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G. S T R ¨ O H M E R (Iowa City, Ia.) W. M. Z A J A ¸ C Z K O W S K I (Warszawa)

LOCAL EXISTENCE OF SOLUTIONS OF THE FREE BOUNDARY PROBLEM

FOR THE EQUATIONS OF COMPRESSIBLE BAROTROPIC VISCOUS SELF-GRAVITATING FLUIDS

Abstract. Local existence of solutions is proved for equations describing the motion of a viscous compressible barotropic and self-gravitating fluid in a domain bounded by a free surface. First by the Galerkin method and reg- ularization techniques the existence of solutions of the linearized momentum equations is proved, next by the method of successive approximations local existence to the nonlinear problem is shown.

1. Introduction. In this paper we prove the existence of local solutions to equations describing the motion of a viscous compressible barotropic fluid under the self-gravitating force in a bounded domain Ω

t

⊂ R

3

with a free boundary S

t

. Let v = v(x, t) be the velocity of the fluid, ̺ = ̺(x, t) the density, p = p(̺) the pressure, µ and ν the constant viscosity coefficients and p

0

the external constant pressure. Then the problem is described by the following system of equations (see [1], Chs. 1,2):

(1.1)

̺(v

t

+ v · ∇v) − div T(v, p) = ̺∇U in e Ω

T

= S

0≤t≤T

t

× {t},

̺

t

+ div(̺v) = 0 in e Ω

T

,

T · n = −p

0

n on e S

T

= S

0≤t≤T

S

t

× {t},

̺|

t=0

= ̺

0

, v|

t=0

= v

0

in Ω = Ω

0

, v · n = − φ

t

|∇φ| on e S

T

,

1991 Mathematics Subject Classification: 76N10, 35G30.

Key words and phrases: local existence, compressible viscous barotropic self-gravita- ting fluids.

The research of W. M. Zaj¸aczkowski was partially supported by Polish Government grant KBN 2 P301 008 06.

[1]

(2)

where φ(x, t) = 0 describes S

t

, n is the unit outward vector normal to S

t

, n = ∇φ/|∇φ|, Ω

t

is the domain at time t, S

t

= ∂Ω

t

, Ω = Ω

t

|

t=0

= Ω

0

, S = ∂Ω.

By T = T(v, p) we denote the stress tensor of the form (1.2) T (v, p) = {T

ij

}

i,j=1,2,3

= {−pδ

ij

+ D

ij

(v)}

i,j=1,2,3

, where

(1.3) D(v) = {D

ij

}

i,j=1,2,3

= {µ(∂

xi

v

j

+ ∂

xj

v

i

) + (ν − µ)δ

ij

div v}

i,j=1,2,3

is the deformation tensor.

Moreover, U (x, t) is the self-gravitating potential such that

(1.4) U (x, t) = k

\

t

̺(y, t)

|x − y| dy, where k is the gravitation constant.

By the continuity equation (1.1)

2

and the kinematic condition (1.1)

5

the total mass is conserved, so

(1.5)

\

t

̺(x, t) dx =

\

̺

0

(x) dx = M, where M is a given constant.

Let Ω be given. We introduce the Lagrangian coordinates ξ as the initial data for the Cauchy problem

(1.6) dx

dt = v(x, t), x|

t=0

= ξ ∈ Ω, ξ = (ξ

1

, ξ

2

, ξ

3

).

Integrating (1.6) we obtain a transformation between the Eulerian x and the Lagrangian ξ coordinates,

(1.7) x = x(ξ, t) ≡ ξ +

t

\

0

u(ξ, τ ) dτ ≡ x

u

(ξ, t),

where u(ξ, t) = v(x

u

(ξ, t), t) and the index u in x

u

(ξ, t) will be omitted in evident cases.

Then, by (1.1)

5

, Ω

t

= {x ∈ R

3

: x = x(ξ, t), ξ ∈ Ω} and S

t

= {x ∈ R

3

: x = x(ξ, t), ξ ∈ S}.

Let η(ξ, t) = ̺(x(ξ, t), t), q(ξ, t) = p(x(ξ, t), t), ∇

u

= ∂

x

ξ

i

ξi

, ∂

ξi

= ∇

ξi

,

T

u

(u, q) = −qI + D

u

(u), I = {δ

ij

}

i,j=1,2,3

is the unit matrix, D

u

(u) =

{µ(∂

xi

ξ

k

ξk

u

j

+ ∂

xj

ξ

k

ξk

u

i

) + (ν − µ)δ

ij

u

· u}, where ∇

u

· u = ∂

xi

ξ

k

ξk

u

i

and summation over repeated indices is assumed.

(3)

Since S

t

is determined (at least locally) by the equation φ(x, t) = 0, S is described by φ(x(ξ, t), t)|

t=0

= e φ(ξ) = 0. Moreover, we have

n

u

= n(x

u

(ξ, t), t) = ∇

x

φ(x, t)

|∇

x

φ(x, t)|

x=x

u(ξ,t)

,

n

0

= n

u0

(ξ, t) = ∇

ξ

φ(ξ) e

|∇

ξ

φ(ξ)| e .

In Lagrangian coordinates the problem (1.1) takes the form

(1.8)

ηu

t

− div

u

T

u

(u, q) = U

u

(η) in Ω

T

= Ω × (0, T ), η

t

+ η div

u

u = 0 in Ω

T

,

T

u

(u, q) · n

u

= −p

0

n

u

on S

T

= S × (0, T ),

u|

t=0

= v

0

in Ω,

η|

t=0

= ̺

0

in Ω,

where

(1.9) U

u

(η) = k

\

η(ϑ, t)

|x

u

(ξ, t) − x

u

(ϑ, t)| A(x

u

(ϑ, t)) dϑ

and A is the Jacobian determinant of the transformation x = x(ξ, t).

The proof of the existence of solutions of problem (1.8) is divided into a few steps. First we consider the problem

(1.10)

u

t

− div D(u) = f

1

, D (u) · n = b

1

, u|

t=0

= v

0

.

At the second step we examine the problem with a given positive function η(ξ, t):

(1.11)

ηu

t

− div D(u) = f in Ω

T

, D (u) · n = g on S

T

, u|

t=0

= v

0

in Ω.

