Necessary Optimality Conditions for Higher-Order Infinite Horizon

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Apr 15, 2012 - petitiveness (“Programa Operacional Factores de Competitividade”) and by Portuguese funds through the Cen- ter for Research and ...
arXiv:1204.3329v1 [math.OC] 15 Apr 2012

Necessary Optimality Conditions for Higher-Order Infinite Horizon Variational Problems on Time Scales Nat´alia Martins1

Delfim F. M. Torres2

(Communicated by Boris S. Mordukhovich) Abstract We obtain Euler–Lagrange and transversality optimality conditions for higher-order infinite horizon variational problems on a time scale. The new necessary optimality conditions improve the classical results both in the continuous and discrete settings: our results seem new and interesting even in the particular cases when the time scale is the set of real numbers or the set of integers. Keywords: time scales, calculus of variations, infinite horizon problems, Euler–Lagrange equations, transversality conditions. Mathematics Subject Classification 2010: 34N05, 39A12, 49K05.

1

Introduction

We consider infinite horizon variational problems on time scales, which consist in maximizing a delta integral with a Lagrangian involving higher-order delta derivatives on a given unbounded time scale. Problems of the calculus of variations of such type have many applications in economics both in discrete (i.e., when the time scale is the set of integers) and continuous (i.e., when the time scale is the set of real numbers) time settings (see, e.g., [1,2]). Indeed, the dynamic processes of economics are usually described with discrete or continuous models. The time scale approach adopted here puts discrete and continuous models of economics together and, most important, extends them to more realistic situations of unequally spaced points in time (time-varying graininess). Consider a typical situation of a consumer, that has to make decisions concerning how much to consume and how much to spend with the goal to maximize his lifetime utility subject to certain constraints. Problems of the calculus of variations on time scales provide a natural way to model such a consumer, that has an income from different sources, at unequal time intervals, and makes expenditures also at unequal time intervals [3]. The reader interested on the usefulness of the calculus of variations on time scales in economics is referred to [3–6] and references therein. This is a preprint of a paper whose final and definite form will appear in Journal of Optimization Theory and Applications (JOTA). Paper submitted 17-Nov-2011; revised 24-March-2012 and 10April-2012; accepted for publication 15-April-2012. This work was supported by FEDER funds through COMPETE — Operational Programme Factors of Competitiveness (“Programa Operacional Factores de Competitividade”) and by Portuguese funds through the Center for Research and Development in Mathematics and Applications (University of Aveiro) and the Portuguese Foundation for Science and Technology (“FCT — Funda¸ca ˜o para a Ciˆ encia e a Tecnologia”), within project PEstC/MAT/UI4106/2011 with COMPETE number FCOMP-01-0124-FEDER-022690. Torres was also supported by project PTDC/MAT/113470/2009. 1 Assistant Professor in the Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal. Senior Researcher in the Center for Research and Development in Mathematics and Applications, Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal. E-mail: [email protected]. 2 Corresponding author. Associate Professor with Habilitation in the Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal. Coordinator of the research group on Mathematical Theory of Systems and Control, Center for Research and Development in Mathematics and Applications, Department of Mathematics, University of Aveiro, 3810-193 Aveiro, Portugal. E-mail: [email protected].

1

Clearly, for infinite horizon variational problems, the delta integral does not necessarily converge: it may diverge to plus or minus infinity or it may oscillate. In such situations, the extension of the standard definition of optimality used in the time scale setting (see, e.g., [7, 8]) to the unbounded time domain is not useful. Indeed, if, for example, for every admissible function the value of the integral functional is equal to plus infinity, then each admissible path could be called an optimal path. To handle this and similar situations in a rigorous way, several alternative definitions of optimality for problems with unbounded time domain have been proposed in the literature (see, e.g., [9–12]). In this paper, we follow the notion of weakly optimal solution introduced by Brock in the economic literature. In the case when the variational functional converges for all admissible paths, Brock’s notion coincides with the standard definition of optimality. Many results in infinite horizon optimal control with this type of optimality can be found in the book [13]. For the method of discrete approximations, that allows to approximate continuous-time control problems by those associated with discrete dynamics, we refer the reader to [14]. The goal of this paper is to provide necessary optimality conditions to higher-order infinite horizon variational problems on time scales. Our main result is Theorem 3.1. It provides a nontrivial generalization of the recent results of [15, 16]. Moreover, Theorem 3.1 improves the continuous results of Okomura et al. [17] when one chooses the time scale to be the set of real numbers, while, in the particular case when the time scale is the set of integers, it generalizes the discrete-time results of Cai and Nitta [18]. The paper is organized as follows. In Section 2, we present some preliminary results and basic definitions necessary in the sequel. Main results are given in Section 3: in Section 3.1, we prove some fundamental lemmas of the calculus of variations for infinite horizon variational problems; the Euler–Lagrange equation and the transversality conditions for higher-order infinite horizon variational problems are proved in Section 3.2 and discussed in Section 3.3. We end the paper with two illustrative examples (Section 4) and a summary of the major results (Section 5).

2

Preliminaries

A time scale is an arbitrary, nonempty and closed subset T of R (endowed with the topology of a subspace of R). In a time scale T, we consider the following two operators: the forward jump operator σ : T → T, defined by σ(t) := inf {s ∈ T : s > t} if t 6= sup T and σ(sup T) := sup T, and the backward jump operator ρ : T → T, defined by ρ(t) := sup {s ∈ T : s < t} if t 6= inf T and ρ(inf T) := inf T. A point t ∈ T is called right-dense, right-scattered, left-dense or left-scattered if and only if σ(t) = t, σ(t) > t, ρ(t) = t or ρ(t) < t, respectively. We say that t is isolated if and only if ρ(t) < t < σ(t), t is dense if and only if ρ(t) = t = σ(t). The mapping µ : T → [0, +∞[ is defined by µ(t) := σ(t) − t and is called the graininess function. In order to introduce the definition of delta derivative, we define a new set Tκ . If T has a left-scattered maximum M , then Tκ = T \ {M }, otherwise, Tκ = T. Definition 2.1. We say that a function f : T → R is delta differentiable at t ∈ Tκ if and only if there is a number f ∆ (t) such that, for all ε > 0, there exists a neighborhood U of t such that |f (σ(t)) − f (s) − f ∆ (t)(σ(t) − s)| ≤ ε|σ(t) − s| for all s ∈ U . We call f ∆ (t) the delta derivative of f at t. Moreover, we say that f is delta differentiable (or ∆-differentiable) on T provided f ∆ (t) exists for all t ∈ Tκ . Remark 2.1. If T = R, then f ∆ = f ′ , where f ′ denotes the usual derivative on R. If T = Z, then f ∆ = f (t + 1) − f (t), i.e., f ∆ is the usual forward difference. For any time scale T, if f is a constant, then f ∆ = 0; if f (t) = kt for some constant k, then f ∆ = k. In order to simplify expressions, we denote the composition f ◦ σ by f σ . Theorem 2.1. [19] Let T be a time scale, f : T → R, and t ∈ Tκ . The following holds: 1. If f is delta differentiable at t, then f is continuous at t. 2

