Topological Methods in Nonlinear Analysis
Journal of the Juliusz Schauder Center
Volume 10, 1997, 195–223
ON LOCAL MOTION OF A COMPRESSIBLE BAROTROPIC
VISCOUS FLUID WITH THE BOUNDARY SLIP CONDITION
Marek Burnat — Wojciech M. ZajĄczkowski
(Submitted by A. Granas)
Dedicated to O. A. Ladyzhenskaya
1. Introduction
We consider the motion of a compressible barotropic viscous fluid in a bounded domain Ω ⊂ R3 with the boundary slip condition. Let ρ = ρ(x, t) be the
density of the fluid, v = v(x, t) the velocity, p = p(ρ(x, t)) the pressure, f =
f (x, t) the external force field per unit mass. Then the motion is described by
the following problem (see [3]):
(1.1)
ρ(vt + v · ∇v) − div T(v, p) = ρf
in ΩT = Ω × (0, T ),
ρt + div (ρv) = 0
in ΩT ,
ρ|t=0 = ρ0
in Ω,
v|t=0 = v0
τ α · T(v, p) · n + γv · τ α = 0,
α = 1, 2, on S T = S × (0, T ),
on S T ,
v·n=0
where T(v, p) is the stress tensor of the form
(1.2)
T(v, p) = {Tij (v, p)}i,j=1,2,3
= {µ(∂xi vj + ∂xj vi ) + (ν − µ)div vδij − pδij }i,j=1,2,3 ,
1991 Mathematics Subject Classification. 35K50, 35F30, 76N10.
Key words and phrases. Compressible barotropic viscous fluid, local existence, potential
theory, anisotropic Sobolev spaces with fractional derivatives.
The second named author was supported by Polish KBN Grant 2P03A 065 08.
c
1997
Juliusz Schauder Center for Nonlinear Studies
195
196
M. Burnat — W. M. Zajączkowski
n, τ α , α = 1, 2, are unit orthonormal vectors such that n is the outward normal
vector and τ 1 , τ 2 are tangent to S. Finally, γ is a positive constant and µ, ν are
constant viscosity coefficients.
By virtue of (1.1)2 and (1.1)5 the total mass is conserved, so
Z
Z
(1.3)
ρ0 (x) dx =
ρ(x, t) dx = M.
Ω
Ω
Moreover, from thermodynamic considerations we have
ν > µ/3.
(1.4)
Let us introduce the quantity for vectors u, v ∈ H 1 (Ω)
Z
(1.5)
EΩ (u, v) = (∂xi uj + ∂xj ui )(∂xi vj + ∂xj vi ) dx,
Ω
where the summation convention over the repeated indices is used. We assume
that EΩ (u) = EΩ (u, u).
We recall from [7] that the vectors for which EΩ (u) = 0 form a finite dimensional affine space of vectors such that
u = A + B × x,
where A and B are constant vectors.
e = {u : EΩ (u) < ∞, u · n = 0 on S}. If Ω is a region obtained
We define H(Ω)
e
by rotation about a vector B, we denote by H(Ω) the space of functions in H(Ω)
satisfying the condition
Z
u(x)u0 (x) dx = 0,
Ω
e
where u0 = B × x; otherwise we set H(Ω) = H(Ω)
(see [7]).
From Lemmas 2.1, 2.2 from [4] (see also Lemma 4 in [7]) we infer
Lemma 1.1 (Korn inequality). Let S ∈ H 3+α , α ∈ (1/2, 1). Then for any
u ∈ H(Ω),
(1.6)
kuk21,Ω ≤ cEΩ (u),
where c is a positive constant.
We introduce Lagrangian coordinates as the initial data for the Cauchy problem
dx
(1.7)
= v(x, t), x|t=0 = ξ,
dt
then
Z t
(1.8)
x=ξ+
v(ξ, τ ) dτ ≡ xv (ξ, t),
0
and v(ξ, t) = v(xv (ξ, t), t). We sometimes omit the index v in xv .
On Local Motion of a Compressible Barotropic Viscous Fluid
197
To prove the existence of solutions to problem (1.1) we integrate the equation
of continuity using the Lagrangian coordinates (1.8). Therefore we have
(1.9)
ρ(x, t) = ρ0 (ξ) exp
Z
−
t
(∇v · v)(ξ, τ ) dτ ,
0
where ∇v = (∂ξk /∂xi )v i,ξk .
Assuming that v = u in (1.9) and assuming that transformation (1.8) is
generated by vector u, we also consider instead of (1.1) the following linearized
problem
in ΩT ,
ρu v t − divu Tu (v, p(ρu )) = ρu fu
(1.10)
τ uα · Tu (v, p(ρu )) · nu + γv · τ uα = 0, α = 1, 2, on S T ,
v · nu = 0
on S T ,
v|t=0 = v0
in Ω,
where ρu (ξ, t) = ρ(xu (ξ, t), t), fu (ξ, t) = f (xu (ξ, t), t) and see also (6.3), which
assigns the transformation
v = Φ(u).
(1.11)
A fixed point of (1.11) is a solution to problem (1.1).
To prove the existence of solutions to problem (1.10) we first consider the
following problem
ωt − div D(ω) = F
(1.12)
in ΩT ,
τ α · D(ω) · n = Gα , α = 1, 2, on S T ,
ω · n = G3
on S T ,
ω|t=0 = ω0
in Ω.
To consider this problem we use the potential technique developed by V. A. Solonnikov. The existence will be proved in Sobolev–Slobodetskiı̆ spaces.
To prove the existence of solutions to problem (1.10) we need also examine
the following problem
ηωt − div D(ω) = F1
(1.13)
in ΩT ,
τ α · D(ω) · n = G1α , α = 1, 2, on S T ,
ω · n = G13
on S T ,
ω|t=0 = ω0
in Ω,
where η ≥ η0 > 0, η0 is a constant and η is a given function.
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M. Burnat — W. M. Zajączkowski
2. Notation and auxiliary results
l,l/2
We use the anisotropic Sobolev–Slobodetskiı̆ spaces W2 (QT ), l ∈ R+ ,
QT = Q × (0, T ), where Q is either Ω (a domain in R3 ) or S (the boundary
of Ω), with the norm
kuk2W l,l/2 (QT )
2
T
Z
kuk2W l (Q)
2
=
0
≡ kuk2W l,0 (QT )
2
Z
kuk2W l/2 (0,T ) dx
dt +
+
2
Q
2
kukW 0,l/2 (QT ) ,
2
where
X
kuk2W l (Q) =
2
kDα uk2L2 (Q) ,
|α|≤l
for integer l, and
kuk2W l (Q) =
X
2
X Z Z |Dα u(x) − Dα0 u(x0 )|2
x
x
dx dx0 ,
0 |s+2(l−[l])
|x
−
x
Q Q
kDα uk2L2 (Q) +
|α|=[l]
|α|≤[l]
for noninteger l, where s = dim Q, Dxα = ∂xα11 . . . ∂xαss , α = (α1 , . . . , αs ) is a multiindex, [l] is the integer part of l. For Q = S the above norm is introduced by
using local mappings and a partition of unity.
Finally,
X
kuk2W l/2 (0,T ) =
k∂ti uk2L2 (0,T ) ,
2
i≤l/2
for integer l/2, and
kuk2W l/2 (0,T )
2
=
X
k∂ti uk2L2 (0,T )
X Z
+
i≤[l/2]
i=[l/2]
0
T
Z
0
T
|∂ti u(t) − ∂ti0 u(t0 )|2
dt dt0 .
|t − t0 |1+2(l/2−[l/2])
Moreover,
|u|p,Ω = kukLp (Ω) , p ∈ [1, ∞],
Z Z
1/2
|u(x) − u(x0 )|2
0
[u]α,Ω =
dx
dx
,
0 3+2α
Ω Ω |x − x |
Z T Z T
1/2
|u(t) − u(t0 )|2
0
[u]α,(0,T ) =
dt dt
,
|t − t0 |1+2α
0
0
and
Z
T
Z Z
dt
[u]α,ΩT ,x =
0
Ω
Z
Z
T
Ω
Z
dx
[u]α,ΩT ,t =
Ω
0
0
|u(x, t) − u(x0 , t)|2
dx dx0
|x − x0 |3+2α
T
1/2
|u(x, t) − u(x, t0 )|2
dt dt0
|t − t0 |1+2α
,
1/2
.
199
On Local Motion of a Compressible Barotropic Viscous Fluid
To consider the problems with vanishing initial conditions we need a space of
functions which admit a zero extension to t < 0. Therefore, for every γ ≥ 0, we
l,l/2
introduce the space Hγ (QT ) with the norm (see [1], [5], [8])
Z T
2
e−2γt kuk2W l (Q) dt + kuk2H 0,l/2 (QT ) .
kukH l,l/2 (QT ) =
γ
2
0
γ
For l/2 6∈ Z,
kuk2H 0,l/2 (QT )
γ
=γ
l
T
Z
e−2γt kuk2L2 (Q) dt
0
Z
+
T
e−2γt dt
0
Z
∞
k∂tk u0 ( · , t − τ ) − ∂tk u0 ( · , t)k2L2 (Q)
τ 1+2(l/2−k)
0
dτ,
where k = [l/2] < l/2, and u0 (x, t) = u(x, t) for t > 0, u0 (x, t) = 0 for t < 0.
