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+v⋅∇v+∇p=b⋅∇b+hen,∂th+v⋅∇h=0,∂tb+v⋅∇b=b⋅∇v,∇⋅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; D⊂∪0⩽t⩽T{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 x∈D0. | (1.2) |
Let Dt={x∈Rn:(t,x)∈D}, then the free boundary conditions should be
{vN=κ, on ∂Dt,p=0, on ∂Dt,b⋅N=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:
∇Np⩽−ε<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 Lp−Lq 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+vk∂k, then the system (1.1) can be rewritten as:
Dtvj+∂jp=bk∂kbj+δjnh, in D,Dtbj=bk∂kvj, 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 i≠j.
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=bk∂kbj+δjnh0. Thus, let the zero order energy be
E0(t)=12∫Dt(|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)i1⋯ir=Πj1i1⋯Πjrirαj1⋯jr, where Πji=δji−NiNj. |
Let ˉ∂i=Πji∂j be a tangential derivative. In fact, we assume p=0 on ∂Dt, obviously, it follows that ˉ∂ip=0 and
(Π∂2p)ij=θij∇Np, | (1.7) |
where θij=ˉ∂iNj is the second fundamental form of ∂Dt. In fact,
0=ˉ∂iˉ∂jp=Πi′i∂i′Πj′j∂j′p=Πi′iΠj′j∂i′∂j′p−(ˉ∂iNj)Nk∂kp−Nj(ˉ∂iNk)∂kp=(Π∂2p)ij−θij∇Np. |
Next, we define the quadratic form Q of the form:
Q(α,β)=⟨Πα,Πβ⟩=qi1j1⋯qirjrαi1⋯irβj1⋯jr, |
where
qij=δij−η(d)2NiNj,d(x)=dist(x,∂Dt),Ni=−δij∂jd, |
where η is a smooth cutoff function satisfying 0⩽η(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)→{x∈Rn:dist(x,∂Dt)<ς0}, |
given by
(ˉx,ς)→x=ˉx+ςN(ˉx), |
is an injection. Then higher energies for r⩾1 can be denoted by
Er(t)=∫Dtδij(Q(∂rvi,∂rvj)+Q(∂rbi,∂rbj))dx+∫Dt(|∂r−1curlv|2+|∂r−1curlb|2)dx+sgn(r−1)∫∂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
T⩽K(H(0),I(0),E0(0),‖Θ0‖2L2(Ω)⋯,En+1(0),‖∇n+1Θ0‖2L2(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+1∑s=0Es(t)⩽2n+1∑s=0(Es(0)+‖∇sΘ0‖2L2(Ω)),0⩽t⩽T. | (1.11) |
In order to prove Theorem 1.1, we consider the Boussinesq equations with zero viscosity and diffusivity in D:
{∂v∂t+(v⋅∇)v=−∇p+hen,∂h∂t+(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:Ω→D0 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),y∈Ω. | (2.1) |
The Euclidean metric δij in Dt then induces a metric
gab(t,y)=δij∂xi∂ya∂xj∂yb, | (2.2) |
and its inverse
gcd(t,y)=δkl∂yc∂xk∂yd∂xl, | (2.3) |
in Ω for each fixed t. Furthermore, expressed in the y-coordinates, we have
∂i=∂∂xi=∂ya∂xi∂∂ya. | (2.4) |
Let us introduce the notation for the material derivative
Dt=∂∂t|y= const =∂∂t|x= const +vk∂∂xk. |
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=∂xj∂ya(−∂jp+δnjh)+vj∂xk∂ya∂vj∂xk=−∇aP+δnaΘ+uc∇auc. | (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+uc∇auc, in [0,T]×Ω,∇aua=0, in [0,T]×Ω,P=0,u⋅N=κ, on [0,T]×∂Ω. | (2.7) |
where Θ|t=0=Θ0.
