# On blow up of $\displaystyle C^{1}$ solutions of isentropic Euler system

Shyam Sundar Ghoshal<EMAIL_ADDRESS>Animesh Jana
<EMAIL_ADDRESS>Centre for Applicable Mathematics, Tata Institute
of Fundamental Research, Post Bag No 6503, Sharadanagar, Bangalore - 560065,
India.  CEREMADE (CNRS UMR no. 7534), Université Paris Dauphine, PSL Research
University, Place du Maréchal De Lattre De Tassigny, Paris CEDEX 16, 75775
France.

###### Abstract

In this article, we study the break-down of smooth and continuous solutions to
isentropic Euler system in multi dimension. Sideris [Comm. Math. Phys. 1985]
proved the blow up of smooth solutions when initial data satisfies an
‘integral condition’. We show that a $\displaystyle C^{1}$ solution of
isentropic Euler equation breaks down if (i) gradient of initial velocity has
a negative real eigenvalue at some point $\displaystyle
x_{0}\in\mathbb{R}^{d}$ and (ii) Hessian of initial density satisfies a
smallness condition in Sobolev space. Our proof also works for the data which
fails to satisfy the above-mentioned ‘integral condition’. Furthermore, we
prove the global existence of smooth solution when (i) eigenvalues of gradient
of initial velocity have non-negative real-part and (ii) initial density
satisfies a smallness condition. This extends the global existence result of
[Grassin, Indiana Univ. Math. J. 1998]. Another goal of this article is to
study the breakdown of continuous weak solutions of isentropic Euler
equations. We are able to show that the ‘integral condition’ of Sideris can
cause the breakdown of continuous solutions in finite time. This improves the
blow up result of Sideris from $\displaystyle C^{1}$ to continuous space.

###### keywords:

Euler equations , blow up , Sobolev space , Energy inequality

###### MSC:

35L65 , 35Q31 , 76N10

###### Contents

  1. 1 Introduction
  2. 2 Main results
    1. 2.1 Blow up of continuous solution
    2. 2.2 Blow up of $\displaystyle C^{1}$ solution
    3. 2.3 Global existence of smooth solution
    4. 2.4 Comparison with previous results
  3. 3 Preliminaries
  4. 4 Energy estimates
    1. 4.1 Proof of Proposition 4.1
  5. 5 Blow up of $\displaystyle C^{1}$ solutions
    1. 5.1 Proof of Proposition 2.3
    2. 5.2 Proof of Theorem 2.5
  6. 6 Break down of continuity of a solution
    1. 6.1 Proof of Theorem 2.2
  7. 7 Global existence of smooth solution

##  1 Introduction

In this article, we are interested in regularity and breakdown of smooth
solution to the isentropic Euler equations which reads as follows,

$\displaystyle\displaystyle\partial_{t}\varrho+\operatorname{div}_{x}(\varrho\textbf{u})$
$\displaystyle\displaystyle=0\mbox{ for }x\in\mathbb{R}^{d},t>0,$ (1.1)
$\displaystyle\displaystyle\partial_{t}(\varrho\textbf{u})+\operatorname{div}_{x}(\varrho\textbf{u}\otimes\textbf{u})+\nabla_{x}p(\varrho)$
$\displaystyle\displaystyle=0\mbox{ for }x\in\mathbb{R}^{d},t>0,$ (1.2)

where $\displaystyle\varrho,\textbf{u}$ denote the density and velocity of a
compressible fluid and the function $\displaystyle
p:[0,\infty)\rightarrow[0,\infty)$ represents the pressure. Here we consider
$\displaystyle p(\varrho)=\varrho^{\gamma}$ for $\displaystyle\gamma>1$.

Despite of the importance of PDE (1.1)–(1.2) in the physical world, very
little is known about well-posedness. For sufficiently smooth initial data,
the local-in-time existence of smooth solutions is known due to [9, 30, 40].
In one dimension, the $\displaystyle C^{1}$–norm of solution blows up in
finite time for compact support velocity (see [29, 33, 35, 40]). In his
seminal work [45], Sideris proposed a sufficient condition for multi-dimension
to show that the smooth solution breaks down in finite time. See also [7, 12,
13, 28, 31, 38, 46] for results on blow up and lifespan of smooth solution to
isentropic Euler system and related hyperbolic system of conservation laws. In
this article, we first prove that for a certain restriction on initial data,
the continuity of a solution breaks down in finite time if it also satisfies
the entropy inequality in the sense of distribution,

$\partial_{t}\left[\frac{1}{2}\varrho\left|\textbf{u}\right|^{2}+P(\varrho)\right]+\operatorname{div}_{x}\left[\left(\frac{1}{2}\varrho\left|\textbf{u}\right|^{2}+P(\varrho)+p(\varrho)\right)\textbf{u}\right]\leq
0,\mbox{ where
}P(\varrho)=\varrho\int\limits_{0}^{\varrho}\frac{p(r)}{r^{2}}dr.$ (1.3)

We note that the initial data for which the result of Sideris [45] is valid
also satisfies the assumption of our result. Hence, for the initial data
considered in [45], the corresponding solution not only loses its
$\displaystyle C^{1}$ regularity in finite time, it also becomes
discontinuous.

Recently, there has been a growing interest in existence of continuous
solutions for fluid flow equations (see [15, 23, 27]). For isentropic Euler
equation, it has been shown [23] that there are infinitely many Hölder
continuous weak solutions from a specific initial data. It is then natural to
ask whether these solutions are globally defined or not. We show that if a
continuous solution satisfies entropy inequality (1.3) and an ‘integral
condition’ (as in (2.4) below), then their continuity breaks down in finite
time.

On the other hand, global smooth solution exists [33] in 1-D for a class of
initial data. Chemin [9] proved the global existence of smooth solution in
multi-D when initial density has finite second spatial moment. Later, Serre
[43] showed that global-in-time solution exists when the initial density and
velocity, $\displaystyle(\varrho_{0},\textbf{u}_{0})$ are sufficiently close
to $\displaystyle(0,Ax)$ in an appropriate Sobolev norm and $\displaystyle A$
is a positive definite constant matrix. The theory of global existence for
isentropic Euler system has been extended by Grassin [25] for a larger class
of initial data. In the article [25], it has been shown that smooth solution
exists for all time $\displaystyle t>0$ if (i) the initial density is
sufficiently small in Sobolev norm and (ii) eigenvalues of
$\displaystyle\nabla_{x}\textbf{u}_{0}$ belong to a compact set of
$\displaystyle\mathbb{C}\setminus(-\infty,0)$. In this article, we establish a
relation between the eigenvalues of $\displaystyle\nabla_{x}\textbf{u}_{0}$
and the global existence of smooth solution. We consider the background
velocity $\displaystyle\overline{\textbf{u}}$ solving the following Cauchy
problem

$\left.\begin{array}[]{rl}\partial_{t}\overline{\textbf{u}}+\overline{\textbf{u}}\cdot\nabla_{x}\overline{\textbf{u}}&=0\,\,\,\,\mbox{
for }x\in\mathbb{R}^{d},t>0,\\\
\overline{\textbf{u}}(0,x)&=\overline{\textbf{u}}_{0}\mbox{ for
}x\in\mathbb{R}^{d}.\end{array}\right\\}$ (1.4)

For the vectorial Burgers equation (1.4), global smooth solution exists if and
only if the initial data $\displaystyle\overline{\textbf{u}}_{0}$ eigenvalues
of $\displaystyle\nabla_{x}\overline{\textbf{u}}_{0}$ belong to a compact set
of $\displaystyle\mathbb{C}\setminus(-\infty,0)$, equivalently, we have finite
time blow up of $\displaystyle C^{1}$ solutions of (1.4) if and only if
$\displaystyle\nabla_{x}\overline{\textbf{u}}_{0}$ has at least one negative
real eigenvalue. In this article, we want to show an analogous result for
isentropic Euler system (1.1)–(1.2). We first prove the finite time blow up of
smooth solution when $\displaystyle\nabla_{x}{\textbf{u}}_{0}(x_{0})$ has at
least one negative real eigenvalue for some $\displaystyle
x_{0}\in\mathbb{R}^{d}$. In the spirit of [25], we also show that global
smooth solution exists for
$\displaystyle\inf\limits_{x\in\mathbb{R}^{d}}\min\limits_{\xi\in\mathbb{S}^{d-1}}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi\geq
0$ where $\displaystyle\mathbb{S}^{d-1}$ is the unit sphere in
$\displaystyle\mathbb{R}^{d}$ and initial density satisfies a smallness
condition. Combining these two results, we conclude the characterization of
initial velocity providing global smooth solutions.

Though our main concern of this article is to find the optimal
characterization of initial data causing blow up of $\displaystyle C^{1}$ or
continuous solution, it is also an interesting subject to study the
description and stability of the shock and how the solution behaves after the
blow up time. Majda [38, 39] studied the existence and stability of shock for
multi-D system of conservation laws. Description of shock formation has been
studied in [12, 37]. We also refer recent results [7] in this direction. For
spherically symmetric smooth initial data, the blow up of $\displaystyle
L^{\infty}$-norm of density and velocity has been shown in [41].

In a forthcoming paper [22], we obtain analogous results for compressible
Navier-Stokes equation with low regularity.

##  2 Main results

###  2.1 Blow up of continuous solution

Before we state our result on break-down of continuous solution we provide the
definition of weak solution.

######  Definition 2.1.

Let $\displaystyle T^{*}>0$. We say $\displaystyle(\varrho,\textbf{u})\in
C([0,T^{*});L^{1}_{loc}(\mathbb{R}^{d}))\cap
L^{\infty}([0,T^{*})\times\mathbb{R}^{d})$ is a weak solution to (1.1)–(1.2)
if the following holds.

  * •

For all $\displaystyle 0\leq t_{1}<t_{2}<T^{*}$ and $\displaystyle\varphi\in
C^{1}_{c}([t_{1},t_{2}]\times\mathbb{R}^{d})$,

$\left[\int\limits_{\mathbb{R}^{d}}\varrho(t,\cdot)\varphi(t,\cdot)\,dx\right]_{t=t_{1}}^{t=t_{2}}=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\left[\varrho\partial_{t}\varphi+\varrho\textbf{u}\cdot\nabla_{x}\varphi\right]\,dxdt.$
(2.1)

  * •

For all $\displaystyle 0\leq t_{1}<t_{2}<T^{*}$ and $\displaystyle\psi\in
C^{1}_{c}([t_{1},t_{2}]\times\mathbb{R}^{d};\mathbb{R}^{d})$,

$\left[\int\limits_{\mathbb{R}^{d}}\varrho\textbf{u}(t,\cdot)\cdot\psi(t,\cdot)\,dx\right]_{t=t_{1}}^{t=t_{2}}=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\left[\varrho\textbf{u}\cdot\partial_{t}\psi+\varrho\textbf{u}\otimes\textbf{u}:\nabla_{x}\psi+p(\varrho)\operatorname{div}_{x}\psi\right]\,dxdt.$
(2.2)

A weak solution $\displaystyle(\varrho,\textbf{u})$ will be called as
admissible solution if it satisfies the following inequality

$\displaystyle\displaystyle\int\limits_{0}^{T^{*}}\int\limits_{\mathbb{R}^{d}}\left[\left(\frac{\varrho\left|\textbf{u}\right|^{2}}{2}+P(\varrho)\right)\partial_{t}\varphi+\left(\frac{\varrho\left|\textbf{u}\right|^{2}}{2}+P(\varrho)+p(\varrho)\right)\textbf{u}\cdot\nabla_{x}\varphi\right]dxdt$
$\displaystyle\displaystyle+\int\limits_{\mathbb{R}^{d}}\left(\frac{\varrho_{0}\left|\textbf{u}_{0}\right|^{2}}{2}+P(\varrho_{0})\right)\varphi(0,\cdot)\,dx\geq
0,$ (2.3)

for all $\displaystyle 0\leq\varphi\in
C^{1}_{c}([0,T^{*})\times\mathbb{R}^{d})$ where $\displaystyle
P(\varrho)=\frac{1}{\gamma-1}\varrho^{\gamma}$.

The well-posedness of admissible solution to isentropic Euler system is
largely open. In one dimension the well-posedness for admissible weak solution
is known in BV setting [3, 4, 24]. But, in multi-D even for Riemann type
planar initial data it has been shown [10] that infinitely many $\displaystyle
L^{\infty}$ admissible weak solutions exist. In three dimension, infinitely
many continuous admissible weak solution has been constructed [23] so that
they satisfy same initial data. On the other hand, uniqueness has been shown
[17] when solution has sufficient Hölder regularity and the velocity gradient
satisfies a one-sided bound condition. See also [19, 20, 21] for recent
results on uniqueness of Hölder continuous solutions for compressible Euler
system and relation hyperbolic conservation laws.

To study the breakdown of continuity of weak solutions, we impose the
following condition on initial data as in [45],

$\frac{1}{\omega_{d}R^{d+1}}\int\limits_{\mathbb{R}^{d}}\varrho_{0}\textbf{u}_{0}\cdot
x\,dx\geq(d+1)\sigma\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})},$
(2.4)

where
$\displaystyle\sigma=\sqrt{\gamma}(\overline{\varrho})^{\frac{\gamma-1}{2}}$
and $\displaystyle\omega_{d}$ denotes the area of $\displaystyle d$
dimensional sphere. Furthermore, we additionally assume that
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ satisfies

$(\varrho_{0}(x),\textbf{u}_{0}(x))=(\overline{\varrho},\textbf{0})\mbox{ for
}\left|x\right|\geq R\mbox{ and
}\int\limits_{\mathbb{R}^{d}}(\varrho_{0}-\overline{\varrho})\,dx\geq 0,$
(2.5)

for some $\displaystyle\overline{\varrho}>0$ and $\displaystyle R>0$. We show
that a continuous admissible solution breaks down in finite time if the
initial data satisfies above two conditions (2.4) and (2.5). More precisely,
we prove the following result.

######  Theorem 2.2.

Let $\displaystyle(\varrho,\textbf{u})\in C([0,T^{*})\times\mathbb{R}^{d})$ be
an admissible solution to (1.1)–(1.2) with initial data
$\displaystyle(\varrho_{0},\textbf{u}_{0})$according to Definition 2.1. We
assume that the initial data $\displaystyle(\varrho_{0},\textbf{u}_{0})$
satisfies (2.4) and (2.5). Then, $\displaystyle T^{*}$ is finite.

Sideris [45] proved that $\displaystyle C^{1}$ solutions break down in finite
time if the initial data satisfies the conditions (2.4) and (2.5). With a
similar condition, blow up of classical solution has been shown [49] Navier-
Stokes. We refer to [48, 50] and references therein for more results on break-
down of $\displaystyle C^{1}$ solutions with a condition like (2.4).

We note that the entropy condition (1.3) and the continuity of solution plays
key role to prove that characteristics generated from
$\displaystyle\\{\left|x\right|\leq R\\}$ are confined inside the cone
$\displaystyle\\{\left|x\right|\leq R+\sigma t\\}$. For $\displaystyle C^{1}$
solutions we do not additionally need to assume (1.3) as it can be shown by
using (1.1)–(1.2) (see [14]) that equality holds in (1.3) for $\displaystyle
C^{1}$ solutions. For isentropic Euler equations, equality in (1.3) can still
hold even for weak solutions which may not be $\displaystyle C^{1}$ (see [2,
18, 26]). Furthermore, we would like to mention a recent result [23] on
existence of infinitely many continuous admissible solutions of isentropic
Euler equations. For some specific choice of continuous initial data, this
[23] gives an existential results for (1.1)–(1.2). We wish to remark that due
to our result in Theorem 2.2 the solutions constructed in [23] can not be
extended globally (in time) if the initial data satisfies the integral
condition (2.4).

###  2.2 Blow up of $\displaystyle C^{1}$ solution

In the view of blow up result of Sideris [45], it is natural to ask whether
global $\displaystyle C^{1}$ solution exists if a $\displaystyle C^{1}$
initial data does not satisfy the condition (2.4). We answer to this question
in our next result. To prove a finite time blow up of $\displaystyle C^{1}$
solution, we impose the following conditions on the initial data.

(ND)

Negative definiteness of velocity gradient: There exists
$\displaystyle\lambda_{max}>0$ and $\displaystyle x_{0}\in\mathbb{R}^{d}$ such
that

$\nabla_{x}\textbf{u}_{0}(x_{0})\mbox{ is symmetric and
}\nabla_{x}\textbf{u}_{0}(x_{0})\xi_{0}=-\lambda_{max}\xi_{0}\mbox{ for some
}\xi_{0}\in\mathbb{S}^{d-1}.$ (2.6)

We now state our result on blow up of $\displaystyle C^{1}$ solutions of
isentropic Euler system (1.1)–(1.2).

