close

Вход

Забыли?

вход по аккаунту

?

1361-6544%2Faa84c0

код для вставкиСкачать
Nonlinearity
Related content
PAPER
Vanishing pressure limit for compressible
Navier–Stokes equations with degenerate
viscosities
To cite this article: Zhilei Liang 2017 Nonlinearity 30 4173
- Global strong solution to the twodimensional density-dependent nematic
liquid crystal flows with vacuum
Lin Li, Qiao Liu and Xin Zhong
- Global classical solutions to the 3D
isentropic compressible Navier–Stokes
equations in a bounded domain
Haibo Yu and Junning Zhao
- On the non-resistive limit and the magnetic
boundary-layer for one-dimensional
compressible magnetohydrodynamics
Song Jiang and Jianwen Zhang
View the article online for updates and enhancements.
This content was downloaded from IP address 129.16.69.49 on 29/10/2017 at 05:31
London Mathematical Society
Nonlinearity 30 (2017) 4173–4190
Nonlinearity
https://doi.org/10.1088/1361-6544/aa84c0
Vanishing pressure limit for compressible
Navier–Stokes equations with degenerate
viscosities
Zhilei Liang
School of Economic Mathematics, Southwestern University of Finance
and Economics, Chengdu 611130, People’s Republic of China
E-mail: zhilei0592@gmail.com
Received 14 January 2017, revised 27 July 2017
Accepted for publication 8 August 2017
Published 12 October 2017
Recommended by Professor Koji Ohkitani
Abstract
In this paper we study a vanishing pressure process for highly compressible
Navier–Stokes equations as the Mach number tends to infinity. We first prove
the global existence of weak solutions for the pressureless system in the
framework (Li and Xin 2015 arXiv:1504.06826v2), where the weak solutions
are established for compressible Navier–Stokes equations with degenerate
viscous coefficients. Furthermore, a rate of convergence of the density in
L∞ 0, T; L2 (RN ) is obtained, in case when the velocity corresponds to the
gradient of density at initial time.
Keywords: compressible Navier–Stokes, mach number limit, pressureless,
global weak solutions, density-dependent viscosities
Mathematics Subject Classification numbers: 76N10, 35Q35
1. Introduction
The time evolution of a viscous compressible barotropic fluid occupying the whole space
RN (N = 2, 3) is governed by the equations
∂t ρε + div(ρε uε ) = 0,
(1)
∂t (ρε uε ) + div(ρε uε ⊗ uε ) + ∇Pε − divSε = 0,
where the unknown functions ρε and uε are the density and the velocity. The pressure Pε = εργε
with γ > 1 is given and ε > 0 is related to Mach number, and the stress tensor takes the form
1361-6544/17/114173+18$33.00 © 2017 IOP Publishing Ltd & London Mathematical Society Printed in the UK
4173
Z Liang
Nonlinearity 30 (2017) 4173
Sε = h(ρε )∇uε + g(ρε )divuε I,
(2)
in which I is the identical matrix, h and g are functions of ρε satisfying the physical restrictions
h(ρε ) > 0, h(ρε ) + Ng(ρε ) 0.
(3)
For simplicity reasons, in this paper we assume
h(ρε ) = ρα
g(ρε ) = (α − 1)ρα
for α > (N − 1)/N, ε ∈ (0, 1).
ε,
ε,
(4)
The initial functions are imposed as
ρε (x, t = 0) = ρ0 0, ρε uε (x, t = 0) = m0 , x ∈ RN .
(5)
For fixed ε, equation (1) is one of the most important mathematical models describing the
motion of a viscous flow. There are many papers on the existence and asymptotics of solutions
because of its mathematical challenges and wide physical applications; see [3, 6, 7, 16, 18,
20–22] and the references therein. Consider the constant viscosities h and g defined in (3), and
more general symmetric stress tensor
∇uε + (∇uε )tr
Sε = h
+ gdivuε I,
2
Lions [20] first proved the global existence of weak solutions of (1) if the adiabatic index
γ 3N/(N + 2). Later, γ was relaxed by Feireisl et al [7] to γ > N/2 and by Jiang and
Zhang [13] to γ > 1 under some extra spherically symmetry assumptions. While for the case
when viscosities are density-dependent, Bresch and Desjardins [4, 5] developed a new entropy
structure on condition that
g(ρε ) = ρε h (ρε ) − h(ρε ).
This gives an estimate on the gradient of density, and thereby, some further compactness
information on density. Li et al [16] proposed the global entropy weak solution to system (1)
in one-dimensional bounded interval and studied the vacuum vanishing phenomena in finite
time span. Similar results in [16] were extended to the Cauchy problem in [14] by Jiu and Xin.
Guo et al [10] obtained the global existence of weak solution to (1) if some spherically symmetric assumptions are made. However, the problem becomes much more difficult in general
high dimension spaces. Mellet and Vasseur [21] provided a compactness framework which
ensures the existence of weak solutions as a limit of approximation solutions, but leaves such
approximations sequence open in [21]. Until recently, the problem was solved in two impressive papers by Vasseur and Yu [22] and Li and Xin [18], where they constructed separately
appropriate approximations from different approaches. Vasseur and Yu [23] also considered
the compressible quantum Navier–Stokes equations with damping, which helps to understand
the existence of global weak solutions to the compressible Navier–Stokes equations.
Referring to [18, 21], we give the weak solution of system (1) in below
Definition 1.1. For fixed ε > 0, we call (ρε , uε ) a weak solution to the problem (1)–(5), if

0 ρε ∈ L∞ 0, T; L1 (RN ) ∩ Lγ (RN ) ,


 α−1/2 √



∇ρε
, ρε uε ∈ L∞ 0, T; L2 (RN )N ,


 (γ+α−1)/2
∇ρε
∈ L2 0, T; L2 (RN )N ,

−1,1
2
N N×N

h(ρ
0,
T;
W
,
)∇u
∈
L
(R
)

