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International Journal of Differential Equations
Volume 2011 (2011), Article ID 261963, 16 pages
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: Alberto Cabada

Copyright © 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.


We present a constructive approach to establish existence and uniqueness of solution of singular boundary value problem (𝑝(𝑥)𝑦(𝑥))=𝑞(𝑥)𝑓(𝑥,𝑦,𝑝𝑦) for 0<𝑥𝑏,𝑦(0)=𝑎,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾1. Here 𝑝(𝑥)>0 on (0,𝑏) allowing 𝑝(0)=0. Further 𝑞(𝑥) may be allowed to have integrable discontinuity at 𝑥=0, so the problem may be doubly singular.

1. Introduction

Consider the following singular boundary value problem: 𝑀𝑦𝑝(𝑥)𝑦(𝑥)=𝑞(𝑥)𝑓𝑥,𝑦(𝑥),𝑝(𝑥)𝑦(𝑥),0<𝑥𝑏,(1.1)𝑦(0)=𝑎,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾1,(1.2) where 𝛼1>0, 𝛽10, 𝛾1 is any finite constant. We assume that 𝑝(𝑥) and 𝑞(𝑥) satisfy the following conditions:(A1) 𝑝(𝑥)>0 in (0,𝑏], 𝑝𝐶[0,𝑏]𝐶1(0,𝑏) and 𝑏0(𝑑𝑡/𝑝(𝑡))<,(A2) 𝑞(𝑥)>0 in (0,𝑏] and 𝑏0𝑞(𝑡)𝑑𝑡<.

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:𝛼𝑛+1+̃𝑘(𝑥)𝛼𝑛+1+̃𝑙(𝑥)𝛼𝑛+1=𝑓𝑥,𝛼𝑛,𝛼𝑛+̃𝑘(𝑥)𝛼𝑛+̃𝑙(𝑥)𝛼𝑛,𝛼𝑛+1(0)=𝛼𝑛+1(1)=0,(1.3) for the regular boundary value problem 𝑦=𝑓(𝑥,𝑦,𝑦),𝑦(0)=𝑦(1)=0. They also suggest that (1.3) with ̃𝑙(𝑥)=0 or with constant ̃𝑘 and ̃𝑙 does not work for the Dirichlet boundary condition.

Thus, for our problem we consider the following iterative scheme: 𝐿𝑦𝑛+1=𝐹𝑥,𝑦𝑛,𝑝𝑦𝑛𝑦,0<𝑥𝑏,𝑛+1(0)=0,𝛼1𝑦𝑛+1(𝑏)+𝛽1𝑝(𝑏)𝑦𝑛+1(𝑏)=𝛾1,(1.4) where 𝐿𝑦=𝑝(𝑥)𝑦(𝑥)𝜇(𝑥)𝑞(𝑥)𝑝(𝑥)𝑦𝐹(𝑥)𝜆𝑘(𝑥)𝑞(𝑥)𝑦(𝑥),𝑥,𝑦,𝑝𝑦=𝑞(𝑥)𝑓𝑥,𝑦,𝑝𝑦𝜇(𝑥)𝑞(𝑥)𝑝(𝑥)𝑦(𝑥)𝜆𝑘(𝑥)𝑞(𝑥)𝑦(𝑥).(1.5) We assume that 𝑘(𝑥) and 𝜇(𝑥) satisfy the following conditions.(A3)𝑘(𝑥)𝐶[0,𝑏], 𝑚𝑘, 𝑀𝑘𝑅 such that 0<𝑚𝑘𝑘(𝑥)𝑀𝑘.(A4)𝜇(𝑥)𝐿1𝑞(0,𝑏), that is,𝑏0𝑞(𝑥)𝜇(𝑥)𝑑𝑥<.(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 𝑏0(𝑑𝑡/𝑝(𝑡))<, several researchers ([25]) 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, 79]): (i)𝑦𝑓(𝑥,𝑦,0)>0, |𝑦|>𝑀0, where 𝑀0 is a constant ([79]) or(ii)𝑓(𝑥,𝑀1,0)0𝑓(𝑥,𝑀2,0),𝑀1,𝑀20 ([4, 5]).

But such sign restrictions are quite restrictive as the simple differential equation 𝑦=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 𝑥=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 𝑥=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(𝑥)=𝑘(𝑥)𝑞(𝑥) and the inner product defined as𝑢,𝑣=𝑏0𝑠(𝑡)𝑢(𝑡)𝑣(𝑡)𝑑𝑡.(2.1) Conditions on 𝑝, 𝑞, 𝜇, and 𝑘 guarantee that the singular point 𝑥=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<𝜆0<𝜆1<𝜆2<, and let the corresponding eigenfunctions be 𝜓0,𝜓1,𝜓2,, respectively. Next, we transform 𝐿𝑦=0 by changing variable 𝑧=𝑔(𝑥)𝑦, where 𝑔(𝑥)=𝑒𝑥0𝜇(𝑡)𝑞(𝑡)𝑑𝑡, to𝑝(𝑥)𝑧(𝑥)+𝑟(𝑥)𝑧(𝑥)=𝜆𝑠(𝑥)𝑧(𝑥),0<𝑥𝑏,(2.2) where 𝑟(𝑥)=(1/2){𝜇(𝑥)𝑞(𝑥)𝑝(𝑥)}+(1/4){𝜇(𝑥)𝑞(𝑥)}2𝑝(𝑥). Now following the analysis of Theorem 2.7, (i), (ii) and Theorem 2.17 of [11] for the operator (1/𝑠)(𝑟+𝑀), where 𝑀 is defined by (1.1), the following results can be established.