2. If f is continuous at t and t is right-scattered, then f is delta differentiable at t and f ∆ (t) =

f σ (t) − f (t) . µ(t)

3. If t is right-dense, then f is delta differentiable at t if and only if the limit f (t) − f (s) s→t t−s lim

exists as a finite number. In this case, f (t) − f (s) . s→t t−s

f ∆ (t) = lim

4. If f is delta differentiable at t, then f σ (t) = f (t) + µ(t)f ∆ (t). Definition 2.2. Let f, F : T → R. Function F is called a delta antiderivative of f if and only if F ∆ (t) = f (t) for all t ∈ Tκ . In this case we define the delta integral of f from a to b (a, b ∈ T) by Z

b

f (t)∆t := F (b) − F (a).

a

Definition 2.3. A function f : T → R is rd-continuous if and only if it is continuous at the right-dense points and its left-sided limits exist (finite) at all left-dense points. The set of all rd-continuous functions f : T → R is denoted by Crd (T, R). Theorem 2.2. [19] Every rd-continuous function f : T → R has a delta antiderivative. In particular, if a ∈ T, then the function F defined by Z t F (t) := f (τ )∆τ, t ∈ T , a

is a delta antiderivative of f . Theorem 2.3. [19] If a, b, c ∈ T, a ≤ c ≤ b, α ∈ R, and f, g ∈ Crd (T, R), then Z b Z b Z b 1. (f (t) + g(t)) ∆t = f (t)∆t + g(t)∆t; a

2.

a

Z

b

Z

b

αf (t)∆t = α

a

3.

f (t)∆t = −

a

4.

Z

Z

a

b

f (t)∆t; a

a

f (t)∆t;

b

Z

a

Z

b

f (t)∆t = 0;

a

5.

f (t)∆t =

a

Z

c

f (t)∆t +

a

Z

b

f (t)∆t;

c

6. if f (t) > 0 for all a < t ≤ b, then

Z

b

f (t)∆t > 0;

a

7. If f and g are ∆-differentiable, then (a)

Z

a

b σ



f (t)g (t)∆t =

[(f g)(t)]t=b t=a



Z

b

f ∆ (t)g(t)∆t;

a

3

(b)

b

Z



f (t)g (t)∆t =

a

t=b [(f g)(t)]t=a



Z

b

f ∆ (t)g σ (t)∆t.

a

For more definitions, notations, and results concerning the theory of time scales, we refer the reader to the books [19, 20]. In what follows, σ denotes the forward jump operator and ∆ is the k delta derivative of a given time scale T. As usual, for f : T → R we define f σ := f ◦ σ k , where σ k := σ ◦ σ k−1 , k ∈ N, and σ 0 = id. We assume that T is a time scale such that sup T = +∞ 0 and we suppose that a, T, T ′ ∈ T are such that T ′ ≥ T > a. Let r ∈ N and f ∆ := f . The  r−1 ∆ r r r rth delta derivative of f : T → R is the function f ∆ : Tκ → R defined by f ∆ := f ∆ , r−1

provided f ∆ is delta differentiable. By ∂i F we denote the partial derivative of a function F with respect to its ith argument. All intervals are time scales intervals, that is, we simply write [a, b] and [a, +∞[ to denote, respectively, the set [a, b] ∩ T and [a, +∞[ ∩ T. We consider the following higher-order variational problem on T: Z +∞   r r−1 r−1 r L t, xσ (t), xσ ∆ (t), . . . , xσ∆ (t), x∆ (t) ∆t −→ max a

2r x ∈ Crd ([a, +∞[, R)

i

x∆ (a) = αi ,

(1)

i = 0, . . . , r − 1,

where r ∈ N, α0 , . . . , αr−1 are fixed real numbers, (u0 , . . . , ur ) → L(t, u0 , . . . , ur ) is a C 1 (Rr+1 , R) r function for any t ∈ [a, +∞[, and ∂i+2 L ∈ Crd ([a, +∞[, R) for all i = 0, . . . , r. Remark 2.2. The results of this paper are trivially generalized for functions x : [a, +∞[→ Rn (n ∈ N), but for simplicity of presentation we restrict ourselves to the scalar case (n = 1). Definition 2.4. We say that x is an admissible function for problem (1) if and only if i

2r x ∈ Crd ([a, +∞[, R) and x∆ (a) = αi , i = 0, . . . , r − 1.

As optimality criteria, we use the following generalization of Brock’s notion of optimality. Definition 2.5. Function x∗ is weakly maximal to problem (1) if and only if x∗ is admissible and lim

inf ′

T →+∞ T ≥T

Z

a

T′

h   r r−1 r−1 r L t, xσ (t), xσ ∆ (t), . . . , xσ∆ (t), x∆ (t)  i r r−1 r−1 r −L t, xσ∗ (t), x∗σ ∆ (t), . . . , x∗σ∆ (t), x∆ ∆t ≤ 0 ∗ (t)

for all admissible function x.

It is well known that, for certain time scales T, the forward jump operator σ is not delta differentiable. Furthermore, the chain rule, as we know it from the classical calculus, that is, when T = R, is not valid in general. However, if we suppose that the time scale T satisfies the condition (H)

for each t ∈ T, (r − 1) (σ(t) − a1 t − a0 ) = 0 for some a1 ∈ R+ and a0 ∈ R,

then we can deal with these two limitations as noted in Remark 2.3 and Lemma 2.1. Remark 2.3. Condition (H) is equivalent to r = 1 or σ(t) = a1 t + a0 for some a1 ∈ R+ and a0 ∈ R. Thus, for the first order infinite horizon variational problem [15], we impose no restriction on the time scale T. For the higher-order problems (i.e., for r ≥ 2) such restriction on the time scale is necessary. Indeed, for r > 1 we are implicitly assuming in (1) that σ be delta differentiable, which is not true for a general time scale T. Note that, for r > 1, condition (H) implies that σ be delta differentiable and σ ∆ (t) = a1 , t ∈ T. Furthermore, note that condition (H) includes the following important cases: the differential calculus (T = R, a1 = 1, a0 = 0); the difference calculus (T = Z, a1 = 1, a0 = 1); the h-calculus (T = hZ := {hz : z ∈ Z} for some h > 0, a1 = 1, a0 = h); and the q-calculus (T = q N0 := {q k : k ∈ N0 } for some q > 1, a1 = q, a0 = 0). 4

Lemma 2.1. [21] Let T be a time scale satisfying condition (H) and r > 1. If f : T → R is two times delta differentiable, then f σ∆ (t) = a1 f ∆σ (t), t ∈ T. The next lemma will be very useful for the proof of our higher-order fundamental lemmas of the calculus of variations on time scales (more precisely, will be useful for Lemma 3.5 and Lemma 3.7). An analogous nabla version can be found in [22]. 2r Lemma 2.2. Assume that the time scale T satisfies condition (H) and η ∈ Crd ([a, +∞[, R) is i i−1 such that η ∆ (a) = 0 for all i = 0, . . . , r. Then, η σ∆ (a) = 0 for each i = 1, . . . , r.

