# Computation of Positive Solutions for Nonlinear Impulsive Integral Boundary Value Problems withp-Laplacian on Infinite Intervals

Authors: Xingqiu Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper deals with the existence and iteration of positive solutions for nonlinear second-order impulsive integral boundary value problems withp-Laplacian on infinite intervals. Our approach is based on the monotone iterative technique.


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## Body

## 1. Introduction

The theory of impulsive differential equations has been emerging as an important area of investigation in recent years. It has been extensively applied to biology, biologic medicine, optimum control in economics, chemical technology, population dynamics, and so on. It is much richer because all the structure of its emergence has deep physical background and realistic mathematical model and coincides with many phenomena in nature. For an introduction of the basic theory of impulsive differential equations inRn, the reader is referred to see Lakshmikantham et al. [1, 2], Samoĭlenko and Perestyuk [3], and the references therein.Boundary value problems on infinite intervals arise quite naturally in the study of radially symmetric solutions of nonlinear elliptic equations and models of gas pressure in a semi-infinite porous medium; see [4–7], for example. In a recent paper [8], by means of a fixed-point theorem due to Avery and Peterson, Li and Nieto obtained some new results on the existence of multiple positive solutions for the following multipoint boundary value problem with a finite number of impulsive times on an infinite interval:
(1)u′′(t)+q(t)f(t,u(t))=0,∀0<t<∞,t≠tk,k=1,2,…,p,Δu(tk)=Ik(u(tk)),k=1,2,…,p,u(0)=∑i=1m-2‍αiu(ξi),u′(∞)=0,
where f∈C([0,+∞)×[0,+∞),[0,+∞)), Ik∈C([0,+∞),[0,+∞)), u′(∞)=limt→∞u′(t), 0<ξ1<ξ2<⋯<ξm-2<+∞, 0<t1<t2<⋯<tp<+∞, and q∈C([0,+∞),[0,+∞)).Boundary value problems with integral boundary conditions for ordinary differential equations arise in different fields of applied mathematics and physics such as heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics. Moreover, boundary value problems with Riemann-Stieltjes integral conditions constitute a very interesting and important class of problems. They include two-point, three-point, and multipoint boundary value problems as special cases; see [9–14]. For boundary value problems with other integral boundary conditions and comments on their importance, we refer the reader to the papers [11–20] and the references therein.There are relatively few papers available for integral boundary value problems for impulsive differential equations on an infinite interval with an infinite number of impulsive times up to now. In [21], Zhang et al. investigated the existence of minimal nonnegative solution for the following second-order impulsive differential equation
(2)-x′′(t)=f(t,x(t),x′(t))=0,t∈J,t≠tk,Δx|t=tk=Ik(x(tk)),k=1,2,…,Δx′|t=tk=I¯k(x(tk)),k=1,2,…,x(0)=∫0+∞‍g(t)x(t)dt,x′(∞)=0,
where f∈C(J×J×J,J), Ik∈C(J,J), I¯k∈C(J,J), J=[0,+∞), 0<t1<t2<⋯<tk<⋯, tk→∞, and g(t)∈L[J,J] with ∫0+∞‍g(t)dt<1. Δx|t=tk denotes the jump of x(t) at t=tk, that is,
(3)Δx|t=tk=x(tk+)-x(tk-),
where x(tk+) and x(tk-) represent the right-hand limit and left-hand limit of x(t) at t=tk, respectively. Δx′|t=tk has a similar meaning to x′(t).In the past few years, the existence and the multiplicity of bounded or unbounded positive solutions to nonlinear differential equations on infinite intervals have been studied by several different techniques; we refer the reader to [5–8, 21–29] and the references therein. However, most of these papers only considered the existence of positive solutions under various boundary value conditions. Seeing such a fact, a natural question which arises is “how can we find the solutions when they are known to exist?” More recently, Ma et al. [30] and Sun et al. [31, 32] established iterative schemes for approximating the solutions for some boundary value problems defined on finite intervals by virtue of the iterative technique.However, to the author's knowledge, the corresponding theory for impulsive integral boundary value problems withp-Laplacian operator and infinite impulsive times on infinite intervals has not been considered till now. Motivated by previous papers, the purpose of this paper is to obtain the existence of positive solutions and establish a corresponding iterative scheme for the following impulsive integral boundary value problem of second-order differential equation with p-Laplacian on an infinite interval
(4)(φp(x′(t)))′+q(t)f(t,x(t),x′(t))=0,t∈J+′,Δx|t=tk=Ik(x(tk)),k=1,2,…,x(0)=∫0+∞‍g(t)x(t)dt,x′(∞)=x∞,
where φp(s)=|s|p-2s, p>1, J=[0,+∞), J+=(0,+∞), J+′=J+∖{t1,t2,…,tk,…}, 0<t1<t2<⋯<tk<⋯, tk→∞, and g(t)∈L[J,J] with ∫0+∞‍g(t)dt<1, ∫0+∞‍tg(t)dt<+∞, and 0≤x′(∞)=limt→+∞x′(t).It is clear that(5)φp(s+t)≤{2p-1(φp(s)+φp(t)),p≥2,s,t>0,φp(s)+φp(t),1<p<2,s,t>0,(6)φp-1(s+t)≤{21/(p-1)(φp-1(s)+φp-1(t)),p≥2,s,t>0,φp-1(s)+φp-1(t),1<p<2,s,t>0.
