On multipoint resonant problems on the half-line

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(1)López-Somoza and Minhós Boundary Value Problems https://doi.org/10.1186/s13661-019-1153-9. (2019) 2019:38. RESEARCH. Open Access. On multipoint resonant problems on the half-line Lucía López-Somoza1*. and Feliz Minhós2. *. Correspondence: lucia.lopez.somoza@usc.es 1 Departamento de Estatística, Análise Matemática e Optimización Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela, Galicia, Spain Full list of author information is available at the end of the article. Abstract In this work we obtain sufficient conditions for the existence of bounded solutions of a resonant multipoint second-order boundary value problem, with a fully differential equation. The noninvertibility of the linear part is overcome by a new perturbation technique, which allows obtaining an existence result and a localization theorem. Our hypotheses are clearly much less restrictive than those existent in the literature and, moreover, they can be applied to higher-order, resonant or nonresonant, boundary value problems defined on the half-line or even on the real line. Keywords: Multipoint problems; Unbounded domains; Resonant problems. 1 Introduction In this paper, we will prove the existence of bounded solutions for the multipoint boundary value problem ⎧ ⎨u (t) = f (t, u(t), u (t)), ⎩u(0) = 0,. t ∈ [0, ∞), m–1 u (∞) = i=1 αi u (ξi ), . (1). where αi > 0 and 0 = ξ1 < · · · < ξm–1 < ∞. We assume that coefficients αi satisfy the resonance condition m–1 . αi = 1.. (2). i=1. A boundary value problem is said to be resonant when the correspondent homogeneous problem has nontrivial solutions. In fact, under condition (2), the homogeneous boundary value problem related to (1), namely ⎧ ⎨u (t) = 0,. t ∈ [0, ∞),. ⎩u(0) = 0,. u (∞) =. m–1 i=1. αi u (ξi ),. (3). has a nontrivial solution. These resonant problems have been studied for many years by many methods: degree theory has been used in, for instance, [3, 7, 13, 22], Lyapunov–Schmidt arguments in [17], © The Author(s) 2019. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made..

(2) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. a Leggett–Williams theorem in [8, 21], fixed point and fixed point index theories in [2, 9, 10, 25], monotone method together with the upper and lower solutions technique in [20], among others. Boundary value problems on unbounded intervals arise in many models of applied mathematics, such as in combustion theory, in plasma physics, to model the unsteady flow of a gas through semi-infinite porous media, to study the electrical potential of an isolated neutral atom, etc. For more details, techniques and applications in this field we refer, for example, to [11, 14–16, 24], and the monograph [1]. From a theoretical point of view, resonance problems can be formulated by an equation Lx = Nx, where L is a noninvertible operator. Therefore, in particular, the resonance condition (2) implies that the Green’s function related to problem (3) does not exist. This issue is overcome applying several techniques. For instance, in [12] the authors studied the problem ⎧ ⎨u (t) + f (t, u(t)) = 0, ⎩u(0) = 0,. t ∈ [0, ∞), m–1 u (∞) = i=1 αi u (ξi ), . also under condition (2) and, to deal with the resonance problem, they defined some suitable operators and were able to find a solution in the space |u(t)| <∞ , E = u ∈ C [0, ∞), u(0) = 0, sup t∈[0,∞) 1 + t so, clearly, that solution could be unbounded. Our arguments apply a different technique to find bounded solutions for problem (1). Moreover, we note that, on the contrary to [12], we allow the nonlinearity f to depend on the first derivative of u. In [6], a similar third order boundary value problem is considered, namely ⎧ ⎨u (t) = f (t, u(t), u (t), u (t)) = 0, t ∈ [0, ∞), ⎩u(0) = u (0) = 0, u (∞) = m–2 i=1 αi u (ξi ), coupled with the resonance condition m–2 . αi = 1.. i=1. The techniques used in [6] are basically the same as in [12] and, again, the authors are able to find a solution which could be unbounded. On the other hand, they allow the nonlinearity f to depend on all the derivatives up to the highest possible order but, to do that, they asked for the following quite restrictive condition on the nonlinearity: (H0 ) f : [0, ∞) × R3 → R is s2 -Carathéodory, that is, (i) f (·, u, v, w) is measurable for each (u, v, w) fixed. (ii) f (t, ·, ·, ·) is continuous for a.e. t ∈ [0, ∞). (iii) For each r > 0 there exists ψr ∈ L1 [0, ∞) with tψr , t 2 ψr ∈ L1 [0, ∞) such that. f (t, u, v, w) ≤ ψr (t),. ∀(u, v, w) ∈ (–r, r) × (–r, r) × (–r, r), a.e. t ∈ [0, ∞).. Page 2 of 19.