To examine the nonlinear problem (1.8) we need an existence result for the problem

(1.12)

ηu

t

− div

w

D

w

(u) = f

3

, D

w

(u) · n

w

= g

3

, u|

t=0

= u

0

,

where η > 0 and w = w(ξ, t) are given functions.

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Finally, we prove the existence of solutions to (1.8), hence also to (1.1), by the following method of successive approximations:

(1.13)

η

m

t

u

m+1

− div

um

D

u

m

(u

m+1

) = −∇

um

q(η

m

) + η

m

U

um

m

) in Ω

T

, D

u

m

(u

m+1

) · n

um

= −(q(η

m

) − p

0

)n

um

on S

T

,

u

m+1

|

t=0

= v

0

in Ω,

where η

m

and u

m

are treated as given, and

(1.14) ∂

t

η

m

+ η

m

div

um

u

m

= 0 in Ω

T

, η

m

|

t=0

= ̺

0

in Ω, where u

m

is treated as given, and m = 0, 1, . . .

We want to point out that the presented proof of existence uses the Galerkin method and some regularization techniques because the case con- sidered is singular in potential theory. This is related to H

3

(Ω) regularity.

Ordinarily the Galerkin and regularization methods are connected with the energy method which is much more natural for (1.1) than the potential tech- nique. Moreover, this technique is applied in the stability proof for (1.1) in [4]. We have also to emphasize that H

3

(Ω) regularity for v is the lowest pos- sible regularity in spaces with integer derivatives for solutions of nonlinear problems such as (1.1) to exist. As follows from [3] the existence of solutions to (1.1) can be shown in the spaces H

2+α,1+α/2

(Ω

T

), α ∈ (1/2, 1), but the norm of these spaces contains fractional derivatives and is not convenient for our considerations in [4].

In [2] local existence of solutions for the free boundary problem for the equations of a viscous compressible heat-conducting self-gravitating fluid is proved. However, the proof is done in a different way and the regularity obtained is not suitable for our considerations in [4].

2. Notation. To simplify considerations we introduce the following notation:

kuk

s,Q

= kuk

Hs(Q)

, s ∈ N ∪ {0}, Q ∈ {Ω, Ω

t

, S, S

t

}, Ω

t

= Ω × (0, t), S

t

= S × (0, t),

|u|

p,Q

= kuk

Lp(Q)

, p ∈ [1, ∞],

kuk

s,p,q,ΩT

= kuk

Lq(0,T ;Wps(Ω))

, p, q ∈ [1, ∞], 0 ≤ s ∈ Z.

We define the space Γ

lk

(Ω) as part of T

k−l

i=0

C

i

([0, T ]; H

k−i

(Ω)) with the norm kuk

Γk

l(Ω)

= P

k−l

i=0

k∂

ti

uk

k−i,Ω

.

(5)

Then we denote by L

p

(0, T ; Γ

lk

(Ω)) the closure of C

(Ω

T

) with the norm



T\

0

 X

k−l

i=0

k∂

ti

uk

k−i,Ω



p

dt 

1/p

, p ∈ [1, ∞].

Moreover, we introduce

|u|

k,l,p,ΩT

= kuk

Lp(0,T ; Γk

l(Ω))

.

3. Existence of solutions. We prove the existence of solutions to prob- lem (1.1) by the method of successive approximations described by problems (1.13) and (1.14). Therefore, we first consider the following auxiliary prob- lem:

(3.1)

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. Moreover, η is such that (3.2) 0 < ̺

≤ η ≤ ̺

< ∞

and w = w(ξ, t) is such that

(3.3) x = ξ +

t

\

0

w(ξ, τ ) dτ ≡ x

w

(ξ, t) ≡ x(ξ, t), and

(3.4) ∂x

∂ξ , ∂ξ

∂x

are matrices with determinants close to 1 for t ∈ [0, T ].

Definition 3.1. By a weak solution to problem (3.1) we mean a function u which satisfies the integral identity

(3.5)

\

[ηu

t

ϕ + D

w

(u) · D

w

(ϕ) − F · ϕ]J

w

dξ −

\

S

GϕJ

w

s

= 0

for any sufficiently smooth function ϕ, where D

w

(u) · D

w

(ϕ) =

µ2

(∇

wi

u

j

+

wj

u

i

)(∇

wi

ϕ

j

+ ∇

wj

ϕ

i

) + (ν − µ)∇

w

· u∇

w

· ϕ and J

w

is the Jacobian determinant of the transformation x = x

w

(ξ, t).

To obtain the integral formula for (3.1) we use the following integration

by parts:

(6)

\

div

w

D

w

(v(x

w

(ξ, t), t)ϕ(x

w

(ξ, t), t)J

w

=

\

t

div D(v(x, t))ϕ(x, t) dx

= −

\

t

D

(v) · D

(ϕ)dx +

\

St

n · D(v)ϕ ds

= −

\

D

w

(v(x

w

(ξ, t), t)) · D

w

(ϕ(x

w

(ξ, t), t))J

w

dξ +

\

S

n

w

· D

w

(v(x

w

(ξ, t), t))ϕ(x

w

(ξ, t), t)J

w

S

, where D

(v) · D

(ϕ) =

µ2

(∂

xi

v

j

+ ∂

xj

v

i

)(∂

xi

ϕ

j

+ ∂

xj

ϕ

i

) + (ν − µ) div v div ϕ.

Take a basis {ϕ

k

} in L

2

(Ω). Then we are looking for an approximate solution of (3.5) in the form

(3.6) u

n

=

X

n i=1

c

in

(t)ϕ

i

(ξ),

where c

in

, i = 1, . . . , n, are solutions of the following system of ordinary differential equations:

\

[ηu

nt

ϕ

i

+ D

w

(u

n

) · D

w

i

) − F · ϕ

i

]J

w

dξ −

\

S

i

J

w

S

= 0, (3.7)

u

n

|

t=0

= X

n i=1

c

in

(0)ϕ

i

(ξ), c

in

(0) =

\

v

i0

ϕ

n

dξ,

where i = 1, . . . , n, and existence follows from the theory of ordinary differ- ential equations.

Next we have to obtain estimates for solutions of (3.7).

