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Solution formula for generalized two-phase Stokes equations and its applications to maximal regularity: Model problems

  • In this paper, we give a solution formula for the two-phase Stokes equations with and without surface tension and gravity over the whole space with a flat interface. The solution formula has already been considered by Shibata and Shimizu. However, we have reconstructed the formula so that we are able to easily prove resolvent and maximal regularity estimates. The previous work required the assumption of additional conditions on normal components. Here, although we consider normal components, the assumption is weaker than before. The method is based on an H-calculus which has already been applied for the Stokes problems with various boundary conditions in the half-space.

    Citation: Naoto Kajiwara. Solution formula for generalized two-phase Stokes equations and its applications to maximal regularity: Model problems[J]. AIMS Mathematics, 2024, 9(7): 18186-18210. doi: 10.3934/math.2024888

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  • In this paper, we give a solution formula for the two-phase Stokes equations with and without surface tension and gravity over the whole space with a flat interface. The solution formula has already been considered by Shibata and Shimizu. However, we have reconstructed the formula so that we are able to easily prove resolvent and maximal regularity estimates. The previous work required the assumption of additional conditions on normal components. Here, although we consider normal components, the assumption is weaker than before. The method is based on an H-calculus which has already been applied for the Stokes problems with various boundary conditions in the half-space.



    It is known that the motion of viscous incompressible fluids is governed by this Navier-Stokes equations. When we consider two fluids that are separated by a free surface, the analysis is a difficult problem. Mathematically, the free-boundary problem is formulated based on initial-boundary value problems, as follows. Let Ω+(t) and Ω(t) be domains in Rn that contain different fluids, and let them have the same time-dependent boundary Γ(t)=Ω+(t)(=Ω(t)) and Rn=Ω+(t)Ω(t)Γ(t). The unknowns are the boundary Γ(t), the velocity v(x,t)=t(v1,,vn), and pressure θ(x,t) defined on Ω(t)=Ω+(t)Ω(t). The equations are as follows:

    {ρ(tv+(v)v)DivS(v,θ)=0inΩ(t),t>0,divv=0inΩ(t),t>0,[[S(v,θ)νt]]=cσHνt+[[ρ]]cgxnνtonΓ(t),t>0,[[v]]=0onΓ(t),t>0,V=vνtonΓ(t),t>0,v|t=0=v0inΩ(0). (1.1)

    Here, S(v,θ)=μD(v)θI=(μ(ivj+jvi)δijθ)ij is an n×n symmetric stress tensor, V is the normal velocity of Γ(t), νt is the unit outward normal vector pointing from Ω+(t) to Ω(t), H is the mean curvature of Γ(t), which is given by Hνt=ΔΓ(t)x, where ΔΓ(t) denotes the Laplace-Beltrami operator on Γ(t). The letters ρ,μ,cσ, and cg denote the coefficients of density, viscosity, surface tension, and gravity, respectively. Here, ρ and μ are positive constants on each domain Ω±(t). The symbol [[]] denotes a jump across the interface Γ(t). For example, the quantity [[ρ]] means that [[ρ]]=ρ|Ω+(t)ρ|Ω(t) for the piecewise constant density ρ defined on Ω(t).

    It is known that the Hanzawa transformation is a useful technique to solve free boundary problems. In this method, the unknown Γ(t) is given by a height function defined on the boundary of a fixed domain. After applying this transformation, the equations become quasi-linear equations. Therefore, it is important to consider the linearized equations. In addition to the above discussion, maximal regularity for the linearized equations over the whole space with a flat interface is a necessary, as described below;

    {ρtUDivS(U,Θ)=Fin˙Rn(:=Rn+Rn),t>0,divU=Fdin˙Rn,t>0,tY+Un=DonRn0(:=Rn+),t>0,[[S(U,Θ)ν]]([[ρ]]cg+cσΔ)Yν=[[G]]onRn0,t>0,[[U]]=[[H]]onRn0,t>0,(U,Y)|t=0=(0,0)in˙Rn, (1.2)

    where F,Fd,D,G, and H are external forces and ν=(0,,0,1). Moreover, we consider the corresponding resolvent equations and the case that cσ=cg=0:

    {ρλuDivS(u,θ)=fin˙Rn,divu=fdin˙Rn,λη+un=donRn0,[[S(u,θ)ν]]([[ρ]]cg+cσΔ)ην=[[g]]onRn0,[[u]]=[[h]]onRn0, (1.3)
    {ρtUDivS(U,Θ)=Fin˙Rn,t>0,divU=Fdin˙Rn,t>0,[[S(U,Θ)ν]]=[[G]]onRn0,t>0,[[U]]=[[H]]onRn0,t>0,U|t=0=0in˙Rn, (1.4)
    {ρλuDivS(u,θ)=fin˙Rn,divu=fdin˙Rn,[[S(u,θ)ν]]=[[g]]onRn0,[[u]]=[[h]]onRn0. (1.5)

    In this paper, we construct the solution formulas for these four problems. The approach is based on the standard method, which entails the use of partial Fourier transforms and Laplace transforms of the equations. When we solve ordinary differential equations, we need to consider the matrix. In the previous related works [30,33], the authors also derived the solution formulas by analyzing the ordinary differential equations and the matrix. However, our approach will be easier than before. We focus only on the determinant of the matrix and the order of growth of the cofactor matrix. We do not need to calculate the inverse of the 4×4 matrix. Thus, we are able to more effectively obtain the solution formulas. This is one of our main theorems. As an application, we are able to prove the resolvent estimate and maximal regularity estimate. When we obtain the solution formulas with a suitable form, we know that they have these estimates. This strategy has been shown in [17], which considered the Stokes equations with various boundary conditions in the half-space. We remark that the authors of [30,33] had to assume additional conditions for hn and Hn. They had to assume a regularity for the derivative of hn in all directions. On the other hand, we can relax some conditions. We clarify that only nhn is the essential condition. This may be useful for future researchers to consider the external forces which act in the tangential direction. The computational complexity is also much less than before. Moreover, our result on maximal regularity for the problem with surface tension is easy to understand from the perspective of regularity theory. See Theorem 2.4. We expect that our method and analysis will be applied in future works.

    There are several papers on two-phase free boundary problems. The problems can be divided into two cases: one is a compact free surface, and the other is a non-compact one. For simplicity, we only consider the first case. Tanaka [38] proved the global existence theorem in L2 Sobolev-Slobodetskii space. Denisova proved the same results with cσ=0 in both Hölder space [5] and L2-based Sobolev space [6]. Denisova and Solonnikov extended their results to be applicable to capillary fluids, i.e., cσ>0, in both the whole space [4] and bounded domain [7]. Shimizu [24] treated the case in which cσ=0 in Lp-Lq settings. Köhne et al. proved global well-posedness for the capillary fluids in Lp-settings as well as their asymptotic behavior, in [19]. Saito and Shibata [23] considered a comprehensive approach for two-phase problems. Moreover, there are some papers that focus on two-phase problems, e.g., varifold solutions [1] and viscosity solutions [15,37]. For more results on resolvent estimates and maximal regularity, see also [17,20,22,29,30,31,32,33,34].

