1. Introduction - Numdam

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 suitable 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.

Introduction In this paper we are concerned with the optimal control of the following semilinear elliptic equation ;y = ey + u in (1.1) y=0 on ; 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 Lp (), 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 1.

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: [email protected]. O. Kavian: Universit e de Versailles Saint-Quentin et Centre de Math ematiques Appliqu ees, Ecole Polytechnique, 91128 Palaiseau Cedex, France. E-mail: [email protected]. J.-P. Puel: Universit e de Versailles Saint-Quentin et Centre de Math ematiques Appliqu ees, Ecole Polytechnique, 91128 Palaiseau Cedex, France. E-mail: [email protected]. 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 publication July 16, 1998. c Soci et e de Math ematiques Appliqu ees et Industrielles. Typeset by LATEX.

362

E. CASAS, O. KAVIAN, AND J.-P. PUEL

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 = fy 2 H01() : ey 2 L1 ()g (1.2) and it satises the equation in the distribution sense. Then the optimal control problem will be formulated in the following terms

Z Z 8 N > < Minimize J (y u) := L(x y(x))dx + p ju(x)jpdx (P) > : (y u) 2 Y K satises (1:1) where K is a nonempty closed convex subset of Lp (), 2 p < +1, N 0 and L : R ;! R is 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 (x y) a1(x) + 1(jy;j1 + ey ) (1.3) @y L(x y) a2 (x) ; 2(jy ;j2 + ey ) (1.4) for some a1 , a2 2 L1 (), 1 , 2 0, 1 = np=(n ; 2p) if p < n=2, and 0 1 < +1 if p n=2, and 1 2 < p. For instance, yd 2 Y being given, a typical functional J would be Z Z 1 N 2 J (y u) := 2 jy (x) ; yd (x)j dx + p ju(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 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.

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 H01 () 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 L1 () for every z 2 H01() and 2.

Z

Z

ry(x)rz(x)dx = ey(x) + u(x)]z(x)dx:

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

(2.1)

OPTIMAL CONTROL OF AN ILL-POSED ELLIPTIC SEMILINEAR EQUATION 363

Proof. By the denition of a solution of (1.1), for all z 2 Cc1 () we have

Z

Z

ry(x)rz(x)dx = ey(x) + u(x)]z(x)dx:

(2.2)

1 Given z 2 L1 () \ H01(), we can take a sequence fzk g1 k=1 Cc (), with 1 kzk k1 kzk1 + 1, for every k 2 N and zk ! z in H0 () and zk * z in L1 ()-w. Then for all k 1 we can replace z by zk 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 L1 () \ H01 (). Let us take now z 2 H01 () such that z 0 and set Tk (z )(x) := kz(x) ifif z0(x)z>(xk) k: Then Tk (z ) 2 L1 () \ H01() and

Z

Z

ry(x)rTk(z)(x)dx = ey(x) + u(x)]Tk(z)(x)dx

8k 2 N:

Since Tk (z ) ! z in H01(), the only trouble to pass to the limit in this identity comes from the term ey zk . As Tk (z ) 0 and ey > 0, then from the monotone convergence theorem, taking into account that Tk (z )(x) " z (x) for almost all x 2 , we deduce

Z

Z

ey(x) z (x)dx

= k!lim ey(x) Tk (z )(x)dx +1 Z Z = k!lim ry(x)rTk(z)(x)dx ; u(x)Tk (z)(x)dx +1 =

Z

ry(x)rz(x)dx ;

Z

u(x)z (x)dx < +1:

Next, for a general z 2 H01 (), we notice that z = z + ; z ; with z + z ; 2 H01(). This proves that (2.1) is satised. Moreover ey 2 H ;1 () and (2.1) shows that for all z 2 H01 ()

hey zi =

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 L2 () and assume that y 2 H 2() is a solution of (1.1). Then for any x0 2 Rn we have Z n Z 1 2 2 2 ; 1 jry (x)j dx + 2 ; jry ( )j ( ) ( ; x0 )]d (2.3) Z Z y ( x ) = n (e ; 1)dx ; u(x)(x ; x0 ) ry (x)]dx

where ( ) denotes the outward unit normal vector to ; at the point . ESAIM: Cocv, November 1998, Vol. 3, 361{380

364


Proof. Since y 2 H 2(), then ey = ;y ; u 2 L2(). Therefore we can multiply the equation (1.1) by any function of L2 () and make the integration in . In particular we can take (x ; x0) ry (x) 2 L2() as this function:

Z

;y (x ; x0) ry ] =

Z

ey (x ; x

0 ) ry ]dx +

Z

u(x ; x0) ry ]dx:

(2.4)

Let us make an integration by parts in the rst integral

Z

;y (x ; x0 ) ry ]

Z

Z

ryr(x ; x0) ry] ; @ y()( ; x0) ry()]d ; Z @x y@x (xj ; x0j )@x y ]dx ; @ y ( )( ; x0 ) ry ()]d ; ij =1 Z Z n n X X j@x yj2dx + @x y (xj ; x0j )@x2 x y]dx ij =1 Zi=1 @ y ()( ; x0 ) ry ()]d ; n Z n Z X X (xj ; x0j )@x (@x y )2dx j@x yj2dx + 12 ij =1 Zi=1 @ y( )( ; x0 ) ry ()]d:

