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International Journal of Differential Equations
VolumeΒ 2011Β (2011), Article IDΒ 261963, 16 pages
http://dx.doi.org/10.1155/2011/261963
Research Article

On a Constructive Approach for Derivative-Dependent Singular Boundary Value Problems

1Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, India
2Department of Mathematics, BITS Pilani, Rajasthan, Pilani 333031, India

Received 20 May 2011; Revised 24 August 2011; Accepted 13 September 2011

Academic Editor: AlbertoΒ Cabada

Copyright Β© 2011 R. K. Pandey and Amit K. Verma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We present a constructive approach to establish existence and uniqueness of solution of singular boundary value problem βˆ’(𝑝(π‘₯)π‘¦ξ…ž(π‘₯))ξ…ž=π‘ž(π‘₯)𝑓(π‘₯,𝑦,π‘π‘¦ξ…ž) for 0<π‘₯≀𝑏,𝑦(0)=π‘Ž,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾1. Here 𝑝(π‘₯)>0 on (0,𝑏) allowing 𝑝(0)=0. Further π‘ž(π‘₯) may be allowed to have integrable discontinuity at π‘₯=0, so the problem may be doubly singular.

1. Introduction

Consider the following singular boundary value problem: ξ€·π‘€π‘¦β‰‘βˆ’π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žξ€·=π‘ž(π‘₯)𝑓π‘₯,𝑦(π‘₯),𝑝(π‘₯)π‘¦ξ…žξ€Έ(π‘₯),0<π‘₯≀𝑏,(1.1)𝑦(0)=π‘Ž,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾1,(1.2) where 𝛼1>0, 𝛽1β‰₯0, 𝛾1 is any finite constant. We assume that 𝑝(π‘₯) and π‘ž(π‘₯) satisfy the following conditions:(A1) 𝑝(π‘₯)>0 in (0,𝑏], π‘βˆˆπΆ[0,𝑏]∩𝐢1(0,𝑏) and βˆ«π‘0(𝑑𝑑/𝑝(𝑑))<∞,(A2) π‘ž(π‘₯)>0 in (0,𝑏] and βˆ«π‘0π‘ž(𝑑)𝑑𝑑<∞.

In this work we establish existence and uniqueness of solution of the singular problem (1.1)-(1.2). We use monotone iterative method. For this we require an appropriate iterative scheme. In this regard Cherpion et al. [1] suggest the following approximation scheme:βˆ’π›Όξ…žξ…žπ‘›+1+Μƒπ‘˜(π‘₯)π›Όξ…žπ‘›+1+̃𝑙(π‘₯)𝛼𝑛+1ξ€·=𝑓π‘₯,𝛼𝑛,π›Όξ…žπ‘›ξ€Έ+Μƒπ‘˜(π‘₯)π›Όξ…žπ‘›+̃𝑙(π‘₯)𝛼𝑛,𝛼𝑛+1(0)=𝛼𝑛+1(1)=0,(1.3) for the regular boundary value problem βˆ’π‘¦ξ…žξ…ž=𝑓(π‘₯,𝑦,𝑦′),𝑦(0)=𝑦(1)=0. They also suggest that (1.3) with ̃𝑙(π‘₯)=0 or with constant Μƒπ‘˜ and ̃𝑙 does not work for the Dirichlet boundary condition.

Thus, for our problem we consider the following iterative scheme: 𝐿𝑦𝑛+1ξ€·=𝐹π‘₯,𝑦𝑛,π‘π‘¦ξ…žπ‘›ξ€Έπ‘¦,0<π‘₯≀𝑏,𝑛+1(0)=0,𝛼1𝑦𝑛+1(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…žπ‘›+1(𝑏)=𝛾1,(1.4) where 𝐿𝑦=βˆ’π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žβˆ’πœ‡(π‘₯)π‘ž(π‘₯)𝑝(π‘₯)π‘¦ξ…žπΉξ€·(π‘₯)βˆ’πœ†π‘˜(π‘₯)π‘ž(π‘₯)𝑦(π‘₯),π‘₯,𝑦,π‘π‘¦ξ…žξ€Έξ€·=π‘ž(π‘₯)𝑓π‘₯,𝑦,π‘π‘¦ξ…žξ€Έβˆ’πœ‡(π‘₯)π‘ž(π‘₯)𝑝(π‘₯)π‘¦ξ…ž(π‘₯)βˆ’πœ†π‘˜(π‘₯)π‘ž(π‘₯)𝑦(π‘₯).(1.5) We assume that π‘˜(π‘₯) and πœ‡(π‘₯) satisfy the following conditions.(A3)π‘˜(π‘₯)∈𝐢[0,𝑏], βˆƒπ‘šπ‘˜, π‘€π‘˜βˆˆπ‘… such that 0<π‘šπ‘˜β‰€π‘˜(π‘₯)β‰€π‘€π‘˜.(A4)πœ‡(π‘₯)∈𝐿1π‘ž(0,𝑏), that is,βˆ«π‘0π‘ž(π‘₯)πœ‡(π‘₯)𝑑π‘₯<∞.(A5)Further, we assume that the homogeneous boundary value problem 𝐿𝑦=0, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=0 has only trivial solution.

For βˆ«π‘0(𝑑𝑑/𝑝(𝑑))<∞, several researchers ([2–5]) suggest to reduce the singular problem to regular problem by a change of variable. But in [6] it is suggested that a direct consideration of singular problems provide better results.

Further, the following sign restrictions are imposed by several researchers ([4, 5, 7–9]): (i)𝑦𝑓(π‘₯,𝑦,0)>0, |𝑦|>𝑀0, where 𝑀0 is a constant ([7–9]) or(ii)𝑓(π‘₯,𝑀1,0)β‰₯0β‰₯𝑓(π‘₯,βˆ’π‘€2,0),𝑀1,𝑀2β‰₯0 ([4, 5]).

But such sign restrictions are quite restrictive as the simple differential equation π‘¦ξ…žξ…ž=2 fails to satisfy the sign restrictions (i) and (ii) ([7]).

In the present work we consider the singular boundary value problem (SBVP) directly and do not impose any sign restriction. Further, we do not assume that the point π‘₯=0 is a regular singular point as assumed in [6, 9]. We use iterative scheme (1.4) to establish existence and uniqueness of the solution of the problem. With the help of nonnegativity of Green's function, existence uniqueness of linear singular boundary value problem (LSBVP) is established.

This paper is divided in four sections. In Section 2, we show that singular point π‘₯=0 is of limit circle type; hence, spectrum is pure point spectrum with complete set of orthonormal eigenfunctions. In Section 3, we prove the existence uniqueness of the corresponding LSBVP. Finally, in Section 4, using the results of Section 3, we establish the existence uniqueness of solutions of the nonlinear problem (1.1)-(1.2).

