# Existence of Positive Solutions for Some Superlinear Fourth-Order Boundary Value Problems

Authors: Imed Bachar; Habib Mâagli
Journal: Journal of Function Spaces
 (2014)

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## Abstract

We are concerned with the following superlinear fourth-order equationu4t+utφt,−ut=0,t∈0,1;−u0=u1=0,−u′0=a,−u′1=-b,  where a,−b are nonnegative constants such that a+b>0 and φt,−s is a nonnegative continuous function that is required to satisfy some appropriate conditions related to a class K satisfying suitable integrability condition. Our purpose is to prove the existence, uniqueness, and global behavior of a classical positive solution to the above problem by using a method based on estimates on the Green function and perturbation arguments.


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## Body

## 1. Introduction

A natural motivation for studying higher order BVPs lies in their applications. It is well-known (see for instance [1]) that the deformation of an elastic beam in equilibrium state, whose both ends clamped, can be described by fourth-order BVP
(1)u(4)(t)=f(t,u(t)),t∈(0,1),
subject to the boundary conditions
(2)u(0)=u(1)=u′(0)=u′(1)=0.
When the nonlinearity f is nonnegative such problems have been extensively investigated by many researchers, and various forms of the equation and boundary condition have been discussed; see, for example, [1–15] and references therein.In particular, in 2005, Ma and Tisdel [1] studied the existence of positive solutions to the sublinear BVP
(3)u(4)(t)=p(t)uσ(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
where σ∈(0,1) and p:(0,1)→[0,∞) is continuous which may be singular at both ends t=0 and t=1 and satisfying some integrable conditions. Using the method of lower and upper solutions for fourth-order boundary value problems, they have given some necessary and sufficient conditions for the existence of regular positive solutions to the boundary value problem (3).In [8], the authors used fixed-point index results, to prove the existence of a positive solution for the boundary value problem
(4)u(4)(t)=p(t)f(u(t)),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
where f:[0,∞)→[0,∞) is nondecreasing continuous function allowed to be superlinear and p:(0,1)→[0,∞) is continuous which may be singular at both ends t=0 and t=1 satisfying some adequate conditions.Recently, in [4], the author considered problem (3), with σ∈(-1,1) and p is a nonnegative continuous function on (0,1) satisfying some hypotheses related to Karamata regular variation theory. Using the Schauder fixed-point theorem, he established the existence and uniqueness of a positive solution to (3).Here, by using a method based on estimates on the Green function and perturbation arguments, we show that, for negative nonlinearityf, problem (1) has a unique positive classical solution subject to some boundary conditions. More precisely, we are concerned with the following superlinear fourth order problem
(5)u(4)(t)+u(t)φ(t,u(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b,
where a,b are nonnegative constants such that a+b>0. Our goal is to answer the questions of existence, uniqueness, and global behavior of a classical positive solution to problem (5), where the nonnegative nonlinear term φ(t,s) is required to satisfy some appropriate conditions related to the following class K.Definition 1.
A Borel measurable functionq in (0,1) belongs to the class K if q satisfies the following condition:
(6)∫01‍r3(1-r)3|q(r)|dr<∞.
We will often refer in this paper to(7)ω(t)∶=at(1-t)2+bt2(1-t),
the unique solution of the problem
(8)u(4)(t)=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b.
Observe that, fort∈[0,1],
(9)min⁡(a,b)ρ(t)≤ω(t)≤max⁡(a,b)ρ(t),
where ρ(t)∶=t(1-t).
Also we denote byG(t,s) the Green function of the operator u→u(4), with the conditions u(0)=u(1)=u′(0)=u′(1)=0, which can be explicitly given by
(10)G(t,s)=16{t2(1-s)2[3s-t(1+2s)],0≤t≤s≤1,s2(1-t)2[3t-s(1+2t)],0≤s≤t≤1,=16(t∧s)2(1-t∨s)2[3(t∨s)-(t∧s)(1+2(t∨s))],
where t∧s=min⁡(t,s) and t∨s=max⁡(t,s) (see [2]).
