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=v⋅νtonΓ(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;
{ρ∂tU−DivS(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:
{ρλu−DivS(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) |
{ρ∂tU−DivS(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) |
{ρλu−DivS(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)∈Rn∣xn>0},Rn−:={x=(x1,…,xn)∈Rn∣xn<0},Rn0:={x=(x′,0)=(x1,…,xn−1,0)∈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=(∂i∂jf∣i,j=1,…,n),∇f=(∂ifj∣i,j=1,…,n),∇2f=(∂i∂jfk∣i,j,k=1,…,n). |
Even if g=(g1,…,g˜n)∈X˜n for some ˜n, we denote g∈X and ‖g‖X by ∑˜nj=1‖gj‖X for simplicity. Set
ˆW1q(D)={π∈Lq,loc(D)∣∇π∈Lq(D)},ˆW1q,0(D)={π∈ˆW1q(D)∣π|∂D=0} |
and let ˆW−1q(D) denote the dual space of ˆW1q′,0(D), where 1/q+1/q′=1. For π∈ˆW−1q(D)∩Lq(D), we have
‖π‖ˆW−1q(D)=sup{|∫Dπϕdx|∣ϕ∈ˆW1q′,0(D),‖∇ϕ‖Lq′(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:R→X∣e−γ0tf(t)∈Lp(R,X),f(t)=0fort<0},Wmp,0,γ0(R,X):={f∈Lp,0,γ0(R,X)∣e−γ0t∂jtf(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 γ0≥0. Let F and F−1 denote the Fourier transform and its inverse, defined as follows:
F[f](ξ)=Fx[f](ξ)=∫Rne−ix⋅ξf(x)dx,F−1[g](x)=F−1ξ[g](x)=1(2π)n∫Rneix⋅ξg(ξ)dξ. |
Similarly, let L and L−1λ denote the two-sided Laplace transform and its inverse, defined as follows:
L[f](λ)=∫∞−∞e−λtf(t)dt,L−1λ[g](t)=12π∫∞−∞eλtg(λ)dτ, |
where λ=γ+iτ∈C. Given s≥0 and the X-valued function f, we use the following Bessel potential spaces to treat fractional orders:
Hsp,0,γ0(R,X):={f:R→X∣Λsγf:=L−1λ[|λ|sL[f](λ)](t)∈Lp,0,γ(R,X)for anyγ≥γ0},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):={hn∈W2q(˙Rn)∣|∇′|−1∂nhn:=F−1ξ′|ξ′|−1Fx′(∂nhn)(x′,xn)∈Lq(˙Rn)}. |
We remark that this condition is weaker than that described in [33] since they assumed that |∇′|−1hn∈ˆW1q(˙Rn). It will be easier to apply this assumption to handle a difficult term. Let Σε,γ:={λ∈C∖{0}∣|argλ|<π−ε,|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 λ∈Σε and
f∈Lq(˙Rn),fd∈ˆW−1q(Rn)∩W1q(˙Rn),g∈W1q(˙Rn),h∈W2q(˙Rn),hn∈Eq(˙Rn) |
problem (1.5) admits a unique solution (u,θ)∈W2q(˙Rn)׈W1q(˙Rn) with the following resolvent estimate:
‖(λu,λ1/2∇u,∇2u,∇θ)‖Lq(˙Rn)≤Cn,q,ε{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)}. |
Theorem 2.2. Let 1<p,q<∞ and γ0≥0. Then, for any
F∈Lp,0,γ0(R,Lq(˙Rn)),Fd∈W1p,0,γ0(R,ˆW−1q(Rn))∩Lp,0,γ0(R,W1q(˙Rn)),G∈H1/2p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W1q(˙Rn)),H∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Hn∈W1p,0,γ0(R,Eq(˙Rn)), |
problem (1.4) admits a unique solution (U,Θ) such that
U∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Θ∈Lp,0,γ0(R,ˆW1q(˙Rn)) |
with the following maximal Lp-Lq regularity estimate:
‖e−γt(∂tU,γU,Λ1/2γ∇U,∇2U,∇Θ)‖Lp(R,Lq(˙Rn))≤Cn,p,q,γ0{‖e−γt(F,Λ1/2γFd,∇Fd,Λ1/2γG,∇G,∂tH,∇2H,∂t(|∇′|−1∂nHn)‖Lp(R,Lq(˙Rn))+‖e−γt(∂tFd,γFd)‖Lp(R,ˆW−1q(Rn))}, |
for any γ≥γ0.