2. Eigenfunction Expansion

Let 𝐿2𝑠(0,𝑏) be a Hilbert space with weight 𝑠(π‘₯)=π‘˜(π‘₯)π‘ž(π‘₯) and the inner product defined asξ€œβŸ¨π‘’,π‘£βŸ©=𝑏0𝑠(𝑑)𝑒(𝑑)𝑣(𝑑)𝑑𝑑.(2.1) Conditions on 𝑝, π‘ž, πœ‡, and π‘˜ guarantee that the singular point π‘₯=0 is of limit circle type (Weyl's Theorem, [10, page 438]. Thus, we have pure point spectrum ([11, page 125]. Next, from the Lagrange's identity, it is easy to see that all the eigenvalues are real, simple, positive, and eigenfunctions are orthogonal. Let the eigenvalues be 0<πœ†0<πœ†1<πœ†2<β‹―, and let the corresponding eigenfunctions be πœ“0,πœ“1,πœ“2,…, respectively. Next, we transform 𝐿𝑦=0 by changing variable βˆšπ‘§=𝑔(π‘₯)𝑦, where 𝑔(π‘₯)=π‘’βˆ«π‘₯0πœ‡(𝑑)π‘ž(𝑑)𝑑𝑑, toβˆ’ξ€·π‘(π‘₯)π‘§ξ…žξ€Έ(π‘₯)ξ…ž+π‘Ÿ(π‘₯)𝑧(π‘₯)=πœ†π‘ (π‘₯)𝑧(π‘₯),0<π‘₯≀𝑏,(2.2) where π‘Ÿ(π‘₯)=(1/2){πœ‡(π‘₯)π‘ž(π‘₯)𝑝(π‘₯)}β€²+(1/4){πœ‡(π‘₯)π‘ž(π‘₯)}2𝑝(π‘₯). Now following the analysis of Theorem 2.7, (i), (ii) and Theorem 2.17 of [11] for the operator (1/𝑠)(π‘Ÿ+𝑀), where 𝑀 is defined by (1.1), the following results can be established.

Theorem 2.1. Let 𝑓(π‘₯) be the primitive of an absolutely continuous function, and let 1𝑠(π‘Ÿ+𝑀)π‘“βˆˆπΏ2𝑠𝑓(0,𝑏),(𝑏)sinπ›Όβˆ’π‘(𝑏)π‘“ξ…ž(𝑏)cos𝛼=0,where𝛼isreal,limπ‘₯β†’0𝑝(π‘₯)π‘Šπ‘₯[]𝑓,πœ“=0,(2.3) for every nonreal πœ†, where πœ“(π‘₯,πœ†)∈𝐿2𝑠(0,𝑏) is a solution of (2.2) and π‘Š[𝑓,πœ“] is the wronskian of 𝑓 and πœ“. Then 𝑓(π‘₯)=βˆžξ“π‘›=0π‘π‘›πœ“π‘›(π‘₯),(0≀π‘₯≀𝑏),(2.4) being the series absolutely and uniformly convergent on [0,𝑏].

Theorem 2.2. Let π‘“βˆˆπΏ2𝑠(0,𝑏). Then ξ€œπ‘0𝑠(π‘₯){𝑓(π‘₯)}2𝑑π‘₯=βˆžξ“π‘›=0𝑐2𝑛.(2.5)

Theorem 2.3. Let π‘“βˆˆπΏ2𝑠(0,𝑏), and let Ξ¦(π‘₯,πœ†) be the solution of βˆ’ξ€·π‘(π‘₯)π‘§ξ…žξ€Έ(π‘₯)ξ…ž+π‘Ÿ(π‘₯)𝑧(π‘₯)βˆ’πœ†π‘ (π‘₯)𝑧(π‘₯)=𝑠(π‘₯)𝑓(π‘₯),0<π‘₯≀𝑏,(2.6) satisfying 𝛼11𝑧(𝑏)+𝛽1𝑝(𝑏)𝑧′(𝑏)=0, where 𝛼11=𝛼1βˆ’(1/2)𝛽1πœ‡(𝑏)𝑝(𝑏)π‘ž(𝑏). Then for πœ† not equal to any of the values of πœ†π‘›, one has Ξ¦(π‘₯,πœ†)=βˆžξ“π‘›=0π‘π‘›πœ“π‘›πœ†βˆ’πœ†π‘›,(2.7) where the series is absolutely convergent.

Remark 2.4. Since ‖⋅‖𝑠 on 𝐿2𝑠(0,𝑏) is equivalent to β€–β‹…β€–π‘ž on 𝐿2π‘ž(0,𝑏), we can apply Theorems 2.1–2.3 in 𝐿2π‘ž(0,𝑏) also.

3. Linear Singular Sturm-Liouville’s Problem

In this section we apply Theorem 1.1 of   [12] to the differential operator 𝐿 and generate two linearly independent solutions of the linear problem. Further, with the help of these solutions, Green's function is constructed, and nonnegativity of the Green's function is established.

Theorem 3.1. Let 𝑝(π‘₯), π‘ž(π‘₯), π‘˜(π‘₯), and πœ‡(π‘₯) satisfy (A1), (A2), (A3), and (A4), respectively. Then the initial value problems (IVPs) 𝐿𝑦=0,0<π‘₯≀𝑏,𝑦(0)=π‘Ž0,limπ‘₯β†’0+𝑝(π‘₯)π‘¦ξ…ž(π‘₯)=𝑏0,(3.1)𝐿𝑦=0,0<π‘₯≀𝑏,𝑦(𝑏)=𝑐0,𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝑑0(3.2) have a solution in 𝐿2𝑠(0,𝑏) or equivalently in 𝐿2π‘ž(0,𝑏) (Remark 2.4).

3.1. Green's Function

Green's function 𝐺(π‘₯,𝑑,πœ†) for the differential operator 𝐿 can be defined as1𝐺(π‘₯,𝑑,πœ†)=𝑝(π‘₯)π‘Š(πœ“,πœ™)βˆ’πœ“(𝑑)πœ™(π‘₯),if0<𝑑≀π‘₯,βˆ’πœ“(π‘₯)πœ™(𝑑),ifπ‘₯≀𝑑≀𝑏,(3.3) where πœ™=𝑆{𝛼1𝑦1(π‘₯)+𝛽1𝑦2(π‘₯)}, 𝑆=1/𝛼21+𝛽21, 𝑦1(𝑏)=0, 𝑝(𝑏)π‘¦ξ…ž1(𝑏)=βˆ’1, 𝑦2(𝑏)=1, 𝑝(𝑏)π‘¦ξ…ž2(𝑏)=0, and πœ“ is a nontrivial solution of IVP (3.1) with π‘Ž0=0, 𝑏0=1. From (A5), it is easy to conclude that 𝑝(π‘₯)π‘Š(πœ“,πœ™)|π‘₯=𝑏≠0; thus, πœ™ and πœ“ are linearly independent.

Next we establish nonnegativity of Green's function. For this we need to establish following results.

Lemma 3.2. If 𝑦(π‘₯) satisfies 𝐿𝑦(π‘₯)=π‘ž(π‘₯)𝑓(π‘₯)β‰₯0, for 0<π‘₯≀𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=𝛾1β‰₯0, where 𝑝(π‘₯), π‘ž(π‘₯), πœ‡(π‘₯), and π‘˜(π‘₯) satisfy (A1), (A2), (A3), and (A4), respectively, then 𝑦(π‘₯)β‰₯0 provided that πœ†β‰€0.

