Lower bound of decay rate for higher-order derivatives of solution to the compressible fluid models of Korteweg type

Abstract

This paper concerns the lower bound decay rate of global solution for compressible Navier–Stokes–Korteweg system in three-dimensional whole space under the $H^{4}\times H^{3}$ framework. At first, the lower bound of decay rate for the global solution converging to constant equilibrium state (1, 0) in $L^2$-norm is $(1+t)^{-\frac{3}{4}}$ if the initial data satisfy some low-frequency assumption additionally. Furthermore, we also show that the lower bound of the $k(k\in [1, 3])$th-order spatial derivatives of solution converging to zero in $L^2$-norm is $(1+t)^{-\frac{3+2k}{4}}$. Finally, it is proved that the lower bound of decay rate for the time derivatives of density and velocity converging to zero in $L^2$-norm is $(1+t)^{-\frac{5}{4}}$.

1 Introduction

In this paper, we are concerned with the lower bounds of decay rate for the global solution to the compressible Navier–Stokes–Korteweg system in three-dimensional whole space:

$$\begin{aligned} \left\{ \begin{aligned}&\rho _t+\mathrm{div}(\rho u)=0,\\&(\rho u)_t+\mathrm{div}(\rho u\otimes u)-\mu \Delta u-(\mu +\nu )\nabla \mathrm{div}u +\nabla P(\rho )=\kappa \rho \nabla \Delta \rho ,\\ \end{aligned} \right. \end{aligned}$$

(1.1)

where $t\ge 0$ is time, $x\in \mathbb {R}^3$ is spatial coordinate and the unknown functions $\rho =\rho (x,t)$ and $u=(u_1,u_2,u_3)(x,t)$ represent density and velocity, respectively. The pressure $P(\rho )$ is a smooth function in a neighborhood of 1 with $P'(1)>0$. The constant viscosity coefficients $\mu$ and $\nu$ satisfy the following physical conditions: $\mu >0, \, 2\mu +3\nu \ge 0.$ The constant capillary coefficient $\kappa$ satisfies $\kappa >0$. To complete the system (1.1), the initial data are given by

$$\begin{aligned} \left. (\rho , u)(x,t)\right| _{t=0}=(\rho _0(x), u_0(x)). \end{aligned}$$

(1.2)

Furthermore, as the space variable tends to infinity, we assume

$$\begin{aligned} \underset{|x|\rightarrow \infty }{\lim }(\rho _0-1, u_0)(x)=0. \end{aligned}$$

(1.3)

Korteweg-type models, supposing that the energy of the fluid depends on standard variables and the gradient of the density, are based on an extended version of nonequilibrium thermodynamics. The Navier–Stokes–Korteweg model, describing the dynamics of a liquid–vapor mixture with diffuse interphase and being used to model the motion of compressible fluid with capillary effect of materiel, originals from the work of van der Waals [23] and Korteweg [17], and the modern form was derived by Dunn and Serrin [8]. It should be noted that the system (1.1) will reduce to the well-known compressible Navier–Stokes system if the capillary coefficient satisfies $\kappa =0$.

There are many studies on the well-posedness of solutions to the compressible fluid models of Korteweg type. For the one-dimensional case, many researchers have studied extensively; refer to [3, 5] and the references therein. Charve and Haspot [3] obtained the global strong solution in the case of Saint-Venant viscosity coefficients. Chen et al. [5] obtained the global existence of classical solutions with large initial data away from vacuum for the isothermal compressible fluid of Korteweg type under the condition that the viscosity coefficient and capillarity coefficient are dependent on the density. For the multi-dimensional case, Hattori and Li [12] showed the local existence of smooth solution with large initial data for the isothermal compressible fluid models of Korteweg type in $\mathbb {R}^2$. Later, Hattori and Li [13] showed the global existence of smooth solution with small initial perturbation for the isothermal compressible fluid models of Korteweg type in high dimensions in some Sobolev space. Danchin and Desjardins in [7] showed the existence and uniqueness of suitably smooth solutions in critical Besov space. This result was improved by Haspot [11] by showing the global existence of weak solution, while the initial data belong to the energy space. Bresch et al. [2] obtained global weak solutions for the isothermal Korteweg model in a periodic or strip domain without smallness of assumption on the initial data when the viscosity $\mu$ and the capillarity coefficient $\kappa$ are dependent on the density. For more results about the well-posedness result, the readers can refer to [1, 4, 6, 14,15,16, 18, 20, 21] and the references therein.

The study for the asymptotic behavior of solution to the compressible Navier–Stokes–Korteweg equations has attracted many scholars’ attention. First of all, the researchers in [20, 24, 25] established the time decay rates for the global solution under the $H^N(N\ge 3), H^2$ and $H^1$ framework, respectively. More precisely, Wang and Tan [25] established the global existence of solution and built the time convergence rates for the case $k=0,1$,

$$\begin{aligned} \Vert \nabla ^k(\rho -1)(t)\Vert _{H^{N+1-k}}+\Vert \nabla ^k u (t)\Vert _{H^{N-k}}\le C(1+t)^{-\frac{3+2k}{4}}. \end{aligned}$$

(1.4)

Later, Gao et al. [10] proved that the convergence rates (1.4) come true when $k\in [0,N]$ with $N\ge 3$, which will be showed in Theorem 1.1. In order to obtain fast time convergence rates for the higher-order spatial derivatives of solution, Tan and Zhang [22] considered the case of initial data belonging to some negative Sobolev space rather than general $L^1$ space. More precisely, if the initial data $(\rho _0-1, u_0)\in (H^{N+1}\cap {\dot{H}}^{-s}) \times (H^N\cap {\dot{H}}^{-s})$ ($N \ge 3$ and $s\in \left[ 0, \frac{3}{2}\right)$), they established time decay rates

$$\begin{aligned} \Vert \nabla ^k(\rho -1)(t)\Vert _{H^{N+1-k}}+\Vert \nabla ^k u (t)\Vert _{H^{N-k}}\le C(1+t)^{-\frac{s+k}{2}}, \end{aligned}$$

where $k=0,1,2,\ldots , N-1$. On the other hand, Li [18], Wang and Wang [24] studied the time decay rates of smooth solution and strong solutions under the smallness assumption on the potential external force in some Sobolev space, respectively. It should be noted that the decay rate (1.4) is called “optimal” in the sense that this rate of solution for the nonlinear part coincide with the decay rate of linearized one. Thus, the aim of this paper is devoted to providing lower bounds of decay rate (coincide with upper rate) for the global solution itself and its derivatives. In other words, this implies that the decay rate (1.4) obtained in [25] is really optimal.

Notation In this paper, the symbol $\nabla ^k$ with an integer $k \ge 0$ stands for the usual any spatial derivatives of order k. For example, we define $\nabla ^k v= \left\{ \partial _x^\alpha v_i||\alpha |=k, \, i=1,2,3\right\} , \, v=(v_1, v_2, v_3).$ We also denote the Fourier transform $\mathcal {F}(f):=\hat{f}$. Denote by $\Lambda ^s$ the pseudo-differential operator defined by $\Lambda ^s f=\mathcal {F}^{-1}(|\xi |^s \hat{f}(\xi ))$. Denote $L^2(\mathbb {R}^3)$ and $H^s(\mathbb {R}^3)$ as the usual Lebesgue space and Sobolev space. For the sake of simplicity, we write $\int f \mathrm{d}x:=\int _{\mathbb {R}^3} f \mathrm{d}x$ and $\Vert (A, B)\Vert _X:=\Vert A \Vert _X+\Vert B\Vert _X$.

First of all, we recall the main results obtained in [10, 25] in the following:

Theorem 1.1

[10, 25] Assume that the initial data $\rho _0-1\in H^{N+1}$ and $u_0\in H^{N}$ for any integer $N\ge 3$ and there exists a small constant $\delta >0$ such that

$$\begin{aligned} \Vert \rho _0-1\Vert _{H^{N+1}} +\Vert u_0\Vert _{H^N}\le \delta , \end{aligned}$$

(1.5)

then the solution $(\rho , u)$ of (1.1)–(1.3) satisfies for all $t \ge 0$

$$\begin{aligned}&\Vert (\rho -1)(t)\Vert _{H^{N+1}}^2+\Vert u(t)\Vert _{H^{N}}^2+\int \limits _0^t (\Vert \nabla \rho (s)\Vert _{H^{N+1}}^2+\Vert \nabla u (s)\Vert _{H^N}^2)\mathrm{d}s \nonumber \\&\quad \le C(\Vert \rho _0-1\Vert _{H^{N+1}}^2+\Vert u_0\Vert _{H^{N}}^2). \end{aligned}$$

(1.6)

Furthermore, if $\Vert (\rho _0-1, u_0)\Vert _{L^1}$ is finite additionally, then it holds on

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^{\alpha }(\rho -1)(t)\Vert _{H^{N+1-\alpha }}+\Vert \nabla ^{\alpha } u(t)\Vert _{H^{N-\alpha }} \le C(1+t)^{-\frac{3+2\alpha }{4}}, \end{aligned} \end{aligned}$$

(1.7)

for all $\alpha \in [0,N]$. Here, C is a positive constant independent of time.

Next, we establish the lower bound of time decay rates for the global solution of (1.1)–(1.3) only under the $H^4\times H^3$ framework for the sake of simplicity.

Theorem 1.2

Denote $\varrho _0 :=\rho _0-1$ and $m_0 :=\rho _0 u_0$. Assume that the Fourier transform $\mathcal {F}(\varrho _0, m_0)=(\hat{\varrho _0}, \hat{m_0})$ satisfies

$$\begin{aligned} |\hat{\varrho }_0|\ge c_0, \quad \hat{m}_0=0, \ 0 \le |\xi | \ll 1, \end{aligned}$$

(1.8)

with $c_0$ a positive constant. Then, the global solution $(\rho , u)$ obtained in Theorem 1.1 has the decay rates for all $t \ge t_*$

$$\begin{aligned} c_1(1+t)^{-\frac{3+2k}{4}} \le \Vert \nabla ^k(\rho -1)(t)\Vert _{L^2}\le C_1(1+t)^{-\frac{3+2k}{4}}, \quad k=0,1,2,3; \end{aligned}$$

(1.9)

$$\begin{aligned} c_1(1+t)^{-\frac{3+2k}{4}} \le \Vert \nabla ^k u(t)\Vert _{L^{2}}\le C_1(1+t)^{-\frac{3+2k}{4}}, \quad k=0,1,2,3. \end{aligned}$$

(1.10)

Here, $t_*$ is a positive large time, and $c_1$ and $C_1$ are two positive constants independent of time.

