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Research article

Regularity of weak solution of the compressible Navier-Stokes equations with self-consistent Poisson equation by Moser iteration

  • Received: 13 April 2023 Revised: 08 July 2023 Accepted: 13 July 2023 Published: 19 July 2023
  • MSC : 35L65, 35Q35, 76N10

  • In this paper, we consider the regularity of the weak solution to the compressible Navier-Stokes-Poisson equations in period domain ΩR3 provided that the density ρ(t,x) with integrability on the space L(0,T;Lq0(Ω)) where q0 satisfies a certain condition and T>0, by which we could present that supt,xρ(t,x)< and inft,xρ(t,x)>0. Furthermore, we develop the estimate for the velocity uL by the Moser iteration method and Gronwall inequality.

    Citation: Cuiman Jia, Feng Tian. Regularity of weak solution of the compressible Navier-Stokes equations with self-consistent Poisson equation by Moser iteration[J]. AIMS Mathematics, 2023, 8(10): 22944-22962. doi: 10.3934/math.20231167

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  • In this paper, we consider the regularity of the weak solution to the compressible Navier-Stokes-Poisson equations in period domain ΩR3 provided that the density ρ(t,x) with integrability on the space L(0,T;Lq0(Ω)) where q0 satisfies a certain condition and T>0, by which we could present that supt,xρ(t,x)< and inft,xρ(t,x)>0. Furthermore, we develop the estimate for the velocity uL by the Moser iteration method and Gronwall inequality.



    In this article, we will investigate the regularity of weak solutions of the compressible Navier-Stokes-Poisson (NSP) equations. The compressible (NSP) system is used to simulate the motion of charged particles. It consists of the compressible Navier-Stokes (NS) equations with the electrostatic potential controlled by the self-consistent Poisson equation. A carrier type of NSP system is represented as the following form in R3:

    {tρ+div(ρu)=0,t(ρu)+div(ρuu)+p(ρ)μΔu(μ+ν)divuρΦ=ρf,ΔΦ=ρˉρ,   lim|x|Φ(t,x)=0. (1.1)

    Here, ρ=ρ(t,x)>0 is the density, u(t,x)=(u1,u2,u3)(t,x) is the velocity, p(ρ) denotes the pressure and Φ denotes the electrostatic potential. ¯ρ>0 is a constant. μ>0 and ν are constant viscous coefficients that satisfy 2μ+3ν0. The term f denotes a given external force.

    There are many conclusions about NS equations on the condition that Φ=0 in (1.1). In the past, people have studied the existence of weak solution to the compressible NS equations. P. L. Lions [12] proved the global existence of finite energy weak solution to NS equations when p(ρ)=Rργ,γ>95 in 1998; Feireisl [6] claimed the existence of the weak solution when p(ρ)=Rργ,γ>32 in 2001. Afterwards, Choe, Kim and Cho [2,3] showed the local existence of the strong solution to NS equations. In 2011, Sun, Wang and Zhang [19] proved a blow-up criterion for a strong solution to the compressible viscous heat-conductive flows in R3 on condition that μ>17ν in accordance with the upper bound of (ρ,ρ1,θ). Valli [24] proved the uniqueness of the solution if the data ρ0,u0 and f satisfy some condition in bounded smooth domain ΩR3. In 2013, Wen and Zhu [25] established a blow-up criterion for strong solutions to the Cauchy problem of compressible isentropic NS equations with vacuum provided μ>329ν in the light of the integrability of the density. Though Choe [4] proved the regularity to the weak solution, which is defined by D. Hoff [9], it is indeterminate whether the weak solutions is the same to the finite energy weak solution.

    However, there are not many conclusions related to the solution to NSP equation. In a series of articles [14,15,18], M. Okada et al. [17,18] studied the vacuum problem of NSP equations which are spherically symmetrical with solid core. They [17] considered a free boundary problem for the equation of the one-dimensional isentropic motion with density-dependent viscosity and proved that there exists a unique weak solution globally in time; Then, Kobayashi and Suzuki [11] proved the existence of weak solution to NSP equations of which the method is similar to the method of [6]. In 2007, Zhang and Tan [27] proved the existence of the global weak solution to the compressible NSP equations where p(ρ)=aρlogdρ for sufficiently large densities by using suitable Orlicz spaces in R2. We may refer to [1,7,20,21,22,26] for more about NS or NSP equations. As far as we are concerned, to the author's best knowledge, there are not many studies on the regularity for the corresponding weak solution to the compressible NSP equations.

