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Nonlinear Analysis: Real World Applications 12 (2011) 1510–1531

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Nonlinear Analysis: Real World Applications

journal homepage: www.elsevier.com/locate/nonrwa

Solutions of incompressible hydrodynamic flow of liquid crystalsHuanyao Wen, Shijin Ding ∗

School of Mathematical Sciences, South China Normal University, Guangzhou, Guangdong 510631, China

a r t i c l e i n f o

Article history:Received 12 August 2009Accepted 20 October 2010

Keywords:Liquid crystalsIncompressible flowWell-posedness

a b s t r a c t

In this paper, we investigate the incompressible hydrodynamic flow of the nematic liquidcrystals in dimension N (N = 2 or 3). We obtain the local existence and uniqueness of thesolution if the initial density ρ0 ≥ 0. Particularly, if ρ0 has a positive bound from below,and N = 2, we get the global existence and uniqueness of the solution with small initialdata.

Crown Copyright© 2010 Published by Elsevier Ltd. All rights reserved.

1. Introduction

We consider the following incompressible hydrodynamic flow of nematic liquids crystals:

ρt + ∇ · (ρu) = 0; (1.1)

(ρu)t + ∇ ·

ρu

u

+ ∇P = γ1u − λ∇ ·

∇n

∇n

; (1.2)

∇ · u = 0; (1.3)

nt + (u · ∇)n = θ(1n + |∇n|2n) (1.4)

in Ω × (0, +∞), where Ω is a bounded, smooth domain in RN (N = 2 or 3). Here u = (u1, u2, . . . , uN) denotes the velocity,n : Ω × (0, +∞) → S2 represents the director of the molecules. The viscosity constants γ , λ, θ > 0. ρ and P(x, t) arescalar functions denoting the density and the pressure. ∇n

∇n denotes N × N matrix whose (i, j)-th entry is given by

(nxi · nxj), for i, j = 1, 2, . . . ,N .

If |∇n|2 in (1.4) is replaced by 1−|n|

2

ε2, it is the Ginzburg–Landau approximation of the simplified Ericksen–Leslie system,

introduced by Lin in [1], and analyzed by Lin, Liu, Shkoller, Blanca et al.. More precisely, when ρ = constant, Lin andLiu [2] proved the local existence of the classical solutions and the global existence of the weak solutions in dimension twoand three with Dirichlet boundary conditions. For any fixed ε, they also obtained the existence and uniqueness of globalclassical solution either in dimension two or dimension three for large fluid viscosity γ . To continue the work in [2], Lin andLiu [3] proved that the one dimensional space–time Hausdorff measure of the singular set of the ‘‘suitable’’ weak solutionsis zero. For a more general system and other boundary problems, please refer to [4,3,5–14] and references therein. Whenρ = constant, Liu and Zhang obtain the global weak solution in dimension three with the initial density ρ0 ∈ L2(Ω) [10].Recently, Jiang and Tan improve the condition of ρ0 (i.e. ρ0 ∈ Lγ (Ω), and γ > 3

2 ) in [15]. They cannot get the estimatesuniformly for ε, and therefore cannot take the limit ε → 0.

∗ Corresponding author.E-mail addresses: [email protected] (H. Wen), [email protected] (S. Ding).

1468-1218/$ – see front matter Crown Copyright© 2010 Published by Elsevier Ltd. All rights reserved.doi:10.1016/j.nonrwa.2010.10.010


H. Wen, S. Ding / Nonlinear Analysis: Real World Applications 12 (2011) 1510–1531 1511

In this paper, we consider the following initial and boundary conditions:

(ρ, u, n)|t=0 = (ρ0, u0, n0), with ρ0 ≥ 0, |n0| = 1, ∇ · u0 = 0, and u0|∂Ω = 0; (1.5)u(x, t) = 0, n(x, t) = n0(x), for (x, t) ∈ ∂Ω × (0, +∞). (1.6)

Our aim is to study the existence and uniqueness of a solution to (1.1)–(1.6). Compared with the approximate problem,|∇n|2 in (1.4) brings us some new difficulties. The systems (1.1)–(1.4) can be viewed as Navier–Stokes equations couplingthe heat flow of a harmonic map. Since the strong solutions of a harmonic map must be blowing up at finite time [16], wecannot expect that (1.1)–(1.6) have a global strong solution with general initial data. Therefore, we only get the existence ofa local solution for ρ0 ≥ 0 and N = 2 or 3 by iteration compared with the Galerkin method of [2]. If ρ0 ≥ M1 > 0, we obtainanother local solution for N = 2 or 3 by the contraction fixed points theorem, and get a global solution with small initialdata for N = 2 by some a priori estimates. With the regularities of the solution we get, the solution is accurately unique.

Denote QT = Ω × (0, T ). The main results of the paper are as follows:

Theorem 1.1. Assume that ρ0 ≥ 0, ρ0 ∈ H1(Ω)

L∞(Ω), u0 ∈ H2(Ω)

H10 (Ω), n0 ∈ H3(Ω), and the following compatible

conditions are valid

γ1u0 − ∇P0 − λ∇ ·

∇n0

∇n0

= ρ

120 g, and ∇ · u0 = 0, in Ω, (1.7)

for (P0, g) ∈ H1(Ω) × L2(Ω). Then there exists a constant T ⋆ > 0 such that for any T ≤ T ⋆ (1.1)–(1.6) has a unique solutionρ, u, P, n satisfying

ρ ∈ L∞(0, T ;H1(Ω))

L∞(QT ), ρt ∈ L∞(0, T ; L2(Ω)),

u ∈ L∞

0, T ;H2(Ω)

H1

0 (Ω)

L2(0, T ;W 2,6(Ω)),

ut ∈ L2(0, T ;H10 (Ω)),

√ρut ∈ L∞(0, T ; L2(Ω)),

P ∈ L∞(0, T ;H1(Ω))

L2(0, T ;W 1,6(Ω)), |n| = 1, in Q T ,

n ∈ W 4,22 (QT )

L∞(0, T ;H3(Ω)), nt ∈ L∞(0, T ;H1

0 (Ω)).

Remark 1.1.

1. P is unique in the sense of up to a function of t , so does P in the following two theorems.2. For N = 2, u ∈ L2(0, T ;W 2,6(Ω)) and P ∈ L2(0, T ;W 1,6(Ω)) in Theorem 1.1 can be replaced by u ∈ L2(0, T ;W 2,p(Ω))

and P ∈ L2(0, T ;W 1,p(Ω)), for any p ≥ 6. It seems hard to get u ∈ Lp(0, T ;W 2,p(Ω)) and P ∈ Lp(0, T ;W 1,p(Ω)) underthe conditions of Theorem 1.1, while this solution can be obtained in the next two theorems if the initial density has apositive bound from below.

Theorem 1.2. Assume 0 < M1 ≤ ρ0 ≤ M2, u0 ∈ W2− 2

pp (Ω), ρ0 ∈ C1(Ω), p > N (N = 2 or 3) and n0 ∈ C2,α(Ω), for

α = 2−N+2p , if p ∈ (N,N + 2), and α ∈ (0, 1), if p ≥ N + 2. Then there exists a constant T0 > 0 such that for any T ≤ T0 the

problems (1.1)–(1.6) have a unique solution ρ ∈ C1(Q T ), u ∈ W 2,1p (QT ), P ∈ Lp(0, T ;W 1,p(Ω)), n ∈ C2+α,1+ α

2 (Q T ), |n| = 1in Q T .

Theorem 1.3. In addition to the conditions of Theorem 1.2, if N = 2, and there is a sufficiently small constant c0 > 0independent of γ , λ, θ , such that∫

Ω

|∇n0|2dx +

1λ

∫Ω

ρ0|u0|2dx ≤ c0,

then for any T > 0, problems (1.1)–(1.6) have a unique solution ρ ∈ C1(Q T ), u ∈ W 2,1p (QT ), P ∈ Lp(0, T ;W 1,p(Ω)), n ∈

C2+α,1+ α2 (Q T ), |n| = 1 in Q T .

The rest of the paper is organized as follows: in section two, we give some useful inequalities whichwill be used in the proofof the main theorems. In section three, we use the technique of iteration to prove Theorem 1.1. In the last section, we onlygive the outline of the proof of Theorem 1.2 for brevity. Based on the local existence of the solution, we give some a prioriestimates globally in time to get Theorem 1.3.


1512 H. Wen, S. Ding / Nonlinear Analysis: Real World Applications 12 (2011) 1510–1531

2. Preliminaries

In this section, we give some useful lemmas which will be used in the next two sections.

Lemma 2.1 ([9, Lemma 2.1]). Let h ∈ L∞(0, T ), h ≥ 0, a ∈ R, and b ∈ L1(0, T ), b ≥ 0 satisfy the inequality

h(τ ) ≤ a +

∫ τ

0b(t)h(t)dt, for all τ ∈ [0, T ].

Then

h(τ ) ≤ a exp∫ τ

0b(t)dt

, for a.e. τ ∈ [0, T ].

Lemma 2.2 ([8], Ch. III, Section 2). Let X0, X, and X1 be three Banach spaces such that X0 ⊂ X ⊂ X1, the injection of X into X1being continuous, and the injection of X0 into X being compact. Then for each η > 0, there exists some positive constant c(η)depending on η (and on the spaces X0, X, X1) such that

‖v‖X ≤ η‖v‖X0 + c(η)‖v‖X1 , ∀v ∈ X0.

Lemma 2.3. Let Ω ⊂ RN be an arbitrary bounded domain with piecewise smooth boundaries. Then the following inequality isvalid for every function u ∈ W 1,m

0 (Ω) or u ∈ W 1,m(Ω),

Ωu(x)dx = 0:

‖u‖Lq(Ω) ≤ c‖∇u‖αLm(Ω)‖u‖

1−αLr (Ω) (2.1)

where m ≥ 1, r ≥ 1, and α = (1/r − 1/q)(1/r − 1/m+ 1/N)−1; moreover, if m < N and N > 1, then q ∈ [r,mN/(N −m)]for r ≤ mN/(N −m), and q ∈ [mN/(N −m), r] for r ≥ mN/(N −m). if m ≥ N > 1, then q ∈ [r, +∞) is arbitrary; moreover,if m > N, then q ∈ [r, +∞]. If m ≥ N = 1, then q ∈ [r, +∞]. The positive constant c depends on N,m, r, α and the domainΩ but independent of the function u.

This lemma can be found in [17] and the references therein.

Lemma 2.4. Let Ω ⊂ RN be an arbitrary bounded domain with piecewise smooth boundaries. Then the following inequality isvalid for every function v ∈ W 1,m(Ω):

‖v‖Lq(Ω) ≤c(‖v‖L1(Ω) + ‖∇v‖αLm(Ω)‖v‖

1−αLr (Ω)) (2.2)

where N,m, r, q, and α are the same as those in Lemma 2.3. The positive constantc depends on N,m, r, α and the domain Ω

but independent of the function u.

