Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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508 J. Pr¨uss and G. Simonett

for gravity to act on the ﬂuids. The governing equations then are given by the

system

⎧

⎪

⎨

⎪

⎩



∂

u +(u|∇)u



− μΔu + ∇q =0 in Ω(t)

div u =0 in Ω(t)

−[[ S(u, q)ν]] = σκν +[[ρ]] γ

yν on Γ(t)

[[ u]] = 0 o n Γ ( t)

V =(u|ν)onΓ(t)

u(0) = u

in Ω

Γ(0) = Γ

(1.1)

Here ρ and μ are given by

ρ = ρ

(t)

+ ρ

(t)

,μ= μ

(t)

+ μ

(t)

with χ the indicator function, where the constants ρ

and μ

denote the densities

and viscosities of the respective ﬂuids. The constant σ>0 denotes the surface

tension, and γ

is the acceleration of gravity. Moreover, S(u, q) is the stress tensor

deﬁned by

S(u, q)=μ



∇u +(∇u)



− qI in Ω

(t),

where q =˜q + ργ

y denotes the modiﬁed pressure incorporating the potential of

the gravity force, and

[[ v]] = ( v

(t)

−v

(t)



Γ(t)

denotes the jump of the quantity v, deﬁned on the respective domains Ω

(t), across

the interface Γ(t). Finally, κ = κ(t, ·) is the mean curvature of the free boundary

Γ(t), ν = ν(t, ·) is the unit normal ﬁeld on Γ(t), and V = V (t, ·) is the normal

velocity of Γ(t). Here we use the convention that ν(t, ·)pointsfromΩ

(t)into

(t), and that κ(t, x) is negative when Ω

(t) is convex in a neighborhood of

x ∈ Γ(t).

Given is the initial position Γ

=graph(h

) of the interface, and the initial velocity

:Ω

→ R

n+1

, Ω

:= Ω

(0) ∪ Ω

(0).

The unknowns are the velocity ﬁeld u(t, ·):Ω(t) → R

n+1

, the pressure ﬁeld

q(t, ·):Ω(t) → R, and the free boundary Γ(t), where Ω(t):=Ω

(t) ∪ Ω

(t).

Our main result shows that problem (1.1) admits a unique local smooth solution,

provided that ∇h



∞

:= sup

x∈R

|∇h

(x)| is suﬃciently small.

Theorem 1.1. Let p>n+3. Then given β>0,thereexistsη = η(β) > 0 such

that for all initial values

) ∈ W

2−2/p

(Ω

, R

n+1

) ×W

3−2/p

), [[ u

]] = 0 ,

satisfying the compatibility conditions

[[ μD(u

)ν

− μ(ν

|D(u

)ν

]] = 0 , div u

=0 on Ω

Two-phase Navier-Stokes Equations 509

with D(u

):=(∇u

+(∇u

)

), and the smallness-boundedness condition

∇h



∞

≤ η, u



∞

≤ β,

there is t

= t

) > 0 such that problem (1.1) admits a classical solu-

tion (u, q, Γ) on (0,t

). The solution is unique in the function class described in

Theorem 4.2. In addition, Γ(t) is a graph over R

given by a function h(t) and

M =

t∈(0,t

)



{t}×Γ(t)



is a real analytic manifold, and with

O := {(t, x, y): t ∈ (0,t

),x∈ R

,y = h(t, x)},

the function (u, q):O→R

n+2

is real analytic.

Remarks 1.2. (a) More precise statements for the transformed problem will be

given in Section 4. Due to the restriction p>n+3weobtain

h ∈ C(J; BUC

)) ∩ C

(J; BUC

)),

with J =[0,t

], where BUC means bounded and uniformly continuous. In partic-

ular, the normal of Ω

(t), the normal velocity of Γ(t), and the mean curvature of

Γ(t) are well deﬁned and continuous, so that (1.1) makes sense pointwise. For u

we obtain

u ∈ BUC(J ×R

n+1

, R

n+1

), ∇u ∈ BUC(O, R

(n+1)

Also interesting is the fact that the surface pressure jump is analytic on M as

well.

