Second-order discontinuous problems with nonlinear functional boundary conditions on the half-line

We establish sufficient conditions for the existence of extremal solutions for a second-order problem on the half-line with discontinuous right-hand side and nonlinear boundary conditions. Our results are new even in the classical case of continuous nonlinearities. We illustrate their applicability with an example.


Introduction
We study the existence of solutions of the nonlinear equation on the half-line coupled with the functional boundary conditions L(x(0), x (0), x) = 0, x (+∞) := lim where B ∈ R and L : C([0, ∞)) × R 2 → R is a continuous function and it is nonincreasing in the second and third variables.Boundary value problems on unbounded intervals are becoming popular in the literature because of their applications to model real world problems in engineering or chemistry, see [1].For instance, second-order nonlinear differential problems arise in the investigation of radial solutions for elliptic equations.
The functional boundary conditions considered here are quite general and were recently studied in [8].They include the extensively studied Sturm-Liouville conditions [11,12], integral boundary conditions [2], multipoint [18] and other boundary conditions with, for example, maximum or minimum arguments.Observe that the boundary condition at infinity implies that solutions, if they exist, are unbounded if B = 0.
The main novelty here is that we allow the function f to satisfy weaker conditions than usual.In particular, it may be discontinuous in the second variable over a countable number of admissible curves.This more general case wants new fixed point theorems which generalize the classical Schauder's fixed point theorem [9,15].
We shall assume the existence of well ordered upper and lower solutions on unbounded domains and a Nagumo condition to control the first derivative in order to obtain existence results for (1.1)- (1.2).Another interesting point in this paper is that we are able to relax the usual definition of lower and upper solutions, cf.[2,8,[11][12][13], and therefore our main existence result is new even in the classical case of continuous right-hand sides in (1.1).See Remark 3.2 for details.
Furthermore, we also prove the existence of extremal solutions for (1.1)-(1.2) by adapting the arguments in [5] to unbounded domains.This is also a new result even for a continuous function f (t, x, x ).
This paper is organized as it follows: in Section 2 we present the fixed point tools which we will employ later and we introduce the Banach space where we will look for solutions to (1.1)-(1.2) as well as the definition for lower and upper solutions.The integral operator associated to problem (1.1)-(1.2) is also studied here.In Section 3, an existence and localization result is established.Finally, in Section 4, we investigate the existence of extremal solutions between the lower and upper functions, that is, a least and a greatest solution.

Definition and preliminaries
First, we present some definitions and fixed point theorems based on multivalued theory which will be the key to work with discontinuous operators.
Let K be a nonempty subset of a normed space (X, • ) and T : K −→ X an operator, not necessarily continuous.
Definition 2.1.The closed-convex envelope of an operator T : K −→ X is the multivalued mapping T : K −→ 2 X given by where B ε (x) denotes the closed ball centered at x and radius ε, and co means closed convex hull.
Remark 2.2.The closed-convex envelope (cc-envelope, for short) is similar to the Krasovskij regularization (see [10]), but here we are 'convexifying' operators instead of nonlinear parts of differential equations.
Remark 2.3.Note that T is an upper semicontinuous multivalued mapping which assumes closed and convex values (see [4,15]) provided that T K is a relatively compact subset of X.
Theorem 2.4 ([15, Theorem 3.1]).Let K be a nonempty, convex and compact subset of X.Any mapping T : K −→ K has at least one fixed point provided that for every x ∈ K ∩ TK we have where T denotes the closed-convex envelope of T.
Remark 2.5.Condition (2.2) is equivalent to Fix(T) ⊂ Fix(T), where Fix(S) denotes the set of fixed points of the operator S.
Theorem 2.6 ([9, Theorem 2.7]).Let K be a nonempty, closed and convex subset of X and T : K −→ K be a mapping such that T K is a relatively compact subset of X and it satisfies condition (2.2).Then T has a fixed point in K.
Now we present some definitions and results concerning the problem (1.1)-(1.2).Consider the space It is clear that (X, • ) is a Banach space and it was employed in [8] where problem (1.1)-(1.2) was studied.For convenience we denote Our approach is based on lower and upper solutions method [7] and fixed point theory.Thus we define the lower and upper solutions for problem (1.1)-(1.2) and we present a Nagumo condition which gives some a priori bound on the first derivative for all possible solutions of the differential equation (1.1) between the lower and the upper solutions.(i) For any t 0 ∈ (0, ∞), either D − α(t 0 ) < D + α(t 0 ), or there exists an open interval I 0 such that t 0 ∈ I 0 , α ∈ W 2,1 (I 0 ) and (iii) There exists N ∈ N such that α ∈ W Assume there exist a continuous function . Then, there exists R > 0 such that for every function f : E → R satisfying for a.e.t ∈ [0, ∞) and all (x, y) ∈ R 2 with (t, x, y) ∈ E, and for every solution x of (1.1) such that ᾱ ≤ x ≤ β, we have Let x be a solution of (1.1) and t ∈ [0, ∞) such that x (t) > R.