Remark 1.1

The lower bounds of decay rates (1.9), (1.10) for the derivatives of density and velocity to the compressible Navier–Stokes–Korteweg system are obtained for the first time.

Remark 1.2

Although we only establish the time decay rates under the $H^4 \times H^3$ framework in Theorem 1.2, the method used here can be applied to the $H^{N+1} \times H^N(N\ge 3)$-framework. Under the condition (1.8), the global classical solution $(\rho , u)$ of the system (1.1) has the decay rates for all $t \ge t_*$

$$\begin{aligned}&c_1(1+t)^{-\frac{3+2k}{4}}\le \Vert \nabla ^k(\rho -1)(t)\Vert _{L^2} \le C_1(1+t)^{-\frac{3+2k}{4}}, \quad k \in [0, N];\\&c_1(1+t)^{-\frac{3+2k}{4}}\le \Vert \nabla ^k u(t)\Vert _{L^2}\le C_1(1+t)^{-\frac{3+2k}{4}}, \quad k \in [0, N]. \end{aligned}$$

Here, $t_*$ is a positive large time, and $c_1$ and $C_1$ are two positive constants independent of time.

Remark 1.3

It should be pointed out that under the $H^3$-framework, the decay rates (1.9) and (1.10) for the global solution to the compressible Navier–Stokes equations can only be obtained under the condition that $k=0,1$ (see [9]). However, these decay rates for the solution to the Navier–Stokes–Korteweg equations can be obtained when $k=0,1,2,3$. The difference here is the appearance of Korteweg term $\kappa \rho \nabla \Delta \rho$ that will obtain enough dissipation for the density.

Finally, we will establish the lower bounds of decay rates for the time derivatives of solution to the compressible Navier–Stokes–Korteweg system (1.1).

Theorem 1.3

Assume the condition (1.8) holds on, then the global classical solution $(\rho , u)$ obtained in Theorem 1.1 satisfies for all $t \ge t_*$

$$\begin{aligned} c_2(1+t)^{-\frac{5}{4}}&\le \Vert \partial _t u(t)\Vert _{L^2} \le C_2(1+t)^{-\frac{5}{4}}. \end{aligned}$$

(1.11)

Furthermore, if there exists a small constant $\delta _1$ such that $\Vert u_0\Vert _{L^1}\le \delta _1$, it holds on for all $t \ge t_*$

$$\begin{aligned} c_2(1+t)^{-\frac{5}{4}}\le \Vert \partial _t \rho (t)\Vert _{L^2}\le C_2(1+t)^{-\frac{5}{4}}. \end{aligned}$$

(1.12)

Here, $t_*$ is a positive large time, and $c_2$ and $C_2$ are two positive constants independent of time.

Remark 1.4

The lower bounds of decay rates for the time derivatives of density and velocity for the compressible Navier–Stokes–Korteweg system in the $L^2$ norm are obtained for the first time.

Now we make comments on the analysis of this paper. At first, we give the lower bound of decay rate for the higher-order spatial derivative of solution to the compressible Navier–Stokes–Korteweg equations (1.1). Define U and $U_l$ as the solution to the nonlinear and linearized problems, respectively. Define $U_\delta := U-U_l$, then we have for any integer k

$$\begin{aligned} \Vert \nabla ^k U\Vert _{L^2} \ge \Vert \nabla ^k U_l\Vert _{L^2}-\Vert \nabla ^k U_\delta \Vert _{L^2}. \end{aligned}$$

For $\alpha _{l,k} < \alpha _{\delta , k}$, if the solutions $U_l$ and $U_\delta$ satisfy

$$\begin{aligned} \Vert \nabla ^k U_l\Vert _{L^2} \ge C_{l,k} (1+t)^{-\alpha _{l,k}}, \quad \Vert \nabla ^k U_\delta \Vert _{L^2} \le C_{\delta ,k} (1+t)^{-\alpha _{\delta , k}}, \end{aligned}$$

(1.13)

where $C_{l,k}$ and $C_{\delta ,k}$ are positive constants independent of time. For large time t, it holds on

$$\begin{aligned} \Vert \nabla ^k U\Vert _{L^2} \ge C_{l,k} (1+t)^{-\alpha _{l,k}}-C_{\delta , k}(1+t)^{-\alpha _{\delta ,k}} \ge C(1+t)^{-\alpha _{l,k}}, \end{aligned}$$

where C is a positive constant that independent of time. It is easy to obtain the lower bound of decay rate for the linearized part by applying the spectral analysis to the semigroup for the linearized Navier–Stokes–Korteweg system (2.4); see Proposition 2.1 in Sect. 2. Therefore, it is significant to obtain upper bound of decay rate of $\nabla ^k U_\delta$. To achieve this goal, we establish the energy estimates, see (2.21) and (2.22) in Lemma 2.2. Then, with the help of the upper bound decay estimate (1.7), the upper bound of decay rate for $U_\delta$ can be obtained. While taking the case $k\ge 1$ into account, we utilize the Fourier Splitting method developed by Schonbek [19] to obtain the upper bound of decay rate for $\nabla ^k U_\delta$.

Next, the upper and lower bounds of decay rate for the time derivative of velocity can be obtained by using the equation and lower bound of first-order spatial derivative. If we use the transport equation to obtain the lower bound of decay rate for the time derivative of density, we need to get the lower bound for the quantity divergence of velocity(i.e., $\mathrm{div} u$). To achieve this target, we need to assume the smallness for the initial velocity in $L^1$.

The rest of this paper is organized as follows. In Sect. 2, we establish the lower bound of decay rate for the solution itself and derivative, and then, we establish the upper and lower bounds of decay rate for the time derivatives of solution. In Sect. 3, we prove technical estimates used in Sect. 2.

2 Lower bounds of decay for spatial derivative

   In this section, we will address the lower bound of decay rates for the solution itself and its derivative. To this end, the upper decay rates for the difference between the nonlinear and linearized parts will be established. Finally, we address the upper decay rate of solution for the higher-order spatial derivative.

2.1 Lower bounds of decay for spatial derivative

In this subsection, we will establish optimal time decay rates of solution for the compressible Navier–Stokes–Korteweg equations (1.1)-(1.3). Let us denote $\varrho :=\rho -1, m:=\rho u$. For simplicity, we take $P'(1)=1$; then, we rewrite (1.1) in the perturbation form as follows:

$$\begin{aligned} \left\{ \begin{aligned}&\varrho _t+\mathrm{div}m=0, \\&m_t-\mu \Delta m-(\mu +\nu )\nabla \mathrm{div} m+\nabla \varrho -\kappa \nabla \Delta \varrho =-\mathrm{div} S, \end{aligned} \right. \end{aligned}$$

(2.1)

where the function $S=S(\varrho , u)$ is defined as

$$\begin{aligned} S= & {} (\varrho +1)u\otimes u+\mu \nabla (\varrho u)+(\mu +\nu )\mathrm{div}(\varrho u)\mathbb {I}_{3\times 3}\nonumber \\&+(P(1+\varrho )-P(1)-\varrho )\mathbb {I}_{3\times 3}-\kappa \varrho \nabla ^2\varrho +\frac{1}{2}\kappa |\nabla \varrho |^2 I_{3\times 3}. \end{aligned}$$

(2.2)

The initial data are given as

$$\begin{aligned} \left. (\varrho , m)(x,t)\right| _{t=0}=(\varrho _0, m_0)(x) \rightarrow (0,0) \quad {\text{ as } } \quad |x|\rightarrow \infty . \end{aligned}$$

(2.3)

In order to obtain the lower decay estimate, we need to analysis the linearized part:

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t \varrho _l+\mathrm{div} m_l=0, \\&\partial _t m_l-\mu \Delta m_l-(\mu +\nu )\nabla \mathrm{div} m_l+\nabla \varrho _l-\kappa \nabla \Delta \varrho _{l}=0, \end{aligned} \right. \end{aligned}$$

(2.4)

with the initial data

$$\begin{aligned} \left. (\varrho _l, m_l)(x,t)\right| _{t=0}=(\varrho _0, m_0)(x) \rightarrow (0,0) \quad {\text{ as } } \quad |x|\rightarrow \infty . \end{aligned}$$

(2.5)

Here, we assume initial data for the linearized system (2.4) is the same as the original problem (2.1).

Proposition 2.1

Let $\varrho _0\in H^{4}(\mathbb {R}^3)\bigcap L^1(\mathbb {R}^3)$, $m_0\in H^{3}(\mathbb {R}^3)\bigcap L^1(\mathbb {R}^3)$. Assume that the Fourier transform $\mathcal {F}(\varrho _0, m_0)=(\hat{\varrho } _{0}, \hat{m}_0)$ satisfies $|\hat{\varrho } _{0}|\ge c_0, \, |\hat{m}_{0}|=0, \, 0 \le |\xi | \ll 1$ with $c_0$ a positive constant, then there exists a positive large time $t_{*}$ such that for $t\ge t_{*}$, we have for $s=0,1,2,3$,

$$\begin{aligned} \Vert \nabla ^s \varrho _l(t)\Vert _{L^2}\ge & {} c (1+t)^{-(\frac{3}{4}+\frac{s}{2})};\nonumber \\ \Vert \nabla ^s m_l(t)\Vert _{L^2}\ge & {} c (1+t)^{-(\frac{3}{4}+\frac{s}{2})}, \end{aligned}$$

(2.6)

where c is a positive constant independent of time t.

