CUBO A Mathematical Journal Vol.16, No¯ 01, (105–116). March 2014 Upper and Lower Solutions for φ−Laplacian Third-order BVPs on the Half-Line Smäıl Djebali 1 and Ouiza Saifi 1,2 Laboratoire ”Théorie du point Fixe et Applications” 1École Normale Supérieure, P.B. 92, 16050 Kouba. Algiers, Algeria djebali@ens-kouba.dz, djebali@hotmail.com 2Department of Economics, Faculty of Economic and Management Sciences. Algiers University 3, Algeria saifi kouba@yahoo.fr ABSTRACT In this paper, we investigate the existence of positive solution for a class of singular third-order boundary value problem associated with a φ-Laplacian operator and posed on the positive half-line: { (φ(−x′′))′(t) + f(t, x(t)) = 0, t > 0, x(0) = µx′(0), x′(+∞) = x′′(+∞) = 0 where µ ≥ 0. By using the upper and lower solution approach and the fixed point theory, the existence of positive solutions is proved under a monotonic condition on f. The nonlinearity f may be singular at x = 0. An example of application is included to illustrate the main existence result. RESUMEN En este art́ıculo investigamos la existencia de una solución positiva de una clase de problema singular de valores de frontera de tercer-orden asociado con el operador φ- Laplaciano y colocado sobre la semirecta real positiva: { (φ(−x′′))′(t) + f(t, x(t)) = 0, t > 0, x(0) = µx′(0), x′(+∞) = x′′(+∞) = 0 donde µ ≥ 0. Usando la técnica de sub y súper soluciones y la teoŕıa del punto fijo, se prueba la existencia de soluciones positivas bajo una condición de monotonicidad sobre f. La nolinealidad f puede ser singular en x = 0. Se incluye un ejemplo de aplicación para ilustrar el resultado principal de existencia. Keywords and Phrases: Third order, half-line, φ−Laplacian, singular problem, positive solu- tion, fixed point, upper and lower solution. 2010 AMS Mathematics Subject Classification: 34B15, 34B18, 34B40, 47H10.. 106 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) 1 Introduction This paper is concerned with the existence of positive solutions to the following third-order bound- ary value problem posed on the half-line and associated with a φ−Laplacian operator:    (φ(−x′′))′(t) + f(t, x(t)) = 0, t > 0, x(0) = µx′(0), x′(+∞) = x′′(+∞) = 0, (1) where µ ≥ 0 and f = f(t, x) : R+ × (0, +∞) −→ R+ is a continuous function which may have space singularity at x = 0 and R+ = [0, +∞). The map φ : R −→ R is a continuous, increasing homeomorphism such that φ(0) = 0 (for instance the p−Laplacian ϕp(s) = |s| p−1s, p > 1). Boundary value problems (bvps for short) on the half-line appear in many applied problems re- lating to various phenomena in physics, biology, and combustion theory (see, e.g., [1] and references therein). In the last couple of years, the