# Existence of Solutions for Singular Second-Order Ordinary Differential Equations with Periodic and Deviated Nonlocal Multipoint Boundary Conditions

Authors: Ahmed M. A. El-Sayed; Fatma M. Gaafar
Journal: Journal of Function Spaces
 (2018)

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## Abstract

This paper studies the existence of continuous solutions for a class of nonlinear singular second-order ordinary differential equations subject to one of the following boundary conditions: periodic-deviated multipoint boundary conditions, periodic-integral boundary conditions, and periodic-nonlocal integral conditions in the Riemann-Stieltjes sense. An existence result based on the Schauder fixed point theorem and Leray-Schauder continuation principle is used to obtain at least one continuous solution for the singular second-order ordinary differential problems. Two examples are given to show the application of our results.


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## Body

## 1. Introduction

In this paper, we study the existence of at least one continuous solution for the second-order differential equation(1)x′′=ft,xt,x′t,a.e.t∈0,1,where f:[0,1]×R2→R is a given function satisfying Carathéodory’s conditions (i.e., for a.e. t∈[0,1] the function f(t,.,.) is continuous on R2 and for each (u,v)∈R2, the function f(.,u,v) is measurable on [0,1]) and is allowed to be singular at t=0 and t=1.We consider the following nonlocal and multipoint boundary conditionsperiodic-deviated multipoint boundary conditions of the form:(2)x0=x1,∑i=1maixϕξi=0,periodic and integral boundary conditions:(3)x0=x1,∫01xϕθdθ=0,and periodic and nonlocal integral conditions:(4)x0=x1,∫cdxθdηθ=0,0<c<d<1where ϕ:[0,1]→[0,1] is a deviated continuous given function, i.e., ϕ(ξi)≤ξi,ξi∈[0,1], ai∈R with ∑i=1mai≠0,i=1,2,3,…,m, and η(θ) is an increasing function and the integral is meant in the Riemann-Stieltjes sense for 0<c<d<1.For the functionϕ(t) we can take(i)
ϕt=ksin⁡t, t∈[0,π/2], where 0<k≤1.(ii)
ϕ(t)=ktγ, t∈[0,1], k∈(0,1], and γ≥1.The considered problem is to determine sufficient conditions onf guaranteeing that the boundary value problem (1), (2) or (1), (3) or (1), (4) has a solution.Note. One must notice that the following periodic and multipoint boundary conditions are special cases of our periodic and multipoint boundary conditions: (5)x0=x1,∑i=1maixξi=0,ξi∈0,1.x0=x1,xξ=0,ξ∈0,1.x0=x1,xξ=-xη,ξ,η∈0,1.We list the following assumptions which we use in this paper:(H0)ai∈R, ∑i=1mai≠0 and ϕ(ξi)≤ξi, where 0<ξ1<ξ2<⋯<ξm<1.(H1) There exists p(t),r(t)∈Lloc1(0,1) with t(1-t)p(t),t(1-t)r(t),q(t)∈L1[0,1] such that f has a growth condition of the form (6)ft,u,v≤ptuσ1+qtvσ2+rt,a.e.t∈0,1,u,v∈R2and0<σ1,σ2<1,where Lloc1(0,1)={x∣x∈L1[a,b] for every closed interval [a,b]⊂(0,1)}.We point out that our results can not be extended to the cases in [1, 2]. On the one hand the cases σ1 and σ2=1 which are used in them are not included in our hypothesis and on the other hand periodic and deviated multipoint boundary conditions are different from boundary conditions of them. Hence we consider a new class of assumptions and periodic and deviated multipoint boundary conditions.The investigation of multipoint boundary value problems was started by Il’in and Moiseev [3]. From that time, many researchers investigate wide-ranging nonlinear multipoint nonlocal boundary value problems (see Chen et al. [4], Guo et al. [5], Ma [6], Webb [7], Webb and Zima [8], and Zhang and Liu [9] and the references cited therein).The singular ordinary differential equations emerge in the fields of the theory of boundary layer, Newtonian fluid mechanics, gas dynamics, and so on. The theory of singular boundary value problems has turned into an essential region of examination during the last few decades; one may refer to Agarwal and O’Regan [10], Agarwal et al. [11], Cheng [12], Cui and Zou [13], Du and Zhao [14], Erbe and Mathsen [15], Feng and Ge [16], Han and Wu [17], Jiang et al. [18], Kiguradze [19], Li et al. [20], Li and Yan [21], B. Liu et al. [22], L. S. Liu et al. [23], Liua [24], Ma and Thompson [2], Moshinsky [25], Pandey and Barnwal [26], Sun and Zhang [27], Suna [28], Timoshenko [29], Wei [30], Yan et al. [31], Yaoa and Lina [32], Zhao [33], Zhang and Sun [34], and Zhang and Feng [35], and the references therein.The existence of solutions of ordinary differential equation with integral and periodic boundary conditions has been widely considered in recent years; they constitute very important and interesting problems because they have various applications in thermoelasticity, chemical engineering, population dynamics, and underground water flow and include nonlocal and multipoint boundary conditions (see, for example, Feng et al. [36], Feng and Cong [37], Hua et al. [38], Jiang et al. [18], Nanware and Dhaigude [39], Song et al. [40], Staněk [41], Webb and Infante [42], Yan et al. [31], Yang [43], Yao [44], Zhang and Ge [45], and Zhang and Xu [46], and the references therein).


