Global existence of strong solutions of Navier–Stokes equations with non-Newtonian potential for one-dimensional isentropic compressible fluids

Nonlinear Analysis 74 (2011) 5876–5891

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Global existence of strong solutions of Navier–Stokes equations with non-Newtonian potential for one-dimensional isentropic compressible fluids✩ Liu Hongzhi a,b , Yuan Hongjun a,∗ , Qiao Jiezeng b , Li Fanpei b a

Institute of Mathematics, Jilin University, Changchun Jilin, 130012, China College, Huhhot 010051, China

b

Inner Mongolia Finance and Economics College, Huhhot 010051, China

article

abstract

info

Article history: Received 25 June 2010 Accepted 7 May 2011 Communicated by Enzo Mitidieri

The aim of this paper is to discuss the global existence and uniqueness of strong solution for a class of the isentropic compressible Navier–Stokes equations with non-Newtonian in one-dimensional bounded intervals. We prove two global existence results on strong solutions of the isentropic compressible Navier–Stokes equations. The first result shows only the existence, and the second one shows the existence and uniqueness result based on the first result, but the uniqueness requires some compatibility condition. © 2011 Elsevier Ltd. All rights reserved.

MSC: 76N10 76A05 Keywords: Navier–Stokes equations Isentropic compressible fluids Global strong solutions Vacuum Non-Newtonian potential

1. Introduction In this paper, we consider a class compressible Navier–Stokes equations with non-Newtonian potential for onedimensional isentropic compressible fluids

  ρt + (ρ u)x =2 0  (ρ u)t + (ρ u )x + ρ Φx − λuxx + Px = ρ f     ∫  δ(Φ )2 + 1  2−2 p 1 x   Φx = 4π g ρ − ρ dx   (Φx )2 + δ |Ω | Ω

in (0, T ) × (0, 1) in (0, T ) × (0, 1)

(a) (b)

in (0, T ) × (0, 1).

(c)

(1.1)

x

Here the unknown functions ρ = ρ(x, t ) and u = u(x, t ) denote the density and velocity, respectively. P = aρ γ (a > 0, γ > 1) is the pressure. Φ = Φ (x, t ) is the non-Newtonian gravitational potential; 1 < p < 2, 0 < δ < 1. In the rest of this paper, we normalize a = 1. Physically, this system describes the motion of compressible viscous isentropic flow under the non-Newtonian gravitational force.

✩ This work is supported by the China Sciences Foundation (10971080).

∗

Corresponding author. E-mail addresses: [email protected], [email protected] (H. Yuan).

0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.05.026

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

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In this paper, we only consider that Ω is a one-dimensional bounded interval. For simplicity we only consider that Ω = (0, 1), T < +∞. The initial and boundary conditions are

ρ|t =0 = ρ0 (x) ≥ 0, u|t =0 = u0 ∀x ∈ (0, 1) u(0, t ) = u(1, t ) = 0, Φ (0, t ) = Φ (1, t ) = 0,

(1.2)

∀t > 0.

(1.3)

The Navier–stokes equations for compressible fluids have been studied by many authors. Lions [1] used the weak convergence method and first showed the existence of global weak solutions for isentropic flow under the assumption that γ ≥ 32 if n = 2 and γ ≥ 95 if n = 3. In a paper [2], Feireisl et al. extended the Lions global existence result in R3 to the case

γ > 32 . Kim in [3] have showed that the global existence of strong solution for the one-dimensional Navier–Stokes equation.

For Navier–Stokes equations with non-Newtonian potential, in [4], Junping Yin and Zhong Tan proved the existence of strong solution to Navier–Stokes–Poisson equations. For the existence, uniqueness, or other virtues of the strong solutions of Navier–Stokes equations, we may refer to [5–12]. But as for Navier–Stokes equations with non-Newtonian potential, the results of strong solutions are few. 1.1. Main results Theorem 1.1. Assume that the initial conditions satisfy

ρ0 ∈ H 1 (0, 1),

2r

u0 ∈ H01 (0, 1),

f ∈ L2loc (0, ∞, L 2r −1 (0, 1)).

(1.4)

Then there is a global strong solution (ρ, u, Φ ) of (1.1)–(1.3), such that for all T ∈ (0, ∞), we have

ρ ∈ L∞ (0, T ; H 1 (0, 1)), Φ ∈ L∞ (0, T ; H 3 (0, 1)), u ∈ L∞ (0, T ; H01 (0, 1)),

ρt ∈ L∞ (0, T ; L2 (0, 1)) Φt ∈ L∞ (0, T ; H 2 (0, 1)) (ρ u)t , uxx ∈ L2 (0, T ; L2 (0, 1)).

Theorem 1.2. Assume that the initial conditions satisfy

ρ0 ∈ H 1 (0, 1),

2r

u0 ∈ H01 (0, 1) ∩ H 2 (0, 1)

f ∈ L2loc (0, ∞, L 2r −1 (0, 1))

fx , ft ∈ L2loc (0, ∞, L2 (0, 1))

(1.5)

and compatibility condition 1

λ(u0 )xx − (aρ0 )γx = ρ02 g for some g ∈ L2 (0, 1).

(1.6)

Then there is a unique strong solution (ρ, u, Φ ), such that for all T ∈ (0, ∞), we have

ρ ∈ L∞ (0, T ; H 1 (0, 1)) Φ ∈ L∞ (0, T ; H 3 (0, 1))

u ∈ L∞ (0, T ; H 2 (0, 1))

ρt ,

ut , Gx ∈ L2 (0, T ; H 1 (0, 1))

√ ρ ut ∈ L∞ (0, T ; L2 (0, 1))

Φt ∈ L∞ (0, T ; H 2 (0, 1)) where G = λux − P .

2. A priori estimates for smooth solutions To get the existence of strong solutions, obviously, we require some more regular estimates. So we provide that (ρ, u, Φ ) 1 is a smooth solution of (1.1)–(1.3) and T ∈ (0, ∞) is some fixed time. Moreover we may let m0 := 0 ρ0 (x)dx be initial mass and m0 > 0. To simplify, we let λ ≡ 1. In fact, as we can deal with approximate system, we only consider initial nonvacuum. Combining the classical results of (1.1)(c) with our correlated uniform estimates, we may get the existence of strong solutions of our system. To prove uniqueness, we use the classical method. Lemma 2.1. 1

∫

(ρ u2 + P )dx +

sup 0≤t ≤T

0

T

∫ 0

1

∫

u2x dxdt ≤ C

(2.1)

0

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) and |f | 0

Proof. It follows from (1.1)(a) and (b) that

ρ ut + (ρ u)ux + ρ Φx − λuxx + Px = ρ f

2r

L2 (0,T ;L 2r −1 (0,1))

, but is independent of the lower bound of ρ0 .