Firstly, we need to define the zero order energy as
E0(t)=12∫Ω(|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)=12∫ΩDt(gabuaub)dμg=∫Ω(uaDtua)dμg+∫Ω12(Dtgab)(uaub)dμg=∫Ω[−ua∇aP+uaδnaΘ0+uauc∇auc]dμg−∫Ωhab(uaub)dμg=−∫∂ΩNauaPdμγ+∫ΩuaδnaΘ0dμg+∫Ωuauc∇aucdμg−12∫Ωgac(∇cud+∇duc)gdbuaubdμg=∫ΩuaδnaΘ0dμg, |
then, using the Hölder inequality
ddtE0(t)⩽C‖u‖L2(Ω)‖Θ0‖L2(Ω)⩽C(E0(t)+‖Θ0‖2L2(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)+‖Θ0‖2L2(Ω))⩽C(E0(0)+‖Θ0‖2L2(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)+∇b∇aP=[Dt,∇b]ua+∇bDtua+∇b∇aP=−(∇a∇bud)ud+∇bδnaΘ0+∇b(uc∇auc)=−(∇a∇bud)ud+δna∇bΘ0+∇buc∇auc+uc∇b∇auc=∇buc∇auc+δna∇bΘ0. | (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 gbdγae∇aub∇eud,
Dt(gbdγae∇aub∇eud)=(Dtgbd)γae∇aub∇eud+gbd(Dtγae)∇aub∇eud+2gbdγae(Dt∇aub)∇eud=−2gbchcfgfdγae∇aub∇eud−2gbdγachcfγfe∇aub∇eud−2gbdγae∇eud∇a∇bP+2gbdγae∇eudδnb∇aΘ0+2gbdγae∇eud∇auc∇buc=−γae(∇cuf+∇fuc)∇auc∇euf−2γacγfe(∇cuf+∇fuc)∇aud∇eud−2γae∇eub∇a∇bP+2γae∇eubδnb∇aΘ0+2γae∇eub∇auc∇buc=−2γae∇cuf∇auc∇euf−4γaeγfc∇euf∇aud∇cud+2γae∇eub∇auc∇buc−2γae∇eub∇a∇bP+2γae∇eubδnb∇aΘ0=−4γaeγfc∇euf∇aud∇cud−2γae∇eub∇a∇bP+2γae∇eubδbn∇aΘ0=−4γaeγfc∇euf∇aud∇cud+2γae∇eubδnb∇aΘ0−2∇b(γae∇eub∇aP)+2(∇bγae)(∇eub∇aP). | (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(Dt∇aub)(curlu)cd=−2gaegfcgbd(∇euf+∇fue)(curlu)ab(curlu)cd+4gacgbd(curlu)cd∇aue∇bue−4gacgbd(curlu)cd∇a∇bP+4gacgbd(curlu)cdδnb∇aΘ0=−4gaegbd∇euc(curlu)ab(curlu)cd+4gacgbd(curlu)cdδnb∇aΘ0. | (3.7) |
Now, we can define the first order energy as
E1(t)=∫Ωgbdγae∇aub∇euddμg+∫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|+1ς0⩽K,on[0,T]×∂Ω, |
when t ∈ [0, T], we can get
E1(t)⩽2eCMt(E1(0)+‖∇Θ0‖2L2(Ω))+CK2(VolΩ)(eCMt−1). | (3.9) |
Proof.
ddtE1(t)=∫ΩDt(gbdγae∇aub∇eud)dμg+∫ΩDt(|curlu|2)dμg. |
From (3.6)–(3.8), ([15], Lemma 2.1) and Gauss' formula, it yield to
ddtE1(t)=∫Ω(−4γaeγfc∇euf∇aud∇cud+2γae∇eubδnb∇aΘ0)dμg+2∫Ω(∇bγae)(∇eub∇aP)dμg−4∫Ωgaegbd∇euc(curlu)ab(curlu)cddμg+4∫Ωgacgbd(curlu)cdδnb∇aΘ0dμg−2∫∂ΩNb(γae∇eub∇aP)dμγ. |
In fact the integral on the boundary is zero. Since P=0 on ∂Ω, we have ˉ∇P=γae∇aP=0.
