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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

in

y

= 0 on ;

(1.1)

where Rn,

n >

2, is a bounded open set, ; being the boundary, which is assumed to be Lipschitz. The function

u

is the control, that will be taken in some space

L

p(), and

y

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 need

E. 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.

(2)

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 functions

Y

=f

y

2

H

01() : ey 2

L

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

)jp

dx

(

yu

)2

Y

K

satises (1

:

1)

where

K

is a nonempty closed convex subset of

L

p(), 2

p <

+1,

N

0 and

L

: R;!Ris a Caratheodory function of class

C

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(j

y

;j1+ ey)

(1.3)

L

(

xy

)

a

2(

x

);

2(j

y

;j2+ ey)

(1.4) for some

a

1,

a

2 2

L

1(),

1,

2 0,

1 =

np=

(

n

;2

p

) if

p < n=

2, and 0

1

<

+1 if

p

n=

2, and 1

2

< p

. For instance,

y

d2

Y

being given, a typical functional

J

would be

J

(

yu

) := 12Z j

y

(

x

);

y

d(

x

)j2

dx

+

N p

Z

j

u

(

x

)jp

dx:

(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 and

p

= 2, respectively. The case

p

= 2 presents some diculties and we will be able to prove the existence of an optimal control under some additional assumption on the function

L

. We should note that, as it seems to us, the case 1

p <

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

2

Y

be a solution of (1.1), then ey 2

H

;1(), ey

z

2

L

1() for every

z

2

H

01() and

Z

r

y

(

x

)r

z

(

x

)

dx

=

Z

ey(x)+

u

(

x

)]

z

(

x

)

dx:

(2.1)

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(3)

Proof. By the denition of a solution of (1.1), for all

z

2

C

c1() we have

Z

r

y

(

x

)r

z

(

x

)

dx

=

Z

ey(x)+

u

(

x

)]

z

(

x

)

dx:

(2.2) Given

z

2

L

1()\

H

01(), we can take a sequencef

z

kg1k=1

C

c1(), with

k

z

kk1 k

z

k1+ 1, for every

k

2 N and

z

k !

z

in

H

01() and

z

k

* z

in

L

1()-w. Then for all

k

1 we can replace

z

by

z

k in (2.2) and pass to the limit when

k

! 1 to obtain that the identity in (2.2) is also true for any

z

2

L

1()\

H

01().

Let us take now

z

2

H

01() such that

z

0 and set

T

k(

z

)(

x

) :=

k

if

z

(

x

)

> k

z

(

x

) if 0

z

(

x

)

k:

Then

T

k(

z

)2

L

1()\

H

01() and

Z

r

y

(

x

)r

T

k(

z

)(

x

)

dx

=

Z

ey(x)+

u

(

x

)]

T

k(

z

)(

x

)

dx

8

k

2N

:

Since

T

k(

z

) !

z

in

H

01(), the only trouble to pass to the limit in this identity comes from the term ey

z

k. As

T

k(

z

)0 and ey

>

0, then from the monotone convergence theorem, taking into account that

T

k(

z

)(

x

) "

z

(

x

) for almost all

x

2, we deduce

Z

ey(x)

z

(

x

)

dx

= limk

!+1 Z

ey(x)

T

k(

z

)(

x

)

dx

= limk

!+1 Z

r

y

(

x

)r

T

k(

z

)(

x

)

dx

;Z

u

(

x

)

T

k(

z

)(

x

)

dx

=

Z

r

y

(

x

)r

z

(

x

)

dx

;Z

u

(

x

)

z

(

x

)

dx <

+1

:

Next, for a general

z

2

H

01(), we notice that

z

=

z

+;

z

; with

z

+

z

; 2

H

01(). This proves that (2.1) is satised. Moreover ey 2

H

;1() and (2.1) shows that for all

z

2

H

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

2

L

2() and assume that

y

2

H

2() is a solution of (1.1). Then for any

x

0 2Rnwe have

n

2 ;1

Z

jr

y

(

x

)j2

dx

+ 12Z

;

jr

y

(

)j2

(

)(

;

x

0)]

d

=

n

Z (ey(x);1)

dx

;Z

u

(

x

)(

x

;

x

0)r

y

(

x

)]

dx

(2.3) where

(

) denotes the outward unit normal vector to ; at the point

.

