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Error code: DatasetGenerationError
Exception: ArrowInvalid
Message: JSON parse error: Missing a closing quotation mark in string. in row 35
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 145, in _generate_tables
dataset = json.load(f)
File "/usr/local/lib/python3.9/json/__init__.py", line 293, in load
return loads(fp.read(),
File "/usr/local/lib/python3.9/json/__init__.py", line 346, in loads
return _default_decoder.decode(s)
File "/usr/local/lib/python3.9/json/decoder.py", line 340, in decode
raise JSONDecodeError("Extra data", s, end)
json.decoder.JSONDecodeError: Extra data: line 2 column 1 (char 101492)
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1995, in _prepare_split_single
for _, table in generator:
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 148, in _generate_tables
raise e
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/packaged_modules/json/json.py", line 122, in _generate_tables
pa_table = paj.read_json(
File "pyarrow/_json.pyx", line 308, in pyarrow._json.read_json
File "pyarrow/error.pxi", line 154, in pyarrow.lib.pyarrow_internal_check_status
File "pyarrow/error.pxi", line 91, in pyarrow.lib.check_status
pyarrow.lib.ArrowInvalid: JSON parse error: Missing a closing quotation mark in string. in row 35
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1529, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1154, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1027, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1122, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1882, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2038, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
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\section{Introduction}
\quad
In this paper, we hope to investigate the decay estimate of strong solution
to the isentropic compressible Navier-Stokes equations and the full compressible
Navier-Stokes equations with large initial data in the three-dimensional whole space.
The isentropic compressible Navier-Stokes (ICNS) equations are given as:
\begin{equation*}\tag{ICNS}\label{cns}
\left\{\begin{aligned}
&\partial_t\rho+\mathop{\rm div}\nolimits(\rho u)=0,\\
&\partial_t(\rho u)+\mathop{\rm div}\nolimits(\rho u\otimes u)-\mu\Delta u-(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u+\nabla P=0,\\
& \underset{|x|\rightarrow\infty}{\lim}(\rho-1, u)(x, t)=(0,0),
\end{aligned}\right.
\end{equation*}
where $(x, t)\in \mathbb{R}^3 \times \mathbb{R}^+$.
The unknown functions $\rho, u=(u_1, u_2, u_3)$ and $P$ represent
the density, velocity and pressure, respectively.
The pressure $P$ is given by a smooth function $P=P(\rho)=A\rho^\gamma$ with the adiabatic exponent $\gamma \ge 1$, and without loss of generality, we assume $A=1$ in this paper.
The constants $\mu$ and $\lambda$ are the viscosity coefficients,
which satisfy the following conditions:
$\mu>0$, $2\mu+3\lambda \ge0$.
To complete system \eqref{cns}, the initial data is given by
\begin{equation*}
\left.(\rho, u)(x, t)\right|_{t=0}=(\rho_0(x), u_0(x)).
\end{equation*}
There are many interesting works on the isentropic compressible Navier-Stokes equations since its physical importance, let us review some results related to well-posedness theory. In the absence of vacuum, the local well-posedness for the compressible Navier-Stokes equations was proved by Nash \cite{nash}.
In the presence of vacuum, Huang, Li and Xin \cite{huang-li-xin} established the global existence and uniqueness of strong solution to system \eqref{cns} in three-dimensional space in the condition that the initial energy is small.
Recently, Li and Xin obtained similar results for the dimension two in \cite{li-xin}, in addition, they established the large time behavior of the solutions for system \eqref{cns} with small initial data but allowing large oscillations in \cite{li-xin}.
The large time behavior of the solutions to the isentropic compressible Navier-Stokes system has been studied extensively. Matsumura and Nishida \cite{nishida2} first obtained the optimal decay rate for strong solution to the compressible Navier-Stokes system, and Ponce \cite{ponce} obtained the optimal $L^p(p\ge 2)$ decay rate. For more information on the long time behavior of the isentropic compressible Navier-Stokes equations with or without external forces, we refer to \cite{duan1,li-zhang} and the references therein. All these decay results mentioned above are restricted to the perturbation framework, that is, if the initial data is a small perturbation of constant equilibrium in $L^1\cap H^3$, then $L^2$ decay of the solution to \eqref{cns} is
$$\|\rho(t)-1\|_{L^2}+\|u (t)\|_{L^2}\le C(1+t)^{-\frac34}.$$
In the perturbation setting, the approach on proving the decay for the solutions of
compressible Navier-Stokes system rely heavily on the analysis of the linearization of the system. More precisely, most of these decay results proved by combining the linear optimal decay of spectral analysis with the energy method.
From another point of view, under the assumption that the initial perturbation is bounded in $\dot H^{-s}$, Guo and Wang \cite{guo-wang} obtained the optimal decay rates of the higher-order spatial derivatives of the solution for system \eqref{cns} by using pure energy method. Specifically, they combined energy estimates with the interpolation between negative and positive Sobolev norms and obtained the time decay rates for the isentropic compressible Navier-Stokes equations and Boltzmann equation. The new method developed in \cite{guo-wang} has a wide range of applications recently, see \cite{tan-tong,tan-wang,tan-wang-tong, tan-wang-wang, tan-wu-zhou,tan-zhou,wang}. It should be noticed that all of these decay results are established under the assumption that the initial data is a small perturbation of constant equilibrium state. As far as we know, there are no decay results established by using the method coming from \cite{guo-wang} beyond the near-equilibrium regime. Recently, He, Huang and Wang \cite{he-huang-wang} established the global stability of large solution to \eqref{cns} in $\mathbb R^3$, more precisely, they obtained the decay rate
\begin{equation*}
\|\rho(t) -1\|_{H^1}+\|u(t)\|_{H^1}\le C(1+t)^{-\frac34(\frac2p-1)},
\end{equation*}
where $(\rho_0-1,u_0)\in L^p(\mathbb R^3)\cap H^2(\mathbb R^3)$ with $p\in[1,2)$. Later, we improved this decay result in \cite{gao-wei-yao}. More precisely, we not only shown that the upper decay rate of the first order spatial derivative converging to zero in $H^1-$ norm was $(1+t)^{-\frac32(\frac1p-\frac12)-\frac12}$, but also gave the lower bound of decay rate for the global solution of \eqref{cns} converging to constant equilibrium state in $L^2-$ norm was $(1+t)^{-\frac34}$ in the case $p=1$.
\textit{The first purpose in this paper is to investigate the convergence rate of the global large solution studied in \cite{he-huang-wang} when the initial data belongs to some negative Sobolev space $\dot H^{-s}$ instead of Lebesgue space $L^1$ .}
The advantages are that the negative Sobolev norms of the solution are shown to be preserved along time evolution
and the first order derivative of the solution can obtain the optimal convergence rate.
Second, we introduce the full compressible Navier-Stokes equations, which govern the motion of the compressible viscous and heat-conductive gases. The full compressible Navier-Stokes equations are written as follows
\begin{equation*}\tag{FCNS}\label{fullcns0}
\left\{\begin{aligned}
&\partial_t\rho+\mathop{\rm div}\nolimits (\rho u) =0,\\
&\partial_t(\rho u)+\mathop{\rm div}\nolimits (\rho u\otimes u)+\nabla P(\rho T)=\mathop{\rm div}\nolimits \mathbb S(u),\\
&\partial_t(\rho E)+\mathop{\rm div}\nolimits(\rho Eu)+\mathop{\rm div}\nolimits(P(\rho,T)u)=\mathop{\rm div}\nolimits(\mathbb S(u)u)+\Delta T,\\
& \underset{|x|\rightarrow\infty}{\lim}(\rho-1, u,T-1)=0,
\end{aligned}\right.
\end{equation*}
where $\rho$, $u$, $T$ represent the density, the velocity and the absolute temperature of the fluid, respectively. In this paper, we are concerned with the perfect heat conducting and viscous gases, that is, the pressure $P(\rho,T)$ is given by $P=\rho T$. $E$ represents total energy, given by $E=T+\frac{|u|^2}{2}$. $\mathbb S(u)$ is the stress tensor:
$$\mathbb S(u)= \mu(\nabla u+(\nabla u)^{'})+\lambda\mathop{\rm div}\nolimits u \ \mathbb I_{3\times3}.$$
Here $\mu$ and $\lambda$ are the coefficients of viscosity, which are assumed to be constants, and satisfying $\mu>0$, $2\mu+3\lambda\ge0$.
There are extensively literatures on system \eqref{fullcns0}. In the absence of vacuum, the global well-posedness was first established by Matsumura and Nishida \cite{nishida1} in the condition that the initial data was close to constant equilibrium in $H^3$-framework. Matsumura and Nishida \cite{nishida2} first established the $L^2$ time-decay rate of classical solutions for system \eqref{fullcns0}. For an exterior domain in $\mathbb R^3$, Kobayashi and Shibata \cite{kobayashi2} investigated the time decay rate of solutions for \eqref{fullcns0}(see \cite{kobayashi1} for further developments). For the half space in $\mathbb R^3$, the authors investigated the asymptotic behaviors of solutions for the compressible Navier-Stokes equations in \cite{kagei1} and \cite{kagei2}.
For the full compressible Navier-Stokes equations with external force, the authors \cite{duan22}
obtained the convergence rate in various norms for the
solution to the non-trivial stationary profile in the whole space, when the initial perturbation of
the non-trivial stationary solution and the potential force were small in some Sobolev norms.
In the presence of vacuum, Li and his collaborator established the global well-posedness of strong solution for system \eqref{fullcns0} in a series of papers \cite{lijinkai1,lijinkai2,lijinkai3,lijinkai4} for the one dimensional case. In the three dimensional case, Huang and Li \cite{huang-li} established global weak solution when the initial energy is small.
Wen and Zhu \cite{wen-zhu} proved that the strong solution exists globally in time if the initial mass is small for the fixed coefficients of viscosity and heat conduction.
Many other global well-posedness and the large time behavior refer to
\cite{huang-li, danchin,qin, kawashima, jiang2} and the references therein.
However, when the initial data is far away from the equilibrium, there are few results on the well-posedness and large-time behavior. In one-dimensional bounded domains, Kazhikhov and Shelukhin \cite{kazhikhov} first obtained the global existence of solutions to system \eqref{fullcns0} ; Jiang \cite{jiang}, Li and Liang \cite{li-liang} studied long-time behavior of large solution to system \eqref{fullcns0}. Very recently, He, Huang and Wang \cite{he-huang-wang2} proved the global-in-time stability of large solution for system \eqref{fullcns0} under the assumption that
the density is uniformly bounded in $C^\alpha$ for some small $0<\alpha<1$ in the whole space. More precisely, they got the uniform-in-time bound for the global solution by using some techniques about the blow-up criterion come from \cite{sun-wang, wen-zhu2, lijing2}, and they obtained decay rate
\begin{equation}\label{00}
\|u(t)\|_{H^1}+\|\rho(t)-1\|_{H^1}+\|T(t)-1\|_{H^1}\le C(1+t)^{-\frac34}.
\end{equation}
Here the initial data $(\rho_0-1,u_0)\in L^1(\mathbb R^3)\cap H^2(\mathbb R^3)$ and $T_0-1\in L^1(\mathbb R^3)\cap H^1(\mathbb R^3)$. It is obviously that the decay rate of the first order spatial derivative of solution in \eqref{00} is not optimal.
Thus the second aim of this paper is to address the following three problems
for the global solution investigated in \cite{he-huang-wang2}:
\textit{(i)obtain optimal decay rate(at least faster than $(1+t)^{-\frac{3}{4}}$) for the first order
spatial derivative of global solution;
(ii)prove that the second order spatial derivative of global solution will converge to zero;
(iii)establish the decay rate for the global solution when the initial perturbation
belongs to some negative Sobolev space $\dot H^{-s}$ instead of Lebesgue space $L^1$.}
\textbf{Notation:} In this paper, we use $H^s(s\in\mathbb R)$ to denote the usual Sobolev space with norm $\|\cdot\|_{H^s}$ and $L^p(\mathbb{R}^3)$ to denote the usual $L^p$ space with norm $\|\cdot\|_{L^p}$. $\mathcal{F}(f):=\widehat{f}$ represents the usual Fourier transform of the function $f$.
The operator $\Lambda^s, s\in\mathbb R$ defined as $$\Lambda^s f(x)=\int_{\mathbb R^3}|\xi|^s\hat f(\xi)e^{2\pi ix \cdot \xi}d \xi.$$
Then the definition of the Sobolev space is given as
$$\|f\|_{\dot H^s}:=\|\Lambda^s f\|_{L^2}=\||\xi|^s\hat f\|_{L^2}.$$
When the index $s$ is non-positive, we called the space as negative Sobolev space. For convenience, we will change the index to be $``-s"$ with $s\ge0$. For the sake of simplicity, we write $\int f dx:=\int _{\mathbb{R}^3} f dx$
and $\|(A, B)\|_X:=\| A \|_X+\| B\|_X$. We will use the notation $a\sim b$ whenever $a\le C_1b$ and $b\le C_2a$ where $C_1$ and $C_2$ are universal constants. We will employ the notation $a\lesssim b$ to mean that $a\le Cb$ for a universal constant $C>0$ that only depends on the parameters coming from the problem but independent of time, and may change from line to line.
This paper is organized as follows. In Section \ref{2}, we will state the main results of this paper and comment the analysis of the proofs. In Section \ref{3}, we will prove the main theorem of the isentropic compressible Navier-Stokes equations. In Section \ref{4}, we will prove the main two theorems of the full compressible Navier-Stokes equations. In Appendix \ref{appendix}, we present some useful inequalities, which are extensively used in this paper.
\section{Main results}\label{2}
\subsection{Main results for the isentropic compressible Navier-Stokes equations.}
\quad
In this subsection, we are concerned with the decay rate
for large solution of the system \eqref{cns}. First of all, we recall the following results obtained in \cite{he-huang-wang}, which will be used in this paper frequently.
\begin{theo}(see \cite{he-huang-wang})\label{Thm1}
Let $\mu>\frac12\lambda$, and $(\rho,u)$ be a global and smooth solution of \eqref{cns} with initial data $(\rho_0,u_0)$ where $\rho_0\ge c>0$. Suppose the admissible condition holds:
$$ \partial_tu\mid_{t=0}=-u_0\cdot\nabla u_0+\frac{1}{\rho_0}Lu_0-\frac{1}{\rho_0}\nabla \rho_0^\gamma,$$
where operator $L$ is defined by $Lu=-\mathop{\rm div}\nolimits(\mu\nabla u)-\nabla((\lambda+\mu)\mathop{\rm div}\nolimits u)$. Assume that $a:=\rho-1$, and $\sup_{t\ge0}\|\rho(t)\|_{C^\alpha}\le M$ for small $0<\alpha<1$. Then if $a_0,u_0 \in L^{p}(\mathbb{R}^3)\cap H^2(\mathbb{R}^3)$ with $p\in[1,2]$, we have\\
(1)\textbf{(Lower bound of the density)}\\
\quad There exists a positive constant $\underline{\rho}=\underline{\rho}(c,M)$
such that for all $t\ge0$
\begin{equation}\label{Lower}
\rho(t)\ge\underline{\rho}.
\end{equation}
(2)\textbf{(Uniform-in-time bounds for the regularity of the solution)}
\begin{equation}\label{uniform-bound}
\|a\|^2_{L^\infty(H^2)}+\|u\|^2_{L^\infty(H^2)}+\int_0^\infty(\|\nabla a\|^2_{H^1}+\|\nabla u\|^2_{H^2})d\tau\le C(\underline{\rho},M,\|a_0\|_{H^2},\|u_0\|_{H^2}).
\end{equation}
(3)\textbf{(Decay estimate for the solution)}
\begin{equation}\label{decay}
\|u(t)\|_{H^1}+\|a(t)\|_{H^1}\le
C(M,\|a_0\|_{L^{p}\cap H^1},\|u_0\|_{L^{p}\cap H^2})(1+t)^{-\beta(p)},
\end{equation}
where $\beta(p)=\frac34(\frac{2}{p}-1)$.
\end{theo}
Then our first main result is stated in the following theorem:
\begin{theo}\label{Thm2}
Suppose all the conditions in Theorem \ref{Thm1} hold on, except for $a_0:=\rho_0-1\in \dot H^{-s}$, and $u_0\in \dot H^{-s}$ with $s\in(0,3/2)$. Let $(\rho, u)$ be the global solution of \eqref{cns}, then it hold on for all $t\ge 0$,
\begin{equation*}
\|\Lambda^{-s} a(t)\|^2_{L^2}+\|\Lambda^{-s} u(t)\|^2_{L^2}+\|\Lambda^{-s}\dot u(t)\|_{L^2}^2\le C,
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\|a(t)\|^2_{H^1}+\|u(t)\|^2_{H^1}+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}.
\end{aligned}
\end{equation*}
For all $t\ge T_1$, there holds
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2\le C(1+t)^{-(1+s)},
\end{equation*}
where $T_1$ is constant given in Lemma \ref{diss}, and $C$ is a constant independent of time.
\end{theo}
\begin{rema}
The reason for the constraint $s<3/2$ comes from applying Lemma \ref{l2} to estimate the nonlinear terms when doing the negative Sobolev estimate via $\Lambda^{-s}$. For $s\ge 3/2$, the nonlinear estimates would not work.
\end{rema}
\begin{rema}
Compared with the usual $L^p-L^2(p\in[1, 2))$ approach of investing the optimal decay of the solution, an important feature is that the $\dot H^{-s}$ norm of the solution is preserved along time evolution, but it is difficult to prove that the $L^p$ norm of the solution can be preserved along time evolution. It should be noticed that both $\dot H^{-s}$ and $L^p$ norms enhance the decay rate of the solution. When $s\in (0,3/2)$, there is the embedding relation
$\dot H^{-s}\hookrightarrow L^p, \quad p\in(1,2).$
\end{rema}
The proof of Theorem \ref{Thm2} will be given in Section \ref{3}, we now make some comments on the analysis of the proof. The main idea of the proof is combining the energy estimates with the interpolation between negative and positive Sobolev norms as Lemma \ref{l1}. Thus it is important to prove the $\dot H^{-s}$ norm of the solution is preserved along time evolution. When deriving the negative Sobolev estimates, we use different interpolation inequalities to estimate the nonlinear terms in the case $s\in(0,1/2]$ and $s\in(1/2,3/2)$ respectively. Thus we need to split the proof of Theorem \ref{Thm2} into two parts. When $s\in(0,1/2]$, we can check that the quantity $(\Lambda^{-s} a, \Lambda^{-s}(\rho u))$ is uniform bounded respect to time
since $\int_0^t(\|\nabla u\|_{H^1}^2+\|\nabla a\|_{H^1}^2)d\tau$ is uniform bounded respect to time.
