# The Nonsteady Boussinesq System with Mixed Boundary Conditions including Conditions of Friction Type

Authors: Tujin Kim
Journal: International Journal of Differential Equations
 (2020)

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## Abstract

In this paper, we are concerned with the nonsteady Boussinesq system under mixed boundary conditions. The boundary conditions for fluid may include Tresca slip, leak and one-sided leak conditions, velocity, static (or total) pressure, rotation, and stress (or total stress) together, and the boundary conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. Relying on the relations among strain, rotation, normal derivative of velocity, and shape of the boundary surface, we get variational formulation. The formulations consist of a variational inequality for velocity due to the boundary conditions of friction type and a variational equation for temperature. For the case of boundary conditions including the static pressure and stress, we prove that if the data of the problem are small enough and compatibility conditions at the initial instance are satisfied, then there exists a unique solution on the given interval. For the case of boundary conditions including the total pressure and total stress, we prove the existence of a solution without restriction on the data and parameters of the problem.


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## Body

## 1. Introduction

In this paper, we are concerned with the Boussinesq equation for heat convection(1)∂v∂t−2∇⋅μθℰv+v⋅∇v+∇p=1−α0θf,∇⋅v=0,∂θ∂t−∇⋅κθ∇θ+v⋅∇θ=g,v0=v0,θ0=θ0,under mixed boundary conditions. Here, v,p, and θ are, respectively, velocity, pressure, and temperature, and α0 is the parameter for buoyancy effect, f is the body force, g is the heat source, μ is the viscosity, and κ is the thermal conductivity. The strain tensor ℰv is the one with the components εijv=1/2∂xivj+∂xjvi. System (1) is a special case of(2)∂v∂t−2∇⋅μθℰv+v⋅∇v+∇p=1−α0θf,∇⋅v=0,∂θ∂t−∇⋅κθ∇θ+v⋅∇θ−α2μθℰv:ℰv=α1θf⋅v+g,v0=v0,θ0=θ0,which is a mathematical model for nonsteady motion of heat-conducting incompressible Newtonian fluid. Here, α1 is the parameter for dissipation of energy due to expansion, α2 is a positive real number, and for two matrices A=aij and B=bij, A:B=∑ijaijbij and A=∑ijaij21/2. The term μθℰv:ℰv represents the dissipation of energy due to viscosity (the Joule effect). Owing to the dissipation of energy due to viscosity μθℰv:ℰv, study of (2) is usually more difficult than the Boussinesq system.For the papers concerned with (2), we refer to Introduction of [1]. Here, we more mention [2–5] concerned with (2), where α1=0. In [2], the problem under nonhomogeneous Dirichlet boundary conditions for velocity and temperature in the time-dependent domain was studied, and existence of a local-in-time solution or existence of the solution on the given interval for small data was proved. In [3], existence of a strong solution and periodic solution for the 2D problem was studied under the boundary conditions and domain as above. In [4], under homogeneous Dirichlet boundary conditions for velocity and temperature, existence of a strong solution and periodic solution were studied when data of the problem are small enough. Łukaszewicz and Krzyżanowski [5] dealt with the initial boundary value problem on a time-dependent domain with the homogeneous Dirichlet boundary condition for velocity and temperature, and they proved the existence and uniqueness of local weak solutions and the existence of a global weak solution for small initial data.Several papers are concerned with (1). In [6, 7], the existence and uniqueness (for 2D) of a solution to the problem were studied under the homogeneous Dirichlet boundary condition for velocity and mixture of nonhomogeneous Dirichlet and Neumann boundary conditions for temperature. In [8], for the problem with nonhomogeneous Dirichlet boundary conditions for velocity and temperature, the existence of the time periodic solution was proved (see [9]). In [10–13], problem (1) on the time-dependent domain was studied under the nonhomogeneous Dirichlet boundary condition for velocity and temperature. In [14, 15], the problem on exterior domains with the homogeneous Dirichlet boundary condition for velocity and nonhomogeneous Dirichlet boundary condition for temperature was studied. In [16], problem (1) was studied under the mixture of the nonhomogeneous Dirichlet boundary condition and the stress boundary condition for fluid and the mixture of nonhomogeneous Dirichlet, Neumann, and Robin boundary conditions for temperature. They proved the existence of a unique local-in-time solution under a compatibility condition at the initial instance (see (27) and (31) of [16]). In [17], problem (1) in the cylindrical pipe with inflow and outflow was studied under slip boundary conditions for velocity and the Neumann conditions for temperature. In that, it was proved that there exists a solution on the given interval when norms of derivatives in the direction along the cylinder of the initial velocity, initial temperature, and the external force are small enough. In [18], the existence of a solution to problem (1) on the time-dependent domain was studied under the mixture of the Dirichlet condition of velocity, total pressure, and rotation boundary conditions for fluid and the mixture of Dirichlet, Neumann, and Robin boundary conditions for temperature.On the contrary, for movement of fluid (v,p), different kinds of boundary conditions are used, and in practice, we deal with the mixture of some kinds of boundary conditions. On some portions of the boundary, we can use boundary conditions with stress or rotation, whereas when there is flux through a portion of the boundary, we can deal with the static pressure p or the total pressure (Bernoulli’s pressure) 1/2v2+p boundary conditions. There are many literature studies for the Navier–Stokes problem with mixed boundary conditions (see Introduction of [19, 20] and references therein). Recently, Navier–Stokes system with mixed boundary conditions including friction-type conditions was studied (cf. [20, 21]).In [1], problem (2) is studied under mixed boundary conditions, and the boundary conditions for fluid may include Tresca slip, leak and one-sided leak conditions, velocity, total pressure, rotation, and total stress together, and the conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. From the result of [1], we can get results for (1) with the boundary conditions as in [1]; however, the result demands that the parameter for buoyancy effect α0 is small enough in accordance with the data of the problem, and the solution includes “defect measure” as in [22]. Also, for (2) and (1), the problem with a mixed boundary condition including the static pressure (not total pressure) and stress (not total stress) together is not yet considered.When one of static pressure, stress, or the outflow boundary condition is given on a portion of the boundary, for the initial boundary value problems of the Navier–Stokes equations, the existence of a unique local-in-time solution and a unique solution on a given interval for small data of the problem are proved. From the mathematical point of view, the main difficulty of such problems results from the fact that in a priori estimation of solution, the term arising from the nonlinear termv⋅∇v is not canceled (cf. Preface in [23]).In the present paper, we are first concerned with heat convection equation (1) under mixed boundary conditions including the static pressure and stress. The boundary conditions for fluid may include conditions of friction type (Tresca slip, threshold leak, and one-sided leak conditions), velocity, static pressure, rotation, and stress together, and the conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. Due to the boundary conditions of friction type, it is difficult to follow the methods in [16, 20]. The main difficulty of this problem is from the estimate of approximate solutions, and due to simultaneous velocity and temperature, the estimate is more difficult than the case of the Navier–Stokes equations. Also, in this paper, we prove the existence of a solution to (1) with the boundary conditions as in [1] without restriction on the parameter for buoyancy effect α0.This paper consists of 5 sections. In the last part of Section1, we give notations.In Section2, the problems to study and assumptions for future are stated. According to the boundary conditions for fluid, Problems I and II are distinguished. Problem I includes the static pressure and the stress conditions, whereas Problem II includes the total pressure and the total stress boundary conditions. Assumption for Problem I is stronger than the one for Problem II.In Section3, we first get a variational formulation for Problem I which consists of six formulae with six unknown functions, that is, using velocity, tangent stress on slip surface, normal stress on the leak surface, normal stresses on one-sided leak surfaces, and temperature together as unknown functions (Problem I-VE). Then, we get a new variational formulation for Problem I consisting of one variational inequality for velocity and a variational equation for temperature (Problem I-VI).The variational formulation for Problem II is obtained in the same way as in [1], and smoothness of the solution with respect to t is weaker than the one in Problem I. In the end of Section 3, the main results of this paper are stated (Theorems 1 and 2). The main result for Problem I asserts that if the data of the problem are small enough and compatibility conditions at the initial time (conditions 4 and 6 of Theorem 1) hold, then there exists a unique smooth solution. The main result for Problem II asserts the existence of a solution without restriction on the parameter for buoyancy unlike [1].Section4 is devoted to the proof of Theorem 1. To this end, first in Section 4.1, we consider an approximate problem, where the variational inequality for velocity is replaced by an equation with the gradient of the Moreau regularization of the functional due to the boundary conditions of friction type. Developing the method for the proof of Theorem 4.4 of [21], we get existence and estimations of approximate solutions for small data under the compatibility conditions at initial time. In Section 4.2, we complete the proof of the existence and uniqueness of a solution.Section 5 is devoted to the proof of Theorem 2. The existence of solutions to an approximate problem and relative compactness of the set of solutions are studied. Then, passing to limit, we get the conclusion.Throughout this paper, we will use the following notation. LetΩ be a connected bounded open subset of ℝl,l=2,3. ∂Ω∈C0,1,(3)∂Ω=∪i=111Γ¯i=Γ¯D∪Γ¯R,ΓD∩ΓR=∅, Γi∩Γj=∅ for i≠j, and Γi=∪jΓij, where Γij are connected open subsets of ∂Ω, and Γij∈C2,1 for i=2,3,7 and Γij∈C1 for others. When X is a Banach space, X=Xl and X∗ is a dual space of X. Let Wα,pΩ be Sobolev spaces; H1Ω=W1,2Ω, and so H1Ω=H1Ωl. An inner product and norm in the spaces L2Ω and L2Ω are denoted, respectively, by ⋅,⋅ and ⋅, and ⋅,⋅ means the duality pairing between a Sobolev space X and its dual one. Also, ⋅,⋅Γi is an inner product in L2Γi or L2Γi, and ⋅,⋅Γi means the duality pairing between H1/2Γi and H−1/2Γi or between H1/2Γi and H−1/2Γi. The inner product and norms in ℝl, respectively, are denoted by ⋅,⋅ℝl and ⋅. Sometimes, the inner product between a and b in ℝl is denoted by a⋅b.Letnx and τx be, respectively, outward normal and tangent unit vectors at x in ∂Ω. When f∈H−1/2Γi, if f,wΓi≥0≤0∀w∈C0∞Γi with w≥0, then we denote by f≥0≤0 on Γi. For convergence in spaces, ⟶, ⇀, and ⇀∗ mean, respectively, strong, weak, and weak ∗ convergence. Derivative of ft,x with respect to t is denoted by f′. We also assume that 0<T<∞.


## 2. Problems and Assumptions

For temperature, we are concerned with the boundary conditions(4)θΓD=0,κθ∂θ∂n+βxθΓR=gRt,x,βx,gRt,x−given functions onΓR,0,T×ΓR.Stress tensorS is the one with components sij=−pδij+2μεijv, and stress vector on the boundary surface is σv,p=S⋅n. The value of the normal stress vector on the boundary surface is σnv,p=σ⋅n. And στv,p=σv,p−σnv,pn. Total stress tensor St is the one with components sijt=−p+1/2v2δij+2μθεijv, and the total stress vector on the boundary surface is σtθ,v,p=St⋅n. The value of the total normal stress vector on the boundary surface is σntθ,v,p=σt⋅n. And στtθ,v,p=σtθ,v,p−σntθ,v,pn.For Problem I, we assume thatμ and κ are independent of θ. Thus, Problem I is the one with the boundary conditions(5)vΓ1=0,vτΓ2=0,−pΓ2=ϕ2,vnΓ3=0,rotv×nΓ3=ϕ3μ,vτΓ4=0,−p+2μεnnvΓ4=ϕ4,vnΓ5=0,2μεnτv+αvτΓ5=ϕ5,α:amatrix,−pn+2μεnvΓ6=ϕ6,vτΓ7=0,−p+μ∂v∂n⋅nΓ7=ϕ7,vn=0,στv≤gτ,στv⋅vτ+gτvτ=0onΓ8,vτ=0,σnv,p≤gn,σnv,pvn+gnvn=0onΓ9,vτ=0,vn≥0,σnv,p+g+n≥0,σnv,p+g+nvn=0onΓ10,vτ=0,vn≤0,σnv,p−g−n≤0,σnv,p−g−nvn=0onΓ11,and Problem II is the one with the boundary conditions(6)vΓ1=0,vτΓ2=0,−p+12v2Γ2=ϕ2,vnΓ3=0,rotv×nΓ3=ϕ3μθ,vτΓ4=0,−p−12v2+2μθεnnvΓ4=ϕ4,vnΓ5=0,2μθεnτv+αvτΓ5=ϕ5,α:a matrix,−pn−12v2n+2μθεnvΓ6=ϕ6,vτΓ7=0,−p−12v2+μθ∂v∂n⋅nΓ7=ϕ7,vn=0,στtθ,v≤gτ,στtθ,v⋅vτ+gτvτ=0onΓ8,vτ=0,σntθ,v,p≤gn,σntθ,v,pvn+gnvn=0onΓ9,vτ=0,vn≥0,σntθ,v,p+g+n≥0,σntθ,v,p+g+nvn=0onΓ10,vτ=0,vn≤0,σntθ,v,p−g−n≤0,σntθ,v,p−g−nvn=0onΓ11,where εnv=ℰvn, εnnv=ℰvn,nℝl, εnτv=ℰvn−εnnvn, vτ=v−v⋅nn, vn=v⋅n, and hi,ϕi,αjk (components of matrix α) are given functions or vectors of functions of t,x on Σi=0,T×Γi. For convenience in what follows, the problems with boundary conditions (5) and (6) are called, respectively, the case of static pressure and the case of total pressure.Let(7)V=u∈H1Ω:divu=0,uΓ1=0,uτΓ2∪Γ4∪Γ7∪Γ9∪Γ10∪Γ11=0,unΓ3∪Γ5∪Γ8=0,H:completion inL2ΩofV,KΩ=u∈V:unΓ10≥0,unΓ11≤0,HK:closure inL2ΩofKΩ,KQ=u∈L20,T;V:u′∈L20,T;V∗;unΓ10≥0,unΓ11≤0,WΓD1,2Ω=y∈W1,2Ω:yΓD=0.We assume thatgτ∈L2Γ8, gn∈L2Γ9, g+n∈L2Γ10, and that g−n∈L2Γ11, and gτ>0, gn>0, g+n>0, and g−n>0 for a.e. x of the portions of boundary. Also, we use the following assumption.Assumption 1.
(for the case of static pressure). We assume the following:(1)
Γ1≠∅, ΓD≠∅, and(8)ΓR⊂∪i=1,3,5,8Γi.(2)
IfΓi, where i is 10 or 11, is nonempty, then at least one of Γj:j∈2,4,7,9−11\i is nonempty, and there exists diffeomorphism in C1 between Γi and Γj.(3)
For the functions of (1),(9)f∈W1,∞0,T;L3Ω,g∈W1,20,T;WΓD1,2Ω∗,μandκare independent ofθ.(4)
For the functions of (4) and (5),(10)gR∈W1,20,T;L4/3ΓR,β1≥βx≥0,β1−a constant,βx−measurable,ϕi∈W1,∞0,T;H−1/2Γi,i=2,4,7,ϕi∈W1,∞0,T;H−1/2Γi,i=3,5,6,αij∈L∞Γ5.Assumption 2.
(for the case of total pressure). We assume (1) and (2) of Assumption1 and the following:
(3′) For the functions of (1),(11)f∈L∞0,T;L3Ω,g∈L20,T;WΓD1,2Ω∗,μ∈Cℝ,0<μ0≤μξ≤μ1<∞,∀ξ∈ℝ,κ∈Cℝ,0<κ0≤κξ≤κ1<∞,∀ξ∈ℝ.
(4′) For the functions of (4) and (6),(12)gR∈L20,T;L4/3ΓR,β1≥βx≥0,β1−a constant,βx−measurable,ϕi∈L20,T;H−1/2Γi,i=2,4,7,ϕi∈L20,T;H−1/2Γi,i=3,5,6,αij∈L∞Γ5.Remark 1.
OnΓ10Γ11, only outflow (inflow) is possible, and so (2) of Assumption 1 is used to guarantee divv=0. In Theorems 3.3 and 3.5 of [24], for the proof of equivalence of variational formulations to variational inequalities, this assumption was used via Lemma 3.2 of [24]. In this paper, this assumption is also necessary to guarantee equivalence between Problems I-VE and I-VI in Remark 4.