5878

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

multiplying this equation by u, integrating (by parts) over (0, 1), and combining the Eq. (1.1)(a) and (c), we can deal with each term as follows: 1

∫

ρ ut udx +

1

∫

1

(ρ u)ux udx =

1

2 0 ∫ 1 1

0

0

∫

ρ(u2 )t dx +

1

2 0 ∫ 1 1

ρ(u )t dx − 2 0 2 ∫ 1 1 d = ρ u2 dx =

2 dt

1

∫

1

∫

aγ ρ γ −1 ρx udx =

= − 1

∫

d

a

∫

ρ u(u2 )dx (ρ u)x (u )dx =

0

2

1

1

∫

2

ρ(u )t dx + 2

0

1 2

1

∫

ρt u2 dx 0

0

aγ ρ γ −1 (−ρt − ρ ux )dx

0

0

2

1

∫

d dt

1

∫

1

∫

γ

aρ dx + γ

Px udx 0

0

1

ρ γ dx γ − 1 dt 0 ∫ 1 ∫ 1 ∫ 1 ρ uΦx dx = − (ρ u)x Φ dx = ρt Φ dx 0 0 0    ∫ 1  ∫ 1  2−p 2−p δ(Φx )2 + 1  2 δ(Φx )2 + 1  2 1 1 = Φx Φ dx = − Φx Φx dx 4π g 0 (Φx )2 + δ 4π g 0 (Φx )2 + δ xt t ∫ 1 ∫ 1 2−p 2−p δ(Φx )2 + 1  2 2 δ(Φx )2 + 1  2 1 d 1 Φx dx + Φx Φxt dx = − 4π g dt 0 (Φx )2 + δ 4π g 0 (Φx )2 + δ ∫ 1 ∫ ∫ 2 2 − p 2−p   1 Φ x 1 d δ(Φx )2 + 1 2 2 δs + 1 2 1 d = − Φ dsdx dx + x 4π g dt 0 (Φx )2 + δ 8pπ g dt 0 0 s+δ Px udx =

0

combining these estimates, we can conclude that

∫ 1 2−p δ(Φx )2 + 1  2 2 Φx dx 2 dt 0 γ − 1 dt 0 4π g dt 0 (Φx )2 + δ ∫ 1 ∫ 1 ∫ Φx2  2−p δs + 1  2 1 d dsdx ≤ ρ uf dx + 8pπ g dt 0 0 s+δ 0 1 d

∫

1

a

ρ u2 dx +

d

∫

1

1

ρ γ dx −

d

(2.2)

we introduce the energy formula

∫ 1 ∫ 1 ∫ Φx2  2−p 2−p δs + 1  2 δ(Φx )2 + 1  2 2 1 ρu + ρ dx − Φ dx + dsdx E (t ) = x 2 γ −1 4π g 0 (Φx )2 + δ 8pπ g 0 0 s+δ 0 ∫ t∫ 1 ∫ t∫ 1 E (t ) + λ|ux |2 dxdτ ≤ E (0) + ρ|u|f dxdτ . ∫

0

1



0

1

2

a

γ



1

0

0

Multiplying (1.1)(c) by Φ and integrating over (0, 1), we get

 ∫ 1  ∫ 1  ∫ 1 2−p δ(Φx )2 + 1  2 ρ Φ dx − m0 Φ dx Φx Φ dx = 4π g (Φx )2 + δ 0 0 0 x ∫ 1  ∫ 1 ∫ 1 ∫ 1 2−p 2−p δ(Φx )2 + 1  2 2 2 2 δ Φx ≤ Φx dx = −4π g ρ Φ dx − m0 Φ dx (Φx )2 + δ 0 0 0 0 ≤ |Φ |L∞ (0,1) 8π gm0 ≤ |Φx |L2 (0,1) 8π gm0 ≤ C |Φx |L2 (0,1) ≤ ε|Φx |2L2 (0,1) + Cε ∫ 1 |Φx |2 dx ≤ c (m0 ) 1 < p < 2

(2.3)

(2.4)

0

∫ 1 ∫ 1 ∫ 1 2−p δ(Φx )2 + 1  2 2 2 Φ dx ≤ Φ dx + Φxp dx ≤ C x x 2 +δ ( Φ ) x 0 0 0

(2.5)

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

5879

combining these estimates, we can conclude that 1 √ 2

| ρ u(t )|2L2 (0,1) +

a 4(γ − 1)

√

γ

sup (| ρ u(t )|2L2 (0,1) ) + |ρ|Lγ (0,1) +

T

∫

0≤t ≤T

T

∫

γ

|ρ|Lγ (0,1) + λ

1

∫

0

u2x dxds ≤ C 0

1

∫

u2x dxdt ≤ C (m0 ).

(2.6)



0

0

Lemma 2.2. sup |ρ(t )|L∞ (0,1) ≤ C

(2.7)

0≤t ≤T

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) , but is independent of the lower bound of ρ0 . 0

Proof. Consider Lagrangian flow X = X (t , x) of u and define

  ∂X = u(t , X (t , x)), ∂t  X (0, x) = x ∈ [0, 1]. Then we only require to prove ρ(t , X (t , x)) ≤ C , for any (t , x) ∈ (0, T ] × (0, 1). Let t0 ∈ (0, T ] be any fixed time. Combining 1 C −1 ≤ 0 ρ0 (x)dx = m0 ≤ C , the conservative mass, and the means of the Lagrangian flow X = X (t , x), proving by contradiction, we can easily find some x1 ∈ (0, 1), such that C −1 ≤ ρ0 (x1 )

and ρ(t0 , X (t0 , x1 )) ≤ C .

(2.8)

Furthermore, if we take some subset (a, b) ⊂ (0, 1) such that C conservative mass, we get X (t0 ,b)

∫

X ( t0 ,a )

ρ(t0 , x)dx =

−1

≤ ρ(t0 , x), then combining the means of X and the

b

∫

ρ0 (x)dx ≤ C . a

Next, we prove that for any x2 ∈ (0, 1), ρ(t0 , X (t0 , x2 )) ≤ C holds. Let Xj (t ) := X (t , xj ), j = 1, 2 and L(t ) = log ρ(t , X2 (t ))− log ρ(t , X1 (t )) Then using (1.1)(a)–(c), we get dL dt

1

=

ρ(t , X2 (t ))

=

ρ(t , X2 (t ))

1

  dX2 ρt (t , X2 (t )) + ρx − dt

ρ(t , X1 (t ))

X2 (t )

∫

X1 (t )

let U (t ) =

 X2 (t ) X1 (t )

dU (t ) dt

ρt (t , X1 (t )) + ρx



dt

(−ρx (t , X1 )u(t , X1 ) − ρ(t , X1 )ux (t , X1 ) + ρx u(t , X1 ))

= −(ux (t , X2 ) − ux (t , X1 )) = − =−

ρ(t , X1 (t ))

dX1

(−ρx (t , X2 )u(t , X2 ) − ρ(t , X2 )ux (t , X2 ) + ρx u(t , X2 ))

1

−



1

X2 (t )

∫

uxx dx

X1 (t )

(ρ ut + ρ uux + Px + ρ Φx − ρ f )dx

(2.9)

ρ u(t , x)dx. Then

= ρ u(t , X2 (t ))

dX2 dt

dX1

− ρ u(t , X1 (t ))

dt

= ρ u2 (t , X2 (t )) − ρ u2 (t , X1 (t )) + ∫

X2 (t )

= X1 (t )

(ρ u(t , x))ux (t , x)dx +

X2 (t )

∫

X1 (t )

X1 (t )

X2 (t )

+ X1 (t )

X2 (t )

∫

∫

(ρ u(t , x))t dx

[−(ρ u)x u + ρ ut ]dx

(ρ ut (t , x))dx

(2.10)

substituting (2.10) into (2.9), we have dL(t ) dt

+

dU (t ) dt

∫ =−

X2 (t ) X1 (t )

∫ Px dx +

X2 (t ) X1 (t )

ρ Φx dx −

∫

X2 (t ) X1 (t )

 ρ f dx .

(2.11)

5880

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891 P (ρ(t ,X2 ))−P (ρ(t ,X1 )) . L(t )

Let α(t ) =

dL(t ) dt

+

dU (t )

We easily find α(t ) ≥ 0. Thus (2.11) becomes

dt

X2 (t )

∫

= −α(L + U ) + α U −

X1 (t )

(ρ Φx − ρ f )dx.