Therefore, we can establish the decay estimate
$$\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C (1+t)^{-s}, \quad s\in(0,1/2].$$
When $s\in(1/2,3/2)$, in order to derive the propagation of the negative Sobolev norms of the solution,
it is necessary to check that
$\int_0^t\|(a,u)\|_{L^2}^{s-1/2}\|\nabla(a,u)\|_{L^2}^{5/2-s}d\tau$
is uniform bounded respect to time. To achieve that, we need to improve the decay rate for the first order derivative of solution when $s\in(0,1/2]$. According to Lemma \ref{diss}, it is easy to derive that there exists a $T_1>0$, there holds
$$\|\nabla a (t)\|_{L^2}^2+\|\nabla u(t)\|_{L^2}^2
\le C(1+t)^{-(1+s)}, \quad t\ge T_1.$$
Once this decay rate are obtained, the propagation of the negative Sobolev norms of the solution in the case $s\in(1/2,3/2)$ follows by the estimates in Section \ref{3} and the uniform bound \eqref{uniform-bound}.
\subsection{Main results for the full compressible Navier-Stokes equations.}
\quad In this subsection, we state the main results of the full compressible Navier-Stokes equations \eqref{fullcns0}.
Before stating our main results, we need to introduce the following results obtained in \cite{he-huang-wang2}, which will be used frequently.
\begin{theo}(see \cite{he-huang-wang2})\label{0them}
Let $\mu>\frac12\lambda$, and $(\rho,u,T)$ be a global and smooth solution of \eqref{fullcns0} with initial data $(\rho_0,u_0,T_0)$ where $\rho_0\ge c>0$, $T_0\ge c>0$, and satisfying the admissible conditions:
\begin{equation}\label{admissible}
\begin{aligned}
& u_t|_{t=0}=-u_0\cdot\nabla u_0+\frac{1}{\rho_0}\mathop{\rm div}\nolimits(\mathbb S(u_0))-\frac{1}{\rho_0}\nabla(\rho_0 T_0),\\&
T_t|_{t=0}=-u_0\cdot \nabla T_0-T_0\mathop{\rm div}\nolimits u_0+\frac{1}{\rho_0}\mathbb S(u_0):\nabla u_0+\frac{1}{\rho_0}\Delta T_0.
\end{aligned}
\end{equation}
Assume that $(\rho, T)$ satisfies that
\begin{equation}\label{thetarho}
\|\rho,T\|_{L^{\infty}}\le M_1,\quad \|\rho\|_{C^{\alpha}}\le M_2,
\end{equation}
where $\alpha$ is a positive and sufficiently small constant. Denote that $a\overset{def}{=}\rho-1$, $\theta\overset{def}{=}T-1$, then if $(a_0,u_0)\in L^1(\mathbb R^3)\cap H^2(\mathbb R^3)$ and $\theta_0\in L^1(\mathbb R^3)\cap H^1(\mathbb R^3)$, we have \\
(1) \textbf{(Propagation of the lower bounds of the density and the temperature)}\\
\quad There exist two constants $c_1$ and $c_2$ depending on $\alpha$ and $M$ such that
$$\rho(t,x)\ge c_1>0,\quad T(t,x)\ge c_2>0.$$
(2)\textbf{(Uniform-in-time bounds for the regularity)}
\begin{equation}\label{uniform}
\|a,u\|^2_{L^\infty_tH^2}+\|\theta\|^2_{L^\infty_tH^1}+\int_0^\infty(\|(a,\nabla u)\|_{H^2}^2+\|\nabla\theta\|^2_{H^1})d\tau\le C(M,\|a_0\|_{L^1\cap H^2},\|u_0\|_{L^1\cap H^2},\|\theta_0\|_{L^1\cap H^1}),
\end{equation}
(3)\textbf{(Long time behavior of the solution)}\\
\begin{equation}\label{decay22}
\|u(t)\|_{H^1}+\|a(t)\|_{H^1}+\|\theta(t)\|_{H^1}\le C(M,\|a_0\|_{L^1\cap H^2},\|u_0\|_{L^1\cap H^2},\|\theta_0\|_{L^1\cap H^1})(1+t)^{-\frac34},
\end{equation}
where $M=\max\{M_1,M_2\}$.
\end{theo}
Now we are in the position to state our main results for the full compressible Navier-Stokes equations.
\begin{theo}\label{Thm3}
Under the assumptions of Theorem \ref{0them}, assume $(\rho, u,\theta)$ be the global solution of system \eqref{fullcns0}, then we have
\begin{equation}\label{Thm3decay}
\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\partial_ta\|_{L^2}^2+\|\partial_tu\|_{L^2}^2+\|\partial_t\theta\|_{L^2}^2\le C(1+t)^{-\frac52}, \quad t\ge T_3,
\end{equation}
where $T_3$ is a large constant given in Lemma \ref{lemma4}, and $C$ is a constant independent of time.
\end{theo}
\begin{rema}\label{nabla2theta}
Although the initial data $\theta_0\in H^1(\mathbb R^3)$ given in Theorem \ref{Thm3},
it is easy to check that $\theta$ actually belongs to $H^2(\mathbb R^3)$ duing to the
compatibility conditions \eqref{admissible}. In other words, we have
\begin{equation}\label{thetabound}
\|\nabla^2\theta\|_{L^\infty(0,\infty;L^2)}<\infty.
\end{equation}
This proof can be found in Corollary 2.1 in \cite{he-huang-wang2}.
\end{rema}
\begin{rema}
Compared with the decay rate \eqref{decay22}, our decay results not only imply that the second order spatial derivative of solution converges to zero, but also shows that the decay rate for the first order spatial derivative of solution is optimal in the sense that it coincides with the decay rate of the solution to the heat equation.
\end{rema}
\begin{rema}
By the Sobolev interpolation inequality, it is shown that the solution $(\rho,u, T)$ converges to the constant equilibrium state $(1,0,1)$ at the $L^q(2\le q\le 6)$-rate $(1+t)^{-\frac34-\frac{3q-6}{4q}}$.
\end{rema}
\begin{rema}
In our previous work \cite{gao-wei-yao}, we obtained the similar decay results on the isentropic compressible Navier-Stokes equations. In above theorem, Since the pressure $P=\rho\theta$ in the system \eqref{fullcns0}, and we only have the uniform bound $\|\theta\|_{L^\infty_t(H^1)}\le C$ in \eqref{uniform}, thus the energy estimate for the second order spatial derivative of solution is more complicated.
\end{rema}
Let us give some comments on the analysis for the proof of Theorem \ref{Thm3}.
Since the solution itself and its first order spatial derivative decay with the same $L^2-$ rate $(1+t)^{-\frac34}$,
these quantities can be small enough if the time is large.
Thus, we take the strategy of the frame of small initial data (cf.\cite{nishida1}) to establish the energy estimate:
\begin{equation}\label{ana}
\begin{aligned}
\frac{d}{dt}&\mathcal E_2^2(t)+c_0(\|\nabla^2 u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2)\\&
\lesssim Q(t)(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2)
+\|\nabla u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2+\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}\|\nabla^3u\|_{L^2},
\end{aligned}
\end{equation}
where $\mathcal E_2^2(t)$ is equivalent to $\|\nabla(a,u,\theta)\|_{H^1}^2$,
and $Q(t)$ consists of some difficult terms, such as $\|(a,u)\|_{L^\infty}$ and $\|\nabla(a,u)\|_{L^3}$.
To close the estimate, the difficulty is that we can only use the smallness of quantities
in $H^1$ norm rather than $H^2$ norm.
Thus, our idea is to apply the Sobolev interpolation inequality to control these quantities by the product
of solution itself and the second order spatial derivative. Since
the latter one is uniform bounded with respect to time(see \eqref{uniform}), $Q(t)$ is a
small quantity which appears as a prefactor in front of dissipation term.
And hence, $Q(t)$ is a small quantity actually after some large time.
The second and third terms on the righ-thand side of inequality \eqref{ana} can be controlled by the similar
method, see \eqref{df} and \eqref{df2} in detail.
Then, the terms on the right-hand side of \eqref{ana} can be absorbed into the second term
on the left-hand side of inequality \eqref{df} after a fixed large time.
Finally, we hope to perform the upper decay rate \eqref{Thm3decay} by using
the energy inequality \eqref{ana} and the time-frequency splitting method by Schonbek \cite{schonbek1}.
However, unlike the incompressible flow(cf.\cite{schonbek2, Schonbek-Wiegner}), the dissipation of density is weaker than the one of the velocity and temperature for the full compressible Navier-Stokes equations.
To obtain the dissipative estimate of density, we will weaken the coefficients of velocity and temperature dissipation, and thus one part of the dissipation of density will play a role of damping term. Finally, according to time-frequency splitting method, we can derive decay estimate \eqref{Thm3decay}.
Our third result investigates the decay rate for the global solution of system \eqref{fullcns0} in the case that the initial data $(a_0, u_0,\theta_0)\in \dot H^{-s}$ with $s\in(0,3/2)$, stated as follows
\begin{theo}\label{1theo}
Under the assumptions of Theorem \ref{0them}, but $(a_0, u_0,\theta_0)\in \dot H^{-s}$ with $s\in(0,3/2)$. Let $(\rho, u,\theta)$ be the global solution of system \eqref{fullcns0}, then for all $t\ge 0$,
\begin{equation*}
\|\Lambda^{-s} a(t)\|^2_{L^2}+\|\Lambda^{-s} u(t)\|^2_{L^2}+\|\Lambda^{-s}\dot u(t)\|_{L^2}^2+\|\Lambda^{-s}\theta(t)\|_{L^2}^2\le C,
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\|a(t)\|^2_{H^1}+\|u(t)\|^2_{H^1}+\|\theta(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}.
\end{aligned}
\end{equation*}
For $t\ge T_4$, there holds
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2+\|\nabla\theta(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \end{equation*}
where $T_4$ is given in Proposition \ref{prop4}, $C$ is a constant independent of time.
\end{theo}
The proof of Theorem \ref{1theo} is similar as the proof of Theorem \ref{Thm2} in Section \ref{3}, except that additional estimates of the temperature equation are established. And it should be noticed that we only have the uniform bound $\|\theta\|_{L^\infty_t(H^1)}\le C$ in \eqref{uniform}, thus the energy estimate for the second order spatial derivative of solution is more complicated, which is necessary for improving the decay rate for the first order derivative of solution when $s\in(0,1/2]$.
\section{The proof of the results for ICNS}\label{3}
\quad First, we need to derive the evolution of the negative Sobolev norms of the solution for the isentropic compressible Navier-Stokes equations \eqref{cns}.
\textbf{Energy evolution of negative Sobolev norms.} In what follows, we will derive the evolution of the negative Sobolev norms of the solution of system \eqref{cns}. In order to estimate the nonlinear terms, we need to restrict ourselves to that $s\in(0,3/2)$. Recalling $a:=\rho-1$, we rewrite \eqref{cns} as
\begin{equation}\label{cns1}
\left\{\begin{aligned}
&\partial_ta+\mathop{\rm div}\nolimits u=-\mathop{\rm div}\nolimits(au),\\
&\partial_t(\rho u)+\mathop{\rm div}\nolimits(\rho u\otimes u)-\mu\Delta u-(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u+\gamma\nabla a+(\gamma-1)\nabla H(\rho|1)=0,\\
\end{aligned}\right.
\end{equation}
where
\begin{equation}
H(\rho|1)=\left\{
\begin{array}{cl}
&\frac{1}{\gamma-1}(\rho^{\gamma}-1-\gamma(\rho-1)), \quad {\rm when} \quad \gamma>1,\\
&\rho\ln\rho-\rho+1, \qquad {\rm when} \quad \gamma=1.\\
\end{array}\right.
\end{equation}
\begin{lemm}\label{l3}
For $s\in(0, 1/2]$, we have
\begin{equation}\label{est11}
\begin{aligned}
\frac12&\frac{d}{dt}\int(\gamma|\Lambda^{-s} a|^2+|\Lambda^{-s}(\rho u)|^2)dx+C\int|\nabla\Lambda^{-s} u|^2dx\\&
\lesssim (\|\nabla u\|_{H^1}^2+\|\nabla a\|^2_{H^1})(\|\Lambda^{-s}(\rho u)\|_{L^2}+\|\Lambda^{-s} a\|_{L^2}+\|\nabla\Lambda^{-s} u\|_{L^2}) +\|\nabla u\|^2_{H^1}\|\nabla \rho\|_{H^1},
\end{aligned}
\end{equation}
and for $s\in(1/2,3/2)$, we have
\begin{equation}\label{est12}
\begin{aligned}
\frac12&\frac{d}{dt}\int(\gamma|\Lambda^{-s} a|^2+|\Lambda^{-s}(\rho u)|^2)dx+C\int|\nabla\Lambda^{-s} u|^2dx\\&
\lesssim \|(a,u)\|_{L^2}^{s-1/2}\|(\nabla a, \nabla u)\|_{L^2}^{5/2-s}(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s}(\rho u)\|_{L^2})\\&
\quad +\|(a,u)\|_{L^2}^{s-1/2}\|(\nabla a, \nabla u)\|_{L^2}^{3/2-s}\|(a, u)\|_{H^1}\|\nabla\Lambda^{-s} u\|_{L^2}\\&
\quad +\|u\|_{L^2}^{2s}\|\nabla u\|_{L^2}^{3-2s}\|\nabla a\|_{L^2}+\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|u\|_{L^2}^{s+1/2}\|\nabla u\|_{L^2}^{5/2-s}.
\end{aligned}
\end{equation}
\end{lemm}
\begin{proof}
Applying $\Lambda ^{-s}$ to $\eqref{cns1}_1$, $\eqref{cns1}_2$ and multiplying the resulting by $\gamma\Lambda^{-s}a$, $\Lambda^{-s}(\rho u)$ respectively, summing up and then integrating over $\mathbb R^3$ by parts, we obtain
\begin{equation}\label{est1}
\begin{aligned}
&\frac12\frac{d}{dt}\int(\gamma|\Lambda^{-s} a|^2+|\Lambda^{-s}(\rho u)|^2)dx+\mu\int|\nabla\Lambda^{-s} u|^2dx+(\mu+\lambda)\int|\mathop{\rm div}\nolimits \Lambda^{-s} u|^2dx\\&
=-\int\Lambda^{-s}\nabla\left((\gamma-1)H(\rho|1)\right)\cdot \Lambda^{-s}(\rho u)dx-\int\Lambda^{-s}\mathop{\rm div}\nolimits(\rho u\otimes u)\cdot \Lambda^{-s}(\rho u)dx\\&
\quad -\gamma\int \Lambda^{-s}\mathop{\rm div}\nolimits(au)\cdot \Lambda^{-s} adx+\mu\int\Lambda^{-s}(\Delta u)\cdot\Lambda^{-s}(au)dx\\&
\quad +(\mu+\lambda)\int\Lambda^{-s}(\nabla\mathop{\rm div}\nolimits u)\cdot\Lambda^{-s}(au)dx\\&
:=I_1+I_2+I_3+I_4+I_5.
\end{aligned}
\end{equation}
In order to estimate the nonlinear terms in the right-hand side of \eqref{est1}, we restrict the value of $s$. If $s\in(0,1/2]$, then $1/2+s/3<1$ and $3/s\ge 6$. Then thanks to \eqref{a.3}, we obtain
\begin{equation}\label{H0}
\begin{aligned}
I_1&=-\int\Lambda^{-s}\nabla\left((\gamma-1)H(\rho|1)\right)\cdot \Lambda^{-s}(\rho u)dx\\&
\lesssim \|\Lambda^{-s}(\nabla H(\rho|1))\|_{L^2}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|\nabla H(\rho|1)\|_{L^{\frac{1}{1/2+s/3}}}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|a\nabla a\|_{L^{\frac{1}{1/2+s/3}}}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|a\|_{L^{3/s}}\|\nabla a\|_{L^2}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|\nabla a\|_{L^2}^{1/2+s}\|\nabla^2a\|_{L^2}^{1/2-s}\|\nabla a\|_{L^2}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim (\|\nabla a\|_{L^2}^2+\|\nabla ^2a\|_{L^2}^2)\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|\nabla a\|^2_{H^1}\|\Lambda^{-s}(\rho u)\|_{L^2},
\end{aligned}
\end{equation}
where we have used $\underline \rho\le \rho\le M$
and $|\nabla H(\rho|1)|\lesssim |a\nabla a|$, which is not difficult to obtain by Taylor expansion.
Integrating by part and applying the uniform bound \eqref{uniform-bound}, we have
\begin{equation*}
\begin{aligned}
I_2&=\int \Lambda^{-s}(\rho u\otimes u)\cdot \Lambda^{-s}(\nabla\rho u+\rho\mathop{\rm div}\nolimits u)dx\\&
\lesssim \|\Lambda^{-s}(\rho u\otimes u)\|_{L^2}(\|\Lambda^{-s}(\nabla \rho u)\|_{L^2}+\|\Lambda^{-s}(\rho\mathop{\rm div}\nolimits u)\|_{L^2})\\&
\lesssim \|\rho\|_{L^\infty}\|u\|_{L^2}\|u\|_{L^{3/s}}(\|u\|_{L^{3/s}}\|\nabla\rho\|_{L^2}+\|\rho\|_{L^{3/s}}\|\nabla u\|_{L^2})\\&
\lesssim \|\nabla u\|_{L^2}^{1/2+s}\|\nabla^2 u\|^{1/2-s}_{L^2}(\|\nabla u\|_{L^2}^{1/2+s}\|\nabla^2u\|_{L^2}^{1/2-s}\|\nabla\rho\|_{L^2}+\|\nabla\rho\|_{L^2}^{1/2+s}\|\nabla^2\rho\|_{L^2}^{1/2-s}\|\nabla u\|_{L^2})\\&
\lesssim \|\nabla u\|^2_{H^1}\|\nabla\rho\|_{H^1}.
\end{aligned}
\end{equation*}
Similarly, it is easy to check that
\begin{equation*}
\begin{aligned}
I_3\lesssim (\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1})\|\Lambda^{-s} a\|_{L^2}.
\end{aligned}
\end{equation*}
Integrating by part and using the same method as $I_2$, one arrives at
\begin{equation*}
\begin{aligned}
I_4&=\mu\int\Lambda^{-s}(\Delta u)\Lambda^{-s}(au)dx\\&
=-\mu \int\nabla\Lambda^{-s} u\cdot\Lambda^{-s}(\nabla au+a\nabla u)dx\\&
\lesssim \|\nabla\Lambda^{-s} u\|_{L^2}(\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1}).
\end{aligned}
\end{equation*}
Similarly, it holds on
\begin{equation*}
I_5\lesssim \|\nabla\Lambda^{-s} u\|_{L^2}(\|\nabla u\|^2_{H^1}+\|\nabla a\|_{H^1}^2).