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(4)

Proof. Since

y

2

H

2(), then ey = ;

y

;

u

2

L

2(). Therefore we can multiply the equation (1.1) by any function of

L

2() and make the inte- gration in . In particular we can take (

x

;

x

0)r

y

(

x

) 2

L

2() as this function:

Z

;

y

(

x

;

x

0)r

y

] =

Z

ey(

x

;

x

0)r

y

]

dx

+

Z

u

(

x

;

x

0)r

y

]

dx:

(2.4) Let us make an integration by parts in the rst integral

Z

;

y

(

x

;

x

0)r

y

]

=

Z

r

y

r(

x

;

x

0)r

y

];

Z

;

@

y

(

)(

;

x

0)r

y

(

)]

d

= Xn

ij=1

Z

@

xi

y@

xi(

x

j;

x

0j)

@

xj

y

]

dx

;Z

;

@

y

(

)(

;

x

0)r

y

(

)]

d

= Xn

i=1

Z

j

@

xi

y

j2

dx

+ Xn

ij=1

Z

@

xi

y

(

x

j;

x

0j)

@

x2ixj

y

]

dx

; Z

;

@

y

(

)(

;

x

0)r

y

(

)]

d

= Xn

i=1

Z

j

@

xi

y

j2

dx

+ 12 Xn

ij=1

Z

(

x

j;

x

0j)

@

xj(

@

xi

y

)2

dx

; Z

;

@

y

(

)(

;

x

0)r

y

(

)]

d:

Now we can integrate by parts in the last relation, taking into account that for

2 ; we have r

y

(

) = jr

y

(

)j

(

), then (

;

x

0)r

y

(

) =

jr

y

(

)j

(

)(

;

x

0) and

@

y

(

) =r

y

(

)

(

) =jr

y

(

)j, we get

Z

;

y

(

x

;

x

0)r

y

] =Xn

i=1

Z

j

@

xi

y

j2

dx

+ 12ijXn=1

; Z

(

@

xi

y

)2+

Z

;

(

x

j;

x

0j)

j(

x

)(

@

xi

y

)2

d

; Z

;

@

y

(

)(

;

x

0)r

y

(

)]

d

=1;

n

2

n

X

i=1

Z

jr

y

j2

dx

;12Z

;

(

)(

;

x

0)]jr

y

(

)j2

d:

(2.5)

In order to handle the rst term in the right hand side of (2.4), for

k

1 set

T

k(

y

)(

x

) :=

8

<

:

+

k

if

y

(

x

)

>

+

k y

(

x

) if ;

k

y

(

x

)

k

;

k

if

y

(

x

)

<

;

k:

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(5)

One checks easily that

T

k(

y

)!

y

in

H

01() and a.e., eTk(y)!eyin

L

2() and a.e. On the other hand eTk(y)r

T

k(

y

)!eyr

y

in

L

1() but as r(eTk(y)) = eTk(y)r

T

k(

y

), we conclude that eyr

y

=r(ey ;1) in

L

1(). Therefore

Z

eyr

y

dx

=;

Z

(ey;1)r

dx

for all 2(

C

1(Rn))n, and we may write:

Z

ey(

x

;

x

0)r

y

]

dx

=Xn

j=1

Z

(

x

j;

x

0j)ey

@

xj

ydx

=Xn

j=1

Z

(

x

j;

x

0j)

@

xjey;1]

dx

;Xn

j=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

2

L

2(), is star-shaped with respect to some interior point

x

0 (i.e.(

;

x

0)

(

) 0 for

2;) and

y

2

H

01() satises

y

2

L

2(). Then we have

n

2 ;1 Z jr

y

(

x

)j2

dx

Z

y

(

x

;

x

0)r

y

]

dx:

(2.7) Proof. Let us consider a sequence of bounded open sets 12 , with\1j=1j = , with ;j =

@

j of class

C

11 and such that all of them are star-shaped with respect to

x

0. For each

j

we consider the problem

;

z

= (ey+

u

)

in j

z

= 0 on ;j

where

is the characteristic function of . This linear problem has a unique solution

y

j 2

H

01(j)\

H

2(j). We extend each

y

j and

y

to 1 by zero, thus

y

j,

y

2

H

01(1). From the above equation we get

Z

1

jr

y

jj2

dx

=

Z

j

jr

y

jj2

dx

=

Z

ey+

u

]

y

j

dx

key +

u

kL2( )k

y

jkL2( )

which proves the boundedness of f

y

jg1j=1 in

H

01(1). By taking a subse- quence we can assume that

y

j !