Proof. If a is right-dense, then the result is trivial (just use Lemma 2.1 and the fact that σ(a) = a). Suppose that a be right-scattered and fix i ∈ {1, . . . , r}. Since  i−1 σ i−1  i−1 ∆ η∆ (a) − η ∆ (a) i ∆ ∆ , η (a) = η (a) = σ(a) − a  i−1 σ i i−1 η ∆ (a) = 0, and η ∆ (a) = 0, then η ∆ (a) = 0. By Lemma 2.1, η σ∆

i−1

i−1

 i−1 σ (a), (a) = (a1 )i−1 η ∆

proving that η σ∆ (a) = 0.  We end this section recalling a result that will be needed in the proof of our Theorem 3.1. Theorem 2.4. [23] Let S and T be subsets of a normed vector space. Let f be a map defined on T × S, having values in some complete normed vector space. Let v be adherent to S and w adherent to T . Assume that: 1. limx→v f (t, x) exists for each t ∈ T ; 2. limt→w f (t, x) exists uniformly for x ∈ S. Then the limits limt→w limx→v f (t, x), limx→v limt→w f (t, x), and lim(t,x)→(w,v) f (t, x) all exist and are equal.

3

Main Results

We prove a first-order necessary optimality condition for higher-order infinite horizon variational r problems on scales. For simplicity of expressions,  time  we introduce the operator h·i defined by r r−1 r−1 r hxir (t) := t, xσ (t), xσ ∆ (t), . . . , xσ∆ (t), x∆ (t) .

Theorem 3.1 (Euler–Lagrange Equation and Transversality Conditions). Let T be a time scale satisfying condition (H) and such that sup T = +∞. Suppose that x∗ be a maximizer to problem r−1 2r (1) and let η ∈ Crd ([a, +∞[, R) be such that η(a) = 0, η ∆ (a) = 0, . . . , η ∆ (a) = 0. Define T′

Lhx∗ + ǫηir (t) − Lhx∗ ir (t) ∆t, ǫ a Z T′   r r V (ε, T ) := inf Lhx + ǫηi (t) − Lhx i (t) ∆t, ∗ ∗ ′ ′

A(ε, T ) :=

Z

T ≥T

a

V (ε) := lim V (ε, T ). T →+∞

Suppose that V (ε, T ) exists for all T ; ε→0 ε

1. lim

5

2.

lim

T →+∞

V (ε, T ) exists uniformly for ε; ε

3. For every T ′ > a, T > a, and ε ∈ R \ {0}, there exists a sequence (A(ε, Tn′ ))n∈N such that lim A(ε, Tn′ ) = inf A(ε, T ′ ) ′

n→+∞

T ≥T

uniformly for ε. Then x∗ satisfies the Euler–Lagrange equation r X



i

(−1)

i=0

1 a1

 i(i−1) 2

∆i

(∂i+2 L)

hxir (t) = 0

(2)

for all t ∈ [a, +∞[ and the r transversality conditions lim

inf

T →+∞ T ′ ≥T

(

r



∂r+2−(k−1) Lhxi (T ) +

k−1 X

i

(−1) ∂r+2−(k−1)+i L

i=1

k = 1, . . . , r, with

Ψri (k)

∆i

r



× xσ

k−1

hxi (T ) · ∆

!

Ψri (k)

r−k

)

(T ′ )

= 0, (3)

r−(k−1)+(j−1) i  Y 1 = . a1 j=1

The proof of Theorem 3.1 is given in §3.2. Before that we state and prove several useful auxiliary results. In particular, we prove in §3.1 a higher-order integration by parts formula (Lemma 3.1) and three higher-order fundamental lemmas of the calculus of variations on time scales (Lemmas 3.5, 3.6 and 3.7).

3.1

Fundamental Lemmas

In our results we use the standard convention that

Pj

k=1

γ(k) = 0 whenever j = 0.

Lemma 3.1 (Higher-order integration by parts formula). Let r ∈ N, T be a time scale satisfying r 2r condition (H), a, b ∈ T, a < b, f ∈ Crd ([a, σ r (b)], R), and g ∈ Crd ([a, σ r (b)], R). For each i = 1, . . . , r we have Z

a

b

f (t)g σ

r−i



i



(t)∆t = f (t)g σ

r−i



i−1

(t) +

i−1 X

k

(−1)k f ∆ (t)g σ

r−i+k



i−1−k

k=1

 i−j b k  Y 1  (t) · a 1 j=1

a

+ (−1)i

Z

b a



1 a1

 i(i−1) 2

i

r

f ∆ (t)g σ (t)∆t.

Proof. We prove the lemma by mathematical induction. If r = 1, the result is obviously true: it coincides with the usual integration by parts formula on time scales. Assuming that the result

6

holds for an arbitrary r, we will prove it for r + 1. Fix i = 1, . . . , r. By the induction hypotheses, Z b Z b r+1−i i r−i i ∆ f (t)g σ (t)∆t = f (t)(g σ )σ ∆ (t)∆t a

a



= f (t)(g σ )σ

r−i

a



= f (t)g σ

(t) +

i−1 X

k

(−1)k f ∆ (t)(g σ )σ

r−i+k



i−1−k

k=1

Z

+ (−1)i



i−1

r+1−i

b ∆

1 a1

i−1

 i(i−1) 2

(t) +

a

i

r

f ∆ (t)(g σ )σ (t)∆t

i−1 X

k

(−1)k f ∆ (t)g σ

r+1−i+k



i−1−k

k=1

Z

+ (−1)i

a

b

1 a1

 i(i−1) 2

 i−j b k  Y 1  (t) · a 1 j=1

 i−j b k  Y 1  (t) · a 1 j=1

a

i

f ∆ (t)g σ

r+1

(t)∆t.