(3) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 3 of 19. Here, we must point out that, although in this paper we work with the second-order problem, the same techniques could be applied to the third-order problem. In this sense, we allow the nonlinearity f to depend on all the derivatives up to the highest possible order but using either hypothesis (H1 ) or (H2 ) instead of (H0 ). This way, our hypotheses regarding the nonlinearity are clearly much less restrictive than (H0 ) and our method develops a different approach complementing that in [6]. We would also like to mention that our technique of modifying the problem, in order to obtain another with a related Green’s function in L1 [0, ∞) ∩ L∞ [0, ∞), is also applicable to problems without resonance. Thus, if we used this idea in problems like ⎧ ⎨u(4) (t) + ku(t) = f (t, u(t), u (t), u (t), u (t)), ⎩u(±∞) = 0,. t ∈ R,. . u (±∞) = 0,. considered in [19], we could extend the results in that reference to nonlinearities satisfying (H2 ) instead of (H1 ). The same could be said about [18]. The paper is divided into several sections: In Sect. 2, we construct an auxiliary differential problem whose solutions are the same as those of problem (1). In Sect. 3, this auxiliary problem is transformed into an integral one, for which some bounded solutions are found. These solutions are showed to be solutions of the original problem. Finally, Sect. 4 includes an example.. 2 Preliminaries We will construct now a modified problem, which will be shown to be equivalent to (1), for which it is possible to construct the related Green’s function. Indeed, consider the modified problem ⎧ ⎨u (t) + ku (t) + Mu(t) = 0, t ∈ [0, ∞), ⎩u(0) = 0, u (∞) = m–1 i=1 αi u (ξi ),. (4). where k and M are positive numbers such that k 2 – 4M < 0 and m–1 . kξ – 2i. αi e. i=1 √. k – sin (γ ξi ) + γ cos (γ ξi ) = 0, 2 2. . with γ = 4M–k 2 If we denote m–1 hl (s) =. i=l. kξi. αi e– 2 (– k2 sin (γ (s – ξi )) + γ cos (γ (s – ξi ))) , m–1 – kξi k 2 (– sin (γ ξi ) + γ cos (γ ξi )) i=1 αi e 2. then the Green’s function related to problem (4) is given by the following expression:. G(t, s) =. ⎧ ⎪ – sin (γ t)hl (s), ⎪ ⎪ ⎪ ⎪ ⎨0,. 0 ≤ t ≤ s, ξl–1 ≤ s < ξl ,. 0 ≤ t ≤ s, ξm–1 ≤ s, 1 – k(t+s) e 2 ⎪ γ ⎪ – sin (γ t)hl (s) + sin (γ (s – t)), 0 ≤ s < t, ξl–1 ≤ s < ξl , ⎪ ⎪ ⎪ ⎩ sin (γ (s – t)), 0 ≤ s < t, ξm–1 ≤ s..

(4) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. The previous expression has been obtained by some explicit calculations, which will be detailed next. First of all, from the properties of the Green’s function (see [4] for the details), we know that it must have the following form: ⎧ ⎨c1(s) cos(γ t) + c2(s) sin(γ t), 0 ≤ t ≤ s, k G(t, s) = e– 2 t ⎩c3(s) cos(γ t) + c4(s) sin(γ t), 0 ≤ s < t, for certain functions c1, . . . , c4. Moreover, as it must satisfy the boundary conditions, u(0) = 0 implies that c1 ≡ 0. Next, from the fact that lim. s→t –. ∂G ∂G (t, s) = lim+ (t, s) s→t ∂t ∂t. and lim G(t, s) = lim+ G(t, s) + 1,. s→t –. s→t. we obtain c4(s) = c2(s) +. 1 ks e 2 cos(γ s) γ. and 1 k c3(s) = – e 2 s sin(γ s). γ Finally, from the condition u (∞) = m–1 i=1 αi u (ξi ), we deduce that ⎧ ⎨– 1 e– k2 s h (s), ξ ≤ s < ξ , l l–1 l γ c2(s) = ⎩0, ξm–1 ≤ s. The first derivative of the Green’s function is given by. 1 k(t+s) ∂G (t, s) = e– 2 ∂t γ. ⎧ ⎪ 0 ≤ t ≤ s, ξl–1 ≤ s < ξl , ( k2 sin (γ t) – γ cos (γ t))hl (s), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 ≤ t ≤ s, ξm–1 ≤ s, ⎪ ⎨0, k ( 2 sin (γ t) – γ cos (γ t))hl (s) ⎪ ⎪ ⎪ ⎪ ⎪ – k2 sin (γ (s – t)) – cos (γ (s – t)), 0 ≤ s < t, ξl–1 ≤ s < ξl , ⎪ ⎪ ⎪ ⎩ k – 2 sin (γ (s – t)) – cos (γ (s – t)), 0 ≤ s < t, ξm–1 ≤ s.. Remark 1 It is easy to see that there exist two positive constants, C1 and C2 , such that. G(t, s) ≤ C1 e– k(t+s) 2 and. ∂G ≤ C2 e– k(t+s). 2 , (t, s). ∂t. for all (t, s) ∈ [0, ∞) × [0, ∞).. Page 4 of 19.

(5) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. As a consequence, it is clear that both G(t, ·) and for all t ∈ [0, ∞).. Page 5 of 19. ∂G (t, ·) ∂t. belong to L1 [0, ∞) ∩ L∞ [0, ∞). Let us consider . X = u ∈ C 1 [0, ∞) : sup u(t) < ∞, sup u (t) < ∞ t∈[0,∞). t∈[0,∞). equipped with the norm

