ESAIM: Control, Optimisation and Calculus of Variations
URL: http://www.emath.fr/cocv/ November 1998, Vol. 3, 361{380
OPTIMAL CONTROL OF AN ILL-POSED ELLIPTIC SEMILINEAR EQUATION WITH AN EXPONENTIAL NON LINEARITY
E. CASAS, O. KAVIAN, AND J.-P. PUEL
Abstract. We study here an optimal control problem for a semilinear elliptic equation with an exponential nonlinearity, such that we cannot expect to have a solution of the state equation for any given control. We then have to speak of pairs (control, state). After having dened a suit- able functional class in which we look for solutions, we prove existence of an optimal pair for a large class of cost functions using a non standard compactness argument. Then, we derive a rst order optimality system assuming the optimal pair is slightly more regular.
1. Introduction
In this paper we are concerned with the optimal control of the following semilinear elliptic equation
;
y
= ey+u
iny
= 0 on ; (1.1)where Rn,
n >
2, is a bounded open set, ; being the boundary, which is assumed to be Lipschitz. The functionu
is the control, that will be taken in some spaceL
p(), andy
denotes the state in our control problem.The equation (1.1) appears in several contexts: we refer for instance to D.A. Franck-Kamenetskii 5] for combustion theory in chemical reactors, S. Chandrasekhar 3] in the study of stellar structures. The equation (1.1) is ill-posed in the sense that there is no solution for some controls
u
and many solutions can be found for some others (see for instance I.M. Gelfand 7], M.G. Crandall and P.H. Rabinowitz 4], F. Mignot and J.P. Puel 8, 9], Th. Gallouet, F. Mignot and J.P. Puel 6]). Because of the term ey, we needE. Casas: Dpt. Matem atica Aplicada Y Ciencias de la Computaci on, E.T.S.I.I y T., Univiersidad de Cantabria, Av. Los Castros s/n 39005 Santander, Spain. E-mail:
casas@etsiso.macc.unican.es.
O. Kavian: Universit e de Versailles Saint-Quentin et Centre de Math ema- tiques Appliqu ees, Ecole Polytechnique, 91128 Palaiseau Cedex, France. E-mail:
kavian@math.uvsq.fr.
J.-P. Puel: Universit e de Versailles Saint-Quentin et Centre de Math ema- tiques Appliqu ees, Ecole Polytechnique, 91128 Palaiseau Cedex, France. E-mail:
jppuel@cmapx.polytechnique.fr.
This research was partially supported by the European Union, under HCM Project number ERBCHRXCT940471. The rst author was also partially supported by Direcci on General de Investigaci on Cient ca y T ecnica (Spain).
Received by the journal October 28, 1997. Revised July 6, 1998. Accepted for publi- cation July 16, 1998.
c Soci et e de Math ematiques Appliqu ees et Industrielles. Typeset by LATEX.
to explain what we mean by a solution of (1.1). We will say that
y
is a solution of (1.1) if it belongs to the class of functionsY
=fy
2H
01() : ey 2L
1()g (1.2) and it satises the equation in the distribution sense. Then the optimal control problem will be formulated in the following terms(P)
8
>
<
>
:
Minimize
J
(yu
) :=Z
L
(xy
(x
))dx
+N p
Z
j
u
(x
)jpdx
(yu
)2Y
K
satises (1:
1)where
K
is a nonempty closed convex subset ofL
p(), 2p <
+1,N
0 andL
: R;!Ris a Caratheodory function of classC
1 with respect to the second variable and satisfying appropriate growth conditions which will be shown to be the following