For l/2 ∈ Z,
Z T
l/2
e−2γt (γ l kuk2L2 (Q) + k∂t uk2L2 (Q) ) dt,
kuk2H 0,l/2 (QT ) =
γ
0
and we assume that ∂tj u|t=0 = 0, j = 0, . . . , l/2 − 1, so u0 (x, t) has a generalized
l/2
derivative ∂t u0 in Q × (−∞, T ).
For simplicity we write
kukl,QT = kukW l,l/2 (QT ) ,
kukl,Q = kukW2l (Q) ,
kukl,γ,QT = kukH l,l/2 (QT ) ,
kukp,Q = kukLp (Q) .
2
γ
In the above definitions we can use the notation
X
l/2
X
α 2
α 2
kukW2l (Q) =
|Dx u|2,Q +
[Dx u]l−[l],Q
.
|α|≤[l]
|α|=[l]
= Rn × (0, T ), Dn+1
= Rn+ × (0, T ),
We set Rn+ = {x ∈ Rn : xn > 0}, Rn+1
T
T
n = 2, 3. For functions defined in Rn+1
and vanishing sufficiently fast at infinity
∞
we define the Fourier transform with respect to x and the Laplace transform
with respect to t by the formula
Z ∞
Z
−st
e
(2.1)
f (ξ, s) =
e
dt
f (x, t)e−ix·ξ dx,
0
where x · ξ =
Pn
i=1
k|u|k2l,γ,Rn+1 =
∞
Z
Rn
xi ξi . Hence we define the norm
Z ∞
dξ
|e
u(ξ, s)|2 (|s| + |ξ|2 )l dξ0 ,
Rn
s = γ + iξ0 , γ ∈ R+ .
−∞
Similarly, for functions defined in Dn+1
∞ , we have
Z ∞
Z
0 0
0
−st
e
f (ξ , xn , s) =
e
dt
f (x, t)e−ix ·ξ dx0 ,
0
Rn−1
200
M. Burnat — W. M. Zajączkowski
where x0 = (x1 , . . . , xn−1 ), ξ 0 = (ξ1 , . . . , ξn−1 ), and introduce the norm
Z ∞
XZ
2
0
k∂xj n u
e(ξ 0 , xn , s)k20,R1 (|s| + |ξ 0 |2 )l−j dξ0
k|u|kl,γ,Dn+1 =
dξ
∞
j≤[l]
Rn−1
Z
+
dξ 0
Rn−1
+
−∞
Z
∞
−∞
ke
u(ξ 0 , · , s)k2l,R1 dξ0 ,
+
s = γ + iξ0 ,
γ ∈ R+ .
We introduce a partition of unity. Let us define two collections of open subsets
S
S
{ω } and {Ω(k) }, k ∈ M ∪ N , such that ω (k) ⊂ Ω(k) ⊂ Ω, k ω (k) = k Ω(k) =
Ω, Ω(k) ∩ S = ∅ for k ∈ M and Ω(k) ∩ S 6= ∅ for k ∈ N . We assume that
at most N0 of the Ω(k) have nonempty intersection, and supk diam Ω(k) ≤ 2λ
for some λ > 0. Let ζ (k) (x) be a smooth function such that 0 ≤ ζ (k) (x) ≤ 1,
ζ (k) (x) = 1 for x ∈ ω (k) , ζ (k) (x) = 0 for x ∈ Ω\Ω(k) and |Dxν ζ (k) (x)| ≤ c/λ|ν| .
P
Then 1 ≤ k (ζ (k) (x))2 ≤ N0 . Introducing the function
(k)
ζ (k) (x)
,
(l)
2
l (ζ (x))
η (k) (x) = P
P
we have η (k) (x) = 0 for x ∈ Ω\Ω(k) , k η (k) (x)ζ (k) (x) = 1 and |Dxν η (k) (x)| ≤
c/λ|ν| . By ξ (k) we denote the center of ω (k) and Ω(k) for k ∈ M and the center
of ω (k) ∩ S and Ω(k) ∩ S for k ∈ N .
Considering problems invariant with respect to translations and rotations we
can introduce a local coordinates system y = (y1 , y2 , y3 ) with the center at ξ (k)
such that the part Se(k) = S ∩Ω(k) of the boundary is described by y3 = F (y1 , y2 ).
Then we consider new coordinates defined by
zi = y i ,
i = 1, 2,
z3 = y3 − F (y1 , y2 ).
b (k) ⊃ ω
We will denote this transformation by Ω
b (k) 3 z = Φk (y), where y ∈ ω (k) ⊂
b (k) are described in local coordinates at ξ (k)
Ω(k) ; we assume that the sets ω
b (k) , Ω
by the inequalities
|yi | < λ,
i = 1, 2,
0 < y3 − F (y1 , y2 ) < λ,
|yi | < 2λ,
i = 1, 2,
0 < y3 − F (y1 , y2 ) < 2λ,
respectively.
Let y = Yk (x) be a transformation from coordinates x to local coordinates
y, which is a composition of a translation and a rotation. Then we set
−1
u
b(k) (z, t) = u(Φ−1
k ◦ Yk (z), t),
u
e(k) (z, t) = u
b(k) (z, t)ζb(k) (z, t).
l,l/2
From [5] we recall the necessary for us properties of spaces Hγ
l,l/2
Hγ (Dn+1
), l ∈ R+ .
T
(Rn+1
) and
T
On Local Motion of a Compressible Barotropic Viscous Fluid
201
l,l/2
Lemma 2.1. Any function u ∈ Hγ (Rn+1
), T < ∞, can be extended to
T
l,l/2
n+1
R∞ in such a way that the extended function u0 ∈ Hγ (Rn+1
∞ ) and
ku0 kl,γ,Rn+1
≤ ckukl,γ,Rn+1 .
∞
(2.2)
T
l,l/2
Any function u ∈ Hγ (Dn+1
), T < ∞, can be extended to Rn+1
in such a way
T
T
l,l/2
n+1
0
that the extended function u ∈ Hγ (RT ) and
ku0 kl,γ,Rn+1 ≤ ckukl,γ,Dn+1 .
(2.3)
T
T
Lemma 2.2. There exist constants c1 and c2 , which do not depend on u
and γ, such that
(2.4)
c1 k|u|kl,γ,Rn+1
≤ kukl,γ,Rn+1
≤ c2 k|u|kl,γ,Rn+1
.
∞
∞
∞
Lemma 2.3. There exist constants c3 and c4 , which do not depend on u
and γ, such that
(2.5)
c3 k|u|kl,γ,Dn+1
≤ kukl,γ,Dn+1
≤ c4 k|u|kl,γ,Dn+1
.
∞
∞
∞
From [5] we recall
r,r/2
Lemma 2.4. Let u ∈ Hγ
r − |α|,
(2.6)
(ΩT ). Then for every ε ∈ (0, 1) and 0 ≤ q <
kDxα ukq,γ,ΩT ≤ εr−|α|−q kukr,γ,ΩT + cε−q−|α| ke−γt uk0,ΩT
≤ (εr−|α|−q + cγ −r/2 ε−q−|α| )kukr,γ,ΩT .
l,l/2
) and 0 < 2m + |α| < l. Then
Lemma 2.5 (see [5]). Let u ∈ Hγ (Rn+1
T
l1 ,l1 /2
n+1
m α
(RT ), where l1 = l − 2m − |α| and
∂t Dx u ∈ Hγ
(2.7)
k∂tm Dxα ukl1 ,γ,Rn+1 ≤ ckukl,γ,Rn+1 .
T
T
Moreover, for ρ ∈ (0, l1 ) and ε > 0,
(2.8)
k∂tm Dxα ukρ,γ,Rn+1 ≤ εl1 −ρ kukl,γ,Rn+1 + cε−h ke−γt uk0,Rn+1 ,
T
T
T
where h = ρ + 2m + |α|.
l,l/2
) and 0 ≤ 2m + |α| < l − 1/2. Then ∂tm Dxα u|xn =0 ∈
Let u ∈ Hγ (Dn+1
T
l2 ,l2 /2
n
Hγ
(RT ), where l2 = l − 2m − |α| − 1/2 and
(2.9)
k∂tm Dxα u|xn =0 kl2 ,γ,RnT ≤ ckukl,γ,Dn+1 .
T
To obtain the results from Section 6 we need
202
M. Burnat — W. M. Zajączkowski
Lemma 2.6 (see [5]). Let’s assume that a ∈ W2l+1 (Ω), f ∈ W2l (Ω), g ∈
l > 1/2, Ω ⊂ R3 . Then the following estimates hold
W2l+1 (Ω),
kaf kl,Ω ≤ c1 |a|∞,Ω kf kl,Ω + kakl+1,Ω (εkf kl,Ω + c2 (ε)kf k0,Ω ),
(2.10)
kaf kl,Ω ≤ c3 kf kl,Ω (εkakl+1,Ω + c4 (ε)kak0,Ω ),
kagkl+1,Ω ≤ c5 |a|∞,Ω kgkl+1,Ω + kakl+1,Ω (εkgkl+1,Ω + c6 (ε)kgk0,Ω ),
kagkl+1,Ω ≤ c7 kakl+1,Ω kgkl+1,Ω ,
where ε ∈ (0, 1).