(6) u

(7) = max

(8) u

(9) ∞ , u ∞ , where

(10) v

(11) ∞ = supt∈[0,∞) |v(t)|. It is easy to prove that (X,

(12) ·

(13) ) is a Banach space. To deal with the lack of compactness of the set X, we will use the following result: Theorem 2 ([5]) Let Y be a Banach space and C (R, Y ) the space of all bounded continuous functions x : R → Y . For a set D ⊂ C (R, Y ) to be relatively compact, it is necessary and sufficient that: 1. D is uniformly bounded; 2. Functions from D are equicontinuous on every compact subinterval of [0, ∞); 3. Functions from D are equiconvergent at ∞, that is, given ε > 0, there exists T > 0 such that for all t ≥ T, we have that x(t) – lim x(t) < ε. t→∞. To prove the existence of solutions, we will consider two different results. First of all, we will use the very well-known Schauder’s fixed point theorem: Theorem 3 ([26]) Let Y be a nonempty, closed, bounded and convex subset of a Banach space X, and suppose that P : Y → Y is a compact operator. Then P has at least one fixed point in Y . On the other hand, we will also give a result to prove the existence of solutions based on the lower and upper solutions technique. To do that, we need to introduce the following definition: Definition 4 A function α ∈ X is said to be a lower solution of problem (1) if ⎧ ⎨α (t) ≥ f (t, α(t), α (t)), ⎩α(0) ≤ 0,. t ∈ [0, ∞), m–1 α (∞) ≥ i=1 αi α (ξi ). . A function β ∈ X is said to be an upper solution of (1) if the reversed inequalities hold.. 3 Main results Consider the following integral operator T : X → X defined by . ∞. Tu(t) =. G(t, s) f s, u(s), u (s) + ku (s) + Mu(s) ds.. 0. It is clear that solutions of problem (1) are fixed points of operator T.. (5).

(14) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 6 of 19. Moreover, we will assume that at least one of the two following conditions holds: (H1 ) The nonlinearity f : [0, ∞) × R2 → R satisfies L1 -Carathéodory condition, that is, (i) f (·, u, v) is measurable for each (u, v) fixed. (ii) f (t, ·, ·) is continuous for a.e. t ∈ [0, ∞). (iii) For each r > 0 there exists ϕr ∈ L1 [0, ∞) such that. f (t, u, v) ≤ ϕr (t),. ∀(u, v) ∈ (–r, r) × (–r, r), a.e. t ∈ [0, ∞).. (H2 ) The nonlinearity f : [0, ∞) × R2 → R satisfies L∞ -Carathéodory condition, that is, (i) f (·, u, v) is measurable for each (u, v) fixed. (ii) f (t, ·, ·) is continuous for a.e. t ∈ [0, ∞). (iii) For each r > 0 there exists φr ∈ L∞ [0, ∞) such that. f (t, u, v) ≤ φr (t),. ∀(u, v) ∈ (–r, r) × (–r, r), a.e. t ∈ [0, ∞).. Under one of these conditions, we will be able to prove the following result. Lemma 5 Assume that either (H1 ) or (H2 ) holds. Then operator T defined in (5) is completely continuous. Proof The proof will be divided into several steps. Step 1: T is well-defined in X. Given an arbitrary u ∈ X, we will prove that Tu ∈ X. First, we will make the proof in case hypothesis (H1 ) holds. If u ∈ X, then there exists some r > 0 such that

(15) u

(16) < r. Therefore, it holds that . ∞ Tu(t) = . G(t, s) f s, u(s), u (s) + ku (s) + Mu(s) ds. 0 ∞. . G(t, s) f s, u(s), u (s) + k u (s) + M u(s) ds ≤ . 0. . 0. ∞. G(t, s) ϕr (s) + (k + M)r ds. ≤. ∞. ≤. C1 e–. 0. . k(t+s) 2. . ϕr (s) + (k + M)r ds. 2 ks e– 2 ϕr (s) ds + (k + M)r k 0. ∞ . 2M – kt – ks 2 2 r e ϕr (s) ds + 2 + = C1 e k 0 kt. = C1 e– 2. ∞. and, analogously, . (Tu) (t) = . ∂G (t, s) f s, u(s), u (s) + ku (s) + Mu(s) ds ∂t 0. ∞ ∂G . . . ≤ ∂t (t, s) f s, u(s), u (s) + k u (s) + M u(s) ds 0. ∞ ∂G ϕr (s) + (k + M)r ds. (t, s) ≤. ∂t 0 ∞. (6).

(17) López-Somoza and Minhós Boundary Value Problems. ≤. ∞. k(t+s) 2. C2 e–. 0 kt. (2019) 2019:38. . ϕr (s) + (k + M)r ds. . 2M ks r . e– 2 ϕr (s) ds + 2 + k. ∞. = C2 e– 2. Page 7 of 19. 0. ks. (7) ks. Now, since ϕr ∈ L1 [0, ∞) and e– 2 ∈ L∞ [0, ∞), it holds that ϕr (s)e– 2 ∈ L1 [0, ∞). Thus, it is clear that. sup Tu(t) < ∞ and. sup (Tu) (t) < ∞,. t∈[0,∞). t∈[0,∞). that is, Tu ∈ X. On the other hand, if (H2 ) holds instead of (H1 ), following similar steps as in the previous case, we obtain the following upper bounds:. Tu(t) ≤ C1 e– kt2. . ∞. 0. . 2M ks r e– 2 φr (s) ds + 2 + k. and. (Tu) (t) ≤ C2 e– kt2. . ∞. 0. . 2M ks r . e– 2 φr (s) ds + 2 + k ks. ks. In this case φr ∈ L∞ [0, ∞) and, since e– 2 ∈ L1 [0, ∞), we obtain that φr (s)e– 2 ∈ L1 [0, ∞). Therefore we conclude again that Tu ∈ X. Step 2: T is a continuous operator. We will detail the proof for the case in which (H1 ) holds. For (H2 ) the proof will be analogous, with the obvious changes, as done in Step 1. Consider a sequence {un }n∈N and assume that it converges to u in X, that is, lim un (t) = u(t). lim u (t) = u (t) n→∞ n. and. n→∞. for all t ∈ [0, ∞).. Then, since f (t, ·, ·) is continuous for a.e. t ∈ [0, ∞), we deduce that lim f s, un (s), un (s) = f s, u(s), u (s) for a.e. s ∈ [0, ∞).. n→∞. Let us see that {Tun }n∈N converges to Tu. Since {un }n∈N is convergent in X, there exists some r > 0 such that