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 γ0≥1 that depends on ε>0 such that, for any λ∈Σε,γ0 and
f∈Lq(˙Rn),fd∈ˆW−1q(Rn)∩W1q(˙Rn),g∈W1q(˙Rn),h∈W2q(˙Rn),hn∈Eq(˙Rn),d∈W2q(˙Rn) |
problem (1.3) admits a unique solution (u,θ,η)∈W2q(˙Rn)׈W1q(˙Rn)×W3q(˙Rn) with the following resolvent estimate:
‖(λu,λ1/2∇u,∇2u,∇θ)‖Lq(˙Rn)+|λ|‖η‖W2q(˙Rn)+‖η‖W3q(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)}. |
Moreover, we have
|λ|3/2‖η‖W1q(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)+|λ|1/2‖d‖W1q(˙Rn)} |
and
|λ|2‖η‖Lq(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)+|λ|‖d‖Lq(˙Rn)}. |
Theorem 2.4. Let 1<p,q<∞. Then, there exists a constant γ0≥1 such that, for any
F∈Lp,0,γ0(R,Lq(˙Rn)),Fd∈W1p,0,γ0(R,ˆW−1q(Rn))∩Lp,0,γ0(R,W1q(˙Rn)),G∈H1/2p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W1q(˙Rn)),H∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Hn∈W1p,0,γ0(R,Eq(˙Rn)),D∈Lp,0,γ0(R,W2q(˙Rn)), |
problem (1.2) admits a unique solution (U,Θ,Y) such that
U∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Θ∈Lp,0,γ0(R,ˆW1q(˙Rn)),Y∈Lp,0,γ0(R,W3q(˙Rn))∩W1p,0,γ0(R,W2q(˙Rn)) |
with the following maximal Lp-Lq regularity estimate:
‖e−γt(∂tU,γU,Λ1/2γ∇U,∇2U,∇Θ)‖Lp(R,Lq(˙Rn))+‖e−γt(∂tY,γY)‖Lp(R,W2q(˙Rn))+‖e−γtY‖Lp(R,W3q(˙Rn))≤Cn,p,q,γ0{‖e−γt(F,Λ1/2γFd,∇Fd,Λ1/2γG,∇G,∂tH,∇2H,∂t(|∇′|−1∂nHn))‖Lp(R,Lq(˙Rn))+‖e−γt(∂tFd,γFd)‖Lp(R,ˆW−1q(Rn))+‖e−γtD‖Lp(R,W2q(˙Rn))}, |
for any γ≥γ0. Moreover, we see that, if D∈H1/2p,0,γ0(R,W1q(˙Rn)), then Y∈H3/2p,0,γ0(R,W1q(˙Rn)) and
‖e−γtΛ3/2γY‖Lp(R,W1q(˙Rn))≤Cn,p,q,γ0{‖e−γt(F,Λ1/2γFd,∇Fd,Λ1/2γG,∇G,∂tH,∇2H,∂t(|∇′|−1∂nHn))‖Lp(R,Lq(˙Rn))+‖e−γt(∂tFd,γFd)‖Lp(R,ˆW−1q(Rn))+‖e−γtD‖Lp(R,W2q(˙Rn))+‖e−γtΛ1/2γD‖Lp(R,W1q(˙Rn))} |
for any γ≥γ0. Moreover, we see that, if D∈W1p,0,γ0(R,Lq(˙Rn)), then Y∈W2p,0,γ0(R,Lq(˙Rn)) and
‖e−γt∂2tY‖Lp(R,Lq(˙Rn))≤Cn,p,q,γ0{‖e−γt(F,Λ1/2γFd,∇Fd,Λ1/2γG,∇G,∂tH,∇2H,∂t(|∇′|−1∂nHn))‖Lp(R,Lq(˙Rn))+‖e−γt(∂tFd,γFd)‖Lp(R,ˆW−1q(Rn))+‖e−γtD‖Lp(R,W2q(˙Rn))+‖e−γt∂tD‖Lp(R,Lq(˙Rn))} |
for any γ≥γ0.
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 γ0≥0.
(1) For any fd∈ˆW−1q(Rn)∩W1q(˙Rn), there exists a z∈W2q(˙Rn) such that divz=fd in ˙Rn, [[z]]=0 on Rn0, and the following estimates hold:
‖z‖Lq(˙Rn)≤Cn,q‖fd‖ˆW−1q(˙Rn),‖∇j+1z‖Lq(˙Rn)≤Cn,q‖∇jfd‖Lq(˙Rn)(j=0,1). |
(2) For any Fd∈W1p,0,γ0(R,ˆW−1q(Rn))∩Lp,0,γ0(R,W1q(˙Rn)), there exists a
Z∈W1p,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:
‖e−γt(∂tZ,γZ)‖Lp(R,Lq(˙Rn))≤Cn,p,q‖e−γt(∂tFd,γFd)‖Lp(R,ˆW−1q(Rn)),‖e−γtΛ1/2γ∇Z‖Lp(R,Lq(˙Rn))≤Cn,p,q‖e−γtΛ1/2γFd‖Lp(R,Lq(˙Rn)),‖e−γt∇2Z‖Lp(R,Lq(˙Rn))≤Cn,p,q‖e−γt∇Fd‖Lp(R,Lq(˙Rn)) |
for any γ≥γ0.
