Global Low-Energy Weak Solutions of a Fluid–Particle Interaction Model with Vacuum in ℝ³

Abstract

Provided that the initial data $({\rho }_0,v_0,{\eta }_0)$ is of small energy around steady state $({\rho }^{▵},0,0)$, in this work we obtain the global-in-time existence of weak solutions to a fluid particle interaction system. It should be pointed out that vacuum is allowed in this work.

1. Introduction

This work is devoted to the analysis of the compressible Navier–Stokes–Smoluchowski equations in ${\mathbb{R}}^3$, a model for fluid–particle interactions (see [1,2,3,4]):

$$\left\{\begin{array}{c}{\rho }_s+\nabla ·\left(\rho v\right)=0,\hfill \\ \rho v_s+\rho v·\nabla v+\nabla (P+\eta )-\mu \Delta v-\lambda \nabla (\nabla ·v)=-(\eta +\rho )\nabla \Phi ,\hfill \\ {\eta }_s+\nabla ·\left(\eta (v-\nabla \Phi )\right)=\Delta \eta ,\hfill \end{array}\right.$$

(1)

with the initial data

$$\begin{array}{c}\hfill (\rho ,v,\eta )(x,0)=({\rho }_0,v_0,{\eta }_0),in{\mathbb{R}}^3,\end{array}$$

(2)

and

$$\begin{array}{c}\hfill (\rho ,v,\eta )(x,s)\to ({\rho }_{\infty },0,0)as\left|x\right|\to \infty \end{array}$$

(3)

for a constant vector $({\rho }_{\infty },0,0)$ satisfying ${\rho }_{\infty }>0.$ Here, the density of the fluid $\rho \ge 0$, the fluid velocity field $v=(v^1,v^2,v^3)$, and $\eta \ge 0$ indicates the concentration of particles in the mixture. Furthermore, the pressure function of fluid P satisfies

$$P=a{\rho }^{\gamma },a>0,\gamma >1.$$

(4)

And the time-independent external potential force $\Phi =\Phi \left(x\right):{\mathbb{R}}^3\to {\mathbb{R}}_{+}$ is the effect of gravity and buoyancy. The viscosity coefficients $\lambda$ and $\mu$ satisfy

$$\mu >0,\lambda +\frac{2}{3}\mu \ge 0.$$

(5)

The mathematical analysis of the fluid–particle interaction model (1) has garnered significant attention since it was rigorously derived in [3]. Research efforts have been directed towards understanding the well-posedness and dynamic behavior of solutions under various conditions. Early studies focused on establishing the existence of different classes of solutions. For instance, in one spatial dimension, global classical solutions were shown to exist even in the presence of initial vacuum [5]. In higher dimensions, the existence of global weak solutions under physical constraints was first investigated in [4], followed by the construction of suitable weak solutions and the analysis of singular limits in low stratification regimes [6,7]. The regularity and long-time behavior of solutions have also been extensively explored. The global well-posedness of classical solutions for initial data with small energy near a steady state was established in [8], with subsequent works examining the incompressible limit [9] and time-decay rates [10,11]. In two dimensions, results on global strong solutions have been obtained for both large initial data potentially containing a vacuum [12] and for local-in-time strong solutions [13]. A common thread in these higher-dimensional studies is the reliance on specific functional settings, often requiring the density to be bounded away from zero or imposing other restrictive conditions on the initial data or external forces.

When the particle density $\eta$ is zero, system (1.1) simplifies to the compressible Navier–Stokes equations with an external force. The vast literature on this topic, e.g., [14,15,16,17,18,19,20,21,22,23], has provided deep insights. Most relevant to the current work are the results on global weak solutions with small-energy initial data [24], an approach that was later extended to allow for the presence of a vacuum while maintaining a bounded density [25].

Motivated by these developments, particularly the advances in the Navier–Stokes theory with external forces and vacuum [25], the primary goal of this paper is to investigate the existence of low-energy weak solutions for the full fluid–particle interaction model (1)–(3). Our key contribution is the establishment of the global-in-time existence of such weak solutions while explicitly allowing the fluid density $\rho$ and particle density $\eta$ to contain vacuum, a scenario that has not been addressed in the existing literature for this coupled system in three dimensions. This extends the previous results for the Navier–Stokes equations [25] to the more complex fluid–particle interaction model.

In order to state the main results, we first introduce the stationary system, which will be used to overcome the difficulties from the external potential force. Similarly to the stationary solution of Navier–Stokes equations in [26], there exists a stationary solution $({\rho }^{▵},0,0)$ of (1)–(3), given by

$$\left\{\begin{array}{c}\nabla P\left({\rho }^{▵}\right)=-{\rho }^{▵}\nabla \Phi ,\hfill \\ {\rho }^{▵}\left(x\right)\to {\rho }_{\infty },as\left|x\right|\to \infty ,\hfill \end{array}\right.$$

(6)

which gives that

$${\int }_{{\rho }_{\infty }}^{{\rho }^{▵}}\frac{P^{\prime }}{\rho }d\rho +\Phi \left(x\right)=0,as\left|x\right|\to \infty ,$$

(7)

and

$${\rho }^{▵}\left(x\right)={\left({\left({\rho }_{\infty }\right)}^{\gamma -1}-\frac{\gamma -1}{a\gamma }\Phi \right)}^{\frac{1}{\gamma -1}}>0$$

(8)

for $\Phi$, satisfying

$$\underset{x\in {\mathbb{R}}^3}{sup}\Phi <\frac{a\gamma }{\gamma -1}{\left({\rho }_{\infty }\right)}^{\gamma -1}.$$

(9)

Define $\dot{f}\triangleq f_s+v·\nabla f,\int fdx\triangleq {\int }_{{\mathbb{R}}^3}fdx$. The initial energy is defined as:

$$E_0\triangleq \int \left(\frac{1}{2}{\rho }_0{\left|v_0\right|}^2+G\left({\rho }_0\right)+{\eta }_{0}ln{\eta }_0+{\eta }_0\Phi \right)dx,$$

(10)

where $G\left(\rho \right)$ denotes the potential energy density given by

$$G\left(\rho \right)\triangleq {\int }_{{\rho }^{▵}}^{\rho }{\int }_{{\rho }^{▵}}^r\frac{P^{\prime }}{s}dsdr=\rho {\int }_{{\rho }^{▵}}^{\rho }\frac{P\left(s\right)-P\left({\rho }^{▵}\right)}{s^2}ds.$$

(11)

In this paper, we employ some notations in [27] to denote the Sobolev space.

Definition 1.

(Weak solution) A pair of functions $(\rho ,v,\eta )$ is called a global weak solution to (1) if $(\rho -{\rho }_{\infty },\rho v,\eta )\in C([0,\infty );H^{-1}\left({\mathbb{R}}^3\right)),v\in L^2((0,\infty );D^1\left({\mathbb{R}}^3\right))$, $\eta \in L^{\infty }((0,\infty );L^2\left({\mathbb{R}}^3\right))\cap L^2((0,\infty );D^1\left({\mathbb{R}}^3\right))$. Moreover, Equations (12)–(14) hold for any test function $\phi \in \mathcal{D}({\mathbb{R}}^3\times (s_1,s_2))$ with $s_2\ge s_1\ge 0$ and $j=1,2,3$:

$$\int \rho \phi (x,s){dx|}_{s_1}^{s_2}={\int }_{s_1}^{s_2}\int (\rho {\phi }_s+\rho v·\nabla \phi )dxds,$$

(12)

$$\begin{array}{cc}\hfill & \int \rho v^j{\phi (x,s)dx|}_{s_1}^{s_2}+{\int }_{s_1}^{s_2}\int (\mu \nabla v^j·\nabla \phi +\lambda (\nabla ·v){\phi }_{x_j})dxds\hfill \\ \hfill =& {\int }_{s_1}^{s_2}\int (\rho v^j{\phi }_s+\rho v^{j}v·\nabla \phi +(P+\eta ){\phi }_{x_j}-(\eta +\rho ){\Phi }_{x_j}\phi )dxds,\hfill \end{array}$$

(13)

$$\int \eta \phi (x,s){dx|}_{s_1}^{s_2}={\int }_{s_1}^{s_2}\int (\eta {\phi }_s+\eta (v-\nabla \Phi )·\nabla \phi -\nabla \eta ·\nabla \phi )dxds.$$

(14)

We now present the main result of this paper, concerning the global-in-time existence of weak solutions.

Theorem 1.