mathematical investigation of such problems, especially second-order boundary value problems have attracted several authors (see, e.g., [4], [5], [6], [7], [8], [9] and the references therein). However, only some of them were interested in higher-order differential equations on [0, +∞) (see [9], [11], [12]). The aim of this work to study a third-order differential equation with a φ-Laplacian derivative operator and posed on the positive half-line. Our approach is based on the upper and lower solution method adapted to this class of problems combined with the Schauder fixed point theorem. This papers essentially consists of three sections. Section 2 is devoted to some preliminaries facts and basic notions needed in this paper. A fixed point formulation is also provided in this section. In Section 3, we present our existence result of positive solutions when the nonlinearity f is monotonic with respect to x but may be singular at x = 0. The case f is not singular at x = 0 is also considered with less hypotheses. Our existence theorem is illustrated by means of an example of application. A function x is said to be a solution of problem (1) if x ∈ X = {x | x ∈ C2((0, ∞), R)} and φ(−x′′) ∈ C1((0, ∞), R)} (2) and satisfies (1). In addition, x said to be a positive solution if x(t) > 0 for t ∈ (0, +∞). 2 Auxiliary Lemmas A mapping defined on a Banach space is said to be completely continuous if it is continuous and maps bounded sets into relatively compact sets. Let Cl([0, ∞), R) = {x ∈ C([0, ∞), R) | lim t→+∞ x(t) exists}. For x ∈ Cl([0, ∞), R), define ‖x‖l = sup t∈R+ |x(t)|· Then (Cl, ‖x‖l) is a Banach space. CUBO 16, 1 (2014) Upper and Lower Solutions for φ−Laplacian . . . 107 Lemma 2.1. ([3], p. 62) Let M ⊆ Cl(R +, R). Then M is relatively compact in Cl(R +, R) if the following three conditions hold: (a) M is uniformly bounded in Cl(R +, R). (b) The functions belonging to M are almost equicontinuous on R+, i.e., equicontinuous on every compact interval of R+. (c) The functions from M are equiconvergent, that is, for every ε > 0, there exists T(ε) > 0 such that |x(t) − x(+∞)| < ε for any t ≥ T(ε) and x ∈ M. Note that the space E = {x ∈ C([0, ∞), R) | lim t→+∞ x(t) 1 + t exists}. (3) is also a Banach space with the norm ‖x‖ = sup t∈R+ |x(t)| 1+t . From Lemma 2.1, we easily deduce Lemma 2.2. Let M ⊆ E. Then M is relatively compact in E if the following conditions hold: (a) M is bounded in E, (b) the functions belonging to {u| u(t) = x(t) 1+t , x ∈ M} are locally equicontinuous on [0, +∞), (c) the functions belonging to {u| u(t) = x(t) 1+t , x ∈ M} are equiconvergent at +∞. Definition 2.3. Let α, β ∈ X. Then α is called a lower solution of (1) if α satisfies    (φ(−α′′(t)))′ + f(t, α(t)) ≥ 0, t > 0 α(0) ≤ µα′(0), lim t→+∞ α′(t) ≤ 0, lim t→+∞ α′′(t) ≥ 0. β is called an upper solution of (1) if the above inequalities are reversed. Let G(t, s) = { s + µ, 0 ≤ s ≤ t < +∞ t + µ, 0 ≤ t ≤ s < +∞, be the Green function of the linear problem −x′′ = x(0) − µx′(0) = x′(+∞) = 0. The following lemmas are straightforward; the proofs are omitted. Lemma 2.4. Assume that δ ∈ C(R+, R+) satisfies ∫+∞ 0 δ(s)ds < +∞ and let x(t) = ∫+∞ 0 G(t, s)δ(s)ds. Then    x′′(t) + δ(t) = 0, t > 0, x(0) = µx′(0), lim t→+∞ x′(t) = 0. (4) Lemma 2.5. Assume that δ ∈ C(R+, R+) ∩ L1(r, +∞) for all r > 0 and ∫+∞ 0 φ−1 (∫+∞ s δ(τ)dτ ) ds < +∞. 108 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) If x(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s δ(τ)dτ ) ds, then x ∈ X and    (φ(−x′′(t)))′ + δ(t) = 0, t > 0, x(0) = µx′(0), x′(+∞) = x′′(+∞) = 0. (5) Consider the positive cone S = {x ∈ C1[0, +∞) | x(t) ≥ 0, concave on [0, +∞), lim t→+∞ x′(t) = 0}. (6) In addition to the null function, S contains, e.g., ln(1 + t), so S is a nonempty subset; moreover S has the following properties: Lemma 2.6. Let x ∈ S \ {0}. Then there exists a positive constant λx such that (a) for all θ > 1, x(t) ≥ λx 1 θ , ∀ t ∈ [1/θ, θ], (b) If ρ(t) = { t, t ∈ [0, 1] 1, t ≥ 1 (7) then x(t) ≥ λxρ(t), ∀ t ∈ [0, +∞). Proof. (a) Notice that every x ∈ S is nondecreasing and thus by L’Hopital’s rule lim t→+∞ x(t) 1+t = 0; as a consequence, the function x(t) 1+t achieves its maximum at some point tm ∈ [0, +∞); let λx = x(tm) 1+tm = ‖x‖ > 0. By concavity of x, we have for t ∈ [1/θ, θ] x(t) ≥ min t∈[ 1 θ ,θ] x(t) = x( 1 θ ) = x(θ−1+θtm θ+θtm 1 θ−1+θtm + 1 θ+θtm tm) ≥ θ−1+θtm θ+θtm x( 1 θ−1+θtm ) + 1 θ+θtm x(tm) ≥ 1 θ+θtm x(tm) = 1 θ x(tm) 1+tm = λx 1 θ , whence the first part of the lemma. (b) Fix t0 ∈ [0, +∞) and distinguish between four cases. (i) If t0 = 0, then x(0) ≥ 0 = λxρ(0). (ii) If t0 ∈ (0, 1), then 1 t0 ∈ (1, +∞). From Part (a), x(s) ≥ λxt0, ∀ s ∈ [t0, 1t0 ]. In particular for s = t0, x(t0) ≥ λxt0 = λxρ(t0). CUBO 16, 1 (2014) Upper and Lower Solutions for φ−Laplacian . . . 109 (iii) If t0 = 1, let {tn}n be a real sequence such that 0 < tn < 1 and tn → 1, as n → +∞. By (ii), we know that x(tn) ≥ λxtn, ∀ n ≥ 1. Then x(1) = lim n→+∞ x(tn) ≥ λx lim n→+∞ tn = λx = λxρ(1). (iv) Finally, let t0 ∈ (1, +∞), since x is nondecreasing, then x(t0) ≥ x(1) ≥ λx = λxρ(t0), ending the proof of the lemma. 