Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift
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658 S. Shimizu
In fact, by the assumption q ≤ r,choosings in such a way that 1/q =1/r +1/s,
noting that n(1/q − 1/s)=n/r < 1 and using the H¨older inequality and the
Sobolev imbedding theorem we have
ab(t)
L
q
(
˙
R
n
)
≤a
L
r
(
˙
R
n
)
b(t)
L
s
(
˙
R
n
)
≤ Ca
L
r
(
˙
R
n
)
b(t)
W
1
q
(
˙
R
n
)
.
Applying the H¨older inequality and the Sobolev imbedding theorem to F (v, θ)in
(2.6) and using (2.11), we see that
e
−γt
F (v, θ)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
(∂
t
v, ∇
2
v, ∇θ)
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
e
−γt
v
L
p
(R,W
1
q
(
˙
R
n
))
. (2.12)
Here and hereafter, C
1
and C
2
denote generic constants independent of K
1
and
K
2
. By the interpolation inequality, it holds that
e
−γt
∇v
L
q
(
˙
R
n
)
≤ e
−γt
∇
2
v
L
q
(
˙
R
n
)
+
1
4
e
−γt
v
L
q
(
˙
R
n
)
,
which implies
e
−γt
∇v
L
p
(R,L
q
(
˙
R
n
))
≤ e
−γt
∇
2
v
L
p
(R,L
q
(
˙
R
n
))
+
1
4γ
γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
.
(2.13)
Combining (2.12) and (2.13), we have
e
−γt
F (v, θ)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
(∂
t
v, ∇
2
v, ∇θ)
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
e
−γt
∇
2
v
L
p
(R,L
q
(
˙
R
n
))
+
C()+1
γ
γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
. (2.14)
The function F
d
(v) is also expressed by F
d
(v)=div
˜
F
d
(v). By (2.11) and (2.13)
we obtain
e
−γt
∂
t
˜
F
d
(v)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
∂
t
v
L
p
(R,L
q
(
˙
R
n
))
, (2.15)
e
−γt
∇F
d
(v)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
∇
2
v
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
e
−γt
v
L
p
(R,W
1
q
(
˙
R
n
))
+
C()+1
γ
γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
. (2.16)
Applying (2.11) to B(v) in (2.6), we see that
e
−γt
D
t
1
2
B(v)
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
∇B(v)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
(∂
t
v, D
t
1
2
∇v,∇
2
v)
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
e
−γt
D
t
1
2
v
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
v
L
p
(R,W
1
q
(
˙
R
n
))
.
By using the relation (cf. Proposition 2.6 in [21])
e
−γt
D
t
1
2
u
L
p
(R,L
q
(
˙
R
n
))
≤ CR
−
1
2
e
−γt
∂
t
u
L
p
(R,L
q
(
˙
R
n
))
+ CR
1
2
γe
−γt
u
L
p
(R,L
q
(
˙
R
n
))
(2.17)
Local Solvability of Free Boundary Problems 659
for every u ∈ H
1
2
p,(0)
(R,L
q
(
˙
R
n
)), 1 ≤ R<∞ and (2.13), we obtain
e
−γt
D
t
1
2
B(v)
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
∇B(v)
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
(∂
t
v, D
t
1
2
∇v,∇
2
v)
