# Monotonic Positive Solutions of Nonlocal Boundary Value Problems for a Second-Order Functional Differential Equation

Authors: A. M. A. El-Sayed; E. M. Hamdallah; Kh. W. El-kadeky
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the existence of at least one monotonic positive solution for the nonlocal boundary value problem of the second-order functional differential equationx′′(t)=f(t,x(ϕ(t))), t∈(0,1), with the nonlocal condition ∑k=1makx(τk)=x0, x′(0)+∑j=1nbjx′(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 ∫adx(t)dt=x0, x′(0)+x(e)-x(c)=x1 will be considered.


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## Body

## 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 [3–22] and references therein.In most of all these papers, the authors assume that the functionf:[0,1]×R+→R+ is continuous. They all assume that(1.1)lim⁡x→∞f(x)x=0or∞,lim⁡x→0f(x)x=0or∞.
These assumptions are restrictive, and there are many functions that do not satisfy these assumptions.Here we assume that the functionf:[0,1]×R+→R+ is measurable in t∈[0,1] for all x∈R+ and continuous in x∈R+ for almost all t∈[0,1] is and there exists an integrable function a∈L1[0,1] and a constant b>0 such that(1.2)|f(t,x)|≤|a(t)|+b|x|,∀(t,x)∈[0,1]×D.
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 equation(1.3)x′′(t)=f(t,x(ϕ(t))),t∈(0,1),
with the nonlocal condition(1.4)∑k=1makx(τk)=x0,x′(0)+∑j=1nbjx′(ηj)=x1,
where τk∈(a,d)⊂(0,1),ηj∈(c,e)⊂(0,1),andx0,x1>0.As an application, the problem with the integral and nonlocal conditions(1.5)∫adx(t)dt=x0,x′(0)+x(e)-x(c)=x1,
is studied.It must be noticed that the nonlocal conditions(1.6)x(τ)=x0,τ∈(a,d),x′(0)+x′(η)=x1,η∈(c,e),∑k=1makx(τk)=0,τk∈(a,d),x′(0)+∑j=1nbjx′(ηj)=0,ηj∈(c,e),∫adx(t)dt=0,x′(0)+x(e)=x(c)
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 x∈R+ and continuous in x∈R+ for almost all t∈[0,1] and there exists an integrable function a∈L1[0,1], and a constant b>0 such that
(2.1)|f(t,x)|≤|a(t)|+b|x|,∀(t,x)∈[0,1]×D.(ii)
ϕ:(0,1)→(0,1) is continuous.(iii)
b<1/(3-B),B=(∑j=1nbj+1)-1.(iv)
(2.2)∑k=1mak>0,∀k=1,2,…,m,∑j=1nbj>0,∀j=1,2,…,n.
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
(2.3)x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(t-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0t(t-s)f(s,x(ϕ(s)))ds,
where A=(∑k=1mak)-1,B=(∑j=1nbj+1)-1.Proof.
Integrating (1.3), we get
(2.4)x′(t)=x′(0)+∫0tf(s,x(ϕ(s)))ds.
Integrating (2.4), we obtain
(2.5)x(t)=x(0)+x′(0)t+∫0t(t-s)f(s,x(ϕ(s)))ds.
Let t=τk, in (2.5), we get
(2.6)∑k=1makx(τk)=∑k=1nakx(0)+∑k=1nakτkx′(0)+∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds,
and we deduce that
(2.7)x(0)=A{x0-∑k=1makτkx′(0)-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds},A=(∑k=1mak)-1.
Substitute from (2.7) into (2.5), we obtain
(2.8)x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+x′(0)(t-A∑k=1makτk)+∫0t(t-s)f(s,x(ϕ(s)))ds.
Let t=ηj, in (2.4), we obtain
(2.9)∑j=1nbjx′(ηj)=∑j=1nbjx′(0)+∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds,x1-x′(0)=x′(0)∑j=1nbj+∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds, and we deduce that
(2.10)x′(0)=B(x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds),B=(∑j=1nbj+1)-1.
Substitute from (2.10) into (2.8), we obtain
(2.11)x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(t-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds},+∫0t(t-s)f(s,x(ϕ(s)))ds,