    For simplicity, let f0 of (1.1) in the paper. Thus, the problem we are concerned with in our study is the regularity of weak solutions to the isentropic compressible NSP equations in the periodic domain Ω=R3Z3 as

    {tρ+div(ρu)=0,  for  (t,x)(0,T)×Ω,t(ρu)+div(ρuu)μΔu(μ+ν)divu+p(ρ)ρΦ=0,ΔΦ=ρˉρ, (1.2)

    with the following initial conditions:

    {ρ(t,x)|t=0=ρ0>0,u(t,x)|t=0=u0, xΩ,ρ0W1,6(Ω),u0H10(Ω)L(Ω),0<mρ0M<+, (1.3)

    where ¯ρ>0 is a constant, p(ρ)=Rργ,R>0, γ>1 and m and M both are constants.

    Definition 1.1. (ρ,u) is called the finite energy weak solution of the Eq (1.2) for (t,x)ΩT=(0,T)×Ω, if (ρ,u) satisfies

    ρL(0,T;L1(Ω))L(0,T;Lγ(Ω)),uL2(0,T;H1(Ω))

    and the energy inequality

    dE(t)dt+μΩ|u|2dx+(μ+ν)Ω|divu|2dx0,

    holds in D((0,)) with the finite total energy, where

    E(t)=Ω(12ρ|u|2+p(ρ)γ1+12|Φ|2)dx<,fort(0,).

    All the priori estimates of this paper depend on the assumption that ρ and u are C for the time interval as mentioned in [4,5].

    Theorem 1.1. Suppose (ρ,u) is the weak solutions denoted by the Definition 1.1 to the Eqs (1.2) and (1.3). If ρL(0,T;Lq0(Ω)) for time T>0 and for q0 depending only on γ, then

    0<infΩTρρsupΩTρ<+,supΩT|u|<+.

    Furthermore, we show that ρC(0,T;Lq(Ω))L(ΩT) for all q[1,), and ρutL2(ΩT),PuL2(0,T;H2(Ω)),GL2(0,T;H2(Ω)),uL(0,T;L2(Ω)), where Pu refers to the divergence-free part of u, G=(2μ+ν)div up(ρ).

    The paper is structured as follows. In Section 2, we give some preliminaries to study the regularity of the weak solution to the NSP equations. In Section 3, one develops the priori estimate of ρ under some condition by many techniques; In Section 4, we shall perfect the proof of Section 3; Finally, in Section 5, we present a prior L estimate of u by Moser iteration.

    In this section, we collect some auxiliary results which will be used to investigate subsequent studies.

    First, we define the divergence-free part of any vector field u as Pu, and the gradient part of u as Qu which means Qu=Δ1divu. Naturally, the following propositions hold.

    Proposition 2.1.

    curl(Qu)=0,div(Pu)=0,Δu=(divu)curlcurlu,Pu=uQu.

    Proposition 2.2. (Page 67 of [13]). Suppose that fL2(Ω), and

    {Δv=p+f,  xΩ,divv=0,v|Ω=0,

    then vW2,2(Ω), and pL2(Ω). Furthermore,

    vW2,2(Ω)+pL2(Ω)CfL2(Ω).

    Proposition 2.3. Suppose that ρL(0,T;Lp0(Ω)) for p0=max{(2γ1)α+4γ2,4}, where α is a nonnegative real number. Then, we obtain

    Ωρ|u|α+2dxΩρ0|u0|α+2dx+supt[0,T]ρ(2γ1)α+4γ2L(2γ1)α+4γ2(Ω)tα+22+ρ¯ρα+2L4(Ω)supt[0,T]ρL1(Ω)tα+2.

    Proof. Let |u|αu as a test function to (1.2), then one obtain

    1α+2ddtΩρ|u|α+2dx+μ(α+1)Ω|u|α|u|2dx+α(μ+ν)Ω|u|αdivuudx+(μ+ν)Ω|divu|2|u|αdx=ΩRργ|u|αdivudx+αΩRργ|u|αudx+ΩρΦ|u|αudx. (2.1)

    Applying Hölder's inequality and Young's inequality to (2.1), implies that

    ΩRργ|u|αdivudx12Ω|u|α|u|2dx+2R2supt[0,T]ρ(2γ1)α+4γα+2L(2γ1)α+4γ2(Ωρ|u|α+2dx)αα+2. (2.2)

    Similarly, we obtain

    αΩRργ|u|αudx12Ω|u|α|u|2dx+2α2R2supt[0,T]ρ(2γ1)α+4γα+2L(2γ1)α+4γ2(Ωρ|u|α+2dx)αα+2. (2.3)

    Noting that ΔΦ=ρ¯ρ, and using Hölder's inequality and Sobolev inequality, we could have

    ΩρΦ|u|αudxρ¯ρL4(Ω)supt[0,T]ρ1α+2L1(Ω)(Ωρ|u|α+2dx)α+1α+2. (2.4)

    Combining (2.2)–(2.4) with (2.1), and letting U(t)=Ωρ|u|α+2dx, one could have

    ddtU(t)supt[0,T]ρ(2γ1)α+4γα+2L(2γ1)α+4γ2(Ω)[U(t)]αα+2+ρ¯ρL4(Ω)supt[0,T]ρ1α+2L1(Ω)[U(t)]α+1α+2.

