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
= 23) with boundary@
2C
1. More precisely, the problem under study is as follows.Let us consider the nonstationary Navier-Stokes equations
@
tv
(tx
);v
(tx
)+ (v
r)v
+rp
=f
(x
) +!u
in divv
= 0 (1:
1) with initial and boundary conditionsv
j = 0v
jt=0 =v
0(x
) (1:
2) wherev
(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 forx
2!
0 forx
2n!:
Let (^
v
(x
)^p
(x
)) be a steady-state solution of the Navier-Stokes equations;^
v
+ (^v
r)^v
+rp
^=f
(x
) in div^v
= 0v
^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.
close enough to the initial condition
k
v
0;v
^kV1()"
(1:
4) where the parameter"
is suciently small. We want to nd a controlu
such that for givenT >
0, the following equality holds:v
(Tx
) = ^v
(x
):
(1:
5) We assumeCondition 1.1. The boundary
@
= Ki=1;i 2C
1 (;i\;j = fg for alli
6=j
) where;i is an
;1-dimensional connected manifold of classC
1:
For each ;i, there exists a neighborhood Ai Rn and a dieomorphismi
2
C
1(AiRn) such that i(;i) =S
1n.*The main result of this paper is the following Theorem.
Theorem 1.2. Let
v
0 2V
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 thatsuppv
^ . Then for suciently small"
there exists a solution (vpu
)2V
12(Q
)L
2(0T 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@
. Letv
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 pointv
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:
@
tv
(tx
);v
(tx
) =rp
+f
(tx
) +!u
in divv
= 0 (1:
6)v
j = 0v
jt=0=v
0(x
)v
jt=T 0:
(1:
7) We haveTheorem 1.3. Let
v
0 2V
1()f
2L
2(0T V
0()) and there exists" >
0 such thatRQjf
j2e
(T;t)12+"dxdt <
1:
Then there exists a solution (vpu
)2V
12(Q
)L
2(0T 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
rB
r =fx
2Rnjx
j< r
gpoint 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 pressurep:
Then in Theorem 3.1, we obtain a Carleman estimate for the velocityy
of the uid via a weightedL
2-norm of the density of external forcesf
and the pressurep:
Moreover, for the pressure, by Theorem 2.11, one can choose a weightedL
2 norm over (0T
)!:
And nally in Theorem 3.6, we prove an estimate (not of Carleman type) for the velocity wherep
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] whenn
= 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
) =rp
+f
inQ
(2:
1)div
y
= 0y
j@= 0y
(0x
) =y
0(x
) (2:
2) where Rnis a bounded domain with@ C
1Q
= (0T
) and the operatorsB
(^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)Denote
Q
! = (0T
)!
= (0T
)@ :
Let be the outward unit normal to@ :
In this paper we use the following functional spaces. Recall thatW
pk()k
0 1p <
1 is the Sobolev space of functions with nite normk
u
kWpk() = (X
jjk Z
@
jju
(x
)/@x
11:::@x
nnpdx
)1=p where= (12:::
n)jj=1++nW
12(Q
) =fw
(tx
)jw
2L
2(0T W
22())@w @t
2L
2(0T L
2())gV
1() =fv
(x
) = (v
1:::v
n) 2 (W
21())ndivv
= 0v
j@= 0gV
0() =fv
(x
) = (v
1:::v
n) 2 (L
2())ndivv
= 0 (v
)j@= 0gV
;1() = (V
1())V
12(Q
) =fv
(tx
) 2 (W
12(Q
))ndivv
= 0v
j@ = 0gL
2(Q
) =fv
(tx
)Z
Q
v
2dxdt <
1g:
We haveProposition 2.1. (27]) Let be a bounded domain inRnwith
@
2C
2:
Then (L
2())n=V
0()(V
0())?where
(
V
0())?=fv
(x
) = (v
1:::v
n) 2 (L
2())nv
=rpp
2W
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 changex
!x=
jx
j2:
)Set Ui = i;1(f
x
2 Rnj1;" <
jx
j<
1g) where"
2 (01):
For all suciently small"
, the setUi is correctly dened andU
i
\U
j =fg for all
i
6=j @
Ui= ;ii (2:
4) wherei= i;1(S
1;"n ) and(
!