Theorem 2.1. Let 𝑓(𝑥) be the primitive of an absolutely continuous function, and let 1𝑠(𝑟+𝑀)𝑓𝐿2𝑠𝑓(0,𝑏),(𝑏)sin𝛼𝑝(𝑏)𝑓(𝑏)cos𝛼=0,where𝛼isreal,lim𝑥0𝑝(𝑥)𝑊𝑥[]𝑓,𝜓=0,(2.3) for every nonreal 𝜆, where 𝜓(𝑥,𝜆)𝐿2𝑠(0,𝑏) is a solution of (2.2) and 𝑊[𝑓,𝜓] is the wronskian of 𝑓 and 𝜓. Then 𝑓(𝑥)=𝑛=0𝑐𝑛𝜓𝑛(𝑥),(0𝑥𝑏),(2.4) being the series absolutely and uniformly convergent on [0,𝑏].

Theorem 2.2. Let 𝑓𝐿2𝑠(0,𝑏). Then 𝑏0𝑠(𝑥){𝑓(𝑥)}2𝑑𝑥=𝑛=0𝑐2𝑛.(2.5)

Theorem 2.3. Let 𝑓𝐿2𝑠(0,𝑏), and let Φ(𝑥,𝜆) be the solution of 𝑝(𝑥)𝑧(𝑥)+𝑟(𝑥)𝑧(𝑥)𝜆𝑠(𝑥)𝑧(𝑥)=𝑠(𝑥)𝑓(𝑥),0<𝑥𝑏,(2.6) satisfying 𝛼11𝑧(𝑏)+𝛽1𝑝(𝑏)𝑧(𝑏)=0, where 𝛼11=𝛼1(1/2)𝛽1𝜇(𝑏)𝑝(𝑏)𝑞(𝑏). Then for 𝜆 not equal to any of the values of 𝜆𝑛, one has Φ(𝑥,𝜆)=𝑛=0𝑐𝑛𝜓𝑛𝜆𝜆𝑛,(2.7) where the series is absolutely convergent.

Remark 2.4. Since 𝑠 on 𝐿2𝑠(0,𝑏) is equivalent to 𝑞 on 𝐿2𝑞(0,𝑏), we can apply Theorems 2.12.3 in 𝐿2𝑞(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(𝑥), 𝑞(𝑥), 𝑘(𝑥), and 𝜇(𝑥) satisfy (A1), (A2), (A3), and (A4), respectively. Then the initial value problems (IVPs) 𝐿𝑦=0,0<𝑥𝑏,𝑦(0)=𝑎0,lim𝑥0+𝑝(𝑥)𝑦(𝑥)=𝑏0,(3.1)𝐿𝑦=0,0<𝑥𝑏,𝑦(𝑏)=𝑐0,𝑝(𝑏)𝑦(𝑏)=𝑑0(3.2) have a solution in 𝐿2𝑠(0,𝑏) or equivalently in 𝐿2𝑞(0,𝑏) (Remark 2.4).

3.1. Green's Function

Green's function 𝐺(𝑥,𝑡,𝜆) for the differential operator 𝐿 can be defined as1𝐺(𝑥,𝑡,𝜆)=𝑝(𝑥)𝑊(𝜓,𝜙)𝜓(𝑡)𝜙(𝑥),if0<𝑡𝑥,𝜓(𝑥)𝜙(𝑡),if𝑥𝑡𝑏,(3.3) where 𝜙=𝑆{𝛼1𝑦1(𝑥)+𝛽1𝑦2(𝑥)}, 𝑆=1/𝛼21+𝛽21, 𝑦1(𝑏)=0, 𝑝(𝑏)𝑦1(𝑏)=1, 𝑦2(𝑏)=1, 𝑝(𝑏)𝑦2(𝑏)=0, and 𝜓 is a nontrivial solution of IVP (3.1) with 𝑎0=0, 𝑏0=1. From (A5), it is easy to conclude that 𝑝(𝑥)𝑊(𝜓,𝜙)|𝑥=𝑏0; thus, 𝜙 and 𝜓 are linearly independent.

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

Lemma 3.2. If 𝑦(𝑥) satisfies 𝐿𝑦(𝑥)=𝑞(𝑥)𝑓(𝑥)0, for 0<𝑥𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾10, where 𝑝(𝑥), 𝑞(𝑥), 𝜇(𝑥), and 𝑘(𝑥) satisfy (A1), (A2), (A3), and (A4), respectively, then 𝑦(𝑥)0 provided that 𝜆0.

Proof. We divide the proof in two cases as follows.
Case i. When 𝜆<0. On contrary assume that there exists a point 𝑐(0,𝑏) such that 𝑦(𝑐)<0. Then from the continuity of the solutions there exists a point 𝑑(0,𝑏) such that 𝑦(𝑑)<0, 𝑦(0)=0, and 𝑦(𝑑)0. Now at the point 𝑑, we have 𝑝(𝑑)𝑦(𝑑)𝑝(𝑑)𝑦(𝑑)𝜇(𝑑)𝑞(𝑑)𝑝(𝑑)𝑦(𝑑)𝜆𝑘(𝑑)𝑞(𝑑)𝑦(𝑑)<0,(3.4) which is a contradiction. Hence, 𝑦(𝑥)0 when 𝜆<0.
Case ii. When 𝜆=0. Using the same notations as in the previous case, we have 𝑦(𝑥)>0 for 𝑥>𝑑 and 𝑦(𝑥)<0 for 𝑥<𝑑.
Now, we consider the interval [𝑑,𝑥0](0,𝑏) where 𝑦(𝑑)=0, 𝑦<0 in [𝑑,𝑥0], and 𝑦>0 in (𝑑,𝑥0]. Then 𝑃=𝑥0𝑑𝑦𝑝(𝑡)𝑔(𝑡)(𝑡)2𝑠(𝑡)𝑔(𝑡)𝑓(𝑡)𝑦(𝑡)𝑑𝑡>0.(3.5) Integrating the first term by parts, we get 𝑥𝑃=𝑝0𝑔𝑥0𝑦𝑥0𝑦𝑥0<0,(3.6) which is again a contradiction. Thus, 𝑦(𝑥)0 when 𝜆0.

