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Abstract. We consider the motion of a viscous compressible barotropic fluid in R

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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

3

bounded 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

t

of Ω

t

is 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

0

the 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

0

in Ω ,

Tn − σHn = p

0

n 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]

(2)

where φ(x, t) = 0 describes S

t

, e Ω

T

= S

t∈(0,T )

t

× {t}, Ω

t

is 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)

5

means that the free boundary S

t

is 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(∂

xi

v

j

+ ∂

xj

v

i

) + (ν − µ)δ

ij

div 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

t

be 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

sα

g

−1/2

b g

αβ

sβ

= g

−1/2

sα

g

1/2

g

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

= ∂

sα

x, {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)

5

implies that the boundary S

t

is a material surface, so if ξ ∈ S = S

0

then X

u

(ξ, t) ∈ S

t

and S

t

= {x : x = X

u

(ξ, t), ξ ∈ S}.

In virtue of the continuity equation (1.1)

2

and (1.1)

5

the total mass M is conserved and

(1.7) R

t

%(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

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an estimate (see (4.3)) for the linearized problem (1.1)

1,3,4

with 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,0

2 (ΩT)

=  R

T

0

kuk

2Wl

2(Ω)

dt 

1/2

,

kuk

W20,l/2(ΩT)

=

 R

kuk

2

W2l/2((0,T ))

dx



1/2

,

kuk

2Wl

2(Ω)

= X

|α|≤[l]

kD

xα

uk

2L2(Ω)

+ X

|α|=[l]

R

R

|D

αx

u(x, t) − D

yα

u(y, t)|

2

|x − y|

3+2(l−[l])

dx dy

≡ X

|α|≤[l]

|D

αx

u|

22,Ω

+ X

|α|=[l]

[D

xα

u]

2l−[l],Ω

≡ X

|α|≤[l]

|D

αx

u|

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

2

W2l/2((0,T ))

=

[l/2]

X

j=0

k∂

tj

uk

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

|∂

tj

u|

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

(4)

must be omitted, kuk

2

W2l,l/2(ΩT)

= X

|α|≤[l]

kD

αx,t

uk

2L2(ΩT)

+ X

|α|=[l]



T

R

0

R

R

|D

αx,t

u(x, t) − D

αy,t

u(y, t)|

2

|x − y|

3+2(l−[l])

dx dy dt + R

Ω T

R

0 T

R

0

|D

αx,t

u(x, t) − D

x,τα

u(x, τ )|

2

|t − τ |

1+2(l/2−[l/2])

dt dτ dx



≡ X

|α|≤[l]

|D

x,tα

u|

22,ΩT

+ X

|α|=[l]

([D

αx,t

u]

2l−[l],ΩT,x

+ [D

x,tα

u]

2l/2−[l/2],ΩT,t

)

≡ X

|α|≤[l]

|D

x,tα

u|

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

3

and 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)

3

for 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/2

2 (Ω)

if Q = Ω

T

or Q = S

T

, l ≥ 0 , kuk

l,Q

= kuk

Wl

2(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 ;Wl

2(Ω))

, 1 ≤ p ≤ ∞ . Let us introduce a space Γ

rl

(Ω) with the norm

kuk

Γl

r(Ω)

= X

i≤[l/2]

ti

u

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

T

or S

T

we define u

l,r,Q

= kuk

Wl,l/2

r (Q)

. We also define the following norms:

|u|

l,0,∞,ΩT

= sup

t∈[0,T ]

 X

|α|≤[l]

|D

x,tα

u|

22,Ω

+ X

|α|=[l]

[D

αx,t

u]

l−[l],Ω



1/2

,

|u|

l,0,Ω

=

 X

|α|≤[l]

|D

x,tα

u|

22,Ω

+ X

|α|=[l]

[D

x,tα

u]

l−[l],Ω



1/2

, l ∈ R

+

.

