On multi-point resonant problems on the half-line

In this work we obtain sufficient conditions for the existence of bounded solutions of a resonant multi-point 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 to obtain an existence result and a localization theorem. Our hypotheses are clearly much less restrictive than the ones existent in the literature and, moreover, they can be applied to higher order, resonant or non-resonant, boundary value problems defined on the half-line or even on the real line.


Introduction
In this paper, we will prove the existence of bounded solutions for the multi-point boundary value problem        u ′′ (t) = f (t, u(t), u ′ (t)), t ∈ [0, ∞), where α i > 0 and 0 = ξ 1 < · · · < ξ m−1 < +∞. We assume that the coefficients α i satisfy the resonant condition 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), has a nontrivial solution. These resonant problems have been studied for many years under a huge variety of arguments: degree theory has been used in, for instance, [4,7,13,22], Lyapunov-Schmidt arguments, [17], a Leggett-Williams theorem [8,21], fixed point and fixed point index theories, [3,9,10,23], monotone method together with upper and lower solutions technique, [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, ... For more details, techniques and applications in this field we refer, for example, to [11,[14][15][16]24], and the monograph [1].
In a theoretical point of view, resonance problems can be formulated as an equation Lx = N x, where L is a noninvertible operator. Therefore, in particular, the resonant 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, t ∈ [0, ∞), 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 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 The techniques used in [6] are basically the same than 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: (ii) f (t, ·, ·, ·) is continuous for a. e. t ∈ [0, ∞).
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 (H 1 ) or (H 2 ) instead of (H 0 ). This way, our hypotheses are clearly much less restrictive than (H 0 ) so our method improves the results in [6].
We would also like to mention that our technique of modifying the problem, in order to obtain another one with a related Green's function in L 1 [0, ∞) ∩ L ∞ [0, ∞), is also applicable to problems without resonance. Thus, if we used this idea in problems like [19], we could extend the results in that reference to nonlinearities satisfying (H 2 ) instead of (H 1 ). The same could be said about [18].
The paper is divided into several sections: In Section 2, we construct an auxiliary differential problem whose solutions are the same than those of problem (1). In Section 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, Section 4 includes an example which can not be solved with the results in [12].

Preliminaries
We will construct now a modified problem, which will be shown equivalent to (1), for which it is possible to construct the related Green's function.
Indeed, consider the modified problem where k and M are positive numbers such that k 2 − 4 M < 0 and , then the Green's function related to problem (4) is given by the following expression: The first derivative of the Green's function is given by It is easy to see that there exist two positive constants, C 1 and C 2 , such that On the other hand, to deal with the lack of compactness of the set X that we will consider in the following section, we will use the following result: . Let E be a Banach space and C(R, E) the space of all bounded continuous functions x : R → E. For a set D ⊂ C(R, E) to be relatively compact, it is necessary and sufficient that: 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 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: . 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: A function β ∈ X is said to be an upper solution of (1) if the reversed inequalities hold.

Main results
Let us consider It is easy to prove that (X, · ) is a Banach space.
Consider the following integral operator T : X → X defined by It is clear that solutions of problem (1) are fixed points of operator T . Moreover, we will assume that at least one of the two following conditions holds: Under one of these conditions we will be able to prove the following result.
Lemma 5. Assume that either (H 1 ) or (H 2 ) 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 T u ∈ X.
First, we will make the proof in case hypothesis (H 1 ) holds. If u ∈ X, then there exists some r > 0 such that u < r. Therefore, it holds that and, analogously, Now, since ϕ r ∈ L 1 [0, ∞) and e − ks On the other hand, if (H 2 ) holds instead of (H 1 ), following similar steps to the previous case, we obtain the following upper bounds: ∞ 0 e − ks 2 φ r (s) ds + 2 + 2 M k r .
Step 2: T is a continuous operator.
We will detail the proof for the case in which (H 1 ) holds. For (H 2 ) the proof will be analogous, with the obvious changes, as it occurred in Step 1.
Consider the sequence {u n } n∈N and assume that it converges to u in X, that is, Then, since f (t, ·, ·) is continuous for a. e. t ∈ [0, ∞), it is deduced that Let's see that {T u n } n∈N converges to T u.
Since {u n } n∈N is convergent in X, then there exists some r > 0 such that u n < r for all n ∈ N. Now, if (H 1 ) holds, Then, we deduce from Lebesgue's Dominated Convergence Theorem that Analogously, we get that Thus, {T u n } n∈N converges to T u in X.
Step 3: T is compact. Again, we will make the proof only for the case in which (H 1 ) holds, being the other one analogous.
Let B be a bounded subset of X, that is, there exists some r > 0 such that u < 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, ∞), for all u ∈ B, that is, T (B) is uniformly bounded.
• If s ≥ ξ m−1 , then Therefore, we can affirm that for a given ε > 0 there exists some δ > 0 such that if |t 1 − t 2 | < δ then, for s ∈ [0, t 2 ) ∪ (t 1 , ∞), it holds that This implies that the first and third terms of the last part of inequality (8) tend to zero with independence of the function u ∈ B.

(iii) T (B) is equiconvergent at ∞:
Given u ∈ B, it holds that that is, T B is equiconvergent at ∞. Therefore, from Theorem 2, we conclude that T (B) is relatively compact in X.
Now we will see our existence results.
Proof. We will prove the first case, being the second one analogous. Consider D = {u ∈ X : u < R}.
If u ∈ D then, and, since G(t, s) = 0 for s ≥ max{t, ξ m−1 }, If t > ξ m−1 , the previous expression leads to On the other hand, if t ≤ ξ m−1 , we obtain that Therefore, Analogously, it can be seen that Thus, by (9), Therefore, T D ⊂ 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 some value t 0 ∈ [0, ∞) for which f (t 0 , 0, 0) = 0, this solution can not be the trivial one. Now, we will give another existence result based on the lower and upper solutions technique: Theorem 7. Let α, β ∈ X be lower and upper solutions of problem (1), respectively, with Assume that the nonlinearity f (t, x, y) is nondecreasing in y and, suppose that, for C 1 and C 2 given by Remark 1, either • (H 1 ) holds and, moreover, there exists some R > 0 such that

or
• (H 2 ) holds and, moreover, there exists some R > 0 such that Then, problem (1) has a solution u ∈ X such that Proof. We will prove the first case, being the second one analogous. Let ε > 0 be such that Consider the modified problem where the function δ : [0, ∞) × R → R is given by Define now operator T * : X → X by T * u(t) = ∞ 0 G(t, s) f (s, δ(s, u(s)), u ′ (s)) + k u ′ (s) + M u(s) + ε(u(s) − δ(s, u(s))) .
Following the same steps as in Lemma 5, it is easy to prove that if (H 1 ) holds, then T * is well-defined in X and it is a completely continuous operator.
Moreover, it is clear that, by (10), |δ(t, u(t))| ≤ R for all t ∈ [0, ∞). Thus, if we consider D = {u ∈ X : u < R} and u ∈ D then, following analogous steps than in the proof of Theorem 6, it can be deduced that Therefore, T D ⊂ 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 that which implies that it is also a solution of problem (5).
We have that for |x|, |y| < r, it holds that |f (t, x, y)| ≤ 1 1000 (2 + sin t) (r + 1) 2 , so we could take φ r (t) = 1 1000 (2 + sin t) (r + 1) 2 and hypothesis (H 2 ) holds. We note that, since φ r / ∈ L 1 [0, ∞), results in [12] can not 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 7 hold.
Moreover, for M = 0.35 and k = 0.86, we obtain the following approximations for C 1 and C 2 :