Abstract

We study the existence of at least one monotonic positive solution for the nonlocal boundary value problem of the second-order functional differential equation 𝑥(𝑡)=𝑓(𝑡,𝑥(𝜙(𝑡))), 𝑡(0,1), with the nonlocal condition 𝑚𝑘=1𝑎𝑘𝑥(𝜏𝑘)=𝑥0, 𝑥(0)+𝑛𝑗=1𝑏𝑗𝑥(𝜂𝑗)=𝑥1, where 𝜏𝑘(𝑎,𝑑)(0,1), 𝜂𝑗(𝑐,𝑒)(0,1), and 𝑥0,𝑥1>0. As an application the integral and the nonlocal conditions 𝑑𝑎𝑥(𝑡)𝑑𝑡=𝑥0, 𝑥(0)+𝑥(𝑒)𝑥(𝑐)=𝑥1 will be considered.

1. Introduction

The nonlocal boundary value problems of ordinary differential equations arise in a variety of different areas of applied mathematics and physics.

The study of nonlocal boundary value problems was initiated by Il’in and Moiseev [1, 2]. Since then, the non-local boundary value problems have been studied by several authors. The reader is referred to [322] and references therein.

In most of all these papers, the authors assume that the function 𝑓[0,1]×𝑅+𝑅+ is continuous. They all assume thatlim𝑥𝑓(𝑥)𝑥=0or,lim𝑥0𝑓(𝑥)𝑥=0or.(1.1) These assumptions are restrictive, and there are many functions that do not satisfy these assumptions.

Here we assume that the function 𝑓[0,1]×𝑅+𝑅+ is measurable in 𝑡[0,1] for all 𝑥𝑅+ and continuous in 𝑥𝑅+ for almost all 𝑡[0,1] is and there exists an integrable function 𝑎𝐿1[0,1] and a constant 𝑏>0 such that||||||||[]𝑓(𝑡,𝑥)𝑎(𝑡)+𝑏|𝑥|,(𝑡,𝑥)0,1×𝐷.(1.2) Our aim here is to study the existence of at least one monotonic positive solution for the nonlocal problem of the second-order functional differential equation𝑥(𝑡)=𝑓(𝑡,𝑥(𝜙(𝑡))),𝑡(0,1),(1.3) with the nonlocal condition𝑚𝑘=1𝑎𝑘𝑥𝜏𝑘=𝑥0,𝑥(0)+𝑛𝑗=1𝑏𝑗𝑥𝜂𝑗=𝑥1,(1.4) where 𝜏𝑘(𝑎,𝑑)(0,1),𝜂𝑗(𝑐,𝑒)(0,1),and𝑥0,𝑥1>0.

As an application, the problem with the integral and nonlocal conditions𝑑𝑎𝑥(𝑡)𝑑𝑡=𝑥0,𝑥(0)+𝑥(𝑒)𝑥(𝑐)=𝑥1,(1.5) is studied.

It must be noticed that the nonlocal conditions𝑥(𝜏)=𝑥0,𝜏(𝑎,𝑑),𝑥(0)+𝑥(𝜂)=𝑥1,𝜂(𝑐,𝑒),𝑚𝑘=1𝑎𝑘𝑥𝜏𝑘=0,𝜏𝑘(𝑎,𝑑),𝑥(0)+𝑛𝑗=1𝑏𝑗𝑥𝜂𝑗=0,𝜂𝑗(𝑐,𝑒),𝑑𝑎𝑥(𝑡)𝑑𝑡=0,𝑥(0)+𝑥(𝑒)=𝑥(𝑐)(1.6) are special cases of our the nonlocal and integral conditions.

2. Integral Equation Representation

Consider the functional differential equation (1.3) with the nonlocal condition (1.4) with the following assumptions.(i)𝑓[0,1]×𝑅+𝑅+ is measurable in 𝑡[0,1] for all 𝑥𝑅+ and continuous in 𝑥𝑅+ for almost all 𝑡[0,1] and there exists an integrable function 𝑎𝐿1[0,1], and a constant 𝑏>0 such that ||||||||[]𝑓(𝑡,𝑥)𝑎(𝑡)+𝑏|𝑥|,(𝑡,𝑥)0,1×𝐷.(2.1)(ii)𝜙(0,1)(0,1) is continuous.(iii)𝑏<1/(3𝐵),𝐵=(𝑛𝑗=1𝑏𝑗+1)1.(iv)𝑚𝑘=1𝑎𝑘>0,𝑘=1,2,,𝑚,𝑛𝑗=1𝑏𝑗>0,𝑗=1,2,,𝑛.(2.2) Now, we have the following Lemma.