(5)

We introduce

W

l,l/22

(Q

T

) = {u ∈ W

2l,l/2

(Q

T

) : ∂

ti

u|

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,t

u|

22,Ω

t

dt



1/2

, κ ∈ (0, 1) , kuk

(l),ΩT

= kuk

l,ΩT

+ u

[l]+κ,ΩT

,

kuk

l,ΩT

= kuk

l,ΩT

+ T

−κ

X

|α|=[l]

|D

x,tα

u|

2,ΩT

.

We denote by W

2,κl,l/2

(Ω

T

) the space with the norm k k

(l),ΩT

. 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

−κ

hui

[l],ΩT

≤ u

[l]+κ,ΩT

≤ c[u]

[l]+κ,ΩT,t

+ cT

−κ

hui

[l],ΩT

≤ cT

l/2−[l/2]−κ

[u]

l,ΩT,t

+ cT

−κ

hui

[l],ΩT

, where κ ≤ l/2 − [l/2], hui

[l],ΩT

= P

|α|=[l]

|D

αx,t

u|

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

2

Hγl,l/2(ΩT)

=

T

R

0

e

−2γt

kuk

2l,Ω

dt + kuk

2

Hγ0,l/2(ΩT)

, where γ ≥ 0. Here for l/2 6∈ Z,

kuk

2

Hγ0,l/2(ΩT)

= γ

l

T

R

0

e

−2γt

kuk

20,Ω

dt

+

T

R

0

e

−2γt

dt

R

0

dτ k∂

tk

u

0

(·, t − τ ) − ∂

tk

u

0

(·, t)k

20,Ω

τ

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

2

H0,l/2γ (ΩT)

=

T

R

0

e

−2γt

l

kuk

20,Ω

+ k∂

tl/2

uk

20,Ω

) dt ,

and we assume that ∂

tj

u|

t=0

= 0, j = 0, . . . , l/2 − 1, so u

0

(x, t) has a general- ized derivative ∂

tl/2

u

0

in L

2

(Ω × (−∞, T )). Moreover, we introduce the notation kuk

l,2,γ,ΩT

= kuk

Hl,l/2

γ (ΩT)

.

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For the purposes of Section 4 we define [u]

p,%,Ω

=

 R

R

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

αx

u]

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,t

u|

p,Q

, l ∈ N ,

[u]

p,%,ΩT,x

=



T

R

0

dt R

dx R

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

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,t

u]

p,l−[l],ΩT,x

+ [D

x,tα

u]

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,En

dh 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

xα

uii

p,l−[l],En+1(T ),xi

+hh∂

t[l/2]

uii

p,l/2−[l/2],En+1(T ),t

,

(7)

where

hhuii

p,%,En+1(T ),xi

=



R

0

|u(x

1

, . . . , x

i

+ h, . . . , x

n

, t) − u(x, t)|

pp,E

n+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,E

n+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

µx

u|

f,Ω

≤ ckuk

d,Ω

,

3/2 − 3/f + |µ| ≤ d, 0 ≤ d ∈ R, 1 < f ∈ R, |µ| ∈ N ∪ {0} , (2.3) [D

µx

u]

p,%,Ω

≤ ckuk

l,Ω

,

3/2 − 3/p + |µ| + % ≤ l, % ∈ (0, 1), 1 < p ∈ R, |µ| ∈ N ∪ {0} , (2.4) |D

µx

u|

f,Ω

≤ ε

1−κ

hui

[d],Ω

+ cε

−κ

|u|

2,Ω

,

0 < κ = (3/2 − 3/f + |µ|)/[d] < 1, ε ∈ (0, 1) , (2.5) |D

µx

u|

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

µx

u]

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∂

ti

uk

2−2/p,p,Ω

≤ c(k∂

ti

uk

2,p,ΩT

+ k∂

ti

u|

t=0

k

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

r

r,α,E

e

n(T ),t

≤ [u]

r

r,α,D

e

n(T ),t

+ 2

T

R

0

dt |u(·, t)|

rr,En−1

1 t

, 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)