Then, by ([11], (A.3) and (A.5)) and the fact ∇NN=0. we obtain that
θab=(δca−NaNc)∇cNb=∇aNb−Na∇NNb=∇aNb, |
Thus, we have
∇bγae=∇b(gae−NaNe)=−∇b(NaNe)=−(∇bNa)Ne−(∇bNe)Na=−θabNe−θebNa. |
From the Hölder inequality and ([11], (A.5)), it follows that
ddtE1(t)⩽CKM(VolΩ)1/2E1/21(t)+CM(VolΩ)1/2‖∇Θ0‖L2(Ω)+C‖∇u‖L∞(Ω)(‖∇u‖2L2(Ω)+‖curlu‖2L2(Ω))⩽CKM(VolΩ)1/2(E1(t)+‖∇Θ0‖2L2(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
Dt∇rua=Dt∇a1⋯∇arua=Dt(∂xi1∂ya1⋯∂xir∂yar∂xi∂ya∂i1⋯∂irvi)=−∇r∇aP−r−1∑s=1(rs+1)(∇1+su)⋅∇r−sua+δna∇rΘ0+∇auc∇ruc, |
when r⩾2, by moving terms, we have
Dt∇rua+∇r∇aP=(curlu)ac∇ruc+sgn(2−r)r−2∑s=1(rs+1)(∇1+su)⋅∇r−sua+δna∇rΘ0. | (4.1) |
Define the r-th order energy for r⩾2 as
Er(t)=∫ΩgbdγafγAF∇r−1A∇aub∇r−1F∇fuddμg+∫Ω|∇r−1curlu|2dμg+∫∂ΩγafγAF∇r−1A∇aP∇r−1F∇fPϑdμγ, | (4.2) |
where ϑ=1/(−∇NP), then we can get the following estimates.
Theorem 4.1. For integer 2⩽r⩽n+1, there exists a constant T>0 such that, for any smooth solution to system (2.7) for 0⩽t⩽T satisfying
|∇P|⩽M,|∇u|⩽M,|∇Θ0|⩽M,in[0,T]×Ω,|θ|+1/ς0⩽K,on[0,T]×∂Ω,−∇NP⩾ε>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)+‖∇rΘ0‖2L2(Ω))+C2(eC1t−1), | (4.4) |
where the constants C1 and C2 depend on K,M,L,1/ε,VolΩ,E0(0),‖Θ0‖2L2(Ω),E1(0),⋯, Er−1(0) and ‖∇r−1Θ0‖2L2(Ω).
Proof. Appling (4.2), the derivative of Er with respect to t is
ddtEr(t)=∫ΩDt(gbdγafγAF∇r−1A∇aub∇r−1F∇fud)dμg+∫ΩDt|∇r−1curlu|2dμg+∫∂ΩDt(γafγAF∇r−1A∇aP∇r−1F∇fP)ϑdμγ+∫∂ΩγafγAF∇r−1A∇aP∇r−1F∇fP(ϑtϑ−hNN)ϑdμγ. | (4.5) |
From ([15], Lemma 2.1) and (4.1), we directly have
Dt(gbdγafγAF∇r−1A∇aub∇r−1F∇fud)=−2∇cueγafγAF∇r−1A∇auc∇r−1F∇fue−4r∇cueγacγefγAF∇r−1A∇aud∇r−1F∇fud−2γafγAF∇r−1F∇fub∇r−1A∇a∇bP+2γafγAF∇r−1F∇fub(curlu)bc∇r−1A∇auc+2sgn(2−r)γafγAF∇r−1F∇fudr−2∑s=1(rs+1)((∇s+1u)⋅∇r−sud)Aa+2γafγAF∇r−1F∇fubδnb∇r−1A∇aΘ0. |
and
Dt(γafγAF∇r−1A∇aP∇r−1F∇fP)=−2r∇cueγacγefγAF∇r−1A∇aP∇r−1F∇fP+2γafγAF∇r−1A∇aPDt(∇r−1F∇fP). |
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(gbdγafγAF∇r−1A∇aub∇r−1F∇fud)dμg+∫∂ΩDt(γafγAF∇r−1A∇aP∇r−1F∇fP)ϑdμγ⩽C‖∇u‖L∞(Ω)Er(t)+CEr(t)+‖∇rΘ0‖2L2(Ω)+CE1/2r(t)r−2∑s=1‖∇s+1u‖L4(Ω)‖∇r−su‖L4(Ω)+2∫∂ΩγafγAF∇rAaP(Dt∇rFfP−1ϑNb∇rFfub)ϑdμγ+2∫Ω∇b(γafγAF)∇r−1F∇fub∇r−1A∇aPdμg. | (4.6) |
Then, using the Hölder inequality, it yield that
2∫Ω∇b(γafγAF)∇r−1F∇fub∇r−1A∇aPdμg⩽CKE1/2r(t)‖∇rP‖L2(Ω). | (4.7) |