ε
ε
loc




g(ρ )divu ∈ L2 0, T; W −1,1 (RN ) ,

ε
ε
loc
4174
Z Liang
Nonlinearity 30 (2017) 4173
√
( ρε , uε ) satisfy (1)1 in distribution sense, and the integral equality
T
√
√
√ √
m0 φ(x, 0) +
ρε ( ρε uε )∂t φ + ρε uε ⊗ ρε uε : ∇φ + εργε divφ
RN
RN
0
=h(ρε )∇uε , ∇φ + g(ρε )divuε , divφ
holds true for any test functions φ ∈ C0∞ RN × [0, T) N , where
h(ρε )∇uε , ∇φ = −
T
0
RN
√
ρα−1/2
ρε uε φ +
ε
2α √
ρε ujε ∂i ρα−1/2
∂i φj
ε
2α − 1
and
g(ρε )divuε , divφ
T
= −(α − 1)
0
RN
√
ρα−1/2
ρε uε · ∇divφ +
ε
2α √
ρε uε ∇ρα−1/2 divφ .
2α − 1
The following important existence results of weak solutions are obtained in [18] by Li and
Xin.
Proposition 1.1. ([18]) Assume that the initial function in (5) satisfies

α−1/2
1
N
γ
N

∇ρ0
∈ L2 (RN ),
0 ρ0 ∈ L (R ) ∩ L (R ),
m0 ∈ L2γ/(γ+1) (RN ),
ρ0 ≡ 0, m0 = 0 a.e. on {x ∈ RN | ρ0 = 0},

 −(1+η0 )
ρ0
|m0 |2+η0 ∈ L1 (RN )
for some η0 > 0.
(6)
Additionally, assume that for N = 2
α > 1/2, γ > 1, γ 2α − 1,
(7)
and for N = 3

γ ∈ (1, 3),
γ ∈ (1, 6α − 3) if α ∈ [3/4, 1],
(8)

4
1
3
γ ∈ [2α − 1, 3α − 1] and ρ−3
0 |m0 | ∈ L (R ) if α ∈ (1, 2).
Then the problem (1)–(5) has global weak solutions (ρε , uε ) in the sense of definition 1.1.
Remark 1.1. By (8), if we multiply (1)2 by 4|uε |2 uε and compute directly, we infer
4
∞
ρ
0, T; L1 (R3 ) .
(9)
ε |uε | ∈ L
Proof. The rigorous proof is available in the appendix.
□
It seems rather natural to expect that, as ε → 0+, the limit (ρ, u) of (ρε , uε ) satisfies the
corre­sponding pressureless system
∂t ρ + div(ρu) = 0,
(10)
∂t (ρu) + div(ρu ⊗ u) − div (h(ρ)∇u + g(ρ)divuI) = 0.
As in [11], we define the weak solution (called quasi-solution) to system (10)
4175
Z Liang
Nonlinearity 30 (2017) 4173
Definition 1.2. The function (ρ, u) is called a quasi-solution, if

0 ρ ∈ L∞ 0, T; L1 (RN ) ,


 α−1/2 √


, ρu ∈ L∞ 0, T; L2 (RN )N ,
∇ρ
−1,1
h(ρ)∇u ∈ L2 0, T; Wloc
(RN )N×N ,




−1,1

g(ρ)divu ∈ L2 0, T; Wloc
(RN ) ;
√
in addition, ( ρ, u) satisfy (10)1 in distribution sense, and the integral equality
T
√
√
√ √
m0 φ(x, 0) +
ρ( ρu)∂t φ + ρu ⊗ ρu : ∇φ
RN
0
RN
=h(ρ)∇u, ∇φ + g(ρ)divu, divφ
holds true, where the quantities on the right side are defined as the same of h(ρε )∇uε , ∇φ
and g(ρε )divuε , divφ.
Remark 1.2. The quasi-solution in definition 1.2 was first proposed by Haspot to approximate in some sense the compressible Navier–Stokes equations [11].
In this paper, we choose ε = η −2 > 0 with η being the Mach number. The readers can refer
to [9, 11, 19, 20] for more information in this aspect. There are satisfactory results on the incompressible limit when η → 0, we refer readers to the pioneering works by Desjardins et al and
Lions and Masmoudi [2, 3, 19] when the viscous coefficients are constant. Regretfully, results
are rarely available up to publication when η → ∞. One major difficulty is the compactness lack
of the density because its Lγ-bound is no longer conserved for constant viscosities. However,
the case of density-dependent viscosity is much different due to the new BD entropy inequality.
Haspot [11] proved the highly compressible limit (ε → 0) in the sense of distribution in suitable
Lebesgue spaces, and discussed the global existence of quasi-solutions as a convergence limit
from approximation solutions of system (1), although such approx­imations are only a priori
exist. It is worthy mentioning that in [11] the author constructed a family of explicit solutions
(ρ, u) with ρ satisfying the porous medium equation, the heat equation, or the fast diffusion
equation, up to the choice of α. Haspot and Zatorska [12] consider the one-dimensional Cauchy
problem and obtain a rate of convergence of ρε and other related properties.
We are interested in the limit procedure for the weak solutions (ρε , uε ) of (1) as ε tends to
zero, and then get some convergence rate the solutions. In particular, on the basis of existence
results obtained in [18] by Li and Xin, we adopt some ideas in [11] and [21] and first show
the quasi-solutions stability for the solutions (ρε , uε ) of (1). Secondly, in the spirit of [12], we
obtain a convergence rate of ρ − ρε in terms of ε in high dimensions by the argument of duality, as long as the initial velocity associated with the gradient of initial density.
Theorem 1.1. Let the conditions (6)–(8) in proposition 1.1 hold true. Then, for α 1, the
solution (ρε , uε ) of (1) converges to a limit function (ρ, u) which solves (10) in the sense of
definition 1.2. Furthermore,
q1
ρε → ρ in C [0, T]; Lloc
(RN ) ,
(11)
q2
ρε uε → ρu in L2 0, T; Lloc
(RN ) ,
(12)
where q1 ∈ [1, ∞), q2 ∈ [1, 2) if N = 2 ; q1 ∈ [1, 6α − 3), q2 ∈ [1, 12α−6
if N = 3.
6α−1 )
4176
Z Liang
Nonlinearity 30 (2017) 4173
Remark 1.3. The assumptions in proposition 1.1 guarantee the existence of (ρε , uε ) to (1),
whose proof are available in [18]. We allow more general viscosities at the cost of stress tensor
having the form (2), although it seems not appropriate from a physical point of view.
Remark 1.4. In case of α = 1, theorem 1.1 is valid for the symmetric viscous stress tensor
tr
ε)
Sε = div ρε ∇uε +(∇u
, where the existence of (ρε , uε ) are achieved in [18, 22]. Moreover,
2
the case α < 1 can also be discussed by modifying slightly the argument in theorem 1.1.
Theorem 1.2. In addition to the assumptions made in theorem 1.1, let
u0 + αρα−2
∇ρ0 = 0.
(13)
0
Then there is a positive C which may depend on T such that for α 3/2
sup |(ρε − ρ)(·, t)L2 (RN ) Cεσ ,
0tT
where σ <
1
2(2α−1)
if N = 2 , σ =
4α−3
4(2α−1)2
if N = 3.
Remark 1.5. For one-dimensional problem, Haspot and Zatorska [12] first obtained a rate
of convergence of ρε − ρ in suitable Sobolev spaces for 1 < α 3/2. We remark that the
argument in [12] relies heavily on the upper bound of density.
In the rest of this paper, section 2 is for some useful lemmas and sections 3 and 4 are
devoted to proving theorems 1.1 and 1.2 respectively.
2. Preliminaries
Lemma 2.1. (see [8, 15]) Let BR = {x ∈ RN : |x| < R}. For any v ∈ W 1,q (BR ) ∩ Lr (BR ),
it satisfies
γ
1−γ
v
(14)
L p (BR ) C1 vLr (BR ) + C2 ∇vLq (BR ) vLr (BR ) ,
where the constant Ci (i = 1, 2) depends only on p, q, r, γ; and the exponents 0 γ 1,
1 q, r ∞ satisfy 1p = γ( 1q − N1 ) + (1 − γ) 1r and