T

R

0

|f (x)|

p

x

pl

dx



1/p

≤ A(p, l)



T

R

0

dx

T

R

0

dy |f (x) − f (y)|

p

|x − y|

1+pl



1/p

,

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where f is such that the right-hand side is finite and 0 < A = α/[(1 − (pl)

−1/p

a

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

 R

ax

0

|f (y)|

p

dy 

1/p

 R

ax

0

dy 

1/p0

(2.10)

+ 1 ax



ax

R

0

|f (x) − f (y)|

p

|x − y|

1+pl

dy



1/p

 R

ax

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)

p0

x

p0(1+l)

. Therefore, (2.10) becomes

(2.11) |f (x)| ≤  1 ax

ax

R

0

|f (y)|

p

dy



1/p

+ a

−1

α(p, a)x

l



ax

R

0

|f (x) − f (y)|

p

|x − y|

1+pl

dy



1/p

. In view of the Minkowski inequality, (2.11) implies



T

R

0

|f (x)|

p

x

pl

dx



1/p

≤ a

−1/p



T

R

0

dx x

1+pl

ax

R

0

|f (y)|

p

dy



1/p

(2.12)

+ α a



T

R

0

dx

ax

R

0

|f (x) − f (y)|

p

|x − y|

1+pl

dy



1/p

= a

−1/p



aT

R

0

|f (y)|

p

dy

T

R

y/a

dx x

1+pl



1/p

+ α a



T

R

0

dx

ax

R

0

|f (x) − f (y)|

p

|x − y|

1+pl

dy



1/p

≡ I .

(9)

Integrating the second integral in the first term yields



T

R

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/p

a

l−1/p



T

R

0

|f (y)|

p

y

pl

dy



1/p

+ α a



T

R

0

dx

ax

R

0

|f (x) − f (y)|

p

|x − y|

1+pl

dy



1/p

. Since l > 1/p, assuming that



T

R

0

|f (x)|

p

x

pl

dx



1/p

< ∞

(here we use f (0) = 0) and that a is so small that 1 − (pl)

−1/p

a

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,Ω

dt t

τ

≤ c

1

T

R

0

dt

T

R

0

dt

0

|u(·, t) − u(·, t

0

)|

22,Ω

|t − t

0

|

1+τ

+ c

2

T

−τ

T

R

0

|u|

22,Ω

dt , where c

1

, c

2

do 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∇

2u

u − ν∇

u

u

· u + ∇

u

q = 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

0

n(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

= ∂

x

ξ

i

ξi

, ∇

ξi

= ∂

ξi

, T

u

(u, q) = −qδ + D

u

(u), δ = {δ

ij

} and D

u

(u) = {µ(∂

xi

ξ

k

ξk

u

j

+ ∂

xj

ξ

k

ξk

u

i

) + (ν − µ)δ

ij

u

· u}, with ∇

u

· u = ∂

xi

ξ

k

ξk

u

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

t

0

ξj

u

i

(ξ, τ ) dτ . Assuming |∇

ξ

u|

∞,ΩT

≤ M we obtain

(3.2) 0 < c

1

(1 − M t)

3

≤ det{∂

ξj

x

i

} ≤ c

2

(1 + M t)

3

, t ≤ T ,

(10)

where c

1

, c

2

are constants and T is sufficiently small. Moreover, det A = exp( R

t

0

u

· u dτ ) = %

0

/η.

Let S

t

be 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) = ∇

x

φ(x, t)

|∇

x

φ(x, t)|

x=x(ξ,t)

, n

0

(ξ) = ∇

ξ

φ(ξ) e

|∇

ξ

φ(ξ)| e .

To prove the existence of solutions of (3.1) we consider first the following linear problem:

(3.3)

u

t

− µ∆

ξ

u − ν∇

ξ

ξ

· u = f

1

in Ω

T

,

Π

0

D

ξ

(u)n

0

= g

1

on S

T

,

n

0

D

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

0

in Ω ,

where Π

0

, Π are projections defined by Πg = g − (g · n)n, Π

0

g = g − (g · n

0

)n

0

, and D

ξ

(u) = {µ(∂

ξi

u

j

+ ∂

ξj

u

i

) + (ν − µ)δ

ij

ξk

u

k

}.

