On the Solvability of the Periodically Forced Relativistic Pendulum Equation on Time Scales

We study some properties of the range of the relativistic pendulum operator P , that is, the set of possible continuous T -periodic forcing terms p for which the equation P x = p admits a T -periodic solution over a T - periodic time scale T . Writing p ( t ) = p 0 ( t ) + p , we prove the existence of a nonempty compact interval I ( p 0 ), depending continuously on p 0 , such that the problem has a solution if and only if p ∈ I ( p 0 ) and at least two diﬀerent solutions when p is an interior point. Furthermore, we give suﬃcient conditions for nondegeneracy; speciﬁcally, we prove that if T is small then I ( p 0 ) is a neighbourhood of 0 for arbitrary p 0 . The results in the present paper improve the smallness condition obtained in previous works for the continuous case T = R .


Introduction
The T -periodic problem for the forced relativistic pendulum equation on time scales reads Px(t) := (ϕ(x ∆ (t))) ∆ + ax ∆ (t) + b sin x(t) = p 0 (t) + s, t ∈ T, where a, b > 0 and s are real numbers, T is an arbitrary T -periodic nonempty closed subset of R for some T > 0, ϕ : (−c, c) → R is the relativistic operator ϕ(x) := x with c > 0 and p 0 is continuous and T -periodic in T, with zero average. In this work, we are concerned with the set of all possible values of s such that (1) admits a T -periodic solution.
The time scales theory was introduced in 1988, in the PhD thesis of Stefan Hilger [11], as an attempt to unify discrete and continuous calculus. The time scale R corresponds to the continuous case and, hence, yields results for ordinary differential equations. If the time scale is Z, then the results apply to standard difference equations. However, the generality of the set T produces many different situations in which the time scales formalism is useful in several applications. For example, in the study of hybrid discrete-continuous dynamical systems, see [6].
In the past decades, periodic problems involving the relativistic forced pendulum differential equation for the continuous case T = R were studied by many authors, see [3,4,8,13,17,18]. In particular, the works [3,18] are concerned with the so-called solvability set, that is, the set I(p 0 ) of values of s for which (1) has at least one T -periodic solution. We remark that problem (1) is 2πperiodic and, consequently, if x is a T -periodic solution then x + 2kπ is also a T -periodic solution for all k ∈ Z. For this reason, the multiplicity results for (1) usually refer to the existence of geometrically distinct T -periodic solutions, i.e. solutions not differing by a multiple of 2π.
For the standard pendulum equation with a = 0, the solvability set was analyzed in the pioneering work [9], where it was proved that I(p 0 ) ⊂ [−b, b] is a nonempty compact interval containing 0. Moreover, I(p 0 ) depends continuously on p 0 . These results were partially extended to the relativistic case in [8]; however, the method of proof in both works is variational and, consequently, cannot be applied to the case a > 0. This latter situation was studied in [10] for the standard pendulum and in [18] for the relativistic case. An interesting question, stated already in [9] is whether or not the equation may be degenerate, namely: is there any p 0 such that I(p 0 ) reduces to a single point? Many works are devoted to this problem and, for the classical pendulum, nondegeneracy has been proved for an open and dense subset ofC T , the space of zero-average Tperiodic continuous functions. However, the question for arbitrary p 0 remains unsolved. For a survey on the pendulum equation and open problems see for example [14].
The purpose of this work is to extend the results in [3] and [18] to the context of time scales. To this end, we prove in the first place that the set I(p 0 ) is a nonempty compact interval depending continuously on p 0 .
The method of proof is inspired in a simple idea introduced in [10] for the standard pendulum equation, which basically employs the Schauder Theorem and the method of upper and lower solutions. Moreover, by a Leray-Schauder degree argument it shall be proved that if s is an interior point of I(p 0 ), then the problem admits at least two geometrically distinct periodic solutions.
Furthermore, sufficient conditions shall be given in order to guarantee that 0 ∈ I(p 0 ). We recall that, when a = 0, this is not trivial even in the continuous case T = R. For the classical pendulum equation, there exist well known examples with 0 / ∈ I(p 0 ) for arbitrary values of T ; for the relativistic case, it was proved in [3] that, if cT < √ 3π, then 0 ∈ I(p 0 ) • . It is worth noticing that, however, the problem is still open for large values of T . As we shall see, a slight improvement of the previous bound can be deduced from the results in the present paper. Specifically, we shall prove the existence of T * with cT * > π such that if T ≤ T * then 0 ∈ I(p 0 ) and it is an interior point when the inequality is strict. An inferior bound for T * can be characterized as a zero of a real function; for the continuous case T = R, it is shown that cT * > √ 3π and verified numerically that cT * > 6.318. We remark that the computation is independent of p 0 : in other words, if T < T * , then the range of the operator P contains a set of the formC T + [−ε, ε] for some ε > 0.
