Optimal Control of PDE
Eduardo Casas
Dpto. Matemática Aplicada y Ciencias de la Computación
Universidad de Cantabria, Santander (Spain)
CIMPA-UNESCO-GUADELOUPE School
Pointe-à-Pitre
January 3–18, 2009
Introduction.
In a control problem we find the following basic elements.
(1) A control u that we can handle according to our interests,
which can be chosen among a family of feasible controls K.
(2) The state of the system y to be controlled, which depends on
the control. Some limitations can be imposed on the state,
in mathematical terms y ∈ C, which means that not every
possible state of the system is satisfactory.
(3) A state equation that establishes the dependence between the
control and the state. In the next sections this state equation
will be a partial differential equation, y being the solution of
the equation and u a function arising in the equation so that
any change in the control u produces a change in the solution
y. However the origin of control theory was connected with the
control of systems governed by ordinary differential equations
and there is a huge activity in this field; see, for instance, the
classical books Pontriaguine et al. [21] or Lee and Markus
[15].
(4) A function to be minimized, called the objective function or
the cost function, depending on the control and the state (y, u).
The objective is to determine an admissible control that provides a
satisfactory state for us and that minimizes the value of functional J.
It is called the optimal control and the associated state is the optimal
state. The basic questions to study are the existence of a solution and
its computation. However to obtain the solution we must use some numerical methods, arising some delicate mathematical questions in this
numerical analysis. The first step to solve numerically the problem requires the discretization of the control problem, which is made usually
by finite elements. A natural question is how good the approximation
is, of course we would like to have some error estimates of these approximations. In order to derive the error estimates it is essential to have
some regularity of the optimal control, some order of differentiability
is necessary, at least some derivatives in a weak sense. The regularity
of the optimal control can be deduced from the first order optimality
3
4
INTRODUCTION.
conditions. Another key tool in the proof of the error estimates is the
use of the second order optimality conditions. Therefore our analysis
requires to derive the first and second order conditions for optimality.
Once we have a discrete control problem we have to use some numerical algorithm of optimization to solve this problem. When the
problem is not convex, the optimization algorithms typically provides
local minima, the question now is if these local minima are significant
for the original control problem.
The following steps must be followed when we study an optimal
control problem:
(1) Existence of a solution.
(2) First and second order optimality conditions.
(3) Numerical approximation.
(4) Numerical resolution of the discrete control problem.
We will not discuss the numerical analysis, we will only consider
the first two points for a model problem. In this model problem the
state equation will be a semilinear elliptic partial differential equation.
There are no many books devoted to the questions we are going
to study here. Firstly let me mention the book by Profesor J.L. Lions
[17], which is an obliged reference in the study of the theory of optimal
control problems of partial differential equations. In this text, that
has left an indelible track, the reader will be able to find some of
the methods used in the resolution of the two first questions above
indicated. More recent books are X. Li and J. Yong [16], H.O. Fattorini
[11] and F. Tröltzsch [25].
CHAPTER 1
Setting of the Problem and Existence of a Solution
Let Ω be an open, connected and bounded domain in Rn , n = 2, 3,
with a Lipschitz boundary Γ. In this domain we consider the following
state equation
½
Ay + a0 (x, y) = u in Ω
(1.1)
y = 0 on Γ
where a0 : Ω × R −→ R is a Carathéodory function and A denotes a
second-order elliptic operator of the form
n
X
Ay(x) = −
∂xj (aij (x)∂xi y(x))
i,j=1
with the coefficients aij ∈ L∞ (Ω) satisfying
n
X
2
λA kξk ≤
aij (x)ξi ξj ∀ξ ∈ Rn , for a.e. x ∈ Ω
i,j=1
for some λA > 0. In (1.1), the function u denotes the control and we
will denote by yu the solution associated to u. We will state later the
conditions leading to the existence and uniqueness of a solution of (1.1)
in C(Ω̄) ∩ H 1 (Ω).
The optimal control problem is formulated as follows
Z
Z

N


min J(u) =
L(x, yu (x)) dx +
u(x)2 dx


2
Ω
Ω
(P)
1
subject
to
(y
,
u)
∈
(C(
Ω̄)
∩
H
(Ω))
× L2 (Ω),

u



u ∈ K and a(x) ≤ g(x, yu (x)) ≤ b(x) ∀x ∈ K.
We impose the following assumptions on the data of the control
problem.
(A1) In the whole paper N ≥ 0 is a real number and K is a convex
and closed subset of L2 (Ω). We introduce the set
Uad = {u ∈ K : a(x) ≤ g(x, yu (x)) ≤ b(x) ∀x ∈ K}
and we assume that Uad is not empty.
5
6
1. SETTING OF THE PROBLEM AND EXISTENCE OF A SOLUTION
(A2) The mapping a0 : Ω×R −→ R is a Carathéodory function of class
C 2 with respect to the second variable and there exists a real number
p > n/2 such that
a0 (·, 0) ∈ Lp (Ω),
∂a0
(x, y) ≥ 0
∂y
for a.e. x ∈ Ω.
Moreover, for all M > 0, there exists a constant C0,M > 0 such that
¯ ¯ 2
¯
¯
¯ ¯ ∂ a0
¯
¯ ∂a0
¯
¯
¯ ≤ C0,M ,
¯
+
(x,
y)
(x,
y)
¯ ¯ ∂y 2
¯
¯ ∂y
¯ 2
¯
2
¯ ∂ a0
¯
∂
a
0
¯
¯ ≤ C0,M |y2 − y1 |,
(x,
y
)
−
(x,
y
)
2
1
¯ ∂y 2
¯
∂y 2
for a.e. x ∈ Ω and |y|, |yi | ≤ M , i = 1, 2.
(A3) L : Ω × R −→ R is a Carathéodory function of class C 2 with
respect to the second variable, L(·, 0) ∈ L1 (Ω), and for all M > 0 there
exist a constant CL,M > 0 and a function ψM ∈ L1 (Ω) such that
¯
¯
¯
¯ 2
¯ ∂L
¯
¯
¯
¯ (x, y)¯ ≤ ψM (x), ¯ ∂ L (x, y)¯ ≤ CL,M ,
¯ ∂y
¯
¯
¯ ∂y 2
¯
¯ 2
2
¯
¯∂ L
∂
L
¯ ≤ CL,M |y2 − y1 |,
¯
(x,
y
)
−
(x,
y
)
2
1
¯
¯ ∂y 2
∂y 2
for a.e. x ∈ Ω and |y|, |yi | ≤ M , i = 1, 2.
(A4) K is a compact subset of Ω̄ and the function g : K × R −→ R
is continuous, together with its derivatives (∂ j g/∂y j ) ∈ C(K × R) for
j = 0, 1, 2. We also assume that a, b : K −→ [−∞, +∞] are measurable
functions, with a(x) < b(x) for every x ∈ K, such that their domains
Dom(a) = {x ∈ K : −∞ < a(x)} and Dom(b) = {x ∈ K : b(x) < ∞}
are closed sets and a and b are continuous on their respective domains.
Finally, we assume that either K ∩Γ = ∅ or a(x) < g(x, 0) < b(x) holds
for every x ∈ K ∩ Γ. We will denote
Yab = {z ∈ C(K) : a(x) ≤ z(x) ≤ b(x) ∀x ∈ K}.
Let us remark that a (b) can be identically equal to −∞ (+∞),
which means that we only have upper (lower) bounds on the state.
Thus the above framework define a quite general formulation for the
pointwise state constraints.
1. SETTING OF THE PROBLEM AND EXISTENCE OF A SOLUTION
7
Remark 1.1. By using Tietze’s theorem we can extend the functions a and b from their respective domains to continuous functions on
K, denoted by ā and b̄ respectively, such that ā(x) < b̄(x) ∀x ∈ K.
Since K is compact, there exists ρ̄ > 0 such that b̄(x) − ā(x) > ρ̄ for
every x ∈ K. If we define z̄ = (ā + b̄)/2, then Bρ̄/2 (z̄) ⊂ Yab , where
Bρ̄/2 (z̄) denotes the open ball in C(K) with center at z̄ and radius ρ̄/2.
Therefore Yab is a closed convex set with nonempty interior.
Remark 1.2. A typical functional in control theory is
Z
©
ª
1
J(u) =
|yu (x) − yd (x)|2 + N u2 (x) dx,
2 Ω
where
yd ∈ L2 (Ω) denotes the ideal state of the system. The term
R
N u2 (x)dx can be considered as the cost term and it is said that the
Ω
control is expensive if N is big, however the control is cheap if N is
small or zero. From a mathematical point of view the presence of the
term N u2 , with N > 0, has a regularizing effect on the optimal control;
see §2.3.
Remark 1.3. The most frequent choices for the set of controls are
K = L2 (Ω) or
K = Uα,β = {u ∈ L2 (Ω) : α(x) ≤ u(x) ≤ β(x) a.e. in Ω}
where α, β ∈ L2 (Ω).
Given u ∈ Lr (Ω), with r > n/2, the existence of a solution yu in
H01 (Ω) ∩ L∞ (Ω) of (1.1) can be proved as follows. Without lost of
generality we can assume that r ≤ 2. First we truncate a0
a0M (x, t) = a0 (x, Proj[−M,+M ] (t)).
The Assumption (A2) implies that
n
.
2
Now we can define the mapping F : Lr (Ω) −→ Lr (Ω) by F (y) = z
where z ∈ H 1 (Ω) is the unique solution of
½
Az + a0M (x, y) = u in Ω
z = 0 on Γ.
|a0M (x, y(x))| ≤ |a0 (x, 0)| + C0M M ∈ Lp (Ω), with p >
Now it is easy to use the Schauder’s theorem to deduce the existence a
fixed point yM of F . On the other hand, thanks to the monotonicity of
a0 with respect to the second variable we get that {yM }∞
M =1 is uniformly
bounded in L∞ (Ω) (see, for instance, Stampacchia [24]). Consequently
for M large enough a0M (x, yM (x)) = a0 (x, yM (x)) and then yM = yu .
Thus, the monotonicity of a0 implies that yu is the unique solution of
8
1. SETTING OF THE PROBLEM AND EXISTENCE OF A SOLUTION
(1.1) in H01 (Ω) ∩ L∞ (Ω). On the other hand the inclusion Ayu ∈ Lq (Ω),
with q = min{r, p} > n/2, implies that yu belongs to the space of
Hölder functions C θ (Ω̄) for some 0 < θ < 1; see [12].
In the case of Lipschitz coefficients aij and a regular boundary Γ we
have some additional regularity for yu . Indeed, the result by Grisvard
[13] implies that yu ∈ W 2,q (Ω), where q is defined as above. In the case
of convex sets Ω and supposing that p, r ≥ 2 we have H 2 (Ω)-regularity
of the states, see [13] again. The following theorem summarizes these
regularity properties.
Theorem 1.4. For any control u ∈ Lr (Ω), with r > n/2, there
exists a unique solution yu of (1.1) in H01 (Ω) ∩ C θ (Ω̄) for some 0 <
θ < 1. Moreover there exists a constant CA > 0 such that
¡
¢
(1.2)
kyu kH01 (Ω) + kyu kC θ (Ω̄) ≤ CA ka0 (·, 0)kLp (Ω) + kukLr (Ω) .
Moreover if Ω is convex and aij ∈ C 0,1 (Ω̄) for 1 ≤ i, j ≤ n and p, r ≥ 2,
then yu ∈ H 2 (Ω). Finally, if aij ∈ C 0,1 (Ω̄) for 1 ≤ i, j ≤ n and Γ is of
class C 1,1 , then yu ∈ W 2,q (Ω) where q = min{r, p}.
The next theorem states the existence of a solution for the control
problem (P).
Theorem 1.5. Under the assumptions (A1)-(A4), the problem (P)
has at least a solution if one of the following hypotheses holds
(1) Either K is bounded in L2 (Ω)
(2) or L(x, y) ≥ ψ(x)+λy 2 , with 0 < 4|λ|CA2 < N and ψ ∈ L1 (Ω).
Proof. Let {uk } ⊂ K be a minimizing sequence of (P), this means
that J(uk ) → inf(P). Under the first hypothesis of the theorem, we get
2
that {uk }∞
k=1 is a bounded sequence in L (Ω). In the second case, from
(1.2) it is easy to deduce that
µ
¶
Z
N
2
2
2
− 2|λ|CA kuk k2L2 (Ω) .
J(uk ) ≥
ψ(x) dx − 2|λ|CA ka0 (·, 0)kLp (Ω) +
2
Ω
2
Thus {uk }∞
k=1 is a bounded sequence in L (Ω) in any of the two cases.
Let us take a subsequence, again denoted in the same way, converging weakly in L2 (Ω) to an element ū. Since K is convex and closed
in L2 (Ω), then it is weakly closed, hence ū ∈ K. From Theorem 1.4
and the compactness of the embedding C θ (Ω̄) ⊂ C(Ω̄), we deduce that
yuk → yū strongly in H01 (Ω) ∩ C(Ω̄). This along with the continuity of
g and the fact that a(x) ≤ g(x, yuk (x)) ≤ b(x) for every x ∈ K implies
that a(x) ≤ g(x, yū (x)) ≤ b(x) too, hence ū ∈ Uad .
1. SETTING OF THE PROBLEM AND EXISTENCE OF A SOLUTION
9
Finally we have
J(ū) ≤ lim inf J(uk ) = inf (P).
k→∞
An existence theorem can be proved by a quite similar way for a
more general class of cost functional
Z
J(u) =
L(x, yu (x), u(x)) dx,
Ω
where L : Ω × R × R −→ R is a Carathéodory function satisfying the
following assumptions
H1) For every (x, y) ∈ Ω × R, L(x, y, ·) : R −→ R is a convex
function.
H2) For any M > 0 there exists a function ψM ∈ L1 (Ω) and a
constant C1 > 0 such that
L(x, y, u) ≤ ψM (x) + C1 u2 a.e. x ∈ Ω, ∀|y| ≤ M.
H3) Either K is bounded in L2 (Ω) or L(x, y) ≥ ψ(x)+C2 (N u2 −y 2 ),
with 0 < 2CA2 < N , C2 > 0 and ψ ∈ L1 (Ω).
Then problem (P) has at least one solution.
If we assume that K is bounded in L∞ (Ω), then it is enough to
suppose H1) and
H2) For any M > 0 there exists a function ψM ∈ L1 (Ω) such that
|L(x, y, u)| ≤ ψM (x) a.e. x ∈ Ω, ∀|y|, |u| ≤ M.
The convexity with respect to the control of L is a key point in the
proof. If this assumption does not hold, then the existence of a solution
can fail. Let us see an example.
½
−∆y = u in Ω
y = 0 on Γ.