(2.12)

Using the theorem of ODE, we get L(t ) + U (t ) = e−

t 0

α(s)ds

t

∫

(L(0) + U (0)) +

e−

t s

α(τ )dτ

 ∫ α(s)U (s) −

X2 (s)

X1 (s)

0

 (ρ Φx − ρ f )dx ds

combining L(0) < C (|ρ0 (x)|L∞ (0,1) ≤ C ), we have t

∫

L(t ) ≤ C + |U (0)| + |U (t )| +

e−

t s

α(τ )dτ



α(s)|U (s)| +

∫

X1 (s)

0 t

∫

e−

×

t s

α(τ )dτ

α(s)|U (s)|ds ≤ sup |U (s)| 0≤s≤t0

0

= − sup |U (t )|

t

∫

0 ≤ t ≤ t0

e−

t s

α(τ )dτ

0

= sup |U (t )|e−

t

e−

t s

 ρ|Φx |dx ds

α(τ )dτ

α(s)ds

0

 t t   α(τ )dτ = sup |U (t )|e− s α(τ ) 

0

0≤t ≤t0

s

α(τ )

0

t

∫ d

t

∫

X2 (s)

dτ ≤ sup |U (t )|.

0 ≤ t ≤ t0

0≤t ≤t0

Particularly, we let t = t0 . Thus L(t0 ) ≤ C + |U (0)| + sup |U (t )| +

∫

0≤t ≤t0

t0

X2 (s)

∫

X1 (s)

0

(ρ|Φx | − ρ f )dxds.

(2.13)

We deal with the latter terms as follows:

|U (0)| ≤

∫

x2

ρ0 |u(0, x)|dx ≤ C

(2.14)

x1

|U (t )| ≤

∫

X2 (t )

X1 (t )

ρ|u(t , x)|dx ≤

X2 (t )

∫

X1 (t )

ρ dx

 12 ∫

X2 (t ) X1 (t )

ρ u2 dx

 21

.

Consequently sup |U (t )| ≤ sup 0≤t ≤t0

1

∫

0 ≤ t ≤ t0

ρ dx

 21 ∫

0

 12 ρ u2 dx .

1 0

Combining (2.6), we get sup |U (t )| ≤ C

(2.15)

0≤t ≤t0

∫ t∫ 0

∫ t∫ 0

X2 (s) X1 (s) X2 (s) X1 (s)

ρ|f |dxds ≤ C

∫ t ∫ 0

1

γ

ρ dx

 γ1 ∫

0

0

2γ

|f | γ −1 dx

 γ2−γ1

ds ≤ C (T )

0

∫ t ∫

ρ|Φx |dxds ≤ C ∫ t 0∫ ≤

1

1

γ

 γ1 ∫

1

ρ dx

0

2γ

|Φx | γ −1 dx

 γ2−γ1

ds

0

1

|Φxx |dxds ≤ C (m0 , T )

(2.16)

0

from (2.13)–(2.16), we have L(t0 ) ≤ C . Then log ρ(t0 , X (t0 , x2 )) = log ρ(t0 , X (t0 , x1 )) + L(t0 ) ≤ C .

(2.17)

Because to x2 , t0 are arbitrary, Lemma 2.2 is proved. To get the higher estimates, the effective viscous flux is very important. In the proof below, usually we will use the regularity of Gx = uxx − px . 

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

5881

Lemma 2.3. sup (|u|L∞ (0,1) + |ux |L2 (0,1) ) +

T

∫

0≤t ≤T

√ | ρ ut |2L2 (0,1) dt ≤ C

0

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) and |f |

(2.18)

2r

L2 (0,T ;L 2r −1 (0,1))

0

, but is independent of the lower bound of ρ0 .

Proof. Multiplying (1.1)(b) by ut and integrating over (0, 1), we get 1

∫

ρ u2t dx +

1

∫

ρ uux ut dx +

1

∫

ρ Φx ut dx − λ

1

∫

ρ fut dx −

uxx ut dx =

1

∫

Px ut dx 0

0

0

0

0

0

1

∫

Thus 1

∫

ρ

u2t dx

d

+

dt

0

1

∫

1 2

0

∫

u2x dx

1

ρ

≤C

u2 u2x dx

1

∫

Φx2 dx

+

0

1

∫ +

ρ|f | dx . 2

Puxt dx +

0

0



1

∫

(2.19)

0

Next, we deal with each of the above terms as follows: 1

∫

Puxt dx = 0

= = 1

∫ −

d

1

∫

dt 0 ∫ 1 d dt 0 ∫ 1 d dt

1

∫ Pux dx −

Pt ux dx =

∫0 1 Pux dx −

∫0 1 Pux dx −

0

d

1

∫

dt

0

Puuxx dx + (γ − 1)

aγ ρ γ −1 ρt ux dx =

1

0

Pu2x dx 0

Pu(Gx + Px )dx + (γ − 1)

= γ

∫

1

Pu2x dx +

∫0 1

1

∫

P (G + P )2 dx

(2.20)

0

aγ ρ γ −1 ρ ux ux dx +

0

aγ ρ γ −1 ρt ux dx

0

1

∫

0

∫

1

∫ Pux dx −

∫

1

∫

aγ ρ γ −1 ρx uux dx

0

1

Px uux dx

∫0 1

∫ 1 (Pu)x ux dx − Pu2x dx 0 0 0 ∫ 1 ∫ 1 = (γ − 1) Pu2x dx + Puuxx dx

= γ

Pu2x dx +

0 1

∫

Pu(Gx + Px )dx = −

−

0 1

∫

dt

∫ PuGx dx −

0

d

P 2 dx = 2 0

∫

PPt dx = −2 0

1

∫

P 2 ux dx − 2

0 1

Px uPdx = 0

(γ − 1)

1

Px uPdx 0 1

Px uPdx = 2(2γ − 1)

PPx udx − 2 0

d

2(2γ − 1) dt

1

∫

Px uPdx 0

1

∫

P 2 dx 0

1

∫

(γ Pux + Px u)Pdx ∫ 1

∫

0 1

PuPx dx

1

∫

0

1

0

1

= −2γ ∫ = 4γ ∫

0

1

∫

Pu2x dx = (γ − 1)

1

∫

0

P (G + P )2 dx 0

= (γ − 1)

∫

PG2 dx + 2(γ − 1) 0

= (γ − 1)

∫ ∫

1

P 2 Gdx + (γ − 1)

∫

1

1

∫

P 3 dx 0

1

P 2 (ux − P )dx + (γ − 1)

0

PG2 dx − 4(γ − 1) 0

1

∫ 0

PG2 dx + 2(γ − 1) 0

= (γ − 1)

1

∫

1

PPx udx − (γ − 1) 0

1

∫

1

∫

P 3 dx 0

P 3 dx 0

5882

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891 1

∫

d

Puxt dx =

1

∫

dt 0 ∫ 1 d

0

=

dt 0 ∫ 1 d

=

dt 0 ∫ 1 d

=

dt 0 ∫ 1 d

=

dt 0 ∫ 1 d

=

dt

1

∫

Pu(Gx + Px )dx + (γ − 1)

Pux dx − 0

∫

1

PuPx dx + (γ − 1)

∫0 1

∫0 1 PuGx dx −

Pux dx −

∫0 1

∫0 1 PuGx dx −

Pux dx − 0

∫

PuPx dx + (γ − 1)

PuGx dx − (4γ − 3)

Pux dx − 0

∫

1

PuGx dx −

Pux dx − 0

0

PuPx dx + (γ − 1)

0

1

P (G + P )2 dx 0

1

∫ PuGx dx −

Pux dx −

1

∫

d dt

2

P (G2 − P 2 )dx − 4(γ − 1)

0

PuPx dx + (γ − 1)