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 solutionx∈C[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 x∈C[0,1].Proof.
Define the subsetQr⊂C(0,1) by Qr={x∈C:|x(t)|≤r,r=(Ax0+Bx1+(3-B)∥a∥)/(1-(3-B)b),r>0}. Clear the set Qr which is nonempty, closed, and convex.
LetH be an operator defined by(3.1)(Hx)(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(t-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0t(t-s)f(s,x(ϕ(s)))ds.
Letx∈Qr, then(3.2)|(Hx)(t)|≤A{x0+∑k=1mak∫0τk(τk-s)|f(s,x(ϕ(s)))|ds}+B(t-A∑k=1makτk){x1+∑j=1nbj∫0ηj|f(s,x(ϕ(s)))|ds}+∫0t(t-s)|f(s,x(ϕ(s)))|ds≤A{x0+∑k=1mak∫01[|a(s)|+b|x(ϕ(s))|]ds}+B{x1+∑j=1nbj∫01[|a(s)|+b|x(ϕ(s))|]ds}+∫01[|a(s)|+b|x(ϕ(s))|]ds≤Ax0+‖a‖+bsup⁡t∈I|x(ϕ(t))|+Bx1+B∑j=1nbj‖a‖+bB∑j=1nbjsup⁡t∈I|x(ϕ(t))|+‖a‖+bsup⁡t∈I|x(ϕ(t))|≤Ax0+Bx1+2‖a‖+2b‖x‖+(1-B)‖a‖+b(1-B)‖x‖≤Ax0+Bx1+(3-B)‖a‖+(3-B)br≤r,
then H:Qr→Qr and {Hx(t)} is uniformly bounded in Qr.
Also fort1,t2∈[0,1] such that t1<t2, we have(3.3)(Hx)(t2)-(Hx)(t1)=B(t2-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(t)))ds}+∫0t2(t2-s)f(s,x(ϕ(t)))ds-B(t1-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(t)))ds}-∫0t1(t1-s)f(s,x(ϕ(t)))ds=B(t2-t1){x1-∑j=1nbj∫0ηjf(s,x(ϕ(t)))ds}+∫0t1(t2-t1)f(s,x(ϕ(t)))ds+∫t1t2(t2-s)f(s,x(ϕ(t)))ds.
Then
(3.4)|(Hx)(t2)-(Hx)(t1)|≤B|t2-t1|{x1+∑j=1nbj∫0ηj[|a(s)|+b|x(ϕ(s))|]ds}+|t2-t1|∫0t1[|a(s)|+b|x(ϕ(s))|]ds+∫t1t2(t2-s)[|a(s)|+b|x(ϕ(s))|]ds≤B|t2-t1|x1+∑j=1nbj[‖a‖+br]+|t2-t1|[‖a‖+br]+∫t1t2‖a‖ds+br[t2-t1].
The above inequality shows that
(3.5)|(Hx)(t2)-(Hx)(t1)|⟶0ast2⟶t1.
Therefore {Hx(t)} is equicontinuous. By the Arzelà-Ascoli theorem, {Hx(t)} is relatively compact.
Since all conditions of the Schauder theorem hold, thenH has a fixed point in Qr which proves the existence of at least one solution x∈C[0,1] of the integral equation (2.3), where(3.6)lim⁡t→0+x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}-BA∑k=1makτk{x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}=x(0),lim⁡t→1-x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(1-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫01(1-s)f(s,x(ϕ(s)))ds=x(1).
To complete the proof, we prove that the integral equation (2.3) satisfies nonlocal problem (1.3)-(1.4). Differentiating (2.3), we get
(3.7)x′(t)=B{x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0tf(s,x(ϕ(s)))ds,(3.8)x′′(t)=f(t,x(ϕ(t))).
Let t=τk in (2.3), we obtain
(3.9)x(τk)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+∫0τk(τk-s)f(s,x(ϕ(s)))ds,
which proves
(3.10)∑k=1makx(τk)=x0.
Also let t=ηj in (3.7), we obtain
(3.11)x′(ηj)=B{x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0ηjf(s,x(ϕ(s)))ds,
then
(3.12)∑j=1nbjx′(ηj)=B∑j=1nbj{x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds.
Let t=0 in (3.7), we obtain
(3.13)x′(0)=B{x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}.
Adding (3.12) and (3.13), we obtain
(3.14)x′(0)+∑j=1nbjx′(ηj)=x1.
This implies that there exists at least one solution x∈C[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
(3.15)x(t1)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(t1-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0t1(t1-s)f(s,x(ϕ(s)))ds<A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+B(t2-A∑k=1makτk){x1-∑j=1nbj∫0ηjf(s,x(ϕ(s)))ds}+∫0t2(t2-s)f(s,x(ϕ(s)))ds=x(t2),
which proves that the solution x of the problem (1.3)-(1.4) is monotonic nondecreasing.


