P. B. M U C H A and W. Z A J A ¸ C Z K O W S K I (Warszawa)
ON LOCAL EXISTENCE OF SOLUTIONS OF THE FREE BOUNDARY PROBLEM FOR AN INCOMPRESSIBLE
VISCOUS SELF-GRAVITATING FLUID MOTION
Abstract. The local-in-time existence of solutions of the free boundary problem for an incompressible viscous self-gravitating fluid motion is proved.
We show the existence of solutions with lowest possible regularity for this problem such that u ∈ W
r2,1( e Ω
T) with r > 3. The existence is proved by the method of successive approximations where the solvability of the Cauchy–
Neumann problem for the Stokes system is applied. We have to underline that in the L
p-approach the Lagrangian coordinates must be used. We are looking for solutions with lowest possible regularity because this simplifies the proof and decreases the number of compatibility conditions.
1. Introduction. In this paper we consider the motion of a viscous incompressible fluid in a bounded domain Ω
t⊂ R
3with a free boundary S
twhich is under the self-gravitational force. Let v = v(x, t) be the velocity of the fluid, p = p(x, t) the pressure, ν the constant viscosity coefficient and p
0the external pressure. Then the problem is described by the following system:
(1.1)
v
t+ v · ∇v − div T(v, p) = ∇U in e Ω
T,
div v = 0 in e Ω
T,
T(u, p) · n = −p
0n on e S
T,
v|
t=0= v
0in Ω,
Ω
t|
t=0= Ω, S
t|
t=0= S,
v · n = −ϕ
t/|∇ϕ| on e S
T,
2000 Mathematics Subject Classification: 35Q30, 76D05.
Key words and phrases: local existence, Navier–Stokes equations, incompressible vis- cous barotropic self-gravitating fluid, sharp regularity, anisotropic Sobolev space.
Research supported by Polish KBN Grant 2 P03A 038 16.
[319]
where e Ω
T= S
t≤T
Ω
t× {t}, e S
T= S
t≤T
S
t× {t}, ϕ(x, t) = 0 describes S
tat least locally, n is the unit outward vector normal to S
t, n = ∇ϕ/|∇ϕ|, Ω
tis the domain at time t, S
t= ∂Ω
t, t ≤ T . Moreover, the dot · denotes the scalar product in R
3.
By T = T(v, p) we denote the stress tensor of the form (1.2) T(v, p) = {T
ij}
i,j=1,2,3= {−pδ
ij+ D
ij(v)}
i,j=1,2,3where
(1.3) D(v) = {D
ij(v)}
i,j=1,2,3= {ν(v
i,xj+ v
j,xi)}
i,j=1,2,3is the velocity deformation tensor.
Moreover, U (Ω
t, x, t) is the self-gravitational potential
(1.4) U (Ω
t, x, t) = k
\
Ωt
dy
|x − y| ,
where k is the gravitation constant and some arguments of U are omitted in evident cases.
In view of the equation (1.1)
2and the kinematic condition (1.1)
6the total volume is conserved:
(1.5) |Ω
t| =
\
Ωt
dx =
\
Ω
dx = |Ω|.
Let Ω be given. Then we introduce the Lagrangian coordinates ξ as the initial data for the following Cauchy problem:
(1.6) ∂x
∂t = v(x, t), x|
t=0= ξ, ξ = (ξ
1, ξ
2, ξ
3).
Integrating (1.6), we obtain a transformation which connects the Eulerian x and the Lagrangian ξ coordinates,
(1.7) x = x(ξ, t) ≡ ξ +
t
\
0
u(ξ, t
′) dt
′≡ x
u(ξ, t),
where u(ξ, t) = v(x
u(ξ, t), t) and the index u in x
u(ξ, t) will be omitted when no confusion can arise.
Then from (1.1)
6we have Ω
t= {x ∈ R
3: x = x(ξ, t), ξ ∈ Ω} and S
t= {x ∈ R
3: x = x(ξ, t), ξ ∈ S = S
0= ∂Ω}.
Our aim is to prove the local-in-time existence of solutions to problem (1.1) with lowest possible regularity. Therefore we apply the L
p-approach.
The result of the paper is the following theorem.