Lemma 3.2. Assume that ̺

≤ η, η

t

∈ L

2

(0, T ; H

1

(Ω)), F ∈ L

2

(Ω

T

), G ∈ L

2

(S

T

), w ∈ L

2

(0, T ; H

3

(Ω)). Assume that

(3.8) sup

t∈[0,T ]

sup

ξ∈Ω

|I − ξ

x

| ≤ δ,

where δ is sufficiently small and I is the unit matrix. Then for solutions of (3.7) the following inequality holds:

ku

n

k

20,Ω

+ c

0

ku

n

k

21,2,2,Ωt

≤ ψ

1

(1/̺

, t, kη

t

k

1,2,2,Ωt

, a(w, t)) (3.9)

× h

\

̺

0

v

20

dx + kF k

20,Ωt

+ kGk

20,St

i ,

where ψ

1

is an increasing positive function, a(w, t) = t

1/2

kwk

3,2,2,Ωt

, and

t ≤ T .

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P r o o f. Multiplying (3.7) by c

in

and summing over i from 1 to n we get (3.10) 1

2

\

 η d

dt u

2n

+ |D

w

(u

n

)|

2



J

w

dξ =

\

F · u

n

J

w

dξ +

\

S

G · u

n

J

w

S

. Using the Korn inequality

(3.11) kuk

21,Ω

≤ c(kD

′′

(u)k

20,Ω

+ kuk

20,Ω

) and |D

(v)|

2

≥ c

0

|D

′′

(v)|

2

, c

0

= min 

3

4

ν −

13

µ  ,

µ2

, where D

′′

(u) = {µ(∂

xi

u

j

+ ∂

xj

u

i

)}, we have

kuk

21,Ω

≤ c(kD

′′w

(u)k

20,Ω

+ kD

′′

(u) − D

′′w

(u)k

20,Ω

+ kuk

20,Ω

), so in view of (3.8) we get

(3.12) kuk

21,Ω

≤ c(δ)(kD

′′w

(u)k

20,Ω

+ kuk

20,Ω

).

Using (3.12) in (3.10) implies d

dt

\

ηu

2n

J

w

dξ + c

0

ku

n

k

21,Ω

≤ c

\

(|η

t

| + η|div

w

w|)|u

n

|

2

J

w

dξ (3.13)

+ c(ku

n

k

20,Ω

+ kF k

20,Ω

+ kGk

20,S

).

Estimating the first term on the r.h.s. by εku

k

20,Ω

+

 c(ε)

̺

t

k

21,Ω

+ |div

w

w|

∞,Ω



\

ηu

2n

J

w

dξ, ε ∈ (0, 1), from (3.13) we get

(3.14) d dt

\

ηu

2n

J

w

dξ + c

0

ku

n

k

21,Ω

 c

̺

(1 + kη

t

k

21,Ω

) + |div

w

w|

∞,Ω



\

ηu

2n

J

w

dξ + c(kF k

20,Ω

+ kGk

20,S

).

Integrating (3.14) with respect to time yields (3.15)

\

ηu

2n

J

w

dξ + c

0

ku

n

k

21,2,2,Ωt

exp

 c

̺

(t + kη

t

k

21,2,2,Ωt

+ ϕ(a(w, t))) h

\

̺

0

v

20

dξ + kF k

20,Ωt

+ kGk

20,St

i , where ϕ is an increasing positive function. From (3.15) we obtain (3.9).

This concludes the proof.

From (3.9) we can prove the existence of weak solutions such that u ∈

L

(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)). However, we want to obtain more reg-

ular weak solutions simultaneously. Therefore we show

(8)

Lemma 3.3. Assume that ̺

≤ η, F

t

∈ L

2

(Ω

T

), G

t

∈ L

2

(S

T

), F ∈ L

(0, T ; L

2

(Ω)), G ∈ L

(0, T ; L

2

(S)), w ∈ L

2

(0, T ; H

3

(Ω)), η

t

∈ L

2

(0, T ; H

1

(Ω)) and

T

̺

0

u

2t

(0) dξ < ∞. Assume (3.8). Then ku

nt

k

20,Ω

+ c

0

ku

nt

k

21,2,2,Ωt

≤ ψ

2

(1/̺

, a(w, t), t, kη

t

k

1,2,2,Ωt

) (3.16)

× h

\

̺

0

u

2t

(0) dξ + kF

t

k

20,Ωt

+ kG

t

k

20,St

+ sup

t

(ku

n

k

21,Ω

+ kF k

20,Ω

+ kGk

20,S

)

×

t

\

0

ε

1

kwk

23,Ω

+ c(ε

1

)kwk

20,Ω

 dt i

, where ψ

2

is an increasing positive function and ε

1

∈ (0, 1).

P r o o f. Differentiating (3.7) with respect to t, multiplying by ˙c

in

and summing up over i from 1 to n we get

(3.17) d dt

\

ηu

2nt

J

w

dξ + c

0

ku

nt

k

21,Ω

≤ c

\

t

|u

2nt

J

w

dξ + ϕ

1

(a(w, t))

\

ηu

2nt

J

w

|w

ξ

| dξ + c(ku

nt

k

20,Ω

+ kF

t

k

20,Ω

+ kG

t

k

20,S

)

+ ϕ

1

(a(w, t))|w

ξ

|

2∞,Ω

(ku

n

k

21,Ω

+ kF k

20,Ω

+ kGk

20,S

), where ϕ

1

is an increasing positive function, a(w, t) was defined in Lemma 3.2, and the Korn inequality and condition (3.8) were used.

Estimating the first term on the r.h.s. of (3.17) by c

0

2 ku

nt

k

21,Ω

+ c(µ, ν)

̺

t

k

21,Ω

\

ηu

2nt

J

w

dξ we can write (3.17) in the form

(3.18) d dt

\

ηu

2nt

J

w

dξ + c

0

ku

nt

k

21,Ω

≤ ϕ

2

(a(w, t))(1 + 1/̺

)(1 + kη

t

k

21,Ω

+ |w

ξ

|

∞,Ω

)

×

\

ηu

2nt

J

w

dξ + c(kF

t

k

20,Ω

+ kG

t

k

20,S

) + ϕ

1

(a(w, t))(εkwk

23,Ω

+ c(ε)kwk

20,Ω

)

× (ku

n

k

21,Ω

+ kF k

20,Ω

+ kGk

20,S

).