@L @y
(xy
)
a
1(x
) +1(jy
;j1+ ey) (1.3)L
(xy
)a
2(x
);2(jy
;j2+ ey) (1.4) for somea
1,a
2 2L
1(), 1, 2 0, 1 =np=
(n
;2p
) ifp < n=
2, and 01<
+1 ifp
n=
2, and 12< p
. For instance,y
d2Y
being given, a typical functionalJ
would beJ
(yu
) := 12Z jy
(x
);y
d(x
)j2dx
+N p
Z
j
u
(x
)jpdx:
(1.5) We should emphasize on the fact that one of the main diculties of the problem is to choose an appropriate class of solutions such that (P) has a solution in that class.The plan of the paper is as follows. In Section 2 we will analyze the state equation and we will establish the necessary background to study the control problem. The existence of a solution for (P) is studied in Sections 3 and 4 for the cases
p >
2 andp
= 2, respectively. The casep
= 2 presents some diculties and we will be able to prove the existence of an optimal control under some additional assumption on the functionL
. We should note that, as it seems to us, the case 1p <
2 cannot be treated with the techniques we use in this paper. Finally in Section 5 the optimality conditions will be investigated.2. Analysis of the State Equation
We start this section by establishing that any solution of (1.1) in the sense dened in Section 1 is a solution in the variational sense in
H
01() this requires to prove some regularity of the term ey.Lemma 2.1. Let
y
2Y
be a solution of (1.1), then ey 2H
;1(), eyz
2L
1() for everyz
2H
01() andZ
r
y
(x
)rz
(x
)dx
=Z
ey(x)+
u
(x
)]z
(x
)dx:
(2.1)ESAIM: Cocv, November 1998, Vol. 3, 361{380
Proof. By the denition of a solution of (1.1), for all
z
2C
c1() we haveZ
r
y
(x
)rz
(x
)dx
=Z
ey(x)+
u
(x
)]z
(x
)dx:
(2.2) Givenz
2L
1()\H
01(), we can take a sequencefz
kg1k=1C
c1(), withk
z
kk1 kz
k1+ 1, for everyk
2 N andz
k !z
inH
01() andz
k* z
inL
1()-w. Then for allk
1 we can replacez
byz
k in (2.2) and pass to the limit whenk
! 1 to obtain that the identity in (2.2) is also true for anyz
2L
1()\H
01().Let us take now
z
2H
01() such thatz
0 and setT
k(z
)(x
) :=k
ifz
(x
)> k
z
(x
) if 0z
(x
)k:
Then
T
k(z
)2L
1()\H
01() andZ
r
y
(x
)rT
k(z
)(x
)dx
=Z
ey(x)+
u
(x
)]T
k(z
)(x
)dx
8k
2N:
SinceT
k(z
) !z
inH
01(), the only trouble to pass to the limit in this identity comes from the term eyz
k. AsT
k(z
)0 and ey>
0, then from the monotone convergence theorem, taking into account thatT
k(z
)(x
) "z
(x
) for almost allx
2, we deduceZ
ey(x)
z
(x
)dx
= limk!+1 Z
ey(x)
T
k(z
)(x
)dx
= limk
!+1 Z
r
y
(x
)rT
k(z
)(x
)dx
;Zu
(x
)T
k(z
)(x
)dx
=
Z
r
y
(x
)rz
(x
)dx
;Zu
(x
)z
(x
)dx <
+1:
Next, for a general
z
2H
01(), we notice thatz
=z
+;z
; withz
+z
; 2H
01(). This proves that (2.1) is satised. Moreover ey 2H
;1() and (2.1) shows that for allz
2H
01()hey
z
i=Z
ey(x)
z
(x
)dx:
Now we state a very important identity to derive some estimates, in terms of
u
, on the solution of (1.1).Theorem 2.2. Let
u
2L
2() and assume thaty
2H
2() is a solution of (1.1). Then for anyx
0 2Rnwe haven
2 ;1Z
jr
y
(x
)j2dx
+ 12Z;
jr
y
()j2()(;x
0)]d
=
n
Z (ey(x);1)dx
;Zu
(x
)(x
;x
0)ry
(x
)]dx
(2.3) where () denotes the outward unit normal vector to ; at the point.ESAIM: Cocv, November 1998, Vol. 3, 361{380
Proof. Since
y
2H
2(), then ey = ;y
;u
2L
2(). Therefore we can multiply the equation (1.1) by any function ofL
