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International Mathematical Forum, Vol. 6, 2011, no. 7, 303 - 324

Sufficient Variational Conditions for Isoperimetric Control Problems Javier F. Rosenblueth IIMAS-UNAM, Universidad Nacional Aut´onoma de M´exico Apartado Postal 20-726, M´exico DF 01000, M´exico [email protected] Gerardo S´ anchez Licea Facultad de Ciencias, Universidad Nacional Aut´onoma de M´exico Departamento de Matem´aticas, M´exico DF 04510, M´exico [email protected] Abstract For optimal control problems involving isoperimetric constraints, by using a two-norm approach, a new sufficiency theorem for a proper strong minimum is obtained. It is applicable to processes that satisfy the Legendre-Clebsch necessary condition but its strict version is not imposed, that is, the processes may be singular. The conditions are expressed explicitly in terms of the second variation along nonnull admissible variations, as well as the Weierstrass excess functions of the constraints delimiting the problem. The proof we provide corresponds to a generalization, in several respects, of a sufficiency proof due to Hestenes for the fixed-endpoint isoperimetric problem in the calculus of variations.

Mathematics Subject Classification: 49K15 Keywords: Optimal control, sufficient variational conditions, isoperimetric problems, singular extremals

1. Introduction In this paper we shall be concerned with a sufficiency theorem for strong minima for a fixed-endpoint optimal control problem with isoperimetric constraints. The proof we provide corresponds to a generalization, in several respects, of a direct sufficiency proof (in the sense that it deals explicitly with

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the positivity of the second variation along admissible variations) given by Hestenes in [8] for the fixed-endpoint isoperimetric problem in the calculus of variations. For comparison reasons, and to clearly situate the main contributions of this paper, let us first state the sufficiency result given in [8] for the calculus of variations problem. The following summary of the main concepts related to the problem studied by Hestenes is written essentially in the same vein as in the original text. Suppose we are given an interval T := [t0 , t1 ] in R, two points ξ0 , ξ1 in n R , a (relatively) open set A in T × Rn × Rn , constants α1 , . . . , αr in R, and functions L, L1 , . . . , Lr mapping T × Rn × Rn to R. Denote by X := AC(T ; Rn ) the space of absolutely continuous functions mapping T to Rn . Elements of X will be called arcs and an arc x is called admissible if it satisfies a. b. c. d.

L(t, x(t), x(t)), ˙ Li (t, x(t), x(t)) ˙ (i = 1, . . . , r) are integrable on T . x(t0 ) = ξ0 , x(t1 ) = ξ1 . t ˙ = 0 (i = 1, . . . , r). Ii (x) := αi + t01 Li (t, x(t), x(t))dt (t, x(t), x(t)) ˙ ∈ A a.e. in T .

The isoperimetric problem treated in [8], which we label (Q), is that of minimizing  t1 L(t, x(t), x(t))dt ˙ I(x) := t0

over all admissible arcs. It is assumed that the functions L, L1 , . . . , Lr are of class C 2 on A. Before stating the sufficiency theorem proved in [8], let us recall a few concepts. • An admissible arc x0 is called a strong minimum of (Q) if there is a neighborhood F of x0 in tx-space such that the inequality I(x) ≥ I(x0 ) holds for all admissible arcs x = x0 whose elements (t, x(t)) are in F . If the inequality can be replaced by a strict inequality, the minimum is said to be a proper minimum. • Sufficient conditions are given in terms of an integral J of the form  t1 r r   J(x) = I(x) + λi Ii (x) = λi αi + F (t, x(t), x(t))dt ˙ i=1

t0

i=1

where the function F mapping T × Rn × Rn to R is given by F (t, x, x) ˙ = L(t, x, x) ˙ +

r 

λi Li (t, x, x) ˙

i=1

and the multipliers λ1 , . . . , λr are constants in R.

Sufficient conditions for isoperimetric control problems

305

• We shall have occasion to refer to the first variation of J along x given by 



t1

J (x; y) = t0

{Fx y(t) + Fx˙ y(t)}dt ˙

and the second variation of J along x given by  t1  {y(t), Fxxy(t) + 2y(t), Fxx˙ y(t) ˙ + y(t), ˙ Fx˙ x˙ y(t)}dt ˙ J (x; y) = t0

where the partial derivatives of F are evaluated at (t, x(t), x(t)). ˙ 1 • An admissible arc x of class C is called an extremal if it satisfies Euler’s equation, that is,  t Fx˙ (t, x(t), x(t)) ˙ = Fx (s, x(s), x(s))ds ˙ + c (t ∈ T ) t0

for some c ∈ Rn or, equivalently, if J  (x; y) = 0 for all y ∈ X satisfying y(t0 ) = y(t1 ) = 0. • An admissible arc x of class C 1 is called nonsingular if the determinant |Fx˙ x˙ (t, x(t), x(t))| ˙ is different from zero (t ∈ T ). • For any admissible arc x denote by Y (x) the class of arcs y having square integrable derivatives and satisfying y(t0 ) = y(t1 ) = 0 and Ii (x; y) = 0 where Ii (x; y) denotes the first variation of Ii along x. Elements of Y (x) will be called admissible variations along x. • The Weierstrass E-function for F is given by ˙ − x) ˙ E(t, x, x, ˙ u) := F (t, x, u) − F (t, x, x) ˙ − Fx˙ (t, x, x)(u ˙ u) will denote the corresponding Weierstrass E-function for Li and Ei (t, x, x, (i = 1, . . . , r). • An arc x0 is said to satisfy the strengthened Weierstrass condition if there exist h > 0 and a neighborhood N of the elements (t, x, x) ˙ on x0 such that E(t, x, x, ˙ u) ≥ h|Ei(t, x, x, ˙ u)| (i = 1, . . . , r) for all (t, x, x, ˙ u) with (t, x, x) ˙ in N and (t, x, u) ∈ A. This is in contrast with the necessary condition of Weierstrass which states that E(t, x0 (t), x˙ 0 (t), u) ≥ 0 for all (t, u) with (t, x0 (t), u) ∈ A. • For all arcs y let  t1 ϕ(y(t))dt ˙ where ϕ(c) := (1 + |c|2)1/2 − 1 D(y) := t0