=

n Z X

=

i

=

i

j

i

; =

i

i

j

i

;

j

i

;

Now we can integrate by parts in the last relation, taking into account that for 2 ; we have ry ( ) = jry ( )j ( ), then ( ; x0 ) ry ( ) =

jry()j () ( ; x0) and @ y() = ry() () = jry()j, we get

Z

;y (x ; x0) ry ] =

Z

n Z X

Zi=1

j@x yj2dx i

n X + 12 ; (@x y)2 + (xj ; x0j ) j (x)(@x y)2d ; ij =1 i

i

Z

; @ y()( ; x0) ry()]d

(2.5)

;

Z n Z X n 1 2 = 1; 2 jryj dx ; 2 () ( ; x0)]jry()j2d:

i=1

;

In order to handle the rst term in the right hand side of (2.4), for k 1 set 8 < +k if y(x) > +k Tk (y)(x) := : y (x) if ; k y (x) k ;k if y(x) < ;k:



One checks easily that Tk (y ) ! y in H01() and a.e., eT (y) ! ey in L2 () and a.e. On the other hand eT (y) rTk (y ) ! ey ry in L1 () but as r(eT (y) ) = eT (y) rTk (y ), we conclude that ey ry = r(ey ; 1) in L1 (). Therefore k

k

Z

k

k

ey ry dx = ;

Z

(ey ; 1)r dx for all 2 (C 1(Rn))n , and we may write: Z n Z X ey (x ; x0 ) ry ]dx = (xj ; x0j )ey @x ydx j =1 Z n Z n Z X X y y = (xj ; x0j )@x e ; 1]dx ; e ; 1]dx = ;n ey ; 1]dx: j =1 j =1 j

j

(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 L2 (), is star-shaped with respect to some interior point x0 (i.e. ( ; x0) ( ) 0 for 2 ;) and y 2 H01() satises y 2 L2 (). Then we have

n

;1 2

Z

jry(x)j2dx

Z

y (x ; x0) ry ]dx:

(2.7)

Proof. Let us consider a sequence of bounded open sets 1 2 , , with ;j = @ j of class C 11 and such that all of them are with \1 j =1 j = star-shaped with respect to x0 . 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 yj 2 H01(j ) \ H 2 (j ). We extend each yj and y to 1 by zero, thus yj , y 2 H01 (1). From the above equation we get

Z

1

jryj j2dx =

Z

Z jryj j2dx = ey + u]yj dx key + ukL2( )kyj kL2( )

j

1 which proves the boundedness of fyj g1 j =1 in H0 (1). By taking a subsequence we can assume that yj ! y~ weakly in H01(1) and yj (x) ! y~(x) for almost all x 2 1. Obviously we have that y~(x) = 0 for x 2 1 n . Then we have that y~ = 0 on ; and ;~y = lim j !1 ;yj = ey + u in . This leads to the equality y~ = y . Moreover Z Z Z 2 2 jryj dx lim inf jryj j dx lim sup jryj j2 dx j !1 1

Z 1

j !1

1 ey + u]yj dx

Z

jryj j2dx = jlim !1 j !1 Z Z Z y 2 = e + u]ydx = jry j dx = jry j2dx 1 hence limj !1 kyj kH01( 1 ) = ky kH01( 1 ) , which proves the strong convergence yj ! y in H01 (1). = lim sup

j

j


366


Now arguing again as in the proof of Theorem 2.2 and using the fact that ( ; x0 ) j ( ) 0 for every 2 ;j (here j is the outward normal to ;j ), we deduce Z Z (;y )(x ; x0) ry ]dx = jlim (;yj )(x ; x0) ryj ]dx !1

= jlim !1

Z

(;yj )(x ; x0) ryj ]dx

) Z Z n 1 jryj j2dx ; 2 ( ; x0) j ()]jryj j2d = jlim 1; 2 !1 ; n Z Z n 2 jlim 1; 2 jryj j dx = 1 ; 2 jryj2dx: !1 ( j

j

j

Remark. Inequality (2.7) makes sense when we assume only that y 2 Y and u 2 L2() but we do not know whether it holds under these assumptions. Actually (2.7) is true for solutions (y u) 2 Y K such that there is a sequence of solutions (yk uk ) 2 Y K with yk 2 H 2() satisfying (1.1), uk ! u strongly in L2() (indeed this implies that yk * y in H01()-weak and ey ! ey in L1 ()). Corollary 2.4. Let us assume that u 2 L2(), is star-shaped with respect to some interior point x0 and y 2 Y is a solution of (1.1) such that ey 2 L2 (). Then we have k

Z Z Z 2dx n (ey(x) ; 1)dx ; u(x)(x ; x0 ) ry (x)]dx: ; 1 jr y ( x ) j 2

n

(2.8)

Proof. If y 2 H 2() this inequality is an immediate consequence of (2.3). Indeed it is enough to note that ( ) ( ; x0 ) 0 for almost every 2 ; because is assumed to be star-shaped with respect to x0 . 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 L2(). 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 Ci , 1 i 4, independent of u and y such that key ykL1( ) C1kuk2L2( ) + C2 (2.9)

kykH01( ) C3kukL2( ) + C4: Proof. Thanks to Theorem 2.1 we know that

Z

jry(x)j2dx =

Z

ey(x) y (x)dx +


Z

u(x)y(x)dx:

(2.10)

(2.11)


Combining this identity with (2.8) we deduce

Z

n

ey ydx +

n

Z

; 1 uydx 2 ;Z 1 2 Z n ey ; 1]dx ; u(x ; x0] ry]dx:

Which yields

Z

2n Z ey ; 1]dx + c kuk 2 ky k 1 : 1 L ( ) H0 ( ) n;2

ey ydx

Let us set

(2.12)

n = x 2 : y (x) > n4;n 2 :

Then from (2.12) we get Z Z Z 4 1 y y e ydx 2 e ydx + e ;2 dx + c1kukL2 ( ) ky kH01( ) n therefore Z 1 ey ydx c ; 1 Z ey ydx + c1kukL2 ( ) ky kH01( ) : 2 2 2 n

n

n

n

n t e jtj c(n)
n=2 < +1 q = : < +1 if p = n=2 np=(n ; 2p) if p < n=2: Moreover ky ; kL ( ) Cpq ku; kL ( ) for some constant Cpq > 0 independent of u and y if we denote by r] the integer part of r and k := r] + 1 then ky+kL ( ) (k!)1=r(C1kukL2( ) + C2)1=r: q

p

r

Proof. Let us take k = r] + 1. Then Rjy + jr < k!ey 2 L1 (), which proves that y + 2 Lr (). Using the estimate ey dx C1kuk2L2 ( ) + C2 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 H01() satisfying ; = ;u; . We have ;y ;u; = ; ESAIM: Cocv, November 1998, Vol. 3, 361{380

368


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 H01(), then 1 =1;1 kzk C (n s)(kT k ;1 + kzk 2 ) W ( ) L ( ) s s Here we have ; = ;u; 2 Lp() and Lp () W ;1s () if L ( )

s

s

n

1 = 1 ; 1:

s

p n

Consequently one sees that the estimate of the proposition on ky ; kL ( ) holds if q = s , which means 1 = 1 ; 2: q

q

p n

Proposition 2.7. Let u and y be as in Proposition 2.6 and satisfying the

inequality (2.8). Then y 2 Lp () and

kykL ( ) C1pkukL ( ) + C2p for some constants Cip > 0 independent of u and y . Proof. Let us take k = p] + 1. From (2.9), the inequality et jtjet + 1 for every t 2 R and taking into account that p 2, we deduce

Z 1=p ky+kL ( ) k! ey dx c1kuk2L2( ) + c2 p

p

p

p

; c3kuk2L=p2( ) + c4 c3 kukL2( ) + 1 + c4 c5kukL ( ) + c6: On the other hand, from Proposition 2.6 it follows that y ; 2 Lp (). Indeed it is enough to note that np=(n ; 2p) > p if p < n=2. Moreover ky; kL ( ) Cpku; kL ( ): Now it is enough to write y = y + ; y ; and to use the two inequalities p

p

p

obtained above to achieve the desired result.

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(yk uk )g1 k=1 of feasible elements. Assuming that either K is bounded in Lp () or N > 0, we can deduce that fuk g1 k=1 is bounded. The dicult part is to deduce that fyk g1 k=1 is bounded in H01(). If the elements (yk uk ) satisfy the inequality (2.8), then Theorem 2.5 provides the necessary inequalities to deduce the boundedness of fyk g1 k=1 . Unfortunately (2.8) has been proved to hold only for solutions of the state equation with ey 2 L2(). 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 3.



W 2q () \ H01() Y ). Not every element of Y needs to be a solution of (1.1) with ey 2 L2(). In fact we have the following result. p Theorem 3.1. Let us assume that f(yk uk )g1 k=1 Y L () is a sequence of functions satisfying (1.1) and converging weakly to some element (y u) in H01() Lp (), with p > 2. Then ey ! ey in L1 (), y 2 Y and (y u) satises also (1.1). The same result holds if yk * y in H01() and uk ! u strongly in L2 (). k

Proof. The weak convergence yk * y in H01 () implies the strong convergence yk ! y in L2(). Then taking a subsequence if necessary, we can assume that yk (x) ! y (x) for almost every point x 2 . In particular ey (x) ! ey(x) almost everywhere in . Let us use Vitali's theorem to prove the convergence ey ! ey in L1(). First let us note that the boundedness 2 y of fuk g1 k=1 in L () along with (2.9) imply that ke yk kL1 ( ) C for some constant C < +1 and all k 2 N. 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 Z Z ey (x) y (x)dx + em dx ey (x)dx 1 k

k

k

m fx2E :y (x)>mg

k

E

k

k

k

m1

Z

fx2E :y (x)mg k

C + em < 8k 2 N ey (x)jyk (x)jdx + em meas(E ) m k

which allows to conclude the desired convergence. Now it is easy to pass to the limit in the state equation satised by (yk uk ) and to conclude that (u y ) satises (1.1). Let us prove that (y u) satises (2.8). First of all we will prove that there exists a subsequence, that we will denote in the same way, such that ryk (x) ! ry(x) for almost all point x 2 . To achieve this aim, we remark that ey ! ey in L1 () while (uk )k is bounded in Lp (): therefore (yk )k is bounded in L1 (). 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) (yk )k such that yk * y in H 1() weakly and ryk ! ry almost everywhere. ; Now the weak convergence ryk * ry in L2 () n along with the pointwise convergence implies the strong convergence in (Lr ())n for all r < 2 (we use here the fact that the weak convergence in L2() implies that the sequence jryk ; ry jr is equi-integrable and we apply Vitali's 0theorem). In particular the strong convergence of ryk ! ry holds in Lp (), with (1=p) + (1=p0) = 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: k