Suppose that ${\rho }^{▵}$ satisfies (6) and $\Phi \in H^3$ satisfies (9), and that the initial data $({\rho }_0,v_0,{\eta }_0)$ satisfies

$$\left\{\begin{array}{c}{\eta }_{0}ln{\eta }_0\in L^1,0\le inf{\rho }_0\le sup{\rho }_0\le \tilde{\rho },\hfill \\ \parallel \nabla v_0{\parallel }_{L^2}^2\le K_1,\parallel \nabla {\eta }_0{\parallel }_{L^2}^2\le K_2{,\parallel \nabla \Phi \parallel }_{L^2}+\parallel e^{-\frac{\Phi }{4}}{\parallel }_{L^1}+{\parallel {\eta }_0\parallel }_{L^2}^2\le E_0\hfill \end{array}\right.$$

(15)

for given numbers $K_1,K_2>0$ (not necessarily small), $\tilde{\rho }\ge \overline{\rho }+1$. Then, there exists a positive constant ε depending on ${\rho }^{\infty },a,\gamma ,$$\underset{x\in {\mathbb{R}}^3}{inf}\Phi$, ${\parallel \Phi \parallel }_{H^3}$, $\overline{\rho }$, $\tilde{\rho }$, $\mu ,\lambda$, $K_1$ and $K_2$, such that if

$$E_0\le \epsilon ,$$

(16)

the Cauchy problem for (1) admits a weak solution $(\rho ,v,\tilde{\eta })$ in the sense of Definition 1, satisfying

$$0\le \rho (x,s)\le 2\tilde{\rho },forallx\in {\mathbb{R}}^3,s\ge 0.$$

(17)

Remark 1.

As mentioned previously, two kinds of weak solutions to the fluid–particle system are studied in [6,7]. More precisely, in a $C^{2,\alpha }$ spatial domain $\Omega \subset {\mathbb{R}}^3$, equipped with the initial-boundary conditions that

$${v|}_{\partial \Omega }{=(\nabla \eta +\eta \nabla \Phi )·\mathit{n}|}_{\partial \Omega }=0,$$

and

$$\begin{array}{cc}\hfill & 0<{\rho }_0\in L^{\gamma }(\Omega )\cap L_{+}^1(\Omega ),{\rho }_0{v}_0\in L^{\frac{6}{5}}\cap L^1(\Omega ),\hfill \\ \hfill & {\rho }_0{\left|v_0\right|}^2\in L^1(\Omega ),{\eta }_{0}ln{\eta }_0\in L^1(\Omega ),{\eta }_0\in L^2(\Omega )\cap L_{+}^1(\Omega ),\hfill \end{array}$$

with $\mathit{n}$ denoting the outer normal vector to the boundary $\partial \Omega$, then the global existence of dissipative weak solutions without vacuum in the sense of the normalized solutions and suitable weak solutions with a relative entropy inequality and also without vacuum are established, respectively. Motivated by these works, we focus on the global existence of weak solutions to the fluid–particle system with a vacuum in this manuscript.

2. Preliminaries

Using the standard method in [28,29], we can obtain the following existence theorem when $inf{\rho }_0>0$.

Lemma 1.

If $\Phi \in H^3$ and the initial data $({\rho }_0,v_0,{\eta }_0)$ satisfies

$$({\rho }_0,v_0,{\eta }_0)\in H^3,inf{\rho }_0>0.$$

(18)

Then there exist a time $T_0>0$, such that the Cauchy problem (1) has a unique smooth solution $(\rho ,v,\eta )$ on ${\mathbb{R}}^3\times [0,T_0]$, satisfying

$$\rho >0,forallx\in {\mathbb{R}}^3,s\in [0,T_0],$$

(19)

$$\rho (x,s)-{\rho }_{\infty }\in C([0,T_0];H^3\left({\mathbb{R}}^3\right))\cap C^1([0,T_0];H^2\left({\mathbb{R}}^3\right)),$$

(20)

and

$$(v,\eta )\in C([0,T_0];H^3\left({\mathbb{R}}^3\right))\cap C^1([0,T_0];H^2\left({\mathbb{R}}^3\right))\cap L^2([0,T_0];H^4\left({\mathbb{R}}^3\right)).$$

(21)

3. Some Useful Estimates of (1)

Let $T>0$ be a fixed time and $(\rho ,v,\eta )$ be the smooth solution of (1) on ${\mathbb{R}}^3\times (0,T]$. In order to obtain some useful estimates, we denote

$$B_1\left(T\right)\triangleq \underset{s\in [0,T]}{sup}\theta \left({\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2\right)+{\int }_0^T\int \theta \left(\rho |\dot{v}{|}^2+|{\eta }_s{|}^2+{|{\nabla }^2\eta |}^2\right)dxds,$$

(22)

$$B_2\left(T\right)\triangleq \underset{s\in [0,T]}{sup}\int {\theta }^2\left(\rho |\dot{v}{|}^2+|{\eta }_s{|}^2+{|{\nabla }^2\eta |}^2\right)dx+{\int }_0^T\int {\theta }^2\left(|\nabla \dot{v}{|}^2+{|\nabla {\eta }_s|}^2\right)dxds,$$

(23)

and

$$B_3\left(T\right)\triangleq \underset{s\in [0,T]}{sup}{(\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2),$$

(24)

where $\theta \left(s\right)\triangleq min\left\{1,s\right\}$.

Proposition 1.

Let the conditions of Theorem 1 hold and $(\rho ,v,\eta )$ be a smooth solution to (1)–(3) on ${\mathbb{R}}^3\times (0,T]$, and let another constant $K>0$ exist, such that

$$\left\{\begin{array}{c}0\le \rho \le 2\tilde{\rho }forall(x,s)\in {\mathbb{R}}^3\times [0,T],\hfill \\ B_1\left(T\right)+B_2\left(T\right)\le 2E_0^{1/2},B_3\left(\theta \left(T\right)\right)\le 3K,\hfill \end{array}\right.$$

(25)

then the following estimates hold

$$\begin{array}{c}\hfill \left\{\begin{array}{c}0\le \rho \le \frac{7}{4}\tilde{\rho }forall(x,s)\in {\mathbb{R}}^3\times [0,T],\hfill \\ B_1\left(T\right)+B_2\left(T\right)\le E_0^{1/2},B_3\left(\theta \left(T\right)\right)\le 2K,\hfill \end{array}\right.\end{array}$$

(26)

provided $\epsilon \le \overline{\epsilon }.$

In this section, generic constants $C,K>0$ depend on ${\rho }^{\infty }$, a, $\gamma$, $\underset{x\in {\mathbb{R}}^3}{inf}\Phi$, ${\parallel \Phi \parallel }_{H^3}$, $\overline{\rho }$$\tilde{\rho }$, $\mu$, $\lambda$, $K_1$, and $K_2$, but not on the time $T>0$.

Lemma 2.

Under the conditions of Proposition 1, we have

$$\begin{array}{cc}\hfill & \int \left(\frac{{\rho \left|v\right|}^2}{2}+G\left(\rho \right)+\eta |ln\eta |+\eta \Phi \right)dx+{\int }_0^T\int \left({\mu |\nabla v|}^2+{|\sqrt{\eta }\nabla \Phi +2\nabla \sqrt{\eta }|}^2\right)dxds\hfill \\ \hfill & \le C{E}_0,\hfill \end{array}$$

(27)

$$\begin{array}{c}\hfill \int {\left|\eta \right|}^{2}dx+{\int }_0^T\int {|\nabla \eta |}^{2}dxds\le C{E}_0,\end{array}$$

(28)

and

$$\begin{array}{c}\hfill {\int }_0^T{(\parallel \nabla v\parallel }_{L^2}^4+{\parallel \nabla \eta \parallel }_{L^2}^4)ds\le C,\end{array}$$

(29)

provided that $E_0\le {\overline{ϵ}}_1\triangleq min\left\{1,{\left(2k\right)}^{-1}\right\}$.

Proof. 

Using the second equation of (1) and (6), we get

$$\rho v_s+\rho v·\nabla v+\rho (\frac{\nabla P}{\rho }-\frac{\nabla P\left({\rho }^{▵}\right))}{{\rho }^{▵}})+\nabla \eta -\mu \Delta v-\lambda \nabla (\nabla ·v)=-\eta \nabla \Phi .$$

(30)

Multiplying (30) and the third equation of (1) by v and $(ln\eta +1)$, respectively, we get, after integrating it by parts, that

$$\begin{array}{cc}\hfill & \frac{d}{ds}\int \left(G\left(\rho \right)+\frac{{\rho \left|v\right|}^2}{2}\right)dx+\int \nabla \eta ·vdx+\frac{\mu }{2}{\parallel \nabla v\parallel }_{L^2}^2+\frac{\lambda }{2}{\parallel \nabla ·v\parallel }_{L^2}^2\hfill \\ \hfill =& -\frac{d}{ds}\int \eta \Phi dx-\int ({|\sqrt{\eta }\nabla \Phi |}^2+\nabla \eta \nabla \Phi )dx,\hfill \end{array}$$

(31)

and

$$\begin{array}{c}\hfill \frac{d}{ds}\int \eta ln\eta dx-\int \nabla \eta ·vdx+\int (\nabla \eta \nabla \Phi +\frac{{|\nabla \eta |}^2}{\eta })dx=0,\end{array}$$

(32)

thus, we have

$$\begin{array}{cc}\hfill & \int \left(G\left(\rho \right)+\frac{{\rho \left|v\right|}^2}{2}+\eta ln\eta +\eta \Phi \right)dx\hfill \\ \hfill & +{\int }_0^T\left(\frac{\mu }{2}{\parallel \nabla v\parallel }_{L^2}^2+\frac{\lambda }{2}{\parallel \nabla ·v\parallel }_{L^2}^2+{\parallel \sqrt{\eta }\nabla \Phi +2\nabla \sqrt{\eta }\parallel }_{L^2}^2\right)ds=E_0.\hfill \end{array}$$

