Proof. Taking inner product of system $(2.1)_1$ with $\omega$, and system $(2.1)_2$ with $u$, after integration by parts, adding them together, it yields that

(3.5)\begin{align} & \frac{1}{2}\frac{\partial}{\partial t}(\|\omega\|_{L^2}^2+\|u\|_{L^2}^2)+\mu\|u\|_{L^2}^2 ={-}\int_{\mathbb{R}^d}(u\cdot\nabla\omega)\omega+(u\cdot\nabla u)udx\\ & \qquad-\frac{\gamma-1}{2}\int_{\mathbb{R}^d}\omega^2\text{div}u+(\omega\nabla\omega)u dx\nonumber\\ & \quad={-}\frac{1}{2}\int_{\mathbb{R}^d}u\cdot\nabla(\omega^2+u^2)dx+\frac{\gamma-1}{2}\int_{\mathbb{R}^d} (\omega\nabla\omega)u dx\nonumber\\ & \quad\leq C\|(\omega,u)\|_{L^\infty}\|u\|_{L^2}\|(\nabla\omega,\nabla u)\|_{L^2}, \nonumber \end{align}

where we have used

\[ \int_{\mathbb{R}^d}(\bar{\pi}\text{div}u)\omega dx+\int_{\mathbb{R}^d}(\bar{\pi}\nabla \omega)u dx=0, \]

in the first equality, the Hölder inequality in the last inequality, and $\|(\omega,\,u)\|_{X}=\|\omega \|_{X}+\|u\|_{X}$.

Next, we will show inequality (3.4). Applying the operator $\partial \nabla ^\delta$ on both sides of system (2.1), it follows that

(3.6)\begin{align} & \partial_t\partial\nabla^\delta\omega+\partial\nabla^\delta(\bar{\pi}\text{div}u+u\cdot\nabla\omega) ={-}\frac{\gamma-1}{2}\partial\nabla^\delta(\omega\text{div}u), \end{align}

(3.7)\begin{align} & \partial_t\partial\nabla^\delta u+\partial\nabla^\delta(u\cdot\nabla u+\bar{\pi}\nabla\omega+\mu u)={-}\frac{\gamma-1}{2}\partial\nabla^\delta(\omega\nabla\omega). \end{align}

Multiplying equation (3.6) and equation (3.7) by $\partial \nabla ^\delta \omega$ and $\partial \nabla ^\delta u$ respectively, $|\delta |\leq s-1$. Adding them together and integration by parts, one has that

(3.8)\begin{align} & \frac{1}{2}\frac{\partial}{\partial t}\sum_{|\delta|\leq s-1}(\|\partial\nabla^\delta\omega\|_{L^2}^2+\|\partial\nabla^\delta u\|_{L^2}^2)+\mu\sum_{|\delta|\leq s-1}\|\partial\nabla^\delta u\|_{L^2}^2 \\ & \quad={-}\sum_{|\delta|\leq s-1}\int_{\mathbb{R}^d}\partial\nabla^\delta(u\cdot\nabla\omega+\frac{\gamma-1}{2}\omega\text{div}u) \partial\nabla^\delta\omega dx\nonumber\\ & \qquad-\sum_{|\delta|\leq s-1}\int_{\mathbb{R}^d}\partial\nabla^\delta(u\cdot\nabla u+\frac{\gamma-1}{2}\omega\nabla\omega)\partial\nabla^\delta udx\nonumber\\ & \quad=:-\sum_{|\delta|\leq s-1}(I_1+I_2+I_3+I_4). \nonumber \end{align}

where the first equality is guaranteed by

\[ \int_{\mathbb{R}^d}\partial\nabla^\delta(\bar{\pi}\text{div}u)\partial\nabla^\delta\omega dx+\int_{\mathbb{R}^d}\partial\nabla^\delta(\bar{\pi}\nabla\omega)\partial\nabla^\delta u dx=0. \]

For $|\delta |\geq 0$, we first deal with the term $I_1$ as follows

(3.9)\begin{equation} \begin{aligned} |I_1| & =\int_{\mathbb{R}^d}\partial\nabla^\delta(u\cdot\nabla\omega)\partial\nabla^\delta\omega dx\\ & =\int_{\mathbb{R}^d}\left(\sum_{\alpha+\beta=\delta}\partial(\nabla^\alpha u\cdot\nabla^{1+\beta}\omega)\right)\partial\nabla^\delta\omega dx\\ & =\int_{\mathbb{R}^d}\partial(\nabla^\delta u\cdot\nabla\omega+ u\cdot\nabla^\delta\nabla\omega)\partial\nabla^\delta\omega dx\\ & \quad+\int_{\mathbb{R}^d}\left(\sum_{\alpha,\beta\leq\delta-1,\alpha+\beta=\delta}(\partial\nabla^\alpha u\cdot\nabla^{1+\beta}\omega+\nabla^\alpha u\cdot\partial\nabla^{1+\beta}\omega)\right)\partial\nabla^\delta\omega dx\\ & =:I_1^{(1)}+I_1^{(2)}. \end{aligned} \end{equation}

When $\delta =0$, thanks to the Hölder inequality, we have

(3.10)\begin{equation} \begin{aligned} \left|\int_{\mathbb{R}^d}u\partial\nabla\omega\partial\omega dx\right| & =\frac{1}{2}\left|\int_{\mathbb{R}^d}\text{div}u(\partial\omega)^2dx\right|\\ & \leq C\|\text{div}u\|_{L^\infty}\|\partial\omega\|_{L^2}^2. \end{aligned} \end{equation}

As $\delta >0$, by virtue of the Gagliardo–Nirenberg inequality yields that

(3.11)\begin{equation} \begin{aligned} \left|\int_{\mathbb{R}^d}\nabla^\delta u\partial\nabla\omega\partial\nabla^\delta\omega dx\right| & \leq \|\partial\nabla \omega\|_{L^p}\|\nabla^\delta u\|_{L^q}\|\partial\nabla^\delta\omega\|_{L^2}\\ & \leq\|\partial\omega\|_{L^\infty}^\theta\|\partial\nabla^\delta\omega\|_{L^2}^{1-\theta}\|\nabla u\|_{L^\infty}^{1-\theta}\|\partial\nabla^\delta u\|_{L^2}^\theta\|\partial\nabla^\delta\omega\|_{L^2}\\ & \leq C(\|\partial\omega\|_{L^\infty}\|\partial\nabla^\delta u\|_{L^2}+\|\nabla u\|_{L^\infty}\|\partial\nabla^\delta\omega\|_{L^2})\|\partial\nabla^\delta\omega\|_{L^2}, \end{aligned} \end{equation}

where $\frac {1}{p}+\frac {1}{q}=\frac {1}{2}$ for $p,\,q\in [2,\,\infty ]$, and $\theta \in (0,\,1)$, we also have used Young's inequality in the last inequality. In view of (3.10), (3.11) and the Hölder inequality, $I_1^{(1)}$ can be dealt with as