Throughout this paper, we adopt the following assumptions.(H1)
f(t,u,v)∈C(J×J×J,J),f(t,0,0)≢0 on any subinterval of J, and when u,v are bounded, f(t,(1+t)u,v) is bounded on J.(H2)
q(t) is a nonnegative measurable function defined in J+ and q(t) does not identically vanish on any subinterval of J+, and
(7)0<∫0+∞‍q(t)dt<+∞,0<∫0+∞‍φp-1(∫s+∞‍q(τ)dτ)ds<+∞.(H3)
Ik∈C(J,J), and there exist ak≥0,bk≥0 such that
(8)0≤Ik(x)≤ak+bkx,forx∈J(k=1,2,3,…),a*=∑k=1∞‍ak<+∞,b*=∑k=1∞‍bk(1+tk)<+∞,with b*<(1/3)(1-∫0+∞‍g(t)dt).Ifp=2,Ik=0(k=1,2,…), g(t)≡0, x′(∞)=0, then BVP (4) reduces to the following two-point boundary value problem:
(9)-x′′(t)=f(t,x(t),x′(t))=0,t∈J,x(0)=0,x′(∞)=0,
which has been studied in [23].Compared with [8, 21], the main features of the present paper are as follows. Firstly, second-order differential operator is replaced by a more general p-Laplacian operator. Secondly, in this paper, x∞ in boundary value conditions may not be zero which will bring about computational difficulties. Thirdly, by applying monotone iterative techniques, we construct successive iterative schemes starting off with simple known functions. It is worth pointing out that the first terms of our iterative schemes are simple functions. Therefore, the iterative schemes are significant and feasible.The rest of this paper is organized as follows. In Section2, we give some preliminaries and establish several lemmas. The main theorems are formulated and proved in Section 3. Then, in Section 4, an example is presented to illustrate the main results.


## 2. Preliminaries and Several Lemmas

Definition 1.
LetE be a real Banach space. A nonempty closed set P⊂E is said to be a cone provided that (1)
au+bv∈P for all u,v∈P and all a≥0,b≥0,(2)
u,-u∈P implies that u=0.Definition 2.
A mapα:P→[0,+∞) is said to be concave on P, if α(tu+(1-t)v)≥tα(u)+(1-t)α(v) for all u,v∈P and t∈[0,1].
LetPC[J,R]={x:x is a map from J into R such that x(t) is continuous at t≠tk, left continous at t=tk and x(tk+) exists for k=1,2,…,}, PC1[J,R]={x∈PC[J,R]:x′(t) exists and is continuous at t≠tk, left continous at t=tk and x′(tk+) exists for k=1,2,…,}(10)FPC[J,R]={x∈PC[J,R]:supt∈J|x(t)|1+t<∞},E=DPC1[J,R]={x∈PC1[J,R]:supt∈J|x(t)|1+t<∞,supt∈J|x′(t)|<∞}.
Obviously, DPC1[J,R]⊂FPC[J,R]. It is clear that FPC[J,R] is a Banach space with the norm
(11)∥x∥F=supt∈J|x(t)|1+t,
and DPC1[J,R] is also a Banach space with the norm
(12)∥x∥D=max{∥x∥F,∥x′∥B},
where ∥x′∥B=supt∈J|x′(t)|. Let J0=[0,t1], Jk=(tk,tk+1](k=1,2,3,…). Define a cone P⊂E by
(13)P={x∈E:xisconcaveandnondecreasingonJ,x(t)≥0,x′(t)≥0,t∈J}.Remark 3.
