1 Introduction

In this note, we consider the following three-dimensional compressible Navier–Stokes equations with revised Maxwell’s law (see [10] for instance):

$$\begin{aligned} \left\{ \begin{array}{ll} \rho _{t}+\textrm{div}(\rho {\textbf{u}})=0,\\ (\rho {\textbf{u}})_{t}+\textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})+\nabla p(\rho )=\textrm{div}{\mathbb {S}},\end{array}\right. \end{aligned}$$
(1.1)

where the fluid density \(\rho =\rho (x,t)\), the fluid velocity \({\textbf{u}}={\textbf{u}}(x,t)=(u_{1}(x,t),u_{2}(x,t),u_{3}(x,t))\) and the stress tensor \({\mathbb {S}}={\mathbb {S}}(x,t)\) are the unknown functions with \((x,t)\in {\mathbb {R}}^{3}\times {\mathbb {R}}^{+}\). The function \(p(\rho )=a\rho ^{\gamma }\) represents the pressure with \(a>0\) and \(\gamma >1\) being two constants, \(\gamma \) is called the adiabatic exponent. The stress tensor \({\mathbb {S}}\) is assumed to satisfy the revised Maxwell’s law: \({\mathbb {S}}={\mathbb {S}}_{1}+S_{2}{\mathbb {I}}\), where \({\mathbb {S}}_{1}\) and \(S_{2}\) are given by the following equations

$$\begin{aligned}{} & {} \tau _{1}({\mathbb {S}}_{1})_{t}+{\mathbb {S}}_{1}=\mu [(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}], \end{aligned}$$
(1.2)
$$\begin{aligned}{} & {} \tau _{2}(S_{2})_{t}+S_{2}=\lambda \textrm{div}{\textbf{u}}, \end{aligned}$$
(1.3)

where \(\mu \) and \(\lambda \) are positive constants, \(\tau _{1}\) and \(\tau _{2}\) denote the shear relaxation time and the compressible relaxation time, respectively, \({\mathbb {I}}\) represents \(3\times 3\) identity matrix. Equations (1.2) and (1.3) were first proposed by Yong [14] from some mathematical point of view, later Chakraborty and Sader [2] showed the fact that the division of \({\mathbb {S}}\) into two parts has its physical meanings, where the importance of this model for describing high frequency limits is underlined together with the presentation of numerical experiments.

There were some works about the mathematical analysis of the Navier–Stokes equations with revised Maxwell’s law in the literature. Yong proved that system (1.1)–(1.3) is symmetric hyperbolic system, which implies local well-posedness immediately. Again due to the symmetric hyperbolic property of system (1.1)–(1.3), the important property of finite propagation speed is available, which allows one to define some averaged quantities as in [8] and shows finite time blowup of solutions by establishing a Riccati-type inequality, see [10]. Precisely, Wang and Hu [10] considered system (1.1)–(1.3) with initial data

$$\begin{aligned} (\rho ,{\textbf{u}},{\mathbb {S}}_{1},S_{2})(x,0)=(\rho _{0},{\textbf{u}}_{0},{\mathbb {S}}_{10},S_{20})(x), \end{aligned}$$
(1.4)

they showed that if the initial data \((\rho _{0}-{\overline{\rho }},{\textbf{u}}_{0},{\mathbb {S}}_{10},S_{20})\) are compactly supported in \(B_{R}:=\{x\in {\mathbb {R}}^{3}||x|\le R\}\) for some \(R>0\) with \({\overline{\rho }}\) being any positive constant, then there exists a constant \(\sigma \) such that

$$\begin{aligned} (\rho (\cdot ,t)-{\overline{\rho }},{\textbf{u}}(\cdot ,t),{\mathbb {S}}_{1}(\cdot ,t),S_{2}(\cdot ,t))=(0,0,0,0) \end{aligned}$$
(1.5)

on \(D=\{x\in {\mathbb {R}}^{3}||x|>R+\sigma t\}\) for \(C^{1}\) solutions \((\rho (\cdot ,t),{\textbf{u}}(\cdot ,t),{\mathbb {S}}_{1}(\cdot ,t),S_{2}(\cdot ,t))\) to the Cauchy problem (1.1)–(1.4) on [0, T], see Proposition 2.1 of [10]. This guarantees that the following averaged quantities are finite:

$$\begin{aligned}{} & {} m(t):=\int _{{\mathbb {R}}^{3}}(\rho (x,t)-{\overline{\rho }})dx, \end{aligned}$$
(1.6)
$$\begin{aligned}{} & {} A(t):=\int _{{\mathbb {R}}^{3}}S_{2}(x,t)dx, \end{aligned}$$
(1.7)
$$\begin{aligned}{} & {} F(t):=\int _{{\mathbb {R}}^{3}}x\cdot (\rho {\textbf{u}})(x,t)dx. \end{aligned}$$
(1.8)

By establishing a Riccati-type inequality of F(t), the authors of Wang and Hu [10] proved that the life span of any \(C^{1}\) solution to (1.1)–(1.4) must be finite for some special large initial data. Unfortunately, the following three restrictions are required in [10]:

$$\begin{aligned} m(0)\ge 0,\ \ \textrm{Trace}\{{\mathbb {S}}_{10}(x)\}=0,\ \ A(0)\le 0. \end{aligned}$$
(1.9)

In this note, our first aim is to give a blowup result for the \(C^{1}\) solutions to (1.1)–(1.4) without the conditions (1.9).