Obviously, we have to estimate ‖∇rP‖L2(Ω). Firstly, we need to take divergence on the first equation of (2.7), by ([11], Lemma 2.2), we nave
ΔP=−∇aub∇bua+∂nΘ0. | (4.8) |
For r⩾2, we have
∇r−2ΔP=−r−2∑s=0(r−2s)∇s∇aub∇r−2−s∇bua−∇r−2∂nΘ0. |
Then by ([11], (A.12)), for ς1⩾1/K1, it follows that
‖u‖L∞(Ω)⩽C2∑s=0Kn/2−s1‖∇su‖L2(Ω)⩽C(K1)2∑s=0E1/2s(t). | (4.9) |
In view of (4.9), for s⩾0, similarly, we can directly get taht
‖∇su‖L∞(Ω)⩽C2∑ℓ=0Kn/2−ℓ1‖∇ℓ+su‖L2(Ω)⩽C(K1)2∑ℓ=0E1/2s+ℓ(t), | (4.10) |
When r=3,4, By the Hölder inequality, (4.9) and (4.10), we have
‖∇r−2ΔP‖L2(Ω)⩽Cr−2∑s=0‖∇s∇aub∇r−2−s∇bua‖L2(Ω)+C‖∇r−2∂nΘ0‖L2(Ω)⩽C‖∇u‖L∞(Ω)‖∇r−1u‖L2(Ω)+(r−3)C‖∇2u‖L∞(Ω)‖∇2u‖L2(Ω)+C‖∇r−1Θ0‖L2(Ω)⩽C(K1)r−1∑ℓ=1Eℓ(t)+C(K1)E1/22(t)E1/2r(t)+C(K1)‖∇r−1Θ0‖L2(Ω)⩽C(K1)r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+C(K1)E1/22(t)E1/2r(t). | (4.11) |
The last inequality is due to the inequality ‖∇r−1Θ0‖L2(Ω)⩽CE0+C‖∇r−1Θ0‖2L2(Ω). For r=2, we have the following estimate from the assumption of (4.3) and the Hölder inequality, i.e.,
‖ΔP‖L2(Ω)⩽CM(E1(t)+‖∇Θ0‖2L2(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
‖∇rP‖L2(Ω)⩽δr‖Π∇rP‖L2(∂Ω)+C(1/δr,K,VolΩ)∑s⩽r−2‖∇sΔP‖L2(Ω)⩽δr‖Π∇rP‖L2(∂Ω)+C(1/δr,K,K1,M,VolΩ)r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+(r−2)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 r⩾1,
‖Π∇rP‖L2(∂Ω)⩽C(K,K1)(‖θ‖L∞(∂Ω)+(r−2)∑k⩽r−3‖ˉ∇kθ‖L2(∂Ω))∑k⩽r−1‖∇kP‖L2(∂Ω). | (4.14) |
By ([11], (A.7)), we obtain that Π∇2P=θ∇NP, then from (4.3), we get
‖θ‖L2(∂Ω)=‖Π∇2P∇NP‖L2(∂Ω)⩽1ε‖Π∇2P‖L2(∂Ω). | (4.15) |
Then, for r=2,3,4, we need to estimate ‖Π∇rP‖L2(∂Ω) and ‖∇rP‖L2(Ω).
{{When r = 2}}, by using (4.13) and (4.14), we have
‖Π∇2P‖L2(∂Ω)⩽‖θ‖L∞(∂Ω)‖∇P‖L2(∂Ω)⩽C(K,VolΩ)(‖∇2P‖L2(Ω)+‖∇P‖L2(Ω))⩽C(K,VolΩ)δ2‖Π∇2P‖L2(∂Ω)+C(K,VolΩ)(VolΩ)1/2M+C(1/δ2,K,K1,M,VolΩ)(1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))). |
We can take δ2 so small that the first term can be absorbed by the left-hand side. Thus
‖Π∇2P‖L2(∂Ω),‖∇2P‖L2(Ω)⩽C(K,K1,M,VolΩ)(1+1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))), | (4.16) |
‖θ‖L2(∂Ω)⩽C(K,K1,M,VolΩ,1/ε)(1+1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))). | (4.17) |
From Theorem 3.1 and zero order energy estimate, there exists a constant T>0 such that Ei(t)⩽CEi(0)+‖∇iΘ0‖2L2(Ω) for t∈[0,T] and i=0,1.