(33)

Using the result (3.15) in [12], we have

$$\begin{array}{c}\hfill \int \eta |ln\eta |dx\le \int \left(\eta ln\eta +\eta \Phi +C{e}^{-\frac{\Phi }{4}}\right)dx\le C{E}_0.\end{array}$$

(34)

The estimates (33) and (34) lead to (27). We deduce from (25) that

$$\begin{array}{cc}\hfill {\int }_0^T{\parallel \nabla v\parallel }_{L^2}^{4}ds\le & C\underset{0\le s\le \theta \left(T\right)}{sup}{\parallel \nabla v\parallel }_{L^2}^2{\int }_0^{\theta \left(T\right)}{\parallel \nabla v\parallel }_{L^2}^{2}ds+C\underset{\theta \left(T\right)\le s\le T}{sup}{\theta \parallel \nabla v\parallel }_{L^2}^2{\int }_{\theta \left(T\right)}^T{\parallel \nabla v\parallel }_{L^2}^{2}ds\hfill \\ \hfill \le & CK{E}_0+C{E}_0^{3/2}.\hfill \end{array}$$

(35)

Multiplying (1)₃ by $\eta$ and integrating it by parts, we find that

$$\begin{array}{c}\hfill \frac{d}{ds}\int {\left|\eta \right|}^{2}dx+\int {|\nabla \eta |}^{2}dx\le {\parallel \eta \parallel }_{L^2}^2{(\parallel \nabla v\parallel }_{L^2}^4+{\parallel \nabla \Phi \parallel }_{L^2}^2),\end{array}$$

(36)

integrating (36) with respect to s, we obtain (28).

Similarly to (35), we arrive at

$${\int }_0^T{\parallel \nabla \eta \parallel }_{L^2}^{4}ds\le C{E}_0.$$

□

Remark 2.

We deduce from the direct calculations (see [30]) that

$$\parallel \rho -{\rho }^{▵}{\parallel }_{L^2}^2\le C{E}_0.$$

(37)

Define

$$G\triangleq {\left({\rho }^{▵}\right)}^{-1}[(\mu +\lambda )(\nabla ·v)-(P-P\left({\rho }^{▵}\right))-\eta ].$$

(38)

Using the above results, we get Lemma 3 as Lemma 4.3 in [8]. We skip the detailed proof here and present only the main idea.

Lemma 3.

Under the conditions of Proposition 1, we have

$$\begin{array}{c}\hfill {\parallel \nabla G\parallel }_{L^2}+\parallel \nabla \left({\left({\rho }^{▵}\right)}^{-1}curlv\right){\parallel }_{L^2}\le C(\parallel \rho \dot{v}{\parallel }_{L^2}+{\parallel \nabla v\parallel }_{L^3}+{\parallel \eta \parallel }_{H^1}+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^2),\end{array}$$

(39)

$$\begin{array}{c}\hfill {\parallel \nabla v\parallel }_{L^6}\le C\left(\parallel \rho \dot{v}{\parallel }_{L^2}+{\parallel \nabla v\parallel }_{L^2}+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^2+{\parallel \eta \parallel }_{H^1}\right).\end{array}$$

(40)

Lemma 4.

Under the conditions of Proposition 1, we obtain

$$B_3\left(\theta \left(T\right)\right)+{\int }_0^{\theta \left(T\right)}\left(\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\parallel {\eta }_s{\parallel }_{L^2}^2+{\parallel {\nabla }^2\eta \parallel }_{L^2}^2\right)ds\le 2K,$$

(41)

provided $\epsilon \le {\overline{\epsilon }}_2$.

Proof. 

Multiplying (1)₂ by $v_s$, we get from integrating it by parts that

$$\begin{array}{cc}\hfill & \frac{1}{2}\frac{d}{ds}{(\mu \parallel \nabla v\parallel }_{L^2}^2{+\lambda \parallel \nabla ·v\parallel }_{L^2}^2)+\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2\hfill \\ \hfill =& \frac{d}{ds}\int \left((P-P\left({\rho }^{▵}\right))+\eta )(\nabla ·v)-(\eta +\rho )\nabla \Phi v\right)dx-\int (P-P\left({\rho }^{▵}\right)){+\eta )}_s(\nabla ·v)dx\hfill \\ & +\int (\rho v·\nabla v)·\dot{v}dx+\int {(\eta +\rho )}_s\nabla \Phi vdx\triangleq \frac{d}{ds}{\mathrm{I}}_0+\sum _{i=1}^3{\mathrm{I}}_i.\hfill \end{array}$$

(42)

We deal with the term ${\mathrm{I}}_1$, and find that

$$\begin{array}{ccc}\hfill {\mathrm{I}}_1& =& \gamma \int P{(\nabla ·v)}^{2}dx-\frac{1}{\lambda +\mu }\int \nabla Pv\left({\rho }^{▵}G+P-P\left({\rho }^{▵}\right)+\eta \right)dx+\int {\eta }_s(\nabla ·v)dx\hfill \\ & \le & \frac{1}{4}\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\frac{1}{4}\parallel {\eta }_s{\parallel }_{L^2}^2+{C(\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2)+C{E}_0^{\frac{1}{3}},\hfill \end{array}$$

(43)

where we have used Lemma 3 and the following fact that

$$P_s=-\nabla ·\left(Pv\right)-(\rho P^{\prime }-P)(\nabla ·v),$$

(44)

$$\nabla ·v=\frac{1}{\lambda +\mu }\left({\rho }^{▵}G+P-P\left({\rho }^{▵}\right))+\eta \right),$$

(45)

and using (37), we have

$$\begin{array}{c}\hfill {\parallel v\parallel }_{L^2}^2=\int \frac{{\rho }^{▵}-\rho }{{\rho }^{▵}}{|v|}^2+\frac{1}{{\rho }^{▵}}{\int \rho |v|}^2\le \frac{1}{2}{\parallel v\parallel }_{L^2}^2+C{\parallel \nabla v\parallel }_{L^2}^2+C{E}_0.\end{array}$$

For the terms ${\mathrm{I}}_2,{\mathrm{I}}_3$, we have

$$\begin{array}{ccc}\hfill {\mathrm{I}}_2& \le & \parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}{\parallel \nabla v\parallel }_{L^3}{\parallel \nabla v\parallel }_{L^2}\hfill \\ \hfill & \le & C\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}{\parallel \nabla v\parallel }_{L^2}^{\frac{3}{2}}(1+\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^{\frac{1}{2}}+{\parallel \nabla v\parallel }_{L^2}^{\frac{1}{2}}+{\parallel \nabla \eta \parallel }_{L^2}^{\frac{1}{2}}+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6})\hfill \\ & \le & \frac{1}{4}\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+{C(\parallel \nabla v\parallel }_{L^2}^4+{\parallel \nabla v\parallel }_{L^2}^6+{\parallel \nabla \eta \parallel }_{L^2}^2+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^4),\hfill \end{array}$$

(46)

and

$$\begin{array}{c}\hfill {\mathrm{I}}_3\le \frac{1}{4}\parallel {\eta }_s{\parallel }_{L^2}^2+C{\parallel \nabla v\parallel }_{L^2}^2+C{E}_0.\end{array}$$

(47)

Putting (43), (46), and (47) into (42), we get

$$\begin{array}{cc}\hfill & \frac{d}{ds}(\frac{\mu }{2}{\parallel \nabla v\parallel }_{L^2}^2+\frac{\lambda }{2}{\parallel \nabla ·v\parallel }_{L^2}^2)+\frac{1}{2}{\parallel {\rho }^{\frac{1}{2}}\dot{v}\parallel }_{L^2}^2\hfill \\ \hfill \le & \frac{d}{ds}{\mathrm{I}}_0+\frac{1}{2}\parallel {\eta }_s{\parallel }_{L^2}^2+{C(\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2+{\parallel \nabla v\parallel }_{L^2}^6+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^4)+C{E}_0^{\frac{1}{3}}.\hfill \end{array}$$

(48)

Multiplying (1)₃ by $({\eta }_s-\Delta \eta )$, integrating it by parts, we obtain

$$\begin{array}{cc}\hfill & \frac{d}{ds}{\parallel \nabla \eta \parallel }_{L^2}^2+(\parallel {\eta }_s{\parallel }_{L^2}^2+\parallel {\nabla }^2\eta {\parallel }_{L^2}^2)\hfill \\ \hfill =& {\int |\nabla ·\left(\eta (v-\nabla \Phi )\right)|}^{2}dx\hfill \\ \hfill \le & C\left({\parallel \eta \parallel }_{L^{\infty }}^2{(\parallel \nabla v\parallel }_{L^2}^2+\parallel {\nabla }^2{\Phi \parallel }_{L^2}^2{)+(\parallel v\parallel }_{L^6}^2+{\parallel \nabla \Phi \parallel }_{L^6}^2{)\parallel \nabla \eta \parallel }_{L^3}^2\right)\hfill \\ \hfill \le & \frac{1}{2}\parallel {\nabla }^2{\eta \parallel }_{L^2}^2+{C\left(\right(\parallel \nabla v\parallel }_{L^2}^4+{\parallel \nabla \eta \parallel }_{L^2}^2{)\parallel \nabla \eta \parallel }_{L^2}^2+{\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2).\hfill \end{array}$$

