MINIMUM PRINCIPLE OF THE TEMPERATURE IN THREE …

MINIMUM PRINCIPLE OF THE TEMPERATURE IN

THREE-DIMENSIONAL COMPRESSIBLE FLOW WITH VACUUM

GENG CHEN, WEIZHE ZHANG, AND SHENGGUO ZHU

Abstract. In this paper, we consider the initial-boundary value problem to the threedimensional (3-D), compressible Navier-Stokes-Fourier system for the viscous, heat con-ductive, Newtonian polytropic flow, where the heat conductivity depends on temperaturein a power law of Chapman-Enskog. First, based on the mathematical structure of ournonlinear system, via introducing some new variable and one type of initial layer compat-ibility condition, we established the local-in-time well-posedenss of strong solutions withvacuum and arbitrarily large data. Secondly, inspired by De Giorgi’s proof of Holderregularity for the solutions of elliptic equation with discontinuous coefficients, via intro-ducing some interpolation estimates and a special continuity argument, we show thatunder the thermo-insulated boundary conditions in some bounded smooth domain, theminimum of the temperature will not increase within the life span of our strong solu-tions even initial vacuum appears. Our conclusion first verifies the necessity of the thirdlaw of thermodynamics for polytropic flow governed by our nonlinear system from fluidmechanics in 3-D space.

1. Introduction

The motion of compressible viscous, heat-conductive, Newtonian polytropic fluid oc-cupying a bounded smooth domain V ⊂ R3 is governed by the following compressibleNavier-Stokes-Fourier system (CNSF):

ρt + div(ρu) = 0,

(ρu)t + div(ρu⊗ u) +∇P = divT,(ρE)t + div

((ρE + P )u

)= div(uT) + div(κ∇θ).

(1.1)

We consider its initial-boundary value problem (IBVP) with the following initial data

(ρ, u, θ)|t=0 = (ρ0(x), u0(x), θ0(x)), x ∈ V, (1.2)

and the Dirichlet and thermo-insulated boundary conditions for (u, θ):

u|∂V = 0, ∇θ · n|∂V = 0, (1.3)

where n = (n1, n2, n3) is the unit outward normal to ∂V.

Date: Apr. 20, 2017.1991 Mathematics Subject Classification. Primary: 35A01, 35B50, 35D35; Secondary: 35Q36, 35J20.Key words and phrases. Compressible Navier-Stokes-Fourier system, Strong solution, Vacuum, De

Giorgi iteration, Minimum principle.

1

2 GENG CHEN, WEIZHE ZHANG, AND SHENGGUO ZHU

In system (1.1), t ≥ 0 is the time variable; x ∈ V is the spatial coordinate; ρ ≥ 0 is themass density; u ∈ R3 is the velocity of fluids;

E =1

2|u|2 + e

is the specific total energy and e is the specific internal energy; P is the pressure forpolytropic flow satisfying:

P = Rρθ = (γ − 1)ρe, e =R

γ − 1θ, (1.4)

where R is a positive constant, γ > 1 is the adiabatic exponent and θ is the absolutetemperature. T is the viscosity stress tensor given by

T = 2µD(u) + λdivuI3, D(u) =∇u+ (∇u)>

2, (1.5)

where D(u) is the deformation tensor, I3 is the 3 × 3 identity matrix, µ is the shearviscosity coefficient, and λ+ 2

3µ is the bulk viscosity coefficient. κ is the heat conductivitycoefficient.

The mission in this paper is to establish the minimum principle of the temperaturefor strong solutions with large data and possible vacuum to IBVP (1.1)-(1.5) under theassumptions on viscosities and heat conductivity:

µ =α, λ = β, κ(θ) = νθb, (1.6)

where (α, β, ν, b) are all constants satisfying:

α > 0, 2α+ 3β ≥ 0, ν > 0, and b ≥ 0. (1.7)

For this purpose, we are also interested in the local-in-time well-posedness of correspondingstrong solutions with large data and non-negative mass densities to the above IBVP.

In the classical setting that (µ, λ, κ) are all constants, there are a lot of literatures on thewell-posedness of solutions and their hehaviors to the correspinding problem. The mainbreakthrough for the well-posedness of solutions in 3-D space with generic initial dataincluding vacuum is due to Lions [14], where he established the global existence of weaksolutions with finite energy to the compressible isentropic flow provided that γ > 9

5 in R3

(see also Feireisl-Novotny-Petzeltova [5, 6, 7] for γ > 32 and the corresponding conclusion

for the non-isentropic flow). Note that here the weak solutions may contain vacuumthough the spatial measure of the set of vacuum has to be small. However, the uniquenessproblem of these weak solutions is widely open due to their fairly low regularities.

On the other hand, if we consider the solutions with relative higher regularities whichcould make sure that density connects to vacuum continuously, we need to face the de-generacy of time evolution in momentum and energy equations (1.1)2-(1.1)3. It is obviousthat the leading coefficients of ut and Et in momentum and energy equations vanish atvacuum, and this leads to infinitely many ways to define velocity and temperature (if theyexit) when vacuum appears. This degeneracy leads to an essential difficulty that it is hardto find a reasonable way to extend the definitions of velocity and temperature into vacuumregion. In 2006, a remedy was suggested by Cho-Kim [3], where they imposed initially acompatibility condition

Lu0 +∇P 0 =√ρ0f1, −κ4θ0 −Q(u0) =

√ρ0f2, (1.8)

NAVIER-STOKES EQUATIONS 3

for some (f1, f2) ∈ L2, andP0 = Rρ0θ0, Lu0 = −µ4u0 − (µ+ λ)∇divu0,

Q(u0) = µ2 |∇u0 + (∇u0)>|2 + λ|divu0|2.

Very roughly, this condition is equivalent to the L2-integrability of(√ρut(t = 0, x),

√ρθt(t =

0, x))

and plays a key role in deducing that

(√ρut,√ρθt) ∈ L∞([0, T∗];L

2(R3)) and (∇ut,∇θt) ∈ L2([0, T∗];L2(R3))

for a short time T∗ > 0. Then they successfully established the local well-posedness ofsmooth solutions with non-negative density in R3, which, has been extended to be a globalone with small energy but large oscillations by Huang-Li [9] (see Ding-Li-Pan-Xin-Zhu[4, 13, 19, 20] for the corresponding results to compressible degenerate viscous flow). Forthe lower bound estimate of the absolute temperature, recently in one-dimensional (1-D)space, Liang-Li [11] showed us that for the strong solutions with large data to the IBVPfor CNSF in unbounded domains, the temperature is postiviely bounded from below,independent of both time and space. Mellet-Vassuer [15] proved that the temperature isalways positive when t ≥ t0 (for arbitrary t0 > 0) for any 3-D solutions with finite initialentropy. However, unitl now, we know very little on the strict minimum principle of theabsolute termerature for the multi-dimensional problem with large data of the polytropicviscous flow. Our result obtained in this paper has taken a first step toward this directionfor constant viscous and heat conductive flow.

In the classical theory of gas dynamics, when we derive CNSF from the Boltzmannequation through the Chapman-Enskog expansion to the second order, cf. Chapman-Cowling [2] and Li-Qin [12], we find that, under some reasonable physical assumptions,the viscosity coefficients and the heat conductivity coefficient κ are not constants butfunctions of absolute temperature θ. Actually, we see that in [2], for the cut-off inversepower force model, if the intermolecular potential varies as r−a with a ∈ (0,∞), where ris the intermolecular distance, then we have

µ(θ) =αθb, λ(θ) = βθb, κ(θ) = νθb, (1.9)

for b = 12 + 2

a . In particular, for Maxwellian molecules, we have a = 4 and b = 1; while for

elastic spheres model, we have a =∞ and b = 12 .

However, when (µ, λ, κ) are dependent of θ as in (1.9), there are very few results nomatter on weak or strong solutions because of the possible degeneracy caused by the initialvacuum and the strong nonlinearity in viscosities and heat diffusion. As a pioneer paperin this direction, Jenssen-Karper [10] proved the global existence of a weak solution in 1-Dspace to (1.1) under the assumption

µ =α, κ(θ) = νθb for b ∈[0,

3

2

). (1.10)

Under this simplified relation the viscosity coefficient µ is a constant, but the porousmedium type energy equation (1.1)3 still introduces significant difficulties. To overcomethese difficulties and then obtain a global existence of weak solution, some new argumentshave been introduced in [10]. When b ∈ (0,+∞), Pan-Zhang [16] showed that (1.1) admitsa unique global strong solution away from vacuum in 1-D space. Recently, when (1.9) is


satisfied, Pan-Zhang [18] showed the global existence of variational solutions with vacuumon some bounded smooth domain of R3, if the pressure is given as

P = Rρθ + ργ for γ > 3.

Unfortunately, due to the high degeneracy for this system in vacuum domain, it isdifficult to get uniform estimates for the velocity or temperature near vacuum. So, whenthe heat conductivity depends on temperature in a power law of Chapman-Enskog suchas in (1.6), the minimum principle of the absolute temperture as well as the related well-posedness of the strong or classical solutions with vacuum and arbitrarily large data in3-D space are totally open until this paper.

Here and throughout this paper, we adopt the following simplified notations for thestandard homogeneous and inhomogeneous Sobolev spaces:

‖f‖s = ‖f‖Hs(V), |f |p = ‖f‖Lp(V), ‖f‖Wm,r = ‖f‖Wm,r(V),

‖f‖LpLq = ‖f‖Lp([0,T ];Lq(V)), Dk,r = f ∈ L1loc(V) : |f |Dk,r = |∇kf |r < +∞,

Dk = Dk,2, |f |Dk = ‖f‖Dk(V), |f |Dk,r = ‖f‖Dk,r(V),

∫Vfdx =

∫f

D10 = f ∈ L6(V) : |f |D1

0= |∇f |2 <∞, f |∂V = 0, |f |D1

0= ‖f‖D1

0(V).

A detailed study of homogeneous Sobolev spaces could be found in [8].

In order to show our main results, we first introduce the following definition of strongsolutions to IBVP (1.1)-(1.7).

Definition 1.1. Let q ∈ (3, 6]. Functions (ρ, u, θ) are called a strong solution on [0, T ]×Vto IBVP (1.1)-(1.7), if

(1) (ρ, u, θ) satisfy the system (1.1) a.e. in (0, T )× V;(2) (ρ, u, θ) belong to the following class Φ with some regularities:

Φ =(ρ, u, θ)|0 ≤ ρ, ρ ∈ C([0, T ];H1 ∩W 1,q), ρt ∈ C([0, T ];L2 ∩ Lq),

(θ, u) ∈ C([0, T ];D1 ∩D2) ∩ L2([0, T ];D2,q),

(θt, ut) ∈ L2([0, T ];D1), (√ρθt,√ρut) ∈ L∞([0, T ];L2);

(1.11)

(3) (ρ, u, θ) satisfy the corresponding initial conditions a.e. on t = 0 × V, and alsosatisfy the corresponding boundary conditions in the sense of traces.

In the next two subsections, we will introduce our main results in this paper. In Subsec-tion 1.1, we will introduce the local-in-time existence result for strong solutions with largedata and vacuum, which also serves as a preparation for the minimum principle resultwhich will be introduced in Subsection 1.2. The minimum principle result is the mainresult in this paper, which shows that the temperature in (1.1) on the whole life span ofthe corresponding solutions satisfies a minimum principle even when the initial vacuumappears.


1.1. Existence of the unique local strong solution with vacuum. As has beenobserved in [3], the lack of a positive lower bound of ρ0 should be compensated with somecompatibility conditions on the initial data. Similarly, if we denote

V = x ∈ V| ρ0(x) = 0, P 0 = Rρ0θ0,

Q(u) =µ

2|∇u+ (∇u)>|2 + λ|divu|2,

(1.12)

then we have the following existence theorem:

Theorem 1.1. Assume that γ > 1 and θ > 0 is a real constant. Let the initial data(ρ0, u0, θ0) satisfy the following regularity conditions

ρ0 ≥ 0, ρ0 ∈ H1 ∩W 1,q, q ∈ (3, 6],

u0 ∈ D1 ∩D2, θ0 ∈ D1 ∩D2, θ0 ≥ θ,(1.13)

and the initial layer compatibility conditions

−divT(u0) +∇P 0 =√ρ0g1,

− 1

cv

(ν(b+ 1)−14θb+1

0 +Q(u0))

=√ρ0g2

(1.14)

for some (g1, g2) ∈ L2. Then there exists a sufficiently small constant ε0 > 0 only depend-ing on (ρ0, u0, θ0), R, cv, γ, b, ν and q such that if

|V | ≤ ε0, (1.15)

then there exists a positive time T∗ and a unique strong solution (ρ, u, θ) in [0, T∗]× V tothe IBVP (1.1)-(1.7).

