PONTRYAGIN'S PRINCIPLE FOR STATE-CONSTRAINED BOUNDARY CONTROL PROBLEMS OF SEMILINEAR PARABOLIC EQUATIONS

This paper deals with state-constrained optimal control problems governed by semilinear parabolic equations. We establish a minimum principle of Pontryagin's type. To deal with the state constraints, we introduce a penalty problem by using Ekeland's principle. The key tool for the proof is the use of a special kind of spike perturbations distributed in the domain where the controls are defined. Conditions for normality of optimality conditions are given.

to assume a stability condition of the optimal cost functional with respect to small perturbation of the feasible state set.
In this paper, we consider a boundary control problem governed by a parabolic semilinear equation. General state constraints are included in the formulation of the problem. The idea developed in [12] is used here. To deal with the state constraints we penalize them. The lack of convexity of the control set and the noncontinuity with respect to the control of the functions involved in the control problem make it difficult to formulate a penalty problem having a solution converging to the optimal control of the original problem, however. Ekeland's variational principle is the key to obtaining the suitable penalization.
Pontryagin's principle is often established in a nonqualified form, which implies that the cost functional does not appear in the conditions for optimality. In the absence of equality state constraints, we give a condition that leads to a qualified optimality system. This condition was introduced by Bonnans [4] and Bonnans and Casas [6]. It consists of assuming a certain kind of Lipschitz dependence of the optimal cost functional with respect to small perturbations of the state constraint. It is proved that this condition is satisfied "almost everywhere (a.e.)." We will distinguish strong and weak Pontryagin principles, depending on whether the optimality system is qualified or not. To prove the strong principle we make an exact penalization of the state contraints.
One of the difficulties found in the optimality system is the adjoint state equation. This equation can have measures as data in the domain, on the boundary, and as a final condition. There are not many papers written about parabolic equations involving measures. For these equations the reader is referred to Barbu and Precupanu [1], Lasiecka [24], Tröltzsch [32], and Boccardo and Gallouët [3], the last one dealing with quasi-linear equations. Here we use the transposition method to derive a general result of existence and "uniqueness" of solution. Since we do not assume continuity of the coefficients of the state equation, we need to be precise in which sense the solution is unique; see Serrin [30] for a nonuniqueness result in W 1,p 0 (Ω) (p < 2) of an elliptic problem well posed in H 1 (Ω).
The paper is organized as follows. In the next section, the control problem is formulated. The state constraints are presented in an abstract framework. We show through some examples how the usual state constraints are included in the abstract formulation. The weak and strong Pontryagin principles are formulated in sections 3 and 4, respectively. In section 5, the state equation is studied and the spike perturbations are defined. The linear parabolic equations involving measures are analyzed in section 6. All the mentioned papers dealing with control of evolution equations, except [22], followed the semigroup approach to analyze the state and adjoint state equations. Here we will follow the variational approach, which allows us to obtain some pointwise information of the solutions of the PDEs. This information is very important for studying the control problems with pointwise state constraints. Finally, the proofs of weak and strong principles are given in section 7.
As usual, we assume the following hypotheses on A: Once given the state equation, we introduce the cost functional where y u is the solution of (2.3) associated with u; σ denotes the usual (n − 1)dimensional measure on Γ induced by the parametrization (remember that Γ is a Lipschitz manifold); and L : Ω T × R −→ R and l : Σ T × R × K −→ R are of class C 1 with respect to the second variable, L being measurable with respect to the first one, (2.10) The space of controls U is formed by the measurable functions u : Σ T −→ K such that the mapping is measurable for every y ∈ R. In section 5 we will prove that there exists a unique solution of (2.3) Finally we introduce the state constraints. Let Z be a separable Banach space and Q ⊂ Z a closed convex subset with nonempty interior. Given two mappings of class C 1 , G : Y −→ Z and F : C(Ω T ) −→ R s , s ≥ 1, we formulate the optimal control problem as follows: Let us show how the usual examples of state constraints can be handled with this formulation.
