Abstract

We consider the following class of nonlinear singular differential equation −(𝑝(𝑥)𝑦′(𝑥))′+ğ‘ž(𝑥)𝑓(𝑥,𝑦(𝑥),𝑝(𝑥)𝑦′(𝑥))=0,0<𝑥<1 subject to the Neumann boundary condition 𝑦′(0)=𝑦′(1)=0. Conditions on 𝑝(𝑥) and ğ‘ž(𝑥) ensure that 𝑥=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≤𝑢0) or the other (𝑢0≤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, 𝑢0≥𝑣0. As far as reverse-ordered upper and lower solutions are considered, that is, 𝑢0≤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:−𝑝(𝑥)ğ‘¦î…žî€¸(𝑥)+ğ‘ž(𝑥)𝑓𝑥,𝑦(𝑥),𝑝(𝑥)ğ‘¦î…žî€¸ğ‘¦(𝑥)=0,0<𝑥<1,(0)=0,ğ‘¦î…ž(1)=0.(1.1)(A1)Let 𝑝(𝑥) satisfy the following conditions. (i)𝑝(0)=0 and 𝑝>0 in (0,1). (ii)𝑝∈𝐶[0,1]∩𝐶1(0,1). (A2) Let ğ‘ž(𝑥) satisfy the following conditions. (i)ğ‘ž(𝑥)>0 in (0,1) and ğ‘ž(𝑥)∈𝐶(0,1]. (ii)∫10ğ‘ž(𝑥)𝑑𝑥<∞. (iii)lim𝑥→0(ğ‘ž(𝑥))/(ğ‘î…ž(𝑥))=0. (iv)∫10∫(1/𝑝(𝑥))(𝑥0ğ‘ž(𝑠)𝑑𝑠)1/2𝑑𝑥<∞.

In this paper, we consider a computationally simple iterative scheme defined byâˆ’î€·ğ‘ğ‘¦î…žğ‘›î€¸î…ž+ğœ†ğ‘žğ‘¦ğ‘›î€·=âˆ’ğ‘žğ‘“ğ‘¥,𝑦𝑛−1,ğ‘ğ‘¦î…žğ‘›âˆ’1+ğœ†ğ‘žğ‘¦ğ‘›âˆ’1𝑦,0<𝑥<1,î…žğ‘›(0)=0,ğ‘¦î…žğ‘›(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 ℎ(𝑥)∈𝐶[0,1], and let 𝜆∈ℝ0(ℝ0=ℝ⧵{0}), let 𝐴∈ℝ and let 𝐵∈ℝ. Now, consider the following class of linear singular problems:−𝑝(𝑥)ğ‘¦î…žî€¸(𝑥)î…žğ‘¦+ğœ†ğ‘ž(𝑥)𝑦(𝑥)=ğ‘ž(𝑥)ℎ(𝑥),0<𝑥<1,(2.1)(0)=𝐴,ğ‘¦î…ž(1)=𝐵.(2.2) The corresponding homogeneous system (eigenvalue problem) is given by−𝑝(𝑥)ğ‘¦î…žî€¸(𝑥)î…žğ‘¦+ğœ†ğ‘ž(𝑥)𝑦(𝑥)=0,0<𝑥<1,(2.3)(0)=0,ğ‘¦î…ž(1)=0.(2.4) The solution of the nonhomogeneous problem (2.1)-(2.2) can be written as follows:𝑤(𝑥)=𝑧1(𝑥)𝑥0ğ‘ž(𝑡)ℎ(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝐴𝑑𝑡+ğ‘§î…ž1(0)+𝑧0(𝑥)1ğ‘¥ğ‘ž(𝑡)ℎ(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝐵𝑑𝑡+ğ‘§î…ž0(1),(2.5) where 𝑧0(𝑥,𝜆) is the solution of−𝑝(𝑥)ğ‘§î…ž0(𝑥)+ğœ†ğ‘ž(𝑥)𝑧0(𝑥)=0,0<𝑥<1,𝑧0(0)=1,ğ‘§î…ž0(0)=0,(2.6)𝑧1(𝑥,𝜆) is the solution of−𝑝(𝑥)ğ‘§î…ž1(𝑥)+ğœ†ğ‘ž(𝑥)𝑧1(𝑥)=0,0<𝑥<1,𝑧1(1)=1,ğ‘§î…ž1(1)=0,(2.7) and 𝑊𝑝(𝑧1,𝑧0)=𝑝(𝑡)(𝑧1ğ‘§î…ž0âˆ’ğ‘§î…ž1𝑧0). By replacing 𝑥 with 1−𝑥 in (2.6), it is easy to verify that 𝑧1(𝑥)=𝑧0(1−𝑥),(2.8) for both positive and negative values of 𝜆.