Proof

By virtue of the semigroup theory for evolution equation, the solution $(\varrho _l,m_l)$ of the linearized problem (2.4), (2.5) can be expressed by

$$\begin{aligned} U_t=BU, \ U(0)=U_0, \end{aligned}$$

(2.7)

where $U=(\varrho _l,m_l)^{t}$ and

$$\begin{aligned} B=\left[ \begin{array}{ll} 0&{} \quad -\mathrm{div}\\ -\nabla +\kappa \nabla \Delta &{}\quad \mu \Delta +(\mu +\nu )\nabla \mathrm{div}\\ \end{array} \right] . \end{aligned}$$

(2.8)

Applying Fourier transform to Eq. (2.7), it holds on

$$\begin{aligned} \hat{U}_t=A(\xi )\hat{U}, \ \hat{U}(0)=\hat{U}_0, \end{aligned}$$

(2.9)

where

$$\begin{aligned} A(\xi )=\left[ \begin{array}{ll} 0&{}\quad -i\xi ^{t}\\ -i\xi -i\kappa |\xi |^2\xi &{}\quad -\mu |\xi |^2I_{3\times 3}-(\mu +\nu )\xi \xi ^{t}\\ \end{array} \right] . \end{aligned}$$

(2.10)

Then by computation, we obtain the Fourier transform $\hat{G}(\xi ,t)$ of Green’s function $G(x,t)=e^{tB}$ as follows:

$$\begin{aligned} \hat{G}(\xi ,t)=\left[ \begin{array}{ll} \frac{\lambda _{+}e^{\lambda _{-}t}-\lambda _{-}e^{\lambda _{+}t}}{\lambda _{+}-\lambda _{-}}&{}{}\quad - \frac{i\xi ^{t}(e^{\lambda _{+}t}-e^{\lambda _{-}t})}{\lambda _{+}-\lambda _{-}}\\ -\frac{i\xi (1+\kappa |\xi |^2)(e^{\lambda _{+}t}-e^{\lambda _{-}t})}{\lambda _{+}-\lambda _{-}} &{}{}\quad \frac{\lambda _{+}e^{\lambda _{+}t}-\lambda _{-}e^{\lambda _{-}t}}{\lambda _{+}-\lambda _{-}}\cdot \frac{\xi \xi ^t}{|\xi |^2}+e^{-\lambda _0 t}\left( I-\frac{\xi \xi ^t}{|\xi |^2}\right) \\ \end{array} \right] , \end{aligned}$$

(2.11)

where

$$\begin{aligned} \lambda _0= & {} -\mu |\xi |^2 \, (\text {double}),\\ \lambda _{+}= & {} -\frac{1}{2}(2\mu +\nu )|\xi |^2+\frac{1}{2}i\sqrt{4(1+\kappa |\xi |^2)|\xi |^2-(2\mu +\nu )^2|\xi |^4},\\ \lambda _{-}= & {} -\frac{1}{2}(2\mu +\nu )|\xi |^2-\frac{1}{2}i\sqrt{4(1+\kappa |\xi |^2)|\xi |^2-(2\mu +\nu )^2|\xi |^4}. \end{aligned}$$

Thus, we can obtain the expression for $\hat{\varrho _l}(\xi ,t)$ and $\hat{m_l}(\xi ,t)$ as follows,

$$\begin{aligned} \hat{\varrho _l}(\xi ,t)=\frac{\lambda _{+}e^{\lambda _{-}t}-\lambda _{-}e^{\lambda _{+}t}}{\lambda _{+}-\lambda _{-}}\hat{\varrho _l}_0- \frac{i\xi ^t(e^{\lambda _{+}t}-e^{\lambda _{-}t})}{\lambda _{+}-\lambda _{-}}\hat{m_l}_0, \end{aligned}$$

(2.12)

and

$$\begin{aligned} \hat{m_l}(\xi ,t)&=-\frac{i\xi (1+\kappa |\xi |^2)(e^{\lambda _{+}t}-e^{\lambda _{-}t})}{\lambda _{+}-\lambda _{-}}\hat{\varrho _l}_0+ \left[ \frac{\lambda _{+}e^{\lambda _{+}t}-\lambda _{-}e^{\lambda _{-}t}}{\lambda _{+}-\lambda _{-}}\cdot \frac{\xi \xi ^t}{|\xi |^2}+e^{-\lambda _0 t}\left( I-\frac{\xi \xi ^t}{|\xi |^2}\right) \right] \hat{m_l}_0. \end{aligned}$$

(2.13)

Then it is easy to verify that

$$\begin{aligned} \hat{\varrho _l}(\xi ,t)\sim \left\{ \begin{aligned}&e^{-\frac{1}{2}(2\mu +\nu )|\xi |^2 t}\cos (bt)|\hat{\varrho _l}_0|+\mathcal {O}(1)e^{-\frac{1}{2}(2\mu +\nu )|\xi |^2 t} \frac{\sin {(bt)}}{b}|\xi |^2|\hat{\varrho _l}_0|, \ |\xi |\ll 1, \\&\mathcal {O}(1)e^{-R_0 |\xi |^2 t}|\hat{\varrho _l}_0|, \ |\xi |\gg 1, \end{aligned} \right. \end{aligned}$$

(2.14)

and

$$\begin{aligned} \hat{m_l}(\xi ,t)\sim \left\{ \begin{aligned}&e^{-\frac{1}{2}(2\mu +\nu )|\xi |^2 t}\sin (bt)|\hat{\varrho _l}_0|,\ |\xi |\ll 1,\\&\mathcal {O}(1)e^{-R_0 |\xi |^2 t}|\hat{\varrho _l}_0|, \ |\xi |\gg 1, \end{aligned} \right. \end{aligned}$$

(2.15)

where

$$\begin{aligned} b=\frac{1}{2}\sqrt{4(1+\kappa |\xi |^2)|\xi |^2-(2\mu +\nu )^2|\xi |^4}\sim |\xi |+\mathcal {O}(|\xi |^3), \ |\xi |\ll 1, \end{aligned}$$

$R_0$ is a positive constant. By Parseval’s theorem, for $s=0,1,2,3$, it holds on

$$\begin{aligned} \begin{aligned} \Vert \nabla ^s \varrho _l\Vert _{L^2}^2&=\int \limits _{|\xi |\le \eta }|\xi |^{2s}|\hat{\varrho _l}|^2 \mathrm{d}\xi +\int \limits _{|\xi |\ge \eta }|\xi |^{2s}|\hat{\varrho _l}|^2 \mathrm{d}\xi \nonumber \\&\ge \frac{1}{2}\int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}\cos ^2(bt)|\hat{\varrho _l}_0|^2 \mathrm{d}\xi \!-\!C \int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s+4}\frac{\sin ^2(bt)}{b^2}|\hat{\varrho _l}_0|^2 \mathrm{d}\xi \nonumber \\&\quad -Ce^{-2R_0\eta ^2 t}, \end{aligned} \end{aligned}$$

where $\eta$ is a small but fixed positive constant. Due to the fact that $b\sim |\xi |+\mathcal {O}(|\xi |^3)$ as $|\xi |\ll 1$, it holds on

$$\begin{aligned} \int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s+4}\frac{\sin ^2(bt)}{b^2}|\hat{\varrho _l}_0|^2 \mathrm{d}\xi \le C t^{-(s+\frac{5}{2})}. \end{aligned}$$

Employing the mean value formula, we get

$$\begin{aligned} \cos (|\xi |+\mathcal {O}(|\xi |^3))=\cos (|\xi |t)+\mathcal {O}(|\xi |^3 t); \end{aligned}$$

then, we have

$$\begin{aligned} \cos ^2(|\xi |+\mathcal {O}(|\xi |^3))\ge \frac{1}{2}\cos ^2(|\xi |t)-\mathcal {O}([|\xi |^3 t]^2), \end{aligned}$$

which, together with the above inequality, we obtain that

$$\begin{aligned}&\int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}\cos ^2(bt)|\hat{\varrho _l}_0|^2 \mathrm{d}\xi \nonumber \\&\quad \ge C\int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}\cos ^2(|\xi |t) \mathrm{d}\xi - C\int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}|\xi |^6 t^2 \mathrm{d}\xi \nonumber \\&\quad =: I_1-I_2. \end{aligned}$$

(2.16)

By direct computation, we have

$$\begin{aligned} \int \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}|\xi |^6 t^2 \mathrm{d}\xi \le C t^{-(s+\frac{5}{2})}; \end{aligned}$$

(2.17)

therefore, we only need to consider the lower bound of $I_1$. We claim that the following estimate (which will be proved in Sect. 3),

$$\begin{aligned} I_1 \ge C t^{-(\frac{3}{2}+s)}. \end{aligned}$$

(2.18)

Substituting the estimate (2.17) and (2.18) into (2.16), one arrives at

$$\begin{aligned} \int \limits \limits _{|\xi |\le \eta }e^{-(2\mu +\nu )|\xi |^{2}t}|\xi |^{2s}\cos ^2(bt)|\hat{\varrho }_0|^2 \mathrm{d}\xi \ge C t^{-(\frac{3}{2}+s)}. \end{aligned}$$

Hence, we can easily obtain that for $s=0,1,2,3$,

$$\begin{aligned} \Vert \nabla ^s \varrho _l\Vert _{L^2}^2\ge C (1+t)^{-(\frac{3}{2}+s)}, \end{aligned}$$

where C is a positive constant independent of time. Similarly, we can also derive that for $s=0,1,2,3$,

$$\begin{aligned} \Vert \nabla ^s m_l\Vert _{L^2}^2\ge C (1+t)^{-(\frac{3}{2}+s)}, \end{aligned}$$

with C a positive constant independent of time. Therefore, we finish the proof of this proposition. $\square$

In order to obtain the lower bound for the solution of the compressible Navier–Stokes–Korteweg equation (2.1), we need to address the upper decay rate for the difference between the nonlinear and linearized part. Hence, let us denote

$$\begin{aligned} \varrho _\delta :=\varrho -\varrho _l, m_\delta :=m-m_l, \end{aligned}$$

then they satisfy the following system

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t \varrho _\delta +\mathrm{div} m_\delta =0,\\&\partial _t m_\delta -\mu \Delta m_\delta -(\mu +\nu )\nabla \mathrm{div} m_\delta +\nabla \varrho _\delta -\kappa \nabla \Delta \varrho _\delta =-\mathrm{div} S, \end{aligned} \right. \end{aligned}$$

(2.19)

with the zero initial data

$$\begin{aligned} \left. (\varrho _\delta , m_\delta )(x,t)\right| _{t=0}=(0, 0). \end{aligned}$$

(2.20)

Here, the force term is defined in (2.2). Now we will establish the decay rate for the solution $(\varrho _\delta , m_\delta )$ of equation (2.19) in the following.