The outline of the paper is as follows. In Section2, we give some sharp estimates on Green’s function G(t,s), including the following 3G-inequality: for each t,s,r∈(0,1),
(11)G(t,r)G(r,s)G(t,s)≤32[(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)],(ρ(r)ρ(t))2G(t,r)≤12(ρ(r))3,
where ρ(r)=r(1-r).
In particular, we derive from this 3G-inequality that, for eachq∈K, one has
(12)αq∶=sup⁡t,s∈(0,1)∫01‍G(t,r)G(r,s)G(t,s)|q(r)|dr<∞.
In Section3, our purpose is to study the superlinear fourth-order problem (5). The nonlinearity φ is required to satisfy a combination of the following assumptions.(H1)
φ is a nonnegative continuous function in (0,1)×[0,∞).(H2)
There exists a nonnegative functionq∈K∩C(0,1) with αq≤1/2 such that for each t∈(0,1), the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1].(H3)
For eacht∈(0,1), the function s→sφ(t,s) is nondecreasing on [0,∞).
We will first exploit the 3G-inequality to prove that the inverse of fourth-order operators that are perturbed by a zero-order term are positivity preserving. That is, ifψ is a positive measurable function and q is a nonnegative function belonging to the class K∩(C(0,1)) with αq≤1/2, then the following problem
(13)u(4)(t)+q(t)u(t)=ψ(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
has a positive solution. It turns out to prove that problem (13) admits a positive Green function G(t,s).
Based on the construction of this Green function and by using perturbation arguments, we prove the following.Theorem 2.
Assume(H1)-(H2), then problem (5) has a positive solution u in C4((0,1))∩C([0,1]) satisfying
(14)c0ω(t)≤u(t)≤ω(t),t∈[0,1],
where c0 is a constant in (0,1).
Moreover, if hypothesis(H3) is also satisfied, then the solution u to problem (5) satisfying (14) is unique.Corollary 3.
Letf be a nonnegative function in C1([0,∞)) such that the map s→θ(s)=sf(s) is nondecreasing on [0,∞). Let p be a nonnegative continuous function on (0,1) such that the function t→p~(t)∶=p(t)max⁡0≤ξ≤ω(t)⁡θ′(ξ) belongs to the class K. Then for λ∈[0,1/2αp~), the following problem
(15)u(4)(t)+λp(t)u(t)f(u(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b,
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(16)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].Observe that in Theorem2, we obtain a positive classical solution u to problem (5) whose behavior is not affected by the perturbed term. That is, it behaves like the solution ω of the homogeneous problem (8). As typical example of nonlinearity satisfying (H1)–(H3), we quote φ(t,s)=λp(t)sσ, where σ≥0,   p is a positive continuous function on (0,1) such that
(17)∫01‍r3+σ(1-r)3+σp(r)dr<∞,
and q(t)=λp~(t)∶=λ(σ+1)p(t)(ω(t))σ, with λ∈[0,1/2αp~). Note that by using (9) and (17) the function p~ belongs to the class K.As usual, letB((0,1)) be the set of Borel measurable functions in (0,1) and let B+((0,1)) be the set of nonnegative ones.We define the kernelV:B+((0,1))→B+((0,1)) by
(18)Vψ(t)∶=∫01‍G(t,r)ψ(r)dr,t∈[0,1].Letq∈B+((0,1)). We define the kernel V(q·):B+((0,1))→B+((0,1)) by
(19)V(q·)(ψ)(t)∶=∫01‍G(t,r)q(r)ψ(r)dr,t∈[0,1].Remark 4 (see [1]).
Letψ∈B+((0,1)) such that the function r→(ρ(r))2ψ(r) is continuous and integrable on (0,1), then Vψ is the unique solution in C4((0,1))∩C([0,1]) of
(20)u(4)(t)=ψ(t),t∈(0,1);u(0)=u(1)=u′(0)=u′(1)=0.


## 2. Estimates on the Green’s Function

From the explicit expression of Green’s function (10) we derive the following.Proposition 5 (see [4, 7]).
