Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift
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518 J. Pr¨uss and G. Simonett
while the estimates for b
j,k
F
4
(J
k
)
can be shown by approximating b by func-
tions that have better time regularity and by carefully estimating the products
θ
j
(b − b(kδ,x
j
))
F
4
(J
k
)
.
We now concentrate on the first interval J
0
=[0,δ]. Let L ∈L(
0
E
4
(a),
0
F
4
(a))
denote the operator with symbol στk(z), i.e.,
L := (σD
1/2
n
−[[ ρ]] γ
a
D
−1/2
n
)k(GD
−1
n
):=L
1
+ L
2
.
If follows from (3.16) and step (iii) that the operator
S
j
:= G + L +(b(0,x
j
)+b
j,0
|∇):
0
E
4
(δ) →
0
F
4
(δ)
is invertible. Moreover, there is a constant C
0
, depending only on sup
j
|b(0,x
j
)| –
and therefore only on b
BC(J ×R
n
)
– such that S
−1
j
L(
0
F
4
(δ),
0
E
4
(δ))
≤ C
0
,j∈ N.
(vi) Suppose that for a given g ∈
0
F
4
(δ)wehaveasolutionh ∈
0
E
4
(δ)of
Gh + Lh +(b|∇)h = g.
Multiplying this equation by φ
j
,usingthatb∂
α
φ
j
=(b(0,x
j
)+b
j,0
) ∂
α
φ
j
and
rr
c
= I this yields
S
j
φ
j
h − [L, φ
j
]h − (b|∇φ
j
)h =(S
j
−[L, φ
j
]r − (b|∇φ
j
)r) r
c
h = r
c
g,
where [·, ·] denotes the commutator. We now interpret this equation as an equation
in
p
(
0
F
4
(δ)). It follows from step (iv) that (S
j
) ∈ Isom(
p
(
0
E
4
(δ)),
p
(
0
F
4
(δ))) and
(S
−1
j
)
L(
p
(
0
F
4
(δ)),
p
(
0
E
4
(δ)))
≤ C
0
. (3.17)
We shall show below in step (vi) that the commutators satisfy
([L, φ
j
]+(b|∇φ
j
)) ∈L(
0
F(a),
p
(
0
F
4
(a))). (3.18)
Assuming this property, it follows from (3.15) that
(([L, φ
j
]+(b|∇φ
j
))r(h
j
))
p
(
0
F
4
(δ))
≤ C(h
j
)
p
(
0
F
4
(δ))
≤ Cδ
α
(h
j
)
p
(
0
E
4
(δ))
for some α depending only on p and n. Therefore, choosing δ small enough we can
conclude that (S
j
−([L, φ
j
]+(b|∇φ
j
))r) ∈ Isom(
p
(
0
E
4
(δ)),
p
(
0
F
4
(δ))) with
(S
j
−([L, φ
j
]+(b|∇φ
j
))r)
−1
≤2C
0
.
Let T
b
:= r (S
j
− ([L, φ
j
]+(b|∇φ
j
))r)
−1
r
c
.ThenT
b
∈L(
0
F
4
(δ),
0
E
4
(δ)) is a left
inverse of S
b
:= G + L +(b|∇). Hence
h
0
E
4
(δ)
= T
b
S
b
h
0
E
4
(δ)
≤ 2C
0
rr
c
S
b
h
0
E
4
(δ)
,h∈
0
E
4
(δ). (3.19)
Replacing b by ρb, ρ ∈ [0, 1], we have a continuous family {S
ρb
} of operators S
ρb
which all satisfy the a priori estimate (3.19) uniformly in ρ ∈ [0, 1]. Since S
0
is an
isomorphism, we can infer from a homotopy argument that S
b
is an isomorphism
as well. Repeating successively these arguments for the intervals J
k
, including the
reduction from step (i), proves the assertion of the corollary.
(vii) We still have to verify the estimate in (3.18). Since the covering {U
j
: j ∈ N}
has finite multiplicity, one obtains
((∂
α
φ
j
)g)
p
(
0
F
4
(a))
≤ C(α)g
0
F
4
(a)
,g∈
0
F
4
(a). (3.20)
Two-phase Navier-Stokes Equations 519
This together with Proposition 5.5(b) shows that
(b|∇φ
j
)h
p
(
0
F
4
(a)
≤ Cbh
0
F
4
(a)
≤ C
0
(b
∞
+ b
F
4
(a)
)h
0
F
4
(a)
.
