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E. Z A D R Z Y ´ N S K A and W. M. Z A J A ¸ C Z K O W S K I (Warszawa)

LOCAL EXISTENCE OF SOLUTIONS OF A FREE BOUNDARY PROBLEM FOR EQUATIONS OF

COMPRESSIBLE VISCOUS HEAT-CONDUCTING FLUIDS

Abstract. The local existence and the uniqueness of solutions for equa- tions describing the motion of viscous compressible heat-conducting fluids in a domain bounded by a free surface is proved. First, we prove the existence of solutions of some auxiliary problems by the Galerkin method and by regu- larization techniques. Next, we use the method of successive approximations to prove the local existence for the main problem.

1. Introduction. This paper is concerned with the local motion of a drop of a viscous compressible heat-conducting fluid. Let Ω

t

⊂ R

3

be a bounded domain of the drop at time t. Let v = v(x, t) (v = (v

1

, v

2

, v

3

)) be the velocity of the fluid, ̺ = ̺(x, t) the density, θ = θ(x, t) the temperature, f = f (x, t) the external force field per unit mass, r = r(x, t) the heat sources per unit mass, θ = θ(x, t) the heat flow per unit surface, p = p(̺, θ) the pressure, c

v

= c

v

(̺, θ) the specific heat at constant volume, µ and ν the constant viscosity coefficients, κ the constant coefficient of heat conductivity, and p

0

the external (constant) pressure. Then the motion of the drop is described by the following system of equations (see [3], Chs. 2 and 5):

̺[v

t

+ (v · ∇)v] − div T(v, p) = ̺f in e Ω

T

,

̺

t

+ div(̺v) = 0 in e Ω

T

,

̺c

v

t

+ v · ∇θ) − κ∆θ + θp

θ

div v

− µ 2

X

3 i,j=1

(v

ixj

+ v

jxi

)

2

− (ν − µ)(div v)

2

= ̺r in e Ω

T

, (1.1)

1991 Mathematics Subject Classification: 35A05, 35R35, 76N10.

Key words and phrases: free boundary, compressible viscous heat-conducting fluids, local existence.

[179]

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T · n = −p

0

n on e S

T

, v · n = −φ

t

/ |∇φ| on e S

T

,

∂θ

∂n = θ on e S

T

,

̺ |

t=0

= ̺

0

, v |

t=0

= v

0

, θ |

t=0

= θ

0

in Ω, (1.1)

[cont.]

where e Ω

T

= S

t∈(0,T )

t

× {t}, e S

T

= S

t∈(0,T )

S

t

× {t}, S

t

= ∂Ω

t

, φ(x, t) = 0 describes S

t

, n is the unit outward vector normal to the boundary, i.e.

n = ∇φ/|∇φ|, Ω = Ω

t

|

t=0

= Ω

0

. By T = T(v, p) we denote the stress tensor of the form

T(v, p) = {T

ij

}

i,j=1,2,3

= {−pδ

ij

+ D

ij

(v) }

i,j=1,2,3

, where

(1.2) D (v) = {D

ij

(v) }

i,j=1,2,3

= {µ(v

ixj

+ v

jxi

) + (ν − µ)δ

ij

div v }

i,j=1,2,3

is the deformation tensor. Moreover, in view of the thermodynamic consi- derations we assume that c

v

> 0, κ > 0, ν >

13

µ > 0.

Let the domain Ω be given. Then by (1.1)

5

,

t

= {x ∈ R

3

: x = x(ξ, t), ξ ∈ Ω}, where x = x(ξ, t) is the solution of the Cauchy problem

(1.3) ∂x

∂t = v(x, t), x |

t=0

= ξ ∈ Ω, ξ = (ξ

1

, ξ

2

, ξ

3

).

Integrating (1.3) we obtain the following relation between the Eulerian x and Lagrangian ξ coordinates of the same fluid particle:

x = ξ +

t

\

0

u(ξ, t

) dt

≡ X

u

(ξ, t),

where u(ξ, t) = v(X

u

(ξ, t), t). Moreover, by (1.1)

5

, S

t

= {x : x = x(ξ, t), ξ ∈ S = ∂Ω}.

By the continuity equation (1.1)

2

and the kinematic condition (1.1)

5

the total mass is conserved, i.e.

\

t

̺(x, t) dx =

\

̺

0

(ξ) dξ = M, where M is a given constant.

The aim of the paper is to show the local existence theorem for problem

(1.1). In order to prove the local-in-time existence of solutions of (1.1) we

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rewrite it in Lagrangian coordinates as follows:

(1.4)

ηu

t

− div

u

T

u

(u, p) = ηg in Ω

T

= Ω × (0, T ),

η

t

+ η div

u

u = 0 in Ω

T

,

ηc

v

(η, γ)γ

t

− κ∇

2u

γ + γp

γ

(η, γ) div

u

u

− µ 2

X

3 i,j=1

xi

· ∇

ξ

u

j

+ ξ

xj

· ∇

ξ

u

i

)

2

− (ν − µ)(div

u

u)

2

= ηk in Ω

T

,

T

u

(u, p) · n

u

= −p

0

n

u

on S

T

= S × (0, T ),

n

u

· ∇

u

γ = γ on S

T

,

η |

t=0

= ̺

0

, u |

t=0

= v

0

, γ |

t=0

= θ

0

in Ω,

where η(ξ, t) = ̺(X

u

(ξ, t), t), γ(ξ, t) = θ(X

u

(ξ, t), t), p = p(η, γ), g(ξ, t) = f (X

u

(ξ, t), t), k(ξ, t) = r(X

u

(ξ, t), t), γ(ξ, t) = θ(X

u

(ξ, t), t), ∇

u

= ξ

ix

ξi

, T

u

(u, p) = −pI + D

u

(u), I = {δ

ij

}

i,j=1,2,3

is the unit matrix, D

u

(u) = {D

uij

(u) }

i,j=1,2,3

= {µ(∂

xi

ξ

k

ξk

u

j

+ ∂

xj

ξ

k

ξk

u

i

) + (ν − µ)δ

ij

div

u

u }, div

u

u

= ∇

u

· u = ∂

xi

ξ

k

ξk

u

i

, div

u

T(u, p) = {∂

xj

ξ

k

ξk

T

uij

(u, p) }

i=1,2,3

(∂

xi

ξ

k

are the elements of the matrix ξ

x

which is inverse to x

ξ

= I +

Tt

0

u

ξ

(ξ, t

) dt

) and summation over repeated indices is assumed.