Proof. For any v ∈ W 1,m(Ω), we denote u = v −1

|Ω|

Ω

v(x)dx. Then u ∈ W 1,m(Ω), and

Ωu(x)dx = 0, and we get (2.2)

from (2.1).

Lemma 2.5 ([13]). Assume X ⊂ E ⊂ Y are Banach spaces and X →→ E. Then the following embeddings are compact:

(i)ϕ : ϕ ∈ Lq(0, T ; X),

∂ϕ

∂t∈ L1(0, T ; Y )

→→ Lq(0, T ; E), if 1 ≤ q ≤ ∞;

(ii)ϕ : ϕ ∈ L∞(0, T ; X),

∂ϕ

∂t∈ Lr(0, T ; Y )

→→ C([0, T ]; E), if 1 < r ≤ ∞.

3. ρ0 ≥ 0: local existence and uniqueness of the solution

Denote

AT ,c1 = m = (m1,m2,m3)|m(x, 0) = n0(x), |||m|||A ≤ c1,BT ,c2 = v = (v1, v2, . . . , vN)|v(x, 0) = u0(x), v|∂Ω = 0, |||v|||B ≤ c2,

where c1 and c2 are positive constants, and

|||m|||A = ‖mt‖L2(0,T ;H2(Ω)) + ‖mt‖L∞(0,T ;H1(Ω)) + ‖m‖L∞(0,T ;H3(Ω)),

|||v|||B = ‖v‖L2(0,T ;W2,6(Ω)) + ‖v‖L∞(0,T ;H2(Ω)) + ‖vt‖L2(0,T ;H1(Ω)).

Without loss of generality, we assume c1, c2 ≥ 1. To prove Theorem 1.1, we will construct a sequence of approximatesolutions to (1.1)–(1.6) and use the technique of iteration based on the two sets AT ,c1 and BT ,c2 .



In order to construct the approximate solutions, we consider the following auxiliary problem for v ∈ BT ,c2ρt + (v · ∇)ρ = 0,ρut + ∇P = γ1u + ρf1 + f2,∇ · u = 0,

(3.1)

with initial and boundary conditionsρ(x, 0) = ρ0(x), x ∈ Ω, (3.2)u(x, 0) = u0(x), x ∈ Ω, (3.3)u|∂Ω = 0. (3.4)

Lemma 3.1. Assume that ρ0 ∈ H1(Ω)

L∞(Ω), ρ0 ≥ 0, u0 ∈ H2(Ω)

H10 (Ω), fi ∈ L2(0, T ;H1(Ω)), fit ∈ L2(QT ), for

i = 1, 2, and the following compatible conditions are valid

γ1u0(x) − ∇P0(x) + f2(x, 0) = ρ120 g(x), and ∇ · u0(x) = 0, in Ω, (3.5)

for (P0, g) ∈ H1(Ω) × L2(Ω). Then for any T > 0 (3.1)–(3.4) has a unique solution (ρ, u, P) satisfying

ρ ∈ L∞(0, T ;H1(Ω))

L∞(QT ), ρt ∈ L∞(0, T ; L2(Ω))

u ∈ L∞

0, T ;H2(Ω)

H1

0 (Ω)

L2(0, T ;W 2,6(Ω)),

ut ∈ L2(0, T ;H10 (Ω)),

√ρut ∈ L∞(0, T ; L2(Ω)),

P ∈ L∞(0, T ;H1(Ω))

L2(0, T ;W 1,6(Ω)).

Remark 3.1. fi ∈ L2(0, T ;H1(Ω)) and fit ∈ L2(QT ) imply fi ∈ C([0, T ]; L2(Ω)), for i = 1, 2.Proof. Mollifying the initial data ρ0 by ρ0j (j ≥ 1) such that ρ0j ∈ C1(Ω), 1

j ≤ ρ0j ≤ ‖ρ0‖L∞(Ω) +1j , we consider (3.1)–(3.4)

with initial data ρ0 replaced by ρ0j. By [18], (3.1)1 and (3.2), there exists a unique solution ρj ∈ C1(Q T ) for each j. By [11],we have the following estimates

1j

≤ ρj ≤ c, in QT . (3.6)

‖ρj(·, t)‖H1(Ω) ≤ c expc∫ t

0‖∇v(·, s)‖W1,6(Ω)ds

, (3.7)

‖ρjt(·, t)‖L2(Ω) ≤ c‖v(·, t)‖L∞(Ω) expc∫ t

0‖∇v(·, s)‖W1,6(Ω)ds

, (3.8)

for 0 < t < T .Considering (3.1)2–(3.1)3 and (3.3)–(3.4) with ρ replaced by ρj, and using the semi-discrete Galerkin method (cf. [11]

or [19]), we can construct the approximate solutions. To prove the convergence of the sequence, we only need a prioriestimates for them. For simplicity, we omit the subscripts j in the solutions.

Differentiating (3.1)2 with respect to t , multiplying the resulting equation by ut , and integrating over Ω , we have

12

ddt

∫Ω

ρ|ut |2dx + γ

∫Ω

|∇ut |2dx =

∫Ω

f2t · utdx +

∫Ω

ρt f1 · utdx +

∫Ω

ρf1t · utdx −12

∫Ω

ρt |ut |2dx.

Together with (3.1)1, the Cauchy inequality and the Poincare inequality, we get

12

ddt

∫Ω

ρ|ut |2dx + γ

∫Ω

|∇ut |2dx ≤ ‖f2t‖L2(Ω)‖ut‖L2(Ω) + ‖ρt‖L2(Ω)‖f1‖L4(Ω)‖ut‖L4(Ω)

+ ‖√

ρut‖L2(Ω)‖√

ρf1t‖L2(Ω) +12

∫Ω

∇ρ · v|ut |2dx

≤ c(‖f2t‖L2(Ω) + ‖ρt‖L2(Ω)‖f1‖L4(Ω))‖∇ut‖L2(Ω) + c42‖√

ρut‖2L2(Ω)

+cc42

‖√

ρf1t‖2L2(Ω)

−12

∫Ω

ρ∇ · v|ut |2dx −

∫Ω

ρ(v · ∇)ut · utdx

≤γ

2

∫Ω

|∇ut |2dx + c‖f2t‖2

L2(Ω)+ c‖ρt‖

2L2(Ω)

‖f1‖2L4(Ω)

+ c42‖√

ρut‖2L2(Ω)

+ cc−42 ‖ρ‖L∞(Ω)‖f1t‖2

L2(Ω)+ c‖∇v‖L∞(Ω)‖

√ρut‖

2L2(Ω)

+ c∫

Ω

ρ2|v|

2|ut |

2dx.



Thus12

ddt

∫Ω

ρ|ut |2dx +

γ

2

∫Ω

|∇ut |2dx ≤ c(c42 + ‖∇v‖L∞(Ω) + ‖ρ‖L∞(Ω)‖v‖

2L∞(Ω))

∫Ω

ρ|ut |2dx + c‖f2t‖2

L2(Ω)

+ c‖ρt‖2L2(Ω)

‖f1‖2L4(Ω)

+ cc−42 ‖ρ‖L∞(Ω)‖f1t‖2

L2(Ω).

Combining (3.6), we have

ddt

∫Ω

ρ|ut |2dx + γ

∫Ω

|∇ut |2dx ≤ c(c42 + ‖v‖W2,6(Ω) + c22‖ρ0‖L∞(Ω))

∫Ω

ρ|ut |2dx + c‖f2t‖2

L2(Ω)

+ c‖ρt‖2L2(Ω)

‖f1‖2L4(Ω)

+ cc−42 ‖ρ0‖L∞(Ω)‖f1t‖2

L2(Ω). (3.9)

Integrating (3.9) over (τ , t), we have∫Ω

ρ|ut |2(x, t)dx + γ

∫ t

τ

∫Ω

|∇ut(x, ξ)|2dxdξ ≤

∫Ω

ρ|ut |2(x, τ )dx + c

∫ t

τ

(c42 + ‖v(·, ξ)‖W2,6(Ω))

×

∫Ω

ρ|ut |2(x, ξ)dxdξ + c

∫ t

τ

‖f2t(·, ξ)‖2L2(Ω)

dξ

+ c∫ t

τ

‖ρt(·, ξ)‖2L2(Ω)

‖f1(·, ξ)‖2L4(Ω)

dξ + cc−42

×

∫ t

τ

‖f1t(·, ξ)‖2L2(Ω)

dξ . (3.10)

It follows from the equation that∫Ω

ρ|ut |2(x, τ )dx =

∫Ω

[γ1u(x, τ ) + ρ(x, τ )f1(x, τ ) + f2(x, τ ) − ∇P(x, τ )] · utdx

=

∫Ω

[γ1u(x, τ ) + ρ(x, τ )f1(x, τ ) + f2(x, τ ) − ∇P0(x)] · utdx.

By the Cauchy inequality, we have∫Ω

ρ|ut |2(x, τ )dx ≤ c

∫Ω

1ρ

|γ1u(x, τ ) + ρ(x, τ )f1(x, τ ) + f2(x, τ ) − ∇P0(x) |2 dx. (3.11)

Combining (3.10), (3.11), and taking τ → 0+, we get∫Ω


∫ t

0

∫Ω

|∇ut(x, ξ)|2dxdξ

≤ c∫

Ω

1ρ0

|γ1u0(x) + ρ0(x)f1(x, 0) + f2(x, 0) − ∇P0(x) |2 dx + c

∫ t

0(c42 + ‖v(·, ξ)‖W2,6(Ω))

×

∫Ω

ρ|ut |2(x, ξ)dxdξ + c

∫ t

0‖f2t(·, ξ)‖2

L2(Ω)dξ + c

∫ t

0‖ρt(·, ξ)‖2

L2(Ω)‖f1(·, ξ)‖2

L4(Ω)dξ

+ cc−42

∫ t

0‖f1t(·, ξ)‖2

L2(Ω)dξ . (3.12)

Together with (3.8) and the compatible condition (3.5), we have∫Ω


∫ t

0

∫Ω

|∇ut(x, ξ)|2dxdξ

≤ c∫

Ω

|g|2dx + c∫

Ω

|f1(x, 0)|2dx + c∫ t

0(c42 + ‖v(·, ξ)‖W2,6(Ω))

∫Ω

ρ|ut |2(x, ξ)dxdξ

+ c∫ T

0‖f2t(·, ξ)‖2

L2(Ω)dξ + cc22 expcc2T

12

∫ T

0‖f1(·, ξ)‖2

L4(Ω)dξ + cc−4

2

∫ T

0‖f1t(·, ξ)‖2

L2(Ω)dξ . (3.13)

By (3.13) and Lemma 2.1, we have∫Ω

ρ|ut |2dx ≤ cF1 exp

∫ T

0c(c42 + ‖v(·, ξ)‖W2,6(Ω))dξ



≤ cF1 expc(c42T + c2T12 ), (3.14)

where

F1 =

∫Ω

|g|2dx +

∫Ω

|f1(x, 0)|2dx +

∫ T

0‖f2t(·, ξ)‖2

L2(Ω)dξ + c22 expcc2T

12

∫ T

0‖f1(·, ξ)‖2

L4(Ω)dξ

+ c−42

∫ T

0‖f1t(·, ξ)‖2

L2(Ω)dξ .