Remark 2.1. Existence of 𝑧0(𝑥) and 𝑧1(𝑥) 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 𝐿2ğ‘ž(0,1) be a Hilbert space with inner product defined by ⟨𝑓,𝑔⟩=10ğ‘ž(𝑥)𝑓(𝑥)𝑔(𝑥)𝑑𝑥.(2.9) From (A2) (iv), it can easily be verified that 𝑥=0 is a singular point of limit circle type (see [6, Remark (i) page 434]) in 𝐿2ğ‘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 ğ‘§î…ž0(1,𝜆). Since ğ‘§î…ž0(1,𝜆) is an analytic function of 𝜆 so its zeros will be isolated and they all will be negative. Let them be −𝜆0,−𝜆1,−𝜆2,…, where 𝜆𝑖>0 for 𝑖=0,1,2,…. Now, we have −𝜆0 as the first negative zero of ğ‘§î…ž0(1,𝜆) or in other words first negative eigenvalue of (2.3)-(2.4).
Since 𝑧0(𝑥,𝜆) does not change its sign for −𝜆0<𝜆<0 and 𝑧0(0,𝜆)=1; therefore, 𝑧0(𝑥,𝜆)>0 for all 𝑥∈[0,1] and for all −𝜆0<𝜆<0.

Remark 2.4. Using (2.6), 𝑧1(𝑥)=𝑧0(1−𝑥), it is easy to prove that if 𝜆>0 then for all 𝑥∈(0,1], 𝑧0(𝑥)>1, and ğ‘§î…ž0(𝑥)>0 and for all 𝑥∈[0,1), we have 𝑧1(𝑥)>1 and ğ‘§î…ž1(𝑥)<0.

Remark 2.5. Using Remark 2.3, 𝑧1(𝑥)=𝑧0(1−𝑥), and the differential equation (2.6), it is easy to prove that if −𝜆0<𝜆<0 for all 𝑥∈[0,1), 𝑧0(𝑥)>0 and ğ‘§î…ž1(𝑥)>0 and for all 𝑥∈(0,1], we have ğ‘§î…ž0(𝑥)<0 and 𝑧1(𝑥)>0.

Remark 2.6. Let 𝜆>0 and let â„Žâˆˆğ¶[0,1]. If ℎ≥0(orℎ≤0), then 𝑥0ğ‘ž(𝑡)ℎ(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡,1ğ‘¥ğ‘ž(𝑡)ℎ(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡(2.10) are nonnegative (or nonpositive).

Remark 2.7. Let −𝜆0<𝜆<0 and let â„Žâˆˆğ¶[0,1]. If ℎ≥0(orℎ≤0), then 𝑥0ğ‘ž(𝑡)ℎ(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡,1ğ‘¥ğ‘ž(𝑡)ℎ(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡(2.11) are nonpositive (or nonnegative).

Proposition 2.8 (Maximum Principle). Let 𝜆>0. If 𝐴≤0, 𝐵≥0 (or 𝐴≥0,𝐵≤0) and â„Žâˆˆğ¶[0,1] is such that ℎ≥0 (or ℎ≤0), then 𝑤(𝑥)≥0 (or 𝑤(𝑥)≤0), where 𝑤(𝑥) is the solution of (2.1)-(2.2).

Proposition 2.9 (Antimaximum Principle). Let −𝜆0<𝜆<0. If 𝐴≤0, 𝐵≥0 (or𝐴≥0,𝐵≤0) and â„Žâˆˆğ¶[0,1] is such that ℎ≥0(or ℎ≤0), then 𝑤(𝑥)≤0(or 𝑤(𝑥)≥0), where 𝑤(𝑥) is the solution of (2.1)-(2.2).

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

Lemma 2.10. Let 𝑀 and 𝑁∈ℝ+. If 𝜆>0 is such that 𝜆≥𝑀1−𝑁10î‚¶ğ‘ž(𝑥)𝑑𝑥−1,(2.12) then for all 𝑥∈[0,1], (𝑀−𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)ğ‘§î…ž0(𝑥)≤0.(2.13)

Proof. Integrating (2.6) from 0 to 𝑥 and using the fact that ğ‘§î…ž0(𝑥)>0 in (0,1], we get 𝑝(𝑥)ğ‘§î…ž0(𝑥)≤𝜆𝑧0(𝑥)10ğ‘ž(𝑥)𝑑𝑥.(2.14) Therefore, we get (𝑀−𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)ğ‘§î…ž0(𝑥)≤(𝑀−𝜆)𝑧0+𝑁𝜆𝑧0∫(𝑥)10ğ‘ž(𝑥)𝑑𝑥. Hence, (2.13) will hold if ∫(𝑀−𝜆)+𝑁𝜆10ğ‘ž(𝑥)𝑑𝑥≤0. Hence the result.