(18) un

(19) < r for all n ∈ N. Now, if (H1 ) holds,. Tun (t) – Tu(t) ≤. . ∞. G(t, s) f s, un (s), u (s) – f s, u(s), u (s) ds n. 0. . G(t, s) k u (s) – u (s) + M un (s) – u(s) ds n. ∞. + 0. . ∞. ≤ C1. e–. k(t+s) 2. 0. . ∞. + C1 ≤ C1 0. 0 ∞. e–. f s, un (s), u (s) – f s, u(s), u (s) ds n. k(t+s) 2. k un (s) – u (s) + M un (s) – u(s) ds. ks e– 2 2ϕr (s) + 2(k + M)r ds < ∞..

(20) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Then, we deduce from Lebesgue’s Dominated Convergence Theorem that . ∞. lim

(21) Tun – Tu

(22) ∞ ≤ lim C1. n→∞. n→∞. 0. . ks e– 2 f s, un (s), un (s) – f s, u(s), u (s) ds ∞. + lim C1 n→∞. . 0. ks lim e– 2 f s, un (s), un (s) – f s, u(s), u (s) ds. ∞. = C1 0. . ks e– 2 k un (s) – u (s) + M un (s) – u(s) ds. n→∞. ks lim e– 2 k un (s) – u (s) + M un (s) – u(s) ds. ∞. + C1. n→∞. 0. = 0. Analogously, we get that lim (Tun ) – (Tu) ∞ ≤ C2. . n→∞. ks lim e– 2 f s, un (s), un (s) – f s, u(s), u (s) ds. ∞. 0. . n→∞ ∞. + C2. ks lim e– 2 k un (s) – u (s) + M un (s) – u(s) ds. n→∞. 0. = 0. Thus, {Tun }n∈N converges to Tu in X. Step 3: T is compact. Again, we will make the proof only for the case in which (H1 ) holds, the other case being analogous. Let B be a bounded subset of X, that is, there exists some r > 0 such that

(23) u

(24) < r, for all u ∈ B. Let us see that T(B) is relatively compact in X. (i) T(B) is uniformly bounded. If u ∈ B, then, for t ∈ [0, ∞),. . . 2M r e ϕr (s) ds + 2 + k 0. ∞ . 2M ks r =: M1 > 0 e– 2 ϕr (s) ds + 2 + ≤ C1 k 0. Tu(t) ≤ C1 e– kt2. ∞. – ks 2. and. . . 2M r e ϕr (s) ds + 2 + k 0. ∞ . 2M ks r =: M2 > 0. e– 2 ϕr (s) ds + 2 + ≤ C2 k 0. (Tu) (t) ≤ C2 e– kt2. ∞. – ks 2. Thus,

(25) Tu

(26) ≤ max{M1 , M2 }, for all u ∈ B, that is, T(B) is uniformly bounded.. Page 8 of 19.

(27) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 9 of 19. (ii) T(B) is equicontinuous. Let t1 , t2 ∈ [0, L] for some L > 0 and assume that t1 > t2 . Then,. Tu(t1 ) – Tu(t2 ) ∞. . G(t1 , s) – G(t2 , s) f s, u(s), u (s) + k u (s) + M u(s) ds ≤ . 0 ∞. G(t1 , s) – G(t2 , s) ϕr (s) + (k + M)r ds. ≤ . 0. = 0. t2 . G(t1 , s) – G(t2 , s) ϕr (s) + (k + M)r ds. . +. t1 . G(t1 , s) – G(t2 , s) ϕr (s) + (k + M)r ds. t2. +. ∞. G(t1 , s) – G(t2 , s) ϕr (s) + (k + M)r ds.. (8). t1. We will find some suitable upper bounds for the difference |G(t1 , s) – G(t2 , s)|. For 0 ≤ t2 < t1 ≤ s, we have two possibilities: • If ξl–1 ≤ s < ξl for some 2 ≤ l ≤ m – 1, then. kt kt G(t1 , s) – G(t2 , s) = 1 hl (s) e– ks2 –e– 21 sin(γ t1 ) + e– 22 sin(γ t2 ) . γ • If s ≥ ξm–1 , then. G(t1 , s) – G(t2 , s) = 0. On the other hand, for 0 ≤ s ≤ t2 < t1 : • If ξl–1 ≤ s < ξl for some 2 ≤ l ≤ m – 1, then. kt kt G(t1 , s) – G(t2 , s) ≤ 1 hl (s) e– ks2 –e– 21 sin(γ t1 ) + e– 22 sin(γ t2 ) γ kt2 1 ks kt1 + e– 2 e– 2 sin γ (s – t1 ) – e– 2 sin γ (s – t2 ) . γ Last term in the previous sum can be upper-bounded as follows: – kt1 kt e 2 sin γ (s – t1 ) – e– 22 sin γ (s – t2 ) . kt1. kt2 = sin(γ s) e– 2 cos(γ t1 ) – e– 2 cos(γ t2 ). . kt1 kt2 + cos(γ s) –e– 2 sin(γ t1 ) + e– 2 sin(γ t2 ). kt1 kt2 ≤ e– 2 cos(γ t1 ) – e– 2 cos(γ t2 ). kt1 kt2 + –e– 2 sin(γ t1 ) + e– 2 sin(γ t2 ) ..