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. Thus, we have

n

u

= n(X

u

(ξ, t), t) = ∇

x

φ(x, t)

|∇

x

φ(x, t) |

x=Xu(ξ,t)

and n

0

= n

0

(ξ) = ∇

ξ

φ(ξ) e

|∇

ξ

φ(ξ) e | . The proof of the existence of solutions of problem (1.4) is divided into a few steps. First, we examine the auxiliary problem (3.1) and the problem

(1.5)

ηc

v

(η, β)γ

t

− κ∇

2ξ

γ

= K + µ 2

X

3 i,j=1

xi

· ∇

ξ

w

j

+ ξ

xj

· ∇

ξ

w

i

)

2

in Ω

T

,

n · ∇

ξ

γ = γ on S

T

,

γ |

t=0

= θ

0

in Ω,

where η > 0, β > 0 and w are given functions, ξ

xi

= ξ

xi

(w). We prove the existence of solutions of problems (3.1) and (1.5) by the Galerkin method and by some regularization techniques.

Next, by using the Schauder–Tikhonov fixed point theorem we obtain

the local existence of solutions of problems (3.40) and (3.76) (see Lemmas

3.5 and 3.6).

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Finally, applying the method of successive approximations we prove the local existence and the uniqueness of a solution (u, γ, η) of problem (1.4) such that u, γ ∈ A

T

, η ∈ B

T

, where T ≤ T

, T

> 0 is a certain constant;

A

T

and B

T

are given by (2.1) and (2.2) (see Theorem 4.2).

We have already considered problem (1.1) in papers [7]–[11]. In [7] we proved by using potential techniques from [5] the local existence of solu- tions of (1.4) in Sobolev–Slobodetski˘ı spaces, i.e. we obtained (u, γ, η) ∈ W

24,2

(Ω

T

) × W

24,2

(Ω

T

) × C(0, T ; Γ

3,3/2

(Ω)) for T ≤ T

, where T

> 0 is a certain constant. We cannot apply potential theory in the present paper because this theory is singular in the case of H

3

(Ω) regularity (with respect to the space variable ξ) considered in the paper.

Papers [8] and [9] are concerned with conservation laws and a differential inequality, respectively, used in [10]–[11] to prove the global existence the- orem for problem (1.1) in the case of a special form of the internal energy per unit mass ε = ε(̺, θ). The main result of the present paper, i.e. The- orem 4.2, will be used in [12] to examine the global motion of the viscous compressible barotropic fluid in the general case, i.e. without assuming any conditions on the form of the pressure p.

In this paper we use some results of paper [6], which is concerned with the local existence of solutions of a free boundary problem for the equations of compressible barotropic viscous self-gravitating fluids.

Moreover, local existence theorems for free boundary problems for equa- tions of compressible viscous heat-conducting and self-gravitating fluids are proved in [2] and [4].

2. Notation and auxiliary results. We use the following notation:

• kuk

s,Q

= kuk

Hs(Q)

, s ≥ 0, s rational, Q = Ω, S, S = ∂Ω;

• |u|

p,Q

= kuk

Lp(Ω)

, p ∈ [1, ∞];

• kuk

s,p,q,ΩT

= kuk

Lq(0,T ;Wps(Ω))

, p, q ∈ [1, ∞], 0 ≤ s ∈ Z, Ω

T

= Ω × (0, T );

• kuk

s,p,q,ST

= kuk

Lq(0,T ;Wps(S))

, p, q ∈ [1, ∞], s ≥ 0, s rational, S

T

= S × (0, T ).

Moreover, we introduce the spaces:

(2.1) A

T

= {u ∈ C(0, T ; H

2

(Ω)) ∩ L

2

(0, T ; H

3

(Ω)) :

u

t

∈ C(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

2

(Ω)), u

tt

∈ C(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)) } and

(2.2) B

T

= {u ∈ C(0, T ; H

2

(Ω)) : u

t

∈ C(0, T ; H

1

(Ω)) ∩L

2

(0, T ; H

2

(Ω)),

u

tt

∈ C(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)) }

(5)

with the norms

kuk

AT

= ( sup

0≤t≤T

kuk

22,Ω

+ kuk

23,2,2,ΩT

+ sup

0≤t≤T

ku

t

k

21,Ω

(2.3)

+ ku

t

k

22,2,2,ΩT

+ sup

0≤t≤T

ku

tt

k

20,Ω

+ ku

tt

k

21,2,2,ΩT

)

1/2

and

kuk

BT

= ( sup

0≤t≤T

kuk

22,Ω

+ sup

0≤t≤T

ku

t

k

21,Ω

+ ku

t

k

22,2,2,ΩT

(2.4)

+ sup

0≤t≤T

ku

tt

k

20,Ω

+ ku

tt

k

21,2,2,ΩT

)

1/2

. Finally, define

|u|

l,k,Q

= X

0≤i≤l−k

k∂

ti

u k

l−i,2,2,Q

, where l ≥ k, k ∈ Z

+

∪ {0}, Q = Ω

t

, S

t

and

u

l,k,Ω

= X

0≤i≤l−k

k∂

ti

u k

l−i,Ω

, where l ≥ k, k ∈ Z

+

∪ {0}.

We denote all positive constants in estimates by the same letter c. We also use the following lemmas.

Lemma 2.1. The following imbedding holds:

W

rl

(Ω) ⊂ L

αp

(Ω) (Ω ⊂ R

3

, Ω satisfies the cone condition), where either

κ = |α|

l + 3 lr − 3

lp < 1 and 1 ≤ r ≤ p ≤ ∞ or κ = 1 and 1 < r ≤ p < ∞,

and L

αp

(Ω) is the space of functions u such that |D

αξ

u |

p,Ω

< ∞;

W

rl

(Ω) ⊂ L

αq

(S) (S = ∂Ω, Ω ⊂ R

3

), where either

κ = |α|

l + 3 lr − 2

lq < 1 and 1 ≤ r ≤ q ≤ ∞ or κ = 1 and 1 < r ≤ q < ∞,

and L

αq

(S) is the space of functions u such that |D

ξα

u |

q,S

< ∞. Moreover, the following inequalities hold :

|D

αξ

u |

p,Ω

≤ cε

1−κ

|D

lξ

u |

r,Ω

+ cε

−κ

|u|

r,Ω

, where

κ = |α|

l + 3 lr − 3

lp < 1, 1 ≤ r ≤ p ≤ ∞,

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ε is a parameter and c > 0 is a constant independent of u and ε;

|D

ξα

u |

q,S

≤ cε

1−κ

|D

ξl

u |

r,Ω

+ cε

−κ

|u|

r,Ω

, where

κ = |α|

l + 3 lr − 2

lq < 1, 1 ≤ r ≤ q ≤ ∞, ε is a parameter and c > 0 is a constant independent of u and ε.

Lemma 2.1 follows from Theorem 10.2 of [1].

