Resonant Problem for a Class of Bvps on the Half-line

We provide an existence result for a Neumann nonlinear boundary value problem posed on the half-line. Our main tool is the multi-valued version of the Miranda Theorem.

In recent years, the existence of solutions for boundary-value problems received wide attention.We can write down almost any nonlinear differential equation in the form Lx = N(x), where L is a linear and N nonlinear operator in appropriate Banach spaces.If kernel L is nontrivial then the equation is called resonant and one can manage the problem by using the coincidence degree in that case.But, if the domain is unbounded the operator is usually non-Fredholm (like in our case).Here, for instance, we can use the perturbation method (see [2]) or we can work in the space H 2 (R + ) (see [7]) to obtain the existence results.
The method presented below enables us to get the existence result under weaker assumptions than those mentioned in the above cited methods.Consequently, we will need some facts from set-valued analysis and multi-valued version of the Miranda Theorem to get the Theorem about the existence of at least one solution to the resonant problem.Now, we shall set up notation and terminology.
EJQTDE, 2009 No. 53, p. 1 Denote by BC(R + , R) (we write BC) the Banach space of continuous and bounded functions with supremum norm and by BCL(R + , R) (we write BCL) its closed subspace of continuous and bounded functions which have finite limits at +∞.
The following theorem gives a sufficient condition for compactness in the space BC, ( [6]).
By a space we mean a metric space.Let X, Y be spaces.A set-valued map Φ : The theorem below is a generalization of the well known Miranda Theorem [5, p. 214] which gives zeros of single-valued continuous maps.
The result, we shall present in this paper, is a generalization of the sublinear case considered in [8].Therefore, the following Lemma will be of crucial importance for our reason.R) is non-negative and satisfies the following inequality where C ≥ 0 is a real constant and ω(s) is a continuous, positive and nondecreasing function such that W (+∞) = +∞, where .
Then, one has for all t ≥ 0 that

Resonant problem
Let us consider an asymptotic BVP where The following assumption will be needed in this paper: (ii) there exists M > 0 such that xf (t, x, y) ≥ 0 for t ≥ 0, y ∈ R and |x| ≥ M. Now, we can formulate our main result.
Theorem 3.Under assumption (i) and (ii) problem (3) has at least one solution.
The proof will be divided into a sequence of Lemmas.
First, we consider problem for fixed c ∈ R. Observe that ( 4) is equivalent to an initial value problem Since f is continuous, then by the Local Existence Theorem we get that problem (5) has at least one local solution.We can write (4) as Set By (i) and ( 7) we get Now, due to Bihari's Lemma (see Lemma 1), we have Hence, by the Theorem on a Priori Bounds [5, p. 146], (5) has a global solution for t ≥ 0. We obtain that (4) has a global solution for t ≥ 0.Moreover, by assumption (i) and ( 7), we have Hence all global solutions are bounded for t ≥ 0. The function t → f (t, c + t 0 y c (u)du, y c (t)) is absolutely integrable; i.e., In particular, there exists a limit lim t→∞ y c (t), for every c.Thus all solutions of (4) have finite limits at +∞.

EJQTDE, 2009 No. 53, p. 4
Let us consider the nonlinear operator A, R × BCL ∋ (c, x) → A c (x) ∈ BCL, given by It is easy to see that A is well defined.By using the Lebesgue Dominated Convergence Theorem one can prove the continuity of A.
Lemma 2. Under assumption (i) operator A is completely continuous.
Proof.We shall show that the image of Now, we prove condition (2).We show that for any t 0 ≥ 0 and ε > 0 there exists δ > 0 that for each Let δ = min{δ 1 , δ 2 }.Then, for |t − t 0 | < δ, we get It remains to prove condition (3).By assumption (i) for every ε > 0 there exist t 1 , t 2 large enough that The proof is complete.
Note that the solutions of ( 4) are fixed points of operator A defined by (10).Let fix A c (•) denote the set of fixed points of operator A c , where c is given.Let us consider the multifunction g : R ⊸ R given by By (9), we get that the solutions of ( 4) are equibounded for any c.Hence both sequences (x n ) and (c n ) are bounded.Proposition 2 yields that operator A is completely continuous.Then, by ( 14), (x n ) is relatively compact.Hence, passing to a subsequence if necessary, we may assume that x n → x 0 in BCL.The continuity of A implies that Hence, x 0 ∈ Φ(c 0 ) and the proof is complete.Proof.We shall show that fix A c (•) is connected in BCL.On the contrary, suppose that the set is not connected.Since fix A c (•) is compact, there exist compact sets A and B such that By ( 9) there exists T > 0 such that for any y

Theorem 2 .
([8]) Let g be an USC map from [− M , M] into convex and compact subsets of R and satisfying the conditions: if d ∈ g( M), then d ≥ 0(1)

Lemma 4 .
Let assumption (i) hold.Then the set-valued map Φ has connected values.