Pontryagin Maximum Principle for Semilinear and Quasilinear Parabolic Equations with Pointwise State Constraints Bei Hu1 and Jiongmin Yong2
Abstract. This paper studies the rst order necessary conditions for the optimal controls of semilinear and quasilinear parabolic partial di erential equations with pointwise state constraints. Pontryagin type maximum principle is obtained.
Keywords. parabolic equations, pointwise state constraints, optimal control, Pontryagin maximum principle, Ekeland variational principle.
AMS(MOS) subject classi cations. 49K20, 35K10, 35K20. x1. Introduction. In this paper, we are concerned with the following parabolic equation: (1:1)
8 > > > yt > >
y @ = 0; > > > > : y = y0(x);
t=0
in T ;
x 2 ;
where aij and f are some given functions, T = (0; T ) with lRn being a bounded domain and T > 0 being a given time duration. The function u(x; t) is called the control, which takes value in some separable metric space U . The solution y(x; t) to (1.1) (for given y0 (x) and u(x; t)) is called the state of the system and y0 (x) is referred to as the initial state. We set U = fu : T ! U u is measurable g. Under proper conditions (see x2), we have that for any y0 2 C0 ( ) and u 2 U , (1.1) admits a unique solution y(x; t) which is in 1 2
Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556. This author was partially supported by US NSF Grant DMS 90-24986 and DMS 92-24935. IMA, University of Minnesota, Minneapolis, MN 55455; on leave from Department of Mathematics, Fudan University, Shanghai 200433, China. This author was partially supported by NSF of China and Fok Ying Tung Education Foundation. 1
C ( T ) (actually, it is even better, see x2). Then, we may talk about the state constraint of form: G(y) 2 Q;
(1:2)
for some continuously Frechet di erentiable map G : C0( T ) ! Z , where C0( T ) = f 2 C ( T ) @ [0;T ] = 0g, Z is some Banach space and Q Z . Let us take a look of two important examples of the above type constraints. First, let Z = C0( T ), (1:3)
Q = f 2 Z (x; t) 0; (x; t) 2 T g
and G(y) = g(x; t; y(x; t)) for some function g : T lR ! lR. Then, (1.2) reads (1:4)
g(x; t; y(x; t)) 0;
(x; t) 2 T :
Our second example is as follows: We again let Z = C0( T ), and de ne (1:5)
Q = f 2 Z (xi; ti ) = bi ; 1 i mg S
for some given (di erent) points (xi ; ti ) 2 T ( fT g) and numbers bi , and G = I , the identity operator on Z . Then, (1.2) reads (1:6)
1 i m:
y(xi ; ti ) = bi ;
For elliptic equations, similar constraints like (1.4) and (1.6) were considered in [4, 5, 19]. The above two examples all require pointwise behavior of the state y(x; t). There are many other examples covered by (1.2) (see x5). It is seen that our state constraint (1.2) is very general. Now, we introduce the following functional: (1:7)
J (u) =
Z
T
f 0 (x; t; y(x; t); u(x; t))dxdt;
for some function f 0 , where y(x; t) is the solution of (1.1) corresponding to u. This is called the cost functional. Next, we set (1:8)
Uad fu 2 U the corresponding y satis es (1.2) g: 2
Any element u 2 Uad is called an admissible control. In what follows, we assume that Uad 6= ;. Then, we may state our optimal control problem as follows:
Problem C. Find u 2 Uad, such that (1:9)
J (u) = Uinf J (u): ad
Whenever such a u 2 Uad exists, we call it an optimal control; the corresponding state y is called an optimal state and (y; u) is called an optimal pair. Our goal is to obtain a set of rst order necessary conditions for the optimal pairs. This set of conditions is called the Pontryagin maximum principle. In recent papers [3, 4, 5, 20], the Pontryagin maximum principle was derived for semilinear and quasilinear elliptic partial di erential equations with pointwise state constraints (see [2] also). For parabolic equations, abstract evolution equation setting was used a little earlier to obtain similar results ([9, 10, 14, 15, 18]). We note that by using the abstract framework for parabolic equations, people treat the time variable t and the spatial variable x unequally, in the sense that the variable x is \averaged" and actually does not appear explicitly in the whole process. Consequently, some pointwise information of the state y(x; t), like Holder continuity, values at some particular points (x0 ; t0) 2 T , are lost. In particular, the problem with the state constraint (1.6) can not be covered by abstract framework. In this paper, we use the idea of [5] (see [13, 20] also) to discuss the optimal control problem for parabolic equations without using the abstract evolution equations. By this approach, we retain some pointwise behavior of the state y(x; t). Consequently, we can treat general (pointwise) state constraint (1.2), which contains (1.6) as a special case. Due to the fact that U is just a separable metric space, only the spike perturbation of the control is allowed when we derive the necessary conditions. On the other hand, the pointwise state constraint is presented. These two together cause the main diculty in our procedure. The key which overcomes this main diculty is to nd the \Taylor expansion" of the state with respect to the spike variation of the control, in a strong enough topology that is sucient for us to treat the pointwise constraint. We achieve this by improving a technical lemma found in [5] and using proper estimates for parabolic equations. Once 3
this is obtained, we then use the usual procedure of applying Ekeland variational principle to derive the desired conclusion. We refer the readers to [17, 19] for some classical relevant results. The rest of the paper is organized as follows. In Section 2, we give some preliminary result and state the main result. Section 3 is devoted to proving some technical lemmas. The proof of Pontryagin maximum principle is carried out in Section 4. Some applications are given in Section 5. The result corresponding to quasilinear parabolic equations is brie y discussed in Section 6.