It remains to prove that the result is true for i = r + 1. Note that Z b Z b r ∆r+1 f (t)g (t)∆t = f (t)(g ∆ )∆ (t)∆t a

a



= f (t)(g ∆ )∆ + (−1)r

r−1

b

Z



r

= f (t)g ∆ (t) +

b

Z

+ (−1)r

 r−j b k  Y k k r−1−k 1  (−1)k f ∆ (t)(g ∆ )σ ∆ (t) · a 1 j=1

k=1

a



(t) +

r−1 X

a

1 a1

r−1 X

a

 r(r−1) 2

r

r

f ∆ (t)(g ∆ )σ (t)∆t k

(−1)k f ∆ (t)g σ

k

∆r−k

(t) ·

k=1



1 a1

  r(r−1) 2

1 a1

r



r

(by induction hypotheses) k Y k 

1 a1

j=1

1 a1

r

f ∆ (t)(g σ )∆ (t)∆t

r−j

b 

a

(by Lemma 2.1).

Using the standard integration by parts formula in the last delta integral, and taking into account  k Q  r−j Q  r+1−j k k 1 that a11 = j=1 a11 , we conclude that j=1 a1 Z

b

f (t)g ∆

r+1

a







r

(t)∆t = f (t)g ∆ (t) + r

r

+ (−1)r f ∆ (t)g σ (t) r

= f (t)g ∆ (t) + + (−1)r+1

Z

r X



1 a1

r−1 X

k

(−1)k f ∆ (t)g σ

k

k

a

 r+1−j b k  Y 1  (t) · a1 j=1

a

 r(r+1) 2

(−1)k f ∆ (t)g σ





r−k

k=1



k=1 b

k

b

 − (−1)r a

r−k

Z

a

b

1 a1

 r(r+1) 2

f∆

r+1

(t)g σ

r+1

(t)∆t

 r+1−j b k  Y 1  (t) · a1 j=1

a

1 a1

 r(r+1) 2

f∆

r+1

(t)g σ

r+1

(t)∆t,

proving that the result is true for i = r + 1.  Before presenting the higher-order fundamental lemmas of the calculus of variations on time scales, we need the following three preliminary results. 7

Lemma 3.2. Suppose that a ∈ T and f ∈ Crd ([a, +∞[, R) be such that f ≥ 0 on [a, +∞[. If lim

T′

Z

inf ′

T →+∞ T ≥T

f (t)∆t = 0,

a

then f ≡ 0 on [a, +∞[. Proof. Suppose, by contradiction, that there exists t0 ∈ [a, +∞[ such that f (t0 ) > 0. Fix b ∈ T Rb such that a0 ≤ t0 < b. We will prove that a f (t)∆t > 0. If t0 is right-scattered, then b

Z

f (t)∆t =

a



Z

Z

t0

f (t)∆t + a

σ(t0 )

Z

f (t)∆t +

t0

σ(t0 )

Z

b

f (t)∆t σ(t0 )

f (t)∆t = f (t0 ) (σ(t0 ) − t0 ) > 0. t0

If t0 is right-dense, then, by the continuity of f at t0 , there exists δ > 0 such that f (t) > 0 for all t ∈ [t0 , t0 + δ[ and, therefore, Z b Z t0 +δ Z b Z t0 f (t)∆t + f (t)∆t f (t)∆t + f (t)∆t = a

a



Z

t0 +δ

t0

t0 +δ

f (t)∆t > 0.

t0

Then, for any T > b, inf ′

T ≥T

Z

T′

T

Z

f (t)∆t =

a

f (t)∆t ≥

a

and hence lim

inf ′

T →+∞ T ≥T

which is a contradiction.

Z

Z

b

f (t)∆t > 0,

a

T′

Z

f (t)∆ ≥

b

f (t)∆t > 0,

a

a



Lemma 3.3. Let f ∈ Crd ([a, +∞[, R). If lim

inf ′

T →+∞ T ≥T

Z

T′

f (t)η ∆ (t)∆t = 0

a

1 for all η ∈ Crd ([a, +∞[, R) such that η(a) = 0, then f (t) = c for all t ∈ [a, +∞[, where c ∈ R.

Proof. Fix T, T ′ ∈ T such that T ′ ≥ T > a. Let c be a constant defined by the condition Z

T′

(f (τ ) − c) ∆τ = 0,

a

and let η(t) =

Z

t

(f (τ ) − c) ∆τ.

a

1 ([a, +∞[, R), η ∆ (t) = f (t) − c, Clearly, η ∈ Crd

η(a) =

Z

a

(f (τ ) − c) ∆τ = 0,

and



η(T ) =

a

Z

T′

(f (τ ) − c) ∆τ = 0.

a

Observe that Z

T′ ∆

(f (t) − c) η (t)∆t =

Z

a

a

8

T′

(f (t) − c)2 ∆t

and Z

T′ ∆

(f (t) − c) η (t)∆t =

a

Hence,

Z

T′ ∆

f (t)η (t)∆t − c

a

lim

inf

T →+∞ T ′ ≥T

T′

Z



η (t)∆t =

a

T′

Z



f (t)η (t)∆t = lim

inf

T →+∞ T ′ ≥T

a

Z

Z

T′

T′

f (t)η ∆ (t)∆t. a

2

(f (t) − c) ∆t = 0,

a

which shows, by Lemma 3.2, that f (t) − c = 0 for all t ∈ [a, +∞[.



Lemma 3.4. Let f, g ∈ Crd ([a, +∞[, R). If Z T′  lim inf f (t)η σ (t) + g(t)η ∆ (t) ∆t = 0 ′ T →+∞ T ≥T

a

1 for all η ∈ Crd ([a, +∞[, R) such that η(a) = 0, then g is delta differentiable and

g ∆ (t) = f (t)

∀t ∈ [a, +∞[.

Rt Proof. Fix T, T ′ ∈ T such that T ′ ≥ T > a and define A(t) = a f (τ )∆τ . Then A∆ (t) = f (t) for all t ∈ [a, +∞[ and Z T′ Z T′ Z T′ T′ ∆ σ ′ ′ ∆ A (t)η (t)∆t = A(T )η(T ) − f (t)η σ (t)∆t. A(t)η (t)∆t = [A(t)η(t)]a − a

a

a



Restricting η to those such that η(T ) = 0, we obtain Z T′  lim inf f (t)η σ (t) + g(t)η ∆ (t) ∆t = lim ′ T →+∞ T ≥T

inf

T →+∞ T ′ ≥T

a

Z

T′

(−A(t) + g(t)) η ∆ (t)∆t = 0.

a

By Lemma 3.3 we may conclude that there exists c ∈ R such that −A(t) + g(t) = c for all t ∈ [a, +∞[. Therefore, A∆ (t) = g ∆ (t) for all t ∈ [a, +∞[, proving the desired result: g ∆ (t) = f (t) for all t ∈ [a, +∞[.  We are now in conditions to prove the following three fundamental lemmas of the calculus of variations for higher-order infinite horizon variational problems on time scales. Lemma 3.5 (Higher-order fundamental lemma of the calculus of variations I). Let T be a time scale satisfying condition (H) and such that sup T = +∞. Suppose that f0 ∈ Crd ([a, +∞[, R), 1 r f1 ∈ Crd ([a, +∞[, R), . . ., fr ∈ Crd ([a, +∞[, R). If ! Z T′ X r r−i i fi (t)η σ ∆ (t) ∆t = 0 lim inf ′ T →+∞ T ≥T

a

i=0

2r for all η ∈ Crd ([a, +∞[, R) such that η(a) = 0, η ∆ (a) = 0, . . . , η ∆ r X

(−1)i

i=0



1 a1



i(i−1) 2

i

fi∆ (t) = 0

r−1

(a) = 0, then

∀t ∈ [a, +∞[.