## 2. Preliminary Lemmas

We shall use the classical Banach spacesL1[0,1], C[0,1], and Ck[0,1]. We denoted by AC[0,1] the space of absolute continuous functions on [0,1] and denoted ACk[a,b]={x∈Ck[a,b]∣x(k)∈AC[a,b]}. Let ACloc(0,1) be the space consisting of functions that are absolutely continuous on every closed interval [a,b]⊂(0,1).LetE be the Banach space(7)E=y∈Lloc10,1∣t1-tyt∈L10,1,with the norm(8)yE=∫01τ1-τyτdτ<∞.Let X be the Banach space(9)X=x∈C10,1∣x∈C0,1,limt→1⁡1-tx′tandlimt→0⁡tx′texistequipped with the norm(10)xX=x+x1,where x=maxt∈[0,1]⁡xt and x1=maxt∈[0,1]⁡t1-tx′t.For convenience we set the following notation:A=1+∑i=1mai/∑i=1mai.Lemma 1 (see [47]).
Lety∈E. Then(i)
∫0tτy(τ)dτ∫t1(1-τ)y(τ)dτ∈L1[0,1], and(11)∫01∫0tτyτdτdt=∫01∫t11-τyτdτdt=∫01τ1-τyτdτ(ii)
limt→0⁡t∫t1(1-τ)y(τ)dτ=0, limt→1⁡(1-t)∫0tτy(τ)dτ=0.Theorem 2 (Schauder’s fixed point theorem).
LetK be a closed convex subset of a Banach space X. If N:K→K is continuous and N(K) is relatively compact, then N has a fixed point in K.Theorem 3 (Leray-Schauder continuation principle; see, e.g., [48]).
LetX be a Banach space and N:X→X be a compact map. Assume that there exists D>0 so that if u=λNu for λ∈(0,1), then uX≤D. Then u=Nu is solvable.The following lemmas play a pivotal role to define the solutions for the given problems.Lemma 4.
The singular second-order ordinary differential equation with periodic and deviated multipoint boundary conditions (1)-(2) is equivalent to the functional integral equation:(12)xt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ.Proof.
The solution of (1) is given by(13)xt=x0+tx′0+∫0tt-τfτ,xτ,x′τdτ.Condition x(0)=x(1) gives(14)x′0=-∫011-τfτ,xτ,x′τdτ,and we have(15)xt=x0-t∫011-τfτ,xτ,x′τdτ+∫0tt-τfτ,xτ,x′τdτ,=x0-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ.Also condition ∑i=1maixϕξi=0 gives(16)x0=1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ,and substituting it in (15) we get (12).
Conversely, (cf. (12)) x(t)∈ACloc(0,1). Moreover(17)x′t=∫0tτfτ,xτ,x′τdτ-∫t11-τfτ,xτ,x′τdτ.Integrating x′(t) from 0 to t(t∈(0,1)) and interchanging the order of integration to get (18)xt-x0=∫0tx′ρdρ=∫0t∫0ρτfτ,xτ,x′τdτdρ-∫0t∫ρ11-τfτ,xτ,x′τdτdρ,=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ,=xt-1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ-∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ,hence