We highlight that our paper is devoted to equations on time scales that involve a ϕ-laplacian of relativistic type, for which the literature is scarce. For example, in [16], the existence of heteroclinic solutions for a family of equations on time scales that includes the unforced relativistic pendulum is proved. However, to our knowledge there are no papers concerned with periodic solutions and, more precisely, the solvability set for equations with a singular ϕ-laplacian on time scales.
This work is organized as follows. In Section 2, we establish the notation, terminology and preliminary results which will be used throughout the paper. In Section 3 we prove that the set I(p 0 ) is a nonempty compact interval depending continuously on p 0 , and that two geometrically distinct T -periodic solutions exist when s is an interior point. Finally, Section 4 is devoted to find sufficient conditions in order to guarantee that 0 ∈ I(p 0 ) and improve the condition obtained in [3] for the continuous case.

Notation and preliminaries
Fix T > 0 and assume that T is T -periodic, i.e. T + T = T. Let C T = C T (T, R) be the Banach space of all continuous T -periodic functions on T endowed with the uniform norm and letC T be the subspace of those elements of C T having zero average. By we shall denote the Banach space of all continuous T -periodic functions on T that are ∆-differentiable functions with continuous ∆-derivatives, endowed with the usual norm Equation (1) can be written as where f : ) for all t ∈ T. We remark that necessarily x ∞ < c.
For x ∈ C T , the average, the maximum value and the minimum value of x shall be denoted respectively by x, x max and x min , namely For details on time scales theory we refer the reader to [6,7].

Upper and lower solutions and degree
Let us define T -periodic lower and upper solutions for problem (2) as follows.
It is worth recalling the problem of finding T -periodic solutions of (2) over the closure of the set can be reduced to a fixed point equation T is a compact operator that can be defined according to the nonlinear version of the continuation method (see e.g. [15]), namely where N f is the Nemitskii operator associated to f and K :C T →C T is the (nonlinear) compact operator given by Kξ = x, with x ∈ C 1 T the unique solution of the problem (ϕ(x ∆ (t))) ∆ = ξ(t) with zero average. We recall, for the reader's convenience, that the definition of K based upon the existence, easy to prove, of a (unique) completely continuous map c : For the purposes of the present paper, we shall only need the following result, which is an adaptation of Theorem 3.7 in [1]: Theorem 2.2 Suppose that (2) has a T -periodic lower solution α and an upper solution β such that α(t) ≤ β(t) for all t ∈ T. Then problem (1) has at least one T -periodic solution x with α(t) ≤ x(t) ≤ β(t) for all t ∈ T. If furthermore α and β are strict, then deg LS (I − M f , Ω α,β (0), 0) = 1, where deg LS stands for the Leray-Schauder degree.

The solvability set I(p 0 )
In this section, we shall prove that the solution set I(p 0 ) is a nonempty compact set; furthermore, employing the method of upper and lower solutions it shall be verified that I(p 0 ) is an interval depending continuously on p 0 . Finally, the excision property of the degree will be employed to verify that if s is an interior point of I(p 0 ), then the problem has at least 2 geometrically different T -periodic solutions. Proof: For the reader's convenience, we shall proceed in several steps.
Step 1 (An associated integro-differential problem). Observe that if x ∈ C 1 T is a solution of (1), then, ∆-integration over [0, T ] T yields s = b T T 0 sin(x(t))∆t. Therefore, it proves convenient to consider the integro-differential Dirichlet problem with s(x) := b T T 0 sin(x(t))∆t. By Schauder's fixed point theorem, it is straightforward to prove that for each r ∈ R there exists at least one solution x ∈ C([0, T ] T ) of (4) such that x(0) = x(T ) = r.
Step 2 (I(p 0 ) is is nonempty and bounded). Let x be a solution of (4) such that and hence ϕ(x ∆ (T )) = ϕ(x ∆ (0)). It follows that x may be extended in a Tperiodic fashion to a solution of (1) with s = s(x). In other words, x is a solution of (4) for some r ∈ [0, 2π]} = ∅.