2() and make the inte- gration in . In particular we can take (x
;x
0)ry
(x
) 2L
2() as this function:Z
;
y
(x
;x
0)ry
] =Z
ey(
x
;x
0)ry
]dx
+Z
u
(x
;x
0)ry
]dx:
(2.4) Let us make an integration by parts in the rst integral
Z
;
y
(x
;x
0)ry
]=
Z
r
y
r(x
;x
0)ry
];Z
;
@
y
()(;x
0)ry
()]d
= Xn
ij=1
Z
@
xiy@
xi(x
j;x
0j)@
xjy
]dx
;Z;
@
y
()(;x
0)ry
()]d
= Xn
i=1
Z
j
@
xiy
j2dx
+ Xnij=1
Z
@
xiy
(x
j;x
0j)@
x2ixjy
]dx
; Z
;
@
y
()(;x
0)ry
()]d
= Xn
i=1
Z
j
@
xiy
j2dx
+ 12 Xnij=1
Z
(
x
j;x
0j)@
xj(@
xiy
)2dx
; Z
;
@
y
()(;x
0)ry
()]d:
Now we can integrate by parts in the last relation, taking into account that for
2 ; we have ry
() = jry
()j(), then (;x
0)ry
() =jr
y
()j()(;x
0) and@
y
() =ry
()() =jry
()j, we getZ
;
y
(x
;x
0)ry
] =Xni=1
Z
j
@
xiy
j2dx
+ 12ijXn=1; Z
(
@
xiy
)2+Z
;
(
x
j;x
0j)j(x
)(@
xiy
)2d
; Z
;
@
y
()(;x
0)ry
()]d
=1;
n
2n
X
i=1
Z
jr
y
j2dx
;12Z;
()(;x
0)]jry
()j2d:
(2.5)
In order to handle the rst term in the right hand side of (2.4), for
k
1 setT
k(y
)(x
) :=8
<
:
+
k
ify
(x
)>
+k y
(x
) if ;k
y
(x
)k
;
k
ify
(x
)<
;k:
ESAIM: Cocv, November 1998, Vol. 3, 361{380
One checks easily that
T
k(y
)!y
inH
01() and a.e., eTk(y)!eyinL
2() and a.e. On the other hand eTk(y)rT
k(y
)!eyry
inL
1() but as r(eTk(y)) = eTk(y)rT
k(y
), we conclude that eyry
=r(ey ;1) inL
1(). ThereforeZ
eyr
y
dx
=;Z
(ey;1)r
dx
for all 2(C
1(Rn))n, and we may write:Z
ey(
x
;x
0)ry
]dx
=Xnj=1
Z
(
x
j;x
0j)ey@
xjydx
=Xn
j=1
Z
(
x
j;x
0j)@
xjey;1]dx
;Xnj=1
Z
ey;1]
dx
=;n
Z ey;1]dx:
(2.6) Putting together (2.5) and (2.6) in (2.4), we get (2.3).
Lemma 2.3. Let us assume that
u
2L
2(), is star-shaped with respect to some interior pointx
0 (i.e.(;x
0)() 0 for 2;) andy
2H
01() satisesy
2L
2(). Then we haven
2 ;1 Z jr
y
(x
)j2dx
Zy
(x
;x
0)ry
]dx:
(2.7) Proof. Let us consider a sequence of bounded open sets 12 , with\1j=1j = , with ;j =@
j of classC
11 and such that all of them are star-shaped with respect tox
0. For eachj
we consider the problem;
z
= (ey+u
) in jz
= 0 on ;jwhere
is the characteristic function of . This linear problem has a unique solutiony
j 2H
01(j)\H
2(j). We extend eachy
j andy
to 1 by zero, thusy
j,y
2H
01(1). From the above equation we getZ
1
jr
y
jj2dx
=Z
j
jr
y
jj2dx
=Z
ey+
u
]y
jdx
key +u
kL2( )ky
jkL2( ) which proves the boundedness of fy
jg1j=1 inH
01(1). By taking a subse- quence we can assume thaty
j !y
~weakly inH
01(1) andy
j(x
) !y
~(x
) for almost allx
21. Obviously we have that ~y
(x
) = 0 forx
21n . Then we have that ~y
= 0 on ; and ;~y
= limj!1;y
j = ey +u
in . This leads to the equality ~y
=y
. MoreoverZ
1
jr
y
j2dx
liminfj!1 Z
1
jr
y
jj2dx
limsupj!1 Z
1
jr
y
jj2dx
= limsupj
!1 Z
j
jr
y
jj2dx
= limj!1 Z
j
ey +
u
]y
jdx
=
Z
ey+
u
]ydx
=Z
jr
y
j2dx
=Z
1
jr
y
j2dx
hence limj!1ky
jkH01( 1)=ky
kH01(
1
), which proves the strong convergence
y
j !y
inH
01(1).ESAIM: Cocv, November 1998, Vol. 3, 361{380
Now arguing again as in the proof of Theorem 2.2 and using the fact that (
;x
0)j()0 for every 2;j (here j is the outward normal to ;j), we deduceZ(;
y
)(x
;x
0)ry
]dx
= limj!1 Z
(;
y
j)(x
;x
0)ry
j]dx
= limj
!1 Z
j
(;
y
j)(x
;x
0)ry
j]dx
= limj
!1 (
1;
n
2Z
j
jr
y
jj2dx
;12Z;
j
(
;x
0)j()]jry
jj2d
)
jlim
!1
1;
n
2Z
jr
y
jj2dx
=1;n
2Z
jr
y
j2dx:
Remark.
