Optimal pointwise control of semilinear parabolic equations

Nonlinear Analysis 39 (2000) 135 – 156

www.elsevier.nl/locate/na

Optimal pointwise control of semilinear parabolic equations J. Droniou a , J.-P. Raymond b;∗ b

a Ecole Normale Supà erieure, 46 AllÃee d’Italie, 69364 Lyon, Cedex 07, France UniversitÃe Paul Sabatier, UMR CNRS MIP, UFR MIG, 31062 Toulouse Cedex 4, France

Received 22 July 1997; accepted 7 December 1997

Keywords: Optimal control; Pointwise control; Semilinear parabolic equations with measures as data

1. Introduction of class C 2 Let be a bounded open subset in R N (N ≥ 2) with a boundary PN and A be a second order dierential operator de ned by Ay = − i; j=1 Di (aij (x)Dj y)+ a0 (x)y, (Di denotes the partial derivative with respect to xi ). We consider the following boundary value problem: @y + Ay + (x; t; y)=u(t)x0 @t @y =0 (or y =0) @nA

in Q;

on ;

y(0)= y0 in ;

(1)

where Q = × ]0; T [;  = × ]0; T [; x0 denotes the Dirac measure at x0 ∈ , the control variable u belongs to some subset KU of Lq (0; T ),  is a Caratheodory function from Q × R into R. We are interested in the control problem (P ) ∗

inf {I (y; u) | (y; u) ∈ L1 (0; T ; W 1; 1 ( )) × KU ; (y; u) satis es (1)};

Corresponding author.

0362-546X/99/$ - see front matter ? 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 8 ) 0 0 1 7 0 - 9

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

where

Z

I (y; u) = 1

Q

Z

|y − zd |r dx dt + 2



|y(T ) − yd |s dx + 3

Z 0

T

|u|q (t) dt

( i ≥ 0 for i = 1; 2; 3). For a linear equation (when  ≡ 0) and for q = s = 2, 1 = 0, this problem has been studied by Lions in [9, 10]. A characterization of controls u for which yu (T ) (yu is the solution of Eq. (1) corresponding to u) belongs to L2 ( ) is given in [10, 15]. Still in the case when  ≡ 0, this problem has also been studied by Anita [3] for 2 = 3 = 0, q = ∞, r = 1, and KU is a closed convex bounded subset of L∞ (0; T ) (the existence and the characterization of solutions are established). The case of a nonlinear equation corresponding to (x; t; y) ≡ y3 is addressed in [10], an approximate problem is considered (in which the Dirac mass x0 is replaced by the characteristic function of some ball B(x0 ; ”)), but optimality conditions for solutions of (P ) by a passage to the limit are not carried out. Moreover, from [5] we know that if u ≡ 1 and if (x; t; y) ≡ |y| −1 y, then Eq. (1) admits a weak solution if and only if ¡N=(N − 2). Even if what follows can be extended to nonlinearities more general than |y| −1 y, for clarity we consider only this case, with 1 ≤ ¡N=(N − 2), and we treat the case of homogeneous Neumann boundary conditions (the results of this paper can be easily adapted to homogeneous Dirichlet boundary conditions). Thus the state equation is @y + Ay + |y| −1 y = u(t)x0 in Q; @t @y = 0 on ; y(0) = y0 in : @nA

(2)

To well de ne the control problem we must indicate in which space y and y(T ) may be observed. The answer clearly depends on q [10, 15]. We prove that if q¿max(1; 2 = (2 −N +N +2)), then y(T ) can be observed in Ls ( ) for every 1 ≤ s¡Nq0 =(Nq0 − 2) (q0 is the conjugate exponent to q). For simplicity, we only consider a terminal observation. Thus we study the control problem (P)

inf {J (y; u) | (y; u) ∈ L1 (0; T ; W 1; 1 ( )) × KU ; (y; u) satis es (2)};

where J (y; u) =

Z

s



|y(T ) − yd | dx +

Z 0

T

|u|q (t) dt

( ≥ 0 and yd is a given function in Ls ( )). Even if, at least in the case u ∈ L∞ (0; T ), the existence result for (2) can be deduced from [5], the estimates and the regularity results of Theorem 2.1 seems to be new. Moreover our proof is dierent. By using estimates on analytic semigroups 0  for linear equations (Proposi(Lemma 2.1), we rst prove estimates in Lq (0; T ; C( )) tions 2.3 and 2.5). Next with the so-called transposition method and by taking advantage of the structure of the measure u(t)x0 , we obtain estimates for linear equations with right-hand side of the form u(t)x0 (Propositions 2.1 and 2.2). Existence and

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

137

regularity results for Eq. (2) are stated in Theorem 2.1. Some results concerning the adjoint equation are established in Section 3. The existence of solutions for the control problem and optimality conditions are established in Section 4. 2. State equation In all the sequel for 1 ¿0 and 2 nonnegative, we set 1 =2 = ∞ if 2 = 0. Some constants Ci in estimates in propositions depend on dierent exponents. Since the constant Ci may intervene in a proof for dierent exponents, for clarity we sometimes indicate this dependence. The constants K1 and K2 are the ones which intervene in semigroup estimates (Lemma 2.1).  The coecients aij (A1) The coecient a0 of A is positive and belongs to C( ). 1;   belong to C ( ) with 0¡ ≤ 1, aij = aji , and they satisfy (for some m0 ¿0) N X

aij (x)i j ≥ m0 ||2

 for every  ∈ R N and every x ∈ :

i; j=1

(A2) KU is a closed convex subset of Lq (0; T ), max(1; 2 =(2 − N + N + 2))¡q¡∞ and 1¡s¡Nq0 =(Nq0 − 2): Remark 2.1. It is well known that the condition a0 ¿0 is not restrictive. Indeed if y is a solution of Eq. (1), then z = e−t y is the solution of @z + Az + z + e−t (x; t; et z) = e−t u(t)x0 @t @z = 0 (or z = 0) on ; @nA z(0) = y0 in :

in Q;

Remark 2.2. When 1 ≤ ¡N=(N − 2) and 1¡q; we can easily verify that the condition 2 =(2 − N + N + 2)¡q is equivalent to ¡(N + 2)q0 =(Nq0 − 2) and also to q0 ¡2 =(N − N − 2)+ (where (·)+ = Max (0; ·)): Thus the inequality ¡min(N=(N − 2); (N + 2)q0 =(Nq0 − 2)) is assumed throughout the paper. In the sequel we consider equations of the form (1) for (x; t; y) = |y| −1 y or ˜ (x; t; y) = a(x; t)y (where a belongs to some space Lk (0; T ; Lk ( ))), and when ux0 is l replaced by f ∈ L (0; T ; Mb ( )), l ≥ 1. For all these equations, we consider solutions in the sense of the following de nition. Deÿnition 2.1. Let f be in Ll (0; T ; Mb ( )). We shall say that y ∈ L1 (0; T ; W 1; 1 ( )) is a weak solution of the equation @y + Ay + (x; t; y) = f @t

in Q;

@y =0 @nA

on ;

y(0) = y0

in ;

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

if (·; y(·)) belongs to L1 (Q) and if   Z N X @ −y + aij (x)Dj yDi  + (x; t; y) dx dt @t Q i; j=1

Z =

T 0

Z

hf(t); (t)iMb ( ) × Cb ( ) dt +



(0)y0 dx

 (hf(t); (t)iM ( ) × C ( ) denotes the  such that (T ) = 0 on . for every  ∈ C 1 (Q) b b integral over of (t) for the measure f(t):) To prove the existence of a weak solution for Eq. (2), we need some estimates for linear equations established below. 2.1. Estimates for linear equations First recall some results for analytic semigroups. We denote by A˜ the operator de ned by    @y = 0 on ˜ = Ay: ˜ = y ∈ C 2 ( ) ; Ay D(A) @nA For 1 ≤ l¡∞, we denote by Al the closure of A˜ in Ll ( ). The operator −Al is the generator of a strongly continuous analytic semigroup Sl (t)t ≥ 0 in Ll ( ) [1]. For 1¡l¡∞ the domain of Al is D(Al ) = {y ∈ W 2; l ( ) | @y=@nA = 0 on }. For 1 = l, 1 1 RD(A1 ) is the set Rof functions y in L ( ) such that there exists z ∈ L ( ) satisfying ˜ For any 1 ≤ l¡∞, 0 belongs to the z(x)v(x) dx = y(x)Av(x) dx for all v ∈ D(A).