{When r = 3}, by (4.3), (4.14), (4.16) and (4.17), we have
‖Π∇3P‖L2(∂Ω)⩽C(K,K1)(K+‖θ‖L2(∂Ω))∑k⩽2‖∇kP‖L2(∂Ω)⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))‖∇3P‖L2(Ω)+C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω)), | (4.18) |
and by (4.13), it follows that
‖∇3P‖L2(Ω)⩽δ3C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))‖∇3P‖L2(Ω)+δ3C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))+C(1/δ3,K,K1,M,VolΩ)(2∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(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
‖∇3P‖L2(Ω),‖Π∇3P‖L2(∂Ω)⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))×(1+2∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+E1/22(t)E1/23(t)). | (4.20) |
{When r = 4}, since
ˉ∇b∇NP=γdb∇d(Na∇aP)=(δdb−NbNd)((∇dNa)∇aP+Na∇d∇aP)=θab∇aP+Na∇b∇aP−NbNd(θad∇aP+Na∇d∇aP), |
by [[11], (A.31) and (A.8)], (4.14), (4.16), (4.17) and (4.20), we have
‖ˉ∇∇NP‖L2(∂Ω)⩽C‖θ‖L∞(∂Ω)‖∇P‖L2(∂Ω)+C‖∇2P‖L2(∂Ω)⩽C(K,VolΩ)(‖∇3P‖L2(Ω)+‖∇2P‖L2(Ω)+‖∇P‖L2(Ω))⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))×(1+2∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+E1/22(t)E1/23(t)). |
Thus, from [[11], (A.8)], we can get (ˉ∇θ)∇NP=Π∇3P−3θ˜⊗ˉ∇∇NP, and
‖ˉ∇θ‖L2(∂Ω)⩽1ε(‖Π∇3P‖L2(∂Ω)+C‖θ‖L∞(∂Ω)‖ˉ∇∇Np‖L2(∂Ω))⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))×(1+2∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+E1/22(t)E1/23(t)). |
Hence, by (4.14), it follows that
‖Π∇4P‖L2(∂Ω)⩽C(K,K1)(K+‖θ‖L2(∂Ω)+‖ˉ∇θ‖L2(∂Ω))∑k⩽4‖∇kP‖L2(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
‖∇4P‖L2(Ω),‖Π∇4P‖L2(∂Ω)⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))×(1+3∑ℓ=0((Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+E1/22(t)E1/24(t)). | (4.21) |
Therefore, by using (4.16), (4.20) and (4.21), we can get for r⩾2
‖∇rP‖L2(Ω)⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))×(1+r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+(r−2)(E1/22(t)E1/2r(t))), |
from which and (4.7), we have
