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(1)

ON EXACT CONTROLLABILITY

FOR THE NAVIER-STOKES EQUATIONS

O.YU. IMANUVILOV

Abstract. We study the local exact controllability problem for the Navier-Stokes equations that describe an incompressible uid ow in a bounded domain with control distributed in a subdomain ! . The result that we obtain in this paper is as follows. Suppose that

^

v(x) is a given steady-state solution of the Navier-Stokes equations.

Letv0(x) be a given initial condition and k^v();v0k<" where " is small enough. Then there exists a locally distributed control u, with suppu (0T)!such that the solutionv(tx) of the Navier-Stokes equations:

@tv;v+ (v r)v=rp+u+f divv= 0 v j@= 0 v jt=0 =v0 coincides with ^v(x) at instantT,v(Tx)^v(x).

1. Introduction

This paper is concerned with the local exact controllability of the Navier- Stokes equations, dened on a bounded domain Rn (

n

= 2

3) with boundary

@

2

C

1. More precisely, the problem under study is as follows.

Let us consider the nonstationary Navier-Stokes equations

@

t

v

(

tx

);

v

(

tx

)+ (

v

r)

v

+r

p

=

f

(

x

) +

!

u

in

div

v

= 0

(1

:

1) with initial and boundary conditions

v

j = 0

v

jt=0 =

v

0(

x

)

(1

:

2) where

v

(

tx

) = (

v

1(

tx

)

:::v

n(

tx

)) is the uid velocity,

p

the pressure,

f

(

x

) = (

f

1(

x

)

::f

n(

x

)) a density of external forces,

u

(

tx

) a control dis- tributed in an arbitrary xed subdomain

!

of the domain and

! is the characteristic function of the set

!

:

!(

x

) = 1

for

x

2

!

0

for

x

2n

!:

Let (^

v

(

x

)

^

p

(

x

)) be a steady-state solution of the Navier-Stokes equations

;^

v

+ (^

v

r)^

v

+r

p

^=

f

(

x

) in

div^

v

= 0

v

^j@ = 0 (1

:

3)

Korea Institute for Advanced Study, 207-43 Chungryangri-dong Dongdaemoon-ku Seoul 130-012, Korea. E-mail: oleg@kias.k ais t.a c.k r

Research partially supported by KIAS (Grant KIAS-M97003), GARC-KOSEF and KOSEF(K94073).

Submitted February 17, 1997. Revised July 27, 1997. Accepted for publication De- cember 23, 1998.

c

Societe de Mathematiques Appliquees et Industrielles. Typeset by TEX.

(2)

close enough to the initial condition

k

v

0;

v

^kV1()

"

(1

:

4) where the parameter

"

is suciently small. We want to nd a control

u

such that for given

T >

0, the following equality holds:

v

(

Tx

) = ^

v

(

x

)

:

(1

:

5) We assume

Condition 1.1. The boundary

@

= Ki=1;i 2

C

1

(;i\;j = fg for all

i

6=

j

) where;i is a

n

;1-dimensional connected manifold of class

C

1

:

For each ;i, there exists a neighborhood Ai Rn and a dieomorphism

i

2

C

1(Ai

Rn) such that i(;i) =

S

1n.*

The main result of this paper is the following Theorem.

Theorem 1.2. Let

v

0 2

V

1() and pair (^

v p

^) 2 (

V

1()T(

W

11 ())n)

W

21() is a given steady state solution of the Navier-Stokes equations (1.3) such thatsupp

v

^ . Then for suciently small

"

there exists a solution (

vpu

)2

V

12(

Q

)

L

2(0

T W

21())(

L

2(

Q

!))n of problem (1.1), (1.2), (1.4), (1.5).

To explain this result, let us assume that ^

v

j@ = 0 and ^

v

is an unstable singular point of the dynamical system generated by equation (1.1) in the phase space of solenoidal vector elds with adherence conditions on

@

. Let

v

0 be an initial condition in a neighborhood of the function ^

v

. This work shows that one can construct a locally distributed control such that the trajectory goes out of point

v

0 and reaches ^

v

in nite time. In other words, by means of the locally distributed control, one can suppress the generation of turbulence. This result claries the question of the connection between turbulence and controllability (see J.-L. Lions 26]).