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and denote by  ·  the supremum norm in X. With these concepts in mind we are in a position to state the sufficiency theorem, for the fixed-endpoint calculus of variations problem involving isoperimetric constraints, proved by Hestenes in [8]. 1.1 Theorem: Let x0 be an admissible arc. Suppose there exist multipliers λ1 , . . . , λr such that, relative to the function J(x) = I(x) +

r 

λi Ii (x) =

i=1

r  i=1

 λi αi +

t1

F (t, x(t), x(t))dt, ˙ t0

i. x0 is a nonsingular extremal. ii. J  (x0 ; y) > 0 for all nonnull admissible variations y along x0 . iii. x0 satisfies the strengthened Weierstrass condition. Then there exist ρ, δ > 0 such that, for all admissible arcs x with x−x0  < ρ, I(x) ≥ I(x0 ) + δD(x − x0 ). In particular, x0 is a proper strong minimum of (Q). The development of the technique used to prove this result as it appears in [8], as well as its application to more general problems, can be traced back to different papers of the author and McShane (see [1–7, 12]). In this paper we generalize Theorem 1.1 in various directions. To begin with, we shall be concerned with the optimal control problem of minimizing an integral of the form  t1 L(t, x(t), u(t))dt I(x, u) = t0

over all pairs (x, u) with (t, x(t), u(t)) ∈ A, satisfying the dynamics x(t) ˙ = f (t, x(t), u(t)), the fixed-endpoint constraints x(t0 ) = ξ0 , x(t1 ) = ξ1 , and a set of integral constraints of the form  t1 Ii (x, u) := αi + Li (t, x(t), u(t))dt = 0 (i = 1, . . . , r). t0

As we shall note throughout the proof, the assumption of openness of the set A will no longer be required. Moreover, our main sufficiency result does not impose, as in Theorem 1.1, the nonsingularity of the process under consideration due to a way of strengthening the Weierstrass condition which differs from the usual one. When this modification is inserted in the isoperimetric calculus of variations problem, the usual way of strengthening that condition implies the new one and therefore our sufficiency theorem covers, even in a calculus of

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variations context, a wide range of problems that lie beyond the scope of the classical theory. For optimal control problems, the same conclusion is reached when our main result is compared with other sufficiency theorems available in the literature (see, for example, [11, 13] and references therein). The above problem, as one readily verifies, is equivalent to the Lagrange problem of minimizing I(x, u) subject to a. (x(t), ˙ z(t)) ˙ = f˜(t, x(t), u(t)) a.e. in T , b. (x(t0 ), z(t0 )) = (ξ0 , α), (x(t1 ), z(t1 )) = (ξ1 , 0), c. (t, x(t), u(t)) ∈ A (t ∈ T ), where f˜(t, x, u) = (f (t, x, u), L1(t, x, u), . . . , Lr (t, x, u)),

α = (α1 , . . . , αr ).

If A is the whole space T × Rn × Rm , a sufficiency result provided in [11] is expressed in terms of the Hamiltonian (for the unconstrained problem) given by ˜ x, u), H(t, x, u, λ) := L(t, x, u) + λ, f(t, assuming that the control function under consideration is continuous, the strict Legendre-Clebsch condition (and therefore nonsingularity) holds on the whole time interval, and a certain Riccati equation has a bounded solution satisfying some boundary conditions. The same occurs if the set of constraints is defined by a set of mixed inequality constraints of the form gi (t, x, u) ≤ 0 and the conditions are expressed in terms of the augmented Hamiltonian defined by ˜ x, u, λ, μ) := H(t, x, u, λ) + μ, g(t, x, u). H(t, In both cases, the two-norm approach developed in [9, 10] is used by observing that for (x − x0 , u − u0 )∞ < , the deviation of the cost function I(x, u) from I(x0 , u0 ) can be measured in the L2 -norm (x, u)2 :=



t1

t0

(|x(t)|2 + |u(t)|2 )dt

1/2 .

In our case, without the assumptions of continuity of the control function and that of nonsingularity of the process under consideration, a similar conclusion will follow by noting that for x − x0  < , the deviation of the cost functions can be measured in terms of the D-function defined for optimal control problems as 

t1

D(u) := t0

ϕ(u(t))dt where ϕ(c) := (1 + |c|2 )1/2 − 1.

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Moreover, the conditions involved will be expressed in terms of the second variation and the Weierstrass excess function, instead of a bounded solution to a certain Riccati equation. Let us end this introduction by stating explicitly the new sufficiency theorem for the calculus of variations problem derived from the main result of the paper. Before doing it, let us briefly mention a few consequences of the conditions imposed in Theorem 1.1. Suppose first that x0 is a piecewise C 1 admissible arc that satisfies the condition of Weierstrass. Then it also satisfies the condition of Legendre, that is, Lx˙ x˙ (t, x0 (t), x˙ 0 (t)) ≥ 0 (t ∈ T ). If also x0 is nonsingular, then it satisfies the strengthened condition of Legendre where the above inequality is strict. Suppose further that there exist h > 0 and a neighborhood N of the elements (t, x, x) ˙ on x0 such that E(t, x, x, ˙ u) ≥ 0 for all (t, x, x) ˙ in N and (t, x, u) ∈ A. Then, as shown in [8], one can diminish N and h if necessary so that the relation E(t, x, x, ˙ u) ≥ hϕ(u − x) ˙ holds whenever (t, x, x) ˙ is in N and (t, x, u) is in A. Now, the sufficiency theorem we provide in this paper, when treating the calculus of variations problem, corresponds to the following result. The only modifications we make with respect to the previous assumptions are that the set A is any subset of T × Rn × Rn and the functions L, L1 , . . . , Lr are continuous and of class C 2 with respect to x and u on the whole space T × Rn × Rn . 1.2 Theorem: Let x0 be an admissible arc with x˙ 0 ∈ L∞ (T ; Rn ). Suppose there exist multipliers λ1 , . . . , λr such that, relative to the function J(x) = I(x) +

r 

λi Ii (x) =

i=1

r  i=1

 λi αi +

t1

F (t, x(t), x(t))dt, ˙ t0

i. x0 satisfies Euler’s equation and Legendre’s condition. ii. J  (x0 ; y) > 0 for all nonnull admissible variations y along x0 . iii. There exist h > 0 and a neighborhood F of x0 in tx-space such that, for all admissible arcs x whose elements (t, x(t)) are in F , E(t, x(t), x˙ 0 (t), x(t)) ˙ ≥ hϕ(x(t) ˙ − x˙ 0 (t)) a.e. in T, 

t1

t0

  E(t, x(t), x˙ 0 (t), x(t))dt ˙ ≥ h

t1

t0

  Ei(t, x(t), x˙ 0 (t), x(t))dt ˙  (i = 1, . . . , r).

