INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES
WARSZAWA 1992
ON LOCAL MOTION OF A COMPRESSIBLE BAROTROPIC VISCOUS FLUID
BOUNDED BY A FREE SURFACE
W. M. Z A J A ¸ C Z K O W S K I
Institute of Mathematics, Polish Academy of Sciences Sniadeckich 8, 00-950 Warszawa, Poland ´
Abstract. We consider the motion of a viscous compressible barotropic fluid in R
3bounded by a free surface which is under constant exterior pressure, both with surface tension and without it. In the first case we prove local existence of solutions in anisotropic Hilbert spaces with noninteger derivatives. In the case without surface tension the anisotropic Sobolev spaces with integration exponent p > 3 are used to omit the coefficients which are increasing functions of 1/T , where T is the existence time.
1. Introduction. First we consider the motion of a viscous compressible barotropic fluid in a bounded domain Ω
t⊂ R
3, which depends on time t ∈ R
1+. The shape of the (free) boundary S
tof Ω
tis governed by the surface tension.
Let v = v(x, t) be the velocity of the fluid, % = %(x, t) the density, f = f (x, t) the external force field per unit mass, p = p(%) the pressure, µ and ν the viscos- ity coefficients, σ the surface tension coefficient and p
0the external (constant) pressure. Then the problem is described by the following system (see [4], Chs. 1, 2, 7):
(1.1)
%(v
t+ v · ∇v) + ∇p(%) − µ∆v − ν∇ div v = %f in e Ω
T,
%
t+ div(%v) = 0 in e Ω
T,
%|
t=0= %
0, v|
t=0= v
0in Ω ,
Tn − σHn = p
0n on e S
T,
v · n = −φ
t/|∇φ| on e S
T,
1991 Mathematics Subject Classification: 35A07, 35Q35, 35R35, 76N10.
Key words and phrases: free boundary, compressible barotropic viscous fluid, local existence, surface tension, anisotropic Sobolev spaces.
[511]
where φ(x, t) = 0 describes S
t, e Ω
T= S
t∈(0,T )
Ω
t× {t}, Ω
tis the domain of the drop at time t, Ω
0= Ω is its initial domain, e S
T= S
t∈(0,T )
S
t, n is the unit outward vector normal to the boundary (n = ∇φ/|∇φ|), µ, ν, σ are constant coefficients. Moreover, thermodynamical considerations imply ν ≥ 1/(3µ) > 0, σ > 0. The last condition (1.1)
5means that the free boundary S
tis built up of moving fluid particles. Finally, T = T(v, p) denotes the stress tensor of the form (1.2) T
ij= −pδ
ij+ µ(∂
xiv
j+ ∂
xjv
i) + (ν − µ)δ
ijdiv v ≡ −pδ
ij+ D
ij(v) , where i, j = 1, 2, 3, D = D(v) = {D
ij} is the deformation tensor and H is the double mean curvature of S
t, which is negative for convex domains and can be expressed in the form
(1.3) Hn = ∆
St(t)x , x = (x
1, x
2, x
3) ,
where ∆
St(t) is the Laplace–Beltrami operator on S
t. Let S
tbe determined by x = x(s
1, s
2, t), (s
1, s
2) ∈ U ⊂ R
2, where U is an open set. Then
(1.4) ∆
St(t) = g
−1/2∂
sαg
−1/2b g
αβ∂
sβ= g
−1/2∂
sαg
1/2g
αβ∂
sβ, α, β = 1, 2 , where the convention summation over repeated indices is assumed, g = det{g
αβ}
α,β=1,2, g
αβ= x
α· x
β, where x
α= ∂
sαx, {g
αβ} is the inverse matrix to {g
αβ} and { b g
αβ} is the matrix of algebraic complements for {g
αβ}.
Let the domain Ω be given. Then, by (1.1)
5, Ω
t= {x ∈ R
3: x = x(ξ, t), ξ ∈ Ω}, where x = x(ξ, t) is the solution of the Cauchy problem
(1.5) ∂x
∂t = v(x, t), x|
t=0= ξ ∈ Ω, ξ = (ξ
1, ξ
2, ξ
3) .
Therefore the transformation x = x(ξ, t) connects the Eulerian x and Lagrangian ξ coordinates of the same fluid particle. Hence
(1.6) x = ξ +
t
R
0
u(ξ, s) ds ≡ X
u(ξ, t) ,
where u(ξ, t) = v(X
u(ξ, t), t). Moreover, the kinematic boundary condition (1.1)
5implies that the boundary S
tis a material surface, so if ξ ∈ S = S
0then X
u(ξ, t) ∈ S
tand S
t= {x : x = X
u(ξ, t), ξ ∈ S}.
In virtue of the continuity equation (1.1)
2and (1.1)
5the total mass M is conserved and
(1.7) R
Ωt
%(x, t) dx = M , which is a relation between % and Ω
t.
The aim of Section 3 of this paper is to prove local existence of solutions for the problem (1.1). We use spaces W
2l,l/2(Ω
T) with noninteger derivatives.
In Section 4 we show local existence of solutions to (1.1) with σ = 0. Since the
existence for the nonlinear problem was considered in [16] it is sufficient to find
an estimate (see (4.3)) for the linearized problem (1.1)
1,3,4with σ = 0 (see (4.1)) with a constant independent of T for T < ∞. A similar estimate for a scalar parabolic equation in the case of the Neumann boundary condition was found in [14].
Local existence of solutions in the compressible case was considered in [6, 7, 13]. In the incompressible case local existence is proved in [2, 11].
