Using the density equation (1.1), we could change (1.1)-(1.10) into the following equivalent form:
$$ \textstyle\begin{cases} \rho_{t}+\nabla\cdot(\rho u)=0,\\ \rho u_{t}+\rho u\cdot\nabla u-\Delta u-\nabla\operatorname{div} u+\nabla p=-\frac{2}{3}\nabla(\rho k),\\ \rho h_{t}+\rho u\cdot\nabla h-\Delta h=p_{t}+u\cdot\nabla p+S_{k},\\ \rho k_{t}+\rho u\cdot\nabla k-\Delta k=G-\rho\varepsilon,\\ \rho\varepsilon_{t}+\rho u\cdot\nabla\varepsilon-\Delta \varepsilon= \frac{C_{1}G\varepsilon}{k}-\frac{C_{2}\rho\varepsilon^{2}}{k},\\ (\rho,u,h,k,\varepsilon)(x,0)=(\rho _{0}(x),u_{0}(x),h_{0}(x),k_{0}(x),\varepsilon_{0}(x)),\\ (u\cdot\vec{n},h, \frac{\partial k}{\partial\vec{n}},\frac{\partial \varepsilon}{\partial\vec{n}})|_{\partial\Omega}=(0, 0, 0, 0). \end{cases} $$
(2.1)
Then we consider the following linearized problem of (2.1):
$$\begin{aligned}& \rho_{t}+\nabla\cdot(\rho v)=0, \end{aligned}$$
(2.2)
$$\begin{aligned}& \rho u_{t}+\rho v\cdot\nabla u-\Delta u-\nabla\operatorname{div} u+\nabla p=-\frac{2}{3}\nabla(\rho\pi), \end{aligned}$$
(2.3)
$$\begin{aligned}& \rho h_{t}+\rho v\cdot\nabla h-\Delta h=p_{t}+u\cdot \nabla p+S_{k}^{\prime}, \end{aligned}$$
(2.4)
$$\begin{aligned}& \rho k_{t}+\rho v\cdot\nabla k-\Delta k=G^{\prime}-\rho \theta, \end{aligned}$$
(2.5)
$$\begin{aligned}& \rho\varepsilon_{t}+\rho v\cdot\nabla\varepsilon-\Delta \varepsilon= \frac{C_{1}G^{\prime}\theta}{\pi}-\frac{C_{2}\rho\theta^{2}}{\pi }, \end{aligned}$$
(2.6)
$$\begin{aligned}& (\rho,v,h,\pi,\theta) (x,0)=\bigl(\rho _{0}(x),u_{0}(x),h_{0}(x),k_{0}(x), \varepsilon_{0}(x)\bigr), \end{aligned}$$
(2.7)
$$\begin{aligned}& \biggl(v\cdot\vec{n},h, \frac{\partial\pi}{\partial\vec{n}},\frac{\partial \theta}{\partial\vec{n}} \biggr)\bigg|_{\partial\Omega}=(0, 0, 0, 0), \end{aligned}$$
(2.8)
with
$$\begin{aligned}& S_{k}^{\prime}= \biggl[\mu \biggl(\frac{\partial v^{i}}{\partial x_{j}}+ \frac{\partial v^{j}}{\partial x_{i}} \biggr)-\frac{2}{3}\delta_{ij} \frac{\partial v^{k}}{\partial x_{k}} \biggr]\frac{\partial v^{i}}{\partial x_{j}}\ +\frac{\mu_{t}}{\rho^{2}} \frac{\partial p}{\partial x_{j}}\frac {\partial\rho}{\partial x_{j}}, \\& G^{\prime}=\frac{\partial v^{i}}{\partial x_{j}} \biggl[\mu_{e} \biggl( \frac{\partial v^{i}}{\partial x_{j}}+\frac{\partial v^{j}}{ \partial x_{i}} \biggr)-\frac{2}{3} \delta_{ij} \biggl(\rho\pi+ \mu_{e}\frac{\partial v^{k}}{\partial x_{k}} \biggr) \biggr], \end{aligned}$$
where v, π, and θ are known quantities on \((0,T_{1})\times\Omega\) with \(T_{1}>0\).
Here we also impose the following regularity conditions on the initial data:
$$ \textstyle\begin{cases} 0< m< \rho_{0},\quad \rho_{0}\in H^{3}(\Omega),\\ u_{0}\in H^{3}(\Omega),\\ (h_{0}, k_{0}, \varepsilon_{0})\in H^{2}(\Omega),\\ (u_{0}\cdot\vec{n},h_{0}, \frac{\partial k_{0}}{\partial\vec{n}},\frac{\partial \varepsilon_{0}}{\partial\vec{n}} ) |_{\partial \Omega}=(0, 0, 0, 0),\\ 0< m< k_{0}. \end{cases} $$
(2.9)
For the known quantities v, π, θ, we assume that \(v(0)=u_{0}\), \(\pi(0)=k_{0}\), \(\theta(0)=\varepsilon_{0}\), and
$$ \textstyle\begin{cases} \sup _{0\leq t\leq T_{2}}(\Vert v\Vert _{H^{1}} +\Vert \pi \Vert _{H^{1}}+\Vert \theta \Vert _{H^{1}})\\ \quad{}+\int_{0}^{T_{2}}(\Vert \pi \Vert _{H^{3}}^{2} +\Vert v_{t}\Vert _{H^{1}}^{2}+\Vert \pi_{t}\Vert _{H^{1}}^{2}+\Vert \theta_{t}\Vert _{H^{1}}^{2})\,\mathrm{d}t\leq c_{1},\\ \sup _{0\leq t\leq T_{2}}\Vert v\Vert _{H^{2}}\leq c_{2},\\ \sup _{0\leq t\leq T_{2}}\Vert v\Vert _{H^{3}}\leq c_{3},\\ \int_{0}^{T_{2}}\Vert v\Vert _{H^{4}}^{2}\,\mathrm{d}t\leq c_{4},\\ \sup _{0\leq t\leq T_{2}}\Vert \pi \Vert _{H^{2}}\leq c_{5},\\ \sup _{0\leq t\leq T_{2}}\Vert \theta \Vert _{H^{2}}\leq c_{6}, \end{cases} $$
(2.10)
for some fixed constants \(c_{i}\) satisfying \(1< c_{0}< c_{i}\) (\(i=1,2,\ldots,6\)) and some time \(T_{2}>0\). Here
$$c_{0}=2+\bigl\Vert (\rho_{0}, u_{0})\bigr\Vert _{H^{3}}+\bigl\Vert (h_{0},k_{0}, \varepsilon_{0})\bigr\Vert _{H^{2}}. $$
For simplicity, we set another small time T as \(T =\min\{ c_{0}^{-6\gamma-16}c_{1}^{-10}c_{2}^{-8}c_{3}^{-8}c_{4}^{-2} c_{5}^{-2}c_{6}^{-4} , T_{1} , T_{2}\}\) and all of the T in Section 2 are defined as this.
Remark 2.1
Here it should be emphasized that throughout this paper, C denotes a generic positive constant which is only dependent on m, γ, and \(\vert \Omega \vert \), but independent of \(c_{i}\) (\(i=0,1,2,\ldots,6\)).
Remark 2.2
From the physical viewpoint, we assume that the turbulent kinetic energy k has a positive lower bound away from zero, namely, \(0< m< k\) with m a constant. We do not know whether \(0< m< k\) holds afterwards if the initial turbulent kinetic energy \(k_{0}>m\).
In this section we aim to prove the following local existence theorem of the linearized system (2.2)-(2.6).
Theorem 2.1
There exists a unique strong solution
\((\rho,u,h,k,\varepsilon)\)
to the linearized problem (2.2)-(2.8) and (2.9) in
\([0,T]\)
satisfying the estimates (2.99) and (2.100) as well as the regularity
$$\begin{aligned}& \rho\in C\bigl(0,T;H^{3}\bigr),\quad\quad \rho_{t} \in C \bigl(0,T;H^{1}\bigr),\quad\quad u \in C\bigl(0,T; H^{3}\bigr)\cap L^{2}\bigl(0,T; H^{4}\bigr), \\& u_{t}\in L^{2}\bigl(0,T;H^{1}\bigr), \quad\quad k\in C \bigl(0,T; H^{2}\bigr)\cap L^{2}\bigl(0,T;H^{3} \bigr),\quad\quad k_{t}\in L^{2}\bigl(0,T;H^{1}\bigr), \\& \varepsilon\in C\bigl(0,T; H^{2}\bigr),\quad\quad \varepsilon_{t} \in L^{2}\bigl(0,T;H^{1}\bigr),\quad\quad h\in C\bigl(0,T; H^{2}\bigr),\quad\quad h_{t}\in L^{2}\bigl(0,T;H^{1} \bigr), \\& (\sqrt{\rho}u_{t},\sqrt{\rho}k_{t},\sqrt{\rho} \varepsilon _{t},\sqrt{\rho}h_{t})\in L^{\infty} \bigl(0,T;L^{2}\bigr). \end{aligned}$$
In the following part, we decompose the proof of Theorem 2.1 into some lemmas.
Lemma 2.1
There exists a unique strong solution
ρ
to the linear transport problem (2.2) and (2.9) such that
$$ \rho\geq\frac{m}{e},\quad\quad \Vert \rho \Vert _{H^{3}(\Omega)}\leq Cc_{0},\quad\quad \Vert \rho_{t}\Vert _{H^{1}(\Omega )}\leq Cc_{0}c_{2} $$
(2.11)
for
\(0\leq t\leq T\).
