Global Analysis of Compressible Navier–Stokes–Korteweg Equations: Well-Posedness and Gevrey Analyticity

Abstract

This paper investigates the Cauchy problem for the full compressible Navier–Stokes–Korteweg equations, which model fluid dynamics with capillary properties in ${\mathbb{R}}^d(d\ge 3)$. And the global well-posedness and Gevrey analytic of strong solutions for the system are established in the $L^2-L^p$ type critical hybrid Besov space with $2\le p\le \frac{2d}{d-2}$ and $p<d$.

1. Introduction

This work focuses on a multi-dimensional, compressible viscous fluid endowed with capillary, a physically significant model in compressible fluid flows. The governing system takes the following form (see [1,2,3,4,5,6] and relevant references):

$$\left\{\begin{array}{c}{\partial }_t\rho +\mathrm{div}\left(\rho \mathit{u}\right)=0,\hfill \\ {\partial }_t\left(\rho u\right)+\mathrm{div}\left(\rho \mathit{u}\otimes \mathit{u}\right)+\nabla P=\mathrm{div}\left(\mathcal{D}+\mathcal{K}\right),\hfill \\ {\partial }_t\left[\rho \left(e+\frac{1}{2}{\left|\mathit{u}\right|}^2\right)\right]+\mathrm{div}\left[\mathit{u}\left(\rho e+\frac{1}{2}\rho {\left|\mathit{u}\right|}^2+P\right)\right]=\mathrm{div}((\mathcal{D}+\mathcal{K})·\mathit{u}-q).\hfill \end{array}\right.$$

(1)

Here, $\rho =\rho (t,\mathit{x})\in {\mathbb{R}}_{+}$ ($(t,\mathit{x})\in {\mathbb{R}}_{+}\times {\mathbb{R}}^d$, $d\ge 1$) represents the density, $\mathit{u}=\mathit{u}(t,\mathit{x})\in {\mathbb{R}}^d$ denotes the velocity field, and e stands for the internal energy per unit mass fulfilling the Joule law: ${\partial }_{\mathcal{T}}e=C_v$, with $C_v$ being positive constants. The viscous stress tensor $\mathcal{D}$ is defined as $\mathcal{D}=2\mu \left(\rho \right)D\left(\mathit{u}\right)+\lambda \left(\rho \right)\mathrm{div}\mathit{u}\mathrm{Id}$. In this expression, the bulk $\mu \left(\rho \right)$ and the shear $\lambda \left(\rho \right)$ are density-dependent functions, which are supposed to be smooth enough and fulfill $\mu \left(\rho \right)>0$ and $2\mu \left(\rho \right)+\lambda \left(\rho \right)>0$. $D\left(\mathit{u}\right)=\frac{1}{2}(\nabla \mathit{u}+{}^{\top }\nabla \mathit{u})$ denotes the deformation tensor and $\mathrm{Id}$ stands for the unit matrix. Heat conduction is described by $q=-\eta \left(\rho \right)\nabla \mathcal{T}$, where $\eta \left(\rho \right)>0$ stands for a smooth function on $\rho$, and $\mathcal{T}$ is temperature. The Korteweg tensor $\mathcal{K}$ is represented by the following expression:

$$\mathcal{K}=\rho \mathrm{div}(\kappa \left(\rho \right)\nabla \rho )\mathrm{Id}+\frac{1}{2}\left(\kappa \left(\rho \right)-\rho {\kappa }^{\prime }\left(\rho \right)\right){|\nabla \rho |}^2\mathrm{Id}-\kappa \left(\rho \right)\nabla \rho \otimes \nabla \rho .$$

Here, $\kappa \left(\rho \right)>0$ stands for the capillarity function, which depends only on $\rho$, and is supposed to be smooth enough.

In particular, when $\kappa \left(\rho \right)=\frac{\kappa }{\rho }$, system (1) contains the so-called quantum Navier–Stokes equations (see [7]). This specific system has undergone a thorough derivation and in-depth analysis by A Jüngel and J. P. Milisic in [8,9]. In this context, P signifies pressure and takes the following form:

$$P(\rho ,\mathcal{T})={\pi }_0+\mathcal{T}{\pi }_1,$$

with ${\pi }_i(i=1,2)$ being smooth functions on the density $\rho$. This pressure law is very general and encompasses various fluid models. In particular, for ideal fluids, ${\pi }_0$ is zero and ${\pi }_1$ is equal to $R\rho$, with R being a universal positive constant. Meanwhile, for barotropic fluids, ${\pi }_1=0$.

Using thermodynamic Gibbs relations for both internal energy and Helmholtz free energy, one can conclude the following system on the temperature by the Maxwell relation. And finally, it follows from (1) that

$$\left\{\begin{array}{c}{\partial }_t\rho +\mathrm{div}\left(\rho \mathit{u}\right)=0,\hfill \\ {\partial }_t\mathit{u}+\mathit{u}·\nabla \mathit{u}+\frac{1}{\rho }\nabla \left({\pi }_0\left(\rho \right)+\mathcal{T}{\pi }_1\left(\rho \right)\right)=\frac{1}{\rho }\mathrm{div}\left(2\mu \left(\rho \right)D\left(\mathit{u}\right)+\lambda \left(\rho \right)\mathrm{div}\mathit{u}\mathrm{Id}\right)\hfill \\ +\nabla \left(\kappa \left(\rho \right)\Delta \rho +\frac{1}{2}{\kappa }^{\prime }\left(\rho \right){|\nabla \rho |}^2\right),\hfill \\ {\partial }_t\mathcal{T}+\mathit{u}·\nabla \mathcal{T}+\frac{1}{C_v\rho }\mathcal{T}{\pi }_1\left(\rho \right)\mathrm{div}\mathit{u}=\frac{1}{C_v\rho }\left(2\mu \left(\rho \right)D\left(\mathit{u}\right):D\left(\mathit{u}\right)+\lambda \left(\rho \right){\left(\mathrm{div}\mathit{u}\right)}^2\right)\hfill \\ +\frac{1}{C_v\rho }\mathrm{div}(\eta \left(\rho \right)\nabla \mathcal{T})+\frac{1}{C_v}\left(\kappa \left(\rho \right)\Delta \rho +\frac{1}{2}\nabla \kappa \left(\rho \right)·\nabla \rho \right)\mathrm{div}\mathit{u}\hfill \\ +\frac{\kappa \left(\rho \right)}{C_v\rho }\left(\frac{1}{2}{|\nabla \rho |}^2\mathrm{div}\mathit{u}-\nabla \rho \otimes \nabla \rho :\nabla \mathit{u}\right).\hfill \end{array}\right.$$

(2)

Here, the definition of $A:B$ is given by

$$A:B=\mathrm{Tr}AB=\sum _{1\le i,j\le d}{A}_{ij}{B}_{ji},\mathrm{with}A={\left(A_{ij}\right)}_{1\le i,j\le d},B={\left(B_{ij}\right)}_{1\le i,j\le d}.$$

In this paper, we analyze the initial problem of system (2) with its corresponding initial data being

$${(\rho ,\mathit{u},\mathcal{T})|}_{t=0}=({\rho }_0,{\mathit{u}}_0,{\mathcal{T}}_0)$$

(3)

and concentrate on strong solutions that approximate a specific constant equilibrium $({\rho }_{\infty },0,{\mathcal{T}}_{\infty })$, where ${\rho }_{\infty }$ and ${\mathcal{T}}_{\infty }$ are positive and satisfy the following condition:

$${\partial }_{\rho }P({\rho }_{\infty },{\mathcal{T}}_{\infty })>0\mathrm{and}{\partial }_{\mathcal{T}}P({\rho }_{\infty },{\mathcal{T}}_{\infty })>0.$$

(4)

The full compressible Navier–Stokes–Korteweg system was originally established by J. F. Van der Waals [10] and Korteweg [11] during their research on the theory of capillarity involving diffuse interfaces. Subsequently, Dunn and Serrin [12] provided a rigorous derivation of the capillarity theory with diffuse interfaces. In fluid mechanics, especially in the liquid–gas two-phase flow (see [13,14,15]), the Korteweg term is often used to characterize the capillary phenomenon caused by surface tension. Owing to its significant physical implications and the associated mathematical difficulties, numerous studies have been conducted on the well-posedness and time-decay estimates of solutions. Under the high Sobolev regularity framework, Hattori and Li [16] studied the global well-posedness of smooth solutions. Later, Hou, Peng, and Zhu [3] improved the work and investigated the global existence of solutions for the Cauchy problem of the system (1) on the condition that the initial energy is small. Chen, He, and Zhao [1] established the global well-posedness of smooth solutions to the initial problem of the $1D$ model with large initial data. Hou, Yao, and Zhu [4] proved that, as time progresses globally, the unique smooth solutions of the $3D$ Korteweg system tend toward the smooth solutions of the $3D$ full compressible Navier–Stokes system. Kotschote [17] derived the theory of maximal $L^p$-regularity, and based on this, he proved the local existence and uniqueness of strong solutions within a bounded domain whose boundary exhibits $C^3$-smoothness. Later, he investigated the global well-posedness and time-decay estimates of strong solutions in [18]. In [2], Chen and Zhao studied the global existence, uniqueness, and nonlinear stability of stationary solutions for the initial problem with a general form external force. Li [19] and Wang and Wang [20] studied the large-time behavior of strong and smooth solutions under the action of a potential external force, respectively. Li, Liu, and Yin [21] explored the outflow problem and investigated the time-decay estimates of solutions.

This study aims to explore the global well-posedness of strong solutions to Equations (2) and (3) within critical regularity spaces (see [22]). The norms associated with these functional spaces exhibit invariance under the following transformation for any positive value of l:

$$\rho (t,x)⇝\rho (l^{2}t,lx),u(t,x)⇝lu(l^{2}t,lx)\mathrm{and}\mathcal{T}(t,x)⇝l^2\mathcal{T}(l^{2}t,lx).$$

(5)

This definition of criticality relates to the property of scaling invariance (2), provided the pressure laws ${\pi }_0$, ${\pi }_1$ have been changed into $l^2{\pi }_0$, $l^2{\pi }_1$. In what follows, we review some previous studies which motivated us to start this work. For the isentropic case, Wang and Wang [23] investigated the large-time behavior of mild solutions in the nonhomogeneous critical Besov spaces. Danchin and Desjardins [24] proved the global well-posedness of solutions in the $L^2$-type critical Besov spaces. Later, Charve, Danchin, and Xu [25] generalized the works of [24] to more general $L^p$-type critical Besov spaces. And then, Kawashima, Shibata, and Xu [26] further established the large-time behavior of the solutions (constructed in [25]) in the $L^p$-type critical Besov spaces. Chikami and Kobayashi [27] investigated the global existence of solutions for the zero sound speed case in critical Besov spaces, while also deriving optimal temporal decay rates for solutions with non-vanishing sound speed. Zhang [28] extended these results by proving global well-posedness under a smallness assumption on initial data, which holds regardless of whether the sound speed vanishes or remains positive. Notably, Zhang’s work permits the initial velocity to be arbitrarily large in ${\dot{B}}_{p,1}^{d/p-1}$. Recently, Song [29] considered the large data problems and investigated the global existence of solutions in the critical framework. However, as far as we know, the research from the perspective of scaling invariance on the non-isentropic system remains relatively scarce. Drawing inspiration from the scaling characteristics, Haspot [30] extended the findings of [24] to the non-isentropic system scenario. In this extended study, the global well-posedness is established in the $L^2$ framework with the smallness imposed on both low and high frequency. This paper endeavors to apply the Fourier method to achieve the desired global existence and Gevrey analyticity of strong solutions in the general $L^p$ framework with the smallness imposed only on low frequency.

The organization of the rest of this manuscript is outlined below. In Section 2, we briefly explain the notations used in this paper and go over the relevant analysis tools. In Section 3, we give the reformulation of system (2) and show our main works. Finally, the proof of Theorem 1 is given Section 4.

2. Preliminary

Let us briefly go over the definition and some properties of the corresponding space and introduce some nonlinear estimates involving Besov Gevrey regularity.

2.1. An Overview of Besov Spaces and Their Properties

For the convenience of the reader, we recall the definitions and the properties used in this paper of Besov space and Chemin–Lerner spaces (see Chapters 2 and 3 of [22,31] for more details). Based on the Littlewood–Paley decomposition, one can find the following definition.

Definition 1.

Assuming $\sigma \in \mathbb{R}$ and $1\le p,r\le \infty ,$ ${\mathcal{S}}^{\prime }$ being tempered distributions, the Besov spaces ${\dot{B}}_{p,r}^{\sigma }$ are defined by

$${\dot{B}}_{p,r}^{\sigma }\triangleq \{f\in {\mathcal{S}}_0^{\prime }{:\parallel f\parallel }_{{\dot{B}}_{p,r}^{\sigma }}<+\infty \}\mathit{with}{\mathcal{S}}_0^{\prime }=\{f\in {\mathcal{S}}^{\prime }|\underset{j\to -\infty }{lim}\parallel {\dot{\sigma }}_{j}u{\parallel }_{L^{\infty }}=0\}.$$

Here,

$${\parallel f\parallel }_{{\dot{B}}_{p,r}^{\sigma }}\triangleq \parallel (2^{js}\parallel {\dot{\Delta }}_j{f\parallel }_{L^p}{)\parallel }_{{\ell }^r\left(\mathbb{Z}\right)}$$

(6)

with ${\dot{\Delta }}_j$ being dyadic blocks.

When investigating evolution PDEs, considering the Besov spaces with mixed space-time is a reasonable approach. These spaces were initially introduced by J.-Y. Chemin and N. Lerner in [32,33].

Definition 2.

Assuming $T>0,\sigma \in \mathbb{R}$ and $1\le r,\varrho \le \infty$, we define the Chemin–Lerner space ${\tilde{L}}_T^{\varrho }\left({\dot{B}}_{p,r}^{\sigma }\right)$ as follows:

$${\tilde{L}}_T^{\varrho }\left({\dot{B}}_{p,r}^{\sigma }\right)\triangleq \{f\in L^{\varrho }(0,T;{\mathcal{S}}_0^{\prime }){:\parallel f\parallel }_{{\tilde{L}}_T^{\varrho }\left({\dot{B}}_{p,r}^{\sigma }\right)}<+\infty \},$$

with the norm

$${\parallel f\parallel }_{{\tilde{L}}_T^{\varrho }\left({\dot{B}}_{p,r}^{\sigma }\right)}\triangleq \parallel (2^{js}\parallel {\dot{\Delta }}_j{f\parallel }_{L_T^{\varrho }\left(L^p\right)}{)\parallel }_{{\ell }^r\left(\mathbb{Z}\right)}.$$

(7)

We denote by ${\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{p,r}^{\sigma })$ a functional space that is defined as the following:

$${\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{p,r}^{\sigma })\triangleq \{u\in \mathcal{C}({\mathbb{R}}_{+};{\dot{B}}_{p,r}^{\sigma }){\mathrm{s}.\mathrm{t}\parallel f\parallel }_{{\tilde{L}}^{\infty }\left({\dot{B}}_{p,r}^{\sigma }\right)}<+\infty \},$$

where the index T is omitted for $T=+\infty$.