we get(19)x0=1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ.Similarly integrate x′(t) from t(t∈(0,1)) to 1 and interchange the order of integration to get (20)x1-xt=∫t1x′ρdρ=∫t1∫0ρτfτ,xτ,x′τdτdρ-∫t1∫ρ11-τfτ,xτ,x′τdτdρ,=1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ,=-xt+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ.Hence we have(21)x1=1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ.Then from (19) and (21) we have x(0)=x(1); also from (12) we have ∑i=1maix(ϕ(ξi))=0.
This proves that the solution of singular second-order ordinary differential equation with periodic and deviated multipoint boundary conditions (1)-(2) is equivalent to the functional integral equation (12).Define a nonlinear operatorF by(22)Fxt=ft,xt,x′t.From (H1), we have (23)FxE=∫01t1-tft,xt,x′tdt,≤∫01t1-tptxtσ1+qtx′tσ2+rtdt,≤xσ1∫01t1-tptdt+∫01t1-σ21-t1-σ2qttσ21-tσ2x′tσ2dt+∫01t1-trtdt≤pExσ1+qL1x1σ2+rE<∞,and we conclude that F:X→E is well defined.Forx∈X, we define an operator N by(24)Nxt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ.For x∈X and from (H0) and (H1), we have(25)Nxt≤1-t∫0tτfτ,xτ,x′τdτ+t∫t11-τfτ,xτ,x′τdτ+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτfτ,xτ,x′τdτ+∑i=1maiϕξi∫ϕξi11-τfτ,xτ,x′τdτ,≤∫0tτ1-τfτ,xτ,x′τdτ+∫t1τ1-τfτ,xτ,x′τdτ+1∑i=1mai∑i=1mai∫0ϕξiτ1-ϕξifτ,xτ,x′τdτ+∫ϕξi1ϕξi1-τfτ,xτ,x′τdτ,≤1+∑i=1mai∑i=1mai∫01τ1-τfτ,xτ,x′τdτ,=AFxE,≤ApExσ1+qL1x1σ2+rE<∞.Then (Nx):[0,1]→R is well defined and bounded.Lemma 5.
Letf:[0,1]×R2→R be a function satisfying Carathéodory’s conditions and conditions (H0) and (H1) are fulfilled; then Nx∈X and(26)Nx′′t=ft,xt,x′t,a.e.t∈0,1.Proof.
Firstly, to showN:X→C[0,1], we have Nx∈ACloc(0,1) and by Lemma 1 and from (19), (21), and (24) we have limt→0⁡(Nx)(t)=(Nx)(0) and limt→1⁡(Nx)(t)=(Nx)(1); then (Nx)(t)∈C[0,1]; moreover we have(27)Nx′t=∫0tτfτ,xτ,x′τdτ-∫t11-τfτ,xτ,x′τdτ.Now since(28)∫01Nx′tdt≤∫01∫0tτfτ,xτ,x′τdτdt+∫01∫t11-τfτ,xτ,x′τdτdt,=∫01τfτ,xτ,x′τ∫τ1dtdτ+∫011-τfτ,xτ,x′τ∫0τdtdτ,=2∫01τ1-τfτ,xτ,x′τdτ=2FxE≤2pExσ1+qL1x1σ2+rE<∞,then we have Nx∈AC[0,1]. Now we have (27) and condition (H1) together with the fact that τf(τ,x(τ),x′(τ))∈L1[0,r] and 1-τfτ,xτ,x′τ∈L1[r,1] for each r∈(0,1); we have Nx′(t)∈ACloc(0,1) and(29)Nx′′t=ft,xt,x′t,a.e.t∈0,1.Now to prove that limt→1⁡1-tNx′(t) and limt→0⁡tNx′(t) exist, we prove t1-tNx′(t)∈AC[0,1], since (30)∫01t1-tNx′t′dt=∫011-tNx′t-tNx′t+t1-tNx′′tdt≤∫011-tNx′t+tNx′t+t1-tNx′′tdt=∫01Nx′t+t1-tNx′′tdt≤∫01∫0tτfτ,xτ,x′τdτdt+∫01∫t11-τfτ,xτ,x′τdτdt+∫01t1-tft,xt,x′tdt=3∫01τ1-τfτ,xτ,x′τdτ≤3pExσ1+qL1x1σ2+rE<∞.Then Nx∈X which completes the proof of the lemma.Lemma 6.