Fort,s∈[0,1], one has
(21)13(t∧s)2(1-t∨s)2(t∨s)(1-t∧s)≤G(t,s)≤12(t∧s)2(1-t∨s)2(t∨s)(1-t∧s),(22)13(ρ(t))2(ρ(s))2≤G(t,s)≤12ρ(t)ρ(s)(ρ(t)∧ρ(s)).Using (22), we deduce the following.Corollary 6.
Fors,t∈(0,1), one has(i)
(23)(ρ(s)ρ(t))2G(t,s)≤12(ρ(s))3.(ii)
Forψ∈B+((0,1)), the function t→Vψ(t) is continuous on [0,1] if and only if the integral ∫01(ρ(s))2ψ(s)ds converges.Theorem 7 (3G inequality).
For eacht,s,r∈(0,1), one has
(24)G(t,r)G(r,s)G(t,s)≤32[(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)],
where ρ(t)=t(1-t).Proof.
To prove the inequality, we denote byA(t,s)=(ρ(t)ρ(s))2/G(t,s) and we claim that A is a quasi-metric, that is, for each t,s,r∈(0,1),
(25)A(t,s)≤32[A(t,r)+A(r,s)].
By symmetry, we may assume thatt≤s.
Using (21), we deduce that
(26)2s(1-t)≤A(t,s)≤3s(1-t).
To show (25), we separate the proof into three cases.
Case
1
(
r
≤
t
≤
s
). In this case, one has
(27)A(t,r)+A(r,s)≥2(t+s)(1-r)≥2s(1-t)≥23A(t,s).
Case
2
(
t
≤
r
≤
s
). We obtain
(28)A(t,r)+A(r,s)≥2r(1-t)+2s(1-r)≥2s(1-t)≥23A(t,s).
Case
3
(
t
≤
s
≤
r
). One has
(29)A(t,r)+A(r,s)≥2r(1-t)+2r(1-s)≥2s(1-t)≥23A(t,s).
This completes the proof.In the sequel, for anyq∈B((0,1)), we recall that
(30)αq∶=sup⁡t,s∈(0,1)∫01‍G(t,r)G(r,s)G(t,s)|q(r)|dr,
and we denote by
(31)h1(t)=t(1-t)2,h2(t)=t2(1-t),fort∈[0,1].Proposition 8.
Letq be a function in K; then(i)
(32)αq≤32∫01‍(ρ(r))3|q(r)|dr<∞.(ii)
Fort∈[0,1], one has
(33)∫01‍G(t,s)h1(s)|q(s)|ds≤αqh1(t).(iii)
Fort∈[0,1], one has
(34)∫01‍G(t,s)h2(s)|q(s)|ds≤αqh2(t).
In particular, fort∈[0,1], one has
(35)∫01‍G(t,s)ρ(s)|q(s)|ds≤αqρ(t).Proof.
Letq be a function in K.(i) Using (24) and (23), one has, for t,s∈(0,1),
(36)∫01‍G(t,r)G(r,s)G(t,s)|q(r)|dr≤32∫01‍(ρ(r)ρ(t))2G(t,r)+(ρ(r)ρ(s))2G(r,s)|q(r)|dr≤32∫01‍(ρ(r))3|q(r)|dr.Hence(37)αq≤32∫01‍(ρ(r))3|q(r)|dr<∞.(ii) Since, for eacht,s∈(0,1), one has lim⁡r→0⁡(G(s,r)/G(t,r))=h1(s)/h1(t), then we deduce by Fatou’s lemma and (12), that
(38)∫01‍G(t,s)h1(s)h1(t)|q(s)|ds≤liminf⁡r→0∫01‍G(t,s)G(s,r)G(t,r)|q(s)|ds≤αq;which implies that, fort∈[0,1],
(39)∫01‍G(t,s)h1(s)|q(s)|ds≤αqh1(t).(iii) Similarly, we prove inequality (34) by observing that
(40)lim⁡r→1G(s,r)G(t,r)=h2(s)h2(t).Inequality (35) follows from Proposition 8 (ii)-(iii) and the fact that ρ(t)=h1(t)+h2(t). This completes the proof.


