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T. K O B A Y A S H I (Tsukuba)

W. M. Z A J A ¸ C Z K O W S K I (Warszawa)

ON GLOBAL MOTION OF A COMPRESSIBLE BAROTROPIC VISCOUS FLUID WITH

BOUNDARY SLIP CONDITION

Abstract. Global-in-time existence of solutions for equations of viscous compressible barotropic fluid in a bounded domain Ω ⊂ R

3

with the bound- ary slip condition is proved. The solution is close to an equilibrium solution.

The proof is based on the energy method. Moreover, in the L

2

-approach the result is sharp (the regularity of the solution cannot be decreased) be- cause the velocity belongs to H

2+α,1+α/2

(Ω × R

+

) and the density belongs to H

1+α,1/2+α/2

(Ω × R

+

), α ∈ (1/2, 1).

1. Introduction. In this paper we prove the existence of global-in-time solutions to the following problem (see [5, 10]):

(1.1)

̺(v

t

+ v · ∇v) − div T(v, p) = ̺f in Ω

T

= Ω × (0, T ),

̺

t

+ div(̺v) = 0 in Ω

T

,

̺|

t=0

= ̺

0

, v|

t=0

= v

0

in Ω,

n · T(v, p) · τ

α

+ γv · τ

α

= 0, α = 1, 2, on S

T

= S × (0, T ),

v · n = 0 on S

T

,

where v = v(x, t) is the velocity of the fluid, ̺ = ̺(x, t) the density, f = f (x, t) the external force field, n the unit outward vector normal to S, τ

α

, α = 1, 2, tangent vectors to S, 0 < γ the constant slip coefficient, and S = ∂Ω, where Ω ⊂ R

3

is a bounded domain. By T(v, p) we denote the stress tensor of the form

(1.2) T (v, p) = {µ(∂

xi

v

j

+ ∂

xj

v

i

) + (ν − µ)δ

ij

div v − δ

ij

p}

i,j=1,2,3

,

1991 Mathematics Subject Classification: 76N10, 35B35, 35G30.

Key words and phrases: global existence, compressible barotropic viscous fluid, bound- ary slip condition, Hilbert–Besov spaces.

[159]

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where µ, ν are constant viscosity coefficients satisfying the thermodynamic restrictions

(1.3) ν >

13

µ > 0.

Moreover, we consider the barotropic motion,

(1.4) p = p(̺).

From (1.1)

2,5

it follows that the total mass of the fluid in Ω is conserved, (1.5)

\

̺ dx = M =

\

̺

0

dx.

In this paper we prove existence of global-in-time solutions which are close to the equilibrium solution

(1.6) v

e

= 0, ̺

e

= M

|Ω| , where |Ω| = vol Ω.

The proof is based on a local existence result from [3] and on the prolon- gation technique from [12]. To recall the result from [3] we have to introduce the Lagrangian coordinates which are initial data to the Cauchy problem

(1.7) ∂x

∂t = v(x, t), x|

t=0

= ξ ∈ Ω.

Integrating (1.7) we obtain a relation between the Eulerian x and the La- grangian ξ coordinates:

(1.8) x = ξ +

t

\

0

u(ξ, τ ) dτ ≡ x

u

(ξ, t) ≡ x(ξ, t),

where u(ξ, t) = v(x(ξ, t), t). Moreover, we introduce η(ξ, t) = ̺(x(ξ, t), t), q(ξ, t) = p(η(ξ, t)), and g(ξ, t) = f (x(ξ, t), t).

Theorem 1.1. Assume that g ∈ H

α,α/2

(Ω

T

), v

0

∈ H

1+α

(Ω), ̺

0

∈ H

1+α

(Ω), 1/̺

0

∈ L

(Ω), S ∈ C

2+α

, α ∈ (1/2, 1). Then there exists a time T > 0 such that for t ≤ T there exists a solution (u, η) to problem (1.1) with u ∈ H

2+α,1+α/2

(Ω

t

), η ∈ H

1+α,1/2+α/2

(Ω

t

), and the following estimate holds:

kuk

2+α,Ωt

+ kηk

1+α,Ωt

≤ ϕ

1

(T, kSk

C2+α

, γ(T ))[kv

0

k

1+α,Ω

(1.9)

+ k̺

0

k

1+α,Ω

+ kgk

α,Ωt

], t ≤ T,

|1/η|

∞,Ωt

≤ ϕ

2

(T, kSk

C2+α

, γ(T ))|1/̺

0

|

∞,Ω

,

where γ(t) = kv

0

k

1+α,Ω

+ k̺

0

k

1+α,Ω

+ kgk

α,Ωt

and ϕ

1

, ϕ

2

are increasing positive functions of their arguments.

To prove global existence we have to control the variation of the solu-

tion in a neighbourhood of the equilibrium solution. For this purpose we

(3)

introduce

(1.10) ̺

σ

= ̺ − ̺

e

, p

σ

= p − p

e

, where p

e

= p(̺

e

). Then (1.1) implies

(1.11)

̺(v

t

+ v · ∇v) − div T(v, p

σ

) = ̺f in Ω

T

,

̺

σt

+ v · ∇̺

σ

+ ̺ div v = 0 in Ω

T

, n · D(v) · τ

α

+ γv · τ

α

= 0, α = 1, 2, on S

T

,

v · n = 0 on S

T

,

v|

t=0

= v

0

, ̺

σ

|

t=0

= ̺

σ0

≡ ̺

0

− ̺

e

in Ω, where

D (v) = {µ(v

ixj

+ v

jxi

) + (ν − µ)δ

ij

div v}

i,j=1,2,3

.

Sometimes instead of (1.11)

2

we use the following equation for the pressure:

(1.12) p

σt

+ v · ∇p

σ

+ p

̺

̺ div v = 0.

Applying Theorem 1.1 to problem (1.11) yields

Theorem 1.2. Let the assumptions of Theorem 1.1 hold. Let ̺

σ0

∈ H

1+α

(Ω). Then for solutions of the problem (1.11) we have the estimate

kuk

2+α,Ωt

+ kη

σ

k

1+α,Ωt

≤ ϕ

3

(T, kSk

C2+α

, γ(T ))[kv

0

k

1+α,Ω

(1.13)

+ k̺

σ0

k

1+α,Ω

+ kgk

α,Ωt

], t ≤ T, where ϕ

3

is an increasing positive function of its arguments and γ(T ) is defined in Theorem 1.1.