Proof
First, applying the particle trajectory method to the equation (2.3), we easily deduce
$$\rho\geq\rho_{0}\exp \biggl(- \int_{0}^{T}\Vert \nabla v\Vert _{L^{\infty }} \,\mathrm{d}t \biggr) \geq\rho_{0}\exp(-c_{3}T)\geq \frac{\rho_{0}}{e}\geq\frac {m}{e} $$
and thus
$$\frac{1}{\rho}\leq\frac{e}{m}\leq C. $$
Second, by simple calculation, we have
$$\frac{\mathrm{d}}{\mathrm{d}t}\Vert \rho \Vert _{H^{3}}\leq C\Vert v\Vert _{H^{3}}\Vert \rho \Vert _{H^{3}}+C\bigl\Vert \nabla^{4}v\bigr\Vert _{L^{2}}, $$
applying the Gronwall and Hölder’s inequalities, one gets
$$\Vert \rho \Vert _{H^{3}}\leq \biggl[\exp \biggl(C \int_{0}^{t}\Vert v\Vert _{H^{3}}\, \mathrm{d}t \biggr) \biggr] \biggl(\Vert \rho_{0}\Vert _{H^{3}} +C \int_{0}^{t}\Vert v\Vert _{H^{4}}\, \mathrm{d}t \biggr)\leq Cc_{0} $$
for \(0\leq t\leq T\).
Next, from the equation (2.2), one obtains
$$\Vert \rho_{t}\Vert _{H^{1}}=\bigl\Vert \nabla\cdot(\rho v)\bigr\Vert _{H^{1}}\leq C\Vert \rho \Vert _{H^{3}}\Vert v \Vert _{H^{2}}\leq Cc_{0}c_{2} $$
for \(0\leq t\leq T\).
Thus, we complete the proof of Lemma 2.1. □
Next, we estimate the velocity field u.
Lemma 2.2
There exists a unique strong solution u to the initial boundary value problem (2.3) and (2.9) such that
$$\begin{aligned}& \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2} + \Vert u\Vert _{H^{1}}^{2}+ \int_{0}^{t}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s \leq Cc_{0}^{5+2\gamma},\quad\quad \Vert u\Vert _{H^{2}}\leq Cc_{0}^{\frac {5}{2}+3\gamma}c_{1}^{2}, \end{aligned}$$
(2.12)
$$\begin{aligned}& \Vert u\Vert _{H^{3}}\leq Cc_{0}^{\frac{13}{2}+3\gamma}c_{1}^{4}c_{2}c_{5},\quad\quad \int_{0}^{t}\Vert u\Vert _{H^{4}}^{2} \,\mathrm{d}s \leq Cc_{0}^{9+6\gamma }c_{1}^{5}c_{2}^{2} \end{aligned}$$
(2.13)
for
\(0\leq t\leq T\).
Proof
We only need to prove the estimates. Differentiating the equation (2.3) with respect to t, then multiplying both sides of the result by \(u_{t}\) and integrating over Ω, we derive that
$$\begin{aligned}& \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t} \int\rho u_{t}^{2}\,\mathrm{d}x+\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+\Vert \operatorname{div} u_{t}\Vert _{L^{2}}^{2} \\& \quad=- \int\rho_{t}v\cdot\nabla u\cdot u_{t}- \int\rho v_{t}\cdot \nabla u\cdot u_{t}-2 \int\rho v\cdot\nabla u_{t}\cdot u_{t} \\& \quad\quad{}- \int \nabla p_{t}\cdot u_{t}-\frac{2}{3} \int\bigl[\nabla(\rho\pi)\bigr]_{t}\cdot u_{t} \\& \quad=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}, \end{aligned}$$
(2.14)
where we have used the equation (2.2) and integration by parts. We will estimate \(I_{i}\) (\(i=1,2,\ldots,5\)) item by item.
First, because ρ has a lower bound away from zero, we easily deduce \(\Vert u_{t}\Vert _{L^{2}}\leq C\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}\). Therefore, using the Hölder, Sobolev, and Young inequalities and (2.10), we have
$$\begin{aligned}& I_{1}\leq C\Vert v\Vert _{L^{\infty}} \Vert \rho_{t}\Vert _{L^{3}}\Vert \nabla u\Vert _{L^{2}} \Vert u_{t}\Vert _{L^{6}} \leq C\Vert v\Vert _{L^{\infty}} \Vert \rho_{t}\Vert _{L^{3}}\Vert \nabla u \Vert _{L^{2}}\bigl(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+\Vert \nabla u_{t}\Vert _{L^{2}}\bigr) \\& \hphantom{I_{1}}\leq Cc_{0}^{2}c_{2}^{4}\Vert \nabla u\Vert _{L^{2}}^{2}+C\Vert \sqrt{\rho }u_{t}\Vert _{L^{2}}^{2}+\frac{1}{8}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.15)
$$\begin{aligned}& I_{3}\leq C\Vert \rho \Vert _{L^{\infty}}^{\frac{1}{2}}\Vert v \Vert _{L^{\infty}} \Vert \nabla u_{t}\Vert _{L^{2}} \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}} \leq Cc_{0}c_{2}^{2}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}}^{2}+\frac{1}{8}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.16)
$$\begin{aligned}& I_{2} \leq C\Vert \rho \Vert _{L^{\infty}}^{\frac{1}{2}}\Vert v_{t}\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{3}} \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}} \leq C \eta^{-1}c_{0}\Vert \nabla u\Vert _{L^{3}}^{2}+ \eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{ \rho}u_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.17)
where \(\eta>0\) is a small number to be determined later.
Next, to evaluate \(\Vert \nabla u\Vert _{L^{3}}^{2}\) in (2.17), we can first the Sobolev interpolation inequality to get
$$ \Vert \nabla u\Vert _{L^{3}}^{2}\leq C\Vert \nabla u \Vert _{L^{2}}\Vert \nabla u\Vert _{L^{6}}\leq C\Vert \nabla u\Vert _{L^{2}}\Vert \nabla u\Vert _{H^{1}}. $$
(2.18)
Then applying the standard elliptic regularity result to the equation (2.3) and using (2.18), we have
$$\begin{aligned} \Vert \nabla u\Vert _{H^{1}} \leq& Cc_{0}^{\gamma} \bigl(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+\Vert v\Vert _{L^{6}}\Vert \nabla u\Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla u\Vert _{H^{1}}^{\frac{1}{2}} +\Vert \nabla\rho \Vert _{L^{2}} \\ &{}+\Vert \nabla\rho \Vert _{L^{4}}\Vert \pi \Vert _{L^{4}}+\Vert \nabla\pi \Vert _{L^{2}}\bigr), \end{aligned}$$
thus the Young inequality and (2.10) yield
$$ \Vert \nabla u\Vert _{H^{1}}\leq Cc_{0}^{2\gamma} \bigl(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+c_{1}^{2} \Vert \nabla u\Vert _{L^{2}}+c_{0}c_{1} \bigr). $$
(2.19)
Combining (2.17), (2.18), and (2.19), and using the Young inequality, we get
$$ I_{2} \leq C\eta^{-1}c_{0}^{2\gamma+1}\bigl( \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+c_{1}^{2} \Vert \nabla u\Vert _{L^{2}}^{2} +c_{0}^{2}c_{1}^{2} \bigr) +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}. $$
(2.20)
By integration by parts, we have
$$\begin{aligned}& I_{4}= \int p_{t}\operatorname{div} u_{t} \leq Cc_{0}^{\gamma-1}\Vert \rho_{t}\Vert _{L^{2}}\Vert \nabla u_{t}\Vert _{L^{2}}\leq Cc_{0}^{2\gamma}c_{2}^{2}+\frac{1}{8} \Vert \nabla u_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.21)
$$\begin{aligned}& I_{5} =\frac{2}{3} \int\rho_{t}\pi\nabla\cdot u_{t}-\frac {2}{3} \int\pi_{t}\nabla\rho\cdot u_{t} -\frac{2}{3} \int\rho\nabla\pi_{t}\cdot u_{t} \\& \hphantom{I_{5} } \leq C\Vert \rho_{t}\Vert _{L^{3}}\Vert \pi \Vert _{L^{6}}\Vert \nabla u_{t}\Vert _{L^{2}} +Cc_{0}^{\frac{1}{2}}\Vert \nabla\rho \Vert _{L^{3}}\Vert \pi_{t}\Vert _{L^{6}}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}} +Cc_{0}^{\frac{1}{2}}\Vert \nabla \pi_{t}\Vert _{L^{2}}\Vert \sqrt{\rho}u_{t} \Vert _{L^{2}} \\& \hphantom{I_{5} }\leq Cc_{0}^{2}c_{1}^{2}c_{2}^{2}+C \eta^{-1}c_{0}^{3}+C\eta \Vert \pi _{t} \Vert _{H^{1}}^{2}\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2} +\frac{1}{8}\Vert \nabla u_{t} \Vert _{L^{2}}^{2}. \end{aligned}$$
(2.22)
On the other hand, we easily have
$$ \frac{\mathrm{d}}{\mathrm{d}t} \int \vert \nabla u\vert ^{2} =2 \int\nabla u\cdot\nabla u_{t}\leq\frac{1}{8}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+C\Vert \nabla u \Vert _{L^{2}}^{2} $$
(2.23)
and
$$ \frac{\mathrm{d}}{\mathrm{d}t} \int \vert u\vert ^{2}\leq Cc_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}\Vert u\Vert _{L^{2}} \leq Cc_{0}\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+C\Vert u\Vert _{L^{2}}^{2}. $$
(2.24)
Combining (2.14)-(2.16) and (2.20)-(2.24), we get
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\bigl(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+\Vert u\Vert _{H^{1}}^{2} \bigr)+\Vert \nabla u_{t}\Vert _{L^{2}}^{2} \\& \quad\leq C\bigl(c_{0}^{2}c_{2}^{4}+ \eta^{-1}c_{0}^{2\gamma+1}c_{1}^{2}+ \eta \Vert \pi_{t}\Vert _{H^{1}}^{2} +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\bigr) \bigl(\Vert \sqrt{ \rho}u_{t}\Vert _{L^{2}}^{2}+\Vert u\Vert _{H^{1}}^{2}\bigr) \\& \quad\quad{}+C\bigl(c_{0}^{2\gamma}c_{1}^{2}c_{2}^{2}+ \eta^{-1}c_{0}^{2\gamma +3}c_{1}^{2} \bigr), \end{aligned}$$
(2.25)
setting \(\eta=\frac{1}{c_{1}}\) and using the Gronwall inequality, we derive
$$ \Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+\Vert u\Vert _{H^{1}}^{2}+ \int_{0}^{t}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5+2\gamma} $$
(2.26)
for \(0\leq t\leq T\), where we have used the fact that \(\lim _{t\rightarrow0}(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}^{2}+\Vert u\Vert _{H^{1}}^{2})\leq Cc_{0}^{5+2\gamma}\).
