A priori estimates of local solutions to compressible Navier-Stokes equations with free boundary

J. Math. Anal. Appl. 482 (2020) 123506

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A priori estimates of local solutions to compressible Navier-Stokes equations with free boundary Huihui Kong a , Chuangchuang Liang b,∗ a

School of Mathematics and Physics, University of Science and Technology Beijing, Beijing 100083, China b College of Mathematics and Statistic, Chongqing University, Chongqing 400044, China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 22 June 2019 Available online 23 September 2019 Submitted by D. Wang

In this paper, we consider the vacuum free boundary problem of the compressible Navier-Stokes system in two dimension. Under the Lagrangian variables, we present a priori estimates of local-in-time solutions in a proper Sobolev space. © 2019 Elsevier Inc. All rights reserved.

Keywords: Compressible Navier-Stokes Free boundary A priori estimates

1. Introduction For 0 ≤ t ≤ T , the evolution of two dimensional compressible fluid with a moving free-surface can be modeled by the compressible Navier-Stokes system: ⎧ ρt + div(ρu) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ (ρu)t + div(ρu ⊗ u) + ∇P = divS(u), V(Γ(t)) = u · n, ⎪ ⎪ ⎪ −P n + S(u) · n = 0, ⎪ ⎪ ⎩ (ρ, u)|t=0 = (ρ0 , u0 ),

in Ω(t), in Ω(t), on Γ(t), on Γ(t), on Ω(0).

(1.1)

The open, bounded subset Ω(t) ⊆ R2 with Ω(0) = Ω denotes the changing volume occupied by the fluid at time t. Γ(t) := ∂Ω(t) denotes the moving vacuum boundary, V(Γ(t)) denotes the normal velocity of Γ(t) and n denotes the exterior unit normal vector of Γ(t). ρ denotes the density of the fluid and the vector-flied u = (u1 , u2 ) denotes the Eulerian velocity flied. The viscous stress tensor S(u) is expressed as * Corresponding author. E-mail addresses: [email protected] (H. Kong), [email protected] (C. Liang). https://doi.org/10.1016/j.jmaa.2019.123506 0022-247X/© 2019 Elsevier Inc. All rights reserved.

2

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

S(u) = μ(∇u + ∇t u) + (μ − μ)divu,

(1.2)

with the Lame ´ viscosity coefficients satisfying μ > 0, μ ≥ 0. The pressure P (ρ) is given by P (x, t) = κργ for γ > 1,

(1.3)

where κ is a constant which we can set to be unity. The initial density ρ0 satisfies ρ0 > 0 in Ω(0) and ρ0 = 0 on Γ(0),

(1.4)

which, together with (1.1)1 , leads to ρ > 0 in Ω(t) and ρ = 0 on Γ(t).

(1.5)

The free boundary problem of compressible fluid has attracted lots of research interests and been intensively studied in the literature. When the stress tensor is balanced by a surface tensor or an external pressure, the local-in-time results can be tracked back to [16,18] etc. In this case, there is a non-vacuum equilibrium state. The global solutions close to the equilibrium state are obtained in [17–19]. If the free boundary is the interface between fluid and vacuum, the most works in this direction are one dimension or spherically symmetric case. Global existence of a spherically symmetric weak solution to the multi-dimensional case with density-dependent viscosity coefficients was shown by Guo-Li-Xin [4] subject to the stress free boundary condition and positive flow density near/at the free boundary. With different condition on viscosity stress tensor, the global existence of radial symmetric strong solution was investigated by Li-Zhang [9] in two dimension recently. When the density connects to vacuum continuously, Luo-Xin-Yang [13] investigated the regularity and the behavior of solutions near the interfaces between the flow and the vacuum in one dimension, which is further generalized by Zeng in [20]. The spherically symmetric motion was considered in [8] which obtained the global existence of weak solution and showed that the solution became a strong one away from symmetric center. The spherically symmetric solution was also obtained in [3,7] with consideration of self-gravitation. In particular, when the density connects vacuum continuously, one important situation is that the density satisfies the physical vacuum boundary condition (see [10–12]) across the vacuum boundary. That is, the sound speed c satisfies −∞ < ∇n c2  −C < 0, ρ = 0, on the boundary with c2 = P  (ρ). The sound speed is only 1/2-H o¨lder continuous which makes trouble to deal with this kind of problem (see [10]). The local well-posedness was investigated by [1,2,5] for inviscid flows. For the viscous case, important progresses have been made on the local well-posedness [6] and global well-posedness[14,15] (also spherically symmetric case). The main purpose of this work is to study the physical vacuum free surface problem without imposing the symmetry condition. By using the similar method of [1], we transform our original problem to another formulation which is the equations of the velocity with density ρ0 as a coefficient. This paper is organized as follows. In Section 2, we introduce the Lagrangian form of our problem and states the main result of this paper. Section 3, which gives the proof of our main result, contains four subsections. Subsection 3.1 shows the basic a priori assumption and the useful facts under this assumption. Subsection 3.2 and 3.3 gives out tangential derivative estimates and elliptic estimates (normal estimates) respectively. In the final subsection 3.4, we collect these estimates to obtain the final energy estimate (3.119).

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

3

2. Main result 2.1. Fixing the domain by Lagrangian transform In order to transform the changing domain Ω(t) into a fixed domain, we use the Lagrangian coordinates. Set η(x, t) as the “position” of the particle x at time t. Thus, ∂t η = u ◦ η for t > 0 and η(0, x) = x, x ∈ Ω,

(2.6)

where ◦ denotes the composition so that u ◦ η := u(η(x, t), t). We set v = u ◦ η, f = ρ ◦ η, A = (∇η)−1 , J = det∇η, a = JA. Therefore, the Lagrangian version of (1.1) can be written as ⎧ i ⎪ = 0, ft + f Aji v,j ⎪ ⎪ ⎨ f v i + Ak f γ = Al [(S (v)) ] , A ij ,l t i ,k j ⎪ (v) · N = 0, S A ⎪ ⎪ ⎩ (f, v, η)|t=0 = (ρ0 , u0 , e),

in Ω × (0, T ], in Ω × (0, T ], on Γ := ∂Ω, on Γ,

(2.7)

where we have used Einstein’s summation convention, that is the repeated indices i, j, k etc. are summed from ∂F 1 to 2 and the notation F,k denotes ∂x , the kth-partial derivative of F for k = 1, 2. N /|N | is the Lagrangian k form of the exterior unit normal vector n, e(x) = x denotes the identity map of Ω and SA (v) stands for the j i n i viscous stress tensor under the Lagrangian coordinates with (SA (v))ij = μ(Aki v,k +Akj v,k ) +(μ −μ)Am n v,m δj . i i Since Jt = JAji v,j = aji v,j , it follows from solving (2.7)1 that f = ρ0 J −1 . Then, we can rewrite (2.7) as ⎧ i k γ −γ l ⎪ ⎨ ρ0 vt + ai (ρ0 J ),k = aj [(SA (v))ij ],l , SA (v) · N = 0, ⎪ ⎩ (v, η)| t=0 = (u0 , e),

in Ω × (0, T ], on Γ, on Ω.

(2.8)

In this article, we assume the initial domain Ω ⊆ R2 is given by Ω = {(x1 , x2 ) ∈ T × R|0 < x2 < 1}. Here, we assume the domain is horizontally periodic for T = R/N the usual 1-torus which is identified with the unit length and periodic boundary conditions. In this case, the reference vacuum boundary we consider at t = 0 is the top boundary Γ = {x2 = 1}, while the bottom boundary {x2 = 0} is fixed with the boundary condition v = 0 on Γb = {x2 = 0}. The moving vacuum boundary is then given by Γ(t) = η(t)(Γ) = η(x1 , 1, t). We will focus our analysis on the case γ = 2. Thus, the model we consider in this article can be written in Lagrangian coordinates as below:

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

4

⎧ ⎪ ρ0 vti + aki (ρ20 J −2 ),k = alj [(SA (v))ij ],l , in Ω × (0, T ] ⎪ ⎪ ⎨ S (v) · N = 0, on Γ, A ⎪ v = 0, on Γb , ⎪ ⎪ ⎩ (v, η)| on Ω, t=0 = (u0 , e),

(2.9)

2 1 where N = (−η,1 , η,1 ) on Γ. From (2.6), we can obtain that

 ∇η =

 t 1 t 1 dτ v dτ 1 + 0 v,1 1 2 1 2 0 ,2 t 2 , J = det∇η = η,1 η,2 − η,2 η,1 , = t 2 v dτ 1 + 0 v,2 dτ 0 ,1     2 1 2 1 −η,2 −η,2 η,2 η,2 −1 −1 , a = JA = . A = (∇η) = J 2 1 2 1 −η,1 η,1 −η,1 η,1 1 η,1 2 η,1

1 η,2 2 η,2





(2.10)

(2.11)

It is a fact that the column vectors of matrix a are divergence-free, that is aki,k = 0 for i = 1, 2.