4γ − 3

1

P (2GP + 2P 2 )dx

0 ∫

1

Pux Pdx

∫0 1 PuPx dx 0

1

∫

P (G2 − P 2 )dx 0

2(2γ − 1)

0

∫0 1

∫

P (G − P )dx + 2(γ − 1) 2

0

1

∫

P (G2 − P 2 )dx + (γ − 1)

∫0 1

1

∫

1

∫

P 2 dx + (γ − 1)

1

∫

P (G2 − P 2 )dx. 0

Consequently 1

∫

Puxt dx = 0

d dt

1

∫

Pux dx − 0

dt

4γ − 3

1

∫

d

2(2γ − 1)

0

P 2 dx + (γ − 1)

1

∫

P (G2 − P 2 )dx −

1

∫

0

PuGx dx

(2.21)

0

substituting (2.21) into (2.19) and integrating over (0, t ), we get

∫ t∫ 0

1

1

1

2

2

ρ u2t dxds + u2x (t )dx − u2x (0)dx 0

(ρ

≤C+ 0

u2 u2x

Pux (0)dx −

−

0

2(2γ − 1)

ρ u2 u2x dxds ≤

4γ − 3 2(2γ − 1)

P 2 (0)dx − (γ − 1)

2

0

0

0

1

∫

|ux |4L2 (0,1) ds

0

t 0

√ γ−1 |ρ|L∞ (20,1) | ρ u|L2 (0,1) |Gx |L2 (0,1) ds ≤ C

t

∫ 0

|Gx |L2 (0,1) ds

and Gx = uxx − Px = ρ ut + ρ uux + ρ Φx . Consequently,

√ |Gx |L2 (0,1) ≤ C (| ρ ut |L2 (0,1) + |uux |L2 (0,1) + |Φx |L2 (0,1) ). Then (2.23) becomes 1

t

∫ P |u||Gx |dxds ≤ C 0

√ (| ρ ut |L2 (0,1) + |uux |L2 (0,1) + |Φx |L2 (0,1) )ds ∫ t∫

1

≤ C +C 0

0

|ux |4L2 (0,1) ds +

1

∫ t∫

2

combining the above estimates, we get t

∫ 0

√ | ρ ut |2L2 (0,1) ds + |ux |2L2 (0,1) ≤ C + C

using the Grownwall’s inequality, we obtain t

∫ 0

0

P (u2x + P 2 )dxds ≤ C

∫

0

0

P 3 dxds

0

P |u||Gx |dxds ≤ C

0

1

1

∫ ∫

1

∫ t∫

∫ t∫

(2.22)

|ρ|L∞ (0,1) |u|L∞ (0,1) |ux |2L2 (0,1) ds ≤

PG dxds ≤ 2

0

Pux (t )dx 0

t

∫ 0 t

1

∫ t∫

0

0

1

∫

(ρ u2 u2x + PG2 + P |u||Gx |)dxds

0

0

P 2 (t )dx +

0 1

∫

+ ρ f )ds + 2

0

1

∫ t∫

(

Φx2 dx

1

≤C+

0

4γ − 3

1 0

∫ t∫

1

+ PG + P |u||Gx |)dxds +

∫

0

∫ t∫

2

0

1

∫

∫ t∫

1

∫ t∫

√ (| ρ ut |2L2 (0,1) + |ux |2L2 (0,1) )ds ≤ C

t

∫ 0

|ux |4L2 (0,1) ds

0

1

ρ u2t dxds 0

(2.23)

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

5883

|u|L∞ (0,1) ≤ |u|W 1,2 (0,1) = |u|L2 (0,1) + |ux |L2 (0,1) ≤ C |ux |L2 (0,1) ∫ T √ 2 ∞ | ρ ut |2L2 (0,1) dt ≤ C .  sup (|u|L (0,1) + |ux |L2 (0,1) ) + 0≤t ≤T

0

Lemma 2.4. T

∫

(|ux |2L∞ (0,1) + |Gx |2L2 (0,1) )dt ≤ C

0

(2.24)

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) and |f |

2r

L2 (0,T ;L 2r −1 (0,1))

0

, but is independent of the lower bound of ρ0 .

Proof.

|ux |2L∞ (0,1) ≤ 2(|G|2L∞ (0,1) + |P |2L∞ (0,1) ) ≤ 2(|G|2L2 (0,1) + |Gx |2L2 (0,1) + |P |2L∞ (0,1) ) ∫ T ∫ 1  ∫ T ∫ T∫ 1 2 2 2 |G|L2 (0,1) dt ≤ 2 |ux | dxdt + |P | dxdt ≤ C 0

0

T

∫

|Gx |2L2 (0,1) dt ≤ C

0

0

∫

T

∫0

0

0

 √ √ 2 2 2 (C + | ρ ut |L2 (0,1) + |uux |L2 (0,1) + | ρ f |L2 (0,1) ) ds T

≤ C+ 0

(|ux |2L2 (0,1) )ds ≤ C .

From the above estimates, we have sup (|ρ|L∞ (0,1) + |ux |L2 (0,1) ) +

0≤t ≤T

T

∫ 0

√ (| ρ ut |2L2 (0,1) + |ux |2L∞ (0,1) + |Gx |2L2 (0,1) )dt ≤ C

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) .

(2.25)



0

Lemma 2.5. sup |ρx |L2 (0,1) ≤ C (T )

(2.26)

0 ≤t ≤T

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) , but is independent of the lower bound of ρ0 . 0

Proof. Differentiating (1.1)(a) with respect to x gives

(ρx )t + (pu)xx = 0. Multiplying it by ρx and integrating over (0, 1), we get 1

∫

(ρx )t ρx dx + 0

(ρ u)xx ρx dx = 0 0

d 1 dt 2

1

∫ 0

(ρx u)x ρx dx =

[(ρx u) + ρ ux ]x ρx dx = 0

0 1

∫

0

1

∫

(ρx )2 dx +

1

∫

1

∫

(ρxx u + ρx ux )ρx dx =

0

(2.27)

1

∫

ρxx ρx udx + 0

∫

1

ρx2 ux dx.

0

However 1

∫

ρxx ρx udx = −

1

∫

0

ρx (ρx u)x dx = − 0

1

ρxx ρx udx = − 0

1 2

ρxx ρx udx − 0

Thus

∫

1

∫

1

∫

ρx2 ux dx 0

1

∫

ρx2 ux dx. 0

5884

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

but 1

∫

ρ

2 x ux dx

ρx2 dx

≤| |

0

0 1

∫

1

∫ ux L∞ (0,1) 1

∫

1

∫

ρ uxx ρx dx ∫ 1 ρ(Gx + Px )ρx dx ρx2 dx + ≤ |ux |L∞ (0,1) 0 0  ∫ 1  ∫ 1 ∫ 1 ∫ 1 aγ ρ γ ρx2 dx ρx2 dx + G2x dx + ρx2 dx + ≤ C |ux |L∞ (0,1) 0 0 0 0   ∫ 1 ∫ 1 ∫ 1 ρx2 dx G2x dx + ρx2 dx + ≤ C |ux |L∞ (0,1)

(ρ ux )x ρx dx =

ρx2 ux dx + ∫ 1

0

0

0

(2.28)

0

0

0

combining (2.27) and the above estimates, we get d 1

1

∫

dt 2

1

 ∫ (ρx )2 dx ≤ C |ux |L∞ (0,1)

G2x dx +

1

∫

 ρx2 dx .

0

0

0

0

1

∫

ρx2 dx +

From the above lemmas and using Grownwall’s inequality, we have sup |ρx |L2 (0,1) ≤ C (T ). 