Z

Minimize J(u) = [yu (x)2 + (u2 (x) − 1)2 ]dx
(P)
 −1 ≤ u(x) ≤ +1, xΩ ∈ Ω.
Let us take a sequence of controls {uk }∞
k=1 such that |uk (x)| = 1 for
every x ∈ Ω and verifying that uk * 0 ∗ weakly in L∞ (Ω). The reader
can make a construction of such a sequence (include Ω in a n-cube to
simplify the proof). Then, taking into account that yuk → 0 uniformly
in Ω, we have
Z
0≤
inf
J(u) ≤ lim J(uk ) = lim
yuk (x)2 dx = 0.
−1≤u(x)≤+1
k→∞
k→∞
Ω
10
1. SETTING OF THE PROBLEM AND EXISTENCE OF A SOLUTION
But it is obvious that J(u) > 0 for any feasible control, which proves
the non existence of an optimal control.
In the lack of convexity of L, it is necessary to use some compactness
argumentation to prove the existence of a solution. The compactness
of the set of feasible controls has been used to get the existence of a
solution in control problems in the coefficients of the partial differential
operator. These type of problems appear in structural optimization
problems and in the identification of the coefficients of the operator;
see Casas [3] and [4].
To deal with control problems in the absence of convexity and compactness, (P) is sometimes included in a more general problem (P̄), in
such a way that inf(P)= inf(P̄), (P̄) having a solution. This leads to
the relaxation theory; see Ekeland and Temam [10], Warga [26], Young
[27], Roubiček [22], Pedregal [20].
CHAPTER 2
First Order Optimality Conditions
In this chapter we are going to study the first order optimality conditions. They are necessary conditions for local optimality. In the case
of convex problems they become also sufficient conditions for global optimality. In the absence of convexity the sufficiency requires the use of
second order optimality conditions. We will prove sufficient conditions
of second order in the next chapter. The sufficient conditions play a
crucial role in the numerical analysis of the problems. From the first
order necessary conditions we can deduce some regularity properties of
the optimal control as we will prove later.
2.1. First Order Optimality Conditions
The key tool to get the first order optimality conditions is provided
by the next lemma.
Lemma 2.1. Let U and Z be two Banach spaces and K ⊂ U and
C ⊂ Z two convex sets, C having a nonempty interior. Let ū be a local
solution of problem
½
Min J(u)
(Q)
u ∈ K and F (u) ∈ C,
where J : U −→ (−∞, +∞] and F : U −→ Z are Gâteaux differentiable at ū. Then there exist a real number ᾱ ≥ 0 and an element
µ̄ ∈ Z 0 such that
(2.1)
ᾱ + kµ̄kZ 0 > 0;
(2.2)
hµ̄, z − F (ū)i ≤ 0 ∀z ∈ C;
(2.3)
hᾱJ 0 (ū) + [DF (ū)]∗ µ̄, u − ūi ≥ 0 ∀u ∈ K.
Moreover can take ᾱ = 1 if the Salter conditions holds:
(2.4)
o
∃u0 ∈ K such that F (ū) + DF (ū) · (u0 − ū) ∈ C .
Reciprocally, if ū is a feasible control, (Q) is a convex problem and
(2.2) and (2.3) hold with ᾱ = 1, then ū is a global solution of (Q).
11
12
2. FIRST ORDER OPTIMALITY CONDITIONS
We recall that (Q) is said convex if the function J is convex and
the set
Uad = {u ∈ K : F (u) ∈ C}
is also convex.
Proof. First we make the proof for global solutions of (Q). Consider the sets
A = {(z, λ) ∈ Z × R : ∃u ∈ K such that
z = F (ū) + DF (ū) · (u − ū) and λ = J 0 (ū) · (u − ū)}
o
and B = C ×(−∞, 0). It is is obvious that A and B are convex sets.
Moreover they are disjoints. Indeed, suppose that there exists an element u0 ∈ K such that
z0 = F (ū) + DF (ū) · (u0 − ū) =
o
F (ū) + limρ→0 ρ1 [F (ū + ρ(u0 − ū)) − F (ū)] ∈ C
and
1
(J(ū + ρ(u0 − ū)) − J(ū)) < 0.
ρ→0 ρ
Then we can find ρ0 ∈ (0, 1) such that
o
1
zρ = F (ū) + (F (ū + ρ(u0 − ū)) − F (ū)) ∈ C ∀ρ ∈ (0, ρ0 ),
ρ
1
(J(ū + ρ(u0 − ū)) − J(ū)) < 0 ∀ρ ∈ (0, ρ0 ).
ρ
This implies
λ0 = J 0 (ū) · (u0 − ū) = lim
o
F (ū + ρ(u0 − ū)) = ρzρ + (1 − ρ)F (ū) ∈ C
and
J(ū + ρ(u0 − ū)) < J(ū)
for every ρ ∈ (0, ρ0 ), which contradicts the fact that ū is a solution of
(Q).
Now taking into account that B is an open set, from the separation
theorems of convex sets (see, for instance, Brezis [2]) we deduce the
existence of µ̄ ∈ Z 0 and ᾱ ∈ R such that
(2.5)
hµ̄, z1 i + ᾱλ1 > hµ̄, z2 i + ᾱλ2
∀(z1 , λ1 ) ∈ A, ∀(z2 , λ2 ) ∈ B.
Let us prove that ᾱ ≥ 0. If ᾱ < 0, we take λ1 = 0, z1 = F (ū),
o
z2 ∈ C fixed and λ2 = −k in (2.5), with k a positive integer, which
leads to
hµ̄, F (ū)i > hµ̄, z2 i − ᾱk.
2.1. FIRST ORDER OPTIMALITY CONDITIONS
13
Taking k large enough we get a contradiction, hence ᾱ ≥ 0. Moreover,
since (2.5) is a strict inequality, ᾱ = kµ̄k = 0 is not possible, which
proves (2.1).
Since B̄ = C̄ × (−∞, 0], we get from (2.5)
(2.6) hµ̄, z1 i + ᾱλ1 ≥ hµ̄, z2 i + ᾱλ2
∀(z1 , λ1 ) ∈ A, ∀(z2 , λ2 ) ∈ B̄.
Now it is sufficient to take z1 = F (ū), z2 = z ∈ C and λ1 = λ2 = 0
to deduce (2.2). The inequality (2.3) is obtained setting z1 = F (ū) +
DF (ū) · (u − ū), λ1 = J 0 (ū) · (u − ū), with u ∈ K, λ2 = 0 and z2 = F (ū).
Finally, let us prove that ᾱ 6= 0 when the Slater condition holds.
Let us assume that (2.4) holds and ᾱ = 0. As a first consequence of
this assumption we get
(2.7)
o
hµ̄, z − F (ū)i < 0 ∀z ∈ C .
o
To prove this inequality it is enough to suppose that there exists z0 ∈ C
such that hµ̄, z0 − F (ū)i = 0. Using (2.2) we deduce
hµ̄, z + z0 − F (ū)i ≤ 0 ∀z ∈ B² (0),
o
with ² > 0 small enough, in such a way that B² (z0 ) ⊂ C. Then hµ̄, zi ≤
0 for all z ∈ B² (0), hence µ̄ = 0, which contradicts (2.1).
o
Taking now z = F (ū) + DF (ū) · (u0 − ū) ∈ C in (2.7), it follows
h[DF (ū)]∗ µ̄, u0 − ūi = hµ̄, DF (ū) · (u0 − ū)i < 0,
which contradicts (2.3), therefore ᾱ > 0. It is enough to divide (2.2)
and (2.3) by ᾱ and to denote again the quotient µ̄/ᾱ by µ̄ to deduce
the desired result.
Now let us consider the case where ū is only a local solution of (Q).
In this case, ū is a global solution of
½
Min J(u)
(Qε )
u ∈ K ∩ Bε (ū) and F (u) ∈ C,
for some ε > 0 small enough. Then (2.1) and (2.2) hold and (2.3) is
replaced by
hᾱJ 0 (ū) + [DF (ū)]∗ µ̄, u − ūi ≥ 0 ∀u ∈ K ∩ Bε (ū).
Now it is an easy exercise to prove that the previous inequality implies
(2.3). On the other hand let us remark that if the Slater condition
(2.4) holds for some u0 ∈ K, then there exists ρ0 > 0 such that uρ =
ū + ρ(u0 − ū) ∈ K ∩ Bε (ū) for every 0 < ρ < ρ0 . Moreover, for any of
14
2. FIRST ORDER OPTIMALITY CONDITIONS
these functions uρ we have
F (ū) + DF (ū) · (uρ − ū) = F (ū) + ρDF (ū) · (u0 − ū)
o
= (1 − ρ)F (ū) + ρ[F (ū) + DF (ū) · (u0 − ū)] ∈C,
which implies that Slater condition also holds at ū for the problem
(Qε ).
Finally, let us prove the reciprocal implication. We assume that J is
a convex function, Uad is a convex set set and (2.2) and (2.3) hold with
ᾱ = 1. Let us take u ∈ Uad arbitrary and define uρ = ū + ρ(u − ū) for
every ρ ∈ [0, 1], then uρ ∈ Uad . Using (2.2) and the fact that F (uρ ) ∈ C
we get
1
h[DF (ū)]∗ µ̄, u − ūi = lim hµ̄, F (uρ ) − F (ū)i ≤ 0.
ρ&0 ρ
Using this inequality and (2.3) we obtain
0 ≤ hJ 0 (ū) + [DF (ū)]∗ µ̄, u − ūi ≤ J 0 (ū)(u − ū).
Finally from this inequality and the convexity of J we conclude
J(uρ ) − J(ū)
≤ J(u) − J(ū).
ρ&0
ρ
0 ≤ lim
Since u is arbitrary in Uad , we deduce that ū is a global solution.
In order to apply this lemma to the study of problem (P) we need
to analyze the differentiability of the functionals involved in the control
problem.
Proposition 2.2. The mapping G : L2 (Ω) −→ H01 (Ω) ∩ C θ (Ω̄)
defined by G(u) = yu is of class C 2 . Furthermore if u, v ∈ L2 (Ω) and
z = DG(u) · v, then z is the unique solution in H01 (Ω) ∩ C θ (Ω̄) of
Dirichlet problem