Lemma 2.1. The solution of the nonlocal problem (1.3)-(1.4) can be expressed by the integral equation 𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵𝑡𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(2.3) where 𝐴=(𝑚𝑘=1𝑎𝑘)1,𝐵=(𝑛𝑗=1𝑏𝑗+1)1.

Proof. Integrating (1.3), we get 𝑥(𝑡)=𝑥(0)+𝑡0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(2.4) Integrating (2.4), we obtain 𝑥(𝑡)=𝑥(0)+𝑥(0)𝑡+𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(2.5) Let 𝑡=𝜏𝑘, in (2.5), we get 𝑚𝑘=1𝑎𝑘𝑥𝜏𝑘=𝑛𝑘=1𝑎𝑘𝑥(0)+𝑛𝑘=1𝑎𝑘𝜏𝑘𝑥(0)+𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(2.6) and we deduce that 𝑥𝑥(0)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥(0)𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,𝐴=𝑚𝑘=1𝑎𝑘1.(2.7) Substitute from (2.7) into (2.5), we obtain 𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝑥(0)𝑡𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘+𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(2.8) Let 𝑡=𝜂𝑗, in (2.4), we obtain 𝑛𝑗=1𝑏𝑗𝑥𝜂𝑗=𝑛𝑗=1𝑏𝑗𝑥(0)+𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑥𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,1𝑥(0)=𝑥(0)𝑛𝑗=1𝑏𝑗+𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(2.9) and we deduce that 𝑥𝑥(0)=𝐵1𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,𝐵=𝑛𝑗=1𝑏𝑗+11.(2.10) Substitute from (2.10) into (2.8), we obtain 𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵𝑡𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0,+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(2.11) which proves that the solution of the nonlocal problem (1.3)-(1.4) can be expressed by the integral equation (2.3).

3. Existence of Solution

We study here the existence of at least one monotonic nondecreasing solution 𝑥𝐶[0,1] for the integral equation (2.3).

Theorem 3.1. Assume that (i)–(iv) are satisfied. Then the nonlocal problem (1.3)-(1.4) has at least one solution 𝑥𝐶[0,1].

