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Vol. LXXXIV, 1 (2015), pp. 1–12

EXISTENCE OF POSITIVE SOLUTIONS FOR NONLINEAR FRACTIONAL PROBLEM ON THE HALF LINE

S. BELARBI

Abstract. This paper deals with the differential equations of fractional order on the half-line. By Leggett-Williams theorem, we present recent results for the exis-tence of positive solutions for a Caputo fractional problem. An illustrative example is also presented.

1. Introduction

In the last few years, fractional differential equations theory have received in-creasing attention. This theory has been developed very quickly and attracted a considerable interest from researches (see [1, 2, 3, 6, 7]).

The motivation for those works stems from both the development of the theory and the applications of such constructions in various sciences such as physics, mechanics, chemistry, engineering, and so on. For an extensive collection of such results, we refer the readers to the monographs by Kilbas and al [10], Miller and Ross [16] and Podlubn´y [18].

As one of the focal topics in the research, some kinds of fractional differential equation with specific configurations have been presented. More specifically, In [4], the authors investigated the existence and multiplicity of positive solutions of the nonlinear fractional differential equation boundary value problem

(

x(t) + a(t)f (x(t)) = 0 0 < t < 1, 1 < α ≤ 2 x(0) = 0, x0(1) = 0

By using Krasnosel’skii’ s fixed point theorem and Leggett-Williams theorem [17, 11], some sufficient conditions for the existence of positive solutions to the above FBVP are obtained. Moreover, the study of positive solution has been studied in [5, 8, 9, 13, 19].

To the best of our knowledge, there are few papers devoted to the study of fractional differential equations with a Laplacian operator on the half-line [12, 14, 15, 20, 21], where boundary value problems on the half-line have been applied (unsteady flow of gas through a semi-infinite porous medium, the theory of drain flows, etc).

Received September 26, 2013; revised October 24, 2014.

2010 Mathematics Subject Classification. Primary 34A12, 34B18, 34B15.

Key words and phrases. Caputo derivative; fractional differential equation; fixed point; half line; positive solution.

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S. BELARBI

In this article, we are concerned with the following fractional differential equa-tion:             p(t)Φ(Dαx(t))0+ f (t, x(t)) = 0, t ∈ [0, ∞), 0 < α ≤ 1 x(0) = ∞ R 0 g(s)x(s)ds + θ, lim t→+∞p(t)Φ(D αx(t)) = ρ, (1)

where ρ ≥ 0, θ ≥ 0, and Dα denotes Caputo fractional derivative of order α, g : [0, ∞) → [0, ∞) is continuous with R0∞g(s)ds < 1, p : [0, ∞) → (0, ∞) is continuous, and Φ(x) = |x|q−2x with q > 1, and, the inverse function of Φ is Φ−1(x) = |x|q0−2x, where 1

q + 1 q0 = 1.

This paper is organized as follows: In section 2, we prepare some material need to prove our main results. In section 3, we obtain existence results of the positive solutions for (1) using Leggett-Williams theorem. In section 4, we give an example to illustrate our results.

2. Preliminaries

In this section, we give some definitions, lemmas and properties which will be used in the next sections.

Definition 1. The Riemann-Liouville fractional integral operator of order α > 0, for a continuous function f on [0, ∞) is defined as:

Jαf (t) = 1 Γ(α) Z t 0 (t − τ )α−1f (τ )dτ α > 0, t > 0. (2)

Definition 2. The Caputo derivative of order α of f ∈ Cn([0, ∞[) is defined as: Dαf (t) = 1 Γ(n − α) Z t 0 (t − τ )n−α−1f(n)(τ )dτ n − 1 < α, n ∈ N∗. (3)

Definition 3. A function f : [0, ∞) × R → R is called a Carath´eodory function if the following conditions are satisfied

(i) For each u ∈ R, t → f (t, u) is measurable on (0, ∞), (ii) For each t ∈ [0, ∞), u → f (t, u) is continuous on R, (iii) For each r > 0, there exists Br, Br(t) > 0, t ∈ [0, ∞),

R∞

0 Br(s)ds < ∞, such that |u| ≤ r implies |f (t, u)| ≤ Br(t), t ∈ [0, ∞).

Definition 4. Let X be a real Banach space. The nonempty convex closed subset P of X is called a cone in X if

(i) ax ∈ P and x + y ∈ P for all x, y ∈ P and a ≥ 0, (ii) if x ∈ P and −x ∈ P, then x = 0.