If |x (t)| > r for all t ∈ [0, ∞), then for T > 0 big enough we have e θT , a contradiction.Therefore there exist t 0 < t 1 (or a contradiction, so we deduce that x (t) < R. In the same way we prove that x (t) > −R.
Remark 2.9.Observe that condition (2.3) in Proposition 2.8 could just be replaced by condition (2.4).However, the first one is easier to check in practice.
Remark 2.10.Notice that a better condition about N, which allows a quadratic growth with respect to the third variable of the nonlinear term f for the differential equation (1.1), is commonly employed in the literature (see, e.g.[2,11,12,14]), namely, ∞ r s N(s) ds = +∞.
Unfortunately, this type of conditions require harder assumptions about M such as sup 0≤t<∞ In particular, the previous hypothesis avoids that M could be singular at t = 0.
Lemma 2.11.Let h ∈ L 1 ([0, ∞)).Then x ∈ X is the unique solution of the problem For applying the fixed point theorems is necessary to guarantee that certain sets are relatively compact.Nevertheless, the classical Ascoli-Arzelà theorem fails due to the noncompactness of the infinite interval [0, ∞), so this difficulty is overcome by the following result, see [1].Lemma 2.12.Let A ⊂ X.The set A is said to be relatively compact if the following conditions hold: (a) A is uniformly bounded in X; (b) the functions belonging to A are equicontinuous on any compact interval of [0, ∞); (c) the functions f from A are equiconvergent at +∞, i.e., given ε > 0 there exists T(ε) > 0 such that f (t) − f (+∞) < ε for any t > T(ε) and f ∈ A.
Now we shall construct a modified problem for proving the existence of solutions for (1.1)-(1.2) under well-ordered lower and upper solutions.
Suppose that there exist α and β lower and upper solutions for Assume that for f : [0, ∞) × R 2 → R the following conditions hold: (H1) compositions t ∈ I → f (t, x(t), y(t)) are measurable whenever x(t) and y(t) are measurable; (H2) there exist a continuous function Consider the modified problem where and with R given by Proposition 2.8.Notice that for t ∈ [0, ∞), where (u) + (t) = max{u(t), 0}.Hence, we have ϕ( , see [7,17].The operator T : X → X associated to the modified problem (2.5) is defined as In order to achieve an existence result for problem (1.1)-(1.2) we shall prove that the operator T has a fixed point by applying Theorem 2.6.In this direction we present some previous lemmas.
Proof.Given x ∈ X, we shall show that Tx ∈ X.From (H2), (2.6) and (2.7) we have Therefore T is well defined.
We shall allow the function f to be discontinuous in the second variable over some curves satisfying a 'transversality' condition.These curves were introduced in [15], but as far as we know this is the first time that such type of discontinuity conditions were presented for boundary value problems on infinity intervals. or We say that the admissible discontinuity curve γ is inviable for the differential equation if it satisfies (2.9) or (2.10).
Now we state three technical results that we need in the proof of condition (2.2) for the operator T. Their proofs can be lookep up in [15].
In the sequel m denotes the Lebesgue measure in R.
For every measurable set J ⊂ (a, b) such that m(J) > 0 there is a measurable set J 0 ⊂ J such that m (J \ J 0 ) = 0 and for all τ 0 ∈ J 0 we have For every measurable set J ⊂ (a, b) such that m(J) > 0 there is a measurable set J 0 ⊂ J such that m (J \ J 0 ) = 0 and for all τ 0 ∈ J 0 we have Now we present the result which gives the main difference between our existence results and the classical ones.It provides the proof for condition (2.2) what allows to avoid the continuity of the operator T and thus the continuity of f .Lemma 2.19.Assume that conditions (H1), (H2) and (H3) there exist admissible discontinuity curves γ n : Then the operator T satisfies condition (2.2) for all x ∈ X, i.e., Fix(T) ⊂ Fix(T) where T is the cc-envelope of T.