( N x ) ( t ) : X → X is a completely continuous operator.Proof.
By the definition ofN and (H1), let xk⊂X be a convergent sequence to x in X; then limk→∞⁡xk-x=0. Since f satisfies Carathéodory’s conditions, we have(31)limk→∞⁡ft,xkt,xk′t=ft,xt,x′tfor  a.et∈0,1.From conditions (H0) and (H1) together with the fact that τfτ,xτ,x′τ∈L1[0,r] and 1-τfτ,xτ,x′τ∈L1[r,1] for each r∈(0,1), we obtain by applying Lebesgue dominated convergence theorem (cf. (24))(32)Nxkt-Nxt→0ask→∞.Also we have (cf. (27)) (33)Nxk′t-Nx′t→0ask→∞Therefore we obtain limk→∞⁡(Nxk)(t)-(Nx)(t)X=0; that is, N is a continuous operator.
Secondly, letB⊂X be a bounded set. We will show that (Nx)(B) is a relatively compact subset of X. Let xk⊂B; let 0≤t1<t2≤1. From (27) and condition (H1), we have (34)Nxkt2-Nxkt1=∫t1t2Nxk′tdt≤∫t1t2Nxk′tdt,≤∫t1t2∫0tτfτ,xkτ,xk′τdτ+∫t1t2∫t11-τfτ,xkτ,xk′τdτdt,≤∫t1t2∫0tτpτxkτσ1+qτxk′τσ2+rτdτdt+∫t1t2∫t11-τpτxkτσ1+qτxk′τσ2+rτdτdt,≤∫t1t2xkσ1∫0tτpτdτ+xk1σ2∫0tτqττσ21-τσ2dτ+∫0tτrτdτdt+∫t1t2xkσ1∫t11-τpτdτ+xk1σ2∫t11-τqττσ21-τσ2dτ+∫t11-τrτdτdt≤∫t1t2xkσ1∫0tτpτdτ+xk1σ2∫0tτqττ1-τdτ+∫0tτrτdτdt+∫t1t2xkσ1∫t11-τpτdτ+xk1σ2∫t11-τqττ1-τdτ+∫t11-τrτdτdt.Similarly we have(35)t21-t2Nxk′t2-t11-t1Nxk′t1=∫t1t2t1-tNxk′t′dt,=∫t1t2t1-tNxk′′t-tNxk′t+1-tNxk′tdt,≤∫t1t2t1-tNxk′′t+tNxk′t+1-tNxk′tdt,=∫t1t2t1-tNxk′′t+Nxk′tdt.From (26) we obtain (36)t1-tNxk′′t=t1-tft,xkt,xk′t,≤t1-tptxktσ1+qtxk′tσ2+rt,≤t1-tptxkσ1+t1-σ21-t1-σ2qtxk1σ2+t1-trt,≤t1-tptxkσ1+qtxk1σ2+t1-trt.Then from (35) and by similar way as (34) we obtain(37)t21-t2Nxk′t2-t11-t1Nxk′t1≤xkσ1∫t1t2t1-tptdt+xk1σ2∫t1t2qtdt+∫t1t2t1-trtdt+∫t1t2xkσ1∫0tτpτdτ+xk1σ2∫0tτqττ1-τdτ+∫0tτrτdτdt+∫t1t2xkσ1∫t11-τpτdτ+xk1σ2∫t1qττdτ+∫t11-τrτdτdt.Since t(1-t)p(t),q(t),t(1-t)r(t)∈L1[0,1] and by using (i) of Lemma 1, thus from (34) and (37) we have the fact that (Nxk)(t) is equicontinuous on [0,1]. By Arzelá-Ascoli theorem, {Nx(t)} and {t1-tNx′(t)} are relatively compact and we have that fact that N:X→X is a completely continuous operator.