Then there exist ρ, δ > 0 such that, for all admissible arcs x with x−x0  < ρ, I(x) ≥ I(x0 ) + δD(x − x0 ).

Sufficient conditions for isoperimetric control problems

309

In particular, x0 is a proper strong minimum of (Q). 2. Statement of the problem and the main result In this section we shall state the fixed-endpoint optimal control problem with isoperimetric constraints we shall be concerned with, as well as the main result of the paper. Some notions given now coincide with those of the previous section but others have a different meaning. To avoid any confusion, we have included all those needed for the problem under consideration. The problem we shall deal with can be stated as follows. Suppose we are given an interval T := [t0 , t1 ] in R, two points ξ0 , ξ1 in Rn , a set A in T × Rn × Rm , constants α1 , . . . , αr in R, and functions L, L1 , . . . , Lr mapping T × Rn × Rm to R and f mapping T × Rn × Rm to Rn . Let X := AC(T ; Rn ) denote the space of absolutely continuous functions mapping T to Rn , let U := L1 (T ; Rm ), and set Z := X × U. Denote by Ze (A) the set of all (x, u) ∈ Z satisfying a. b. c. d. e.

L(t, x(t), u(t)), Li (t, x(t), u(t)) (i = 1, . . . , r) are integrable on T . x(t) ˙ = f (t, x(t), u(t)) a.e. in T . x(t0 ) = ξ0 , x(t1 ) = ξ1 . t Ii (x, u) := αi + t01 Li (t, x(t), u(t))dt = 0 (i = 1, . . . , r). (t, x(t), u(t)) ∈ A (t ∈ T ).

The problem we shall deal with, which we label (P), is that of minimizing I over Ze (A), where  t1 L(t, x(t), u(t))dt. I(x, u) := t0

Elements of Z will be called processes and a process is admissible if it belongs to Ze (A). Thus, admissible processes are couples (x, u) comprising functions x ∈ X and u ∈ U which satisfy the constraints of the problem. For any C ⊂ T × Rn × Rm consider the set S(C) := {(x, u) ∈ Ze (C) | I(x, u) ≤ I(y, v) for all (y, v) ∈ Ze (C)} where Ze (C) is defined as Ze (A) but replacing A with C, that is, condition (e) is replaced by (t, x(t), u(t)) ∈ C (t ∈ T ). A process (x, u) is called a (global) solution of (P) if it belongs to S(A), and a strong minimum of (P) if there exists  > 0 such that (x, u) belongs to S(T0 (x; ) ∩ A), where T0 (x; ) := {(t, y) ∈ T × Rn : |x(t) − y| < } × Rm denotes the tube of radius  > 0 centred on the trajectory x ∈ X. A strong minimum (x, u) of (P) is proper if the inequality in the definition of S(T0 (x; )∩ A) is strict for processes in Ze (T0 (x; ) ∩ A) different than (x, u).

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We shall assume throughout the paper that the functions L, L1 , . . . , Lr and f are continuous and of class C 2 with respect to x and u on T × Rn × Rm , and there exists a continuous function ψ: T × Rn → R such that |fu (t, x, u)| ≤ ψ(t, x) for all (t, x, u) ∈ T × Rn × Rm . Given r real numbers λ1 , . . . , λr consider the functional  t1 r r   λi Ii (x, u) = λi αi + L0 (t, x(t), u(t))dt, I0 (x, u) := I(x, u) + i=1

i=1

t0

where the function L0 mapping T × Rn × Rm to R is given by L0 (t, x, u) := L(t, x, u) +

r 

λi Li (t, x, u).

i=1

For the theory to follow we shall find convenient to introduce the following definitions. • For all (t, x, u, p) ∈ T × Rn × Rm × Rn , let H(t, x, u, p) := p, f (t, x, u) − L0 (t, x, u). • A triple (x, u, p) will be called an extremal if (x, u) is an admissible process, p ∈ X, p(t) ˙ = −Hx∗ (t, x(t), u(t), p(t)) (a.e. in T ) and Hu (t, x(t), u(t), p(t)) = 0 (t ∈ T ) where ‘∗ ’ denotes transpose. • For a given p ∈ X define, for all (t, x, u) ∈ T × Rn × Rm , ˙ x. F0 (t, x, u) := L0 (t, x, u) − p(t), f (t, x, u) − p(t), With respect to F0 , define the functional J as  t1 F0 (t, x(t), u(t))dt J(x, u) := β + t0

r

where β = p(t1 ), ξ1 −p(t0 ), ξ0 + 1 λi αi . Note that, if (x, u) is an admissible process, we have J(x, u) = I0 (x, u) = I(x, u). Consider the first variations of J and Ii with respect to (x, u) ∈ X ×L∞ (T ; Rm ) over (y, v) ∈ Z which are given, respectively, by  t1  {F0x (t, x(t), u(t))y(t) + F0u (t, x(t), u(t))v(t)}dt, J ((x, u); (y, v)) := t0

Sufficient conditions for isoperimetric control problems

Ii ((x, u); (y, v))



t1

:= t0

311

{Lix (t, x(t), u(t))y(t) + Liu (t, x(t), u(t))v(t)}dt.

The second variation of J with respect to (x, u) ∈ X ×L∞ (T ; Rm ) over (y, v) ∈ X × L2 (T ; Rm ) is given by 



t1

J ((x, u); (y, v)) :=

2Ω(t, y(t), v(t))dt, t0

where, for all (t, y, v) ∈ T × Rn × Rm , 2Ω(t, y, v) := y, F0xx(t, x(t), u(t))y + 2y, F0xu (t, x(t), u(t))v + v, F0uu (t, x(t), u(t))v. Denote by E0 the Weierstrass excess function with respect to F0 , that is, E0 (t, x, u, v) := F0 (t, x, v) − F0 (t, x, u) − F0u (t, x, u)(v − u) for all (t, x, u, v) ∈ T × Rn × Rm × Rm . Also, with respect to Li (i = 1, . . . , r), denote by Ei the corresponding Weierstrass excess functions given by Ei(t, x, u, v) = Li (t, x, v) − Li (t, x, u) − Liu (t, x, u)(v − u). • A process (x, u) is nonsingular if the determinant |F0uu (t, x(t), u(t))| is different from zero (t ∈ T ). • For all (x, u) ∈ X × L∞ (T ; Rm ) denote by Y (x, u) the class of all (y, v) ∈ X × L2 (T ; Rm ) satisfying y(t) ˙ = A(t)y(t) + B(t)v(t), a.e. in T,