Since the local existence of solutions to problem (1.1) was proved in spaces such that u ∈ H

2+α,1+α/2

(Ω

T

), η ∈ H

1+α,1/2+α/2

(Ω

T

), we have to formulate problem (1.11) and (1.12) in Lagrangian coordinates. Hence we have

(1.14)

ηu

t

− div

u

T

u

(u

,

q

σ

) = ηg in Ω

T

, q

σt

+ p

η

η div

u

u = 0 in Ω

T

, n

u

· D

u

(u) · τ

+ γu · τ

= 0, α = 1, 2, on S

T

,

u · n

u

= 0 on S

T

,

u|

t=0

= v

0

, q

σ

|

t=0

= q

σ0

≡ q

0

− p

e

in Ω, where q

σ

= q − p

e

, and

(1.15) η

t

+ η div

u

u = 0 in Ω

T

, η|

t=0

= v

0

in Ω,

where ∇

u

= ξ

ix

ξi

, T

u

(u, q

σ

) = −q

σ

I + D

u

(u), I is the unit matrix and

the operators D

u

, div

u

are obtained from D and div by replacing ∇ by ∇

u

.

Finally n

u

(ξ, t) = n(x(ξ, t)) and τ

u,α

(ξ, t) = τ

α

(x(ξ, t)), α = 1, 2.

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Moreover, in view of the local solvability the Jacobi matrix with de- terminant J

x(ξ,t)

of the transformation x = x(ξ, t), with elements a

ij

= δ

ij

+

Tt

0

ξj

u

i

(ξ, τ ) dτ , is invertible.

We have to underline that the use of the Lagrangian coordinates is not natural for problem (1.1) because the domain Ω is fixed. However in [3]

and also in this paper the Lagrangian coordinates have to be used because otherwise the nonlinear term v · ∇̺ in the continuity equation cannot be estimated in the proofs for v and ̺ such that v ∈ H

2+α,1+α/2

(Ω

T

), ̺ ∈ H

1+α,1/2+α/2

(Ω

T

), α ∈ (1/2, 1).

Moreover, we have not been able to show the invariance of the norm of H

2+α,1+α/2

(Ω

T

), α ∈ (1/2, 1), under the transformation (1.8), where u ∈ H

2+α,1+α/2

(Ω

T

). More precisely we examine the following problem. Let the transformation (1.8) be generated by w ∈ H

2+α,1+α/2

(Ω

T

), α ∈ (1/2, 1).

Then we have x = x

w

(ξ, t). Let v = v(x, t) and u(ξ, t) = v(x

w

(ξ, t), t). We want both u, v to belong to H

2+α,1+α/2

(Ω

T

). We can easily show that (1.16) c

1

kvk

L2(0,T ;H2+α(Ω))

≤ kuk

L2(0,T ;H2+α(Ω))

≤ c

2

kvk

L2(0,T ;H2+α(Ω))

, where c

1

, c

2

are positive, nonvanishing and depend on kwk

2+α,ΩT

.

However, we are not able to show (1.16) for v and u in H

2+α,1+α/2

(Ω

T

) under the assumption that c

1

, c

2

depend on kwk

2+α,ΩT

only. We can only show this if w is more regular but this is not appropriate for us.

Now we formulate the main result.

Theorem 1.3. Assume that the equilibrium solution is defined by (1.6) and the total mass of the fluid is determined by (1.5). Assume the thermo- dynamic restrictions (1.3) and that the barotropic motion with state equa- tion (1.4) is considered. Assume that the bounded domain Ω is not ro- tationally symmetric and S ∈ C

2+α

, ̺

0

, v

0

∈ H

1+α

(Ω), 1/̺

0

∈ L

(Ω), g ∈ H

α,α/2

(Ω

T

), p ∈ C

2

, α ∈ (1/2, 1), g = g(ξ, t) = f (x

u

(ξ, t), t) and x = x

u

(ξ, t) defined by (1.8) determines the transformation between the Eu- lerian and Lagrangian coordinates. Assume that kv

0

k

1+α,Ω

, k̺

0

− ̺

e

k

1+α,Ω

, kgk

α,Ωt

, t ∈ R

+

are sufficiently small , where ̺

e

= M/|Ω|. Then there exists a global solution to problem (1.1) such that

u ∈ H

2+α,1+α/2

(Ω

t

), η ∈ H

1+α,1/2+α/2

(Ω

t

),

t ∈ R

+

, where u(ξ, t) = v(x

u

(ξ, t), t), η(ξ, t) = ̺(x

u

(ξ, t), t) and u deter- mines the transformation (1.8).

Finally, we want to add some historical remarks and comments concern-

ing global-in-time existence for viscous compressible fluids. The first results

concerning global-in-time existence for the Cauchy problem for equations

of viscous compressible and heat conducting fluids were obtained by Mat-

sumura and Nishida [6, 7]. The corresponding result for an initial boundary

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value problem for the same equations was also shown by Matsumura and Nishida [8, 9]. They considered only a half space and they assumed that the initial density, velocity and temperature are from H

3

and the external force has a gradient form. They assumed the Dirichlet boundary conditions for velocity and the Neumann boundary conditions for temperature. Valli [15] improved the results for the barotropic case showing global-in-time exis- tence for an arbitrary bounded domain Ω ⊂ R

3

, general small external force field and for initial density and velocity from H

2

(Ω). The result of [15] was extended to general viscous compressible heat-conducting fluid equations by Valli–Zaj¸aczkowski [16]. The global-in-time existence results from [6–9, 15, 16] follow from some a priori estimate whose proof depends heavily on the L

2

-approach. However, the results are not sharp in the L

2

-framework, where sharpness means that the existence result cannot be proved with less regular data. In the L

p

-framework global-in-time existence of solutions for equations of viscous compressible heat-conducting fluids with Dirichlet boundary conditions was proved by Str¨ ohmer in [13, 14]. However his results are not sharp because he imposed too much regularity on data functions.

Our result is sharp for the L

2

-approach and this is the reason why the fractional derivative spaces have been used. However this forced us to use the Lagrangian coordinates which are not appropriate for problems in fixed domains. Unfortunately in general our result is not sharp because then the L

p

-approach is necessary.

Our result can be extended to the Dirichlet and Neumann problems and also to equations of viscous compressible heat-conducting fluids.