Lemma 2.11. Let 𝑀 and 𝑁∈ℝ+. If −𝜆0<𝜆<0 is such that ∫−(10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−1<𝜆≤−𝑀 and (𝑀+𝜆)1+𝜆101𝑝(𝑥)𝑥0î‚¶î€œğ‘ž(𝑡)𝑑𝑡𝑑𝑥−𝑁𝜆10ğ‘ž(𝑥)𝑑𝑥≤0,(2.15) then for all 𝑥∈[0,1], (𝑀+𝜆)𝑧0(𝑥)−𝑁𝑝(𝑥)ğ‘§î…ž0(𝑥)≤0.(2.16)

Proof. Using (2.6) and Remark 2.5, it can be deduced that 𝑧0(𝑥) and 𝑝(𝑥)ğ‘§î…ž0(𝑥) are decreasing functions of 𝑥 for −𝜆0<𝜆<0, thus (𝑀+𝜆)𝑧0(𝑥)−𝑁𝑝(𝑥)ğ‘§î…ž0(𝑥)≤(𝑀+𝜆)𝑧0(1)−𝑁𝑝(1)ğ‘§î…ž0(1).(2.17) Now using (2.6), we get −𝑝(1)ğ‘§î…ž0∫(1)≤(−𝜆)10ğ‘ž(𝑥)𝑑𝑥 and 𝑧0∫(1)>1+𝜆10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥. This completes the proof.

Note 2. In Lemma 2.11, we arrive at the integral ∫10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥 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 𝑢0∈𝐶[0,1]∩𝐶2(0,1] is an upper solution of (1.1) if âˆ’î€·ğ‘ğ‘¢î…ž0+ğ‘žğ‘“ğ‘¥,𝑢0,ğ‘ğ‘¢î…ž0≥0,0<𝑥<1,ğ‘¢î…ž0(0)≤0â‰¤ğ‘¢î…ž0(1).(3.1)

Definition 3.2. A function 𝑣0∈𝐶[0,1]∩𝐶2(0,1] is a lower solution of (1.1) if âˆ’î€·ğ‘ğ‘£î…ž0+ğ‘žğ‘“ğ‘¥,𝑣0,ğ‘ğ‘£î…ž0≤0,0<𝑥<1,ğ‘£î…ž0(0)≥0â‰¥ğ‘£î…ž0(1).(3.2) Now, for every 𝑛, the problem (1.2) has a unique solution 𝑦𝑛+1 given by (2.5) with ℎ(𝑥)=−𝑓(𝑥,𝑦𝑛,ğ‘ğ‘¦î…žğ‘›)+𝜆𝑦𝑛, 𝐴=0, and 𝐵=0.

In this section, we show that for the proposed scheme (1.2) a good choice of 𝜆 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 𝜆>0. If 𝑢𝑛 is an upper solution of (1.1) and 𝑢𝑛+1 is defined by (1.2), then 𝑢𝑛+1≤𝑢𝑛.

Proof. Let 𝑤𝑛=𝑢𝑛+1−𝑢𝑛, then âˆ’î€·ğ‘ğ‘¤î…žğ‘›î€¸î…ž+ğœ†ğ‘žğ‘¤ğ‘›=î€·ğ‘ğ‘¢î…žğ‘›î€¸î…žî€·âˆ’ğ‘žğ‘“ğ‘¥,𝑢𝑛,ğ‘ğ‘¢î…žğ‘›î€¸ğ‘¤â‰¤0,î…žğ‘›(0)≥0,𝑤𝑛′(1)≤0,(3.3) and using Proposition 2.8, we have 𝑢𝑛+1≤𝑢𝑛.

