POSITIVE SOLUTIONS FOR FIRST ORDER NONLINEAR FUNCTIONAL BOUNDARY VALUE PROBLEMS ON INFINITE INTERVALS

1. IntroductionBoundaryvalueproblemsonin nite intervalsappearinmanyproblemsofpracti-cal interest, for example in linear elasticity problems, nonlinear uid ow problemsand foundation engineering (see e.g. [1,10,16] and the references therein). Thisis the reason why these problems have been studied quite extensively in the lit-erature, especially the ones involving second order di erential equations. Secondorder boundary value problems on in nite intervals are treated with various meth-ods, such as xed point theorems (see e.g. [1,5,6,10,14-16,23]), upper and lowersolutions method (see e.g. [2,8,9]), diagonalization method (see e.g. [1,18,20]) andothers. An interesting overview on in nite interval problems, including real worldexamples, history and various methods of solvability, can be found in the recentbook of Agarwal and O’ Regan [1].A rather less extensive study has been done on rst order boundary value prob-lems on in nite intervals. One of the major ways to deal with such problems is touse numerical methods, see for example [10,16]. In short, the basic idea in this caseis to build a nite interval problem such that its solution approximates the solutionof the in nite problemquite well on the nite interval. The diculty of this methodlies in setting the nite interval boundary value problem in such a way that theapproximation is accurate. Another way to deal with boundary value problems onin nite intervals is to use xed point theorems. See for example [7,9]. Using thisapproach one will have to reformulate the boundary value problem to an operatorequation and to use an appropriate compactness criterion on in nite intervals forthe corresponding operator (see Lemma 2.2 below).In the recent years a growing interest has arisen for positive solutions of bound-ary value problems. See for example [10,11,17,18,23]. Also, nowadays, functionalboundary value problems are extensively investigated, usually via xed point the-orems (see [6,12,21,22] and the references therein).


Introduction
Boundary value problems on infinite intervals appear in many problems of practical interest, for example in linear elasticity problems, nonlinear fluid flow problems and foundation engineering (see e.g.[1,10,16] and the references therein).This is the reason why these problems have been studied quite extensively in the literature, especially the ones involving second order differential equations.Second order boundary value problems on infinite intervals are treated with various methods, such as fixed point theorems (see e.g.[1,5,6,10,[14][15][16]23]), upper and lower solutions method (see e.g.[2,8,9]), diagonalization method (see e.g.[1,18,20]) and others.An interesting overview on infinite interval problems, including real world examples, history and various methods of solvability, can be found in the recent book of Agarwal and O' Regan [1].
A rather less extensive study has been done on first order boundary value problems on infinite intervals.One of the major ways to deal with such problems is to use numerical methods, see for example [10,16].In short, the basic idea in this case is to build a finite interval problem such that its solution approximates the solution of the infinite problem quite well on the finite interval.The difficulty of this method lies in setting the finite interval boundary value problem in such a way that the approximation is accurate.Another way to deal with boundary value problems on infinite intervals is to use fixed point theorems.See for example [7,9].Using this approach one will have to reformulate the boundary value problem to an operator equation and to use an appropriate compactness criterion on infinite intervals for the corresponding operator (see Lemma 2.2 below).
In the recent years a growing interest has arisen for positive solutions of boundary value problems.See for example [10,11,17,18,23].Also, nowadays, functional boundary value problems are extensively investigated, usually via fixed point theorems (see [6,12,21,22] and the references therein).This paper is motivated by [10][11][12]14,18].We will deal with a first order functional boundary value problem on an infinite interval, seeking positive non-zero solutions.In order to do that, we will use two fixed point theorems, one of them being the well known Krasnoselkii's fixed point theorem.
Let R be the set of real numbers, R + := {x ∈ R : x ≥ 0} and J := [−r, 0] for some r ≥ 0. If I is an interval in R we denote by C(I) the set of all continuous real functions ψ : I → R. Also, we denote by BC(I) the Banach space of all ψ ∈ C(I) such that sup{|ψ(s)| : s ∈ I} < +∞ endowed with the usual sup-norm If x ∈ C(J ∪ R + ) and t ∈ R + , then we denote by x t the element of C(J) defined by The boundary value problem, which we will study, consists of the equation along with the nonlinear condition where f : R + × C(J) → R, φ : J → R are continuous functions, x(+∞) := lim t→+∞ x(t) and it holds that As indicated in [13], since x(+∞) exists in R, we must suppose that the following is true lim However, this is a direct consequence of the forthcoming assumption (H 3 ), so we do not state it as a separate assumption.The paper is organized as follows.In Section 1 we state the boundary value problem and in Section 2 we present the fixed point theorems, formulate the corresponding operator and prove that it is compact.Then in Sections 3 and 4 we prove our new results for the functional and the ordinary case, respectively, and in Section 5 we give an application.We must notice that even for the ordinary boundary value problem, which corresponds to the case r = 0, the results we present in Section 4, are new.