Lemma 2.2. Assume that η ∈ C(0, T ; H

2

(Ω)), η

t

∈ C(0, T ; H

1

(Ω)), η

tt

∈ L

2

(Ω

T

), η > 0, β ∈ A

T

, β > 0, c

v

∈ C

2

(R

2+

), c

v

> 0. Then ηc

v

(η, β) ∈ C(0, T ; H

2

(Ω)), ∂

t

[ηc

v

(η, β)] ∈ C(0, T ; H

1

(Ω)), ∂

t2

[ηc

v

(η, β)] ∈ C(0, T ; L

2

(Ω)), 1/(ηc

v

(η, β)) ∈ C(Ω × [0, T ]) and

(2.5) sup

t

kηc

v

(η, β) k

22,Ω

≤ ckc

v

k

2C2( ¯V )

sup

t

kηk

22,Ω

f

1

(sup

t

kηk

22,Ω

, sup

t

kβk

22,Ω

), where f

1

(x

1

, x

2

) = 1 + x

1

+ x

2

+ x

21

+ x

22

+ x

1

x

2

and c > 0 is a constant;

(2.6) sup

t

k∂

t

[ηc

v

(η, β)] k

21,Ω

≤ ckc

v

k

2C2( ¯V )

f

2

(sup

t

kηk

22,Ω

, sup

t

kβk

22,Ω

, sup

t

t

k

21,Ω

, sup

t

t

k

21,Ω

), where f

2

(x

1

, x

2

, x

3

, x

4

) = x

1

(x

3

+x

1

x

3

+x

2

x

3

+x

1

x

4

+x

2

x

4

)+x

3

(1+x

2

+x

4

) and c > 0 is a constant;

(2.7) sup

t

k∂

t2

[ηc

v

(η, β)] k

20,Ω

≤ ckc

v

k

2C2( ¯V )

f

3

(sup

t

kηk

22,Ω

, sup

t

t

k

21,Ω

, sup

t

t

k

21,Ω

, sup

t

tt

k

20,Ω

, sup

t

tt

k

20,Ω

), where f

3

(x

1

, x

2

, x

3

, x

4

, x

5

) = x

1

(x

22

+ x

2

x

3

+ x

4

+ x

23

+ x

5

) + x

2

x

3

+ x

4

and c > 0 is a constant;

sup

T

[ηc

v

(η, β)] ≤ kc

v

k

C( ¯V )

sup

T

η;

(2.8)

sup

T

1 ηc

v

(η, β) ≤

1 c

v

C( ¯V )

sup

T

1 η . (2.9)

In (2.5)–(2.9), V ⊂ R

2

is a bounded domain such that (η(ξ, t), β(ξ, t)) ∈ V for any (ξ, t) ∈ Ω

T

.

The proof of the above lemma is obtained using Lemma 2.1.

Now, consider the continuity equation (1.4)

2

. Integrating it we have (2.10) η(ξ, t) = ̺

0

(ξ) exp h

t

\

0

div

u

u dt

i

.

By direct calculations we obtain the following lemma.

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Lemma 2.3. Let ̺

0

∈ H

2

(Ω), ̺

0

> 0, u ∈ L

(0, T ; H

2

(Ω)) ∩ L

2

(0, T ; H

3

(Ω)), u

t

∈ L

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

2

(Ω)). Then η given by (2.10) belongs to B

T

and the following estimates hold :

sup

t

η ≤ k̺

0

k

2,Ω

φ

1

(a(u, t)), sup

t

kηk

22,Ω

≤ k̺

0

k

22,Ω

φ

2

(a(u, t)), sup

t

t

k

21,Ω

≤ k̺

0

k

22,Ω

φ

3

(a(u, t), a

0

(u

t

, t), ku(0)k

22,Ω

), sup

t

tt

k

20,Ω

≤ k̺

0

k

22,Ω

φ

4

(a(u, t), a

0

(u

t

, t), ku(0)k

22,Ω

, ku

t

(0) k

21,Ω

), kη

t

k

21,2,2,Ωt

≤ tk̺

0

k

22,Ω

φ

3

(a(u, t), a

0

(u

t

, t), ku(0)k

22,Ω

),

tt

k

20,Ωt

≤ tk̺

0

k

22,Ω

φ

4

(a(u, t), a

0

(u

t

, t), ku(0)k

22,Ω

, ku

t

(0) k

21,Ω

), kη

t

k

22,2,2,Ωt

≤ k̺

0

k

22,Ω

kuk

23,2,2,Ωt

φ

5

(t, t

a1

kuk

23,2,2,Ωt

),

tt

k

21,2,2,Ωt

≤ k̺

0

k

22,Ω

φ

6

(a(u, t))[φ

7

(a(u, t), b(t, u, ε

3

)) + ku

t

k

22,2,2,Ωt

], where t ≤ T , φ

i

(i = 1, . . . , 7) are positive increasing continuous functions of their arguments, a(u, t) = t

Tt

0

kuk

23,Ω

dt

, a

0

(u

t

, t) = t

Tt

0

ku

t

k

22,Ω

dt

, b is given by (3.46) and a

1

> 0 is a constant. Moreover , 1/η ∈ B

T

and

sup

t

1 η +

1

η

2

Bt

≤ φ

8

( kuk

23,2,2,Ωt

, sup

t

kuk

22,Ω

, ku

t

k

22,2,2,Ωt

, sup

t

ku

t

k

22,Ω

), where t ≤ T and φ

8

is a positive increasing continuous function of its argu- ments.

3. Existence of solutions of auxiliary problems. In order to prove the local-in-time solvability of problem (1.4) we have to consider a few aux- iliary problems. First, we consider the problem

(3.1)

ηu

t

− div D(u) = F in Ω

T

, D (u) · n

0

= G on S

T

, u |

t=0

= v

0

in Ω,

where D(u) is defined by (1.2) and η is a given function. Moreover, (3.2) 0 < ̺

≤ η ≤ ̺

< ∞,

where ̺

and ̺

are constants.

Definition 3.1. By a weak solution of problem (3.1) we mean a function u ∈ C(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)) with u

t

∈ L

2

(Ω

T

) which satisfies the integral identity

(3.3)

\



ηu

t

φ + µ

2 S (u)S(φ) + (ν − µ) div u div φ − F φ

 dξ −

\

S

Gφ dξ

s

= 0

(8)

for all φ ∈ H

1

(Ω) and the initial condition u |

t=0

= v

0

; here f φ = P

3

i=1

f

i

φ

i

, f = u

t

, F , G, S(u) = {u

j

+ u

i

}

i,j=1,2,3

and S (u)S(φ) =

X

3 i,j=1

(u

j

+ u

i

)(φ

j

+ φ

i

).

In order to prove the existence of a weak solution to problem (3.1) we shall apply a Galerkin procedure. Choose a sequence of functions φ

1

, φ

2

, . . . such that: φ

i

∈ H

1

(Ω) for all i; φ

1

, . . . , φ

n

are linearly independent for each n; the set of all linear combinations of the functions φ

i

is dense in H

1

(Ω).