    This reduces to

    U(t)U(0)+supt[0,T]ρ(2γ1)α+4γ2L(2γ1)α+4γ2(Ω)tα+22+ρ¯ρα+2L4(Ω)supt[0,T]ρL1(Ω)tα+2.

    Now, applying the operator "divergence-free fields" P to (1.2), we consequently obtain that

    P(ρut+ρuuρΦ)μΔPu=0. (3.1)

    Denote H=ρut+ρuuρΦ, then (3.1) will be reformulated equivalently as

    P(H)μΔPu=0. (3.2)

    Then, by taking divergence operator to the Eq (1.2), one deduces that

    div(ρut+ρuuρΦ)Δ((2μ+ν)divup(ρ))=0. (3.3)

    Denote G=(2μ+ν)divup(ρ), and we have again by formula (3.3) that

    G=Δ1div(ρut+ρuuρΦ). (3.4)

    Note that the definition of Qu=Δ1divu, then (3.4) will be reformulated equivalently as

    Q(H)G=0. (3.5)

    By virtue of (3.2) and (3.5), we get

    {μΔPu+G=P(H)+Q(H),div(Pu)=0.

    Note that H=P(H)+Q(H), and from Proposition 2.2 and the Stokes problem (see 2.2 The stokes problem and the operator A of Chapter 2 in Temam [23]), it follows that

    2PuLr(Ω)+GLr(Ω)HLr(Ω),for1<r<. (3.6)

    Observe that H=ρut+ρuuρΦ, then clearly we have

    HLr(Ω)c(ρutLr(Ω)+ρuuLr(Ω)+ρΦLr(Ω)),for1<r<, (3.7)

    where c=c(N), and N denotes the dimension.

    By virtue of Proposition 2.1, we can get

    uLrPuLr+divuLr, for1<r<. (3.8)

    From the Gagliardo-Nirenberg-Sobolev (G-N-S) inequality, we have

    uL3r3ruLr,for1<r<3. (3.9)

    From G=(2μ+ν)divup(ρ), we obtain

    divuLr12μ+ν(GLr+RργLγr). (3.10)

    Combining (3.8)–(3.10), immediately we have

    uL3r3r(Ω)2PuLr(Ω)+12μ+ν(GLr(Ω)+RργL3rγ3r). (3.11)

    Lemma 3.1. Suppose that ρ0L(Ω) and u0H1(Ω), and assume that 2Pu and G is bounded, and ρL(0,T;Lp1(Ω)), where p1=max{p0,39,5γ,40γ19}. Then we could obtain that

    supΩTρ(x,t)<,infΩTρ(x,t)>0.

    Proof. From ρt+div(ρu)=0, we get the following relations

    dρ(x,t)dt=ρ(x,t)divu(x,t).

    We can rewrite it in differential notation and integrate the equality above from 0 to t, then we have

    lnρ(x,t)=lnρ0t0divu(x,s)ds.

    Using the equation G=(2μ+ν)divup(ρ), one deduces that

    lnρ(x,t)=lnρ012μ+νt0(G+Rργ)ds. (3.12)

    Note that G=Δ1div((ρu)t+div(ρuu)ρΦ). Now, considering the term with respect to G of (3.12), we can infer that

    t0Gds=t0Δ1div((ρu)s+div(ρuu)ρΦ)ds=t0(Δ1div((ρu)s)ds+t0Δ1div(div(ρuu))dst0Δ1div(ρΦ)ds.

    Note that

    t0(Δ1div(ρu)s)ds=Δ1div(ρu)Δ1div(ρ0u0)t0uΔ1div(ρu)ds.