0≤t ≤T

Lemma 2.6. sup (|ρ|H 1 (0,1) + |ρt |L2 (0,1) + |ux |L2 (0,1) ) +

T

∫

0≤t ≤T

(|(ρ u)t |2L2 (0,1) + |uxx |2L2 (0,1) )ds ≤ C (T )

0

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) , but is independent of the lower bound of ρ0 . 0

Proof. From Eq. (1.1)(a), we have ρ = [−(ρx u + ρ ux )]2 integrating it over (0, 1), we get 2 t

1

∫

1

∫

ρt2 dx = 0

1

∫

ρx2 u2 dx + 2 0

0

ux 2L∞ (0,1)

1

∫

ρx ρ uux dx +

ρ 2 u2x dx 0

1

∫

ρ

2 x dx

≤| |

2 L∞ (0,1)

+ 2|ρ|L∞ (0,1) |u|L∞ (0,1) |ρx |L2 (0,1) |ux |L2 (0,1) + |ρ|

0

1

∫

u2x dx. 0

Thus 1

∫

ρ

sup 0≤t ≤T

2 t dx

2 L∞ (0,1)

≤ sup |u|

sup

0≤t ≤T

0≤t ≤T

0

2 L∞ (0,1)

+ sup |ρ| 0≤t ≤T

1

∫

ρx2 dx + 2 sup |ρ|L∞ (0,1) sup |u|L∞ (0,1) sup |ρx |L2 (0,1) sup |ux |L2 (0,1) 0 ≤t ≤T

0

ux 2L2 (0,1)

sup | | 0≤t ≤T

≤C

0≤t ≤T

0≤t ≤T

0 ≤t ≤T

(2.29)

consequently T

∫ 0

∫ 0

1

(ρ u)

2 t dxdt

T

∫

1

∫

(ρt u) dxdt + 2 2

≤2 0

0

0 2 L∞ (0,1)

∫

T

1

≤ 2|ρ 2 |2L∞ (0,T )×(0,1)

0

1

∫

(ρ ut )2 dxdt 0

1

∫

(ρt )2 dxdt

≤ 2 sup |u| 0≤t ≤T

T

∫

0 T

∫ 0

∫

1

√ ( ρ ut )2 dxdt ≤ C

(2.30)

0

combining the momentum equation, we have uxx = (ρ u)t + (ρ u2 )x + ρ Φx + Px − ρ f

(2.31)

combining |u|L∞ (0,T )×(0,1) ≤ C , |ρ|L∞ (0,T )×(0,1) ≤ C , |ρx |L∞ (0,T ;L2 (0,1)) ≤ C , |ux |L∞ (0,T ;L2 (0,1)) ≤ C we see that (ρ u2 )x ∈ L2 (0, T ) × (0, 1) ≤ C , in fact, we have

(ρ u2 )x ∈ L∞ (0, T ; L2 (0, 1))

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

5885

|(ρ u2 )x |L2 = |(ρx u2 ) + ρ uux |L2 = |u|2L∞ |ρx |L2 + |ρ|L∞ |u|L∞ |ux |L2 ≤ C ∫ T∫ 1 ∫ T∫ 1 |Φx |2 dxdt ≤ C |ρ Φx |2 dxdt ≤ |ρ|2L∞ (0,T )×(0,1) 0

0

0

0

ρ f ∈ L2 ((0, T ) × (0, 1)) combining the above estimates and (2.31), we have T

∫

sup (|ρ|H 1 (0,1) + |ρt |L2 (0,1) + |ux |L2 (0,1) ) +

0≤t ≤T

(|(ρ u)t |2L2 (0,1) + |uxx |2L2 (0,1) )ds ≤ C (T ).

0

To prove Theorem 1.2, we must deal with the following estimate.

(2.32)



Lemma 2.7.

√ | ρ ut (t )|2L2 (0,1) +

t

∫

utx 2L2 (0,1) ds

| τ

|

where C is dependent on |ρ0 |H 1 (1,0) , |u0 |H 1 (1,0) , |f | lower bound of ρ0 .

2r

L2 (0,T ;L 2r −1 (0,1))

0

t

∫

√ ≤ C + | ρ ut (τ )|2L2 (0,1) + C

0

√ |ux |L∞ (0,1) | ρ ut |2L2 (0,1) ds

(2.33)

, |fx |L2 (0,T ;L2 (0,1)) and |ft |L2 (0,T ;L2 (0,1)) , but is independent of the

Proof. Combining the momentum equation (1.1)(b) and the mass equation (1.1)(a), we can easily get

ρ ut + ρ uux + ρ Φx − uxx + Px = ρ f . Differentiating it with respect to time, we have

ρ utt + ρt ut + ρ uuxt + ρt uux + ρ ut ux + ρt Φx + ρ Φxt + Pxt − uxxt = ρt f + ρ ft that is

ρ utt + ρ uuxt + pxt − uxxt = −ρt (ut + uux + Φx − f ) − ρ ut ux − ρ Φxt + ρ ft . Multiplying it by ut and integrating over (0, 1), we have 1

∫

d dt

1

0

2

∫

1

ρ

u2t dx

1

∫ −

1 2

0

ρ

2 t ut ddx

1

∫ +

1 2

0

(ρ

)

uu2t x dx

1

1 2

0

1

∫ −

2

(ρ u)

ρ u2t ux dx −

2 x ut dx

1

∫ −

u2xt dx

Pt uxt dx +

1

∫

1

∫

0

0

0 1

∫

ρ Φxt dx +

ρ ft ut dx

0

0

1

∫

(ρ uu2t )x dx = 0 1 2

0

1

1

∫

0

∫

− 0

(ρ u)x (u2t + uux ut + Φx ut − fut )dx −

=

1

∫

1

∫

ρt u2t dx =

1 2

0

(ρ u)x u2t dx.

From Eq. (1.1)(a), we have 1

∫

d dt

1 2

0

ρ u2t dx +

1

∫

u2xt dx − 0

1

∫

Pt uxt dx 0

1

∫

ρ u(

u2t

=−

+ uux ut + Φx ut − fut )x −

1

∫

0

ρ

u2t ux dx

∫

ρ Φxt ut dx +

−

0

0

1

ρ ft ut dx.

(2.34)

0

We deal with each term of (2.34) as follows: 1

∫ −

1

∫

aγ ρ

Pt uxt dx = 0

γ −1

(ρ u)x uxt dx =

0

= = = =

d

dt

1

1

1 0

γ 2

0

∫

γ 2

0

∫

γ 2

0

dt d

1

∫

dt d

(Px u + γ Pux )uxt dx 0

∫

dt d

1

∫

γ 2

Pu2x dx

1

∫ −

2

0

Pu2x dx +

γ

Pt u2x dx

+

1

∫

Px uuxt dx + 0

Pu2x dx

1

∫ +

Px uuxt dx + 0

Pu2x dx +

1

∫

Px uuxt dx + 0

1

∫

Px uuxt dx 0

γ 2

γ

2

(Px u + γ Pux )u2x dx 0 1

∫

2

γ

1

∫

γ Pu3x dx − 0

γ 2

1

∫

Pu3x dx − 0

γ 2

1

∫

(−Pu(u2x )x + (γ − 1)Pu3x )dx. 0

1

∫

Pu(u2x )x dx 0

5886

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

Substituting it into (2.34), we have 1

∫

d dt



1

ρ

2

0

u2t

+

γ 2

Pu2x



Px |u||uxt |dx + γ

1

∫

P |ux ||u||uxx |dx +

γ

0

0 1

∫

ρ u(Φx ut )x dx +

− 1

∫

(γ − 1)P |ux |3 dx 0

ρ|ut ||u||ux | dx + 2

1

∫

ρ|fx ||u||ut |dx +

ρ|ut | |ux |dx − 2

1

∫

ρ|uxt ||u|2 |ux |dx 0

ρ Φxt ut dx 0

ρ|f ||u||uxt |dx +

1

∫

ρ|ft ||ut |dx ≡

13 −

0

0

0

ρ|ut ||u| |uxx |dx + 2

1

∫

1

∫

1

∫ 0

0

0

+

2

0

0 1

∫

1

∫

1

∫

ρ|ut ||u||uxt |dx +

+2

Ij

j =1

1

∫

|Px ||u||uxt |dx ≤ ε|uxt |2L2 (0,1) + C (ε)|Px |2L2 (0,1)