Setting u=v+z,˜f=f−(ρλz−Δz) and U=V+Z,˜F=F−(ρ∂tZ−ΔZ), we would like to find (v,θ),(V,Θ) such that
{ρλv−DivS(v,θ)=˜fin˙Rn,divv=0in˙Rn,[[S(v,θ)ν]]=[[g−μD(z)ν]]onRn0,[[v]]=[[h]]onRn0. | (3.1) |
and
{ρ∂tV−DivS(V,Θ)=˜Fin˙Rn,t>0,divV=0in˙Rn,t>0,[[S(V,Θ)ν]]=[[G−μD(Z)ν]]onRn0,t>0,[[V]]=[[H]]onRn0,t>0,V|t=0=0inRn−1. | (3.2) |
Let ˜g:=g−μD(z)ν and ˜G:=G−μD(Z)ν. We see that
‖˜f‖Lq(˙Rn)≤‖f‖Lq(˙Rn)+Cn,q(|λ|‖fd‖ˆW−1q(Rn)+‖∇fd‖Lq(˙Rn)),‖e−γt˜F‖Lp(R,Lq(˙Rn))≤‖e−γtF‖Lp(R,Lq(˙Rn))+Cn,p,q(‖e−γt∂tFd‖Lp(R,ˆW−1q(Rn))+‖e−γt∇Fd‖Lp(R,Lq(˙Rn))),‖(λ1/2˜g,∇˜g)‖Lq(˙Rn)≤C‖(λ1/2fd,∇fd,λ1/2g,∇g)‖Lq(˙Rn),‖e−γt(Λ1/2γ˜G,∇˜G)‖Lp(R,Lq(˙Rn))≤C‖e−γt(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=(δjk−ξjξk|ξ|−2)jk be the Helmholtz decomposition. Then, the functions
ψ±(x)=F−1ξ[P(ξ)Fxf(ξ)ρ±λ+μ±|ξ|2](x),ϕ±(x)=−F−1ξ[iξ⋅Fxf(ξ)|ξ|2](x),Ψ±(x,t)=LλF−1ξ[P(ξ)FxLF(ξ,λ)ρ±λ+μ±|ξ|2](x,t),Φ±(x,t)=−LλF−1ξ[iξ⋅FxLF(ξ,λ)|ξ|2](x,t) |
satisfy
(ψ±,ϕ±)∈W2q(Rn)׈W1q(Rn),ρ±λψ±−μ±Δψ±+∇ϕ±=f,divψ±=0inRn,‖(λψ±,λ1/2∇ψ±,∇2ψ±,∇ϕ±)‖Lq(Rn±)≤Cn,q,ε‖f‖Lq(˙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,‖e−γt(∂tΨ±,γΨ±,Λ1/2γ∇Ψ,∇2Ψ±,∇Φ±)‖Lp(R,Lq(Rn±))≤Cn,p,q,γ0‖e−γtF‖Lp(R,Lq(˙Rn)), |
for any 1<p,q<∞, γ≥γ0≥0,f∈Lq(˙Rn),F∈Lp,0,γ0(R,Lq(˙Rn)), and λ∈Σε with 0<ε<π/2, according to (3.19) in [34]. We define
(ψ,ϕ,Ψ,Φ):={(ψ+,ϕ+,Ψ+,Φ+)forx∈Rn+,(ψ−,ϕ−,Ψ−,Φ−)forx∈Rn−. |
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
{ρλw−μΔw+∇κ=0in˙Rn,divw=0in˙Rn,[[S(w,κ)ν]]=[[g−μD(ψ)ν]]onRn0,[[w]]=[[h−ψ]]onRn0. | (3.3) |
and
{ρ∂tW−μΔW+∇Ξ=0in˙Rn,t>0,divW=0in˙Rn,t>0,[[S(W,Ξ)ν]]=[[G−μD(Ψ)ν]]onRn0,t>0,[[W]]=[[H−Ψ]]onRn0,t>0,W|t=0=0inRn−1. | (3.4) |
Let
˜g:=g−μD(ψ)ν,˜h:=h−ψ,˜G:=G−μD(Ψ)ν,˜H:=H−Ψ. |
Since we have the estimates
‖(λ1/2˜g,∇˜g,λ˜h,∇2˜h,λ|∇′|−1∂n˜hn)‖Lq(˙Rn)≤C‖(f,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn),‖e−γt(Λ1/2γ˜G,∇˜G,∂t˜H,∇2˜H,∂t(|∇′|−1∂n˜Hn))‖Lp(R,Lq(˙Rn))≤C‖e−γt(F,Λ1/2γG,∇G,∂tH,∇2H,∂t(|∇′|−1∂nHn))‖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|∇′|−1∂nψn‖Lq(˙Rn)≤C‖f‖Lq(˙Rn)‖e−γt∂t(|∇′|−1∂nΨn)‖Lp(R,Lq(˙Rn))≤C‖e−γtF‖Lp(R,Lq(˙Rn)) |
in the Appendix.
We shall give a solution of the resolvent problem (1.5) with f=fd=0 and λ∈Σε. We apply a partial Fourier transform with respect to the tangential direction x′∈Rn−1. We apply notation
ˆv(ξ′,xn):=Fx′v(ξ′,xn):=∫Rn−1e−ix′⋅ξ′v(x′,xn)dx′,F−1ξ′w(x′,xn)=1(2π)n−1∫Rn−1eix′⋅ξ′w(ξ′,xn)dξ′ |
for the functions v,w:Rn±→C. Let u±=t(u±1,…,u±(n−1),u±n). Here and hereafter, the index j runs from 1 to n−1 unless stated otherwise.