2. Notation and auxiliary results. First we introduce a partition of unity ({ e Ω

i

}, {ζ

i

}), Ω = S

i

Ω e

i

. Let e Ω be one of the e Ω

i

’s and ζ(ξ) = ζ

i

(ξ) the corresponding function. If e Ω is an interior subdomain, then let e ω be such that e ω ⊂ e Ω and ζ(ξ) = 1 for ξ ∈ e ω. Otherwise we assume that e Ω ∩ S 6= ∅, e

ω ∩ S 6= ∅, e ω ⊂ e Ω. Let β ∈ e ω ∩ S ⊂ e Ω ∩ S, e S ≡ S ∩ e Ω. Introduce local coordinates {y} connected with {ξ} by

(2.1) y

k

= α

kl

l

− β

l

), α

3k

= n

k

(β), k = 1, 2, 3,

where α

kl

is a constant orthogonal matrix, such that e S is determined by y

3

= F (y

1

, y

2

), F ∈ H

3/2+α

and

Ω = {y : |y e

i

| < d, i = 1, 2, F (y

) < y

3

< F (y

) + d, y

= (y

1

, y

2

)}.

Next we introduce functions u

and q

by

(2.2) u

i

(y) = α

ij

u

j

(ξ)|

ξ=ξ(y)

, q

(y) = q(ξ)|

ξ=ξ(y)

,

where ξ = ξ(y) is the inverse transformation to (2.1). Further we introduce

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new variables by

(2.3) z

i

= y

i

, i = 1, 2, z

3

= y

3

− e F (y), y ∈ e Ω,

which will be denoted by z = Φ(y), where e F is an extension of F to e Ω with F ∈ H e

2+α

( e Ω). Let b Ω = Φ( e Ω) = {z : |z

i

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

3

< d} and S = Φ( e b S). Define

(2.4) u(z) = u b

(y)|

y=Φ−1(z)

, q(z) = q b

(y)|

y=Φ−1(z)

.

Introduce b ∇

k

= ξ

lxk

z

l

zi

|

ξ=χ−1(z)

, where χ(ξ) = Φ(ψ

0

(ξ)) and y = ψ

0

(ξ) is described by (2.1). Introduce also the notation

e

u(ξ) = u(ξ)ζ(ξ), q e

σ

(ξ) = q

σ

(ξ)ζ(ξ), ξ ∈ e Ω, e Ω ∩ S = ∅, (2.5)

e

u(z) = b u(z)b ζ(z), q e

σ

(z) = b q

σ

(z)b ζ(z), z ∈ b Ω = Φ( e Ω), e Ω ∩ S 6= ∅, (2.6)

where b ζ(z) = ζ(ξ)|

ξ=χ−1(z)

.

Under the above notation problem (1.14) has the following form in an interior subdomain:

(2.7)

ηe u

it

− ∇

j

T

ij

(e u, e q

σ

) = ηeg

i

− ∇

uj

B

uij

(u, ζ) − T

uij

(u, q

σ

)∇

uj

ζ

− (∇

j

T

ij

(e u, e q

σ

) − ∇

uj

T

uij

(e u, e q

σ

)) ≡ ηeg

i

+ k

1i

, i = 1, 2, 3, e

q

σt

+ p

1

∇ · e u = p

1

u · ∇

u

ζ + (p

1

− Ψ (η))∇

u

· uζ + p

1

(∇ · e u − ∇

u

· e u) ≡ k

2

,

where Ψ (η) = p

η

(η)η, p

1

= Ψ (̺

e

), and in a boundary subdomain:

(2.8) b

η e u

it

− ∇

j

T

ij

(e u, e q

σ

) = b η eg

i

− b ∇

j

B b

ij

(b u, b ζ) − b T

ij

(b u, b q

σ

) b ∇

j

ζ b

− (∇

j

T

ij

(e u, e q

σ

) − b ∇

j

T b

ij

(e u, e q

σ

)) ≡ b η eg

i

+ k

3i

, i = 1, 2, 3, e

q

σt

+ p

1

∇ · e u = p

1

b u · b ∇b ζ + (p

1

− Ψ (b η)) b ∇ · e ub ζ + p

1

(∇ · e u − b ∇ · e u) ≡ k

4

, b

n · b D (e u) · b τ

α

= b n · b B (b u, b ζ) · b τ

α

− γb u · b τ

α

≡ k

, α = 1, 2, b

n · e u = 0,

where the summation convention is assumed,

(2.9) B

uij

(u, ζ) = µ(u

i

uj

ζ + u

j

ui

ζ) + (ν − µ)δ

ij

u · ∇

u

ζ,

and b T , b B indicate that the operator ∇

u

is replaced by b ∇. Finally the dot · denotes the scalar product in R

3

.

Let ψ(y) ≡ y

3

− e F (y) = 0. Then b n = ∇ψ/|∇ψ|, τ

1

= (0, 1, 0) × b n,

τ

2

= (1, 0, 0) × b n. Moreover, in the next considerations we denote z

1

, z

2

by τ , and z

3

by n. From [11] we recall the Korn inequality. For vectors

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u, v ∈ H

1

(Ω) we introduce (2.10) E(u, v) =

\

(∂

xi

u

j

+ ∂

xj

u

i

)(∂

xi

v

j

+ ∂

xj

v

i

) dx.

The vectors which satisfy E(u, u) = 0 have the form

(2.11) u = A + B × x,

where A, B are arbitrary constant vectors.

We define e H(Ω) = {u : E(u, u) < ∞, u · n = 0 on S}. If Ω is a domain obtained by rotation about the vector B we set H(Ω) = {u ∈ e H(Ω) :

T

u · B × x dx = 0}. Otherwise we set H(Ω) = e H(Ω).

Lemma 2.1 (see [11]). Let S ∈ C

2

. Then for u ∈ H(Ω), (2.12) kuk

21,Ω

≤ c

0

E(u, u).

Finally, we introduce the notation. By H

k+α,k/2+α/2

(Ω

T

), k ∈ N ∪ {0}, α ∈ (1/2, 1), we denote the Hilbert space with the norm

kuk

2Hk+α,k/2+α/2(ΩT)

= X

|β|+2i≤k

k∂

xβ

ti

uk

2L

2(ΩT)

+ X

|β|=k T

\

0

\

\

|∂

xβ

u(x, t) − ∂

xβ

u(x

, t)|

2

|x − x

|

3+2α

dx dx

dt

+

\

Ω T

\

0 T

\

0

|∂

t[k/2]

u(x, t) − ∂

t[k/2]

u(x, t

)|

2

|t − t

|

1+α+k−2[k/2]

dx dt dt

. Similarly we introduce the spaces H

k+α

(Ω), H

k+α,k/2+α/2

(S

T

). Moreover, we use the notation

kuk

Hk+α,k/2+α/2(ΩT)

= kuk

k+α,ΩT

,

kuk

Hk+α(Ω)

= kuk

k+α,Ω

, kuk

Hk+α(0,T )

= kuk

k+α,(0,T )

, kuk

Lp(Ω)

= |u|

p,Ω

, p ∈ [1, ∞],

kuk

Hs(ΩT)

= kuk

s,ΩT

, s ∈ N ∪ {0}.