Lemma 2.2

For any smooth solution $(\varrho _\delta , m_\delta )$ of the equation (2.19), it holds on

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}(\Vert \nabla ^l (\varrho _\delta ,m_\delta )\Vert _{H^{3-l}}^2+\kappa \Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{3-l}}^2)+C\Vert \nabla ^{l+1} m_\delta \Vert _{H^{3-l}}^2 \nonumber \\&\quad \le C \Vert \nabla ^l(\varrho ,u)\Vert _{H^{3-l}}^2 (\Vert \nabla \varrho \Vert _{H^{2}}^2+\Vert \nabla u\Vert _{H^{2}}^2)+C\Vert \nabla \varrho \Vert _{H^1}^2(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2), \end{aligned}$$

(2.21)

and

$$\begin{aligned}&\sum _{k=l}^3\frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^l m_\delta \cdot \nabla ^{l+1} \varrho _\delta \mathrm{d}x +\Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{3-l}}^2+\frac{1}{2}\kappa \Vert \nabla ^{l+2} \varrho _\delta \Vert _{H^{3-l}}^2 \nonumber \\&\quad \le C\Vert \nabla ^{l+1} m_\delta \Vert _{H^{3-l}}^2 + C \Vert \nabla ^l(\varrho ,u)\Vert _{H^{3-l}}^2 (\Vert \nabla \varrho \Vert _{H^{2}}^2+\Vert \nabla u\Vert _{H^{2}}^2) \nonumber \\&\qquad +\,C\Vert \nabla \varrho \Vert _{H^1}^2(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2), \end{aligned}$$

(2.22)

where $l=0,1,2,3$.

The above inequalities (2.21) and (2.22) in Lemma 2.2 will be proved later in Sect. 3. Multiplying inequality (2.22) by a small constant $\delta$ and adding with (2.21), together with the decay rates (1.7), we get for all $t \ge 0$,

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_l^3(t) +\frac{\delta }{2}\Vert \nabla ^{l+1}\varrho _\delta \Vert _{H^{3-l}}^2 +\frac{\delta }{2}\kappa \Vert \nabla ^{l+2}\varrho _\delta \Vert _{H^{3-l}}^2 +C\Vert \nabla ^{l+1}m_\delta \Vert _{H^{3-l}}^2\nonumber \\&\quad \le C(1+t)^{-(4+l)}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2), \end{aligned}$$

(2.23)

where $l=0,1,2,3$. Here, the energy $\mathcal {E}_l^3(t)$ is defined by

$$\begin{aligned} \mathcal {E}_l^3(t):= \Vert \nabla ^l (\varrho _\delta ,m_\delta )\Vert _{H^{3-l}}^2+\kappa \Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{3-l}}^2 +\delta \sum _{k=l}^3 \int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1}\varrho _\delta \mathrm{d}x. \end{aligned}$$

(2.24)

Due to the smallness of $\delta$, there are two constants $C_*$ and $C^*$(independent of time) such that

$$\begin{aligned} C_* (\Vert \nabla ^l \varrho _\delta (t)\Vert _{H^{4-l}}^2+\Vert \nabla ^l m_\delta (t)\Vert _{H^{3-l}}^2) \le \mathcal {E}_l^3(t) \le C^* (\Vert \nabla ^l \varrho _\delta (t)\Vert _{H^{4-l}}^2+\Vert \nabla ^l m_\delta (t)\Vert _{H^{3-l}}^2). \end{aligned}$$

(2.25)

Now we establish the upper bound decay rate of solution $(\varrho _\delta , m_\delta )$ for the equation (2.19).

Lemma 2.3

Under the assumptions in Theorem 1.1, the smooth solution $(\varrho _\delta , m_\delta )$ of equation (2.19) satisfies

$$\begin{aligned} \Vert \nabla ^l \varrho _\delta (t)\Vert _{H^{4-l}}+\Vert \nabla ^l m_\delta (t)\Vert _{H^{3-l}}\le C(1+t)^{-\frac{5+2l}{4}}, \end{aligned}$$

(2.26)

where $l=0,1,2,3$. Here, C is a constant independent of time.

Proof

We will take the strategy of induction to give the proof of estimate (2.26). Taking $l=0$ in (2.23), then we have

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_0^3(t) +C(\Vert \nabla \varrho _\delta \Vert _{H^{4}}^2+\Vert \nabla m_\delta \Vert _{H^{3}}^2) \le C(1+t)^{-4}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2). \end{aligned}$$

Obviously, the dissipation term $\Vert \nabla \varrho _\delta \Vert _{H^{4}}^2+\Vert \nabla m_\delta \Vert _{H^{3}}^2$ cannot control the energy term $\mathcal {E}_0^3(t)$ in above inequality. Thus, we add both sides of the above inequality with term $\Vert (\varrho _\delta , m_\delta )\Vert _{L^2}^2$ and get

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_0^3(t) +C(\Vert \varrho _\delta \Vert _{H^{5}}^2+\Vert m_\delta \Vert _{H^{4}}^2)\nonumber \\&\quad \le \Vert (\varrho _\delta , m_\delta )\Vert _{L^2}^2+C(1+t)^{-4}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2). \end{aligned}$$

(2.27)

By virtue of the Duhamel principle formula and estimate (1.7), we have

$$\begin{aligned} \Vert (\varrho _\delta , m_\delta )(t)\Vert _{L^2}\le & {} \int \limits _0^t (1+t-\tau )^{-\frac{5}{4}}(\Vert |\xi |^{-1}\mathcal {F}(\mathrm{div} S)\Vert _{L^\infty } +\Vert \mathrm{div} S\Vert _{L^2})\mathrm{d}\tau \nonumber \\\le & {} \int \limits _0^t (1+t-\tau )^{-\frac{5}{4}}(\Vert S\Vert _{L^1}+\Vert \mathrm{div} S \Vert _{L^2})\mathrm{d}\tau \nonumber \\\le & {} C\int \limits _0^t (1+t-\tau )^{-\frac{5}{4}}(1+\tau )^{-\frac{3}{2}}\mathrm{d}\tau \nonumber \\\le & {} C(1+t)^{-\frac{5}{4}}, \end{aligned}$$

(2.28)

where we have used Sobolev inequality to get

$$\begin{aligned}&\Vert S\Vert _{L^1}+\Vert \mathrm{div} S \Vert _{L^2}\nonumber \\&\quad \le \Vert \varrho +1\Vert _{L^\infty }\Vert u\Vert _{L^2}\Vert u\Vert _{L^2}+\Vert \nabla \varrho \Vert _{L^2}\Vert u\Vert _{L^2}+\Vert \varrho \Vert _{L^2}\Vert \nabla u\Vert _{L^2}+\Vert \varrho \Vert _{L^2}\Vert \varrho \Vert _{L^2}\\&\qquad +\Vert \varrho \Vert _{L^2}\Vert \nabla ^2\varrho \Vert _{L^2} +\Vert \nabla \varrho \Vert _{L^2}\Vert \nabla \varrho \Vert _{L^2}+\Vert 1+\varrho \Vert _{L^\infty }\Vert u\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2} +\Vert u\Vert _{L^\infty }^2\Vert \nabla \varrho \Vert _{L^2}\\&\qquad +\Vert \nabla ^2 (\varrho u)\Vert _{L^2}+\Vert \nabla \mathrm{div}(\varrho u)\Vert _{L^2}+ \Vert \nabla (\varrho \nabla ^2\varrho )\Vert _{L^2}+\Vert \varrho \nabla \varrho \Vert _{L^2}\nonumber \\&\quad \le C \Vert \nabla (\varrho ,u)\Vert _{H^1}\Vert (\varrho ,u)\Vert _{L^2}+ C\Vert \varrho \Vert _{L^2}^2+ C\Vert \nabla \varrho \Vert _{L^2}^2+ C\Vert \nabla (\varrho ,u)\Vert _{H^1}^2+ C\Vert \nabla \varrho \Vert _{H^1}\Vert \nabla ^3\varrho \Vert _{L^2}\nonumber \\&\quad \le C(1+t)^{-\frac{3}{2}}. \end{aligned}$$

Using (2.27), (2.28) and equivalent relation (2.25), one arrives at

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_0^3(t)+\frac{C}{C^{*}}\mathcal {E}_0^3(t) \le C(1+t)^{-\frac{5}{2}}+ C (1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2), \end{aligned}$$

which implies

$$\begin{aligned} \mathcal {E}_0^3(t)\le C\int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau + C \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\mathrm{d}\tau . \end{aligned}$$

It is easy to obtain that

$$\begin{aligned} \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau \le C(1+t)^{-\frac{5}{2}}, \end{aligned}$$

where we have used the fact that

$$\begin{aligned} e^{-\frac{C}{C^{*}}t}\le C (1+t)^{-\frac{5}{2}}, \end{aligned}$$

for some constant $C>0$, and

$$\begin{aligned}&\int \limits _0^t e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau \nonumber \\&\quad \le C\int \limits _0^t (1+t-\tau )^{-\frac{5}{2}}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau \nonumber \\&\quad = C \int \limits _0^{\frac{t}{2}}+\int \limits _{\frac{t}{2}}^t (1+t-\tau )^{-\frac{5}{2}}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau \nonumber \\&\quad \le C \left( 1+\frac{t}{2}\right) ^{-\frac{5}{2}}\int \limits _0^{\frac{t}{2}}(1+\tau )^{-\frac{5}{2}} \mathrm{d}\tau +\left( 1+\frac{t}{2}\right) ^{-\frac{5}{2}}\int \limits _{\frac{t}{2}}^t (1+t-\tau )^{-\frac{5}{2}} \mathrm{d}\tau \nonumber \\&\quad \le C (1+t)^{-\frac{5}{2}}. \end{aligned}$$

(2.29)

In the sequel, we only need to deal with the term

$$\begin{aligned} \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\mathrm{d}\tau . \end{aligned}$$

We claim the estimate(which will be proved in Sect. 3),

$$\begin{aligned} \begin{aligned} \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\mathrm{d}\tau \le C(1+t)^{-\frac{5}{2}}. \end{aligned} \end{aligned}$$

(2.30)

Hence, we can easily obtain that

$$\begin{aligned} \Vert \varrho _\delta (t)\Vert _{H^4}^2+\Vert m_\delta (t)\Vert _{H^3}^2 \le C(1+t)^{-\frac{5}{2}}. \end{aligned}$$

(2.31)

We now assume that the decay rate (2.26) holds on for the case $k=l$, i.e.,

$$\begin{aligned} \Vert \nabla ^l \varrho _\delta (t)\Vert _{H^{4-l}}+\Vert \nabla ^l m_\delta (t)\Vert _{H^{3-l}}\le C(1+t)^{-\frac{5+2l}{4}}, \end{aligned}$$