I1 = 0

1

∫

I2 = γ

1

∫

|ux ||uxx |dx ≤ C (|ux |2L∞ (0,1) + |uxx |2L2 (0,1) )

P |ux ||u||uxx |dx ≤ C 0

γ

I3 =

u2xt dx 0

1

∫ ≤

1

∫ dx +

0 1

∫

2

(γ − 1)P |ux |3 dx ≤ C |ux |2L2 (0,1) |ux |2L∞ (0,1) ≤ C |ux |2L∞ (0,1)

0 1

∫

√ √ ρ|ut ||u||uxt |dx ≤ C | ρ ut |L2 (0,1) |uxt |L2 (0,1) ≤ ε|uxt |2L2 (0,1) + C (ε)| ρ ut |2L2 (0,1)

I4 = 2 0 1

∫ I5 = 0

1

∫

√ √ ρ|ut ||u||ux |2 dx ≤ C | ρ ut |L2 (0,1) |ux |L∞ (0,1) |ux |L2 (0,1) ≤ C | ρ ut |2L2 (0,1) + C |ux |2L∞ (0,1) √ √ ρ|ut ||u|2 |uxx |dx ≤ C | ρ ut |L2 (0,1) |uxx |L2 (0,1) ≤ C (| ρ ut |2L2 (0,1) + |uxx |2L2 (0,1) )

I6 = 0

1

∫

ρ|uxt ||u|2 |ux |dx ≤ C

I7 = 0

1

∫

|uxt ||ux |dx ≤ ε|uxt |2L2 (0,1) + C (ε)|ux |2L2 (0,1)

0

combining Eq. (1.1)(c) we can get|Φxx |L∞ (0,1) ≤ C

 ∫ 1 ∫ 1  ρ u(Φx ut )x dx ≤ ρ|u||Φx ||uxt |dx + ρ|u||Φxx ||ut |dx 0 0 0 √ ≤ C |Φx |2L2 (0,1) + C | ρ ut |2L2 (0,1) + ε|uxt |2L2 (0,1) ∫ 1 √ I9 = ρ|ut |2 |ux |dx ≤ C |ux |L∞ (0,1) | ρ ut |2L2 (0,1) 0 ∫ 1  ∫ 1    √ 2 2   I10 =  ρ Φxt ut dx ≤ C | ρ ut |L2 (0,1) + C |Φxt | dx ∫  I8 = 

1

0

1

∫ I11 = 0

0

√ ρ|fx ||u||ut |dx ≤ C | ρ ut |2L2 (0,1) + C |fx |2L2 (0,1)

1

∫ I12 = 0

ρ|f ||u||uxt |dx ≤ C |f |2L2 (0,1) + ε|uxt |2L2 (0,1) .

We deal with the estimate of Φxt . Differentiating (1.1)(c) with respect to time, multiplying it by Φt and integrating over (0, 1), we get p−2

µ0 2

1

∫

Φxt2 dx ≤

1

∫

0

(Φx2 + µ0 )

p−4 2

((p − 1)Φx2 + µ0 )Φxt2 dx

0 1

∫

((Φx2 + µ0 )

=

p−2 2

Φx )t Φxt dx = −4π g

0

≤ C |ρt |L2 (0,1) |Φt |L2 (0,1) ≤ C |ρt |L2 (0,1) |Φxt |L2 (0,1) ≤ C |ρt |2L2 (0,1) + ε|Φxt |2L2 (0,1) .

1

∫

ρt Φt dx 0

(2.35)

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

5887

Thus 1

∫

Φxt2 dx ≤ C |ρt |2L2 (0,1)

0

√

I10 ≤ C (| ρ ut |2L2 (0,1) + |ρt |2L2 (0,1) ) 1

∫ I13 = 0

√ ρ|ft ||ut |dx ≤ C |ft |2L2 (0,1) + C | ρ ut |2L2 (0,1)

from the estimates of I1 –I13 and (2.35) we obtain 1

∫

d dt



1 2

0

ρ

u2t

γ

+

2

Pu2x



1

∫

u2xt dx

dx + 0

√ ≤ C (|Px |2L2 (0,1) + |ux |2L∞ (0,1) + |uxx |2L2 (0,1) + |ft |2L2 (0,1) + |fx |2L2 (0,1) + | ρ ut |2L2 (0,1) + |ρt |2L2 (0,1) ) √ + C |ux |L∞ (0,1) | ρ ut |2L2 (0,1)

(2.36)

integrating it over (τ , t ) ⊂ (0, T ), we conclude that 1

∫

ρ u2t (t )dx + 0

≤C +C

∫ t∫ τ

1

u2xt dxds 0

1

∫

ρ u2t (τ )dx + 0

1

∫



Pu2x (τ )dx + C 0

t

∫ τ

√ |ux |L∞ (0,1) | ρ us |2L2 (0,1) ds.

Using Lemma 2.3, we get 1

∫

ρ u2t (t )dx +

∫ t∫ τ

0

1

√

u2xt dxds ≤ C + C | ρ ut (τ )|2L2 (0,1) + C

0

t

∫ τ

√ |ux |L∞ (0,1) | ρ us |2L2 (0,1) ds.

(2.37)

Combining the momentum equation (1.1)(b) and compatibility condition, we have

ρ ut + ρ uux + ρ Φx − uxx + Px = ρ f 1

1

1

1

1

1

ρ 2 ut = −(ρ 2 uux + ρ 2 Φx − ρ − 2 uxx + ρ − 2 px ) + ρ 2 f consequently 1 1 √ √ √ √ | ρ ut (τ )|2L2 (0,1) ≤ (C (| ρ f |2L2 (0,1) ) + | ρ Φx |2L2 (0,1) ) + (| ρ uux |2L2 (0,1) + |ρ − 2 px − ρ − 2 uxx |2L2 (0,1) ).

γ

1

Combining the above estimates |ft | ∈ L2 (0, T ; L2 (0, 1)), sup0≤t ≤T |f |2L2 (0,1) ≤ C , |Φx |2L2 (0,1) ≤ C |ρ|Lγ (0,1) + C , |ρ 2 uux |2L2 (0,1) ≤ C |ux |2L2 (0,1) and sup0≤t ≤T |ux |2L2 (0,1) ≤ C , we obtain

1 1 √ | ρ ut (τ )|2L2 (0,1) ≤ C + C |ρ − 2 Px (τ ) − ρ − 2 uxx (τ )|2L2 (0,1) .

Substituting it into (2.37) and letting τ → 0, we get 1

∫

1 2

0

ρ u2t (t )dx +

∫ t∫ 0

1

u2xt dxds ≤ C + C 0

0

Using Grownwall’s inequality and 1

∫

ρ u2t (t )dx + 0

∫ t∫ τ

t

∫

T 0

√ |ux |L∞ (0,1) | ρ us |2L2 (0,1) ds.

|ux |L∞ (0,1) dt ≤ C , we get

1

u2xt dxds ≤ C .

(2.38)

0

Next, we construct the approximate systems to deal with the existence.