Proof. We prove the lemma by mathematical induction. If r = 1, the result is true by Lemma 3.4. Assume now that the result is true for some r. We will prove that the result is also true for r + 1. Suppose that ! Z T′ X r+1 σr+1−i ∆i fi (t)η (t) ∆t = 0 lim inf ′ T →+∞ T ≥T

for all η ∈ that

2(r+1) Crd

a

i=0

r

([a, +∞[, R) such that η(a) = 0, η ∆ (a) = 0, . . . , η ∆ (a) = 0. We want to prove r+1 X i=0

i

(−1)



1 a1

 i(i−1) 2

i

fi∆ (t) = 0

9

∀t ∈ [a, +∞[.

Note that Z T′ X r+1 a

fi (t)η

σ

r+1−i



i

!

(t) ∆t =

i=0

r X

T′

Z

a

fi (t)η

σ

r+1−i



i

!

(t) ∆t +

i=0

Z

T′

a

 r ∆ (t)∆t. fr+1 (t) η ∆

Using the integration by parts formula in the last integral, we obtain that Z T′  r ∆ h iT ′ Z T ′  r σ r ∆ fr+1 (t) η ∆ (t)∆t = fr+1 (t)η ∆ (t) − fr+1 (t) η ∆ (t)∆t. a

a

Since η



r

a

r

(a) = 0 and we can restrict ourselves to those η such that η ∆ (T ′ ) = 0, then Z T′ Z T′  r σ  r ∆ ∆ ∆ (t)∆t fr+1 (t) η ∆ (t)∆t = − fr+1 (t) η a

a

and, by Lemma 2.1, Z T′



fr+1 (t) η

a

Hence, Z

r+1 X

T′ a

fi (t)η σ

r+1−i



∆r

i

T′

=

T′

=

Z

r X

(t)∆t = −

Z

T′ ∆ (t) fr+1

a



1 a1

r

r

η σ∆ (t)∆t.

!

(t) ∆t

i=0

Z

∆

!

r r 1 η σ∆ (t)∆t a1 a i=0 !   r  r−1 X r−i i 1 ∆ σ σ ∆ σ ∆r fi (t) (η ) (t) + fr (t) − fr+1 (t) (η ) (t) ∆t a1 i=0

a

a

fi (t)η

σr+1−i ∆i

(t) ∆t −

Z

T′

∆ fr+1 (t)



and, therefore, lim

inf

T →+∞ T ′ ≥T

= lim

Z

r+1 X

T′

a

inf

T →+∞ T ′ ≥T

fi (t)η σ

r+1−i



i

!

(t) ∆t

i=0

Z

a

T′

"r−1 X

r−i i ∆ σ σ

fi (t) (η )

i=0

#   r  1 σ ∆r ∆ (η ) (t) ∆t = 0. (t) + fr (t) − fr+1 (t) a1 r−1

By Lemma 2.2, η σ (a) = 0, (η σ )∆ (a) = 0, . . . , (η σ )∆ (a) = 0. Then, by the induction hypothesis, we conclude that   i(i−1)   r(r−1)   r ∆r r−1 2 2 X 1 1 1 i ∆ ∆i r (−1) (t) = 0 fr (t) − fr+1 (t) fi (t) + (−1) a a a 1 1 1 i=0 for all t ∈ [a, +∞[, which is equivalent to r+1 X i=0

i

(−1)



1 a1

 i(i−1) 2

i

fi∆ (t) = 0

∀t ∈ [a, +∞[. 

Lemma 3.6 (Higher-order fundamental lemma of the calculus of variations II). Let T be a time scale satisfying condition (H) and such that sup T = +∞. Suppose that f0 ∈ Crd ([a, +∞[, R) and r fi ∈ Crd ([a, +∞[, R) for all i = 1, . . . , r. If ! Z T′ X r r−i i lim inf fi (t)η σ ∆ (t) ∆t = 0 ′ T →+∞ T ≥T

a

i=0

10

r−1

2r for all η ∈ Crd ([a, +∞[, R) such that η(a) = 0, η ∆ (a) = 0, . . . , η ∆ (a) = 0, then n o ′ ∆r−1 ′ f (T ) · η (T ) = 0. lim inf r ′ T →+∞ T ≥T

Proof. Note that ! Z T′ X r σr−i ∆i fi (t)η (t)∆t a

i=0

= =

Z Z

T′

σr

f0 (t)η (t)∆t + a

Z r X

fi (t)η

σr−i ∆i

(t)∆t

a

i=1

T′

T′

!

r

f0 (t)η σ (t)∆t a

 ′ i−j T k  Y r−i+k i−1−k r−i i−1 k 1 ∆ fi (t)η σ ∆ (t) +  (−1)k fi∆ (t)η σ (t) · + a 1 j=1 i=1 k=1 a   i(i−1) ′   Z r T 2 X i r 1 (−1)i fi∆ (t)η σ (t)∆t + (by Lemma 3.1) a 1 a i=1     i(i−1) Z T′ r 2 X ∆i r 1 f0 (t) +  · η σ (t)∆t (−1)i fi (t) = a 1 a i=1 r X



i−1 X

T ′ i−j k  Y k r−i+k i−1−k r−i i−1 1 ∆  fi (t)η σ ∆ (t) + (−1)k fi∆ (t)η σ (t) · + a 1 j=1 i=1 k=1 a   i(i−1)   Z T′ X r 2 r ∆i  · η σ (t)∆t  (−1)i fi (t) 1 = a 1 a i=0 r X



r−1 X

+

i=1



i−1 X



fi (t)η σ

+ fr (t)η ∆

r−1

r−i

∆i−1

(t) +

i−1 X

k=1 r−1 X

(t) +

T ′ i−j k  Y k r−i+k i−1−k 1 ∆  (−1)k fi∆ (t)η σ (t) · a 1 j=1 a

k

(−1)k fr∆ (t)η σ

k

∆r−1−k

(t) ·

k=1

a

T′

r X

fi (t)η

σr−i ∆i

(t)∆t

i=0

=

r−1 X i=1



fi (t)η σ 

+ fr (t)η ∆

r−i



r−j k  Y 1  a1 j=1

a

and, by Lemma 3.5, we get Z

T ′

!

i−1

(t) +

i−1 X

k

(−1)k fi∆ (t)η σ

r−i+k



i−1−k

k=1

r−1

(t) +

r−1 X

T ′ i−j k  Y 1  (t) · a 1 j=1 a

k

(−1)k fr∆ (t)η σ

k

∆r−1−k

(t) ·

k  Y

j=1

k=1

1 a1

r−j

Therefore, restricting the variations η to those such that ησ

r−k

∆k−1

(T ′ ) = η σ

r−k

∆k−1

(a) = 0,

11

k = 1, . . . , r − 1,

T ′ 

a

.