The result we obtain in Theorem 1.2 is local. On the other hand, for the linearized Navier-Stokes system, we can prove global zero-controllability, see Theorem 4.3.

One important special case is the following controllability problem for the Stokes system:

@

t

v

(

tx

);

v

(

tx

) =r

p

+

f

(

tx

) +

!

u

in

div

v

= 0

(1

:

6)

v

j = 0

v

jt=0=

v

0(

x

)

v

jt=T 0

:

(1

:

7) We have

Theorem 1.3. Let

v

0 2

V

1()

f

2

L

2(0

T V

0()) and there exists

" >

0 such thatRQj

f

j2

e

(T;t)12+"

dxdt <

1

:

Then there exists a solution (

vpu

)2

V

12(

Q

)

L

2(0

T W

21())(

L

2(

Q

!))n to problem (1.6),(1.7).

This paper is organized as follows. To prove Theorem 1.2 we use a vari- ant of the implicit function theorem. The only nontrivial condition to be checked is to show that the derivative of the corresponding mapping at some

*

S

rn=

@B

r

B

r =f

x

2Rnj

x

j

< r

g

(3)

point is an epimorphism. In our case, this problem is equivalent to the zero controllability of the linearization of the Navier-Stokes equations at point ^

v:

(see problem (4.1)-(4.3).) Sections 2-4 are devoted to this problem. One of the usual ways to solve the controllability problem for evolution equations is to reduce it to an observability problem for the adjoint equation. Thus, in section 2 we introduce a linear operator (see equation (2.1)) which after the change

t

! ;

t

is formally adjoint to the derivative of the Navier-Stokes equations at point ^

v

(

x

)

:

The observability problem for this operator is solved in three steps. First in Theorem 2.11, we get an appropriate estimate for the pressure

p:

Then in Theorem 3.1, we obtain a Carleman estimate for the velocity

y

of the uid via a weighted

L

2-norm of the density of external forces

f

and the pressure

p:

Moreover, for the pressure, by Theorem 2.11, one can choose a weighted

L

2 norm over (0

T

)

!:

And nally in Theorem 3.6, we prove an estimate (not of Carleman type) for the velocity where

p

and an initial condition are absent from the right-hand side. In section 4, this observability estimate is converted into a controllability result in The- orem 4.3. In section 5, all conditions for the implicit function theorem are checked.

We close this section by mentioning some of the previous results regarding our problems. The solvability of (1.1), (1.2),(1.5) was rst proved in A.V.

Fursikov, O.Yu. Imanuvilov 12] in the case when (1.1) is Burgers' equation.

For a control distributed in a domain

!

such that

@ !

, this problem was studied in the case of the Navier-Stokes equations and ^

v

0 in A.V.

Fursikov, O.Yu. Imanuvilov 13] in dimension

n

= 2 and in A.V. Fursikov 10] when

n

= 3

:

The case of the Navier-Stokes equations and ^

v

6= 0 has been studied in A.V. Fursikov, O.Yu. Imanuvilov 14], 17], O.Yu. Imanuvilov 21] and for the Boussinesq system in 15] (see also 16]). On the other hand, in pioneering works 3]-5], J.-M. Coron proved the global approximate con- trollability for the 2-D Euler equations and the 2-D Navier-Stokes equations with slip boundary conditions. In 6], combining results on global approxi- mate and local exact controllability results, J.-M. Coron and A.V. Fursikov obtained the global exact controllability for the Navier-Stokes system on a 2-D manifold without boundary.

In 7], C. Fabre obtained an approximate controllability for \cut o"

Navier-Stokes equations.

2. Estimate for the pressure Let us consider the system

@y @t

;

y

+

B

(

y

^

v

) +

B

(^

vy

) =r

p

+

f

in

Q

(2

:

1)

div

y

= 0

y

j@= 0

y

(0

x

) =

y

0(

x

)

(2

:

2) where Rnis a bounded domain with

@ C

1

Q

= (0

T

) and the operators

B

(^

v

)

B

(

^

v

) are dened by the formulas:

B

(

y

^

v

) = ((

y @

^

v

@x

1)

:::

(

y @

^

v

@x

n))

B

(^

vy

) =;(^

v

r)

y:

(2

:

3)

(4)

Denote

Q

! = (0

T

)

!