The spaces H

k+α,k/2+α/2

(Ω

T

) are Sobolev–Slobodetski˘ı spaces and are de- noted also by W

k+α,k/2+α/2

2

(Ω

T

). They are particular cases of Besov spaces.

First we recall some properties of isotropic Besov spaces which are fre- quently used in this paper; next we define anisotropic Besov spaces and formulate some imbedding theorems which we need.

Let us introduce the differences

i

(h)u(x) = u(x + he

i

) − u(x),

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where x ∈ R

n

and e

i

, i = 1, . . . , n, are the standard unit vectors. Then we define inductively the m-difference

mi

(h)u(x) = ∆

i

(h)(∆

m−1i

(h)u(x)) = X

m j=0

(−1)

m−j

c

jm

u(x + jhe

i

),

where c

jm

=

mj



=

j!(m−j)!m!

. Moreover, we introduce the difference

∆(y)f (x) = f (x + y) − f (x), x, y ∈ R

n

, and inductively

m

(y)f (x) = ∆(y)(∆

m−1

(y)f (x)).

Since

∆(x − y)f (y) = f (x) − f (y) we have

m

(x − y)f (y) = X

n

i=1

m

((x − y) · e

i

)f (y) = X

n i=1

mi

(h)f (y), where the last equality holds for (x − y) · e

i

= h.

We define the isotropic Besov spaces by introducing the norm (see [2, Sect. 18])

(2.13) kuk

Bpl(Rn)

= kuk

Lp(Rn)

+ X

n i=1



h\0

0

dh

\

Rn

dx |∆

mi

(h)∂

xki

u|

p

h

1+(l−k)p



1/p

, where m > l − k, m, k ∈ N ∪ {0}, l ∈ R

+

and l 6∈ Z.

It was shown in [4] that the Besov spaces defined by (2.13) all coincide and have equivalent norms for all m, k satisfying m > l − k.

Next we define the L

p

-scale of Sobolev–Slobodetski˘ı spaces by introduc- ing the norm

(2.14) kuk

Wpl(Rn)

= kuk

Lp(Rn)

+ X

n i=1



h\0

0

dh

\

Rn

dx |∆

i

(h)∂

x[l]i

u|

p

h

1+p(l−[l])



1/p

, where l 6∈ Z, [l] is the integral part of l. We frequently write l = k + α, k ∈ N ∪ {0}, α ∈ (0, 1), so k = [l] and α = l − [l].

By the Golovkin theorem [4] the norms of the spaces B

pl

(R

n

) and W

pl

(R

n

) are equivalent.

We also define the norms (2.15) kuk

Bel

p(Rn)

= kuk

Lp(Rn)

+



\

Rn

dx

\

Rn

dy |∆

m

(x − y)∂

yk

u(y)|

p

|x − y|

n+p(l−k)



1/p

(9)

for any m > l − k (here ∂

yk

u = P

|α|=k

D

yα

u) and (2.16) kuk

Wfl

p(Rn)

= kuk

Lp(Rn)

+



\

Rn

dx

\

Rn

dy |∆(x − y)∂

y[l]

u(y)|

p

|x − y|

n+p(l−[l])



1/p

.

Using Lemma 7.44 of [1] we can show that the spaces e B

pl

(R

n

) and B

pl

(R

n

) (with h

0

= ∞) and f W

pl

(R

n

) and W

pl

(R

n

) (with h

0

= ∞) all coincide and have equivalent norms.

Therefore all norms (2.13)–(2.16) are equivalent. Now we recall some imbedding theorems for Besov spaces (see [2, Sect. 18]):

(2.17) D

σ

B

pl

(R

n

) ⊂ B

̺q

(R

n

) for n/p − n/q + σ + ̺ ≤ l.

For κ = 1 −

1l np

nq

+ σ + ̺) > 0 we have the interpolation inequality (2.18) k∂

σx

uk

B̺q(Rn)

≤ ε

1−κ

kuk

Bpl(Rn)

+ cε

−κ

kuk

Lp0(Rn)

,

with p

0

≥ 1, ε ∈ (0, 1). In the above notation B

pl

(R

n

) with l ∈ Z

+

is a Sobolev space and L

p

(R

n

) = W

p0

(R

n

).

All the above remarks can be applied to spaces of functions defined on a bounded domain Ω ⊂ R

n

. Moreover by using a partition of unity we can define spaces of traces on the boundary of Ω and formulate the corresponding trace theorems.

Now we introduce some anisotropic Sobolev–Slobodetski˘ı and Besov spaces which are necessary for our considerations. First we define

kuk

Lp1,p2(ΩT)

= 

T\

0

dt 

\

dx |u(x, t)|

p1



p2/p1



1/p2

and

kuk

Lp1,p2(ΩT)

= 

\

dx 

T\

0

dt |u(x, t)|

p1



p2/p1



1/p2

, where p

i

∈ [1, ∞], i = 1, 2.

We need imbedding theorems between the above spaces and W

2l,l/2

(Ω

T

), l ∈ R

+

, Ω ⊂ R

n

. We apply the results from [2, Sect. 18]. Since W

2l,l/2

(Ω

T

) is isotropic with respect to the power of integration we have, for 2 ≤ p

1

, p

2

< ∞,

(2.19) ∂

xα1

tα2

W

2l,l/2

(Ω

T

) ⊂ L

p1,p2

(Ω

T

) if n 2 − n

p

1

+ 2 2 − 2

p

2

+ α

1

+ 2α

2

≤ l, (2.20) ∂

xα1

tα2

W

2l,l/2

(Ω

T

) ⊂ L

p1,p2

(Ω

T

) if n

2 − n p

2

+ 2 2 − 2

p

1

+ α

1

+ 2α

2

≤ l,

(10)

(2.21) ∂

xα1

tα2

W

2l,l/2

(Ω

T

) ⊂ L

p

(0, T ; B

qσ

(Ω)) if n

2 − n q + 2

2 − 2

p + α

1

+ 2α

2

+ σ ≤ l, and finally

(2.22) ∂

xα1

αt2

W

2l,l/2

(Ω

T

) ⊂ L

p

(Ω; B

qσ

(0, T )) if n

2 − n p + 2

2 − 2

q + α

1

+ 2α

2

+ 2σ ≤ l, 2 ≤ p, q < ∞.