Let ψ be a nonnegative functional on a cone P of a real Banach space X. We define the sets

Pr= {y ∈ P : |y| < r},

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We need also the Leggett-Williams fixed point theorem [8]

Theorem 5. Let a < b < d < c be positive numbers, T : Pc → Pc be a com-pletely continuous operator, and ψ be a nonnegative continuous concave functional on P, such that ψ(y) ≤ kyk for all y ∈ Pc. Suppose that

– {y ∈ P (ψ; b, d) : ψ(y) > b} 6= ∅, ψ(T y) > b for y ∈ P (ψ; b, d), – kT yk < a, for y ∈ P, with kyk ≤ a,

– ψ(T y) > a for y ∈ P (ψ; b, c) with kT yk > d.

Then T has at least three fixed points y1, y2 and y3 such that ky1k < a, ψ(y2) > b and ky3k > a with ψ(y3) < b.

We cite also the following three lemmas:

Lemma 6. For α > 0, the general solution of the fractional differential equation Dαx = 0 is given by

x(t) = c0+ c1t + c2t2+ · · · + cn−1tn−1, (4)

where ci ∈ R, i = 0, 1, 2, .., n − 1, n = [α] + 1. Lemma 7. Let α > 0. Then we have

JαDαx(t) = x(t) + c0+ c1t + c2t2+ · · · + cn−1tn−1, (5)

for some ci∈ R, i = 0, 1, 2, . . . , n − 1, n = [α] + 1. Lemma 8. Let β > α > 0. Then the formula

DαJβf (t) = Jβ−αf (t), t ∈ [a, b] (6)

is valid.

3. Main Results We introduce the following quantities:

λ := λ(t) = Z t 0 Φ−1 1 p(s)  ds. λ∗:= Z ∞ 0 Φ−1 1 p(s)  ds. For k > 1 large enough, we take λ(1k) < 1.

We take also µ := Z 1k 0 Φ−1 1 p(s)  ds 1 1 +R0∞Φ−1( 1 p(s))ds . It is clear that 0 < µ < 1.

Now, we consider the following Banach space X = {x ∈ C[0, ∞) : lim

t→∞x(t) < ∞}, with the norm

kxk = sup t∈[0,∞)

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S. BELARBI

and we define the cone P by

P :=nx ∈ X : x(t) ≥ 0, x(t) is non-decreasing on [0, ∞), min t∈[1 k,k] x(t) ≥ µ sup t∈[0,∞) x(t)o. On P, we define the functional

ψ(y) := min t∈[1

k,k]

y(t).

It is easy to see that ψ is a nonnegative continuous concave functional on P satisfies

ψ(y) ≤ kyk , y ∈ P. Now, we introduce the following hypotheses:

(H111) The constant λ∗ < ∞, and the function p : [0, ∞) → (0, ∞) is continuous and satisfies Z ∞ 0 g(t) Z t 0 Φ−1 1 p(u)  dudt < ∞.

(H222) The function f : [0, ∞) × [0, ∞) → [0, ∞) is a Carath´eodory function with f (t, 0) 6= 0 on each subinterval of [0, ∞).

(H333) There exist A, c > 0 such that

ρ + Z ∞ 0 Ac (1 + t)2ds !" R∞ 0 g(u) h θ + JαΦ−1( 1 p(t)) i du 1 −R∞ 0 g(s)ds + JαΦ−1 1 p(t)  # + θ ≤ c. (H444) There exist b > 0, B > 0, such that

R∞ 0 g(u) h θ + Γ(α)1 Rk1 0 ( 1 k − τ )Φ −1( ρ p(τ ) + 1 p(τ ) R∞ τ Bb (1+s)2ds)dτ i du 1 −R∞ 0 g(s)ds ≥ b. We prove the following two lemmas:

Lemma 9. Let 0 < α ≤ 1 and suppose that (H111) and (H222) hold. A solution of the problem (1) is given by

x(t) = 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u) " θ + JαΦ−1(ρ + R∞ t f (s, x(s))ds p(t) ) # du + θ + JαΦ−1 " ρ +Rt∞f (s, x(s))ds p(t) # (7)

Proof. Since f is a Carath´eodory function, then we can write p(t)Φ(Dαx(t)) = ρ + Z ∞ t f (s, x(s))ds. Therefore, Dαx(t) = Φ−1 " ρ +R∞ t f (s, x(s))ds p(t) # .