Nq
Nq

min{r, N−q } p max{r, N−q }, if q < N;
r p < ∞,
if q = N;

r p ∞,
if q > N.
The following L p -bound estimate is taken from [17, lemma 2.4], whose proof is available by
adopting [6, lemma 12] and the elliptic theory due to Agmon et al [1].
Lemma 2.2. ([17, lemma 2.4]) Let p ∈ (1, +∞) and k ∈ N. Then for all v ∈ W 2+k,p (BR )
with 0-Dirichlet boundary condition, it holds that
∇2+k vL p (BR ) CvW k,p (BR ) ,
where the C relies only on p and k.
Lemma 2.3. Assume that f is increasing and convex in R+ = [0, +∞) with f (0) = 0 . Then,
|x − y| f (|x − y|) (x − y)( f (x) − f (y)),
4177
∀ x, y ∈ R+ .
Z Liang
Nonlinearity 30 (2017) 4173
Proof. Define F(x) = f (x) − f (y) − f (x − y). Since f is convex, then F (x) 0 for
x y 0 . This and F(0) = 0 deduce F(x) 0. Hence,
f (|x − y|) = f (x − y) f (x) − f (y) = | f (x) − f (y)|.
Repeating the argument when y x 0 , we obtain
f (|x − y|) | f (x) − f (y)|,
∀ x, y ∈ R+ .
This, along with the monotonicity of f, leads to
|x − y| f (|x − y|) |x − y|| f (x) − f (y)| = (x − y)( f (x) − f (y)),
□
the required.
3. Proof of theorem 1.1
In what follows, the operations are based on hypotheses imposed in proposition 1.1, and the
generic constant C > 0 is ε independent.
Firstly, for all existing time t 0, we have
ρ
ε (·, t)L1 (RN ) = ρ0 L1 (RN )
(15)
and
sup
0tT
RN
RN
2
ρε |uε | +
|m0 |2
+ε
ρ0
RN
εργε
ργ0 .
+
T
0
RN
2
α
2
ρα
ε |∇uε | + (α − 1)ρε (divuε )
(16)
Following in [21], a straight calculation shows
T
α−2
2
γ
sup
ρε |uε + αρε ∇ρε | + ερε + αγε
ρα+γ−3
|∇ρε |2
ε
0tT RN
0
RN
2
ρ0 |u0 + αρα−2
∇ρ
|
+
ε
ργ0 .
0
0
RN
(17)
RN
The initial condition (6) and (15)–(17) guarantee
2 sup
ρε + ∇ρα−1/2
dx C.
(18)
ε
0tT
RN
We claim that
ρε ∈ L∞ 0, T; L1 (RN ) ∩ Lq (RN ) , ∇ρα−1/2
∈ L∞ 0, T; L2 (RN ) ,
(19)
ε
where q < ∞ if N = 2, and q = 6α − 3 if N = 3.
Proof. [Proof of (19)]. If α 3/2, by (14) we have
ρα−1/2
L p (RN ) C ρα−1/2
L(α−1/2)−1 (RN ) + ∇ρα−1/2
L2 (RN )
ε
ε
ε
α−1/2
2 (RN )
C ρε L1 (RN ) + ∇ρα−1/2
,
L
ε
4178
(20)
Z Liang
Nonlinearity 30 (2017) 4173
where p (α − 1/2)−1 for N = 2 and p = 6 for N = 3. While for α > 3/2, by (14) and
interpolation theorem, one has
α−1/2
α−1/2
p (RN ) C
1 (RN ) + C∇ρ
2 (RN )
ρα−1/2
ρ
L
L
L
ε
ε
ε
(1−θ)(α−1/2)
α−1/2 θ
C ρε L1 (RN )
ρε
L p (RN ) + ∇ρα−1/2
L2 (RN ) ,
ε
(21)
N = 2 and p = 6 if N = 3. The (19) thus follows from (18),
where θ = p(α−1/2)−p
p(α−1/2)−1 , p > 1 if
(20) and (21).
The key issue in proving theorem 1.1 is to get the ε -independent estimates and take ε -limit
in definition 1.1. In terms of (8) and (19), one has
T
ε
ργε divφ → 0 as ε → 0.
0
RN
Besides, we also need to justify (11) and (12) and the strong convergence of
purpose it suffices to prove the lemmas 3.1–3.3 below.
√
ρε uε . For that
□
Lemma 3.1. Upon to some subsequence, it satisfies
q1
ρε → ρ in C [0, T]; Lloc
(RN ) ,
(22)
where q1 ∈ [1, ∞) if N = 2 and q1 ∈ [1, 6α − 3) if N = 3.
Proof. By (19) and Hölder inequality, we have
1/2
α
α−1/2
∇ρ
L2 (RN ) C.
(23)
ε L1 (RN ) Cρε L1 (RN ) ∇ρε
Since 1 α 2α − 1 < 6α − 3, from (19) and (16) we deduce
T
1
N
ρα
u
+
ρα
ε ε L (R )
ε divuε L1 (RN )
0
√
α−1/2
ρε
L2 (RN ) ρε uε L2 (RN )
T
α/2
+ sup ρε L2 (RN )
ρα/2
ε divuε L2 (RN )
0tT
0
C.
This, along with
α
α
∂ t ρα
(24)
ε = (1 − α)ρε divuε − div(ρε uε ),
−1,1
2
0, T; Wloc
(RN ) . By the Aubin–Lions lemma, we get
ensures that ∂t ρα
ε ∈L
β
α
N
ρα
→
ρ
in
C
[0,
T];
L
(R
)
for β ∈ [1, 3/2).
ε
loc
Therefore, up to some subsequence,
α
α
ρ
almost everywhere.
(25)
ε →ρ ,
4179
Z Liang
Nonlinearity 30 (2017) 4173
q
So, (19) and (25) guarantee the strong convergence of ρε to ρ in L∞ 0, T; Lloc (RN )