Proposition 3.4. Assume that (𝐻1) there exist upper solution (𝑢0) and lower solution (𝑣0) in 𝐶[0,1]∩𝐶2(0,1] such that 𝑣0≤𝑢0 for all 𝑥∈[0,1],(𝐻2)the function 𝑓∶𝐷→ℝ is continuous on𝐷∶=𝑥,𝑦,ğ‘ğ‘¦î…žî€¸âˆˆ[]0,1×𝑅×𝑅∶𝑣0≤𝑦≤𝑢0,(3.4)(𝐻3)there exists 𝑀≥0 such that for all (𝑥,𝜏,ğ‘ğ‘£î…ž),(𝑥,ğœŽ,ğ‘ğ‘£î…ž)∈𝐷,𝑓𝑥,𝜏,ğ‘ğ‘£î…žî€¸î€·âˆ’ğ‘“ğ‘¥,ğœŽ,ğ‘ğ‘£î…žî€¸â‰¥ğ‘€(ğœâˆ’ğœŽ),(ğœâ‰¤ğœŽ),(3.5)(𝐻4)there exist 𝑁≥0 such that for all (𝑥,𝑢,ğ‘ğ‘£î…ž1)(𝑥,𝑢,ğ‘ğ‘£î…ž2)∈𝐷,||𝑓𝑥,𝑢,ğ‘ğ‘£î…ž1−𝑓𝑥,𝑢,ğ‘ğ‘£î…ž2||||â‰¤ğ‘ğ‘ğ‘£î…ž2âˆ’ğ‘ğ‘£î…ž1||.(3.6) Let 𝜆>0 be such that ∫𝜆≥𝑀(1−𝑁10ğ‘ž(𝑥)𝑑𝑥)−1. Then the functions 𝑢𝑛+1 defined recursively by (1.2) are such that, for all 𝑛∈ℕ, (i)𝑢𝑛 is an upper solution of (1.1). (ii)𝑢𝑛+1≤𝑢𝑛.

Proof. We prove the claims by the principle of mathematical induction. Since 𝑢0 is an upper solution and by Lemma 3.3  𝑢0≥𝑢1; therefore, both the claims are true for 𝑛=0.
Further, let the claims be true for 𝑛−1, that is, 𝑢𝑛−1 is an upper solution and 𝑢𝑛−1≥𝑢𝑛. Now, we are required to prove that 𝑢𝑛 is an upper solution and 𝑢𝑛+1≤𝑢𝑛. To prove this, let 𝑤=𝑢𝑛−𝑢𝑛−1, then we have âˆ’î€·ğ‘ğ‘¢î…žğ‘›î€¸î…žî€·+ğ‘žğ‘“ğ‘¥,𝑢𝑛,ğ‘ğ‘¢î…žğ‘›î€¸î€ºî€·â‰¥ğ‘(𝑀−𝜆)𝑤−𝑁signğ‘¤î…žî€¸ğ‘ğ‘¤î…žî€».(3.7) Thus, to prove that 𝑢𝑛 is an upper solution, we are required to prove that (𝑀−𝜆)𝑤−𝑁signğ‘¤î…žî€¸ğ‘ğ‘¤î…žâ‰¥0.(3.8) Now, since 𝑤 satisfies âˆ’î€·ğ‘ğ‘¤î…žî€¸î…žî€·+ğœ†ğ‘žğ‘¤=ğ‘ğ‘¢î…žğ‘›âˆ’1î€¸î…žî€·âˆ’ğ‘žğ‘“ğ‘¥,𝑢𝑛−1,ğ‘ğ‘¢î…žğ‘›âˆ’1≤0,ğ‘¤î…ž(0)≥0,𝑤′(1)≤0,(3.9) from Proposition 2.8, we have 𝑤≤0 for 𝜆>0. Now, putting the value of 𝑤 from (2.5) in (3.8), and in view of ℎ=(ğ‘ğ‘¢î…žğ‘›âˆ’1)î…žâˆ’ğ‘žğ‘“(𝑥,𝑢𝑛−1,ğ‘ğ‘¢î…žğ‘›âˆ’1)≤0, we deduce that to prove (3.8) it is sufficient to prove that (𝑀−𝜆)𝑧0−𝑁signğ‘¤î…žî€¸ğ‘ğ‘§î…ž0≤0,(𝑀−𝜆)𝑧1−𝑁signğ‘¤î…žî€¸ğ‘ğ‘§î…ž1,≤0(3.10) for all 𝑥∈[0,1]. Since 𝑧1=𝑧0(1−𝑥), using Remark 2.6, the above inequalities will be true if for all 𝑥∈[0,1] we have (𝑀−𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)ğ‘§î…ž0(𝑥)≤0.(3.11) Which is true (Lemma 2.10). Therefore, (3.8) holds, and hence 𝑢𝑛 is an upper solution.
Now applying Lemma 3.3, we deduce that 𝑢𝑛+1≤𝑢𝑛. 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 𝜆>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 𝜆>0 be such that ∫𝜆≥𝑀(1−𝑁10ğ‘ž(𝑥)𝑑𝑥)−1. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that for all 𝑛∈ℕ, (i)𝑣𝑛 is a lower solution of (1.1).(ii)𝑣𝑛≤𝑣𝑛+1.

