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A priori estimates for the free boundary problem of incompressible inviscid Boussinesq and MHD-Boussinesq equations without heat diffusion

  • For all physical spatial dimensions n=2 and 3, we establish a priori estimates of Sobolev norms for free boundary problem of inviscid Boussinesq and MHD-Boussinesq equations without heat diffusion under the Taylor-type sign condition on the initial free boundary. It is different from MHD equations because the energy of the system is not conserved.

    Citation: Wei Zhang. A priori estimates for the free boundary problem of incompressible inviscid Boussinesq and MHD-Boussinesq equations without heat diffusion[J]. AIMS Mathematics, 2023, 8(3): 6074-6094. doi: 10.3934/math.2023307

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  • For all physical spatial dimensions n=2 and 3, we establish a priori estimates of Sobolev norms for free boundary problem of inviscid Boussinesq and MHD-Boussinesq equations without heat diffusion under the Taylor-type sign condition on the initial free boundary. It is different from MHD equations because the energy of the system is not conserved.



    The Boussinesq equations for magnetohydrodynamic convection (Boussinesq-MHD) model are of relevance to study a number of models coming from atmospheric or oceanographic turbulence. Let us consider the following inviscid MHD-Boussinesq equations without heat diffusion in D:

    {tv+vv+p=bb+hen,th+vh=0,tb+vb=bv,u=0,b=0, (1.1)

    where v, b, p and h denote the velocity, the magnetic field, the fluid pressure and the temperature, respectively; D0tT{t}×Rn is an unknowns to be determined for n{2,3}. We want to find a set D, v, h and b solving (1.1) and satisfying the initial conditions:

    {x:(0,x)D}=D0,(v,b,h)|t=0=(v0(x),b0(x),h0(x)) for xD0. (1.2)

    Let Dt={xRn:(t,x)D}, then the free boundary conditions should be

    {vN=κ, on Dt,p=0, on Dt,bN=0, on Dt, (1.3)

    for each t[0,T], where N is the exterior unit normal to Dt, vN=Nivi, κ is the normal velocity of Dt. To get a priori bounds for (1.1) and (1.3) in Sobolev spaces, we need assume:

    Npε<0 on Dt, (1.4)

    where ε is positive constant. On the other hand, the condition (1.4) can hold initially and it will hold true within some time. In fact, the pressure is larger in the interior than on the boundary. Moreover, (1.4) is called Taylor sign condition for the Euler Eq (1.4) also plays the same role, for the Boussinesq equations.

    The free boundary problems of incompressible Euler equations and ideal incompressible MHD have been studied by many people in recent decades. Wu [20,21] first made great contributions and proved well-posedness for incompressible and irrotational water wave problems. On the motion of the free surface of a liquid, Christodoulou-Lindblad [8] obtaind a priori energy estimate. In [9], Gu and Wang had benn proved local well-posedness of free-surface incompressible ideal magnetohydrodynamic equation. Hao and Luo obtain well-posedness for the linearized free boundary problem of incompressible ideal magnetohydrodynamics equations in [10]. More works for flows has been extensively studied in [13,14,16].

    On the other hand, there have been some results for Boussinesq equations. Chae and Nam [6] proved local existence and blow-up criterion for the Boussinesq equations, and established [7] for initial data in Hölder spaces Cr with r>1. Another local well-posedness was recently obtained [17] for the critical Besov spaces B2/p+1p,1. Numerical investigations on similar data for the 2-D inviscid Boussinesq equations appeared to indicate that there is no finite time singularity formation [19]. Miao and Zheng [18] obtained the global well-posedness for the Boussinesq system with horizontal dissipation. An analytical work on the inviscid Boussinesq equation can be found in [5]. The global well-posedness for weak or strong solutions was obtained by the stability and instability for a fully nonlinear 2-D MHD-Boussinesq equations in [1,2,3]. Liu, Bian and Pu [4] obtained global well-posedness of the 3D Boussinesq-MHD system without heat diffusion. There are few results about two-phase fluid motion in the Oberbeck-Boussinesq approximation. Hao and Zhang established the maximal LpLq regularity for the two-phase fluid motion of the linearized Oberbeck-Boussinesq approximation in [12]. In recently, they obtained Local well-posedness for two-phase fluid motion in the Oberbeck-Boussinesq approximation.

    In this paper, we adopt a geometrical point of view used in [8,11], and estimate quantities such as the second fundamental form. We denote the material derivative by Dt=t+vkk, then the system (1.1) can be rewritten as:

    Dtvj+jp=bkkbj+δjnh, in D,Dtbj=bkkvj, in D,Dth=0, in D,jvj=0,jbj=0, in D,vN=κ,bjNj=0, on [0,T]×Dt,p=0, on D,Np<0, on {t=0}×D0, (1.5)

    where δij is the Kronecker delta symbol such that δii=1 and δij=0 for ij.