Inequality (2.7) makes sense when we assume only thaty
2Y
andu
2L
2() but we do not know whether it holds under these assump- tions. Actually (2.7) is true for solutions (yu
)2Y
K
such that there is a sequence of solutions (y
ku
k) 2Y
K
withy
k 2H
2() satisfying (1.1),u
k !u
strongly inL
2() (indeed this implies thaty
k* y
inH
01()-weak and eyk !ey inL
1()).Corollary 2.4. Let us assume that
u
2L
2(), is star-shaped with re- spect to some interior pointx
0 andy
2Y
is a solution of (1.1) such that ey 2L
2(). Then we haven
2 ;1Z
jr
y
(x
)j2dx
n
Z (ey(x);1)dx
;Zu
(x
)(x
;x
0)ry
(x
)]dx:
(2.8) Proof. If
y
2H
2() this inequality is an immediate consequence of (2.3).Indeed it is enough to note that
()(;x
0)0 for almost every 2; because is assumed to be star-shaped with respect tox
0. Since we have not assumed ; to be of classC
11 or to be convex, we cannot deduce theH
2()-regularity ofy
from the fact that ey +u
2L
2(). However (2.6) is still valid, therefore the result follows from the previous lemma.As a consequence of this corollary, we deduce some estimates for
y
in terms ofu
.Theorem 2.5. Let us assume that
y
is a solution of (1.1) that satises the inequality (2.8). Then there exist positive constantsC
i, 1i
4, independent ofu
andy
such thatkey
y
kL1( )C
1ku
k2L2( )+C
2 (2.9)k
y
kH01 ( )C
3ku
kL2( )+C
4:
(2.10) Proof. Thanks to Theorem 2.1 we know thatZ
jr
y
(x
)j2dx
=Z
ey(x)
y
(x
)dx
+Z
u
(x
)y
(x
)dx:
(2.11)ESAIM: Cocv, November 1998, Vol. 3, 361{380
Combining this identity with (2.8) we deduce
n
2 ;1Z
ey
ydx
+n
2 ;1Z
uydx
n
Z ey;1]dx
;Zu
(x
;x
0]ry
]dx:
Which yields
Z
ey
ydx
2n n
;2Z
ey;1]
dx
+c
1ku
kL2( )ky
kH01( )
:
(2.12) Let us setn=
x
2 :y
(x
)>
4n n
;2
:
Then from (2.12) we getZ
ey
ydx
12Zn
ey
ydx
+Z
n n
en;24n
dx
+c
1ku
kL2( )ky
kH01 ( ) therefore12
Z
ey
ydx
c
2;12Zn n
ey
ydx
+c
1ku
kL2( )ky
kH01 ( ):
Taking into account that etj
t
jc
(n
)<
+1 for any real numbert
n4;2n , we get from the previous inequalityZ
eyj
y
jdx
c
3+c
1ku
kL2( )ky
kH01( )
:
(2.13)Using this inequality in (2.11), we obtain (2.10). Finally using (2.10) in (2.13) we get the estimate (2.9).