Definition. A solution of the boundary value problem
Searching for positive solutions of the boundary value problem (1. Proof.From (1.1), we have (2.1) 2) and (2.2) we get Using (2.1) and (2.4) we have Also, combining (2.3) and (2.4) we get Also, for any s ∈ J we have Finally, it is clear that Rx is continuous at zero.The proof is complete.
and consider the following assumptions.
By (H 2 ), we conclude that for every s ∈ R + and m > 0, the sup y J ∈[0,m] f (s, y) exists in R + .So, we set and we assume that (H 3 ) For every m > 0, it holds Then for every m > 0 set (H 4 ) Assume that there exists ρ > 0 such that endowed with the usual norm The following compactness criterion, which is due to Avramescu [4], will be used to prove that R is completely continuous.Note that the classical Arzela -Ascoli Theorem cannot be applied here, since the domain of R is a space of functions with unbounded domain.
Then M is relatively compact in Y (t 0 ).
We can now prove the following. where Therefore, from Lebesgue's Dominated Convergence Theorem, it follows that where m > 0. We will prove that R(V ) is relatively compact in Y (−r).Indeed for every t ∈ J ∪ R + we have Also, for arbitrary t 1 , t 2 ∈ J ∪ R + , we have Therefore, since φ is uniformly continuous, in any of the above cases and for any > 0 there exists δ( , φ, F ) > 0 such that Furthermore, for t 0 ∈ R + we have So, having in mind assumption (H 3 ) we conclude that for every > 0, there exists for every t 0 ≥ T ( , F ) and x ∈ V .Now, we can apply Lemma 2.2 and get that R(V ) is relatively compact in Y (−r).This completes the proof.
Definition.Let E be a real Banach space.A cone in E is a nonempty, closed set P ⊂ E such that (i) κu + λv ∈ P for all u, v ∈ P and all κ, λ ≥ 0 (ii) u, −u ∈ P implies u = 0.
Let P be a cone in a Banach space E.Then, for any b > 0, we denote by P b the set and by ∂P b the boundary of P b in P, i.e. the set Part of our results are based on the following theorem, which is an application of the fixed point theory in a cone.Its proof can be found in [3].Also, we will use the well known Krasnoselskii's fixed point theorem (see [13] Then g has a fixed point x ∈ P ∩ (Ω 2 \Ω 1 ).
In this paper, we will use the following Theorem 2.7, which is a corollary of Theorem 2.6, for the special case when the sets Ω 1 and Ω 2 are balls (with common center at point zero and positive, nonequal radium).