3. Existence of solutions to problem (1.12) with
vanishing initial data and in the half space
Considering the problem (1.12) in the half–space x3 > 0 and with vanishing
initial conditions we have
ωt − div D(ω) = 0
x3 > 0,
∂ωi
∂ω3
µ
+
= bi , i = 1, 2, x3 = 0,
∂x3
∂xi
ω 3 = b3
x3 = 0,
(3.1)
ω|t=0 = 0
x3 > 0.
By applying the Fourier–Laplace transformation
Z ∞
Z
0 0
fe(ξ 0 , x3 , s) =
e−st dt
f (x, t)e−ix ·ξ dx0 ,
(3.2)
0
Res > 0,
R2
s = γ + iξ0 ,
where ξ 0 = (ξ1 , ξ2 ), x0 = (x1 , x2 ), x0 · ξ 0 = x1 ξ1 + x2 ξ2 , we obtain problem (3.1)
in the form
(3.3)
(3.4)
de
ω3
d2 ω
ek
+ νiξk
− (s + µξ 2 )e
ωk − νξk ξj ω
ej = 0, k = 1, 2, x3
2
dx3
dx3
d2 ω
e3
de
ωj
(µ + ν) 2 + νiξj
− (s + µξ 2 )e
ω3 = 0,
x3
dx3
dx3
de
ωk
µ
+ µiξk ω
e3 = b̃k ,
k = 1, 2, x3
dx3
ω
e3 = eb3 ,
x3
µ
> 0,
> 0,
= 0,
= 0,
ω
e → 0 as x3 → ∞, where ξ = (ξ1 , ξ2 ), ξ 2 = ξ12 +ξ22 and the summation convention
over the repeated indices is assumed.
Every solution to (3.3) vanishing at infinity has the form
(3.5)
ω
e = Φ(ξ, s)e−τ1 x3 + ψ(ξ, s)(ξ1 , ξ2 , iτ2 )e−τ2 x3 ,
p
where Φ(ξ, s) = (φ1 , φ2 , (i/τ1 )ξ · φ), φj = φj (ξ, s), j = 1, 2, τ1 = s/µ + ξ 2 , τ2 =
p
s/(µ + ν) + ξ 2 , arg τj ∈ (−π/4, π/4), j = 1, 2, ξ ·φ = ξ1 φ1 +ξ2 φ2 , φ = (φ1 , φ2 ).
On Local Motion of a Compressible Barotropic Viscous Fluid
203
Inserting (3.5) into (3.4) yields
τ1 φk + τ2 ξk ψ = ieb3 ξk − ebk /µ ,
ξ · φ + τ1 τ2 ψ = −iτ1eb3 .
(3.6)
k = 1, 2,
From (3.6) we have
(3.7)
τ1 φ · ξ + τ2 ξ 2 ψ = ieb3 ξ 2 − eb · ξ/µ,
Solving (3.7), we obtain
1e
τ1
2 e
2ξ
i
b
−
b·ξ
,
(3.8) φ·ξ = 2
3
τ1 − ξ 2
µ
ψ=
ξ · φ + τ1 τ2 ψ = −iτ1eb3 .
1e
1
2
2 e
(τ
−ξ
)i
b
+
b·ξ
,
3
1
τ2 (τ12 − ξ 2 )
µ
where
τ12 − ξ 2 = s/µ.
(3.9)
Using (3.8) and (3.9) in (3.6) implies
(3.10)
1 e
1 e
b · ξξk −
bk ,
φk = −
sτ1
µτ1
k = 1, 2,
1 e
ψ=
ib3 +
τ2
1e
b·ξ .
s
Let
ei = e−τi x3 ,
(3.11)
i = 1, 2,
e0 =
e1 − e2
.
τ1 − τ2
Then we write (3.5) in the form




ξ1
φ1 + ψξ1
 e1 ≡ V e0 + W e1 ,
(3.12)
ω
e = ψ  ξ2  (e2 − e1 ) + 
φ2 + ψξ2
iτ2
iφ · ξ/τ1 + iτ2 ψ
where
(3.13)
1 e
1e
Vj =
ib3 + b · ξ (τ2 − τ1 )ξj
τ2
s
c0
=−
(iseb3 + eb · ξ)ξj , j = 1, 2,
τ2 (τ1 + τ2 )
ic0
1e
e
V3 = i ib3 + b · ξ (τ2 − τ1 ) = −
(iseb3 + eb · ξ),
s
τ1 + τ2
where we used
τ12 − τ22 =
ν
s ≡ c0 s,
µ(µ + ν)
and
(3.14)
1 1
1 e
1 e
i
Wj = φj + ψξj =
−
b · ξξj −
bj + eb3 ξj
s τ2
τ1
µτ1
τ2
c0
eb · ξξj − 1 ebj + i eb3 ξj , j = 1, 2,
=
τ1 τ2 (τ1 + τ2 )
µτ1
τ2
W3 = eb3 .
204
M. Burnat — W. M. Zajączkowski
We need the following result (see [10], [5], [8]).
Lemma 3.1. For ξ ∈ R2 , s = γ + iξ0 , γ ∈ R+ , ξ0 ∈ R, γ > 0, and for any
nonnegative integer j and κ ∈ (0, 1),
k∂xj 3 ei k20,R+ ≤ c1 |τi |2j−1 ,
(3.15)
∞
∞Z
Z
0
0
|∂xj 3 ei (x3 + z) − ∂xj 3 ei (x3 )|2
i = 1, 2,
dx3 dz
≤ c2 |τi |2(j+κ)−1 ,
z 1+2κ
i = 1, 2,
where c1 , c2 do not depend on τi , i = 1, 2 and |ξ|.
Moreover,
|τ1 |2j−1 + |τ2 |2j−1
,
|τ1 |2
Z ∞Z ∞
dx3 dz
|∂xj 3 e0 (x3 + z) − ∂xj 3 e0 (x3 )|2 1+2κ
z
0
0
k∂xj 3 e0 k20,R+ ≤ c3
(3.16)
≤ c4
|τ1 |2(j+κ)−1 + |τ2 |2(j+κ)−1
,
|τ1 |2
where c3 , c4 do not depend on τi , i = 1, 2 and |ξ|.
Using Lemma 3.1, we obtain
Lemma 3.2. We assume that
bi ∈ Hγ1/2+α,1/4+α/2 (R3∞ ),
i = 1, 2,
b3 ∈ Hγ3/2+α,3/4+α/2 (R3∞ ),
γ > 0, α > 0.
2+α,1+α/2
(D4∞ ) and the folloThen there exists a solution to problem (3.1) in Hγ
wing estimate holds
X
3
2
X
(3.17)
k|ωi |k2+α,γ,D4∞ ≤ c(γ)
k|bi |k1/2+α,γ,R3∞ + k|b3 |k3/2+α,γ,R3∞ ,
i=1
i=1
where c(γ) remains bounded for γ ≥ γ0 > 0.
Proof. The existence follows from constructions (3.12)–(3.14). To obtain
(3.17) we consider
Z ∞
XZ
k|ω|k22+α,γ,D4∞ =
dξ
k∂xj 3 ω
e (ξ, x3 , s)k20,R1 (|s| + |ξ|2 )2+α−j dξ0
R2
j≤2
Z
R2
≤
XZ
j≤2
∞
ke
ω (ξ, x3 , s)k22+α,R1 dξ0
dξ
+
R2
+
−∞
Z
+
−∞
∞
Z
dξ
−∞
(|V |2 k∂xj 3 e0 k20,R1 + |W |2 k∂xj 3 e1 k20,R1 )
+
+
· (|s| + |ξ|2 )2+α−j dξ0
Z
Z ∞
+
dξ
(|V |2 ke0 k22+α,R1 + |W |2 ke1 k22+α,R1 ) dξ0 ≡ I.
R2
−∞
+
+
On Local Motion of a Compressible Barotropic Viscous Fluid
205
Using the inequalities (see [10], [5])
|τ1 |
≤ c2 ,
|τ2 |
1
|τ1 + τ2 | ≥ √ (|τ1 | + |τ2 |),
2
c3 (|s| + ξ 2 ) ≤ |τi |2 ≤ c4 (|s| + ξ 2 ), i = 1, 2,
c1 ≤
we obtain
|V | ≤ c(|τ | |eb3 | + |eb0 |2 ),
2
(3.18)
2
1 e0 2 e
|W | ≤ c
|b | + |b3 | ,
|τ |2
2
where τ replaces either τ1 or τ2 , and eb0 = (eb1 , eb2 ).
In view of (3.18) and Lemma 3.1 we obtain
Z
Z ∞
I ≤c
dξ
(|eb0 |2 (1 + |τ |1+2α ) + |eb3 |2 (1 + |τ |3+2α )) dξ0
≤
−∞
R2
0 2
c(kb k1/2+α,γ,R3∞
+ kb3 k23/2+α,γ,R3∞ ).
This concludes the proof.
Now we consider the nonhomogeneous problem
ωt − div D(ω) = f
x3 > 0,
∂ωi
∂ω3
µ
+
= bi , i = 1, 2, x3 = 0,
∂x3
∂xi
ω 3 = b3
x3 = 0,
(3.19)
ω|t=0 = 0
x3 > 0.
α,α/2
(D4T ), b0 =
Lemma 3.3. Let α > 0, γ > 0. If we assume that f ∈ Hγ
α+3/2,α/2+3/4
α+1/2,α/2+1/4
3
3
(RT ), T > 0, then the pro(RT ), b3 ∈ Hγ
(b1 , b2 ) ∈ Hγ
2+α,1+α/2
(D4T )
blem (3.19) has a unique solution for ν > µ/3 such that ω ∈ Hγ
and
(3.20)
kωk2+α,γ,D4T ≤ c(kf kα,γ,D4T + kb0 kα+1/2,γ,R3T + kb3 kα+3/2,γ,R3T ).