with q ∈ [1, ∞) if N = 2 and q ∈ [1, 6α − 3) if N = 3. Choosing α = 1 implies
−1,1
∂t ρε ∈ L2 0, T; Wloc
(RN ) , we conclude (22) by the Aubin–Lions lemma.
□
√
Consequently, the (19) and (25) implies that
√
2
ρε ρ in L2 0, T; Lloc
(RN ) , ρεα−1/2 ρα−1/2
1
in L2 0, T; Hloc
(RN ) .
Lemma 3.2. Upon to some subsequence, it satisfies
√
√
2
ρε uε → ρu in L2 0, T; Lloc
(RN ) .
(26)
Proof. The process is divided into several steps.
□
(1+α)/2
uε with χ(x) being smooth and
Step 1. Define mε = χ(ρε )ρα
ε + (1 − χ(ρε ))ρε
satisfying χ(x) = 1 if |x| 1 and χ(x) = 0 if |x| 2.
We claim that, for p ∈ [1, 3/2),
p
mε → χ(ρ)ρα + (1 − χ(ρ))ρ(1+α)/2 u in L2 0, T; Lloc
(RN ) .
Consequently,
mε → χ(ρ)ρα + (1 − χ(ρ))ρ(1+α)/2 u, almost everywhere.
(27)
In fact, we deduce from (19) and (16) that
T
0
2
∇(χ(ρε )ρα
ε uε )L1 (RN )
C
C
C
and
T
0
T
0
T
0
√
α−1/2
α
2
2
ρα
ρ
|u
|∇ρ
|
+
χρ
|∇u
|
|u
||∇ρ
|
+
χ
1
N
1
ε ε
ε L (R )
ε L ({1ρε 2})
ε
ε
ε ε
√
2
α/2
2
ρε uε 2L2 ∇ρα−1/2
2L2 (RN ) + ρα/2
ε
ε L2 (RN ) ρε ∇uε L2 (RN )
∇((1 − χ(ρε ))ρ(1+α)/2
uε )2L1 (RN ) dt
ε
C
0
+C
C
T
C,
0
T
ρ(1+α)/2
|uε ||∇ρε |2L1 ({1ρε 2})
ε
T
0
(1 − χ(ρε ))(ρ(2−α)/2
|uε |∇ρα−1/2
| + ρ(1+α)/2
|∇uε |)2L1 ({1ρε })
ε
ε
ε
√
√
2
ρε uε 2L2 (RN ) ∇ρα−1/2
2L2 (RN ) + ρε 2L2 (RN ) ρα/2
ε
ε ∇uε L2 (RN )
(28)
4180
Z Liang
Nonlinearity 30 (2017) 4173
1/2
(2−α)/2
where we have used 1{ρε 1} ρε
1{ρε 1} ρε
guarantees
∇mε ∈ L2 0, T; L1 (RN ) .
since α 1. The last two inequalities
Furthermore, if
−1,1
2
∂
0, T; Wloc
(RN ) .
(29)
t mε ∈ L
3/2
Then, the Aubin–Lions lemma shows there is a m ∈ L2 0, T; Lloc (RN ) such that
mε → m, almost everywhere.
(30)
√
This combining with (25) and ρε uε ∈ L∞ 0, T; L2 (RN ) provides
2
m2
(ρα
ε uε )
=
lim
inf
lim
inf
ρε |uε |2 C.
2α−1
ε→0
ρ2α−1
{ρε 1} ρ
{ρε 1} ε→0
RN
ε
So, m = 0 on vacuum sets. We define u = m χ(ρ)ρα + (1 − χ(ρ))ρ(1+α)/2 −1 if ρ > 0 and
u = 0 if ρ = 0 . The proof is thus completed.
We need to check (29). Let us first prove
∂t (1 − χ(ρε ))ρ(1+α)/2
uε ∈ L2 0, T; W −1,1 (RN ) .
(31)
ε
By (1) and (24), a careful calculation shows
∂t (1 − χ(ρε ))ρ(1+α)/2
u
ε
ε
2
χ ρε uε ∂t ρ(α+1)/2
+ (1 − χ(ρε ))∂t (ρ(1+α)/2
uε )
ε
ε
α+1
2
α − 1 (α+1)/2
(α+1)/2
=
χ ρε uε
ρε
divuε + div(ρε
uε )
α+1
2
3 − α (α−1)/2
(α−1)/2
+ (1 − χ(ρε ))ρε uε
ρε
divuε − div(ρε
uε )
2
=−
α
γ
[div(ρα
+ (1 − χ(ρε ))ρ(α−1)/2
ε
ε ∇uε + (α − 1)ρε divuε I) − ε∇ρε − div(ρε uε ⊗ uε )] .
(32)
The terms in (32) are dealt with as follows: firstly,
α − 1 (α+1)/2
(α+1)/2
ρε
χ ρε uε
divuε + div(ρε
uε )
2
α + 1 (α+3)/2
α+1
χ ρε
uε uε · ∇χ(ρ(α+3)/2
=
uε divuε +
)
ε
2
α+3
α+1
α + 1 (α+3)/2
χ ρε
(1 − χ(ρ(α+3)/2
uε divuε +
))(uε · ∇uε + uε divuε )
=
ε
2
α+3
α+1
∂j ((1 − χ(ρ(α+3)/2
−
))ukε uεj )
ε
α+3
∈ L2 0, T; W −1,1 (RN ) ,
where we have used
4181
Z Liang
Nonlinearity 30 (2017) 4173
(1 − χ(ρ(α+3)/2
))ukε uεj 2L1 (RN )
ε
T
+
ρ(α+3)/2
χ (ρε )uε divuε + (1 − χ(ρ(α+3)/2
))(uε · ∇uε + uε divuε )2L1 (RN )
ε
ε
0
|uε |2 2L1 ({1ρ(α+3)/2 })
ε
T
|uε ||divuε |2L1 ({1ρε 2}) + |uε ||∇uε |2L1 ({1ρ(α+3)/2 })
+
ε
0
Cρε |uε |2 2L1 (RN ) +
C,
T
0
√
2
ρε uε 2L2 (RN ) ρα/2
ε ∇uε L2 (RN )
owes to (16).
Secondly, by virtue of (16) and (19),
(1 − χ(ρε ))ρ(α−1)/2
div(ρα
ε
ε ∇uε )
ρα/2
= div((1 − χ(ρε ))ρα−1/2
ε
ε ∇uε ) −
+ χ ρ(3α−1)/2
∇ρε · ∇uε
ε
α−1
α−1/2
(1 − χ(ρε ))ρα/2
ε ∇uε ∇ρε
2α − 1
(33)
∈ L2 (0, T; W −1,1 (RN )).
By similar argument, we receive
Next,
2
−1,1
(1 − χ(ρε ))ρ(α−1)/2
div(ρα
(RN )).
ε
ε divuε I) ∈ L (0, T; W
(1 − χ(ρε ))ρε uε
3 − α (α−1)/2
ρε
divuε − div(ρ(α−1)/2
uε )
ε
2
− (1 − χ(ρε ))ρ(α−1)/2
div(ρε uε ⊗ uε )
ε
2
1−α
(1 − χ(ρε ))ρ(α+1)/2
(1 − χ(ρ(α+3)/2
uε divuε −
))(uε divuε + uε · ∇uε )
=
ε
ε
2
α+3
2
(α+3)/2
k j
k j (α+1)/2