Proof. We divide the proof in two cases as follows.
Case i. When πœ†<0. On contrary assume that there exists a point π‘βˆˆ(0,𝑏) such that 𝑦(𝑐)<0. Then from the continuity of the solutions there exists a point π‘‘βˆˆ(0,𝑏) such that 𝑦(𝑑)<0, 𝑦′(0)=0, and π‘¦ξ…žξ…ž(𝑑)β‰₯0. Now at the point 𝑑, we have βˆ’π‘(𝑑)π‘¦ξ…žξ…ž(𝑑)βˆ’π‘ξ…ž(𝑑)π‘¦ξ…ž(𝑑)βˆ’πœ‡(𝑑)π‘ž(𝑑)𝑝(𝑑)π‘¦ξ…ž(𝑑)βˆ’πœ†π‘˜(𝑑)π‘ž(𝑑)𝑦(𝑑)<0,(3.4) which is a contradiction. Hence, 𝑦(π‘₯)β‰₯0 when πœ†<0.
Case ii. When πœ†=0. Using the same notations as in the previous case, we have 𝑦′(π‘₯)>0 for π‘₯>𝑑 and 𝑦′(π‘₯)<0 for π‘₯<𝑑.
Now, we consider the interval [𝑑,π‘₯0]βŠ‚(0,𝑏) where 𝑦′(𝑑)=0, 𝑦<0 in [𝑑,π‘₯0], and 𝑦′>0 in (𝑑,π‘₯0]. Then ξ€œπ‘ƒ=π‘₯0𝑑𝑦𝑝(𝑑)𝑔(𝑑)ξ…žξ€Έ(𝑑)2ξ‚‡βˆ’π‘ (𝑑)𝑔(𝑑)𝑓(𝑑)𝑦(𝑑)𝑑𝑑>0.(3.5) Integrating the first term by parts, we get ξ€·π‘₯𝑃=𝑝0𝑔π‘₯0ξ€Έπ‘¦ξ…žξ€·π‘₯0𝑦π‘₯0ξ€Έ<0,(3.6) which is again a contradiction. Thus, 𝑦(π‘₯)β‰₯0 when πœ†β‰€0.

Lemma 3.3. Consider the following differential equation: 𝐿𝑦(π‘₯)=0,0<π‘₯≀𝑏,(3.7) where 𝑝(π‘₯), π‘ž(π‘₯), πœ‡(π‘₯), and π‘˜(π‘₯) satisfy (A1), (A2), (A3), and (A4), respectively, with the boundary conditions: 𝑦(0)=0,𝛼1𝑝(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾1.(3.8) Then LSBVP (3.7)-(3.8) has a unique solution given by 𝛾𝑦(π‘₯)=1πœ“(π‘₯)𝛼1πœ“(𝑏)+𝛽1𝑝(𝑏)πœ“ξ…ž(,𝑏)(3.9) provided that πœ† is none of the eigenvalues of the corresponding eigenvalue problem and πœ“ satisfies (3.1). Moreover, 𝑦(π‘₯)β‰₯0 if 𝛾1β‰₯0 and 0<πœ†<πœ†0, where πœ†0 is the first positive zero of 𝛼1πœ“(𝑏,πœ†)+𝛽1𝑝(𝑏)πœ“β€²(𝑏,πœ†).

Proof. From Theorem 3.1, it is easy to see that the unique solution of (3.7)-(3.8) can be written as 𝛾𝑦(π‘₯)=1πœ“(π‘₯)𝛼1πœ“(𝑏)+𝛽1𝑝(𝑏)πœ“ξ…ž(,𝑏)(3.10) provided that 𝛼1πœ“(𝑏,πœ†)+𝛽1𝑝(𝑏)πœ“β€²(𝑏,πœ†)β‰ 0; that is, πœ† is none of the eigenvalue of the corresponding eigenvalue problem (Section 2). Since πœ“(0)=0, limπ‘₯β†’0+𝑝(π‘₯)πœ“β€²(π‘₯)=1, and πœ“(π‘₯,πœ†) does not change sign for 0<πœ†<πœ†0, we get that 𝑦(π‘₯)β‰₯0 for 0<πœ†<πœ†0, provided that 𝛾1β‰₯0.

Lemma 3.4. For the linear differential operator associated with 𝐿𝑦(π‘₯)=π‘ž(π‘₯)𝑓(π‘₯),0<π‘₯≀𝑏,(3.11)𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾1,(3.12) with π‘“βˆˆπΏ2π‘ž(0,𝑏), the generalized Green's function for the corresponding homogeneous boundary value problem is given by 𝐺(π‘₯,𝑑,πœ†)=βˆžξ“π‘›=0πœ“ξ€·π‘₯,πœ†π‘›ξ€Έπœ“ξ€·π‘‘,πœ†π‘›ξ€Έπœ†π‘›,βˆ’πœ†(3.13) where πœ“(π‘₯,πœ†π‘–) are the normalized eigenfunctions corresponding to the eigenvalue πœ†π‘–. 𝐺(π‘₯,𝑑,πœ†) satisfies the homogeneous boundary condition provided that πœ†β‰ πœ†0,πœ†1,…. Solution of the non-homogeneous LSBVP (3.11)-(3.12) is 𝛾𝑦(π‘₯)=1πœ“(π‘₯,πœ†)𝛼1πœ“(𝑏)+𝛽1𝑝(𝑏)πœ“ξ…ž+ξ€œ(𝑏)𝑏0π‘ž(𝑑)𝑓(𝑑)𝐺(π‘₯,𝑑,πœ†)𝑑𝑑.(3.14) The series on the right is absolutely convergent.

Proof. The solution 𝑦(π‘₯) of (3.11)-(3.12) can be written as sum of the solution of (3.11) with boundary condition 𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=0 and solution of (3.7) with boundary condition 𝑦(0)=0, 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=𝛾1, 𝛾𝑦(π‘₯)=1πœ“(π‘₯,πœ†)𝛼1πœ“(𝑏,πœ†)+𝛽1𝑝(𝑏)πœ“ξ…ž+ξ€œ(𝑏,πœ†)𝑏0π‘ž(𝑑)𝑓(𝑑)𝐺(π‘₯,𝑑,πœ†)𝑑𝑑,(3.15) where 𝐺(π‘₯,𝑑,πœ†) is Green's function defined by (3.3). Now using the analysis of ([11, page 38], it is easy to show that the generalized Green's function is given by 𝐺(π‘₯,𝑑,πœ†)=βˆžξ“π‘›=0πœ“ξ€·π‘₯,πœ†π‘›ξ€Έπœ“ξ€·π‘‘,πœ†π‘›ξ€Έπœ†π‘›,βˆ’πœ†(3.16) and absolute convergence of the series on the right-hand side follows from the analysis of ([11, page 38]. This completes the proof.

Lemma 3.5. If π‘“βˆˆπΏ2π‘ž(0,𝑏), 𝛾1β‰₯0 and 𝑓β‰₯0, then solution of (3.11)-(3.12) is nonnegative provided that 0<πœ†<πœ†0.

Proof. We first show that 𝐺(π‘₯,𝑑)β‰₯0 for all 0≀π‘₯,𝑑≀𝑏 if 0<πœ†<πœ†0. Fixing 𝑑, 𝐺(π‘₯,𝑑) satisfies 𝐿𝐺(π‘₯,𝑑)=0,0<π‘₯β‰€π‘‘βˆ’, where β€²β‰‘πœ•/πœ•π‘₯. Since 𝐺(0,𝑑)=0, 𝛼1𝐺(π‘‘βˆ’,𝑑)+𝛽1𝑝(π‘‘βˆ’)πΊξ…ž(π‘‘βˆ’,𝑑)β‰₯0 for 0<πœ†<πœ†0, from Lemma 3.3  𝐺(π‘₯,𝑑)β‰₯0 for 0≀π‘₯β‰€π‘‘βˆ’, provided that 0<πœ†<πœ†0. By the symmetry, continuity, and 𝐺(𝑑,𝑑)β‰₯0 for 0<πœ†<πœ†0, it follows that 𝐺(π‘₯,𝑑)β‰₯0 for 0≀π‘₯,𝑑≀𝑏, provided that 0<πœ†<πœ†0. The result follows.