Proof. We extend f to a function f 0 on R4∞ in such a way that f 0 ∈
(see (2.2), (2.3)) and
α,α/2
Hγ
(R4∞ )
(3.21)
kf 0 kα,γ,R4∞ ≤ ckf kα,γ,D4T .
Let ω 0 be a solution of the problem
(3.22)
L(∂x , ∂t )ω 0 ≡ ωt0 − div D(ω 0 ) = f 0
in R4∞ .
Applying the Fourier–Laplace transformation (2.1) to (3.22) yields
(3.23)
L(iξ, s)e
ω 0 = fe0 ,
206
M. Burnat — W. M. Zajączkowski
so ω
e 0 = L−1 (iξ, s)fe0 . Since det L(iξ, s) = (s+µξ 2 )2 (s+(µ+ν)ξ 2 ) and s = γ +iξ0
with γ ≥ γ0 > 0, one obtains easily kω 0 k2+α,γ,R4∞ ≤ ckf 0 kα,γ,R4∞ , and so
kωk2+α,γ,D4T ≤ ckf kα,γ,D4∞ .
(3.24)
Now v = ω − ω 0 is a solution to the problem
(3.25)
vt − div D(v) = 0,
∂vi
∂v3
µ
+
= hi ,
∂x3
∂xi
v3 = h3 ,
i = 1, 2,
v|t=0 = 0,
where
(3.26)
0
∂ωi
∂ω30
hi = b i − µ
+
,
∂x3
∂xi
h3 = b3 − ω30 .
i = 1, 2,
Using (3.26) we have
(3.27)
khi k1/2+α/2,γ,R3T ≤ c(kbi k1/2+α/2,γ,R3T + kω 0 k2+α,γ,D4T ),
i = 1, 2,
0
kh3 k3/2+α/2,γ,R3T ≤ c(kb3 k3/2+α/2,γ,R3T + kω k2+α,γ,D4T ),
so applying Lemma 3.2 to problem (3.25) we see that ω = v + ω 0 is a solution to
(3.19) and the estimate (3.20) holds.
4. Existence of solutions to problem (1.12)
First we consider the problem (1.12) with vanishing initial data
L(∂x , ∂t )u ≡ ut − div D(u) = f
(4.1)
in Ω × (−∞, T ),
B1i (x, ∂x )u ≡ τ i · D(u) · n = bi , i = 1, 2, on S × (−∞, T ),
B2 (x)u ≡ u · n = b3
on S × (−∞, T ).
We write shortly B(x, ∂x )u = (B1 (x, ∂x )u, B2 (x)u).
Let f (k) (x, t) = ζ (k) (x, t)f (x, t). We denote by R(k) , k ∈ M, the operator
(4.2)
u(k) (x, t) = R(k) f (k) (x, t),
where u(k) (x, t) is a solution of the Cauchy problem
(4.3)
L(∂x , ∂t )u(k) (x, t) = f (k) (x, t).
For k ∈ N we define R(k) to be the operator
(4.4)
u
b(k) (z, t) = R(k) (fb(k) (z, t), bb(k) (z, t)),
where u
b(k) (z, t) is a solution to the boundary value problem
(4.5)
L(∂z , ∂t )b
u(k) (z, t) = fb(k) (z, t),
B(z, ∂z )b
u(k) (z, t) = bb(k) (z, t),
On Local Motion of a Compressible Barotropic Viscous Fluid
207
where u
b(k) (z, t) = Zk−1 u(k) (x, t) and Zk describes the transformation between
u
b(k) (z, t) and u(k) (x, t). Then we define an operator R (called a regularizer) by
the formula (see [2], [6])
X
(4.6)
Rh =
η (k) (x)u(k) (x, t),
k
where
(
h(k) (x, t) =
f (k) (x, t)
k ∈ M,
Zk {fb(k) (z, t), bb(k) (z, t)} k ∈ N ,
and
(
u(k) (x, t) =
R(k) f (k) (x, t)
Zk R
(k)
k ∈ M,
(Zk−1 f (k) (x, t), Zk−1 b(k) (x, t))
k ∈ N.
Lemma 3.3 implies the existence of solutions of problems (4.3), (4.5) and the
estimates
ku(k) k2+α,γ,R4∞ ≤ ckf (k) kα,γ,R4∞ ,
(4.7)
k ∈ M,
and
(4.8)
(k)
kb
u(k) k2+α,γ,R4∞ ≤ c(kfb(k) kα,γ,D4T + kbb k1/2+α/2,γ,R3∞
(k)
+ kbb3 k3/2+α/2,γ,R3∞ ),
k ∈ N.
Let
h = (f, b0 , b00 ) ∈ Hγα,α/2 (ΩT ) × Hγ1/2+α,1/4+α/2 (S T ) × Hγ3/2+α,3/4+α/2 (S T ) ≡ Hγα
2+α,1+α/2
and let Vγα = Hγ
(ΩT ). Inequalities (4.7) and (4.8) imply
Lemma 4.1 (see [10], [11]). Let S ∈ C 2+α , h ∈ Hγα , α > 1/2 and let γ
be sufficiently large. Then there exists a bounded linear operator R : Hγα → Vγα
(defined by (4.6)) such that
(4.9)
kRhkVγα ≤ ckhkHγα ,
where c does not depend on γ or h.
We write problem (4.1) in the following short form
(4.10)
Au = h ,
A = (L, B).
Lemma 4.2. Let S ∈ H 3+α , h ∈ Hγα with γ sufficiently large and α > 1/2.
Then
(4.11)
ARh = h + T h,
208
M. Burnat — W. M. Zajączkowski
where T is a bounded operator in Hγα with a small norm for small λ and large γ.
Proof. We have
X
LRh =
(L(∂x , ∂t )η (k) u(k) − η (k) L(∂x , ∂t )u(k) )
k∈M∪N
+
X
η (k) Zk (L(∂z − ∇F ∂z3 , ∂t ) − L(∂z , ∂t ))Zk−1 u(k) (x, t)
k∈N
+
X
X
η (k) L(∂x , ∂t )u(k) (x, t) +
k∈M
η (k) Zk L(∂z , ∂t )Zk−1 u(k) (x, t)
k∈N
= f + T1 h
and
X
BRh =
(B(x, ∂x )η (k) u(k) − η (k) B(x, ∂x )u(k) )
k∈N
+
X
η (k) (B(x, ∂x ) − B(ξ (k) , ∂x ))u(k)
k∈N
+
X
η (k) Zk (B(ξ (k) , ∂z − ∇F ∂z3 ) − B(ξ (k) , ∂z ))Zk−1 u(k) (x, t)
k∈N
+
X
η (k) Zk B(ξ (k) , ∂z )Zk−1 u(k) (x, t) = b + T2 h.
k∈N
Now, we estimate operators T1 and T2 . By Lemmas 2.4, 2.5 and Lemma 3.3
the first term in T1 h is estimated in the following way
X
X
(k) (k)
(k)
(k) ≤c
ku(k) k1+α,γ,Q(k)
(Lη
u
−
η
Lu
)
T
α,γ,Ω
k∈M∪N
≤ c(ε
δ1
+ c0 (ε)γ
−δ2
)
X
k∈M∪N
ku
(k)
k2+α,γ,Q(k)
k∈M∪N
≤ c(εδ1 + c0 (ε)γ −δ2 )khkHγα ,
where δi > 0, i = 1, 2, Q(k) = Ω(k) × (0, T ) and c0 (ε) is a decreasing function.
The second term in T1 h is bounded by
c
X
(k∇Fe∇2 Fe∇b
u(k) |z=Φk (y(x)) kα,γ,Q(k)
k∈N
+ k(∇Fe(1 + ∇Fe)∇2 u
b(k) |z=Φk (y(x)) kα,γ,Q(k)
+ k(∇2 Fe∇b
u(k) |z=Φk (y(x)) kα,γ,Q(k) )
X
≤c
(p(k∇Fek1+α,Q(k) )ku(k) k1+α,γ,Q(k)
k∈N
+ |∇Fe|∞,Q(k) (1 + |∇Fe|∞,Q(k) )ku(k) k2+α,γ,Q(k) ) ≡ I,
On Local Motion of a Compressible Barotropic Viscous Fluid
209
where p is an increasing function. Using |∇Fe|∞,Ω(k) ≤ cλa k∇Fek1+α,Ω(k) , a > 0,
the interpolation inequalities from Lemma 2.4 and Lemma 3.3, we have
I ≤ c(εδ1 + c0 (ε)(λδ2 + γ −δ3 ))khkHγα ,
δi > 0, i = 1, 2, 3,
and c0 (ε) is a decreasing function.
Now, we consider the second term in T2 h. Therefore, we have to estimate the
expression by
X
(kη (k) (B1 (x, ∂x ) − B1 (ξ (k) , ∂x ))u(k) k1/2+α,γ,Q(k) ∩S
k∈N
+ kη (k) (B2 (x, ∂x ) − B2 (ξ (k) , ∂x ))u(k) k3/2+α,γ,Q(k) ∩S )
X
≤
p(k∇Fek1+α,Ω(k) )k∇Fek2+α,Ω(k) ku(k) k2+α,γ,Q(k)
k∈N
≤c(εδ1 + c0 (ε)λδ2 )khkHγα ,
δi > 0, i = 1, 2,
where p = p( · ) is a polynomial and c0 (ε) is a decreasing function.
