Sign of the solutions of linear fractional differential equations and some applications
aa r X i v : . [ m a t h . C A ] F e b SIGN OF THE SOLUTIONS OF LINEAR FRACTIONALDIFFERENTIAL EQUATIONS AND SOME APPLICATIONS
RUI A. C. FERREIRA
Abstract.
In this work we wish to highlight some consequences of a recentresult proved in [N. D. Cong and H. T. Tuan, Generation of nonlocal fractionaldynamical systems by fractional differential equations, J. Integral EquationsAppl. (2017), no. 4, 585–608]. Particular emphasis will be given to itsapplication on fractional variational problems of Herglotz type. Preamble
Very recently, in 2017, Cong et. al. [7] proved the following result (we refer thereader to Section 2 for the definitions appearing below):
Theorem 1.1.
Consider the following fractional differential equation (1.1) C D αa + [ x ]( t ) = f ( t, x ( t )) , < α ≤ , where f : [ a, ∞ ) × R → R is a continuous function satisfying the Lipschitz condition (1.2) | f ( t, x ) − f ( t, y ) | ≤ L ( t ) | x − y | , t ∈ [ a, ∞ ) , x, y ∈ R , L ∈ C ([ a, ∞ ) , R ) . Then, for any two different initial values x a = x a in R , the solutions x and x of (1.1) starting from x a = x ( a ) and x a = x ( a ) verify x ( t ) = x ( t ) for all t ∈ [ a, ∞ ) .Remark . The existence and uniqueness of (continuous) solutions for (1.1) withthe given initial conditions is guaranteed by, e.g., [3, Theorem 2].
Remark . Theorem 1.1 was conjectured in 2008 by Diethelm [9] and solvedpartially therein. However, it was only in 2017 that a complete and correct proofof it was given (see [7] for the historical developments regarding this result).In particular, for 0 < α < C D αa + [ x ]( t ) = g ( t ) x ( t ) , x ( a ) = x a > , for which f ( t, x ) = g ( t ) x (with g ∈ C ([ a, ∞ ) , R )) obviosuly satisfies (1.2). Thenwe may conclude from Theorem 1.1 that the solution of (1.3) is positive on [ a, ∞ ).We find this result of much interest and, to the best of our knowledge, it was notsufficiently highlighted in the literature so far; therefore, we shall write it in thefollowing: Mathematics Subject Classification.
Primary 26A33, 34A30; Secondary 49K30.
Key words and phrases.
Linear equation, fractional derivative, Herglotz variational problem.The author was supported by the “Funda¸c˜ao para a Ciˆencia e a Tecnologia (FCT)” throughthe program “Stimulus of Scientific Employment, Individual Support-2017 Call” with referenceCEECIND/00640/2017.
Theorem 1.4.
Let < α < and g ∈ C ([ a, ∞ ) , R ) . Then the solution of the IVP (1.3) is positive on [ a, ∞ ) . We may extract several interesting consequences of Theorem 1.4 and we will refersome in Section 3 of this work. But before we go into it we want to oberve that,following the same steps as those of the proof of Theorem 1.1, we may prove ananalogous result where in the differential equation (1.1) we use the left Riemann–Liouville fractional derivative and the initial conditions are given by I − αa + [ x ]( a ) = x a (again we refer the reader to Section 2 to understand the meaning of the symbolsused in this section). Then, we may prove the following result: Theorem 1.5.
Let < α < and g ∈ C (( a, ∞ ) , R ) . Then the solution of the IVP (1.4) D αa + [ x ]( t ) = g ( t ) x ( t ) , I − αa + [ x ]( a ) = x a > , is positive on ( a, ∞ ) . However, in this work, we will be especially interested in the analogous theoremto Theorem 1.5 but using the right fractional derivative (cf. the proof of Theorem3.11). In order to accomplish it we may appeal to the duality results (for left andright fractional operators) presented and proved in [6]. We, therefore, have:
Theorem 1.6.
Let < α < and g ∈ C (( −∞ , b ) , R ) . Then the solution of theIVP (1.5) D αb − [ x ]( t ) = g ( t ) x ( t ) , I − αb − [ x ]( b ) = x b > , is positive on ( −∞ , b ) .Remark . It is clear that, if x b <
0, then the solution in Theorem 1.6 is negativeon ( −∞ , b ).The remaining of this manuscript is organized as follows: In Section 2 we providethe reader with the definitions and results of fractional calculus needed in this work.In Section 3 we present interesting applications of Theorems 1.4 and 1.6 enunciatedabove. 2. Preliminaries on fractional calculus
Let I be an interval of R and n ∈ N . Suppose that E ( I, R n ) is a space offunctions. We denote by E loc ( I, R n ) the space of functions x : I → R n such that x ∈ E ( J, R n ) for every compact subinterval J ⊂ I .We now introduce the left and right fractional integrals and derivatives used inthis work. Definition 2.1.