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Thanks to Lemma 7, we deduce that x(t) = x(0) + JαΦ−1 " ρ +R∞ t f (s, x(s))ds p(t) # . The boundary conditions in (1) imply that

x(t) = Z ∞ 0 g(s)x(s)ds + θ + JαΦ−1 " ρ +R∞ t f (s, x(s))ds p(t) # . (8) Since Z ∞ 0 g(s)ds < 1, then it follows that

Z ∞ 0 g(s)x(s)ds = 1 1 −R∞ 0 g(s)ds · Z ∞ 0 g(u) " θ + JαΦ−1(ρ + R∞ t f (s, x(s))ds p(t) ) # du. (9)

Combining (8) and (9), we get Lemma 9. 

Lemma 10. Suppose that (H111) and (H222) hold and x is a solution of (1). Then we have:

(a) Dαx(t) ≥ 0; t ∈ [0, ∞[.

(b) The function x is concave with respect to λ on [0, ∞), and it is positive with respect to t on [0, ∞).

Proof. (a) Since x is a solution of (1), then for all t ∈ [0, ∞) we have [p(t)Φ(Dαx(t))]0≤ 0.

Thanks to (H222),we have

ρ − p(t)Φ(Dαx(t)) ≤ 0, t ∈ [0, ∞) which satisfies

p(t)Φ(Dαx(t)) ≥ 0, since ρ ≥ 0. Thus

Dαx(t) ≥ 0 for all t ∈ [0, ∞).

(b) Since Dαx ≥ 0, then to prove that x > 0, it suffices to show that x(0) ≥ 0. We have x(0) = Z ∞ 0 g(s)x(s)ds + θ ≥ x(0) Z ∞ 0 g(s)ds. SinceR∞

0 g(s)ds < 1, it follows then that x(0) ≥ 0. Hence

x(t) ≥ 0 for t ∈ [0, ∞). With help of (H222) it follows that

x(t) > 0 for all t ∈ (0, ∞).

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S. BELARBI

Thanks to (H111), we have λ∗< ∞, and then λ ∈ C([0, ∞ ), [0, λ∗[). On the other hand, we have

dx dλ = dx dt 1 Φ−1( 1 p(t)  ≥ 0, p(t)Φdx dt  = Φdx dλ  , and d2x dλ2 =  p(t)Φdx dt 0 Φ0dx dλ dλ dt .

Using the fact that [p(t)Φ(Dαx(t))]0 ≤ 0, α = 1, Φ0(x) > 0, x > 0 and

dt > 0, we obtain d2x

dλ2 ≤ 0.

Hence x(t) is concave with respect to λ on [0, ∞). The proof is complete.  Now, we define the following nonlinear operator T : P → X by:

T x(t) = 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u) " θ + JαΦ−1 ρ + R∞ t f (s, x(s))ds p(t) !# du + θ + JαΦ−1 " ρ +R∞ t f (s, x(s))ds p(t) # . (10)

Then, we prove the following result:

Lemma 11. Suppose that (H111) and (H222) hold. We have (i*) For x ∈ P , T x satisfies

(p(t)Φ(DαT x(t)))0+ f (t, x(t)) = 0, t ∈ [0, ∞), 0 < α ≤ 1, T x(0) = Z ∞ 0 g(s)T x(s)ds + θ, θ ≥ 0, lim t→+∞p(t)Φ(D αT x(t)) = ρ, ρ ≥ 0. (ii*) T x ∈ P for each x ∈ P.

(iii*) x is a bounded positive solution of (1) if and only if x is a solution the equation x = T x in P .

