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International Journal of Differential Equations
VolumeΒ 2011, Article IDΒ 986948, 13 pages
http://dx.doi.org/10.1155/2011/986948
Research Article

Nonlinear Singular BVP of Limit Circle Type and the Presence of Reverse-Ordered Upper and Lower Solutions

1Department of Mathematics, BITS Pilani, Pilani, Rajasthan 333031, India
2Department of Mathematics and Astronomy, University of Lucknow, Lucknow 226007, India

Received 27 May 2011; Accepted 3 July 2011

Academic Editor: Mohamed A.Β El-Gebeily

Copyright Β© 2011 Amit K. Verma and Lajja 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 consider the following class of nonlinear singular differential equation βˆ’(𝑝(π‘₯)𝑦′(π‘₯))β€²+π‘ž(π‘₯)𝑓(π‘₯,𝑦(π‘₯),𝑝(π‘₯)𝑦′(π‘₯))=0,0<π‘₯<1 subject to the Neumann boundary condition 𝑦′(0)=𝑦′(1)=0. Conditions on 𝑝(π‘₯) and π‘ž(π‘₯) ensure that π‘₯=0 is a singular point of limit circle type. A simple approximation scheme which is iterative in nature is considered. The initial iterates are upper and lower solutions which can be ordered in one way (𝑣0≀𝑒0) or the other (𝑒0≀𝑣0).

1. Introduction

The upper and lower solution technique is the most promising technique as far as singular boundary value problems, are concerned [1]. Recently, lot of activities are there regarding upper and lower solutions technique (see [2, 3] and the references therein). To see the application of the similar kind of problems, one should see the references of [3]. In most of the results, upper and lower solutions are well ordered, that is, 𝑒0β‰₯𝑣0. As far as reverse-ordered upper and lower solutions are considered, that is, 𝑒0≀𝑣0, the literature is not that rich. Though references are there for nonsingular boundary value problem, but singular boundary value problems require further exploration. The details of the work done for the nonsingular problem when upper and lower solutions are in reverse order can be seen in [4, 5]. To fill this gap in the present paper, we consider the following singular BVP:βˆ’ξ€·π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žξ€·+π‘ž(π‘₯)𝑓π‘₯,𝑦(π‘₯),𝑝(π‘₯)π‘¦ξ…žξ€Έπ‘¦(π‘₯)=0,0<π‘₯<1,ξ…ž(0)=0,π‘¦ξ…ž(1)=0.(1.1)(A1)Let 𝑝(π‘₯) satisfy the following conditions. (i)𝑝(0)=0 and 𝑝>0 in (0,1). (ii)π‘βˆˆπΆ[0,1]∩𝐢1(0,1). (A2) Let π‘ž(π‘₯) satisfy the following conditions. (i)π‘ž(π‘₯)>0 in (0,1) and π‘ž(π‘₯)∈𝐢(0,1]. (ii)∫10π‘ž(π‘₯)𝑑π‘₯<∞. (iii)limπ‘₯β†’0(π‘ž(π‘₯))/(π‘ξ…ž(π‘₯))=0. (iv)∫10∫(1/𝑝(π‘₯))(π‘₯0π‘ž(𝑠)𝑑𝑠)1/2𝑑π‘₯<∞.

In this paper, we consider a computationally simple iterative scheme defined byβˆ’ξ€·π‘π‘¦ξ…žπ‘›ξ€Έξ…ž+πœ†π‘žπ‘¦π‘›ξ€·=βˆ’π‘žπ‘“π‘₯,π‘¦π‘›βˆ’1,π‘π‘¦ξ…žπ‘›βˆ’1ξ€Έ+πœ†π‘žπ‘¦π‘›βˆ’1𝑦,0<π‘₯<1,ξ…žπ‘›(0)=0,π‘¦ξ…žπ‘›(1)=0.(1.2) Starting with upper and lower solutions, we generate monotone sequences. To generate these monotonic sequences, we need the existence of some differential inequalities. To prove these differential inequalities, we analyze the corresponding singular IVP and extract properties of the solutions and their derivative.

We have arranged the paper in four sections. In Section 2, we discuss some elementary results, for example, maximum principles and existence of two differential inequalities. Then using these elementary results, we establish existence results for well-ordered upper and lower solutions in Section 3 and for reverse-ordered upper and lower solutions in Section 4. In Section 5, we conclude this paper with some remarks.