By (3.13), (3.14), and the Poincare inequality, we have∫ T

0‖ut‖

2H1(Ω)

dt ≤ cF1[(c42T + c2T12 ) expc(c42T + c2T

12 ) + 1]. (3.15)

Since

−γ1u + ∇P = ρf1 + f2 − ρut ,

we have from the estimates for the stationary Stokes equations (cf. [7]), and (3.14)

‖u‖2H2(Ω)

+ ‖P‖2H1(Ω)

≤ c‖ρf1‖2L2(Ω)

+ c‖f2‖2L2(Ω)

+ c‖ρut‖2L2(Ω)

≤ c‖f1‖2L2(Ω)

+ c‖f2‖2L2(Ω)

+ cF1 expc(c42T + c2T12 ), (3.16)

‖u‖2W2,6(Ω)

+ ‖P‖2W1,6(Ω)

≤ c‖ρf1‖2L6(Ω)

+ c‖f2‖2L6(Ω)

+ c‖ρut‖2L6(Ω)

≤ c‖ρ0‖2L∞(Ω)‖f1‖

2L6(Ω)

+ c‖f2‖2L6(Ω)

+ c‖ρ0‖2L∞(Ω)‖ut‖

2L6(Ω)

,

here we have normalized P as

ΩP(x, t)dx = 0.

Using the Sobolev inequality, and integrating over (0, T ), we have∫ T

0‖u‖2

W2,6(Ω)dt +

∫ T

0‖P‖

2W1,6(Ω)

dt ≤ c∫ T

0‖f1‖2

H1(Ω)dt + c

∫ T

0‖f2‖2

H1(Ω)dt

+ cF1[(c42T + c2T12 ) expc(c42T + c2T

12 ) + 1]. (3.17)

The constant c in (3.14)–(3.17) depends on the initial data and some other known constants but does not depend on c2, Tand j. Since the above estimates are uniform for the lower-bound of ρj, we can take the limit of the approximate solutionsto get a solution (ρ, u, P) to (3.1)–(3.4). By the lower semi-continuity, (3.14)–(3.17) is also valid for the limits (ρ, u, P).

The uniqueness of the solution can be proved by the standardmethod (cf. [11]). For simplicity, we omit it here. The proofof Lemma 3.1 is completed.

(1.1)–(1.3) can be rewritten as follows:

ρt + (u · ∇)ρ = 0; (1.1′)

ρut + ρ(u · ∇)u + ∇P = γ1u − λ∇ ·

∇n

∇n

; (1.2′)

∇ · u = 0. (1.3′)

In the section, γ , λ, and θ play no role. Without loss of generality, we assume γ , λ, θ = 1. Since (1.1)–(1.3) is equivalentto (1.1′), (1.2′) and (1.3′) if the regularities of the solutions are as those in Theorem 1.1 (or Theorems 1.2–1.3), we linearize(1.1′), (1.2′) and (1.3′) and (1.4)–(1.6) for (v,m) ∈ BT ,c2 × AT ,c1 as follows:

ρt + (v · ∇)ρ = 0,ρut + ρ(v · ∇)v + ∇P = 1u − ∇ ·

∇n

∇n

,

∇ · u = 0,nt − 1n = |∇m|

2m − (v · ∇)m

(3.18)

with initial and boundary conditions

(ρ, u, n)|t=0 = (ρ0, u0, n0), x ∈ Ω,

(u, n)|∂Ω = (0, n0).

Since (|∇m|2m − (v · ∇)m) ∈ W 2,1

2 (QT ), we obtain from [12] that (3.18)4 with the corresponding initial and boundaryconditions has a unique solution

n ∈ W 4,22 (QT )

L∞(0, T ;H3(Ω)), and nt ∈ L∞(0, T ;H1

0 (Ω)). (3.19)



By (3.19), and Lemma 3.1, (3.18)1–(3.18)3 with the corresponding initial-boundary conditions has a unique solution (ρ, u, P)with the regularities like that in Lemma 3.1.

Therefore, we can get a solution (ρ1, u1, P1, n1) to (3.18) with (v,m) replaced by (u0, n0), where (u0, n0) ∈ BT ,c2 × AT ,c1 .Assuming (uk−1, nk−1) ∈ BT ,c2 × AT ,c1 for k ≥ 1, we can construct an approximate solution (ρk, uk, Pk, nk) satisfying

ρkt + (uk−1

· ∇)ρk= 0,

ρkukt + ρk(uk−1

· ∇)uk−1+ ∇Pk

= 1uk− ∇ ·

∇nk

∇nk

,

∇ · uk= 0,

nkt − 1nk

= |∇nk−1|2nk−1

− (uk−1· ∇)nk−1

(3.20)


(ρk, uk, nk)|t=0 = (ρ0, u0, n0), x ∈ Ω,

(uk, nk)|∂Ω = (0, n0).

In step 1 and step 2, we denote the letter c to be the constants independent of c1, c2, and T .Step 1: We will prove

|||nk|||A ≤ c1. (3.21)

It follows from (3.20)4, the Cauchy inequality, and integration by parts that

12

ddt

∫Ω

|∇nkt |

2dx +

∫Ω

|1nkt |

2dx = −2∫

Ω

(∇nk−1· ∇nk−1

t )nk−1· 1nk

tdx −

∫Ω

|∇nk−1|2nk−1

t · 1nktdx

+

∫Ω

(uk−1· ∇)nk−1

t · 1nktdx +

∫Ω

(uk−1t · ∇)nk−1

· 1nktdx

≤12

∫Ω

|1nkt |

2dx + c∫

Ω

|∇nk−1|2|∇nk−1

t |2|nk−1

|2dx + c

∫Ω

|∇nk−1|4|nk−1

t |2dx

+ c∫

Ω

|uk−1|2|∇nk−1

t |2dx −

∫Ω

(∇uk−1t · ∇)nk−1

· ∇nktdx

−

∫Ω

(uk−1t · ∇)∇nk−1

· ∇nktdx. (3.22)

Applying the Cauchy inequality to the last two terms of (3.22), we get

ddt

∫Ω

|∇nkt |

2dx +

∫Ω

|1nkt |

2dx ≤ cc61 + cc21c22 + cc21c

22

∫Ω

|∇nkt |

2dx + cc−21 c−2

2

∫Ω

|∇uk−1t |

2|∇nk−1

|2dx

+ cc−21 c−2

2

∫Ω

|uk−1t |

2|∇

2nk−1|2dx

≤ cc61 + cc21c22 + cc21c

22

∫Ω

|∇nkt |

2dx + cc−21 c−2

2 ‖∇nk−1‖2L∞(Ω)

∫Ω

|∇uk−1t |

2dx

+ cc−21 c−2

2 ‖uk−1t ‖

2L4(Ω)

‖∇2nk−1

‖2L4(Ω)

.

Together with the Sobolev inequalities, we have

ddt

∫Ω

|∇nkt |

2dx +

∫Ω

|1nkt |

2dx ≤ cc61 + cc21c22 + cc21c

22

∫Ω

|∇nkt |

2dx + cc−21 c−2

2 ‖∇nk−1‖2H2(Ω)

∫Ω

|∇uk−1t |

2dx

+ cc−21 c−2

2 ‖uk−1t ‖

2H1(Ω)

‖∇2nk−1

‖2H1(Ω)

≤ cc61 + cc21c22 + cc21c

22

∫Ω

|∇nkt |

2dx + cc−22 ‖uk−1

t ‖2H1(Ω)

. (3.23)

Integrating (3.23) over (0, t), and using (3.20)4, we have∫Ω

|∇nkt |

2dx +

∫ t

0

∫Ω

|1nkt |

2dxds ≤ cc21c22

∫ t

0

∫Ω

|∇nkt |

2dxds + c(c61 + c21c22 )T + c. (3.24)

By Lemma 2.1, we get∫Ω

|∇nkt |

2dx ≤ [c(c61 + c21c22 )T + c] expcc21c

22T .



Taking T ⋆1 =

1c61+c21 c

22, and letting T ≤ T ⋆

1 , we have∫Ω

|∇nkt |

2dx ≤ c.

Since nkt |∂Ω = 0, we have

‖nkt (·, t)‖H1(Ω) ≤ c, (3.25)

for 0 < t < T . By (3.24) and (3.25), we have∫ T

0

∫Ω

|1nkt (x, t)|

2dxdt ≤ c. (3.26)

We use the elliptic estimate (cf. [17] or [12]) to get

‖nkt (·, t)‖H2(Ω) ≤ c‖1nk

t (·, t)‖L2(Ω), for any t ∈ (0, T ). (3.27)

(3.26) and (3.27) give

‖nkt ‖L2(0,T ;H2(Ω)) ≤ c. (3.28)

Using (3.20)4 and the elliptic estimate, we obtain

‖nk(·, t)‖H3(Ω) ≤ c‖n0‖H3(Ω) + c‖nkt (·, t)‖H1(Ω) + c‖uk−1

· ∇nk−1‖H1(Ω) + c‖|∇nk−1

|2nk−1

‖H1(Ω),

for 0 < t < T . Together with (3.25), we have

‖nk(·, t)‖H3(Ω) ≤ c + c‖uk−1· ∇nk−1

‖H1(Ω) + c‖|∇nk−1|2nk−1

‖H1(Ω). (3.29)

‖uk−1· ∇nk−1

‖2H1(Ω)

≤ c∫

Ω

|uk−1|2|∇nk−1

|2dx + c

∫Ω

|∇uk−1|2|∇nk−1

|2dx + c

∫Ω

|uk−1|2|∇

2nk−1|2dx

≤ c∫

Ω

|uk−1− u0|

2|∇nk−1

|2dx + c

∫Ω

|∇nk−1|2dx + c

∫Ω

|∇uk−1|2|∇(nk−1

− n0)|2dx

+ c∫

Ω

|∇uk−1|2dx + c

∫Ω

|uk−1− u0|

2|∇

2nk−1|2dx + c

∫Ω

|∇2nk−1

|2dx

≤ cc21

∫Ω

∫ t

0uk−1t (x, s)ds

2 dx + c∫

Ω

|∇(nk−1− n0)|

2dx+

× c‖∇uk−1‖2L4(Ω)

‖∇(nk−1− n0)‖

2L4(Ω)

+ c∫

Ω

|∇(uk−1− u0)|

2dx+

× c∫

Ω

∫ t

0uk−1t (x, s)ds

2 |∇2nk−1

|2dx + c

∫Ω

|∇2(nk−1

− n0)|2dx + c.