3 Main Existence Results First we list some assumptions: (H1) f ∈ C(R + × (0, +∞), R+) and f(t, x) is a nonincreasing relatively to the second argument. (H2) For every λ > 0, ∫+∞ 0 f(τ, λρ(τ))dτ < +∞, ∫+∞ 0 φ−1 (∫+∞ s f(τ, λρ(τ))dτ ) ds < +∞. (H3) There exists a function a ∈ S \ {0} such that for t ≥ 0    b(t) := ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f(τ, a(τ))dτ ) ds ≥ a(t), ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f(τ, b(τ))dτ ) ds ≥ a(t). For x ∈ S \ {0}, define a fixed point operator T by Tx(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f(τ, x(τ))dτ ) ds. We have Lemma 3.1. Assume (H1)-(H2) holds. Then the operator T maps S \ {0} into X ∩ S. In addition    (φ(−(Tx)′′))′(t) + f(t, x(t)) = 0, t > 0, (Tx)(0) = µ(Tx)′(0), (Tx)′(+∞) = (Tx)′′(+∞) = 0. (8) Proof. (a) For λ > 0, let Fλ(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f(τ, λρ(τ))dτ ) ds, then lim t→+∞ Fλ(t) 1 + t = 0. 110 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) Indeed, using the convergence of the second integral in (H2), we get lim t→+∞ F′λ(t) = lim t→+∞ ∫+∞ t φ−1 (∫+∞ s f(τ, λρ(τ))dτ ) ds = 0, then Fλ is nondecreasing and lim t→+∞ Fλ(t) 1 + t =    0, if lim t→+∞ Fλ(t) < ∞, lim t→+∞ F′λ(t) = 0, if lim t→+∞ Fλ(t) = ∞. (b) Given x ∈ S \ {0}, by Lemma 2.6, there exists λx > 0 such that x(t) ≥ λxρ(t), t ∈ R +. By (H1), (H2), and Part (a), we have Tx(t) 1+t = ∫ +∞ 0 G(t,s)φ −1( ∫ +∞ s f(τ,x(τ))dτ)ds 1+t ≤ ∫ +∞ 0 G(t,s)φ −1( ∫ +∞ s f(τ,λxρ(τ))dτ)ds 1+t = Fλx (t) 1+t . Hence lim t→+∞ Tx(t) 1+t = 0. Then Tx ∈ E and even Tx ∈ X ∩ S. Indeed Tx(t) ≥ 0, (Tx)′(t) = ∫+∞ t φ−1 (∫+∞ s f(τ, x(τ))dτ ) ds =⇒ lim t→+∞ (Tx)′(t) = 0, (Tx)′′(t) = −φ−1 (∫+∞ t f(τ, x(τ))dτ ) ≤ 0, and thus (8) is satisfied. Now we state and prove our main existence result: Theorem 3.2. Assume that Assumptions (H1) − (H3) hold. Then the boundary value problem (1) has at least one positive solution x ∈ X which satisfies x(t) ≥ λ0ρ(t) for some λ0 > 0. Proof. The proof is be split into three steps. Step 1. We first determine appropriate upper and lower solution for the bvp (1). Since a ∈ S\{0} and b(t) = Ta(t), then by (H3) and Lemma 3.1, we have b, Tb ∈ S \ {0}. Moreover T being nonincreasing relatively to x, we have a ≤ b ⇒ a ≤ Tb ≤ Ta = b. (9) Therefore, for all t > 0    (φ(−(Tb)′′))′(t) + f(t, Tb(t)) ≥ (φ(−(Tb)′′))′(t) + f(t, b(t)) = 0 (Tb)(0) = µ(Tb)′(0), (Tb)′(+∞) = 0, (Tb)′′(+∞) = 0 (10) CUBO 16, 1 (2014) Upper and Lower Solutions for φ−Laplacian . . . 111 and    (φ(−(Ta)′′))′(t) + f(t, Ta(t)) ≤ (φ(−(Ta)′′))′(t) + f(t, a(t)) = 0, (Ta)(0) = µ(Ta)′(0), (Ta)′(+∞) = 0, (Ta)′′(+∞) = 0. (11) The functions α(t) = Tb(t) and β(t) = Ta(t) are lower and upper solution of the bvp (1), respectively with α ≤ β. Step 2. We claim that the following regular modified boundary value problem    (φ(−x′′))′(t) + f∗(t, x(t)) = 0, t > 0, x(0) = µx′(0), x′(+∞) = x′′(+∞) = 0 (12) has a positive solution, where f∗(t, x) =    f(t, α), x < α(t), f(t, x), α(t) ≤ x ≤ β(t), f(t, β), x > β(t), (13) To see this, consider the operator A : E → E defined