The estimates for the commutators [L, φ
j
] are more involved. The operator A =
GD
−1
n
with canonical domain is sectorial and admits a bounded H
∞
-calculus with
angle π/2in
0
H
s
p
(J; K
r
p
(R
n
)), for K ∈{H, W },andalsoin
0
W
s
p
(J; K
r
p
(R
n
)) by
real interpolation. Hence fixing θ ∈ (0,π/2), the following resolvent estimate holds
in these spaces:
z(z − A)
−1
≤M, for all z ∈−Σ
θ
.
The function k(z) is holomorphic in C \(−∞, −2δ
0
]forsomeδ
0
> 0 and behaves
like 1/z as |z|→∞. Choose the contour
Γ=(∞,δ
0
]e
iψ
∪δ
0
e
i[ψ,2π−ψ]
∪ [δ
0
, ∞)e
−iψ
,
where π>ψ>π− θ. Then we have the Dunford integral
k(A)=
1
2πi
Γ
k(z)(z −A)
−1
dz,
which is absolutely convergent. This shows that k(A) is bounded, as is Ak(A)
thanks to A ∈H
∞
,thusA
1/2
k(A) is bounded as well. Therefore the identity
k(A)D
−1/2
n
= G
−1/2
A
1/2
k(A)showsthatL
2
is bounded since G
−1/2
is, and (3.18)
follows for [L
2
,φ
j
]. For the commutator [L
1
,φ
j
]weobtain
[L
1
,φ
j
]=σ[k(A)D
1/2
n
,φ
j
]=σ[k(A),φ
j
]D
1/2
n
+ σk(A)[D
1/2
n
,φ
j
].
Using the Dunford integral for k(A) this yields
[k(A),φ
j
]=
1
2πi
Γ
k(z)[(z−A)
−1
,φ
j
]dz =
1
2πi
Γ
k(z)(z−A)
−1
[A, φ
j
](z−A)
−1
dz,
hence with
[A, φ
j
]=GD
−1
n
[φ
j
,D
n
]D
−1
n
= A(Δφ
j
+2(∇φ
j
|∇))D
−1
n
= A(Δφ
j
D
−1
n
+2i(∇φ
j
|R)D
−1/2
n
),
we have
[k(A),φ
j
]D
1/2
n
=
1
2πi
Γ
k(z)A(z−A)
−1
{−Δφ
j
G
−1/2
A
1/2
+2i(∇φ
j
|R)}(z−A)
−1
dz.
Let h ∈
0
F
4
be given. Then we obtain from
k(z)A(z − A)
−1
L(
0
F
4
)
≤ C/|z|, A
1/2
(z − A)
−1
L(
0
F
4
)
≤ C/|z|
1/2
,z∈ Γ,
from (3.20), and from Minkowski’s inequality for integrals
#
#
#
Γ
k(z)A(z −A)
−1
Δφ
j
G
−1/2
A
1/2
(z − A)
−1
hdz
#
#
#
p
(
0
F
4
)
≤ C
Γ
1
|z|
(Δφ
j
G
−1/2
A
1/2
(z − A)
−1
h)
p
(
0
F
4
)
|dz|
≤ C
Γ
1
|z|
3/2
h
0
F
4
|dz|≤Ch
0
F
4
520 J. Pr¨uss and G. Simonett
wherewealsousedthatG
−1/2
is bounded on compact intervals. In the same way
we can estimate the second term in the integral representation of [k(A),φ
j
]D
1/2
,
this time using the fact that R is bounded.
To estimate the commutators [D
1/2
n
,φ
j
]in
0
F
4
note that
(D
n
)
1/2
= D
n
(D
n
)
−1/2
=
1
√
π
D
n
∞
0
e
−D
n
t
t
−
1
2
dt
=
1
√
π
D
n
1
0
e
−D
n
t
t
−
1
2
dt + D
n
∞
1
e
−D
n
t
t
−
1
2
dt
=:
1
√
π
(T
1
+ T
2
),
with e
−D
n
t
denoting the bounded analytic semigroup generated by the Laplacian
in H
s
p
(R
n
) which extends by real interpolation to W
s
p
(R
n
), and then canonically
to
0
F
4
. Thus by (3.20) there is a constant C>0 such that for h ∈
0
F
4
we have
(φ
j
T
2
h)
p
(
0
F
4
)
=
#
#
#
#
(φ
j
∞
1
D
n
e
−D
n
t
t
−
1
2
hdt)
#
#
#
#
p
(
0
F
4
)
≤ C
#
#
#
#
∞
1
D
n
e
−D
n
t
t
−
1
2
hdt
#
#
#
#
0
F
4
≤ C
∞
1
t
−
3
2
dt h
0
F
4
≤ Ch
0
F
4
,
(T
2
φ
j
h)
p
(
0
F
4
)
=
#
#
#
#
∞
1
D
n
e
−D
n
t
t
−
1
2
(φ
j
h) dt
#
#
#
#
p
(
0
F
4
)
≤ C(φ
j
h)
p
(
0
F
4
)
≤ Ch
0
F
4
.