######  Proposition 2.3.

Let $\displaystyle(\varrho,\textbf{u})\in
C^{1}([0,T^{*}]\times\mathbb{R}^{d})$ be a solution of isentropic Euler system
(1.1)–(1.2) corresponding to the initial data
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ satisfying  (ND) and
$\displaystyle\textbf{u}_{0}(x_{0}+r\xi_{0})-\textbf{u}_{0}(x_{0})=-\lambda_{0}r\xi_{0}$
for some $\displaystyle r>0,\lambda_{0}>0,\xi_{0}\in\mathbb{S}^{d-1}$ and
$\displaystyle x_{0}\in\mathbb{R}^{d}$. Define $\displaystyle
M:=\left|\left|\nabla_{x}\textbf{u}\right|\right|_{L^{\infty}([0,T^{*}]\times\mathbb{R}^{d})}$.
If $\displaystyle\varrho,\textbf{u}$ and $\displaystyle\lambda_{0}$ satisfy
the following estimate,

$\left|\lambda_{max}-\lambda_{0}\right|+\left|\left|\pi(\cdot)\nabla_{x}\pi(\cdot)\right|\right|_{L^{\infty}([0,1/\lambda_{max}]\times\mathbb{R}^{d})}\leq\epsilon_{0}:=\frac{\lambda_{0}r}{2}\left[\frac{rM}{\lambda_{max}}+\frac{2e^{\frac{M}{\lambda_{max}}}}{M}\right]^{-1},$
(2.7)

where
$\displaystyle\pi=\sqrt{\frac{\gamma-1}{4\gamma}}\varrho^{\frac{\gamma-1}{2}}$,
then we have $\displaystyle T^{*}\leq 1/\lambda_{max}$.

######  Remark 2.4.

  1. 1.

We would like to mention that in order to prove the blow up result, the
smallness assumption on $\displaystyle\pi\nabla_{x}\pi$ as in (2.7) is not
essentially required on whole $\displaystyle\mathbb{R}^{d}$. One can possibly
replace it by a locally smallness assumption on the domain of dependence.

  2. 2.

We note that the condition on $\displaystyle\textbf{u}_{0}$ of having a
direction such that
$\displaystyle\textbf{u}_{0}(x_{0}+r\xi_{0})-\textbf{u}_{0}(x_{0})=-\lambda_{0}r\xi_{0}$
holds can be replaced by assuming that
$\displaystyle\nabla_{x}\textbf{u}_{0}(x)$ has $\displaystyle d$ real
eigenvalues in a neighborhood of $\displaystyle x_{0}\in\mathbb{R}^{d}$. When
$\displaystyle\nabla_{x}\textbf{u}_{0}$ has $\displaystyle d$ real eigenvalues
then it is guaranteed that there exists a direction such that
$\displaystyle\textbf{u}_{0}(x_{0}+r\xi_{0})-\textbf{u}_{0}(x_{0})=-\lambda_{0}r\xi_{0}$
holds. This can be shown in a similar manner as in Lax theory for solving
Riemann problem of $\displaystyle n\times n$ hyperbolic system in 1-D (see [5,
32]).

Since the smallness condition (2.7) in Proposition 2.3 is given on the
solution, it does not directly serve our purpose to classify the initial data
causing finite time blow up of $\displaystyle C^{1}$. We want to replace (2.7)
by considering smallness on $\displaystyle H^{m}$-norm of
$\displaystyle\nabla^{2}_{x}\varrho^{\frac{\gamma-1}{2}}_{0}$ and
$\displaystyle\nabla^{2}_{x}\textbf{u}_{0}$ and the blow up result holds true.

######  Theorem 2.5.

Let $\displaystyle(\varrho,\textbf{u})\in
C^{1}([0,T^{*})\times\mathbb{R}^{d})$ be a solution of isentropic Euler system
(1.1)–(1.2) corresponding to the initial data
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ satisfying

$\left(\left|\left|\nabla^{2}_{x}\varrho^{\frac{\gamma-1}{2}}_{0}\right|\right|_{H^{m}(\mathbb{R}^{d})}+\left|\left|\nabla^{2}_{x}\textbf{u}_{0}\right|\right|_{H^{m}(\mathbb{R}^{d})}\right)\cdot\left|\left|\varrho^{\frac{\gamma-1}{2}}_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\leq\epsilon_{0}\mbox{
for some integer }m>1+\frac{d}{2},\epsilon_{0}>0.$ (2.8)

We assume that $\displaystyle\textbf{u}_{0}$ satisfies  (ND) for some
$\displaystyle\lambda_{max}>0$. Then, $\displaystyle T^{*}$ is finite provided
$\displaystyle\epsilon_{0}<\frac{\lambda^{2}_{max}}{5(\gamma-1)}$.

######  Remark 2.6.

  1. 1.

We remark that there exists initial data
$\displaystyle\varrho_{0},\textbf{u}_{0}$ such that it satisfies  (ND) and
(2.8) but it fails to satisfy (2.4). See Example 5.2 in section 5.2.
Therefore, for the $\displaystyle C^{1}$ solutions arsing from initial data as
in Example 5.2, we can not invoke blow up result of Sideris [45] but Theorem
2.3 is still applicable as they satisfies (2.8).

  2. 2.

Similar to (1) of Remark 2.4, we again mention that one can obtain the blow up
result of Theorem 2.5 by using a local smallness condition with a proper use
of finite speed of propagation.

We prove Theorem 2.5 in a similar technique as Proposition 2.3 with the help
of an energy estimate of Kato [30] for $\displaystyle C^{1}$ solutions.

###  2.3 Global existence of smooth solution

We remark that the result of Theorem 2.5 complements the global existence
theory developed in [9, 25, 43]. We recall that if initial data satisfies (i)
$\displaystyle\min\limits_{\xi\in\mathbb{S}^{d-1}}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi\geq
c>0$ for all $\displaystyle x\in\mathbb{R}^{d}$ and (ii)
$\displaystyle\pi_{0}$ is sufficiently small then the global smooth solution
exists (see [25]). As Theorem 2.5 rules out the possibility of global solution
when velocity gradient $\displaystyle\nabla_{x}\textbf{u}_{0}$ has at least
one negative real eigenvalue, now it is natural to ask whether global solution
when $\displaystyle\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi\geq 0$ for all
$\displaystyle\xi\in\mathbb{S}^{d-1}$. We answer this question in our next
result.

######  Proposition 2.7.

Let $\displaystyle m>1+\frac{d}{2}$ and
$\displaystyle\varrho_{0},\textbf{u}_{0}$ be satisfying

(G-1)

$\displaystyle\nabla_{x}^{2}\textbf{u}_{0}\in H^{m-1}(\mathbb{R}^{d})$ and
$\displaystyle\nabla_{x}\textbf{u}_{0}\in L^{\infty}(\mathbb{R}^{d})$.

(G-2)

$\displaystyle\inf\left\\{\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi;\,\xi\in\mathbb{S}^{d-1},x\in\mathbb{R}^{d}\right\\}\geq
0$ where
$\displaystyle\mathbb{S}^{d-1}=\\{x\in\mathbb{R}^{d};\,\left|x\right|=1\\}$.

(G-3)

$\displaystyle\varrho_{0}$ has compact support and furthermore,

$\mbox{supp}(\varrho_{0})\subset\left\\{x\in\mathbb{R}^{d};\,\inf\limits_{\xi\in\mathbb{R}^{d},\left|\xi\right|=1}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi\geq\alpha\right\\}\mbox{
for some }\alpha>0.$ (2.9)

Then there exists $\displaystyle\epsilon_{0}>0$ such that the following holds:
for
$\displaystyle\left|\left|\varrho^{\frac{\gamma-1}{2}}\right|\right|_{H^{m}(\mathbb{R}^{d})}<\epsilon_{0}$,
there exists a global smooth solution $\displaystyle(\varrho,\textbf{u})$ of
(1.1)–(1.2) corresponding to the initial data
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ and it satisfies,

$\left(\varrho^{\frac{\gamma-1}{2}},\textbf{u}-\textbf{v}\right)\in
C^{1}([0,\infty);H^{m-1}((\mathbb{R}^{d}))\cap
C([0,\infty);H^{m}(\mathbb{R}^{d})),$ (2.10)

where v is the global solution to (1.4) with initial data
$\displaystyle\textbf{u}_{0}$.

######  Remark 2.8.

We note that the hypothesis  (G-3) restricts support of density
$\displaystyle\varrho$ in a set where
$\displaystyle\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi$ is strictly away from
$\displaystyle 0$ for all $\displaystyle\xi\in\mathbb{S}^{d-1}$. Therefore, if
$\displaystyle\nabla_{x}u_{0}\equiv\textbf{0}_{d}$ or
$\displaystyle\nabla_{x}u_{0}(x)$ is anti-symmetric matrix for all
$\displaystyle x\in\mathbb{R}^{d}$, then the hypothesis  (G-3) forces
$\displaystyle\varrho_{0}\equiv 0$ and in this case the conclusion of
Proposition 2.7 holds true trivially.

###  2.4 Comparison with previous results

As we mentioned earlier Grassin [25] proved the global existence of smooth
solutions if the initial data satisfies (i)
$\displaystyle\inf\left\\{\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi;\,\xi\in\mathbb{S}^{d-1},x\in\mathbb{R}^{d}\right\\}\geq
c_{0}$ for some $\displaystyle c_{0}>0$ and (ii)
$\displaystyle\varrho_{0}^{\frac{\gamma-1}{2}}$ is small in Sobolev space
$\displaystyle H^{m}$ for some $\displaystyle m>1+\frac{d}{2}$. When (i)
holds, global solution $\displaystyle\overline{\textbf{u}}$ exists for the
Cauchy problem (1.4). The proof is based on the energy estimate on
$\displaystyle\textbf{u}-\overline{\textbf{u}}$ and
$\displaystyle\varrho^{\frac{\gamma-1}{2}}$. This extends the previous results
of Serre [43] and Chemin [9]. We also refer to [42] where author studied the
smooth solution in negative time as well for finite mass and the solution is
surrounded by vacuum. Global existence smooth solutions have been shown in
[44] when initial velocity is close to a linear function in
$\displaystyle\mathbb{R}^{d}$. In all these articles, it has been crucially
used the strict positiveness of
$\displaystyle\inf\limits_{\xi\in\mathbb{S}^{d-1},x\in\mathbb{R}^{d}}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi$.
In Theorem 2.7, we relax this positiveness by non-negativeness. We could do
this by restricting the support of initial density in the set when
$\displaystyle\inf\limits_{\xi\in\mathbb{S}^{d-1}}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi>0$.

Sideris [45] showed the finite time breakdown of smooth solution when initial
data satisfies the integral condition (2.4). As in Theorem 2.2 we show that
finite time blow up happens even if we replace the $\displaystyle C^{1}$
solution by a continuous one. Note that when we talk about continuous solution
it is no longer satisfying the equation in the classical way. We consider a
weak solution which is continuous and satisfying the entropy inequality (1.3).
Theorem 2.2 not only extends the result of Sideris [45] for continuous entropy
solutions, it also tells us that the $\displaystyle C^{1}$ solution considered
in [45] loses its smoothness in finite time and besides, it fails to be
continuous as well. Our result also tells that the continuous solutions
constructed in [23] can not be global if it the data satisfies (2.4).

In the spirit of Sideris [45], blow up phenomenon holds for compressible
Navier-Stokes [49] equations and other hyperbolic systems [46] as well. When
the solution is containing vacuum then blow up of $\displaystyle C^{3}$
solution has been proved in [8] for irrotational and compressive initial
velocity. It has been shown [47] that for class of data
$\displaystyle\varrho_{0}=\overline{\varrho}+\epsilon\tilde{\varrho}_{0},\textbf{u}_{0}=\epsilon\tilde{\textbf{u}}_{0}$
with small enough $\displaystyle\epsilon>0$ it has been shown the life span of
the smooth solution $\displaystyle T^{*}>C/\epsilon$ for a given
$\displaystyle C>0$.

In [7], authors have shown the shock formation for smooth solution to
(1.1)–(1.2) in dimension two at time of order $\displaystyle\epsilon$ when
initial data is bounded but it has slope $\displaystyle-1/\epsilon$ for some
small enough $\displaystyle\epsilon>0$. They have also shown the stability of
the shock formation. In literature, Majda [39, 38] started the study of
stability of shock formation. Christodoulou [12] has proved the shock
formation for irrotational initial velocity for relativistic Euler equations.
We refer to [11] for the non-relativistic case. For 2-D isentropic Euler
equations, shock formation has been studied in [37] for velocity with
vorticity. We note that Theorem 2.5 and [7] can not be directly compared as
(ND) put minimal restriction on initial velocity but Theorem 2.5 requires the
smallness assumption (2.8) whereas the results of [7] need large enough
negative slope but there is no restriction of type (2.8).

Rest of the article is organized as follows. In the next section, we recall
some basic results and notations. We establish energy estimates in section 4.
We prove blow up of $\displaystyle C^{1}$ and continuous solutions in section
5 and 6 respectively. In the last section, we prove the result of global-in-
time existence.

##  3 Preliminaries

In this article, we use the following notations.

  * •

The notation $\displaystyle\partial_{\alpha}^{n}$ stands for
$\displaystyle\frac{d^{n}}{dx^{\alpha_{1}}_{1}dx^{\alpha_{2}}_{2}\cdots
dx^{\alpha_{d}}_{d}}$ where
$\displaystyle\alpha=(\alpha_{1},\cdots,\alpha_{d})$ and
$\displaystyle\alpha_{1}+\alpha_{2}+\dots+\alpha_{d}=n$.

  * •

For multi index $\displaystyle\alpha\in\mathbb{N}^{n}$,
$\displaystyle\alpha=(\alpha_{1},\cdots,\alpha_{n})$, we use
$\displaystyle\left|\alpha\right|=\alpha_{1}+\cdots+\alpha_{n}$.

  * •

For
$\displaystyle\alpha=(\alpha_{1},\cdots_{n}),\beta=(\beta_{1},\cdots,\beta_{1}),\tilde{\beta}=(\tilde{\beta}_{1},\cdots,\tilde{\beta}_{1})\in\mathbb{N}^{n}$,
we denote $\displaystyle c(\beta,\tilde{\beta};\alpha)$ as follows

$c(\beta,\tilde{\beta};\alpha)=\frac{\alpha_{1}!\alpha_{2}!\cdots\alpha_{n}!}{\beta_{1}!\beta_{2}!\cdots\beta_{n}!\tilde{\beta}_{1}!\tilde{\beta}_{2}!\cdots\tilde{\beta}_{n}!}\mbox{
where }\alpha_{i}=\beta_{i}+\tilde{\beta}_{i}\mbox{ for }1\leq i\leq n.$ (3.1)

  * •

We often use the following inequality which we refer as
“$\displaystyle\epsilon$-inequality”

$ab\leq\epsilon a^{2}+\frac{1}{4\epsilon}b^{2}\mbox{ for
}a,b\in\mathbb{R}\mbox{ and }\epsilon>0.$ (3.2)

  * •

Let $\displaystyle p>1,q>1$ and $\displaystyle a,b\geq 0$, Then the following
inequality holds,

$ab\leq\frac{a^{p}}{p}+\frac{b^{q}}{q}\quad\mbox{ if $\displaystyle p,q$
satisfy }\frac{1}{p}+\frac{1}{q}=1.$ (3.3)

This is known as ‘Young’s inequality’.

  * •

By $\displaystyle W^{m,p}(\mathbb{R}^{d})$ we denote the standard Sobolev
spaces for $\displaystyle m\in\mathbb{N}$ and $\displaystyle p\geq 1$. We use
the notation $\displaystyle H^{m}=W^{m,2}$.

In the following lemma, we recall the equivalence of various norms in finite
dimension.

######  Lemma 3.1.

Let $\displaystyle m\geq 1,n\geq 1$ and $\displaystyle d\geq 1$. There exists
$\displaystyle C_{m,n,d}>1$ such that the following holds,

$\frac{1}{C_{m,n,d}}\sum\limits_{k=1}^{d}\left|x_{k}\right|^{n}\leq\left(\sum\limits_{k=1}^{d}\left|x_{k}\right|^{m}\right)^{\frac{n}{m}}\leq
C_{m,n,d}\sum\limits_{k=1}^{d}\left|x_{k}\right|^{n}\mbox{ for all
}x=(x_{1},\cdots,x_{d})\in\mathbb{R}^{d}.$ (3.4)

For proof of Lemma 3.1 we refer to [34]. Next, we recall a classical result on
embedding of Sobolev spaces.

######  Lemma 3.2.

[Morrey] Let $\displaystyle q>N$. Then we have the following continuous
injection $\displaystyle W^{1,q}(\mathbb{R}^{N})\subset
L^{\infty}(\mathbb{R}^{N})$. Furthermore, for all $\displaystyle g\in
W^{1,q}(\mathbb{R}^{N})$,

$\left|g(z_{1})-g(z_{2})\right|\leq
C\left|z_{1}-z_{2}\right|^{\alpha}\left|\left|\nabla_{x}g\right|\right|_{L^{q}(\mathbb{R}^{N})}\mbox{
for a.e. }z_{1},z_{2}\in\mathbb{R}^{N},$ (3.5)

where $\displaystyle C=C(q,N)$, $\displaystyle\alpha=1-\frac{1}{q}$ and it
does not depend on $\displaystyle g$.

