \mathcal{F}(e^{{a\gamma |.|}^{\gamma }}\Delta u)\left(z\right){\parallel }_{L^1}dz+\alpha {\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}dz\hfill \\ & & \le \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}+\left(\beta +{\displaystyle \frac{1}{2\mu }}\right){\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{3}dz\hfill \\ & & \le \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}+(\beta +{\mu }^{-1}){\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{3}dz.\hfill \end{array}$$

Using the Gronwall Lemma, for $0\le t\le T<T_{a,\gamma }^{*}$ we obtain

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(T\right){\parallel }_{L^1}& \le & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}exp\left((\beta +{\mu }^{-1}){\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{2}dz\right).\hfill \end{array}$$

This implies

$$\begin{array}{ccc}\hfill \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(T\right){\parallel }_{L^1}^{2}exp\left(-2(\beta +{\mu }^{-1}){\int }_t^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{2}dz\right)& \le & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right){\parallel }_{L^1}^2.\hfill \end{array}$$

Integrating from $[t_0,T]\subset [0,T_{a,\gamma }^{*})$, we get

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1-exp\left(-2(\beta +{\mu }^{-1}){\int }_{t_0}^T{\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(z\right)\parallel }_{L^1}^{2}dz\right)}{2(\beta +{\mu }^{-1})(T-t_0)}}& \le & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t_0\right){\parallel }_{L^1}^2.\hfill \end{array}$$

Consequently, if $T\to T_{a,\gamma }^{*}$ applying Equation (17), we conclude

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t_0)}}& \le & \parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t_0\right){\parallel }_{L^1}^2,\forall t_0\in [0,T_{a,\gamma }^{*}).\hfill \end{array}$$

(20)

This completes the proof of Theorem 2 $\left(iii\right)$.

4.4. Proof of Theorem 2 (iv)

Let $0<a^{\prime }=a{\gamma }^{\prime }=a\left({\displaystyle \frac{1+\gamma }{2}}\right)<a$. According to the injection property of the Sobolev spaces

$${\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right)\hookrightarrow {\dot{H}}_{a^{\prime },\gamma }^1\left({\mathbb{R}}^3\right)$$

we ensure that

$$u\in C([0,T_{a,\gamma }^{*}),{\dot{H}}_{a^{\prime },\gamma }^1\left({\mathbb{R}}^3\right)).$$

Further, if $u\in C([0,T_{a^{\prime },\gamma }^{*}),{\dot{H}}_{a^{\prime },\gamma }^1\left({\mathbb{R}}^3\right))$ and $u\in C([0,T_{a,\gamma }^{*}),{\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right))$. The fact that

$${\parallel u\parallel }_{{\dot{H}}_{a^{\prime },\gamma }^1\left({\mathbb{R}}^3\right)}\le {\parallel u\parallel }_{{\dot{H}}_{a,\gamma }^1\left({\mathbb{R}}^3\right)}$$

implies that

$$\begin{array}{c}\hfill T_{a^{\prime },\gamma }^{*}=T_{a{\gamma }^{\prime },\gamma }^{*}=T_{a\left({\displaystyle \frac{1+\gamma }{2}}\right),\gamma }^{*}\ge T_{a,\gamma }^{*}.\end{array}$$

(21)

On the other hand, From Equation (20), we can write for all $t\in [0,T_{a,\gamma }^{*})$

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{{a\gamma |.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2.$$

Equation (4) implies

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le C_{{\gamma }^{\prime }}^2{\parallel u\left(t\right)\parallel }_{{\dot{H}}_{a{\gamma }^{\prime },\gamma }^1}^2.$$

Taking the limit superior, as $t\nearrow T_{a,\gamma }^{*}$, we conclude

$$lim\underset{t\nearrow T_{a,\gamma }^{*}}{sup}{\parallel u\left(t\right)\parallel }_{{\dot{H}}_{a{\gamma }^{\prime },\gamma }^1}=\infty$$

and

$$\begin{array}{c}\hfill T_{a,\gamma }^{*}\ge T_{a^{\prime },\gamma }^{*}=T_{a{\gamma }^{\prime },\gamma }^{*}=T_{a\left({\displaystyle \frac{1+\gamma }{2}}\right),\gamma }^{*}.\end{array}$$

(22)

As a consequence, from Equations (21) and (22), we obtain

$$T_{a,\gamma }^{*}=T_{a^{\prime },\gamma }^{*}=T_{a{\gamma }^{\prime },\gamma }^{*}=T_{a\left({\displaystyle \frac{1+\gamma }{2}}\right),\gamma }^{*}$$

and similarly, we infer

$$T_{a{\gamma }^{\prime },\gamma }^{*}=T_{a\left({\displaystyle \frac{1+\gamma }{2}}\right),\gamma }^{*}=T_{a\left({\displaystyle \frac{1+\gamma }{2}}\right)\left({\displaystyle \frac{1+\gamma }{2}}\right),\gamma }^{*}=...=T_{a{\left({\displaystyle \frac{1+\gamma }{2}}\right)}^n,\gamma }^{*}.$$