(1 − χ(ρε
+ ∂j
))uε uε − (1 − χ(ρε ))uε uε ρε
α+3
∈ L2 (0, T; W −1,1 (RN )),
where the following inequality has been used
T
(1 − χ(ρε ))ρ(α+1)/2
|uε |2 L1 (RN )
ε
0
T
(α+1)/2
∇((1 − χ(ρε ))ρε
|uε |)2L1 (RN ) + uε 2L2 ({ρε 1}) ,
0
C T 1/2
α/2
ρε |uε |2 2L2 + ρε 2L2 ,
0
C,
due to (9), (19) and (28).
4182
N=2
N=3
Z Liang
Nonlinearity 30 (2017) 4173
Finally, since (19) and (2γ + α − 1)/2 ∈ (1, 6α − 3), it has
(1 − χ(ρε ))ρ(α−1)/2
∇ργε
ε
2γ
∇((1 − χ(ρε ))ρ(2γ+α−1)/2
=
) + ρ(2γ+α−1)/2
χ∇ρε
ε
ε
2γ + α − 1
∈ L∞ (0, T; W −1,1 (RN )).
A similar argument yields
2
−1,1
∂t (χ(ρε )ρα
(RN ) ,
ε uε ) ∈ L 0, T; W
which combining with (31) gives the desired (29).
Step 2. It satisfies
√
ρu ln1/2 (e + |u|2 ) ∈ L∞ 0, T; L2 (RN ) .
(34)
To this end, let us first check
√
ρε uε ln1/2 (e + |uε |2 ) ∈ L∞ 0, T; L2 (RN ) .
(35)
Clearly, the (35) follows directly from (9) and (16) in case of N = 3. Now let us pay attention
to N = 2. Following in [11, 21], we have for any δ ∈ (0, 2)
d
2
2
ρε (1 + |uε | ) ln(1 + |uε | ) +
ρε [1 + ln(1 + |uε |2 )]|∇uε |2
dt RN
RN
δ/2
2−δ
2
2
α
2
2
(4γ−2α−δ)/(2−δ)
ρε (1 + |uε | )
.
C
ρε |∇uε | + Cε
ρε
(36)
RN
RN
RN
Thus, using (15) and (16), integration of (36) in time conclude the (35), so long as
2−δ
T 2
ε2
ρ(4γ−2α−δ)/(2−δ)
C,
ε
RN
0
which is fulfilled because of (19). Making use of (25), (27), (35), the Fatou lemma, we get (34).
√
√
Step 3. Given constant M > 1, the (25) and (27) ensure that ρε uε |uε M → ρu|uM
√
almost everywhere when ρ > 0 . If we also define ρu|uM on sets {ρ = 0}, then
√
√
√
ρε uε |uε M M ρε → 0 = ρu|uM for ρ = 0.
Recalling (19), it satisfies for q > 2
√
ρε uε |uε M ∈ L∞ (0, T; Lq ) ,
and therefore,
T
0
RN
√
√
| ρε uε |uε M − ρu|uM |2 → 0
On the other hand, it follows from (35) and (34) that
T
√
√
2
( ρε uε |uε >M + ρu|u>M )
0
RN
as ε → 0.
T
C
ρε |uε |2 ln(1 + |uε |2 ) + ρ|u|2 ln(1 + |u|2 )
2
ln(1 + M ) 0 RN
→ 0 as M → ∞.
4183
Z Liang
Nonlinearity 30 (2017) 4173
In conclusion, sending ε → 0 first and then M → ∞ yields
T
T
√
√ 2
√
√
| ρε uε − ρu| 2
| ρε uε |uε M − ρu|uM |2
0
RN
0
+2
0
RN
T
RN
√
√
| ρε uε |uε >M |2 + | ρu|u>M |2 → 0.
Lemma 3.3. It satisfies
q2
ρε uε → ρu in L2 0, T; Lloc
(RN ) ,
(37)
if N = 3.
where q2 ∈ [1, 2) if N = 2 and q2 ∈ [1, 12α−6
6α−1 )
Proof. Making use of lemma 3.2, (19) and (34), and the inequality
√
√
| x − y| |x − y|, ∀ x 0, y 0,
we conclude the (37) from the following
ρε uε − ρuLq2 (RN )
√ √
√
√
√ √
ρε ( ρε uε − ρu)Lq2 (RN ) + ( ρε − ρ) ρuLq2 (RN )
√
√
√
√
√
√
ρε Lq (RN ) ( ρε uε − ρu)L2 (RN ) + ρε − ρLq (RN ) ρuL2 (RN )
√
√
1/2
C ( ρε uε − ρu)L2 (RN ) + ρε − ρLq/2 (RN ) ,
(38)
where q2 = (1/q + 1/2)−1 with q 2 if N = 2 and q ∈ [2, 6α − 3) if N = 3.
□
4. Proof of theorem 1.2
Utilizing (4), we deduce from (10) that
ρ(·,
t)L1 (RN ) = ρε (·, t)L1 (RN ) = ρ0 L1 (RN ) ,
(39)
and
T
α
|m0 |2
sup
ρ|u|2 +
ρ |∇u|2 + (α − 1)ρα (divu)2 (40)
t∈[0,T] RN
RN
RN ρ0
0
t∈[0,T]
RN
2
sup
ρ|u + αρ
∇ρ| ρ0 |u0 + αρα−2
∇ρ0 |2 .
(41)
0
α−2
RN
In addition, the same method as (19) runs
ρ ∈ L∞ 0, T; L1 (RN ) ∩ Lq (RN ) , ∇ρα−1/2 ∈ L∞ 0, T; L2 (RN ) ,
(42)
where q < ∞ if N = 2 and q = 6α − 3 if N = 3.
Set z = ρε − ρ. Subtracting (10)1 from (1)1 receives
α
α−2
∇ρε ) − ρ(u + αρα−2 ∇ρ) ,
∂t z = (ρα
ε − ρ ) − div ρε (uε + αρε
z(x, t = 0) = 0.
4184
Z Liang
Nonlinearity 30 (2017) 4173
Multiplying the above by ϕ(x, t) ∈ C0∞ (RN × [0, +∞)) and integrating the expression by
parts give rise to
t
zϕ(x, t) =
z (ϕs + aϕ)
RN
0
RN
t
ρε (uε + αρα−2
∇ρε ) − ρ(u + αρα−2 ∇ρ) · ∇ϕ,
+
ε
0
RN
(43)
α
where a = (ρα
ε − ρ )/z if z = 0 and a = 0 if z = 0.
To be continued, consider the following backward parabolic equation

