Abstract

We consider the following class of nonlinear singular differential equation (𝑝(𝑥)𝑦(𝑥))+𝑞(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 𝑞(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 (𝑣0𝑢0) or the other (𝑢0𝑣0).

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, 𝑢0𝑣0. As far as reverse-ordered upper and lower solutions are considered, that is, 𝑢0𝑣0, 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)𝑦(𝑥)+𝑞(𝑥)𝑓𝑥,𝑦(𝑥),𝑝(𝑥)𝑦𝑦(𝑥)=0,0<𝑥<1,(0)=0,𝑦(1)=0.(1.1)(A1)Let 𝑝(𝑥) satisfy the following conditions. (i)𝑝(0)=0 and 𝑝>0 in (0,1). (ii)𝑝𝐶[0,1]𝐶1(0,1). (A2) Let 𝑞(𝑥) satisfy the following conditions. (i)𝑞(𝑥)>0 in (0,1) and 𝑞(𝑥)𝐶(0,1]. (ii)10𝑞(𝑥)𝑑𝑥<. (iii)lim𝑥0(𝑞(𝑥))/(𝑝(𝑥))=0. (iv)10(1/𝑝(𝑥))(𝑥0𝑞(𝑠)𝑑𝑠)1/2𝑑𝑥<.

In this paper, we consider a computationally simple iterative scheme defined by𝑝𝑦𝑛+𝜆𝑞𝑦𝑛=𝑞𝑓𝑥,𝑦𝑛1,𝑝𝑦𝑛1+𝜆𝑞𝑦𝑛1𝑦,0<𝑥<1,𝑛(0)=0,𝑦𝑛(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 (𝑥)𝐶[0,1], and let 𝜆0(0={0}), let 𝐴 and let 𝐵. Now, consider the following class of linear singular problems:𝑝(𝑥)𝑦(𝑥)𝑦+𝜆𝑞(𝑥)𝑦(𝑥)=𝑞(𝑥)(𝑥),0<𝑥<1,(2.1)(0)=𝐴,𝑦(1)=𝐵.(2.2) The corresponding homogeneous system (eigenvalue problem) is given by𝑝(𝑥)𝑦(𝑥)𝑦+𝜆𝑞(𝑥)𝑦(𝑥)=0,0<𝑥<1,(2.3)(0)=0,𝑦(1)=0.(2.4) The solution of the nonhomogeneous problem (2.1)-(2.2) can be written as follows:𝑤(𝑥)=𝑧1(𝑥)𝑥0𝑞(𝑡)(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝐴𝑑𝑡+𝑧1(0)+𝑧0(𝑥)1𝑥𝑞(𝑡)(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝐵𝑑𝑡+𝑧0(1),(2.5) where 𝑧0(𝑥,𝜆) is the solution of𝑝(𝑥)𝑧0(𝑥)+𝜆𝑞(𝑥)𝑧0(𝑥)=0,0<𝑥<1,𝑧0(0)=1,𝑧0(0)=0,(2.6)𝑧1(𝑥,𝜆) is the solution of𝑝(𝑥)𝑧1(𝑥)+𝜆𝑞(𝑥)𝑧1(𝑥)=0,0<𝑥<1,𝑧1(1)=1,𝑧1(1)=0,(2.7) and 𝑊𝑝(𝑧1,𝑧0)=𝑝(𝑡)(𝑧1𝑧0𝑧1𝑧0). By replacing 𝑥 with 1𝑥 in (2.6), it is easy to verify that 𝑧1(𝑥)=𝑧0(1𝑥),(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 𝐿2𝑞(0,1) be a Hilbert space with inner product defined by 𝑓,𝑔=10𝑞(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 𝐿2𝑞(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 𝑧0(1,𝜆). Since 𝑧0(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 𝑧0(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(1𝑥), it is easy to prove that if 𝜆>0 then for all 𝑥(0,1], 𝑧0(𝑥)>1, and 𝑧0(𝑥)>0 and for all 𝑥[0,1), we have 𝑧1(𝑥)>1 and 𝑧1(𝑥)<0.