## 3. Existence Results

Theorem 7.
Letf:[0,1]×R2→R satisfy the Carathéodory conditions. Assume that assumptions (H0) and (H1) hold. Then problem (1)-(2) has at least one solution.Proof.
In order to apply Theorem2, we define a ball Bk={x∈X:xX≤k,t∈[0,1]}, where(38)k≥max⁡31+ApE1/1-σ1,31+AqL11/1-σ2,31+ArE.Firstly, we prove that N:Bk→Bk. For x∈Bk, and from (25), we have (39)Nxt≤ApExσ1+qL1x1σ2+rE,≤ApEkσ1+qL1kσ2+rE.Also from (27) (40)t1-tNx′t≤t1-t∫0tτfτ,xτ,x′τdτ+t1-t∫t11-τfτ,xτ,x′τdτ,≤t∫0tτ1-τfτ,xτ,x′τdτ+1-t∫t1τ1-τfτ,xτ,x′τdτ,≤∫01τ1-τfτ,xτ,x′τdτ≤pExσ1+qL1x1σ2+rE≤pEkσ1+qL1kσ2+rE.Then (41)NxX=maxt∈0,1⁡Nxt+maxt∈0,1⁡t1-tNx′t,≤2+∑i=1mai∑i=1maipEkσ1+qL1kσ2+rE,=1+ApEkσ1+qL1kσ2+rE≤k3+k3+k3=k.Thus we have N:Bk→Bk and Lemma 6 provides that (Nx)(t):X→X is a completely continuous operator. Hence Theorem 2 implies the existence of a solution of problem (1)-(2) in Bk.Theorem 8.
Letf:[0,1]×R2→R be a function satisfying the Carathéodory conditions. Assume that assumptions (H0) and (H1) hold. Then problem (1)-(2) has at least one solution in X.Proof.
From Lemma4, we have the fact that x(t) is a solution of problem (1)-(2) if and only if x(t) satisfies the integral equation (12). We apply Theorem 3 to obtain the existence of at least one solution of (1)-(2) in X.
To do this, by Lemma6, it suffices to verify that the set of all possible solutions of the family of equations(42)x′′=λft,xt,x′t,a.e.t∈0,1,(43)u0=u1,∑i=1maixϕξi=0,is a priori bounded in X by a constant independent of λ∈(0,1).
Lety(t)∈X be a solution of (42)-(43) for some λ∈(0,1). Then, for t∈[0,1], we have (44)yt=-1-t∫0tτλfτ,yτ,y′τdτ-t∫t11-τλfτ,yτ,y′τdτ+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτλfτ,yτ,y′τdτ+∑i=1maiϕξi∫ϕξi11-τλfτ,yτ,y′τdτ,=-1-t∫0tτy′′τdτ-t∫t11-τy′′τdτ+1∑i=1mai∑i=1mai1-ϕξi∫0ϕξiτy′′τdτ+∑i=1maiϕξi∫ϕξi11-τy′′τdτ,≤∫0tτ1-τy′′τdτ+∫t1τ1-τy′′τdτ+1∑i=1mai∑i=1mai∫0ϕξiτ1-ϕξiy′′τdτ+∫ϕξi1ϕξi1-τy′′τdτ,=1+∑i=1mai∑i=1mai∫01τ1-τy′′τdτ,which implies that(45)y≤Ay′′E.Similarly (cf. (17)), (46)t1-ty′t=t1-t∫0tτλfτ,yτ,y′τdτ-t1-t∫t11-τλfτ,yτ,y′τdτ,=t1-t∫0tτy′′τdτ-t1-t∫t11-τy′′τdτ,≤t∫0tτ1-τy′′τdτ+1-t∫t1τ1-τy′′τdτ,≤∫01τ1-τy′′τdτ≤y′′E,which implies that(47)y1≤y′′E.Now from (42), (45), and (47) and condition (H1), it follows that (48)t1-ty′′t=λt1-tft,yt,y′t≤t1-tptytσ1+qty′tσ2+rt.And we have(49)∫01t1-ty′′tdt≤∫01t1-tptytσ1+qty′tσ2+rtdt,≤pEyσ1+qL1y1σ2+rE,y′′E≤pEAσ1y′′Eσ2+qL1y′′Eσ2+rE.Since σ1,σ2∈(0,1), from the above inequality, there exists a constant B, such that y′′E≤B which implies from (45) that(50)y≤AB.Also from (47) we have(51)y1≤B.Therefore(52)yX≤2+∑i=1mai∑i=1maiB=1+AB.Then all conditions of Theorem 3 are satisfied and then problem (1)-(2) has at least one solution in X.Corollary 9.
Under the assumptions of Theorem8, the singular second-order ordinary differential equation with periodic and multipoint nonlocal boundary conditions(53)x′′t=ft,xt,x′t,a.e.t∈0,1,x0=x1,∑i=1maixξi=0,ξi∈0,1,has at least one solution in X represented by(54)xt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+1∑i=1mai∑i=1mai1-ξi∫0ξiτfτ,xτ,x′τdτ+∑i=1maiξi∫ξi11-τfτ,xτ,x′τdτ.Proof.
Lettingϕ(ξi)=ξi∈(0,1) in problem (1)-(2), then from Theorem 8 we derive that singular second-order ordinary differential problem (53) has at least one solution given by (54).Example 10.
Consider the singular second-order ordinary differential problem(55)x′′=λ1etsin⁡πtt1-tπ+xx1/2+1+cos⁡x2+x′33+etx′1/6+λ2e-2tt1-t1+t2,a.e.t∈0,1,x0=x1,2xsin⁡π8-xsin⁡π6+xsin⁡π4=0,where λ1,λ2∈R. Here, f(t,u,v)=λ1etsin⁡πt/t(1-t)π+uu1/2+1+cos⁡(u2+v3)/3+etv1/6+λ2e-2t/t(1-t)(1+t2). Clearly(56)ft,u,v≤ptu1/2+qtv1/6+rt,where p(t)=λ1et/t(1-t), q(t)=1/2, r(t)=λ2/t(1-t)(1+t2), and ϕ(ξi)=sin⁡ξiforξi∈[π/8,π/4]⊂(0,1). We can see that p(t),r(t)∈E and q(t)∈L1. Indeed we have pE=λ1(e-1), qL1=1/2, rE=|λ2|π/4, ϕ(ξi)≤ξi, and ∑i=13ai=2≠0. Assumptions (H0) and (H1) hold; therefore, by Theorem 8, singular boundary value problem (55) has a continuous solution in X.