(2.12)

2.2. The energy function and compatibility conditions Define the following high-order energy function as: E(t) =

2



ρ0 |∂tk v|2 dx +

k=0 Ω



+

ρ20 J −1 dx +

Ω

D(t) = ∂t v 2H 3 (Ω) +

2

{

k=1 Ω



|∂t ∇v|2 dx + v 2H 3 (Ω) Ω

ρ20 J −3 |∂t J|2 dx +

Ω

ρ20 J −3 |∂t ∂1 J|2 dx.

Ω



ρ0 |∂tk v|2 dxt +



ρ0 |∂t ∂1k v|2 dx} + Ω

2



|∂tk ∇v|2 dx.

(2.13)

(2.14)

k=0 Ω

Assume the initial data (ρ0 , u0 ) is smooth enough such that the following data are well-defined: 1 l 1 {a [(SA (v))ij ],l − aki (ρ20 J −2 ),k }|t=0 = (μ  ui0 + μ ∂i divu0 − ∂i ρ20 ), ρ0 j ρ0 1 ∂t2 v i |t=0 := ∂t { {alj [(SA (v))ij ],l − aki (ρ20 J −2 ),k }}|t=0 , ρ0 1 ∂t ∇v i |t=0 := ∇{ {alj [(SA (v))ij ],l − aki (ρ20 J −2 ),k }}|t=0 . ρ0 ∂t v i |t=0 :=

(2.15) (2.16) (2.17)

2.3. Main result Theorem 2.1. Assume that ρ0 ∈ H 4 (Ω), ρ0 > 0 for x ∈ Ω, ρ0 = 0 for x on Γ, and ρ0 satisfies physical vacuum boundary condition. Furthermore, suppose u0 is given such that E0 < +∞. Then, for T > 0 which taken small enough, the solution (η, v) to (2.6), (2.9) on [0, T ] satisfies E(t) < +∞, t ∈ [0, T ].

(2.18)

Remark 2.1. Using the similar method, it can also give a priori estimates of local solutions in three dimension, which may be needed higher derivative estimates. In order to show the key point clearly, we only state the result in two dimension case for simplicity.

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

5

3. A priori estimates 3.1. A priori assumption The basic a priori assumption is that,

∇v L∞ ≤ C, x,t

≤ J ≤ 32 . Thus, for t ∈ [0, T ], it is easy to check the useful facts:

1 2

and T ( 1) is small enough such that

(3.19)

(i) |aji | ≤ T ∇v L∞ ≤ CT, f or i = j; |aii | ≤ 1 + CT, f or i = 1, 2. x,t

(3.20)

i (ii) |∂τ aji | ≤ ∇v L∞ ≤ C, |∂t J| = |aji v,j | ≤ C, f or i, j = 1, 2. x,t

t (iii)

|∂lm aji |

≤|

t ∂lm ∇vdτ |;

|∂l J| ≤ C|

0

|∂lm aji |2 dx

≤

Ω



t |

∂lm ∇vdτ |2 dx

|∂lm ∇v|2 dτ dx

t 0

Ω

T



|∂lm ∇v|2 dxdτ ≤ T sup t∈[0,T ]

0 Ω



|∂l J| dx ≤ C Ω

|∂lm ∇v|2 dx.

∂l ∇vdτ | dx ≤ C 2

0

Ω

t |∂l ∇v|2 dτ dx

t Ω

T

≤ CT

(3.23)

Ω

t |

2

(v)

(3.22)

t

≤

0

Ω

≤T

∂l ∇vdτ | f or l = 1, 2 and m = 1, 2, . . . 0

(iv)

(3.21)

0

|∂l ∇v| dxdτ ≤ CT sup

|∂l ∇v|2 dx.

2

t∈[0,T ]

0 Ω

(3.24)

Ω

3.2. The tangential derivative estimates of energy and dissipation In this subsection, we will take the tangential derivative estimates up to the third order under the basic assumption (3.19). Lemma 3.1. Under the assumption (3.19), it holds that

sup { t∈[0,T ]

ρ0 |v| dx + 2

Ω

ρ20 J −1 dx}

T

T

|∇v| dxdt ≤ CE0 + CT

+ 0 Ω

Ω

|∇v|2 dxdt.

2

(3.25)

0 Ω

Proof. Multiplying (2.9)1 by v and integrating the resulting equation with respect to x, we have 1 d 2 dt



ρ0 |v|2 dx + Ω

Ω

aki (ρ20 J −2 ),k v i dx −

alj (SA (v))ij,l v i dx = 0.

(3.26)

Ω

Integrating by part and using the boundary conditions, we can deal with the second term of (3.26) as follows:

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

6





aki (ρ20 J −2 ),k v i dx

i ρ20 J −2 aki v,k dx

=−

Ω

+

Ω

Γ



i ρ20 J −2 aki v,k dx = −

=−

ρ20 J −2 a2i v i dx1

Ω

−

ρ20 J −2 a2i v i dx1

Γb



ρ20 J −2 ∂t Jdx =

d dt

Ω



ρ20 J −1 dx.

(3.27)

Ω

The third term of (3.26) can be rewritten in the following form below with help of integrating by part and the boundary condition:

−



Ω

(SA (v))ij alj v,li dx −

alj (SA (v))ij,l v i dx =

a2j (SA (v))ij v i dx

Ω

Γ



(SA (v))ij alj v,li dx −

=

(SA (v) · N ) · vdSx

Ω

Γ



(SA (v))ij alj v,li dx.

=

(3.28)

Ω

Since aij = δji +

t 0

∂τ aij (x, τ )dτ and ∂τ aij L∞ ≤ ∇v L∞ ≤ C, then x,t x,t

j l i j l i i l i (SA (v))ij alj v,li = μJ −1 akj v,k aj v,l + μJ −1 aki v,k aj v,l + (μ − μ)J −1 akj v,k ai v,l j i i i j = μJ −1 v,j v,j + μJ −1 v,j v,i + (μ − μ)J −1 v,ii v,j + R,

(3.29)

with

R =μJ

+μJ

−1

−1

t

t

i i [v,j v,l

∂τ alj dτ

+

i i v,j v,k 0

0

t

t

∂τ alj dτ

+

i j v,j v,k

−1

∂τ aki dτ

+

j i v,k v,l

0

t ∂τ aki dτ

+

j v,ii v,k

0

∂τ alj dτ ] 0

t ∂τ ali dτ

∂τ akj ] 0

0

t j i [v,j v,l

t ∂τ alj dτ

t 0

+(μ − μ)J

+

i v,li v,k

0

[v,ij v,li



t ∂τ akj dτ

t ∂τ akj dτ

+

j v,li v,k

0

t ∂τ ali dτ

0

∂τ akj ] 0

and |R| ≤ CT |∇v|2 . Since v|Γb = 0, by Korn’s inequality, we have



|∇v| dx ≤ C 2

Ω

j i i i j (μJ −1 v,j v,j + μJ −1 v,j v,i + (μ − μ)J −1 v,ii v,j )dx.

(3.30)

Ω

Integrating (3.26) with respect to t, together with (3.27)-(3.30), we obtain



ρ0 |v| dx + 2

Ω

ρ20 J −1 dx

≤ C( Ω

|∇v|2 dxdt

+ 0 Ω

Ω



T



ρ0 |u0 | dx + 2

T

ρ20 dx)

Ω

|∇v|2 dxdt. 2

+ CT 0 Ω

(3.31)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

7

Lemma 3.2. Under the assumption (3.19), it holds that

|∇v| dx +

sup t∈[0,T ]

T

ρ0 |∂t v|2 dxdt

2

0 Ω

Ω



|∇v|2 dx + CT sup

≤ P (E0 ) + (δ + CT ) sup t∈[0,T ]

t∈[0,T ]

Ω

ρ20 J −3 |∂t J|2 dx.

(3.32)

Ω

Proof. Multiplying (2.9)1 by ∂t v and integrating the resulting equation with respect to x, we obtain



aki (ρ20 J −2 ),k ∂t v i dx −

ρ0 |∂t v|2 dx + Ω

Ω

alj (SA (v))ij,l ∂t v i dx = 0.

(3.33)

Ω

i Integrating by part and using the fact ∂t J = aji v,j , it holds that



aki (ρ20 J −2 ),k ∂t v i dx = −

Ω

d =− dt





i ρ20 J −2 aki ∂t v,k dx

Ω i ρ20 J −2 aki v,k dx

Ω



+

i ρ20 J −2 ∂t aki v,k dx

−2

Ω

ρ20 J −3 |∂t J|2 dx.