2∫Ω∇b(γafγAF)∇r−1F∇fub∇r−1A∇aPdμg⩽C(K,K1,M,VolΩ,1/ε,E0(0),‖Θ0‖2L2(Ω),E1(0),‖∇Θ0‖2L2(Ω))E1/2r(t)×(1+r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω))+(r−2)E1/22(t)E1/2r(t)). |
Now, since P=0 on ∂Ω, we have γab∇aP=0 on ∂Ω. Thus we also get
−ϑ−1Nb=∇bP. | (4.22) |
By using the Hölder inequality and (4.22), we obtain
∫∂ΩγafγAF∇rAaP(Dt∇rFfP−1ϑNb∇rFfub)ϑdμγ⩽C‖ϑ‖1/2L∞(∂Ω)E1/2r(t)‖Π(Dt(∇rP)−ϑ−1Nb∇rub)‖L2(∂Ω)=C‖ϑ‖1/2L∞(∂Ω)E1/2r(t)‖Π(Dt(∇rP)+∇ru⋅∇P)‖L2(∂Ω), | (4.23) |
hence, we will estimate ΠDt(∇rP) and ∇ru⋅∇P, from ([11], Lemma 2.3), it yields that
Dt∇rP+∇ru⋅∇P=[Dt,∇r]P+∇rDtP+∇ru⋅∇P=sgn(2−r)r−2∑s=1(rs+1)(∇s+1u)⋅∇r−sP+∇rDtP. | (4.24) |
For 2⩽r⩽4, From ([11], (A.18), (A.31) and (A.17)), we can get
‖Π∇rDtP‖L2(∂Ω)⩽C(K,K1,VolΩ)(‖θ‖L∞(∂Ω)+(r−2)∑k⩽r−3‖ˉ∇kθ‖L2(∂Ω))×∑k⩽r‖∇kDtP‖L2(Ω), | (4.25) |
then since by ([11], (A.17)), we get that
‖∇rDtP‖L2(Ω)⩽δ‖Π∇rDtP‖L2(∂Ω)+C(1/δ,K,VolΩ)∑s⩽r−2‖∇sΔDtP‖L2(Ω). | (4.26) |
Now, by ([11], Lemmas 2.1 and 2.3), (2.7), (3.4), (3.5) and (4.8), we have
ΔDtP=2hab∇a∇bP+(Δue)∇eP−Dt(△P)=2hab∇a∇bP+(Δue)∇eP−2Dt(gbd)∇aud∇bua−2gbdDt(∇aud)∇bua+Dt(∂nΘ0)=2hab∇a∇bP+(Δue)∇eP+4hbd∇aud∇bua+2gbd∇bua∇a∇dP−2gbd∇bua∇auc∇duc−2gbd∇duaδnb∇aΘ0+Dt(∂nΘ0)=4gac∇cub∇a∇bP+(Δue)∇eP+2∇eub∇bua∇aue−2gbd∇buaδnb∇aΘ0. |
For s=2 (similarly for s=0,1), From (4.12), (4.15) and ([11], Lemma A.12), it follows that,
‖∇2ΔDtP‖L2(Ω)⩽C‖∇3u∇2P+∇2u∇3P+∇u∇4P+∇P∇4u‖L2(Ω)+C‖∇3u∇u∇u+∇2u∇2u∇u‖L2(Ω)+C‖∇3u∇Θ0+∇2u∇2Θ0+∇u∇3Θ0‖L2(Ω)⩽C‖∇u‖L∞(Ω)‖∇4P‖L2(Ω)+C‖∇3u‖L2(Ω)‖∇2P‖L∞(Ω)+C‖∇2u‖L4(Ω)‖∇3P‖L4(Ω)+C‖∇4u‖L2(Ω)‖∇P‖L∞(Ω)+C‖∇u‖L∞(Ω)‖∇u‖L∞(Ω)‖∇3u‖L2(Ω)+C‖∇u‖L∞(Ω)‖∇2u‖L4(Ω)‖∇2u‖L4(Ω)+C‖∇u‖L∞(Ω)‖∇3Θ0‖L2(Ω)+C‖∇3u‖L2(Ω)‖∇Θ0‖L∞(Ω)+C‖∇2u‖L4(Ω)‖∇2Θ0‖L4(Ω). |
By ([11], (A.11)) and (4.10), we have
‖∇s+1u‖L4(Ω)⩽C‖∇su‖1/2L∞(Ω)(2∑ℓ=0‖∇s+ℓu‖L2(Ω)K2−ℓ1)1/2⩽C(K1)2∑ℓ=0E1/2s+ℓ(t). | (4.27) |
Obviously, we can get
‖∇s+1Θ0‖L4(Ω)⩽C‖∇sΘ0‖1/2L∞(Ω)(2∑ℓ=0‖∇s+ℓΘ0‖L2(Ω)K2−ℓ1)1/2⩽C(K1)2∑ℓ=0‖∇s+ℓΘ0‖L2(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
‖∇sΔDtP‖L2(Ω)⩽C(K,K1,M,L,1/ε,VolΩ,E0(0),‖Θ0‖L2(Ω))×(1+r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω)))(1+(Er(t)+‖∇rΘ0‖2L2(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
