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Review Topical Sections

Molecular mechanisms involved in taste learning and memory

  • Received: 22 July 2017 Accepted: 24 September 2017 Published: 26 September 2017
  • Taste learning, and particularly conditioned taste aversion (CTA), is an adaptive learning involving complex brain mechanisms and molecular pathways. Taste learning and CTA are critical behaviors for survival, and the knowledge of the molecular bases involved in the acquisition, retention and extinction of CTA can help to understand the brain mechanisms of normal and altered taste learning. The aim of this review is to describe recent findings on the molecular mechanisms of taste learning, from the genetic, receptors, and intracellular and extracellular signaling biological levels. We can conclude that some molecular pathways and processes for the acquisition of taste learning and the formation of taste memories are well identified. However, new molecular, neurobiological and behavioral studies are needed to thoroughly elucidate the complexity of the taste system and the neural mechanisms of CTA.

    Citation: Andrés Molero-Chamizo, Guadalupe Nathzidy Rivera-Urbina. Molecular mechanisms involved in taste learning and memory[J]. AIMS Molecular Science, 2017, 4(4): 389-398. doi: 10.3934/molsci.2017.4.389

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  • Taste learning, and particularly conditioned taste aversion (CTA), is an adaptive learning involving complex brain mechanisms and molecular pathways. Taste learning and CTA are critical behaviors for survival, and the knowledge of the molecular bases involved in the acquisition, retention and extinction of CTA can help to understand the brain mechanisms of normal and altered taste learning. The aim of this review is to describe recent findings on the molecular mechanisms of taste learning, from the genetic, receptors, and intracellular and extracellular signaling biological levels. We can conclude that some molecular pathways and processes for the acquisition of taste learning and the formation of taste memories are well identified. However, new molecular, neurobiological and behavioral studies are needed to thoroughly elucidate the complexity of the taste system and the neural mechanisms of CTA.


    In this paper, we are concerned with the sharp decay rates of solutions to the Cauchy problem for the isentropic Navier-Stokes equations:

    {tρ+div(ρu)=0,(t,x)R+×R3,t(ρu)+div(ρuu)+p(ρ)=divT,(t,x)R+×R3,lim|x|ρ=ˉρ,lim|x|u=0,tR+,(ρ,u)|t=0=(ρ0,u0),xR3, (1.1)

    which governs the motion of a isentropic compressible viscous fluid. The unknown functions ρ and u represent the density and velocity of the fluid respectively. The pressure p=p(ρ) is a smooth function in a neighborhood of a positive constant ˉρ s.t. p(ˉρ)>0. T is the viscosity stress tensor given by T=μ(u+(u)t)+ν(divu)I with I the identity matrix. We assume that the constant viscosity coefficients μ>0 and ν satisfy ν+23μ>0. Throughout this article, by optimal time decay rate, we refer to the best possible decay rate in upper bound as many literatures, and the sharp time decay rate includes the best possible upper and lower bounds.

    Using the classical spectral method, the optimal time decay rate (upper bound) of the linearized equations of the isentropic Navier-Stokes equations are well known. One may then expect that the small solution of the nonlinear equations (1.1) have the same decay rate as the linear one. Our work is devoted to proving the sharp time decay rate (for both upper and lower bound) for the nonlinear system.

    In the case of one space dimension, Zeng [24] and Liu-Zeng [15] offered a detailed analysis of the solution to a class of hyperbolic-parabolic system through point-wise estimate, including the isentropic Navier-Stokes system. For multi-dimensional Navier-Stokes equations (and/or Navier-Stokes-Fourier system), the Hs global existence and time-decay rate of strong solutions with the initial perturbation small in HsL1 are obtained in whole space first by A. Matsumura and T. Nishida [17], [18]. When the small initial perturbation belongs to H3 only, using a weighted energy method, A. Matsumura [16] showed the time-decay rate (1+t)34 of upper bound in L-norm. Since then, there are concrete development on the upper bound time-decay estimates: the optimal Lp (with 2p) upper bound decay rate was proved by G. Ponce [19], combining the spectral analysis on linearized system and the energy method for small initial perturbation in L1. For the isentropic Navier-Stokes equations with artificial viscosity, D. Hoff and K. Zumbrun [6], [7] studied the Green's function and derived the Lp (1p) upper bound time decay rate of diffusive waves for the small initial perturbation belongs to HmL1 with m4. Liu and Wang [14] studied the point-wise estimates of the Green function of the linearized isentropic Navier-Stokes system in 3D and then analyzed the coupling of nonlinear diffusion waves, obtained the optimal (upper bound) decay rate. These results were further extended to the exterior problem [12], [11], or the half space problem [9], [10], [8]. Recently, Guo and Wang in [5] developed a new general energy method for proving the optimal (upper bound) time decay rates of the solutions to the dissipative equations in the whole space, using a family of scaled energy estimates with minimum derivative counts and interpolations among them without linear decay analysis.

    When additional external force is taken into account, the external force does affect the long time behavior of solutions. The upper bound of time decay rates were studied intensively, see for instance [1] and [2] on unbounded domain, [22], [23] on the convergence of the non-stationary flow to the corresponding steady flow when the initial date are small in H3L65, and [4], [3], on the optimal LpLq upper bound decay rates for potential forces.

    The main goal of current paper is to establish the sharp decay rate, on both upper and lower bounds, to the solutions of (1.1) using relatively simple energy method. We remark that similar results had been pursued by M. Schonbek [20], [21] for incompressible Navier-Stokes equations, and by Li, Matsumura-Zhang [13] for isentropic Navier-Stokes-Poisson system. Although they share the same spirit in obtaining the lower bound decay rates, the feature of the spectrum near zero exhibits quite different behaviors, leading to different analysis. For instance, we explored the elegant structure of the higher order nonlinear terms of Navier-Stokes, when choosing conservative variables: density and momentum. The conservative form of the sharp equations provided a natural derivative structure in these terms, leading to the possibility of a faster decay rate estimate. We will make a more detailed comparison later in this paper.

    Define n=ρˉρ, and let m=ρu=(n+ˉρ)u be the momentum. We rewrite (1.1) as

    {tn+divm=0,(t,x)R+×R3,tm+c2nˉμm(ˉμ+ˉν)divm=F,(t,x)R+×R3,lim|x|n=0,lim|x|m=0,tR+,(n,m)|t=0=(ρ0ˉρ,ρ0u0),xR3, (1.2)

    where ˉμ=μˉρ, ˉν=νˉρ, c=p(ˉρ)>0 is the sound speed, and

    F=div{mmn+ˉρ+ˉμ(nmn+ˉρ)}{(ˉμ+ˉν)div(nmn+ˉρ)+(p(n+ˉρ)p(ˉρ)c2n)}.

    It is this structure of F that plays an important role in our analysis.

    Our aim is to obtain a clear picture of the large time behavior of U=(n,m) in L2(R3) when U0=(ρ0ˉρ,ρ0u0) is sufficiently smooth and small. We introduce the following initial value problem of the linearized Navier-Stokes system corresponding to (1.2):

    {t˜n+div˜m=0,(t,x)R+×R3,t˜m+c2˜nˉμ˜m(ˉμ+ˉν)div˜m=0,(t,x)R+×R3,lim|x|˜n=0,lim|x|˜m=0,tR+,(˜n,˜m)|t=0=(ρ0ˉρ,ρ0u0),xR3, (1.3)

    where ˉμ=μˉρ, ˉν=νˉρ, c=p(ˉρ). It is known that the L2-norm of ˜U=(˜n,˜m) decays at the optimal upper bound rate (1+t)34 for generic small initial data, see for instance [18]. A detailed proof on the optimal lower and upper bound rate will be given in the section 3 of this paper. In section 4, we prove that (U˜U)(,t)L2 decays at a faster rate than ˜U(,t)L2, under some reasonable conditions on the initial data. Therefore, U(,t)L2 shares the sharp decay rate of (1+t)34.

    Notation. For a, we mean that there is a uniform constant C , which may be different on different lines, such that a \leq Cb . And a \approx b stands for a \lesssim b and b\lesssim a .

    We now state our main result.

    Theorem 1.1. Assume that (n_0, m_0)\in L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)\cap H^3({\mathop{\mathbb R\kern 0pt}\nolimits}^3) , \delta_0 = : \|(n_0,m_0)\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)\cap H^3({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} is sufficiently small, and

    \begin{equation} \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}(n_0,m_0)dx\neq 0, \end{equation} (1.4)

    then there is a unique global classical solution \widetilde U = (\widetilde{n}, \widetilde{m})\in \mathcal C([0,\infty); H^3({\mathop{\mathbb R\kern 0pt}\nolimits}^3)) of the linearized system (1.3) satisfying for some positive constant C

    \begin{eqnarray*} \begin{split} C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq\|{\nabla}^k \widetilde n(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \quad k = 0,1,2,3,\\ C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq\|{\nabla}^{k} \widetilde m(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \quad k = 0,1,2,3, \end{split} \end{eqnarray*}

    and the initial value problem (1.2) has a unique solution U = (n, m)\in \mathcal C([0,\infty); H^3({\mathop{\mathbb R\kern 0pt}\nolimits}^3)) . Moreover, let n_h = n-\widetilde{n} and m_h = m-\widetilde{m} , then it holds that

    \begin{eqnarray*} \begin{split} &\|{\nabla}^{k} ( n_h,m_h)(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0^2(1+t)^{-\frac54-\frac{k}2},\quad k = 0,1,2,\\ &\|{\nabla}^{3} m_h(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0^2(1+t)^{-\frac{11}4},\quad \|{\nabla}^{3} n_h(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0(1+t)^{-\frac74}. \end{split} \end{eqnarray*}

    As a consequence, there exists a positive constant C_1 such that

    \begin{eqnarray*} \begin{split} C_1^{-1}(1+t)^{-\frac34-\frac{k}2}\leq&\| {\nabla}^{k} n (t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C_1(1+t)^{-\frac34-\frac{k}2},\quad k = 0,1, 2,\\ C_1^{-1}(1+t)^{-\frac34-\frac{k}2}\leq&\| {\nabla}^{k} m (t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C_1(1+t)^{-\frac34-\frac{k}2},\quad k = 0,1, 2,3.\\ \end{split} \end{eqnarray*}

    Remark 1.1. We remark that this theorem is valid under the condition (1.4) which is important in the lower bound estimate to the linearized problem. When (1.4) fails, the decay rate of the linearized system (1.3) depends on the order of the degeneracy of moments. Assume (n_0, m_0)\in L^1\cap H^3 and belong to certain appropriate weighted L^p spaces, similar situation happened also in the incompressible Navier-Stokes equations, c.f. [20], [21]. We also note that our condition (1.4) is weaker than those in most of previous results where the differentiability of Fourier transform of initial disturbance is required in general.