Theorem 1.1. Let r > 3, v
0∈ W
r2−2/r(Ω), S ∈ W
r2−1/r. Then there exists T
0> 0 such that for all T ≤ T
0there exists a unique solution (u, p) of (1.1) such that u ∈ W
r2,1( e Ω
T), p ∈ W
r1,0( e Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
( e S
T)
and the following estimate holds : (1.8) kuk
W2,1r ( eΩT)
+ kpk
W1,0r ( eΩT)
+ kpk
W1−1/r,1/2−1/(2r)r ( eST)
≤ c(T )kv
0k
W2−2/r r (Ω). To prove Theorem 1.1 we need solvability of the Cauchy–Neumann prob- lem for the Stokes system from [3]. To recall the result we formulate the problem
(1.9)
u
t− div T(u, p) = F in Ω
T,
div u = G in Ω
T,
n · T(u, p) = H on S
T, u|
t=0= u
0in Ω, where Ω
T= Ω × [0, T ] and S
T= S × [0, T ].
Theorem 1.2 (see [3]). Let r > 3, F ∈ L
r(Ω
T), G ∈ W
r1,0(Ω
T), G
t− div F = div B + A, A, B ∈ L
r(Ω
T),
H ∈ W
1−1/r,1/2−1/(2r)r
(S
T), u
0∈ W
r2−2/r(Ω), S ∈ W
r2−2/r, and assume the compatibility conditions
(1.10) div u
0= G(x, 0), n · T(u
0, p
0)|
S= H(x, 0),
where p
0= p|
t=0. Then there exists a unique solution (u, p) to problem (1.9) such that
u ∈ W
r2,1(Ω
T), p ∈ W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T)
and the following estimate holds : (1.11) kuk
W2,1r (ΩT)
+ kpk
W1,0r (ΩT)
+ kpk
W1−1/r,1/2−1/(2r)r (ST)
≤ C(T )[kF k
Lr(ΩT)+ kGk
W1,0r (ΩT)
+ kBk
Lr(ΩT)+ kAk
Lr(ΩT)+ kHk
W1−1/r,1/2−1/(2r)r (ST)
+ ku
0k
W2−2/r r (Ω)],
where C(T ) is a constant increasing with T which does not depend on the solution (u, p).
Problem (1.1) without the self-gravitation force is considered in [4].
Moreover we recall that the local existence of solutions to problem (1.1) with surface tension is shown in [5].
2. Notation. We need the anisotropic Sobolev spaces W
rm,n(Q
T) where
m, n ∈ R
+∪ {0}, r ≥ 1 and Q
T= Q × (0, T ), with the norm
(2.1) kuk
rWm,nr (QT)
=
T\
0
\
Q
|u(x, t)|
rdx dt
+ X
0≤|m′|≤[|m|]
T
\
0
\
Q
|D
xm′u(x, t)|
rdx dt
+ X
|m′|=[|m|]
T
\
0
dt
\
Q
\
Q
|D
xm′u(x, t) − D
mx′u(x
′, t)|
r|x − x
′|
s+r(|m|−[|m|])dx dx
′+ X
0≤|n′|≤[|n|]
T\
0
\
Q
|D
tn′u(x, t)|
rdx dt
+
\
Q
dx
T
\
0 T
\
0
|D
t[n]u(x, t) − D
t[n](x, t
′)|
r|t − t
′|
1+r(n−[n])dt dt
′,
where s = dim Q, [α] is the integral part of α, D
xl= ∂
xl11. . . ∂
xlsswhere l = (l
1, . . . , l
s) is a multiindex.
In the proof we will use the following results.
Proposition 2.1 (see [1]). Let u ∈ W
rm,n(Ω
T), m, n ∈ R
+. If q ≥ r and
κ = X
3 i=1α
i+ 1
r − 1 q
1 m +
β + 1
r − 1 q
1 n < 1 then
kD
βtD
αxuk
Lq(ΩT)≤ ε
1−κkuk
Wrm,n(ΩT)+ cε
−κkuk
Lr(ΩT)for all ε ∈ (0, 1).
Proposition 2.2 (see [1, 2]). Let u ∈ W
r2m,m(Ω
T), m ∈ R
+. If 2m − 1/r > 0 then u = u|
STis well defined as a function in W
2m−1/r,m−1/(2r)r
(S
T)
and
kuk
W2m−1/r,m−1/(2r)r (ST)
≤ ckuk
W2m,mr (ΩT)
. Proposition 2.3 (see [1, 2]). Let u ∈ W
2m−1/r,m−1/(2r)r
(S
T), m ∈ R
+. If 2m − 1/r > 0 then there exists a function e u ∈ W
r2m,m(Ω
T) such that e
u|
ST= u and the following estimate holds:
ke uk
W2m,mr (ΩT)
≤ ckuk
W2m−1/r,m−1/(2r)r (ST)
.