(3.12)\begin{equation} I_1^{(1)}\leq C\|(\partial\omega,\partial u)\|_{L^\infty}\|(\partial\nabla^\delta\omega,\partial\nabla^\delta u)\|_{L^2}\|\partial\nabla^\delta\omega\|_{L^2}. \end{equation}

On the other hand, in order to deal with $I_1^{(2)}$, in view of the Gagliardo–Nirenberg inequality, for $\alpha,\,\beta \leq \delta -1,\,\;\alpha +\beta =\delta$, one has that

(3.13)\begin{equation} \begin{aligned}& \int_{\mathbb{R}^d}\Big{(}\partial\nabla^\alpha u\cdot\nabla^{1+\beta}\omega\Big{)}\partial\nabla^\delta\omega dx \leq C\|\partial\nabla^\alpha u\|_{L^p}\|\nabla^{1+\beta}\omega\|_{L^q}\|\partial\nabla^\delta\omega\|_{L^2}\\ & \quad\leq C\|\partial u\|_{L^\infty}^\vartheta\|\partial\nabla^\delta u\|_{L^2}^{1-\vartheta}\|\nabla\omega\|_{L^\infty}^{1-\vartheta}\|\nabla^{\delta+1}\omega\|_{L^2}^{\vartheta}\|\partial\nabla^\delta\omega\|_{L^2}\\ & \quad\leq C(\|\partial u\|_{L^\infty}\|\nabla^{\delta+1}\omega\|_{L^2}+\|\nabla\omega\|_{L^\infty}\|\partial\nabla^\delta u\|_{L^2})\|\partial\nabla^\delta\omega\|_{L^2}, \end{aligned} \end{equation}

where $\frac {1}{p}+\frac {1}{q}=\frac {1}{2}$ for $p,\,q\in [2,\,\infty ]$, and $\vartheta \in (0,\,1)$, we also have used Young's inequality in the last inequality. Similarly, we can show that

(3.14)\begin{equation} \begin{aligned}\int_{\mathbb{R}^d}(\nabla^\alpha u\cdot\partial\nabla^{1+\beta}\omega)\partial\nabla^\delta\omega dx & \leq C\|\nabla^\alpha u\|_{L^p}\|\partial\nabla^{1+\beta}\omega\|_{L^q}\|\partial\nabla^\delta\omega\|_{L^2}\\ & \leq C(\|\nabla u\|_{L^\infty}\|\partial\nabla^{\delta}\omega\|_{L^2}\\ & \quad +\|\partial\omega\|_{L^\infty}\|\nabla^{\delta+1} u\|_{L^2})\|\partial\nabla^\delta\omega\|_{L^2}. \end{aligned} \end{equation}

Combining (3.13) with (3.14), it yields that

(3.15)\begin{equation} I_1^{(2)}\leq C\|(\partial\omega,\partial u)\|_{L^\infty}\|(\partial\nabla^\delta\omega,\partial\nabla^\delta u)\|_{L^2}\|\partial\nabla^\delta\omega\|_{L^2}. \end{equation}

In view of (3.9), (3.12) and (3.15), it gives that

(3.16)\begin{equation} I_1=:I_1^{(1)}+I_1^{(2)}\leq C\|(\partial\omega,\partial u)\|_{L^\infty}(\|\partial\nabla^\delta\omega\|_{L^2}^2+\|\partial\nabla^\delta u\|_{L^2}^2). \end{equation}

Similar to the process of proving (3.16), we can estimate $I_2,\,\;I_3$ and $I_4$ to derive

(3.17)\begin{equation} I_2+I_3+I_4\leq C\|(\partial\omega,\partial u)\|_{L^\infty}(\|\partial\nabla^\delta\omega\|_{L^2}^2+\|\partial\nabla^\delta u\|_{L^2}^2). \end{equation}

Substituting (3.16) and (3.17) into (3.8), by virtue of (3.1) and (3.2), thus we consequently obtain (3.4).

Proof. Multiplying the operator $\nabla ^\delta$ on both sides of system (2.1), one has that

(3.20)\begin{align} & \partial_t\nabla^\delta\omega={-}\nabla^\delta(\bar{\pi}\text{div}u+u\cdot\nabla\omega) -\frac{\gamma-1}{2}\nabla^\delta(\omega\text{div}u), \end{align}

(3.21)\begin{align} & \nabla^{\delta+1}\omega=\frac{-1}{\bar{\pi}}[\partial_t\nabla^\delta u+\nabla^\delta(u\cdot\nabla u+\mu u)]-\frac{\gamma-1}{\bar{\pi}}\nabla^\delta(\omega\nabla\omega). \end{align}

Taking $L^2$ norm of the equation (3.20) and (3.21) for $|\delta |\leq s-1$, adding them together yields that

(3.22)\begin{equation} \begin{aligned} \sum_{|\delta|\leq s-1}\|\partial\nabla^{\delta}\omega\|_{L^2} & \leq C\sum_{|\delta|\leq s-1}\|\nabla^\delta(\text{div}u+u\cdot\nabla\omega)+\nabla^\delta(\omega\text{div}u)\|_{L^2} \\ & \quad +C\sum_{|\delta|\leq s-1}\|\partial_t\nabla^\delta u+\nabla^\delta(u\cdot\nabla u+u)+\nabla^\delta(\omega\nabla\omega)\|_{L^2}\\ & \leq C\sum_{|\delta|\leq s-1}\|\partial\nabla^{\delta}u\|_{L^2}+\|\nabla^{\delta}(u\cdot\nabla\omega\!+\!\omega\text{div}u\!+\!u\cdot\nabla u+\omega\nabla\omega)\|_{L^2}. \end{aligned} \end{equation}

Similar to the method of dealing with (3.9), one can easily check that

(3.23)\begin{align} & \begin{aligned} \|\nabla^{\delta}(u\cdot\nabla\omega)\|_{L^2} & \leq C(\|\nabla\omega\|_{L^\infty}\|\nabla^\delta u\|_{L^2}+\|u\|_{L^\infty}\|\nabla^{\delta+1} \omega\|_{L^2})\\ & \leq C(I(\omega)I(u))^{\frac{1}{2}}, \end{aligned} \end{align}

(3.24)\begin{align} & \begin{aligned} \|\nabla^{\delta}(\omega\text{div}u)\|_{L^2} & \leq C(\|\nabla u\|_{L^\infty}\|\nabla^\delta\omega\|_{L^2}+\|\omega\|_{L^\infty}\|\nabla^{\delta+1} u\|_{L^2})\\ & \leq C(I(\omega)I(u))^{\frac{1}{2}}, \end{aligned} \end{align}

(3.25)\begin{align} & \begin{aligned} \|\nabla^{\delta}(u\cdot\nabla u)\|_{L^2} & \leq C(\|\nabla u\|_{L^\infty}\|\nabla^\delta u\|_{L^2}+\|u\|_{L^\infty}\|\nabla^{\delta+1} u\|_{L^2})\\ & \leq C(I(u)Q(u))^{\frac{1}{2}}, \end{aligned} \end{align}