n

Z

2 ;1

n Z jryj2dx lim inf ; 1 jryk j2dx k !1 2 Z Z y lim inf n e ; 1]dx ; uk (x ; x0 ) ryk dx k!1 Z Z = n ey ; 1]dx ; u(x ; x0 ) rydx k


370


which concludes the proof when (yk uk ) * (y u) in H01() Lp () and p > 2. On the other hand it is clear that when p = 2 and uk ! u strongly in L2 (), in the above inequalities one has

Z

uk (x ; x0 ) ryk dx !

Z

u(x ; x0) rydx

as k ! 1. Now we reformulate the control problem as follows Z Z 8 N > < Minimize J (y u) := L(x y(x))dx + p ju(x)jpdx (P ) > : (y u) 2 Y K satises (1:1): The fact of taking (y u) 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 x0 2 and that (1.1) has a solution for some control u 2 Lp (), p 2. Then there exists a solution z of (1.1) corresponding to the same control u and belonging to the class Y . Before proving this proposition, we state and prove the following lemma: Lemma 3.3. For k 1 and t 2 R let us denote fk (t) := minfek etg. Then there is zk 2 H01() such that ;z = fk (z) + u in (3.1) zk = 0 on ;: 1 Moreover zk zk+1 and the sequence fzk g1 k=1 is bounded in H0 (). Proof. We denote with y a solution of (1.1) associated to the control u. Let us take y0 2 H01() such that ;y0 = u. Then we have the following three inequalities: ;y0 fk (y0) + u ;y fk (y) + u ;(y ; y0) = ey > 0: From the last relation we deduce that y0 y in . From the two rst relations it follows that y0 is a subsolution and y is a supersolution of (3.1). Combining all these, by the now classical techniques introduced by D.H. Sattinger 10] (see also H. Amann 1]) we get the existence of a solution zk of (3.1) such that y0 zk y . Moreover we have zk zk+1 and

Z

jrzk j2dx

=

Z

ez + juj]jzk jdx k

y y e + juj]jzkjdx ke + jujkH ;1( )kzk kH01( ) 1 which implies that fzk gk=1 is bounded in H01(). Proof of Proposition 3.2. The sequence fzk g1 k=1 being given by Lemma 3.3,

Z

fk (zk ) + u]zk dx

Z

taking a subsequence, denoted in the same way, we infer that there exists an element z 2 H01() such that zk ! z weakly in H01() zk (x) ! z(x) a.e. x 2 : (3.2) ESAIM: Cocv, November 1998, Vol. 3, 361{380


Therefore fk (zk (x)) ! ez(x) for almost all x 2 and fk (zk ) ez ey . Then we can apply the dominated convergence theorem to obtain that ez ! ez in L1 (). Now it is easy to pass to the limit and to deduce that z satises (1.1). We are going to prove that z 2 Y . First of all let us remark that fk (zk ) + u 2 L2(), hence zk 2 L2 (). Therefore we can multiply (3.1) by (x ; x0 ) rzk and integrate over and get: k

k

Z ; zk (x ; x0) rzk ]dx Z Z = fk (zk )(x ; x0 ) rzk ]dx + u(x ; x0) rzk ]dx: (3.3) Let us dene the function Fk : R ;! R by t k Fk (t) = ek (t ;e k;+11) ; 1 ifif tt > k: Then Fk is the primitive of fk (which is dened in Lemma 3.3) satisfying Fk (0) = 0 and Fk (t) et ; 1. Arguing as in the proof of Theorem 2.2 we obtain Z Z Z fk (zk )(x ; x0) rzk ]dx = ;n Fk (zk )dx ;n ez ; 1]dx: (3.4)

Now from Lemma 2.3 we get

Z

n

jrzk j2dx

k

Z

(zk )(x ; x0 ) rzk ]dx: (3.5) 2 ;1 Combining (3.3), (3.4) and (3.5), the following inequality is obtained

n

Z

2 ;1

Using (3.2) and

n

Z Z jrzk j2dx n ez ; 1]dx ; u(x ; x0) rzk ]dx: k

the convergence ez

Z

jrzj2dx n

Z

k

! ez

in

1 L (), it comes

ez ; 1]dx ;

Z

u(x ; x0 ) rz ]dx 2 ;1 which proves that z 2 Y as desired. In the next proposition we state some interesting properties of the set of feasible controls. Proposition 3.4. Let p 2 and let be star-shaped with respect to one of its interior points. Then the set of controls u 2 Lp() for which there exists a solution y 2 Y is non empty, convex and closed in Lp (). Proof. It is enough to take any function y 2 Cc1 () and to put u = ;y ; ey to deduce that the set of feasible controls is non empty. Let us take a p sequence of feasible controls fuk g1 u. k=1 converging in L () to some function Thanks to Theorem 2.5, we know that the corresponding states fyk g1 k=1 Y 1 are bounded in H0 (). Then Theorem 3.1 claims that any weak limit of fyk g1 k=1 is an element of Y satisfying (1.1) along with the control u, which proves that the set of feasible controls is closed. Now let us prove the convexity. If u1 and u2 are two feasible controls, with associated states y1 and y2 , respectively, and 2 (0 1), we set u = ESAIM: Cocv, November 1998, Vol. 3, 361{380