3. Proof of the existence Our method that constructed approximate systems is similar to that in [11], we take a semidiscrete Galerkin scheme. We take our basic function space as X = H01 (0, 1) ∩ H 2 (0, 1) and the finite-dimensional subspaces as X m = span {ϕ 1 , ϕ 2 , . . . , ϕ m } ⊂ X ∩ C 2 ([0, 1]). Here ϕ m is the mth eigenfunction of the strongly elliptic operator A = − ∂∂x2 2

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H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

defined on X . Let ρ0 , u0 satisfy the hypotheses of Theorem 1.1 or Theorem 1.2. Assume for the moment that ρ0δ ∈ C 1 ([0, 1]) and ρ0δ ≥ δ in (0, 1) (for some constant δ > 0). We may construct an approximate solution for any v ∈ X m , ϕ ∈ C 2 ([0, 1])

∫ 1 ∫ 1  m m m m m m m m m  ρ m f δ v dx  (ρ ut + ρ u · ux + Au + ρ Φx + Px ) · v dx =   0 0  ∫ ∫  1 1 ρtm ϕ dx + (ρ m um )x ϕ dx = 0  0   ∫0 1 ∫ 1   p−2 m0    ((|Φxm |2 + µ0 ) 2 Φxm )x ϕ dx = 4π g ρm − ϕ dx |Ω | 0 0 2r

where f δ ∈ C 1 ((0, T ) × (0, 1)) and f δ → f in L2 (0, T ; L 2r −1 (0, 1)). The initial and boundary conditions are um 0 ≡

m −

(u0 , ϕ k )L2 [0,1] ϕ k and ρ m (0) = ρ0δ > δ,

ρ δ (0) < |ρ0 |L∞ + 1,

k=1

|ρ0δ − ρ0 |H 1 (0,1) → 0,

um (0, x) = um (1, x) = 0,

Φ m (0, x) = Φ m (1, x) = 0.

Under the hypotheses of Theorem 1.1, similarly, for any fixed δ > 0, we may get the similar estimate of Lemmas 2.1–2.6 2 sup (|ρδm |L∞ (0,1) + |um xδ |L2 (0,1) ) +

T

∫

0≤t ≤T

0

 2 m m 2 (| ρδm um δ t |L2 (0,1) + |uxδ |L2 (0,1) + |Gxδ |L2 (0,1) )dt ≤ C (T ) T

∫

2 sup (|ρδm |H 1 (0,1) + |ρδmt |L2 (0,1) + |um δ x |L2 (0,1) ) +

0≤t ≤T

0

2 m 2 (|(ρδm um δ )t |L2 (0,1) + |uxxδ |L2 (0,1) )ds ≤ C (T ).

Combining the course of estimates and the initial condition of approximate system, we can easily deduce that C is dependent on T , ρ0 , u0 . Moreover, because the constants C of Lemmas 2.1–2.6 are independent of the lower bound of ρ0 . Here, C (T ) does not depend on δ and m (for any m ≥ M , M is dependent on the approximate velocity of initial condition). Thus, we can deduce from the two above estimates that (ρ m , um , Φ m ) converges up to an extraction of subsequences to some limit (ρδ , uδ , Φδ ) in the obvious weak sense, and there are estimates sup (|ρδ |L∞ (0,1) + |uxδ |2L2 (0,1) ) +

T

∫

0≤t ≤T

0

√ (| ρδ uδt |2L2 (0,1) + |uxδ |L2 (0,1) + |Gxδ |2L2 (0,1) )dt ≤ C (T )

sup (|ρδ |H 1 (0,1) + |ρδ t |L2 (0,1) + |uδ x |2L2 (0,1) ) +

T

∫

0≤t ≤T

0

(|(ρδ uδ )t |2L2 (0,1) + |uxxδ |2L2 (0,1) )ds ≤ C (T ).

Because C (T ) is independent of δ , when δ → 0, we can deduce that (ρδ , uδ , Φδ ) converges, up to an extraction of subsequences, to some limit (ρ, u, φ) in weak sense and sup (|ρ| 0≤t ≤T

L∞ (0,1)

ux 2L2 (0,1)

+| |

)+

T

∫ 0

√ (| ρ ut |2L2 (0,1) + |ux |2L2 (0,1) + |Gx |2L2 (0,1) )dt ≤ C (T )

sup (|ρ|H 1 (0,1) + |ρt |L2 (0,1) + |ux |L2 (0,1) ) +

T

∫

0≤t ≤T

(|(ρ u)t |2L2 (0,1) + |uxx |2L2 (0,1) )dt ≤ C (T ).

0

From the Lp -strong estimates of the Eq. (1.1)(c), we can easily get the regularity in Theorem 1.1. As for Theorem 1.2, we can deal with it similarly, but the initial and outer power conditions are

δ ≤ ρ δ (0) ≤ |ρ0 |L∞ + 1,

ρ0δ ∈ C 2 ([0, 1]),

g δ ∈ Cc2 (0, 1),

|g0δ − g |L2 (0,1) → 0.

|ρ0δ − ρ0 |H 1 (0,1) → 0. f ε ∈ Cc2 ((0, ∞) × (0, 1))

(f δ , fxδ , ftδ ) → (f , fx , ft ) in L2loc (0, ∞; L2 (0, 1)). Because we have compatibility condition, we let uδ0 ∈ C 3 [0, 1] be the solution of the following elliptic equation 1

uδ0xx = (pδ )x + ρ02 g δ ,

0 < x < 1,

uδ0 (0) = uδ0 (1) = 0.

Combining the classical stableness results of the elliptic equation and the compatibility condition of Theorem 1.2, we deduce that uδ0 → u0 in H 2 (0, 1), and u0 satisfies the compatibility of Theorem 1.2. For any fixed δ > 0, similarly, we may get the similar results with Lemmas 2.1–2.7 and we have estimates sup (|ρ δ |H 1 (0,1) + |ρtδ |L2 (0,1) + |uδx |2L2 (0,1) ) +

0≤t ≤T

∫  | ρ δ uδt (t )|2L2 (0,1) +

T 0

|uδxt |2L2 (0,1) dt ≤ C (T )

∫ 0

T

(|uδx |2L∞ (0,1) + |uδxx |2L2 (0,1) )dt ≤ C (T )

H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

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where C is independent of δ . Thus when δ → 0, we can deduce from the above estimates that (ρδ , uδ , Φδ ) converges to some limit (ρ, u, φ) and we have estimates sup (|ρ|H 1 (0,1) + |ρt |L2 (0,1) + |ux |L2 (0,1) ) +

0≤t ≤T

√ | ρ ut (t )|2L2 (0,1) +

T

∫

(|ux |2L∞ (0,1) + |uxx |2L2 (0,1) )dt ≤ C (T )

0

T

∫ 0

|uxt |2L2 (0,1) dt ≤ C (T ).