x2. Preliminary and the Main Result. Let us rst give some assumptions and preliminary results. We let lRn be a S bounded domain with @ 2 C 2, T = (0; T ), @p T = ( f0g) (@ [0; T ]) be the parabolic boundary of T , U be a separable metric space. We use j j or k k (some times with a subscript) as the norm in various spaces, which can be identi ed from the context. For any measurable set S lRn, we use jS j to denote the Lebesgue measure of the set S . In what follows, we will denote by C0( T ) C ( T ) the set of all continuous function on
T which vanish on @ [0; T ], by C ; =2( T ) the set of all continuous functions on T which are Holder continuous in (x; t) with the exponent in x and =2 in t ( 2 (0; 1)) and by C0( ) the set of all continuous functions on which vanish on @ . The following assumptions will be assumed throughout of the paper. (A1) The function aij : T ! lR is measurable, aij = aji and there exist constants > > 0, such that (2:1)
for a.e. (x; t) 2 T ; 2 lRn:
jj2 aij (x; t)i j jj2;
(A2) The function f : [0; T ] lR U ! lR has the following properties: f (; ; y; u) is measurable on [0; T ], f (x; t; ; u) is in C 1(lR) with f (x; t; ; ) and fy (x; t; ; ) being continuous on lR U . There exists a constant C > 0, such that (2:2)
f (x; t; y; u)y C (jyj2 + 1);
8(x; t; y; u) 2 [0; T ] lR U: 4
Moreover, for any R > 0, there exists an MR > 0, such that (2:3)
jf (x; t; y; u)j + jfy (x; t; y; u)j MR ;
8(x; t; u) 2 [0; T ] U; jyj R:
The same conditions, except (2.2), hold for the function f 0 : [0; T ] lR U ! lR. (A3) Z is a Banach space with the dual Z being strictly convex. Q Z is convex and closed, and is of nite codimension in Z (see below or [15, De nition 2.2]). The map G : C0( T ) ! Z is continuously Frechet di erentiable. Let us make some remarks on (A3). First, a set Q Z is said to be nite codimensional in Z if for some z0 2 Q, the space Z0 spanned by Q z0 fz z0 z 2 Qg is a nite codimensional subspace of Z and the convex hull co(Q z0) of Q z0 has a nonempty relative interior in Z0 . It is not hard to see that the set Q de ned by (1.3) has a nonempty interior in Z and hence is of codimension 0 in Z ; the set Q de ned by (1.5) is of codimension m in Z . Next, we consider the case Z = C0( T ). Then, by Hahn-Banach Theorem, for any 2 Z C0( T ) , there exists a e 2 C ( T ) = M( T ) (the set of all Radon measures on T ), such that = e C0( T ) and (2:4)
h e; i =
Z
T
Then, for any 2 C0( T ) (note @ = 0) (2:5)
h ; i =
Z
T
8 2 C ( T ):
de;
d =
Z S
T ( f0;T g)
d:
In what follows, we let M0 ( T ) be the set of all Radon measures on T with the support S contained in T ( f0; T g). Clearly, M0 ( T ) = C0( T ) with the identi cation being (2.5). It is known that if we use the usual norm in C0( T ), the dual C0( T ) of it is not strictly convex. However, since the space C0( T ) is a separable Banach space, by [7, p.167], there exists a norm, denoted by j j0, which is equivalent to the norm k kC0( T ), such that the dual of (C0( T ); jj0 ) is strictly convex. It is clear that any element 2 (C0( T ); jj0 ) can still be identi ed with an element of M0( T ), such that (2.5) holds. This will be useful when we discuss the case with Z = C ( T ) (see x5). 5
Next, we de ne
dQ () = inf jz j; 2Q
(2:6)
8 2 Z:
Then, dQ : Z ! lR is convex and Lipschitz continuous (with the Lipschitz constant being 1). From [6], we know that the Clarke's generalized gradient, denoted by @dQ , which coincides with the subdi erential in the sense of the convex analysis in this case [6, Proposition 2.2.7], is convex and weak -compact. Therefore, given 2 @dQ (), we have that
h ; z i +dQ () dQ (z);
(2:7)
8z 2 Z:
This implies that j h ; z i j jz j, for all z 2 Z , since dQ () is Lipschitz continuous with Lipschitz constant 1. Thus, kkZ 1. The identity kkZ = 1 being true whenever 2= Q; see [15, Lemma 3.4]. Since Z is strictly convex, @dQ () is a singleton for every 2= Q [15, Corollary 3.5]. Furthermore, dQ : Z ! lR is G^ateaux di erentiable at every point 2= Q and frdQ()g = @dQ () [6, Proposition 2.2.4], where rdQ() is the G^ateaux derivative of dQ () at . Hence
krdQ()kZ = 1;
(2:8)
8 2= Q:
The following result is basic.
Proposition 2.1. Let (A1){(A2) hold. Then, for any u 2 U and y0 2 C ( ) T C0( ) (0 < < 1), there exists a 2 (0; 1), such that (1.1) has a unique solution y y(; ; u) 2 T
C ; =2( T ) L2(0; T ; H01 ( )). Furthermore, there exists a constant C > 0, independent of u 2 U , such that (2:9)
ky(; ; u)kC ; =2 ( T ) C;
8u 2 U :
Sketch of the Proof. Uniqueness follows immediately from the energy estimates. For the existence, it suces to establish the a-priori estimates for the solution. The assumption (2.2) immediately gives us the L1( T ) estimates. Then the standard energy inequality gives L2(0; T ; H 1 ( )) estimates, and the existence of the solution follows. The estimates (2.9) is standard and can be found in [12, Chapter III, section 10].
6
T
In what follows, any pair (y; u) 2 C ; =2( T ) C0( T ) U satisfying (1.1) is called a feasible pair and we refer to the corresponding y and u as feasible state and control, respectively. Clearly, under (A1){(A2), U coincides with the set of all feasible controls and for each feasible control u 2 U there corresponds a unique feasible state. Also, we see that the cost functional J (u) is well-de ned for each u 2 U and the state constraint (1.2) clearly makes sense. Now, we assume that the set Uad de ned in (1.8) is nonempty and there exists an optimal pair (y; u) to Problem C. Our main result then can be stated as follows:
Theorem 2.2. (Maximum Principle) Let (A1){(A3) hold and let the following compatibility condition, for the set Q, the map G and the initial state y0 , holds: [
supp G0 () @dQ (G()) T ( fT g); (2:10) 8 2 C0( T ) with G() 2 Q; t=0 = y0(x): Let (y; u) be an optimal pair of Problem C. Then, there exists a constant 0 0, a function +2 ), and a ' 2 @dQ (G(y )) Z , such that 2 Lq (0; T ; W01;q ( )) (1 < q < nn+1
j 0 j + k'kZ > 0;
(2:11)
(2:12)
(2:13) (2:14) where (2:15)
8 > > > > > > > > > < > > > > > > > > > :
t+
n X
(aij (x; t) xj )xi = fy (x; t; y (x; t); u(x; t))
i;j =1
0 f 0 (x; t; y (x; t); u(x; t)) + (G0 (y ) ') ; y
T
@ = 0;
0
in T ;
')
t=T = (G (y )
fT g ;
h z G(y); ' i 0;
8z 2 Q:
H (x; t; y (x; t);u(x; t); 0 ; (x; t)) = max H (x; t; y (x; t); v; 0 ; (x; t)); v2U a.e. (x; t) 2 [0; T ]: H (x; t; y; u; 0 ; ) = 0f 0 (x; t; y; u) + f (x; t; y; u); 8(x; t; y; u; 0; ) 2 [0; T ] lR U lR lR: 7
In the above, (2.12) is called the adjoint equation, (2.13) is called the transversality condition and (2.14) is called the maximum condition. It will be seen that in the proof, we only need (2.10) holds for = y . Also, in x5, we will give some examples for which such a compatible condition holds.