When \(\tau _{1}=\tau _{2}=0\), the system (1.1)–(1.3) is reduced to the classical compressible Navier–Stokes system. Many authors have studied the blowup of smooth solutions to the classical compressible Navier–Stokes system, see [1, 3, 5,6,7, 9, 12, 13] and the references therein. Particularly, the authors of Jiu [5] and Wang et al. [9] investigated the blowup of classical solutions to the classical compressible Navier–Stokes system when the density and the velocity decay at far fields. Motivated by Jiu [5] and Wang et al. [9], we can present a blowup result of classical solutions to (1.1)–(1.4) when \(\tau _{1}=0\) and the density and the velocity decay at far fields. This is our second aim of this note. Different from Jiu [5] and Wang et al. [9], for the problem (1.1)–(1.4) we cannot obtain the conversation of the total energy or the decrease with time of the total energy. This makes us difficult to get the upper bound of the momentum of inertia, which is crucial to establish the Riccati-type inequality of a weighted momentum. To overcome this difficulty, we introduce a new averaged quantity J(t) [see (3.3) below], which decreases with time and is related to the total energy.

Remark 1.1

For the results about the non-isentropic Navier–Stokes equations with revised Maxwell’s law, we can refer to [4, 11].

2 Improvement on the Blowup Result of Wang and Hu [10]

Our result in this section is stated as follows.

Theorem 2.1

Let \((\rho ,u,{\mathbb {S}}_{1},S_{2})\) be a \(C^{1}\) solution to the Cauchy problem (1.1)–(1.4) for \(0\le t\le T_{1}\) with the initial data \((\rho _{0}-{\overline{\rho }},{\textbf{u}}_{0},{\mathbb {S}}_{10},S_{20})\) being compactly supported in \(B_{R}:=\{x\in {\mathbb {R}}^{3}||x|\le R\}\). For any fixed \(t_{1}^{*}>0\), if

$$\begin{aligned} F(0)>\sqrt{2[m(0)+{\overline{\rho }}|B(t_{1}^{*})|][3{\overline{p}}|B(t_{1}^{*})|+3|A(0)|+|G(0)|]}(R+\sigma t_{1}^{*}) \end{aligned}$$
(2.1)

and

$$\begin{aligned} F(0)>\left\{ \int _{0}^{t_{1}^{*}}\frac{dt}{2(R+\sigma t)^{2}[m(0)+{\overline{\rho }}|B(t)|]}\right\} ^{-1}, \end{aligned}$$
(2.2)

then \(T_{1}<t_{1}^{*}\), where

$$\begin{aligned} |B(t_{1}^{*})|=\frac{4}{3}\pi (R+\sigma t_{1}^{*})^{3},\quad {\overline{p}}=p({\overline{\rho }}),\quad G(0)=\int _{{\mathbb {R}}^{3}}\mathrm {Trace\{{\mathbb {S}}_{10}\}}dx. \end{aligned}$$
(2.3)

Proof

By (1.6), (1.1)\(_{1}\) and (1.5), we get

$$\begin{aligned} m'(t)=\int _{{\mathbb {R}}^{3}}\rho _{t}dx=-\int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}})dx=0, \end{aligned}$$
(2.4)

which implies that

$$\begin{aligned} m(t)=m(0). \end{aligned}$$
(2.5)

In view of (1.7), (1.3) and (1.5), one has

$$\begin{aligned} A'(t)=-\frac{1}{\tau _{2}}\int _{{\mathbb {R}}^{3}}S_{2}dx+\frac{\lambda }{\tau _{2}}\int _{{\mathbb {R}}^{3}}\textrm{div}{\textbf{u}}dx=-\frac{1}{\tau _{2}}\int _{{\mathbb {R}}^{3}}S_{2}dx=-\frac{1}{\tau _{2}}A(t), \end{aligned}$$
(2.6)

so we obtain

$$\begin{aligned} |A(t)|=|A(0)|e^{-\frac{t}{\tau _{2}}}\le |A(0)|. \end{aligned}$$
(2.7)

By (1.2), we know that

$$\begin{aligned} \tau _{1}\partial _{t}\textrm{Trace}\{{\mathbb {S}}_{1}\}+\textrm{Trace}\{{\mathbb {S}}_{1}\}=0 \end{aligned}$$
(2.8)

due to

$$\begin{aligned} \textrm{Trace}\{(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}\}=0. \end{aligned}$$
(2.9)