(3.34)

Ω

The third term of (3.33) can be expressed as below



−

alj (SA (v))ij,l ∂t v i dx Ω



=μ

= Ω

j l J −1 aki v,k aj ∂t v,li dx + μ

Ω

=



μ 2

j i J −1 aki alj ∂t (v,k v,l )dx +

d μ { dt 2



−

μ 2

(SA (v) · N ) · ∂t v i dx

i l J −1 akj v,k aj ∂t v,li dx + (μ − μ)



μ 2

j l i J −1 aki v,k aj v,l dx +

μ 2

j i ∂t (J −1 aki alj )v,k v,l dx −

Ω



i i J −1 akj alj ∂t (v,k v,l )dx +

μ 2

i l J −1 aki v,k aj ∂t v,lj dx

i l i J −1 akj v,k aj v,l dx +



μ − μ 2



μ −μ 2

Ω



Ω

Ω

Ω



− Ω

Ω

Ω

=

alj (SA (v))ij ∂t v,li dx



Ω i l j J −1 aki v,k aj v,l dx }

Ω i i ∂t (J −1 akj alj )v,k v,l dx −



μ −μ 2

Ω

i j J −1 aki alj ∂t (v,k v,l )dx

(i)

Ω

i j ∂t (J −1 aki alj )v,k v,l dx,

(3.35)

(ii)

where (i) indeed can be written as 1 (i) = 2

(SA (v))ij alj v,li dx, Ω

which can be estimated the same as Lemma 3.1.

(3.36)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

8

Thus, one has 1 d 2 dt

(SA (v))ij alj v,li dx

d − dt

Ω



=−





i ρ20 J −2 aki v,k dx

Ω i (ρ20 J −2 )∂t aki v,k dx + 2

Ω



ρ0 |∂t v|2 dx

+ Ω

ρ20 J −3 |∂t J|2 dx + (ii).

(3.37)

Ω

After integrating (3.37) with respect to t, we have 1 2

t



(SA (v))ij alj v,li dx Ω

=

1 2



ρ0 |∂τ v|2 dxdτ

+ 0 Ω

μ ( |∇u0 + ∇t u0 |2 + (μ − μ)|divu0 |2 )dx − 2

Ω

t



−



ρ20 divu0 dx +

Ω i ρ20 J −2 ∂τ aki v,k dxdτ + 2

0 Ω

t



i ρ20 J −2 aki v,k dx

Ω

ρ20 J −3 |∂τ J|2 dxdτ +

0 Ω

t (ii)dτ.

(3.38)

|∇v|2 dx

(3.39)

0

Similarly as Lemma 3.1, one can obtain that



|∇v| dx ≤ C 2

Ω

(SA (v))ij alj v,li dx

+ CT

Ω

Ω

The right hand terms of (3.38) can be estimated as follows:

i ρ20 J −2 aki v,k dx|

| Ω

t

|

≤ Cδ

−1



ρ40 dx Ω

i ρ20 J −2 ∂τ aki v,k dxdτ | ≤ CT

0 Ω

|∇v|2 dx,

+δ

(3.40)

Ω



|∇v|2 dx.

ρ40 dx + CT Ω

(3.41)

Ω

Since |∂t (J −1 akj ali )| can be bounded by P ( ∇v L∞ ), it is easy to get x,t

t |

(ii)dτ | ≤ CT sup

|∇v|2 dx.

t∈[0,T ]

0

(3.42)

Ω

Therefore, it follows from (3.38)-(3.42) that

|∇v| dx +

sup t∈[0,T ]



T

ρ0 |∂t v|2 dxdt ≤ C

2

Ω



Ω



|∇v|2 dx + CT sup

+ (δ + CT ) sup t∈[0,T ]

0 Ω

Ω

|∇u0 |2 dx + C(1 + δ)

t∈[0,T ]

Ω

ρ40 dx Ω

ρ20 J −3 |∂t J|2 dx. 2

(3.43)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

9

Lemma 3.3. Under the assumption (3.19), it holds that

sup { t∈[0,T ]



ρ20 J −3 |∂t J|2 dx}

ρ0 |∂t v| dx + 2

Ω

T

0 Ω

Ω

≤ P (E0 ) + CT sup t∈[0,T ]

+ CT (1 + δ −1 ) sup t∈[0,T ]

|∂t ∇v|2 dxdt

+

ρ20 J −3 |∂t J|2 dx

T

|∂t ∇v|2 dxdt

+ (δ + CT ) 0 Ω

Ω



|∇v|2 dx.

(3.44)

Ω

Proof. Differentiating (2.9) with respect to t and then taking the inner product with ∂t v, there is 1 d 2 dt



ρ0 |∂t v|2 dx +

Ω

∂t (aki ρ20 J −2 ),k ∂t v i dx −

Ω

∂t {alj (SA (v))ij,l }∂t v i dx = 0.

(3.45)

Ω

Integrating by part and using the boundary conditions, we can obtain that

∂t (aki ρ20 J −2 ),k ∂t v i dx

Ω

Ω



i ρ20 J −3 ∂t Jaki ∂t v,k dx

=2





ρ20 J −3 ∂t J∂t2 Jdx

=2 Ω

−



i ρ20 J −2 ∂t aki v,k dx

−

Ω

d = dt

i ∂t (aki ρ20 J −2 )∂t v,k dx

=−



Ω i ρ20 J −3 ∂t J∂t aki v,k dx

−2 Ω

ρ20 J −3 |∂t J|2 dx + 3

Ω



ρ20 J −4 (∂t J)3 dx − 2

Ω

−



i ρ20 J −2 ∂t aki v,k dx

Ω i ρ20 J −3 ∂t J∂t aki v,k dx

Ω

i ρ20 J −2 ∂t aki v,k dx,

Ω

(3.46)

(i)

and

−

∂t {alj (SA (v))ij,l }∂t v i dx Ω



∂t {alj (SA (v))ij }∂t v,li dx −

= Ω

∂t {a2j (SA (v))ij }∂t v i dx Γ



∂t {alj (SA (v))ij }∂t v,li dx −

= Ω

∂t (SA (v) · N ) · ∂t vdSx Γ



(SA (∂t v))ij alj ∂t v,li dx + μ

= Ω



+μ Ω

j ∂t (J −1 aki alj )v,k ∂t v,li dx

Ω i ∂t (J −1 akj alj )v,k ∂t v,li dx





+ (μ − μ) Ω

i ∂t (J −1 aki alj )v,k ∂t v,lj dx, (ii)

(3.47)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

10

with the estimates

(i) ≤ C

ρ20 J −3 |∂t J|2 dx

|∇v|2 dx,

+C

Ω

Ω



|∂t ∇v|2 dx + Cδ −1

(ii) ≤ δ Ω



|∇v|2 dx. Ω

Using the same method as Lemma 3.1, it holds that





|∂t ∇v| dx ≤ C 2

Ω

(SA (∂t v))ij alj ∂t v,li dx

|∂t ∇v|2 dx.

+ CT

Ω

(3.48)

Ω

After integrating (3.45) with respect to time t, we obtain from (3.46)-(3.48) that

sup { t∈[0,T ]

≤C

ρ0 |∂t v| dx + 2

Ω

ρ20 J −3 |∂t J|2 dx}

Ω

|∂t ∇v|2 dxdt

+ 0 Ω



t∈[0,T ]

Ω

T





ρ20 |divu0 |2 dx + CT sup

ρ0 (∂t v|t=0 )2 dx + C Ω

T

|∂t ∇v|2 dxdt + CT (1 + δ −1 ) sup

+ (δ + CT )

t∈[0,T ]

0 Ω

ρ20 J −3 |∂t J|2 dx

Ω

|∇v|2 dx. 2

(3.49)

Ω

Lemma 3.4. Under the assumption (3.19), it holds that

|∂t ∇v| dx +

sup t∈[0,T ]

T



ρ0 |∂t2 v|2 dxdt

2

≤ CE0 + CT sup t∈[0,T ]

0 Ω

Ω

+ Cδ −1 T sup t∈[0,T ]

T



|∂t ∇v|dx Ω



|∇v|dx + δ

|∂t2 ∇v|2 dxdt.

(3.50)

0 Ω

Ω

Proof. Differentiate (2.9)1 with respect to t and then take the inner product with ∂t2 v to get



ρ0 |∂t2 v|2 dx Ω

+

∂t {aki (ρ20 J −2 ),k }∂t2 v i dx

−

Ω

∂t {alj (SA (v))ij,l }∂t2 v i dx = 0.

(3.51)

Ω

The pressure term can be estimated as follows:

∂t {aki (ρ20 J −2 ),k }∂t2 v i dx = −

Ω

≤ Cδ −1

i ρ20 ∂t (aki J −2 )∂t2 v,k dx

Ω

|∂t2 ∇v|2 dx.

ρ40 dx + δ Ω

Similarly, one gets







Ω

(3.52)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506



∂t {alj (SA (v))ij,l }∂t2 v i dx

−

∂t {alj (SA (v))ij }∂t2 v,li dx

=

Ω

Γ





∂t {alj (SA (v))ij }∂t2 v,li dx =

= Ω

alj (SA (∂t v))ij ∂t2 v,li dx + μ Ω



i 2 i ∂t (J −1 akj alj )v,k ∂t v,l dx + (μ − μ)

+μ

∂t (SA (v) · N ) · ∂t2 vdSx

+

Ω



11

Ω

j 2 i ∂t (J −1 aki alj )v,k ∂t v,l dx

Ω



i 2 j ∂t (J −1 aki alj )v,k ∂t v,l dx,

Ω

(3.53)

(i)

with

alj (SA (∂t v))ij ∂t2 v,li dx = Ω 

+

μ −μ 2



d μ { dt 2

−

μ 2

j l J −1 aki ∂t v,k aj ∂t v,li dx +

Ω i l J −1 aki ∂t v,k aj ∂t v,lj dx

Ω





}−

μ 2



i l J −1 akj ∂t v,k aj ∂t v,li dx

Ω j ∂t (J −1 aki alj )∂t v,k ∂t v,li dx

Ω

(ii)

i ∂t (J −1 akj alj )∂t v,k ∂t v,li dx −



μ 2

μ − μ 2

Ω



i ∂t (J −1 aki alj )∂t v,k ∂t v,lj dx.