Let a < b be two real numbers and [ a, b ] ⊂ I . The left and rightRiemann–Liouville fractional integrals of order α > f ∈ L loc ( I, R n )are defined, respectively by I αa + [ f ]( t ) = 1Γ( α ) Z ta ( t − s ) α − f ( s ) ds, t ≥ a, and I αb − [ f ]( t ) = 1Γ( α ) Z bt ( s − t ) α − f ( s ) ds, t ≤ b, provided that the right-hand side exists. For α = 0 we set I a + [ f ]( t ) = I b − [ f ]( t ) = f ( t ). IGN OF THE SOLUTIONS OF LINEAR FRACTIONAL DIFFERENTIAL EQUATIONS AND SOME APPLICATIONS3
Definition 2.2.
The left and right Riemann–Liouville (RL) fractional derivativesof order 0 < α ≤ f ∈ L loc ( I, R n ) such that I − αa + [ f ] ∈ AC loc ( I, R n ),respectively I − αb − [ f ] ∈ AC loc ( I, R n ), are defined by D αa + [ f ]( t ) = ddt (cid:2) I − αa + [ f ] (cid:3) ( t ) , respectively, D αb − [ f ]( t ) = − ddt (cid:2) I − αb − [ f ] (cid:3) ( t ) . We denote by AC αa + ( I, R n ), respectively AC αb − ( I, R n ), the set of all functions f ∈ L loc ( I, R n ) possessing a left, respectively right, RL fractional derivative of order0 < α ≤ Definition 2.3.
The left and right Caputo fractional derivatives of order 0 < α ≤ f ∈ C ( I, R n ) such that f − f ( a ) ∈ AC αa + ( I, R n ), respectively f − f ( b ) ∈ AC αb − ( I, R n ), are defined by C D αa + [ f ]( t ) = D αa + [ f − f ( a )]( t ) , respectively, C D αb − [ f ]( t ) = D αb − [ f − f ( b )]( t ) . We denote by C AC αa + ( I, R n ), respectively C AC αb − ( I, R n ), the set of all functions f ∈ C ( I, R n ) possessing a left, respectively right, Caputo fractional derivative oforder 0 < α ≤ E α,β ( z ) = ∞ X k =0 z k Γ( αk + β ) , α > , β ∈ C , and we put E α ( z ) = E α, ( z ). It satisfies (cf. [18]),(2.1) E α,β ( t ) ≥ < α ≤ , β ≥ α. The following formulas may be found in [15].
Lemma 2.4.
Suppose that α, β, γ are positive real numbers and λ ∈ R . Then, (2.2) I α + [ s γ − E β,γ ( λs β )]( t ) = t α + γ − E β,α + γ ( λt β ) , t ≥ , and (2.3) E α,β ( t ) = 1Γ( β ) + tE α,α + β ( t ) , t ∈ R . The following result may be consulted in, e.g., [17, Theorem 5.15].
Theorem 2.5 (Variation of constants formula) . The solution of the IVP C D αa + [ x ]( t ) = λx ( t ) + f ( t ) , λ ∈ R , < α ≤ , t ≥ a, (2.4) x ( a ) = x a , (2.5) can be given by (2.6) x ( t ) = x a E α ( λ ( t − a ) α ) + Z ta ( t − s ) α − E α,α ( λ ( t − s ) α ) f ( s ) ds. We will also use the following result:
RUI A. C. FERREIRA
Theorem 2.6.
Let f : ( −∞ , b ] → R be a continuous function. Then, the solution x ∈ C (( −∞ , b ) , R ) of the IVP D αb − [ x ]( t ) = f ( t ) x ( t ) , < α ≤ , t < b,I − αb − [ x ]( b ) = x b , can be given by (2.7) x ( t ) = x b Γ( α ) ∞ X k =0 T kf [( b − s ) α − ]( t ) , t < b, where T f [ φ ] = φ and T k +1 f [ φ ] = T f [ T kf φ ] ( k ∈ N ) , with T f being the operator definedby T f [ φ ] = I αb − [ f φ ] .Proof. We only need to invoke the duality results of [6] and apply them to [8,Theorem 2.3]. (cid:3) Applications
Two direct consequences of Theorem 1.4 and a related result.