Proof. The proof of (i* ) follows from the definition of T and is omitted. To show (ii* ), we note from (i∗) that T x is a solution of (1). Then, Lemma 10 implies that

T x(t) ≥ 0, T0x(t) ≥ 0 for all t ∈ [0, ∞), and T x(t) is concave with respect to λ.

To complete the proof of T P ⊆ P, it suffices to prove that min t∈[1/k,k] T x(t) ≥ µ sup t∈[0,∞) T x(t). (11)

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Since x ∈ X and f is a Carath´eodory function, then T x(t) ≤ 1 1 −R0∞g(s)ds Z ∞ 0 g(u) " θ + JαΦ−1(ρ + R∞ t Br(s)ds p(t) ) # du + θ + 1 Γ(α) Z ∞ 0 (t − τ )α−1Φ−1 " ρ +R∞ τ Br(s)ds p(τ ) # dτ < ∞. This means that supt∈[0,∞)T x(t) exists. So, we shall consider two cases:

Case A. Suppose T x(t) achieves its maximum at σ ∈ [0, ∞) For t ∈ [1/k, k], we can write

T x(t) ≥ T x(1/k) = T x(t(λ(1/k))) = T x  t 1 − λ(1/k) + λ(σ) 1 + λ(σ) λ(1/k) 1 − λ(1/k) + λ(σ)+ λ(1/k) 1 + λ(σ)λ(σ) 

Thanks to the concavity of T x with respect to λ, we have: T x(t) ≥ 1 − λ(1/k) + λ(σ) 1 + λ(σ) T x  t  λ(1/k) 1 − λ(1/k) + λ(σ)  + λ(1/k) 1 + λ(σ)T x(t(λ(σ))) ≥ λ(1/k) 1 + λ(σ)T x(t(λ(σ))) = Z 1/k 0 Φ−1  1 p(s)  ds 1 1 + λ(σ)T x(t(λ(σ))) ≥ Z 1/k 0 Φ−1  1 p(s)  ds 1 1 +R∞ 0 Φ−1( 1 p(s))ds sup t∈[0,∞) T x(t) = µ sup t∈[0,∞) T x(t).

Case B. Now, suppose T x(t) achieves its maximum at ∞ : Choose σ0∈ [0, ∞), then, with the same arguments as before, we get for t ∈ [1/k, k] that

T x(t) ≥ µT x(σ0). Let σ0 → ∞, then for t ∈ [1/k, k], we can write

T x(t) ≥ µ sup t∈[0,∞)

T x(t). Thanks to A and B, we deduce that T x ∈ P .

The proof of (iii*) is based on Lemma 8 and it can be omitted.  Our main result is given by:

Theorem 12. Assume that the hypothesis (H111) and (H222) hold, and there exist constants a, b and c such that 0 < a < b < c.

(D1) f (t, x) < Aa

(1 + t)2 for all t ∈ (0, ∞) and x ∈ [0, a]; (D2) f (t, x) >

Bb

(1 + t)2 for all t ∈ [1/k, k] and x ∈ [b, c]; (D3) f (t, x) ≤

Ac

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S. BELARBI

Then, the (1) has at least three bounded positive solutions x1, x2 and x3satisfying kx1k < a, b < ψ(x2), a < x3 with ψ(x3) < b.

Proof. We shall prove that the operator T is completely continuous on P. It is easy to verify that T : P → P is well defined. We prove that T is continuous and maps bounded sets into pre-compact sets: Let xn→ x0as n → ∞ in P , then there exists r0 such that

sup n≥0 kxnk < r0. Hence, we have Z ∞ 0 |f (s, xn(s)) − f (s, x0(s))| ds ≤ Z ∞ 0 Br0(s)ds.

By the Lebesgue dominated convergence theorem, we obtain Z ∞

t

f (u, xn(u))du → Z ∞

t

f (u, x0(u))du uniformly as n → ∞. Let ε > 0. For all n, we have

Φ(ρ) + Z ∞ s f (u, xn(u))du ≤ Φ(ρ) + Z ∞ 0 Br0(s)ds ≡ r.