Lemma 3.1. Let f

1

∈ W

2,κl,l/2

(Ω

T

), g

1

, g

2

∈ W

l+1/2,l/2+1/4

2,κ

(S

T

), h

1

W

l−1/2,l/2−1/4

2,κ

(S

T

), u

0

∈ W

2l+1

(Ω), S ∈ W

2l+2

and 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

≤ c

1

(X

1

+ |u(0)|

l+1,0,Ω

) ≤ c

2

(X

1

+ X

2

+ ku

0

k

l+1,Ω

) , where

X

1

= kf

1

k

l,ΩT

+

2

X

i=1

kg

i

k

l+1/2,ST

+ kh

1

k

l−1/2,ST

,

X

2

=

[l/2−1/2]

X

i=0

k∂

ti

f

1

(0)k

l−1−2i,Ω

,

|u(0)|

l+1,0,Ω

=

[l/2+1/2]

X

i=0

k∂

ti

u(0)k

l+1−2i

≤ c

3

(X

2

+ ku

0

k

l+1,Ω

) , and the constants c

1

, c

2

, c

3

do not depend on T for T < ∞.

P r o o f. Let ϕ

i

= ∂

ti

u|

t=0

∈ W

2l+1−2i

(Ω) be calculated from (3.3)

1

inductively:

(3.5) ϕ

i+1

= µ∇

2ξ

ϕ

i

+ ν∇

ξ

ξ

· ϕ

i

+ ∂

ti

f

1

(0) , i ≤ [l/2 − 1/2] .

(11)

They satisfy the following compatibility conditions:

(3.6)

Π

0

D

ξ

i

)n

0

= ∂

ti

g

1

(0) , i ≤ [l/2 + 1/4 − 1/2] , n

0

D

ξ

i

)n

0

− σn

0

S

(0)ϕ

i−1

= ∂

ti

g

2

(0) + σ∂

ti−1

h

1

(0) ,

1 ≤ i ≤ [l/2 − 1/4 − 1/2] , n

0

D

ξ

0

)n

0

= g

2

(0) , for i = 0 .

We extend the functions ϕ

i

, i = 0, . . . , [l/2 + 1/2], to R

3

in 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

i

k

l+1−2i,R3

≤ ckϕ

i

k

l+1−2i,Ω

. Now we construct a function e v such that

(3.7) ∂

ti

e 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+

≤ ck e vk

l+2,R3×R+

≤ c

[l/2+1/2]

X

i=0

k ϕ e

i

k

l+1−2i,R3

≤ c

[l/2+1/2]

X

i=0

i

k

l+1−2i,Ω

= c|u(0)|

l+1,0,Ω

≤ c(X

2

+ ku

0

k

l+1,Ω

) . Let v = e v|

T

. Then introducing the function

(3.9) w = u − v ,

we see that it is a solution of the problem

(3.10)

w

t

− µ∇

2ξ

w − ν∇

ξ

ξ

· w = f

10

in Ω

T

,

Π

0

D

ξ

(w)n

0

= g

10

on S

T

,

n

0

D

ξ

(w)n

0

− σn

0

S

(0)

t

R

0

w(τ ) dτ = g

02

+ σ

t

R

0

h

01

dτ on S

T

,

w|

t=0

= 0 in Ω ,

where

(3.11)

f

10

= f

1

− (v

t

− µ∇

2ξ

v − ν∇

ξ

ξ

· v) ∈ W

l,l/22

(Ω

T

) , g

10

= g

1

− Π

0

D

ξ

(v)n

0

∈ W

l+1/2,l/2+1/4

2

(S

T

) ,

g

20

= g

2

− n

0

D

ξ

(v)n

0

∈ W

l+1/2,l/2+1/4

2

(S

T

) ,

h

01

= h

1

− σn

0

S

(0)v ∈ W

l−1/2,l/2−1/4

2

(S

T

) .