We nish this section by proving two propositions that will be very useful in the next sections. Here we will use the notation
y
+ = maxfy
0g andy
;= maxf;y
0g.Proposition 2.6. Let
y
be a solution of (1.1) corresponding to a functionu
2L
p(). Theny
+2L
r() for all 1r <
+1y
; 2L
q() withq
=8
<
:
+1 if
p > n=
2<
+1 ifp
=n=
2np=
(n
;2p
) ifp < n=
2:
Moreoverk
y
;kLq( )C
pqku
;kLp( )for some constantC
pq>
0 independent ofu
andy
if we denote byr
] the integer part ofr
andk
:=r
] + 1 thenk
y
+kLr( )(k
!)1=r(C
1ku
kL2( )+C
2)1=r:
Proof. Let us take
k
=r
] + 1. Then jy
+jr< k
!ey 2L
1(), which proves thaty
+ 2L
r(). Using the estimate R eydx
C
1ku
k2L2( )+C
2 one gets the last estimate of the proposition.Now assume that
p < n=
2 (the casep
n=
2 may be treated in the same way). Consider 2H
01() satisfying;=;u
;. We have;
y
;u
;=;ESAIM: Cocv, November 1998, Vol. 3, 361{380
and by the maximum principle we conclude that
y
in . As<
0, we conclude that 0y
; ; =jj. Now recall that by a classical result of G. Stampacchia 12] (theorems 4.1 and 4.2), if for somes
2 (ands < n
) one has ;z
=T
2W
;1s() andz
2H
01(), thenk
z
kLs( )C
(ns
)(kT
kW;1s( )+kz
kL2( ))s
1 = 1s
;n
1 Here we have ;=;u
; 2L
p() andL
p()W
;1s() if1
s
= 1p
;n:
1Consequently one sees that the estimate of the proposition on k
y
;kLq( )holds if
q
=s
, which means1
q
= 1p
;n:
2Proposition 2.7. Let
u
andy
be as in Proposition 2.6 and satisfying the inequality (2.8). Theny
2L
p() andk
y
kLp( )C
1pku
kLp( )+C
2p for some constantsC
ip>
0 independent ofu
andy
.Proof. Let us take
k
=p
] + 1. From (2.9), the inequality et jt
jet+ 1 for everyt
2Rand taking into account thatp
2, we deducek
y
+kLp( )
k
!Z
ey
dx
p
c
1ku
k2L2( )+c
2 1=pc
3ku
k2L=p2( )+c
4c
3;ku
kL2( )+ 1+c
4c
5ku
kLp( )+c
6:
On the other hand, from Proposition 2.6 it follows that
y
; 2L
p().Indeed it is enough to note that
np=
(n
;2p
)> p
ifp < n=
2. Moreoverk
y
;kLp( )C
pku
;kLp( ):
Now it is enough to write
y
=y
+;y
; and to use the two inequalities obtained above to achieve the desired result.3. Existence of an Optimal Control. Case p>2 The aim of this section is to study the existence of a solution for the optimal control problem. As usual, to prove the existence of such a solution, we take a minimizing sequence f(
y
ku
k)g1k=1 of feasible elements. Assuming that eitherK
is bounded inL
p() orN >
0, we can deduce that fu
kg1k=1is bounded. The dicult part is to deduce that f
y
kg1k=1 is bounded inH
01(). If the elements (y
ku
k) satisfy the inequality (2.8), then Theorem 2.5 provides the necessary inequalities to deduce the boundedness offy
kg1k=1. Unfortunately (2.8) has been proved to hold only for solutions of the state equation with ey 2L
2(). This leads us to consider the following class of states: we denote by Y the subset ofY
formed by the functions which satisfy (1.1) and (2.8) for some controlu
(note that forq > n=
2 one hasESAIM: Cocv, November 1998, Vol. 3, 361{380
W
2q()\H
01() Y). Not every element of Y needs to be a solution of (1.1) with ey 2L
2(). In fact we have the following result.Theorem 3.1. Let us assume thatf(
y
ku
k)g1k=1YL
p() is a sequence of functions satisfying (1.1) and converging weakly to some element (yu
) inH
01()L
p(), withp >
2. Then eyk ! ey inL
1(),y
2 Y and (yu
) satises also (1.1). The same result holds ify
k* y
inH
01() andu
k !u
strongly inL
2().Proof. The weak convergence
y
k* y
inH
01() implies the strong conver- gencey
k !y
inL
2(). Then taking a subsequence if necessary, we can assume thaty
k(x
) !y
(x
) for almost every pointx
2 . In particular eyk(x)!ey(x) almost everywhere in . Let us use Vitali's theorem to prove the convergence eyk !ey inL
1(). First let us note that the boundedness of fu
kg1k=1 inL