    Then, one obtains

    lnρ(t,x)=lnρ012μ+ν(Δ1div(ρu)(x,t)Δ1div(ρ0u0))12μ+νt0Rργds+12μ+νt0Δ1div(ρΦ)ds12μ+νt0(Δ1div2(ρuu)uΔ1div(ρu))dslnρ012μ+ν(t0Rργds+(A0a0)+t0(A1+A2)ds), (3.13)

    where A0=Δ1div(ρu),a0=Δ1div(ρ0u0),A1=Δ1div(ρΦ),A2=Δ1div2(ρuu)uΔ1div(ρu), and div2 is an operator defined by div2M=ijMij for a 3×3 matrix M=(Mij).

    Therefore, we have

    lnρ(x,t)lnρ0L(Ω)+A0a0L(Ω)+t0(A1(,s)L(Ω)+A2(,s)L(Ω))ds, (3.14)
    lnρ(x,t)ln(infΩρ0)t0RργL(Ω)dsA0a0L(Ω)t0(A1(,s)L(Ω)+A2(,s)L(Ω))ds. (3.15)

    Using (3.14), one could obtain that

    lnρ(x,t)lnρ0L(Ω)A0a0L(Ω)+t0(A1(,s)L(Ω)+A2(,s)L(Ω))dsA(t).

    Hence, one yields that

    ρ(t,x)ρ0L(Ω)exp{A}.

    From above inequality, (3.15) implies that

    lnρ(t,x)ln(infΩρ0)At0Rρ0γL(Ω)exp{γA}ds.

    Therefore

    ρ(infΩρ0)exp{At0Rρ0γL(Ω)exp(γA)ds}.

    Obviously, once one proves supt[0,T]A(t)<, we may obtain

    supΩTρ(x,t)<,andinfΩTρ(x,t)>0.

    Therefore, it's necessary to study the estimates of A(t).

    First, observe A0=Δ1div(ρu),A1=Δ1div(ρΦ),A2=Δ1div2(ρuu)uΔ1div(ρu), then we obtain

    ΔA0=div(ρu),ΔA1=div(ρΦ).

    Meanwhile, we have

    ΔA2=div(ΔA0u+ρ(u)u+(uA0)).

    Using the Calderon-Zygmund theorem (Theorem 9.9 of [8]), we get

    A0Lp(Ω)cρuLp(Ω),A1Lp(Ω)cρΦLp(Ω),A2Lp(Ω)c(ρ|u|+|A0|)|u|Lp(Ω),

    where 1<p<, c depends on p. For all vW1,4(Ω), one yields

    vL(Ω)cvL4(Ω).

    Hence, we have

    A0L(Ω)A0L4(Ω)ρuL4(Ω)ρ78L7(Ω)ρ18uL8(Ω). (3.16)

    Applying the same method of estimating A0L(Ω) to estimate A1L(Ω), we obtain that

    A1L(Ω)A1L4(Ω)ρΦL4(Ω)ρL8(Ω)ΦL8(Ω).

    According to ΔΦ=ρ¯ρ and the Calderon-Zygmung inequality [16], we have

    ΦL3β3β(Ω)ρ¯ρLβ,  (β>1).

    Moreover, we estimate that

    A1L(Ω)ρL8(Ω)ρ¯ρL2411(Ω)ρ2L8(Ω)+ρL8(Ω). (3.17)

    Similarly for A2, we could obtain

    A2L(Ω)A2L4(Ω)(ρ|u|+|A0|)|u|L4(Ω)(ρ|u|+|A0|)L20(Ω)uL5(Ω)ρu2L20(Ω)+u2L5(Ω)B21+B22. (3.18)

    Applying Hölder's inequality to B1, we obtain

    B1ρuL20(Ω)ρ3940L39(Ω)ρ140uL40(Ω).

    Similarly, by using the Calderon-Zygmung inequality and Hölder's inequality, one has

    B2PuL5(Ω)+divuL5(Ω)PuL5(Ω)+(GL5(Ω)+RργL5γ(Ω))2PuL158(Ω)+GL158(Ω)+RργL5γ(Ω).

    Hence, we get the estimate of A2 by Young inequality,

    A2L(Ω)ρ3910L39(Ω)+ρ140u4L40(Ω)+2Pu2L158(Ω)+G2L158(Ω)+Rρ2γL5γ(Ω). (3.19)

    Therefore, combining (3.16), (3.17) and (3.19), one could have

    At0ρL8(Ω)ρ¯ρL2411(Ω)ds+t0{ρ140u4L40(Ω)+ρ3910L39(Ω)+2Pu2L158(Ω)+G2L158(Ω)+Rρ2γL5γ(Ω)}ds+ρ78L7(Ω)ρ18uL8(Ω). (3.20)

    Observe Proposition 2.3 and that if 2Pu and G is bounded which we will prove in the Lemma 4.3 of the next section, then we could obtain that supt[0,T]A(t)<.