For proof of Lemma 3.2 we refer to [6, Theorem 9.12].

##  4 Energy estimates

The blow up of $\displaystyle C^{1}$ solution for a class of initial condition
relies on an energy estimate in the spirit of [30]. Here, we reproduce the
regularity result of Kato [30] for integer exponents.

######  Proposition 4.1.

Let $\displaystyle(\varrho,\textbf{u})\in
C^{1}([0,T^{*})\times\mathbb{R}^{d})$ be a solution of isentropic Euler system
(1.1)–(1.2) corresponding to the initial data
$\displaystyle(\varrho_{0},\textbf{u}_{0})$. We further assume that
$\displaystyle(\nabla_{x}^{k}\varrho_{0}^{\frac{\gamma-1}{2}},\nabla_{x}^{k}\textbf{u}_{0})\in
H^{m}(\mathbb{R}^{d})$ for some $\displaystyle m\in\mathbb{N}$ and
$\displaystyle k\in\mathbb{N}\cup\\{0\\}$. Then, we have

$(\nabla_{x}^{k}\varrho^{\frac{\gamma-1}{2}}(t,\cdot),\nabla_{x}^{k}\textbf{u}(t,\cdot))\in
H^{m}(\mathbb{R}^{d})\mbox{ for all }t\in[0,T^{*}).$ (4.1)

We note that the exponent of regularity in Proposition 4.1 are considered to
be positive integers. We expect that it can be extended to the positive real
numbers in a similar spirit as in [30].

The proof of Proposition 4.1 can be done in a similar manner as in [30]. For
sake of completeness, we provide here some explicit estimates which will help
us in section 5. To prove the energy estimates we need the following results
on interpolation of $\displaystyle W^{1,\infty}$ and $\displaystyle W^{3,2}$.
We refer to [1] for more on interpolation of spaces.

######  Lemma 4.2.

Let $\displaystyle\psi\in C^{3}_{c}(\mathbb{R}^{d})$. Then the following
holds,

$\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{2}\psi\right|^{4}\,dx\leq
C_{d}\left|\left|\nabla_{x}\psi\right|\right|^{2}_{L^{\infty}(\mathbb{R}^{d})}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\psi\right|^{2}\,dx.$
(4.2)

###### Proof.

Throughout the proof we use $\displaystyle C_{d}$ as a generic constant
depending only on dimension $\displaystyle d$. By using integration by parts,
we can see that

$\displaystyle\displaystyle\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{4}\,dx$
$\displaystyle\displaystyle=\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\partial^{2}_{ij}\psi(\partial^{2}_{ij}\psi)^{3}\,dx$
$\displaystyle\displaystyle=-C_{d}\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\partial_{i}\psi(3\partial^{2}_{ij}\psi)^{2}\partial^{3}_{ijj}\psi\,dx.$

Applying $\displaystyle\epsilon$-inequality (3.2) with
$\displaystyle\epsilon=1/2$ we obtain,

$\displaystyle\displaystyle\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{4}\,dx$
$\displaystyle\displaystyle\leq\frac{1}{2}\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{4}\,dx+C_{d}\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\left|\partial_{i}\psi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}^{2}\left|\partial^{3}_{ijj}\psi\right|^{2}\,dx$
$\displaystyle\displaystyle\leq\frac{1}{2}\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{4}\,dx+C_{d}\left|\left|\nabla_{x}\psi\right|\right|^{2}_{L^{\infty}(\mathbb{R}^{d})}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\psi\right|^{2}\,dx.$

Hence, we have

$\sum\limits_{i,j=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{4}\,dx\leq
C_{d}\left|\left|\nabla_{x}\psi\right|\right|^{2}_{L^{\infty}(\mathbb{R}^{d})}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\psi\right|^{2}\,dx.$

By using the inequality (3.4) we conclude the result (4.2). ∎

In the next result, we estimate
$\displaystyle\left|\left|\nabla_{x}^{2}\phi\nabla_{x}^{3}\psi\right|\right|_{L^{2}}$
in terms of
$\displaystyle\left|\left|\nabla_{x}^{4}\psi\right|\right|_{L^{2}},\left|\left|\nabla_{x}^{4}\phi\right|\right|_{L^{2}}$
and the $\displaystyle W^{1,\infty}$ norm of $\displaystyle\psi,\phi$. It
plays an important role in derivation of energy estimates.

######  Lemma 4.3.

Let $\displaystyle\phi,\psi\in C^{4}_{c}(\mathbb{R}^{d}),\,d\geq 1$. Then the
following holds,

$\displaystyle\displaystyle\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{2}\psi\right|^{2}\left|\nabla_{x}^{3}\psi\right|^{2}+\left|\nabla_{x}^{2}\phi\right|^{2}\left|\nabla_{x}^{3}\phi\right|^{2}\right]\,dx+\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{2}\psi\right|^{2}\left|\nabla_{x}^{3}\phi\right|^{2}+\left|\nabla_{x}^{2}\phi\right|^{2}\left|\nabla_{x}^{3}\psi\right|^{2}\right]\,dx$
$\displaystyle\displaystyle\leq
C_{d}\left(1+\left|\left|\nabla_{x}\phi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\psi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)^{3}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{4}\phi\right|^{2}+\left|\nabla_{x}^{4}\psi\right|^{2}\,dx.$
(4.3)

###### Proof.

Applying integration by part, we get

$\displaystyle\displaystyle\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{2}\left|\partial^{3}_{klr}\phi\right|^{2}\,dx$
$\displaystyle\displaystyle=-\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\partial_{j}\psi\partial^{3}_{iij}\psi\left|\partial^{3}_{klr}\phi\right|^{2}\,dx$
$\displaystyle\displaystyle-2\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\partial_{j}\psi\partial^{2}_{ij}\psi\partial^{3}_{klr}\phi\partial^{4}_{iklr}\phi\,dx.$
(4.4)

By using the relation between finite dimensional norms (3.4), we see

$\frac{1}{C_{d}}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{3/2}\,dx\leq\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\psi\right|^{3}\,dx\leq{C_{d}}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{3/2}\,dx,$
(4.5)

where $\displaystyle C_{d}$ is a generic constant depending only on dimension
$\displaystyle d$. With the help of (4.5), Young’s inequality (3.3) and
$\displaystyle\epsilon$-inequality, from (4.4), we obtain

$\displaystyle\displaystyle\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{2}\left|\partial^{3}_{klr}\phi\right|^{2}\,dx\leq$
$\displaystyle\displaystyle
C_{d}\left(1+\left|\left|\nabla_{x}\psi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)^{2}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{3}\psi\right|^{3}+\left|\nabla_{x}^{3}\phi\right|^{3}\right]\,dx$
$\displaystyle\displaystyle+C_{d}\left(1+\left|\left|\nabla_{x}\psi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)^{2}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{4}\psi\right|^{2}\,dx.$
(4.6)

Using integration by parts, we have,

$\displaystyle\displaystyle\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{3/2}\,dx$
$\displaystyle\displaystyle=\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{1/2}(\partial^{3}_{i^{\prime}j^{\prime}k^{\prime}}\psi)^{2}\,dx$
$\displaystyle\displaystyle=-\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{1/2}\partial^{2}_{j^{\prime}k^{\prime}}\psi\partial^{4}_{i^{\prime}i^{\prime}j^{\prime}k^{\prime}}\psi\,dx$
$\displaystyle\displaystyle-\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\sum\limits_{i,j,k=1}^{d}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{-1/2}\partial^{2}_{j^{\prime}k^{\prime}}\psi\partial^{3}_{i^{\prime}j^{\prime}k^{\prime}}\psi\partial^{4}_{i^{\prime}ijk}\psi\partial^{3}_{ijk}\psi\,dx.$
(4.7)

Now, with the help of $\displaystyle\epsilon$-inequality (3.2), we get

$\displaystyle\displaystyle\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{3/2}\,dx$
$\displaystyle\displaystyle\leq\epsilon\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\sum\limits_{i,j,k=1}^{d}\int\limits_{\mathbb{R}^{d}}(\partial^{3}_{ijk}\psi)^{2}(\partial^{2}_{j^{\prime}k^{\prime}}\psi)^{2}\,dx+C_{\epsilon}\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{4}_{i^{\prime}i^{\prime}j^{\prime}k^{\prime}}\psi\right|^{2}\,dx$
$\displaystyle\displaystyle+C_{d}\sum\limits_{i^{\prime},j^{\prime},k^{\prime},i=1}^{d}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{1/2}\left|\partial^{2}_{j^{\prime}k^{\prime}}\psi\right|\left|\partial^{4}_{i^{\prime}ijk}\psi\right|\,dx.$
(4.8)

Again by similar argument, we have

$\displaystyle\displaystyle\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)^{3/2}\,dx$
$\displaystyle\displaystyle\leq\epsilon\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\sum\limits_{i,j,k=1}^{d}\int\limits_{\mathbb{R}^{d}}(\partial^{3}_{ijk}\psi)^{2}(\partial^{2}_{j^{\prime}k^{\prime}}\psi)^{2}\,dx+C_{\epsilon}\sum\limits_{i^{\prime},j^{\prime},k^{\prime}=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{4}_{i^{\prime}i^{\prime}j^{\prime}k^{\prime}}\psi\right|^{2}\,dx$
$\displaystyle\displaystyle+\epsilon
C_{d}\int\limits_{\mathbb{R}^{d}}\left(\sum\limits_{i,j,k=1}^{d}(\partial^{3}_{ijk}\psi)^{2}\right)\left(\sum\limits_{i,j=1}^{d}(\partial^{2}_{ij}\psi)^{2}\right)\,dx+C_{\epsilon,d}\int\limits_{\mathbb{R}^{d}}\left|\nabla^{4}_{x}\psi\right|\,dx,$

or equivalently,

$\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\psi\right|^{3}\,dx\leq\epsilon
C_{d}\int\limits_{\mathbb{R}^{d}}\left|\nabla^{3}_{x}\psi\right|^{2}\left|\nabla^{2}_{x}\psi\right|^{2}\,dx+C_{\epsilon}C_{d}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{4}\psi\right|^{2}\,dx.$
(4.9)

Similarly, we have

$\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{3}\phi\right|^{3}\,dx\leq\epsilon
C_{d}\int\limits_{\mathbb{R}^{d}}\left|\nabla^{3}_{x}\phi\right|^{2}\left|\nabla^{2}_{x}\phi\right|^{2}\,dx+C_{\epsilon}C_{d}\int\limits_{\mathbb{R}^{d}}\left|\nabla_{x}^{4}\phi\right|^{2}\,dx.$
(4.10)

Let’s define $\displaystyle\mathcal{I}_{i},\,1\leq i\leq 4$, as follows,

$\displaystyle\displaystyle\mathcal{I}_{1}$
$\displaystyle\displaystyle:=\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{2}\left|\partial^{3}_{klr}\phi\right|^{2}\,dx,\quad\quad\mathcal{I}_{2}:=\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\phi\right|^{2}\left|\partial^{3}_{klr}\psi\right|^{2}\,dx,$
$\displaystyle\displaystyle\mathcal{I}_{3}$
$\displaystyle\displaystyle:=\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\psi\right|^{2}\left|\partial^{3}_{klr}\psi\right|^{2}\,dx,\quad\quad\mathcal{I}_{4}:=\sum\limits_{i,j,k,l,r=1}^{d}\int\limits_{\mathbb{R}^{d}}\left|\partial^{2}_{ij}\phi\right|^{2}\left|\partial^{3}_{klr}\phi\right|^{2}\,dx.$

We also set $\displaystyle
M:=1+\left|\left|\nabla_{x}\psi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\phi\right|\right|_{L^{\infty}(\mathbb{R}^{d})}$.
Combining (4.9) and (4.10) with (4.6), we obtain

$\mathcal{I}_{1}\leq\epsilon
C_{d}M^{2}\left[\mathcal{I}_{3}+\mathcal{I}_{4}\right]+C_{\epsilon}C_{d}M^{2}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{4}\psi\right|^{2}+\left|\nabla_{x}^{4}\phi\right|^{2}\right]\,dx.$
(4.11)

By a similar argument, we have

$\mathcal{I}_{k}\leq\epsilon
C_{d}M^{2}\left[\mathcal{I}_{3}+\mathcal{I}_{4}\right]+C_{\epsilon}C_{d}M^{2}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{4}\psi\right|^{2}+\left|\nabla_{x}^{4}\phi\right|^{2}\right]\,dx\mbox{
for }k=2,3,4.$ (4.12)

Finally, we combine (4.11), (4.12) and then choosing appropriate
$\displaystyle\epsilon$, we obtain (4.3). ∎

###  4.1 Proof of Proposition 4.1

With the help of Lemma 4.2 and 4.3, we are going to prove Proposition 4.1.
Proof also relies on the symmetrization of (1.1)–(1.2) in the spirit of [40].
The method is similar to the one in [30]. Here, we can reduce the regularity
exponent even smaller than $\displaystyle 1+\frac{d}{2}$ due to our assumption
on $\displaystyle C^{1}$ of $\displaystyle\varrho$ and u.

###### Proof of Proposition 4.1.

We consider the following symmetrization of (1.1)–(1.2) in the spirit of [40]

$\displaystyle\displaystyle\partial_{t}\pi+\textbf{u}\cdot\nabla_{x}\pi+C_{1}\pi\operatorname{div}_{x}\textbf{u}$
$\displaystyle\displaystyle=0,$ (4.13)
$\displaystyle\displaystyle\partial_{t}\textbf{u}+\textbf{u}\cdot\nabla_{x}\textbf{u}+C_{1}\pi\nabla_{x}\pi$
$\displaystyle\displaystyle=0,$ (4.14)

where $\displaystyle C_{1}=\frac{\gamma-1}{2}$ and
$\displaystyle\pi=\sqrt{\frac{\gamma-1}{4\gamma}}\varrho^{\frac{\gamma-1}{2}}$.
Applying derivative $\displaystyle\partial_{\alpha}^{n}$ for
$\displaystyle\alpha=(\alpha_{1},\cdots,\alpha_{d})$ with
$\displaystyle\alpha_{1}+\dots+\alpha_{d}=n$ we obtain,

$\displaystyle\displaystyle\partial_{t}\partial^{n}_{\alpha}\pi+\sum\limits_{\beta_{1}+\beta_{2}=\alpha}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\textbf{u}\cdot\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\pi+C_{1}\sum\limits_{\beta_{1}+\beta_{2}=\alpha}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\pi\operatorname{div}_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\textbf{u}$
$\displaystyle\displaystyle=0,$ (4.15)
$\displaystyle\displaystyle\partial_{t}\partial^{n}_{\alpha}\textbf{u}+\sum\limits_{\beta_{1}+\beta_{2}=\alpha}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\textbf{u}\cdot\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\textbf{u}+C_{1}\sum\limits_{\beta_{1}+\beta_{2}=\alpha}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\pi\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\pi$
$\displaystyle\displaystyle=0,$ (4.16)

where $\displaystyle c(\beta_{1},\beta_{2};\alpha)$ is as in (3.1). We
rearrange (4.15) as follows

$\displaystyle\displaystyle\partial_{t}\partial^{n}_{\alpha}\pi$
$\displaystyle\displaystyle=-(\textbf{u}\cdot\nabla_{x}\partial^{n}_{\alpha}\pi+\partial^{n}_{\alpha}\textbf{u}\cdot\nabla_{x}\pi)-\sum\limits_{\beta_{1}+\beta_{2}=\alpha,\left|\beta_{1}\right|\geq
1,\left|\beta_{2}\right|\geq
1}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\textbf{u}\cdot\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\pi$
$\displaystyle\displaystyle-
C_{1}(\pi\operatorname{div}_{x}\partial^{n}_{\alpha}\textbf{u}+\partial^{n}_{\alpha}\pi\operatorname{div}_{x}\textbf{u})-C_{1}\sum\limits_{\beta_{1}+\beta_{2}=\alpha,\left|\beta_{1}\right|\geq
1,\left|\beta_{2}\right|\geq
1}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\pi\operatorname{div}_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\textbf{u}.$
(4.17)

Multiplying (4.17) with $\displaystyle\partial^{n}_{\alpha}\pi$ and then
integrating over $\displaystyle\mathbb{R}^{d}$ we get

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left|\partial^{n}_{\alpha}\pi\right|^{2}$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\right]\,dx-
C_{1}\int\limits_{\mathbb{R}^{d}}\pi\operatorname{div}_{x}\partial^{n}_{\alpha}\textbf{u}\partial^{n}_{\alpha}\pi.$
(4.18)