\end{equation*}
Collecting all above estimates, we deduce \eqref{est11}.
Now if $s\in (1/2,3/2)$, we shall estimate the right-hand side of \eqref{est1} in a different way. Since $s\in(1/2,3/2)$, we have $1/2+s/3<1$ and $2<3/s<6$. Then using the different Sobolev interpolation, we have
\begin{equation*}
\begin{aligned}
I_1&=\int\Lambda^{-s}(\nabla(\gamma-1)H(\rho|1))\cdot\Lambda^{-s}(\rho u)dx\\&
\lesssim \|a\|_{L^{3/s}}\|\nabla a\|_{L^2}\|\Lambda^{-s}(\rho u)\|_{L^2}\\&
\lesssim \|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2}\|\Lambda^{-s}(\rho u)\|_{L^2}.
\end{aligned}
\end{equation*}
Integrating by part, it holds on
\begin{equation*}
\begin{aligned}
I_2&=\int\Lambda^{-s}(\rho u\otimes u)\cdot\Lambda^{-s}(\nabla\rho u+a\mathop{\rm div}\nolimits u+\mathop{\rm div}\nolimits u)dx\\&
\lesssim \|\Lambda^{-s}(\rho u\otimes u)\|_{L^2}(\|\Lambda^{-s}(\nabla\rho u)\|_{L^2}+\|\Lambda^{-s}(a\mathop{\rm div}\nolimits u)\|_{L^2}+\|\Lambda^{-s}(\mathop{\rm div}\nolimits u)\|_{L^2})\\&
\lesssim \|\rho\|_{L^\infty}\|u\|_{L^{3/s}}\|u\|_{L^2}(\|u\|_{L^{3/s}}\|\nabla \rho\|_{L^2}+\|a\|_{L^{3/s}}\|\nabla u\|_{L^2}+\|\nabla\Lambda^{-s} u\|_{L^2}).
\end{aligned}
\end{equation*}
Using the Sobolev interpolation
\begin{equation*}
\|f\|_{L^{3/s}}\lesssim \|f\|_{L^2}^{s-1/2}\|\nabla f\|_{L^2}^{3/2-s}, \quad s\in(1/2,3/2)
\end{equation*}
again, it is easy to derive
\begin{equation*}
\begin{aligned}
I_2&\lesssim \|u\|_{L^2}^{2s}\|\nabla u\|_{L^2}^{3-2s}\|\nabla\rho\|_{L^2}+\|u\|_{L^2}^{s+1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla\Lambda^{-s} u\|_{L^2}\\&
\quad +\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|u\|_{L^2}^{s+1/2}\|\nabla u\|_{L^2}^{5/2-s}.
\end{aligned}
\end{equation*}
Similarly, we have
\begin{equation*}
I_3\lesssim (\|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2}+\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2})\|\Lambda^{-s} a\|_{L^2},
\end{equation*}
\begin{equation*}
I_4\lesssim \|\Lambda^{-s}\nabla u\|_{L^2}(\|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2}+\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}),
\end{equation*}
\begin{equation*}
I_5\lesssim \|\Lambda^{-s}\nabla u\|_{L^2}(\|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2}+\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}).
\end{equation*}
Collecting all above estimates together, we obtain \eqref{est12}. Thus we complete the proof of lemma.
\end{proof}
Before giving decay result when $s\in(0,1/2]$, we need to introduce the dissipation inequality obtained in \cite{he-huang-wang}.
\begin{prop}(see \cite{he-huang-wang})\label{pro1}
Under the assumptions of Theorem \ref{Thm1}, then there holds
\begin{equation}\label{est2}
\frac{d}{dt}X_1(t)+C(\|\nabla^2u\|^2_{L^2}+\|\nabla u\|^2_{L^2}+\|\nabla a\|^2_{L^2}+\|\nabla\dot u\|^2_{L^2})\le 0,
\end{equation}
where $X_1(t)\sim \|u\|_{H^1}^2+\|a\|^2_{H^1}+\|\dot u\|^2_{L^2}$.
\end{prop}
\begin{prop}\label{prop1}
Under the assumptions of Theorem \ref{Thm2}, when $s\in(0,1/2]$, then for all $t\ge 0$, we have
\begin{equation}\label{prop1est}
\begin{aligned}
&\|\Lambda^{-s} a(t)\|_{L^2}^2+ \|\Lambda^{-s} u(t)\|_{L^2}^2+ \|\Lambda^{-s} \dot u(t)\|_{L^2}^2\le C,\\&
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s},
\end{aligned}
\end{equation}
where $C$ is a constant independent of time.
\end{prop}
\begin{proof}
Integrating \eqref{est11} over $[0,t)$ and using Cauchy inequality, we get
\begin{equation}\label{331}
\begin{aligned}
\frac12&\sup_{\tau\in[0,t)}(\|\Lambda^{-s} a(\tau)\|^2_{L^2}+\|\Lambda^{-s}(\rho u)(\tau)\|^2_{L^2}) +\frac C2\int_0^t\|\nabla\Lambda^{-s} u\|^2_{L^2}d\tau\\&
\le \frac12(\|\Lambda^{-s} a_0\|^2_{L^2}+\|\Lambda^{-s}(\rho_0u_0)\|^2_{L^2})\\&
\quad +\sup_{\tau\in[0,t)}(\|\Lambda^{-s} a(\tau)\|_{L^2}+\|\Lambda^{-s}(\rho u)(\tau)\|_{L^2})\int_0^t(\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1})d\tau\\&
\quad +\epsilon(\|\nabla u\|_{L^\infty_\tau(H^1)}^2+\|\nabla a\|_{L^\infty_\tau(H^1)}^2)\int_0^t \|\Lambda^{-s}\nabla u\|_{L^2}^2d\tau\\&
\quad +C_\epsilon\int_0^t(\|\nabla u\|_{H^1}^2+\|\nabla a\|_{H^1}^2)d\tau+\|\nabla\rho\|_{L^\infty_\tau(H^1)}\int_0^t\|\nabla u\|_{H^1}^2d\tau.
\end{aligned}
\end{equation}
Notice that $\underline \rho\le\rho_0\le M$, and using Hausdorff-Young inequality \eqref{hausdorff}, we have
\begin{equation*}
\begin{aligned}
\|\rho_0u_0\|_{\dot H^{-s}}=\||\xi|^{-s}\widehat{\rho_0u_0}\|_{L^2}&\le \||\xi|^{-s}\hat\rho_0*\hat u_0\|_{L^2}\le\|\hat\rho_0\|_{L^1}\||\xi|^{-s}\hat u_0\|_{L^2}
\le \|\rho_0\|_{L^\infty}\|u_0\|_{\dot H^{-s}}
\le C\|u_0\|_{\dot H^{-s}},
\end{aligned}
\end{equation*}
this together with $u_0\in\dot H^{-s}$ implies $\|\Lambda^{-s}(\rho_0 u_0)\|_{L^2}\le C$.
Then according to the uniform bound \eqref{uniform-bound}
and choosing $\epsilon$ small enough in inequality \eqref{331}, we obtain
\begin{equation}\label{propa}
\|\Lambda^{-s} a(t)\|^2_{L^2}+\|\Lambda^{-s}(\rho u)\|^2_{L^2}\le C, \quad s\in(0,1/2].
\end{equation}
Since $\underline\rho\le\rho\le M$, it is easy to check $\widehat{\rho u}\sim \hat u$, hence we have
\begin{equation}\label{propu}
\|\Lambda^{-s} u(t)\|_{L^2}^2\le \|\Lambda^{-s}(\rho u)(t)\|_{L^2}^2\le C.
\end{equation}
Next, let us prove $\|\Lambda^{-s}\dot u(t)\|\le C$.
We have known $u\in \dot H^{-s}\cap H^2$, thus we have $u\in \dot H^{2-s}$ by Sobolev interpolation inequality. Then we have
\begin{equation}\label{prop1est1}
\||\xi|^{-s}\widehat {\Delta u}\|_{L^2}\le \||\xi|^{2-s}\hat u\|_{L^2}\le \|u\|_{\dot H^{2-s}}\le C,
\end{equation}
which implies
\begin{equation}\label{lapu}
\Delta u\in \dot H^{-s}.
\end{equation}
Similarly, we have $a\in\dot H^{1-s}$. In follows, we set ${\bm a}\overset{\Delta}{=} \rho^\gamma-1$. Observing thet ${\bm a}=(\int_0^1\gamma(t\rho+(1-t))^{\gamma-1}dt)a$. Since $\underline \rho\le \rho\le M$, it is easy to see
\begin{equation*}
\bm a\le \gamma(M+1)^{\gamma-1}a,
\end{equation*}
which implies $\hat{\bm a}\lesssim \hat a$. Thus we have
\begin{equation*}
\||\xi|^{-s}\widehat {\nabla{\bm a}}\|_{L^2}\le \||\xi|^{1-s}\hat{\bm a}\|_{L^2}\le C \||\xi|^{1-s}\hat a\|_{L^2}\le C\|a\|_{\dot H^{1-s}}\le C,
\end{equation*}
which means
\begin{equation}\label{aa}
\nabla {\bm a}\in \dot H^{-s}.
\end{equation}
Since $\rho\dot u+\nabla {\bm a}=\mu\Delta u+(\lambda+\mu)\nabla\mathop{\rm div}\nolimits u$, combing \eqref{lapu} and \eqref{aa}, we have $\rho\dot u\in \dot H^{-s}$. Using $\underline\rho\le\rho\le M$ again, we deduce that
\begin{equation*}
\dot u\in \dot H^{-s}.
\end{equation*}
So far, we have
\begin{equation}\label{dotu}
\|\Lambda^{-s} a(t)\|_{L^2}+\|\Lambda^{-s} u(t)\|_{L^2}+\|\Lambda^{-s}\dot u(t)\|_{L^2}\le C, \quad s\in(0,1/2].
\end{equation}
According to \eqref{a.2}, we have
\begin{equation}\label{001}
\|\nabla^{l+1}f\|_{L^2}\ge C\|\Lambda^{-s} f\|_{L^2}^{-\frac{1}{l+s}}\|\nabla^lf\|_{L^2}^{1+\frac{1}{l+s}}.
\end{equation}
By this fact and the uniform bound \eqref{uniform-bound} and \eqref{dotu}, we deduce
\begin{equation*}
\begin{aligned}
\|\nabla a\|_{L^2}^2+\|\nabla u\|_{L^2}^2+\|\nabla^2u\|_{L^2}^2+\|\nabla \dot u\|_{L^2}^2
\ge C(\|a\|_{H^1}^2+\|u\|_{H^1}^2+\|\dot u\|_{L^2}^2)^{1+\frac1s}.
\end{aligned}
\end{equation*}
Thus, thanks to Proposition \ref{pro1} and notice that the estimate \eqref{est2} has nothing to do with the initial data, we obtain
\begin{equation*}
\frac{d}{dt}(\|a\|_{H^1}^2+\|u\|_{H^1}^2+\|\dot u\|_{L^2}^2)+C(\|a\|_{H^1}^2+\|u\|_{H^1}^2+\|\dot u\|_{L^2}^2)^{1+\frac1s}\le 0.
\end{equation*}
Solving this inequality directly, and using \eqref{uniform-bound}, we get
\begin{equation*}
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s},
\end{equation*}
which completes the proof.
\end{proof}
Before deriving the propagation of the negative Sobolev norms of the solution for $s\in(1/2,3/2)$, we need to improve the decay estimate of the first order spatial derivative of the solution for $s\in(0,1/2]$.
And hence, we need to introduce the following lemma.
\begin{lemm}\label{diss}
Under the assumptions of Theorem \ref{Thm2}, and $s\in(0,1/2]$, we define
\begin{equation*}
\mathcal{E}^2_1(t)\stackrel{\Delta}{=}\|\nabla u\|^2_{H^1}+P'(1)\|\nabla a\|^2_{H^1}+2\delta_0\int\nabla u\cdot\nabla^2a dx.
\end{equation*}
Then there exists a large time $T_1$, such that
\begin{equation}\label{1111}
\frac{d}{dt}\mathcal{E}^2_1(t)+c_* (\|\nabla^2 u\|_{H^1}^2+\|\nabla^2 a\|_{L^2}^2) \le 0
\end{equation}
holds on for all $t\ge T_1$.
Here $c_*=\min{\{\mu,\delta_0P'(1)}\}$, and $\delta_0$ is a small constant.
\end{lemm}
\begin{proof}
To obtain \eqref{1111}, we need to introduce the following energy estimate coming from Lemma 2.4 in \cite{gao-wei-yao}
\begin{equation*}
\begin{split}
&\frac{d}{dt}\left\{\frac12\|\nabla u\|_{H^1}^2+\frac{P'(1)}{2}\|\nabla a\|_{H^1}^2
+\delta_0 \int\nabla u\cdot\nabla^2a dx \right\}
+\frac{3\mu}{4}\|\nabla^2 u\|_{H^1}^2 +\frac{3\delta_0}{4}P'(1)\|\nabla^2 a\|_{L^2}^2\\
&\le C(\|a\|^{\frac14}_{L^2}+\|u\|^{\frac14}_{L^2}+\|\nabla u\|^{\frac14}_{L^2}+\|a \|_{H^1}+\|u \|_{H^1})
(\|\nabla^2 u\|^2_{H^1}+\|\nabla^2 a\|^2_{L^2}).
\end{split}
\end{equation*}
According to the decay result \eqref{prop1est} obtained in Proposition \ref{prop1}, one may conclude that
\begin{equation*}
\|a\|^{\frac14}_{L^2}+\|u\|^{\frac14}_{L^2}+\|\nabla u\|^{\frac14}_{L^2}+\|a \|_{H^1}+\|u \|_{H^1}\le C(1+t)^{-\frac s8},
\end{equation*}
and thus, there exists a large time $T_1>0$ such that
\begin{equation*}
\|a\|^{\frac14}_{L^2}+\|u\|^{\frac14}_{L^2}+\|\nabla u\|^{\frac14}_{L^2}+\|a \|_{H^1}+\|u \|_{H^1}\le \frac14\min\{\mu,\delta_0P'(1)\}
\end{equation*}
holds on for all $t\ge T_1$. Therefore, we obtain
\begin{equation*}
\frac{d}{dt}\{\|\nabla u\|_{H^1}^2+P'(1)\|\nabla a\|_{H^1}^2
+2\delta_0\int\nabla u\cdot\nabla^2a dx \}+\mu\|\nabla^2 u\|_{H^1}^2 +\delta_0P'(1)\|\nabla^2 a\|_{L^2}^2\le 0.
\end{equation*}
Taking $c_*=\min{\{\mu,\delta_0P'(1)}\}$, it holds on
\begin{equation*}
\frac{d}{dt}\mathcal{E}^2_1(t)+c_* (\|\nabla^2 u\|_{H^1}^2+\|\nabla^2 a\|_{L^2}^2) \le 0.
\end{equation*}
By virtue of the smallness of $\delta_0$, there exist two constants $c_3$ and $C_3$(independent of time), such that
\begin{equation}\label{E1}
c_3(\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1})\le\mathcal{E}^2_1(t)\le C_3(\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1}).
\end{equation}
Therefore, we complete the proof of this lemma.
\end{proof}
With Lemma \ref{diss} at hand, we can improve the decay estimate of the first derivative of the solution, and also establish the decay estimate of the second derivative of the solution when $s\in(0,1/2]$.
\begin{prop}\label{prop5}
Under the assumptions of Theorem \ref{Thm2}, and $s\in(0,1/2]$, then, it holds on for all $t\ge T_1$
\begin{equation}\label{hidecay}
\|\nabla u(t)\|_{H^1}^2+\|\nabla a(t)\|_{H^1}^2\le C(1+t)^{-(1+s)},
\end{equation}
where $C$ is a constant independent of time, and $T_1$ is given in Lemma \ref{diss}.
\end{prop}
\begin{proof}
According to \eqref{a.2}, it is easy to check that
\begin{equation*}
\|\nabla^{l+1}f\|_{L^2}\ge C\|\Lambda^{-s} f\|_{L^2}^{-\frac{1}{l+s}}\|\nabla^lf\|_{L^2}^{1+\frac{1}{l+s}},
\end{equation*}
this together with \eqref{uniform-bound} and \eqref{dotu} gives rise to
\begin{equation*}
\|\nabla^2 u\|^2_{H^1}+\|\nabla^2 a\|^2_{L^2}\ge C(\|\nabla u\|^2_{H^1}+\|\nabla a\|^2_{H^1})^{1+\frac{1}{1+s}}.
\end{equation*}
By Lemma \ref{diss} and the equivalent condition \eqref{E1}, we have
\begin{equation*}
\frac{d}{dt}\mathcal{E}^2_1(t)+C(\mathcal{E}^2_1(t))^{1+\frac{1}{1+s}}\le 0 \quad for \quad t\ge T_1.
\end{equation*}
Solving this inequality directly and using \eqref{uniform-bound}, it holds on
\begin{equation}\label{decay2}
\mathcal{E}^2_1(t)\le \left(\mathcal{E}^2_1(T_1)^{-\frac{2}{1+s}}+\frac{C(t-T_1)}{1+s}\right)^{-(1+s)}\le C(1+t)^{-(1+s)}, \quad t\ge T_1.
\end{equation}
Consequently, the combination of the equivalent condition \eqref{E1} and \eqref{decay2}, yields that for $s\in(0,1/2]$,
\begin{equation*}
\|\nabla u(t)\|_{H^1}^2+\|\nabla a(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \quad t\ge T_1,
\end{equation*}
which completes the proof.
\end{proof}
Now we are in the position to establish the decay estimate for the solution of \eqref{cns} with initial data $(a_0,u_0)\in \dot H^{-s}(s\in(1/2,3/2))$.
\begin{prop}\label{prop6}
Under the assumptions of Theorem \ref{Thm2}, when $s\in(1/2,3/2)$, then for all $t\ge0$, we have
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+ \|\Lambda^{-s} u(t)\|_{L^2}^2+ \|\Lambda^{-s} \dot u(t)\|_{L^2}^2\le C,
\end{equation*}
and
\begin{equation*}
\|a(t)\|^2_{H^1}+\|u(t)\|^2_{H^1}+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}.
\end{equation*}
For $t\ge T_1$, there holds
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2\le C(1+t)^{-(1+s)},
\end{equation*}
where $T_1$ is a fixed large time given in Lemma \ref{diss}, and $C$ is a constant independent of time.
\end{prop}
\begin{proof}
Notice that $a_0, u_0\in\dot H^{-s}\cap L^2 \subset\dot H^{-1/2}$ when $s\in(1/2,3/2)$. Then we derive from what we have proved in \eqref{prop1est} and \eqref{hidecay} with $s=1/2$ that the following decay results:
\begin{equation}\label{decay3}
\begin{aligned}
&\|a(t)\|_{L^2}+\|u(t)\|_{L^2}\le C(1+t)^{-\frac14}, \quad t\ge0,\\&
\|\nabla a(t)\|_{H^1}+\|\nabla u(t)\|_{H^1}\le C(1+t)^{-\frac34}, \quad t\ge T_1.