Remark 1.1. In principle, we will first apply the basic argument laid out in [3] for constantheat conductivity (b=0) to deal with the degeneracy of the time evolution appearing in themomentum and energy equations. Secondly, we need to pay extra attention via introducingsome new variable to solve another two new significant difficulties from the porous mediumtype diffusion in the energy equation (1.1)3 : the possible degeneracy and strong nonlinearityin heat diffusion. Taking the higher order terms of θ as an example, from the classicalregularity theory for elliptic equation and (1.1)3, we only have

|θ|D2 ≤ C∣∣∣ 1

θb|∇θ|2

∣∣∣2

+ C||∇u|2|2 + ....

Then it is obvious that we need to study the positivity of θ and deal with the quadraticterms |∇θ|2 and |∇u|2. This is one of essential reasons that why we need to assume thesize of the initial vacuum domain is sufficiently small.

The explanation for the compatibility between (1.13) and (1.14) can be shown as follows:

Theorem 1.2. Let conditions supposed in Theorem 1.1 hold. We assume that one of thefollowing two conditions hold:

(1). |V | = 0, i.e., the initial vacuum set V has zero 3-D Lebesgue measure; or


(2). |V | > 0 and the elliptic system−divT(u) = 0 in V,

− 1

cv

(ν(b+ 1)−14ψ +Q(u)

)= 0 in V

(1.16)

only has a zero solution (u, ψ) in D10(V ) ∩D2(V ).

Then there exists a unique (local) strong solution (ρ, u, θ) such that

‖ρ(t)− ρ0‖H1∩W 1,q(V) + ‖(u(t)− u0, θ(t)− θ0)‖D1∩D2(V) → 0, as t→ 0, (1.17)

if and only if the initial data (ρ0, u0, θ0) satisfy the compatibility conditions (1.14).

Remark 1.2. (ρ, u, θ) satisfies IBVP (1.1)-(1.7) in the sense of distribution, so we onlyhave

ρu(t = 0, x) = ρ0u0, ρθ(t = 0, x) = ρ0θ0.

In the vacuum domain V , the relations

u(t = 0, x) = u0, and θ(t = 0, x) = θ0

maybe not hold. Theorem 1.2 tells us that if the initial vacuum domain V has a sufficientlysimple geometry, for instance, the Lipschitz continuous domain, we have

u(t = 0, x) = u0, and θ(t = 0, x) = θ0 a.e. in V.

1.2. Minimum principle on the absolute temperature . It is well-known that thelinear heat equation with thermo-insulated boundary condition

θt = ∆θ in V,

θ(t = 0, x) = θ0(x) in V,

∇θ · n = 0 in ∂V

(1.18)

satisfies the minimum principle. Particularly if the initial data

θ0(x) ≥ θ > 0, for any x ∈ V, (1.19)

then for the solution θ we have,

θ(t, x) ≥ θ > 0 for any (t, x) ∈ (0,+∞)× V. (1.20)

This fundamental property could be generalized to some more complicated models. Inthis paper, we will show a minimum principle on temperature θ for the system (1.1).

According to system (1.1), the relations E = 12 |u|

2 + e and (1.4), the time evolution ofthe temperature can be governed by the following equation:

ρθt + ρu · ∇θ +1

cv

(Pdivu− νdiv(θb∇θ)

)=

1

cvQ(u). (1.21)

It is obvious that there are several obstacles in obtaining a minimum principle on θ.

(1) The temperature is in a multi-dimensional non-linear hyperbolic-parabolic coupledsystem (1.1) instead of a scalar linear parabolic equation as shown in (1.18);


(2) The pressure P may do work so that the specific internal energy of the fluid thatwe consider may decrease. We need to pay extra attention to deal with the termPdivu in our mathematical analysis;

(3) The leading coefficient of θt in (1.1)3 vanishes when ρ = 0, and as a consequence,the temperature equation loses its classical parabolic structure as in (1.18). Moreprecisely, the equation is degenerated into a non-linear elliptic equations with somequadratic terms in the form of

div(uT)− u · ∇T =µ

2|∇u+ (∇u)>|2 + λ|divu|2

in vacuum domain.

The main ideas used in this paper to conquer these difficulties include: i, study thesolutions by the De Giorgi’s proof of Holder regularity for the solutions of elliptic equationwith discontinuous coefficients; ii, use some interpolation estimates; iii. introduce a newcontinuity argument.

Our conclusion is that the minimum of the temperature will not increase as t increases,under the thermo-insulated boundary conditions in any bounded smooth domain, as statedin the following theorem.

Theorem 1.3. Assume that γ > 1, θ > 0 is a real constant and

2α+ 3β > 0.

Let (ρ0, u0, θ0) satisfy (1.13)-(1.14) and in particularly, θ0 ≥ θ. If (ρ, u, θ) on [0, T ] × Vis the strong solution to IBVP (1.1)-(1.7), then

θ(t, x) ≥ θ, for all (t, x) ∈ [0, T ]× V. (1.22)

Remark 1.3. This minimum principle verifies the necessity of the third law of thermo-dynamics in our nonlinear system, which is an important open problem no matter for themathematical analysis or the physical understanding of the compressible Navier-Stokes-Fourier system in the muti-dimensional space.

Moreover, we need to point out that this minimum principle in the temperature holds inthe whole life span of the corresponding solution.

We now outline the organization of the rest of this paper. In Section 2, we list someimportant lemmas that will be used frequently in our proof. In Section 3, the existenceof the unique local strong solution with vacuum will be proved. Section 4 is devoted toproving the necessity and sufficiency of the compatibility conditions shown in Theorem1.2. The proof of the minimum principle in Theorem 1.3 is given in Section 5, which isachieved in four steps:

(1) Constructing proper functionals that will be used in De Giorgi iteration based onthe mathematical structure of the CNSF (§5.1);

(2) Establishing the desired interpolation estimates for the cases 0 ≤ b < 1, especiallyfor the term Rρθdivu (§5.2);


(3) Establishing the desired interpolation estimates for the cases 1 ≤ b <∞, especiallyfor the term Rρθdivu (§5.3);

(4) Proving the minimum principle of the absolute temperature via introducing aspecial and new continuity argument (§5.4).

Finally, we will also give an appendix in order to prove some important lemmas that willbe shown or used in Sections 2-3.

2. Preliminary

In this section, we give some important lemmas which will be used frequently in ourproof. The first one comes from the well-known Gagliardo-Nirenberg inequality:

Lemma 2.1. [8] For q ∈ (3, 6], there exists a constant C > 0 depending on q such thatfor

f ∈ D10(V), g ∈ D1

0 ∩D2(V), h ∈W 1,q(V),

the following inequalities hold:

|f |6 ≤ C|f |D10, |g|∞ ≤ C|g|

126 |∇g|

126 ≤ |g|D1

0∩D2 , |h|∞ ≤ C‖h‖W 1,q . (2.1)

Secondly, we introduce some Poincare type inequality (see Chapter 8 in [14]):

Lemma 2.2. [14] There exists a constant C depending only on V, |ρ|r with r ≥ 1, whereρ ≥ 0 is a real function satisfying |ρ|1 > 0, such that for every F ≥ 0 satisfying

ρF ∈ L1(V),√ρF ∈ L2, ∇F ∈ L2(V),

the following inequality holds:

|F |6 ≤ C(|ρF |1 + (1 + |ρ|2)|∇F |2) ≤ C(|√ρF |2 + (1 + |ρ|2)|∇F |2).

Next we consider a homogeneous Dirichlet boundary value problem: let U = (U1, U2, U3),F = (F 1, F 2, F 3) and −α4U − (α+ β)∇divU = F in V;

U = 0 for x ∈ ∂V.(2.2)

If F ∈ W−1,2(V), then there exists a unique weak solution U ∈ H10 (V). We begin from

recalling various estimates for this system in the Ll(V) space with l ∈ (1,+∞). Theseresults can be found in [1].

Lemma 2.3. [1] Let (1.7) hold, and let V be a bounded, smooth domain and set l ∈(1,+∞). There exists a constant C depending only on α, β, l and V such that: if F ∈Ll(V), then we have

‖U‖W 2,l ≤ C|F |l. (2.3)

Moreover, if4U = F, ∇U(x) · n|∂V = 0,

then for any weak solution U ∈ H1, (2.3) also holds.

Finally, using Aubin-Lions Lemma, one has (c.f. [17]),


Lemma 2.4. [17] Let X0, X and X1 be three Banach spaces satisfying X0 ⊂ X ⊂ X1.Suppose that X0 is compactly embedded in X and that X is continuously embedded in X1.

I) Let G be bounded in Lp(0, T ;X0) with 1 ≤ p < +∞, and ∂G∂t be bounded in L1(0, T ;X1).

Then G is relatively compact in Lp(0, T ;X).

II) Let F be bounded in L∞(0, T ;X0) and ∂F∂t be bounded in Lp(0, T ;X1) with p > 1.

Then F is relatively compact in C(0, T ;X).

3. Well-posedness of strong solutions with vacuum

In this section, we always assume that γ > 1, θ > 0 is a real constant and (1.7) holds.The goal of this section is to prove Theorem 1.1. To this end, we first reformulate oursystem (1.1) into a new form via introducing a new variable ψ.

3.1. Reformulation. According to the equation (1.21) and (1.4), one quickly has

ρθt + ρu · ∇θ +1

cv

(Rρθdivu− νdiv(θb∇θ)

)=

1

cvQ(u). (3.1)

Then via introducing the new variable ψ = θb+1, system (1.1) can be rewritten as

ρt + div(ρu) = 0,

ρut + ρu · ∇u+∇(Rρψ

1b+1

)= divT,

ρψt + ρu · ∇ψ +1

cvR(b+ 1)ρψdivu− 1

cvνψ

bb+14ψ =

1

cv(b+ 1)ψ

bb+1Q(u).

(3.2)

The initial data are given by

(ρ, u, ψ)|t=0 = (ρ0(x), u0(x), ψ0(x)) = (ρ0(x), u0(x), θb+10 (x)), x ∈ V, (3.3)

and the boundary conditions can be shown as: Dirichlet and thermo-insulated boundaryconditions for (u, ψ): V ⊂ R3 is a bounded smooth domain and

u|∂V = 0, ∇ψ · n|∂V = 0. (3.4)

To prove Theorem 1.1, our first step is to establish the following existence result for thereformulated problem (3.2)-(3.4).

Theorem 3.1. Let the initial data (ρ0, u0, ψ0) satisfy the following regularities:

ρ0 ≥ 0, ρ0 ∈ H1 ∩W 1,q, q ∈ (3, 6],

u0 ∈ D1 ∩D2, ψ0 ∈ D1 ∩D2, ψ0 ≥ ψ = θb+1,(3.5)

and the initial layer compatibility conditions

−divT(u0) +∇(Rρ0(ψ0)

1b+1

)=√ρ0g1,

− 1

cv(b+ 1)ψ

bb+1

0

(ν(b+ 1)−14ψ0 +Q(u0)

)=√ρ0g2

(3.6)


for some (g1, g2) ∈ L2. Then there exists a sufficiently small constant ε0 > 0 only depend-ing on (ρ0, u0, ψ0), R, cv, γ, b, ν and q such that if

|V | ≤ ε0, (3.7)

then there exists a positive time T∗ and a unique strong solution (ρ, u, ψ) on [0, T∗]×V toIBVP (3.2)-(3.4) satisfying

0 ≤ ρ ∈ C([0, T∗];H1 ∩W 1,q), (ψ, u) ∈ C([0, T∗];D

1 ∩D2) ∩ L2([0, T∗];D2,q),

(ψt, ut) ∈ L2([0, T∗];D1), (

√ρψt,√ρut) ∈ L∞([0, T∗];L

2).(3.8)

We will prove this existence theorem in the Subsections 3.2-3.5, and at the end of thissection, we will show that this theorem indeed implies the Theorem 1.1. For simplicity, inthe following subsections, we denote

a1 =1

cvR(b+ 1), a2 =

1

cvν, a3 =

1

cv(b+ 1), P ∗ = Rρψ

1b+1 .