Westart by stating a generalization of Theorem 1.4.
Theorem 3.1.
Let < α < and f ∈ C ([ a, ∞ ) × R , R ) be such that f ( t,
0) = 0 and it satisfies the Lipschitz condition (1.2) . Then the solution of the IVP C D αa + [ x ]( t ) = f ( t, x ( t )) , x ( a ) = x a > , is positive on [ a, ∞ ) .Proof. Just observe that the trivial solution x ( t ) = 0 solves the IVP C D αa [ x ]( t ) = f ( t, x ( t )) , x ( a ) = 0 on [ a, ∞ ) and apply Theorem 1.1. (cid:3) Example 3.2.
Consider f ( t, x ) = g ( t ) ln( x + 1) with g ∈ C ( R +0 , R ) and x ∈ R .We have, | f ( t, x ) − f ( t, y ) | = | g ( t ) || ln( x + γ ) − ln( y + γ ) | ≤ | g ( t ) || x − y | , x, y ∈ R , where we have used the mean value theorem. Since f ( t,
0) = 0, it follows fromTheorem 3.1 that the solution of C D α + [ x ]( t ) = g ( t ) ln( x ( t ) + 1) , x ( a ) = 1 , t ≥ , is positive.The following result seems to be new in the literature. Theorem 3.3.
For g ∈ C ([ a, ∞ ] , R ) and < α < define k ( t, s ) = 1Γ( α ) ( t − s ) α − g ( s ) , and the j th iterated kernel k j for j = 1 , , . . . via the recurrence relation k ( t, s ) = k ( t, s ) , k j ( t, s ) = Z ts k ( t, τ ) k j − ( τ, s ) dτ, j = 2 , , . . . . Then the function x ( t ) = x a (cid:18) Z ta R ( t, s ) ds (cid:19) , t ∈ [ a, ∞ ) , where R ( t, s ) = P ∞ j =1 k j ( t, s ) is positive. IGN OF THE SOLUTIONS OF LINEAR FRACTIONAL DIFFERENTIAL EQUATIONS AND SOME APPLICATIONS5
Proof.
This result follows from the representation for the solution of (1.3) given in[10, Theorem 7.10]. (cid:3)
Remark . Observe that the previous result is by no means obvious as the function g may be negative.We end this section with a lateral but nevertheless interesting result, namely,we deduce a Bernoulli-type inequality using the theory of fractional calculus . Theproof follows the same lines as the one in [11] (see also [12]), which was done usingdiscrete fractional operators. Theorem 3.5 (Fractional Bernoulli’s inequality) . Let λ ∈ R and < α ≤ . Thenthe following inequality holds: (3.1) E α ( λt α ) ≥ λt α Γ( α + 1) + 1 , t ≥ . Proof.
The formula trivially holds for t = 0. Let x ( t ) = λ t α Γ( α + 1) , t ≥ . We have x (0) = 0 and C D αa x ( t ) = λ (cf. [17, Property 2.1]). Therefore, λx ( t ) + λ = λ t α Γ( α + 1) + λ ≥ C D α x ( t ) . Define m ( t ) = λx ( t ) + λ − C D α x ( t ), which is nonnegative. Then, using Theorem2.5, we get x ( t ) = Z t ( t − s ) α − E α,α ( λ ( t − s ) α )[ λ − m ( s )] ds = λ Z t ( t − s ) α − E α,α ( λ ( t − s ) α ) ds − Z t ( t − s ) α − E α,α ( λ ( t − s ) α ) m ( s ) ds ≤ λ Z t ( t − s ) α − E α,α ( λ ( t − s ) α ) ds = λt α E α,α +1 ( λt α ) , where we have used (2.1) and (2.2). Therefore, λ t α Γ( α + 1) ≤ λt α E α,α +1 ( λt α ) , which for t > λ Γ( α + 1) ≤ λE α,α +1 ( λt α ) , and upon using (2.3) and some rearrangements furnishes (3.1). (cid:3) Remark . It is clear that, for α = 1, inequality (3.1) reads as e λt ≥ λt + 1. Thisinequality is the continuous version of the Bernoulli inequality (cf. [1]), hence thename given in Theorem 3.5. We have never seen such type of result in the literature. That is, (1 + x ) n ≥ nx for x > − n ∈ N . RUI A. C. FERREIRA
Herglotz’s variational problem.