Ω

(3.54)

(iii)

Since |∂t (J −1 akj ali )| can be bounded by P ( ∇v L∞ ), it is obvious that x,t

(i) ≤ δ

|∂t2 ∇v|2 dx + Cδ −1

Ω

|∇v|2 dx. Ω



|∂t ∇v|2 dx.

(iii) ≤ C Ω

With the help aij = δji +

t 0

∂τ aij (x, τ )dτ , we use the same technique as above to get



|∂t ∇v| dx ≤ C(ii) + CT

|∂t ∇v|2 dx

2

Ω

(3.55)

Ω

Integrating (3.51) with respect to time t, the resulted equation together with (3.53)-(3.55) leads to

|∂t ∇v| dx +

sup t∈[0,T ]

T

2

ρ0 |∂t2 v|2 dxdt ≤ CT δ −1

0 Ω

Ω

+ CT sup t∈[0,T ]



Ω

|∂t ∇v|dx + Cδ −1 T sup t∈[0,T ]

Ω



Ω

T



|∇v|dx + δ Ω

(∂t ∇v|t=0 )2 dx

ρ40 dx + C

|∂t2 ∇v|2 dxdt. 2 0 Ω

Lemma 3.5. Under the assumption (3.19), it holds that

|∂1 ∇v| dx +

sup t∈[0,T ]

T

2

Ω

0 Ω

∂1 ρ0 + Cδ −1 T √ L∞ sup ρ0 x t∈[0,T ]

ρ0 |∂t ∂1 v|2 dxdt ≤ P (E0 ) + Cδ −1 T

Ω

|ρ0 ∂1 ρ0 |2 dx Ω

ρ0 |∂t v|2 dx + C(δ −1 + 1)T sup t∈[0,T ]



|∂1 ∇v|2 dx Ω

(3.56)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

12

T

T

|∂t ∂1 v| dxdt + δ

|∂t ∂1 ∇v|2 dxdt.

2

+δ 0 Ω

(3.57)

0 Ω

Proof. Act ∂1 on (2.9)1 and then take the inner product with ∂t ∂1 v to obtain



ρ0 |∂t ∂1 v| dx + Ω

Ω



−

∂1 ρ0 ∂t v · ∂t ∂1 vdx +

2

∂1 (aki ρ20 J −2 ),k ∂t ∂1 v i dx

Ω

∂1 (alj (SA (v))ij,l )∂t ∂1 v i dx = 0.

(3.58)

Ω

It is easy to see that

Ω

∂1 ρ0 ∂1 ρ0 ∂t v i ∂t ∂1 v i dx ≤ Cδ −1 T √ L∞ sup ρ0 x t∈[0,T ]



ρ0 |∂t v|2 dx + δ Ω

|∂t ∂1 v|2 dx.

(3.59)

Ω

Integrate by part and use Hölder’s inequality and the facts (3.23)-(3.24) to get

∂1 (aki ρ20 J −2 ),k ∂t ∂1 v i dx = −

Ω



≤δ

|∂t ∂1 ∇v|2 dx + Cδ −1

Ω





i ∂1 (aki ρ20 J −2 )∂t ∂1 v,k dx

Ω



|ρ0 ∂1 ρ0 |2 dx + Cδ −1 T sup

|∂1 ∇v|2 dx.

t∈[0,T ]

Ω

(3.60)

Ω

After integrating by part, the viscosity term can be split into two terms which we can estimate separately:

−

∂1 (alj (SA (v))ij,l )∂t ∂1 v i dx

=

Ω

Ω

+μ

alj (SA (∂1 v)ij )∂t ∂1 v,li dx

= Ω

i ∂1 (J −1 alj ajk )v,k ∂t ∂1 v,li dx + μ

Ω

∂1 (SA (v) · N ) · ∂t ∂1 vdSx Ω

∂1 (alj (SA (v))ij )∂t ∂1 v,li dx



−

Ω



=

∂1 (alj (SA (v))ij )∂t ∂1 v,li dx





+μ

j ∂1 (J −1 alj aik )v,k ∂t ∂1 v,li dx

Ω

(i) j ∂1 (J −1 akj ail )v,k ∂t ∂1 v,li dx.

Ω

(3.61)

(ii)

The term (i) can be rewritten in the form 1 d (i) = 2 dt

alj (SA (∂1 v)ij )∂1 v,li dx + R,

(3.62)

Ω

where |R| = | − −



μ −μ 2

μ 2

Ω

and



j ∂t (J −1 aki alj )∂1 v,k ∂1 v,li dx −

μ 2

Ω i ∂t (J −1 aki alj )∂1 v,k ∂1 v,lj dx| ≤ C



i ∂t (J −1 akj alj )∂1 v,k ∂1 v,li dx

Ω



|∂1 ∇v|2 dx, Ω

(3.63)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506







|∂1 ∇v| dx ≤ C 2

Ω

13

alj (SA (∂1 v)ij )∂1 v,li dx

|∂1 ∇v|2 dx.

+ CT

Ω

(3.64)

Ω

By Hölder’s inequality and the facts (3.23)-(3.24), the second term is bounded as below

(ii) ≤ δ



|∂t ∂1 ∇v|2 dx + Cδ −1 T sup

|∂1 ∇v|2 dx.

t∈[0,T ]

Ω

(3.65)

Ω

Integrate (3.58) with respect to time, then (3.57) can be deduced from the resulted equation and above estimates. 2 Lemma 3.6. Under the assumption (3.19), it holds that



ρ0 |∂t ∂1 v| dx + 2

Ω

ρ20 J −3 |∂t ∂1 J|2 dx

T

0 Ω

Ω

T

≤ P (E0 ) + δ 0 Ω

t∈[0,T ]

t∈[0,T ]

∂1 ρ0 |∂t ∂1 v|2 dxdt + CT δ −1 √ 2L∞ sup ρ0 t∈[0,T ]



+ CT δ −1 sup

+ CT sup

|ρ0 ∂t2 v|2 dx Ω



t∈[0,T ]

Ω

|∂1 ∇v|2 dx + CT δ −1

T

T

ρ20 J −3 |∂t ∂1 J|2 dx



Ω

|∂1 ∇v|2 dx 0 Ω

T

|∂1 ∇v|2 dx sup



|∂1 ∇2 v|2 dx

t∈[0,T ]

Ω



∂t ∇v 2H 2 dt 0

t∈[0,T ]



{ρ0 |∂1 J|2 + |∂1 ρ20 |2 + |∂1 ∇v|2 }dx + CT sup

Ω

+ CT 2 sup

|∂t ∂1 ∇v|2 dxdt

+

|∂t ∇2 v|2 dx.

(3.66)

0 Ω

Ω

Proof. Act ∂t ∂1 on (2.9)1 and then take the inner product with ∂t ∂1 v to obtain 1 d 2 dt

−



ρ0 |∂t ∂1 v|2 dx + Ω

∂1 ρ0 ∂t2 v · ∂t ∂1 vdx +

Ω

∂t ∂1 (aki ρ20 J −2 ),k ∂t ∂1 v i dx

Ω

∂t ∂1 (alj (SA (v))ij,l )∂t ∂1 v i dx Ω

= 0.

(3.67)

(i)

The second term of (3.67) is estimated by

∂1 ρ0 ∂t2 v

· ∂t ∂1 vdx ≤ Cδ

−1

∂1 ρ0

√ 2L∞ ρ0

Ω



|ρ0 ∂t2 v|2 dx

|∂t ∂1 v|2 dx.