3.2. Linearization. Let V be a bounded smooth domain. Now we consider the followinglinearized problem:

ρt + div(ρw) = 0,

ρψt + ρw · ∇ψ + a1ρψdivw − a2φbb+14ψ = a3φ

bb+1Q(w),

ρut + ρw · ∇u+∇P ∗ − divT = 0,

(ρ, u, ψ)|t=0 = (ρδ0(x), u0(x), ψ0(x)),

u(t, x)|∂V = 0, ∇ψ(t, x) · n|∂V = 0, for t ≥ 0,

(3.9)

where ρδ0 = ρ0 + δ for some constant δ > 0, w(t, x) ∈ R3 is a known vector, and φ(t, x) isa known function. Assume that w|∂V = 0, ∇φ · n|∂V = 0, and

(w, φ) ∈ C([0, T ];H2) ∩ L2([0, T ];W 2,q), (wt, φt) ∈ L2([0, T ];H1),

φ ≥ 1

2ψ > 0, (w(t = 0, x), φ(t = 0, x)) = (u0(x), ψ0(x)), for x ∈ R3.

(3.10)

We easily have the global existence of the unique strong solution (ρδ, uδ, ψδ) to (3.9)-(3.10) by the standard methods for every δ > 0.

Lemma 3.1. Assume that (ρ0, u0, ψ0) satisfies (3.5)-(3.6). Then there exists a uniquestrong solution (ρδ, uδ, ψδ) to IBVP (3.9)-(3.10) for every δ > 0 satisfying

ρδ ∈ C([0, T ];W 1,q), (ψδ, uδ) ∈ C([0, T ];H2) ∩ L2([0, T ];W 2,q), (3.11)

and ρδ ≥ δ for some positive constant δ > 0.

Proof. First, the existence of a unique solution ρδ to (3.9)1 can be obtained by the standardtheory of transport equation, and ρδ can be written as

ρδ(t, x) = ρδ0(U(0; t, x)) exp(−∫ t

0divw(s, U(s; t, x))ds

), (3.12)


where U ∈ C([0, T ]× [0, T ]× V) is the solution to the initial value problemd

dsU(s; t, x) = w(s, U(s; t, x)), 0 ≤ s ≤ T,

U(t; t, x) = x, x ∈ V, 0 ≤ t ≤ T.(3.13)

So we can get the lower bound of ρδ easily.Secondly, from the regularity properties of ρδ, w and φ, it is not difficult to solve (ψδ, uδ)

from the linear parabolic equationsψδt + w · ∇ψδ + a1ψ

δdivw − a2

ρδφ

bb+14ψδ =

a3

ρδφ

bb+1Q(w),

uδt + w · ∇uδ + (ρδ)−1∇(Rρδ(ψδ)

1b+1

)− (ρδ)−1divT(uδ) = 0,

(3.14)

to complete the proof of this lemma. Here we omit the details.

3.3. A priori estimates independent of δ. In this subsection, we will get some a prioriestimates for the solution (ρδ, uδ, ψδ) obtained in Lemma 3.1, independent of δ.

We first fix a positive constant c0 that

2 + ‖ρ0‖W 1,q + ‖(ψ0, u0)‖2 + |(g1, g2)|2 ≤ c0. (3.15)

Then we rewrite (3.6) into

−divT(u0) +∇(Rρδ0ψ

1b+1

0

)=√ρδ0g

δ1,

− 1

cv(b+ 1)ψ

bb+1

0

(ν(b+ 1)−14ψ0 +Q(u0)

)=√ρδ0g

δ2,

(3.16)

where

gδ1 =(ρ0

ρδ0

) 12g1 +Rδ

∇ψ1b+1

0

(ρδ0)12

, gδ2 =(ρ0

ρδ0

) 12g2.

Then according to (3.15), for any δ > 0 small enough, we have

1 + ‖ρδ0‖W 1,q + ‖(ψ0, u0)‖2 + |(gδ1, gδ2)|2 ≤ c0. (3.17)

Secondly, for w and ψ, let positive constants ci (i = 1, 2, 3, 4, 5) satisfy

sup0≤t≤T ∗

‖w(t)‖21 +

∫ T ∗

0

(|w|2D2,q + ‖wt‖21

)dt ≤ c2

1, sup0≤t≤T ∗

|w(t)|2D2 ≤c22,

sup0≤t≤T ∗

‖φ(t)‖21 ≤ c23,

∫ T ∗

0

(|φ|2D2,q + ‖φt‖21

)dt ≤ c2

4, sup0≤t≤T ∗

|φ(t)|2D2 ≤c25,

(3.18)

for some time T ∗ ∈ (0, T ), where constants ci (i = 1, 2, 3, 4, 5) satisfy

2 < c0 < c1 < c2 < c3 < c4 < c5.

Constants ci (i = 1, 2, 3, 4, 5) and T ∗ will be determined later and they depend only on c0

and the fixed constants µ, λ, ν, q, R, cv, |V| and T (see (3.51)).In this paper, we denote C ≥ 1 a generic constant depending only on fixed constants µ,

λ, ν, q, R, cv, |V| and T . Is this definition for the whole paper?

For simplicity, we denote a4 = a2

(12ψ) bb+1

.


For simplicity, in the rest of Subsection 3.3, we will denote (ρδ, uδ, ψδ) by (ρ, u, ψ) ifthere is no confusion.

Now we start with the a priori estimates for ρ.

Lemma 3.2. For 0 ≤ t ≤ T1 = min(T ∗, (1 + c22)−1), we have

1 + |ρ(t)|2∞ + ‖ρ(t)‖2W 1,q ≤ Cc20, |ρt(t)|q ≤ Cc0c2.

Proof. From standard energy estimate for the transport equation, we have

‖ρ(t)‖W 1,q ≤ ‖ρδ0‖W 1,q exp(C

∫ t

0‖∇w(s)‖W 1,qds

)≤ C(c2t+ c2t

12 ) ≤ C, (3.19)

for 0 ≤ t ≤ T1 = min(T ∗, (1 + c22)−1), where we have used the fact w · n|∂V = 0.

For the term ρt, from the continuity equation (3.9)1, we get

|ρt|q ≤ C(|ρ|∞|∇w|q + |w|∞|∇ρ|q) ≤ Cc0||w||2 ≤ Cc0c2.

Next we will study the evolution of the initial vacuum domain for ρ0. We denoteVR0 ⊂ V a neighborhood containing the initial vacuum region V :

V ⊂ VR0 = x ∈ V|dist(x, V ) ≤ R0,

where R0 > 0 is a sufficiently small constant. We first give two lemmas.

Lemma 3.3. For every sufficiently small R0 << 1, there exists a time TR0 ∈ (0, T ∗) smallenough and a constant aR0 > 0 such that

ρ(t, x) ≥ aR0 +1

2δ > 0, ∀ (t, x) ∈ [0, TR0 ]× (V/VR0),

where TR0 = min(T1, (ln 2)2(Cc2)−2, 2R0(6c2)−1) and aR0 are both independent of δ.

Lemma 3.4.

ψ(t, x) ≥ 1

2ψ, for (t, x) ∈ [0, T2 = min(T1, (ln 2)2(Cc2)−2)]× V.

The proof of these two lemmas can be found in the appendix.

Next we give the a priori estimates for ψ.

Lemma 3.5.

|√ρψ(t)|22 + ‖ψ(t)‖21 ≤ Cc30, |√ρψt(t)|22 +

∫ t

0|ψt(s)|2D1ds ≤M(c3),∫ t

0|ψ(s)|2D2,qds ≤M(c3), |ψ(t)|D2 ≤M(c3)c

bb+1

5 ,

for 0 ≤ t ≤ T3 = min(T2, TR0 ,M−1(c5)), where R0 satisfies |VR0 | ≤

(a4/(20CcK5

))3for

a constant K ≥ 18, and M = M(·) : [2,+∞) → [1,+∞) denotes a strictly increasingcontinuous function depending only on fixed constants µ, λ, ν, q, R, cv, |V| and T .


Proof. Step 1. Multiplying (3.9)2 by ψ and integrating over V, we have

1

2

d

dt

∫ρ|ψ|2 + a2

∫Vφ

bb+1 |∇ψ|2

≤C∫ (|∇φ

bb+1 · ∇ψψ|+ ρ|ψ|2|divw|+ φ

bb+1 |∇w|2|ψ|

)≤CΛ1 + C

(|ρ|

12∞|√ρψ|2|∇w|3 + |φ|

bb+1∞ |∇w|2|∇w|3

)|ψ|6

≤CΛ1 +a4

20c20

(|√ρψ|22 + c20|∇ψ|22) + Cc52|

√ρψ|22 + Cc85,

(3.20)

where we have used the Poincare type inequality for ψ in Lemma 2.2.For the term Λ1, from Lemmas 3.2-3.4, we have

Λ1 =

∫VR0

|∇φbb+1 · ∇ψψ|dx+

∫V/VR0

|∇φbb+1 · ∇ψψ|dx

≤C|∇ψ|2|ψ|6|∇φ|6|VR0 |16 + C|∇ψ|2|

√ρψ|

122 |√ρψ|

126 |∇φ|6

≤a4

10|∇ψ|22 + C(c2

5|VR0 |13 + c7

5)|√ρψ|22 + Cc45|VR0 |13 |∇ψ|22,

(3.21)

which, along with (3.20), from Gronwall’s inequality, implies

|√ρψ|22 +

∫ t

0|∇ψ|22ds ≤ C(c3

0 + c85t) exp(Cc7

5t) ≤ Cc30 (3.22)

for 0 ≤ t ≤ T ′′′ = min(T2, TR0 , (1 + c85)−1) and R0 satisfying |VR0 | ≤

(a4/(20Cc45)

)3.

Step:2. Differentiating (3.9)2 with respect to t, we have

ρψtt − a2φbb+14ψt − a2φ

bb+1

t 4ψ

=− ρtψt − (ρw · ∇ψ)t − a1(ρψdivw)t + a3

(φ

bb+1Q(w)

)t.

(3.23)

Multiplying (3.23) by ψt and integrating over V, we have

1

2

d

dt

∫ρ|ψt|2 + a2

∫φ

bb+1 |∇ψt|2

≤C∫ (|ρtw · ∇ψψt|+ |∇φ

bb+1 · ∇ψtψt|+ |φ

bb+1

t 4ψψt|+ |ρwt · ∇ψψt|)

+ C

∫ (|ρw · ∇ψtψt|+ |(ρψ)tdivwψt|+ |ρψdivwtψt|

)+ C

∫ (∣∣φ bb+1Q(w)tψt

∣∣+∣∣φ b

b+1

t Q(w)ψt∣∣) ≡: C

9∑i=1

Ii.