In this section we will deduce necessaryoptimality conditions for a fractional variational problem of Herglotz type [14, 16]and, in particular, show the usefulness of Theorem 1.6. Let us first state what wemean here by the Herglotz variational problem in the classical case: Following [19,Problem P H ], we consider: z ( b ) −→ minsubject to ˙ z ( t ) = L ( t, x ( t ) , ˙ x ( t ) , z ( t )) , t ∈ [ a, b ] , (3.2) x ( a ) = x a , z ( a ) = z a , x a , z a ∈ R . The aim is to find a couple ( x, z ) in an appropriate space of functions that solves(3.2), i.e. that satisfy all the conditions in (3.2).In our work we will consider the following fractional version of (3.2): z ( b ) −→ minsubject to C D αa [ z ]( t ) = L ( t, x ( t ) , C D αa [ x ]( t ) , z ( t )) , t ∈ [ a, b ] , (3.3) x ( a ) = x a , z ( a ) = z a , x a , z a ∈ R . Before proceeding let us just note that, if the function L does not depend on itsfourth variable, then z is immediately determined and we may write (3.3) as1Γ( α ) Z ba ( b − t ) α − (cid:20) L ( t, x ( t ) , C D αa [ x ]( t )) + z a ( b − t ) − α b − a (cid:21) dt −→ minsubject to x ( a ) = x a , x a ∈ R , in view of I αa + [ C D αa [ z ]]( t ) = z ( t ) − z ( a ). So, if we defineˆ L ( t, x, v ) = L ( t, x, v ) + z a ( b − t ) − α b − a , we get the problem L ( x ) = 1Γ( α ) Z ba ( b − t ) α − ˆ L ( t, x ( t ) , C D αa [ x ]( t )) dt −→ minsubject to x ( a ) = x a , x a ∈ R . This is the basic problem of the fractional calculus of variational with fixed initialcondition. The same problem with fixed initial and final conditions was recentlystudied in [13].To the best of our knowledge the first work considering Herglotz-type problemsinvolving fractional derivatives is [2]. However, the problem we consider here isdifferent in nature from the one considered in [2] (when 0 < α <
1) as the authorsconsidered the differential equation˙ z ( t ) = L ( t, x ( t ) , C D αa [ x ]( t ) , z ( t )) , instead of C D αa [ z ]( t ) = L ( t, x ( t ) , C D αa [ x ]( t ) , z ( t )) considered above in (3.3). More-over, the proofs of the necessary optimality conditions are quite different.In this work we will obtain first and second order necessary otimality conditionsfor (3.3) by using a recent result of [4], namely, the fractional version of the cel-ebrated Pontryagin Maximum Principle (PMP). For the benefit of the reader werecall here the main result of [4]. IGN OF THE SOLUTIONS OF LINEAR FRACTIONAL DIFFERENTIAL EQUATIONS AND SOME APPLICATIONS7
Consider the Optimal Control Problem (OCP) of Bolza type given by ϕ ( x ( a ) , x ( b )) + I αa + [ F ( · , x, u )]( b ) −→ minsubject to x ∈ C AC αa + ([ a, b ] , R n ) , u ∈ L ∞ ([ a, b ] , R m ) , C D αa [ x ]( t ) = f ( t, x ( t ) , u ( t )) a.e. t ∈ [ a, b ] ,g ( x ( a ) , x ( b )) ∈ C,u ( t ) ∈ U a.e. t ∈ [ a, b ] . A couple ( x ∗ , u ∗ ) is said to be an optimal solution to OCP if it satisfies all theabove constraints and it minimizes the cost among all couples ( x, u ) satisfying thoseconstraints. Obviously, the functions involved in the OCP satisfy some regularityconditions, that we will skip here and refer the reader to [4, Page 8]. Under thesehypothesis (regularity conditions), we have the following theorem. Theorem 3.7 (PMP) . Assume that ( x ∗ , u ∗ ) ∈ C AC αa + ([ a, b ] , R n ) × L ∞ ([ a, b ] , R m ) is an optimal solution to the OCP. Then, there exists a nontrivial couple ( p, p ) ,where p ∈ AC αb − ([ a, b ] , R n ) (called adjoint vector) and p ≤ , such that the followingconditions hold:(i) Fractional Hamiltonian system: C D αa [ x ∗ ]( t ) = ∂ H ( t, x ∗ ( t ) , u ∗ ( t ) , p ( t ) , p ) D αb − [ p ]( t ) = ∂ H ( t, x ∗ ( t ) , u ∗ ( t ) , p ( t ) , p ) , for almost every t ∈ [ a, b ] , where the Hamiltonian H : [ a, b ) × R n × R m × R n × R → R associated to Problem (OCP) is defined by H ( t, x, u, p, p ) = h p, f ( t, x, u ) i R n + p ( b − t ) α − Γ( α ) F ( t, x, u ) . (ii) Hamiltonian maximization condition: u ∗ ( t ) = arg max u ∈ U H ( t, x ∗ ( t ) , u, p ( t ) , p ) a.e. t ∈ [ a, b ] . (iii) Transversality conditions on the adjoint vector: if in addition g is submer-sive at ( x ∗ ( a ) , x ∗ ( b ) , then the couple ( p, p ) satisfy I − αb − [ p ]( a ) = − p ∂ ϕ ( x ∗ ( a ) , x ∗ ( b )) − ∂ g ( x ∗ ( a ) , x ∗ ( b )) T × Ψ ,I − αb − [ p ]( b ) = p ∂ ϕ ( x ∗ ( a ) , x ∗ ( b )) + ∂ g ( x ∗ ( a ) , x ∗ ( b )) T × Ψ , where Ψ ∈ N C [ g ( x ∗ ( a ) , x ∗ ( b ))] , with N C [ x ] = { z ∈ R j : ∀ x ′ ∈ C, h z, x ′ − x i R j } ≤ } . A series of remarks is in order.
Remark . If U = R , i.e., there is no control constraint in the OCP, and theHamiltonian is differentiable with respect to its third variable, then the maximiza-tion condition (ii) in Theorem 3.7 implies (cf. [4, Remark 3.18]) ∂ H ( t, x ∗ ( t ) , u ∗ ( t ) , p ( t ) , p ) = 0 a.e. t ∈ [ a, b ] . In the notation of [4] we considering here β = α . A function g : R n × R n → R j is said to be submersive at a point ( x a , x b ) ∈ R n × R n if itsdifferential at this point is surjective. RUI A. C. FERREIRA
Moreover, if H is twice differentiable with respect to its third variable, we easillysee that(3.4) ∂ H ( t, x ∗ ( t ) , u ∗ ( t ) , p ( t ) , p ) ≤ a.e. t ∈ [ a, b ] . Remark . If the initial point is fixed and if the final point is free in the OCP,then we may take (cf. [4, Remark 3.17]) I − αb − [ p ]( b ) = − ∂ ϕ ( x ∗ ( a ) , x ∗ ( b )) . Remark . It is mentioned in [4, pag. 15] and shown in [5, Theorem 5.3] that p ∈ ( C [ a, b ) , R n ).It follows the main result of this section: Theorem 3.11.
Consider the function L ( t, x, u, z ) in (3.3) to have continuous par-tial derivatives with respect to x, u and z . Suppose that ( x ⋆ , z ⋆ ) ∈ C AC αa + ([ a, b ] , R ) × C AC αa + ([ a, b ] , R ) , with C D αa [ x ⋆ ] ∈ C ([ a, b ] , R ) , solves the Herglotz problem (3.3) .Then (3.5) I αb − [ p∂ L ( · , x ⋆ , C D αa [ x ⋆ ] , z ⋆ )]( t ) + p ( t ) ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )) = 0 , for all t ∈ [ a, b ) , where p is the solution of D αb − [ p ]( t ) = p ( t ) ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )) , t ∈ [ a, b ) , I − αb − [ p ]( b ) = − . Moreover, if L is twice continuously differentiable with respect to u , then the Le-gendre necessary optimality condition holds: (3.6) ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )) ≥ , t ∈ [ a, b ] . Proof.