    This paper is organized as follows. First, we introduce some notation and state our main theorems in Section 2. The main objective of this work was to shorten the proofs of estimates and weaken the assumption on the normal components relative to that in the previous work [33]. In Section 3, we cite some theorems from [33], which is the standard way to consider solution formulas. This implies that it is enough to consider the cases that f=fd=0 and F=Fd=0. The solution formula derived from the boundary data is the most important part. This is demonstrated in Section 4 for (1.4) and (1.5). Then, in Section 5, we prove the resolvent Lq estimate and maximal Lp-Lq estimate as based on Theorem 6.1 in [17]. Analysis of (1.2) and (1.3) is given in Section 6. The solution formulas and the estimates depend on the results for (1.4) and (1.5).

    In this section, we provide and describe some notation and function spaces and give main theorems. Let Rn+, Rn, Rn0 be the upper and lower half-spaces and the corresponding flat boundary, and let Q+, Q, Qn0 be the corresponding time-space domains, as follows:

    Rn+:={x=(x1,,xn)Rnxn>0},Rn:={x=(x1,,xn)Rnxn<0},Rn0:={x=(x,0)=(x1,,xn1,0)Rn},Q+:=Rn+×(0,),Q:=Rn×(0,),Q0:=Rn0×(0,).

    Given a domain D, Lebesgue and Sobolev spaces are denoted by Lq(D) and Wmq(D) with the norms Lq(D) and Wmq(D). This is similar for the X-valued spaces Lp(R,X) and Wmp(R,X). For a scalar function f and n-vector f=(f1,,fn), we use the following symbols:

    f=(1f,,nf),2f=(ijfi,j=1,,n),f=(ifji,j=1,,n),2f=(ijfki,j,k=1,,n).

    Even if g=(g1,,g˜n)X˜n for some ˜n, we denote gX and gX by ˜nj=1gjX for simplicity. Set

    ˆW1q(D)={πLq,loc(D)πLq(D)},ˆW1q,0(D)={πˆW1q(D)π|D=0}

    and let ˆW1q(D) denote the dual space of ˆW1q,0(D), where 1/q+1/q=1. For πˆW1q(D)Lq(D), we have

    πˆW1q(D)=sup{|Dπϕdx|ϕˆW1q,0(D),ϕLq(D)=1}.

    Although we usually consider the time interval R+ for initial-value problems, we consider the functions on R to enable use of the Fourier transform. Thus, and to consider Laplace transforms as Fourier transforms, we introduce some function spaces:

    Lp,0,γ0(R,X):={f:RXeγ0tf(t)Lp(R,X),f(t)=0fort<0},Wmp,0,γ0(R,X):={fLp,0,γ0(R,X)eγ0tjtf(t)Lp(R,X),j=1,,m},Lp,0(R,X):=Lp,0,0(R;X),Wmp,0(R,X):=Wmp,0,0(R;X)

    for some γ00. Let F and F1 denote the Fourier transform and its inverse, defined as follows:

    F[f](ξ)=Fx[f](ξ)=Rneixξf(x)dx,F1[g](x)=F1ξ[g](x)=1(2π)nRneixξg(ξ)dξ.

    Similarly, let L and L1λ denote the two-sided Laplace transform and its inverse, defined as follows:

    L[f](λ)=eλtf(t)dt,L1λ[g](t)=12πeλtg(λ)dτ,

    where λ=γ+iτC. Given s0 and the X-valued function f, we use the following Bessel potential spaces to treat fractional orders:

    Hsp,0,γ0(R,X):={f:RXΛsγf:=L1λ[|λ|sL[f](λ)](t)Lp,0,γ(R,X)for anyγγ0},Hsp,0(R,X):=Hsp,0,0(R,X).

    Since we need to consider the n-th component of the velocity, we introduce the following function space:

    Eq(˙Rn):={hnW2q(˙Rn)||1nhn:=F1ξ|ξ|1Fx(nhn)(x,xn)Lq(˙Rn)}.

    We remark that this condition is weaker than that described in [33] since they assumed that ||1hnˆW1q(˙Rn). It will be easier to apply this assumption to handle a difficult term. Let Σε,γ:={λC{0}|argλ|<πε,|λ|γ} and Σε:=Σε,0. Throughout this paper, let ρ,μ be positive constants on each domain Rn±, denoted by ρ± and μ±. Next, we shall state our main results.

    Theorem 2.1. Let 0<ε<π/2 and 1<q<. Then, for any λΣε and

    fLq(˙Rn),fdˆW1q(Rn)W1q(˙Rn),gW1q(˙Rn),hW2q(˙Rn),hnEq(˙Rn)

    problem (1.5) admits a unique solution (u,θ)W2q(˙Rn)׈W1q(˙Rn) with the following resolvent estimate:

    (λu,λ1/2u,2u,θ)Lq(˙Rn)Cn,q,ε{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)}.

    Theorem 2.2. Let 1<p,q< and γ00. Then, for any

    FLp,0,γ0(R,Lq(˙Rn)),FdW1p,0,γ0(R,ˆW1q(Rn))Lp,0,γ0(R,W1q(˙Rn)),GH1/2p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W1q(˙Rn)),HW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),HnW1p,0,γ0(R,Eq(˙Rn)),

    problem (1.4) admits a unique solution (U,Θ) such that

    UW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),ΘLp,0,γ0(R,ˆW1q(˙Rn))

    with the following maximal Lp-Lq regularity estimate:

    eγt(tU,γU,Λ1/2γU,2U,Θ)Lp(R,Lq(˙Rn))Cn,p,q,γ0{eγt(F,Λ1/2γFd,Fd,Λ1/2γG,G,tH,2H,t(||1nHn)Lp(R,Lq(˙Rn))+eγt(tFd,γFd)Lp(R,ˆW1q(Rn))},

    for any γγ0.

    We can extend the above theorems to the problems (1.2) and (1.3). Let cσ>0 and cg>0.

    Theorem 2.3. Let 0<ε<π/2 and 1<q<. Then, there exists a constant γ01 that depends on ε>0 such that, for any λΣε,γ0 and

    fLq(˙Rn),fdˆW1q(Rn)W1q(˙Rn),gW1q(˙Rn),hW2q(˙Rn),hnEq(˙Rn),dW2q(˙Rn)

    problem (1.3) admits a unique solution (u,θ,η)W2q(˙Rn)׈W1q(˙Rn)×W3q(˙Rn) with the following resolvent estimate:

    (λu,λ1/2u,2u,θ)Lq(˙Rn)+|λ|ηW2q(˙Rn)+ηW3q(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)+dW2q(˙Rn)}.