= (0

T

)

@ :

Let

be the outward unit normal to

@ :

In this paper we use the following functional spaces. Recall that

W

pk()

k

0

1

p <

1 is the Sobolev space of functions with nite norm

k

u

kWpk

() = (X

jjk Z

@

jj

u

(

x

)/

@x

11

:::@x

nnp

dx

)1=p

where

= (

1

2

:::

n)

j

j=

1++

n

W

12(

Q

) =f

w

(

tx

)j

w

2

L

2(0

T W

22())

@w @t

2

L

2(0

T L

2())g

V

1() =f

v

(

x

) = (

v

1

:::v

n) 2 (

W

21())ndiv

v

= 0

v

j@= 0g

V

0() =f

v

(

x

) = (

v

1

:::v

n) 2 (

L

2())ndiv

v

= 0

(

v

)j@= 0g

V

;1() = (

V

1())

V

12(

Q

) =f

v

(

tx

) 2 (

W

12(

Q

))ndiv

v

= 0

v

j@ = 0g

L

2(

Q

) =f

v

(

tx

)

Z

Q

v

2

dxdt <

1g

:

We have

Proposition 2.1. (27]) Let be a bounded domain inRnwith

@

2

C

2

:

Then (

L

2())n=

V

0()(

V

0())?

where

(

V

0())?=f

v

(

x

) = (

v

1

:::v

n) 2 (

L

2())n

v

=r

pp

2

W

21()g

:

Here and below we assume that the pair (^

v p

^) satises (1.3) and

(^

v p

^)2(

V

1()\(

W

11())n)

W

21() supp^

v :

Let

!

be an arbitrary xed subdomain and ibe the mapping from Condition 1.1.

Without loss of generality we can assume that i(Ai\)

B

1. (Otherwise we can make the change

x

!

x=

j

x

j2

:

)

Set Ui = i;1(f

x

2 Rnj1;

" <

j

x

j

<

1g)

where

"

2 (0

1)

:

For all suciently small

"

, the setUi is correctly dened and

U

i

\U

j =fg for all

i

6=

j @

Ui= ;i

i

(2

:

4) where

i= i;1(

S

1;"n ) and

(

!

supp ^

v

)\Ui=fg8

i

= 1

:::K:

(2

:

5) Let

G

Rnbe a domain which satises the following condition:

(5)

Condition 2.2. The domain

G

is dieomorphic to the cylinder;0

T

0]

where

T

0

>

0 is a number and ; Rn is a closed (

n

;1)-dimensional manifold of class

C

1

:

This condition implies immediately that

@G

2

C

1 and

@G

=

0

1

where

i is a (

n

;1)-dimensional connected manifold of class

C

1

:

Let

w

(

x

) be a harmonic function in

G

:

w

= 0 in

G

(2

:

6)

such that

@w

@

j1 = 0

(2

:

7)

Z

0

j

w

j2

d

(2

M

)2

Z

1

j

w

j2

d

(2

"

)2

:

(2

:

8) Let

(

x

)2

C

1(

G

) be a function satisfying the conditions

0

< C

1jr

(

x

)j8

x

2

G

j1 = 0 j0 = 1

:

(2

:

9) Then the set

t=f

x

2

G

(

x

) =

t

g

t

20

1] (2

:

10) is a smooth manifold dieomorphic to

0 and

1. We have:

Theorem 2.3.(28]) There exist a constant

C

2

>

0 and a function

(

x

) 2

C

1(

G

), satisfying condition (2.9), such that for any function

w

2

W

212(

G

) for which (2.6)-(2.8) are satised

Z

t

j

w

j2

d

C

2k

w

k2(1;t)L2(1 )

k

w

k2tL2(0)

C

2(2

"

)2(1;t)(2

M

)2t

(2

:

11) where

t is the manifold (2.10).

Obviously the domain Ui satises Condition 2.2 for all

i

2 f1

:::K

g

:

Denote by

i(

x

) the function from Theorem 2.3 which corresponds to the domainUi

:

Similar to (2.10) we set

t(

i

) =f

x

2Ui

i(

x

) =

t

g

:

(2

:

12) For all

r

2(0

1) we introduce the auxiliary domains

O

i(

r

) =f

x

i2Uij

r <

i(

x

)

<

1g

O(

r

) =Ki=1Oi(

r

)

:

(2

:

13) Let

w

0b

w

be an arbitrary subdomain. We have:

Lemma 2.4. There exists a function

(

x

)2

C

1() such that

(

x

) = 1;

i(

x

) 8

x

2Ui (

x

)

>

0 in

jr

(

x

)j

>

08

x

2n

!