2. Preliminaries

Let β„Ž(π‘₯)∈𝐢[0,1], and let πœ†βˆˆβ„0(ℝ0=ℝ⧡{0}), let π΄βˆˆβ„ and let π΅βˆˆβ„. Now, consider the following class of linear singular problems:βˆ’ξ€·π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žπ‘¦+πœ†π‘ž(π‘₯)𝑦(π‘₯)=π‘ž(π‘₯)β„Ž(π‘₯),0<π‘₯<1,(2.1)ξ…ž(0)=𝐴,π‘¦ξ…ž(1)=𝐡.(2.2) The corresponding homogeneous system (eigenvalue problem) is given byβˆ’ξ€·π‘(π‘₯)π‘¦ξ…žξ€Έ(π‘₯)ξ…žπ‘¦+πœ†π‘ž(π‘₯)𝑦(π‘₯)=0,0<π‘₯<1,(2.3)ξ…ž(0)=0,π‘¦ξ…ž(1)=0.(2.4) The solution of the nonhomogeneous problem (2.1)-(2.2) can be written as follows:𝑀(π‘₯)=𝑧1ξƒ¬ξ€œ(π‘₯)π‘₯0π‘ž(𝑑)β„Ž(𝑑)𝑧0(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0𝐴𝑑𝑑+π‘§ξ…ž1ξƒ­(0)+𝑧0ξƒ¬ξ€œ(π‘₯)1π‘₯π‘ž(𝑑)β„Ž(𝑑)𝑧1(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0𝐡𝑑𝑑+π‘§ξ…ž0ξƒ­(1),(2.5) where 𝑧0(π‘₯,πœ†) is the solution ofβˆ’ξ€·π‘(π‘₯)π‘§ξ…ž0ξ€Έ(π‘₯)ξ…ž+πœ†π‘ž(π‘₯)𝑧0(π‘₯)=0,0<π‘₯<1,𝑧0(0)=1,π‘§ξ…ž0(0)=0,(2.6)𝑧1(π‘₯,πœ†) is the solution ofβˆ’ξ€·π‘(π‘₯)π‘§ξ…ž1ξ€Έ(π‘₯)ξ…ž+πœ†π‘ž(π‘₯)𝑧1(π‘₯)=0,0<π‘₯<1,𝑧1(1)=1,π‘§ξ…ž1(1)=0,(2.7) and π‘Šπ‘(𝑧1,𝑧0)=𝑝(𝑑)(𝑧1π‘§ξ…ž0βˆ’π‘§ξ…ž1𝑧0). By replacing π‘₯ with 1βˆ’π‘₯ in (2.6), it is easy to verify that 𝑧1(π‘₯)=𝑧0(1βˆ’π‘₯),(2.8) for both positive and negative values of πœ†.

Remark 2.1. Existence of 𝑧0(π‘₯) and 𝑧1(π‘₯) satisfying the IVP (2.6) and IVP (2.7), respectively, is an immediate consequence of the result due to O'Regan [6, Theorem  2.1, page 432].

Remark 2.2. Let 𝐿2π‘ž(0,1) be a Hilbert space with inner product defined by ξ€œβŸ¨π‘“,π‘”βŸ©=10π‘ž(π‘₯)𝑓(π‘₯)𝑔(π‘₯)𝑑π‘₯.(2.9) From (A2) (iv), it can easily be verified that π‘₯=0 is a singular point of limit circle type (see [6, Remark (i) page 434]) in 𝐿2π‘ž(0,1). Thus, we have pure point spectrum [7, page 125]. It is easy to show that the eigenvalues are real, simple, and negative.

Remark 2.3. Since 𝑧0 and 𝑧1 are two linearly independent solutions of (2.3), the eigenvalues of the eigenvalue problem (2.3)-(2.4) will be the zeros of π‘§ξ…ž0(1,πœ†). Since π‘§ξ…ž0(1,πœ†) is an analytic function of πœ† so its zeros will be isolated and they all will be negative. Let them be βˆ’πœ†0,βˆ’πœ†1,βˆ’πœ†2,…, where πœ†π‘–>0 for 𝑖=0,1,2,…. Now, we have βˆ’πœ†0 as the first negative zero of π‘§ξ…ž0(1,πœ†) or in other words first negative eigenvalue of (2.3)-(2.4).
Since 𝑧0(π‘₯,πœ†) does not change its sign for βˆ’πœ†0<πœ†<0 and 𝑧0(0,πœ†)=1; therefore, 𝑧0(π‘₯,πœ†)>0 for all π‘₯∈[0,1] and for all βˆ’πœ†0<πœ†<0.

Remark 2.4. Using (2.6), 𝑧1(π‘₯)=𝑧0(1βˆ’π‘₯), it is easy to prove that if πœ†>0 then for all π‘₯∈(0,1], 𝑧0(π‘₯)>1, and π‘§ξ…ž0(π‘₯)>0 and for all π‘₯∈[0,1), we have 𝑧1(π‘₯)>1 and π‘§ξ…ž1(π‘₯)<0.

Remark 2.5. Using Remark 2.3, 𝑧1(π‘₯)=𝑧0(1βˆ’π‘₯), and the differential equation (2.6), it is easy to prove that if βˆ’πœ†0<πœ†<0 for all π‘₯∈[0,1), 𝑧0(π‘₯)>0 and π‘§ξ…ž1(π‘₯)>0 and for all π‘₯∈(0,1], we have π‘§ξ…ž0(π‘₯)<0 and 𝑧1(π‘₯)>0.

Remark 2.6. Let πœ†>0 and let β„ŽβˆˆπΆ[0,1]. If β„Žβ‰₯0(orβ„Žβ‰€0), then ξ€œπ‘₯0π‘ž(𝑑)β„Ž(𝑑)𝑧0(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0ξ€Έξ€œπ‘‘π‘‘,1π‘₯π‘ž(𝑑)β„Ž(𝑑)𝑧1(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0𝑑𝑑(2.10) are nonnegative (or nonpositive).

Remark 2.7. Let βˆ’πœ†0<πœ†<0 and let β„ŽβˆˆπΆ[0,1]. If β„Žβ‰₯0(orβ„Žβ‰€0), then ξ€œπ‘₯0π‘ž(𝑑)β„Ž(𝑑)𝑧0(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0ξ€Έξ€œπ‘‘π‘‘,1π‘₯π‘ž(𝑑)β„Ž(𝑑)𝑧1(𝑑)π‘Šπ‘ξ€·π‘§1,𝑧0𝑑𝑑(2.11) are nonpositive (or nonnegative).