Define

$$\begin{aligned} G(t)=\int _{{\mathbb {R}}^{3}}\mathrm {Trace\{{\mathbb {S}}_{1}(x,t)\}}dx, \end{aligned}$$
(2.10)

then by (2.8), we have

$$\begin{aligned} |G(t)|=|G(0)|e^{-\frac{t}{\tau _{1}}}\le |G(0)|. \end{aligned}$$
(2.11)

It follows from (1.8), (1.1)\(_{2}\), (1.5), (2.7) and (2.11) that

$$\begin{aligned} F'(t)= & {} \int _{{\mathbb {R}}^{3}}x\cdot (\rho {\textbf{u}})_{t}dx\nonumber \\= & {} -\int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})dx-\int _{{\mathbb {R}}^{3}}x\cdot \nabla (p(\rho )-{\overline{p}})dx+\int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}{\mathbb {S}}dx\nonumber \\= & {} \int _{{\mathbb {R}}^{3}}\rho |{\textbf{u}}|^{2}dx+3\int _{{\mathbb {R}}^{3}}(p(\rho )-{\overline{p}})dx-3\int _{{\mathbb {R}}^{3}}S_{2}dx-\int _{{\mathbb {R}}^{3}}\mathrm {Trace\{{\mathbb {S}}_{1}\}}dx\nonumber \\= & {} \int _{B(t)}\rho |{\textbf{u}}|^{2}dx+3\int _{B(t)}(p(\rho )-{\overline{p}}))dx-3A(t)-G(t)\nonumber \\\ge & {} \int _{B(t)}\rho |{\textbf{u}}|^{2}dx-3{\overline{p}}|B(t)|-3|A(0)|-|G(0)|, \end{aligned}$$
(2.12)

where

$$\begin{aligned} B(t)=\{x\in {\mathbb {R}}^{3}||x|\le R+\sigma t\},\quad |B(t)|=\frac{4}{3}\pi (R+\sigma t)^{3}. \end{aligned}$$
(2.13)

By the Schwarz inequality, it holds

$$\begin{aligned} F(t)^{2}=\left( \int _{B(t)}x\cdot \rho {\textbf{u}}dx\right) ^{2}\le \int _{B(t)}|x|^{2}\rho dx\cdot \int _{B(t)}\rho |{\textbf{u}}|^{2}dx. \end{aligned}$$
(2.14)

By (1.6) and (2.5), we know that

$$\begin{aligned} \int _{B(t)}|x|^{2}\rho dx\le & {} (R+\sigma t)^{2}\int _{B(t)}\rho dx =(R+\sigma t)^{2}\left[ \int _{B(t)}(\rho -{\overline{\rho }})dx+\int _{B(t)}{\overline{\rho }}dx\right] \nonumber \\= & {} (R+\sigma t)^{2}\left[ m(t)+{\overline{\rho }}|B(t)|\right] =(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] .\nonumber \\ \end{aligned}$$
(2.15)

Combining (2.12), (2.14) and (2.15), we obtain

$$\begin{aligned} F'(t)\ge \frac{F(t)^{2}}{(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] }-3{\overline{p}}|B(t)|-3|A(0)|-|G(0)|. \end{aligned}$$
(2.16)

For any fixed \(t_{1}^{*}>0\), when \(0\le t\le t_{1}^{*}\), it follows from (2.16) that

$$\begin{aligned} F'(t)\ge & {} \frac{F(t)^{2}}{2(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] }\nonumber \\{} & {} +\,\left\{ \frac{F(t)^{2}}{2(R+\sigma t_{1}^{*})^{2}\left[ m(0)+{\overline{\rho }}|B(t_{1}^{*})|\right] }-3{\overline{p}}|B(t_{1}^{*})|-3|A(0)|-|G(0)|\right\} .\nonumber \\ \end{aligned}$$
(2.17)

By (2.1), we know that

$$\begin{aligned} \frac{F(0)^{2}}{2(R+\sigma t_{1}^{*})^{2}\left[ m(0)+{\overline{\rho }}|B(t_{1}^{*})|\right] }-3{\overline{p}}|B(t_{1}^{*})|-3|A(0)|-|G(0)|>0, \end{aligned}$$
(2.18)

which together with (2.17) imply that \(F'(0)>0\), so \(F(t)>F(0)>0\) holds at least for a small time \(t>0\), then by (2.17) we derive that

$$\begin{aligned} F'(t)\ge \frac{F(t)^{2}}{2(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] } \end{aligned}$$
(2.19)

and

$$\begin{aligned} \frac{F(t)^{2}}{2(R+\sigma t_{1}^{*})^{2}\left[ m(0)+{\overline{\rho }}|B(t_{1}^{*})|\right] }-3{\overline{p}}|B(t_{1}^{*})|-3|A(0)|-|G(0)|>0 \end{aligned}$$
(2.20)

hold whenever F(t) exists and \(0<t\le t_{1}^{*}\). We divide (2.19) by \(F(t)^{2}\) and integrate the resultant inequality over \([0,t_{1}^{*}]\) to have

$$\begin{aligned} \frac{1}{F(0)}>\frac{1}{F(0)}-\frac{1}{F(t_{1}^{*})}\ge \int _{0}^{t_{1}^{*}}\frac{dt}{2(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] }. \end{aligned}$$
(2.21)