(28) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 10 of 19. As a consequence,. . kt kt G(t1 , s) – G(t2 , s) ≤ 1 hl (s) + 1 e– ks2 –e– 21 sin(γ t1 ) + e– 22 sin(γ t2 ) γ. kt2 1 ks kt1 + e– 2 e– 2 cos(γ t1 ) – e– 2 cos(γ t2 ) . γ • If s ≥ ξm–1 , then kt. kt G(t1 , s) – G(t2 , s) = 1 e– ks2 e– 21 sin γ (s – t1 ) – e– 22 sin γ (s – t2 ) γ. kt2 1 ks kt1 ≤ e– 2 e– 2 cos(γ t1 ) – e– 2 cos(γ t2 ) γ. kt1 kt2 1 ks + e– 2 –e– 2 sin(γ t1 ) + e– 2 sin(γ t2 ) . γ Therefore, we can affirm that for a given ε > 0 there exists some δ > 0 such that if |t1 – t2 | < δ then, for s ∈ [0, t2 ) ∪ (t1 , ∞), it holds that. G(t1 , s) – G(t2 , s) ≤ εe– ks2 . This implies that the first and third terms of the last part of inequality (8) tend to zero regardless of the function u ∈ B. On the other hand, for 0 ≤ t2 ≤ s ≤ t1 : • If ξl–1 ≤ s < ξl for some 2 ≤ l ≤ m – 1, then. . kt kt G(t1 , s) – G(t2 , s) ≤ 1 e– ks2 hl (s) –e– 21 sin(γ t1 ) + e– 22 sin(γ t2 ) γ 1 ks kt1 + e– 2 e– 2 sin γ (s – t1 ) . γ • If s ≥ ξm–1 , then kt. G(t1 , s) – G(t2 , s) = 1 e– ks2 e– 21 sin γ (s – t1 ) . γ Thus, when s ∈ [t2 , t1 ], it holds that. G(t1 , s) – G(t2 , s) ≤ Ce– ks2 , for some constant C. This implies that |G(t1 , ·) – G(t2 , ·)|(ϕr (·) + (k + M)r) ∈ L1 [t1 , t2 ] for any t1 , t2 ∈ [0, ∞). Then it is clear that . t1 . t2. G(t1 , s) – G(t2 , s) ϕr (s) + (k + M)r ds −−−→ 0 t1 →t2. regardless of the function u ∈ B. Thus we conclude that given ε > 0 there exists δ(ε) > 0 such that if |t1 – t2 | < δ, then |Tu(t1 ) – Tu(t2 )| < ε for all u ∈ B..

(29) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 11 of 19. In a completely analogous way, finding suitable upper bounds for (t , s) – | ∂G ∂t 1. ∂G (t , s)|, ∂t 2. it is possible to prove that given ε > 0 there exists δ(ε) > 0 such. that if |t1 – t2 | < δ, then |(Tu) (t1 ) – (Tu) (t2 )| < ε for all u ∈ B. Therefore, T(B) is equicontinuous.. (iii) T(B) is equiconvergent at ∞. Given u ∈ B, it holds that. . – kt 2 Tu(t) – lim Tu(t) ≤ C1 e t→∞. . 2M r e ϕr (s) ds + 2 + k 0 . ∞ 2M kt ks r e– 2 ϕr (s) ds + 2 + + lim C1 e– 2 t→∞ k 0 . ∞ 2M – kt – ks 2 2 r −−−→ 0 e ϕr (s) ds + 2 + ≤ C1 e t→∞ k 0 ∞. – ks 2. and . 2M r e ϕr (s) ds + 2 + k 0 . ∞ 2M kt ks r e– 2 ϕr (s) ds + 2 + + lim C2 e– 2 t→∞ k 0. ∞ . 2M kt ks ≤ C2 e– 2 r −−−→ 0, e– 2 ϕr (s) ds + 2 + t→∞ k 0. . – kt 2 (Tu) (t) – lim (Tu) (t) ≤ C2 e t→∞. ∞. – ks 2. . that is, TB is equiconvergent at ∞. Therefore, from Theorem 2, we conclude that T(B) is relatively compact in X. Now we will see our existence results.. Theorem 6 Let f : [0, ∞) × R2 → R be such that there exists t0 ∈ [0, ∞) for which f (t0 , 0, 0) = 0. Moreover, suppose that, for C1 and C2 given in Remark 1, either • (H1 ) holds and, moreover, there exists some R > 0 such that t kt ks max{C1 , C2 } max sup e– 2 e– 2 ϕR (s) ds, t>ξm–1. 0. ξm–1. ks. . e– 2 ϕR (s) ds. 0. . kξm–1 M 1 + max{C1 , C2 } max , 2 1 – e– 2 1+ R < R, 2 k. (9). or • (H2 ) holds and, moreover, there is R > 0 such that t kt ks e– 2 φR (s) ds, max{C1 , C2 } max sup e– 2 t>ξm–1. 0. 0. ξm–1. ks. e– 2 φR (s) ds. . kξm–1 M 1 + max{C1 , C2 } max , 2 1 – e– 2 1+ R < R. 2 k Then problem (1) has at least one nontrivial solution.. .