Example 2.1. Given a continuous function g :Ω T × R −→ R of class C 1 in respect to the second variable, the constraint g(x, t, y u (x, t)) ≤ δ for all (x, t) ∈Ω T , with δ > 0 being a given number, can be written in the above framework by putting Z = C(Ω T ), G : Y −→ C(Ω T ), defined by G(y) = g(·, y(·)), and then we can include the equality constraints y u (x j , t j ) = δ j , 1 ≤ j ≤ s, in the above formulation. Indeed, it is enough to define the functions F j : C(Ω T ) −→ R given by F j (y) = y(x j )−δ j and to take F = (F 1 , . . . , F s ) T . Then F is of class C 1 .
Example 2.3. Let g : Ω × [0, T ] × R −→ R be a function measurable with respect to the first variable, continuous with respect to the second, of class C 1 with respect to the third, and such that ∂g/∂y is also continuous in the last two variables. Moreover, it is assumed that for every M > 0 there exists a function ψ M ∈ L 1 (Ω) such that is included in the above formulation by taking Z = C[0, T ], , with G(y) = y, and Q the closed ball in L 1 (Ω) of center at 0 and radius δ.
Example 2.5. For every 1 ≤ j ≤ k let g j : Ω T × R −→ R be a measurable function of class C 1 with respect to the second variable such that for each M > 0 there exists a function η j M ∈ L 1 (Ω T ) satisfying Then the constraints are included in the formulation of (P) by choosing G = (G 1 , . . . , G k ) T , with Example 2.6. The equality constraints can also be included in problem (P) in the obvious way by assuming the same hypotheses as in Example 2.5. Example 2.7. Integral constraints on the gradient of the state can be considered within our formulation of problem (P): In this case we can take Z = R and Q = (−∞, δ]. Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php 3. The weak Pontryagin principle. Before formulating the weak Pontryagin principle, we introduce some notation. Given α ≥ 0, we define the Hamiltonian H α : Σ T × R × K × R −→ R as follows: H α (x, t, y, u, ϕ) = αl(x, t, y, u) + ϕf (x, t, y, u). Now we can establish Pontryagin's principle. THEOREM 3.1. Ifū ∈ U is a solution of (P), then there existᾱ ≥ 0,ȳ ∈ C(Ω T ) ∩ L 2 ([0, T ], H 1 (Ω)), andφ ∈ L r ([0, T ], W 1,p (Ω)) for all p, r ∈ [1, 2) with (2/r)+(n/p) > n + 1,μ ∈ Z andλ ∈ R s such that Hᾱ(x, t,ȳ(x, t), u(x, t),φ(x, t))dσ(x)dt; (3.5) where A * denotes the formal adjoint operator of A. Moreover, if one of the following assumptions is satisfied, (A1) Functions f and l are continuous with respect to the third variable on (K, d) and this space is separable; T for every (y, u) ∈ R × K, then the following pointwise relation holds: where M (Ω T ) is the space of the real and regular Borel measures inΩ T . Let us assume that [DG(ȳ)] * μ =φ +ν, withφ ∈ L 2 ([0, T ], H 1 (Ω) ) andν ∈ M(Ω T ), then we can write Analogous considerations can be made for [DF (ȳ)] * λ .
Let us apply the above principle to the examples given in section 2. Example 3.3. In Example 2.1, Z = C(Ω T ); therefore, the Lagrange multiplierμ whose existence is established in Theorem 3.1 is a measure inΩ T . In this case the transversality condition (3.4) is written as follows: From this relation we can deduce thatμ is a positive measure concentrated in the set of points (x, t) ∈Ω T , where g(x, t,ȳ(x, t)) = δ. In particular, it could be a Dirac measure or a combination of Dirac measures; see Casas [7].