Together with the Hölder inequality and Lemma 2.2, we have

‖uk−1· ∇nk−1

‖2H1(Ω)

≤ cc21T∫ T

0

∫Ω

|uk−1t |

2dxdt + δ‖nk−1− n0‖

2H2(Ω)

+ c(δ)∫

Ω

|nk−1− n0|

2dx

+ c‖uk−1‖2H2(Ω)

‖nk−1− n0‖

2H2(Ω)

+ δ‖uk−1− u0‖

2H2(Ω)

+ c(δ)∫

Ω

|uk−1− u0|

2dx

+ cT∫ T

0

∫Ω

|uk−1t (x, s)|2|∇2nk−1(x, t)|2dxds + δ‖nk−1

− n0‖2H3(Ω)

+ c(δ)∫

Ω

|nk−1− n0|

2dx + c

≤ cc21c22T + 2δc21 + c(δ)T

∫ T

0

∫Ω

|nk−1t |

2dxdt + cc22δ‖nk−1

− n0‖2H3(Ω)

+ cc22c(δ)‖nk−1

− n0‖2L2(Ω)

+ 2δc22 + c(δ)T∫ T

0

∫Ω

|uk−1t |

2dxdt

+ cT∫ T

0‖uk−1

t (·, s)‖2L4(Ω)

‖∇2nk−1(·, t)‖2

L4(Ω)ds + 2δc21

+ c(δ)T∫ T

0

∫Ω

|nk−1t |

2dxdt + c



≤ cc21c22T + 2δc21 + c(δ)c21T + cc22 (c

21 + 1)δ + cc21c

22c(δ)T + 2δc22

+ c(δ)c22T + cc21c22T + 2δc21 + c(δ)c21T + c

≤ cc21c22δ + c[c21c

22 + c21c

22c(δ)]T + c.

Take δ =1

c21 c22, T ⋆

2 = minT ⋆1 , 1

c21 c22+c21 c

22 c(δ)

, and let T ≤ T ⋆

2 , we have

‖uk−1· ∇nk−1

‖H1(Ω) ≤ c. (3.30)

‖|∇nk−1|2nk−1

‖H1(Ω) ≤ c‖|∇nk−1|2nk−1

‖L2(Ω) + c‖(∇nk−1· ∇∇nk−1)nk−1

‖L2(Ω) + c‖∇nk−1‖3L6(Ω)

.

Together with the Sobolev inequalities and the Hölder inequality, we get

‖|∇nk−1|2nk−1

‖H1(Ω) ≤ c‖nk−1‖H2(Ω)‖∇nk−1

‖2L4(Ω)

+ c‖nk−1‖H2(Ω)‖∇nk−1

‖L4(Ω)‖∇2nk−1

‖L4(Ω) + c‖∇nk−1‖3H1(Ω)

≤ c‖nk−1‖3H2(Ω)

+ c‖nk−1‖2H2(Ω)

‖∇2nk−1

‖L4(Ω)

≤ c‖nk−1‖3W2,4(Ω)

≤ c‖nk−1− n0‖

3W2,4(Ω)

+ c.

Together with Lemma 2.2 and the Hölder inequalities, we have

‖|∇nk−1|2nk−1

‖H1(Ω) ≤ δ‖nk−1− n0‖

3H3(Ω)

+ c(δ)‖nk−1− n0‖

3L2(Ω)

+ c

≤ cc31δ + c(δ)∫ t

0nk−1t (·, s)ds

3

L2(Ω)

+ c

≤ cc31δ + c(δ)c31T32 + c.

Take δ =1c31

, T ⋆3 = min

T ⋆2 ,

[1

c(δ)c31

] 23, and let T ≤ T ⋆

3 , we have

‖|∇nk−1|2nk−1

‖H1(Ω) ≤ c. (3.31)

By (3.29)–(3.31), we get

‖nk‖L∞(0,T ;H3(Ω)) ≤ c. (3.32)

It follows from (3.25), (3.28) and (3.32) that

‖nkt ‖L2(0,T ;H2(Ω)) + ‖nk

t ‖L∞(0,T ;H1(Ω)) + ‖nk‖L∞(0,T ;H3(Ω)) ≤ c.

Choosing c1 ≥ c , we get (3.21). Step 1 is complete.Step 2: We will prove

0 ≤ ρk≤ c, in QT , (3.33)

‖ρk‖L∞(0,T ;H1(Ω)) ≤ c, (3.34)

‖ρkt ‖L∞(0,T ;L2(Ω)) ≤ cc2. (3.35)

It follows from (3.20)1, (3.6)–(3.8) that

0 ≤ ρk≤ c, in QT ,

‖ρk‖L∞(0,T ;H1(Ω)) ≤ c exp

cc2T

12

,

‖ρkt ‖L∞(0,T ;L2(Ω)) ≤ cc2 expcc2T

12 .

Take T ⋆4 = minT ⋆

3 , c−22 , and let T ≤ T ⋆

4 , we get (3.33)–(3.35).Step 3: We will prove

|||uk|||B ≤ c2. (3.36)

‖Pk‖L∞(0,T ;H1(Ω)) + ‖Pk

‖L2(0,T ;W1,6(Ω)) ≤ c. (3.37)

‖

ρkuk

t ‖L∞(0,T ;L2(Ω)) ≤ c. (3.38)



We obtain from (3.14)–(3.15) with (v, ρ, u) replaced by (uk−1, ρk, uk)∫Ω

ρk|uk

t |2dx ≤ cF1 expc(c42T + c2T

12 ), (3.39)∫ T

0‖uk

t ‖2H1(Ω)

dt ≤ cF1[(c42T + c2T12 ) expc(c42T + c2T

12 ) + 1]. (3.40)

F1 ≤ c + c∫ T

0

∫Ω

|∇nkt |

2|∇

2nk|2dxdt + c

∫ T

0

∫Ω

|∇nk|2|∇

2nkt |

2dxdt

+ cc22 expcc2T12

∫ T

0‖uk−1

· ∇uk−1‖2L4(Ω)

dt

+ cc−42

∫ T

0

∫Ω

|uk−1t · ∇uk−1

|2dxdt + cc−4

2

∫ T

0

∫Ω

|uk−1· ∇uk−1

t |2dxdt

≤ c + cc41 + cc62T expcc2T12 + cc−4

2

∫ T

0‖uk−1

t ‖2L4(Ω)

‖∇uk−1‖2L4(Ω)

dt + cc−22

∫ T

0

∫Ω

|∇uk−1t |

2dxdt

≤ c + cc41 + cc62T expcc2T12 + cc−4

2

∫ T

0‖uk−1

t ‖2H1(Ω)

‖∇uk−1‖2H1(Ω)

dt

≤ c + cc41 + cc62T expcc2T12 .

Take T ⋆5 = minT ⋆

4 , c−62 , and let T ≤ T ⋆

5 , we have

F1 ≤ c. (3.41)

By (3.39) and (3.41), we get (3.38) for T ≤ T ⋆5 . (3.40) and (3.41) give∫ T

0‖uk

t ‖2H1(Ω)

dt ≤ c[(c42T + c2T12 ) expc(c42T + c2T

12 ) + 1].

Since T ≤ T ⋆5 ≤ c−6

2 , we have

‖ukt ‖L2(0,T ;H1(Ω)) ≤ c. (3.42)

We obtain from (3.17) for T ≤ T ⋆5∫ T

0‖uk

‖2W2,6(Ω)

dt +

∫ T

0‖Pk

‖2W1,6(Ω)

dt ≤ c∫ T

0‖uk−1

· ∇uk−1‖2H1(Ω)

dt + c∫ T

0‖∇ ·

∇nk

∇nk

‖2H1(Ω)

dt

+ cF1[(c42T + c2T12 ) expc(c42T + c2T

12 ) + 1]

≤ c∫ T

0‖uk−1

· ∇uk−1‖2H1(Ω)

dt + c∫ T

0‖∇ ·

∇nk

∇nk

‖2H1(Ω)

dt + c

≤ c(c42 + c41 )T + c,

here we have used (3.32) and the Sobolev inequalities.Since T ≤ T ⋆

5 ≤ T ⋆1 , we have T ≤

1c42+c41

. Therefore,

‖uk‖L2(0,T ;W2,6(Ω)) + ‖Pk

‖L2(0,T ;W1,6(Ω)) ≤ c. (3.43)

It follows from (3.16) that for T ≤ T ⋆5

‖uk‖2H2(Ω)

+ ‖Pk‖2H1(Ω)

≤ c‖uk−1· ∇uk−1

‖2L2(Ω)

+ c‖∇ ·

∇nk

∇nk

‖2L2(Ω)

+ cF1 expc(c42T + c2T12 )

≤ c∫

Ω

|uk−1|2|∇uk−1

|2dx + c

∫Ω

|∇nk|2|∇

2nk|2dx + c

≤ c∫

Ω

|uk−1− u0|

2|∇uk−1

|2dx + c

∫Ω

|∇uk−1|2dx + c

∫Ω

|∇(nk− n0)|

2|∇

2nk|2dx

+ c∫

Ω

|∇2nk

|2dx + c.



By the Hölder inequality, the Sobolev inequality and Lemma 2.2, we get

‖uk‖2H2(Ω)

+ ‖Pk‖2H1(Ω)

≤ c‖uk−1− u0‖

2L4(Ω)

‖∇uk−1‖2L4(Ω)

+ c∫

Ω

|∇(uk−1− u0)|

2dx

+ c‖∇(nk− n0)‖

2L4(Ω)

‖∇2nk

‖2L4(Ω)

+ c∫

Ω

|∇2(nk

− n0)|2dx + c

≤ c22δ‖uk−1

− u0‖2H1(Ω)

+ c22c(δ)‖uk−1

− u0‖2L2(Ω)

+ δ‖uk−1− u0‖

2H2(Ω)

+ c(δ)∫

Ω

|uk−1− u0|

2dx + c21δ‖nk− n0‖

2H2(Ω)

+ c21c(δ)‖nk− n0‖

2L2(Ω)

+ δ‖nk− n0‖

2H3(Ω)

+ c(δ)∫

Ω

|nk− n0|

2dx + c

≤ cc42δ + c22c(δ)∫ t

0uk−1t (·, s)ds

2

L2(Ω)

+ cc22δ + c(δ)∫

Ω

∫ t

0uk−1t (·, s)ds

2 dx+ cc41δ + c21c(δ)

∫ t

0nkt (·, s)ds

2

L2(Ω)

+ cc21δ + c(δ)∫

Ω

∫ t

0nkt (·, s)ds

2 dx + c

≤ c(c41 + c42 )δ + (c41 + c42 )c(δ)T + c.

Take δ =1

c41+c42, T ⋆

6 = minT ⋆5 , 1

(c41+c42 )c(δ)

, and let T ≤ T ⋆

6 , we have

‖uk(·, t)‖L∞(0,T ;H2(Ω)) + ‖Pk‖L∞(0,T ;H1(Ω)) ≤ c. (3.44)

By (3.42)–(3.44), we get (3.37) and

‖uk‖L2(0,T ;W2,6(Ω)) + ‖uk

‖L∞(0,T ;H2(Ω)) + ‖ukt ‖L2(0,T ;H1(Ω)) ≤ c.