+δ

Ω

Ω

After integrating by part, the third term can be written in the following form

Ω

∂t ∂1 (aki ρ20 J −2 ),k ∂t ∂1 v i dx = −

= Ω



i ∂t ∂1 (aki ρ20 J −2 )∂t ∂1 v,k dx

Ω i 2ρ20 J −3 ∂t ∂1 Jaki ∂t ∂1 v,k dx



+ Ω

{−6ρ20 J −4 ∂t J∂1 Jaki − ρ20 ∂t J −2 ∂1 aki

(3.68)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

14

i −∂1 ρ20 ∂t J −2 aki − ∂1 (∂t aki ρ20 J −2 )}∂t ∂1 v,k dx R1



d i = ρ20 J −3 |∂t ∂1 J|2 dx + 3 ρ20 J −4 ∂t J|∂t ∂1 J|2 dx− 2ρ20 J −3 ∂t ∂1 J{∂t ∂1 aki v,k dt Ω

Ω

i i +∂t aki ∂1 v,k + ∂t v,k ∂1 aki }dx

R2

Ω

+ R1 ,

(3.69)

with the estimates of R1 , R2 below |R1 | ≤ Cδ

−1

|R2 | ≤ C



{ρ0 |∂1 J| + 2

|∂1 ρ20 |2

Ω

ρ20 J −3 |∂t ∂1 J|2 dx + C

Ω

+ |∂1 ∇v| }dx + δ Ω

|∂1 ∇v|2 dx Ω

t∈[0,T ]

|∂1 ∇2 v|2 dx

t∈[0,T ]

Ω





|∂1 ∇v|2 dx sup

+ CT 2 sup

|∂t ∂1 ∇v|2 dx,

2

Ω

|∂t ∇2 v|2 dx. Ω

For the last term of (3.67), we integrate by part to get



∂t ∂1 (alj (SA (v))ij )∂t ∂1 v,li dx −

(i) = Ω

∂t ∂1 (SA (v) · N ) · ∂t ∂1 vdSx Γ



alj (SA (∂t ∂1 v))ij ∂t ∂1 v,li + R3 ,

=

(3.70)

Ω

where R3 is written as

j j i R3 = {μ∂1 (J −1 aki alj )∂t v,k + μ∂1 (J −1 alj akj )∂t v,k + (μ − μ)∂1 (J −1 ali akj )∂t v,k }∂t ∂1 v,li dx Ω

+

R31 j j i {μ∂t (J −1 aki alj )∂1 v,k + μ∂t (J −1 alj akj )∂1 v,k + (μ − μ)∂t (J −1 ali akj )∂1 v,k }∂t ∂1 v,li dx

Ω



+

R32 j j i {μ∂t ∂1 (J −1 aki alj )v,k + μ∂t ∂1 (J −1 alj akj )v,k + (μ − μ)∂t ∂1 (J −1 ali akj )v,k }∂t ∂1 v,li dx.

Ω

R33

These terms can be estimated as

|R21 | ≤ C

t |

Ω

≤ CT δ

∂1 ∇vdτ ||∂t ∇v||∂t ∂1 ∇v|dx 0

−1



T

∂t ∇v 2L∞ x

≤ CT δ −1 ∂t ∇v 2H 2 |R22 + R23 | ≤ Cδ −1

|∂1 ∇v| dx + δ

|∂t ∂1 ∇v|2 dx

2

0 Ω

Ω

T





|∂1 ∇v|2 dx + δ 0 Ω



|∂1 ∇v|2 dx + δ Ω

|∂t ∂1 ∇v|2 dx, Ω

|∂t ∂1 ∇v|2 dx. Ω

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

15

Then, similarly as above, it holds that





|∂t ∂1 ∇v|2 dx ≤ C Ω

|∂t ∂1 ∇v|2 dx.

alj (SA (∂t ∂1 v))ij ∂t ∂1 v,li + CT Ω

(3.71)

Ω

After integrating (3.67) with respect to time t, (3.66) holds from (3.68)-(3.71). 2 Lemma 3.7. Under the assumption (3.19), it holds that



T

|∂12 ∇v|2 dx

sup t∈[0,T ]

+

≤ P (E0 ) + δ

0 Ω

Ω

T

0 Ω

∂12 ρ0 2

√ L∞ sup ρ0 t∈[0,T ] Ω

+ Cδ −1 T



ρ0 |∂t2 v|2 dx}

+ CT (1 + δ

|∂12 ρ20 |2 dx + Cδ −1 T { ∂1 ρ20 2L∞ sup

ρ0 |∂t ∂1 v|2 dx Ω



|∂1 ∇v|2 dx)2 + ( sup



Ω

|∂1 ∇v|2 dx Ω

|∂1 ∇2 v|2 dx)2 }.

t∈[0,T ]

Ω

|∂12 ∇v|2 dx

) sup

t∈[0,T ]

Ω

t∈[0,T ]

−1



t∈[0,T ]

+ ( sup

|∂t ∂12 v|2 dxdt 0 Ω

∂1 ρ0 |∂t ∂12 ∇v|2 dxdt + Cδ −1 T { √ 2L∞ sup ρ0 t∈[0,T ]

+δ

+

T

ρ0 |∂t ∂12 v|2 dxdt

(3.72)

Ω

Proof. Act ∂12 on (2.9)1 and then take the inner product with ∂t ∂12 v to obtain



ρ0 |∂t ∂12 v|2 dx

∂1 ρ0 ∂t ∂1 v∂t ∂12 vdx

+2

Ω

Ω

+

∂12 ρ0 ∂t v∂t ∂12 vdx

+ Ω

∂12 (aki ρ20 J −2 ),k ∂t ∂12 v i dx

Ω

(i)

−

(ii)

∂12 (alj SA (v)ij ),l ∂t ∂12 v i dx Ω

= 0.

(3.73)

(iii)

The terms (i)-(iii) can be estimated in the following way: (i) ≤Cδ

−1

∂1 ρ0

√ 2L∞ ρ0

Ω

−1

∂ 2 ρ0

√1 2L∞ ρ0

ρ0 |∂t v|2 dx Ω

|∂t ∂12 v|2 dx,

+δ

ρ0 |∂t ∂1 v| dx + Cδ 2

Ω



(ii) = −



(3.74)

Ω i ∂12 (aki ρ20 J −2 )∂t ∂12 v,k dx



≤δ

|∂t ∂12 ∇v|2 dx + Cδ −1 sup t∈[0,T ]

Ω

+ Cδ −1 T sup t∈[0,T ]

Ω

|∂12 ρ20 |2 dx Ω

|∂12 ∇v|2 dx + Cδ −1 T ∂1 ρ20 2L∞ sup

t∈[0,T ]

|∂1 ∇v|2 dx Ω

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

16

+ Cδ

−1

|∂1 ∇v| dx) + Cδ 2

T ( sup t∈[0,T ]



2

−1

Ω

Ω

|∂1 ∇2 v|2 dx)2 ,

T ( sup t∈[0,T ]



∂12 (alj (SA (v))ij )∂t ∂12 v,li dx −

(iii) =



∂12 (SA (v) · N ) · ∂t ∂12 vdSx Γ



alj (SA (∂12 v))ij ∂t ∂12 v,li + R,

=

(3.75)

Ω

(3.76)

Ω

where R is the remainder term which can be written as

j j i R = {μ∂1 (J −1 aki alj )∂1 v,k + μ∂1 (J −1 alj akj )∂1 v,k + (μ − μ)∂1 (J −1 ali akj )∂1 v,k }∂t ∂12 v,li dx Ω



+

j j i {μ∂12 (J −1 aki alj )v,k + μ∂12 (J −1 alj akj )v,k + (μ − μ)∂12 (J −1 ali akj )v,k }∂t ∂12 v,li dx,

Ω

and with the help of (3.19)-(3.24) R can be bounded by

|∂t ∂12 ∇v|2 dx + Cδ −1 T ( sup

|R| ≤δ

|∂1 ∇v|2 dx)2

t∈[0,T ]

Ω



+ Cδ −1 T ( sup t∈[0,T ]

Ω



|∂1 ∇2 v|2 dx)2 + Cδ −1 T sup t∈[0,T ]

Ω

|∂12 ∇v|2 dx. Ω

Furthermore, there is

alj (SA (∂12 v))ij ∂t ∂12 v,li =

1 d 2 dt

Ω

(SA (∂12 v))ij alj ∂12 v,li + R1 ,

(3.77)

Ω

with

μ |R1 | = | − 2 −



μ −μ 2



j 2 i ∂t (J −1 aki alj )∂12 v,k ∂1 v,l dx

μ − 2

Ω i 2 j ∂t (J −1 aki alj )∂12 v,k ∂1 v,l dx| ≤ C

Ω





i 2 i ∂t (J −1 akj alj )∂12 v,k ∂1 v,l dx

Ω

|∂12 ∇v|2 dx,

(3.78)

Ω

and



|∂12 ∇v|2 dx ≤ C Ω

|∂12 ∇v|2 dx.

(SA (∂12 v)ij )alj ∂12 v,li dx + CT Ω

(3.79)

Ω

Integrating (3.73) with respect to time and putting these estimates (3.75)-(3.79) together lead to (3.72). 2 Lemma 3.8. Under the assumption (3.19), it holds that

sup t∈[0,T ]

T

ρ0 |∂t ∂12 v|2 dx

Ω

T

|∂t ∂12 ∇v|2 dxdt

+ 0 Ω

≤ P (E0 ) + (δ + CT )

|∂t ∂12 ∇v|2 dt 0 Ω

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

T

T

|∂t ∂12 v|2 dxdt

+ δ(

|∂t2 ∂1 v|2 dxdt)

+

0 Ω

0 Ω

∂1 ρ0 + Cδ −1 T ( √ 2L∞ sup ρ0 t∈[0,T ]

+ Cδ

−1

t∈[0,T ]

Ω

|∂12 ∇v|2 dx

+ 1) sup t∈[0,T ]

Ω

t∈[0,T ]

Ω



T



∂t ∇v 2Hx2 dt){

T (1 +

|∂1 ∇2 v|2 dx)2 } + Cδ −1 T ( sup t∈[0,T ]

Ω

|∂12 ∇v|2 dx

sup t∈[0,T ]



Ω

|∂1 ∇v|2 dx Ω

T

|∂1 ∇ v| dx)(

|∂t ∂1 ∇v| dxdt +

2

|∂t ∂1 ∇2 v|2 dxdt).