(9)

Integrating (3.18) with respect to t we get (3.19)

\

ηu

2nt

J

w

dξ + c

0

ku

nt

k

21,2,2,Ωt

≤ exp[ϕ

2

(a(w, t))(1 + 1/̺

)

× (t + kη

t

k

21,2,2,Ωt

+ a(w, t))]

× h

\

̺

0

u

2t

(0) dξ + kF

t

k

20,Ωt

+ kG

t

k

20,St

+ ϕ

1

(a(w, t)) sup

t

(ku

n

k

21,Ω

+ kF k

20,Ω

+ kGk

20,S

)

×

t

\

0

(εkwk

23,Ω

+ c(ε)kwk

20,Ω

) dt i . From (3.19) we have (3.16). This concludes the proof.

To estimate the expression sup

t

ku

n

k

21,Ω

on the r.h.s. of (3.16) we need the following result.

Lemma 3.4. Let the assumptions of Lemma 3.3 be satisfied. Then (3.20) ku

nt

k

20,Ωt

+ c

0

ku

n

k

21,Ω

≤ ψ

3

 t, 1/̺

, a(w, t),

t

\

0

1

kwk

23,Ω

+ c(ε

1

)kwk

20,Ω

) dt 

× h

ku

0

k

21,Ω

+

\

̺

0

v

20

dx + kF k

20,Ωt

+ c(ε

2

)kGk

20,St

+ ε

2

ku

nt

k

21,2,2,Ωt

i , where ψ

3

is an increasing positive function and ε

1

, ε

2

∈ (0, 1).

P r o o f. Multiplying (3.7) by ˙c

in

and summing over i from 1 to n we get (3.21)

\

ηu

2nt

J

w

dξ +

\

D

w

(u

n

) · D

w

(u

nt

)J

w

=

\

F · u

nt

J

w

dξ +

\

S

G · u

nt

J

w

S

. From (3.21) in view of the H¨older and Young inequalities we obtain (3.22)

\

ηu

2nt

J

w

dξ + d dt

\

|D

w

(u

n

)|

2

J

w

≤ c(|w

ξ

|

2∞,Ω

+ 1)

\

|D

w

(u

n

)|

2

J

w

dξ + c

\

|∇

ξ

u

n

|

2

J

w

+ εku

nt

k

21,Ω

+ c(ε)kGk

20,S

+ c(̺

)kF k

20,Ω

.

(10)

Integrating (3.22) with respect to time, using the Korn inequality and (3.8) we get

(3.23)

\

t

ηu

2nt

J

w

dξ + c

0

ku

n

k

21,Ω

≤ exp h c 

\t

0

1

kwk

23,Ω

+ c(ε

1

)kwk

20,Ω

) dt + t i

× [ku

0

k

21,Ω

+ ku

n

k

21,2,2,Ωt

+ ε

2

ku

nt

k

21,2,2,Ωt

+ c(ε

2

)kGk

20,St

+ c(̺

)kF k

20,Ωt

] + cku

n

k

20,Ω

. Using (3.9) in (3.23) yields

(3.24) ku

nt

k

20,Ωt

+ c

0

ku

n

k

21,Ω

≤ exp h c 

t\

0

1

kwk

23,Ω

+ c(ε

1

)kwk

20,Ω

) dt + t i

× h

ku

0

k

21,Ω

+ ψ

1



\

̺

0

v

02

dx + kF k

20,Ωt

+ kGk

20,St



+ ε

2

ku

nt

k

21,2,2,Ωt

+ c(ε

2

)kGk

20,St

+ c(̺

)kF k

20,Ωt

i .

From (3.24) we obtain (3.20). This concludes the proof.

Inserting the estimate for ku

n

k

21,Ω

from (3.20) into the r.h.s. of (3.16) and assuming that

ε

2

ψ

3

c

0

ψ

2

b(t, ε

1

, w) = c

0

2 , where

(3.25) b(t, ε

1

, w) =

t

\

0

1

kwk

23,Ω

+ c(ε

1

)kwk

20,Ω

) dt, we obtain

(3.26) ku

nt

k

20,Ω

+ c

0

2 ku

nt

k

21,2,2,Ωt

≤ ψ

2

h

\

̺

0

u

2t

(0) dξ + kF

t

k

20,Ωt

+ kG

t

k

20,St

+ sup

t

(kF k

20,Ω

+ kGk

20,S

)b(t

1

, ε

1

, w) i + ψ

2

ψ

3

µ b



ku

0

k

21,Ω

+

\

̺

0

v

02

dξ + kF k

20,Ωt

+ c

 ψ

2

ψ

3

b µ

2



kGk

20,St



.

Simplifying the expression we get

(11)

Lemma 3.5.

(3.27) ku

nt

k

20,Ω

+ c

0

ku

nt

k

21,2,2,Ωt

≤ ψ

4

(t, 1/̺

, a(w, t), b(t, ε

1

, w), kη

t

k

1,2,2,Ωt

)

× h

\

̺

0

v

02

dξ +

\

̺

0

u

2t

(0) dξ + ku

0

k

21,Ω

+ kF

t

k

20,Ωt

+ kG

t

k

20,St

+ kF k

20,Ωt

+ kGk

20,St

+ sup

t

(kF k

20,Ω

+ kGk

20,S

) i . From (3.27), (3.20) and (3.9) we get

Lemma 3.6. Let the assumptions of Lemmas 3.2–3.4 be satisfied. Then (3.28) ku

n

k

20,Ω

+ ku

n

k

21,Ω

+ ku

nt

k

20,Ω

+ ku

n

k

21,2,2,Ωt

+ ku

nt

k

21,2,2,Ωt

≤ ψ

5

(t, 1/̺

, a(w, t), b(t, ε

1

, w), kη

t

k

1,2,2,Ωt

)

× h

\

̺

0

v

02

dx +

\

̺

0

u

2t

(0) dx + ku

0

k

21,Ω

+ kF

t

k

20,Ωt

+ kG

t

k

20,Ωt

+ sup

t

(kF k

20,Ω

+ kGk

20,S

) i , where ψ

5

is an increasing positive function of its arguments.