In this paper, a crucial step involves limiting the norms defined in (6) and (7) to the low- or high-frequency components of distributions. To illustrate, let us select an integer $j_0$.

$${\parallel f\parallel }_{{\dot{B}}_{p,1}^{\sigma }}^{\ell }\triangleq \sum _{j\le j_0}{2}^{j\sigma }\parallel {\dot{\Delta }}_j{f\parallel }_{L^p}{\mathrm{and}\parallel f\parallel }_{{\dot{B}}_{p,1}^{\sigma }}^h\triangleq \sum _{j\ge j_0-1}{2}^{j\sigma }{\parallel {\dot{\Delta }}_{j}f\parallel }_{L^p},$$

$${\parallel f\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\sigma }\right)}^{\ell }\triangleq \sum _{j\le j_0}{2}^{j\sigma }\parallel {\dot{\Delta }}_j{f\parallel }_{L_T^{\infty }\left(L^p\right)}{\mathrm{and}\parallel f\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\sigma }\right)}^h\triangleq \sum _{j\ge j_0-1}{2}^{j\sigma }{\parallel {\dot{\Delta }}_{j}f\parallel }_{L_T^{\infty }\left(L^p\right)}.$$

What we need to clarify is that the technical constraints necessitate a small overlap in the low- and high-frequency ranges.

Let us review the fundamental properties of Besov spaces, and recall the product law and composite function estimates.

Proposition 1.

(Embedding properties)

• Assuming $p\in [1,\infty ],$ then ${\dot{B}}_{p,1}^0\hookrightarrow L^p\hookrightarrow {\dot{B}}_{p,\infty }^0.$
• Assuming $\sigma \in \mathbb{R}$, $1\le p_1\le p_2\le \infty ,$ and $1\le r_1\le r_2\le \infty ,$ then ${\dot{B}}_{p_1,r_1}^{\sigma }\hookrightarrow {\dot{B}}_{p_2,r_2}^{\sigma -d(\frac{1}{p_1}-\frac{1}{p_2})}$.
• The space ${\dot{B}}_{p,1}^{\frac{d}{p}}$ admits a continuous embedding into the space of bounded continuous functions (which decay at infinity if further $p<\infty$).

Proposition 2.

Assume $1\le p,r,r_1,r_2\le \infty$.

• If $f\in {\dot{B}}_{p,r_1}^{{\sigma }_1}\cap {\dot{B}}_{p,r_2}^{{\sigma }_2}$ and ${\sigma }_1\ne {\sigma }_2$, then $f\in {\dot{B}}_{p,r}^{\varrho {\sigma }_1+(1-\varrho ){\sigma }_2}$ for all $\varrho \in (0,1)$ and

  $${\parallel f\parallel }_{{\dot{B}}_{p,r}^{\varrho {\sigma }_1+(1-\varrho ){\sigma }_2}}\le {\parallel f\parallel }_{{\dot{B}}_{p,r_1}^{{\sigma }_1}}^{\varrho }{\parallel f\parallel }_{{\dot{B}}_{p,r_2}^{{\sigma }_2}}^{1-\varrho }\mathit{with}\frac{1}{r}=\frac{\varrho }{r_2}+\frac{1-\varrho }{r_2}.$$
• Assuming $f\in {\dot{B}}_{p,\infty }^{{\sigma }_1}\cap {\dot{B}}_{p,\infty }^{{\sigma }_2}$ and ${\sigma }_1<{\sigma }_2$, then $f\in {\dot{B}}_{p,1}^{\varrho {\sigma }_1+(1-\varrho ){\sigma }_2}$ for any $\varrho \in (0,1)$ and

  $${\parallel f\parallel }_{{\dot{B}}_{p,1}^{\varrho {\sigma }_1+(1-\varrho ){\sigma }_2}}\le \frac{C}{\varrho (1-\varrho )({\sigma }_2-{\sigma }_1)}{\parallel f\parallel }_{{\dot{B}}_{p,\infty }^{{\sigma }_1}}^{\varrho }{\parallel f\parallel }_{{\dot{B}}_{p,\infty }^{{\sigma }_2}}^{1-\varrho }.$$

In addition, we should also bring to mind the following Bernstein inequality

$$\parallel D^k{f\parallel }_{L^b}\le C^{1+k}{\lambda }^{k+d(\frac{1}{a}-\frac{1}{b})}{\parallel f\parallel }_{L^a}$$

(8)

for all $k\in \mathbb{N}$ and $1\le a\le b\le \infty$.

Let us establish a generalized frequency-localized estimate: for any function f with Fourier support satisfying $\mathrm{Supp}\mathcal{F}f\subset \{\xi \in {\mathbb{R}}^d:R_1\lambda \le |\xi |\le R_2\lambda \}$ (where $0<R_1<R_2$ and $\lambda >0$), and for any smooth homogeneous function A of degree m defined on ${\mathbb{R}}^d\setminus \left\{0\right\},$ the following inequality holds for $1\le a\le \infty$ (see, e.g., Lemma 2.2 in [22]):

$${\parallel A\left(D\right)f\parallel }_{L^a}\lesssim {\lambda }^m{\parallel f\parallel }_{L^a}.$$

(9)

This immediately yields the equivalence relation between derivatives in Besov spaces: $\parallel D^k{f\parallel }_{{\dot{B}}_{p,r}^{\sigma }}\approx {\parallel f\parallel }_{{\dot{B}}_{p,r}^{\sigma +k}}$ for all $k\in \mathbb{N}.$

The following fundamental product estimates play a crucial role in controlling nonlinear terms and are used repeatedly throughout our analysis.

Proposition 3

([22,34]). Let $1\le p,r\le \infty$, then the following product laws hold:

(a) Assuming $\sigma >0$, we have

$${\parallel fg\parallel }_{{\dot{B}}_{p,r}^{\sigma }}\lesssim {\parallel f\parallel }_{L^{\infty }}{\parallel g\parallel }_{{\dot{B}}_{p,r}^{\sigma }}+{\parallel g\parallel }_{L^{\infty }}{\parallel f\parallel }_{{\dot{B}}_{p,r}^{\sigma }}.$$

(b) Assuming ${\sigma }_1,{\sigma }_2\le \frac{d}{p}$ and ${\sigma }_1+{\sigma }_2>dmax(0,\frac{2}{p}-1)$, we have

$${\parallel fg\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_1+{\sigma }_2-\frac{d}{p}}}\lesssim {\parallel f\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_1}}{\parallel g\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_2}}.$$

(c) Assuming ${\sigma }_1\le \frac{d}{p},{\sigma }_2<\frac{d}{p},$ and ${\sigma }_1+{\sigma }_2\ge dmax(0,\frac{2}{p}-1)$, we have

$${\parallel fg\parallel }_{{\dot{B}}_{p,\infty }^{{\sigma }_1+{\sigma }_2-\frac{d}{p}}}\lesssim {\parallel f\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_1}}{\parallel g\parallel }_{{\dot{B}}_{p,\infty }^{{\sigma }_2}}.$$

To establish proper Lebesgue norm compatibility across different frequency regimes, we must develop modified product estimates beyond classical results. Specifically,

Proposition 4

([35,36]). Assume ${\sigma }_1$, ${\sigma }_2$, $p_1$, and $p_2$ are the real numbers satisfying

$${\sigma }_1+{\sigma }_2>0,{\sigma }_1\le \frac{d}{p_1},{\sigma }_2\le \frac{d}{p_2},{\sigma }_1\ge {\sigma }_2,\mathit{and}\frac{1}{p_1}+\frac{1}{p_2}\le 1.$$

Then the following product laws hold:

$${\parallel fg\parallel }_{{\dot{B}}_{q,1}^{{\sigma }_2}}\lesssim {\parallel f\parallel }_{{\dot{B}}_{p_1,1}^{{\sigma }_1}}{\parallel g\parallel }_{{\dot{B}}_{p_2,1}^{{\sigma }_2}}\mathit{with}\frac{1}{q}=\frac{1}{p_1}+\frac{1}{p_2}-\frac{{\sigma }_1}{d}.$$

The product estimates above can lead to the following inequalities.

Corollary 1

([26,37]). Let $1-\frac{d}{2}<{\sigma }_1\le {\sigma }_0=\frac{2d}{p}-\frac{d}{2}$ and p satisfy (14). It holds that

$${\parallel fg\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }\lesssim {\parallel fg\parallel }_{{\dot{B}}_{2,\infty }^{-{\sigma }_0}}^{\ell }\lesssim {\parallel f\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}{\parallel g\parallel }_{{\dot{B}}_{p,1}^{1-\frac{d}{p}}}.$$

In order to establish the proof of Theorem 1, the following nonstandard product estimate (see for example [36,38]) is additionally required.

Proposition 5

([39]). Assuming $j_0\in \mathbb{Z}$ and set $z^{\ell }\triangleq {\dot{S}}_{j_0}z$, $z^h\triangleq z-z^{\ell }$, then for all $\sigma \in \mathbb{R}$,

$${\parallel z\parallel }_{{\dot{B}}_{2,\infty }^{\sigma }}^{\ell }\triangleq \underset{j\le j_0}{sup}{2}^{j\sigma }{\parallel {\dot{\Delta }}_{j}z\parallel }_{L^2}.$$

One can find a positive integer $N_0$ such that for every $2\le p\le 4$ and arbitrary $\sigma >0$, it holds that

$$\begin{array}{ccc}\hfill & & \parallel f{g}^h{\parallel }_{{\dot{B}}_{2,\infty }^{-{\sigma }_0}}^{\ell }\le C\left({\parallel f\parallel }_{{\dot{B}}_{p,1}^{\sigma }}+{\parallel {\dot{\sigma }}_{j_0+N_0}f\parallel }_{L^{p^{*}}}\right){\parallel g^h\parallel }_{{\dot{B}}_{p,\infty }^{-\sigma }},\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill & & \parallel f^h{g\parallel }_{{\dot{B}}_{2,\infty }^{-{\sigma }_0}}^{\ell }\le C\left(\parallel f^h{\parallel }_{{\dot{B}}_{p,1}^{\sigma }}+{\parallel {\dot{\sigma }}_{j_0+N_0}{f}^h\parallel }_{L^{p^{*}}}\right){\parallel g\parallel }_{{\dot{B}}_{p,\infty }^{-\sigma }}\hfill \end{array}$$

(11)

with ${\sigma }_0\triangleq \frac{2d}{p}-\frac{d}{2}$ and $\frac{1}{p^{*}}+\frac{1}{p}=\frac{1}{2}$, and C depending only on $j_0$, d, and σ.

To bound the nonlinear compositions appearing in (13), we establish the following estimates:

Proposition 6

([26,35,37]). Assume $F:\mathbb{R}\to \mathbb{R}$ is smooth and satisfying $F\left(0\right)=0$. Then for any $1\le p,r\le \infty$, $\sigma >0$, it holds that $F\left(u\right)\in {\dot{B}}_{p,r}^{\sigma }\cap L^{\infty }$ for $u\in {\dot{B}}_{p,r}^{\sigma }\cap L^{\infty }$, and

$${\parallel F\left(u\right)\parallel }_{{\dot{B}}_{p,r}^{\sigma }}\le C{\parallel u\parallel }_{{\dot{B}}_{p,r}^{\sigma }}$$

with C depending only on ${\parallel u\parallel }_{L^{\infty }}$, σ, p, and d.

When $\sigma >-min\left(\frac{d}{p},\frac{d}{p^{\prime }}\right)$, we have $u\in {\dot{B}}_{p,r}^{\sigma }\cap {\dot{B}}_{p,1}^{\frac{d}{p}}$ that implies $F\left(u\right)\in {\dot{B}}_{p,r}^{\sigma }\cap {\dot{B}}_{p,1}^{\frac{d}{p}}$, and we have

$${\parallel F\left(u\right)\parallel }_{{\dot{B}}_{p,r}^{\sigma }}\le C(1+{\parallel u\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}){\parallel u\parallel }_{{\dot{B}}_{p,r}^{\sigma }}.$$

2.2. Some Nonlinear Estimates Involving Besov Gevrey Regularity

This subsection focuses on deriving essential nonlinear estimates in Besov Gevrey spaces for controlling the right-hand terms of (13).

Following standard harmonic analysis techniques, we employ the Bony decomposition for products of tempered distributions:

$$fg=T_{f}g+R(f,g)+T_{g}f,$$

(12)

where we define the paraproduct and remainder operators, respectively, as

$$T_{f}g\triangleq \sum _{j\in \mathbb{Z}}{\dot{\sigma }}_{j-1}f{\dot{\Delta }}_{j}g\mathrm{and}R(f,g)\triangleq \sum _{j\in \mathbb{Z}}\sum _{|j^{\prime }-j|\le 1}{\dot{\Delta }}_{j}f{\dot{\Delta }}_{j^{\prime }}g.$$

Throughout this subsection, we shall fix some real number $c>0$ and argue that $F\left(t\right)\triangleq e^{\sqrt{ct}{\Lambda }_1}f$ and $G\left(t\right)\triangleq e^{\sqrt{ct}{\Lambda }_1}g$ for $t>0$. Our investigation commences with the analysis of these operators for Lebesgue exponents $p\in (1,\infty )$.

Proposition 7.

Let $s\in \mathbb{R}$, $1<p,p_1,p_2<\infty$ and $1\le r,r_1,r_2\le \infty$ with $1/p=1/p_1+1/p_2$ and $1/r=1/r_1+1/r_2$. One can find a constant C such that for any f, g and $\sigma >0$ (or $\sigma \ge 0$ if $r_1=1$),

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{T}_f{g\parallel }_{{\dot{B}}_{p,r}^{s-\sigma }}\le {C\parallel F\parallel }_{{\dot{B}}_{p_1,r_1}^{-\sigma }}{\parallel G\parallel }_{{\dot{B}}_{p_2,r_2}^s},$$

and for any ${\sigma }_1,{\sigma }_2\in \mathbb{R}$ with ${\sigma }_1+{\sigma }_2>0$,

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{R(f,g)\parallel }_{{\dot{B}}_{p,r}^{{\sigma }_1+{\sigma }_2}}\le {C\parallel F\parallel }_{{\dot{B}}_{p_1,r_1}^{{\sigma }_1}}{\parallel G\parallel }_{{\dot{B}}_{p_2,r_2}^{{\sigma }_2}}.$$

Proposition 7 can lead to the following inequalities.

Corollary 2.

Let p satisfy (14) and $d\ge 3$, then there exists a constant $C>0$ such that

$$\begin{array}{cc}\hfill \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}\left(T_{f}g\right){\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-3}}& \le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}},\hfill \\ \hfill \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}\left(T_{f}g\right){\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-2}}& \le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}},\hfill \\ \hfill \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}{R(f,g)\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-2}}& \le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}.\hfill \end{array}$$

Corollary 3.

If p satisfies (14) and $d\ge 3$, then there exists a constant $C>0$ such that

$$\begin{array}{cc}\hfill \parallel e^{\sqrt{ct}{\Lambda }_1}{\left(fg\right)\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }& \le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}+{\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}},\hfill \\ \hfill \parallel e^{\sqrt{ct}{\Lambda }_1}{\left(fg\right)\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }& \le C({\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}+{\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}){\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}},\hfill \\ \hfill \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}{\left(fg\right)\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-2}}& \le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}.\hfill \end{array}$$

To prove our main results, we also need the following estimates corresponding to the case $p_2=p$.

Proposition 8.

