# Classical Solvability of a Free Boundary Problem for an Incompressible Viscous Fluid with a Surface Density Equation

Authors: Yoshiaki Kusaka
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We investigate a mathematical model introduced
by Shikhmurzaev to remove singularities that arise when classical hydrodynamic models are applied to certain physical phenomena. The
model is described as a free boundary problem consisting of the Navier-Stokes equations and a surface mass balance equation. We prove the
local-in-time solvability in Hölder spaces.


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## Body

## 1. Introduction

Let a time-dependent bounded domainΩt⊂R3 with the outer boundary Γt≡∂Ωt be filled with an incompressible viscous fluid, and let Γt represent the interface. In Ωt, we assume that the flow is governed by the Navier-Stokes equations:
(1)∂v∂t+(v·∇)v+∇p-νΔv=0,∇·v=0,
where v is the velocity, p is the pressure, and ν, which is assumed to be a positive constant, is the kinematic viscosity.OnΓt, we assume the following equations:
(2)ΠT(v,p)n=∇-σ,n·T(v,p)n=σH,(3)DρsDt+ρs∇-·vs=ρ(v-vs)·n,(4)(v-vs)·n=-ρes-ρsρτ,Π(v-vs)=-χ∇-σ,(5)σ=γ(ρ--ρs).
Here vs and ρs are the velocity and the density of surface layer, respectively. T(v,p)=νD(v)-pI is the stress tensor, where D(v)=((∂vi/∂xj)+(∂vi/∂xj))i,j=1,2,3 is the velocity deformation tensor. H is the twice mean curvature of Γt at the point x, which is negative if Ωt is convex in the neighborhood of x. n is the unit outward normal to Γt at the point x. Π is the projection operator onto the tangent plane at the point x on Γt. D/Dt denotes the derivative along the trajectory of particle on Γt. ∇- is the gradient restricted to the surface. ρ, ρes, τ, χ, γ, ρ- are positive constants; in particular, ρ is the density of the bulk and τ is the characteristic time scale over which the surface density ρs relaxes to its equilibrium value ρes.Finally, to complete the problem, we give the initial conditions:(6)v|t=0=v0onΩ-≡Ω-0,ρs|t=0=ρ0onΓ≡Γ0.It is known that singularities arise when the the classical hydrodynamic equations and modeling assumptions are applied to certain physical phenomena. For example, the application of the classical no-slip boundary condition to the spreading of a drop on a plate gives rise to a nonintegrable shear stress, and the application of the classical kinematic condition at the free boundary to the formation of a cusp on a free surface of a viscous fluid leads to an infinite energy dissipation in the fluid (e.g., refer to [1] and the references therein).To remove the above mentioned singularities, we are required to modify classical boundary conditions by taking into account molecular interaction near interfaces. The molecule in the liquid region which is very close to another phases experiences an asymmetric force due to the presence of another materials. This gives rise to the variation in the density in the liquid region near to the adjacent phase, and the surface tension occurs as a result of this variation in density. The thin layer in the liquid region in which the above mentioned density variation occurs is called the surface layer.Through [1–4], Shikhmurzaev developed a theory to remove the above mentioned singularities by introducing a surface layer which is treated as a separate phase. In this theory, the no-slip condition assumed in classical models for dynamic wetting processes is modified as the Navier-slip condition through thermodynamic considerations on the surface layer (refer to [2, 3]). The formation of a free surface cusp associated with fluid flow is also investigated in [4]. In [4], the cusp formation is modeled as an interface disappearance process. In this model, an internal surface stretching from the cusp, which is referred to as “the surface-tension-relaxation tail”, is introduced. The above mentioned singularity associated with the modeling of cusp formation arises owing to the absence of viscous stress at the cusp with which the surface tension acting from the liquid surface is balanced. In this model, the surface tension at the cusp can be balanced by shear stresses acting on this tail.The problem (1)–(6) is a model describing the behavior of an isolated liquid drop in which the interface is modeled as a surface layer based on Shikhmurzaev’s theory. The dynamics of the liquid in this layer are governed by (3) which represents conservation of mass. The right-hand side of (3) represents the source consisting of a flow of molecules from the bulk. Equations in (4) are conditions that minimize the rate of entropy production in the surface layer. Equation (5) represents a linearized state equation in the surface layer (refer to [1] for details). In Shikhmurzaev’s theory, the surface layer is modeled as a sharp interface as a result of a continuum approximation. Thus, in the above problem, the surface layer is described by the equations given in (3)–(5) defined on a geometric surface, and the behavior of the surface layer is related to (1) in the bulk through the boundary conditions given in (2).In the present paper, we prove the local-in-time classical solvability of problem (1)–(6). As is mentioned above, this model is important as a basic model to describe the above mentioned physical phenomena; however, as far as the author knows, any rigorous proofs on its solvability have not been given. In the present paper, we consider the case where the mass exchange between the interface and the bulk does not occur. As is seen in Section 2, under such an assumption, we can reformulate our problem as a problem defined in a domain with a fixed known boundary by introducing Lagrangian coordinates, and in Section 3, we construct a unique solution of the reformulated problem in Hölder spaces with the aid of the method of successive approximations.