## 3. Variational Formulations and Main Results

SinceΓ1≠∅ and ΓD≠∅, by the Korn inequality and Poincaré inequality, we use(13)v,uV=ℰv,ℰu,y,zWΓD1,2Ω=∇y,∇z.


### 3.1. Variational Formulations: The Case of Static Pressure

By Theorems 2.1 and 2.2 of [19], for v∈H2Ω∩V and u∈V, we have(14)−2∇⋅μℰv,u=2μℰv,ℰu−2μℰvn,u∪i=211Γi=2μℰv,ℰu+2μkxv,uΓ2−μrotv×n,uΓ3+2μSv˜,u˜Γ3−2μεnnv,unΓ4=−2μεnτv,uΓ5−2μεnv,uΓ6−μ∂v∂n,uΓ7+μkxv,uΓ7−2μεnτv,uΓ8−2μεnnv,unΓ9−2μεnnv,unΓ10−2μεnnv,unΓ11,where S is the shape operator of the boundary surface (cf. (A.1) in [19]), v˜,u˜ are expressions of v,u in a local coordinate system on Γ3, and kxΓi=2×mean curvature ofΓi.Forp∈H1Ω and u∈V, we have(15)∇p,u=−p,divu+p,un∪i=211Γi=p,unΓ2∪Γ4∪Γ7∪Γ9∪Γ10∪Γ11+pn,uΓ6,where unΓ3∪Γ5∪Γ8=0 was used. By (4), for θ∈H2Ω and φ∈WΓD1,2Ω, we have(16)−∇⋅κ∇θ,φ=κ∇θ,∇φ−κ∂θ∂n,φΓR=κ∇θ,∇φ+βθ−gR,φΓR.Taking into account (8) and vnΓ3∪Γ5∪Γ8=0, for v∈V, θ∈H1Ω, and φ∈WΓD1,2Ω, we have(17)v⋅∇θ,φ=vnθ,φΓR−θv,∇φ=−θv,∇φ.By (14)–(17), we can see that smooth solutions v,p,θ of problems (1), (4), and (5) satisfy(18)∂v∂t,u+2μℰv,ℰu+v⋅∇v,u+2μkxv,uΓ2+2μSv˜,u˜Γ3+2αxv,uΓ5+μkxv,uΓ7−2μεnτv,uΓ8+p−2μεnnv,unΓ9∪Γ10∪Γ11=1−α0θf,u+∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V,∂θ∂t,φ+κ∇θ,∇φ−θv,∇φ+βθ,φΓR=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω,στθ,v≤gτ,στθ,v⋅vτ+gτvτ=0,onΓ8,σnv,p≤gn,σnv,pvn+gnvn=0,onΓ9,σnv,p+g+n≥0,σnv,p+g+nvn=0,onΓ10,σnv,p−g−n≤0,σnv,p−g−nvn=0,onΓ11,θΓD=0.Definea01⋅,⋅,a11⋅,⋅,⋅, and f1t∈V∗ by(19)a01w,u=2μℰw,ℰu+2μkxw,uΓ2+2μSw˜,u˜Γ3+2αxw,uΓ5+μkxw,uΓ7,∀w,u∈V,a11v,u,w=v⋅∇u,w,∀v,u,w∈V,f1t,u=∑i=2,4,7ϕit,unΓi+∑i=3,5,6ϕit,uΓi,∀u∈V.Defineb1⋅,⋅ and g1t∈WΓD1,2Ω∗ by(20)b1θ,φ=κ∇θ,∇φ+βxθ,φΓR,∀θ∈W1,2Ω,φ∈WΓD1,2Ω,g1t,φ=gRt,φΓR+gt,φ,∀φ∈WΓD1,2Ω.Remark 2.
Under (4) of Assumption1, the duality product f1,u of (19) has a meaning (cf. Remark 3.1 in [24]). By (9) and (10),(21)f1∈W1,∞0,T;V∗,g1∈W1,20,T;WΓD1,2Ω∗.
Then, taking into account(22)στθ,v=2μεnτv,σnθ,v,p=−p+2μεnnv,and (18), we introduce the following variational formulation for problems (1), (4), and (5).Problem (I-VE).
Findv∈KQ, θ∈L20,T;WΓD1,2Ω, and στ,σn,σ+n,σ−n∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11 a.e. t∈0,T such that v0=v0, θ0=θ0, and(23)∂v∂t,u+a01v,u+a11v,v,u−στ,uτΓ8−σn,unΓ9−σ+n,unΓ10−σ−n,unΓ11−f−α0θf,u=f1,u,∀u∈L20,T;V,∂θ∂t,φ+b1θ,φ−θv,∇φ=g1,φ,∀φ∈L20,T;WΓD1,2Ω,στ≤gτ,στ⋅vτ+gτvτ=0,onΓ8,σn≤gn,σnvn+gnvn=0,onΓ9,σ+n+g+n≥0,σ+n+g+n,vnΓ10=0onΓ10,σ−n−g−n≤0,σ−n−g−n,vnΓ11=0,onΓ11,where Lτ2Γ8 is the subspace of L2Γ8 consisting of functions such that u,nL2Γ8=0.Remark 3.
Under Assumption1, if a solution is smooth enough, (v∈L20,T;H2Ω, v′∈L20,T;L2Ω, θ∈L20,T;W2,2Ω,θ′∈L20,T;L2Ω), then Problem I-VE is equivalent to problems (1), (4), and (5) in the following sense.
By Theorem 3.4 of [24], at a.e., there exists pt satisfying the first equation of (1), and v,p satisfies boundary condition (5). As given in Section 1, ch. 2 of [25], it is proved that θ satisfies the third equation of (1) and boundary condition (4).
We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE if the solution is smooth enough (cf. Remark4).
For fixedθ, let us consider the problem(24)∂v∂t,u+a01v,u+a11v,v,u−στ,uτΓ8−σn,unΓ9−σ+n,unΓ10−σ−n,unΓ11−f−α0θf,u=f1,u,∀u∈L20,T;V,στ≤gτ,στ⋅vτ+gτvτ=0onΓ8,σn≤gn,σnvn+gnvn=0onΓ9,σ+n+g+n≥0,σ+n+g+n,vnΓ10=0onΓ10,σ−n−g−n≤0,σ−n−g−n,vnΓ11=0onΓ11.
Subtracting the first formula of (21) with u=v from the first formula of (21), we get(25)∂v∂t,u−v+a01v,u−v+a11v,v,u−v−στ,uτ−vτΓ8−σn,un−vnΓ9,−σ+n,un−vnΓ10−σ−n,un−vnΓ11−f−α0θf,u−v=f1,u−v,∀u∈V.
Define the functionalsϕτ,ϕn,ϕ+,ϕ−, respectively, by(26)ϕτη=∫Γ8gτηdx,∀η∈Lτ2Γ8,ϕnη=∫Γ9gnηdx,∀η∈L2Γ9,ϕ+η=∫Γ10g+nηdx,∀η∈L2Γ10,ϕ−η=−∫Γ11g−nηdx,∀η∈L2Γ11.
Since ifu∈KΩ, then uΓ8∈Lτ2Γ8, unΓ9∈L2Γ9, unΓ10∈L2Γ10, and unΓ11∈L2Γ11, in what follows, for convenience, we use the notation(27)ϕτu=ϕτuΓ8,ϕnu=ϕnunΓ9,ϕ+u=ϕ+unΓ10,ϕ−u=ϕ−unΓ11,∀u∈KΩ.
Define a functionalΦ:V⟶ℝ¯≡ℝ∪+∞ by(28)Φu=ϕτu+ϕnu+ϕ+u+ϕ−u,∀u∈KΩ,+∞,∀u∉KΩ.
Note thatΦu≥0 since unΓ10≥0,unΓ11≤0,∀u∈KΩ. Then, the functional Φ∈V⟶ℝ¯ is proper (cf. Definition A.1 of [21]), convex, lower semicontinuous, and(29)Φu≥0,∀u∈V,Φ0V=0.
Define a functionalΨu by(30)Ψu=∫0TΦutdt,ifΦut∈L10,T,+∞,otherwise.
In the same way as Problem I in [24], from (25), we get(31)∂v∂t,u−v+a01v,u−v+a11v,v,u−v+Φu−Φv≥1−α0θf,u−v+f1,u−v.
Define operatorsA1∈V⟶V∗, B1∈V×V⟶V∗, and C1∈WΓD1,2Ω⟶WΓD1,2Ω∗, respectively, by(32)A1v,u=a01v,u,∀v,u∈V,B1v,u,w=a11v,u,w,∀v,u,w∈V,C1θ,φ=b01θ,φ,∀θ,φ∈WΓD1,2Ω.
Ifv is a solution to (24), then we can see that the solution satisfies (cf. (3 20) of [21])(33)∫0Tv′t+A1vt+B1vt,vt−1−α0θtft−f1t,ut−vtdt+Ψu−Ψv≥0,∀u∈L40,T;V.
Therefore, we have the following variational formulation for problems (1), (4), and (5).Problem (I-VI).
Findv,θ∈L∞0,T;H∩L20,T;V×L20,T;WΓD1,2∩L∞0,T;L2 such that(34)∫0Tv′+A1vt+B1vt,vt−1−α0θf−f1,ut−vtdt+Ψu−Ψv≥0,∀u∈L40,T;V,∫0T∂θ∂t,φ+C1θt,φ−θv,∇φ−g1,φdt=0,∀φ∈L20,T;WΓD1,2,v0=v0,θ0=θ0.Remark 4.
If the solutions to Problem I-VI are smooth as much asv∈L20,T;V and v′∈L20,T;V∗, then the first one of (34) is equivalent to(35)v′t+A1vt+B1vt,vt−1−α0θf−f1,u−vt+Φt−Φvt≥0,for a.e.t∈0,T,∀u∈KΩ,(Remark, pp. 114 in [26]). In (31), putting F1,u−v=−∂v/∂t−1−α0θf−f1,u−v, by Theorem 3.5 of [24], we can see existence of στ,σn,σ+n,σ−n∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11 for a.e. t∈0,T such that v,θ,στ,σn,σ+n,σ−n is a solution to Problem I-VE.


### 3.2. Variational Formulations: The Case of Total Pressure

Takingv⋅∇v=rotv×v+1/2gradv2 into account, by (14)–(17) with μ,κ depending on θ, we can see that smooth solutions v,p,θ of problems (1), (4), and (6) satisfy the following:(36)∂v∂t,u+2μθℰv,ℰu+rotv×v,u+2μθkxv,uΓ2+2μθSv˜,u˜Γ3+2αxv,uΓ5+μθkxv,uΓ7−2μθεnτv,uΓ8+p+12v2−2μθεnnv,unΓ9∪Γ10∪Γ11=1−α0θf,u+∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V,∂θ∂t,φ+κθ∇θ,∇φ−θv,∇φ+βθ,φΓR=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω,στtθ,v≤gτ,στtθ,v⋅vτ+gτvτ=0onΓ8,σntθ,v,p≤gn,σntθ,v,pvn+gnvn=0onΓ9,σntθ,v,p+g+n≥0,σntθ,v,p+g+nvn=0onΓ10,σntθ,v,p−g−n≤0,σntθ,v,p−g−nvn=0onΓ11,θΓD=0.Definea02θ˜;⋅,⋅,a12⋅,⋅,⋅, and f2∈V∗ by(37)a02θ˜;w,u=2μθ˜ℰw,ℰu+2μθ˜kxw,uΓ2+2μθ˜Sw˜,u˜Γ3+2αxw,uΓ5+μθ˜kxw,uΓ7,∀w,u∈V,θ˜∈WΓD1,2Ω,a12v,u,w=rotv×u,w,∀v,u,w∈V,f2,u=∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V.Defineb2θ˜;⋅,⋅ and g2∈WΓD1,2Ω∗ by(38)b2θ˜;θ,φ=κθ˜∇θ,∇φ+βxθ,φΓR,∀θ˜,θ∈WΓD1,2Ω,φ∈WΓD1,2Ω,g2,φ=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω.By (11) and (12),(39)f2∈L20,T;V∗,g2∈L20,T;WΓD1,2∗.Then, taking into account(40)στtθ,v=2μθεnτv,σntθ,v,p=−p+12v2+2μθεnnv,and (36), we introduce the following variational formulation for problems (1), (4), and (6).Problem (II-VE).
Findv∈KQ, θ∈L∞0,T;L2Ω∩L20,T;WΓD1,2Ω, and στt,σnt,σ+nt,σ−nt∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11, in a.e., t∈0,T, such that v0=v0, θ0=θ0, and(41)∂v∂t,u+a02θ;v,u+a12v,v,u−στt,uτΓ8−σnt,unΓ9−σ+nt,unΓ10−σ−nt,unΓ11−f−α0θf,u=f2,u,∀u∈V,∂θ∂t,φ+b2θ;θ,φ−θv,∇φ=g2,φ,∀φ∈WΓD1,2Ω,στt≤gτ,στt⋅vτ+gτvτ=0onΓ8,σnt≤gn,σntvn+gnvn=0onΓ9,σ+nt+g+n≥0,σ+nt+g+n,vnΓ10=0onΓ10,σ−nt−g−n≤0,σ−nt−g−n,vnΓ11=0onΓ11,where Lτ2Γ8 is the subspace of L2Γ8 consisting of functions such that u,nL2Γ8=0.
Define operatorsA2θ∈V⟶V∗ and B2∈V×V⟶V∗, respectively, by(42)A2θ˜v,u=a02θ˜;v,u,∀v,u∈V,θ˜∈WΓD1,2Ω,B2v,u,w=a12v,u,v,∀v,u,w∈V.
Let functionalΨ be defined by (25)–(28). Then, in the same way as Problem I-VI of [21], we find a variational inequality for velocity. Then, we get another variational formulation consisting of a variational inequality for velocity and a variational equation for temperature, which is equivalent to Problem II-VE if the solution is smooth enough.Problem (II-VI).
Findv,θ∈L∞0,T;H∩L20,T;V×L∞0,T;L2Ω∩L20,T;WΓD1,2Ω such that(43)∫0Tu′+A2θvt+B2vt,vt−1−α0θf−f2,ut−vtdt+Ψu−Ψv≥−12v0−u02,∀u∈L40,T;Vwithu′∈L20,T;V∗,∫0T−θ,∂φ∂t+b2θ;θ,φ−θv,∇φ−g2,φdt=θ0x,φx,0,∀φ∈C10,T;WΓD1,2Ωwithφ⋅,T=0.