    Remark 1.1. Different from the fixed boundary problem, in this paper, we do not need to assume the condition of temperature on the boundary in (1.5). In the following proof, we can find that the higher order energy of temperature is actually controlled by its initial data, and is not affected by the boundary condition.

    Remark 1.2. Owe to Dth=0, the first equation of Eq (1.5) can transform to Dtvj+jp=bkkbj+δjnh0. Thus, let the zero order energy be

    E0(t)=12Dt(|v(t,x)|2+|b(t,x)|2)dx. (1.6)

    A direct computation yields

    ddtE0(t)0.

    Different from MHD equations, the energy of the system is not conserved, but it can be controlled by the initial data and time T.

    Let's introduce some knowledge of Riemannian geometry such as second fundamental form of the free surface and tensor products by [8]. In order to get energies, we introduce orthogonal projection Π to the tangent space of the boundary of a (0, r) tensor α is defined to be the projection of each component along the normal:

    (Πα)i1ir=Πj1i1Πjrirαj1jr, where Πji=δjiNiNj.

    Let ˉi=Πjij be a tangential derivative. In fact, we assume p=0 on Dt, obviously, it follows that ˉip=0 and

    (Π2p)ij=θijNp, (1.7)

    where θij=ˉiNj is the second fundamental form of Dt. In fact,

    0=ˉiˉjp=ΠiiiΠjjjp=ΠiiΠjjijp(ˉiNj)NkkpNj(ˉiNk)kp=(Π2p)ijθijNp.

    Next, we define the quadratic form Q of the form:

    Q(α,β)=Πα,Πβ=qi1j1qirjrαi1irβj1jr,

    where

    qij=δijη(d)2NiNj,d(x)=dist(x,Dt),Ni=δijjd,

    where η is a smooth cutoff function satisfying 0η(d)1, η(d)=1 when d<d0/4, and η(d)=0 when d>d0/2. d0 is a fixed number that is smaller than the injectivity radius of the normal exponential map ς0, defined to be the largest number ς0 such that the map

    Dt×(ς0,ς0){xRn:dist(x,Dt)<ς0},

    given by

    (ˉx,ς)x=ˉx+ςN(ˉx),

    is an injection. Then higher energies for r1 can be denoted by

    Er(t)=Dtδij(Q(rvi,rvj)+Q(rbi,rbj))dx+Dt(|r1curlv|2+|r1curlb|2)dx+sgn(r1)DtQ(rp,rp)ϑdS, (1.8)

    where

    ϑ=(Np)1.

    In this paper, we prove the following main theorem.

    Theorem 1.1. Let

    H(0)=max(θ(0,)L(D0),1/ς0(0)),I(0)=1/(Np(0,))L(D0)=1/ε(0)>0. (1.9)

    There exists a continuous function K>0 such that if

    TK(H(0),I(0),E0(0),Θ02L2(Ω),En+1(0),n+1Θ02L2(Ω),VolD0), (1.10)

    then any smooth solution of the free boundary problem for inviscid MHD-Boussinesq Eq (1.5) without heat diffusion satisfies

    n+1s=0Es(t)2n+1s=0(Es(0)+sΘ02L2(Ω)),0tT. (1.11)

    In order to prove Theorem 1.1, we consider the Boussinesq equations with zero viscosity and diffusivity in D:

    {vt+(v)v=p+hen,ht+(v)h=0,v=0, (1.12)

    with the following conditions on the free boundary

    {vN=κ, on Dt,p=0, on Dt. (1.13)

    We will prove a priori bounds for (1.12) and (1.13) in Sobolev spaces under the assumption (1.4).

    Let us now outline the proof of Theorem 1.1. Firstly, for the Boussinesq equations, we transform the free boundary problem to a fixed initial boundary problem in the Lagrangian coordinates in Section 2. In Section 3, we prove the zero order and the first order energy estimates. Section 4 is devoted to the higher order energy estimates by using the identities derived in Section 2, then, for the MHD-Boussinesq equations, we can get a similar conclusion in Section 5. Finally, we justify the a priori assumptions in Section 6.

    As we all know, the method to deal with the free boundary problem is to transform the unknown region into a fixed region. We can introduce Lagrangian coordinates or co-moving coordinates to transform the free boundary problem to a fixed boundary problem. Let f0:ΩD0 is a diffeomorphism, where Ω be bounded domain in Rn. Then the Lagrangian coordinates in (t,y) where x=x(t,y)=ft(y) are given by solving

    dxdt=v(t,x(t,y)),x(0,y)=f0(y),yΩ. (2.1)

    The Euclidean metric δij in Dt then induces a metric

    gab(t,y)=δijxiyaxjyb, (2.2)

    and its inverse

    gcd(t,y)=δklycxkydxl, (2.3)

    in Ω for each fixed t. Furthermore, expressed in the y-coordinates, we have

    i=xi=yaxiya. (2.4)

    Let us introduce the notation for the material derivative

    Dt=t|y= const =t|x= const +vkxk.