\end{aligned}
\end{equation}
Substituting \eqref{decay3} into \eqref{est12}, it is easy to derive that for $s\in(1/2,3/2)$,
\begin{equation*}
\begin{aligned}
&\sup_{\tau\in[0,t]}(\|\Lambda^{-s} a(\tau)\|^2_{L^2}+\|\Lambda^{-s}(\rho u)(\tau)\|^2_{L^2})+\int_0^t\|\nabla\Lambda^{-s} u(\tau)\|^2_{L^2}d\tau\\&
\quad \lesssim \|\Lambda^{-s} a_0\|_{L^2}^2+\|\Lambda^{-s}(\rho_0u_0)\|_{L^2}^2\\&
\qquad +\int_0^{T_1}(1+\tau)^{-\frac12}(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s}(\rho u)\|_{L^2}+\|\nabla \Lambda^{-s} u\|_{L^2})d\tau+\int_0^{T_1}(1+\tau)^{-1}d\tau\\&
\qquad +\int_{T_1}^t(1+\tau)^{\frac12s-\frac74}(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s}(\rho u)\|_{L^2})d\tau+\int_{T_1}^t(1+\tau)^{\frac12s-\frac54}\|\nabla\Lambda^{-s} u\|_{L^2}d\tau\\&
\qquad +\int_{T_1}^t(1+\tau)^{s-3}d\tau.
\end{aligned}
\end{equation*}
Notice that we can fix the value of $T_1$ according to Lemma \ref{diss}, then direct calculation derives that for all $t\ge0$, there holds
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+\|\Lambda^{-s}(\rho u)(t)\|_{L^2}^2\le C \quad for \quad s\in(1/2,3/2).
\end{equation*}
Then repeating the progress in the proof of Proposition \ref{prop1} for $s\in(1/2,3/2)$, we obtain that for all $t\ge0$
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+ \|\Lambda^{-s} u(t)\|_{L^2}^2+ \|\Lambda^{-s} \dot u(t)\|_{L^2}^2\le C, \quad s\in(1/2,3/2),
\end{equation*}
and
\begin{equation*}
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}, \quad s\in(1/2,3/2).
\end{equation*}
Similar to Proposition \ref{prop5} for $s\in(1/2,3/2)$, it is easy to derive that for all $t\ge T_1$
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \quad s\in(1/2, 3/2).
\end{equation*}
The combination of above two estimates completes the proof.
\end{proof}
\textbf{Proof of Theorem \ref{Thm2}:}
With the uniform bounds and decay estimates stated in Propositions \ref{prop1}, \ref{prop5} and \ref{prop6},
we can complete the the proof of Theorem \ref{Thm2}.
\section{The proof of the main results for FCNS}\label{4}
Denoting $a:=\rho-1$, $\theta:=T-1$, we rewrite \eqref{fullcns0} in the perturbation form as follows
\begin{equation}\label{fullcns}
\left\{\begin{aligned}
&\partial_ta+\mathop{\rm div}\nolimits u =S_1,\\
&\partial_tu-\mu\Delta u-(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u+\nabla a+\nabla\theta=S_2,\\
&\partial_t\theta-\Delta\theta+\mathop{\rm div}\nolimits u=S_3,
\end{aligned}\right.
\end{equation}
where the nonlinear terms $S_1$, $S_2$ and $S_3$ are defined by
\begin{equation}\label{S}
\left\{\begin{aligned}
&S_1:=-a\mathop{\rm div}\nolimits u-u\cdot\nabla a,\\
&S_2:=-u\cdot\nabla u-h(a)[\mu\Delta u+(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u]+h(a)(\nabla a+\nabla \theta)-g(a)\nabla(a\theta),\\
&S_3:=-\mathop{\rm div}\nolimits(\theta u)+g(a)(2\mu|D u|^2+\lambda(\mathop{\rm div}\nolimits u)^2)-h(a)\Delta\theta,
\end{aligned}\right.
\end{equation}
where
\begin{equation*}
h(a):=\frac{a}{1+a}, \quad g(a):=\frac{1}{1+a}.
\end{equation*}
\subsection{The proof of Theorem \ref{Thm3}.}
\quad
First, we shall establish the estimate for the first order spatial derivative of the solution as follows.
\begin{lemm}\label{l4}
Under the assumptions of Theorem \ref{0them}, the global solution $(a, u, \theta)$ of Cauchy problem \eqref{fullcns} has the estimate
\begin{equation}\label{l4est}
\begin{aligned}
\frac12&\frac{d}{dt}\int(|\nabla a|^2+|\nabla u|^2+|\nabla\theta|^2)dx+\mu\int|\nabla^2u|^2dx+(\mu+\lambda)\int|\nabla\mathop{\rm div}\nolimits u|^2dx+\int|\nabla^2\theta|^2dx\\&
\le C(\|(a,u,\theta)\|_{H^1}+\|\nabla(a, u)\|_{L^2}^{\frac12})(\|\nabla^2u\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2).
\end{aligned}
\end{equation}
\end{lemm}
\begin{proof}
Applying $\nabla$ operator to $\eqref{fullcns}_1$, multiplying by $\nabla a$ and integrating over $\mathbb R^3$, we have
\begin{equation*}
\frac12\frac{d}{dt}\int |\nabla a|^2dx+\int\nabla\mathop{\rm div}\nolimits u\cdot\nabla adx=\int\nabla S_1\cdot\nabla adx.
\end{equation*}
Integrating by part and applying H\"{o}lder inequality, we obtain
\begin{equation}\label{m1}
\frac12\frac{d}{dt}\int |\nabla a|^2dx-\int\nabla u\cdot\nabla ^2adx\le \|S_1\|_{L^2}\|\nabla^2a\|_{L^2}.
\end{equation}
Direct calculation gives
\begin{equation}\label{n8}
\|S_1\|_{L^2}\le \|a\|_{L^3}\|\mathop{\rm div}\nolimits u\|_{L^6}+\|u\|_{L^3}\|\nabla a\|_{L^6}\le C(\|a\|_{H^1}+\|u\|_{H^1})(\|\nabla^2u\|_{L^2}+\|\nabla^2a\|_{L^2}),
\end{equation}
this together with \eqref{m1} yields
\begin{equation}\label{m7}
\frac12\frac{d}{dt}\int |\nabla a|^2dx-\int\nabla u\cdot\nabla ^2adx\le C(\|a\|_{H^1}+\|u\|_{H^1})(\|\nabla^2u\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2).
\end{equation}
Second, applying $\nabla$ operator to $\eqref{fullcns}_2$, and then multiplying by $\nabla u$ and integrating over $\mathbb R^3$, we have
\begin{equation*}
\frac12
\frac{d}{dt}\int|\nabla u|^2dx+\mu\int|\nabla^2u|^2dx
+\!(\mu\!+\lambda)\!\!\int|\nabla\mathop{\rm div}\nolimits u|^2dx
\!+\!\!\int\nabla^2a\cdot\nabla udx\!+\!\!\int\nabla^2\theta\cdot\nabla udx
=\!\int\nabla S_2\cdot\nabla udx,
\end{equation*}
which yields directly
\begin{equation*}
\begin{aligned}
\frac12&\frac{d}{dt}\!\!\int|\nabla u|^2dx+\mu\!\!\int|\nabla^2u|^2dx
\!+(\mu+\lambda)\!\!\int|\nabla\mathop{\rm div}\nolimits u|^2dx+\!\!\int\nabla^2a\cdot\nabla udx
\!+\!\!\int\nabla^2\theta\cdot\nabla udx
\le \|S_2\|_{L^2}\|\nabla^2u\|_{L^2}.
\end{aligned}
\end{equation*}
Using Sobolev inequality, it is easy to check that
\begin{equation}\label{m2}
\|u\cdot\nabla u\|_{L^2}\le\|u\|_{L^3}\|\nabla u\|_{L^6}\le C\|u\|_{H^1}\|\nabla^2u\|_{L^2}.
\end{equation}
Recalling the lower bound of density, and using Sobolev inequality and \eqref{uniform}, we obtain
\begin{equation}
\|\frac{a}{1+a}[\mu\Delta u+(\mu+\lambda)\mathop{\rm div}\nolimits u]\|_{L^2}
\le C\|a\|_{L^\infty}\|\nabla^2u\|_{L^2}
\le C\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}
\le C\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}.
\end{equation}
Similarly, it holds on
\begin{equation}\label{m3}
\|\frac{a}{1+a}(\nabla a+\nabla\theta)\|_{L^2}\le C\|a\|_{L^3}(\|\nabla a\|_{L^6}+\|\nabla\theta\|_{L^6})\le C
\|a\|_{H^1}(\|\nabla^2a\|_{L^2}+\|\nabla^2\theta\|_{L^2}),
\end{equation}
and
\begin{equation}\label{m4}
\|\frac{1}{1+a}\nabla(a\theta)\|_{L^2}
\le C(\|a\|_{L^3}\|\nabla\theta\|_{L^6}+\|\theta\|_{L^3}\|\nabla a\|_{L^6})
\le C(\|a\|_{H^1}+\|\theta\|_{H^1})(\|\nabla^2\theta\|_{L^2}+\|\nabla^2a\|_{L^2}).
\end{equation}
The combination of \eqref{m2}-\eqref{m4} gives
\begin{equation*}
\|S_2\|_{L^2}\|\nabla^2u\|_{L^2}
\le C(\|u\|_{H^1}+\|a\|_{H^1}+\|\theta\|_{H^1}
+\|\nabla a\|_{L^2}^{\frac12})(\|\nabla^2\theta\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2u\|_{L^2}^2),
\end{equation*}
which implies
\begin{equation}\label{m8}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla u|^2dx+\mu\int|\nabla^2u|^2dx+(\mu+\lambda)\int|\nabla\mathop{\rm div}\nolimits u|^2dx+\int\nabla^2a\cdot\nabla udx+\int\nabla^2\theta\cdot\nabla udx\\&
\le C(\|u\|_{H^1}+\|a\|_{H^1}+\|\theta\|_{H^1}+\|\nabla a\|_{L^2}^{\frac12})(\|\nabla^2\theta\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2u\|_{L^2}^2).
\end{aligned}
\end{equation}
Last, applying $\nabla$ operator to $\eqref{fullcns}_3$,
multiplying by $\nabla \theta$ and integrating over $\mathbb R^3$, we have
\begin{equation*}
\frac12\frac{d}{dt}\int|\nabla\theta|^2dx+\int|\nabla^2\theta|^2dx+\int\nabla\mathop{\rm div}\nolimits u\cdot\nabla\theta dx=\int\nabla S_3\cdot\nabla\theta dx.
\end{equation*}
which integrating by part yields directly
\begin{equation}\label{m5}
\frac12\frac{d}{dt}\int|\nabla\theta|^2dx+\int|\nabla^2\theta|^2dx-\int\nabla u\cdot\nabla^2\theta dx\le \|S_3\|_{L^2}\|\nabla^2\theta\|_{L^2}.
\end{equation}
According to \eqref{uniform} and routine calculation, it is easy to check that
\begin{equation}\label{p2}
\begin{aligned}
\|S_3\|_{L^2}&\le\|\theta\|_{L^3}\|\mathop{\rm div}\nolimits u\|_{L^6}+\|u\|_{L^3}\|\nabla\theta\|_{L^6}+\|\nabla u\|_{L^3}\|\nabla u\|_{L^6}+\|a\|_{L^\infty}\|\nabla^2\theta\|_{L^2}\\&
\le C(\|\theta\|_{H^1}+\|u\|_{H^1})(\|\nabla^2u\|_{L^2}+\|\nabla^2\theta\|_{L^2})+C\|\nabla u\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}\\&
\quad +C\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12}\|\nabla^2\theta\|_{L^2}\\&
\le C(\|\theta\|_{H^1}+\|u\|_{H^1}+\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12})(\|\nabla^2u\|_{L^2}+\|\nabla^2\theta\|_{L^2}),
\end{aligned}
\end{equation}
which implies that
\begin{equation}\label{m6}
\|S_3\|_{L^2}\|\nabla^2\theta\|_{L^2}\le C(\|\theta\|_{H^1}+\|u\|_{H^1}+\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12})(\|\nabla^2u\|_{L^2}^2+\|\nabla^2\theta\|^2_{L^2}).
\end{equation}
The combination of \eqref{m5} and \eqref{m6} gives immediately
\begin{equation}\label{m9}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla\theta|^2dx+\int|\nabla^2\theta|^2dx-\int\nabla u\cdot\nabla^2\theta dx\\&
\le C(\|\theta\|_{H^1}+\|u\|_{H^1}+\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12})(\|\nabla^2u\|^2_{L^2}+\|\nabla^2\theta\|^2_{L^2}).
\end{aligned}
\end{equation}
Summing \eqref{m7}, \eqref{m8} and \eqref{m9}, we obtain
\begin{equation*}
\begin{aligned}
\frac12&\frac{d}{dt}\int(|\nabla a|^2+|\nabla u|^2+|\nabla\theta|^2)dx+\mu\int|\nabla^2u|^2dx+(\mu+\lambda)\int|\nabla\mathop{\rm div}\nolimits u|^2dx+\int|\nabla^2\theta|^2dx\\&
\le C(\|(a,u,\theta)\|_{H^1}
+\|\nabla(a, u)\|_{L^2}^{\frac12})(\|\nabla^2u\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2).
\end{aligned}
\end{equation*}
Therefore, we completes the proof.
\end{proof}
Next, we shall establish the energy estimate for the second order spatial derivative of the solution.
\begin{lemm}\label{l5}
Under the assumptions of Theorem \ref{0them}, the global solution $(a, u, \theta)$ of Cauchy problem \eqref{fullcns} has the estimate
\begin{equation}\label{l5est}
\begin{aligned}
\frac12&\frac{d}{dt}\int(|\nabla^2a|^2+|\nabla^2u|^2+|\nabla^2\theta|^2)dx+\mu\int|\nabla^3u|^2dx+(\mu+\lambda)\int|\nabla^2\mathop{\rm div}\nolimits u|^2dx+\int|\nabla^3\theta|^2dx\\&
\le C(\|\nabla u\|_{L^2}^{\frac14}+\|\nabla (u,a)\|_{L^2}^{\frac12}+\|(u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14})(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2).
\end{aligned}
\end{equation}
\end{lemm}
\begin{proof}
First, applying $\nabla^2$ operator to $\eqref{fullcns}_1$, and then multiplying by $\nabla^2a$ and integrating over $\mathbb R^3$, we get
\begin{equation}\label{s1}
\frac12\frac{d}{dt}\int|\nabla^2a|^2dx+\int\nabla^2\mathop{\rm div}\nolimits u\cdot\nabla^2adx\le \|\nabla^2S_1\|_{L^2}\|\nabla^2a\|_{L^2}.
\end{equation}
Recall that $S_1=-a\mathop{\rm div}\nolimits u-u\cdot\nabla a$, direct calculation shows
$$\nabla^2(a\mathop{\rm div}\nolimits u)=a\nabla^2\mathop{\rm div}\nolimits u+2\nabla a\nabla\mathop{\rm div}\nolimits u+\nabla^2a\mathop{\rm div}\nolimits u,$$
it follows that
\begin{equation}\label{s2}
\begin{aligned}
\|&\nabla^2(a\mathop{\rm div}\nolimits u)\|_{L^2}\|\nabla^2a\|_{L^2}\\&
\le (\|a\|_{L^\infty}\|\nabla^2\mathop{\rm div}\nolimits u\|_{L^2}+\|\nabla a\|_{L^3}\|\nabla\mathop{\rm div}\nolimits u\|_{L^6})\|\nabla^2a\|_{L^2}+\|\mathop{\rm div}\nolimits u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2\\&
\le C(\|a\|_{L^\infty}+\|\nabla a\|_{L^3})(\|\nabla^3u\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2)+\|\mathop{\rm div}\nolimits u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2.
\end{aligned}
\end{equation}
By routine checking, it holds on
$$\nabla^2(u\cdot\nabla a)=u\cdot\nabla(\nabla^2a)+2\nabla u\cdot\nabla^2a+\nabla^2u\cdot\nabla a.$$
Integrating by part, we get
$$\int u\cdot\nabla(\nabla^2a)\cdot\nabla^2adx=\int u\cdot\nabla(\frac12|\nabla^2a|^2)dx=-\frac12\int(\mathop{\rm div}\nolimits u)|\nabla^2a|^2dx,$$
and hence, we obtain
\begin{equation}\label{s3}
\begin{aligned}
|\int\nabla^2(u\cdot\nabla a)\cdot\nabla^2adx|&\le C(\|\nabla u\|_{L^\infty}\|\nabla^2a\|_{L^2}+\|\nabla^2u\|_{L^6}\|\nabla a\|_{L^3})\|\nabla^2a\|_{L^2}\\&\
\quad +C\|\mathop{\rm div}\nolimits u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2\\&
\le C\|\nabla u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2+C\|\nabla a\|_{L^3}\|\nabla^3u\|_{L^2}\|\nabla^2a\|_{L^2}.
\end{aligned}
\end{equation}
The combination of \eqref{s1}, \eqref{s2} and \eqref{s3} yields
\begin{equation*}
|\int\nabla^2S_1\cdot\nabla^2adx|\le C(\|\nabla a\|_{L^3}+\|a\|_{L^\infty}+\|\nabla u\|_{L^\infty})\|\nabla^2a\|_{L^2}^2+C(\|\nabla a\|_{L^3}+\|a\|_{L^\infty})\|\nabla^3u\|_{L^2}^2.
\end{equation*}
Thanks to the Sobolev inequality and the uniform estimate \eqref{uniform}, it follows that
$$\|a\|_{L^\infty}+\|\nabla a\|_{L^3}\le \|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12}\le C\|\nabla a\|_{L^2}^{\frac12},$$
and
\begin{equation}\label{df}
\|\nabla u\|_{L^\infty}\|\nabla^2a\|_{L^2}^2\le C\|\nabla u\|_{L^2}^{\frac14}\|\nabla^3u\|_{L^2}^{\frac34}\|\nabla^2a\|_{L^2}^{\frac54}\|\nabla^2a\|_{L^2}^{\frac34}\le C\|\nabla u\|_{L^2}^{\frac14}(\|\nabla^3u\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2).
\end{equation}
The combination of above estimates yields
\begin{equation}\label{n5}
\frac12\frac{d}{dt}\int |\nabla^2a|^2dx-\int\nabla^2u\cdot\nabla^3adx\le (\|\nabla a\|_{L^2}^{\frac12}+\|\nabla u\|_{L^2}^{\frac14})(\|\nabla^2a\|_{L^2}^2+\|\nabla^3u\|_{L^2}^2).