resolvent of −Al and there exists ¿0 such Re (Al ) ≥  (it is a consequence of (A1) and of the fact that (Al ) is independent of l). Therefore, for ¿0, there exists a constant K0 = K0 (l; ) such that kAl Sl (t)’kLl ( ) ≤ K0 t − k’kLl ( ) ; for every t¿0 and every ’ ∈ Ll ( ) (see [8, 12], Al is the -power of Al ). Thanks to this result the following lemma can be established. The rst part of Lemma 2.1 is stated in [1], the second part is established in [14]. Lemma 2.1. For every 1 ≤ l ≤  ≤ ∞ with l¡∞; there exists a constant K1 = K1 (; l) such that N

1

1

kSl (t)’kL ( ) ≤ K1 t − 2 ( l −  ) k’kLl ( )

(3)

for every ’ ∈ Ll ( ) and every t¿0. For every 1 ≤ l ≤  ≤ ∞ with l¡∞; and every ¿0; there exists a constant K2 = K2 (; l; ) such that N

1

1

kAl Sl (t)’kL ( ) ≤ K2 t − 2 ( l −  )− k’kLl ( ) for every ’ ∈ Ll ( ) and every t¿0.

(4)

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139

We consider the linear equation @y + Ay + ay = f @t

in Q;

@y =0 @nA

on ;

y(0) = y0

in ;

(5)

˜ ˜ k) where f belongs to Lq (0; T ; Mb ( )), and a belongs to Lk (0; T ; Lk ( )) for every (k; satisfying

N q ˜ ; ≤ k¡∞; 1 ≤ k¡

−1 (N − 2)( − 1) 1 1 N N ¡ + + if ¿1; 2 q0 2( − 1)k ( − 1)k˜

(6)

and ˜ 1 ≤ k¡∞;

1 ≤ k¡∞

if = 1:

We look for estimates for the solution y of Eq. (5) in Lr˜(0; T ; Lr ( )) with q ≤ r˜ ≤ ∞;

1¡r¡

N N −2

and

N 1 1 N ¡ + + : 2 q0 2r r˜

(7)

The following lemma will be often used in calculations throughout the paper. ˜

˜ k) satisfying (6), Lemma 2.2. If a function a belongs to Lk (0; T ; Lk ( )) for every (k; ˜ r) satisfying (7), then ay belongs to and if y belongs to Lr˜(0; T ; Lr ( )) for every (r; L (Q) for every 1 ≤ ¡ inf (N= (N − 2); (N + 2)q0 =[ (Nq0 − 2)]). If a sequence (an )n ˜ ˜ k) satisfying (6), and if (yn )n is bounded is bounded in Lk (0; T ; Lk ( )) for every (k; in Lr˜(0; T ; Lr ( )) for every (r; ˜ r) satisfying (7), then (an yn )n is bounded in L (Q) for every 1 ≤ ¡ inf (N=[ (N − 2)]; (N + 2)q0 [ (Nq0 − 2)]). Proof. If q ≥ N=(N − 2), then for k˜ = q=( − 1), r˜ = q, we can verify that ay belongs to L˜ (0; T ; L ( )) for ˜ = q= and for every 1 ≤ ¡N=[ (N − 2)]. Therefore the rst part of the lemma is proved in this case. If q¡N=(N − 2), then we can verify that ay belongs to L (Q) for every 1 ≤ ¡[(N + 2)q0 ]=[ (Nq0 − 2)]. The second part of the lemma can be proved in the same way. ˜

Proposition 2.1. If a is a nonnegative function belonging to Lk (0; T ; Lk ( )) for ˜ k) satisfying (6), if f belongs to Lq (0; T ; Mb ( )) and if y0 belongs to every (k; Nq0 =(Nq0 −2) ( ); then Eq. (5) admits a unique weak solution in L1 (0; T ; W 1; 1 ( )); this L ˜ r) satisfying (7) and there exists a solution belongs to Lr˜(0; T ; Lr ( )) for every (r; ˜ r; q); not depending on the function a; such that constant C1 = C1 (r;   kykLr˜(0; T ; Lr ( )) ≤ C1 kfkLq (0; T ; Mb ( )) + ky0 kLNq0 =(Nq0 −2) ( ) :

(8)

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

In particular for r˜ = ∞ and for every 1¡r¡Nq0 =(Nq0 − 2); y belongs to C([0; T ]; Lrw ( )) and we have   (9) kykL∞ (0; T ; Lr ( )) ≤ C1 (∞; r; q) kfkLq (0; T ; Mb ( )) + ky0 kLNq0 =(Nq0 −2) ( ) : (C([0; T ]; Lrw ( )) denotes the space of continuous functions from [0; T ] into Lr ( ) endowed with its weak topology.) Proposition 2.2. Let f be in Ll (0; T ; Mb ( )). The equation @y + Ay = f @t

@y =0 @nA

in Q;

on ;

y(0) = 0

in

0

(10) 0

admits a unique solution in L1 (0; T ; W 1; 1 ( )); it belongs to L (0; T ; W 1; d ( )) for every (; d; l) satisfying 1 ≤  ≤ l0 ≤ ∞;

N 1 1 1 + ¡ + 0; 2d  2 l

N ¡d¡∞;

(11)

and there exists a constant C2 = C2 (; d; l) such that kykL0 (0;T ;W 1; d0 ( )) ≤ C2 kfkLl (0; T ; Mb ( )) : 0

(12)

0

For every y0 in LNq =(Nq −2) ( ); the equation @y + Ay = 0 @t

@y =0 @nA

in Q;

on ;

y(0) = y0

in ; 0

(13) 0

admits a unique solution in L1 (0; T ; W 1; 1 ( )); it belongs to L (0; T ; W 1; d ( )) for every (; d) satisfying 2¡¡∞;

1¡d ≤

Nq0 ; 2

N 1 1 1 + ¡ + 0; 2d  2 q

(14)

and there exists a constant C3 = C3 (; d; q) such that kykL0 (0; T ; W 1; d0 ( )) ≤ C3 ky0 kLNq0 =(Nq0 −2) ( ) :

(15)

Propositions 2.1 and 2.2 are proved thanks to the so-called transposition method. For this, we prove some estimates in the propositions below. Proposition 2.3. Consider the following terminal boundary value problem: −

@z + Az + az = g @t

in Q;

@z =0 @nA

on ;

z(T ) = 0 in :

(16)

˜

We suppose that a is a nonnegative function belonging to Lk (0; T ; Lk ( )) for every ˜ k) satisfying (6), and g belongs to L (0; T ; Ld ( )). Then the weak solution of (k; 0  for every (; d; l) satisfying Eq. (16) belongs to Ll (0; T ; C( )) 1 ≤  ≤ l0 ¡∞;

N ¡d¡∞; 2

N 1 1 + ¡1 + 0 ; 2d  l

(17)

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141

and there exists a constant C4 = C4 (l; ; d) not depending on the function a such that kzkLl0(0; T ; L∞ ( )) ≤ C4 kgkL (0; T ; Ld ( )) :

(18)

Proof. (1) We rst consider the case when a = 0 and when g is regular. We denote by zˆ the solution of Eq. (16) corresponding to a = 0. As in [13, 14], we use a duality method. Let  be in D( ) and y be the solution of the Cauchy problem −

@y + Ay = 0 @t

@y =0 @nA

in Q;

on ;

y(T ) =  in :