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
R
−
1
2
e
−γt
∂
t
v
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
∇
2
v
L
p
(R,L
q
(
˙
R
n
))
+
C()+1
γ
γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
. (2.18)
Obviously we have
D
t
1
2
H
0
L
q
(
˙
R
n
)
+ ∇H
0
L
p
(R,L
q
(
˙
R
n
))
≤ C(D
t
1
2
˜
h, ∇
˜
h)
L
p
(R,L
q
(
˙
R
n
))
. (2.19)
For notational simplicity we set
[(v, θ,
¯
θ)]
p,q,γ
= e
−γt
v
W
2,1
q,p
(
˙
R
n
×R)
+ e
−γt
∇θ
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
¯
θ
H
1,1/2
q,p
(
˙
R
n
×R)
+ γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
for (v, θ,
¯
θ) ∈D
p,q
(
˙
R
n
×R)andforanyγ ≥ 0. Applying Theorem 2.2 to (2.7) and
using (2.9), (2.10), (2.14)–(2.16), (2.18) and (2.19), we have
[(w, κ, ¯κ)]
p,q,γ
(2.20)
≤ (C
1
K
1
+ C
2
K
2
+ C
2
K
2
R
−
1
2
+ C
2
K
2
(R
1
2
+ C()+1)γ
−1
)[(v, θ,
¯
θ)]
p,q,γ
+ CI
γ
,
where
I
γ
= e
−γt
˜
f
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
(1 + ∂
n
A)
˜
f
d
W
1
p
(R,
ˆ
W
−1
q
(R
n
))
+ e
−γt
(1 + ∂
n
A)
˜
f
d
L
p
(R,W
1
q
(
˙
R
n
))
+ e
−γt
(D
t
1
2
f
d
, ∇
˜
h)
L
p
(R,L
q
(
˙
R
n
))
for any γ ≥ 0. We define the map Φ on D
p,q
(
˙
R
n
× R)byΦ(v, θ,
¯
θ)=(w, κ, ¯κ).
Since the equation (2.7) is linear, from (2.20) we have
[Φ(v
1
,θ
1
,
¯
θ
1
) − Φ(v
2
,θ
2
,
¯
θ
2
)]
p,q,γ
≤ (C
1
K
1
+ C
2
K
2
+ C
2
K
2
R
−
1
2
+ C
2
K
2
(R
1
2
+ C()+1)γ
−1
0
)
× [(v
1
,θ
1
,
¯
θ
1
) − (v
2
,θ
2
,
¯
θ
2
)]
p,q,γ
(2.21)
for any γ>γ
0
.AfterchoosingK
1
, and R in such a way that
C
1
K
1
≤ 1/8,C
2
K
2
≤ 1/8,C
2
K
2
R
−
1
2
≤ 1/8,
we take γ
0
in such a way that
C
2
K
2
(R
1
2
+ C()+1)γ
−1
0
≤ 1/8.
Then (2.21) shows that Φ is a contraction map on D
p,q
(
˙
R
n
× R), and therefore
by the fixed point theorem of S. Banach, we see that the map Φ admits a unique
660 S. Shimizu
fixed point (v, θ,
¯
θ) ∈D
p,q
(
˙
R
n
×R), which solves the equation (2.5). Moreover from
(2.20) with (w, κ, ¯κ)=(v, θ,
¯
θ)andC
1
K
1
+C
2
K
2
+C
2
K
2
R
−
1
2
+C
2
K
2
(R
1
2
+C()+
1)γ
−1
0
≤ 1/2, we have [(v,θ,
¯
θ)]
p,q,γ
≤ 2CI
γ
for γ ≥ γ
0
, from which Theorem 2.1
follows immediately.
3. Analysis in a bent space for a problem
with surface tension and gravity
In this section we consider the problem with surface tension and gravity in a bent
space:
ρ∂
t
u −Div S(u, π)=0 inΩ,t>0,
div u =0 inΩ,t>0,
∂
t
η − ν · u = d on Γ,t>0,
[[ S(u, π)ν]] − (σΔ
Γ
+[[ρ]] c
g
)ην=0 onΓ,t>0,
[[ u]] = 0 o n Γ ,t>0,
u|
t=0
=0 inΩ,η|
t=0
=0 onΓ. (3.1)
We set
E
p,q,γ
0
(Ω × R)={(u, π, ¯π, η) ∈ W
2,1
q,p,γ
0
,(0)
(Ω × R)
n
×L
p,γ
0
,(0)
(R,
ˆ
W
1
q
(Ω))
×H
1,1/2
q,p,γ
0
,(0)
(Ω ×R) ×(W
1
p,γ
0
,(0)
(R,W
2−1/q
q
(Γ)) ∩ L
p,γ
0
,(0)
(R,W
3−1/q
q
(Γ)))
| [[ ¯π]] = [[ π]] }.