### 3.3. Main Results

The main results of this paper are as follows.Theorem 1 (the case of static pressure).
Let Assumption1 be satisfied. Suppose that(1)
The norms off, ϕi,i=2−6,g,gR in the spaces they belong to are small enough(2)
v0∈V and Φv0=0(3)
θ0∈WΓD1,2Ω(4)
A1v0+B1v0,v0−f10∈H (compatibility condition at initial time for velocity)(5)
v0V and A1v0+B1v0,v0−1−α0θ0f0−f10 are small enough(6)
C1θ0+v0⋅∇θ0−g10∈L2Ω (compatibility condition at initial time for temperature)(7)
θ0WΓD1,2Ω and C1θ0+v0⋅∇θ0−g10L2Ω are small enough
Then, there exists a solutionv,θ to (34) such that(44)v∈C0,T;V,v′∈L20,T;V∩L∞0,T;H,θ∈C0,T;WΓD1,2Ω,θ′∈L20,T;WΓD1,2Ω∩L∞0,T;L2Ω.The solution satisfyingvV≤c and θWΓD1,2≤c for a constant c>0 small enough is unique.Theorem 2 (the case of total pressure).
Let Assumption2 be satisfied; v0∈HK and θ0∈L2Ω. Then, there exists a solution v,θ∈L∞0,T;H∩L20,T;V×L∞0,T;L2Ω∩L20,T;WΓD1,2Ω to (43).


## 3.1. Variational Formulations: The Case of Static Pressure

By Theorems 2.1 and 2.2 of [19], for v∈H2Ω∩V and u∈V, we have(14)−2∇⋅μℰv,u=2μℰv,ℰu−2μℰvn,u∪i=211Γi=2μℰv,ℰu+2μkxv,uΓ2−μrotv×n,uΓ3+2μSv˜,u˜Γ3−2μεnnv,unΓ4=−2μεnτv,uΓ5−2μεnv,uΓ6−μ∂v∂n,uΓ7+μkxv,uΓ7−2μεnτv,uΓ8−2μεnnv,unΓ9−2μεnnv,unΓ10−2μεnnv,unΓ11,where S is the shape operator of the boundary surface (cf. (A.1) in [19]), v˜,u˜ are expressions of v,u in a local coordinate system on Γ3, and kxΓi=2×mean curvature ofΓi.Forp∈H1Ω and u∈V, we have(15)∇p,u=−p,divu+p,un∪i=211Γi=p,unΓ2∪Γ4∪Γ7∪Γ9∪Γ10∪Γ11+pn,uΓ6,where unΓ3∪Γ5∪Γ8=0 was used. By (4), for θ∈H2Ω and φ∈WΓD1,2Ω, we have(16)−∇⋅κ∇θ,φ=κ∇θ,∇φ−κ∂θ∂n,φΓR=κ∇θ,∇φ+βθ−gR,φΓR.Taking into account (8) and vnΓ3∪Γ5∪Γ8=0, for v∈V, θ∈H1Ω, and φ∈WΓD1,2Ω, we have(17)v⋅∇θ,φ=vnθ,φΓR−θv,∇φ=−θv,∇φ.By (14)–(17), we can see that smooth solutions v,p,θ of problems (1), (4), and (5) satisfy(18)∂v∂t,u+2μℰv,ℰu+v⋅∇v,u+2μkxv,uΓ2+2μSv˜,u˜Γ3+2αxv,uΓ5+μkxv,uΓ7−2μεnτv,uΓ8+p−2μεnnv,unΓ9∪Γ10∪Γ11=1−α0θf,u+∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V,∂θ∂t,φ+κ∇θ,∇φ−θv,∇φ+βθ,φΓR=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω,στθ,v≤gτ,στθ,v⋅vτ+gτvτ=0,onΓ8,σnv,p≤gn,σnv,pvn+gnvn=0,onΓ9,σnv,p+g+n≥0,σnv,p+g+nvn=0,onΓ10,σnv,p−g−n≤0,σnv,p−g−nvn=0,onΓ11,θΓD=0.Definea01⋅,⋅,a11⋅,⋅,⋅, and f1t∈V∗ by(19)a01w,u=2μℰw,ℰu+2μkxw,uΓ2+2μSw˜,u˜Γ3+2αxw,uΓ5+μkxw,uΓ7,∀w,u∈V,a11v,u,w=v⋅∇u,w,∀v,u,w∈V,f1t,u=∑i=2,4,7ϕit,unΓi+∑i=3,5,6ϕit,uΓi,∀u∈V.Defineb1⋅,⋅ and g1t∈WΓD1,2Ω∗ by(20)b1θ,φ=κ∇θ,∇φ+βxθ,φΓR,∀θ∈W1,2Ω,φ∈WΓD1,2Ω,g1t,φ=gRt,φΓR+gt,φ,∀φ∈WΓD1,2Ω.Remark 2.
Under (4) of Assumption1, the duality product f1,u of (19) has a meaning (cf. Remark 3.1 in [24]). By (9) and (10),(21)f1∈W1,∞0,T;V∗,g1∈W1,20,T;WΓD1,2Ω∗.
Then, taking into account(22)στθ,v=2μεnτv,σnθ,v,p=−p+2μεnnv,and (18), we introduce the following variational formulation for problems (1), (4), and (5).Problem (I-VE).
Findv∈KQ, θ∈L20,T;WΓD1,2Ω, and στ,σn,σ+n,σ−n∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11 a.e. t∈0,T such that v0=v0, θ0=θ0, and(23)∂v∂t,u+a01v,u+a11v,v,u−στ,uτΓ8−σn,unΓ9−σ+n,unΓ10−σ−n,unΓ11−f−α0θf,u=f1,u,∀u∈L20,T;V,∂θ∂t,φ+b1θ,φ−θv,∇φ=g1,φ,∀φ∈L20,T;WΓD1,2Ω,στ≤gτ,στ⋅vτ+gτvτ=0,onΓ8,σn≤gn,σnvn+gnvn=0,onΓ9,σ+n+g+n≥0,σ+n+g+n,vnΓ10=0onΓ10,σ−n−g−n≤0,σ−n−g−n,vnΓ11=0,onΓ11,where Lτ2Γ8 is the subspace of L2Γ8 consisting of functions such that u,nL2Γ8=0.Remark 3.
Under Assumption1, if a solution is smooth enough, (v∈L20,T;H2Ω, v′∈L20,T;L2Ω, θ∈L20,T;W2,2Ω,θ′∈L20,T;L2Ω), then Problem I-VE is equivalent to problems (1), (4), and (5) in the following sense.
By Theorem 3.4 of [24], at a.e., there exists pt satisfying the first equation of (1), and v,p satisfies boundary condition (5). As given in Section 1, ch. 2 of [25], it is proved that θ satisfies the third equation of (1) and boundary condition (4).
We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE if the solution is smooth enough (cf. Remark4).
For fixedθ, let us consider the problem(24)∂v∂t,u+a01v,u+a11v,v,u−στ,uτΓ8−σn,unΓ9−σ+n,unΓ10−σ−n,unΓ11−f−α0θf,u=f1,u,∀u∈L20,T;V,στ≤gτ,στ⋅vτ+gτvτ=0onΓ8,σn≤gn,σnvn+gnvn=0onΓ9,σ+n+g+n≥0,σ+n+g+n,vnΓ10=0onΓ10,σ−n−g−n≤0,σ−n−g−n,vnΓ11=0onΓ11.
Subtracting the first formula of (21) with u=v from the first formula of (21), we get(25)∂v∂t,u−v+a01v,u−v+a11v,v,u−v−στ,uτ−vτΓ8−σn,un−vnΓ9,−σ+n,un−vnΓ10−σ−n,un−vnΓ11−f−α0θf,u−v=f1,u−v,∀u∈V.
Define the functionalsϕτ,ϕn,ϕ+,ϕ−, respectively, by(26)ϕτη=∫Γ8gτηdx,∀η∈Lτ2Γ8,ϕnη=∫Γ9gnηdx,∀η∈L2Γ9,ϕ+η=∫Γ10g+nηdx,∀η∈L2Γ10,ϕ−η=−∫Γ11g−nηdx,∀η∈L2Γ11.
Since ifu∈KΩ, then uΓ8∈Lτ2Γ8, unΓ9∈L2Γ9, unΓ10∈L2Γ10, and unΓ11∈L2Γ11, in what follows, for convenience, we use the notation(27)ϕτu=ϕτuΓ8,ϕnu=ϕnunΓ9,ϕ+u=ϕ+unΓ10,ϕ−u=ϕ−unΓ11,∀u∈KΩ.
Define a functionalΦ:V⟶ℝ¯≡ℝ∪+∞ by(28)Φu=ϕτu+ϕnu+ϕ+u+ϕ−u,∀u∈KΩ,+∞,∀u∉KΩ.
Note thatΦu≥0 since unΓ10≥0,unΓ11≤0,∀u∈KΩ. Then, the functional Φ∈V⟶ℝ¯ is proper (cf. Definition A.1 of [21]), convex, lower semicontinuous, and(29)Φu≥0,∀u∈V,Φ0V=0.
Define a functionalΨu by(30)Ψu=∫0TΦutdt,ifΦut∈L10,T,+∞,otherwise.
In the same way as Problem I in [24], from (25), we get(31)∂v∂t,u−v+a01v,u−v+a11v,v,u−v+Φu−Φv≥1−α0θf,u−v+f1,u−v.
Define operatorsA1∈V⟶V∗, B1∈V×V⟶V∗, and C1∈WΓD1,2Ω⟶WΓD1,2Ω∗, respectively, by(32)A1v,u=a01v,u,∀v,u∈V,B1v,u,w=a11v,u,w,∀v,u,w∈V,C1θ,φ=b01θ,φ,∀θ,φ∈WΓD1,2Ω.
Ifv is a solution to (24), then we can see that the solution satisfies (cf. (3 20) of [21])(33)∫0Tv′t+A1vt+B1vt,vt−1−α0θtft−f1t,ut−vtdt+Ψu−Ψv≥0,∀u∈L40,T;V.
Therefore, we have the following variational formulation for problems (1), (4), and (5).Problem (I-VI).
Findv,θ∈L∞0,T;H∩L20,T;V×L20,T;WΓD1,2∩L∞0,T;L2 such that(34)∫0Tv′+A1vt+B1vt,vt−1−α0θf−f1,ut−vtdt+Ψu−Ψv≥0,∀u∈L40,T;V,∫0T∂θ∂t,φ+C1θt,φ−θv,∇φ−g1,φdt=0,∀φ∈L20,T;WΓD1,2,v0=v0,θ0=θ0.Remark 4.
If the solutions to Problem I-VI are smooth as much asv∈L20,T;V and v′∈L20,T;V∗, then the first one of (34) is equivalent to(35)v′t+A1vt+B1vt,vt−1−α0θf−f1,u−vt+Φt−Φvt≥0,for a.e.t∈0,T,∀u∈KΩ,(Remark, pp. 114 in [26]). In (31), putting F1,u−v=−∂v/∂t−1−α0θf−f1,u−v, by Theorem 3.5 of [24], we can see existence of στ,σn,σ+n,σ−n∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11 for a.e. t∈0,T such that v,θ,στ,σn,σ+n,σ−n is a solution to Problem I-VE.


## 3.2. Variational Formulations: The Case of Total Pressure

Takingv⋅∇v=rotv×v+1/2gradv2 into account, by (14)–(17) with μ,κ depending on θ, we can see that smooth solutions v,p,θ of problems (1), (4), and (6) satisfy the following:(36)∂v∂t,u+2μθℰv,ℰu+rotv×v,u+2μθkxv,uΓ2+2μθSv˜,u˜Γ3+2αxv,uΓ5+μθkxv,uΓ7−2μθεnτv,uΓ8+p+12v2−2μθεnnv,unΓ9∪Γ10∪Γ11=1−α0θf,u+∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V,∂θ∂t,φ+κθ∇θ,∇φ−θv,∇φ+βθ,φΓR=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω,στtθ,v≤gτ,στtθ,v⋅vτ+gτvτ=0onΓ8,σntθ,v,p≤gn,σntθ,v,pvn+gnvn=0onΓ9,σntθ,v,p+g+n≥0,σntθ,v,p+g+nvn=0onΓ10,σntθ,v,p−g−n≤0,σntθ,v,p−g−nvn=0onΓ11,θΓD=0.Definea02θ˜;⋅,⋅,a12⋅,⋅,⋅, and f2∈V∗ by(37)a02θ˜;w,u=2μθ˜ℰw,ℰu+2μθ˜kxw,uΓ2+2μθ˜Sw˜,u˜Γ3+2αxw,uΓ5+μθ˜kxw,uΓ7,∀w,u∈V,θ˜∈WΓD1,2Ω,a12v,u,w=rotv×u,w,∀v,u,w∈V,f2,u=∑i=2,4,7ϕi,unΓi+∑i=3,5,6ϕi,uΓi,∀u∈V.Defineb2θ˜;⋅,⋅ and g2∈WΓD1,2Ω∗ by(38)b2θ˜;θ,φ=κθ˜∇θ,∇φ+βxθ,φΓR,∀θ˜,θ∈WΓD1,2Ω,φ∈WΓD1,2Ω,g2,φ=gR,φΓR+g,φ,∀φ∈WΓD1,2Ω.By (11) and (12),(39)f2∈L20,T;V∗,g2∈L20,T;WΓD1,2∗.Then, taking into account(40)στtθ,v=2μθεnτv,σntθ,v,p=−p+12v2+2μθεnnv,and (36), we introduce the following variational formulation for problems (1), (4), and (6).Problem (II-VE).
Findv∈KQ, θ∈L∞0,T;L2Ω∩L20,T;WΓD1,2Ω, and στt,σnt,σ+nt,σ−nt∈Lτ2Γ8×L2Γ9×H−1/2Γ10×H−1/2Γ11, in a.e., t∈0,T, such that v0=v0, θ0=θ0, and(41)∂v∂t,u+a02θ;v,u+a12v,v,u−στt,uτΓ8−σnt,unΓ9−σ+nt,unΓ10−σ−nt,unΓ11−f−α0θf,u=f2,u,∀u∈V,∂θ∂t,φ+b2θ;θ,φ−θv,∇φ=g2,φ,∀φ∈WΓD1,2Ω,στt≤gτ,στt⋅vτ+gτvτ=0onΓ8,σnt≤gn,σntvn+gnvn=0onΓ9,σ+nt+g+n≥0,σ+nt+g+n,vnΓ10=0onΓ10,σ−nt−g−n≤0,σ−nt−g−n,vnΓ11=0onΓ11,where Lτ2Γ8 is the subspace of L2Γ8 consisting of functions such that u,nL2Γ8=0.
Define operatorsA2θ∈V⟶V∗ and B2∈V×V⟶V∗, respectively, by(42)A2θ˜v,u=a02θ˜;v,u,∀v,u∈V,θ˜∈WΓD1,2Ω,B2v,u,w=a12v,u,v,∀v,u,w∈V.
Let functionalΨ be defined by (25)–(28). Then, in the same way as Problem I-VI of [21], we find a variational inequality for velocity. Then, we get another variational formulation consisting of a variational inequality for velocity and a variational equation for temperature, which is equivalent to Problem II-VE if the solution is smooth enough.Problem (II-VI).
Findv,θ∈L∞0,T;H∩L20,T;V×L∞0,T;L2Ω∩L20,T;WΓD1,2Ω such that(43)∫0Tu′+A2θvt+B2vt,vt−1−α0θf−f2,ut−vtdt+Ψu−Ψv≥−12v0−u02,∀u∈L40,T;Vwithu′∈L20,T;V∗,∫0T−θ,∂φ∂t+b2θ;θ,φ−θv,∇φ−g2,φdt=θ0x,φx,0,∀φ∈C10,T;WΓD1,2Ωwithφ⋅,T=0.