The adjoint state equation (3.2) now becomes Since ∂g/∂y is a continuous function inΩ T , then the product (∂g/∂y)μ is well defined and can be identified again with a measure. If the points (x j , t j ) are all of them included in Ω T , then the adjoint state equation is If some points x j are in Γ, then the corresponding termλ j δ (xj ,tj ) should appear on the Neumann condition. Analogously, if t j = T for some index j, thenλδ (xj ,T ) should be included in the final condition. Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Then we have the following equation forφ: So, in particular, we have thatφ(T ) = 0 if the state constraint is not active in T . This type of state constraints has been studied by many authors; see Barbu and Precupanu [1], Lasiecka [24], and Tröltzsch [32]. All of them consider the semigroup theory approach to deal with the state and adjoint state equations. They prove some regularity of the adjoint stateφ; see section 6. Example 3.6. In Example 2.4, the Lagrange multiplierμ is an element of Z = L ∞ (Ω T ); therefore, (3.2) reduces in this case to In this case, assuming more regularity for the functions ψ dM and ψ bM given in (2.8)-(2.9), we can obtain additional regularity forφ. For instance, if we take function ψ bM ∈ Lp([0, T ], Lq(Ω)), thenφ ∈ Y . H 2,1 (Ω)-regularity is also obtained provided that Γ is of class C 2 and the coefficients a ij of A are Lipschitz in the variable x. Example 3.7. The Lagrange multipliers in Example 2.5 are positive real numbers {μ j } k j=1 . The positivity is a consequence of the transversality condition (3.3). The adjoint state equation can be written as follows: By increasing the regularity of functions η j , we can improve the regularity ofφ such as it was described in Example 3.6. For the equality constraints considered in Example 2.6 the adjoint state equation is similar to the above one. The only difference is that the Lagrange multipliers can be negative. Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Example 3.8. In Example 2.7, the Lagrange multiplierμ is a nonnegative real number,φ ∈ Y , and the adjoint state equation is

The strong Pontryagin principle.
In this section we will prove that, in the absence of equality constraints, Theorem 3.1 holds withᾱ = 1 for "almost all" control problems. We will be precise about this term later. The key to achieving this result is the introduction of a stability assumption of the optimal cost functional with respect to small perturbations of the set of feasible controls. This stability allows us to accomplish an exact penalization of the state constraints. First of all let us formulate the following control problem: with the same notation and assumptions of section 2 and setting Q δ = Q +B δ (0) for every δ > 0. DEFINITION 4.1. We say that (P δ ) is strongly stable if there exist > 0 and C > 0 such that This concept was first introduced in relation with optimal control problems by Bonnans [4]; see also Bonnans and Casas [6]. A weaker stability concept was used by Casas [8] to analyze the convergence of the numerical discretizations of optimal control problems. The following proposition states that almost all problems (P δ ) are strongly stable. PROPOSITION 4.2. Let δ 0 ≥ 0 be the smallest number such that (P δ ) has feasible controls for every δ > δ 0 . Then (P δ ) is strongly stable for all δ > δ 0 except at most a zero Lebesgue measure set.
Proof. It is enough to consider the function h : and remark that it is a nonincreasing monotone function and, consequently, differentiable at every point of (δ 0 , +∞) except at a zero measure set. Now it is obvious to check that (P δ ) is strongly stable at every point where h is differentiable. Now we state the strong Pontryagin principle. THEOREM 4.3. If (P δ ) is strongly stable andū is a solution of this problem, then Theorem 3.1 remains to be true withᾱ = 1.

Analysis of the state equation. In this section we will see that (2.3) is well posed in
, H 1 (Ω)) for every control u ∈ U. Also we will study the variations of the state with respect to some pointwise perturbations of the control which are the crucial point in the proof of Pontryagin's principle. In U we consider Ekeland's distance where m Σ T is the measure on Σ T obtained as the product of σ and the Lebesgue measure in the interval (0, T ). It is easy to check that (U , d E ) is a complete metric space. Indeed the proof given by Ekeland [16] can be repeated in this framework.
The uniqueness of the solution in Y can be proved by using the Gronwall inequality in the standard way along with the monotonicity of the nonlinear terms. Let us prove the existence.