We arrange (4.16) to have

$\displaystyle\displaystyle\partial_{t}\partial^{n}_{\alpha}\textbf{u}$
$\displaystyle\displaystyle=-(\textbf{u}\cdot\nabla_{x}\partial^{n}_{\alpha}\textbf{u}+\partial^{n}_{\alpha}\textbf{u}\cdot\nabla_{x}\textbf{u})-\sum\limits_{\beta_{1}+\beta_{2}=\alpha,\left|\beta_{1}\right|\geq
1,\left|\beta_{2}\right|\geq
1}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\textbf{u}\cdot\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\textbf{u}$
$\displaystyle\displaystyle-
C_{1}(\pi\nabla_{x}\partial^{n}_{\alpha}\pi+\partial^{n}_{\alpha}\pi\nabla_{x}\pi)-C_{1}\sum\limits_{\beta_{1}+\beta_{2}=\alpha,\left|\beta_{1}\right|\geq
1,\left|\beta_{2}\right|\geq
1}c(\beta_{1},\beta_{2};\alpha)\partial^{\left|\beta_{1}\right|}_{\beta_{1}}\pi\nabla_{x}\partial^{\left|\beta_{2}\right|}_{\beta_{2}}\pi,$
(4.19)

where $\displaystyle c(\alpha_{1},\beta_{2};\alpha)$ is as in (3.1).
Multiplying (4.19) with $\displaystyle\partial^{n}_{\alpha}\textbf{u}$ and
then integrating over $\displaystyle\mathbb{R}^{d}$ we obtain

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left|\partial^{n}_{\alpha}\textbf{u}\right|^{2}$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\right]\,dx-
C_{1}\int\limits_{\mathbb{R}^{d}}\pi\nabla_{x}\partial^{n}_{\alpha}\pi\cdot\partial^{n}_{\alpha}\textbf{u}.$
(4.20)

Applying integration by parts we get

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left|\partial^{n}_{\alpha}\textbf{u}\right|^{2}$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{1}\int\limits_{\mathbb{R}^{d}}\partial^{n}_{\alpha}\pi\nabla_{x}\pi\cdot\partial^{n}_{\alpha}\textbf{u}\,dx+C_{1}\int\limits_{\mathbb{R}^{d}}\pi\operatorname{div}_{x}\partial^{n}_{\alpha}\textbf{u}\partial^{n}_{\alpha}\pi\,dx.$

We further estimate,

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left|\partial^{n}_{\alpha}\textbf{u}\right|^{2}$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{1}\int\limits_{\mathbb{R}^{d}}\pi\partial^{n}_{\alpha}\pi\operatorname{div}_{x}\partial^{n}_{\alpha}\textbf{u}\,dx.$
(4.21)

Combining (4.18), (4.21) along with the help of (3.4) we have

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\textbf{u}\right|^{2}+\left|\nabla_{x}^{n}\pi\right|^{2}\right]\,dx$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\right]\,dx$
$\displaystyle\displaystyle+C_{d,n}\sum\limits_{k=1}^{n-1}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k}\pi\right|^{2}\left|\nabla_{x}^{n-k+1}\textbf{u}\right|^{2}\,dx+\left|\nabla_{x}^{k}\textbf{u}\right|^{2}\left|\nabla_{x}^{n-k+1}\pi\right|^{2}\right]\,dx.$
(4.22)

By using Lemma 4.2 and 4.3 with appropriate choice of
$\displaystyle\psi,\phi$, we obtain

$\displaystyle\displaystyle\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\textbf{u}\right|^{2}+\left|\nabla_{x}^{n}\pi\right|^{2}\right]\,dx$
$\displaystyle\displaystyle\leq
C_{d,n}\left(1+\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\right)\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\pi\right|^{2}+\left|\nabla_{x}^{n}\textbf{u}\right|^{2}\right]\,dx.$

By using Grönwall’s inequality, we get

$\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\textbf{u}\right|^{2}+\left|\nabla_{x}^{n}\pi\right|^{2}\right]\,dx\leq
e^{M(t)}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{n}\textbf{u}_{0}\right|^{2}+\left|\nabla_{x}^{n}\pi_{0}\right|^{2}\right]\,dx,$
(4.23)

where $\displaystyle M(t)$ is defined as

$M(t):=\int\limits_{0}^{t}C_{d,n}(1+\left|\left|\nabla_{x}\textbf{u}(s,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}+\left|\left|\nabla_{x}\pi(s,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})})\,ds.$
(4.24)

This completes the proof of Proposition 4.1.

∎

##  5 Blow up of $\displaystyle C^{1}$ solutions

###  5.1 Proof of Proposition 2.3

In this subsection, we prove the blow up result of Proposition 2.3. Here, we
do not derive the Riccati equation directly. Rather, we compare the velocity
$\displaystyle\textbf{u}(t,\cdot)$ with the solution of vectorial Burgers
equation (1.4) with initial data $\displaystyle\textbf{u}_{0}$.

###### Proof Proposition 2.3:.

We argue by contradiction. Suppose that $\displaystyle
T^{*}>\frac{1}{\lambda_{max}}$. Let us consider v to be the solution to the
following Cauchy problem,

$\displaystyle\displaystyle\partial_{t}\textbf{v}+\textbf{v}\cdot\nabla_{x}\textbf{v}$
$\displaystyle\displaystyle=0,\mbox{ for }x\in\mathbb{R}^{d},t>0,$ (5.1)
$\displaystyle\displaystyle\textbf{v}(0,x)$
$\displaystyle\displaystyle=\textbf{u}_{0}(x)\mbox{ for }x\in\mathbb{R}^{d}.$
(5.2)

Set
$\displaystyle\pi:=\sqrt{\frac{\gamma-1}{4\gamma}}\varrho^{\frac{\gamma-1}{2}}$
and we consider the following symmetrization of the system (1.1)–(1.2),

$\displaystyle\displaystyle\partial_{t}\pi+\textbf{u}\cdot\nabla_{x}\pi+C_{1}\pi\operatorname{div}_{x}\textbf{u}$
$\displaystyle\displaystyle=0,$ (5.3)
$\displaystyle\displaystyle\partial_{t}\textbf{u}+\textbf{u}\cdot\nabla_{x}\textbf{u}+C_{1}\pi\nabla_{x}\pi$
$\displaystyle\displaystyle=0,$ (5.4)

where $\displaystyle C_{1}=\frac{\gamma-1}{2}$. From (5.1) and (5.4) we have

$\partial_{t}(\textbf{u}-\textbf{v})+\textbf{u}\cdot\nabla_{x}\textbf{u}-\textbf{v}\cdot\nabla_{x}\textbf{v}+C_{1}\pi\nabla_{x}\pi=0.$

Let $\displaystyle\boldsymbol{\varphi}=\textbf{u}-\textbf{v}$. Then, we get

$\partial_{t}\boldsymbol{\varphi}+\boldsymbol{\varphi}\cdot\nabla\textbf{u}+\textbf{v}\cdot\nabla_{x}\boldsymbol{\varphi}+C_{1}\pi\nabla_{x}\pi=0.$

Multiply by $\displaystyle\boldsymbol{\varphi}$ to get

$\frac{1}{2}\partial_{t}\left|\boldsymbol{\varphi}\right|^{2}+\nabla\textbf{u}:\boldsymbol{\varphi}\otimes\boldsymbol{\varphi}+\frac{1}{2}\textbf{v}\cdot\nabla_{x}\left|\boldsymbol{\varphi}\right|^{2}+C_{1}\pi\nabla_{x}\pi\cdot\boldsymbol{\varphi}=0,$
(5.5)

or equivalently,

$(\partial_{t}+\textbf{v}\cdot\nabla_{x})\left(\frac{\left|\boldsymbol{\varphi}\right|^{2}}{2}\right)=-\nabla\textbf{u}:\boldsymbol{\varphi}\otimes\boldsymbol{\varphi}-C_{1}\pi\nabla_{x}\pi\cdot\boldsymbol{\varphi}.$
(5.6)

Now we consider $\displaystyle X(t,x_{0})$ defined as

$\frac{dX(t,x_{0})}{dt}=\textbf{v}(t,X(t,x_{0}))\mbox{ and }X(0,x_{0})=x_{0}.$

We define $\displaystyle
g(t)=\frac{1}{2}\left|\boldsymbol{\varphi}(t,X(t,x_{0}))\right|^{2}$ to get

$\displaystyle\displaystyle\frac{dg(t)}{dt}$
$\displaystyle\displaystyle=-\nabla_{x}\textbf{u}(t,X(t,x_{0})):\boldsymbol{\varphi}(t,X(t,x_{0}))\otimes\boldsymbol{\varphi}(t,X(t,x_{0}))$
$\displaystyle\displaystyle-
C_{1}\pi(t,X(t,x_{0}))\nabla_{x}\pi(t,X(t,x_{0}))\cdot\boldsymbol{\varphi}(t,X(t,x_{0}))$
$\displaystyle\displaystyle\leq
2\left|\nabla_{x}\textbf{u}(t,X(t,x_{0}))\right|a(t)+C_{1}\pi(t,X(t,x_{0}))\left|\nabla_{x}\pi(t,X(t,x_{0}))\right|\sqrt{g(t)}.$

Since $\displaystyle T^{*}>1/\lambda_{max}$, we have $\displaystyle
M\geq\left|\left|\nabla_{x}\textbf{u}\right|\right|_{L^{\infty}([0,1/\lambda_{max}]\times\mathbb{R}^{d})}$
and from (2.7) it follows that
$\displaystyle\epsilon_{0}\geq\left|\left|\pi\nabla_{x}\pi\right|\right|_{L^{\infty}([0,1/\lambda_{max}]\times\mathbb{R}^{d})}$.
Then, we can write

$g^{\prime}(t)\leq Mg(t)+\epsilon_{0}\sqrt{g(t)}.$

Let $\displaystyle f(t)=\sqrt{g}(t)$. We have

$f^{\prime}(t)\leq Mf(t)+\epsilon_{0}\mbox{ or equivalently,
}\left(f(t)e^{-Mt}\right)^{\prime}\leq\epsilon_{0}e^{-Mt}.$

Subsequently, we get

$f(t)\leq
e^{Mt}\left[f(0)+\frac{\epsilon_{0}}{M}\left(1-e^{Mt}\right)\right]=\frac{\epsilon_{0}}{M}\left(e^{Mt}-1\right).$

From the assumption on $\displaystyle\textbf{u}_{0}$ we have
$\displaystyle\textbf{u}_{0}(x_{0}+r\xi)-\textbf{u}_{0}(x_{0})=-\lambda_{0}r\xi$.
Let $\displaystyle x_{\xi}=x_{0}+r\xi$, $\displaystyle
X(t)=x_{\xi}+t\textbf{u}_{0}(x_{\xi})$ and $\displaystyle
Y(t)=x_{0}+t\textbf{u}_{0}(x_{0})$. Then we have

$\displaystyle\displaystyle\left|X(t)-Y(t)\right|^{2}$
$\displaystyle\displaystyle=\left|x_{\xi}-x_{0}+t(\textbf{u}_{0}(x_{\xi})-\textbf{u}_{0}(x_{0}))\right|^{2}$
$\displaystyle\displaystyle=\left|x_{\xi}-x_{0}\right|^{2}+2t(\textbf{u}_{0}(x_{\xi})-\textbf{u}_{0}(x_{0}))\cdot(x_{\xi}-x_{0})+t^{2}\left|\textbf{u}_{0}(x_{\xi})-\textbf{u}_{0}(x_{0})\right|^{2}$
$\displaystyle\displaystyle=\left|x_{\xi}-x_{0}\right|^{2}-2\lambda_{0}t\left|x_{\xi}-x_{0}\right|^{2}+\lambda_{0}^{2}t^{2}\left|x_{\xi}-x_{0}\right|^{2}$
$\displaystyle\displaystyle=(1-t\lambda_{0})^{2}\left|x_{\xi}-x_{0}\right|^{2}.$

Since $\displaystyle\textbf{v}=\textbf{u}-\boldsymbol{\varphi}$, we note that

$\displaystyle\displaystyle(x_{\xi}-x_{0})\cdot(\textbf{u}_{0}(x_{\xi})-\textbf{u}_{0}(x_{0}))$
$\displaystyle\displaystyle=(x_{\xi}-x_{0})\cdot(\textbf{v}(X(t))-\textbf{v}(Y(t)))$
$\displaystyle\displaystyle=(x_{\xi}-x_{0})\cdot(\textbf{u}(X(t))-\textbf{u}(Y(t)))-(x_{\xi}-x_{0})\cdot(\boldsymbol{\varphi}(X(t))-\boldsymbol{\varphi}(Y(t))).$

As
$\displaystyle\left|\left|\boldsymbol{\varphi}(t)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\leq
e^{Mt}\epsilon_{0}$ for all $\displaystyle t\in[0,1/\lambda_{max}]$, we get

$\left|(x_{\xi}-x_{0})\cdot(\boldsymbol{\varphi}(X(t))-\boldsymbol{\varphi}(Y(t)))\right|\leq\frac{2e^{Mt}}{M}\left|x_{\xi}-x_{0}\right|\epsilon_{0}\leq\frac{2e^{\frac{M}{\lambda_{max}}}}{M}r\epsilon_{0}.$
(5.7)

Similarly, we have

$\displaystyle\displaystyle\left|(x_{\xi}-x_{0})\cdot(\textbf{u}(X(t))-\textbf{u}(Y(t)))\right|$
$\displaystyle\displaystyle=\left|\int\limits_{0}^{1}(X(t)-Y(t))\cdot\nabla_{x}\textbf{u}(t,\theta
X(t)+(1-\theta)Y(t))\cdot(x_{\xi}-x_{0})\,d\theta\right|$
$\displaystyle\displaystyle\leq\left|X(t)-Y(t)\right|\left|x_{\xi}-x_{0}\right|\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}$
$\displaystyle\displaystyle=(1-\lambda_{0}t)\left|x_{\xi}-x_{0}\right|^{2}\left|\left|\nabla_{x}\textbf{u}(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}$
$\displaystyle\displaystyle=(1-\lambda_{0}t)r^{2}M.$

Subsequently,

$\displaystyle\displaystyle\lambda_{0}r^{2}=\lambda_{0}\left|x_{\xi}-x_{0}\right|^{2}$
$\displaystyle\displaystyle=\left|(x_{\xi}-x_{0})\cdot(\textbf{v}_{0}(x_{\xi})-\textbf{v}_{0}(x_{0}))\right|$
$\displaystyle\displaystyle\leq(1-\lambda_{0}t)r^{2}M+\frac{2e^{\frac{M}{\lambda_{max}}}}{M}r\epsilon_{0}.$

Therefore, for $\displaystyle t=\frac{1}{\lambda_{max}}$ we obtain

$\displaystyle\displaystyle\lambda_{0}r\leq\left(1-\frac{\lambda_{0}}{\lambda_{max}}\right)rM+\frac{2e^{\frac{M}{\lambda_{max}}}}{M}\epsilon_{0}$
$\displaystyle\displaystyle=\frac{\lambda_{max}-\lambda_{0}}{\lambda_{max}}rM+\frac{2e^{\frac{M}{\lambda_{max}}}}{M}\epsilon_{0}$
$\displaystyle\displaystyle\leq\left[\frac{rM}{\lambda_{max}}+\frac{2e^{\frac{M}{\lambda_{max}}}}{M}\right]\epsilon_{0}.$

This contradicts with the assumption (2.7) on $\displaystyle\epsilon_{0}$.
Hence, we prove Proposition 2.3. ∎

###  5.2 Proof of Theorem 2.5

Before we prove the blow up result of Theorem 2.5 we construct an example of
data $\displaystyle\textbf{u}_{0}$ in dimension $\displaystyle 2$ such that it
satisfies (2.8).