Lemma 3.3. Consider the following differential equation: 𝐿𝑦(𝑥)=0,0<𝑥𝑏,(3.7) where 𝑝(𝑥), 𝑞(𝑥), 𝜇(𝑥), and 𝑘(𝑥) satisfy (A1), (A2), (A3), and (A4), respectively, with the boundary conditions: 𝑦(0)=0,𝛼1𝑝(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾1.(3.8) Then LSBVP (3.7)-(3.8) has a unique solution given by 𝛾𝑦(𝑥)=1𝜓(𝑥)𝛼1𝜓(𝑏)+𝛽1𝑝(𝑏)𝜓(,𝑏)(3.9) provided that 𝜆 is none of the eigenvalues of the corresponding eigenvalue problem and 𝜓 satisfies (3.1). Moreover, 𝑦(𝑥)0 if 𝛾10 and 0<𝜆<𝜆0, where 𝜆0 is the first positive zero of 𝛼1𝜓(𝑏,𝜆)+𝛽1𝑝(𝑏)𝜓(𝑏,𝜆).

Proof. From Theorem 3.1, it is easy to see that the unique solution of (3.7)-(3.8) can be written as 𝛾𝑦(𝑥)=1𝜓(𝑥)𝛼1𝜓(𝑏)+𝛽1𝑝(𝑏)𝜓(,𝑏)(3.10) provided that 𝛼1𝜓(𝑏,𝜆)+𝛽1𝑝(𝑏)𝜓(𝑏,𝜆)0; that is, 𝜆 is none of the eigenvalue of the corresponding eigenvalue problem (Section 2). Since 𝜓(0)=0, lim𝑥0+𝑝(𝑥)𝜓(𝑥)=1, and 𝜓(𝑥,𝜆) does not change sign for 0<𝜆<𝜆0, we get that 𝑦(𝑥)0 for 0<𝜆<𝜆0, provided that 𝛾10.

Lemma 3.4. For the linear differential operator associated with 𝐿𝑦(𝑥)=𝑞(𝑥)𝑓(𝑥),0<𝑥𝑏,(3.11)𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾1,(3.12) with 𝑓𝐿2𝑞(0,𝑏), the generalized Green's function for the corresponding homogeneous boundary value problem is given by 𝐺(𝑥,𝑡,𝜆)=𝑛=0𝜓𝑥,𝜆𝑛𝜓𝑡,𝜆𝑛𝜆𝑛,𝜆(3.13) where 𝜓(𝑥,𝜆𝑖) are the normalized eigenfunctions corresponding to the eigenvalue 𝜆𝑖. 𝐺(𝑥,𝑡,𝜆) satisfies the homogeneous boundary condition provided that 𝜆𝜆0,𝜆1,. Solution of the non-homogeneous LSBVP (3.11)-(3.12) is 𝛾𝑦(𝑥)=1𝜓(𝑥,𝜆)𝛼1𝜓(𝑏)+𝛽1𝑝(𝑏)𝜓+(𝑏)𝑏0𝑞(𝑡)𝑓(𝑡)𝐺(𝑥,𝑡,𝜆)𝑑𝑡.(3.14) The series on the right is absolutely convergent.

Proof. The solution 𝑦(𝑥) 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(𝑥)=1𝜓(𝑥,𝜆)𝛼1𝜓(𝑏,𝜆)+𝛽1𝑝(𝑏)𝜓+(𝑏,𝜆)𝑏0𝑞(𝑡)𝑓(𝑡)𝐺(𝑥,𝑡,𝜆)𝑑𝑡,(3.15) where 𝐺(𝑥,𝑡,𝜆) 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(𝑥,𝑡,𝜆)=𝑛=0𝜓𝑥,𝜆𝑛𝜓𝑡,𝜆𝑛𝜆𝑛,𝜆(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 𝑓𝐿2𝑞(0,𝑏), 𝛾10 and 𝑓0, then solution of (3.11)-(3.12) is nonnegative provided that 0<𝜆<𝜆0.

Proof. We first show that 𝐺(𝑥,𝑡)0 for all 0𝑥,𝑡𝑏 if 0<𝜆<𝜆0. Fixing 𝑡, 𝐺(𝑥,𝑡) satisfies 𝐿𝐺(𝑥,𝑡)=0,0<𝑥𝑡, where 𝜕/𝜕𝑥. Since 𝐺(0,𝑡)=0, 𝛼1𝐺(𝑡,𝑡)+𝛽1𝑝(𝑡)𝐺(𝑡,𝑡)0 for 0<𝜆<𝜆0, from Lemma 3.3  𝐺(𝑥,𝑡)0 for 0𝑥𝑡, provided that 0<𝜆<𝜆0. By the symmetry, continuity, and 𝐺(𝑡,𝑡)0 for 0<𝜆<𝜆0, it follows that 𝐺(𝑥,𝑡)0 for 0𝑥,𝑡𝑏, provided that 0<𝜆<𝜆0. The result follows.