by Ax(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f∗(τ, x(τ))dτ ) ds. It is clear that a fixed point of the operator A is a solution of the boundary value problem (12). Since α ∈ S \ {0}, by lemma 2.6 (b), there exists a positive constant λα such that α(t) ≥ λαρ(t), ∀ t ∈ R +. Moreover f(t, x) being nonincreasing in x, we have f∗(t, x) ≤ f(t, α(t)) ≤ f(t, λαρ(t)) (14) for all positive t. (a) A(E) ⊆ E. For x ∈ E and t ∈ R+, we have, using (14) Ax(t) 1+t = ∫ +∞ 0 G(t,s)φ −1 ( ∫ +∞ s f ∗ (τ,x(τ))dτ)ds 1+t ≤ ∫ +∞ 0 G(t,s)φ −1 ( ∫ +∞ s f(τ,λαρ(τ))dτ)ds 1+t = Fλα (t) 1+t , hence lim t→+∞ Ax(t) 1+t = 0 and A(E) ⊆ E. (b) A is continuous. Let some sequence {xn}n≥1 ⊆ E be such that lim n→+∞ xn = x0 ∈ E. Then 112 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) we have ‖Axn − Ax0‖ = sup t∈R+ |Axn(t)−Ax0(t)| 1+t = sup t∈R+ ∫+∞ 0 G(t,s) 1+t |φ−1( ∫+∞ s f∗(τ, xn(τ))dτ) − φ −1( ∫+∞ s f∗(τ, x0(τ))dτ)|ds ≤ max(1, µ) ∫+∞ 0 φ−1( ∫+∞ s f∗(τ, xn(τ))dτ) − φ −1( ∫+∞ s f∗(τ, x0(τ))dτ)|ds. Since ∣ ∣ ∣ φ−1 (∫+∞ s f∗(τ, xn(τ))dτ ) − φ−1 (∫+∞ s f∗(τ, x0(τ))dτ ) ∣ ∣ ∣ ≤ 2φ−1 (∫+∞ 0 f(τ, λαρ(τ) ) dτ, then the continuity of f∗, φ−1, (H2) and the Lebesgue dominated convergence theorem, we deduce ‖Axn − Ax0‖ −→ 0, as n −→ +∞ (c) A(E) is relatively compact. Indeed (i) A(E) is uniformly bounded. For x ∈ E, we have ‖Ax‖ = sup t∈R+ |Ax(t)| 1+t ≤ sup t∈R+ ∫+∞ 0 G(t,s) 1+t φ−1 (∫+∞ s f∗(τ, x(τ))dτ) ) ds ≤ max(1, µ) ∫+∞ 0 φ−1 (∫+∞ s f∗(τ, x(τ))dτ) ) ds ≤ max(1, µ) ∫+∞ 0 φ−1 (∫+∞ s f(τ, λαρ(τ))dτ) ) ds < +∞. (ii) { A(E) 1+t } is almost equicontinuous. For a given T > 0, x ∈ E, and t, t′ ∈ [0, T] (t > t′), we have ∣ ∣ ∣ Ax(t) 1+t − Ax(t ′ ) 1+t′ ∣ ∣ ∣ ≤ ∫+∞ 0 ∣ ∣ ∣ G(t,s) 1+t − G(t ′ ,s) 1+t′ ∣ ∣ ∣ φ−1 (∫+∞ s f∗(τ, x(τ))dτ) ) ds ≤ ∫T 0 ∣ ∣ ∣ G(t,s) 1+t − G(t ′ ,s) 1+t′ ∣ ∣ ∣ φ−1 (∫+∞ s f(τ, λαρ(τ))dτ) ) ds + ∣ ∣ ∣ t+µ 1+t − t ′ +µ 1+t′ ∣ ∣ ∣ ∫+∞ T φ−1 (∫+∞ s f(τ, λαρ(τ))dτ) ) ds, CUBO 16, 1 (2014) Upper and Lower Solutions for φ−Laplacian . . . 113 then by (H2), for any ε > 0 and T > 0, there exists δ > 0 such that ∣ ∣ ∣ Ax(t) 1+t − Ax(t ′ ) 1+t′ ∣ ∣ ∣ < ε for all t, t′ ∈ [0, T] with |t − t′| < δ. Hence { A(E) 1+t } are almost equicontiuous. (iii) { A(E) 1+t } is equiconvergent at +∞. Since lim t→+∞ Ax(t) 1+t = 0, then by (H2) we have lim t→+∞ sup x∈E | Ax(t) 1+t − lim t→+∞ Ax(t) 1+t | = lim t→+∞ sup x∈E ∫ +∞ 0 G(t,s)φ −1( ∫ +∞ s f ∗ (τ,x(τ))dτ))ds 1+t ≤ lim t→+∞ ∫ +∞ 0 G(t,s)φ −1( ∫ +∞ s f(τ,λαρ(τ))dτ))ds 1+t = lim t→+∞ Fλα (t) 1+t = 0. Lemma 2.2 guarantees that A(E) is relatively compact. Finally by the Schauder fixed point theorem (see, e.g., [2]), the operator A has at least one fixed point x ∈ E, which is further in X by Lemma 3.1, solution of the bvp (12). Step 3. Next we will prove that the boundary