######  Example 5.1.

We consider
$\displaystyle\textbf{u}_{0}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ defined
as $\displaystyle\textbf{u}_{0}(x_{1},x_{2})=(u_{1},u_{2})$

$\displaystyle\displaystyle u_{1}(x_{1},x_{2})$
$\displaystyle\displaystyle=-x_{1}\left(1+\frac{x_{1}^{2}}{R^{2}}\right)^{-n}\varphi\left(\left(1+\frac{x_{1}^{2}}{R^{2}}\right)\frac{x_{2}}{R}\right),$
(5.8) $\displaystyle\displaystyle u_{2}(x_{1},x_{2})$
$\displaystyle\displaystyle=-x_{2}\left(1+\frac{x_{2}^{2}}{R^{2}}\right)^{-n}\varphi\left(\left(1+\frac{x_{2}^{2}}{R^{2}}\right)\frac{x_{1}}{R}\right),$
(5.9)

for some $\displaystyle\varphi\in
C^{\infty}_{c}(\mathbb{R}),R>0,n\in\mathbb{N}$ such that

$\left.\begin{array}[]{rl}\varphi(y)=1&\mbox{ if }\left|y\right|\leq 1,\\\
0\leq\varphi(y)\leq 1&\mbox{ if }1\leq\left|y\right|\leq 2,\\\
\varphi(y)=0&\mbox{ if }\left|y\right|\geq 2.\end{array}\right\\}$ (5.10)

We calculate for $\displaystyle j=k+l\geq 1$,

$\displaystyle\displaystyle\partial_{x_{1}}^{k}\partial_{x_{2}}^{l}u_{1}=\frac{1}{R^{j-1}}\sum\limits_{i=0}^{j}\mathcal{P}_{i}^{k,l}\left(\frac{x_{1}}{R},\frac{x_{2}}{R}\right)\left(1+\frac{x_{1}^{2}}{R^{2}}\right)^{-n-i}\varphi^{(j-i)}\left(\left(1+\frac{x_{1}^{2}}{R^{2}}\right)\frac{x_{2}}{R}\right)$
(5.11)

where $\displaystyle\mathcal{P}^{k,l}_{i}$ is a polynomial of two variables.
Subsequently, we obtain

$\left|\left|\nabla_{x}^{j}u_{1}\right|\right|_{L^{2}(\mathbb{R}^{2})}=\mathcal{O}(1)R^{-j+\frac{3}{2}}\mbox{
for }1\leq j\leq 5.$

Hence, we have
$\displaystyle\left|\left|\nabla^{2}_{x}u_{1}\right|\right|_{H^{3}(\mathbb{R}^{2})}=\mathcal{O}(1)R^{-\frac{1}{2}}$.
By a similar argument we can also have
$\displaystyle\left|\left|\nabla^{2}_{x}u_{2}\right|\right|_{H^{3}(\mathbb{R}^{2})}=\mathcal{O}(1)R^{-\frac{1}{2}}$.
We also note that

$\nabla_{x}\textbf{u}(0,0)=\begin{pmatrix}\partial_{x_{1}}u_{1}(0,0)&\partial_{x_{2}}u_{1}(0,0)\\\
\partial_{x_{1}}u_{2}(0,0)&\partial_{x_{2}}u_{2}(0,0)\end{pmatrix}=\begin{pmatrix}-1&0\\\
0&-1\end{pmatrix}.$ (5.12)

Choosing large enough $\displaystyle R>1$, we can see that
$\displaystyle\textbf{u}_{0}$ satisfies (2.8) and (2.6) with
$\displaystyle\lambda_{max}=1$.

Next, we construct another example of data such that it satisfies (2.8) but it
fails to satisfy the integral condition (2.4).

######  Example 5.2.

Let us consider $\displaystyle\textbf{u}_{0}$ defined as in Example 5.1. We
now define $\displaystyle\tilde{\textbf{u}}_{0}$ as
$\displaystyle\psi_{\lambda}\textbf{u}_{0}$ where $\displaystyle\psi\in
C_{c}^{\infty}$ having the property

$\left.\begin{array}[]{ll}\psi_{\lambda}(x)=1&\mbox{ for }x\in B(0,2R),\\\
0\leq\psi_{\lambda}(x)\leq 1&\mbox{ for }2R\leq\left|x\right|\leq\lambda,\\\
\psi_{\lambda}(x)=0&\mbox{ for
}\left|x\right|\geq\lambda,\end{array}\right\\}$ (5.13)

for some $\displaystyle\lambda>2R$ where $\displaystyle R$ is as Example 5.1.
We set $\displaystyle\varrho_{0}(x)=\overline{\varrho}$ for all $\displaystyle
x\in\mathbb{R}^{2}$ and some constant $\displaystyle\overline{\varrho}$. We
choose $\displaystyle\overline{\varrho}$ large enough so that we have

$\frac{1}{\omega_{2}\lambda^{3}}\int\limits_{\mathbb{R}^{2}}\varrho_{0}(x)\textbf{u}_{0}(x)\cdot
x\,dx\leq\overline{\varrho}\frac{C_{1}R^{4}}{\omega_{2}\lambda^{3}}\mbox{ for
some }C_{1}>0.$ (5.14)

Choosing $\displaystyle\lambda$ large enough and
$\displaystyle\overline{\varrho}\geq 1$, we can have
$\displaystyle\overline{\varrho}\frac{C_{1}R^{4}}{\omega_{2}\lambda^{3}}\leq
3\sqrt{\gamma}(\overline{\varrho})^{\frac{\gamma-1}{2}}(\overline{\varrho})$.
This shows that $\displaystyle(\varrho_{0},\tilde{\textbf{u}}_{0})$ does not
satisfy the integral condition (2.4).

In the following example, we consider a particular radially symmetric initial
velocity $\displaystyle\textbf{u}_{0}$ such that it satisfies smallness
condition (2.8) and having a decreasing direction as in (2.6).

######  Example 5.3 (Radially symmetric data).

We consider initial velocity
$\displaystyle\textbf{u}_{0}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ defined
as follows

$\textbf{u}_{0}(x)=h(r)\frac{x}{\left|x\right|}\mbox{ where
}r=\left|x\right|.$

By an elementary calculation we get

$\nabla_{x}\textbf{u}_{0}(x)=\frac{h^{\prime}(r)}{r^{2}}x\otimes
x+\frac{h(r)}{r^{3}}\begin{pmatrix}x_{2}^{2}&x_{1}x_{2}\\\
x_{1}x_{2}&x_{1}^{2}\end{pmatrix}.$

We write $\displaystyle
a_{ij}=\partial_{i}u_{j}(x)=\frac{h^{\prime}(r)}{r^{2}}\mathcal{P}^{1}_{2}(x_{1},x_{2})+\frac{h(r)}{r^{3}}\mathcal{P}^{2}_{2}(x_{1},x_{2})$
where $\displaystyle\mathcal{P}^{j}_{2}(x_{1},x_{2}),j=1,2$ are polynomials of
degree 2. Then we have

$\displaystyle\displaystyle\partial_{k}a_{ij}$
$\displaystyle\displaystyle=\frac{h^{(2)}(r)}{r^{3}}\mathcal{P}^{1}_{3}(x_{1},x_{2})+\frac{h^{(1)}(r)}{r^{4}}\mathcal{P}^{2}_{3}(x_{1},x_{2})+\frac{h(r)}{r^{5}}\mathcal{P}^{3}_{3}(x_{1},x_{2}),$
$\displaystyle\displaystyle\partial_{l}(\partial_{k}a_{ij})$
$\displaystyle\displaystyle=\frac{h^{(3)}(r)}{r^{4}}\mathcal{P}^{1}_{4}(x_{1},x_{2})+\frac{h^{(2)}(r)}{r^{5}}\mathcal{P}^{2}_{4}(x_{1},x_{2})+\frac{h^{(1)}(r)}{r^{6}}\mathcal{P}^{3}_{4}(x_{1},x_{2})+\frac{h(r)}{r^{7}}\mathcal{P}^{4}_{4}(x_{1},x_{2}),$
$\displaystyle\displaystyle\partial^{2}_{lm}(\partial_{k}a_{ij})$
$\displaystyle\displaystyle=\frac{h^{(4)}(r)}{r^{5}}\mathcal{P}^{1}_{5}(x_{1},x_{2})+\frac{h^{(3)}(r)}{r^{6}}\mathcal{P}^{2}_{5}(x_{1},x_{2})+\frac{h^{(2)}(r)}{r^{7}}\mathcal{P}^{3}_{5}(x_{1},x_{2})+\frac{h^{(1)}(r)}{r^{8}}\mathcal{P}^{4}_{5}(x_{1},x_{2})$
$\displaystyle\displaystyle+\frac{h(r)}{r^{9}}\mathcal{P}^{5}_{5}(x_{1},x_{2}),$
$\displaystyle\displaystyle\partial^{3}_{lmn}(\partial_{k}a_{ij})$
$\displaystyle\displaystyle=\frac{h^{(5)}(r)}{r^{6}}\mathcal{P}^{1}_{6}(x_{1},x_{2})+\frac{h^{(4)}(r)}{r^{7}}\mathcal{P}^{2}_{6}(x_{1},x_{2})+\frac{h^{(3)}(r)}{r^{8}}\mathcal{P}^{3}_{6}(x_{1},x_{2})+\frac{h^{(2)}(r)}{r^{9}}\mathcal{P}^{4}_{6}(x_{1},x_{2})$
$\displaystyle\displaystyle+\frac{h^{(1)}(r)}{r^{10}}\mathcal{P}^{5}_{6}(x_{1},x_{2})+\frac{h(r)}{r^{11}}\mathcal{P}^{6}_{6}(x_{1},x_{2}).$

Now, we choose $\displaystyle h(z)=Re^{-\frac{z}{R}}\varphi(z-R)$ for
$\displaystyle R>1$ and $\displaystyle\varphi$ is defined as

$\left.\begin{array}[]{ll}\varphi(z)=1&\mbox{ for }\left|z\right|\leq 1,\\\
0\leq\varphi(z)\leq 1&\mbox{ for }1\leq\left|z\right|\leq 2,\\\
\varphi(z)=0&\mbox{ for }\left|z\right|\geq 2.\end{array}\right\\}$

Then we have

$\left|\left|\nabla_{x}^{l}(\partial_{k}a_{ij})\right|\right|_{L^{2}(\mathbb{R}^{2})}=\mathcal{O}(1)R^{-1-l}\mbox{
for }l=0,1,2,3.$

Further, we note that
$\displaystyle\nabla_{x}\textbf{u}_{0}(R,0)=\begin{pmatrix}-1&0\\\
0&1\end{pmatrix}$. Hence, the initial velocity satisfies (2.6) with
$\displaystyle\lambda_{max}=1$ and we can choose an appropriate
$\displaystyle\varrho_{0}$ and $\displaystyle R>1$ such that (2.8) is
satisfied.

See [36, 41] for studies on blow up of spherically symmetric solutions.

Now, we prove the blow up result for $\displaystyle C^{1}$ solution when
initial data satisfies (2.6) and (2.8). The proof is based on Kato type energy
estimate derived as in section 4 and deduction of a Riccati equation.

###### Proof of Theorem 2.5:.

We argue by contradiction. Suppose that $\displaystyle(\varrho,\textbf{u})\in
C^{1}([0,T]\times\mathbb{R}^{d};[0,\infty)\times\mathbb{R}^{d}))$ for some
$\displaystyle T>2/\lambda_{max}$. Similar to Proposition 2.3, we consider the
following symmetrization of (1.1)–(1.2) in the spirit of [40],

$\displaystyle\displaystyle\partial_{t}\pi+\textbf{u}\cdot\nabla_{x}\pi+C_{1}\pi\operatorname{div}_{x}\textbf{u}$
$\displaystyle\displaystyle=0,$ (5.15)
$\displaystyle\displaystyle\partial_{t}\textbf{u}+\textbf{u}\cdot\nabla_{x}\textbf{u}+C_{1}\pi\nabla_{x}\pi$
$\displaystyle\displaystyle=0,$ (5.16)

where $\displaystyle C_{1}=\frac{\gamma-1}{2}$ and
$\displaystyle\pi=\sqrt{\frac{\gamma-1}{4\gamma}}\varrho^{\frac{\gamma-1}{2}}$.
Let us consider the flow $\displaystyle X(t,x_{0})$ defined as follows,

$\frac{dX(t,x_{0})}{dt}=\textbf{u}(t,X(t,x_{0}))\mbox{ and }X(0,x_{0})=x_{0}.$

By using energy estimate (4.23) we obtain

$\sum\limits_{k=0}^{m}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k+2}\pi(t,\cdot)\right|^{2}+\left|\nabla_{x}^{k+2}\textbf{u}(t,\cdot)\right|^{2}\right]\,dx\leq
e^{C_{2}t}\sum\limits_{k=0}^{m}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k+2}\pi_{0}\right|^{2}+\left|\nabla_{x}^{k+2}\textbf{u}_{0}\right|^{2}\right]\,dx.$
(5.17)

Then from assumption (2.8) and Lemma 3.2 we have the smallness of
$\displaystyle\left|\left|\nabla^{2}_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}$
as

$\left|\left|\nabla^{2}_{x}\pi(t,\cdot)\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\leq
e^{C_{2}t}\kappa\mbox{ where
}\kappa=\sum\limits_{k=0}^{m}\int\limits_{\mathbb{R}^{d}}\left[\left|\nabla_{x}^{k+2}\pi_{0}\right|^{2}+\left|\nabla_{x}^{k+2}\textbf{u}_{0}\right|^{2}\right]\,dx.$
(5.18)

Subsequently, from (5.15), we obtain

$0\leq\pi(t,X(t,x_{0}))\leq e^{C_{2}t}\pi(x_{0}).$ (5.19)

From (2.6), there exists $\displaystyle x_{0}\in\mathbb{R}^{d}$ and
$\displaystyle\xi_{0}\in\mathbb{S}^{d-1}$ such that

$\sup\limits_{\xi\in\mathbb{S}^{d-1}}\left[-\nabla_{x}\textbf{u}_{0}(x_{0}):\xi\otimes\xi\right]=-\nabla_{x}\textbf{u}_{0}(x_{0}):\xi_{0}\otimes\xi_{0}=\lambda_{max}.$

Applying $\displaystyle\nabla_{x}$ on both side of (5.16), we have

$\partial_{t}\nabla\textbf{u}+\textbf{u}\cdot\nabla^{2}_{x}\textbf{u}+\nabla_{x}\textbf{u}\nabla_{x}\textbf{u}+C_{1}\pi\nabla^{2}_{x}\pi+C_{1}\nabla_{x}\pi\otimes\nabla_{x}\pi=0.$
(5.20)

First we observe that

$\displaystyle\displaystyle\partial_{t}[\nabla_{x}\textbf{u}-\nabla_{x}^{T}\textbf{u}]+\textbf{u}\cdot\nabla_{x}[\nabla_{x}\textbf{u}-\nabla_{x}^{T}\textbf{u}]$
$\displaystyle\displaystyle=\nabla_{x}^{T}\textbf{u}\nabla_{x}^{T}\textbf{u}-\nabla_{x}\textbf{u}\nabla_{x}\textbf{u}$
(5.21)
$\displaystyle\displaystyle=\nabla_{x}^{T}\textbf{u}[\nabla_{x}^{T}\textbf{u}-\nabla_{x}\textbf{u}]+[\nabla^{T}_{x}\textbf{u}-\nabla_{x}\textbf{u}]\nabla_{x}\textbf{u}.$
(5.22)

Along the curve $\displaystyle X(t,x_{0})$ we have
$\displaystyle\frac{d}{dt}\left|[\nabla_{x}\textbf{u}-\nabla_{x}^{T}\textbf{u}](t,X(t))\right|\leq
C_{1}\left|[\nabla_{x}\textbf{u}-\nabla_{x}^{T}\textbf{u}](t,X(t))\right|$. By
Grönwall’s lemma we conclude that
$\displaystyle\nabla_{x}\textbf{u}(t,X(t))=-\nabla_{x}^{T}\textbf{u}(t,X(t))$
since
$\displaystyle\nabla_{x}\textbf{u}_{0}(x_{0})=-\nabla_{x}^{T}\textbf{u}_{0}(x_{0})$.
Now, for any $\displaystyle\xi\in\mathbb{S}^{d-1}$ we have,

$\displaystyle\displaystyle\partial_{t}[-\nabla\textbf{u}:\xi\otimes\xi]+\textbf{u}\cdot\nabla_{x}[-\nabla\textbf{u}:\xi\otimes\xi]$
$\displaystyle\displaystyle=[\nabla_{x}\textbf{u}\nabla_{x}\textbf{u}]:\xi\otimes\xi+C_{1}\pi\nabla^{2}_{x}\pi:\xi\otimes\xi+C_{1}\left|\xi\cdot\nabla_{x}\pi\right|^{2}$
$\displaystyle\displaystyle\geq[\nabla_{x}\textbf{u}\nabla_{x}\textbf{u}]:\xi\otimes\xi+C_{1}\pi\nabla^{2}_{x}\pi:\xi\otimes\xi.$
(5.23)

We note that
$\displaystyle[\nabla_{x}\textbf{u}_{0}(x_{0})\nabla_{x}\textbf{u}_{0}(x_{0})]:\xi_{0}\otimes\xi_{0}=\lambda_{max}^{2}$
since $\displaystyle-\nabla\textbf{u}_{0}(x_{0})\xi_{0}=\lambda_{max}\xi_{0}$.
Let $\displaystyle\delta>0$. Since $\displaystyle\textbf{u}\in
C^{1}([0,T)\times\mathbb{R}^{d})$ there exists $\displaystyle t_{1}>0$ such
that the following holds

$[\nabla_{x}\textbf{u}(t,X(t,x_{0}))\nabla_{x}\textbf{u}(t,X(t,x_{0}))]:\xi_{0}\otimes\xi_{0}\geq\left|\nabla_{x}\textbf{u}(t,X(t,x_{0})):\xi_{0}\otimes\xi_{0}\right|^{2}-\delta^{2}.$
(5.24)