On the other hand, by (2.2) we know that

$$\begin{aligned} \frac{1}{F(0)}<\int _{0}^{t_{1}^{*}}\frac{dt}{2(R+\sigma t)^{2}\left[ m(0)+{\overline{\rho }}|B(t)|\right] }, \end{aligned}$$
(2.22)

which together with (2.21) imply that \(T_{1}<t_{1}^{*}\). We complete the proof of Theorem 2.1. \(\square \)

Remark 2.1

In Theorem 2.1 above, we show the blowup of classical solutions to the problem (1.1)–(1.4) without the three restrictions \(m(0)\ge 0\), \(\textrm{Trace}\{{\mathbb {S}}_{10}(x)\}=0\) and \(A(0)\le 0\), so we have improved the blowup result of [10].

3 Blowup for Solutions Decay at Far Fields

In this section, we need the following averaged quantities:

$$\begin{aligned} I(t):= & {} \frac{1}{2}\int _{{\mathbb {R}}^{3}}|x|^{2}\rho (x,t)dx, \end{aligned}$$
(3.1)
$$\begin{aligned} E(t):= & {} \frac{1}{2}\int _{{\mathbb {R}}^{3}}\rho |{\textbf{u}}|^{2}dx+\frac{a}{\gamma -1}\int _{{\mathbb {R}}^{3}}\rho ^{\gamma }dx=E_{k}(t)+E_{i}(t), \end{aligned}$$
(3.2)
$$\begin{aligned} J(t):= & {} E(t)+\frac{\tau _{2}}{2\lambda }\int _{{\mathbb {R}}^{3}}S_{2}^{2}dx, \end{aligned}$$
(3.3)

where \(I(t),E(t),E_{k}(t)\) and \(E_{i}(t)\) represent the momentum of inertia, the total energy, the kinetic energy and the internal energy, respectively, which have been used in [5, 9]. We remark that the averaged quantity J(t) is new in the literature. In this section, we only consider the classical solutions with decay at far fields. Precisely, for any \(T>0\), we require that the solutions \((\rho ,{\textbf{u}},{\mathbb {S}}_{1},S_{2})\) satisfy that

$$\begin{aligned} \rho |{\textbf{u}}|^{3},\rho ^{\gamma }|{\textbf{u}}|,|S_{2}{\textbf{u}}|,|{\textbf{u}}\nabla {\textbf{u}}|,|x|\rho |{\textbf{u}}|^{2},|x|\rho ^{\gamma },|x\nabla {\textbf{u}}|,|xS_{2}|,|{\textbf{u}}|\in L^{\infty }(0,T;L^{1}({\mathbb {R}}^{3})).\nonumber \\ \end{aligned}$$
(3.4)

We should remark that the condition (3.4) guarantees that the integration by parts in our calculations makes sense (see also [5, 9]).

The main result of this section is stated as follows.

Theorem 3.1

Let \((\rho ,{\textbf{u}},{\mathbb {S}}_{1},S_{2})\) be a \(C^{1}\) solution to the Cauchy problem (1.1)–(1.4) with \(\tau _{1}=0\) satisfying (3.4) for \(0\le t\le T_{2}\) with the initial data satisfying \(0<|A(0)|,I(0),J(0)<+\infty \). For any fixed \(t_{2}^{*}>0\), if

$$\begin{aligned} F(0)>\sqrt{6|A(0)|\{[\max \{2,3(\gamma -1)\}J(0)+3|A(0)|]t_{2}^{*2}+2F(0)t_{2}^{*}+2I(0)\}}\nonumber \\ \end{aligned}$$
(3.5)

and

$$\begin{aligned} F(0)>\left\{ \int _{0}^{t_{2}^{*}}\frac{dt}{2\{[\max \{2,3(\gamma -1)\}J(0)+3|A(0)|]t^{2}+2F(0)t+2I(0)\}}\right\} ,\nonumber \\ \end{aligned}$$
(3.6)

then \(T_{2}<t_{2}^{*}\).

Proof

We multiply (1.1)\(_{2}\) by \({\textbf{u}}\) and integrate it over \({\mathbb {R}}^{3}\) to have

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}^{3}}(\rho {\textbf{u}})_{t}\cdot {\textbf{u}}dx+\int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})\cdot {\textbf{u}}dx+\int _{{\mathbb {R}}^{3}}\nabla p(\rho )\cdot {\textbf{u}}dx\nonumber \\{} & {} \quad =\mu \int _{{\mathbb {R}}^{3}}\textrm{div}[(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}]\cdot {\textbf{u}}dx+\int _{{\mathbb {R}}^{3}}\nabla S_{2}\cdot {\textbf{u}}dx, \end{aligned}$$
(3.7)

where we have used \({\mathbb {S}}={\mathbb {S}}_{1}+S_{2}{\mathbb {I}}\) and (1.2) with \(\tau _{1}=0\). We use integration by part and (1.1)\(_{1}\) to obtain