    Remark 1.2. In [13], Li, Matsumura-Zhang proved the lower bound decay rate of the linearized isentropic Navier-Stokes-Poisson system, they only require |\widehat n_0(\xi)|>c_0>0 for |\xi|\ll 1 with c_0 a constant due to the special structure of the spectrum from the help of the Poisson term. This condition is proposed in Fourier space, similar to (1.4) in some sense. In our case, the spectrum is different and the different structure leads to different sharp decay rates.

    In what follows, we will set n = \rho-\bar \rho , u = u-0 . We rewrite (1.1) in the perturbation form as

    \begin{equation} \left\{\begin{array}{l} \partial_t n + \bar \rho{\mathop{{\rm{div}}}\nolimits} u = -n{\mathop{{\rm{div}}}\nolimits} u - u\cdot {\nabla} n,\\ \partial_t u + \gamma\bar \rho{\nabla} n -\bar \mu△ u-(\bar \mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} u\\ \qquad = -u\cdot {\nabla} u- \bar \mu f(n)△ u-(\bar \mu+\bar\nu)f(n){\nabla}{\mathop{{\rm{div}}}\nolimits} u-g(n){\nabla} n, \\ \lim\limits_{|x|\to\infty}n = 0, \quad\lim\limits_{|x|\to\infty}u = 0,\\ (n,u)\big|_{t = 0} = (\rho_0-\bar \rho,u_0), \end{array}\right. \end{equation} (2.1)

    where \bar \mu = \frac{\mu}{ \bar \rho } , \bar \nu = \frac{\nu}{ \bar \rho } , \gamma = \frac{p'(\bar \rho)}{\bar \rho^2} , and the nonlinear functions f and g are defined by

    \begin{equation} f(n): = \frac n{n+\bar \rho}, \quad \quad g(n): = \frac{p'(n+\bar \rho)}{n+\bar \rho}-\frac{p'(\bar \rho)}{\bar \rho}. \end{equation} (2.2)

    We assume that there exist a time of existence T>0 and sufficiently small \delta>0 , such that a priori estimate

    \begin{equation} \|n(t)\|_{H^3}+\|u(t)\|_{H^3}\leq \delta, \end{equation} (2.3)

    holds for any t\in[0,T] . First of all, by (2.3) and Sobolev's inequality, we obtain that

    \begin{eqnarray*} \frac{\bar \rho}2\leq n+\bar \rho\leq 2\bar \rho. \end{eqnarray*}

    Hence, we immediately have

    \begin{equation} \left|f(n)\right|,\left|g(n)\right|\leq C|n|, \quad \big|{\nabla}^kf(n)\big|,\big|{\nabla}^kg(n)\big|\leq C \quad \forall k\in{\mathop{\mathbb N\kern 0pt}\nolimits}^+, \end{equation} (2.4)

    where f(n) and g(n) are nonlinear functions of n defined by (2.2).

    Next, we begin with the energy estimates including n and u themselves. The following results is essentially due to A. Matsumura and T. Nishida [17], [18].

    Theorem 2.1. Assume that (n_0, u_0)\in H^3({\mathop{\mathbb R\kern 0pt}\nolimits}^3) , then there exists a constant \delta_0>0 such that if

    \begin{eqnarray*} \|n_0\|_{H^3}+\|u_0\|_{H^3} \leq \delta_0, \end{eqnarray*}

    then the problem (2.1) admits a unique global solution (n(t), u(t)) satisfying that for all t\geq0 ,

    \begin{eqnarray*} \|n(t)\|_{H^3}^2+\|u(t)\|_{H^3}^2+\int_0^t\left (\|{\nabla} n(\tau)\|_{H^2}^2+\|{\nabla} u(\tau)\|_{H^3}^2 \right)d\tau \leq C\left(\|n_0\|^2_{H^3}+\|u_0\|^2_{H^3}\right), \end{eqnarray*}

    where C is a positive constant independent of time.

    The proof of this theorem is divided into several subsections.

    For k = 0 , multiplying the first equation in (2.1) by \gamma n and the second equation in (2.1) by u , summing up and then integrating the result over {\mathop{\mathbb R\kern 0pt}\nolimits}^3 by parts. By virtue of Hölder's inequality, Sobolev's inequality and the fact (2.4), we obtain that

    \begin{equation} \begin{split} &\frac12 \frac{d}{d t}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left (\gamma|n|^2+|u|^2\right)dx+ \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left(\bar\mu|{\nabla} u|^2+(\bar\mu+\bar\nu)|{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ = &\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\gamma\left(-n{\mathop{{\rm{div}}}\nolimits} u - u\cdot {\nabla} n\right)n-\big(u\cdot {\nabla} u+\bar\mu f(n)△ u\\ &\quad+(\bar \mu+\bar\nu)f(n){\nabla}{\mathop{{\rm{div}}}\nolimits} u+g(n){\nabla} n\big)\cdot u dx\\ \lesssim&\|n\|_{L^3}\|{\nabla} u\|_{L^2}\| n\|_{L^6}+\left(\|u\|_{L^3}\|{\nabla} u\|_{L^2}+\|n\|_{L^3}\|{\nabla} n\|_{L^2}\right)\| u\|_{L^6}\\ &\quad +\left(\|u\|_{L^\infty}\|{\nabla} n\|_{L^2}+\|n\|_{L^\infty}\|{\nabla} u\|_{L^2}\right)\|{\nabla} u\|_{L^2}\\ \lesssim&\left(\|n\|_{L^3}+\|u\|_{L^3}+\|n\|_{L^\infty}+\|u\|_{L^\infty}\right)\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|^2_{L^2}\right). \end{split} \end{equation} (2.5)

    Now for 1\leq k \leq 3 , applying {\nabla}^k to (2.1) and then multiplying the first equation by \gamma{\nabla}^k n and the second equation by {\nabla}^k u , summing up and integrating over {\mathop{\mathbb R\kern 0pt}\nolimits}^3 . For k = 1 we have

    \begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla} n|^2+|{\nabla} u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left( \bar\mu|{\nabla}^2u|^2 +(\bar\mu+\bar\nu)|{\nabla} {\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim&\left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right). \end{split} \end{equation} (2.6)

    For k = 2 we have

    \begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla}^2 n|^2+|{\nabla}^2 u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\bar\mu|{\nabla}^3 u|^2 +(\bar\mu+\bar\nu)|{\nabla}^2{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim& \left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right).\\ \end{split} \end{equation} (2.7)

    For k = 3 we have

    \begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla}^3 n|^2+|{\nabla}^3 u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left(\bar\mu |{\nabla}^4 u|^2 +(\bar\mu+\bar\nu)|{\nabla}^3{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim& \left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla}^3 n\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^4 u\|_{L^2}^2\right)\\ &\quad+\|{\nabla} n\|_{L^3}\|{\nabla}^4 u\|_{L^2}\|{\nabla}^2 n\|_{L^6} +\|{\nabla} u\|_{L^3}\|{\nabla}^4 u\|_{L^2}\|{\nabla}^2 u\|_{L^6}\\ &\quad +\|{\nabla}^2 n\|_{L^3}\left(\|{\nabla}^3 n\|_{L^2}+\|{\nabla}^4 u\|_{L^2}\right)\|{\nabla}^2 u\|_{L^6}. \end{split} \end{equation} (2.8)

    Summing up the above estimates, noting that \delta>0 is small, we obtain that

    \begin{equation} \begin{split} \frac{d}{dt}\sum\limits_{0\leq k\leq3}\left(\gamma\|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right) + C_1\sum\limits_{1\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2} \leq C_2\delta\sum\limits_{1\leq k\leq3}\|{\nabla}^{k} n\|^2_{L^2}. \end{split} \end{equation} (2.9)

    For 0\leq k \leq 2 , applying {\nabla} ^k to the second equation in (2.1) and then multiplying by {\nabla}^{k+1} n . The key idea is to integrate by parts in the t -variable and to use the continuity equation. Thus integrating the results by parts for both the t - and x -variables, we obtain for k = 0 that

    \begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} u\cdot {\nabla} n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla} n|^2 dx\\ \lesssim &\|{\nabla} u\|_{L^2}^2+\|{\nabla} n\|_{L^2}\|{\nabla}^2 u\|_{L^2}+ (\|n\|_{L^\infty}+ \|u\|_{L^\infty})\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|_{L^2}^2\right), \end{split} \end{equation} (2.10)

    for k = 1 , we get

    \begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla} u\cdot {\nabla}^2 n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla}^2 n|^2 dx\\ \lesssim&\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^2 n\|_{L^2}\|{\nabla}^3 u\|_{L^2}+ (\|(n, u)\|_{L^\infty}+\|({\nabla} n, {\nabla} u)\|_{L^\infty})\\ &\quad\times\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right), \end{split} \end{equation} (2.11)

    and for k = 2 we have

    \begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^3 n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla}^3 n|^2 dx\\ \lesssim &\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}\|{\nabla}^4 u\|_{L^2}+ (\|(n, u)\|_{L^\infty}+\|({\nabla} n, {\nabla} u)\|_{L^\infty})\\ &\quad\times\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right). \end{split} \end{equation} (2.12)

    Plugging the above estimates, using the smallness of \delta>0 , we obtain that

    \begin{equation} \begin{split} \frac{d}{dt}\sum\limits_{0\leq k\leq 2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^k u\cdot {\nabla}^{k+1} n dx +C_3\sum\limits_{1\leq k\leq 3}\|{\nabla}^{k} n\|_{L^2}^2 \leq C_4\sum\limits_{1\leq k\leq 4}\|{\nabla}^{k} u\|_{L^2}^2. \end{split} \end{equation} (2.13)