Next, by (2.19) and (2.26), we deduce
$$ \Vert \nabla u\Vert _{H^{1}}\leq Cc_{0}^{\frac{5}{2}+3\gamma}c_{1}^{2}, $$
(2.27)
which implies (2.12) by (2.26).
Next, we will estimate \(\int_{0}^{t}\Vert u\Vert _{H^{4}}^{2}\,\mathrm{d}t\). By the standard elliptic regularity result of the equation (2.3), we have
$$ \bigl\Vert \nabla^{4}u\bigr\Vert _{L^{2}}\leq \Vert \rho u_{t}\Vert _{H^{2}}+\Vert \rho v\cdot \nabla u\Vert _{H^{2}}+\Vert \nabla p\Vert _{H^{2}}+\biggl\Vert \frac{2}{3}\nabla(\rho\pi )\biggr\Vert _{H^{2}}. $$
(2.28)
By simple calculation, the first term of the right-hand side of (2.28) can be controlled as
$$ \Vert \rho u_{t}\Vert _{H^{2}}\leq C\bigl(\Vert \rho u_{t}\Vert _{L^{2}}+\Vert \rho \Vert _{H^{2}} \Vert u_{t}\Vert _{H^{2}}\bigr)\leq Cc_{0}\Vert u_{t}\Vert _{H^{2}}. $$
(2.29)
In order to estimate \(\Vert \nabla^{2}u_{t}\Vert _{L^{2}}\), differentiating the equation (2.3) with respect to t yields
$$\begin{aligned} \Delta u_{t}+\nabla\operatorname{div} u_{t} =&\rho_{t}u_{t}+\rho u_{tt}+ \rho_{t}v\cdot\nabla u+\rho v_{t}\cdot \nabla u +\rho v\cdot \nabla u_{t}+\nabla p_{t} \\ &{}+\frac{2}{3}(\nabla \rho_{t}\pi +\rho_{t}\nabla\pi+\nabla\rho\pi_{t}+ \rho\nabla\pi _{t}), \end{aligned}$$
(2.30)
applying the standard elliptic regularity result to (2.30) and using (2.26), one obtains
$$\begin{aligned} \bigl\Vert \nabla^{2}u_{t}\bigr\Vert _{L^{2}} \leq& C\bigl(\Vert \rho_{t}\Vert _{L^{4}}\Vert u_{t}\Vert _{L^{4}}+\Vert \rho u_{tt}\Vert _{L^{2}}+\Vert \rho_{t}\Vert _{L^{4}}\Vert v \Vert _{L^{\infty}} \Vert \nabla u\Vert _{L^{4}} \\ &{}+\Vert \rho \Vert _{L^{\infty}} \Vert v_{t}\Vert _{L^{4}}\Vert \nabla u\Vert _{L^{4}} +\Vert v\Vert _{L^{\infty}} \Vert u_{t}\Vert _{H^{1}} +\Vert \rho \Vert _{H^{2}}^{\gamma} \Vert \rho_{t}\Vert _{H^{1}} +\Vert \pi \Vert _{L^{\infty}} \Vert \rho_{t}\Vert _{H^{1}} \\ &{}+\Vert \rho_{t} \Vert _{L^{4}}\Vert \nabla\pi \Vert _{L^{4}} +\Vert \nabla\rho \Vert _{L^{4}}\Vert \pi_{t}\Vert _{L^{4}} +\Vert \rho \Vert _{L^{\infty}} \Vert \nabla \pi_{t}\Vert _{L^{2}}\bigr) \\ \leq& C\bigl(\Vert \rho u_{tt}\Vert _{L^{2}}+c_{0}^{\frac{7}{2}+3\gamma }c_{1}^{2}c_{2}^{2}c_{5}+c_{0}^{\frac{7}{2}+3\gamma}c_{1}^{2} \Vert v_{t}\Vert _{H^{1}} \\ &{} +c_{0}c_{2} \Vert u_{t}\Vert _{H^{1}}+c_{0}\Vert \pi_{t}\Vert _{H^{1}}\bigr), \end{aligned}$$
(2.31)
therefore, the key point is to estimate \(\Vert \rho u_{tt}\Vert _{L^{2}}\). Because we have the fact \(\Vert \rho u_{tt}\Vert _{L^{2}}\leq C\Vert \sqrt{\rho} u_{tt}\Vert _{L^{2}}\), we could first estimate \(\Vert \sqrt{\rho} u_{tt}\Vert _{L^{2}}\) as follows.
Multiplying both sides of (2.30) by \(u_{tt}\) and integrating the result over Ω yield
$$\begin{aligned}& \int\rho u_{tt}^{2}\,\mathrm{d}x+\frac{1}{2} \frac{\mathrm{d}}{\mathrm {d}t}\Vert \nabla u_{t}\Vert _{L^{2}}^{2} +\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \operatorname{div} u_{t} \Vert _{L^{2}}^{2} \\& \quad=- \int\rho_{t}u_{t}\cdot u_{tt}- \int\rho_{t}v\cdot\nabla u\cdot u_{tt}- \int\rho v_{t}\cdot\nabla u\cdot u_{tt} - \int\rho v\cdot\nabla u_{t}\cdot u_{tt}- \int\nabla p_{t}\cdot u_{tt} \\& \quad\quad{}-\frac{2}{3} \int(\pi\nabla\rho_{t}+\rho_{t}\nabla\pi+\pi _{t}\nabla\rho+\rho\nabla\pi_{t})\cdot u_{tt} \\& \quad=J_{1}+J_{2}+J_{3}+J_{4}+J_{5}+J_{6}. \end{aligned}$$
(2.32)
Using the Hölder, Sobolev, and Young inequalities and (2.10) and (2.26), we get
$$\begin{aligned}& J_{1} \leq Cc_{0}^{\frac{1}{2}}\Vert \rho_{t}\Vert _{L^{3}}\Vert u_{t}\Vert _{L^{6}}\Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}} \leq Cc_{0}^{\frac{1}{2}}\Vert \rho_{t}\Vert _{L^{3}}\bigl(\Vert \sqrt{\rho}u_{t}\Vert _{L^{2}}+ \Vert \nabla u_{t}\Vert _{L^{2}}\bigr)\Vert \sqrt{ \rho}u_{tt}\Vert _{L^{2}} \\& \hphantom{ J_{1}}\leq Cc_{0}^{3}c_{2}^{2}\Vert \nabla u_{t}\Vert _{L^{2}}^{2}+Cc_{0}^{8+2\gamma}c_{2}^{2}+ \frac{1}{18}\Vert \sqrt{\rho }u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.33)
$$\begin{aligned}& J_{2}\leq Cc_{0}^{\frac{1}{2}}\Vert \sqrt{ \rho}u_{tt}\Vert _{L^{2}}\Vert \rho _{t}\Vert _{L^{3}} \Vert v\Vert _{L^{\infty}} \Vert \nabla u\Vert _{L^{6}} \leq Cc_{0}^{8+6\gamma}c_{1}^{4}c_{2}^{4}+ \frac{1}{18}\Vert \sqrt{\rho }u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.34)
$$\begin{aligned}& J_{3}\leq Cc_{0}^{\frac{1}{2}}\Vert \sqrt{ \rho}u_{tt}\Vert _{L^{2}}\Vert v_{t}\Vert _{L^{3}}\Vert \nabla u\Vert _{L^{6}} \leq Cc_{0}^{6+6\gamma}c_{1}^{4} \Vert v_{t}\Vert _{H^{1}}^{2}+\frac{1}{18} \Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.35)
$$\begin{aligned}& J_{4}\leq Cc_{0}^{\frac{1}{2}}\Vert v\Vert _{L^{\infty}} \Vert \sqrt{\rho }u_{tt}\Vert _{L^{2}}\Vert \nabla u_{t}\Vert _{L^{2}} \leq Cc_{0}c_{2}^{2} \Vert \nabla u_{t}\Vert _{L^{2}}^{2}+ \frac{1}{18}\Vert \sqrt {\rho}u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.36)
$$\begin{aligned}& J_{5}\leq Cc_{0}^{\frac{1}{2}}\Vert \sqrt{ \rho}u_{tt}\Vert _{L^{2}}\Vert \nabla p_{t}\Vert _{L^{2}} \leq Cc_{0}^{2\gamma+1}c_{2}^{2}+ \frac{1}{18}\Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.37)
$$\begin{aligned}& J_{6} \leq Cc_{0}^{\frac{1}{2}}\Vert \pi \Vert _{L^{\infty}} \Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}\Vert \nabla\rho_{t}\Vert _{L^{2}}+ Cc_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}\Vert \nabla\pi \Vert _{L^{4}}\Vert \rho_{t}\Vert _{L^{4}} \\& \hphantom{ J_{6}=}{} +Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}\Vert \nabla\rho \Vert _{L^{\infty}} \Vert \pi_{t}\Vert _{L^{2}} +Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}\Vert \nabla \pi_{t}\Vert _{L^{2}} \\& \hphantom{ J_{6}} \leq Cc_{0}^{3}c_{2}^{2}c_{5}^{2}+Cc_{0}^{3} \Vert \pi_{t}\Vert _{H^{1}}^{2}+\frac{2}{9} \Vert \sqrt{\rho}u_{tt}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.38)
inserting (2.33)-(2.38) to (2.32), then integrating the result over \((0,t)\), we derive
$$ \int_{0}^{t} \int_{\Omega}\rho u_{tt}^{2}\,\mathrm{d}x\,\mathrm{d}t+ \Vert \nabla u_{t}\Vert _{L^{2}}^{2}\leq Cc_{0}^{6+6\gamma}c_{1}^{5}c_{2}^{2}, $$
(2.39)
where we have used the equation (2.3) to get \(\lim_{t\rightarrow 0}\Vert \nabla u_{t}(t)\Vert _{L^{2}}^{2}\leq Cc_{0}^{2\gamma+4}\).