If a 0 and f are bounded functions, then the existence and uniqueness of a solution ) is a consequence of the monotonicity of f imposed in (2.1) and the condition on a 0 given in (2.6); see Lions [29] or Ladyzhenskaya, Solonnikov, and Ural'tseva [23] for a proof based in Galerkin's approximation of the problem. If f is not bounded, we can consider the usual truncation of the function Thus hypothesis (2.2) implies the boundedness of f m .
An analogous modification can be made on a 0 . Then we deduce the existence and uniqueness of a solution y m ∈ L ∞ ([0, T ], L 2 (Ω)) ∩ L 2 ([0, T ], H 1 (Ω)) for problem (2.3) with a 0 and f replaced by a 0m and f m , respectively. Now thanks to the assumptions (2.1)-(2.8), we can apply the procedure of Ladyzhenskaya, Solonnikov, and Ural'tseva [23] to deduce the existence of a constant M > 0 independent of m and u ∈ U such that (5.2) holds for y u replaced by y m . This implies that Consequently, the uniqueness of a solution of (2.3) lets us obtain the identity y m = y u and the inequality (5.2).
In order to prove the continuity of y u , we first suppose that y 0 ∈ C θ (Ω T ) for some constant θ ∈ (0, 1]. Then, by applying the results of di Benedetto [2], we deduce Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php that y u ∈ C β,β/2 (Ω T ) for some β ∈ (0, θ]. When y 0 is not a Hölder function, we can take a sequence {y 0k } ∞ k=1 ⊂ C θ (Ω T ) converging uniformly to y 0 inΩ T . Then the corresponding solutions of (2.3), denoted by y k , are Hölder functions. Now, by applying the methods of [23] is easy to deduce the convergence y k → y u in L ∞ (Ω T ), which proves the continuity of y u .
Finally, the convergence The uniform convergence is obtained again by using the arguments of [23].
The rest of the section is devoted to the proof of the following theorem and if we denote by y ρ and y the states corresponding to u ρ and u, respectively, then the following equalities hold: and The first step is the proof of the following result The functions a r : (−Λ Γ , +Λ Γ ) n−1 −→ R are Lipschitz, and for some coordinate system (x r , x r,n ) = (x r,1 , . . . , x r,n ) in R n we have that and for every set E = {(x r , a r (x r )) : x r ∈ F }, with F ⊂ (−Λ Γ , +Λ Γ ) n−1 Lebesgue measurable, the following identity holds: For every k ∈ N we decompose the interval [−Λ Γ , +Λ Γ ] into k closed subintervals of length 2Λ Γ /k and disjoint interiors. Now we make all possible Cartesian products of these subintervals and obtain a family of cubes {Q k,i } k n−1 i=1 of equal Lebesgue measure, covering [−Λ Γ , +Λ Γ ] n−1 and with disjoint interiors. For every r = 1, . . . , d and every Let us see that such an F r k,j exists. For every t ∈ [0, 1] we define Q k,j (t) as the cube with the same center as Q k,j and the length of each side being equal to t times the length of the sides of Q k,j . So Q k,j (1) = Q k,j and Q k,j (0) is reduced to one point: the center of Q k,j . Let us consider the function g : [0, 1] −→ R defined by Then it is obvious that g is continuous and Therefore there exists 0 < t 0 < 1 such that Thus we can choose F r k,j = Q k,j (t 0 ). Now we set We are going to prove that Once this is proved, the convergence ( If A ⊂ Γ is an open set, then Thus (5.7) holds for every open subset of Γ. Let us take a closed set K ⊂ Γ, Finally, let A ⊂ Γ be a σ-measurable set. Given > 0 arbitrary, we can take K ⊂ Γ closed and V ⊂ Γ open such that K ⊂ A ⊂ V and Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php which concludes the proof of (5.7).
Step 2. The sets J k .
To construct the sets J k , we decompose the interval [0, T ] into k closed intervals I k j of length T/k and disjoint interiors. For each j = 1, . . . , k we take a subinterval J k j ⊂ 0 I k j of length √ ρT /k and the same center as I k j . Finally, we define J k as the union of the intervals {J k j } k j=1 . Then |J k | = √ ρT and the convergence (1/ √ ρ)χ J k → 1 * weakly in L ∞ (0, T ) can be proved following the same ideas as in the previous step.