y

~weakly in

H

01(1) and

y

j(

x

) !

y

~(

x

) for almost all

x

21. Obviously we have that ~

y

(

x

) = 0 for

x

21n . Then we have that ~

y

= 0 on ; and ;~

y

= limj!1;

y

j = ey +

u

in . This leads to the equality ~

y

=

y

. Moreover

Z

1

jr

y

j2

dx

liminfj

!1 Z

1

jr

y

jj2

dx

limsupj

!1 Z

1

jr

y

jj2

dx

= limsupj

!1 Z

j

jr

y

jj2

dx

= limj

!1 Z

j

ey +

u

]

y

j

dx

=

Z

ey+

u

]

ydx

=

Z

jr

y

j2

dx

=

Z

1

jr

y

j2

dx

hence limj!1k

y

jkH01( 1)=k

y

kH01

(

1

), which proves the strong convergence

y

j !

y

in

H

01(1).

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(6)

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)r

y

]

dx

= limj

!1 Z

(;

y

j)(

x

;

x

0)r

y

j]

dx

= limj

!1 Z

j

(;

y

j)(

x

;

x

0)r

y

j]

dx

= limj

!1 (

1;

n

2

Z

j

jr

y

jj2

dx

;12Z

;

j

(

;

x

0)

j(

)]jr

y

jj2

d

)

jlim

!1

1;

n

2

Z

jr

y

jj2

dx

=1;

n

2

Z

jr

y

j2

dx:

Remark.

Inequality (2.7) makes sense when we assume only that

y

2

Y

and

u

2

L

2() but we do not know whether it holds under these assump- tions. Actually (2.7) is true for solutions (

yu

)2

Y

K

such that there is a sequence of solutions (

y

k

u

k) 2

Y

K

with

y

k 2

H

2() satisfying (1.1),

u

k !

u

strongly in

L

2() (indeed this implies that

y

k

* y

in

H

01()-weak and eyk !ey in

L

1()).

Corollary 2.4. Let us assume that

u

2

L

2(), is star-shaped with re- spect to some interior point

x

0 and

y

2

Y

is a solution of (1.1) such that ey 2

L

2(). Then we have

n

2 ;1

Z

jr

y

(

x

)j2

dx

n

Z (ey(x);1)

dx

;Z

u

(

x

)(

x

;

x

0)r

y

(

x

)]

dx:

(2.8) Proof. If

y

2

H

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 to

x

0. Since we have not assumed ; to be of class

C

11 or to be convex, we cannot deduce the

H

2()-regularity of

y

from the fact that ey +

u

2

L

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 of

u

.

Theorem 2.5. Let us assume that

y

is a solution of (1.1) that satises the inequality (2.8). Then there exist positive constants

C

i, 1

i

4, independent of

u

and

y

such that

key

y

kL1( )

C

1k

u

k2L2( )+

C

2

(2.9)

k

y

kH01 ( )

C

3k

u

kL2( )+

C

4

:

(2.10) Proof. Thanks to Theorem 2.1 we know that

Z

jr

y

(

x

)j2

dx

=

Z

ey(x)

y

(

x

)

dx

+

Z

u

(

x

)

y

(

x

)

dx:

(2.11)

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(7)

Combining this identity with (2.8) we deduce

n

2 ;1

Z

ey

ydx

+

n

2 ;1

Z

uydx

n

Z ey;1]

dx

;Z

u

(

x

;

x

0]r

y

]

dx:

Which yields

Z

ey

ydx

2

n n

;2

Z

ey;1]

dx

+

c

1k

u

kL2( )k

y

kH01

( )

:

(2.12) Let us set

n=

x

2 :

y

(

x

)

>

4

n n

;2

:

Then from (2.12) we get

Z

ey

ydx

12Z

n

ey

ydx

+

Z

n n

en;24n

dx

+

c

1k

u

kL2( )k

y

kH01 ( )

therefore

12

Z

ey

ydx

c

2;12Z

n n

ey

ydx

+

c

1k

u

kL2( )k

y

kH01 ( )

:

Taking into account that etj

t

j

c

(

n

)

<

+1 for any real number

t

n4;2n , we get from the previous inequality

Z

eyj

y

j

dx

c

3+

c

1k

u

kL2( )k

y

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

+ = maxf

y

0g and

y

;= maxf;

y

0g.