Hence ([φ
j
,T
2
]h)
p
(
0
F
4
)
≤ Ch
0
F
4
.
We consider next the commutator [T
1
,φ
j
]. Let k
t
(x)=(2πt)
−n/2
exp(−|x|
2
/4t)
denote the Gaussian kernel. Then for fixed t>0, the operator D
n
e
−D
n
t
is the con-
volution with kernel −Δk
t
(x), which is of class C
∞
. It is not difficult to see that
there are constants C, c > 0 such that
|Δk
t
(x)|≤Ct
−(n+2)/2
e
−c|x|
2
/t
,x∈ R
n
,t>0. (3.21)
Choosing a cut-off function χ ∈ C
∞
(R
n
)withχ ≡ 1inB
ρ
(0), supp (χ) ⊂ B
2ρ
(0)
and 0 ≤ χ ≤ 1 elsewhere, we set
−Δk
t
(x)=−(1 −χ(x))Δk
t
(x) −χ(x)Δk
t
(x)=:k
3,t
(x)+k
4,t
(x),x∈ R
n
,t>0,
and we denote by T
l
the convolution operators with kernels
1
0
k
l,t
t
−1/2
dt, l =3, 4.
For the kernel of T
3
we obtain from (3.21) the estimate
|
1
0
k
3,t
(x)t
−1/2
dt|≤C
1
0
e
−c|x|
2
/t
t
−(n+3)/2
dt
≤ Ce
−c
1
|x|
2
∞
1
e
−c
2
|x|
2
s
s
(n−1)/2
ds ≤ Ce
−c
1
|x|
2
,
Two-phase Navier-Stokes Equations 521
as k
3,t
(x)=0for|x|≤ρ. Thus this kernel is in L
1
(R
n
) and hence we may estimate
the commutator [T
3
,φ
j
]inthesamewayas[T
2
,φ
j
].
For the remaining commutator [T
4
,φ
j
]notethat
∂
α
[T
4
,φ
j
]=
!
β≤α
α
β
[T
4
,∂
β
φ
j
]∂
α−β
.
This shows that it is enough to estimate the commutator [T
4
,φ
j
]inL
p
(R
n
), as it
then extends to H
m
p
(R
n
) and by interpolation to W
s
p
(R
n
), and then canonically
to
0
F
4
. Next we observe that for x, y ∈ R
n
∂
α
φ
j
(y) − ∂
α
φ
j
(x)=∂
α
φ
j
(x)(y − x)+r
j,α
(x, y),
where |r
j,α
(x, y)|≤C|x −y|
2
, with some constant C independent of j and |α|≤2.
Therefore
[T
4
,φ
j
]h(x)=
1
0
R
n
(φ
j
(y) − φ
j
(x))k
4,t
(x −y)h(y) dy t
−
1
2
dt
= −φ
j
(x)
R
n
1
0
(y −x)k
4,t
(x − y)t
−
1
2
dt h(y) dy+
+
R
n
1
0
r
j
(x, y)k
4,t
(x − y)t
−
1
2
dt h(y) dy
=: T
5,j
h(x)+T
6,j
h(x).
We observe that the support of the kernel k
4,t
is contained in B
2ρ
(0), and conse-
quently we may replace h by ψ
j
h,whereψ
j
is a cut-off function which equals 1
on supp(φ
j
)+B
2ρ
(0). In the following we fix a smooth cut-off function ψ which
equals 1 on supp(φ)+B
2ρ
(0) and then set ψ
j
:= τ
x
j
ψ.Wethenhave
(T
l,j
h)
p
(L
p
)
= (T
l,j
ψ
j
h)
p
(L
p
)
≤ sup
k
T
l,k
L(L
p
)
(ψ
j
h)
p
(L
p
)
≤ Ch
L
p
,
provided we can show that the operators T
l,k
are L
p
-bounded with bound inde-
pendent of k ∈ N for l =5, 6.