(3.24)


For terms I1-I9, according to Lemmas 3.2-3.4, 2.1-2.2 and Young’s inequality, we have

I1 =

∫|ρtw · ∇ψψt|

≤C|ρt|3|w|∞|∇ψ|2|ψt|6 ≤ Cc82|∇ψ|22 +a4

20c20

(|√ρψt|22 + c20|∇ψt|22),

I2 =

∫VR0

|∇φbb+1 · ∇ψtψt|dx+

∫V/VR0

|∇φbb+1 · ∇ψtψt|dx

≤C|∇ψt|2|ψt|6|∇φ|6|VR0 |16 + C|∇ψt|2|

√ρψt|

122 |√ρψt|

126 |∇φ|6

≤a4

10|∇ψt|22 + C(c2

5|VR0 |13 + c7

5)|√ρψt|22 + Cc45|VR0 |13 |∇ψt|22,

I3 =

∫VR0

|φbb+1

t 4ψψt|dx+

∫V/VR0

|φbb+1

t 4ψψt|dx

≤C|4ψ|2|ψt|6|φt|6|VR0 |16 + C|4ψ|2|

√ρψt|

122 |√ρψt|

126 |φt|6

≤ a4

10c20

(|√ρψt|22 + c20|∇ψt|22) + C|VR0 |

13 |φt|26|4ψ|22

+ η|φt|26 + η−2c30|√ρψt|22|4ψ|42,

I4 + I7 =

∫|ρwt · ∇ψψt|+

∫|ρψdivwtψt|

≤C|ρ|12∞(|wt|6|∇ψ|2 + |∇wt|2|ψ|6)|√ρψt|

122 |√ρψt|

126

≤C|∇wt|22 + C(c110 + c9

0|∇ψ|42)|√ρψt|22 +a4

20c20

(|√ρψt|22 + c20|∇ψt|22),

I5 =

∫|ρw · ∇ψtψt| ≤ C|ρ|

12∞|w|∞|∇ψt|2|

√ρψt|2 ≤ Cc32|

√ρψt|22 +

a4

20|∇ψt|22,

I6 =

∫|P ∗t divwψt|

≤C(|ρt|2|ψ|6|∇w|6|ψt|6 + |ρ|12∞|√ρψt|2|∇w|3|ψt|6)

≤Cc52|√ρψt|22 +

a4

20c20

(|√ρψt|22 + c20|∇ψt|22) + Cc10

2 |∇ψ|22 + Cc112 ,

I8 =

∫ ∣∣φ bb+1Q(w)tψt

∣∣≤C|φ|6|ψt|6|∇wt|2|∇w|6 ≤

a4

20c20

(|√ρψt|22 + c20|∇ψt|22) + Cc6

3|∇wt|22,

I9 =

∫VR0

∣∣φ bb+1

t Q(w)ψt∣∣dx+

∫V/VR0

∣∣φ bb+1

t Q(w)ψt∣∣dx

≤C|∇w|3|∇w|6| ψt|6|φt|6|VR0 |16 + C|∇w|3|∇w|6|φt|6|

√ρψt|

122 |√ρψt|

126

≤ a4

10c20

(|√ρψt|22 + c20|∇ψt|22) + Cc62|VR0 |

13 |φt|26 + η|φt|26 + Cc11

2 η−2|√ρψt|22,

(3.25)

Then combining (3.24)-(3.25) and denoting L(t) = 1+ |∇ψ|22 + |√ρψt|22, we quickly have


1

2

d

dt

∫ρ|ψt|2 + a2

∫φ

bb+1 |∇ψt|2

≤(a4

2+ Cc45|VR0 |

13

)|∇ψt|22 + Cc11

2 + Cc63|∇wt|22 + C(|VR0 |

13 (|4ψ|22 + c6

2) + η)|φt|26

+ C(c11

5 + c25|VR0 |

13 + η−2c11

2 + c90|∇ψ|42 + η−2c3

0|4ψ|42)L(t),

(3.26)

where η > 0 is a constant, and we have used Lemma 2.2.Now we need to consider the term |ψ|D2 . From equation (3.9)2, we know:

−a2φbb+14ψ = −(ρψt + ρw · ∇ψ + a1ρψdivw) + a3φ

bb+1Q(w). (3.27)

Then from Lemmas 2.1-2.3, we have

|ψ|D2 ≤C(|ρ|

12∞|√ρψt|2 + |ρ|∞|w|∞|∇ψ|2 + |ρ|6|ψ|6|divw|6 + c

bb+1

5 |∇w|6|∇w|3)

≤Cc32(|∇ψ|2 + |√ρψt|2) + Cc

723 c

bb+1

5 .

(3.28)

On the other hand,

d

dt|∇ψ(t)|22 ≤ C|∇ψ|2|∇ψt|2 ≤ C|∇ψ|22 +

a4

20|∇ψt|22. (3.29)

Then letting R0 sufficiently small such that |VR0 | ≤ min(

a420CcK5

)3for a sufficiently large

constant K ≥ 18, and letting η = c−K5 from (3.26) and (3.28)-(3.29) we have

d

dtL(t) + a4

∫|∇ψt|2 ≤ Cc63|∇wt|22L(t) + C(c2K+12

5 + c9−K5 |φt|26)L3(t). (3.30)

Denote

H(t) = L(t) exp(−∫ t

0Cc63|∇wt|22ds

),

then from (3.30), we have

d

dtH(t) ≤ C(c2K+12

5 + c9−K5 |φt|26)H3(t). (3.31)

Next we need to solve this inequality. From (3.9)2, we have

|√ρψt|22 ≤ |ρ|∞‖∇w‖21|∇ψ|22 +

∫|Φ|2/ρ, (3.32)

where

Φ = − 1

cv(b+ 1)ψ

bb+1

0

(ν(b+ 1)−14ψ0 +Q(u0)

).

Via Lemma 3.1, we easily have

limt7→0

∫ ( |Φ(t)|2

ρ− |Φ(0)|2

ρ0

)≤ lim

t7→0

(1

δ

∫|Φ(t)− Φ(0)|2 +

1

δδ|ρ(t)− ρ0|∞

∫|Φ(0)|2

)= 0.

According to the compatibility conditions (3.6) and equation (3.9)2, we have

lim supτ→0

|√ρψt(τ)|22 ≤ |ρ0|∞‖∇w0‖21|∇ψ0|22 + |g2|22 ≤ Cc50, (3.33)


which implies that

lim supτ→0

H(τ) ≤ Cc50.

Now integrating (3.31) over [τ, t] for any τ ∈ (0, t), letting τ → 0, and solving theresulting inequality, we have

H(t) ≤Cc50 +

Cc50(1− Cc50(c2K+12

5 t+ c11−K5 )

) 12

≤ Cc50,

when 0 < t ≤ T′′′′

= min(T2, TR0 , (1 + Cc5)−4K

)for constant K sufficiently large. Then

we immediately have

L(t) ≤M(c3), for 0 ≤ t ≤ T ′′′′ , (3.34)

where M = M(·) : [2,+∞) → [1,+∞) denotes a strictly increasing continuous function,and depends only on fixed constants µ, λ, ν, q, R, cv, |V| and T .

Therefore, from (3.30) and (3.34), we quickly have

L(t) + a4

∫ t

0|ψt|2D1ds ≤M(c3), for 0 ≤ t ≤ T ′′′′ . (3.35)

From (3.28), we quickly have

|ψ|D2 ≤Cc22(|∇ψ|2 + |√ρψt|2) + Cc33c

bb+1

5 ≤M(c3)cbb+1

5 . (3.36)

For the term |ψ|D2,q , similarly, via Lemma 2.3 and (3.27), we also have∫ t

0|ψ|2D2,qds ≤C

∫ t

0

∣∣ρψt + ρw · ∇ψ + a1ρψdivw − a3φbb+1Q(w)

∣∣2qds ≤M(c3) (3.37)

for 0 ≤ t ≤ T3 = min(T′′′′,M−1(c5)

).

According to P ∗ = Rρψ1b+1 and Lemma 3.4, for 0 ≤ t ≤ T3, we easily obtain that

|∇P ∗|2 ≤M(c3), |∇P ∗|q ≤M(c3)cbb+1

5 , |P ∗t |2 ≤M(c3). (3.38)

Step 3. Multiplying (3.9)2 by ψt and integrating over V, we have

a2

2

d

dt

∫φ

bb+1 |∇ψ|2 +

∫ρ|ψt|2

≤C∫ (|φ

bb+1

t ||∇ψ|2 +(|∇φ

bb+1 ||∇ψ|+ |ρ|ψ||divw|+ ρ|w||∇ψ|+ |φ

bb+1Q(w)|

)ψt|)

≤C|φt|6|∇ψ|2|∇ψ|3 + C|∇φ|3|∇ψ|2|ψt|6 + C|√ρψt|2|ρ|12∞|ψ|6|∇w|3

+ C|√ρψt|2|ρ|12∞|∇ψ|2|w|∞ + C|φ|

bb+1∞ |∇w|2|∇w|3|ψt|6

≤η(|φt|26 + |ψt|26) +1

2|√ρψt|22 + η−1M(c5),

(3.39)

where η = M−1(c5). Then integrating (3.39) over (0, t), we have

|∇ψ|22 +

∫ t

0|√ρψt|22ds ≤ Cc30, for 0 ≤ t ≤ T3 = min

(T′′′′,M−1(c5)

). (3.40)


Next we give the a priori estimates for the velocity u.

Lemma 3.6. For 0 ≤ t ≤ T3, we have

‖u(t)‖21 + |√ρut(t)|22 +

∫ t

0|ut|2D1ds ≤ Cc50,

∫ t

0|u|2D2,qds ≤ Cc70, |u(t)|2D2 ≤Cc13

1 ,

Proof. Differentiating (3.9)3 with respect to t, we have

ρutt − divTt = −ρtut − (ρw · ∇u)t −∇P ∗t . (3.41)

Then multiplying (3.41) by ut and integrating over V, via (3.4) we have

1

2

d

dt

∫ρ|ut|2 +

∫ ((µ+ λ)|divut|2 + µ(∇ut)2

)≤C

∫ (|ρtw · ∇u · ut|+ |ρw · ∇ut · ut|+ |ρwt · ∇u · ut|+ | P ∗t divut|

)≤C(|w|∞|∇ut|2 + |wt|6|∇u|3

)|ρ|

12∞|√ρut|2 + C

(|ρt|3|w|∞|∇u|2 + |P ∗t |2

)|∇ut|2

≤M(c3) + C(c62 + η|∇wt|22)|√ρut|22 + Cη−1c0‖∇u‖21.

(3.42)

Now we have to estimate |u|D2 , due to

divT = ρut + ρw · ∇u+∇P ∗, (3.43)

and Lemma 2.3, we have

|u|D2 ≤C(|ρ|12∞|√ρut|2 + |ρ|2|w|6|∇u|

122 |∇u|

123 + |ρ|∞|∇ψ|2 + |ψ|6|∇ρ|3)

≤Cc41(|√ρut|2 + |∇u|2) +

1

2|u|D2 + Cc

520 .

(3.44)

On the other hand, we also have

d

dt|∇u(t)|22 ≤ 2|∇u|2|∇ut|2 ≤ C|∇u|22 +

µ

20|∇ut|22, (3.45)

which, along with (3.41)-(3.44), implies that

1

2

d

dt(|√ρut|22 + |∇u|22) +

µ

2

∫|∇ut|2

≤M(c3) + η−1c60 + C(c6

2 + η|∇wt|22 + η−1c91)(|√ρut|22 + |∇u|22

).

(3.46)

Similarly to the proof of (3.33), via (3.6) and equations (3.9)2, we have

lim supτ→0

|√ρut(τ)|22 ≤ |ρ0|∞‖∇w0‖21|∇u0|22 + |g1|22 ≤ Cc50. (3.47)

Then from Gronwall’s inequality and (3.47), via letting η = (c22)−1, we have

|∇u(t)|22 + |√ρut(t)|22 +

∫ t

0|ut|2D1ds

≤(Cc50 +M(c3)t) exp(∫ t

0(c11

2 + c−22 |∇wt|

22)ds

)≤ Cc50

(3.48)


for 0 ≤ t ≤ min(T3,M

−1(c5)), which together with (3.44), implies that

|u|D2 ≤ Cc612

1 ,

∫ t

0|u|2D2,qds ≤

∫ t

0

(|ρut + ρw · ∇u+∇P |2q

)ds ≤ Cc7

0,

for 0 ≤ t ≤ T4 = min(T3,M

−1(c5)).

Then based on Lemmas 3.2-3.6, we have chosen M = M(·) : [2,+∞) → [1,+∞) as astrictly increasing continuous function depending only on fixed constants µ, λ, ν, q, R, cv,|V| and T . When 0 ≤ t ≤ T3 = min(T2, TR0 ,M

−1(c5)), with R0 satisfying

|VR0 | ≤(a4/(20CcK5 )

)3, (3.49)

for a sufficiently large constant K ≥ 18, the following a priori estimates hold

|ρ(t)|2∞ + ‖ρ(t)‖2W 1,q ≤ Cc20, |ρt(t)|q ≤ Cc0c2, ψ(t, x) ≥ 1

2ψ, ‖ψ(t)‖21 ≤Cc30,

|√ρψt(t)|22 +

∫ t

0(|ψt(s)|2D1 + |ψ(s)|2D2,q)ds ≤M(c3), |ψ(t)|2D2 ≤M(c3)c

2bb+1

5 ,

‖u(t)‖21 + |√ρut(t)|22 +

∫ t

0

(|u(s)|2D2,q + |ut(s)|2D1

)ds ≤ Cc70, |u(t)|2D2 ≤Cc13

1 .