Remark 2.5. Using Remark 2.3, 𝑧1(𝑥)=𝑧0(1𝑥), 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 𝑧1(𝑥)>0 and for all 𝑥(0,1], we have 𝑧0(𝑥)<0 and 𝑧1(𝑥)>0.

Remark 2.6. Let 𝜆>0 and let 𝐶[0,1]. If 0(or0), then 𝑥0𝑞(𝑡)(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡,1𝑥𝑞(𝑡)(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡(2.10) are nonnegative (or nonpositive).

Remark 2.7. Let 𝜆0<𝜆<0 and let 𝐶[0,1]. If 0(or0), then 𝑥0𝑞(𝑡)(𝑡)𝑧0(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡,1𝑥𝑞(𝑡)(𝑡)𝑧1(𝑡)𝑊𝑝𝑧1,𝑧0𝑑𝑡(2.11) are nonpositive (or nonnegative).

Proposition 2.8 (Maximum Principle). Let 𝜆>0. If 𝐴0, 𝐵0 (or 𝐴0,𝐵0) and 𝐶[0,1] is such that 0 (or 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 𝐴0, 𝐵0 (or𝐴0,𝐵0) and 𝐶[0,1] is such that 0(or 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+. If 𝜆>0 is such that 𝜆𝑀1𝑁10𝑞(𝑥)𝑑𝑥1,(2.12) then for all 𝑥[0,1], (𝑀𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)𝑧0(𝑥)0.(2.13)

Proof. Integrating (2.6) from 0 to 𝑥 and using the fact that 𝑧0(𝑥)>0 in (0,1], we get 𝑝(𝑥)𝑧0(𝑥)𝜆𝑧0(𝑥)10𝑞(𝑥)𝑑𝑥.(2.14) Therefore, we get (𝑀𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)𝑧0(𝑥)(𝑀𝜆)𝑧0+𝑁𝜆𝑧0(𝑥)10𝑞(𝑥)𝑑𝑥. Hence, (2.13) will hold if (𝑀𝜆)+𝑁𝜆10𝑞(𝑥)𝑑𝑥0. Hence the result.

Lemma 2.11. Let 𝑀 and 𝑁+. If 𝜆0<𝜆<0 is such that (10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)1<𝜆𝑀 and (𝑀+𝜆)1+𝜆101𝑝(𝑥)𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥𝑁𝜆10𝑞(𝑥)𝑑𝑥0,(2.15) then for all 𝑥[0,1], (𝑀+𝜆)𝑧0(𝑥)𝑁𝑝(𝑥)𝑧0(𝑥)0.(2.16)

Proof. Using (2.6) and Remark 2.5, it can be deduced that 𝑧0(𝑥) and 𝑝(𝑥)𝑧0(𝑥) are decreasing functions of 𝑥 for 𝜆0<𝜆<0, thus (𝑀+𝜆)𝑧0(𝑥)𝑁𝑝(𝑥)𝑧0(𝑥)(𝑀+𝜆)𝑧0(1)𝑁𝑝(1)𝑧0(1).(2.17) Now using (2.6), we get 𝑝(1)𝑧0(1)(𝜆)10𝑞(𝑥)𝑑𝑥 and 𝑧0(1)>1+𝜆10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥. This completes the proof.

Note 2. In Lemma 2.11, we arrive at the integral 10(1/𝑝(𝑥))𝑥0𝑞(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 𝑢0𝐶[0,1]𝐶2(0,1] is an upper solution of (1.1) if 𝑝𝑢0+𝑞𝑓𝑥,𝑢0,𝑝𝑢00,0<𝑥<1,𝑢0(0)0𝑢0(1).(3.1)

Definition 3.2. A function 𝑣0𝐶[0,1]𝐶2(0,1] is a lower solution of (1.1) if 𝑝𝑣0+𝑞𝑓𝑥,𝑣0,𝑝𝑣00,0<𝑥<1,𝑣0(0)0𝑣0(1).(3.2) Now, for every 𝑛, the problem (1.2) has a unique solution 𝑦𝑛+1 given by (2.5) with (𝑥)=𝑓(𝑥,𝑦𝑛,𝑝𝑦𝑛)+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+1𝑢𝑛.