x3. Some Technical Lemmas. In order to derive the rst order necessary conditions for optimal pairs, we need some sort of \directional derivatives" of the state y and the cost functional J (u) in the control variable u. However, since the control domain U is just a metric space and there is no convexity in general, the perturbation of the control variable is restricted to be of \spike" type. Thus, to nd the \directional derivative" is not obvious. In this section, we will present some technical lemmas which will give us exactly the \directional derivatives" we need in the proof of the maximum principle in x4. The results of this section is comparable with those in [5] for elliptic equations (see [13{15, 20] also).
Lemma 3.1. Let h0 2 L1( ) and h 2 Lp( ), 1 < p < 1. For any 2 (0; 1), we de ne (3:1)
E = fE E measurable and jE j = j j g:
Let the embedding Y ,! Lp0 ( ) be compact (p0 = p p 1 ). Then, (3:2)
Z inf 1 E 2E
1 (x) h0(x)dx +
1 1 h
= 0: E E Y
Proof. Let 2 (0; 1) be given and let > 0 be arbitrary. We let B be the closed unit ball in Y . This set is compact in Lp0 ( ) by our assumption. Thus, we can nd a set of nitely many step functions fi ; 1 i rg, such that for any y 2 B, there exists a i 2 satisfying
ky i kLp0 ( ) < :
(3:3)
Since is a nite set, we may let fKj gmj=1 be a partition of with jKj j > 0 for each 1 j m, such that (3:4)
i (x) =
m X j =1
cij Kj (x); 8
x 2 1 i r:
Then, for any y 2 B, by choosing i 2 with the property (3.3), we have (3:5) Z Z m p0 X 1 y ( ) d y (x) dx jK j j
j =1 Kj
Kj
m nZ 0 1X p 3 jy(x) K j j =1 Z
i (x)jp0 dx +
Z
Z
Kj
i (x)
1 Z ()d p0 dx jKj j Kj i
p0 o 1 + ( y ( ) ( )) d dx i Kj jKj j Kj ) (Z Z m 1 Z X 0=p 0 0 1 0 p p p p 3 jy(x) i (x)j dx + jK jp0 jKj j jy() i ()j ddx
Kj Kj i=1 j
3p0 1
p0 +
Z
jy()
i ()jp0 d
2 3p0 1 p0 :
Here, we have used the fact that i (x) is a constant on each set Kj . We have seen that the above estimate is uniform for y 2 B. Now, for any y 2 B, let us de ne ye : ! lR to be the following: Z 1 (3:6) ye(x) = jK j y()d; x 2 Kj ; 1 j m: j Kj Then, (3.5) can be written as (by setting "p0 = 2 3p0 1p0 , which is still arbitrary)
ky yekLp0 ( ) < ";
(3:7)
y 2 B:
Next, on each Kj , we approximate the function h0 and h by step functions:
rj rj X X 0 (3:8) hj (x) = ij Fij (x); hj (x) = ij Fij (x); x 2 ; i=1 i=1 j being a partition of K , jF j > 0, and such that with ij ; ij 2 lR, fFij gri=1 j ij
(3:9)
Z
Kj
jh0(x)
h0j (x)jdx +
Z
jh(x) hj (x)jdx < "jKj j;
Kj Let us take Eij Fij , such that jEij j = jFij j.
are simple functions, we have (3:10)
8Z > > > < Kj Z > > > :
Sj Set Ei = ri=1 Eij . Since h0j (x) and hj (x)
Z
1 j (x)h0 (x)dx; j Kj E Z 1 j (x)h (x)dx; hj (x)dx = j E Kj Kj
h0j (x)dx =
1 j m:
9
1 j m:
S Finally, we take E = mj=1 Ej . Then, E 2 E , and for any y 2 B, Z 1
(3:11)
1 h(x)y(x)dx Z 1 1 h(x)ye(x)dx E E
Z + 1 + 1 E jh(x)j jy(x) ye(x)jdx
m Z X 1 j =1 Kj
From (3.7) we see that
1 j (x) E
h(x)ye(x)dx + 1 +
1 khk p ky yek 0 : L ( ) Lp ( )
(3:12) 1 + 1 khkLp ( )ky yekLp0 ( ) " 1 + 1 khkLp( ): On the other hand, we notice that ye(x) is a constant on each set Kj (we denote this constant by ye(Kj )). Thus, by (3.9){(3.10) and (3.6), we have m Z X 1 j =1 Kj
(3:13)
m X j =1
1 j (x) E
jye(Kj )j
Z
Kj
h(x)ye(x)dx
1 + 1 Ej (x) jh(x) hj (x)jdx
m Z X + ye(Kj ) 1 Kj j =1 X m
1 j h (x)dx E j
1 + 1 "jKj j jye(Kj )j " 1 + 1 kykL1 ( ) " 1 + 1 C: j =1 Here, kykL1( ) C kykY C , since y 2 B. Thus, (3.11){(3.13) imply that
1 1
1
(3:14)
E h Y " 1 + (C + khkLp ( )): On the other hand by (3.9){(3.10) again, we have Z 1
(3:15)
1 (x) E
0 h (x)dx
m Z X 1 Kj j =1
1 j (x) E
0 h (x)dx
m Z X 1 0 0 1+ jh (x) hj (x)jdx + 1 1 Ej h0j (x)dx Kj j =1 j =1 Kj < " 1 + 1 : m X
Z
10
Therefore, our conclusion follows. The above result was proved for Y = W01;p( ) in [5] using di erent methods. The proof given here is inspired by a personal communication of the second author with E. Casas. Now we consider the equation (3:16)
8 > < t
(aij (x; t)xi )xj + c(x; t) (x; t) = 1 1 E (x; t) h(x; t); in T ; > : @p T = 0: S
where aij satis es (A1) and @p T = (@ [0; T ]) ( f0g) is the parabolic boundary of
T . It is clear that the solution = E (x; t) is uniquely determined by the choice of the coecient c and the set E.