Proposition 2.8 (Maximum Principle). Let πœ†>0. If 𝐴≀0, 𝐡β‰₯0 (or 𝐴β‰₯0,𝐡≀0) and β„ŽβˆˆπΆ[0,1] is such that β„Žβ‰₯0 (or β„Žβ‰€0), then 𝑀(π‘₯)β‰₯0 (or 𝑀(π‘₯)≀0), where 𝑀(π‘₯) is the solution of (2.1)-(2.2).

Proposition 2.9 (Antimaximum Principle). Let βˆ’πœ†0<πœ†<0. If 𝐴≀0, 𝐡β‰₯0 (or𝐴β‰₯0,𝐡≀0) and β„ŽβˆˆπΆ[0,1] is such that β„Žβ‰₯0(or β„Žβ‰€0), then 𝑀(π‘₯)≀0(or 𝑀(π‘₯)β‰₯0), where 𝑀(π‘₯) is the solution of (2.1)-(2.2).

Now, we derive conditions on πœ† which will help us to prove the monotonicity of the solutions generated by the iterative scheme (1.2).

Lemma 2.10. Let 𝑀 and π‘βˆˆβ„+. If πœ†>0 is such that ξ‚΅ξ€œπœ†β‰₯𝑀1βˆ’π‘10ξ‚Άπ‘ž(π‘₯)𝑑π‘₯βˆ’1,(2.12) then for all π‘₯∈[0,1], (π‘€βˆ’πœ†)𝑧0(π‘₯)+𝑁𝑝(π‘₯)π‘§ξ…ž0(π‘₯)≀0.(2.13)

Proof. Integrating (2.6) from 0 to π‘₯ and using the fact that π‘§ξ…ž0(π‘₯)>0 in (0,1], we get 𝑝(π‘₯)π‘§ξ…ž0(π‘₯)β‰€πœ†π‘§0ξ€œ(π‘₯)10π‘ž(π‘₯)𝑑π‘₯.(2.14) Therefore, we get (π‘€βˆ’πœ†)𝑧0(π‘₯)+𝑁𝑝(π‘₯)π‘§ξ…ž0(π‘₯)≀(π‘€βˆ’πœ†)𝑧0+π‘πœ†π‘§0∫(π‘₯)10π‘ž(π‘₯)𝑑π‘₯. Hence, (2.13) will hold if ∫(π‘€βˆ’πœ†)+π‘πœ†10π‘ž(π‘₯)𝑑π‘₯≀0. Hence the result.

Lemma 2.11. Let 𝑀 and π‘βˆˆβ„+. If βˆ’πœ†0<πœ†<0 is such that βˆ«βˆ’(10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’1<πœ†β‰€βˆ’π‘€ and ξ‚΅ξ€œ(𝑀+πœ†)1+πœ†101ξ€œπ‘(π‘₯)π‘₯0ξ‚Άξ€œπ‘ž(𝑑)𝑑𝑑𝑑π‘₯βˆ’π‘πœ†10π‘ž(π‘₯)𝑑π‘₯≀0,(2.15) then for all π‘₯∈[0,1], (𝑀+πœ†)𝑧0(π‘₯)βˆ’π‘π‘(π‘₯)π‘§ξ…ž0(π‘₯)≀0.(2.16)

Proof. Using (2.6) and Remark 2.5, it can be deduced that 𝑧0(π‘₯) and 𝑝(π‘₯)π‘§ξ…ž0(π‘₯) are decreasing functions of π‘₯ for βˆ’πœ†0<πœ†<0, thus (𝑀+πœ†)𝑧0(π‘₯)βˆ’π‘π‘(π‘₯)π‘§ξ…ž0(π‘₯)≀(𝑀+πœ†)𝑧0(1)βˆ’π‘π‘(1)π‘§ξ…ž0(1).(2.17) Now using (2.6), we get βˆ’π‘(1)π‘§ξ…ž0∫(1)≀(βˆ’πœ†)10π‘ž(π‘₯)𝑑π‘₯ and 𝑧0∫(1)>1+πœ†10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯. This completes the proof.

Note 2. In Lemma 2.11, we arrive at the integral ∫10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯ which is an improper integral, and it should be convergent. Using the assumption (A2) (iv) and Remark (𝑖) and (𝑖𝑖) at [6, page 434] its convergence can be established.

3. Well-Ordered Upper and Lower Solutions

Let us define upper and lower solutions.

Definition 3.1. A function 𝑒0∈𝐢[0,1]∩𝐢2(0,1] is an upper solution of (1.1) if βˆ’ξ€·π‘π‘’ξ…ž0ξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑒0,π‘π‘’ξ…ž0ξ€Έβ‰₯0,0<π‘₯<1,π‘’ξ…ž0(0)≀0β‰€π‘’ξ…ž0(1).(3.1)

Definition 3.2. A function 𝑣0∈𝐢[0,1]∩𝐢2(0,1] is a lower solution of (1.1) if βˆ’ξ€·π‘π‘£ξ…ž0ξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑣0,π‘π‘£ξ…ž0≀0,0<π‘₯<1,π‘£ξ…ž0(0)β‰₯0β‰₯π‘£ξ…ž0(1).(3.2) Now, for every 𝑛, the problem (1.2) has a unique solution 𝑦𝑛+1 given by (2.5) with β„Ž(π‘₯)=βˆ’π‘“(π‘₯,𝑦𝑛,π‘π‘¦ξ…žπ‘›)+πœ†π‘¦π‘›, 𝐴=0, and 𝐡=0.