    Moreover, we have

    |λ|3/2ηW1q(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)+dW2q(˙Rn)+|λ|1/2dW1q(˙Rn)}

    and

    |λ|2ηLq(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)+dW2q(˙Rn)+|λ|dLq(˙Rn)}.

    Theorem 2.4. Let 1<p,q<. Then, there exists a constant γ01 such that, for any

    FLp,0,γ0(R,Lq(˙Rn)),FdW1p,0,γ0(R,ˆW1q(Rn))Lp,0,γ0(R,W1q(˙Rn)),GH1/2p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W1q(˙Rn)),HW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),HnW1p,0,γ0(R,Eq(˙Rn)),DLp,0,γ0(R,W2q(˙Rn)),

    problem (1.2) admits a unique solution (U,Θ,Y) such that

    UW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),ΘLp,0,γ0(R,ˆW1q(˙Rn)),YLp,0,γ0(R,W3q(˙Rn))W1p,0,γ0(R,W2q(˙Rn))

    with the following maximal Lp-Lq regularity estimate:

    eγt(tU,γU,Λ1/2γU,2U,Θ)Lp(R,Lq(˙Rn))+eγt(tY,γY)Lp(R,W2q(˙Rn))+eγtYLp(R,W3q(˙Rn))Cn,p,q,γ0{eγt(F,Λ1/2γFd,Fd,Λ1/2γG,G,tH,2H,t(||1nHn))Lp(R,Lq(˙Rn))+eγt(tFd,γFd)Lp(R,ˆW1q(Rn))+eγtDLp(R,W2q(˙Rn))},

    for any γγ0. Moreover, we see that, if DH1/2p,0,γ0(R,W1q(˙Rn)), then YH3/2p,0,γ0(R,W1q(˙Rn)) and

    eγtΛ3/2γYLp(R,W1q(˙Rn))Cn,p,q,γ0{eγt(F,Λ1/2γFd,Fd,Λ1/2γG,G,tH,2H,t(||1nHn))Lp(R,Lq(˙Rn))+eγt(tFd,γFd)Lp(R,ˆW1q(Rn))+eγtDLp(R,W2q(˙Rn))+eγtΛ1/2γDLp(R,W1q(˙Rn))}

    for any γγ0. Moreover, we see that, if DW1p,0,γ0(R,Lq(˙Rn)), then YW2p,0,γ0(R,Lq(˙Rn)) and

    eγt2tYLp(R,Lq(˙Rn))Cn,p,q,γ0{eγt(F,Λ1/2γFd,Fd,Λ1/2γG,G,tH,2H,t(||1nHn))Lp(R,Lq(˙Rn))+eγt(tFd,γFd)Lp(R,ˆW1q(Rn))+eγtDLp(R,W2q(˙Rn))+eγttDLp(R,Lq(˙Rn))}

    for any γγ0.

    Remark 2.5. (ⅰ) In Theorems 2.1 and 2.3, the uniqueness implies that, if f=fd=[[g]]=[[h]]=0 and [[d]]=0, then u=0, θ=0 with [[θ]]=0, and η|Rn0=0. In Theorems 2.2 and 2.4, the uniqueness has a similar implication.

    (ⅱ) By interpolation theory, we have

    W1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn))H1/2p,0,γ0(R,W1q(˙Rn)),W2p,0,γ0(R,Lq(˙Rn))W1p,0,γ0(R,W2q(˙Rn))H3/2p,0,γ0(R,W1q(˙Rn)).

    In this section, we follow the method in [33]; thus, it is enough to consider the case that f=fd=0 and F=Fd=0 by subtracting solutions of inhomogeneous data.

    We start with whole-space problems.

    Lemma 3.1. ([33, Lemma 2.1]) Let 1<p,q< and γ00.

    (1) For any fdˆW1q(Rn)W1q(˙Rn), there exists a zW2q(˙Rn) such that divz=fd in ˙Rn, [[z]]=0 on Rn0, and the following estimates hold:

    zLq(˙Rn)Cn,qfdˆW1q(˙Rn),j+1zLq(˙Rn)Cn,qjfdLq(˙Rn)(j=0,1).

    (2) For any FdW1p,0,γ0(R,ˆW1q(Rn))Lp,0,γ0(R,W1q(˙Rn)), there exists a

    ZW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn))

    such that divZ=Fd in ˙Rn×R, [[Z(t)]]=0 on Rn0×R, and the following estimates hold:

    eγt(tZ,γZ)Lp(R,Lq(˙Rn))Cn,p,qeγt(tFd,γFd)Lp(R,ˆW1q(Rn)),eγtΛ1/2γZLp(R,Lq(˙Rn))Cn,p,qeγtΛ1/2γFdLp(R,Lq(˙Rn)),eγt2ZLp(R,Lq(˙Rn))Cn,p,qeγtFdLp(R,Lq(˙Rn))

    for any γγ0.

    Setting u=v+z,˜f=f(ρλzΔz) and U=V+Z,˜F=F(ρtZΔZ), we would like to find (v,θ),(V,Θ) such that

    {ρλvDivS(v,θ)=˜fin˙Rn,divv=0in˙Rn,[[S(v,θ)ν]]=[[gμD(z)ν]]onRn0,[[v]]=[[h]]onRn0. (3.1)

    and

    {ρtVDivS(V,Θ)=˜Fin˙Rn,t>0,divV=0in˙Rn,t>0,[[S(V,Θ)ν]]=[[GμD(Z)ν]]onRn0,t>0,[[V]]=[[H]]onRn0,t>0,V|t=0=0inRn1. (3.2)

    Let ˜g:=gμD(z)ν and ˜G:=GμD(Z)ν. We see that

    ˜fLq(˙Rn)fLq(˙Rn)+Cn,q(|λ|fdˆW1q(Rn)+fdLq(˙Rn)),eγt˜FLp(R,Lq(˙Rn))eγtFLp(R,Lq(˙Rn))+Cn,p,q(eγttFdLp(R,ˆW1q(Rn))+eγtFdLp(R,Lq(˙Rn))),(λ1/2˜g,˜g)Lq(˙Rn)C(λ1/2fd,fd,λ1/2g,g)Lq(˙Rn),eγt(Λ1/2γ˜G,˜G)Lp(R,Lq(˙Rn))Ceγt(Λ1/2γFd,Fd,Λ1/2γG,G)Lp(R,Lq(˙Rn)).

    Therefore, we can reduce the problem as follows: fd=0,Fd=0.