Now choosing a subsequence and passing with n to infinity we get Lemma 3.7. Assume that ̺

≤ η ≤ ̺

, w ∈ L

2

(0, T ; H

3

(Ω)), η

t

∈ L

2

(0, T ; H

1

(Ω)), v

0

∈ H

1

(Ω), u

t

(0) ∈ L

2

(Ω), F

t

∈ L

2

(Ω

T

), G

t

∈ L

2

(S

T

), F ∈ L

(0, T ; L

2

(Ω)) and G ∈ L

(0, T ; L

2

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

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)), u

t

∈ L

2

(0, T ; H

1

(Ω)) ∩ L

(0, T ; L

2

(Ω)), and

(3.29) kuk

21,Ω

+ ku

t

k

20,Ω

+ kuk

21,2,2,Ωt

+ ku

t

k

21,2,2,Ωt

≤ ψ

5

(t, 1/̺

, a(w, t), b(t, ε

1

, w), kη

t

k

1,2,2,Ωt

)

× h

\

̺

0

v

20

dx +

\

̺

0

u

2t

(0) dx + kv

0

k

21,Ω

+ kF

t

k

20,Ωt

+ kG

t

k

20,St

+ sup

t

(kF k

20,Ω

+ kGk

20,S

) i

.

Having proved the existence of weak solutions to problem (3.1) expressed

by Lemma 3.7 we obtain by regularization techniques (see Appendix, The-

orem 4.1 and Remark 4.2) the following result:

(12)

Lemma 3.8. Let the assumptions of Lemma 3.7 be satisfied. Let v

0

∈ H

2

(Ω), w ∈ L

2

(0, T ; H

3

(Ω)), F ∈ L

2

(0, T ; H

1

(Ω)), G ∈ L

2

(0, T ; H

3/2

(S)), η ∈ L

(0, T ; H

2

(Ω)), and S ∈ H

5/2

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

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

3

(Ω)), u

t

∈ L

(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)), and

(3.30) ku

t

k

20,Ω

+ kuk

21,Ω

+ kuk

23,2,2,Ωt

+ ku

t

k

21,2,2,Ωt

≤ ψ

6

(t, 1/̺

, a(w, t), b(t, ε

1

, w), kη

t

k

1,2,2,Ωt

, kηk

2,2,∞,Ωt

)

× h

\

̺

0

v

20

dx +

\

̺

0

u

2t

(0) dx + kv

0

k

22,Ω

+ kF k

21,Ωt

+ kG

t

k

20,St

+ kGk

23/2,2,2,St

+ kF k

20,2,∞,Ωt

+ kGk

20,2,∞,St

i , where ψ

6

is an increasing positive function.

Now we prove the existence of solutions of (1.1) by the method of suc- cessive approximations determined by problems (1.13) and (1.14). First we show the boundedness of the sequence described by (1.13) and (1.14) in the norm defined in Lemma 3.8.

To simplify considerations let us introduce

(3.31) α

m

(t) = ku

mt

k

20,Ω

+ ku

m

k

21,Ω

+ ku

m

k

23,2,2,Ωt

+ ku

mt

k

21,2,2,Ωt

. Lemma 3.9. Assume that v

0

∈ H

2

(Ω), ̺

0

∈ H

2

(Ω), and there exist two positive constants ̺

and ̺

, ̺

< ̺

and ̺

≤ ̺

0

≤ ̺

.

T (v

0

, p(̺

0

))n = −p

0

n on S.

Then for A such that G(0, 0, F

0

) < A, α

m

(0) ≤ A, where F

0

= kv

0

k

22,Ω

+ ku

t

(0)k

20,Ω

+ k̺

0

k

22,Ω

and G is defined by the r.h.s. of (3.41), there exists T

such that for t ≤ T

,

(3.32) α

m

(t) ≤ A, m = 1, 2, . . . Moreover , in view of (1.1)

1

we have

ku

t

(0)k

0,Ω

≤ ckv

0

k

22,Ω

+ p

)

̺

0

k

1,Ω

+ µ + ν

̺

kv

0

k

2,Ω

+ ck̺

0

k

0,Ω

. P r o o f. First we obtain estimates for solutions of (1.14). Integrating (1.14) we get

(3.33) η

m

(ξ, t) = ̺

0

(ξ) exp h

t

\

0

div

um

u

m

dτ i

.

(13)

From (3.33) we have

(3.34)

sup

t

η

m

+ sup

t

1 η

m

≤ k̺

0

k

2,Ω

ϕ

1

(a(u

m

, t)), sup

t

m

k

2,Ω

≤ k̺

0

k

2,Ω

ϕ

1

(a(u

m

, t))ϕ

2

(a(u

m

, t)), where a(u

m

, t) = t

1/2

(

Tt

0

ku

m

k

23,Ω

dt)

1/2

. Moreover,

η

mt

= ̺

0

(ξ) exp h

t

\

0

div

um

u

m

dτ i

(− div

um

u

m

).

Therefore

(3.35) kη

mt

k

1,2,2,Ωt

≤ k̺

0

k

2,Ω

ϕ

3

(a(u

m

, t))b(t, ε, u

m

), where b(t, ε, u

m

) is defined by (3.25).

Comparing (3.1) with (1.13) we have

(3.36) F = −∇

um

q(η

m

) + η

m

U

um

m

), G = −(p(η

m

) − p

0

)n

um

. From (3.36) we have

(3.37) kF k

21,2,2,Ωt

+ kF k

20,2,∞,Ωt

≤ ϕ

4

(t, a(u

m

, t), k̺

0

k

2,Ω

) and

(3.38) kF

t

k

20,Ωt

≤ ϕ

5

(a(u

m

, t), sup

t

m

k

2,Ω

)b(t, ε, u

m

).

Moreover,

(3.39) kGk

23/2,2,2,St

+ kGk

20,2,∞,Ωt

≤ ϕ

6

(a(u

m

, t), t, sup

t

m

k

2,Ω

) and

(3.40) kG

t

k

20,St

≤ ϕ

7

(a(u

m

, t), sup

t

m

k

2,Ω

)b(t, ε, u

m

).

Using the fact that

a

2

(u

m

, t) ≤ tα

m

, b(t, ε, u

m

) ≤ t

a

α

m

+ cF

0

, a > 0, and inserting all the above estimates into (3.30) we get

(3.41) α

m+1

(t) ≤ G(t, t

a

α

m

(t), F

0

),

where a > 0, F

0

= kv

0

k

22,Ω

+ ku

t

(0)k

20,Ω

+ k̺

0

k

22,Ω

, and G is an increasing positive function.

Let A be such that G(0, 0, F

0

) < A. Since G is a continuous increasing

function of its arguments there exists T

> 0 such that for t ≤ T

we have

(3.42) G(t, t

a

A, F

0

) ≤ A.