(12)

Therefore, we have (3.12) kf

10

k

l,ΩT

+

2

X

i=1

kg

i0

k

l+1/2,ST

+ kh

01

k

l−1/2,ST

≤ c 

kf

1

k

l,ΩT

+

2

X

i=1

kg

i

k

(l+1/2),ST

+ kh

1

k

l−1/2,ST

+ kvk

l+2,ΩT

 + D

ξ

(v)n

0 [l+1/2]+κ,ST

+ n

0

D

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

0

D

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

01

can 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/2−1/4

0

(S

T

) and by Lemmas 2.3 and 2.4 that g

100

, g

200

∈ H

l+1/2,l/2+1/4

0

(S

T

)

and

(3.14)

kf

100

k

l,2,0,ΩT

≤ ckf

10

k

l,ΩT

,

kg

i00

k

l+1/2,2,0,ST

≤ ckg

i0

k

(l+1/2),ST

, i = 1, 2 , kh

001

k

l−1/2,2,0,ST

≤ ckh

01

k

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+1/4

γ

(S

T

), h

001

∈ H

l−1/2,l/2−1/4

γ

(S

T

) and there ex- ists a constant c(γ) such that

(3.15)

kf

100

k

l,2,γ,ΩT

≤ c(γ)kf

100

k

l,2,0,ΩT

,

kg

i00

k

l+1/2,2,γ,ST

≤ c(γ)kg

i00

k

l+1/2,2,0,ST

, i = 1, 2, kh

001

k

l−1/2,2,γ,ST

≤ c(γ)kh

001

k

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∇

2ξ

w − ν∇ e

ξ

ξ

· w = f e

100

in Ω × (−∞, T ) , Π

0

D

ξ

( w)n e

0

= g

100

on S × (−∞, T ) , n

0

D

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 ) ,

(13)

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

0

k

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

0

k

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

≤ 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

0

k

l+2,Ω

) , where c

1

, c

2

, c

3

are 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∆

ξ

u − ν∇

ξ

ξ

· u = F in Ω

T

,

Π

0

D

ξ

(u)n

0

= G

1

on S

T

,

n

0

D

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

0

in Ω .

Lemma 3.2. Assume that F ∈ W

2,κl,l/2

(Ω

T

), G

i

∈ W

l+1/2,l/2+1/4

2,κ

(S

T

), i = 1, 2, H ∈ W

l−1/2,l/2−1/4

2,κ

(S

T

), 1/η ∈ L

(Ω

T

), η ∈ L

(0, T ; Γ

2l+1

(Ω)) ∩ W

l+1,l/2+1/2

2

(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

≤ ϕ

1

(T, |1/η|

∞,ΩT

, |η|

l+1,0,∞,ΩT

, kηk

l+1,ΩT

)

× h

kF k

l,ΩT

+

2

X

i=1

kG

i

k

l+1/2,ST

+ kHk

l−1/2,ST

+ |u(0)|

l+1,0,Ω

+ kuk

l,ΩT

i

, where ϕ

1

is 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

N

k=1

ζ

k(λ)

(ξ, t) = 1, (ξ, t) ∈ Ω

T

, λ =

(14)

max

k

diam Q

(λ)k

, ζ

k(λ)

≥ 0, 0 ≤ µ

0

≤ P

N

k=1

k(λ)

)

2

≤ N

0

and |D

ξ,tα

ζ

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∆

ξ

u

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

0

D

ξ

(u

k

)n

0

= Π

0

D

ξ

k(λ)

)n

0

u + G

1k

≡ G

01k

+ G

1k

≡ e G

1k

,

n

0

D

ξ

(u

k

)n

0

− σn

0

S

(0)

t

R

0

u

k

(τ ) dτ

= n

0

D

ξ

k(λ)

)n

0

u − n

0

D

ξ

k(λ)

