OPTIMAL CONTROL OF LINEAR BOTTLENECK
PROBLEMS
M. BERGOUNIOUX AND F. TROLTZSCH
Abstract. The regularity of Lagrange multipliers for state-constrained optimal control problems belongs to the basic questions of control the- ory. Here, we investigate bottleneck problems arising from optimal con- trol problems for PDEs with certain mixed control-state inequality con- straints. We show how to obtain Lagrange multipliers inLp-spaces for linear problems and give an application to linear parabolic optimal con- trol problems.
1. Introduction
The regularity of Lagrange multipliers for state-constrained optimal con- trol problems belongs to the basic questions of control theory. This is known for control systems governed by ODEs and refers to PDEs as well. In many cases, Lagrange multipliers have to be measures, for instance, if pointwise state-constraints are given in distributed parameter systems. Unfortunately, several methods cannot be applied for measures as Lagrange multipliers. In particular, standard second order sucient optimality conditions cannot be established, and the convergence of some numerical techniques is not en- sured.
It is well known that Lagrange multipliers can be obtained for nonlinear programming problems in Banach spaces (including control problems) by considering the associated linearized problem (see Zowe and Kurcyusz 15]).
In this way, the theory of linear programming problems in Banach spaces may serve as a tool to prove the existence of Lagrange multipliers.
For linear programming problems inLp-spaces with pointwise constraints of bottleneck-type it has been known for a long time that associated La- grange multipliers may exist in Lp-spaces, too. We mention for instance Grinold 7],8], Levinson 10], and Tyndall 14]. Moreover, we refer to An- derson 1] and the monography by Anderson and Nash 2].
Bottleneck problems appear in optimal control problems for ODEs with certain mixed control-state inequality constraints. However, it seems that this kind of problems has not been studied since many years nevertheless we may refer to a paper by Anderson and Philpott 3] which deals with
Matine Bergounioux, Dept. de Mathematiques, UMR 6628, Universite d'Orleans, 45067 Orleans, France. E-mail: [email protected].
Fredi Troltzsch, Technical University of Chemnitz, Faculty of Mathematics, D-09107 Chemnitz, Germany. E-mail: [email protected].
This work was supported by EEC, HCM Contract CHRX-CT94-0471.
Received by the journal September 23, 1997. Revised April 7, 1998. Accepted for publication May 6, 1998.
c Societe de Mathematiques Appliquees et Industrielles. Typeset by LATEX.
a ODE problem close to the PDE problem we are going to investigate.
Indeed, the associated theory can be extended also to the control of PDEs, see Bergounioux and Tiba 4] and the short note of Troltzsch 13]. Here, we aim to extend those ideas to make them applicable to state-constrained control problems governed by PDEs instead of ODEs.
In contrast to the results on linear continuous programming, where the corresponding integral operators have bounded and measurable kernels, we shall have to deal with weakly singular operators. Moreover, we have to consider Lebesgue spacesLp(Q)1<p<1, to get the maximum regularity for Lagrange multipliers. In the early papers we mentioned above, only spaces of L1-type appeared. To our knowledge, bottleneck problems have not been studied yet from this extended point of view.
We develop the theory for linear problems with application to a linear par- abolic optimal distributed control problem as an example. In a forthcoming paper, a class of nonlinear parabolic control problems with pointwise mixed control-state inequality constraints will be discussed using these results.
In this way we hope to contribute with a class of problems, where dif- ferent methods such as augmented Lagrangian methods, Lagrange-Newton methods, or second-order sucient optimality conditions can be justied by having suciently regular multipliers in Lp-spaces.
The paper is organized as follows: rst we introduce a class of linear programming problems in L1(Q) where QRN+1is a certain measurable set. On extending the problem to Lp(Q) (p>N =2 + 1) (without changing the feasible set, which remains a subset of L1(Q)), the associated dual problem is dened. In sections 4 and 5 the existence of solutions to the dual problems is shown for dierent choices of primal problems. In section 6 we investigate duality relations. Finally, linear control problems for PDEs are discussed in section 7 and 8.