Suppose that $1<p,q<\infty$ and $1\le r,r_1,r_2\le \infty$ meet the condition $1/r=1/r_1+1/r_2$. Then we can find a constant C such that for any f, g and $\sigma >0$ (or $\sigma \ge 0$ if $r_1=1$),

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{T}_f{g\parallel }_{{\dot{B}}_{p,r}^{s-\sigma }}\le {C\parallel F\parallel }_{{\dot{B}}_{q,r_1}^{\frac{d}{q}-\sigma }}{\parallel G\parallel }_{{\dot{B}}_{p,r_2}^s},$$

and for any ${\sigma }_1,{\sigma }_2\in \mathbb{R}$ with ${\sigma }_1+{\sigma }_2>0$,

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{R(f,g)\parallel }_{{\dot{B}}_{p,r}^{{\sigma }_1+{\sigma }_2}}\le {C\parallel F\parallel }_{{\dot{B}}_{q,r_1}^{{\sigma }_1+\frac{d}{q}}}{\parallel G\parallel }_{{\dot{B}}_{p,r_2}^{{\sigma }_2}}.$$

By integrating the preceding proposition with functional embedding results and Bony’s decomposition technique, we obtain the following crucial Gevrey product estimates, which will play a fundamental role in our subsequent analysis.

Proposition 9.

Assume $1<p<\infty$, ${\sigma }_1,{\sigma }_2\le \frac{d}{p}$ such that

$${\sigma }_1+{\sigma }_2>dmax\left(0,-1+\frac{2}{p}\right).$$

We can establish the following uniform bound with some positive constant C:

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{\left(fg\right)\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_1+{\sigma }_2-\frac{d}{p}}}\le {C\parallel F\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_1}}{\parallel G\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_2}}.$$

Remark 1.

Proposition 9 implies the space $\{f\in {\dot{B}}_{p,1}^{\frac{d}{p}}|e^{\sqrt{ct}{\Lambda }_1}f\in {\dot{B}}_{p,1}^{\frac{d}{p}}\}$ is an algebra whenever $1<p<\infty$.

To complete the proof of our main result, our analysis requires both (i) bilinear estimates in Besov Gevrey spaces and (ii) composition estimates for real analytic functions.

Proposition 10.

Assume Φ is a real analytic function in a neighborhood of 0, such that $\Phi \left(0\right)=0$. Assume $1<p<\infty$ and satisfies $-min\left(\frac{d}{p},\frac{d}{p^{\prime }}\right)<\sigma \le \frac{d}{p}$ with $\frac{1}{p^{\prime }}+\frac{1}{p}=1$. Then one can find two constants $R_0$ and D depending only on p, d, and Φ such that if, for some $T>0$,

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{z\parallel }_{{\tilde{L}}_T^{\infty }({\dot{B}}_{p,1}^{\frac{d}{p}})}\le R_0,$$

then we have for all $q\in [1,\infty ]$

$$\parallel e^{\sqrt{ct}{\Lambda }_1}{\Phi \left(z\right)\parallel }_{{\tilde{L}}_T^q\left({\dot{B}}_{p,1}^{\sigma }\right)}\le D{\parallel z\parallel }_{{\tilde{L}}_T^q\left({\dot{B}}_{p,1}^{\sigma }\right)}.$$

Remark 2.

In proving our principal theorems, we employ the above proposition with the parameter choices $\sigma =\frac{d}{p}$ (requiring $d\ge 2$ and $1<p<2d$) or $\sigma =\frac{d}{p}-1$.

Finally, let us end this section with the endpoint maximal regularity property of the heat system with a complex diffusion coefficient (see [25] for details).

Proposition 11.

Let $T>0$, $\sigma \in \mathbb{R}$ and $1\le {\varrho }_2,p,r\le \infty$. Let u fulfill

$$\left\{\begin{array}{c}{\partial }_{t}u-\nu \Delta u=f,\hfill \\ u_{|t=0}=u_0.\hfill \end{array}\right.$$

Here, $\nu \in \mathbb{C}$ is a complex number and its real part $Re\nu$ is positive. Then, one can find a positive constant C depending only on d such that for every ${\varrho }_1\in [{\varrho }_2,\infty ]$, it holds that

$${\left(Re\nu \right)}^{\frac{1}{{\varrho }_1}}{\parallel u\parallel }_{{\tilde{L}}_T^{{\varrho }_1}({\dot{B}}_{p,r}^{\sigma +\frac{2}{{\varrho }_1}})}\le C(\parallel u_0{\parallel }_{{\dot{B}}_{p,r}^{\sigma }}+{\left(Re\nu \right)}^{\frac{1}{{\varrho }_2}-1}{\parallel f\parallel }_{{\tilde{L}}_T^{{\varrho }_2}({\dot{B}}_{p,r}^{\sigma -2+\frac{2}{{\varrho }_2}})}).$$

In the entire paper, C stands for a generic positive constant that may change from line to line. $A\lesssim B$ or $A\gtrsim B$ means that $A\le CB$ or $A\ge CB$, and we also use $A\sim B$ to denote that $B\lesssim A\lesssim B$. For any Banach space X and $A,B$ within it, we define the norm ${\parallel (A,B)\parallel }_X\triangleq {\parallel A\parallel }_X+{\parallel B\parallel }_X$. For all $T>0$ and $\varrho \in [1,+\infty ]$, the notation $L_T^{\varrho }\left(X\right)\triangleq L^{\varrho }(0,T;X)$ or $L_T^{\varrho }\left(X\right)$ designates the set of measurable functions $f:[0,T]\to X$ with $t\mapsto {\parallel f\left(t\right)\parallel }_X$ in $L^{\varrho }(0,T)$, endowed with the norm ${\parallel ·\parallel }_{L_T^{\varrho }\left(X\right)}{\triangleq \parallel \parallel ·\parallel }_X{\parallel }_{L^{\varrho }(0,T)}$. Let $\mathcal{F}\left(f\right)=\widehat{f}$ be the Fourier transform of f and correspondingly we use ${\mathcal{F}}^{-1}\left(f\right)=\check{f}$ to denote its inverse, and further, ${\Lambda }^{\sigma }f\triangleq {\mathcal{F}}^{-1}{\left(\right|\xi |}^{\sigma }\mathcal{F}\left(f\right))$$(\sigma \in \mathbb{R})$.

3. Reformulation of System (2) and Main Results

Before annunciating the central assertion of this paper, we shall transform Equation (2) into a nonlinear perturbation format with respect to the constant equilibrium state $({\rho }_{\infty },0,{\mathcal{T}}_{\infty })$ with ${\rho }_{\infty }>0$ and ${\mathcal{T}}_{\infty }>0$, dealing with the nonlinear terms as source terms. Assuming $\rho ={\rho }_{\infty }(1+a)$ and $\mathcal{T}=\theta +{\mathcal{T}}_{\infty }$, it follows from (2) that whenever $a>-1$, the triplet $(a,\mathit{u},\theta )$ fulfills

$$\left\{\begin{array}{c}{\partial }_{t}a+\mathrm{div}\mathit{u}=f,\hfill \\ {\partial }_t\mathit{u}-\mathcal{A}\mathit{u}+{\alpha }_1\nabla a-\delta \nabla \Delta a+{\gamma }_1\nabla \theta =g,\hfill \\ {\partial }_t\theta -{\alpha }_2\Delta \theta +{\gamma }_2\mathrm{div}\mathit{u}=h.\hfill \end{array}\right.$$

(13)

Here, $\mathcal{A}\triangleq \overline{\mu }\Delta +(\overline{\mu }+\overline{\lambda })\nabla \mathrm{div}=\frac{{\mu }_{\infty }}{{\rho }_{\infty }}\Delta +\frac{{\mu }_{\infty }+{\lambda }_{\infty }}{{\rho }_{\infty }}\nabla \mathrm{div}$, $\delta \triangleq {\rho }_{\infty }{\kappa }_{\infty }$, (${\mu }_{\infty }\triangleq \mu \left({\rho }_{\infty }\right),{\lambda }_{\infty }\triangleq \lambda \left({\rho }_{\infty }\right)$ and ${\kappa }_{\infty }\triangleq \kappa \left({\rho }_{\infty }\right)$), ${\alpha }_1\triangleq {\pi }_0^{\prime }\left({\rho }_{\infty }\right)+{\mathcal{T}}_{\infty }{\pi }_1^{\prime }\left({\rho }_{\infty }\right)={\partial }_{\rho }P({\rho }_{\infty },{\mathcal{T}}_{\infty })$, ${\gamma }_1\triangleq \frac{{\pi }_1\left({\rho }_{\infty }\right)}{{\rho }_{\infty }}$, ${\alpha }_2\triangleq \frac{{\eta }_{\infty }}{C_v{\rho }_{\infty }}({\eta }_{\infty }\triangleq \eta \left({\rho }_{\infty }\right))$, ${\gamma }_2\triangleq \frac{{\mathcal{T}}_{\infty }{\pi }_1\left({\rho }_{\infty }\right)}{C_v{\rho }_{\infty }}$. And the nonlinear terms on the right side of the density equation and velocity equation are accordingly expressed as follows:

$$\begin{array}{ccc}\hfill f& \triangleq & -\mathrm{div}\left(a\mathit{u}\right),\hfill \\ \hfill g& \triangleq & -\mathit{u}·\nabla \mathit{u}+\frac{1}{{\rho }_{\infty }(1+a)}\mathrm{div}\left(2\tilde{\mu }\left(a\right)D\left(\mathit{u}\right)+\tilde{\lambda }\left(a\right)\mathrm{div}\mathit{u}\mathrm{Id}\right)-I\left(a\right)\mathcal{A}\mathit{u}\hfill \\ & & -k_1\left(a\right)\nabla a+{\rho }_{\infty }\nabla \left(\tilde{\kappa }\left(a\right)\Delta a+\frac{1}{2}\nabla \tilde{\kappa }\left(a\right)·\nabla a\right)-\theta \nabla k_2\left(a\right)-k_3\left(a\right)\nabla \theta ,\hfill \end{array}$$

where $I\left(a\right)\triangleq \frac{a}{1+a}$, $\tilde{\mu }\left(a\right)\triangleq \mu \left({\rho }_{\infty }(1+a)\right)-\mu \left({\rho }_{\infty }\right)$, $\tilde{\lambda }\left(a\right)\triangleq \lambda \left({\rho }_{\infty }(1+a)\right)-\lambda \left({\rho }_{\infty }\right)$, $k_1\left(a\right)\triangleq \frac{{\partial }_{\rho }P({\rho }_{\infty }(1+a),{\mathcal{T}}_{\infty })}{1+a}-{\partial }_{\rho }P({\rho }_{\infty },{\mathcal{T}}_{\infty })$, $k_2\left(a\right)\triangleq {\int }_0^a\frac{{\pi }_1^{\prime }(1+z)}{1+z}dz$, $k_3\left(a\right)\triangleq \frac{{\pi }_1\left({\rho }_{\infty }(1+a)\right)}{{\rho }_{\infty }(1+a)}-\frac{{\pi }_1\left({\rho }_{\infty }\right)}{{\rho }_{\infty }}$, $\tilde{\kappa }\left(a\right)\triangleq \kappa \left({\rho }_{\infty }(1+a)\right)-\kappa \left({\rho }_{\infty }\right)$. And also, the nonlinear term on the right side of the energy equation, which is where the difficulty lies compared to the isentropic case, is accordingly expressed as follows:

$$\begin{array}{ccc}\hfill h& \triangleq & -\mathit{u}·\nabla \theta -\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)\Delta \theta -\frac{{\mathcal{T}}_{\infty }}{C_v}{k}_3\left(a\right)\mathrm{div}\mathit{u}\hfill \\ & & +\frac{1}{C_v{\rho }_{\infty }(1+a)}({\tilde{\eta }}^{\prime }\left(a\right)\nabla a·\nabla \theta -{\pi }_1\left({\rho }_{\infty }(1+a))\theta \mathrm{div}\mathit{u}\right)\hfill \\ & & +\frac{1}{C_v{\rho }_{\infty }(1+a)}\left(2\mu \left({\rho }_{\infty }(1+a)\right)D\left(u\right):D\left(u\right)+\lambda \left({\rho }_{\infty }(1+a)\right){\left(\mathrm{div}\mathit{u}\right)}^2\right)\hfill \\ & & +\frac{{\rho }_{\infty }}{C_v}\kappa \left({\rho }_{\infty }(1+a)\right)\Delta a\mathrm{div}\mathit{u}+\frac{{\rho }_{\infty }}{2C_v}\nabla \tilde{\kappa }\left(a\right)·\nabla a\mathrm{div}\mathit{u}\hfill \\ & & +\frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a)\right)}{C_v(1+a)}\left(\frac{1}{2}{|\nabla a|}^2\mathrm{div}\mathit{u}-\nabla a\otimes \nabla a:\nabla \mathit{u}\right)\hfill \end{array}$$

with $\tilde{\eta }\left(a\right)\triangleq \eta \left({\rho }_{\infty }(1+a)\right)-\eta \left({\rho }_{\infty }\right)$.

Next, we state our main work of our paper as follows:

Theorem 1.

Assume ${\rho }_{\infty }$ and ${\mathcal{T}}_{\infty }$ are positive constants satisfying (4). And suppose that κ, μ, λ, P, and η have real analyticity. (The further assumption of real analyticity near zero is used to establish the global evolution of Gevrey.) If d and p satisfy the following condition:

$$2\le p<3\mathit{for}d=3\mathit{and}2\le p\le \frac{2d}{d-2}\mathit{for}d\ge 4,$$

(14)

one can identify a positive small constant $\epsilon =\epsilon (p,d,\kappa ,\mu ,\lambda ,\eta ,P,C_v,{\rho }_{\infty },{\mathcal{T}}_{\infty })$ and an integer $j_0\in \mathbb{Z}$ such that if $a_0=\frac{{\rho }_0-{\rho }_{\infty }}{{\rho }_{\infty }}\in {\dot{B}}_{p,1}^{\frac{d}{p}}$, $u_0\in {\dot{B}}_{p,1}^{\frac{d}{p}-1}$, ${\theta }_0={\mathcal{T}}_0-{\mathcal{T}}_{\infty }\in {\dot{B}}_{p,1}^{\frac{d}{p}-2}$ and if, in addition, $(a_0^{\ell },u_0^{\ell },{\theta }_0^{\ell })\in {\dot{B}}_{2,1}^{\frac{d}{2}-1}$ with

$${\mathcal{X}}_{p,0}\triangleq \parallel (a_0,{\mathit{u}}_0,{\theta }_0){\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }+\parallel (a_0,{\mathit{u}}_0){\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^{\ell }+{\parallel {\theta }_0\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^{\ell }\le \epsilon ,$$

(15)

subsequently, a unique global-in-time solution $(a,\mathit{u},\theta )$ to Equation (13) exists in the space $X_p$, whose definition is given by

$$\begin{array}{ccc}\hfill X_p& \triangleq & {\left\{(a,\mathit{u},\theta )\right|(a,\mathit{u},\theta )}^{\ell }\in {\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{2,1}^{\frac{d}{2}-1})\cap L^1({\mathbb{R}}_{+};{\dot{B}}_{2,1}^{\frac{d}{2}+1}),\hfill \\ & & a^h\in {\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}})\cap L^1({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}+2}),{\mathit{u}}^h\in {\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}-1})\cap L^1({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}+1})\hfill \\ & & {\theta }^h\in {\tilde{\mathcal{C}}}_b({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}-2})\cap L^1({\mathbb{R}}_{+};{\dot{B}}_{p,1}^{\frac{d}{p}})\}.\hfill \end{array}$$

And also, there exists a constant C depending on $p,d,\kappa ,\mu ,\lambda ,\eta ,P,C_v,{\rho }_{\infty }$, and ${\mathcal{T}}_{\infty }$, such that

$${\mathcal{X}}_p\left(t\right)\le C{\mathcal{X}}_{p,0}\mathit{for}\mathit{all}t\ge 0,$$

where

$$\begin{array}{ccc}\hfill {\mathcal{X}}_p\left(t\right)& \triangleq & {\parallel (a,\mathit{u},\theta )\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (a,\mathit{u},\theta )\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}+1}\right)}^{\ell }+{\parallel (\nabla a,\mathit{u})\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & & +{\parallel (\nabla a,\mathit{u})\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel \theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

(16)

Moreover, one can find a positive constant $c_0$ depending on $p,d,\kappa ,\mu ,\lambda ,\eta ,P,C_v,{\rho }_{\infty }$, and ${\mathcal{T}}_{\infty }$ such that $(a,\mathit{u},\theta )$ belongs to the space

$$Y_p\triangleq \left\{(a,\mathit{u},\theta )\in X_p|e^{\sqrt{c_{0}t}{\Lambda }_1}(a,\mathit{u},\theta )\in X_p\right\}$$

with ${\Lambda }_1$ being the Fourier multiplier with symbol ${\left|\xi \right|}_1=\sum _{i=1}^d\left|{\xi }_i\right|$.