    Let u(t,y), Θ(t, y), P(t,y) represent the velocity, temperature, pressure in the Lagrangian coordinates, respectively, then from ([15], Lemma 2.1) and (1.12), we can obtain

    Dtua=xjya(jp+δnjh)+vjxkyavjxk=aP+δnaΘ+ucauc. (2.5)

    Obviously, because the temperature Θ is a scalar, so we directly obtain

    DtΘ=0. (2.6)

    Thus, the temperature can be directly expressed by the initial temperature in Lagrangian coordinates, then the system (1.12) and (1.13) can be rewritten in the Lagrangian coordinates as

    Dtua+aP=δnaΘ0+ucauc, in [0,T]×Ω,aua=0, in [0,T]×Ω,P=0,uN=κ, on [0,T]×Ω. (2.7)

    where Θ|t=0=Θ0.

    Firstly, we need to define the zero order energy as

    E0(t)=12Ω(|u(t,y)|2)dy. (3.1)

    Owe to ([15], Lemma 2.1) and (2.7), Gauss' formula and Dtdμg=0, we can get

    ddtE0(t)=12ΩDt(gabuaub)dμg=Ω(uaDtua)dμg+Ω12(Dtgab)(uaub)dμg=Ω[uaaP+uaδnaΘ0+uaucauc]dμgΩhab(uaub)dμg=ΩNauaPdμγ+ΩuaδnaΘ0dμg+Ωuaucaucdμg12Ωgac(cud+duc)gdbuaubdμg=ΩuaδnaΘ0dμg,

    then, using the Hölder inequality

    ddtE0(t)CuL2(Ω)Θ0L2(Ω)C(E0(t)+Θ02L2(Ω)). (3.2)

    when t[0,T] with a constant T>0, by the Gronwall inequality, it follows that

    E0(t)C(T)(E0(0)+Θ02L2(Ω))C(E0(0)+Θ02L2(Ω)). (3.3)

    Zero order energy is controlled by initial value and time T. By ([11], Lemma 2.3), (2.5) and (2.7), we can get

    Dt(bua)+baP=[Dt,b]ua+bDtua+baP=(abud)ud+bδnaΘ0+b(ucauc)=(abud)ud+δnabΘ0+bucauc+ucbauc=bucauc+δnabΘ0. (3.4)

    Then by (2.6) and ([11], Lemma 2.3), we have that

    Dt(Θ)=[Dt,]Θ+DtΘ=0. (3.5)

    In fact, the estimates of Θ and Θ are only related to their initial data. The difficulty is that the velocity term will appear in the higher derivatives, but we can still control them.

    Now, we can calculate the first order energy estimates. By (3.4), ([15], Lemma 2.1) and ([11], (A.13)), we can get the material derivative of gbdγaeaubeud,

    Dt(gbdγaeaubeud)=(Dtgbd)γaeaubeud+gbd(Dtγae)aubeud+2gbdγae(Dtaub)eud=2gbchcfgfdγaeaubeud2gbdγachcfγfeaubeud2gbdγaeeudabP+2gbdγaeeudδnbaΘ0+2gbdγaeeudaucbuc=γae(cuf+fuc)auceuf2γacγfe(cuf+fuc)audeud2γaeeubabP+2γaeeubδnbaΘ0+2γaeeubaucbuc=2γaecufauceuf4γaeγfceufaudcud+2γaeeubaucbuc2γaeeubabP+2γaeeubδnbaΘ0=4γaeγfceufaudcud2γaeeubabP+2γaeeubδbnaΘ0=4γaeγfceufaudcud+2γaeeubδnbaΘ02b(γaeeubaP)+2(bγae)(eubaP). (3.6)

    Similarly, we can easily calculate the material derivative of |curlu|2. It follows that

    Dt|curlu|2=Dt(gacgbd(curlu)ab(curlu)cd)=2(Dtgac)gbd(curlu)ab(curlu)cd+4gacgbd(Dtaub)(curlu)cd=2gaegfcgbd(euf+fue)(curlu)ab(curlu)cd+4gacgbd(curlu)cdauebue4gacgbd(curlu)cdabP+4gacgbd(curlu)cdδnbaΘ0=4gaegbdeuc(curlu)ab(curlu)cd+4gacgbd(curlu)cdδnbaΘ0. (3.7)

    Now, we can define the first order energy as

    E1(t)=Ωgbdγaeaubeuddμg+Ω|curlu|2dμg. (3.8)

    Finally, we have the following theorem.