Lemma 3.2. Suppose that n+2 2 < p < 1. Then, there exists a 2 (0; 1) such that for each h0 2 L1( T ), h 2 Lp( T ), K = fc(x; t); kckL1 ( T ) K g (K > 0), and any 2 (0; 1), (3:17)
inf sup
E 2E c 2K
Z
T
1 1 E (x; t) h0(x; t)dxdt + E C;=2 ( T ) = 0:
Proof. By the assumption, h 2 Lp( T ), p > n+2 2 . Therefore the right-hand-side of the equation (3.16) is in Lp( T ), (although the Lp( T ) norm may blow up as ! 0). It follows from the parabolic estimates [12, Chapter III, section 10] that there exists 2 (0; 1), such that
sup sup kE kC ; =2 ( T ) C:
(3:18)
E 2E c 2K
We claim that, for any 0 < < , (3.17) hods. In fact, we rst note that the identity mapping C ; =2( T ) ,! Lp0 ( T ) is compact. Therefore if we take Y = C ; =2( T ) in Lemma 3.1, then (3:19)
Z inf sup 1 E 2E c 2K T Z inf 1 E 2E
T
1 h0dxdt + Z 1 1 h dxdt E E E
T
1 h0dxdt + C
1 1 h
= 0: E E Y 11
Using a change of variable (x; t) = (x; t)eKt if necessary, we may assume without loss of generality that c(x; t) 0. Multiplying equation (3.16) with E and integrating over
T , we immediately obtain Z
(3:20)
(E )2 (x; T )dx +
Z
Z Z T C 1 0
T
jrE (x; )j2 dxd
1 h E
E dxdt :
Notice that E = 0 on @ ftg, for each t 2 (0; T ). Therefore, Z
jrE j2(x; t)dx,
Z
(E )2 (x; t)dx
C by Poincare's inequality. Integrating over t 2 [0; T ], and taking
(3.20) into account, yields (3:21)
Z
Z
T
(E )2 (x; t)dxdt C
T
1 1 E (x; t) h(x; t)E (x; t)dxdt :
By the interpolation theorem (see Lemma 3.4 below), for any " > 0, there exists C" > 0, such that (3:22)
k kC;=2( T ) "k kC ; =2 ( T ) + C"k kL2 ( T ); 8 2 C ; =2( T ):
Using (3.19), (3.21) and (3.22), we obtain inf sup
E 2E c 2K
Z
T
1 1 E (x; t) h0(x; t)dxdt + E C;=2( T ) "C:
Since " can be arbitrarily small, the lemma follows. Now, for any feasible pair (y; u), we de ne 8 < c(x; t)
= fy (x; t; y(x; t); u(x; t)); : c0(x; t) = f 0 (x; t; y (x; t); u(x; t)); y
(3:23) and for given v 2 U , (3:24)
(
h(x; t) = f (x; t; y(x; t); v (x; t)) f (x; t; y(x; t); u(x; t)); h0 (x; t) = f 0 (x; t; y(x; t); v(x; t)) f 0 (x; t; y(x; t); u(x; t)): 12
Consider the following problem (3:25)
8 > > > :z
(aij (x; t)zxi )xj + c(x; t)z = h(x; t);
i;j =1
in T ;
@p T = 0: T
Clearly, since h 2 L1( T ), this problem admits a unique solution z 2 C ; =2( T ) L2(0; T ; H01 ( )), as in Proposition 2.1. Our main result of this section is the following.
Theorem 3.3. Let (y; u) be a given feasible pair and v 2 U be xed. Then, for any 2 (0; 1), there exists a measurable set E T , with property jEj = j T j, such that if we de ne u by (3:26)
u(x; t) =
(
if (x; t) 2 T n E; if (x; t) 2 E;
u(x; t); v(x; t);
and let y be the state corresponding to u, then the following hold: 8 < y
= y + z + r ; 1 : lim kr kC ;=2 ( ) = 0; T !0
(3:27) for some 2 (0; 1), and
8 < J (u )
= J (u) + z0 + r0 ; 1 : lim jr0 j = 0; !0
(3:28)
where z is the solution of (3.25) and z0 is given by (3:29)
z0
=
Z
[c0(x; t)z(x; t) + h0 (x; t)]dx:
Proof. First, we recall the so-called Ekeland distance. For any u; v 2 U , we let
(3:30)
d(u; v) = jf(x; t) 2 u(x; t) 6= v(x; t)gj:
It is standard that (U ; d(; )) is a complete metric space (see [8]). Clearly, d(u; v) jEj. 13
Now, we set
z(x; t) = y(x; t) y(x; t) ;
(3:31)
x 2 :
Then, z satis es the following (3:32)
8 > > < (z
)t
> > :z
(aij (x; t)(z )xi )xj + c(x; t)z = 1 E (x; t)h(x; t);
i;j =1
@p T = 0;
where (3:33)
n X
c(x; t) =
Z
1 0
fy (x; t; y(x; t) + (y (x; t) y(x; t)); u (x; t))d :
We see that (note (2.9) and (2.3)), c(x; t) and h(x; t) are uniformly bounded (with the bounds independent of E, the controls u and v). The function h(x; t) is actually independent of the set E; we shall use this fact when we apply Lemma 3.2. Since h 2 L1( T ) Lp( T ) for any p > 1, the parabolic Holder's estimates implies that y y = z satis es, for xed p > n+2 2 , (3:34)
ky ykC ; =2 ( T ) C k E kLp ( T ) !(C) ! 0;
as ! 0;
where the constant C is independent of E, and ! is a modulus of continuity. It follows that (3:35)
c(x; t) ! c(x; t) fy (x; t; y(x; t); u(x; t));
in Lp( T ); 1 p < 1:
By recalling z, the solution of (3.25), we have the following: (3:36)
8 > > > (z > > > >
> > > > > > : (z
= (c(x; t) c(x; t))z z) @p T = 0;
n X
i;j =1
(aij (x; t)(z z)xi )xj + c(x; t)(z z)
1 1 E (x; t) h(x; t);
We note that the above equation is linear in (z z). Thus, we may write z z = + with and satisfying the following: (3:37)
8 > > < (
)t
> > :
n X
i;j =1
aij (x; t)( )xi
xj + c (x; t) =
@p T = 0;
14
(c(x; t) c(x; t))z
and (3:38)
8 > > < (
)t
n X
(aij (x; t)( )xi )xj + c(x; t) =
> > :
i;j =1
1 1 E (x; t) h(x; t);
@p T = 0;