    Second, we would like to reduce the case of f=0, F=0. Let P(ξ)=(Pj,k)jk=(δjkξjξk|ξ|2)jk be the Helmholtz decomposition. Then, the functions

    ψ±(x)=F1ξ[P(ξ)Fxf(ξ)ρ±λ+μ±|ξ|2](x),ϕ±(x)=F1ξ[iξFxf(ξ)|ξ|2](x),Ψ±(x,t)=LλF1ξ[P(ξ)FxLF(ξ,λ)ρ±λ+μ±|ξ|2](x,t),Φ±(x,t)=LλF1ξ[iξFxLF(ξ,λ)|ξ|2](x,t)

    satisfy

    (ψ±,ϕ±)W2q(Rn)׈W1q(Rn),ρ±λψ±μ±Δψ±+ϕ±=f,divψ±=0inRn,(λψ±,λ1/2ψ±,2ψ±,ϕ±)Lq(Rn±)Cn,q,εfLq(˙Rn)

    and

    Ψ±W1p,0,γ0(R,Lq(Rn))Lp,0,γ0(R,W2q(Rn)),Φ±Lp,0,γ0(R,ˆW1q(Rn)),ρ±tΨ±μ±ΔΨ±+Φ±=F,divΨ±=0inRn×(0,),Ψ±|t=0=0,eγt(tΨ±,γΨ±,Λ1/2γΨ,2Ψ±,Φ±)Lp(R,Lq(Rn±))Cn,p,q,γ0eγtFLp(R,Lq(˙Rn)),

    for any 1<p,q<, γγ00,fLq(˙Rn),FLp,0,γ0(R,Lq(˙Rn)), and λΣε with 0<ε<π/2, according to (3.19) in [34]. We define

    (ψ,ϕ,Ψ,Φ):={(ψ+,ϕ+,Ψ+,Φ+)forxRn+,(ψ,ϕ,Ψ,Φ)forxRn.

    Then, we have that [[ϕ]]=0 on Rn0 and [[Φ(t)]]=0 on Q0.

    Setting u:=ψ+w, θ:=ϕ+κ in (1.5) with fd=0 and U:=Ψ+W, Θ=Φ+Ξ in (1.4) with Fd=0, respectively, we have

    {ρλwμΔw+κ=0in˙Rn,divw=0in˙Rn,[[S(w,κ)ν]]=[[gμD(ψ)ν]]onRn0,[[w]]=[[hψ]]onRn0. (3.3)

    and

    {ρtWμΔW+Ξ=0in˙Rn,t>0,divW=0in˙Rn,t>0,[[S(W,Ξ)ν]]=[[GμD(Ψ)ν]]onRn0,t>0,[[W]]=[[HΨ]]onRn0,t>0,W|t=0=0inRn1. (3.4)

    Let

    ˜g:=gμD(ψ)ν,˜h:=hψ,˜G:=GμD(Ψ)ν,˜H:=HΨ.

    Since we have the estimates

    (λ1/2˜g,˜g,λ˜h,2˜h,λ||1n˜hn)Lq(˙Rn)C(f,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn),eγt(Λ1/2γ˜G,˜G,t˜H,2˜H,t(||1n˜Hn))Lp(R,Lq(˙Rn))Ceγt(F,Λ1/2γG,G,tH,2H,t(||1nHn))Lp(R,Lq(˙Rn)),

    we conclude that f=fd=0 and F=Fd=0 are sufficient for Theorems 2.1 and 2.2, where we prove that

    λ||1nψnLq(˙Rn)CfLq(˙Rn)eγtt(||1nΨn)Lp(R,Lq(˙Rn))CeγtFLp(R,Lq(˙Rn))

    in the Appendix.

    We shall give a solution of the resolvent problem (1.5) with f=fd=0 and λΣε. We apply a partial Fourier transform with respect to the tangential direction xRn1. We apply notation

    ˆv(ξ,xn):=Fxv(ξ,xn):=Rn1eixξv(x,xn)dx,F1ξw(x,xn)=1(2π)n1Rn1eixξw(ξ,xn)dξ

    for the functions v,w:Rn±C. Let u±=t(u±1,,u±(n1),u±n). Here and hereafter, the index j runs from 1 to n1 unless stated otherwise.

    We need to solve the following second-order ordinary differential equations:

    {(ρ±λ+μ±|ξ|2μ±2n)ˆu±j+iξjˆθ±=0inxn0,(ρ±λ+μ±|ξ|2μ±2n)ˆu±n+nˆθ±=0inxn0,n1j=1iξjˆu±j+nˆu±n=0inxn0,[[μ(iξjˆun+nˆuj)]]=[[ˆgj]]onxn=0,[[2μnˆunˆθ]]=[[ˆgn]]onxn=0,[[ˆu]]=[[ˆh]]onxn=0. (4.1)

    Set

    A:=n1j=1ξ2j,B±:=ρ±(μ±)1λ+A2

    with positive real parts. Here, we consider ξ to have complex values, as follows:

    ξj˜Ση:={zC{0}|argz|<η}{zC{0}πη<|argz|}

    for η(0,π/4). The details are given in Lemma 5.2.

    We find the solution of the form

    ˆu±j(ξ,xn)=α±j(eB±xneAxn)+β±jeB±xn(j=1,,n),ˆθ±(ξ,xn)=γ±eAxn.

    Then, the equations become

    {μ±(B2±A2)α±j+iξjγ±=0,μ±(B2±A2)α±nAγ±=0,iξα±±Aα±n=0,iξ(α±+β±)B±(α±n+β±n)=0, (4.2)

    and

    {μ+(B2+A2)α+n+μ+(B2++A2)β+nμ(B2A2)αnμ(B2+A2)βn=iξ[[ˆg]],μ+(B+A)2α+n2μ+AB+β+n+μ(BA)2αn2μABβn=A[[ˆgn]],(B+A)α+n+B+β+n+(BA)αn+Bβn=iξ[[ˆh]],β+nβn=[[ˆhn]].

    This means that

    [μ+(B++A)μ+(B2++A2)μ(B+A)μ(B2+A2)μ+(B+A)2μ+AB+μ(BA)2μAB1B+1B0101][(B+A)α+nβ+n(BA)αnβn]=[iξ[[ˆg]]A[[ˆgn]]iξ[[ˆh]][[ˆhn]]]=[iξT0000A0000iξT00001][[[ˆg1]][[ˆgn]][[ˆh1]][[ˆhn]]],

    where ξT is the transpose of ξ. We define

    L:=[μ+(B++A)μ+(B2++A2)μ(B+A)μ(B2+A2)μ+(B+A)2μ+AB+μ(BA)2μAB1B+1B0101],R:=(rij):=[iξT0000A0000iξT00001](1i4,1j2n).

    We have

    detL=(μ+μ)2A3{(3μ+μ)μ+B++(3μμ+)μB}A2{(μ+B++μB)2+μ+μ(B++B)2}A(μ+B++μB)(μ+B2++μB2)

    and it is known, from [33, Lemma 5.5], that the determinant is not zero for λΣε, ξRn1.

    We introduce some new notation:

    M+=M+(A,B+,xn)=eB+xneAxnB+A,M=M(A,B,xn)=eBxneAxnBA,(ai,j):=(L1R)ij=(detL)1(4s=1Lisrsj)ij(1i4,1j2n),

    where we use the cofactor matrix of L, denoted by Cof(L)=(Lij).