## 3. Proofs of Main Results

In this section, we aim at proving Theorem2 and Corollary 3. So, we need the following preliminaries results.For a nonnegative functionq in K such that αq<1, we define the function G(t,s) on [0,1]×[0,1], by
(41)G(t,s)=∑n=0∞(-1)nGn(t,s),
where G0(t,s)=G(t,s) and
(42)Gn(t,s)=∫01‍G(t,r)Gn-1(r,s)q(r)dr,n≥1.
Next, we establish some inequalities on Gn(t,s). In particular, we deduce that G(t,s) is well defined.Lemma 9.
Letq be a nonnegative function in K such that αq<1; then, for each n≥0 and t,s∈[0,1], one has(i)
Gn(t,s)≤αqnG(t,s).In particular,G(t,s) is well defined in [0,1]×[0,1].(ii)
(43)Ln(ρ(t))2(ρ(s))2≤Gn(t,s)≤Rnρ(t)ρ(s)(ρ(t)∧ρ(s));whereLn=(1/3n+1)(∫01‍(ρ(r))4q(r)dr)n and Rn=(1/2n+1)(∫01‍(ρ(r))3q(r)dr)n.(iii)
Gn+1(t,s)=∫01Gn(t,r)G(r,s)q(r)dr and Gn(t,s)=Gn(s,t).(iv)
∫01G(t,r)G(r,s)q(r)dr=∫01G(t,r)G(r,s)q(r)dr.Proof.
(i) The assertion is clear forn=0.
Assume that inequality in (i) holds for somen≥0; then by using (42) and (12), we obtain
(44)Gn+1(t,s)≤αqn∫01‍G(t,r)G(r,s)q(r)dr≤αqn+1G(t,s).
Now, sinceGn(t,s)≤αqnG(t,s), it follows that G(t,s) is well defined in [0,1]×[0,1].
(ii) Using (22) and (42), we obtain (43) by induction.
(iii) SinceG(t,s)=G(s,t), then equalities in (iii) are clear for n=0.
Assume that for a given integern≥1 and t,s∈[0,1], one has
(45)Gn(t,s)=∫01‍Gn-1(t,r)G(r,s)q(r)dr,=Gn-1(s,t).
Using (42) and Fubini-Tonelli theorem, we obtain
(46)Gn+1(t,s)=∫01‍G(t,r)(∫01‍Gn-1(r,ξ)G(ξ,s)q(ξ)dξ)q(r)dr=∫01‍(∫01‍G(t,r)Gn-1(r,ξ)q(r)dr)G(ξ,s)q(ξ)dξ=∫01‍Gn(t,ξ)G(ξ,s)q(ξ)dξ.
On the other hand, by (42) and (45), we deduce that
(47)Gn(t,s)=∫01‍G(t,r)Gn-1(r,s)q(r)dr=∫01‍Gn-1(s,r)G(r,t)q(r)dr=Gn(s,t).
(iv) Letn≥0 and t,r,s∈[0,1]. By Lemma 9 (i) and (22), one has
(48)0≤Gn(t,r)G(r,s)q(r)≤αqnG(t,r)G(r,s)q(r).
Hence the series∑n≥0∫01‍Gn(t,r)G(r,s)q(r)dr converges.
So we deduce by the dominated convergence theorem and Lemma9 (iii) that
(49)∫01G(t,r)G(r,s)q(r)dr=∑n=0∞∫01(-1)nGn(t,r)G(r,s)q(r)dr=∑n=0∞∫01(-1)nG(t,r)Gn(r,s)q(r)dr=∫01G(t,r)G(r,s)q(r)dr.Proposition 10.
Letq be a nonnegative function in K such that αq<1. Then the function (t,s)→G(t,s) is continuous on [0,1]×[0,1].Proof.
We claim that forn≥0, the function Gn(t,s) is continuous on [0,1]×[0,1].
The assertion is clear forn=0.
Now assume that for a given integern≥1, the function (t,s)→Gn-1(t,s) is continuous on [0,1]×[0,1]. By (42), one has
(50)Gn(t,s)=∫01‍G(t,r)Gn-1(r,s)q(r)dr.
Since the function(t,s,r)→G(t,r)Gn-1(r,s) is continuous on [0,1]×[0,1]×[0,1] and by Lemma 9 (i) and (22)
(51)G(t,r)Gn-1(r,s)q(r)≤αqn-1G(t,r)G(r,s)q(r)≤(ρ(r))3q(r),
we deduce by the dominated convergence theorem that(t,s)→Gn(t,s) is continuous on [0,1]×[0,1].
