Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift
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668 S. Shimizu
In order to treat (5.3), we discuss an analytic semigroup to the initial boundary
value problem
ρ∂
t
u −Div S(u, π)=0 divu =0 inΩ,t>0,
[[ S(u, π)ν]] = 0 [[ u]] = 0 o n Γ ,t>0,
u|
t=0
= u
0
in Ω. (5.4)
A corresponding resolvent problem to (5.4) is
λu −
1
ρ
Div S(u, π)=u
0
div u =0 inΩ,
[[ S(u, π)ν]] = 0 [[ u]] = 0 o n Γ . (5.5)
Let us introduce the Helmholtz decomposition
L
q
(Ω)
n
= J
q
(Ω) ⊕ G
q
(Ω) (5.6)
for 1 <q<∞,where⊕ is the direct sum and
J
q
(Ω) = {u ∈ C
∞
0
(Ω)
n
| div u =0inΩ}
·
L
q
(Ω)
,
G
q
(Ω) = {∇π | π ∈
ˆ
W
1
q
(Ω), [[ π]] = 0 }.
Let us define the solution operator K from W
2
q
(Ω)
n
into
ˆ
W
1
q
(Ω) by the formula
K(u)=π:
Δπ =0 inΩ,
[[ π]] = [[ ν · μD(u)ν −div u]] o n Γ ,
[[ ρ
−1
∂
ν
π]] = [[ ρ
−1
ν · (μDiv D(u) −∇div u)]] on Γ.
We introduce an operator A
q
and the domain D(A
q
):
A
q
u = −Div S(u, K(u)) for u ∈D(A
q
),
D(A
q
)={u ∈ J
q
(Ω) ∩ W
2
q
(Ω)
n
| [[ S(u, K(u))ν]] = 0 , [[ u]] = 0 }.
By a similar argument as §3 in [21] we obtain the following.
Theorem 5.1.
(1) The operator A
q
generates an analytic semigroup {e
−tA
q
}
t≥0
on J
q
(Ω).
(2) If u
0
∈ (J
q
(Ω), D(A
q
))
1−1/p,p
,then(5.4) admits a unique solution (u, π, ¯π) ∈
D
p,q,γ
0
(Ω × R) which satisfies 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)
≤ Cu
0
(J
q
(Ω),D(A
q
))
1−1/p,p
for any γ ≥ γ
0
.
Local Solvability of Free Boundary Problems 669
If the initial data u
0
∈ (L
q
(Ω),W
2
q
(Ω))
1−1/p,p
= B
2(1−1/p)
q,p
(Ω) satisfies the
compatibility condition (1.9), then by Theorem 5.1 we know that the problem
(5.3) admits a unique solution
u
2
∈ W
2,1
q,p
(Ω × (0, 2)),π
2
∈ L
p
((0, 2),
ˆ
W
1
q
(Ω)).
Moreover there exists ¯π
2
∈ H
1,1/2
q,p
(Ω × (0, 2)) such that [[¯π]] = [[ π
2
]] . ( u
2
, π
2
,¯π
2
)
satisfies the estimate
u
2
W
2,1
q,p
(Ω×(0,2))
+ π
2
L
p
((0,2),
ˆ
W
1
q
(Ω))
+ ¯π
2
H
1,1/2
q,p
(Ω×(0,2))
≤ C
2
u
0
+ u
1
B
2(1−1/p)
q,p
(Ω))
. (5.7)
If we set z = u
1
+ u
2
and τ = π
1
+ π
2
,then(z, τ) satisfies the time-dependent
linear equation in the time interval (0, 2):
ρ∂
t
z − Div S(z, τ)=0 inΩ× (0, 2),
div z =0 inΩ× (0, 2),
[[ S(z,τ)ν]] = ( σH
0
(Γ) + [[ρ]] c
g
ξ
n
)ν on Γ × (0, 2),
[[ z]] = 0 o n Γ × (0, 2),
z|
t=0
= u
0
in Ω. (5.8)
z is our initial flow.