In this section, we show that for the proposed scheme (1.2) a good choice of πœ† is possible so that the solutions generated by the approximation scheme converge monotonically to solutions of (1.1). We require a number of results.

Lemma 3.3. Let πœ†>0. If 𝑒𝑛 is an upper solution of (1.1) and 𝑒𝑛+1 is defined by (1.2), then 𝑒𝑛+1≀𝑒𝑛.

Proof. Let 𝑀𝑛=𝑒𝑛+1βˆ’π‘’π‘›, then βˆ’ξ€·π‘π‘€ξ…žπ‘›ξ€Έξ…ž+πœ†π‘žπ‘€π‘›=ξ€·π‘π‘’ξ…žπ‘›ξ€Έξ…žξ€·βˆ’π‘žπ‘“π‘₯,𝑒𝑛,π‘π‘’ξ…žπ‘›ξ€Έπ‘€β‰€0,ξ…žπ‘›(0)β‰₯0,𝑀𝑛′(1)≀0,(3.3) and using Proposition 2.8, we have 𝑒𝑛+1≀𝑒𝑛.

Proposition 3.4. Assume that (𝐻1) there exist upper solution (𝑒0) and lower solution (𝑣0) in 𝐢[0,1]∩𝐢2(0,1] such that 𝑣0≀𝑒0 for all π‘₯∈[0,1],(𝐻2)the function π‘“βˆΆπ·β†’β„ is continuous on𝐷∢=ξ€½ξ€·π‘₯,𝑦,π‘π‘¦ξ…žξ€Έβˆˆ[]0,1Γ—π‘…Γ—π‘…βˆΆπ‘£0≀𝑦≀𝑒0ξ€Ύ,(3.4)(𝐻3)there exists 𝑀β‰₯0 such that for all (π‘₯,𝜏,π‘π‘£ξ…ž),(π‘₯,𝜎,π‘π‘£ξ…ž)∈𝐷,𝑓π‘₯,𝜏,π‘π‘£ξ…žξ€Έξ€·βˆ’π‘“π‘₯,𝜎,π‘π‘£ξ…žξ€Έβ‰₯𝑀(πœβˆ’πœŽ),(πœβ‰€πœŽ),(3.5)(𝐻4)there exist 𝑁β‰₯0 such that for all (π‘₯,𝑒,π‘π‘£ξ…ž1)(π‘₯,𝑒,π‘π‘£ξ…ž2)∈𝐷,||𝑓π‘₯,𝑒,π‘π‘£ξ…ž1ξ€Έξ€·βˆ’π‘“π‘₯,𝑒,π‘π‘£ξ…ž2ξ€Έ||||β‰€π‘π‘π‘£ξ…ž2βˆ’π‘π‘£ξ…ž1||.(3.6) Let πœ†>0 be such that βˆ«πœ†β‰₯𝑀(1βˆ’π‘10π‘ž(π‘₯)𝑑π‘₯)βˆ’1. Then the functions 𝑒𝑛+1 defined recursively by (1.2) are such that, for all π‘›βˆˆβ„•, (i)𝑒𝑛 is an upper solution of (1.1). (ii)𝑒𝑛+1≀𝑒𝑛.

Proof. We prove the claims by the principle of mathematical induction. Since 𝑒0 is an upper solution and by Lemma 3.3  𝑒0β‰₯𝑒1; therefore, both the claims are true for 𝑛=0.
Further, let the claims be true for π‘›βˆ’1, that is, π‘’π‘›βˆ’1 is an upper solution and π‘’π‘›βˆ’1β‰₯𝑒𝑛. Now, we are required to prove that 𝑒𝑛 is an upper solution and 𝑒𝑛+1≀𝑒𝑛. To prove this, let 𝑀=π‘’π‘›βˆ’π‘’π‘›βˆ’1, then we have βˆ’ξ€·π‘π‘’ξ…žπ‘›ξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑒𝑛,π‘π‘’ξ…žπ‘›ξ€Έξ€Ίξ€·β‰₯𝑝(π‘€βˆ’πœ†)π‘€βˆ’π‘signπ‘€ξ…žξ€Έπ‘π‘€ξ…žξ€».(3.7) Thus, to prove that 𝑒𝑛 is an upper solution, we are required to prove that ξ€·(π‘€βˆ’πœ†)π‘€βˆ’π‘signπ‘€ξ…žξ€Έπ‘π‘€ξ…žβ‰₯0.(3.8) Now, since 𝑀 satisfies βˆ’ξ€·π‘π‘€ξ…žξ€Έξ…žξ€·+πœ†π‘žπ‘€=π‘π‘’ξ…žπ‘›βˆ’1ξ€Έξ…žξ€·βˆ’π‘žπ‘“π‘₯,π‘’π‘›βˆ’1,π‘π‘’ξ…žπ‘›βˆ’1≀0,π‘€ξ…ž(0)β‰₯0,𝑀′(1)≀0,(3.9) from Proposition 2.8, we have 𝑀≀0 for πœ†>0. Now, putting the value of 𝑀 from (2.5) in (3.8), and in view of β„Ž=(π‘π‘’ξ…žπ‘›βˆ’1)ξ…žβˆ’π‘žπ‘“(π‘₯,π‘’π‘›βˆ’1,π‘π‘’ξ…žπ‘›βˆ’1)≀0, we deduce that to prove (3.8) it is sufficient to prove that (π‘€βˆ’πœ†)𝑧0ξ€·βˆ’π‘signπ‘€ξ…žξ€Έπ‘π‘§ξ…ž0≀0,(π‘€βˆ’πœ†)𝑧1ξ€·βˆ’π‘signπ‘€ξ…žξ€Έπ‘π‘§ξ…ž1,≀0(3.10) for all π‘₯∈[0,1]. Since 𝑧1=𝑧0(1βˆ’π‘₯), using Remark 2.6, the above inequalities will be true if for all π‘₯∈[0,1] we have (π‘€βˆ’πœ†)𝑧0(π‘₯)+𝑁𝑝(π‘₯)π‘§ξ…ž0(π‘₯)≀0.(3.11) Which is true (Lemma 2.10). Therefore, (3.8) holds, and hence 𝑒𝑛 is an upper solution.
Now applying Lemma 3.3, we deduce that 𝑒𝑛+1≀𝑒𝑛. This completes the proof.