Step 3. The sets E k .
Proof. Let us take a sequence {C k } ∞ k=1 of closed cubes with sides parallel to the axes and [31, pp. 167-170]. Fixed r, for each cube C l , it is obvious that From here it follows Since > 0 is arbitrary, the previous relations conclude the proof. Finally, we are ready to prove Theorem 5.2.
Proof of Theorem 5.2. Let ρ ∈ (0, 1) be fixed. Applying Proposition 5.3, we deduce the existence of measurable sets t) if (x, t) ∈ E k , Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php and we denote by y k and y the states corresponding to u k and u, respectively. Now, subtracting the equations satisfied by y k and y, and putting z k = (y k −y)/ρ we obtain
To conclude the proof it is enough to note that

y(x, t), u(x, t))z(x, t) dσ(x)dt
and to take into account the convergences previously established and (5.18).

Linear parabolic equations involving measure data.
Let µ be a regular Borel measure inΩ T . We can write µ = µ Ω T + µ Σ T + µ T + µ 0 , where µ Ω T = µ| Ω T , µ Σ T = µ| Σ T , µ T = µ|Ω ×{T } , and µ 0 = µ|Ω ×{0} . The aim of this section is the study of the following problem: The reader is referred to Boccardo and Gallouët [3] for the study of a quasi-linear parabolic equation with a measure in Ω T as a datum. Here we improve the results of [3] by exploiting the linearity of the equation.

is the space of bounded and continuous functions in Ω T and ν T (x, t) is the outward unit normal vector to ∂Ω T at the point (x, t).
By applying this theorem to the function w defined above and using (6.2) and (6.3), we have for all y ∈ Y 0 ∩ C 1 (Ω T ) From the identity and taking into account that we can identify Now we have the following result of existence and uniqueness of solution for problem (6.1). THEOREM 6.3. There exists a unique function ϕ ∈ L r ([0, T ], W 1,p (Ω)) ∀r, p ∈ [1, 2) with (2/r) + (n/p) > n + 1 such that it is a solution of (6.1) and Moreover, there exists a constant C r,p > 0 independent of µ such that Now for every ψ = (ψ 0 , ψ 1 , . . . , ψ n ) ∈ D(Ω T ) n+1 , we denote by y ψ the solution in Then Using (6.8) and the properties of f k , g k , and h k , we deduce from (6.8) ψ j L r ([0,T ],L p (Ω)) , (6.11) the last inequality being a consequence of the estimates for the solution of (6.9); see di Benedetto [2] and Ladyzhenskaya, Solonnikov, and Ural'tseva [23]. From the density of the space {ψ 0 − n j=1 ∂ xj ψ j : ψ ∈ D(Ω T ) n+1 } in L r ([0, T ], W 1,p (Ω) ) and estimate (6.11) follows the boundedness of {ϕ k } k in the space L r ([0, T ], W 1,p (Ω)). Moreover, by taking a subsequence if necessary, we can assume that ϕ k → ϕ weakly in L r ([0, T ], W 1,p (Ω)) and (6.7) is satisfied.