∂a0

(x, yu (x))z = v in Ω,
Az +
(2.8)
∂y

z = 0 on Γ.
Finally, for every v1 , v2 ∈ L2 (Ω), zv1 v2 = G00 (u)v1 v2 is the solution of

∂a0
∂ 2 a0
 Az
(x, yu )zv1 v2 +
(x, yu )zv1 zv2 = 0 in Ω
v1 v2 +
∂y
∂y 2
(2.9)

zv1 v2 = 0 on Γ,
where zvi = G0 (u)vi , i = 1, 2.
2.1. FIRST ORDER OPTIMALITY CONDITIONS
15
Proof. Let us fix q = min{p, 2} > n/2. To prove the differentiability of G we will apply the implicit function theorem. Let us consider
the Banach space
V (Ω) = {y ∈ H01 (Ω) ∩ C θ (Ω̄) : Ay ∈ Lq (Ω)},
endowed with the norm
kykV (Ω) = kykH01 (Ω) + kykC θ (Ω̄) + kAykLq (Ω) .
Now let us take the function
F : V (Ω) × Lq (Ω) −→ Lq (Ω)
defined by
F(y, u) = Ay + a0 (·, y) − u.
It is obvious that F is of class C 2 , yu ∈ V (Ω) for every u ∈ Lq (Ω),
F(yu , u) = 0 and
∂F
∂a0
(y, u) · z = Az +
(x, y)z
∂y
∂y
is an isomorphism from V (Ω) into Lq (Ω). By applying the implicit
function theorem we deduce that G is of class C 2 and z = DG(u) · v
is given by (2.8). Finally (2.9) follows by differentiating twice with
respect to u in the equation
AG(u) + a0 (·, G(u)) = u.
As a consequence of this result we get the following proposition.
Proposition 2.3. The function J : L2 (Ω) → R is of class C 2 .
Moreover, for every u, v, v1 , v2 ∈ L2 (Ω)
Z
0
(2.10)
J (u)v = (ϕu + N u)v dx
Ω
and
(2.11)
J 00 (u)v1 v2
¸
Z · 2
∂ 2 a0
∂ L
(x, yu )zv1 zv2 + N v1 v2 − ϕu 2 (x, yu )zv1 zv2 dx
=
2
∂y
Ω ∂y
where zvi = G0 (u)vi , i = 1, 2, and ϕu ∈ W01,s (Ω) for every 1 ≤ s <
n/(n − 1) is the unique solution of problem

 A∗ ϕ + ∂a0 (x, yu )ϕ = ∂L (x, yu ) in Ω
∂y
∂y
(2.12)

ϕ =
0
on Γ,
A∗ being the adjoint operator of A.
16
2. FIRST ORDER OPTIMALITY CONDITIONS
Proof. First of all, let us remark that the right hand side of (2.12)
belongs to L1 (Ω) 6⊂ H −1 (Ω); see Assumption (A3). This implies that
the solution of (2.12) does not belong to H01 (Ω), but it is just an element
of W01,s (Ω) for all 1 ≤ s < n/(n−1); see Stampacchia [24]. Since n ≤ 3,
then W 1,s (Ω) ⊂ L2 (Ω) for 2n/(n + 2) ≤ s < n/(n − 1).
From Assumption (A3), Proposition 2.2 and the chain rule we get
¸
Z ·
∂L
0
J (u) · v =
(x, yu (x))zv (x) + N u(x)v(x) dx,
Ω ∂y
where zv = G0 (u)v. Using (2.12) in this expression we obtain
¾
Z ½
∂a0
∗
0
J (u) · v =
[A ϕu +
(x, yu )ϕu ]zv + N u(x)v(x) dx
∂y
Ω
¾
Z ½
∂a0
=
(x, yu )zv ]ϕu + N u(x)v(x) dx
[Azv +
∂y
Ω
Z
= [ϕu (x) + N u(x)]v(x) dx,
Ω
which proves (2.10). Finally (2.11) follows in a similar way by application of the chain rule and Proposition 2.2.
The next result is an immediate consequence of Proposition 2.2.
Proposition 2.4. The mapping F : L2 (Ω) → C(K), defined by
F (u) = g(·, yu (·)), is of class C 2 . Moreover, for all u, v, v1 , v2 ∈ L2 (Ω)
F 0 (u)v =
(2.13)
∂g
(·, yu (·))zv (·)
∂y
and
(2.14)
F 00 (u)v1 v2 =
∂ 2g
∂g
(·, yu (·))zv1 (·)zv2 (·) +
(·, yu (·))zv1 v2 (·)
2
∂y
∂y
where zv = G0 (u)v, zvi = G0 (u)vi , i = 1, 2, and zv1 v2 = G00 (u)v1 v2 .
Before stating the first order necessary optimality conditions, let
us fix some notation. We denote by M (K) the Banach space of all
real and regular Borel measures in K, which is identified with the dual
space of C(K). The norm in M (K) is given by
½Z
¾
kµkM (K) = |µ|(K) = sup
z(x) dµ(x) : z ∈ C(K), kzkC(K) ≤ 1 ,
K
where |µ|(K) is the total variation of the measure µ.
Combining Lemma 2.1 with the previous propositions we get the
first order optimality conditions.
2.1. FIRST ORDER OPTIMALITY CONDITIONS
17
Theorem 2.5. Let ū be a local minimum of (P). Then there exist
ᾱ ≥ 0, ȳ ∈ H01 (Ω) ∩ C θ (Ω̄), ϕ̄ ∈ W01,s (Ω) for all 1 ≤ s < n/(n − 1), and
µ̄ ∈ M (Ω), with (ᾱ, µ̄) 6= (0, 0), such that the following relationships
hold
½
Aȳ + a0 (x, ȳ) = ū in Ω,
(2.15)
ȳ = 0 on Γ,

∂a0
∂L
∂g
 ∗
A ϕ̄ +
(x, ȳ)ϕ̄ = ᾱ (x, ȳ) +
(x, ȳ)µ̄ in Ω,
(2.16)
∂y
∂y
∂y

ϕ̄ =
0
on Γ,
Z
(2.17)
(z(x) − g(x, ȳ(x))dµ̄(x) ≤ 0 ∀z ∈ Yab ,
K
Z
(2.18)
[ϕ̄(x) + ᾱN ū(x)](u(x) − ū(x))dx ≥ 0 ∀u ∈ K.
Ω
Furthermore, if there exists u0 ∈ K such that
(2.19)
a(x) < g(x, ȳ(x)) +
∂g
(x, ȳ(x))z0 (x) < b(x) ∀x ∈ K,
∂y
where z0 ∈ H01 (Ω) ∩ C θ (Ω̄) is the unique solution of

 Az + ∂a0 (x, ȳ)z = u0 − ū in Ω
∂y
(2.20)

z =
0
on Γ,
then (2.15)-(2.18) hold with ᾱ = 1. Reciprocally, if ū ∈ Uad , the functions a0 and g are constant or linear with respect to y, L is convex
with respect to y and (2.15)-(2.18) hold with ᾱ = 1, then ū is a global
solution of (P).
Proof. The previous theorem is a consequence of Lemma 2.1 and
the Propositions 2.3 and 2.4. Indeed it is enough to take Z = C(K),
C = Yab , F (u) = g(·, yu ) and J as the cost functional of (P). In this
framework (2.2) is equivalent to (2.17). Then it is immediate the existence of µ̄ and ᾱ, with (ᾱ, µ̄) 6= (0, 0), such that (2.17) holds. Let us
check that (2.3) is equivalent to (2.18). First denote the state associate
to ū by ȳ, then (2.15) holds. Now we define ϕū , ψµ̄ ∈ W01,s (Ω) satisfying

 A∗ ϕū + ∂a0 (x, ȳ)ϕū = ᾱ ∂L (x, ȳ) in Ω
∂y
∂y
(2.21)