(49)

Putting (48) and (49) together, we arrive at

$$\begin{array}{cc}\hfill & \frac{d}{ds}(\frac{\mu }{2}{\parallel \nabla v\parallel }_{L^2}^2+\frac{\lambda }{2}{\parallel \nabla ·v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2)+\frac{1}{2}(\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\parallel {\eta }_s{\parallel }_{L^2}^2+\parallel {\nabla }^2\eta {\parallel }_{L^2}^2)\hfill \\ \hfill \le & \frac{d}{ds}{\mathrm{I}}_0+C\left({\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2+{\parallel \nabla v\parallel }_{L^2}^6+{(\parallel \nabla v\parallel }_{L^2}^4+{\parallel \nabla \eta \parallel }_{L^2}^2{)\parallel \nabla \eta \parallel }_{L^2}^2+{\parallel \rho -{\rho }^{▵}\parallel }_{L^6}^4\right)\hfill \\ \hfill & +C{E}_0^{\frac{1}{3}},\hfill \end{array}$$

(50)

Integrating it over $[0,\theta (T\left)\right]$ and using (25), we have

$$\begin{array}{cc}\hfill & B_3(\theta (T))+{\int }_0^{\theta (T)}\left(\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\parallel {\eta }_s{\parallel }_{L^2}^2+{\parallel {\nabla }^2\eta \parallel }_{L^2}^2\right)ds\hfill \\ \hfill \le & C(E_0^{\frac{1}{3}}+K_1+K_2)+C_1{E}_0{B}_3^2(\theta (T))\hfill \\ \hfill \le & K+C_1{E}_0{B}_3^2(\theta (T)),\hfill \end{array}$$

(51)

where we have used

$$|{\mathrm{I}}_0|\le \frac{\mu }{4}{\parallel \nabla v\parallel }_{L^2}^2+C{E}_0,$$

and

$$\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^4\le {\left(\parallel \rho -{\rho }^{▵}{\parallel }_{L^{\infty }}^4{\parallel \rho -{\rho }^{▵}\parallel }_{L^2}^2\right)}^{\frac{2}{3}}\le C{E}_0^{\frac{2}{3}}.$$

Therefore, Lemma 4 is proved by setting $E_0\le {\overline{\epsilon }}_2\triangleq {\left(9K{C}_1\right)}^{-1}$. □

Applying the same processes as in Lemma 4.4 of paper [8], we get Lemma 5.

Lemma 5.

Under the conditions of Proposition 1, we obtain

$$\begin{array}{cc}\hfill & B_1\left(T\right)\le C{E}_0+C{\int }_0^T\theta {\parallel \nabla v\parallel }_{L^3}^{3}ds,\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill & B_2\left(T\right)\le C{E}_0+C{B}_1\left(T\right)+C{\int }_0^T{\theta }^2{\parallel \nabla v\parallel }_{L^4}^{4}ds,\hfill \end{array}$$

(53)

provided $\epsilon \le {\overline{\epsilon }}_2$.

Lemma 6.

Under the conditions of Proposition 1, we have

$${\int }_0^T{\theta }^2\left(\parallel \rho -{\rho }^{▵}{\parallel }_{L^4}^4+{\parallel \nabla v\parallel }_{L^4}^4+{\parallel G\parallel }_{L^4}^4\right)ds\le C{E}_0,$$

(54)

provided $E_0\le {\overline{\epsilon }}_3$.

Proof. 

In view of (1) and (38), we get

$${(\rho -{\rho }^{▵})}_s+\frac{{\rho }^{▵}}{\mu +\lambda }(P-P\left({\rho }^{▵}\right))=-\nabla ·\left(v(\rho -{\rho }^{▵})\right)-v·\nabla {\rho }^{▵}-\frac{{\left({\rho }^{▵}\right)}^{2}G}{\mu +\lambda }-\frac{{\rho }^{▵}\eta }{\mu +\lambda }.$$

(55)

Multiplying (55) by $4{(\rho -{\rho }^{▵})}^3$, integrating it by parts, we deduce

$$\begin{array}{cc}\hfill \frac{d}{ds}\int {(\rho -{\rho }^{▵})}^{4}dx+\frac{C}{\mu +\lambda }\int {(\rho -{\rho }^{▵})}^{4}dx\le & \frac{C}{2(\mu +\lambda )}{\parallel \rho -{\rho }^{▵}\parallel }_{L^4}^4\hfill \\ \hfill & +C(\parallel \nabla v{\parallel }_{L^2}^2+\parallel G{\parallel }_{L^4}^4+\parallel \eta {\parallel }_{L^4}^4).\hfill \end{array}$$

(56)

Multiplying (56) by ${\theta }^2$ and integrating it over $[0,T]$, we arrive at

$$\begin{array}{c}\hfill {\int }_0^T{\theta }^2\parallel \rho -{\rho }^{▵}{\parallel }_{L^4}^{4}ds\le C{E}_0+C{\int }_0^T{\theta }^2{\parallel G\parallel }_{L^4}^{4}ds.\end{array}$$

(57)

From (25), (38), and (37), we get that

$$\begin{array}{cc}\hfill {\parallel \nabla v\parallel }_{L^4}& \le C(\parallel G{\parallel }_{L^4}+\parallel \rho -{\rho }^{▵}{\parallel }_{L^4}+\parallel \eta {\parallel }_{L^4}+\parallel {\left({\rho }^{▵}\right)}^{-1}curlv{\parallel }_{L^4}),\hfill \end{array}$$

(58)

and

$$\begin{array}{c}\hfill {\parallel G\parallel }_{L^2}+{\parallel {\left({\rho }^{▵}\right)}^{-1}curlv\parallel }_{L^2}\le C\left({\parallel \nabla v\parallel }_{L^2}+1\right)\le C,\end{array}$$

(59)

and

$$\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^6\le \parallel \rho -{\rho }^{▵}{\parallel }_{L^{\infty }}^3\parallel \rho -{\rho }^{▵}{\parallel }_{L^3}^3\le C\parallel \rho -{\rho }^{▵}{\parallel }_{L^2}\parallel \rho -{\rho }^{▵}{\parallel }_{L^4}^2\le C{E}_0^{\frac{1}{2}}{\parallel \rho -{\rho }^{▵}\parallel }_{L^4}^2,$$

(60)

which, together with (25), (38), and (39), yields

$$\begin{array}{cc}\hfill & {\int }_0^T{\theta }^2\left(\parallel \rho -{\rho }^{▵}{\parallel }_{L^4}^4+{\parallel \nabla v\parallel }_{L^4}^4+{\parallel G\parallel }_{L^4}^4\right)ds\hfill \\ \hfill \le & C{\int }_0^T{\theta }^2\left(\parallel {\left({\rho }^{▵}\right)}^{-1}{curlv\parallel }_{L^4}^4+{\parallel G\parallel }_{L^4}^4\right)ds+C{E}_0\hfill \\ \hfill \le & C{\int }_0^T{\theta }^2\left(\parallel {\left({\rho }^{▵}\right)}^{-1}{curlv\parallel }_{L^2}+{\parallel G\parallel }_{L^2}\right)\left(\parallel \nabla \left({\left({\rho }^{▵}\right)}^{-1}curlv\right){\parallel }_{L^2}^3+{\parallel \nabla G\parallel }_{L^2}^3\right)ds+C{E}_0\hfill \\ \hfill \le & C{\int }_0^T\theta \parallel \rho \dot{v}{\parallel }_{L^2}\theta {\parallel \rho \dot{v}\parallel }_{L^2}^{2}ds+C{\int }_0^T\left({\theta }^2{\parallel \nabla v\parallel }_{L^2}{\parallel \nabla v\parallel }_{L^4}^2+{\theta }^2{E}_0^{\frac{1}{2}}{\parallel \nabla v\parallel }_{L^2}{\parallel \rho -{\rho }^{▵}\parallel }_{L^4}^2\right)ds+C{E}_0\hfill \\ \hfill \le & \frac{1}{2}{\int }_0^T{\theta }^2{\parallel \nabla v\parallel }_{L^4}^{4}ds+C_2{E}_0^{\frac{1}{2}}{\int }_0^T{\theta }^2{\parallel \rho -{\rho }^{▵}\parallel }_{L^4}^{4}ds+C{E}_0.\hfill \end{array}$$

(61)

Taking $E_0\le {\overline{\epsilon }}_3\triangleq min\left\{{\overline{\epsilon }}_2,{\left(2C_2\right)}^{-2}\right\}$, we obtain (54) from (61). □

Lemma 7.