2. Notation and auxiliary results. In Section 3 of this paper we use the anisotropic Sobolev–Slobodetski˘ı spaces W
2l,l/2(Ω
T), l ∈ R
+(see [3], Ch. 18) of functions defined in Ω
T= Ω × (0, T ). In fact W
2l,l/2, l 6∈ Z, are Besov spaces; the equivalence between W
2l,l/2, l 6∈ Z, and Besov spaces follows from considerations in [1], Ch. 7. In the case of noninteger l we introduce the following norms (Ω ⊂ R
3) for functions defined in Ω
T:
(2.1)
kuk
Wl,02 (ΩT)
= RT
0
kuk
2Wl2(Ω)
dt
1/2,
kuk
W20,l/2(ΩT)=
R
Ω
kuk
2W2l/2((0,T ))
dx
1/2,
kuk
2Wl2(Ω)
= X
|α|≤[l]
kD
xαuk
2L2(Ω)+ X
|α|=[l]
R
Ω
R
Ω
|D
αxu(x, t) − D
yαu(y, t)|
2|x − y|
3+2(l−[l])dx dy
≡ X
|α|≤[l]
|D
αxu|
22,Ω+ X
|α|=[l]
[D
xαu]
2l−[l],Ω≡ X
|α|≤[l]
|D
αxu|
22,Ω+ [u]
2l,Ω,
where D
xα= ∂
xα11∂
xα22∂
αx33, |α| = α
1+ α
2+ α
3, ∂
xi= ∂/∂x
i, [l] is the integer part of l, % = l − [l] ∈ (0, 1) and in the case when l is an integer the last term must be omitted,
kuk
2W2l/2((0,T ))
=
[l/2]
X
j=0
k∂
tjuk
2L2((0,T ))+
T
R
0 T
R
0
|∂
t[l/2]u(x, t) − ∂
τ[l/2]u(x, τ )|
2|t − τ |
1+2(l/2−[l/2])dt dτ
≡
[l/2]
X
j=0
|∂
tju|
22,(0,T )+ [∂
t[l/2]u]
2l/2−[l/2],(0,T ),
where % = l/2 − [l/2] ∈ (0, 1) and in the case when l/2 is an integer the last term
must be omitted, kuk
2W2l,l/2(ΩT)
= X
|α|≤[l]
kD
αx,tuk
2L2(ΩT)+ X
|α|=[l]
TR
0
R
Ω
R
Ω
|D
αx,tu(x, t) − D
αy,tu(y, t)|
2|x − y|
3+2(l−[l])dx dy dt + R
Ω T
R
0 T
R
0
|D
αx,tu(x, t) − D
x,ταu(x, τ )|
2|t − τ |
1+2(l/2−[l/2])dt dτ dx
≡ X
|α|≤[l]
|D
x,tαu|
22,ΩT+ X
|α|=[l]
([D
αx,tu]
2l−[l],ΩT,x+ [D
x,tαu]
2l/2−[l/2],ΩT,t)
≡ X
|α|≤[l]
|D
x,tαu|
22,ΩT+ [u]
2l,ΩT,x+ [u]
2l,ΩT,t,
where α = (α
0, α
1, α
2, α
3) is a multiindex, D
αx,t= ∂
tα0∂
αx11∂
xα22∂
xα33, |α| = 2α
0+ α
1+ α
2+ α
3and we use generalized (Sobolev) derivatives. Similarly using local coordinates and a partition of unity we introduce the norm in the space W
2l,l/2(S
T) of functions defined on S
T= S × (0, T ), where S = ∂Ω. We also use W
2l(Ω) with the norm (2.1)
3for functions defined in Ω. We do not distinguish norms of scalar and vector-valued functions. To simplify notation we write
kuk
l,Q= kuk
Wl,l/22 (Ω)
if Q = Ω
Tor Q = S
T, l ≥ 0 , kuk
l,Q= kuk
Wl2(Q)
if Q = Ω or Q = (0, T ), l ≥ 0 , and W
20,0(Q) = W
20(Q) = L
2(Q). Moreover,
kuk
Lp(Q)= |u|
p,Q, kuk
l,p,ΩT= kuk
Lp(0,T ;Wl2(Ω))
, 1 ≤ p ≤ ∞ . Let us introduce a space Γ
rl(Ω) with the norm
kuk
Γlr(Ω)
= X
i≤[l/2]
∂
tiu
l−2i,r,Ω,
where u
l,r,Q= kuk
Wrl(Q), for Q either Ω or S. In the case when Q is either Ω
Tor S
Twe define u
l,r,Q= kuk
Wl,l/2r (Q)
. We also define the following norms:
|u|
l,0,∞,ΩT= sup
t∈[0,T ]
X
|α|≤[l]
|D
x,tαu|
22,Ω+ X
|α|=[l]
[D
αx,tu]
l−[l],Ω1/2,
|u|
l,0,Ω=
X
|α|≤[l]
|D
x,tαu|
22,Ω+ X
|α|=[l]
[D
x,tαu]
l−[l],Ω 1/2, l ∈ R
+.
We introduce
W
◦l,l/22(Q
T) = {u ∈ W
2l,l/2(Q
T) : ∂
tiu|
t=0= 0, i ≤ [l/2 − 1/2]} , where Q is either Ω or S.
We also need the notation u
[l]+κ,ΩT=
X
|α|=[l]
T
R
0
|D
αx,tu|
22,Ωt
2κdt
1/2, κ ∈ (0, 1) , kuk
(l),ΩT,κ= kuk
l,ΩT+ u
[l]+κ,ΩT,
kuk
l,ΩT,κ= kuk
l,ΩT+ T
−κX
|α|=[l]
|D
x,tαu|
2,ΩT.
We denote by W
2,κl,l/2(Ω
T) the space with the norm k k
(l),ΩT,κ. For T finite and κ ≤ l/2 − [l/2] the above norms are equivalent because in view of Lemma 2.6 below we have
T
−κhui
[l],ΩT≤ u
[l]+κ,ΩT≤ c[u]
[l]+κ,ΩT,t+ cT
−κhui
[l],ΩT≤ cT
l/2−[l/2]−κ[u]
l,ΩT,t+ cT
−κhui
[l],ΩT, where κ ≤ l/2 − [l/2], hui
[l],ΩT= P
|α|=[l]
|D
αx,tu|
2,ΩT.