On the other hand, we know that Φ−1 is uniformly continuous on [0, r]. It follows then that there exists δ > 0, such that for x, y ∈ [0, r], |x − y| < δ, we have

Φ−1(x) − Φ−1(y) < ε. So for the above δ > 0, there exists N > 0, such that

ρ + Z ∞ t f (u, xn(u))du − (ρ + Z ∞ t f (u, x0(u))du) < δ, where n > N , t ∈ [0, ∞).

Then for n > N , we can write Φ−1(ρ + Z ∞ t f (u, xn(u))du) − Φ−1(ρ + Z ∞ t f (u, x0(u))du) < ε. Hence, for t ∈ [0, ∞), and n > N, yields

|T xn− T x0(t)| = 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)hJαhΦ−1ρ + R∞ t f (s, xn(s))ds p(t)  − Φ−1ρ + R∞ t f (s, x0(s))ds p(t) i dui + JαhΦ−1ρ + R∞ t f (s, xn(s))ds p(t)  − Φ−1ρ + R∞ t f (s, x0(s))ds p(t) i ≤ ε 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)JαΦ−1 1 p(t)  du + εJαΦ−1 1 p(t)  ≤ εh 1 1 −R0∞g(s)ds Z ∞ 0 g(u)JαΦ−1 1 p(t)  du + JαΦ−1 1 p(t) i .

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It follows then that

kT xn− T x0k → 0 uniformly as n → ∞. So, T is continuous.

Let Ω be an arbitrary bounded subset of P . First, we shall prove that T is bounded: Since Ω is bounded, then there exists r > 0, such that kxk ≤ r, for all x ∈ Ω. Thanks to Definition 3, we obtain

0 ≤ T x(t) = 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)hθ + JαΦ−1ρ + R∞ t f (s, x(s))ds p(t) i du + θ + JαΦ−1hρ + R∞ t f (s, x(s))ds p(t) i ≤ 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)hθ + JαΦ−1ρ + R∞ s Br(s)ds p(t) i du + θ + 1 Γ(α) Z ∞ 0 (t − τ )α−1Φ−1hρ + R∞ τ Br(s)ds p(τ ) i dτ. This means that T Ω is bounded.

Now we shall prove the equicontinuity of T (Br). Let us take t1, t2 ∈ (0, ∞), t1< t2 and x ∈ Ω. We have |T x(t2) − T x(t1)| ≤ 1 1 −R∞ 0 g(s)ds × Z ∞ 0 g(u)h Z t2 0 (t2− τ )α−1− (t1− τ )α−1 Γ(α)  Φ−1ρ + R∞ τ Br(s)ds p(τ )  dτidu + 1 Γ(α) Z t2 0 ((t2− τ )α−1− (t1− τ )α−1)Φ−1 hρ +R ∞ τ Br(s)ds p(τ ) i dτ.

We see that the function ϕ(x) = xα−1 − (α − 1)x is decreasing on [0, 1] , and increasing on (1, ∞). Consequently we can write

(t2− τ )α−1− (t1− τ )α−1≤ (α − 1)(t2− t1), on [0, 1] and

(t1− τ )α−1− (t2− τ )α−1≤ (α − 1)(t1− t2), on (1, ∞). We deduce from these two inequalities that if t1,t2∈ [0, 1], then

|T x(t2) − T x(t1)| ≤ 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)h(α − 1)(t2− t1) Γ(α) Z t2 0 Φ−1ρ + R∞ τ Br(s)ds p(τ )  dτidu +(α − 1)(t2− t1) Γ(α) Z t2 0 Φ−1hρ + R∞ τ Br(s)ds p(τ ) i dτ

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S. BELARBI and |T x(t2) − T x(t1)| ≤ 1 1 −R0∞g(s)ds Z ∞ 0 g(u)h(α − 1)(t1− t2) Γ(α) Z t2 0 Φ−1ρ + R∞ τ Br(s)ds p(τ )  dτidu +(α − 1)(t1− t2) Γ(α) Z t2 0 Φ−1hρ + R∞ τ Br(s)ds p(τ ) i dτ, provided that t1, t2∈ (1, ∞).

When t1 → t2, in the above tow inequalities, we claim that |T x(t2) − T x(t1)| tend to 0. Consequently T Ω is equicontinuous. According to the Ascoli-Arzela theorem we deduce that T is completely continuous operator.