The operators T
l,j
satisfy
T
5,j
h = φ
j
(q ∗ h)withq(x)=χ(x)
1
0
xΔk
t
(x)t
−
1
2
dt, x ∈ R
n
|T
6,j
h|≤r ∗|h| with r(x)=Cχ(x)
1
0
|x|
2
|Δk
t
(x)|t
−
1
2
dt, x ∈ R
n
.
The Fourier transform of q is given by q(ξ)=C χ ∗
1
0
∇
ξ
(|ξ|
2
e
−t|ξ|
2
)t
−1/2
dt and
we verify that
sup
α≤(1,...,1)
sup
ξ∈R
n
|ξ|
|α|
|∂
α
q(ξ)|≤M
for some M<∞. It thus follows from Mikhlin’s multiplier theorem that
T
5,j
h
L
p
≤ Cφ
j
∞
h
L
p
≤ Ch
L
p
.
522 J. Pr¨uss and G. Simonett
Finally, in order to estimate T
6,j
we infer from (3.21) that
r(x) ≤ C
1
0
|x|
2
e
−c|x|
2
/t
t
−
n+3
2
dt ≤ Ce
−c
1
|x|
2
|x|
−(n−1)
∞
1
e
−c
2
s
s
(n−1)/2
ds
for x ∈ R
n
. It follows that r ∈ L
1
(R
n
) which implies by Young’s inequality
T
6,j
h
p
≤ Ch
p
with a uniform constant C.
Remarks 3.2. (a) We mention that the proof for the estimate of [D
1/2
n
,φ
j
] follows
the ideas of [15, Lemma 6.4].
(b) If ρ
2
≤ ρ
1
, i.e., the light fluid lies above the heavy one, then the estimate (3.8)
can be improved in the following sense: for every β>0andλ
0
> 0thereare
positive constants δ, η = η(β)andc
j
= c
j
(β,λ
0
,δ,η) such that
c
0
5
|λ| + |τ|
6
≤ ˜s(λ, τ, ζ) ≤ c
1
5
|λ|+ |τ|
6
(3.22)
for all (λ, τ, ζ) ∈ Σ
π/2+η
× Σ
η
× U
β,δ
and |λ|≥λ
0
. For this we observe that
estimates (3.9) and (3.10) certainly also hold in case that ρ
2
≤ ρ
1
. On the other
hand, given M>0 we conclude as in (3.11) that L ≤ Re ((ρ
1
− ρ
2
)γ
a
k(z)) ≤ R
and |Im ((ρ
1
− ρ
2
)γ
a
k(z))|≤H for |z|≤M , with appropriate positive constants
L, R, H. This shows that there exists α = α(M,η) ∈ (0,π/2) such that
(ρ
1
− ρ
2
)γ
a
k(z)/τ ∈ Σ
α
, (λ, τ ) ∈ Σ
π/2+η
× Σ
η
, |z|≤M (3.23)
with η ∈ (0,η
0
) chosen small enough, where we can assume that α coincides with
the angle in (3.12). Combining (3.12) and (3.23) yields
|˜s(λ, τ, ζ)|≥c(ψ)
5
|λ| + |τ(σk(z)+iζ)+(ρ
1
− ρ
2
)γ
a
k(z)/τ|
6
≥ c(ψ)c(α)
5
|λ|+ |τ(σk(z)+iζ)| + |(ρ
1
− ρ
2
)γ
a
k(z)/τ|
6
≥ c
0
(M,β,δ,η)
5
|λ| + |τ|]
provided (λ, τ, ζ) ∈ Σ
π/2+η
× Σ
η
× U
β,δ
and |λ|≤M|τ|
2
. Noting again that the
curves |λ| = m|τ| and |λ| = M |τ|
2
intersect at (m/M, m
2
/M ) we obtain (3.22) by
choosing M big enough.
(c) If ρ
2
≤ ρ
1
we can conclude from the lower estimate in (3.22) that the function
˜s does not have zeros in Σ
π/2
×R
+
×[−β,β]. This holds in particular true for the
symbol s(λ, τ):=˜s(λ, τ, 0), indicating that there are no instabilities in case that
the light fluid lies on top of the heavy one.
(d) If ρ
2
>ρ
1
then it is shown in [29] that the symbol s has for each τ ∈ (0,τ
∗
)with
τ
∗
:= ((ρ
2
− ρ
1
)γ
a
/σ)
1/2
a zero λ = λ(τ) > 0, pertinent to the Rayleigh-Taylor
instability.