## 2. Reformulation of the Problem

In this section, we reformulate our problem in Lagrangian coordinates. By Lagrangian coordinates we mean the initial coordinates of the fluid particles. In the case where no exchange of molecules occurs between the surface and the bulk, (4)1 is reduced to v·n=vs·n. This relation indicates that the following kinematic condition at the interface is satisfied: the interface consists of the particles located at the interface at the initial time. This circumstance enables us to relate each point x∈Ω-t to its initial point ξ∈Ω- by relation (10) given below.Before reformulating our problem, we rewrite (3) as a nonlinear parabolic equation on Γt with the time derivative D~ρs/D~t, where D~ρs/D~t denotes the derivative along the trajectory of particle on the interface with velocity v. Noting the following relation (e.g., see [5]):
(7)(DρsDt)n=DρsDt-vs·∇-ρs=D~ρsD~t-v·∇-ρs,
where (Dρs/Dt)n represents the derivative along the trajectory which is normal to the interface, (3) can be written as
(8)D~ρsD~t+ρs∇-·vs=(v-vs)·∇-ρs.
Then eliminating vs from the above equation with the aid of the relation (4)2, we obtain the following equation:
(9)D~ρsD~t-χγρ0s∇-2ρs=-ρs∇-·v+χγ{(ρs-ρ0s)∇-2ρs+∇-ρs·∇-ρs}.Now let us reformulate our problem. The Lagrangian and Eulerian coordinates are related by(10)x=Xu(ξ,t)≡ξ+∫0t‍u(ξ,τ)dτ,
where u(ξ,τ) is the velocity at time t of the particle which was located at ξ at t=0. By changing the variables from x to ξ by relation (10), problem (1)–(6) is reformulated as the following problem defined in the cylindrical domain Ω0T=Ω×(0,T) with the lateral boundary Γ0T≡Γ×(0,T):
(11)∂u∂t-ν∇u2u+∇uq=0,∇u·u=0inΩ0T,(12)ΠΠuTu(u,q)nu=Π∇Γtθ,n·Tu(u,q)nu=θ∇Γt2Xu(ξ,t)|ξ∈Γ·n+n·∇Γtθ,∂rs∂t-χγρ0s∇Γt2rs=-rs∇Γt·u+χγ{(rs-ρ0s)∇Γt2rs+∇Γtrs·∇Γtrs},θ=γ(ρ--rs)onΓ0T,(13)u|t=0=v0onΩ-,rs|t=0=ρ0sonΓ.
In (11)–(13), u, q and rs are v(Xu(ξ,t),t),p(Xu(ξ,t),t), and ρs(Xu(ξ,t),t), respectively. Consider ∇u=(𝒥u-1)t∇≡𝒥u*∇; here 𝒥u denotes the Jacobian matrix of Xu, and the notation At means the transpose of the matrix A. n is the outward unit normal to Γ at the point ξ, nu=𝒥u*n/|𝒥u*n|, and Π and Πu are the operators defined by Πf=f-(f·n)f and Πuf=f-(f·nu)f, respectively. Tu(u,q) is the tensor with the elements ν∑k=13‍(Ajk(∂ui/∂ξk)+Aik(∂uj/∂ξk))-pδij, where Aij is the (i,j)-element of 𝒥u*, and δij is Kronecker’s delta. The operators ∇Γtf and ∇Γt·A are defined by
(14)∇Γtf=∑α,β=1,2‍gαβ∂f∂sβ∂Xu(s)∂sα,∇Γt·A=1g∑α=1,2‍∂∂sαgAα,
where g=det(gαβ)α,β=1,2, gαβ=(∂Xu(s)/∂sα)·(∂Xu(s)/∂sβ), gαβ denote the components of the inverse matrix of (gαβ)α,β=1,2, Xu(s)=Xu(ξ(s),t), s=(s1,s2) denotes the local coordinates on Γ, and Aα denotes the components of the vector A with respect to the basis (∂Xu(s)/∂sα), α=1,2. Finally, the operator ∇Γt2f is defined as