In our considerations we will use well known imbedding theorems for
Sobolev spaces. All constants are denoted by the same letter c.
3. Proof of Theorem 1.1. To prove local existence of solutions to problem (1.1) we write it in the Lagrangian coordinates:
(3.1)
u
t− div
uT
u(u, q) = ∇
uU
uin Ω
T,
div
uu = 0 in Ω
T,
n
u· T
u(u, q) = −p
0n
uon S
T,
u|
t=0= v
0on Ω,
where u(ξ, t) = v(x(ξ, t), t), q(ξ, t) = p(x(ξ, t), t), ∇
u= ξ
i,x∇
ξi, T
u(u, q) = D
u(u) − qI, D
u(u) = ν{ξ
k,xiu
j,ξk+ ξ
k,xju
i,ξk}
i,j=1,2,3,
U
u(ξ, t) =
\
Ω
kJ
y(ξ′,t)dξ
′|x(ξ, t) − y(ξ
′, t)| ,
where J
x(ξ,t)is the Jacobian of the transformation x = x(ξ, t), div
uu = ξ
k,xiu
i,xk, n
u(ξ, t) = n(x(ξ, t), t), I is the unit matrix and the summation convention over repeated indices is used.
To prove the existence of solutions to (3.1) we use the following method of successive approximations:
(3.2)
u
m+1,t− div
umT
um(u
m+1, q
m+1) = ∇
umU
umin Ω
T,
div
umu
m+1= 0 in Ω
T,
n
um· T
um(u
m+1, q
m+1) = −p
0n
umon S
T,
u
m+1|
t=0= v
0on Ω,
where m = 0, 1, 2, . . . and u
mis treated as a given function. Assume u
0= 0 and q
0= 0.
To apply Theorem 1.2 we write (3.2) in the form
(3.3)
u
m+1,t− div T(u
m+1, q
m+1)
= div
umT
um(u
m+1, q
m+1)
− div T(u
m+1, q
m+1) + ∇
umU
umin Ω
T, div u
m+1= div u
m+1− div
umu
m+1in Ω
T, n
0· T(u
m+1, q
m+1)
= n
0· T(u
m+1, q
m+1)
− n
um· T
um(u
m+1, q
m+1) − p
0n
umon S
T,
u
m+1|
t=0= v
0on Ω,
where the operators without index contain derivatives with respect to ξ and
n
0is the unit outward vector normal to S.
First we obtain a uniform bound for the sequence {u
m}
∞m=0determined by (3.3).
Lemma 3.1. Assume that S ∈ W
r2−1/r, v
0∈ W
r2−2/r(Ω). Then (3.4) ku
mk
W2,1r (ΩT)
+ kq
mk
W1,0r (ΩT)
+ kq
mk
W1−1/r,1/2−1/(2r)r (ST)
≤ c(kv
0k
W2−2/rr (Ω)
, kSk
W2−2/rr
)
if T is small enough.
P r o o f. Applying Theorem 1.2 to problem (3.3) yields (3.5) ku
m+1k
W2,1r (ΩT)
+ kq
m+1k
W1,0r (ΩT)
+ kq
m+1k
W1−1/r,1/2−1/(2r)r (ST)
≤ ckdiv T(u
m+1, q
m+1) − div
umT
um(u
m+1, q
m+1)k
Lr(ΩT)+ ck∇
umU
umk
Lr(ΩT)+ ckdiv u
m+1− div
umu
m+1k
W1,0r (ΩT)
+ ckn
0· T(u
m+1, q
m+1)
− n
um· T
um(u
m+1, q
m+1)k
W1−1/r,1/2−1/(2r)r (ST)
+ ckn
umk
W1−1/r,1/2−1/(2r)r (ST)
+ ckv
0k
W2−2/rr (Ω)
+ ck((I − A
∗(u
m))u
m+1)
tk
Lr(ΩT),
where ((I − A
∗(u
m))u
m+1)
tis treated as e B from Theorem 1.2 ( e A = 0) and A
ij(u
m) = δ
ij+
Tt
0
u
mi,ξjdτ , A
∗kl(u
m) = A
−1lk(u
m). Here we note that div
umu
m+1= A
−1kl∂
ξlu
lm+1= div
ξ(A
∗u
m+1),
which follows from P
3 k=1 ∂∂ξk
A
lk(u
m)(ξ, t) = 0. All the above relations hold under the assumption that div
um−1u
m= 0.