(0))n

0

u

0

+ G

2k

+

t

R

0

[−σn

0

[∆

S

(0), ζ

k(λ)

]u + n

0

D

ξ

(u)ζ

k,τ(λ)

n

0

] dτ

+

t

R

0

(−G

2

ζ

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)

3

we differentiate (3.18)

3

with respect to time, multiply by ζ

k(λ)

and integrate the result with respect to time to get

n

0

D

ξ

(u

k

)n

0

− σn

0

S

(0)

t

R

0

u

k

(τ ) dτ

=

t

R

0

[n

0

D

ξ

k(λ)

)n

0

u

τ

− σn

0

[∆

S

(0), ζ

k(λ)

]u + n

0

D

ξ

(uζ

k,τ(λ)

)n

0

] dτ

+ G

2k

t

R

0

(G

2

ζ

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−1

t, η

k

= η

k

k

, t

k

), and applying Lemma 3.1 yields (3.21) ku

k

k

l+2,ΩT /ηk

≤ c 

k e F

k

k

l,ΩT /ηk

+

2

X

i=1

k e G

ik

k

l+1/2,ST /ηk

+ k e H

k

k

l−1/2,ST /ηk

+ ku

0k

k

l+1,Ω



.

(15)

Going back to the variable t in (3.21) implies (3.22) ku

k

k

l+2,ΩT

≤ ϕ

1

(1/min η, max η)

× 

k e F

k

k

l,ΩT

+

2

X

i=1

k e G

ik

k

l+1/2,ST

+ k e H

k

k

l−1/2,ST

+ ku

0k

k

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

kt

k

l,ΩT

≤ cλ

α

|η|

Cα(ΩT)

ku

k

k

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

0+|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



t

R

0

dt R

R

dξ dξ

0

X

|α|=[l]

|α|

X

|β|=1

|D

βξ,t

ηD

ξ,tα−β

u

kt

− D

ξβ0,t

ηD

α−βξ0,t

u

kt

|

2

|ξ − ξ

0

|

3+2(l−[l])



1/2



T

R

0

dt R

R

dξ dξ

0

X

|α|=[l]

|α|

X

|β|=1

 |D

βξ,t

η − D

ξβ0,t

η|

2

|D

ξ,tα−β

u

kt

|

2

|ξ − ξ

0

|

3+2(l−[l])

+ |D

ξβ0,t

η|

2

|D

ξ,tα−β

u

kt

− D

ξα−β0,t

u

kt

|

2

|ξ − ξ

0

|

3+2(l−[l])



1/2



T

R

0

dt X

|α|=[l]

|α|

X

|β|=1



[D

βξ,t

η]

2l−[l]+ε/2,2pβ,Ω



R

R

dξ dξ

0

|D

α−βξ,t

u

kt

|

2qβ

|ξ − ξ

0

|

3−εqβ



1/qβ

+



R

R

dξ dξ

0

|D

ξβ0,t

η|

2qβ0

|ξ − ξ

0

|

2−εqβ0



1/qβ0

[D

α−βξ,t

u

kt

]

2l−[l]+ε/2,2p0β,Ω



1/2

≤ c  R

T

0

dt X

|α|=[l]

|α|

X

|β|=1

([D

ξ,tβ

η]

2l−[l]+ε/2,2pβ,Ω

|D

ξ,tα−β

u

kt

|

22qβ,Ω

+ |D

βξ,t

η|

22q0

β,Ω

[D

α−βξ,t

u

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.