(b) It is possible to relax the assumption p>n+3.In fact, p>(n +3)/2 turns

out to be suﬃcient. In order to keep the arguments simple, we impose here the

stronger condition p>n+3.

immiscible ﬂuids – with the heavier ﬂuid lying above a ﬂuid of lesser density – leads

to an instability, the Rayleigh-Taylor instability. In this case, small disturbances

of the equilibrium situation (u, h)=(0, 0) can cause instabilities, where the heavy

ﬂuid moves down under the inﬂuence of gravity, and the light material is displaced

upwards, leading to vortices. Our results show that problem (1.1) is also well

posed in this case, provided ∇h



∞

is small enough, yielding smooth solutions

for a short time. In the forthcoming publication [29] we will give a rigorous proof

showing that the equilibrium solution (u, h)=(0, 0) is L

-unstable. To the best

of our knowledge these are the ﬁrst rigorous results concerning the Navier-Stokes

equations subject to the Rayleigh-Taylor instability.

(d) If γ

= 0 then it is shown in [28] that problem (1.1) admits a solution with

the same regularity properties on an arbitrary ﬁxed time interval [0,t

], provided

that u



2−2/p

(Ω

)

and h



3−2/p

)

are suﬃciently small (depending on t

(e) We point out that in Theorem 1.1 we only need a smallness condition on the

sup-norm of ∇h

(relative to the vertical component of the velocity). In case of a

more general geometry, this condition can always be achieved by a judicious choice

of a reference manifold.

510 J. Pr¨uss and G. Simonett

The motion of a layer of viscous, incompressible ﬂuid in an ocean of inﬁ-

nite extent, bounded below by a solid surface and above by a free surface which

includes the eﬀects of surface tension and gravity (in which case Ω

is a strip,

bounded above by Γ

andbelowbyaﬁxedsurfaceΓ

) has been considered by

Allain [1], Beale [7], Beale and Nishida [8], Tani [35], by Tani and Tanaka [36], and

by Shibata and Shimizu [32]. If the initial state and the initial velocity are close

to equilibrium, global existence of solutions is proved in [7] for σ>0, and in [36]

for σ ≥ 0, and the asymptotic decay rate for t →∞is studied in [8]. We also refer

to [9], where in addition the presence of a surfactant on the free boundary and in

one of the bulk phases is considered.

In case that Ω

(t) is a bounded domain, γ

=0,andΩ

(t)=∅, one obtains the

one-phase Navier-Stokes equations with surface tension, describing the motion of

an isolated volume of ﬂuid. For an overview of the existing literature in this case

we refer to the recent publications [28, 31, 32, 33].

Results concerning the two-phase problem (1.1) with γ

=0inthe3D-case are

obtained in [11, 12, 13, 34]. In more detail, Densiova [12] establishes existence and

uniqueness of solutions (of the transformed problem in Lagrangian coordinates)

with v ∈ W

s,s/2

for s ∈ (5/2, 3) in case that one of the domains is bounded. Tanaka

[34] considers the two-phase Navier-Stokes equations with thermo-capillary con-

vection in bounded domains, and he obtains existence and uniqueness of solutions

with (v, θ) ∈ W

s,s/2

for s ∈ (7/2, 4), with θ denoting the temperature.

In order to prove our main result we transform problem (1.1) into a problem on

a ﬁxed domain. The transformation is expressed in terms of the unknown height

function h describing the free boundary. Our analysis proceeds with establish-

ing maximal regularity results for an associated linear problem. relying on the

powerful theory of maximal regularity, in particular on the H

∞

-calculus for sec-

torial operators, the Dore-Venni theorem, and the Kalton-Weis theorem, see for

instance [2, 14, 16, 22, 23, 26, 30].

Based on the linear estimates we can solve the nonlinear problem by the

contraction mapping principle. Analyticity of solutions is obtained as in [28] by

the implicit function theorem in conjunction with a scaling argument, relying on

an idea that goes back to Angenent [4, 5] and Masuda [24]; see also [17, 18, 20].

The plan for this paper is as follows. Section 2 contains the transformation of

the problem to a half-space and the determination of the proper underlying linear

problem. In Section 3 we analyze this linearization and prove the crucial maxi-

mal regularity result in an L

-setting. Section 4 is then devoted to the nonlinear

problem and contains the proof of our main result. Finally we collect and prove in

an appendix some of the technical results used in order to estimate the nonlinear

terms.