We consider the so-called Primal Problem
(P)
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
max
Z
Q
a(xt)u(xt)dx dt
c
1(xt)u(xt);x
Z
t
0 Z
G(xts)u(s)d dsc2(xt) a.e. in Q
d
1(xt)u(xt)d2(xt) a.e. in Q:
In this setting, a bounded, measurable subset RNN 1Q :=
(0T) and functions a 2 L1(Q)cidi 2 L1(Q)i= 12 are given, where
T >0 is xed. Dene D=f(ts)2R2 j0 s<tTg and a continuous function G: 2D!Rsupposed to satisfy
jG(xts)jk1(t;s);N2 exp
;k
2 jx;j
2
t;s
(1.1)
where k1k2 are positive real numbers (we have in mind Green's functions
G for parabolic partial dierential equations).
The main part of our analysis is devoted to the case c1 =;1, c2 <1,
d
1 = 0d2=1, and toc1 =;1, c2 <1,d1 = 0d2<1. The cases c1 =
;1, c2 = +1 and d1 =;1, d2 = +1 are more or less trivial. However,
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we were not able to show the existence of regular Lagrange multipliers for the general case of (P), where all bounds really appear.
2. An integral equation We would like to discuss rst the integral equation
u(xt) =c(xt) +
Z
t
0 Z
G(xts)u(s)d ds (2.1) for given c 2L1(Q). Equations of this type are basic for our theory. Let us introduce the integral operator K:
(Ku)(xt) =Z t
0 Z
G(xts)u(s)dds: (2.2) It is easy to show thatKis continuous in L1(Q), and fromLp(Q) toL1(Q) for N 2p > N =2 + 1 and N = 1p > 2. We refer to Troltzsch 11], p.
138, Lemma 5.6.6.
For convenience, we endow the spaceL1(Q) with the equivalent normkuk,
kuk
= sup essje;tu(xt)j
(xt)2Q (2.3)
where >0 (it holds e;TkukL1(Q)kukkukL1(Q)).
Lemma 2.1. K is a contraction in L1(Q), for suciently large >o. Proof. In view of the denition ofkuk and the estimate (1.1) we get
kKuk
= sup
(xt)2Q
ess
e
;t t
Z
0 Z
G(xts)ese;su(s)dds
sup
(xt)2Q
ess
Z
t
0 Z
R N
k
1(t;s);N2e;(t;s) exp
;k
2 jx;j
2
t;s
ddskuk
= sup
(xt)2Q
essk
Z
t
0 e
;(t;s)
dskuk
k
kuk
where k=k1k;2 N2 N2 (note that R;11 e;k 2d=p =kk>0).
Corollary 2.2. For all c2L1(Q), the equation (2.1) has a unique solu- tion u2L1(Q). The mapping c7!u is continuous in L1(Q).
Proof. We have
u= (I;K);1c= (X1
n=0 K
n)c (2.4)
by well known results on Neumann series. Moreover,k(I;K);1k1=(1;
kKk
), where we have used for convenience the symbolkkto denote also the norm of the operator K induced by kk. The result follows from the equivalence of the norms kk andkkL1(Q).
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Corollary 2.3. If G(xts)0 on 2D and c(xt) 0 a.e. in Q,
c2L
1(Q), then u(xt)0 a.e. inQ.
Proof. The Neumann series representation (2.4) yields this result, asc0 implies Knc0 8n2N.
Corollary 2.4. (Comparison principle). Suppose that G(xts)0 on 2 D and ui 2 L1(Q) i = 12 are solutions of (2.1) associated to
c
i 2L
1(Q) i= 12. If c1(xt) c2(xt) holds a.e. in Q, then u1(xt)
u
2(xt) a.e. in Q.
This result follows immediately from Corollary 2.3.
In the next corollary we consider Kas an operator fromLp(Q) toL1(Q).
Corollary 2.5. (Extended comparison principle). Suppose p N =2 + 1.
Then the integral equation (2.1) has for each c 2 Lp(Q) a unique solution
u 2L
p(Q), and the mapping c7!u is continuous inLp(Q).