(2.32)

for $l=1,2$. Then, we should verify that the estimate (2.26) holds on for the case $k=l+1$. Indeed, we replace l by $l+1$ in (2.23) to obtain that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_{l+1}^3(t) +C(\Vert \nabla ^{l+2}\varrho _\delta \Vert _{H^{3-l}}^2 +\Vert \nabla ^{l+2}m_\delta \Vert _{H^{2-l}}^2) \nonumber \\&\quad \le C(1+t)^{-(5+l)}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2). \end{aligned}$$

(2.33)

For some constant R defined below, denoting the time sphere (see [19])

$$\begin{aligned} S_0:=\left\{ \left. \xi \in \mathbb {R}^3\right| |\xi |\le \left( \frac{R}{1+t}\right) ^\frac{1}{2}\right\} , \end{aligned}$$

it follows immediately

$$\begin{aligned}&\Vert \nabla ^{l+2} \varrho _\delta \Vert _{H^{3-l}}^2 \ge \frac{R}{1+t}\Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{3-l}}^2-\frac{R^2}{(1+t)^2}\Vert \nabla ^l\varrho _\delta \Vert _{H^{3-l}}^2; \nonumber \\&\quad \Vert \nabla ^{l+2} m_\delta \Vert _{H^{2-l}}^2 \ge \frac{R}{1+t}\Vert \nabla ^{l+1} m_\delta \Vert _{H^{2-l}}^2-\frac{R^2}{(1+t)^2}\Vert \nabla ^l m_\delta \Vert _{H^{2-l}}^2. \end{aligned}$$

(2.34)

By substituting (2.34) into (2.33), we can easily get

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_{l+1}^3(t) +\frac{CR}{1+t}(\Vert \nabla ^{l+1}\varrho _\delta \Vert _{H^{3-l}}^2 +\Vert \nabla ^{l+1}m_\delta \Vert _{H^{2-l}}^2)\\&\quad \le \frac{C R^2}{(1+t)^2}(\Vert \nabla ^l\varrho _\delta \Vert _{H^{3-l}}^2\!+\!\Vert \nabla ^l m_\delta \Vert _{H^{2-l}}^2)+C(1+t)^{-(5+l)}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2)\\&\quad \le C R^2(1+t)^{-\frac{9+2l}{2}}+C(1+t)^{-(5+l)}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2), \end{aligned}$$

where we have used the assumption (2.32). Notice that the term $\mathcal {E}_{l+1}^3(t)$ is equivalent to the norm $\Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{4-l}}^2+\Vert \nabla ^{l+1} m_\delta \Vert _{H^{3-l}}^2$; hence, we obtain

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\mathcal {E}_{l+1}^3(t) +\frac{CR}{C^{*}(1+t)}\mathcal {E}_{l+1}^3(t)\\&\quad \le C R^2(1+t)^{-\frac{9+2l}{2}}+C(1+t)^{-(5+l)}+C(1+t)^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2). \end{aligned}$$

Choosing $R=C^{*}(l+4)/C$ and multiplying the above inequality by $(1+t)^{l+4}$, we obtain that

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}[(1+t)^{l+4}\mathcal {E}_{l+1}^3(t)] \le C (1+t)^{-\frac{1}{2}}+C(1+t)^{l+\frac{3}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^{4} u\Vert _{L^2}^2). \end{aligned}$$

(2.35)

We claim that the following estimate holds on (which will be proved in Sect. 3),

$$\begin{aligned} \int \limits _0^t (1+\tau )^{l+\frac{3}{2}}(\Vert \nabla ^4 \varrho (\tau )\Vert _{H^1}^2+\Vert \nabla ^{4} u(\tau )\Vert _{L^2}^2) \mathrm{d}\tau \le C, \end{aligned}$$

(2.36)

which together with (2.35) yields directly

$$\begin{aligned} \mathcal {E}_{l+1}^3(t)\le C (1+t)^{-\frac{7+2l}{2}}. \end{aligned}$$

Then, due to the fact that the term $\mathcal {E}_{l+1}^3(t)$ is equivalent to the norm $\Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{4-l}}^2+\Vert \nabla ^{l+1} m_\delta \Vert _{H^{3-l}}^2$, we obtain

$$\begin{aligned} \Vert \nabla ^{l+1} \varrho _\delta \Vert _{H^{4-l}}^2+\Vert \nabla ^{l+1} m_\delta \Vert _{H^{3-l}}^2\le C (1+t)^{-\frac{7+2l}{2}}. \end{aligned}$$

Thus, by the general step of induction, we have given the proof for (2.26). $\square$

Finally, we establish the lower bound estimates.

Lemma 2.4

Under all the assumptions of Theorem 1.2, the solution $(\rho , u)$ of equation (1.1) has the following estimate

$$\begin{aligned} \begin{aligned}&\Vert \nabla ^k (\rho -1)(t)\Vert _{L^2}\ge C(1+t)^{-\frac{3+2k}{4}}, \quad k=0,1,2,3;\\&\Vert \nabla ^k u(t)\Vert _{L^2}\ge C(1+t)^{-\frac{3+2k}{4}}, \quad k=0,1,2,3, \end{aligned} \end{aligned}$$

(2.37)

for all $t \ge t_*$, where $t_*$ is a positive constant.

Proof

Recall the definition

$$\begin{aligned} \varrho _\delta =\varrho -\varrho _l,\ m_\delta =m-m_l, \end{aligned}$$

we have

$$\begin{aligned} \Vert \varrho _l\Vert _{L^2}=\Vert \varrho -\varrho _\delta \Vert _{L^2}\le \Vert \varrho \Vert _{L^2}+\Vert \varrho _\delta \Vert _{L^2}, \end{aligned}$$

which, together with estimates (2.6) and (2.26), yields directly

$$\begin{aligned} \begin{aligned} \Vert \varrho (t) \Vert _{L^2}&\ge \Vert \varrho _l(t)\Vert _{L^2}-\Vert \varrho _\delta (t)\Vert _{L^2}\\&\ge C_1(1+t)^{-\frac{3}{4}}-C_2(1+t)^{-\frac{5}{4}}\\&\ge C_1(1+t)^{-\frac{3}{4}}-\frac{C_2}{(1+t)^{\frac{1}{2}}}(1+t)^{-\frac{3}{4}}. \end{aligned} \end{aligned}$$

Choosing $t \ge \frac{4C_2^2-C_1^2}{C_1^2}$, it holds on

$$\begin{aligned} \Vert \varrho (t) \Vert _{L^2} \ge C(1+t)^{-\frac{3}{4}}. \end{aligned}$$

Similarly, using estimates (2.6) and (2.26), we also have

$$\begin{aligned} \Vert \nabla ^k \varrho (t)\Vert _{L^2}\ge & {} C(1+t)^{-\frac{3+2k}{4}},\quad k=1,2,3; \nonumber \\ \Vert \nabla ^{k} m (t)\Vert _{L^2}\ge & {} C(1+t)^{-\frac{3+2k_1}{4}},\quad k=0,1,2,3. \end{aligned}$$

(2.38)

Finally, we establish the lower decay rate for the velocity. Using decay estimate (1.7) and Morse inequality, we get

$$\begin{aligned} \Vert \nabla ^k m\Vert _{L^2}\le & {} \Vert \nabla ^k u\Vert _{L^2}+\Vert \nabla ^k (\varrho u)\Vert _{L^2}\\\le & {} \Vert \nabla ^k u\Vert _{L^2}+C\Vert \varrho \Vert _{L^\infty }\Vert \nabla ^k u\Vert _{L^2}+C\Vert u\Vert _{L^\infty }\Vert \nabla ^k \varrho \Vert _{L^2}\\\le & {} \Vert \nabla ^k u\Vert _{L^2}+C(1+t)^{-\frac{9+2k}{4}}, \end{aligned}$$

which, together with (2.38), yields directly

$$\begin{aligned} \Vert \nabla ^k u\Vert _{L^2} \ge \Vert \nabla ^k m\Vert _{L^2}-C(1+t)^{-\frac{9+2k}{4}} \ge C(1+t)^{-\frac{3+2k}{4}}. \end{aligned}$$

Therefore, we complete the proof of lemma. $\square$

2.2 Upper and lower bounds of decay for time derivative

In this subsection, we will establish the upper and lower bounds for the time derivatives of density and velocity. In [25], Wang and Tan have rewriten (1.1) in the perturbation form as

$$\begin{aligned} \left\{ \begin{aligned}&\varrho _t+\mathrm{div} u=G_1,\\&u_t-\mu \Delta u-(\mu +\nu )\nabla \mathrm{div} u+\nabla \varrho -\kappa \nabla \Delta \varrho =G_2, \end{aligned} \right. \end{aligned}$$

(2.39)

where $\varrho =\rho -1$, and the function $G_i(i=1,2)$ is defined as

$$\begin{aligned} \left\{ \begin{aligned}&G_1=-\varrho \mathrm{div} u-u\cdot \nabla \varrho ,\\&G_2=-u \cdot \nabla u-\frac{\varrho }{\varrho +1}(\mu \Delta u+(\mu +\nu )\nabla \mathrm{div} u) -\left( \frac{P'(\varrho +1)}{\varrho +1}-1\right) \nabla \varrho . \end{aligned} \right. \end{aligned}$$

The initial data are given as

$$\begin{aligned} \left. (\varrho , u)(x,t)\right| _{t=0}=(\varrho _0, u_0)(x) \rightarrow (0,0) \quad {\text{ as } } \quad |x|\rightarrow \infty . \end{aligned}$$

(2.40)

Now, we establish the upper and lower bounds of decay rate for the time derivative of solution in the $L^2$ norm. The lower decay rate estimate for the time derivative of density and velocity can be obtained by using the method in [9]. However, we still give the estimate in detail due to the appearance of the Korteweg term.

Lemma 2.5

Under the assumptions in Theorem 1.2, the global solution $(\varrho , u)$ of equation (2.39) has the following estimate

$$\begin{aligned} C(1+t)^{-\frac{5}{4}}\le & {} \Vert \partial _t \varrho (t)\Vert _{L^2}\le C(1+t)^{-\frac{5}{4}}; \nonumber \\ C(1+t)^{-\frac{5}{4}}\le & {} \Vert \partial _t u(t)\Vert _{L^2}\le C(1+t)^{-\frac{5}{4}}, \end{aligned}$$

(2.41)

for all $t \ge t_*$ with $t_*$ a positive constant. Here, C is a positive constant independent of time.