To continue the induction we need to have div
umu
m+1= 0, but this is given by (3.3)
2.
Now we estimate the particular terms from the r.h.s. of (3.5). Define a
m= T
(r−1)/rku
mk
Wr2,1(ΩT), α
m(t) = {α
ij(u
m)} = {
Tt
0
u
mi,ξjdτ }.
For r > 3 we have kα
mk
L∞(Ω)≤ ca
m.
To estimate the first term on the r.h.s. of (3.5) we calculate div
umT
um(u
m+1, q
m+1) − div T(u
m+1, q
m+1)
= {ν(ξ
lxjξ
kxjxsx
sξlδ
σi+ ξ
lxjξ
kxixsx
sξlδ
σj)u
m+1σ,ξk+ ν(ξ
lxiξ
kxj− δ
jkδ
jl)u
m+1i,ξlξk+ ν(ξ
lxjξ
kξi− δ
ljδ
ki)u
m+1j,ξlξk− (ξ
lxj− δ
ljq
m+1,ξl)},
where the matrix ξ
,xdepends on u
m.
Since x
iξj= δ
ij+
Tt
0
u
iξj(τ ) dτ = δ
ij+ α
ijand ξ
jxiis the inverse matrix to x
iξj, we have
ξ
jxi= δ
ij+ φ
ij(α),
where φ
ijis a polynomial matrix-valued function which contains terms of α and α
2(α = {α
ij}). Then ξ
j,xixk= φ
ij,αrsα
rs,ξσξ
σ,xk, where α
rs,ξσ=
Tt
0
u
r,ξsξσ(τ ) dτ .
Then we write the first term of the r.h.s. of (3.5) in the form
kψ
1(α
m)(I − A(u
m))(u
m+1,ξξ+ q
m+1,ξ) + ψ
2(α
m)A(u
m)
,ξu
m+1,ξk
Lr(ΩT)≤ φ(a
m)a
m(ku
m+1k
W2,1r (ΩT)
+ kq
m+1k
W1,0r (ΩT)
), where ψ
iare some functions with ψ
i(0) 6= 0 and φ always denotes an in- creasing positive function.
We estimate the third term by the same quantity.
The fourth term can be expressed in the form
kψ
3(α
m)α
mu
m+1,ξ+ ψ
4(α
m)α
mq
m+1k
W1−1/r,1/2−1/(2r)r (ST)
≤ kψ
3(α
m)α
mu
m+1,ξk
W1−1/r,1/2−1/(2r)r (ST)
+ kψ
4(α
m)α
mq
m+1k
W1−1/r,1/2−1/(2r)r (ST)
≡ I + J, where
I ≤
T\0
kψ
3(α
m)α
mu
m+1,ξk
rW1r(Ω)
dτ
1/r+
\Ω
kψ
3(α
m)α
mu
m+1,ξk
rW1/2 r (0,T ) 1/r≡ L + K.
Next we have L ≤
T\0
kψ
3(α
m)α
mu
m+1,ξk
rLr(Ω)dτ
1/r+
T\0
ψ
3,αm(α
m)
t
\
0
u
m,ξξdτ α
mu
m+1,ξr
Lr(Ω)
dt
1/r+
T\0
ψ
3(α
m)
t
\
0
u
m,ξξdτ u
m+1,ξr
Lr(Ω)
dt
1/r+
T\0
kψ
3(α
m)α
mu
m+1,ξξk
rLr(Ω)dτ
1/r≡ L
1+ L
2+ L
3+ L
4.