2

0 Ω

Ω

+ Cδ

−1

ρ0 |∂t2 v|2 dx)

0



T

2

|∂12 ρ20 |2 dx

T

∂12 ρ0 2

√ L∞ sup ρ0 t∈[0,T ] Ω

ρ0 |∂t ∂12 v|2 dx +

|∂1 ∇v|2 dx)2 + ( sup

+ ( sup

Ω



T ( ∂1 ρ20 2L∞

+

+ Cδ

−1

17

(3.80)

0 Ω

Proof. Act ∂t ∂12 on (2.9)1 and take the inner product with ∂t ∂12 v to get 1 d 2 dt



ρ0 |∂t ∂12 v|2 dx + 2 Ω

+

∂1 ρ0 ∂t2 ∂1 v∂t ∂12 vdx +

Ω

∂12 ρ0 ∂t2 v∂t ∂12 vdx Ω

∂t ∂12 (aki ρ20 J −2 ),k ∂t ∂12 v i dx

Ω

(i)

−

(ii)

∂t ∂12 (alj SA (v)ij ),l ∂t ∂12 v i dx Ω

= 0.

(3.81)

(iii)

Then, it holds that |(i)| ≤Cδ

−1

∂1 ρ0

√ 2L∞ ρ0

ρ0 |∂t ∂12 v|2 dx Ω

Ω

Ω

−1

Ω

|∂12 ρ20 |2 dx

+ Cδ

−1

≤δ

|∂t ∂12 ∇v|2 dx

( ∂1 ρ20 2L∞

+ Cδ −1 ( ∂1 ρ20 2L∞ + 1){( sup t∈[0,T ]



Ω



|∂12 ∇v|2 dx

+ 1) sup 0∈[0,T ]

Ω



|∂1 ∇v|2 dx)2 + ( sup Ω

t∈[0,T ]



∂t ∂12 (alj SA (v)ij )∂t ∂12 v,li dx −

(iii) =

ρ0 |∂t2 v|2 dx (3.82)

Ω





|∂t2 ∂1 v|2 dx,

Ω



∂ 2 ρ0

√1 2L∞ ρ0

Ω

i ∂t ∂12 (aki ρ20 J −2 )∂t ∂12 v,k dx|

|(ii)| =| −

+ Cδ

−1



|∂t ∂12 v|2 dx + δ

+δ

+ Cδ



|∂1 ∇2 v|2 dx)2 }, Ω

∂t ∂12 (SA (v) · N ) · ∂t ∂12 vdSx Γ

alj SA (∂t ∂12 v)ij ∂t ∂12 v,li dx + R,

=

(3.84)

Ω

with

R= Ω

(3.83)

j j i {μ∂t ∂12 (J −1 aki alj )v,k + μ∂t ∂12 (J −1 alj akj )v,k + (μ − μ)∂t ∂12 (J −1 ali akj )v,k }∂t ∂12 v,li dx R1

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

18

+

j j i {μ∂12 (J −1 aki alj )∂t v,k + μ∂12 (J −1 alj akj )∂t v,k + (μ − μ)∂12 (J −1 ali akj )∂t v,k }∂t ∂12 v,li dx

Ω



+

R2 j j i {μ∂t ∂1 (J −1 aki alj )∂1 v,k + μ∂t ∂1 (J −1 alj akj )∂1 v,k + (μ − μ)∂t ∂1 (J −1 ali akj )∂1 v,k }∂t ∂12 v,li dx

Ω



+

R3 j j i {μ∂1 (J −1 aki alj )∂t ∂1 v,k + μ∂1 (J −1 alj akj )∂t ∂1 v,k + (μ − μ)∂1 (J −1 ali akj )∂t ∂1 v,k }∂t ∂12 v,li dx

Ω



+

R4 j j i {μ∂t (J −1 aki alj )∂12 v,k + μ∂t (J −1 alj akj )∂12 v,k + (μ − μ)∂t (J −1 ali akj )∂12 v,k }∂t ∂12 v,li dx.

Ω

Similarly, there is

R5



|∂t ∂12 ∇v|2 dx

≤C

Ω

(SA (∂t ∂12 v)ij )alj ∂t ∂12 v,li dx

|∂t ∂12 ∇v|2 dx,

+ CT

Ω

(3.85)

Ω

and each term of R can be bounded in the following way:



2 2 −1 |R1 | ≤δ |∂t ∂1 ∇v| dx + Cδ { sup |∂12 ∇v|2 dx t∈[0,T ]

Ω

t∈[0,T ]

Ω

Ω

t∈[0,T ]

|∂1 ∇v|2 dx)2 + ( sup

|∂t ∂12 ∇v|2 dx + Cδ −1 ( sup

t∈[0,T ]

+ Cδ −1 ( sup t∈[0,T ]

Ω

Ω

Ω

Ω

|∂1 ∇2 v|2 dx)(

Ω

|∂t ∂12 ∇v|2 dx + Cδ −1

(3.87)

Ω

|∂1 ∇2 v|2 dx)2 ,

t∈[0,T ]

|∂12 ∇v|2 dx

|∂1 ∇v|2 dx)2

t∈[0,T ]

+ sup |R5 | ≤δ





|∂t ∂12 ∇v|2 dx + Cδ −1 T ( sup

Ω





|∂1 ∇2 v|2 dx)2 },

t∈[0,T ]

Ω

|R4 | ≤δ

Ω



Ω

Ω

|∂1 ∇2 v|2 dx)2 }

t∈[0,T ]

t∈[0,T ]



t∈[0,T ]



+ ( sup

|∂12 ∇v|2 dx

sup



Ω

(3.86)



T ∂t ∇v 2L∞ {

|∂t ∂12 ∇v|2 dx + Cδ −1 T ∂t ∇v 2Hx2 { sup

Ω

|R3 | ≤δ

Ω

|∂1 ∇v|2 dx)2 + ( sup

t∈[0,T ]



+ Cδ

−1



+ ( sup ≤δ

|∂1 ∇2 v|2 dx)2 },

2

t∈[0,T ]

|∂t ∂12 ∇v|2 dx





|∂1 ∇v| dx) + ( sup 2

+ ( sup |R2 | ≤δ

Ω



(3.88)

|∂1 ∇v|2 dx Ω



|∂t ∂1 ∇v|2 dx +

Ω

|∂t ∂1 ∇2 v|2 dx),

(3.89)

Ω

|∂12 ∇v|2 dx.

Ω

After integrating (3.81) with respect to time t, we obtain (3.80) from (3.83)-(3.90). 2

(3.90)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

19

Lemma 3.9. Under the assumption (3.19), it holds that



T

ρ0 |∂t2 v|2 dx

T

|∂t2 ∇v|2 dxdt

+

≤ P (E0 ) + (δ + CT )

0 Ω

Ω



+ Cδ −1 T ( sup t∈[0,T ]

0 Ω



|∇v|2 dx + sup

|∂t ∇v|2 dx).

t∈[0,T ]

Ω

|∂t2 ∇v|2 dxdt

(3.91)

Ω

Proof. Act ∂t2 on (2.9) and take the inner product with ∂t2 v to get 1 d 2 dt



ρ0 |∂t2 v|2 dx Ω

+

∂t2 (aki ρ20 J −2 ),k ∂t2 v i dx

−

Ω

∂t2 (alj SA (v)ij ),l ∂t2 v i dx = 0.

(3.92)

Ω

Integrating by part and using the boundary conditions and Hölder’s inequality, we obtain that

|

∂t2 (aki ρ20 J −2 ),k ∂t2 v i dx| = | −

Ω



≤δ



|∂t2 ∇v|2 dx + Cδ −1 (

Ω



i ∂t2 (aki ρ20 J −2 )∂t2 v,k dx|

Ω



|∇v|2 dx +

Ω

|∂t ∇v|2 dx),

(3.93)

Ω

and



− Ω

∂t2 (alj SA (v)ij )∂t2 v,li dx −

∂t2 (alj SA (v)ij ),l ∂t2 v i dx = Ω



∂t2 (SA (v) · N ) · ∂t2 vdSx Γ

alj SA (∂t2 v)ij ∂t2 v,li dx + R,

=

(3.94)

Ω

with the remainder term

j j i R = {μ∂t (J −1 aki alj )∂t v,k + μ∂t (J −1 alj akj )∂t v,k + (μ − μ)∂t (J −1 ali akj )∂t v,k }∂t2 v,li dx Ω



+

j j i {μ∂t2 (J −1 aki alj )v,k + μ∂t2 (J −1 alj akj )v,k + (μ − μ)∂t2 (J −1 ali akj )v,k }∂t2 v,li dx,

(3.95)

Ω

which can be bounded by

|R| ≤ δ Ω

|∂t2 ∇v|2 dx + Cδ −1

|∂t ∇v|2 dx.