(3.26)\begin{align} & \begin{aligned} \|\nabla^{\delta}(\omega\nabla\omega)\|_{L^2} & \leq C(\|\nabla\omega\|_{L^\infty}\|\nabla^\delta \omega\|_{L^2}+\|\omega\|_{L^\infty}\|\nabla^{\delta+1} \omega\|_{L^2})\\ & \leq C(I(\omega)Q(\omega))^{\frac{1}{2}}, \end{aligned} \end{align}

where we have used for $s>1+\frac {d}{2}$ that

\begin{align*} & \|f\|_{L^\infty}\leq C\|f\|_{H^{s-1}}\leq CI^{\frac{1}{2}}(f),\\ & \|\nabla f\|_{L^\infty}\leq C\|\nabla f\|_{H^{s-1}}\leq CQ^{\frac{1}{2}}(f). \end{align*}

Combining (3.22)–(3.25) and (3.26), it yields that

\[ Q(\omega)(t)\leq CI(\omega)(t)Q(\omega)(t)+C(1+I(\omega))I(u). \]

If the solution $\omega$ satisfies $I(\omega )(t)\ll 1$ for any time $t>0$, choosing $CI(\omega )(t)\leq \frac {1}{3}$ then we have

\[ 2Q(\omega)(t)\leq(3C+1)I(u)(t), \]

which derives the inequality (3.19).

Proof. Combining (3.3) and (3.4) in lemma 3.1 yields that

(3.27)\begin{equation} \begin{aligned} \frac{\partial}{\partial t}I(\omega,u)(t)+2\mu I(u)(t) & \leq C(\|(\omega,u)\|_{L^\infty}\|u\|_{L^2}\|\nabla(\omega, u)\|_{L^2}\\ & \quad+\|(\partial\omega,\partial u)\|_{L^\infty}(Q(\omega,u)(t))). \end{aligned} \end{equation}

Note that

(3.28)\begin{align} & \begin{aligned} \|(\omega,u)\|_{L^\infty}\|u\|_{L^2}\|\nabla(\omega, u)\|_{L^2} & \leq\|(\omega,u)\|_{H^s}\|u\|_{H^s}\|\nabla(\omega, u)\|_{H^{s-1}}\\ & \leq I^{\frac{1}{2}}(\omega,u)I^{\frac{1}{2}}(u)Q^{\frac{1}{2}}(\omega,u), \end{aligned} \end{align}

(3.29)\begin{align} & \begin{aligned} \|(\partial\omega,\partial u)\|_{L^\infty}Q(\omega,u)(t) & \leq\|\partial(\omega,u)\|_{H^{s-1}}Q(\omega,u)(t)\\ & \leq I^{\frac{1}{2}}(\omega,u)(t)Q(\omega,u)(t), \end{aligned} \end{align}

where we have used $s>1+\frac {d}{2}$, which guarantees

\[ \|f\|_{L^\infty}< C\|f\|_{H^{s-1}},\;\quad\qquad\|\nabla f\|_{L^\infty}< C\|f\|_{H^{s}}. \]

Inserting (3.28) and (3.29) into (3.27), one has that

(3.30)\begin{equation} \begin{aligned}\frac{\partial}{\partial t}I(\omega,u)(t)+2\mu I(u)(t) & \leq CI^{\frac{1}{2}}(\omega,u)[Q(\omega)+I(u)]+CI^{\frac{1}{2}}(\omega,u)Q(\omega,u)\\ & \leq CI^{\frac{1}{2}}(\omega,u)[Q(\omega)+I(u)]+CI^{\frac{1}{2}}(\omega,u)[Q(\omega)+I(u)]\\ & \leq CI^{\frac{1}{2}}(\omega,u)[Q(\omega)+I(u)]. \end{aligned} \end{equation}

Suppose the solution $(\omega,\,u)$ satisfies $I(\omega,\,u)(t)=\epsilon _0\ll 1$ for any $t\in \mathbb {R}^+$, choosing $CI(\omega,\,u)(t)\leq \frac {1}{3}$, then it follows from lemma 3.2 that the solution satisfies (3.19), in view of (3.30) we have

\begin{align*} \frac{\partial}{\partial t}I(\omega,u)(t)+2\mu I(u)(t)& \leq CI^{\frac{1}{2}}(\omega,u)[Q(\omega)+I(u)]\\ & \leq\frac{C}{2}(\epsilon_0)^{\frac{1}{2}}(3C+2)I(u)(t). \end{align*}

Since $\epsilon _0$ is sufficiently small, we can choose $\epsilon _0$ such that

\[ \mu_0=:2\mu-\frac{C}{2}(\epsilon_0)^{\frac{1}{2}}(3C+2)>0. \]

Consequently, we have

(3.31)\begin{equation} \frac{\partial}{\partial t}I(\omega,u)(t)+\mu_0 I(u)(t)\leq 0. \end{equation}

Integrating the inequality (3.31) with respect to the time variable on interval $[0,\,t]$, it follows that

(3.32)\begin{equation} I(\omega,u)(t)+\mu_0 \int_0^tI(u)(\tau)d\tau\leq I(\omega_0,u_0). \end{equation}

Thus if the initial data $I(\omega _0,\,u_0)=\epsilon _0$ is small enough, then the inequality (3.32) guarantees for all $t>0$ that

\[ I(\omega,u)(t)\leq\epsilon_0\ll1, \]

which completes the proof of theorem 3.3.

Proof. Note that the linear system (4.1) is equivalent to

(4.3)\begin{equation} \begin{aligned} \frac{\partial}{\partial t}\left(\begin{array}{@{}c@{}cc}\omega \\ u \end{array}\right) & =\left(\begin{array}{@{}ccc@{}} 0 & -\bar{\pi}\nabla \\- \bar{\pi}\nabla^\top & {-}\mu \mathbb{I}_d \end{array}\right) \left(\begin{array}{@{}c@{}cc} \omega \\ u \end{array}\right),\\ & =:\mathcal{A}\left(\begin{array}{@{}c@{}cc} \omega \\ u \end{array}\right),\quad\;(t,x)\in\mathbb{R}^{+}\times\mathbb{R}^d \end{aligned} \end{equation}

with initial data $v_0(x)\in L^1(\mathbb {R}^d)\cap H^s(\mathbb {R}^d)$, where $\top$ denotes the transposition of vector, $\bar {\pi }$ and $\mu$ are positive constants, and $\mathbb {I}_d$ is $d\times d$ unit matrix. In view of the Fourier transformation and $v=(\omega,\,u)^\top$ we have

\[ \partial_t\widehat{v}(t,\xi)=\mathcal {A}(\xi)\widehat{v}(t,\xi) \]

with the $(d+1)\times (d+1)$ matrix

\[ \mathcal {A}(\xi)=\left(\begin{array}{ccc} 0 & -i\bar{\pi}\xi \\- i\bar{\pi}\xi^\top & -\mu \mathbb{I}_d \end{array}\right). \]