Now we introduce spaces convenient for proving the existence of solutions to the linearized parabolic problem (1.1)
1,3,4(see [8, 15]):
kuk
2Hγl,l/2(ΩT)
=
T
R
0
e
−2γtkuk
2l,Ωdt + kuk
2Hγ0,l/2(ΩT)
, where γ ≥ 0. Here for l/2 6∈ Z,
kuk
2Hγ0,l/2(ΩT)
= γ
lT
R
0
e
−2γtkuk
20,Ωdt
+
T
R
0
e
−2γtdt
∞
R
0
dτ k∂
tku
0(·, t − τ ) − ∂
tku
0(·, t)k
20,Ωτ
1+2(l/2−k)and k = [l/2], u
0(x, t) = u(x, t) for t > 0, u
0(x, t) = 0 for t ≤ 0. For l/2 ∈ Z kuk
2H0,l/2γ (ΩT)
=
T
R
0
e
−2γt(γ
lkuk
20,Ω+ k∂
tl/2uk
20,Ω) dt ,
and we assume that ∂
tju|
t=0= 0, j = 0, . . . , l/2 − 1, so u
0(x, t) has a general- ized derivative ∂
tl/2u
0in L
2(Ω × (−∞, T )). Moreover, we introduce the notation kuk
l,2,γ,ΩT= kuk
Hl,l/2γ (ΩT)
.
For the purposes of Section 4 we define [u]
p,%,Ω=
R
Ω
R
Ω
dx dx
0|u(x, t) − u(x
0, t)|
p|x − x
0|
3+p% 1/p, [u]
2,%,Ω≡ [u]
%,Ω, [u]
l,p,Ω= X
|α|=[l]
[D
αxu]
p,l−[l],Ω.
Next we introduce the notation used by V. A. Solonnikov (see [10]):
E
n+1= {x
1, . . . , x
n, t} ≡ R
n+1, E
n= {x
1, . . . , x
n} ≡ R
n, E e
n= {(x, t) ∈ E
n+1: x
n= 0} , D
n+1= {(x, t) ∈ E
n+1: t > 0} , D e
n+1= {(x, t) ∈ E
n+1: x
n> 0} , D
n= {x ∈ E
n: x
n> 0} ≡ R
n+,
D e
n= {(x
0, t) ∈ e E
n: t > 0}, x
0= (x
1, . . . , x
n−1) ,
R = {(x, t) ∈ E
n+1: x
n> 0, t > 0} , E
n−1= {x ∈ E
n: x
n= 0} ≡ R
n−1. If G is one of the sets E
n+1, e E
n, D
n+1, e D
n+1, e D
n, R, we denote by G(T ) the set of all points in G with t ≤ T , and G(∞) = G.
Moreover, we introduce the following norms and seminorms:
hui
l,p,Q= X
|α|=l
|D
αx,tu|
p,Q, l ∈ N ,
[u]
p,%,ΩT,x=
TR
0
dt R
Ω
dx R
Ω
dx
0|u(x, t) − u(x
0, t)|
p|x − x
0|
n+p% 1/p, Ω ⊂ R
n, % ∈ (0, 1) ,
[u]
p,%,ΩT,t=
R
Ω
dx
T
R
0
dt
T
R
0
dt
0|u(x, t) − u(x, t
0)|
p|t − t
0|
1+p% 1/p, % ∈ (0, 1), p ∈ (1, ∞) , [u]
p,%,ΩT= [u]
p,%,ΩT,x+ [u]
p,%,ΩT,t,
[u]
l,p,ΩT= X
|α|=[l]
([D
αx,tu]
p,l−[l],ΩT,x+ [D
x,tαu]
p,l/2−[l/2],ΩT,t)
≡ [u]
l,p,ΩT,x+ [u]
l,p,ΩT,t,
for l ≥ 1, l − [l] ∈ (0, 1), l/2 − [l/2] ∈ (0, 1), p ∈ (1, ∞).
Finally, we define hhuii
p,%,En,xi=
∞R
0
|u(x
1, . . . , x
i+h, . . . , x
n)−u(x)|
pp,Endh h
1+p% 1/p, % ∈ (0, 1) .
In the case of E
n+1(T ) we have hhuii
l,p,En+1(T )=
n
X
i=1
X
|α|=[l]
hhD
xαuii
p,l−[l],En+1(T ),xi+hh∂
t[l/2]uii
p,l/2−[l/2],En+1(T ),t,
where
hhuii
p,%,En+1(T ),xi=
∞R
0
|u(x
1, . . . , x
i+ h, . . . , x
n, t) − u(x, t)|
pp,En+1(T )
dh h
1+p% 1/p, i = 1, . . . , n ,
hhuii
p,%,En+1(T ),t=
∞R
0
|u(x, t − h) − u(x, t)|
pp,En+1(T )
dh h
1+p% 1/p, % ∈ (0, 1) . We need the following imbedding theorems and interpolation inequalities:
Lemma 2.1 (see [3]). For a sufficiently smooth u on Ω we have (2.2) |D
µxu|
f,Ω≤ ckuk
d,Ω,
3/2 − 3/f + |µ| ≤ d, 0 ≤ d ∈ R, 1 < f ∈ R, |µ| ∈ N ∪ {0} , (2.3) [D
µxu]
p,%,Ω≤ ckuk
l,Ω,
3/2 − 3/p + |µ| + % ≤ l, % ∈ (0, 1), 1 < p ∈ R, |µ| ∈ N ∪ {0} , (2.4) |D
µxu|
f,Ω≤ ε
1−κhui
[d],Ω+ cε
−κ|u|
2,Ω,
0 < κ = (3/2 − 3/f + |µ|)/[d] < 1, ε ∈ (0, 1) , (2.5) |D
µxu|
f,Ω≤ ε
1−κ1[u]
d,2,Ω+ cε
−κ1|u|
2,Ω,
0 < κ
1= (3/2 − 3/f + |µ|)/d < 1, (2.6) [D
µxu]
p,%,Ω≤ ε
1−κ2[u]
l,2,Ω+ cε
−κ2|u|
2,Ω,
0 < κ
2= (3/2 − 3/p + |µ| + %)/l < 1 . Lemma 2.2 (see [9, 12]). For a sufficiently regular u we have
(2.7) k∂
tiuk
2−2/p,p,Ω≤ c(k∂
tiuk
2,p,ΩT+ k∂
tiu|
t=0k
2−2/p,p,Ω) , where the constant c does not depend on T .