The following theorem is the main result in this section.
Theorem 3.1. Let 1 <p,q<∞, n<r<∞ and q ≤ r. Assume that
α
W
3−1/r
r
(R
n−1
)
≤ M . Then there exists constant K
0
with 0 <K
0
≤ 1 and γ
0
> 1
depending on M , p, q and n such that if ∇
α
L
∞
(R
n−1
)
≤ K
0
, then the following
assertion holds: For d ∈ L
p,γ
0
,(0)
(R,W
2−1/q
q
(Γ)), the problem (3.1) admits a unique
solution (u, π, ¯π, η) ∈E
p,q,γ
0
(Ω × R) satisfying the estimate:
e
−γt
u
W
2,1
q,p
(Ω×R)
+ e
−γt
∇π
L
p
(R,L
q
(Ω))
+e
−γt
¯π
H
1,1/2
q,p
(Ω×R)
+e
−γt
(∂
t
η, ∇
η)
L
p
(R,W
2−1/q
q
(R
n−1
))
≤ Ce
−γt
d
L
p
(R,W
2−1/q
q
(R
n−1
))
for any γ ≥ γ
0
,whereC is a constant depending on M , γ
0
, p, q and n.
Local Solvability of Free Boundary Problems 661
A proof of Theorem 1.2
Combining Theorem 2.1 and Theorem 3.1, we obtain Theorem 1.2.
In order to prove Theorem 3.1, we based our argument on the maximal L
p
-L
q
regularity result of the problem with surface tension and gravity with plainer
interface in R
n
:
ρ∂
t
u − Div S(u, π)=0 in
˙
R
n
,t>0,
div u =0 in
˙
R
n
,t>0,
∂
t
η − ν
0
·u = d on R
n
0
,t>0,
[[ S(u, π)ν
0
]] − (σΔ
+[[ρ]] c
g
)ην
0
=[[g]] o n R
n
0
,t>0,
[[ u]] = 0 o n R
n
0
,t>0,
u|
t=0
=0 in
˙
R
n
,η|
t=0
=0 onR
n−1
, (3.2)
where ν
0
=
T
(0,...,0, −1). We set
E
p,q,γ
0
(
˙
R
n
× R)={(u, π, ¯π, η) ∈ W
2,1
q,p,γ
0
,(0)
(
˙
R
n
×R)
n
×L
p,γ
0
,(0)
(R,
ˆ
W
1
q
(
˙
R
n
))
×H
1,1/2
q,p,γ
0
,(0)
(
˙
R
n
×R) × (W
1
p
(R,W
2−1/q
q
(R
n−1
)) ∩L
p
(R,W
3−1/q
q
(R
n−1
)))
| [[ ¯π]] = [[ π]] }.
We obtained the following result (cf. Theorem 1.4 in [23]).
Theorem 3.2. Let 1 <p,q<∞. There exists a constant γ
0
> 1 depending on p, q
and n such that the following assertion holds: For d ∈ L
p,γ
0
,(0)
(R,W
2−1/q
q
(R
n−1
))
and g ∈H
1,1/2
q,p,γ
0
(0)
(
˙
R
n
×R)
n
, the problem (3.2) admits a unique solution (u,π,¯π,η)∈
E
p,q,γ
0
(
˙
R
n
×R) satisfying the estimate:
e
−γt
u
W
2,1
q,p
(
˙
R
n
×R)
+ e
−γt
∇π
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
¯π
H
1,1/2
q,p
(
˙
R
n
×R)
+ γe
−γt
u
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
(∂
t
η, ∇
η)
L
p
(R,W
2−1/q
q
(R
n−1
))
≤ C
e
−γt
d
L
p
(R,W
2−1/q
q
(R
n−1
))
+ e
−γt
g
H
1,1/2
q,p
(
˙
R
n
×R)
for any γ ≥ γ
0
,whereC is a constant depending on M , γ
0
, p, q and n.