### 3.1. Positive Solution

Letbj=0,j=1,2,…n and x1=0, then the nonlocal problem condition (1.4) will be(3.16)∑k=1makx(τk)=x0,x′(0)=0.Theorem 3.3.
Let the assumptions (i)–(iv) of Theorem3.1 be satisfied. Then the solution of the nonlocal problem (1.3)–(3.16) is positive t∈[d,1].Proof.
Letbj=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
(3.17)x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+∫0t(t-s)f(s,x(ϕ(s)))ds,
where A=(∑k=1mak)-1.
Lett∈[d,1], then(3.18)∫0τk(τk-s)f(s,x(ϕ(s)))ds≤∫0t(t-s)f(s,x(ϕ(s)))ds,τk≤t,∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds≤∑k=1mak∫0t(t-s)f(s,x(ϕ(s)))ds.
Multiplying by A=(∑k=1mak)-1, we obtain
(3.19)A∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds≤A∑k=1mak∫0t(t-s)f(s,x(ϕ(s)))ds=∫0t(t-s)f(s,x(ϕ(s)))ds,
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
(3.20)x′(0)=0,x(η)=x0,η∈(a,d)⊂(0,1).
Applying our results here, we deduce that the two-point boundary value problem (1.3)–(3.20) has at least one monotonic nondecreasing solution x∈C[0,1] represented by the integral equation
(3.21)x(t)=x0-∫0η(η-s)f(s,x(ϕ(s)))ds+∫0t(t-s)f(s,x(ϕ(s)))ds.
This the solution is positive with t>η.


## 3.1. Positive Solution

Letbj=0,j=1,2,…n and x1=0, then the nonlocal problem condition (1.4) will be(3.16)∑k=1makx(τk)=x0,x′(0)=0.Theorem 3.3.
Let the assumptions (i)–(iv) of Theorem3.1 be satisfied. Then the solution of the nonlocal problem (1.3)–(3.16) is positive t∈[d,1].Proof.
Letbj=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
(3.17)x(t)=A{x0-∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds}+∫0t(t-s)f(s,x(ϕ(s)))ds,
where A=(∑k=1mak)-1.
Lett∈[d,1], then(3.18)∫0τk(τk-s)f(s,x(ϕ(s)))ds≤∫0t(t-s)f(s,x(ϕ(s)))ds,τk≤t,∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds≤∑k=1mak∫0t(t-s)f(s,x(ϕ(s)))ds.
Multiplying by A=(∑k=1mak)-1, we obtain
(3.19)A∑k=1mak∫0τk(τk-s)f(s,x(ϕ(s)))ds≤A∑k=1mak∫0t(t-s)f(s,x(ϕ(s)))ds=∫0t(t-s)f(s,x(ϕ(s)))ds,
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
(3.20)x′(0)=0,x(η)=x0,η∈(a,d)⊂(0,1).
Applying our results here, we deduce that the two-point boundary value problem (1.3)–(3.20) has at least one monotonic nondecreasing solution x∈C[0,1] represented by the integral equation
(3.21)x(t)=x0-∫0η(η-s)f(s,x(ϕ(s)))ds+∫0t(t-s)f(s,x(ϕ(s)))ds.
This the solution is positive with t>η.


## 4. Nonlocal Integral Condition

Letx∈C[0,1] be the solution of the nonlocal problem (1.3) and (1.4).Letak=tk-tk-1,τk∈(tk-1,tk)⊂(a,d)⊂(0,1) and let bj=ξj-ξj-1,ηj∈(ξj-1,ξj)⊂(c,e)⊂(0,1), then(4.1)∑k=1m(tk-tk-1)x(τk)=x0,x′(0)+∑j=1n(ξj-ξj-1)x′(ηj)=x1.
From the continuity of the solution x of the nonlocal problem (1.3) and (1.4), we obtain(4.2)lim⁡m→∞∑k=1m(tk-tk-1)x(τk)=∫adx(s)ds,x′(0)+lim⁡n→∞∑j=1n(ξj-ξj-1)x′(ηj)=x′(0)+∫cex′(s)ds,
and the nonlocal condition (1.4) transformed to the integral condition(4.3)∫adx(s)ds=x0,x′(0)+x(e)-x(c)=x1,
and the solution of the integral equation (2.3) will be(4.4)x(t)=(d-a)-1{x0-∫ad∫0t(t-s)f(s,x(ϕ(s)))dsdt}+((b-c)+1)-1(t-1){x1-∫ce∫0tf(s,x(ϕ(s)))dsdt}+∫0tf(s,x(ϕ(s)))ds.
Now, we have the following theorem.Theorem 4.1.
Let the assumptions (i)–(iv) of Theorem3.1 be satisfied. Then the nonlocal problem
(4.5)x′′(t)=f(t,x(ϕ(t))),t∈(0,1),∫adx(s)ds=x0,x′(0)+x(e)-x(c)=x1
has at least one monotonic nondecreasing solution x∈C[0,1] represented by (4.4).



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