Let c2 ≥ c , we get (3.36). Step 3 is complete.Step 4: Taking limits.

In this step, we denotec to be the constants depending on c1, c2 and other known constants but T . Denote ρk+1=

ρk+1− ρk, uk+1

= uk+1− uk, nk+1

= nk+1− nk, P

k+1= Pk+1

− Pk.We have from (3.21)

ρk+1t + (uk

· ∇)ρk+1= −uk

· ∇ρk,

ρk+1uk+1t + ∇P

k+1= 1uk+1

− ρk+1[uk

t + (uk· ∇)uk

− (uk−1· ∇)uk

] − ρk(uk· ∇)uk

−ρk+1(uk−1· ∇)uk

− ∇ ·∇nk+1

∇nk+1− ∇ ·

∇nk

∇nk+1 ,

∇ · uk+1= 0,

nk+1t − 1nk+1

= [∇nk· (∇nk

+ ∇nk−1)]nk+ |∇nk−1

|2nk

− (uk· ∇)nk

+ (uk−1· ∇)nk,

(3.45)

with the initial and boundary conditions

(ρk+1, uk+1, nk+1)|t=0 = (0, 0, 0), x ∈ Ω,

(uk+1, nk+1)|∂Ω = (0, 0).

Multiplying (3.45)2 by uk+1, integrating over Ω , using integration by parts, and combining (3.21), we have

12

ddt

∫Ω

ρk+1|uk+1

|2dx +

∫Ω

|∇uk+1|2dx = −

∫Ω

ρk+1[uk

t + (uk· ∇)uk

− (uk−1· ∇)uk

] · uk+1dx

+

∫Ω

ρk+1(uk· ∇)uk+1

· uk+1dx −

∫Ω

ρk(uk· ∇)uk

· uk+1dx

−

∫Ω

ρk+1(uk−1· ∇)uk

· uk+1dx

+

∫Ω

∇nk+1

∇nk+1

+ ∇nk

∇nk+1

· ∇uk+1dx.

Together with the Hölder inequality, and (3.21), (3.33) and (3.36), we have

12

ddt

∫Ω

ρk+1|uk+1

|2dx +

∫Ω

|∇uk+1|2dx ≤c‖ρk+1

‖L32 (Ω)

‖ukt + (uk

· ∇)uk− (uk−1

· ∇)uk‖L6(Ω)‖u

k+1‖L6(Ω)

+c‖ρk+1uk+1‖L2(Ω)‖∇uk+1

‖L2(Ω) +c‖ρkuk‖L2(Ω)‖∇uk

‖L4(Ω)‖uk+1

‖L4(Ω)



+c‖ρk+1uk+1‖L2(Ω)‖∇uk

‖L2(Ω) +c‖∇nk+1‖L2(Ω)‖∇uk+1

‖L2(Ω)

≤c‖ρk+1‖L32 (Ω)

‖ukt + (uk

· ∇)uk− (uk−1

· ∇)uk‖H1(Ω)‖∇uk+1

‖L2(Ω)

+c‖ρk+1uk+1‖L2(Ω)‖∇uk+1

‖L2(Ω) +c‖ρkuk‖L2(Ω)‖∇uk+1

‖L2(Ω)

+c‖ρk+1uk+1‖L2(Ω)‖∇uk

‖L2(Ω) +c‖∇nk+1‖L2(Ω)‖∇uk+1

‖L2(Ω)

≤12‖∇uk+1

‖2L2(Ω)

+c‖ρk+1‖2

L32 (Ω)

‖ukt + (uk

· ∇)uk− (uk−1

· ∇)uk‖2H1(Ω)

+c‖ρk+1uk+1‖2L2(Ω)

+c‖ρkuk‖2L2(Ω)

+c‖∇nk+1‖2L2(Ω)

+cc(δ)‖ρk+1uk+1‖2L2(Ω)

+ δ‖∇uk‖2L2(Ω)

.

Thusddt

∫Ω

ρk+1|uk+1

|2dx +

∫Ω

|∇uk+1|2dx ≤ cc(δ) ∫

Ω

ρk+1|uk+1

|2dx +cAk(t)‖ρk+1

‖2

L32 (Ω)

+c‖ρkuk‖2L2(Ω)

+ δ‖∇uk‖2L2(Ω)

+c‖∇nk+1‖2L2(Ω)

, (3.46)

where Ak(t) = ‖[ukt + (uk

· ∇)uk− (uk−1

· ∇)uk](·, t)‖2

H1(Ω).

We have from (3.45)1ddt

∫Ω

|ρk+1|32 dx ≤ −

∫Ω

∇|ρk+1|32 · ukdx +c ∫

Ω

|ρk+1|12 |∇ρk

||uk|dx

≤ c‖ρk+1‖

12

L32 (Ω)

‖∇ρk‖L2(Ω)‖u

k‖L6(Ω)

≤ c‖ρk+1‖

12

L32 (Ω)

‖∇uk‖L2(Ω)

ddt

‖ρk+1‖2

L32 (Ω)

=43‖ρk+1

‖

12

L32 (Ω)

ddt

‖ρk+1‖

32

L32 (Ω)

≤ c‖ρk+1‖L32 (Ω)

‖∇uk‖L2(Ω)

≤ δ‖∇uk‖2L2(Ω)

+cc(δ)‖ρk+1‖2

L32 (Ω)

. (3.47)

It follows from (3.45)4 that

ddt

∫Ω

|∇nk+1|2dx +

∫Ω

|1nk+1|2dx ≤c ∫

Ω

|∇nk|2dx + 2

∫Ω

(uk· ∇)nk

· 1nk+1dx.

Using integration by parts, we have

ddt

∫Ω

|∇nk+1|2dx +

∫Ω

|1nk+1|2dx ≤ c ∫

Ω

|∇nk|2dx −

∫Ω

(uk· ∇)∇nk

· ∇nk+1dx −

∫Ω

(∇uk· ∇)nk

· ∇nk+1dx

≤ c ∫Ω

|∇nk|2dx +c‖∇nk+1

‖L2(Ω)‖uk‖L4(Ω)‖∇

2nk‖L4(Ω)

+c‖∇nk+1‖L2(Ω)‖∇uk

‖L2(Ω)‖∇nk‖L∞(Ω).

Together with the Poincare inequality, the Sobolev inequality, and (3.21), we have

ddt

∫Ω

|∇nk+1|2dx +

∫Ω

|1nk+1|2dx ≤ c ∫

Ω

|∇nk|2dx +c‖∇nk+1

‖L2(Ω)‖∇uk‖L2(Ω)

≤ c ∫Ω

|∇nk|2dx + δ‖∇uk

‖2L2(Ω)

+cc(δ)‖∇nk+1‖2L2(Ω)

. (3.48)

Denote Bk(t) = ‖

ρkuk(·, t)‖2L2(Ω)

+ ‖ρk(·, t)‖2

L32 (Ω)

+ ‖∇nk(·, t)‖2L2(Ω)

, and Ck(t) = ‖∇uk(·, t)‖2L2(Ω)

+ ‖1nk(·, t)‖2L2(Ω)

.

By (3.46)–(3.48), we get

ddt

Bk+1(t) + Ck+1(t) ≤c[c(δ) + Ak(t)]Bk+1(t) +cBk(t) + 3δCk(t).

Integrating over (0, t), we have

Bk+1(t) +

∫ t

0Ck+1(s)ds ≤c ∫ t

0[c(δ) + Ak(s)]Bk+1(s)ds +c ∫ T

0Bk(s)ds + 3δ

∫ T

0Ck(s)ds. (3.49)



By Lemma 2.1, (3.36) and (3.49), we have

sup0≤t≤T

Bk+1(t) ≤

[c ∫ T

0Bk(s)ds + 3δ

∫ T

0Ck(s)ds

]expcc(δ)T +c. (3.50)

We have from (3.49)–(3.50)∫ T

0Ck+1(s)ds ≤

[c ∫ T

0Bk(s)ds + 3δ

∫ T

0Ck(s)ds

][(cc(δ)T +c) · expcc(δ)T +c + 1]. (3.51)

By (3.50)–(3.51), we have

sup0≤t≤T

Bk+1(t) +

∫ T

0Ck+1(s)ds ≤

[cT sup0≤t≤T

Bk(t) + 3δ∫ T

0Ck(s)ds

][(cc(δ)T +c) · expcc(δ)T +c + 1].

Take δ =1

24c exp2c+12 , T⋆= min

T ⋆6 , 1

c(δ) ,1

4c(2c exp2c+1)

, and let T ≤ T ⋆, we have

sup0≤t≤T

Bk+1(t) +

∫ T

0Ck+1(s)ds ≤

14

[sup

0≤t≤TBk(t) +

∫ T

0Ck(s)ds

],

for any k ≥ 1. By iteration, we have

sup0≤t≤T

Bk+1(t) +

∫ T

0Ck+1(s)ds ≤

14k−1

[sup

0≤t≤TB2(t) +

∫ T

0C2(s)ds

].

Together with the Poincare inequality and the elliptic estimate, we have

‖ρk+1‖L∞(0,T ;L

32 (Ω))

+ ‖nk+1‖L∞(0,T ;H1(Ω)) + ‖uk+1

‖L2(0,T ;H1(Ω)) + ‖nk+1‖L2(0,T ;H2(Ω)) ≤

c2k−1

.

Therefore,∞−k=2

‖ρk‖L∞(0,T ;L

32 (Ω))

< ∞

∞−k=2

‖uk‖L2(0,T ;H1(Ω)) < ∞

∞−k=2

(‖nk‖L∞(0,T ;H1(Ω)) + ‖nk

‖L2(0,T ;H2(Ω))) < ∞.