    Theorem 3.1. For any smooth solution of system (2.7) satisfying the following assumptions

    |P|M,|u|M,in[0,T]×Ω,|θ|+|u|+1ς0K,on[0,T]×Ω,

    when t [0, T], we can get

    E1(t)2eCMt(E1(0)+Θ02L2(Ω))+CK2(VolΩ)(eCMt1). (3.9)

    Proof.

    ddtE1(t)=ΩDt(gbdγaeaubeud)dμg+ΩDt(|curlu|2)dμg.

    From (3.6)–(3.8), ([15], Lemma 2.1) and Gauss' formula, it yield to

    ddtE1(t)=Ω(4γaeγfceufaudcud+2γaeeubδnbaΘ0)dμg+2Ω(bγae)(eubaP)dμg4Ωgaegbdeuc(curlu)ab(curlu)cddμg+4Ωgacgbd(curlu)cdδnbaΘ0dμg2ΩNb(γaeeubaP)dμγ.

    In fact the integral on the boundary is zero. Since P=0 on Ω, we have ˉP=γaeaP=0.

    Then, by ([11], (A.3) and (A.5)) and the fact NN=0. we obtain that

    θab=(δcaNaNc)cNb=aNbNaNNb=aNb,

    Thus, we have

    bγae=b(gaeNaNe)=b(NaNe)=(bNa)Ne(bNe)Na=θabNeθebNa.

    From the Hölder inequality and ([11], (A.5)), it follows that

    ddtE1(t)CKM(VolΩ)1/2E1/21(t)+CM(VolΩ)1/2Θ0L2(Ω)+CuL(Ω)(u2L2(Ω)+curlu2L2(Ω))CKM(VolΩ)1/2(E1(t)+Θ02L2(Ω))1/2+CME1(t).

    By the Gronwall inequality, it yields the desired estimate.

    Remark 3.1. In fact, we can find that the integral involving P is zero, so pressure does not affect boundary integral in E1. But for the higher order estimates, we have to introduce boundary integrals for P.

    In this section, we will get the higher order energy estimates. By ([11], Lemma 2.2) and (1.12), we obtain that

    Dtrua=Dta1arua=Dt(xi1ya1xiryarxiyai1irvi)=raPr1s=1(rs+1)(1+su)rsua+δnarΘ0+aucruc,

    when r2, by moving terms, we have

    Dtrua+raP=(curlu)acruc+sgn(2r)r2s=1(rs+1)(1+su)rsua+δnarΘ0. (4.1)

    Define the r-th order energy for r2 as

    Er(t)=ΩgbdγafγAFr1Aaubr1Ffuddμg+Ω|r1curlu|2dμg+ΩγafγAFr1AaPr1FfPϑdμγ, (4.2)

    where ϑ=1/(NP), then we can get the following estimates.

    Theorem 4.1. For integer 2rn+1, there exists a constant T>0 such that, for any smooth solution to system (2.7) for 0tT satisfying

    |P|M,|u|M,|Θ0|M,in[0,T]×Ω,|θ|+1/ς0K,on[0,T]×Ω,NPε>0,on[0,T]×Ω,|2P|+|NDtP|L,on[0,T]×Ω, (4.3)

    when t[0,T], we obtain

    Er(t)eC1t(Er(0)+rΘ02L2(Ω))+C2(eC1t1), (4.4)

    where the constants C1 and C2 depend on K,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω),E1(0),, Er1(0) and r1Θ02L2(Ω).

    Proof. Appling (4.2), the derivative of Er with respect to t is

    ddtEr(t)=ΩDt(gbdγafγAFr1Aaubr1Ffud)dμg+ΩDt|r1curlu|2dμg+ΩDt(γafγAFr1AaPr1FfP)ϑdμγ+ΩγafγAFr1AaPr1FfP(ϑtϑhNN)ϑdμγ. (4.5)

    From ([15], Lemma 2.1) and (4.1), we directly have

    Dt(gbdγafγAFr1Aaubr1Ffud)=2cueγafγAFr1Aaucr1Ffue4rcueγacγefγAFr1Aaudr1Ffud2γafγAFr1Ffubr1AabP+2γafγAFr1Ffub(curlu)bcr1Aauc+2sgn(2r)γafγAFr1Ffudr2s=1(rs+1)((s+1u)rsud)Aa+2γafγAFr1Ffubδnbr1AaΘ0.

    and

    Dt(γafγAFr1AaPr1FfP)=2rcueγacγefγAFr1AaPr1FfP+2γafγAFr1AaPDt(r1FfP).