(14)

From (3.42) we see that if α

m

(t) ≤ A then α

m+1

(t) ≤ A for t ≤ T

. Here A must be so large that α

m

(0) ≤ A.

To end the proof we have to construct the zero approximation function u

0

. We use the solution of the problem

u

0t

− div D(u

0

) = 0 in Ω

T

, n · D(u

0

) = (p(̺

0

) − p

0

)n on S

T

, u

0

|

t=0

= v

0

in Ω.

The existence of solutions to the above problem follows from the Galerkin method and can be proved in the classes determined by α

0

(t) < ∞. More- over, the compatibility condition is satisfied. Finally, A must be so large that α

0

(t) ≤ A, t ≤ T

. This concludes the proof.

Now we prove the convergence of the sequence {u

m

, η

m

}.

To show this we obtain from (1.13) and (1.14) the following system of problems for the differences U

m

= u

m

− u

m−1

and H

m

= η

m

− η

m−1

:

(3.43)

η

m

t

U

m+1

− div

um

D

u

m

U

m+1

= − H

m

t

u

m

− (div

um

D

u

m

(u

m

) − div

um−1

D

u

m−1

(u

m

))

− (∇

um

− ∇

um−1

)q(η

m

) − ∇

um−1

(q(η

m

) − q(η

m−1

)) + H

m

U

um

m

) + η

m−1

U

um

(H

m

)

+ η

m−1

(U

um

m−1

) − U

um−1

m−1

))

≡ X

7 i=1

F

i

≡ e F , D

u

m

(U

m+1

) · n

um

= − (D

um

(u

m

) · n

um

− D

um−1

(u

m

) · n

um−1

)

− q(η

m

)(n

um

− n

um−1

)

− (q(η

m

) − q(η

m−1

))n

um−1

+ p

0

(n

um

− n

um−1

)

≡ X

4 i=1

G

i

≡ e G, U

m+1

|

t=0

= 0,

and

(3.44) ∂

t

H

m

+ H

m

div

um

u

m

= −η

m−1

(div

um

u

m

− div

um−1

u

m−1

),

H

m

|

t=0

= 0.

(15)

Now we write the expressions on the r.h.s. of (3.43)

1

in qualitative forms:

(3.45)

F

2

= f

1 t

\

0

U

dτ u

mξξ

+ f

2 t

\

0

U

t

\

0

dτ u

mξξ

u

+ f

3 t

\

0

U

t

\

0

u

m−1,ξξ

dτ u

+ f

4 t

\

0

U

mξξ

dτ u

,

F

3

= f

5

f

1

t

\

0

U

dτ η

,

F

4

= f

6

f

2

+ η

m−1ξ

)H

m

+ f

7

f

3

H

, G

1

= f

8

t

\

0

U

dτ u

,

G

2

= f

9

f

4

t

\

0

U

dτ,

G

3

= p

0

f

10 t

\

0

U

dτ, G

4

= f

11

f

5

H

m

, where f

i

= f

i

(I+

Tt

0

u

dτ, I+

Tt

0

u

m−1,ξ

dτ ), i = 1, . . . , 11, f

j

= f

j

m

, η

m−1

), j = 1, . . . , 5, are C

functions of their arguments and I is the unit matrix.

Moreover, we have the estimates

(3.46) |f

i

| ≤ ϕ

1

(A), |f

j

| ≤ ϕ

2

(A)

where ϕ

1

, ϕ

2

are increasing positive functions, for i, j as above.

Therefore we have

Lemma 3.10. Let the assumptions of Lemma 3.9 be satisfied. Then there exists 0 < T

∗∗

sufficiently small such that

(3.47) kU

m+1

k

21,Ω

+ kU

m+1,t

k

20,Ωt

+ kU

m+1

k

22,2,2,Ωt

≤ δkU

m

k

22,2,2,Ωt

, where δ = δ(t) < 1 for t ≤ T

∗∗

.

P r o o f. To show (3.47) we multiply (3.43) by U

m+1

J

um

and integrate over Ω. Therefore after integration by parts we get

(3.48) 1 2

\

η

m

d

dt U

m+12

J

um

dξ +

\

|D

um

(U

m+1

)|

2

J

um

=

\

F U e

m+1

J

um

dξ +

\

S

GU e

m+1

J

um

S

.

(16)

First we estimate all terms on the r.h.s.:

\

t

H

m

u

mt

U

m+1

J

um

dξ dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)t sup

t

ku

mt

k

20,Ω

sup

t

kH

m

k

21,Ω

,

\

t

F

2

U

m+1

J

um

dξ dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)t

t

\

0

kU

m

k

22,Ω

dt,

\

t

F

3

U

m+1

J

um

dξ dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)t

t

\

0

kU

m

k

22,Ω

dt,

\

t

F

4

U

m+1

J

um

dξ dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)t sup

t

kH

m

k

21,Ω

,

\

t

(F

5

+ F

6

)U

m+1

J

um

dξ dt ≤ εkU

m+1

k

20,Ωt

+ c(ε)ϕ(A)t sup

t

kH

m

k

20,Ω

,

\

t

F

7

U

m+1

J

um

dξ dt ≤ εkU

m+1

k

20,Ωt

+ c(ε)ϕ(A)tkU

m

k

21,2,2,Ωt

. Next we estimate the boundary term (3.48):

\

St

(G

1

+ G

2

+ G

3

)U

m+1

J

um

s

dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)tkU

m

k

22,2,2,Ωt

,

\

St

G

4

U

m+1

J

um

s

dt ≤ εkU

m+1

k

21,2,2,Ωt

+ c(ε)ϕ(A)t sup

t

kH

m

k

21,Ω

. Using the Korn inequality in (3.48), integrating with respect to time, using the above estimates and taking ε sufficiently small we obtain

(3.49) kU

m+1

k

20,Ω

+kU

m+1

k

21,2,2,Ωt

≤ ϕ(A)t(kU

m

k

22,2,2,Ωt

+sup

t

kH

m

k

21,Ω

).

(17)

Multiplying (3.43)

1

by U

m+1,t

J

um

, integrating over Ω and by parts we have (3.50)

\

η

m

|U

m+1,t

|

2

J

um

dξ +

\

D

u

m

(U

m+1

) · D

um

(U

m+1,t

)J

um

=

\

S

GU e

m+1,t

J

um

dξ +

\

F · U e

m+1,t

J

um

dξ.