Now, we look for a solution (u, π) of the equation (4.13) of the form: u = z+w
and π = τ + κ in the time interval (0,T)with0<T≤ 1. Setting ¯τ = π
1
+¯π
2
,we
see that w, κ and η should satisfy the equations:
ρ∂
t
w − Div S(w, κ)=DivQ(z + w)+R(z + w)∇(τ + κ)inΩ× (0,T),
div w = E(z + w)=div
˜
E(z + w)inΩ× (0,T),
∂
t
η − ν ·w = G(z + w)+ν · z on Γ × (0,T),
[[ ΠμD(w)ν]] = [[ H
t
(z + w)]] on Γ × (0,T),
[[ ν · S(w, κ)ν]] + σ(m −Δ
Γ
)η =[[H
n
(z + w, ¯τ + κ)]] on Γ × (0,T),
[[ w]] = 0 o n Γ × (0,T),
w|
t=0
=0inΩ,η|
t=0
=0onΓ. (5.9)
6. The nonlinear problem
In this section we solve (5.9), namely we shall prove the following theorem.
Theorem 6.1. Let 2 <p<∞ and n<q<∞.Let(u
1
,π
1
) be a solution of
(5.1) and (u
2
,π
2
,η
2
) be a solution of (5.3). Then there exists T>0 depending on
u
0
B
2(1−1/p)
q,p
(Ω)
and α
W
3−1/q
q
(R
n−1
)
such that (5.9) admits a unique solution
w ∈ W
2,1
q,p
(Ω × (0,T))
n
,κ∈ L
p
((0,T),
ˆ
W
1
q
(Ω)),
η ∈ W
1
p
((0,T),W
2−1/q
q
(Γ)) ∩ L
p
((0,T),W
3−1/q
q
(Γ)).
670 S. Shimizu
We prove Theorem 6.1 by using the maximal L
p
-L
q
regularity theorem. If
e
−γ
0
t
u ∈ L
p
(R,L
q
(Ω)) for γ
0
> 1, then u ∈ L
p
((0,T),L
q
(Ω)) for any T with
0 <T <∞, because it holds that
T
0
u
p
L
q
(Ω)
dt =
T
0
e
pγ
0
t
e
−γ
0
t
u
p
L
q
(Ω)
dt
≤ e
pγ
0
T
∞
0
e
−γ
0
t
u
p
L
q
(Ω)
dt
≤ Ce
−γ
0
t
u
L
p
(R,L
q
(Ω))
.
Therefore as a corollary of Theorem 1.2 with q = r, it holds that the maximal
regularity of (1.10) is local in time.
Theorem 6.2. Let 1 <p,q<∞. If the right-hand members f , f
d
, d and h of
(1.10) satisfy the conditions
f ∈ L
p
((0,T),L
q
(Ω))
n
,f
d
∈ L
p
((0,T),W
1
q
(Ω)) ∩ W
1
p
((0,T),
ˆ
W
−1
q
(R
n
)),
d ∈ L
p
((0,T),W
2−1/q
q
(Γ)),h∈ H
1,1/2
q,p
(Ω × (0,T))
n
and compatibility conditions
¯
f
d
|
t=0
=0,h|
t=0
=0,
then (1.10) admits a unique solution (u, π, η):
u ∈ W
2,1
q,p
(Ω × (0,T)),π∈ L
p
((0,T),
ˆ
W
1
q
(Ω)),
η ∈ W
1
p
((0,T),W
2−1/q
q
(Γ)) ∩ L
p
((0,T),W
3−1/q
q
(Γ)).