Continuing, we have
L
1≤ φ(a
m(T ))a
m(T )ku
m+1k
W2,1r (ΩT)
, L
2+ L
3≤ φ(a
m(T ))a
m(T )
T\0
t
\
0
u
m,ξξdτ u
m+1,ξr
Lr(Ω)
dt
1/r≤ φ(a
m(T ))a
m(T )a
2m(T )ku
m+1k
Wr2,1(ΩT), L
4≤ φ(a
m(T ))a
m(T )ku
m+1k
W2,1r (ΩT)
. Next we examine
K ≤
\Ω
dξ
T\
0
dt
T\
0
dt
′|ψ
3(α
m(t)) − ψ
3(α
m(t
′))|
r|α
m(t)|
r|u
m+1,ξ(t)|
r|t − t
′|
1+r/2 1/r+
\Ω
dξ
T\
0
dt
T\
0
dt
′|ψ
3(α
m(t
′))|
r|α
m(t) − α
m(t
′)|
r|u
m+1,ξ(t)|
r|t − t
′|
1+r/2 1/r+
\Ω
dξ
T
\
0
dt
T
\
0
dt
′|ψ
3(α
m(t))|
r|α
m(t)|
r|u
m+1,ξ(t)−u
m+1,ξ(t
′)|
r|t − t
′|
1+r/2 1/r≡ K
1+ K
2+ K
3. Using the formula
(3.6) ψ
3(α
m(t)) − ψ
3(α
m(α
m(t
′))
= (α
m(t) − α
m(t
′))
1
\
0
ψ
3,αm(t′)(α
m(t
′) + s(α
m(t) − α
m(t
′)) ds we obtain
K
1+ K
2≤ φ(a
m(T ))
\Ω
dξ
T
\
0
dt
T
\
0
dt
′|
Tt
t′
u
m,ξdτ |
r|u
m+1,ξ(t)|
r|t − t
′|
1+r/2 1/r≤ φ(a
m(T ))
\Ω
dξ
T
\
0
dt
T
\
0
dt
′|t − t
′|
r/2−2t
\
t′
|u
m,ξ|
rdτ |u
m+1,ξ(t)|
r1/r≤ φ(a
m(T ))ku
mk
Wr2,1(ΩT) \Ω
dξ
T\
0
dt
T\
0
dt
′|t − t
′|
r/2−2|u
m+1,ξ|
r1/r≡ K
4. Integrating with respect to t
′we get (r/2 > 1)
K
4≤ φ(a
m(T ))T
r/2−1ku
mk
Wr2,1(ΩT)ku
m+1k
Wr2,1(ΩT).
Finally
K
3≤ φ(a
m(T ))a
m(T )ku
m+1k
W2,1r (ΩT)
. Summarizing the above considerations we obtain
I ≤ φ(a
m(T ))a
m(T )ku
m+1k
W2,1r (ΩT)
+ φ(a
m(T ))T
r/2−1ku
mk
W2,1r (ΩT)
ku
m+1k
W2,1r (ΩT)
. Similarly, we obtain
J ≤ φ(a
m(T ))a
m(T )ke q
m+1k
W1,1/2 r (ΩT)+ φ(a
m(T ))T
r/2−1ku
mk
W2,1r (ΩT)
ke q
m+1k
W1,1/2 r (ΩT), where e q
m+1is an extension of q
m+1∈ W
1−1/r,1/2−1/(2r)r
(S
T).
The fifth term on the r.h.s. of (3.5) is estimated by kψ
5(α
m)k
W1−1/r,1/2−1/(2r)r (ST)
≤
T\0
kψ
5(α
m(t))k
W1−1/rr (S)
dt
1/r+
\S
kψ
5(α
m(t))k
W1/2−1/(2r)r (0,T )
dξ
1/r≡ M
1+ M
2, where
M
1≤ φ(a
m(T ))
T\0
dt
t
\
0
u
m,ξdτ
r Wr1(Ω)
1/r≤ φ(a
m(T ))T
1/ra
m(T ),
M
2≤
\S
dξ
T
\
0
dt
T
\
0
dt
′|ψ
5(α
m(t)) − ψ
5(α
m(t
′))|
r|t − t
′|
1+r(1/2−1/(2r)) 1/r; using (3.6) we have
φ(a
m(T ))
\S
dξ
T\
0
dt
T\
0
dt
′|
Tt
t′
u
m,ξ(ξ, τ ) dτ |
r|t − t
′|
1+r(1/2−1/(2r)) 1/r≤
\S
dξ
T\
0
|u
m,ξ(ξ, τ )|
rdτ
1/rT\0
dt
T\
0
|t − t
′|
r/2−3/2dt
′1/r≤ φ(a
m(T ))T
1/2+r/2ku
mk
Wr2,1(ΩT).