    Proof of Theorem 2.1. Multiplying (2.13) by \frac{2C_2\delta}{C_3} , adding it with (2.9), with the help of smallness of \delta>0 , we deduce that there exists a constant C_5>0 such that

    \begin{equation} \begin{split} \frac{d}{dt}&\bigg\{\sum\limits_{0\leq k\leq3}\left(\gamma\|{\nabla}^k n\|^2_{L^2}+ \|{\nabla}^k u\|^2_{L^2} \right)+\frac{2C_2\delta}{C_3}\sum\limits_{0\leq k\leq 2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^k u\cdot {\nabla}^{k+1} n dx \bigg\}\\ &\qquad+ C_5\bigg\{\sum\limits_{1\leq k\leq 3}\|{\nabla}^{k} n\|_{L^2}^2+\sum\limits_{1\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2}\bigg\} \leq 0. \end{split} \end{equation} (2.14)

    Next, we define \mathcal E(t) to be C_5^{-1} times the expression under the time derivative in (2.14). Then we may write (2.14) as

    \begin{equation} \frac{d}{dt}\mathcal E(t)+\|{\nabla} n(t)\|_{H^2}^2+\|{\nabla} u(t)\|^2_{H^3} \leq 0. \end{equation} (2.15)

    Observe that since \delta is small, then there exists a constant C_6>0 such that

    \begin{eqnarray*} C_6^{-1}\left(\| n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right) \leq\mathcal E(t)\leq C_6\left(\| n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right). \end{eqnarray*}

    Then integrating (2.15) directly in time, we get

    \begin{eqnarray*} \begin{split} &\sup\limits_{0\leq t \leq T}\left (\|n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right)+C_6\int_0^T \left(\|{\nabla} n(\tau)\|_{H^2}^2+\|{\nabla} u(\tau)\|^2_{H^3}\right)d\tau\\ \leq & C_6^2\left(\|n_0\|^2_{H^3}+\|u_0\|^2_{H^3}\right). \end{split} \end{eqnarray*}

    Using a standard continuity argument along with classical local wellposedness theory, this closes the a priori assumption (2.3) if we assume \|n_0\|_{H^3}+\|u_0\|_{H^3}\leq\delta_0 is sufficiently small. We can then extend the solution globally in time and complete the proof of Theorem 2.1.

    In this section, we consider the initial value problem for the linearized Navier-Stokes system

    \begin{equation} \left\{\begin{array}{l} \partial_t \widetilde{n} + {\mathop{{\rm{div}}}\nolimits} \widetilde{m} = 0, \qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \partial_t \widetilde{m} + c^2 {\nabla} \widetilde{n} - \bar\mu △ \widetilde{m}-(\bar \mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} \widetilde{m} = 0,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \lim\limits_{|x|\to\infty}\widetilde n = 0, \quad\lim\limits_{|x|\to\infty}\widetilde m = 0,\qquad t\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+,\\ (\widetilde{n} ,\widetilde{m})\big|_{t = 0} = (\rho_0-\bar\rho, \rho_0u_0),\qquad x\in{\mathop{\mathbb R\kern 0pt}\nolimits}^3, \end{array}\right. \end{equation} (3.1)

    where \bar\mu = \frac{\mu}{\bar\rho} , \bar\nu = \frac{\nu}{\bar\rho} , c = \sqrt{p'(\bar \rho)} .

    In terms of the semigroup theory for evolutionary equations, the solution (\widetilde{n}, \widetilde{m}) of the linearized Navier-Stokes problem (3.1) can be expressed for \widetilde U = (\widetilde{n}, \widetilde{m})^ t as

    \begin{eqnarray*} \widetilde U_t = B \widetilde U,\quad t \geq 0, \qquad \widetilde U(0) = \widetilde U_0, \end{eqnarray*}

    which gives rise to

    \begin{eqnarray*} \widetilde U(t) = S(t)\widetilde U_0 = e^{tB}\widetilde U_0, \quad t \geq 0, \end{eqnarray*}

    where B is defined as

    B = {\left( \begin{matrix} 0 & -{\mathop{{\rm{div}}}\nolimits} \\ -c^2 \nabla & \bar\mu △angle+(\bar\mu+\bar\nu)\nabla{\mathop{{\rm{div}}}\nolimits} \end{matrix} \right).}

    What left is to analyze the differential operator B in terms of its Fourier expression A(\xi) and show the long time properties of the semigroup S(t) . Applying the Fourier transform to system (3.1), we have

    \begin{eqnarray*} \partial_t \widehat{\widetilde U}(t,\xi) = A(\xi)\widehat{\widetilde U}(t,\xi), \quad t \geq 0, \qquad \widehat{\widetilde U}(0,\xi) = \widehat{\widetilde U}_0(\xi), \end{eqnarray*}

    where \xi = (\xi_1,\xi_2,\xi_3)^t , and A(\xi) is defined as

    \begin{eqnarray*} A(\xi) = {\left( \begin{matrix} 0 & -i\xi^t \\ -c^2 i\xi & -\bar\mu |\xi|^2I_{3\times3}-(\bar\mu+\bar\nu)\xi\otimes\xi \end{matrix} \right).} \end{eqnarray*}

    The eigenvalues of the matrix A can be computed by

    \begin{eqnarray*} \det(A(\xi)-\lambda I) = -(\lambda+\bar\mu|\xi|^2)^2(\lambda^2+(2\bar\mu+\bar\nu) |\xi|^2\lambda + c^2|\xi|^2) = 0, \end{eqnarray*}

    which implies

    \begin{eqnarray*} \lambda_0 = -\bar\mu|\xi|^2 (\text{double}), \quad \lambda_1 = \lambda_1(|\xi|), \quad \lambda_2 = \lambda_2(|\xi|). \end{eqnarray*}

    The semigroup e^{tA} is expressed as

    \begin{eqnarray*} e^{tA} = e^{\lambda_0 t}P_0+e^{\lambda_1 t}P_1+e^{\lambda_2 t}P_2, \end{eqnarray*}

    where the project operators P_i can be computed as

    \begin{eqnarray*} P_i = \prod\limits_{i\ne j}\frac{A(\xi)-\lambda_j I}{\lambda_i-\lambda_j}. \end{eqnarray*}

    By a direct computation, we can verify the exact expression for the Fourier transform \widehat G(t,\xi) of Green's function G(t,x) = e^{tB} as

    \begin{eqnarray*} \begin{split} \widehat G(t,\xi)& = e^{tA} = {\left( \begin{matrix} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}& -\frac{i\xi^t(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} & e^{-\lambda_0 t}(I-\frac{\xi \otimes \xi} {|\xi|^2})+\frac{\xi \otimes \xi} {|\xi|^2}\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} \end{matrix} \right)} = {\left( \begin{matrix} \widehat N \\ \widehat M \end{matrix} \right).} \end{split} \end{eqnarray*}

    Indeed, we can make the following decomposition for (\widetilde n, \widetilde m) = G \ast \widetilde U_0 as

    \begin{eqnarray*} \begin{split} \widehat {\widetilde n} = \widehat N \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal N} +\widehat {\mathfrak N} )\cdot \widehat{\widetilde U}_0,\quad \widehat {\widetilde m} = \widehat M \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal M} +\widehat {\mathfrak M} )\cdot \widehat{\widetilde U}_0, \end{split} \end{eqnarray*}

    where

    \begin{eqnarray*} \begin{split} &\widehat {\mathcal N} = \left( \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} \quad 0\right),\quad \widehat {\mathfrak N} = \left(0 \quad -\frac{i\xi^t(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\right),\\ &\widehat {\mathcal M} = \left(-\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \quad 0\right),\quad \widehat {\mathfrak M} = \left(0 \quad e^{-\lambda_0 t}(I-\frac{\xi \otimes \xi} {|\xi|^2})+\frac{\xi \otimes \xi} {|\xi|^2}\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}\right). \end{split} \end{eqnarray*}

    We further decompose the Fourier transform \widehat N , \widehat M into low frequency term and high frequency term below.

    Define

    \begin{eqnarray*} \begin{split} \widehat {\mathcal N} = \widehat {\mathcal N}_1+\widehat {\mathcal N}_2,\quad \widehat {\mathfrak N} = \widehat {\mathfrak N}_1+\widehat {\mathfrak N}_2,\quad \widehat {\mathcal M} = \widehat {\mathcal M}_1+\widehat {\mathcal M}_2,\quad \widehat {\mathfrak M} = \widehat {\mathfrak M}_1+\widehat {\mathfrak M}_2, \end{split} \end{eqnarray*}

    where (\cdot)_1 = \chi(\xi)(\cdot) , (\cdot)_2 = (1-\chi(\xi))(\cdot) , and \chi(\xi) is a smooth cut off function such that

    \begin{eqnarray*} \begin{split} \chi(\xi) = { \left\{\begin{array}{l} 1,\quad |\xi|\leq R,\\ 0,\quad |\xi|\geq R+1. \end{array}\right.} \end{split} \end{eqnarray*}

    Then we have the following decomposition for (\widetilde n, \widetilde m) = G \ast \widetilde U_0 as

    \begin{equation} \begin{split} &\widehat {\widetilde n} = \widehat N \cdot \widehat{\widetilde U}_0 = \widehat {N}_1 \cdot \widehat{\widetilde U}_0+\widehat {N}_2 \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal N}_1+\widehat {\mathfrak N}_1)\cdot \widehat{\widetilde U}_0 +(\widehat {\mathcal N}_2+\widehat {\mathfrak N}_2)\cdot \widehat{\widetilde U}_0,\\ &\widehat {\widetilde m} = \widehat M \cdot \widehat{\widetilde U}_0 = \widehat {M}_1 \cdot \widehat{U}_0+\widehat {M}_2 \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal M}_1+\widehat {\mathfrak M}_1)\cdot \widehat{\widetilde U}_0 +(\widehat {\mathcal M}_2+\widehat {\mathfrak M}_2)\cdot \widehat{\widetilde U}_0. \end{split} \end{equation} (3.2)

