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498 T. Nau and J. Saal

Taking L

-norm with respect to t and applying the contraction principle of Kahane

therefore yields

j=1

(·)bD

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

≤ Cb

∞

j=1

(·)



+ δ



1−

|β|

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

Thanks to (4.8) this implies

j=1

(·)bD

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

≤ Cb

∞

−(1−

|β|

)

j=1

(·)f

([0,1],X)

Thus for δ

> (Cb

∞

/ε)

1/(1−|β|/2m)

the assertion follows.

In case that b ∈ L

n−k

), H¨older’s inequality and the Gagliardo-Nirenberg

inequality imply for τ(2m − ν)=

n−k

and arbitrary t ∈ [0, 1] that

j=1

(t)bD

(λ

+ δ

+ A

+ δ)

−1

≤b

j=1

(t)D

(λ

+ δ

+ A

+ δ)

−1



≤ Cb



|α|=2m

j=1

(t)D

(λ

+ δ

+ A

+ δ)

−1



τ/p

j=1

(t)(λ

+ δ

+ A

+ δ)

−1

1−τ

Taking L

-norm with respect to t and applying once more H¨older’s inequality we

deduce

j=1

(·)bD

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

≤ Cb



|α|=2m

j=1

(·)D

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)



τ/p

j=1

(·)(λ

+ δ

+ A

+ δ)

−1

1−τ

([0,1],X)

R-sectoriality of Cylindrical Boundary Value Problems 499

The contraction principle of Kahane then gives us

j=1

(·)bD

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

≤ Cb



|α|=2m

j=1

(·)D

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)



τ/p

j=1

(·)

+ δ

(λ

+ δ

+ A

+ δ)

−1

1−τ

([0,1],X)

Taking into account (4.8) we arrive at

j=1

(·)bD

(λ

+ δ

+ A

+ δ)

−1

([0,1],X)

≤ Cb

τ−1

j=1

(·)f

([0,1],X)

Choosing δ

> (Cb

/ε)

1/(1−τ)

proves the lemma. 

Proof of Theorem 2.3. We denote by

(x, D):=

|α|=2m

(x)D

the principal part of A

(x, D)andbyA

its realization in X with domain D(A

2m,p

n−k

,E). Recall that A

(x, D)=A

,D) does not depend on x

∈ V .

Freezing the coeﬃcients at some arbitrary x

∈ R

n−k

∪{∞}, Proposition 4.5

applies to A

(D):=A

,D).

So, we ﬁrst choose a large ball B

(0) ⊂ R

n−k

with a ﬁxed radius r

> 0such

that

) − a

(∞)|≤η/M

, for all |x

|≥r

, |α

| =2m,

and set U

:= R

n−k

(0). Here M

{α

∈ N

n−k

; |α

| =2m, a

=0}

and η = η(∞) is the constant given in Corollary 4.7 for the principal part of

the ’limiting operator’ A

(∞,D)=

|α|=2m

(∞)D

. For every x

∈ B

(0)

let η = η(x

) be the constant given in Corollary 4.7 for the ’frozen coeﬃcients

operator’ A

(D):=A

,D). By our continuity assumptions on the coeﬃcients

then there exists a radius r = r(x

) such that

) −a

)|≤η(x

)/M

, for all |x

−x

|≤r(x

), |α

| =2m.

500 T. Nau and J. Saal

Obviously the collection {B

r(x

)

)) : x

∈ B

(0)} represents an open covering

of B

(0). Thus, by compactness we have

(0) ⊆

;

j=1

r(x

)

for a certain ﬁnite set (x

)

j=1

For simplicity we set x

:= (x

, 0), r

:= r(x

), and U

:= B

)forj =

1,...,N,aswellasx

:= ∞.Foreachj =0,...,N we deﬁne coeﬃcients of

(x, D)-localizations

1,loc

(x, D):=

|α|=2m

j,α

(x)D

by reﬂection, i.e., we set

0,α

(x)=



(x) ,x

/∈ B

(0),

(

|x|

x),x

∈ B

(0),

and

j,α

(x)=



(x) ,x

∈ B

|x−x

(x − x

)),x

/∈ B

Then by deﬁnition we have

|α

|=2m

j,α

(x) − a

)|≤η(x

)

for x =(x

, 0) ∈ R

n−k

× R

and j =0,...,N,thatis,A

1,loc

(x, D)isasmall

variation of A

,D):=A

,D)+A

(x, D). Hence Corollary 4.7 applies to

loc

:= A

1,loc

+ A

In other words, for each φ>ϕ

(A,B)

there exists δ = δ(φ) > 0 such that A

loc

+ δ ∈

RS(X) and we have

R({λ



(λ + A

loc

+ δ)

−1

; λ ∈ Σ

π−φ

, 0 ≤  + |β| + |γ|≤2m}) ≤ C

< ∞

(4.9)

for j =0,...,N.