Therefore,

ρk→ ρ1

+

∞−i=2

ρ i, in L∞(0, T ; L32 (Ω)),

uk→ u1

+

∞−i=2

ui, in L2(0, T ;H1(Ω)), (3.52)

nk→ n1

+

∞−i=2

ni, in L∞(0, T ;H1(Ω))

L2(0, T ;H2(Ω)), (3.53)

as k → ∞.By Lemma 2.5, (3.34)–(3.38), we can extract a subsequence denoted by (ρki , uki , nki) such that as i → ∞

ρki ρ, weak-∗ in L∞(QT ), (3.54)

(∇ρki , ρkit ) (∇ρ, ρt), weak-∗ in L∞(0, T ; L2(Ω)), (3.55)

ρki → ρ, strongly in C([0, T ]; L2(Ω)), (3.56)

uki → u, strongly in C(Q T ), (3.57)

(∇uki , ∇2uki) (∇u, ∇2u), weak-∗ in L∞(0, T ; L2(Ω)), (3.58)

∇2uki ∇

2u, weakly in L2(0, T ; L6(Ω)), (3.59)

ukit ut , weakly in L2(0, T ;H1

0 (Ω)), (3.60)


H. Wen, S. Ding / Nonlinear Analysis: Real World Applications 12 (2011) 1510–1531 1523ρkiuki

t √

ρut , weak-∗ in L∞(0, T ; L2(Ω)), (3.61)

(Pki , ∇Pki) (P, ∇P), weak-∗ in L∞(0, T ; L2(Ω)), (3.62)

∇Pki ∇P, weakly in L2(0, T ; L6(Ω)), (3.63)

nki → n, strongly in C(0, T ;H2(Ω)), (3.64)

(∇nki , ∇2nki , ∇3nki) (∇n, ∇2n, ∇3n), weak-∗ in L∞(0, T ; L2(Ω)), (3.65)

nkit nt , weak-∗ in L∞(0, T ;H1

0 (Ω)), (3.66)

∇2nki

t ∇2nt , weakly in L2(0, T ; L2(Ω)). (3.67)

By lower semi-continuity, we have

|||n|||A + |||u|||B + ‖√

ρut‖L∞(0,T ;L2(Ω)) + ‖P‖L∞(0,T ;H1(Ω)) + ‖P‖L2(0,T ;W1,6(Ω))

+ ‖ρ‖L∞(QT ) + ‖ρ‖L∞(0,T ;H1(Ω)) + ‖ρt‖L∞(0,T ;L2(Ω)) ≤c.By the uniqueness of the limits, we get ρ = ρ1

+∑

∞

i=2 ρ i, u = u1+

∑∞

i=2 ui and n = n1

+∑

∞

i=2 ni. (3.52) and (3.53) imply

uki−1→ u, in L2(0, T ;H1(Ω)), (3.68)

nki−1→ n, in L∞(0, T ;H1(Ω))

L2(0, T ;H2(Ω)), (3.69)

as i → ∞. Replacing k in (3.20) with the corresponding initial and boundary conditions by ki, and taking the limits, we canobtain that (ρ, u, P, n) is accurately a solution to problems (1.1)–(1.6) with regularities like in Theorem 1.1.

Since |||n|||A ≤c , we have from (1.4) and the parabolic theory n ∈ W 4,22 (QT ).Multiplying (1.4) by n, we get an equation for

|n|2 − 1. By the maximum principle, we get |n| = 1 in Q T . The uniqueness of the solution can be obtained by the standardmethod similar to step 4. The proof of Theorem 1.1 is complete.

4. ρ0 ≥ M1 > 0: local solution, and global solution with small initial data

Theorem 1.2 can be proved by iteration technique like in the last section or contraction fixed points theorem. Forsimplicity, we sketch the proof with the contraction fixed points theorem as follows.

Denote

GT ,c3 = m = (m1,m2,m3)|m(x, 0) = n0(x), |||m|||G ≤ c3, andHT ,c4 = v = (v1, v2, . . . , vN)|m(x, 0) = n0(x), v|∂Ω = 0, |||v|||H ≤ c4, where|||m|||G = ‖m‖W2,1

p+2(QT )+ sup

0≤t≤T‖m‖

W2− 2

p+2p+2 (Ω)

,

and

|||v|||H = ‖v‖W2,1p (QT )

+ sup0≤t≤T

‖v‖W

2− 2p

p (Ω)

.

Obviously, HT ,c4 ×GT ,c3 with norm ||| · ||| = ||| · |||G +||| · |||H is a complete space. Consider (3.18) with (v,m) ∈ HT ,c4 ×GT ,c3 .By the Lp theory for parabolic equation (cf. [17]) and the Navier–Stokes equation (cf. [18]), we can get a unique solution

ρ ∈ C1(Q T ), u ∈ W 2,1p (QT ), ∇P ∈ Lp(QT ), n ∈ W 2,1

p+2(QT ).

We normalize P such that

ΩPdx = 0. By the Poincare inequality, we get P ∈ Lp(0, T ;W 1,p(Ω)). Therefore, we can define a

mapM(v,m) = (u, n). With the estimates in [18,17], we can easily obtain thatM is a contraction map from HT ,c4 × GT ,c3 toHT ,c4 × GT ,c3 . Hence, there is a unique (u, n) ∈ HT ,c4 × GT ,c3 such thatM(u, n) = (u, n). Together with the Schauder theoryfor parabolic equations (cf. [17]), we have n ∈ C2+α,1+ α

2 (Q T ), where α is as that in Theorem 1.2. The proof of Theorem 1.2is completed.

Based on Theorem 1.2, we only need some a priori estimates globally in time to get Theorem 1.3. In the followingestimates, we denote the letter c to be the constant dependent on γ , λ, θ, T , and some other known constants, butindependent of ρ, u, n and P . We have the following priori estimates for any T > 0.

Lemma 4.1. We have the identity∫Ω

(ρ|u|2 + λ|∇n|2)dx + 2γ∫ T

0

∫Ω

|∇u|2 ≤

∫Ω

(ρ0|u0|2+ λ|∇n0|

2)dx, (4.1)

and

0 < M1 ≤ ρ ≤ M2. (4.2)



Proof. Multiplying (1.2′) by u, and integrating the resulting equation with respect to the space variables over Ω , we have

12

ddt

∫Ω

ρ|u|2dx + γ

∫Ω

|∇u|2dx = −λ

2−i,j=1

∫Ω

(nxixi .nxj + nxi .nxixj)ujdx

= −λ

2−i,j=1

∫Ω

nxixi .nxjujdx − λ

2−i=1

∫Ω

|∇n|2

2

xi

uidx

= −λ

2−i,j=1

∫Ω

nxixi .nxjujdx, (4.3)

here we have used integration by parts, (1.1′) and (1.3′).Since |n| = 1, we change (1.4) into this:

nt + (u · ∇)n = −θn × (n × 1n). (4.4)

Multiplying (4.4) by 1n, integrating the resulting equation over Ω , and using integration by parts, we obtain

12

ddt

∫Ω

|∇n|2dx + θ

∫Ω

|n × 1n|2dx =

2−i,j=1

∫Ω

nxixi · nxjujdx. (4.5)

(4.3) + (4.5) × λ gives

ddt

∫Ω

ρ|u|2

2+

λ|∇n|2

2

dx + λθ

∫Ω

|n × 1n|2dx + γ

∫Ω

|∇u|2dx = 0. (4.6)

Integrating (4.6) with respect to the time variable over (0, T ), we get (4.1).Using the characteristic method (cf. [18]), we obtain

M1 ≤ infΩ

ρ0 ≤ ρ ≤ supΩ

ρ0 ≤ M2.

The proof of the lemma is complete.

(4.1) and (4.2) immediately give the following corollary:

Corollary 4.1.

sup0≤t≤T

∫Ω

|u|2dx ≤ c.

Lemma 4.2. we have the inequality∫ T

0

∫Ω

|∇2n|2 + |nt |

2dxdt ≤ c.

Proof. Multiplying (1.4) by 1n, integrating the resulting equation over Ω , and using integration by parts, we have

θ

∫Ω

|1n|2dx +12

ddt

∫Ω

|∇n|2dx = θ

∫Ω

|∇n|4dx +

∫Ω

(u.∇)n.1ndx. (4.7)

(4.3) + (4.7) × λ gives

ddt

∫Ω

ρ|u|2

2+

λ|∇n|2

2

dx + λθ

∫Ω

|1n|2dx + γ

∫Ω

|∇u|2dx = θλ

∫Ω

|∇n|4dx

≤ θλ(c‖∇n‖L1(Ω) +c‖∇2n‖12L2(Ω)

‖∇n‖12L2(Ω)

)4

≤ 8θλc4‖∇n‖4L1(Ω)

+ 8θλc4‖∇2n‖2L2(Ω)

‖∇n‖2L2(Ω)

≤ c + 8c4θλc1‖1n‖2L2(Ω)

‖∇n‖2L2(Ω)

≤ c + 8c4θλc1‖1n‖2L2(Ω)

∫Ω

|∇n0|2dx +

1λ

∫Ω

ρ0|u0|2dx

≤ c + 8c4θλc1c0‖1n‖2

L2(Ω),



where we have used Lemmas 2.4 and 4.1, the conditions of Theorem 1.2 and the elliptic estimate

‖∇2n‖2

L2(Ω)≤c1(‖1n‖2

L2(Ω)+ 1). (4.8)

Choosing c0 > 0 sufficiently small such that

8c4c1c0 <13, (4.9)

we have

ddt

∫Ω

ρ|u|2

2+

λ|∇n|2

2

dx +

2λθ

3

∫Ω

|1n|2dx + γ

∫Ω

|∇u|2dx ≤ c.

Together with (4.8), we get∫ T

0

∫Ω

|∇2n|2dxdt ≤ c. (4.10)

It follows from (1.4) and |n| = 1 that∫Ω

|nt |2dx ≤ c

∫Ω

|1n|2dx + c∫

Ω

|∇n|4dx + c∫

Ω

|u|2|∇n|2dx.

Using Lemmas 2.3, 2.4 and 4.1 and Corollary 4.1, we have∫Ω

|nt |2dx ≤ c

∫Ω

|1n|2dx + c(‖∇n‖L1(Ω) + ‖∇2n‖

12L2(Ω)

‖∇n‖12L2(Ω)

)4 + c‖u‖2L4(Ω)

‖∇n‖2L4(Ω)

≤ c∫

Ω

|1n|2dx + c + c‖∇2n‖2L2(Ω)

+ c‖u‖L2(Ω)‖∇u‖L2(Ω)(‖∇n‖L1(Ω) + ‖∇2n‖

12L2(Ω)

‖∇n‖12L2(Ω)

)2

≤ c∫

Ω

|∇2n|2dx + c + c‖∇u‖L2(Ω)(1 + ‖∇

2n‖L2(Ω))

≤ c∫

Ω

|∇2n|2dx + c + c

∫Ω

|∇u|2dx.

Together with (4.10) and Lemma 4.1, we have∫ T

0

∫Ω

|nt |2dxdt ≤ c.

The proof of the lemma is complete.

Lemma 4.3. we have the inequality∫Ω

|∇2n|2dx ≤ c + c

∫Ω

|∇u|2dx + c∫

Ω

|nt |2dx,

for any 0 ≤ t ≤ T .

Proof. It follows from (1.4), Lemmas 2.3 and 2.4 that

θ2∫

Ω

|1n|2dx ≤ 3θ2∫

Ω

|∇n|4dx + 3∫

Ω

|nt |2dx + 3

∫Ω

|u|2|∇n|2dx

≤ 3θ2(c‖∇2n‖12L2(Ω)

‖∇n‖12L2(Ω)

+c‖∇n‖L1(Ω))4+ 3

∫Ω

|nt |2dx + 3‖u‖2

L4(Ω)‖∇n‖2

L4(Ω)

≤ 3θ2(c‖∇2n‖12L2(Ω)

‖∇n‖12L2(Ω)

+c‖∇n‖L1(Ω))4

+ 3∫

Ω

|nt |2dx + c‖u‖L2(Ω)‖∇u‖L2(Ω)(‖∇n‖L1(Ω) + ‖∇

2n‖12L2(Ω)

‖∇n‖12L2(Ω)

)2.