For any n we define an approximate solution of problem (3.1) by

(3.4)

u

n

= X

n i=1

c

in

(t)φ

i

(ξ),

\



ηu

nt

φ

i

+ µ

2 S (u

n

)S(φ

i

) + (ν − µ) div u

n

div φ

i

− F φ

i

 dξ

\

S

i

s

= 0, u

n

(0) = u

n0

,

where u

n0

→ v

0

in H

1

(Ω), u

nt

(0) → u

t

(0) in H

1

(Ω), u

ntt

(0) → u

tt

(0) in L

2

(Ω) and ku

n0

k

1,Ω

≤ ckv

0

k

1,Ω

, ku

nt

(0) k

1,Ω

≤ cku

t

(0) k

1,Ω

, |u

ntt

(0) |

2,Ω

≤ c |u

tt

(0) |

2,Ω

; u

t

(0) and u

tt

(0) are calculated from (3.1); c > 0 is a constant.

Lemma 3.2. Let assumption (3.2) be satisfied. Let η ∈ C(Ω×[0, T ]), η

t

∈ L

2

(0, T ; H

1

(Ω)), η

tt

∈ L

2

(Ω

T

), F ∈ H

2

(0, T ; L

2

(Ω)), G ∈ H

2

(0, T ; L

2

(S)), v

0

∈ H

1

(Ω), u

t

(0) ∈ H

1

(Ω), u

tt

(0) ∈ L

2

(Ω). Then there exists a unique weak solution of problem (3.1) such that u ∈ L

(0, T ; H

1

(Ω)), u

t

∈L

(0, T ; H

1

(Ω)), u

tt

∈ L

(0, T ; L

2

(Ω)) ∩L

2

(0, T ; H

1

(Ω)) and the following estimate is satisfied:

(3.5) kuk

21,Ω

+ ku

t

k

21,Ω

+ ku

tt

k

20,Ω

+ kuk

21,2,2,Ωt

+ ku

t

k

21,2,2,Ωt

+ ku

tt

k

21,2,2,Ωt

≤ Ψ

1

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

, kη

tt

k

20,Ωt

)[ kF k

20,Ωt

+ kF

t

k

20,Ωt

+ ε

1

kF

tt

k

20,Ωt

+ kGk

20,St

+ kG

t

k

20,St

+ ε

1

kG

tt

k

20,St

+ kv

0

k

21,Ω

+ ku

t

(0) k

21,Ω

+ ku

tt

(0) k

20,Ω

],

where t ≤ T , Ψ

1

is a positive increasing continuous function of its arguments

and ε

1

∈ (0, 1) is a sufficiently small constant.

(9)

P r o o f. First, multiply (3.4) by c

in

and sum up over i from 1 to n.

Using the Korn inequality and Lemma 2.1 we get d

dt

\

ηu

2n

dξ + c ku

n

k

21,Ω

≤ c

̺

(1 + kη

t

k

21,Ω

)

\

ηu

2n

+ c( kF k

20,Ω

+ kGk

20,S

) + ε ku

n

k

21,Ω

, where we have also used the fact that

µ 2

\

|S(u

n

) |

2

dξ + (ν − µ)kdiv u

n

k

20,Ω

≥ c

\

|S(u

n

) |

2

dξ.

Hence, integrating with respect to time, taking ε > 0 sufficiently small and using the Gronwall inequality we have

\

ηu

2n

dξ + ku

n

k

21,2,2,Ωt

≤ Ψ

2

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

) (3.6)

× 

\

η(0)v

02

dξ + kF k

20,Ωt

+ kGk

20,St

 , where Ψ

2

is a positive increasing continuous function.

Next, multiplying (3.4) by ˙c

in

and summing up over i we obtain (3.7)

\



ηu

2nt

+ 1 2

d dt

 µ

2 |S(u

n

) |

2

+ (ν − µ)(div u

n

)

2



− F u

nt

 dξ

\

S

Gu

nt

s

= 0.

Integrating (3.7) with respect to t and using the Korn inequality yields

\

t

ηu

2nt

dξ dt

+ ku

n

k

21,Ω

≤ εku

nt

k

21,2,2,Ωt

+ c ku

n

k

20,Ω

(3.8)

+ c( kF k

20,Ωt

+ kGk

20,St

+ kv

0

k

21,Ω

).

Differentiating (3.4) with respect to t, multiplying by ˙c

in

and summing up over i from 1 to n we get

d dt

\

ηu

2nt

dξ + c ku

nt

k

21,Ω

≤ εku

nt

k

21,Ω

+ c

̺

(1 + kη

t

k

21,Ω

)

\

ηu

2nt

dξ (3.9)

+ c( kF

t

k

20,Ω

+ kG

t

k

20,S

).

Integrating (3.9) with respect to time gives (3.10)

\

ηu

2nt

dξ + ku

nt

k

21,2,2,Ωt

≤ Ψ

2

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

) 

\

η(0)u

2t

(0) dξ + kF

t

k

20,Ωt

+ kG

t

k

20,St

 .

(10)

Next, differentiate (3.4) with respect to t, multiply the result by ¨ c

in

and sum up over i from 1 to n. We obtain

\

ηu

2ntt

dξ + 1 2

d dt

\

 µ

2 |S(u

n

t) |

2

+ (ν − µ)(div u

nt

)

2

 dξ

≤ εku

ntt

k

21,Ω

+ c( kη

t

k

21,Ω

ku

nt

k

20,Ω

+ kF

t

k

20,Ω

+ kG

t

k

20,S

).

Hence (3.11)

\

t

ηu

2ntt

dξ dt

+ ku

nt

k

21,Ω

≤ εku

ntt

k

21,2,2,Ωt

+ c( kη

t

k

21,2,2,Ωt

sup

t

ku

nt

k

20,Ω

+ kF

t

k

20,Ωt

+ kG

t

k

20,St

+ ku

nt

k

20,Ωt

+ ku

ntξ

(0) k

20,Ω

).

Finally, differentiating twice (3.4) with respect to t, multiplying by ¨ c

in

and summing up over i we have d

dt

\

ηu

2ntt

dξ + c ku

ntt

k

21,Ω

≤ c

̺

( kη

t

k

21,Ω

+ 1)

\

ηu

2ntt

dξ (3.12)

+ c( kη

tt

k

20,Ω

ku

nt

k

21,Ω

+ ε

1

kF

tt

k

20,Ω

+ ε

1

kG

tt

k

20,S

).