2() along with (2.9) imply thatkeyky
kkL1( )C
for some constantC <
+1 and allk
2N. Now given>
0, let us takem >
0 such thatC=m < =
2 and ==
(2em). Then for every measurable setE
, with meas(E
)<
, we haveZ
Eeyk(x)
dx
m
1Z
fx2E:yk(x)>mgeyk(x)
y
k(x
)dx
+Z
fx2E:yk(x)mgem
dx
m
1Z
eyk(x)j
y
k(x
)jdx
+ emmeas(E
)C
m
+ em<
8k
2N which allows to conclude the desired convergence. Now it is easy to pass to the limit in the state equation satised by (y
ku
k) and to conclude that (uy
) satises (1.1).Let us prove that (
yu
) satises (2.8). First of all we will prove that there exists a subsequence, that we will denote in the same way, such thatr
y
k(x
) ! ry
(x
) for almost all pointx
2 . To achieve this aim, we remark that eyk ! ey inL
1() while (u
k)k is bounded inL
p(): therefore (y
k)k is bounded inL
1(). Now by a result due to L. Boccardo and F. Murat 2] (theorem 2.1) one may conclude that there exists a subsequence (still denoted by) (y
k)k such thaty
k* y
inH
1() weakly andry
k !ry
almost everywhere.Now the weak convergencer
y
k*
ry
in ;L
2()n along with the point- wise convergence implies the strong convergence in (L
r())n for allr <
2 (we use here the fact that the weak convergence inL
2() implies that the sequence jry
k ;ry
jr is equi-integrable and we apply Vitali's theorem).In particular the strong convergence of r
y
k ! ry
holds inL
p0(), with (1=p
) + (1=p
0) = 1 (this is the only place where we need the assumptionp >
2). Then we can pass to the limit in the inequality (2.8) to obtain:n
2 ;1Z
jr
y
j2dx
liminfk!1
n
2 ;1
Z
jr
y
kj2dx
liminfk
!1
n
Z eyk ;1]dx
;Zu
k(x
;x
0)ry
kdx
=
n
Z ey;1]dx
;Zu
(x
;x
0)rydx
ESAIM: Cocv, November 1998, Vol. 3, 361{380
which concludes the proof when (
y
ku
k)*
(yu
) inH
01()L
p() andp >
2. On the other hand it is clear that whenp
= 2 andu
k !u
strongly inL
2(), in the above inequalities one hasZ
u
k(x
;x
0)ry
kdx
!Zu
(x
;x
0)rydx
ask
!1.Now we reformulate the control problem as follows (P)
8
>
<
>
:
Minimize
J
(yu
) :=Z
L
(xy
(x
))dx
+N p
Z
j
u
(x
)jpdx
(yu
)2YK
satises (1:
1):
The fact of taking (
yu
) 2 YK
imposes a restriction on the class of solutions of (1.1), but it is not restrictive with respect to the controls. More precisely, we have the following resultProposition 3.2. Let us assume that is star-shaped with respect to some point
x
0 2 and that (1.1) has a solution for some controlu
2L
p(),p
2. Then there exists a solutionz
of (1.1) corresponding to the same controlu
and belonging to the classY.Before proving this proposition, we state and prove the following lemma:
Lemma 3.3. For
k
1 andt
2Rlet us denotef
k(t
) := minfeketg. Then there isz
k 2H
01() such that;
z
=f
k(z
) +u
inz
k = 0 on;:
(3.1)Moreover
z
kz
k+1 and the sequencefz
kg1k=1 is bounded inH
01().Proof. We denote with
y
a solution of (1.1) associated to the controlu
. Let us takey
02H
01() such that;y
0 =u
. Then we have the following three inequalities:;
y
0f
k(y
0) +u
;y
f
k(y
) +u
;(y
;y
0) = ey>
0:
From the last relation we deduce thaty
0y
in . From the two rst re- lations it follows thaty
0 is a subsolution andy
is a supersolution of (3.1).Combining all these, by the now classical techniques introduced by D.H. Sat- tinger 10] (see also H. Amann 1]) we get the existence of a solution
z
k of (3.1) such thaty
0z
ky
. Moreover we havez
kz
k+1 andZ
jr
z
kj2dx
=Z
f
k(z
k) +u
]z
kdx
Z ezk+ju
j]jz
kjdx
Z
ey+j
u
j]jz
kjdx
key+ju
jkH;1( )kz
kkH01( ) which implies that fz
kg1k=1 is bounded inH
01().Proof of Proposition 3.2. The sequence f
z
kg1k=1 being given by Lemma 3.3, taking a subsequence, denoted in the same way, we infer that there exists an elementz
2H
01() such thatz
k !z
weakly inH
01()z
k(x
)!z
(x
) a.e.x
2:
(3.2)ESAIM: Cocv, November 1998, Vol. 3, 361{380