ησ

k

∆r−1−k

(T ′ ) = η σ

k

∆r−1−k

(a) = 0,

k = 1, . . . , r − 1,

we get lim

Z

inf ′

T →+∞ T ≥T

T′

a

r X

fi (t)η

σr−i ∆i

!

(t) ∆t = 0 ⇒ lim

i=0

inf ′

T →+∞ T ≥T

n

fr (T ′ )η ∆

r−1

o (T ′ ) = 0,

proving the desired result.



Lemma 3.7 (Higher-order fundamental lemma of the calculus of variations III). Let T be a time scale satisfying condition (H) and such that sup T = +∞. Suppose that f0 ∈ Crd ([a, +∞[, R) and r fi ∈ Crd ([a, +∞[, R) for all i = 1, . . . , r. If ! Z T′ X r σr−i ∆i fi (t)η (t) ∆t = 0 lim inf ′ T →+∞ T ≥T

a

i=0

2r for all η ∈ Crd ([a, +∞[, R) such that η(a) = 0, η ∆ (a) = 0, . . . , η ∆

lim

inf

T →+∞ T ′ ≥T

(



fr−(k−1) (T ) +

k−1 X

i

(−1) fr−(k−1)+i

i=1

for all k = 1, . . . , r, where

Ψri (k)

∆i



(T ) ·

r−1

(a) = 0, then !

Ψri (k)

·η

σk−1 ∆r−k



)

(T )

=0

r−(k−1)+(j−1) i  Y 1 = . a1 j=1

Proof. We do the proof by induction. Let r = 1. Using the integration by parts formula and Lemma 3.5, we obtain limT →+∞ inf T ′ ≥T f1 (T ′ )η(T ′ ) = 0, showing that the result is true for r = 1. Assuming now that the result holds for an arbitrary r, we will prove it for r + 1. Suppose that ! Z T′ X r+1 σr+1−i ∆i fi (t)η (t) ∆t = 0 lim inf ′ T →+∞ T ≥T

2(r+1)

for all η ∈ Crd lim

inf

T →+∞ T ′ ≥T

(

a

i=0 i

([a, +∞[, R) such that η ∆ (a) = 0, i = 0, . . . , r. We want to prove that ′

fr+1−(k−1) (T ) +

k−1 X

i

(−1) fr+1−(k−1)+i

i=1

∆i



(T ) ·

!

Ψr+1 (k) i ×η

σk−1 ∆r+1−k



)

(T )

=0

(4)

for k = 1, . . . , r + 1. Fix some k = 2, . . . , r + 1. The main idea of the proof is that the kth transversality condition for the variational problem of order r + 1 is obtained from the (k − 1)th transversality condition for the variational problem of order r. Using the same techniques as in Lemma 3.5, we prove that ! Z T′ X r+1 σr+1−i ∆i lim inf fi (t)η (t) ∆t = 0 ′ T →+∞ T ≥T

a

i=0

implies lim

inf

T →+∞ T ′ ≥T

(Z

a

T′

r−1 X i=0

r−i i ∆ σ σ

fi (t) (η )

! )   r  1 σ ∆r ∆ (η ) (t) ∆t = 0. (t) + fr (t) − fr+1 (t) a1

12

Since, by Lemma 2.2, η σ (a) = 0, (η σ )∆ (a) = 0, . . . , (η σ )∆ hypothesis for k − 1, we conclude that lim

inf ′

T →+∞ T ≥T

(

k−3 X i=1

r−1

(a) = 0, then, by the induction

 ∆i  ∆k−2 (T ′ ) · Ψrk−2 (k − 1) (−1)i fr−(k−2)+i (T ′ ) · Ψri (k − 1) + (−1)k−2 fr k−1



+ fr−(k−2) (T ) + (−1)

∆k−1  1 r  (T ′ ) · Ψrk−2 (k − 1) fr+1 a1

!

σ σk−2 ∆r−(k−1)

· (η )



)

= 0,

)

= 0.

(T )

which is equivalent to lim

inf

T →+∞ T ′ ≥T

(



fr−(k−2) (T ) +

k−1 X

i

(−1) fr−(k−2)+i

i=1

∆i



(T ) ·

Ψri (k

!

− 1)

× ησ

k−1



r−(k−1)

(T ′ )

This proves equation (4) for k = 2, . . . , r + 1. It remains to prove that equation (4) is also true for k = 1. This condition follows from Lemma 3.6. 

3.2

Proof of Theorem 3.1

Using our notion of weak maximality, if x∗ is optimal, then V (ε) ≤ 0 for every ε ∈ R. Since V (0) = 0, then 0 is an extremal point of V . We will prove that V is differentiable at t = 0, and hence V ′ (0) = 0. Note that V (ε) V (ε, T ) = lim lim ε→0 T →+∞ ε ε V (ε, T ) (by hypothesis 1 and 2 and Theorem 2.4) = lim lim T →+∞ ε→0 ε = lim lim inf A(ε, T ′ ) ′

V ′ (0) = lim

ε→0

T →+∞ ε→0 T ≥T

= lim lim lim A(ε, Tn′ )

(by hypothesis 3 )

T →+∞ ε→0 n→+∞

= lim

lim lim A(ε, Tn′ )

= lim

inf lim A(ε, T ′ )

= lim

inf lim ′

(by hypothesis 3 and Theorem 2.4)

T →+∞ n→+∞ ε→0

(by hypothesis 3 )

T →+∞ T ′ ≥T ε→0

T →+∞ T ≥T ε→0

Z

T′

a

Lhx∗ + ǫηir (t) − Lhx∗ ir (t) ∆t ε

T′

Lhx∗ + ǫηir (t) − Lhx∗ ir (t) ∆t ε→0 T →+∞ T ≥T a ε ! Z T′ X r r−i i ∂i+2 Lhx∗ ir (t) · η σ ∆ (t) ∆t. = lim inf ′ = lim

inf ′

Z

T →+∞ T ≥T

lim

a

i=0

Therefore, lim

inf

T →+∞ T ′ ≥T

Z

T′

a

Using Lemma 3.5, we conclude that r X i=0

i

(−1)



1 a1

r X

r

∂i+2 Lhx∗ i (t) · η

i=0

 i(i−1) 2

∆i

(∂i+2 L)

13

σr−i ∆i

!