Continuing, we have (3.51)

\

η

m

|U

m+1,t

|

2

J

um

dξ +

\

D

u

m

(U

m+1

) · d dt D

u

m

(U

m+1

)J

um

\

D

u

m

(U

m+1

) · ∂

t

(D

um

)(U

m+1

)J

um

= d dt

\

S

G · U e

m+1

J

um

S

\

S

G e

t

· U

m+1

J

um

S

\

S

G · U e

m+1

J

um

div

um

u

m

S

+

\

F · U e

m+1,t

J

um

dξ.

In view of the H¨older and Young inequalities we get (3.52)

\

η

m

|U

m+1,t

|

2

J

um

dξ + d dt

\

|D

um

(U

m+1

)|

2

J

um

≤ d dt

\

S

G · U e

m+1

J

um

S

+ cku

m

k

23,Ω

·

\

|D

um

(U

m+1

)|

2

J

um

+ c

\

|∇

ξ

U

m+1

|

2

J

um

dξ + ck e F k

20,Ω

+ ε

1

(k e G

t

k

20,S

+ ϕ(A)ku

m

k

23,Ω

k e Gk

20,S

) + c(ε

1

)kU

m+1

k

21,Ω

,

where ε

1

∈ (0, 1).

Integrating with respect to time and using the Korn inequality we obtain from (3.52)

(3.53) kU

m+1,t

k

20,Ω

+ kU

m+1

k

21,Ω

≤ [ε

2

k e Gk

20,S

+ c(ε

2

)(ε

3

kU

m+1,ξ

k

20,Ω

+ c(ε

3

)kU

m+1

k

20,Ω

) + ckU

m+1

k

21,2,2,Ωt

+ ck e F k

20,Ωt

+ ε

1

ϕ(A)(k e G

t

k

20,St

+ sup

t

k e Gk

20,S

)]e

A

+ ckU

m+1

k

20,Ω

,

(18)

where we used the facts that

\

S

G · U e

m+1

J

um

dξ ≤ ε

2

k e Gk

20,S

+ c(ε

2

)kU

m+1

k

20,S

, kU

m+1

k

20,S

≤ ε

3

kU

m+1ξ

k

20,Ω

+ c(ε

3

)kU

m+1

k

20,Ω

,

t

\

0

ku

m

k

23,Ω

k e Gk

20,S

dt ≤ sup

t

k e Gk

20,S

t

\

0

ku

m

k

23,Ω

dt ≤ A sup

t

k e Gk

20,S

. Using (3.49) in (3.53) implies

(3.54) kU

m+1

k

21,Ω

+ kU

m+1,t

k

20,Ωt

+ kU

m+1

k

21,2,2,Ωt

≤ ϕ(A)[ε(k e G

t

k

20,St

+ sup

t

k e Gk

20,S

) + k e F k

20,Ωt

] + ϕ(A)t(kU

m

k

22,2,2,Ωt

+ sup

t

kH

m

k

21,Ω

).

Now from the regularity result for the parabolic problem η

m

U

m+1,t

− div

um

D

u

m

(U

m+1

) = e F in Ω

T

, D

um

(U

m+1

) · n

um

= e G on S

T

,

U

m+1

|

t=0

= 0 in Ω,

we obtain (see Theorem 4.1 and Remark 4.2)

(3.56) kU

m+1

k

22,2,2,Ωt

≤ c(k e F k

20,Ωt

+ k e Gk

21/2,2,2,St

) + ckU

m+1

k

20,Ωt

. Now collecting (3.49), (3.54) and (3.56) together, we get

(3.57) kU

m+1

k

21,Ω

+ kU

m+1,t

k

20,Ωt

+ kU

m+1

k

22,2,2,Ωt

≤ ϕ(A)[ε sup

t

k e Gk

20,S

+ εk e G

t

k

20,St

+ k e F k

20,Ωt

+ k e Gk

21/2,2,2,St

] + ϕ(A)t(kU

m

k

22,2,2,Ωt

+ sup

t

kH

m

k

21,Ω

).

Using the form of e F and e G we estimate the terms in the first bracket on the r.h.s. of (3.57):

(3.58)

k e Gk

20,S

≤ ϕ(A) h t

t

\

0

kU

m

k

22,Ω

dt + sup

t

kH

m

k

21,Ω

i ,

k e G

t

k

20,St

≤ ϕ(A) h

t\

0

kU

m

k

22,Ω

dt +

t

\

0

kH

mt

k

21,Ω

dt i ,

k e Gk

21/2,2,2,St

+ k e F k

20,Ωt

≤ ϕ(A)t h

t\

0

kU

m

k

22,Ω

dt + sup

t

kH

m

k

21,Ω

i

.

(19)

Using the equation (3.47) we have the estimate (3.59)

t

\

0

kH

mt

k

21,Ω

dt ≤ ϕ(A)(t sup

t

kH

m

k

21,Ω

+ kU

m

k

22,2,2,Ωt

).

From (3.57)–(3.59) it follows that

(3.60) kU

m+1

k

21,Ω

+ kU

m+1,t

k

20,Ωt

+ kU

m+1

k

22,2,2,Ωt

≤ ϕ(A)(c(ε)t + ε)(kU

m

k

22,2,2,Ωt

+ sup

t

kH

m

k

21,Ω

).

Integrating (3.44) we respect to time yields (3.61) H

m

(ξ, t)

= − exp h

t

\

0

div

um

u

m

dτ i

t\

0

h η

m−1

(div

um

u

m

− div

um−1

u

m−1

)

× exp 

t\

0

div

um

u

m

dt

′′

i dt

, hence one has

(3.62) sup

t

kH

m

k

21,Ω

≤ ϕ(A)tkU

m

k

22,2,2,Ωt

. Using (3.62) in (3.60) yields

(3.63) kU

m+1

k

21,Ω

+ kU

m+1,t

k

20,Ωt

+ kU

m+1

k

22,2,2,Ωt

≤ ϕ(A)(c(ε)t + ε)kU

m

k

22,2,2,Ωt

. Therefore for t so small that

(3.64) (c(ε)t + ε)ϕ(A) < 1

we have convergence of the sequence {u

m

, η

m

} to a solution. Assume that (3.64) holds for t ≤ T

∗∗

. This concludes the proof.