We need to solve the following second-order ordinary differential equations:
{(ρ±λ+μ±|ξ′|2−μ±∂2n)ˆu±j+iξjˆθ±=0inxn≠0,(ρ±λ+μ±|ξ′|2−μ±∂2n)ˆu±n+∂nˆθ±=0inxn≠0,n−1∑j=1iξjˆu±j+∂nˆu±n=0inxn≠0,[[μ(iξjˆun+∂nˆuj)]]=−[[ˆgj]]onxn=0,[[2μ∂nˆun−ˆθ]]=−[[ˆgn]]onxn=0,[[ˆu]]=[[ˆh]]onxn=0. | (4.1) |
Set
A:=√n−1∑j=1ξ2j,B±:=√ρ±(μ±)−1λ+A2 |
with positive real parts. Here, we consider ξ′ to have complex values, as follows:
ξj∈˜Ση:={z∈C∖{0}∣|argz|<η}∪{z∈C∖{0}∣π−η<|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(e∓B±xn−e∓Axn)+β±je∓B±xn(j=1,…,n),ˆθ±(ξ′,xn)=γ±e∓Axn. |
Then, the equations become
{−μ±(B2±−A2)α±j+iξjγ±=0,−μ±(B2±−A2)α±n∓Aγ±=0,−iξ′⋅α′±±Aα±n=0,iξ′⋅(α′±+β′±)∓B±(α±n+β±n)=0, | (4.2) |
and
{μ+(B2+−A2)α+n+μ+(B2++A2)β+n−μ−(B2−−A2)α−n−μ−(B2−+A2)β−n=iξ′⋅[[ˆg′]],μ+(B+−A)2α+n−2μ+AB+β+n+μ−(B−−A)2α−n−2μ−AB−β−n=−A[[ˆg′n]],(B+−A)α+n+B+β+n+(B–A)α−n+B−β−n=iξ′⋅[[ˆh′]],β+n−β−n=[[ˆh′n]]. |
This means that
[μ+(B++A)μ+(B2++A2)−μ−(B−+A)−μ−(B2−+A2)μ+(B+−A)−2μ+AB+μ−(B–A)−2μ−AB−1B+1B−010−1][(B+−A)α+nβ+n(B–A)α−nβ−n]=[iξ′⋅[[ˆg′]]−A[[ˆgn]]iξ′⋅[[ˆh′]][[ˆh′n]]]=[iξ′T0000−A0000iξ′T00001][[[ˆg1]]⋮[[ˆgn]][[ˆh1]]⋮[[ˆhn]]], |
where ξ′T 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−(B–A)−2μ−AB−1B+1B−010−1],R:=(rij):=[iξ′T0000−A0000iξ′T00001](1≤i≤4,1≤j≤2n). |
We have
detL=(μ+−μ−)2A3−{(3μ+−μ−)μ+B++(3μ−−μ+)μ−B−}A2−{(μ+B++μ−B−)2+μ+μ−(B++B−)2}A−(μ+B++μ−B−)(μ+B2++μ−B2−) |
and it is known, from [33, Lemma 5.5], that the determinant is not zero for λ∈Σε, ξ′∈Rn−1.
We introduce some new notation:
M+=M+(A,B+,xn)=e−B+xn−e−AxnB+−A,M−=M−(A,B−,xn)=eB−xn−eAxnB–A,(ai,j):=(L−1R)ij=(detL)−1(4∑s=1Lisrsj)ij(1≤i≤4,1≤j≤2n), |
where we use the cofactor matrix of L, denoted by Cof(L)=(Lij).
From these observations, we have
ˆu+n(ξ′,xn)=n∑k=1{(a1,kM++a2,ke−B+xn)[[ˆgk]]+(a1,n+kM++a2,n+ke−B+xn)[[ˆhk]]},ˆu−n(ξ′,xn)=n∑k=1{(a3,kM−+a4,keB−xn)[[ˆgk]]+(a3,n+kM−+a4,n+keB−xn)[[ˆhk]]}. |
To simplify, we define the following symbols for k=1,…,n:
ϕk,+n(λ,ξ′,xn)=a1,kM++a2,ke−B+xn,ψk,+n(λ,ξ′,xn)=a1,n+kM++a2,n+ke−B+xn,ϕk,−n(λ,ξ′,xn)=a3,kM−+a4,keB−xn,ψk,−n(λ,ξ′,xn)=a3,n+kM−+a4,n+keB−xn, |
which yields the solution formulas for u±n:
ˆu+n=ˆu+n(ξ′,xn)=n∑k=1(ϕk,+n[[ˆgk]]+ψk,+n[[ˆhk]]),xn>0,ˆu−n=ˆu−n(ξ′,xn)=n∑k=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,ke−Axn,ωk,+(λ,ξ′,xn)=−μ+(B++A)Aa1,n+ke−Axn,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)=n∑k=1(χk,±[[ˆgk]]+ωk,±[[ˆhk]]),xn≷0. |