Corollary 3.6. If 𝑦(π‘₯) satisfies 𝐿𝑦(π‘₯)=π‘ž(π‘₯)𝑓(π‘₯)β‰₯0 for 0<π‘₯≀𝑏 and 𝑦(0)=0, 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=𝛾1β‰₯0, then 𝑦(π‘₯)β‰₯0, provided that πœ†<πœ†0.

Proof. The proof follows from Lemmas 3.2 and 3.5.

Corollary 3.7. The solution of the boundary value problem in Lemma 3.4 is unique.

Proof. The proof follows from Corollary 3.6.

4. Nonlinear Sturm-Liouville’s Problem

In this section, we establish the existence uniqueness of solution of the nonlinear problem (1.1)-(1.2). For this, first we prove that the sequences generated by (1.4) are monotonic sequences (Lemmas 4.2 and 4.3). Then using the bound for 𝑝𝑦′ (Lemmas 4.9 and 4.10), the uniform convergence of these sequences to a solution of the nonlinear problem is established (Theorem 4.11). Finally the uniqueness of the solution is established in Theorem 4.14.

The nonlinear boundary value problem βˆ’ξ€·π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žξ€·=π‘ž(π‘₯)𝑓π‘₯,𝑦,π‘π‘¦ξ…žξ€Έ,0<π‘₯≀𝑏,𝑦(0)=π‘Ž,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾2(4.1) can be transformed to βˆ’ξ€·π‘(π‘₯)π‘’ξ…žξ€Έ(π‘₯)ξ…žξ€·=π‘ž(π‘₯)𝑓π‘₯,𝑒+π‘Ž,π‘π‘’ξ…žξ€Έ,0<π‘₯≀𝑏,𝑒(0)=0,𝛼1𝑒(𝑏)+𝛽1𝑝(𝑏)π‘’ξ…ž(𝑏)=𝛾2βˆ’π‘Žπ›Ό1,(4.2) with 𝑒=π‘¦βˆ’π‘Ž. Further, the functions 𝑓(π‘₯,𝑒+π‘Ž,𝑝𝑒′) and 𝑓(π‘₯,𝑦,𝑝𝑦′) satisfy the same Lipschitz condition, so we may work with the boundary value problemβˆ’ξ€·π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žξ€·=π‘ž(π‘₯)𝑓π‘₯,𝑦,π‘π‘¦ξ…žξ€Έ,0<π‘₯≀𝑏,𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)=𝛾1.(4.3) Next, we define upper solution 𝑒0(π‘₯) and lower solution 𝑣0(π‘₯) such that 𝑒0β‰₯𝑣0, which work as initial iterates for our constructive approach.

Definition 4.1. A function 𝑒0(π‘₯)∈𝐢[0,𝑏]∩𝐢2(0,𝑏] is an upper solution if βˆ’ξ€·π‘(π‘₯)π‘’ξ…ž0ξ€Έ(π‘₯)ξ…žξ€·β‰₯π‘ž(π‘₯)𝑓π‘₯,𝑒0,π‘π‘’ξ…ž0𝑒,0<π‘₯≀𝑏,0(0)=0,𝛼1𝑒0(𝑏)+𝛽1𝑝(𝑏)π‘’ξ…ž0(𝑏)β‰₯𝛾1,(4.4) and a function 𝑣0(π‘₯)∈𝐢[0,𝑏]∩𝐢2(0,𝑏] is a lower solution if βˆ’ξ€·π‘(π‘₯)π‘£ξ…ž0ξ€Έ(π‘₯)ξ…žξ€·β‰€π‘ž(π‘₯)𝑓π‘₯,𝑣0,π‘π‘£ξ…ž0𝑣,0<π‘₯≀𝑏,0(0)=0,𝛼1𝑣0(𝑏)+𝛽1𝑝(𝑏)π‘£ξ…ž0(𝑏)≀𝛾1.(4.5)

Lemma 4.2. If πœ†<0, πœ†π‘˜(π‘₯)≀𝐾1, |πœ‡(π‘₯)|≀𝐿1, 𝐿𝑦β‰₯0 for 0<π‘₯≀𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)β‰₯0, then 𝐾1ξ€Έξ€·πœ‡βˆ’πœ†π‘˜(π‘₯)π‘¦βˆ’(π‘₯)+𝐿1ξ€·signπ‘¦ξ…žξ€Έξ€Έπ‘π‘¦β€²β‰₯0,0<π‘₯≀𝑏,(4.6) provided that ξ€œ1+πœ†π‘0π‘‘π‘‘ξ€œπ‘(𝑑)𝑔(𝑑)𝑏0𝑠(𝑑)𝑔(𝑑)π‘‘π‘‘βˆ’supπ‘π‘”π‘ξ‚ξ€œπ‘0𝑑𝑑𝐾𝑝(𝑑)𝑔(𝑑)>0,0<π‘₯≀𝑏,1ξ€Έβˆ’ξ€·πΏβˆ’πœ†π‘˜(π‘₯)1βˆ’||πœ‡||ξ€ΈΞ¦(π‘₯)(𝑝,π‘ž,𝑠,𝑔)β‰₯0,0<π‘₯≀𝑏,(4.7) hold. Here, ξ‚€Ξ¦(𝑝,π‘ž,𝑠,𝑔)=supπ‘π‘”π‘ξ‚ξ€œπΌ(𝑝,π‘ž,𝑠,𝑔)βˆ’πœ†π‘0ξ‚΅ξ€œπ‘ (𝑑)𝑔(𝑑)𝑑𝑑,(4.8)𝐼(𝑝,π‘ž,𝑠,𝑔)=1+πœ†π‘0π‘‘π‘‘ξ€œπ‘(𝑑)𝑔(𝑑)𝑏0𝑠(𝑑)𝑔(𝑑)π‘‘π‘‘βˆ’supπ‘π‘”π‘ξ‚ξ€œπ‘0𝑑𝑑𝑝(𝑑)𝑔(𝑑)βˆ’1.(4.9)