## 3.3. Main Results

The main results of this paper are as follows.Theorem 1 (the case of static pressure).
Let Assumption1 be satisfied. Suppose that(1)
The norms off, ϕi,i=2−6,g,gR in the spaces they belong to are small enough(2)
v0∈V and Φv0=0(3)
θ0∈WΓD1,2Ω(4)
A1v0+B1v0,v0−f10∈H (compatibility condition at initial time for velocity)(5)
v0V and A1v0+B1v0,v0−1−α0θ0f0−f10 are small enough(6)
C1θ0+v0⋅∇θ0−g10∈L2Ω (compatibility condition at initial time for temperature)(7)
θ0WΓD1,2Ω and C1θ0+v0⋅∇θ0−g10L2Ω are small enough
Then, there exists a solutionv,θ to (34) such that(44)v∈C0,T;V,v′∈L20,T;V∩L∞0,T;H,θ∈C0,T;WΓD1,2Ω,θ′∈L20,T;WΓD1,2Ω∩L∞0,T;L2Ω.The solution satisfyingvV≤c and θWΓD1,2≤c for a constant c>0 small enough is unique.Theorem 2 (the case of total pressure).
Let Assumption2 be satisfied; v0∈HK and θ0∈L2Ω. Then, there exists a solution v,θ∈L∞0,T;H∩L20,T;V×L∞0,T;L2Ω∩L20,T;WΓD1,2Ω to (43).


## 4. Proof of Theorem 1

### 4.1. Existence and Estimation of Solutions to an Approximate Problem

We first consider a problem approximating (34).For every0<ε<1, define a functional Φε by(45)Φεy=infy−uV22ε+Φu;u∈V,y∈V,which is called the Moreau regularization of Φ. When ∂Φ:V⟶2V is the subdifferential of Φ in the Hilbert space V, let Jε=I+ε∂Φ−1 and ∂Φε:=ε−1I−Jε(the Yosida approximation of ∂ϕ) for all ε>0. Then, the functional Φε is convex, continuous, Fréchet differentiable, and ∇Φε=∂Φε≡ε−1I−Jε for all 1>ε>0. Moreover,(46)Φεy=y−JεyV22ε+ΦJεy,∀y∈V,(47)limε⟶0Φεy=Φy,ΦJεy≤Φεy≤Φy,∀y∈V,(cf. Theorem 2.9 in [27]). The operator ∇Φε is Lipschitz continuous with the constant 2ε−1(cf. Proposition 2.3 in [27]) and monotone (cf. Lemma 4.10 of ch. III in [25]).By the fact thatΓ2j,Γ3j,andΓ7j are in C2.1Γij and 4 of Assumption 1, there exists a constant M such that(48)Sx∞,kx∞,αL∞Γ5≤M.Thus, there existsc∗ such that(49)2μkxz,zΓ2+2μSz˜,z˜Γ3+αxz,zΓ5+μkxz,zΓ7≤μ4zV2+c∗z2,∀z∈V,(cf. .5.1.10 of [28]). Thus,(50)A1u,u≥7μ4uV2−c∗u2,∀u∈V,A1u,v≤c1uVvV,∃c1>0,∀u,v∈V,(51)B1v,u,w≤c2vVuVwV,∀u,v∈V,where the operators A1,B1 are the ones in (32).Letuj,j=1,2,… and φj,j=1,2,… be, respectively, bases of the space V and WΓD1,2Ω. Without loss of generality, we assume that u1=v0 and φ1=θ0 as in [26]. We find a solution vm=∑j=1mgjmtuj and θm=∑j=1mrjmtφj to the problem(52)∂vm∂t,uj+2μℰvm,ℰuj+vm⋅∇vm,uj+2μkxvm,ujΓ2+2μSv˜m,u˜jΓ3+2αxvm,ujΓ5+μkxvm,ujΓ7+∇Φεvmt,uj=1−α0θmf,uj+f1,uj,∂θm∂t,φj+κ∇θm,∇φj+βxθm,φjΓR−vmθm,∇φj=g1,φj,vm0=v0,θm0=θ0,which gives us a system for gjmt and rjmt,j=1−m. The solutions to (52) depend on ε, but for convenience of notation, here and in what follows, we use subindex m instead of subindex mε. For tm, there exist absolute continuous functions gjmt and rjmt on 0,tm. Since f∈W1,∞0,T;L3Ω, f1∈W1,∞0,T;V∗, g1∈W1,20,T;WΓD1,2Ω∗, and ∇Φε is Lipschitz continuous, gjm′t and rjm′t are in fact absolute continuous. If vmt and θmt are bounded and vmt,θmt are integrable, then gjmt and rjmt are prolonged over tm. Under smallness of the data of the problem and the compatibility condition of the data at the initial instant, we will find estimates for vmt and θmt in the following, by which we obtain (111) and see that tm=T.Multiplying the first and the second equation of (52), respectively, by gjmt and φjmt and adding for i=1,…,m, we get(53)∂vm∂t,vm+2μℰvm,ℰvm+vm⋅∇vm,vm+2μkxvm,vmΓ2+2μSv˜m,v˜mΓ3+2αxvm,vmΓ5+μkxvm,vmΓ7+∇Φεvmt,vm=1−α0θmf,vm+f1,vm,∂θm∂t,θm+κ∇θm,∇θm+βxθm,θmΓR−vmθm,∇θm=g1,θm,vm0=v0,θm0=θ0.We will find a priori estimates for(54)It:=vmt2+vm′t2+θmt2+θm′t2.SinceΦε is convex, continuous, and Fréchet differentiable, we have(55)Φεu−Φεvmt≥∇Φεvmt,u−vmt,∀u∈V,and so by Φε0V=0,(56)0≤Φεvmt≤∇Φεvmt,vmt.Also,(57)2−α0θmtf,vmt≤cα0θmtWΓD1,22fL32+μ4vmtV2.By virtue of (50), (51), (56), and (57), we have from the first equation of (53)(58)ddtvmt2+7μ2vmtV2−2c2vmtV3+2Φεvmt≤cα0θmtWΓD1,22fL32+cfL32+cf1V∗2+μ2vmtV2+2c∗vmt2,where c∗ and c2 are, respectively, the ones in (49) and (51), and so(59)ddtvmt2+3μ−2c2vmtVvmtV2+2Φεvmt≤cα0θmtWΓD1,22fL32+cfL32+cf1V∗2+2c∗vmt2.Here and in what follows are the constants independent of the data of problem which are denoted byc with the exceptions of c∗andc2.Settingt=0 in the first equation of (52) and multiplying the resulting equation by gjm′0 and adding for j=1,…,m, we get(60)vm′02+A1vm0,vm′0+B1vm0,vm0,vm′0+∇Φεv0,vm′0=1−α0θ0f0,vm′0+f10,vm′0.By condition (2) of Theorem1, for any u∈V, we have Φu≥Φv0=0, which by (47), it implies that Φεv0=0 and ∇Φεv0=0. Then, from (60), we have(61)vm′0≤A1v0+B1v0,v0−1−α0θ0f0−f10,which is valid by the compatibility condition at the initial time for velocity (condition (4)) and the conditions for θ0,f. On the contrary, taking into account (50), (51), and (56), we have from the first equation of (53)(62)7μ2vmtV2≤2c2vmV3+21−α0θmtft,vmt+2f1t,vmt+2c∗vm2−2vmt,vm′t,and so(63)3μvmtV≤2c2vmtV2+cftL3+cα0θmtftL3+cf1tV∗+2δvm′t+2c∗δvmt,where δ is such that ⋅≤δ⋅V. Since vmθm,∇θm=0 by (17), we get from the second equation of (53)(64)ddtθmt2+2κθmtWΓD1,22+2βxθm,θmΓR=2g1,θmt.From (64), we have(65)ddtθmt2+κθmtWΓD1,22+βxθm,θmΓR≤1κg1WΓD1,2∗2,(66)θmt2+∫0t∫Ωκ∇θms2dxds≤θ02+1κ∫0tg1sWΓD1,2∗2ds.Settingt=0 in the second equation of (52) and multiplying the resulting equation by rjm′0 and adding for j=1,…,m, we get(67)θm′02+b1θ0,θm′0+v0⋅∇θ0,θm′0=g0,θm′0,where −v0θ0,∇θm′0=v0⋅∇θ0,θm′0 was used. From (67), we have(68)θm′0≤C1θ0+v0⋅∇θ0−g10,which is valid by the compatibility condition at the initial time for temperature (condition (6)).On the contrary, taking into accountvmθm,∇θm=0, from the second equation of (53), we have(69)κθmtWΓD1,22≤g1WΓD1,2∗θmtWΓD1,2+δ1θmtWΓD1,2θm′t,where δ1 is such that ⋅≤δ1⋅WΓD1,2, and so(70)θmtWΓD1,2≤1κg1WΓD1,2∗+δ1θm′t.Taking into account (66), we have from (63)(71)3μvmtV≤c2vmtV2+cfL∞0,T;L3+cα0θ02+1κ∫0Tg1WΓD1,2∗2ds1/2fL∞0,T;L3+f1L∞0,T;V∗+max2δ,2c∗δvm′t+vmt.Differentiating the first equality of (52) with respect to t, we have that(72)vm″t,vj+A1vm′t,vj+B1vmt,vmt′,vj+∇Φεvm′,vj=−α0θm′tf,vj−1−α0θmtf′,vj+f1′,vj.Multiplying (72) by gjm′t and adding for j, we have(73)vm″t,vm′t+A1vm′t,vm′t+B1vmt,vmt′,vm′t+∇Φεvm′,vm′t=−α0θm′tf,vm′t−1−α0θmtf′,vm′t+f1′,vm′t.CalculatingBvmt,vmt′, we have(74)B1vmt,vmt′,vm′t=B1vm′,vm,vm′+B1vm,vm′,vm′≤2c2vmVvm′V2,where c2 is the one in (51). Also, by the Hölder inequality and the Young inequality, we have(75)2α0θm′tf,vm′t≤cα0θm′tWΓD1,22fL32+μ8vm′tV2,2α0θmtf′,vm′t≤cα0θmtWΓD1,22f′L32+μ8vm′tV2,2f′t,vm′t≤cf′tL32+μ8vm′tV2,2f1′t,vm′t≤cf1′tV∗2+μ8vm′tV2.Taking into account (50), (74), and (75) and the fact that ∇Φεvm′,vm′≥0, which is owing to monotonicity of r from ∇Φε (cf. [26], pp. 116), from (73), we have(76)ddtvm′t2+3μ−4c2vmtVvm′tV2+μ2vm′tV2≤cα0θm′tWΓD1,22fL32+cα0θmtWΓD1,22f′L32+cf′L32+cf1′V∗2+2c∗vm′t2+μ2vm′tV2,that is,(77)ddtvm′t2+3μ−4c2vmtVvm′tV2≤cα0θm′tWΓD1,22fL32+cα0θmtWΓD1,22f′L32+cf′L32+cf1′V∗2+2c∗vm′t2.Differentiating the second equality of (52) with respect to t, we have that(78)θm″t,φj+C1θm′t,φj−vm′θ,∇φj−vmθ′,∇φj=g1′,φj.Multiplying (78) by rjm′t and adding for j, we have that(79)θm″t,θm′t+C1θm′t,θm′t−vm′θm,∇θm′t−vmθ′,∇θm′t=g1′,θm′t.On the contrary, we get(80)2vm′θ,∇θm′t≤cκvm′V2θWΓD1,22+κθm′tWΓD1,22.Taking into accountvmθ′,∇θm′t=0 (see (17) and (80)), we have from (79)(81)ddtθm′t2+2κθm′tWΓD1,22≤cκvm′V2θWΓD1,22+κθm′tWΓD1,22+cg1′WΓD1,2∗2+κ4θm′tWΓD1,22.We have from (81)(82)ddtθm′t2+3κ4θm′tWΓD1,22≤cκvm′V2θWΓD1,22+cg1′WΓD1,2∗2.Adding (59), (77), (65), and (82), we have(83)ddtIt+2μ−4c2vmtVvmtV2+vm′tV2+μ−cκθmWΓD1,22vm′tV2+κ−cα0fL32−cα0f′L32θmWΓD1,22+3κ4−cα0fL32θm′WΓD1,22≤cfL32+f′L32+cf1V∗2+f1′V∗2+cg1WΓD1,2∗2+g1′WΓD1,2∗2+2c∗vmt2+vm′t2.Integrating (83), we have(84)It+∫0t2μ−4c2vmsVvmtV2+vm′sV2+μ−cκθmWΓD1,22vm′sV2+κ−cα0fL32−cα0f′L32θmWΓD1,22+3κ4−cα0fL32θm′WΓD1,22ds≤I0+Ft+2c∗∫0tvms2+vm′s2ds,where(85)Ft:=ctfW1,∞0,T;L32+f1W1,∞0,T;V∗2+cg1W1,20,t;WΓD1,2∗2.By (61) and (68), we have(86)I0≤v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102.By the condition of theorem, letfW1,∞0,T;L3 be so small that(87)κ−cα0ftL32−cα0f′tL32≥0,at a.e.t∈0,T,3κ4−cα0fL32≥0,at a.e.t∈0,T.If(88)vm0V=v0V<μ2c2,(89)θm0WΓD1,2=θ0WΓD1,2<μκc,are valid, then there exists tm such that(90)2μ−4c2vmtV≥0,μ−cκθmtWΓD1,2≥0,on 0,tm. Therefore, taking into account (86), by the Gronwall inequality, we have(91)It≤v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102+FTe2c∗t,on all intervals of t satisfying (90).Using the estimate, we will obtain a quadratic inequality satisfied byvmtV.Put(92)β:=v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102+FTe2c∗T.Note thatβ depends only on the data of the problem.Then, whenf satisfies (87), we can see from (91) that(93)vmt+vm′t≤2vmt2+vm′t21/2≤2β,θm′t≤β,on 0,tm, where (90) holds. Let the data of the problem be so small that(94)cκ2g1W1,20,T;WΓD1,2∗+δ1β≤μ2.By (70) and (93), for the small data of the problem, we have(95)cκθmtWΓD1,2≤cκ2g1W1,20,T;WΓD1,2∗+δ1β≤μ2,on 0,tm, which implies(96)μ−cκθmtWΓD1,2≥μ2,∀t∈0,tm.Therefore, for such small data of the problem that (94) is valid, if(97)2μ−4c2vmtV≥0,∀t∈0,tm+γ,γ>0,tm+γ≤T,then owing to (96), step by step, we have(98)μ−cκθmtWΓD1,2≥μ2,∀t∈0,tm+γ.From the above, we see that, for the small data of the problem satisfying (87)–(89) and (94),(99)μ−cκθmtWΓD1,2≥μ2,is valid on the interval where the first inequality of (90) is valid.Put(100)γ:=ftL∞0,T;L3+cα0θ02+1κ∫0Tg1WΓD1,2∗2ds1/2ftL∞0,T;L3+f1tL∞0,T;V∗+max2δ,2c∗δ2β.By (66), (93), and (63), for the small data satisfying (87)–(89) and (94), we have a quadratic inequality for vmtV, which is the one we want,(101)0≤γ−3μvmtV+2c2vmtV2,on the intervals where the first inequality of (90) is satisfied.By the conditions of the theorem, we can assume that the data of the problem are so small that (87)–(89) and (94) are valid, and γ satisfies the following inequality:(102)9μ2−8c2γ>4μ2.Now, let us prove that if(103)v0V≤3μ−9μ2−8c2γ4c2<μ4c2,then for any m,(104)2μ−4c2vmtV≥μ,∀t∈0,T.Since2μ−4c2v0V>μ, on an interval 0,tm,(105)2μ−4c2vmtV≥μ.Let us first prove that if the first inequality of (90) is valid on an interval 0,t¯m, then more stronger(106)2μ−4c2vmtV≥μ,∀t∈0,t¯m,is valid. Putting y=vmtV in (101) (which is valid on the interval where the first inequality of (90) holds when (87)–(89) and (94) are valid), we get(107)0≤γ−3μy+2c2y2on0,t¯m.By virtue of (102), there exist two real roots of z=γ−3μy+c2y2:(108)y1=3μ−9μ2−8c2γ4c2,y2=3μ+9μ2−8c2γ4c2,and on the intervals 0,y1 and y2,+∞, (107) holds. Thus, by continuity of vmtV with respect to t from v0V∈0,y1, we have that vmtV∈0,y1∀t∈0,t¯m, that is,(109)vmtV≤3μ−9μ2−8c2γ4c2<μ4c2,∀t∈0,t¯m.Thus,(110)2μ−4c2vmtV>μ,∀t∈0,t¯m,which shows (106). Thus, by step by step, we see that the fist inequality of (90) is valid on 0,T, and (104) is valid.If (103) is valid, then so is (88). Therefore, for the small data satisfying (87), (89), (94), (102), and (103), we also have (99) on 0,T. By (104) and (99), we have(111)vmtV≤μ4c2,∀t∈0,T,∀m,∀ε>0;cκθmtWΓD1,2≤μ2,∀t∈0,T,∀m,∀ε>0.Then, by (91) and (83), we have(112)vm′t≤const,∀t∈0,T,∀m,∀ε>0;vm′L20,T;V≤const,∀m,∀ε>0;θm′t≤const,∀t∈0,T,∀m,∀ε>0;θm′L20,T;WΓD1,2≤const∀m,∀ε>0.By (111),(113)∫0TΦεvmtdt≤const,∀m,∀ε>0,and so by (46) and (47),(114)∫0Tvmt−JεvmtV2dt≤cε,∀m,∀ε>0.