The seventh term of the r.h.s. of (3.5) will be considered in the from k((I − A
∗(u
m))u
m+1)
tk
Lr(ΩT)≤ kψ
6(α
m)α
mu
m+1,tk
Lr(ΩT)+ kψ
7(α
m)u
m,ξu
m+1k
Lr(ΩT)≡ N
1+ N
2and we have
N
1≤ φ(a
m(T ))|α
m| ku
m+1k
W2,1r (ΩT)
, N
2≤ φ(a
m(T ))
(u
m,ξ− v
0,ξ+ v
0,ξ) v
0+
t
\
0
u
m+1,tdt
Lr(ΩT)
(3.7)
≤ φ(a
m(T ))T
1/rkv
0k
2W2−2/rr (Ω)
+ φ(a
m(T ))ku
mk
W2,1r (ΩT)
a
m+1+ φ(a
m(T ))kv
0k
W2−2/r r (Ω)T
β× (ku
mk
W2,1r (ΩT)
+ kv
0k
W2−2/rr (Ω)
).
In the last term of the r.h.s. of (3.7)
2we have applied the imbedding W
r1,1/2(Ω
T) ⊂ C
β(0, T ; L
r(Ω)) with 0 < β < 1/2 − 1/r. This enables us to get
ku
m,ξ− v
0,ξk
Lr≤ T
β(ku
mk
W2,1r (ΩT)
+ kv
0k
W2−2/r r (Ω)).
Finally we consider the second term of the r.h.s. of (3.5). We have
∇
umU
um= ∇
um\
Ω
J
xum(ξ′,t)|x
um(ξ, t) − x
um(ξ
′, t)| dξ
′= −
\
Ω
∇
umx
um(ξ, t) · (x
um(ξ, t) − x
um(ξ
′, t))
|x
um(ξ, t) − x
um(ξ
′, t)|
3J
xum(ξ′,t)dξ
′= −
\
Ω
∇
umx
um(ξ, t) · (ξ −ξ
′)(1+
T1 0
ds
Tt
0
∂
su
m(ξ
′+s(ξ − ξ
′), τ ) dτ )
|ξ −ξ
′|
3|1 +
T1 0
ds
Tt
0
∂
su
m(ξ
′+ s(ξ − ξ
′), τ ) dτ |
3× J
xum(ξ′,t)dξ
′. Assuming that
t
\
0
ku
mξ(·, τ )k
L∞(Ω)dτ < 1, we obtain
k∇
umU
umk
Lr(ΩT)≤ φ(a
m(T ))T
1/r\
Ω
dξ
′|ξ − ξ
′|
2Lr(Ω)
≤ φ(a
m(T ))T
1/r.
For simplicity we introduce X
k= ku
kk
W2,1r (ΩT)
+ kq
kk
W1,0r (ΩT)
+ kq
mk
W1−1/r,1/2−1/(2r)r (ST)
.
Summing up the estimates for all terms of the r.h.s. of (3.5) we get X
m+1≤ a
mφ(a
m)X
m+1+ φ(a
m)T
r/2−1X
mX
m+1+ φ(a
m)a
mT
1/r+ φ(a
m)T
1/2+r/2X
m+ φ(a
m)X
ma
m+1(T ) + (T
1/r+ T
β)φ(a
m) + φ(a
m(T ))T
βX
m.
Putting
a = min
r − 1 r , r − 2
2 , 1 r , 1
2 + r 2 , β
we have
X
m+1≤ T
aφ(a
m)X
mX
m+1+ T
aφ(a
m)X
m(3.8)
+ φ(a
m)T
aX
mX
m+1+ T
aφ(a
m).
By induction we prove that X
k≤ 1 (X
0= 0). Taking T such that T ≤ 1 and T
aφ(1) ≤ 1/4, inserting X
m≤ 1 in (3.8), we obtain
X
m+1≤ 1
4 X
m+1+ 1 4 + 1
4 X
m+1+ 1 4 , which gives X
m+1≤ 1.
The proof of the lemma is complete.
Lemma 3.2. Assume that S ∈ W
r2−1/r, v
0∈ W
r2−2/r(Ω). Then there exist u ∈ W
r2,1(Ω
T) and p ∈ W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T) such that
u
m→ u in W
r2,1(Ω
T) and q
m→ p in W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T)
as m → ∞ for T small enough.