(e) Further mapping properties of the boundary symbol s(λ, τ ):=˜s(λ, τ, 0) and the
associated operator S in case that γ
a
= 0 have been derived in [27]. In particular,
we have investigated the singularities and zeros of s, and we have studied the
mapping properties of S in case of low and high frequencies, respectively.
Two-phase Navier-Stokes Equations 523
4. The nonlinear problem
In this section we prove existence and uniqueness of solutions for the nonlinear
problem (2.1), and we show additionally that solutions immediately regularize
and are real analytic in space and time. In order to facilitate this task, we first
introduce some notation. We set
E
1
(a):={u ∈ H
1
p
(J; L
p
(R
n+1
, R
n+1
)) ∩ L
p
(J; H
2
p
(
˙
R
n+1
, R
n+1
)) : [[u]] = 0 },
E
2
(a):=L
p
(J;
˙
H
1
p
(
˙
R
n+1
)),
E
3
(a):=W
1/2−1/2p
p
(J; L
p
(R
n
)) ∩ L
p
(J; W
1−1/p
p
(R
n
)),
E
4
(a):=W
2−1/2p
p
(J; L
p
(R
n
)) ∩ H
1
p
(J; W
2−1/p
p
(R
n
))
∩W
1/2−1/2p
p
(J; H
2
p
(R
n
)) ∩L
p
(J; W
3−1/p
p
(R
n
)),
E(a):={(u, π, q, h) ∈ E
1
(a) × E
2
(a) × E
3
(a) × E
4
(a): [[π]] = q}.
(4.1)
The space E(a) is given the natural norm
(u, π, q, h)
E(a)
= u
E
1
(a)
+ π
E
2
(a)
+ q
E
3
(a)
+ h
E
4
(a)
which turns it into a Banach space. Moreover, we set
F
1
(a):=L
p
(J; L
p
(R
n+1
, R
n+1
)),
F
2
(a):=H
1
p
(J; H
−1
p
(R
n+1
)) ∩ L
p
(J; H
1
p
(
˙
R
n+1
)),
F
3
(a):=W
1/2−1/2p
p
(J; L
p
(R
n
, R
n+1
)) ∩ L
p
(J; W
1−1/p
p
(R
n
, R
n+1
)),
F
4
(a):=W
1−1/2p
p
(J; L
p
(R
n
)) ∩ L
p
(J; W
2−1/p
p
(R
n
)),
F(a):=F
1
(a) × F
2
(a) × F
3
(a) × F
4
(a).
(4.2)
The generic elements of F(a) are the functions (f, f
d
,g,g
h
).
Let b ∈ F
4
(a)
n
be a given function. Then we define the nonlinear mapping
N
b
(u, π, q, h):=
F (u, π, h),F
d
(u, h),G(u, q, h), (b −γv|∇h)
(4.3)
for (u, π, q, h) ∈ E(a), where, as before, u =(v, w), F =(F
v
,F
w
)andG =
(G
v
,G
w
). We will now study the mapping properties of N
b
and we will derive
estimates for the Fr´echet derivative of N
b
. In the following, the notion C
ω
means
real analytic.
Proposition 4.1. Suppose p>n+3and b ∈ F
4
(a)
n
.Then
N
b
∈ C
ω
(E(a) , F(a)),a>0. (4.4)
Let DN
b
(u, π, q, h) denote the Fr´echet derivative of N
b
at (u, π, q, h) ∈ E(a).Then
DN
b
(u, π, q, h) ∈L(
0
E(a),
0
F(a)), and for any number a
0
> 0 there is a positive
524 J. Pr¨uss and G. Simonett
constant M
0
= M
0
(a
0
,p) such that
DN
b
(u, π, q, h)
L(
0
E(a),
0
F(a))
≤ M
0
5
b − γv
BC(J;BC) ∩F
4
(a)
+ (u, π, q, h)
E(a)
6
+ M
0
5
∇h
BC(J ;BC
1
)
+ h
E
4
(a)
+ u
BC(J;BC)
u
E
1
(a)
6
+ M
0
5
P (∇h
BC(J ;BC)
)∇h
BC(J;BC)
+ Q
∇h
BC(J ;BC
1
)
, h
E
4
(a)
h
E
4
(a)
6
for all (u, π, q, h) ∈ E(a) and all a ∈ (0,a
0
]. Here, P and Q are fixed polynomials
with coefficients equal to one.
Proof. The proof of the proposition is relegated to the end of the appendix.