ϕū =
0
on Γ
18
2. FIRST ORDER OPTIMALITY CONDITIONS
and

 A∗ ψµ̄ + ∂a0 (x, ȳ)ψµ̄ = ∂g (x, ȳ)µ̄
∂y
∂y

ψµ̄ =
0
(2.22)
in Ω
on Γ
respectively. Setting ϕ̄ = ϕµ̄ + ψµ̄ , then we have that ϕ̄ satisfies (2.16).
From (2.10) and (2.13) we get for every v ∈ L2 (Ω)
0
∗
hᾱJ
Z (ū) + [DF (ū)] µ̄, viZ
∂g
= (ϕū + ᾱN ū)v dx +
(x, ȳ)zv dµ̄(x)
K ∂y
ZΩ
∂a0
(x, ȳ)ψµ̄ , zv iM (K),C(K)
= (ϕū + ᾱN ū)v dx + hA∗ ψµ̄ +
∂y
Ω
Z
Z
∂a0
= (ϕū + ᾱN ū)v dx + (Azv +
(x, ȳ)zv )ψµ̄ dx
∂y
Ω
Ω
Z
= (ϕ̄ + ᾱN ū)v dx.
Ω
which proves the desired equivalence.
Finally, we observe that (2.20) is the formulation of the Slater hypothesis (2.4) corresponding to the problem (P). The rest is obvious.
Let us prove some properties of the Lagrange multiplier µ̄. Define
Ka = {x ∈ K : g(x, ȳ(x) = a(x)}
and
Kb = {x ∈ K : g(x, ȳ(x)) = b(x)}.
Now we set K0 = Ka ∪ Kb . Then we have the following decomposition
of µ̄.
Corollary 2.6. The multiplier µ̄ has the following Lebesgue decomposition µ̄ = µ̄+ − µ̄− , where µ̄− and µ̄+ are positive measures on
K whose supports are contained in Ka and Kb respectively. Moreover
we have
kµ̄kM (K) = |µ̄(K0 )| = µ̄− (Ka ) + µ̄+ (Kb ).
Proof. Let us prove that the support of µ̄ is concentrated in K0 .
Take a sequence of compact subsets of K, {Kj }∞
j=1 , such that
o
o
Kj+1 ⊂Kj ⊂ Kj = Kj ∀j ≥ 1 and
∞
\
j=1
Kj = K0 .
2.1. FIRST ORDER OPTIMALITY CONDITIONS
19
o
For every j we have a(x) < g(x, ȳ(x)) < b(x) for all x ∈ K\ Kj , then
we deduce the existence of a sequence of strict positive numbers {εj }∞
j=1
such that
o
a(x) + εj < g(x, ȳ(x)) < b(x) − εj ∀x ∈ K\ Kj , ∀j ≥ 1.
o
For every function z ∈ C(K) with support in K\ Kj and z 6≡ 0 we
define
z(x)
zj (x) = εj
+ g(x, ȳ(x)) ∀x ∈ K.
kzkC(K)
It is clear that zj (x) = g(x, ȳ(x)) for every x ∈ Kj and zj ∈ Yab . Then
(2.17) leads to
Z
Z
kzkC(K)
z(x) dµ̄(x) =
(zj (x) − g(x, ȳ(x))) dµ̄(x) ≤ 0
εj
K\Kj
K
o
for every z ∈ C0 (K\ Kj ), which implies that µ̄|K\Kj = 0, therefore
|µ̄|(K \ K0 ) = lim |µ̄|(K \ Kj ) = 0,
j→∞
hence the support of µ̄ is included in K0 .
Finally, since Ka and Kb are disjoint, any function z ∈ C(Ka ) can be
extended to a continuous function z̃ in K0 by setting z̃(x) = g(x, ȳ(x))
for x ∈ Kb . Using Tietze’s Theorem and assuming that a(x) ≤ z(x) ≤
b(x) for all x ∈ Ka , we also extend z̃ to a continuous function in K
such that z̃ ∈ Yab . Therefore, using once again (2.17) we get
Z
Z
(z(x) − a(x)) dµ̄(x) =
(z̃(x) − g(x, ȳ(x))) dµ̄(x) ≤ 0.
Ka
K
This inequality implies that µ̄ is nonpositive in Ka . Analogously we
prove that µ̄ is nonnegative in Kb , which concludes the proof.
Remark 2.7. Sometimes the state constraints are active at a finite
set of points K0 = {xj }m
j=1 ⊂ K. Then the previous corollary implies
that
½
m
X
≥ 0 if g(xj , ȳ(xj )) = b(xj ),
(2.23) µ̄ =
λ̄j δxj , with λ̄j =
≤ 0 if g(xj , ȳ(xj )) = a(xj ),
j=1
where δxj denotes the Dirac measure centered at xj . If we denote by
ϕ̄j , 1 ≤ j ≤ m, and ϕ̄0 the solutions of

 A∗ ϕ̄j + ∂a0 (x, ȳ(x))ϕ̄j = δx in Ω
j
∂y
(2.24)

ϕ̄j = 0
on Γ
20
and
(2.25)
2. FIRST ORDER OPTIMALITY CONDITIONS

 A∗ ϕ̄0 + ∂a0 (x, ȳ(x))ϕ̄0 = ∂L (x, ȳ) in Ω
∂y
∂y

ϕ̄0 = 0
on Γ
then the adjoint state defined by (2.16) is given by
m
X
∂g
(2.26)
ϕ̄ = ᾱϕ̄0 +
λ̄j (xj , ȳ(xj ))ϕ̄j .
∂y
j=1
2.2. Regularity of the Optimal Controls
From the optimality conditions we can deduce some properties on
the behavior of the local minima. In the following theorem we consider
some different situations.
Theorem 2.8. Let us assume that ū is a local minimum of (P).
Then the following statements hold.
• If K = L2 (Ω) and N > 0, then
1
n
(2.27)
ᾱ = 1 and ū = − ϕ̄ ∈ W01,s (Ω) for all 1 ≤ s <
.
N
n−1
• If K = Uα,β = {u ∈ L2 (Ω) : α(x) ≤ u(x) ≤ β(x) a.e. in Ω}
and N > 0 and ᾱ = 1, then ū is given by the expression
µ
¶
ϕ̄(x)
(2.28)
ū(x) = Proj[α(x),β(x)] −
.
N
Moreover, ū ∈ W 1,s (Ω) for all 1 ≤ s < n/(n − 1) if α, β ∈
W 1,s (Ω).
• If K = Uα,β and ᾱN = 0, then
½
α(x) if ϕ̄(x) > 0
(2.29)
ū(x) =
β(x) if ϕ̄(x) < 0.
The proof follows easily from the inequality (2.18). Now we can
improve the regularity results if we assume more regularity on the data
of the control problem. Let us start supposing that K0 is a finite set.
The following result was proved in [5].
Theorem 2.9. Assume p > n in Assumption (A2) and ψM ∈ Lp (Ω)
in (A3). Suppose also that N > 0, K = Uαβ , ai,j , α, β ∈ C 0,1 (Ω̄),
1 ≤ i, j ≤ n and that Γ is of class C 1,1 . Let (ȳ, ū, ϕ̄, µ̄) ∈ H01 (Ω) ∩
C θ (Ω̄) × L∞ (Ω) × W01,s (Ω) × M (K), for all 1 ≤ s < n/(n − 1), satisfy
the optimality system (2.15)-(2.18) with ᾱ = 1. If the active set consists
2.2. REGULARITY OF THE OPTIMAL CONTROLS
21
0,1
of finitely many points, i.e. K0 = {xj }m
(Ω̄)
j=1 , then ū belongs to C
2,p
and ȳ to W (Ω).
Proof. From (2.24) and (2.25) we deduce that ϕ̄0 ∈ W 2,p (Ω) ⊂
C (Ω̄) and ϕ̄j ∈ W 2,p (Ω \ B̄ε (xj )) ⊂ C 1 (Ω̄ \ Bε (xj )), 1 ≤ j ≤ m for
any ε > 0. Furthermore it is well known, see for instance [1], that the
asymptotic behavior of ϕ̄j (x) when x → xj is of the type
 µ
¶
1


if n = 2

 C1 log |x − x | + C2
j
(2.30)
ϕ̄j (x) ≈
¶
µ

1

 C1
+ C2
if n > 2,

|x − xj |n−2
1
with C1 > 0. In particular we have that ϕ̄ ∈ C 1 (Ω̄ \ {xj }j∈Iū ), where
Iū = {j ∈ [1, m] : λ̄j 6= 0}.
Let us take
(2.31)
Cαβ = kαkL∞ (Ω) + kβkL∞ (Ω) + 1.
Thanks to (2.30) we can take εαβ > 0 such that
[
|ϕ̄(x)| > N Cαβ ∀x ∈
Bεαβ (xj ),
j∈Iū
the sign in Bεαβ (xj ) being equal to the sign of λ̄j (∂g/∂y)(xj , ȳ(xj )).
Finally, (2.28) implies that

∂g


 α(x) if λ̄j (xj , ȳ(xj )) > 0
∂y
ū(x) =
∀x ∈ Bεαβ (xj ).
∂g


 β(x) if λ̄j (xj , ȳ(xj )) < 0
∂y
Finally, the Lipschitz regularity of ϕ̄ in Ω̄ \ ∪j∈Iū Bεαβ (xj ) and α and β
in Ω̄ respectively, along with (2.28) implies that ū ∈ C 0,1 (Ω̄). Indeed,
it is enough to observe that for any numbers t1 , t2 ∈ R and elements
x1 , x2 ∈ Ω̄ it is satisfied
¯
¯
¯Proj[α(x ),β(x )] (t2 ) − Proj[α(x ),β(x )] (t1 )¯
2
2
1
1
= | max{α(x2 ), min{t2 , β(x2 )}} − max{α(x1 ), min{t1 , β(x1 )}}|
≤ |α(x2 ) − α(x1 )| + |β(x2 ) − β(x1 )| + |t2 − t1 |.
Surprisingly, the bound constraints on the controls contribute to
increase the regularity of ū. Now the question arises if this Lipschitz
property remains also valid for an infinite number of points where the
pointwise state constraints are active. Unfortunately, the answer is
22
2. FIRST ORDER OPTIMALITY CONDITIONS
negative. In fact, the optimal control can even fail to be continuous if
K0 is an infinite and numerable set. Let us present a counterexample.
√
Counterexample. We set Ω = {x ∈ R2 : kxk < 2},
½
1
if kxk ≤ 1 √
ȳ(x) =
1 − (kxk2 − 1)4 if 1 < kxk ≤ 2,
∞
k
∞
= (0, 0), and
K = {xk }∞
k=1 ∪ {x }, where x = (1/k, 0) and x
∞
X
1
µ̄ =
δ k.
k2 x
k=1
Now we define ϕ̄ ∈ W01,s (Ω) for all 1 ≤ s < n/(n − 1) as the solution
of the equation
(
−∆ϕ̄ = ȳ + µ̄ in Ω,
(2.32)
ϕ̄ = 0
on Γ.
The function ϕ̄ can be decomposed in the following form
∞
X
1
ϕ̄(x) = ψ̄(x) +
[ψk (x) + φ(x − xk )],
2
k
k=1
where φ(x) = −(1/2π) log kxk is the fundamental solution of −∆ and
the functions ψ̄, ψk ∈ C 2 (Ω̄) satisfy
(
(
−∆ψk (x) = 0
in Ω,
−∆ψ̄(x) = ȳ(x) in Ω,
ψ̄(x) = 0
on Γ,
ψk (x) = −φ(x − xk )
on Γ.
Finally we set

¯
¯
∞
∞
¯
¯ X
X

1
1

¯
 M = ¯¯ψ̄(0) +
ψk (0)¯ +
φ(xk ) + 1,
2
2
¯
¯
k
k
(2.33)
k=1
k=1


 ū(x) = Proj
[−M,+M ] (−ϕ̄(x))
and a0 (x) = ū(x) + ∆ȳ(x). Then ū is the unique global solution of the
control problem