Corollary 3.6. If 𝑦(𝑥) satisfies 𝐿𝑦(𝑥)=𝑞(𝑥)𝑓(𝑥)0 for 0<𝑥𝑏 and 𝑦(0)=0, 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾10, then 𝑦(𝑥)0, provided that 𝜆<𝜆0.

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 𝑝(𝑥)𝑦(𝑥)=𝑞(𝑥)𝑓𝑥,𝑦,𝑝𝑦,0<𝑥𝑏,𝑦(0)=𝑎,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾2(4.1) can be transformed to 𝑝(𝑥)𝑢(𝑥)=𝑞(𝑥)𝑓𝑥,𝑢+𝑎,𝑝𝑢,0<𝑥𝑏,𝑢(0)=0,𝛼1𝑢(𝑏)+𝛽1𝑝(𝑏)𝑢(𝑏)=𝛾2𝑎𝛼1,(4.2) with 𝑢=𝑦𝑎. Further, the functions 𝑓(𝑥,𝑢+𝑎,𝑝𝑢) and 𝑓(𝑥,𝑦,𝑝𝑦) satisfy the same Lipschitz condition, so we may work with the boundary value problem𝑝(𝑥)𝑦(𝑥)=𝑞(𝑥)𝑓𝑥,𝑦,𝑝𝑦,0<𝑥𝑏,𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾1.(4.3) Next, we define upper solution 𝑢0(𝑥) and lower solution 𝑣0(𝑥) such that 𝑢0𝑣0, which work as initial iterates for our constructive approach.

Definition 4.1. A function 𝑢0(𝑥)𝐶[0,𝑏]𝐶2(0,𝑏] is an upper solution if 𝑝(𝑥)𝑢0(𝑥)𝑞(𝑥)𝑓𝑥,𝑢0,𝑝𝑢0𝑢,0<𝑥𝑏,0(0)=0,𝛼1𝑢0(𝑏)+𝛽1𝑝(𝑏)𝑢0(𝑏)𝛾1,(4.4) and a function 𝑣0(𝑥)𝐶[0,𝑏]𝐶2(0,𝑏] is a lower solution if 𝑝(𝑥)𝑣0(𝑥)𝑞(𝑥)𝑓𝑥,𝑣0,𝑝𝑣0𝑣,0<𝑥𝑏,0(0)=0,𝛼1𝑣0(𝑏)+𝛽1𝑝(𝑏)𝑣0(𝑏)𝛾1.(4.5)

Lemma 4.2. If 𝜆<0, 𝜆𝑘(𝑥)𝐾1, |𝜇(𝑥)|𝐿1, 𝐿𝑦0 for 0<𝑥𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)0, then 𝐾1𝜇𝜆𝑘(𝑥)𝑦(𝑥)+𝐿1sign𝑦𝑝𝑦0,0<𝑥𝑏,(4.6) provided that 1+𝜆𝑏0𝑑𝑡𝑝(𝑡)𝑔(𝑡)𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡sup𝑝𝑔𝑏𝑏0𝑑𝑡𝐾𝑝(𝑡)𝑔(𝑡)>0,0<𝑥𝑏,1𝐿𝜆𝑘(𝑥)1||𝜇||Φ(𝑥)(𝑝,𝑞,𝑠,𝑔)0,0<𝑥𝑏,(4.7) hold. Here, Φ(𝑝,𝑞,𝑠,𝑔)=sup𝑝𝑔𝑏𝐼(𝑝,𝑞,𝑠,𝑔)𝜆𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡,(4.8)𝐼(𝑝,𝑞,𝑠,𝑔)=1+𝜆𝑏0𝑑𝑡𝑝(𝑡)𝑔(𝑡)𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡sup𝑝𝑔𝑏𝑏0𝑑𝑡𝑝(𝑡)𝑔(𝑡)1.(4.9)

Proof. The solution of the equation 𝐿𝑦=𝑞𝑓0, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)=𝛾10 is given by (3.15) where 𝐺(𝑥,𝑡,𝜆) is defined by (3.3). Substituting 𝑦(𝑥) from (3.15) into (4.6), it is easy to see that we require the following inequalities in order to complete the proof 𝐾1𝜇𝜆𝑘(𝑥)𝜓(𝑥)+𝐿1sign𝑦𝑝𝜓𝐾0,0<𝑥𝑏,(4.10)1𝜆𝑘(𝑥)𝜙𝜇(𝑥)+𝐿1sign𝑦𝑝𝜙0,0<𝑥𝑏.(4.11)
Here 𝜓 satisfies the IVP at 𝑥=0; that is, 𝐿𝜓=0, 𝜓(0)=0 and lim𝑥0+𝑝(𝑥)𝜓(𝑥)=1, and 𝜙 satisfies the IVP at 𝑥=𝑏; that is, 𝐿𝜙=0, 𝜙(𝑏)=𝛽1 and 𝑝(𝑏)𝜙(𝑏)=𝛼1. The solutions 𝜓 and 𝜙 cannot have either point of maxima (at the point of maxima the 𝐿𝜓=0 or 𝐿𝜙=0 will be contradicted) or point of minima (since to have minima, maxima is bound to occur). So, finally we have 𝜓(𝑥)0 and 𝜙(𝑥)0 on [0,𝑏]. As |𝜇(𝑥)|𝐿1, it is enough to prove the following inequalities: 𝐾1𝐿𝜆𝑘(𝑥)𝜓1||𝜇||(𝑥)𝑝𝜓𝐾0,0<𝑥𝑏,(4.12)1𝐿𝜆𝑘(𝑥)𝜙+1||||𝜇(𝑥)𝑝𝜙0,0<𝑥𝑏.(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 𝜓(𝑏)=𝑏𝜓(𝜏). Writing 𝐿𝜓=0 in the following form: 𝑝(𝑥)𝑔(𝑥)𝜓(𝑥)𝜆𝑠(𝑥)𝑔(𝑥)𝜓(𝑥)=0,(4.14) and integrating it first from 𝜏 to 𝑥 and then 𝑥 to 𝑏, we get that 𝑝(𝑥)𝜓(𝑥)𝜓(𝑥)Φ(𝑝,𝑞,𝑠,𝑔)on0<𝑥𝑏.(4.15) Here Φ(𝑝,𝑞,𝑠,𝑔) is given by (4.8). Now, the result follows from (4.7), (4.12), and (4.15).