Proof. The solution of the equation 𝐿𝑦=π‘žπ‘“β‰₯0, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦′(𝑏)=𝛾1β‰₯0 is given by (3.15) where 𝐺(π‘₯,𝑑,πœ†) is defined by (3.3). Substituting 𝑦(π‘₯) from (3.15) into (4.6), it is easy to see that we require the following inequalities in order to complete the proof 𝐾1ξ€Έξ€·πœ‡βˆ’πœ†π‘˜(π‘₯)πœ“βˆ’(π‘₯)+𝐿1ξ€·signπ‘¦ξ…žξ€Έξ€Έπ‘πœ“ξ…žξ€·πΎβ‰₯0,0<π‘₯≀𝑏,(4.10)1ξ€Έξ€·βˆ’πœ†π‘˜(π‘₯)πœ™βˆ’πœ‡(π‘₯)+𝐿1ξ€·signπ‘¦ξ…žξ€Έξ€Έπ‘πœ™ξ…žβ‰₯0,0<π‘₯≀𝑏.(4.11)
Here πœ“ satisfies the IVP at π‘₯=0; that is, πΏπœ“=0, πœ“(0)=0 and limπ‘₯β†’0+𝑝(π‘₯)πœ“(π‘₯)=1, and πœ™ satisfies the IVP at π‘₯=𝑏; that is, πΏπœ™=0, πœ™(𝑏)=𝛽1 and 𝑝(𝑏)πœ™ξ…ž(𝑏)=βˆ’π›Ό1. The solutions πœ“ and πœ™ cannot have either point of maxima (at the point of maxima the πΏπœ“=0 or πΏπœ™=0 will be contradicted) or point of minima (since to have minima, maxima is bound to occur). So, finally we have πœ“ξ…ž(π‘₯)β‰₯0 and πœ™ξ…ž(π‘₯)≀0 on [0,𝑏]. As |πœ‡(π‘₯)|≀𝐿1, it is enough to prove the following inequalities: 𝐾1ξ€Έξ€·πΏβˆ’πœ†π‘˜(π‘₯)πœ“βˆ’1βˆ’||πœ‡||ξ€Έ(π‘₯)π‘πœ“ξ…žξ€·πΎβ‰₯0,0<π‘₯≀𝑏,(4.12)1ξ€Έξ€·πΏβˆ’πœ†π‘˜(π‘₯)πœ™+1βˆ’||||ξ€Έπœ‡(π‘₯)π‘πœ™ξ…žβ‰₯0,0<π‘₯≀𝑏.(4.13) Next, we prove the inequality (4.12), and the other one can be proved in a similar manner.
By the mean value theorem, there exist 𝜏∈(0,𝑏) such that πœ“(𝑏)=π‘πœ“β€²(𝜏). Writing πΏπœ“=0 in the following form: βˆ’ξ€·π‘(π‘₯)𝑔(π‘₯)πœ“(π‘₯)ξ…žξ€Έξ…žβˆ’πœ†π‘ (π‘₯)𝑔(π‘₯)πœ“(π‘₯)=0,(4.14) and integrating it first from 𝜏 to π‘₯ and then π‘₯ to 𝑏, we get that 𝑝(π‘₯)πœ“ξ…ž(π‘₯)β‰€πœ“(π‘₯)Ξ¦(𝑝,π‘ž,𝑠,𝑔)on0<π‘₯≀𝑏.(4.15) Here Ξ¦(𝑝,π‘ž,𝑠,𝑔) is given by (4.8). Now, the result follows from (4.7), (4.12), and (4.15).

Lemma 4.3. If 0<πœ†<πœ†0, πœ†π‘˜(π‘₯)≀𝐾1, |πœ‡(π‘₯)|≀𝐿1, 𝐿𝑦β‰₯0 for 0<π‘₯≀𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)β‰₯0, then 𝐾1ξ€Έξ€·πœ‡βˆ’πœ†π‘˜(π‘₯)π‘¦βˆ’(π‘₯)+𝐿1ξ€·signπ‘¦ξ…žξ€Έξ€Έπ‘π‘¦ξ…žβ‰₯0,0<π‘₯≀𝑏,(4.16) provided that ξ‚€1βˆ’supπ‘π‘”π‘ξ‚ξ€œπ‘0𝑑𝑑𝐾𝑝𝑔>0,0<π‘₯≀𝑏,1ξ€Έξ‚΅ξ‚€βˆ’πœ†π‘˜(π‘₯)1βˆ’supπ‘π‘”π‘ξ‚ξ€œπ‘0π‘‘π‘‘ξ‚Άβˆ’ξ€·πΏπ‘π‘”1βˆ’||πœ‡||ξ€Έξ‚€(π‘₯)sup𝑝𝑔𝑏β‰₯0,0<π‘₯≀𝑏(4.17) or ξ€œ1βˆ’πœ†π‘0ξ€œπ‘ (𝑑)𝑔(𝑑)𝑑𝑑𝑏0𝑑𝑑𝐾𝑝(𝑑)𝑔(𝑑)>0,1ξ€Έξ‚΅ξ€œβˆ’πœ†π‘˜(π‘₯)1βˆ’πœ†π‘0ξ€œπ‘ π‘”π‘‘π‘‘π‘0π‘‘π‘‘ξ‚Άβˆ’ξ€·πΏπ‘π‘”1βˆ’||πœ‡||ξ€Έπœ†ξ€œ(π‘₯)𝑏0𝑠𝑔𝑑𝑑β‰₯0,0<π‘₯≀𝑏,(4.18) hold.

Proof. Similar to the proof of Lemma 4.2, we need to establish two inequalities (4.10)-(4.11) for 0<πœ†<πœ†0. Here πœ“ and πœ™ cannot have the point of minima in (0,𝑏), because at the point of minima, the differential equation πΏπœ“=0 or πΏπœ™=0 will be contradicted. So either πœ“β€²(π‘₯)β‰₯0 and πœ™ξ…ž(π‘₯)≀0 or πœ“(π‘₯) and πœ™(π‘₯) both are concave downwards on [0,𝑏]. Thus we can divide the proof in two cases:
Case i. πœ“ and πœ™ both are concave downwards.
We prove for πœ“ as similar analysis provides result for πœ™. Let the point of maxima be π‘₯0∈(0,𝑏). Then πœ“β€²(π‘₯)>0 for π‘₯<π‘₯0 and πœ“β€²(π‘₯)<0 for π‘₯>π‘₯0. On both sides of π‘₯0, the inequality (4.10) will be reduced into the following two inequalities: 𝐾1ξ€Έξ€·πΏβˆ’πœ†π‘˜(π‘₯)πœ“βˆ’1ξ€Έ+πœ‡(π‘₯)π‘πœ“β€²β‰₯0,πœ“ξ…žξ€·πΎ(π‘₯)β‰₯0,1ξ€Έξ€·πΏβˆ’πœ†π‘˜(π‘₯)πœ“+1ξ€Έβˆ’πœ‡(π‘₯)π‘πœ“β€²β‰₯0,πœ“ξ…ž(π‘₯)≀0.(4.19) For a point π‘₯ on the left side of π‘₯0, we integrate (4.14) from π‘₯ to π‘₯0 twice and get 𝑝(π‘₯)πœ“ξ…žβˆ«(π‘₯)β‰€πœ†πœ“(π‘₯)𝑏0𝑠(𝑑)𝑔(𝑑)π‘‘π‘‘βˆ«1βˆ’πœ†π‘0π‘ βˆ«(𝑑)𝑔(𝑑)𝑑𝑑𝑏0(1/𝑝(𝑑)𝑔(𝑑))𝑑𝑑.(4.20) Similarly for any point π‘₯ on the right side of π‘₯0, we get βˆ’π‘(π‘₯)πœ“ξ…žβˆ«(π‘₯)β‰€πœ†πœ“(π‘₯)𝑏0𝑠(𝑑)𝑔(𝑑)π‘‘π‘‘βˆ«1βˆ’πœ†π‘0π‘ βˆ«(𝑑)𝑔(𝑑)𝑑𝑑𝑏0(1/𝑝(𝑑)𝑔(𝑑))𝑑𝑑.(4.21) Now, the result follows from the fact that |πœ‡(π‘₯)|≀𝐿1 and from (4.18) to (4.21).

Case ii. When πœ“ξ…ž(π‘₯)β‰₯0 and πœ™ξ…ž(π‘₯)≀0.
To establish the inequality (4.16), we require to establish the inequalities (4.12)-(4.13). We prove the inequality (4.12), and the proof for (4.13) is quite similar. By the mean value theorem, there exists 𝜏∈(0,𝑏) such that πœ“(𝑏)=π‘πœ“β€²(𝜏). Integrating (4.14) first from 𝜏 to π‘₯ and then from π‘₯ to 𝑏, we get 𝑝(π‘₯)πœ“β€²(π‘₯)β‰€πœ“(π‘₯)sup(𝑝𝑔/𝑏)∫1βˆ’sup(𝑝𝑔/𝑏)𝑏0,(𝑑𝑑/𝑝𝑔)(4.22) and the result follows from (4.12), (4.17), and (4.22). This completes the proof.