Therefore, integrating (3.12) with respect to time yields (3.13)

\

ηu

2ntt

dξ + ku

ntt

k

21,2,2,Ωt

≤ Ψ

2

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

) 

\

η(0)u

2tt

(0) dξ + ε

1

kF

tt

k

20,Ωt

+ ε

1

kG

tt

k

20,St

+ sup

t

ku

nt

k

21,Ω

tt

k

20,Ωt

 .

Using inequality (3.11) in (3.13) and assuming that εΨ

2

tt

k

20,Ωt

is suffi- ciently small we obtain

(3.14)

\

ηu

2ntt

dξ + ku

ntt

k

21,2,2,Ωt

≤ Ψ

3

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

, kη

tt

k

20,Ωt

) 

\

η(0)u

2tt

(0) dξ + ku

(0) k

20,Ω

+ ε

1

kF

tt

k

20,Ωt

+ ε

1

kG

tt

k

20,St

+ kF

t

k

20,Ωt

+ kG

t

k

20,St

+ ku

nt

k

20,Ωt

+ sup

t

ku

nt

k

20,Ω

 ,

where Ψ

3

is a positive increasing continuous function of its arguments.

(11)

Now, taking into account inequalities (3.6), (3.8), (3.10), (3.11) and (3.14) we get the estimate

ku

n

k

21,Ω

+ ku

nt

k

21,Ω

+ ku

ntt

k

20,Ω

+ ku

n

k

21,2,2,Ωt

+ ku

nt

k

21,2,2,Ωt

+ ku

ntt

k

21,2,2,Ωt

≤ Ψ

1

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

, kη

tt

k

20,Ωt

) h

kF k

20,Ωt

+ kF

t

k

20,Ωt

+ ε

1

kF

tt

k

20,Ωt

+ kGk

20,St

+ kG

t

k

20,St

+ ε

1

kG

tt

k

20,St

+

\

η(0)v

20

dξ +

\

η(0)u

2t

(0) dξ +

\

η(0)u

2tt

(0) dξ + kv

0

k

21,Ω

+ ku

(0) k

20,Ω

i .

Choosing a subsequence and letting n → ∞ we obtain the existence of a solution of (3.1) and estimate (3.5). Uniqueness follows from (3.5). This concludes the proof.

Lemma 3.3. Let assumption (3.2) be satisfied. Let η ∈ C(0, T ; H

2

(Ω)), η

t

∈ C(0, T ; H

1

(Ω)), 1/η ∈ C(0, T ; H

2

(Ω)), (1/η)

t

∈ C(0, T ; H

1

(Ω)), η

tt

∈ L

2

(Ω

T

), F ∈ H

2

(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)), G ∈ L

2

(0, T ; H

3/2

(S)), G

t

∈ L

2

(0, T ; H

1/2

(S)), G

tt

∈ L

2

(S

T

)), S ∈ H

5/2

, v

0

∈ H

2

(Ω), u

t

(0) ∈ H

1

(Ω), u

tt

(0) ∈ L

2

(Ω) (where u

t

(0) and u

tt

(0) are calculated from (3.1)).

Moreover , let the compatibility condition be satisfied:

D (v

0

) · n

0

= G(0) on S.

Then the solution u of problem (3.1) belongs to A

T

and for t ≤ T the following estimate holds :

kuk

2At

≤ Ψ

4

(1/̺

, ̺

, t, h(t, η, ε

2

))[ kF k

21,2,2,Ωt

+ kF

t

k

20,Ωt

(3.15)

+ ε

1

kF

tt

k

20,Ωt

+ sup

t

kF k

20,Ω

+ kGk

23/2,2,2,St

+ kG

t

k

21/2,2,2,St

+ ε

1

kG

tt

k

20,St

+ u(0)

22,0,Ω

],

where ε

i

∈ (0, 1) (i = 1, 2) are sufficiently small constants, Ψ

4

is a positive increasing continuous function of its arguments depending also on kΦk

23,Ω

(where Φ is a transformation which straightens locally the boundary of Ω) and

h(t, η, ε

2

) = kη

t

k

21,2,2,Ωt

+ kη

tt

k

20,Ωt

+ ε

2

sup

t

kηk

22,Ω

(3.16)

+ c(ε

2

) sup

t

kηk

20,Ω

+ ε

2

sup

t

t

k

21,Ω

+ c(ε

2

) sup

t

t

k

20,Ω

.

P r o o f. In [6] it is proved that the solution u of problem (3.1) be-

longs to L

2

(0, T ; H

3

(Ω)), so in view of Lemma 3.2 it suffices to prove that

u ∈ C(0, T ; H

2

(Ω)) with u

t

∈ C(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

2

(Ω)), u

tt

C(0, T ; L

2

(Ω)) and that the estimate (3.15) holds. In order to do it con-

(12)

sider as in [6] a covering {Ω

j

}

nj=1

of Ω and associate with this covering a partition of unity {ζ

j

}

nj=1

, i.e. P

n

j=1

ζ

j

= 1, supp ζ

j

⊂ Ω

j

, ζ

j

∈ C

0

(R

3

).

Denote by e Ω an arbitrary set of the covering {Ω

j

} such that e Ω ∩ S = ∅.

Denote by ζ a function of the partition of unity {ζ

j

} such that supp ζ ⊂ e Ω.

Since the identity (3.3) is satisfied for any test function φ it is also fulfilled for ζφ. Then we have

(3.17)

\

Ωe



ηu

t

ζφ + µ

2 S (u)S(ζφ) + (ν − µ) div u div(ζφ) − F ζφ

 dξ

\

S∩ eΩ= eS

Gζφ dξ

s

= 0.

Now, apply the transformation Φ : e Ω → b Ω which straightens locally the boundary of Ω. Then (3.17) takes the form

(3.18)

\

Ωb

 b

η u e

t

φ + b µ

2 [b S ( u)b e S (b φ) + b S ( e u)b D

1

(b ζ, b φ) − b D

1

( u, b b ζ)b S (b φ)]

 J dz +

\

Ωb

{(ν − µ)[ c div e u c div b φ + c div b ub φ · b ∇b ζ − b u · b ∇b ζ c div b φ] − b F b ζ b φ }J dz

\

Sb

Gb b ζ b φ √ g dz

s

= 0,

where e Ω ∋ ξ → Φ(ξ) = z ∈ b Ω, b u = u ◦ Φ

−1

, e u = b ub ζ, J is the Jacobian of the transformation ξ = Φ

−1

(z) = (z

1

, z

2

, z

3

+ e ψ(z)), e ψ is an extension to b Ω of a function ψ such that e S is described by ξ

3

= ψ(ξ

1

, ξ

2

), g = 1 + ψ

2z1

+ ψ

z22

,

ξ

in S, b D

1

, etc. is replaced by b ∇ = ∇

ξ

Φ(ξ) |

ξ=Φ−1(z)

· ∇

z

, and b D

1

(b ζ, w) = b { b w

i

∇ b

j

ζ + b w b

j

∇ b

i

ζ b }

i,j=1,2,3

(w = φ, u; b ∇ = ( b ∇)

i=1,2,3

).