0

:

(2

:

14) Proof. First we construct an auxiliary function

(

x

)2

C

1() such that

(

x

) = 1;

i(

x

) 8

x

2Ui (

x

)

>

0 in

:

(2

:

15)

(6)

To do this, we consider the sequence of domains ~Ui Rn

i

=f1

:::K

g with the following properties

@

U~i= ;i

~i2

C

1

Ui U~i

i U~i

U~i\U~j =fgfor

i

6=

j

where ~

iis a connected (

n

;1)-dimensional

C

1 manifold. (For example we can choose ~Ui as ~Ui = i;1(f

x

2Rnj1;

" <

~ j

x

j

<

1g) for some ~

":

)

Since by assumption

iis a

C

1surface, one can extend the function

i(

x

) to a smooth function of

C

1(~Ui) such that

i0 in some neighborhood of

~

i

:

Set ~

(

x

)jU~i 1;

i(

x

) and ~

0 in nKi=1U~i

:

Let

(

x

)2

C

01(nKi=1Ui) be a nonnegative function such that

(

x

)

>

0 8

x

2f

x

2nKi=1Uij

~(

x

) = 0g

:

Then the function

(

x

) = ~

+

s

satises (2.15) for all

s >

0 suciently large.

Now let us show that the function

(

x

) which satises (2.15) can be chosen as a Morse function. Since by (2.9), (2.15)

jr

(

x

)j

>

0 8

x

2Ki=1Ui there exists a sequence of domains ~~Ui Rnsuch that

@

U~~i= ;i

~~i2

C

1

Ui U~~i U~i

i U~~i

U~~i\U~~j =fgfor

i

6=

j

where ~~

i is a connected (

n

;1)-dimensional

C

1 manifold, and

jr

(

x

)j

>

0 8

x

2Ki=1U~~i

:

Let

2

C

01(nKi=1Ui) such that

(

x

) = 18

x

2nKi=1U~~i (

x

) = 0 8

x

2Ki=1Ui

:

For every

" >

0 there exists a Morse function

"such thatk

;

"kC2()

":

Set

"(

x

) = (1;

(

x

))

(

x

)+

(

x

)

"(

x

)

:

Obviously

"(

x

) = 1;

i(

x

) 8

x

2Ui

and for all suciently small

" >

0

"(

x

)

>

0 in

:

Let us show that for all small

"

,

" is a Morse function. Actually, in

K

i=1 U

i function

" has no critical points. In nKi=1U~~i,

" coincides with the Morse function

". Short calculations give the inequality

jr

"(

x

)j=jr

(

x

);r

(

x

)(

;

")(

x

);

(

x

)(r

;r

")(

x

)j

C

;2k

kC1()k

;

"kC2()

C

;2k

kC1( )

"

(7)

where

C >

0

x

2ki=1U~~i.

Since for all suciently small

"

, the right-hand side of this inequality is positive, the function

" has no critical points in Ki=1U~~i

:

Denote by M the set of critical points of the function

"

:

Exactly in the same way as it was done in 2], 20], one can construct a dieomorphism

r

: ! such that

r

(

x

) =

x

8

x

2Ki=1Ui

r

;1(M)

!

0

:

Thus, the function

(

x

) =

"(

r

(

x

)) satises all the conditions of our

lemma. tu

We set

'

(

tx

) =

e

(x)

=

(

t

(

T

;

t

))2

(2

:

16)

(

tx

) = (

e

;

e

2jj jjC())

=

(

t

(

T

;

t

))2

(

t

) =

(

tx

0)

'

(

t

) =

'

(

tx

0)

(2

:

17) where

>

1, function

from Lemma 2.4 and

x

0 2

@ :

By (2.9), (2.14) the functions

'

are independent of the selection of

x

0 2

@ :

Note that (2.9), (2.14) imply the obvious inequality

0

>

(

tx

)

(

t

)

'

(

tx

)

'

(

t

) 8(

tx

)2

Q:

Let us introduce a function

`

(

t

)2

C

10

T

] by the formula

`

(

t

)

>

0

t

2(0

T

)

`

(

t

) = 1

t

2(0

14

T

)

t t

2(34

TT

)

:

(2

:

18) In this section, our aim is to get an estimate for the function

p

using the trace of

p

on

@

and the restriction of

p

on 0

T

]

!