So, combining (2.29), (2.31), and (2.39), we obtain
$$ \int_{0}^{t}\Vert \rho u_{t}\Vert _{H^{2}}^{2}\leq Cc_{0}^{9+6\gamma }c_{1}^{5}c_{2}^{2}. $$
(2.40)
In the following, we shall estimate the rest terms of the inequality (2.28).
For the second term of the inequality (2.28), direct calculation yields
$$ \Vert \rho v\cdot\nabla u\Vert _{H^{2}}\leq C\Vert \rho \Vert _{H^{2}}\Vert v\Vert _{H^{2}}\Vert u\Vert _{H^{3}} \leq Cc_{0}c_{2}\Vert u\Vert _{H^{3}}, $$
(2.41)
therefore, we have to evaluate \(\Vert u\Vert _{H^{3}}\). In fact, applying the standard elliptic regularity result to the equation (2.3), we obtain
$$ \bigl\Vert \nabla^{3}u\bigr\Vert _{L^{2}}\leq C\bigl( \Vert \rho u_{t}\Vert _{H^{1}}+\Vert \rho v\cdot \nabla u \Vert _{H^{1}}+\Vert \nabla p\Vert _{H^{1}}+\bigl\Vert \nabla(\rho\pi)\bigr\Vert _{H^{1}}\bigr), $$
(2.42)
we could estimate the right-hand side of (2.42) item by item.
First, from (2.26), we have \(\Vert u_{t}\Vert _{L^{2}}\leq Cc_{0}^{\frac{5}{2}+\gamma}\), thus
$$\begin{aligned} \Vert \rho u_{t}\Vert _{H^{1}} \leq& Cc_{0} \Vert u_{t}\Vert _{L^{2}}+\Vert \nabla\rho \Vert _{L^{\infty}} \Vert u_{t}\Vert _{L^{2}}+Cc_{0} \Vert \nabla u_{t}\Vert _{L^{2}} \\ \leq& Cc_{0}^{\frac{7}{2}+\gamma}+Cc_{0} \Vert \nabla u_{t}\Vert _{L^{2}}. \end{aligned}$$
(2.43)
Second, using the Sobolev interpolation inequality and the Young inequality, we get
$$\begin{aligned}& \Vert \rho v\cdot\nabla u\Vert _{H^{1}} \\& \quad\leq C\bigl(\Vert \rho v \cdot\nabla u\Vert _{L^{2}}+\bigl\Vert \nabla(\rho v\cdot\nabla u)\bigr\Vert _{L^{2}}\bigr) \\& \quad\leq C\bigl(c_{0}\Vert v\Vert _{L^{\infty}} \Vert \nabla u \Vert _{L^{2}}+\Vert \nabla\rho \Vert _{L^{\infty}} \Vert v\Vert _{L^{\infty}} \Vert \nabla u\Vert _{L^{2}} +c_{0}\Vert \nabla v\Vert _{L^{2}}\Vert \nabla u\Vert _{L^{2}}^{\frac{1}{4}} \bigl\Vert \nabla ^{3} u\bigr\Vert _{L^{2}}^{\frac{3}{4}} \\& \quad\quad{}+c_{0}\Vert v\Vert _{L^{\infty}}\bigl\Vert \nabla^{2}u\bigr\Vert _{L^{2}}\bigr) \\& \quad \leq Cc_{0}^{\frac{13}{2}+3\gamma}c_{1}^{4}c_{2}+ \frac{3}{4}\Vert u\Vert _{H^{3}}. \end{aligned}$$
(2.44)
Third, due to (2.11), we easily derive
$$ \Vert \nabla p\Vert _{H^{1}}\leq Cc_{0}^{2}. $$
(2.45)
Last, by simple calculation, one gets
$$ \bigl\Vert \nabla(\rho\pi)\bigr\Vert _{H^{1}} \leq C\Vert \rho \Vert _{H^{3}}\Vert \pi \Vert _{H^{2}} \leq Cc_{0}c_{5}. $$
(2.46)
Combining (2.39) and (2.42)-(2.46), we deduce
$$ \Vert u\Vert _{H^{3}} \leq Cc_{0}^{\frac{13}{2}+3\gamma}c_{1}^{4}c_{2}c_{5}. $$
(2.47)
Next, by simple calculation, the third and fourth terms on the right-hand side of (2.28) can be estimated as
$$ \Vert \nabla p\Vert _{H^{2}}\leq Cc_{0}^{3}, \quad\quad \bigl\Vert \nabla(\rho\pi)\bigr\Vert _{H^{2}}\leq Cc_{0} \Vert \pi \Vert _{H^{3}}. $$
(2.48)
Combining (2.26), (2.28), (2.40), (2.41), and (2.47)-(2.48), one deduces
$$ \int_{0}^{t}\Vert u\Vert _{H^{4}}^{2} \,\mathrm{d}t\leq Cc_{0}^{9+6\gamma }c_{1}^{5}c_{2}^{2}, $$
(2.49)
for \(0\leq t\leq T\).
Thus, we complete the proof of Lemma 2.2. □
In the following part, we estimate the turbulent kinetic energy k.
Lemma 2.3
There exists a unique strong solution k to the initial boundary value problem (2.5) and (2.9) such that
$$\begin{aligned}& \Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2} + \Vert k\Vert _{H^{1}}^{2} + \int_{0}^{t}\Vert \nabla k_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5}, \end{aligned}$$
(2.50)
$$\begin{aligned}& \Vert k\Vert _{H^{2}}\leq Cc_{0}^{\frac{7}{2}}c_{1}c_{2}^{2},\quad\quad \int_{0}^{t}\Vert k\Vert _{H^{3}}^{2} \,\mathrm{d}s\leq Cc_{0}^{7}, \end{aligned}$$
(2.51)
for
\(0\leq t\leq T\).
Proof
We only need to prove the estimates. Differentiating the equation (2.5) with respect to t, then multiplying both sides of the resulting equation by \(k_{t}\) and integrating over Ω, we get
$$\begin{aligned} \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2} +\Vert \nabla k_{t}\Vert _{L^{2}}^{2} =&- \int\rho_{t}v\cdot\nabla k\cdot k_{t} - \int\rho v_{t}\cdot\nabla k\cdot k_{t} -2 \int\rho v\cdot\nabla k_{t}\cdot k_{t} \\ &{} + \int G_{t}^{\prime}\cdot k_{t} - \int\rho_{t}\theta\cdot k_{t} - \int\rho\theta_{t}\cdot k_{t} \\ =& \sum _{i=1}^{6}K_{i}, \end{aligned}$$
(2.52)
we could evaluate \(K_{i}\) (\(i=1,\ldots,6\)) as follows.