(3.96)

Ω

Integrating (3.92) with respect to time t, we obtain (3.91) from the resulted equation and above estimates (3.92)-(3.96). 2 3.3. Elliptic estimates In this subsection, based on the tangential derivative estimates obtained in the last subsection, we will make the full space derivative estimates through elliptic estimates.

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

20

Lemma 3.10. Under the assumption (3.19), it holds that





|∇2 v|2 dx + Ω

Ω

Ω

ρ0 |∂t v|2 dx + C

Ω

|∂t ∇v|2 dx

|∇3 v|2 dx)2

t∈[0,T ]

Ω

|∂1 ρ20 |2 dx

Ω



|∇2 v|2 dx + sup

+ CT ( sup

|∂12 ∇v|2 + C

Ω



Ω

t∈[0,T ]

|∂1 ∇v|2 dx + C Ω



ρ0 |∂t ∂1 v|2 dx + C

+C



|∇3 v|2 dx ≤ C

(3.97)

Ω

Proof. The second order derivative estimate can be made as follows. Recall the equation (2.9)1 alj (SA (v)ij ),l = ρ0 vti + aki (ρ20 J −2 ),k and the right hand viscosity term can be expanded as j j i alj (SA (v)ij ),l = μJ −1 aki alj v,kl + μJ −1 alj akj v,kl + (μ − μ)J −1 ali akj v,kl j j i + μalj ∂l (J −1 aki )v,k + μalj ∂l (J −1 akj )v,k + (μ − μ)ali ∂l (J −1 akj )v,k

Ri

.

(3.98)

For i = 1, it is reduced from (2.9)1 and (3.98) that

1 2 {(μ + μ )(a21 )2 + μ(a22 )2 }J −1 v,22 = −μ a21 a22 v,22 −

j 1 {μJ −1 ak1 alj v,kl + μJ −1 alj akj v,kl

k=1 or l=1 

+ (μ −

j μ)J −1 al1 akj v,kl }

− R1 +

ρ0 vt1

+

ak1 (ρ20 J −2 ),k ,

(3.99)

which, together with the properties (3.19)-(3.24), leads to





1 2 |v,22 | dx ≤CT Ω

2 2 |v,22 | dx + C Ω

Ω



|∂1 ∇v|2 dx + CT sup t∈[0,T ]



|∂1 ρ20 |2 dx + C

+C Ω

≤C



|∂1 ∇v|2 dx + CT sup Ω

|∂1 ρ20 |2 dx + C

+C

|∇2 v|2 dx

t∈[0,T ]



Ω

Ω

ρ0 |∂t v|2 dx Ω



|∇2 v|2 dx

Ω

ρ0 |∂t v|2 dx

(3.100)

Ω

Similarly, for i = 2, there is

2 1 {(μ + μ )(a22 )2 + μ(a21 )2 }J −1 v,22 = −μ a21 a22 v,22 −

j 2 {μJ −1 ak2 alj v,kl + μJ −1 alj akj v,kl

k=1 or l=1 

+ (μ −

j μ)J −1 al2 akj v,kl }

− R2 +

ρ0 vt2

+

ak2 (ρ20 J −2 ),k ,

(3.101)

which, also together with the properties (3.19)-(3.24), gives



2 2 |v,22 | dx ≤CT Ω

1 2 |v,22 | dx + C

Ω



|∂1 ∇v|2 dx + CT sup

Ω

t∈[0,T ]



|∂1 ρ20 |2 dx + C

+C Ω

ρ0 |∂t v|2 dx Ω

|∇2 v|2 dx Ω

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

≤C

|∂1 ∇v| dx + CT sup

|∇2 v|2 dx

2

t∈[0,T ]

Ω



Ω

|∂1 ρ20 |2 dx + C

+C

21

Ω

ρ0 |∂t v|2 dx.

(3.102)

Ω

The combination of (3.100) and (3.101) leads to



|∇ v| dx ≤C 2

Ω

|∂1 ∇v| dx + CT sup

2

|∇2 v|2 dx

2

Ω

t∈[0,T ]



|∂1 ρ20 |2 dx + C

+C Ω

Ω

ρ0 |∂t v|2 dx.

(3.103)

Ω

Now, we will make the third order derivative estimates. Taking the tangential derivative of (2.9), there is ∂1 [alj (SA (v)ij ),l ] = ρ0 ∂t ∂1 v i + ∂1 ρ0 ∂t v i + ∂1 (aki ρ20 J −2 ),k .

(3.104)

The right hand term is written as j i ∂1 [alj (SA (v)ij ),l ] = alj (SA (∂1 v)ij ),l + μ∂1 (J −1 aki alj )v,kl + μ∂1 (J −1 alj akj )v,kl j j i + (μ − μ)∂1 (J −1 ali akj )v,kl + μ∂1 ∂l (J −1 alj aki )v,k + μ∂1 ∂l (J −1 alj akj )v,k j + (μ − μ)∂1 ∂l (J −1 ali akj )v,k

(3.105)

with j j i alj (SA (∂1 v)ij ),l = μJ −1 aki alj ∂1 v,kl + μJ −1 alj akj ∂1 v,kl + (μ − μ)J −1 ali akj ∂1 v,kl j j i + μalj ∂l (J −1 aki )∂1 v,k + μalj ∂l (J −1 akj )∂1 v,k + (μ − μ)ali ∂l (J −1 akj )∂1 v,k .

(3.106)

Consequently, we obtain the following estimate from (3.104)-(3.106) with help of (3.19)-(3.24):





|∂1 ∇ v| dx ≤CT (1 + 2

2

∇ρ20 L∞ )

Ω

sup t∈[0,T ]

t∈[0,T ]

+C

Ω

Ω



|∇3 v|2 dx)2

t∈[0,T ]

Ω

|∂12 ∇v|2

2

|∇2 v|2 dx + sup

+ CT ( sup

|∂1 ∇ v| dx + C 2

Ω

∂1 ρ0 ρ0 |∂t ∂1 v|2 dx + C √ L∞ ρ0

Ω

ρ0 |∂t v|2 dx.

(3.107)

Ω

Act ∂2 on (2.9)1 to obtain ∂2 [alj (SA (v)ij ),l ] = ρ0 ∂t ∂2 v i + ∂2 ρ0 ∂t v i + ∂2 (aki ρ20 J −2 ),k , and it holds that





|∂2 ∇2 v|2 dx ≤CT (1 + ∇ρ20 L∞ ) sup t∈[0,T ]

Ω



Ω



|∇2 v|2 dx + sup

+ CT ( sup t∈[0,T ]

|∂2 ∇2 v|2 dx + C

Ω

t∈[0,T ]

|∂1 ∇2 v|2 Ω

|∇3 v|2 dx)2 Ω

(3.108)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

22



∂2 ρ0 |∂t ∇v| dx + C √ L∞ ρ0

ρ0 |∂t v|2 dx.

2

+C Ω

(3.109)

Ω

Then, (3.107) and (3.109) lead to



|∇3 v|2 dx ≤CT (1 + ∇ρ20 L∞ ) sup |∇3 v|2 dx + C |∂12 ∇v|2 t∈[0,T ]

Ω



+C

+C Ω

|∇3 v|2 dx)2

t∈[0,T ]

Ω

Ω

∂1 ρ0 ρ0 |∂t ∂1 v|2 dx + C √ L∞ ρ0

Ω



Ω



|∇2 v|2 dx + sup

+ CT ( sup t∈[0,T ]

Ω

∂2 ρ0 |∂t ∇v|2 dx + C √ L∞ ρ0



ρ0 |∂t v|2 dx Ω

ρ0 |∂t v|2 dx.

(3.110)

Ω

Using the similar method as the last lemma, we will give the high order derivative estimates of dissipation in next lemma. 2 Lemma 3.11. Under the assumption (3.19), it holds that

T



T

|∂t ∇ v| dxdt + 2

|∂t ∇3 v|2 dxdt ≤ CT

2

0 Ω

0 Ω

Ω

T

T

T

|∂t ∇ v| dxdt + CT 2

+ CT

|∂t ∇ v| dxdt + C

2

0 Ω

3

0 Ω

T

T

|∂t ∂12 ∇v|2 dxdt + C 0 Ω

|∂t2 ∇v|2 dxdt + C 0 Ω

|∇ v| dx + sup

+ CT ( sup t∈[0,T ]

Ω

T

|∇ v| dx)(

2

t∈[0,T ]

3

|∂t ∇v|2 dxdt

2

0 Ω

Ω

T



|∂t ∇2 v|2 dxdt) + CT ( sup

+

ρ0 |∂t2 v|2 dxdt 0 Ω

2

|∂t ∂1 ∇v|2 dxdt

2

0 Ω

T

+C

|∇2 ρ20 |2 dx

t∈[0,T ]

0 Ω

|∇2 v|2 dx + sup

+ CT ( sup t∈[0,T ]

t∈[0,T ]

Ω

|∇2 v|2 dx + sup

Ω

t∈[0,T ]

|∇3 v|2 dx)2 Ω

|∇3 v|2 dx).