\end{equation}
Second, applying $\nabla^2$ operator to $\eqref{fullcns}_2$, multiplying by $\nabla^2u$
and integrating over $\mathbb R^3$, we get
\begin{equation}\label{n1}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla^2u|^2dx+\mu\int|\nabla^3u|^2dx+(\mu+\lambda)\int|\nabla^2\mathop{\rm div}\nolimits u|^2dx+\int\nabla^3a\cdot\nabla^2udx+\int\nabla^3\theta\cdot\nabla^2udx\\&
=\int\nabla^2S_2\cdot\nabla^2udx=-\int\nabla S_2\cdot\nabla^3udx.
\end{aligned}
\end{equation}
By routine checking, it is easy to see
\begin{equation*}
\begin{aligned}
\nabla S_2&=-\nabla u\cdot\nabla u-u\cdot\nabla^2 u-\frac{a}{1+a}[\mu\nabla\Delta u+(\mu+\lambda)\nabla^2\mathop{\rm div}\nolimits u]+\frac{\nabla a}{(1+a)^2}[\mu\Delta u+(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u]\\&
\quad -\frac{a}{1+a}(\nabla^2a+\nabla^2\theta)+\frac{\nabla a}{(1+a)^2}(\nabla a+\nabla\theta)-\frac{1}{1+a}\nabla^2(a\theta)+\frac{\nabla a}{(1+a)^2}\nabla(a\theta).
\end{aligned}
\end{equation*}
By using Sobolev inequality, it holds on
\begin{equation*}
\begin{aligned}
\|\nabla S_2\|_{L^2}&\le\|\nabla u\|_{L^3}\|\nabla u\|_{L^6}+\|u\|_{L^3}\|\nabla^2u\|_{L^6}+\|a\|_{L^\infty}\|\nabla^3u\|_{L^2}+\|\nabla a\|_{L^3}\|\nabla^2u\|_{L^6}\\&
\quad +\|a\|_{L^\infty}(\|\nabla^2a\|_{L^2}+\|\nabla^2\theta\|_{L^2})+\|\nabla a\|_{L^3}(\|\nabla a\|_{L^6}+\|\nabla\theta\|_{L^6})+\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}\\&
\quad +\|a\|_{L^\infty}\|\nabla^2\theta\|_{L^2}+\|\theta\|_{L^6}\|(\nabla a)^2\|_{L^3}+\|a\|_{L^\infty}\|\nabla a\|_{L^3}\|\nabla\theta\|_{L^6}\\&
\le C(\|\nabla (u,a)\|_{L^3}+\|u\|_{L^3}+\|a\|_{L^\infty}+\|a\|_{L^\infty}\|\nabla a\|_{L^3})(\|\nabla ^2u\|_{H^1}+\|\nabla^2a\|_{L^2}+\|\nabla^2\theta\|_{L^2})\\&
\quad +\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}+C\|\nabla\theta\|_{L^2}\|\nabla^2a\|_{L^2}^2.
\end{aligned}
\end{equation*}
By virtue of Sobolev inequality and the uniform bound \eqref{uniform}, we see
\begin{equation*}
\begin{aligned}
\|&\nabla (u,a)\|_{L^3}+\|u\|_{L^3}+\|a\|_{L^\infty}+\|a\|_{L^\infty}\|\nabla a\|_{L^3}\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}^{\frac12}+\|u\|_{H^1}+\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12}\|\nabla a\|_{H^1})\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}+\|u\|_{H^1}+\|\nabla a\|_{L^2}^{\frac12}),
\end{aligned}
\end{equation*}
and
$$\|\nabla\theta\|_{L^2}\|\nabla^2a\|_{L^2}^2\le C\|\nabla\theta\|_{L^2}\|\nabla^2a\|_{L^2}.$$
Hence, we have
\begin{equation}\label{n2}
\|\nabla S_2\|_{L^2}
\le C(\|\nabla u\|_{L^2}^{\frac12}\!+\!\|u\|_{H^1}
\!+\!\|\nabla a\|_{L^2}^{\frac12}\!+\!\|\nabla\theta\|_{L^2})
(\|\nabla ^2u\|_{H^1}
\!+\!\|\nabla^2a\|_{L^2}+\|\nabla^2\theta\|_{L^2})
\!+\!\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}.
\end{equation}
The combination of \eqref{n1} and \eqref{n2} gives rise to
\begin{equation}\label{n3}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla^2u|^2dx+\mu\int|\nabla^3u|^2dx+(\mu+\lambda)\int|\nabla^2\mathop{\rm div}\nolimits u|^2dx+\int\nabla^3a\cdot\nabla^2udx+\int\nabla^3\theta\cdot\nabla^2udx\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}\!+\!\|u\|_{H^1}+\|\nabla a\|_{L^2}^{\frac12}\!+\!\|\nabla\theta\|_{L^2})
(\|\nabla ^2u\|_{H^1}^2\!+\!\|\nabla^2a\|_{L^2}^2\!+\!\|\nabla^2\theta\|_{L^2}^2)
\!+\!\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}\|\nabla^3u\|_{L^2}.
\end{aligned}
\end{equation}
For the last term on the right-hand side of above inequality, using Cauchy inequality,
Sobolev inequality, and the uniform bound \eqref{uniform}, we have
\begin{equation}\label{df2}
\begin{aligned}
\|\theta\|_{L^\infty}\|\nabla^2a\|_{L^2}\|\nabla^3u\|_{L^2}&\le\|\theta\|_{L^2}^{\frac14}\|\nabla^2\theta\|_{L^2}^{\frac34}\|\nabla^2a\|_{L^2}^{\frac14}\|\nabla^2a\|_{L^2}^{\frac34}\|\nabla^3u\|_{L^2}\\&
\le \|\theta\|_{L^2}^{\frac14}(\|\nabla^2\theta\|_{L^2}+\|\nabla^2a\|_{L^2})\|\nabla^3u\|_{L^2}\\&
\le \|\theta\|_{L^2}^{\frac14}(\|\nabla^2\theta\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^3u\|_{L^2}^2),
\end{aligned}
\end{equation}
this together with \eqref{n3} gives
\begin{equation}\label{n6}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla^2u|^2dx+\mu\int|\nabla^3u|^2dx+(\mu+\lambda)\int|\nabla^2\mathop{\rm div}\nolimits u|^2dx+\int\nabla^3a\cdot\nabla^2udx+\int\nabla^3\theta\cdot\nabla^2udx\\&
\le C(\|\nabla(u,a)\|_{L^2}^{\frac12}+\|u\|_{H^1}+\|\nabla\theta\|_{L^2}+\|\theta\|_{L^2}^{\frac14})(\|\nabla ^2u\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2).
\end{aligned}
\end{equation}
Last, applying $\nabla^2$ operator to $\eqref{fullcns}_3$, and then multiplying by $\nabla^2\theta$ and integrating over $\mathbb R^3$, we get
\begin{equation*}
\frac12\frac{d}{dt}\int|\nabla^2\theta|^2dx+\int|\nabla^3\theta|^2dx+\int\nabla^2\mathop{\rm div}\nolimits u\cdot\nabla^2\theta dx=\int\nabla^2S_3\cdot\nabla^2\theta dx.
\end{equation*}
Integrating by part, we get
\begin{equation}\label{n4}
\frac12\frac{d}{dt}\int|\nabla^2\theta|^2dx+\int|\nabla^3\theta|^2dx-\int\nabla^2 u\cdot\nabla^3\theta dx\le \|\nabla S_3\|_{L^2}\|\nabla^3\theta\|_{L^2}.
\end{equation}
It is easy to compute
\begin{equation*}
\begin{aligned}
\|\nabla S_3\|_{L^2}&\le\|\nabla u\|_{L^3}\|\nabla\theta\|_{L^6}+\|\theta\|_{L^3}\|\nabla^2u\|_{L^6}+\|u\|_{L^3}\|\nabla^2\theta\|_{L^6}+\|\nabla u\|_{L^3}\|\nabla^2u\|_{L^6}\\&
\quad +\|\nabla a\|_{L^6}\|(\nabla u)^2\|_{L^3}+\|a\|_{L^\infty}\|\nabla^3\theta\|_{L^2}+\|\nabla a\|_{L^3}\|\nabla^2\theta\|_{L^6}\\&
\le C(\|\nabla u\|_{L^3}+\|\theta\|_{L^3}+\|u\|_{L^3}+\|a\|_{L^\infty}+\|\nabla a\|_{L^3})(\|\nabla^2\theta\|_{H^1}+\|\nabla^3 u\|_{L^2})\\&
\quad +\|\nabla u\|_{L^\infty}\|\nabla u\|_{L^3}\|\nabla a\|_{L^6}.
\end{aligned}
\end{equation*}
Using Sobolev inequality and the uniform bound \eqref{uniform}, we have
\begin{equation*}
\begin{aligned}
\|\nabla& u\|_{L^3}+\|\theta\|_{L^3}+\|u\|_{L^3}+\|a\|_{L^\infty}+\|\nabla a\|_{L^3}\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}\|\nabla^2u\|_{L^2}^{\frac12}+\|\theta\|_{H^1}+\|u\|_{H^1}+\|\nabla a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac12})\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12}+\|\theta\|_{H^1}+\|u\|_{H^1}),
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\|\nabla u\|_{L^\infty}\|\nabla u\|_{L^3}\|\nabla a\|_{L^6}&\le \|\nabla^2u\|_{L^2}^{\frac12}\|\nabla^3u\|_{L^2}^{\frac12}\|\nabla u\|_{L^2}^{\frac12}|\nabla^2 u\|_{L^2}^{\frac12}\|\nabla^2 a\|_{L^2}\\&
\le C\|\nabla u\|_{L^2}^{\frac12}(\|\nabla^2u\|_{L^2}+\|\nabla^3u\|_{L^2}).
\end{aligned}
\end{equation*}
Thus, it follows that
\begin{equation*}
\begin{aligned}
\|\nabla S_3\|_{L^2}\|\nabla^3\theta\|_{L^2}&\le C(\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12}+\|\theta\|_{H^1}+\|u\|_{H^1})(\|\nabla^2\theta\|^2_{H^1}+\|\nabla^2u\|_{H^1}^2).
\end{aligned}
\end{equation*}
Substituting above two estimates into \eqref{n4}, we obtain
\begin{equation}\label{n7}
\begin{aligned}
\frac12&\frac{d}{dt}\int|\nabla^2\theta|^2dx+\int|\nabla^3\theta|^2dx-\int\nabla^2 u\cdot\nabla^3\theta dx\\&
\le C(\|\nabla u\|_{L^2}^{\frac12}+\|\nabla a\|_{L^2}^{\frac12}+\|\theta\|_{H^1}+\|u\|_{H^1})(\|\nabla^2\theta\|^2_{H^1}+\|\nabla^2u\|_{H^1}^2).
\end{aligned}
\end{equation}
Finally, the combination of \eqref{n5}, \eqref{n6} and \eqref{n7} gives rise to
\begin{equation*}
\begin{aligned}
\frac12&\frac{d}{dt}\int(|\nabla^2a|^2+|\nabla^2u|^2+|\nabla^2\theta|^2)dx+\mu\int|\nabla^3u|^2dx+(\mu+\lambda)\int|\nabla^2\mathop{\rm div}\nolimits u|^2dx+\int|\nabla^3\theta|^2dx\\&
\le C(\|\nabla u\|_{L^2}^{\frac14}+\|\nabla (u,a)\|_{L^2}^{\frac12}+\|(u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14})(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2),
\end{aligned}
\end{equation*}
which completes the proof of this lemma.
\end{proof}
In order to close the estimate, we need to establish the dissipation estimate for $\nabla^2a$.
\begin{lemm}\label{l6}
Under the assumptions of Theorem \ref{0them}, the global solution $(a, u, \theta)$ of Cauchy problem \eqref{fullcns} has the estimate
\begin{equation}\label{l6est}
\begin{aligned}
\frac{d}{dt}&\int\nabla u\cdot\nabla^2adx+\frac34\int|\nabla^2a|^2dx\\&
\le C(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{L^2}^2)+C(\|(a,u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac12})(\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2).
\end{aligned}
\end{equation}
\end{lemm}
\begin{proof}
Applying $\nabla$ operator to $\eqref{fullcns}_2$,
multiplying by $\nabla^2a$ and integrating over $\mathbb R^3$, we get
\begin{equation*}
\int\partial_t\nabla u\cdot\nabla^2adx-\int[\mu\nabla\Delta u+(\mu+\lambda)\nabla^2\mathop{\rm div}\nolimits u]\cdot\nabla^2adx
+\int|\nabla^2a|^2dx+\int\nabla^2\theta\cdot\nabla^2adx=\int \nabla S_2\cdot\nabla^2adx.
\end{equation*}
According to $\eqref{fullcns}_1$, it holds on
\begin{equation*}
\begin{aligned}
\int\partial_t\nabla u\cdot\nabla^2adx&=\int\partial_t(\nabla u\cdot\nabla^2a)dx-\int\nabla u\cdot\nabla^2a_tdx\\&
=\frac{d}{dt}\int\nabla u\cdot\nabla^2adx+\int\nabla\mathop{\rm div}\nolimits u\cdot\partial_t(\nabla a)dx\\&
=\frac{d}{dt}\int\nabla u\cdot\nabla^2adx+\int\nabla\mathop{\rm div}\nolimits u\cdot(\nabla S_1-\nabla\mathop{\rm div}\nolimits u)dx.
\end{aligned}
\end{equation*}
Then using Cauchy and H\"{o}lder inequalities, we obtain
\begin{equation*}
\begin{aligned}
\frac{d}{dt}&\int\nabla u\cdot\nabla^2adx+\int|\nabla^2a|^2dx\\&
=\int[\mu\nabla\Delta u+(\mu+\lambda)\nabla^2\mathop{\rm div}\nolimits u]\cdot\nabla^2adx+\int \nabla S_2\cdot\nabla^2adx\\&
\quad -\int\nabla\mathop{\rm div}\nolimits u\cdot(\nabla S_1-\nabla\mathop{\rm div}\nolimits u)dx-\int\nabla^2\theta\cdot\nabla^2adx\\&
\le \frac14\|\nabla^2a\|_{L^2}^2+C(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{L^2}^2)+C(\|\nabla S_2\|_{L^2}^2+\|S_1\|_{L^2}^2).
\end{aligned}
\end{equation*}
According to \eqref{n8}, \eqref{n2} and the uniform bound \eqref{uniform}, it is easy to get
\begin{equation}\label{s5}
\begin{aligned}
\frac{d}{dt}&\int\nabla u\cdot\nabla^2adx+\frac34\int|\nabla^2a|^2dx\\&
\le C(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{L^2}^2)+C(\|a\|_{H^1}+\|u\|_{H^1}+\|\theta\|_{H^1})\|\nabla^2a\|_{L^2}^2+\|\theta\|_{L^\infty}^2\|\nabla^2a\|_{L^2}^2.
\end{aligned}
\end{equation}
Using the Sobolev interpolation and the uniform bound \eqref{uniform} , we have
\begin{equation}\label{s4}
\|\theta\|_{L^\infty}^2\|\nabla^2a\|_{L^2}^2\le \|\theta\|_{L^2}^{\frac12}\|\nabla^2\theta\|_{L^2}^{\frac32}\|\nabla^2a\|_{L^2}^{\frac12}\|\nabla^2a\|_{L^2}^{\frac32}
\le \|\theta\|_{L^2}^{\frac12}(\|\nabla^2\theta\|_{L^2}^2+\|\nabla^2a\|_{L^2}^2).
\end{equation}
Substituting \eqref{s4} into \eqref{s5} derives
\begin{equation*}
\begin{aligned}
\frac{d}{dt}&\int\nabla u\cdot\nabla^2adx+\frac34\int|\nabla^2a|^2dx\\&
\le C(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{L^2}^2)+C(\|a\|_{H^1}+\|u\|_{H^1}+\|\theta\|_{H^1}+\|\theta\|_{L^2}^{\frac12})(\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2).
\end{aligned}
\end{equation*}
Thus we complete the proof.
\end{proof}
Combining all the estimates obtained in Lemmas \ref{l4}-\ref{l6}, we derive the following energy estimate.
\begin{lemm}\label{l5}
Under the assumptions of Theorem \ref{0them}, we define
$$\mathcal E_2^2(t)\overset{\Delta}{=}\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla \theta\|_{H^1}^2+\delta\int\nabla u\cdot\nabla^2adx.$$
Then there exists a large time $T_2$, such that
\begin{equation}\label{b1}
\frac{d}{dt}\mathcal E_2^2(t)+c_0\left(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2\right)\le 0, \quad t\ge T_2.
\end{equation}
Here $c_0$ is a positive constant, and $\delta$ is a small constant.
\end{lemm}
\begin{proof}
Adding \eqref{l4est} with \eqref{l5est}, we have
\begin{equation}\label{p1}
\begin{aligned}
&\frac12\frac{d}{dt}(\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla \theta\|_{H^1}^2)+\mu\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2\\&
\quad \le C(\|\nabla u\|_{L^2}^{\frac14}+\|\nabla(u,a)\|_{L^2}^{\frac12}+\|(a,u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14})\left(\|\nabla^2u\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{H^1}^2\right).
\end{aligned}
\end{equation}
Multiplying $\delta$ to \eqref{l6est} and adding with \eqref{p1}, we can choose $\delta$ being small enough to get
\begin{equation}\label{pp11}
\begin{aligned}
\frac12&\frac{d}{dt}\left(\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla \theta\|_{H^1}^2+2\delta\int\nabla u\cdot\nabla^2adx\right)
+\frac{3}{4}(\mu\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2)+\frac{3}{4}\delta\|\nabla^2a\|_{L^2}^2\\&
\le C\left(\|\nabla u\|_{L^2}^{\frac14}+\|\nabla(u,a)\|_{L^2}^{\frac12}+\|(a,u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14}+\|\theta\|_{L^2}^{\frac12}\right)\left(\|\nabla^2u\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2+\|\nabla^2\theta\|_{H^1}^2\right).
\end{aligned}
\end{equation}
Thanks to the decay rate \eqref{decay22} obtained in Theorem \ref{0them}, we obtain
\begin{equation*}
\|\nabla u\|_{L^2}^{\frac14}+\|\nabla(u,a)\|_{L^2}^{\frac12}+\|(a,u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14}+\|\theta\|_{L^2}^{\frac12}\le C(1+t)^{-\frac{3}{16}}.