Positive solutions for the functional problem
Before we state our main results we set , Λ := Φ + φ(−r) A Theorem 3.1.Suppose that conditions (H 1 ) − (H 4 ) hold.Also suppose that if φ = 0, there exists t 0 ∈ R + such that f (t 0 , 0) = 0. Then the boundary value problem (1.1) − (1.2) has at least one positive nonzero solution x such that More precisely we have Proof.We will first justify why any positive solution x of the boundary value problem (1.1) − (1.2), if one exists, is nonzero.By hypothesis, if φ = 0, then there exists EJQTDE, 2004 No. 8, p. 8 Then it is clear that equation (1.1) does not have the zero solution.On the other hand, if φ(t 1 ) = 0 for some t 1 ∈ J, then by (1.2) we get that Ax(t 1 ) − x(+∞) = 0, which also means that x = 0. Now, set and observe that P is a cone in BC(J ∪ R + ).Also, we notice that for every x ∈ P and t ∈ R + we have x t ∈ C + (J), so by (H 2 ) f (t, x t ) ≥ 0 and , taking into account (H 1 ), we easily obtain Rx(t) ≥ 0. This means that R(P) ⊂ P, so since we are looking for a positive solution of the boundary value problem (1.1) − (1.2), it is enough to find a fixed point of the operator R : P → P.
Let ρ be the constant introduced by (H 4 ).Then, obviously R(P ρ ) ⊆ P and, from Lemma 2.3 it follows that R is a completely continuous operator.
Furthermore, we will show that Rx J∪R + < x J∪R + , for every x ∈ ∂P ρ .Assume that this is not true.Then, for some x ∈ ∂P ρ , it holds x J∪R + ≤ Rx J∪R + .Also, observe that, from the formula of Rx and assumptions (H 2 ), (H 3 ), for every t ∈ R + , we have Additionally, for every t ∈ J we have

So, we have
which contradicts (H 4 ).
We can now apply Theorem 2.5 to obtain that the boundary value problem (1.1) − (1.2) has at least one positive nonzero solution x, such that (3.1) 0 < x J∪R + < ρ.EJQTDE, 2004 No. 8, p. 9 If x is a positive solution of the boundary value problem (1.1) − (1.2), then, taking into account the formula of R and the fact that A > 1, we conclude that Now it is easy to see that Φ ≤ Q(ρ) and so, by (H 4 ), Φ < ρ.Then, taking into account (3.1), we obtain Φ ≤ x J∪R + < ρ and the proof is complete.
Theorem 3.2.Suppose that (H 1 ) − (H 5 ) hold and φ ∈ C(J) is a nondecreasing function.Also suppose that there exists γ > 0 such that where w, v are the functions involved in (H 5 ).Then the boundary value problem (1.1) − (1.2) has at least one positive solution x, with where ρ is the constant involved in (H 4 ) and ρ = τ := Aγ.More precisely we have Proof.Define the set Notice that K is a cone in BC(J ∪ R + ).It is clear that, by (H 1 ) and (H 2 ), for any x ∈ K d , where d = max{ρ, τ }, we have Rx(t) ≥ 0 for every t ∈ J ∪ R + .Also, since x ≥ 0 we have, also by (H 2 ), that (Rx) (t) = f (t, x t ) ≥ 0, t ∈ R + .Namely Rx|R + is a nondecreasing function.Also, taking into account the formula of Rx|J and the fact that φ is nondecreasing, we get that Rx|J is also nondecreasing.Since Rx is continuous at zero, we conclude that Rx is also nondecreasing on J ∪ R + .Moreover it is clear that ARx(0) − Rx(+∞) = φ(0).By (H 1 ), we have φ(0) A ≥ 0 and thus Rx(0) Furthermore, as we did in Theorem 3.1, we can prove that Rx J∪R + < x J∪R + for x ∈ ∂K ρ .
EJQTDE, 2004 No. 8, p. 10 Now we will prove that Rx J∪R + ≥ x J∪R + for every x ∈ ∂K τ .For this purpose it suffices to prove that Rx ≥ τ for every x ∈ ∂K τ .By (H 1 ) we have So, using (H 5 ) and the fact that x is nondecreasing, we obtain However, x(0) ≥ 1 A x(+∞) and x(+∞) = x J∪R + , since x is nondecreasing.Therefore, taking into account (3.2) and the fact that x J∪R + = τ = Aγ, we get Hence, since Rx is nondecreasing, we have Therefore, for every x ∈ ∂K τ we have Rx ≥ x J∪R + and so Rx J∪R + ≥ x J∪R + .Thus, we can apply Theorem 2.7 to get that the boundary value problem (1.1) − (1.2) has at least one positive solution x, such that (3.3) min{τ, ρ} < x J∪R + < max{τ, ρ}.