Given h
0
∈ W
3−2/p
p
(R
n
) we define
Θ
h
0
(x, y):=(x, y + h
0
(x)), (x, y) ∈ R
n
× R. (4.5)
Letting Ω
h
0
,i
:= {(x, y) ∈ R
n
×R :(−1)
i
(y−(h
0
(x)) > 0} and Ω
h
0
:= Ω
h
0
,1
∪Ω
h
0
,2
we obtain from Sobolev’s embedding theorem that
Θ
h
0
∈ Diff
2
(
˙
R
n+1
, Ω
h
0
) ∩ Diff
2
(R
n+1
−
, Ω
h
0
,1
) ∩Diff
2
(R
n+1
+
, Ω
h
0
,2
),
i.e., Θ
h
0
yields a C
2
-diffeomorphism between the indicated domains. The inverse
transformation obviously is given by Θ
−1
h
0
(x, y)=(x, y − h
0
(x)). It then follows
from the chain rule and the transformation rule for integrals that
Θ
∗
h
0
∈ Isom(H
k
p
(
˙
R
n+1
),H
k
p
(Ω
h
0
)), [Θ
∗
h
0
]
−1
=Θ
h
0
∗
,k=0, 1, 2,
whereweusethenotation
Θ
∗
h
0
u := u ◦ Θ
h
0
,u:Ω
h
0
→ R
m
,
Θ
h
0
∗
v := v ◦ Θ
−1
h
0
,v:
˙
R
n+1
→ R
m
.
We are now ready to prove our main result of this section.
Theorem 4.2. (Existence of solutions for the nonlinear problem (2.1)).
(a) For every β>0 there exists a constant η = η(β) > 0 such that for all initial
values
(u
0
,h
0
) ∈ W
2−2/p
p
(
˙
R
n+1
, R
n+1
) ×W
3−2/p
p
(R
n
) with [[ u
0
]] = 0 ,
satisfying the compatibility conditions
[[ μD(Θ
h
0
∗
u
0
)ν
0
− μ(ν
0
|D(Θ
h
0
∗
u
0
)ν
0
)ν
0
]] = 0 ,div(Θ
h
0
∗
u
0
)=0, (4.6)
and the smallness-boundedness condition
∇h
0
∞
≤ η, u
0
∞
≤ β, (4.7)
there is a number t
0
= t
0
(u
0
,h
0
) such that the nonlinear problem (2.1) admits
auniquesolution(u, π, [[ π]] ,h) ∈ E
1
(t
0
).
(b) The solution has the additional regularity properties
(u, π) ∈ C
ω
((0,t
0
) ×
˙
R
n+1
, R
n+2
), [[ π]] ,h ∈ C
ω
((0,t
0
) × R
n
). (4.8)
In particular, M =
:
t∈(0,t
0
)
{t}×Γ(t)
is a real analytic manifold.
Two-phase Navier-Stokes Equations 525
Proof. The proof of this result proceeds in a similar way as the proof of Theo-
rem 6.3 in [28].
For a given function b ∈ F
4
(a)
n
we consider the nonlinear problem
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
ρ∂
t
u −μΔu + ∇π = F (u, π, h)in
˙
R
n+1
div u = F
d
(u, h)in
˙
R
n+1
−[[ μ∂
y
v]] − [[ μ∇
x
w]] = G
v
(u, [[ π]] ,h)onR
n
−2[[μ∂
y
w]] + [[ π]] − σΔh = G
w
(u, h)onR
n
[[ u]] = 0 o n R
n
∂
t
h − γw +(b|∇h)=(b − γv|∇h)onR
n
u(0) = u
0
,h(0) = h
0
,
(4.9)
which clearly is equivalent to (2.1).
In order to economize our notation we set z := (u, π, q, h)for(u, π, q, h) ∈ E(a).
With this notation, the nonlinear problem (2.1) can be restated as
L
b
z = N
b
(z), (u(0),h(0)) = (u
0
,h
0
), (4.10)
where L
b
denotes the linear operator on the left-hand side of (4.9), and N
b
corre-
spondingly denotes the nonlinear mapping on the right-hand site of (4.9).
It is convenient to first introduce an auxiliary function z
∗
= z
∗
b
∈ E(a)which
resolves the compatibility conditions and the initial conditions in (4.10), and then
to solve the resulting reduced problem
L
b
z = N
b
(z + z
∗
) − L
b
z
∗
=: K
b
(z),z∈
0
E(a), (4.11)
by means of a fixed point argument.