0. To prove this estimate, we need to recall some previous results on Carleman inequalities for the Laplace operator.

Let us consider the analog of problem (2.6)-(2.8) in the domain Ui:

w

= 0 in Ui

(2

:

19)

@w @

ji = 0

(2

:

20)

Set

A

=

A

(

) = max

2 1

4 1]

e

(1;);1

1;

:

(2

:

21)

By (2.9), (2.14) there exists

0

>

1 such that

1 +

A

(

)

(

x

)

e

(x) 8

>

0

x

2O(14)

e

= (

e

(x))ji

>

1 +

A

(

)

(

x

)

> e

(x)

8

>

0

x

2O(12)nO(34)

:

(2

:

22)

(8)

We have:

Lemma 2.5. Let the function

w

2

W

212(Ui) be a solution of problem (2.19), (2.20). Then there exist ^

2(0

1) and ^

>

1 such that for

>

^

((

T

;

s t

)

t

)2

Z

O

i (

1

4 )

j

w

j2

e

2s((T1+ A;t)t)2

dx

C

4(

Z

@

j

w

j2

e

2s'

d

+

Z

i

j

w

j2

e

2s '^

d

)

(2

:

23) where the domainOi(

r

) is dened in (2.13),

i

2f1

:::K

g

:

Proof. Set

(2

M

)2=

Z

;i

j

w

j2

d

(2

"

)2=

Z

i

j

w

j2

d:

Let

>

1 and

(

i

) be the manifold dened by (2.12). Note that, by (2.14)

A

= max

2 1

4 1]

e

(1;);1

1;

= maxx2Oi(14)

e

(x);1

(

x

)

e

34=

(

e

ji)34

:

(2

:

24) Using inequality (2.11), we obtain

Z

Oi( 1

4 )

j

w

j2

e

2sA

dx

C

6Z 1

1

4 Z

j

w

j2

e

2sA(1;)

dd

C

7Z 1

1

4

k

e

sA

w

k2(1;L2(i) )

k

w

k2L2(;i)

d

C

7Z 1

0

k

e

sA

w

k2(1;)L2(i )

k

w

k2L2(;i)

d

C

8Z 1

0

(

"e

sA)2(1;)

M

2

d

=

I:

(2.25) Short calculations give the equality

I

=

M

2 2

"

"e

sA

M

2

;1

#

=ln

"e

sA

M

:

(2

:

26) Obviously, by (2.24) there exist ^

2 (0

1) and ^

>

1 such that for all

> >

^ 1

A

+ 1

e

34+ 1^

min

x2

i

e

(x);1 = ^

e

;1

:

(2

:

27) Let us consider two cases.

A) Let

"e

s(A+1)

M

1

:

(2

:

28)

Thus

ln

("eMsA)0 and

;

ln

(

"e

sA

M

) =

s

;

ln

(

"e

s(A+1)

M

)

s:

(2

:

29)

(9)

Inequalities (2.26), (2.28) and (2.29) imply

I

M

2

2

s :

By this inequality and (2.25), keeping in mind that

e

j@= 1

we obtain

s

Z

O

i (

1

4 )

j

w

j2

e

2s(1+A )

dx

C

10Z

;

i

j

w

j2

e

2se

d:

(2

:

30) B) Let

"e

s(A+1)

M

1

:

Then by (2.25), (2.27)

I

C

11Z 1

0

"

2

e

2sA

e

2s

d

C

11

"

2

e

2s(e34+1)

C

12Z

i

w

2

e

2s( e^ ;1)

d

where

> >

^ 1

:

From this inequality, increasing the parameter ^

2(0

1) if necessary, we obtain

s

Z

O

i (

1

4 )

j

w

j2

e

2s(1+A )

dx

C

13Z

i

j

w

j2

e

2s e^

d:

(2

:

31) Inequalities (2.30), (2.31) after the change

s

!

s=

(

t

(

T

;

t

))2 imply

(2.23). tu

We have:

Lemma 2.6. Let

p

2

W

212(Ui) be a harmonic function in Ui. Then there exists ^

>

1 such that for

>

^

there exists

s

0(

)