Next we choose a partition of unity (ϕ

)

j=0

⊂ C

∞

n−k

)ofR

n−k

sub-

ordinate to the open covering (U

)

j=0

such that 0 ≤ ϕ

≤ 1. In addition, we

ﬁx ψ

∈ C

∞

n−k

) such that ψ

≡ 1 on supp ϕ

and supp ψ

⊂ U

.Weset

B(x, D):=A(x, D) − A

(x, D) and pick λ ∈ Σ

π−φ

.Then

λu + A(·,D)u = f

holds if and only if

λu + A

(·,D)u = f −B(·,D)u.

R-sectoriality of Cylindrical Boundary Value Problems 501

Multiplying the line above by ϕ

we obtain

λϕ

u + A

(·,D)ϕ

u = ϕ

f +[A

(·,D),ϕ

]u −ϕ

B(·,D)u,

where the commutators

(·,D),ϕ

]:=A

(·,D)ϕ

− ϕ

(·,D)=[A

(·,D),ϕ

]

do only depend on A

(·,D). Applying the resolvent of A

loc

to the localized equa-

tions we deduce

u =(λ + A

loc

+ δ)

−1

f +(λ + A

loc

+ δ)

−1

([A

(·,D),ϕ

]u − ϕ

B(·,D)u).

By multiplying with ψ

and by summing up over j we gain the representation

u =

j=0

(λ+A

loc

+δ)

−1

f +

j=0

(λ+A

loc

+δ)

−1

([A

(·,D),ϕ

]u−ϕ

B(·,D))u.

Hence we obtain

(I −

j=0

(λ + A

loc

+ δ)

−1

(·,D))u =

j=0

(λ + A

loc

+ δ)

−1

where

(·,D):=[A

(·,D),ϕ

] − ϕ

B(·,D)

is a diﬀerential operator in X of lower order whose coeﬃcients fulﬁll the assump-

tions of Lemma 4.8. We set

(λ, δ):=

j=0

(λ + A

loc

+ δ)

−1

(4.10)

and

(λ, δ):=

j=0

(λ + A

loc

+ δ)

−1

(·,D).

Relation (4.9) and Lemma 4.8(a) now imply that

R

(λ, δ



+ δ

)u

2m,p

(Ω,F )

+ δ

R

(λ, δ



+ δ

)u

≤ C



R

(λ + δ

,δ



)u

2m,p

(Ω,F )

+ |λ + δ

|R

(λ + δ

,δ



)u



≤ CC

(·,D)u

≤ C



εu

2m,p

n−k

,E)

+ C(ε)u



≤



u

2m,p

n−k

,E)

+ δ

u



≤



u

2m,p

(Ω,F )

+ δ

u



(λ ∈ Σ

π−φ

)

for some δ



> 0 and provided that δ

> 0 is suﬃciently large. Setting δ := δ



+ δ

we see that then

:= (I −R

(λ, δ))

−1

(λ, δ):L

n−k

,E) → D(A)

502 T. Nau and J. Saal

is a left inverse of λ + A + δ which admits an estimate

|λ|L

f

≤ Cf

(λ ∈ Σ

π−φ

Thus, if we can prove that there exists a right inverse as well, we obtain A + δ ∈

S(X)andφ

A+δ

≤ φ.

To this end, let f ∈ X be arbitrary. Then

(λ + A(·,D)+δ)R

(λ, δ)f =(λ + A

(·,D)+δ)R

(λ, δ)f + B(·,D)R

(λ, δ)f

=(λ + A

(·,D)+δ)

j=0

(λ + A

loc

+ δ)

−1

+ B(·,D)

j=0

(λ + A

loc

+ δ)

−1

j=0

(λ + A

(·,D)+δ)(λ + A

loc

+ δ)

−1

j=0

D(·,D)(λ + A

loc

+ δ)

−1

where

D(·,D):=[A

(·,D),ψ

]+B(·,D)ψ

is again a diﬀerential operator in X of lower order whose coeﬃcients fulﬁll the

assumptions of Lemma 4.8. Since supp ψ

⊂ U

and ψ ≡ 1 on supp ϕ

,weobtain

(λ + A(·,D)+δ)R

(λ, δ)f = f + R

(λ, δ)f

with

(λ, δ):=

j=0

D(·,D)(λ + A

loc

+ δ)

−1

Lemma 4.8(b) implies R

(λ, δ)

L(X)

≤ 1/2 for large enough δ>0. Consequently,

:= R

(λ, δ)(I + R

(λ, δ))

−1

is a right inverse of λ + A + δ.

With the help of the Leibniz rule and the contraction principle of Kahane,

from representation (4.10) and relation (4.9) we obtain that

R({λ



(λ, δ)}) ≤ C(N +1).