Proof.Without loss of generality, suppose that R > 0 as given by Proposition 2.8 satisfies (2.11) Consider the closed and convex subset of X, where M(t) = max s∈[0,R] N(s)M(t).It is clear that TX ⊂ K and then TX ⊂ K.
The proof follows the ideas of [15,Theorem 4.4].Thus, we fix x ∈ K and consider the following three cases.
The assumption implies that for a.a.
as one can easily check by considering all possible combinations of the cases for a.a.t ∈ [0, ∞), hence Tx k − Tx → 0, by Lebesgue's dominated convergence theorem.
Case 2: m({t ∈ I n : x(t) = γ n (t)}) > 0 for some n ∈ N such that γ n is inviable.In this case we can prove that x ∈ Tx.We shall show that there exist ρ, ε > 0 such that for every finite family x i ∈ B ε(x) and First, we fix some notation.Let us assume that for some n ∈ N we have m({t ∈ I n : x(t) = γ n (t)}) > 0 and there exist ε > 0 and ψ ∈ L 1 (I n ), ψ(t) > 0 for a.a.t ∈ I n , such that (2.10) holds with γ replaced by γ n .(The proof is similar if we assume (2.9) instead of (2.10), so we omit it.) We denote J = {t ∈ I n : x(t) = γ n (t)}, and we observe that m({t ) on a set of positive measure, which is a contradiction with the definition of upper solution.Now we distinguish between two sub-cases.

.15)
Let us now fix a point τ 0 ∈ J 1 .From (2.14) and (2.15) we deduce that there exist t − < τ 0 and t + > τ 0 , t ± sufficiently close to τ 0 so that the following inequalities are satisfied: ) Finally, we define a positive number ρ = min 1 4 and we are now in a position to prove that x ∈ Tx.It is sufficient to prove the following claim.
Let x i and λ i be as in the Claim and, for simplicity, denote y = ∑ λ i Tx i .For a.a. (2.21) On the other hand, for every i ∈ {1, 2, . . ., m} and for a.a.t ∈ J we have for a.a.t ∈ A.
Therefore the assumptions on γ n ensure that for a.a.t ∈ A we have The set {t ∈ J : γ n (t) ∈ (α(t), β(t))} can be written as the following countable union so there exists some n 0 ∈ N such that m({t ∈ J : Hence, from (2.10) it follows that for a.a.t ∈ A and all x i ∈ B ε(x).Now the proof of the Claim follows exactly as in Case 2.1.
Case 3: m({t ∈ I n : x(t) = γ n (t)}) > 0 only for some of those n ∈ N such that γ n is viable.Let us prove that in this case the relation x ∈ Tx implies x = Tx.Note first that x ∈ Tx implies that x satisfies the boundary conditions in (2.5), because every element in Tx is, roughly speaking, a limit of convex combinations of functions y satisfying and L is continuous.Now it only remains to show that x ∈ Tx implies that x satisfies the ODE in (2.5).
Let us consider the subsequence of all viable admissible discontinuity curves in the conditions of Case 3, which we denote again by {γ n } n∈N to avoid overloading notation.We have m(J n ) > 0 for all n ∈ N, where For each n ∈ N and for a.a.t ∈ J n we have γ n (t) = f (t, γ n (t), γ n (t)) and from α ≤ γ n ≤ β and |γ n (t)| < R it follows that γ n (t) = f (t, ϕ(t, γ n (t)), δ R ((ϕ(t, γ n (t))) )), so γ n is viable for (2.5).Then for a.a.t ∈ J n we have and therefore (2.26) Now we assume that x ∈ Tx and we prove that it implies that i=1 λ k,i Tx k,i , and notice that y k → x uniformly in [0, ∞) and x k,i − x ≤ 1/k for all k ∈ N and all i ∈ {1, 2, . . ., m(k)}.Note also that (2.27)For a.a.t ∈ [0, ∞)\ J we have that either x(t) ∈ [α(t), β(t)], and then f (t, ϕ(t, •), δ R ((ϕ(t, •)) )) is continuous at x(t), so for any ε > 0 there is some Combining this result with (2.26), we see that x solves (2.5), which implies that x is a fixed point of T.
Remark 2.20.Admissible discontinuity curves may be defined in infinite intervals as the union of that in Definition 2.15.
Proof.For simplicity, we divide the proof in several steps.First, we will prove that the modified problem (2.5) has at least a solution, that is, we will ensure that the operator T defined as in (2.8) has a fixed point.Later, we will show that it is a solution for problem (1.1)-(1.2).