$$\begin{aligned} \int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})\cdot {\textbf{u}}dx= & {} \int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}})|{\textbf{u}}|^{2}dx +\int _{{\mathbb {R}}^{3}}\rho {\textbf{u}}\cdot \nabla {\textbf{u}}\cdot {\textbf{u}}dx\nonumber \\= & {} \int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}})|{\textbf{u}}|^{2}dx +\frac{1}{2}\int _{{\mathbb {R}}^{3}}\rho {\textbf{u}}\cdot \nabla (|{\textbf{u}}|^{2})dx\nonumber \\= & {} \frac{1}{2}\int _{{\mathbb {R}}^{3}}\textrm{div}(\rho {\textbf{u}})|{\textbf{u}}|^{2}dx\nonumber \\= & {} -\frac{1}{2}\int _{{\mathbb {R}}^{3}}\rho _{t}|{\textbf{u}}|^{2}dx. \end{aligned}$$
(3.8)

Similarly,

$$\begin{aligned} \int _{{\mathbb {R}}^{3}}\nabla p(\rho )\cdot {\textbf{u}}dx= & {} a\gamma \int _{{\mathbb {R}}^{3}}\rho ^{\gamma -1}\nabla \rho \cdot {\textbf{u}}dx\nonumber \\= & {} \frac{a\gamma }{\gamma -1}\int _{{\mathbb {R}}^{3}}\nabla (\rho ^{\gamma -1})\cdot (\rho {\textbf{u}})dx\nonumber \\= & {} -\frac{a\gamma }{\gamma -1}\int _{{\mathbb {R}}^{3}}\rho ^{\gamma -1}\textrm{div}(\rho {\textbf{u}})dx\nonumber \\= & {} \frac{a\gamma }{\gamma -1}\int _{{\mathbb {R}}^{3}}\rho ^{\gamma -1}\rho _{t}dx\nonumber \\= & {} \frac{d}{dt}\int _{{\mathbb {R}}^{3}}\frac{\rho ^{\gamma }}{\gamma -1}dx, \end{aligned}$$
(3.9)
$$\begin{aligned} \int _{{\mathbb {R}}^{3}}\nabla S_{2}\cdot {\textbf{u}}dx= & {} -\int _{{\mathbb {R}}^{3}}S_{2}\textrm{div}{\textbf{u}}dx. \end{aligned}$$
(3.10)

We estimate the integral \(\mu \int _{{\mathbb {R}}^{3}}\textrm{div}[(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}]\cdot {\textbf{u}}dx\) as follows. Noticing that

$$\begin{aligned} \textrm{div}(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})= & {} \left( 2\frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ,\right. \nonumber \\{} & {} \quad \frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) +2\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{2}}{\partial x_{2}}\right) +\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ,\nonumber \\{} & {} \quad \left. \frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) +2\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{3}}{\partial x_{3}}\right) \right) ,\nonumber \\ \end{aligned}$$
(3.11)

one has

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}^{3}}\textrm{div}(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})\cdot {\textbf{u}}dx\nonumber \\{} & {} \quad =\int _{{\mathbb {R}}^{3}}u_{1}\left[ 2\frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) \right] dx\nonumber \\{} & {} \qquad +\,\int _{{\mathbb {R}}^{3}}u_{2}\left[ \frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) +2\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{2}}{\partial x_{2}}\right) +\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) \right] dx\nonumber \\{} & {} \qquad +\,\int _{{\mathbb {R}}^{3}}u_{3}\left[ \frac{\partial }{\partial x_{1}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) +\frac{\partial }{\partial x_{2}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) +2\frac{\partial }{\partial x_{3}}\left( \frac{\partial u_{3}}{\partial x_{3}}\right) \right] dx\nonumber \\{} & {} \quad =-2\int _{{\mathbb {R}}^{3}}\sum \limits _{i=1}^{3}\left( \frac{\partial u_{i}}{\partial x_{i}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left[ \left( \frac{\partial u_{1}}{\partial x_{2}}\right) ^{2}+\left( \frac{\partial u_{1}}{\partial x_{3}}\right) ^{2}\right] dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \nonumber \\{} & {} \qquad -\,\int _{{\mathbb {R}}^{3}}\left[ \left( \frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}+\left( \frac{\partial u_{2}}{\partial x_{3}}\right) ^{2}\right] dx-\int _{{\mathbb {R}}^{3}}\left[ \left( \frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}+\left( \frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}\right] dx\ \ \ \ \ \ \ \nonumber \\{} & {} \qquad -\,2\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}\cdot \frac{\partial u_{2}}{\partial x_{1}}+\frac{\partial u_{1}}{\partial x_{3}}\cdot \frac{\partial u_{3}}{\partial x_{1}}+\frac{\partial u_{2}}{\partial x_{3}}\cdot \frac{\partial u_{3}}{\partial x_{2}}\right) dx\ \ \ \ \nonumber \\{} & {} \quad =-2\int _{{\mathbb {R}}^{3}}\sum \limits _{i=1}^{3}\left( \frac{\partial u_{i}}{\partial x_{i}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}dx\nonumber \\{} & {} \qquad -\,\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}dx. \end{aligned}$$
(3.12)
$$\begin{aligned}{} & {} \frac{2}{3}\int _{{\mathbb {R}}^{3}}\textrm{div}(\textrm{div}{\textbf{u}}{\mathbb {I}})\cdot {\textbf{u}}dx=-\frac{2}{3}\int _{{\mathbb {R}}^{3}}(\sum \limits _{i=1}^{3}\frac{\partial u_{i}}{\partial x_{i}})^{2}dx. \end{aligned}$$
(3.13)