(15)∇Γt2f=∇Γt·∇Γtf=1g∑α,β=1,2‍∂∂sαggαβ∂f∂sβ.Note that in derivation of (12)2, we have used the formula Hn=∇-2x≡∇-2Xu(ξ,t). Note also that although (12)1,2 are different from the following formulas which are obtained directly from (2):
(16)ΠuTu(u,q)nu=∇Γtθ,nu·Tu(u,q)nu=θ∇Γt2Xu(ξ,t)|ξ∈Γ·nu,
problem (11)–(13) is equivalent to problem (1)–(6) as far as the condition n·nu>0, which is valid for sufficiently small t, is satisfied.We now introduce some function spaces. LetD be a domain in Rn, let T be a positive constant, let DT be a cylindrical domain D×(0,T), let l be a nonnegative integer, and let α, γ∈(0,1).ByCl+α(D), we define the space of functions f(x), x∈D, with the norm
(17)|f|D(l+α)≡∑|m|≤l‍|∂xmf|D+[f]D(l+α),|f|D≡supx∈D|f(x)|,[f]D(l+α)≡∑|m|=l‍[∂xmf]D(α)≡supx,y∈D,x≠y∑m=|l|‍|∂xmf(x)-∂ymf(y)||x-y|α,|m|=∑i=1n‍mi,∂xm=∂|m|∂x1m1⋯∂xnmn,
for a multi-index m=(mi)(mi≥0,i=1,…,n).ByCl+α,((l+α)/2)(DT) we denote an anisotropic Hölder space of functions whose norm is defined by
(18)|f|DT(l+α,((l+α)/2))≡∑2r+|m|=0l‍|∂tr∂xmf|DT+[f]DT(l+α,((l+α)/2)),
where
(19)|f|DT≡sup(x,t)∈DT|f(x,t)|,[f]DT(l+α,((l+α)/2))≡∑2r+|m|=l-1l‍[∂tr∂xmf]DT(0,((l+α-(2r+|m|))/2))+∑2r+|m|=l‍[∂tr∂xmf]DT(α,0).
Here,
(20)[f]DT(0,(α/2))≡sup(x,t),(x,t′)∈DT,t≠t′|f(x,t)-f(x,t′)||t-t′|(α/2),[f]DT(α,0)≡sup(x,t),(x′,t)∈DT,x≠x′|f(x,t)-f(x′,t)||x-x′|α.Finally, we introduce the function spaceC1+α,γ(DT) equipped with the norm
(21)∥f∥DT(1+α,γ)≡|f|DT+|∇f|DT(α,(α/2))+|f|DT(1+α,γ),
where
(22)|f|DT(1+α,γ)≡supτ,t∈(0,T),τ≠t|f(·,t)-f(·,τ)|D(γ)|t-τ|((1+α-γ)/2).Now, let us state our main result.Theorem 1.
Letα, γ be constants satisfying 0<α, γ<1. Assume that
(23)v0∈C2+α(Ω-),ρ0s∈C2+α(Γ),Γ∈C3+α.
Assume that there exist positive constants Δ1 and Δ2 such that ρ--ρ0s≥Δ1>0 and ρ0s≥Δ2>0 on Γ. In addition, assume that the following compatibility conditions are satisfied:
(24)∇·v0=0,(νΠD(v0)n+γ∇Γρ0s)|Γ=0,
where ∇Γ is the operator corresponding to ∇Γt with t=0; namely, ∇Γ is given by the formula in (14) with Xu(s)=ξ(s).
Then, for a positive constantT, problem (11)–(13) has a unique solution (u,q,rs) with the following smoothness:
(25)u∈C2+α,1+(α/2)(Ω0T),q∈C1+α,γ(Ω0T)∩C1+α,((1+α)/2)(Γ0T),rs∈C2+α,1+(α/2)(Γ0T).


## 3. Proof of the Main Result

In this section, we will prove Theorem1.We begin with preparing estimates of solutions to some linear problems. For the following problem:(26)∂u∂t-ν∇w2u+∇wq=F1,∇w·u=F2inΩ0T,νΠΠwDw(u)nw=F3,n·Tw(u,q)nw-Θ-n·∇Γ~t2∫0t‍udτ=b+∫0t‍BdτonΓ0T,u|t=0=v0onΩ-,
the following result is given in [6]. In (26), ∇w, Πw, Tw(u,q)(=νDw(u)-qI), nw are defined for a given vector w in the same manner as ∇u, Πu, Tu(u,q), nu are defined, and ∇Γ~t is defined by (14) with Xu=Xw.Theorem 2.
LetT>0, and let α, γ be positive constants satisfying 0<α, γ<1. Assume that
(27)Γ∈C2+α,v0∈C2+α(Ω-),F1∈Cα,(α/2)(Ω0T),F2∈C1+α,((1+α)/2)(Ω0T),F3∈C1+α,((1+α)/2)(Γ0T),b∈C1+α,((1+α)/2)(Γ0T),B∈Cα,(α/2)(Γ0T),Θ-∈Cα(Γ).