(3.50)

Therefore, if we define the constants ci (i = 1, 2, 3, 4, 5) and T ∗ by

c1 = C12 c

720 , c2 = C

12 c

132

1 = C154 c

914

0 , c3 = c2 = C154 c

914

0 ,

c4 =√M(c3) =

√M(C

154 c

914

0 ), c5 = Mb+12 (c3) = M

b+12 (C

154 c

914

0 ),

T ∗ = min(T, (12R0c2)−1,M−1(c5)),

(3.51)

then we deduce that

sup0≤t≤T ∗

‖u(t)‖21 + ess sup0≤t≤T ∗

|√ρut(t)|22 +

∫ T ∗

0

(|u(s)|2D2,q + |ut(s)|2D1

)ds ≤c2

1,

sup0≤t≤T ∗

|u(t)|2D2 ≤ c22, sup

0≤t≤T ∗|√ρψ(t)|22 + sup

0≤t≤T ∗‖ψ(t)‖21 ≤c2

3,

ψ(t, x) ≥ 1

2ψ, ess sup

0≤t≤T ∗|√ρψt(t)|22 +

∫ T ∗

0(|ψt(s)|2D1 + |ψ(s)|2D2,q)ds ≤c2

4,

sup0≤t≤T ∗

|ψ(t)|2D2 ≤ c25, sup

0≤t≤T ∗

(|ρ(t)|2∞ + ‖ρ(t)‖2W 1,q + |ρt(t)|q) ≤c2

2.

(3.52)

Moreover, for sufficiently small R0 > 0 satisfying (3.49), we also have

ρ(t, x) ≥ aR0 +1

2δ > 0, ∀ (t, x) ∈ [0, T ∗]× (V/VR0), (3.53)

where aR0 is a positive constant independent of δ.


3.4. Passing limit from nonvacuum to vacuum. In this subsection, we will give theexistence of the strong solution with vacuum to our linear problem:

ρt + div(ρw) = 0,

ρψt + ρw · ∇ψ + a1ρψdivw − a2φbb+14ψ = a3φ

bb+1Q(w),

ρut + ρw · ∇u+∇P ∗ − divT = 0,

(ρ, u, ψ)|t=0 = (ρ0(x), u0(x), ψ0(x)),

u|∂V = 0, ∇ψ · n|∂V = 0.

(3.54)

Lemma 3.7. Assume (3.5)-(3.6) hold. Then there exists a unique strong solution (ρ, u, ψ)on [0, T ∗]× V to IBVP (3.54) satisfying

ρ ∈ C([0, T ∗];W 1,q), (ψ, u) ∈ C([0, T ∗];H2) ∩ L2([0, T ∗];W 2,q),

ψ ≥ 1

2ψ, (

√ρψt,√ρut) ∈ L∞([0, T ∗];L2), (ψt, ut) ∈ L2([0, T ∗];H1).

(3.55)

Moreover, the a priori estimates (3.52) also hold for our solution (ρ, u, ψ), and for suffi-ciently small R0 > 0 there exists a constant aR0 independent of δ such that

ρ(t, x) ≥ aR0 > 0, ∀ (t, x) ∈ [0, T ∗]× (V/VR0). (3.56)

Proof. We divide the proof into three steps.Step 1. Existence. For ρδ0 = ρ0 + δ with δ ∈ (0, 1), from Lemma 3.1, there exists a unique

strong solution (ρδ, uδ, ψδ) on [0, T ∗]× V satisfying (3.52)-(3.53), where the constants c1-c5, C, R0, T ∗ and aR0 are independent of δ. Then there exists a subsequence of solutions(ρδ, uδ, ψδ) converging to a limit (ρ, u, ψ) in weak or weak* sense:

ρδ ρ weakly* in L∞([0, T ∗];W 1,q(V)),

(ψδ, uδ) (ψ, u) weakly* in L∞([0, T ∗];H2(V)),

(ψδt , uδt ) (ψt, ut) weakly in L2([0, T ∗];H1(V)).

(3.57)

Moreover, due to the compactness property in [17], there exists a subsequence of solutions(ρδ, uδ, ψδ) satisfying:

(ρδ, uδ, ψδ)→ (ρ, u, ψ) in C([0, T ∗];H1(K)), (3.58)

where K is any compact subset of V.From the lower semi-continuity of norms, we know that (ρ, u, ψ) also satisfies the es-

timates (3.52)-(3.53). Then it is easy to show (ρ, u, ψ) is a weak solution in the sense ofdistribution and satisfies:

ρ ∈ L∞([0, T ∗];W 1,q), ρt ∈ L∞([0, T ∗];Lq),

(ψ, u) ∈ L∞([0, T ∗];H2) ∩ L2([0, T ∗];W 2,q), ψ ≥ 1

2ψ,

(√ρψt,√ρut) ∈ L∞([0, T ∗];L2), (ψt, ut) ∈ L2([0, T ∗];H1).

(3.59)


Step 2. Uniqueness. Let (ρ1, u1, ψ1) and (ρ2, u2, ψ2) be two solutions obtained above. Then

ρ1 = ρ2 can be obtained by the same method used in Lemma 3.1. Let ψ = ψ1−ψ2. Fromequation (3.54)2, we have

ρψt + ρw · ∇ψ + a1ρψdivw − a2φbb+14ψ = 0.

Then multiplying the above equations by ψ and integrating over V, we have

1

2

d

dt|√ρψ|22 + a4

∫|∇ψ|2 ≤ C|ψ|22, (3.60)

which, along with ∇ψ · n|∂V = 0, immediately means that ψ1 = ψ2. Via the similarargument, we can show that u1 = u2.Step 3. The time-continuity. The continuity of ρ can be obtained via the same method asin Lemma 3.1. Similarly, from (3.59), we have

(ψ, u) ∈ C([0, T ∗];H1) ∩ C([0, T ∗];D2 − weak).

From equations (3.54) and (3.59), we know that

(ρψt, ρut) ∈ L2([0, T ∗];L2), and ((ρψt)t, (ρut)t) ∈ L2([0, T ∗];H−1).

Thus from the Aubin-Lions lemma, we have (ρψt, ρut) ∈ C([0, T ∗];L2). Due to equations(3.27) and (3.43), and Lemma 2.3, we have (u, ψ) ∈ C([0, T ∗];D2).

3.5. Strong convergence in L2 space. In this subsection, we will give the proof forTheorem 3.1 based on some classical iteration scheme. Let us assume as in Section 3.3:

2 + ‖ρ0‖W 1,q + ‖(u0, ψ0)‖2 + |(g1, g2)|2 ≤ c0.

Next, let (u0, ψ0) be the solutions to the following linear problemsu0t −4u0 = 0; u0(0) = u0 in V; u|∂V = 0,

ψ0t −4ψ0 = 0; ψ0(0) = ψ0 in V; ∇ψ · n|∂V = 0.

Then we can choose a time T ∗∗ ∈ (0, T ∗) such that (u0, ψ0) satisfies (3.18).

Proof. Step 1. Existence. Let (w, φ) = (u0, ψ0). We can get (ρ1, u1, ψ1) as a strong

solution to (3.54). Then we construct approximate solutions (ρk+1, uk+1, ψk+1) inductivelyas follows: assume (uk, ψk) is defined for k ≥ 1, and let (ρk+1, uk+1, ψk+1) be the solutionto (3.54) with (w, φ) replaced by (uk, ψk) as following:

ρk+1t + div(ρk+1uk) = 0,

ρk+1ψk+1t + ρk+1uk · ∇ψk+1 + a1ρ

k+1ψk+1divuk

= a2(ψk)bb+14ψk+1 + a3(ψk)

bb+1Q(uk),

ρk+1uk+1t + ρk+1uk · ∇uk+1 +∇(P ∗)k+1 − divT(uk+1) = 0,

(ρk+1, uk+1, ψk+1)|t=0 = (ρ0(x), u0(x), ψ0(x)), x ∈ R3,

uk+1|∂V = 0, ∇ψk+1 · n|∂V = 0,

(3.61)


where (P ∗)k+1 = Rρk+1(ψk+1)1b+1 . Then from Subsection 3.4, we know that the solution

sequences (ρk, uk, ψk) also satisfy the a priori estimates (3.52) and (3.56).Next, we show that (ρk, uk, ψk) converges to a limit in a strong sense. Denote

ρk+1 = ρk+1 − ρk, uk+1 = uk+1 − uk, ψk+1= ψk+1 − ψk,

then we have

ρk+1t + div(ρk+1uk) + div(ρkuk) = 0,

ρk+1ψk+1t + ρk+1uk · ∇ψk+1 − a2(ψk)

bb+14ψk+1

=a2

((ψk)

bb+1 − (ψk−1)

bb+1)4ψk + a3

((ψk)

bb+1 − (ψk−1)

bb+1)Q(uk)

+ a3(ψk−1)bb+1 (Q(uk)−Q(uk−1))− ρk+1(ψkt + uk−1 · ∇ψk)

− a1ρk+1ψkdivuk−1 − ρk+1

(uk · ∇ψk + a1ψ

k+1divuk + a1ψ

kdivuk),

ρk+1uk+1t + ρk+1uk · ∇uk+1 − divT(uk+1)

=ρk+1(−ukt − uk−1 · ∇uk)− ρk+1uk · ∇uk

−R∇(ρk+1

((ψk+1)

1b+1 − (ψk)

1b+1)

+ ρk+1(ψk)1b+1

).

(3.62)

First, multiplying (3.62)1 by ρk+1 and integrating over V, for 0 < η ≤ 110 , we have

d

dt|ρk+1|22 ≤ C

(|uk|2D2,q + η−1 + 1

)|ρk+1|22 + η|∇uk|22, (3.63)

Secondly, multiplying (3.62)2 by ψk+1

and integrating over V, we have

1

2

d

dt|√ρk+1ψ

k+1|22 + a2

∫(ψk)

bb+1 |∇ψk+1|2

=

∫ (− a2∇(ψk)

bb+1 · ∇ψk+1

+ a2

((ψk)

bb+1 − (ψk−1)

bb+1)4ψk

)ψk+1

+

∫a3

(((ψk)

bb+1 − (ψk−1)

bb+1)Q(uk) + (ψk−1)

bb+1(Q(uk)−Q(uk−1)

))ψk+1

−∫ρk+1

(ψkt + uk−1 · ∇ψk + a1ψ

kdivuk−1)ψk+1

−∫ρk+1

(uk · ∇ψk + a1ψ

k+1divuk + a1ψ

kdivuk)ψk+1 ≡:

23∑i=14

Ii.

According to Holder’s inequality, Lemma 2.1 and Young’s inequality, we have

I14 =− a2

∫∇(ψk)

bb+1 · ∇ψk+1

ψk+1

≤C∫VR0

|∇ψk||∇ψk+1||ψk+1|dx+ C

∫V/VR0

|∇ψk||∇ψk+1||ψk+1|dx

≤C|∇ψk|6|∇ψk+1|2

(|ψk+1|6|VR0 |

16 + |√ρk+1ψ

k+1|122 |√ρk+1ψ

k+1|126

),


and

I15 =

∫a2

((ψk)

bb+1 − (ψk−1)

bb+1)4ψkψk+1

≤C∫VR0

|4ψk||ψk||ψk+1|dx+ C

∫V/VR0

|4ψk||ψk||ψk+1|dx

≤C|4ψk|2|ψk|6(|ψk+1|6|VR0 |

16 + |√ρk+1ψ

k+1|122 |√ρk+1ψ

k+1|126

),

I16 =a3

∫ ((ψk)

bb+1 − (ψk−1)

bb+1 )

)Q(uk)ψ

k+1

≤C∫VR0

|ψk||ψk+1||∇uk|2dx+ C

∫V/VR0

|ψk||ψk+1||∇uk|2dx

≤C|∇uk|3|∇uk|6|ψk|6(|ψk+1|6|VR0 |

16 +√ρk+1ψ

k+1|122 |√ρk+1ψ

k+1|126

),

I17 =a3

∫(ψk−1)

bb+1

(Q(uk)−Q(uk−1)

)ψk+1

≤C|ψk−1|bb+1∞ |∇uk +∇uk−1|3|∇uk|2|ψ

k+1|6,

I18+I19 + I20 = −∫ρk+1

(ψkt + uk−1 · ∇ψk + a1ψ

kdivuk−1)ψk+1

≤C|ρk+1|2|ψkt |3|ψk+1|6 + C|ρk+1|2|ψ

k+1|6‖ψk‖2‖uk−1‖2,

I21 =−∫ρk+1uk · ∇ψkψk+1 ≤ C|√ρk+1ψ

k+1|2|ρk+1|12∞|uk|6‖∇ψk‖1,

I22+I23 = −a1

∫ρk+1

(ψk+1

divuk + ψkdivuk)ψk+1

≤C|divuk|W 1,q |√ρk+1ψk+1|22 + C|√ρk+1ψ

k+1|2|ρk+1|12∞|∇uk|2|ψk|∞.