First, using a similar method to deriving (2.15), (2.20), (2.16), respectively, one has
$$\begin{aligned}& K_{1} \leq Cc_{0}^{2}c_{2}^{4} \Vert \nabla k\Vert _{L^{2}}^{2}+C\Vert \sqrt{ \rho}k_{t}\Vert _{L^{2}}^{2}+\frac{1}{10} \Vert \nabla k_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.53)
$$\begin{aligned}& K_{2}\leq C\eta^{-1}c_{0}^{2\gamma+1}\bigl( \Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+c_{1}^{2} \Vert \nabla k\Vert _{L^{2}}^{2} +c_{0}^{2}c_{1}^{2}c_{2}^{4} \bigr) +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.54)
$$\begin{aligned}& K_{3} \leq Cc_{0}c_{2}^{2}\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+ \frac{1}{10}\Vert \nabla k_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.55)
Next, differentiating \(G^{\prime}\) with respect to t and inserting the result thus obtained into \(K_{4}\) yield
$$\begin{aligned} K_{4} \leq& C \int \vert \nabla v_{t}\vert \vert \nabla v\vert \vert k_{t}\vert +C \int \vert \rho \vert \vert \pi \vert \vert \nabla v_{t}\vert \vert k_{t}\vert +C \int \vert \rho_{t}\vert \vert \pi \vert \vert \nabla v\vert \vert k_{t}\vert \\ &{}+C \int \vert \rho \vert \vert \pi_{t}\vert \vert \nabla v\vert \vert k_{t}\vert \\ \leq& Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho}k_{t} \Vert _{L^{2}}\Vert \nabla v_{t}\Vert _{L^{2}} \Vert \nabla v\Vert _{L^{\infty}} +Cc_{0}^{\frac{1}{2}}\Vert \pi \Vert _{L^{\infty}} \Vert \nabla v_{t}\Vert _{L^{2}} \Vert \sqrt{\rho}k_{t}\Vert _{L^{2}} \\ &{}+C\Vert \pi \Vert _{L^{\infty}} \Vert \rho_{t}\Vert _{L^{3}}\Vert \nabla v\Vert _{L^{2}}\Vert k_{t} \Vert _{L^{6}} +Cc_{0}^{\frac{1}{2}}\Vert \sqrt{ \rho}k_{t}\Vert _{L^{2}}\Vert \pi_{t}\Vert _{L^{6}}\Vert \nabla v\Vert _{L^{3}} \\ \leq& C\eta^{-1}c_{0}c_{3}^{2}c_{5}^{2}+Cc_{0}^{2}c_{1}^{2}c_{2}^{2}c_{5}^{2} +C\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2} +C\eta \bigl(\Vert v_{t}\Vert _{H^{1}}^{2} +\Vert \pi_{t}\Vert _{H^{1}}^{2}\bigr)\Vert \sqrt{ \rho}k_{t}\Vert _{L^{2}}^{2} \\ &{} +\frac{1}{10} \Vert \nabla k_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.56)
Last, direct calculation leads to
$$\begin{aligned}& K_{5}\leq \Vert \rho_{t}\Vert _{L^{3}}\Vert \theta \Vert _{L^{2}}\Vert k_{t}\Vert _{L^{6}} \leq Cc_{0}^{2}c_{1}^{2}c_{2}^{2} +C\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2} + \frac{1}{10}\Vert \nabla k_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.57)
$$\begin{aligned}& K_{6}\leq Cc_{0}^{\frac{1}{2}}\Vert \sqrt{ \rho}k_{t}\Vert _{L^{2}}\Vert \theta _{t}\Vert _{L^{2}} \leq C\eta^{-1}c_{0}+\eta \Vert \theta_{t}\Vert _{L^{2}}^{2}\Vert \sqrt{\rho }k_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.58)
On the other hand, we easily get
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla k\Vert _{L^{2}}^{2}\leq \frac {1}{10}\Vert \nabla k_{t}\Vert _{L^{2}}^{2} +C\Vert \nabla k\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.59)
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\Vert k\Vert _{L^{2}}^{2} \leq Cc_{0}\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+C \Vert k\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.60)
Combining (2.52)-(2.60), we obtain
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\bigl(\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+\Vert k\Vert _{H^{1}}^{2} \bigr)+\Vert \nabla k_{t}\Vert _{L^{2}}^{2} \\& \quad\leq C\bigl(c_{0}^{2}c_{2}^{4}+ \eta^{-1}c_{0}^{2\gamma+1}c_{1}^{2}+ \eta \Vert v_{t}\Vert _{H^{1}}^{2}+\eta \Vert \pi_{t}\Vert _{H^{1}}^{2} +\eta \Vert \theta_{t}\Vert _{L^{2}}^{2}\bigr) \bigl(\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2} +\Vert k\Vert _{H^{1}}^{2}\bigr) \\& \quad\quad{}+C\bigl(\eta ^{-1}c_{0}^{2}c_{1}^{2}c_{2}^{4} c_{3}^{2}c_{5}^{2}+c_{0}^{2}c_{1}^{2}c_{2}^{2}c_{5}^{2} \bigr), \end{aligned}$$
(2.61)
setting \(\eta=c_{1}^{-1}\) and using the Gronwall inequality, we deduce
$$ \Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+\Vert k\Vert _{H^{1}}^{2}+ \int_{0}^{t}\Vert \nabla k_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s \leq Cc_{0}^{5} $$
(2.62)
for \(0\leq t\leq T\), where we have used the fact that \(\lim _{t\rightarrow0}(\Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}^{2}+\Vert k\Vert _{H^{1}}^{2})\leq Cc_{0}^{5}\).
Then, by the standard elliptic regularity result of the equation (2.5) and using (2.62), we have
$$\begin{aligned} \Vert \nabla k\Vert _{H^{1}} \leq& Cc_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}k_{t}\Vert _{L^{2}}+Cc_{0} \Vert v\Vert _{L^{\infty}} \Vert \nabla k\Vert _{L^{2}}+ C\Vert \nabla v\Vert _{L^{4}}^{2} \\ &{}+Cc_{0}\Vert \pi \Vert _{L^{4}}\Vert \nabla v\Vert _{L^{4}}+Cc_{0}\Vert \theta \Vert _{L^{2}} \\ \leq& Cc_{0}^{\frac{7}{2}}c_{1}c_{2}^{2} \end{aligned}$$
(2.63)
and
$$ \bigl\Vert \nabla^{2} k\bigr\Vert _{H^{1}}\leq C\bigl( \Vert \rho k_{t}\Vert _{H^{1}}+\Vert \rho v\cdot \nabla k \Vert _{H^{1}}+\bigl\Vert G^{\prime}\bigr\Vert _{H^{1}}+ \Vert \rho\theta \Vert _{H^{1}}\bigr). $$
(2.64)
To evaluate \(\int_{0}^{t}\Vert k\Vert _{H^{3}}^{2}\,\mathrm{d}t\), we will estimate the right-hand side of (2.64) item by item.
In fact, we derive by using (2.62) and (2.63) that
$$\begin{aligned}& \Vert \rho k_{t}\Vert _{H^{1}}\leq C\bigl(\Vert \rho k_{t}\Vert _{L^{2}}+\bigl\Vert \nabla(\rho k_{t})\bigr\Vert _{L^{2}}\bigr) \\& \hphantom{\Vert \rho k_{t}\Vert _{H^{1}}} \leq Cc_{0}^{\frac{7}{2}}+Cc_{0} \Vert \nabla k_{t}\Vert _{L^{2}}, \end{aligned}$$
(2.65)
$$\begin{aligned}& \Vert \rho v\cdot\nabla k\Vert _{H^{1}}\leq C\bigl(\Vert \rho v \cdot\nabla k\Vert _{L^{2}}+\bigl\Vert \nabla(\rho v\cdot\nabla k)\bigr\Vert _{L^{2}}\bigr) \\& \hphantom{\Vert \rho v\cdot\nabla k\Vert _{H^{1}}}\leq C\bigl(c_{0}\Vert v\Vert _{L^{\infty}} \Vert \nabla k \Vert _{L^{2}}+\Vert \nabla\rho \Vert _{L^{\infty}} \Vert v\Vert _{L^{\infty}} \Vert \nabla k\Vert _{L^{2}} \\& \hphantom{\Vert \rho v\cdot\nabla k\Vert _{H^{1}}\leq}{} +c_{0}\Vert \nabla v\Vert _{L^{4}} \Vert \nabla k \Vert _{L^{4}}+c_{0}\Vert v\Vert _{L^{\infty }}\bigl\Vert \nabla^{2}k\bigr\Vert _{L^{2}}\bigr) \\& \hphantom{\Vert \rho v\cdot\nabla k\Vert _{H^{1}} }\leq Cc_{0}^{\frac{9}{2}}c_{1}c_{2}^{3}, \end{aligned}$$
(2.66)
$$\begin{aligned}& \bigl\Vert G^{\prime}\bigr\Vert _{H^{1}}\leq C\bigl(\Vert \nabla v\Vert _{L^{4}}^{2}+\Vert \nabla v\cdot\rho \cdot\pi \Vert _{L^{2}}+\bigl\Vert \nabla v\cdot\nabla^{2}v\bigr\Vert _{L^{2}} +\bigl\Vert \nabla(\nabla v\cdot\rho\cdot\pi)\bigr\Vert _{L^{2}}\bigr) \\& \hphantom{\bigl\Vert G^{\prime}\bigr\Vert _{H^{1}} }\leq C\bigl(\Vert \nabla v\Vert _{L^{4}}^{2}+c_{0} \Vert \pi \Vert _{L^{\infty}} \Vert \nabla v\Vert _{L^{2}}+\Vert \nabla v\Vert _{L^{4}}\bigl\Vert \nabla^{2} v\bigr\Vert _{L^{4}} +c_{0}\Vert \pi \Vert _{L^{\infty}}\bigl\Vert \nabla^{2} v\bigr\Vert _{L^{2}} \\& \hphantom{\bigl\Vert G^{\prime}\bigr\Vert _{H^{1}}\leq}{} +\Vert \pi \Vert _{L^{\infty}} \Vert \nabla\rho \Vert _{L^{\infty}} \Vert \nabla v\Vert _{L^{2}}+c_{0}\Vert \nabla v\Vert _{L^{\infty}} \Vert \nabla\pi \Vert _{L^{2}}\bigr) \\& \hphantom{\bigl\Vert G^{\prime}\bigr\Vert _{H^{1}} } \leq Cc_{0}c_{1}c_{2}^{2}c_{3}c_{5}, \end{aligned}$$
(2.67)
and
$$ \Vert \rho\theta \Vert _{H^{1}}\leq C\Vert \rho \Vert _{H^{3}}\Vert \theta \Vert _{H^{1}}\leq Cc_{0}c_{1}. $$
(2.68)
Therefore, inserting (2.65)-(2.68) to (2.64) and integrating the result thus obtained over \((0,t)\), one gets
$$ \int_{0}^{t}\Vert k\Vert _{H^{3}}^{2} \,\mathrm{d}t\leq Cc_{0}^{7} $$
(2.69)
for \(0\leq t\leq T\).