Similar considerations can be applied to the other terms of T1 and T2 . Summarizing we have
kT hkHγα ≤ c[εδ1 + c0 (ε)(λδ2 + γ −δ3 )]khkHγα .
(4.12)
This concludes the proof.
Lemma 4.3. Let S ∈ H 3+α , α > 1/2. Then for every v ∈ Vγα ,
RAv = v + W v,
(4.13)
where W is a bounded operator in Vγα whose norm can be made small for small λ
and large γ, because
kW vkVγα ≤ c(εδ1 + c0 (ε)(λδ2 + γ −δ3 ))kvkVγα ,
(4.14)
ε ∈ (0, 1), δi , i = 1, 2, 3, and c0 (ε) has the same properties as before.
Proof. We have
X
(4.15)
RAv =
η (k) Zk R(k) [Zk−1 ζ (k) L(∂x , ∂t )v, Zk−1 ζ (k) B(x, ∂x )v|S ]
k∈N
+
X
η (k) R(k) ζ (k) L(∂x , ∂t )v
k∈M
=
X
η (k) Zk R(k) [Zk−1 (ζ (k) L(∂x , ∂t )v − L(∂x , ∂t )ζ (k) v),
k∈N
Zk−1 (ζ (k) B(x, ∂x )v − B(x, ∂x )ζ (k) v)|S ]
X
+
η (k) Zk R(k) [0, Zk−1 (B(x, ∂x ) − B(ξ (k) , ∂x ))ζ (k) v|S ]
k∈N
+
X
k∈N
η (k) Zk R(k) [Zk−1 L(∂x , ∂t )ζ (k) v, Zk−1 B(ξ (k) , ∂x )ζ (k) v|S ]
210
M. Burnat — W. M. Zajączkowski
+
X
η (k) R(k) (ζ (k) L(∂x , ∂t )v − L(∂x , ∂t )ζ (k) v)
k∈M
+
X
η (k) R(k) L(∂x , ∂t )ζ (k) v.
k∈M
By uniqueness, for the Cauchy problem (4.3) we have
R(k) L(∂x , ∂t )ζ (k) v = ζ (k) v,
(4.16)
k ∈ M,
so the last term on the r.h.s. of (4.15) is equal to v. The third term on the r.h.s.
of (4.15) has the form
X
(4.17)
η (k) Zk R(k) [L(∂z − ∇Fe(z)∂z3 , ∂t )e
v (k) ,
k∈N
B (k) (ξ (k) , ∂z − ∇Fe(z)∂z3 )e
v (k) |z3 =0 ]
X
=
η (k) (Zk R(k) [(L(∂z − ∇Fe(z)∂z3 , ∂t ) − L(∂z , ∂t ))e
v (k) ,
k∈N
(B (k) (ξ (k) , ∂z − ∇Fe(z)∂z3 ) − B (k) (ξ (k) , ∂z ))e
v (k) |z3 =0 ] + Zk ve(k) ),
where ve(k) is a solution to problem (4.5), so
Zk R(k) [L(∂z , ∂t )e
v (k) , B (k) (ξ (k) , ∂z )e
v (k) |z3 =0 ] = Zk ve(k) = ζ (k) (x)v(x, t),
for k ∈ N and B (k) is obtained from B by applying Zk−1 . Therefore, the operator W is determined by the first, second and fourth expressions on the r.h.s.
of (4.15) and the first term on the r.h.s. of (4.17).
The fourth term on the r.h.s. of (4.15) is estimated in the following way
(4.18)
kη (k) R(k) (ζ (k) L(∂x ,∂t )v − L(∂x , ∂t )ζ (k) v)k2+α,γ,ΩT
≤ c(k∇ζ (k) ∇vkα,γ,ΩT + k∇2 ζ (k) vkα,γ,ΩT )
≤ ckvk1+α,γ,Q(k)
≤ c(εδ1 + c0 (ε)γ −δ2 )kvk2+α,γ,Q(k) ,
where (2.6) has been used and δi > 0, i = 1, 2.
Next, we consider the first term on the r.h.s. of (4.17). The first part of this
term is bounded by
X
(4.19) c
(k∇z Fe∇2z Fe∇z ve(k) kα,γ,Q(k) + k∇z Fe(1 + ∇z Fe)∇2z ve(k) kα,γ,Q(k)
k∈N
≤ cp(kFek2+α,ΩT )(εδ1 + c0 (ε)(λδ2 + γ −δ3 ))kvkVγα .
Continuing we obtain (4.14). This concludes the proof.
Summarizing we have
On Local Motion of a Compressible Barotropic Viscous Fluid
211
α,α/2
Theorem 4.4. Let’s assume that S ∈ H 3+α , f ∈ Hγ
(Ω × (−∞, T )),
1/2+α,1/4+α/2
3/2+α,3/4+α/2
0
00
b ∈ Hγ
(S × (−∞, T )), b ∈ Hγ
(S × (−∞, T )), α > 1/2
and γ is sufficiently large. Then there exists a unique solution to problem (4.1)
2+α,1+α/2
such that u ∈ Hγ
(Ω × (−∞, T )) and
(4.20)
kuk2+α,γ,ΩT ≤ c(kf kα,γ,ΩT + kb0 k1/2+α,γ,S T + kb00 k3/2+α,γ,S T ),
where c does not depend on u and γ.
Now, we consider problem (4.1) with nonvanishing initial data. Therefore,
we formulate it in the form
(4.21)
ut − div D(u) = f1
in ΩT ,
τ · D(u) · n = b01
on S T ,
u · n = b001
on S T ,
u|t=0 = u0
in Ω.
We have
α,α/2
1/2+α/2,1/4+α/4
Lemma 4.5. We assume that f1 ∈ W2
(ΩT ), b01 ∈ W2
(S T ),
3/2+α/2,3/4+α/4
b001 ∈ W2
(S T ), u0 ∈ W21+α (Ω), S ∈ W23+α , α ∈ (1/2, 1), and use
the following compatibility conditions
(4.22)
b001 |t=0 − u0 · n = 0.
b01 |t=0 − τ · D(u0 ) · n = 0,
2+α,1+α/2
Then there exists a solution to problem (4.21) such that u ∈ W2
and the estimate holds
(ΩT )
(4.23) kuk2+α,ΩT ≤ c(kf1 kα,ΩT +kb01 k1/2+α/2,S T +kb001 k3/2+α/2,S T +ku0 k1+α,Ω ).
2+α,1+α/2
Proof. Since u0 ∈ W21+α (Ω) there exists a function u
e0 ∈ W2
(ΩT )
such that u
e0 |t=0 = u0 and ke
u0 k2+α,ΩT ≤ cku0 k1+α,Ω . Introducing the function
v =u−u
e0 we see that it is a solution to the problem
(4.24)
vt − div D(v) = f2
in ΩT ,
τ · D (v) · n = b02
on S T ,
v · n = b002
on S T ,
v|t=0 = 0
in Ω,
where
α,α/2
f2 = f1 − (e
u0t − div D(e
u0 )) ∈ W2
(4.25)
(ΩT ),
1/2+α/2,1/4+α/4
b02 = b01 − τ · D(e
u0 ) · n ∈ W2
3/2+α/2,3/4+α/4
b002 = b001 − u
e0 · n ∈ W2
(S T ),
(S T ),
212
M. Burnat — W. M. Zajączkowski
and we have the following estimates
kf2 kα,ΩT ≤ c(kf1 kα,ΩT + ku0 k1+α,Ω ),
(4.26)
kb02 k1/2+α/2,S T
kb002 k3/2+α/2,S T
≤ c(kb01 k1/2+α/2,S T + ku0 k1+α,Ω ),
≤ c(kb001 k3/2+α/2,S T + ku0 k1+α,Ω ).
Since the compatibility conditions (4.22) are satisfied the functions f2 , b02 , b002 can
be extended by zero for t < 0, and the extended functions fe2 , eb02 , eb002 are such
that
α,α/2
fe2 ∈ H0
(ΩT ),
eb0 ∈ H 1/2+α/2,1/4+α/4 (S T ),
2
0
eb00 ∈ H 3/2+α/2,3/4+α/4 (S T ),
(4.27)
2
0
and
kfe2 kα,0,ΩT ≤ ckf2 kα,ΩT ,
(4.28)
keb02 k1/2+α/2,0,S T ≤ ckb02 k1/2+α/2,S T ,
keb002 k3/2+α/2,0,S T ≤ ckb002 k3/2+α/2,S T .
α,α/2
α,α/2
Since T < ∞ the norms Hγ
(ΩT ) and H0
larly for boundary norms) we have that
(ΩT ) are equivalent (and simi-
fe2 ∈ Hγα,α/2 (ΩT ),
eb0 ∈ H 1/2+α/2,1/4+α/4 (S T ),
2
γ
(4.29)
eb00 ∈ H 3/2+α/2,3/4+α/4 (S T ),
2
γ
and
(4.30)
kfe2 kα,γ,S T ≤ c(γ)kfe2 kα,0,ΩT ,
keb02 k1/2+α/2,γ,S T ≤ c(γ)keb02 k1/2+α/2,0,S T ,
keb00 k3/2+α/2,γ,S T ≤ c(γ)keb00 k3/2+α/2,0,S T .