Using (5.18), (5.23) and (5.24) we get

$\frac{d}{dt}[-\nabla\textbf{u}(t,X(t,x_{0})):\xi_{0}\otimes\xi_{0}]\geq\left|\nabla_{x}\textbf{u}(t,X(t,x_{0})):\xi_{0}\otimes\xi_{0}\right|^{2}-\delta^{2}-C_{1}\kappa\pi(x_{0})e^{2C_{2}t}.$
(5.25)

Define $\displaystyle
a(t):=[-\nabla\textbf{u}(t,X(t,x_{0})):\xi_{0}\otimes\xi_{0}]$. For
sufficiently small $\displaystyle\delta>0$ we have

$a^{\prime}(t)\geq a^{2}(t)-2C_{1}\kappa\pi(x_{0})e^{2C_{2}t}.$ (5.26)

From the assumption (2.8), we have
$\displaystyle\kappa\pi_{0}(x_{0})\leq\epsilon_{0}$. Hence,

$a^{\prime}(t)\geq a^{2}(t)-2C_{1}\epsilon_{0}e^{2C_{2}t}.$ (5.27)

Next, we consider $\displaystyle b(t)=a^{m}(t)$. Then we have

$\displaystyle\displaystyle b^{\prime}(t)=ma^{m-1}(t)a^{\prime}(t)$
$\displaystyle\displaystyle\geq
ma^{m-1}(t)[a^{2}(t)-2C_{1}\epsilon_{0}e^{2C_{2}t}]$
$\displaystyle\displaystyle=ma^{m+1}(t)-2C_{1}m\epsilon_{0}e^{2C_{2}t}a^{m-1}$
$\displaystyle\displaystyle=mb^{1+\frac{1}{m}}(t)-2C_{1}m\epsilon_{0}e^{2C_{2}t}b^{1-\frac{1}{m}}(t).$

We define $\displaystyle\Lambda(t)=b(t)e^{-At}$. Then we have

$\displaystyle\displaystyle\Lambda^{\prime}(t)$
$\displaystyle\displaystyle=b^{\prime}(t)e^{-At}-Ab(t)e^{-At}$
$\displaystyle\displaystyle\geq
mb^{1+\frac{1}{m}}(t)e^{-At}-Ab(t)e^{-At}-2C_{1}m\epsilon_{0}e^{2C_{2}t}b^{1-\frac{1}{m}}(t)e^{-At}$
$\displaystyle\displaystyle=mb^{1+\frac{1}{m}}(t)e^{-At-\frac{At}{m}}e^{\frac{At}{m}}-A\Lambda(t)-2C_{1}m\epsilon_{0}e^{2C_{2}t}b^{1-\frac{1}{m}}(t)e^{-At}$
$\displaystyle\displaystyle=m\Lambda^{1+\frac{1}{m}}(t)e^{\frac{At}{m}}-A\Lambda(t)-2C_{1}m\epsilon_{0}e^{2C_{2}t}b^{1-\frac{1}{m}}(t)e^{-At}.$

We choose large $\displaystyle A$ such that $\displaystyle A\geq 2C_{2}$. Then
we have

$\Lambda^{\prime}(t)\geq
m\Lambda^{1+\frac{1}{m}}(t)e^{\frac{At}{m}}-A\Lambda(t)-2C_{1}m\epsilon_{0}b^{1-\frac{1}{m}}(t).$

Using $\displaystyle b(t)=\Lambda(t)e^{At}$, we get

$\displaystyle\displaystyle\Lambda^{\prime}(t)$
$\displaystyle\displaystyle\geq
m\Lambda^{1+\frac{1}{m}}(t)e^{\frac{At}{m}}-A\Lambda(t)-2C_{1}m\epsilon_{0}\Lambda^{1-\frac{1}{m}}(t)e^{At-\frac{At}{m}}$
$\displaystyle\displaystyle=m\Lambda^{1+\frac{1}{m}}(t)e^{\frac{At}{m}}\left[1-2C_{1}\epsilon_{0}\Lambda^{-\frac{2}{m}}(t)e^{-\frac{At}{m}}\right]-A\Lambda(t).$

We note that
$\displaystyle\Lambda^{\frac{2}{m}}(t)=b^{\frac{2}{m}}(t)e^{-\frac{2At}{m}}=a^{2}(t)e^{-\frac{2At}{m}}$.
We choose large enough $\displaystyle m>1$ such that $\displaystyle
e^{\frac{At}{m}}\leq\frac{5}{4}$ for all $\displaystyle t\in[0,T]$. From the
assumption on initial data, we have

$2C_{1}\epsilon_{0}\Lambda^{-\frac{2}{m}}(t)e^{-\frac{At}{m}}=2C_{1}\epsilon_{0}a^{-2}(t)e^{\frac{At}{m}}\leq\frac{5C_{1}}{2}\epsilon_{0}a^{-2}(t).$

Since $\displaystyle a^{\prime}(t)\geq 0$, we have $\displaystyle a(t)\geq
a(0)=\lambda_{max}-\delta$, from the choice of $\displaystyle\epsilon_{0}$ (as
in Theorem 2.5) we get

$2C_{1}\epsilon_{0}\Lambda^{-\frac{2}{m}}(t)e^{-\frac{At}{m}}\leq\frac{5C_{1}}{2}\epsilon_{0}(\lambda_{max}-\delta)^{-2}\leq\frac{1}{4}\mbox{
for small enough }\delta>0.$

Therefore, we have

$\displaystyle\displaystyle\Lambda^{\prime}(t)$
$\displaystyle\displaystyle\geq\frac{3m}{4}\Lambda^{1+\frac{1}{m}}(t)-A\Lambda(t)$
$\displaystyle\displaystyle\geq\frac{3m}{4}\Lambda^{1+\frac{1}{m}}(t)\left[1-\frac{4A}{3m}\Lambda^{-\frac{1}{m}}(t)\right].$

For large enough $\displaystyle m>1$, we have
$\displaystyle\frac{A}{m}\Lambda^{-\frac{1}{m}}(t)\leq\frac{1}{4}$. Hence,

$\Lambda^{\prime}(t)\geq\frac{m}{2}\Lambda^{1+\frac{1}{m}}(t).$ (5.28)

Therefore we have

$m\left[\frac{1}{\Lambda^{\frac{1}{m}}(0)}-\frac{1}{\Lambda^{\frac{1}{m}}(t)}\right]\geq\frac{mt}{2},$
(5.29)

or equivalently,

$\frac{1}{\Lambda^{\frac{1}{m}}(t)}\leq\frac{1}{\Lambda^{\frac{1}{m}}(0)}-\frac{t}{2}.$
(5.30)

Recall that
$\displaystyle\Lambda^{\frac{1}{m}}(t)=(b(t)e^{-At})^{\frac{1}{m}}=b^{\frac{1}{m}}(t)e^{-\frac{At}{m}}=a(t)e^{-\frac{At}{m}}$
and $\displaystyle\Lambda^{\frac{1}{m}}(0)=b^{\frac{1}{m}}(0)=a(0)$.
Therefore,

$\frac{1}{a(t)e^{-\frac{At}{m}}}\leq\frac{1}{a(0)}-\frac{t}{2}\mbox{ for
}t\in[0,t_{1}].$ (5.31)

Since (5.31) remains true for all $\displaystyle m$ sufficiently large, we
obtain

$\frac{1}{a(t)}\leq\frac{1}{a(0)}-\frac{t}{2}\mbox{ for all }t\in[0,t_{1}].$
(5.32)

Therefore,

$\frac{1}{a(t_{1})}\leq\frac{1}{a(0)}-\frac{t_{1}}{2}.$ (5.33)

Let $\displaystyle\nu(t)$ be defined as
$\displaystyle\nu(t):=\sup\limits_{\xi\in\mathbb{S}^{d-1}}[-\nabla\textbf{u}(t,X(t,x_{0})):\xi\otimes\xi]$.
We observe that $\displaystyle\nu(0)=a(0)$ and $\displaystyle\nu(t_{1})\geq
a(t_{1})$. Therefore,

$\frac{1}{\nu(t_{1})}\leq\frac{1}{a(t_{1})}\leq\frac{1}{a(0)}-\frac{t_{1}}{2}=\frac{1}{\nu(0)}-\frac{t_{1}}{2}.$
(5.34)

Now, we apply the same argument at time $\displaystyle t_{1}$ and
$\displaystyle x_{1}=X(t_{1},x_{0})$. In this case, we reuse the previous
method for direction $\displaystyle\xi_{1}\in\mathbb{S}^{d-1}$ satisfying
$\displaystyle\nu(t_{1})=[-\nabla\textbf{u}(t,X(t,x_{0})):\xi_{1}\otimes\xi_{1}]$.
We note that $\displaystyle\nu(t_{1})\geq\nu(0)$. Hence, there exists
$\displaystyle t_{2}\geq 2t_{1}$ such that (5.24) holds. We repeat the same
argument for time $\displaystyle t_{1},t_{2}$ we can obtain

$\frac{1}{\nu(t_{2})}\leq\frac{1}{\nu(t_{1})}-\frac{t_{2}-t_{1}}{2}.$ (5.35)

Since $\displaystyle\textbf{u},\pi$ belong to $\displaystyle
C^{1}([0,T]\times\mathbb{R}^{d})$, for any $\displaystyle t^{*}\in(0,T)$, we
have finitely many points $\displaystyle
0=t_{0}<t_{1}<t_{2}<\cdots<t_{k}=t^{*}$ such that

$\frac{1}{\nu(t_{j})}\leq\frac{1}{\nu(t_{j-1})}-\frac{t_{j+1}-t_{j}}{2}\mbox{
for all }1\leq j\leq k.$ (5.36)

Hence, we get

$\frac{1}{\nu(t_{*})}\leq\frac{1}{\nu(0)}-\sum\limits_{j=1}^{k}\frac{t_{j}-t_{j-1}}{2}=\frac{1}{\nu(0)}-\frac{t^{*}}{2},$
(5.37)

or equivalently,

$\nu(t^{*})\geq\frac{2\nu(0)}{2-\nu(0)t^{*}}.$ (5.38)

Hence, $\displaystyle\nu(t)\rightarrow\infty$ as $\displaystyle t\rightarrow
2/\nu(0)$. This completes the proof of Theorem 2.5.

∎

##  6 Break down of continuity of a solution

Adapting [14, Theorem 4.1.1], we have the following result.

######  Lemma 6.1.

Let $\displaystyle(\varrho,\textbf{u})\in C([0,T^{*})\times\mathbb{R}^{d})$ be
an admissible solution to (1.1)–(1.2) (as in Definition 2.1) corresponding to
initial data $\displaystyle(\varrho_{0},\textbf{u}_{0})$. Suppose
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ satisfies

$(\varrho_{0}(x),\textbf{u}_{0}(x))=(\overline{\varrho},\textbf{0})\mbox{ for
}\left|x\right|\geq R\mbox{ for some }R>0.$ (6.1)

Then we have

$(\varrho(t,\cdot),\textbf{u}(t,\cdot))=(\overline{\varrho},\textbf{0})\mbox{
for }\left|x\right|\geq R+\sigma t\mbox{ for }t\in[0,T^{*})\mbox{ and
}\sigma=\sqrt{\gamma}(\overline{\varrho})^{\frac{\gamma-1}{2}}.$ (6.2)

Though in [14, Theorem 4.1.1], the result has been proved for classical
$\displaystyle C^{1}$ solution but we can repeat the proof with the assumption
of entropy inequality (1.3).

######  Remark 6.2.

  1. 1.

We remark that if a weak solutions having Hölder regularity $\displaystyle
C^{\frac{1}{3}}$ or if it is continuous and BV the equality holds in (1.3). We
refer to [18] for details in this regards.

  2. 2.

We also remark that the result of finite speed of propagation remains valid
even for $\displaystyle L^{\infty}$ solutions but in that case we must take
$\displaystyle\sigma$ as follows

$\sigma=\sup\limits_{\xi\in\mathbb{S}^{d-1},(t,x)\in[0,T^{*})\times\mathbb{R}^{d}}\left|\textbf{u}(t,x)\cdot\xi\pm\sqrt{\gamma}(\varrho(t,x))^{\frac{\gamma-1}{2}}\right|.$
(6.3)

  3. 3.

We also note that the proof of Theorem 2.2 can also be replicated for
$\displaystyle L^{\infty}$ solutions but $\displaystyle L^{\infty}$–data
$\displaystyle\varrho_{0},\textbf{u}_{0}$ fails to satisfy the condition
(2.4). Indeed, we have $\displaystyle\left|\textbf{u}_{0}\right|\leq\sigma$
where $\displaystyle\sigma$ is defined as in (6.3). Subsequently,

$\frac{1}{\omega_{d}R^{d+1}}\int\limits_{B(\textbf{0},R)}\varrho_{0}\textbf{u}_{0}\cdot
x\,dx\leq\frac{\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\sigma}{\omega_{d}R^{d+1}}\int\limits_{B(\textbf{0},R)}\left|x\right|\,dx=\frac{\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\sigma}{d+1}.$

###### Proof of Lemma 6.1:.

Consider $\displaystyle\eta(\varrho,\textbf{m})$ and $\displaystyle
Q(\varrho,\textbf{m})$ as follows

$\displaystyle\displaystyle\eta(\varrho,\textbf{m})$
$\displaystyle\displaystyle=\frac{\left|\textbf{m}\right|^{2}}{2\varrho}+P(\varrho)-P^{\prime}(\overline{\varrho})(\varrho-\overline{\varrho})-P(\overline{\varrho}),$
(6.4) $\displaystyle\displaystyle Q(\varrho,\textbf{m})$
$\displaystyle\displaystyle=\frac{\textbf{m}}{\varrho}\left[\frac{\left|\textbf{m}\right|^{2}}{2\varrho}+P(\varrho)-P^{\prime}(\overline{\varrho})(\varrho-\overline{\varrho})-P(\overline{\varrho})\right].$
(6.5)

For $\displaystyle s\in\mathbb{R}$ and $\displaystyle\xi\in\mathbb{S}^{d-1}$
we define
$\displaystyle\Phi(s,\xi;\varrho,\textbf{m})=s\eta(\varrho,\textbf{m})-\xi\cdot
Q(\varrho,\textbf{m})$. We also set

$F_{\xi}(\varrho,\textbf{m})=\left(\textbf{m}\cdot\xi,\frac{\textbf{m}\cdot\xi}{\varrho^{2}}\textbf{m}+p(\varrho)\xi\right).$

We note that

$\displaystyle\displaystyle\Phi(s,\xi;\overline{\varrho},\textbf{0})=0,\quad\nabla_{\varrho,\textbf{m}}\Phi(s,\xi;\overline{\varrho},\textbf{0})=\textbf{0},$
(6.6)
$\displaystyle\displaystyle\nabla_{\varrho,\textbf{m}}^{2}\Phi(s,\xi;\overline{\varrho},\textbf{0})=\nabla^{2}\eta(\overline{\varrho},\textbf{0})\left[s\mathbb{I}_{d+1}-\nabla_{\varrho,\textbf{m}}F_{\xi}(\overline{\varrho},\textbf{0})\right].$
(6.7)

Let eigenvalues of $\displaystyle F_{\xi}(\overline{\varrho},\textbf{0})$ be
$\displaystyle\lambda_{1}(\overline{\varrho},\textbf{0})=-\sqrt{p^{\prime}(\overline{\varrho})},\lambda_{d+1}(\overline{\varrho},\textbf{0})=\sqrt{p^{\prime}(\overline{\varrho})}$
and $\displaystyle\lambda_{j}(\overline{\varrho},\textbf{0})=0,\,2\leq j\leq
d$. Let $\displaystyle\zeta_{j},1\leq j\leq d+1$ be eigenvectors corresponding
to $\displaystyle\lambda_{j},1\leq j\leq d+1$ respectively. Subsequently, we
have

$\nabla_{\varrho,\textbf{m}}^{2}\Phi(s,\xi;\overline{\varrho},\textbf{0}):\zeta_{j}\otimes\zeta_{j}=(s-\lambda_{j}(\overline{\varrho},\textbf{0}))\nabla^{2}\eta(\overline{\varrho},\textbf{0}):\zeta_{j}\otimes\zeta_{j}$
(6.8)

Hence, for $\displaystyle s=\sqrt{p^{\prime}(\overline{\varrho})}+\epsilon$
with $\displaystyle\epsilon>0$ we get

$\nabla_{\varrho,\textbf{m}}^{2}\Phi(\sqrt{p^{\prime}(\overline{\varrho})}+\epsilon,\xi;\overline{\varrho},\textbf{0}):\zeta_{j}\otimes\zeta_{j}\geq
C_{0}\epsilon\mbox{ for some }C_{0}>0.$

From the smoothness of
$\displaystyle(\varrho,\textbf{m})\mapsto\Phi(s,\xi;\varrho,\textbf{m})$ there
exists $\displaystyle\delta=\delta(\epsilon)>0$ such that