Now, for $0<a^{\prime }<a$, there exist $n\in {\mathbb{N}}^{*}$ such that $a{\left({\displaystyle \frac{1+\gamma }{2}}\right)}^n<a^{\prime }<a$, then

$$T_{a^{\prime },\gamma }^{*}=T_{a,\gamma }^{*}$$

and for $a^{\prime }>0$, we have

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a^{\prime },\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{a^{\prime }\gamma {|.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2.$$

As $T_{a^{\prime },\gamma }^{*}=T_{a,\gamma }^{*}$, then

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{a^{\prime }\gamma {|.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2.$$

This implies

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{a^{\prime }\left({\displaystyle \frac{1+\gamma }{2}}\right){|.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2.$$

Consequently, for $a^{\prime }=a{\left({\displaystyle \frac{1+\gamma }{2}}\right)}^n$ we can deduce

$${\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \mathcal{F}\left(e^{a{\left({\displaystyle \frac{1+\gamma }{2}}\right)}^n{|.|}^{\gamma }}u\right)\left(t\right)\parallel }_{L^1}^2.$$

The Dominated Convergence Theorem implies, if $n\to \infty$

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)}}\le {\parallel \widehat{u}\left(t\right)\parallel }_{L^1}^2\end{array}$$

(23)

and $T_{a,\gamma }^{*}=T_{0,\gamma }^{*}$.

Therefore, by using Lemma 1, for $k\in \mathbb{N}$ such that $1+{\displaystyle \frac{k}{2}}\gamma \ge 2$, Equation (23) becomes

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{{\left(2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)\right)}^{1/2}}}& \le & \parallel \widehat{u}{\left(t\right)\parallel }_{L^1}\hfill \\ & \le & {B\left(2\right)\parallel u\left(t\right)\parallel }_{L^2}^{1-{\displaystyle \frac{3}{2(1+\frac{k}{2}\gamma )}}}{\parallel u\left(t\right)\parallel }_{{\dot{H}}^{1+{\displaystyle \frac{k}{2}}\gamma }}^{{\displaystyle \frac{3}{2(1+\frac{k}{2}\gamma )}}},\forall t\in [0,T_{a,\gamma }^{*}).\hfill \end{array}$$

The energy estimate Equation (2) yields

$${\displaystyle \frac{1}{{\left(2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)\right)}^{1/2}}}\le B\left(2\right)\parallel u_0{\parallel }_{L^2}^{1-{\displaystyle \frac{3}{2(1+\frac{k}{2}\gamma )}}}{\parallel u\left(t\right)\parallel }_{{\dot{H}}^{1+{\displaystyle \frac{k}{2}}\gamma }}^{{\displaystyle \frac{3}{2(1+\frac{k}{2}\gamma )}}}$$

or equivalently

$${\left({\displaystyle \frac{1}{{\left(2(\beta +{\mu }^{-1})(T_{a,\gamma }^{*}-t)\right)}^{1/2}}}\right)}^{{\displaystyle \frac{2(1+\frac{k}{2}\gamma )}{3}}}{\left(B\left(2\right)\right)}^{{\displaystyle \frac{-2(1+\frac{k}{2}\gamma )}{3}}}\parallel u_0{\parallel }_{L^2}^{1-{\displaystyle \frac{2(1+\frac{k}{2}\gamma )}{3}}}\le {\parallel u\left(t\right)\parallel }_{{\dot{H}}^{1+{\displaystyle \frac{k}{2}}\gamma }}.$$

Then,

$${\displaystyle \frac{C_1}{{(T_{a,\gamma }^{*}-t)}^{\frac{2}{3}}}}{\left({\displaystyle \frac{C_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)}^k\le {\parallel u\left(t\right)\parallel }_{{\dot{H}}^{1+{\displaystyle \frac{k}{2}}\gamma }}^2$$

with

$$C_1={\left({\displaystyle \frac{\parallel u_0{\parallel }_{L^2}}{2(\beta +{\mu }^{-1})B\left(2\right)}}\right)}^{2/3},C_2={\left({\displaystyle \frac{1}{2(\beta +{\mu }^{-1})B{\left(2\right)}^2{\parallel u_0\parallel }_{L^2}^2}}\right)}^{{\displaystyle \frac{\gamma }{3}}}.$$

This implies

$${\displaystyle \frac{1}{k!}}{\displaystyle \frac{C_1}{{(T_{a,\gamma }^{*}-t)}^{\frac{2}{3}}}}{\left({\displaystyle \frac{2a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)}^k\le {\int }_{\xi }{\displaystyle \frac{{\left(2a\right)}^k}{k!}}{\left|\xi \right|}^{k\gamma }{\left|\xi \right|}^2{\left|\widehat{u}(t,\xi )\right|}^{2}d\xi .$$