Proposition 2.6. Let

y

be a solution of (1.1) corresponding to a function

u

2

L

p(). Then

y

+2

L

r() for all 1

r <

+1

y

; 2

L

q() with

q

=

8

<

:

+1 if

p > n=

2

<

+1 if

p

=

n=

2

np=

(

n

;2

p

) if

p < n=

2

:

Moreoverk

y

;kLq( )

C

pqk

u

;kLp( )for some constant

C

pq

>

0 independent of

u

and

y

if we denote by

r

] the integer part of

r

and

k

:=

r

] + 1 then

k

y

+kLr( )(

k

!)1=r(

C

1k

u

kL2( )+

C

2)1=r

:

Proof. Let us take

k

=

r

] + 1. Then j

y

+jr

< k

!ey 2

L

1(), which proves that

y

+ 2

L

r(). Using the estimate R ey

dx

C

1k

u

k2L2( )+

C

2 one gets the last estimate of the proposition.

Now assume that

p < n=

2 (the case

p

n=

2 may be treated in the same way). Consider

2

H

01() satisfying;

=;

u

;. We have

;

y

;

u

;=;

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(8)

and by the maximum principle we conclude that

y

in . As

<

0, we conclude that 0

y

;

; =j

j. Now recall that by a classical result of G. Stampacchia 12] (theorems 4.1 and 4.2), if for some

s

2 (and

s < n

) one has ;

z

=

T

2

W

;1s() and

z

2

H

01(), then

k

z

kLs( )

C

(

ns

)(k

T

kW;1s( )+k

z

kL2( ))

s

1 = 1

s

;

n

1 Here we have ;

=;

u

; 2

L

p() and

L

p()

W

;1s() if

1

s

= 1

p

;

n:

1

Consequently one sees that the estimate of the proposition on k

y

;kLq( )

holds if

q

=

s

, which means

1

q

= 1

p

;

n:

2

Proposition 2.7. Let

u

and

y

be as in Proposition 2.6 and satisfying the inequality (2.8). Then

y

2

L

p() and

k

y

kLp( )

C

1pk

u

kLp( )+

C

2p

for some constants

C

ip

>

0 independent of

u

and

y

.

Proof. Let us take

k

=

p

] + 1. From (2.9), the inequality et j

t

jet+ 1 for every

t

2Rand taking into account that

p

2, we deduce

k

y

+kLp( )

k

!

Z

ey

dx

p

c

1k

u

k2L2( )+

c

2 1=p

c

3k

u

k2L=p2( )+

c

4

c

3;k

u

kL2( )+ 1+

c

4

c

5k

u

kLp( )+

c

6

:

On the other hand, from Proposition 2.6 it follows that

y

; 2

L

p().

Indeed it is enough to note that

np=

(

n

;2

p

)

> p

if

p < n=

2. Moreover

k

y

;kLp( )

C

pk

u

;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

k

u

k)g1k=1 of feasible elements. Assuming that either

K

is bounded in

L

p() or

N >

0, we can deduce that f

u

kg1k=1

is bounded. The dicult part is to deduce that f

y

kg1k=1 is bounded in

H

01(). If the elements (

y

k

u

k) satisfy the inequality (2.8), then Theorem 2.5 provides the necessary inequalities to deduce the boundedness off

y

kg1k=1. Unfortunately (2.8) has been proved to hold only for solutions of the state equation with ey 2

L

2(). This leads us to consider the following class of states: we denote by Y the subset of

Y

formed by the functions which satisfy (1.1) and (2.8) for some control

u

(note that for

q > n=

2 one has

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(9)

W

2q()\

H

01() Y). Not every element of Y needs to be a solution of (1.1) with ey 2

L

2(). In fact we have the following result.

Theorem 3.1. Let us assume thatf(

y

k

u

k)g1k=1Y

L

p() is a sequence of functions satisfying (1.1) and converging weakly to some element (

yu

) in

H

01()

L

p(), with

p >

2. Then eyk ! ey in

L

1(),

y

2 Y and (

yu

) satises also (1.1). The same result holds if

y

k

* y

in

H

01() and

u

k !

u

strongly in

L

2().