(3.111)

Ω

Proof. Act ∂t on (2.9)1 to get ∂t [alj (SA (v)ij ),l ] = ρ0 ∂t2 v i + ∂t (aki ρ20 J −2 ),k

(3.112)

The right hand term is expanded as j i ∂t [alj (SA (v)ij ),l ] = alj (SA (∂t v)ij ),l + μ∂t (J −1 aki alj )v,kl + μ∂t (J −1 alj akj )v,kl j j i + (μ − μ)∂t (J −1 ali akj )v,kl + μ∂t ∂l (J −1 alj aki )v,k + μ∂t ∂l (J −1 alj akj )v,k j + (μ − μ)∂t ∂l (J −1 ali akj )v,k ,

(3.113)

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

23

with j j i alj (SA (∂t v)ij ),l = μJ −1 aki alj ∂t v,kl + μJ −1 alj akj ∂t v,kl + (μ − μ)J −1 ali akj ∂t v,kl j j i + μalj ∂l (J −1 aki )∂t v,k + μalj ∂l (J −1 akj )∂t v,k + (μ − μ)ali ∂l (J −1 akj )∂t v,k .

(3.114)

Using the similar method as above, we obtain from (3.112)-(3.114) that

T

T

|∂t ∇ v| dxdt ≤ CT 2

0 Ω

T

|∂t ∇ v| dxdt + C

2

2

0 Ω

T

0 Ω



ρ0 |∂t2 v|2 dxdt + CT ( sup

+





T

Ω

Ω



|∇3 v|2 dx)(

t∈[0,T ]

|∇3 v|2 dx)

t∈[0,T ]

Ω

|∇2 v|2 dx + sup

+ CT ( sup t∈[0,T ]

|∇2 v|2 dx + sup

t∈[0,T ]

0 Ω

T

|∂t ∂1 ∇v|2 dxdt

2

|∂t ∇v|2 dxdt 0 Ω

Ω

|∂t ∇2 v|2 dxdt).

+

(3.115)

0 Ω

Furthermore, acting ∂t ∂m on (2.9) for m = 1, 2, there is ∂t ∂m [alj (SA (v)ij ),l ] = ρ0 ∂t2 ∂m v i + ∂m ρ0 ∂t2 v i + ∂t ∂m (aki ρ20 J −2 ),k ,

(3.116)

from which, we obtain that

T

T

|∂t ∇ v| dxdt ≤ CT 3

0 Ω

T

|∂t ∇ v| dxdt + C

2

3

0 Ω

T

+ ρ0 L∞

|∇ v| dx + sup

t∈[0,T ]

ρ0 |∂t2 v|2 dxdt 0 Ω

t∈[0,T ]

Ω

3

Ω

t∈[0,T ]

Ω

t∈[0,T ]



|∇2 v|2 dx + sup

+ CT ( sup

t∈[0,T ]

Ω

Ω

T

|∂t ∇v|2 dxdt +

0 Ω



|∇2 v|2 dx

2

T

|∇3 v|2 dx)(



|∇ v| dx) + CT ( sup

2

+ sup

T



2

+ CT ( sup t∈[0,T ]

0 Ω

∂1 ρ0 |∂t2 ∇v|2 dxdt + √ L∞ ρ0

0 Ω

|∂t ∂12 ∇v|2 dxdt

2

|∂t ∇2 v|2 dxdt) 0 Ω



|∇3 v|2 dx)2 + CT Ω

|∇2 ρ20 |2 dx.

2

(3.117)

Ω

3.4. The combination of tangential estimates and elliptic estimates Summing the inequalities provided by our energy estimates in subsections 3.2-3.3 and taking δ small enough, we obtain a polynomial-type inequality

T t∈[0,T ]

T D(t)dt ≤ P (E0 ) + T P ( sup E(t) +

sup E(t) + 0

t∈[0,T ]

D(t)dt) 0

(3.118)

24

H. Kong, C. Liang / J. Math. Anal. Appl. 482 (2020) 123506

where P (·) denotes a polynomial function. Take M0 big enough such that P (E0 ) ≤ M0 . Taking T sufficiently small, we have

T D(t)dt ≤ 2M0 .

sup E(t) + t∈[0,T ]

(3.119)

0

Thus, the basic a priori assumption (3.19) can be bounded as follows, 3 ≤

∇v L∞ ≤ C v L∞ sup E(t) ≤ C. x,t t (H )

(3.120)

t∈[0,T ]

Then, the a priori estimates for the local solution can be closed at this point. Acknowledgments The authors are grateful to the anonymous referees for their insightful comments and suggestions to improve this paper. Kong’s work is supported by the Fundamental Research Funds for the Central Universities (FRF-TP-18-078A1) and NSFC grant No. 11901029. Liang’s work is supported by NSFC grant No. 11701053. References [1] D. Coutand, H. Lindblad, S. Shkoller, A priori estimates for the free-boundary 3D compressible Euler equations in physical vacuum, Comm. Math. Phys. 296 (2) (2010) 559–587. [2] D. Coutand, S. Shkoller, Well-posedness in smooth function spaces for the moving-boundary three-dimensional compressible Euler equations in physical vacuum, Arch. Ration. Mech. Anal. 206 (2) (2012) 515–616. [3] S. Gao, Global Solutions to the Navier-Stokes-Poisson Equations for Self-Gravitating Gaseous Stars, Thesis (Ph.D.), Northwestern University, 2010. [4] Z.H. Guo, H.L. Li, Z.P. Xin, Lagrange structure and dynamics for spherically symmetric compressible Navier-Stokes equations, Comm. Math. Phys. 309 (2) (2012) 371–412. [5] C.C. Hao, Remarks on the free boundary problem of compressible Euler equations in physical vacuum with general initial densities, Discrete Contin. Dyn. Syst. Ser. B 20 (9) (2015) 2885–2931. [6] J. Jang, Local well-posedness of dynamics of viscous gaseous stars, Arch. Ration. Mech. Anal. 195 (2010) 797–863. [7] H.H. Kong, H.-L. Li, Free boundary value problem to 3D spherically symmetric compressible Navier-Stokes-Poisson equations, Z. Angew. Math. Phys. 68 (2017) 21, 34 pp. [8] H.H. Kong, H.-L. Li, C.C. Liang, Global solutions to 3D isentropic compressible Navier-Stokes equations with free boundary, Bull. Inst. Math. Acad. Sin. (N.S.) 10 (2015) 575–613. [9] H.-L. Li, X.W. Zhang, Global strong solutions to radial symmetric compressible Navier-Stokes equations with free boundary, J. Differential Equations 261 (11) (2016) 6341–6367. [10] T.-P. Liu, Compressible flow with damping and vacuum, Jpn. J. Ind. Appl. Math. 13 (1996) 25–32. [11] T.-P. Liu, T. Yang, Compressible Euler equations with vacuum, J. Differential Equations 140 (1997) 223–237. [12] T.-P. Liu, J. Smoller, On the vacuum state for isentropic gas dynamics equations, Adv. Math. 1 (1980) 345–359. [13] T. Luo, Z.P. Xin, T. Yang, Interface behavior of compressible Navier-Stokes equations with vacuum, SIAM J. Math. Anal. 31 (2000) 1175–1191. [14] T. Luo, Z.P. Xin, H.H. Zeng, Nonlinear asymptotic stability of the Lane-Emden solutions for the viscous gaseous star problem with degenerate density dependent viscosities, Comm. Math. Phys. 347 (3) (2016) 657–702. [15] T. Luo, Z.P. Xin, H.H. Zeng, On nonlinear asymptotic stability of the Lane-Emden solutions for the viscous gaseous star problem, Adv. Math. 291 (2016) 90–182. [16] V.A. Solonnikov, A. Tani, Free boundary problem for a viscous compressible flow with a surface tension, in: Constantin Carathéodory: An International Tribute, Vol. I, II, World Sci. Publ., Teaneck, NJ, 1991, pp. 1270–1303. [17] V.A. Solonnikov, A. Tani, Evolution free boundary problem for equations of motion of viscous compressible barotropic liquid, in: The Navier-Stokes Equations II-Theory and Numerical Methods, Oberwolfach, 1991, in: Lecture Notes in Math., vol. 1530, Springer, Berlin, 1992, pp. 30–55. [18] W.M. Zajaczkowski, On nonstationary motion of a compressible barotropic viscous fluid bounded by a free surface, Dissertationes Math. (Rozprawy Mat.) 324 (1993), 101 pp. [19] W.M. Zajaczkowski, On nonstationary motion of a compressible barotropic viscous capillary fluid bounded by a free surface, SIAM J. Math. Anal. 25 (1) (1994) 1–84. [20] H.H. Zeng, Global-in-time smoothness of solutions to the vacuum free boundary problem for compressible isentropic Navier-Stokes equations, Nonlinearity 28 (2) (2015) 331–345.