    Lemma 4.1. Let 1<r<2, ρL(0,T;Lp2(Ω)), and p2(r)=max{p0,p1,5r22r,r2r,4}, then

    T0(2Pu2Lr(Ω)+G2Lr(Ω))dtsupt[0,T]ρLr2r(Ω)ΩTρ|ut|2dxdt+supt[0,T]{ρ5r22rL5r22r(Ω)(Ωρ|u|4r2r)2r2ru2L2(Ω)}+supt[0,T]ρLr(Ω)ρ¯ρL4(Ω).

    Proof. Using (3.6) and (3.7), we obtain

    2PuLr(Ω)+GLr(Ω)c(ρutLr(Ω)+ρuuLr(Ω)+ρΦLr(Ω)),

    where 1<r<, c=c(N) and N is the dimension. For 1<r<2, it follows that

    ρutLr(Ω)ρ12Lr2r(Ω)ρutL2(Ω),ρuuLr(Ω)ρ5r24rL5r22r(Ω)ρ2r4ruL4r2r(Ω)uL2(Ω).

    According to ΔΦ=ρ¯ρ, we obtain

    ρΦLr(Ω)ρLr(Ω)ΦL(Ω)ρLr(Ω)ΔΦL4(Ω)ρLr(Ω)ρ¯ρL4(Ω).

    Hence, if 1<r<2, we obtain

    2Pu2Lr(Ω)+G2Lr(Ω)ρLr2r(Ω)ρut2L2(Ω)+ρ5r22rL5r22r(Ω)ρ2r4ruL4r2r(Ω)u2L2(Ω)+ρ2Lr(Ω)ρ¯ρ2L4(Ω). (4.1)

    Hence, integrating (4.1) over [0,T], we could get

    T0(2Pu2Lr(Ω)+G2Lr(Ω))dtT0{ρLr2r(Ω)ρut2L2(Ω)+ρ5r22rL5r22r(Ω)ρ2r4ru2L4r2r(Ω)u2L2(Ω)+ρ2Lr(Ω)ρ¯ρ2L4(Ω)}dt.

    Lemma 4.2. Let 65<r<2, and p3(r)=max{4γ,(2γ1)r+1r1,6γr65r,3γr3r,32r3,2γ1r1+2γ,3(2γ1)2r3+2γ}. Here we assume ρ0L(0,T;Lp3(Ω)). Then, one chould obtain that

    T0Ωρ|ut|2(t,x)dxdt+μsupt[0,T]Ω|u(t,x)|2dx+1μ2T0Ω(R3(γ1)ρ3γ)dxdt3εT0(2Pu2Lr(Ω)+G2Lr(Ω))dt+C,for any ε>0,

    where C=C(μ,ν,R,u0H1(Ω),ρ0L2γ(Ω),supt[0,T]ρLp3(Ω)).

    Proof. Multiply ut to (1.2)2 and integrate over ΩT, then we could obtain that

    T0Ω|ρu2t|dxdt+μ2supt[0,T]Ω|u|2dx+μ+ν2supt[0,T]Ω|divu|2dx+T0Ωp(ρ)utdxdtT0Ω|ρΦut|dxdt+μ2Ω|u0|2dx+μ+ν2Ω|divu0|2dx+T0Ω|ρ(uu)ut|dxdtT0Ω|ρΦut|dxdt+μ2Ω|u0|2dx+μ+ν2Ω|divu0|2dx+12T0Ω|ρu2t|dxdt+12T0Ω|ρu2(u)2|dxdt. (4.2)

    Therefore, we have

    T0Ω|ρu2t|dxdt+μsupt[0,T]Ω|u|2dx+(μ+ν)supt[0,T]Ω|divu|2dx+2T0Ωp(ρ)utdxdt2T0Ω|ρΦut|dxdt+μΩ|u0|2dx+(μ+ν)Ω|divu0|2dx+T0Ω|ρu2|u|2|dxdt. (4.3)

    Set

    I=Ωp(ρ)utdx,J=ΩρΦutdx.

    Noting that ρt+div(ρu)=0, one obtains that

    (Rργ)t=Rγργ1ρt=Rγργ1div(ρu)=u(Rργ)Rγργdivu.