Let us prove that ϕ does not depend on r and p. Indeed, passing to the limit in (6.10) and remembering that y ψ (0) = 0, we get It is obvious that there is at most one function ϕ in L 1 ([0, T ], W 1,1 (Ω)) satisfying (6.12), which proves that ϕ is independent of r and p. Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Given y ∈ Y 0 ∩ C 1 (Ω T ), multiplying (6.8) by y and integrating by parts, it follows that Now passing to the limit we deduce (6.2) and consequently ϕ is a solution of (6.1). Let us prove (6.6). Given y ∈ Y ∞ 0 , multiplying (6.8) by y and integrating by parts, we deduce Now (6.6) is obtained by passing to the limit. Finally, the uniqueness of ϕ follows from (6.6). Indeed, the regularity results for the Neumann problem associated with the operator (∂/∂t) + A (see [2] or [23]) prove the surjectivity of the mapping This along with (6.6) implies that the zero function of L r ([0, T ], W 1,p (Ω)) is the only one satisfying This shows the uniqueness of ϕ. An interesting case arises when µ = gω, with g ∈ C([0, T ], L 2 (Ω)) and ω ∈ M [0, T ] see Example 3.5. In this particular case we have the following result. THEOREM 6.4. With the above notation, there exists a unique function ϕ in the space in Ω. (6.14) Given f ∈ D(Ω T ), let us denote by y f the solution in Y of the problem Then From (6.15) it follows by using the classical arguments that From the first inequality and (6.16) we deduce the boundedness of the sequence {ϕ k } k in the space L ∞ ([0, T ], L 2 (Ω)). The second inequality leads to the boundedness of the same sequence in L 2 ([0, T ], H 1 (Ω)). The rest of the proof is easy.

Proof of Pontryagin principle.
In this section we prove Theorems 3.1 and 4.3. A crucial point in the proofs is the use of Ekeland's variational principle that we state now. LEMMA 7.1 (see Ekeland [16] Proof of Theorem 3.1. Since Z is separable, we can take in Z a norm · Z such that Z endowed with the dual norm · Z is strictly convex. Then the function is convex, Lipschitz and Gâteaux differentiable at every point z ∈ Q, with ∂d Q (z) = {∇d Q (z)}, where the Clarke's generalized gradient and the subdifferential in the sense of the convex analysis coincide for this function. Therefore, given ξ ∈ ∂d Q (y), we have that Moreover, ∇d Q (z) Z = 1 for every z ∈ Q; see Clarke [15] and Casas and Yong [14].
Let us take J : U −→ R defined by It is obvious that J (u) > 0 for every u ∈ U and J (ū) = . On the other hand, thanks to Theorem 5.1 we have that J is continuous in (U, d E ), with d E defined by (5.1). Therefore we can apply Ekeland's variational principle and deduce the existence of u ∈ U such that Given v ∈ U arbitrary, let us take E ρ and u ρ as in Theorem 5.2, Then with the help of (5.3) and (5.4) we get where y and y ρ are the states associated with u and u ρ , respectively, and z ∈ Y By using Theorem 6.3, we can take a function ϕ ∈ L r ([0, T ], W 1,p (Ω)) ∀r, p ∈ [1, 2) with (2/r) + (n/p) > n + 1 such that Thanks to the assumptions (2.2) and (2.7), we have that z ∈ Y ∞ 0 . Then we can apply (6.6) with y = z and deduce from (7.3)-(7.5) and the definition of H α given in section 3 the inequality Now we pass to the limit when → 0. To do this, let us remark that α 2 + µ 2 Z + |λ | 2 = 1. (7.10) Then we take subsequences, denoted in the same way, satisfying α →ᾱ in R, λ →λ in R n , µ →μ in the * weak topology of Z . (7.11) On the other hand, the convergence y →ȳ in Y follows from Theorem 5.1. The boundedness of {ϕ } in L r ([0, T ], W 1,p (Ω)) follows from (6.7) and (7.10). Then, using (7.11), it is easy to pass to the limit in (7.8) and (7.9) and to deduce (3.3) and (3.5). Now remembering the definition of µ and ξ and (7.1), we deduce µ , z − G(y ) ≤ 0 ∀z ∈ Q. (7.12) Passing to the limit in this expression we obtain (3.4). Let us prove (3.1). To do this, let us suppose thatᾱ = |λ| = 0; then from (7.10) it follows µ Z → 1 as → 0. Let us take z 0 ∈ o Q and ρ > 0 such thatB ρ (z 0 ) ⊂ o Q. Then (7.12) implies that µ , z + z 0 − G(y ) ≤ 0 ∀z ∈B ρ (0). Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Passing to the limit which proves thatμ = 0.