>

0 such that

((

T

;

s t

)

t

)2

Z

O(

1

4 )

j

p

j2

e

2s((T(1+A );t)t)2

dx

C

;Z

@

j

p

j2

e

2s'

d

+

Z

i

(jr

p

j2+

p

2)

e

2s '

d

8

s

s

0(

)

(2

:

32) for some

2(0

1)

:

Proof. Let ^

be dened in Lemma 2.5 and ^

be the maximum of the cor- responding parameter from Lemma 2.5 and

0 from (2.22). We are looking for a function

p

of the form:

p

=

z

1 +

z

2

z

1 = 0 in Ui

z

1j;i = 0

@z @~n

(1

i

)ji =

@p

@~n

(

i

)ji (2

:

33) and

z

2 = 0 in Ui

z

2j;i =

p @z @~n

(2

i

)ji = 0

(2

:

34)

(10)

where

~n

(

i

) is a unit outward normal toUi

:

Note that

'

ji =

e

((

T

;

t

)

t

)2 8

i

=f1

:::K

g

:

Also by (2.22) there exists

2(^

1)

>

34 such that

1 +

A

maxx2Oi(14 )

(

x

)

((

T

;

t

)

t

)2

< e

((

T

;

t

)

t

)2 =

'

ji

:

From this inequality, (2.14) and (2.33) we obtain

Z

i

j

z

1j2

e

2s '

d

+

s

((

T

;

t

)

t

)2

Z

O

i (

1

4 )

j

z

1j2

e

2s((T(1+A );t)t)2

dx

C

13Z

i

@p

@~n

(

i

)

2

e

2s '

dx:

(2

:

35)

Then, by Lemma 2.5, the function

z

2 satises the estimate ((

T

;

s t

)

t

)2

Z

O

i (

1

4 )

j

z

2j2

e

2s((T1+A;t)t)2

dx

C

13(

Z

@

j

p

j2

e

2s'

d

+

Z

i

j

z

2j2

e

2s '

d

)

C

14(

Z

@

j

p

j2

e

2s'

d

+

Z

i

(j

z

1j2+

p

2)

e

2s '

d

)

:

(2.36) By (2.35), (2.36) we have

((

T

;

s t

)

t

)2

Z

O

i (

1

4 )

j

z

2j2

e

2s((T1+A;t)t)2

dx

C

15(

Z

@

j

p

j2

e

2s'

d

+

Z

i

jr

p

j2

e

2s '

d

)

C

15(

Z

j

p

j2

e

2s'

d

+

Z

i

(jr

p

j2+

p

2)

e

2s '

d

)

:

(2

:

37)

Inequalities (2.35), (2.37) imply (2.32). tu

Let us consider the Dirichlet boundary value problem for the Laplace operator

z

(

x

) =

f

(

x

)

x

2

z

j@ = 0

:

(2

:

38) We have:

Lemma 2.7. Let

f

(

x

) 2

L

2(). There exists ^

>

1 such that for

>

^, there exists

s

0(

)

>

0 such that for any

s > s

0 the solution

z

(

x

) 2

W

22() of problem (2.38) satises the estimate

Z

0

@ 1

s'

n

X

ij=1

@

2

z

@x

i

@x

j

2+

s

2

'

jr

z

j2+

s

3

4

'

3

z

2

1

A

e

2s

dx

C

16(

Z

f

2

e

2s

dx

+

Z

!0

s

3

4

'

3

z

2

e

2s

dx

)

(2

:

39)

(11)

where

C

1

>

0 does not depend on

st

.

The Carleman inequality (2.39) can be proved in the same way as the corresponding Carleman estimate for a parabolic equation in 20], 17]. Note that for the case

@ !

0 this estimate was proved in 19].

Now to continue estimating the pressure

p

, we have to use equations (2.1), (2.2). Applying the operatordivto both parts of equation (2.1) we obtain

p

=div(

B

(^

vy

) +

B

(

y v

^)) in (2

:

40) for a.e.

t

20

T

].

The following lemma gives an estimate of the

L

2-norm of the pressure

p

. Lemma 2.8. Let

f

2

L

2(0

T V

1()) and

p

(

t

) 2

L

2() satises (2.40).