$\Phi(\sqrt{p^{\prime}(\overline{\varrho})}+\epsilon,\xi;\varrho,\textbf{m})\geq\frac{\epsilon}{4}\left[\left|\textbf{m}\right|^{2}+\left|\varrho-\overline{\varrho}\right|^{2}\right]\mbox{
for
}\left|\textbf{m}\right|+\left|\varrho-\overline{\varrho}\right|\leq\delta\mbox{
for all }\xi\in\mathbb{S}^{d-1}.$ (6.9)

######  Claim 6.3.

Let $\displaystyle(\varrho,\textbf{u})\in C([0,T^{*})\times\mathbb{R}^{d})$ be
an admissible solution to (1.1)–(1.2) (as in Definition 2.1) corresponding to
initial data $\displaystyle(\varrho_{0},\textbf{u}_{0})$. Suppose
$\displaystyle(\varrho_{0},\textbf{u}_{0})$ satisfies

$(\varrho_{0}(x),\textbf{u}_{0}(x))=(\overline{\varrho},\textbf{0})\mbox{ if
}x\cdot\xi\geq R\mbox{ for some }\xi\in\mathbb{S}^{d-1}\mbox{ and }R>0.$
(6.10)

Then we have

$(\varrho(t,\cdot),\textbf{u}(t,\cdot))=(\overline{\varrho},\textbf{0})\mbox{
for }x\cdot\xi\geq R+\sigma t\mbox{ for }t\in[0,T^{*})\mbox{ and
}\sigma=\sqrt{\gamma}(\overline{\varrho})^{\frac{\gamma-1}{2}}.$ (6.11)

###### Proof of Claim 6.3:.

We are going to prove this by contradiction. Fix $\displaystyle\epsilon>0$.
Suppose there exists a point
$\displaystyle(t_{0},x_{0})\in(0,T^{*})\times\mathbb{R}^{d}$ such that (i)
$\displaystyle x_{0}\cdot\xi\geq R+\sigma t_{0}$ and (ii)
$\displaystyle(\varrho(t_{0},x_{0}),\textbf{u}(t_{0},x_{0}))\neq(\overline{\varrho},\textbf{0})$.
Without loss of generality we can assume that $\displaystyle
t_{0}<\delta=\delta(\epsilon)$ if required we choose the minimal time
$\displaystyle t^{*}$ as $\displaystyle t_{0}$ when (i) and (ii) holds and we
replace $\displaystyle R$ by $\displaystyle
R+\sigma\left(t^{*}-\frac{\delta}{2}\right)$. Now consider the following set

$\mathcal{J}_{t_{0},x_{0}}:=\left\\{(t,x)\in[0,\infty)\times\mathbb{R}^{d});x\cdot\xi\geq
x_{0}\cdot\xi-\sigma_{\epsilon}(t_{0}-t),\left|x-x_{0}\right|\leq\sigma_{\epsilon}(t-t_{0}),t\in[0,t_{0}]\right\\},$
(6.12)

where $\displaystyle\sigma_{\epsilon}=\sigma+\epsilon$. We define
$\displaystyle\mathcal{J}^{n}:=\left\\{(t,x)\in\mathcal{J}_{t_{0},x_{0}};\mbox{dist}((t,x),\partial\mathcal{J}_{t_{0},x_{0}})\geq
1/n\right\\}$ where $\displaystyle\partial\mathcal{J}_{t_{0},x_{0}}$ denotes
the boundary of $\displaystyle\mathcal{J}_{t_{0},x_{0}}$. Consider
$\displaystyle\psi_{n}:[0,\infty)\times\mathbb{R}^{d}\rightarrow\mathbb{R}$
such that $\displaystyle\psi\in
C^{\infty}_{c}([0,\infty)\times\mathbb{R}^{d})$ satisfying

$\left.\begin{array}[]{ll}\psi_{n}(t,x)=1&\mbox{ if
}(t,x)\in\mathcal{J}^{n},\\\ 0\leq\psi_{n}(t,x)\leq 1&\mbox{ if
}(t,x)\in\mathcal{J}_{t_{0},x_{0}}\setminus\mathcal{J}^{n},\\\
\psi_{n}(t,x)=0&\mbox{ if
}(t,x)\in\mathcal{J}^{c}_{t_{0},x_{0}},\end{array}\right\\}$ (6.13)

where $\displaystyle A^{c}$ denotes the complement of set $\displaystyle A$ in
whole space. Taking $\displaystyle\varphi=\psi_{n}$ in (1.3) we obtain

$\int\limits_{\partial\mathcal{J}_{t_{0},x_{0}}}\Phi(s(t,y),\nu(t,y);\varrho(t,y),\textbf{m}(t,y))d\mathcal{S}_{t,y}\leq
0,$ (6.14)

where $\displaystyle d\mathcal{S}_{t,y}$ is the surface measure on
$\displaystyle\partial\mathcal{J}_{t_{0},x_{0}}$ and
$\displaystyle(s(t,y),\nu(t,y))$ is the outward normal at
$\displaystyle(t,y)$. We observe that
$\displaystyle\partial\mathcal{J}_{t_{0},x_{0}}=\mathcal{B}_{1}\cup\mathcal{B}_{2}\cup\mathcal{B}_{3}$
where

$\displaystyle\displaystyle\mathcal{B}_{1}$
$\displaystyle\displaystyle=\\{(0,x);x\cdot\xi\geq
x_{0}\cdot\xi-\sigma_{\epsilon}t_{0},\left|x-x_{0}\right|\leq\sigma_{\epsilon}t_{0}\\},$
$\displaystyle\displaystyle\mathcal{B}_{2}$
$\displaystyle\displaystyle=\\{(0,x);x\cdot\xi=x_{0}\cdot\xi-\sigma_{\epsilon}(t_{0}-t)\\},$
$\displaystyle\displaystyle\mathcal{B}_{3}$
$\displaystyle\displaystyle=\\{(0,x);x\cdot\xi\geq
x_{0}\cdot\xi-\sigma_{\epsilon}(t_{0}-t),\left|x-x_{0}\right|=\sigma_{\epsilon}(t_{0}-t)\\}.$

Therefore, we have

$(s(t,y),\nu(t,y))=\left\\{\begin{array}[]{rl}(-1,0)&\mbox{ if
}(t,y)\in\mathcal{B}_{1},\\\ (\sigma_{\epsilon},-\xi)&\mbox{ if
}(t,y)\in\mathcal{B}_{2},\\\
\left(\sigma_{\epsilon},\frac{y-x_{0}}{\sigma_{\epsilon}(t_{0}-t)}\right)&\mbox{
if }(t,y)\in\mathcal{B}_{3}.\end{array}\right.$

We also observe that $\displaystyle(t_{0},x_{0})$ is chosen such a way that
$\displaystyle\left|\varrho(t,x)-\overline{\varrho}\right|+\left|\textbf{m}(t,x)\right|\leq\delta$
for all $\displaystyle(t,x)\in\mathcal{J}_{t_{0},x_{0}}$. Hence, by using
(6.9) we get a contradiction. This concludes the proof of Claim 6.3. ∎

To complete the proof of Lemma 6.1 let us consider a point $\displaystyle
x_{0}\in\mathbb{R}^{d}$ such that $\displaystyle\left|x_{0}\right|=R$. Set
$\displaystyle\xi=x_{0}/\left|x_{0}\right|$. By Claim 6.3 we have
$\displaystyle(\varrho(t,x),\textbf{m}(t,x))=(\overline{\varrho},\textbf{0})$
for $\displaystyle x\cdot\xi\geq x_{0}\cdot\xi+\sigma t$. Note that
$\displaystyle x_{0}\cdot\xi=\left|x_{0}\right|=R$. We further observe that

$\\{(t,x)\in[0,\infty)\times\mathbb{R}^{d};\left|x-x_{0}\right|\geq R+\sigma
t\\}=\bigcup\limits_{\xi\in\mathbb{S}^{d-1}}\\{(t,x)\in[0,\infty)\times\mathbb{R}^{d};x\cdot\xi\geq
R+\sigma t\\}$ (6.15)

This completes the proof of Lemma 6.1. ∎

###  6.1 Proof of Theorem 2.2

###### Proof of Theorem 2.2:.

Let us consider $\displaystyle F(t)$ and $\displaystyle M(t)$ defined as
follows

$M(t):=\int\limits_{\mathbb{R}^{d}}(\varrho(t,x)-\bar{\varrho})\,dx\mbox{ and
}F(t):=\int\limits_{\mathbb{R}^{d}}x\cdot\varrho\textbf{u}\,dx.$

Let $\displaystyle\\{\eta_{\epsilon}\\}_{\epsilon>0}$ be Friedrich mollifiers.
Then, we consider $\displaystyle F_{\epsilon},M_{\epsilon}(t)$ defined as
follows

$\displaystyle\displaystyle
F_{\epsilon}(t)=\int\limits_{\mathbb{R}^{d}}x\cdot(\varrho\textbf{u})_{\epsilon}(t,\cdot)\,dx\mbox{
and
}M_{\epsilon}(t)=\int\limits_{\mathbb{R}^{d}}(\varrho_{\epsilon}(t,x)-\bar{\varrho})\,dx$
(6.16)

Then by Fundamental Theorem of Calculus we have

$\displaystyle\displaystyle
F_{\epsilon}(t_{2})-F_{\epsilon}(t_{1})=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}x\cdot\partial_{t}(\varrho\textbf{u})_{\epsilon}(t,\cdot)\,dxdt\mbox{
for }t_{2}>t_{1}>2\epsilon.$ (6.17)

We can mollify the system (1.1)–(1.2) with $\displaystyle\eta_{\epsilon}$ to
obtain

$\displaystyle\displaystyle\partial_{t}\varrho_{\epsilon}+\operatorname{div}_{x}(\varrho\textbf{u})_{\epsilon}$
$\displaystyle\displaystyle=0$ (6.18)
$\displaystyle\displaystyle\partial_{t}(\varrho\textbf{u})_{\epsilon}+\operatorname{div}_{x}\left(\varrho\textbf{u}\otimes\textbf{u}\right)_{\epsilon}+\nabla_{x}p(\varrho)_{\epsilon}$
$\displaystyle\displaystyle=0.$ (6.19)

By using (6.19) we get

$\displaystyle\displaystyle F_{\epsilon}(t_{2})-F_{\epsilon}(t_{1})$
$\displaystyle\displaystyle=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{N}}x\cdot\partial_{t}(\varrho\textbf{u})_{\epsilon}(t,\cdot)\,dxdt$
$\displaystyle\displaystyle=-\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}x\cdot\left[\operatorname{div}_{x}\left(\varrho\textbf{u}\otimes\textbf{u}\right)_{\epsilon}+\nabla_{x}p(\varrho)_{\epsilon}\right]\,dx$
$\displaystyle\displaystyle=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\left[\mathbb{I}_{d}:\left(\varrho\textbf{u}\otimes\textbf{u}\right)_{\epsilon}+d(p(\varrho)_{\epsilon}-p(\bar{\varrho}))\right]\,dx$
$\displaystyle\displaystyle=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\left[\left(\varrho\left|\textbf{u}\right|^{2}\right)_{\epsilon}+d(p(\varrho)_{\epsilon}-p(\bar{\varrho}))\right]\,dx.$
(6.20)

Similarly, with the help of (6.18) we obtain

$M_{\epsilon}(t_{2})-M_{\epsilon}(t_{1})=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\partial_{t}(\varrho_{\epsilon}-\bar{\varrho})\,dxdt=-\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\operatorname{div}_{x}(\varrho\textbf{u})_{\epsilon}\,dxdt=0.$
(6.21)

Passing to the limit in (6.20), (6.21) as $\displaystyle\epsilon\rightarrow
0$, we obtain

$F(t_{2})-F(t_{1})=\int\limits_{t_{1}}^{t_{2}}\int\limits_{\mathbb{R}^{d}}\left[\left(\varrho\left|\textbf{u}\right|^{2}\right)+d(p(\varrho)-p(\bar{\varrho}))\right]\,dxdt\mbox{
and }M(t_{2})=M(t_{1})\mbox{ for }t_{2}>t_{1}\geq 0$ (6.22)

Subsequently, we have

$\int\limits_{B(t)}(\varrho(t,\cdot)-\bar{\varrho})\,dx=\int\limits_{\mathbb{R}^{d}}(\varrho(t,\cdot)-\bar{\varrho})\,dx=M(t)=M(0)=\int\limits_{\mathbb{R}^{d}}(\varrho_{0}(\cdot)-\bar{\varrho})\,dx.$

Note that

$\int\limits_{B(t)}\varrho(t,\cdot)\,dx=M(t)+\bar{\varrho}\left|B(t)\right|=M(0)+\bar{\varrho}\left|B(t)\right|.$
(6.23)

Then, by using Jensen’s inequality we obtain

$\displaystyle\displaystyle\int\limits_{B(t)}p(\varrho)\,dx=\kappa\int\limits_{B(t)}\varrho^{\gamma}(t,\cdot)\,dx$
$\displaystyle\displaystyle\geq\kappa\left|B(t)\right|^{1-\gamma}\left(\int\limits_{B(t)}\varrho(t,\cdot)\,dx\right)^{\gamma}$
$\displaystyle\displaystyle=\kappa\left|B(t)\right|^{1-\gamma}\left(M(0)+\left|B(t)\right|\bar{\varrho}\right)^{\gamma}$
$\displaystyle\displaystyle\geq\kappa\left|B(t)\right|^{1-\gamma}\left(\left|B(t)\right|\bar{\varrho}\right)^{\gamma}$
$\displaystyle\displaystyle=\int\limits_{B(t)}p(\bar{\varrho})\,dx.$ (6.24)

Since, $\displaystyle\varrho,\textbf{u}\in C([0,T^{*})\times\mathbb{R}^{d})$,
we can pass $\displaystyle t_{1}\rightarrow 0$ and put $\displaystyle t_{2}=t$
to obtain

$F(t)=F(0)+\int\limits_{0}^{t}\int\limits_{\mathbb{R}^{d}}\left[\left(\varrho\left|\textbf{u}\right|^{2}\right)+d(p(\varrho)-p(\bar{\varrho}))\right]\,dxds.$
(6.25)

Note that supports of u and $\displaystyle p(\varrho)-p(\bar{\varrho})$ is
contained in $\displaystyle B(t)$. Hence, the following function,
$\displaystyle\Psi$ is well-defined,

$\Psi(s):=\int\limits_{\mathbb{R}^{d}}\left[\left(\varrho\left|\textbf{u}\right|^{2}\right)+d(p(\varrho)-p(\bar{\varrho}))\right](s,\cdot)\,dx.$
(6.26)

By our assumption $\displaystyle\Psi$ is continuous as well. Therefore, we
have $\displaystyle F\in C^{1}(0,T^{*})$ and

$F^{\prime}(t)=\int\limits_{\mathbb{R}^{d}}\left[\left(\varrho\left|\textbf{u}\right|^{2}\right)+d(p(\varrho)-p(\bar{\varrho}))\right](t,\cdot)\,dx\mbox{
for }t\in(0,T^{*}).$ (6.27)

From (6.24) we further obtain

$F^{\prime}(t)\geq\int\limits_{\mathbb{R}^{d}}\varrho\left|\textbf{u}\right|^{2}(t,\cdot)\,dx\mbox{
for }t\in(0,T^{*}).$ (6.28)

By using finite speed of propagation, Lemma 6.1, we observe that

$F(t)=\int\limits_{\mathbb{R}^{d}}x\cdot\varrho\textbf{u}\,dx=\int\limits_{B(t)}x\cdot\varrho\textbf{u}\,dx.$
(6.29)

By using Hölder’s inequality, we get

$F^{2}(t)\leq\left(\int\limits_{B(t)}\left|x\right|^{2}\varrho\,dx\right)\left(\int\limits_{B(t)}\varrho\left|\textbf{u}\right|^{2}\,dx\right).$
(6.30)

We note that

$\displaystyle\displaystyle\int\limits_{B(t)}\left|x\right|^{2}\varrho\,dx\leq(R+\sigma
t)^{2}\int\limits_{B(t)}\varrho(t,\cdot)\,dx$
$\displaystyle\displaystyle=(R+\sigma
t)^{2}\left[\int\limits_{B(t)}\bar{\varrho}\,dx+M(0)\right]$
$\displaystyle\displaystyle=(R+\sigma
t)^{2}\left[\int\limits_{B_{0}}\varrho_{0}\,dx-\int\limits_{B_{0}}\bar{\varrho}\,dx+\int\limits_{B(t)}\bar{\varrho}\,dx\right]$
$\displaystyle\displaystyle\leq\omega_{d}(R+\sigma
t)^{d+2}\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}.$
(6.31)

Combining (6.28), (6.30) and (6.31), we get

$F^{2}(t)\leq\omega_{d}(R+\sigma t)^{d+2}F^{\prime}(t).$

Therefore, by Fundamental Theorem of Calculus, we have

$-\frac{1}{d+1}\left[\frac{1}{(R+\sigma
t)^{d+1}}-\frac{1}{R^{d+1}}\right]\leq\omega_{d}\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\left[\frac{1}{F(0)}-\frac{1}{F(t)}\right],$

or equivalently,

$\frac{\omega_{d}\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}}{F(0)}\geq\frac{\omega_{d}\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}}{F(t)}+\frac{1}{(d+1)R^{d+1}}-\frac{1}{d+1}\frac{1}{(R+\sigma
t)^{d+1}}.$