Using Lemma 4.1, Corollary 4.1, (4.8) and the conditions of Theorem 1.3, we get

θ2∫

Ω

|1n|2dx ≤ 24θ2c4c1‖1n‖2L2(Ω)

‖∇n‖2L2(Ω)

+ c + 3∫

Ω

|nt |2dx + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖∇

2n‖L2(Ω)

≤ 24θ2c4c1‖1n‖2L2(Ω)

∫Ω

|∇n0|2dx +

1λ

∫Ω

ρ0|u0|2dx

+ c



+ 3∫

Ω

|nt |2dx + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖1n‖L2(Ω)

≤ 24θ2c4c1c0‖1n‖2L2(Ω)

+ c + 3∫

Ω

|nt |2dx + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖1n‖L2(Ω).

Together with (4.9) and the Cauchy inequality, we have∫Ω

|1n|2dx ≤ c + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖1n‖L2(Ω) + c∫

Ω

|nt |2dx

≤ c + c‖∇u‖2L2(Ω)

+12‖1n‖2

L2(Ω)+ c

∫Ω

|nt |2dx.

Hence,∫Ω

|1n|2dx ≤ c + c‖∇u‖2L2(Ω)

+ c∫

Ω

|nt |2dx. (4.11)

By (4.8) and (4.11), we complete the proof of the lemma.

Lemma 4.4. We have the inequality∫Ω

(|nt |2+ |∇u|2)dx +

∫ t

0

∫Ω

|ut |2+ |∇

2u|2dxdt ≤ c, for any 0 ≤ t ≤ T .

Proof. Some of the following procedures are formal because the regularities of the solution is insufficient to validate severalof the steps. Rigorous derivations of the inequalities can be achieved easily by using mollifiers, or by using differencequotients and taking the limits. Step 1:

ddt

∫Ω

|nt |2dx = 2

∫Ω

nt · nttdx

= 2∫

Ω

nt · [θ1nt + θ |∇n|2nt + 2θ(∇n · ∇nt)n − (ut · ∇)n − (u · ∇)nt ]dx.

Together with integration by parts and the Cauchy inequality, we have

ddt

∫Ω

|nt |2dx ≤ −2θ

∫Ω

|∇nt |2dx + c

∫Ω

|nt |2|∇n|2dx +

θ

2

∫Ω

|∇nt |2dx +

θ

16

∫Ω

|ut |2dx − 2

∫Ω

(u · ∇)nt · ntdx.

Since ∇ · u = 0, we have∫Ω

(u · ∇)nt · ntdx =

∫Ω

u · ∇

|nt |

2

2

dx

= 0.

Therefore,ddt

∫Ω

|nt |2dx ≤ −

3θ2

∫Ω

|∇nt |2dx + c

∫Ω

|nt |2|∇n|2dx +

θ

16

∫Ω

|ut |2dx

≤ −3θ2

∫Ω

|∇nt |2dx + c‖nt‖

2L4(Ω)

‖∇n‖2L4(Ω)

+θ

16

∫Ω

|ut |2dx

≤ −3θ2

∫Ω

|∇nt |2dx + c(‖nt‖

2L2(Ω)

+ ‖nt‖L2(Ω)‖∇nt‖L2(Ω))(‖∇n‖2L2(Ω)

+ ‖∇n‖L2(Ω)‖∇2n‖L2(Ω))

+θ

16

∫Ω

|ut |2dx

≤ −3θ2

∫Ω

|∇nt |2dx + c‖nt‖

2L2(Ω)

+ c‖nt‖2L2(Ω)

‖∇2n‖L2(Ω) + c‖nt‖L2(Ω)‖∇nt‖L2(Ω)

+ c‖nt‖L2(Ω)‖∇nt‖L2(Ω)‖∇2n‖L2(Ω) +

θ

16

∫Ω

|ut |2dx

≤ −3θ2

∫Ω

|∇nt |2dx +

θ

2

∫Ω

|∇nt |2dx + c

∫Ω

|nt |2dx + c

∫Ω

|nt |2dx

∫Ω

|∇2n|2dx +

θ

16

∫Ω

|ut |2dx

= −θ

∫Ω

|∇nt |2dx + c

∫Ω

|nt |2dx + c

∫Ω

|nt |2dx

∫Ω

|∇2n|2dx +

θ

16

∫Ω

|ut |2dx,



here we have used Lemmas 2.4 and 4.1 and the Cauchy inequality. Together with the Cauchy inequality and Lemma 4.3, wehave

ddt

∫Ω

|nt |2dx + θ

∫Ω

|∇nt |2dx ≤ c‖nt‖

4L2(Ω)

+ c‖∇2n‖4L2(Ω)

+θ

16

∫Ω

|ut |2dx + c

≤ c‖nt‖4L2(Ω)

+ c‖∇u‖4L2(Ω)

+θ

16

∫Ω

|ut |2dx + c. (4.12)

Step 2: It follows from (1.4), Lemmas 2.3 and 2.4 that

‖∇1n‖L2(Ω) ≤ ‖∇nt‖L2(Ω) + c‖∇u.∇n‖L2(Ω) + c‖u.∇∇n‖L2(Ω) + c‖∇∇n.∇n‖L2(Ω) + c‖∇n‖3L6(Ω)

≤ ‖∇nt‖L2(Ω) + c‖∇u‖L4(Ω)‖∇n‖L4(Ω) + c‖u‖L4(Ω)‖∇2n‖L4(Ω) + c‖∇2n‖L4(Ω)‖∇n‖L4(Ω) + c‖∇n‖3

L6(Ω)

≤ ‖∇nt‖L2(Ω) + c(‖∇u‖L1(Ω) + ‖∇2u‖

12L2(Ω)

‖∇u‖12L2(Ω)

)(‖∇n‖L1(Ω) + ‖∇2n‖

12L2(Ω)

‖∇n‖12L2(Ω)

)

+ c‖u‖12L2(Ω)

‖∇u‖12L2(Ω)

(1 + ‖1n‖L4(Ω)) + c(1 + ‖1n‖L4(Ω))(‖∇n‖L1(Ω) + ‖∇2n‖

12L2(Ω)

‖∇n‖12L2(Ω)

)

+ c(‖∇n‖3L1(Ω)

+ ‖∇2n‖2

L2(Ω)‖∇n‖L2(Ω)).

Together with Lemma 4.1, we have

‖∇1n‖L2(Ω) ≤ ‖∇nt‖L2(Ω) + c(‖∇u‖L2(Ω) + ‖∇2u‖

12L2(Ω)

‖∇u‖12L2(Ω)

)(1 + ‖∇2n‖

12L2(Ω)

)

+ c‖∇u‖12L2(Ω)

(1 + ‖1n‖L1(Ω) + ‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

)

+ c(1 + ‖1n‖L1(Ω) + ‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

)(1 + ‖∇2n‖

12L2(Ω)

) + c + c‖∇2n‖2L2(Ω)

≤ ‖∇nt‖L2(Ω) + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖∇2n‖

12L2(Ω)

+ c‖∇2u‖12L2(Ω)

‖∇u‖12L2(Ω)

+ c‖∇2u‖12L2(Ω)

‖∇u‖12L2(Ω)

‖∇2n‖

12L2(Ω)

+ c‖∇u‖12L2(Ω)

‖1n‖L2(Ω) + c‖∇u‖12L2(Ω)

‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

+ c + c‖∇2n‖2L2(Ω)

+ c‖1n‖L1(Ω) + c‖1n‖L1(Ω)‖∇2n‖

12L2(Ω)

+ c‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

+ c‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

‖∇2n‖

12L2(Ω)

.

By the Cauchy inequality, we have

‖∇1n‖L2(Ω) ≤ ‖∇nt‖L2(Ω) + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖∇2n‖L2(Ω) + c‖∇2u‖

12L2(Ω)

‖∇u‖12L2(Ω)

+ c‖∇2u‖12L2(Ω)

‖∇u‖12L2(Ω)

‖∇2n‖

12L2(Ω)

+12‖∇1n‖L2(Ω) + c‖∇2n‖2

L2(Ω)+ c.

Thus

‖∇1n‖L2(Ω) ≤ 2‖∇nt‖L2(Ω) + c‖∇u‖L2(Ω) + c‖∇u‖L2(Ω)‖∇2n‖L2(Ω) + c‖∇2u‖

12L2(Ω)

‖∇u‖12L2(Ω)

+ c‖∇2u‖12L2(Ω)

‖∇u‖12L2(Ω)

‖∇2n‖

12L2(Ω)

+ c‖∇2n‖2L2(Ω)

+ c.

Therefore,

‖∇1n‖2L2(Ω)

≤ 8‖∇nt‖2L2(Ω)

+ c‖∇u‖2L2(Ω)

+ c‖∇u‖2L2(Ω)

‖∇2n‖2

L2(Ω)+ c‖∇2u‖L2(Ω)‖∇u‖L2(Ω)

+ c‖∇2u‖L2(Ω)‖∇u‖L2(Ω)‖∇2n‖L2(Ω) + c‖∇2n‖4

L2(Ω)+ c

≤ 8‖∇nt‖2L2(Ω)

+ c‖∇u‖4L2(Ω)

+ c‖∇2n‖4L2(Ω)

+14‖∇

2u‖2L2(Ω)

+ c.

By Lemma 4.3, we have

‖∇1n‖2L2(Ω)

≤ 8‖∇nt‖2L2(Ω)

+ c‖∇u‖4L2(Ω)

+ c‖nt‖4L2(Ω)

+14‖∇

2u‖2L2(Ω)

+ c. (4.13)

Step 3: Multiplying (1.2′) by ut , integrating the resulting equation over Ω , and employing integration by parts, we have∫Ω

ρ|ut |2dx +

γ

2ddt

∫Ω

|∇u|2dx = −

∫Ω

ρ(u · ∇)u · utdx − λ

∫Ω

[∇ ·

∇n

∇n

] · utdx.



Since

∇ ·

∇n

∇n

= 1n · ∇n + ∇

|∇n|2

2

.

By integration by parts and ∇ · u = 0, we have∫Ω

ρ|ut |2dx +

γ

2ddt

∫Ω

|∇u|2dx = −

∫Ω

ρ(u · ∇)u · utdx − λ

∫Ω

1n · ∇n · utdx.

Together with (4.2), we get∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx ≤ c‖ut‖L2(Ω)‖u‖L4(Ω)‖∇u‖L4(Ω) + c‖1n‖L4(Ω)‖∇n‖L4(Ω)‖ut‖L2(Ω)

≤12

∫Ω

|ut |2dx + c‖u‖2

L4(Ω)‖∇u‖2

L4(Ω)+ c‖1n‖2

L4(Ω)‖∇n‖2

L4(Ω).

Thus ∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx ≤ c‖u‖2L4(Ω)

‖∇u‖2L4(Ω)

+ c‖1n‖2L4(Ω)

‖∇n‖2L4(Ω)

.