Then combining the estimates for Ii (i = 14, ..., 23), for t ∈ [0, T ∗∗], we have

d

dt|√ρk+1ψ

k+1|22 + a2|(ψk)b

2(b+1)∇ψk+1|22≤ Ek1η(t)|

√ρk+1ψ

k+1|22 + Ek2 (t)|ρk+1|22 + η(|∇ψk|22 + |√ρkψk|22)

+(a4

2+ Cη−1c6

5|VR0 |13 )|∇ψk+1|22 + C|∇uk|22,

Ek1η(t) = C(

1 + η−2 + |uk|D2,q + η−1|VR0 |13

), Ek2 (t) = C(1 + |ψkt |23).

(3.64)

Finally, multiplying (3.62)3 by uk+1 and integrating over V, similarly we have

1

2

d

dt|√ρk+1uk+1|22 +

1

2

∫ (µ|∇uk+1|2 + (λ+ µ)|divuk+1|22

)≤C(1 + η−1

)(|√ρk+1uk+1|22 + |√ρk+1ψ

k+1|22) + F k(t)|ρk+1|22 + η|∇uk|22,(3.65)

where the term F k(t) = C(1 + |ukt |23).Now, let ε > 0 be a sufficiently small constant and denote

Λk+1(T ∗∗, η, ε) = sup0≤t≤T ∗∗

|ρk+1(t)|22 + ε sup0≤t≤T ∗∗

|√ρk+1ψk+1

(t)|22 + sup0≤t≤T ∗∗

|√ρk+1uk+1(t)|22,


then let |VR0 |13 ≤ a4η(Cc6

5)−1, from (3.63)-(3.65), so by the Gronwall’s inequality, we have

Λk+1(T ∗∗, η, ε) +

∫ T ∗∗

0

(a4

4ε|∇ψk+1|22 +

µ

2|∇uk+1|22

)dt

≤∫ T ∗∗

0Gkη,εΛ

k+1(s, η, ε)dt+

∫ T ∗∗

0

(ηε|∇ψk|22 + ηε|√ρkψk|22 + (η + Cε)|∇uk|22

)dt

for some Gkη such that∫ t

0Gkη(s)ds ≤ C

(1 + ε+ η−2t+ η−1|VR0 |

13 t)

= f(C, t, ε, η, R0), for 0 ≤ t ≤ T ∗∗.

Then from Gronwall’s inequality, we have

Λk+1(T ∗∗, η, ε) +

∫ T ∗∗

0

(a4

4ε|∇ψk+1|22 +

µ

2|∇uk+1|22

)dt

≤(ηεT ∗∗ sup

0≤t≤T ∗∗|√ρkψk(t)|22 +

∫ T ∗∗

0

(ηε|∇ψk|22 + (η + Cε)|∇uk|22

)dt)

exp f(C, t, ε, η, R0).

First, we can choose 0 < ε = ε0 < 1 small enough such that

(1 + C)ε0 exp(C + Cε0) ≤ min( µ

32,

1

32

);

secondly, we can choose 0 < η = η0 small enough such that

(1 + C)(η0 + η0ε0)exp(C + Cε0

)≤(a4ε0

32,µ

32,ε032,

1

32

);

thirdly, we can choose T ∗∗ = T∗ small enough such that

(1 + η0ε0T∗)exp(Cη−2

0 T∗)≤ 2;

at last, we can choose R0 sufficiently small such that

exp(Cη−1

0 |VR0 |13T∗)≤ 2.

So, when Λk+1 = Λk+1(T∗, η0, ε0), we have

∞∑k=1

(1

2Λk+1 +

∫ T∗

0

(a4ε08|∇ψk+1|22 +

µ

8|∇uk+1|22

)ds)≤ C < +∞.

Thus we know that the full consequence (ρk, uk, ψk) converges to a limit (ρ, u, ψ) in thefollowing strong sense:

ρk → ρ in L∞([0, T∗];L2(V)), (ψk, uk)→ (ψ, u) in L2([0, T∗];D

1(V)). (3.66)

Due to the local uniform estimates (3.52) and (3.56), and the strong convergence in (3.66),it is easy to see that (ρ, u, ψ) is a weak solution in the sense of distribution. Via the lowersemi-continuity of norms, we also have that (ρ, u, ψ) satisfies the regularities in (3.59).

Step 2. The uniqueness. Let (ρ1, u1, ψ1) and (ρ2, u2, ψ2) be two strong solutions toIBVP(3.2)-(3.3) with (3.4) satisfying the regularity (3.59). We denote that

ρ = ρ1 − ρ2, u = u1 − u2, ψ = ψ1 − ψ2.


Similarly to the derivations of (3.63)-(3.65), let

Λ(t) = |ρ|22 + |√ρ1ψ|22 + C|√ρ1u|

22,

where C > 0 is a sufficiently large constant, then

d

dtΛ(t) +

1

2Cµ|∇u|22 + |∇ψ|22 ≤ Ψ(t)Λ(t), with

∫ t

0Ψ(s)ds ≤ C for t ∈ [0, T∗].

Then from the Gronwall’s inequality and ∇ψ · n|∂V = 0, u · n = 0, we deduce thatρ = u = ψ = 0.

Step 3. Time-continuity. The time-continuity can be obtained by the same method asin the proof of Lemma 3.7. Here we omit its details.

3.6. Proof of Theorem 1.1. Now we give the proof for Theorem 1.1.

Proof. First, from (3.1), IBVP (1.1)-(1.7) can be written into

ρt + div(ρu) = 0,

ρut + ρu · ∇u+∇P = divT,

ρθt + ρu · ∇θ +1

cv

(Rρθdivu− νdiv(θb∇θ)

)=

1

cvQ(u),

(ρ, u, θ)|t=0 = (ρ0(x), u0(x), θ0(x)), x ∈ V,

u|∂V = 0, ∇θ · n|∂V = 0.

(3.67)

Secondly, from Theorem 3.1, we know that IBVP (3.2)-(3.4) has a unique strong solution

(ρ, u, ψ1

1+b ) ∈ Φ, which quickly implies that

(ρ, u, θ) = (ρ, u, ψ1

1+b )

is our desired unique strong solution for (3.67), and also IBVP (1.1)-(1.7).

4. necessity and sufficiency of the compatibility condition

In this section, we need to prove Theorem 1.2.

Proof. Step 1. Necessity. Let (ρ, u, θ) be a strong solution on [0, T∗] × R3 to IBVP (1.1)-(1.7) with the regularities shown in Definition 1.1. Then due to (1.1), we have

−divT(u) +∇P =√ρG1,

− 1

cv

(ν(b+ 1)−14θb+1 +Q(u)

)=√ρG2

(4.1)

for 0 ≤ t ≤ T∗, where

G1(t) =√ρ(−ut − u · ∇u), G2(t) =

√ρ(−θt − u · ∇θ −Rθdivu).

Since(√ρut,√ρθt,√ρu · ∇u,√ρu · ∇θ,√ρθdivu) ∈ L∞([0, T∗];L

2),

we have(G1, G2) ∈ L∞([0, T∗];L

2).


So there exists a sequence tk (tk → 0) such that

(G(tk), G2(tk)) (f, g) weakly* in L2 for some (f, g) ∈ L2.

So, let t = tk → 0 in (4.1), then we obtain

divT(u(0, x)) +∇P (0, x) =√ρ(0, x)f,

− 1

cv

(ν(b+ 1)−14θb+1(0, x) +Q(u(0, x))

)=√ρ(0, x)g.

(4.2)

Combining with the strong convergence (1.17) and (4.2), we know that the necessity ofthe compatibility conditions are obtained. Moreover, from the construction of our strongsolutions in Section 3, we easily deduce that

f = g1, and g = g2.

Step 2. Sufficiency. To prove the sufficiency. Let (ρ0, u0, θ0) be the initial data satisfying(1.13)-(1.14). Then there exists a unique solution (ρ, u, θ) to (1.1)- (1.3):

ρ(t, x) ∈ C([0, T∗];W1,q), (θ, u)(t, x) ∈ C([0, T∗];H

2(V)).

Then we only need to make sure that

ρ(0, x) = ρ0, u(0, x) = u0, θ(0, x) = θ0, x ∈ V.

From the weak formulation of the strong solution, we easily have

ρ(0, x) = ρ0, ρ(0, x)u(0, x) = ρ0u0, ρ(0, x)θ(0, x) = ρ0θ0, x ∈ V.

It remains to prove that

u(0, x) = u0(x), and θ(0, x) = θ0(x)

when x ∈ V . Let

u0 = u0 − u(0, x), ψ0 = θb+10 − θb+1(0, x).

According to the proof of the necessity, we know that (ρ(0, x), u(0, x), θ(0, x)) also satisfiesthe relation (1.14) for (g1, g2) ∈ L2. Then we quickly know that

(ψ0, u0) ∈ D10(V ) ∩D2(V )

is the unique solution of the elliptic problem (1.16), and thus

u0 = 0, ψ0 = 0, in V,

which implies that

u(0, x) = u0(x), θ(0, x) = θ0(x), x ∈ V.


5. Minimum principle.

Based on the solution class obtained in Section 3, now in this section, we will give theproof for the minimum principle of θ shown in Theorem 1.3. Let (ρ, u, θ) on [0, T ]×V bethe strong solution to IBVP (1.1)-(1.7), then we have

sup0≤t≤T

‖u(t)‖22 + ess sup0≤t≤T

|√ρut(t)|22 +

∫ T

0

(|u(s)|2D2,q + |ut(s)|2D1

)ds ≤C,

sup0≤t≤T

‖θ(t)‖22 + ess sup0≤t≤T

|√ρθt(t)|22 +

∫ T

0

(|θ(s)|2D2,q + |θt(s)|2D1

)ds ≤C,

sup0≤t≤T

(|ρ(t)|2∞ + ‖ρ(t)‖2W 1,q + |ρt(t)|q) ≤C,

(5.1)

where C is a generic constant depending only on (ρ0, u0, θ0) and fixed constants µ, λ, ν,q, R, cv, |V| and T .

Because the initial temperature θ0 ≥ θ > 0 and by the continuity of θ, we know thatthere exists a time T = T (C, θ) ∈ (0, T ] such that

θ(t, x) > 0 for [0, T ]× V. (5.2)

Then based on (5.2), we will prove that

θ(t, x) ≥ θ for [0, T ]× V. (5.3)

Hence the desired minimum principle follows from the uniform choice of T and (5.3).Next we will prove (5.3) in four steps:

5.1. Definition of functional Uk. Let m > b be a constant and Mo > 1 be a constantlarge enough. Here m and Mo can be chosen as

θ =( 1

Mo

) 2m−b

. (5.4)

We define the ratio

K = 2m/(m− b) ≥ 2. (5.5)

Without lose of generality, we consider the case of 0 < θ ≤ 1 small enough, and the othercases could be obtained by scaling arguments.

For the De Giorgi iteration technique used in our following proof, we need to introducea sequence Nk:

N0 = Mo, Nk+1 = Nk +Mo

zk+1, z ≥ max

(Mo, 2

), (5.6)

which implies that

M0 ≤ Nk ≤1

1− 1z

Mo, N∞ = limk→∞

Nk =1

1− 1z

Mo. (5.7)

We also define

ξk(θ) =[ 1

θmK

−Nk

]+, ϕk(θ) =

[ 1

θm−NK

k

]+.


One can see that ξk(θ) and ϕk(θ) have the same supports, which can be denoted as

Vk =

(x, t)∣∣θm < 1/NK

k

. (5.8)

Here we have some relations between ξk and ϕk,

ξk(θ) ≤ ϕk(θ), and ξk(θ)K ≤ ϕk(θ), (5.9)

when θ =(

1Mo

) 2m−b

such that Vk is not an empty set.