    Note that G=(2μ+ν)divup(ρ), we deduce that

    I=Ω(Rργ)utdx=ddtΩRργdivudx+Ω(Rργ)tdivudx. (4.4)

    Further, one obtains

    Ω(Rργ)tdivudx=Ωu(Rργ)divuΩRγργ(divu)2dx=Ω(u(Rργ)divu+Rργ(divu)2)dxR(γ1)Ωργ(divu)2dx=Ωdiv(Rργu)divudxR(γ1)Ωργ(divu)2dx=R2μ+νΩργu(G+Rργ)dxR(γ1)(2μ+ν)2Ωργ(G2R2ρ2γ+2(2μ+ν)Rργdivu)dx=R2μ+νΩργuGdx+R2μ+νΩργu(Rργ)dxR(γ1)(2μ+ν)2ΩργG2dx+R3(γ1)(2μ+ν)2Ωργρ2γdx2R2(γ1)2μ+νΩρ2γdivudx=R2μ+νΩργuGdxR22(2μ+ν)Ωρ2γdivudxR(γ1)(2μ+ν)2ΩργG2dx+R3(γ1)(2μ+ν)2Ωρ3γdx2R2(γ1)2μ+νΩρ2γdivudx=R2μ+νΩργuGdxR2(4γ3)2(2μ+ν)Ωρ2γdivudxR(γ1)(2μ+ν)2ΩργG2dx+R3(γ1)(2μ+ν)2Ωρ3γdx.

    Hence, we have

    I=ddtΩRργdivudx+12μ+νΩRργuGdx4γ32(2μ+ν)ΩR2ρ2γdivudx1(2μ+ν)2ΩR(γ1)ργG2dx+1(2μ+ν)2ΩR3(γ1)ρ3γdx.

    Integrate I over [0,T], then one could obtain

    T0I(t)dt=ΩRργdivu(T,x)dx+ΩRργ0divu0dx+12μ+νT0Ω(RργuG)dxdt4γ32(2μ+ν)T0Ω(R2ρ2γdivu)dxdt1(2μ+ν)2T0Ω(R(γ1)ργG2)dxdt+1(2μ+ν)2T0Ω(R3(γ1)ρ3γ)dxdt. (4.5)

    Observing that ΔΦ=ρˉρ, estimate the term J, then one obtains

    JΦL(Ω)ρutL2(Ω)ρL2(Ω)ρ¯ρL4(Ω)ρutL2(Ω)ρL2(Ω)14ερ¯ρ4L4(Ω)+ε2ρut2L2(Ω)+14ερ4L2(Ω).

    Hence, we obtain

    T0J(t)dt14εT0{ρ¯ρ4L4(Ω)+ρ2L1(Ω)}dt+ε2T0Ωρu2tdxdt. (4.6)

    Combining (4.5) and (4.6), (4.3) can be written as

    T0Ωρ|ut|2dxdt+supt[0,T]Ω|u|2dx+supt[0,T]Ω|divu|2dx+T0Ω(R3(γ1)ρ3γ)dxdtΩ|u0|2dx+Ω|divu0|2dx+Ω|Rργ0divu0|dx+supt[0,T]ΩR|ργdivu|dx+T0Ω|RργuG|dxdt+T0Ω|R(γ1)ργG2|dxdt+T0Ω|ρu2(u)2|dxdt+T0Ω|R2ρ2γdivu|dxdt+T0{ρ¯ρ4L4(Ω)+ρ2L1(Ω)}dt9k=1ik. (4.7)

    In what follows, estimate i1i9. First, based on Cauchy's inequality, we have

    i4supt[0,T]Ω|Rργdivu|dxc(ε)R2supt[0,T]ρ2γL2γ+εsupt[0,T]Ω|divu|2dx.

    Similarly, we obtain

    i5T0Ω|RργuG|dxdtT0(GLr(Ω)RργuLrr1(Ω))dtεT0G2Lr(Ω)dt+14εR2supt[0,T]((Ωρ|u|2rr1dx)r1rρ(2γ1)r+1rL(2γ1)r+1r1)T.

    Observe that ρ2γL2γ(Ω)ρ2γL6γr65r(Ω), and make use of the same method as i4, then one obtains that

    i6T0Ω|(R(γ1)ργG2)|dxdtT0Ω(εG2+14εR2(γ1)2ρ2γG2)dxdtεT0G2L2(Ω)dt+14εR2(γ1)2supt[0,T]ρ2γL6γr65rT0G2L2(Ω)dt.