2
2
By virtue of (4.23)–(4.30) and Theorem 4.4 we obtain the existence of solutions
2+α,1+α/2
(ΩT ) and the following estimate
to problem (4.24) such that v ∈ Hγ
holds
(4.31)
kvk2+α,γ,ΩT ≤ c(kf1 kα,ΩT + kb01 k1/2+α/2,S T
+ kb001 k3/2+α/2,S T + ku0 k1+α,Ω ).
Now, in view of the definition of v, we obtain the existence of solutions to (4.21)
2+α,1+α/2
such that u ∈ W2
(ΩT ) and by (4.31) we have
(4.32) kuk2+α,ΩT ≤ c(kvk2+α,ΩT + ku0 k1+α,Ω ≤ c(kvk2+α,γ,ΩT + ku0 k1+α,Ω ),
hence, (4.23) holds. This concludes the proof.
On Local Motion of a Compressible Barotropic Viscous Fluid
213
5. Existence of solutions to (1.13)
Using a partition of unity in ΩT we have (see [12]):
Lemma 5.1. If we assume that
η ∈ C β (ΩT ),
1/η ∈ L∞ (ΩT ),
β > 0,
1/2+α,1/4+α/2
G1j ∈ W2
(S T ),
j = 1, 2,
w0 ∈ W21+α (Ω),
η > 0,
α,α/2
F1 ∈ W2
(ΩT ),
G13 ∈ W 3/2+α,3/4+α/2 (S T ),
S ∈ H 3+α .
2+α,1+α/2
Then there exists a solution to problem (1.13) such that w ∈ W2
and the estimate holds
(ΩT )
(5.1) kwk2+α,ΩT ≤ c(kηkC β (ΩT ) , |1/η|∞,ΩT )
2
X
· kF1 kα,ΩT +
kG1j k1/2+α,S T + kG13 k3/2+α,S T + kw0 k1+α,Ω .
j=1
6. Existence of solutions to problem (1.1)
To prove the existence of solutions to problem (1.1) we formulate it in Lagrangian coordinates. To simplify the considerations we introduce them once
again using another notation. By Lagrangian coordinates we mean the initial
data for the following Cauchy problem
(6.1)
dx
= v(x, t),
dt
x|t=0 = ξ ∈ Ω,
ξ = (ξ1 , ξ2 , ξ3 ).
Integrating (6.1) we obtain the following transformation between the Eulerian x
and the Lagrangian ξ coordinates of the same fluid particle. Hence
Z t
(6.2)
x=ξ+
u(ξ, s) ds ≡ x(ξ, t),
0
where u(ξ, t) = v(x(ξ, t), t).
The condition (1.1)5 implies that
Ω 3 x = x(ξ, t)
for ξ ∈ Ω,
S 3 x = x(ξ, t)
for ξ ∈ S.
Using the Lagrangian coordinates, we can formulate the problem (1.1) in the
form
(6.3)
ηut − divu Du (u) + ∇u q = ηg
in ΩT ,
ηt + η∇u · u = 0
in ΩT ,
τ uα · Du (u) · nu + γu · τ uα = 0, α = 1, 2, on S T ,
on S T ,
nu · u = 0
u|t=0 = v0 ,
η|t=0 = ρ0
in Ω,
214
M. Burnat — W. M. Zajączkowski
where η(ξ, t) = ρ(x(ξ, t), t), q(ξ, t) = p(x(ξ, t), t), g(ξ, t) = f (x(ξ, t), t), ∇u =
∂x ξi ∇ξi , ∂ξi = ∇ξi , Du (u) = {µ(∇ui uj + ∇uj ui ) + (ν − µ)δij ∇u · u}, ∇u · u =
∂xi ξj ∇ξj ui , ∇ui = ∂xi ξk ∂ξk and the summation convention over the repeated
indices is assumed. Moreover, nu (ξ, t) = n(x(ξ, t), t), τ u (ξ, t) = τ (x(ξ, t), t).
Let Ax(ξ,t) be the Jacobi matrix of the transformation x = x(ξ, t) with eleRt
ments aij (ξ, t) = δij + 0 uiξj dτ and let Jx(ξ,t) = det{aij }i,j=1,2,3 be the Jacobian. Then
∂Jx(ξ,t)
∂aij
∂ui
=
Aij = Aij
, Jx(ξ,0) = 1,
∂t
∂t
∂ξj
where Aij are the algebraic complements of aij and A = {Aij }i,j=1,2,3 .
Hence
Z t
Z t
∂ui
(6.4)
Jx(ξ,t) = 1 +
Aij
dτ = 1 +
A · ∇ · u dτ.
∂ξj
0
0
Moreover, since Aij ∂ui /∂ξj = Aij akj ∂vi /∂xk = ∇·v(x, t)|x=x(ξ,t) Jx(ξ,t) it follows
that
Z t
Z t
∇u · u dτ ,
∇ · v|x(ξ,t) dτ = exp
Jx(ξ,t) = exp
0
0
−1
Jx(ξ,t)
A
where ∇u =
· ∇.
Therefore from (6.3)2,5 we have
η(ξ, t) = ρ0 (ξ) exp
(6.5)
Z
t
−
∇u · u dτ
0
−1
= ρ0 (ξ)Jx(ξ,t)
(ξ, t).
From (6.4) and (6.5) we obtain also
−1
Z t
η(ξ, t) = ρ0 (ξ) 1 +
A · ∇ · u dτ
.
(6.6)
0
Lemma 6.1. We assume that ρ0 ∈ W21+α (Ω), α ∈ (1/2, 1), ρ0 (ξ) ≥ ρ∗ > 0,
and
T 1/2 kuk2+α,ΩT ≤ δ,
(6.7)
where δ is sufficiently small. Then solutions of problem (6.3)2,5 satisfy
kη( · , t)k1+α,Ω ≤ φ(a)kρ0 k1+α,Ω ,
[ηξ ]α/2,ΩT ,t ≤ kρ0 k1+α,Ω φ(a, T )a,
(6.8)
sup
t
T
|ηξ (t) − ηξ (t0 )|2
dξ dt0
|t − t0 |1+α
Ω 0
Z T
Z
2
1−α
2
≤ kρ0 k1+α,Ω φ(a)T
kuk2+α,Ω dt · 1 +
Z Z
0
0
T
kuk22+α,Ω
dt ,
215
On Local Motion of a Compressible Barotropic Viscous Fluid
where a = T a kuk2+α,ΩT , a > 0 and φ is an increasing positive function.
Proof. (6.8)1 follows easily from (6.5). To show (6.8)2 we get from (6.5)
Ω
T
T
|ηξ (t) − ηξ (t0 )|2
dξ dt dt0
|t − t0 |1+α
0
0
Z Z T Z T R t0
| t uξ dτ |2
2
≤ kρ0 k1+α,Ω φ(a)
|t − t0 |1+α
0
Ω 0
R t0
R
R t0
t
| t uξ dτ |2 | 0 uξξ dτ |2
| t uξξ dτ |2
+
+
dξ dt dt0
|t − t0 |1+α
|t − t0 |1+α
Z T
Z
Z T
Z
≤ kρ0 k21+α,Ω φ(a)T 2−α
(u2ξ + u2ξξ ) dτ +
dξ
u2ξ dτ
Z Z
Z
0
Ω
T
≤ kρ0 k21+α,Ω φ(a)T 2−α
Z
0
kuk22+α,Ω dτ 1 +
Z
0
T
u2ξξ dτ
0
T
kuk22+α,Ω dτ .
0
Therefore, (6.8)2 has been proved.
A proof of (6.8)3 is similar to the proof of (6.8)2 .
To prove the existence of solutions to problem (1.1) we use the following
method of successive approximations
ηm um+1t − divum Dum (um+1 ) + ∇um q(ηm ) = ηm g
(6.9)
in ΩT ,
τ um α · Dum (um+1 ) · num = −γum · τ um α , α = 1, 2, on S T ,
num · um+1 = 0
on S T ,
um+1 |t=0 = v0
in Ω,
where ηm , um are given, and
ηmt + ηm divum um = 0 in ΩT ,
(6.10)
ηm |t=0 = ρ0
in Ω,
where um is given.
To apply Lemma 5.1 we write (6.9) in the form
ηm um+1t − divD(um+1 ) = −(div D(um+1 )
− divum Dum (um+1 )) − ∇um q(ηm ) + ηm g ≡ l1 + l2 + l3
(6.11)
in ΩT ,
τ α · D(um+1 ) · n = (τ α · D(um+1 ) · n
− τ um α Dum (um+1 ) · num ) − γum · τ um α ≡ l4 + l5 , α = 1, 2, on S T ,
n · um+1 = (n − num ) · um+1 ≡ l6
on S T ,
um+1 |t=0 = v0
in Ω,
where n = n(ξ, t), τ α = τ α (ξ, t), α = 1, 2. First we show
216
M. Burnat — W. M. Zajączkowski
Lemma 6.2. We assume that S ∈ W23+α , ρ0 ∈ W21+α (Ω), v0 ∈ W21+α (Ω),
2+α,1+α/2
α ∈ (1/2, 1), ρ0 (ξ) ≥ ρ∗ > 0, where ρ∗ is a constant, and um ∈ W2
(ΩT )
and (6.7) holds with δ sufficiently small. Then for solutions of the problem (6.11)
2+α,1+α/2
such that um+1 ∈ W2
(ΩT ) the following inequality holds
(6.12)
kum+1 k2+α,Ωt ≤ G(γ, ta kum k2+α,Ωt , t),
t ≤ T, a > 0,
where G is an increasing positive function of its arguments and
γ = kρ0 k1+α,Ω + kgkα,ΩT + kv0 k1+α,Ω ,
and G(γ, 0, 0) = G0 (γ) > 0.