Z
1


(yu2 (x) + u2 (x)) dx
min J(u) =


2 Ω



subject to (yu , u) ∈ (C(Ω̄) ∩ H 1 (Ω)) × L∞ (Ω),
(P)



−M ≤ u(x) ≤ +M for a.e. x ∈ Ω,




−1 ≤ yu (x) ≤ +1 ∀x ∈ K,
2.2. REGULARITY OF THE OPTIMAL CONTROLS
23
where yu is the solution of
½
−∆y + a0 (x) = u in Ω,
(2.34)
y = 0 on Γ.
As a first step to prove that ū is a solution of problem, we verify
2
that M is a real number: Since {φ(x − xk )}∞
k=1 is bounded in C (Γ),
2
∞
the sequence {ψk }k=1 is bounded in C (Ω̄). Therefore, the convergence
of the first series of (2.33) is obvious. The convergence of the second
one is also clear,
∞
∞
X
1
1 X 1
k
φ(x ) =
log k < ∞.
k2
2π k=1 k 2
k=1
Problem (P) is strictly convex and ū is a feasible control with associated state ȳ satisfying the state constraints. Therefore, there exists
a unique solution characterized by the optimality system. In other
words, the first order optimality conditions are necessary and sufficient
for a global minimum. Let us check that (ȳ, ū, ϕ̄, µ̄) ∈ H01 (Ω) ∩ C(Ω̄) ×
L∞ (Ω) × W01,s (Ω) × M (K) satisfies the optimality system (2.15)-(2.18)
with ᾱ = 1. First, in view of the definition of a0 , it is clear that ȳ is the
state associated to ū. On the other hand, ϕ̄ is the solution of (2.32),
which is the same as (2.16) for our example. Relation (2.18) follows
directly from the definition of ū given in (2.33). Finally, because of the
definition of µ̄ and K, (2.17) can be written in the form
∞
∞
X
X
1
1
k
z(x ) ≤
∀z ∈ C(K) such that −1 ≤ z(x) ≤ +1 ∀x ∈ K,
2
k
k2
k=1
k=1
which obviously is satisfied.
Now we prove that ū is not continuous at x = 0. Notice that
ϕ̄(xk ) = +∞ for every k ∈ N, because φ(0) = +∞. Therefore, (2.33)
implies that ū(xk ) = −M for every k. Since xk → 0, the continuity of
ū at x = 0 requires that u(x) → −M as x → 0. However, we have for
ξ j = (xj + xj+1 )/2 that
(2.35)
!
Ã
∞
∞
X
X
1
1
lim ū(ξ j ) = Proj[−M,+M ] ψ̄(0) +
ψ (0) +
φ(xk ) > −M.
2 k
2
j→∞
k
k
k=1
k=1
Therefore ū is not continuous at (0, 0). Nevertheless, we are able to
improve the regularity result of Theorem 2.8.
Theorem 2.10. Suppose that ū is a strict local minimum of (P).
We also assume that Assumptions (A1)-(A4) and (2.19) hold, N > 0,
24
2. FIRST ORDER OPTIMALITY CONDITIONS
K = Uαβ , α, β ∈ L∞ (Ω) ∩ H 1 (Ω), aij ∈ C(Ω̄) (1 ≤ i, j ≤ n), p > n/2,
and ψM ∈ Lp (Ω) in (A3). Then ū ∈ H 1 (Ω).
Proof. Fix εū > 0 such that ū is a strict global minimum of (P)
in the closed ball B̄εū (ū) ⊂ L2 (Ω). This implies that ū is the unique
global solution of the problem

min J(u)




 subject to (yu , u) ∈ (H 1 (Ω) ∩ C θ (Ω̄)) × L2 (Ω),
0
(P0 )

α(x) ≤ u(x) ≤ β(x) for a.e. x ∈ Ω, ku − ūkL2 (Ω) ≤ εū




a(x) ≤ g(x, yu (x)) ≤ b(x) ∀x ∈ K,
where yu is the solution of (1.1).
Now we select a sequence {xk }∞
k=1 being dense in Dom(a) ∪ Dom(b)
and consider the family of control problems

min J(u)




 subject to (yu , u) ∈ (C(Ω̄) ∩ H 1 (Ω)) × L2 (Ω),
(Pk )

α(x) ≤ u(x) ≤ β(x) for a.e. x ∈ Ω, ku − ūkL2 (Ω) ≤ εū




a(xj ) ≤ g(xj , yu (xj )) ≤ b(xj ), 1 ≤ j ≤ k.
Obviously, ū is a feasible control for every problem (Pk ). Therefore, the existence of a global minimum uk of (Pk ) is a consequence of
Theorem 1.5.
The proof of the theorem is split into three steps: First, we show
2
that the sequence {uk }∞
k=1 converges to ū strongly in L (Ω). In a second
step, we will check that the linearized Slater condition corresponding
to problem (Pk ) holds for all sufficiently large k. Finally, we confirm
1
the boundedness of {uk }∞
k=1 in H (Ω).
∞
Step 1 - Convergence of {uk }k=1 . By taking a subsequence, if necessary, we can suppose that uk * ũ weakly in L2 (Ω). This implies that
yk = yuk → ỹ = yũ strongly in H01 (Ω) ∩ C(Ω̄). Because of the density
of {xk }∞
k=1 in Dom(a) ∪ Dom(b) and the fact that
a(xj ) ≤ g(xj , ỹ(xj )) = lim g(xj , yk (xj )) ≤ b(xj ) ∀ j ≥ 1,
k→∞
it holds that a(x) ≤ g(x, ỹ(x)) ≤ b(x) for every x ∈ K. The control
constraints define a closed and convex subset of L2 (Ω), hence ũ satisfies
all the control constraints. Therefore ũ is a feasible control for problem
(P0 ). Since ū is the solution of (P0 ), uk is a solution of (Pk ), and ū is
feasible for every problem (Pk ), we have J(uk ) ≤ J(ū) and further
J(ū) ≤ J(ũ) ≤ lim inf J(uk ) ≤ lim sup J(uk ) ≤ J(ū).
k→∞
k→∞
2.2. REGULARITY OF THE OPTIMAL CONTROLS
25
Since ū is the unique solution of (P0 ), this implies ū = ũ and J(uk ) →
J(ū), hence the strong convergence uk → ū in L2 (Ω) follows from the
fact J(uk ) → J(ū) along with the uniform convergence yk → ȳ.
Step 2 - The linearized Slater condition for (Pk ) holds at uk . Assumption (2.19) ensures the existence of a number ρ > 0 such that the
following inequalities hold
∂g
(x, ȳ(x))z0 (x) < b(x) − ρ ∀x ∈ K.
∂y
(2.36) a(x) + ρ < g(x, ȳ(x)) +
Given 0 < ε < 1, we multiply the previous inequality by ε and the
inequality a(x) ≤ g(x, ȳ(x)) ≤ b(x) by 1 − ε. Next, we add both
inequalities to get
(2.37) a(x)+ερ < g(x, ȳ(x))+ε
We fix
∂g
(x, ȳ(x))z0 (x) < b(x)−ερ ∀ x ∈ K.
∂y
½
0 < ε < min 1,
εū
ku0 − ūkL2 (Ω)
¾
and define u0,ε = ε(u0 −ū)+ū. It is obvious that u0,ε satisfies the control
constraints of problem (Pk ) for any k. Consider now the solutions zk
of the boundary value problem

 Az + ∂a0 (x, yk )z = u0,ε − uk in Ω
∂y

z =
0
on Γ.
In view of uk → ū in L2 (Ω) and yk → ȳ in C(Ω̄), we obtain the
convergence zk → εz0 in H01 (Ω) ∩ C(Ω̄). Finally, from (2.37) we deduce
the existence of k0 > 0 such that
ρ
∂g
ρ
(2.38) a(xj ) + ε ≤ g(xj , yk (xj )) + (xj , yk (xj ))zk (xj ) ≤ b(xj ) − ε
2
∂y
2
for 1 ≤ j ≤ k and every k ≥ k0 .
1
Step 3 - {uk }∞
k=1 is bounded in H (Ω). The strong convergence
uk → ū in L2 (Ω) implies that kuk − ūkL2 (Ω) < εū for k large enough.
Consequently, the additional constraint kuk − ūkL2 (Ω) ≤ εū is not active,
and hence uk is a local minimum of the problem

min J(u)



 subject to (yu , u) ∈ (C(Ω̄) ∩ H 1 (Ω)) × L∞ (Ω),
(Qk )
α(x) ≤ u(x) ≤ β(x) for a.e. x ∈ Ω,




a(xj ) ≤ g(xj , yu (xj )) ≤ b(xj ) 1 ≤ j ≤ k.
26
2. FIRST ORDER OPTIMALITY CONDITIONS
From Theorem 2.8-(2.28) we get
µ
(2.39)
uk (x) = Proj[α(x),β(x)]
¶
1
− ϕk (x) ,
N
with
(2.40)
ϕk = ϕk,0 +
k
X
λk,j
j=1
∂g
(xj , yk (xj ))ϕk,j .
∂y
Above, {λk,j }kj=1 are the Lagrange multipliers, more precisely
(2.41)
µk =
k
X
λk,j δxj ,
j=1
with
½
λk,j =
≥ 0 if g(xj , yk (xj )) = b(xj )
≤ 0 if g(xj , yk (xj )) = a(xj )
see Remark 2.7. Finally, ϕk,0 and {ϕk,j }kj=1 are given by
(2.42)

 A∗ ϕk,0 + ∂a0 (x, yk (x))ϕk,0 = ∂L (x, yk ) in Ω
∂y
∂y

ϕk,0 =
0
on Γ
(2.43)

 A∗ ϕk,j + ∂a0 (x, yk (x))ϕk,j = δx in Ω
j
∂y

ϕk,j = 0 on Γ.
Let us prove the following boundedness property:
(2.44)
∃C > 0 such that kµk kM (K) =
k
X
|λk,j | ≤ C ∀k.
j=1
From (2.38) we get

ρ
∂g


 λk,j > 0 ⇒ g(xj , yk (xj )) = b(xj ) ⇒ ∂y (xj , yk (xj ))zk (xj ) ≤ −ε 2

ρ
∂g

 λk,j < 0 ⇒ g(xj , yk (xj )) = a(xj ) ⇒
(xj , yk (xj ))zk (xj ) ≥ +ε .
∂y
2
2.2. REGULARITY OF THE OPTIMAL CONTROLS
27
Next, in view of (2.18) with u = u0,ε = ε(u0 − ū) + ū we obtain
Z
0 ≤ (ϕ̄k + N uk )(u0,ε − uk )
Ω
0
= J (uk )(u0,ε − uk ) +
k
X
λk,j
j=1
∂g
(xj , yk (xj ))zk (xj )
∂y
k
≤ J 0 (uk )(u0,ε − uk ) − ε
ρX
|λk,j |.
2 j=1
This implies that
k
X
|λk,j | ≤
j=1
2 0
2
J0 (uk )(u0,ε − uk ) → J 0 (ū)(u0 − ū) when k → ∞,
ερ
ρ
hence (2.44) holds. Now (2.40), (2.42) and (2.43) lead to