supp ^v
)\Ui=fg8i
= 1:::K:
(2:
5) LetG
Rnbe a domain which satises the following condition:Condition 2.2. The domain
G
is dieomorphic to the cylinder;0T
0] whereT
0>
0 is a number and ; Rn is a closed (n
;1)-dimensional manifold of classC
1:
This condition implies immediately that
@G
2C
1 and@G
= 0 1 wherei is a (n
;1)-dimensional connected manifold of classC
1:
Let
w
(x
) be a harmonic function inG
:w
= 0 inG
(2:
6)such that
@w
@
j1 = 0 (2:
7)Z
0
j
w
j2d
(2M
)2 Z
1
j
w
j2d
(2"
)2:
(2:
8) Let (x
)2C
1(G
) be a function satisfying the conditions0
< C
1jr(x
)j8x
2G
j1 = 0 j0 = 1:
(2:
9) Then the set t=fx
2G
(x
) =t
gt
201] (2:
10) is a smooth manifold dieomorphic to0 and1. We have:Theorem 2.3.(28]) There exist a constant
C
2>
0 and a function (x
) 2C
1(G
), satisfying condition (2.9), such that for any functionw
2W
212(G
) for which (2.6)-(2.8) are satisedZ
t
j
w
j2d
C
2kw
k2(1;t)L2(1 )k
w
k2tL2(0)C
2(2"
)2(1;t)(2M
)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
) =fx
2Uii(x
) =t
g:
(2:
12) For allr
2(01) we introduce the auxiliary domainsO
i(
r
) =fx
i2Uijr <
i(x
)<
1g O(r
) =Ki=1Oi(r
):
(2:
13) Letw
0bw
be an arbitrary subdomain. We have:Lemma 2.4. There exists a function
(x
)2C
1() such that (x
) = 1;i(x
) 8x
2Ui (x
)>
0 in jr(x
)j>
08x
2n!
0:
(2:
14) Proof. First we construct an auxiliary function (x
)2C
1() such that (x
) = 1;i(x
) 8x
2Ui (x
)>
0 in:
(2:
15)To do this, we consider the sequence of domains ~Ui Rn
i
=f1:::K
g with the following properties@
U~i= ;i~i2C
1 Ui U~ii U~i U~i\U~j =fgfori
6=j
where ~iis a connected (n
;1)-dimensionalC
1 manifold. (For example we can choose ~Ui as ~Ui = i;1(fx
2Rnj1;" <
~ jx
j<
1g) for some ~":
)Since by assumption
iis aC
1surface, one can extend the functioni(x
) to a smooth function ofC
1(~Ui) such thati0 in some neighborhood of~
i:
Set ~
(x
)jU~i 1;i(x
) and ~0 in nKi=1U~i:
Let
(x
)2C
01(nKi=1Ui) be a nonnegative function such that (x
)>
0 8x
2fx
2nKi=1Uij~(x
) = 0g:
Then the function
(x
) = ~ +s
satises (2.15) for alls >
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 8x
2Ki=1Ui there exists a sequence of domains ~~Ui Rnsuch that@
U~~i= ;i~~i2C
1 Ui U~~i U~ii U~~i U~~i\U~~j =fgfori
6=j
where ~~i is a connected (n
;1)-dimensionalC
1 manifold, andjr
(x
)j>
0 8x
2Ki=1U~~i:
Let 2C
01(nKi=1Ui) such that (x
) = 18x
2nKi=1U~~i (x
) = 0 8x
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
) 8x
2Ui and for all suciently small" >
0 "(x
)>
0 in:
Let us show that for all small
"
, " is a Morse function. Actually, inK
i=1 U
i function
" has no critical points. In nKi=1U~~i," coincides with the Morse function ". Short calculations give the inequalityjr
"(x
)j=jr(x
);r(x
)(;")(x
);(x
)(r;r")(x
)jC
;2kkC1()k;"kC2()C
;2kkC1( )"
where
C >
0x
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 dieomorphismr
: ! such thatr
(x
) =x
8x
2Ki=1Uir
;1(M)!
0:
Thus, the function
(x
) = "(r
(x
)) satises all the conditions of ourlemma. 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 andx
0 2@ :
By (2.9), (2.14) the functions'
are independent of the selection ofx
0 2@ :
Note that (2.9), (2.14) imply the obvious inequality0
>
(tx
)(t
)'
(tx
)'
(t
) 8(tx
)2Q:
Let us introduce a function
`
(t
)2C
10T
] by the formula`
(t
)>
0t
2(0T
)`
(t
) = 1t
2(014T
)t t
2(34TT
):
(2:
18) In this section, our aim is to get an estimate for the functionp
using the trace ofp
on@
and the restriction ofp
on 0T
]!