Step 1. Problem (2.5) has at least a solution x ∈ X.
By Lemma 2.13, the operator T is well defined.Consider the closed and convex set D defined as where For x ∈ D, we have and Therefore, TD ⊂ D. In addition, TD is relatively compact, by Lemma 2.14, and T satisfies condition (2.2), by Lemma 2.19.Then Theorem 2.6 guarantees that the operator T has at least a fixed point x ∈ D.
Let x ∈ X be a solution of (2.5).Suppose that there exists t ∈ [0, ∞) such that α(t) > x(t).Then inf It does not happen at 0, since If the infimum is attained as t tends to infinity, there exists T > 0 such that and α ∈ W 2,1 ((T, ∞)).Hence, and so x − α is a concave function on [T, ∞).
Then there are two options: either there exists t 0 > T such that t 0 is a relative minimum (in this case the reasoning is analogous to that we do below when the infimum is attained at t 0 ∈ (0, ∞)) or there exists T > T such that (x − α) ( T) < 0 and, by (3.1), However, by the definition of α, so, by the definition of lower solution, there exists an open interval I 0 such that t 0 ∈ I 0 and α (t) ≥ f (t, α(t), α (t)) for a.a.t ∈ I 0 .
On the other hand, by the continuity of x − α, there exists ε > 0 such that for all t ∈ (t 0 − ε, t 0 + ε) we have x(t) − α(t) < 0.Then, It is an immediate consequence of the Nagumo condition, see Proposition 2.8.
Moreover, in the recent paper [13, Remark 3.3], the authors observe that it is unknown if this strict inequality can be weakened.Observe that this difficulty was overcome in our previous theorem.
To finish this section, we illustrate our existence result with an example which shows its applicability.
Consider the problem (1.1)-(1.2) with the following functional boundary conditions where 0 < η ≤ 1, and the function which implies that S is a closed subset of X due to T is upper semicontinuous and {0} is a closed set, see [3,Lemma 17.4].Hence, since TX is relatively compact and S ⊂ TX, S is a compact set.Define x min (t) = inf{x(t) : x ∈ S} for t ∈ [0, ∞).The evaluation map δ t : X → R given by δ t (x) = x(t) is a continuous map and then δ t (S) = {x(t) : x ∈ S} is compact.Thus, for each t 0 ∈ [0, ∞), there exists a function x 0 ∈ S such that x 0 (t 0 ) = x min (t 0 ) and x min is a continuous function on [0, ∞).
Let us see that x min is a solution of (2.5).It is clear that x min will be the least solution.By the upper semicontinuity of the operator T and the condition Fix(T) = Fix(T), the limit in X of a sequence of elements in S must be a solution of (2.5).
Given T > 0 and ε > 0 we shall prove that there exists v ∈ S such that |v(t) − x min (t)| ≤ ε for all t ∈ [0, T].
Then, a sequence of elements in S converges pointwise to x min and by the compactness of S, up to a subsequence, it converges in S.
Following [5,Theorem 4.1], the idea is to construct an upper solution for problem (4.1) and to apply Theorem 3.1 in order to obtain the function v ∈ S looked for.
Then β n−1 (0) = x 0 (0) and β n−1 (0) ≤ x 0 (0), so from the monotonicity hypothesis about L and the fact that x 0 ∈ S, we have L(β n−1 (0), β n−1 (0)) ≥ L(x 0 (0), x 0 (0)) = 0, and it is immediate to check that β n−1 is an upper solution for problem (4.1).From Theorem 3.1 it follows that there exists v ∈ S such that α(t) ≤ v(t) ≤ β n−1 (t) for t ∈ [0, ∞) and, by the construction of β n−1 and the definition of x min , we have that v(t i ) = x min (t i ) for i = 0, 1, . . ., n − 1.Hence, for each t ∈ [0, T] there is i ∈ {0, A similar reasoning shows that problem (4.1) has the greatest solution between α and β.Remark 4.2.We note that the existence of extremal solutions in addition to information about the set of solutions for problem (4.1) it provides a method to achieve new existence results for problems where the nonlinearity f has a functional dependence, see [6].

Definition 2 . 7 .
A function α ∈ Y is said to be a lower solution for the problem (1.1)-(1.2) if the following conditions are satisfied.

17 .
Let a, b ∈ R, a < b, and let h ∈ L 1 (a, b) be such that h > 0 a.e. on (a, b).