Proof. Let 𝑤𝑛=𝑢𝑛+1𝑢𝑛, then 𝑝𝑤𝑛+𝜆𝑞𝑤𝑛=𝑝𝑢𝑛𝑞𝑓𝑥,𝑢𝑛,𝑝𝑢𝑛𝑤0,𝑛(0)0,𝑤𝑛(1)0,(3.3) and using Proposition 2.8, we have 𝑢𝑛+1𝑢𝑛.

Proposition 3.4. Assume that (𝐻1) there exist upper solution (𝑢0) and lower solution (𝑣0) in 𝐶[0,1]𝐶2(0,1] such that 𝑣0𝑢0 for all 𝑥[0,1],(𝐻2)the function 𝑓𝐷 is continuous on𝐷=𝑥,𝑦,𝑝𝑦[]0,1×𝑅×𝑅𝑣0𝑦𝑢0,(3.4)(𝐻3)there exists 𝑀0 such that for all (𝑥,𝜏,𝑝𝑣),(𝑥,𝜎,𝑝𝑣)𝐷,𝑓𝑥,𝜏,𝑝𝑣𝑓𝑥,𝜎,𝑝𝑣𝑀(𝜏𝜎),(𝜏𝜎),(3.5)(𝐻4)there exist 𝑁0 such that for all (𝑥,𝑢,𝑝𝑣1)(𝑥,𝑢,𝑝𝑣2)𝐷,||𝑓𝑥,𝑢,𝑝𝑣1𝑓𝑥,𝑢,𝑝𝑣2||||𝑁𝑝𝑣2𝑝𝑣1||.(3.6) Let 𝜆>0 be such that 𝜆𝑀(1𝑁10𝑞(𝑥)𝑑𝑥)1. Then the functions 𝑢𝑛+1 defined recursively by (1.2) are such that, for all 𝑛, (i)𝑢𝑛 is an upper solution of (1.1). (ii)𝑢𝑛+1𝑢𝑛.

Proof. We prove the claims by the principle of mathematical induction. Since 𝑢0 is an upper solution and by Lemma 3.3  𝑢0𝑢1; therefore, both the claims are true for 𝑛=0.
Further, let the claims be true for 𝑛1, that is, 𝑢𝑛1 is an upper solution and 𝑢𝑛1𝑢𝑛. Now, we are required to prove that 𝑢𝑛 is an upper solution and 𝑢𝑛+1𝑢𝑛. To prove this, let 𝑤=𝑢𝑛𝑢𝑛1, then we have 𝑝𝑢𝑛+𝑞𝑓𝑥,𝑢𝑛,𝑝𝑢𝑛𝑝(𝑀𝜆)𝑤𝑁sign𝑤𝑝𝑤.(3.7) Thus, to prove that 𝑢𝑛 is an upper solution, we are required to prove that (𝑀𝜆)𝑤𝑁sign𝑤𝑝𝑤0.(3.8) Now, since 𝑤 satisfies 𝑝𝑤+𝜆𝑞𝑤=𝑝𝑢𝑛1𝑞𝑓𝑥,𝑢𝑛1,𝑝𝑢𝑛10,𝑤(0)0,𝑤(1)0,(3.9) from Proposition 2.8, we have 𝑤0 for 𝜆>0. Now, putting the value of 𝑤 from (2.5) in (3.8), and in view of =(𝑝𝑢𝑛1)𝑞𝑓(𝑥,𝑢𝑛1,𝑝𝑢𝑛1)0, we deduce that to prove (3.8) it is sufficient to prove that (𝑀𝜆)𝑧0𝑁sign𝑤𝑝𝑧00,(𝑀𝜆)𝑧1𝑁sign𝑤𝑝𝑧1,0(3.10) for all 𝑥[0,1]. Since 𝑧1=𝑧0(1𝑥), using Remark 2.6, the above inequalities will be true if for all 𝑥[0,1] we have (𝑀𝜆)𝑧0(𝑥)+𝑁𝑝(𝑥)𝑧0(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+1𝑢𝑛. 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 𝑣𝑛𝑣𝑛+1.