Moreover, we need that b Ω = {z ∈ R

3

: |z

i

| < d, i = 1, 2, 0 < z

3

< d }, S = Φ( e b S) = {z ∈ R

3

: |z

i

| < d, i = 1, 2, z

3

= 0 }. Since the first integral in (3.18) vanishes on ∂ b Ω \ b S it can be extended by zero onto R

3+

= {z ∈ R

3

: z

3

> 0 }.

Assume in (3.18) (as in [6]) b φ = δ

−1n

δ

n

u, where e δ

h

u(z) = 1

h [u(z

+ h, z

3

) − u(z)], δ

h−1

u(z) = 1

h [u(z

− h, z

3

) − u(z)], z

= (z

1

, z

2

).

Then the first term in (3.18) can be rewritten as (see (A.7) in [6])

\

Ωb

b

η e u

t

φJ dz = b −

\

Ωb

δ

h

η b u e

t

δ

h

uJ dz e −

\

Ωb

b

η u e

t

δ

h

uδ e

h

J dz (3.19)

− 1 2

d dt

\

Ωb

b

η |δ

h

u e |

2

J dz + 1 2

\

Ωb

b

η

t

h

u e |

2

J dz

(13)

and by Lemma 2.1 the first two terms in (3.19) are bounded by ε kδ

h

e u k

20, b

+ c(ε)(ε

2

kb η k

22, b

+ c(ε

2

) kb η k

20, b

) ke u

t

k

21, b

,

where ε ∈ (0, 1) and ε

2

∈ (0, 1). Hence in the same way as in [6] we obtain (cf. inequality (A.8) of [6]) the estimate

(3.20) 1 2

d dt

\

Ωb

b

η |δ

h

u e |

2

J dz + µ

2 kδ

h

u e k

21, b

≤ c(ε

2

kb η k

22, b

+ c(ε

2

) kb η k

20, b

) ke u

t

k

21, b

+ c kb η

t

k

21, b

h

u e k

20, b

+ ε kδ

h

b u k

21, b

+ c kb u k

21, b

+ c( k e F k

20, b

+ k e G k

21/2, bS

).

Integrating (3.20) with respect to time, going back to the old variables, summing over all neighborhoods of the partition of unity, using the fact that ε is sufficiently small and letting h tend to 0 we get

(3.21)

\

ηu

2τ

dξ + µ ku

τ

k

21,2,2,Ωt

\

η(0)v

2

dξ + (ε

2

sup

t

kηk

22,Ω

+ c(ε

2

) sup

t

kηk

20,Ω

) ku

t

k

21,2,2,Ωt

+ c kη

t

k

21,2,2,Ωt

sup

t

kuk

21,Ω

+ c( kuk

21,2,2,Ωt

+ kF k

20,Ωt

+ kGk

1/2,2,2,St

), where u

τ

denotes the tangent derivatives to the boundary and the constant c depends on kΦk

23,Ω

.

To calculate the normal derivatives we use the equation

− div D(u) = F − ηu

t

. Hence we have

(3.22) ku

n

k

21,Ω

≤ c(kF k

20,Ω

+ ku

τ

k

21,Ω

) + c(ε

2

kηk

21,Ω

c(ε

2

) kηk

20,Ω

) ku

t

k

21,Ω

. Now, inequalities (3.21) and (3.22) imply

(3.23)

\

ηu

2τ

dξ + kuk

22,2,2,Ωt

\

η(0)v

2

dξ + (ε

2

sup

t

kηk

22,Ω

+ c(ε

2

) sup

t

kηk

20,Ω

) ku

t

k

21,2,2,Ωt

+ c kη

t

k

21,2,2,Ωt

sup

t

kuk

21,Ω

+ c( kuk

21,2,2,Ωt

+ kF k

20,Ωt

+ kGk

21/2,2,2,St

), where t ≤ T and the r.h.s. of (3.23) is bounded in terms of the estimates for the weak solution (see Lemma 3.2).

Now, we obtain estimates for sup

t

kuk

22,Ω

and ku

t

k

22,2,2,Ωt

. To do this we

put b φ = δ

−1h

δ

h

e u

t

in (3.18).

(14)

Using the H¨older and Young inequalities and Lemma 2.1 we obtain (3.24)

\

Ωb

b

η |δ

h

u e

t

|

2

J dz + 1 2

d dt

\

Ωb

 µ

2 |bS(δ

n

e u) |

2

+ (ν − µ)( c divδ

n

u) e

2

 J dz

≤ εkδ

h

u e

t

k

21, b

+ c[ ke u

t

k

21, b

2

h

b η k

21, b

+ c(ε

2

) kδ

h

η b k

20, b

) + ke u

t

k

21, b

2

kb η k

22, b

+ c(ε

2

) kb η k

20, b

) kδ

h

J k

21, b

+ kb u k

21, b

+ ke u k

22, b

h

J k

21, b

+ k e F k

20, b

+ k e G k

21/2, bS

].

Integrating (3.24) with respect to time and using the Korn inequality yields (3.25)

\

Ωbt

b

η |δ

h

u e

t

|

2

J dz dt

+ kδ

h

e u k

21, b

≤ εkδ

h

u e

t

k

21,2,2, bt

+ c[ ke u

t

k

21,2,2, bt

2

sup

t

h

η b k

21, b

+ c(ε

2

) sup

t

h

b η k

20, b

+ c(ε

2

) sup

t

kb η k

20, b

) + ke u

t

k

21,2,2, bt

kΦk

23, b

2

sup

t

kb η k

22, b

+ c(ε

2

) sup

t

kηk

20,Ω

) + kb u k

21,2,2, bt

+ kΦk

23, b

ke u k

22,2,2, bt

+ k e F k

20, bt

+ k e G k

21/2,2,2, bSt

+ kv

0

k

22, b

].

Now, since by Lemma 3.2, (ηu

t

)

,t

= η

t

u

t

+ ηu

tt

∈ L

2

(Ω

t

) and F

t

∈ L

2

(Ω

t

) we have [div D(u)]

,t

= div D(u

t

) ∈ L

2

(Ω

t

). Therefore, differentiating (3.1)

1

and (3.1)

2

with respect to t we obtain the problem

(3.26)

ηu

tt

− div D(u

t

) = F

t

− η

t

u

t

in Ω

T

, D (u

t

) · n

0

= G

t

on S

T

, u

t

|

t=0

= u

t

(0) in Ω.