Lemma 4.3. If 0<𝜆<𝜆0, 𝜆𝑘(𝑥)𝐾1, |𝜇(𝑥)|𝐿1, 𝐿𝑦0 for 0<𝑥𝑏, 𝑦(0)=0, and 𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)0, then 𝐾1𝜇𝜆𝑘(𝑥)𝑦(𝑥)+𝐿1sign𝑦𝑝𝑦0,0<𝑥𝑏,(4.16) provided that 1sup𝑝𝑔𝑏𝑏0𝑑𝑡𝐾𝑝𝑔>0,0<𝑥𝑏,1𝜆𝑘(𝑥)1sup𝑝𝑔𝑏𝑏0𝑑𝑡𝐿𝑝𝑔1||𝜇||(𝑥)sup𝑝𝑔𝑏0,0<𝑥𝑏(4.17) or 1𝜆𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡𝑏0𝑑𝑡𝐾𝑝(𝑡)𝑔(𝑡)>0,1𝜆𝑘(𝑥)1𝜆𝑏0𝑠𝑔𝑑𝑡𝑏0𝑑𝑡𝐿𝑝𝑔1||𝜇||𝜆(𝑥)𝑏0𝑠𝑔𝑑𝑡0,0<𝑥𝑏,(4.18) hold.

Proof. Similar to the proof of Lemma 4.2, we need to establish two inequalities (4.10)-(4.11) for 0<𝜆<𝜆0. Here 𝜓 and 𝜙 cannot have the point of minima in (0,𝑏), because at the point of minima, the differential equation 𝐿𝜓=0 or 𝐿𝜙=0 will be contradicted. So either 𝜓(𝑥)0 and 𝜙(𝑥)0 or 𝜓(𝑥) and 𝜙(𝑥) both are concave downwards on [0,𝑏]. Thus we can divide the proof in two cases:
Case i. 𝜓 and 𝜙 both are concave downwards.
We prove for 𝜓 as similar analysis provides result for 𝜙. Let the point of maxima be 𝑥0(0,𝑏). Then 𝜓(𝑥)>0 for 𝑥<𝑥0 and 𝜓(𝑥)<0 for 𝑥>𝑥0. On both sides of 𝑥0, the inequality (4.10) will be reduced into the following two inequalities: 𝐾1𝐿𝜆𝑘(𝑥)𝜓1+𝜇(𝑥)𝑝𝜓0,𝜓𝐾(𝑥)0,1𝐿𝜆𝑘(𝑥)𝜓+1𝜇(𝑥)𝑝𝜓0,𝜓(𝑥)0.(4.19) For a point 𝑥 on the left side of 𝑥0, we integrate (4.14) from 𝑥 to 𝑥0 twice and get 𝑝(𝑥)𝜓(𝑥)𝜆𝜓(𝑥)𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡1𝜆𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡𝑏0(1/𝑝(𝑡)𝑔(𝑡))𝑑𝑡.(4.20) Similarly for any point 𝑥 on the right side of 𝑥0, we get 𝑝(𝑥)𝜓(𝑥)𝜆𝜓(𝑥)𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡1𝜆𝑏0𝑠(𝑡)𝑔(𝑡)𝑑𝑡𝑏0(1/𝑝(𝑡)𝑔(𝑡))𝑑𝑡.(4.21) Now, the result follows from the fact that |𝜇(𝑥)|𝐿1 and from (4.18) to (4.21).

Case ii. When 𝜓(𝑥)0 and 𝜙(𝑥)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 𝜓(𝑏)=𝑏𝜓(𝜏). Integrating (4.14) first from 𝜏 to 𝑥 and then from 𝑥 to 𝑏, we get 𝑝(𝑥)𝜓(𝑥)𝜓(𝑥)sup(𝑝𝑔/𝑏)1sup(𝑝𝑔/𝑏)𝑏0,(𝑑𝑡/𝑝𝑔)(4.22) and the result follows from (4.12), (4.17), and (4.22). This completes the proof.

Lemma 4.4. If 𝑢𝑛 is an upper solution of (4.3) and 𝑢𝑛+1 is defined by (1.4)–(1.5), then 𝑢𝑛𝑢𝑛+1 for 𝜆<𝜆0.

Proof. Let 𝑤=𝑢𝑛𝑢𝑛+1. 𝑤 satisfies 𝐿𝑤=(𝑝𝑢𝑛)𝑞𝑓(𝑥,𝑢𝑛,𝑝𝑦𝑛)0,0<𝑥𝑏,𝑤(0)=0,𝛼1𝑤(𝑏)+𝛽1𝑝(𝑏)𝑤(𝑏)0, and the result follows from Corollary 3.6.