In what follows, we prove Theorem 3.1. We use the same notation as in §2.
A Laplace-Beltrami operator on Γ is defined by
Δ
Γ
η =
1
√
g
n−1
!
j,k=1
∂
j
(
√
gg
jk
∂
k
η), g
jk
=
(1 + |∇
α|
2
)δ
jk
− ∂
j
α∂
k
α
g
, g =1+|∇
α|
2
.
662 S. Shimizu
We derive an equivalent equation to (3.1) in
˙
R
n
.Ifweset
v(y)=u ◦ Φ
−1
(y)=u(x),θ(y)=p ◦ Φ
−1
(y)=p(x),
then by (2.4) we have
ρ∂
t
v − Div S(v, θ)=F (v, θ)in
˙
R
n
,t>0,
div v = F
d
(v)in
˙
R
n
,t>0,
∂
t
η + v
n
= d + D(v)onR
n
0
,t>0,
[[ S(v, θ)ν
0
]] − (σΔ
+[[ρ]] c
g
)ην
0
=[[H(v)]] + T (η)ν
0
on R
n
0
,t>0,
[[ v]] = 0 o n R
n
0
,t>0,
v|
t=0
=0 in
˙
R
n
,η|
t=0
=0 onR
n−1
, (3.3)
where F (v, θ), F
d
(v)andH(v) are same as in (2.6), D(v)andthenth component
of T (η) expressed by T
n
(η)aregivenby
D(v)=(
√
g)
−1
(∇
α,
1+|∇
α|
2
− 1) ·v,
T
n
(η)=−σg
−1
|∇
α|
2
Δ
η −σg
−2
n−1
!
j,k=1
∂
j
α∂
k
α∂
j
∂
k
η. (3.4)
We seek the solution of (3.3) by setting v = w + z and θ = κ + τ such that (w, κ)
satisfies the Neumann problem:
ρ∂
t
w − Div S(w, κ)=F(w, κ)in
˙
R
n
,t>0,
div w = F
d
(w)in
˙
R
n
,t>0,
[[ S(w, κ)ν
0
]] = [[ H(w)]] on R
n
0
,t>0,
[[ w]] = 0 o n R
n
0
,t>0,
w|
t=0
=0 in
˙
R
n
, (3.5)
and (z,τ,η) satisfies the equation:
ρ∂
t
z − Div S(z,τ)=0 in
˙
R
n
,t>0,
div z =0 in
˙
R
n
,t>0,
∂
t
η + z
n
= d + D(z)onR
n
0
,t>0,
[[ S(z,τ)ν
0
]] − (σΔ
+[[ρ]] c
g
)ην
0
= T (η)ν
0
on R
n
0
,t>0,
[[ z]] = 0 o n R
n
0
,t>0,
z|
t=0
=0 in
˙
R
n
,η|
t=0
=0 inR
n−1
. (3.6)
Since we know the unique solvability of (3.5) because (3.5) is the same problem as
(2.5), we solve (3.6) by the contradiction mapping principle based on Theorem 3.2.
Given (v, θ,
¯
θ, η) ∈E
p,q,γ
0
(
˙
R
n
× R), let (w, κ, ¯κ, ζ) ∈E
p,q,γ
0
(
˙
R
n
× R) be a unique
Local Solvability of Free Boundary Problems 663
solution to the equation:
ρ∂
t
w − Div S(w, κ)=0 in
˙
R
n
,t>0,
div w =0 in
˙
R
n
,t>0,
∂
t
η + w
n
= d + D(v)onR
n
0
,t>0,
[[ S(w, κ)ν
0
]] − (σΔ
+[[ρ]] c
g
)ζν
0
= T (η)ν
0
on R
n
0
,t>0,
[[ w]] = 0 o n R
n
0
,t>0,
w|
t=0
=0 inR
n
0
,η|
t=0
=0 inR
n−1
. (3.7)
The next lemma is proved in the same way as Lemma 2.3.