Moreover there exists [[ ¯π]] = [[ π]] such that ¯π ∈ H
1,1/2
q,p
(Ω × (0,T)). The solutions
satisfy the estimate
u
W
2,1
q,p
(Ω×(0,T ))
+ ∇π
L
p
((0,T ),L
q
(Ω))
+ ¯π
H
1,1/2
q,p
(Ω×(0,T ))
+ ∂
t
η
L
p
((0,T ),W
2−1/q
q
(Γ))
+ η
L
p
((0,T ),W
3−1/q
q
(Γ))
≤ C
f
L
p
((0,T ),L
q
(Ω))
+ d
L
p
((0,T ),W
2−1/q
q
(Γ))
+ f
d
L
p
((0,T ),W
1
q
(Ω))
+ ∂
t
f
d
L
p
((0,T ),
ˆ
W
−1
q
(R
n
))
+ h
H
1,1/2
q,p
(Ω×(0,T ))
. (6.1)
In what follows we construct a solution of (5.9) by the contraction mapping
principle based on Theorem 6.2.
Step 1. We set
M =max(C
1
,C
2
)
σH
0
(Γ)
W
1−1/q
q
(Γ)
+[[ρ]] c
g
α
W
1−1/q
q
(Γ)
+ u
0
B
2(1−1/p)
q,p
(Ω)
.
(6.2)
Local Solvability of Free Boundary Problems 671
By (5.2) and (5.7) we define the underlying space I
R,T
by
I
R,T
={(v, θ,
¯
θ, η) |
v ∈ W
2,1
q,p
(Ω × (0,T))
n
,θ∈ L
p
((0,T),
ˆ
W
1
q
(Ω)),
¯
θ ∈ H
1,
1
2
q,p,0
(Ω × (0,T)),
∂
t
η ∈ L
p
((0,T),W
2−1/q
q
(Γ)),η∈ L
p
((0,T),W
3−1/q
q
(Γ)),
θ =
¯
θ on Γ ×(0,T), (As.1)
v(ξ,0) = 0 in Ω,η(ξ
, 0) = 0 on Γ, (As.2)
v
W
2,1
q,p
(Ω×(0,T ))
+ θ
L
p
((0,T ),W
1
q
(Ω))
+
¯
θ
H
1,1/2
q,p
(Ω×(0,T ))
+ ∂
t
η
L
p
((0,T ),W
2−1/q
q
(Γ))
+ η
L
p
((0,T ),W
3−1/q
q
(Γ))
≤ R}, (As.3)
where T is a positive number and R ≥ 1 is a large number determined later. Since
we consider the local in time solvability of (5.9), we may assume that 0 <T≤ 1
in the course of our proof of Theorem 6.1.
Given (v, θ,
¯
θ, η, ) ∈I
R,T
,let(V,Θ,Y) be a solution to the linear equation
ρ∂
t
V − Div S(V,Θ)
=DivQ(u
1
+ u
2
+ v)+R(u
1
+ u
2
+ v)∇(π
1
+ π
2
+ θ)inΩ,t>0,
div V = E(u
1
+ u
2
+ v)=div
˜
E(u
1
+ u
2
+ v)inΩ,t>0,
∂
t
Y − ν ·V = G(u
1
+ u
2
+ v)+ν · (u
1
+ u
2
)onΓ,t>0,
[[ ΠμD(V )ν]] = [[ H
t
(u
1
+ u
2
+ v)]] on Γ,t>0,
[[ ν · S(V,Θ)ν]] + σ(m − Δ
Γ
)Y =[[H
n
(u
1
+ u
2
+ v, π
1
+¯π
2
+
¯
θ)]] on Γ,t>0,
[[ V ]] = 0 o n Γ ,t>0,
V |
t=0
=0inΩ,Y|
t=0
=0onΓ. (6.3)
Our task is to show that if we define the map Φ(v, θ,
¯
θ, η)=(V, Θ,
¯
Θ,Y), then
Φ is a contraction map from I
R,T
into itself. To do so we check the condition on
the right-hand members in (6.3), to apply Theorem 6.2 to (6.3). All the constants
independent of R and T are denoted by C in what follows. To estimate L
∞
norm
of functions and the multiplication of several functions we use the following fact:
W
1
q
(Ω) ⊂ L
∞
(Ω), f
L
∞
(Ω)
≤ Cf
W
1
q
(Ω)
for f ∈ W
1
q
(Ω),
#
#
#
#
#
#
f
N
F
j=1
g
j
#
#
#
#
#
#
L
q
(Ω)
≤ Cf
L
q
(Ω)
N
F
j=1
g
j
W
1
q
(Ω)
for f ∈ L
q
(Ω),g
j
∈ W
1
q
(Ω),
#
#
#
#
#
#
N
F
j=1
f
j
#
#
#
#
#
#
W
1
q
(Ω)
≤ C
N
F
j=1
f
j
W
1
q
(Ω)
for f
j
∈ W
1
q
(Ω) (j =1,...,N), (6.4)
which follows from the Sobolev imbedding theorem and the assumption: n<q<
∞. Here and hereafter, p
denotes the dual exponent of p,thatisp
= p/(p − 1).