Since g is regular, by a straightforward calculation we obtain Z  Z T Z d (x)ˆz (x; t) dx = − y(x; T + t − )ˆz (x; ) dx d d

t Z =

Z

T



t

Z + Z =

T

Z

t T

t

(Ay(x; T + t − )ˆz (x; ) − y(x; T + t − )Aˆz (x; )) dx d y(x; T + t − )g(x; ) dx d

Z

y(x; T + t − )g(x; ) dx d:

(19)

Thanks to (3) we have ky(t)kLd0 ( ) ≤ K1 (d0 ; 1)(T − t)−N=2d kkL1 ( )

(20)

for all 1 ≤ d ≤ ∞ and all  ∈ D( ). From Eqs. (19) and (20), it follows Z  (x)ˆz (x; t) dx | kkL1 ( ) = 1 kˆz (t)kL∞ ( ) = Sup

Z ≤ K1

t

T

(T + t − )−N=2d kg()kLd ( ) d:

Notice that t → kg(t)kLd ( ) belongs to L (0; T ) and t → t −N=2d belongs to Li (0; T ) for ˆ = (T + t)−N=2d ]−∞;T ] every 1 ≤ i¡2d=N . If we set g(t) ˆ = kg(t)kLd ( ) ]−∞;T ] (t), h(t) RT (T +t) (where ]−∞;T ] is the characteristic function of ]−∞; T ]), then t (T +t−)−N=2d RT 0 ˆ kg()kLd ( ) d = gˆ ∗ h(t). Thus t → t (T + t − )−N=2d kg()kLd ( ) d belongs to Ll (0; T ) 0 if 1=l0 = (1=)+(1=i)−1. Therefore, zˆ belongs to Ll (0; T ; L∞ ( )) for every l0 satisfying (17) and we have ˜ kˆz kLl0(0; T ; L∞ ( )) ≤ Ckgk L (0; T ; Ld ( )) ˜ ; d). The same estimate can be obtained if g is not regular. for some constant C˜ = C(l; For that it is sucient to use an approximation process.

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

(2) Now we suppose that a is regular and nonnegative. We set g+ = max(0; g), g = max(0; −g), we denote by zˆ 1 (resp. zˆ 2 ) the solution of Eq. (16) corresponding to a = 0 and to g+ (resp. g− ) and by z1 (resp. z2 ) the solution of Eq. (16) corresponding to g+ (resp. g− ). The function w = z1 − zˆ 1 is the solution of −

−

@w + Aw + aw = −aˆz 1 @t

@w =0 @nA

in Q;

on ;

w(T ) = 0

in :

Since aˆz 1 belongs to L1 (Q) and is nonnegative, thanks to a classical comparison theorem, we obtain 0 ≤ z1 ≤ zˆ 1 a.e. on Q, and with Step 1 we have ˜ + kL (0; T ; Ld ( )) : kz1 kLl0(0; T ; L∞ ( )) ≤ kˆz 1 kLl0(0; T ; L∞ ( )) ≤ Ckg We can prove a similar estimate for z2 and we have ˜ kzkLl0(0; T ; L∞ ( )) ≤ 2Ckgk L (0; T ; Ld ( )) : Notice that C˜ is the constant of Step 1 and it is independent of a. Moreover we can 0  Indeed if g is regular the result is obeasily verify that z belongs to Ll (0; T ; C( )). vious, if not, we can consider a sequence (gn )n of regular functions converging to g in L (0; T ; Ld ( )). If zn is the solution of Eq. (16) corresponding to gn , then with the 0  previous estimate we see that (zn )n is a Cauchy sequence in Ll (0; T ; C( )), converging l0  to z (the solution of Eq. (16) corresponding to g) in L (0; T ; C( )). ˜ (3) We suppose that g belongs to L∞ ( ) and that a belongs to Lk (0; T ; Lk ( )) for ˜ k) satisfying (6). Let (an )n be a sequence of nonnegative regular functions every (k; ˜ ˜ k) satisfying (6). Denote by z n the converging to a in Lk (0; T ; Lk ( )) for every (k; solution of −

@z + Az + an z = g @t

@z =0 @nA

in Q;

on

;

z(T ) = 0:

(21)

0

The sequence (zn )n is bounded in Ll (0; T ; L∞ ( )). The sequence (zn )n is bounded in ˜ L∞ (Q) (because g ∈ L∞ (Q)), the sequence (an zn )n is bounded in Lk (0; T ; Lk ( )) for ˜ k) satisfying (6). The sequence (zn )n is also bounded in Ls (0; T ; W 1; r ( )) every (k; for (N=2) + 12 ¡(1=s) + (N=2r) [13], therefore the vector distribution dzn =dt is bounded in L (0; T ; (W 1;  ( ))0 ) for ¿1 big enough and some ¿1. Thus from Aubin Theorem [11, Theorem 1.5.1], we deduce that (zn )n is relatively compact in L1 (Q). Thus we can pass to the limit in the variational formulation satis ed by zn and we see that (zn )n converges to the weak solution z of Eq. (16) for the weak-star topology of 0 ˜ To prove that z belongs to Ll (0; T ; L∞ ( )), and that z satis es (18) with C4 = 2C. l0 ∞  L (0; T ; C( )), we notice that z ∈ L (Q) (because g ∈ L∞ (Q)) and that −

@z + Az = −az + g @t

in Q;

@z =0 @nA

on ;

z(T ) = 0:

(22)

 for every  ≥ 1 satisfying (N=2k) + Thanks to Step 1, z belongs to L (0; T ; C( )) ˜ ˜ ˜ k; ) exists be(1= k)¡1 + (1=), where (k; k) satis es (6) and k˜ ≤  (the triplet (k; 0 0 0 cause ¡N=(N − 2) and ¡(N + 2)q =(Nq − 2) ≤ ((N + 4)q − 2)=(Nq0 − 2)). We

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

143

 ∩ Ll0(0; T ; L∞ ( )), therefore z belongs to have proved that z belongs to L (0; T ; C( )) 0  Ll (0; T ; C( )). ˜ (4) We suppose that g belongs to L (0; T ; Ld ( )) and that a belongs to Lk (0; T ; k ˜ k) satisfying (6). Let (gn )n be a sequence of regular functions L ( )) for every (k; converging to g in L (0; T ; Ld ( )) for every (; d; l) satisfying (17). Let z n be the solution of Eq. (16) corresponding to gn . Thanks to estimate (18) proved in Step 3 for 0  that it converges z n , we can prove that (z n )n is a Cauchy sequence in Ll (0; T ; C( )), l0  to the weak solution z of Eq. (16) and that (18) is satis ed. in L (0; T ; C( )) Proposition 2.4. Consider the following equation: −

@z + Az = −div  @t

in Q;

@z =0 @nA

on ;

z(T ) = 0 in :

(23) 0

Suppose that  ∈ (D(Q)) N . Then the weak solution of Eq. (23) belongs to Ll (0; T ;  C( )) for every (; d; l) satisfying (11), and there exists a constant C6 = C6 (q; ; d) not depending on a such that kzkLl0(0; T ; L∞ ( )) ≤ C6 kkL (0; T ; (Ld ( )) N )

(24)

for every  ∈ (D(Q)) N . Proof. We denote by y the solution of the Cauchy problem −

@y + Ay = 0 @t

@y =0 @nA

in Q;

on ;

y(0) = 

in :

As in the proof of Proposition 2.3, by a direct calculation we have Z

Z

(x)z(x; t) =

T

Z

t

Dy(x; T + t − )(x; ) dx d:

(25)

From [8], Theorem 1.6.1, we know that the domain of Ad0 is continuously imbedded 0 in W 1; d ( ) if ¿ 12 . Thus from (4) we deduce ˜ d0 y(t)kLd0( ) ≤ CK ˜ 2 (d0 ; 1)(T − t)(−N=2d)− kkL1 ( ) ky(t)kW 1; d0( ) ≤ CkA

(26)

for every 21 ¡¡1. From Eqs. (25) and (26), it follows that ˜ 2 kz(t)kL∞ ( ) ≤ CK

Z t

T

(T + t − )(−N=2d)− k()k(Ld ( )) N d:

Notice that t → k(t)k(Ld ( )) N belongs to L (0; T ) and that, for every 1 ≤ i¡2d=(N + d), we can choose ¿ 12 such that t → t (−N=2d)− belongs to Li (0; T ). Therefore, still by

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

using integrability results for convolution products (as in the proof of Proposition 2.3), RT 0 we claim that t → t (T + t − )(−N=2d)− kg()k(Ld ( )) N d belongs to Ll (0; T ) for every l0 ≥  satisfying 12 + (1=l0 )¿(1=) + (N=2d) and that kzkLl0(0; T ; L∞ ( )) ≤ C6 kkL (0; T ; (Ld ( )) N ) 0  for C6 = C6 (l; ; d). Since  is regular, it is clear that z belongs to Ll (0; T ; C( )).