372


u1 + (1 ; )u2 and y = y1 + (1 ; )y2 . Let us take 2 H01() such that ; = u. Then ; e + u and ; y = ey1 + (1 ; )ey2 + u ey + u: Therefore is a subsolution of (1.1) for the control u and y is a supersolution. On the other hand, ;(y ; ) ey > 0, then y . Therefore we deduce the existence of a solution z of (1.1) associated to the control u, with z y. Finally, the conclusion follows from Proposition 3.2. So far we have studied the properties of the feasible pairs (y u) 2 Y p L () satisfying (1.1). Let us say something about the action of functional J on these pairs. For each one of these pairs (y u), J is well dened and ;1 < J (y u) +1. Indeed the only trouble can come from the integral of L(x y (x)). With the notation of (1.4), let us set f (x) = L(x y (x)) + 2 jy ;(x)j2 + ey(x) ; a2(x): Then f is a nonnegative measurable function and consequently its integral is well dened as a number in 0 +1]. On the other hand, it is enough to use Theorem 2.5 and Proposition 2.6 and the assumptions on 2 to deduce that jyj2 and ey are integrable functions. Therefore the integral of L(x y(x)) is well dened, though it could be +1 in some cases. Finally we establish our result of existence of a solution to (P ). Theorem 3.5. Let us assume that p > 2 an (i) There exists a pair (y u) 2 Y K satisfying (1.1). (ii) Either K is bounded or N > 0. Then problem (P ) has at least one solution. Proof. Let us assume that := inf(P )< +1 (otherwise the theorem is obvious). Let f(yk uk )g1 k=1 be a minimizing sequence for problem (P ): J (yk uk) # . Thus we can suppose that J (yk uk ) + 1. The main p point in the proof is to establish the boundedness of fuk g1 k=1 in L (). This is obvious Z Let us assumeZthat K is not bounded. Z We have Z if K is bounded. N a2 (x)dx + p juk (x)jpdx ; 2 jyk;(x)j2 dx ; 2 ey (x)dx J (yk uk) + 1: Using the fact that the exponent q dened in proposition 2.6 satises q > p we conclude that kyk; k22 C kuk kp2 . In the same way we observe that Z ey (x)dx C (kuk k22 + 1) k

and nally we obtain:

k

kuk kpp C + C kuk k2p + C kuk kp2 : As 2 < p and p > 2, this implies that (uk )k is bounded in Lp (). 1 The boundedness of fyk g1 k=1 in H0 () is a consequence of (2.10). Therefore, taking a subsequence if necessary, we can assume that (yk uk ) * (y u) weakly in H01 () Lp (). Theorem 3.1 asserts that y 2 Y and (y u) satises (1.1). Moreover the convexity and closedness of K in Lp() implies that u 2 K . Thus (y u) is a feasible pair for problem (P ). Let us prove that it ESAIM: Cocv, November 1998, Vol. 3, 361{380


is a solution. Since yk ! y strongly in L2 () we can take a subsequence in such a way that yk (x) ! y (x) for almost all x 2 . Let us set fk (x) = L(x yk (x)) + 2 jyk;(x)j2 + ey (x) ; a2(x) and f (x) = L(x y (x)) + 2 jy (x)j2 + ey(x) ; a2(x): Then fk (x) ! f (x) almost everywhere and fk 0. Therefore we can apply Fatou's Lemma and the convergences ey ! ey (Theorem 3.1) and jyk j2 ! jyj2 (Proposition 2.7) in L1() to derive k

k

Z

Z n o J (y u) = f (x)dx ; 2 jy ;(x)j2 + ey(x) ; a2(x) dx Z Z N + p ju(x)jpdx lim inf fk (x)dx k!1 Z Z h i N p ; y ( x ) 2 ; 2 jyk (x)j + e ; a2(x) dx + p juk (x)j dx Z Z = lim inf L(x yk (x))dx + Np juk (x)jpdx k!1 = lim inf J ( y u ) = inf ( P ) : k k k!1 k

This concludes the proof. We conclude this section by studying the uniqueness of the solution. Theorem 3.6. Let p 2 and assume that is star-shaped with respect to some x0 2 and that the set of admissible pairs (y u) 2 Y K satisfying (1.1) is not empty. Assume also that the function t 7! L(x t) is monotone on R, non decreasing and convex for almost all x 2 . Then problem (P ) has at most one solution if one of the following conditions holds: (i) N > 0 (ii) L(x ) is strictly increasing (iii) L(x ) is strictly convex. Proof. Let us assume that (y1 u1) and (y2 u2) are two di#erent solutions of (P ). We note in particular that y1 6= y2 and we set u = (u1 + u2 )=2. Looking at the proof of Proposition 3.4, we remark that one may prove the existence of a solution y 2 Y of (1.1) corresponding to the control u, with y (y1 + y2)=2. In fact we have that this inequality is strict in . Indeed ;(y1 + y2)=2 ; y] 12 (ey1 + ey2 ) ; ey e(y1 +y2 )=2 ; ey 0 in and due to the fact that y1 6= y2 , we have 12 (ey1 + ey2 ) ; ey 6 0 on . Now by the strong maximum principle we conclude that (y1 + y2 )=2 > y in . Therefore Z Z L(x y(x))dx L(x (y1(x) + y2(x))=2)dx Z