Two-phase Navier-Stokes Equations 511

2. The transformed problem

The nonlinear problem (1.1) can be transformed to a problem on a ﬁxed domain

by means of the transformations

v(t, x, y):=(u

,...,u

)(t, x, y + h(t, x)),

w(t, x, y):=u

n+1

(t, x, y + h(t, x)),

π(t, x, y):=q(t, x, y + h(t, x)),

where t ∈ J =[0,a], x ∈ R

, y ∈ R, y = 0. With a slight abuse of notation we will

in the sequel denote the transformed velocity again by u, that is, we set u =(v, w).

With this notation we obtain the transformed problem

⎧

⎪

⎨

⎪

⎩

ρ∂

u −μΔu + ∇π = F (u, π, h)in

n+1

div u = F

(u, h)in

n+1

−[[ μ∂

v]] − [[ μ∇

w]] = G

(u, [[ π]] ,h)onR

−2[[μ∂

w]] + [[ π]] − σΔh − [[ ρ]] γ

h = G

(u, h)onR

[[ u]] = 0 o n R

∂

h − γw = −(γv|∇h)onR

u(0) = u

,h(0) = h

(2.1)

for t>0, where

n+1

= {(x, y) ∈ R

× R : y =0}.

The nonlinear functions have been computed in [28] and are given by:

(v, w, π, h)=μ{−2(∇h|∇

)∂

v + |∇h|

∂

v − Δh∂

v} + ∂

π∇h

+ ρ{−(v|∇

)v +(∇h|v)∂

v − w∂

v} + ρ∂

h∂

(v, w, h)=μ{−2(∇h|∇

)∂

w + |∇h|

∂

w − Δh∂

+ ρ{−(v|∇

)w +(∇h|v)∂

w − w∂

w} + ρ∂

h∂

(v, h)=(∇h|∂

(2.2)

and

(v, w, [[ π]] ,h)= − [[ μ(∇

v +(∇

)]]∇h + |∇h|

[[ μ∂

v]] + ( ∇h|[[ μ∂

v]] ) ∇h

− [[ μ∂

w]] ∇h + {[[ π]] − σ(Δh − G

(h))}∇h,

(v, w, h)= − (∇h|[[ μ∇

w]] ) − (∇h|[[ μ∂

v]] ) + |∇h|

[[ μ∂

w]] − σG

(h)

(2.3)

with

(h)=

|∇h|

Δh

(1 +



1+|∇h|

)



1+|∇h|

(∇h|∇

h∇h)

(1 + |∇h|

)

3/2

, (2.4)

where ∇

h denotes the Hessian matrix of all second-order derivatives of h.

Before studying solvability results for problem (2.1) let us ﬁrst introduce

suitable function spaces. Let Ω ⊆ R

be open and X be an arbitrary Banach

512 J. Pr¨uss and G. Simonett

space. By L

(Ω; X)andH

(Ω; X), for 1 ≤ p ≤∞, s ∈ R,wedenotetheX-

valued Lebesgue and the Bessel potential spaces of order s, respectively. We will

also frequently make use of the fractional Sobolev-Slobodeckij spaces W

(Ω; X),

1 ≤ p<∞, s ∈ R \ Z,withnorm

g

(Ω;X)

= g

[s]

(Ω;X)

|α|=[s]





∂

g(x) − ∂

g(y)

|x −y|

m+(s−[s])p

dx dy



1/p

, (2.5)

where [s] denotes the largest integer smaller than s.Leta ∈ (0, ∞]andJ =[0,a].

We set

(J; X):=

⎧

⎪

⎨

⎪

⎩

{g ∈ W

(J; X):g(0) = g



(0) = ···= g

(k)

(0) = 0},

if k +

<s<k+1+

,k∈ N ∪{0},

(J; X), if s<

The spaces

(J; X) are deﬁned analogously. Here we remind that H

= W

for

k ∈ Z and 1 <p<∞,andthatW

= B

for s ∈ R \ Z.