Proposition 3.6. Assume that (𝐻1), (𝐻2), (𝐻3), and (𝐻4) hold, and let 𝜆>0 be such that 𝜆𝑀(1𝑁10𝑞(𝑥)𝑑𝑥)1. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that for all 𝑛, (i)𝑣𝑛 is a lower solution of (1.1).(ii)𝑣𝑛𝑣𝑛+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 𝜆𝑀(1𝑁10𝑞(𝑥)𝑑𝑥)1and for all 𝑥[0,1]𝑓𝑥,𝑣0,𝑝𝑣0𝑓𝑥,𝑢0,𝑝𝑢0𝑢+𝜆0𝑣00.(3.12) Then for all 𝑛, the functions 𝑢𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛𝑢𝑛.

Proof. We define a function 𝑖(𝑥)=𝑓𝑥,𝑣𝑖𝑝𝑣𝑖𝑓𝑥,𝑢𝑖,𝑝𝑢𝑖𝑢+𝜆𝑖𝑣𝑖,𝑖.(3.13) It is easy to see that for all 𝑖, 𝑤𝑖=𝑢𝑖𝑣𝑖 satisfies the following differential equation: 𝑝𝑤𝑖+𝜆𝑞𝑤𝑖𝑓=𝑞𝑥,𝑣𝑖1,𝑝𝑣𝑖1𝑓𝑥,𝑢𝑖1,𝑝𝑢𝑖1𝑢+𝜆𝑖1𝑣𝑖1=𝑞𝑖1.(3.14) Now to prove this proposition again, we use the principle of mathematical induction. For 𝑖=1, we have 00, and 𝑤1 is the solution of (2.1)-(2.2) with 𝐴=0 and 𝐵=0. Using Proposition 2.8, we deduce that 𝑤10, that is, 𝑢1𝑣1.
Now, let 𝑛2, let 𝑛20, and let 𝑢𝑛1𝑣𝑛1, then we are required to prove that 𝑛10 and 𝑢𝑛𝑣𝑛. First, we show that for all 𝑥[0,1] the function 𝑛1 is nonnegative. Indeed, we have 𝑛1=𝑓𝑥,𝑣𝑛1,𝑝𝑣𝑛1𝑓𝑥,𝑢𝑛1,𝑝𝑢𝑛1𝑢+𝜆𝑛1𝑣𝑛1(𝑀𝜆)𝑤𝑛1+𝑁sign𝑤𝑛1𝑝𝑤𝑛1.(3.15) Here 𝑤𝑛1 is a solution of (2.1) with (𝑥)=𝑛20, 𝐴=0, and 𝐵=0. Arguments similar to Proposition 3.4 can be used to prove that 𝑛10. Now, we have 𝑛10, 𝑤𝑛(0)=0, and 𝑤𝑛(1)=0, thus from Proposition 2.8, we deduce that 𝑤𝑛0, that is, 𝑢𝑛𝑣𝑛.

Lemma 3.8. If 𝑓(𝑥,𝑢,𝑝𝑢) satisfies (𝐻1), (𝐻2), and (𝐻5)for all (𝑥,𝑢,𝑝𝑢)𝐷, |𝑓(𝑥,𝑢,𝑝𝑢)|𝜑(|𝑝𝑢|), where 𝜑[0,)(0,) is continuous and satisfies.0𝑑𝑠>𝜑(𝑠)10𝑞(𝑥)𝑑𝑥,(3.16) then there exists 𝑅0>0 such that any solution of 𝑝𝑢+𝑞𝑓𝑥,𝑢,𝑝𝑢0,0<𝑥<1,𝑢(0)=0=𝑢(1)(3.17) with 𝑢[𝑣0,𝑢0], for all 𝑥[0,1], satisfies 𝑝𝑢<𝑅0.