    Note that divuLq12μ+ν(GLq+R||ρ||γLγq) on account of G=(2μ+ν)divup(ρ). Make use of (3.11), Cauchy inequality and the interpolation inequality, then one yields

    i7T0u2L2r3r(Ω)ρu2L2r2r3(Ω)dtT0u32L3r3r(Ω)u12Lr3r(Ω)ρu2L2r2r3(Ω)dt3ε4T0u2L3r3r(Ω)dt+14ε3T0ρu8L2r2r3(Ω)u2Lr3r(Ω)dt3ε4T0(2Pu2Lr(Ω)+12μ+ν(G2Lr(Ω)+Rρ2γL3rγ3r(Ω)))dt+18ε3T0(ρu16L2r2r3(Ω)+u4Lr3r(Ω))dt3ε4T0(2Pu2L2(Ω)+12μ+ν(G2L2(Ω)+Rρ2γL3rγ3r(Ω)))dt+18ε3T0((Ωρu4r2r3dx)4(2r3)rρ12rL32r3(Ω)+u4L2(Ω))dt,

    where 32<r<2. Similarly, we have

    i84γ3μT0Ω|(R2ρ2γdivu)|dxdt4γ3μR2(T0Ω|divu|2dxdt)12(T0Ωρ4γdxdt)124γ3μR2(εT0Ω|u|2dxdt+1εT0Ωρ4γdxdt).

    Combining i1i9 with (4.7), we arrive at the conclusion of the lemma easily.

    Combining Lemma 4.1 with Lemma 4.2, let ε be small sufficiently, then the following lemma is obtained naturally.

    Lemma 4.3. Suppose that ρ0 and u0 satisfy the assumptions of Lemma 4.2. Assume ρL(0,T;Lp4(Ω)), then we obtain that

    ΩTρu2tdxdt+sup0<≤tTΩ|u|2dxC,T02Pu2Lr(Ω)+G2Lr(Ω)dtC,

    where C=C(μ,R,T,u0H1(Ω),ρ0L2γ(Ω),supt[0,T]ρLp4(Ω)), and p4(r)=max{p0(r),p1(r),p2(r),p3(r)}.

    Moreover, combining the results of the Lemma 4.3 with (3.20), we could obtain sup0<t<TA(t)< easily, which is obvious that we prove that Lemma 3.1 completely.

    In the following, we shall present the L bound of u by the Moser iteration [4,10] provided that the result of the Lemma 3.1 holds on. Furthermore, we could get the local L estimate of u by Gronwall inequality.

    Lemma 5.1. Let all conditions satisfy the assumptions of Proposition 2.3 and Lemma 3.1, then we have

    sup(t,x)ΩT|u(x,t)|C,

    where the constant C=C(T,supΩTρ(x,t),infΩTρ(x,t),u0L(Ω))>0.

    Proof. Integrating the equality (2.1) on (0,T), one could arrive at

    1α+2sup0tTΩρ|u|α+2dx+μ(α+1)T0Ω|u|2|u|αdxdt+α(μ+ν)T0Ω|u|αdivuudxdt+(μ+ν)T0Ω|divu|2|u|αdxdt1α+2Ωρ0|u0|α+2dx+T0ΩRργ|u|αdivudxdt+αT0ΩRργ|u|αudxdt+T0ΩρΦ|u|αudxdt1α+2Ωρ0|u0|α+2dx+T0(j1+j2+j3)(t)dt. (5.1)

    Now, we consider the estimates of the right terms of (5.1).

    T0j1(t)dtc1T0Ω|u|α|divu|dxdtc1T0|u|α2divuL2(Ω)|u|α2L2(Ω)dtc1T0|u|α2divuL2(Ω)uα2Lα+2(Ω)dt12ΩT|u|α|u|2dxdt+c212T0uαLα+2(Ω)dt12ΩT|u|α|u|2dxdt+c212(T0uα+2Lα+2(Ω)dt)αα+2T2α+2=12ΩT|u|α|u|2dxdt+c212(ΩT|u|α+2dxdt)αα+2T2α+2,

    where c1=p(ˆρ) and ˆρ=supΩTρ(t,x). Similarly, we have

    T0j2(t)dt12ΩT|u|α|u|2dxdt+c212(ΩT|u|α+2dxdt)αα+2T2α+2.

    Note that ΔΦ=ρ¯ρ and (2.4), then we have

    T0j3(t)dtˆρΩT|Φdiv(|u|αu)|dxdt(ˆρ2+ˆρ)T0Ω(|u|α|divu|+|u|α|u|)dxdt12ΩT|u|α|u|2dxdt+ˆρ4+ˆρ22(ΩT|u|α+2dxdt)αα+2T2α+2.

    Hence, one could get

    ˜ρα+2supt[0,T]Ω|u|α+2dx+c(μ,ν,α)ΩT(|u|2|u|α)dxdtMα+2u0α+2L(Ω)+(p2(ˆρ)+ˆρ2+ˆρ4)(ΩT|u|α+2dxdt)αα+2T2α+2, (5.2)

    where ˜ρ=infΩTρ(x,t). Observe

    (|a|+|b|)p{|a|p+|b|p,0p1,2p1(|a|p+|b|p),p>1.