Z 1 2 L(x y1(x))dx + L(x y2(x))dx ESAIM: Cocv, November 1998, Vol. 3, 361{380

374


and the inequality

Z

L(x y(x))dx 12

Z

L(x y1(x))dx +

Z

L(x y2(x))dx

is strict if the strict convexity or monotonicity of L(x ) is assumed. If the previous inequality is strict or if N > 0 we deduce J (y u) < 12 J (y1 u1) + J (y2 u2)] = inf (P ) which is a contradiction with the fact that (u y ) is feasible for problem (P ). Existence of an Optimal Control. Case p = 2 As we noticed before, when p = 2 the conclusions of Theorem 3.1 holds only when we know that uk ! u strongly in L2(). The diculty comes from the fact that the strong convergence ryk ! ry can only be proved in Lr (), with r < 2. Hence we cannot pass to the limit in the inequality (2.8) satised for every (yk uk ). Therefore we cannot deduce that (y u) satises (2.8), and consequently we are not able to prove the existence of a solution in Y K . Since we have estimates on the state (see Theorem 2.5) only for elements of Y , we cannot deduce, in general, the boundedness in Y K of a minimizing sequence of problem (P). In this section we will show that, under some additional assumptions on the function L, it is possible to have 1 a minimizing sequence f(yk uk )g1 k=1 of (P) with fyk gk=1 Y . In this way, we can deduce the boundedness of the states and prove the existence of an optimal solution. Theorem 4.1. Let us assume that is star-shaped with respect to one of its interior points. We also make the hypotheses (i) The function t 7! L(x t) dened on R is monotone non decreasing for almost every x 2 . (ii) There exists a pair (y u) 2 Y K satisfying (1.1). (iii) Either K is bounded or N > 0 and 2 < 2 in (1.4). Then problems (P) and (P ) have at least one solution. For each solution (y u) of (P), we can nd y~ 2 Y such that (~y u) is a solution of (P) and (P ). Moreover, if L(x ) is strictly increasing for almost all x 2 , then any optimal solution of (P) is also a solution of (P ). Proof. Let us take a minimizing sequence f(yk uk )g1 k=1 Y K . From Proposition 3.2 we deduce the existence of elements zk 2 Y such that (zk uk ) satises (1.1). Moreover, by looking at the proof of the mentioned proposition, we know that zk yk in . Now using the monotonicity of L(x ), we get that J (zk uk ) J (yk uk ). Therefore f(zk uk )g1 k=1 is also a minimizing sequence of (P). Arguing as in the proof of Theorem 3.5, we can obtain a subsequence, denoted in the same way, converging to an element (y u) 2 Y K solution of Problem (P). If y 62 Y , we can apply again Proposition 3.2 to deduce the existence of an element y~ 2 Y , with y~ y, such that (~y u) satises (1.1). Again the monotonicity of L(x ) leads to J (~y u) J (y u). So (~y u) is a solution of (P), and consequently of (P ) too. 4.



Finally, if L(x ) is strictly increasing and y 62 Y , then J (~y u) < J (y u), which contradicts the optimality of (y u). The Optimality Conditions The aim of this section is to derive some optimality conditions for the control problem. We will prove two theorems corresponding to the cases K = Lp() and K Lp () with K 6= Lp (). Let us start with the rst case. Theorem 5.1. Assume that is star-shaped with respect to one of its interior points, p 2 and K = Lp (). If (y u) is a solution of problem (P ) (resp. (P)) with ey 2 Lp(), then ;y = ey + u in (5.1) y = 0 on ; and setting := ;N jujp;2u, one has ' 2 W01s () for every s < n=(n ; 1) and 5.

8 < ;' = ey' + @L (x y) in (5.2) @y : ' = 0 on ;: Moreover, if p > n=2, n 5 and a1 2 L2n=(n+2) () in (1.3), then ' 2 H01(). Proof. Let us take z 2 H01() \ L1 () such that z 2 Lp (). For every 2 R, = 6 0, we set y = y + z and u = u ; z + ey ; ey+z : Then Corollary 2.4 implies that y 2 Y , with ;y = ey + u . On the other hand u 2 Lp (). Then (y u) is a feasible point for (P ) (resp. (P)),

consequently, using Lebesgue's convergence theorem along with assumption (1.3), we get 0 lim J (y u) ; J (y u)

Z ju (x)jp ; ju(x)jp L ( x y (x)) ; L(x y(x)) dx + lim N = lim dx !0 !0 p Z @L Z p;2 u ;;z ; eyz dx: = ( x y ( x )) z ( x ) dx + N j u j @y From the linearity of the previous relation with respect to z we deduce that Z @L Z ; (x y(x))z (x)dx + N jujp;2u ;z ; eyz dx = 0 @y 1 1 for every z 2 H0 () \ L () such that z 2 Lp (). Let us set ' := ;N jujp;2u, then Z ; Z ' ;z ; eyz dx = @L (x y(x))z (x)dx (5.3) @y !0 Z