Lemma 2.3 (see [10], Sect. 12, p. 72; [12], Th. 5). The norms [u]
l,p,En+1(T )and hhuii
l,p,En+1(T )are equivalent.
Lemma 2.4 (see [10]). Let u = 0 for t ≤ 0. Then (2.8) hhuii
rr,α,E
e
n(T ),t≤ [u]
rr,α,D
e
n(T ),t+ 2
rα
T
R
0
dt |u(·, t)|
rr,En−11 t
rα, where α ∈ (0, 1).
Lemma 2.5. Let f (0) = 0 and supp f ⊂ [0, T ]. Let p ∈ (1, ∞), 1/p < l < 1.
Then there exists a constant A(l, p) such that (2.9)
TR
0
|f (x)|
px
pldx
1/p≤ A(p, l)
TR
0
dx
T
R
0
dy |f (x) − f (y)|
p|x − y|
1+pl 1/p,
where f is such that the right-hand side is finite and 0 < A = α/[(1 − (pl)
−1/pa
l−1/p)a] ,
α = ((1 − (1 − a)
p0(1+l))/(p
0(1 + l))
1/p0) , a ∈ (0, 1) , 1/p + 1/p
0= 1 . P r o o f. It is sufficient to prove the lemma for smooth functions vanishing near 0. We consider the identity
f (x) = 1 ax
ax
R
0
f (y) dy + 1 ax
ax
R
0
[f (x) − f (y)] dy , a < 1 .
By the H¨ older inequality we have
|f (x)| ≤ 1 ax
Rax
0
|f (y)|
pdy
1/pRax
0
dy
1/p0(2.10)
+ 1 ax
axR
0
|f (x) − f (y)|
p|x − y|
1+pldy
1/pRax
0
|x − y|
p0(1/p+l)dy
1/p0, where 1/p + 1/p
0= 1. We calculate the integral
ax
R
0
|x − y|
p0(1/p+l)dy =
ax
R
0
(x − y)
p0(1/p+l)dy
= 1 − (1 − a)
p0(1+l)p
0(1 + l) x
p0(1+l)≡ α(p, a)
p0x
p0(1+l). Therefore, (2.10) becomes
(2.11) |f (x)| ≤ 1 ax
ax
R
0
|f (y)|
pdy
1/p+ a
−1α(p, a)x
l axR
0
|f (x) − f (y)|
p|x − y|
1+pldy
1/p. In view of the Minkowski inequality, (2.11) implies
TR
0
|f (x)|
px
pldx
1/p≤ a
−1/p TR
0
dx x
1+plax
R
0
|f (y)|
pdy
1/p(2.12)
+ α a
TR
0
dx
ax
R
0
|f (x) − f (y)|
p|x − y|
1+pldy
1/p= a
−1/p aTR
0
|f (y)|
pdy
T
R
y/a
dx x
1+pl 1/p+ α a
TR
0
dx
ax
R
0
|f (x) − f (y)|
p|x − y|
1+pldy
1/p≡ I .
Integrating the second integral in the first term yields
TR
y/a
dx x
1+pl 1/p= 1
pl [(y/a)
−pl− T
−pl]
1/p≤ 1 pl
1/p(y/a)
−l.
Hence we have I ≤ (pl)
−1/pa
l−1/p TR
0
|f (y)|
py
pldy
1/p+ α a
TR
0
dx
ax
R
0
|f (x) − f (y)|
p|x − y|
1+pldy
1/p. Since l > 1/p, assuming that
TR
0
|f (x)|
px
pldx
1/p< ∞
(here we use f (0) = 0) and that a is so small that 1 − (pl)
−1/pa
l−1/p> 0 we obtain (2.9). This concludes the proof.
We recall Lemma 6.3 from [8].
Lemma 2.6. Let τ ∈ (0, 1). Then for u ∈ W
20,τ /2(Ω
T) (2.13)
T
R
0
|u|
22,Ωdt t
τ≤ c
1T
R
0
dt
T
R
0
dt
0|u(·, t) − u(·, t
0)|
22,Ω|t − t
0|
1+τ+ c
2T
−τT
R
0
|u|
22,Ωdt , where c
1, c
2do not depend on T and u.
For T = ∞ the last term in (2.13) vanishes. The above result was shown in [12], Lemma 2, p. 138.
3. Local existence. To prove the local existence of solutions to (1.1) we write it in the Lagrangian coordinates introduced by (1.5) and (1.6):
(3.1)
ηu
t− µ∇
2uu − ν∇
u∇
u· u + ∇
uq = ηg in Ω
T,
η
t+ η∇
u· u = 0 in Ω
T,
T
u(u, q)n(ξ, t) − σ∆
St(t)X
u(ξ, t) = −p
0n(ξ, t) on S
T,
u|
t=0= v
0(ξ) in Ω ,
η|
t=0= %
0(ξ) in Ω ,
where η(ξ, t) = %(X
u(ξ, t), t), q(ξ, t) = p(X
u(ξ, t), t), g(ξ, t) = f (X
u(ξ, t), t),
∇
u= ∂
xξ
i∇
ξi, ∇
ξi= ∂
ξi, T
u(u, q) = −qδ + D
u(u), δ = {δ
ij} and D
u(u) = {µ(∂
xiξ
k∇
ξku
j+ ∂
xjξ
k∇
ξku
i) + (ν − µ)δ
ij∇
u· u}, with ∇
u· u = ∂
xiξ
k∇
ξku
i. Let A be the Jacobi matrix of the transformation x = x(ξ, t) ≡ X
u(ξ, t) with elements a
ij= δ
ij+ R
t0
∂
ξju
i(ξ, τ ) dτ . Assuming |∇
ξu|
∞,ΩT≤ M we obtain
(3.2) 0 < c
1(1 − M t)
3≤ det{∂
ξjx
i} ≤ c
2(1 + M t)
3, t ≤ T ,
where c
1, c
2are constants and T is sufficiently small. Moreover, det A = exp( R
t0
∇
u· u dτ ) = %
0/η.