Ifx satisfies (4), then (φp(x′(t)))′=-q(t)f(t,x(t),x′(t))≤0, and t∈[0,+∞) which implies that φp(x′(t)) is nonincreasing on J; that is, x′(t) is also nonincreasing on J. Thus, x is concave on [0,+∞). Moreover, if x′(∞)=x∞≥0, then x′(t)≥0, t∈[0,+∞), and so x is monotone increasing on [0,+∞).Lemma 4.
Let conditions(H1)–(H3) hold. Then, x∈P with (φp(x′(t)))′∈C(0,+∞) is a solution of BVP (4) if and only if x∈C[0,+∞) is a fixed point of the following operator equation:
(14)(Ax)(t)=11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk)).Proof.
Suppose thatx∈P with (φp(x′(t)))′∈C(0,+∞) is a solution of BVP (4). For t∈J, integrating (4) from t to +∞, we have
(15)∫t+∞‍φp(x′(τ))′dτ=-∫t+∞‍q(τ)f(τ,x(τ),x′(τ))dτ.
That is
(16)φp(x′(∞))-φp(x′(t))=-∫t+∞‍q(τ)f(τ,x(τ),x′(τ))dτ,
which implies that
(17)x′(t)=φp-1(∫t+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)).
If t1<t≤t2, integrating (17) from 0 to t1, we get that
(18)x(t1)-x(0)=∫0t1‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds.
Integrating (17) from t1 to t, we obtain
(19)x(t)-x(t1+)=∫t1t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds.
Adding (18) and (19) together, we have
(20)x(t)=x(0)+∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+I1(x(t1)),t1<t≤t2.
Repeating previous process, we get that
(21)x(t)=x(0)+∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(x(tk))=∫0+∞‍g(t)x(t)dt+∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(x(tk)).
It follows that
(22)∫0+∞‍g(t)x(t)dt=11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt.
Substituting (22) into (21), we get that
(23)x(t)=11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(x(tk)).
Forx∈P, there exists r0 such that ∥x∥D<r0. Set Br0=sup{f(t,(1+t)u,v)∣(t,u,v)∈J×[0,r0]×[0,r0]}, and we have by (H1) and (H3) that
(24)∫0+∞‍q(s)f(s,x(s),x′(s))ds≤∫0+∞‍q(s)ds·Br0,∑tk<t‍Ik(x(tk))≤∑k=1∞‍Ik(x(tk))≤a*+b*r0<+∞.
By (6), (24), we have
(25)x(t)=11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(x(tk))≤11-∫0+∞‍g(t)dt×∫0+∞‍tg(t)dt·φp-1×(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+tφp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+11-∫0+∞‍g(t)dt(a*+b*r0)≤21/(p-1)(11-∫0+∞‍g(t)dt·∫0+∞‍tg(t)dt+t)×[φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ)+x∞]+11-∫0+∞‍g(t)dt(a*+b*r0).
It follows from (24) and (25) that the right term in (23) is well defined. Thus, we have proved that x is a fixed point of the operator A defined by (14).
Conversely, suppose thatx∈C[0,+∞) is a fixed point of the operator equation (14). Evidently,
(26)Δx|t=tk=Ik(x(tk))(k=1,2,…).
Direct differentiation of (14) implies that, for t≠tk,
(27)x′(t)=φp-1(∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞)),Δx′|t=tk=0(k=1,2,…),(φp(x′(t)))′=-q(t)f(t,x(t),x′(t)),
which means that (φp(x′(t)))′∈C(J′). It is easy to verify that x(0)=∫0+∞‍g(t)x(t)dt,  x′(∞)=x∞. The proof of Lemma 4 is complete.To obtain the complete continuity ofA, the following lemma is still needed.Lemma 5 (see [33, 34]).
LetW be a bounded subset of P. Then, P is relatively compact in E if {W(t)/(1+t)} and {W′(t)} are both equicontinuous on any finite subinterval Jk∩[0,T](k=1,2,…) for any T>0, and for any ε>0, there exists N>0 such that
(28)|x(t′)1+t′-x(t′′)1+t′′|<ε,|x′(t′)-x′(t′′)|<ε,∀t′,t′′≥N,
uniformly with respect to x∈W as t′,t′′≥N, where W(t)={x(t)∣x∈W}, W′(t)={x′(t)∣x∈W}, t∈[0,+∞).This lemma is a simple improvement of the Corduneanu theorem in [33, 34].Lemma 6.