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

Proposition 3.7. Assume that (𝐻1), (𝐻2), (𝐻3), and (𝐻4) hold, and let 𝜆>0 such that ∫𝜆≥𝑀(1−𝑁10ğ‘ž(𝑥)𝑑𝑥)−1and for all 𝑥∈[0,1]𝑓𝑥,𝑣0,ğ‘ğ‘£î…ž0−𝑓𝑥,𝑢0,ğ‘ğ‘¢î…ž0𝑢+𝜆0−𝑣0≥0.(3.12) Then for all 𝑛∈ℕ, the functions 𝑢𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛≤𝑢𝑛.

Proof. We define a function â„Žğ‘–î€·(𝑥)=𝑓𝑥,ğ‘£ğ‘–ğ‘ğ‘£î…žğ‘–î€¸î€·âˆ’ğ‘“ğ‘¥,𝑢𝑖,ğ‘ğ‘¢î…žğ‘–î€¸î€·ğ‘¢+𝜆𝑖−𝑣𝑖,𝑖∈ℕ.(3.13) It is easy to see that for all 𝑖∈ℕ, 𝑤𝑖=𝑢𝑖−𝑣𝑖 satisfies the following differential equation: âˆ’î€·ğ‘ğ‘¤î…žğ‘–î€¸î…ž+ğœ†ğ‘žğ‘¤ğ‘–î€½ğ‘“î€·=ğ‘žğ‘¥,𝑣𝑖−1,ğ‘ğ‘£î…žğ‘–âˆ’1−𝑓𝑥,𝑢𝑖−1,ğ‘ğ‘¢î…žğ‘–âˆ’1𝑢+𝜆𝑖−1−𝑣𝑖−1=ğ‘žâ„Žğ‘–âˆ’1.(3.14) Now to prove this proposition again, we use the principle of mathematical induction. For 𝑖=1, we have ℎ0≥0, and 𝑤1 is the solution of (2.1)-(2.2) with 𝐴=0 and 𝐵=0. Using Proposition 2.8, we deduce that 𝑤1≥0, that is, 𝑢1≥𝑣1.
Now, let 𝑛≥2, let â„Žğ‘›âˆ’2≥0, and let 𝑢𝑛−1≥𝑣𝑛−1, then we are required to prove that â„Žğ‘›âˆ’1≥0 and 𝑢𝑛≥𝑣𝑛. First, we show that for all 𝑥∈[0,1] the function â„Žğ‘›âˆ’1 is nonnegative. Indeed, we have â„Žğ‘›âˆ’1=𝑓𝑥,𝑣𝑛−1,ğ‘ğ‘£î…žğ‘›âˆ’1−𝑓𝑥,𝑢𝑛−1,ğ‘ğ‘¢î…žğ‘›âˆ’1𝑢+𝜆𝑛−1−𝑣𝑛−1(≥−𝑀−𝜆)𝑤𝑛−1+𝑁signğ‘¤î…žğ‘›âˆ’1î€¸ğ‘ğ‘¤î…žğ‘›âˆ’1.(3.15) Here 𝑤𝑛−1 is a solution of (2.1) with ℎ(𝑥)=â„Žğ‘›âˆ’2≥0, 𝐴=0, and 𝐵=0. Arguments similar to Proposition 3.4 can be used to prove that â„Žğ‘›âˆ’1≥0. Now, we have â„Žğ‘›âˆ’1≥0, ğ‘¤î…žğ‘›(0)=0, and ğ‘¤î…žğ‘›(1)=0, thus from Proposition 2.8, we deduce that 𝑤𝑛≥0, that is, 𝑢𝑛≥𝑣𝑛.

Lemma 3.8. If 𝑓(𝑥,𝑢,ğ‘ğ‘¢î…ž) satisfies (𝐻1), (𝐻2), and (𝐻5)for all (𝑥,𝑢,ğ‘ğ‘¢î…ž)∈𝐷, |𝑓(𝑥,𝑢,ğ‘ğ‘¢î…ž)|≤𝜑(|ğ‘ğ‘¢î…ž|), where 𝜑∶[0,∞)→(0,∞) is continuous and satisfies.∞0𝑑𝑠>𝜑(𝑠)10ğ‘ž(𝑥)𝑑𝑥,(3.16) then there exists 𝑅0>0 such that any solution of âˆ’î€·ğ‘ğ‘¢î…žî€¸î…žî€·+ğ‘žğ‘“ğ‘¥,𝑢,ğ‘ğ‘¢î…žî€¸â‰¥0,0<𝑥<1,ğ‘¢î…ž(0)=0=ğ‘¢î…ž(1)(3.17) with 𝑢∈[𝑣0,𝑢0], for all 𝑥∈[0,1], satisfies â€–ğ‘ğ‘¢î…žâ€–âˆž<𝑅0.