Lemma 3.3. Let 1 <r<∞, 0 <<1 and α ∈ W
3−1/r
r
(R
n−1
). Then there exists
A
±
(x) ∈ W
3
r
(R
n
±
) which satisfies A
±
|
x
n
=±0
= α, ∂
n
A
±
L
∞
(R
n
±
)
≤ and
A
±
L
r
(R
n
±
)
≤ Cα
L
r
(R
n−1
)
,
∇
k
A
±
L
r
(R
n
±
)
≤ Cα
W
k−1/r
r
(R
n−1
)
,k=1, 2, 3. (3.8)
Let φ(x
n
) be a function in C
∞
(R) such that φ(x
n
)=1whenx
n
< 1and
φ(x
n
)=0whenx
n
> 2. We extend η to the function Y
±
defined on R
n
±
by using
the formula
Y
+
(x, t)=φ(x
n
)L
−1
λ
F
−1
ξ
[e
−(1+|ξ
|
2
)
1/2
x
n
LF
ξ
[η](ξ
,λ)](x
,t) x
n
> 0,
Y
−
(x, t)=0 x
n
< 0,
where F
x
and F
−1
ξ
denote the Fourier transform and its inversion transform with
respect to x
,andL and L
−1
λ
denote the Laplace transform and its inversion
transform defined by (1.7). We know that
[[ Y ]] = η, Y
W
q
(
˙
R
n
)
≤ Cη
W
−1/q
q
(R
n−1
)
,=1, 2, 3. (3.9)
For the estimate of D(v)andT (η) in (3.7), we use extended functions A and
Y instead of α and η:
D(v)=(
√
g
A
)
−1
(∇
A,
1+|∇
A|
2
− 1) ·v, g
A
=1+|∇
A|
2
,
T
n
(η)=−σg
−1
A
|∇
A|
2
Δ
Y − σg
−2
A
n−1
!
j,k=1
∂
j
A∂
k
A∂
j
∂
k
Y.
We set
K
1
= ∇A
L
∞
(
˙
R
n
)
,K
2
= A
W
3
r
(
˙
R
n
)
.
By the Sobolev imbedding theorem under the assumption n<r<∞ we have
A
W
2
∞
(
˙
R
n
)
≤ CK
2
. (3.10)
664 S. Shimizu
We assume that 0 <K
1
< 1 a priori, and therefore K
1
≤ K
1
for ≥ 1. Using
(2.11) and (3.10) we have
e
−γt
D(v)
L
p
(R,W
2
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
∇
2
v
L
p
(R,L
q
(
˙
R
n
))
+ C
2
K
2
e
−γt
v
L
p
(R,W
1
q
(
˙
R
n
))
, (3.11)
e
−γt
(D
t
1
2
T (Y ), ∇T (Y ))
L
p
(R,L
q
(
˙
R
n
))
≤ C
1
K
1
e
−γt
(D
t
1
2
Y,∇
Y )
L
p
(R,W
2
q
(R
n−1
))
+ C
2
K
2
e
−γt
Y
L
p
(R,W
2
q
(R
n−1
))
(3.12)
for γ ≥ γ
0
. For notational simplicity we set
[(v, θ,
¯
θ, η)]
p,q,γ
0
= e
−γt
v
W
2,1
q,p
(
˙
R
n
×R)
+ e
−γt
∇θ
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
¯
θ
H
1,1/2
q,p
(
˙
R
n
×R)
+ γe
−γt
v
L
p
(R,L
q
(
˙
R
n
))
+ e
−γt
(∂
t
η, ∇η)
L
p
(R,W
2−1/q
q
(
˙
R
n−1
))
for (v, θ,
¯
θ, η) ∈E
p,q,γ
0
(
˙
R
n
× R)andγ ≥ γ
0
. Applying Theorem 3.2 to (3.7) and
using (3.11) and (3.12), for any γ ≥ γ
0
with γ
0
≥ 1wehave