By computing the determinant $|\lambda I-\mathcal {A}(\xi )|=0$, we derive that the eigenvalues of the matrix $\mathcal {A}(\xi )$ are $\lambda _1=\cdots =\lambda _{d-1}=-\mu$, and

(4.4)\begin{equation} \left\{\begin{array}{@{}ll} \lambda_{d}={-}\dfrac{1}{2}(\mu+\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2}),\\ \lambda_{d+1}={-}\dfrac{1}{2}(\mu-\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2}). \end{array}\right. \end{equation}

By virtue of the eigenvalue $\lambda _i=-\mu$ of matrix $\mathcal {A}(\xi )$, $i=1,\,\cdots,\,d-1$, one has that the unit orthonormal eigenvectors $b_i=(0,\,y_i)^\top =(0,\,y_{i1},\,y_{i2},\,\cdots,\,y_{id})^\top$, such that for every $i=1,\,\cdots,\,d-1$

\[ \xi\cdot y_i=\xi_1y_{i1}+\xi_2y_{i2}+\cdots\xi_dy_{id}=0. \]

Similarly, it is easy to obtain the unit eigenvectors $b_j$ of the eigenvalue $\lambda _j$, ${j=d,\,\;d+1}$ satisfying

\[ b_d=\frac{(i\lambda_{d+1},\bar{\pi}\xi)^\top}{\sqrt{\lambda_{d+1}^2+|\bar{\pi}\xi|^2}}\qquad\text{and}\qquad b_{d+1}=\frac{(i\lambda_{d},\bar{\pi}\xi)^\top}{\sqrt{\lambda_{d}^2+|\bar{\pi}\xi|^2}},\;i^2={-}1. \]

Thus we can choose the unitary matrix $\mathcal {B}(\xi )=(b_1,\,\cdots,\,b_d,\, \widetilde {b}_{d+1})$ such that

\[ \mathcal {A}(\xi)\mathcal {B}(\xi)=\mathcal {B}(\xi)\left(\begin{array}{@{}ccc@{}cc} -\mu \mathbb{I}_{d-1} & 0 & 0 & \\ 0 & \lambda_d & \eta \\ 0 & 0 & \lambda_{d+1} \end{array}\right), \]

where $b_1,\,\cdots,\,b_d$ and $\widetilde {b}_{d+1}$ are unit orthonormal eigenvectors in $\mathbb {R}^{d+1}$, the function $\eta$ satisfies

(4.5)\begin{equation} \eta=\left\{\begin{array}{@{}ll} -(\mu+\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2}), & \text{if}\quad \mu^2-4\bar{\pi}^2|\xi|^2\geq0,\\ - \mu, & \text{if}\quad \mu^2-4\bar{\pi}^2|\xi|^2<0. \end{array}\right. \end{equation}

Consequently, the solutions of system (4.1) are given by $v(t,\,x)=S(t)v_0(x)$, where

(4.6)\begin{equation} \begin{aligned} \widehat{S}(t)=\exp(t\mathcal {A}(\xi)) & =\mathcal {B}(\xi)\mathcal {D}(t,\xi)\mathcal {B}^{{-}1}(\xi)\\ & =:\mathcal {B}(\xi)\left(\begin{array}{@{}ccc@{}cc} e^{-\mu t} \mathbb{I}_{d-1} & 0 & 0 & \\ 0 & e^{\lambda_dt} & \kappa\eta \\ 0 & 0 & e^{\lambda_{d+1}t} \end{array}\right)\mathcal {B}^{{-}1}(\xi) \end{aligned} \end{equation}

with the function

\[ \kappa=\frac{e^{\lambda_{d}t}-e^{\lambda_{d+1}t}}{\lambda_{d}-\lambda_{d+1}}. \]

Next, in order to show (4.2), we first estimate every element of the matrix $\mathcal {D}(t,\,\xi )$.

Case 1: if $\mu ^2-4\bar {\pi }^2|\xi |^2<0$, then we have

\begin{align*} \kappa& =\frac{e^{\lambda_{d}t}-e^{\lambda_{d+1}t}}{\lambda_{d}-\lambda_{d+1}}=e^{-\frac{1}{2}\mu t}\frac{2\sin(\frac{t}{2}\sqrt{4\bar{\pi}^2|\xi|^2-\mu^2})}{\sqrt{4\bar{\pi}^2|\xi|^2-\mu^2}}\\ & \leq \left\{\begin{array}{@{}ll} Cte^{-\dfrac{1}{2}\mu t}\leq Ce^{-\dfrac{1}{3}\mu t}, & \quad \text{if}\quad t\sqrt{4\bar{\pi}^2|\xi|^2-\mu^2}\ll1,\\ Ce^{-\dfrac{1}{3}\mu t}, & \quad\text{if}\quad t\sqrt{4\bar{\pi}^2|\xi|^2-\mu^2}>1. \end{array}\right. \end{align*}

Case 2: if $\mu ^2-4\bar {\pi }^2|\xi |^2\geq 0$, then we have

\[ \kappa=\frac{e^{\lambda_{d}t}-e^{\lambda_{d+1}t}}{\lambda_{d}-\lambda_{d+1}}=e^{-\frac{1}{2}\mu t}\frac{2\sinh(\frac{t}{2}\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2})}{\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2}}. \]

As $0\leq \sqrt {\mu ^2-4\bar {\pi }^2|\xi |^2}\ll 1$, one has that

\[ \kappa\leq Cte^{-\frac{1}{2}\mu t}\leq Ce^{-\frac{1}{3}\mu t}. \]

Otherwise, if $\sqrt {\mu ^2-4\bar {\pi }^2|\xi |^2}\in (\delta _0,\,\frac {\mu }{2})$, for some $\delta _0>0$, then it is easy to check that

\[ \frac{t}{2}\delta_0<\frac{t}{2}\sqrt{\mu^2-4\bar{\pi}^2|\xi|^2}<\frac{t}{4}\mu, \]

thus we have

\[ \kappa\leq Ce^{-\frac{1}{5}\mu t}. \]

On the other hand, if $\sqrt {\mu ^2-4\bar {\pi }^2|\xi |^2}\in [\frac {\mu }{2},\,\mu ]$, then one can easily check that

\[ -\frac{3}{16}\mu\leq{-}\frac{|\bar{\pi}\xi|^2}{\mu}\leq0, \]

this implies that

\[ \kappa\leq Ce^{-\frac{|\bar{\pi}\xi|^2}{\mu} t}. \]

Note that

\[ |\eta|\leq\max\left\{\mu,2|\lambda_{d-1}|\right\}\leq2\mu. \]

Thus it follows that

(4.7)\begin{equation} |\kappa\eta|\leq \left\{\begin{array}{@{}ll} Ce^{-\dfrac{|\bar{\pi}\xi|^2}{\mu} t}, & \quad \text{if}\quad |\bar{\pi}\xi|\leq\dfrac{\sqrt{3}}{4}\mu,\\ Ce^{-\dfrac{1}{5}\mu t}, & \quad\text{if}\quad |\bar{\pi}\xi|>\dfrac{\sqrt{3}}{4}\mu. \end{array}\right. \end{equation}

In fact, in a similar way, the bound of all diagonal elements of the matrix $\mathcal {D}(t,\,\xi )$ satisfies (4.7).