Step 3 (I(p 0 ) is connected). Assume that s 1 , s 2 ∈ I(p 0 ) are such that s 1 < s 2 , and let x 1 and x 2 be T -periodic solutions of (1) for s 1 and s 2 , respectively. Then for any s ∈ (s 1 , s 2 ) it is verified that x 1 and x 2 are strict upper and a lower solutions of (1), respectively. Replacing x 1 by x 1 + 2kπ, with k the first integer such that x 2 < x 1 + 2kπ and applying Theorem 2.2 with α = x 2 and β = x 1 +2kπ, we conclude that problem (1) has at least one T -periodic solution, whence s ∈ I(p 0 ).
Step 4 (I(p 0 ) is closed). Let {s n } ⊂ I(p 0 ) converge to some s, and let x n ∈ C 1 T be a solution of (1) for s n . Without loss of generality, we may assume that x n (0) ∈ [0, 2π]. Because x ∆ n ∞ < c, by Arzelà-Ascoli theorem there exists a subsequence (still denoted {x n }) that converges uniformly to some x. Furthermore, from (1) we deduce the existence of a constant C independent of n such that |(ϕ(x ∆ n (t))) ∆ | ≤ C for all t. We claim that ϕ(x ∆ n ) is also uniformly bounded, that is, x ∆ n ∞ is bounded away from c. Indeed, otherwise passing to a subsequence we may suppose for example that ϕ( CT and, consequently, ϕ(x ∆ n ) min → +∞. This implies that (x ∆ n ) min → c, which contradicts the fact that x ∆ n has zero average. Using Arzelà-Ascoli again, we may assume that ϕ(x ∆ n ) converges uniformly to some function v and, from the identity x n (t) = x n (0) + t 0 x ∆ n (ξ)∆ξ we deduce that x ∈ C 1 T and x ∆ = ϕ −1 (v). Now integrate the equation for each n and take limit for n → ∞ to obtain In turn, this implies that x is a solution of (4) with s(x) = s; hence, I(p 0 ) is closed and the proof is complete.
Step 5 (continuous dependence on p 0 ). Let {p n 0 } n∈N ⊂C T be a sequence that converges to some p 0 . We shall prove that D(p n 0 ) → D(p 0 ); the proof for d is analogous. Similarly to Step 4, it is seen that if a subsequence of {D(p n 0 )} converges to some D, then the problem for p 0 with s = D admits a solution and, consequently, D ≤ D(p 0 ). Thus, it suffices to prove that lim inf n→∞ D(p n 0 ) ≥ D(p 0 ). Indeed, otherwise, passing to a subsequence we may suppose that D(p n 0 ) → D < D(p 0 ). Fix η > 0 such that D + η < D(p 0 ) and let x be a T -periodic solution of (1) for s = D(p 0 ). Take n large enough such that and let x n be a T -periodic solution of (1) for p n 0 and s = D(p n 0 ). The previous inequalities imply that x and x n are respectively a lower and an upper solution of the problem for p n 0 and s = D+η and, without loss of generality, we may assume that x < x n . Thus, (1) has a T -periodic solution for p n 0 and s = D + η > D(p n 0 ), a contradiction.
The following theorem establishes the existence of at least two geometrically different T -periodic solutions to problem (1) when s is an interior point.
Theorem 3.2 Assume that p 0 ∈ C T has zero average. If s ∈ (d(p 0 ), D(p 0 )), then the problem (1) has at least two geometrically different T -periodic solutions.
Proof: For s ∈ (d(p 0 ), D(p 0 )), let s 1 := d(p 0 ) < s < D(p 0 ) := s 2 and let x 1 , x 2 be as in Step 3 of the previous proof. Then x 1 and x 2 are strict upper and lower solutions for s, respectively. Due to the 2π-periodicity of (1), we may assume that x 2 < x 1 and x 2 + 2π ≤ x 1 and, consequently, Ω x2,x1 and Ω x2+2π,x1+2π are disjoint open subsets of Ω x2,x1+2π . From Theorem 2.2 and the excision property of the Leray-Schauder degree, we deduce the existence of three different solutions y 1 , y 2 , y 3 ∈ C 1 T such that for all t ∈ T. If y 2 = y 1 + 2π, then y 3 = y 1 , y 1 + 2π and the conclusion follows.