x ∈ BR , 0 s < t,
∂s ϕR + an ϕR = 0,
ϕR = 0,
x ∈ ∂BR , 0 s < t,
(44)

ϕR (x, t) = θ(x) ∈ H01 (BR ), x ∈ BR ,
where an = η1/n ∗ aK,ε ∈ [2−1 ε, 2K] and η1/n being the standard Friedrichs’ mollifier such that
as n → ∞,

K, a > K,
a, ε a K,
a
→
a
=
(45)
n
K,ε

ε, a < ε.
The classical linear parabolic theory (see [15]) ensures that (44) has a unique solution
ϕR ∈ L∞ 0, t; H01 ∩ L2 0, t; H 2 . If we multiply (44) by ϕR , we infer for τ ∈ [0, t]
t
1
1
2
2
|∇ϕR | (x, τ ) −
|∇ϕR | (x, t) +
an |ϕR |2 = 0,
2 BR
2 BR
BR
τ
and thus,
t
2
2
|∇ϕ
|
(x,
τ
)
+
a
|ϕ
|
|∇θ|2 .
(46)
R
n
R
BR
τ
BR
BR
Define a smooth cut-off function ξR satisfying
ξR = 1 in BR/2 , ξR = 0 in RN \ BR , |∇k ξR | CR−k , (k = 1, 2).
(47)
If we extend ϕR to RN by zero and replace ϕR in (43) with ϕ = ξR ϕR, the first term on the
right-hand side of (43) satisfies
t
z (ϕs + aϕ)
0
RN
t
t
α
zξR (∂s ϕR + aϕR ) +
(ρα
=
ε − ρ ) (2∇ξR ∇ϕR + ϕR ξR )
N
N
R
R
0
0
t
t
α
=
zξR (a − an ) ϕR +
(ρα
ε − ρ ) (2∇ξR ∇ϕR + ϕR ξR )
0
RN
0
I1 + I2 ,
where in the second equality we used (44).
RN
(48)
4185
Z Liang
Nonlinearity 30 (2017) 4173
By (45) and (46), one has
t
|I1 | BR
0
z2 (a − an )
an
2
1/2 t
0
BR
an |ϕR |
2
1/2
2 1/2
z2 (a − an )
∇θL2 (BR )
an
BR
0
1/2
t
√
2
2
2ε−1 ∇θL2 (BR )
z2 (a − aK,ε ) + z2 (aK,ε − an )
t
BR
0
Cε1/2 ∇θL2 (BR ) ,
where the last inequality owes to (19) and (42), and the following two inequalities:
t
T
2
z2 (aK,ε − an ) C
z2L6α−3 (aK,ε − an )2 12α−6 → 0 (n → ∞)
BR
0
L
0
6α−5
and
t
0
2
2
BR
2
z (a − aK,ε ) Cε sup
0tT
Cε
2
z2L2
+
T
0
BR
∩{x∈RN :a>K}
α
(ρα
ε −ρ )
2
(K → ∞).
Therefore, (48) is estimated as
t
z (ϕs + aϕ) Cε1/2 ∇θL2 (BR ) + I2 .
(49)
RN
0
The second term on the right-hand side of (43) satisfies for ϕ = ξR ϕR
t
ρε (uε + αρα−2
∇ρε ) − ρ(u + αρα−2 ∇ρ) · ∇(ξR ϕR )
ε
RN
0
C
+
T
0
T
0
BR
BR
J1 + J2 .
ρε |uε + αρα−2
∇ρε | + ρ|u + αρα−2 ∇ρ| |∇ϕR |
ε
ρε |uε + αρα−2
∇ρε | + ρ|u + αρα−2 ∇ρ| |∇ξR ||ϕR |
ε
(50)
Owing to (13) and (17),
√
(51)
ρε |uε + αρα−2
∇ρε |2L2 (RN ) ερ0 γLγ (RN ) Cε.
ε
This together with (13) and (41) shows for q̄ > 2
T
√
√
J1 C
ρε |uε + αρα−2
∇ρε |L2 (BR ) ρε L(1/2−1/q̄)−1 (BR ) ∇ϕR Lq̄ (BR )
ε
0
1/2
Cε
T
0
We discuss J1 in two cases.
√
ρε L(1/2−1/q̄)−1 (BR ) ∇ϕR Lq̄ (BR ) .
4186
(52)
Z Liang
Nonlinearity 30 (2017) 4173
• Let q̄ = 2 + δ with δ > 0 small in case of N = 2 .
By (19) and (14),
J1 Cε1/2
Cε1/2
Cε1/2
1
Cε 2+δ
Cε
1
2+δ
T
0
√
ρε L(1/2−1/(2+δ))−1 (BR ) ∇ϕR L2+δ (BR )
T
∇ϕR L2+δ (BR )
T
2
δ
−δ
√
2+δ
2+δ
2(2+δ)
∇ϕR L2 (BR ) an ϕR L2 (BR )
∇ϕR L2 (BR ) + ε
0
1/2 T
√
2
an ϕR L2 (BR )
sup ∇ϕR L2 (BR ) +
0
t∈[0,T]
0
∇θL2 (BR ) ,
(53)
where in the third inequality we have used
√
2
∇
ϕR L2 (BR ) CϕR L2 (BR ) ε−1/2 an ϕR L2 (BR ) ,
(54)
owing to (45) and lemma 2.2.
•Let q̄ =
6α−3
3α−2
in case of N = 3.
Similar to (53), we deduce
T
√
J1 Cε1/2
ρε L12α−6 (BR ) ∇ϕR Cε1/2
0
T
0
4α−3
4(2α−1)
6α−3
L 3α−2 (BR )
−1
−1
√
(1 + ε 4(2α−1) )∇ϕR L2 (BR ) + ε 4(2α−1) an ϕR L2 (BR )
∇θL2 (BR ) .
Cε
With the aid of (52), (53) and (55), the (50) satisfies
t
ρε (uε + αρα−2
∇ρε ) + ρ(u + αρα−2 ∇ρ) · ∇(ξR ϕR )
ε
0
RN
1
N = 2,
ε 2+δ ,
J2 + C∇θL2 (BR )
4α−3
4(2α−1)
ε
, N = 3,
(55)
which, along with (43) and (49), implies
1
ε 2+δ ,
N = 2,
zξR ϕR (x, t) I2 + J2 + C∇θL2 (BR )
(56)
4α−3
4(2α−1)
N
ε
, N = 3.
R
Next, by (19), (42), (47), the Poincaré inequality, we deduce from (48) that
t
α
I2 =
(ρα