y(t0) = y(t1) = 0,

Ii ((x0 , u0 ); (y, v)) = 0 (i = 1, . . . , r) where A(t) := fx (t, x(t), u(t)), B(t) := fu (t, x(t), u(t)) (t ∈ T ). Elements of Y (x, u) will be called admissible variations along (x, u). • For all u ∈ U let  t1 D(u) := ϕ(u(t))dt where ϕ(c) := (1 + |c|2 )1/2 − 1 t0

and denote by  ·  the supremum norm in X. Let us now state the main theorem of the paper. It corresponds to a sufficiency result for a proper strong minimum of problem (P) assuming, with respect to a given extremal, the Legendre-Clebsch condition, the positivity of the second variation on the set of nonnull admissible variations, a condition related to the Weierstrass excess function with respect to F0 , and a condition

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related to the Weierstrass excess functions which delimit the problem under consideration. 2.1 Theorem: Let (x0 , u0) be an admissible process with u0 ∈ L∞ (T ; Rm ). Suppose there exist p ∈ X and multipliers λ1 , . . . , λr such that (x0 , u0 , p) is an extremal satisfying i. F0uu (t, x0 (t), u0 (t)) ≥ 0 a.e. in T . ii. J  ((x0 , u0 ); (y, v)) > 0 for all (y, v) nonnull admissible variations along (x0 , u0 ). iii. There exist h,  > 0 such that, for all (x, u) ∈ Ze (T0 (x0 ; ) ∩ A), E0 (t, x(t), u0 (t), u(t)) ≥ hϕ(u(t) − u0 (t)) a.e. in T,  t1  t1    E0 (t, x(t), u0 (t), u(t))dt ≥ h Ei (t, x(t), u0 (t), u(t))dt (i = 1, . . . , r). t0

t0

Then there exist ρ, δ > 0 such that, for all admissible processes (x, u) with x − x0  < ρ, J(x, u) ≥ J(x0 , u0) + δD(u − u0 ). In particular, (x0 , u0) is a proper strong minimum of (P). Before proving this theorem, let us show that it implies another sufficiency result for a proper strong minimum appropriate for nonsingular extremals. To do so, let us first define a restricted tube of radius  > 0 centred on a given process (x, u) as T1 ((x, u); ) := {(t, y, v) ∈ T0 (x; ) : |u(t) − v| < } The following result is a simple consequence of Lemma 3.2 of [14]. 2.2 Lemma: Suppose F0 is C 2 with respect to u and (x0 , u0 ) ∈ Z with u0 continuous satisfies i. F0uu (t, x0 (t), u0 (t)) > 0 (t ∈ T ). ii. There exists  > 0 such that, for all (t, x, u) ∈ T1 ((x0 , u0); ) and (t, x, v) ∈ A, E0 (t, x, u, v) ≥ 0 . Then there exist δ, h > 0 such that, for all (x, u) ∈ Ze (T0 (x0 ; δ) ∩ A), E0 (t, x(t), u0 (t), u(t)) ≥ hϕ(u(t) − u0 (t)) a.e. in T. By an application of Theorem 2.1 and Lemma 2.2 we obtain the following result. 2.3 Theorem: Let (x0 , u0 ) be an admissible process with u0 continuous. Suppose there exist p ∈ X and multipliers λ1 , . . . , λr such that (x0 , u0 , p) is an extremal satisfying

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i. F0uu (t, x0 (t), u0 (t)) > 0 (t ∈ T ). ii. J  ((x0 , u0 ); (y, v)) > 0 for all nonnull admissible variations (y, v) along (x0 , u0 ). iii. There exist h,  > 0 such that E0 (t, x, u, v) ≥ 0 for all (t, x, u) ∈ T1 ((x0 , u0); ) and (t, x, v) ∈ A, and for all (x, u) ∈ Ze (T0 (x0 ; ) ∩ A),  t1  t1    E0 (t, x(t), u0 (t), u(t))dt ≥ h Ei (t, x(t), u0 (t), u(t))dt (i = 1, . . . , r). t0

t0

Then (x0 , u0 ) is a proper strong minimum of (P). 3. Proof of Theorem 2.1 In this section we shall prove Theorem 2.1. We first state an auxiliary result (proved in [15]) on which the proof of the theorem is strongly based. Implicit in the statement of the result we have included a generalization of the notion of a directionally convergent sequence of trajectories, first introduced in a calculus of variations context by Hestenes in [8]. 3.1 Lemma: Let {zq := (xq , uq )} be a sequence in Z, (x0 , u0) ∈ Z, and suppose that lim D(uq − u0 ) = 0 and dq := [2D(uq − u0 )]1/2 > 0 (q ∈ N).

q→∞

For all q ∈ N and t ∈ T define

1/2 1 wq (t) := 1 + ϕ(uq (t) − u0 (t)) , 2 yq (t) :=

xq (t) − x0 (t) dq

and vq (t) :=

uq (t) − u0 (t) . dq

Then the following hold: a. For some v0 ∈ L2 (T ; Rm ) and some subsequence of {zq } (we do not relabel), {vq } converges weakly to v0 in L1 (T ; Rm ). b. Let Aq ∈ L∞ (T ; Rn×n ) and Bq ∈ L∞ (T ; Rn×m ) be matrix functions for which there exist constants m0 , m1 > 0 such that Aq ∞ ≤ m0 and Bq ∞ ≤ m1 (q ∈ N) and suppose that ζq satisfies the system ˙ = Aq (t)ζ(t) + Bq (t)vq (t) (a.e. in T ), ζ(t)

ζ(t0 ) = 0.