Under the conditions of Proposition 1, we arrive at

$$\begin{array}{c}\hfill B_1\left(T\right)+B_2\left(T\right)\le E_0^{1/2},\end{array}$$

(62)

provided $E_0\le {\overline{\epsilon }}_4$.

Proof. 

It holds from (52), (53), (27), and (54) that

$$\begin{array}{cc}\hfill B_1\left(T\right)+B_2\left(T\right)\le & C{E}_0+C{\int }_0^T\theta {\parallel \nabla v\parallel }_{L^3}^{3}ds\hfill \\ \hfill \le & C{E}_0+C{\int }_0^T{\theta \parallel \nabla v\parallel }_{L^2}{\parallel \nabla v\parallel }_{L^4}^{2}ds\hfill \\ \hfill \le & C_3{E}_0\le E_0^{1/2},\hfill \end{array}$$

(63)

provided that $E_0\le {\overline{\epsilon }}_4=min\left\{{\overline{\epsilon }}_3,C_3^{-2}\right\}$. □

With Lemmas 2–7 at hand, we can obtain Lemma 8 by the same processes as in Lemma 4.8 of [8]. We skip the detailed proof here and present only the main idea.

Lemma 8.

Under the conditions of Proposition 1, we have

$$\underset{s\in [0,T]}{sup}{(\parallel \nabla v\parallel }_{L^2}^2+{\parallel \nabla \eta \parallel }_{L^2}^2)+{\int }_0^T(\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\parallel {\eta }_s{\parallel }_{L^2}^2+\parallel {\nabla }^2\eta {\parallel }_{L^2}^2)ds\le C,$$

(64)

and

$$\underset{s\in [0,T]}{sup}\theta \parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+{\int }_0^T\theta {\parallel \nabla \dot{v}\parallel }_{L^2}^{2}ds\le C,$$

(65)

provided $E_0\le {\overline{\epsilon }}_4$.

Proof. 

Applying (41) and (62), we obtain (64). The proof of (65) is due to the same processes of Lemma 4.8 in [8]. For completeness, we sketch it here. Using (1) and (6), we get

$$\begin{array}{c}\hfill \rho \dot{v}+\nabla (P-P\left({\rho }^{▵}\right)+\eta )-\mu \Delta v-\lambda \nabla (\nabla ·v)=-(\eta +\rho -{\rho }^{▵})\nabla \Phi .\end{array}$$

(66)

Operating $\theta {\dot{v}}^j[\partial /\partial s+\mathrm{div}(v·)]$ to (66)^j, making use of Lemma 4.8 in [8], we arrive at

$$\begin{array}{cc}\hfill & \underset{0\le s\le T}{sup}\theta \parallel {\rho }^{1/2}\dot{v}{\parallel }_{L^2}^2+{\int }_0^T\theta {\parallel \nabla \dot{v}\parallel }_{L^2}^{2}ds\hfill \\ \hfill \le & C{E}_0+C{\int }_0^T\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^{2}ds+C{\int }_0^T\theta {\parallel \nabla v\parallel }_{L^4}^{4}ds\hfill \\ \hfill \le & C+C{\int }_0^{\theta \left(T\right)}{\theta \parallel \nabla v\parallel }_{L^2}{\parallel \nabla v\parallel }_{L^6}^{3}ds\hfill \\ \hfill \le & C+C{\int }_0^{\theta \left(T\right)}\theta \left(\parallel \rho \dot{v}{\parallel }_{L^2}^3+{\parallel \nabla v\parallel }_{L^2}^3+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^3+\parallel \rho -{\rho }^{▵}{\parallel }_{L^6}^6+{\parallel \eta \parallel }_{L^2}^3+{\parallel \nabla \eta \parallel }_{L^2}^3\right)ds\hfill \\ \hfill \le & C+C\underset{s\in [0,T]}{sup}\left({\theta }^{\frac{1}{2}}{\parallel {\rho }^{\frac{1}{2}}\dot{v}\parallel }_{L^2}\right),\hfill \end{array}$$

(67)

which, implies (65). □

By Zlotnik inequality in [31], we have Lemma 9.

Lemma 9.

Assuming the conditions of Proposition 1, we obtain

$$\underset{0\le s\le T}{sup}{\parallel \rho \parallel }_{L^{\infty }}\le \frac{7}{4}\tilde{\rho },$$

(68)

provided $\epsilon \le \overline{\epsilon }\triangleq min\left\{{\overline{\epsilon }}_6,{\left(\frac{\tilde{\rho }}{4C}\right)}^2\right\}$.

Proof. 

We use (1)₁ and (1)₂ to get that

$$D_s\rho =g\left(\rho \right)+b^{\prime }\left(s\right)$$

Here,

$$g\left(\rho \right)\triangleq -\frac{a\rho }{\mu +\lambda }({\rho }^{\gamma }-{\left({\rho }^{▵}\right)}^{\gamma }),b\left(s\right)\triangleq -\frac{1}{\mu +\lambda }{\int }_0^T\rho (\tilde{G}+\eta )ds,$$

(69)

and

$$\tilde{G}\triangleq \left(\mu +\lambda \right)(\nabla ·v)-P+P\left({\rho }^{▵}\right)-\eta .$$

(70)

From (69), (1)₂, and (6), we get that

$$\underset{\rho \to \infty }{lim}g\left(\rho \right)=-\infty ,$$

and

$$▵\tilde{G}=\nabla ·\left(\rho \dot{v}+(\eta +(\rho -{\rho }^{▵}))\nabla \Phi \right),$$

(71)

which, used by the $L^p$-estimate, the Hölder inequality, and Lemma 8, leads to

$$\begin{array}{c}\hfill \parallel \tilde{G}{\parallel }_{L^2}\le C\parallel \left(\mu +\lambda \right)(\nabla ·v)-P+P\left({\rho }^{▵}\right){)-\eta \parallel }_{L^2}\le C,\end{array}$$

(72)

and

$$\begin{array}{c}\hfill \parallel \nabla \tilde{G}{\parallel }_{L^4}\le \tilde{C}\parallel \rho \dot{v}{\parallel }_{L^2}^{\frac{1}{4}}\parallel \rho \dot{v}{\parallel }_{L^6}^{\frac{3}{4}}+{C\parallel \eta \parallel }_{L^4}+C\parallel \rho -{\rho }^{▵}{\parallel }_{L^{12}}{\parallel \nabla \Phi \parallel }_{L^6}\le C{\theta }^{-\frac{1}{8}}{\parallel \nabla \dot{v}\parallel }_{L^2}^{\frac{3}{4}}+C{E}_0^{\frac{1}{12}},\end{array}$$

(73)

and

$$\begin{array}{c}\hfill \parallel \tilde{G}{\parallel }_{L^{\infty }}\le C\parallel \tilde{G}{\parallel }_{L^2}^{\frac{1}{7}}\parallel \nabla \tilde{G}{\parallel }_{L^4}^{\frac{6}{7}}\le C{\theta }^{-\frac{3}{28}}{\parallel \nabla \dot{v}\parallel }_{L^2}^{\frac{9}{14}}+C{E}_0^{\frac{1}{14}}.\end{array}$$

(74)

For $0\le s_1<s_2\le \theta \left(T\right)\le 1$, it follows from (62), (65), and (74) that

$$\begin{array}{cc}\hfill |b\left(s_2\right)-b\left(s_1\right)|\le & C{\int }_0^{\theta \left(T\right)}(\parallel \tilde{G}{\parallel }_{L^{\infty }}+{\parallel \eta \parallel }_{L^{\infty }})ds\hfill \\ \hfill \le & C{E}_0^{\frac{1}{14}}+C{\int }_0^{\theta \left(T\right)}{\theta }^{-\frac{1}{2}}{\left(\theta \parallel \nabla \dot{v}{\parallel }_{L^2}^2\right)}^{\frac{1}{4}}{\left({\theta }^2{\parallel \nabla \dot{v}\parallel }_{L^2}^2\right)}^{\frac{1}{14}}ds\hfill \\ \hfill \le & C{E}_0^{\frac{1}{14}}+C{\left(B_2\left(T\right)\right)}^{\frac{1}{14}}\le C{E}_0^{\frac{1}{28}}.\hfill \end{array}$$

(75)

For $T\in [0,\theta (T\left)\right]$, we choose $N_0,N_1$ and ${\xi }^{*}$ as follows:

$$N_0=C{E}_0^{\frac{1}{28}},N_1=0,\overline{\zeta }=\overline{\rho }.$$

It holds that

$$g\left(\xi \right)=-\frac{a\xi }{\mu +\lambda }\left({\xi }^{\gamma }-{\overline{\rho }}^{\gamma }\right)\le -N_1=0,\forall \xi \ge \overline{\zeta }=\overline{\rho },$$

Thus, we have

$$\underset{0\le s\le \theta \left(T\right)}{sup}{\parallel \rho \parallel }_{L^{\infty }}\le max\{\overline{\rho },\tilde{\rho }\}+N_0\le \tilde{\rho }+C{E}_0^{\frac{1}{28}}\le \frac{3}{2}\tilde{\rho },$$