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+μ−B−1−1][β+jβ−j]=[[[ˆgj]][[ˆhj]]]+[−μ+(B+−A)α+j+μ+iξjβ+n−μ−(B−−A)α−j−μ−iξjβ−n0]=[[[ˆgj]][[ˆhj]]]+iξjA[μ+μ+A−μ−−μ−A0000][(B+−A)α+nβ+n(B–A)α−nβ−n], |
[β+jβ−j]=1μ+B++μ−B−{[[[ˆgj]]+μ−B−[[ˆhj]][[ˆgj]]−μ+B+[[ˆhj]]]+iξjA[μ+μ+A−μ−−μ−Aμ+μ+A−μ−−μ−A][(B+−A)α+nβ+n(B–A)α−nβ−n]}=1μ+B++μ−B−[[[ˆgj]]+μ−B−[[ˆhj]]+iξjA∑nk=1{μ+(a1,k+Aa2,k)[[ˆgk]]−μ−(a3,n+k+Aa4,n+k)[[ˆhk]]}[[ˆgj]]−μ+B+[[ˆhj]]+iξjA∑nk=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±je∓B±xn=:n∑k=1(ϕk,±j[[ˆgk]]+ψk,±j[[ˆhk]]),xn≷0, |
where
ϕk,+j(λ,ξ′,xn)=−iξjAa1,kM++1μ+B++μ−B−(δk,j+iμ+ξjA(a1,k+Aa2,k))e−B+xn,ψk,+j(λ,ξ′,xn)=−iξjAa1,n+kM++1μ+B++μ−B−(μ−B−δk,j−iμ−ξjA(a3,n+k+Aa4,n+k))e−B+xn,ϕk,−j(λ,ξ′,xn)=iξjAa3,kM−+1μ+B++μ−B−(δk,j+iμ+ξjA(a1,k+Aa2,k))eB−xn,ψk,−j(λ,ξ′,xn)=iξjAa3,n+kM−−1μ+B++μ−B−(μ+B+δk,j+iμ−ξjA(a3,n+k+Aa4,n+k))eB−xn. |
Here, we introduce a new notation, i.e., [[[f]]](x′,xn):=f(x′,xn)−f(x′,−xn), for f:˙Rn→C. 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)=∓n∑k=1{∫R±F−1ξ′[∂nϕk,±j(λ,ξ′,xn+yn)Fx′[[[gk]]]](x,yn)dyn+∫R±F−1ξ′[ϕk,±j(λ,ξ′,xn+yn)Fx′[[[∂ngk]]]](x,yn)dyn+∫R±F−1ξ′[∂nψk,±j(λ,ξ′,xn+yn)Fx′[[[hk]]]](x,yn)dyn+∫R±F−1ξ′[ψk,±j(λ,ξ′,xn+yn)Fx′[[[∂nhk]]]](x,yn)dyn},(j=1,…,n), | (4.3) |
θ±(x)=∓n∑k=1{∫R±F−1ξ′[∂nχk,±(λ,ξ′,xn+yn)Fx′[[[gk]]]](x,yn)dyn+∫R±F−1ξ′[χk,±(λ,ξ′,xn+yn)Fx′[[[∂ngk]]]](x,yn)dyn+∫R±F−1ξ′[∂nωk,±(λ,ξ′,xn+yn)Fx′[[[hk]]]](x,yn)dyn+∫R±F−1ξ′[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)=∓L−1λn∑k=1{∫R±F−1ξ′[∂nϕk,±j(λ,ξ′,xn+yn)Fx′L[[[Gk]]]](x,yn)dyn+∫R±F−1ξ′[ϕk,±j(λ,ξ′,xn+yn)Fx′L[[[∂nGk]]]](x,yn)dyn+∫R±F−1ξ′[∂nψk,±j(λ,ξ′,xn+yn)Fx′L[[[Hk]]]](x,yn)dyn+∫R±F−1ξ′[ψk,±j(λ,ξ′,xn+yn)Fx′L[[[∂nHk]]]](x,yn)dyn},(j=1,…,n), | (4.5) |
Θ±(x,t)=∓L−1λn∑k=1{∫R±F−1ξ′[∂nχk,±(λ,ξ′,xn+yn)Fx′L[[[Gk]]]](x,yn)dyn+∫R±F−1ξ′[χk,±(λ,ξ′,xn+yn)Fx′L[[[∂nGk]]]](x,yn)dyn+∫R±F−1ξ′[∂nωk,±(λ,ξ′,xn+yn)Fx′L[[[Hk]]]](x,yn)dyn+∫R±F−1ξ′[ωk,±(λ,ξ′,xn+yn)Fx′L[[[∂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[F−1ξ′m(ξ′,xn+yn)Fx′f](x,yn)dyn,˜Tγ[mλ]g(x,t)=L−1λ∫∞0[F−1ξ′mλ(ξ′,xn+yn)Fx′Lg](x,yn,λ)dyn,=[eγtF−1τ→tT[mλ]Ft→τ(e−γtg)](x,t), |
where λ=γ+iτ∈Σε, m,mλ are C-valued functions, f:Rn+→C, and g:Rn+×R→C. 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∞(˜Σn−1η).
(b) There exist η∈(0,π/2) and C>0 such that supξ′∈˜Σn−1η|m(ξ′,xn)|≤Cx−1n for all xn>0.
Then, T[m] is a bounded linear operator on Lq(Rn+) for every 1<q<∞.
(ⅱ) Let γ0≥0 and mλ satisfy the following two conditions:
(c) There exists η∈(0,π/2−ε) such that, for each xn>0 and γ≥γ0,
˜Σnη∋(τ,ξ′)↦mλ(ξ′,xn)∈C |
is bounded and holomorphic.