\end{equation*}
Thus, there exists a large time $T_2>0$, such that
\begin{equation}\label{cdelta}
C\left(\|\nabla u\|_{L^2}^{\frac14}+\|\nabla(u,a)\|_{L^2}^{\frac12}+\|(a,u,\theta)\|_{H^1}+\|\theta\|_{L^2}^{\frac14}+\|\theta\|_{L^2}^{\frac12}\right)\le \frac14\min\{\mu,1,\delta\}
\end{equation}
holds on for all $t\ge T_2$. Therefore we obtain the following estimate
\begin{equation*}
\frac{d}{dt}\left(\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2
+\|\nabla \theta\|_{H^1}^2+2\delta\int\nabla u\cdot\nabla^2adx\right)
+\mu\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\delta\|\nabla^2a\|_{L^2}^2\le 0
\end{equation*}
Taking $c_0=\min\{\mu,1,\delta\}$, then we get
\begin{equation*}
\frac{d}{dt} \left(\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2
+\|\nabla \theta\|_{H^1}^2+\delta\int\nabla u\cdot\nabla^2adx\right)
+c_0\left(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2\right)\le 0.
\end{equation*}
By virtue of the smallness of $\delta$, there are two constants $c_4$ and $C_4$ (independent of time) such that
\begin{equation}\label{b4}
c_4(\|\nabla u\|_{H^1}^2+\|\nabla a\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2)\le \mathcal E_2^2(t)\le C_4(\|\nabla u\|_{H^1}^2+\|\nabla a\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2).
\end{equation}
Thus, we complete the proof of this lemma.
\end{proof}
Next, let us establish the following decay estimate, which will give the proof for the Theorem \ref{Thm3}.
\begin{lemm}\label{lemma4}
Under the assumptions of Theorem \ref{0them}, there exists a large time $T_3$, such that
\begin{equation}\label{41}
\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\partial_ta\|_{L^2}^2+\|\partial_tu\|_{L^2}^2+\|\partial_t\theta\|_{L^2}^2\le C(1+t)^{-\frac52}
\end{equation}
holds on all $t\ge T_3$, here $C$ is a constant independent of time.
\end{lemm}
\begin{proof}
In order to obtain the time decay rate \eqref{41}, our method
here is to use the Fourier splitting method (by Schonbek\cite{schonbek1}),
which has been applied to obtain decay rate for the incompressible Navier-Stokes equations
in higher order derivative norm (cf.\cite{{schonbek2},{Schonbek-Wiegner}}).
The difficulty, arising from the compressible Navier-Stokes equations, is the appearance of density that obeys the
transport equation rather than diffusive one. To get rid of
this difficulty, our idea is to rewrite the inequality \eqref{b1} as follows
\begin{equation}\label{b2}
\frac{d}{dt}\mathcal E_2^2(t)+\frac{c_0}{2}\left(\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2\right)+\frac{c_0}{2}\|\nabla^2a\|_{L^2}^2\le 0, \quad t\ge T_2.
\end{equation}
Define $S_0:=\{\xi\in\mathbb R^3||\xi|\le(\frac{R}{1+t})^{\frac12}\}$, then we split the phase space $\mathbb R^3$ into two time-dependent regions, $R$ is a constant defined below. It is easy to see
\begin{equation*}
\begin{aligned}
\int_{\mathbb R^3}|\nabla^3u|^2dx&\ge\int_{\mathbb R^3/S_0}|\xi|^6|\hat u|^2d\xi
\ge\frac{R}{1+t}\int_{\mathbb R^3}|\xi|^4|\hat u|^2d\xi\\&
\ge \frac{R}{1+t}\int_{\mathbb R^3/S_0}|\xi|^4|\hat u|^2d\xi-\frac{R}{1+t}\int_{S_0}|\xi|^4|\hat u|^2d\xi\\&
\ge \frac{R}{1+t}\int_{\mathbb R^3/S_0}|\xi|^4|\hat u|^2d\xi-\frac{R^2}{(1+t)^2}\int_{\mathbb R^3/S_0}|\xi|^2|\hat u|^2d\xi,
\end{aligned}
\end{equation*}
which means
$$\|\nabla^3u\|_{L^2}^2\ge\frac{R}{1+t}\|\nabla^2u\|_{L^2}^2-\frac{R^2}{(1+t)^2}\|\nabla u\|_{L^2}^2.$$
Hence, in the same way, we obtain
\begin{equation}\label{b3}
\begin{aligned}
&\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2\\
&\ge \frac{R}{1+t}(\|\nabla u\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\nabla a\|_{L^2}^2)-\frac{R^2}{(1+t)^2}(\|u\|_{H^1}^2+\|\theta\|_{H^1}^2+\|a\|_{L^2}^2).
\end{aligned}
\end{equation}
Substituting \eqref{b3} into \eqref{b2}, we have
\begin{equation*}
\frac{d}{dt}\mathcal E_2^2(t)+\frac{c_0R}{2(1+t)}(\|\nabla u\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\nabla a\|_{L^2}^2)+\frac{c_0}{2}\|\nabla^2a\|_{L^2}^2
\le \frac{c_0R^2}{2(1+t)^2}(\|u\|_{H^1}^2+\|\theta\|_{H^1}^2+\|a\|_{L^2}^2).
\end{equation*}
The term $\|\nabla^2a\|_{L^2}^2$ on the left-hand side of above inequality plays a role of damping term, hence, it holds on
\begin{equation*}
\frac{d}{dt}\mathcal E_2^2(t)+\frac{c_0R}{2(1+t)}(\|\nabla u\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\nabla a\|_{H^1}^2)
\le \frac{c_0R^2}{2(1+t)^2}(\|u\|_{H^1}^2+\|\theta\|_{H^1}^2+\|a\|_{L^2}^2),
\end{equation*}
for all $t\ge T_3:=\max\{T_2, R-1\}$.
By the equivalent relation \eqref{b4} and the decay result \eqref{decay2} in Theorem \ref{0them}, we have
\begin{equation*}
\frac{d}{dt}\mathcal E_2^2(t)+\frac{c_0R}{2C_2(1+t)}\mathcal E_2^2(t)\le \frac{c_0R^2}{2}C(1+t)^{-\frac32-2}.
\end{equation*}
Choosing $R=\frac{6C_4}{c_0}$ and multiplying the resulting inequality by $(1+t)^3$, it follows that
\begin{equation*}
\frac{d}{dt}\left[(1+t)^3\mathcal E_2^2(t)\right]\le C(1+t)^{-\frac12}.
\end{equation*}
Integrating the above inequality over $[T_3,t]$, we obtain
\begin{equation*}
\mathcal E_2^2(t)\le (1+t)^{-3}\left[(1+T_3)^3\mathcal E_2^2(T_3)+C(1+t)^{\frac12}-C(1+T_3)^{\frac12}\right],
\end{equation*}
which, together with the uniform bound \eqref{uniform}
and estimate \eqref{thetabound} in Remark \ref{nabla2theta}, it holds on
$$\mathcal E_2^2(t)\le C(1+t)^{-\frac52}.$$
Using the equivalent relation \eqref{b4} again, we have for all $t\ge T_3:=\max\{T_2,\frac{6C_4}{c_0}-1\}$
\begin{equation}\label{p22}
\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla \theta\|_{H^1}^2\le C(1+t)^{-\frac52}.
\end{equation}
Last, according to the equation \eqref{fullcns} and the decay results obtained before, we get
\begin{equation}\label{p3}
\begin{aligned}
\|\partial_ta\|_{L^2}&\le\|\mathop{\rm div}\nolimits u\|_{L^2}+\|S_1\|_{L^2}
\le \|\nabla u\|_{L^2}+C(\|a\|_{H^1}+\|u\|_{H^1})(\|\nabla^2u\|_{L^2}+\|\nabla^2a\|_{L^2})
\le C(1+t)^{-\frac54}.
\end{aligned}
\end{equation}
By \eqref{m2}-\eqref{m4}, we get
\begin{equation}\label{p4}
\begin{aligned}
\|\partial_tu\|_{L^2}&\le \mu\|\Delta u\|_{L^2}+(\mu+\lambda)\|\nabla\mathop{\rm div}\nolimits u\|_{L^2}+\|\nabla a\|_{L^2}+\|\nabla \theta\|_{L^2}+\|S_2\|_{L^2}\\&
\le C(\|\nabla^2u\|_{L^2}+\|\nabla a\|_{L^2}+\|\nabla\theta\|_{L^2})\\&
\quad +C(\|u\|_{H^1}+\|a\|_{L^2}^{\frac14}+\|a\|_{H^1}+\|\theta\|_{H^1})(\|\nabla^2u\|_{L^2}+\|\nabla^2a\|_{L^2}+\|\nabla^2\theta\|_{L^2})\\&
\le C(1+t)^{-\frac54}.
\end{aligned}
\end{equation}
According to \eqref{p2}, we get
\begin{equation}\label{p5}
\begin{aligned}
\|\partial_t\theta\|_{L^2}&\le \|\Delta\theta\|_{L^2}+\|\mathop{\rm div}\nolimits u\|_{L^2}+\|S_3\|_{L^2}\\&
\le \|\nabla^2\theta\|_{L^2}+\|\nabla u\|_{L^2}+C(\|\theta\|_{H^1}+\|u\|_{H^1}+\|\nabla u\|_{L^2}^{\frac12}+\|a\|_{L^2}^{\frac14})(\|\nabla^2u\|_{L^2}+\|\nabla^2\theta\|_{L^2})\\&
\le C(1+t)^{-\frac54}.
\end{aligned}
\end{equation}
The combination of \eqref{p22}-\eqref{p5} completes the proof of this lemma.
\end{proof}
\subsection{The proof of Theorem \ref{1theo}.}
\quad
First, we derive the evolution of the negative Sobolev norms of the solution of system \eqref{fullcns0}.
Similar to the estimates in Section \ref{3}, we need to restrict $s\in(0,3/2)$ to estimate the nonlinear terms.
\begin{lemm}
For $s\in(0,1/2]$, we have
\begin{equation}\label{p7}
\begin{aligned}
\frac{d}{dt}&\int (|\Lambda^{-s} a^2|+|\Lambda^{-s} u|^2+|\Lambda^{-s}\theta|^2)dx+C(\|\nabla\Lambda^{-s} u\|_{L^2}^2+\|\nabla\Lambda^{-s}\theta\|_{L^2}^2)\\&\lesssim (\|\nabla a\|_{H^1}^2+\|\nabla\theta\|_{H^1}^2+\|\nabla u\|_{H^2}^2)(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s} u\|_{L^2}+\|\Lambda^{-s}\theta\|_{L^2}),
\end{aligned}
\end{equation}
and for $s\in(1/2,3/2)$, we have
\begin{equation}\label{p8}
\begin{aligned}
&\frac{d}{dt}\int (|\Lambda^{-s} a^2|+|\Lambda^{-s} u|^2+|\Lambda^{-s}\theta|^2)dx+C(\|\nabla\Lambda^{-s} u\|_{L^2}^2+\|\nabla\Lambda^{-s}\theta\|_{L^2}^2)\\&
\quad\lesssim \|(a,u,\theta)\|_{L^2}^{s-1/2}\|\nabla(a,u,\theta)\|_{L^2}^{3/2-s}(\|\nabla(u,\theta)\|_{H^1}+\|\nabla a\|_{L^2})(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s} u\|_{L^2}+\|\Lambda^{-s}\theta\|_{L^2})\\&
\qquad +\|\nabla u\|_{L^2}^{s+1/2}\|\nabla^2u\|_{L^2}^{3/2-s}\|\Lambda^{-s}\theta\|_{L^2}.
\end{aligned}
\end{equation}
\end{lemm}
\begin{proof}
Applying $\Lambda^{-s}$ to $\eqref{fullcns}_1$, $\eqref{fullcns}_2$, $\eqref{fullcns}_3$ and multiplying the resulting by $\Lambda^{-s} a$, $\Lambda^{-s} u$, $\Lambda^{-s}\theta$ respectively, summing up and integrating over $\mathbb R^3$ by parts, we obtain
\begin{equation}\label{p6}
\begin{aligned}
&\frac12\frac{d}{dt}\int(|\Lambda^{-s} a|^2+|\Lambda^{-s} u|^2+|\Lambda^{-s}\theta|^2)dx+\mu\int|\nabla\Lambda^{-s} u|^2dx+(\mu+\lambda)\int|\mathop{\rm div}\nolimits\Lambda^{-s} u|^2dx+\int|\nabla\Lambda^{-s}\theta|^2dx\\&
\quad=\int\Lambda^{-s}(-a\mathop{\rm div}\nolimits u-u\cdot\nabla a)\cdot\Lambda^{-s} adx+\int\Lambda^{-s}(-u\cdot\nabla u)\cdot\Lambda^{-s} udx\\&
\qquad -\int\Lambda^{-s}(h(a)[\mu\Delta u+(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u])\cdot\Lambda^{-s} udx+\int\Lambda^{-s}(h(a)(\nabla a+\nabla\theta))\cdot\Lambda^{-s} u\\&
\qquad -\int\Lambda^{-s}(g(a)\nabla(a\theta))\cdot\Lambda^{-s} udx-\int\Lambda^{-s}(\mathop{\rm div}\nolimits(\theta u))\cdot\Lambda^{-s}\theta dx\\&
\qquad +\int\Lambda^{-s}(g(a)(2\mu|Du|^2+\lambda(\mathop{\rm div}\nolimits u)^2))\cdot\Lambda^{-s}\theta dx-\int\Lambda^{-s}(h(a)\Delta\theta)\cdot\Lambda^{-s}\theta dx
\overset{\Delta}{=}\sum_{i=1}^8 J_i.
\end{aligned}
\end{equation}
Similar to the estimates in Lemma \ref{l3}, when $s\in(0,1/2]$,
using \eqref{a.3}, Sobolev interpolation inequality, H\"{o}lder inequality and Young inequality, we obtain
\begin{equation*}
\begin{aligned}
&J_1\lesssim (\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2)\|\Lambda^{-s} a\|_{L^2},\\&
J_2\lesssim \|\nabla u\|_{H^1}^2\|\Lambda^{-s} u\|_{L^2},\\&
J_3\lesssim (\|\nabla a\|_{H^1}^2+\|\nabla^2u\|_{L^2}^2)\|\Lambda^{-s} u\|_{L^2},\\&
J_4\lesssim (\|\nabla a\|_{H^1}^2+\|\nabla\theta\|_{L^2}^2)\|\Lambda^{-s} u\|_{L^2},\\&
J_5\lesssim (\|\nabla\theta\|_{H^1}^2+\|\nabla a\|_{H^1}^2)\|\Lambda^{-s} u\|_{L^2},\\&
J_6\lesssim (\|\nabla \theta\|_{H^1}^2+\|\nabla u\|_{H^1}^2)\|\Lambda^{-s}\theta\|_{L^2},\\&
J_7\lesssim \|\nabla u\|_{H^2}^2\|\Lambda^{-s}\theta\|_{L^2},\\&
J_8\lesssim (\|\nabla a\|_{H^1}^2+\|\nabla^2\theta\|_{L^2}^2)\|\Lambda^{-s}\theta\|_{L^2}.
\end{aligned}
\end{equation*}
Substituting above estimates from $J_1$ to $J_8$ into \eqref{p6}, we obtain the estimate \eqref{p7}.
Next, if $s\in(1/2,3/2)$, we should deal with $J_1\sim J_8$ in a different way,
using the different Sobolev interpolation, it is easy to check that
\begin{equation*}
\begin{aligned}
&J_1\lesssim (\|a\|_{L^{3/s}}\|\nabla u\|_{L^2}+\|u\|_{L^{3/s}}\|\nabla a\|_{L^2})\|\Lambda^{-s} a\|_{L^2}\\&
\quad \lesssim(\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}+\|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2})\|\Lambda^{-s} a\|_{L^2},\\&
J_2\lesssim \|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}\|\Lambda^{-s} u\|_{L^2},\\&
J_3\lesssim \|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla^2 u\|_{L^2}\|\Lambda^{-s} u\|_{L^2},\\&
J_4\lesssim \|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}(\|\nabla a\|_{L^2}+\|\nabla\theta\|_{L^2})\|\Lambda^{-s} u\|_{L^2},\\&
J_5\lesssim (\|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla \theta\|_{L^2}+\|\theta\|_{L^2}^{s-1/2}\|\nabla \theta\|_{L^2}^{3/2-s}\|\nabla a\|_{L^2})\|\Lambda^{-s} u\|_{L^2},\\&
J_6\lesssim (\|\theta\|_{L^2}^{s-1/2}\|\nabla \theta\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}+\|u\|_{L^2}^{s-1/2}\|\nabla u\|_{L^2}^{3/2-s}\|\nabla \theta\|_{L^2})\|\Lambda^{-s} \theta\|_{L^2},\\&
J_7\lesssim \|\nabla u\|_{L^2}^{s-1/2}\|\nabla^2 u\|_{L^2}^{3/2-s}\|\nabla u\|_{L^2}\|\Lambda^{-s}\theta\|_{L^2},\\&
J_8\lesssim \|a\|_{L^2}^{s-1/2}\|\nabla a\|_{L^2}^{3/2-s}\|\nabla^2 \theta\|_{L^2}\|\Lambda^{-s}\theta\|_{L^2}.
\end{aligned}
\end{equation*}
Plugging the estimates from $J_1$ to $J_8$ into \eqref{p6}, we deduce \eqref{p8}.
Thus we complete the proof of this lemma.
\end{proof}
Next, let us prove the decay result when the initial data $(a_0,u_0,\theta_0)\in\dot H^{-s}$, $s\in (0,3/2)$.
Similar to the proof in section \ref{3}, we split the proof into two parts. Before that, we need to introduce the following proposition first, which comes from \cite{he-huang-wang2}.
\begin{prop}\label{prop2}
Under the assumptions of Theorem \ref{0them}, there holds
\begin{equation}\label{prop2est}
\frac{d}{dt}X_2(t)+C\left(\|\nabla^2u\|_{L^2}^2+\|\nabla u\|_{L^2}^2+\|\nabla a\|_{L^2}^2+\|\nabla\dot u\|_{L^2}^2+\|\nabla\theta\|_{L^2}^2+\|\nabla^2\theta\|_{L^2}^2\right)\le 0,
\end{equation}
where $$X_2(t)\sim \|u\|_{H^1}^2+\|a\|_{H^1}^2+\|\dot u\|_{L^2}^2+\|\theta\|_{H^1}^2.$$
\end{prop}
With the Proposition \ref{prop2} at hand, we obtain the following decay result.
\begin{prop}\label{prop3}
Under the sssumptions of Theorem \ref{1theo}, when $s\in (0,1/2]$, then for all $t\ge 0$, we have
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+ \|\Lambda^{-s} u(t)\|_{L^2}^2+\|\Lambda^{-s}\theta(t)\|_{L^2}^2+\|\Lambda^{-s} \dot u(t)\|_{L^2}^2\le C,
\end{equation*}
\begin{equation}\label{prop30}
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\theta(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s},
\end{equation}
where $C$ is a constant independent of time.