## 4. Problems with Periodic and Integral Boundary Conditions

Following exactly the same arguments used in Sections2 and 3, we can prove analogous results for the problems (1), (3) and (1), (4), so we omit the proof of main results of this section.Firstly, we deal with the problem (1), (3). By analogous method as in Lemma 4, we have the following result.Lemma 11.
The singular second-order ordinary differential equation with periodic and integral boundary conditions (1), (3) is equivalent to the functional integral equation:(57)xt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+∫011-ϕθ∫0ϕθτfτ,xτ,x′τdτdθ+∫01ϕθ∫ϕθ11-τfτ,xτ,x′τdτdθ.
We define an operatorN1 associated with the boundary value problem (1), (3) as follows: for x∈X(58)N1xt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+∫011-ϕθ∫0ϕθτfτ,xτ,x′τdτdθ+∫01ϕθ∫ϕθ11-τfτ,xτ,x′τdτdθ.From (H1), we have (59)N1xt≤1-t∫0tτfτ,xτ,x′τdτ+t∫t11-τfτ,xτ,x′τdτ+∫011-ϕθ∫0ϕθτfτ,xτ,x′τdτdθ+∫01ϕθ∫ϕθ11-τfτ,xτ,x′τdτdθ,≤∫0tτ1-τfτ,xτ,x′τdτ+∫t1τ1-τfτ,xτ,x′τdτ+∫01∫0ϕθτ1-ϕθfτ,xτ,x′τdτ+∫ϕθ1ϕθ1-τfτ,xτ,x′τdτdθ,=∫01τ1-τfτ,xτ,x′τdτ+∫01∫01τ1-τfτ,xτ,x′τdτdθ,=2FxE≤2pExσ1+qL1x1σ2+rE<∞.Then (N1x):[0,1]→R is well defined and bounded.Lemma 12.
Letf:[0,1]×R2→R be a function satisfying Carathéodory’s conditions and (H1) is fulfilled; then (N1x)(t):X→X is a completely continuous operator and (60)Nx′′t=ft,xt,x′t,a.e.t∈0,1.Proof.
The proof is similar to that of Lemmas5 and 6.Using the operator (58) and the method of proof for the results obtained in Section 3, we can establish the following result for the problem (1), (3).Theorem 13.
Letf:[0,1]×R2→R be a function satisfying the Carathéodory conditions. Assume that the assumption (H1) holds. Then there exists at least one solution x∈X of the nonlinear singular ordinary differential equation with periodic and integral conditions (1), (3) represented by (57).The existence of solutions for the problem (1), (4) can be proved by a similar way.Theorem 14.
Letf:[0,1]×R2→R be a function satisfying the Carathéodory conditions. Assume that the assumption (H1) holds. Then there exists at least one solution x∈X of the nonlinear singular ordinary differential equation with periodic and nonlocal integral conditions(61)x′′t=ft,xt,x′t,a.e.t∈0,1,x0=x1,∫cdxθdηθ=0,0<c<d<1,represented by (62)xt=-1-t∫0tτfτ,xτ,x′τdτ-t∫t11-τfτ,xτ,x′τdτ+1ηd-ηc∫cd1-θ∫0θτfτ,xτ,x′τdτdηθ+∫cdθ∫θ11-τfτ,xτ,x′τdτdηθ,where η(θ) is an increasing function and the integral is meant in the Riemann-Stieltjes sense for 0<c<d<1.Example 15.
Consider the singular second-order ordinary differential problem(63)x′′=cos⁡t24t1-tsin⁡x1/42+t+1x′3/231+x′+1+cos2⁡xett1-t,a.e.t∈0,1,under periodic and integral conditions(64)x0=x1,∫01x23θ5dθ=0,or under periodic and nonlocal integral conditions(65)x0=x1,∫1/31/2xθsin⁡πθdθ=0.Here, f(t,u,v)=cos⁡t2/4t1-tsin⁡(|u|1/4/2)+(t+1)v3/2/3(1+v)+(1+cos2⁡x)et/t(1-t). Clearly(66)ft,u,v≤ptu1/4+qtv1/2+rt,where p(t)=1/8t(1-t), q(t)=(t+1)/3, and r(t)=2et/t(1-t), and we have ϕ(θ)=2/3θ5 and dη(θ)=sin⁡πθdθ. Clearly, pE=1/8, qL1=5/9, and rE=2(e-1). Assumption (H1) holds; therefore, by Theorems 13 and 14, singular boundary value problem (63) under the condition (64) or (65) has a continuous solution in X.



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