Proof. Consider an interval [𝑥,𝑥0]⊂[0,1] such that ∀𝑠∈𝑥,𝑥0,ğ‘¢î…ž(𝑠)<0,ğ‘¢î…žî€·ğ‘¥0=0.(3.18) Now using (𝐻5), we have î€·ğ‘ğ‘¢î…žî€¸î…žî€·||â‰¤ğ‘žğœ‘ğ‘ğ‘¢î…ž||,(3.19) and after integrating it from 𝑥 to 𝑥0 and using (𝐻5), we have âˆ’ğ‘ğ‘¢î…žâ‰¤ğ‘…0.(3.20) Similarly for the interval [𝑥0,𝑥], we have ğ‘ğ‘¢î…žâ‰¤ğ‘…0.(3.21) Thus â€–â€–ğ‘ğ‘¢î…žâ€–â€–âˆžâ‰¤ğ‘…0.(3.22) In the same way, we can prove the following result for lower solutions.

Lemma 3.9. If 𝑓(𝑥,𝑣,ğ‘ğ‘£î…ž) satisfies (𝐻1), (𝐻2), and (𝐻5), then there exists 𝑅0>0 such that any solution of âˆ’î€·ğ‘ğ‘£î…žî€¸î…žî€·+ğ‘žğ‘“ğ‘¥,𝑣,ğ‘ğ‘£î…žî€¸â‰¤0,0<𝑥<1,ğ‘£î…ž(0)=0=ğ‘£î…ž(1)(3.23) with 𝑣∈[𝑣0,𝑢0], for all 𝑥∈[0,1], satisfies â€–ğ‘ğ‘£î…žâ€–âˆž<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 𝜆>0 be such that 𝜆≥𝑀1−𝑁10î‚¶ğ‘ž(𝑥)𝑑𝑥−1,(3.24) and for all 𝑥∈[0,1], 𝑓𝑥,𝑣0,ğ‘ğ‘£î…ž0−𝑓𝑥,𝑢0,ğ‘ğ‘¢î…ž0𝑢+𝜆0−𝑣0≥0.(3.25) Then the sequences {𝑢𝑛} and {𝑣𝑛} defined by (1.2) converge monotonically to solutions ̃𝑢(𝑥) and ̃𝑣(𝑥) of (1.1). Any solution 𝑧(𝑥) of (1.1) in 𝐷 satisfies ̃𝑣(𝑥)≤𝑧(𝑥)≤̃𝑢(𝑥).(3.26)

Proof. Using Lemma 3.3 to Lemma 3.9 and Proposition 3.4 to Proposition 3.7, we deduce that the sequences {𝑢𝑛} and {𝑣𝑛} are monotonic (𝑢0≥𝑢1≥𝑢2⋯≥𝑢𝑛≥𝑣𝑛⋯≥𝑣2≥𝑣1≥𝑣0) and are bounded by 𝑣0 and 𝑢0 in 𝐶[0,1], and by Dini's theorem, they converge uniformly to ̃𝑢 and ̃𝑣 (say). We can also deduce that the sequences {ğ‘ğ‘¢î…žğ‘›} and {ğ‘ğ‘£î…žğ‘›} are uniformly bounded and equicontinuous in 𝐶[0,1], and by Arzela-Ascoli theorem, there exists uniformly convergent subsequences {ğ‘ğ‘¢î…žğ‘›ğ‘˜} and {ğ‘ğ‘£î…žğ‘›ğ‘˜} in 𝐶[0,1]. It is easy to observe that 𝑢𝑛→̃𝑢 and 𝑣𝑛→̃𝑣 imply ğ‘ğ‘¢î…žğ‘›â†’ğ‘Ìƒğ‘¢î…ž and ğ‘Ìƒğ‘£î…žğ‘›Ìƒğ‘£â†’ğ‘î…ž.
Solution of (1.2) is given by (2.5) where ℎ(𝑥)=−𝑓(𝑥,𝑦𝑛−1,ğ‘ğ‘¦î…žğ‘›âˆ’1)+𝜆𝑦𝑛−1. Since the sequences are uniformly convergent taking limit as ğ‘›â†’âˆž, we get ̃𝑢 and ̃𝑣 as the solutions of the nonlinear boundary value problem (1.1). Any solution 𝑧(𝑥) in 𝐷 plays the role of 𝑢0. Hence, ̃𝑧(𝑥)≥𝑣(𝑥). Similarly 𝑧(𝑥)≤̃𝑢(𝑥).