There exists ^

>

1 such that for all

>

^, there exists

s

0(

) such that

Z

s

2

2

'

2

p

2

e

2s

dx

C

10(

)

Z

!

0

s

2

2

'

2

p

2

e

2s

dx

+

Z

@

s'p

2

e

2s

d

+

Z

jr

y

j2

s

2

'

+

s

2

'

j

y

j2

e

2s

dx

8

s > s

0(

)

(2.41) where the constant

C

10 is independent of

s

and

t

.

Proof. Let ^

be a maximum of the corresponding parameters from Lemma 2.6 and Lemma 2.7 and

0 from (2.22). Note that suppdiv(

B

(^

vy

) +

B

(

y

^

v

)) supp ^

v:

By (2.5)

supp ^

v

\Ui=fgfor all

i

2f1

:::K

g

:

So there exists a neighborhood of

i, a domain

G

isuch that

p

is the harmonic function in

G

iand Ki=1

G

i nKi=1Oi(18) =fg

:

Note also that

'

ji =

e

((

T

;

t

)

t

)2 8

i

=f1

:::K

g

:

Thus, by the properties of interior regularity of solutions of elliptic equations, there exists a constant

C

such that

K

X

i=1 Z

i

jr

p

j2

e

2s '

d

C

Z

K

i=1 G

i

j

p

j2

e

2s'

dx

(2

:

42) where

2(0

1) is dened in Lemma 2.6.

By (2.13), (2.14) lim

"!+0

min

x2O(

3

4

;")nO(

3

4 )

(

x

) = lim

"!+0

max

x2O(

3

4

;")nO(

3

4 )

(

x

) = 14

:

Thus, by (2.22), there exists

"

0 2(0

1001 ) such that

1 +

A

inf

x2O(

3

4

;"0)nO(

3

4 )

(

x

)

>

sup

x2O(

3

4

;"

0 )nO(

3

4 )

e

:

(2

:

43)

(12)

Again, using the property of interior regularity of harmonic functions, by (2.43) we have

Z

O(

3

4 (1;"

0 ))nO(

3

4

; 3

8

"

0 )

(jr

p

j2+

p

2)

e

2s'

dx

C

Z

O(

3

4

;"0)nO(

3

4 )

p

2

e

2s((T1+A;t)t)2

dx

8

s >

1

:

(2

:

44) Let

(

x

) be a function such that

(

x

) = 18

x

2nO(34(1;

"

0))

0 in jO(34

; 3

8

"0) 0

:

Then by (2.40) the function

z

=

p

satises the equation

z

= 2(r

r

p

)+

p

+

div(

B

(^

vy

)+

B

(

y

^

v

)) in

z

j@= 0

:

(2

:

45) Note that

supp(2(r

r

p

)+

p

) O(34(1;

"

0))nO(34;3 8

"

0)

:

Short calculations give the inequality:

jdiv(

B

(^

vy

) +

B

(

y

^

v

))(

tx

)j

C

(jr

v

^(

x

)j+j

v

^(

x

)j)(jr

v

(

tx

)j+j

v

(

tx

)j) in

Q:

(2

:

46) By (2.44), (2.46), (2.39) we obtain from (2.45)

s

2Z

'

2

2

p

2

e

2s'

dx

s

2Z

nO(

3

4 (1;"

0 ))

'

2

p

2

e

2s'

dx

C

(

)(

Z

s'

1 (jr

v

j2+j

v

j2)

e

2s'

dx

+Z

!

0

s

2

'

2

p

2

e

2s'

dx

+ 1

s'

Z

O(

3

4

;"

0 )nO(

3

4 )

p

2

e

2s((T1+A;t)t)2

dx

) 8

s > s

0(

)

:

(2.47) On the other hand, by (2.32), (2.41) we have

s

2Z

O(

1

4 )

'

2

p

2(

e

2s'+

e

2s((T1+A;t)t)2)

dx

C

(

Z

@

s'p

2

e

2s'

d

+

Z

K

i=1 Gi

s'

j

p

j2

e

2s'

dx

)

:

(2

:

48) SinceKi=1

G

i nO(34(1;

"

0)) and O(34;

"

0)nO(34) O(14) inequalities

(2.47), (2.48) imply (2.41). tu

Now in right-hand side of (2.41), we have to estimate the integral on containing pressure

p:

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