Proof of Proposition 2.1. (1) We rst prove the estimate in Lr˜(0; T ; Lr ( )) in the  and y0 ≡ 0. Let z be the solution of case when f is regular (for example f ∈ C(Q)) r˜0 r0 (16) corresponding to g ∈ L (0; T ; L ( )). If (r; ˜ r) satis es (7), we see that r˜0 ≤ q0 , 0 0 0 0 r ¿(N=2) and 1 + (1=q )¿(N=2r ) + (1= r˜ ), and Proposition 2.3 yields kzkLq0 (0; T ; L∞ ( )) ≤ C4 kgkLr˜0 (0; T ;Lr0 ( )) : Let y be the solution of Eq. (5) corresponding to y0 ≡ 0, by a Green formula, we have Z Z yg dx dt = fz dx dt: Q

Q

For every (r; ˜ r) verifying (7), we have  Z  kykLr˜(0; T ; Lr ( )) = sup yg dx dt| kgkLr˜0 (0; T ; Lr0 ( )) = 1 ≤ C4 kfkLq (0; T ; L1 ( )) Q

 In particular for r˜ = ∞ and every r¡Nq0 =(Nq0 − 2) we obtain for every f ∈ C(Q). kykL∞ (0; T ; Lr ( )) ≤ C4 kfkLq (0; T ; L1 ( )) : (2) Suppose that f belongs to Lq (0; T ; Mb ( )), and let (fn )n be a sequence of regular functions converging to f in the following sense Z

Z limn

Q

fn  dx dt =

T

0

hf(t); (t)iMb ( ) × Cb ( ) dt

 for every  ∈ C(Q);

limn kfn kLq (0; T ; L1 ( )) = kfkLq (0; T ; Mb ( )) and kfn (t)kL1 ( ) ≤ kf(t)kMb ( )

on [0; T ]:

Let yn be the solution of @y + Ay + ay = fn @t

in Q;

@y =0 @nA

on ;

y(0) = 0

in :

(27)

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

145

Thanks to the estimate in Step 1, we obtain kyn kLr˜(0; T ; Lr ( )) ≤ C4 kfn kLq (0; T ; L1 ( )) :

(28)

From (yn )n , we can extract a subsequence, still indexed by n, such that (yn )n con˜ r) satisfying verges to y for the weak-star topology of Lr˜(0; T ; Lr ( )) for every (r; (7). By passing to the limit in the variational formulation of Eq. (27), we can easily verify that y is the solution of Eq. (5) corresponding to y0 ≡ 0. By passing to the limit in (28), we obtain (8). Since the solution of Eq. (5) is unique, the original ˜ r) sequence (yn )n converges to y weakly-star in Lr˜(0; T ; Lr ( )), for every (r; satisfying (7). (3) Now we prove that the solution y of Eq. (5) corresponding to y0 ≡ 0 belongs to C([0; T ]; Lrw ( )). Since y is the weak-star limit of (yn )n in L∞ (0; T ; Lr ( )) for every 1¡r¡Nq0 =(Nq0 − 2), and since kfn (t)kL1 ( ) ≤ kf(t)kMb ( ) for almost every t ∈ [0; T ] and every n, thanks to [4], Theorem 2, we can prove that y ∈ C([0; T ]; L1 ( )). We already know that y belongs to L∞ (0; T ; Lr ( )) for every 1¡r¡Nq0 =(Nq0 − 2), therefore y belongs to C([0; T ]; Lrw ( )) for every 1¡r¡Nq0 = (Nq0 − 2) (see, e.g., [7], Chapter 18, Lemma 5.6). (4) If yˆ is the solution of @y + Ay = 0 @t then

in Q;

 ky(t)k ˆ Lr ( ) ≤ K1 r;

@y =0 @nA

Nq0 Nq0 − 2



on ;

0

y(0) = y0

in ;

(29)

0

t −N=2((Nq −2)=Nq −1=r) ky0 kLNq0 =(Nq0 −2) ( )

for every Nq0 =(Nq0 − 2) ≤ r¡∞. Therefore yˆ belongs to Lr˜(0; T ; Lr ( )) for every (r; ˜ r) satisfying together Nq0 =(Nq0 − 2) ≤ r¡∞ and (7), and we have kyk ˆ Lr˜(0; T ; Lr ( )) ≤ Cky0 kLNq0 =(Nq0 −2) ( ) 0

0

for some C = C(r; ˜ r; q). Moreover yˆ belongs to C([0; T ]; LNq =(Nq −2) ( )). Thus the estimate kyk ˆ Lr˜(0; T ; Lr ( )) ≤ Cky0 kLNq0 =(Nq0 −2) ( ) is true for all (r; ˜ r) satisfying (7). (5) The estimate of Step 4 can be proved in the case when a 6= 0 and a is regular, by using a comparison principle, as in Step 2 of the proof of Proposition 2.3. ˜ ˜ k) satisfying (6) Now, we suppose that a belongs to Lk (0; T ; Lk ( )) for every (k;  Eq. (6), and that y0 belongs to C( ). Let (an )n be a sequence of nonnegative regular ˜ k) satisfying (6). Let yn be the solution of functions, converging to a for every (k; Eq. (5) corresponding to an and to f ≡ 0, and let y be the solution of Eq. (5) corresponding to a and to f ≡ 0. By arguing as in Step 3 of the proof of Proposition 2.3,

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we can verify that (yn )n converges to y for the weak-star topology of Lr˜(0; T ; Lr ( )) for every (r; ˜ r) satisfying (7), and that (15) is satis ed. Therefore y belongs to 0 0 L∞ (0; T ; LNq =(Nq −2) ( )). We notice that (yn )n obeys @yn + Ayn + ayn = (a − an )yn @t

in Q;

@yn =0 @nA

on ;

yn (0) = y0

in

and that ((a − an )yn )n tends to zero in L (Q) for some ¿1 (it is a consequence of Lemma 2.2). As in Step 3, we can prove that y belongs to C([0; T ]; L1 ( )). Therefore 0 0 y belongs to C([0; T ]; LwNq =(Nq −2) ( )). Remark 2.3. With the notation of step 6 in the previous proof; we can write @(yn − ym ) + A(yn − ym ) + an (yn − ym ) = (am − an )ym @t @(yn − ym ) =0 @nA

on ;

(yn − ym )(0) = 0

in Q;

in :