    From these observations, we have

    ˆu+n(ξ,xn)=nk=1{(a1,kM++a2,keB+xn)[[ˆgk]]+(a1,n+kM++a2,n+keB+xn)[[ˆhk]]},ˆun(ξ,xn)=nk=1{(a3,kM+a4,keBxn)[[ˆgk]]+(a3,n+kM+a4,n+keBxn)[[ˆhk]]}.

    To simplify, we define the following symbols for k=1,,n:

    ϕk,+n(λ,ξ,xn)=a1,kM++a2,keB+xn,ψk,+n(λ,ξ,xn)=a1,n+kM++a2,n+keB+xn,ϕk,n(λ,ξ,xn)=a3,kM+a4,keBxn,ψk,n(λ,ξ,xn)=a3,n+kM+a4,n+keBxn,

    which yields the solution formulas for u±n:

    ˆu+n=ˆu+n(ξ,xn)=nk=1(ϕk,+n[[ˆgk]]+ψk,+n[[ˆhk]]),xn>0,ˆun=ˆun(ξ,xn)=nk=1(ϕk,n[[ˆgk]]+ψk,n[[ˆhk]]),xn<0.

    Since

    γ±=μ±(B±+A)A(B±A)α±n

    from the second equation of (4.2), by letting

    χk,+(λ,ξ,xn)=μ+(B++A)Aa1,keAxn,ωk,+(λ,ξ,xn)=μ+(B++A)Aa1,n+keAxn,χk,(λ,ξ,xn)=μ(B+A)Aa3,keAxn,ωk,(λ,ξ,xn)=μ(B+A)Aa3,n+keAxn,

    we have

    ˆθ±=ˆθ±(ξ,xn)=nk=1(χk,±[[ˆgk]]+ωk,±[[ˆhk]]),xn0.

    From the first equation of (4.2), α±j=(iξj/A)α±n. From the fourth and the sixth equations of (4.1), β±j satisfies

    [μ+B+μB11][β+jβj]=[[[ˆgj]][[ˆhj]]]+[μ+(B+A)α+j+μ+iξjβ+nμ(BA)αjμiξjβn0]=[[[ˆgj]][[ˆhj]]]+iξjA[μ+μ+AμμA0000][(B+A)α+nβ+n(BA)αnβn],
    [β+jβj]=1μ+B++μB{[[[ˆgj]]+μB[[ˆhj]][[ˆgj]]μ+B+[[ˆhj]]]+iξjA[μ+μ+AμμAμ+μ+AμμA][(B+A)α+nβ+n(BA)αnβn]}=1μ+B++μB[[[ˆgj]]+μB[[ˆhj]]+iξjAnk=1{μ+(a1,k+Aa2,k)[[ˆgk]]μ(a3,n+k+Aa4,n+k)[[ˆhk]]}[[ˆgj]]μ+B+[[ˆhj]]+iξjAnk=1{μ+(a1,k+Aa2,k)[[ˆgk]]μ(a3,n+k+Aa4,n+k)[[ˆhk]]}].

    Therefore,

    ˆu±j=ˆu±j(ξ,xn)=(B±A)α±jM±+β±jeB±xn=:nk=1(ϕk,±j[[ˆgk]]+ψk,±j[[ˆhk]]),xn0,

    where

    ϕk,+j(λ,ξ,xn)=iξjAa1,kM++1μ+B++μB(δk,j+iμ+ξjA(a1,k+Aa2,k))eB+xn,ψk,+j(λ,ξ,xn)=iξjAa1,n+kM++1μ+B++μB(μBδk,jiμξjA(a3,n+k+Aa4,n+k))eB+xn,ϕk,j(λ,ξ,xn)=iξjAa3,kM+1μ+B++μB(δk,j+iμ+ξjA(a1,k+Aa2,k))eBxn,ψk,j(λ,ξ,xn)=iξjAa3,n+kM1μ+B++μB(μ+B+δk,j+iμξjA(a3,n+k+Aa4,n+k))eBxn.

    Here, we introduce a new notation, i.e., [[[f]]](x,xn):=f(x,xn)f(x,xn), for f:˙RnC. Then, we have that [[f]](x)=limxn+0[[[f]]](x,xn) and [[a]](ξ)=R±[[[na]]](ξ,yn)dyn for a function a with a(,xn)0 as xn±. So, we obtain the solution formulas of integral form;

    u±j(x)=nk=1{R±F1ξ[nϕk,±j(λ,ξ,xn+yn)Fx[[[gk]]]](x,yn)dyn+R±F1ξ[ϕk,±j(λ,ξ,xn+yn)Fx[[[ngk]]]](x,yn)dyn+R±F1ξ[nψk,±j(λ,ξ,xn+yn)Fx[[[hk]]]](x,yn)dyn+R±F1ξ[ψk,±j(λ,ξ,xn+yn)Fx[[[nhk]]]](x,yn)dyn},(j=1,,n), (4.3)
    θ±(x)=nk=1{R±F1ξ[nχk,±(λ,ξ,xn+yn)Fx[[[gk]]]](x,yn)dyn+R±F1ξ[χk,±(λ,ξ,xn+yn)Fx[[[ngk]]]](x,yn)dyn+R±F1ξ[nωk,±(λ,ξ,xn+yn)Fx[[[hk]]]](x,yn)dyn+R±F1ξ[ωk,±(λ,ξ,xn+yn)Fx[[[nhk]]]](x,yn)dyn}. (4.4)

    Since the Laplace-transformed non-stationary Stokes equations of (1.4) with F=Fd=0 on R are equivalent to the resolvent problem (1.5) with f=fd=0, we have the following formula:

    U±j(x,t)=L1λnk=1{R±F1ξ[nϕk,±j(λ,ξ,xn+yn)FxL[[[Gk]]]](x,yn)dyn+R±F1ξ[ϕk,±j(λ,ξ,xn+yn)FxL[[[nGk]]]](x,yn)dyn+R±F1ξ[nψk,±j(λ,ξ,xn+yn)FxL[[[Hk]]]](x,yn)dyn+R±F1ξ[ψk,±j(λ,ξ,xn+yn)FxL[[[nHk]]]](x,yn)dyn},(j=1,,n), (4.5)
    Θ±(x,t)=L1λnk=1{R±F1ξ[nχk,±(λ,ξ,xn+yn)FxL[[[Gk]]]](x,yn)dyn+R±F1ξ[χk,±(λ,ξ,xn+yn)FxL[[[nGk]]]](x,yn)dyn+R±F1ξ[nωk,±(λ,ξ,xn+yn)FxL[[[Hk]]]](x,yn)dyn+R±F1ξ[ωk,±(λ,ξ,xn+yn)FxL[[[nHk]]]](x,yn)dyn}. (4.6)

    We decompose the solutions (4.3) and (4.4) to obtain the independent variables on the right-hand side of the resolvent estimates. Analysis of the solutions (4.5) and (4.6) are based on the same analysis for the resolvent problems.