‖Π∇rDtP‖L2(∂Ω)⩽C(K,K1,M,L,1/ε,VolΩ,E0(0),‖Θ0‖L2(Ω))×(1+r−1∑ℓ=0(Eℓ(t)+‖∇ℓΘ0‖2L2(Ω)))(1+(Er(t)+‖∇rΘ0‖2L2(Ω))1/2). | (4.30) |
Finally, for the case r=3,4 and s=r−2, we need to estimate the remaining term Π((∇s+1u)⋅∇r−sP), by ([11], Lemma A.14) and (4.3), we get
‖Π((∇r−1u)⋅∇2P)‖L2(∂Ω)⩽‖∇r−1u‖L2(∂Ω)‖∇2P‖L∞(∂Ω)⩽CL‖∇2u‖L2(n−1)/(n−2)(∂Ω)⩽C(K,VolΩ)L(‖∇ru‖L2(Ω)+‖∇r−1u‖L2(Ω))⩽C(K,L,VolΩ)(E1/2r−1(t)+E1/2r(t)). |
For n=3, r=4 and s=1, we have similarly
‖Π((∇2u)⋅∇3P)‖L2(∂Ω)=‖Π∇2u⋅Π∇3p+Π(∇2u⋅N)˜⊗Π(N⋅∇3P)‖L2(∂Ω)⩽C‖Π∇2u‖L4(∂Ω)‖Π∇3P‖L4(∂Ω)+C‖Π(Na∇2ua)‖L4(∂Ω)‖Π(∇N∇2P)‖L4(∂Ω)⩽C‖∇2u‖L4(∂Ω)‖∇3P‖L4(∂Ω)⩽C(K,VolΩ)(‖∇3u‖L2(Ω)+‖∇2u‖L2(Ω))(‖∇4P‖L2(Ω)+‖∇3P‖L2(Ω))⩽C(K,K1,VolΩ)(E1/23(t)+E1/22(t))(3∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω))+(4∑ℓ=0E1/2ℓ(t))E1/22(t))⩽C(K,K1,VolΩ)3∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω))4∑ℓ=0E1/2ℓ(t). |
Then, we can get
|(4.23)|⩽C(K,K1,M,L,1/ε,VolΩ,E0(0),‖Θ0‖2L2(Ω))×(1+r−1∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω)))(1+Er(t)+‖∇rΘ0‖2L2(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),‖Θ0‖2L2(Ω))×(1+r−1∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω)))(1+Er(t)+‖∇rΘ0‖2L2(Ω))). |
Now, we calculate the material derivatives of |∇r−1curlu|2. By ([15], Lemma 2.1) and (4.1), we can get
Dt(|∇r−1curlu|2)=Dt(gacgbdgAF∇r−1A(curlu)ab∇r−1F(curlu)cd)=(r+1)Dt(gac)gbdgAF∇r−1A(curlu)ab∇r−1F(curlu)cd+4gacgbdgAFDt(∇r−1A∇aub)∇r−1F(curlu)cd=−2(r+1)gae∇eucgbdgAF∇r−1A(curlu)ab∇r−1F(curlu)cd−4gacgbdgAF∇r−1F(curlu)cd∇rAa∇bP+4gacgbdgAF∇r−1F(curlu)cdδbn∇rAaΘ0+4gacgbdgAF∇r−1F(curlu)cd(curlu)be∇rAaue+4sgn(2−r)gacgAF∇r−1F(curlu)cdr−2∑s=1(rs+1)((∇1+su)⋅∇r−sud)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(|∇r−1curlu|2)dμg⩽C(K,K1,M,L,1/ε,VolΩ,E0(0),‖Θ0‖2L2(Ω))×(1+r−1∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω)))(1+Er(t)+‖∇rΘ0‖2L2(Ω)). |
Hence, we need to estimate the last term in (4.5). From ([11], (A.12)), we get
Dt(∇Np)=−2hadNd∇ap+hNN∇Np+∇NDtp, |
and
ϑtϑ=2hadNd∇ap∇Np−hNN+∇NDtp∇Np. |
Thus, the integrals can be controlled by C(K,M,L,1/ε)(Er(t)+‖∇rΘ0‖2L2(Ω)).