[(w, κ, ¯κ, ζ)]
p,q,γ
0
≤ (C
1
K
1
+ C
2
K
2
+ C
2
K
2
(C()+1)γ
−1
0
)[(v, θ,
¯
θ,η)]
p,q,γ
0
+ Cd
L
p
(R,W
2−1/q
q
(R
n−1
))
. (3.13)
We define the map Φ on E
p,q,γ
0
(
˙
R
n
× R)byΦ(v, θ,
¯
θ, η)=(w, κ, ¯κ, ζ). Since the
equation (3.7) is linear, from (3.13) we have
[Φ(v
1
,θ
1
,
¯
θ
1
,η
1
) −Φ(v
2
,θ
2
,
¯
θ
2
,η
2
)]
p,q,γ
0
≤ (C
1
K
1
+ C
2
K
2
+ C
2
K
2
(C()+1)γ
−1
0
)[(v
1
,θ
1
,
¯
θ
1
,η
1
) −(v
2
,θ
2
,
¯
θ
2
,η
2
)]
p,q,γ
0
(3.14)
for any γ ≥ γ
0
.AfterchoosingK
1
and in such a way that
C
1
K
1
≤ 1/8,C
2
K
2
≤ 1/8,
we take γ
0
in such a way that
C
2
K
2
(C()+1)γ
−1
0
≤ 1/4.
Then (3.14) shows that Φ is a contraction map on E
p,q,γ
0
(
˙
R
n
×R), and therefore by
the fixed point theorem of S. Banach, we see that the map Φ admits a unique fixed
point (v,θ,
¯
θ, η) ∈E
p,q,γ
0
(
˙
R
n
×R), which solves the equation (3.6). Moreover from
(3.13) with (w, κ, ¯κ, ζ)=(v, θ,
¯
θ, η)andC
1
K
1
+C
2
K
2
+C
2
K
2
(C()+1)γ
−1
0
≤ 1/2,
we have [(v, θ,
¯
θ, η)]
p,q,γ
0
≤ 2Ce
−γt
d
L
p
(R,W
2−1/q
q
(R
n−1
))
, from which Theorem 3.1
follows immediately.
Local Solvability of Free Boundary Problems 665
4. Reduction of the boundary condition to linearized problems
In this section, we shall discuss the reduction of the boundary condition
[[ ( S(u, π)+Q(u))ν
tu
]] − σHν
tu
−[[ ρ]] c
g
X
u,n
ν
tu
=0, (4.1)
which is the first key step in our proof of Theorem 1.1. Let Π
t
and Π be projections
to tangent hyperplanes of Γ(t) and Γ, which are defined by
Π
t
d = d − (d, ν
tu
)ν
tu
, Πd = d −(d, ν)ν (4.2)
for an arbitrary vector field d defined on Γ(t) and Γ, respectively. We know the
following fact (cf. Solonnikov [24] and Shibata-Shimizu [22, Appendix]).
Lemma 4.1. If ν
t
· ν =0, then for arbitrary vector d, d =0is equivalent to
ΠΠ
t
d =0,ν·d =0. (4.3)
We apply Lemma 4.1 for (4.1). Since we obtain
[[ Π
t
(μD(u)+Q(u))ν
tu
]] = 0 ( 4 . 4 )
by applying Π
t
to the left-hand side of (4.1), the first equation of (4.3) for (4.1)
is given by
[[ ΠμD(u)ν]] = −[[ Π(Π
t
−Π)(μD(u)ν
tu
)+Π(μD(u)(ν
tu
− ν)) + ΠΠ
t
(Q(u)ν
tu
)]],
(4.5)
wherewehaveusedΠΠ = Π.