Thanks to the method developed in the proof of Proposition 2:3; we can prove that kyn − ym kL˜(0; T ; L ( )) ≤ Ck(an − am )ym kL˜(0; T ; L ( )) with  ≤; ˜ ≤ ˜ and (N=2) + (1= )¡(1= ˜ ) ˜ + (N=2) + 1. In particular for ˜ =∞; ˜ + (1= r)=(1= ˜ k) obeys (6) and (1=k) + (1=r)=(1=); (1= k) ˜ ); ˜ ¡N=(N − 2) ; where (k; (r; ˜ r) obeys (7), we obtain kyn − ym kC([0;T ]; L ( )) ≤ Ck(an − am )kLk˜ (0; T ; Lk ( )) kym kLr˜(0; T ; Lr ( )) ; ˜ + 1. Therefore (yn )n is a Cauchy sewith (N =2) − ( =q0 )¡(N=2) + (1= )¡(N=2) quence in C([0; T ]; L ( )); when (N =2) − ( =q0 )¡(N=2) + 1. This condition is in particular satisÿed for  = 1 (because ¡q0 (N + 2)=(Nq0 − 2)) but not in general for  = Nq0 =(Nq0 − 2). Proof of Proposition 2.2. (1) We rst prove the estimate (12) in the case when  Let z be the solution of Eq. (23). If (; d; l) satis es (11), from f ∈ C(Q). Proposition 2.4, we obtain kzkLl0(0; T ; L∞ ( )) ≤ C6 kkL (0; T ; (Ld ( )) N ) for every  ∈ (D(Q)) N . Let y be the solution of Eq. (10), with a Green formula we have Z Z Dy  dx dt = fz dx dt: Q

Q

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

It follows kDykL0 (0; T ; (Ld0 ( )) N ) =

X i

147

 Z  sup Di y i dx dt ki kL (0; T ; Ld ( )) = 1 Q

≤ NC6 kfkLl (0; T ; L1 ( ))  for every f ∈ C(Q). We nally obtain (12) with this estimate and with Proposition 2.1. (2) Suppose that f belongs to Ll (0; T ; Mb ( )). Let (fn )n be a sequence of regular functions converging to f in the following sense: Z T Z  fn  dx dt = hf(t); (t)iMb ( ) × Cb ( ) dt for every  ∈ C(Q); limn Q

0

limn kfn kLl (0; T ; L1 ( )) = kfkLl (0; T ;Mb ( )) : Let yn be the solution of @y + Ay = fn @t

in Q;

@y =0 @nA

on ;

y(0) = 0

in :

(30)

Thanks to estimate (12) proved in Step 1 for regular functions, we obtain kyn kL0 (0; T ; W 1; d0 ( )) ≤ C2 kfn kLl (0; T ; L1 ( )) :

(31)

From (yn )n , we can extract a subsequence, still indexed by n, such that (yn )n con0 0 verges to y for the weak-star topology of L (0; T ; W 1; d ( )) for every (; d) satisfying (11). By passing to the limit in the variational formulation of Eq. (30), we can easily verify that y is the solution of Eq. (10). We obtain estimate (12) by passing to the limit in (31). (3) Let y be the solution of @y + Ay = 0 @t

in Q;

@y =0 @nA

on ;

y(0) = y0

in : 0

From [8] Theorem 1.6.1, the domain of Ad0 is continuously imbedded in W 1; d ( ) if ¿ 12 . With Lemma 2.1, we obtain 0

0

0

ky(t)kW 1; d0 ( ) ≤ CkAd0 S(t)y0 kLd0 ( ) ≤ CK2 t −N=2(((Nq −2)=Nq )−1=d )− ky0 kLNq0 =(Nq0 −2) ( ) if Nq0 =(Nq0 − 2) ≤ d0 ¡∞. For every (d; ) satisfying (14), we can choose ¿ 12 to have N N 1 1 1 1 N + − 0 ¡ + − 0 ¡ 0 + 0 : 2 2 q 2 q 2d  Therefore kykL0 (0; T ;W 1; d0 ( )) ≤ Cky0 kLNq0 =(Nq0 −2) ( ) : This completes the proof.

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2.2. State equation Theorem 2.4. The state equation (2) admits a unique solution; it belongs to Lr˜(0; T ; ˜ r) satisfying (7) and Lr ( )) for every (r;   kykLr˜(0; T ;Lr ( )) ≤ C1 kukLq (0; T ) + ky0 kLNq0 =(Nq0 −2) ( ) ˜ r; q) is the constant in estimate (8)). The solution of Eq. (2) also (where C1 = C1 (r; belongs to C([0; T ]; Lrw ( )) for every 1 ≤ r¡Nq0 =(Nq0 − 2) and we have   kykL∞ (0; T ;Lr ( )) ≤ C1 (∞; r; q) kukLq (0; T ) + ky0 kLNq0 =(Nq0 −2) ( ) : 0

0

0

0

It also belongs to L1 (0; T ; W 1; d1 ( ))+L2 (0; T ; W 1; d2 ( )) for every (1 ; d1 ) satisfying 1¡1 ≤

q ; q−

N ¡d1 ¡∞;

N 1 3 N ( − 1)

+ ¡ − + 0; 2d1 1 2 2 q

(32)

and every (2 ; d2 ) satisfying (14). There exists a constant C7 = C7 (1 ; d1 ; q) such that + kk 02 ky − k 01 1; d0 1; d0 L (0;T ;W 1 ( )) L (0;T ;W 2 ( ))   ≤ C7 kukLq (0; T ) + kuk Lq (0; T ) + ky0 k LNq0 =(Nq0 −2) ( ) + C3 ky0 kLNq0 =(Nq0 −2) ( ) ; where C3 is the constant in estimate (15) and  is the solution of @ + A = 0 in Q; @t

@ = 0 on ; @nA

(0) = y0

in :

Remark 2.4. Since ¡N=(N − 2)¡(N + 1)=(N − 2); we have (q − )=q¡ 32 − [N ( − 1)]=2 + ( =q0 ). Therefore the set of pairs (1 ; d1 ) satisfying (32) is nonempty. Proof. First notice that the uniqueness can be proved as in [6]. Indeed, since the coecients of the operator A are regular, if y is a solution of (2) in the sense of De nition 2.1, it also satis es  Z T Z Z  @ + yA + |y| −1 y dx dt = −y u(t)(x0 ; t) dt + (0)y0 dx @t Q

0  and @=@nA = 0 on .  such that (T ) = 0 on

for every  ∈ C 2; 1 (Q) Let (un )n be a sequence of regular functions converging to u in Lq (0; T ) such that kun kLq (0; T ) ≤ kukLq (0; T ) , let (vn )n be a sequence of nonnegative regular functions converging to x0 for the narrow topology of Mb ( ) and such that kvn kL1 ( ) = 1. We denote by yn the solution of (2) corresponding to un vn . (1) Estimate in Lr˜(0; T ; Lr ( )). By setting an = |yn | −1 , Proposition 2.1 yields   kyn kLr˜(0; T ;Lr ( )) ≤ C1 kukLq (0; T ) + ky0 kLNq0 =(Nq0 −2) ( ) for every n and every (r; r) ˜ satisfying (7). In particular, the sequence (yn )n is bounded in Lr (Q) for all 1 ≤ r¡inf((N + 2)q0 =(Nq0 − 2); N=(N − 2)) (see Lemma 2.2). Since

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

149

¡inf((N + 2)q0 =(Nq0 − 2); N=(N − 2)), for every 1 ≤ ¡inf((N + 2)q0 =[ (Nq0 − 2)]; N=[ (N − 2)]), (yn )n is bounded in L (Q) and   ˜ 1 ( ; ; q) kukLq (0;T ) + ky0 kLNq0 =(Nq0 −2) ( ) ; (33) kyn kL (Q) ≤ CC where the constant C˜ only depends on the measure of Q. Moreover, still with Proposition 2.1, for every l ≥ 1 satisfying l ≥ q and (1=q0 ) + (1= l)¿N ( − 1)=2 , we have   (34) kyn kLl (0; T ; L1 ( )) ≤ C1 ( l; ; q) kukLq (0;T ) + ky0 kLNq0 =(Nq0 −2) ( ) : 0

0

0

0

(2) Estimate in L1 (0; T ; W 1; d1 ( )) + L2 (0; T ; W 1; d2 ( )). Let n be the solution of @n + An = un vn − |yn | −1 yn @t

in Q;

@n =0 @nA

on ;

n (0) = 0

in :

Then we have yn = n + . With Proposition 2.2, for every (1 ; d1 ; l) satisfying (11), we obtain
≤ C2 (1 ; d1 ; l) kukLl (0; T ) + kyn kLl (0; T ;L1 ( )) (35) kn k 01 1; d0 1 L (0; T ;W