    To derive the long time decay rate of solution, we need to use accurate approximation to the Fourier transform \widehat G(t,x) of Green's function for both lower frequency and high frequency. In terms of the definition of the eigenvalues, we are able to obtain that it holds for |\xi|\leq\eta for some small positive constant \eta that

    \begin{equation} \begin{split} \lambda_1 = -\frac{2\bar\mu+\bar\nu}2 |\xi|^2 + \frac i 2\sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} = a+bi,\\ \lambda_2 = -\frac{2\bar\mu+\bar\nu}2 |\xi|^2 - \frac i 2 \sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} = a-bi, \end{split} \end{equation} (3.3)

    and we have

    \begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(bt)+\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(bt)}{b}\right]\\ &\sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(bt)-\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(bt)}{b}\right]\\ & \sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split} \frac{e^{\lambda_1 t}-e^{\lambda_2 t}}{\lambda_1-\lambda_2} = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\frac{\sin(bt)}{b}\sim O(1)\frac 1{|\xi|}e^{-\frac12(2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*}

    where

    \begin{eqnarray*} b = \frac12 \sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} \sim c|\xi|+O(|\xi|^3), \quad |\xi|\leq\eta. \end{eqnarray*}

    For the high frequency |\xi|\geq\eta , we are also able to obtain that it holds for |\xi|\geq\eta that

    \begin{equation} \begin{split} \lambda_1 = -\frac{2\bar\mu+\bar\nu} 2 |\xi|^2 - \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} = a-b,\\ \lambda_2 = -\frac{2\bar\mu+\bar\nu} 2 |\xi|^2 + \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} = a+b, \end{split} \end{equation} (3.4)

    and we have

    \begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} & = \frac12 e^{(a+b)t}[1+e^{-2bt}] - \frac{a}{2b} e^{(a+b)t}[1-e^{-2bt}] \sim O(1)e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} & = \frac{a+b}{2b} e^{(a+b)t}[1-e^{-2bt}] +e^{(a-b)t} \sim O(1)e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split} \frac{e^{\lambda_1 t}-e^{\lambda_2 t}}{\lambda_1-\lambda_2} = \frac{1}{2b}e^{(a+b)t}[1- e^{-2bt}] \sim O(1)\frac1{|\xi|^2}e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*}

    where

    \begin{eqnarray*} b = \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} \sim \frac12 (2\bar\mu+\bar\nu) |\xi|^2-\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}), \quad |\xi|\geq \eta. \end{eqnarray*}

    Here R_0 , \eta are some fixed positive constants.

    In this section, we apply the spectral analysis to the semigroup for the linearized Navier-Stokes system. We will establish the L^2 and L^p ( 2\leq p \leq \infty ) time decay rate of the global solutions for the linearized Navier-Stokes system.

    With the help of the formula for Green's function in Fourier space and the asymptotic analysis on its elements, we are able to establish the L^2 time decay rate. Indeed, we have the L^2 -time decay rate of the global strong solution to the problem for the linearized Navier-Stokes system as follows.

    Proposition 4.1. Let U_0 = (n_0, m_0)\in L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)\cap H^l({\mathop{\mathbb R\kern 0pt}\nolimits}^3) with l\geq3 , then (\widetilde n,\widetilde m) solves the linearized Navier-Stokes system (3.1) and satisfies for 0\leq k\leq l that

    \begin{eqnarray*} \|{\nabla}^{k} (\widetilde n,\widetilde m)(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}\big(\|U_0\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^k U_0\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big), \end{eqnarray*}

    where C is a positive constant independent of time.

    Proof. A straightforward computation together with the formula of the Green's function \widehat G(t,\xi) gives

    \begin{eqnarray*} \begin{split} \widehat {\widetilde n}(t,\xi)& = \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}\widehat{n}_0-\frac{i\xi\cdot\widehat{m}_0(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ &\sim \left\{\begin{array}{l} O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\leq\eta,\\ O(1)e^{-R_0 t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\geq\eta, \end{array}\right.\\ \widehat {\widetilde m}(t,\xi)& = -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\widehat{n}_0+e^{-\lambda_0 t}\widehat{m}_0+\left(\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}-e^{-\lambda_0 t}\right)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2} \\ &\sim \left\{\begin{array}{l} O(1)e^{- \bar\mu|\xi|^2t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\leq\eta,\\ O(1)e^{-R_0 t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\geq\eta, \end{array}\right. \end{split} \end{eqnarray*}

    here and below, R_0 , \eta are some fixed positive constants. Therefore, we have the L^2 -decay rate for (\widetilde n, \widetilde m) as

    \begin{eqnarray*} \begin{split} &\quad\|(\widehat {\widetilde n},\widehat {\widetilde m})(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}^2\\ & = \int_{|\xi|\leq \eta}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2\bar\mu|\xi|^2t}(|\widehat{n}_0|^2+|\widehat{m}_0|^2) d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}(|\widehat{n}_0|^2+|\widehat{m}_0|^2)d\xi\\ &\lesssim(1+t)^{-\frac32}\|({n}_0,{m}_0)\|^2_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3) \cap L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}. \end{split} \end{eqnarray*}

    And the L^2 -decay rate on the derivatives of (\widetilde n,\widetilde m) as

    \begin{eqnarray*} \begin{split} &\quad\|(\widehat {{\nabla}^{k}\widetilde n},\widehat {{\nabla}^{k}\widetilde m})(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}^2\\ & = \int_{|\xi|\leq \eta}|\xi|^{2k}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|\xi|^{2k}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2\bar\mu|\xi|^2t}|\xi|^{2k}(|\widehat{n}_0|^2+|\widehat{m}_0|^2) d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k}(|\widehat{n}_0|^2+|\widehat{m}_0|^2)d\xi\\ &\lesssim(1+t)^{-\frac32-k}\big(\|({n}_0,{m}_0)\|^2_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|({\nabla}^{k}{n}_0,{\nabla}^{k}{m}_0)\|^2_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big). \end{split} \end{eqnarray*}

    The proof of the Proposition 4.1 is completed.

    It should be noted that the L^2 -time decay rates derived above are optimal.

    Proposition 4.2. Let U_0 = ( n_0, m_0)\in L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)\cap H^l({\mathop{\mathbb R\kern 0pt}\nolimits}^3) with l\geq3 , assume that M_n = \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}n_0(x) d x and M_m = \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} m_0(x) dx satisfies that M_n , M_m are at least not all zeros, then the solution (\widetilde n,\widetilde m) of the linearized Navier-Stokes system (3.1) given by Proposition 4.1 satisfies for 0\leq k\leq l

    \begin{eqnarray*} \begin{split} C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq \|{\nabla}^{k} \widetilde n(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2},\\ C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq \|{\nabla}^{k} \widetilde m(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \end{split} \end{eqnarray*}

    where C is a positive constant independent of time.

    Proof. We only show the case of k = 0 for simplicity, the argument applies to the other orders of derivatives. From the formula of the Green's function \widehat G(t,\xi) , we deduce that

    \begin{eqnarray*} \begin{split} &\quad\widehat {\widetilde n}(t,\xi)\\& = \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}\widehat{n}_0-\frac{i\xi\cdot\widehat{m}_0(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(b t)\widehat{n}_0-i\xi\cdot\widehat{m}_0\frac{\sin(b t)}{b}\right]+e^{-\frac12(2\bar\mu+\bar\nu)|\xi|^2t}\left[\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(b t)}{b}\widehat{n}_0\right]\\ & = T_1+T_2, \quad \text{for}\quad|\xi|\leq \eta,\\ \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split} &\quad\widehat {\widetilde m}(t,\xi)\\ & = -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\widehat{n}_0+e^{-\lambda_0 t}\widehat{m}_0+\left(\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}-e^{-\lambda_0 t}\right)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2} \\ & = \bigg[e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(b t)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}-c^2i\xi\frac{\sin(b t)}{b}\widehat{n}_0\right]+e^{- \bar\mu|\xi|^2t}\left[\widehat{m}_0-\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}\right]\bigg]\\ &\quad\quad-e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(b t)}{b}\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}\right]\\ & = S_1+S_2, \quad \text{for}\quad|\xi|\leq \eta,\\ \end{split} \end{eqnarray*}

    here and below, \eta is a sufficiently small but fixed constant.

    It is easy to check that

    \begin{equation} \begin{split} \|\widehat {\widetilde n}(t,\xi)\|^2_{L^2}& = \int_{|\xi|\leq \eta}|\widehat {\widetilde n}(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|\widehat {\widetilde n}(t,\xi)|^2 d\xi\\ &\geq \int_{|\xi|\leq \eta}|T_1+T_2|^2 d\xi \geq \int_{|\xi|\leq \eta}\frac12|T_1|^2-|T_2|^2 d\xi. \end{split} \end{equation} (4.1)

    We then calculate that

    \begin{equation} \begin{split} \int_{|\xi|\leq \eta}|T_2|^2 d\xi &\lesssim \|\widehat{n}_0\|_{L^\infty}^2\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}|\xi|^4\left(\frac{\sin(b t)}{b}\right)^2 d\xi \\ &\lesssim\|\widehat{n}_0\|_{L^\infty}^2\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}|\xi|^2 d\xi\lesssim(1+t)^{-\frac52}\|{n}_0\|_{L^1}^2. \end{split} \end{equation} (4.2)

    Since n_0(x) \in L^1 implies \widehat{n}_0(\xi) \in C({{\mathop{\mathbb R\kern 0pt}\nolimits}^3}) . If \widehat{n}_0(0) = \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}n_0(x) d x\neq 0 , we deduce that \widehat{n}_0(\xi)\neq 0 for |\xi|\leq\eta when \eta is sufficiently small. One finds that, when M_n\neq 0 ,

    \begin{eqnarray*} \begin{split} |\widehat{n}_0(\xi)|^2\geq \frac1C\left|\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}n_0(x) d x\right|^2 \geq \frac{M_n^2}C,\quad \text{for}\quad |\xi|\leq\eta. \end{split} \end{eqnarray*}

    For \widehat{m}_0 , a similar argument yields that, when M_m\neq 0 , we have

    \begin{eqnarray*} \begin{split} \frac{\left|\xi\cdot\widehat{m}_0(\xi)\right|^2}{|\xi|^2} \geq \frac{\left|\xi\cdot M_m\right|^2}{C|\xi|^2},\quad \text{for}\quad |\xi|\leq\eta. \end{split} \end{eqnarray*}