Proof. The weak convergence

y

k

* y

in

H

01() implies the strong conver- gence

y

k !

y

in

L

2(). Then taking a subsequence if necessary, we can assume that

y

k(

x

) !

y

(

x

) for almost every point

x

2 . In particular eyk(x)!ey(x) almost everywhere in . Let us use Vitali's theorem to prove the convergence eyk !ey in

L

1(). First let us note that the boundedness of f

u

kg1k=1 in

L

2() along with (2.9) imply thatkeyk

y

kkL1( )

C

for some constant

C <

+1 and all

k

2N. Now given

>

0, let us take

m >

0 such that

C=m < =

2 and

=

=

(2em). Then for every measurable set

E

, with meas(

E

)

<

, we have

Z

Eeyk(x)

dx

m

1

Z

fx2E:yk(x)>mgeyk(x)

y

k(

x

)

dx

+

Z

fx2E:yk(x)mgem

dx

m

1

Z

eyk(x)j

y

k(

x

)j

dx

+ emmeas(

E

)

C

m

+ em

<

8

k

2N

which allows to conclude the desired convergence. Now it is easy to pass to the limit in the state equation satised by (

y

k

u

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 that

r

y

k(

x

) ! r

y

(

x

) for almost all point

x

2 . To achieve this aim, we remark that eyk ! ey in

L

1() while (

u

k)k is bounded in

L

p(): therefore (

y

k)k is bounded in

L

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 that

y

k

* y

in

H

1() weakly andr

y

k !r

y

almost everywhere.

Now the weak convergencer

y

k

*

r

y

in ;

L

2()n along with the point- wise convergence implies the strong convergence in (

L

r())n for all

r <

2 (we use here the fact that the weak convergence in

L

2() implies that the sequence jr

y

k ;r

y

jr is equi-integrable and we apply Vitali's theorem).

In particular the strong convergence of r

y

k ! r

y

holds in

L

p0(), with (1

=p

) + (1

=p

0) = 1 (this is the only place where we need the assumption

p >

2). Then we can pass to the limit in the inequality (2.8) to obtain:

n

2 ;1

Z

jr

y

j2

dx

liminfk

!1

n

2 ;1

Z

jr

y

kj2

dx

liminfk

!1

n

Z eyk ;1]

dx

;Z

u

k(

x

;

x

0)r

y

k

dx

=

n

Z ey;1]

dx

;Z

u

(

x

;

x

0)r

ydx

ESAIM: Cocv, November 1998, Vol. 3, 361{380

(10)

which concludes the proof when (

y

k

u

k)

*

(

yu

) in

H

01()

L

p() and

p >

2. On the other hand it is clear that when

p

= 2 and

u

k !

u

strongly in

L

2(), in the above inequalities one has

Z

u

k(

x

;

x

0)r

y

k

dx

!Z

u

(

x

;

x

0)r

ydx

as

k

!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

)jp

dx

(

yu

)2Y

K

satises (1

:

1)

:

The fact of taking (

yu

) 2 Y

K

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 result

Proposition 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 control

u

2

L

p(),

p

2. Then there exists a solution

z

of (1.1) corresponding to the same control

u

and belonging to the classY.

Before proving this proposition, we state and prove the following lemma:

Lemma 3.3. For

k

1 and

t

2Rlet us denote

f

k(

t

) := minfek

etg. Then there is

z

k 2

H

01() such that

;

z

=

f

k(

z

) +

u

in

z

k = 0 on;

:

(3.1)

Moreover

z

k

z

k+1 and the sequencef

z

kg1k=1 is bounded in

H

01().

Proof. We denote with

y

a solution of (1.1) associated to the control

u

. Let us take

y

02

H

01() such that;

y

0 =

u

. Then we have the following three inequalities:

;

y

0

f

k(

y

0) +

u

;

y

f

k(

y

) +

u

;(

y

;

y

0) = ey

>

0

:

From the last relation we deduce that

y

0

y

in . From the two rst re- lations it follows that

y

0 is a subsolution and

y

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 that

y

0

z

k

y

. Moreover we have

z

k

z

k+1 and

Z

jr

z

kj2

dx

=

Z

f

k(

z

k) +

u

]

z

k

dx

Z ezk+j

u

j]j

z

kj

dx

Z

ey+j

u

j]j

z

kj

dx

key+j

u

jkH;1( )k

z

kkH01( )

which implies that f

z

kg1k=1 is bounded in

H

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 element

z

2

H

01() such that

z

k !

z

weakly in

H

01()

z

k(

x

)!

z

(

x

) a.e.

x

2

:

(3.2)

ESAIM: Cocv, November 1998, Vol. 3, 361{380

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