    We shall provide a theorem to prove the main theorems. Let us respectively define the operators T and ˜Tγ by

    T[m]f(x)=0[F1ξm(ξ,xn+yn)Fxf](x,yn)dyn,˜Tγ[mλ]g(x,t)=L1λ0[F1ξmλ(ξ,xn+yn)FxLg](x,yn,λ)dyn,=[eγtF1τtT[mλ]Ftτ(eγtg)](x,t),

    where λ=γ+iτΣε, m,mλ are C-valued functions, f:Rn+C, and g:Rn+×RC. The following theorem was taken from [17]. See also [18], where R-boundedness and the difference from previous works are written.

    Theorem 5.1. ([17, Theorem 6.1]) (ⅰ) Let m satisfy the following two conditions:

    (a) There exists η(0,π/2) such that {m(,xn),xn>0}H(˜Σn1η).

    (b) There exist η(0,π/2) and C>0 such that supξ˜Σn1η|m(ξ,xn)|Cx1n for all xn>0.

    Then, T[m] is a bounded linear operator on Lq(Rn+) for every 1<q<.

    (ⅱ) Let γ00 and mλ satisfy the following two conditions:

    (c) There exists η(0,π/2ε) such that, for each xn>0 and γγ0,

    ˜Σnη(τ,ξ)mλ(ξ,xn)C

    is bounded and holomorphic.

    (d) There exist η(0,π/2ε) and C>0 such that sup{|mλ(ξ,xn)|(τ,ξ)˜Σnη}Cx1n for all γγ0 and xn>0.

    Then, ˜Tγ[mλ] satisfies

    eγt˜Tγ[mλ]gLp(R,Lq(Rn+))CeγtgLp(R,Lq(Rn+))

    for every γγ0 and 1<p,q<.

    By using the identities given by

    B2±=ρ±(μ±)1λ+n1m=1ξ2m,1=B2±B2±=ρ±(μ±)1λ1/2B2±λ1/2n1m=1iξmB2±(iξm),

    we have

    u±j(x)=nk=1{R±F1ξ[ρ±(μ±)1λ1/2B2±nϕk,±j(λ,ξ,xn+yn)Fx[[[λ1/2gk]]]](x,yn)dynn1m=1R±F1ξ[iξmB2±nϕk,±j(λ,ξ,xn+yn)Fx[[[mgk]]]](x,yn)dyn+R±F1ξ[ϕk,±j(λ,ξ,xn+yn)Fx[[[ngk]]]](x,yn)dyn+R±F1ξ[B2±nψk,±j(λ,ξ,xn+yn)Fx[[[(ρ±(μ±)1λΔ)hk]]]](x,yn)dynn1m=1R±F1ξ[iξmB2±ψk,±j(λ,ξ,xn+yn)Fx[[[mnhk]]]](x,yn)dyn}n1k=1R±F1ξ[ρ±(μ±)1λ1/2B2±ψk,±j(λ,ξ,xn+yn)Fx[[[λ1/2nhk]]]](x,yn)dyn+R±F1ξ[ρ±(μ±)1AB2±ψk,±j(λ,ξ,xn+yn)Fx[[[λ||1nhn]]]](x,yn)dynθ±(x)=nk=1{R±F1ξ[ρ±(μ±)1λ1/2B2±nχk,±(λ,ξ,xn+yn)Fx[[[λ1/2gk]]]](x,yn)dynn1m=1R±F1ξ[iξmB2±nχk,±(λ,ξ,xn+yn)Fx[[[mgk]]]](x,yn)dyn+R±F1ξ[χk,±(λ,ξ,xn+yn)Fx[[[ngk]]]](x,yn)dyn+R±F1ξ[B2±nωk,±(λ,ξ,xn+yn)Fx[[[(ρ±(μ±)1λΔ)hk]]]](x,yn)dynn1m=1R±F1ξ[iξmB2±ωk,±(λ,ξ,xn+yn)Fx[[[mnhk]]]](x,yn)dyn}n1k=1R±F1ξ[ρ±(μ±)1λ1/2B2±ωk,±(λ,ξ,xn+yn)Fx[[[λ1/2nhk]]]](x,yn)dyn+R±F1ξ[ρ±(μ±)1AB2±ωk,±(λ,ξ,xn+yn)Fx[[[λ||1nhn]]]](x,yn)dyn

    for j=1,,n.

    Let Suj±(λ,ξ,xn) and Sθ±(λ,ξ,xn) be considered as follows;

    Suj±(λ,ξ,xn):={ρ±(μ±)1λ1/2B2±nϕk,±j(λ,ξ,xn)k{1,,n},iξmB2±nϕk,±j(λ,ξ,xn)k{1,,n},m{1,,n1},ϕk,±j(λ,ξ,xn)k{1,,n},B2±nψk,±j(λ,ξ,xn)k{1,,n},iξmB2±ψk,±j(λ,ξ,xn)k{1,,n},m{1,,n1},ρ±(μ±)1λ1/2B2±ψk,±j(λ,ξ,xn)k{1,,n1},ρ±(μ±)1AB2±ψn,±j(λ,ξ,xn),Sθ±(λ,ξ,xn):={ρ±(μ±)1λ1/2B2±nχk,±(λ,ξ,xn)k{1,,n},iξmB2±nχk,±(λ,ξ,xn)k{1,,n},m{1,,n1},χk,±(λ,ξ,xn)k{1,,n},B2±nωk,±(λ,ξ,xn)k{1,,n},iξmB2±ωk,±(λ,ξ,xn)k{1,,n},m{1,,n1},ρ±(μ±)1λ1/2B2±ωk,±(λ,ξ,xn)k{1,,n1},ρ±(μ±)1AB2±ωn,±(λ,ξ,xn).

    We shall confirm that all of the symbols are bounded in the sense that

    sup(λ,ξ)ΣεטΣn1η,=1,,n1{(|λ|+|λ|1/2|ξ|+|ξ||ξ|)|Suj±|+(|λ|1/2+|ξ|)|nSuj±|+|2nSuj±|+|ξ||Sθ±|+|nSθ±|}<C(±xn)1 (5.1)

    for suitable ε,η; thus, we can prove Theorem 2.1 with f=fd=0.

    Following the method in [17], we have some identities:

    nM±(A,B±,xn)=eB±xnAM±(A,B±,xn),2nM±(A,B±,xn)=(A+B±)eB±xn+A2M±(A,B±,xn),3nM±(A,B±,xn)=(A2+AB±+B2±)eB±xnA3M±(A,B±,xn)

    as well as the following useful lemma, where we let ˜A:=n1j=1|ξj|2.