    When M_n\neq 0 , M_m\neq 0 , with the help of the above analysis, using b\sim c|\xi|+O(|\xi|^3) for |\xi|\leq \eta , we obtain that

    \begin{equation} \label{optimal3} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\nonumber\\ &\geq \frac{M_n^2}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi+\frac1C\int_{|\xi|\leq \eta}\frac{\left|\xi\cdot M_m\right|^2}{b^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi \nonumber \end{split} \end{equation}
    \begin{equation} \begin{split}&\geq \frac{\min\{{M_n^2},\frac{M_m^2}{3c^2}\}}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\big(\cos^2(b t)+\sin^2(b t)\big)d\xi\\ &\geq C_1\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}d\xi\\ &\geq C^{-1}(1+t)^{-\frac32}. \end{split} \end{equation} (4.3)

    If M_n\neq 0 , M_m = 0 , and by the conituinity of \widehat{m}_0 near \xi = 0 , there exists a small enough constant \epsilon such that \epsilon\to0 as \xi\to 0 , and

    \begin{eqnarray*} |\widehat{m}_0(\xi)|^2 < \epsilon,\quad \text{for}\quad |\xi|\leq\eta. \end{eqnarray*}

    We thus use the help of spherical coordinates and the change of variables r = |\xi|\sqrt{t} to obtain that

    \begin{equation} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\\ &\geq \frac{M_n^2}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi-\frac{\epsilon}{Cc^2}\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\geq \frac{M_n^2}Ct^{-\frac32}\int_0^{\eta\sqrt t}e^{-(2\bar\mu+\bar\nu)r^2}\cos^2(cr\sqrt t)r^2dr-\frac{\epsilon}{Cc^2}t^{-\frac32}\int_0^{\eta\sqrt t}e^{-(2\bar\mu+\bar\nu)r^2}\sin^2(cr\sqrt t)r^2dr\\ &\geq \frac{M_n^2}Ct^{-\frac32}\sum\limits_{k = 0}^{[\frac{c\eta t}\pi]-1}\int_{\frac{k\pi}{c\sqrt t}}^{\frac{k\pi+\frac\pi4}{c\sqrt t}}e^{-(2\bar\mu+\bar\nu)r^2}\cos^2(cr\sqrt t)r^2dr-\frac{\epsilon}{Cc^2}(1+t)^{-\frac32}\\ &\geq \frac{M_n^2}{2C}t^{-\frac32}\sum\limits_{k = 0}^{[\frac{c\eta t}\pi]-1}\int_{\frac{k\pi}{c\sqrt t}}^{\frac{k\pi+\frac\pi4}{c\sqrt t}}e^{-(2\bar\mu+\bar\nu)r^2}r^2dr-\frac{\epsilon}{Cc^2}(1+t)^{-\frac32}\\ &\geq C_1^{-1}(1+t)^{-\frac32}-C_2^{-1}\epsilon(1+t)^{-\frac32}\\ .&\geq C^{-1}(1+t)^{-\frac32} \end{split} \end{equation} (4.4)

    In the case of M_n = 0 , M_m\neq0 , we can use a similar argument to obtain that

    \begin{equation} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\\ &\geq -\frac\epsilon C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi+\frac{M_m^2}{3Cc^2}\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\geq C^{-1}(1+t)^{-\frac32}. \end{split} \end{equation} (4.5)

    Combining the above estimates (4.1), (4.2), (4.3), (4.4) and (4.5), we obtain the lower bound of the time decay rate for {\widetilde n}(t,x) as

    \begin{eqnarray*} \| {\widetilde n}(t,x)\|^2_{L^2} = \|\widehat {\widetilde n}(t,\xi)\|^2_{L^2}\geq C^{-1}(1+t)^{-\frac32}. \end{eqnarray*}

    The lower bound of the time decay rate for {\widetilde m}(t,x) can be shown in a similar fashion. It is not difficult to derive that

    \begin{equation} \begin{split} \|\widehat {\widetilde m}(t,\xi)\|^2_{L^2}\geq \int_{|\xi|\leq \eta}\frac12|S_1|^2-|S_2|^2 d\xi, \end{split} \end{equation} (4.6)

    then we find that

    \begin{equation} \begin{split} \int_{|\xi|\leq \eta}|S_2|^2 d\xi \lesssim(1+t)^{-\frac52}\|{m}_0\|_{L^1}^2. \end{split} \end{equation} (4.7)

    We then calculate that

    \begin{eqnarray*} \begin{split} &\quad\int_{|\xi|\leq \eta}|S_1|^2 d\xi\\ &\geq \bigg\{\frac{c^4M_n^2}C\int_{|\xi|\leq \eta}\frac{|\xi|^2}{b^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\qquad+\frac{1}C\int_{|\xi|\leq \eta}\frac{\left|\xi\cdot M_m\right|^2}{|\xi|^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(bt)d\xi\bigg\}\\ &\qquad +\bigg\{\int_{|\xi|\leq \eta}e^{-\frac12(4\bar\mu+\bar\nu)|\xi|^2t}\cos(bt)\frac{\xi(\xi\cdot{\widehat m}_0)}{|\xi|^2}\left({\widehat m}_0-\frac{\xi(\xi\cdot{\widehat m}_0)}{|\xi|^2}\right)d\xi\bigg\}\\ & = J_1+J_2. \end{split} \end{eqnarray*}

    A direct computation gives rise to

    \begin{equation} J_1\geq C^{-1}(1+t)^{-\frac32},\qquad J_2 = 0. \end{equation} (4.8)

    Combining the above estimates (4.6), (4.7) and (4.8), we obtain the lower bound of the time decay rate for {\widetilde m}(t,x) as

    \begin{eqnarray*} \| {\widetilde m}(t,x)\|^2_{L^2} = \|\widehat {\widetilde m}(t,\xi)\|^2_{L^2}\geq C^{-1}(1+t)^{-\frac32}. \end{eqnarray*}

    Then the proof of Proposition 4.2 is completed.

    In this subsection, we establish the following L^p -time decay rate of the global strong solution to the linearized Navier-Stokes system with p \in [2,+\infty] .

    Proposition 4.3. Let U_0 = (n_0,m_0)\in L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)\cap W^{l,p}({\mathop{\mathbb R\kern 0pt}\nolimits}^3) with l\geq 3 , then (\widetilde n,\widetilde m) solves the linearized Navier-Stokes system (3.1) and satisfies for 0\leq k\leq l and p \in [2,+\infty] that

    \begin{eqnarray*} \|{\nabla}^{k} (\widetilde n,\widetilde m)(t)\|_{L^p({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac32(1-\frac1p)-\frac{k}2}\big(\|U_0\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^k U_0\|_{L^p({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big), \end{eqnarray*}

    where C is a positive constant independent of time.

    To prove Proposition 4.3, the following two lemmas in [6] are helpful.

    Lemma 4.1. Let n\geq 1 and assume that \hat f(\xi) \in L^\infty \cap C^{n+1}({\mathop{\mathbb R\kern 0pt}\nolimits}^n/\{0\}) , with

    \begin{eqnarray*} \begin{split} |{\nabla}^\alpha _\xi \hat f(\xi)| \leq C'{ \left\{\begin{array}{l} |\xi|^{-|\alpha|+\sigma_1},\quad |\xi|\leq R, |\alpha| = n,\\ |\xi|^{-|\alpha|-\sigma_2},\quad |\xi|\geq R, |\alpha| = n-1,n,n+1, \end{array}\right.} \end{split} \end{eqnarray*}

    where \sigma_1, \sigma_2>0 and n>2-2\sigma_2 . Then \hat f(\xi) is continuous at 0 and \infty , and

    \begin{eqnarray*} f = m_1+m_2\delta, \end{eqnarray*}

    where m_1\in L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^n) satisfies \|m_1\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^n)} \leq C(C') , m_2 is the constant

    \begin{eqnarray*} m_2 = (2\pi)^{-\frac n 2}\lim\limits_{|\xi| \to \infty} \hat f(\xi), \end{eqnarray*}

    and \delta is the Dirac distribution. In particular, \hat f(\xi) is a strong L^p multiplier, 1\leq p \leq \infty , in the sense that, for any g\in L^p ,

    \begin{eqnarray*} \|f\ast g\|_{L^p} \leq C\| g\|_{L^p},\quad 1\leq p \leq \infty, \end{eqnarray*}

    where C depends only on |m_2|\leq \| \hat f\|_{L^\infty} and the constant C' above.

    Lemma 4.2. Let \hat g(t,\xi) = \hat K(t,\xi)\hat f(\xi) , where \hat K(t,\xi) = e^{-\vartheta|\xi|^2 t} , \hat f(\xi)\in L^\infty\cap C^{n+1}({\mathop{\mathbb R\kern 0pt}\nolimits}^n) , and

    \begin{eqnarray*} |{\nabla}_\xi ^\beta \hat f(\xi)|\leq C'|\xi|^{-|\beta|},\quad |\beta|\leq n+1. \end{eqnarray*}

    Then {\nabla}_x ^\alpha g(t,\cdot)\in L^p for t>0 , and for all \alpha , 1\leq p \leq \infty , we have

    \begin{eqnarray*} \|{\nabla}_x ^\alpha g(t,\cdot)\|_{L^p}\leq C(|\alpha|)t^{-\frac n 2(1-\frac 1 p)-\frac{|\alpha|}{2}}. \end{eqnarray*}

    In particular, \widehat{{\nabla}_x ^\alpha g(t,x)} = (i\xi)^\alpha \hat g(t,\xi) is a strong L^p multiplier, with norm bounded by C(|\alpha|,\vartheta)C't^{-\frac{|\alpha|}2} , where the constant C(|\alpha|,\vartheta) depends only on |\alpha| and \vartheta .

    Now let us turn to the proof of Proposition 4.3.