\end{prop}
\begin{proof}
Integrating \eqref{p7} over $[0,t]$, then using the uniform bound \eqref{uniform}, we have
\begin{equation}
\begin{aligned}
&\sup_{\tau\in[0,t]}(\|\Lambda^{-s} a(\tau)\|_{L^2}^2+\|\Lambda^{-s} u(\tau)\|_{L^2}^2+\|\Lambda^{-s} \theta(\tau)\|_{L^2}^2)\\&
\quad \le (\|\Lambda^{-s} a_0\|_{L^2}^2+\|\Lambda^{-s} u_0\|_{L^2}^2+\|\Lambda^{-s} \theta_0\|_{L^2}^2)\\&
\qquad +C\int_0^t(\|\nabla (a,\theta)(\tau)\|_{H^1}^2+\|\nabla u(\tau)\|_{H^2}^2)(\|\Lambda^{-s} a(\tau)\|_{L^2}+\|\Lambda^{-s} u(\tau)\|_{L^2}+\|\Lambda^{-s}\theta(\tau)\|_{L^2})d\tau\\&
\quad \le C+C\sup_{\tau\in[0,t]}(\|\Lambda^{-s} a(\tau)\|_{L^2}+\|\Lambda^{-s} u(\tau)\|_{L^2}+\|\Lambda^{-s}\theta(\tau)\|_{L^2}),
\end{aligned}
\end{equation}
which implies
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+\|\Lambda^{-s} u(t)\|_{L^2}^2+\|\Lambda^{-s} \theta(t)\|_{L^2}^2\le C \quad {\rm for}~s\in(0,1/2].
\end{equation*}
Similar as \eqref{prop1est1} in Proposition \ref{prop1}, we have
\begin{equation}\label{prop3est1}
\Delta u\in \dot H^{-s}.
\end{equation}
Recall that $a\in \dot H^{-s}\cap H^2$, $\theta\in\dot H^{-s}\cap H^1$,
we get $a\in \dot H^{1-s}$ and $\theta\in\dot H^{1-s}$, and hence
\begin{equation*}
\begin{aligned}
\||\xi|^{-s}\widehat{\nabla(a\theta)}\|_{L^2}&\lesssim \||\xi|^{1-s}\widehat {a\theta}\|_{L^2}\lesssim \||\xi|^{1-s}\hat a*\hat\theta\|_{L^2}\lesssim \||\xi|^{1-s}\hat a\|_{L^2}\|\hat\theta\|_{L^1}
\lesssim\||\xi|^{1-s}\hat a\|_{L^2}\|\theta\|_{L^\infty}
\le C,
\end{aligned}
\end{equation*}
where we used the condition \eqref{thetarho} in the last inequality. Hence, it is easy to check that
\begin{equation}\label{prop3est2}
\nabla P=\nabla(a\theta)+\nabla a+\nabla\theta\in \dot H^{-s}.
\end{equation}
Since $\rho\dot u+\nabla P=\mu\Delta u+(\mu+\lambda)\nabla\mathop{\rm div}\nolimits u$, combining \eqref{prop3est1} and \eqref{prop3est2}, we have $\rho\dot u\in \dot H^{-s}$. Using $\underline\rho\le \rho\le M$, we deduce that
$$\dot u\in \dot H^{-s}.$$
So far, we have
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+\|\Lambda^{-s} u(t)\|_{L^2}^2+\|\Lambda^{-s} \theta(t)\|_{L^2}^2+\|\Lambda^{-s}\dot u(t)\|_{L^2}^2\le C \quad for \quad s\in(0,1/2].
\end{equation*}
Thanks to \eqref{a.2}, it holds on
\begin{equation*}
\|\nabla^{l+1}f\|_{L^2}\ge C\|\Lambda^{-s} f\|_{L^2}^{-\frac{1}{l+s}}\|\nabla^lf\|_{L^2}^{1+\frac{1}{l+s}}.
\end{equation*}
By this fact and the uniform bound \eqref{uniform}, we deduce that
\begin{equation*}
\|\nabla u\|_{H^1}^2+\|\nabla a\|_{L^2}^2+\|\nabla\dot u\|_{L^2}^2+\|\nabla\theta\|_{H^1}^2\ge C(\|u\|_{H^1}^2+\|a\|_{H^1}^2+\|\dot u\|_{L^2}^2+\|\theta\|_{H^1}^2)^{1+\frac1s}.
\end{equation*}
Substituting above inequality into \eqref{prop2est}, we get
\begin{equation*}
\frac{d}{dt}X_2(t)+CX_2(t)^{1+\frac1s}\le 0.
\end{equation*}
Solving this ODE directly, and using the uniform bound \eqref{uniform}, we have
\begin{equation*}
X_2(t)\le C(1+t)^{-s}, \quad s\in(0,1/2].
\end{equation*}
Therefore we complete the proof of this lemma.
\end{proof}
Similar to the analysis in section \ref{3}, before we derive the propagation of the negative Sobolev norms of the solution for $s\in(1/2,3/2)$, it is important to improve the decay estimate of the first order spatial derivative of the solution for $s\in(0,1/2]$.
\begin{prop}\label{prop4}
Under the assumptions of Theorem \ref{1theo}, and $s\in(0,1/2]$, there exists a large time $T_4$ such that
\begin{equation*}
\|\nabla u(t)\|_{H^1}^2+\|\nabla a(t)\|_{H^1}^2+\|\nabla\theta(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}
\end{equation*}
holds on for all $t\ge T_4$,
where $C$ is a constant independent of time.
\end{prop}
\begin{proof}
Thanks to \eqref{a.2}, we have
\begin{equation*}
\|\nabla^{l+1}f\|_{L^2}\ge C\|\Lambda^{-s} f\|_{L^2}^{-\frac{1}{l+s}}\|\nabla^lf\|_{L^2}^{1+\frac{1}{l+s}},
\end{equation*}
this together with the uniform bound \eqref{uniform}, we obtain
\begin{equation}\label{ooo1}
\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\|\nabla^2a\|_{L^2}^2\ge C(\|\nabla u\|^2_{H^1}+\|\nabla\theta\|_{H^1}^2+\|\nabla a\|_{H^1}^2)^{1+\frac{1}{1+s}}.
\end{equation}
On the other hand, similar as the proof of Lemma \ref{diss}, the combination of \eqref{pp11} in Lemma \ref{l5} and the decay rate \eqref{prop30} in Proposition \ref{prop3} easily derives that there exits a large time $T_4$ such that
\begin{equation*}
\frac{d}{dt}\left(\|\nabla a\|_{H^1}^2+\|\nabla u\|_{H^1}^2+\|\nabla \theta\|_{H^1}^2+2\delta\int\nabla u\cdot\nabla^2adx\right)
+\mu\|\nabla^2u\|_{H^1}^2+\|\nabla^2\theta\|_{H^1}^2+\delta\|\nabla^2a\|_{L^2}^2\le 0
\end{equation*}
holds on for all $t\ge T_4$.
Thus, by virtue of the equivalent relation \eqref{b4} and \eqref{ooo1}, there holds for all $t\ge T_4$
\begin{equation}\label{ee2}
\frac{d}{dt}\mathcal E_2^2(t)+C(\mathcal E_2^2(t))^{1+\frac{1}{1+s}}\le 0.
\end{equation}
Solving the inequality \eqref{ee2} directly, and according to \eqref{uniform} and $\|\nabla^2\theta\|_{L^\infty_tL^2}< \infty$, we obtain
\begin{equation*}
\mathcal E_2^2(t)\le C(1+t)^{-(1+s)}, \quad t\ge T_4,
\end{equation*}
which implies
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2+\|\nabla\theta(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \quad t\ge T_4.
\end{equation*}
Therefore we complete the proof of this lemma.
\end{proof}
\begin{prop}\label{prop7}
Under the assumptions of Theorem \ref{1theo}, when $s\in(1/2,3/2)$, then for all $t\ge 0$, we have
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+\|\Lambda^{-s} u(t)\|_{L^2}^2+\|\Lambda^{-s} \theta(t)\|_{L^2}^2+\|\Lambda^{-s}\dot u(t)\|_{L^2}^2\le C,\end{equation*}
and
\begin{equation*}
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\theta(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}.
\end{equation*}
For $t\ge T_4$, there holds
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2+\|\nabla\theta(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \end{equation*}
where $T_4$ is given in Proposition \ref{prop4}, and $C$ is a constant independent of time.
\end{prop}
\begin{proof}
Notice that $(a_0,u_0,\theta_0)\in \dot H^{-s}\cap L^2\subset \dot H^{-1/2}$ when $s\in(1/2,3/2)$, then we derive from what we have provn in Propositions \ref{prop3} and \ref{prop4} with $s=1/2$ that
\begin{equation}\label{p88}
\begin{aligned}
&\|a(t)\|_{L^2}+\|u(t)\|_{L^2}+\|\theta(t)\|_{L^2}\le C(1+t)^{-\frac14}, \quad t\ge0,\\&
\|\nabla a(t)\|_{H^1}+\|\nabla u(t)\|_{H^1}+\|\nabla \theta(t)\|_{H^1}\le C(1+t)^{-\frac34},\quad t\ge T_4.
\end{aligned}
\end{equation}
Substituting \eqref{p88} into \eqref{p8}, it is easy to obtain that for $s\in(1/2,3/2)$,
\begin{equation*}
\begin{aligned}
&\sup_{\tau\in[0,t]}\left(\|\Lambda^{-s} a(\tau)\|_{L^2}^2+\|\Lambda^{-s} u(\tau)\|_{L^2}^2+\|\Lambda^{-s} \theta(\tau)\|_{L^2}^2\right)+C\int_0^t(\|\nabla\Lambda^{-s} u\|_{L^2}^2+\|\nabla\Lambda^{-s}\theta\|_{L^2}^2)d\tau\\&
\quad \lesssim \|\Lambda^{-s} a_0\|_{L^2}^2+ \|\Lambda^{-s} u_0\|_{L^2}^2+ \|\Lambda^{-s} \theta_0\|_{L^2}^2+\int_0^{T_4}(1+\tau)^{-\frac14(s+\frac12)}\|\Lambda^{-s}\theta\|_{L^2}d\tau\\&
\qquad+\int_0^{T_4}(1+\tau)^{-\frac14}(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s} u\|_{L^2}+\|\Lambda^{-s}\theta\|_{L^2})d\tau
+\int_{T_4}^t(1+\tau)^{-\frac32}\|\Lambda^{-s}\theta\|_{L^2}d\tau\\&
\qquad +\int_{T_4}^t(1+\tau)^{-\frac74+\frac12s}(\|\Lambda^{-s} a\|_{L^2}+\|\Lambda^{-s} u\|_{L^2}+\|\Lambda^{-s}\theta\|_{L^2})d\tau.
\end{aligned}
\end{equation*}
This implies
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}+\|\Lambda^{-s} u(t)\|_{L^2}+\|\Lambda^{-s} \theta(t)\|_{L^2}\le C \quad {\rm for} ~s\in(1/2,3/2).
\end{equation*}
According to Proposition \ref{prop2}, repeat the progress in Proposition \ref{prop3} for $s\in(1/2,3/2)$,
there hold for all $t\ge 0$
\begin{equation*}
\|\Lambda^{-s} a(t)\|_{L^2}^2+\|\Lambda^{-s} u(t)\|_{L^2}^2+\|\Lambda^{-s} \theta(t)\|_{L^2}^2+\|\Lambda^{-s}\dot u(t)\|_{L^2}^2\le C, \quad s\in(1/2,3/2),
\end{equation*}
and
\begin{equation*}
\|a(t)\|_{H^1}^2+\|u(t)\|_{H^1}^2+\|\theta(t)\|_{H^1}^2+\|\dot u(t)\|_{L^2}^2\le C(1+t)^{-s}, \quad s\in(1/2,3/2).
\end{equation*}
Similarly, repeat the progress in Proposition \ref{prop4} for $s\in(1/2,3/2)$,
it holds for all $t\ge T_4$
\begin{equation*}
\|\nabla a(t)\|_{H^1}^2+\|\nabla u(t)\|_{H^1}^2+\|\nabla\theta(t)\|_{H^1}^2\le C(1+t)^{-(1+s)}, \quad s\in(1/2,3/2).
\end{equation*}
The combination of above two estimates completes the proof of this lemma.
\end{proof}
\textbf{Proof of Theorem \ref{1theo}:}
With the uniform bounds and decay estimates stated in Propositions \ref{prop3}, \ref{prop4} and \ref{prop7},
we can complete the the proof of Theorem \ref{1theo}.
|
{
"timestamp": "2020-07-28T02:36:41",
"yymm": "2007",
"arxiv_id": "2007.13450",
"language": "en",
"url": "https://arxiv.org/abs/2007.13450"
}
|
\section{Introduction}
The double orthogonal projection is a four-dimensional generalization of Monge's projection. In Monge's projection, an object in the 3-space is orthogonally projected into two mutually perpendicular planes (horizontal and vertical, i.e. top and front view). One of these planes is chosen to be the drawing (or picture) plane and the second is rotated about their intersecting line to the drawing plane. Therefore, each point in the 3-space has two conjugated images in the drawing plane. In the double orthogonal projection, an object is in the 4-space, and we project it orthogonally into two mutually perpendicular 3-spaces. Let $x,y,z$, and $w$ be the orthogonal system of coordinate axes of the 4-space, and $\Xi(x,y,z)$ and $\Omega(x,y,w)$ be the 3-spaces of projection. If $\Omega$ is chosen to be the modeling 3-space (instead of the drawing plane), then $\Xi$ is rotated about their common plane $\pi(x,y)$ such that $z$ and $w$ have opposite orientations (usually $w$ up and $z$ down). Analogically, each point in the 4-space has two conjugated images in the modeling 3-space. Elementary constructions and principles of the double orthogonal projection were described in \cite{Zamboj2018a}, sections, and lighting of polytopes in \cite{Zamboj2018b}, intersections of lines, planes, and 3-spaces with a 3-sphere in \cite{Zamboj2019b}, and regular quadric sections as intersections of 4-dimensional cones with 3-spaces in \cite{Zamboj2019a}. Similarly to Monge's projection, in which an observer can reach any point of the drawing plane, in the four-dimensional case an observer can reach any point of the modeling 3-space. For this purpose, our constructions are supplemented by online interactive models \cite{Zamboj2020icggGGB} created in \emph{GeoGebra~5}.
A projection between an $n$-sphere embedded in an $(n+1)$-space from its point $N$ and an $n$-dimensional hyperplane supplemented with a point at infinity $\{\infty\}$, not passing through $N$ is called stereographic projection. Due to its angle-preserving property, it is a convenient tool to visualize spheres. The 3-dimensional case of a stereographic projection of a 2-sphere from the North pole~$N$ to a tangent plane at the South pole and its Monge's projection is depicted in Figure~\ref{fig:sp3d}. For a brief overview see \cite{Snyder1987}, pp. 154--163, and for more general view with the construction of a spherical inversion used in Section~\ref{sec:sphinv} see \cite{Odehnal2020}, pp. 368--378.
\begin{figure}[!htb]
\centering
\includegraphics[height=4.5cm]{127-Figure1a_Zamboj}\hfill\includegraphics[height=4.5cm]{127-Figure1b_Zamboj}
\caption{A stereographic projection of the southern hemisphere from the North pole $N$ to the tangent plane at the South pole (i.e. the horizontal plane of projection), and its Monge's projection. Images of a quadrilateral with its sides along parallels and meridians are highlighted.\\
Interactive model available at \texttt{https://www.geogebra.org/m/mfctqfcs}}
\label{fig:sp3d}
\end{figure}
A stereographic projection of a 3-sphere onto a 3-space was used to create 3-D printed models and study properties of four-dimensional polytopes in \cite{Schleimer2012} and later on in \cite{Hart2014} to visualize symmetries of the quaternionic group on a 3-D printed models of monkeys in a hypercube (limbs, head, and tail of the monkey represent faces of cubical cells). Well designed online applets with stereographic projections of a hypercube are at \cite{Sirius14000} and including other polytopes, sections, and tori in a 3-sphere at \cite{Cervone}. A stereographic projection of a 3-sphere is often used to visualize objects usually studied in topology. The topology of a 3-sphere is studied in \cite{Kocak1987}. Several animations and four-dimensional stereographic projection are described in \cite{Banchoff1996}, Chapters 6 and 8. The same author in \cite{Banchoff1988} used computer graphics to visualize stereographic images of Pinkall's tori corresponding in the Hopf fibration to simple closed curves on a 2-sphere (see also \cite{Pinkall1985} and \cite{3dxm}). An interactive visualization of a stereographic projection of a Clifford torus on a 3-sphere is at \cite{Balmens2012}. Videos and animations of the Hopf fibration and its stereographic projection are at \cite{Johnson2011, Dimensions, Chinyere2012}. Commented videos with interactive environment discussing quaternions and stereographic projection are at \cite{Eater}. A stereographic and double-stereographic projection of an arbitrary object in the 4-space was described in \cite{Ohori2017}. As an example of a recent application, a stereographic projection to a 3-sphere is used in \cite{Hosseinbor2015} to analyze multiple disconnected anatomical structures mathematically represented as a composition of compact finite three-dimensional surfaces.
\begin{figure}[!b]
\centering
\includegraphics[height=7cm, trim=50 50 50 50, clip]{127-Figure2a_Zamboj}\hfill\includegraphics[height=7cm, trim=50 50 50 50, clip]{127-Figure2b_Zamboj}
\caption{The system of concentric spheres, which are stereographic images of parallel sections of the half of the 3-sphere split by the 3-space parallel to $\Xi$ from the point $N$ with the maximal $w$-coordinate to the tangent 3-space $\Xi$ at the antipodal point $S$, and its double orthogonal projection. On the right side are conjugated images and stereographic image of a hyperspherical hexahedron, as an analogy to a spherical quadrilateral.\\
Interactive model available at \texttt{https://www.geogebra.org/m/sz7ykk54}}
\label{fig:sp4d}
\end{figure}
The previous references were based on the analytic representation of points in the fourth dimension. With the use of the double orthogonal projections, we can construct images of four-dimensional objects synthetically and use the modeling 3-space to be also the projecting 3-space of a stereographic projection. A construction of a hyperspherical tetrahedron in a special position and its images in a stereographic and the double orthogonal projection is in \cite{Zamboj2020a}. A synthetic construction and animation of Hopf fibers in the double orthogonal projection and their stereographic projection are discussed in \cite{Zamboj2020b}. In this paper, we will extend these results in several aspects.