From Lp -strong estimates of Eq. (1.1)(c), we can easily get the regularity in Theorem 1.2. 4. Proof of the uniqueness

¯ ) be two solutions that satisfy the same initial condition. Then combining (1.1)(a) and (b), we Let (ρ, u, Φ ) and (ρ, ¯ u¯ , Φ have ρ ut + ρ uux + ρ Φx − uxx + Px = ρ f ¯ x − u¯ xx + P¯x = ρ¯ f . ρ¯ u¯ t + ρ¯ u¯ u¯x + ρ¯ Φ Thus

ρ(u − u¯ )t + ρ u(u − u¯ )x − (u − u¯ )xx ¯ x ) − (P − P¯ )x − ρ(Φ − Φ ¯ )x − ρ(u − u¯ )¯ux . = (ρ − ρ)( ¯ f − u¯ t − u¯ u¯ x − Φ

(4.1)

Multiplying it by (u − u¯ ) and integrating over (0, 1), we have 1

∫

d dt

1

0

2

∫

1

ρ(u − u¯ )2 dx +

1

∫

1 2

(ρ u)x (u − u¯ )2 dx + 0

¯ x ||u − u¯ |dx + |ρ − ρ|| ¯ f − u¯ t − u¯ u¯ x − Φ

≤

1

∫ 0 1

∫

0

ρ u(u − u¯ )x (u − u¯ )dx +

1

∫

(u − u¯ )2x dx 0

|P − P¯ ||(u − u¯ )x |dx 0

1

∫

¯ )x ||u − u¯ |dx + ρ|(Φ − Φ

+ 0

1

∫

ρ|u − u¯ |2 |¯ux |dx 0

that is 1

∫

d dt

1 2

0

ρ(u − u¯ )2 dx +

1

∫

(u − u¯ )2x dx

0

¯ x |L2 (0,1) |u − u¯ |L∞ (0,1) + |P − P¯ |L2 (0,1) |(u − u¯ )x |L2 (0,1) ≤ |ρ − ρ| ¯ L2 (0,1) |f − u¯ t − u¯ u¯ x − Φ √ ¯ )x |L2 (0,1) |u − u¯ |L2 (0,1) + | ρ|u − u¯ ||22 + C |(Φ − Φ |¯u | ∞ L (0,1) x L (0,1) √ 2 2 2 ≤ ε|(u − u¯ )x |L2 (0,1) + |ρ − ρ| ¯ L2 (0,1) (C + C |¯ut |L2 (0,1) ) + |P − P¯ |2L2 (0,1) + | ρ|u − u¯ ||2L2 (0,1) |¯ux |L∞ (0,1) consequently 1

∫

d dt

1 2

0

ρ(u − u¯ )2 dx +

1

∫

(u − u¯ )2x dx

0

√ ≤ |ρ − ρ| ¯ L2 (0,1) (C + C |¯ut |2L2 (0,1) ) + |P − P¯ |2L2 (0,1) + | ρ|u − u¯ ||2L2 (0,1) |¯ux |L∞ (0,1) . Moreover, from the conservative mass equation, we have

ρt + ρx u + ρ ux = 0,

ρ¯ t + ρ¯ x u¯ + ρ¯ u¯ x = 0.

Then

(ρ − ρ) ¯ t + (ρ − ρ) ¯ x u + ρ¯ x u − ρ¯ x u¯ + (ρ − ρ) ¯ ux + ρ¯ ux − ρ¯ u¯ x = 0 that is

(ρ − ρ) ¯ t + (ρ − ρ) ¯ x u + ρ¯ x (u − u¯ ) + (ρ − ρ) ¯ ux + ρ( ¯ ux − u¯ x ) = 0. Multiplying it by (ρ − ρ) ¯ and integrating over (0, 1), we get 1

∫

d dt

1

0

2

∫

1

(ρ − ρ) ¯ 2 dx −

(ρ − ρ) ¯ 2 ux dx +

+ 0

1

∫

1

0

2

∫

1

(ρ − ρ) ¯ 2 ux dx +

1

∫

ρ¯ x (u − u¯ )(ρ − ρ) ¯ dx 0

ρ( ¯ u − u¯ )x (ρ − ρ) ¯ dx. 0

(4.2)

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H. Liu et al. / Nonlinear Analysis 74 (2011) 5876–5891

Thus 1

∫

d dt

1 2

0

≤C

(ρ − ρ) ¯ 2 dx 1

∫

(ρ − ρ) ¯ 2 |ux |dx +

1

∫

|ρ¯ x ||(u − u¯ )||(ρ − ρ)| ¯ dx +

∫

ρ|( ¯ u − u¯ )x ||(ρ − ρ)| ¯ dx



0

0

0

1

that is 1

∫

d dt

1 2

0

(ρ − ρ) ¯ 2 dx

 ∫ 1 ∫  2 ρ − ρ¯ dx + ≤ C |ux |L∞ (0,1) 0

+

ρ¯ x2 dx

 12

|(u − u¯ )|L∞ (0,1)

0

1

∫

1

|(u − u¯ )x |2 dx

 12 ∫

1

∫

(ρ − ρ) ¯ 2 dx

 12

0

 12 

1

|(ρ − ρ)| ¯ 2 dx 0

0

 ≤ C |ux |L∞ (0,1) + C (ε) + C (ε)

1

∫

ρ¯ x2 dx

∫

0

1

(ρ − ρ) ¯ 2 dx + ε|(u − u¯ )x |L2 (0,1) .

0

(4.3)

Moreover, multiplying (1.1)(a) by aγ ρ γ −1 , we get Pt + Px u + γ Pux = 0,

P¯ t + P¯ x u¯ + γ P¯ u¯ x = 0.

Similarly, we get

(P − P¯ )t + (P − P¯ )x u¯ + P¯x (u − u¯ ) + γ (P − P¯ )ux + γ P¯ (u − u¯ )x = 0. Multiplying it by (P − P¯ ) and integrating over (0, 1), we get 1

∫

d dt

1 2

0

+γ

(P − P¯ )2 dx −

1

∫ 0

1

∫

(P − P¯ )2 ux dx + 0

1 2

∫

(P − P¯ )2 ux dx +

1

∫

P¯ x (u − u¯ )(P − P¯ )dx 0

1

P¯ (u − u¯ )x (P − P¯ )dx.

0

Then 1

∫

d dt

(P − P¯ )2 dx ≤ C |ux |L∞ (0,1) |P − P¯ |2L2 (0,1) + |P¯x |L2 (0,1) |u − u¯ |L∞ (0,1) |P − P¯ |L2 (0,1)

0

+ ε|(u − u¯ )x |2L2 (0,1) + C (ε)|P − P¯ |2L2 (0,1) ≤ C (|ux |L∞ (0,1) + |P¯x |L2 (0,1) + 1)|P − P¯ |2L2 (0,1) + 2ε|(u − u¯ )x |2L2 (0,1) .

(4.4)

From (4.2)–(4.4), we obtain 1

∫

d dt

(ρ(u − u¯ ) + (ρ − ρ) ¯ + (P − P¯ )2 )dx + 2

0

2

1

∫

(u − u¯ )2x dx 0

≤ C (1 + |¯ut |2L2 (0,1) + |¯ux |2L∞ (0,1) + |ρx |2L2 (0,1) + |ux |L∞ (0,1) + |P¯x |2L2 (0,1) ) √ × (| ρ(u − u¯ )|2L2 (0,1) + |ρ − ρ| ¯ 2L2 (0,1) + |P − P¯ |2L2 (0,1) ). Using Grownwall’s inequality and the above regularity of strong solution, we have

√ | ρ(u − u¯ )|2L2 (0,1) + |ρ − ρ| ¯ 2L2 (0,1) + |P − P¯ |2L2 (0,1) = 0 ¯ |W 2,2 (0,1) . Finally, we get the uniqueness. that is, u = u¯ , ρ = ρ¯ . From the classical theorems of Eq. (1.1)(c), we get |Φ = Φ References [1] P.L. Lions, Mathematical Topics in Fluids Mechanics, Vol. 2, in: Oxford Lecture Series in Mathematics and its Applications, vol. 10, Clarendon Press, Oxford, 1998. [2] E. Feireisl, A. Novotný, H. Petzeltová, On the existence of globally defined weak solution to the Navier–Stokes equations, J. Math. Fluid Mech. 3 (2001) 358–392. [3] H. Kim, Global existence of strong solutions of the Navier–Stokes equations for one-dimensional isentropic compressible fluids, Available at: http://com2mac.postech.ac.kr/papers/2001/01-38.pdf.

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