### 4.2. Existence and Uniqueness of a Solution

Let us prove the existence of a solution. Owing to (111) and (112), we can extract subsequences, which are denoted with the subindex as before, such that(115)vm⟶vinC0,T;V,vm′⇀v′inL20,T;V,vm′⇀∗v′inL∞0,T;H,θm⟶θinC0,T;WΓD1,2,θm′⇀θ′inL20,T;WΓD1,2,θm′⇀∗θ′inL∞0,T;L2Ω,when m⟶∞ and ε⟶0.Puttingu=∑j=1Mkjtuj, where kjt∈C10,T and M is the positive integer, multiply the first equation of (52) by kjt and add for j=1,…,M. Then, multiply the first equation of (52) by gjmt and add for j=1,…,m. Substituting the resulting equations, we have(116)vm′t+A1vmt+B1vmt,vmt+∇Φεvm,ut−vmt=1−α0θf+f1,ut−vmt.SinceΦε is convex, continuous, and Fréchet differentiable, we have(117)Φεut−Φεvmt≥∇Φεvmt,u−vmt.Taking into account (117), we have from (116)(118)∫0Tvm′t+A1vmt+B1vmt,vmt−1−α0θmf−f1,ut−vmtdt+∫0TΦεut−Φεvmtdt≥0.SinceΦu≥Φεu and ΦJεwmt≤Φεwmt(see (47)), we have from (118)(119)∫0Tvm′t+A1vmt+B1vmt,vmt−1−α0θmf−f1,ut−vmtdt+Ψudt−∫0TΦJεvmtdt≥0.By (114), Jεvm⟶v in L20,T;V as m⟶∞ and ε⟶0, and by lower semicontinuity of Φ,(120)limm⟶∞,ε⟶0∫0TΦJεvmtdt≥∫0TΦvtdt.In the routine way, we can prove that(121)∫0TB1vmt,vmt,vtdt⟶∫0TB1vt,vt,vtdt,as m⟶∞ and ε⟶0.Sincelimmk⟶∞,ε⟶0A1vmt,vmt=A1vt,vt, by (120) and (121), we have from (119)(122)∫0Tv′t+A1vt+B1vt,vt−1−α0θf−f1,ut−vtdt+Ψu−Ψv≥0.Sincevm0=v0, by (115), it is obvious that v0=v0.B1vt,vt∈L∞0,T;V∗, and the set u=∑j=1Mkjtuj;kjt∈C10,T,M:positive intiger is dense in L40,T;V, and so (122) is valid for all u∈L40,T;V.By (115), we can get from the second equation of (52) that(123)∫0T∂θm∂t,φ+κ∇θm,∇φ+βxθm,φΓR−vmθm,∇φdt=∫0Tg1,φdt,φ∈L20,T;WΓD1,2.Easily, we see that(124)κ∇θm,∇φ⟶κ∇θ,∇φ,for a.e.t∈0,T.Also,(125)∫0Tvmθm,∇φ−vθ,∇φdt≤∫0Tvm−vL6θmL3∇φL2dt+∫0Tvθm−θ,∇φdt≤vm−vL∞0,T;VθmL20;T;L3φL20,T;WΓD1,2+∫0Tvθm−θ,∇φdt.Sincev∇φ∈L20,T;L6/5Ω, by (115), ∫0Tvθm−θ,∇φdt⟶0. Thus,(126)∫0Tvmθm,∇φdt⟶∫0Tvθ,∇φdt.It is easy to prove that(127)∫0Tβxθm,φΓRdt⟶∫0Tβxθ,φΓRdt.Therefore, from (123), we have(128)∫0T∂θ∂t,φ+κ∇θ,∇φ+βxθ,φΓR−vθ,∇φdt=∫0Tg1,φdt.Sinceθm0=θ0, by (115), it is obvious that θ0=θ0. Therefore, we proved the existence of a solution.Let us prove uniqueness of a solution. Letv1,θ1andv2,θ2 be the two solutions to Problem I-VI satisfying inequality (111) instead of approximate solutions. Then, taking into account (44) (Remark 4), from (35), we have(129)v1′t+A1v1t+B1v1t,v1t−1−α0θ1f−f1,v2t−v1t+Φv2t−Φv1t≥0,v2′t+A1v2t+B1v2t,v2t−1−α0θ2f−f1,v1t−v2t+Φv1t−Φv2t≥0,which imply(130)v1′t−v2′t,v1t−v2t+A1v1t−v2t,v1t−v2t≤α0θ1−θ2f,v1t−v2t+B1v1t,v1t−B1v2t,v2t,v1t−v2t.By virtue of (50) and (51), we have(131)ddtv1t−v2t2+7μ2v1t−v2tV2≤2c∗v1t−v2t2+μ2v1t−v2tV2+cfL32θ1−θ22+2Bv1t−v2t,v1t,v1t−v2t+2Bv2t,v1t−v2t,v1t−v2t≤2c∗v1t−v2t2+μ2v1t−v2tV2+cfL32θ1−θ22+2c2v1tV+v2tVv1t−v2tV2,where c2 is the one in (51). By (111),(132)2c2v1tV+v2tV≤μ,and so we have(133)dv1t−v2t2dt+2μv1t−v2tV2≤2c∗v1t−v2t2+cfL32θ1−θ22.Also, from(134)∂θ1∂t,φ+κ∇θ1,∇φ+βxθ1,φΓR−v1θ1,∇φ1=g1,φ,∂θ2∂t,φ+κ∇θ2,∇φ+βxθ2,φΓR−v2θ2,∇φ=g1,φ,we have(135)∂θ1−θ2∂t,θ1−θ2+κ∇θ1−∇θ2,∇θ1−∇θ2+βxθ1−θ2,θ1−θ2ΓR−v1θ1−θ2,∇θ1−θ2−v1−v2θ2,∇θ1−θ2=0.Taking into accountv1θ1−θ2,∇θ1−θ2=0(see (17)), by (80), we have(136)ddtθ1−θ22+2κ∇θ1−∇θ22≤cκv1t−v2tV2θ2tWΓD1,22+κ∇θ1−∇θ22,and so(137)ddtθ1t−θ2t2≤cκv1t−v2tV2θ2tWΓD1,22.By (111),(138)cκθ2tWΓD1,22≤μ2.Therefore, adding (133) and (137), we get(139)ddtv1t−v2t2+θ1t−θ2t2≤2c∗+cfL32v1t−v2t2+θ1t−θ2t2.We have from (139)(140)v1t−v2t2+θ1t−θ2t2≤∫0t2c∗+2cfL32v1s−v2s2+θ1s−θ2s2ds,which implies v1t=v2t and θ1t=θ2t for all t∈0,T.