value problem (1) has at least one positive solution. For this, we only need to check that α(t) ≤ x(t) ≤ β(t), ∀t ∈ R+. Since x is a solution of the bvp (12) x(0) = µx′(0), lim t→+∞ x′(t) = lim t→+∞ x′′(t) = 0 (15) In addition, f(t, x) is nonincreasing in x f(t, β(t)) ≤ f∗(t, x) ≤ f(t, α(t)), ∀ t ∈ R+. (16) It follows from (9) and (H3) that f(t, b(t)) ≤ f∗(t, x) ≤ f(t, a(t)), ∀ t ∈ R+. (17) Since a ∈ S \ {0}, by Lemma 3.1 (φ(−β′′(t)))′ = (φ(−Ta)′′(t)))′ = −f(t, a(t)), ∀ t ∈ R+. These, together with Lemma 3.1 (9), (15)-(17) yield    (φ(−β′′(t)))′ − (φ(−x′′(t)))′ = −f(t, a(t)) + f∗(t, x(t)) ≤ 0, t ∈ R+ (β − x)(0) = µ(β − x)′(0), (β − x)′(+∞) = 0, (β − x)′′(+∞) = 0 (18) This implies that the function z defined by z(t) = (φ(−β′′(t))) − (φ(−x′′(t))) is a non- increasing function in R+. Moreover z(+∞) = 0 implies z(t) ≥ 0, ∀ t ≥ 0 and then (β − x)′′(t) ≤ 0, ∀ t ∈ R+ which means that (β − x)′ is nonincreasing in R+. Now (β − x)′(+∞) = 0 then (β − x)′(t) ≥ 0, ∀t ∈ R+ and so β − x is nondecreasing on R+. Finally the boundary condition (β − x)(0) = µ(β − x)′(0) ≥ 0 implies that x(t) ≤ β(t), for 114 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) all t ∈ R+. In a similar way, we can prove that x(t) ≥ α(t), for all t ∈ R+. Therefore, x is a solution of the bvp (1). In addition, there existence of a positive constant λ0 = λα such that x(t) ≥ α(t) ≥ λ0ρ(t), ∀ t ∈ R +. The proof of Theorem 3.2 is completed. However, when f(t, x) is nonsingular at x = 0, i.e. f : R+ × R+ −→ R+ is a continuous function, then for all x ≥ 0, f(t, x) ≤ f(t, 0). In this case, we have Theorem 3.3. Assume that assumption (H1) holds and (H2) ′ 0 < ∫+∞ 0 f(τ, 0)dτ < +∞ and ∫+∞ 0 φ−1 (∫+∞ s f(τ, 0)dτ ) ds < +∞. Then the bvp (1) has at least one positive solution x ∈ X such that x(t) ≥ λ0ρ(t) for some λ0 > 0. The proof is similar to that of Theorem 3.2. We only check that T(S) ⊂ S ∩ X and if we take a(t) = 0, ∀ t ≥ 0, then condition (H3) holds. Finally the condition (H2) ′ implies that β = Ta = b, α = Tb belong to S \ {0}. Example 3.4. Consider the singular boundary value problem    (φ(−x′′(t)))′ + e−tm(t)g(x(t)) = 0, x(0) = µx′(0), lim t→+∞ x′(t) = lim t→+∞ x′′(t) = 0, (19) where 0 ≤ µ ≤ 8 3 , φ(x) = x 1 3 , f(t, x) = e−tm(t)g(x), g(x) = { 1 x , x ∈ (0, 1] 1, x ≥ 1, and m(t) = { t3, t ∈ [0, 1] 1 t2 , t ≥ 1, Then, we have (H1) f ∈ C((0, +∞) × R+, R+) and f(t, x) is a nonincreasing with respect to x for every positive t. (H2) For all λ > 0, ∫+∞ 0 f(τ, λρ(τ))dτ ≤ max{1, 1 λ } < +∞, and ∫+∞ 0 φ−1 (∫+∞ s f(τ, λρ(τ))dτ ) ds < +∞. CUBO 16, 1 (2014) Upper and Lower Solutions for φ−Laplacian . . . 