Similarly, we can prove the following two results (Lemma 3.5, Proposition 3.6) for lower solutions.

Lemma 3.5. Let πœ†>0. If 𝑣𝑛 is a lower solution of (1.1) and 𝑣𝑛+1 is defined by (1.2), then 𝑣𝑛≀𝑣𝑛+1.

Proposition 3.6. Assume that (𝐻1), (𝐻2), (𝐻3), and (𝐻4) hold, and let πœ†>0 be such that βˆ«πœ†β‰₯𝑀(1βˆ’π‘10π‘ž(π‘₯)𝑑π‘₯)βˆ’1. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that for all π‘›βˆˆβ„•, (i)𝑣𝑛 is a lower solution of (1.1).(ii)𝑣𝑛≀𝑣𝑛+1.

In the next result, we prove that upper solution 𝑒𝑛 is larger than lower solution 𝑣𝑛 for all 𝑛.

Proposition 3.7. Assume that (𝐻1), (𝐻2), (𝐻3), and (𝐻4) hold, and let πœ†>0 such that βˆ«πœ†β‰₯𝑀(1βˆ’π‘10π‘ž(π‘₯)𝑑π‘₯)βˆ’1and for all π‘₯∈[0,1]𝑓π‘₯,𝑣0,π‘π‘£ξ…ž0ξ€Έξ€·βˆ’π‘“π‘₯,𝑒0,π‘π‘’ξ…ž0𝑒+πœ†0βˆ’π‘£0ξ€Έβ‰₯0.(3.12) Then for all π‘›βˆˆβ„•, the functions 𝑒𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛≀𝑒𝑛.

Proof. We define a function β„Žπ‘–ξ€·(π‘₯)=𝑓π‘₯,π‘£π‘–π‘π‘£ξ…žπ‘–ξ€Έξ€·βˆ’π‘“π‘₯,𝑒𝑖,π‘π‘’ξ…žπ‘–ξ€Έξ€·π‘’+πœ†π‘–βˆ’π‘£π‘–ξ€Έ,π‘–βˆˆβ„•.(3.13) It is easy to see that for all π‘–βˆˆβ„•, 𝑀𝑖=π‘’π‘–βˆ’π‘£π‘– satisfies the following differential equation: βˆ’ξ€·π‘π‘€ξ…žπ‘–ξ€Έξ…ž+πœ†π‘žπ‘€π‘–ξ€½π‘“ξ€·=π‘žπ‘₯,π‘£π‘–βˆ’1,π‘π‘£ξ…žπ‘–βˆ’1ξ€Έξ€·βˆ’π‘“π‘₯,π‘’π‘–βˆ’1,π‘π‘’ξ…žπ‘–βˆ’1𝑒+πœ†π‘–βˆ’1βˆ’π‘£π‘–βˆ’1ξ€Έξ€Ύ=π‘žβ„Žπ‘–βˆ’1.(3.14) Now to prove this proposition again, we use the principle of mathematical induction. For 𝑖=1, we have β„Ž0β‰₯0, and 𝑀1 is the solution of (2.1)-(2.2) with 𝐴=0 and 𝐡=0. Using Proposition 2.8, we deduce that 𝑀1β‰₯0, that is, 𝑒1β‰₯𝑣1.
Now, let 𝑛β‰₯2, let β„Žπ‘›βˆ’2β‰₯0, and let π‘’π‘›βˆ’1β‰₯π‘£π‘›βˆ’1, then we are required to prove that β„Žπ‘›βˆ’1β‰₯0 and 𝑒𝑛β‰₯𝑣𝑛. First, we show that for all π‘₯∈[0,1] the function β„Žπ‘›βˆ’1 is nonnegative. Indeed, we have 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’1𝑒+πœ†π‘›βˆ’1βˆ’π‘£π‘›βˆ’1ξ€Έξ€Ί(β‰₯βˆ’π‘€βˆ’πœ†)π‘€π‘›βˆ’1ξ€·+𝑁signπ‘€ξ…žπ‘›βˆ’1ξ€Έπ‘π‘€ξ…žπ‘›βˆ’1ξ€».(3.15) Here π‘€π‘›βˆ’1 is a solution of (2.1) with β„Ž(π‘₯)=β„Žπ‘›βˆ’2β‰₯0, 𝐴=0, and 𝐡=0. Arguments similar to Proposition 3.4 can be used to prove that β„Žπ‘›βˆ’1β‰₯0. Now, we have β„Žπ‘›βˆ’1β‰₯0, π‘€ξ…žπ‘›(0)=0, and π‘€ξ…žπ‘›(1)=0, thus from Proposition 2.8, we deduce that 𝑀𝑛β‰₯0, that is, 𝑒𝑛β‰₯𝑣𝑛.