P r o o f. We show that {(u
m, q
m)}
∞n=1is convergent. For this purpose we consider v
m= u
m+1− u
m, r
m= q
m+1− q
mwhich satisfy the system
(3.9)
v
m,t− div T(v
m, r
m) = div
umT
um(u
m+1, q
m+1)
− div
um−1T
um−1(u
m, q
m) − div T(v
m, q
m)
+∇
umU
um− ∇
um−1U
um−1≡ I in Ω
T, div v
m= div v
m− div
umu
m+1+ div
um−1u
m≡ J in Ω
T, n
0· T(v
m, r
m) = n
0T(v
m, r
m) − n
umT(u
m+1, q
m+1)
+n
um−1(u
m, q
m) − p
0n
um+ p
0n
um−1≡ K on S
T,
v
m|
t=0= 0 on Ω.
By Theorem 1.2 we obtain an estimate on solutions of (3.9):
(3.10) kv
mk
W2,1r (ΩT)
+ kr
mk
W1,0r (ΩT)
+ kr
mk
W1−1/r,1/2−1/(2r)r (ST)
≤ c(kIk
Lr(ΩT)+ kJk
W1,0r (ΩT)
+ kKk
W1−1/r,1/2−1/(2r)r (ST)
+ kBk
Lr(ΩT)),
where B is defined by the relation J
t= div B.
First we estimate the terms of the r.h.s. of (3.9)
1in L
r(Ω
T). Let I = I
1+ I
2. We examine
I
1= div
umT
um(u
m+1, q
m+1) − div
um−1T
um−1(u
m, q
m) − div T(v
m, q
m)
= (div
umT
um− div T)(v
m, r
m)
+ (div
umT
um− div
um−1T
um−1)(u
m, q
m) ≡ I
11+ I
12.
I
11is estimated in the same way as the first term of the r.h.s. of (3.5):
kI
11k
Lr(ΩT)≤ φ(a
m)a
m(kvk
Wr2,1(ΩT)+ kr
mk
Wr1,0(ΩT)).
For the second term we have
kI
12k
Lr(ΩT)= k(div
um(T
um− T
um−1)
+ (div
um− div
um−1)T
um−1)(u
m, q
m)k
Lr(ΩT)≤ φ(a
m)T
(r−1)/rkv
m−1k
Wr2,1(ΩT). Next we consider
I
2= ∇
umU
um− ∇
um−1U
um−1= ∇
um(U
um− U
um−1) + (∇
um− ∇
um−1)U
um−1≡ I
21+ I
22. The first term is
I
21= − A(u
m)
\
Ω
x
m(ξ, t) − y
m(ξ
′, t)
|x
m(ξ, t) − y
m(ξ
′, t)|
3J
ym(ξ′,t)− x
m−1(ξ, t) − y
m−1(ξ
′, t)
|x
m−1(ξ, t) − y
m−1(ξ
′, t)|
3J
ym−1(ξ′,t)dξ
′= − A(u
m)
\
Ω
x
m(ξ, t) − y
m(ξ
′, t)
|x
m(ξ, t) − y
m(ξ
′, t)|
3(J
ym(ξ′,t)− J
ym−1(ξ′,t)) dξ
′−A(u
m)
\
Ω
x
m(ξ, t) − y
m(ξ
′, t)
|x
m(ξ, t) − y
m(ξ
′, t)|
3− x
m−1(ξ, t) − y
m−1(ξ
′, t)
|x
m−1(ξ, t) − y
m−1(ξ
′, t)|
3J
ym−1(ξ′,t)dξ
′≡ I
211+ I
212, where x
k(ξ, t) = ξ +
Tt
0
u
k(ξ, t
′) dt
′and y
k(ξ
′, t) = ξ
′+
Tt
0
u
k(ξ
′, t
′) dt
′. Since
|J
ym(ξ′,t)− J
ym−1(ξ′,t)| ≤ cku
m− u
m−1k
W2,1r (ΩT)
, we have
kI
211k
Lr(ΩT)≤ φ(a
m)T
1/rkv
m−1k
W2,1r (ΩT)
.
The same holds for I
212. Since
\
x
m(ξ, t) − y
m(ξ
′, t)
|x
m(ξ, t) − y
m(ξ
′, t)|
3− x
m−1(ξ, t) − y
m−1(ξ
′, t)
|x
m−1(ξ, t) − y
m−1(ξ
′, t)|
3dξ
′≤ cku
m− u
m−1k
W2,1r (ΩT)
, we obtain
kI
212k
Lr(ΩT)≤ φ(a
m)T
1/rkv
m−1k
W2,1r (ΩT)
. We estimate J in W
r1,0(Ω
T) by the same quantity.