Then there exist σ0 ∈ L2 (T ; Rn ) and a subsequence of {ζq } (we do not relabel) such that {ζ˙q } converges weakly in L1 (T ; Rn ) to σ0 . Moreover, if we define  t y0 (t) := σ0 (s)ds (t ∈ T ), t0

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then ζq (t) → y0 (t) uniformly on T . c. There exists some subsequence of {zq } (we do not relabel) such that uq (t) → u0 (t) almost uniformly on T (that is, for every  > 0 there exists S ⊂ T with m(S ) <  such that uq (t) → u0 (t) uniformly on T \ S ) and hence wq (t) → 1 almost uniformly on T . d. Suppose S ⊂ T is measurable and wq (t) → 1 uniformly on S. Let Rq , R0 be quadratic forms with m × m associated matrices Rq (·) measurable on S, R0 (·) ∈ L∞ (S; Rm×m ), Rq (t) → R0 (t) uniformly on S, and R0 (t) ≥ 0 (t ∈ S). Then, for some subsequence of {zq } (we do not relabel),   R0 (t; v0 (t))dt. lim inf Rq (t; vq (t))dt ≥ q→∞

S

S

Proof of Theorem 2.1: Assume that, for all ρ, δ > 0, there exists (x, u) ∈ Ze (A) with x − x0  < ρ such that J(x, u) < J(x0 , u0 ) + δD(u − u0 ). (1) We are going to show that this contradicts (ii) of Theorem 2.1 and the first statement will follow. The second conclusion is a consequence of the first since J(x, u) = I(x, u) for all (x, u) ∈ Ze (A). Let z0 := (x0 , u0). Note that, for all z = (x, u) ∈ Ze (A), J(z) = J(z0 ) + J  (z0 ; z − z0 ) + K0 (z) + E˜0 (z) 

where E˜0 (x, u) :=  K0 (x, u) :=

t1 t0

t1

t0

(2)

E0 (t, x(t), u0 (t), u(t))dt,

{M0 (t, x(t)) + u(t) − u0 (t), N0 (t, x(t))}dt,

and the functions M0 and N0 are given by M0 (t, y) := F0 (t, y, u0(t)) − F0 (t, x0 (t), u0 (t)) − F0x (t, x0 (t), u0 (t))(y − x0 (t)), ∗ ∗ (t, y, u0(t)) − F0u (t, x0 (t), u0 (t)). N0 (t, y) := F0u

By Taylor’s theorem we have 1 M0 (t, y) = y − x0 (t), P0 (t, y)(y − x0 (t)), 2

N0 (t, y) = Q0 (t, y)(y − x0 (t)),

where  P0 (t, y) := 2

0

1

(1 − λ)F0xx (t, x0 (t) + λ(y − x0 (t)), u0 (t))dλ,

Sufficient conditions for isoperimetric control problems

 Q0 (t, y) :=

1 0

315

F0ux (t, x0 (t) + λ(y − x0 (t)), u0(t))dλ.

Let us begin by proving the existence of α0 , δ > 0 such that, for all z = (x, u) ∈ Ze (A) with x − x0  < δ, E˜0 (x, u) ≥ hD(u − u0 ),

(3)

|K0 (x, u)| ≤ α0 x − x0 [1 + D(u − u0 )].

(4)

By assumption (iii) of the theorem,  t1 ˜ E0 (t, x(t), u0 (t), u(t))dt E0 (z) = t0  t1 ϕ(u(t) − u0(t))dt = hD(u − u0 ) ≥ h t0

for all z ∈ Ze (A) satisfying x − x0  < . Choose α, μ > 0 such that, for all z ∈ Ze (A) with x − x0  < μ and all t ∈ T , |M0 (t, x(t)) + u(t) − u0(t), N0 (t, x(t))| ≤ α|x(t) − x0 (t)|[1 + |u(t) − u0(t)|2 ]1/2 . Set α0 := max{α, α(t1 − t0 )}. Then, for all z ∈ Ze (A) with x − x0  < μ,  t1 [1 + ϕ(u(t) − u0 (t))]dt ≤ α0 x − x0 [1 + D(u − u0 )] |K0 (z)| ≤ αx − x0  t0

and hence (3) and (4) hold with α0 given above and δ = min{, μ}. Now, by (1), for all q ∈ N there exists zq = (xq , uq ) ∈ Ze (A) such that xq − x0  < δ,

1 xq − x0  < , q

1 J(zq ) − J(z0 ) < D(uq − u0 ). q

(5)

Observe that the last inequality implies that uq (t) = u0 (t) on a set of positive measure and so D(uq − u0 ) > 0 (q ∈ N). Since J  (z0 ; w) = 0 for all w ∈ Z, it follows by (2), (3) and (4) that J(zq ) − J(z0 ) = K0 (zq ) + E˜0(zq ) ≥ −α0 xq − x0  + D(uq − u0 )(h − α0 xq − x0 ). By (5) we obtain

1 α0 α0 D(uq − u0 ) h − − < q q q and consequently D(uq − u0 ) → 0, q → ∞. Define dq , wq , yq , and vq as in Lemma 3.1. By Lemma 3.1(a) there exist v0 ∈ L2 (T ; Rm ) and some subsequence of {zq } (without relabelling) such that {vq } converges weakly in L1 (T ; Rm ) to v0 . By Taylor’s theorem, for all q ∈ N, y˙ q (t) = Aq (t)yq (t) + Bq (t)vq (t) a.e. in T

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where

 Aq (t) =

1 0

 Bq (t) =

1

0

fx (t, x0 (t) + λ[xq (t) − x0 (t)], u0 (t))dλ, fu (t, xq (t), uq (t) + λ[u0 (t) − uq (t)])dλ.

By continuity of fx there exists m0 > 0 such that Aq ∞ ≤ m0 (q ∈ N). Moreover, for all t ∈ T and q ∈ N,  1 |Bq (t)| ≤ |fu (t, xq (t), uq (t) + λ[u0 (t) − uq (t)])|dλ ≤ ψ(t, xq (t)). 0

Since ψ is continuous, there exists m1 > 0 such that Bq ∞ ≤ m1 (q ∈ N). By Lemma 3.1(b), there exist σ0 ∈ L2 (T ; Rn ) and a subsequence of {zq } (we do not relabel) such that, if  y0 (t) :=

t t0

σ0 (s)ds (t ∈ T ),

then yq (t) → y0 (t) uniformly on T . The theorem will be proved if we show that J  (z0 ; (y0, v0 )) ≤ 0, (y0 , v0 ) ∈ Y (z0 ), and (y0 , v0 ) = (0, 0). The fact that y0 (t0 ) = y0 (t1 ) = 0 follows by Lemma 3.1(b). Now, by definition of the functional K0 , for all q ∈ N,  t1   M0 (t, xq (t)) N0 (t, xq (t)) K0 (zq ) = + , vq (t) dt. d2q d2q dq t0 In view of Lemma 3.1(b), M0 (t, xq (t)) 1 = y0(t), F0xx (t, x0 (t), u0 (t))y0 (t), 2 q→∞ dq 2 lim