On the other hand, we consider the inner product of the boundary condition
with ν. Using the fact that Hν
tu
=Δ
Γ(t)
X
u
and substituting (1.4) for (4.1), we
obtain
[[ ν · (S(u, π)+Q(u))ν
tu
]] − σν · (Δ
Γ(t)
−Δ
Γ
)
ξ +
t
0
u(ξ, τ) dτ
− σν · Δ
Γ
ξ +
t
0
u(ξ, τ) dτ
− [[ ρ]] c
g
ν ·
ξ
n
+
t
0
u
n
(ξ,τ) dτ
ν
tu
=0. (4.6)
Taking a commutator between Δ
Γ
and ν·,wehave
ν ·Δ
Γ
t
0
udτ =Δ
Γ
t
0
ν ·udτ
−(Δ
Γ
ν)·
t
0
udτ−2
∇
Γ
·
t
0
udτ
∇
Γ
·ν. (4.7)
By (4.6) and (4.7), we obtain
[[ ν · S(u, π)ν]] − σΔ
Γ
t
0
ν ·udτ
+ σ
Δ
Γ
ν ·
t
0
udτ + ν · (Δ
Γ
−Δ
Γ(t)
)
t
0
udτ + ν · (Δ
Γ
− Δ
Γ(t)
)ξ
=[[ν · S(u, π)(ν − ν
tu
) − ν ·Q(u)ν
tu
]] + σH
0
(Γ) + [[ρ]] c
g
ξ
n
+[[ρ]] c
g
t
0
u
n
dτ ν ·ν
tu
+[[ρ]] c
g
ξ
n
ν ·(ν
tu
− ν) − 2σ
∇
Γ
·
t
0
udτ
∇
Γ
· ν,
(4.8)
666 S. Shimizu
wherewehaveusedν · Δ
Γ
ξ = H
0
(Γ). We denote the terms in the bracket of the
left-hand side of (4.8) by F (u), that is
F (u)=Δ
Γ
ν ·
t
0
udτ + ν · (Δ
Γ
− Δ
Γ(t)
)
t
0
udτ + ν · (Δ
Γ
− Δ
Γ(t)
)ξ.
In (4.8), since Δ
Γ(t)
and Δ
Γ
contain the second-order tangential derivatives of X
u
,
in order to avoid the loss of regularity we apply the inverse operator (m − Δ
Γ
)
−1
with sufficiently large number m to F (u). Namely, we proceed as follows:
[[ ν · S(u, π)ν]] + σ(m − Δ
Γ
)
t
0
ν · udτ +(m − Δ
Γ
)
−1
F (u)
−σm
t
0
ν ·udτ
=[[ν · S(u, π)(ν − ν
tu
) − ν ·Q(u)ν
tu
]] + σH
0
(Γ) + [[ρ]] c
g
ξ
n
+[[ρ]] c
g
t
0
u
n
dτ ν ·ν
tu
+[[ρ]] c
g
ξ
n
ν · (ν
tu
−ν) −2σ
∇
Γ
·
t
0
udτ
∇
Γ
·ν.
(4.9)
We introduce a function η by the formula
η =
t
0
ν · udτ +(m − Δ
Γ
)
−1
F (u)onΓ. (4.10)
Then, from (4.9) and (4.10), we obtain the two equations on the boundary Γ as
follows:
[[ ν · S(u, π)ν]] + σ(m − Δ
Γ
)η
=[[ν · S(u, π)(ν − ν
tu
) − ν ·Q(u)ν
tu
]] + σH
0
(Γ) + [[ρ]] c
g
ξ
n
+[[ρ]] c
g
t
0
u
n
dτ ν ·ν
tu
+[[ρ]] c
g
ξ
n
ν ·(ν
tu
− ν) − 2σ
∇
Γ
·
t
0
udτ
∇
Γ
· ν
+ σm
t
0
ν · udτ, (4.11)
∂
t
η − ν ·u =(m − Δ
Γ
)
−1
˙
F (u), (4.12)
where
˙
F (u) denotes the derivative of F(u) with respect to t.