(d) There exist η∈(0,π/2−ε) and C>0 such that sup{|mλ(ξ′,xn)|∣(τ,ξ′)∈˜Σnη}≤Cx−1n for all γ≥γ0 and xn>0.
Then, ˜Tγ[mλ] satisfies
‖e−γt˜Tγ[mλ]g‖Lp(R,Lq(Rn+))≤C‖e−γtg‖Lp(R,Lq(Rn+)) |
for every γ≥γ0 and 1<p,q<∞.
By using the identities given by
B2±=ρ±(μ±)−1λ+n−1∑m=1ξ2m,1=B2±B2±=ρ±(μ±)−1λ1/2B2±λ1/2−n−1∑m=1iξmB2±(iξm), |
we have
u±j(x)=∓n∑k=1{∫R±F−1ξ′[ρ±(μ±)−1λ1/2B−2±∂nϕk,±j(λ,ξ′,xn+yn)Fx′[[[λ1/2gk]]]](x,yn)dyn−n−1∑m=1∫R±F−1ξ′[iξmB−2±∂nϕk,±j(λ,ξ′,xn+yn)Fx′[[[∂mgk]]]](x,yn)dyn+∫R±F−1ξ′[ϕk,±j(λ,ξ′,xn+yn)Fx′[[[∂ngk]]]](x,yn)dyn+∫R±F−1ξ′[B−2±∂nψk,±j(λ,ξ′,xn+yn)Fx′[[[(ρ±(μ±)−1λ−Δ′)hk]]]](x,yn)dyn−n−1∑m=1∫R±F−1ξ′[iξmB−2±ψk,±j(λ,ξ′,xn+yn)Fx′[[[∂m∂nhk]]]](x,yn)dyn}∓n−1∑k=1∫R±F−1ξ′[ρ±(μ±)−1λ1/2B−2±ψk,±j(λ,ξ′,xn+yn)Fx′[[[λ1/2∂nhk]]]](x,yn)dyn+∫R±F−1ξ′[ρ±(μ±)−1AB−2±ψk,±j(λ,ξ′,xn+yn)Fx′[[[λ|∇′|−1∂nhn]]]](x,yn)dynθ±(x)=∓n∑k=1{∫R±F−1ξ′[ρ±(μ±)−1λ1/2B−2±∂nχk,±(λ,ξ′,xn+yn)Fx′[[[λ1/2gk]]]](x,yn)dyn−n−1∑m=1∫R±F−1ξ′[iξmB−2±∂nχk,±(λ,ξ′,xn+yn)Fx′[[[∂mgk]]]](x,yn)dyn+∫R±F−1ξ′[χk,±(λ,ξ′,xn+yn)Fx′[[[∂ngk]]]](x,yn)dyn+∫R±F−1ξ′[B−2±∂nωk,±(λ,ξ′,xn+yn)Fx′[[[(ρ±(μ±)−1λ−Δ′)hk]]]](x,yn)dyn−n−1∑m=1∫R±F−1ξ′[iξmB−2±ωk,±(λ,ξ′,xn+yn)Fx′[[[∂m∂nhk]]]](x,yn)dyn}∓n−1∑k=1∫R±F−1ξ′[ρ±(μ±)−1λ1/2B−2±ωk,±(λ,ξ′,xn+yn)Fx′[[[λ1/2∂nhk]]]](x,yn)dyn+∫R±F−1ξ′[ρ±(μ±)−1AB−2±ωk,±(λ,ξ′,xn+yn)Fx′[[[λ|∇′|−1∂nhn]]]](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±)−1λ1/2B−2±∂nϕk,±j(λ,ξ′,xn)k∈{1,…,n},iξmB−2±∂nϕk,±j(λ,ξ′,xn)k∈{1,…,n},m∈{1,…,n−1},ϕk,±j(λ,ξ′,xn)k∈{1,…,n},B−2±∂nψk,±j(λ,ξ′,xn)k∈{1,…,n},iξmB−2±ψk,±j(λ,ξ′,xn)k∈{1,…,n},m∈{1,…,n−1},ρ±(μ±)−1λ1/2B−2±ψk,±j(λ,ξ′,xn)k∈{1,…,n−1},ρ±(μ±)−1AB−2±ψn,±j(λ,ξ′,xn),Sθ±(λ,ξ′,xn):={ρ±(μ±)−1λ1/2B−2±∂nχk,±(λ,ξ′,xn)k∈{1,…,n},iξmB−2±∂nχk,±(λ,ξ′,xn)k∈{1,…,n},m∈{1,…,n−1},χk,±(λ,ξ′,xn)k∈{1,…,n},B−2±∂nωk,±(λ,ξ′,xn)k∈{1,…,n},iξmB−2±ωk,±(λ,ξ′,xn)k∈{1,…,n},m∈{1,…,n−1},ρ±(μ±)−1λ1/2B−2±ωk,±(λ,ξ′,xn)k∈{1,…,n−1},ρ±(μ±)−1AB−2±ωn,±(λ,ξ′,xn). |
We shall confirm that all of the symbols are bounded in the sense that
sup(λ,ξ′)∈ΣεטΣn−1ηℓ,ℓ′=1,…,n−1{(|λ|+|λ|1/2|ξℓ|+|ξℓ||ξℓ′|)|Suj±|+(|λ|1/2+|ξℓ|)|∂nSuj±|+|∂2nSuj±|+|ξℓ||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)=∓e∓B±xn∓AM±(A,B±,xn),∂2nM±(A,B±,xn)=(A+B±)e∓B±xn+A2M±(A,B±,xn),∂3nM±(A,B±,xn)=∓(A2+AB±+B2±)e∓B±xn∓A3M±(A,B±,xn) |
as well as the following useful lemma, where we let ˜A:=√∑n−1j=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)∈ΣεטΣn−1η×R±, we have
c˜A≤ReA≤|A|≤˜A,c(|λ|1/2+˜A)≤ReB±≤|B±|≤C(|λ|1/2+˜A),|∂mne∓Axn|≤C˜A−1+m(±xn)−1,forxn≷0,|∂mne∓B±xn|≤C(|λ|1/2+˜A)−1+m(±xn)−1,forxn≷0,|M±(A,B±,xn)|≤C(|λ|1/2+˜A)−1˜A−1(±xn)−1,forxn≷0,|∂mnM±(A,B±,xn)|≤C(|λ|1/2+˜A)−2+m(±xn)−1,forxn≷0,m≠0,c(|λ|1/2+˜A)≤|μ+B++μ−B−| |
for m=0,1,2,3, with positive constants c and C, which are independent of λ,ξ′,xn.