( ))

for every n. If ¡N=(N − 2), we have 1+( =q0 )−[N ( − 1)]=2 ≥ (q − )=q. Therefore, if (32) is satis ed, there exists l ≥ q= such that 32 +( =q0 )−N ( − 1)=2 ≥ 32 −(1=l) = 12 + (1=l0 )¿(N=2d1 )+(1=1 ) and 1¡1 ≤ l0 ≤ q=(q − ). Thanks to (34) and (35), for every (1 ; d1 ) satisfying (32), there exists a constant C7 = C7 (1 ; d1 ; q) such that  

q (0; T ) + kuk q kuk ≤ C + ky k kn k 01 0 =(Nq0 −2) 1; d0 7 L 0 Nq (0; T ) L 1 ( ) L L (0; T ;W

( ))

for every n. With Proposition 2.2, we also have kk

0

L 2 (0; T ;W

1; d0 2 ( ))

≤ C3 ky0 kLNq0 =(Nq0 −2) ( )

for every (2 ; d2 ) satisfying (14). (3) Estimates of dyn =dt. From the previous estimates we can prove that dyn =dt is a distribution with values in (W 1;  ( ))0 and that (dyn =dt)n is bounded in Ll (0; T ; (W 1;  ( ))0 ) for some l¿1 and for  big enough. Thus, from Aubin Theorem [11] Theorem 1.5.1, the sequence (yn )n is relatively compact in L1 (Q). (4) Passage to the limit. From (yn )n we can extract a subsequence, still denoted by ˜ satisfying (7), and (yn )n , weakly-star convergent to y in Lr˜(0; T ; Lr ( )) for every (r; r) 0 0 such that (yn − )n weakly converges to y −  in L1 (0; T ; W 1; d1 ( )) for every (1 ; d1 ) satisfying (32). Since (yn )n is relatively compact in L1 (Q), we can also suppose that (yn )n converges to y almost everywhere in Q. Because of (33), we can easily prove (with Holder’s inequality and Vitali’s Theorem) that (|yn | −1 yn )n converges to |y| −1 y in L (Q) for all 1 ≤ ¡inf((N + 2)q0 =[ (Nq0 − 2)]; N=[ (N − 2)]). Now by passing to the limit in the variational formulation satis ed by yn , we see that y is a solution of Eq. (2). The estimates on y clearly follow from estimates on yn proved in Step 1 and Step 2.

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3. Adjoint equation 3.1. Adjoint equation We consider the following terminal boundary value problem −

@p + Ap + ap = 0 @t

in Q;

@p =0 @nA

on ;

p(T ) = pT

in :

(36)

If u is a solution of (P), if yu is the solution of Eq. (2) corresponding to u, if we set a = |yu | −1 and pT = s|yu (T ) − yd |s−2 (yu (T ) − yd ), then Eq. (36) corresponds to the adjoint equation for (P) associated with (u; yu ). Since yu (T ) belongs to Lr ( ) for all 1 ≤ r¡Nq0 =(Nq0 − 2), we have to study Eq. (36) for pT ∈ L ( ) with ¡ Nq0 =[(Nq0 − 2)(s − 1)]. ˜

Theorem 3.1. Suppose that a is nonnegative and belongs to Lk (0; T ; Lk ( )) for every ˜ k) satisfying (6). If pT belongs to L ( ) for all 1 ≤ ¡Nq0 =[(Nq0 − 2)(s − 1)]; (k;  for every l¡2q0 =[(Nq0 − 2) the weak solution of Eq. (36) belongs to Ll (0; T ; C( )) (s − 1)]. Moreover; for every l¡(2=N ); there exists a constant C8 = C8 (l; ); not depending on the function a such that kpkLl (0; T ;C( ))  ≤ C8 kpT kL ( ) :

(37)

Proof. (1) We rst prove the result for a ≡ 0 by using estimates on analytic semigroups. We denote by pˆ the weak solution of Eq. (36) corresponding to a = 0. From Lemma 2.1, we deduce −N=2 kpT kL ( ) : kp(t)k ˆ L∞ ( ) ≤ K1 (∞; )(T − t)

Therefore pˆ belongs to Ll (0; T ; L∞ ( )) if l¡2=N and we have kpk ˆ Ll (0; T ; L∞ ( )) ≤ C8 kpT kL ( ) for some C8 = C8 (l; ). Thanks to this estimate and by using approximation arguments,  we can easily prove that pˆ belongs to Ll (0; T ; C( )). (2) We suppose that a 6= 0 and a is regular. If p is the weak solution of Eq. (36), then w = p − pˆ satis es −

@w + Aw + aw = −apˆ @t

in Q;

@w =0 @nA

on ;

w(T ) = 0:

(38)

 for every l¡2=N and a belongs to L∞ (Q), then Since pˆ belongs to Ll (0; T ; C( )) 1 apˆ belongs to L (Q). As in the proof of Proposition 2.3, we can use a comparison principle to prove that p belongs to Ll (0; T ; L∞ ( )) and that (37) is satis ed. ˜ ˜ k) satisfy(3) Now, we suppose that a belongs to Lk (0; T ; Lk ( )) for every (k;  ing (6), and that pT belongs to C( ). Let (an )n be a sequence of regular func˜ k) satisfying (6). Let pn be the solution of (36) tions, converging to a for every (k; corresponding to an and let p be the solution of Eq. (36) corresponding to a. By

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

151

arguing as in Step 3 of the proof of Proposition 2.3, we can verify that, for every 1¡l¡2q0 =[(Nq0 − 2)(s − 1)], (pn )n converges to p for the weak-star topology of Ll (0; T ; L∞ ( )) and that kpkLl (0; T ;L∞ ( )) ≤ C8 kpT kL ( ) is satis ed if l¡(2=N ) ¡  we notice that 2q0 =[(Nq0 − 2)(s − 1)]. To prove that p belongs to Ll (0; T ; C( )),  and p ∈ L∞ (Q) (because pT belongs to C( )) −

@p + Ap = −ap @t

in Q;

@p =0 @nA

on ;

p(T ) = pT

in :

(39)

 Thanks to Step 1 and to Proposition 2.3, we can prove that p belongs to L (0; T ; C( )) ˜ k) obeys ˜ where k˜ is chosen so that N + 1 ¡1 + 1 and so that (k; for every  ≥ k, 2k  k˜ (6) (since ¡N=(N − 2) such a  exists). We already know that p belongs to Ll (0; T ;  for every 1 ≤ l¡2q0 =[(Nq0 − 2)(s − 1)]. L∞ ( )), therefore p belongs to Ll (0; T ; C( )) (4) Now, we suppose that pT belongs to L ( ). We denote by (pTn )n a sequence of bounded functions converging to pT in L ( ). Let pn be the solution of Eq. (36) corresponding to pTn . Thanks to the estimate (37) established for pn in Step 3, we  that it converges to p can prove that (pn )n is a Cauchy sequence in Ll (0; T ; C( )), l  in L (0; T ; C( )) and that (37) is satis ed by p. 3.2. Green formula Lemma 3.1. Let (an )n be a sequence of nonnegative functions converging to a in ˜ ˜ k) satisfying (6). Let zn be the solution of Lk (0; T ; Lk ( )) for every (k; @z + Az + an z = vx0 @t

in Q;

@z = 0 on ; @nA

z(0) = 0

in ;

(40)

and let z be the solution of @z + Az + az = vx0 @t

in Q;

@z = 0 on ; @nA

z(0) = 0 in :

(41)