N0 (t, xq (t)) = F0ux (t, x0 (t), u0(t))y0 (t) q→∞ dq lim

both uniformly on T and, since {vq } converges weakly to v0 in L1 (T ; Rm ), K0 (zq ) 1 1  J (z0 ; (y0 , v0 )) = lim + q→∞ 2 d2q 2



t1

t0

v0 (t), F0uu (t, x0 (t), u0(t))v0 (t)dt. (6)

Let us now show that, for some subsequence of {zq } (we do not relabel),  E˜0 (zq ) 1 t1 lim inf ≥ v0 (t), F0uu (t, x0 (t), u0(t))v0 (t)dt. q→∞ d2q 2 t0

(7)

Sufficient conditions for isoperimetric control problems

317

By Lemma 3.1(c), there exist some subsequence of {zq } (without relabelling) and S ⊂ T measurable such that uq (t) → u0 (t) uniformly on S. By Taylor’s theorem, for all t ∈ S and all q ∈ N, we have 1 1 E0 (t, xq (t), u0 (t), uq (t)) = vq (t), Rq (t)vq (t) 2 dq 2 where  Rq (t) := 2

0

1

(1 − λ)F0uu (t, xq (t), u0 (t) + λ[uq (t) − u0 (t)])dλ.

Clearly, lim Rq (t) = R0 (t) := F0uu (t, x0 (t), u0(t)) uniformly on S.

q→∞

By assumption (i) of the theorem, R0 (t) ≥ 0 (t ∈ S). Moreover, by assumption (iii) and by Lemma 3.1(d), for some subsequence of {zq } (without relabelling),  1 E˜0 (zq ) ≥ v0 (t), F0uu (t, x0 (t), u0 (t))v0 (t)dt. lim inf q→∞ d2q 2 S Since S can be chosen to differ from T by a set of an arbitrary small measure and since the function t → v0 (t), F0uu (t, x0 (t), u0 (t))v0 (t) belongs to L1 (T ; R), this inequality holds when S = T . This establishes (7). Now, by (5), (6) and (7), we obtain E˜0 (zq ) 1  K0 (zq ) J(zq ) − J(z0 ) J (z0 ; (y0, v0 )) ≤ lim + lim inf = lim inf ≤ 0. 2 2 q→∞ q→∞ q→∞ 2 dq dq d2q In addition, if (y0 , v0 ) = (0, 0), then lim

q→∞

K0 (zq ) =0 d2q

and so, by (3), E˜0 (zq ) 1 h ≤ lim inf ≤0 q→∞ 2 d2q contradicting the positivity of h. To show that (y0 , v0 ) ∈ Y (z0 ), we know by Lemma 3.1(c) that there exists S ⊂ T measurable such that Aq (t) → A0 (t) := fx (t, x0 (t), u0 (t)) and Bq (t) → B0 (t) := fu (t, x0 (t), u0 (t))

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both uniformly on S. Since yq (t) → y0 (t) uniformly on S and {vq } converges weakly to v0 in L1 (S; Rm ), it follows that {y˙ q } converges weakly in L1 (S; Rn ) to A0 y0 + B0 v0 . By Lemma 3.1(b), {y˙q } converges weakly in L1 (S; Rn ) to σ0 = y˙0 . Hence, y˙0 (t) = A0 (t)y0 (t) + B0 (t)v0 (t) (t ∈ S). Since S can be chosen to differ from T by a set of an arbitrary small measure, there cannot exist a subset of T of positive measure on which the functions y0 and v0 do not satisfy the differential equation y˙ 0 (t) = A0 (t)y0 (t) + B0 (t)v0 (t). Consequently, y˙ 0 (t) = A0 (t)y0 (t) + B0 (t)v0 (t) a.e. in T. It remains to show that Ii (z0 ; (y0 , v0 )) = 0 (i = 1, . . . , r). To do so, observe that for all z = (x, u) ∈ Ze (A) and all i = 1, . . . , r, Ii (z) = Ii (z0 ) + Ii (z0 ; z − z0 ) + Ki (z) + E˜i(z) 

where E˜i (x, u) :=  Ki (x, u) :=

t1

t0

t1

t0

(8)

Ei (t, x(t), u0 (t), u(t))dt,

{Mi (t, x(t)) + u(t) − u0 (t), Ni (t, x(t))}dt,

and the functions Mi and Ni are given by Mi (t, y) := Li (t, y, u0(t)) − Li (t, x0 (t), u0(t)) − Lix (t, x0 (t), u0 (t))(y − x0 (t)), Ni (t, y) := L∗iu (t, y, u0(t)) − L∗iu (t, x0 (t), u0 (t)). By Taylor’s theorem we have 1 Mi (t, y) = y − x0 (t), Pi(t, y)(y − x0 (t)), 2

Ni (t, y) = Qi (t, y)(y − x0 (t)),

where  Pi (t, y) := 2

0

1

(1 − λ)Lixx (t, x0 (t) + λ(y − x0 (t)), u0 (t))dλ,

 Qi (t, y) :=

0

1

Liux (t, x0 (t) + λ(y − x0 (t)), u0 (t))dλ.

Clearly, for all i = 0, 1, . . . , r, Mi (t, xq (t)) = 0 and q→∞ dq lim

lim Ni (t, xq (t)) = 0

q→∞

Sufficient conditions for isoperimetric control problems

319

all uniformly on T . Thus, since {vq } converges weakly to v0 in L1 (T ; Rm ),  t1  t1 Ki (zq ) Mi (t, xq (t)) = lim dt + lim vq (t), Ni (t, xq (t))dt = 0 (9) lim q→∞ q→∞ t q→∞ t dq dq 0 0 for all i = 0, 1, . . . , r. Since J  (z0 ; w) = 0 for all w ∈ Z, by (2), (5), and (9), 0 ≥ lim sup q→∞

E˜0 (zq ) J(zq ) − J(z0 ) = lim sup . dq dq q→∞

By the first part of (iii) of the theorem, E˜0 (zq ) ≥ 0 (q ∈ N), and so E˜0 (zq ) = 0. q→∞ dq lim

By the second part of (iii), E˜i (zq ) = 0 (i = 1, . . . , r). q→∞ dq lim

(10)

Since Ii (zq ) = Ii (z0 ) = 0 (i = 1, . . . , r), we have by (8), (9) and (10), 0 = lim Ii (z0 ; (yq , vq )) (i = 1, . . . , r). q→∞