    Hence, we could obtain that

    ΩT|u(t,x)|53(α+2)dxdtT0u2(α+2)3L32(Ω)uα+2L3(Ω)dt=T0(Ω|u|α+2dx)23uα+222L6(Ω)dt(supt[0,T]Ω|u|α+2dx)23T0uα+222H1(Ω)dt(supt[0,T]Ω|u|α+2dx)23(ΩT(|u|α+22)2dxdt+ΩT|u|α+2dxdt)(supt[0,T]Ω|u|α+2dx)23((α+2)24ΩT|u|α|u|2dxdt+ΩT|u|α+2dxdt). (5.3)

    Combining (5.2) and (5.3), we could get

    ΩT|u(t,x)|53(α+2)dxdtC(α+2)83(ΩT|u|α+2dxdt)53+C(α+2)83, (5.4)

    where C=C(T,M,supΩTρ,infΩTρ,u0L(Ω)).

    Let r=53, rk=2+αk, where  k2, then the inequality (5.3) is rearranged to be

    ΩT|u|rk+1dxdtC1rkC2(ΩT|u|rkdxdt)r+C1rkC2, (5.5)

    where C1=C1(T,M,supΩTρ,infΩTρ,u0L(Ω))>1, and C2=C2(T,M,supΩTρ,infΩTρ,u0L(Ω))>1.

    Make use of the inequality (5.5), then we obtain

    ΩT|u(t,x)|53(αk+2)dxdt=ΩT|u|rk+1dxdtC1rkC2(ΩT|u|rkdxdt)r+C1rkC22r1C1rkC2[C1rr(k1)C2(ΩT|u|rk1dxdt)r2+C1rr(k1)C2]+C1rkC2=223C1+r1rC2(k+r(k1))(ΩT|u|rk1dxdt)r2+223C1+r1rC2(k+r(k1))+C1rkC2223C1+r1rC2(k+r(k1))[C1r(k2)C2(ΩT|u|rk2dxdt)r+C1r(k2)C2]r3+223C1+r1rC2(k+r(k1))+C1rkC22(r1)+(r21)++(rk11)C1+r++rk21rC2(k+r(k1)++rk22)(ΩT|u|r2dxdt)rk1+2(r1)+(r21)++(rk11)C1+r+r2++rk21rC2(k+r(k1)++rk22)+2(r1)+(r21)++(rk21)C1+r+r2++rk31rC2(k+r(k1)++rk33)++C1rC2k.

    Next, we will consider the following inequality

    ΩT|u|rk+12(r1)+(r21)++(rk11)Ck2l=0rl1rC2k2l=0(kl)rl(ΩT|u|r2dxdt)rk1+(k1)2(r1)+(r21)++(rk11)Ck2l=0rl1rC2k2l=0(kl)rl.

    Denote a=2(r1)+(r21)++(rk11), then we have the inequality

    uLrk+1(ΩT)a1rk+1Ck2l=0rlk11rC2k2l=0(kl)rlk1(ΩT|u|r2dxdt)r2+(k1)r(k+1)a1rk+1Ck2l=0rlk11rC2k2l=0(kl)rlk1. (5.6)

    Observe that b=k1l=1rlk2<+, c=k1l=1(kl)rlk2<+, d=(k1)r(k+1)<+. Moreover, we have 2((r1)++(rk21))r(k+1)<.

    Obviously when k goes to , we could obtain that

    u(x,t)L(ΩT)2Cb1rcC2uLr2(ΩT)+2dCb1rcC2.

    Hence, we obtain the inequality

    supΩT|u|2Cb1rcC2uLr2(ΩT)+2dCb1rcC2.

    Combining the above result and Proposition 2.3, we could obtain that

    supΩT|u(x,t)|C,

    where C=C(T,supΩTρ,infΩTρ,u0L(Ω)).

    Finally, let us give the proof of Theorem 1.1.

    Proof. Under the assumptions of Lemmas 4.1–4.3, we employ the result in Lemma 3.1, Proposition 2.3 and the a prior estimates in Lemma 5.1. Therefore, we could achieve the results of Theorem 1.1 easily.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by the National Natural Science Foundation of China (42274166), the Fundamental Research Funds for the Central Universities (3132021195, 3132022202), Basic Scientific Research Project of The Educational Department of Liaoning Province (LJKMZ20220368).

    The authors declare that they have no conflict of interest.



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