Moreover, the corresponding interpolation inequalities hold.

3. Inequality for global existence. First we obtain an energy type inequality:

Lemma 3.1. Assume that (v, ̺

σ

) is the solution to problem (1.11) deter- mined by Theorem 1.2. Let the assumptions of Lemma 2.1 hold. Then (3.1) d

dt

\

 1

2 ̺v

2

+ p

2σ

2p

̺

̺



dx + c

0

kvk

21,Ω

+ γ X

2 α=1

\

S

|v · τ

α

|

2

dS

≤ ψ

1

(|̺|

∞,ΩT

, |1/̺|

∞,ΩT

)kp

σ

k

41,Ω

+

\

̺f · v dx, where ψ

1

is an increasing positive function.

P r o o f. Multiplying (1.11)

1

by v, integrating over Ω, using the boundary conditions and the continuity equation we obtain

(3.2) d dt

1 2

\

̺v

2

dx +

\

D (v) · ∇v dx

\

p

σ

div v dx + γ X

2 α=1

\

S

|v · τ

α

|

2

dS =

\

̺f · v dx.

In view of the Korn inequality (see Lemma 2.1), using (1.12) in the third term on the l.h.s. of (3.2) yields

(3.3) d dt

1 2

\

̺v

2

dx + c

0

kvk

21,Ω

+

\

p

σ

p

̺

̺ (p

σt

+ v · ∇p

σ

) dx + γ

X

2 α=1

\

S

|v · τ

α

|

2

dS ≤

\

̺f · v dx,

where c

0

is the constant from the Korn inequality.

(11)

Continuing calculations in the third term on the l.h.s. of (3.3) we get

\

1

2p

̺

̺ (p

2σ,t

+ v · ∇p

2σ

) dx =

\



̺

 p

2σ

2p

̺

̺

2



t

+ ̺v · ∇

 p

2σ

2p

̺

̺

2



dx

\

1 2 ̺p

2σ

 1 p

̺

̺

2



p

(p

σt

+ v · ∇p

σ

) dx

= d dt

\

p

2σ

2p

̺

̺ dx + N

1

, where

N

1

= 1 2

\

p

̺

̺

2

 1 p

̺

̺

2



p

p

2σ

div v dx, so

|N

1

| ≤ εk div vk

20,Ω

+ ϕ

1

(1/ε, |̺|

∞,ΩT

, |1/̺|

∞,ΩT

)kp

σ

k

41,Ω

,

where ϕ

1

is an increasing positive function. Using the above considerations in (3.3) we obtain (3.1). This concludes the proof.

Now we show

Lemma 3.2. For the local solution determined by Theorems 1.1, 1.2 we have

ϕ(t, Ω) + kuk

22+α,Ωt

+ kq

σ

k

21+α,Ωt

≤ c(kuk

20,Ωt

+ kq

σ

k

20,Ωt

) (3.4)

+ ckgk

2α,Ωt

+ cX

12

+ ϕ(0, Ω), where t ≤ T , T is the time of local existence, and ϕ(t, Ω) and X

1

are defined by (3.62) and (3.64), respectively.

P r o o f. First we obtain the inequality for interior subdomains. Assume that ξ

0

∈ e Ω, where e Ω is an interior subdomain, and ζ ∈ C

0

(R

3

) is the corresponding function from the partition of unity such that ζ(ξ) = 1 for ξ ∈ B

λ

0

) = {ξ ∈ R

3

: |ξ − ξ

0

| < λ} and ζ(ξ) = 0 for ξ ∈ R

3

\ B

0

).

Denote by ∆

s

(z)f (x) the sth finite difference of f such that

s

(z)f (x) = X

s k=0

c

ks

(−1)

s−k

f (x + kz), where c

ks

=

sk



=

k!(s−k)!s!

.

In this proof, functions ϕ

j

, ψ

j

, j ∈ N ∪ {0}, are increasing continuous

positive functions of their arguments. Applying ∆

s

(z) to (2.7)

1

, multiplying

the result by ∆

s

(z)e u

i

, integrating over B

0

) and integrating with respect

to z over R

3

with the weight 1/|z|

3+2(1+α)

we obtain

(12)

(3.5)

\

R3

dz

\

B0)

dξ η∆

s

(z)e u

it

s

(z)e u

i

|z|

3+2(1+α)

\

R3

dz

\

B0)

dξ ∇

ξj

(∆

s

(z)T

ij

(e u, e q

σ

))∆

s

(z)e u

i

|z|

3+2(1+α)

= −

\

R3

dz

\

B0)

dξ (∆

s

(z)(ηe u

it

) − η∆

s

(z)e u

it

)∆

s

(z)e u

i

|z|

3+2(1+α)

+

\

R3

dz

\

B0)

dξ ∆

s

(z)(ηeg

i

+ k

1i

)∆

s

(z)e u

i

|z|

3+2(1+α)

≡ I

1

+ I

2

. Using the continuity equation (1.15)

1

in the first term and integrating by parts in the second with the use of the boundary condition e u|

∂B0)

= 0 we obtain

(3.6) 1 2

d dt

\

R3

dz

\

B0)

dξ η|∆

s

(z)e u|

2

|z|

3+2(1+α)

+

\

R3

dz

\

B0)

dξ D (∆

s

(z)e u) · ∇∆

s

(z)e u

|z|

3+2(1+α)

\

R3

dz

\

B0)

dξ ∆

s

(z)e q

σ

div(∆

s

(z)e u)

|z|

3+2(1+α)

= I

1

+ I

2

− 1 2

\

R3

dz

\

B0)

dξ η div

u

u|∆

s

(z)e u|

2

|z|

3+2(1+α)

≡ X

3 i=1

I

i

.

To examine the last term on the l.h.s. of (3.6) we use (2.7)

2

. Applying

s

(z) to (2.7)

2

, multiplying the result by ∆

s

(z)e q

σ

, integrating over B

0

) and integrating with respect to z with the weight 1/|z|

3+2(1+α)

we obtain (3.7) 1

2 d dt

\

R3

dz

\

B0)

dξ |∆

s

(z)e q

σ

|

2

|z|

3+2(1+α)

+ p

1

\

R3

dz

\

B0)

dξ ∆

s

(z)e q

σ

div ∆

s

(z)e u

|z|

3+2(1+α)

=

\

R3

dz

\

B0)

dξ ∆

s

(z)k

2

s

(z)e q

σ

|z|

3+2(1+α)

≡ I

4

.