Let(H1)–(H3) hold. Then A:P→P is completely continuous.Proof.
For anyx∈P, by (14), we have
(29)φp((Ax)′)(t)=∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞),(φp(Ax)′(t))′=-q(t)f(t,x(t),x′(t)).
It follows from (14), (29), and (H1) that (Ax)(t)≥0, (Ax)′(t)≥x∞≥0, (Ax)′′(t)≤0, that is, A(P)⊂P. Now, we prove that A is continuous and compact respectively. Let xn∈P, xn→x as n→∞, then there exists r0 such that supn∈N∖{0}∥xn∥<r0. Let Br0=sup{f(t,(1+t)u,v)∣(t,u,v)∈J×[0,r0]×[0,r0]}. By (H1) and (H2), we have
(30)∫0+∞‍q(τ)|f(τ,xn(τ),xn′(τ))-f(τ,x(τ),x′(τ))|dτ≤2Br0·∫0+∞‍q(s)ds<+∞.
It follows from (30) and dominated convergence theorem that
(31)∫0+∞‍q(τ)|f(τ,xn(τ),xn′(τ))-f(τ,x(τ),x′(τ))|dτ→∫0+∞‍q(τ)|f(τ,x(τ),x′(τ))-f(τ,x(τ),x′(τ))|dτ,
which implies that
(32)|φp-1(∫0+∞‍q(τ)f(τ,xn(τ),xn′(τ))dτ+φp(x∞))ds-φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds|→0,n→∞.
By (30)–(32), (H3) and dominated convergence theorem, we get that
(33)∥Axn-Ax∥F=supt∈J{11-∫0+∞‍g(t)dt11+t×|11-∫0+∞‍g(t)dt×[∑tk<t‍∫0+∞‍g(t)×∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,xn(τ),xn′(τ))dτ+φp(x∞)∫s+∞‍)dsdt-∫0+∞‍g(t)×∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)dsdt+∫0+∞‍g(t)·∑tk<t‍(Ik(xn(tk))-Ik(x(tk)))]dt+∫0t‍[φp-1(∫s+∞‍q(τ)f(τ,xn(τ),xn′(τ))dτ+φp(x∞)∫s+∞‍∫s+∞‍)-φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)]ds+∑tk<t‍(Ik(xn(tk))-Ik(x(tk)))11-∫0+∞‍g(t)dt|}≤11-∫0+∞‍g(t)dt×∫0+∞‍tg(t)dt·|φp-1(∫0+∞‍q(τ)f(τ,xn(τ),xn′(τ))dτ+φp(x∞))-φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))|+|φp-1(∫0+∞‍q(τ)f(τ,xn(τ),xn′(τ))dτ+φp(x∞))-φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))|+11-∫0+∞‍g(t)dt×∫0+∞‍g(t)·∑tk<t‍|Ik(xn(tk))-Ik(x(tk))|dt+∑tk<t‍|Ik(xn(tk))-Ik(x(tk))|→0(n→∞),∥(Axn)′-(Ax)′∥B=supt∈J{|φp-1(∫t+∞‍q(s)f(s,xn(s),xn′(s))ds+φp(x∞))=supt∈J-φp-1(∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞))|}→0(n→∞).It follows from (33) that A is continuous.
LetΩ⊂P be any bounded subset. Then, there exists r>0 such that ∥x∥D≤r for any x∈Ω. Obviously,
(34)∥Ax∥F=supt∈J{11-∫0+∞‍g(t)dt11+t×|11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))11-∫0+∞‍g(t)dt|}≤11-∫0+∞‍g(t)dt×∫0+∞‍tg(t)dt·φp-1×(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+11-∫0+∞‍g(t)dt(a*+b*r)≤21/(p-1)[11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1]×[φp-1(Br)·φp-1(∫0+∞‍q(τ)dτ)+x∞]+11-∫0+∞‍g(t)dt(a*+b*r),∥(Ax)′∥B=supt∈J{|φp-1(∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞))|}≤21/(p-1)[φp-1(∫0+∞‍q(s)ds)·φp-1(Br)+x∞].From (34), (H2), and (H3), we know that AΩ is bounded.