ε − ρ ) (2∇ξR ∇ϕR + ϕR ξR )
0
RN
T
(57)
α
−1
CR
ρα
ε − ρ L2 (BR \BR/2 ) ∇ϕR L2 (BR )
0
CR−1 ∇θL2 (BR ) → 0
(R → ∞).
4187
Z Liang
Nonlinearity 30 (2017) 4173
By (13), (41), (19), (46), (47), (51) and (54), the Poincaré inequality, we deduce from (50) that
T
J2 C
ρε |uε + αρα−2
∇ρε | + ρ|u + αρα−2 ∇ρ| |∇ξR ||ϕR |
ε
0
CR−1
BR \BR/2
T
0
√
√
ρε L4 (BR \BR/2 ) ρε |uε + αρα−2
∇ρε |L2 (RN ) ϕR L4 (BR )
ε
1/2
Cε1/2 sup ρε L2 (BR \BR/2 )
t∈[0,T]
T
T
0
∇ϕR L4 (BR )
√
∇ϕR L2 (BR ) + an ϕR L2 (BR )
C sup
1/2
ρε L2 (BR \BR/2 )
C sup
1/2
ρε L2 (BR \BR/2 ) ∇θL2 (BR )
t∈[0,T]
t∈[0,T]
0
→0
(R → ∞).
(58)
Particularly, if we replace the function ϕR (x, t) in (56) with
1
α− 12
ϕ
− ρα− 2 (x, t),
(59)
R (x, t) = ξR ρε
we conclude from (56)–(58) that by sending R → ∞ ,
1
ε 2+δ ,
N = 2,
α− 12
α− 12
(ρε − ρ)(ρε
−ρ
)C
4α−3
ε 4(2α−1) , N = 3.
RN
In terms of lemma 2.3, it satisfies for α 32
1
ε 2+δ ,
N = 2,
α+ 12
|ρε − ρ|
C
4α−3
ε 4(2α−1) , N = 3.
RN
The proof of theorem 1.2 is complete by exploiting (19) and (42), and interpolation
inequalities.
Acknowledgments
The author is supported by NNSFC Grant No. 11301422. The author is grateful for the anonymous reviewers’ helpful comments and suggestions which improved both the mathematical
results and the way to present them.
Appendix. Proof of (9)
Multiplying equations (1)2 by 4|uε |2 uε yields
d
2
2
ρε |uε |4 + 4
ρα
ε |uε | |∇uε |
dt R3
R3
α
2
2
+8
ρε |uε | |∇|uε || + 4(α − 1)
ρα
ε divuε |uε | (divuε |uε | + 2uε · ∇|uε |)
R3
R3
= 4ε
∇ργε div(|uε |2 uε ).
R3
By Young’s inequality, it satisfies for all α > 1/2
4188
Z Liang
Nonlinearity 30 (2017) 4173
2
2
ρα
|u
|
|∇|u
||
+
4(α
−
1)
ρα
ε
ε ε
ε divuε |uε | (divuε |uε | + 2uε · ∇|uε |)
R3
R3
5
5
2
2
ρα
ρα |∇uε |2 |uε |2 .
−
ε (divuε ) |uε | −
2 R3
2 R3 ε
8
Next,
∇ργε div(|uε |2 uε )
2
2
ε
ρα
|u
|
|∇u
|
+
Cε
ρ2γ−α
|uε |2
ε
ε
ε
ε
R3
R3
2
2
2γ−α−1/2
2 2
ε
ρα
L2 (R3 ) 1 + ρ1/2
ε |uε | |∇uε | + Cερε
ε |uε | L2 (R3 ) .
ε
R3
R3
For small ε 1/2, the above three inequalities ensure that
d
2
2
ρε |uε |4 +
ρα
ε |uε | |∇uε |
dt R3
R3
2 2
Cερ2γ−α−1/2
L2 (R3 ) 1 + ρ1/2
ε
ε |uε | L2 (R3 ) .
The proof can be done by means of the Gronwall inequality, provided
T
ε
ρ2γ−α−1/2
L2 (R3 ) dt C.
ε
0
In fact, since (7) and (8) implies 1 4γ − 2α − 1 2γ + 4α − 3, it has
2γ−α−1/2 2
ρε
L2 (R3 ) =
+
ρ4γ−2α−1
ε
{ρε 1}
{ρε 1}
ρε +
{ρε 1}
{ρε 1}
ρ2γ+4α−3
C + C∇ρ(γ+α−1)/2
4L2 (R3 ) .
ε
ε
where the last inequality owes to (19), Sobolev inequality and the following
ρ2γ+4α−3
Cρε 2α−1
ρ(γ+α−1)/2
4L6 (R3 ) C∇ρ(γ+α−1)/2
4L2 (R3 ) .
ε
ε
ε
L6α−3 (R3 )
R3
Therefore,
ε
T
0
ρ2γ−α−1/2
L2 (R3 )
ε
Cε + ε
T
0
∇ρ(γ+α−1)/2
2L2 (R3 ) C,
ε
where the last inequality owes to (17) and the C is independent of ε .
References
[1] Agmon S, Douglis A and Nirenberg L 1959 Estimates near the boundary for solutions of elliptic
partial differential equations satisfying general boundary conditions I Commun. Pure Appl.
Math. 12 623–727
4189
Z Liang
Nonlinearity 30 (2017) 4173
Agmon S, Douglis A and Nirenberg L 1964 Estimates near the boundary for solutions of elliptic
partial differential equations satisfying general boundary conditions II Commun. Pure Appl.
Math. 17 35–92
[2] Desjardins B and Grenier E 1999 Low Mach number limit of viscous compressible flows in the
whole space Proc. R. Soc. A 455 2271–9