Combining (2.62), (2.63), and (2.69), we complete the proof of Lemma 2.3. □
In the next part, we estimate the viscous dissipation rates of the turbulent flows ε.
Lemma 2.4
There exists a unique strong solution
ε
to the initial boundary value problem (2.6) and (2.9) such that
$$\begin{aligned}& \Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2} +\Vert \varepsilon \Vert _{H^{1}}^{2} + \int_{0}^{t}\Vert \nabla\varepsilon_{t} \Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5}, \end{aligned}$$
(2.70)
$$\begin{aligned}& \Vert \varepsilon \Vert _{H^{2}}\leq Cc_{0}^{\frac {9}{2}}c_{1}^{2}c_{2}^{2} \end{aligned}$$
(2.71)
for
\(0\leq t\leq T\).
Proof
We only need to prove the estimates. Differentiating the equation (2.6) with respect to t, then multiplying both sides of the result by \(\varepsilon_{t}\) and integrating over Ω, one obtains
$$\begin{aligned}& \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\Vert \sqrt{\rho}\varepsilon _{t}\Vert _{L^{2}}^{2} +\Vert \nabla \varepsilon_{t}\Vert _{L^{2}}^{2} \\ & \quad= - \int\rho_{t}v\cdot\nabla\varepsilon\cdot\varepsilon_{t} - \int\rho v_{t}\cdot\nabla\varepsilon\cdot\varepsilon_{t} -2 \int\rho v\cdot\nabla\varepsilon_{t}\cdot\varepsilon _{t} \\ & \quad\quad{}+ \int \biggl(\frac{C_{1}G^{\prime}\theta}{\pi} \biggr)_{t}\cdot \varepsilon_{t} - \int \biggl(\frac{C_{2}\rho\theta^{2}}{\pi} \biggr)_{t}\cdot \varepsilon_{t} \\ & \quad= \sum _{i=1}^{5}E_{i}. \end{aligned}$$
(2.72)
We could evaluate \(E_{4}\) and \(E_{5}\) in the first place. Because π has an upper and a lower bound away from zero, direct calculation yields
$$\begin{aligned} E_{4} \leq& C \int\bigl(\bigl\vert G^{\prime}_{t}\theta\bigr\vert +\bigl\vert G^{\prime}\theta_{t}\bigr\vert +\bigl\vert G^{\prime}\theta \pi_{t}\bigr\vert \bigr)\vert \varepsilon_{t}\vert \\ \leq& C \int\bigl(\vert \nabla v_{t}\cdot\nabla v\vert +\vert \rho_{t}\pi\nabla v\vert +\vert \rho\pi_{t}\nabla v \vert +\vert \rho\pi\nabla v_{t}\vert \bigr)\vert \theta \vert \vert \varepsilon_{t}\vert \\ &{}+C \int\bigl(\vert \nabla v\vert ^{2}+\vert \rho\pi\nabla v \vert \bigr)\vert \theta_{t}\vert \vert \varepsilon_{t} \vert +C \int\bigl(\vert \nabla v\vert ^{2}+\vert \rho\pi\nabla v \vert \bigr)\vert \theta \vert \vert \pi _{t}\vert \vert \varepsilon_{t}\vert \\ \leq &Cc_{0}^{\frac{1}{2}}\Vert \theta \Vert _{L^{\infty}} \Vert \nabla v\Vert _{L^{\infty}} \Vert \nabla v_{t}\Vert _{L^{2}}\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}} +Cc_{0}^{\frac{1}{2}}\Vert \pi \Vert _{L^{\infty}} \Vert \sqrt{\rho}\varepsilon _{t}\Vert _{L^{2}}\Vert \rho_{t}\Vert _{L^{6}}\Vert \nabla v \Vert _{L^{6}}\Vert \theta \Vert _{L^{6}} \\ &{}+Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho}\varepsilon_{t} \Vert _{L^{2}}\Vert \pi _{t}\Vert _{L^{6}}\Vert \nabla v\Vert _{L^{6}}\Vert \theta \Vert _{L^{6}} +Cc_{0}\Vert \pi \Vert _{L^{\infty}} \Vert \theta \Vert _{L^{\infty}} \Vert \nabla v_{t}\Vert _{L^{2}}\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}} \\ &{}+C\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}\Vert \theta_{t}\Vert _{L^{2}}\Vert \nabla v\Vert _{L^{\infty}}^{2} +Cc_{0}\Vert \pi \Vert _{L^{\infty}} \Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}\Vert \theta_{t}\Vert _{L^{2}}\Vert \nabla v\Vert _{L^{\infty}} \\ &{}+Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho}\varepsilon_{t} \Vert _{L^{2}}\Vert \pi _{t}\Vert _{L^{6}}\Vert \nabla v\Vert _{L^{6}}^{2}\Vert \theta \Vert _{L^{\infty}} +Cc_{0}^{\frac{1}{2}}\Vert \pi \Vert _{L^{\infty}} \Vert \sqrt{ \rho}\varepsilon _{t}\Vert _{L^{2}}\Vert \pi_{t} \Vert _{L^{6}}\Vert \nabla v\Vert _{L^{6}}\Vert \theta \Vert _{L^{6}} \\ \leq& C\eta ^{-1}c_{0}c_{1}^{2}c_{2}^{4}c_{6}^{2}c_{3}^{4}c_{5}^{2}+Cc_{0}^{4}c_{1}^{2}c_{2}^{4}c_{5}^{2} +C\eta\bigl(\Vert \nabla v_{t}\Vert _{L^{2}}^{2}+ \Vert \pi_{t}\Vert _{L^{6}}^{2}+\Vert \theta _{t}\Vert _{L^{2}}^{2}\bigr) \Vert \sqrt{\rho} \varepsilon_{t}\Vert _{L^{2}}^{2} \\ &{}+C\Vert \sqrt{ \rho }\varepsilon_{t}\Vert _{L^{2}}^{2} \end{aligned}$$
(2.73)
and
$$\begin{aligned} E_{5} \leq& C \int\bigl\vert \rho_{t}\theta^{2} \varepsilon_{t}\bigr\vert +C \int \vert \theta\theta_{t}\rho\varepsilon_{t} \vert +C \int\bigl\vert \rho\theta^{2}\pi_{t} \varepsilon_{t}\bigr\vert \\ \leq& C\Vert \rho_{t}\Vert _{L^{3}}\Vert \theta \Vert _{L^{4}}^{2}\Vert \varepsilon_{t}\Vert _{L^{6}} +Cc_{0}^{\frac{1}{2}}\Vert \sqrt{\rho} \varepsilon_{t}\Vert _{L^{2}}\Vert \theta _{t} \Vert _{L^{2}}\Vert \theta \Vert _{L^{\infty}} +Cc_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}\Vert \pi _{t}\Vert _{L^{2}}\Vert \theta \Vert _{L^{\infty}}^{2} \\ \leq& C\eta^{-1}c_{0}c_{6}^{4}+Cc_{0}^{2}c_{1}^{4}c_{2}^{2}+C \Vert \sqrt {\rho}\varepsilon_{t}\Vert _{L^{2}}^{2}+C \eta\bigl(\Vert \theta_{t}\Vert _{L^{2}}^{2} + \Vert \pi_{t}\Vert _{L^{2}}^{2}\bigr)\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2} \\ &{}+ \frac{1}{8}\Vert \nabla\varepsilon_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.74)
Next, using an argument similar to that used in deriving (2.53), (2.54), (2.55), (2.60), and (2.59), respectively, one gets
$$\begin{aligned}& E_{1} \leq Cc_{0}^{2}c_{2}^{4} \Vert \nabla\varepsilon \Vert _{L^{2}}^{2}+C\Vert \sqrt { \rho}\varepsilon_{t}\Vert _{L^{2}}^{2}+ \frac{1}{10}\Vert \nabla\varepsilon _{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.75)
$$\begin{aligned}& E_{2}\leq C\eta^{-1}c_{0}^{2\gamma+1}\bigl( \Vert \sqrt{\rho}\varepsilon _{t}\Vert _{L^{2}}^{2}+c_{1}^{2} \Vert \nabla\varepsilon \Vert _{L^{2}}^{2} +c_{1}^{4}c_{2}^{4}\bigr) +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{\rho} \varepsilon_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.76)
$$\begin{aligned}& E_{3} \leq Cc_{0}c_{2}^{2}\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2}+ \frac {1}{10}\Vert \nabla\varepsilon_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.77)
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\Vert \varepsilon \Vert _{L^{2}}^{2} \leq C\Vert \varepsilon \Vert _{L^{2}}^{2}+Cc_{0} \Vert \sqrt{\rho}\varepsilon _{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.78)
and finally
$$ \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla\varepsilon \Vert _{L^{2}}^{2} \leq\frac{1}{8}\Vert \nabla\varepsilon_{t}\Vert _{L^{2}}^{2} +C\Vert \nabla\varepsilon \Vert _{L^{2}}^{2}. $$
(2.79)
Combining (2.72)-(2.79), one obtains
$$\begin{aligned}& \frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\bigl(\Vert \sqrt{\rho}\varepsilon _{t}\Vert _{L^{2}}^{2}+ \Vert \varepsilon \Vert _{H^{1}}^{2}\bigr) +\Vert \nabla\varepsilon_{t} \Vert _{L^{2}}^{2} \\& \quad\leq C\bigl(c_{0}^{2}c_{2}^{4} + \eta^{-1}c_{0}^{2\gamma+1}c_{1}^{2} + \eta \Vert v_{t}\Vert _{H^{1}}^{2} +\eta \Vert \theta_{t}\Vert _{H^{1}}^{2} +\eta \Vert \pi_{t}\Vert _{H^{1}}^{2} \bigr) \bigl(\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2}+ \Vert \varepsilon \Vert _{H^{1}}^{2}\bigr) \\& \quad\quad{}+C\eta^{-1}c_{0}c_{1}^{4}c_{2}^{4}c_{6}^{4}c_{3}^{4}c_{5}^{2} +Cc_{0}^{4}c_{1}^{4}c_{2}^{4}c_{5}^{2}, \end{aligned}$$
(2.80)
setting \(\eta=c_{1}^{-1}\) and using the Gronwall inequality, one obtains
$$ \Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2} +\Vert \varepsilon \Vert _{H^{1}}^{2} + \int_{0}^{t}\Vert \nabla\varepsilon_{t} \Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5} $$
(2.81)
for \(0\leq t\leq T\), where we have used the fact that \(\lim _{t\rightarrow0}(\Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}}^{2}+\Vert \varepsilon \Vert _{H^{1}}^{2})\leq Cc_{0}^{5}\).