Proposition 4.5. Let 𝑢0 be an upper solution of (4.3), and let 𝑓(𝑥,𝑦,𝑝𝑦) satisfy the following(F1) 𝑓(𝑥,𝑦,𝑝𝑦) is continuous on𝐷0=𝑥,𝑦,𝑝𝑦[]×𝑣0,𝑏0,𝑢0×,(4.23)(F2) 𝐾1𝐾1(𝐷0) such that for all (𝑥,𝑦,𝑣),(𝑥,𝑤,𝑣)𝐷0,𝐾1(𝑦𝑤)𝑓(𝑥,𝑦,𝑣)𝑓(𝑥,𝑤,𝑣)for𝑦𝑤,and(4.24)(F3)0𝐿1𝐿1(𝐷0) such that for all (𝑥,𝑦,𝑣1),(𝑥,𝑦,𝑣2)𝐷0,||𝑓𝑥,𝑦,𝑣1𝑓𝑥,𝑦,𝑣2||𝐿1||𝑣1𝑣2||,(4.25) and (4.7), (4.17), or (4.18) hold. Then the functions 𝑢𝑛 defined by (1.4)–(1.5) are such that, for all 𝑛, (i) 𝑢𝑛 is upper solution of (4.3) and (ii) 𝑢𝑛𝑢𝑛+1.

Proof. Since 𝑢0 is an upper solution from Lemma 4.4, we have 𝑢0𝑢1. Assume that the claim is true for 𝑛1; that is, 𝑢𝑛1 is an upper solution and 𝑢𝑛1𝑢𝑛.
Let 𝑤=𝑢𝑛1𝑢𝑛. We have 𝑝𝑢𝑛𝑞𝑓𝑥,𝑢𝑛,𝑝𝑢𝑛𝐾𝑞1𝜆𝑘(𝑥)𝑤𝜇(𝑥)+𝐿1sign𝑤𝑝𝑤,(4.26) and from Lemmas 4.2 and 4.3 we get (𝑝𝑢𝑛)𝑞𝑓(𝑥,𝑢𝑛,𝑝𝑢𝑛)0,0<𝑥𝑏.
Thus, 𝑢𝑛 is an upper solution for all 𝑛. From Lemma 4.4 we have 𝑢𝑛𝑢𝑛+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 𝑣𝑛𝑣𝑛+1 for 𝜆<𝜆0.

Proposition 4.7. Let 𝑣0 be a lower solution of (4.3), let 𝑓(𝑥,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 𝑛, (i) 𝑣𝑛 is lower solution of (4.3) and (ii) 𝑣𝑛𝑣𝑛+1.

Proposition 4.8. If 𝑓(𝑥,𝑦,𝑝𝑦) satisfies(F4) 𝑓(𝑥,𝑢0,𝑝𝑢0)𝑓(𝑥,𝑣0,𝑝𝑣0)𝜇(𝑥)(𝑝𝑢0𝑝𝑣0)𝜆𝑘(𝑥)(𝑢0𝑣0)0 for 0<𝑥𝑏 such that 𝜆𝑘(𝑥)𝐾1 and |𝜇(𝑥)|𝐿1,and in addition let (F1)–(F3) and (4.7), (4.17), or (4.18) hold, then for all 𝑛 the functions 𝑢𝑛 and 𝑣𝑛 defined by (1.4)–(1.5) satisfy 𝑣𝑛𝑢𝑛.

Proof. Let 𝑤𝑖=𝑢𝑖𝑣𝑖, then 𝑤𝑖 satisfies 𝐿𝑤𝑖=𝑞(𝑥)𝑖1 for all 𝑖𝑁 such that 𝑖(𝑥)=𝑓𝑥,𝑢𝑖,𝑝𝑢𝑖𝑓𝑥,𝑣𝑖,𝑝𝑣𝑖𝜇(𝑥)𝑝𝑢𝑖𝑝𝑣𝑖𝑢𝜆𝑘(𝑥)𝑖𝑣𝑖,0<𝑥𝑏.(4.27) Since 𝑣0𝑢0, we prove that 𝑣1𝑢1. Since 𝑤1 is solution of 𝐿𝑤1=𝑞00, 𝑤1(0)=0 and 𝛼1𝑤1(0)+𝛽1𝑝(𝑏)𝑤1(𝑏)=0, from Corollary 3.6 we have 𝑤10. Let 𝑛2, let 𝑛20, and 𝑢𝑛1𝑣𝑛1, then we prove that 𝑛10 and 𝑢𝑛𝑣𝑛. Consider 𝑛1=𝑓𝑥,𝑢𝑛1,𝑝𝑢𝑛1𝑓𝑥,𝑣𝑛1,𝑝𝑣𝑛1𝜇(𝑥)𝑝𝑤𝑛1𝜆𝑘(𝑥)𝑤𝑛1𝐾1𝑤𝜆𝑘(𝑥)𝑛1𝜇+𝐿sign𝑤𝑛1𝑝𝑤𝑛1.(4.28) Since 𝑤𝑛1 is a solution of 𝐿𝑤𝑛1=𝑞𝑛20, 𝑤𝑛1(0)=0, and 𝛼1𝑤𝑛1(0)+𝛽1𝑝(𝑏)𝑤𝑛1(𝑏)=0; hence, from Lemmas 4.2 and 4.3, we have 𝑛10. Thus, from Corollary 3.6 on 𝐿𝑤𝑛=𝑞𝑛10, 𝑤𝑛(0)=0 and 𝛼1𝑤𝑛(0)+𝛽1𝑝(𝑏)𝑤𝑛(𝑏)=0, we have 𝑤𝑛0, that is, 𝑢𝑛𝑣𝑛. This completes the proof.