By Holder estimates again, (notice that z 2 C ; =2( T ) L1( T ), and p is xed with p > n+2 2 ), we have (3:39)
k kC ; =2 ( T ) C k(c c)zkLp ( T ) = o(1) as ! 0;
all the constants involved in the above are independent of the choices of E. Now we x 2 (0; ) as in Lemma 3.2. Then we can choose E T with the property jEj = j T j, such that the solution of (3.38) satis es (3:40)
Z
T
1 1 E (x; t) h0(x; t)dxdt + kkC;=2 ( T ) :
This proves (3.27). The proof of (3.28) is similar but simpler. The above result will play a very important role in the proof of our main result (Theorem 2.2). Conclusion (3.27) gives a \Taylor expansion" (of rst order) in the space C ;=2( T ) C ( T ). This will be sucient for us to deal with the pointwise state constraint. It is not hard for us to see that the stronger the topology under which (3.27) holds, the harder for us to prove it. Thus, for example, if in (3.27), C ;=2( T ) is replaced by Lp( T ), then it will be much easier to prove it. In another word, an Lp( T ) constraint of the state is much easier to treat than a C ( T ) constraint. We now give a proof for the interpolation theorem used in the proof of Lemma 3.2. The identity mappings C ; =2( T ) ,! C ;=2( T ) ,! L2( T ) are continuous and compact. Therefore the interpolation follows from a compactness argument. However, the compactness argument does not give us the exact form of the the constants C". Lemma 3.4 below is a stronger statement. The interpolation involves di erent type of spaces. Nonetheless, the proof is similar to that in, for example, [11]. 15
Lemma 3.4. Suppose that @ is Lipschitz continuous, 0 < < 1 and 0 < p 1. Then there exists a constant C , depending only on and T , such that (3:41)
1=p
k kC;=2( T ) 4"[ ]C ; =2 ( T ) + 3C" k kLp( T ); 8 2 C ; =2( T ); 80 < " 1;
where = ( ) + ( n+2)p , and (3:42)
(
8 > > > [ ] ;=2( T ) > > < C
)
= sup pj (x; t) 2 (x; t)j ; (x; t) 6= (x; t) 2 T ; ( jx xj + jt tj ) k kC;=2( T ) = k kC( T ) + [ ]C;=2( T ) when 0 < < 1; > > > > > : k kC;=2( T ) = k kC( T ) when = 0:
(The k kLp ( T ) should be understood in the usual sense, it should be noted that it is not a norm when 0 < p < 1.) Proof. We let = "1=( ), then 0 < 1. Splitting the sup in (3.43) into two set p p f jx xj2 + jt tj g and f jx xj2 + jt tj > g immediately gives us
(3:43)
k kC;=2( T ) [ ]C ; =2( T ) +
1 + 2 k kC( T );
from which the case p = 1 follows. Now consider the case 0 < p < 1. Since is continuous on T , k kC( T ) = j (xe; et)j, q for some (xe; et) 2 T . Now let B = f(x; t) 2 T ; jx xej2 + jt etj g, then by the mean value Theorem, 1 T
jB T j
!1=p
Z
B
T
T
j (x; t)jp dxdt
= j (x ; t )j;
T
T
for some (x ; t ) 2 B T . Since @ is Lipschitz continuous, jB T j n+2=C for some generic constant C > 0. It follows that (3:44)
k kC( T ) = j (xe; et)j j (xe; et) (x ; t )j + j (x ; t )j [ ]C ; =2 ( T ) + 16
C
n+2
1=p
k kLp( T );
the case = 0 follows immediately. Now substituting (3.45) into (3.44), we obtain
k kC;=2( T ) 4 [ ]C ; =2( T ) + 3
and the general case 0 < < < 1, 0 < p < 1 follows.
C
1=p
n+2
k kLp ( T );
x4. Proof of Theorem 2.2. In this section, we are going to prove our main theorem of this paper. Proof of Theorem 2.2. Let (y; u) be an optimal pair. For any u 2 U , let y(; ; u) be the corresponding state, emphasizing the dependence of it on the control. For any " > 0, we de ne
(4:1)
J"(u) = f[(J (u) J (u) + ")+ ]2 + dQ(G(y(; ; u)))2 g1=2:
Clearly, this functional is continuous on the (complete) metric space (U ; d) (recall that d is the Ekeland distance, see (3.30)). Also, we have (4:2) (4:3)
8u 2 U ;
J"(u) > 0;
J"(u) = " inf J (u) + ": U "
Hence, by Ekeland's variational principle ([6]), we can nd a u" 2 U , such that
p
(4:4)
d(u; u") ";
(4:5)
J"(u") J"(u);
(4:6)
J"(ub) J"(u")
p" d(ub; u");
8ub 2 U :
We let v 2 U and " > 0 be xed and let y" = y(; ; u"). By Theorem 3.3, we know that for any 2 (0; 1), there exists a measurable set E" T with the property jE"j = j T j, such that if we de ne 8 < u" (x; t); if (x; t) 2 T n E"; " (4:7) u(x; t) = : v(x; t); if (x; t) 2 E"; 17
and let y" = y(; ; u") be the corresponding state, then 8 < y"
= y" + z" + r" ; : J (u" ) = J (u" ) + z 0;" + r0;" ;
(4:8)
where z" and z0;" satisfy the following: (4:9)
8 > > < z"
n X
t
> > :z
i;j
(aij (x; t)zx"i )xj fy (x; t; y" (x; t); u" (x; t))z" = h"(x; t);
in T ;
@p T = 0:
z0;" =
(4:10)
Z
T
[fy0 (x; t; y" (x; t); u" (x; t))z" (x; t) + h0;"(x; t)]dx;
with (4:11)
(
h"(x; t) = f (x; t; y" (x; t); v(x; t)) f (x; t; y" (x; t); u" (x; t)); h0;"(x; t) = f 0 (x; t; y" (x; t); v(x; t)) f 0 (x; t; y" (x; t); u" (x; t)):