376


Given f 2 C01 (), let z 2 H01() be the solution of Dirichlet problem ;z = f in (5.4) z=0 on ;: Then z 2 H01() \ L1 () and for s < n=(n ; 1) and (1=s) + (1=s0) = 1 we have (see G. Stampacchia 12], Theorems 4.1 and 4.2) kzkL1( ) Cskf kW ;1 0 ( ): (5.5) s

Combining (5.3), (5.4) and (5.5) we get

Z

Z

Z

@L (x y(x))z (x) + eyz dx @y !

'(;z)dx = y @L @y (x y)L1( ) + ke kL1( ) kzkL1( ) ! @L Cs @y (x y) + keykL1 ( ) kf kW ;1 L1 ( )

'fdx =

0 ( ) :

s

Taking into account that C01 () is dense in W ;1s0 (), we deduce from the 0 above inequality that ' 2 W ;1s0 () = W01s () for every s < n=(n ; 1). The fact that (5.2) follows from (5.3) is a straightforward consequence of the denition of '. Finally, if p > n=2, from the fact that ;y = ey + u 2 Lp () and using again the above mentioned results of G. Stampacchia 12], it follows that y 2 L1 (). On the other hand, W01s () Lns=(n;s) () H ;1() if s is close enough to n=(n ; 1) and n 5. Therefore the right hand side of (5.2) belongs to H ;1(), which allows to conclude that ' 2 H01(). In case of a problem with control constraints, we have the following result. Theorem 5.2. Let us assume that is star-shaped with respect to some x0 2 , p 2 and p > n=2. If (y u) is a solution of problem (P ) (resp. (P)) with ey 2 Lp (), then there exist a real number 0 and a function ' 2 W01s() for every s < n=(n ; 1) such that + k'kW01 ( ) > 0 (5.6) s

;y = ey + u in y = 0 on ;

8 < ;' = ey' + @L (x y) in @y : ' = 0 on ; Z ; ' + N jujp;2 u (u ; u)dx 0 8u 2 K:

Moreover, if n 5 and a1 2 L2n=(n+2) () in (1.3), then ' 2 H01().


(5.7) (5.8) (5.9)


The proof of this theorem requires some previous lemmas. First of all, let us remark that y 2 L1 (). Indeed ;y = ey + u 2 Lp () with p > n=2, thus it is enough to use again the mentioned results of G. Stampacchia 12] to deduce the boundedness of y. In particular we have that if (y u) is a solution of (P), then it is also a solution of (P ) because y 2 Y see Corollary 2.4. Given > 0 we dene J : Lp () Lp() ;! R by Z 1 jey( ) ; wjpdx J (u w) := J (y u) +

where y(uw) problem

(uw)

p Z Z 1 + p ju ; ujpdx + 1p jey ; wjpdx 1 is the unique element of H0 () solving the boundary value uw

;y = u + w in y=0 on ;:

(5.10)

Once again using the mentioned results of G. Stampacchia 12] we obtain that y(uw) 2 L1 () and therefore J is well dened. Now we consider the following control problem J (u w) (P ) Minimize (u w) 2 K \ B1 (u) B1 (ey) where B1 (u) (resp. B1 (ey)) denotes the closed unit ball of Lp () with center at u (resp. ey). We have the following result. Lemma 5.3. Problem (P ) has at least one solution (u w ). Moreover we have 1 key ; w kp = 0 yk lim k u ; u k = lim k w ; e = lim L ( ) L ( ) L ( ) !0 !0 !0 (5.11) p

p

p

lim ky ; ykL1 ( ) = lim ky ; ykH01( ) = 0 !0

!0

(5.12)

where y is the solution of (5.10) corresponding to (u w ). Proof. The existence of a solution is obvious because of the convexity, boundedness and closedness of the set of feasible controls as well as the weak lower semicontinuity of J . Furthermore f(u w )g0< 0 such that < 0 : ku ; u kL ( ) < 1 and ku ; u kL ( ) < 1: (5.13) Furthermore, there exists ' 2 W01s () for every s < n=(n ; 1) such that ;y = u + w in (5.14) y = 0 on ; p

p

8 < ;' = ey ' + @L (x y ) + g in (5.15) @y : ' = 0 on ; Z ' + N ju jp;2u + ju ; ujp;2(u ; u) (u ; u )dx 0

8u 2 K

(5.16)

where g ! 0 in Lp0 () as & 0. Proof. The existence of a sequence f(u w )g0< < 0 of solutions satisfying (5.13) is a consequence of Lemma 5.3. Given h 2 Lp () we denote by zh 2 H01() \ L1() the solution of ;z = h in z=0 on ;: Let us set y = y(u w ) and ' = 1 jey ; w jp;2(ey ; w ) + jey ; w jp;2(ey ; w ):

From the optimality of (u w ) we deduce that for all u 2 K and w 2 Lp ():

@J (u w )(u ; u ) 0 and @J (u w )w = 0: (5.17) @u @w We compute the second derivative for each w 2 Lp () @J (u w )w = Z @L (x y ) + ey ' + g z dx ; Z ' wdx = 0 w @w @y

(5.18)

where

g = ;ey jey ; w jp;2 (ey ; w ): From (5.11) and (5.12) we get that g ! 0 strongly in Lp0 ().