Together with Lemmas 2.3, 2.4 and 4.1, Corollary 4.1, and the Cauchy inequality, we have∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx ≤ c‖u‖L2(Ω)‖∇u‖L2(Ω)(‖∇u‖2L2(Ω)

+ ‖∇u‖L2(Ω)‖∇2u‖L2(Ω)) + c(‖1n‖2

L2(Ω)

+ ‖∇1n‖L2(Ω)‖1n‖L2(Ω))(‖∇n‖2L2(Ω)

+ ‖∇2n‖L2(Ω)‖∇n‖L2(Ω))

≤ c‖∇u‖3L2(Ω)

+ c‖∇u‖2L2(Ω)

‖∇2u‖L2(Ω) + c‖∇2n‖3

L2(Ω)

+ c + c‖∇1n‖L2(Ω)(‖∇2n‖2

L2(Ω)+ 1). (4.14)

It follows from [7], (4.2), Lemmas 2.3 and 2.4 that

‖∇2u‖L2(Ω) ≤ c‖ρu · ∇u‖L2(Ω) + c‖∇ ·

∇n

∇n

‖L2(Ω) + c‖ρut‖L2(Ω)

≤ c‖∇u‖L4(Ω)‖u‖L4(Ω) + c‖∇2n‖L4(Ω)‖∇n‖L4(Ω) + c‖ut‖L2(Ω)

≤ c(‖∇u‖L1(Ω) + ‖∇2u‖

12L2(Ω)

‖∇u‖12L2(Ω)

)‖∇u‖12L2(Ω)

‖u‖12L2(Ω)

+ c‖1n‖L4(Ω)‖∇n‖L4(Ω)

+ c‖∇n‖L4(Ω) + c‖ut‖L2(Ω),

here we have used the elliptic estimate

‖∇2n‖L4(Ω) ≤ c‖1n‖L4(Ω) + c.

Together with Lemmas 2.4 and 4.1, Corollary 4.1 and the Cauchy inequality, we have

‖∇2u‖L2(Ω) ≤ c‖∇u‖

32L2(Ω)

+ c‖∇2u‖12L2(Ω)

‖∇u‖L2(Ω) + c(‖1n‖L1(Ω) + ‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

)

× (‖∇n‖L1(Ω) + ‖∇n‖12L2(Ω)

‖∇2n‖

12L2(Ω)

) + c(‖∇n‖L1(Ω) + ‖∇n‖12L2(Ω)

‖∇2n‖

12L2(Ω)

) + c‖ut‖L2(Ω)

≤ c‖∇u‖32L2(Ω)

+12‖∇

2u‖L2(Ω) + c‖∇u‖2L2(Ω)

+ c‖1n‖L2(Ω) + c‖1n‖12L2(Ω)

‖∇1n‖12L2(Ω)

+ c‖∇2n‖32L2(Ω)

+ c‖∇2n‖L2(Ω)‖∇1n‖12L2(Ω)

+ c + c‖ut‖L2(Ω).

Combining the Cauchy inequality, we have

‖∇2u‖L2(Ω) ≤ c‖∇u‖2

L2(Ω)+ c + c‖∇2n‖L2(Ω)‖∇1n‖

12L2(Ω)

+ c‖∇1n‖12L2(Ω)

+ c‖∇2n‖32L2(Ω)

+ c‖ut‖L2(Ω)

= c‖∇u‖2L2(Ω)

+ c + c(1 + ‖∇2n‖L2(Ω))‖∇1n‖

12L2(Ω)

+ c‖∇2n‖32L2(Ω)

+ c‖ut‖L2(Ω). (4.15)

By (4.14) and (4.15), and the Cauchy inequality, we have∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx +

∫Ω

|∇2u|2dx ≤ c‖∇u‖2

L2(Ω)‖∇

2u‖L2(Ω) + c‖∇1n‖L2(Ω)(‖∇2n‖2

L2(Ω)+ 1)

+ c‖∇u‖4L2(Ω)

+ c‖∇2n‖3L2(Ω)

+ c



≤12

∫Ω

|∇2u|2dx + c‖∇u‖4

L2(Ω)+ c‖∇1n‖L2(Ω)(‖∇

2n‖2L2(Ω)

+ 1)

+ c‖∇2n‖4L2(Ω)

+ c.

Thus ∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx +12

∫Ω

|∇2u|2dx ≤ c‖∇u‖4

L2(Ω)+ c‖∇1n‖L2(Ω)(‖∇

2n‖2L2(Ω)

+ 1)

+ c‖∇2n‖4L2(Ω)

+ c. (4.16)

By (4.16), the Cauchy inequality and Lemma 4.3, we have∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx +12

∫Ω

|∇2u|2dx ≤ c‖∇u‖4

L2(Ω)+

12‖∇1n‖2

L2(Ω)+ c‖∇2n‖4

L2(Ω)+ c

≤ c‖∇u‖4L2(Ω)

+12‖∇1n‖2

L2(Ω)+ c‖nt‖

4L2(Ω)

+ c. (4.17)

We have from (4.13) and (4.17)∫Ω

|ut |2dx +

ddt

∫Ω

|∇u|2dx +14

∫Ω

|∇2u|2dx +

12

∫Ω

|∇1n|2dx

≤ 8‖∇nt‖2L2(Ω)

+ c‖∇u‖4L2(Ω)

+ c‖nt‖4L2(Ω)

+ c. (4.18)

By (4.12) and (4.18), we have

ddt

∫Ω

|∇u|2dx +8θ

∫Ω

|nt |2dx

+

12

∫Ω

|ut |2dx +

14

∫Ω

|∇2u|2dx ≤ c‖∇u‖4

L2(Ω)+ c‖nt‖

4L2(Ω)

+ c

≤ c∫

Ω

|∇u|2dx +

∫Ω

|nt |2dx

2

+ c.

Integrating over (τ , t), we have∫Ω

|∇u(x, t)|2dx +

∫Ω

|nt(x, t)|2dx +

∫ t

τ

∫Ω

|ut(x, s)|2dx +

∫Ω

|∇2u(x, s)|2dx

ds

≤ c∫

Ω

|∇u(x, τ )|2dx +

∫Ω

|nt(x, τ )|2dx

+ c∫ t

τ

∫Ω

|∇u(x, s)|2dx +

∫Ω

|nt(x, s)|2dx2

ds + c. (4.19)

It follows from (1.4) that∫Ω

|nt(x, τ )|2dx ≤ c∫

Ω

|1n(x, τ )|2dx + c∫

Ω

|∇n(x, τ )|4dx + c∫

Ω

|u(x, τ )|4dx. (4.20)

By (4.19) and (4.20), we have∫Ω

|∇u(x, t)|2dx +

∫Ω

|nt(x, t)|2dx +

∫ t

τ

∫Ω

|ut(x, s)|2dx +

∫Ω

|∇2u(x, s)|2dx

ds

≤ c∫

Ω

|∇u(x, τ )|2dx +

∫Ω

|1n(x, τ )|2dx +

∫Ω

|∇n(x, τ )|4dx +

∫Ω

|u(x, τ )|4dx

+ c∫ t

τ

∫Ω

|∇u(x, s)|2dx +

∫Ω

|nt(x, s)|2dx2

ds + c.

Let τ → 0+, together with the initial conditions, we have∫Ω

|∇u(x, t)|2dx +

∫Ω

|nt(x, t)|2dx +

∫ t

0

∫Ω

|ut(x, s)|2dx +

∫Ω

|∇2u(x, s)|2dx

ds

≤ c∫ t

0

∫Ω

|∇u(x, s)|2dx +

∫Ω

|nt(x, s)|2dx2

ds + c. (4.21)

By Lemmas 2.1, 4.1 and 4.2 and (4.21), we complete the proof of the lemma.



Lemma 4.5. We have the inequality∫Ω

|∇2n|2dx ≤ c, for any 0 ≤ t ≤ T .

Proof. It can be obtained directly by Lemmas 4.3 and 4.4.

Lemma 4.6. We have the inequality

‖ρ‖C1(Q T ) + ‖u‖W2,1p (QT )

+ ‖P‖Lp(0,T ;W1,p(Ω)) + ‖n‖C2+α, 2+α

2 (Q T )≤ c.

Proof. Since sup0≤t≤T ‖u(·, t)‖H1(Ω) ≤ c and sup0≤t≤T ‖n(·, t)‖H2(Ω) ≤ c , we have from the Sobolev inequality

‖u‖Lq1 (QT ) ≤ c, (4.22)

‖n‖Lq1 (QT ) + ‖∇n‖Lq1 (QT ) ≤ c, (4.23)

for any q1 > 0.Combining (4.22) and (4.23), we obtain

‖(u · ∇)n‖Lp+2(QT ) ≤ c, and ‖ |∇n|2n‖Lp+2(QT ) ≤ c. (4.24)

By the standard parabolic estimates [17] and (1.4), we have

‖n‖W2,1p+2(QT )

≤ c. (4.25)

Denote

F = −λ∇ ·

∇n

∇n

ρ

,

(1.1)–(1.3) can be changed as follows:ρt + (u · ∇)ρ = 0,ρut + ρ(u · ∇)u + ∇P = γ1u + ρF ,∇ · u = 0,

(4.26)


(ρ, u)|t=0 = (ρ0, u0), (4.27)u|∂Ω = 0. (4.28)

We have from (4.2) and (4.25)

‖F‖Lp(QT ) ≤ c. (4.29)

Since (4.26) is the Navier–Stokes equation, and the dimension N = 2, we obtain from (4.29) and [18]

‖ρ‖C1(Q T ) + ‖u‖W2,1p (QT )

+ ‖P‖Lp(0,T ;W1,p(Ω)) ≤ c. (4.30)

By the Sobolev inequality, (4.25) and (4.30), we have

‖∇n‖Cα2,

α22 (Q T )

≤ c, (4.31)

‖u‖Cα1,

α12 (Q T )

≤ c, (4.32)

where α1 = 2 −N+2p , and α2 = 1 −

N+2p+2 . p > N and N ≥ 2 imply α2 < α1. It follows from (1.4), (4.31) and (4.32) and the

Schauder theory that

‖n‖C2+α2,

2+α22 (Q T )

≤ c.

Thus ‖∇n‖Cα, α

2 (Q T )≤ c , where the α is defined in Theorem 1.2. Since α1 ≥ α, we use the Schauder theory again to get

‖n‖C2+α, 2+α

2 (Q T )≤ c.

Based on the local existence of the solution and the global priori estimates, the proof of the existence of the global solutionis completed by standard arguments (cf. [18]). The uniqueness of the global solution is based on that of the local solution.Therefore, Theorem 1.3 is obtained.



Acknowledgement

The authorswould like to thank Professor ChangyouWang for his suggestions and discussions. The authors are supportedby the National Basic Research Program of China (973 Program) No. 2011CB808002, and by the National Natural ScienceFoundation of China No. 11071086.

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