A∗ ϕk + ∂a0 (x, yk )ϕk = ∂L (x, yk ) + ∂g (x, yk )µk
∂y
∂y
∂y
(2.45)

ϕk =
0
in Ω
on Γ.
Let us set
µ
with Cαβ
that
¶
1
vk (x) = Proj[−Cα,β ,+Cα,β ] − ϕk (x) ,
N
defined by (2.31). From the last relation and (2.39) it follows
uk (x) = Proj[α(x),β(x)] (vk (x)).
Notice that the trace of uk on Γ is not necessarily zero, therefore it is a
delicate question to multiply the equation (2.45) by uk and to integrate
by parts. However, vk vanishes on Γ and hence the previous operation
can be done without difficulty and we will do it later.
1
The goal is to prove that {vk }∞
k=1 is bounded in H (Ω), which yields
the boundedness of {uk }∞
k=1 in the same space. The last claim is an
immediate consequence of
|∇uk (x)| ≤ |∇vk (x)| + |∇α(x)| + |∇β(x)| a.e. Ω.
1
1
If {uk }∞
k=1 is bounded in H (Ω), then ū ∈ H (Ω) obviously.
1
Let us prove the boundedness of {vk }∞
k=1 in H0 (Ω). The solution of
a Dirichlet problem associated with an elliptic operator with coefficients
aij ∈ C(Ω̄) and Lipschitz boundary Γ belongs to W01,r (Ω), if the right
hand side is in W −1,r (Ω) for any n < r < n + εn , where εn > 0 depends
on n and n ∈ {2, 3}; cf. Jerison and Kenig [14] and Mateos [18]. If
28
2. FIRST ORDER OPTIMALITY CONDITIONS
p > n/2, then Lp (Ω) ⊂ W −1,2p (Ω) and consequently ϕk,0 ∈ W01,r (Ω)
holds for all r in the range indicated above with r < 2p.
In view of this, we have ϕk ∈ W01,s (Ω)∩W 1,r (Ω\Sk ), where Sk is the
set of points xj such that λk,j (∂g/∂y)(xj , yk (xj )) 6= 0. Notice that by
(2.40) only these ϕk,j appear in the representation of ϕk . Taking into
account that vk is constant in a neighborhood of every point xj ∈ Sk ,
we deduce that vk ∈ W01,r (Ω) ⊂ C(Ω̄). Therefore, we are justified to
multiply equation (2.45) by −vk and to integrate by parts. We get
!
Z ÃX
n
∂a0
−
aij (x)∂xi vk ∂xj ϕk +
(x, yk )vk ϕk dx
∂y
Ω i,j=1
(2.46)
Z
k
X
∂L
∂g
=−
(x, yk )vk dx −
λk,j (xj , yk (xj ))vk (xj ).
∂y
Ω ∂y
j=1
From the definition of vk we obtain for a.a. x ∈ Ω

 − 1 ∇ϕ (x) if −C ≤ − 1 ϕ (x) ≤ +C
k
α,β
k
α,β
N
N
(2.47) ∇vk (x) =

0
otherwise.
Invoking this property in (2.46) along with the boundedness of {yk }∞
k=1
in C(Ω̄), the estimate kvk kL∞ (Ω) ≤ Cα,β , and the assumptions (A3) and
(A4), we get
Z
2
λA N
|∇vk | dx ≤ kψM kL2 (Ω) kvk kL2 (Ω) + C
Ω
k
X
|λk,j | kvk kL∞ (Ω) ≤ C 0 .
j=1
1
Clearly, this implies that {vk }∞
k=1 is bounded in H (Ω) as required.
2.3. On the uniqueness of the Lagrange multiplier µ̄
In this section, we provide a sufficient condition for the uniqueness
of the Lagrange multiplier associated with the state constraints. We
also analyze some situations where these conditions are satisfied. It is
known that a non-uniqueness of Lagrange multipliers may lower the
efficiency of numerical methods, e.g. primal-dual active set methods.
Moreover, some other theoretical properties of optimization problems
depend on the uniqueness of multipliers. Therefore, this is a desirable
property.
In this section we will assume that K = Uαβ .
2.3. ON THE UNIQUENESS OF THE LAGRANGE MULTIPLIER µ̄
29
Theorem 2.11. Assume (A1)–(A4), (2.19) and the existence of
some ε > 0 such that the following property holds
(2.48) T : L2 (Ωε ) −→ C(K0 ), T v =
∂g
(x, ȳ(x))zv , R(T ) = C(K0 )
∂y
where R(T ) denotes the range of T and
Ωε = {x ∈ Ω : α(x) + ε < ū(x) < β(x) − ε},
zv ∈ H01 (Ω) ∩ C θ (Ω̄) satisfies

 Azv + ∂a0 (x, ȳ)zv = v in Ω
∂y
(2.49)

zv = 0 on Γ,
and v is extended by zero to the whole domain Ω. Then there exists
a unique Lagrange multiplier µ̄ ∈ M (K) such that (2.15)-(2.18) hold
with ᾱ = 1.
Proof. Let us assume to the contrary that µ̄i , i = 1, 2, are two
Lagrange multipliers associated to the state constraints corresponding
to the optimal control ū. Let us denote by ϕ̄i the corresponding adjoint
states. Then (2.18) leads to
Z
0 ≤ (ϕ̄i + N ū)(u − ū) dx = J 0 (ū)(u − ū)+
Ω
Z
∂g
(x, ȳ(x))zu−ū (x) dµ̄i (x), i = 1, 2, ∀u ∈ Uαβ .
K ∂y
Taking v ∈ L∞ (Ωε ) \ {0} arbitrarily, we have for a.e. x ∈ Ω
ε
α(x) ≤ uρ (x) = ū(x) + ρv(x) ≤ β(x) ∀|ρ| <
,
kvkL∞ (Ωε )
where v is extended by zero to the whole domain Ω. Inserting u = uρ
in the above inequality, with positive and negative ρ and remembering
that suppµ̄i ⊂ K0 , we deduce
Z
∂g
0
(x, ȳ(x))zv (x) dµ̄i (x) = 0, i = 1, 2,
J (ū)v +
K0 ∂y
which leads to
hµ̄1 , T vi = −J 0 (ū)v = hµ̄2 , T vi ∀v ∈ L∞ (Ωε ).
Since L∞ (Ωε ) is dense in L2 (Ωε ) and T (L2 (Ωε )) is dense in C(K0 ) we
obtain from the above identity that µ̄1 = µ̄2 .
30
2. FIRST ORDER OPTIMALITY CONDITIONS
Remark 2.12. For a finite set K = {xj }nj=1 , assumption (2.48) is
equivalent to the independence of the gradients {Fj0 (ū)}j∈I0 in L2 (Ωε ),
where Fj : L2 (Ωε ) −→ R is defined by Fj (u) = g(xj , yu (xj )) and I0 is
the set of indexes j corresponding to active constraints. It is a regularity assumption on the control problem at ū. This type of assumption
was introduced by the authors in [8] to analyze control constrained
problems with finitely many state constraints. The first author proved
in [5] that, under very general hypotheses, this assumption is equivalent to the Slater condition in the case of a finite number of pointwise
state constraints.
We show finally that (2.48) holds under some more explicit assumptions on ū and on the set of points K0 , where the state constraint is
active.
Theorem 2.13. Assume that (A1)-(A4) and (2.29) hold and that
the coefficients aij belong to C 0,1 (Ω̄) (1 ≤ i, j ≤ n). We also suppose
the following properties:
(1) The Lebesgue measure of K0 is zero.
(2) There exists ε > 0 such that, for every open connected component A of Ω \ K0 , the set A ∩ Ωε has a nonempty interior.
(3) (∂g/∂y)(x, ȳ(x)) 6= 0 for every x ∈ K0 .
Then the regularity assumption (2.48) is satisfied.
Remark 2.14. If α, β ∈ C(Ω̄), then ū ∈ C(Ω̄ \ K0 ); cf. Theorem
2.8. Hence property 2 of the theorem is fulfilled, if ū is not identically
equal to α or β in any open connected component A ⊂ Ω \ K0 . Indeed,
since ū ∈ C(A) and ū 6≡ α and ū 6≡ β in A, there exists x0 ∈ A such
that α(x0 ) < ū(x0 ) < β(x0 ). Consequently, the continuity of ū implies
the existence of ε > 0 such that A ∩ Ωε contains a ball Bρ (x0 ).
Let us also mention that property 3 of the theorem is trivially satisfied if the state constraint is a(x) ≤ y(x) ≤ b(x) for every x ∈ K.
Proof of Theorem 2.13. Fix ε > 0 as in property (2). We will argue
by contradiction. If R(T ) 6= C(K0 ), then there exists µ ∈ C(K0 )0 =
M (K0 ), µ 6= 0, such that
Z
(2.50)
0 = hµ, T vi =
K0
∂g
(x, ȳ(x))zv (x) dµ(x) ∀v ∈ L2 (Ωε ).
∂y
2.3. ON THE UNIQUENESS OF THE LAGRANGE MULTIPLIER µ̄
31
We take the function ψ ∈ W01,s (Ω) for all 1 ≤ s < n/(n − 1) satisfying

 A∗ ψ + ∂a0 (x, ȳ(x))ψ = ∂g (x, ȳ(x))µ in Ω
∂y
∂y
(2.51)