Note that ψ(x) ≤ kxk for x ∈ Pc. We will show that the conditions of Theorem 5 are satisfied: Put x ∈ Pc. Then kxk ≤ c, we find

kT x(t)k = sup t∈[0,∞) T x(t) = 1 1 −R∞ 0 g(s)ds Z ∞ 0 g(u)hθ + JαΦ−1ρ + R∞ t f (s, x(s))ds p(t) i du + θ + JαΦ−1hρ + R∞ t f (s, x(s))ds p(t) i . Thanks to (D3), we have kT x(t)k ≤ ρ + R∞ 0 Ac (1+s)2ds  1 −R∞ 0 g(s)ds Z ∞ 0 g(u)hθ + JαΦ−1 1 p(t) i du + θ +ρ + Z ∞ 0 Ac (1 + s)2ds  JαΦ−1 1 p(t)  . Using (H333), we get kT x(t)k ≤ c This implies that T : Pc→ Pc.

By the same way, if x ∈ Pa, then with help of (D1), we obtain kT xk < a, and therefore (C2) is satisfied.

Let d be a fixed constant such that b < d ≤ c. Then ψ(d) ≥ d > b and kdk = d, it means P (ψ, b, d) 6= ∅.

For any x ∈ P (ψ, b, d), it holds that kxk ≤ d and ψ(x) = mint∈[1

k,k]x(t). Then we have ψ(T x) = min t∈[1 k,k] T x(t) = T x1 k  = R∞ 0 g(u) h θ + 1 Γ(α) Rk1 0 ( 1 k − τ )Φ −1 ρ+Rτ∞f (s,x(s))ds p(τ ) dτ i du 1 −R∞ 0 g(s)ds + θ + 1 Γ(α) Z k1 0 1 k− τ  Φ−1ρ + R∞ τ f (s, x(s))ds p(τ )  dτ

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In view of (D2) and (H444), we obtain ψ(T x) ≥ R∞ 0 g(u)  θ +Γ(α)1 R1k 0  1 k − τ  Φ−1ρ+ R∞ τ Bb (1+s)2ds p(τ )  dτ  du 1 −R∞ 0 g(s)ds ≥ b. Thus (C1) is satisfied.

Finally, for any x ∈ P (ψ, b, c) with kT xk > d, then kxk ≤ c and mint∈[1

k,k]x(t) ≥ b,

by the same method, we can also show that ψ(T x) > b easily, which means that (C3) holds.

Therefore, by the conclusion of Theorem 5, the operator T has at least three fixed points. This implies that (1) has at least three solutions. 

4. Example Let us consider the problem

 1 16exp t 2  D12x(t) 40 +2x + 1 9 + t2 = 0, t ∈ [0, ∞), x(0) = 1 8 Z ∞ 0 exp(−2s)x(s)ds + 1, 1 16t→+∞lim exp  t 2   D12x(t) 4 = 1. (12) We have p(t) = 1 16exp t 2, Φ(x) = x 4, g(t) = exp(−2t) 8 , f (t, x) = 2x + 1 9 + t2, θ = ρ = 1. It is clear that λ = λ(t) = 16  1 − exp  −t 8  and Z ∞ 0 g(t)dt = 1 8 Z ∞ 0 exp(−2t)dt = 1 16 < 1. Through a simple calculation, we get

λ∗= 16 < ∞, and Z ∞ 0 g(t) Z t 0 Φ−1  1 p(u)  dudt = 1 17 < ∞, The function f is a Carath´eodory function and f (t, 0) 6= 0 on each subinterval of [0, ∞). So (H111) and (H222) hold. Taking k = 1010, then λ(1 k) < 1 and 0 < µ = 16 17(1 − exp(− 1 8×1010)) < 1.

Next, in order to demonstrate our main result obtained, we choose a, b, c, A, B and C such that (H333) and (H444) be satisfied.

By Theorem 5, we conclude that the example (12) has at least three positive solutions.

Acknowledgment. The author would like to thank Professor Zoubir Dah-mani for his helpful suggestions

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S. BELARBI

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