## 4.1. Existence and Estimation of Solutions to an Approximate Problem

We first consider a problem approximating (34).For every0<ε<1, define a functional Φε by(45)Φεy=infy−uV22ε+Φu;u∈V,y∈V,which is called the Moreau regularization of Φ. When ∂Φ:V⟶2V is the subdifferential of Φ in the Hilbert space V, let Jε=I+ε∂Φ−1 and ∂Φε:=ε−1I−Jε(the Yosida approximation of ∂ϕ) for all ε>0. Then, the functional Φε is convex, continuous, Fréchet differentiable, and ∇Φε=∂Φε≡ε−1I−Jε for all 1>ε>0. Moreover,(46)Φεy=y−JεyV22ε+ΦJεy,∀y∈V,(47)limε⟶0Φεy=Φy,ΦJεy≤Φεy≤Φy,∀y∈V,(cf. Theorem 2.9 in [27]). The operator ∇Φε is Lipschitz continuous with the constant 2ε−1(cf. Proposition 2.3 in [27]) and monotone (cf. Lemma 4.10 of ch. III in [25]).By the fact thatΓ2j,Γ3j,andΓ7j are in C2.1Γij and 4 of Assumption 1, there exists a constant M such that(48)Sx∞,kx∞,αL∞Γ5≤M.Thus, there existsc∗ such that(49)2μkxz,zΓ2+2μSz˜,z˜Γ3+αxz,zΓ5+μkxz,zΓ7≤μ4zV2+c∗z2,∀z∈V,(cf. .5.1.10 of [28]). Thus,(50)A1u,u≥7μ4uV2−c∗u2,∀u∈V,A1u,v≤c1uVvV,∃c1>0,∀u,v∈V,(51)B1v,u,w≤c2vVuVwV,∀u,v∈V,where the operators A1,B1 are the ones in (32).Letuj,j=1,2,… and φj,j=1,2,… be, respectively, bases of the space V and WΓD1,2Ω. Without loss of generality, we assume that u1=v0 and φ1=θ0 as in [26]. We find a solution vm=∑j=1mgjmtuj and θm=∑j=1mrjmtφj to the problem(52)∂vm∂t,uj+2μℰvm,ℰuj+vm⋅∇vm,uj+2μkxvm,ujΓ2+2μSv˜m,u˜jΓ3+2αxvm,ujΓ5+μkxvm,ujΓ7+∇Φεvmt,uj=1−α0θmf,uj+f1,uj,∂θm∂t,φj+κ∇θm,∇φj+βxθm,φjΓR−vmθm,∇φj=g1,φj,vm0=v0,θm0=θ0,which gives us a system for gjmt and rjmt,j=1−m. The solutions to (52) depend on ε, but for convenience of notation, here and in what follows, we use subindex m instead of subindex mε. For tm, there exist absolute continuous functions gjmt and rjmt on 0,tm. Since f∈W1,∞0,T;L3Ω, f1∈W1,∞0,T;V∗, g1∈W1,20,T;WΓD1,2Ω∗, and ∇Φε is Lipschitz continuous, gjm′t and rjm′t are in fact absolute continuous. If vmt and θmt are bounded and vmt,θmt are integrable, then gjmt and rjmt are prolonged over tm. Under smallness of the data of the problem and the compatibility condition of the data at the initial instant, we will find estimates for vmt and θmt in the following, by which we obtain (111) and see that tm=T.Multiplying the first and the second equation of (52), respectively, by gjmt and φjmt and adding for i=1,…,m, we get(53)∂vm∂t,vm+2μℰvm,ℰvm+vm⋅∇vm,vm+2μkxvm,vmΓ2+2μSv˜m,v˜mΓ3+2αxvm,vmΓ5+μkxvm,vmΓ7+∇Φεvmt,vm=1−α0θmf,vm+f1,vm,∂θm∂t,θm+κ∇θm,∇θm+βxθm,θmΓR−vmθm,∇θm=g1,θm,vm0=v0,θm0=θ0.We will find a priori estimates for(54)It:=vmt2+vm′t2+θmt2+θm′t2.SinceΦε is convex, continuous, and Fréchet differentiable, we have(55)Φεu−Φεvmt≥∇Φεvmt,u−vmt,∀u∈V,and so by Φε0V=0,(56)0≤Φεvmt≤∇Φεvmt,vmt.Also,(57)2−α0θmtf,vmt≤cα0θmtWΓD1,22fL32+μ4vmtV2.By virtue of (50), (51), (56), and (57), we have from the first equation of (53)(58)ddtvmt2+7μ2vmtV2−2c2vmtV3+2Φεvmt≤cα0θmtWΓD1,22fL32+cfL32+cf1V∗2+μ2vmtV2+2c∗vmt2,where c∗ and c2 are, respectively, the ones in (49) and (51), and so(59)ddtvmt2+3μ−2c2vmtVvmtV2+2Φεvmt≤cα0θmtWΓD1,22fL32+cfL32+cf1V∗2+2c∗vmt2.Here and in what follows are the constants independent of the data of problem which are denoted byc with the exceptions of c∗andc2.Settingt=0 in the first equation of (52) and multiplying the resulting equation by gjm′0 and adding for j=1,…,m, we get(60)vm′02+A1vm0,vm′0+B1vm0,vm0,vm′0+∇Φεv0,vm′0=1−α0θ0f0,vm′0+f10,vm′0.By condition (2) of Theorem1, for any u∈V, we have Φu≥Φv0=0, which by (47), it implies that Φεv0=0 and ∇Φεv0=0. Then, from (60), we have(61)vm′0≤A1v0+B1v0,v0−1−α0θ0f0−f10,which is valid by the compatibility condition at the initial time for velocity (condition (4)) and the conditions for θ0,f. On the contrary, taking into account (50), (51), and (56), we have from the first equation of (53)(62)7μ2vmtV2≤2c2vmV3+21−α0θmtft,vmt+2f1t,vmt+2c∗vm2−2vmt,vm′t,and so(63)3μvmtV≤2c2vmtV2+cftL3+cα0θmtftL3+cf1tV∗+2δvm′t+2c∗δvmt,where δ is such that ⋅≤δ⋅V. Since vmθm,∇θm=0 by (17), we get from the second equation of (53)(64)ddtθmt2+2κθmtWΓD1,22+2βxθm,θmΓR=2g1,θmt.From (64), we have(65)ddtθmt2+κθmtWΓD1,22+βxθm,θmΓR≤1κg1WΓD1,2∗2,(66)θmt2+∫0t∫Ωκ∇θms2dxds≤θ02+1κ∫0tg1sWΓD1,2∗2ds.Settingt=0 in the second equation of (52) and multiplying the resulting equation by rjm′0 and adding for j=1,…,m, we get(67)θm′02+b1θ0,θm′0+v0⋅∇θ0,θm′0=g0,θm′0,where −v0θ0,∇θm′0=v0⋅∇θ0,θm′0 was used. From (67), we have(68)θm′0≤C1θ0+v0⋅∇θ0−g10,which is valid by the compatibility condition at the initial time for temperature (condition (6)).On the contrary, taking into accountvmθm,∇θm=0, from the second equation of (53), we have(69)κθmtWΓD1,22≤g1WΓD1,2∗θmtWΓD1,2+δ1θmtWΓD1,2θm′t,where δ1 is such that ⋅≤δ1⋅WΓD1,2, and so(70)θmtWΓD1,2≤1κg1WΓD1,2∗+δ1θm′t.Taking into account (66), we have from (63)(71)3μvmtV≤c2vmtV2+cfL∞0,T;L3+cα0θ02+1κ∫0Tg1WΓD1,2∗2ds1/2fL∞0,T;L3+f1L∞0,T;V∗+max2δ,2c∗δvm′t+vmt.Differentiating the first equality of (52) with respect to t, we have that(72)vm″t,vj+A1vm′t,vj+B1vmt,vmt′,vj+∇Φεvm′,vj=−α0θm′tf,vj−1−α0θmtf′,vj+f1′,vj.Multiplying (72) by gjm′t and adding for j, we have(73)vm″t,vm′t+A1vm′t,vm′t+B1vmt,vmt′,vm′t+∇Φεvm′,vm′t=−α0θm′tf,vm′t−1−α0θmtf′,vm′t+f1′,vm′t.CalculatingBvmt,vmt′, we have(74)B1vmt,vmt′,vm′t=B1vm′,vm,vm′+B1vm,vm′,vm′≤2c2vmVvm′V2,where c2 is the one in (51). Also, by the Hölder inequality and the Young inequality, we have(75)2α0θm′tf,vm′t≤cα0θm′tWΓD1,22fL32+μ8vm′tV2,2α0θmtf′,vm′t≤cα0θmtWΓD1,22f′L32+μ8vm′tV2,2f′t,vm′t≤cf′tL32+μ8vm′tV2,2f1′t,vm′t≤cf1′tV∗2+μ8vm′tV2.Taking into account (50), (74), and (75) and the fact that ∇Φεvm′,vm′≥0, which is owing to monotonicity of r from ∇Φε (cf. [26], pp. 116), from (73), we have(76)ddtvm′t2+3μ−4c2vmtVvm′tV2+μ2vm′tV2≤cα0θm′tWΓD1,22fL32+cα0θmtWΓD1,22f′L32+cf′L32+cf1′V∗2+2c∗vm′t2+μ2vm′tV2,that is,(77)ddtvm′t2+3μ−4c2vmtVvm′tV2≤cα0θm′tWΓD1,22fL32+cα0θmtWΓD1,22f′L32+cf′L32+cf1′V∗2+2c∗vm′t2.Differentiating the second equality of (52) with respect to t, we have that(78)θm″t,φj+C1θm′t,φj−vm′θ,∇φj−vmθ′,∇φj=g1′,φj.Multiplying (78) by rjm′t and adding for j, we have that(79)θm″t,θm′t+C1θm′t,θm′t−vm′θm,∇θm′t−vmθ′,∇θm′t=g1′,θm′t.On the contrary, we get(80)2vm′θ,∇θm′t≤cκvm′V2θWΓD1,22+κθm′tWΓD1,22.Taking into accountvmθ′,∇θm′t=0 (see (17) and (80)), we have from (79)(81)ddtθm′t2+2κθm′tWΓD1,22≤cκvm′V2θWΓD1,22+κθm′tWΓD1,22+cg1′WΓD1,2∗2+κ4θm′tWΓD1,22.We have from (81)(82)ddtθm′t2+3κ4θm′tWΓD1,22≤cκvm′V2θWΓD1,22+cg1′WΓD1,2∗2.Adding (59), (77), (65), and (82), we have(83)ddtIt+2μ−4c2vmtVvmtV2+vm′tV2+μ−cκθmWΓD1,22vm′tV2+κ−cα0fL32−cα0f′L32θmWΓD1,22+3κ4−cα0fL32θm′WΓD1,22≤cfL32+f′L32+cf1V∗2+f1′V∗2+cg1WΓD1,2∗2+g1′WΓD1,2∗2+2c∗vmt2+vm′t2.Integrating (83), we have(84)It+∫0t2μ−4c2vmsVvmtV2+vm′sV2+μ−cκθmWΓD1,22vm′sV2+κ−cα0fL32−cα0f′L32θmWΓD1,22+3κ4−cα0fL32θm′WΓD1,22ds≤I0+Ft+2c∗∫0tvms2+vm′s2ds,where(85)Ft:=ctfW1,∞0,T;L32+f1W1,∞0,T;V∗2+cg1W1,20,t;WΓD1,2∗2.By (61) and (68), we have(86)I0≤v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102.By the condition of theorem, letfW1,∞0,T;L3 be so small that(87)κ−cα0ftL32−cα0f′tL32≥0,at a.e.t∈0,T,3κ4−cα0fL32≥0,at a.e.t∈0,T.If(88)vm0V=v0V<μ2c2,(89)θm0WΓD1,2=θ0WΓD1,2<μκc,are valid, then there exists tm such that(90)2μ−4c2vmtV≥0,μ−cκθmtWΓD1,2≥0,on 0,tm. Therefore, taking into account (86), by the Gronwall inequality, we have(91)It≤v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102+FTe2c∗t,on all intervals of t satisfying (90).Using the estimate, we will obtain a quadratic inequality satisfied byvmtV.Put(92)β:=v02+A1v0+B1v0,v0−1−α0θ0f0−f102+θ02+C1θ0+v0⋅∇θ0−g102+FTe2c∗T.Note thatβ depends only on the data of the problem.Then, whenf satisfies (87), we can see from (91) that(93)vmt+vm′t≤2vmt2+vm′t21/2≤2β,θm′t≤β,on 0,tm, where (90) holds. Let the data of the problem be so small that(94)cκ2g1W1,20,T;WΓD1,2∗+δ1β≤μ2.By (70) and (93), for the small data of the problem, we have(95)cκθmtWΓD1,2≤cκ2g1W1,20,T;WΓD1,2∗+δ1β≤μ2,on 0,tm, which implies(96)μ−cκθmtWΓD1,2≥μ2,∀t∈0,tm.Therefore, for such small data of the problem that (94) is valid, if(97)2μ−4c2vmtV≥0,∀t∈0,tm+γ,γ>0,tm+γ≤T,then owing to (96), step by step, we have(98)μ−cκθmtWΓD1,2≥μ2,∀t∈0,tm+γ.From the above, we see that, for the small data of the problem satisfying (87)–(89) and (94),(99)μ−cκθmtWΓD1,2≥μ2,is valid on the interval where the first inequality of (90) is valid.Put(100)γ:=ftL∞0,T;L3+cα0θ02+1κ∫0Tg1WΓD1,2∗2ds1/2ftL∞0,T;L3+f1tL∞0,T;V∗+max2δ,2c∗δ2β.By (66), (93), and (63), for the small data satisfying (87)–(89) and (94), we have a quadratic inequality for vmtV, which is the one we want,(101)0≤γ−3μvmtV+2c2vmtV2,on the intervals where the first inequality of (90) is satisfied.By the conditions of the theorem, we can assume that the data of the problem are so small that (87)–(89) and (94) are valid, and γ satisfies the following inequality:(102)9μ2−8c2γ>4μ2.Now, let us prove that if(103)v0V≤3μ−9μ2−8c2γ4c2<μ4c2,then for any m,(104)2μ−4c2vmtV≥μ,∀t∈0,T.Since2μ−4c2v0V>μ, on an interval 0,tm,(105)2μ−4c2vmtV≥μ.Let us first prove that if the first inequality of (90) is valid on an interval 0,t¯m, then more stronger(106)2μ−4c2vmtV≥μ,∀t∈0,t¯m,is valid. Putting y=vmtV in (101) (which is valid on the interval where the first inequality of (90) holds when (87)–(89) and (94) are valid), we get(107)0≤γ−3μy+2c2y2on0,t¯m.By virtue of (102), there exist two real roots of z=γ−3μy+c2y2:(108)y1=3μ−9μ2−8c2γ4c2,y2=3μ+9μ2−8c2γ4c2,and on the intervals 0,y1 and y2,+∞, (107) holds. Thus, by continuity of vmtV with respect to t from v0V∈0,y1, we have that vmtV∈0,y1∀t∈0,t¯m, that is,(109)vmtV≤3μ−9μ2−8c2γ4c2<μ4c2,∀t∈0,t¯m.Thus,(110)2μ−4c2vmtV>μ,∀t∈0,t¯m,which shows (106). Thus, by step by step, we see that the fist inequality of (90) is valid on 0,T, and (104) is valid.If (103) is valid, then so is (88). Therefore, for the small data satisfying (87), (89), (94), (102), and (103), we also have (99) on 0,T. By (104) and (99), we have(111)vmtV≤μ4c2,∀t∈0,T,∀m,∀ε>0;cκθmtWΓD1,2≤μ2,∀t∈0,T,∀m,∀ε>0.Then, by (91) and (83), we have(112)vm′t≤const,∀t∈0,T,∀m,∀ε>0;vm′L20,T;V≤const,∀m,∀ε>0;θm′t≤const,∀t∈0,T,∀m,∀ε>0;θm′L20,T;WΓD1,2≤const∀m,∀ε>0.By (111),(113)∫0TΦεvmtdt≤const,∀m,∀ε>0,and so by (46) and (47),(114)∫0Tvmt−JεvmtV2dt≤cε,∀m,∀ε>0.


## 4.2. Existence and Uniqueness of a Solution

Let us prove the existence of a solution. Owing to (111) and (112), we can extract subsequences, which are denoted with the subindex as before, such that(115)vm⟶vinC0,T;V,vm′⇀v′inL20,T;V,vm′⇀∗v′inL∞0,T;H,θm⟶θinC0,T;WΓD1,2,θm′⇀θ′inL20,T;WΓD1,2,θm′⇀∗θ′inL∞0,T;L2Ω,when m⟶∞ and ε⟶0.Puttingu=∑j=1Mkjtuj, where kjt∈C10,T and M is the positive integer, multiply the first equation of (52) by kjt and add for j=1,…,M. Then, multiply the first equation of (52) by gjmt and add for j=1,…,m. Substituting the resulting equations, we have(116)vm′t+A1vmt+B1vmt,vmt+∇Φεvm,ut−vmt=1−α0θf+f1,ut−vmt.SinceΦε is convex, continuous, and Fréchet differentiable, we have(117)Φεut−Φεvmt≥∇Φεvmt,u−vmt.Taking into account (117), we have from (116)(118)∫0Tvm′t+A1vmt+B1vmt,vmt−1−α0θmf−f1,ut−vmtdt+∫0TΦεut−Φεvmtdt≥0.SinceΦu≥Φεu and ΦJεwmt≤Φεwmt(see (47)), we have from (118)(119)∫0Tvm′t+A1vmt+B1vmt,vmt−1−α0θmf−f1,ut−vmtdt+Ψudt−∫0TΦJεvmtdt≥0.By (114), Jεvm⟶v in L20,T;V as m⟶∞ and ε⟶0, and by lower semicontinuity of Φ,(120)limm⟶∞,ε⟶0∫0TΦJεvmtdt≥∫0TΦvtdt.In the routine way, we can prove that(121)∫0TB1vmt,vmt,vtdt⟶∫0TB1vt,vt,vtdt,as m⟶∞ and ε⟶0.Sincelimmk⟶∞,ε⟶0A1vmt,vmt=A1vt,vt, by (120) and (121), we have from (119)(122)∫0Tv′t+A1vt+B1vt,vt−1−α0θf−f1,ut−vtdt+Ψu−Ψv≥0.Sincevm0=v0, by (115), it is obvious that v0=v0.B1vt,vt∈L∞0,T;V∗, and the set u=∑j=1Mkjtuj;kjt∈C10,T,M:positive intiger is dense in L40,T;V, and so (122) is valid for all u∈L40,T;V.By (115), we can get from the second equation of (52) that(123)∫0T∂θm∂t,φ+κ∇θm,∇φ+βxθm,φΓR−vmθm,∇φdt=∫0Tg1,φdt,φ∈L20,T;WΓD1,2.Easily, we see that(124)κ∇θm,∇φ⟶κ∇θ,∇φ,for a.e.t∈0,T.Also,(125)∫0Tvmθm,∇φ−vθ,∇φdt≤∫0Tvm−vL6θmL3∇φL2dt+∫0Tvθm−θ,∇φdt≤vm−vL∞0,T;VθmL20;T;L3φL20,T;WΓD1,2+∫0Tvθm−θ,∇φdt.Sincev∇φ∈L20,T;L6/5Ω, by (115), ∫0Tvθm−θ,∇φdt⟶0. Thus,(126)∫0Tvmθm,∇φdt⟶∫0Tvθ,∇φdt.It is easy to prove that(127)∫0Tβxθm,φΓRdt⟶∫0Tβxθ,φΓRdt.Therefore, from (123), we have(128)∫0T∂θ∂t,φ+κ∇θ,∇φ+βxθ,φΓR−vθ,∇φdt=∫0Tg1,φdt.Sinceθm0=θ0, by (115), it is obvious that θ0=θ0. Therefore, we proved the existence of a solution.Let us prove uniqueness of a solution. Letv1,θ1andv2,θ2 be the two solutions to Problem I-VI satisfying inequality (111) instead of approximate solutions. Then, taking into account (44) (Remark 4), from (35), we have(129)v1′t+A1v1t+B1v1t,v1t−1−α0θ1f−f1,v2t−v1t+Φv2t−Φv1t≥0,v2′t+A1v2t+B1v2t,v2t−1−α0θ2f−f1,v1t−v2t+Φv1t−Φv2t≥0,which imply(130)v1′t−v2′t,v1t−v2t+A1v1t−v2t,v1t−v2t≤α0θ1−θ2f,v1t−v2t+B1v1t,v1t−B1v2t,v2t,v1t−v2t.By virtue of (50) and (51), we have(131)ddtv1t−v2t2+7μ2v1t−v2tV2≤2c∗v1t−v2t2+μ2v1t−v2tV2+cfL32θ1−θ22+2Bv1t−v2t,v1t,v1t−v2t+2Bv2t,v1t−v2t,v1t−v2t≤2c∗v1t−v2t2+μ2v1t−v2tV2+cfL32θ1−θ22+2c2v1tV+v2tVv1t−v2tV2,where c2 is the one in (51). By (111),(132)2c2v1tV+v2tV≤μ,and so we have(133)dv1t−v2t2dt+2μv1t−v2tV2≤2c∗v1t−v2t2+cfL32θ1−θ22.Also, from(134)∂θ1∂t,φ+κ∇θ1,∇φ+βxθ1,φΓR−v1θ1,∇φ1=g1,φ,∂θ2∂t,φ+κ∇θ2,∇φ+βxθ2,φΓR−v2θ2,∇φ=g1,φ,we have(135)∂θ1−θ2∂t,θ1−θ2+κ∇θ1−∇θ2,∇θ1−∇θ2+βxθ1−θ2,θ1−θ2ΓR−v1θ1−θ2,∇θ1−θ2−v1−v2θ2,∇θ1−θ2=0.Taking into accountv1θ1−θ2,∇θ1−θ2=0(see (17)), by (80), we have(136)ddtθ1−θ22+2κ∇θ1−∇θ22≤cκv1t−v2tV2θ2tWΓD1,22+κ∇θ1−∇θ22,and so(137)ddtθ1t−θ2t2≤cκv1t−v2tV2θ2tWΓD1,22.By (111),(138)cκθ2tWΓD1,22≤μ2.Therefore, adding (133) and (137), we get(139)ddtv1t−v2t2+θ1t−θ2t2≤2c∗+cfL32v1t−v2t2+θ1t−θ2t2.We have from (139)(140)v1t−v2t2+θ1t−θ2t2≤∫0t2c∗+2cfL32v1s−v2s2+θ1s−θ2s2ds,which implies v1t=v2t and θ1t=θ2t for all t∈0,T.