Assume that there exists a positive constant Δ1 such that Θ-≥Δ1>0 on Γ. Assume that the following compatibility conditions are satisfied:
(28)∇·v0=F2(ξ,0),νΠD(v0)n|Γ=F3(ξ,0).
Assume that there exist functions h∈Cα,(α/2)(Ω0T), Hk, k=1,2,3, with a finite norm |Hk|Ω0T(1+α,γ) satisfying the relation
(29)∂F2∂t-∇w·F1=∇·h,h=∑k=13‍∂ξkHk,
in the sense of distribution. Furthermore, assume that w∈C2+α,1+(α/2)(Ω0T) satisfies the inequality
(30)(T+T1/2)|w|Ω0T(2+α,1+(α/2))+T((1-α+γ)/2)|∂ξw|Ω0T≤δ,
for a sufficiently small positive constant δ.
Then problem (26) has a unique solution (u,q) satisfying the following inequality:
(31)|u|Ω0T(2+α,1+(α/2))+∥q∥Ω0T(1+α,γ)+|q|Γ0T(1+α,((1+α)/2))≤C-1(T){∑k=13‍|F1|Ω0T(α,(α/2))+|F2|Ω0T(1+α,((1+α)/2))mmmmmmmm+|F3|Γ0T(1+α,((1+α)/2))+|b|Γ0T(1+α,((1+α)/2))mmmmmmmm+|B|Γ0T(α,(α/2))+|v0|Ω(2+α)+|h|Ω0T(α,(α/2))mmmmmmmm+∑k=13‍|Hk|Ω0T(1+α,γ)+PT(w)(|v0|Ω(1)+|b(·,0)|Γ)},
where PT(w)=T((1-α)/2)|w|Ω0T(1,0)+|∂ξw|Ω0T(α,(α/2))+[∂ξw]Ω0T(0,((1+α-γ)/2)) and C-1(T) is a nondecreasing function of T.For the following problem:(32)∂rs∂t-χγρ0s∇Γ~t2rs=GonΓ0T,rs|t=0=ρ0sonΓ,
we have the following theorem. The assertion of the theorem immediately follows from the Hölder estimates for linear parabolic equations (e.g., see [7]).Theorem 3.
LetT>0, and let α be a positive constant satisfying 0<α<1. Assume that
(33)Γ∈C2+α,ρ0s∈C2+α(Γ),G∈Cα,(α/2)(Γ0T).
Assume that there exists a positive constant Δ2 such that ρ0s≥Δ2>0 on Γ. Further assume the same assumptions for w stated in Theorem 2.
Then, problem (32) has a unique solution rs satisfying the following inequality:
(34)|rs|Γ0T(2+α,1+(α/2))≤C-2(T)(|G|Γ0T(α,(α/2))+|ρ0s|Γ(2+α)),
where C-2(T) is a nondecreasing function of T.Combining the above results, we can easily obtain Theorem4 given below for the following problem:
(35)∂u∂t-ν∇w2u+∇wq=F1,∇w·u=F2inΩ0T,νΠΠwDw(u)nw+γΠ∇Γ~trs=F3,n·Tw(u,q)nw-Θ-n·∇Γ~t2∫0t‍udτ+γn·∇Γ~trs=b+∫0t‍BdτonΓ0T,u|t=0=v0onΩ-,∂rs∂t-χγρ0s∇Γ~t2rs=GonΓ0T,rs|t=0=ρ0sonΓ.
The estimate given in the theorem will be essentially used in the later argument to prove Theorem 1.Theorem 4.
Under the same assumptions given in Theorem2 where only compatibility condition is replaced by
(36)∇·v0=F2(ξ,0),(νΠD(v0)n+γΠ∇Γρ0s)|Γ=F3(ξ,0),
and Theorem 3, problem (35) has a unique solution (u,q,rs) satisfying the following inequality:
(37)|u|Ω0T(2+α,1+(α/2))+∥q∥Ω0T(1+α,γ)+|q|Γ0T(1+α,((1+α)/2))+|rs|Γ0T(2+α,1+(α/2))≤C-3(T){|F1|Ω0T(α,(α/2))+|F2|Ω0T(1+α,((1+α)/2))mmmmmmmn+|F3|Γ0T(1+α,((1+α)/2))+|b|Γ0T(1+α,((1+α)/2))mmmmmmmn+|B|Γ0T(α,(α/2))+|h|Ω0T(α,(α/2))+∑k=13‍|Hk|Ω0T(1+α,γ)mmmmmmmn+|v0|Ω(2+α)+|G|Γ0T(α,(α/2))+|ρ0s|Γ(2+α)mmmmmmmn+PT(w)(|v0|Ω(1)+|b(·,0)|Γ)},
where C-3(T) is a nondecreasing function of T.In addition, we prepare estimates for𝒥u*, which are used later.Lemma 5.