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
() = max2 1
4 1]
e
(1;);11;
:
(2:
21)By (2.9), (2.14) there exists
0>
1 such that1 +
A
()(x
)e
(x) 8>
0x
2O(14)e
= (e
(x))ji>
1 +A
()(x
)> e
(x)8
>
0x
2O(12)nO(34):
(2
:
22)We have:
Lemma 2.5. Let the function
w
2W
212(Ui) be a solution of problem (2.19), (2.20). Then there exist ^ 2(01) and ^>
1 such that for>
^((
T
;s t
)t
)2Z
O
i (
1
4 )
j
w
j2e
2s((T1+ A;t)t)2dx
C
4(Z
@
j
w
j2e
2s'd
+Z
i
j
w
j2e
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
j2d
(2"
)2=Z
i
j
w
j2d:
Let
>
1 and (i
) be the manifold dened by (2.12). Note that, by (2.14)A
= max2 1
4 1]
e
(1;);11;
= maxx2Oi(14)e
(x);1 (x
)e
34=(e
ji)34:
(2:
24) Using inequality (2.11), we obtainZ
Oi( 1
4 )
j
w
j2e
2sAdx
C
6Z 11
4 Z
j
w
j2e
2sA(1;)dd
C
7Z 11
4
k
e
sAw
k2(1;L2(i) )k
w
k2L2(;i)d
C
7Z 10
k
e
sAw
k2(1;)L2(i )k
w
k2L2(;i)d
C
8Z 10
(
"e
sA)2(1;)M
2d
=I:
(2.25) Short calculations give the equalityI
=M
2 2"
"e
sAM
2
;1
#
=ln
"e
sAM
:
(2:
26) Obviously, by (2.24) there exist ^ 2 (01) and ^>
1 such that for all> >
^ 1A
+ 1e
34+ 1^minx2
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
sAM
) =s
;ln
("e
s(A+1)M
)s:
(2:
29)Inequalities (2.26), (2.28) and (2.29) imply
I
M
22
s :
By this inequality and (2.25), keeping in mind that
e
j@= 1we obtains
ZO
i (
1
4 )
j
w
j2e
2s(1+A )dx
C
10Z;
i
j
w
j2e
2sed:
(2:
30) B) Let"e
s(A+1)M
1:
Then by (2.25), (2.27)I
C
11Z 10
"
2e
2sAe
2sd
C
11"
2e
2s(e34+1)C
12Z
i
w
2e
2s( e^ ;1)d
where> >
^ 1:
From this inequality, increasing the parameter ^ 2(01) if necessary, we obtains
ZO
i (
1
4 )
j
w
j2e
2s(1+A )dx
C
13Z
i
j
w
j2e
2s e^d:
(2:
31) Inequalities (2.30), (2.31) after the changes
!s=
(t
(T
;t
))2 imply(2.23). tu
We have:
Lemma 2.6. Let
p
2W
212(Ui) be a harmonic function in Ui. Then there exists ^>
1 such that for>
^there existss
0()>
0 such that((
T
;s t
)t
)2Z
O(
1
4 )
j
p
j2e
2s((T(1+A );t)t)2dx
C
;Z@
j
p
j2e
2s'd
+Z
i
(jr
p
j2+p
2)e
2s 'd
8s
s
0() (2:
32) for some 2(01):
Proof. Let ^
be dened in Lemma 2.5 and ^ be the maximum of the cor- responding parameter from Lemma 2.5 and0 from (2.22). We are looking for a functionp
of the form:p
=z
1 +z
2z
1 = 0 in Uiz
1j;i = 0@z @~n
(1i
)ji =@p
@~n
(i
)ji (2:
33) andz
2 = 0 in Uiz
2j;i =p @z @~n
(2i
)ji = 0 (2:
34)where
~n
(i
) is a unit outward normal toUi:
Note that'
ji =e
((
T
;t
)t
)2 8i
=f1:::K
g:
Also by (2.22) there exists 2(^1)>
34 such that1 +
A
maxx2Oi(14 )(x
)((
T
;t
)t
)2< e
((
T
;t
)t
)2 ='
ji:
From this inequality, (2.14) and (2.33) we obtainZ
i
j
z
1j2e
2s 'd
+s
((T
;t
)t
)2Z
O
i (
1
4 )
j
z
1j2e
2s((T(1+A );t)t)2dx
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
)2Z
O
i (
1
4 )
j