Remark 3.

When $p=2$, our result on global well-posedness reduces to the $L^2$ framework (see [30]) with the smallness assumption imposed only on low frequency.

4. The Proof of Theorem 1

This section is devoted to the proof of Theorem 1 by means of the classical fixed-point theorem and the suitable effective velocity.

4.1. The Linearized System

Within this subsection, supposing that f, g, and h are given, we aim to illustrate the smoothing features of (13). Given a function $z\in \mathcal{C}({\mathbb{R}}_{+};\mathcal{S}\left({\mathbb{R}}^d\right))$, we denote its Fourier transform with respect to the spatial variable as $\widehat{z}$. The following lemma holds the key to understanding this remarkable property.

Lemma 1.

One can find two constant $C,c_0>0$ depending on ${\kappa }_{\infty },{\mu }_{\infty },{\lambda }_{\infty },{\eta }_{\infty },{\rho }_{\infty }$, and ${\mathcal{T}}_{\infty }$ such that for all $\xi \in {\mathbb{R}}^d$ and $t\ge 0$, it holds that

$$\begin{array}{ccc}\hfill \left|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\mathit{u}},\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right)\right|& \le & C\left(e^{-c_0{\left|\xi \right|}^{2}t}\right|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\mathit{u}},\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right)(0,\xi )|\hfill \\ & & +{\int }_0^t{e}^{-c_0{\left|\xi \right|}^2(t-\tau )}\left|\left(\widehat{f},\left|\xi \right|\widehat{g},\widehat{f},\frac{1}{1+\left|\xi \right|}\widehat{h}\right)(\tau ,\xi )\right|d\tau ).\hfill \end{array}$$

(17)

Proof. 

The proof of Lemma 1 primarily relies on energy estimates for system (13). Through the Duhamel principle, the analysis can be simplified to the homogeneous case where $f=0$, $g=0$, and $h=0$. Introducing the variables $\omega ={\Lambda }^{-1}\mathrm{div}\mathit{u}$ and $\Omega ={\Lambda }^{-1}\mathrm{curl}\mathit{u}$, we derive from system (13) that

$$\left\{\begin{array}{c}{\partial }_{t}a+\Lambda \omega =0,\hfill \\ {\partial }_t\omega -\nu \Delta \omega -{\alpha }_1\Lambda a-\delta {\Lambda }^{3}a-{\gamma }_1\Lambda \theta =0,\hfill \\ {\partial }_t\Omega -\overline{\mu }\Delta \Omega =0,\hfill \\ {\partial }_t\theta -{\alpha }_2\Delta \theta +{\gamma }_2\Lambda \omega =0\hfill \end{array}\right.$$

(18)

with $\overline{\mu }=\frac{{\mu }_{\infty }}{{\rho }_{\infty }}$ and $\nu =2\overline{\mu }+\overline{\lambda }=\frac{2{\mu }_{\infty }+{\lambda }_{\infty }}{{\rho }_{\infty }}$. Applying the Fourier transform of both sides of (18) yields

$$\left\{\begin{array}{c}{\partial }_t\widehat{a}+\left|\xi \right|\widehat{\omega }=0,\hfill \\ {\partial }_t\widehat{\omega }+{\nu \left|\xi \right|}^2\widehat{\omega }-{\alpha }_1\left|\xi \right|\widehat{a}-{\delta \left|\xi \right|}^3\widehat{a}-{\gamma }_1\left|\xi \right|\widehat{\theta }=0,\hfill \\ {\partial }_t\widehat{\Omega }-\overline{\mu }\Delta \widehat{\Omega }=0,\hfill \\ {\partial }_t\widehat{\theta }+{\alpha }_2{\left|\xi \right|}^2\widehat{\theta }+{\gamma }_2\left|\xi \right|\widehat{\omega }=0.\hfill \end{array}\right.$$

(19)

Then we obtain from the third equation of (19) that

$$|\widehat{\Omega }\left(t\right)|\le e^{-\overline{\mu }{\left|\xi \right|}^{2}t}\left|\widehat{\Omega }\left(0\right)\right|.$$

(20)

By multiplying the first equation in (19) by the complex conjugate $\overline{\widehat{a}}$, the second equation by $\overline{\widehat{\omega }}$, and the fourth equation by $\overline{\widehat{\theta }}$, we have

$$\begin{array}{ccc}\hfill & & \frac{1}{2}\frac{d}{dt}|\widehat{a}{|}^2+\left|\xi \right|Re\left(\widehat{\omega }\overline{\widehat{a}}\right)=0,\hfill \end{array}$$

(21)

$$\begin{array}{ccc}\hfill & & \frac{1}{2}\frac{d}{dt}|\widehat{\omega }{|}^2+{\nu \left|\xi \right|}^2|\widehat{\omega }{|}^2-({\alpha }_1+{\delta \left|\xi \right|}^2\left)\right|\xi |Re\left(\widehat{a}\overline{\widehat{\omega }}\right)-{\gamma }_1\left|\xi \right|Re\left(\widehat{\theta }\overline{\widehat{\omega }}\right)=0,\hfill \end{array}$$

(22)

$$\begin{array}{ccc}\hfill & & \frac{1}{2}\frac{d}{dt}|\widehat{\theta }{|}^2+{\alpha }_2{\left|\xi \right|}^2{\left|\theta \right|}^2+{\gamma }_2\left|\xi \right|Re\left(\widehat{\omega }\overline{\widehat{\theta }}\right)=0.\hfill \end{array}$$

(23)

Multiplying (21) by ${\alpha }_1+\delta {\left|\xi \right|}^2$, multiplying (23) by $\frac{{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}$ ($\alpha >0$, will be determined later), then combining the results with (22), we obtain

$$\begin{array}{ccc}\hfill & & \frac{1}{2}\frac{d}{dt}\left(({\alpha }_1+{\delta \left|\xi \right|}^2\left)\right|\widehat{a}{|}^2+|\widehat{\omega }{|}^2+\frac{{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}{\left|\widehat{\theta }\right|}^2\right)+{\nu \left|\xi \right|}^2{\left|\widehat{\omega }\right|}^2\hfill \\ & & +\frac{{\alpha }_2{\gamma }_1{\left|\xi \right|}^2}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}{\left|\widehat{\theta }\right|}^2-\frac{2\alpha {\gamma }_1{\left|\xi \right|}^2+{\gamma }_1{\alpha }^2{\left|\xi \right|}^3}{{(1+\alpha |\xi \left|\right)}^2}Re\left(\widehat{\omega }\overline{\widehat{\theta }}\right)=0.\hfill \end{array}$$

(24)

Using the identity $Re\left(\widehat{\omega }\overline{\widehat{a}}\right)=Re\left(\widehat{a}\overline{\widehat{\omega }}\right)$, we derive the dissipation for a by multiplying the first equation in (19) by $-\left|\xi \right|\overline{\widehat{\omega }}$ and the second equation by $-\left|\xi \right|\overline{\widehat{a}}$. Combining these yields

$$\begin{array}{ccc}\hfill & & \frac{d}{dt}(-|\xi |Re\left(\widehat{a}\overline{\widehat{\omega }}\right){)-|\xi |}^2|\widehat{\omega }{|}^2-{\nu \left|\xi \right|}^{3}Re\left(\widehat{\omega }\overline{\widehat{a}}\right)+{\delta \left|\xi \right|}^4{\left|\widehat{a}\right|}^2\hfill \\ & & +{\alpha }_1{\left|\xi \right|}^2|\widehat{a}{|}^2+{\gamma }_1{\left|\xi \right|}^{2}Re\left(\widehat{\theta }\overline{\widehat{a}}\right)=0.\hfill \end{array}$$

(25)

Multiplying (25) by $\alpha >0$, multiplying (21) by ${\alpha \nu \left|\xi \right|}^2$, and adding them to (24), we obtain

$$\begin{array}{ccc}& & \frac{1}{2}\frac{d}{dt}{\mathcal{L}}^2\left(t\right)+{(\nu -\alpha )\left|\xi \right|}^2|\widehat{\omega }{|}^2+{\alpha \delta \left|\xi \right|}^4|\widehat{a}{|}^2+\alpha {\alpha }_1{\left|\xi \right|}^2|\widehat{a}{|}^2+\frac{{\alpha }_2{\gamma }_1{\left|\xi \right|}^2}{{\gamma }_2{(1+\alpha |\xi |)}^2}{\left|\widehat{\theta }\right|}^2\hfill \\ & & -\frac{2\alpha {\gamma }_1{\left|\xi \right|}^2+{\gamma }_1{\alpha }^2{\left|\xi \right|}^3}{{(1+\alpha |\xi \left|\right)}^2}Re\left(\widehat{\omega }\overline{\widehat{\theta }}\right)-\alpha {\gamma }_1{\left|\xi \right|}^{2}Re\left(\widehat{\theta }\overline{\widehat{a}}\right)=0\hfill \end{array}$$

with

$${\mathcal{L}}^2\left(t\right)={\alpha }_1|\widehat{a}{|}^2+{(\delta +\alpha \nu )\left|\xi \right|}^2|\widehat{a}{|}^2+|\widehat{\omega }{|}^2+\frac{{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi |)}^2}|\widehat{\theta }{|}^2-2\alpha \left|\xi \right|Re\left(\widehat{a}\overline{\widehat{\omega }}\right).$$

Using the Cauchy–Schwarz and Young inequalities, one can obtain that

$$\begin{array}{ccc}\hfill \alpha |\xi |Re\left(\widehat{a}\overline{\widehat{\omega }}\right)& \le & \frac{\alpha \nu }{2}{\left|\xi \right|}^2|\widehat{a}{|}^2+\frac{\alpha }{2\nu }{\left|\widehat{\omega }\right|}^2,\hfill \\ \hfill \frac{2\alpha {\gamma }_1}{{(1+\alpha |\xi \left|\right)}^2}Re\left(\widehat{\omega }\overline{\widehat{\theta }}\right)& \le & \frac{\alpha {\alpha }_2{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{\alpha {\gamma }_1{\gamma }_2}{{\alpha }_2{(1+\alpha |\xi \left|\right)}^2}{\left|\widehat{\omega }\right|}^2\hfill \\ & \le & \frac{\alpha {\alpha }_2{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{\alpha {\gamma }_1{\gamma }_2}{{\alpha }_2}{\left|\widehat{\omega }\right|}^2,\hfill \\ \hfill \frac{{\alpha }^2{\gamma }_1\left|\xi \right|}{{(1+\alpha |\xi \left|\right)}^2}Re\langle \widehat{\omega }\overline{\widehat{\theta }}\rangle & \le & \frac{\alpha {\alpha }_2{\gamma }_1}{4{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{{\alpha }^3{\gamma }_1{\gamma }_2{\left|\xi \right|}^2}{{\alpha }_2{(1+\alpha |\xi \left|\right)}^2}{\left|\widehat{\omega }\right|}^2\hfill \\ & \le & \frac{\alpha {\alpha }_2{\gamma }_1}{4{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{\alpha {\gamma }_1{\gamma }_2}{{\alpha }_2}{\left|\widehat{\omega }\right|}^2,\hfill \\ \hfill \alpha {\gamma }_{1}Re\left(\widehat{\theta }\overline{\widehat{a}}\right)& \le & \frac{\alpha {\gamma }_1^2}{{\alpha }_1{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{\alpha {\alpha }_1{(1+\alpha |\xi \left|\right)}^2}{4}{\left|\widehat{a}\right|}^2\hfill \\ & \le & \frac{\alpha {\gamma }_1^2}{{\alpha }_1{(1+\alpha |\xi \left|\right)}^2}|\widehat{\theta }{|}^2+\frac{\alpha {\alpha }_1(1+{\alpha }^2{\left|\xi \right|}^2)}{2}{\left|\widehat{a}\right|}^2.\hfill \end{array}$$

Then we deduce that $\alpha$ fulfills

$$1-\frac{\alpha }{\nu }>0,\nu -\alpha -\frac{2\alpha {\gamma }_1{\gamma }_2}{{\alpha }_2}>0,1-\frac{5\alpha }{4}-\frac{\alpha {\gamma }_1{\gamma }_2}{{\alpha }_1{\alpha }_2}>0\mathrm{and}\delta -\frac{{\alpha }^2{\alpha }_1}{2}>0.$$

Choosing $0<\alpha <min\left(\frac{{\alpha }_2\nu }{{\alpha }_2+2{\gamma }_1{\gamma }_2},\frac{4{\alpha }_1{\alpha }_2}{5{\alpha }_1{\alpha }_2+4{\gamma }_1{\gamma }_2},\sqrt{\frac{2\delta }{{\alpha }_1}}\right)$, we arrive at

$${\mathcal{L}}_{\left|\xi \right|}^2\left(t\right)\approx |\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right){|}^2$$

and

$$\begin{array}{ccc}& & (\nu -\alpha )|\widehat{\omega }{|}^2+{\alpha \delta \left|\xi \right|}^2|\widehat{a}{|}^2+\alpha {\alpha }_1|\widehat{a}{|}^2+\frac{{\alpha }_2{\gamma }_1}{{\gamma }_2{(1+\alpha |\xi \left|\right)}^2}{\left|\widehat{\theta }\right|}^2\hfill \\ & & -\frac{2\alpha {\gamma }_1+{\gamma }_1{\alpha }^2\left|\xi \right|}{{(1+\alpha |\xi \left|\right)}^2}Re\left(\widehat{\omega }\overline{\widehat{\theta }}\right)-\alpha {\gamma }_{1}Re\left(\widehat{\theta }\overline{\widehat{a}}\right)\approx |\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right){|}^2.\hfill \end{array}$$

Therefore, one can conclude that there exists a constant $c_0>0$ such that

$$\frac{d}{dt}|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right){|}^2+c_0{\left|\xi \right|}^2|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right){|}^2\le 0,$$

which leads, after time integration, to

$$\left|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right)\left(t\right)\right|\le C{e}^{-c_0{\left|\xi \right|}^{2}t}\left|\left(\widehat{a},\left|\xi \right|\widehat{a},\widehat{\omega },\frac{1}{1+\left|\xi \right|}\widehat{\theta }\right)\left(0\right)\right|.$$

(26)

Putting (26) together with (21) yields the proof of the lemma in the case $f=0$, $g=0$, and $h=0$. For the general case $f\ne 0$, $g\ne 0$, and $h\ne 0$, according to the Duhamel formula, we arrive at (17). This completes the proof of Lemma 1. □

The proof requires two additional technical lemmas from [25].