and for some 2 (0; 1), 1 jr0;" j = 0: 1 kr" k lim ;= 2 ( T ) = lim C !0 !0
(4:12)
Now, we take ub = u" in (4.6). Then, it follows that
p"j j J"(u") J"(u") T (
(4:13)
[(J (u") J (u) + ")+]2 [(J (u") J (u) + ")+ ]2 = J (u" ) +1 J (u") " " ) d (G(y" ))2 d (G(y" ))2 + Q Q "
"
+
! (J (u ) J (Ju("u)) + ") z0;" + h dQ (JG((uy")))" ; G0 (y" )z" i; "
"
where (4:14)
" =
(
rdQ (G(y" )); 0; 18
if G(y" ) 2= Q; if G(y" ) 2 Q:
( ! 0);
We note that since G : C0( T ) ! Z , to obtain the convergence in (4.13), the expansion (4.8) in the space C0( T ) is necessary. Next, we de ne ('0;"; '") 2 [0; 1] M( T ) as follows: 8 0;" > > > zt > > > < > > > > > :
i;j =1
+ f (x; t; y (x; t); v(x; t)) f (x; t; y (x; t); u(x; t));
in T ;
z @p T = 0:
and (4:24)
(aij (x; t)zxi )xj = fy (x; t; y (x; t); u(x; t))z
z0
=
Z
ZT
+
fy0 (x; t; y (x; t); u(x; t))z(x; t)dxdt
T
[f 0 (x; t; y (x; t); v(x; t)) f 0 (x; t; y (x; t); u(x; t))]dxdt:
We note that the solution z of (4.23) and the quantity z0 de ned by (4.24) depend on the choice of v 2 U . Thus, we denote them by z(; ; v) and z0 (v). respectively. Then, taking limits in (4.16), we obtain (4:25)
'0z0 (v) + h '; G0 (y)z(; ; v) i 0;
8v 2 U :
Now, we let 0
(4:26)
= '0 2 [ 1; 0]:
Then, (2.11) follows from (4.21). Also, we obtain (2.13) by taking limits in (4.20) (along the above-mentioned subsequence). Furthermore, (4.25) can be written as (4:27)
0 z 0 (v )
h G0 (y) '; z(; ; v) i 0;
8v 2 U :
We note that G0 (y) ' 2 M( T ) and by our compatible condition (2.10), we see that
(4:28)
h G0 (y ) '; z i = h(G0 (y ) ') T ; z i M( T );C( T ) + h(G0 (y) ') fT g ; z t=T i M( );C( ):
+2 . By [1], we know that (2.12) admits a solution in Lq (0; T; W01;q ( )) for any 1 < q < nn+1 However, unlike the elliptic equations, the function z(x; t; v) is not smooth enough (in t direction) to be a test function for the equation (2.12). We shall get around this problem by approximating the equations for z and the equations (2.12).
20
We consider the following approximation for z(x; t; v): (4:29)
n X
8 > > zt > > > < > > > > > :
(aij (x; t)zx i )xj = fy (x; t; y (x; t); u(x; t))z
i;j =1
+ f (x; t; y (x; t); v(x; t)) f (x; t; y (x; t); u(x; t));
in T ;
z @p T = 0;
where aij 2 C 2;1( T ), aij satis es (A1), and
aij ! aij
(4:30)
in Lp( T ); as ! 0;
for any 1 < p < 1. By Lp estimates for the parabolic equations, z = z (; ; v) 2 Wp2;1( T ) for any 1 < p < 1. As before, we have the estimates
kz kC ; =2 ( T ) + krxz kL2 ( T ) C:
(4:31)
where the constants C and are independent of and v 2 U . Thus, by compactness and the uniqueness of the equation (4.23), one can easily derive that
kz zkC( T ) ! 0;
(4:32)
as ! 0:
Clearly, (4.27) and (4.32) imply that (4:33)
lim !0
0 z 0 (v )
h G0 (y ) '; z (; ; v) i 0;
8v 2 U :
By [1], if we replace aij with aij , then (2.12) has a solution 2 Lq (0; T; W01;q ( )), where +2 . ( is actually unique.) Furthermore, 1 < q < nn+1
k kLq (0;T;W01;q ( )) C;
(4:34)
where the constant C is independent of . Clearly, v is not involved in the de nition of . By passing to a subsequence if necessary, we have, as ! 0, (4:35)
w *
w xj * xj
in Lq ( T );
in Lq ( T );
for some function . It follows that is a solution of (2.12). 21
Since z 2 Wp2;1( T ) for p > q=(q 1), we can use z as a test function in the equation for . Then, by some direct computation, we can reduce (4.33) to the following: Z
T
(4:36)
0 [f 0 (x; t; y (x; t); u(x; t))
f 0 (x; t; y (x; t); v(x; t))]
+ (x; t)[f (x; t; y (x; t); u(x; t)) f (x; t; y (x; t); v(x; t))]gdx o(1); as ! 0:
Now letting ! 0 and recalling (4.35), we obtain, Z
T
(4:37)
0 [f 0 (x; t; y (x; t); u(x; t))
f 0 (x; t; y (x; t); v(x; t))]
+ (x; t)[f (x; t; y (x; t); u(x; t)) f (x; t; y (x; t); v(x; t))]gdx [H (x; t; y (x; t); u(x; t); 0 ; (x; t)) H (x; t; y (x; t); v(x; t); 0 ; (x; t))]dx;
0; 8v 2 U : Z
Then, by the separability of U and the continuity of the Hamiltonian H in the variable v, noticing also that v 2 U is arbitrary, we obtain the maximum condition (2.14) (see [5]).
x5. Applications. In this section, we would like to discuss some special cases which are covered by our main result. We rst consider the following case. Let Z = C0( T ) with some norm j j0 which is equivalent to k kC( T ) and the dual of it, still denoted by M0( T ), is strictly convex. We let Q Z be de ned as in (1.3) and g : T lR ! lR be continuous with gy (x; t; y) being also continuous. Moreover, (5:1)
(
g(x; t; 0) < 0; 8(x; t) 2 @ [0; T ]; g(x; 0; y0 (x)) < 0; 8x 2 ;
where y0 2 C0( ). We let G() = g(x; t; (x; t)), for any 2 C0( T ). Then, the following result holds.