Consequently,

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}^{3}}\textrm{div}\left[ (\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}\right] \cdot {\textbf{u}}dx\nonumber \\{} & {} \quad =-2\int _{{\mathbb {R}}^{3}}\sum \limits _{i=1}^{3}\left( \frac{\partial u_{i}}{\partial x_{i}}\right) ^{2}dx+\frac{2}{3}\int _{{\mathbb {R}}^{3}}(\sum \limits _{i=1}^{3}\frac{\partial u_{i}}{\partial x_{i}})^{2}dx\nonumber \\{} & {} \qquad -\,\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}dx\nonumber \\{} & {} \quad \le -\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}dx\nonumber \\{} & {} \qquad -\,\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}dx. \end{aligned}$$
(3.14)

Combining (3.7)–(3.10), (3.14) and (3.2), we obtain

$$\begin{aligned} \frac{d}{dt}E(t)\le & {} -\int _{{\mathbb {R}}^{3}}S_{2}\textrm{div}{\textbf{u}}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}dx\nonumber \\{} & {} -\,\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}dx. \end{aligned}$$
(3.15)

We multiply (1.3) by \(\frac{S_{2}}{\lambda }\) and integrate the resultant equation over \({\mathbb {R}}^{3}\) to have

$$\begin{aligned} \frac{\tau _{2}}{2\lambda }\frac{d}{dt}\int _{{\mathbb {R}}^{3}}S_{2}^{2}dx=-\frac{1}{\lambda }\int _{{\mathbb {R}}^{3}}S_{2}^{2}dx+\int _{{\mathbb {R}}^{3}}S_{2}\textrm{div}{\textbf{u}}dx. \end{aligned}$$
(3.16)

In view of (3.15), (3.16) and (3.3), it holds

$$\begin{aligned} \frac{d}{dt}J(t)\le & {} -\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{2}}+\frac{\partial u_{2}}{\partial x_{1}}\right) ^{2}dx-\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{1}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{1}}\right) ^{2}dx\nonumber \\{} & {} -\,\int _{{\mathbb {R}}^{3}}\left( \frac{\partial u_{2}}{\partial x_{3}}+\frac{\partial u_{3}}{\partial x_{2}}\right) ^{2}dx\le 0, \end{aligned}$$
(3.17)

which implies that

$$\begin{aligned} J(t)\le J(0),\quad 0\le t\le T_{2}. \end{aligned}$$
(3.18)

By (3.1), (1.1)\(_{1}\) and (1.8), we know that

$$\begin{aligned} I'(t)=\frac{1}{2}\int _{{\mathbb {R}}^{3}}|x|^{2}\rho _{t}dx=-\frac{1}{2}\int _{{\mathbb {R}}^{3}}|x|^{2}\textrm{div}(\rho {\textbf{u}})dx=\int _{{\mathbb {R}}^{3}}x\cdot \rho {\textbf{u}}dx=F(t).\nonumber \\ \end{aligned}$$
(3.19)

In view of (1.8) and (1.1)\(_{2}\), we get

$$\begin{aligned} F'(t)= & {} \int _{{\mathbb {R}}^{3}}x\cdot (\rho {\textbf{u}})_{t}dx\nonumber \\= & {} -\int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})dx-\int _{{\mathbb {R}}^{3}}x\cdot \nabla p(\rho )dx\nonumber \\{} & {} +\,\mu \int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}[(\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}]dx+\int _{{\mathbb {R}}^{3}}x\cdot \nabla S_{2}dx, \end{aligned}$$
(3.20)

where we have used \({\mathbb {S}}={\mathbb {S}}_{1}+S_{2}{\mathbb {I}}\) and (1.2) with \(\tau _{1}=0\). Using (3.4) and integration by part, we obtain