On the other hand, we have

ξk−1 ≥M0/zk, in the region Vk. (5.10)

Now we define

Uk =

∫ρϕk(θ)(t, x) +

1

cv

∫ t

0

∫Vm

(Q(u)

θm+1+

(m+ 1)κ(θ)|∇θ|2

θm+2

)1Vkdxds. (5.11)

Then multiplying both sides of the temperature equation (3.1) by ϕ′k(θ), one has

Uk =− 1

cv

∫ t

0

∫V

m

θm+1Rρθ1Vkdivudxds ≤ C

∫ t

0

∫V

ρθ

θm+1|divu|1Vkdxds, (5.12)

where we have used the continuity equation (1.1)1, definition of ϕk(θ) and (5.4).We have the following bounds from the definition of Uk:

‖ρϕk‖L∞L1 ≤ CUk, ‖∇ξk‖L2L2 ≤ CU12k .

(5.13)

Next we need to control the right hand side term of (5.12) by some power of Uk−1 suchthat we can use the De Giorgi iteration technique to get

limk→∞

Uk = 0. (5.14)

So the main task is to give a sufficiently good interpolation estimates for the right hand-side of (5.12) .

Here is a useful lemma in helping us prove (5.14) in the future.

Lemma 5.1. We assume that U0 is a small enough positive number,

γ1 > 0, γ2 > 1, γ3 > 0, M > 1 and z > 1

are all real constants. If the following conditions hold:

Uk+1 ≤ P1

( zkM

)γ1Uγ2k M

γ3 , (5.15)

where P1 > 0 is a constant independent of γi (i = 1, ..., 3), M and z, then we have

limk→∞

Uk = 0.

The proof for this lemma could be seen in the appendix of this paper.


5.2. Estimates for Iterations I. In this section, we will give the estimates needed forLemma 5.1 in the case when 0 ≤ b < 1.

The term in equation (5.12) could be controlled in this way:∫ t

0

∫V

ρθ

θm+1|divu|1Vkdxds ≤

∫ t

0

(∫V

(δ1|divu|2

θm+1+ C(δ1)ρ2θ−m+1

)1Vkdxds, (5.16)

where δ1 > 0 is a constant small enough and C(δ1) > 0 is a constant depending on δ1 andthe generic constant C introduced in (5.1). In fact, we only need to control the secondterm for future use.

First using Holder’s inequality, we have(∫ t

0

(∫V

(ρµ1ξλ1Kk−1 )q1dx) p1q1 ds

) 1p1

=(∫ t

0

(∫Vρµ1q1ξµ1q1Kk−1 · ξ(λ1−µ1)q1K

k−1 dx) p1q1 ds

) 1p1 ≤ ‖ρξKk−1‖

µ1L∞L1‖ξk−1‖

(λ1−µ1)KL2L6 ,

(5.17)

where the parameters satisfy

(λ1 − µ1)q1K

6+ µ1q1 = 1,

(λ1 − µ1)p1K

2= 1. (5.18)

Now we need to control the term ‖ξk−1‖(λ1−µ1)KL2L6 . From Lemma 2.2, we have

|ξk−1|6 ≤ C(|ρξk−1|1 + |∇ξk−1|2

), (5.19)

which, along with (5.13), means that

‖ξk−1‖(λ1−µ1)KL2L6 ≤C

(‖ρξk−1‖

(λ1−µ1)KL2L1 + ‖∇ξk−1‖

(λ1−µ1)KL2L2

)≤C(U

(λ1−µ1)Kk−1 + U

(λ1−µ1)K2

k−1

).

(5.20)

Secondly, we can choose m satisfying b < m ≤ 1. Thus we have∫ t

0

∫Vρ2θ−m+11Vkdxds ≤ C

∫ t

0

∫Vρ2

ξλ1Kk−1

(Mo/zk)λ1Kdxds

≤C 1

(Mo/zk)λ1K‖ρ2−µ1‖Lp2Lq2

(∫ t

0

(∫V


) 1p1

≤C 1

(Mo/zk)λ1K‖ρ2−µ1‖Lp2Lq2‖ρξKk−1‖

µ1L∞L1‖ξk−1‖

(λ1−µ1)KL2L6

≤C 1


(Uµ1+(λ1−µ1)Kk−1 + U

µ1+(λ1−µ1)K

2k−1

),

(5.21)

where the positive indices also need to satisfy


(λ1 − µ1)q1K

6+ µ1q1 = 1,

(λ1 − µ1)p1K

2= 1,

λ1 > µ1, µ1 < 2,1

p1+

1

p2= 1,

1

q1+

1

q2= 1,

λ1K > 0, µ1 +(λ1 − µ1)K

2> 1.

(5.22)

Denoting

x =(λ1 − µ1)K

6, y = µ1, (5.23)

due to (5.18) and (5.22), we need to consider the following stronger requirements:

0 < x+ y < 1, 3x+ y > 1, 3x < 1. (5.24)

We just need to choose x < 13 but very close to 1

3 , and y a positive small enough number.Thus, we have

Uk ≤C(zk/M0)λ1K(Uµ1+(λ1−µ1)Kk−1 + U

µ1+(λ1−µ1)K

2k−1

)for b ∈ [0, 1). (5.25)

Moreover, from the defintion of the functional Uk, b < m ≤ 1 and (5.16), we also havethe following estimate on Uk:

Uk ≤C(δ1)

∫ t

0

(∫Vρ2θ−m+11Vkdx

)ds ≤ C(δ1)

∫ t

0

∫Vρ21Vkdxds, (5.26)

where δ1 > 0 is a small enough constant, and C(δ1) > 0 is a constant independent of kand z.

5.3. Estimates for Iterations II. In this section, we will finish the estimates neededfor Lemma 5.1 in the case when 1 ≤ b <∞.

The negative power of θ could be bounded in the following way:

1

θm−11Vk ≤ C

(ξm−1m/K

k−1 + (Mo)m−1m/K

). (5.27)

So we have the estimates∫ t

0

∫Vρ2θ−m+11Vkdxds

≤C∫ t

0

∫Vρ2

ξλ1K+m−1

m/K

k−1

(Mo/zk)λ1K+

ξλ2Kk−1

(Mo/zk)λ2K(Mo)−m−1m/K

dxds.

(5.28)

Let’s define the constant

K1 = K +(m− 1)

λ1m/K.


Then, from (5.13) and (5.20), the first term in inequality (5.28) could be estimated as

∫ t

0

∫Vρ2

ξλ1K+m−1

m/K

k−1

(Mo/zk)λ1Kdxds =

∫ t

0

∫Vρ2

ξλ1K1k−1

(Mo/zk)λ1Kdxds

≤C 1


(∫ t

0

(∫V

(ρµ1ξλ1K1

k−1

)q1dx) p1q1 ds

) 1p1

≤C 1

(Mo/zk)λ1K‖ρ2−µ1‖Lp2Lq2‖ρξK1

k−1‖µ1L∞L1‖ξk−1‖

(λ1−µ1)K1

L2L6

≤C 1


(Uµ1+(λ1−µ1)K1

k−1 + Uµ1+

(λ1−µ1)K12

k−1

),

(5.29)

where the positive index need to satisfy (5.22). While we can choose the parameterssimilarly as in Section 5.2. If (x, y) is still defined in (5.23), then we can also choose x < 1

3

but x is very close to 13 , and y a positive number small enough.

Similarly, the second term in inequality (5.28) could be estimated as

∫ t

0

∫Vρ2

ξλ2Kk−1


dxds

≤C 1


‖ρ2−µ2‖Lp4Lq4(∫ t

0

(∫V


) 1p3

≤C 1


‖ρ2−µ2‖Lp4Lq4‖ρξK‖µ2L∞L1‖ξk−1‖(λ2−µ2)KL2L6

≤C 1


‖ρ2−µ2‖Lp4Lq4(Uµ2+(λ2−µ2)Kk−1 + U

µ2+(λ2−µ2)K

2k−1

),

(5.30)

where the positive index need to satisfy

(λ2 − µ2)q4K

6+ µ1q2 = 1,

(λ2 − µ2)p4K

2= 1,

λ2 > µ2, µ2 < 2,1

p3+

1

p4= 1,

1

q3+

1

q4= 1,

λ2K > 0, µ2 +(λ2 − µ2)K

2> 1.

(5.31)

While we can choose the parameters similarly as in the above case. If

x =(λ2 − µ2)K

6, y = µ2,

we need to choose x < 13 but x is very close to 1

3 , and y a positive number small enough.


In summary, we have

Uk ≤C(zk/Mo)s1(Uµ1+(λ1−µ1)K1

k−1 + Uµ1+

(λ1−µ1)K12

k−1

+ Uµ2+(λ2−µ2)Kk−1 + U

µ2+(λ2−µ2)K

2k−1

)(Mo)

(m−1)m/K for b ∈ [1,∞),

(5.32)

where s1 = max(λ1K,λ2K

).

5.4. Continuity argument. In this section, we will fix t ∈ (0, Tm), and prove

limk→∞

Uk(t) = 0, (5.33)

then using which to prove the minimum principle.As a preparation, first we have the following lemma:

Lemma 5.2. Let t ∈ (0, Tm) and (5.33) hold. One has

θ(t, x) ≥(

1− 1

z

)Kmθ for x ∈ V. (5.34)

Its proof can be found in the appendix.

Secondly, we consider the following function

LG(t) = − sup0<τ<t

∣∣∣ 1

θ(τ)

∣∣∣mK∞. (5.35)

Notice that LG is monotone decreasing, and define

t1 = supLG(t)>−N∞

t, t2 = infLG(t)<−N∞

t,

i.e. for any ε > 0 small enough, there exists a constant δ(ε) > 0, such thatLG(t2 + ε) < −N∞ − δ(ε),

LG(t1 − ε) > −N∞ + δ(ε),

where N∞ is defined in (5.7).Obviously, t1 ≤ t2, as a fact, we could choose z so that they are the same:

Lemma 5.3. We can also choose a zs ∈ (s, s+ 1) for any s ≥ 2 such that t2 = t1.

The proof of this lemma can also be found in the appendix.

The final step of the proof for Theorem 1.3 will be divided into two parts: 0 < b ≤ 1and b > 1, as before.

5.4.1. Case 0 ≤ b ≤ 1. Now we first consider the case when 0 < b ≤ 1, and we are readyto use the continuity argument to show the uniform bound for θ.

First, we want to show t1 > 0. Go back to the estimate (5.26):

Uk ≤ C(δ1)

∫ t

0

∫Vρ21Vkdxds. (5.36)


So if t ≤ 1C1

with C1 = 1C(δ1)C2C2

(here C1 is independent of k and z), then we have for

arbitrarily k = 0, 1, 2, ...,

Uk ≤1

C2< 1, when C2 is large enough and independent of k and z.

Thus from (5.25), we have

Uk ≤C(zk/M0)λ1KUµ1+

(λ1−µ1)K2

k−1 for b ∈ [0, 1) and t ∈[0,

1

C1

]. (5.37)

Then, due to Lemma 5.1, for some C2 large enough, one has

limk→∞

Uk = 0 for t ∈[0,

1

C1

], (5.38)

which, along with Lemma 5.2, implies that

θ ≥(

1− 1

z

)Kmθ for t ≥

[0,

1

C1

].

According to the definition of t1, it is obvious that t1 ≥ 1C1

.

Secondly, we want show that t1 = T , i.e.

θ ≥(

1− 1

z

)Kmθ for t ∈ [0, T ). (5.39)

Here we recall that T is defined in (5.2).We prove it by contradiction. Assume t1 < T , then we will find a contradiction. We

can define

s2 = t1 −1

2C1.

By the definition of t1, we have LG(s2) > −N∞. So when k large enough, say whenk > k1(s2), we have

LG(s2) > −Nk1(s2) ≥ −Nk > −N∞, (5.40)

which means that:

θ(t) ≥(

1− 1

z

)Kmθ for t ∈ (0, s2).

Then we have

Uk ≤ C(δ1)

∫ t

s2

∫Vρ21Vkdxdτ. (5.41)

As long as

t ≤ mins2 +1

C1, T,

via the completely same argument for proving (5.38), one has

limk→∞

Uk = 0.

Thus

t2 ≥ mins2 +

1

C1, T

= mint1 +

1

2C1, T> t1, (5.42)

which contradicts with Lemma 5.3. So t1 = T .


At last, replacing (s, s+ 1) in Lemma 5.3 with (n, n+ 1) for sufficiently large constantn, we can also choose an increasing sequence of zn such that

zn ∈ (n, n+ 1).

So we have

limn→∞

zn =∞, as n→∞.

Therefore the desired minimum principle can be obtained via letting zn →∞ in (5.34).