    Proof of Proposition 4.3. We first analyze above higher frequency terms denoted by \widehat {(\cdot)}_2 . Recall that

    \begin{eqnarray*} \begin{split} \lambda_1 = -(2\bar\mu+\bar\nu) |\xi|^2+\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}),\quad \lambda_2 = -\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}),\quad |\xi|\geq \eta. \end{split} \end{eqnarray*}

    We shall prove that the higher frequency terms are L^p Fourier multipliers with an exponential time decay coefficient C e^{-c_1t} for some constants c_1>0 . For simplicity, we only show that \widehat {\mathcal N}_2 is an L^p Fourier multiplier at higher frequency as follows. It holds

    \begin{eqnarray*} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} = e^{\lambda_2 t}+\frac{\lambda_2 e^{\lambda_2 t}}{\lambda_1-\lambda_2}-\frac{\lambda_2 e^{\lambda_1 t}}{\lambda_1-\lambda_2}. \end{eqnarray*}

    By a direct computation, it is easy to verify

    \begin{eqnarray*} |{\nabla}_\xi^k \lambda_2|\lesssim|\xi|^{-2-k},\quad |\xi|\geq\eta, \end{eqnarray*}

    which gives rise to

    \begin{eqnarray*} \begin{split} \bigg|{\nabla}_\xi^k \Big[(1-\chi(\cdot))e^{\lambda_2 t}\Big]\bigg|, \left|{\nabla}_\xi^k \Big[(1-\chi(\cdot))\frac{\lambda_2 e^{\lambda_2 t}}{\lambda_1-\lambda_2}\Big]\right|\lesssim{ \left\{\begin{array}{l} 0, \quad |\xi|\leq R,\\ e^{-c_1t}|\xi|^{-2-k},\quad |\xi|\geq R, \end{array}\right.} \end{split} \end{eqnarray*}

    here and below, R>0 is a given constant. Thus, from Lemma 4.1 it follows that the inverse Fourier transform of the term (1-\chi(\cdot))\left(e^{\lambda_2 t}+\frac{\lambda_2 e^{\lambda_2 t}}{\lambda_1-\lambda_2}\right) is an L^p multiplier with the coefficient Ce^{-c_1t} . The other part of \widehat {\mathcal N}_2 at higher frequency can be written as

    \begin{eqnarray*} (1-\chi(\cdot))\frac{\lambda_2 e^{\lambda_1 t}}{\lambda_1-\lambda_2} \sim e^{-\frac12(2\bar\mu+\bar\nu) |\xi|^2 t}\Big[(1-\chi(\cdot))\frac{e^{(-\lambda_2-\frac12(2\bar\mu+\bar\nu) |\xi|^2)t}}{\lambda_1-\lambda_2}\Big]. \end{eqnarray*}

    We can regard e^{-\frac12 (2\bar\mu+\bar\nu) |\xi|^2 t} as the function K(t,\xi) of Lemma 4.2, and the rest term satisfies the condition. Thus, the inverse Fourier transform of (1-\chi(\cdot))\frac{\lambda_2 e^{\lambda_1 t}}{\lambda_1-\lambda_2} is also an L^p multiplier with the coefficient Ce^{-c_1t} . These facts imply that \widehat {\mathcal N}_2 at higher frequency is an L^p multiplier with the coefficient Ce^{-c_1t} . Applying the similar analysis to the terms \widehat {\mathfrak N}_2 , \widehat {\mathcal M}_2 , and \widehat {\mathfrak M}_2 , we can show that their inverse Fourier transform are all L^p multiplier with the constant coefficient Ce^{-c_1t} . Then

    \begin{equation} \|({\nabla}_x^k({\mathcal N}_2 \ast f),{\nabla}_x^k({\mathfrak N}_2 \ast f),{\nabla}_x^k( {\mathcal M}_2 \ast f),{\nabla}_x^k( {\mathfrak M}_2 \ast f))(t)\|_{L^p} \leq Ce^{-c_1t}\|{\nabla}_x^k f\|_{L^p}, \end{equation} (4.9)

    for all integer k\geq 0 , and p\in[2,\infty] .

    We also need to deal with the corresponding lower frequency terms denoted by \widehat {(\cdot)}_1 . Recall that

    \begin{eqnarray*} \begin{split} &\frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}, \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2},\frac{|\xi|(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*}

    which imply that for |\xi|\leq\eta that

    \begin{eqnarray*} |\widehat{\mathcal N}_1|\sim O(1)e^{-c_2|\xi|^2t},\quad |\widehat{\mathfrak N}_1|\sim O(1)e^{-c_2|\xi|^2t},\\ |\widehat{\mathcal M}_1|\sim O(1)e^{-c_2|\xi|^2t},\quad |\widehat{\mathfrak M}_1|\sim O(1)e^{-c_2|\xi|^2t}, \end{eqnarray*}

    for some constants c_2>0 . Thus, by Hausdroff-Young's inequality with p\in[2, +\infty] , we can obtain

    \begin{equation} \begin{split} \|({\nabla}^k{\mathcal N}_1,{\nabla}^k{\mathfrak N}_1,{\nabla}^k {\mathcal M}_1,{\nabla}^k {\mathfrak M}_1)(t)\|_{L^p} \leq& C\left(\int_{|\xi|\leq \eta}\big||\xi|^k e^{-c_2|\xi|^2t}\big|^q d\xi\right)^{\frac 1q}\\ \leq& C(1+t)^{-\frac32(1-\frac1p)-\frac k2}. \end{split} \end{equation} (4.10)

    Combining (4.9) and (4.10), we finally have for t>0 that

    \begin{eqnarray*} \begin{split} \|({\nabla}^k(N \ast f),{\nabla}^k(M \ast f))(t)\|_{L^p}& = \|({\nabla}^k((N_1+N_2) \ast f),{\nabla}^k((M_1+M_2) \ast f))(t)\|_{L^p}\\ &\leq C(1+t)^{-\frac32(1-\frac1p)-\frac k2}\|f\|_{L^1}+Ce^{-c_1t}\|{\nabla}^k f\|_{L^p}\\ &\leq C(1+t)^{-\frac32(1-\frac1p)-\frac k2}(\|f\|_{L^1}+\|{\nabla}^k f\|_{L^p}). \end{split} \end{eqnarray*}

    The proof of Proposition 4.3 is completed.

    We are ready to prove Theorem 1.1 on the sharp time decay rate of the global solution to the initial value problem for the nonlinear Navier-Stokes system.

    In what follows, we will set n_h = n-\widetilde n and m_h = m-\widetilde m , then we have

    \begin{equation} \left\{\begin{array}{l} \partial_t n_h + {\mathop{{\rm{div}}}\nolimits} m_h = 0,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \partial_t m_h + c^2 {\nabla} n_h- \bar\mu △ m_h- (\bar\mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} m_h = F,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \lim\limits_{|x|\to\infty}n_h = 0, \quad\lim\limits_{|x|\to\infty} m_h = 0,\qquad t\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+,\\ (n_h,m_h)\big|_{t = 0} = (0,0),\qquad x\in{\mathop{\mathbb R\kern 0pt}\nolimits}^3, \end{array}\right. \end{equation} (5.1)

    where \bar\mu = \frac{\mu}{\bar\rho} , \bar\nu = \frac{\nu}{\bar\rho} , c = \sqrt{p'(\bar \rho)} , and

    \begin{eqnarray*} \begin{split} F = &-{\mathop{{\rm{div}}}\nolimits}\Big\{ \frac{(m_h+\widetilde m)\otimes (m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}+\bar\mu{\nabla}\big(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}\big)\Big\}\\ & - {\nabla}\Big\{(\bar\mu+\bar\nu){\mathop{{\rm{div}}}\nolimits}(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho})+\big(p(n_h+\widetilde n+\bar\rho)-p(\bar\rho)-c^2(n_h+\widetilde n)\big)\Big\}. \end{split} \end{eqnarray*}

    Denote U_h = (n_h, m_h)^t , we have the equivalent form of system (5.1) in vector form

    \begin{eqnarray*} \partial_t U_h = BU_h+H,\quad t\geq0,\qquad U_h(0) = 0, \end{eqnarray*}

    where the nonlinear term H(\widetilde U, U_h) = (0, F(\widetilde U, U_h))^t . Thus, we can represent the solution in term of the semigroup

    \begin{eqnarray*} U_h(t) = S(t)\ast U_{h}(0)+\int_0^t S(t-\tau)\ast H(\widetilde U, U_h)(\tau)d\tau, \end{eqnarray*}

    which (n_h, m_h) can be decomposed as

    \begin{equation} n_h = N\ast U_{h}(0)+\int_0^t \mathfrak N (t-\tau)\ast H(\tau)d\tau, \end{equation} (5.2)
    \begin{equation} m_h = M\ast U_{h}(0)+\int_0^t \mathfrak M (t-\tau)\ast H(\tau)d\tau. \end{equation} (5.3)

    Furthermore, in view of the above definition for \widehat{\mathfrak N}(\xi) and \widehat{\mathfrak M}(\xi) , it is easy to verify for some constants c_3>0 , c_4>0 , R_0>0 , we discover that

    \begin{eqnarray*} |\widehat{\mathfrak N}(\xi)|\sim O(1)e^{-c_3|\xi|^2t}, \quad |\widehat{\mathfrak M}(\xi)|\sim O(1)e^{-c_3|\xi|^2t}, \quad|\xi|\leq\eta, \end{eqnarray*}
    \begin{eqnarray*} |\widehat{\mathfrak N}(\xi)|\sim O(1)\frac1{|\xi|}e^{-R_0t}, \quad |\widehat{\mathfrak M}(\xi)|\sim O(1)\frac1{|\xi|^2}e^{-R_0t}+O(1)e^{-c_4|\xi|^2t}, \quad |\xi|\geq\eta. \end{eqnarray*}

    Thus, applying a similar argument as in the proof of Proposition 4.1, we have

    \begin{equation} \|({\nabla}^k {\mathfrak N}\ast H, {\nabla}^k {\mathfrak M}\ast H)(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k+1} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} (5.4)
    \begin{equation} \|({\nabla}^k {\mathfrak N}\ast H, {\nabla}^k {\mathfrak M}\ast H)(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} (5.5)
    \begin{equation} \|{\nabla}^k {\mathfrak M}\ast H(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k-1} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} (5.6)

    for any non-negative integer k and

    \begin{equation} \begin{split} Q = &\Big|\frac{(m_h+\widetilde m)\otimes (m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}+\bar\mu{\nabla}\big(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}\big)\Big|\\ &+\Big|(\bar\mu+\bar\nu){\mathop{{\rm{div}}}\nolimits}(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho})+\big(p(n_h+\widetilde n+\bar\rho)-p(\bar\rho)-c^2(n_h+\widetilde n)\big)\Big|. \end{split} \end{equation} (5.7)

    For readers' convenience, we show how to estimate \|{\nabla}^k {\mathfrak M}\ast H(t)\|_{L^2} as an example. The other two estimates can be obtained by the similar argument. Indeed,

    \begin{eqnarray*} \begin{split} &\quad\|{\nabla}^k {\mathfrak M}\ast H(t)\|_{L^2}^2\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2c_3|\xi|^2t}|\xi|^{2k}|\widehat H|^2d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k-4}|\widehat H|^2d\xi\\ &\quad+\int_{|\xi|\geq \eta}e^{-2c_4|\xi|^2 t}|\xi|^{2k}|\widehat H|^2d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2c_3|\xi|^2t}|\xi|^{2k+2}|\widehat Q|^2d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k-2}|\widehat Q|^2d\xi\\ &\quad+\int_{|\xi|\geq \eta}e^{-2c_4|\xi|^2t}|\xi|^{2k+2}|\widehat Q|^2d\xi\\ &\lesssim(1+t)^{-3(\frac1q-\frac12)-1-k}\big(\|Q\|^2_{L^q({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^{\tilde k} Q\|^2_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big),\quad q = 1,2,\quad k-1\leq\tilde k\in{\mathop{\mathbb N\kern 0pt}\nolimits}^+. \end{split} \end{eqnarray*}

    In this subsection, we establish the faster decay rate for (n_h, m_h) . We will start with an a priori assumption on a carefully chosen quantity \Lambda(t) defined in (5.8), and then later prove a better estimate with the help of the smallness of initial data.