$$\begin{aligned} -\int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}(\rho {\textbf{u}}\otimes {\textbf{u}})dx= & {} -\int _{{\mathbb {R}}^{3}}x_{1}\left[ \frac{\partial }{\partial x_{1}}(\rho u_{1}^{2})+\frac{\partial }{\partial x_{2}}(\rho u_{2}u_{1})+\frac{\partial }{\partial x_{3}}(\rho u_{3}u_{1})\right] dx\nonumber \\{} & {} -\,\int _{{\mathbb {R}}^{3}}x_{2}\left[ \frac{\partial }{\partial x_{1}}(\rho u_{1}u_{2})+\frac{\partial }{\partial x_{2}}(\rho u_{2}^{2})+\frac{\partial }{\partial x_{3}}(\rho u_{3}u_{2})\right] dx\nonumber \\{} & {} -\,\int _{{\mathbb {R}}^{3}}x_{3}\left[ \frac{\partial }{\partial x_{1}}(\rho u_{1}u_{3})+\frac{\partial }{\partial x_{2}}(\rho u_{2}u_{3})+\frac{\partial }{\partial x_{3}}(\rho u_{3}^{2})\right] dx\nonumber \\= & {} \int _{{\mathbb {R}}^{3}}\rho (u_{1}^{2}+u_{2}^{2}+u_{3}^{2})dx=\int _{{\mathbb {R}}^{3}}\rho |{\textbf{u}}|^{2}dx, \end{aligned}$$
(3.21)
$$\begin{aligned} -\int _{{\mathbb {R}}^{3}}x\cdot \nabla p(\rho )dx= & {} 3a\int _{{\mathbb {R}}^{3}}\rho ^{\gamma }dx, \end{aligned}$$
(3.22)
$$\begin{aligned}{} & {} \mu \int _{{\mathbb {R}}^{3}}x\cdot \textrm{div}\left[ (\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}\right] dx\nonumber \\{} & {} \quad =-\mu \int _{{\mathbb {R}}^{3}}\textrm{Trace}\left[ (\nabla {\textbf{u}}+\nabla {\textbf{u}}^{T})-\frac{2}{3}\textrm{div}{\textbf{u}}{\mathbb {I}}\right] dx=0, \end{aligned}$$
(3.23)
$$\begin{aligned} \int _{{\mathbb {R}}^{3}}x\cdot \nabla S_{2}dx=-3\int _{{\mathbb {R}}^{3}}S_{2}dx. \end{aligned}$$
(3.24)

It follows from (3.20) to (3.24) that

$$\begin{aligned} F'(t)=\int _{{\mathbb {R}}^{3}}\rho |{\textbf{u}}|^{2}dx+3a\int _{{\mathbb {R}}^{3}}\rho ^{\gamma }dx-3\int _{{\mathbb {R}}^{3}}S_{2}dx, \end{aligned}$$
(3.25)

which together with (3.2) lead to

$$\begin{aligned} F'(t)=2E_{k}(t)+3(\gamma -1)E_{i}(t)-3\int _{{\mathbb {R}}^{3}}S_{2}dx. \end{aligned}$$
(3.26)

We integrate (1.3) over \({\mathbb {R}}^{3}\) and use the condition \(|{\textbf{u}}|\in L^{\infty }(0,T;L^{1}({\mathbb {R}}^{3}))\) in (3.4) to have

$$\begin{aligned} \tau _{2}\frac{d}{dt}\int _{{\mathbb {R}}^{3}}S_{2}dx+\int _{{\mathbb {R}}^{3}}S_{2}dx=\lambda \int _{{\mathbb {R}}^{3}}\textrm{div}{\textbf{u}}dx=0, \end{aligned}$$
(3.27)

which together with (1.7) and (3.26) imply that

$$\begin{aligned} F'(t)= & {} 2E_{k}(t)+3(\gamma -1)E_{i}(t)-3A(t)\nonumber \\= & {} 2E_{k}(t)+3(\gamma -1)E_{i}(t)-3A(0)e^{-\frac{t}{\tau _{2}}}. \end{aligned}$$
(3.28)

By (3.19), (3.28), (3.2), (3.3) and (3.18), we know that

$$\begin{aligned} I''(t)= & {} F'(t)\le \max \{2,3(\gamma -1)\}J(t)+3|A(0)|\nonumber \\\le & {} \max \{2,3(\gamma -1)\}J(0)+3|A(0)|. \end{aligned}$$
(3.29)

We integrate (3.29) over \({\mathbb {R}}^{3}\) twice and use (3.19) to obtain

$$\begin{aligned} I(t)\le \frac{1}{2}[\max \{2,3(\gamma -1)\}J(0)+3|A(0)|]t^{2}+F(0)t+I(0). \end{aligned}$$
(3.30)

By the Schwarz inequality, (3.1) and (3.2), it holds

$$\begin{aligned} F(t)^{2}=\left( \int _{{\mathbb {R}}^{3}}x\cdot \rho {\textbf{u}}dx\right) ^{2}\le \int _{{\mathbb {R}}^{3}}|x|^{2}\rho dx\cdot \int _{{\mathbb {R}}^{3}}\rho |{\textbf{u}}|^{2}dx=4I(t)E_{k}(t), \end{aligned}$$
(3.31)

which together with (3.28) and (3.30) imply that

$$\begin{aligned} F'(t)\ge & {} 2E_{k}(t)-3|A(0)|\ge \frac{F(t)^{2}}{2I(t)}-3|A(0)|\nonumber \\\ge & {} \frac{F(t)^{2}}{[\max \{2,3(\gamma -1)\}J(0)+3|A(0)|]t^{2}+2F(0)t+2I(0)}-3|A(0)|.\nonumber \\ \end{aligned}$$
(3.32)

This corresponds to (2.16), so the rest proof is similar to the one of Theorem 2.1, we omit the details. \(\square \)

In fact, if we assume \(A(0)\le 0\), we can obtain a refined blowup result for the Cauchy problem (1.1)–(1.4) with solutions decay at far fields.