Problem (3.26) is, with respect to v = u

t

, of analogous form to problem (3.1), so to get an estimate for kvk

22,2,2,Ωt

= ku

t

k

22,2,2,Ωt

we use the same argument as in the case of the estimate for kuk

22,2,2,Ωt

. Thus, (3.20) is now replaced by

(3.27) 1 2

d dt

\

Ωb

b

η |δ

h

u e

t

|

2

J dz + µ

2 kδ

h

e u

t

k

21, b

≤ c(ε

2

kb η k

22, b

+ c(ε

2

) kb η k

20, b

) ke u

tt

k

21, b

+ c kb η

t

k

21, b

h

u e

t

k

20, b

+ ε kδ

h

e u

t

k

21,Ω

+ c kb u

t

k

21, b

+ c( k e F

t

k

20, b

+ k e G

t

k

21/2, bS

+ kb η

t

k

21, b

ke u

t

k

21, b

).

Integrating (3.27) with respect to time gives

(15)

(3.28)

\

Ωb

b

η |δ

h

u e

t

|

2

J dz +

t

\

0

h

e u

t

k

21, b

dt

≤ c(ε

2

sup

t

kb η k

22, b

+ c(ε

2

) sup

t

kb η k

20, b

) ke u

tt

k

21,2,2, bt

+ c kb η

t

k

21,2,2, bt

(sup

t

h

e u

t

k

20, b

+ sup

t

ke u

t

k

21, b

) + c kb u

t

k

21,2,2, bt

+ c 

k e F

t

k

20, bt

+ k e G

t

k

21/2,2,2, bSt

+

\

Ωb

η(0) |δ

h

u

t

(0) |

2

dξ  . Adding inequalities (3.25) and (3.28), next going back to the old variables, summing over all neighbourhoods of the partition of unity, using the fact that ε is sufficiently small and letting h tend to 0 we get

(3.29)

\

t

η |u

|

2

dξ dt

+

\

η |u

|

2

dξ + ku

τ

k

21,Ω

+ ku

k

1,2,2,Ωt

≤ c(kΦk

23,Ω

) h

ku

t

k

21,2,2,Ωt

1

sup

t

kηk

22,Ω

+ c(ε

1

) sup

t

kηk

20,Ω

) + ku

tt

k

21,2,2,Ωt

2

sup

t

kηk

22,Ω

+ c(ε

2

) sup

t

kηk

20,Ω

) + kuk

22,2,2,Ωt

+ ku

t

k

21,2,2,Ωt

+ kη

t

k

21,2,2,Ωt

sup

t

ku

t

k

21,Ω

+ kF k

20,Ωt

+ kF

t

k

20,Ωt

+ kGk

21/2,2,2,St

+ kG

t

k

21/2,2,2,St

+ kv

0

k

22,Ω

+

\

η(0) |u

(0) |

2

dξ i . In order to calculate the normal derivatives of u

t

we use the equation

− div D(u

t

) = F

t

− ηu

tt

− η

t

u

t

. Hence, we have

ku

tn

k

21,2,2,Ωt

≤ c[kF

t

k

20,Ωt

+ ku

k

21,2,2,Ωt

+ kη

t

k

21,2,2,Ωt

sup

t

ku

t

k

21,Ω

(3.30)

+ (ε

2

sup

t

kηk

22,Ω

+ c(ε

2

) sup

t

kηk

20,Ω

) ku

tt

k

20,Ωt

].

Taking into account inequalities (3.29), (3.30) and the inequality sup

t

ku

n

k

21,Ω

≤ c(sup

t

kF k

20,Ω

+ sup

t

ku

τ

k

21,Ω

) + c(ε

1

sup

t

kηk

21,Ω

+ c(ε

1

) sup

t

kηk

20,Ω

) sup

t

ku

t

k

21,Ω

, which follows from (3.22), and using estimate (3.5) we find that u ∈ L

(0, T ; H

2

(Ω)), u

t

∈ L

2

(0, T ; H

2

(Ω)) and

(3.31) kuk

22,Ω

+ ku

t

k

22,2,2,Ωt

≤ Ψ

5

(1/̺

, ̺

, t, kη

t

k

21,2,2,Ωt

, kη

tt

k

20,Ωt

, ε

2

sup

t

kηk

22,Ω

+ c(ε

2

) sup

t

kηk

20,Ω

)

× [kF k

20,Ωt

+ kF

t

k

20,Ωt

+ ε

1

kF

tt

k

20,Ωt

+ sup

t

kF k

20,Ω

+ kGk

21/2,2,2,St

+ kG

t

k

21/2,2,2,St

+ ε

1

kG

tt

k

20,St

+ u(0)

22,0,Ω

],

where Ψ

5

is a positive increasing continuous function.

(16)

It remains to find an estimate for kuk

23,2,2,Ωt

. Using the same argument as in [6] and after similar calculations to those for (A.16) of [6] we get (3.32) 1

2 d dt

\

Ωb

b

η |∂

τ2

j

δ

u |

2

J dz + µ

2 k∂

τ2

j

δ

u e k

21, b

≤ c[kb η k

22, b

kj

δ

u e

t

k

21, b

+ (ε

2

kb η

t

k

21, b

+ c(ε

2

) kb η

t

k

20, b

) k∂

τ2

j

δ

u e k

20, b

+ kb η k

22, b

ke u

t

k

21, b

+ kb u k

22, b

+ k e F k

21, b

+ k e G k

23/2, bS

+ ε k∂

n2

τ

j

δ

u e k

20, b

], where j

δ

is the Friedrichs mollifier operator. Integrating (3.32) with respect to time yields

(3.33)

\

Ωb

b

η |∂

τ2

j

δ

e u |

2

J dz + µ k∂

τ2

j

δ

e u k

21,2,2, bt

≤ c h

kb η k

22,2,2, bt

sup

t

ke u

t

k

21, b

+ (ε

2

sup

t

kb η

t

k

21, b

+ c(ε

2

) sup

t

kb η

t

k

20, b

) kb u k

22,2,2, bt

+ kb u k

22,2,2, bt

+ k e F k

21,2,2, bt

+ k e G k

23/2,2,2, bSt

+ ε k∂

n2

τ

j

δ

e u k

20, bt

+

\

Ωb

b

η(0) |∂

τ2

u(0) e |

2

J dz i . Next, (A.18) of [6] is now replaced by

(3.34) ∂

τ

n2

\

ω

δ

(x − y)

 1 − 1

Φ

23,x

23,x

− Φ

23,y

)

 e u(y) dy

2

0, bΩt

≤ ckb η k

22,2,2, bt

sup

t

ke u

t

k

21, b

+ c

1

k∂

τ2

j

δ

e u k

21,2,2, bt

+ c kb u k

22,2,2, bt

+ c k e F k

21,2,2, bt

, where ω

δ

is the smooth kernel of the mollifier operator j

δ

. Estimates (3.33) and (3.34) imply

(3.35) 2c

1

µ

\

Ωb

b

η |∂

τ2

j

δ

u e |

2

J dz + c

1

k∂

τ2

j

δ

u e k

21,2,2, bt

+ ∂

τ

n2

\

ω

δ

(x − y)

 1 − 1

Φ

23,x

23,x

− Φ

23,y

) − 2εc

1

µ

 e u(y) dy

2

0, bΩt

≤ c h

kb η k

22,2,2, bt

sup

t

ke u

t

k

21, b

+ (ε

2

sup

t

kb η

t

k

21, b

+ c(ε

2

) sup

t

kb η

t

k

20, b

) kb u k

22,2,2, bt

+ kb u k

22,2,2, bt

+ k e F k

21,2,2, bt

+ k e G k

23/2,2,2, bSt

+ ε k∂

n2

τ

j

δ

e u k

20, bt

+

\

Ωb

b

η(0) |∂

τ2

u(0) e |

2

J dz i .