We recall that
L−1=[μ+(B++A)μ+(B2++A2)−μ−(B−+A)−μ−(B2−+A2)μ+(B+−A)−2μ+AB+μ−(B–A)−2μ−AB−1B+1B−010−1]−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,
|4∑s=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),…,(2,2n−1),(3,1),…,(3,n),(4,n+1),…,(4,2n−1),C˜A(|λ|1/2+˜A)3for(i,j)=(1,n+1),…,(1,2n−1),(3,n+1),…,(3,2n−1),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(λ∈Σε,ξ′∈˜Σn−1η). |
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+e−iθ,B−e−iθ)|≥c(|λe−2iθ|+|A|2)3/2≥c(|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||∂mne∓B±xn|≤|detL|−1(|4∑s=1L1srsk||∂mnM±|+|4∑s=1L2srsk||∂mne∓B±xn|)≤C(|λ|1/2+˜A)−2+m(±xn)−1,|∂mnϕk,±j|≤C(|a1,k||∂mnM±|+|μ+B++μ−B−|−1(1+|a1,k|+˜A|a2,k|)|∂mne∓B±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,…,n−1}C(|λ|1/2+˜A)−1˜Am(±xn)−1ifk=n,|∂mnχk,±|≤C˜A−1+m(±xn)−1,k∈{1,…,n},|∂mnωk,±|≤{C(|λ|1/2+˜A)˜A−1+m(±xn)−1ifk∈{1,…,n−1}C(|λ|1/2+˜A)2˜A−2+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, λ1/2∂ℓu, ∂ℓ∂ℓ′u, λ1/2∂nu, ∂ℓ∂nu, ∂2nu, ∂ℓθ, and ∂nθ.
We also see that the new symbols Suj± and Sθ±, multiplied by λ, ξℓ, 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+)≤‖f‖Lq(˙Rn).
Theorem 5.4. Let 0<ε<π/2 and 1<q<∞. Then, for any λ∈Σε,g∈W1q(˙Rn), h∈W2q(˙Rn), and hn∈Eq(˙Rn), the problem (1.5) with f=fd=0 admits a solution (u,θ)∈W2q(˙Rn)׈W1q(˙Rn) with the following resolvent estimate:
‖(λu,λ1/2∇u,∇2u,∇θ)‖Lq(˙Rn)≤C‖(λ1/2g,∇g,λh,λ1/2∇h,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)≤C‖(λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖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 γ0≥0. Then, for any
H∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Hn∈W1p,0,γ0(R,Eq(˙Rn)), |
the problem (1.4) with F=Fd=0 admits a solution (U,Π) such that
U∈W1p,0,γ0(R,Lq(˙Rn))∩Lp,0,γ0(R,W2q(˙Rn)),Π∈Lp,0,γ0(R,ˆW1q(˙Rn)) |
with the following maximal Lp-Lq regularity:
‖e−γt(∂tU,γU,Λ1/2γ∇U,∇2U,∇Π)‖Lp(R,Lq(˙Rn))≤C‖e−γt(∂tH,Λ1/2γ∇H,∇2H,∂t(|∇′|−1∂nHn))‖Lp(R,Lq(˙Rn))≤C‖e−γt(∂tH,∇2H,∂t(|∇′|−1∂nHn))‖Lp(R,Lq(˙Rn)) |
for any γ≥γ0 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:
{ρλv−DivS(v,τ)=fin˙Rn,divv=fdin˙Rn,[[S(v,τ)ν]]=[[g]]onRn0,[[v]]=[[h]]onRn0, | (6.1) |
{ρ∂tV−DivS(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
{ρλw−DivS(w,κ)=0in˙Rn,divw=0in˙Rn,λη+wn=d−vn=:˜donRn0,[[S(w,κ)ν]]−([[ρ]]cg+cσΔ′)ην=0onRn0,[[w]]=0onRn0, | (6.3) |
{ρ∂tW−DivS(W,Ξ)=0in˙Rn,t>0,divW=0in˙Rn,t>0,∂tY+Wn=D−Vn=:˜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]]cg−cσA2)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′)=detLλdetL−A{μ+(B++A)+μ−(B−+A)}([[ρ]]cg−cσA2)ˆ˜d,ˆw±j(λ,ξ′,xn)=−ϕn,±j([[ρ]]cg−cσA2)ˆη(j=1,⋯,n),ˆκ±(λ,ξ′,xn)=−χn,±([[ρ]]cg−cσA2)ˆη |
with the following estimate:
L(λ,ξ′):=λdetL−A{μ+(B++A)+μ−(B−+A)}([[ρ]]cg−cσA2)|L(λ,ξ′)|≥c(|λ|+˜A)(|λ|1/2+˜A)3 |
for (λ,ξ′)∈Σε,γ0טΣn−1η with 0<ε<π/2, 0<η<ε/2, and γ0≥1. The proof for ξ′∈Rn−1 is in [33, Lemma 6.1]. However, the proof for complex values is almost the same.