Proof

At first, we establish upper bound time decay rate for $\partial _t \varrho$ and $\partial _t u$ in the $L^2$ norm. With the help of the equation (2.39), we can easily obtain

$$\begin{aligned} \Vert \partial _t \varrho \Vert _{L^2}\le C \Vert \mathrm{div}u \Vert _{L^2}+\Vert G_1\Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _t u \Vert _{L^2}\le C \Vert \Delta u \Vert _{L^2}+C\Vert \nabla \mathrm{div}u \Vert _{L^2}+C\Vert \nabla \varrho \Vert _{L^2}+C\Vert \nabla \Delta \varrho \Vert _{L^2}+\Vert G_2\Vert _{L^2}. \end{aligned}$$

By virtue of Sobolev’s inequality and time decay rate (1.7), we have

$$\begin{aligned} \Vert G_1\Vert _{L^2}\le & {} \Vert \varrho \mathrm{div}u\Vert _{L^2}+\Vert u\cdot \nabla \varrho \Vert _{L^2} \nonumber \\\le & {} C\Vert \varrho \Vert _{L^\infty }\Vert \mathrm{div}u\Vert _{L^2}+C\Vert u\Vert _{L^\infty }\Vert \nabla \varrho \Vert _{L^2} \nonumber \\\le & {} C(1+t)^{-\frac{5}{2}}, \end{aligned}$$

(2.42)

and

$$\begin{aligned} \Vert G_2\Vert _{L^2}\le & {} C \Vert u\cdot \nabla u\Vert _{L^2} + C \left\| \frac{\varrho }{1+\varrho }\Delta u \right\| _{L^2}+C \left\| \frac{\varrho }{1+\varrho }\nabla \mathrm{div} u \right\| _{L^2}+ C \left\| \left( \frac{P'(1+\varrho )}{1+\varrho }-1\right) \nabla \varrho \right\| _{L^2} \nonumber \\\le & {} \Vert u\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2}+\left\| \frac{1}{1+\varrho }\Vert _{L^\infty } \right\| \varrho \Vert _{L^\infty }\Vert \nabla ^2 u\Vert _{L^2}\!+\! \left\| \frac{1}{1+\varrho } \right\| _{L^\infty } \Vert \nabla \varrho \Vert _{L^6}\Vert \nabla \varrho \Vert _{L^3}\nonumber \\&+\left\| \frac{1}{1+\varrho }\Vert _{L^\infty } \right\| \varrho \Vert _{L^\infty }\Vert \nabla \varrho \Vert _{L^2}\nonumber \\\le & {} C(1+t)^{-\frac{5}{2}}. \end{aligned}$$

(2.43)

Then, we can easily derive that

$$\begin{aligned} \Vert \partial _t \varrho \Vert _{L^2}\le C (1+t)^{-\frac{5}{4}}, \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _t u \Vert _{L^2}\le C(1+t)^{-\frac{5}{4}}. \end{aligned}$$

Next, we establish lower bound time decay rate for $\partial _t u$ in the $L^2$ norm. Using the momentum equation in (2.39), we have

$$\begin{aligned} \Vert \nabla \varrho \Vert _{L^2}\le \Vert \partial _t u \Vert _{L^2} +C\Vert \nabla ^2 u\Vert _{L^2} +C\Vert \nabla \Delta \varrho \Vert _{L^2}+\Vert G_2\Vert _{L^2}. \end{aligned}$$

With the help of the inequality (2.43), we can get for all $t\ge t_{*}$,

$$\begin{aligned} \Vert \partial _t u\Vert _{L^2} \ge C(1+t)^{-\frac{5}{4}}-C(1+t)^{-\frac{7}{4}}-C(1+t)^{-\frac{9}{4}}-C(1+t)^{-\frac{5}{2}} \ge C(1+t)^{-\frac{5}{4}}. \end{aligned}$$

(2.44)

Finally, we establish lower bound time decay rate for $\partial _t \varrho$ in the $L^2$ norm. To achieve this target, we use the transport equation in (2.39) to obtain

$$\begin{aligned} \Vert \mathrm{div} u\Vert _{L^2}\le \Vert \partial _t \varrho \Vert _{L^2}+\Vert G_1\Vert _{L^2}; \end{aligned}$$

hence, together with the inequality (2.42), we obtain

$$\begin{aligned} \Vert \partial _t \varrho \Vert _{L^2} \ge \Vert \mathrm{div} u\Vert _{L^2}-C(1+t)^{-\frac{5}{2}}. \end{aligned}$$

(2.45)

Now, we need to establish the lower bound decay rate for $\Vert \mathrm{div} u\Vert _{L^2}$. Notice the differential relation $\Delta =\nabla \mathrm{div}-\nabla \times \nabla \times$, we get

$$\begin{aligned} \Vert \nabla u\Vert _{L^2}^2=\Vert \mathrm{div} u\Vert _{L^2}^2+\Vert \nabla \times u\Vert _{L^2}^2. \end{aligned}$$

And hence, one arrives at

$$\begin{aligned} \Vert \mathrm{div} u\Vert _{L^2} \ge C\Vert \nabla u\Vert _{L^2}-C\Vert \nabla \times u\Vert _{L^2} \ge C(1+t)^{-\frac{5}{4}}-C\Vert \nabla \times u\Vert _{L^2}, \end{aligned}$$

(2.46)

which implies that we need to establish upper bound decay rate for $\Vert \nabla \times u\Vert _{L^2}$. To this end, applying the operator $\nabla \times$ to both sides of the equation (2.39), we arrive at

$$\begin{aligned} \partial _t(\nabla \times u)-\mu \Delta (\nabla \times u)=\nabla \times G_2. \end{aligned}$$

Using Sobolev’s inequality, uniform bound (1.6) and decay rate (1.7), we have

$$\begin{aligned}&\Vert G_2\Vert _{L^1}+\Vert G_2\Vert _{L^2}\nonumber \\&\quad \le C \Vert u\cdot \nabla u\Vert _{L^1} + C \left\| \frac{\varrho }{1+\varrho }\Delta u \right\| _{L^1}+C\left\| \frac{\varrho }{1+\varrho }\nabla \mathrm{div} u \right\| _{L^1}+ C\left\| \left( \frac{P'(1+\varrho )}{1+\varrho }-1\right) \nabla \varrho \right\| _{L^1}\nonumber \\&\qquad +C \Vert u\cdot \nabla u\Vert _{L^2} + C\left\| \frac{\varrho }{1+\varrho }\Delta u \right\| _{L^2}+C \left\| \frac{\varrho }{1+\varrho }\nabla \mathrm{div} u \right\| _{L^2}+ C\left\| \left( \frac{P'(1+\varrho )}{1+\varrho }-1\right) \nabla \varrho \right\| _{L^2}\nonumber \\&\quad \le C \Vert u\Vert _{L^2}\Vert \nabla u\Vert _{L^2}+C\left\| \frac{\varrho }{1+\varrho } \right\| _{L^\infty }\Vert \varrho \Vert _{L^2}\Vert \nabla ^2 u\Vert _{L^2}+C \left\| \frac{\varrho }{1+\varrho }\right\| _{L^\infty }\Vert \varrho \Vert _{L^2}\Vert \nabla ^2 \varrho \Vert _{L^2}\nonumber \\&\qquad + C\Vert u\Vert _{L^3}\Vert \nabla u\Vert _{L^6}+C \left\| \frac{\varrho }{1+\varrho }\right\| _{L^\infty }\Vert \varrho \Vert _{L^3}\Vert \nabla ^2 u\Vert _{L^6}+C \left\| \frac{\varrho }{1+\varrho } \right\| _{L^\infty }\Vert \varrho \Vert _{L^3}\Vert \nabla ^2 \varrho \Vert _{L^6} \nonumber \\&\quad \le C \Vert (\varrho ,u)\Vert _{H^1}\Vert \nabla (\varrho ,u)\Vert _{H^2}\nonumber \\&\quad \le C\delta (1+t)^{-\frac{5}{4}}. \end{aligned}$$

(2.47)

By virtue of the Duhamel principle formula and (2.47), we get

$$\begin{aligned} \Vert \nabla \times u\Vert _{L^2}\le & {} C(1+t)^{-\frac{5}{4}}(\Vert \Lambda ^{-1}\mathcal {F}(\nabla \times u_0)\Vert _{L^\infty } +\Vert \Lambda ^{-1}\mathcal {F}(\nabla \times u_0)\Vert _{L^2})\\&+\,C\int \limits _0^t (1+t-\tau )^{-\frac{5}{4}} (\Vert \Lambda ^{-1}\mathcal {F}(\nabla \times G_2)\Vert _{L^\infty } +\Vert \Lambda ^{-1}\mathcal {F}(\nabla \times G_2)\Vert _{L^2})\mathrm{d}\tau \\\le & {} C(1+t)^{-\frac{5}{4}}(\Vert u_0\Vert _{L^1}+\Vert u_0\Vert _{L^2}) +C\int \limits _0^t (1+t-\tau )^{-\frac{5}{4}}(\Vert G_2\Vert _{L^1}+\Vert G_2\Vert _{L^2})\mathrm{d}\tau \\\le & {} C(\delta +\delta _1)(1+t)^{-\frac{5}{4}}+C\delta \int \limits _0^t(1+t-\tau )^{-\frac{5}{4}}(1+\tau )^{-\frac{5}{4}}\mathrm{d}\tau \\\le & {} C(\delta +\delta _1)(1+t)^{-\frac{5}{4}}, \end{aligned}$$

which, together with estimates (2.45) and (2.46), yields directly

$$\begin{aligned} \Vert \partial _t \varrho \Vert _{L^2} \ge C(1+t)^{-\frac{5}{4}}-C(\delta +\delta _1)(1+t)^{-\frac{5}{4}}-C(1+t)^{-\frac{5}{2}}, \end{aligned}$$

where the positive constant C on right-hand side of the above inequality depends on $c_{0}$ given in Proposition 2.1, but not on $\delta$ and $\delta _1$. Then, by virtue of the smallness of $\delta$ and $\delta _1$, we have for $t\ge t_{*}$,

$$\begin{aligned} \Vert \partial _t \varrho \Vert _{L^2} \ge C(1+t)^{-\frac{5}{4}}. \end{aligned}$$

Therefore, we complete the proof of this lemma. $\square$

3 Proof of some technical estimates

In this section, we will establish the claim estimates that have been used in Sect. 2. That is to say, we will establish the claim estimates (2.18), (2.21), (2.22), (2.30) and (2.36).