(30) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 12 of 19. Proof We will prove the first case, the second being analogous. Consider D = u ∈ X :

(31) u

(32) < R . If u ∈ D then. Tu(t) ≤. . ∞. G(t, s) ϕR (s) + (k + M)R ds, ∀t ∈ [0, ∞),. 0. and, since G(t, s) = 0 for s ≥ max{t, ξm–1 },. Tu(t) ≤. . max{t,ξm–1 } . G(t, s) ϕR (s) + (k + M)R ds, ∀t ∈ [0, ∞).. 0. If t > ξm–1 , the previous expression leads to. Tu(t) ≤. . t. G(t, s) ϕR (s) + (k + M)R ds. 0 kt. . t. ≤ C1 e– 2. ks e– 2 ϕR (s) + (k + M)R ds. 0. t M kt ks kt kt R e– 2 ϕR (s) ds + 2e– 2 1 – e– 2 1 + ≤ C1 e– 2 k 0. t M 1 – kt – ks 2 2 ≤ C1 e 1+ R . e ϕR (s) ds + 2 k 0 On the other hand, if t ≤ ξm–1 , we obtain that. Tu(t) ≤. . ξm–1 . G(t, s) ϕR (s) + (k + M)R ds. 0 kt. . ξm–1. ≤ C1 e– 2. ks e– 2 ϕR (s) + (k + M)R ds. 0. M 1–e 1+ R e ϕR (s) ds + 2e k 0. . ξm–1 kξ M – ks – m–1 2 2 1+ ≤ C1 R . e ϕR (s) ds + 2 1 – e k 0. kt ≤ C1 e– 2. ξm–1. – ks 2. – kt 2. . kξ – m–1 2. . Therefore, t . ks Tu(t) ≤ C1 max sup e– kt2 e– 2 ϕR (s) ds, t>ξm–1. 0. ξm–1. ks. . e– 2 ϕR (s) ds. 0. . kξm–1 M 1 + C1 max , 2 1 – e– 2 1+ R, 2 k. ∀t ∈ [0, ∞)..

(33) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 13 of 19. Analogously, it can be seen that t. ks (Tu) (t) ≤ C2 max sup e– kt2 e– 2 ϕR (s) ds, t>ξm–1. ξm–1. . ks. e– 2 ϕR (s) ds. 0. 0. . kξm–1 M 1 + C2 max , 2 1 – e– 2 1+ R, 2 k. ∀t ∈ [0, ∞).. Thus, by (9), t kt ks

(34) Tu

(35) ≤ max{C1 , C2 } max sup e– 2 e– 2 ϕR (s) ds, t>ξm–1. ξm–1. ks. . e– 2 ϕR (s) ds. 0. 0. . kξm–1 M 1 + max{C1 , C2 } max , 2 1 – e– 2 1+ R < R, 2 k . that is, Tu ∈ D. Therefore, TD ⊂ D and, from Theorem 3, the operator T has at least one fixed point in D, which is a solution of problem (1). Moreover, since there exists at least one value t0 ∈ [0, ∞) for which f (t0 , 0, 0) = 0, this solution cannot be the trivial one. Now, we will give another existence result based on the lower and upper solutions technique. The proof will follow the lines of [23]. Before formulating the theorem, we will give a lemma that we will use in the proof. Lemma 7 ([23, Lemma 2]) Let u ∈ C 1 (I), v, w ∈ W 2,1 (I) and define. p t, u(t) =. ⎧ ⎪ ⎪ ⎨v(t), ⎪ ⎪ ⎩. u(t),. u(t) < v(t), v(t) ≤ u(t) ≤ w(t),. w(t), u(t) > w(t).. Then, the two following properties hold: 1. dtd p(t, u(t)) exists for a.e. t ∈ I. 2. If u, um ∈ C 1 (I) and um converges to u in C 1 (I), then lim. m→∞. d d p t, um (t) = p t, u(t) , dt dt. a.e. t ∈ I.. Theorem 8 Let α, β ∈ X be lower and upper solutions of problem (1), respectively, with α(t) ≤ β(t),. ∀t ∈ [0, ∞),. and denote R = max

(36) α

(37) ∞ ,

(38) β

(39) ∞ , α ∞ , β ∞ . Assume that, for C1 and C2 given by Remark 1, either. (10).

(40) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 14 of 19. • (H1 ) holds and, moreover, there exists some R > 0 such that t kt ks e– 2 ϕmax{R,R} (s) ds, max{C1 , C2 } max sup e– 2 t>ξm–1. 0. ξm–1. . ks. e– 2 ϕmax{R,R} (s) ds. 0. . kξm–1 M 1 1+ R < R, + max{C1 , C2 } max , 2 1 – e– 2 2 k or • (H2 ) holds and, moreover, there exists some R > 0 such that t kt ks e– 2 φmax{R,R} (s) ds, max{C1 , C2 } max sup e– 2 t>ξm–1. ξm–1. – ks 2. e. φmax{R,R} (s) ds. 0. 0. . kξ M 1 – m–1 2 + max{C1 , C2 } max , 2 1 – e 1+ R < R. 2 k Then, problem (1) has a solution u ∈ X such that α(t) ≤ u(t) ≤ β(t),. ∀t ∈ [0, ∞).. Proof We will prove the first case, the second being analogous. Let ε > 0 be such that t kt ks e– 2 ϕmax{R,R} (s) ds, max{C1 , C2 } max sup e– 2 t>ξm–1. 0. ξm–1. . ks. e– 2 ϕmax{R,R} (s) ds. 0. . kξm–1 M ε 1 1+ R + (R + R) < R. + max{C1 , C2 } max , 2 1 – e– 2 2 k k Consider the modified problem ⎧ d ⎪ ⎪ ⎨u (t) + ku (t) + Mu(t) = f (t, δ(t, u(t)), dt δ(t, u(t))) ⎪ ⎪ ⎩. + ku (t) + Mu(t) + ε(u(t) – δ(t, u(t))), u (∞) = m–1 i=1 αi u (ξi ),. u(0) = 0,. where the function δ : [0, ∞) × R → R is given by ⎧ ⎪ ⎪ ⎨β(t), u(t) > β(t),. δ t, u(t) = u(t), ⎪ ⎪ ⎩ α(t),. α(t) ≤ u(t) ≤ β(t), u(t) < α(t).. Define now operator T ∗ : X → X by. d G(t, s) f s, δ s, u(s) , δ s, u(s) T u(t) = ds 0. ds. + ku (s) + Mu(s) + ε u(s) – δ s, u(s) ∗. . ∞. t ∈ [0, ∞),. (11).