## 5. Proof of Theorem 2

### 5.1. Existence and Estimation of a Solution to an Approximate Problem

We first consider a problem approximating (43). For every 0<ε<1, let a functional Φε be defined by (45).Letuj,j=1,2,… and φj,j=1,2,… be, respectively, bases of the space V and WΓD1,2Ω. Without loss of generality, we assume that u1=v0 and φ1=θ0 as in [26]. We find a solution vm=∑j=1mgjmtuj and θm=∑j=1mrjmtφj to the problem(141)∂vm∂t,uj+2μθmℰvm,ℰuj+rotvm×vm,uj+2μθmkxvm,ujΓ2+2μθmSv˜m,u˜jΓ3+2αxvm,ujΓ5+μθmkxvm,ujΓ7+∇Φεvmt,uj=1−α0θmf,uj+f2,uj,∂θm∂t,φj+κθm∇θm,∇φj+βxθm,φjΓR−vmθm,∇φj=g2,φj,vm0=v0,θm0=θ0,which gives us a system for gjmt and rjmt,j=1−m. The solutions to (141) depend on ε, but for convenience of notation, here and in what follows, we use subindex m. For tm, there exist absolute continuous functions gjmt and rjmt on 0,tm. If vmt and θmt are bounded and vmtandθmt are integrable, then gjmt and rjmt are prolonged over tm. We will find estimates (157) in the following, by which we see that tm=T.Multiplying the first and the second equation of (141), respectively, by gjmt and φjmt and adding for i=1,…,m, we get(142)∂vm∂t,vm+2μθmℰvm,ℰvm+rotvm×vm,vm+2μθmkxvm,vmΓ2+2μθmSv˜m,v˜mΓ3+2αx,tvm,vmΓ5+μθmkxvm,vmΓ7+∇Φεvmt,vm+α0θmf,vm=f,vm+f2,vm,∂θm∂t,θm+κθm∇θm,∇θm+βxθm,θmΓR−vmθm,∇θm=g2,θm,vm0=v0,θm0=θ0.Let us estimate terms on the left-hand side above. It is easy to see(143)2μθmℰvm,ℰvmdt≥2μ0vmV2.By the fact thatΓ2j,Γ3j,Γ7j are in C2.1Γij and Assumption 2, there exists a constant M such that(144)Sx∞,kx∞,αL∞Γ5≤M.Thus,(145)2μθmkxv,vΓ2+2μθmSv˜m,v˜mΓ3+2αxvm,vmΓ5+μθmkxvm,vmΓ7≤μ02vmV2+k11vm2,(cf. Theorem 1.5.1.10 of [28]). Obviously,(146)rotvm×vm,vm=0.SinceΦε is convex, continuous and Fréchet differentiable, we have(147)Φεy−Φεx≥∇Φεx,y−x,∀x,y∈V.Thus,(148)∇Φεvmt,0V−vmt≤Φε0V−Φεvmt≤−Φεvmt,(149)∇Φεvmt,vmt≥Φεvmt.Also, by the Hölder inequality, we have(150)α0θmf,vmdt≤k12vmt2+κ04θmtWΓD1,22,where k12=cα0fL∞0,T;L32.(151)f,wm+f2,wm≤μ02wmV2+cfL32+f2v∗2.Also, we have(152)κθm∇θm,∇θm≥κ0θmWΓD1,22,βxθm,θmΓR≥0.By (17), we have(153)wmθm,∇θm=0.(154)g2,θm≤κ04θmtWΓD1,22+cg2WΓD1,2∗2.Taking(155)k1=k11+k12,by (143)–(154), we have from (142)(156)ddtvmt2+ddtθmt2+2μ0vmV2+κ0θmWΓD1,22+Φεvmt≤cftL32+f2tv∗2+g2tWΓD1,2∗2+2k1vmt2.Applying the Gronwall inequality, we have from (156)(157)vmt2+θmt2≤v02+θ02+∫0tfsL32+f2sv∗2+g2sWΓD1,2∗2dse2k1t,vmL20,T;V2+θmL20,t;WΓD1,22≤cv02+θ02+∫0TftL32+f2tv∗2+g2tWΓD1,2∗2dt,∫0TΦεvmtdt≤cv02+θ02+∫0TftL32+f2tv∗2+g2tWΓD1,2∗2dt.Note thatc in (157) depends on T and (via k12) f but independent of m and ε.By (46) and (29) and the third inequality of (157), we have(158)∫0Tvmt−JεvmtV2dt≤cε,with c independent of ε. Multiplying the first equation of (141) by gjmt−gjms, summing for j, and taking into account (147), we have(159)12dvmt−vms2dt+A2θmvmt+B2vmt,vmt+α0θmtft−ft−f2t,vmt−vms=∇Φεvmt,vms−vmt≤Φεvms−Φεvms≤Φεvms,where the operators A2θm,B2 are the ones in (42). By (145) and (144), we have(160)A2θmvmt,vmt≥3μ02vmtV2−k11vmt2,A2θmvmt,vms≤cvmtVvmsV,where k11 is the one in (145). Taking into account (160) and the fact that B2vmt,vmt,vmt=0, we have from (159)(161)12dvmt−vms2dt≤Φεvms+A2θmvmt,vms+B2vmt,vmt,vms+−α0θmtft+ft+f2t,vmt−vms+k11vmt2.Let us integrate every term of (161) first with respect to t from s to s+h and then with respect to s from 0 to T, where vmt=0 when t∈T,T+h.(162)∫0T∫ss+hdvmt−vms2dtdtds=∫0Tvms+h−vms2ds.By the third inequality of (157),(163)∫0T∫ss+hΦεvmsdtds≤h∫0TΦεvmsds≤c1h.By (157) and (161), we have(164)∫0T∫ss+hA2θmvmt,vmsdtds≤c∫0TvmsV∫ss+hvmtVdtds≤c∫0TvmsVhvmL20,T;Vds≤c2h.SincewL3≤KwL21/2‖w‖L61/2,(165)B2v,w,z=rotv×w,z≤KrotvL2wL3zL6≤KvVw1/2wV1/2zV,and so by (157), we have(166)∫0T∫ss+hB2vmt,vmtvmsdtds≤K∫0T∫ss+hvmtV3/2vmt1/2vmsVdtds≤K∫0TvmsV∫ss+hvmtV2dt3/4∫ss+hvmt2dt1/4ds≤c3h1/4.Also, by (157), we have(167)∫0T∫ss+hf+f2t,vmtdtds≤∫0Tf+f2t,vmt∫t−htdsdt≤c4h,∫0T∫ss+hf+f2t,−vmsdtds≤K∫0TvmsV∫ss+hf+f2t‖V∗dtds≤c5h.In the same way, we get(168)∫0T∫ss+hα0θmf,vmt−vmsdtds≤c6h+c7h,∫0T∫ss+hk11vmt2dtds≤c8h.Note that constantsci,i=1−8, are independent of m,ε. By virtue of (162)–(168), uniformly with respect to m,ε,(169)∫0Tvms+h−vms2ds≤Oh1/4,and the set vm is relatively compact in L20,T;W9/10,2Ω(see Theorem 5 of [29]). Also, we have(170)κθm∇θm,∇φ≤κ1∇θmL2φWΓD1,2Ω,βxθm,φΓR≤cθmWΓD1,2ΩφWΓD1,2Ω,vmθm,∇φ≤cvmVθmL3φWΓD1,2Ω.By (170), from the second equation of (141), we have(171)∂θm∂t,φ≤cθmWΓD1,2Ω+vmV2+θmWΓD1,2Ω2+g2WΓD1,2∗φWΓD1,2,∀φ∈WΓD1,2Ω.Hence, by (157), we know that(172)θm′∈L10,T;WΓD1,τ∗,θm′L10,T;WΓD1,τ∗≤c,where c is independent of m,ε. Thus, the set θε is relatively compact in L20,T;W9/10,2Ω(see Corollary 5 of [29]).


### 5.2. Existence of a Solution

We can extract subsequences, which are denoted as before, such that(173)vm⇀vinL20,T;V,vm⇀∗vinL∞0,T;H,vm⟶vinL20,T;W9/10,2Ω,θm⇀θinL20,T;WD1,2Ω,θm⇀∗θinL∞0,T;L2Ω,θm⟶θinL20,T;W9/10,2Ω,as m⟶∞ and ε⟶0.On the contrary, puttingu=∑j=1Mkjtuj, where kjt∈C10,T and is a positive integer, multiply the first equation of (141) by kjt and add for j=1,…,M. Then, multiply the first equation of (141) by gjmt and add for j=1,…,m. Substituting the resulting equations, we have(174)∫0T∂vm∂t+A2θmvmt+B2vmt,vmt+∇Φεvmt,ut−vmtdt=∫0T−α0θmf+f+f2,ut−vmtdt.Since(175)∫0Tvm′t,ut−vmtdt=∫0Tu′t,ut−vmtdt−12vmT−uT2+12vm0−u02,taking into account (147), we have from (174)(176)∫0Tu′t+A2θmvmt+B2vmt,vmt,ut−vmtdt−∫0T−α0θmf+f+f2,ut−vmtdt+∫0TΦεut−Φεvmtdt≥−12vm0−u02.SinceΦεu≤Φu and ΦJεvmt≤Φεvmt(cf. (47)), we have from (176)(177)∫0Tu′t+A2θmvmt+B2vmt,vmt,ut−vmtdt−∫0T−α0θmf+f+f2,ut−vmtdt+Ψu−∫0TΦJεvmtdt≥−12vm0−u02.By (173) and Corollary Appendix B.2 of [1], taking a subsequence if necessary, we have(178)∫0TA2θmvmt,utdt≡∫0Ta02θmt;vmt,utdt⟶∫0TA2θvt,utdt.Owing to (173),(179)vm⟶vinL20,T;L2∂Ω.Thus, taking a subsequence if necessary, we have (see Lemma Appendix B.1 of [1])(180)2μθmkxvm,vmΓ2+2μθmSv˜m,v˜mΓ3+2αxvm,vmΓ5+μθmkxvm,vmΓ7⟶2μθkxv,vΓ2+2μθSv˜,v˜Γ3+2αxv,vΓ5+μθkxv,vΓ7.Therefore, taking into account(181)liminf2μθmℰvm,ℰvm≥2μθℰv,ℰv(see Corollary Appendix B.3 of [1]), we have(182)liminfA2θmvmt,vmt≥A2θvt,vt.By (178) and (182), we have(183)liminfm⟶∞ε⟶0∫0TA2θmvmt,ut−vmtdt≤∫0TA2θvt,ut−vtdt.By (158) and (173), Jεvm⇀v in L20,T;V as m⟶∞ and ε⟶0. Since the functional Φ:V⟶ℝ¯ is lower weak semicontinuous, we have(184)liminfm⟶∞ε⟶0∫0TΦJεvmtdt≥∫0TΦvtdt≡Ψv.In a rather routine way, we can prove that(185)∫0TB2vmt,vmt,utdt⟶∫0TB2vt,vt,utdtas m⟶∞,ε⟶0. For convenience of readers, we give a proof in the following:(186)∫0TB2vmt,vmt,ut−B2vt,vt,utdt=∫0TB2vmt,vmt−vt,utdt+∫0TB2vmt−vt,vt,utdt≡I1+I2.By (165), the Hölder inequality with exponents 2,4,4, and (173), we have(187)I1≤Ksupt∈0,TutH1∫0TvmtH1vmt−vtH11/2vmt−vt1/2dt≤Ksupt∈0,TutH1vmL20,T;H1vm−vL20,T;H11/2vm−vL20,T;L21/2⟶0.By (165),(188)∫0TB2zt,vt,utdt≤Ksupt∈0,TutH1ztL20,T;H1vL20,T;H11/2vL20,T;L21/2≤K1ztL20,T;H1,which means that the mapping(189)zt∈L20,T;V⟶∫0TB2zt,vt,utdt,is continuous and linear on L20,T;V, that is, there exists f∈L20,T;V∗ such that(190)∫0TB2zt,vt,utdt=∫0Tzt,ftdt.Thus, by (173),(191)I2⟶0,asm⟶∞,ε⟶0.By (186)–(191), we get (185).Let us prove that(192)∫0Tα0θmf,ut−vmtdt⟶∫0Tα0θf,ut−vtdt⟶0,asm⟶∞,ε⟶0.By (173), we have(193)∫0Tα0θm−α0θf,utdt⟶0,(194)∫0Tα0θmf,vmt−α0θf,vtdt=∫0Tα0θm−θf,vmtdt+∫0Tα0θf,vmt−vtdt⟶0.By (193) and (194), we get (192).Therefore, by (183)–(185) and (192), from (177), we have(195)∫0Tu′t+A2θvt+B2vt,vt−1−α0θf−f2,ut−vtdt+Ψu−Ψv≥−12v0−u02.SinceB2vt,vtV∗≤KvtV3/2vt1/2(cf. (165)) and v∈L∞0,T;L2, B2v,v∈L4/30,T;V∗. Therefore, by density of the set u=∑j=1Mkjtuj,kjt∈C10,T,M=1,2,… in L40,T;V:u′∈L20,T;V∗, (195) is valid for all u∈L40,T;V:u′∈L20,T;V∗.Thus, the first formula of (43) is valid.Puttingφt=∑j=1Mkjtφj, where kjt∈C10,T,kjT=0, multiply the second equation of (141) by kjt and add for j=1,…,M. Thus, we have(196)θmt,φt−∫0tθm,∂φ∂tds+∫0tκθm∇θm,∇φds+∫0tβxθm,φΓRds−∫0tvmθm,∇φds=θm0,φ0+∫0tg2,φds,∀t∈0,T.By Corollary Appendix B.2 of [1], we have(197)∫0tκθm∇θm,∇φds⟶∫0tκθ∇θ,∇φds,asm⟶∞,ε⟶0.SinceW9/10,2Ω⊂L4Ω, we have(198)vm⟶vinL20,T;L4Ω,θm⟶θinL20,T;L4Ω.By (198), we have(199)∫0tvmθm,∇φ−vθ,∇φds≤∫0tvm−vθm,∇φds+∫0tvθm−θ,∇φds≤vm−vL20,T;L4θmL20,T;L4+vL20,T;L4θm−θL20,T;L4∇φL∞0,T;L2⟶0,which implies(200)∫0tvmθm,∇φds⟶∫0tvθ,∇φds,asm⟶∞,ε⟶0.Therefore, taking into account (197) and (200), from (196), we have(201)−∫0Tθ,∂φ∂tdt+∫0Tκθ∇θ,∇φdt+∫0Tβxθ,φΓRdt−∫0Tvθ,∇φdt=θ0,φ0+∫0tg2,φds.Since the setφt=∑j=1Mkjtφj:kjt∈C10,T,kjT=0,M=1,2,… is dense in φ∈C10,T;WΓD1,2:φT=0, from (201), we have the second equation of (43). □