For Ω ⊂ R

open and 1 ≤ p<∞, the homogeneous Sobolev spaces

(Ω) of

order 1 are deﬁned as

(Ω) := ({g ∈ L

1,loc

(Ω) : ∇g

(Ω)

< ∞}, ·

(Ω)

)

g

(Ω)



j=1

∂

g

(Ω)



1/p

(2.6)

Then

(Ω) is a Banach space, provided we factor out the constant functions

and equip the resulting space with the corresponding quotient norm, see for in-

stance [21, Lemma II.5.1]. We will in the sequel always consider the quotient space

topology without change of notation. In case that Ω is locally Lipschitz, it is known

that

(Ω) ⊂ H

p,loc

(Ω), see [21, Remark II.5.1], and consequently, any function

(Ω) has a well-deﬁned trace on ∂Ω.

For s ∈ R and 1 <p<∞ we also consider the homogeneous Bessel-potential

spaces

)oforders, deﬁned by

):=({g ∈S



g ∈ L

)}, ·

)

g

)

:= 

g

)

(2.7)

where S



) denotes the space of all tempered distributions, and

is the Riesz

potential given by

g := (−Δ)

s/2

g := F

−1

(|ξ|

Fg),g∈S



By factoring out all polynomials,

) becomes a Banach space with the natural

quotient norm. For s ∈ R\Z, the homogeneous Sobolev-Slobodeckij spaces

)

of fractional order can be obtained by real interpolation as

):=(

k+1

))

s−k,p

,k<s<k+1,

Two-phase Navier-Stokes Equations 513

where (·, ·)

θ,p

is the real interpolation method. It follows that

∈ Isom(

t+s

)) ∩ Isom(

t+s

)),s,t∈ R, (2.8)

with

for k ∈ Z. We refer to [6, Section 6.3] and [37, Section 5] for

more information on homogeneous functions spaces. In particular, it follows from

parts (ii) and (iii) in [37, Theorem 5.2.3.1] that the deﬁnitions (2.6) and (2.7) are

consistent if Ω = R

, s =1,and1<p<∞. We note in passing that





|g(x) −g(y)|

|x −y|

n+sp

dx dy



1/p





∞

(1−s)p



P (t)g

)



1/p

(2.9)

deﬁne equivalent norms on

)for0<s<1, where P (·) denotes the Poisson

semigroup, see [37, Theorem 5.2.3.2 and Remark 5.2.3.4]. Moreover,

∈L(

n+1

1−1/p

)), (2.10)

where γ

denotes the trace operators, see for instance [21, Theorem II.8.2].

3. The linearized two-phase Stokes problem with free boundary

It turns out that, unfortunately, the nonlinear term (γv|∇h) occurring in (2.1)

cannot be made small in the norm of F

(a), deﬁned below in (4.2), by merely taking

∇h

∞

small. This can, however, be achieved for the modiﬁed term (b − γv|∇h),

provided b is properly chosen so that b(0) = γv

. As a consequence, we now need

to consider the modiﬁed linear problem

⎧

⎪

⎨

⎪

⎩

ρ∂

u − μΔu + ∇π = f in

n+1

div u = f

n+1

−[[ μ∂

v]] − [[ μ∇

w]] = g

on R

−2[[μ∂

w]] + [[ π]] = g

+ σΔh +[[ρ]] γ

h on R

[[ u]] = 0 o n R

∂

h − γw +(b(t, x)|∇)h = g

on R

u(0) = u

,h(0) = h

(3.1)

Here we mention that the simpler case where b =0andγ

= 0 was studied in [28,

Theorem 5.1]. We obtain the following maximal regularity result.

Theorem 3.1. Let p>n+3 be ﬁxed, and assume that ρ

and μ

are positive

constants for j =1, 2,andsetJ =[0,a].Suppose

∈ R

∈ W

1−1/2p

(J; L

, R

)) ∩ L

(J; W

2−1/p

, R

)),

514 J. Pr¨uss and G. Simonett

and set b(·)=b

+ b

(·). Then the Stokes problem with free boundary (3.1) admits

auniquesolution(u, π, h) with regularity

u ∈ H

(J; L

n+1

, R

n+1

)) ∩ L

(J; H

(

n+1

, R

n+1

)),

π ∈ L

(J;