Proof. Define the subset 𝑄𝑟𝐶(0,1) by 𝑄𝑟={𝑥𝐶|𝑥(𝑡)|𝑟,𝑟=(𝐴𝑥0+𝐵𝑥1+(3𝐵)𝑎)/(1(3𝐵)𝑏),𝑟>0}. Clear the set 𝑄𝑟 which is nonempty, closed, and convex.
Let 𝐻 be an operator defined by𝑥(𝐻𝑥)(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑s+𝐵𝑡𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(3.1)
Let 𝑥𝑄𝑟, then||||𝑥(𝐻𝑥)(𝑡)𝐴0+𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘||||𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵𝑡𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1+𝑛𝑗=1𝑏𝑗𝜂𝑗0||||+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0||||𝑥(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝐴0+𝑚𝑘=1𝑎𝑘10||||||||𝑥𝑎(𝑠)+𝑏𝑥(𝜙(𝑠))𝑑𝑠+𝐵1+𝑛𝑗=1𝑏𝑗10||||||||+𝑎(𝑠)+𝑏𝑥(𝜙(𝑠))𝑑𝑠10||𝑎||||𝑥||(𝑠)+𝑏(𝜙(𝑠))𝑑𝑠𝐴𝑥0+𝑎+𝑏sup𝑡𝐼||||𝑥(𝜙(𝑡))+𝐵𝑥1+𝐵𝑛𝑗=1𝑏𝑗𝑎+𝑏𝐵𝑛𝑗=1𝑏𝑗sup𝑡𝐼||||+𝑥(𝜙(𝑡))𝑎+𝑏sup𝑡𝐼||||𝑥(𝜙(𝑡))𝐴𝑥0+𝐵𝑥1+2𝑎+2𝑏𝑥+(1𝐵)𝑎+𝑏(1𝐵)𝑥𝐴𝑥0+𝐵𝑥1+(3𝐵)𝑎+(3𝐵)𝑏𝑟𝑟,(3.2) then 𝐻𝑄𝑟𝑄𝑟 and {𝐻𝑥(𝑡)} is uniformly bounded in 𝑄𝑟.
Also for 𝑡1,𝑡2[0,1] such that 𝑡1<𝑡2, we have𝑡(𝐻𝑥)2𝑡(𝐻𝑥)1𝑡=𝐵2𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠𝑡20𝑡2𝑡𝑠𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠𝐵1𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠𝑡10𝑡1𝑡𝑠𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠=𝐵2𝑡1𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(s,𝑥(𝜙(𝑡)))𝑑𝑠𝑡10𝑡2𝑡1+𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠𝑡2𝑡1𝑡2𝑠𝑓(𝑠,𝑥(𝜙(𝑡)))𝑑𝑠.(3.3) Then ||𝑡(𝐻𝑥)2𝑡(𝐻𝑥)1||||𝑡𝐵2𝑡1||𝑥1+𝑛𝑗=1𝑏𝑗𝜂𝑗0||||||||+||𝑡𝑎(𝑠)+𝑏𝑥(𝜙(𝑠))𝑑𝑠2𝑡1||𝑡10||||||||+𝑎(𝑠)+𝑏𝑥(𝜙(𝑠))𝑑𝑠𝑡2𝑡1𝑡2||𝑎||||𝑥||||𝑡𝑠(𝑠)+𝑏(𝜙(𝑠))𝑑𝑠𝐵2𝑡1||𝑥1+𝑛𝑗=1𝑏𝑗[]+||𝑡𝑎+𝑏𝑟2𝑡1||[]+𝑎+𝑏𝑟𝑡2𝑡1𝑡𝑎𝑑𝑠+𝑏𝑟2𝑡1.(3.4) The above inequality shows that ||𝑡(𝐻𝑥)2𝑡(𝐻𝑥)1||0as𝑡2𝑡1.(3.5) Therefore {𝐻𝑥(𝑡)} is equicontinuous. By the Arzelà-Ascoli theorem, {𝐻𝑥(𝑡)} is relatively compact.
Since all conditions of the Schauder theorem hold, then 𝐻 has a fixed point in 𝑄𝑟 which proves the existence of at least one solution 𝑥𝐶[0,1] of the integral equation (2.3), wherelim𝑡0+𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝐵𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠=𝑥(0),lim𝑡1𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵1𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠10(1𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠=𝑥(1).(3.6) To complete the proof, we prove that the integral equation (2.3) satisfies nonlocal problem (1.3)-(1.4). Differentiating (2.3), we get 𝑥𝑥(𝑡)=𝐵1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0𝑥𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(3.7)(𝑡)=𝑓(𝑡,𝑥(𝜙(𝑡))).(3.8) Let 𝑡=𝜏𝑘 in (2.3), we obtain 𝑥𝜏𝑘𝑥=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘+𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(3.9) which proves 𝑚𝑘=1𝑎𝑘𝑥𝜏𝑘=𝑥0.(3.10) Also let 𝑡=𝜂𝑗 in (3.7), we obtain 𝑥𝜂𝑗𝑥=𝐵1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(3.11) then 𝑛𝑗=1𝑏𝑗𝑥𝜂𝑗=𝐵𝑛𝑗=1𝑏𝑗𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑛𝑗=1𝑏𝑗𝜂𝑗0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(3.12) Let 𝑡=0 in (3.7), we obtain 𝑥𝑥(0)=𝐵1𝑛𝑗=1𝑏𝑗𝜂𝑗0.𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠(3.13) Adding (3.12) and (3.13), we obtain 𝑥(0)+𝑛𝑗=1𝑏𝑗𝑥𝜂𝑗=𝑥1.(3.14) This implies that there exists at least one solution 𝑥𝐶[0,1] of the nonlocal problem (1.3) and (1.4). This completes the proof.

Corollary 3.2. The solution of the problem (1.3)-(1.4) is monotonic nondecreasing.

Proof. Let𝑡1<𝑡2, we deduce from (2.3) that 𝑥𝑡1𝑥=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑡𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵1𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡10𝑡1𝑥𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠<𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑡𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝐵2𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘𝑥1𝑛𝑗=1𝑏𝑗𝜂𝑗0+𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡20𝑡2𝑡𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠=𝑥2,(3.15) which proves that the solution 𝑥 of the problem (1.3)-(1.4) is monotonic nondecreasing.