Using the fact that Φ

23,x

is close to one and Φ

23,x

− Φ

23,y

is close to zero for

x, y ∈ b Ω and for b Ω sufficiently small, we obtain from (3.35), after taking the

(17)

limit as δ → 0, (3.36)

\

Ωb

b

η |∂

τ2

e u |

2

J dz + k∂

τ2

u e k

21,2,2, bt

+ k∂

2n

τ

u e k

20, bt

≤ c h

kb η k

22,2,2, bt

sup

t

ke u

t

k

21, b

+ (ε

2

sup

t

kb η

t

k

21, b

+ c(ε

2

) sup

t

kb η

t

k

20, b

) kb u k

22,2,2, bt

+ kb u k

22,2,2, bt

+ k e F k

21,2,2, bt

+ k e G k

23/2,2,2, bSt

+

\

Ωb

b

η(0) |∂

τ2

u(0) e |

2

J dz i . Next, from equation (A.13) of [6] we have

k∂

n3

e u k

20, bt

≤ c(k∂

n2

τ

u e k

20, bt

+ k∂

n

τ2

u e k

20, bt

(3.37)

+ kb u k

22,2,2, bt

+ kb η k

22,2,2, bt

sup

t

ke u

t

k

21, bt

+ k e F k

21,2,2, bt

).

Taking into account (3.36) and (3.37) we see that u ∈ L

2

(0, T ; H

3

(Ω)) and

kuk

23,2,2,Ωt

≤ c h

kηk

22,2,2,Ωt

sup

t

ku

t

k

21,Ω

(3.38)

+ (ε

2

sup

t

t

k

21,Ω

+ c(ε

2

) sup

t

t

k

20,Ω

) kuk

22,2,2,Ωt

+ kuk

22,2,2,Ωt

+ kF k

21,2,2,Ωt

+ kGk

23/2,2,2,St

+

\

η(0) |∂

τ2

u(0) |

2

dz i .

Adding inequalities (3.31), (3.38) and (3.5) we obtain estimate (3.15).

To complete the proof notice that from u ∈ L

(0, T ; H

2

(Ω)) ∩ L

2

(0, T ; H

3

(Ω)) and u

t

∈ L

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

2

(Ω)) it follows that u ∈ C(0, T ; H

2

(Ω)). Next, since u

t

∈ L

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

2

(Ω)) and u

tt

∈ L

(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)) we have u

t

∈ C(0, T ; H

1

(Ω)).

In order to prove that u

tt

∈ C(0, T ; L

2

(Ω)) differentiate (3.1)

1

with re- spect to t, multiply by φ ∈ H

1

(Ω) and integrate over Ω, using the boundary condition (3.1)

2

, to get

\

u

tt

φ dξ =

\



− η

t

u

t

η + F

t

η − ∇( 1 η )D(u

t

)

 φ dξ

\

D (u

t

)

η ∇φ dξ +

\

S

G

t

η φ dξ

s

∀φ ∈ H

1

(Ω), where D(u

t

) ∇φ = P

3

i,j=1

D

ij

(u

t

ixj

. Hence,

(18)

d dt

\

u

tt

φ dξ =

\



− ∇

 1 η



,t



D (u

t

) − ∇

 1 η

 D (u

tt

) (3.39)

− η

tt

u

t

η − η

t

u

tt

η −

 1 η



,t

η

t

u

t

+ F

tt

η + F

t

 1 η



,t

 φ dξ

\

 D (u

tt

)

η − D(u

t

)

 1 η



,t



∇φ dξ +

\

S

 G

tt

η + G

t

 1 η



,t

 φ dξ

s

≡ hg

1

, φ i

+ hg

2

, φ i

S

∀φ ∈ H

1

(Ω).

Identity (3.39) implies ku

ttt

k

2L2(0,T ;(H1(Ω)))

≤ c

T\

0

sup

kφk1,Ω≤1

|hg

1

, φ i

|

2

dt + c

T\

0

sup

kφk1,Ω≤1

|hg

2

, φ i

S

|

2

dt

≤ Ψ

6

( kηk

2BT

, k1/ηk

2BT

, ku

t

k

22,2,2,ΩT

, ku

tt

k

21,2,2,ΩT

, kG

tt

k

20,ST

,

kG

t

k

1/2,2,2,ST

, kF

tt

k

20,ΩT

, kF

t

k

20,ΩT

), where Ψ

6

is a positive increasing continuous function of its arguments. Hence u

tt

∈ C(0, T ; L

2

(Ω)). This completes the proof of the lemma.

Now, consider the problem (3.40)

ηu

t

− div

w

D

w

(u) = F in Ω

T

, D

w

(u) · n

w

= G on S

T

, u |

t=0

= v

0

in Ω,

where η and w are given functions, n

w

= n(X

w

(ξ, t), t), D

w

(u) = µS

w

(u) + (ν −µ)I div

w

u, S

w

(u) = {∂

xi

ξ

k

ξk

u

j

+∂

xj

ξ

k

ξk

u

i

}

i,j=1,2,3

, I = {δ

ij

}

i,j=1,2,3

. We assume that η satisfies (3.2) and w = w(ξ, t) is such that

x = ξ +

t

\

0

w(ξ, s) ds ≡ X

w

(ξ, t) = x(ξ, t)

and x

ξ

= {∂x

i

/∂ξ

j

}

i,j=1,2,3

, ξ

x

= {∂ξ

i

/∂x

j

}

i,j=1,2,3

are matrices with de- terminants close to 1 for t ∈ [0, T ].

In order to prove the existence of solutions of (3.40) consider first the problem

(3.41)

ηu

t

− div D(u) = div

w

D

w

(u

) − div D(u

) + F in Ω

T

, D (u) · n

0

= D(u

) · n

0

− D

w

(u

) · n

w

+ G on S

T

,

u |

t=0

= v

0

in Ω,

where u

is a given function.

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