Proposition 5.1. For the above Q, G and y0 , condition (2.10) holds. 22
Proof. By (5.1), we see that for any " > 0, there exists a > 0, such that
(5:2)
g(x; t; y) "; if t 2 [; T ]; jyj < ; d(x; @ ) < ; or t 2 [0; ]; x 2 ; jy y0 (x)j < :
Thus, for any 2 C0( T ) with G() 2 Q and t=0 = y0 (x), we have the following: For any " > 0, there exists a > 0, such that (5:3)
g(x; t; (x; t)) ";
if d((x; t); @p T ) < :
Now, for any ' 2 C ( T ), if for some > 0, it holds (5:4)
supp ' f(x; t) 2 T d((x; t); @p T ) g;
then, for all small enough > 0, we have (5:5)
g(x; t; (x; t)) + '(x; t) 0;
(x; t) 2 T :
This means G() + ' 2 Q for all small > 0. Hence, by the de nition of the generalized gradient, we obtain (5:6)
h ; ' i = 0;
8 2 @dQ (G()) M( T ):
In another word, we have (for the above ) (5:7)
supp f(x; t) 2 T d((x; t); @p T ) g:
Since G0 () = gy (x; t; (x; t))I , with I being the identity on Z , we see that (2.10) holds. We have already know that Q has a nonempty interior in Z , hence it is of codimension 0 in Z . Then, our main result is applicable to this case. Let us state the corresponding result.
Theorem 5.2. Let (y; u) be an optimal pair. Then, there exists a constant +2 ) and a ' 2 M0 ( T ), such that function 2 Lq (0; T ; W01;q ( )) (q < nn+1 (5:8)
j 0j + k'kM0( T ) > 0; 23
0
0, a
(5:9)
8 > > > > > > > > > < > > > > > > > > > :
t+
n X
i;j =1
0 f 0 (x; t; y (x; t); u(x; t)) + g (x; t; y (x; t)) ' ; y y
T
@ = 0;
in T ;
t=T = gy (x; T; y (x; T ))' fT g ; Z
(5:10) (5:11)
(aij (x; t) xj )xi = fy (x; t; y (x; t); u(x; t))
i
h
T
z(x; t) g(x; t; y (x; t)) d'(x; t) 0;
8z 2 Q:
H (x; t; y (x; t);u(x; t); 0 ; (x; t)) = max H (x; t; y (x; t); v; 0 ; (x; t)); v2U a.e. (x; t) 2 [0; T ]:
where H is the Hamiltonian de ned by (2.15). Let us make some further remark on the above result. We set (5:12)
0T = f(x; t) 2 T g(x; t; y (x; t)) = 0g:
Then, by our condition (5.1), we see that (5:13)
[
0T T ( fT g):
The set 0T is called the active set for the optimal state y. We have that supp ' 0T :
(5:14)
In fact, for any 2 C ( T ) with supp T n 0T , g(; ; y (; )) " 2 Q if " is small enough. By the transversality condition (5.10), we see immediately that (5:15)
Z
T
(x; t)d'(x; t) = 0:
This gives (5.14). The above situation is comparable with the case discussed in [5] for quasilinear elliptic equations. 24
Next, let us look at another important case. Let Z = C0( T ) as before and let S (xi ; ti ) 2 T ( fT g) (1 i m) be given m (di erent) points and also let bi 2 lR; 1 i m. We de ne Q as in (1.5). Then, we see that Q is a nite codimensional convex and closed subset of Z . Also, it is not hard to see that for any 2 Q and any 2 @dQ (), we have (5:16)
=
m X i=1
i(xi ti);
where i 2 lR and (xi;ti ) is the Dirac measure concentrated at point (xi ; ti ) with mass 1. Thus, we see that condition (2.10) holds (in the present case G = I the identity). Hence, our result applies to this situation. Let us state the corresponding result below.
Theorem 5.3. Let (y; u) be an optimal pair. Then, there exists a constant 0 0, a +2 ) and real numbers i ; 1 i m, such that function 2 Lq (0; T ; W01;q ( )) (q < nn+1 j
(5:17)
(5:18)
(5:19)
8 > > > > > > > > > > > < > > > > > > > > > > > :
t+
n X
(aij (x; t) xj )xi = fy (x; t; y (x; t); u(x; t))
i;j =1
@ = 0;
t=T =
m 0 j + X ji j > 0; i=1
X
ti =T
0 f 0 (x; t; y (x; t); u(x; t)) + X i (xi ;ti ) ; y ti :
= fz 2 Lp( T )
Z
T
z(x; t)dxdt 0g;
G()(x; t) = F (x; t; (x; t)); 25
8 2 C0( T ):
Then, the corresponding state constraint is Z
(5:21)
F (x; t; (x; t))dxdt 0:
T
2. Let Z = Lp( T )m , 1 < p < 1, Fi : T lR ! lR, 1 i m, (5:22)
8 > : G( )(x; t)
Z
T
z(x; t)dxdt = 0g;
= (F1 (x; t; (x; t)); ; Fm (x; t; (x; t));
8 2 C0( T ):
Then, the state constraint is Z
(5:23)
T
1 i m:
Fi (x; t; y(x; t))dxdt = 0;
3. Let Z = C0( T ), F : T lR ! lR such that (5:24)
F (xi ; ti ; y) = gi (y);
1 i m;
S
where (xi ; ti ) 2 T ( fT g) are given di erent points. Let (5:26)
8 0 being very small. Physically, this means, for example, that the temperature at point (xi ; ti ) has to be controlled near bi with an accuracy ".