    Next we will deal with the integration of the higher derivatives of P on the boundary. It is the difficulty in this paper. By the Hölder inequality, we get

    ΩDt(gbdγafγAFr1Aaubr1Ffud)dμg+ΩDt(γafγAFr1AaPr1FfP)ϑdμγCuL(Ω)Er(t)+CEr(t)+rΘ02L2(Ω)+CE1/2r(t)r2s=1s+1uL4(Ω)rsuL4(Ω)+2ΩγafγAFrAaP(DtrFfP1ϑNbrFfub)ϑdμγ+2Ωb(γafγAF)r1Ffubr1AaPdμg. (4.6)

    Then, using the Hölder inequality, it yield that

    2Ωb(γafγAF)r1Ffubr1AaPdμgCKE1/2r(t)rPL2(Ω). (4.7)

    Obviously, we have to estimate rPL2(Ω). Firstly, we need to take divergence on the first equation of (2.7), by ([11], Lemma 2.2), we nave

    ΔP=aubbua+nΘ0. (4.8)

    For r2, we have

    r2ΔP=r2s=0(r2s)saubr2sbuar2nΘ0.

    Then by ([11], (A.12)), for ς11/K1, it follows that

    uL(Ω)C2s=0Kn/2s1suL2(Ω)C(K1)2s=0E1/2s(t). (4.9)

    In view of (4.9), for s0, similarly, we can directly get taht

    suL(Ω)C2=0Kn/21+suL2(Ω)C(K1)2=0E1/2s+(t), (4.10)

    When r=3,4, By the Hölder inequality, (4.9) and (4.10), we have

    r2ΔPL2(Ω)Cr2s=0saubr2sbuaL2(Ω)+Cr2nΘ0L2(Ω)CuL(Ω)r1uL2(Ω)+(r3)C2uL(Ω)2uL2(Ω)+Cr1Θ0L2(Ω)C(K1)r1=1E(t)+C(K1)E1/22(t)E1/2r(t)+C(K1)r1Θ0L2(Ω)C(K1)r1=0(E(t)+Θ02L2(Ω))+C(K1)E1/22(t)E1/2r(t). (4.11)

    The last inequality is due to the inequality r1Θ0L2(Ω)CE0+Cr1Θ02L2(Ω). For r=2, we have the following estimate from the assumption of (4.3) and the Hölder inequality, i.e.,

    ΔPL2(Ω)CM(E1(t)+Θ02L2(Ω))1/2, (4.12)

    which is a lower order energy term. Thus, from ([8], (A.17)), (4.11) and (4.12), we get for any δr>0

    rPL2(Ω)δrΠrPL2(Ω)+C(1/δr,K,VolΩ)sr2sΔPL2(Ω)δrΠrPL2(Ω)+C(1/δr,K,K1,M,VolΩ)r1=0(E(t)+Θ02L2(Ω))+(r2)C(1/δr,K,K1,M,VolΩ)(E1/22(t)E1/2r(t)). (4.13)

    Due to ([11], (A.18)) and P=0 on Ω, we have for r1,

    ΠrPL2(Ω)C(K,K1)(θL(Ω)+(r2)kr3ˉkθL2(Ω))kr1kPL2(Ω). (4.14)

    By ([11], (A.7)), we obtain that Π2P=θNP, then from (4.3), we get

    θL2(Ω)=Π2PNPL2(Ω)1εΠ2PL2(Ω). (4.15)

    Then, for r=2,3,4, we need to estimate ΠrPL2(Ω) and rPL2(Ω).

    {{When r = 2}}, by using (4.13) and (4.14), we have

    Π2PL2(Ω)θL(Ω)PL2(Ω)C(K,VolΩ)(2PL2(Ω)+PL2(Ω))C(K,VolΩ)δ2Π2PL2(Ω)+C(K,VolΩ)(VolΩ)1/2M+C(1/δ2,K,K1,M,VolΩ)(1=0(E(t)+Θ02L2(Ω))).

    We can take δ2 so small that the first term can be absorbed by the left-hand side. Thus

    Π2PL2(Ω),2PL2(Ω)C(K,K1,M,VolΩ)(1+1=0(E(t)+Θ02L2(Ω))), (4.16)
    θL2(Ω)C(K,K1,M,VolΩ,1/ε)(1+1=0(E(t)+Θ02L2(Ω))). (4.17)

    From Theorem 3.1 and zero order energy estimate, there exists a constant T>0 such that Ei(t)CEi(0)+iΘ02L2(Ω) for t[0,T] and i=0,1.

    {When r = 3}, by (4.3), (4.14), (4.16) and (4.17), we have

    Π3PL2(Ω)C(K,K1)(K+θL2(Ω))k2kPL2(Ω)C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))3PL2(Ω)+C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω)), (4.18)

    and by (4.13), it follows that

    3PL2(Ω)δ3C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))3PL2(Ω)+δ3C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))+C(1/δ3,K,K1,M,VolΩ)(2=0(E(t)+Θ02L2(Ω)))+C(1/δ3,K,K1,M,VolΩ)(E1/22(t)E1/23(t)). (4.19)

    Then, we can choose a sufficiently small δ3>0, and by (4.18) and (4.19), it yields that