This proves our claim.
Now by Lemma9 (i) and (22), one has, for each t,s∈[0,1],
(52)Gn(t,s)≤αqnG(t,s)≤12ρ(t)(ρ(s))2αqn≤αqn.
This implies that the series∑n≥0(-1)nGn(t,s) is uniformly convergent on [0,1]×[0,1] and therefore the function (t,s)→G(t,s) is continuous on [0,1]×[0,1].
The proof is completed.Lemma 11.
Letq be a nonnegative function in K such that αq≤1/2. Then for t,s in [0,1], one has
(53)(1-αq)G(t,s)≤G(t,s)≤G(t,s).Proof.
Sinceαq≤1/2, we deduce from Lemma 9 (i) that
(54)|G(t,s)|≤∑n=0∞(αq)nG(t,s)=11-αqG(t,s).
On the other hand, from the expression ofG, one has
(55)G(t,s)=G(t,s)-∑n=0∞(-1)nGn+1(t,s).
Since the series∑n≥0∫01‍G(t,r)Gn(r,s)q(r)dr is convergent, we deduce by (55) and (42) that
(56)G(t,s)=G(t,s)-∑n=0∞(-1)n∫01‍G(t,r)Gn(r,s)q(r)dr=G(t,s)-∫01‍G(t,r)(∑n=0∞(-1)nGn(r,s))q(r)dr.
That is,(57)G(t,s)=G(t,s)-V(qG(·,s))(t).
Now from (54) and Lemma 9 (i) (with n=1), we obtain
(58)V(qG(·,s))(t)≤11-αqV(qG(·,s))(t)=11-αqG1(t,s)≤αq1-αqG(t,s).
This implies that(59)G(t,s)≥G(t,s)-αq1-αqG(t,s)=1-2αq1-αqG(t,s)≥0.
So it follows thatG(t,s)≤G(t,s) and by (57) and Lemma 9 (i) (with n=1), one has
(60)G(t,s)≥G(t,s)-V(qG(·,s))(t)≥(1-αq)G(t,s).In the sequel, for a given nonnegative functionq∈K such that αq≤1/2, we define the operator Vq:B+((0,1))→B+((0,1)) by
(61)Vqψ(t)=∫01‍G(t,s)ψ(s)ds,t∈[0,1].Using (53) and (22), we obtain the following.Corollary 12.
Letq be a nonnegative function in K such that αq≤1/2 and ψ∈B+((0,1)); then the following statements are equivalent.(i)
The functiont→Vqψ(t) is continuous on [0,1].(ii)
The integral∫01(ρ(s))2ψ(s)ds converges.Next, we will prove that the kernelVq satisfies the following resolvent equation.Lemma 13.
Letq be a nonnegative function in K such that αq≤1/2 and ψ∈B+((0,1)). Then Vqψ satisfies the following resolvent equation:
(62)Vψ=Vqψ+Vq(qVψ)=Vqψ+V(qVqψ).
In particular, ifV(qψ)<∞, one has
(63)(I-Vq(q·))(I+V(q·))ψ=(I+V(q·))(I-Vq(q·))ψ=ψ.Proof.
Lett,s∈[0,1], then by (57) one has
(64)G(t,s)=G(t,s)+V(qG(·,s))(t),
which implies by Fubini-Tonelli theorem that, for ψ∈B+((0,1)),
(65)Vψ(t)=∫01‍(G(t,s)+V(qG(·,s))(t))ψ(s)ds=Vqψ(t)+V(qVqψ)(t).
On the other hand, by Lemma9 (iii) and Fubini-Tonelli theorem, we obtain for ψ∈B+((0,1)) and t∈[0,1](66)∫01∫01G(t,r)G(r,s)ψ(s)q(r)drds=∫01∫01G(t,r)G(r,s)ψ(s)q(r)drds;
that is,
(67)Vq(qVψ)(t)=V(qVqψ)(t).