Let S
tbe determined (at least locally) by the equation φ(x, t) = 0. Then S is described by φ(x(ξ, t), t)|
t=0≡ e φ(ξ) = 0. Moreover, we have
n(x(ξ, t), t) = ∇
xφ(x, t)
|∇
xφ(x, t)|
x=x(ξ,t)
, n
0(ξ) = ∇
ξφ(ξ) e
|∇
ξφ(ξ)| e .
To prove the existence of solutions of (3.1) we consider first the following linear problem:
(3.3)
u
t− µ∆
ξu − ν∇
ξ∇
ξ· u = f
1in Ω
T,
Π
0D
ξ(u)n
0= g
1on S
T,
n
0D
ξ(u)n
0− σn
0∆
S(0)
t
R
0
u(τ ) dτ = g
2+ σ
t
R
0
h
1(τ ) dτ on S
T,
u|
t=0= u
0in Ω ,
where Π
0, Π are projections defined by Πg = g − (g · n)n, Π
0g = g − (g · n
0)n
0, and D
ξ(u) = {µ(∂
ξiu
j+ ∂
ξju
i) + (ν − µ)δ
ij∂
ξku
k}.
Lemma 3.1. Let f
1∈ W
2,κl,l/2(Ω
T), g
1, g
2∈ W
l+1/2,l/2+1/42,κ
(S
T), h
1∈
W
l−1/2,l/2−1/42,κ
(S
T), u
0∈ W
2l+1(Ω), S ∈ W
2l+2and T < ∞. Let 0 < l 6∈ Z be such that l/2 = n + 3/4 + κ, n ∈ N ∪ {0}, κ ∈ (0, 1/4). Then there exists a solution of problem (3.3) such that u ∈ W
2,κl+2,l/2+1(Ω
T) and
(3.4) kuk
l+2,ΩT,κ≤ c
1(X
1+ |u(0)|
l+1,0,Ω) ≤ c
2(X
1+ X
2+ ku
0k
l+1,Ω) , where
X
1= kf
1k
l,ΩT,κ+
2
X
i=1
kg
ik
l+1/2,ST,κ+ kh
1k
l−1/2,ST,κ,
X
2=
[l/2−1/2]
X
i=0
k∂
tif
1(0)k
l−1−2i,Ω,
|u(0)|
l+1,0,Ω=
[l/2+1/2]
X
i=0
k∂
tiu(0)k
l+1−2i≤ c
3(X
2+ ku
0k
l+1,Ω) , and the constants c
1, c
2, c
3do not depend on T for T < ∞.
P r o o f. Let ϕ
i= ∂
tiu|
t=0∈ W
2l+1−2i(Ω) be calculated from (3.3)
1inductively:
(3.5) ϕ
i+1= µ∇
2ξϕ
i+ ν∇
ξ∇
ξ· ϕ
i+ ∂
tif
1(0) , i ≤ [l/2 − 1/2] .
They satisfy the following compatibility conditions:
(3.6)
Π
0D
ξ(ϕ
i)n
0= ∂
tig
1(0) , i ≤ [l/2 + 1/4 − 1/2] , n
0D
ξ(ϕ
i)n
0− σn
0∆
S(0)ϕ
i−1= ∂
tig
2(0) + σ∂
ti−1h
1(0) ,
1 ≤ i ≤ [l/2 − 1/4 − 1/2] , n
0D
ξ(ϕ
0)n
0= g
2(0) , for i = 0 .
We extend the functions ϕ
i, i = 0, . . . , [l/2 + 1/2], to R
3in such a way that the extended functions ϕ e
i∈ W
2l+1−2i(R
3), 0 ≤ i ≤ [l/2 + 1/2], satisfy
k ϕ e
ik
l+1−2i,R3≤ ckϕ
ik
l+1−2i,Ω. Now we construct a function e v such that
(3.7) ∂
tie v|
t=0= ϕ e
i, i ≤ [l/2 + 1/2] .
By Theorem 3 of [12] there exists a function e v ∈ W
2,κl+2,l/2+1(R
3× R
+), κ ≤ l/2 − [l/2], satisfying (3.7) and in view of Lemma 2.6 we have
(3.8) k e vk
l+2,R3×R+,κ≤ ck e vk
l+2,R3×R+≤ c
[l/2+1/2]
X
i=0
k ϕ e
ik
l+1−2i,R3≤ c
[l/2+1/2]
X
i=0
kϕ
ik
l+1−2i,Ω= c|u(0)|
l+1,0,Ω≤ c(X
2+ ku
0k
l+1,Ω) . Let v = e v|
ΩT. Then introducing the function
(3.9) w = u − v ,
we see that it is a solution of the problem
(3.10)
w
t− µ∇
2ξw − ν∇
ξ∇
ξ· w = f
10in Ω
T,
Π
0D
ξ(w)n
0= g
10on S
T,
n
0D
ξ(w)n
0− σn
0∆
S(0)
t
R
0
w(τ ) dτ = g
02+ σ
t
R
0
h
01dτ on S
T,
w|
t=0= 0 in Ω ,
where
(3.11)
f
10= f
1− (v
t− µ∇
2ξv − ν∇
ξ∇
ξ· v) ∈ W
◦l,l/22(Ω
T) , g
10= g
1− Π
0D
ξ(v)n
0∈ W
◦l+1/2,l/2+1/42
(S
T) ,
g
20= g
2− n
0D
ξ(v)n
0∈ W
◦l+1/2,l/2+1/42
(S
T) ,
h
01= h
1− σn
0∆
S(0)v ∈ W
◦l−1/2,l/2−1/42
(S
T) .