(

n+1

)),

[[ π]] ∈ W

1/2−1/2p

(J; L

)) ∩ L

(J; W

1−1/p

)),

h ∈ W

2−1/2p

(J; L

)) ∩ H

(J; W

2−1/p

)) ∩ L

(J; W

3−1/p

))

(3.2)

if and only if the data (f,f

,g,g

) satisfy the following regularity and com-

patibility conditions:

(a) f ∈ L

(J; L

n+1

, R

n+1

)),

(b) f

∈ H

(J;

−1

n+1

)) ∩ L

(J; H

(

n+1

)),

) ∈ W

1/2−1/2p

(J; L

, R

n+1

)) ∩ L

(J; W

1−1/p

, R

n+1

)),

(d) g

∈ W

1−1/2p

(J; L

)) ∩ L

(J; W

2−1/p

)),

(e) u

∈ W

2−2/p

(

n+1

, R

n+1

), h

∈ W

3−2/p

(f) div u

= f

(0) in

n+1

and [[ u

]] = 0 on R

if p>3/2,

(g) −[[ μ∂

]] − [[ μ∇

]] = g

(0) on R

if p>3.

The solution map [(f, f

,g,g

) → (u, π, h)] is continuous between the cor-

responding spaces.

If b

≡ 0 then the result is true for all p ∈ (1, ∞), p =3/2, 3.

Proof. (i) Since F

(a), deﬁned by

(a):=W

1−1/2p

(J; L

)) ∩ L

(J; W

2−1/p

)),

is a multiplication algebra for p>n+3, the operator [h → (b|∇)h] maps the space

(a):=W

2−1/2p

(J; L

)) ∩ H

(J; W

2−1/p

)) ∩ L

(J; W

3−1/p

))

continuously into F

(a) with bound |b

| + C

b



(a)

, see Lemma 5.5(a).

As in the proof of [28, Theorem 5.1] it suﬃces to consider the reduced problem

⎧

⎪

⎨

⎪

⎩

ρ∂

u − μΔu + ∇π =0 in

n+1

div u =0 in

n+1

−[[ μ∂

v]] − [[ μ∇

w]] = 0 o n R

−2[[μ∂

w]] + [[ π]] = σΔh +[[ρ]] γ

h on R

[[ u]] = 0 o n R

∂

h − γw +(b(t, x)|∇)h =˜g

on R

u(0) = 0,h(0) = 0,

(3.3)

where the function ˜g

∈

(a) is deﬁned in a similar way as in formula (5.5) in

[28]. This can be accomplished by choosing h

:= h

1,b

∈ E

(a) such that

(0) = h

,∂

(0) = g

(0) + γw

− (b(0)|∇h

Two-phase Navier-Stokes Equations 515

and then setting ˜g

:= ˜g

h,b

:= g

+ γw

−(b|∇h

) −∂

, where w

has the same

meaning as in step (i) of the proof of [28, Theorem 5.1].

(ii) We ﬁrst consider the reduced problem (3.3) for the case where b ≡ b

constant. The corresponding boundary symbol s

(λ, ξ)isgivenby

(λ, ξ)=λ +



σ|ξ|−[[ ρ]] γ

/|ξ|



k(z)+i(b

|ξ), (3.4)

where we use the same notation as in the proof of [28, Theorem 5.1]. Here we

remind that k has the following properties: k is holomorphic in C \ R

−

and

k(0) =

2(μ

+ μ

)

,zk(z) →

+ ρ

for |z|→∞, (3.5)

uniformly in z ∈

for ϑ ∈ [0,π) ﬁxed. In particular there is a constant N = N(ϑ)

such that

|k(z)| + |zk(z)|≤N, z ∈ Σ

. (3.6)