(76)

where

$$\epsilon \le {\overline{\epsilon }}_5\triangleq min\left\{{\overline{\epsilon }}_4,{\left(\frac{\tilde{\rho }}{2C}\right)}^{28}\right\}.$$

For $\theta \left(T\right)\le s_1<s_2\le T$, similarly to (75), we deduce that

$$\begin{array}{cc}\hfill |b\left(s_2\right)-b\left(s_1\right)|\le & C{\int }_{s_1}^{s_2}{\left(\parallel \rho \dot{v}{\parallel }_{L^2}^{\frac{1}{4}}\parallel \rho \dot{v}{\parallel }_{L^6}^{\frac{3}{4}}+{\parallel \eta \parallel }_{L^4}+{\parallel \rho -{\rho }^{▵}\parallel }_{L^{12}}\right)}^{\frac{6}{7}}ds\hfill \\ \hfill & +C{\int }_{s_1}^{s_2}\theta {\parallel {\nabla }^2\eta \parallel }_{L^2}^{2}ds+\frac{a}{4(\mu +\lambda )}(s_2-s_1)\hfill \\ \hfill \le & C{\int }_{\theta \left(T\right)}^T{\theta }^2(\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+\parallel \nabla \dot{v}{\parallel }_{L^2}^2)ds+\left(C{E}_0^{\frac{1}{14}}+\frac{a}{2(\mu +\lambda )}\right)(s_2-s_1)+C{E}_0^{\frac{1}{2}}\hfill \\ \hfill \le & C{E}_0^{\frac{1}{2}}+\frac{a}{\mu +\lambda }(s_2-s_1),\hfill \end{array}$$

(77)

where from $E_0$ in the last inequality of (77), it is given that

$$\epsilon \le {\overline{\epsilon }}_6\triangleq min\left\{{\overline{\epsilon }}_5,{\left(\frac{a}{2C(\mu +\lambda )}\right)}^{14}\right\}.$$

For $s\in \left[\theta \right(T),T]$, we choose $N_0,N_1$ and ${\xi }^{*}$ as follows:

$$N_0=C{E_0}^{1/2},N_1=\frac{a}{\mu +\lambda },\overline{\zeta }=\overline{\rho }+1.$$

Noticing that

$$g\left(\xi \right)=-\frac{a\xi }{\mu +\lambda }\left({\xi }^{\gamma }-{\overline{\rho }}^{\gamma }\right)\le -N_1=-\frac{a}{\mu +\lambda },\forall \xi \ge \overline{\zeta }=\overline{\rho }+1,$$

we obtain

$$\underset{\theta \left(T\right)\le s\le T}{sup}{\parallel \rho \parallel }_{L^{\infty }}\le max\left\{\frac{3\tilde{\rho }}{2},\overline{\rho }+1\right\}+N_0\le \frac{3\tilde{\rho }}{2}+C{E}_0^{1/2}\le \frac{7}{4}\overline{\rho },$$

(78)

here,

$$\epsilon \le \overline{\epsilon }\triangleq min\left\{{\overline{\epsilon }}_6,{\left(\frac{\tilde{\rho }}{4C}\right)}^2\right\}.$$

Hence, we get (68) from (76) and (78). □

4. Proof of Theorem 1

Theorem 2.

Suppose that $({\rho }_0,v_0,{\eta }_0)$ satisfy (18) and $\Phi \in H^3$. Then, for any $0<T<\infty$, the system (1) has a unique smooth solution $(\rho ,v,\eta )$ on ${\mathbb{R}}^3\times [0,T]$, satisfying (19)–(21), where T is substituted for $T_0$, and the initial energy $E_0$ satisfies (16).

Proof. 

From Lemma 1, the Cauchy problem (1) has a unique local smooth solution $(\rho ,v,\eta )$ on ${\mathbb{R}}^3\times (0,T_0]$, where $T_0>0$ may depend on $inf{\rho }_0$. Thus, it follows from (22)–(24) that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{B}_1\left(0\right)=B_2\left(0\right)=0,\hfill \\ B_3\left(0\right)=K_1+K_2\le 3K,0\le {\rho }_0\le \tilde{\rho }.\hfill \end{array}\right.\end{array}$$

Hence, there exists a $s_1\in (0,T_0]$, such that (25) holds for $T=s_1$. Define

$$\overline{T}=sup\left\{T\left|\left(25\right)\mathrm{holds}\right.\right\},$$

(79)

then $\overline{T}\ge s_1>0$.

Next, we will prove

$$\overline{T}=\infty .$$

(80)

Otherwise, $\overline{T}<\infty$. By Proposition 1, we get that (26) holds for $T=\overline{T}$. Thus, from Proposition 2 and Lemmas 1, we get that there exists some $\overline{\overline{T}}>\overline{T}$, such that (25) holds for $T=\overline{\overline{T}}$, which contradicts (79), and we get (80). Hence, by Proposition 2, we obtain Theorem 2. □

Proposition 2.

Let $(\rho ,v,\eta )$ be a smooth solution of (1) on ${\mathbb{R}}^3\times [0,T]$, with $\Phi \in H^3$ satisfying (9) and (15), the initial data $({\rho }_0,v_0,{\eta }_0)$ satisfying (18), and $E_0$ satisfying the smallness condition (16). Then,

$$\rho >0,forallx\in {\mathbb{R}}^3,s\in [0,T],$$

(81)

and

$$\underset{0\le s\le T}{sup}\parallel (\rho -{\rho }_{\infty },v,\eta ){\parallel }_{H^3}+{\int }_0^T{\parallel (v,\eta )\parallel }_{H^4}^{2}ds\le \tilde{C}.$$

(82)

Here, $\tilde{C}>0$ depends on ${\mu ,\lambda ,\parallel \Phi \parallel }_{H^3},a,\gamma ,\overline{\rho },\tilde{\rho },{\parallel ({\rho }_0-{\rho }_{\infty },v_0,{\eta }_0)\parallel }_{H^3},inf{\rho }_0$ and T.

Proof. 

Estimate (81) is a direct consequence of (82) and depends only on the bound of ${\parallel divv\parallel }_{L^1(0,T;L^{\infty })}$. To prove (82), as observed in [26], the essential step is to estimate ${\parallel \nabla v\parallel }_{L^1(0,T;L^{\infty })}$ and ${\parallel \nabla \rho \parallel }_{L^1(0,T;L^l)}$ for $l\in [2,6]$; this is achieved by employing the Beale–Kato–Majda-type inequality developed in [30,32].

Firstly, it follows from (18) and Proposition 1 that

$$\begin{array}{c}\hfill \dot{v}(·,0)={\rho }_0^{-1}\left(\mu ▵v_0+\lambda \nabla (\nabla ·v_0)-\nabla (P\left({\rho }_0\right)+{\eta }_0)-{\eta }_0\nabla \Phi \right)-\nabla \Phi ,\end{array}$$

(83)

and

$$\begin{array}{c}\hfill \rho \le C<\infty forall(x,s)\in {\mathbb{R}}^3\times [0,T],\end{array}$$

(84)

which, together with (64) and (65), gives

$$\underset{s\in [0,T]}{sup}\parallel {\rho }^{\frac{1}{2}}\dot{v}{\parallel }_{L^2}^2+{\int }_0^T{\parallel \nabla \dot{v}\parallel }_{L^2}^{2}ds\le \tilde{C}.$$

(85)

Thus, the estimate for ${\parallel \nabla v\parallel }_{L^1(0,T;L^{\infty })}$ follows.

Secondly, for $2\le l\le 6$, we deduce from (1) that

$$\begin{array}{cc}\hfill & {\left(\right|\nabla \rho |}^l{)}_s+\nabla ·{\left(\right|\nabla \rho |}^l{v)+(l-1)|\nabla \rho |}^l(\nabla ·v)\hfill \\ \hfill & +{l|\nabla \rho |}^{l-2}{(\nabla \rho )}^s\nabla v(\nabla \rho )+l\rho {|\nabla \rho |}^{l-2}\nabla \rho ·\nabla (\nabla ·v)=0.\hfill \end{array}$$

(86)

Multiplying (86) by ${|\nabla \rho |}^{l-2}\nabla \rho$ with $l\ge 2$, we have

$$\begin{array}{c}\hfill \frac{d}{ds}{\parallel \nabla \rho \parallel }_{L^l}\le \tilde{C}\left(1+{\parallel \nabla v\parallel }_{L^{\infty }}\right){\parallel \nabla \rho \parallel }_{L^l}+\tilde{C}\left(1+\parallel \rho \dot{v}{\parallel }_{L^l}\right),\end{array}$$

(87)

where we have used

$$\begin{array}{c}\hfill {\parallel {\nabla }^{2}v\parallel }_{L^l}\le \tilde{C}\left(\parallel \rho \dot{v}{\parallel }_{L^l}+{\parallel \nabla \rho \parallel }_{L^l}+1\right).\end{array}$$

(88)