Assume G(xts)0 on 2D, ci2Lp(Q) i= 12 c1(xt)c2(xt) a.e.in Q and let ui 2 Lp(Q) i= 12 be the associated solutions of (2.1).
Then u1(xt)u2(xt) a.e. in Q.
Proof. We putv:=u;c, then (2.1) reads
v=Kc + Kv: (2.5)
As K: Lp(Q) ! L1(Q), continuously, the right hand side Kc belongs to L1(Q) and depends continuously onc. By Corollary 2.2, equation (2.5) admits a unique solution in v2L1(Q) depending continuously on Kcand hence on c. Clearly, u=c+v is a solution of (2.1) in Lp(Q).
The uniqueness of u in Lp(Q) is seen as follows: If u1u2 are solutions in Lp(Q) associated to the same c, then u = u1 ;u2 satises u = Ku. Since u 2 Lp(Q), the smoothing property of K implies even u 2 L1(Q).
Uniqueness in L1(Q) yields u= 0. Therefore,u1 =u2 must hold inLp(Q).
Finally, we derive the comparison principle assuming G 0. Consider once again the auxiliary function v := u;c. If c 0, then Kc 0, too.
From equation (2.5) and the rst comparison principle we get v0 a.e. on
Q. Therefore,
u=c+vc0
holds true. The comparison principle is an easy consequence (put c :=
c
1
;c
2).
So far, we have considered the integral operator K. This operator will appear in our primal problem. Our theory is focused on the associated dual problem, where the (formal) adjoint operator K>
(K>)(xt) =
Z
T
t Z
G(xst)(s) d ds (2.6) plays a crucial role. K> is positive and has the same singularity as K. Therefore, K> behaves like K. In particular, it is a contraction for>o in the normkk. It will be useful to prove the following result:
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Lemma 2.6. The equation
(xt) = max
0a(xt)+
Z
T
t Z
G(xst)(s)d ds (2.7) has exactly one solution 2L1(Q).
Proof. Dene by the projection on the positive cone of L1(Q), (z)(xt) = maxf0z(xt)g=:z(xt)+:
Obviously
j(z1;z2)(xt)jj(z1;z2)(xt)j a.e. in Q therefore, we get
kz1;z2kkz1;z2k
by multiplying with e;t and taking the supremum over Q. The operator
T, expressed by the right hand side of (2.7),
T= a+K>]+
is a contraction in L1(Q) endowed with kk forlarge enough:
kT
1
;T
2 k
= k(a+K>1);(a+K>2)k
kK
>
1
;K
>
2 k
<
k
k
1
;
2 k
is a contraction for >k.
Equation (2.7) reads=T. According to the Banach xed point theo- rem, this equation has a unique solution in L1(Q).
3. Two basic primal problems
From now, we assume that G(xts)0 on 2D and pN =2 + 1.
Let us discuss two main choices of (P).
We consider the problems (P1), (P2), (P1)
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:
max
Z
Q
a(xt)u(xt)dx dt
u(xt)c+Ku](xt) a.e. in Q
u(xt)0 a.e. in Q
(P2)
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:
max
Z
Q
a(xt)u(xt)dx dt
u(xt)c+Ku](xt) a.e. in Q 0u(xt)d(xt) a.e. in Q:
where a c and dareL1(Q) functions.
We start with (P1). Let us dene uc by
u
c =c+Kuc: (3.1)
Note thatuc is bounded and measurable according to Corollary 2.2.
Theorem 3.1. (P1) has a solution u if and only if uc 0.
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Proof. Ifuc 0, then the feasible domain for (P1) is non empty. All feasible elements u 2 Lp(Q) satisfy u uc by the second comparison principle of corollary 2.5, and we know uc 2L1(Q). Moreover,u 0 follows from the constraints, too. This implies the L1-boundedness ofu by kuckL1(Q). So the feasible set is bounded, closed and convex in all (reexive)Lp-spaces for 1 +N =2<p<1. The existence of u 2Lp(Q) is an immediate conclusion.
Of course, u belongs toL1(Q).
On the other hand, if uc is not equal or greater than 0, then the feasible set is empty. This follows again from the comparison principle.