Proof. Applying Lemma (5.1) to (6.11) we have
(6.13) kum+1 k2+α,ΩT ≤ φ1 (kηm kC β (ΩT ) , |1/ηm |∞,ΩT )
X
3
5
X
·
kli kα,ΩT +
kli k1/2+α,S T + kl6 k3/2+α,S T + kv0 k1+α,Ω ,
i=1
i=4
where φ1 is an increasing positive function and β < α − 1/2.
Now we have to estimate the norms from the r.h.s. of (6.13). To estimate l1
we write it in the qualitative form
l1 ≈ ψ1 (b)um+1ξξ + ψ2 (b)bξ um+1ξ ,
Rt
where b = {bij } = { 0 uiξj (ξ, τ ) dτ }, and ψ1 , ψ2 are matrix functions such that
ψ1 = ξx ξx , ψ2 = ξx ξx,b , where the products are the matrix products.
We shall restrict our considerations to estimate the highest derivatives in the
considered norms only.
Z T Z Z Rt
| 0 (umξ − umξ0 ) dτ |2
2
|um+1ξξ |2
[l1 ]α,ΩT ,x ≤ φ2 (am )
|ξ − ξ 0 |3+2α
0
Ω Ω
Rt
|um+1ξξ − um+1ξ0 ξ0 |2 | 0 umξ dτ |2
+
|ξ − ξ 0 |3+2α
Rt
Rt
| 0 (umξ − umξ0 ) dτ |2 | 0 umξξ dτ |2 |um+1ξ |2
+
|ξ − ξ 0 |3+2α
Rt
| (umξξ − umξ0 ξ0 ) dτ |2 |um+1ξ |2
+ 0
|ξ − ξ 0 |3+2α
Z t
2
5
X
|um+1ξ − um+1ξ0 |2
0
dξ
dξ
dt
≡
φ
Ii ,
umξξ dτ + 2
|ξ − ξ 0 |3+2α
0
i=1
where
am = T
a
Z
0
1/2
T
kum k22+αΩ
dτ
,
a > 0.
On Local Motion of a Compressible Barotropic Viscous Fluid
217
Now, we estimate the expression Ii , i = 1, . . . , 5.
T
|umξ − umξ0 |2p1
dξ dξ 0
0
3+2p1 (3µ1 /2−3/2p1 +α)
|ξ
−
ξ
|
0
Ω Ω
1/p2
Z T Z Z
|um+1ξξ (ξ)|2p2
0
dt ≡ I1 ,
·
dξ
dξ
0 3µ2 p2
0
Ω Ω |ξ − ξ |
I1 ≤ T a
Z
Z Z
1/p1
dt
where a > 0, 1/p1 + 1/p2 = 1, µ1 + µ2 = 1 and µ2 p2 < 1. Using the imbeddings
1+α+3µ1 /2−3/2p1
W22+α (Ω) ⊂ W2p1
(Ω) and W22+α (Ω) ⊂ L2p2 (Ω), which holds simultaneously because 3/2 − 3/2p1 + 3µ1 /2 − 3/2p1 + α ≤ 1 + α, 3/2 − 3/2p2 ≤ α
are valid for α > 1/2, we obtain I10 ≤ T a kum k2+α,ΩT kum+1 k2+α,ΩT . Moreover,
we have
I2 ≤ T [um+1ξξ ]2α,ΩT ,x
T
Z
|umξ |2∞,Ω dτ ≤ T
T
Z
0
kum k22+α,Ω dτ [um+1ξξ ]2α,ΩT ,x .
0
Continuing we see that I3 , I5 can be estimated in the similar way as I1 and I4
as I2 . Summarizing we have [l1 ]2α,ΩT ,x ≤ φ3 (am )T a kum k2+α,ΩT kum+1 k2+α,ΩT ,
where a > 0. Next, we estimate
[l1 ]2α/2,ΩT ,t
R t0
| t umξ dτ |2
|um+1ξξ |2
≤ φ4 (am )
|t − t0 |1+α
Ω 0
0
2
Z
|um+1ξξ (t) − um+1ξξ (t0 )|2 t
umξ dτ +
0
1+α
|t − t |
0
R t0
Rt
R t0
2
2
| t umξ dτ | | 0 umξξ dτ | |um+1ξ |2
| t umξξ dτ |2 |um+1ξ |2
+
+
|t − t0 |1+α
|t − t0 |1+α
Z t
2
|um+1ξ (t) − um+1ξ (t0 )|2
dξ dξ 0 dt
+ umξξ dτ |t − t0 |1+α
0
Z
Z T
Z T
≤ φ4 (am ) T 1−α
dξ
|umξ |2 dτ
|um+1ξξ |2 dτ
T
Z Z
Z
T
Ω
0
+ T [um+1ξξ ]2α/2,ΩT ,t
Z
0
T
|umξ |2∞,Ω dτ
0
+T
2−α
Z
Z
0
Z
Z
dξ
Ω
Z
+T
≡
10
X
i=6
Ii2 ,
u2mξξ dτ
0
Z
dξ
Ω
T
0
u2mξξ dτ
0
T
Z
Z
T
u2m+1ξ dτ
0
u2m+1ξ dτ
0
T
u2mξξ
T
Z
|umξ |2 dτ
dξ
Ω
+ T 1−α
T
Z
T
Z
dτ
0
0
T
|um+1ξ (t) − um+1ξ (t0 )|2
dt dt0
|t − t0 |1+α
218
M. Burnat — W. M. Zajączkowski
where I6 + I8 + I9 ≤ φ5 (am ), am kum+1 k2+α,ΩT , and
1/2
Z T
2
1/2
|umξξ |2p1 ,Ω dt
I10 ≤ φ6 (am )T
0
Z
T
Z
·
0
0
T
|um+1ξ (t) − um+1ξ (t0 )|22p2 ,Ω
|t − t0 |1+α
1/2
≤ φ6 (am )T 1/2 kum k2+α,ΩT kum+1 k2+α,ΩT ,
where we used imbeddings D2 W22+α (Ω) ⊂ L2p1 (Ω), D2 W22+α (Ω) ⊂ L2p2 (Ω),
which hold because 3/2 − 3/2p1 ≤ α, 3/2 − 3/2p2 ≤ 1, 1/p1 + 1/p2 = 1 are
satisfied together for α > 1/2.
Summarizing, we have shown
[l1 ]α/2,ΩT ,t ≤ φ7 (am )T a kum k2+α,ΩT kum+1 k2+α,ΩT .
Now, we consider l2 = ψ3 (b)q̇(ηm )ηmξ , where the dot denotes the derivative
with respect to the argument,
Z T Z Z Rt
| 0 (umξ − umξ0 ) dτ |2 |ηmξ |2
[l2 ]α,ΩT ,x ≤ φ8 (am , sup kηm k1+α,Ω )
|ξ − ξ 0 |3+2α
t
0
Ω Ω
0 2
2
|ηm (ξ) − ηm (ξ )| |ηmξ |
0
2
dt dξ dξ + [ηmξ ]α,ΩT ,x
+
|ξ − ξ 0 |3+2α
≤ φ9 (am , sup kηm k1+α,Ω )T a sup kηm k1+α,Ω ,
t
t
where to estimate the first two integrals on the r.h.s of the first inequality we
used the same method as in the case of I1 . Next,
Z Z T Z T R t0
| t umξ dτ |2 |ηmξ |2
[l2 ]α/2,ΩT ,t ≤ φ10 (am , sup kηm k1+α,Ω )
|t − t0 |1+α
t
Ω 0
0
|ηm (t) − ηm (t0 )|2
|ηmξ (t) − ηmξ (t0 )|2
2
+
|η
|
+
dξ dt dt0
mξ
|t − t0 |1+α
|t − t0 |1+α
≡ φ10 I11 .
Continuing,
I11 ≤ T 2−α
Z
T
kum k22+α,Ω dτ sup kηm k21+α,Ω
t
0
Z
T
Z
+
0
0
T
|ηm (t) − ηm (t0 )|22p1 ,Ω
|ηmξ |22p2 ,Ω dt dt0 + [ηmξ ] α2 ,ΩT ,t ≡ I12 ,
|t − t0 |1+α
where 1/p1 +1/p2 = 1. To estimate the middle term in I12 we use the imbeddings
W21 (Ω) ⊂ L2p1 (Ω), DW21+α (Ω) ⊂ L2p2 (Ω), which hold together because the
relations 3/2 − 3/2p1 ≤ 1, 3/2 − 3/2p2 ≤ 1 + α are satisfied for α > 1/2. Using
(6.8) we obtain
[l2 ]α/2,ΩT ,t ≤ φ11 (am , sup kηm k1+α,Ω )am .
t
On Local Motion of a Compressible Barotropic Viscous Fluid
219
In view of Lemma 2.6 we have
kl3 kα,ΩT ≤ sup kηm k1+α,Ω kgkα,ΩT .
t
Next we have
kl4 k1/2+α,S T ≤ ckl4 k1+α,ΩT ,
which can be estimated in the same way as l1 . Next
kl5 k1/2+α,S T ≤ ckl5 k1+α,ΩT ≤ ckum k1+α,ΩT ≤ εkum k2+α,ΩT + c(ε)kum k0,ΩT
≤ εkum k2+α,ΩT + c(ε)T 1/2 (kum k2+α,ΩT + kv0 k1+α,Ω ).