Since yq (t) → y0 (t) uniformly on T and vq → v0 weakly in L1 (T ; Rm ), 0 = lim Ii (z0 ; (yq , vq )) = Ii (z0 ; (y0, v0 )) (i = 1, . . . , r) q→∞

and this completes the proof. 4. Examples In this section we provide three examples of problems with isoperimetric constraints for which a singular proper strong minimum is exhibited. The first two examples, which deal with an admissibility set having an isolated point, illustrate how Theorem 2.1 can be applied in a straightforward way. The third example is of a different nature. For this example, the space of mixed constraints corresponds to the whole space and, though one of the sufficient conditions of the theorem is not fulfilled, the proof of the theorem is used to establish the desired result for the specific singular process under consideration. 4.1 Example: Let a > 0, b > 1 and consider the problem of minimizing  b I(x, u) = {x2 (t) + tx(t)u(t) − x(t)}dt 1

subject to

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a. x(t) ˙ = sin(x(t)u(t)) a.e. in [1, b]. b. x(1) = x(b) = 1. b c. I1 (x, u) = 1 − b + 1 {1 + sin2 (x(t)u(t))}1/2 dt = 0. d. (t, x(t), u(t)) ∈ [1, b] × R × Ba (t ∈ [1, b]) where Ba := {0} ∪ [a, ∞). For this case n = m = r = 1, T = [1, b], A = T × R × Ba , ξ0 = ξ1 = 1, α1 = 1 − b, L(t, x, u) = x2 +txu−x,

L1 (t, x, u) = {1+sin2 (xu)}1/2 ,

f (t, x, u) = sin(xu).

Note first that the assumptions on L, L1 , and f are satisfied. In particular, since fu (t, x, u) = x cos(xu), there exists a continuous function ψ: T × R → R such that |fu (t, x, u)| ≤ ψ(t, x) for all (t, x, u) ∈ T × R × R. For this problem we have H(t, x, u, p) = p sin(xu) − x2 − txu + x − λ1 [1 + sin2 (xu)]1/2 , Hx (t, x, u, p) = pu cos(xu) − 2x − tu + 1 − λ1 u sin(xu) cos(xu)[1 + sin2 (xu)]−1/2 , Hu (t, x, u, p) = px cos(xu) − tx − λ1 x sin(xu) cos(xu)[1 + sin2 (xu)]−1/2 . Clearly (x0 , u0) ≡ (1, 0) is admissible and, in view of the above relations, (x0 , u0 , p) with p(t) = t (t ∈ T ) and λ1 = 0 is an extremal. Now, since F0 (t, x, u) = x2 − 2x + txu − t sin(xu) we have

F0uu (t, x, u) = tx2 sin(xu)

and hence F0uu (t, x0 (t), u0 (t)) = 0 (t ∈ T ) so that (x0 , u0 ) is singular and condition 2.1(i) holds. Note that (y, v) ∈ Y (x0 , u0 ) implies that y(t) ˙ = v(t) a.e. in T . Hence  b J  ((x0 , u0); (y, v)) = 2 y 2(t)dt > 0 1

for all (y, v) ∈ Y (x0 , u0), (y, v) = (0, 0) which corresponds to condition 2.1(ii). Now, the Weierstrass excess function of F0 is given by E0 (t, x, u, v) = −t sin(xv) + t sin(xu) + txv cos(xu) − txu cos(xu). Therefore for all x ≥ 1/2 and some ha > 0 sufficiently small, E0 (t, x, 0, u) = t(xu − sin(xu)) ≥ xu − sin(xu) ≥ ha |xu| ≥

ha ha |u| ≥ ϕ(u) 2 2

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Sufficient conditions for isoperimetric control problems

for all u ∈ Ba . Hence, the first part of 2.1(iii) holds with h = ha and 0 <  < 1/2. Finally, to show that the second part of 2.1(iii) holds observe that, for all (x, u) ∈ Ze (T0 (x0 ; ) ∩ A), 



b

b

E0 (t, x(t), 0, u(t))dt = t[x(t)u(t) − sin(x(t)u(t))]dt 1 1  b  b  b ≥ [x(t)u(t) − sin(x(t)u(t))]dt = x(t)u(t)dt ≥ | sin(x(t)u(t))|dt 1 1 1  b  b    ϕ(sin(x(t)u(t)))dt ≥ ha  E1 (t, x(t), 0, u(t))dt. ≥ 1

1

By Theorem 2.1, (x0 , u0) is a proper strong minimum of the problem in hand. 4.2 Example: Let a > 0 and consider the problem of minimizing  1 I(x, u) = {u4 (t) − tx(t)u(t) − x4 (t)}dt 0

subject to a. x(t) ˙ = tx(t) cos u(t) + sin u(t) a.e. in [0, 1]. b. x(0) = x(1) = 0. 1 c. I1 (x, u) = 0 {tx5 (t) + tx(t)u(t)}dt = 0. d. (t, x(t), u(t)) ∈ [0, 1] × R × Ba (t ∈ [0, 1]) where Ba := (−∞, −a] ∪ {0} ∪ [a, ∞). For this case n = m = r = 1, T = [0, 1], A = T × R × Ba , ξ0 = ξ1 = 0, α1 = 0, L(t, x, u) = u4 −txu−x4 ,

L1 (t, x, u) = tx5 +txu,

f (t, x, u) = tx cos u+sin u.

As in the previous example, the assumptions on L, L1 , and f are clearly satisfied. In particular, since fu (t, x, u) = −tx sin u + cos u, there exists a continuous function ψ: T × R → R such that |fu (t, x, u)| ≤ ψ(t, x) for all (t, x, u) ∈ T × R × R. For this problem we have H(t, x, u, p) = ptx cos u + p sin u − u4 + txu + x4 − λ1 tx5 − λ1 txu, Hx (t, x, u, p) = pt cos u + tu + 4x3 − 5λ1 tx4 − λ1 tu, Hu (t, x, u, p) = −ptx sin u + p cos u − 4u3 + tx − λ1 tx.