Lemma 2.

For the operator $M_1\triangleq e^{-[\sqrt{t-\tau }+\sqrt{\tau }-\sqrt{t}]{\Lambda }_1}$ with $0<\tau <t$, the kernel is $L^1$-integrable, and its $L^1$ norm admits uniform bounds independent of τ and t.

Lemma 3.

For the operator $M_2\triangleq e^{\frac{1}{2}a\Delta +\sqrt{a}{\Lambda }_1}$, the kernel mapping boundedly $L^p$ to $L^p$ for all $1<p<\infty$ is a Fourier multiplier. Moreover, it has operator norms uniformly bounded for all $a\ge 0$.

4.2. Global Estimates in $X_p$ for System (13)

In this subsection, our aim is to prove the first part of Theorem 1. Our proof employs a fixed-point argument in the framework of complete spaces.

Proposition 12.

Assuming ${\mathcal{X}}_p$ and ${\mathcal{X}}_{p,0}$ are defined as in Theorem 1, then one can find a constant C to make the following inequality hold for all $t\ge 0$:

$${\mathcal{X}}_p\left(t\right)\le C({\mathcal{X}}_{p,0}+(1+{\mathcal{X}}_p\left(t\right)+{\mathcal{X}}_p^2\left(t\right)){\mathcal{X}}_p^2\left(t\right)).$$

(27)

Proof. 

Let us first give the low-frequency estimates. First, we apply the operator ${\dot{\Delta }}_j$ to Equation (13) and then replicate the steps used to derive Lemma 1. For arbitrary $j_0\in \mathbb{Z}$ (where we will focus on high-frequency values), there are two positive constants $c_0$ and C such that we have for all $t\ge 0$ and $j\le j_0$,

$$\left|(\widehat{a_j},\widehat{{\mathit{u}}_j},\widehat{{\theta }_j})\left(t\right)\right|\le C\left(e^{-c_0{2}^{2j}t}|(\widehat{a_{0,j}},\widehat{{\mathit{u}}_{0,j}},\widehat{{\theta }_{0,j}})|+{\int }_0^t{e}^{-c_0{2}^{2j}(t-\tau )}\left|(\widehat{f_j},\widehat{g_j},\widehat{h_j})\left(\tau \right)\right|d\tau \right),$$

(28)

where $z_j\triangleq {\dot{\Delta }}_{j}z$ for z in ${\mathcal{S}}_0^{\prime }$. Taking the $L^2$ norm, owing to the Fourier–Plancherel theorem, one can obtain

$$\begin{array}{ccc}\hfill \parallel (a_j,{\mathit{u}}_j,{\theta }_j)\left(t\right){\parallel }_{L^2}& \lesssim & e^{-c_0{2}^{2j}t}{\parallel (a_{0,j},{\mathit{u}}_{0,j},{\theta }_{0,j})\parallel }_{L^2}\hfill \\ & & +{\int }_0^t{e}^{-c_0{2}^{2j}(t-\tau )}{\parallel (f_j,g_j,h_j)\left(\tau \right)\parallel }_{L^2}d\tau ,\hfill \end{array}$$

(29)

which leads, for all $T\ge 0$, after multiplying by $2^{j(\frac{d}{2}-1)}$ and summing on $j\le j_0$, to

$${\parallel (a,\mathit{u},\theta )\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (a,\mathit{u},\theta )\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}+1}\right)}^{\ell }\lesssim \parallel (a_0,{\mathit{u}}_0,{\theta }_0){\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }+{\parallel (f,g,h)\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }.$$

(30)

Next, we shall show the high-frequency estimates. Let $\mathcal{P}=\mathrm{Id}+{(-\Delta )}^{-1}\nabla \mathrm{div}$ be the Leray projector onto a divergence-free vector field. Now, using the Leray projector $\mathcal{P}$ in the second equation of (13) yields

$${\partial }_t\mathcal{P}\mathit{u}-\overline{\mu }\Delta \mathcal{P}\mathit{u}=\mathcal{P}g\mathrm{with}\overline{\mu }=\frac{{\mu }_{\infty }}{{\rho }_{\infty }}.$$

(31)

With the aid of Proposition 7, we arrive at

$${\parallel \mathcal{P}\mathit{u}\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \mathcal{P}\mathit{u}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\lesssim \parallel \mathcal{P}{\mathit{u}}_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel \mathcal{P}g\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h.$$

(32)

Let $\mathcal{Q}=\mathrm{Id}-\mathcal{P}=-(-\Delta )\nabla \mathrm{div}$. Following the methodology developed by Haspot [40], which extends Hoff’s viscous effective flux framework [41], we define

$$v\triangleq \mathcal{Q}\mathit{u}+\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla a\mathrm{so}\mathrm{that}\mathrm{div}v=\mathrm{div}\mathit{u}-\frac{{\alpha }_1}{\nu }a,$$

and deduce that, thanks to $\nu =2\overline{\mu }+\overline{\lambda }=\frac{2{\mu }_{\infty }+{\lambda }_{\infty }}{{\rho }_{\infty }}$,

$$\left\{\begin{array}{c}{\partial }_t\nabla a+\nabla a+\Delta v=\nabla f,\hfill \\ {\partial }_{t}v-\nu \Delta v-\delta \nabla \Delta a=\tilde{g}\hfill \end{array}\right.$$

with $\tilde{g}=\mathcal{Q}g+\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla f+\frac{{\alpha }_1}{\nu }v-\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla a-{\gamma }_1\nabla \theta$. Following from [25,26], we set

$$w=v+\beta \nabla a\mathrm{with}\beta \mathrm{fulfilling}\beta =\frac{\delta }{\nu -\beta },$$

(33)

which reveals that

$${\partial }_{t}w-(\nu -\beta )\Delta w=-\beta \nabla a+\beta \nabla f+\tilde{g}.$$

(34)

Hence, one can choose

$$\beta =\frac{1}{2}(\nu +\sqrt{{\nu }^2-4\delta })\mathrm{so}\mathrm{that}\nu -\beta =\frac{1}{2}(\nu -\sqrt{{\nu }^2-4\delta }),$$

and it is clear that $Re(\nu -\beta )>0$. Consequently, by using Proposition 11, we have

$${\parallel w\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel w\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\lesssim \parallel w_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel \tilde{g}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h.$$

(35)

We note that the operator $\nabla {(-\Delta )}^{-1}$ acts as a degree $-1$ homogeneous Fourier multiplier

$$\parallel \tilde{g}{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim {\parallel g\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel v\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel \theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.$$

According to the high-frequency cutoff, we observe that

$${\parallel v\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim (2^{-4j_0}+2^{-2j_0}){\parallel (\nabla a,v)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h$$

and

$$\begin{array}{c}\hfill {\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim 2^{-2j_0}{\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\end{array}$$

Consequently, we conclude that

$$\begin{array}{ccc}\hfill & & {\parallel w\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel w\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & & \lesssim \parallel w_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+(1+2^{-2j_0}){\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel g\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & & +(2^{-4j_0}+2^{-2j_0})\parallel (\nabla a,v){\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

(36)

To obtain the estimate of v, we use the relation

$$\nabla a=\frac{w-v}{\beta }$$

(37)

to recast the governing equation of v:

$${\partial }_{t}v-\frac{\nu \beta -\delta }{\beta }\Delta v=\frac{\delta }{\beta }\Delta w+\tilde{g}.$$

(38)

We note that

$$\frac{\nu \beta -\delta }{\beta }=\frac{\delta }{\nu -\beta }=\beta ,$$

which ensures that $Re\frac{\nu \beta -\delta }{\beta }=Re\beta >0$. Then we obtain, according to Proposition 11, that

$${\parallel v\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel v\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\lesssim \parallel v_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel \Delta w\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \tilde{g}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h,$$

where

$$\begin{array}{ccc}\hfill & & {\parallel v\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel v\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & & \lesssim \parallel v_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel (\nabla w,\theta )\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+2^{-2j_0}{\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h\hfill \\ & & +{\parallel g\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+(2^{-4j_0}+2^{-2j_0}){\parallel (\nabla a,v)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h.\hfill \end{array}$$

(39)

Similarly, as $\theta$ fulfills

$${\partial }_t\theta -{\alpha }_2\Delta \theta =h-{\gamma }_2\mathrm{div}v-\frac{{\alpha }_1{\gamma }_2}{\nu }a,$$

(40)

one can use Proposition 11 again and obtain

$$\begin{array}{ccc}\hfill {\parallel \theta \parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel \theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h& \lesssim & \parallel {\theta }_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^h+{\parallel h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+2^{-2j_0}{\parallel v\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & & +2^{-4j_0}{\parallel \nabla a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h.\hfill \end{array}$$

(41)

Combining (36), (39), and (41), using (37) and making $j_0$ large enough, we obtain

$$\begin{array}{ccc}& & {\parallel (\nabla a,v)\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \theta \parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel (\nabla a,v)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h\hfill \\ & \lesssim & \parallel (\nabla a_0,v_0){\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+\parallel {\theta }_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^h+{\parallel f\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel g\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h.\hfill \end{array}$$

Finally, owing to $u=v-\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla a+\mathcal{P}u$, due to (33), we can arrive at

$$\begin{array}{ccc}& & {\parallel (\nabla a,\mathit{u})\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \theta \parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel (\nabla a,\mathit{u})\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+2}\right)}^h\hfill \\ & \lesssim & \parallel (\nabla a_0,{\mathit{u}}_0){\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+\parallel {\theta }_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^h+\parallel (\nabla f,g){\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h.\hfill \end{array}$$

Therefore, together with (30), we end up with

$${\mathcal{X}}_p\left(t\right)\lesssim {\mathcal{X}}_{p,0}+{\parallel (f,g,h)\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (\nabla f,g)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h.$$

(42)

Next, we handle those nonlinear terms. To do this, we decompose $g=g_1+g_2$ with

$$\begin{array}{ccc}\hfill g_1& =& -\mathit{u}·\nabla \mathit{u}+\frac{1}{{\rho }_{\infty }(1+a)}\mathrm{div}\left(2\tilde{\mu }\left(a\right)D\left(\mathit{u}\right)+\tilde{\lambda }\left(a\right)\mathrm{div}\mathit{u}\mathrm{Id}\right)-I\left(a\right)\mathcal{A}\mathit{u}\hfill \\ & & -k_1\left(a\right)\nabla a+{\rho }_{\infty }\nabla \left(\tilde{\kappa }\left(a\right)\Delta a+\frac{1}{2}\nabla \tilde{\kappa }\left(a\right)·\nabla a\right),\hfill \\ \hfill g_2& =& -\theta \nabla k_2\left(a\right)-k_3\left(a\right)\nabla \theta .\hfill \end{array}$$

Owing to [25], one can obtain

$$\parallel (f,g_1){\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (\nabla f,g_1)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim (1+{\mathcal{X}}_p\left(T\right)+{\mathcal{X}}_p^2\left(T\right)){\mathcal{X}}_p^2\left(T\right).$$

(43)

We shall use the following inequalities repeatedly, according to embedding and ${\mathcal{X}}_p\left(t\right)$:

$$\parallel (\nabla a,a,\mathit{u},{\theta }^{\ell }){\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+\parallel {\theta }^{\ell }{\parallel }_{L_T^2\left({\dot{B}}_{2,1}^{\frac{d}{2}}\right)}+{\parallel {\theta }^h\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}\lesssim {\mathcal{X}}_p\left(T\right),$$

(44)

$$\parallel (\nabla a,a,\mathit{u},{\theta }^{\ell }){\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}+{\parallel (\nabla a,\nabla \mathit{u},\nabla {\theta }^{\ell },{\theta }^h)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim {\mathcal{X}}_p\left(T\right).$$

(45)

Now, we give the high-frequency estimates of the nonlinear terms $g_2$ and h. We write that

$$g_2=-{\theta }^{\ell }\nabla k_2\left(a\right)-k_3\left(a\right)\nabla {\theta }^{\ell }-{\theta }^h\nabla k_2\left(a\right)-k_3\left(a\right)\nabla {\theta }^h.$$

Propositions 3 and 6 and Equations (44) and (45) ensure that

$$\begin{array}{c}\hfill \parallel {\theta }^{\ell }\nabla k_2\left(a\right)+k_3\left(a\right)\nabla {\theta }^{\ell }{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim \parallel {\theta }^{\ell }{\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel a\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right),\\ \hfill \parallel {\theta }^h\nabla k_2\left(a\right)+k_3\left(a\right)\nabla {\theta }^h{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim {\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel {\theta }^h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right).\end{array}$$

To handle the nonlinear term h and make our proof clearer, we decompose h into 5 parts that are $h={\displaystyle \sum _{j=1}^5}{h}_j$, where

$$\begin{array}{ccc}\hfill h_1& =& -\mathit{u}·\nabla \theta +\frac{{\tilde{\eta }}^{\prime }\left(a\right)}{C_v{\rho }_{\infty }(1+a)}\nabla a·\nabla \theta -\frac{{\pi }_1\left({\rho }_{\infty }(1+a)\right)}{C_v{\rho }_{\infty }(1+a)}\theta \mathrm{div}\mathit{u},\hfill \\ \hfill h_2& =& -\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)\Delta \theta ,\hfill \\ \hfill h_3& =& \frac{1}{C_v{\rho }_{\infty }(1+a)}\left(2\mu \left({\rho }_{\infty }(1+a)\right)D\left(\mathit{u}\right):D\left(\mathit{u}\right)+\lambda \left({\rho }_{\infty }(1+a)\right){\left(\mathrm{div}\mathit{u}\right)}^2\right),\hfill \\ \hfill h_4& =& \frac{{\rho }_{\infty }}{C_v}\kappa \left({\rho }_{\infty }(1+a)\right)\Delta a\mathrm{div}\mathit{u}+\frac{{\rho }_{\infty }}{2C_v}\nabla \tilde{\kappa }\left(a\right)·\nabla a\mathrm{div}\mathit{u},\hfill \\ \hfill h_5& =& \frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a)\right)}{C_v(1+a)}\left(\frac{1}{2}{|\nabla a|}^2\mathrm{div}\mathit{u}-\nabla a\otimes \nabla a:\nabla \mathit{u}\right).\hfill \end{array}$$