Theorem 3.2. If uc 0 (that is, if the feasible set is non empty) and if
d0, then (P2) has a solution.
The result is shown in the same way as for the preceding one.
4. Dual Problem to (P1)
Let us extend the feasible set of (P1) fromL1(Q) toLp(Q) (p>N =2+1).
Any feasible solution u of (P1) satises 0 u uc, whereuc is dened in (3.1). Therefore, all feasible solutions of Lp(Q) belong automatically to
L
1(Q), and the feasible set is independent of this extension to Lp(Q).
Using standard techniques for constructing dual problems (cf. our dis- cussion in section 6), we consider (P1) in Lp(Q) and get the following dual problem in Lq(Q).
(D1)
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:
minZ
Q
c(xt) (xt) dxdt
(xt)a+K>](xt) a.e. in Q
(xt)0 a.e. in Q
where 2 Lq(Q), p;1+q;1 = 1, and K> is the integral operator dened by (2.6).
The kernel ofK>satises the estimate (1.1), henceK>:Lp(Q)!L1(Q) forp>N =2+1, too. However, this is not true fromLq(Q) toL1(Q). On the other hand, K> represents the adjoint operator of K :Lp(Q)! L1(Q)
L
p(Q). Therefore, K>is continuous in Lq(Q)Lp(Q), as well. In view of (1.1) and Lemma 2.1, K> is a contraction in L1(Q), hence the equation
= + K> (4.1)
has for each 2 L1(Q) a unique solution 2 L1(Q). Moreover, we have uniqueness for (4.1) in Lq(Q). Suppose that 2 Lq(Q) solves the homogeneous equation ;K>= 0 we take an arbitraryv2Lp(Q), then
0 =
Z
Q
v(;K>)dxdt=
Z
Q
(v;Kv)dxdt:
In view of the proof of Corollary 2.5, the range of v;Kv is Lp(Q), hence
= 0. This yields uniqueness for (4.1) in Lq(Q).
Remark 4.1. It is easy to see that the solution of
(xt) =ja(xt)j+ (K>)(xt) a.e. in Q (4.2)
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is a feasible and essentially bounded solution of (D1) (this follows from the corollaries of section 2.)
Theorem 4.2. Ifc0, then (D1) has at least one optimal solution given by equation (2.7).
Proof. Let be the unique solution of (2.7). Suppose that2Lq(Q) is any other feasible element for (D1) (dierent from ). Then
a+K> a.e. in Q
and, of course, 0. Next, we construct a sequence 1 2 as follows: we put 1 =2Lq(Q) and dene
2 = maxf0 a+K>1g:
A simple discussion yields 2 1 a.e. in Q. Then the positivity of K>
implies K>2K>1and we get
2
a+K>1 a+K>2 a.e. onQ:
Thus 2 is feasible, and 2 1 holds on Q. Repeating this process, one constructs a non-increasing feasible sequence fng, which has to be point- wisely convergent towards some ~0, i.e.
lim
n!+1
n(xt) = ~(xt) a.e. onQ:
Therefore, fn(xt);~(xt)gq tends pointwisely to zero, too. Moreover, this sequence is bounded by (1(xt))q 2L1(Q). The Lebesgue dominated convergence theorem yields that (;~)q tends to zero in L1(Q), hence n tends to ~ in Lq(Q).
K
> is continuous in Lq(Q) so that a+K>n ! a+K>~ in Lq(Q).
Passing to the limit in
n = maxf0a+K>n;1g
(the right hand side is a continuous mapping in Lq(Q)) gives
~
= maxf0a+K> g:~
We are able to conclude ~2L1(Q). This is seen from the equation
~
(xt) =(xt)a(xt)+(xt)(K>~)(xt)
where (xt) = 0 for ~(xt) = 0 and (xt) = 1 for ~(xt) > 0. This equation is of the type discussed in section 2, where c=a2L1(Q) (the new kernel Ge = G satises the estimate (1.1) too and Ge 0.) Hence
~
2L
1(Q). By uniqueness in L1(Q) we have ~= moreover~. Consequently holds for all feasible solutions, and
Z
Q
c(xt)(xt) dxdt
Z
Q
c(xt) (xt)dx dt follows from c0.