˜ r) Then (zn )n converges to z for the weak-star topology of Lr˜(0; T ; Lr ( )) for all (r; satisfying (7), and (zn (T ))n converges to z(T ) for the weak topology of Lr ( ) for all 1¡r¡Nq0 =(Nq0 − 2). ˜ r) Proof. Thanks to Proposition 2.1, (zn )n is bounded in Lr˜(0; T ; Lr ( )) for all (r; satisfying (7), and (zn (T ))n is bounded in Lr ( ) for all 1¡r¡Nq0 =(Nq0 − 2). The sequence (an zn )n is bounded in L (Q) for some ¿1 (see Lemma 2.2). Therefore (zn )n 0 0 is also bounded in L (0; T ; W 1; d ( )) for every ¿1, d¿1 such that (N=2d)+(1=)¡ 12 ([13], Theorem 4.2). From (zn )n , we can extract a subsequence, still indexed by n, weakly star converging to some z in Lr˜(0; T ; Lr ( )) for all (r; ˜ r) satisfying (7), and 0 0 weakly star converging to z in L (0; T ; W 1; d ( )) for every ¿1, d¿1 such that (N=2d) + (1=)¡ 12 . Moreover, we can suppose that (zn (T ))n converges to some z˜ for the weak topology of Lr ( ) for all 1¡r¡Nq0 =(Nq0 − 2). By passing to the limit in

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

the Green formula ([13], Theorem 4.2)   Z N X @ −zn + aij (x)Dj zn Di  + an zn  dx dt @t Q i; j=1

Z =

T 0

Z

v(t)(x0 ; t) dt −



(T )zn (T ) dx

 we see that z is the solution of Eq. (41) and that z˜ = z(T ). satis ed for every  ∈ C 1 (Q), Theorem 3.2. Let p be the solution of Eq. (36) (with pT ∈ L ( ) for every 1 ≤ ¡ Nq0 =[(Nq0 − 2)(s − 1)]) and z be the solution of @z + Az + az = vx0 @t

in Q;

@z = 0 on ; @nA

z(0) = 0 in :

(42)

˜

˜ k) satisfying (6) and that v Suppose that a belongs to Lk (0; T ; Lk ( )) for every (k; belongs to Lq (0; T ); then we have the Green formula Z Z T p(x0 ; t)v(t) dt = pT (x)z(x; T ) dx: (43) 0



 the result can be deduced from Theorem 4.2 Proof. (1) If a is regular and pT ∈ C( ), in [13].  and v ∈ L∞ (0; T ). Let (an )n be a sequence of (2) First suppose that pT ∈ C( ) ˜ k) satisfying (6). Let pn (resp. p) be regular functions converging to a for every (k; the solution of Eq. (36) corresponding to an (resp. a). Thanks to Proposition 2.3 we  and that (pn )n converges can prove that (pn )n is a Cauchy sequence in L1 (0; T ; C( ))  If zn (resp. z) is the solution of Eq. (42) corresponding to an to p in L1 (0; T ; C( )). (resp. a), then (zn (T ))n converges to z(T ) for the weak-star topology of Lr ( ) for all 1¡r¡Nq0 =(Nq0 − 2) (Lemma 3.1). Therefore we can pass to the limit in the Green formula satis ed by (pn ; zn ) and (43) is satis ed by (p; z). (3) Let us still suppose that v ∈ L∞ (0; T ). Let (pTn )n be a sequence of regular functions converging to pT in L ( ) for all 1 ≤ ¡Nq0 =[(Nq0 − 2)(s − 1)], and let pn be the solution of Eq. (36) corresponding to pn (T ) = pTn . From Theorem 3.1, we de for all 1 ≤ duce that (pn )n converges to the solution p of Eq. (36) in Ll (0; T ; C( )) 0 l¡2q0 =[(Nq0 − 2)(s − 1)], thus (pn (x0 ))n converges to p(x0 ) in Lq (0; T ) (indeed 1 ≤ s¡Nq0 =(Nq0 − 2)). Thanks to Proposition 2.1, z(T ) belongs to Lr ( ) for every 1 ≤ r¡ Nq0 =(Nq0 − 2). Since 1 ≤ s¡Nq0 =(Nq0 − 2), (z(T )pTn )n converges to z(T )pT in L1 ( ). Thus, to complete the proof in the case when a is not regular but when v ∈ L∞ (0; T ), we can pass to the limit in the Green formula satis ed by pn . (4) Now we consider a sequence (vn )n of regular functions converging to v in Lq (0; T ). If zn (resp. z) is the solution of Eq. (42) corresponding to vn (resp. v), thanks to Proposition 2.1, we see that (zn (T ))n converges to z(T ) in Lr ( ) for every 1 ≤ r¡Nq0 =(Nq0 − 2). We can pass to the limit in the Green formula satis ed by zn and the proof is complete.

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153

4. The control problem 4.1. Existence of solutions ˜ r) Lemma 4.1. If (yn )n is a sequence bounded in Lr˜(0; T ; Lr ( )) for every (r; satisfying (7), and if (yn )n converges to y in L1 (Q); then (yn )n converges to y ˜ r) satisfying together (7) and r¡∞. ˜ in Lr˜(0; T ; Lr ( )) for every (r; Proof. Let ¿0 and A be a measurable subset of Q with meas(A) = . Let (r; ˜ r) satisfying together (7) and r¡∞. ˜ Let (r˜1 ; r1 ) satisfying (7) and such that r¡ ˜ r˜1 ;

r¡r1 ;

1 1 1 1 − = − : r r1 r˜ r˜1

Let us set A(t) = A ∩ ( × {t}), |A(t)| = LN (A(t)), where LN is the N -dimensional Lebesgue measure, and denote by A the characteristic function of A. Due to Holder’s inequality and to the equality satis ed by r, r, ˜ r1 , r˜1 , we have r=r r=r Z T Z Z T Z ˜ ˜ |yn (t) − y(t)|r A (t) dx dt = |yn (t) − y(t)|r dx dt 0

Z ≤



T 0

Z ≤

˜ r=r ˜ 1 |A(t)|r=r− T

0

Z

1−r=˜ r˜1 |A(t)| dt

0

|yn (t) − y(t)|r1 dx Z

T

Z

0



A(t)

r=r ˜ 1

dt r1

|yn (t) − y(t)| dx

r˜1 =r1

!r=˜ r˜1 dt

≤ 1−r=˜ r˜1 kyn − ykrL˜r˜1 (0;T ; Lr1 ( )) : With this estimate, the lemma can be proved with Egorov’s theorem. Theorem 4.1. Let us suppose that either ¿0 or KU is bounded in Lq (0; T ). Then the control problem (P) admits solutions. Proof. Let (un )n be a minimizing sequence in KU . It is clear that (un )n is bounded in Lq (0; T ). We can suppose that (un )n converges to some u weakly-star in Lq (0; T ). Since KU is convex and closed in Lq (0; T ), then u ∈ KU . Due to Theorem 2.1, (yun )n (the sequence of solutions of (2) corresponding to un ) is bounded in Lr˜(0; T ; Lr ( )) for every (r; ˜ r) satisfying (7) and (yun − )n (where  is the function de ned in Theorem 2.1) 0 0 is bounded in in L1 (0; T ; W 1; d1 ( )) for every (1 ; d1 ) satisfying (32). By arguments similar to those in Step 3 in the proof of Theorem 2.1, we can also prove that (yun )n is relatively compact in L1 (Q). Thanks to Lemma 4.1 (yun )n converges to yu (the solution ˜ r) satisfying (7) and of Eq. (2) corresponding to u) in Lr˜(0; T ; Lr ( )) for every (r; 0 0 r¡∞ ˜ and (yun − )n converges to yu −  weakly-star in L1 (0; T ; W 1; d1 ( )) for every (1 ; d1 ) satisfying (32). Since for all 1¡s¡Nq0 (Nq0 − 2)   kyun (T )kLs ( ) ≤ C1 (∞; s; q) kun kLq (0; T ) + ky0 kLNq0 =(Nq0 −2) ( ) ;

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

we can also prove that (yun (T ))n (or at least a subsequence) converges to some yT for the weak-star topology of Ls ( ). To prove that yT = yu (T ), we use the Green formula of Theorem 3.2. We introduce the solution pn of −

@p + Ap + an p = 0 @t

in Q;