Let𝒥u and 𝒥u′ be the Jacobian matrices of the mappings Xu and Xu′, respectively. Let us assume that u and u′ satisfy condition (30) for sufficiently small δ>0. Then, the following inequalities hold:
(38)|𝒥u*|Ω0T(1+α,((1+α)/2)),|𝒥u*|Ω0T(1+α,γ)≤C,
where C is a positive constant independent of δ, and
(39)|𝒥u*-𝒥u′*|Ω0T(1+α,((1+α)/2)),|𝒥u*-𝒥u′*|Ω0T(1+α,γ)≤ϵ|u-u′|Ω0T(2+α,1+(α/2))+C(ϵ)∫0T‍|u-u′|Ω0τ(2+α,1+(α/2))dτ,
for arbitrary 0<ϵ<1, where C(ϵ) is a positive constant depending only on ϵ.Proof.
In the following proof,c1, c2, and c3 are positive constants independent of δ and ϵ. Let aij be the (i,j)-component of 𝒥u.
Then, we have(40)δij-|∫0t‍∂ξjuidτ|≤aij≤δij+|∫0t‍∂ξjuidτ|,
where δij denotes Kronecker’s delta. Then, using the inequality
(41)|∫0t‍∂ξjuidτ|Ω0T≤T|u|Ω0T(1,0)≤δ,
from (40), we have
(42)δij-δ≤|aij|Ω0T≤δij+δ.
This inequality implies that det𝒥u>0 holds for sufficiently small δ>0.
Now, letaij′ be the (i,j)-components of 𝒥u′, and let Aij and Aij′ be the cofactors of aij and aij′, respectively.
Then from the inequalities(43)|aij|Ω0T(1+α,((1+α)/2)),|aij′|Ω0T(1+α,((1+α)/2))≤c1(1+δ),
we have
(44)|Aij|Ω0T(1+α,((1+α)/2)),|Aij|Ω0T(1+α,γ)≤c2.
On the other hand, from the inequality(45)|aij-aij′|Ω0T(1+α,((1+α)/2))≤ϵ|u-u′|Ω0T(2+α,1+(α/2))+c3(1+ϵ-((1+α)/(2-α)))∫0T‍|u-u′|Ω0τ(2+α,1+(α/2))dτ,
which are obtained with the aid of the following inequality which holds for arbitrary 0<ϵ<1:
(46)|∫0t‍f(ξ,τ)dτ|Ω0T(1+α,((1+α)/2))≤∫0T‍|f|Ω0τ(1+α,0)dτ+sup|t′-t′′|<ϵ(2/(1-α))(|t′-t′′|-((1+α)/2)∫t′′t′‍|f(ξ,τ)|Ωdτmmmmmmmmmmmm+|t′-t′′|-(α/2)∫t′′t′‍|∂ξf(ξ,τ)|Ωdτ)+supϵ(2/(1-α))<|t′-t′′|≤T(|t′-t′′|-((1+α)/2)∫t′′t′‍|f(ξ,τ)|Ωdτmmmmmmmmmmmm+|t′-t′′|-(α/2)∫t′′t′‍|∂ξf(ξ,τ)|Ωdτ)≤ϵ|f|Ω0T(1+α,((1+α)/2))+c3(1+ϵ-((1+α)/(1-α)))∫0T‍|f|Ω0τ(1+α,((1+α)/2))dτ,
we have
(47)|Aij-Aij′|Ω0T(1+α,((1+α)/2)),|Aij-Aij′|Ω0T(1+α,γ)≤ϵ|u-u′|Ω0T(2+α,1+(α/2))+C1(ϵ)∫0T‍|u-u′|Ω0τ(2+α,1+(α/2))dτ,
where C1(ϵ) is a positive constant depending only on ϵ. From (44) and (47), estimates (38) and (39) immediately follow. Thus, the proof is completed.Now, let us prove Theorem1 by the method of successive approximations. We take (u0,q0,r0s)=(0,0,0), and, for the known nth approximation, we define the (n+1)th approximation by the solutions of the following problem:
(48)∂un+1∂t-ν∇un2un+1+∇unqn+1=0,∇un·un+1=0inΩ0T,νΠΠunDun(un+1)nun+γΠ∇Γn,trn+1s=0,n·Tun(un+1,qn+1)nun-Θ-n·∇Γn,t2∫0t‍un+1dτ+γn·∇Γn,trn+1s=b(un,rns)+∫0t‍B(un,rns)dτonΓ0T,un+1|t=0=v0onΩ-,∂rn+1s∂t-χγρ0s∇Γn,t2rn+1s=G(rns,un)onΓ0T,rn+1s|t=0=ρ0sonΓ,
where
(49)b(un,rns)=θn(H0+n·∫0t‍(∇Γn,τ2)τξdτ),H0=n·∇Γ2ξB(un,rns)={∂∂τ(θn-Θ-)}n·∇Γn,τ2∫0τ‍un(ξ,s)ds+(θn-Θ-)n·(∇Γn,τ2)τ∫0τ‍un(ξ,s)ds+(θn-Θ-)n·∇Γn,τ2un,θn=γ(ρ--rns),Θ-=γ(ρ--ρ0s),G(un,rns)≡-rns∇Γn,t·un+χγ{(rns-ρ0s)∇Γn,t2rns+∇Γn,trns·∇Γn,trns}.