672 S. Shimizu
Lemma 6.3 (Lemma 2.2 in [20]). (1) Let 0 <T ≤ 1 and set
W
2,1
q,p,0
(Ω × (0,T)) = {v ∈ W
2,1
q,p
(Ω × (0,T)) : v(ξ,0) = 0}.
Then, there exists a bounded linear operator E : W
2,1
q,p,0
(Ω ×(0,T)) → W
2,1
q,p
(Ω ×R)
such that Ev = v on Ω ×(0,T), Ev =0for t ∈ [0, 2T ] and
Ev
W
2,1
q,p
(Ω×R)
≤ Cv
W
2,1
q,p
(Ω×(0,T ))
for any v ∈ W
2,1
q,p,0
(Ω × (0,T)),whereC is independent of T .
(2) There exists a bounded linear operator E
2
: W
2,1
q,p
(Ω × (0, 2)) → W
2,1
q,p
(Ω × R)
such that E
2
u
2
= u
2
on Ω × (0, 1), E
2
u
2
=0for t ∈ [−2, 2] and
E
2
u
2
W
2,1
q,p
(Ω×R)
≤ Cu
2
W
2,1
q,p
(Ω×(0,2))
for any u
2
∈ W
2,1
q,p
(Ω × (0, 2)).
Lemma 6.4. For 0 <t≤ T<1 we obtain
(1)
#
#
#
t
0
(u
1
+ u
2
+ v) dτ
#
#
#
L
∞
((0,T ),W
2
q
(Ω))
≤ T
1/p
(2M + R), (6.5)
(2)
#
#
#
t
0
(u
1
+ u
2
+ v) dτ
#
#
#
L
p
((0,T ),W
2
q
(Ω))
≤ T (2M + R), (6.6)
(3)
#
#
#
t
0
(u
1
+ E
2
u
2
+ Ev) dτ
#
#
#
H
1,1/2
q,p
(Ω×R)
≤ T
1
2
(2M + R), (6.7)
(4) u
1
+ u
2
+ v
L
∞
((0,T ),W
1
q
(Ω))
≤ C(2M + T
1
2
−
1
p
R), (6.8)
(5) u
1
+ u
2
+ v
L
p
((0,T ),W
1
q
(Ω))
≤ CT
1
p
(2M + R). (6.9)
Proof. The first inequality follows from
#
#
#
t
0
∇(u
1
+ u
2
+ v) dτ
#
#
#
W
1
q
(Ω)
≤
t
0
u
1
+ u
2
+ v
W
2
q
(Ω)
dt
≤u
1
W
2
q
(Ω)
t
0
dt +
t
0
dt
1/p
(u
2
L
p
((0,T ),W
2
q
(Ω))
+ v
L
p
((0,T ),W
2
q
(Ω))
)
≤ t
1/p
(2M + R),
where we have used (5.2), (5.7) and (As.3). The second inequality is easily derived.