Now, we prove the inequality (4.2). Let $(2l+d)/2<((2l+d)/2)^+\leq (2l+d)/ 2+\varepsilon$ for any $\varepsilon >0$. In view of the Fourier transformation and (4.7), for ${((2l+d)/2)^+\leq s}$, it follows that

(4.8)\begin{equation} \begin{aligned} \left\|\nabla^lS(t)v_0\right\|_{L^\infty} & \leq C\int_{\mathbb{R}^d}\left|e^{ix\cdot\xi}(i\xi)^l\widehat{S}(t)\widehat{v}_0(\xi)\right|d\xi\\ & \leq C\int_{|\bar{\pi}\xi|\leq\frac{\sqrt{3}}{4}\mu}e^{-\frac{|\bar{\pi}\xi|^2}{\mu} t}|\xi|^l|\widehat{v}_0(\xi)|d\xi+C\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}e^{-\frac{t}{5}\mu }|\xi|^l|\widehat{v}_0(\xi)|d\xi\\ & \leq C\|\widehat{v}_0\|_{L^\infty}\int_{0}^{\frac{\sqrt{3}}{4\bar{\pi}}\mu}r^{l+d-1}e^{-\frac{(\bar{\pi}r)^2}{\mu} t}dr+Ce^{-\frac{t}{5}\mu }\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}|\xi|^l|\widehat{v}_0(\xi)|d\xi\\ & \leq C(1+t)^{-(l+d)/2}\|v_0\|_{L^1}+Ce^{-\frac{1}{5}\mu t}\left\|\nabla^{((2l+d)/2)^+}v_0\right\|_{L^2}, \end{aligned} \end{equation}

where the last inequality is guaranteed by

\begin{align*} \left|\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}|\xi|^l|\widehat{v}_0(\xi)|d\xi\right|^2 & \leq\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}|\xi|^{-(d)^+}d\xi\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu} |\xi|^{2l+(d)^+}|\widehat{v}_0(\xi)|^2d\xi\\ & \leq C\left\|\nabla^{((2l+d)/2)^+}v_0\right\|_{L^2}^2. \end{align*}

On the other hand, by virtue of the Fourier transformation and (4.7), for $0\leq k\leq s$, one has that

(4.9)\begin{equation} \begin{aligned} \left\|\nabla^kS(t)v_0\right\|_{L^2}^2 & =\left\||\xi|^k\widehat{S}(t)\widehat{v}_0\right\|_{L^2}^2\\ & \leq C\int_{|\bar{\pi}\xi|\leq\frac{\sqrt{3}}{4}\mu}e^{{-}2t\frac{|\bar{\pi}\xi|^2}{\mu} }|\xi|^{2k}|\widehat{v}_0(\xi)|^2d\xi+C\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}e^{-\frac{2}{5}\mu t}|\xi|^{2k}|\widehat{v}_0(\xi)|^2d\xi\\ & \leq C\|\widehat{v}_0\|_{L^\infty}^2\int_{0}^{\frac{\sqrt{3}}{4\bar{\pi}}\mu}e^{{-}2t\frac{(\bar{\pi}r)^2}{\mu} }r^{2k+d-1}dr+Ce^{-\frac{2}{5}\mu t}\int_{|\bar{\pi}\xi|>\frac{\sqrt{3}}{4}\mu}|\xi|^{2k}|\widehat{v}_0(\xi)|^2d\xi\\ & \leq C(1+t)^{-(k+\frac{d}{2})}\|v_0\|_{L^1}^2+Ce^{-\frac{2}{5}\mu t}\left\|\nabla^{k}v_0\right\|_{L^2}^2, \end{aligned} \end{equation}

which concludes the proof of lemma 4.1.

Proof. In view of the linear system (4.1) and $v=(\omega,\,u)^\top$, the system (2.1) can be transformed into

(4.12)\begin{equation} \begin{aligned} \frac{\partial}{\partial t}\left(\begin{array}{@{}c@{}cc} \omega \\ u \end{array}\right) & =\left(\begin{array}{ccc} 0 & -\bar{\pi}\nabla\\ - \bar{\pi}\nabla^\top & -\mu \mathbb{I}_d \end{array}\right) \left(\begin{array}{@{}c@{}cc} \omega \\ u \end{array}\right)-\left(\begin{array}{@{}c@{}cc} u\cdot\nabla\omega+\dfrac{\gamma-1}{2}\omega\text{div}u \\ u\cdot\nabla u+\dfrac{\gamma-1}{2}\omega\nabla\omega \end{array}\right),\\ & =:\mathcal {A}v+F(v,\nabla v),\qquad\;(t,x)\in\mathbb{R}^{+}\times\mathbb{R}^d \end{aligned} \end{equation}

with initial data $v_0(x)\in L^1(\mathbb {R}^d)\cap H^s(\mathbb {R}^d)$. By the Duhamel principle, the solutions of system (4.12) are given by

(4.13)\begin{equation} v(t,x)=S(t)v_0(x)+\int_0^tS(t-\tau)F(v,\nabla v)(\tau,x)d\tau. \end{equation}

By virtue of the assumption in theorem 4.2, inequality (4.2) and equation (4.13), for $(l+\frac {d}{2})^+\leq s$ we have

(4.14)\begin{equation} \begin{aligned} \left\|\nabla^lv(t,\cdot)\right\|_{L^\infty} & \leq\left\|\nabla^lS(t)v_0\right\|_{L^\infty}+\int_0^t\left\|\nabla^lS(t-\tau)F(v,\nabla v)(\tau)\right\|_{L^\infty}d\tau\\ & \leq C(1+t)^{-(l+d)/2}\|v_0\|_{L^1}+Ce^{-\beta t}\left\|\nabla^{((2l+d)/2)^+}v_0\right\|_{L^2}\\ & \quad+C\int_0^t(1\!+\!t-\tau)^{-(l\!+\!d)/2}\|F\|_{L^1}\!+\!Ce^{-\beta (t-\tau)}\left\|\nabla^{((2l+d)/2)^+}F\right\|_{L^2}d\tau\\ & \leq C(1+t)^{-(l+d)/2}\epsilon_0+C\int_0^t(1+t-\tau)^{-(l+d)/2}\|F(\tau,\cdot)\|_{L^1}d\tau\\ & \quad+C\int_0^te^{-\beta (t-\tau)}\left\|\nabla^{((2l+d)/2)^+}F(\tau,\cdot)\right\|_{L^2}d\tau. \end{aligned} \end{equation}