Computing now the rst derivative of (5.17)

@J (u w )(u ; u ) = Z @L (x y ) + ey ' + g z dx u;u @u @y Z + N ju jp;2 u + ju ; ujp;2 (u ; u) (u ; u )dx 0:

Taking w = u ; u in (5.18), one sees that (5.16) follows from this inequality. Finally, from (5.18) we get that

Z

' (;z)dx =

Z

@L (x y ) + ey ' + g zdx @y

for every z 2 H01 () with ;z 2 Lp (). Arguing as in the proof of Theorem 5.1, we deduce from here that ' 2 W01s () for every s < n=(n ; 1) and (5.15) holds. Now we are ready to conclude the proof of Theorem 5.2. Proof of Theorem 5.2. If f' g0< < 0 is bounded in L1 (), then f' g0< < 0 is also bounded in W01s () and it is easy to pass to the limit in (5.14){(5.16) with the aid of Lemma 5.3 and to deduce (5.7){(5.9) with = 1. Otherwise we take 1 =

k' kL1( )

and we redene ' as ' . Then (5.15) and (5.16) can be written

8 < ;' = ey ' + @L (x y ) + g in @y : ' = 0 on ;

and

Z 8u 2 K ' + N ju jp;2u + ju ; ujp;2 (u ; u) (u ; u )dx 0

respectively. Now f' g0< < 0 is bounded in L1 () and ! 0, then we can pass to the limit in the previous relations and to obtain (5.8) and (5.9) with = 0. It remains to prove that (5.6) holds, or equivalently that ' 6= 0. From the equation satised by ' we deduce that f' g0< < 0 is bounded in W01s() for every s < n=(n ; 1). Then we can take a subsequence, denoted in the same way, such that ' ! ' weakly in W01s (), hence also strongly in L1 (). Now the equality k' kL1 ( ) = 1 leads to k'kL1 ( ) = 1. The H01()-regularity of ' claimed in the theorem follows as in the proof of Theorem 5.1. In some cases we can prove that can be chosen equal to one in the system (5.7)-(5.9). Corollary 5.5. Under the assumptions of Theorem 5.2, if furthermore there exists an open set ! and a neighborhood W of zero in C01 (! ) such that u + w 2 K for every w 2 W , then (5.7)-(5.9) holds with = 1. ESAIM: Cocv, November 1998, Vol. 3, 361{380

380


Proof. If = 0 in (5.7)-(5.9), then from the hypothesis of the corollary and (5.9) we deduce that '(x) = 0 for almost all x 2 ! . Now (5.8) is ;' = ey' in : ' = 0 on ;: Then we deduce that ' = 0 in (see J.C. Saut and B. Scheurer 11]), which contradicts (5.6). References 1] H. Amann: On the number of solutions of nonlinear equations in ordered Banach spaces. J. Func. Anal., 11 1972, 346{384. 2] L. Boccardo, F. Murat: Almost everywhere convergence of the gradients of solutions to elliptic and parabolic equations. Nonlinear Anal., Theory, Methods, Appl., 19 1992, 581{597. 3] S. Chandrasekhar: An introduction to the study of stellar structures . Dover Publishing Inc., 1985. 4] M.G. Crandall, P.H. Rabinowitz: Some continuation and variational methods for positive solutions of nonlinear elliptic eigenvalue problems. Arch. Rational Mech. Anal., 58 1975, 207{218. 5] D.A. Franck-Kamenetskii: Diusion and heat transfer in chemical kinetics . Second edition, Plenum Press, 1969. 6] Th. Gallouet, F. Mignot, J.P. Puel: Quelques r esultats sur le probleme ;u = e . C. R. Acad. Sci. Paris, 307, s erie I, 1988, 289{292. 7] I.M. Gelfand: Some problems in the theory of quasi-linear equations. Uspekhi Mat. Nauk, (N.S.), 14 (86), 1959, 87{158 (in russian) Amer. Math. Soc. Transl., (Ser. 2), 29, 1963, 289{292. 8] F. Mignot, J.P. Puel: Sur une classe de problemes non lin eaires avec nonlin earit e positive, croissante, convexe. Comm. PDE, 5 (8), 1980, 791{836. 9] F. Mignot, J.P. Puel: Solution singuliere radiale de ;u = e . C. R. Acad. Sci. Paris, 307, s erie I, 1988, 379{382. 10] D.H. Sattinger: Monotone methods in nonlinear elliptic and parabolic boundary value problems. Indiana Univ. Math. J., 21 1972, 979{1000. 11] J.C. Saut, B. Scheurer: Sur l'unicit e du probleme de cauchy et le prolongement unique pour des equations elliptiques a coecients non localement born es. J. Di. Eq., 43 1982, 28{43. 12] G. Stampacchia: Le probleme de Dirichlet pour les equations elliptiques du second ordre a coecients discontinus. Ann. Inst. Fourier, 15 1965, 189{258. u

u