    Lemma 5.2. ([17, Lemma 6.3]) Let 0<ε<π/2 and 0<η<ε/2. Then, for any (λ,ξ,xn)ΣεטΣn1η×R±, we have

    c˜AReA|A|˜A,c(|λ|1/2+˜A)ReB±|B±|C(|λ|1/2+˜A),|mneAxn|C˜A1+m(±xn)1,forxn0,|mneB±xn|C(|λ|1/2+˜A)1+m(±xn)1,forxn0,|M±(A,B±,xn)|C(|λ|1/2+˜A)1˜A1(±xn)1,forxn0,|mnM±(A,B±,xn)|C(|λ|1/2+˜A)2+m(±xn)1,forxn0,m0,c(|λ|1/2+˜A)|μ+B++μB|

    for m=0,1,2,3, with positive constants c and C, which are independent of λ,ξ,xn.

    We recall that

    L1=[μ+(B++A)μ+(B2++A2)μ(B+A)μ(B2+A2)μ+(B+A)2μ+AB+μ(BA)2μAB1B+1B0101]1=(detL)1Cof(L)=(detL)1(Lij)ij.

    From the cofactor expansion, we have

    |Lis|{C(|λ|1/2+˜A)for(i,s)=(2,1),(2,2),(4,1),(4,2),C(|λ|1/2+˜A)2for(i,s)=(1,1),(1,2),(2,3),(3,1),(3,2),(4,3),C(|λ|1/2+˜A)3for(i,s)=(1,3),(2,4),(3,3),(4,4),C(|λ|1/2+˜A)4for(i,s)=(1,4),(3,4).

    Then,

    |4s=1Lisrsj|{C˜A(|λ|1/2+˜A)for(i,j)=(2,1),,(2,n),(4,1),,(4,n),C˜A(|λ|1/2+˜A)2for(i,j)={(1,1),,(1,n),(2,n+1),,(2,2n1),(3,1),,(3,n),(4,n+1),,(4,2n1),C˜A(|λ|1/2+˜A)3for(i,j)=(1,n+1),,(1,2n1),(3,n+1),,(3,2n1),C(|λ|1/2+˜A)3for(i,j)=(2,2n),(4,2n),C(|λ|1/2+˜A)4for(i,j)=(1,2n),(3,2n).

    We need to derive the boundedness for detL.

    Lemma 5.3. Let 0<ε<π/2 and 0<η<ε/2. Then, there exist positive constants c and C such that

    c(|λ|1/2+˜A)3|detL|C(|λ|1/2+˜A)3(λΣε,ξ˜Σn1η).

    Proof. Let the angle of A be θ(0,η), i.e., A=|A|eiθ. Since the function of detL=detL(A,B+,B) is homogeneous function of A,B±, it follows that

    |detL(A,B+,B)|=|detL(|A|,B+eiθ,Beiθ)|c(|λe2iθ|+|A|2)3/2c(|λ|1/2+˜A)3

    from the previous results on real values in [30, Lemma 5.5], where we have chosen small η such that 0<2η<ε. It is easy to check the estimate from above.

    We calculate the estimates of ϕk,±j and ψk,±j by combining all of the estimates above, as follows:

    |mnϕk,±n||a1,k||mnM±|+|a2,k||mneB±xn||detL|1(|4s=1L1srsk||mnM±|+|4s=1L2srsk||mneB±xn|)C(|λ|1/2+˜A)2+m(±xn)1,|mnϕk,±j|C(|a1,k||mnM±|+|μ+B++μB|1(1+|a1,k|+˜A|a2,k|)|mneB±xn|)C(|λ|1/2+˜A)2+m(±xn)1,

    for m=0,1,2,3 and j,k=1,,n; similarly,

    |mnψk,±j|{C(|λ|1/2+˜A)1+m(±xn)1ifk{1,,n1}C(|λ|1/2+˜A)1˜Am(±xn)1ifk=n,|mnχk,±|C˜A1+m(±xn)1,k{1,,n},|mnωk,±|{C(|λ|1/2+˜A)˜A1+m(±xn)1ifk{1,,n1}C(|λ|1/2+˜A)2˜A2+m(±xn)1ifk=n,

    for m=0,1,2,3 and j=1,,n.

    We remark that ψn,±j and ωn,± are the coefficients of hn. They are different from ψk,±j and ωk,±. Therefore, we need to assume additional assumptions on normal components.

    These estimates lead to the inequality (5.1), which encompasses the estimates λu, λ1/2u, u, λ1/2nu, nu, 2nu, θ, and nθ.

    We also see that the new symbols Suj± and Sθ±, multiplied by λ, ξ, and n, are holomorphic in (τ,ξ)˜Σnη. Therefore, we are able to apply Theorem 5.1, where we employ a change of variables from xn to xn, also, note that [[[f]]]Lq(Rn+)fLq(˙Rn).

    Theorem 5.4. Let 0<ε<π/2 and 1<q<. Then, for any λΣε,gW1q(˙Rn), hW2q(˙Rn), and hnEq(˙Rn), the problem (1.5) with f=fd=0 admits a solution (u,θ)W2q(˙Rn)׈W1q(˙Rn) with the following resolvent estimate:

    (λu,λ1/2u,2u,θ)Lq(˙Rn)C(λ1/2g,g,λh,λ1/2h,2h,λ||1nhn)Lq(˙Rn)C(λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)

    for some positive constant C.

    This theorem and the estimates in Section 3 can be applied to derive the existence in Theorem 2.1. The uniqueness has been derived in [30,33], where they considered the homogeneous equation and the dual problem.

    For the non-stationary Stokes equations, we have the following theorem according to Theorem 5.1:

    Theorem 5.5. Let 1<p,q< and γ00. Then, for any

    HW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),HnW1p,0,γ0(R,Eq(˙Rn)),

    the problem (1.4) with F=Fd=0 admits a solution (U,Π) such that

    UW1p,0,γ0(R,Lq(˙Rn))Lp,0,γ0(R,W2q(˙Rn)),ΠLp,0,γ0(R,ˆW1q(˙Rn))

    with the following maximal Lp-Lq regularity:

    eγt(tU,γU,Λ1/2γU,2U,Π)Lp(R,Lq(˙Rn))Ceγt(tH,Λ1/2γH,2H,t(||1nHn))Lp(R,Lq(˙Rn))Ceγt(tH,2H,t(||1nHn))Lp(R,Lq(˙Rn))

    for any γγ0 with some positive constant C=Cn,p,q,γ0, depending only on n,p,q, and γ0.