(t) ∆t = 0.

hx∗ ir (t) = 0 ∀t ∈ [a, +∞[ ,

proving that x∗ satisfies the Euler–Lagrange equation (2). By Lemma 3.7, we conclude that lim

inf

T →+∞ T ′ ≥T

(

r



∂r+2−(k−1) Lhx∗ i (T ) +

k−1 X

i

(−1) ∂r+2−(k−1)+i L

i=1

∆i

r

!

Ψri (k)



hx∗ i (T ) ·

×η

σk−1 ∆r−k



)

(T )

=0

(5)

for k = 1, . . . , r. Consider η defined by η(t) = α(t)x∗ (t) for all t ∈ [a, +∞[, where α : [a, +∞[→ R r−1 2r is a Crd function satisfying α(a) = 0, α∆ (a) = 0, . . . , α∆ (a) = 0, and there exists T0 ∈ T r−1 such that α(t) = β ∈ R \ {0} for all t > T0 . Note that η(a) = 0, η ∆ (a) = 0, . . . , η ∆ (a) = 0. Substituting η in equation (5), we conclude that lim

inf

T →+∞ T ′ ≥T

(

r



∂r+2−(k−1) Lhx∗ i (T ) +

k−1 X

i

(−1) ∂r+2−(k−1)+i L

i=1

∆i

r

!

Ψri (k)



hx∗ i (T ) · × x∗σ

k−1



r−k

)

(T ′ )

= 0,

proving that x∗ satisfies the transversality condition (3) for all k = 1, . . . , r.

3.3

Remarks and Corollaries

Note that we have actually proved that, for an infinite horizon variational problem of order r, one has r transversality conditions and that, for each k = 1, . . . , r, the kth transversality condition has exactly k terms. To the best of our knowledge, even for the classical calculus of variations (i.e., when T = R or T = Z) our explicit formulas for the transversality conditions are new. Similarly to the special case when T = R (see [17]) and when T = Z (see [18]), hypotheses 1, 2, and 3 of Theorem 3.1 are impossible to be verified a priori because x∗ is unknown. In practical terms, such hypotheses are assumed to be true and conditions (2) and (3) are applied heuristically to obtain a candidate. If such a candidate is, or not, a solution to the problem is a different question that always require further analysis (see Examples 4.1 and 4.2 in Section 4). For the convenience of the reader, we present the particular cases of Theorem 3.1 for r=1, r = 2, and r = 3. Corollary 3.1. [15] Assuming hypotheses of Theorem 3.1 for r = 1, if x∗ is a maximizer to problem (1), then x∗ satisfies the Euler–Lagrange equation (∂3 L)∆ (t, xσ (t), x∆ (t)) = ∂2 L(t, xσ (t), x∆ (t))

∀t ∈ [a, +∞[ ,

and the transversality condition lim

inf

T →+∞ T ′ ≥T

 ∂3 L(T ′ , xσ (T ′ ), x∆ (T ′ )) · x(T ′ ) = 0.

Corollary 3.2. Assuming hypotheses of Theorem 3.1 for r = 2, if x∗ is a maximizer to problem (1), then x∗ satisfies the Euler–Lagrange equation ∆

∂2 Lhxi2 (t) − (∂3 L) hxi2 (t) +

1 ∆2 (∂4 L) hxi2 (t) = 0 a1

∀t ∈ [a, +∞[ ,

and the two transversality conditions lim

inf ′

T →+∞ T ≥T

 ∂4 Lhxi2 (T ′ ) · x∆ (T ′ ) = 0,

   1 ∆ 2 ′ σ ′ 2 ′ (∂ L) hxi (T ) · x (T ) = 0. lim inf ∂ Lhxi (T ) − 4 3 T →+∞ T ′ ≥T a1 14

Corollary 3.3. Assuming hypotheses of Theorem 3.1 for r = 3, if x∗ is a maximizer to problem (1), then x∗ satisfies the Euler–Lagrange equation ∆

3

3

∂2 Lhxi (t) − (∂3 L) hxi (t) +



1 a1



∆2

(∂4 L)

3

hxi (t) −



1 a1

3

∆3

(∂5 L)

hxi3 (t) = 0

∀t ∈ [a, +∞[ ,

and the three transversality conditions lim

inf ′

T →+∞ T ≥T

lim

inf

T →+∞ T ′ ≥T

lim

inf

T →+∞ T ′ ≥T

(

(

3

n o 2 ∂5 Lhxi3 (T ′ ) · x∆ (T ′ ) = 0,



∂4 Lhxi (T ) −

∂3 Lhxi3 (T ′ ) −



1 a1





1 a1

2





3



!

(∂5 L) hxi (T )

(∂4 L) hxi3 (T ′ ) +



1 a1

3

σ∆

·x

∆2

(∂5 L)



)

(T )

= 0, !

2

)

hxi3 (T ′ ) xσ (T ′ )

= 0.

Considering T = R in Theorem 3.1, we get the following result that improves the results of [17]. Corollary 3.4. Consider the problem Z +∞   L t, x(t), x′ (t), . . . , x(r) (t) dt −→ max a

x ∈ C r ([a, +∞[, R)

x(a) = α0 ,

(6)

x(r−1) (a) = αr−1 ,

...,

where (u0 , . . . , ur ) → L(t, u0 , . . . , ur ) is a C 1 (Rr+1 , R) function for any t ∈ [a, +∞[, α0 , . . . , αr−1 are fixed real numbers, and ∂i+2 L ∈ C r ([a, +∞[, R) for all i = 1, . . . , r and all x ∈ C r ([a, +∞[, R). Suppose that the maximizer to problem (6) exists and is given by x∗ . Let η ∈ C r ([a, +∞[, Rn ) be such that η(a) = 0, . . . , η (r−1) (a) = 0. Define     (r) (r) Z T ′ L t, x∗ (t) + εη(t), . . . , x(r) (t) − L t, x∗ (t), . . . , x∗ (t) ∗ (t) + εη dt, A(ε, T ′ ) := ε a Z T′ h   i (r) (r) (r) V (ε, T ) := inf L t, x (t) + εη(t), . . . , x (t) + εη (t) − L(t, x (t), . . . , x (t)) dt, ∗ ∗ ∗ ∗ ′ T ≥T

a

V (ε) := lim V (ε, T ). T →+∞

Suppose that 1. lim

ε→0

2.