In view of Lemma 4.8(b) and Lemma 3.5 the representation

(λ + A + δ)

−1

= R

(λ, δ)

∞

i=0

(λ, δ)

R-sectoriality of Cylindrical Boundary Value Problems 503

as a Neumann series ﬁnally gives us

R({λ



(λ + A + δ)

−1

; λ ∈ Σ

π−φ

, 0 ≤  + |β| + |γ|≤2m})

≤R({λ



(λ, δ)})R({

∞

i=0

(λ, δ)

})

≤ (N +1)C

∞

i=0

(N +1)

(Cε)

(N +1)C

1 −(N +1)Cε

< ∞.

Hence the proof of Theorem 2.3 is complete. 

5. Mixed orders

All parts of the proof can easily be adjusted to the situation when the diﬀerential

operators A

(·,D)andA

(·,D) have diﬀerent orders, say 2m

and 2m

respec-

tively. Then a cylindrical boundary value problem is given as

λu + A(x, D)u = f in Ω,

(x, D)u =0 on∂Ω(j =1,...,m),

(5.1)

with

A(x, D)=A

,D)+A

,D)

|α

|≤2m

(α

,0)

|α

|≤2m

(0,α

)

and

(x, D)=B

2,j

,D)

|β

|≤m

2,j

j,β

(0,β

)

2,j

< 2m

,j=1,...,m

However, then the notion of parameter-ellipticity for the entire cylindrical bound-

ary value problem is no longer appropriate. Instead we assume the diﬀerential

operator A

(·,D) to be parameter-elliptic in R

n−k

as well as the boundary value

problem

λu + A

(x, D)u = f in V,

2,j

(x, D)u =0 on∂V (j =1,...,m),

(5.2)

to be parameter-elliptic in the cross-section V of Ω with a joint angle of parameter-

ellipticity ϕ ∈ [0,π/2). The exact same proof as the one of Theorem 2.3 can be

used to show the following result.

Theorem 5.1. Given the assumptions of Theorem 2.3,letA

(·,D) in R

n−k

as well

as the boundary value problem (5.2) in V be parameter-elliptic with a joint angle

of parameter-ellipticity ϕ ∈ [0,π/2).ForΩ=R

n−k

× V we deﬁne the L

(Ω,F)-

504 T. Nau and J. Saal

realization of the cylindrical boundary value problem (5.1) by

D(A)=



u ∈ L

(Ω,F); D

u ∈ L

(Ω,F)

for

|α

≤ 1 and B

(·,D)u =0 (j =1,...,m)



Au = A(·,D)u, u ∈ D(A).

Then for each φ>ϕthere exists δ = δ(φ) > 0 such that A + δ ∈RS(L

(Ω,F))

with φ

A+δ

≤ φ. Moreover, for α =(α

,α

) ∈ N

n−k

× N

we have



1−(

|α

)

(λ + A + δ)

−1

; λ ∈ Σ

π−φ

, 0 ≤

|α

≤ 1



< ∞.

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Tobias Nau

University of Konstanz

Department of Mathematics and Statistics

Box D 187

D-78457 Konstanz, Germany

e-mail: tobias.nau@uni-konstanz.de

J¨urgen Saal

Technische Universit¨at Darmstadt

Center of Smart Interfaces

Petersenstraße 32

D-64287 Darmstadt, Germany

e-mail: saal@csi.tu-darmstadt.de

Progress in Nonlinear Diﬀerential Equations

and Their Applications, Vol. 80, 507–540

 2011 Springer Basel AG

Analytic Solutions for the

Two-phase Navier-Stokes Equations

with Surface Tension and Gravity

Jan Pr¨uss and Gieri Simonett

Dedicated to Herbert Amann on the oc casion of his 70th birthday

Abstract. We consider the motion of two superposed immiscible, viscous, in-

compressible, capillary ﬂuids that are separated by a sharp interface which

needs to be determined as part of the problem. Allowing for gravity to act

on the ﬂuids, we prove local well-posedness of the problem. In particular,

we obtain well-posedness for the case where the heavy ﬂuid lies on top of the

light one, that is, for the case where the Rayleigh-Taylor instability is present.

Additionally we show that solutions become real analytic instantaneously.

Mathematics Subject Classiﬁcation (2000). Primary: 35R35. Secondary: 35Q10,

76D03, 76D45, 76T05.

Keywords. Navier-Stokes equations, free boundary problem, surface tension,

gravity, Rayleigh-Taylor instability, well-posedness, analyticity.

1. Introduction and main results

We consider a free boundary problem describing the motion of two immiscible,

viscous, incompressible capillary ﬂuids, ﬂuid

and ﬂuid

, occupying the regions

(t)={(x, y) ∈ R

×R :(−1)

(y −h(t, x)) > 0,t≥ 0},i=1, 2.

The ﬂuids, thus, are separated by the interface

Γ(t):={(x, y) ∈ R

×R : y = h(t, x):x ∈ R

,t≥ 0},

called the free boundary, which needs to be determined as part of the problem.

The motion of the ﬂuids is governed by the incompressible Navier-Stokes equations

where surface tension on the free boundary is included. In addition, we also allow

The research of GS was partially supported by NSF, Grant DMS-0600870.