Let us describe the visualization of a 3-sphere and stereographic images in the double orthogonal projection in Figure~\ref{fig:sp4d} (left) in an analogy to the three-dimensional case in Figure~\ref{fig:sp3d} (right). In Monge's projection, the conjugated images of a \mbox{2-sphere} are disks, in the double orthogonal projection, the conjugated images of a 3-sphere are balls. In 3-D, sections of a \mbox{2-sphere} with planes parallel to the horizontal plane of projection (parallels) are circles, and their images are segments in the vertical plane and circles in the horizontal plane. The stereographic projection from the North pole $N$ to the tangent plane at the South pole $S$ projects these parallels to the system of concentric circles in the drawing plane. In 4-D, sections of a 3-sphere with 3-spaces parallel to the 3-space of projection $\Xi$ are 2-spheres, their $\Omega$-images are disks and $\Xi$-images are 2-spheres. The stereographic projection from the point $N$ with the maximal $w$-coordinate to the tangent 3-space $\Xi$ at the antipodal point $S$ projects the parallel 2-spheres to the system of concentric 2-spheres in the modeling 3-space. In both cases, the point $N$ is projected stereographically into the point at infinity $\{\infty\}$, and hence images of circles through $N$ become lines. In Figure~\ref{fig:sp4d} (right) (see also animated model) is a hyperspherical hexahedron along hyperspherical coordinates.
\section{A synthetic construction of a stereographic projection in the double orthogonal projection}
\begin{figure}[!htb]
\centering
\includegraphics[height=7cm, trim=100 50 50 30, clip]{127-Figure3a_Zamboj}\hfill\includegraphics[height=7cm, trim=100 50 50 30, clip]{127-Figure3b_Zamboj}
\caption{(left) Construction of the stereographic image $A_s$ of the point $A$ on the 3-sphere given by its $\Omega$-image $A_4$. (right) Construction of the conjugated images $A_3$ and $A_4$ of a point $A$ from its stereographic image $A_s$.\\
Interactive model available at \texttt{https://www.geogebra.org/m/xdypddf9}}
\label{fig:sp4dpoint}
\end{figure}
The stereographic image of a point on a 3-sphere is the intersection of the projecting ray with the plane of projection (Figure~\ref{fig:sp4dpoint}, left). Let us have conjugated images of a 3-sphere with a center $Z$ in the double orthogonal projection and the $\Omega$-image $A_4$ of a point $A$ on the 3-sphere. The $\Xi$-image $A_3$ lies on the perpendicular to $\pi$ through $A_4$, i.e. ordinal line of the point $A$. Furthermore, the section of the 3-space through $A$ parallel to the 3-space $\Omega$ with a 3-sphere is a 2-sphere. Its $\Omega$-image is a 2-sphere with the center $Z_4$ through $A_4$, and its $\Xi$-image is a disk with the same radius in a plane parallel to $\pi$. Let $N$ be the abovementioned center of the sterographic projection to the 3-space $\Xi$. The stereographic image $A_s$ of the point $A$ on the 3-sphere lies on the line $N_3A_3$ and also on the perpendicular to $\pi$ through the intersection $A_0$ of $\pi$ and the line $N_4A_4$. Oppositely, the inverse construction of the conjugated images $A_3$ and $A_4$ from the stereographic image~$A_s$ is in Figure~\ref{fig:sp4dpoint} (right). In this case, we need to find the intersection~$A$ of the projecting ray $NA_s$ with the 3-sphere. For such construction, we can use a third orthogonal projection into the plane perpendicular to $\Xi$ through the projecting ray, and rotate it into the modeling 3-space (see \cite{Zamboj2019b} for more details). In the third view, the image of the 3-sphere is a circle with the center $(Z)$, and its intersection with the rotated line $(N)A_s$ is the rotated point $(A)$. Then, the point $A_3$ is constructed with the reverse rotation.
\begin{figure}[!htb]
\centering
\includegraphics[height=7cm, trim=100 50 50 30, clip]{127-Figure4a_Zamboj}\hfill\includegraphics[height=7cm, trim=100 50 50 30, clip]{127-Figure4b_Zamboj}
\caption{(left) The double orthogonal projection and stereographic image of a hyperspherical tetrahedron $ABCD$. (right) The same situation with the circumscribed 2-sphere around $ABCD$.\\
Interactive model available at \texttt{https://www.geogebra.org/m/xdypddf9}}
\label{fig:sp4dtetrahedron}
\end{figure}
In Figure~\ref{fig:sp4dtetrahedron} (left), the same method is used to construct the stereographic images $A_s, B_s, C_s, D_s$ of the vertices of a hyperspherical tetrahedron $ABCD$ (generalization of a spherical triangle). The edges of the tetrahedron $ABCD$ are circular arcs and faces are spherical triangles.\footnote{The faces are not depicted due to insufficient possibilities of the surface parametrization in GeoGebra, but the reader can turn on the visibility of the corresponding spheres in the stereographic projection in the online model.} These properties are preserved due to the conformity of stereographic projection. To construct the stereographic images of the circular edges, we can conveniently use stereographic images of the antipodal points. For example, the edge $A_sB_s$ lie on the circle $A_sB_sA'_s$, where $A'_s$ is the stereographic image of the point $A'$ antipodal to $A$, and so its conjugated images $A'_3$ and $A'_4$ are the mirror images of $A_3$ and $A_4$ about $Z_3$ and $Z_4$, respectively. The conjugated images of the edges are constructed point-by-point from their stereographic images and create elliptical arcs in a general position. Note that we could construct them, however, more laboriously, as the intersections of the 3-sphere and planes.
Further on, a 2-sphere circumscribed around a tetrahedron $ABCD$ is visualized in Figure~\ref{fig:sp4dtetrahedron} (right). While in the stereographic projection it is simply a 2-sphere in the true shape, the conjugated images in the double orthogonal projection are ellipsoids. They can be constructed as the intersections of a 3-sphere with the 3-space defined by the noncospatial points $A,B,C$, and $D$ (see \cite{Zamboj2019b} for details), or, as highlighted in the $\Omega$-image, as the intersections of the conical hypersurface with the vertex $N$ with a 3-sphere.
\begin{figure}[!htb]
\centering
\includegraphics[height=7cm]{127-Figure5_Zamboj}
\caption{My drawing was not a picture of a hat. It was a picture of a boa constrictor digesting an elephant on a 3-sphere in the double orthogonal projection (slightly modified from \cite{Exupery1943}).\\
Interactive model available at \texttt{https://www.geogebra.org/m/mjpeaud4}}
\label{fig:freehand}
\end{figure}
The interactive environment, in combination with the construction of a stereographic projection from a 3-space onto a 3-sphere (Figure~\ref{fig:sp4dpoint}, right), brings a possibility of freehand drawing on a 3-sphere. With the use of the trace tool in GeoGebra (or similar point-by-point construction in other software), we move the point in the stereographic projection, and its dependent conjugated images draw their traces. This way, we can draw a stereographic image (its $\Xi$-image) of a curve, or any other picture, in the modeling 3-space, and simultaneously its double orthogonal projection. A four-dimensional curve on a 3-sphere drawn by hand on a 2-sphere in a stereographic projection is in Figure~\ref{fig:freehand}.
\subsection{Spherical inversion}
\label{sec:sphinv}
\begin{figure}[!htb]
\centering
\includegraphics[width=0.9\textwidth, trim= 40 0 0 0, clip]{127-Figure6_Zamboj}
\caption{Spherical inversion of a tetrahedron $ABCD$ and its circumscribed 2-sphere as a composition of stereographic projections from antipodal poles between a \mbox{3-sphere} and its equatorial 3-space.\\
Interactive model available at \texttt{https://www.geogebra.org/m/uqvek39f}}
\label{fig:sphinv}
\end{figure}
A spherical inversion may be obtained as a composition of two stereographic projections (Figure~\ref{fig:sphinv}). For this purpose, we choose the projecting 3-space $\Sigma$ of the stereographic projection through the center $Z$ of the 3-sphere and parallel to $\Xi$. A point~$A$ in the 3-space $\Sigma\cup\{\infty\}$ is stereographically projected in the projection~$p_N$ from the center $N$ to the point $A^\circ$ on the 3-sphere. The second stereographic projection $p_S$ from the center $S$ projects the point $A^\circ$ to the point $A'$ in the 3-space~\mbox{$\Sigma\cup\{\infty\}$}. Since $\Sigma$ is parallel to $\Xi$, all the $\Xi$-images of objects in the 3-space~$\Sigma$ are in the true shape. The section $\gamma$ of a 3-sphere and the 3-space~$\Sigma$ is a 2-sphere overlapping with the $\Xi$-image of the 3-sphere with the center $Z_3$ in the modeling 3-space. The composition of $p_N$ and $p_S$ is the spherical inversion, and $p_S(p_N(A))=A'$. Especially, all the points on $\gamma$ are fixed. Moreover, $p_S(p_N(\infty))=p_S(N)=Z$, and $p_S(p_N(Z))=p_S(S)=\infty$.
\begin{figure}[!htb]
\centering
\includegraphics[height=7cm]{127-Figure7_Zamboj}
\caption{Spherical inversion of a point $A$ in the orthogonal projection to the projecting plane of the line $\overline{NA}$.}
\label{fig:circinv}
\end{figure}
Note, that for each point $A\neq Z,\{\infty\}$ in $\Sigma$, we could choose a projecting plane of the line $AZ$ perpendicular to $\Xi$ which cuts the 2-sphere $\gamma$ in a circle $c$, and the final composition would be a circle inversion. The situation is depicted in Monge's projection in Figure~\ref{fig:circinv}, in which the circle inversion is in the orthogonal projection into the horizontal plane. Triangles $A_2Z_2N_2$ and $S_2Z_2A'_2$ are in their true shape in the front view, and they are apparently similar. Therefore,
\begin{equation*}
\frac{|A_2Z_2|}{|Z_2N_2|}=\frac{|S_2Z_2|}{|Z_2A'_2|},
\end{equation*} and so
\begin{equation*}
\label{eq:st}
|A_2Z_2||A'_2Z_2|=|Z_2N_2||Z_2S_2|=r^2,
\end{equation*}
where $r$ is the radius of $c,\gamma$, and also the 3-sphere. From the given construction, it also holds that $|A_2Z_2|=|A_1Z_1|=|AZ|$ and $|A'_2Z_2|=|A'Z|$, and we have
\begin{equation*}
|AZ||A'Z|=r^2.
\end{equation*}
The last formula leads to the standard definition of the spherical inversion about the 2-sphere $\gamma$ with the center $Z$ and radius $r$.
Figure~\ref{fig:sphinv} also shows a tetrahedron $ABCD$ with a circumscribed sphere and their inversion. The conformity of stereographic projections is inherited in their composition to the spherical inversion. The lines not passing through the center of the inversion become circles. Consequently, the edges of the image tetrahedron $A'B'C'D'$ are circular arcs with preserved mutual angles. The image of the sphere circumscribed to $ABCD$ is a sphere circumscribed to $A'B'C'D'$.
\subsection{Hopf tori of Clelia curves}
\begin{figure}[!htb]
\centering
\includegraphics[width=0.31\textwidth]{127-Figure8a1_Zamboj} \hspace{1pt} \includegraphics[width=0.31\textwidth]{127-Figure8b1_Zamboj} \hspace{1pt}\includegraphics[width=0.31\textwidth]{127-Figure8c1_Zamboj}\\
\includegraphics[width=0.31\textwidth]{127-Figure8a2_Zamboj} \hspace{1pt} \includegraphics[width=0.31\textwidth]{127-Figure8b2_Zamboj} \hspace{1pt}\includegraphics[width=0.31\textwidth]{127-Figure8c2_Zamboj}
\caption{(Up) Conjugated images of the Hopf tori generated by the Clelia curves for $s=0.5$ (Left), $s=1$ Viviani curve (Center), $s=2$ (Right). Highlighted is a point (blue) on the Clelia curve (black), conjugated images of its fiber (red) and conjugated images of the Hopf torus (green). (Down) Stereographic image of a fiber (purple) and Hopf torus (gray).\\
Interactive model available at \texttt{https://www.geogebra.org/m/sbf4xadp}}
\label{fig:clelia}
\end{figure}
The Hopf fibration is a mapping between a 3-sphere and a 2-sphere. In particular, it sends a point $$\displaystyle P(x_P,y_P,z_P,w_P)\in\mathbb{R}^4$$ on a 3-sphere to the point $$\displaystyle P'(2(x_Pz_P+y_Pw_P),2(-x_Pw_P+y_Pz_P),x_P^2-y_P^2-z_P^2-w_P^2)\in\mathbb{R}^3$$ on a 2-sphere. Oppositely, a point on a 2-sphere in spherical coordinates $$P'(\sin\psi_P \cos\varphi_P, \sin\psi_P\sin\varphi_P,\cos\psi_P)$$ for $\psi_P\in\langle0,\pi\rangle, \varphi_P\in\langle0,2\pi)$ corresponds to a set of points
$$c_P(\cos\frac{\psi_P}{2}\cos(\varphi_P+\beta),\cos\frac{\psi_P}{2}\sin(\varphi_P+\beta),\sin\frac{\psi_P}{2}\cos\beta,\sin\frac{\psi_P}{2}\sin\beta)$$
for $\beta\in \langle0,2\pi)$, which forms a great circle (fiber) on a 3-sphere. If $P'$ lies on a closed curve on the 2-sphere, its corresponding Hopf fibers form a Hopf torus on the 3-sphere. For the visualization of Hopf tori, it is convenient to use the stereographic projection that preserves circles. This way we can construct and study tori on the 3-sphere in the 4-space given by a curve on the 2-sphere in the 3-space, and vice-versa.
For the sake of visualization in the double orthogonal projection, we perform several adjustments. First, we swap the reference axes $y$ and $z$, and hence the 3-spaces of projection will be $\Xi(x,y,z)$ (upper) and $\Omega(x,z,w)$ (lower) with the common plane $\pi(x,z)$.\footnote{This choice reflects the possibility of a definition of the Hopf fibration in the complex number plane, and so the common plane $\pi(x,z)$ corresponds to the real parts of coordinates $(x+\i y, z+\i w)$ of points.} To avoid overlapping of the conjugated images, we translate the center of the abovementioned unit 3-sphere from the origin to the point $(0,1,0,1)$. The preimage 3-sphere and the image 2-sphere are not strictly related in the definition of the Hopf fibration, and topologists often describe them as separated objects. However, in our interpretation, the 2-sphere is chosen to be the equatorial section of the 3-sphere with the 3-space parallel to $\Omega(x,z,w)$, so its $\Xi$-image is in the true shape. Consequently, points of the 3-sphere are stereographically projected from the center $(0,2,0,1)$ to the 3-space $\Omega(x,z,w)$. A synthetic construction of the Hopf fiber of any point on the 2-sphere and a Hopf torus of a circle on a 2-sphere and their stereographic images in this setting are described in \cite{Zamboj2020b}.
We extend these results and highlight interactive possibilities for the so-called Clelia curves given parametrically on the (translated) 2-sphere in the form $$k(\psi)=(\sin(s\psi)\cos\psi, \sin(s\psi)\sin\psi+1,\cos(s\psi)),$$ where $s\in\mathbb{R}$ defines the specific curve, for $\psi\in I\subset\mathbb{R}$ (see \cite{Goemans2016} for details and further generalizations). Conjugated images and stereographic projections of the Hopf tori corresponding to the Clelia curves for $s=\frac{1}{2}, 1,$ and $2$ are visualized in Figure~\ref{fig:clelia}. In the interactive model, the user can choose the parameter $s$ of the curve and change the corresponding Hopf torus, and also move with a point~$P$ on $k(\psi)$ and its corresponding fiber on the 3-sphere by manipulating the parameter~$\psi$. An example of a simple straightforward observation is, that a point, in which the curve on the 2-sphere intersects itself, becomes a circle (or line) in which the torus on the 3-sphere intersects itself.
\section{Conclusion}
The combination of the double orthogonal projection and stereographic projection is an accessible tool for the investigation of a 3-sphere in the 3-space. The simplicity of the synthetic construction of the stereographic image of a point and the use of interactive 3-D modeling software imply that we can actually draw sketches and shapes on the 3-sphere embedded in the 4-space. We have also shown a construction of the well-known relationship between the stereographic projection and the spherical inversion. Our results were applied to the construction of Hopf tori on a 3-sphere generated by Clelia curves on a 2-sphere. The visualizations were presented on interactive 3-D models, which are easily extendible for further theoretical and practical applications in various software.
\bibliographystyle{spmpsci}
|
{
"timestamp": "2020-07-28T02:37:13",
"yymm": "2007",
"arxiv_id": "2007.13470",
"language": "en",
"url": "https://arxiv.org/abs/2007.13470"
}
|
"\\section{Introduction}\\label{sec:Intro}\n\nIn this paper, we obtain optimal (in terms of scaling (...TRUNCATED)
| {"timestamp":"2020-07-28T02:29:59","yymm":"2007","arxiv_id":"2007.13297","language":"en","url":"http(...TRUNCATED)
|
"\\section{Introduction}\n\nCNe are cataclysmic variables (CVs) with 6 to 19 mag brightness increase(...TRUNCATED)
| {"timestamp":"2020-07-28T02:32:34","yymm":"2007","arxiv_id":"2007.13337","language":"en","url":"http(...TRUNCATED)
|
"\\section{Introduction}\n\\label{sec:introd}\nWe undertake the first comprehensive analysis of the (...TRUNCATED)
| {"timestamp":"2020-07-28T02:29:20","yymm":"2007","arxiv_id":"2007.13275","language":"en","url":"http(...TRUNCATED)
|
"\\section{Introduction and Preliminaries}\nLet $H$ be an invertible $n\\times n$ Hermitian matrix. (...TRUNCATED)
| {"timestamp":"2020-10-14T02:17:11","yymm":"2007","arxiv_id":"2007.13396","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction\\label{section:introduction}}\nUnderstanding the nature of the flavor struct(...TRUNCATED)
| {"timestamp":"2020-10-06T02:29:57","yymm":"2007","arxiv_id":"2007.13254","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction}\nThe properties of dense nuclear matter still remain one of the unsolved pr(...TRUNCATED)
| {"timestamp":"2020-11-10T02:33:44","yymm":"2007","arxiv_id":"2007.13399","language":"en","url":"http(...TRUNCATED)
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"\\section{Introduction} \\label{Section I}\nA fundamental bottleneck in operating large-dimensiona(...TRUNCATED)
| {"timestamp":"2020-07-28T02:30:01","yymm":"2007","arxiv_id":"2007.13299","language":"en","url":"http(...TRUNCATED)
|
"\\section{Introduction}\n\nThe end of Inflation set the density fluctuations which dictated the gro(...TRUNCATED)
| {"timestamp":"2020-07-28T02:33:16","yymm":"2007","arxiv_id":"2007.13353","language":"en","url":"http(...TRUNCATED)
|
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