115 (H3) Let a0(t) = 1, then a0 ∈ S and if we put a(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)g(a0(τ))dτ ) ds, then a(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)dτ ) ds. Moreover for all t ∈ R+ a(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)dτ ) ≤ ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τdτ ) ≤ ∫+∞ 0 (s + 8/3)φ−1(e−s)ds ≤ 1 = a0(t). Hence b(t) = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)g(a(τ))dτ ) ds ≥ ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)g(a0(τ))dτ ) ds = a(t). Finally, since g ≥ 1, we have ∫+∞ 0 G(t, s)φ−1 (∫+∞ s f(τ, b(τ))dτ ) ds = ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)g(b(τ))dτ ) ds ≥ ∫+∞ 0 G(t, s)φ−1 (∫+∞ s e−τm(τ)dτ ) ds = a(t). Then all conditions of Theorem 3.2 are fulfilled which guarantees that the bvp (19) has at least one positive solution. Received: May 2013. Accepted: February 2014. References [1] R.P. Agarwal and D. O’Regan, Infinite Interval Problems for Differential, Difference and Integral Equations, Kluwer Academic Publisher, Dordrecht, 2001. [2] R.P. Agarwal, M. Meehan, and D. O’Regan, Fixed Point Theory and Applications, Cambridge Tracts in Mathematics, Vol. 141, Cambridge University Press, 2001. [3] C. Corduneanu, Integral Equations and Stability of Feedback Systems, Academic Press, New York, 1973. 116 Smäıl Djebali & Ouiza Saifi CUBO 16, 1 (2014) [4] S. Djebali and K. Mebarki, Multiple positive solutions for singular BVPs on the positive half- line, Comput. Math. Appl. 55(12) (2008) 2940–2952. [5] S. Djebali and K. Mebarki, Multiple unbounded positive solutions for three-point bvps with sign-changing nonlinearities on the positive half-line, Acta Appl. Math. 109(2) (2010) 361– 388 [6] S. Djebali and O. Saifi, Positive solutions for singular φ−Laplacian BVPs on the positive half-line, E.J.Q.T.D.E. 56 (2009) 1–24. [7] S. Djebali and O. Saifi, Upper and lower solution method for singular φ−Laplacian BVPs with derivative depending nonlinearity on [0, +∞), Comm. Appl. Anal. 14(4) (2010) 463–480 [8] Y. Guo, C. Yu, and J. Wang, Existence of three positive solutions for m−point boundary value problems on infinite intervals, Nonlin. Anal., T.M.A. 71 (2009) 717–722. [9] G. Han and F. Li, Multiple solutions for some fourth-order boundary value problems, Nonl. Anal., T.M.A. 66(11) (2007) 2591–2603. [10] S. Liang and J. Zhang, Positive solutions for singular third-order boundary-value problem with dependence on the first order derivative on the half-line, Acta. Appl. Math. 111 (2010) 27–43. [11] Y. Yang and J. Zhang, Existence of solutions for some fourth-order boundary value problems with parameters, Nonl. Anal. T.M.A. 69(4) (2008) 1364–1375. [12] Y. Yang and J. Zhang, Nontrivial solutions for some fourth-order boundary value problems with parameters, Nonl. Anal. T.M.A. 70(11) (2009) 3966–3977.