In addition, we infer from (64), (88) and Beale-Kato-Majda-type inequality in [32] that

$$\begin{array}{cc}\hfill {\parallel \nabla v\parallel }_{L^{\infty }}\le & C+C\left({\parallel \nabla ·v\parallel }_{L^{\infty }}+{\parallel \omega \parallel }_{L^{\infty }}\right)ln\left(e+\parallel \nabla \dot{v}{\parallel }_{L^2}+{\parallel \nabla \rho \parallel }_{L^6}\right)\hfill \\ \hfill & +C\left({\parallel \nabla ·v\parallel }_{L^{\infty }}+{\parallel \omega \parallel }_{L^{\infty }}\right)ln\left(e+{\parallel \nabla \rho \parallel }_{L^6}\right).\hfill \end{array}$$

(89)

Thus, we arrive at

$$\rho \dot{v}-\nabla \tilde{G}+\mu curl\left(curlv\right)=-\left(\eta +(\rho -{\rho }^{▵})\right)\nabla \Phi ,$$

where, when applied by the $L^p$-estimate, it holds that

$$\begin{array}{cc}\hfill \parallel \nabla \tilde{G}{\parallel }_{L^6}+{\parallel curl\left(curlv\right)\parallel }_{L^6}& \le C\left(\parallel \rho \dot{v}{\parallel }_{L^6}+{\parallel \left(\eta +(\rho -{\rho }^{▵})\right)\nabla \Phi \parallel }_{L^6}\right)\hfill \\ \hfill & \le C\left(\parallel \nabla \dot{v}{\parallel }_{L^2}+1\right).\hfill \end{array}$$

(90)

And we use (64), (85), and (90) to get that

$$\begin{array}{cc}\hfill {\parallel \nabla ·v\parallel }_{L^{\infty }}+{\parallel \omega \parallel }_{L^{\infty }}\le & C\left(\parallel \nabla \dot{v}{\parallel }_{L^2}+1\right).\hfill \end{array}$$

(91)

Setting

$$\Phi \left(s\right)\triangleq e+{\parallel \nabla \rho \parallel }_{L^6},\Psi \left(s\right)\triangleq \left(1+\parallel \nabla \dot{v}{\parallel }_{L^2}\right)ln\left(e+\parallel \nabla \dot{v}{\parallel }_{L^2}\right),$$

it then follows from (87), (89), and (91) with $l=6$ that

$${\Phi }^{\prime }\left(s\right)\le \tilde{C}\Psi \left(s\right)\Phi \left(s\right)ln\Phi \left(s\right),$$

which leads to

$$\frac{d}{ds}ln\Phi \left(s\right)\le \tilde{C}\Psi \left(s\right)ln\Phi \left(s\right),$$

(92)

due to $\Phi \left(T\right)\ge e$. By (85), we get

$$\begin{array}{c}\hfill {\int }_0^T\Psi ds\le \tilde{C}+\tilde{C}{\int }_0^T{\parallel \nabla \dot{v}\parallel }_{L^2}^{2}ds\le \tilde{C}.\end{array}$$

(93)

It gives that

$$\underset{0\le s\le T}{sup}{\parallel \nabla \rho \left(s\right)\parallel }_{L^6}\le \tilde{C}.$$

(94)

Also, we have

$${\int }_0^T{\parallel \nabla v\parallel }_{L^{\infty }}ds\le \tilde{C}.$$

(95)

Taking $l=2$ in (87), using (95), and the previous Lemma, we get

$$\underset{0\le s\le T}{sup}{\parallel \nabla \rho \left(s\right)\parallel }_{L^2}\le \tilde{C},$$

(96)

which, together with (88), (85), and (96), gives

$$\underset{0\le s\le T}{sup}{\parallel {\nabla }^{2}v\parallel }_{L^2}\le \tilde{C}.$$

Finally, in view of all the estimates above, applying a similar method to that in [26], we can obtain the higher-order derivatives of $(\rho ,v,\eta )$. We omit the details for simplicity here. □

Proof of Theorem 1.

Let $j_{\delta }=j_{\delta }\left(x\right)$ be a standard mollifier, and denote

$${\rho }_0^{\delta ,\alpha }=\frac{j_{\delta }\ast {\rho }_0+\alpha }{1+\alpha },v_0^{\delta ,\alpha }=j_{\delta }\ast v_0,{\eta }_0^{\delta ,\alpha }=j_{\delta }\ast {\eta }_0,{\Phi }^{\delta ,\alpha }=j_{\delta }\ast \Phi ,$$

where $0<\delta ,\alpha <1$ and ∗ indicate the usual convolution operation. Furthermore, let ${\rho }^{▵\delta ,\alpha }={\rho }^{▵\delta ,\alpha }\left(x\right)$ be the unique solution of (6), associated with the potential ${\Phi }^{\delta ,\alpha }$. Furthermore, we get

$$\left\{\begin{array}{c}0<\frac{\alpha }{1+\alpha }\le {\rho }_0^{\delta ,\alpha }\le \frac{\tilde{\rho }+\alpha }{1+\alpha }\le \tilde{\rho }<\infty ,\parallel \nabla v_0^{\delta ,\alpha }{\parallel }_{L^2}\le K_1,{\parallel \nabla {\eta }_0^{\delta ,\alpha }\parallel }_{L^2}\le K_2,\hfill \\ ({\rho }_0^{\delta ,\alpha }-{\rho }_{\infty },v_0^{\delta ,\alpha },{\eta }_0^{\delta ,\alpha })\in H^{\infty }\subseteq C^{\infty },\hfill \\ {\displaystyle \underset{\alpha \to 0}{lim}}{\displaystyle \underset{\delta \to 0}{lim}}\left(\parallel {\rho }_0^{\delta ,\alpha }-{\rho }_0{\parallel }_{L^2}+\parallel v_0^{\delta ,\alpha }-v_0{\parallel }_{H^1}+{\parallel {\eta }_0^{\delta ,\alpha }-{\eta }_0\parallel }_{H^1}\right)=0.\hfill \end{array}\right.$$

(97)

The initial energy of the mollifier data $({\rho }_0^{\delta ,\alpha },v_0^{\delta ,\alpha },{\eta }_0^{\delta ,\alpha })$ now becomes

$$E_0^{\delta ,\alpha }=\int \left(\frac{1}{2}{\rho }_0^{\delta ,\alpha }{\left|v_0^{\delta ,\alpha }\right|}^2+G\left({\rho }_0^{\delta ,\alpha }\right)+{\eta }_0^{\delta ,\alpha }ln{\eta }_0^{\delta ,\alpha }+{\eta }_0^{\delta ,\alpha }{\Phi }^{\delta ,\alpha }\right)dx.$$

(98)

Using (97), the results (5.34) in paper [33] for the first two terms of (98), and the estimates that

$$\underset{\alpha \to 0}{lim}\underset{\delta \to 0}{lim}\int {\eta }_0^{\delta ,\alpha }ln{\eta }_0^{\delta ,\alpha }dx\le \int {\eta }_{0}ln{\eta }_{0}dx,$$

(99)

and

$$\underset{\alpha \to 0}{lim}\underset{\delta \to 0}{lim}\int {\eta }_0^{\delta ,\alpha }{\Phi }^{\delta ,\alpha }dx\le \int {\eta }_0\Phi dx,$$

(100)

we have

$$\underset{\alpha \to 0}{lim}\underset{\delta \to 0}{lim}{E}_0^{\delta ,\alpha }\le E_0,$$

(101)

this implies the existence of ${\alpha }_0\in (0,1)$ and ${\delta }_0\left(\alpha \right)$, satisfying

$$E_0^{\delta ,\alpha }\le E_0+\frac{\overline{\epsilon }}{2}\le \overline{\epsilon }$$

(102)

for $0<\alpha <{\alpha }_0$ and $0<\delta <{\delta }_0\left(\alpha \right).$ We use Theorem 2 to get a global smooth solution $({\rho }^{\delta ,\alpha }-{\rho }_{\infty },v^{\delta ,\alpha },{\eta }^{\delta ,\alpha })$ of (1) with the initial data $({\rho }_0^{\delta ,\alpha }-{\rho }_{\infty },v_0^{\delta ,\alpha },{\eta }_0^{\delta ,\alpha })$, satisfying (25) for all $s>0$ uniformly in $\alpha ,\delta$.

Throughout, ${\langle ·\rangle }^{\kappa }$ represents the Hölder norm with exponent $\kappa \in (0,1)$. The proof begins with establishing the uniform Hölder continuity of the families $\left\{v^{\delta ,\alpha }(·,s)\right\}$ on intervals bounded away from $s=0$. □

Lemma 10.

Given $\tau >0$, it holds that

$${\langle v^{\delta ,\alpha }(·,s)\rangle }_{R^3\times [\tau ,\infty )}^{\frac{1}{2},\frac{1}{4}}\le C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho }){|s_2-s_1|}^{\frac{1}{8}},for0<\tau \le s_1<s_2<\infty ,$$

(103)

$${\langle {\eta }^{\delta ,\alpha }(·,s)\rangle }_{R^3\times [\tau ,\infty )}^{\frac{1}{2},\frac{1}{4}}\le C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho }){|s_2-s_1|}^{\frac{1}{8}},for0<\tau \le s_1<s_2<\infty .$$

(104)

Proof. 