(i) Suppose that (u
0
,h
0
) satisfies the (first) compatibility condition in (4.6), and
let
[[ π
0
]] : = θ
∗
h
0
{[[ μ(ν
0
|D(Θ
h
0
∗
u
0
)ν
0
)]] + σκ},
where θ
h
0
:= Θ
h
0
|
R
n+1
×{0}
.Hereweobservethatθ
∗
h
0
[[ ω]] = [[ Θ
∗
h
0
ω]] for any function
ω :Ω
h
0
→ R
m
which has one-sided limits. It is then clear from the definition in
(2.3)–(2.4) that the following compatibility conditions hold:
−[[ μ∂
y
v
0
]] − [[ μ∇
x
w
0
]] = G
v
(u
0
, [[ π
0
]] ,h
0
)onR
n
−2[[μ∂
y
w
0
]] + [[ π
0
]] − σΔh
0
= G
w
(u
0
,h
0
)onR
n
(4.12)
where, as before, u
0
=(v
0
,w
0
). Next we introduce special functions (0,f
∗
d
,g
∗
,g
∗
h
) ∈
F(a) which resolve the necessary compatibility conditions. First we set
c
∗
(t):=
R
+
e
−tD
n+1
E
+
(v
0
|∇h
0
)inR
n+1
+
,
R
−
e
−tD
n+1
E
−
(v
0
|∇h
0
)inR
n+1
−
,
(4.13)
where E
±
∈L(W
2−2/p
p
(R
n+1
±
),W
2−2/p
p
(R
n+1
)) is an appropriate extension oper-
ator and R
±
is the restriction operator. Due to (v
0
|∇h
0
) ∈ W
2−2/p
p
(
˙
R
n+1
)we
526 J. Pr¨uss and G. Simonett
obtain
c
∗
∈ H
1
p
(J; L
p
(R
n+1
)) ∩ L
p
(J; H
2
p
(
˙
R
n+1
)).
Consequently,
f
∗
d
:= ∂
y
c
∗
∈ F
2
(a)andf
∗
d
(0) = F
d
(v
0
,h
0
). (4.14)
Next we set
g
∗
(t):=e
−D
n
t
G(u
0
, [[ π
0
]] ,h
0
).g
∗
h
(t):=e
−D
n
t
(b(0) − γv
0
|∇h
0
).
(4.15)
It then follows from (4.14) and [19, Lemma 8.2] that (0,f
∗
d
,g
∗
,g
∗
h
) ∈ F(a). (4.12)
and the second condition in (4.6) show that the necessary compatibility conditions
of Theorem 3.1 are satisfied and we can conclude that the linear problem
L
b
z
∗
=(0,f
∗
d
,g
∗
,g
∗
h
), (u
∗
(0),h
∗
(0)) = (u
0
,h
0
), (4.16)
has a unique solution z
∗
= z
∗
b
∈ E(a). With the auxiliary function z
∗
now deter-
mined, we can focus on the reduced equation (4.11), which can be converted into
the fixed point equation
z = L
−1
b
K
b
(z),z∈
0
E(a). (4.17)
Due to the choice of (f
∗
d
,g
∗
,g
∗
h
)wehaveK
b
(z) ∈
0
F(a) for any z ∈
0
E(a), and it
follows from Proposition 4.1 that
K
b
∈ C
ω
(
0
E(a),
0
F(a)).
Consequently, L
−1
b
K
b
:
0
E(a) →
0
E(a) is well defined and smooth.
(ii) An inspection of the proof of Theorem 3.1 shows that given β>0 we can find
a positive number δ
0
= δ
0
(b) such that
L
−1
b
∈L(
0
F(a),
0
E(a)), L
−1
b
L(
0
F(a),
0
E(a))
≤ M, a ∈ [0,δ
0
], (4.18)
whenever b ∈ F
4
(a)
n
and b
BC[0,a],BC(R
n
))
≤ β. It should be pointed out that
the bound M is universal for all functions b ∈ F
4
(a)
n
with b
∞
≤ β,whereasthe
number δ
0
= δ(b) may depend on b.
(iii) We will now fix a pair of initial values (u
0
,h
0
) ∈ W
2−2/p
p
(
˙
R
n+1
)×W
3−2/p
p
(R
n
)
satisfying (4.6) and (4.7) with
η := 1/(16M
0
M), (4.19)
where the constants M
0
and M are given in (4.1) and (4.18), respectively. We
choose
b(t):=e
−D
n
t
γv
0
,t≥ 0. (4.20)
Then b ∈ F
4
(a)
n
and b
BC([0,a];BC(R
n
))
≤γv
0
BC(R
n
)
≤ β for any a>0, as
{e
−D
n
t
: t ≥ 0} is a contraction semigroup on BUC(R
n
). Hence the estimate
(4.18) holds true for this (and any other choice) of initial values. It should be
pointed out once more that the bound M is universal for all initial values u
0
with
v
0
∞
≤ β – and hence for b(t):=e
−D
n
t
γv
0
– whereas the number δ
0
may depend
on γv
0
.