In order to derive the results, introducing the following four functions

\begin{align*} & \mathcal {H}(t)=\sup_{\tau\in[0,t]}\|v(\tau,\cdot)\|_{H^s},\\ & J_0(t)=\sup_{\tau\in[0,t]}(1+\tau)^{\frac{d}{4}}\|v(\tau,\cdot)\|_{L^2},\\ & J_1(t)=\sup_{\tau\in[0,t]}(1+\tau)^{\frac{d+2}{4}}\|\nabla v(\tau,\cdot)\|_{L^2},\\ & J_\infty^h(t)=\sup_{\tau\in[0,t]}(1+\tau)^{\frac{d+h}{2}}\|\nabla ^hv(\tau,\cdot)\|_{L^\infty}. \end{align*}

Then for all $\tau \in [0,\,t]$ and $(l+1+\frac {d}{2})^+\leq s$, the nonlinear term $F(v,\,\nabla v)$ in (4.14) can be dealt with

(4.15)\begin{align} & \begin{aligned} \|F(v,\nabla v)(\tau)\|_{L^1} & \leq C\|(\omega,u)\|_{L^2}\|\nabla(\omega,u)\|_{L^2}\\ & =: C\|v\|_{L^2}\|\nabla v\|_{L^2}\\ & \leq C(1+\tau)^{-\frac{d+1}{2}} J_0(t)J_1(t), \end{aligned} \end{align}

(4.16)\begin{align} & \begin{aligned} \|F(v,\nabla v)(\tau)\|_{L^2} & \leq C\|v\|_{L^\infty}\|\nabla v\|_{L^2}\\ & \leq C\|v\|_{L^\infty}\|v\|_{H^s}\\ & \leq C(1+\tau)^{-\frac{d}{2}} J_\infty^0(t)\mathcal {H}(t), \end{aligned} \end{align}

(4.17)\begin{align} & \begin{aligned} \|\nabla^{(l+\frac{d}{2})^+}F(v,\nabla v)(\tau)\|_{L^2} & \leq C\|v\|_{L^\infty}\|\nabla^{(l+1+\frac{d}{2})^+} v\|_{L^2}\\ & \leq C(1+\tau)^{-\frac{d}{2}} J_\infty^0(t)\mathcal {H}(t). \end{aligned} \end{align}

Plugging (4.15) and (4.17) into (4.14) yields that

(4.18)\begin{equation} \begin{aligned} \left\|v(t,\cdot)\right\|_{L^\infty} & \leq C(1+t)^{{-}d/2}\epsilon_0+CJ_\infty^0(t)\mathcal {H}(t)\int_0^te^{-\beta(t-\tau)}(1+\tau)^{-\frac{d}{2}}d\tau \\ & \quad+CJ_0(t)J_1(t)\int_0^t(1+t-\tau)^{-\frac{d}{2}}(1+\tau)^{-\frac{d+1}{2}}d\tau\\ & \leq C(1+t)^{{-}d/2}(\epsilon_0+J_\infty^0(t)\mathcal {H}(t)) \\ & \quad+CJ_0(t)J_1(t)\int_0^t(1+t-\tau)^{-\frac{d}{2}}(1+\tau)^{-\frac{d+1}{2}}d\tau\\ & \leq C(1+t)^{-\frac{d}{2}}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)\mathcal {H}(t)), \end{aligned} \end{equation}

where we have used for $t\gg 1$

\[ \int_0^t(1+t-\tau)^{-\frac{d}{2}}(1+\tau)^{-\frac{d+1}{2}}d\tau \leq C(1+t)^{-\frac{d}{2}}. \]

On the other hand, thanks to (4.2), one can easily check that

(4.19)\begin{equation} \begin{aligned} \|v(t,\cdot)\|_{L^2} & \leq\|S(t)v_0\|_{L^2}+\int_0^t\|S(t-\tau)F(v,\nabla v)(\tau)\|_{L^2}d\tau\\ & \leq C(1+t)^{-\frac{d}{4}}\epsilon_0+C\int_0^te^{-\beta (t-\tau)}\|F(\tau,\cdot)\|_{L^2}d\tau\\ & \quad+C\int_0^t(1+t-\tau)^{-\frac{d}{4}}\|F(\tau,\cdot)\|_{L^1}d\tau\\ & \leq C(1+t)^{-\frac{d}{4}}\epsilon_0+CJ_\infty^0(t)\mathcal {H}(t)\int_0^te^{-\beta(t-\tau)}(1+\tau)^{-\frac{d}{2}}d\tau \\ & \quad+CJ_0(t)J_1(t)\int_0^t(1+t-\tau)^{-\frac{d}{4}}(1+\tau)^{-\frac{d+1}{2}}d\tau\\ & \leq C(1+t)^{-\frac{d}{4}}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)\mathcal {H}(t)), \end{aligned} \end{equation}

where the last inequality comes from for $d\geq 2$ and $t\gg 1$

\[ \Phi(t)=:\int_0^t(1+t-\tau)^{-\frac{d}{4}}(1+\tau)^{-\frac{d+1}{2}}d\tau\leq C(1+t)^{-\frac{d}{4}}, \]

which is guaranteed by

\begin{align*} \Phi(t)& \leq(1+t)^{-\frac{d}{4}}\int_0^t\left(\frac{1}{1+t-\tau}+\frac{1}{1+\tau}\right)^{\frac{d}{4}}(1+\tau)^{-\frac{d+2}{4}}d\tau\\ & \leq C(1+t)^{-\frac{d}{4}}\int_0^t[(1+t-\tau)^{-\frac{d}{4}}+(1+\tau)^{-\frac{d}{4}}](1+\tau)^{-\frac{d+2}{4}}d\tau\\ & \leq C(1+t)^{-\frac{d}{4}}. \end{align*}

Similarly, we have

(4.20)\begin{equation} \begin{aligned} \left\|\nabla v(t,\cdot)\right\|_{L^2} & \leq\|\nabla S(t)v_0\|_{L^2}+\int_0^t\left\|\nabla S(t-\tau)F(v,\nabla v)(\tau)\right\|_{L^2}d\tau\\ & \leq C(1+t)^{-(\frac{d+2}{4})}\epsilon_0+C\int_0^te^{-\beta (t-\tau)}\|\nabla F(\tau,\cdot)\|_{L^2}d\tau\\ & \quad+C\int_0^t(1+t-\tau)^{-\frac{d+2}{4}}\|F(\tau,\cdot)\|_{L^1}d\tau\\ & \leq C(1+t)^{-\frac{d+2}{4}}\epsilon_0+CJ_\infty^0(t)\mathcal {H}(t)\int_0^te^{-\beta(t-\tau)}(1+\tau)^{-\frac{d}{2}}d\tau \\ & \quad+CJ_0(t)J_1(t)\int_0^t(1+t-\tau)^{-\frac{d+2}{4}}(1+\tau)^{-\frac{d+1}{2}}d\tau\\ & \leq C(1+t)^{-\frac{d+2}{4}}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)\mathcal {H}(t)), \end{aligned} \end{equation}

Combining (4.18), (4.19) with (4.20), by the definition of $J_i(t),\,\;i=0,\,1,\,\infty$, we end up with

