UV-Enhanced Ordering

New results about the existence of periodic solutions for second order differential equations are provided. The method of proof relies on the Schauder's fixed point theorem. Some examples are presented to illustrate the main results.

The main purpose of this article is to discuss the existence of periodic solutions of equation (1.1) by means of Schauder's fixed point theorem.The method of proof is in a simple idea and is composed of two steps: The first step is to transform the original equation into a first order integro-differential equation through a linear integral operator and the second step is an application of the Schauder's fixed point theorem.The existence of single periodic solution for (1.1) has been established under suitable behavior of g and f on some closed sets.So some information on the location of periodic solution is also obtained, leading to multiplicity results.Our results are new for (1.2)-(1.5)(see Corollary 3.2, Corollary 3.3, Theorem 3.3), which seems not be found in the literature.

Preliminaries
Clearly, X is a Banach space.
Let p, q ∈ X and consider the following two differential equations Proof.Here we only consider (2.1).Obviously, the periodic solution of (2.1) is unique if ω 0 p(t)dt = 0 and we show that x(t) is the periodic solution of (2.1).Differentiating x(t), we obtain that Hence, x(t) is unique ω-periodic solution of (2.1).
The following well-known Schauder's fixed point theorem is crucial in our arguments.where p > 0 is a constant which is determined later.For any u ∈ X, Ju ∈ X ∩ C 1 (R) and (2.4) We transform (1.1) to that is By Lemma 2.1, we obtain that if u is a periodic solution of (2.5), u satisfies In order to put more emphasis on the above facts, we summarize them in the following lemma.
Lemma 2.3.Define an operator T on X by here ω 0 (p + a(r))dr = 0. Then the fixed point u of T on X is the periodic solution of (2.5) and Ju is the periodic solution of (1.1).
Proof.Since T u = u and we obtain that u is the periodic solution of (2.5).In order to prove that Ju is the periodic solution of (1.1), we only show that Ju satisfies (1.1).Form (2.3)-(2.5),this result follows immediately.
The following theorems are the main results of this paper.Theorem 3.1.Assume that there exist constants m < M, p > 0 such that Using (H 2 ), we obtain that for any u ∈ Ω, which imply that p + a(t) ≥ 0 for all t ∈ R. If p + a(t) ≡ 0, according to (H 1 ) and (H 2 ), we easily to check that g(t, u) Thus any constant C ∈ [m, M ] is the periodic solution of (1.1).We assume that p + a(t) ≥ 0 and p + a(t) ≡ 0. The operator T is well defined.Now we show that T satisfies all conditions of Lemma 2.2.Noting that we obtain that for any u ∈ Ω, Next, we show that T : Ω → Ω is completely continuous.Obviously, T (Ω) is a uniformly bounded set and T is continuous on Ω, so it suffices to show T (Ω) is equicontinuous by Ascoli-Arzela theorem.For any u ∈ Ω, we have Since T (Ω) is bounded and f, g, a are continuous, there exists ρ > 0 such that which implies that T (Ω) is equicontinuous.So T is a completely continuous operator on Ω.By Lemma 2.2, there exists u ∈ Ω with T u = u.Moreover, Ju ∈ [m, M ] is the periodic solution of (1.1).The proof is complete.
Analogously, we have the following theorem.Theorem 3.2.Assume that there exist constants m < M, p > 0 such that Then (1.1) has at least one ω-periodic solution x with m ≤ x ≤ M .Remark 3.1 Let c, µ > 0 and h be ω-periodic continuous function.In [12], by using critical point theorem, authors proved that (1.3) has at least two periodic solutions if h < µ and one periodic solutions if h = µ.Proof.Here we only consider (1.3).If µ = 0, then h ≡ 0 and any constant Λ is the periodic solution of (1.3).Now, we assume that µ = 0. Let a(t) = −c, g(u) = µ sin u and f (t, u, v) = h(t).Put p 1 = (|c| + 1)(|µ| + 1), then p 1 (p Hence, (1.3) has at least one periodic solution EJQTDE, 2013 No. 41, p. 5 Corollary 3.3.Let e be ω-periodic continuous function and f a bounded continuous function on R. Further suppose that there exist constants m < M such that g ∈ C 1 ([m, M ], R) and where α = sup u∈R f (u) and β = inf u∈R f (u).Then (1.4) has at least one ω-periodic solution x with m ≤ x ≤ M .
Assume that e is ω-periodic continuous function, for (1.5), we have the following result.
Then (1.5) has at least one ω-periodic solution x with m ≤ x ≤ M .
Proof.Define the function where p > 0 is sufficiently large which is determined later.By computation, we have Similar to (2.5), using the operator Ju, we transform (1.5) to the equation Define an operator K on the closed, convex set Ω by Hence, the fact that K has fixed point on Ω implies that (1.5) has the periodic solution.
EJQTDE, 2013 No. 41, p. 6 The rest proof is similar to that of Theorem 3.1.Remark 3.2 Using the same idea, for the following differential equation we have the following result.

Some examples
In this section, three examples are provided to highlight our results obtained in previous section.
In recent paper [7], R. Hakl and P J. Torres discussed the existence of positive periodic solution of (4.1) when G, H ∈ L + and F ∈ L, where L is the Banach space of ω-periodic Lebesgue integrable function and L + = {p ∈ L : p ≥ 0 for a.e.t ∈ [0, ω]}.By using method of upper and lower functions, authors established several existence results.By Corollary 3.1, we obtain the following new result.
EJQTDE, 2013 No. 41, p. 7 Proof.By the condition (4.2), we have On the other hand, for any u ∈ [m, M ], Hence for any u ∈ [m, M ], the inequality holds.By Corollary 3.1, (4.1) has at least one positive ω-periodic solution.
The condition (4.2) admits that G changes sign, which is different with those of [7] .For example, using (4.2), one can check that the differential equation has one 2π-periodic solution x with 0.1 ≤ x ≤ 10.When G ≡ 0, (4.2) reduces to which guarantee that the equation
Lemma 2.2.Let X be a Banach space with D ⊂ X closed and convex.Assume that T : D → D is a completely continuous operator, then T has a fixed point in D.