Lemma 4.4. If 𝑒𝑛 is an upper solution of (4.3) and 𝑒𝑛+1 is defined by (1.4)–(1.5), then 𝑒𝑛β‰₯𝑒𝑛+1 for πœ†<πœ†0.

Proof. Let 𝑀=π‘’π‘›βˆ’π‘’π‘›+1. 𝑀 satisfies 𝐿𝑀=βˆ’(π‘π‘’ξ…žπ‘›)ξ…žβˆ’π‘žπ‘“(π‘₯,𝑒𝑛,π‘π‘¦ξ…žπ‘›)β‰₯0,0<π‘₯≀𝑏,𝑀(0)=0,𝛼1𝑀(𝑏)+𝛽1𝑝(𝑏)𝑀′(𝑏)β‰₯0, and the result follows from Corollary 3.6.

Proposition 4.5. Let 𝑒0 be an upper solution of (4.3), and let 𝑓(π‘₯,𝑦,𝑝𝑦′) satisfy the following(F1) 𝑓(π‘₯,𝑦,𝑝𝑦′) is continuous on𝐷0=ξ€½ξ€·π‘₯,𝑦,π‘π‘¦ξ…žξ€ΈβˆΆ[]×𝑣0,𝑏0,𝑒0×ℝ,(4.23)(F2) βˆƒπΎ1≑𝐾1(𝐷0) such that for all (π‘₯,𝑦,𝑣),(π‘₯,𝑀,𝑣)∈𝐷0,𝐾1(π‘¦βˆ’π‘€)≀𝑓(π‘₯,𝑦,𝑣)βˆ’π‘“(π‘₯,𝑀,𝑣)for𝑦β‰₯𝑀,and(4.24)(F3)βˆƒ0≀𝐿1≑𝐿1(𝐷0) such that for all (π‘₯,𝑦,𝑣1),(π‘₯,𝑦,𝑣2)∈𝐷0,||𝑓π‘₯,𝑦,𝑣1ξ€Έξ€·βˆ’π‘“π‘₯,𝑦,𝑣2ξ€Έ||≀𝐿1||𝑣1βˆ’π‘£2||,(4.25) and (4.7), (4.17), or (4.18) hold. Then the functions 𝑒𝑛 defined by (1.4)–(1.5) are such that, for all π‘›βˆˆβ„•, (i) 𝑒𝑛 is upper solution of (4.3) and (ii) 𝑒𝑛β‰₯𝑒𝑛+1.

Proof. Since 𝑒0 is an upper solution from Lemma 4.4, we have 𝑒0β‰₯𝑒1. Assume that the claim is true for π‘›βˆ’1; that is, π‘’π‘›βˆ’1 is an upper solution and π‘’π‘›βˆ’1β‰₯𝑒𝑛.
Let 𝑀=π‘’π‘›βˆ’1βˆ’π‘’π‘›. We have βˆ’ξ€·π‘π‘’ξ…žπ‘›ξ€Έξ…žξ€·βˆ’π‘žπ‘“π‘₯,𝑒𝑛,π‘π‘’ξ…žπ‘›ξ€ΈπΎβ‰₯π‘žξ€½ξ€·1ξ€Έξ€·βˆ’πœ†π‘˜(π‘₯)π‘€βˆ’πœ‡(π‘₯)+𝐿1ξ€·signπ‘€ξ…žξ€Έξ€Έπ‘π‘€ξ…žξ€Ύ,(4.26) and from Lemmas 4.2 and 4.3 we get βˆ’(π‘π‘’ξ…žπ‘›)ξ…žβˆ’π‘žπ‘“(π‘₯,𝑒𝑛,π‘π‘’ξ…žπ‘›)β‰₯0,0<π‘₯≀𝑏.
Thus, 𝑒𝑛 is an upper solution for all π‘›βˆˆβ„•. From Lemma 4.4 we have 𝑒𝑛β‰₯𝑒𝑛+1. Hence, the result follows.
Similar results (Lemma 4.6, Proposition 4.7) follow for lower solutions.

Lemma 4.6. If 𝑣𝑛 is a lower solution of (4.3) and 𝑣𝑛+1 is defined by (1.4)–(1.5) then 𝑣𝑛≀𝑣𝑛+1 for πœ†<πœ†0.

Proposition 4.7. Let 𝑣0 be a lower solution of (4.3), let 𝑓(π‘₯,𝑦,𝑝𝑦′) satisfies (F1)–(F3) and (4.7), (4.17), or (4.18) hold. Then the functions 𝑣𝑛 defined by (1.4)–(1.5) are such that, for all π‘›βˆˆβ„•, (i) 𝑣𝑛 is lower solution of (4.3) and (ii) 𝑣𝑛≀𝑣𝑛+1.

Proposition 4.8. If 𝑓(π‘₯,𝑦,𝑝𝑦′) satisfies(F4) 𝑓(π‘₯,𝑒0,π‘π‘’ξ…ž0)βˆ’π‘“(π‘₯,𝑣0,π‘π‘£ξ…ž0)βˆ’πœ‡(π‘₯)(π‘π‘’ξ…ž0βˆ’π‘π‘£ξ…ž0)βˆ’πœ†π‘˜(π‘₯)(𝑒0βˆ’π‘£0)β‰₯0 for 0<π‘₯≀𝑏 such that πœ†π‘˜(π‘₯)≀𝐾1 and |πœ‡(π‘₯)|≀𝐿1,and in addition let (F1)–(F3) and (4.7), (4.17), or (4.18) hold, then for all π‘›βˆˆβ„• the functions 𝑒𝑛 and 𝑣𝑛 defined by (1.4)–(1.5) satisfy 𝑣𝑛≀𝑒𝑛.

Proof. Let 𝑀𝑖=π‘’π‘–βˆ’π‘£π‘–, then 𝑀𝑖 satisfies 𝐿𝑀𝑖=π‘ž(π‘₯)β„Žπ‘–βˆ’1 for all π‘–βˆˆπ‘ such that β„Žπ‘–ξ€·(π‘₯)=𝑓π‘₯,𝑒𝑖,π‘π‘’ξ…žπ‘–ξ€Έξ€·βˆ’π‘“π‘₯,𝑣𝑖,π‘π‘£ξ…žπ‘–ξ€Έξ€·βˆ’πœ‡(π‘₯)π‘π‘’ξ…žπ‘–βˆ’π‘π‘£ξ…žπ‘–ξ€Έξ€·π‘’βˆ’πœ†π‘˜(π‘₯)π‘–βˆ’π‘£π‘–ξ€Έ,0<π‘₯≀𝑏.(4.27) Since 𝑣0≀𝑒0, we prove that 𝑣1≀𝑒1. Since 𝑀1 is solution of 𝐿𝑀1=π‘žβ„Ž0β‰₯0, 𝑀1(0)=0 and 𝛼1𝑀1(0)+𝛽1𝑝(𝑏)π‘€ξ…ž1(𝑏)=0, from Corollary 3.6 we have 𝑀1β‰₯0. Let 𝑛β‰₯2, let β„Žπ‘›βˆ’2β‰₯0, and π‘’π‘›βˆ’1β‰₯π‘£π‘›βˆ’1, then we prove that β„Žπ‘›βˆ’1β‰₯0 and 𝑒𝑛β‰₯𝑣𝑛. Consider β„Žπ‘›βˆ’1ξ€·=𝑓π‘₯,π‘’π‘›βˆ’1,π‘π‘’ξ…žπ‘›βˆ’1ξ€Έξ€·βˆ’π‘“π‘₯,π‘£π‘›βˆ’1,π‘π‘£ξ…žπ‘›βˆ’1ξ€Έβˆ’πœ‡(π‘₯)π‘π‘€ξ…žπ‘›βˆ’1βˆ’πœ†π‘˜(π‘₯)π‘€π‘›βˆ’1β‰₯𝐾1ξ€Έπ‘€βˆ’πœ†π‘˜(π‘₯)π‘›βˆ’1βˆ’ξ€·ξ€·πœ‡+𝐿signπ‘€ξ…žπ‘›βˆ’1ξ€Έξ€Έπ‘π‘€ξ…žπ‘›βˆ’1.(4.28) Since π‘€π‘›βˆ’1 is a solution of πΏπ‘€π‘›βˆ’1=π‘žβ„Žπ‘›βˆ’2β‰₯0, π‘€π‘›βˆ’1(0)=0, and 𝛼1π‘€π‘›βˆ’1(0)+𝛽1𝑝(𝑏)π‘€ξ…žπ‘›βˆ’1(𝑏)=0; hence, from Lemmas 4.2 and 4.3, we have β„Žπ‘›βˆ’1β‰₯0. Thus, from Corollary 3.6 on 𝐿𝑀𝑛=π‘žβ„Žπ‘›βˆ’1β‰₯0, 𝑀𝑛(0)=0 and 𝛼1𝑀𝑛(0)+𝛽1𝑝(𝑏)π‘€ξ…žπ‘›(𝑏)=0, we have 𝑀𝑛β‰₯0, that is, 𝑒𝑛β‰₯𝑣𝑛. This completes the proof.