x6. Quasilinear Parabolic Equations. Finally, let us remark that the result extends to the quasilinear parabolic equations as those considered in [5] for elliptic cases. The Holder estimates for the gradient (in x direction) of the solution is required, which is available [16], when the leading coecients is assumed to be Holder continuous in x direction. After a careful examination of the proof of Lemma 3.2, we conclude that aij can actually be allowed to depend on . We consider the equations 8 > :
aij (x; t)xi xj + c(x; t) (x; t) =
1 1 E (x; t) h(x; t);
= 0; (x; t) 2 @p T : It is clear that the solution = E (x; t) is uniquely determined by the choice of the coecients aij , c and the set E. Let K > 0, 0 < < and K = f(aij (x; t); c(x; t)); kckL1 ( T ) K; (6:2) jj2 aij (x; t)i j jj2; 8(x; t) 2 T ; 2 lRng
Lemma 6.1. Suppose that n+2 2 < p < 1. Then, there exists 2 (0; 1) such that for each h0 2 L1( T ), h 2 Lp( T ), K > 0, and any 2 (0; 1) (6:3) inf E 2E
( Z Z T sup 1 (a ;c )2K 0
ij
)
1 (x; t) h0(x; t)dxdt +
E C ;=2 ( [0;T ]) = 0: E 27
We now consider the following parabolic equation: 8 yt rx (a(x; t; rx y)) = f (x; t; y; u(t; x)); in T ; > > > < y @ = 0; (6:4) > > > : y t=0= y0 (x); x 2 ; where a satis es @ai : lRn ! lR are continuous. (A1) The functions a : T lRn ! lRn and @p T j There exist > > 0, 1 > 0, 2 (0; 1) and m > 1 such that 8 @ai m 2 j j2 ; > > 8(x; t; p) 2 [0; T ] lRn; < @p (x; t; p)i j (1 + jpj) j (6:5) @ai > > (1 + jpj)m 2 ; : ( x; t; p ) 8(x; t; p) 2 [0; T ] lRn; @pj and ja(x; t; p) a(xe; t; p)j 1(1 + jpj)m 1 jx xej ; (6:6) 8(x; t); (xe; t) 2 [0; T ]; p 2 lRn: Under the assumptions (A1) and (A2), (6.4) has a unique solution y(x; t) such that y; yxj 2 T C ; =m( T ), for any y0 2 C 1+ ( ) C0( ) (see [16]). The a-priori estimates for y and yxj in the space C ; =m( T ) are valid uniformly for u 2 U . If u; ub 2 U and y; yb are the corresponding states, then by using an interpolation theorem of the C k+ spaces, we have, for any 0 < < , (6:7)
ky ybkC( T ) + krx(y yb)kC;=m( T ) ! 0;
uniformly as d(u; ub) ! 0:
Lemma 6.1 and (6.7) are the essential estimates needed to establish the variation theorem similar to Theorem 3.3. We have
Lemma 6.2. The same statement of Theorem 3.3 is still valid for the quasilinear case, @ai (x; t; r y(x; t)). except that the aij (x; t) in (3.25) should be replaced by aij (x; t) @p x j
Sketch of Proof. The proof ofZTheorem 3.3 also goes through, where we shall replace 1 @a aij (x; t) in (3.32) with aij (x; t) @p i (x; t; rx y(x; t) + rx(y (x; t) y(x; t))d . Using j 0 (6.7), we can easily obtain
(6:8)
aij ! aij ;
in C ( T ): 28
Write z z (z zb) + (zb z), where zb satis es the equation (6:9)
8 > > < (zb
)t
> > : zb
n X
(aij (x; t)(zb )xi )xj + c(x; t)zb = h(x; t);
i;j =1
in T ;
@p T = 0:
We can treat z zb the same way as before, apply Lemma 6.1 instead of Lemma 3.2, and then obtain
kz zb kC;=2( T ) ! 0
(6:10)
as ! 0;
for a special choice of E with jEj = j T j. Clearly, by Holder's estimates [12, Chapter III, section 10], for b = zb z,
kbkC ; =2 ( T ) kzb kC ; =2 ( T ) + kzkC ; =2 ( T ) C;
(6:11)
and b satis es the equation (6:12)
8 > > < (b )t > > : b
n X i;j =1
aij (x; t)(b )xi
+ c(x; t)b = [aij (x; t) aij (x; t)]zxi (x; t) x xj j
@p T = 0:
Multiplying the above equation with b and integrating over T give us the usual energy estimates: (6:13) kbkL2 ( T ) krxbkL2 ( T ) C kaij aij kC( T )krxzkL2 ( T ) ! 0;
as ! 0;
where we have used that fact that krxzkL2 ( T ) C , which is an easy consequence of the energy estimates for the solution z(x; t). Using the interpolation (Lemma 3.4), the estimates (6.11) and (6.13), we immediately obtain, for any 0 < < , (6:14)
kzb zkC;=2 ( T ) = kbkC;=2 ( T ) ! 0;
as ! 0:
Combining (6.10) and (6.14), we get an variation theorem like Theorem 3.3. 29
Therefore, with the same argument as in section 4 (where we shall use (6.7) for the convergence of the leading coecients), we have the following maximum principle:
Theorem 6.3. Consider the same problem except that the governing equation (1.1) is reT placed by (6.4). Let (A1), (A2) and (A3) be in force. Suppose that y0 2 C 1+ ( ) C0( )
and that (2.10) holds. Let (y; u) be an optimal pair of Problem C corresponding to the state equation (6.4). Then, there exists a constant 0 0, a function 2 Lq (0; T ; W01;q ( )) (1 < q < nn+2 +1 ), and a ' 2 @dQ (G(y )) Z , such that
j 0 j + k'kZ > 0;
(6:15)
(6:16)
(6:17) (6:18)
8 > > > > > > > > > < > > > > > > > > > :
t+
n X i;j =1
@ = 0;
@ai (x; t; r y (x; t)) = f (x; t; y (x; t); u(x; t)) x xj y @pj xi 0 f 0 (x; t; y (x; t); u(x; t)) + (G0 (y ) ') ; in T ; y
T
0 t=T = (G (y ) ') fT g ;
h z G(y); ' i 0;
8z 2 Q:
0 ; (x; t)); H ( x; t; y ( x; t ) ; v; H (x; t; y (x; t);u(x; t); 0 ; (x; t)) = max v2U a.e. (x; t) 2 [0; T ]:
where (6:19)
H (x; t; y; u; 0 ; ) = 0f 0 (x; t; y; u) + f (x; t; y; u); 8(x; t; y; u; 0; ) 2 [0; T ] lR U lR lR:
Remark 6.4. The semilinear case is not a special case of the quasilinear case, since the Holder continuity of aij (in x direction) is not assumed in the semilinear case. Acknowledgment. The authors thank Dr. Hong-Ming Yin for a stimulating discussion on this problem.
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