3.1. Positive Solution

Let 𝑏𝑗=0,𝑗=1,2,𝑛 and 𝑥1=0, then the nonlocal problem condition (1.4) will be𝑚𝑘=1𝑎𝑘𝑥𝜏𝑘=𝑥0,𝑥(0)=0.(3.16)

Theorem 3.3. Let the assumptions (i)–(iv) of Theorem 3.1 be satisfied. Then the solution of the nonlocal problem (1.3)–(3.16) is positive 𝑡[𝑑,1].

Proof. Let 𝑏𝑗=0,𝑗=1,2,𝑛 and 𝑥1=0 in the integral equation (2.3) and the nonlocal condition (1.4), then the solution of the nonlocal problem (1.3)–(3.16) will be given by the integral equation 𝑥𝑥(𝑡)=𝐴0𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘+𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(3.17) where 𝐴=(𝑚𝑘=1𝑎𝑘)1.
Let 𝑡[𝑑,1], then𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,𝜏𝑘𝑡,𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑚𝑘=1𝑎𝑘𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(3.18) Multiplying by 𝐴=(𝑚𝑘=1𝑎𝑘)1, we obtain 𝐴𝑚𝑘=1𝑎𝑘𝜏𝑘0𝜏𝑘𝑠𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝐴𝑚𝑘=1𝑎𝑘𝑡0=(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠,(3.19) and the solution 𝑥 of the nonlocal problem (1.3) and (3.16), given by the integral equation (3.17), is positive for 𝑡[𝑑,1]. This complete the proof.

Example 3.4. Consider the nonlocal problem of the second-order functional differential equation (1.3) with two-point boundary condition 𝑥(0)=0,𝑥(𝜂)=𝑥0,𝜂(𝑎,𝑑)(0,1).(3.20) Applying our results here, we deduce that the two-point boundary value problem (1.3)–(3.20) has at least one monotonic nondecreasing solution 𝑥𝐶[0,1] represented by the integral equation 𝑥(𝑡)=𝑥0𝜂0(𝜂𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠+𝑡0(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(3.21) This the solution is positive with 𝑡>𝜂.

4. Nonlocal Integral Condition

Let 𝑥𝐶[0,1] be the solution of the nonlocal problem (1.3) and (1.4).

Let 𝑎𝑘=𝑡𝑘𝑡𝑘1,𝜏𝑘(𝑡𝑘1,𝑡𝑘)(𝑎,𝑑)(0,1) and let 𝑏𝑗=𝜉𝑗𝜉𝑗1,𝜂𝑗(𝜉𝑗1,𝜉𝑗)(𝑐,𝑒)(0,1), then𝑚𝑘=1𝑡𝑘𝑡𝑘1𝑥𝜏𝑘=𝑥0,𝑥(0)+𝑛𝑗=1𝜉𝑗𝜉𝑗1𝑥𝜂𝑗=𝑥1.(4.1) From the continuity of the solution 𝑥 of the nonlocal problem (1.3) and (1.4), we obtainlim𝑚𝑚𝑘=1𝑡𝑘𝑡𝑘1𝑥𝜏𝑘=𝑑𝑎𝑥𝑥(𝑠)𝑑𝑠,(0)+lim𝑛𝑛𝑗=1𝜉𝑗𝜉𝑗1𝑥𝜂𝑗=𝑥(0)+𝑒𝑐𝑥(𝑠)𝑑𝑠,(4.2) and the nonlocal condition (1.4) transformed to the integral condition𝑑𝑎𝑥(𝑠)𝑑𝑠=𝑥0,𝑥(0)+𝑥(𝑒)𝑥(𝑐)=𝑥1,(4.3) and the solution of the integral equation (2.3) will be𝑥(𝑡)=(𝑑𝑎)1𝑥0𝑑𝑎𝑡0+(𝑡𝑠)𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑑𝑡((𝑏𝑐)+1)1𝑥(𝑡1)1𝑒𝑐𝑡0𝑓+(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠𝑑𝑡𝑡0𝑓(𝑠,𝑥(𝜙(𝑠)))𝑑𝑠.(4.4) Now, we have the following theorem.

Theorem 4.1. Let the assumptions (i)–(iv) of Theorem 3.1 be satisfied. Then the nonlocal problem 𝑥,(𝑡)=𝑓(𝑡,𝑥(𝜙(𝑡))),𝑡(0,1)𝑑𝑎𝑥(𝑠)𝑑𝑠=𝑥0,𝑥(0)+𝑥(𝑒)𝑥(𝑐)=𝑥1(4.5) has at least one monotonic nondecreasing solution 𝑥𝐶[0,1] represented by (4.4).