By using the complex interpolation relation
H
1/2
p
(R,L
q
(Ω)) = [L
p
(R,L
q
(Ω)),W
1
p
(R,L
q
(Ω))]
1/2
,
between
#
#
#
#
t
0
(u
1
+ E
2
u
2
+ Ev) dτ
#
#
#
#
W
1
p
(R,L
q
(Ω))
≤u
1
+ E
2
u
2
+ Ev
W
1
p
(R,L
q
(Ω))
(6.10)
and
#
#
#
#
t
0
(u
1
+ E
2
u
2
+ Ev) dτ
#
#
#
#
L
p
(R,L
q
(Ω))
≤ T u
1
+ E
2
u
2
+ Ev
L
p
(R,L
q
(Ω))
, (6.11)
Local Solvability of Free Boundary Problems 673
we obtain
#
#
#
#
t
0
(u
1
+ E
2
u
2
+ Ev) dτ
#
#
#
#
H
1
2
p
(R,L
q
(Ω))
≤ T
1
2
u
1
+ E
2
u
2
+ Ev
H
1
2
p
(R,L
q
(Ω))
≤ T
1
2
(2M + R). (6.12)
Combining (6.6) and (6.12) we have the inequality (3). To obtain the fourth in-
equality, we use the relation
H
1
2
p
((0,T),W
1
q
(Ω)) ⊂ Lip
1
2
−
1
p
([0,T],W
1
q
(Ω)), (6.13)
and we have
u
2
(t)
W
1
q
(Ω)
≤u
0
W
1
q
(Ω)
+ Cu
2
H
1
2
p
((0,2),W
1
q
(Ω))
t
1
2
−
1
p
(6.14)
v(t)
W
1
q
(Ω)
≤ Cv
H
1
2
p
((0,1),W
1
q
(Ω))
t
1
2
−
1
p
(6.15)
for 0 <t<1. For the last inequality, we use (6.14) and (6.15).
Remark 6.5. In order to simplify the estimate of polynomials of
t
0
∇(u
1
+u
2
+v) dτ ,
we choose T>0sosmallthat
#
#
#
t
0
∇(u
1
+ u
2
+ v) dτ
#
#
#
W
1
q
(Ω)
≤ 1 (6.16)
for 0 <t≤ T .
Step 2 (Estimate of G). Nonlinear terms Q, R, E and
˜
E are the same terms as
that of a free boundary problem without surface tension and gravity estimated as
in §2 in [20]. Therefore in this paper we estimate nonlinear terms G, H
t
and H
n
.
We consider the term
G(u)=(m − Δ
Γ
)
−1
˙
F (u),
where
˙
F (u) denotes the derivative of F(u) with respect to t,andgivenby
˙
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.
By (6.2), we have
α
W
3−1/q
q
(Γ)
≤ M. (6.17)
Since ν =(∇
α, −1)/
1+|∇
α|
2
,wehave
ν
W
2−1/q
q
(Γ)
≤ M. (6.18)
674 S. Shimizu
The radius vectors of surfaces Γ and Γ(t)for(1.1)areequaltor
0
=(ξ
,α(ξ
)) and
r
t
=
3
ξ
1
+
t
0
u
1
(ξ
,α(ξ
),τ) dτ, ..., ξ
n−1
+
t
0
u
n−1
(ξ
,α(ξ
),τ) dτ,
α(ξ
)+
t
0
u
n
(ξ
,α(ξ
),τ) dτ
4
.
Δ
Γ
and Δ
Γ(t)
are defined by
Δ
Γ
=
1
√
g
n−1
!
j,k=1
∂
∂ξ
j
ˆ
g
jk
√
g
∂
∂ξ
k
, Δ
Γ(t)
=
1
√
g
t
n−1
!
j,k=1
∂
∂ξ
j
ˆ
g
tjk
√
g
t
∂
∂ξ
k
(6.19)
where
g
jk
=
∂r
0
∂ξ
j
·
∂r
0
∂ξ
k
, g
tjk
=
∂r
t
∂ξ
j
·
∂r
t
∂ξ
k
.