It remains to prove (3.6); see Bonnans and Casas [5] or Casas [11] for the study of analogous situations. To do this we consider the coordinate system {(Γ r , a r )} d r=1 of Γ introduced in the proof of Proposition 5.3. Given a point x 0 ∈ o Γr for some 1 ≤ r ≤ d we denote for each > 0 small enough where B (x 0r ) is the ball in R n−1 centered at x 0r and having radius . Now given 0 < t 0 < T , we set The following lemma is used in this proof.
Proof. Let us denote for all (x r , t) ∈ (0, 1) n−1 × (0, T ) Since ω r and f r are Lebesgue integrable functions in (0, 1) n−1 and (0, 1) n−1 × (0, T ), respectively, we know that the set of Lebesgue points of these functions U r and V r , respectively, have measure equal to 1 and T, respectively. Let us define Then m Σ T (S) = m Σ T (Σ T ) and S r ⊂ o Γr ×(0, T ), 1 ≤ r ≤ d. Let us take (x 0 , t 0 ) = (x 0r , a(x 0r ), t 0 ) ∈ S r . Then x 0j and (x 0 , t 0 ) are Lebesgue points of ω r and f r ; consequently, |f r (x r , t) − f r (x 0r , t)|dx r dt , Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php where |B (x 0r )| denotes the (n − 1)-measure of B (x 0r ). The set points of S will be called the Lebesgue points of f . This set depends on the system of coordinates {(Γ r , a r )} d r=1 , but this dependence only affects a set of σ-measure equal to zero.
The first two terms converge to Hᾱ(x 0 , t 0 ,ȳ(x 0 , t 0 ), v,φ(x 0 , t 0 )) because of the continuity of the integrands in (x 0 , t 0 ) ∈ Σ 0 T . Let us prove that the last term goes to zero.
|φ(x, t) −φ(x 0 , t 0 )|dσ(x)dt −→ 0, thanks to (7.14) and the fact that (x, t) → f (x, t,ȳ(x, t), v) is bounded in Σ T because of the assumption (2.2) and the boundedness ofȳ. Now we will prove Theorem 4.3. The key to achieving this result is to carry out an exact penalization of the state constraint. To do this, we will use the distance function d Q δ associated with the set Q δ and defined in the same way as in the proof of Theorem 4.3. PROPOSITION 7.3. If (P δ ) is strongly stable andū is a solution of this problem, then there exists q 0 > 0 such thatū is also a solution of inf u∈U J q (u) = J(u) + qd Q δ (G(y u )) (7.15) for every q ≥ q 0 .
Proof. Let us suppose that it is false. Then there exists a sequence {q k } ∞ k=1 of real numbers, with q k → +∞ and elements {u k } ∞ k=1 ⊂ U such that where y k is the state corresponding to u k . From here we obtain that d Q δ (G(y k )) < J(ū) − J(u k ) q k −→ 0 when k → +∞ and G(y k ) ∈ Q δ . Let δ k > δ be the smallest number such that G(y k ) ∈ Q δ k . Since δ k → δ, we can use (4.1) to deduce > q k d Q δ (G(y k )) = q k (δ k − δ) ∀k ≥ k , which is not possible. Downloaded 05/24/13 to 193.144.185.28. Redistribution subject to SIAM license or copyright; see http://www.siam.org/journals/ojsa.php Since J q is not Gâteaux differentiable on Q δ , we are going to modify slightly this functional to attain the differentiability necessary for the proof. PROPOSITION 7.4. Let us take q ≥ q 0 and for every > 0 let us consider the problem (P δ, ) inf u∈U J q, (u) = J(u) + q d Q δ (G(y u )) 2 + 2 1/2 .
Proof. It is an immediate consequence of the inequality J q (u) ≤ J q, (u) ≤ J q (u) + q ∀u ∈ U.
Finally we are ready to prove the strong Pontryagin principle.
Therefore we have µ Z ≤ q for every > 0. Now we can take a subsequence that converges weakly * to an elementμ ∈ Z . The rest is as in the proof of Theorem 3.1, taking α = 1.