Lemma 4.9. If 𝑓(𝑥,𝑦,𝑝𝑦) satisfies(F5) for all (𝑥,𝑦,𝑣)𝐷0, |𝑓(𝑥,𝑦,𝑣)|𝜑(|𝑣|) where 𝜑[0,)(0,) is continuous and satisfies 𝑏0𝑞(𝑠)𝑑𝑠<𝑙0𝑑𝑠,𝜑(𝑠)(4.29) where 𝑙0=sup[0,𝑏]|𝑝(𝑥)𝑢0(𝑥)/𝑏|, then there exists 𝑅0>0 such that any solution of 𝑝𝑦𝑞𝑓𝑥,𝑦,𝑝𝑦,0<𝑥𝑏,(4.30)𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)𝛾1,(4.31) with 𝑦[𝑣0,𝑢0] for all 𝑥[0,𝑏], satisfies 𝑝𝑦<𝑅0.

Proof. We divide the proof in three parts.
Case i. If solution is not monotone throughout the interval, then we consider the interval (𝑥0,𝑥](0,𝑏) such that 𝑦(𝑥0)=0 and 𝑦(𝑥)>0 for 𝑥>𝑥0. Integrating (4.30) from 𝑥0 to 𝑥 we get 𝑝𝑦0𝑑𝑠𝜑(𝑠)𝑏0𝑞(𝑠)𝑑𝑠.(4.32) From (F5) we can choose 𝑅0>0 such that 𝑝𝑦0𝑑𝑠𝜑(𝑠)𝑏0𝑞(𝑠)𝑑𝑠𝑅0𝑙0𝑑𝑠𝜑(𝑠)𝑅00𝑑𝑠,𝜑(𝑠)(4.33) which gives 𝑝(𝑥)𝑦(𝑥)𝑅0.(4.34) Now we consider the case in which 𝑦(𝑥)<0 for 𝑥<𝑥0, 𝑦(𝑥0)=0, and proceeding in the similar way we get 𝑝(𝑥)𝑦(𝑥)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 𝑦(𝜏)=𝑦(𝑏)𝑦(0)𝑏|||𝑢0𝑏|||.(4.36) Now, integrating (4.30) from 𝜏 to 𝑥, we get 𝑝𝑦0𝑑𝑠𝜑(𝑠)𝑏0𝑞(𝑡)𝑑𝑡+𝑙00𝑑𝑠.𝜑(𝑠)(4.37) Further, from (F5) we can choose 𝑅0 such that 𝑝𝑦0𝑑𝑠𝜑(𝑠)𝑏0𝑞(𝑠)𝑑𝑠+𝑙00𝑑𝑠𝜑(𝑠)𝑅00𝑑𝑠.𝜑(𝑠)(4.38) which gives 𝑝(𝑥)𝑦(𝑥)𝑅0.Case iii. If 𝑦 is monotonically decreasing in (0,𝑏); that is, 𝑦<0 in (0,𝑏), then argument similar to Case  ii yields 𝑝𝑦0𝑑𝑠𝜑(𝑠)𝑏0𝑞(𝑠)𝑑𝑠+𝑙00𝑑𝑠𝜑(𝑠)𝑅00𝑑𝑠,𝜑(𝑠)(4.39) and we get 𝑝(𝑥)𝑦(𝑥)𝑅0,(4.40) and the result follows.

Lemma 4.10. If 𝑓(𝑥,𝑦,𝑝𝑦) satisfies (F5), then there exists 𝑅0>0 such that any solution of 𝑝𝑦𝑞𝑓𝑥,𝑦,𝑝𝑦,0<𝑥𝑏,𝑦(0)=0,𝛼1𝑦(𝑏)+𝛽1𝑝(𝑏)𝑦(𝑏)𝛾1,(4.41) with 𝑦[𝑣0,𝑢0] for all 𝑥[0,𝑏], satisfies 𝑝𝑦<𝑅0.

Proof. Proof follows from the analysis of Lemma 4.9.

Theorem 4.11. Let 𝑢0 and 𝑣0 be upper and lower solutions. Let 𝑓(𝑥,𝑦,𝑝𝑦) 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 𝜆<𝜆0 is chosen such that 𝜆𝑘(𝑥)𝐾1 and |𝜇(𝑥)|𝐿1, where 𝜆0 is the first positive eigenvalue of the corresponding eigenvalue problem, then the sequences {𝑢𝑛} and {𝑣𝑛} generated by (1.4)–(1.5) with initial iterate 𝑢0 and 𝑣0 converge monotonically and uniformly towards solutions ̃𝑢(𝑥) and ̃𝑣(𝑥) of (4.3). Any solution 𝑧(𝑥) in 𝐷0 must satisfy ̃𝑣(𝑥)𝑧(𝑥)̃𝑢(𝑥).