(13)

Multiplying (3.7) by 1/p

1

, adding to (3.6) and using the Korn inequality we get

(3.8) 1 2

d dt

\

R3

dz

\

B0)

 η|∆

s

(z)e u|

2

|z|

3+2(1+α)

+ 1

p

1

· |∆

s

(z)e q

σ

|

2

|z|

3+2(1+α)



+ 1 c

0

\

R3

dz k∆

s

(z)e uk

21,B

0)

|z|

3+2(1+α)

≤ X

3 i=1

I

i

+ 1 p

1

I

4

. Now we estimate the terms on the r.h.s. of (3.8). Since α ∈ (1/2, 1) and s must be such that s > 1 + α (see [4]) we assume that s = 2. First we consider

I

2

=

\

R3

dz

\

B0)

dξ ∆

2

(z)(ηeg

i

+ k

1i

) · ∆

2

(z)e u

i

|z|

3+2(1+α)

= −

\

R3

dz

\

B0)

dξ ∆(z)(ηeg

i

+ k

1i

) · ∆

3

(z)e u

i

|z|

3+2(1+α)

≤ ε

1

\

R3

dz

\

B0)

dξ |∆

3

(z)e u|

2

|z|

3+2(2+α)

+ c(ε

1

)

\

R3

dz

\

B0)

dξ |∆(z)(ηeg + k

1

)|

2

|z|

3+2α

≡ I

21

+ I

22

, where

I

22

≤ c

\

R3

dz

\

B0)

dξ 1

|z|

3+2α

(|∆(z)(ηeg)|

2

+ |∆(z)∇

u

B

u

(u, ζ)|

2

+ |∆(z)T

u

(u, q

σ

)∇

u

ζ|

2

+ |∆(z)(∇T(e u, e q

σ

) − ∇

u

T

u

(e u, e q

σ

))|

2

) ≡ X

4 i=1

I

5i

. Next we estimate

I

51

≤ c

\

R3

dz

\

B0)

dξ |∆(z)ηeg|

2

+ |η∆(z)eg|

2

|z|

3+2α

≤ c



\

R3

dz

\

B0)

dξ |∆(z)η|

2p

|z|

3+2p(α+ε/2)



1/p



\

R3

dz

\

B0)

dξ |eg(ξ, t)|

2p

|z|

3−pε



1/p

+ c|η|

∞,Ω

\

R3

dz

\

B0)

dξ |∆(z)eg(ξ, t)|

2

|z|

3+2α

,

for all p, p

with 1/p+1/p

= 1 and ε > 0. Using the imbeddings B

21+α

(Ω) ⊂

B

2pα+ε/2

(Ω), Ω ⊂ R

3

, valid if 3/2 − 3/(2p) + α + ε/2 < 1 + α, and B

2α

(Ω) ⊂

L

2p

(Ω), Ω ⊂ R

3

, valid if 3/2 − 3/(2p

) < α, which both hold for suitable p,

(14)

ε because 3/2 + ε/2 < 1 + α can be satisfied for α > 1/2, we obtain I

51

≤ ckηk

21+α,Ω

kegk

2α,Ω

,

where we used the fact that B

21+α

(Ω) ⊂ L

(Ω) for α > 1/2.

Introducing the quantity α(t) =

Tt

0

u

ξ

(ξ, τ ) dτ we see that ξ

x

− I = αψ

0

(α), ξ

x2

− I = αψ

1

(α),

ξ

x

ξ

ξ

x

= ∇

ξ

αψ

2

(α) + α∇

ξ

αψ

3

(α), where ξ

x

, I are matrices, I is the unit matrix,

|α(t)|

2

≤ T

t

\

0

|u

ξ

|

2

dτ ≤ cT

t

\

0

kuk

22+α,Ω

dτ, and

a ≡ cT

1/2



T\

0

kuk

22+α,Ω

dτ 

1/2

so

x

− I|

2

, |ξ

x2

− I|

2

≤ a

2

ψ

4

(a), (3.8

)

x

ξ

ξ

x

|

2

≤ ψ

5

(a)

T

\

0

|u

ξξ

|

2

dτ.

Using the H¨older inequality and the imbeddings B

2α

(Ω) ⊂ L

2p

(Ω), B

21+α

(Ω) ⊂ B

α2p

(Ω), Ω ⊂ R

3

, holding for 3/2 − 3/(2p) ≤ α, 3/2 − 3/(2p

) + α ≤ 1 + α, 1/p + 1/p

= 1, we have

I

52

= c

\

R3

dz

\

B0)

|∆(z)∇

u

B

u

(uϕ)|

2

|z|

3+2α

≤ c

\

R3

dz

\

B0)

× 1

|z|

3+2α

[|∆(z)∇u|

2

+ ξ

x4

|∇ζ|

2

+ |∇u|

2

(|∆(z)ξ

x2

|

2

|∇ζ|

2

+ ξ

x2

|∆(z)∇ζ|

2

) + |∆(z)u|

2

(|ξ

x

· ∇ξ

x

|

2

|∇ζ|

2

+ ξ

x2

|∇

2

ζ|

2

) + |u|

2

(|∆(z)(ξ

x

· ∇ξ

x

)|

2

|∇ζ|

2

+ |ξ

x

· ∇ξ

x

|

2

|∆(z)∇ζ|

2

+ |∆(z)ξ

x2

|

2

|∇

2

ζ|

2

+ ξ

4x

|∆(z)∇

2

ζ|

2

)]

≤ ψ

6

(a) h

kuk

21+α,B0)

+ (ku

ξ

k

2Bα

2p(B0))

+ kuk

2Bα

2p(B0))

)

t

\

0

ku

ξξ

k

2L2p(B0))

dτ i + ψ

7

(a)(|∇u|

2∞,B

0)

+ |u|

2∞,B

0)

)

(15)

× h

\t

0

kuk

22+α,B0)

dτ +

t

\

0

ku

ξ

k

2Bα

2p(B0))

t

\

0

ku

ξξ

k

2L2p(B0))

dτ i

≤ cϕ

0

(a)kuk

21+α,Ω

+ ϕ

1

(a)kuk

22+α,Ω

t

\

0

kuk

22+α,Ω

dτ.

Similarly, we have

I

53

≤ ϕ

2

(a)(kuk

21+α,Ω

+ kq

σ

k

2α,Ω

).