Regarding the term with $h_1$, let us write that $h_1=h_{11}+h_{12}$ with

$$\begin{array}{ccc}\hfill h_{11}& =& -\mathit{u}·\nabla {\theta }^{\ell }+\frac{{\tilde{\eta }}^{\prime }\left(a\right)}{C_v{\rho }_{\infty }(1+a)}\nabla a·\nabla {\theta }^{\ell }-\frac{{\pi }_1\left({\rho }_{\infty }(1+a)\right)}{C_v{\rho }_{\infty }(1+a)}{\theta }^{\ell }\mathrm{div}\mathit{u},\hfill \\ \hfill h_{12}& =& -\mathit{u}·\nabla {\theta }^h+\frac{{\tilde{\eta }}^{\prime }\left(a\right)}{C_v{\rho }_{\infty }(1+a)}\nabla a·\nabla {\theta }^h-\frac{{\pi }_1\left({\rho }_{\infty }(1+a)\right)}{C_v{\rho }_{\infty }(1+a)}{\theta }^h\mathrm{div}\mathit{u}.\hfill \end{array}$$

It follows from Propositions 3 and 6, the Bernstein inequality, and Equation (45) that

$$\begin{array}{ccc}\hfill \parallel h_{11}{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h& \lesssim & {(\parallel \mathit{u}\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^2)\parallel {\theta }^{\ell }{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}\hfill \\ & & {+(1+\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)})\parallel {\theta }^{\ell }{\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel \mathit{u}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right),\hfill \\ \hfill \parallel h_{12}{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h& \lesssim & {(1+\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\left)\right(\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+{\parallel \mathit{u}\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)})\parallel {\theta }^h{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).\hfill \end{array}$$

The term $h_2=-\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)\Delta \theta$ may be treated along the same lines. Therefore, we have

$$\parallel h_2{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim {\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}(\parallel {\theta }^{\ell }{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}+\parallel {\theta }^h{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)})\lesssim {\mathcal{X}}_p^2\left(T\right).$$

Using Propositions 3 and 6 and Equations (44) and (45), we obtain

$$\begin{array}{c}\hfill \parallel h_3{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim {(1+\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^2\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).\end{array}$$

Similarly,

$$\begin{array}{c}\hfill \parallel h_4{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim {(1+\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel \mathit{u}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right),\\ \hfill \parallel h_5{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim {(1+\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^2{\parallel \mathit{u}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^3\left(T\right).\end{array}$$

So, we conclude that

$$\parallel g_2{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim (1+{\mathcal{X}}_p\left(T\right)+{\mathcal{X}}_p^2\left(T\right)){\mathcal{X}}_p^2\left(T\right).$$

(46)

We claim that if p satisfies (14), then we obtain the following for $T\ge 0$:

$$\parallel (g_2,h){\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).$$

(47)

Next, let us prove claim (47). In order to handle the term $\theta \nabla k_2\left(a\right)$ of $g_2$, we note that $\theta \nabla k_2\left(a\right)=\frac{{\pi }_1^{\prime }(1+a)}{1+a}{\theta }^{\ell }\nabla a+\frac{{\pi }_1^{\prime }(1+a)}{1+a}{\theta }^h\nabla a$. It follows from Propositions 4 and 6 and Equation (45) that

$$\begin{array}{ccc}\hfill \parallel \frac{{\pi }_1^{\prime }(1+a)}{1+a}{\theta }^{\ell }\nabla a{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right)\parallel {\theta }^{\ell }{\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}{\parallel \nabla a\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).\hfill \end{array}$$

(48)

Using Corollary 1, Proposition 6, and Equation (45) yields

$$\begin{array}{ccc}\hfill \parallel \frac{{\pi }_1^{\prime }(1+a)}{1+a}{\theta }^h\nabla a{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel {\theta }^h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right),\hfill \end{array}$$

(49)

where we used the relation $1-\frac{d}{p}<\frac{d}{p}-1<\frac{d}{p}$. The term with $k_3\left(a\right)\nabla \theta$ is similar: from Propositions 4 and 6, Corollary 1, the Bernstein inequality, and Equations (44) and (45), we have

$$\begin{array}{ccc}\hfill \parallel k_3\left(a\right)\nabla {\theta }^{\ell }{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {\parallel a\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel {\theta }^{\ell }\parallel }_{L_T^2\left({\dot{B}}_{2,1}^{\frac{d}{2}}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right),\hfill \\ \hfill \parallel k_3\left(a\right)\nabla {\theta }^h{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {\parallel a\parallel }_{L_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel {\theta }^h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right).\hfill \end{array}$$

Hence, we eventually obtain

$$\parallel g_2{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }={\parallel \theta \nabla k_2\left(a\right)+k_3\left(a\right)\nabla \theta \parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).$$

Regarding the terms with $h_{11}$ and $h_{12}$, we have by Propositions 4 and 6, Corollary 1, and Equations (44) and (45),

$$\begin{array}{ccc}\hfill \parallel h_{11}{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\parallel {\theta }^{\ell }{\parallel }_{L_T^2\left({\dot{B}}_{2,1}^{\frac{d}{2}}\right)}+\left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel a\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel {\theta }^{\ell }\parallel }_{L_T^2\left({\dot{B}}_{2,1}^{\frac{d}{2}}\right)}\hfill \\ & & +\left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right)\parallel {\theta }^{\ell }{\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}{\parallel \mathit{u}\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel h_{12}{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {\parallel \mathit{u}\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}\parallel {\theta }^h{\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+\left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel {\theta }^h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & & +\left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right)\parallel {\theta }^h{\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & \lesssim & (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).\hfill \end{array}$$

To handle the term with $h_2$, we write that $h_2=h_{21}+h_{22}$ with

$$h_{21}=-\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)\Delta {\theta }^{\ell }\mathrm{and}{h}_{22}=-\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)\Delta {\theta }^h.$$

It follows from Propositions 4 and 6 and Equation (45) that

$$\parallel h_{21}{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim {\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel {\theta }^{\ell }\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}+1}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right).$$

For the term with $h_{22}$, we have to consider the case $2\le p\le \frac{2d}{3}$ and $p>\frac{2d}{p}$ separately. If $2\le p\le \frac{2d}{3}$, then we obtain, taking advantage of Corollary 1,

$$\parallel h_{22}{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }\lesssim {\parallel a\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}\parallel \Delta {\theta }^h{\parallel }_{{\dot{B}}_{p,1}^{1-\frac{d}{p}}}\lesssim {\parallel a\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}{\parallel {\theta }^h\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}.$$

(50)

If $\frac{2d}{3}<p<d$, then applying (10) with $2-\frac{d}{p}>\frac{1}{2}>0$ yields

$$\parallel F{G}^h{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }\lesssim \parallel F{G}^h{\parallel }_{{\dot{B}}_{2,\infty }^{-{\sigma }_0}}^{\ell }\lesssim {(\parallel F\parallel }_{{\dot{B}}_{p,1}^{2-\frac{d}{p}}}+\parallel F^{\ell }{\parallel }_{L^{p^{*}}})\parallel G^h{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}\mathrm{with}\frac{1}{p^{*}}=\frac{1}{2}-\frac{1}{p}.$$

(51)

Additionally, through the composition inequality in $L^p$ spaces and the continuous embeddings ${\dot{B}}_{2,1}^{\frac{d}{p}}\hookrightarrow L^{p^{*}}$ together with ${\dot{B}}_{p,1}^{{\sigma }_0}\hookrightarrow L^{p^{*}}$ yields

$$\parallel {\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)}^{\ell }{\parallel }_{L^{p^{*}}}\lesssim {\parallel a\parallel }_{L^{p^{*}}}\lesssim \parallel a^{\ell }{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{p}}}+\parallel a^h{\parallel }_{{\dot{B}}_{p,1}^{{\sigma }_0}}\lesssim \parallel a^{\ell }{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}+{\parallel a^h\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}},$$

(52)

where the relations $\frac{d}{2}-1\le \frac{d}{p}$ and ${\sigma }_0\le \frac{d}{p}$. It follows from Proposition 6, the embeddings ${\dot{B}}_{2,1}^{2-{\sigma }_0}\hookrightarrow {\dot{B}}_{p,1}^{2-\frac{d}{p}}$ and the relations $2-{\sigma }_0>\frac{d}{2}-1(p>\frac{2d}{3})$ and $2-\frac{d}{p}<\frac{d}{p}(p<d)$ that

$$\begin{array}{ccc}\hfill \parallel \left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right){\parallel }_{{\dot{B}}_{p,1}^{2-\frac{d}{p}}}& \lesssim & {\parallel a\parallel }_{{\dot{B}}_{p,1}^{2-\frac{d}{p}}}\lesssim \parallel a^{\ell }{\parallel }_{{\dot{B}}_{2,1}^{2-{\sigma }_0}}+{\parallel a^h\parallel }_{{\dot{B}}_{p,1}^{2-\frac{d}{p}}}\hfill \\ & \lesssim & \parallel a^{\ell }{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}+{\parallel a^h\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}.\hfill \end{array}$$

(53)

Then we obtain from (51), (52), and (53) that

$$\parallel h_{22}{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }\lesssim \left(\parallel a^{\ell }{\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}+{\parallel a^h\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}\right){\parallel {\theta }^h\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}}}$$

(54)

for all $\frac{2d}{3}<p<d$. Hence, we conclude (thanks to (50) and (54)) that

$$\parallel h_{22}{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim \left(\parallel a^{\ell }{\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}+{\parallel a^h\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel {\theta }^h\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim {\mathcal{X}}_p^2\left(T\right).$$

Regarding the term with $h_3$, we obtain from the Bernstein inequality, Corollary 1, Propositions 3 and 6, and Equations (44) and (45) that

$$\parallel h_3{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim \left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim (1+{\mathcal{X}}_p\left(T\right)){\mathcal{X}}_p^2\left(T\right).$$

With the aid of Corollary 1, Propositions 3 and 6, Equations (44) and (45), and the Bernstein inequality, bounding $h_4$ and $h_5$ follows from essentially the same arguments, and we obtain

$$\begin{array}{ccc}\hfill \parallel h_4{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel a\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\lesssim \left(1+{\mathcal{X}}_p\left(T\right)\right){\mathcal{X}}_p^2\left(T\right),\hfill \\ \hfill \parallel h_5{\parallel }_{L_T^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \left(1+{\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel a\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel a\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel \mathit{u}\parallel }_{L_T^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & \lesssim & \left(1+{\mathcal{X}}_p\left(T\right)\right){\mathcal{X}}_p^3\left(T\right).\hfill \end{array}$$

Combining with these estimates, we conclude that (47) is fulfilled. From (42), (43), (46), and (47), it is not difficult to prove Proposition 12. □

4.3. The First Part of the Proof of Theorem 1

Let $W\left(t\right)$ be the semi-group associated with the left-hand side of (13). It follows from the standard Duhamel formula that

$$\left(\begin{array}{c}a\left(t\right)\\ \mathit{u}\left(t\right)\\ \theta \left(t\right)\end{array}\right)=\left(\begin{array}{c}{a}_L\left(t\right)\\ {\mathit{u}}_L\left(t\right)\\ {\theta }_L\left(t\right)\end{array}\right)+{\int }_0^{t}W\left(t-\tau \right)\left(\begin{array}{c}f\left(\tau \right)\\ g\left(\tau \right)\\ h\left(\tau \right)\end{array}\right)d\tau$$

(55)

with

$$\left(\begin{array}{c}{a}_L\left(t\right)\\ {\mathit{u}}_L\left(t\right)\\ {\theta }_L\left(t\right)\end{array}\right)=W\left(t\right)\left(\begin{array}{c}{a}_0\\ {\mathit{u}}_0\\ {\theta }_0\end{array}\right).$$

Let us rephrase the definition of the functional ${\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}$ within the vicinity of the origin in the space $X_p$ by

$$\begin{array}{c}{\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}(\overline{a},\overline{\mathit{u}},\overline{\theta })\end{array}={\int }_0^{t}W(t-\tau )\left(\begin{array}{c}f(a_L+\overline{a},{\mathit{u}}_L+\overline{\mathit{u}},{\theta }_L+\overline{\theta })\\ g(a_L+\overline{a},{\mathit{u}}_L+\overline{\mathit{u}},{\theta }_L+\overline{\theta })\\ h(a_L+\overline{a},{\mathit{u}}_L+\overline{\mathit{u}},{\theta }_L+\overline{\theta })\end{array}\right)d\tau .$$

(56)

Regarding the first part of Theorem 1, the key lies in demonstrating that the functional ${\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}$ has a fixed point in $X_p$. The proof process is carried out into two steps: stability of some closed ball $B(0,R)$ of $X_p$ by ${\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}$, then contraction in that ball.

First step: Stability of $B(0,R)$. Assuming that the radius R is small enough, we would prove the ball $B(0,R)$ of $X_p$ is stable under ${\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}$. Let $a=a_L+\overline{a}$, $\mathit{u}={\mathit{u}}_L+\overline{\mathit{u}}$ and $\theta ={\theta }_L+\overline{\theta }$. If the initial data fulfill (15), following (42), we have

$$\parallel (a_L,{\mathit{u}}_L,{\theta }_L){\parallel }_{X_p}\le C{\mathcal{X}}_{p,0}\le C\epsilon$$

and

$$\parallel {\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}(\overline{a},\overline{\mathit{u}},\overline{\theta }){\parallel }_{Xp}\le C\left({\parallel (f,g,h)\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (\nabla f,g)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\right).$$

Assuming that $C\epsilon \le R\le \frac{1}{2}$, we obtain from Proposition 12 that

$$\begin{array}{ccc}\hfill \parallel {\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}(\overline{a},\overline{\mathit{u}},\overline{\theta }){\parallel }_{X_p}& \le & C{(C\epsilon +R)}^2\left(1+(C\epsilon +R)+{(C\epsilon +R)}^2\right)\hfill \\ & \le & 4C{R}^2(1+2R+4R^2).\hfill \end{array}$$

(57)

We choose $(R,\epsilon )$ satisfying

$$\begin{array}{c}\hfill R\le \min \left(\frac{1}{2},\frac{1}{80C}\right)\mathrm{and}\epsilon \le \frac{R}{C},\end{array}$$

(58)

then we obtain

$${\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}\left(B(0,R)\right)\subset B(0,R).$$

Second step: The contraction property. Set $({\overline{a}}_i,{\overline{\mathit{u}}}_i,{\overline{\theta }}_i)$ for $i=1,2$ being in the ball $B(0,R)$. And let $a_i=a_L+{\overline{a}}_i$, ${\mathit{u}}_i={\mathit{u}}_L+{\overline{\mathit{u}}}_i$ and ${\theta }_i={\theta }_L+{\overline{\theta }}_i$ for $i=1,2$. According to (56) and (42), we have

$$\begin{array}{ccc}& & \parallel {\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_2,{\overline{\mathit{u}}}_2,{\overline{\theta }}_2)-{\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_1,{\overline{\mathit{u}}}_1,{\overline{\theta }}_1){\parallel }_{X_p}\hfill \\ & \le & C(\parallel f(a_2,{\mathit{u}}_2)-f(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel g(a_2,{\mathit{u}}_2,{\theta }_2)-g(a_1,{\mathit{u}}_1,{\theta }_1)\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\hfill \\ & & +\parallel h(a_2,{\mathit{u}}_2,{\theta }_2)-h(a_1,{\mathit{u}}_1,{\theta }_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel f(a_2,{\mathit{u}}_2)-f(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h\hfill \\ & & +\parallel g(a_2,{\mathit{u}}_2,{\theta }_2)-g(a_1,{\mathit{u}}_1,{\theta }_1){\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel h(a_2,{\mathit{u}}_2,{\theta }_2)-h(a_1,{\mathit{u}}_1,{\theta }_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h),\hfill \end{array}$$

where f, g, and h are defined in (13). We note that $g=g_1+g_2$. Using the computations in Danchin [25], one can arrive at

$$\begin{array}{ccc}& & \parallel f(a_2,{\mathit{u}}_2)-f(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel f(a_2,{\mathit{u}}_2)-f(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h\hfill \\ & \lesssim & \left(\parallel (a_1,{\mathit{u}}_1){\parallel }_E+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1)\parallel }_{X_p};\hfill \\ & & \parallel g_1(a_2,{\mathit{u}}_2)-g_1(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel g_1(a_2,{\mathit{u}}_2)-g_1(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & \lesssim & \left(1+\parallel (a_1,{\mathit{u}}_1){\parallel }_E+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right)\left(\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1)\parallel }_{X_p}.\hfill \end{array}$$