## 5.1. Existence and Estimation of a Solution to an Approximate Problem

We first consider a problem approximating (43). For every 0<ε<1, let a functional Φε be defined by (45).Letuj,j=1,2,… and φj,j=1,2,… be, respectively, bases of the space V and WΓD1,2Ω. Without loss of generality, we assume that u1=v0 and φ1=θ0 as in [26]. We find a solution vm=∑j=1mgjmtuj and θm=∑j=1mrjmtφj to the problem(141)∂vm∂t,uj+2μθmℰvm,ℰuj+rotvm×vm,uj+2μθmkxvm,ujΓ2+2μθmSv˜m,u˜jΓ3+2αxvm,ujΓ5+μθmkxvm,ujΓ7+∇Φεvmt,uj=1−α0θmf,uj+f2,uj,∂θm∂t,φj+κθm∇θm,∇φj+βxθm,φjΓR−vmθm,∇φj=g2,φj,vm0=v0,θm0=θ0,which gives us a system for gjmt and rjmt,j=1−m. The solutions to (141) depend on ε, but for convenience of notation, here and in what follows, we use subindex m. For tm, there exist absolute continuous functions gjmt and rjmt on 0,tm. If vmt and θmt are bounded and vmtandθmt are integrable, then gjmt and rjmt are prolonged over tm. We will find estimates (157) in the following, by which we see that tm=T.Multiplying the first and the second equation of (141), respectively, by gjmt and φjmt and adding for i=1,…,m, we get(142)∂vm∂t,vm+2μθmℰvm,ℰvm+rotvm×vm,vm+2μθmkxvm,vmΓ2+2μθmSv˜m,v˜mΓ3+2αx,tvm,vmΓ5+μθmkxvm,vmΓ7+∇Φεvmt,vm+α0θmf,vm=f,vm+f2,vm,∂θm∂t,θm+κθm∇θm,∇θm+βxθm,θmΓR−vmθm,∇θm=g2,θm,vm0=v0,θm0=θ0.Let us estimate terms on the left-hand side above. It is easy to see(143)2μθmℰvm,ℰvmdt≥2μ0vmV2.By the fact thatΓ2j,Γ3j,Γ7j are in C2.1Γij and Assumption 2, there exists a constant M such that(144)Sx∞,kx∞,αL∞Γ5≤M.Thus,(145)2μθmkxv,vΓ2+2μθmSv˜m,v˜mΓ3+2αxvm,vmΓ5+μθmkxvm,vmΓ7≤μ02vmV2+k11vm2,(cf. Theorem 1.5.1.10 of [28]). Obviously,(146)rotvm×vm,vm=0.SinceΦε is convex, continuous and Fréchet differentiable, we have(147)Φεy−Φεx≥∇Φεx,y−x,∀x,y∈V.Thus,(148)∇Φεvmt,0V−vmt≤Φε0V−Φεvmt≤−Φεvmt,(149)∇Φεvmt,vmt≥Φεvmt.Also, by the Hölder inequality, we have(150)α0θmf,vmdt≤k12vmt2+κ04θmtWΓD1,22,where k12=cα0fL∞0,T;L32.(151)f,wm+f2,wm≤μ02wmV2+cfL32+f2v∗2.Also, we have(152)κθm∇θm,∇θm≥κ0θmWΓD1,22,βxθm,θmΓR≥0.By (17), we have(153)wmθm,∇θm=0.(154)g2,θm≤κ04θmtWΓD1,22+cg2WΓD1,2∗2.Taking(155)k1=k11+k12,by (143)–(154), we have from (142)(156)ddtvmt2+ddtθmt2+2μ0vmV2+κ0θmWΓD1,22+Φεvmt≤cftL32+f2tv∗2+g2tWΓD1,2∗2+2k1vmt2.Applying the Gronwall inequality, we have from (156)(157)vmt2+θmt2≤v02+θ02+∫0tfsL32+f2sv∗2+g2sWΓD1,2∗2dse2k1t,vmL20,T;V2+θmL20,t;WΓD1,22≤cv02+θ02+∫0TftL32+f2tv∗2+g2tWΓD1,2∗2dt,∫0TΦεvmtdt≤cv02+θ02+∫0TftL32+f2tv∗2+g2tWΓD1,2∗2dt.Note thatc in (157) depends on T and (via k12) f but independent of m and ε.By (46) and (29) and the third inequality of (157), we have(158)∫0Tvmt−JεvmtV2dt≤cε,with c independent of ε. Multiplying the first equation of (141) by gjmt−gjms, summing for j, and taking into account (147), we have(159)12dvmt−vms2dt+A2θmvmt+B2vmt,vmt+α0θmtft−ft−f2t,vmt−vms=∇Φεvmt,vms−vmt≤Φεvms−Φεvms≤Φεvms,where the operators A2θm,B2 are the ones in (42). By (145) and (144), we have(160)A2θmvmt,vmt≥3μ02vmtV2−k11vmt2,A2θmvmt,vms≤cvmtVvmsV,where k11 is the one in (145). Taking into account (160) and the fact that B2vmt,vmt,vmt=0, we have from (159)(161)12dvmt−vms2dt≤Φεvms+A2θmvmt,vms+B2vmt,vmt,vms+−α0θmtft+ft+f2t,vmt−vms+k11vmt2.Let us integrate every term of (161) first with respect to t from s to s+h and then with respect to s from 0 to T, where vmt=0 when t∈T,T+h.(162)∫0T∫ss+hdvmt−vms2dtdtds=∫0Tvms+h−vms2ds.By the third inequality of (157),(163)∫0T∫ss+hΦεvmsdtds≤h∫0TΦεvmsds≤c1h.By (157) and (161), we have(164)∫0T∫ss+hA2θmvmt,vmsdtds≤c∫0TvmsV∫ss+hvmtVdtds≤c∫0TvmsVhvmL20,T;Vds≤c2h.SincewL3≤KwL21/2‖w‖L61/2,(165)B2v,w,z=rotv×w,z≤KrotvL2wL3zL6≤KvVw1/2wV1/2zV,and so by (157), we have(166)∫0T∫ss+hB2vmt,vmtvmsdtds≤K∫0T∫ss+hvmtV3/2vmt1/2vmsVdtds≤K∫0TvmsV∫ss+hvmtV2dt3/4∫ss+hvmt2dt1/4ds≤c3h1/4.Also, by (157), we have(167)∫0T∫ss+hf+f2t,vmtdtds≤∫0Tf+f2t,vmt∫t−htdsdt≤c4h,∫0T∫ss+hf+f2t,−vmsdtds≤K∫0TvmsV∫ss+hf+f2t‖V∗dtds≤c5h.In the same way, we get(168)∫0T∫ss+hα0θmf,vmt−vmsdtds≤c6h+c7h,∫0T∫ss+hk11vmt2dtds≤c8h.Note that constantsci,i=1−8, are independent of m,ε. By virtue of (162)–(168), uniformly with respect to m,ε,(169)∫0Tvms+h−vms2ds≤Oh1/4,and the set vm is relatively compact in L20,T;W9/10,2Ω(see Theorem 5 of [29]). Also, we have(170)κθm∇θm,∇φ≤κ1∇θmL2φWΓD1,2Ω,βxθm,φΓR≤cθmWΓD1,2ΩφWΓD1,2Ω,vmθm,∇φ≤cvmVθmL3φWΓD1,2Ω.By (170), from the second equation of (141), we have(171)∂θm∂t,φ≤cθmWΓD1,2Ω+vmV2+θmWΓD1,2Ω2+g2WΓD1,2∗φWΓD1,2,∀φ∈WΓD1,2Ω.Hence, by (157), we know that(172)θm′∈L10,T;WΓD1,τ∗,θm′L10,T;WΓD1,τ∗≤c,where c is independent of m,ε. Thus, the set θε is relatively compact in L20,T;W9/10,2Ω(see Corollary 5 of [29]).


## 5.2. Existence of a Solution

We can extract subsequences, which are denoted as before, such that(173)vm⇀vinL20,T;V,vm⇀∗vinL∞0,T;H,vm⟶vinL20,T;W9/10,2Ω,θm⇀θinL20,T;WD1,2Ω,θm⇀∗θinL∞0,T;L2Ω,θm⟶θinL20,T;W9/10,2Ω,as m⟶∞ and ε⟶0.On the contrary, puttingu=∑j=1Mkjtuj, where kjt∈C10,T and is a positive integer, multiply the first equation of (141) by kjt and add for j=1,…,M. Then, multiply the first equation of (141) by gjmt and add for j=1,…,m. Substituting the resulting equations, we have(174)∫0T∂vm∂t+A2θmvmt+B2vmt,vmt+∇Φεvmt,ut−vmtdt=∫0T−α0θmf+f+f2,ut−vmtdt.Since(175)∫0Tvm′t,ut−vmtdt=∫0Tu′t,ut−vmtdt−12vmT−uT2+12vm0−u02,taking into account (147), we have from (174)(176)∫0Tu′t+A2θmvmt+B2vmt,vmt,ut−vmtdt−∫0T−α0θmf+f+f2,ut−vmtdt+∫0TΦεut−Φεvmtdt≥−12vm0−u02.SinceΦεu≤Φu and ΦJεvmt≤Φεvmt(cf. (47)), we have from (176)(177)∫0Tu′t+A2θmvmt+B2vmt,vmt,ut−vmtdt−∫0T−α0θmf+f+f2,ut−vmtdt+Ψu−∫0TΦJεvmtdt≥−12vm0−u02.By (173) and Corollary Appendix B.2 of [1], taking a subsequence if necessary, we have(178)∫0TA2θmvmt,utdt≡∫0Ta02θmt;vmt,utdt⟶∫0TA2θvt,utdt.Owing to (173),(179)vm⟶vinL20,T;L2∂Ω.Thus, taking a subsequence if necessary, we have (see Lemma Appendix B.1 of [1])(180)2μθmkxvm,vmΓ2+2μθmSv˜m,v˜mΓ3+2αxvm,vmΓ5+μθmkxvm,vmΓ7⟶2μθkxv,vΓ2+2μθSv˜,v˜Γ3+2αxv,vΓ5+μθkxv,vΓ7.Therefore, taking into account(181)liminf2μθmℰvm,ℰvm≥2μθℰv,ℰv(see Corollary Appendix B.3 of [1]), we have(182)liminfA2θmvmt,vmt≥A2θvt,vt.By (178) and (182), we have(183)liminfm⟶∞ε⟶0∫0TA2θmvmt,ut−vmtdt≤∫0TA2θvt,ut−vtdt.By (158) and (173), Jεvm⇀v in L20,T;V as m⟶∞ and ε⟶0. Since the functional Φ:V⟶ℝ¯ is lower weak semicontinuous, we have(184)liminfm⟶∞ε⟶0∫0TΦJεvmtdt≥∫0TΦvtdt≡Ψv.In a rather routine way, we can prove that(185)∫0TB2vmt,vmt,utdt⟶∫0TB2vt,vt,utdtas m⟶∞,ε⟶0. For convenience of readers, we give a proof in the following:(186)∫0TB2vmt,vmt,ut−B2vt,vt,utdt=∫0TB2vmt,vmt−vt,utdt+∫0TB2vmt−vt,vt,utdt≡I1+I2.By (165), the Hölder inequality with exponents 2,4,4, and (173), we have(187)I1≤Ksupt∈0,TutH1∫0TvmtH1vmt−vtH11/2vmt−vt1/2dt≤Ksupt∈0,TutH1vmL20,T;H1vm−vL20,T;H11/2vm−vL20,T;L21/2⟶0.By (165),(188)∫0TB2zt,vt,utdt≤Ksupt∈0,TutH1ztL20,T;H1vL20,T;H11/2vL20,T;L21/2≤K1ztL20,T;H1,which means that the mapping(189)zt∈L20,T;V⟶∫0TB2zt,vt,utdt,is continuous and linear on L20,T;V, that is, there exists f∈L20,T;V∗ such that(190)∫0TB2zt,vt,utdt=∫0Tzt,ftdt.Thus, by (173),(191)I2⟶0,asm⟶∞,ε⟶0.By (186)–(191), we get (185).Let us prove that(192)∫0Tα0θmf,ut−vmtdt⟶∫0Tα0θf,ut−vtdt⟶0,asm⟶∞,ε⟶0.By (173), we have(193)∫0Tα0θm−α0θf,utdt⟶0,(194)∫0Tα0θmf,vmt−α0θf,vtdt=∫0Tα0θm−θf,vmtdt+∫0Tα0θf,vmt−vtdt⟶0.By (193) and (194), we get (192).Therefore, by (183)–(185) and (192), from (177), we have(195)∫0Tu′t+A2θvt+B2vt,vt−1−α0θf−f2,ut−vtdt+Ψu−Ψv≥−12v0−u02.SinceB2vt,vtV∗≤KvtV3/2vt1/2(cf. (165)) and v∈L∞0,T;L2, B2v,v∈L4/30,T;V∗. Therefore, by density of the set u=∑j=1Mkjtuj,kjt∈C10,T,M=1,2,… in L40,T;V:u′∈L20,T;V∗, (195) is valid for all u∈L40,T;V:u′∈L20,T;V∗.Thus, the first formula of (43) is valid.Puttingφt=∑j=1Mkjtφj, where kjt∈C10,T,kjT=0, multiply the second equation of (141) by kjt and add for j=1,…,M. Thus, we have(196)θmt,φt−∫0tθm,∂φ∂tds+∫0tκθm∇θm,∇φds+∫0tβxθm,φΓRds−∫0tvmθm,∇φds=θm0,φ0+∫0tg2,φds,∀t∈0,T.By Corollary Appendix B.2 of [1], we have(197)∫0tκθm∇θm,∇φds⟶∫0tκθ∇θ,∇φds,asm⟶∞,ε⟶0.SinceW9/10,2Ω⊂L4Ω, we have(198)vm⟶vinL20,T;L4Ω,θm⟶θinL20,T;L4Ω.By (198), we have(199)∫0tvmθm,∇φ−vθ,∇φds≤∫0tvm−vθm,∇φds+∫0tvθm−θ,∇φds≤vm−vL20,T;L4θmL20,T;L4+vL20,T;L4θm−θL20,T;L4∇φL∞0,T;L2⟶0,which implies(200)∫0tvmθm,∇φds⟶∫0tvθ,∇φds,asm⟶∞,ε⟶0.Therefore, taking into account (197) and (200), from (196), we have(201)−∫0Tθ,∂φ∂tdt+∫0Tκθ∇θ,∇φdt+∫0Tβxθ,φΓRdt−∫0Tvθ,∇φdt=θ0,φ0+∫0tg2,φds.Since the setφt=∑j=1Mkjtφj:kjt∈C10,T,kjT=0,M=1,2,… is dense in φ∈C10,T;WΓD1,2:φT=0, from (201), we have the second equation of (43). □



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*Source: 6096531-2020-09-03.xml*