Lemma 3.8. If 𝑓(π‘₯,𝑒,π‘π‘’ξ…ž) satisfies (𝐻1), (𝐻2), and (𝐻5)for all (π‘₯,𝑒,π‘π‘’ξ…ž)∈𝐷, |𝑓(π‘₯,𝑒,π‘π‘’ξ…ž)|β‰€πœ‘(|π‘π‘’ξ…ž|), where πœ‘βˆΆ[0,∞)β†’(0,∞) is continuous and satisfies.ξ€œβˆž0𝑑𝑠>ξ€œπœ‘(𝑠)10π‘ž(π‘₯)𝑑π‘₯,(3.16) then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘’ξ…žξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑒,π‘π‘’ξ…žξ€Έβ‰₯0,0<π‘₯<1,π‘’ξ…ž(0)=0=π‘’ξ…ž(1)(3.17) with π‘’βˆˆ[𝑣0,𝑒0], for all π‘₯∈[0,1], satisfies β€–π‘π‘’ξ…žβ€–βˆž<𝑅0.

Proof. Consider an interval [π‘₯,π‘₯0]βŠ‚[0,1] such that ξ€Ίβˆ€π‘ βˆˆπ‘₯,π‘₯0ξ€Έ,π‘’ξ…ž(𝑠)<0,π‘’ξ…žξ€·π‘₯0ξ€Έ=0.(3.18) Now using (𝐻5), we have ξ€·π‘π‘’ξ…žξ€Έξ…žξ€·||β‰€π‘žπœ‘π‘π‘’ξ…ž||ξ€Έ,(3.19) and after integrating it from π‘₯ to π‘₯0 and using (𝐻5), we have βˆ’π‘π‘’ξ…žβ‰€π‘…0.(3.20) Similarly for the interval [π‘₯0,π‘₯], we have π‘π‘’ξ…žβ‰€π‘…0.(3.21) Thus β€–β€–π‘π‘’ξ…žβ€–β€–βˆžβ‰€π‘…0.(3.22) In the same way, we can prove the following result for lower solutions.

Lemma 3.9. If 𝑓(π‘₯,𝑣,π‘π‘£ξ…ž) satisfies (𝐻1), (𝐻2), and (𝐻5), then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘£ξ…žξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑣,π‘π‘£ξ…žξ€Έβ‰€0,0<π‘₯<1,π‘£ξ…ž(0)=0=π‘£ξ…ž(1)(3.23) with π‘£βˆˆ[𝑣0,𝑒0], for all π‘₯∈[0,1], satisfies β€–π‘π‘£ξ…žβ€–βˆž<𝑅0.

Now we are in a situation to prove our final result for the case when upper and lower solutions are well ordered.

Theorem 3.10. Assume (𝐻1), (𝐻2), (𝐻3), (𝐻4), and (𝐻5) are true. Let πœ†>0 be such that ξ‚΅ξ€œπœ†β‰₯𝑀1βˆ’π‘10ξ‚Άπ‘ž(π‘₯)𝑑π‘₯βˆ’1,(3.24) and for all π‘₯∈[0,1], 𝑓π‘₯,𝑣0,π‘π‘£ξ…ž0ξ€Έξ€·βˆ’π‘“π‘₯,𝑒0,π‘π‘’ξ…ž0𝑒+πœ†0βˆ’π‘£0ξ€Έβ‰₯0.(3.25) Then the sequences {𝑒𝑛} and {𝑣𝑛} defined by (1.2) converge monotonically to solutions ̃𝑒(π‘₯) and ̃𝑣(π‘₯) of (1.1). Any solution 𝑧(π‘₯) of (1.1) in 𝐷 satisfies ̃𝑣(π‘₯)≀𝑧(π‘₯)≀̃𝑒(π‘₯).(3.26)

Proof. Using Lemma 3.3 to Lemma 3.9 and Proposition 3.4 to Proposition 3.7, we deduce that the sequences {𝑒𝑛} and {𝑣𝑛} are monotonic (𝑒0β‰₯𝑒1β‰₯𝑒2β‹―β‰₯𝑒𝑛β‰₯𝑣𝑛⋯β‰₯𝑣2β‰₯𝑣1β‰₯𝑣0) and are bounded by 𝑣0 and 𝑒0 in 𝐢[0,1], and by Dini's theorem, they converge uniformly to ̃𝑒 and ̃𝑣 (say). We can also deduce that the sequences {π‘π‘’ξ…žπ‘›} and {π‘π‘£ξ…žπ‘›} are uniformly bounded and equicontinuous in 𝐢[0,1], and by Arzela-Ascoli theorem, there exists uniformly convergent subsequences {π‘π‘’ξ…žπ‘›π‘˜} and {π‘π‘£ξ…žπ‘›π‘˜} in 𝐢[0,1]. It is easy to observe that 𝑒𝑛→̃𝑒 and 𝑣𝑛→̃𝑣 imply π‘π‘’ξ…žπ‘›β†’π‘Μƒπ‘’ξ…ž and π‘Μƒπ‘£ξ…žπ‘›Μƒπ‘£β†’π‘ξ…ž.
Solution of (1.2) is given by (2.5) where β„Ž(π‘₯)=βˆ’π‘“(π‘₯,π‘¦π‘›βˆ’1,π‘π‘¦ξ…žπ‘›βˆ’1)+πœ†π‘¦π‘›βˆ’1. Since the sequences are uniformly convergent taking limit as π‘›β†’βˆž, we get ̃𝑒 and ̃𝑣 as the solutions of the nonlinear boundary value problem (1.1). Any solution 𝑧(π‘₯) in 𝐷 plays the role of 𝑒0. Hence, ̃𝑧(π‘₯)β‰₯𝑣(π‘₯). Similarly 𝑧(π‘₯)≀̃𝑒(π‘₯).