Proof of inequality (2.18)

Let $\zeta =\xi \sqrt{t}$, we obtain

$$\begin{aligned} I_1= & {} C\int \limits _{|\xi |\le \eta }e^{-(\mu +\nu )|\xi |^{2}t}|\xi |^{2s}\cos ^2(|\xi |t) \mathrm{d}\xi \\= & {} Ct^{-(\frac{3}{2}+s)}\int \limits _{|\zeta |\le \eta t^{\frac{1}{2}}}e^{-(\mu +\nu )|\zeta |^{2}}|\zeta |^{2s}\cos ^2(|\zeta |t^{\frac{1}{2}}) \mathrm{d}\zeta \\= & {} Ct^{-(\frac{3}{2}+s)}\int \limits _0^ {\eta t^{\frac{1}{2}}}e^{-(\mu +\nu )R^{2}}R^{2s+2}\cos ^2(Rt^{\frac{1}{2}}) \mathrm{d}R\\\ge & {} Ct^{-(\frac{3}{2}+s)}\sum _{k=0}^{[\frac{\eta t}{\pi }]-1}\int \limits _{\frac{k\pi +\frac{\pi }{6}}{\sqrt{t}}}^ {\frac{k\pi +\frac{\pi }{3}}{\sqrt{t}}}e^{-(\mu +\nu )R^{2}}R^{2s+2}\cos ^2(Rt^{\frac{1}{2}}) \mathrm{d}R\\\ge & {} Ct^{-(\frac{3}{2}+s)}\sum _{k=[\frac{\eta \sqrt{t}}{\pi }]+1}^{[\frac{2\eta \sqrt{t}}{\pi }]+1}\int \limits _{\frac{k\pi +\frac{\pi }{6}}{\sqrt{t}}}^ {\frac{k\pi +\frac{\pi }{3}}{\sqrt{t}}}e^{-(\mu +\nu )R^{2}}R^{2s+2} \mathrm{d}R\\\ge & {} Ct^{-(\frac{3}{2}+s)}, \end{aligned}$$

where there exits a positive large time $t_{*}$ such that for $t\ge t_{*}$, we can obtain

$$\begin{aligned} e^{-(\mu +\nu )R^{2}}\ge e^{-(\mu +\nu )\left( \frac{k\pi +\frac{\pi }{3}}{\sqrt{t}}\right) ^{2}}\ge e^{-(\mu +\nu )\left( \frac{([\frac{2\eta \sqrt{t}}{\pi }]+1)\pi +\frac{\pi }{3}}{\sqrt{t}}\right) ^{2}}\ge C, \end{aligned}$$

with C a positive constant that independent of time. $\square$

Proof of inequality (2.21)

Multiplying the first and second equations of (2.19) by $\varrho _\delta$ and $m_\delta$, respectively, it holds on

$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \limits (|\varrho _\delta |^2+|m_\delta |^2)\mathrm{d}x +\mu \int \limits |\nabla m_\delta |^2 \mathrm{d}x +(\mu +\nu )\int \limits |\mathrm{div} m_\delta |^2 \mathrm{d}x-\kappa \int \limits \nabla \Delta \varrho _\delta \cdot m_\delta \mathrm{d}x=\int \limits S \cdot \nabla m_\delta \mathrm{d}x. \end{aligned}$$

Due to the fact that $P'(1)=1$, we can use the Taylor expression formula to get

$$\begin{aligned} P(1+\varrho )-P(1)-\varrho \sim \varrho ^2, \end{aligned}$$

which, together with Sobolev’s inequality, yields directly

$$\begin{aligned} \Vert S\Vert _{L^2}\le C\Vert (\varrho , u)\Vert _{H^2}(\Vert \nabla \varrho \Vert _{H^1}+\Vert \nabla u\Vert _{L^2}), \end{aligned}$$

where the symbol $\sim$ represents the equivalent relation. By virtue of integrating by parts and using the transport equation, it is easy to get

$$\begin{aligned} \int \limits \nabla \Delta \varrho _\delta \cdot m_\delta \mathrm{d}x=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \limits |\nabla \varrho _\delta |^2\mathrm{d}x. \end{aligned}$$

Then, we get

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}(\Vert (\varrho _\delta , m_\delta )\Vert _{L^2}^2+\kappa \Vert \nabla \varrho _\delta \Vert _{L^2}^2) +\mu \Vert \nabla m_\delta \Vert _{L^2}^2 +\nu \Vert \mathrm{div} m_\delta \Vert _{L^2}^2 \mathrm{d}x \le C\Vert (\varrho , u)\Vert _{H^2}^2(\Vert \nabla \varrho \Vert _{H^1}^2+\Vert u\Vert _{L^2}^2). \end{aligned}$$

Applying the equation (2.19), it is easy to obtain for $k=1,2,3$,

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla ^k (\varrho _\delta , m_\delta )\Vert _{L^2}^2 +\mu \Vert \nabla ^{k+1} m_\delta \Vert _{L^2}^2+(\mu +\nu )\Vert \nabla ^k \mathrm{div} m_\delta \Vert _{L^2}^2-\kappa \int \limits \nabla ^{k+1}\Delta \varrho _\delta \cdot \nabla ^k m_\delta \mathrm{d}x\\&\quad \le \Vert \nabla ^k S\Vert _{L^2} \Vert \nabla ^{k+1} m_\delta \Vert _{L^2}. \end{aligned}$$

At first, we estimate the term $\int \nabla ^{k+1}\Delta \varrho _\delta \cdot \nabla ^k m_\delta \mathrm{d}x$ for $k=1,2,3$. Using integration by parts and transport equation, we can obtain

$$\begin{aligned} \int \limits \nabla ^{k+1}\Delta \varrho _\delta \cdot \nabla ^k m_\delta \mathrm{d}x=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\int \limits |\nabla ^{k+1}\varrho _\delta |^2 \mathrm{d}x. \end{aligned}$$

Now we give the estimates for $\Vert \nabla ^k S\Vert _{L^2}^2, k=1,2,3$. Indeed, when $k=1$, we apply Sobolev’s inequality to obtain

$$\begin{aligned} \Vert \nabla ((1+\varrho )u\otimes u)\Vert _{L^2}\le & {} C\Vert 1+\varrho \Vert _{L^\infty }\Vert u\Vert _{L^\infty }\Vert \nabla u\Vert _{L^2} +C\Vert u\Vert _{L^\infty }^2\Vert \nabla \varrho \Vert _{L^2}\\\le & {} C(1+\Vert \nabla u\Vert _{H^1})\Vert \nabla u\Vert _{H^1}\Vert \nabla (\varrho , u)\Vert _{L^2}\\\le & {} C\Vert \nabla u\Vert _{H^1}\Vert \nabla (\varrho , u)\Vert _{L^2}. \end{aligned}$$

Similarly, we also have for $k=1$,

$$\begin{aligned} \Vert \nabla ^2 (\varrho u)\Vert _{L^2}+\Vert \nabla \mathrm{div}(\varrho u)\Vert _{L^2} \le C\Vert \nabla (\varrho , u)\Vert _{H^1}\Vert \nabla ^2 (\varrho , u)\Vert _{L^2}. \end{aligned}$$

By the same way, we have

$$\begin{aligned} \Vert \nabla (\varrho \nabla ^2\varrho )\Vert _{L^2}\le C\Vert \nabla ^2 \varrho \Vert _{L^2}\Vert \nabla ^2 \varrho \Vert _{H^1}+ C\Vert \nabla \varrho \Vert _{H^1}\Vert \nabla ^3 \varrho \Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla |\nabla \varrho |^2\Vert _{L^2}\le C\Vert \nabla ^2 \varrho \Vert _{L^2}\Vert \nabla ^2 \varrho \Vert _{H^1}. \end{aligned}$$

By virtue of the Taylor expression formula, we get

$$\begin{aligned} \nabla (P(1+\varrho )-P(1)-\varrho )=P'(1+\varrho )\nabla \varrho -P'(1)\nabla \varrho \sim \varrho \nabla \varrho , \end{aligned}$$

where we have used the fact that $P'(1)=1$. Then, we use Sobolev’s inequality to obtain

$$\begin{aligned} \Vert \nabla (P(1+\varrho )-P(1)-\varrho )\Vert _{L^2} \le C\Vert \nabla \varrho \Vert _{H^1}\Vert \nabla \varrho \Vert _{L^2}. \end{aligned}$$

Thus, it holds on for $k=1$,

$$\begin{aligned} \Vert \nabla S\Vert _{L^2}\le C\Vert \nabla (\varrho , u)\Vert _{H^1}(\Vert \nabla \varrho \Vert _{H^2}+\Vert \nabla u\Vert _{H^1}). \end{aligned}$$

Then, we use Cauchy inequality to get

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}(\Vert \nabla (m_\delta , \varrho _\delta )\Vert _{L^2}^2+\kappa \Vert \nabla ^2 \varrho _\delta \Vert _{L^2}^2)+\mu \Vert \nabla ^{2}m_\delta \Vert _{L^2}^2 \le C\Vert \nabla (\varrho , u)\Vert _{H^1}^2 (\Vert \nabla ^2 \varrho \Vert _{H^1}^2+\Vert \nabla u\Vert _{H^1}^2). \end{aligned}$$

Notice that

$$\begin{aligned} \nabla ^2 (P(1+\varrho )-P(1)-\varrho )= & {} P''(1+\varrho )\nabla \varrho \nabla \varrho +P'(1+\varrho )\nabla ^2 \varrho -P'(1)\nabla ^2 \varrho \\\sim & {} \nabla \varrho \nabla \varrho +\varrho \nabla ^2 \varrho , \end{aligned}$$

and

$$\begin{aligned}&\nabla ^3 (P(1+\varrho )-P(1)-\varrho )\\&\quad = 2P'''(1+\varrho )\nabla \varrho \nabla \varrho \nabla \varrho +2P''(1+\varrho )\nabla ^2 \varrho \nabla \varrho +P'(1+\varrho )\nabla ^3 \varrho -P'(1)\nabla ^3 \varrho \\&\quad \sim \nabla \varrho \nabla \varrho \nabla \varrho +\nabla \varrho \nabla ^2 \varrho +\varrho \nabla ^3 \varrho . \end{aligned}$$