Two-phase Navier-Stokes Equations 527
We note in passing that g
∗
h
= 0 for this particular choice of the function b.
Without loss of generality we can assume that M
0
,M ≥ 1. We shall show that
L
−1
b
K
b
is a contraction on a properly defined subset of
0
E(a)fora ∈ (0,δ
0
]chosen
sufficiently small. For r>0anda ∈ (0,δ
0
]weset
0
B
E(a)
(z
∗
,r):={z ∈ E
1
(a):z − z
∗
∈
0
E(a), z − z
∗
E(a)
<r}.
We remark that a and r are independent parameters that can be chosen as we
please. Let then r
0
> 0 be fixed. It is not difficult to see that there exists a number
R
0
= R
0
(u
0
,h
0
,δ
0
,r
0
) such that
∇(h + h
∗
)
BC(J;BC
1
)
+ h + h
∗
E
4
(a)
+ u + u
∗
BC(J;BC)
+ Q
∇(h + h
∗
)
BC(J;BC
1
)
, h + h
∗
E
4
(a)
≤ R
0
for all u ∈
0
B
E
1
(a)
(0,r)andh ∈
0
B
E
4
(a)
(0,r), with a ∈ (0,δ
0
]andr ∈ (0,r
0
]
arbitrary, where z
∗
=(u
∗
,π
∗
,q
∗
,h
∗
) is the solution of equation (4.16) and where
Q is defined in Proposition 4.1. Let M
1
:= M
0
(1 + R
0
). It then follows from
Proposition 4.1 and (4.18) that
D(L
−1
b
K
b
)(z)
0
E(a)
≤ M
1
M
5
b − γ(v + v
∗
)
BC(J ;BC) ∩ F
4
(a)
+ z + z
∗
E(a)
6
+ M
0
M
5
P
∇(h + h
∗
)
BC(J ;BC)
∇(h + h
∗
)
BC(J ;BC)
6
(4.21)
for all z ∈
0
B
E(a)
(0,r)anda ∈ (0,δ
0
].
(iv) For (u
0
,h
0
)fixed,thenormofz
∗
in E(a) (which involves various integral
expressions evaluated over the interval (0,a)) can be made as small as we like by
choosing a ∈ (0,δ
0
] small. Let then a
1
∈ (0,δ
0
]befixedsothat
∇h
∗
BC([0,a
1
],BC)
≤ 2η,
M
1
M
b − γv
∗
BC([0,a
1
];BC) ∩ F
4
(a
1
)
+ z
∗
E(a
1
)
≤ 1/8.
(4.22)
Since (∇h
∗
,b−γv
∗
) ∈
0
BC([0,δ
0
],BC(R
n
)) and ∇h
∗
(0)
∞
= ∇h
0
∞
≤ η,the
estimates in (4.22) certainly hold for a
1
sufficiently small.
In a next step we choose 2r
1
∈ (0,r
0
]sothat
∇h
0
BC([0,a
1
],BC(R
n
)
≤ η,
M
1
M
γv
0
BC([0,a
1
];BC) ∩
0
F
4
(a
1
)
+ z
0
E(a
1
)
≤ 1/8,
(4.23)
for all h ∈
0
B
E(a
1
)
(0, 2r
1
), v ∈
0
B
E(a
1
)
(0, 2r
1
), and z ∈
0
B
E(a
1
)
(0, 2r
1
). It follows
from Proposition 5.1 that (4.23) can indeed be achieved. Combining (4.19)–(4.23)
gives
D(L
−1
b
K
b
)(z)
0
E(a)
≤ 1/2,z∈
0
B
E(a
1
)
(0, 2r
1
) (4.24)
showing that L
−1
b
K
b
:
0
B
E(a
1
)
(0,r
1
) →
0
E(a
1
) is a contraction, where
0
B
E(a
1
)
(0,r
1
)
denotes the closed ball in
0
E(a
1
) with center at 0 and radius r
1
.
It remains so show that L
−1
b
K
b
maps
0
B
E(a
1
)
(0,r
1
) into itself. From (4.24) and the