Remark 3.11. When the source function is derivative independent, that is, 𝑁=0, in this case we can choose πœ†=𝑀.

4. Upper and Lower Solutions in Reverse Order

In this section, we consider the case when the upper and lower solutions are in reverse order, that is,𝑒0(π‘₯)≀𝑣0(π‘₯).(4.1) For this, we require opposite one-sided Lipschitz condition, and we assume that (𝐹1)there exists upper solution (𝑒0) and lower solution (𝑣0) in 𝐢[0,1]∩𝐢2(0,1] such that 𝑒0≀𝑣0 for all π‘₯∈[0,1],(𝐹2)the function π‘“βˆΆπ·0→ℝ is continuous on𝐷0∢=ξ€½ξ€·π‘₯,𝑦,π‘π‘¦ξ…žξ€Έβˆˆ[]0,1Γ—π‘…Γ—π‘…βˆΆπ‘’0≀𝑦≀𝑣0ξ€Ύ,(4.2)(𝐹3)there exists 𝑀β‰₯0 such that for all (π‘₯,Μƒπœ,π‘π‘£ξ…ž),(π‘₯,ξ‚πœŽ,π‘π‘£ξ…ž)∈𝐷0,𝑓π‘₯,ξ‚πœŽ,π‘π‘£ξ…žξ€Έξ€·βˆ’π‘“π‘₯,Μƒπœ,π‘π‘£ξ…žξ€Έξ€·ξ€Έ,ξ€·ξ€Έβ‰₯βˆ’π‘€ξ‚πœŽβˆ’ΜƒπœΜƒπœβ‰€ξ‚πœŽ,(4.3)(𝐹4)there exist 𝑁β‰₯0 such that for all (π‘₯,𝑒,π‘π‘£ξ…ž1)(π‘₯,𝑒,π‘π‘£ξ…ž2)∈𝐷0,||𝑓π‘₯,𝑒,π‘π‘£ξ…ž1ξ€Έξ€·βˆ’π‘“π‘₯,𝑒,π‘π‘£ξ…ž2ξ€Έ||||β‰€π‘π‘π‘£ξ…ž2βˆ’π‘π‘£ξ…ž1||.(4.4) Here again we define the approximation scheme by (1.2) and use the Antimaximum principle. We make a good choice of πœ† so that the sequences thus generated converge to the solution of the nonlinear problem. Similar to Section 3, we require the following lemmas and propositions.

Lemma 4.1. Let βˆ’πœ†0<πœ†<0. If 𝑒𝑛 is an upper solution of (1.1) and 𝑒𝑛+1 is defined by (1.2), then 𝑒𝑛+1β‰₯𝑒𝑛.

Proof. Let 𝑀𝑛=𝑒𝑛+1βˆ’π‘’π‘›, then βˆ’ξ€·π‘π‘€ξ…žπ‘›ξ€Έξ…ž+πœ†π‘žπ‘€π‘›=ξ€·π‘π‘’ξ…žπ‘›ξ€Έξ…žξ€·βˆ’π‘žπ‘“π‘₯,𝑒𝑛,π‘π‘’ξ…žπ‘›ξ€Έπ‘€β‰€0,ξ…žπ‘›(0)β‰₯0,π‘€ξ…žπ‘›(1)≀0,(4.5) and using Proposition 2.8, we have 𝑒𝑛+1β‰₯𝑒𝑛.

Proposition 4.2. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let βˆ’πœ†0<πœ†<0 be such that βˆ«βˆ’(10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’1<πœ†β‰€βˆ’π‘€ and ∫(𝑀+πœ†)(1+πœ†10∫(1/𝑝(π‘₯))π‘₯0βˆ«π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’π‘πœ†10π‘ž(π‘₯)𝑑π‘₯≀0. Then the functions 𝑒𝑛+1 defined recursively by (1.2) are such that, for all π‘›βˆˆβ„•, (i)𝑒𝑛 is an upper solution of (1.1);(ii)𝑒𝑛+1β‰₯𝑒𝑛.

Proof. Using Remarks 2.5 and 2.7, Lemmas 2.11 and 4.1, and on the lines of the proof of Proposition 3.4, this proposition can be deduced easily.
In the same way, we can prove the following results for the lower solutions.

Lemma 4.3. Let βˆ’πœ†0<πœ†<0. If 𝑣𝑛 is a lower solution of (1.1) and 𝑣𝑛+1 is defined by (1.2), then 𝑣𝑛β‰₯𝑣𝑛+1.

Proposition 4.4. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let βˆ’πœ†0<πœ†<0 be such that βˆ«βˆ’(10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’1<πœ†β‰€βˆ’π‘€ and ∫(𝑀+πœ†)(1+πœ†10∫(1/𝑝(π‘₯))π‘₯0βˆ«π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’π‘πœ†10π‘ž(π‘₯)𝑑π‘₯≀0. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that, for all π‘›βˆˆβ„•, (i)𝑣𝑛 is a lower solution of (1.1);(ii)𝑣𝑛β‰₯𝑣𝑛+1.

In the next result, we prove that lower solution 𝑣𝑛 is larger than upper solution 𝑒𝑛 for all 𝑛.