    3PL2(Ω),Π3PL2(Ω)C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))×(1+2=0(E(t)+Θ02L2(Ω))+E1/22(t)E1/23(t)). (4.20)

    {When r = 4}, since

    ˉbNP=γdbd(NaaP)=(δdbNbNd)((dNa)aP+NadaP)=θabaP+NabaPNbNd(θadaP+NadaP),

    by [[11], (A.31) and (A.8)], (4.14), (4.16), (4.17) and (4.20), we have

    ˉNPL2(Ω)CθL(Ω)PL2(Ω)+C2PL2(Ω)C(K,VolΩ)(3PL2(Ω)+2PL2(Ω)+PL2(Ω))C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))×(1+2=0(E(t)+Θ02L2(Ω))+E1/22(t)E1/23(t)).

    Thus, from [[11], (A.8)], we can get (ˉθ)NP=Π3P3θ˜ˉNP, and

    ˉθL2(Ω)1ε(Π3PL2(Ω)+CθL(Ω)ˉNpL2(Ω))C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))×(1+2=0(E(t)+Θ02L2(Ω))+E1/22(t)E1/23(t)).

    Hence, by (4.14), it follows that

    Π4PL2(Ω)C(K,K1)(K+θL2(Ω)+ˉθL2(Ω))k4kPL2(Ω).

    In the end, by (4.15), we choose a sufficiently small δ4>0 which can absorb the highest order term in the right-hand side, and get

    4PL2(Ω),Π4PL2(Ω)C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))×(1+3=0((E(t)+Θ02L2(Ω))+E1/22(t)E1/24(t)). (4.21)

    Therefore, by using (4.16), (4.20) and (4.21), we can get for r2

    rPL2(Ω)C(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))×(1+r1=0(E(t)+Θ02L2(Ω))+(r2)(E1/22(t)E1/2r(t))),

    from which and (4.7), we have

    2Ωb(γafγAF)r1Ffubr1AaPdμgC(K,K1,M,VolΩ,1/ε,E0(0),Θ02L2(Ω),E1(0),Θ02L2(Ω))E1/2r(t)×(1+r1=0(E(t)+Θ02L2(Ω))+(r2)E1/22(t)E1/2r(t)).

    Now, since P=0 on Ω, we have γabaP=0 on Ω. Thus we also get

    ϑ1Nb=bP. (4.22)

    By using the Hölder inequality and (4.22), we obtain

    ΩγafγAFrAaP(DtrFfP1ϑNbrFfub)ϑdμγCϑ1/2L(Ω)E1/2r(t)Π(Dt(rP)ϑ1Nbrub)L2(Ω)=Cϑ1/2L(Ω)E1/2r(t)Π(Dt(rP)+ruP)L2(Ω), (4.23)

    hence, we will estimate ΠDt(rP) and ruP, from ([11], Lemma 2.3), it yields that

    DtrP+ruP=[Dt,r]P+rDtP+ruP=sgn(2r)r2s=1(rs+1)(s+1u)rsP+rDtP. (4.24)

    For 2r4, From ([11], (A.18), (A.31) and (A.17)), we can get

    ΠrDtPL2(Ω)C(K,K1,VolΩ)(θL(Ω)+(r2)kr3ˉkθL2(Ω))×krkDtPL2(Ω), (4.25)

    then since by ([11], (A.17)), we get that

    rDtPL2(Ω)δΠrDtPL2(Ω)+C(1/δ,K,VolΩ)sr2sΔDtPL2(Ω). (4.26)

    Now, by ([11], Lemmas 2.1 and 2.3), (2.7), (3.4), (3.5) and (4.8), we have

    ΔDtP=2hababP+(Δue)ePDt(P)=2hababP+(Δue)eP2Dt(gbd)audbua2gbdDt(aud)bua+Dt(nΘ0)=2hababP+(Δue)eP+4hbdaudbua+2gbdbuaadP2gbdbuaaucduc2gbdduaδnbaΘ0+Dt(nΘ0)=4gaccubabP+(Δue)eP+2eubbuaaue2gbdbuaδnbaΘ0.

    For s=2 (similarly for s=0,1), From (4.12), (4.15) and ([11], Lemma A.12), it follows that,

    2ΔDtPL2(Ω)C3u2P+2u3P+u4P+P4uL2(Ω)+C3uuu+2u2uuL2(Ω)+C3uΘ0+2u2Θ0+u3Θ0L2(Ω)CuL(Ω)4PL2(Ω)+C3uL2(Ω)2PL(Ω)+C2uL4(Ω)3PL4(Ω)+C4uL2(Ω)PL(Ω)+CuL(Ω)uL(Ω)3uL2(Ω)+CuL(Ω)2uL4(Ω)2uL4(Ω)+CuL(Ω)3Θ0L2(Ω)+C3uL2(Ω)Θ0L(Ω)+C2uL4(Ω)2Θ0L4(Ω).