V (ε, T ) exists for all T ; ε

lim

T →+∞

V (ε, T ) exists uniformly for ε; ε

3. For every T ′ > a, T > a, and ε ∈ R \ {0}, there exists a sequence (A(ε, Tn′ ))n∈N such that lim A(ε, Tn′ ) = inf A(ε, T ′ ) ′

n→+∞

T ≥T

uniformly for ε. Then x∗ satisfies the Euler–Lagrange equation r X i=0

(i)

(−1)i (∂i+2 L)



 t, x(t), x′ (t), . . . , x(r) (t) = 0 15

∀t ∈ [a, +∞[ ,

and the r transversality conditions lim

inf ′

T →+∞ T ≥T

(

  ∂r+2−(k−1) L t, x(t), x′ (t), . . . , x(r) (t) +

k−1 X

i

(−1) ∂r+2−(k−1)+i L

i=1

(i)

(r)



(t, x(t), x (t), . . . , x

)  (r−k) ′ (t)) · x (T ) = 0,

k = 1, . . . , r.

4

Illustrative Examples

The following two examples illustrate the usefulness of Theorem 3.1. Example 4.1. Let T be a time scale satisfying condition (H) and such that sup T = +∞ and 0 ∈ T. Consider the problem Z +∞  2 2 − x∆ (t) ∆t −→ max 0

(7)

4 x ∈ Crd ([0, +∞[, R)

x(0) = 0,

x∆ (0) = 1.

By Theorem 3.1, if x∗ is a maximizer to problem (7), then x∗ satisfies the Euler–Lagrange equation ∂2 Lhxi2 (t) − (∂3 L)∆ hxi2 (t) +

2 1 (∂4 L)∆ hxi2 (t) = 0 a1

∀t ∈ [0, +∞[ .

Since ∂2 Lhxi2 (t) = 0,

∂3 Lhxi2 (t) = 0,

2

∂4 Lhxi2 (t) = −2x∆ (t),

then the Euler–Lagrange equation is 4

x∆ (t) = 0

∀t ∈ [0, +∞[.

(8)

Clearly, x∗ (t) = c1 t3 + c2 t2 + c3 t + c4 , where c1 , c2 , c3 , c4 ∈ R, is the solution of (8). Using the initial conditions x∗ (0) = 0 and x∆ ∗ (0) = 1, we get c4 = 0 and c3 = 1 − c2 a0 − c1 a20 . Using the two transversality conditions, we will determine the value of c1 and c2 . Since    1 ∆ 2 ′ σ ′ 2 ′ (∂4 L) hx∗ i (T ) · x∗ (T ) = 0 ∂3 Lhx∗ i (T ) − lim inf T →+∞ T ′ ≥T a1 is equivalent to lim

inf ′

T →+∞ T ≥T

n

 c1 (a1 T ′ + a0 )3 + c2 (a1 T ′ + a0 )2 + (1 − c2 a0 − c1 a20 )(a1 T ′ + a0 )

o × c1 (1 + a21 + a1 )(1 + a1 ) = 0,

then we conclude that c1 = 0. Using the transversality condition  ′ lim inf ∂4 Lhx∗ i2 (T ′ ) · x∆ ∗ (T ) = 0, ′ T →+∞ T ≥T

that is,

lim

inf {−2c2 (1 + a1 ) · (c2 (T ′ + a1 T ′ + a0 ) + 1 − c2 a0 )} = 0,

T →+∞ T ′ ≥T

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we conclude that c2 = 0. Hence, x∗ (t) = t is a candidate to be a maximizer to problem (8). Since lim

inf ′

T →+∞ T ≥T

Z

0

T′

h i 2 2 2 ∆2 L(t, xσ (t), xσ∆ (t), x∆ (t)) − L(t, xσ∗ (t), xσ∆ ∗ (t), x∗ (t)) ∆t = lim

inf ′

T →+∞ T ≥T

Z

0

T′

i h 2 −(x∆ (t))2 ∆t ≤ 0

for every admissible function x, then x∗ is indeed a solution to problem (8). In what follows, we use the standard notation of quantum calculus (see, e.g., [24]): Dq [y](t) :=

y(qt) − y(t) (q − 1)t

and Dq2 [y](t) := Dq [Dq [y]](t).

Example 4.2. Fix q > 1 and let T = q N0 . Consider the following non-autonomous problem: Z +∞  2  dq t −→ max −t 1 + Dq2 [x](t) 1

x(1) = α ,

(9)

Dq [x](1) = β,

where α and β are fixed real numbers. By Theorem 3.1, if x∗ is a maximizer to problem (9), then x∗ must satisfy the Euler–Lagrange equation 1 ∂2 Lhxi2 (t) − Dq [∂3 L]hxi2 (t) + Dq2 [∂4 L]hxi2 (t) = 0 q

∀t ∈ [1, +∞[ .

Since ∂2 Lhxi2 (t) = 0,

∂3 Lhxi2 (t) = 0,

then the Euler–Lagrange equation takes the form   Dq2 2t Dq2 [x] (t) = 0

∂4 Lhxi2 (t) = −2tDq2 [x](t),

∀t ∈ [1, +∞[ .

(10)

It is easy to see that x∗ (t) = k1 t2 + k2 t + k3 t ln t + k4 is the general solution of equation (10). Using the initial conditions we obtain k2 = β − k1 (1 + q) − k3 and k4 = −β + k1 q + k3

q ln q q−1

q ln q + α. q−1

Using the transversality condition    1 2 ′ 2 ′ ′ ∂3 Lhx∗ i (T ) − Dq [∂4 L]hx∗ i (T ) · x∗ (qT ) = 0 lim inf T →+∞ T ′ ≥T q we get k3 = 0. The transversality condition  lim inf ∂4 Lhx∗ i2 (T ′ ) · Dq [x∗ ](T ′ ) = 0 ′ T →+∞ T ≥T

implies that k1 = 0. Hence, x∗ (t) = βt − β + α is a candidate to be a maximizer. Using the definition of weak maximality, we conclude that x∗ (t) = βt − β + α is indeed a maximizer to problem (9).

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5

Concluding Remarks

We have established Euler–Lagrange and transversality optimality conditions for higher-order infinite horizon variational problems on a time scale. The results were obtained for weakly optimal solutions in the Brock sense. The main result is Theorem 3.1, which generalizes the recent results of [15] and [16]. Moreover, the new necessary optimality conditions improve the classical results both in the continuous and discrete settings: if one chooses the time scale to be the set of real numbers, then Theorem 3.1 improves the continuous results of [17]; if one chooses the time scale to be the set of integers, then Theorem 3.1 improves the discrete-time results of [18].

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