The argument is identical to that in the work of Hoff [22].

By Sobolev-embedding Theorem (40) and (64), we obtain

$$\begin{array}{cc}\hfill \langle v^{\delta ,\alpha }(·,s)\rangle \frac{1}{2}\le & C(1+\parallel \nabla v^{\delta ,\alpha }{\parallel }_{L^6})\hfill \\ \hfill \le & C\left(1+\parallel {\rho }^{\delta ,\alpha }{\dot{v}}^{\delta ,\alpha }{\parallel }_{L^2}+\parallel \nabla v^{\delta ,\alpha }{\parallel }_{L^2}+\parallel {\rho }^{\delta ,\alpha }-{\rho }^{▵\delta ,\alpha }{\parallel }_{L^6}^2+\parallel {\eta }^{\delta ,\alpha }{\parallel }_{L^2}+{\parallel \nabla {\eta }^{\delta ,\alpha }\parallel }_{L^2}\right))\hfill \\ \hfill \le & C\left(1+\parallel {\rho }^{\delta ,\alpha }{\dot{v}}^{\delta ,\alpha }{\parallel }_{L^2}\right)\hfill \\ \hfill \le & C\left(\tau \right),s\ge \tau >0.\hfill \end{array}$$

(105)

For $s\ge \tau$ and $x\in {\mathbb{R}}^3$, the bound (105) shows that

$$\left|v^{\delta ,\alpha }(x,s)-\frac{1}{|B_R\left(x\right)|}{\int }_{B_R\left(x\right)}{v}^{\delta ,\alpha }(y,s)dy\right|\le C\left(\tau \right)R^{\frac{1}{2}}.$$

(106)

Let $s_2>s_1\ge \tau >0$; then,

$$\begin{array}{cc}\hfill & \left|v^{\delta ,\alpha }(x,s_2)-v^{\delta ,\alpha }(x,s_1)\right|\hfill \\ \hfill \le & \frac{1}{|B_R\left(x\right)|}{\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}|v_s^{\delta ,\alpha }(y,s)|dyds+C\left(\tau \right)R^{\frac{1}{2}}\hfill \\ \hfill \le & C{R}^{\frac{-3}{2}}{|s_2-s_1|}^{\frac{1}{2}}{\left({\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}{|v_s^{\delta ,\alpha }(y,s)|}^{2}dyds\right)}^{\frac{1}{2}}+C\left(\tau \right)R^{\frac{1}{2}}\hfill \\ \hfill \le & C{R}^{\frac{-3}{2}}{|s_2-s_1|}^{\frac{1}{2}}{\left({\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}(|{\dot{v}}^{\delta ,\alpha }{|}^2+|v^{\delta ,\alpha }{|}^2|\nabla v^{\delta ,\alpha }{|}^2)dyds\right)}^{\frac{1}{2}}+C\left(\tau \right)R^{\frac{1}{2}},\hfill \end{array}$$

(107)

and the following estimates hold

$$\begin{array}{cc}\hfill {\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}{\left|{\dot{v}}^{\delta ,\alpha }\right|}^{2}dyds\le & C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho }){\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}\left({\rho }^{\delta ,\alpha }{\left|{\dot{v}}^{\delta ,\alpha }\right|}^2+|{\rho }^{\delta ,\alpha }-{{\rho }^{▵\delta ,\alpha }|}^2{\left|{\dot{v}}^{\delta ,\alpha }\right|}^2\right)dyds\hfill \\ \hfill \le & C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho })+C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho }){\int }_{s_1}^{s_2}{\parallel \nabla {\dot{v}}^{\delta ,\alpha }\parallel }_{L_2}^2{\parallel {\rho }^{\delta ,\alpha }-{\rho }^{▵\delta ,\alpha }\parallel }_{L^3}^{2}ds\hfill \\ \hfill \le & C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho }),\hfill \end{array}$$

and

$$\begin{array}{c}\hfill {\int }_{s_1}^{s_2}{\int }_{B_R\left(x\right)}|v^{\delta ,\alpha }{|}^2|\nabla v^{\delta ,\alpha }{|}^{2}dyds\le C(\overline{\rho },\tilde{\rho })\underset{s\ge \tau }{sup}\parallel v^{\delta ,\alpha }{\parallel }_{L^{\infty }}^2{\int }_{s_1}^{s_2}{\parallel \nabla v^{\delta ,\alpha }\parallel }_{L^2}^{2}ds\le C(\tau ,\overline{\rho },\tilde{\rho }),\end{array}$$

hence, for $0<\tau \le s_1<s_2<\infty$, the estimate (107) gives

$$\begin{array}{cc}\hfill \left|v^{\delta ,\alpha }(x,s_2)-v^{\delta ,\alpha }(x,s_1)\right|\le & C(\tau ,\underline{\rho },\overline{\rho },\tilde{\rho })R^{\frac{-3}{2}}{|s_2-s_1|}^{\frac{1}{2}}+C\left(\tau \right)R^{\frac{1}{2}},\hfill \end{array}$$

which, chosen by $R=|s_2-s_1{|}^{\frac{1}{4}}$, leads to (103). Similarly, we also prove (104).

Consequently, the compactness of the family $({\rho }^{\delta ,\alpha }-{\rho }_{\infty },v^{\delta ,\alpha },{\eta }^{\delta ,\alpha })$ is established. □

Lemma 11.

There is a sequence ${\alpha }_k,{\delta }_k\to 0$ and functions $\rho ,v,\eta$, such that as $k\to 0$,

$$\begin{array}{c}\hfill v^{{\delta }_k,{\alpha }_k}\to v,{\eta }^{{\delta }_k,{\alpha }_k}\to \eta uniformlyoncompactsetsin{\mathbb{R}}^3\times (0,\infty ),\end{array}$$

(108)

$$\nabla v^{{\delta }_k,{\alpha }_k}(·,s),\nabla {\eta }^{{\delta }_k,{\alpha }_k}(·,s)\rightharpoonup \nabla v(·,s),\nabla \eta (·,s)weaklyin{L}^2({\mathbb{R}}^3\times (0,\infty )),$$

(109)

$${\theta }^{\frac{1}{2}}{\dot{v}}^{{\delta }_k,{\alpha }_k}(·,s),{\theta }^{\frac{3}{2}}\nabla {\dot{v}}^{{\delta }_k,{\alpha }_k}(·,s)\rightharpoonup {\theta }^{\frac{1}{2}}\dot{v},{\theta }^{\frac{3}{2}}\nabla \dot{v}weaklyin{L}^2({\mathbb{R}}^3\times [0,\infty )),$$

(110)

$${\theta }^{\frac{1}{2}}{\eta }_s^{{\delta }_k,{\alpha }_k}(·,s),{\theta }^{\frac{3}{2}}\nabla {\eta }_s^{{\delta }_k,{\alpha }_k}(·,s)\rightharpoonup {\theta }^{\frac{1}{2}}{\eta }_s(·,s),{\theta }^{\frac{3}{2}}\nabla {\eta }_s(·,s)weaklyin{L}^2({\mathbb{R}}^3\times [0,\infty )),$$

(111)

and

$${\rho }^{\delta ,\alpha }(·,s)\to \rho (·,s)stronglyin{L}_{loc}^2({\mathbb{R}}^3\times [0,\infty )).$$

(112)

Proof. 

The uniform convergence (108) is an immediate result of the Ascoli–Arzela theorem. Using the compactness provided by those bounds in (25), these weak forms (109)–(111) follow from the equality of weak-$L^2$ derivatives and distribution derivatives. Referring to Lions [34] and Feireisl [19], the convergence of approximate densities (112) was obtained. □

To explain how the nonlinear coupling terms involving $\eta$ and pressure term pass to the limit, for any test function $\phi \in \mathcal{D}({\mathbb{R}}^3\times (s_1,s_2))$ with $s_2\ge s_1\ge 0$, using (25), (108), and (112), we have

$$\begin{array}{cc}\hfill |{\int }_{s_1}^{s_2}\int \left((P\left({\rho }^{\delta ,\alpha }\right)-P)\right){\phi }_{x_j}dxds|\le & {\int }_{s_1}^{s_2}\parallel a\gamma {\xi }^{\gamma -1}{\phi }_{x_j}{\parallel }_{L_{loc}^2}{\parallel {\rho }^{\delta ,\alpha }-\rho \parallel }_{L_{loc}^2}ds\to 0,\hfill \end{array}$$

and

$$\begin{array}{c}\hfill |{\int }_{s_1}^{s_2}\int ({\eta }^{\delta ,\alpha }{v}^{\delta ,\alpha }-\eta v){\phi }_{x_j}dxds|\le {\int }_{s_1}^{s_2}\int |({\eta }^{\delta ,\alpha }-\eta )v{\phi }_{x_j}+(v^{\delta ,\alpha }-v)\eta {\phi }_{x_j}|dxds\to 0.\end{array}$$

Hence, by virtue of the statements above, we get that the limited function $(\rho -{\rho }_{\infty },v,\eta )$ is a weak solution in the sense of Definition (1) satisfying (25) for all $T\ge 0$.