Proposition 4.5. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let βˆ’πœ†0<πœ†<0 be such that βˆ«βˆ’(10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’1<πœ†β‰€βˆ’π‘€ and ξ‚΅ξ€œ(𝑀+πœ†)1+πœ†101ξ€œπ‘(π‘₯)π‘₯0ξ‚Άξ€œπ‘ž(𝑑)𝑑𝑑𝑑π‘₯βˆ’π‘πœ†10π‘ž(π‘₯)𝑑π‘₯≀0,(4.6) and for all π‘₯∈[0,1], 𝑓π‘₯,𝑣0,π‘π‘£ξ…ž0ξ€Έξ€·βˆ’π‘“π‘₯,𝑒0,π‘π‘’ξ…ž0𝑒+πœ†0βˆ’π‘£0ξ€Έβ‰₯0.(4.7) Then, for all π‘›βˆˆβ„•, the functions 𝑒𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛β‰₯𝑒𝑛.

Now similar to Lemmas 3.8 and 3.9, we state the following two results. These results establish a bound on 𝑝(π‘₯)π‘’ξ…ž(π‘₯) and 𝑝(π‘₯)π‘£ξ…ž(π‘₯).

Lemma 4.6. If 𝑓(π‘₯,𝑒,π‘π‘’ξ…ž) satisfies (𝐹1), (𝐹2), and (𝐹5) for all (π‘₯,𝑒,π‘π‘’ξ…ž)∈𝐷0, |𝑓(π‘₯,𝑒,π‘π‘’ξ…ž)|β‰€πœ‘(|π‘π‘’ξ…ž|), where πœ‘βˆΆ[0,∞)β†’(0,∞) is continuous and satisfiesξ€œβˆž0𝑑𝑠>ξ€œπœ‘(𝑠)10π‘ž(π‘₯)𝑑π‘₯,(4.8) then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘’ξ…žξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑒,π‘π‘’ξ…žξ€Έβ‰₯0,0<π‘₯<1,π‘’ξ…ž(0)=0=π‘’ξ…ž(1)(4.9) with π‘’βˆˆ[𝑒0,𝑣0], for all π‘₯∈[0,1], satisfies β€–π‘π‘’ξ…žβ€–βˆž<𝑅0.

Lemma 4.7. If 𝑓(π‘₯,𝑣,π‘π‘£ξ…ž) satisfies (𝐹1), (𝐹2), and (𝐹5), then there exists 𝑅0>0 such that any solution of βˆ’ξ€·π‘π‘£ξ…žξ€Έξ…žξ€·+π‘žπ‘“π‘₯,𝑣,π‘π‘£ξ…žξ€Έβ‰€0,0<π‘₯<1,π‘£ξ…ž(0)=0=π‘£ξ…ž(1)(4.10) with π‘£βˆˆ[𝑒0,𝑣0], for all π‘₯∈[0,1], satisfies β€–π‘π‘£β€²β€–βˆž<𝑅0.

Finally we arrive at the theorem similar to Theorem 3.10.

Theorem 4.8. Assume (𝐹1), (𝐹2), (𝐹3), (𝐹4), and (𝐹5) are true. Let βˆ’πœ†0<πœ†<0 be such that βˆ«βˆ’(10∫(1/𝑝(π‘₯))π‘₯0π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’1<πœ†β‰€βˆ’π‘€ and ∫(𝑀+πœ†)(1+πœ†10∫(1/𝑝(π‘₯))π‘₯0βˆ«π‘ž(𝑑)𝑑𝑑𝑑π‘₯)βˆ’π‘πœ†10π‘ž(π‘₯)𝑑x≀0, and for all π‘₯∈[0,1], 𝑓π‘₯,𝑣0,π‘π‘£ξ…ž0ξ€Έξ€·βˆ’π‘“π‘₯,𝑒0,π‘π‘’ξ…ž0𝑒+πœ†0βˆ’π‘£0ξ€Έβ‰₯0.(4.11)
Then the sequences {𝑒𝑛} and {𝑣𝑛} defined by (1.2) converge monotonically to solutions ̃𝑒(π‘₯) and ̃𝑣(π‘₯) of (1.1). Any solution 𝑧(π‘₯) of (1.1) in 𝐷0 satisfies ̃̃𝑒(π‘₯)≀𝑧(π‘₯)≀𝑣(π‘₯).(4.12)

Proof. Using Lemma 4.1 to Lemma 4.7 and Proposition 4.2 to Proposition 4.5, we deduce that 𝑒0≀𝑒1≀𝑒2≀⋯≀𝑒𝑛≀𝑣𝑛⋯≀𝑣1≀𝑣0.(4.13) Now similar to the proof of Theorem 3.10, the result of this theorem can be deduced.

Remark 4.9. When the source function is derivative independent, that is, 𝑁=0, in this case we can choose πœ†=βˆ’π‘€.

5. Conclusion

We establish some existence results under quite general conditions on 𝑝(π‘₯), π‘ž(π‘₯), and 𝑓(π‘₯,𝑦,π‘π‘¦ξ…ž). We prove some fundamental differential inequalities which enables us to prove the monotonicity of the sequences {𝑒𝑛} and {𝑣𝑛}. For this we have analyzed the singular differential equation βˆ’(π‘π‘¦ξ…ž)ξ…ž+πœ†π‘žπ‘¦=0 and derived properties of the solutions and their derivatives. This work generalizes our previous work [3]. Lot of exploration is still left. For example, one can consider different type of boundary conditions, and one can also try to remove the Lipschitz condition.

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