In the above formulas, ∇Γn,t is the operator corresponding to ∇Γt with Xu=Xun, and (∇Γn,τ2)τ denotes the operator obtained by differentiating the coefficients of ∇Γn,τ2 with respect to τ.Now, let us verify that all terms of the sequence{(un,qn,rns)} are defined on some time interval independent of n. We begin with the following lemma.Lemma 6.
LetTn be a constant satisfying 0<Tn≤1. Then there exist positive constants δ and β such that if un and rns satisfy the following conditions:
(50)(Tn+Tn1/2)|un|Ω0Tn(2+α,1+(α/2))+Tn((1-α+γ)/2)|∂ξun|Ω0Tn≤δ,(Tn+Tnβ)(|un|Ω0Tn(2+α,1+(α/2))+|rns|Γ0Tn(2+α,1+(α/2)))≤δ,
then the following inequality holds for a positive constant C- independent of un, rns and Tn:
(51)|b(un,rns)|Γ0Tn(1+α,((1+α)/2))+|B(un,rns)|Γ0Tn(α,(α/2))+|G(un,rns)|Γ0Tn(α,(α/2))+PTn(un)≤C-.Proof.
In the proof,c1,…,c9 are positive constants independent of un, rns, and Tn.
ChoosingTn((1-α)/2α) as ϵ in the following interpolation inequality:
(52)|θn-(θn|t=0)|Γ0Tn(1+α,((1+α)/2))≤ϵ|rns-ρ0s|Γ0Tn(2,1)+c1ϵ-(α/(1-α))|rns-ρ0s|Γ0Tn≤ϵ|rns-ρ0s|Γ0Tn(2,1)+c1ϵ-(α/(1-α))∫0Tn‍|∂τ(rns-ρ0s)|Γ0τdτ,
we have the following estimate:
(53)|θn|Γ0Tn(1+α,((1+α)/2))≤c2(|ρ0s|Γ(1+α)+|θn-(θn|t=0)|Γ0Tn(1+α,((1+α)/2)))≤c3{|ρ0s|Γ(1+α)+(Tn((1-α)/2α)+Tn(1/2))|rns-ρ0s|Γ0Tn(2,1)}≤c4(1+δ).
With the aid of the above estimate, we can easily obtain the desired estimate for b(un,rns).
B
(
u
n
,
r
n
s
) is estimated as follows. With the aid of the inequality
(54)|∫0t‍∂ξ2f(ξ,τ)dτ|Γ0T(α,(α/2))≤c5(T+T1-(α/2))|f|Γ0T(2+α,1+(α/2)),
we have
(55)|{∂∂t(θn-Θ-)}n·∇Γn,t2∫0t‍un(ξ,τ)dτ+(θn-Θ-)n·(∇Γn,t2)t∫0t‍un(ξ,τ)dτ|Γ0Tn(α,(α/2))≤c6(Tn+Tn1-(α/2))|rns-ρ0s|Γ0Tn(2+α,1+(α/2))|un|Γ0Tn(2+α,1+(α/2))≤c7(1+δ2).
On the other hand, with the aid of
(56)|θn-Θ-|Γ0Tn(α,(α/2))≤∫0Tn‍|∂τ(θn-Θ-)|Γ0τ(α,(α/2))dτ≤Tn|rns-ρ0s|Γ0Tn(2+α,1+(α/2)),
we have
(57)|(θn-Θ-)n·∇Γn,t2un|Γ0Tn(α,(α/2))≤c8Tn|rns-ρ0s|Γ0Tn(2+α,1+(α/2))|un|Γ0Tn(2+α,1+(α/2))≤c9(1+δ2).
From (55) and (57), we have the desired estimate for B(un,rsn). G(un,rns) and PTn(un) are estimated in a similar manner. Thus, we have proved the lemma.From this lemma, ifun and rns satisfy conditions (50), by applying Theorem 4 to problem (48) we have the following estimate of un+1 and rn+1s:
(58)|un+1|Ω0Tn(2+α,1+(α/2))+|rn+1s|Γ0Tn(2+α,1+(α/2))≤C-3(1)C-,
where C-3(·) is the function given in Theorem 4.Now, let us takeT satisfying the following conditions:
(59)0<T≤1,C-3(1)C-(T+Tβ)≤δ.