To bound the term with $g_2$, we decompose it into

$$\begin{array}{ccc}\hfill g_2(a_2,{\theta }_2)-g_2(a_1,{\theta }_1)& =& ({\theta }_2-{\theta }_1)\nabla k_2\left(a_2\right)+{\theta }_1\nabla (k_2\left(a_2\right)-k_2\left(a_1\right))+k_3\left(a_2\right)\nabla ({\theta }_2-{\theta }_1)\hfill \\ & & +(k_3\left(a_2\right)-k_3\left(a_1\right))\nabla {\theta }_1.\hfill \end{array}$$

Applying Propositions 4 and 6 and Corollary 1 yields

$$\begin{array}{ccc}& & \parallel g_2(a_2,{\theta }_2)-g_2(a_1,{\theta }_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel g_2(a_2,{\theta }_2)-g_2(a_1,{\theta }_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & \lesssim & \left(\parallel (a_1,{\theta }_1){\parallel }_{X_p}+{\parallel (a_2,{\theta }_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1)\parallel }_{X_p}.\hfill \end{array}$$

For the term containing $h={\displaystyle \sum _{i=1}^5}{h}_i$, we shall use the following decompositions for $h_1$

$$\begin{array}{ccc}& & h_1(a_2,{\mathit{u}}_2,{\theta }_2)-h_1(a_1,{\mathit{u}}_1,{\theta }_1)\hfill \\ & =& {\mathit{u}}_2·\nabla ({\theta }_2-{\theta }_1)+({\mathit{u}}_2-{\mathit{u}}_1)·\nabla {\theta }_1+\frac{{\tilde{\eta }}^{\prime }\left(a_2\right)}{C_v{\rho }_{\infty }(1+a_2)}\nabla a_2·\nabla ({\theta }_2-{\theta }_1)\hfill \\ & & +\frac{{\tilde{\eta }}^{\prime }\left(a_2\right)}{C_v{\rho }_{\infty }(1+a_2)}\nabla (a_2-a_1)·\nabla {\theta }_1+\frac{1}{C_v{\rho }_{\infty }}\left(\frac{{\tilde{\eta }}^{\prime }\left(a_2\right)}{1+a_2}-\frac{{\tilde{\eta }}^{\prime }\left(a_1\right)}{1+a_1}\right)\nabla a_1·\nabla {\theta }_1,\hfill \\ & & +\frac{{\pi }_1\left({\rho }_{\infty }(1+a_2)\right)}{C_v{\rho }_{\infty }(1+a_2)}({\theta }_2\mathrm{div}({\mathit{u}}_2-{\mathit{u}}_1)+({\theta }_2-{\theta }_1)\mathrm{div}{\mathit{u}}_1)\hfill \\ & & +\frac{1}{C_v{\rho }_{\infty }}\left(\frac{{\pi }_1\left({\rho }_{\infty }(1+a_2)\right)}{1+a_2}-\frac{{\pi }_1\left({\rho }_{\infty }(1+a_1)\right)}{1+a_1}\right){\theta }_1\mathrm{div}{\mathit{u}}_1.\hfill \end{array}$$

And similarly, we have for $h_2$

$$\begin{array}{ccc}& & h_2(a_2,{\theta }_2)-h_2(a_1,{\theta }_1)\hfill \\ & =& -\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a_2\right)-\frac{\tilde{\eta }\left(a_2\right)}{1+a_2}\right)\Delta ({\theta }_2-{\theta }_1)\hfill \\ & & -\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a_2\right)-\frac{\tilde{\eta }\left(a_2\right)}{1+a_2}-{\eta }_{\infty }I\left(a_1\right)+\frac{\tilde{\eta }\left(a_1\right)}{1+a_1}\right)\Delta {\theta }_1.\hfill \end{array}$$

Next, we decompose $h_3$ as follows:

$$\begin{array}{ccc}& & h_3(a_2,{\mathit{u}}_2)-h_3(a_1,{\mathit{u}}_1)\hfill \\ & =& \frac{2\mu \left({\rho }_{\infty }(1+a_2)\right)}{C_v{\rho }_{\infty }(1+a_2)}D\left({\mathit{u}}_2\right):(D\left({\mathit{u}}_2\right)-D\left({\mathit{u}}_1\right))+\frac{\lambda \left({\rho }_{\infty }(1+a_2)\right)}{C_v{\rho }_{\infty }(1+a_2)}\mathrm{div}{\mathit{u}}_2(\mathrm{div}{\mathit{u}}_2-\mathrm{div}{\mathit{u}}_1)\hfill \\ & & +\frac{2}{C_v{\rho }_{\infty }}\left(\frac{\mu \left({\rho }_{\infty }(1+a_2)\right)}{1+a_2}-\frac{\mu \left({\rho }_{\infty }(1+a_1)\right)}{1+a_1}\right)D\left({\mathit{u}}_2\right):D\left({\mathit{u}}_1\right)\hfill \\ & & +\frac{1}{C_v{\rho }_{\infty }}\left(\frac{\lambda \left({\rho }_{\infty }(1+a_2)\right)}{1+a_2}-\frac{\lambda \left({\rho }_{\infty }(1+a_1)\right)}{1+a_1}\right)\mathrm{div}{\mathit{u}}_2\mathrm{div}{\mathit{u}}_1\hfill \\ & & +\frac{2\mu \left({\rho }_{\infty }(1+a_1)\right)}{C_v{\rho }_{\infty }(1+a_1)}(D\left({\mathit{u}}_2\right)-D\left({\mathit{u}}_1\right)):D\left({\mathit{u}}_1\right)+\frac{\lambda \left({\rho }_{\infty }(1+a_1)\right)}{C_v{\rho }_{\infty }(1+a_1)}(\mathrm{div}{\mathit{u}}_2-\mathrm{div}{\mathit{u}}_1)\mathrm{div}{\mathit{u}}_1.\hfill \end{array}$$

For the term $h_4$, we have

$$\begin{array}{ccc}& & h_4(a_2,{\mathit{u}}_2)-h_4(a_1,{\mathit{u}}_1)\hfill \\ & =& \frac{{\rho }_{\infty }}{C_v}\kappa \left({\rho }_{\infty }(1+a_2)\right)\Delta (a_2-a_1)\mathrm{div}{\mathit{u}}_2+\frac{{\rho }_{\infty }}{2C_v}\nabla \tilde{\kappa }\left(a_2\right)·\nabla (a_2-a_1)\mathrm{div}{\mathit{u}}_2\hfill \\ & & +\frac{{\rho }_{\infty }}{C_v}\kappa \left({\rho }_{\infty }(1+a_2)\right)\Delta a_1\mathrm{div}({\mathit{u}}_2-{\mathit{u}}_1)+\frac{{\rho }_{\infty }}{2C_v}\nabla \tilde{\kappa }\left(a_2\right)·\nabla a_1\mathrm{div}({\mathit{u}}_2-{\mathit{u}}_1)\hfill \\ & & +\frac{{\rho }_{\infty }}{C_v}(\kappa \left({\rho }_{\infty }(1+a_2)\right)-\kappa \left({\rho }_{\infty }(1+a_1)\right))\Delta a_1\mathrm{div}{\mathit{u}}_1+\frac{{\rho }_{\infty }}{2C_v}\nabla (\tilde{\kappa }\left(a_2\right)-\tilde{\kappa }\left(a_1\right))·\nabla a_1\mathrm{div}{\mathit{u}}_1.\hfill \end{array}$$

Finally, for the term $h_5$ we can obtain

$$\begin{array}{ccc}& & h_5(a_2,{\mathit{u}}_2)-h_5(a_1,{\mathit{u}}_1)\hfill \\ & =& \frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a_2)\right)}{C_v(1+a_2)}\left(\frac{1}{2}\nabla a_2·\nabla (a_2-a_1)\mathrm{div}{\mathit{u}}_2-\nabla a_2\otimes \nabla (a_2-a_1):\nabla {\mathit{u}}_2\right)\hfill \\ & & +\frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a_2)\right)}{C_v(1+a_2)}\left(\frac{1}{2}\nabla (a_2-a_1)·\nabla a_1\mathrm{div}{\mathit{u}}_2-\nabla (a_2-a_1)\otimes \nabla a_1:\nabla {\mathit{u}}_2\right)\hfill \\ & & +\frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a_2)\right)}{C_v(1+a_2)}\left(\frac{1}{2}{|\nabla a_1|}^2\mathrm{div}({\mathit{u}}_2-{\mathit{u}}_1)-\nabla a_1\otimes \nabla a_1:\nabla ({\mathit{u}}_2-{\mathit{u}}_1)\right)\hfill \\ & & +\left(\frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a_2)\right)}{C_v(1+a_2)}-\frac{{\rho }_{\infty }\kappa \left({\rho }_{\infty }(1+a_1)\right)}{C_v(1+a_1)}\right)\left(\frac{1}{2}{|\nabla a_1|}^2\mathrm{div}{\mathit{u}}_1-\nabla a_1\otimes \nabla a_1:\nabla {\mathit{u}}_1\right).\hfill \end{array}$$

Hence, using Propositions 4 and 6 and Corollary 1, we discover that

$$\begin{array}{ccc}& & \parallel h_1(a_2,{\mathit{u}}_2,{\theta }_2)-h_1(a_1,{\mathit{u}}_1,{\theta }_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel h_1(a_2,{\mathit{u}}_2,{\theta }_2)-h_1(a_1,{\mathit{u}}_1,{\theta }_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\hfill \\ & \lesssim & \left(1+\parallel (a_1,{\mathit{u}}_1,{\theta }_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2,{\theta }_2)\parallel }_{X_p}\right)\left(\parallel (a_1,{\mathit{u}}_1,{\theta }_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2,{\theta }_2)\parallel }_{X_p}\right)\hfill \\ & & \times \parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1){\parallel }_{X_p},\hfill \\ & & \parallel h_2(a_2,{\theta }_2)-h_2(a_1,{\theta }_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel h_2(a_2,{\theta }_2)-h_2(a_1,{\theta }_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\hfill \\ & \lesssim & \left(\parallel (a_2,{\theta }_1){\parallel }_{X_p}+{\parallel (a_2,{\theta }_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1)\parallel }_{X_p},\hfill \\ & & \parallel h_3(a_2,{\mathit{u}}_2)-h_3(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel h_3(a_2,{\mathit{u}}_2)-h_3(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\hfill \\ & \lesssim & \left(1+\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right)\left(\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1)\parallel }_{X_p},\hfill \\ & & \parallel h_4(a_2,{\mathit{u}}_2)-h_4(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel h_4(a_2,{\mathit{u}}_2)-h_4(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\hfill \\ & \lesssim & \left(1+\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right)\left(\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1)\parallel }_{X_p},\hfill \\ & & \parallel h_5(a_2,{\mathit{u}}_2)-h_5(a_1,{\mathit{u}}_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel h_5(a_2,{\mathit{u}}_2)-h_5(a_1,{\mathit{u}}_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\hfill \\ & \lesssim & \left(1+\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right){\left(\parallel (a_1,{\mathit{u}}_1){\parallel }_{X_p}+{\parallel (a_2,{\mathit{u}}_2)\parallel }_{X_p}\right)}^2{\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1)\parallel }_{X_p}.\hfill \end{array}$$

These estimates lead to the following inequality:

$$\begin{array}{ccc}& & \parallel {\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_2,{\overline{\mathit{u}}}_2,{\overline{\theta }}_2)-{\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_1,{\overline{\mathit{u}}}_1,{\overline{\theta }}_1){\parallel }_{X_p}\hfill \\ & \le & C{\left(1+\parallel ({\overline{a}}_1,{\overline{\mathit{u}}}_1,{\overline{\theta }}_1){\parallel }_{X_p}+\parallel ({\overline{a}}_2,{\overline{\mathit{u}}}_2,{\overline{\theta }}_2{\parallel }_{X_p}+2{\parallel (a_L,{\mathit{u}}_L,{\theta }_L)\parallel }_{X_p}\right)}^2\hfill \\ & & \times (\parallel ({\overline{a}}_1,{\overline{\mathit{u}}}_1,{\overline{\theta }}_1){\parallel }_{X_p}+\parallel ({\overline{a}}_2,{\overline{\mathit{u}}}_2,{\overline{\theta }}_2{\parallel }_E+2\parallel (a_L,{\mathit{u}}_L,{\theta }_L){\parallel }_{X_p})\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1){\parallel }_{X_p}\hfill \\ & \le & 8C{(1+C\epsilon +R)}^2(C\epsilon +R){\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1)\parallel }_{X_p}.\hfill \end{array}$$

Now, if $(R,\epsilon )$ satisfies (58) (for a greater constant C if needed), one can arrive at

$$\parallel {\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_2,{\overline{\mathit{u}}}_2,{\overline{\theta }}_2)-{\Psi }_{(a_L,{\mathit{u}}_L,{\theta }_L)}({\overline{a}}_1,{\overline{\mathit{u}}}_1,{\overline{\theta }}_1){\parallel }_E\le \frac{4}{5}{\parallel ({\overline{a}}_2-{\overline{a}}_1,{\overline{\mathit{u}}}_2-{\overline{\mathit{u}}}_1,{\overline{\theta }}_2-{\overline{\theta }}_1)\parallel }_E.$$

Hence, combing the two steps, a standard fixed-point argument (cf. [CDX]) yields the first part of Theorem 1.

4.4. A Priori Estimates for Gevrey Regularity

The current subsection focuses on deriving Gevrey regularity bounds in the $L^p$ Besov setting, relying on the subsequent proposition.

Proposition 13.

Let $(a,\mathit{u},\theta )$ satisfy (13). Denote $A\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}a$, $\mathit{u}\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}\mathit{u}$ and $\Theta \triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}\theta$, where $c_0$ is the constant of Lemma 1. If ${\parallel A\parallel }_{{\tilde{L}}^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}$ is small enough, then the a priori estimate

$$\begin{array}{c}\hfill {\mathcal{Y}}_p\left(t\right)\le C({\mathcal{X}}_{p,0}+(1+{\mathcal{Y}}_p\left(t\right)){\mathcal{Y}}_p^2\left(t\right))\end{array}$$

holds for all $t\ge 0$, with

$$\begin{array}{ccc}\hfill {\mathcal{Y}}_p\left(t\right)& \triangleq & {\parallel (A,U,\Theta )\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (A,U,\Theta )\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}+1}\right)}^{\ell }+{\parallel (\nabla A,U)\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & & +{\parallel (\nabla A,U)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

Proof. 