Next, applying the standard elliptic regularity result to the equation (2.6) and using (2.81), we have
$$\begin{aligned} \Vert \nabla\varepsilon \Vert _{H^{1}} \leq& C\bigl(c_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}\varepsilon_{t}\Vert _{L^{2}} +c_{0}\Vert v\Vert _{L^{6}}\Vert \nabla\varepsilon \Vert _{L^{3}} +\Vert \nabla v\Vert _{L^{6}}^{2}\Vert \theta \Vert _{L^{6}} \\ &{}+c_{0}\Vert \nabla v\Vert _{L^{6}}\Vert \theta \Vert _{L^{6}}\Vert \pi \Vert _{L^{6}} +c_{0}\Vert \theta \Vert _{L^{4}}^{2} \bigr) \\ \leq& C\bigl(c_{0}^{3}c_{2}^{2}c_{1}^{2}+c_{0}c_{1} \Vert \nabla\varepsilon \Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla \varepsilon \Vert _{L^{6}}^{\frac {1}{2}}\bigr), \end{aligned}$$
(2.82)
therefore, by the Young inequality and (2.81), one deduces
$$\Vert \varepsilon \Vert _{H^{2}}\leq Cc_{0}^{\frac{9}{2}}c_{1}^{2}c_{2}^{2}. $$
Thus, we complete the proof of Lemma 2.4. □
Finally, we estimate the total enthalpy h.
Lemma 2.5
There exists a unique strong solution h to the initial boundary value problem (2.4) and (2.9) such that
$$\begin{aligned}& \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2} + \Vert h\Vert _{H^{1}}^{2} + \int_{0}^{t}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5}, \end{aligned}$$
(2.83)
$$\begin{aligned}& \Vert h\Vert _{H^{2}}\leq Cc_{0}^{\frac{7}{2}+\gamma }c_{1}^{2}c_{2}^{2} \end{aligned}$$
(2.84)
for
\(0\leq t\leq T\).
Proof
We only need to prove the estimates. Differentiating equation (2.4) with respect to t, multiplying both sides of the result equation by \(h_{t}\) and integrating over Ω, one obtains
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\bigl(\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+\Vert h\Vert _{H^{1}}^{2} \bigr) +\Vert \nabla h_{t}\Vert _{L^{2}}^{2} \\ & \quad=- \int\rho_{t}v\cdot\nabla h\cdot h_{t} - \int\rho v_{t}\cdot\nabla h\cdot h_{t} -2 \int\rho v\cdot\nabla h_{t}\cdot h_{t} + \int p_{tt}\cdot h_{t} \\ & \quad\quad{}+ \int u_{t}\cdot\nabla p\cdot h_{t} + \int u\cdot\nabla p_{t}\cdot h_{t} + \int S_{kt}^{\prime}\cdot h_{t} \\ & \quad = \sum _{i=1}^{7}H_{i}. \end{aligned}$$
(2.85)
First of all, using similar methods of deriving the estimates (2.15), (2.20), and (2.16), respectively, one has
$$\begin{aligned}& H_{1} \leq Cc_{0}^{2}c_{2}^{4} \Vert \nabla h\Vert _{L^{2}}^{2}+C\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}}^{2}+\frac{1}{20} \Vert \nabla h_{t}\Vert _{L^{2}}, \end{aligned}$$
(2.86)
$$\begin{aligned}& H_{2}\leq C\eta^{-1}c_{0}^{2\gamma+1} \bigl(c_{0}^{7}c_{2}^{4}+\Vert \sqrt {\rho}h_{t}\Vert _{L^{2}}^{2}+c_{1}^{2} \Vert \nabla h\Vert _{L^{2}}^{2}\bigr) +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}}^{2}, \end{aligned}$$
(2.87)
$$\begin{aligned}& H_{3} \leq Cc_{0}c_{2}^{2}\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+ \frac{1}{20}\Vert \nabla h_{t}\Vert _{L^{2}}. \end{aligned}$$
(2.88)
Second, differentiating the equation (2.2) with respect to t yields
$$ \rho_{tt}=-\rho_{t}\nabla\cdot v +\rho\nabla\cdot v_{t} +v_{t}\cdot\nabla\rho +v\cdot\nabla \rho_{t}. $$
(2.89)
Therefore, by direct calculation and using (2.89), we derive
$$\begin{aligned} H_{4} =& \int\bigl[\gamma(\gamma-1)\rho^{\gamma-2}\rho_{t}^{2} -\gamma\rho^{\gamma-1}(\rho_{t}\nabla\cdot v +\rho\nabla\cdot v_{t} +v_{t}\cdot\nabla\rho +v\cdot\nabla \rho_{t})\bigr]\cdot h_{t} \\ \leq& Cc_{0}^{\gamma-\frac{3}{2}}\Vert \rho_{t}\Vert _{L^{4}}^{2}\Vert \sqrt {\rho}h_{t}\Vert _{L^{2}} +Cc_{0}^{\gamma-\frac{1}{2}}\Vert \rho_{t} \Vert _{L^{3}}\Vert \nabla v\Vert _{L^{6}}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}} \\ &{}+Cc_{0}^{\gamma-\frac{1}{2}} \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}\Vert \nabla v_{t}\Vert _{L^{2}} +Cc_{0}^{\gamma-\frac{1}{2}}\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}\Vert v_{t}\Vert _{L^{6}}\Vert \nabla\rho \Vert _{L^{3}} \\ &{}+Cc_{0}^{\gamma-\frac{1}{2}}\Vert \nabla \rho_{t}\Vert _{L^{2}}\Vert v\Vert _{L^{\infty}} \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}} \\ \leq& C\bigl(c_{0}^{2\gamma+1}c_{2}^{4}+ \eta^{-1}c_{0}^{2\gamma}+\Vert \sqrt { \rho}h_{t}\Vert _{L^{2}}^{2} +\eta \Vert v_{t}\Vert _{H^{1}}^{2}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}}^{2}\bigr)+ \frac {1}{20}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.90)
Third, simple calculation and (2.26) lead to
$$\begin{aligned} H_{5} \leq& Cc_{0}^{\gamma-\frac{1}{2}}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}}\Vert \nabla\rho \Vert _{L^{3}}\Vert u_{t}\Vert _{L^{6}} \leq Cc_{0}^{\gamma-\frac{1}{2}}\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}\Vert \nabla\rho \Vert _{L^{3}}\bigl(\Vert u_{t}\Vert _{L^{2}}+\Vert \nabla u_{t}\Vert _{L^{2}}\bigr) \\ \leq &Cc_{0}^{2\gamma+1}\Vert \sqrt{\rho}h_{t} \Vert _{L^{2}}^{2}+Cc_{0}^{2\gamma+5} +C\Vert \nabla u_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.91)
Next, by direct calculation, we know that \(\nabla p_{t}= \gamma(\gamma -1)\rho^{\gamma-2}\rho_{t}\nabla\rho +\gamma\rho^{\gamma-1}\nabla\rho_{t}\). Therefore,
$$\begin{aligned} H_{6} \leq& Cc_{0}^{\gamma-2} \int \vert \rho_{t}\vert \vert u\vert \vert \nabla \rho \vert \vert h_{t}\vert +Cc_{0}^{\gamma-1} \int \vert u\vert \vert \nabla\rho_{t}\vert \vert h_{t}\vert \\ \leq& Cc_{0}^{\gamma-2}\Vert \nabla\rho \Vert _{L^{\infty}} \Vert \rho_{t}\Vert _{L^{3}}\Vert u\Vert _{L^{2}} \bigl(\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}} +\Vert \nabla h_{t}\Vert _{L^{2}}\bigr) \\ &{}+Cc_{0}^{\gamma-1}\Vert u\Vert _{L^{3}}\Vert \nabla\rho_{t}\Vert _{L^{2}}\bigl(\Vert \sqrt { \rho}h_{t}\Vert _{L^{2}}+\Vert \nabla h_{t} \Vert _{L^{2}}\bigr) \\ \leq& Cc_{0}^{7+2\gamma}c_{2}^{2}+C\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+ \frac{1}{20}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}. \end{aligned}$$