So we obtain(68)Vψ=Vqψ+V(qVqψ)=Vqψ+Vq(qVψ)(t).
This completes the proof.Proposition 14.
Letq be a nonnegative function in K∩C(0,1) such that αq≤1/2 and let ψ∈B+((0,1)) such that t→(ρ(t))2ψ(t) is continuous and integrable on (0,1). Then Vqψ is the unique nonnegative solution in C4((0,1))∩C([0,1]) of the perturbed fourth-order equation (13) satisfying
(69)(1-αq)Vψ≤Vqψ≤Vψ.Proof.
It is clear by Corollary12 that the function t→q(t)Vqψ(t) is continuous on (0,1). Using (62) and (22), there exists a nonnegative constant c such that
(70)Vqψ(t)≤Vψ(t)≤12∫01‍ρ(t)(ρ(s))2ψ(s)ds≤cρ(t).
So we deduce that(71)∫01‍(ρ(s))2q(s)Vqψ(s)ds≤c∫01‍(ρ(s))3q(s)ds<∞.
Hence by using Remark4, the function u=Vqψ=Vψ-V(qVqψ) satisfies the equation
(72)u(4)(t)=ψ(t)-q(t)u(t),t∈(0,1),u(0)=u(1)=u′(0)=u′(1)=0,
and by integration inequalities (53), we obtain (69).
It remains to prove the uniqueness. Assume thatv is another nonnegative solution in C4((0,1))∩C([0,1]) of problem (13) satisfying (69).
Since the functiont→q(t)v(t) is continuous on (0,1) and by (69) and (70), the function t→(ρ(t))2q(t)v(t) is integrable on (0,1); then it follows by Remark 4 that the function v~∶=v+V(qv) satisfies
(73)v~(4)(t)=ψ(t),t∈(0,1),v~(0)=v~(1)=v~′(0)=v~′(1)=0.
From the uniqueness in Remark4, we deduce that
(74)v~∶=v+V(qv)=Vψ.
Hence(75)(I+V(q·))(v-u)=0.
Now since by (69), (70), and (35), one has
(76)V(q|v-u|)≤2cV(qρ)≤2cαqρ<∞,
then by (63), we deduce that u=v. This completes the proof.Proof of Theorem2.
Leta≥0 and b≥0 with a+b>0 and recall that
(77)ω(t)=at(1-t)2+bt2(1-t)=ah1(t)+bh2(t).
Sinceφ satisfies (H2), then there exists a positive function q in K∩C(0,1) such that αq≤1/2 and for each t∈(0,1), the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1].
Let(78)Λ∶={u∈B+((0,1)):(1-αq)ω≤u≤ω},
and define the operator T on Λ by
(79)Tu=ω-Vq(qω)+Vq((q-φ(·,u))u).
By (62) and Proposition 8, one has
(80)Vq(qω)≤V(qω)≤αqω≤ω,
and by (H2), we obtain
(81)0≤φ(·,u)≤q,∀u∈Λ.
So we claim thatΛ is invariant under T. Indeed, using (81) and (80), one has for u∈Λ(82)Tu≤ω-Vq(qω)+Vq(qu)≤ω,Tu≥ω-Vq(qω)≥(1-αq)ω.
Next, we will prove that the operatorT is nondecreasing on Λ. Indeed, let u,v∈Λ be such that u≤v. Since the map s→s(q(t)-φ(t,sω(t))) is nondecreasing on [0,1], for t∈(0,1), we obtain
(83)Tv-Tu=Vq([v(q-φ(·,v))-u(q-φ(·,u))])≥0.
Now, we consider the sequence(un) defined by u0=(1-αq)ω and un+1=Tun, for n∈N. Since Λ is invariant under T, one has u1=Tu0≥u0 and by the monotonicity of T, we deduce that
(84)(1-αq)ω=u0≤u1≤⋯≤un≤un+1≤ω.
Hence by dominated convergence theorem and hypotheses(H1)-(H2), we conclude that the sequence (un) converges to a function u∈Λ satisfying
(85)u=(I-Vq(q·))ω+Vq((q-φ(·,u))u).
That is,(86)(I-Vq(q·))u=(I-Vq(q·))ω-Vq(uφ(·,u)).