Proof. Consider an interval [𝑥,𝑥0][0,1] such that 𝑠𝑥,𝑥0,𝑢(𝑠)<0,𝑢𝑥0=0.(3.18) Now using (𝐻5), we have 𝑝𝑢||𝑞𝜑𝑝𝑢||,(3.19) and after integrating it from 𝑥 to 𝑥0 and using (𝐻5), we have 𝑝𝑢𝑅0.(3.20) Similarly for the interval [𝑥0,𝑥], we have 𝑝𝑢𝑅0.(3.21) Thus 𝑝𝑢𝑅0.(3.22) In the same way, we can prove the following result for lower solutions.

Lemma 3.9. If 𝑓(𝑥,𝑣,𝑝𝑣) satisfies (𝐻1), (𝐻2), and (𝐻5), then there exists 𝑅0>0 such that any solution of 𝑝𝑣+𝑞𝑓𝑥,𝑣,𝑝𝑣0,0<𝑥<1,𝑣(0)=0=𝑣(1)(3.23) with 𝑣[𝑣0,𝑢0], for all 𝑥[0,1], satisfies 𝑝𝑣<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 𝜆𝑀1𝑁10𝑞(𝑥)𝑑𝑥1,(3.24) and for all 𝑥[0,1], 𝑓𝑥,𝑣0,𝑝𝑣0𝑓𝑥,𝑢0,𝑝𝑢0𝑢+𝜆0𝑣00.(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 (𝑢0𝑢1𝑢2𝑢𝑛𝑣𝑛𝑣2𝑣1𝑣0) 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 {𝑝𝑢𝑛} and {𝑝𝑣𝑛} are uniformly bounded and equicontinuous in 𝐶[0,1], and by Arzela-Ascoli theorem, there exists uniformly convergent subsequences {𝑝𝑢𝑛𝑘} and {𝑝𝑣𝑛𝑘} in 𝐶[0,1]. It is easy to observe that 𝑢𝑛̃𝑢 and 𝑣𝑛̃𝑣 imply 𝑝𝑢𝑛𝑝̃𝑢 and 𝑝̃𝑣𝑛̃𝑣𝑝.
Solution of (1.2) is given by (2.5) where (𝑥)=𝑓(𝑥,𝑦𝑛1,𝑝𝑦𝑛1)+𝜆𝑦𝑛1. Since the sequences are uniformly convergent taking limit as 𝑛, 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 𝑢0𝑣0 for all 𝑥[0,1],(𝐹2)the function 𝑓𝐷0 is continuous on𝐷0=𝑥,𝑦,𝑝𝑦[]0,1×𝑅×𝑅𝑢0𝑦𝑣0,(4.2)(𝐹3)there exists 𝑀0 such that for all (𝑥,̃𝜏,𝑝𝑣),(𝑥,𝜎,𝑝𝑣)𝐷0,𝑓𝑥,𝜎,𝑝𝑣𝑓𝑥,̃𝜏,𝑝𝑣,𝑀𝜎̃𝜏̃𝜏𝜎,(4.3)(𝐹4)there exist 𝑁0 such that for all (𝑥,𝑢,𝑝𝑣1)(𝑥,𝑢,𝑝𝑣2)𝐷0,||𝑓𝑥,𝑢,𝑝𝑣1𝑓𝑥,𝑢,𝑝𝑣2||||𝑁𝑝𝑣2𝑝𝑣1||.(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+1𝑢𝑛.

Proof. Let 𝑤𝑛=𝑢𝑛+1𝑢𝑛, then 𝑝𝑤𝑛+𝜆𝑞𝑤𝑛=𝑝𝑢𝑛𝑞𝑓𝑥,𝑢𝑛,𝑝𝑢𝑛𝑤0,𝑛(0)0,𝑤𝑛(1)0,(4.5) and using Proposition 2.8, we have 𝑢𝑛+1𝑢𝑛.