4 Sufficient conditions for 0 ∈ I(p 0 ) In this section, we shall obtain conditions guaranteeing that 0 belongs to the solvability set. Even in the continuous case, this is not clear when a = 0 since, as it is well known, counter-examples exist for the classical pendulum equation for arbitrary periods. In the relativistic case, however, it was proved that 0 ∈ I(p 0 ) when T is sufficiently small and counter-examples for large values of T are not yet known. Here, as mentioned in the introduction, we shall improve the bounds for T obtained in previous works for T = R. The results shall be expressed in terms of k(T), the optimal constant of the inequality T . For instance, for arbitrary T it is readily seen that k(T) ≤ T 2 , because x ∆ has zero average and hence, due to periodicity, We recall that, in the continuous case, the (optimal) Sobolev inequality 3 . The main result of this section reads as follows. If ψ(δ) ≥ 0 for some δ ∈ (0, π 2 ), then 0 ∈ I(p 0 ). Furthermore, if the previous inequality is strict, then 0 ∈ I(p 0 ) • .
Before proceeding to the proof, it is worth to recall that, from Theorem 4.1 and Example 5.3 in [2], in order to prove the existence of T -periodic solutions for s = 0 it suffices to verify that the equation has no T -periodic solutions with average ± π 2 . For example, if x ∈ C 1 T is a solution of (5) such that x = π 2 , then it follows from the definition of k(T) that, for all t ∈ T, In particular, if ck(T) ≤ π 2 , then x(t) ∈ [0, π] for all t ∈ T and, upon integration of equation (5), we deduce: The same contradiction is obtained also if x = − π 2 . For example, the condition cT ≤ π is sufficient for arbitrary T and, in the continuous case, the condition cT ≤ √ 3π is retrieved. However, the previous bound ck(T) ≤ π 2 can be improved, as we shall see in the following proof.
Proof of Theorem 4.1: From the preceding discussion, it may be assumed that π 2 < ck(T) < π. Suppose that x is a solution of (5) such that x = π 2 , then for all t ∈ T and hence Fix δ ∈ (0, π 2 ) and consider the set where m(C δ ) is the measure of the set C δ associated to the ∆-integral, namely m(C δ ) = C δ ∆t. Clearly, a contradiction is obtained when the latter term of (6) is positive. Moreover, notice that if x(t 0 ) ≤ π 2 and t 1 > t 0 is such that x(t 1 ) ≥ π 2 + δ, then In the same way, if t 0 < t 1 are such that x(t 0 ) ≥ π 2 and x(t 1 ) ≤ π 2 − δ, then c(t 1 − t 0 ) > δ. Thus, by periodicity, we deduce that m(C δ ) > 2δ c . The same conclusions are obtained if x = − π 2 ; hence, a sufficient condition for the existence of at least one T -periodic solution is that, for some δ ∈ (0, π 2 ), or, equivalently, that ψ(δ) ≥ 0. Note, furthermore, that if the inequality is strict, then a contradiction is still obtained as in (6) if we add a small parameter s to the function p 0 in (5).
Thus, replacing (7) in ψ, a somewhat explicit condition on T reads: An immediate corollary is the following: There exists a constant T * with cT * > π such that 0 ∈ I(p 0 ) for all p 0 ∈C T if T ≤ T * and it is an interior point if T < T * . For the particular case T = R, it is verified that cT * > √ 3π.
Analogous conclusions are obtained when T = R using Ψ cont instead of Ψ.

Numerical examples and final remarks
As shown in Corollary 4.3, the bound thus obtained always improves the simpler one ck(T) ≤ π 2 and, in particular, it guarantees that if the latter inequality is satisfied then 0 is in fact an interior point of I(p 0 ). In the continuous case, an easy numerical computation gives the sufficient condition cT ≤ 6.318, slightly better than the bound cT < √ 3π obtained in [4] (see Figure 1). For arbitrary T, numerical experiments show that 0 ∈ I(p 0 ) • for cT ≤ 4.19, as shown in Figure  2. Remark 4.5 An estimation of the constant k(T) could be obtained analogously to the continuous case as shown for example in [12]. Let {e n } n∈Z ⊂ C T be an orthonormal basis of L 2 (0, T ) T with e 0 ≡ 1 √ T and E n be a primitive of e n such that E n = 0. Writing x ∆ = n =0 a n e n , it follows that When T = R, taking the usual Fourier basis one has that E n ∞ = √ T 2πn and the value k(R) ≤ T 2 √ 3 is obtained from the well known equality n∈N 1 n 2 = π 2 6 .
Remark 4.6 As mentioned in the introduction, Theorem 4.1 allows to compute an inferior bound for the length of the solvability interval which does not depend on p 0 , provided that T is small enough. In some obvious cases, inferior bounds are obtained for arbitrary T : for example, if p 0 ∞ < b then [−ε, ε] ⊂ I(p 0 ) for ε = b − p 0 ∞ . This is readily verified taking α = π 2 and β = 3π 2 as lower and upper solutions.