(41) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 15 of 19. d From Lemma 7, we know that ds δ(s, u(s)) exists for a.e. s ∈ [0, ∞). Thus, Lemma 7, together with the Carathéodory condition on the nonlinearity, implies that the integral in the latter expression is well-defined. Following the same steps as in Lemma 5, it is easy to prove that if (H1 ) holds, then T ∗ is well-defined in X and it is a completely continuous operator. Moreover, by (10), it is clear that. d δ t, u(t) ≤ max

(42) u

(43) , R for all t ∈ [0, ∞).. dt. δ t, u(t) ≤ R and Thus, if we consider D = u ∈ X :

(44) u

(45) < R. and u ∈ D then, following analogous steps as in the proof of Theorem 6, it can be deduced that t ∗ ks T u ≤ max{C1 , C2 } max sup e– kt2 e– 2 ϕmax{R,R} (s) ds, t>ξm–1. 0. ξm–1. ks. . e– 2 ϕmax{R,R} (s) ds. 0. . . kξm–1 M 1 ε 1+ + max{C1 , C2 } max , 2 1 – e– 2 R + (R + R) < R, 2 k k that is, T ∗ u ∈ D. Therefore, TD ⊂ D and, from Theorem 3, T ∗ has at least one fixed point in D, which corresponds to a solution of problem (11). Finally, we will prove that this solution u of problem (11) satisfies α(t) ≤ u(t) ≤ β(t),. ∀t ∈ [0, ∞),. which implies that it is also a solution of problem (5). Define v(t) = u(t) – β(t) and consider t0 ∈ [0, ∞) ∪ {∞} such that v(t0 ) := sup v(t) : t ∈ [0, ∞) . Suppose that v(t0 ) > 0. Then, since v(0) = –β(0) ≤ 0, necessarily t0 = 0. Thus, there exists t ∈ [0, ∞) such that v(t) > 0. for t ∈ (t, t0 ) and. v (t) ≥ 0.. Now, using the facts that u is a solution of (11) and β is an upper solution of problem (1), we obtain for t ∈ (t, t0 ) that. d u (t) = f t, δ t, u(t) , δ t, u(t) + ε u(t) – δ t, u(t) dt = f t, β(t), β (t) + ε u(t) – β(t) ≥ β (t) + ε u(t) – β(t) ..

(46) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 16 of 19. Thus, we deduce that v (t) = u (t) – β (t) ≥ ε u(t) – β(t) = εv(t) > 0,. t ∈ (t, t0 ),. which implies that v is strictly increasing on (t, t0 ). In particular, since v (t) ≥ 0, it must be true that v > 0 on (t, t0 ). Now, the fact that v > 0 on (t, t0 ) implies that t0 = ∞ and so lim v (t) > 0.. t→∞. On the other hand, since v ∈ X, v(∞) = sup v(t) = C ∈ R t∈[0,∞). and, using L’Hôpital’s Rule, et v(t) et v(t) + et v (t) = lim = lim v(t) + v (t) = C + lim v (t), t t→∞ e t→∞ t→∞ t→∞ et. C = lim v(t) = lim t→∞. hence we deduce that lim v (t) = 0,. t→∞. which is a contradiction. Therefore sup v(t) : t ∈ [0, ∞) ≤ 0, that is, u(t) ≤ β(t),. t ∈ [0, ∞).. Analogously, it can be seen that u(t) ≥ α(t),. t ∈ [0, ∞).. This way, we conclude that u is a solution of problem (1).. . 4 Example Let us consider the following boundary value problem: ⎧ ⎨u (t) =. |1–u(t)| 1 (2 + sin t)e–|u(t)| (u(t)) 2 +1 (u (t) – 1), 1000 . ⎩u(0) = 0,. t ∈ [0, ∞),. (12). u (∞) = 0.11u (0) + 0.89u (0.11).. This problem is a particular case of (1) with f (t, x, y) = α1 = 0.11, α2 = 0.89, ξ1 = 0 and ξ2 = 0.11.. 1 (2 + sin t)e–|x| |1–x| (y – 1), 1000 x2 +1. m = 3,.

(47) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Page 17 of 19. For |x|, |y| < r, it holds that. f (t, x, y) ≤. 1 (2 + sin t)(r + 1)2 , 1000. 1 (2 + sin t)(r + 1)2 and hence hypothesis (H2 ) holds. We note so we could take φr (t) = 1000 1 that, since φr ∈/ L [0, ∞), results in [12] cannot be applied to solve this problem. We will look for a pair of lower and upper solutions of problem (1) and suitable values for k and M for which the hypotheses in Theorem 8 hold. As lower and upper solutions we will take. α(t) =. t2 – t 3 –(t + 1)e–t + 2 400 t +1. and. ∀t ∈ [0, ∞).. β(t) = 1,. It can be checked that