Remark 3.11. When the source function is derivative independent, that is, 𝑁=0, in this case we can choose 𝜆=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,𝑢0(𝑥)≤𝑣0(𝑥).(4.1) For this, we require opposite one-sided Lipschitz condition, and we assume that (𝐹1)there exists upper solution (𝑢0) and lower solution (𝑣0) in 𝐶[0,1]∩𝐶2(0,1] such that 𝑢0≤𝑣0 for all 𝑥∈[0,1],(𝐹2)the function 𝑓∶𝐷0→ℝ is continuous on𝐷0∶=𝑥,𝑦,ğ‘ğ‘¦î…žî€¸âˆˆ[]0,1×𝑅×𝑅∶𝑢0≤𝑦≤𝑣0,(4.2)(𝐹3)there exists 𝑀≥0 such that for all (𝑥,̃𝜏,ğ‘ğ‘£î…ž),(𝑥,î‚ğœŽ,ğ‘ğ‘£î…ž)∈𝐷0,𝑓𝑥,î‚ğœŽ,ğ‘ğ‘£î…žî€¸î€·âˆ’ğ‘“ğ‘¥,̃𝜏,ğ‘ğ‘£î…žî€¸î€·î€¸,î€·î€¸â‰¥âˆ’ğ‘€î‚ğœŽâˆ’ÌƒğœÌƒğœâ‰¤î‚ğœŽ,(4.3)(𝐹4)there exist 𝑁≥0 such that for all (𝑥,𝑢,ğ‘ğ‘£î…ž1)(𝑥,𝑢,ğ‘ğ‘£î…ž2)∈𝐷0,||𝑓𝑥,𝑢,ğ‘ğ‘£î…ž1−𝑓𝑥,𝑢,ğ‘ğ‘£î…ž2||||â‰¤ğ‘ğ‘ğ‘£î…ž2âˆ’ğ‘ğ‘£î…ž1||.(4.4) Here again we define the approximation scheme by (1.2) and use the Antimaximum principle. We make a good choice of 𝜆 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 −𝜆0<𝜆<0. If 𝑢𝑛 is an upper solution of (1.1) and 𝑢𝑛+1 is defined by (1.2), then 𝑢𝑛+1≥𝑢𝑛.

Proof. Let 𝑤𝑛=𝑢𝑛+1−𝑢𝑛, then âˆ’î€·ğ‘ğ‘¤î…žğ‘›î€¸î…ž+ğœ†ğ‘žğ‘¤ğ‘›=î€·ğ‘ğ‘¢î…žğ‘›î€¸î…žî€·âˆ’ğ‘žğ‘“ğ‘¥,𝑢𝑛,ğ‘ğ‘¢î…žğ‘›î€¸ğ‘¤â‰¤0,î…žğ‘›(0)≥0,ğ‘¤î…žğ‘›(1)≤0,(4.5) and using Proposition 2.8, we have 𝑢𝑛+1≥𝑢𝑛.

Proposition 4.2. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let −𝜆0<𝜆<0 be such that ∫−(10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−1<𝜆≤−𝑀 and ∫(𝑀+𝜆)(1+𝜆10∫(1/𝑝(𝑥))𝑥0âˆ«ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−𝑁𝜆10ğ‘ž(𝑥)𝑑𝑥≤0. Then the functions 𝑢𝑛+1 defined recursively by (1.2) are such that, for all 𝑛∈ℕ, (i)𝑢𝑛 is an upper solution of (1.1);(ii)𝑢𝑛+1≥𝑢𝑛.

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 −𝜆0<𝜆<0. If 𝑣𝑛 is a lower solution of (1.1) and 𝑣𝑛+1 is defined by (1.2), then 𝑣𝑛≥𝑣𝑛+1.

Proposition 4.4. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let −𝜆0<𝜆<0 be such that ∫−(10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−1<𝜆≤−𝑀 and ∫(𝑀+𝜆)(1+𝜆10∫(1/𝑝(𝑥))𝑥0âˆ«ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−𝑁𝜆10ğ‘ž(𝑥)𝑑𝑥≤0. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that, for all 𝑛∈ℕ, (i)𝑣𝑛 is a lower solution of (1.1);(ii)𝑣𝑛≥𝑣𝑛+1.