    In this section, we consider the problems (1.2) and (1.3). We shall prove Theorems 2.3 and 2.4. Let (v,τ) and (V,Υ) be solutions to the following problems:

    {ρλvDivS(v,τ)=fin˙Rn,divv=fdin˙Rn,[[S(v,τ)ν]]=[[g]]onRn0,[[v]]=[[h]]onRn0, (6.1)
    {ρtVDivS(V,Υ)=Fin˙Rn,t>0,divV=Fdin˙Rn,t>0,[[S(V,Υ)ν]]=[[G]]onRn0,t>0,[[V]]=[[H]]onRn0,t>0,V|t=0=0in˙Rn. (6.2)

    We shall find the solutions (w,κ,η) and (W,Ξ,Y) satisfying

    {ρλwDivS(w,κ)=0in˙Rn,divw=0in˙Rn,λη+wn=dvn=:˜donRn0,[[S(w,κ)ν]]([[ρ]]cg+cσΔ)ην=0onRn0,[[w]]=0onRn0, (6.3)
    {ρtWDivS(W,Ξ)=0in˙Rn,t>0,divW=0in˙Rn,t>0,tY+Wn=DVn=:˜DonRn0,t>0,[[S(V,Ξ)ν]]([[ρ]]cg+cσΔ)Yν=0onRn0,t>0,[[V]]=0onRn0,t>0,(V,Y)|t=0=(0,0)in˙Rn. (6.4)

    Then (u,θ,η)=(v+w,τ+κ,η) and (U,Θ,Y)=(V+W,Υ+Ξ,Y) are the solutions of (1.3) and (1.2). To solve the equations in (6.3), it is enough to consider that

    ([[ˆh]],[[ˆg]],[[ˆgn]])=(0,0,([[ρ]]cgcσA2)ˆη)

    in (4.1), and that

    {λˆη+ˆwn=ˆ˜donRn0,ˆw±j=ϕn,±j[[ˆgn]]inRn±(j=1,,n). (6.5)

    Note that ϕn,+n(λ,ξ,0)=ϕn,n(λ,ξ,0)=a2,n=a4,n=(detL)1A{μ+(B++A)+μ(B+A)}. Therefore we have the following solution formulas:

    ˆη(λ,ξ)=detLλdetLA{μ+(B++A)+μ(B+A)}([[ρ]]cgcσA2)ˆ˜d,ˆw±j(λ,ξ,xn)=ϕn,±j([[ρ]]cgcσA2)ˆη(j=1,,n),ˆκ±(λ,ξ,xn)=χn,±([[ρ]]cgcσA2)ˆη

    with the following estimate:

    L(λ,ξ):=λdetLA{μ+(B++A)+μ(B+A)}([[ρ]]cgcσA2)|L(λ,ξ)|c(|λ|+˜A)(|λ|1/2+˜A)3

    for (λ,ξ)Σε,γ0טΣn1η with 0<ε<π/2, 0<η<ε/2, and γ01. The proof for ξRn1 is in [33, Lemma 6.1]. However, the proof for complex values is almost the same.

    Let η extend suitably from Rn1 to ˙Rn. Since we have the estimate

    sup(λ,ξ)Σε,γ0טΣn1η=1,,n1{(|λ|+|ξ|)detLL}<C

    and holomorphy, we are able to prove, by applying Fourier multiplier theory as in [22, Proposition 4.3.10, Theorem 4.3.3], that

    (λη,η)Lq(˙Rn)C˜dLq(˙Rn),(λη,2η)Lq(˙Rn)C˜dLq(˙Rn),(λ2η,3η)Lq(˙Rn)C2˜dLq(˙Rn),

    for λΣε,γ0. Then, from the results in the previous section, it follows that

    (λw,λ1/2w,2w,κ)Lq(˙Rn)C(λ1/2gn,gn)Lq(˙Rn)C(λ1/2η,λ1/22η,η,3η)Lq(˙Rn)C˜dW2q(˙Rn),

    where we have used |λ|1/2|λ| when λΣε,γ0; also, C is dependent on γ0, as well as the constants [[ρ]], cg, and cσ. This yields that

    (λu,λ1/2u,2u,θ)Lq(˙Rn)+|λ|ηW2q(˙Rn)+ηW3q(˙Rn)(λv,λ1/2v,2v,λw,λ1/2w,2w,τ,κ)Lq(˙Rn)+|λ|ηW2q(˙Rn)+ηW3q(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)+˜dW2q(˙Rn)}Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)+dW2q(˙Rn)}

    since

    ˜dW2q(˙Rn)Cn,q,ε(dW2q(˙Rn)+vW2q(˙Rn))Cn,q,ε,γ0(dW2q(˙Rn)+|λ|vLq(˙Rn)+2vLq(˙Rn))Cn,q,ε,γ0(dW2q(˙Rn)+(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|fdˆW1q(˙Rn)).

    In addition, we have

    |λ|3/2ηW1q(˙Rn)|λ|1/2˜dW1q(˙Rn)|λ|1/2dW1q(˙Rn)+|λ|1/2vW1q(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|gˆW1q(˙Rn)+dW2q(˙Rn)+|λ|1/2dW1q(˙Rn)}

    and

    |λ|2ηLq(˙Rn)|λ|˜dLq(˙Rn)|λ|dLq(˙Rn)+|λ|vLq(˙Rn)Cn,q,ε,γ0{(f,λ1/2fd,fd,λ1/2g,g,λh,2h,λ||1nhn)Lq(˙Rn)+|λ|gˆW1q(˙Rn)+dW2q(˙Rn)+|λ|dLq(˙Rn)}.

    The proof of Theorem 2.4 is the same as above.

    In this paper, the solution formulas for generalized two-phase Stokes equations have been derived. By using the solution formulas, we have proved the existence of resolvent Lq estimates and maximal Lp-Lq estimates. The method is based on H-calculus, whereas the previous works were based on R-boundedness. The complexity of the calculation is comparatively less, and the conditions on the normal component is relaxed. Our method does not require the estimates of derivatives of the Fourier symbols. Although we were able to obtain an explicit form of the solution, we only applied the order of the coefficient of boundary source terms. Thus, we have not only considered the standard free-boundary condition, we have also considered the problem with surface tension and gravity. This strategy will be useful for future works when we consider other terms.

    Proof of the estimate for normal components

    Proof of the estimate λ||1nψnLq(˙Rn)CfLq(˙Rn). We see that

    λ||1nψ±n=n1k=1F1ξ(λiξn|ξ|1ρ±λ+μ±|ξ|2(ξnξk|ξ|2))Fxfk+F1ξ(λiξn|ξ|1ρ±λ+μ±|ξ|2(1ξ2n|ξ|2))Fxfn.

    All symbols denoted by

    λiξn|ξ|1ρ±λ+μ±|ξ|2ξnξk|ξ|2,λiξn|ξ|1ρ±λ+μ±|ξ|2(1ξ2n|ξ|2)=λiξn|ξ|1ρ±λ+μ±|ξ|2|ξ|2|ξ|2

    are bounded and holomorphic in λΣε, ξ˜Σnη for small ε,η, where we regard |ξ|=n1j=1ξ2j=A and |ξ|2=A2+ξ2n as complex functions. Therefore, by Fourier multiplier theory, we have

    λ||1nψnLq(˙Rn)±λ||1nψ±nLq(Rn)CfLq(˙Rn).

    The other estimate follows similarly.

    The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.

    The research was supported by JSPS KAKENHI Grants No. 19K23408 and No. 20K14350.

    The author declares that he has no competing interests.



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