ψ = 0
on Γ.
From (2.50) and (2.51), it follows for every v ∈ L2 (Ω) vanishing outside
Ωε
¸
Z
Z
Z ·
∂a0
ψv dx =
ψv dx =
Azv +
(x, ȳ(x))zv ψ dx
∂y
ΩZ
Ω
Ω
ε
∂g
(x, ȳ(x))zv dµ = hµ, T vi = 0,
=
K0 ∂y
which implies that ψ = 0 in Ωε . Consider now an open connected
component A of Ω \ K0 . It holds ψ = 0 in the interior of A ∩ Ωε and
A∗ ψ +
∂a0
(x, ȳ(x))ψ = 0 in A,
∂y
therefore ψ = 0 in A; see Saut and Scheurer [23]. Thus we have that
ψ = 0 in Ω \ K0 , but K0 has zero Lebesgue measure, hence ψ = 0 in Ω
and consequently (2.51) along with the property (3) imply that µ = 0
in contrary to our previous assumption.
We conclude our paper by proving that the regularity condition
(2.48) is stronger that the linearized Slater assumption (2.19).
Theorem 2.15. Under the assumptions (A1)-(A4) the regularity
condition (2.48) implies the linearized Slater condition (2.19).
Proof. We define the set
B = {z ∈ C(K0 ) : a(x) < z(x) + g(x, ȳ(x)) < b(x) ∀x ∈ K0 }.
From our assumptions it follows that B is a non empty open set of
C(K0 ), hence condition (2.48) implies the existence of v ∈ L2 (Ωε ) such
that T v ∈ B. By density of L∞ (Ωε ) in L2 (Ωε ) we can assume that
v ∈ L∞ (Ωε ). Outside Ωε , we extend v by zero. The inclusion T v ∈ B
can be expressed by
(2.52)
a(x) < g(x, ȳ(x)) +
∂g
(x, ȳ(x))zv (x) < b(x) ∀x ∈ K0 ,
∂y
where zv ∈ H01 (Ω) ∩ C(Ω̄) is the solution of (2.49). From here, we
deduce the existence of ρ1 > 0 such that
(2.53) a(x) + ρ1 < g(x, ȳ(x)) +
∂g
(x, ȳ(x))zv (x) < b(x) − ρ1 ∀x ∈ K0 .
∂y
32
2. FIRST ORDER OPTIMALITY CONDITIONS
Given ρ ∈ (0, 1) arbitrary, multiplying the inequalities
(2.54)
a(x) ≤ g(x, ȳ(x)) ≤ b(x) ∀x ∈ K
by 1 − ρ, (2.53) by ρ, and adding the resulting inequalities we get for
every x ∈ K0 and all ρ ∈ (0, 1)
∂g
(2.55) a(x) + ρρ1 < g(x, ȳ(x)) + ρ (x, ȳ(x))zv (x) < b(x) − ρρ1 .
∂y
Define now, for any δ > 0,
K0,δ = {x ∈ K : dist(x, K0 ) < δ}.
Taking δ small enough, we get from (2.55) for all x ∈ K0,δ and every
ρ ∈ (0, 1)
ρ1
∂g
ρ1
(2.56) a(x) + ρ < g(x, ȳ(x)) + ρ (x, ȳ(x))zv (x) < b(x) − ρ .
2
∂y
2
On the other hand, since the state constraint is not active in the compact set K \ K0,δ , we deduce the existence of 0 < ρ2 < 1 such that
(2.57)
a(x) + ρ2 < g(x, ȳ(x)) < b(x) − ρ2 ∀x ∈ K \ K0,δ .
If we select a ρ ∈ (0, 1) that satisfies
¯
¯
¯ ∂g
¯ ρ2
ρ ¯¯ (x, ȳ(x))zv (x)¯¯ <
∀x ∈ K,
∂y
2
we obtain from (2.57)
ρ2
∂g
ρ2
(2.58)
a(x) +
< g(x, ȳ(x)) + ρ (x, ȳ(x))zv (x) < b(x) −
2
∂y
2
for every x ∈ K \ K0,δ . Finally, taking 0 < ρ < ε/kvkL∞ (Ωε ) , recalling
the definition of Ωε and that v vanishes in Ω \ Ωε , we deduce that
u0 = ū + ρv ∈ K. Moreover, (2.56) and (2.58) imply that (2.19) holds,
which concludes the proof.
CHAPTER 3
Second Order Optimality Conditions
In this chapter the goal is to derive the sufficient conditions for
local optimality. We will follow the ideas developed in [6]. In the
whole chapter we will assume that N > 0 and K = Uαβ .
Let us introduce some notation. The Lagrange function associated
with problem (P) is defined by
L : L2 (Ω) × M (K) −→ R
Z
L(u, µ) = J(u) +
g(x, yu (x)) dµ(x).
K
Using Proposition 2.2, (2.10) and (2.13) we find that
Z
∂L
(3.1)
(u, µ)v =
(ϕu (x) + N u(x)) v(x) dx,
∂u
Ω
where ϕu ∈ W01,s (Ω), for all 1 ≤
Dirichlet problem

 A∗ ϕ + ∂a0 (x, yu )ϕ =
∂y
(3.2)

ϕ =
s < n/(n − 1), is the solution of the
∂L
∂g
(x, yu ) +
(x, yu (x))µ in Ω
∂y
∂y
0
on Γ.
From the expression (3.1) we get that the inequality (2.18) can be
written as follows
∂L
(3.3)
(ū, µ̄)(u − ū) ≥ 0 ∀u ∈ Uα,β
∂u
assuming that ᾱ = 1.
Before we set up the sufficient second order optimality conditions,
we evaluate the expression of the second derivative of the Lagrangian
with respect to the control. From (2.14) we get
∂ 2L
(u, µ)v1 v2 = J 00 (u)v1 v2
∂u2
¸
Z · 2
∂ g
∂g
+
(x, yu (x))zv1 (x)zv2 (x) +
(x, yu (x))zv1 v2 (x) dµ(x).
2
∂y
K ∂y
33
34
3. SECOND ORDER OPTIMALITY CONDITIONS
By (2.8), (2.9) and (2.11), this is equivalent to
(3.4)
∂2L
(u, µ)v1 v2
∂uZ2 ·
¸
∂ 2L
∂ 2 a0
=
(x, yu )zv1 zv2 + N v1 v2 − ϕu 2 (x, yu )zv1 zv2 dx
2
∂y
Ω ∂y
Z
2
∂ g
+
(x, yu (x))zv1 (x)zv2 (x) dµ(x),
2
K ∂y
where ϕu is the solution of (3.2).
Associated with ū, we define the cone of critical directions by
Cū = {v ∈ L2 (Ω) : v satisfies (3.5), (3.6) and (3.7) below},
(3.5)
(3.6)
(3.7)

 ≥ 0 if ū(x) = α(x),
≤ 0 if ū(x) = β(x),
v(x) =
 = 0 if ϕ̄(x) + N ū(x) 6= 0,
½
∂g
≥ 0 if x ∈ Ka
(x, ȳ(x))zv (x) =
≤ 0 if x ∈ Kb ,
∂y
Z
∂g
(x, ȳ(x))zv (x) dµ̄(x) = 0,
K ∂y
where zv ∈ H01 (Ω) ∩ C θ (Ω̄) satisfies