Let η extend suitably from Rn−1 to ˙Rn. Since we have the estimate
sup(λ,ξ′)∈Σε,γ0טΣn−1ηℓ=1,…,n−1{(|λ|+|ξℓ|)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‖˜d‖Lq(˙Rn),‖(λ∇η,∇2η)‖Lq(˙Rn)≤C‖∇˜d‖Lq(˙Rn),‖(λ∇2η,∇3η)‖Lq(˙Rn)≤C‖∇2˜d‖Lq(˙Rn), |
for λ∈Σε,γ0. Then, from the results in the previous section, it follows that
‖(λw,λ1/2∇w,∇2w,∇κ)‖Lq(˙Rn)≤C‖(λ1/2gn,∇gn)‖Lq(˙Rn)≤C‖(λ1/2η,λ1/2∇2η,∇η,∇3η)‖Lq(˙Rn)≤C‖˜d‖W2q(˙Rn), |
where we have used |λ|1/2≤|λ| when λ∈Σε,γ0; also, C is dependent on γ0, as well as the constants [[ρ]], cg, and cσ. This yields that
‖(λu,λ1/2∇u,∇2u,∇θ)‖Lq(˙Rn)+|λ|‖η‖W2q(˙Rn)+‖η‖W3q(˙Rn)≤‖(λv,λ1/2∇v,∇2v,λw,λ1/2∇w,∇2w,∇τ,∇κ)‖Lq(˙Rn)+|λ|‖η‖W2q(˙Rn)+‖η‖W3q(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)+‖˜d‖W2q(˙Rn)}≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)} |
since
‖˜d‖W2q(˙Rn)≤Cn,q,ε(‖d‖W2q(˙Rn)+‖v‖W2q(˙Rn))≤Cn,q,ε,γ0(‖d‖W2q(˙Rn)+|λ|‖v‖Lq(˙Rn)+‖∇2v‖Lq(˙Rn))≤Cn,q,ε,γ0(‖d‖W2q(˙Rn)+‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖fd‖ˆW−1q(˙Rn)). |
In addition, we have
|λ|3/2‖η‖W1q(˙Rn)≤|λ|1/2‖˜d‖W1q(˙Rn)≤|λ|1/2‖d‖W1q(˙Rn)+|λ|1/2‖v‖W1q(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖g‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)+|λ|1/2‖d‖W1q(˙Rn)} |
and
|λ|2‖η‖Lq(˙Rn)≤|λ|‖˜d‖Lq(˙Rn)≤|λ|‖d‖Lq(˙Rn)+|λ|‖v‖Lq(˙Rn)≤Cn,q,ε,γ0{‖(f,λ1/2fd,∇fd,λ1/2g,∇g,λh,∇2h,λ|∇′|−1∂nhn)‖Lq(˙Rn)+|λ|‖g‖ˆW−1q(˙Rn)+‖d‖W2q(˙Rn)+|λ|‖d‖Lq(˙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|∇′|−1∂nψn‖Lq(˙Rn)≤C‖f‖Lq(˙Rn). We see that
λ|∇′|−1∂nψ±n=n−1∑k=1F−1ξ(λiξn|ξ′|1ρ±λ+μ±|ξ|2(−ξnξk|ξ|2))Fxfk+F−1ξ(λiξn|ξ′|1ρ±λ+μ±|ξ|2(1−ξ2n|ξ|2))Fxfn. |
All symbols denoted by
λiξn|ξ′|1ρ±λ+μ±|ξ|2−ξnξk|ξ|2,λiξn|ξ′|1ρ±λ+μ±|ξ|2(1−ξ2n|ξ|2)=λiξn|ξ′|1ρ±λ+μ±|ξ|2|ξ′|2|ξ|2 |
are bounded and holomorphic in λ∈Σε, ξ∈˜Σnη for small ε,η, where we regard |ξ′|=√∑n−1j=1ξ2j=A and |ξ|2=A2+ξ2n as complex functions. Therefore, by Fourier multiplier theory, we have
‖λ|∇′|−1∂nψn‖Lq(˙Rn)≤∑±‖λ|∇′|−1∂nψ±n‖Lq(Rn)≤C‖f‖Lq(˙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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