Therefore, we have (3.12) kf
10k
l,ΩT+
2
X
i=1
kg
i0k
l+1/2,ST+ kh
01k
l−1/2,ST≤ c
kf
1k
l,ΩT+
2
X
i=1
kg
ik
(l+1/2),ST,κ+ kh
1k
l−1/2,ST+ kvk
l+2,ΩT+ D
ξ(v)n
0 [l+1/2]+κ,ST+ n
0D
ξ(v)n
0 [l+1/2]+κ,ST≤ c(X
1+ kvk
l+2,ΩT) , because
(3.13) D
ξ(v)n
0 [l+1/2]+κ,ST+ n
0D
ξ(v)n
0 [l+1/2]+κ,ST≤ c(T )kvk
l+2,ΩT. The restrictions imposed on l in the assumptions of the lemma imply that l/2 − [l/2] = 3/4 + κ > 1/2, l/2 − 1/4 − [l/2 − 1/4] = 1/2 + κ > 1/2, l/2 + 1/4 − [l/2 + 1/4] = κ < 1/4. Therefore, f
10, g
i0, i = 1, 2, h
01can be extended by zero into t < 0 without losing regularity. Denote the extended functions by f
100, g
100, g
002, h
001, respectively. Moreover, in view of Lemma 2.5 we find that f
100∈ H
0l,l/2(Ω
T), h
001∈ H
l−1/2,l/2−1/40
(S
T) and by Lemmas 2.3 and 2.4 that g
100, g
200∈ H
l+1/2,l/2+1/40
(S
T)
and
(3.14)
kf
100k
l,2,0,ΩT≤ ckf
10k
l,ΩT,
kg
i00k
l+1/2,2,0,ST≤ ckg
i0k
(l+1/2),ST, i = 1, 2 , kh
001k
l−1/2,2,0,ST≤ ckh
01k
l−1/2,ST,
where the constant does not depend on T .
Since T < ∞, the norms of H
γl,l/2(Ω
T) and H
0l,l/2(Ω
T) (and similarly for boundary norms) are equivalent. Therefore, from [8] we deduce that f
100∈ H
γl,l/2(Ω
T), g
001, g
002∈ H
l+1/2,l/2+1/4γ
(S
T), h
001∈ H
l−1/2,l/2−1/4γ
(S
T) and there ex- ists a constant c(γ) such that
(3.15)
kf
100k
l,2,γ,ΩT≤ c(γ)kf
100k
l,2,0,ΩT,
kg
i00k
l+1/2,2,γ,ST≤ c(γ)kg
i00k
l+1/2,2,0,ST, i = 1, 2, kh
001k
l−1/2,2,γ,ST≤ c(γ)kh
001k
l−1/2,2,0,ST.
Performing the above extension on the right-hand sides of (3.10) we obtain the following problem:
(3.16)
w e
t− µ∇
2ξw − ν∇ e
ξ∇
ξ· w = f e
100in Ω × (−∞, T ) , Π
0D
ξ( w)n e
0= g
100on S × (−∞, T ) , n
0D
ξ( w)n e
0− σn
0∆
S(0)
t
R
0
w(τ ) dτ = g e
002+ σ
t
R
0
h
001(τ ) dτ
on S × (−∞, T ) ,
where w is zero for t ≤ 0. By [15] there exists a solution of (3.16) such that e w ∈ H e
γl+2,l/2+1(Ω × (−∞, T )) and (3.12), (3.14), (3.15) imply
k wk e
l+2,2,γ,Ω×(−∞,T )≤ c(γ)(X
1+ X
2+ ku
0k
l+1Ω) . For T < ∞ we have
k wk e
l+2,2,0,Ω×(−∞,T )≤ c(γ, T )k wk e
l+2,2,γ,Ω×(−∞,T )(3.17)
≤ c(γ, T )(X
1+ X
2+ ku
0k
l+1,Ω) , where c(γ, T ) is an increasing function of T .
Let w = w| e
[0,T ]. Then (3.17) yields
kwk
l+2,ΩT,κ≤ c
1(T )kwk
l+2,ΩT,l/2−[l/2]≤ c
2(T )k wk e
l+2,2,0,Ω×(−∞,T )≤ c
3(γ, T )(X
1+ X
2+ ku
0k
l+2,Ω) , where c
1, c
2, c
3are increasing functions of T .
From the above inequality, (3.8) and (3.9) we get (3.4). This concludes the proof.
Now we consider the following problem with η > 0 (we use here the consider- ations from [9, 16]):
(3.18)
ηu
t− µ∆
ξu − ν∇
ξ∇
ξ· u = F in Ω
T,
Π
0D
ξ(u)n
0= G
1on S
T,
n
0D
ξ(u)n
0− σn
0∆
S(0)
t
R
0
u(τ ) dτ = G
2+
T
R
0
H(τ ) dτ on S
T,
u|
t=0= u
0in Ω .
Lemma 3.2. Assume that F ∈ W
2,κl,l/2(Ω
T), G
i∈ W
l+1/2,l/2+1/42,κ
(S
T), i = 1, 2, H ∈ W
l−1/2,l/2−1/42,κ
(S
T), 1/η ∈ L
∞(Ω
T), η ∈ L
∞(0, T ; Γ
2l+1(Ω)) ∩ W
l+1,l/2+1/22
(Ω
T), η ∈ C
α(Ω
T), α ∈ (0, 1) (where C
α(Ω
T) is the H¨ older space (see [10])), u
0∈ W
2l+1(Ω), S ∈ W
2l+2, l > 3/2 satisfies the assumptions of Lemma 3.1 and κ ∈ (0, 1/4). Then there exists a solution of problem (3.18) such that u ∈ W
2,κl+2,l/2+1(Ω
T) and
(3.19) kuk
l+2,ΩT,κ≤ ϕ
1(T, |1/η|
∞,ΩT, |η|
l+1,0,∞,ΩT, kηk
l+1,ΩT)
× h
kF k
l,ΩT,κ+
2
X
i=1
kG
ik
l+1/2,ST,κ+ kHk
l−1/2,ST,κ+ |u(0)|
l+1,0,Ω+ kuk
l,ΩT,κi
, where ϕ
1is an increasing positive function.