z
2j2e
2s((T1+A;t)t)2dx
C
13(Z
@
j
p
j2e
2s'd
+Z
i
j
z
2j2e
2s 'd
)C
14(Z
@
j
p
j2e
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
)2Z
O
i (
1
4 )
j
z
2j2e
2s((T1+A;t)t)2dx
C
15(Z
@
j
p
j2e
2s'd
+Z
i
jr
p
j2e
2s 'd
)C
15(Z
j
p
j2e
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
2z
j@ = 0:
(2:
38) We have:Lemma 2.7. Let
f
(x
) 2L
2(). There exists ^>
1 such that for>
^, there existss
0()>
0 such that for anys > s
0 the solutionz
(x
) 2W
22() of problem (2.38) satises the estimateZ
0
@ 1
s'
n
X
ij=1
@
2z
@x
i@x
j
2+
s
2'
jrz
j2+s
34'
3z
21
A
e
2sdx
C
16(Z
f
2e
2sdx
+Z
!0
s
34'
3z
2e
2sdx
) (2:
39)where
C
1>
0 does not depend onst
.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 obtainp
=div(B
(^vy
) +B
(y v
^)) in (2:
40) for a.e.t
20T
].The following lemma gives an estimate of the
L
2-norm of the pressurep
. Lemma 2.8. Letf
2L
2(0T V
1()) andp
(t
) 2L
2() satises (2.40).There exists ^
>
1 such that for all>
^, there existss
0() such thatZ
s
22'
2p
2e
2sdx
C
10()Z
!
0
s
22'
2p
2e
2sdx
+Z
@
s'p
2e
2sd
+Z
jr
y
j2s
2'
+s
2'
jy
j2e
2sdx
8s > s
0() (2.41) where the constantC
10 is independent ofs
andt
.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 alli
2f1:::K
g:
So there exists a neighborhood of
i, a domainG
isuch thatp
is the harmonic function inG
iand Ki=1G
i nKi=1Oi(18) =fg:
Note also that'
ji =e
((
T
;t
)t
)2 8i
=f1:::K
g:
Thus, by the properties of interior regularity of solutions of elliptic equations, there exists a constant
C
such thatK
X
i=1 Z
i
jr
p
j2e
2s 'd
C
ZK
i=1 G
i
j
p
j2e
2s'dx
(2:
42) where2(01) 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(01001 ) such that1 +
A
infx2O(
3
4
;"0)nO(
3
4 )
(x
)>
supx2O(
3
4
;"
0 )nO(
3
4 )
e
:
(2:
43)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
ZO(
3
4
;"0)nO(
3
4 )
p
2e
2s((T1+A;t)t)2dx
8s >
1:
(2:
44) Let (x
) be a function such that (x
) = 18x
2nO(34(1;"
0)) 0 in jO(34; 3
8
"0) 0
:
Then by (2.40) the functionz
=p
satises the equationz
= 2(rrp
)+p
+div(B
(^vy
)+B
(y
^v
)) inz
j@= 0:
(2:
45) Note thatsupp(2(r
rp
)+p
) O(34(1;"
0))nO(34;3 8"
0):
Short calculations give the inequality:jdiv(
B
(^vy
) +B
(y
^v
))(tx
)jC
(jrv
^(x
)j+jv
^(x
)j)(jrv
(tx
)j+jv
(tx
)j) inQ:
(2:
46) By (2.44), (2.46), (2.39) we obtain from (2.45)s
2Z
'
22p
2e
2s'dx
s
2ZnO(
3
4 (1;"
0 ))
'
2p
2e
2s'dx
C
()(Z
s'
1 (jrv
j2+jv
j2)e
2s'dx
+Z!
0
s
2'
2p
2e
2s'dx
+ 1s'
Z
O(
3
4
;"
0 )nO(
3
4 )
p
2e
2s((T1+A;t)t)2dx
) 8s > s
0():
(2.47) On the other hand, by (2.32), (2.41) we haves
2ZO(
1
4 )
'
2p
2(e
2s'+e
2s((T1+A;t)t)2)dx
C
(Z
@
s'p
2e
2s'd
+Z
K
i=1 Gi
s'
jp
j2e
2s'dx
):
(2:
48) SinceKi=1G
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