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Let (x0 , u0 ) ≡ (0, 0) which is admissible. Moreover, (x0 , u0, p) with p ≡ 0 and λ1 = 0 is an extremal. Now, since F0 (t, x, u) = u4 − txu − x4 we have F0uu (t, x0 (t), u0(t)) = 0 (t ∈ T ) so that (x0 , u0) is a singular process and 2.1(i) holds. ˙ = ty(t) + v(t) a.e. in T . Note that (y, v) ∈ Y (x0 , u0 ) implies that y(t) Hence  1  1  2ty(t)v(t)dt = − 2ty(t)[y(t) ˙ − ty(t)]dt J ((x0 , u0 ); (y, v)) = − 0 0  1  1 ≥ − 2ty(t)y(t)dt ˙ = y 2(t)dt > 0 0

0

for all (y, v) ∈ Y (x0 , u0 ), (y, v) = (0, 0), and 2.1(ii) holds. Finally, for some sufficiently small ha > 0, E0 (t, x, 0, u) = u4 ≥ ha |u| ≥ ha ϕ(u) for all u ∈ Ba and E1 (t, x, u, v) = 0 for all (t, x, u, v) ∈ T × R × R × R. Hence 2.1(iii) holds with h = ha and any  > 0. By Theorem 2.1, (x0 , u0) is a proper strong minimum. 4.3 Example: Consider the same problem of Example 4.2 except for the admissibility set now given by the whole space, that is, A = T × R × R. Proceeding as in the previous problem, it follows that conditions 2.1(i), 2.1(ii) and the second part of 2.1(iii) hold. However, since E0 (t, x, 0, u) = u4 , the first part of 2.1(iii) fails to hold since there do not exist h,  > 0 such that, for all (x, u) ∈ Ze (T0 (x0 ; )), E0 (t, x(t), 0, u(t)) ≥ hϕ(u(t)) a.e. in T. Nevertheless, as we shall show next, the proof of the theorem can be used to prove that (x0 , u0) ≡ (0, 0) is indeed a proper strong minimum of the problem in hand. We proceed essentially as in the proof of the theorem. Let us suppose the contrary. Thus, for all q ∈ N, there exists zq := (xq , uq ) in Ze (T0 (x0 ; 1/q)) with zq = (x0 , u0) such that I(xq , uq ) ≤ I(x0 , u0 ) = 0. Observe first that, in view of the above inequality and the fact that zq ∈ Ze (T0 (x0 ; 1/q)),  1  1  1 4 4 uq (t)dt ≤ {txq (t)uq (t) + xq (t)}dt = {−tx5q (t) + x4q (t)}dt 0

0

0

Sufficient conditions for isoperimetric control problems

323

which tends to zero as q → ∞. Hence uq 4 → 0, q → ∞, and therefore D(uq ) → 0, q → ∞. Since zq ∈ Ze (A) and zq = (x0 , u0 ), we have uq = u0 and so D(uq ) > 0 (q ∈ N). For all q ∈ N define dq , wq , yq and vq as in Lemma 3.1 and let v0 and y0 be any pair of functions given by Lemma 3.1(a) and 3.1(b). Following the proof of Theorem 2.1, one readily verifies that (y0 , v0 ) ∈ Y (x0 , u0 ) and, moreover, since F0uu (t, x0 (t), u0 (t)) = 0 (t ∈ T ) and E0 (t, x(t), u0 (t), u(t)) = u4 (t) ≥ 0 for all t ∈ T and (x, u) ∈ Ze (A), we have J  ((x0 , u0 ); (y0, v0 )) ≤ 0. Since 2.1(ii) is satisfied, this implies that (y0 , v0 ) = (0, 0). Let us show that this yields a contradiction. Indeed, since yq (t) → y0 (t) uniformly on T , the relations I(zq ) ≤ 0 and I1 (zq ) = 0 imply that  1  1  1  1 4  1 vq (t) txq (t)uq (t) 4 4 dt+ yq (t)dt = − txq (t)yq (t)dt+ yq4(t)dt dt ≤ 4 (t) 4 w d 0 0 0 0 0 q q which tends to zero as q → ∞. Hence  lim

q→∞

0

1

vq2 (t) dt = 0. wq2 (t)

But this contradicts the fact that  1 2 vq (t) dt = 1 for all q ∈ N 2 0 wq (t) and the claim is proved. Therefore (x0 , u0 ) ≡ (0, 0) is a proper strong minimum. References [1] Hestenes MR (1934) Sufficient conditions for the problem of Bolza in the calculus of variations, Transactions of the American Mathematical Society, 36: 793-818 [2] Hestenes MR (1936) On sufficient conditions in the problems of Lagrange and Bolza, The Annals of Mathematics, 37: 543-551 [3] Hestenes MR (1937) A direct sufficiency proof for the problem of Bolza in the calculus of variations, Transactions of the American Mathematical Society, 42: 141-154 [4] Hestenes MR (1946) Sufficient conditions for the isoperimetric problem of Bolza in the calculus of variations, Transactions of the American Mathematical Society, 60: 93-118

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[5] Hestenes MR (1947) The Weierstrass E-function in the calculus of variations, Transactions of the American Mathematical Society, 60: 51-71 [6] Hestenes MR (1947) An indirect sufficiency proof for the problem of Bolza in nonparametric form, Transactions of the American Mathematical Society, 62: 509-535 [7] Hestenes MR (1948) Sufficient conditions for multiple integral problems in the calculus of variations, American Journal of Mathematics, 70: 239276 [8] Hestenes MR (1966) Calculus of Variations and Optimal Control Theory, John Wiley & Sons, New York [9] Maurer H, Zowe J (1979) First and second order necessary and sufficient optimality conditions for infinite-dimensional programming problems, Mathematical Programming, 16: 98-110 [10] Maurer H (1981) First and second order sufficient optimality conditions in mathematical programming and optimal control, Mathematical Programming Study, 14: 163-177 [11] Maurer H, Pickenhain S (1995) Second order sufficient conditions for control problems with mixed control-state constraints, Journal of Optimization Theory and Applications, 86: 649-667 [12] McShane EJ (1942) Sufficient conditions for a weak relative minimum in the problem of Bolza, Transactions of the American Mathematical Society, 52: 344-379 [13] Milyutin AA, Osmolovskiˇı (1998) Calculus of Variations and Optimal Control, Translations of Mathematical Monographs 180, American Mathematical Society, Providence, Rhode Island [14] Rosenblueth JF (1999) Variational conditions and conjugate points for the fixed-endpoint control problem, IMA Journal of Mathematical Control & Information, 16: 147-163 [15] Rosenblueth JF, S´anchez Licea G (submitted) Sufficiency and singularity in optimal control, SIAM Journal on Control & Optimization Received: August, 2010