We set G as a matrix whose jk elements are denoted by g
jk
,andset
g =detG =det(g
jk
)=1+|∇
α|
2
.
ˆ
g
jk
are elements of a cofactor matrix of G and
G
−1
=
ˆ
g
jk
g
=
(1 + |∇
α|
2
)δ
jk
− ∂
j
α∂
k
α
g
.
We set G
t
as a matrix whose jk elements are denoted by g
tjk
.
ˆ
g
tjk
are elements
of a cofactor matrix of G
t
,
g
t
=detG
t
=det(g
tjk
),G
−1
t
=
ˆ
g
tjk
g
t
.
If we set
β
ij
=
t
0
∂u
i
∂ξ
j
(ξ
,α(ξ
),τ) dτ,
then we have
∂r
ti
∂ξ
j
= δ
ij
+ β
ij
+ ∂
j
αβ
in
,i=1,...,n−1,
∂r
tn
∂ξ
j
= ∂
j
α + β
nj
+ ∂
j
αβ
nn
,
and
g
tjk
= δ
jk
+ β
jk
+ β
kj
+
n
!
=1
β
j
β
k
+ ∂
k
α
3
β
jn
+ β
nj
+
n
!
=1
β
j
β
n
4
+ ∂
j
α(β
kn
+ β
nk
+
n
!
=1
β
n
β
k
)+∂
j
α∂
k
α
3
1+2β
nn
+
n
!
=1
β
2
n
4
.
Local Solvability of Free Boundary Problems 675
Therefore g
tjk
,
ˆ
g
tjk
and g
t
are expressed by
g
tjk
= g
jk
+ J
1
(∇
α)K
1
t
0
∇
(u
1
+ u
2
+ v) dτ
, (6.20)
ˆ
g
tjk
=
ˆ
g
jk
+ J
2
(∇
α)K
2
t
0
∇
(u
1
+ u
2
+ v) dτ
, (6.21)
g
t
= g + J
3
(∇
α)K
3
t
0
∇
(u
1
+ u
2
+ v) dτ
, (6.22)
where J
1
(·), J
2
(·)andJ
3
(·)arepolynomialsof∇α,andK
1
(·), K
2
(·)andK
3
(·)
are polynomials of
t
0
∇(u
1
+ u
2
+ v) dτ , respectively, such as K
j
(0) = 0.
Proposition 6.6 (cf. Amann [4]). Let 1 <q<∞. For every f ∈ W
−1/q
q
(Γ),there
exists m>>1 such that
(m − Δ
Γ
)
−1
f
W
2−1/q
q
(Γ)
≤ C
q,Γ
f
W
−1/q
q
(Γ)
(6.23)
holds.
Proposition 6.7. Let 1 <q<∞.
(1) For every f ∈ W
−1/q
q
(Γ) and g ∈ W
1
q
(Ω) we have
fg
W
−1/q
q
(Γ)
≤ C
q,Γ
f
W
−1/q
q
(Γ)
g
W
1
q
(Ω)
. (6.24)
(2) For f ∈ W
−1/q
q
(Γ) and g
j
∈ W
1−1/q
q
(Γ) (j =1,...,N), we have
#
#
#
#
#
#
f
N
F
j=1
g
j
#
#
#
#
#
#
W
−1/q
q
(Γ)
≤ Cf
W
−1/q
q
(Γ)
N
F
j=1
g
j
W
1−1/q
q
(Γ)
.
(3) For f
j
∈ W
1−1/q
q
(Γ) (j =1,...,N), we have
#
#
#
#
#
#
N
F
j=1
g
j
#
#
#
#
#
#
W
1−1/q
q
(Γ)
≤ C
N
F
j=1
f
j
W
1−1/q
q
(Γ)
.