Now, if
x is a positive solution of the boundary value problem (1.1) − (1.2), then, taking into account the formula of R and the facts that A > 1 and x is nondecreasing, we conclude that Also, taking into account (3.3), we obtain (3.4) max{min{τ, ρ}, Λ} < x J∪R + < max{τ, ρ}.
Now we observe that for every θ > 0 we have Λ ≤ Q(θ).So, since Q(ρ) < ρ, we have Λ < ρ and if τ > ρ, then max{min{τ, ρ}, Λ} = max{ρ, Λ} = ρ.On the other hand, if τ < ρ, then max{min{τ, ρ}, Λ} = max{τ, Λ}.Therefore (3.4)  where D is defined in Theorem 3.2, ρ is the constant involved in (H 4 ) and ρ = τ .More precisely we have Proof.Define the set The proof is similar to the one we used in Theorem 3.2.Furthermore, as we did in Theorem 3.1, we can prove that R(x) J∪R + < x J∪R + for x ∈ ∂K ρ .Now we will prove that Rx J∪R + ≥ x J∪R + for every x ∈ ∂K τ .For this purpose it suffices to prove that Rx ≥ τ for every x ∈ ∂K τ .As in Theorem 3.2, using (H 1 ) and (H 6 ), we obtain Therefore, taking into account the fact that w is nonincreasing and (3.5) we have Hence, since Rx is nondecreasing, we have Therefore, for every x ∈ ∂K τ we have Rx ≥ x J∪R + and hence Rx J∪R + ≥ x J∪R + .So, applying Theorem 2.7 we get that there exists at least one positive solution x of the boundary value problem (1.1) − (1.2), such that (3.3) holds.Finally, as in the previous Theorem 3.2, we can prove that (3.3) takes the form of (3.4) and the proof is complete.
In this, ordinary, case assumptions (H 2 )-(H 4 ) are replaced by the following: Also, observe that the analogues of assumptions (H 5 ), (H 6 ), for the ordinary case, can be unified in the following: ( H 5 ) There exist E ⊆ R + , with measE > 0, and functions v : E → R + continuous, with sup{v(t) : t ∈ E} > 0 and monotonous w : R + → R + such that f (t, y) ≥ v(t)w(y), (t, y) ∈ E × R.
Finally we set Φ := C A − 1 and then we have the following theorems, which correspond to Theorems 3.1 − 3.5, respectively, for the ordinary case.The proofs of these theorems are omitted, since they can be easily derived from the proofs of Theorems 3.1−3.5,with some obvious modifications.Also it is easy to see that the constant Λ in the present ordinary case is equal to the constant Φ.

An application
Consider the boundary value problem (5.1) x (t) = x 2 t (−

Theorem 2 . 4 .
Let g : P b → P be a completely continuous map such that g(x) = λx for all x ∈ ∂P b and λ ≥ 1.Then g has a fixed point in P b .It easy to see that the condition g(x) = λx for all x ∈ ∂P b and λ ≥ 1, can be replaced by the following stricter assumption g(x) < x for all x ∈ ∂P b So, we get the following corollary of Theorem 2.4.Theorem 2.5.Let g : P b → P be a compact map such that g(x) < x for all x ∈ ∂P b .Then g has a fixed point in P b .