Lemma 4.9. If 𝑓(π‘₯,𝑦,π‘π‘¦ξ…ž) satisfies(F5) for all (π‘₯,𝑦,𝑣)∈𝐷0, |𝑓(π‘₯,𝑦,𝑣)|β‰€πœ‘(|𝑣|) where πœ‘βˆΆ[0,∞)β†’(0,∞) is continuous and satisfies ξ€œπ‘0ξ€œπ‘ž(𝑠)𝑑𝑠<βˆžπ‘™0𝑑𝑠,πœ‘(𝑠)(4.29) where 𝑙0=sup[0,𝑏]|𝑝(π‘₯)𝑒0(π‘₯)/𝑏|, then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘¦ξ…žξ€Έξ…žξ€·β‰₯π‘žπ‘“π‘₯,𝑦,π‘π‘¦ξ…žξ€Έ,0<π‘₯≀𝑏,(4.30)𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)β‰₯𝛾1,(4.31) with π‘¦βˆˆ[𝑣0,𝑒0] for all π‘₯∈[0,𝑏], satisfies β€–π‘π‘¦β€²β€–βˆž<𝑅0.

Proof. We divide the proof in three parts.
Case i. If solution is not monotone throughout the interval, then we consider the interval (π‘₯0,π‘₯]βŠ‚(0,𝑏) such that 𝑦′(π‘₯0)=0 and 𝑦′(π‘₯)>0 for π‘₯>π‘₯0. Integrating (4.30) from π‘₯0 to π‘₯ we get ξ€œπ‘π‘¦β€²0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑏0π‘ž(𝑠)𝑑𝑠.(4.32) From (F5) we can choose 𝑅0>0 such that ξ€œπ‘π‘¦β€²0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑏0ξ€œπ‘ž(𝑠)𝑑𝑠≀𝑅0𝑙0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑅00𝑑𝑠,πœ‘(𝑠)(4.33) which gives 𝑝(π‘₯)π‘¦ξ…ž(π‘₯)≀𝑅0.(4.34) Now we consider the case in which 𝑦′(π‘₯)<0 for π‘₯<π‘₯0, 𝑦′(π‘₯0)=0, and proceeding in the similar way we get βˆ’π‘(π‘₯)π‘¦ξ…ž(π‘₯)≀𝑅0,(4.35) and the result follows.Case ii. If 𝑦 is monotonically increasing in (0,𝑏), that is, 𝑦′>0 in (0,𝑏), then by the mean value theorem there exists a point 𝜏∈(0,𝑏) such that π‘¦ξ…ž(𝜏)=𝑦(𝑏)βˆ’π‘¦(0)𝑏≀|||𝑒0𝑏|||.(4.36) Now, integrating (4.30) from 𝜏 to π‘₯, we get ξ€œπ‘π‘¦β€²0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑏0ξ€œπ‘ž(𝑑)𝑑𝑑+𝑙00𝑑𝑠.πœ‘(𝑠)(4.37) Further, from (F5) we can choose 𝑅0 such that ξ€œπ‘π‘¦β€²0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑏0ξ€œπ‘ž(𝑠)𝑑𝑠+𝑙00π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑅00𝑑𝑠.πœ‘(𝑠)(4.38) which gives 𝑝(π‘₯)𝑦′(π‘₯)≀𝑅0.Case iii. If 𝑦 is monotonically decreasing in (0,𝑏); that is, π‘¦ξ…ž<0 in (0,𝑏), then argument similar to Case  ii yields ξ€œβˆ’π‘π‘¦β€²0π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑏0ξ€œπ‘ž(𝑠)𝑑𝑠+𝑙00π‘‘π‘ β‰€ξ€œπœ‘(𝑠)𝑅00𝑑𝑠,πœ‘(𝑠)(4.39) and we get βˆ’π‘(π‘₯)π‘¦ξ…ž(π‘₯)≀𝑅0,(4.40) and the result follows.

Lemma 4.10. If 𝑓(π‘₯,𝑦,π‘π‘¦ξ…ž) satisfies (F5), then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘¦ξ…žξ€Έξ…žξ€·β‰€π‘žπ‘“π‘₯,𝑦,π‘π‘¦ξ…žξ€Έ,0<π‘₯≀𝑏,𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…ž(𝑏)≀𝛾1,(4.41) with π‘¦βˆˆ[𝑣0,𝑒0] for all π‘₯∈[0,𝑏], satisfies β€–π‘π‘¦β€²β€–βˆž<𝑅0.

Proof. Proof follows from the analysis of Lemma 4.9.

Theorem 4.11. Let 𝑒0 and 𝑣0 be upper and lower solutions. Let 𝑓(π‘₯,𝑦,π‘π‘¦ξ…ž) satisfy (F1) to (F5) and (4.7), (4.17), or (4.18) hold. Then, boundary value problem (4.3) has at least one solution in the region 𝐷0. If πœ†<πœ†0 is chosen such that πœ†π‘˜(π‘₯)≀𝐾1 and |πœ‡(π‘₯)|≀𝐿1, where πœ†0 is the first positive eigenvalue of the corresponding eigenvalue problem, then the sequences {𝑒𝑛} and {𝑣𝑛} generated by (1.4)–(1.5) with initial iterate 𝑒0 and 𝑣0 converge monotonically and uniformly towards solutions ̃𝑒(π‘₯) and ̃𝑣(π‘₯) of (4.3). Any solution 𝑧(π‘₯) in 𝐷0 must satisfy ̃𝑣(π‘₯)≀𝑧(π‘₯)≀̃𝑒(π‘₯).