From Lemmas 3.9 and 3.10 we have

Theorem 3.11. Let the assumptions of Lemmas 3.9 and 3.10 be satisfied.

Then there exists T

∗∗

sufficiently small such that for T ≤ T

∗∗

there exists a solution to problem (1.1) such that

u ∈ L

(0, T ; H

1

(Ω)) ∩ L

2

(0, T ; H

3

(Ω)), u

t

∈ L

(0, T ; L

2

(Ω)) ∩ L

2

(0, T ; H

1

(Ω)) and

(3.65) kuk

1,2,∞,ΩT

+ kuk

3,2,2,ΩT

+ ku

t

k

0,2,∞,ΩT

+ ku

t

k

1,2,2,ΩT

≤ A,

where A is defined in Lemma 3.9. Moreover, η, 1/η ∈ L

(Ω

T

) ∩

L

(0, T ; H

2

(Ω)), η

t

, (1/η)

t

∈ L

(0, T ; L

2

(Ω))∩L

2

(0, T ; H

2

(Ω)), η

tt

, (1/η)

tt

(20)

∈ L

2

(Ω

T

), and

(3.66) kχk

2,2,∞,ΩT

+ kχ

t

k

0,2,∞,ΩT

+ kχ

t

k

2,2,2,ΩT

+ kχ

tt

k

0,ΩT

≤ ϕ(A), where χ replaces either η or 1/η and ϕ is some positive function.

P r o o f. We only have to show the last statement and the estimate (3.66).

They follow from the expression for η, (3.67) η(ξ, t) = ̺

0

(ξ) exp 

t

\

0

div

u

u(ξ, τ ) dτ  .

The most difficult part is to estimate η

tt

. Taking the second derivative of η with respect to time we obtain

η

tt

= ̺

0

exp 

t

\

0

div

u

u(ξ, τ ) dτ 

(−(div

u

u)

t

+ (div

u

u)

2

).

Since the first two factors are bounded we only consider the last bracket.

Qualitatively,

div

u

u = f

1



t\

0

u

ξ

dτ  u

2ξ

, where f

1

is a smooth function and f

1

(

Tt

0

u

ξ

dτ ) is bounded. Next in view of (3.65),

|u

2ξ

|

2,Ωt

≤ 

\t

0

|u

ξ

|

2∞,Ω

dt 

1/2

sup

t

|u

ξ

|

2,Ω

≤ A

2

. Similarly,

(div

u

u)

t

= f

2



t\

0

u

ξ

dτ 

u

ξt

+ f

3



t\

0

u

ξ

dτ  u

2ξ

,

where f

2

, f

3

are smooth functions and the same considerations as above can be applied. This concludes the proof.

4. Appendix. In this section we show the regularity of solutions to problem (1.12). First we consider the problem

(4.1)

ηu

t

− div D(u) = F in Ω

T

, n · D(u) = G on S

T

, u|

t=0

= v

0

in Ω.

We examine (4.1) using the following weak formulation:

(4.2)

\

ηu

t

ϕ dx +

\

D

(u) · D

(ϕ) dx =

\

F · ϕ dx +

\

S

Gϕ ds.

(21)

To examine regularity we only have to consider the integral

(4.3) K(ϕ) :=

\

D

(u) · D

(ϕ) dx.

Set ϕ = ζϕ

1

, where ζ is a smooth function with a support in e Ω ⊂ Ω and ϕ

1

is a test function. Then we get

(4.4) K(ζϕ

1

) =

\

Ωe

[D

(u) · D

(ζ)ϕ

1

+ D

(u

) · D

1

) − uD

(ζ) · D

1

)] dx,

where u

= uζ.

In further considerations we choose e Ω such that e Ω ∩ S 6= ∅. Therefore we apply the transformation Φ : e Ω → b Ω which straightens locally the boundary of Ω. Hence (4.4) takes the form

(4.5) K(b ζ e ϕ

1

) =

\

Ωb

[D

Φ

(b u)·D

Φ

(b ζ) b ϕ

1

+D

Φ

(e u)·D

Φ

( b ϕ

1

)−b uD

Φ

(b ζ)·D

Φ

( b ϕ

1

)]J

Φ

dz,

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

−1

, e u = b ub ζ, D

Φ

is such that ∇

x

in D are replaced by ∇

x

Φ(x)|

x=Φ−1(z)

· ∇

z

and J

Φ

is the Jacobian determinant of the transformation z = Φ(x). We also need the fact that b Ω = {z ∈ R

3

:

|z

i

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

3

< d}, b S = Φ( e S) = {z ∈ R

3

: |z

i

| < d, i = 1, 2, z

3

= 0}, and e S = e Ω ∩ S. Since the integrand in (4.5) vanishes on ∂ b Ω \ b S it can be extended by zero on R

3+

= {z ∈ R

3

: z

3

> 0}. Therefore, we assume b ϕ

1

= δ

h−1

δ

h

e u, where δ

h

u(z) =

1h

(u(z

+ h, z

3

) − u(z)), z

= (z

1

, z

2

), corresponds only to the tangent directions, which will also be denoted by τ . Then from (4.5) under the assumption that S and hence Φ are smooth, we have the estimate

(4.6) K(b ζδ

h−1

δ

h

u) ≥ e c

0

2 kδ

h

uk e

21, b

− εkD

Φ

h

b u)k

20, b

− ckb uk

21, b

, where ε ∈ (0, 1).

Now we consider the first term in (4.2):

\

ηu

t

ϕ dx =

\

Ωe

ηu

t

ϕ

1

dx =

\

Ωb

b

ηe u

t

ϕ b

1

J

Φ

dz (4.7)

=

\

Ωb

δ

h

ηJ b

Φ

u e

t

δ

h

u dz + b

\

Ωb

b

ηδ

h

J

Φ

e u

t

δ

h

b u dz +

\

Ωb

b

ηδ

h

u e

t

δ

h

uJ b

Φ

dz,

where the last term is equal to 1

2

\

Ωb

b η d

dt |δ

h

e u|

2

J

Φ

dz = 1 2

d dt

\

Ωb

b

η|δ

h

e u|

2

J

Φ

dz − 1 2

\

Ωb

b

η

t

h

e u|

2

J

Φ

dz,

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