Proof. Let 1/q +1/q
=1.Forϕ ∈ W
1−1/q
q
(Γ) there exists ψ ∈ W
1
q
(Ω) such that
ψ|
Γ
= ϕ, ψ
W
1
q
(Ω)
≤ Cϕ
W
1−1/q
q
(Γ)
.
It holds that
gϕ
W
1−1/q
q
(Γ)
≤ Cgψ
W
1
q
(Ω)
≤ C
∇gψ
L
q
(Ω)
+ g
L
∞
(Ω)
ψ
W
1
q
(Ω)
.
(6.25)
Let 1/q
=1/q +1/r.Sincen(
1
q
−
1
r
) < 1, by the embedding relation, W
1
q
(Ω) ⊂
L
r
(Ω) holds. Therefore we have
∇gψ
L
q
(Ω)
≤ Cg
W
1
q
(Ω)
ψ
L
r
(Ω)
≤ Cg
W
1
q
(Ω)
ψ
W
1
q
(Ω)
≤ Cg
W
1
q
(Ω)
ϕ
W
1−
1
q
q
(Γ)
. (6.26)
676 S. Shimizu
Combining (6.25) with (6.26), we obtain
"
"
"
"
Γ
fgϕdσ
"
"
"
"
≤ Cf
W
−
1
q
q
(Γ)
gϕ
W
1−
1
q
q
(Γ)
≤ Cf
W
−
1
q
q
(Γ)
g
W
1
q
(Ω)
ϕ
W
1−
1
q
q
(Γ)
,
which implies that
fg
W
−
1
q
q
(Γ)
=sup
ϕ∈C
∞
0
(Γ), ϕ
W
1−
1
q
q
(Γ)
=1
"
"
"
"
Γ
fgϕdσ
"
"
"
"
≤ Cf
W
−
1
q
q
(Γ)
g
W
1
q
(Ω)
.
Assertions (2) and (3) are corollaries of (1).
First we consider the term: (m −Δ
Γ
)
−1
Δ
Γ
ν · (u
1
+ u
2
+ v). By Propositions
6.6 and 6.7, we have
(m −Δ
Γ
)
−1
Δ
Γ
ν · (u
1
+ u
2
+ v)
W
2−1/q
q
(Γ)
≤ C
q,Γ
Δ
Γ
ν · (u
1
+ u
2
+ v)
W
−1/q
q
(Γ)
≤ C
q,Γ
Δ
Γ
ν
W
−1/q
q
(Γ)
u
1
+ u
2
+ v
W
1
q
(Ω)
.
By (6.18) and (6.9),
(m − Δ
Γ
)
−1
Δ
Γ
ν · (u
1
+ u
2
+ v)
L
p
((0,T ),W
2−1/q
q
(Γ))
≤ C
q,Γ
Δ
Γ
ν
W
−1/q
q
(Γ)
u
1
+ u
2
+ v
L
p
((0,T ),W
1
q
(Ω))
≤ CM(2M + R)T
1
p
. (6.27)
Next we consider the term: (m − Δ
Γ
)
−1
ν ·
˙
Δ
Γ(t)
t
0
(u
1
+ u
2
+ v) dτ.
˙
Δ
Γ(t)
is
given by
˙
Δ
Γ(t)
=
n−1
!
j,k=1
∂
t
ˆ
g
tjk
g
t
∂
2
∂ξ
j
∂ξ
k
+
n−1
!
k=1
⎡
⎣
n−1
!