[3] Desjardins B, Grenier E, Lions P and Masmoudi N 1999 Incompressible limit for solutions of the
isentropic Navier–Stokes equations with Dirichlet boundary conditions J. Math. Pures Appl.
78 461–71
[4] Bresch D and Desjardins B 2004 Some diffusive capillary models of Korteweg type C. R. Mec.
332 881–6
[5] Bresch D and Desjardins B 2003 Existence of global weak solution for 2D viscous shallow water
equations and convergence to the quasi-geostrophic model Commun. Math. Phys. 238 211–23
[6] Cho Y, Choe H and Kim H 2004 Unique solvability of the initial boundary value problems for
compressible viscous fluid J. Math. Pures Appl. 83 243–75
[7] Feireisl E, Novotny A and Petzeltov H 2001 On the existence of globally defined weak solutions to
the Navier–Stokes equations J. Math. Fluid Mech. 3 358–92
[8] Gagliardo E 1959 Ulteriori proprietà di alcune classi di funzioni in più variabili Ric. Mat. Napoli.
8 24–51
[9] Goudon T and Junca S 2000 Vanishing pressure in gas dynamics equations Z. Angew. Math. Phys.
51 143–8
[10] Guo Z, Jiu Q and Xin Z 2008 Spherically symmetric isentropic compress- ible flows with densitydependent viscosity coefficients SIAM J. Math. Anal. 39 1402–27
[11] Haspot B 2016 From the highly compressible Navier–Stokes equations to fast diffusion and porous
media equations, existence of global weak solution for the quasi-solutions J. Math. Fluid Mech.
18 243–91
[12] Haspot B and Zatorska E 2016 From the highly compressible Navier–Stokes equations to the
porous medium equation—rate of convergence Discrete Contin. Dyn. Syst. A 36 3107–23
[13] Jiang S and Zhang P 2001 Global spherically symmetry solutions of the compressible isentropic
Navier–Stokes equations Commun. Math. Phys. 215 559–81
[14] Jiu Q and Xin Z 2008 The Cauchy problem for 1D compressible flows with density-dependent
viscosity coefficients Kinet. Relat. Mod. 1 313–30
[15] Ladyzenskaja O, Solonnikov V and Ural’ceva N 1968 Linear and Quasilinear Equations of
Parabolic Type (Providence, RI: American Mathematical Society)
[16] Li H, Li J and Xin Z 2008 Vanishing of vacuum states and blow-up phenomena of the compressible
Navier–Stokes equations Commun. Math. Phys. 281 401–44
[17] Li J and Liang Z 2014 Local well-posedness of strong and classical solutions to Cauchy problem
of the two-dimensional barotropic compressible Navier–Stokes equations with vacuum J. Math.
Pures Appl. 102 640–71
[18] Li J and Xin Z 2015 Global existence of weak solutions to the barotropic compressible Navier–
Stokes flows with degenerate viscosities (http://arxiv.org/abs/1504.06826v2)
[19] Lions P and Masmoudi N 1998 Incompressible limit for a viscous compressible fluid J. Math.
Pures Appl. 77 585–627
[20] Lions P 1998 Mathematical Topics in Fluid Mechanics (Compressible Models vol 2) (Oxford:
Oxford University Press)
[21] Mellet A and Vasseur A 2007 On the barotropic compressible Navier–Stokes equations Commun.
PDE 32 431–52
[22] Vasseur A and Yu C 2016 Existence of global weak solutions for 3D degenerate compressible
Navier–Stokes equations Invent. Math. 206 935–74
[23] Vasseur A and Yu C 2016 Global weak solutions to compressible quantum Navier–Stokes
equations with damping SIAM J. Math. Anal. 48 1489–511
4190
Документ
Категория
Без категории
Просмотров
8
Размер файла
1 138 Кб
Теги
2faa84c0, 6544, 1361
1/--страниц
Пожаловаться на содержимое документа