Since qualitatively

∇T(e u, e q

σ

) − ∇

u

T

u

(e u, e q

σ

) = ϕ

3

(α)

t

\

0

u

ξξ

(τ ) dτ e u

ξ

+ αϕ

4

(α)(e u

ξξ

+ e q

σξ

), where ϕ

i

(0) 6= 0, i = 3, 4, we obtain

I

54

≤ ϕ

5

(a)

t

\

0

kuk

22+α,Ω

dτ (ke uk

22+α,Ω

+ ke q

σ

k

21+α,Ω

).

Summarizing the above considerations we have

|I

2

| ≤ ε

1

ke uk

22+α, e

+ ϕ

6

(1/ε

1

, a)(kuk

21+α, e

+ kq

σ

k

2α, e

) (3.9)

+ ϕ

7

(1/ε

1

, a)

t

\

0

kuk

22+α,Ω

dτ (ke uk

22+α, e

+ ke q

σ

k

21+α, e

), where ε

1

∈ (0, 1).

Now we calculate I

1

= −

\

R3

dz

\

B0)

dξ (∆

2

(z)ηe u

t

+ 2∆(z)η∆(z)e u

t

)∆

2

(z)e u

|z|

3+2(1+α)

= −

\

R3

dz

\

B0)

dξ (∆

2

(z)η

σ

e u

t

+ 2∆(z)η

σ

∆(z)e u

t

)∆

2

(z)e u

|z|

3+2(1+α)

.

We have

|I

1

| ≤ c



\

R3

dz

\

B0)

dξ |∆

2

(z)η

σ

|

2

|z|

3+2(1+α)



1/2



\

R3

dz

\

B0)

dξ |e u

t

|

p

|z|

3−εp



1/p

×



\

R3

dz

\

B0)

dξ |∆

2

(z)e u|

p

|z|

3+p(1+α)+εp



1/p

+ c



\

R3

dz

\

B0)

dξ |∆(z)η

σ

|

p1

|z|

3+(p1/2)(1+α)



1/p1

(16)

×



\

R3

dz

\

B0)

dξ |∆(z)e u

t

|

p2

|z|

3+(p2/2)(1+α)



1/p2

×



\

R3

dz

\

B0)

dξ |∆

2

(z)e u|

p3

|z|

3+p3(1+α)



1/p3

≡ I

11

+ I

12

,

for all p, p

i

with 1/p + 1/p

= 1/2 and 1/p

1

+ 1/p

2

+ 1/p

3

= 1.

Using the imbeddings B

2α

(Ω) ⊂ L

p

(Ω), B

22+α

(Ω) ⊂ B

p1+α+ε

(Ω), valid if 3/2 − 3/p ≤ α, 3/2 − 3/p

+ 1 + α + ε ≤ 2 + α, which both hold for suitable p, ε because 3/2 + ε ≤ 1 + α can be satisfied for α > 1/2, we obtain

I

11

≤ ckη

σ

k

1+α,Ω

ke u

t

k

α, e

ke uk

2+α, e

≤ ε

2

ke uk

22+α, e

+ c(ε

2

)kη

σ

k

21+α,Ω

ke u

t

k

2α, e

.

Using the imbeddings B

21+α

(Ω) ⊂ B

p(1+α)/21

(Ω), B

2α

(Ω) ⊂ B

p(1+α)/22

(Ω), B

22+α

(Ω) ⊂ B

1+αp3

(Ω) valid for 3/2 − 3/p

1

+ (1 + α)/2 ≤ 1 + α, 3/2 − 3/p

2

+ (1 + α)/2 ≤ α, 3/2 − 3/p

3

+ 1 + α ≤ 2 + α, which all hold for suitable p

i

because 3/2 ≤ 1 + α, we get the same bound as for I

11

, I

12

≤ ε

2

ke uk

22+α, e

+ c(ε

2

)kη

σ

k

21+α,Ω

ke u

t

k

2α, e

. Next we have

|I

3

| ≤ ε

3

ke uk

22+α, e

+ c(ε

3

)kuk

22+α,Ω

ke uk

21+α, e

. Finally, we have

|I

4

| ≤ ε

4

\

R3

dz

\

B0)

dξ |∆

2

(z)e q

σ

|

2

|z|

3+2(1+α)

+ ϕ

8

(1/ε

4

, a) h

kuk

21+α, e

+ kuk

22+α, e

t

\

0

kuk

22+α,Ω

dτ + kq

σ

k

21+α, e

kuk

22+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

t

\

0

kuk

22+α,Ω

dτ + kq

σ

k

21+α, e

t

\

0

kuk

22+α,Ω

dτ i .

Summarizing the above considerations we obtain (3.10) d

dt

\

R3

dz

\

B0)

 η

2 · |∆

2

(z)e u|

2

|z|

3+2(1+α)

+ 1

2p

1

· |∆

2

(z)e q

σ

|

2

|z|

3+2(1+α)



+ 1 c

0

\

R3

dz k∆

2

(z)e uk

21,B

0)

|z|

3+2(1+α)

≤ ε

5

(ke uk

22+α, e

+ ke q

σ

k

21+α, e

)

(17)

+ ϕ

9

(1/ε

5

, a)(kuk

21+α, e

+ kq

σ

k

2α, e

)

+ ϕ

10

(1/ε

5

, a)

t

\

0

kuk

22+α,Ω

dτ (kuk

22+α, e

+kq

σ

k

21+α, e

+kq

σ

k

21+α, e

kuk

22+α, e

) + ϕ

11

(1/ε

5

, a)(kq

σ

k

21+α, e

ke u

t

k

2α, e

+ kuk

22+α, e

ke uk

21+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

+ kegk

2α, e

).

Now we want to obtain an energy type estimate for the local solutions satisfying (2.7). Multiplying (2.7)

1

by e u

i

, summing over i, integrating over Ω and using the boundary condition e

(3.11) e u|

∂ e

= 0,

and employing the continuity equation, we obtain (3.12)

\

e Ω

η 1 2

d

dt e u

2

dξ +

\

e Ω

D (e u)∇e u dξ −

\

e Ω

e

q

σ

∇ · e u dξ =

\

e Ω

(ηeg + k

1

)e u dξ.

Using the continuity equation (1.15)

1

and the Korn inequality we get (3.13) d

dt

\

Ωe

1

2 ηe u

2

dξ + 1

c

0

ke uk

21, e

\

Ωe

e

q

σ

∇ · e u dξ

≤ − 1 2

\

Ωe

η div

u

ue u

2

dξ +

\

Ωe

(ηeg + k

1

)e u dξ.