We apply ${\dot{\Delta }}_j$ to (13) and then repeat the procedure leading to Lemma 1. Multiplying by the factor $e^{\sqrt{c_{0}t}{\left|\xi \right|}_1}$, we obtain

$$\begin{array}{ccc}\hfill \left|\right(\widehat{A_j},|\xi |{\widehat{A}}_j,\widehat{U_j},\frac{1}{1+\left|\xi \right|}\widehat{{\Theta }_j}\left)\left(t\right)\right|& \lesssim & e^{\sqrt{c_{0}t}{\left|\xi \right|}_1}{e}^{-c_0{\left|\xi \right|}^{2}t}\left|\right(\widehat{a_{0,j}},\left|\xi \right|\widehat{a_{0,j}},\widehat{{\mathit{u}}_{0,j}},\frac{1}{1+\left|\xi \right|}\widehat{{\theta }_{0,j}}\left)\right|\hfill \\ & & +e^{\sqrt{c_{0}t}{\left|\xi \right|}_1}{\int }_0^t{e}^{-c_0{\left|\xi \right|}^2(t-\tau )}\left|\right(\widehat{f_j},\left|\xi \right|\widehat{f_j},\widehat{g_j},\frac{1}{1+\left|\xi \right|}\widehat{h_j}\left)\left(\tau \right)\right|d\tau .\hfill \end{array}$$

Applying the $L^2$ norm and invoking the Fourier–Plancherel theorem yields, for all $t\ge 0$,

$$\begin{array}{ccc}& & \parallel (A_j,\Lambda A_j,U_j,\frac{1}{1+\Lambda }{\Theta }_j){\parallel }_{L^2}\hfill \\ & \lesssim & \parallel e^{\sqrt{c_{0}t}{\Lambda }_1+\frac{1}{2}{c}_{0}t\Delta }{e}^{\frac{1}{2}{c}_{0}t\Delta }(a_{0,j},\Lambda a_{0,j},{\mathit{u}}_{0,j},\frac{1}{1+\Lambda }{\theta }_{0,j}){\parallel }_{L^2}\hfill \\ & & +{\int }_0^t\parallel e^{(\sqrt{c_0(t-\tau )}{\Lambda }_1+\frac{1}{2}{c}_0(t-\tau )\Delta )}{e}^{-\sqrt{c_0}(\sqrt{t-\tau }+\sqrt{\tau }-\sqrt{t}){\Lambda }_1}{e}^{\frac{1}{2}{c}_0(t-\tau )\Delta }\hfill \\ & & \times (F_j,\Lambda F_j,G_j,\frac{1}{1+\Lambda }{H}_j){\parallel }_{L^2}d\tau ,\hfill \end{array}$$

with $F\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}f$, $G\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}g$, and $H\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}h$. Due to Lemmas 2 and 3, and to the properties of localization of ${\dot{\Delta }}_j$, we have, denoting $c_1=\frac{9}{32}{c}_0$,

$$\begin{array}{ccc}\hfill \parallel (A_j,2^j{A}_j,U_j,\frac{1}{1+2^j}{\Theta }_j){\parallel }_{L^2}& \lesssim & \parallel e^{-c_{1}t{2}^{2j}}(a_{0,j},2^j{a}_{0,j},{\mathit{u}}_{0,j},\frac{1}{1+2^j}{\theta }_{0,j}){\parallel }_{L^2}\hfill \\ & & +{\int }_0^t{\parallel e^{-c_1(t-\tau )2^{2j}}(F_j,2^j{F}_j,G_j,\frac{1}{1+2^j}{H}_j)\parallel }_{L^2}d\tau .\hfill \end{array}$$

Multiplying by $2^{j(\frac{d}{2}-1)}$ and summing on $j\le j_0\in \mathbb{Z}$, we conclude that, for all $t\ge 0$,

$$\begin{array}{ccc}\hfill & & \parallel (A,U,\Theta ){\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+\parallel (A,U,\Theta ){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}+1}\right)}^{\ell }\hfill \\ & \lesssim & \parallel (a_0,{\mathit{u}}_0,{\theta }_0){\parallel }_{{\dot{B}}_{2,1}^{\frac{d}{2}-1}}^{\ell }+{\parallel (F,G,H)\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }.\hfill \end{array}$$

(59)

In the high-frequency regime, let us plan to repeat the computations of the previous section after introducing $e^{\sqrt{c_{0}t}{\Lambda }_1}$ everywhere. Now, applying again the auxiliary functions

$$v=\mathcal{Q}\mathit{u}+\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla a\mathrm{and}w=v+\beta \nabla a\mathrm{with}\beta =\frac{1}{2}(\nu +\sqrt{{\nu }^2-4\delta })$$

and setting $\tilde{\beta }\triangleq \nu -\beta$ and $\tilde{g}=\mathcal{Q}g+\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla f+\frac{{\alpha }_1}{\nu }v-\frac{{\alpha }_1}{\nu }{(-\Delta )}^{-1}\nabla a-{\gamma }_1\nabla \theta$, due to (34), we conclude that

$$w\left(t\right)=e^{\tilde{\beta }t\Delta }{w}_0+{\int }_0^t{e}^{\tilde{\beta }(t-\tau )\Delta }(-\beta \nabla a+\beta \nabla f+\tilde{g})d\tau .$$

Hence, $W\left(t\right)\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}w\left(t\right)$ fulfills

$$W\left(t\right)=e^{\sqrt{c_{0}t}{\Lambda }_1+\tilde{\beta }t\Delta }{w}_0+{\int }_0^t{e}^{[\sqrt{c_0}(\sqrt{t}-\sqrt{\tau }){\Lambda }_1+\tilde{\beta }(t-\tau )\Delta ]}(-\beta \nabla A+\beta \nabla F+\tilde{G})d\tau$$

with $\tilde{G}\left(t\right)\triangleq e^{\sqrt{c_{0}t}{\Lambda }_1}\tilde{g}\left(t\right)$. Arguing as in the proof of Proposition 13, it follows from Lemmas 2 and 3 that the same threshold $j_0$ as in (35) and (36) yields

$$\begin{array}{ccc}& & {\parallel W\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel W\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & \lesssim & \parallel w_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel A\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel F\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \tilde{G}\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & \lesssim & \parallel w_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+(2^{-2j_0}+2^{-4j_0}){\parallel A\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+2}\right)}^h+2^{-2j_0}{\parallel V\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h\hfill \\ & & +(1+2^{-2j_0}){\parallel F\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel G\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h.\hfill \end{array}$$

Then, one can revert to v as in (37), applying $e^{\sqrt{c_{0}t}{\Lambda }_1}$ to (38). Denoting $V\triangleq e^{\sqrt{c_0\tau }{\Lambda }_1}v$ and following the procedure leading to (39), we have

$$\begin{array}{ccc}& & {\parallel V\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel V\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & \lesssim & \parallel v_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel W\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \tilde{G}\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\hfill \\ & \lesssim & \parallel v_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel W\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+2^{-4j_0}{\parallel A\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+2}\right)}^h+2^{-2j_0}{\parallel V\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & & +{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+(1+2^{-2j_0}){\parallel F\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel G\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h.\hfill \end{array}$$

Similarly, as the temperature $\theta$ satisfies (40), we have, denoting $\Theta \triangleq e^{\sqrt{c_0\tau }{\Lambda }_1}\theta$ and following the procedure leading to (41), that

$$\begin{array}{ccc}\hfill {\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h& \lesssim & \parallel {\theta }_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^h+{\parallel H\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+2^{-2j_0}{\parallel V\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\hfill \\ & & +2^{-4j_0}{\parallel A\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+2}\right)}^h.\hfill \end{array}$$

Applying $e^{\sqrt{c_0\tau }{\Lambda }_1}$ to (31) yields

$${\parallel \mathcal{P}U\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \mathcal{P}U\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h\lesssim \parallel \mathcal{P}{\mathit{u}}_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+{\parallel G\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h.$$

Consequently, selecting the same sufficiently large $j_0$ as before and applying (37), we deduce that

$$\begin{array}{ccc}& & {\parallel (\nabla A,U)\parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel \Theta \parallel }_{{\tilde{L}}_T^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h+{\parallel (\nabla A,U)\parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^h+{\parallel \Theta \parallel }_{L_T^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+2}\right)}^h\hfill \\ & \lesssim & \parallel (\nabla a_0,{\mathit{u}}_0){\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-1}}^h+\parallel {\theta }_0{\parallel }_{{\dot{B}}_{p,1}^{\frac{d}{p}-2}}^h+\parallel (\nabla F,G){\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel H\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h.\hfill \end{array}$$

Putting this together with (59), we end up with

$${\mathcal{Y}}_p\left(t\right)\lesssim {\mathcal{X}}_{p,0}+{\parallel (F,G,H)\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (\nabla F,G)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel H\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h.$$

(60)

All that remains to do is to bound F, G, and H in terms of ${\mathcal{Y}}_p\left(t\right)$. Note that $G_1=e^{\sqrt{c_0\tau }{\Lambda }_1}{g}_1$, owing to [25], so we can obtain

$$\parallel (F,G_1){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel (\nabla F,G_1)\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim {\mathcal{Y}}_p^2\left(t\right).$$

(61)

Regarding the low frequencies of $G_2$ and H, they will be strongly based on Corollaries 2 and 3 and Propositions 9 and 10. For the term containing $G_2=e^{\sqrt{c_0\tau }{\Lambda }_1}{g}_2$ with $g_2=-\theta \frac{{\pi }^{\prime }(1+a)}{1+a}\nabla a-k_3\left(a\right)\nabla \theta$, using the fact that ${\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}$ is small, Corollary 3 and Propositions 9 and 10 allow us to obtain that

$$\begin{array}{ccc}\hfill \parallel G_2{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}(\theta \frac{{\pi }^{\prime }(1+a)}{1+a}\nabla a){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+{\parallel e^{\sqrt{c_0\tau }{\Lambda }_1}(k_3\left(a\right)\nabla \theta )\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\hfill \\ & \lesssim & {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^{\ell }{\parallel A\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^{\ell }\hfill \\ & & {+(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel \Theta \parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h{\parallel A\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}+{\parallel \Theta \parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^{\ell }{\parallel A\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\hfill \\ & & +{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

Next, let us bound the term with $H={\displaystyle \sum _{j=1}^5}{e}^{\sqrt{c_0\tau }{\Lambda }_1}{h}_j$. For the term with $e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_1$, we note that $e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_1=e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_{11}+e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_{22}$. It follows from Corollary 3 and Propositions 9 and 10 that

$$\begin{array}{ccc}\hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_{11}{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel (\nabla A,U)\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel \Theta \parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^{\ell }\hfill \\ & +& {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^{\ell }{\parallel (\nabla A,U)\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}\hfill \\ & +& {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel U\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^{\ell },\hfill \\ \hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_{22}{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel (\nabla A,U)\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

For the term containing $e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_2$, we write that

$$e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_2=e^{\sqrt{c_0\tau }{\Lambda }_1}L\left(a\right)\Delta {\theta }^{\ell }+e^{\sqrt{c_0\tau }{\Lambda }_1}L\left(a\right)\Delta {\theta }^h$$

with $L\left(a\right)=-\frac{1}{C_v{\rho }_{\infty }}\left({\eta }_{\infty }I\left(a\right)-\frac{\tilde{\eta }\left(a\right)}{1+a}\right)$. Using Corollary 3 and Proposition 10 yields

$$\parallel e^{\sqrt{c_0\tau }{\Lambda }_1}L\left(a\right)\Delta {\theta }^{\ell }{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim \left({\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^{\ell }.$$

To bound the term with $L\left(a\right)\Delta {\theta }^h$, we write that $L\left(a\right)\Delta {\theta }^h=T_{\Delta {\theta }^h}L\left(a\right)+R(\Delta {\theta }^h,L\left(a\right))+T_{L\left(a\right)}\Delta {\theta }^h$. Then we obtain from Corollary 2 and Proposition 10 that

$$\begin{array}{ccc}\hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}L\left(a\right)\Delta {\theta }^h{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }& \lesssim & \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}(T_{\Delta {\theta }^h}L\left(a\right)+R(\Delta {\theta }^h,L\left(a\right))){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-2}\right)}^{\ell }\hfill \\ & +& \parallel e^{\sqrt{c_{0}t}{\Lambda }_1}{T}_{L\left(a\right)}\Delta {\theta }^h{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-3}\right)}^{\ell }\hfill \\ & \lesssim & {(\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}{)\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h.\hfill \end{array}$$

For the term with $e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_3$, due to Corollary 3 and Propositions 9 and 10, we obtain

$$\begin{array}{c}\hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_3{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }\lesssim \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_3{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-2}\right)}^{\ell }\lesssim (1+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)\parallel U\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^2.\end{array}$$

Similarly,

$$\begin{array}{ccc}\hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_4{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-2}\right)}^{\ell }& \lesssim & (1+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}){\parallel A\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel U\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)},\hfill \\ \hfill \parallel e^{\sqrt{c_0\tau }{\Lambda }_1}{h}_5{\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-2}\right)}^{\ell }& \lesssim & \left(1+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel A\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}{\parallel U\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}.\hfill \end{array}$$

The completion of Proposition 13’s proof requires treatment of the high frequencies in $G_2$ and H. This analysis is elementary, requiring only Propositions 9 and 10. In particular, we obtain

$$\begin{array}{c}\hfill \parallel G_2{\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h\lesssim {\parallel \Theta \parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^{\ell }{\parallel A\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}+{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\end{array}$$

and

$$\begin{array}{ccc}\hfill {\parallel H\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h& \lesssim & \left(1+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}\right){\parallel (A,\nabla A,U)\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}({\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}^{\ell }+{\parallel \Theta \parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^h)\hfill \\ & +& {(1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\left)\right(\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^{\ell }+{\parallel \Theta \parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h{)\parallel U\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)},\hfill \\ & +& \left({1+\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{)}^2\right({\parallel U\parallel }_{L_t^2\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}^2+{\parallel A\parallel }_{{\tilde{L}}_t^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}{\parallel U\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}+1}\right)}).\hfill \end{array}$$

Owing to ${\parallel A\parallel }_{{\tilde{L}}^{\infty }\left({\dot{B}}_{p,1}^{\frac{d}{p}}\right)}$ being small, one can conclude that

$$\parallel (G_2,H){\parallel }_{L_t^1\left({\dot{B}}_{2,1}^{\frac{d}{2}-1}\right)}^{\ell }+\parallel G_2{\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-1}\right)}^h+{\parallel H\parallel }_{L_t^1\left({\dot{B}}_{p,1}^{\frac{d}{p}-2}\right)}^h\lesssim {\mathcal{Y}}_p^2\left(t\right).$$

(62)

Hence, combining with (60)–(62), we conclude that Proposition 13 holds.

In the end, similar to the approach taken in the preceding subsection, by applying an appropriate contraction mapping principle, we can finalize the proof of Theorem 1. Since the steps follow a familiar pattern, the in-depth details are omitted and left for the reader to explore independently. Regarding the uniqueness of the solution, it is guaranteed by ([24], Theorem 5). □