(48) α

(49) ∞ ≈ 0.0087,

(50) α

(51) ∞ ≈ 0.0065 and

(52) β

(53) ∞ = 1,

(54) β

(55) ∞ = 0. Therefore, we obtain that R given in (10) is R = 1. Moreover, for M = 0.35 and k = 0.86, we obtain the following approximations for C1 and C2 : C1 ≈ 1.2305,. C2 ≈ 1.3395.. Therefore, . kξm–1 M 1 max{C1 , C2 } max , 2 1 – e– 2 1+ ≈ 0.9423. 2 k On the other hand, . ξm–1. 2 ks e– 2 φmax{R,R} (s) ds ≈ 0.00022 max{R, R} + 1. 0. and kt. . t. sup e– 2 t>ξm–1. 2 ks e– 2 φmax{R,R} (s) ds ≈ 0.00174 max{R, R} + 1 .. 0. Therefore, we can approximate t – kt – ks 2 2 max{C1 , C2 } max sup e e φmax{R,R} (s) ds, t>ξm–1. 0. ξm–1. – ks 2. e. φmax{R,R} (s) ds. 0. . kξ M 1 – m–1 2 + max{C1 , C2 } max , 2 1 – e 1+ R 2 k 2 ≈ 0.00233 max{R, 1} + 1 + 0.9423R, and it can be seen that for R ∈ (R0 , R1 ), with R0 ≈ 0.1615 and R1 ≈ 22.7199, it holds that 2 0.00233 max{R, 1} + 1 + 0.9423R < R..

(56) López-Somoza and Minhós Boundary Value Problems. (2019) 2019:38. Therefore, we have proved the existence of a solution u of problem (12) such that. t2 – t 3 –t –(t + 1)e + 2 ≤ u(t) ≤ 1, 400 t +1. ∀t ∈ [0, ∞),. which implies that this solution is nontrivial.. Acknowledgements Not applicable. Funding First author was partially supported by Xunta de Galicia (Spain), project EM2014/032, AEI Spain and FEDER, grants MTM2013-43014-P, MTM2016-75140-P, and FPU scholarship, Ministerio de Educación, Cultura y Deporte, Spain. Availability of data and materials Not applicable. Competing interests The authors declare that there is no conflict of interest regarding the publication of this article. Authors’ contributions Each author equally contributed to this paper, and read and approved the final manuscript. Author details 1 Departamento de Estatística, Análise Matemática e Optimización Instituto de Matemáticas, Facultade de Matemáticas, Universidade de Santiago de Compostela, Galicia, Spain. 2 Departamento de Matemática, Escola de Ciências e Tecnologia, Centro de Investigação em Matemática e Aplicações (CIMA), Instituto de Investigação e Formação Avançada, Universidade de Évora, Évora, Portugal.. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 21 October 2018 Accepted: 7 February 2019 References 1. Agarwal, R.P., O’Regan, D.: Infinite Interval Problems for Differential, Difference and Integral Equations (2001) 2. Bai, C., Fang, J.: Existence of positive solutions for three-point boundary value problems at resonance. J. Math. Anal. Appl. 291, 538–549 (2004) 3. Bebernes, J., Gaines, R., Schmitt, K.: Existence of periodic solutions for third and fourth order ordinary differential equations via coincidence degree. Ann. Soc. Sci. Brux. Sér. I 88, 25–36 (1974) 4. Cabada, A.: Green’s Functions in the Theory of Ordinary Differential Equations. Springer Briefs in Math. (2014) 5. Corduneanu, C.: Integral Equations and Stability of Feedback Systems. Academic Press, New York (1973) 6. Djebali, S., Guedda, L.: A third order boundary value problem with nonlinear growth at resonance on the half-axis. Math. Methods Appl. Sci. 40(8), 2859–2871 (2017) 7. Feng, W., Webb, J.R.L.: Solvability of three-point boundary value problems at resonance. Nonlinear Anal. 30, 3227–3238 (1997) 8. Franco, D., Infante, G., Zima, M.: Second order nonlocal boundary value problems at resonance. Math. Nachr. 284, 875–884 (2011) 9. Han, X.: Positive solutions for a three-point boundary value problem at resonance. J. Math. Anal. Appl. 336, 556–568 (2007) 10. Infante, G., Pietramala, P., Tojo, F.A.F.: Nontrivial solutions of local and nonlocal Neumann boundary value problems. Proc. R. Soc. Edinb. A 146, 337–369 (2016) 11. Jiang, W.: Solvability for p-Laplacian boundary value problem at resonance on the half-line. Bound. Value Probl. 2013, 207 (2013) 12. Jiang, W., Yang, C.: The existence of positive solutions for multi-point boundary value problem at resonance on the half-line. Bound. Value Probl. 2016, 13, 12 (2016) 13. Kaufmann, E.R.: Existence of solutions for a nonlinear second-order equation with periodic boundary conditions at resonance. Commun. Appl. Anal. 18, 163–174 (2014) 14. Kim, C.-G.: Existence and iteration of positive solutions for multi-point boundary value problems on a half-line. Comput. Math. Appl. 61(7), 1898–1905 (2011) 15. Kosmatov, N.: Multi-point boundary value problems on an unbounded domain at resonance. Nonlinear Anal. 68, 2158–2171 (2008) 16. Lian, H., Ge, W.: Solvability for second-order three-point boundary value problems on a half-line. Appl. Math. Lett. 19(10), 1000–1006 (2006) 17. Ma, R., Yang, Y.: Existence result for a singular nonlinear boundary value problem at resonance. Nonlinear Anal. 68, 671–680 (2008) 18. Minhós, F., Carrasco, H.: Existence of homoclinic solutions for nonlinear second-order problems. Mediterr. J. Math. 13, 3849 (2016). Page 18 of 19.

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