(4.21)\begin{equation} J_0(t)+J_1(t)+J_\infty^0(t)\leq C(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)\mathcal {H}(t)). \end{equation}

If the initial data satisfy $\|v_0\|_{L^1}+I(v_0)=\epsilon _0\ll 1$, thanks to theorem 3.3, then it implies that

\[ \mathcal {H}(t)\leq I(v)(t)\leq\epsilon_0\ll1. \]

Define $f(t)=J_0(t)+J_1(t)+J_\infty ^0(t)$, in view of inequality (4.21), one has for all $t\in \mathbb {R}^+$ that

(4.22)\begin{equation} f(t)\leq C\epsilon_0+Cf^2(t). \end{equation}

If we choose $\epsilon _0\ll 1$ such that $4C^2\epsilon _0<1$, then the equation

\[ Cy^2-y+C\epsilon_0=0 \]

has two differential roots

\[ 0< y_1=\frac{1-\sqrt{1-4C^2\epsilon_0}}{2C}< y_2=\frac{1+\sqrt{1-4C^2\epsilon_0}}{2C}. \]

Thanks to

\[ f(0)=J_0(0)+J_1(0)+J_\infty^0(0)\leq C\epsilon_0\ll1, \]

in order to ensure inequality (4.22) hold for all $t\geq 0$, thus we deduce that $f(t)< y_1$ is bound, this implies that

(4.23)\begin{equation} \begin{aligned} & \|v(t,\cdot)\|_{L^2}\leq C(1+t)^{-\frac{d}{4}},\\ & \|\nabla v(t,\cdot)\|_{L^2}\leq C(1+t)^{-\frac{d+2}{4}},\\ & \|v(t,\cdot)\|_{L^\infty}\leq C(1+t)^{-\frac{d}{2}}. \end{aligned} \end{equation}

In view of the above inequality (4.23), we can consequently estimate

(4.24)\begin{align} \left\|\nabla^k v(t,\cdot)\right\|_{L^2} & \leq\left\|\nabla^k S(t)v_0\right\|_{L^2}+\int_0^t\left\|\nabla^k S(t-\tau)F(v,\nabla v)(\tau)\right\|_{L^2}d\tau\nonumber\\ & \leq C(1+t)^{-(\frac{d+2k}{4})}\epsilon_0+C\int_0^te^{-\beta (t-\tau)}\left\|\nabla^k F(\tau,\cdot)\right\|_{L^2}d\tau\nonumber\\ & \quad+C\int_0^t(1+t-\tau)^{-\frac{d+2k}{4}}\|F(\tau,\cdot)\|_{L^1}d\tau \nonumber\\ & \leq C(1+t)^{-\frac{d+2k}{4}}\epsilon_0+CJ_\infty^0(t)\mathcal {H}(t)\int_0^te^{-\beta(t-\tau)}(1+\tau)^{-\frac{d}{2}}d\tau \nonumber\\ & \quad+CJ_0(t)J_1(t)\int_0^t(1+t-\tau)^{-\frac{d+2k}{4}}(1+\tau)^{-\frac{d+1}{2}}d\tau\nonumber\\ & \leq C\max\left\{(1+t)^{-\frac{d+2k}{4}},(1+t)^{-\frac{d}{2}}\right\}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t) \mathcal {H}(t))\nonumber\\ & \leq C\max\left\{(1+t)^{-\frac{d+2k}{4}},(1+t)^{-\frac{d}{2}}\right\},\end{align}

where we have used

\[ \left\|\nabla^k F(\tau,\cdot)\right\|_{L^2}\leq C\|v\|_{L^\infty}\left\|\nabla^{k+1}v\right\|_{L^2}\leq C(1+\tau)^{-\frac{d}{2}}J_\infty^0(t)\mathcal {H}(t), \]

and for $0\leq k\leq 1+\frac {d}{2}$

\[ \left|\int_0^t(1+t-\tau)^{-\frac{d+2k}{4}}(1+\tau)^{-\frac{d+1}{2}}d\tau\right|\leq C (1+t)^{-\frac{d+2k}{4}}. \]

On the other hand, thanks to (4.2) and (4.23), similar to estimate (4.18) it follows that

(4.25)\begin{equation} \begin{aligned}\|\nabla v(t,\cdot)\|_{L^\infty} & \leq C(1+t)^{-\frac{d}{2}}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)\mathcal {H}(t))\\ & \leq C(1+t)^{-\frac{d}{2}}. \end{aligned} \end{equation}

This completes the proof of theorem 4.2.

Proof. Taking advantage of the Gagliardo–Nirenberg inequality, it follows that

(4.26)\begin{equation} \begin{aligned}\left\|\nabla^{((2+d)/2)^+}F(\tau,\cdot)\right\|_{L^2} & \leq\|v(\tau,\cdot)\|_{L^\infty}\left\|\nabla^{((4+d)/2)^+} v(\tau,\cdot)\right\|_{L^2} \\ & \leq\|v(\tau,\cdot)\|_{L^\infty}\|\nabla v(\tau,\cdot)\|_{L^\infty}^{1-\theta_1}\|\nabla^s v(\tau,\cdot)\|_{L^2}^{\theta_1}\\ & \leq C(1+\tau)^{-(\frac{d}{2}+(1-\theta_1)\frac{d+1}{2})}J_\infty^0(t)(J_\infty^1(t))^{1-\theta_1}\mathcal {H}^{\theta_1}(t)\\ & \leq C(1+\tau)^{-\frac{d+1}{2}}J_\infty^0(t)(J_\infty^1(t))^{1-\theta_1}\mathcal {H}^{\theta_1}(t), \end{aligned} \end{equation}

where the last inequality is guaranteed by $\frac {d}{2}+2+\frac {1}{d}< s$, the constant

\[ \theta_1=\frac{(1)^+}{s-(1+\frac{d}{2})}\in (0,1). \]

Using inequality (4.26), we can estimate

(4.27)\begin{equation} \begin{aligned}\left\|\nabla v(t,\cdot)\right\|_{L^\infty} & \leq C(1+t)^{-(d+1)/2}\epsilon_0+C\int_0^t(1+t-\tau)^{-(d+1)/2}\|F(\tau,\cdot)\|_{L^1}d\tau\\ & \quad+C\int_0^te^{-\beta (t-\tau)}\left\|\nabla^{((2+d)/2)^+}F(\tau,\cdot)\right\|_{L^2}d\tau\\ & \leq C(1+t)^{-(d+1)/2}(\epsilon_0+J_0(t)J_1(t)+J_\infty(t)^0(J_\infty^1(t))^{1-\theta_1}\mathcal {H}(t)^{\theta_1}). \end{aligned} \end{equation}

Note that $J_0(t),\,\;J_1(t)$ and $J_\infty ^0(t)$ are bounded, in view of Young's inequality yields that

\[ J_\infty^1(t)\leq C, \]

which implies for $s>\frac {d}{2}+2+\frac {1}{d}$ that

\[ \left\|\nabla v(t,\cdot)\right\|_{L^\infty}\leq C(1+t)^{-(d+1)/2}. \]