Proof. From Lemmas 4.24.10, Propositions 4.54.8, and we get two monotonic sequences {𝑢𝑛} and {𝑣𝑛} which are bounded by 𝑢0 and 𝑣0; respectively, and by Dini's Theorem their uniform convergence is assured. Let {𝑢𝑛} and {𝑣𝑛} converge uniformly to ̃𝑢 and ̃𝑣.
By Lemmas 4.9 and 4.10, it is easy to see that the sequences {𝑝𝑢𝑛} and {𝑝𝑣𝑛} are uniformly bounded. Now, from||𝑝𝑦𝑛𝑥1𝑝𝑦𝑛𝑥2||=||||𝑥2𝑥1𝑝𝑦𝑛||||𝑑𝑡,(4.42) uniform convergence of {𝑦𝑛}, properties (A1)–(A4), and (F1), it is easy to prove that {𝑝𝑦𝑛} is equicontinuous. Hence, by Arzela-Ascoli's Theorem there exist a uniform convergent subsequence {𝑝𝑦𝑛𝑘} of {𝑝𝑦𝑛}. Since limit is unique so original sequence will also converge uniformly to the same limit say 𝑝𝑦. It is easy to see that, if 𝑦𝑛̃𝑦, then 𝑝𝑦𝑛𝑝̃𝑦. Therefore sequences {𝑝𝑢𝑛} and {𝑝𝑣𝑛} converge uniformly to 𝑝̃𝑢 and 𝑝̃𝑣, respectively.
Let 𝐺(𝑥,𝑡) be Green's function for the linear boundary value problem 𝐿𝑦𝑛=0, 𝑦𝑛(0)=0, and 𝛼1𝑦𝑛(𝑏)+𝛽1𝑝(𝑏)𝑦𝑛(𝑏)=0. Then solution of (1.4)–(1.5) can be written as𝑦𝑛=𝐶𝑥2+𝑏0𝐹𝐺(𝑥,𝑡)𝑡,𝑦𝑛1,𝑝𝑦𝑛1+𝐻(𝑡)𝑑𝑡,(4.43) where 𝐻(𝑡)=2𝐶(𝑡𝑝(𝑡)+𝑝(𝑡))+2𝐶𝑡𝜇(𝑡)𝑞(𝑡)𝑝(𝑡)+𝜆𝐶𝑡2𝑠(𝑡) and 𝐶=𝛾1/(𝛼1𝑏2+2𝛽1𝑏𝑝(𝑏)).
Now, uniform convergence of {𝑦𝑛}, {𝑝𝑦𝑛} and continuity of 𝑓(𝑥,𝑦,𝑝𝑦) imply that {(1/𝑞)𝐹(𝑥,𝑦𝑛,𝑝𝑦𝑛)} converges uniformly in [0,𝑏]. Hence, {(1/𝑞)𝐹(𝑥,𝑦𝑛,𝑝𝑦𝑛)} converges in the sense of mean in 𝐿2𝑞(0,𝑏). Taking limit as 𝑛 and using Lemma 2.4 ([11, page 27]), we get𝑦=𝐶𝑥2+𝑏0𝐹𝐺(𝑥,𝑡)𝑡,𝑦,𝑝𝑦+𝐻(𝑡)𝑑𝑡,(4.44) which is the solution of the boundary value problem (4.3).
Any solution 𝑧(𝑥) in 𝐷0 plays the role of 𝑢0(𝑥). Hence ̃𝑧(𝑥)𝑣(𝑥). Similarly, 𝑧(𝑥)̃𝑢(𝑥). This completes the proof.

Remark 4.12. The case when 𝜆=0 corresponds to the case when 𝑓(𝑥,𝑦,𝑝𝑦)𝑓(𝑥,𝑝𝑦). In such cases the boundary value problem (4.3) can be reduced to two initial value problems 𝑧=𝑞𝑓(𝑥,𝑧), 𝑧(0)=𝛼1 and 𝑝𝑦=𝑧, 𝑦(0)=𝛽1. From the assumptions on 𝑝(𝑥), 𝑞(𝑥), and 𝑓(𝑥,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(𝑥,𝑦,𝑝𝑦)|𝑁0 in 𝐷0. Then lower solution 𝑣0 and upper solution 𝑢0 may be obtained as solution of the following linear boundary value problems: 𝑝𝑣0+𝑁0𝑞𝑣(𝑥)=0,0<𝑥𝑏,0(0)=0,𝛼1𝑣0(𝑏)+𝛽1𝑝(𝑏)𝑣0(𝑏)=𝛾1,𝑝𝑢0𝑁0𝑢𝑞(𝑥)=0,0<𝑥𝑏,0(0)=0,𝛼1𝑢0(𝑏)+𝛽1𝑝(𝑏)𝑢0(𝑏)=𝛾1.(4.45)

Theorem 4.14. Suppose that 𝑓(𝑥,𝑦,𝑝𝑦) satisfies (F1), (F3), and constants 𝐾1(𝐷0)<𝜆0 such that 𝐾1(𝑢𝑣)𝑓𝑥,𝑢,𝑝𝑦𝑓𝑥,𝑣,𝑝𝑦.(4.46) Then the boundary value problem (4.3) has unique solution.

Proof. Let 𝑢 and 𝑣 be two solutions of (4.3), then we get 𝑝(𝑢𝑣)𝑓=𝑞(𝑥)𝑥,𝑢,𝑝𝑢𝑓𝑥,𝑣,𝑝𝑣𝑝,0<𝑥𝑏,or(𝑢𝑣)+𝐿1𝑞(𝑥)𝑝𝑢𝑝𝑣𝐾1𝑞(𝑥)(𝑢𝑣)0,0<𝑥𝑏,(𝑢𝑣)(0)=0,𝛼1(𝑢𝑣)(𝑏)+𝛽1𝑝(𝑏)(𝑢𝑣)(𝑏)=0.(4.47) Since 𝐾1<𝜆0, from Corollary 3.6 we get 𝑢𝑣0 or 𝑢𝑣. Similarly 𝑣𝑢. Therefore, the solution of (4.3) is unique.


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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