In general, we have
ddtEr(t)⩽C(K,K1,M,L,1/ε,VolΩ,E0(0),‖Θ0‖2L2(Ω))×(1+r−1∑s=0(Es(t)+‖∇sΘ0‖2L2(Ω)))(1+Er(t)+‖∇rΘ0‖2L2(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=uc∇auc+βd∇dβa+δnaΘ0, in [0,T]×Ω,Dtβa=βd∇dua+βc∇auc, 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)=12∫Ω(|u(t,x)|2+|β(t,x)|2)dμg⩽CE0(0)+‖Θ0‖2L2(Ω). |
Similarly, we can define the first order energy as
E1(t)=∫Ω(gbdγae∇aub∇eud+gbdγae∇aβb∇eβd)dμg+∫Ω(|curlu|2+|curlβ|2)dμg. |
Theorem 5.1. For 0⩽t⩽T for any smooth solution of system (5.1) satisfying
|∇P|⩽M,|∇u|⩽M,in[0,T]×Ω, | (5.2) |
|θ|+|∇u|+1ς0⩽K,on[0,T]×∂Ω, | (5.3) |
when t ∈ [0, T], we can get
E1(t)⩽2eCMt(E1(0)+‖∇Θ0‖2L2(Ω))+CK2(VolΩ)(eCMt−1). | (5.4) |
Define the r-th order energy for r⩾2 as
Er(t)=∫ΩgbdγafγAF∇r−1A∇aub∇r−1F∇fuddμg+∫Ω|∇r−1curlu|2dμg+∫ΩgbdγafγAF∇r−1A∇aβb∇r−1F∇fβddμg+∫Ω|∇r−1curlβ|2dμg+∫∂ΩγafγAF∇r−1A∇aP∇r−1F∇fPϑdμγ. | (5.5) |
Theorem 5.2. For the integer 2⩽r⩽n+1, there exists a constant T>0 such that, for any smooth solution to system (5.1) for 0⩽t⩽T satisfying
|β|⩽M1,forr=2,in[0,T]×Ω,|∇P|⩽M,|∇u|⩽M,|∇β|⩽M,|∇Θ0|⩽M,in[0,T]×Ω,|θ|+1/ς0⩽K,on[0,T]×∂Ω,−∇NP⩾ε>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)+‖∇rΘ0‖2L2(Ω))+C2(eC1t−1), | (5.7) |
where the constants C1 and C2 depend on K,M,L,1/ε,VolΩ,E0(0),‖Θ0‖2L2(Ω),E1(0),⋯, Er−1(0) and ‖∇r−1Θ0‖2L2(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 K1⩾1/ς1(t), then there are continuous functions Fj, j = 1, 2, 3, 4, such that
‖∇u‖L∞(Ω)+‖β,∇β‖L∞(Ω)+‖∇Θ0‖L∞(Ω)⩽F1(K1,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω)),‖∇P‖L∞(Ω)+‖∇2P‖L∞(Ω)⩽F2(K1,I,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω),VolΩ),‖∇DtP‖L∞(∂Ω)⩽F4(K1,I,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω),VolΩ),‖θ‖L∞(∂Ω)⩽F3(K1,I,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω)VolΩ),|ddtI|⩽Cr(K1,I,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω),VolΩ),|ddtEr|⩽Cr(K1,I,E0,‖Θ0‖2L2(Ω),⋯,En+1,‖∇n+1Θ0‖2L2(Ω),VolΩ)r∑s=0(Es+‖∇sΘ0‖2L2(Ω)). |
Lemma 6.2. There exists a continuous function K>0 depending on K1, E0(0), ‖Θ0‖2L2(Ω), E1(0), ⋯, En+1(0), ‖∇n+1Θ0‖2L2(Ω) and VolΩ such that for
0⩽t⩽K(K1,E0(0),‖Θ0‖2L2(Ω),E1(0),⋯,En+1(0),‖∇n+1Θ0‖2L2(Ω),VolΩ) |
the following statements hold
Es(t)⩽2(Es(0)+‖∇sΘ0‖2L2(Ω)),0⩽s⩽n+1,I(t)⩽2I(0). |
Furthermore,
12gab(0,y)YaYb⩽gab(t,y)YaYb⩽2gab(0,y)YaYb, |
and
|N(x(t,ˉy))−N(x(0,ˉy))|⩽ε116,ˉy∈∂Ω,|x(t,y)−x(t,y)|⩽ς116,y∈Ω,|∂x(t,ˉy)∂y−∂(0,ˉy)∂y|⩽ε116,ˉy∈∂Ω. |
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 t⩽K, 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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