Finally, thanks to theorem 4.2, we only need to show the last inequality holds for $\frac {d}{2}< k\leq 1+\frac {d}{2}$. Similar to estimate (4.26), one has that

(4.28)\begin{equation} \begin{aligned}\left\|\nabla^{k}F(\tau,\cdot)\right\|_{L^2} & \leq\|v(\tau,\cdot)\|_{L^\infty}\left\|\nabla^{k+1}v(\tau,\cdot)\right\|_{L^2} \\ & \leq\|v(\tau,\cdot)\|_{L^\infty}\|\nabla v(\tau,\cdot)\|_{L^\infty}^{\vartheta_1}\|\nabla^s v(\tau,\cdot)\|_{L^2}^{1-\vartheta_1}\\ & \leq C(1+\tau)^{-(\frac{d}{2}+\vartheta_1\frac{d+1}{2})}J_\infty^0(t)(J_\infty^1(t))^{\vartheta_1}\mathcal {H}^{1-\vartheta_1}(t)\\ & \leq C(1+\tau)^{-\frac{d+2k}{4}}J_\infty^0(t)(J_\infty^1(t))^{\vartheta_1}\mathcal {H}^{1-\vartheta_1}(t), \end{aligned} \end{equation}

where the last inequality is guaranteed by $k\leq [(\frac {3d}{2}+1)(s-1)-\frac {d^2}{4}]/(s+\frac {d}{2})$, the constant

\[ \vartheta_1=\frac{s-(k+1)}{s-(1+\frac{d}{2})}\in (0,1). \]

By virtue of $J_\infty ^1(t)\leq C$ and (4.28), it follows that

(4.29)\begin{align} \left\|\nabla^k v(t,\cdot)\right\|_{L^2} & \leq C(1+t)^{-(\frac{d+2k}{4})}\epsilon_0+C\int_0^te^{-\beta (t-\tau)}\left\|\nabla^k F(\tau,\cdot)\right\|_{L^2}d\tau\nonumber\\ & \quad+C\int_0^t(1+t-\tau)^{-\frac{d+2k}{4}}\|F(\tau,\cdot)\|_{L^1}d\tau\nonumber\\ & \leq C(1+t)^{-\frac{d+2k}{4}}(\epsilon_0+J_0(t)J_1(t))\nonumber\\ & \quad+CJ_\infty^0(t)(J_\infty^1(t))^{\vartheta_1}(\mathcal {H}(t))^{1-\vartheta_1}\int_0^te^{-\beta(t-\tau)}(1+\tau)^{-\frac{d+2k}{4}}d\tau\nonumber\\ & \leq C(1+t)^{-\frac{d+2k}{4}}(\epsilon_0+J_0(t)J_1(t)+J_\infty^0(t)(J_\infty^1(t))^{\vartheta_1}(\mathcal {H}(t))^{1-\vartheta_1})\nonumber\\ & \leq C(1+t)^{-\frac{d+2k}{4}}, \end{align}

where we have applied $0< k\leq 1+\frac {d}{2}$ and $\frac {d}{2}+2+\frac {1}{d}< s$, which is equivalent to

\[ k<[(\frac{3d}{2}+1)(s-1)-\frac{d^2}{4}]/(s+\frac{d}{2}). \]

This completes the proof of corollary 4.4.

Proof. In view of interpolation inequality and theorem 4.2 yields that

\begin{align*} \|v(t,\cdot)\|_{L^p}& \leq\|v(t,\cdot)\|_{L^2}^{\frac{2}{p}}\|v(t,\cdot)\|_{L^\infty}^{1-\frac{2}{p}}\\ & \leq C(1+t)^{-\frac{d}{2}(1-\frac{1}{p})}. \end{align*}

By the Gagliardo–Nirenberg inequality and corollary 4.1, for $0\leq \alpha <1+\frac {d}{2}$ one shows that

\begin{align*} \|\nabla^\alpha v(t,\cdot)\|_{L^p}& \leq\|v(t,\cdot)\|_{L^\infty}^{\theta}\left\|\nabla^{(1+\frac{d}{2})}v(t,\cdot)\right\|_{L^2}^{1-\theta}\\ & \leq C(1+t)^{-\frac{1}{2}(d+\alpha-\frac{d}{p})}, \end{align*}

where $1-\theta =\alpha -\frac {d}{p}$ and $\theta \in (0,\,1)$. Note that

\begin{align*} \left\|\nabla^{((1+\frac{d}{p})} v(t,\cdot)\right\|_{L^p}& \leq\|\nabla v(t,\cdot)\|_{L^\infty}^{1/2}\left\|\nabla^{(1+\frac{d}{2})}v(t,\cdot)\right\|_{L^2}^{1/2}\\ & \leq C(1+t)^{-\frac{d+1}{2}}, \end{align*}

which includes the proof of corollary 4.7.

Proof. If we define the vorticity $\Omega =Du-\nabla u$, where $Du$ stands for the Jacobian matrix of velocity $u$, and $\nabla u$ stands for its transposed matrix, then the vorticity plays a fundamental role in the compressible fluid mechanics. Indeed, by system (2.1), $\Omega$ takes the form of a quasi-linear evolution equation of hyperbolic type

(4.31)\begin{equation} \partial_t\Omega+u\cdot\nabla\Omega+\Omega\cdot Du+\nabla u\cdot\Omega+\mu\Omega=0. \end{equation}

Multiplying $\Omega$ on both sides of equation (4.31), integration by parts, it follows that

(4.32)\begin{equation} \begin{aligned}\frac{\partial}{\partial_t}\int_{\mathbb{R}^d}|\Omega|^2dx+2\mu\int_{\mathbb{R}^d}|\Omega|^2dx & \leq C\|\nabla u\|_{L^\infty}\int_{\mathbb{R}^d}|\Omega|^2dx\\ & \leq\mu\int_{\mathbb{R}^d}|\Omega|^2dx, \end{aligned} \end{equation}

where we have applied $I(v_0)\ll 1$, which guarantees that

\[ C\|\nabla u\|_{L^\infty}\leq CI^{\frac{1}{2}}(v)(t)\leq CI^{\frac{1}{2}}(v_0)\leq\mu. \]

In view of Gronwall's inequality to (4.32) one has that

\[ \|\Omega(t,\cdot)\|_{L^2}\leq\|\Omega_0\|_{L^2}e^{-\frac{\mu}{2}t}. \]

Thanks to $\|\Omega \|_{L^2}\leq C\|\nabla u\|_{L^2}$, and $\|\nabla u\|_{L^2}\leq C\|\Omega \|_{L^2}$ (see proposition 7.5 on page 294 in [Reference Bahouri, Chemin and Danchin1]), therefore, we have

\[ \|\nabla u(t,\cdot)\|_{L^2}\leq C\|\nabla u_0\|_{L^2}e^{-\frac{\mu}{2}t}. \]