On the other hand, since by (80), one has V(qu)≤V(qω)≤ω<∞, then by applying the operator (I+V(q·)) on both sides of the above equality and using (62) and (63), we conclude that u satisfies
(87)u=ω-V(uφ(·,u)).
Next we aim at proving thatu is a solution of problem (5). To this end, we remark by (81) and (9) that
(88)uφ(·,u)≤qω≤max⁡(a,b)qρ.
This implies by Corollary6 (ii) that the function t→V(uφ(.,u))(t) is continuous on [0,1] and so by (87), u is continuous on [0,1].
Now, since by(H1) and (88), the function t→(ρ(t))2u(t)φ(t,u(t)) is continuous and integrable on (0,1), we conclude by Remark 4 that u is the required solution.
It remains to prove that under condition(H3),u is the unique solution to problem (5) satisfying (14). Assume that v is another nonnegative solution in C4((0,1))∩C([0,1]) to problem (5) satisfying (14). Since v≤ω, we deduce by (88) that
(89)0≤vφ(·,v)≤qω≤max⁡(a,b)qρ.
So the functiont→(ρ(t))2v(t)φ(t,v(t)) is continuous and integrable on (0,1) and by Remark 4, we conclude that the function v~∶=v+V(vφ(.,v)) satisfies
(90)v~(4)(t)=0,t∈(0,1),v~(0)=v~(1)=0,v~′(0)=a,v~′(1)=-b.
From the uniqueness in problem (8), we deduce that
(91)v=ω-V(vφ(·,v)).
Now leth be the function defined on (0,1) by
(92)h(t)={v(t)φ(t,v(t))-u(t)φ(t,u(t))v(t)-u(t)ifv(t)≠u(t),0ifv(t)=u(t).
Then by(H3),  h∈B+((0,1)) and by (87) and (91), one has
(93)(I+V(h·))(v-u)=0.
On the other hand, by(H2), we remark that h≤q and by (80) we deduce that
(94)V(h|v-u|)≤2V(qω)≤2αqω<∞.
Hence by (63), we conclude that u=v. This completes the proof.Proof of Corollary3.
Letφ(t,s)=λp(t)f(s),  θ(s)=sf(s), and p~(t)=p(t)max⁡0≤ξ≤ω(t)⁡θ′(ξ). It is clear that hypotheses (H1) and (H3) are satisfied. Since the function q(t)∶=λp~(t) belongs to the class K, one has αq≤1/2 for λ∈[0,1/2αp~). Moreover, by a simple computation, for s∈[0,1] and t∈(0,1), we obtain
(95)dds[s(q(t)-φ(t,sω(t)))]=q(t)-λp(t)θ′(sω(t))≥0.
This implies that the functionφ satisfies hypothesis (H2). So the result follows by Theorem 2.Example 15.
Leta≥0 and b≥0 with a+b>0. Let σ≥0, and p be a positive continuous function on (0,1) such that
(96)∫01‍r3+σ(1-r)3+σp(r)dr<∞.
Since the functionp~(t)∶=(σ+1)p(t)(ω(t))σ belongs to the class K, then, for λ∈[0,1/2αp~), the problem
(97)u(4)(t)+λp(t)uσ+1(t)=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(98)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].Example 16.
Leta≥0 and b≥0 with a+b>0. Let σ≥0,γ>0, and p be a positive continuous function on (0,1) such that
(99)∫01‍r3+σ+γ(1-r)3+σ+γp(r)dr<∞.
Letθ(s)=sσ+1log⁡(1+sγ). Since the function p~(t)∶=p(t)max⁡0≤ξ≤ω(t)⁡θ′(ξ)∈K, then, for λ∈[0,1/2αp~), the problem
(100)u(4)(t)+λp(t)uσ+1(t)log⁡(1+uγ(t))=0,t∈(0,1),u(0)=u(1)=0,u′(0)=a,u′(1)=-b
has a unique positive solution u in C4((0,1))∩C([0,1]) satisfying
(101)(1-λαp~)ω(t)≤u(t)≤ω(t),t∈[0,1].



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*Source: 384958-2014-08-11.xml*