Finally we arrive at the equivalent equation to (1.5) as follows:
ρ∂
t
u −Div S(u, π)=DivQ(u)+R(u)∇π in Ω,t>0,
div u = E(u)=div
˜
E(u)inΩ,t>0,
∂
t
η − ν ·u = G(u)onΓ,t>0,
[[ ΠμD(u)ν]] = [[ H
t
(u)]] on Γ,t>0,
[[ ν · S(u, π)ν]] + σ(m − Δ
Γ
)η =[[H
n
(u, π)]] + σH
0
(Γ) + [[ρ]] c
g
ξ
n
on Γ,t>0,
[[ u]] = 0 o n Γ ,t>0,
u|
t=0
= u
0
(ξ)inΩ,η|
t=0
=0onΓ, (4.13)
Local Solvability of Free Boundary Problems 667
where
G(u)=(m − Δ
Γ
)
−1
˙
F (u),
˙
F (u)=Δ
Γ
ν · u − ν ·
˙
Δ
Γ(t)
t
0
udτ + ν · (Δ
Γ
− Δ
Γ(t)
)u − ν ·
˙
Δ
Γ(t)
ξ,
[[ H
t
(u)]] = −[[ Π(Π
t
− Π)(μD(u)ν
tu
)+Π(μD(u)(ν
tu
− ν)) + ΠΠ
t
(Q(u)ν
tu
)]],
[[ H
n
(u, π)]] = [[ν · S(u, π)(ν − ν
tu
) − ν ·(Q(u)ν
tu
)]]
+[[ρ]] c
g
t
0
u
n
dτ ν ·ν
tu
+[[ρ]] c
g
ξ
n
ν · (ν
tu
−ν)
−2σ
∇
Γ
·
t
0
udτ
∇
Γ
· ν + σm
t
0
ν ·udτ,
and Q(u), R(u), E(u)and
˜
E(u) are nonlinear terms defined by (1.6).
5. Initial flow
In this section, we shall discuss the initial flow to reduce the problem (4.13) to the
case where u
0
(ξ)=0andσH
0
(Γ) + [[ρ]] c
g
ξ
n
= 0. We study the problem in two
steps.
Step 1. Let (u
1
,π
1
) be a solution to the problem:
ρλu
1
− Div S(u
1
,π
1
)=0 inΩ,
div u
1
=0 inΩ,
[[ S(u
1
,π
1
)ν]] = ( σH
0
(Γ) + [[ρ]] c
g
ξ
n
)ν on Γ,
[[ u
1
]] = 0 o n Γ . (5.1)
If positive number λ is large enough, then we know that (5.1) admits a unique
solution
(u
1
,π
1
) ∈ W
2
q
(Ω) ×
ˆ
W
1
q
(Ω)
which satisfies the estimate
|λ|u
1
L
q
(Ω)
+ ∇
2
u
1
L
q
(Ω)
+ ∇π
1
L
q
(Ω)
≤ C
1
(σH
0
(Γ)
W
1−1/q
q
(Γ)
+ c
g
α
W
1−1/q
q
(Γ)
). (5.2)
Step 2. We consider the linear time-dependent problem in the time interval (0, 2):
ρ∂
t
u
2
−Div S(u
2
,π
2
)=−ρλu
1
in Ω ×(0, 2),
div u
2
=0 inΩ×(0, 2),
[[ S(u
2
,π
2
)ν]] = 0 o n Γ × (0, 2),
[[ u
2
]] = 0 o n Γ × (0, 2),
u
2
|
t=0
= u
0
(ξ) − u
1
(ξ)inΩ. (5.3)