(16)

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

1

ku

k

k

l+2,ΩT

+ cku

k

k

l,ΩT

) . Next we consider

[ηu

kt

]

l,ΩT,t



R

T

R

0 T

R

0

dt dt

0

X

|α|=[l]

|α|

X

|β|=1

|D

βξ,t

ηD

ξ,tα−β

u

kt

− D

ξ,tβ 0

ηD

α−βξ,t0

u

kt0

|

2

|t − t

0

|

1+2(l/2−[l/2])



1/2



R

T

R

0 T

R

0

dt dt

0

X

|α|=[l]

|α|

X

|β|=1

 |D

ξ,tβ

η − D

ξ,tβ 0

η|

2

|D

ξ,tα−β

u

kt

|

2

|t − t

0

|

1+2(l/2−[l/2])

+ |D

ξ,tβ 0

η|

2

|D

ξ,tα−β

u

kt

− D

ξ,tα−β0

u

kt0

|

2

|t − t

0

|

1+2(l/2−[l/2])



1/2



T

R

0 T

R

0

dt dt

0

X

|α|=[l]

|α|

X

|β|=1

 |D

ξ,tβ

η − D

βξ,t0

η|

22pβ,Ω

|D

α−βξ,t

u

kt

|

22q

β,Ω

|t − t

0

|

1+2(l/2−[l/2])

+

|D

ξ,tβ 0

η|

22p0

β,Ω

|D

ξ,tα−β

u

kt

− D

ξ,tα−β0

u

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



T

R

0 T

R

0

dt dt

0

X

|β|=1

|D

ξβ

η(t) − D

βξ

η(t

0

)|

22p1,Ω

|t − t

0

|

1+2(l/2−[l/2])

X

|α|=[l]−1 T

R

0

|D

αξ,t

u

kt

|

22q1,Ω

dt



1/2

 sup

t

X

|α|=[l]−1

|D

ξ,tα

u

kt

|

22q1,Ω

T

R

0 T

R

0

dt dt

0

kη(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

5

we consider two cases: [l] = 2s + 1 and

[l] = 2s, s ∈ N ∪ {0}. In the first case the highest derivative ∂

ts+1

u

k

appears 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 ∂

ts

u

k

.

(17)

Therefore to estimate the expression I

6

≡ sup

t

|∂

ts+1

u

k

|

2q1,Ω

we use

(3.25) ∂

t

u

k

= 1

η (µ∆

ξ

u + ν∇

ξ

ξ

· u)ζ

k

+ uζ

kt

+ 1 η F

k

. We obtain

I

6

s

X

i=1

c

is

sup

t

ts−i

1 η

2q1ri,Ω

X

|α|=2

|∂

ti

D

ξα

u|

2q1ri,Ωk

+ {lower order terms} + sup

t

ts

 1 η F

k



2q1,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,∞,ΩT

k

1−κ1

kuk

l+2,ΩT

k

+ ε

−κ1

kuk

l,ΩT

+ kF

k

k

l,ΩT

+ |u(0)|

l+1,0,Ω

) where

(3.26) 3/2 − 3/(2q

1

r

i

) + 2(s − i) ≤ l + 1 , κ

1

= (3/2 − 3/(2q

1

r

i0

) + 2i + 2)/(l + 2) < 1 .

In the case [l] = 2s the highest t-derivative in the first factor in I

5

is ∂

ts

u

k

. Therefore the factor can be estimated by

ε

1−κ2

ku

k

k

l+2,ΩT

+ cε

−κ2

ku

k

k

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,ΩT

k

+ c(ε)kuk

l,ΩT

k

+ kF

k

k

l,ΩT

+ |u(0)|

l+1,0,Ω

) .

Now we examine I

3

for 2 ≤ |β| ≤ [l]. Then in view of (2.2) and (2.5) we have I

3

≤ X

β

1−κ

|u

k

|

l+1,0,∞,ΩT

+ cε

−κ

|u

k

|

l,0,∞,ΩT

)kηk

l+1,ΩT

≡ I

6

,

where 3/2 − 3/(2p

β

) + |β| ≤ [l] + 1, κ

= (3/2 − 3/(2q

β

) + |α − β| + 2)/(l + 1) < 1,

2 ≤ |β| ≤ [l], |α| = [l]. Finally, from Lemma 2.2 the first factor in I

6

is estimated

by εku

k

k

l+2,ΩT

+ c(ε)|u

k

|

l+1,0,Ω

|

t=0

.

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