On the Existence of Positive Solutions of a Nonlinear Differential Equation

We study some existence results for the nonlinear equation (1/A)(Au′)′ = uψ(x,u) for x ∈ (0,ω) with different boundary conditions, where ω ∈ (0,∞], A is a continuous function on [0,ω), positive and differentiable on (0,ω), and ψ is a nonnegative function on (0,ω)× [0,∞) such that t → tψ(x, t) is continuous on [0,∞) for each x ∈ (0,ω). We give asymptotic behavior for positive solutions using a potential theory approach.


Introduction
In this paper, we study the following nonlinear equation: where ω ∈ (0,∞] and A is a continuous function on [0,ω), which is positive and differentiable on (0, ω).
Our aim in this paper is to study (1.1) with a nonlinear term f (x,u) = uψ(x,u) and two boundary conditions.More precisely, we assume that x → 1/A(x) is integrable in the neighborhood of ω and the integral ω 0 (1/A(t))dt may diverges and we search a positive continuous solution u of (1.1).
Our paper is organized as follows.In Section 2, we give some properties of the Green's function G(x, y) of the operator u → −(1/A)(Au ) with Au (0) = 0 and u(ω) = 0, which will be used later.We recall (see [4]) that for x, y in [0,ω), we have We refer in this paper to V f , the potential of a measurable nonnegative function f defined on (0,ω) by Note that V f is a lower semicontinuous function on (0, ω).Moreover, for two nonnegative measurable functions f and g with f ≤ g and Vg is continuous, we have V f is continuous.
In order to simplify our statements, we adopt the following notation.

Properties of Green's function
In the sequel, we denote dt , for x ∈ (0,ω). (2.1) Let a ∈ (0,ω), then for each x ∈ (0,ω), we have Indeed, the result follows from the following inequalities: First, we give the following version of a comparison principle.
(2) It follows from the fact that x → Γ(x,0) is nonincreasing and (3) For x, y ∈ (0,ω), we distinguish the following cases: and this completes the proof.
Next, we will give some inequalities satisfied by Green's function.
On the other hand, using Lebesgue's theorem and Fatou's lemma, we obtain that (2.15) Now, by (2.10) we deduce that α q ≤ 2 q . (2.16) This completes the proof.
For each x ∈ (0,ω), we obtain by Fubini-Tonelli's theorem that (2.32) Now using that ϕ is the solution of the problem (Q) and ρ is differentiable on (0, ω), we obtain by integrating by parts that Hence, from the lower inequality of (2.26), we deduce that This completes the proof of (2.28).
Sonia Ben Othman et al. 9

Proofs of the main results
In this section, we aim at proving Theorems 1.1 and 1.2.We recall that ρ(x) = ω x (1/A(t))dt.Proof of Theorem 1.1.Let c > 0 and q ∈ K satisfying (H 2 ).We denote by the nonempty convex set of Ꮾ + ((0,ω)), and we define the operator T on Λ by We claim that TΛ ⊂ Λ.Indeed, for u ∈ Λ we have by (H 2 ), On the other hand, by using (2.28), we obtain that Hence TΛ ⊂ Λ. Next, we prove that the operator T is nondecreasing on Λ.Let u 1 ,u 2 ∈ Λ such that u 1 ≤ u 2 , then from (H 2 ), we have for each x ∈ (0,ω), Now, we consider the sequence (u j ) j defined by u 0 (x)=c(ρ(x)−V q (qρ)(x)) and u j+1 (x)= Tu j (x), for j ∈ N and x ∈ (0,ω).Then since Λ is invariant under T, we have obviously u 1 = Tu 0 ≥ u 0 and so from the monotonicity of T, we deduce that Hence, the sequence (u j ) converges on (0, ω) to a function u ∈ Λ.Now, using (H 0 ), (H 2 ), and the dominated convergence theorem, we deduce that (Tu j ) j converges to Tu on (0,ω).Consequently, we have or equivalently Applying the operator (I + V (q.)) on both sides of the above equality and using (2.25), we deduce that u satisfies Finally, we need to verify that u is a positive continuous solution for the problem (P 1 ).Indeed, by (H 2 ) we have uψ(•,u) ≤ cqρ, then using the fact that q ∈ K and ρ is bounded on each interval [x 0 ,ω) with x 0 > 0, we deduce the continuity of V (cqρ), which implies the continuity of u on (0, ω).Now since q ∈ K and for each x, y ∈ (0,ω),we have Then the following problem: In the next, we will give the proof of Theorem 1.2.
Proof of Theorem 1.2.Let c > 0, then by hypothesis (H 3 ), there exists q ∈ K such that the function t → t(ψ(x,t) − q(x)) is nonincreasing on [0,c].We consider the nonempty closed convex set Λ given by Tu := c 1 − V q (q) + V q q − ψ(•,u) u , (3.17) and Tu(ω):=lim x→ω Tu(x)=c.Now, by similar arguments as in the proof of Theorem 1.1, we obtain that TΛ ⊂ Λ and T is an increasing operator on Λ.Let (u n ) n be the sequence of functions defined by u 0 = c 1 − V q (q) , u n+1 = Tu n , for n ∈ N. (3.21)
.16) Sonia Ben Othman et al. 11 and we define the operator T on Λ by