In the following we ﬁx β>0. For further analysis it will be convenient to introduce

the related extended symbol

˜s(λ, τ, ζ):=λ + στk(z)+iτζ − [[ ρ]] γ

k(z)/τ, (3.7)

where (λ, τ) ∈ Σ

π/2+η

× Σ

with η suﬃciently small, z := λ/τ

,andζ ∈ U

β,δ

with U

β,δ

:= {ζ ∈ C : |Re ζ| <β+1, |Im ζ| <δ} and δ ∈ (0, 1]. Clearly

˜s(λ, |ξ|, (b

|ξ/|ξ|)) = s

(λ, ξ)for(λ, ξ) ∈ Σ

We are going to show that for every ﬁxed β>0 there are positive constants λ

δ, η = η(β), and c

= c

(β,λ

,δ,η) such that

|λ| + |τ|

≤|˜s(λ, τ, ζ)|≤c

|λ| + |τ|

, (3.8)

for all (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

with |λ|≥λ

. The upper estimate is easy

to obtain: ﬁxing ϑ ∈ (π/2,π)andλ

> 0, it follows from (3.6) and the identity

k(z)/τ = zk(z)τ/λ that

|˜s(λ, τ, ζ)|≤|λ| +



σN +(β +2)+|[[ ρ]] |γ

N/λ



|τ|≤c

|λ|+ |τ|

(3.9)

for all (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

,where|λ|≥λ

and η ∈ (0,η

)withη

(ϑ − π/2)/3.

In order to obtain a lower estimate we proceed as follows. Suppose ﬁrst that

β,λ

> 0areﬁxedandη

is as above. Then we obtain

|˜s(λ, τ, ζ)|≥|λ|−



σN +(β +2)+|[[ ρ]] |γ

N/λ



|τ|

≥ (1/2)|λ|+(m/4)|τ| = c

(β,λ

)[|λ| + |τ|],

(3.10)

provided (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

, η ∈ (0,η

), and |λ|≥λ

as well as

|λ|≥m|τ| with

(m/4) ≥ σN +(β +2)+|[[ ρ]] |γ

N/λ

Next we will derive an estimate from below in case that |λ|≤M|τ|

with M apos-

itive constant. From (3.5) follows that there are constants H, L, R > 0, depending

on M, such that

L ≤ Re (σk(z)) ≤ R, |Im (σk(z))|≤H, (3.11)

516 J. Pr¨uss and G. Simonett

whenever (λ, τ ) ∈ Σ

π/2+η

× Σ

,forη ∈ (0,η

)and|λ|≤M|τ|

,wherez = λ/τ

By choosing δ small enough we obtain from (3.11) and the deﬁnition of U

β,δ

0 <L− δ ≤ Re (σk(z)+iζ) ≤ R + δ, |Im (σk(z)+iζ)|≤H +(β +1)

provided (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

, η ∈ (0,η

)and|λ|≤M|τ|

,where

z = λ/τ

.Bychoosingη small enough we conclude that there is α = α(M, β, δ, η) ∈

(0,π/2) such that

τ(σk(z)+iζ) ∈ Σ

(3.12)

whenever (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

and |z|≤M with z = λ/τ

.Wecan

additionally assume that η is chosen so that ψ := π/2 −α −η>0. This implies

|˜s(λ, τ, ζ)|≥c(ψ)

|λ| + |τ||σk(z)+iζ|

−|τ||[[ ρ]] |γ

N/λ

≥ c(ψ)min(1,L− δ)

|λ|+ |τ|

−|τ||[[ ρ]] |γ

N/λ

≥ c

(M,β,λ

)

|λ| + |τ|

(3.13)

provided (λ, τ, ζ) ∈ Σ

π/2+η

× Σ

× U

β,δ

, |λ|≤M|τ|

and |λ|≥λ

,whereλ

chosen big enough.

Noting that the curves |λ| = m|τ| and |λ| = M |τ|

intersect at (m/M, m

/M )

we obtain (3.8) by choosing λ

:= max(λ

/M ).

(iii) In the following, we ﬁx β>0 and we assume that b

∈ R

with |b

|≤β.