Summing the above inequality over the set $\{k\in \mathbb{N},k\ge {\displaystyle \frac{2}{\gamma }}\}$, one gets

$$\begin{array}{ccc}\hfill {\displaystyle \frac{C_1}{{(T_{a,\gamma }^{*}-t)}^{\frac{2}{3}}}}\left(e^{{\displaystyle \frac{2a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}}--\sum _{0\le k\le {\displaystyle \frac{2}{\gamma }}}{\displaystyle \frac{{\left(\frac{2a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}\right)}^k}{k!}}\right)& \le & {\int }_{\xi }\left(e^{{2a\left|\xi \right|}^{\gamma }}--\sum _{0\le k\le {\displaystyle \frac{2}{\gamma }}}{\displaystyle \frac{{\left(2a\right|\xi |}^{\gamma }{)}^k}{k!}}\right){\left|\xi \right|}^2{\left|\widehat{u}(t,\xi )\right|}^{2}d\xi \hfill \\ & \le & {\int }_{\xi }{e}^{{2a\left|\xi \right|}^{\gamma }}{\left|\xi \right|}^2{\left|\widehat{u}(t,\xi )\right|}^{2}d\xi .\hfill \end{array}$$

Finally, if we put the function

$$E\left(x\right)=\left(e^x-\sum _{k=0}^{{\displaystyle \frac{2}{{\gamma }_0}}}{\displaystyle \frac{x^k}{k!}}\right)\left(x^{{\displaystyle \frac{2}{{\gamma }_0}}+1}{e}^{{\displaystyle \frac{x}{2}}}\right),\forall x\in (0,\infty )$$

where ${\displaystyle \frac{2}{{\gamma }_0}}$ is the integer part of ${\displaystyle \frac{2}{\gamma }}$.

Further, for all $x\in (0,\infty )$ the function E is continuous, and $E>0$. Besides, E is bounded below as $x\to \infty$, i.e., ${lim}_{x\to \infty }E\left(x\right)=\infty$ also bounded below as $x\to 0^{+}$, i.e., ${lim}_{x\to 0^{+}}E\left(x\right)={\displaystyle \frac{1}{(\frac{2}{\gamma }+1)!}}$. Consequently, there is a positive constant $C_{{\gamma }_0}>0$ such that $E\left(x\right)\ge C_{{\gamma }_0}$ for all $x>0$. As a result, we can write

$${\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}\ge {\displaystyle \frac{C_1{C}_{{\gamma }_0}}{{(T_{a,\gamma }^{*}-t)}^{\frac{2}{3}}}}{\left({\displaystyle \frac{2a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)}^{{\displaystyle \frac{2}{{\gamma }_0}}+1}exp\left({\displaystyle \frac{a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)$$

or equivalently

$${\parallel u\left(t\right)\parallel }_{H_{a,\gamma }^1}\ge {\displaystyle \frac{C_3}{{(T_{a,\gamma }^{*}-t)}^{\frac{2({\gamma }_0+\gamma )+\gamma {\gamma }_0}{3{\gamma }_0}}}}exp\left({\displaystyle \frac{a{C}_2}{{(T_{a,\gamma }^{*}-t)}^{\frac{\gamma }{3}}}}\right)$$

with $C_3=C_1{C}_{{\gamma }_0}{\left(2a{C}_2\right)}^{{\displaystyle \frac{2}{{\gamma }_0}}+1}$. This concludes the proof of the inequality $\left(iv\right)$ and the proof of Theorem 2 is completes.

Remark 2.

The blow-up Equations (20)–(23) provide a quantitative description of the growth rate of the velocity field near the singularity. This can be viewed as a practical example illustrating how the energy of the flow increases in finite time within the Sobolev–Gevrey framework.

5. Conclusions

The main contributions of this paper can be summarized as follows:

• We established local existence and uniqueness of solutions for the modified three-dimensional convective Brinkman–Forchheimer system in Sobolev–Gevrey spaces.
• We derived several symmetry-preserving blow-up criteria, showing that the singularity thresholds depend mainly on the Gevrey regularity rather than on any loss of isotropic symmetry.
• These results generalize and refine previous studies on the Navier–Stokes and classical Brinkman–Forchheimer equations, providing a more precise description of the exponential-type blow-up in Sobolev–Gevrey frameworks.
• The analysis clarifies the role of analytic regularity in controlling nonlinear instabilities and symmetry behavior in viscous porous-media flows.
• Future work will focus on the decay and stability of global solutions under similar symmetry constraints.

These findings contribute to a deeper theoretical understanding of blow-up mechanisms in nonlinear fluid models governed by Gevrey regularity.