@p =0 @nA

on ;

p(T ) =  in ;

where an = |yun | −1 and  ∈ D( ), and we introduce the solution p of −

@p + Ap + ap = 0 @t

in Q;

@p =0 @nA

on ;

p(T ) =  in ;

where a = |yu | −1 . Due to previous convergence results, (an )n converges to a in ˜ ˜ k) satisfying (6). We notice that w = pn − p satis es Lk (0; T ; Lk ( )) for all (k; −

@w + Aw + an w = (a − an )p @t

in Q;

@w =0 @nA

on ;

w(T ) = 0

in :

If q ≤ , with estimate (18), we have kpn − pkLq0 (0; T ; C( ))  ≤ C4 (q; ; d)k(a − an )pkL (0;T ; Ld ( )) ≤ C4 ka − an kL (0; T ; Ld ( )) kpkL∞ (Q) for  ≤ q0 and (N=2d) + (1=)¡1 + (1=q0 ). Since ¡(N + 2)q0 =(Nq0 − 2), we have ˜ k) = (; d) = (q0 ; d) satisfying (N=2 − 1=q0 )( − 1)¡1 + (1=q0 ) and we can choose (k; 0 0 ((N=2) − (1=q ))( − 1)¡(N=2d) + (1=)¡1 + (1=q ) = 1 + (1=). If q¿ , still with estimate (18), we have kpn − pkLq0 (0; T ; C( ))  ≤ Ckpn − pkLq=( −1) (0; T ; C( ))    q ≤ CC4 ; ; d ka − an kL (0; T ; Ld ( )) kpkL∞ (Q) ; q− +1 ˜ k) = (; d) satisfying if  = q=( − 1) and N=2d¡1. Since ¡N=(N − 2), we choose (k; N=(2d)¡1 and ((N=2) − (1=q0 ))( − 1)¡(N=2d) + (1=). Thus (pn )n converges to p 0  With the Green formula (43) for pn and yun we have in Lq (0; T ; C( )). Z Z T pn (x0 ; t)un (t) dt = (x)yun (x; T ) dx:

0

By passing to the limit, we obtain Z Z T p(x0 ; t)u(t) dt = (x)yT dx: 0



We also have Z Z (x)yu (x; T ) dx =

0

T

p(x0 ; t)u(t) dt

RT for every  ∈ D( ). Thus we obtain yu (T ) = yT . Since the mapping u → 0 |u|q dt is lower semicontinuous for the weak-star topology in Lq (0; T ), and the mapping

J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

155

R y → |y − yd |s dx is lower semicontinuous for the weak-star topology in Ls ( ), by classical arguments, we prove that u is a solution of (P). 4.2. Optimality conditions Theorem 4.2. If u is a solution of (P); then Z T (p(x0 ; t) + q|u(t)|q−2 u(t))(v − u)(t) dt ≥ 0 0

(44)

for every v ∈ KU ; where p is the solution of @p + Ap + |yu | −1 p = 0 in Q; @t @p = 0 on ; p(T ) = s|yu (T ) − yd |s−2 (yu (T ) − yd ) in ; @nA

−

(45)

and yu is the solution of Eq. (2) corresponding to u. Proof. Let v be in KU , ¿0, denote by y the solution of Eq. (1) corresponding to u + (v − u). The function w = y − yu satis es @w @w + Aw + a w = (v − u)x0 in Q; = 0 on ; w(0) = 0 in ; @t @nA R1 where a = 0 |yu + (y − yu )| −1 d. From estimate (8), still true for the above equation, we obtain ˜ r; q)kv − ukLq (0; T ) kwkLr˜(0; T ; Lr ( )) ≤ C1 (r; for every (r; ˜ r) satisfying (7). Therefore, when  tends to zero, y tends to yu in ˜ r) satisfying (7)) and y (T ) tends to yu (T ) in Lr ( ) for Lr˜(0; T ; Lr ( )) (for all (r; all 1 ≤ r¡Nq0 =(Nq0 − 2). In particular y (T ) tends to yu (T ) in Ls ( ). It also follows ˜ ˜ k) satisfying (6). Now that a tends to a = |yu | −1 in Lk (0; T ; Lk ( )) for every (k; we set z = (y − yu )= and we denote by z the solution of @z + Az + |yu | −1 z = (v − u)x0 in Q; @t @z = 0 on ; z(0) = 0 in : @nA topology of Lr ( ), for all Due to Lemma 3.1, z (T ) tends to z(T R ) for the weak 0 0 s 1 ≤ r¡Nq =(NqR − 2). If we set I (u) = |yu (T ) − yd | dx, from the convexity of the mapping y → |y − yd |s dx, it follows Z I (u + (v − u)) − I (u) s |yu (T ) − yd |s−2 (yu (T ) − yd )z (T ) dx ≤ 

Z ≤ s |y (T ) − yd |s−2 (y (T ) − yd )z (T ) dx:

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J. Droniou, J.-P. Raymond / Nonlinear Analysis 39 (2000) 135 – 156

We set F(u) = J (yu ; u), thanks to the above calculations we obtain Z s|yu (T ) − yd |s−2 (yu (T ) − yd )z(T ) dx F 0 (u)(v − u) =

Z

+

0

T

q|u(t)|q−2 u(t)(v − u)(t) dt: ˜

Now, if p is the solution of Eq. (45), we notice that |yu | −1 belongs to Lk (0; T ; Lk ( )) ˜ k) satisfying (6), and that s|y(T ) − yd |s−2 (y(T ) − yd ) belongs to L ( ) for every (k; for 1 ≤ ¡Nq0 =[(Nq0 − 2)(s − 1)]. Therefore we can use the Green formula of Theorem 3.2 to complete the proof. References [1] H. Amann, Dual semigroups and second order linear elliptic boundary value problems, Is. J. Math., 45 (1983) 225 – 254. [2] S. Anita, Optimal impulse-control of population dynamics with diusion, in: V. Barbu (Ed.), Dierential Equations and Control Theory, Longman, Harlow, 1991, pp. 1– 6. [3] S. Anita, Optimal control of parameter distributed systems with impulses, Appl. Math. Optim. 29 (1994) 93 –107. [4] P. Baras, Compacite de l’operateur f → u solution d’une e quation non lineaire f ∈ (du= dt) + Au, C. R. Acad. Sci. Paris, Serie A, 286 (1978) 1113–1116. [5] P. Baras, M. Pierre, Problemes paraboliques semilineaires avec donnees mesures, in: Thesis of P. Baras, Paris VI, 1985. [6] H. Brezis, A. Friedman, Nonlinear parabolic equations involving measures as initial conditions, J. Math. Pures Appl., 62 (1983) 73 – 97. [7] R. Dautray, J.-L. Lions, Analyse Mathematique et Calcul Numerique, Tome 8, Masson, Paris, 1987. [8] D. Henry, Geometric Theory of Semilinear Parabolic Equations, Springer, Berlin, 1981. [9] J.-L. Lions, Some Methods in the Mathematical Analysis of Systems and their Control, Science Press, Beijing and Gordon and Breach, New York, 1981. [10] J.-L. Lions, Pointwise control for distributed systems, in: H.T. Banks (Ed.), Control and Estimation in Distributed Parameters Sytems, SIAM, Philadelphia, 1992, pp. 1–39. [11] J.-L. Lions, Quelques Methodes de Resolution des Problemes aux Limites non Lineaires, Dunod, Paris, 1969. [12] A. Pazy, Semigroups of Linear Operators and Applications to Partial Dierential Equations, Springer, Berlin, 1983. [13] J.P. Raymond, Nonlinear boundary control of semilinear parabolic equation, Discrete Contin. Dyn. System 3 (1997) 341–370. [14] J.P. Raymond, H. Zidani, Hamiltonian Pontryagin’s principles for control problems governed by semilinear parabolic equations, Appl. Math. Optim., to appear [15] J. Simon, Caracterisation d’un espace fonctionnel intervenant en contrˆole optimal, Ann. Fac. Sciences Toulouse 5 (1983) 149 –169.