Finally,
kl6 k3/2+α,S T ≤ ckl6 k2+α,ΩT ≤ T a kum k2+α,ΩT kum+1 k2+α,ΩT .
Summarizing the above considerations we obtain (6.12). This concludes the
proof.
Let
αm (t) = kum k2+α,ΩT .
(6.14)
Let A > 0 be sufficiently large such that G0 (γ) < A, αm (t) ≤ A. Then there
exists a time T∗ such that for t ≤ T∗ we have
αm+1 (t) ≤ G(γ, ta A, t) ≤ A.
(6.15)
Hence we obtain
(6.16)
αm (t) ≤ A,
for m = 0, 1, . . . , and t ≤ T∗ .
Finally, we define the zero approximation function u0 as a solution to the
following problem
in ΩT ,
u0t − div D(u0 ) = 0
τ α · D(u0 ) · n + γu0 · τ α = 0,
α = 1, 2, on S T ,
u0 · n = 0
on S T ,
u0 |t=0 = v0
in Ω.
Therefore, we have proved
Lemma 6.3. If we assume that data are such that γ < ∞, and that α > 1/2,
then the sequence constructed by (6.9) and (6.10) is bounded for T ≤ T∗ .
Now we show that the sequence {ηm , um } converges. To this end we consider
the differences
(6.17)
Hm = ηm − ηm−1 ,
Um = um − um−1 ,
220
M. Burnat — W. M. Zajączkowski
which are solutions of the following problems
ηm Um+1t − div D(Um+1 )
= − (div D(Um+1 ) − divum Dum (Um+1 ))
+ (divum Dum (um ) − divum−1 Dum−1 (um ))
− q̇(ηm )∇um Hm + q̇(ηm )(∇um − ∇um−1 )(ηm−1 )
− (q̇(ηm ) − q̇(ηm−1 ))∇um−1 ηm−1 + Hm g − Hm umt ≡
7
X
Li ,
i=1
(6.18)
τ α · D(Um+1 ) · n
= (τ α · D(Um+1 ) · n − τ um α · Dum (Um+1 ) · num )
− (τ um α · Dum (um ) · num − τ um−1 α · Dum−1 (um ) · num−1 )
− γ(um · τ um α − um−1 · τ um−1 α ) =
10
X
Li ,
α = 1, 2,
i=8
n · Um+1 = (n − num ) · Um+1 − (num − num−1 ) · um ≡
12
X
Li ,
i=11
Um+1 |t=0 = 0,
and
(6.19)
Hmt + Hm divum um = − ηm−1 (divum um − divum−1 um ),
Hm |t=0 = 0.
To show the convergence of the sequence {um , ηm } we need
Lemma 6.4. Let the assumptions of Lemma 6.3 be satisfied. Then
(6.20)
kUm+1 k2+α,ΩT ≤ c(A)T a kUm k2+α,ΩT ,
where a > 0 and A is the bound from (6.16).
Proof. From Lemma 5.1 we have
(6.21) kUm+1 k2+α,ΩT ≤ φ1 (kηm kC β (ΩT ) , k1/ηm kL∞ (ΩT ) )
X
10
12
7
X
X
kLi k1/2+α,S T +
kLi k3/2+α,S T .
kLi kα,ΩT +
·
i=1
i=8
i=11
Since L1 = ψ1 (bm )bm Um+1ξξ + ψ2 (bm )bmξ Um+1ξ we have
kL1 kα,ΩT ≤ φ2 (A, T )T a kUm+1 k2+α,ΩT .
Rt
Rt
Since L2 = ψ3 (bm , bm−1 ) 0 Umξ dτ umξξ +ψ4 (bm , bm−1 ) 0 Umξξ dτ umξ we obtain
kL2 kα,ΩT ≤ φ3 (A, T )T a kUm k2+α,ΩT .
On Local Motion of a Compressible Barotropic Viscous Fluid
221
Next, we have that L3 = q̇(ηm )ψ5 (bm )Hmξ . Therefore
Z TZ Z |ηm (ξ) − ηm (ξ 0 )|2 |Hmξ |2
[L3 ]2α,ΩT ,x ≤ φ4 (A, T )
|ξ − ξ 0 |3+2α
0
Ω Ω
Rt
| 0 (umξ − umξ0 ) dτ |2 |Hmξ |2
|Hmξ − Hmξ0 |2
+
+
dt dξ dξ 0
|ξ − ξ 0 |3+2α
|ξ − ξ 0 |3+2α
≤ φ5 (A, T )T sup kHm k1+α,Ω ,
t
and
[L3 ]2α,ΩT ,t
T
Z Z
Z
T
≤ φ6 (A, T )
Ω
0
0
R t0
|ηm (t) − ηm (t0 )|2 |Hmξ |2
|t − t0 |1+α
umξ dτ | |Hmξ |2
|Hmξ (t) − Hmξ (t0 )|2
+
+
|t − t0 |1+α
|t − t0 |1+α
a
≤ φ7 (A, T )T sup kHm k1+α,Ω
|
2
t
dξ dt dt0
t
Z Z
T
T
Z
+ φ6 (A, T )
Ω
0
0
|Hmξ (t) − Hmξ (t0 )|2
dξ dt dt0 .
|t − t0 |1+α
To estimate the last expression we integrate (6.19) to get
Z t
(6.22) Hm (ξ, t) = − exp −
divum um dτ
0
Z
·
t
Z
divum um dτ
exp
0
t0
· ηm−1 (divum um − divum−1 um−1 ) dt0 .
0
Using the fact that Hm has the following qualitative form
Z t
Z t
Hm = ψ6 (bm (t))
[ψ7 (bm (τ ))Umξ (τ ) + ψ8 (bm (τ ))
Umτ (τ 0 ) dτ um−1 (τ )] dτ
0
0
we obtain that
Z Z T
Z T
|Hmξ (t) − Hmξ (t0 )|2
0
a
dξ
dt
≤
φ
(A,
T
)T
kUm k22+α,Ω dt.
(6.23)
7
|t − t0 |1+α
0
Ω 0
Similarly, we have
(6.24)
sup kHm k21+α,Ω ≤ φ8 (A, T )T a
t
Z
T
kUm k22+α,Ω dt.
0
Using (6.23) and (6.24) in the estimation of L3 we obtain
kL3 kα,ΩT ≤ φ9 (A, T )T a kUm k2+α,ΩT .
Similarly L3 we estimate L5 . Moreover,
kL4 kα,ΩT ≤ φ10 (A, T )T a kUm k2+α,ΩT .
222
M. Burnat — W. M. Zajączkowski
Let us consider L6 . We have
Z T
Z
[L6 ]2α,ΩT ,x ≤
kL6 k2α,Ω dt ≤
T
kHm k21+α,Ω kgk2α,Ω dt
0
≤
0
sup kHm k21+α,Ω kgk2α,ΩT ,
t
and
[L6 ]2α/2.ΩT ,t
T
T
|Hm (t) − Hm (t0 )|2 |g(t)|2
|t − t0 |1+α
0
Ω 0
0 2
|Hm (t )| |g(t) − g(t0 )|2
+
dξ dt dt0
|t − t0 |1+α
Z TZ T
kHm (t) − Hm (t0 )k21,Ω
≤
kgk21+α,Ω
0 |1+α
|t
−
t
0
0
kHm (t0 )k21+α,Ω |g(t) − g(t0 )|22,Ω
+
dt dt0
|t − t0 |1+α
Z T
kHm (t) − Hm (t0 )k21,Ω
2
2
dt0
≤ sup
+
sup
kH
k
m
1+α,Ω kgkα,ΩT .
|t − t0 |1+α
t
t
0
Z Z
Z
≤
Using (6.23) and (6.24) we have the estimate
kL6 kα,ΩT ≤ φ11 (A, T )T a kUm k2+α,ΩT kgkα,ΩT .
Similarly, we have
kL7 kα,ΩT ≤ φ12 (A, T )T a kUm k2+α,ΩT kumt kα,ΩT .
Continuing the considerations we prove the lemma.
Summarizing we have
Theorem 6.5. Let the assumptions of Lemma 6.3 be satisfied. Then there
exists a time T∗∗ ≤ T∗ such that for t ≤ T∗∗ there exists a solution to problem
2+α,1+α/2
1+α,1/2+α/2
(1.1) such that v ∈ W2
(Ωt ), ρ ∈ W2
(Ωt ).
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[1]
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[2]
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[6]
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[9]
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[10]
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[11]
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Manuscript received April 27, 1997
Marek Burnat
Institute of Applied Mathematics and Mechanics
Warsaw University
Warsaw, POLAND
Wojciech ZajĄczkowski
Institute of Mathematics
Polish Academy of Sciences
Śniadeckich 8
00-950 Warsaw, POLAND
E-mail address: [email protected]
TMNA : Volume 10 – 1997 – No 2