    We begin with following Lemma.

    Lemma 5.1. Let r_1, r_2>0 be real, one has

    \begin{eqnarray*} \begin{split} \int_0^{\frac t 2}(1+t-\tau)^{-r_1}(1+\tau)^{-r_2} d\tau = &\int_0^{\frac t 2}(1+\frac t 2+\tau)^{-r_1}(1+\frac t 2-\tau)^{-r_2} d\tau\\ \lesssim&{ \left\{\begin{array}{l} (1+t)^{-r_1}, \quad \mathit{\text{for}} \quad r_2 > 1,\\ (1+t)^{-(r_1-\epsilon)},\quad \mathit{\text{for}} \quad r_2 = 1,\\ (1+t)^{-(r_1+r_2-1)},\quad \mathit{\text{for}} \quad r_2 < 1, \end{array}\right.} \end{split} \end{eqnarray*}

    and

    \begin{eqnarray*} \begin{split} \int_{\frac t 2}^t(1+t-\tau)^{-r_1}(1+\tau)^{-r_2} d\tau = &\int_0^{\frac t 2}(1+t-\tau)^{-r_2}(1+\tau)^{-r_1} d\tau\\ \lesssim&{ \left\{\begin{array}{l} (1+t)^{-r_2}, \quad \mathit{\text{for}} \quad r_1 > 1,\\ (1+t)^{-(r_2-\epsilon)},\quad \mathit{\text{for}} \quad r_1 = 1,\\ (1+t)^{-(r_1+r_2-1)},\quad \mathit{\text{for}} \quad r_1 < 1, \end{array}\right.} \end{split} \end{eqnarray*}

    where \epsilon>0 is a small but fixed constant.

    Proposition 5.1. Under the assumptions of Theorem 1.1, the solution (n_h, m_h) of the nonlinear system (5.1) satisfies for k = 0,1,2 that

    \begin{eqnarray*} \begin{split} &\|({\nabla} ^k n_h,{\nabla}^{k} m_h)\|_{L^2}\leq C\delta_0^2(1+t)^{-\frac54-\frac {k} 2},\\ &\|{\nabla} ^3 m_h\|_{L^2}\leq C\delta_0^2(1+t)^{-\frac{11}4},\quad \|{\nabla} ^3 n_h\|_{L^2}\leq C\delta_0(1+t)^{-\frac74}, \end{split} \end{eqnarray*}

    where C is a positive constant independent of time.

    From (5.7), we deduce

    \begin{eqnarray*} Q(\widetilde U, U_h) = Q_1+Q_2+Q_3+Q_4, \end{eqnarray*}

    which implies for a smooth solution (n,m) satisfying \|(n,m)\|_{H^3}<\infty that

    \begin{eqnarray*} \begin{split} &Q_1 = Q_1(\widetilde U, U_h)\sim O(1)\left(n_h^2+m_h\otimes m_h+\widetilde n^2+\widetilde m\otimes\widetilde m \right),\\ &Q_2 = Q_2(\widetilde U, U_h)\sim O(1)\left(\widetilde n n_h+\widetilde m\otimes m_h\right),\\ &Q_3 = Q_3(\widetilde U, U_h)\sim O(1)\left({\nabla}(n_h\cdot m_h)+{\nabla}(\widetilde n\cdot\widetilde m)\right),\\ &Q_4 = Q_4(\widetilde U, U_h)\sim O(1)\left({\nabla}(\widetilde n\cdot m_h)+{\nabla}( n_h\cdot\widetilde m) \right). \end{split} \end{eqnarray*}

    Define

    \begin{equation} \begin{split} \Lambda(t) = :&\sup\limits_{0\leq s \leq t}\bigg\{\sum\limits_{k = 0}^2(1+s)^{\frac54+\frac k 2}{\delta_0}^{-\frac34}\|({\nabla} ^k n_h,{\nabla}^{k}m_h)(s)\|_{L^2}\\ &\quad+(1+s)^{\frac74}\|({\nabla}^3 n_h, {\nabla}^3 m_h)(s)\|_{L^2}\bigg \}. \end{split} \end{equation} (5.8)

    Proposition 5.2. Under the assumptions of Theorem 1.1, if for some T>0 , \Lambda(t) \leq \delta_0^{\frac12} for any t\in[0,T] , then it holds that

    \begin{eqnarray*} \begin{split} \Lambda(t)\leq C\delta_0^{\frac34},\quad t\in[0,T], \end{split} \end{eqnarray*}

    where C is a positive constant independent of time.

    The proof of this Proposition 5.2 consists of following three steps.

    Starting with (5.4), (5.5), (5.6) and (5.8), we have after a complicate but straightforward computation that

    \begin{equation} \begin{split} \|(n_h, m_h)\|_{L^2}&\lesssim\int_0^t \|(\mathfrak N (t-\tau)\ast H(\tau), \mathfrak M (t-\tau)\ast H(\tau))\|_{L^2}d\tau\\ &\lesssim\int_0^{t} (1+t-\tau)^{-\frac54}\big(\|Q(\tau)\|_{L^1}+\| Q(\tau)\|_{L^2}\big)d\tau\\ &\lesssim\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right)\int_0^{t} (1+t-\tau)^{-\frac54}(1+\tau)^{-\frac32}d\tau\\ &\lesssim(1+t)^{-\frac54}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{equation} (5.9)

    It is easy to verify that

    \begin{eqnarray*} \begin{split} \|Q(t)\|_{L^1}\lesssim&\|Q_1\|_{L^1}+\|Q_2\|_{L^1}+\|Q_3\|_{L^1}+\|Q_4\|_{L^1}\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^2}^2+\|( n_h, m_h)\|_{L^2}^2+ \|( n_h, m_h)\|_{L^2}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}\\ &+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|(\widetilde n,\widetilde m)\|_{L^2}\left(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2} +\| ({\nabla} n_h,{\nabla} m_h)\|_{L^2}\right)\\ \lesssim & (1+t)^{-\frac32}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*}

    Indeed, by virtue of Hölder's inequality and Gagliardo-Nirenberg's inequality, we obtain that

    \begin{eqnarray*} \|u\|_{L^\infty}\lesssim \|{\nabla} u\|_{L^2}^{\frac12}\|{\nabla} ^2 u\|_{L^2}^{\frac12}, \end{eqnarray*}

    which implies that

    \begin{eqnarray*} \begin{split} &\|Q(t)\|_{L^2}\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^\infty}\big(\|(\widetilde n,\widetilde m)\|_{L^2}+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|(n_h,m_h)\|_{L^2}\\ &+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|( n_h,m_h)\|_{L^\infty}\left(\|( n_h,m_h)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\right)\\ &+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^\infty}\|(n_h,m_h)\|_{L^2}\\ \lesssim &(1+t)^{-\frac94}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*}

    Furthermore, exactly as in the estimate of the high order derivatives, we have

    \begin{equation} \begin{split} &\|({\nabla} n_h, {\nabla} m_h)\|_{L^2}\\ \lesssim &\int_0^{\frac t 2} \|({\nabla}\mathfrak N , {\nabla}\mathfrak M)(t-\tau)\ast H(\tau)\|_{L^2}d\tau+\int_{\frac t 2}^t \|(\mathfrak N, \mathfrak M )(t-\tau)\ast {\nabla} H(\tau)\|_{L^2}d\tau\\ \lesssim &\int_0^{\frac t 2} (1+t-\tau)^{-\frac74}\big(\| Q(\tau)\|_{L^1}+\|{\nabla} Q(\tau)\|_{L^2}\big)d\tau +\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} Q(\tau)\|_{L^2}d\tau\\ \lesssim&\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right)\Bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac74}(1+\tau)^{-\frac32}d\tau+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{11}4}d\tau\Big)\\ \lesssim&(1+t)^{-\frac74}\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right), \end{split} \end{equation} (5.10)

    Similarly, it holds that

    \begin{eqnarray*} \begin{split} &\|{\nabla} Q(t)\|_{L^2}\nonumber\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^\infty}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|({\nabla}^2\widetilde n,{\nabla}^2\widetilde m)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\nonumber \end{split} \end{eqnarray*}
    \begin{eqnarray*} \begin{split}&\quad+\|( {\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2}\big)+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^\infty}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|(n_h,m_h)\|_{L^2}\\ &\quad+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|( n_h,m_h)\|_{L^\infty}\big(\|( {\nabla}^2\widetilde n,{\nabla}^2\widetilde m)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\\ &\quad+\|({\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2}\big)+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^\infty}\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\\ \lesssim & (1+t)^{-\frac{11}4}\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right). \end{split} \end{eqnarray*}