Next, employing Sobolev’s inequality, it is easy to get

$$\begin{aligned} \Vert \nabla ^2 S\Vert _{L^2}\le C\Vert \nabla ^2 (\varrho , u)\Vert _{H^1}(\Vert \nabla \varrho \Vert _{H^2}+\Vert \nabla u\Vert _{H^1})+C \Vert \nabla \varrho \Vert _{H^1}\Vert \nabla ^4 \varrho \Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla ^3 S\Vert _{L^2}\le C\Vert \nabla ^3 (\varrho ,u)\Vert _{L^2}(\Vert \nabla \varrho \Vert _{H^2}+\Vert \nabla u\Vert _{H^1})+ C\Vert \nabla \varrho \Vert _{H^1}(\Vert \nabla ^5 \varrho \Vert _{L^2}+\Vert \nabla ^4 u\Vert _{L^2}). \end{aligned}$$

Then, we obtain the following estimates

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}(\Vert \nabla ^2(m_\delta , \varrho _\delta )\Vert _{L^2}^2+\kappa \Vert \nabla ^3 \varrho _\delta \Vert _{L^2}^2)+\mu \Vert \nabla ^{3}m_\delta \Vert _{L^2}^2\\&\quad \le C\Vert \nabla ^2 (\varrho , u)\Vert _{H^1}^2 (\Vert \nabla \varrho \Vert _{H^2}^2+\Vert \nabla u\Vert _{H^1}^2)+ C\Vert \nabla \varrho \Vert _{H^1}^2\Vert \nabla ^4 \varrho \Vert _{L^2}^2, \end{aligned}$$

and

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}(\Vert \nabla ^3 (m_\delta , \varrho _\delta )\Vert _{L^2}^2+\kappa \Vert \nabla ^4 \varrho _\delta \Vert _{L^2}^2)+\mu \Vert \nabla ^{4}m_\delta \Vert _{L^2}^2\\&\quad \le C\Vert \nabla ^3 (\varrho ,u)\Vert _{L^2}^2(\Vert \nabla \varrho \Vert _{H^2}^2+\Vert \nabla u\Vert _{H^1}^2)+ C\Vert \nabla \varrho \Vert _{H^1}^2(\Vert \nabla ^5 \varrho \Vert _{L^2}^2+\Vert \nabla ^4 u\Vert _{L^2}^2). \end{aligned}$$

Therefore, we complete the proof of claim estimate (2.21). $\square$

Proof of inequality (2.22)

Taking $k(k=0,1,2,3)$th spatial derivatives to the second equation of (2.19) and multiplying the equation by $\nabla ^{k+1} \varrho _\delta$, then we have

$$\begin{aligned}&\int \limits \partial _t \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x+\int \limits |\nabla ^{k+1} \varrho _\delta |^2 \mathrm{d}x-\kappa \int \limits \nabla ^{k+1} \Delta \varrho _\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x\\&\quad =\int \limits (\mu \nabla ^k \Delta m_\delta +(\mu +\nu )\nabla ^{k+1} \mathrm{div} m_\delta )\cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x -\int \limits \nabla ^k \mathrm{div} S\cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x. \end{aligned}$$

Using the first equation of (2.19), it holds on

$$\begin{aligned} \int \limits \partial _t \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x= & {} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x -\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \partial _t \varrho _\delta \mathrm{d}x\\= & {} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x +\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \mathrm{div} m_\delta \mathrm{d}x\\= & {} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x -\int \limits |\nabla ^k \mathrm{div} m_\delta |^2 \mathrm{d}x. \end{aligned}$$

After integration by part, it holds on

$$\begin{aligned} \int \limits \nabla ^{k+1} \Delta \varrho _\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x=-\int \limits |\nabla ^{k+2} \varrho _\delta |^2 \mathrm{d}x. \end{aligned}$$

Thus, we combine the above three equalities to obtain

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x+\int \limits |\nabla ^{k+1} \varrho _\delta |^2 \mathrm{d}x+\kappa \int \limits |\nabla ^{k+2} \varrho _\delta |^2 \mathrm{d}x\\&\quad = \int \limits |\nabla ^k \mathrm{div}m_\delta |^2 \mathrm{d}x-\int \limits \nabla ^k \mathrm{div} S\cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x+\int \limits (\mu \nabla ^k \Delta m_\delta +(\mu +\nu )\nabla ^{k+1} \mathrm{div} m_\delta )\cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x, \end{aligned}$$

which, together with integration by parts and Cauchy inequality, yields directly

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int \limits \nabla ^k m_\delta \cdot \nabla ^{k+1} \varrho _\delta \mathrm{d}x +\int \limits |\nabla ^{k+1} \varrho _\delta |^2 \mathrm{d}x+\frac{1}{2}\kappa \int \limits |\nabla ^{k+2} \varrho _\delta |^2 \mathrm{d}x \le C(\Vert \nabla ^{k+1} m_\delta \Vert _{L^2}^2+\Vert \nabla ^{k}S\Vert _{L^2}^2), \end{aligned}$$

which implies (2.22). Therefore, we complete proof of claim estimate (2.22). $\square$

Proof of inequality (2.30)

It is easy to see that

$$\begin{aligned} \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{5}{2}}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\mathrm{d}\tau \le \int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\mathrm{d}\tau . \end{aligned}$$

In [10], the following estimate holds on

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} \mathcal {F}_l^{3}(t)+C(\Vert \nabla ^{l+1} \varrho \Vert _{H^{4-l}}^2+\Vert \nabla ^{l+1} u\Vert _{H^{3-l}}^2)\le 0, \quad l=0,1,2,3, \end{aligned}$$

(3.1)

where

$$\begin{aligned} \mathcal {F}_l^{3}(t):=\Vert \nabla ^{l} \varrho \Vert _{H^{4-l}}^2+\Vert \nabla ^{l} u\Vert _{H^{3-l}}^2+\delta \sum _{k=l}^3 \int \limits \nabla ^k u\cdot \nabla ^{k+1}\varrho \mathrm{d}x, \end{aligned}$$

and $\delta$ is a small positive number. By multiplying the inequality (3.1) with $l=3$ by $e^{\frac{C}{C^{*}}t}$, it holds on

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}(e^{\frac{C}{C^{*}}t}\mathcal {F}_3^{3}(t))+Ce^{\frac{C}{C^{*}}t}(\Vert \nabla ^4 \varrho \Vert _{H^1}^2+\Vert \nabla ^4 u\Vert _{L^2}^2)\\&\quad \le Ce^{\frac{C}{C^{*}}t}(\Vert \nabla ^3 \varrho \Vert _{H^{1}}^2+\Vert \nabla ^3 u\Vert _{L^2}^2); \end{aligned}$$

then, by integrating about time over [0, t], similar to (2.29), one arrives at

$$\begin{aligned}&\mathcal {F}_3^{3}(t)+\int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(\Vert \nabla ^4 \varrho (\tau )\Vert _{H^1}^2+\Vert \nabla ^4 u(\tau )\Vert _{L^2}^2) \mathrm{d}\tau \\&\quad \le Ce^{-\frac{C}{C^{*}}t}\mathcal {F}_3^{3}(0)+C\int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(\Vert \nabla ^3 \varrho (\tau )\Vert _{H^1}^2+\Vert \nabla ^3 u(\tau )\Vert _{L^2}^2) \mathrm{d}\tau \\&\quad \le Ce^{-\frac{C}{C^{*}}t}(\Vert \nabla ^3 \varrho _0\Vert _{H^1}^2+\Vert \nabla ^3 u_0\Vert _{L^2}^2)+C\int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(\Vert \nabla ^3 \varrho (\tau )\Vert _{H^1}^2+\Vert \nabla ^3 u(\tau )\Vert _{L^2}^2) \mathrm{d}\tau \\&\quad \le C e^{-\frac{C}{C^{*}}t}+C\int \limits _0^{t}e^{-\frac{C}{C^{*}}(t-\tau )}(1+\tau )^{-\frac{9}{2}}\mathrm{d}\tau \\&\quad \le C e^{-\frac{C}{C^{*}}t}+(1+t)^{-\frac{9}{2}}\\&\quad \le C (1+t)^{-\frac{9}{2}}, \end{aligned}$$

where we utilize decay estimate (1.7). Therefore, we complete proof of claim estimate (2.30). $\square$

Proof of inequality (2.36)

Replacing l by $l+1$ in (3.1), and then multiplying both sides by $(1+t)^{l+\frac{3}{2}}$, one arrives at

$$\begin{aligned}&\frac{\mathrm{d}}{\mathrm{d}t}[(1+t)^{l+\frac{3}{2}}\mathcal {F}_{l+1}^{3}(t)]+C(1+t)^{l+\frac{3}{2}}(\Vert \nabla ^{l+2} \varrho \Vert _{H^{3-l}}^2+\Vert \nabla ^{l+2} u\Vert _{H^{2-l}}^2)\\&\quad \le C (1+t)^{l+\frac{1}{2}}\mathcal {F}_{l+1}^{3}(t). \end{aligned}$$

The integration of the above inequality with respect to time over [0, t] implies that

$$\begin{aligned}&\int \limits _0^t (1+\tau )^{l+\frac{3}{2}}(\Vert \nabla ^{l+2} \varrho (\tau )\Vert _{H^{3-l}}^2+\Vert \nabla ^{l+2} u(\tau )\Vert _{H^{2-l}}^2) \mathrm{d}\tau \\&\quad \le C \mathcal {F}_{l+1}^{3}(0)+C\int \limits _0^t (1+\tau )^{l+\frac{1}{2}}\mathcal {E}_{l+1}^{3}(\tau ) \mathrm{d}\tau \\&\quad \le C+C\int \limits _0^t (1+\tau )^{l+\frac{1}{2}} (1+\tau )^{-\frac{5+2l}{2}}\mathrm{d}\tau \\&\quad \le C, \end{aligned}$$

where we used the estimate $\mathcal {F}_{l}^{3}(t)\le C (1+t)^{-\frac{3+2l}{2}}$ for $l=0,1,2,3$. Therefore, we complete proof of claim estimate (2.36). $\square$