Corollary 3.1

Denote

$$\begin{aligned} c_{1}=\max \{2,3(\gamma -1)\},\ \ c_{2}=\sqrt{2[c_{1}J(0)-3A(0)]I(0)-F(0)^{2}}. \end{aligned}$$
(3.33)

Under the assumptions of Theorem 3.1, if we further assume that \(A(0)\le 0\), \(F(0)>0\) and

$$\begin{aligned} \frac{c_{2}}{F(0)}+\textrm{arctan}\frac{F(0)}{c_{2}}<\frac{\pi }{2}, \end{aligned}$$
(3.34)

then the life span \(T_{2}\) of the classical solutions to (1.1)–(1.4) satisfies that

$$\begin{aligned} T_{2}<\frac{c_{2}}{c_{1}J(0)-3A(0)}\tan \left[ \frac{c_{2}}{F(0)}+\textrm{arctan}\frac{F(0)}{c_{2}}\right] -\frac{F(0)}{c_{1}J(0)-3A(0)} \end{aligned}$$
(3.35)

Proof

By the condition \(A(0)\le 0\) and (3.28)–(3.30), we know that

$$\begin{aligned} F'(t)\ge 2E_{k}(t)+3(\gamma -1)E_{i}(t) \end{aligned}$$
(3.36)

and

$$\begin{aligned} I(t)\le \frac{1}{2}[\max \{2,3(\gamma -1)\}J(0)-3A(0)]t^{2}+F(0)t+I(0). \end{aligned}$$
(3.37)

Then (3.32) becomes

$$\begin{aligned} F'(t)\ge \frac{F(t)^{2}}{[\max \{2,3(\gamma -1)\}J(0)-3A(0)]t^{2}+2F(0)t+2I(0)}. \end{aligned}$$
(3.38)

The inequality (3.36) implies that F(t) increases with time, which together with \(F(0)>0\) lead to the fact that \(F(t)>0\) for \(t\in [0,T_{2}]\). We use (3.33) to rewrite (3.38) as

$$\begin{aligned} F'(t)\ge \frac{F(t)^{2}}{[c_{1}J(0)-3A(0)]\left\{ \left( t+\frac{F(0)}{c_{1}J(0)-3A(0)}\right) ^{2}+\frac{c_{2}^{2}}{[c_{1}J(0)-3A(0)]^{2}}\right\} }. \end{aligned}$$
(3.39)

Noticing that

$$\begin{aligned} c_{2}^{2}= & {} 2[c_{1}J(0)-3A(0)]I(0)-F(0)^{2}\ge 2c_{1}J(0)I(0)\nonumber \\{} & {} -\,F(0)^{2}>4E_{k}(0)I(0)-F(0)^{2}\ge 0, \end{aligned}$$
(3.40)

we have \(c_{2}>0\). So we divide (3.39) by \(F(t)^{2}\) and integrate the resultant inequality over \([0,T_{2}]\) to obtain

$$\begin{aligned}{} & {} \frac{1}{F(0)}>\frac{1}{F(0)}-\frac{1}{F(T_{2})}\nonumber \\{} & {} \quad \ge \frac{1}{c_{2}}\left[ \textrm{arctan}\frac{[c_{1}J(0)-3A(0)]\left( T_{2}+\frac{F(0)}{c_{1}J(0)-3A(0)}\right) }{c_{2}}-\textrm{arctan}\frac{F(0)}{c_{2}}\right] ,\qquad \quad \end{aligned}$$
(3.41)

which can be solved as (3.35). The proof of Corollary 3.1 is finished. \(\square \)

Remark 3.1

The condition (3.34) in Corollary 3.1 is used to ensure that the upper bound in (3.35) is positive. In fact, by using (3.34) and the monotonicity of the function \(\tan x\) in \((0,\frac{\pi }{2})\), we have

$$\begin{aligned}{} & {} \frac{c_{2}}{c_{1}J(0)-3A(0)}\tan \left( \frac{c_{2}}{F(0)}+\textrm{arctan}\frac{F(0)}{c_{2}}\right) >\frac{c_{2}}{c_{1}J(0)-3A(0)}\cdot \frac{F(0)}{c_{2}}\nonumber \\{} & {} \quad =\frac{F(0)}{c_{1}J(0)-3A(0)}. \end{aligned}$$
(3.42)

Remark 3.2

If \(\tau _{1}>0\), we do not know how to treat the integral \(\int _{{\mathbb {R}}^{3}}\textrm{div}{\mathbb {S}}_{1}\cdot {\textbf{u}}dx\), which is crucial to obtain the inequality like (3.18). Perhaps we need to construct some new averaged quantities for the case of \(\tau _{1}>0\).