If $\displaystyle t\rightarrow\infty$, we have

$\frac{\omega_{d}\left|\left|\varrho_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}}{F(0)}\geq\frac{1}{(d+1)R^{d+1}},$

which contradicts with (2.4). ∎

##  7 Global existence of smooth solution

In this section, we are going to prove the global existence of smooth solution
to (1.1)–(1.2) when initial data satisfying the condition  (G-1)– (G-3). Our
proof is relying on the technique of [25]. We first consider the following
vectorial Burgers equation,

$\left.\begin{array}[]{rl}\partial_{t}\textbf{v}+\textbf{v}\cdot\nabla_{x}\textbf{v}&=0,\mbox{
for }x\in\mathbb{R}^{d},t>0,\\\ \textbf{v}(0,x)&=\textbf{u}_{0}\mbox{ for
}x\in\mathbb{R}^{d}.\end{array}\right\\}$ (7.1)

By method of characteristics, we can solve the equation (7.1) for small enough
time $\displaystyle t_{0}$ and till the time of existence we have

$\textbf{v}(t,X(t,x_{0}))=\textbf{u}_{0}(x_{0})\mbox{ where
}X(t,x_{0})=x_{0}+t\textbf{u}_{0}(x_{0}).$ (7.2)

We can calculate

$\nabla_{x}\textbf{v}(t,X(t,x_{0}))=(\mathbb{I}_{d}+t\nabla_{x}\textbf{u}_{0}(x_{0}))^{-1}\nabla_{x}\textbf{u}_{0}(x_{0})\mbox{
for }t\in[0,t_{0}).$

Due to the assumption  (G-2), we can see that
$\displaystyle\left|\left|\nabla_{x}\textbf{v}(t,X(t,\cdot))\right|\right|_{L^{\infty}(\mathbb{R})}\leq
C_{1}\left|\left|\nabla_{x}\textbf{u}_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})}$
for some $\displaystyle C_{1}>0$ independent of $\displaystyle t$. Hence, the
solution $\displaystyle\textbf{v}(t,\cdot)$ exists for all time $\displaystyle
t>0$. We further have the following estimate from [25],

######  Proposition 7.1 (Grassin, [25]).

Let $\displaystyle\alpha>0$, $\displaystyle m>1+\frac{d}{2}$ and
$\displaystyle\textbf{u}_{0}:\mathbb{R}^{d}\rightarrow\mathbb{R}^{d}$ such
that the following holds

(A-1)

$\displaystyle\nabla_{x}^{2}\textbf{u}_{0}\in H^{m-1}(\mathbb{R}^{d})$ and
$\displaystyle\nabla_{x}\textbf{u}_{0}\in L^{\infty}(\mathbb{R}^{d})$.

(A-2)

$\displaystyle\inf\left\\{\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi;\,\xi\in\mathbb{S}^{d-1},x\in\mathbb{R}^{d}\right\\}\geq
0$ where
$\displaystyle\mathbb{S}^{d-1}=\\{x\in\mathbb{R}^{d};\,\left|x\right|=1\\}$.

We define

$\Omega_{\alpha}:=\\{x\in\mathbb{R}^{d};\,\inf\limits_{\xi\in\mathbb{S}^{d-1}}\nabla_{x}\textbf{u}_{0}(x):\xi\otimes\xi\geq\alpha\\}.$
(7.3)

Let $\displaystyle\overline{\textbf{u}}$ be the unique smooth solution to
(7.1) with initial data $\displaystyle\textbf{u}_{0}$. Then the following
holds,

  1. 1.

$\displaystyle\nabla_{x}\overline{\textbf{u}}=\frac{1}{1+t}\mathbb{I}_{d}+\frac{1}{(1+t)^{2}}\mathcal{R}(t,x)$
where $\displaystyle\mathcal{R}(t,x)$ is bounded on every compact subset of
$\displaystyle\Omega_{\alpha}(t)$ defined as

$\Omega_{\alpha}(t):=\left\\{x;\,x=x_{0}+t\textbf{u}_{0}(x_{0}),x_{0}\in\Omega_{\alpha}\right\\}.$
(7.4)

  2. 2.

$\displaystyle\left|\left|\nabla_{x}^{2}\overline{\textbf{u}}(t,\cdot)\right|\right|_{L^{2}(\Omega_{\alpha}(t))}\leq
c_{j}(1+t)^{\frac{d}{2}-j-1}$ for $\displaystyle j=2,\dots,m+1$.

  3. 3.

$\displaystyle\left|\left|\nabla_{x}^{2}\overline{\textbf{u}}(t,\cdot)\right|\right|_{L^{\infty}(\Omega_{\alpha}(t))}\leq
c_{0}(1+t)^{-3}$.

As in [25], following [40] we consider the quantity $\displaystyle\pi$ defined
as
$\displaystyle\pi=\sqrt{\frac{\gamma-1}{4\gamma}}\varrho^{\frac{\gamma}{2}}$
and the symmetrization of (1.1)–(1.2) as below,

$\displaystyle\displaystyle(\partial_{t}+\textbf{u}\cdot\nabla_{x})\pi+C_{1}\pi\operatorname{div}_{x}\textbf{u}$
$\displaystyle\displaystyle=0,$ (7.5)
$\displaystyle\displaystyle(\partial_{t}+\textbf{u}\cdot\nabla_{x})\textbf{u}+C_{1}\pi\nabla_{x}\pi$
$\displaystyle\displaystyle=0,$ (7.6)

with $\displaystyle C_{1}=\frac{\gamma-1}{2}$. The system (7.5)–(7.6) can be
written as quasi-linear form,

$\partial_{t}U+\sum\limits_{j=1}^{d}A_{j}(U)\partial_{j}U=0\mbox{ where
}U=(\varrho,\textbf{u})^{T},$ (7.7)

and $\displaystyle A_{j}(U)\in\mathbb{M}_{d+1}(\mathbb{R})$ is defined as

$A_{j}(U)=u_{j}\mathbb{I}_{d+1}+C_{1}\pi e_{j}\otimes e_{j}\mbox{ for
}j=1,\dots,d$ (7.8)

and $\displaystyle\\{e_{0},e_{1},\cdots,e_{d}\\}$ is the standard basis of
$\displaystyle\mathbb{R}^{d+1}$. Note that $\displaystyle\textbf{u}_{0}$ is
not integrable. Therefore, we approximate $\displaystyle\textbf{u}_{0}$ by
$\displaystyle\textbf{u}_{0}^{\lambda}$ as
$\displaystyle\textbf{u}^{\lambda}_{0}=\psi_{\lambda}\textbf{u}_{0}$ where
$\displaystyle\psi\in C_{0}^{\infty}(\mathbb{R}^{d})$ such that

$\psi_{\lambda}(x)=1\mbox{ for }\left|x\right|\leq\lambda.$

We choose $\displaystyle\lambda>0$ large enough such that $\displaystyle
supp(\varrho_{0})\subset\\{x\in\mathbb{R}^{d};\left|x\right|\leq\lambda\\}$.
For system (7.7) with initial data
$\displaystyle(\pi_{0},\textbf{u}_{0}^{\lambda})$, smooth solution exists in
$\displaystyle[0,t_{0})$ for small enough $\displaystyle t_{0}>0$ (see [9]).
Let us consider $\displaystyle\mathcal{Q}$ defined as

$\mathcal{Q}_{\alpha}=\\{(t,x)\in[0,\infty)\times\mathbb{R}^{d};x=y+t\nabla_{x}\textbf{u}_{0}(y),y\in\Omega_{\alpha},\left|y\right|\leq\lambda\\}.$

Similar to [25], we conclude that
$\displaystyle(\pi(t,x),\textbf{u}(t,x))=(0,\textbf{v}(t,x))$ for
$\displaystyle(t,x)\in[0,\infty)\times\left(\mathbb{R}^{d}\setminus\mathcal{Q}_{\alpha}\right)$
by using the following uniqueness result.

######  Proposition 7.2 (Grassin ,[25]).

Let $\displaystyle T_{0}>0$ and $\displaystyle W=(\pi,\textbf{u})\in
C^{1}([0,T_{0}];H^{m-1}(\mathbb{R}^{d}))\cap
C([0,T_{0}]H^{m}(\mathbb{R}^{d}))$ be a classical solution to (7.5)–(7.6)
corresponding to initial data $\displaystyle W_{0}\in H^{m}(\mathbb{R}^{d})$
for some $\displaystyle m>1+\frac{d}{2}$. Let $\displaystyle\overline{W}\in
C^{1}([0,T_{0}]\times\mathbb{R}^{d})$ also be a solution to (7.5)–(7.6)
corresponding to initial data $\displaystyle\overline{W}_{0}$ such that
$\displaystyle W_{0}(x)=\overline{W}_{0}(x)$ for $\displaystyle x\in
B(x_{0},r)$ for some $\displaystyle x_{0}\in\mathbb{R}^{d},r>0$. Consider
$\displaystyle M>0$ and $\displaystyle Q_{T}>0$ defined as

$\displaystyle\displaystyle M$
$\displaystyle\displaystyle=\sup\left\\{(C_{1}\left|\pi\right|+\left|\textbf{u}\right|)(t,x);\,(t,x)\in
D\right\\}\mbox{ where }D=B(x_{0},r)\times[0,T_{0}],$
$\displaystyle\displaystyle Q_{T}$
$\displaystyle\displaystyle=\left\\{(t,x);\,t\in[0,T],x\in
B(x_{0},r-Mt)\right\\}\mbox{ for }T\in[0,T_{1}]\mbox{ with
}T_{1}=\min\left\\{T_{0},r/M\right\\}.$

Then we have

$W(t,x)=\overline{W}(t,x)\mbox{ for }(t,x)\in Q_{T}.$ (7.9)

Now, from system (7.1) and (7.5)–(7.6) we obtain

$\left.\begin{array}[]{rl}(\partial_{t}+\textbf{w}\cdot\nabla_{x})\pi+C_{1}\pi\operatorname{div}_{x}\textbf{w}&=-\nabla_{x}\textbf{v}\cdot\pi-
C_{1}\pi\operatorname{div}_{x}\textbf{v},\\\
(\partial_{t}+\textbf{w}\cdot\nabla_{x})\textbf{w}+C_{1}\pi\nabla_{x}\pi&=-(\textbf{v}\cdot\nabla_{x})\textbf{w}-(\textbf{w}\nabla_{x})\textbf{v},\end{array}\right\\}$
(7.10)

where $\displaystyle\textbf{w}=\textbf{u}-\textbf{v}$. Let us denote
$\displaystyle V=(0,\textbf{v})$ and $\displaystyle U=(\pi,\textbf{w})$. Then
$\displaystyle U$ satisfies

$\partial_{t}U+\sum\limits_{j=1}^{d}F_{j}(U)\partial_{j}U=-G(\nabla_{x}V,U)-\sum\limits_{j=1}^{d}v_{j}\partial_{j}U,$
(7.11)

where $\displaystyle
G(\nabla_{x}V,U)=\begin{pmatrix}C_{1}\pi\operatorname{div}_{x}\textbf{v}\\\
\textbf{w}\cdot\nabla_{x}\textbf{v}\end{pmatrix}$. We define

$\Gamma_{k}(t):=\left[\int\limits_{\mathbb{R}^{d}}\nabla_{x}^{k}U(t,x)\cdot\nabla_{x}^{k}U(t,x)\,dx\right]^{\frac{1}{2}}.$
(7.12)

In order to get the decay estimate for w we consider,

$\Gamma(t)=\sum\limits_{k=0}^{m}(1+t)^{\delta_{k}}\Gamma_{k}(t)\mbox{ where
}\delta_{k}=k+b-a$ (7.13)

and $\displaystyle b$ is defined as below

$b=\left\\{\begin{array}[]{ll}1-\frac{d}{2}&\mbox{ if
}\delta\geq\delta_{c}=1+\frac{2}{d},\\\ \frac{\delta-1}{2}d-\frac{d}{2}&\mbox{
if }1<\delta<\delta_{c}.\end{array}\right.$ (7.14)

We will choose $\displaystyle a$ later. By using (7.11) we can obtain

$\frac{1}{2}\frac{d}{dt}\int\limits_{\mathbb{R}^{d}}\nabla_{x}^{k}U(t,x)\cdot\nabla_{x}^{k}U(t,x)\,dx=\int\limits_{\mathbb{R}^{d}}\mathcal{K}_{k}(U)(t,x)\,dx+\int\limits_{\mathbb{R}^{d}}\mathcal{A}_{k}(U,V)(t,x)\,dx,$

where

$\displaystyle\displaystyle\mathcal{K}_{k}(U)$
$\displaystyle\displaystyle=-\nabla_{x}^{k}U\cdot\left[\nabla_{x}^{k}\left(\sum\limits_{j=1}^{d}A_{j}(U)\partial_{j}U\right)-\sum\limits_{j=1}^{d}A_{j}(U)\partial_{j}\nabla_{x}^{k}U\right]+\frac{1}{2}\sum\limits_{j=1}^{d}\nabla_{x}^{k}U\cdot\partial_{j}A_{j}(U)\nabla_{x}^{k}U,$
$\displaystyle\displaystyle\mathcal{A}_{k}(U,V)$
$\displaystyle\displaystyle=-\nabla_{x}^{k}U\cdot\nabla_{x}^{k}\left(B(\nabla_{x}V,U)\right)-\nabla_{x}^{k}U\cdot\left[\nabla_{x}^{k}\left(\textbf{v}\cdot\nabla_{x}U\right)-\textbf{v}\cdot\nabla_{x}\nabla_{x}^{k}U\right]$
$\displaystyle\displaystyle+\frac{1}{2}\operatorname{div}_{x}(\textbf{v})\nabla_{x}^{k}U\cdot\nabla_{x}^{k}U.$

By using the similar argument as in [25, Proposition 3], we show that

######  Proposition 7.3.

Let $\displaystyle m>1+\frac{d}{2}$. Then there exists $\displaystyle C>0$
depending on $\displaystyle m,d$ and $\displaystyle\bar{C}>0$ depending on
$\displaystyle
m,d,\alpha,\left|\left|\nabla_{x}\textbf{u}_{0}\right|\right|_{L^{\infty}(\mathbb{R}^{d})},\left|\left|\nabla_{x}^{2}\textbf{u}_{0}\right|\right|_{H^{m-1}(\mathbb{R}^{d})}$
such that the following holds,

$\displaystyle\displaystyle\left|\int\limits_{\mathbb{R}^{d}}\mathcal{K}_{k}(t,x)\,dx\right|$
$\displaystyle\displaystyle\leq
C\left|\left|\nabla_{x}U\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\Gamma_{k}^{2}(t),$
$\displaystyle\displaystyle\frac{k+s}{1+t}\Gamma_{k}^{2}(t)+\int\limits_{\mathbb{R}^{d}}\mathcal{A}_{k}(U,V)(t,x)\,dx$
$\displaystyle\displaystyle\leq\bar{C}\Gamma_{k}(t)\Gamma(t)(1+t)^{-\delta_{k}-2}.$

Now, with the help of above estimates, we are going to prove Proposition 2.7.
The argument is similar to the proof of [25, Theorem 1].

###### Proof of Proposition 2.7:.

By using Proposition 7.3 we have

$\frac{1}{2}\frac{d}{dt}\Gamma_{k}^{2}(t)+\frac{k+s}{1+t}\Gamma_{k}^{2}\leq
C\left|\left|\nabla_{x}U\right|\right|_{L^{\infty}(\mathbb{R}^{d})}\Gamma_{k}^{2}(t)+\bar{C}\Gamma_{k}(t)\Gamma(t)(1+t)^{-\delta_{k}-2}.$
(7.15)

Setting $\displaystyle a=1+s+\frac{d}{2}>1$, we get

$\frac{d\Gamma(t)}{dt}+\frac{a}{1+t}\Gamma(t)\leq
C(\Gamma(t))^{2}+\frac{\bar{C}}{(1+t)^{2}}\Gamma(t).$ (7.16)

By [25, Proposition 4], there exists $\displaystyle\epsilon_{0}>0$ such that
$\displaystyle\Gamma(t)\leq C_{3}(1+t)^{-a}$ for some $\displaystyle C_{3}>0$
provided $\displaystyle\Gamma(0)<\epsilon_{0}$. This completes the proof of
Proposition 2.7. ∎

Acknowledgement. SSG would like to express thanks to the Department of Atomic
Energy, Government of India, under project no. 12-R&D-TFR-5.01-0520 for
support. AJ acknowledges the support of Fondation Sciences Mathématiques de
Paris-FSMP.

Disclosure statement: On behalf of all authors, the corresponding author
states that there is no conflict of interest.

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