Proof. From Lemmas 4.2–4.10, Propositions 4.5–4.8, and we get two monotonic sequences {𝑒𝑛} and {𝑣𝑛} which are bounded by 𝑒0 and 𝑣0; respectively, and by Dini's Theorem their uniform convergence is assured. Let {𝑒𝑛} and {𝑣𝑛} converge uniformly to ̃𝑒 and ̃𝑣.
By Lemmas 4.9 and 4.10, it is easy to see that the sequences {π‘π‘’ξ…žπ‘›} and {π‘π‘£ξ…žπ‘›} are uniformly bounded. Now, from||π‘π‘¦ξ…žπ‘›ξ€·π‘₯1ξ€Έβˆ’π‘π‘¦ξ…žπ‘›ξ€·π‘₯2ξ€Έ||=||||ξ€œπ‘₯2π‘₯1ξ€·π‘π‘¦ξ…žπ‘›ξ€Έξ…ž||||𝑑𝑑,(4.42) uniform convergence of {𝑦𝑛}, properties (A1)–(A4), and (F1), it is easy to prove that {π‘π‘¦ξ…žπ‘›} is equicontinuous. Hence, by Arzela-Ascoli's Theorem there exist a uniform convergent subsequence {π‘π‘¦ξ…žπ‘›π‘˜} of {π‘π‘¦ξ…žπ‘›}. Since limit is unique so original sequence will also converge uniformly to the same limit say 𝑝𝑦′. It is easy to see that, if 𝑦𝑛→̃𝑦, then π‘π‘¦ξ…žπ‘›β†’π‘Μƒπ‘¦ξ…ž. Therefore sequences {π‘π‘’ξ…žπ‘›} and {π‘π‘£ξ…žπ‘›} converge uniformly to π‘Μƒπ‘’ξ…ž and π‘Μƒπ‘£ξ…ž, respectively.
Let 𝐺(π‘₯,𝑑) be Green's function for the linear boundary value problem 𝐿𝑦𝑛=0, 𝑦𝑛(0)=0, and 𝛼1𝑦𝑛(𝑏)+𝛽1𝑝(𝑏)π‘¦ξ…žπ‘›(𝑏)=0. Then solution of (1.4)–(1.5) can be written as𝑦𝑛=𝐢π‘₯2+ξ€œπ‘0𝐹𝐺(π‘₯,𝑑)𝑑,π‘¦π‘›βˆ’1,π‘π‘¦ξ…žπ‘›βˆ’1ξ€Έξ€Ύ+𝐻(𝑑)𝑑𝑑,(4.43) where 𝐻(𝑑)=2𝐢(𝑑𝑝′(𝑑)+𝑝(𝑑))+2πΆπ‘‘πœ‡(𝑑)π‘ž(𝑑)𝑝(𝑑)+πœ†πΆπ‘‘2𝑠(𝑑) and 𝐢=𝛾1/(𝛼1𝑏2+2𝛽1𝑏𝑝(𝑏)).
Now, uniform convergence of {𝑦𝑛}, {π‘π‘¦ξ…žπ‘›} and continuity of 𝑓(π‘₯,𝑦,𝑝𝑦′) imply that {(1/π‘ž)𝐹(π‘₯,𝑦𝑛,𝑝𝑦′𝑛)} converges uniformly in [0,𝑏]. Hence, {(1/π‘ž)𝐹(π‘₯,𝑦𝑛,π‘π‘¦ξ…žπ‘›)} converges in the sense of mean in 𝐿2π‘ž(0,𝑏). Taking limit as π‘›β†’βˆž and using Lemma 2.4 ([11, page 27]), we get𝑦=𝐢π‘₯2+ξ€œπ‘0𝐹𝐺(π‘₯,𝑑)𝑑,𝑦,π‘π‘¦ξ…žξ€Έξ€Ύ+𝐻(𝑑)𝑑𝑑,(4.44) which is the solution of the boundary value problem (4.3).
Any solution 𝑧(π‘₯) in 𝐷0 plays the role of 𝑒0(π‘₯). Hence ̃𝑧(π‘₯)β‰₯𝑣(π‘₯). Similarly, 𝑧(π‘₯)≀̃𝑒(π‘₯). This completes the proof.

Remark 4.12. The case when πœ†=0 corresponds to the case when 𝑓(π‘₯,𝑦,𝑝𝑦′)≑𝑓(π‘₯,𝑝𝑦′). In such cases the boundary value problem (4.3) can be reduced to two initial value problems βˆ’π‘§ξ…ž=π‘žπ‘“(π‘₯,𝑧), 𝑧(0)=βˆ’π›Ό1 and π‘π‘¦ξ…ž=𝑧, 𝑦(0)=𝛽1. From the assumptions on 𝑝(π‘₯), π‘ž(π‘₯), and 𝑓(π‘₯,𝑦,𝑝𝑦′), one can easily conclude existence uniqueness of solutions of the nonlinear boundary value problem.

Remark 4.13. Suppose, in addition to the hypothesis of Theorem 4.11, |𝑓(π‘₯,𝑦,𝑝𝑦′)|≀𝑁0 in 𝐷0. Then lower solution 𝑣0 and upper solution 𝑒0 may be obtained as solution of the following linear boundary value problems: βˆ’ξ€·π‘π‘£ξ…ž0ξ€Έβ€²+𝑁0π‘žπ‘£(π‘₯)=0,0<π‘₯≀𝑏,0(0)=0,𝛼1𝑣0(𝑏)+𝛽1𝑝(𝑏)π‘£ξ…ž0(𝑏)=𝛾1,βˆ’ξ€·π‘π‘’ξ…ž0ξ€Έξ…žβˆ’π‘0π‘’π‘ž(π‘₯)=0,0<π‘₯≀𝑏,0(0)=0,𝛼1𝑒0(𝑏)+𝛽1𝑝(𝑏)π‘’ξ…ž0(𝑏)=𝛾1.(4.45)

Theorem 4.14. Suppose that 𝑓(π‘₯,𝑦,𝑝𝑦′) satisfies (F1), (F3), and βˆƒ constants 𝐾1(𝐷0)<πœ†0 such that 𝐾1ξ€·(π‘’βˆ’π‘£)≀𝑓π‘₯,𝑒,π‘π‘¦ξ…žξ€Έξ€·βˆ’π‘“π‘₯,𝑣,π‘π‘¦ξ…žξ€Έ.(4.46) Then the boundary value problem (4.3) has unique solution.

Proof. Let 𝑒 and 𝑣 be two solutions of (4.3), then we get βˆ’ξ€·π‘(π‘’βˆ’π‘£)ξ…žξ€Έξ…žξ€½π‘“ξ€·=π‘ž(π‘₯)π‘₯,𝑒,π‘π‘’ξ…žξ€Έξ€·βˆ’π‘“π‘₯,𝑣,π‘π‘£ξ…žξ€·π‘ξ€Έξ€Ύ,0<π‘₯≀𝑏,orβˆ’(π‘’βˆ’π‘£)ξ…žξ€Έξ…ž+𝐿1π‘žξ€·(π‘₯)π‘π‘’ξ…žβˆ’π‘π‘£ξ…žξ€Έβˆ’πΎ1π‘ž(π‘₯)(π‘’βˆ’π‘£)β‰₯0,0<π‘₯≀𝑏,(π‘’βˆ’π‘£)(0)=0,𝛼1(π‘’βˆ’π‘£)(𝑏)+𝛽1𝑝(𝑏)(π‘’βˆ’π‘£)ξ…ž(𝑏)=0.(4.47) Since 𝐾1<πœ†0, from Corollary 3.6 we get π‘’βˆ’π‘£β‰₯0 or 𝑒β‰₯𝑣. Similarly 𝑣β‰₯𝑒. Therefore, the solution of (4.3) is unique.

Acknowledgments

The authors are grateful to the reviewers for their critical comments and valuable suggestions. This work is supported by Council of Scientific and Industrial Research (CSIR) and DST, New Delhi, India.

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