Multiplying (2.7)

2

by e q

σ

and integrating over e Ω yields

(3.14) 1

2 d dt

\

Ωe

e

q

σ2

dξ + p

1

\

Ωe

e

q

σ

∇ · e u dξ =

\

Ωe

k

2

e q

σ

dξ.

From (3.13) and (3.14) we obtain (3.15) d

dt

\

Ωe

1 2



ηe u

2

+ 1 p

1

q e

σ2



dξ + 1

c

0

ke uk

21, e

≤ 1 p

1

\

Ωe

k

2

q e

σ

dξ − 1 2

\

Ωe

η div

u

ue u

2

dξ +

\

Ωe

(ηeg + k

1

)e u dξ

≡ J

1

+ J

2

+ J

3

, where

|J

1

| ≤ εke q

σ

k

20, e

+ ϕ

12

(1/ε, a) 

kuk

20, e

+ kq

σ

k

21, e

kuk

22, e

+

t

\

0

kuk

22+α,Ω

dτ ke uk

21, e



,

(18)

|J

2

| ≤ εke uk

21, e

+ ϕ

13

(1/ε, a)kuk

21, e

ke uk

21, e

,

|J

3

| ≤ ε(ke uk

21, e

+ kq

σ

k

20, e

)

+ ϕ

14

(1/ε, a) h

kuk

21, e

+

t

\

0

kuk

22, e

dτ (ke uk

22, e

+ ke q

σ

k

21, e

) i

+ ckegk

20, e

. Using the above estimates in (3.15) gives

(3.16) d dt

\

Ωe



ηe u

2

+ 1 p

1

q e

σ2

 dξ + 1

c

0

ke uk

21, e

≤ εkq

σ

k

20, e

+ ϕ

15

(1/ε, a) h

kegk

20, e

+ kuk

21, e

+ kuk

21, e

ke uk

21, e

+ kq

σ

k

21, e

kuk

22, e

+

t

\

0

kuk

22+α,Ω

dt (kuk

22, e

+ kq

σ

k

21, e

) i . From (3.10) and (3.16) after using the equivalence of norms for fractional spaces (see [4]) we obtain

(3.17) d dt

\

e Ω



ηe u

2

+ 1 p

1

q e

σ2

 dξ

+ d dt

\

R3

dz

\

B0)



η |∆

2

(z)e u|

2

|z|

3+2(1+α)

+ 1 p

1

· |∆

2

(z)e q

σ

|

2

|z|

3+2(1+α)



+ ke uk

22+α, e

≤ εke q

σ

k

21+α, e

+ c(kuk

21+α, e

+ kq

σ

k

2α, e

)

+ c[kegk

2α, e

+ kq

σ

k

21+α, e

ke u

t

k

2α, e

+ kuk

22+α, e

ke uk

21+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

]

+ c

t

\

0

kuk

22+α,Ω

dτ [kuk

22+α, e

+ kq

σ

k

21+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

].

To examine the second term on the r.h.s. of (3.17) we use the interpolation inequality

kuk

21+α, e

+ kq

σ

k

2α, e

≤ ε(kuk

22+α, e

+ kq

σ

k

21+α, e

) (3.18)

+ c(ε)(kuk

20, e

+ kq

σ

k

20, e

).

From (2.7)

1

we have

ke q

σ

k

21+α, e

≤ c(ke u

t

k

2α, e

+ ke uk

22+α, e

+ kegk

2α, e

) (3.19)

+ c

t

\

0

kuk

22+α,Ω

dτ (ke uk

22+α, e

+ ke q

σ

k

21+α, e

)

+ ε(kuk

22+α, e

+ kq

σ

k

21+α, e

) + c(ε)(kuk

20, e

+ kq

σ

k

20, e

).

(19)

From (3.17)–(3.19) we obtain (3.20) d

dt

\

Ωe



ηe u

2

+ 1 p

1

q e

σ2

 dξ

+ d dt

\

R3

dz

\

B0)

 η|∆

2

(z)e u|

2

|z|

3+2(1+α)

+ 1 p

1

· |∆

2

(z)e q

σ

|

2

|z|

3+2(1+α)



+ ke uk

22+α, e

+ ke q

σ

k

21+α, e

≤ εke u

t

k

2α, e

+ ckegk

2α, e

+ ε(kuk

22+α, e

+ kq

σ

k

21+α, e

)c(ε)(kuk

20, e

+ kq

σ

k

20, e

) + c[kq

σ

k

21+α, e

ke u

t

k

2α, e

+ kuk

22+α, e

ke uk

21+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

] + c

t

\

0

kuk

22+α,Ω

dτ [kuk

22+α, e

+ kq

σ

k

21+α, e

+ kq

σ

k

21+α, e

kuk

22+α, e

].

To examine the norm ke u

t

k

α, e

from the r.h.s. of (3.20) we use the inter- polation inequality (see [12])

(3.21)

t

\

0

t

k

2α, e

dt ≤ ε

t

\

0

kwk

22+α, e

dt + c(ε)

\

Ωe

kw

t

k

2α/2,(0,t)

dξ.

To estimate the last term on the r.h.s. of (3.21) we consider the time differ- ences

kt

(h)f (ξ, t) = X

k j=0

c

jk

(−1)

k−j

f (ξ, t + jh), k > α.

Applying ∆

kt

(h) to (2.7)

1

we obtain

(3.22) η∆

kt

e u

it

− ∇

j

T

ij

(∆

kt

u, ∆ e

kt

q e

σ

) = η∆

kt

u e

it

− ∆

kt

(ηe u

it

) + ∆

kt

(ηeg

i

+ k

1i

).

From (3.22) we have (3.23)

h\0

0

dh h

1+α

T −kh\0

0

ψ(t) dt

\

Ωe

η|∆

kt

e u

t

|

2

h\0

0

dh h

1+α

T −kh\0

0

ψ(t) dt

\

Ωe

∇T(∆

kt

e u, ∆

kt

q e

σ

)∆

kt

u e

t

=

h0

\

0

dh h

1+α

T −kh0

\

0

ψ(t) dt

\

e Ω

[(η∆

kt

u e

t

− ∆

kt

(ηe u

t

))∆

kt

u e

t

+ ∆

kt

(ηeg + k

1

)∆

kt

e u

t

] dξ ≡ I

1

+ I

2

,

where ψ(t) is a smooth function vanishing for t ≤ t

0

and equal to 1 for

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