P r o o f. We introduce a partition of unity {ζ
k(λ)(ξ, t), Q
(λ)k} (see [9, 16]), Q
(λ)k= supp ζ
k(λ), k = 1, . . . , N , such that P
Nk=1
ζ
k(λ)(ξ, t) = 1, (ξ, t) ∈ Ω
T, λ =
max
kdiam Q
(λ)k, ζ
k(λ)≥ 0, 0 ≤ µ
0≤ P
Nk=1
(ζ
k(λ))
2≤ N
0and |D
ξ,tαζ
k(λ)(ξ, t)| ≤ c|λ|
−|α|. Set u
k= uζ
k(λ)and h
k= hζ
k(λ). Therefore instead of (3.18) we consider
(3.20)
η(ξ
k, t
k)u
kt− µ∆
ξu
k− ν∇
ξ∇
ξ· u
k= [η(ξ
k, t
k) − η(ξ, t)]u
kt+ ηuζ
kt(λ)− µ[∆
ξ, ζ
k(λ)]u − ν[∇
ξ∇
ξ· , ζ
k(λ)]u + F
k≡ F
k0+ F
k≡ e F
k, Π
0D
ξ(u
k)n
0= Π
0D
ξ(ζ
k(λ))n
0u + G
1k≡ G
01k+ G
1k≡ e G
1k,
n
0D
ξ(u
k)n
0− σn
0∆
S(0)
t
R
0
u
k(τ ) dτ
= n
0D
ξ(ζ
k(λ))n
0u − n
0D
ξ(ζ
k(λ)(0))n
0u
0+ G
2k+
t
R
0
[−σn
0[∆
S(0), ζ
k(λ)]u + n
0D
ξ(u)ζ
k,τ(λ)n
0] dτ
+
t
R
0
(−G
2ζ
k,τ(λ)+ H
k(τ )) dτ
≡ G
02k+ G
2k+
t
R
0
(H
k0(τ ) + H
k(τ )) dτ ≡ e G
2+
t
R
0
H(τ ) dτ , e u
k|
t=0= u
0k,
where (ξ
k, t
k) ∈ Q
(λ)k, [L, u]w = L(uw) − uL(w), L is an operator. To obtain the boundary condition (3.20)
3we differentiate (3.18)
3with respect to time, multiply by ζ
k(λ)and integrate the result with respect to time to get
n
0D
ξ(u
k)n
0− σn
0∆
S(0)
t
R
0
u
k(τ ) dτ
=
t
R
0
[n
0D
ξ(ζ
k(λ))n
0u
τ− σn
0[∆
S(0), ζ
k(λ)]u + n
0D
ξ(uζ
k,τ(λ))n
0] dτ
+ G
2k−
t
R
0
(G
2ζ
k,τ(λ)+ H
k) dτ .
Integrating by parts in the first term on the right-hand side we get (3.20)
3. Changing variables τ = η
k−1t, η
k= η
k(ξ
k, t
k), and applying Lemma 3.1 yields (3.21) ku
kk
l+2,ΩT /ηk,κ≤ c
k e F
kk
l,ΩT /ηk,κ+
2
X
i=1
k e G
ikk
l+1/2,ST /ηk,κ+ k e H
kk
l−1/2,ST /ηk,κ+ ku
0kk
l+1,Ω.
Going back to the variable t in (3.21) implies (3.22) ku
kk
l+2,ΩT,κ≤ ϕ
1(1/min η, max η)
×
k e F
kk
l,ΩT,κ+
2
X
i=1
k e G
ikk
l+1/2,ST,κ+ k e H
kk
l−1/2,ST,κ+ ku
0kk
l+1,Ω. Now we employ the explicit forms of F
k0, G
0ik, H
k0, i = 1, 2, which are given by the right-hand sides of (3.20). We only consider the W
2l,l/2(Ω
T) norms, because the remaining part of the W
2,κl,l/2(Ω
T) norm is easier. First we estimate the first term in the last brackets on the right-hand side of (3.22):
(3.23) k(η(ξ
k, t
k) − η(ξ, t))u
ktk
l,ΩT≤ cλ
α|η|
Cα(ΩT)ku
kk
l+2,ΩT+ [ηu
kt]
l,ΩT,x+ [ηu
kt]
l,ΩT,t+ {lower order terms} . Now we estimate the second term. Let
D
ξ,tα= X
2α0+|α0|=|α|
D
αξ0∂
tα0, α
0= (α
1, α
2, α
3) , |α
0| = α
1+ α
2+ α
3, |α| = [l] .
Then [ηu
kt]
l,ΩT,x≤
tR
0
dt R
Ω
R
Ω
dξ dξ
0X
|α|=[l]
|α|
X
|β|=1
|D
βξ,tηD
ξ,tα−βu
kt− D
ξβ0,tηD
α−βξ0,tu
kt|
2|ξ − ξ
0|
3+2(l−[l]) 1/2≤
TR
0
dt R
Ω
R
Ω
dξ dξ
0X
|α|=[l]
|α|
X
|β|=1
|D
βξ,tη − D
ξβ0,tη|
2|D
ξ,tα−βu
kt|
2|ξ − ξ
0|
3+2(l−[l])+ |D
ξβ0,tη|
2|D
ξ,tα−βu
kt− D
ξα−β0,tu
kt|
2|ξ − ξ
0|
3+2(l−[l]) 1/2≤
TR
0
dt X
|α|=[l]
|α|
X
|β|=1
[D
βξ,tη]
2l−[l]+ε/2,2pβ,ΩR
Ω
R
Ω
dξ dξ
0|D
α−βξ,tu
kt|
2qβ|ξ − ξ
0|
3−εqβ 1/qβ+
R
Ω
R
Ω
dξ dξ
0|D
ξβ0,tη|
2qβ0|ξ − ξ
0|
2−εqβ0 1/qβ0[D
α−βξ,tu
kt]
2l−[l]+ε/2,2p0β,Ω 1/2≤ c RT
0
dt X
|α|=[l]
|α|
X
|β|=1
([D
ξ,tβη]
2l−[l]+ε/2,2pβ,Ω|D
ξ,tα−βu
kt|
22qβ,Ω+ |D
βξ,tη|
22q0β,Ω
[D
α−βξ,tu
kt]
2l−[l]+ε/2,2p0β,Ω)
1/2≡ I
1,
where α ≥ β, ε > 0, 1/p
β+ 1/q
β= 1, 1/p
0β+ 1/q
0β= 1.