    Thus, we also get that

    \begin{equation} \begin{split} &\|({\nabla}^2 n_h, {\nabla}^2 m_h)(t)\|_{L^2}\\ \lesssim& \int_0^{\frac t 2} \|({\nabla}^2 \mathfrak N, {\nabla}^2 \mathfrak M) (t-\tau)\ast H(\tau)\|_{L^2}d\tau\\ &\quad+\int_{\frac t 2}^t \|(\mathfrak N, \mathfrak M) (t-\tau)\ast {\nabla}^2 H(\tau)\|_{L^2}d\tau\\ \lesssim &\int_0^{\frac t 2}(1+t-\tau)^{-\frac94}\big(\|Q(\tau)\|_{L^1}+\|{\nabla}^2 Q(\tau)\|_{L^2}\big)d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} ^2Q(\tau)\|_{L^2}d\tau\\ \lesssim& \left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right)\bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac94}(1+\tau)^{-\frac32}d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{13}4}d\tau\bigg)\\ \lesssim&(1+t)^{-\frac94}\left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right). \end{split} \end{equation} (5.11)

    Finally, we have

    \begin{eqnarray*} \begin{split} &\|{\nabla}^2 Q(t)\|_{L^2}\\ \lesssim &(\|(\widetilde n,\widetilde m)\|_{L^\infty}+\|(n_h,m_h)\|_{L^\infty})(\|({\nabla}^3\widetilde n,{\nabla}^3 \widetilde m)\|_{L^2}+\|({\nabla}^3n_h,{\nabla}^3 m_h)\|_{L^2})\\ &\quad+(\|({\nabla} \widetilde n,{\nabla}\widetilde m)\|_{L^\infty}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^\infty}) (\|({\nabla}\widetilde n,{\nabla} \widetilde m)\|_{L^2}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2})\\ &\quad+(\|(\widetilde n,\widetilde m)\|_{L^\infty}+\|(n_h,m_h)\|_{L^\infty}+\|({\nabla} \widetilde n,{\nabla}\widetilde m)\|_{L^\infty}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^\infty})\\ &\quad\times(\|({\nabla}^2\widetilde n,{\nabla}^2 \widetilde m)\|_{L^2}+\|({\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2})\\ \lesssim&(1+t)^{-\frac{13}4}\left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right). \end{split} \end{eqnarray*}

    In this subsection, we will close the a priori estimates and complete the proof of Proposition 5.2. For this purpose, we need to derive the time decay rate of higher order derivatives of (n_h,m_h) . We will establish the following lemma.

    Lemma 5.2. Under the assumption of Theorem 1.1, one has

    \begin{eqnarray*} \|{\nabla}^2 n(t)\|_{H^1}+\|{\nabla}^2 u(t)\|_{H^1}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*}

    In particular, it holds that

    \begin{eqnarray*} \|{\nabla}^3 (n_h, m_h)(t)\|_{L^2}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*}

    Proof. First of all, in view of (2.12), recovering the dissipation estimate for n , we see that

    \begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^3 n dx +C_1\|{\nabla}^3 n\|_{L^2}^2 dx\\ \leq &C_2\left(\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^4 u\|_{L^2}^2\right)+C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\\ &\quad\times\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right). \end{split} \end{equation} (5.12)

    Summing up (2.7) and (2.8) in the energy estimate for (n,u) , we can directly derive

    \begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma |{\nabla}^2 n|^2+|{\nabla}^2 u|^2 + \gamma|{\nabla}^3 n|^2 +|{\nabla}^3 u|^2 \right)dx + C_3\left(\|{\nabla}^3 u|^2 _{L^2}+\|{\nabla}^4 u\|^2 _{L^2}\right) \\ \leq &C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}^2\right). \end{split} \end{equation} (5.13)

    Multiplying (5.12) by \epsilon_1\frac{C_3}{C_2} with \epsilon_1>0 a small but fixed constant, adding it with (5.13), we deduce that there exists a constant C_4>0 such that

    \begin{eqnarray*} \begin{split} &\frac{d}{dt}\bigg\{\sum\limits_{2\leq k\leq3}\left(\gamma \|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right)+\epsilon_1\frac{C_3}{C_2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx \bigg\}\\ &\quad+ C_4\Big(\|{\nabla}^{3} n\|_{L^2}^2+\sum\limits_{3\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2}\Big)\\ \leq &C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right). \end{split} \end{eqnarray*}

    Next, we define

    \mathcal E_1(t) = \bigg\{\sum\limits_{2\leq k\leq3}\left(\gamma \|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right)+\epsilon_1\frac{C_3}{C_2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx \bigg\}.

    Observe that since \epsilon_1\frac{C_3}{C_2} is small, then there exists a constant C_5>0 such that

    \begin{eqnarray*} C_5^{-1}\left(\|{\nabla}^2 n(t)\|^2_{H^1}+\|{\nabla}^2 u(t)\|^2_{H^1}\right) \leq\mathcal E_1(t)\leq C_5\left(\|{\nabla}^2 n(t)\|^2_{H^1}+\|{\nabla}^2 u(t)\|^2_{H^1}\right). \end{eqnarray*}

    Then we arrive at

    \begin{eqnarray*} \frac{d}{dt}\mathcal E_1(t)+C_4\Big(\|{\nabla}^{3} n(t)\|_{L^2}^2+\|{\nabla}^3 u(t)\|^2_{H^1}\Big) \leq C(1+t)^{-5}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{eqnarray*}

    Denote S(t) = \Big\{\xi\big| |\xi| \leq \sqrt{\frac{3(1+\gamma)}{C_4}}(1+t)^{-\frac12}\Big\} the time-dependent n -dimensional sphere. This decomposition allows us to estimate L^2 time decay depend on (\widehat {n}, \widehat {u}) for frequency values \xi \in S(t) , then we obtain that

    \begin{eqnarray*} \begin{split} &\frac{C_4}{3}\|{\nabla}^{3} (n, u)(x)\|_{L^2}^2 \geq\frac{C_4}{3}\int_{S(t)^c} |\xi|^6|(\widehat{n}, \widehat{u})(\xi)|^2d\xi\\ \geq&(1+\gamma)(1+t)^{-1}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi-(1+\gamma)(1+t)^{-1}\int_{S(t)} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi. \end{split} \end{eqnarray*}

    Hence we have

    \begin{eqnarray*} \begin{split} &\frac{d}{dt}\mathcal E_1(t)+(1+t)^{-1}\mathcal E_1(t)+\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\\ \lesssim&(1+t)^{-5}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right)+(1+t)^{-1}\int_{S(t)} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi\\ &\quad+(1+t)^{-1}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx. \end{split} \end{eqnarray*}

    Multiplying the above equation by (1+t)^5 , we obtain that

    \begin{eqnarray*} \begin{split} &\frac{d}{dt}\Big\{(1+t)^5\mathcal E_1(t)\Big\}+(1+t)^5\Big(\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\Big) \lesssim(1+t)^{\frac12}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*}

    Integrating it with respect to time from 0 to T , then we have

    \begin{eqnarray*} \begin{split} &(1+t)^5\mathcal E_1(t)+\int_0^T(1+t)^5\Big(\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\Big)dt\\ \lesssim& \mathcal E_1(0)+(1+t)^{\frac32}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right), \end{split} \end{eqnarray*}

    which implies that

    \begin{eqnarray*} \|{\nabla}^3 n\|^2_{L^2}+\|{\nabla}^3 u\|^2_{L^2}\lesssim\mathcal E_1(t)\lesssim (1+t)^{-5}\delta_0^2+(1+t)^{-\frac72}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{eqnarray*}

    Finally, we have

    \begin{eqnarray*} \|{\nabla}^3 n_h\|_{L^2}+\|{\nabla}^3 m_h\|_{L^2}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*}

    This completes the proof of this Lemma.

    In this subsection, we first combine the above a priori estimates of (5.8), (5.9), (5.10), (5.11) and Lemma 5.2 together to give the proof of the Proposition 5.2. In deed, for any t\in[0,T] , we have shown that

    \begin{equation} \Lambda(t)\leq C\left(\delta_0+\delta_0^{\frac14}\Lambda(t)+\Lambda^2(t)\right) \leq C\delta_0^{\frac34}. \end{equation} (5.14)

    With the help of standard continuity argument, Proposition 5.2 and the smallness of \delta_0>0 , implies that \Lambda(t)\leq C\delta_0^{\frac34} for any t>0 . Moreover, we deduce the time decay estimate for (n_h, m_h) from (5.9), (5.10), (5.11), Lemma 5.2 and (5.14) that

    \begin{eqnarray*} \begin{split} &\|({\nabla}^k n_h, {\nabla}^k m_h)\|_{L^2}\lesssim \delta_0^2(1+t)^{-\frac54-\frac k2},\quad k = 0,1,\\ &\|{\nabla}^2 (n_h, m_h)\|_{L^2}\lesssim\delta_0^{\frac74}(1+t)^{-\frac94},\quad \|{\nabla}^3 (n_h, m_h)\|_{L^2}\lesssim\delta_0(1+t)^{-\frac74}. \end{split} \end{eqnarray*}

    Consequently, for any t\in[0,T] we have

    \begin{equation} \Lambda(t)\leq C\delta_0. \end{equation} (5.15)

    From (5.11) and (5.15), thus we also get that

    \begin{eqnarray*} \|{\nabla}^2 (n_h, m_h)\|_{L^2}\lesssim\delta_0^2(1+t)^{-\frac94}. \end{eqnarray*}

    For {\nabla}^3 m_h , in view of the (5.6), we see that

    \begin{eqnarray*} \begin{split} &\|{\nabla}^3 m_h(t)\|_{L^2} \\\lesssim &\int_0^{\frac t 2}(1+t-\tau)^{-\frac{11}4}\big(\|Q(\tau)\|_{L^1}+\|{\nabla}^2 Q(\tau)\|_{L^2}\big)d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} ^2Q(\tau)\|_{L^2}d\tau\\ \lesssim& \delta_0^2\bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac{11}4}(1+\tau)^{-\frac32}d\tau+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{13}4}d\tau\bigg)\\ \lesssim&\delta_0^2(1+t)^{-\frac{11}4}. \end{split} \end{eqnarray*}

    Hence, we finish the proof of the Proposition 5.1. Theorem 1.1 follows.

    Y. Chen is partially supported by the China Postdoctoral Science Foundation under grant 2019M663198, Guangdong Basic and Applied Basic Research Foundation under grant 2019A1515110733, NNSF of China under grants 11801586, 11971496 and China Scholarship Council. The research of R. Pan is partially supported by National Science Foundation under grants DMS-1516415 and DMS-1813603, and by National Natural Science Foundation of China under grant 11628103. L. Tong's research is partially supported by China Scholarship Council.

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