Let K = K
1+ K
2, where
K
1= n
0T(v
m, r
m) − n
umT(u
m+1, q
m+1) + n
um−1(u
m, q
m)
= (n
0T(v
m, r
m) − n
umT(v
m, r
m))
+ (n
umT
um− n
um−1T
um−1)(u
m, q
m) ≡ K
11+ K
12and
K
2= −p
0(n
um− n
um−1).
The term K
11is estimated just as the fourth term of the r.h.s. of (3.5):
kK
11k
W1−1/r,1/2−1/(2r)r (ST)
≤ φ(a
m(T ))a
m(T )kv
mk
Wr2,1(ΩT)+ φ(a
m(T ))T
r/2−1ku
mk
Wr2,1(ΩT)kv
mk
Wr2,1(ΩT). The second term is
K
12= (n
um(T
um− T
um−1) + (n
um− n
um−1)T
um−1)(u
m, q
m) ≡ K
121+ K
122. For K
121we have
kK
121k
W1−1/r,1/2−1/(2r)r (ST)
≤ φ(a
m(T ))T
(r−1)/rkv
m−1k
W2,1r (ΩT)
. Since
kn
um− n
um−1k
W1−1/r,1/2−1/(2r)r (ST)
≤ φ(a
m(T ))T
(r−1)/rkv
m−1k
W2,1r (ΩT)
, we conclude that
kK
122k
W1−1/r,1/2−1/(2r)r (ST)
+ kK
2k
W1−1/r,1/2−1/(2r)r (ST)
≤ φ(a
m(T ))T
(r−1)/rkv
m−1k
W2,1r (ΩT)
. Finally we have to examine B which is defined by J
t= div B. We have
J = div v
m− div
umu
m−1+ div
um−1u
m= (div v
m− div
umv
m) + (div
um−1u
m− div
umu
m) ≡ J
1+ J
2. To examine J
1we proceed as in the case of the seventh term of the r.h.s.
of (3.5). By the same argument as in Lemma 3.1 we have J
2= (div
um−1− div
um)u
m= div ·((A
∗m−1− A
∗m)u
m).
Hence we put B
2= ((A
∗m−1− A
∗m)u
m)
tand we get, just as for N
2in the
proof of Lemma 3.1, the following estimate:
kB
2k
Lr(ΩT)≤ φ(a
m(T ))(T
(r−1)/r+ T
β)kv
m−1k
W2,1r (ΩT)
, where 0 < β < 1/2 − 1/r.
Define
Y
m= kv
mk
Wr2,1(ΩT)+ kr
mk
Wr1,0(ΩT)+ kr
mk
W1−1/r,1/2−1/(2r)r (ST)
.
Summing up the estimates for all terms of the r.h.s. of (3.10) we obtain Y
m≤ φ(a
m(T ))(T
(r−1)/r+T
r/2−1+T
β)Y
m+φ(a
m(T ))(T
(r−1)/r+T
β)Y
m−1. Taking T so small that φ(a
m(T ))(T
(r−1)/r+ T
r/2−1+ T
β) ≤ 1/2 we get
Y
m≤ φ(a
m(T ))(T
(r−1)/r+ T
β)Y
m−1.
Thus if φ(a
m(T ))(T
(r−1)/r+ T
β) < 1 we have a contraction, hence Y
m→ 0 as m → ∞. This yields the existence of u ∈ W
r2,1(Ω
T) and p ∈ W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T) such that
u
m→ u in W
r2,1(Ω
T),
q
m→ p in W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T).
The proof of the lemma is complete.
By Lemma 3.2 we see that system (3.1) has a unique solution (u, q) in W
r2,1(Ω
T) × W
r1,0(Ω
T) ∩ W
1−1/r,1/2−1/(2r)r
(S
T). By Lemma 3.1 we get the estimate
(3.11) kuk
Wr2,1(ΩT)+ kqk
Wr1,0(ΩT)+ kqk
W1−1/r,1/2−1/(2r)r (ST)
≤ c(kv
0k
W2−2/rr (Ω)
, kSk
W2−2/rr