    By ([11], (A.11)) and (4.10), we have

    s+1uL4(Ω)Csu1/2L(Ω)(2=0s+uL2(Ω)K21)1/2C(K1)2=0E1/2s+(t). (4.27)

    Obviously, we can get

    s+1Θ0L4(Ω)CsΘ01/2L(Ω)(2=0s+Θ0L2(Ω)K21)1/2C(K1)2=0s+Θ0L2(Ω). (4.28)

    From (4.27) and (4.28), we will estimate all terms with L4(Ω) norms and the similar estimate of P by the assumptions. Thus, we have

    sΔDtPL2(Ω)C(K,K1,M,L,1/ε,VolΩ,E0(0),Θ0L2(Ω))×(1+r1=0(E(t)+Θ02L2(Ω)))(1+(Er(t)+rΘ02L2(Ω))1/2). (4.29)

    By (4.25), (4.26) and (4.29), for small δ independent of Er(t), by induction argument for r, we have

    ΠrDtPL2(Ω)C(K,K1,M,L,1/ε,VolΩ,E0(0),Θ0L2(Ω))×(1+r1=0(E(t)+Θ02L2(Ω)))(1+(Er(t)+rΘ02L2(Ω))1/2). (4.30)

    Finally, for the case r=3,4 and s=r2, we need to estimate the remaining term Π((s+1u)rsP), by ([11], Lemma A.14) and (4.3), we get

    Π((r1u)2P)L2(Ω)r1uL2(Ω)2PL(Ω)CL2uL2(n1)/(n2)(Ω)C(K,VolΩ)L(ruL2(Ω)+r1uL2(Ω))C(K,L,VolΩ)(E1/2r1(t)+E1/2r(t)).

    For n=3, r=4 and s=1, we have similarly

    Π((2u)3P)L2(Ω)=Π2uΠ3p+Π(2uN)˜Π(N3P)L2(Ω)CΠ2uL4(Ω)Π3PL4(Ω)+CΠ(Na2ua)L4(Ω)Π(N2P)L4(Ω)C2uL4(Ω)3PL4(Ω)C(K,VolΩ)(3uL2(Ω)+2uL2(Ω))(4PL2(Ω)+3PL2(Ω))C(K,K1,VolΩ)(E1/23(t)+E1/22(t))(3s=0(Es(t)+sΘ02L2(Ω))+(4=0E1/2(t))E1/22(t))C(K,K1,VolΩ)3s=0(Es(t)+sΘ02L2(Ω))4=0E1/2(t).

    Then, we can get

    |(4.23)|C(K,K1,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω))×(1+r1s=0(Es(t)+sΘ02L2(Ω)))(1+Er(t)+rΘ02L2(Ω)).

    By combining (4.27) with (4.28), it yields that

    |(4.5)|+|(4.6)|C(K,K1,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω))×(1+r1s=0(Es(t)+sΘ02L2(Ω)))(1+Er(t)+rΘ02L2(Ω))).

    Now, we calculate the material derivatives of |r1curlu|2. By ([15], Lemma 2.1) and (4.1), we can get

    Dt(|r1curlu|2)=Dt(gacgbdgAFr1A(curlu)abr1F(curlu)cd)=(r+1)Dt(gac)gbdgAFr1A(curlu)abr1F(curlu)cd+4gacgbdgAFDt(r1Aaub)r1F(curlu)cd=2(r+1)gaeeucgbdgAFr1A(curlu)abr1F(curlu)cd4gacgbdgAFr1F(curlu)cdrAabP+4gacgbdgAFr1F(curlu)cdδbnrAaΘ0+4gacgbdgAFr1F(curlu)cd(curlu)berAaue+4sgn(2r)gacgAFr1F(curlu)cdr2s=1(rs+1)((1+su)rsud)Aa.

    The higher order term involving pressure P will vanish by symmetry. For other terms, we can apply the Hölder inequality and the Gauss formula to obtain that

    ΩDt(|r1curlu|2)dμgC(K,K1,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω))×(1+r1s=0(Es(t)+sΘ02L2(Ω)))(1+Er(t)+rΘ02L2(Ω)).

    Hence, we need to estimate the last term in (4.5). From ([11], (A.12)), we get

    Dt(Np)=2hadNdap+hNNNp+NDtp,

    and

    ϑtϑ=2hadNdapNphNN+NDtpNp.

    Thus, the integrals can be controlled by C(K,M,L,1/ε)(Er(t)+rΘ02L2(Ω)).

    In general, we have

    ddtEr(t)C(K,K1,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω))×(1+r1s=0(Es(t)+sΘ02L2(Ω)))(1+Er(t)+rΘ02L2(Ω)), (4.31)

    which implies the desired result by Gronwall's inequality and the induction argument for r{2,,n+1}.