Here β is assumed to be chosen, so that β<1/2 and β<(1-α+γ)/2. Since (u0,r0s)=(0,0), the zeroth approximation (u0,r0s) obviously satisfies conditions (50) for the above T. Hence, from (58), (u1,r1s) satisfies
(60)|u1|Ω0T(2+α,1+(α/2))+|r1s|Γ0T(2+α,1+(α/2))≤C-3(1)C-.
From (59), this inequality indicates that (u1,r1s) satisfies conditions (50), and hence we can obtain the same estimate as (60) for (u2,r2s). Thus, repeating this argument, we can construct a sequence {(un,qn,rns)} such that each term is defined on (0,T).Let us proceed to the proof of the convergence of the sequence{(un,qn,rns)}. In the following argument, C1,…,C10 denote positive constants independent of n and C(ϵ) represents various positive constants depending only on ϵ.Let us set(61)U(n+1)≡un+1-un,Q(n+1)≡qn+1-qn,R(n+1)≡rn+1s-rns.
Subtracting (48) with index n from that with index n+1, we have
(62)∂U(n+1)∂t-ν∇un2U(n+1)+∇unQ(n+1)=ℱ1(n),∇un·U(n+1)=ℱ2(n)inΩ0T,νΠΠunDun(U(n+1))nun+γΠ∇Γn,tR(n+1)=ℱ3(n),n·Tun(U(n+1),Q(n+1))nun-Θ-n·∇Γn,t2∫0t‍U(n+1)dτ+γn·∇Γn,tR(n+1)=b(n)+∫0t‍ℬ(n)dτonΓ0T,U(n+1)|t=0=0onΩ-,∂R(n+1)∂t-χγρ0s∇Γn,t2R(n+1)=𝒢(n)inΓ0T,R(n+1)|t=0=0onΓ,
where
(63)ℱ1(n)≡ν(∇un2-∇un-12)un-(∇un-∇un-1)qn,ℱ2(n)≡-(∇un-∇un-1)·un,ℱ3(n)≡-νΠ{ΠunDun(un)nun-Πun-1Dun-1(un)nun-1}-γΠ(∇Γn,t-∇Γn-1,t)rns,b(n)≡-n·{Tun(un,qn)nun-Tun-1(un,qn)nun-1}-γn·(∇Γn,t-∇Γn-1,t)rns+b(un,rns)-b(un-1,rn-1s),ℬ(n)≡Θ-n·(∇Γn,τ2-∇Γn-1,τ2)τ∫0τ‍un(ξ,s)ds+Θ-n·(∇Γn,τ2-∇Γn-1,τ2)un+B(un,rns)-B(un-1,rn-1s),𝒢(n)≡χγρ0s(∇Γn,t2-∇Γn-1,t2)rns+G(un,rns)-G(un-1,rn-1s).Now, noting that the relations∑j=13‍∂ξjAij=0 hold for the cofactors Aij of the Jacobian matrix of any transformation from ξ to x, by direct calculations, we can verify that the following relations hold:
(64)∂ℱ2(n)∂t-∇un·ℱ1(n)=∇·h(n),h(n)=∑k=13‍∂ξkHk(n),
where
(65)Hk(n)=-(𝒥un-1-𝒥un-1-1)Lk(n)-𝒥un-1Mk(n)+14π∂ξk∫Ω‍Nk(n)(η,t)|ξ-η|dη,
with
(66)Lk(n)≡νAik(n-1)Ail(n-1)∂ξlun-𝒥un-1*ekqn,Mk(n)≡ν(Aik(n)Ail(n)-Aik(n-1)Ail(n-1))∂ξlun-(𝒥un*-𝒥un-1*)ekqn,Nk(n)≡{∂t(𝒥un-1-𝒥un-1-1)}un-{∂ηk(𝒥un-1-𝒥un-1-1)}Lk(n)-(∂ηk𝒥un-1)Mk(n).