Let then S

be the operator corresponding to the symbol s

. It is clear that

is bounded from

(a)to

(a)=:X and it remains to prove that it is

boundedly invertible. For this we use the H

∞

-calculus and similar arguments as

in [27, Section 4] and [28, Section 5]. First we note that D

admits an R-bounded

∞

-calculus in X with angle 0; this follows from [14, Theorem 4.11]. Therefore

by the estimates obtained in (3.8), the operator family

{(λ + D

1/2

)˜s

−1

(λ, D

1/2

,ζ):(λ, ζ) ∈ Σ

π/2+η

× U

β,δ

, |λ|≥λ

}

is R-bounded. Since G = ∂

is in H

∞

(X) with angle π/2, the theorem of Kalton

and Weis [22, Theorem 4.4] implies that the operator family

{(G + D

1/2

)˜s

−1

(G, D

1/2

,ζ):ζ ∈ U

β,δ

}

is bounded and holomorphic on U

β,δ

. Finally, we employ the Dunford calculus for

the bounded linear operator R

:= (b

|R), where R denotes the Riesz operator

with symbol ξ/|ξ|, ξ ∈

. The operator R

is bounded and its spectrum is

σ(R

)=[−|b

|, |b

|], as, e.g., the Mikhlin theorem shows. Since the operator

family

{(G + D

1/2

)˜s

−1

(G, D

1/2

,ζ):ζ ∈ U

β,δ

}

is bounded and holomorphic in a neighborhood of σ(R

), the classical Dunford

calculus shows that the operator

(G + D

1/2

)˜s

−1

(G, D

1/2

)

is bounded in X, uniformly for all b

∈ R

with |b

|≤β. This shows that S

(a) →

(a) is boundedly invertible, uniformly for all b

∈ R

with |b

|≤β.

Two-phase Navier-Stokes Equations 517

We emphasize that the bound for the operator S

−1

(a) →

(a)de-

pends only on the parameters ρ

, μ

, σ, γ

, p and β,for|b

|≤β.

(iv) By means of a perturbation argument the result for constant b can be extended

to variable b = b

+ b

(t, x). In fact, given β>0thereexistsanumberη>0such

that the solution operator S

−1

exists and is bounded uniformly, provided |b

|≤β

and b



∞

+ b



(a)

≤ 2η. This follows easily from the estimate

(b

|∇h)

(a)

≤ c

(b



∞

+ b



(a)

)h

(a)

see Lemma 5.5(c).

(v) In the general case we use a localization technique, similar to [3, Section 9].

For this purpose we ﬁrst decompose J into subintervals J

=[kδ,(k +1)δ]of

length δ>0 and solve the problem successively on these subintervals. Since b ∈

BUC(J; C

, R

)), i.e., b is bounded and uniformly continuous, given any η>0

we may choose δ>0andε>0 such that

|b(t, x) − b(s, y)|≤η for all (t, x), (s, y) ∈ J ×R

with |t−s|≤δ and |x−y|

∞

≤ ε.Let{U

:= x

+(ε/2)Q : j ∈ N} be an enumeration

of the open covering {(ε/2)(z/2+Q):z ∈ Z

} of R

,whereQ =(−1, 1)

. Clearly,

|b(t, x) − b(s, y)|≤η, s,t ∈ J

,x,y∈ U

. (3.14)

Let φ be a smooth cut-oﬀ function with support contained in (ε/2)Q such that

φ ≡ 1on(ε/4)Q. Deﬁne

:= (τ

φ)



k∈N

(τ

φ)



−1/2

,j∈ N,

where (τ

φ)(x):=φ(x − x

). Consequently, φ

is a smooth cut-oﬀ function with

supp(φ

) ⊂ U

and

≡ 1. For a function space F(J; R

) ⊂ L

(J; L

)) we

deﬁne

r(h

):=

, (h

) ∈ F(J; R

)

h := (φ

h),h∈ F(J; R

Similarly as in [3, Section 9] one shows that

r ∈L(

(F(J; R

)), F(J; R

)),r

∈L(F(J; R

),

(F(J; R

))),rr

= I, (3.15)

for F(J; R

) ∈{F

(a), E

(a)}.

Let θ be a smooth cut-oﬀ function with supp(θ) ⊂ (ε/2)Q such that θ ≡ 1

on supp(φ)andletθ

:= τ

θ. Deﬁne

j,k

(t, x):=θ

(x)(b(t, x) − b(kδ, x

)) , (t, x) ∈ J × R

It follows that

b

j,k



BC(J

×R

)

+ b

j,k



)

≤ c

η, k =0,...,m, j ∈ N, (3.16)

provided δ is chosen small enough, where BC stands for bounded and contin-

uous. Indeed, the estimates for b

j,k



BC(J

)

follow immediately from (3.14),