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5. Dual Problem to (P2) In the case of (P2) we obtain the dual problem
(D2)
8
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:
min
Z
Q
c(xt)1(xt) +d(xt) 2(xt)]dxdt
1(xt) +2(xt)a+K>1](xt) a.e. in Q
1(xt)0 2(xt)0 a.e. in Q where the unknown functions i i= 12 belong to Lq(Q).
From now, we assume c0 andd0.
Lemma 5.1. For any feasible pair of (D2) having the objective functional value v, there exists another feasible pair (12) such that
1 + 2 = maxf0a+K>1g (5.1) and the associated value is not greater than v.
Proof. Let (12) be a feasible pair with valuevfor the objective functional.
Then we know
1+2 maxf0a+K>1g:
Once again, we dene decreasing sequencesfn1gfn2g as follows:
1
j :=jj= 12
n+1
1 +n+12 = maxf0a+K>n1gn= 12:::
More precisely, assume that n1 0 and n2 0 have been constructed as above and satisfy
n
1 +n2 maxf0a+K>n1g=n0: We set n+1i =nnii= 12where
n(xt) =
8
<
:
0 ifn(xt) = 0
n(xt)
n
1(xt) +n2(xt) otherwise:
Thus 0 n(xt)1 and we have 0 n+1i ni:The very choice of n givesn+1i 0 andn+11 +n+12 =n = maxf0a+K>n1g. The positivity of
K and 0n+11 n1 yield that maxf0a+K>n1gmaxf0a+K>n+11 g and
n+1
1 +n+12 maxf0a+K>n+11 g=n+1 0: Passing to the limit along the lines of the last proof, we get
lim
n!1
n
i = i in the sense ofLq(Q)
and lim
n!1
n = lim
n!1
maxf0a+K>n1g
= maxf0a+K>1g in the sense ofLq(Q): Moreover,
1+ 2 = maxf0a+K>1g:
Clearly, v v holds for the value of the objective functional as c 0 and
d0.
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So far, we have shown comparison principles in L (Q)p>N =2 + 1. Let us now add a third conclusion for Lq(Q).
Lemma 5.2. If 0 is given inLq(Q) and 2Lq(Q) is the solution of
=+K> (5.2)
then 0 holds a.e. onQ.
Proof. This is shown as follows: dene
M
;=f(xt)2Q j(xt)<0g and
=M;=
1 if (xt)2M; 0 otherwise.
Let v 2 L1(Q) be the unique solution of v;Kv = . By the extended comparison principle we havev 0. Now multiply (5.2) byv and integrate overQ. Then
Z
Q
(xt)v(xt)dx dt = Z
Q
((xt);K>(xt))v(xt)dx dt
=
Z
Q
(v(xt);Kv(xt))(xt) dxdt
=
Z
M
;
(xt)dxdt:
Fromv0 we conclude that (M;) has a null measure hence 0 a.e.
on Q.
Theorem 5.3. Problem(D2) admits at least one solution (12)2L1(Q)2. Proof. Owing to Lemma 5.1 we may restrict ourselves to the set of all pairs (12)2Lq(Q)2 such that
1+2 = maxf0a+K>1gand j 0j= 12: Therefore,
1
1+2 jaj+K>1: So we have 1 jaj+K>1. Consider the solution a of
a(xt) =ja(xt)j+ (K>a)(xt) a.e. in Q: (5.3) This solution belongs toL1(Q) ( applying Lemma 2.1 toK>as an operator from Lp(Q) toL1(Q)) and is unique in Lq(Q) ( see the discussion of (4.1)).
We have
1 =+K>1
where(xt)ja(xt)jand2Lq(Q). The comparison principle of Lemma 5.2 yields 1a, hence 01 implies 1 2L1(Q).
Therefore2 is uniformly bounded as well. In this way we may restrict to a weakly compact subset ofLq(Q), which is uniformly bounded. We complete the proof by standard arguments.
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