In (66), the Einstein summation convention is used, Aij(n) denotes the (i,j)-component of 𝒥n*, and ek, k=1,2,3, denote fundamental unit vectors in R3.For the terms in (63), h(n), and Hk(n), we have the following estimate for arbitrary 0<ϵ<1:
(67)|ℱ1(n)|Ω0T(α,(α/2))+|ℱ2(n)|Ω0T(1+α,((1+α)/2))+|ℱ3(n)|Γ0T(1+α,((1+α)/2))+|b(n)|Γ0T(1+α,((1+α)/2))+|ℬ(n)|Γ0T(α,(α/2))+|h(n)|Ω0T(α,(α/2))+∑k=13‍|Hk(n)|Ω0T(1+α,γ)+|𝒢(n)|Γ0T(α,(α/2))≤C1{ϵ(|U(n)|Ω0T(2+α,1+(α/2))+|R(n)|Γ0T(2+α,1+(α/2)))mmmmn+C(ϵ)∫0T‍(|U(n)|Ω0τ(2+α,1+(α/2))mmmmmmmmmmmn+|R(n)|Γ0τ(2+α,1+(α/2)))dτ}.We will derive here the estimate only ofHk(n) because the other terms can be estimated in a similar and simpler manner. Noting that the following estimates hold for un and qn:
(68)|un|Ω0T(2+α,1+(α/2))+|qn|Ω0T(1+α,γ)≤C2,
and with the aid of estimates (38) and (39), we have
(69)|(𝒥un-1-𝒥un-1-1)Lk(n)|Ω0T(1+α,γ),|𝒥un-1Mk(n)|Ω0T(1+α,γ),|Nk(n)|Ω0T(0,((1+α)/2))≤C3(∫0T‍ϵ|U(n)|Ω0T(2+α,1+(α/2))+C(ϵ)∫0T‍|U(n)|Ω0τ(2+α,1+(α/2))dτ).Noting also the following inequality:(70)|∂ξk∫Ω‍Nk(n)(η,t)|ξ-η|dη|Ω0T(1+α,γ)≤C4|Nk(n)|Ω0T(0,((1+α-γ)/2)),
from estimate (69) for Nk(n), we have
(71)|∂ξk∫Ω‍Nk(n)(η,t)|ξ-η|dη|Ω0T(1+α,γ)≤C5(∫0T‍ϵ|U(n)|Ω0T(2+α,1+(α/2))+C(ϵ)∫0T‍|U(n)|Ω0τ(2+α,1+(α/2))dτ).Thus, from estimates (69) and (71), we have the desired estimate for Hk(n).Now, applying Theorem4 to problem (62), we have the following estimate:
(72)|U(n+1)|Ω0T(2+α,1+(α/2))+∥Q(n+1)∥T+|R(n+1)|Γ0T(2+α,1+(α/2))≤C6{ϵ(|U(n)|Ω0T(2+α,1+(α/2))+|R(n)|Γ0T(2+α,1+(α/2)))mmmmn+C(ϵ)∫0T(|U(n)|Ω0τ(2+α,1+(α/2))+|R(n)|Γ0τ(2+α,1+(α/2)))dτ},‍
where the norm ∥f∥T is defined as ∥f∥T≡|f|Ω0T(1+α,γ)+|∇f|Ω0T(α,(α/2))+|f|Γ0T(1+α,((1+α)/2)). Fixing ϵ so that C6ϵ≤1/2 and summing the above inequalities from n=1 to n=m, we have
(73)Sm(T)≤C7(S1(T)+∫0T‍Sm(τ)dτ),
where
(74)Sm(T)≡∑n=1m‍(|U(n)|Ω0T(2+α,1+(α/2))+∥Q(n)∥T+|R(n)|Γ0T(α,(α/2))).
Then, using Gronwall’s inequality, from the above inequality, we have
(75)Sm(T)≤C8(1+TeC9T).Noting that the right-hand side in (75) is independent of m, we can conclude that the sequence {(un,qn,rns)} is convergent in C2+α,1+(α/2)(Ω0T)×(C1+α,γ(Ω0T)∩C1+α,((1+α)/2)(Γ0T))×C2+α,1+(α/2)(Γ0T).Now, let us prove Theorem1. Taking the limit as n tends to infinity in problem (48), we can easily see that the limit of the sequence {(un,qn,rns)} is a solution of problem (11)–(13). The uniqueness can be proved as follows. Let (u,q,rs) and (u~,q~,r~s) be two solutions of problem (11)–(13). By subtracting one equation from the other, we obtain the equations for the differences V≡u-u~, P≡q-q~, and R≡rs-r~s, the form of which is similar to (62). Then, in a similar manner to obtain (72), we can obtain the following estimate:
(76)|V|Ω0T(2+α,1+(α/2))+|R|Γ0T(2+α,1+(α/2))≤C10{ϵ(|V|Ω0T(2+α,1+(α/2))+|R|Γ0T(2+α,1+(α/2)))mmmmmn+C(ϵ)∫0T(|V|Ω0τ(2+α,1+(α/2))+|R|Γ0τ(2+α,1+(α/2)))dτ}.‍
This inequality implies that V=0 and R=0, and as a consequence, P=0 follows. Thus, the proof of Theorem 1 is completed.



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*Source: 495408-2013-09-30.xml*