j=1
∂
t
∂
j
ˆ
g
tjk
g
t
−
ˆ
g
tjk
∂
j
g
t
2g
2
t
⎤
⎦
∂
∂ξ
k
. (6.28)
Since
g
t
L
∞
((0,T ),W
1−1/q
q
(Γ))
≤ CM,
∂
t
ˆ
g
tjk
L
p
((0,T ),W
−1/q
q
(Γ))
≤ C∇
α
W
1−1/q
q
(Γ)
∇
(u
1
+ u
2
+ v)
L
p
((0,T ),W
1
q
(Ω))
≤ C(2M + R)M,
∂
t
ˆ
g
tjk
∇
g
t
L
∞
((0,T ),W
−1/q
q
(Γ))
≤ C∇
α
W
1−1/q
q
(Γ)
∇
(u
1
+ u
2
+ v)
L
p
((0,T ),W
1
q
(Ω))
×∇
g + ∇
(J
3
K
3
)
L
∞
((0,T ),W
1−1/q
q
(Γ))
≤ C(2M + R)M
2
,
Local Solvability of Free Boundary Problems 677
we have
#
#
#
#
∂
t
ˆ
g
tjk
g
t
#
#
#
#
L
p
((0,T ),W
1−1/q
q
(Γ))
(6.29)
≤ C(∂
t
ˆ
g
tjk
L
p
((0,T ),W
1−1/q
q
(Γ))
+ (∂
t
ˆ
g
tjk
)∇
g
t
L
p
((0,T ),W
−1/q
q
(Γ))
)
≤ C(2M + R)M
2
, (6.30)
#
#
#
#
∂
t
ˆ
g
tjk
g
t
#
#
#
#
L
p
((0,T ),W
1−1/q
q
(Γ))
≤ C(2M + R)M
2
, (6.31)
#
#
#
#
∂
t
∂
j
ˆ
g
tjk
g
t
,∂
t
ˆ
g
tjk
∂
j
g
t
2g
2
t
#
#
#
#
L
p
((0,T ),W
−1/q
q
(Γ))
≤ C(2M + R)M
2
. (6.32)
Therefore we obtain from (6.18), (6.9), (6.31) and (6.32),
#
#
#
#
(m −Δ
Γ
)
−1
ν ·
˙
Δ
Γ(t)
t
0
(u
1
+ u
2
+ v) dτ
#
#
#
#
L
p
((0,T ),W
2−1/q
q
(Γ))
≤ C
q,Γ
#
#
#
#
ν ·
˙
Δ
Γ(t)
t
0
(u
1
+ u
2
+ v) dτ
#
#
#
#
L
p
((0,T ),W
−1/q
q
(Γ))
≤ C
q,Γ
ν
W
1−1/q
q
(Γ))
#
#
#
#
∂
t
ˆ
g
tjk
g
t
#
#
#
#
L
p
((0,T ),W
1−1/q
q
(Γ))
×
#
#
#
#
t
0
∇
2
(u
1
+ u
2
+ v) dτ
#
#
#
#
L
∞
((0,T ),L
q
(Ω))
+ C
q,Γ
ν
W
1−1/q
q
(Γ)
3
#
#
#
#
∂
t
∂
j
ˆ
g
tjk
g
t
#
#
#
#
L
p
((0,T ),W
−1/q
q
(Γ))
+
#
#
#
#
∂
t
ˆ
g
tjk
∂
j
g
t
2g
2
t
#
#
#
#
L
p
((0,T ),W
1−1/q
q
(Γ))
4
#
#
#
#
t
0
∇
(u
1
+ u
2
+ v) dτ
#
#
#
#
L
∞
((0,T ),W
1
q
(Ω))
≤ C
q,Γ
T
1/p
(2M + R)
2
M
3
. (6.33)
Next we consider the term: (m − Δ
Γ
)
−1
ν · (Δ
Γ
− Δ
Γ(t)
)(u
1
+ u
2
+ v). By
(6.19),
(Δ
Γ
− Δ
Γ(t)
)u
=
n−1
!
j,k=1
ˆ
g
jk
g
−
ˆ
g
tjk
g
t
∂
2
u
∂ξ
j
∂ξ
k
+
n−1
!
k=1
⎡
⎣
n−1
!
j=1
∂
j
ˆ
g
jk
g
−
∂
j
ˆ
g
tjk
g
t
−
ˆ
g
jk
∂
j
g
2g
2
−
ˆ
g
tjk
∂
j
g
t
2g
2
t
⎤
⎦
∂u
∂ξ
k