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

Proposition 4.5. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let −𝜆0<𝜆<0 be such that ∫−(10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−1<𝜆≤−𝑀 and (𝑀+𝜆)1+𝜆101𝑝(𝑥)𝑥0î‚¶î€œğ‘ž(𝑡)𝑑𝑡𝑑𝑥−𝑁𝜆10ğ‘ž(𝑥)𝑑𝑥≤0,(4.6) and for all 𝑥∈[0,1], 𝑓𝑥,𝑣0,ğ‘ğ‘£î…ž0−𝑓𝑥,𝑢0,ğ‘ğ‘¢î…ž0𝑢+𝜆0−𝑣0≥0.(4.7) Then, for all 𝑛∈ℕ, the functions 𝑢𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛≥𝑢𝑛.

Now similar to Lemmas 3.8 and 3.9, we state the following two results. These results establish a bound on 𝑝(𝑥)ğ‘¢î…ž(𝑥) and 𝑝(𝑥)ğ‘£î…ž(𝑥).

Lemma 4.6. If 𝑓(𝑥,𝑢,ğ‘ğ‘¢î…ž) satisfies (𝐹1), (𝐹2), and (𝐹5) for all (𝑥,𝑢,ğ‘ğ‘¢î…ž)∈𝐷0, |𝑓(𝑥,𝑢,ğ‘ğ‘¢î…ž)|≤𝜑(|ğ‘ğ‘¢î…ž|), where 𝜑∶[0,∞)→(0,∞) is continuous and satisfies∞0𝑑𝑠>𝜑(𝑠)10ğ‘ž(𝑥)𝑑𝑥,(4.8) then there exists 𝑅0>0 such that any solution of âˆ’î€·ğ‘ğ‘¢î…žî€¸î…žî€·+ğ‘žğ‘“ğ‘¥,𝑢,ğ‘ğ‘¢î…žî€¸â‰¥0,0<𝑥<1,ğ‘¢î…ž(0)=0=ğ‘¢î…ž(1)(4.9) with 𝑢∈[𝑢0,𝑣0], for all 𝑥∈[0,1], satisfies â€–ğ‘ğ‘¢î…žâ€–âˆž<𝑅0.

Lemma 4.7. If 𝑓(𝑥,𝑣,ğ‘ğ‘£î…ž) satisfies (𝐹1), (𝐹2), and (𝐹5), then there exists 𝑅0>0 such that any solution of âˆ’î€·ğ‘ğ‘£î…žî€¸î…žî€·+ğ‘žğ‘“ğ‘¥,𝑣,ğ‘ğ‘£î…žî€¸â‰¤0,0<𝑥<1,ğ‘£î…ž(0)=0=ğ‘£î…ž(1)(4.10) with 𝑣∈[𝑢0,𝑣0], for all 𝑥∈[0,1], satisfies â€–ğ‘ğ‘£â€²â€–âˆž<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 −𝜆0<𝜆<0 be such that ∫−(10∫(1/𝑝(𝑥))𝑥0ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−1<𝜆≤−𝑀 and ∫(𝑀+𝜆)(1+𝜆10∫(1/𝑝(𝑥))𝑥0âˆ«ğ‘ž(𝑡)𝑑𝑡𝑑𝑥)−𝑁𝜆10ğ‘ž(𝑥)𝑑x≤0, and for all 𝑥∈[0,1], 𝑓𝑥,𝑣0,ğ‘ğ‘£î…ž0−𝑓𝑥,𝑢0,ğ‘ğ‘¢î…ž0𝑢+𝜆0−𝑣0≥0.(4.11)
Then the sequences {𝑢𝑛} and {𝑣𝑛} defined by (1.2) converge monotonically to solutions ̃𝑢(𝑥) and ̃𝑣(𝑥) of (1.1). Any solution 𝑧(𝑥) of (1.1) in 𝐷0 satisfies ̃̃𝑢(𝑥)≤𝑧(𝑥)≤𝑣(𝑥).(4.12)

Proof. Using Lemma 4.1 to Lemma 4.7 and Proposition 4.2 to Proposition 4.5, we deduce that 𝑢0≤𝑢1≤𝑢2≤⋯≤𝑢𝑛≤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 𝜆=−𝑀.

5. Conclusion

We establish some existence results under quite general conditions on 𝑝(𝑥), ğ‘ž(𝑥), and 𝑓(𝑥,𝑦,ğ‘ğ‘¦î…ž). We prove some fundamental differential inequalities which enables us to prove the monotonicity of the sequences {𝑢𝑛} and {𝑣𝑛}. For this we have analyzed the singular differential equation −(ğ‘ğ‘¦î…ž)+ğœ†ğ‘žğ‘¦=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.