    The system (1.1) can be written in the Lagrangian coordinates as

    Dtua+aP=ucauc+βddβa+δnaΘ0, in [0,T]×Ω,Dtβa=βddua+βcauc, in [0,T]×Ω,aua=0,aβa=0, in [0,T]×Ω,βaNa=0,P=0, on [0,T]×Ω. (5.1)

    Thus, by (5.1) and ([15], Lemma 2.1), we also get the zero order energy

    E0(t)=12Ω(|u(t,x)|2+|β(t,x)|2)dμgCE0(0)+Θ02L2(Ω).

    Similarly, we can define the first order energy as

    E1(t)=Ω(gbdγaeaubeud+gbdγaeaβbeβd)dμg+Ω(|curlu|2+|curlβ|2)dμg.

    Theorem 5.1. For 0tT for any smooth solution of system (5.1) satisfying

    |P|M,|u|M,in[0,T]×Ω, (5.2)
    |θ|+|u|+1ς0K,on[0,T]×Ω, (5.3)

    when t [0, T], we can get

    E1(t)2eCMt(E1(0)+Θ02L2(Ω))+CK2(VolΩ)(eCMt1). (5.4)

    Define the r-th order energy for r2 as

    Er(t)=ΩgbdγafγAFr1Aaubr1Ffuddμg+Ω|r1curlu|2dμg+ΩgbdγafγAFr1Aaβbr1Ffβddμg+Ω|r1curlβ|2dμg+ΩγafγAFr1AaPr1FfPϑdμγ. (5.5)

    Theorem 5.2. For the integer 2rn+1, there exists a constant T>0 such that, for any smooth solution to system (5.1) for 0tT satisfying

    |β|M1,forr=2,in[0,T]×Ω,|P|M,|u|M,|β|M,|Θ0|M,in[0,T]×Ω,|θ|+1/ς0K,on[0,T]×Ω,NPε>0,on[0,T]×Ω,|2P|+|NDtP|L,on[0,T]×Ω, (5.6)

    when t[0,T], we can get

    Er(t)eC1t(Er(0)+rΘ02L2(Ω))+C2(eC1t1), (5.7)

    where the constants C1 and C2 depend on K,M,L,1/ε,VolΩ,E0(0),Θ02L2(Ω),E1(0),, Er1(0) and r1Θ02L2(Ω).

    Theorems 5.1 and 5.2 can be proved similarly. For the case involving the magnetic field, one can refer to [11]. We omit the details of the proof.

    Since some a priori assumptions are made, them will be justified in this section, at time t, we denote the following values

    H(t)=max(θ(t,)L(Ω),1/ς0(t)),I(t)=1/(NP(t,))L(Ω),ε(t)=1I(t). (6.1)

    Our judgment is very similar to those in [8,11], so we only state the results and omit their proofs as follows.

    Lemma 6.1. Let K11/ς1(t), then there are continuous functions Fj, j = 1, 2, 3, 4, such that

    uL(Ω)+β,βL(Ω)+Θ0L(Ω)F1(K1,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω)),PL(Ω)+2PL(Ω)F2(K1,I,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω),VolΩ),DtPL(Ω)F4(K1,I,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω),VolΩ),θL(Ω)F3(K1,I,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω)VolΩ),|ddtI|Cr(K1,I,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω),VolΩ),|ddtEr|Cr(K1,I,E0,Θ02L2(Ω),,En+1,n+1Θ02L2(Ω),VolΩ)rs=0(Es+sΘ02L2(Ω)).

    Lemma 6.2. There exists a continuous function K>0 depending on K1, E0(0), Θ02L2(Ω), E1(0), , En+1(0), n+1Θ02L2(Ω) and VolΩ such that for

    0tK(K1,E0(0),Θ02L2(Ω),E1(0),,En+1(0),n+1Θ02L2(Ω),VolΩ)

    the following statements hold

    Es(t)2(Es(0)+sΘ02L2(Ω)),0sn+1,I(t)2I(0).

    Furthermore,

    12gab(0,y)YaYbgab(t,y)YaYb2gab(0,y)YaYb,

    and

    |N(x(t,ˉy))N(x(0,ˉy))|ε116,ˉyΩ,|x(t,y)x(t,y)|ς116,yΩ,|x(t,ˉy)y(0,ˉy)y|ε116,ˉyΩ.

    Lemma 6.3. Let K be as in Lemma 6.2. There exists some ε1>0 such that, if

    |N(x(0,y1))N(x(0,y2))|ε12,

    then for tK, it holds

    |N(x(t,y1))N(x(t,y2))|ε1.

    Consequently, Lemmas 6.2 and 6.3 yield immediately Theorem 1.1.

    The authors were supported by National Natural Science Foundation of China (Grant numbers: 12171460 and 11971014).

    The authors declare that there is no conflict of interest.



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