 Azv + ∂a0 (x, ȳ)zv = v in Ω
∂y

zv = 0 on Γ.
The relation (3.6) expresses the natural sign conditions, which must be
fulfilled for feasible directions at active points x ∈ Ka or Kb , respectively. On the other hand, (3.7) states that the derivative of the state
constraint in the direction v must be zero whenever the corresponding Lagrange multiplier is non-vanishing. This restriction is needed for
second-order sufficient conditions. Compared with the finite dimensional case, this is exactly what we can expect. Therefore the relations
(3.6)-(3.7) provide a convenient extension of the usual conditions of the
finite-dimensional case.
Condition (3.7) was proved for the first time in the context of
infinite-dimensional optimization problems in [6]. In earlier papers on
this subject, other extensions to the infinite-dimensional case were suggested. For instance, Maurer and Zowe [19] used first-order sufficient
conditions to account for the strict positivity of Lagrange multipliers.
Inspired by their approach, in [9] an application to state-constrained
3. SECOND ORDER OPTIMALITY CONDITIONS
35
elliptic boundary control was suggested by the authors. In terms of
our problem, equation (3.7) was relaxed by
Z
Z
∂g
(x, ȳ(x))zv (x) dµ̄(x) ≥ −ε
|v(x)| dx
K ∂y
{x:|ϕ̄(x)+N ū(x)|≤τ }
for some ε > 0 and τ > 0, cf. [9, (5.15)]. In the next theorem, which
was proven in [6, Theorem 4.3], we will see that this relaxation is not
necessary. We obtain a smaller cone of critical directions that seems
to be optimal. However, the reader is referred to Theorem 3.2 below,
where we consider the possibility of relaxing the conditions defining the
cone Cū .
Theorem 3.1. Assume that (A1)-(A4) hold. Let ū be a feasible
control of problem (P), ȳ the associated state and (ϕ̄, µ̄) ∈ W01,s (Ω) ×
M (K), for all 1 ≤ s < n/(n − 1), satisfying (2.16)-(2.18) with ᾱ = 1.
Assume further that
∂ 2L
(ū, µ̄)v 2 > 0 ∀v ∈ Cū \ {0}.
∂u2
Then there exist ε > 0 and δ > 0 such that the following inequality
holds:
δ
(3.9) J(ū) + ku − ūk2L2 (Ω) ≤ J(u) if ku − ūkL2 (Ω) ≤ ε and u ∈ Uad .
2
(3.8)
The condition (3.8) seems to be natural. In fact, under some regularity assumption, we can expect the inequality
∂ 2L
(ū, µ̄)v 2 ≥ 0 ∀v ∈ Cū
∂u2
to be necessary for local optimality. At least, this is the case when
the state constraints are of integral type, see [7] and [8], or when K
is a finite set of points, see [5]. In the general case of (P), to our best
knowledge, the necessary second order optimality conditions are still
open.
Proof of Theorem 3.1. We argue by contradiction. Suppose that ū does
not satisfy the quadratic growth condition (3.9). Then there exists a
2
sequence {uk }∞
k=1 ⊂ L (Ω) of feasible controls of (P) such that uk → ū
in L2 (Ω) and
(3.10)
1
J(ū) + kuk − ūk2L2 (Ω) > J(uk ) ∀k.
k
36
3. SECOND ORDER OPTIMALITY CONDITIONS
Let us take
ρk = kuk − ūkL2 (Ω)
and hk =
1
(uk − ū).
ρk
Since khk kL2 (Ω) = 1, we can extract a subsequence, denoted in the
same way, such that hk * h weakly in L2 (Ω). Now we split the proof
in several steps.
(ū, µ̄)h = 0. In the following, we write yk = yuk . Since uk
Step 1: ∂L
∂u
is feasible, it holds that a(x) ≤ g(x, yk (x)) ≤ b(x) for every x ∈ K. By
using (2.17) and (3.10) we obtain
1
L(ū, µ̄) + kuk − ūk2L2 (Ω)
k
Z
(3.11) > J(uk ) +
g(x, yk (x)) dµ̄(x) = L(uk , µ̄).
K
From the mean value theorem we know that
∂L
L(uk , µ̄) = L(ū, µ̄) + ρk
(vk , µ̄)hk ,
∂u
with vk a point between ū and uk . This identity and (3.11) imply
∂L
1
1
(vk , µ̄)hk <
kuk − ūk2L2 (Ω) = kuk − ūkL2 (Ω) .
∂u
kρk
k
Since hk * h weakly in L2 (Ω), vk → ū in L2 (Ω), yvk → ȳ in C(Ω̄) ∩
H01 (Ω) and ϕvk → ϕ̄ in W01,s (Ω) ⊂ L2 (Ω) for s close to n/(n − 1), we
deduce from the above inequality and the expression of the derivative
of the Lagrangian given by (3.1) that
∂L
∂L
(ū, µ̄)h = lim
(vk , µ̄)hk ≤ 0.
k→∞ ∂u
∂u
On the other hand, since α(x) ≤ uk (x) ≤ β(x) holds for almost all
x ∈ Ω, we deduce from the variational inequality (3.3)
(3.12)
∂L
1 ∂L
(ū, µ̄)hk =
(ū, µ̄)(uk − ū) ≥ 0,
∂u
ρk ∂u
which implies
∂L
∂L
(ū, µ̄)h = lim
(vk , µ̄)hk ≥ 0.
k→∞ ∂u
∂u
This inequality, along with (3.12), leads to
(3.13)
∂L
(ū, µ̄)h = 0.
∂u
3. SECOND ORDER OPTIMALITY CONDITIONS
37
Step 2: h ∈ Cū . We have to confirm (3.5)–(3.7). The set of functions
of L2 (Ω) that are nonnegative if ū(x) = α(x) and nonpositive if ū(x) =
β(x), almost everywhere, is convex and closed. Therefore, it is weakly
closed. Moreover uk − ū obviously belongs to this set, thus every hk
also does. Consequently, h belongs to the same set. On the other hand
from (2.28), we deduce
(
ū(x) = α(x) if ϕ̄(x) + N ū(x) > 0,
ū(x) = β(x) if ϕ̄(x) + N ū(x) < 0.
This property along with (3.13) and the sign of h(x) when ū(x) = α(x)
or β(x) imply
Z
|ϕ̄(x) + N ū(x)||h(x)| dx
ΩZ
∂L
= (ϕ̄(x) + N ū(x))h(x) dx =
(ū, µ̄)h = 0,
∂u
Ω
hence h(x) = 0 if ϕ̄(x)+N ū(x) 6= 0, which concludes the proof of (3.5).
Let us prove (3.6). From Proposition 2.2 we have
(yū+ρk hk − ȳ)
in C(Ω̄) ∩ H01 (Ω),
k→∞
ρk
zh = G0 (ū)h = lim
which implies for every x ∈ Ka
(3.14)
[g(x, yū+ρk hk (x)) − g(x, ȳ(x))]
∂g
(x, ȳ(x))zh (x) = lim
≥ 0.
k→∞
∂y
ρk
The last inequality follows from the fact that uk is a feasible control
of (P), ū + ρk hk = uk , and consequently a(x) ≤ g(x, yū+ρk hk (x)) =
g(x, yuk (x)) ≤ b(x) for every x ∈ K. Analogously we prove that
∂g
(x, ȳ(x))zh (x) ≤ 0 ∀x ∈ Kb .
∂y
Finally, we prove (3.7). Taking z = g(·, yuk (·)) in (2.17), we get
Z
∂g
(x, ȳ(x))zh (x) dµ̄(x)
K ∂y
Z
1
= lim
[g(x, yū+ρk hk (x) − g(x, ȳ(x))] dµ̄(x)
k→∞ ρk K
Z
1
= lim
(3.15)
[g(x, yuk (x)) − g(x, ȳ(x))] dµ̄(x) ≤ 0.
k→∞ ρk K
38
3. SECOND ORDER OPTIMALITY CONDITIONS
On the other hand, from (3.10) we find
J(ū + ρk hk ) − J(ū)
k→∞
ρk
J(uk ) − J(ū)
ρk
= lim
≤ lim
= 0.
k→∞
k→∞ k
ρk
J 0 (ū)h = lim
(3.16)
Then (3.13), (3.15), (3.16) and the fact that
Z
∂L
∂g
0
(ū, µ̄)h = J (ū)h +
(x, ȳ(x))zh (x) dµ̄(x),
∂u
K ∂y
imply that
Z
0
J (ū)h =
K
∂g
(x, ȳ(x))zh (x) dµ̄(x) = 0.
∂y
Thus (3.7) holds and we know h ∈ Cū .
Step 3: h = 0. Taking into account (3.8), it is enough to prove that
∂ 2L
(ū, µ̄)h ≤ 0.
∂u2
(3.17)
For this purpose, we evaluate the Lagrangian. By a second-order Taylor
expansion, we derive
(3.18)
L(uk , µ̄) = L(ū, µ̄) + ρk
∂L
ρ2 ∂ 2 L
(ū, µ̄)hk + k 2 (wk , µ̄)h2k ,
∂u
2 ∂u
wk being an intermediate point between ū and uk . From here we get
∂L
ρ2 ∂ 2 L
(ū, µ̄)hk + k 2 (ū, µ̄)h2k
∂u
2 ∂u
·
¸
ρ2k ∂ 2 L
∂ 2L
= L(uk , µ̄) − L(ū, µ̄) +
(ū, µ̄) −
(wk , µ̄) h2k .
2 ∂u2
∂u2
ρk
(3.19)
Now (3.11) can be written
(3.20)
ρ2k
L(uk , µ̄) − L(ū, µ̄) ≤ .
k
On the other hand, taking into account the expression (3.4) of the second derivative of the Lagrangian, the assumptions (A1)-(A3), Theorem
1.4, Proposition 2.2, the fact that uk → ū in L2 (Ω) and khk kL2 (Ω) = 1,
3. SECOND ORDER OPTIMALITY CONDITIONS
we obtain
(3.21)
39
¯· 2
¸ ¯
¯ ∂ L
¯
∂ 2L
2¯
¯
¯ ∂u2 (ū, µ̄) − ∂u2 (wk , µ̄) hk ¯
° 2
°
2
°∂ L
°
∂
L
°
≤°
(ū,
µ̄)
−
(w
,
µ̄)
khk k2L2 (Ω)
k
° ∂u2
°
2
∂u
B(L2 (Ω))
° 2
°
2
°∂ L
°
∂ L
°
→ 0 when k → ∞,
=°
(ū,
µ̄)
−
(w
,
µ̄)
k
° ∂u2
° 2
∂u2
B(L (Ω))
where B(L2 (Ω)) is the space of quadratic forms in L2 (Ω).
Using once again (3.3) along with the identity hk = (uk − ū)/ρk and
(3.19)–(3.21) we deduce
° 2
°
2
°
°
∂ 2L
∂
L
2
∂
L
2
°
°
→ 0.
(ū,
µ̄)h
≤
+
(ū,
µ̄)
−
(w
,
µ̄)
k
k
° 2
∂u2
k ° ∂u2
∂u2
B(L (Ω))
Taking into account the expression of the second derivative of the Lagrangian given by (3.4), we deduce from the above inequality that
∂ 2L
∂2L
2
(ū,
µ̄)h
≤
lim
inf
(ū, µ̄)h2k ≤ 0,
k→∞ ∂u2
∂u2
which along with (3.8) and the fact that h ∈ Cū implies that h = 0.
Step 4: Final Contradiction. We have already proved that hk * 0
weakly in L2 (Ω), therefore zhk → 0 strongly in C(Ω̄) ∩ H01 (Ω). Since
khk kL2 (Ω) = 1, we obtain from (3.4)
∂ 2L
(ū, µ̄)h2k ≤ 0.
k→∞ ∂u2
Thus we have got the contradiction.
We finish this chapter by establishing an equivalent condition to
(3.8) that is more convenient for the numerical analysis of problem
(P). Let us introduce a cone Cūτ of critical directions that is bigger
than Cū . Given τ > 0, we denote by Cūτ the set of elements v ∈ L2 (Ω)
satisfying
N ≤ lim inf
(3.22)
(3.23)
(3.24)

 ≥ 0 if ū(x) = α(x),
≤ 0 if ū(x) = β(x),
v(x) =

= 0 if |ϕ̄(x) + N ū(x)| > τ,
½
∂g
≥ −τ kvkL2 (Ω) if x ∈ Ka
(x, ȳ(x))zv (x) =
≤ +τ kvkL2 (Ω) if x ∈ Kb ,
∂y
Z
∂g
(x, ȳ(x))zv (x) dµ̄(x) ≥ −τ kvkL2 (Ω) .
K ∂y
40
3. SECOND ORDER OPTIMALITY CONDITIONS
Theorem 3.2. Under the assumptions (A1)-(A4), relation (3.8)
holds if and only if there exist τ > 0 and λ > 0 such that
(3.25)
∂ 2L
(ū, µ̄)v 2 ≥ λkvk2L2 (Ω) ∀v ∈ Cūτ .
∂u2
Proof. Since Cū ⊂ Cūτ , it is clear that (3.25) implies (3.8). Let
us prove by contradiction that (3.25) follows from (3.8). Assume that
(3.8) holds but not (3.25). Then, for any positive integer k, there exists
1/k
an element vk ∈ Cū such that
(3.26)
∂L
1
(ū, µ̄)vk2 < kvk k2L2 (Ω) .
∂u
k
Redefining vk by vk /kvk kL2 (Ω) and selecting a subsequence, if necessary,
denoted in the same way, we can assume that
(3.27) kvk kL2 (Ω) = 1, vk * v weakly in L2 (Ω) and
∂L
1
(ū, µ̄)vk2 < ,
∂u
k
and from (3.22)-(3.24)

 ≥ 0 if ū(x) = α(x),
≤ 0 if ū(x) = β(x),
(3.28)
vk (x) =
 = 0 if |ϕ̄(x) + N ū(x)| > 1/k,
½
∂g
≥ −1/k if x ∈ Ka
(3.29)
(x, ȳ(x))zvk (x) =
≤ +1/k if x ∈ Kb ,
∂y
Z
∂g
(x, ȳ(x))zvk (x) dµ̄(x) ≥ −1/k.
(3.30)
K ∂y
Since zvk → zv strongly in H01 (Ω) ∩ C(Ω̄), we can pass to the limit
in (3.28)-(3.30) and get that v ∈ Cū and
(3.31)
∂L
(ū, µ̄)v 2 ≤ 0.
∂u
This is only possible if v = 0; see (3.8). Let us note that the only
delicate point to prove that v ∈ Cū is to establish (3.7). Indeed, (3.5)
and (3.6) follow easily from (3.28) and (3.29). Passing to the limit in
(3.30) we get
Z
∂g
(x, ȳ(x))zv (x) dµ̄(x) ≥ 0.
K ∂y
This inequality, along with (3.6) and the structure of µ̄ established in
Corollary 2.6, implies (3.7).
3. SECOND ORDER OPTIMALITY CONDITIONS
41
Therefore, we have that vk * 0 weakly in L2 (Ω) and zvk → 0
strongly in H01 (Ω) ∩ C(Ω̄). Hence, using the expression (3.4) of the
second derivative of the Lagrangian we get
∂L
(ū, µ̄)vk2 ≤ 0,
N = lim inf N kvk k2L2 (Ω) ≤ lim inf
k→∞
k→∞ ∂u
which is a contradiction.
In [6] the second order optimality conditions are proved for a more
general cost functional
Z
J(u) =
L(x, yu (x), u(x)) dx.
Ω
The sufficient conditions are formulated as follows
¯
¯
¯
¯
∂L
∂2L
¯
(3.32)
(x, ȳ(x), ū(x)) ≥ ω if ¯ϕ̄(x) +
(x, ȳ(x), ū(x))¯¯ ≤ τ,
2
∂u
∂u
2
∂ L
(3.33)
(ū, µ̄)h2 > 0 ∀h ∈ Cū \ {0},
∂u2
for some ω > 0 and τ > 0. The proof follows the same ideas but it
is technically more complicated. However the consequence of these
assumptions are weaker than those given in Theorem 3.1. Indeed,
(3.32) and (3.33) are sufficient for ū to be a strict local minimum in
the sense of L∞ (Ω). More precisely, there exist ε > 0 and δ > 0 such
that
δ
J(ū) + ku − ūk2L2 (Ω) ≤ J(u) if ku − ūkL∞ (Ω) < ε and u ∈ Uad .
2
The difference with respect to Theorem 3.1 is motivated by the fact
that the general functional J defined above is not of class C 2 in L2 (Ω),
but it is C 2 in L∞ (Ω). Here we find the so-called two-norm discrepancy.
The sufficient conditions (3.25) is satisfied with respect to one norm and
the differentiability holds with respect to a different norm. Typically
these two norms are L2 (Ω) and L∞ (Ω). Only in the cases where L is
quadratic with respect to u and u appears linearly in the state equation,
then we can get the result provided in Theorem 3.1, which is the most
useful for the numerical analysis.
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