5.4.2. Case b > 1. From (5.12), we know

Uk ≤C∫ t

0

∫V

ρθ

θm+1|divu|1Vkdxds

≤ m

2cv

∫ t

0

∫V

|divu|2

θm+11Vkdxds+ C

∫ t

0

∫Vρ2θ(ϕk +NK

k )dxds

≤ m

2cv

∫ t

0

∫V

|divu|2

θm+11Vkdxds+ C

∫ t

0Uk(s)ds+ CtNK

k |V|,

(5.43)

which, along with (5.11), implies that∫Vρϕk(θ)(t, x)dx+

1

2cv

∫ t

0

∫Vm

(Q(u)

θm+1+

(m+ 1)κ(θ)|∇θ|2

θm+2

)1Vkdxds

≤C∫ t

0Uk(s)ds+ CtNK

k |V|.(5.44)

Then from Gronwall’s inequality, we have

Uk(t) ≤ (CNKk |V|t) exp(Ct). (5.45)

Noticing that Mo ≤ Nk ≤ 2Mo, then if t ≤ 1C3

with

C3 =1

C((2Mo)K |V|C4 + 1),

where C3 is independent of k and z, then we have for abitrarily k = 0, 1, 2, ...,

Uk ≤2

C4< 1, when C4 is large enough and independent of k and z.

Thus from (5.32), we have

Uk ≤C(zk/Mo)s1UBk−1(Mo)(m−1)m/K for b ∈ [1,∞), (5.46)

where

B = min(µ1 +

(λ1 − µ1)K1

2, µ2 +

(λ2 − µ2)K

2

)> 1.

Then from Lemma 5.1, we have

limk→∞

Uk = 0,

which implies that t1 ≥ 1C3

.Finally, via the similar argument used in Subsection 5.4.1, we will obtain the desired

conclusions for the case when 1 ≤ b < +∞.


6. Appendix

In this appendix, we will give the proof for some important lemmas that are used inour proof.

6.1. Proof of Lemma 2.2. In this subsection, we give the proof for the Poincare typeinequality stated in Section 2.

Proof. We first denote that

F =1

|Ω|

∫ΩF (y)dy,

then via the classical Poincare inequality, we quickly deduce that

F

∫Ωρdx =

∫Ωρ(F − F )dx+

∫ΩρFdx

≤C (|ρF |1 + |ρ|2|∇F |2) ≤ C(|ρ|

121 |√ρF |2 + |ρ|2|∇F |2

),

which implies thatF ≤ C

(|√ρF |2 + |ρ|2|∇F |2

). (6.1)

Second, we consider that

‖F‖1 =|∇F |2 + |F |2 ≤ |∇F |2 + |F − F |2 + F |Ω|12

≤C(|√ρF |2 + (1 + |ρ|2)|∇F |2

),

(6.2)

then according to (6.1)-(6.2) and the classical Sobolev imbedding theorem, we easily obtainthe following inequality:

|F |6 ≤ C‖F‖1 ≤ C(|√ρF |2 + (1 + |ρ|2)|∇F |2

).

6.2. Proof of Lemma 3.3. In order to prove Lemma 3.3, we first denote X ∈ C([0, T1]×[0, T1]× V) is the solution to the initial value problem

d

dtX(t; 0, x0) = w(t,X(t; 0, x0)), 0 ≤ t ≤ T1;

X(0; 0, x0) = x0, x0 ∈ V.(6.3)

Second, let A(t, R0) and B(t, R0) be closed regions that are the images of VR0 and (V/VR0)respectively under the flow map (6.3):A(t, R0) = X(t; 0, x0)|x0 ∈ VR0,

B(t, R0) = X(t; 0, x0)|x0 ∈ (V/VR0).

Proof. Due to the definition of V and the continuity of ρ0, it is obvious to see that for anysufficiently small R′ > 0, there exists a constant aR′ independent of δ such that

ρδ0(x) ≥ aR′ + δ > 0, ∀ x ∈ (V/VR′). (6.4)

From the continuity equation (3.9)1, we have

ρ(t, x) = ρδ0(X(0; 0, x0)) exp(−∫ t

0divw(s;X(s; 0, x0))ds

). (6.5)


It is easy to see that∫ t

0|divw(t,X(t; 0, x0))|ds ≤

∫ t

0|∇w|∞ds ≤ c2t

1/2 ≤ ln 2, (6.6)

for 0 ≤ t ≤ T ′ = min(T1, (ln 2)2(Cc2)−2).So via (6.4) and (6.6), we easily know that for 0 ≤ t ≤ T ′,

ρ(t, x) ≥ 1

2(aR′ + δ) > 0, ∀ x ∈ B(t, R′). (6.7)

From the ODE problem (6.3), we get

|X(0; 0, x0)− x| =|X(0; 0, x0)−X(t; 0, x0)|

≤∫ t

0|w(τ,X(τ ; 0, x0))|dτ ≤ c2t ≤ R′/2,

for all (t, x) ∈ [0, TR′ ]× V, and TR′ = min(T ′, R′(2c2)−1), which means,

V/V3R′/2 ⊂ B(t, R′). (6.8)

Thus we can choose that

R0 =3

2R′, aR0 =

1

2aR′ and TR0 = min

(T1, (ln 2)2(Cc2)−2, 2R0(6c2)−1

).

6.3. Proof of Lemma 3.4. Now we give the proof of the postivity of ψ for the corre-sponding strong solutions to the linear problem.

Proof. From equation (3.9)2, it is easy to have

ρψt + ρw · ∇ψ − a2φbb+14ψ ≥ −a1ρψdivw, (6.9)

where we have used the fact that φbb+1Q(w) ≥ 0. We define

T ′′ = inf t ∈ (0, T ]∣∣ ψ(t, x) = 0, for some x ∈ V.

From Lemma 3.1, we know that ρ ≥ δ > 0. Thus (6.9) implies that

ψt + w · ∇ψ − a2

ρφ

bb+14ψ ≥ −a1ψ|divw|∞ for (t, x) ∈ [0, T ′′]× V. (6.10)

Denote

ψ∗ = ψ exp(a1

∫ t

0|divw(τ, x)|∞dτ

),

then along curve X(t; 0, x0), we have

d

dtψ∗ − a2

ρφ

bb+14ψ∗ ≥ 0. (6.11)

Then, from ψ0(x) ≥ ψ and the classical minimum principle, we have

ψ(t, x) ≥ infx∈V

ψ0(x) exp(− (γ − 1)

∫ t

0|divw(τ, x)|∞dτ

)> 0, (6.12)

for t ∈ [0, T ′′], which is contradictory with the definition of T ′′. Thus we have (6.12) holdsfor t ∈ [0, T1].


Moreover, we have

ψ(t, x) ≥ 1

2ψ for 0 ≤ t ≤ T2 = min(T1, (ln 2)2(Cc2)−2).

6.4. Proof of Lemma 5.1. In this subsection, we specify some recursion relations, andthe conditions for the sequence of functionals Uk to converge to 0. Moreover, we will givethe proof for Lemma 5.1. First, we have

Lemma 6.1. If Uk is a nonnegative sequence such that

Uk+1 ≤ P2zβ1kUβ2k , (6.13)

where P2 is a positive constant independent of β1, β2, k and z.

β1 > 0, β2 > 1 and z > 1

are all constants. If U0 is small enough (specified in equation (6.17)), we have

limk→∞

Uk = 0.

Proof. Let’s define xk = logUk. Then we have the recursion relation

xk+1 ≤ logP2 + β1k log z + β2xk. (6.14)

We want to choose xk going to negative infinity by geometric speed, so we want to show

xk ≤ x0

(β2 + 1

2

)k. (6.15)

We will prove equation (6.15) by induction, for k = 0, the equation (6.15) is obvious, thenif

xk ≤ x0

(β2 + 1

2

)k. (6.16)

And we specify the x0 by assuming

x0 < −P3 ≤ −maxk≥0

logP2 + β1k log zβ2−1

2 (β2+12 )k

, (6.17)

where the constant P3 depends only P2, β1, β2 and z. Then

xk+1 ≤ x0

(β2 + 1

2

)kβ2 + logP2 + β1k log z

≤ x0

(β2 + 1

2

)k+1+β2 − 1

2x0

(β2 + 1

2

)k+ logP2 + β1k log z

≤ x0

(β2 + 1

2

)k+1.

(6.18)

So equation (6.15) is true for any positive integer k, and by letting k → ∞ we finish theproof of lemma 6.1. One can notice that the constant P3(P2, β1, β2, z) in the requiment ofU0 could be chosen continuously depending on z.

Then we have the following variant:


Lemma 6.2. If Uk is a nonnegative sequence such that

Uk+1 ≤ P4zβ1kUβ2k Mβ3 , (6.19)

where P4 is a positive constant independent of β1, β2, β3, k,M and z.

β1 > 0, β2 > 1 β3 > 0, M > 1 and z > 1

are all constants. Then there exists a constant P5 which is independent of M , and if

U0 ≤ P5

( 1

M

) β3β2−1

, (6.20)

we have

limk→∞

Uk = 0.

Proof. We can define

yk = Mβ3β2−1Uk,

then we have

yk+1 ≤ P4zβ1kyβ2k .

Therefore, the proof of this lemma could be quickly finished based on the conclusion ofLemma 6.1.

Finally, it is easy to see that Lemma 5.1 is a direct consequence of Lemmas 6.2-6.3.

6.5. Proof of Lemma 5.2.

Proof. First, from (5.6)-(5.7) and the monotonicity of ϕk, Nk, one has

limk→∞

ϕk =[ 1

θm−( 1

1− 1z

)K 1

θm

]+, lim

k→∞Vk =

(x, t)

∣∣θm <(

1− 1

z

)Kθm. (6.21)

Then it follows from (5.33), (6.21) and the definition of Uk that in the domain

Ω+ =

x ∈ V

∣∣∣ ρ(x, t) > 0, or

(Q(u)

θm+1+

(m+ 1)κ(θ)|∇θ|2

θm+2

)(x, t) > 0

, (6.22)

the fact (5.34) holds in Ω+.Assuming that there exists some domain Ω0 ⊂ V (|Ω0| > 0) given via

Ω0 =

x ∈ V

∣∣∣ ρ(x, t) = 0, and

(Q(u)

θm+1+

(m+ 1)κ(θ)|∇θ|2

θm+2

)(x, t) = 0

, (6.23)

then from the above definition, one has

|∇θ| = 0 for x ∈ Ω0, (6.24)

which, along with

θ(t) ≥(

1− 1

z

)Kmθ for x ∈ V/Ω0,

implies that the fact (5.34) holds in Ω0.


6.6. Proof of Lemma 5.3.

Proof. If for any zi, there exists t1(zi) ≤ t2(zi), then by the monotonicity of LG,

LG(t) = −(

1− 1

zi

)Mo, for any t ∈ (t1, t2). (6.25)

We want to show that (t1(za), t2(za)) and (t1(zb), t2(zb)) are two non-intersecting intervalsfor za 6= zb. As a fact, if za 6= zb, τ ∈ (t1(za), t2(za)), and τ ∈ (t1(zb), t2(zb)),

LG(τ) = −(

1− 1

za

)Mo = −

(1− 1

zb

)Mo, (6.26)

which is impossible, and implies that

(t1(za), t2(za)) and (t1(zb), t2(zb))

are two non-intersecting intervals. So the total length of all these non-intersecting intervalsis not larger than T , ∑

s<zi<s+1

(t2(zi)− t1(zi)) ≤ T . (6.27)

As a fact zi is from uncountable set (2, 3), so we have the existence of zi.

Acknowledgement: We sincerely appreciate Professors Boling Guo and RonghuaPan for their very helpful suggestions and discussions on the problem solved in this paperwhen Chen, Zhang and Zhu were in Georgia Tech, and when Zhu worked in The ChineseUniversity of Hong Kong. The research of G. Chen was partially supported by NSF, withgrant DMS-1715012. The research of S. Zhu was supported by Natural Science Foundationof Shanghai under grant 14ZR1423100, Nation Natural Science Foundation of China undergrants 11231006 and 11571232, and China Scholarship Council 201206230030.

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(G. Chen) School of Mathematics, University of Kansas, Lawrence, KS 66045, U.S.A.E-mail address: [email protected]

(W. Z. Zhang) School of Mathematics, Georgia Tech, Atlanta 30332, U.S.A.E-mail address: [email protected]

(S. G. Zhu) School of Mathematical Sciences, Monash University, Melbourne, VIC 3800,Australia

E-mail address: [email protected]

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