Proposition 4.2. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let 𝜆0<𝜆<0 be such that (10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)1<𝜆𝑀 and (𝑀+𝜆)(1+𝜆10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)𝑁𝜆10𝑞(𝑥)𝑑𝑥0. Then the functions 𝑢𝑛+1 defined recursively by (1.2) are such that, for all 𝑛, (i)𝑢𝑛 is an upper solution of (1.1);(ii)𝑢𝑛+1𝑢𝑛.

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 𝑣𝑛𝑣𝑛+1.

Proposition 4.4. Assume that (𝐹1), (𝐹2), (𝐹3), and (𝐹4) hold. Let 𝜆0<𝜆<0 be such that (10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)1<𝜆𝑀 and (𝑀+𝜆)(1+𝜆10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)𝑁𝜆10𝑞(𝑥)𝑑𝑥0. Then the functions 𝑣𝑛+1 defined recursively by (1.2) are such that, for all 𝑛, (i)𝑣𝑛 is a lower solution of (1.1);(ii)𝑣𝑛𝑣𝑛+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))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)1<𝜆𝑀 and (𝑀+𝜆)1+𝜆101𝑝(𝑥)𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥𝑁𝜆10𝑞(𝑥)𝑑𝑥0,(4.6) and for all 𝑥[0,1], 𝑓𝑥,𝑣0,𝑝𝑣0𝑓𝑥,𝑢0,𝑝𝑢0𝑢+𝜆0𝑣00.(4.7) Then, for all 𝑛, the functions 𝑢𝑛 and 𝑣𝑛 defined recursively by (1.2) satisfy 𝑣𝑛𝑢𝑛.

Now similar to Lemmas 3.8 and 3.9, we state the following two results. These results establish a bound on 𝑝(𝑥)𝑢(𝑥) and 𝑝(𝑥)𝑣(𝑥).

Lemma 4.6. If 𝑓(𝑥,𝑢,𝑝𝑢) satisfies (𝐹1), (𝐹2), and (𝐹5) for all (𝑥,𝑢,𝑝𝑢)𝐷0, |𝑓(𝑥,𝑢,𝑝𝑢)|𝜑(|𝑝𝑢|), where 𝜑[0,)(0,) is continuous and satisfies0𝑑𝑠>𝜑(𝑠)10𝑞(𝑥)𝑑𝑥,(4.8) then there exists 𝑅0>0 such that any solution of 𝑝𝑢+𝑞𝑓𝑥,𝑢,𝑝𝑢0,0<𝑥<1,𝑢(0)=0=𝑢(1)(4.9) with 𝑢[𝑢0,𝑣0], for all 𝑥[0,1], satisfies 𝑝𝑢<𝑅0.

Lemma 4.7. If 𝑓(𝑥,𝑣,𝑝𝑣) satisfies (𝐹1), (𝐹2), and (𝐹5), then there exists 𝑅0>0 such that any solution of 𝑝𝑣+𝑞𝑓𝑥,𝑣,𝑝𝑣0,0<𝑥<1,𝑣(0)=0=𝑣(1)(4.10) with 𝑣[𝑢0,𝑣0], for all 𝑥[0,1], satisfies 𝑝𝑣<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))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)1<𝜆𝑀 and (𝑀+𝜆)(1+𝜆10(1/𝑝(𝑥))𝑥0𝑞(𝑡)𝑑𝑡𝑑𝑥)𝑁𝜆10𝑞(𝑥)𝑑x0, and for all 𝑥[0,1], 𝑓𝑥,𝑣0,𝑝𝑣0𝑓𝑥,𝑢0,𝑝𝑢0𝑢+𝜆0𝑣00.(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 𝑢0𝑢1𝑢2𝑢𝑛𝑣𝑛𝑣1𝑣0.(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), 𝑞(𝑥), and 𝑓(𝑥,𝑦,𝑝𝑦). 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 (𝑝𝑦)+𝜆𝑞𝑦=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.