Using the imbedding theorems (2.2), (2.3) and the interpolation inequalities (2.5), (2.6) with 3/2−3/(2p
β)+|β|+l−[l]+ε/2 ≤ l+1, 3/2−3/(2q
β)+|α|−|β|+2 <
l + 2 and 3/2 − 3/q
β0+ |β| ≤ l + 1, 3/2 − 3/(2p
0β) + |α| − |β| + l − [l] + ε/2 + 2 < l + 2, respectively, we have
I
1≤ c|η|
l+1,0,2,∞,ΩT(ε
1ku
kk
l+2,ΩT+ cku
kk
l,ΩT) . Next we consider
[ηu
kt]
l,ΩT,t≤
R
Ω
dξ
T
R
0 T
R
0
dt dt
0X
|α|=[l]
|α|
X
|β|=1
|D
βξ,tηD
ξ,tα−βu
kt− D
ξ,tβ 0ηD
α−βξ,t0u
kt0|
2|t − t
0|
1+2(l/2−[l/2]) 1/2≤
R
Ω
dξ
T
R
0 T
R
0
dt dt
0X
|α|=[l]
|α|
X
|β|=1
|D
ξ,tβη − D
ξ,tβ 0η|
2|D
ξ,tα−βu
kt|
2|t − t
0|
1+2(l/2−[l/2])+ |D
ξ,tβ 0η|
2|D
ξ,tα−βu
kt− D
ξ,tα−β0u
kt0|
2|t − t
0|
1+2(l/2−[l/2]) 1/2≤
TR
0 T
R
0
dt dt
0X
|α|=[l]
|α|
X
|β|=1
|D
ξ,tβη − D
βξ,t0η|
22pβ,Ω|D
α−βξ,tu
kt|
22qβ,Ω
|t − t
0|
1+2(l/2−[l/2])+
|D
ξ,tβ 0η|
22p0β,Ω
|D
ξ,tα−βu
kt− D
ξ,tα−β0u
kt0|
22q0 β,Ω|t − t
0|
1+2(l/2−[l/2]) 1/2≡ I
2≡ I
3+ I
4, where 1/p
β+ 1/q
β= 1, 1/p
0β+ 1/q
0β= 1. First we estimate I
3. Let |β| = 1. Then
I
3≤
TR
0 T
R
0
dt dt
0X
|β|=1
|D
ξβη(t) − D
βξη(t
0)|
22p1,Ω|t − t
0|
1+2(l/2−[l/2])X
|α|=[l]−1 T
R
0
|D
αξ,tu
kt|
22q1,Ωdt
1/2≤
sup
t
X
|α|=[l]−1
|D
ξ,tαu
kt|
22q1,ΩT
R
0 T
R
0
dt dt
0kη(t) − η(t
0)k
2[l]+1,Ω|t − t
0|
1+2(l/2−[l/2]) 1/2≡ I
5,
where we used the imbedding (2.2) with
(3.24) 3/2 − 3/(2p
1) ≤ [l] .
To estimate the first factor in I
5we consider two cases: [l] = 2s + 1 and
[l] = 2s, s ∈ N ∪ {0}. In the first case the highest derivative ∂
ts+1u
kappears in
the factor which is the highest t-derivative in the W
2l+2,l/2+1(Ω
T) norm but it
does not appear in the W
2l,l/2(Ω
T) norm where the highest t-derivative is ∂
tsu
k.
Therefore to estimate the expression I
6≡ sup
t
|∂
ts+1u
k|
2q1,Ωwe use
(3.25) ∂
tu
k= 1
η (µ∆
ξu + ν∇
ξ∇
ξ· u)ζ
k+ uζ
kt+ 1 η F
k. We obtain
I
6≤
s
X
i=1
c
issup
t
∂
ts−i1 η
2q1ri,ΩX
|α|=2
|∂
tiD
ξαu|
2q1ri,Ωk+ {lower order terms} + sup
t
∂
ts1 η F
k2q1,Ωk
≡ I
7, where 1/r
i+ 1/r
0i= 1, i = 1, . . . , s, and Ω
k= Ω ∩ supp ζ
k.
Now in view of (2.5) and (2.7) we have I
7≤ c|1/η|
l+1,0,∞,ΩTk
(ε
1−κ1kuk
l+2,ΩTk
+ ε
−κ1kuk
l,ΩT+ kF
kk
l,ΩT+ |u(0)|
l+1,0,Ω) where
(3.26) 3/2 − 3/(2q
1r
i) + 2(s − i) ≤ l + 1 , κ
1= (3/2 − 3/(2q
1r
i0) + 2i + 2)/(l + 2) < 1 .
In the case [l] = 2s the highest t-derivative in the first factor in I
5is ∂
tsu
k. Therefore the factor can be estimated by
ε
1−κ2ku
kk
l+2,ΩT+ cε
−κ2ku
kk
l,ΩT+ c|u(0)|
l+1,0,Ω, where
(3.27) κ
2= (3/2 − 3/(2q
1) + [l] + 1)/(l + 2) < 1 .
Summarizing, (3.24), (3.26), (3.27) are satisfied for l > 3/2, where for l ∈ (3/2, 2) we have to assume q
1= 1, p
1= ∞. Therefore
I
5≤ ϕ
01(T, |1/η|
∞,ΩT, |η|
l+1,0,∞,ΩT)kηk
l+1,ΩT× (εkuk
l+2,ΩTk
+ c(ε)kuk
l,ΩTk
+ kF
kk
l,ΩT+ |u(0)|
l+1,0,Ω) .
Now we examine I
3for 2 ≤ |β| ≤ [l]. Then in view of (2.2) and (2.5) we have I
3≤ X
β