(2.92)
Last, simple calculation yields \(\vert S_{kt}^{\prime} \vert \leq C\vert \nabla v\vert \vert \nabla v_{t}\vert +C\rho^{\gamma-1}\vert \rho_{t}\vert \vert \nabla\rho \vert ^{2} +C\rho^{\gamma-1}\vert \nabla\rho_{t}\vert \vert \nabla\rho \vert \), thus
$$\begin{aligned} H_{7} \leq& C \int \vert \nabla v_{t}\vert \vert \nabla v\vert \vert h_{t}\vert +Cc_{0}^{\gamma-1} \int \vert \rho _{t}\vert \vert \nabla\rho \vert ^{2}\vert h_{t}\vert +Cc_{0}^{\gamma-1} \int \vert \nabla\rho_{t}\vert \vert \nabla\rho \vert \vert h_{t}\vert \\ \leq& Cc_{0}^{\frac{1}{2}}\Vert \nabla v\Vert _{L^{\infty}} \Vert \nabla v_{t}\Vert _{L^{2}}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}} +Cc_{0}^{\gamma-\frac{1}{2}} \Vert \rho_{t}\Vert _{L^{6}}\Vert \nabla\rho \Vert _{L^{6}}^{2}\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}} \\ &{}+Cc_{0}^{\gamma-\frac{1}{2}}\Vert \nabla\rho \Vert _{L^{\infty}} \Vert \nabla \rho_{t}\Vert _{L^{2}}\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}} \\ \leq& C\bigl(\eta^{-1}c_{0}c_{3}^{2}+c_{0}^{5+2\gamma}c_{2}^{2} +\eta \Vert \nabla v_{t}\Vert _{L^{2}}^{2}\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2} +\Vert \sqrt{ \rho}h_{t}\Vert _{L^{2}}^{2}\bigr). \end{aligned}$$
(2.93)
Furthermore, we easily have
$$ \frac{\mathrm{d}}{\mathrm{d}t}\Vert h\Vert _{L^{2}}^{2}\leq Cc_{0}\Vert \sqrt{\rho }h_{t}\Vert _{L^{2}}^{2}+C \Vert h\Vert _{L^{2}}^{2} $$
(2.94)
and
$$ \frac{\mathrm{d}}{\mathrm{d}t}\Vert \nabla h\Vert _{L^{2}}^{2}\leq C \Vert \nabla h\Vert _{L^{2}}^{2}+\frac{1}{10}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}. $$
(2.95)
Consequently, combining (2.85)-(2.95), one deduces
$$\begin{aligned}& \frac{\mathrm{d}}{\mathrm{d}t}\bigl(\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+\Vert h\Vert _{H^{1}}^{2} \bigr) +\Vert \nabla h_{t}\Vert _{L^{2}}^{2} \\& \quad\leq C\bigl(c_{0}^{2\gamma+1}c_{2}^{4}+ \eta^{-1}c_{0}^{2\gamma +1}c_{1}^{2}+ \eta \Vert v_{t}\Vert _{H^{1}}^{2}\bigr) \bigl( \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+\Vert h \Vert _{H^{1}}^{2}\bigr) \\& \quad\quad{}+C\bigl(c_{0}^{7+2\gamma}c_{2}^{4}+ \eta^{-1}c_{0}^{8+2\gamma }c_{2}^{4}c_{3}^{2} \bigr), \end{aligned}$$
(2.96)
setting \(\eta=c_{1}^{-1}\) and using the Gronwall inequality, we get
$$ \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+\Vert h\Vert _{H^{1}}^{2} + \int_{0}^{t}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s \leq Cc_{0}^{5} $$
(2.97)
for \(0\leq t \leq T\), where we have used the fact that \(\lim _{t\rightarrow0}(\Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2}+\Vert h\Vert _{H^{1}}^{2})\leq Cc_{0}^{5}\).
Next, using (2.97) and the standard elliptic regularity result of the equation (2.4), one obtains
$$\begin{aligned} \Vert \nabla h\Vert _{H^{1}} \leq& C\bigl(c_{0}^{\frac{1}{2}} \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}} +c_{0} \Vert v\Vert _{L^{6}}\Vert \nabla h\Vert _{L^{3}} +c_{0}^{\gamma-1}\Vert \rho_{t}\Vert _{L^{2}}+c_{0}^{\gamma-1}\Vert u\Vert _{L^{6}} \Vert \nabla\rho \Vert _{L^{3}} \\ &{}+\Vert \nabla v\Vert _{L^{4}}^{2}+c_{0}^{\gamma-1} \Vert \nabla\rho \Vert _{L^{4}}^{2}\bigr) \\ \leq& Cc_{0}^{\frac{5}{2}+\gamma}c_{2}^{2}+Cc_{0}c_{1} \Vert \nabla h\Vert _{L^{2}}^{\frac{1}{2}}\Vert \nabla h\Vert _{H^{1}}^{\frac{1}{2}}, \end{aligned}$$
(2.98)
then the Young inequality and (2.97) yield
$$\Vert h\Vert _{H^{2}}\leq Cc_{0}^{\frac{7}{2}+\gamma }c_{1}^{2}c_{2}^{2}. $$
Thus, we have finished the proof of Lemma 2.5. □
Next, let us define \(c_{i}\) (\(i=1,\ldots,6\)) as follows:
$$\begin{aligned}& c_{1}=Cc_{0}^{7+2\gamma},\quad\quad c_{2}=Cc_{0}^{\frac{5}{2}+3\gamma}c_{1}^{2},\quad\quad c_{5}=Cc_{0}^{\frac{7}{2}}c_{1}c_{2}^{2},\\& c_{6}=Cc_{0}^{\frac{9}{2}}c_{1}^{2}c_{2}^{2},\quad\quad c_{3}=Cc_{0}^{\frac{13}{2}+3\gamma}c_{1}^{4}c_{2}c_{5},\quad\quad c_{4}=Cc_{0}^{9+6\gamma}c_{1}^{5}c_{2}^{2}, \end{aligned}$$
then we conclude from Lemma 2.1 to Lemma 2.5 that
$$ \textstyle\begin{cases} \sup _{0\leq t\leq T}(\Vert u\Vert _{H^{1}} +\Vert k\Vert _{H^{1}}+\Vert \varepsilon \Vert _{H^{1}})\\ \quad{}+\int_{0}^{T}(\Vert k\Vert _{H^{3}}^{2} +\Vert u_{t}\Vert _{H^{1}}^{2}+\Vert k_{t}\Vert _{H^{1}}^{2}+\Vert \varepsilon_{t}\Vert _{H^{1}}^{2})\,\mathrm{d}t\leq c_{1},\\ \sup _{0\leq t\leq T}\Vert u\Vert _{H^{2}}\leq c_{2},\quad\quad \sup _{0\leq t\leq T}\Vert u\Vert _{H^{3}}\leq c_{3},\quad\quad \int_{0}^{T}\Vert u\Vert _{H^{4}}^{2}\,\mathrm{d}t\leq c_{4},\\ \sup _{0\leq t\leq T}\Vert k\Vert _{H^{2}}\leq c_{5},\quad\quad \sup _{0\leq t\leq T}\Vert \varepsilon \Vert _{H^{2}}\leq c_{6}, \end{cases} $$
(2.99)
and
$$ \textstyle\begin{cases} \Vert \rho \Vert _{H^{3}(\Omega)}\leq Cc_{0}, \quad\quad \Vert \rho_{t}\Vert _{H^{1}(\Omega )}\leq Cc_{0}c_{2},\\ \Vert \sqrt{\rho}h_{t}\Vert _{L^{2}}^{2} +\Vert h\Vert _{H^{1}}^{2} +\int_{0}^{t}\Vert \nabla h_{t}\Vert _{L^{2}}^{2}\,\mathrm{d}s\leq Cc_{0}^{5},\\ \Vert h\Vert _{H^{2}}\leq Cc_{0}^{\frac{7}{2}+\gamma}c_{1}^{2}c_{2}^{2}, \end{cases} $$
(2.100)
for \(0\leq t\leq T\).
Using a standard proof as that in [13], we complete the proof of Theorem 2.1. □