Suppose that ( x ⋆ , z ⋆ ) is a solution of (3.3). Then, by letting x ⋆ ( t ) = x ⋆ ( t ), x ⋆ ( t ) = z ⋆ ( t ) and u ⋆ ( t ) = C D αa [ x ⋆ ]( t ), we conclude that ( x ⋆ , x ⋆ , u ⋆ ) solves thefollowing OCP x ⋆ ( b ) −→ minsubject to C D αa [ x ⋆ ]( t ) = u ⋆ ( t ) , t ∈ [ a, b ] , C D αa [ x ⋆ ]( t ) = L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) , t ∈ [ a, b ] ,x ⋆ ( a ) = x a , x ⋆ ( a ) = z a , x a , z a ∈ R . It follows from Theorem 3.7 and Remarks 3.8, 3.9 and 3.10 the existence of a vector( p , p ) ∈ C ([ a, b ) , R ) satisfying(3.7) p ( t ) + p ( t ) ∂ L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) = 0 , a.e. t ∈ [ a, b ] , and D αb − [ p ]( t ) = p ( t ) ∂ L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) , a.e. t ∈ [ a, b ] , I − αb − [ p ]( b ) = 0 , (3.8) D αb − [ p ]( t ) = p ( t ) ∂ L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) , a.e. t ∈ [ a, b ] , I − αb − [ p ]( b ) = − . (3.9)Observe that the continuity of p and p on [ a, b ), together with the assumptionson L and ( x ⋆ , u ⋆ ), imply that (3.7), (3.8) and (3.9) hold on [ a, b ). Moreover, since p ( t ) = I αb − [ p ∂ L ( · , x ⋆ , u ⋆ , x ⋆ )]( t ), then (3.5) follows from immediately (3.7). IGN OF THE SOLUTIONS OF LINEAR FRACTIONAL DIFFERENTIAL EQUATIONS AND SOME APPLICATIONS9
Suppose now that L is twice continuously differentiable with respect to u . Then,by (3.4), we get(3.10) p ( t ) ∂ L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) ≤ , a.e. t ∈ [ a, b ] , and, upon using Theorem 1.6, Remark 1.7 and the continuity of p on [ a, b ), ∂ L ( t, x ⋆ ( t ) , u ⋆ ( t ) , x ⋆ ( t )) ≥ , a.e. t ∈ [ a, b ] . The previous inequality holds on [ a, b ] from the hypothesis on L and ( x ⋆ , u ⋆ ). Theproof is done. (cid:3) We call (3.5) the
Euler–Lagrange equation in integral form for the Herglotz vari-ational problem (3.3).
Remark . The function p of Theorem 3.11 has the representation (cf. Theorem2.6): p ( t ) = − α ) ∞ X k =0 T kf [( b − s ) α − ]( t ) , t < b, where f ( t ) = ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )). Remark . We emphasize the importance of Theorem 1.6 in order to obtainthe Legendre necessary condition (3.6). Because of it we were able to remove thedependence on the function p in the inequality (3.10).Also, suppose that L ( t, x, u, z ) does not depend on x . Then, the Euler–Lagrangeequation (3.5) becomes p ( t ) ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )) = 0 , t ∈ [ a, b ) . Again, we may use Theorem 1.6 and the continuity of the involved functions toconclude that ∂ L ( t, x ⋆ ( t ) , C D αa [ x ⋆ ]( t ) , z ⋆ ( t )) = 0 , t ∈ [ a, b ] . If we let α = 1 in Theorem 3.11, it follows the following: Corollary 3.14.
The first and second order optimality conditions for the varia-tional problem given by (3.2) are, respectively,(3.11) Z bt e R bs ∂ L [ τ ] dτ ∂ L [ s ] ds + e R bt ∂ L [ s ] ds ∂ L [ t ] = 0 , t ∈ [ a, b ] , where [ s ] = ( s, x ⋆ ( s ) , ˙ x ⋆ ( s ) , z ⋆ ( s )), and ∂ L ( t, x ⋆ ( t ) , ˙ x ∗ ( t ) , z ⋆ ( t )) ≥ , t ∈ [ a, b ] . Proof.
Just let α = 1 in Theorem 3.11 and note that, in this case, p ( t ) = − e R bt ∂ L [ s ] ds ,for all t ∈ [ a, b ]. (cid:3) We end this work by noting that we can obtain a differential form for equa-tion (3.11). Indeed, since the integral on the left hand side of (3.11) and f ( t ) = e R bt ∂ L [ s ] ds > a, b ], then ∂ L is also differentiable and, hence,we easilly obtain ∂ L ( t, x ⋆ ( t ) , ˙ x ⋆ ( t ) , z ⋆ ( t )) + ∂ L ( t, x ⋆ ( t ) , ˙ x ⋆ ( t ) , z ⋆ ( t )) ∂ L ( t, x ⋆ ( t ) , ˙ x ⋆ ( t ) , z ⋆ ( t )) − ddt ∂ L ( t, x ⋆ ( t ) , ˙ x ⋆ ( t ) , z ⋆ ( t )) = 0 , t ∈ [ a, b ] . References
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