For anyT>0,x∈Ω,t′, t′′∈Jk∩[0,T] with t′<t′′, by the absolute continuity of the integral, we have
(35)|(Ax)(t′)1+t′-(Ax)(t′′)1+t′′|≤1(1+t′′)(1-∫0+∞‍g(t)dt)·∫0+∞‍g(t)dt·∫t′t′′‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+11-∫0+∞‍g(t)dt×∫0+∞‍g(t)dt·∫0t′‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds·|11+t′′-11+t′|+11+t′′×∫t′t′′‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds+∫0t′‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds·|11+t′′-11+t′|+a*+b*r1-∫0+∞‍g(t)dt·|11+t′′-11+t′|≤21/(p-1)1-∫0+∞‍g(t)dt×[∫t′t′′‍(φp-1(∫0+∞‍q(τ)dτ)·φp-1(Br)+x∞)ds+∫0t′‍(φp-1(∫0+∞‍q(τ)dτ)·φp-1(Br)+x∞)ds·|11+t′′-11+t′|∫t′t′′‍]+a*+b*r1-∫0+∞‍g(t)dt·|11+t′′-11+t′|→0uniformlyast′→t′′,|φp((Ax)′(t′))-φp((Ax)′(t′′))|=|∫t′t′′‍q(s)f(s,x(s),x′(s))ds|≤Br·|∫t′t′′‍q(s)ds|→0uniformlyast′→t′′.
Thus, we have proved that AΩ is equicontinuous on any Jk∩[0,T].
Next, we prove that for anyε>0,x∈Ω, there exits sufficiently large N>0 such that
(36)|(Ax)(t′)1+t′-(Ax)(t′′)1+t′′|<ε,|(Ax)′(t′)-(Ax)′(t′′)|<ε,∀t′,t′′≥N.
For any x∈Ω, we have
(37)limt→+∞11+t·[11-∫0+∞‍g(t)dt×∫0+∞‍g(t)∑tk<t‍Ik(x(tk))dt+∑tk<t‍Ik(x(tk))]≤limt→+∞11+t·a*+b*r1-∫0+∞‍g(t)dt=0,limt→+∞11+t·∫0+∞‍g(t)·[∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds]dt≤limt→+∞11+t21/(p-1)×(11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt)×[φp-1(Br)·φp-1(∫0+∞‍q(τ)dτ)+x∞]=0,limt→+∞11+t·∫0t‍φp-1(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))ds=limt→+∞φp-1(∫t+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))=x∞.
It follows from (37) that
(38)limt→∞|(Ax)(t)1+t|=limt→∞11+t×{11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))}=x∞.
On the other hand, we arrive at
(39)limt→∞|(Ax)′(t)|=limt→∞φp-1(∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞))=x∞.
Thus, (36) can be easily derived from (38) and (39). So, by Lemma 5, we know that AΩ is relatively compact. Thus, we have proved that A:P→P is completely continuous.


## 3. Main Results

For notational convenience, we denote that(40)m=21/(p-1)(11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1)·φp-1(∫0+∞‍q(τ)dτ),(41)m′=(11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1)·φp-1(∫0+∞‍q(τ)dτ),(42)n=21/(p-1)(11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1)x∞,(43)n′=(11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1)x∞,(44)Λ=max{a*1-∫0+∞‍g(t)dt-3b*,n},Λ′=max{a*1-∫0+∞‍g(t)dt-3b*,n′}.Theorem 7.
Assume that(H1)–(H3) hold, and there exists
(45)d>{3Λ,asp≥2,3Λ′,as1<p<2
such that (A1)
f(t,x1,y1)≤f(t,x2,y2) for any 0≤t<+∞,0≤x1≤x2≤d,0≤y1≤y2≤d,(A2)
(46)f(t,(1+t)u,v)≤{φp(d3m),asp≥2,φp(d3m′),as1<p<2,(t,u,v)∈[0,+∞)×[0,d]×[0,d],(A3)
Ik(x1)≤Ik(x2)(k=1,2,…,), for any 0≤x1≤x2.
Then, the boundary value problem (4) admits positive, nondecreasing on [0,+∞) and concave solutions w* and v* such that 0<∥w*∥D≤d, and limn→∞wn=limn→∞Anw0=w*, where
(47)w0(t)=d+dt,t∈J,
and 0<∥v*∥D≤d, limn→∞vn=limn→∞Anv0=v*, where v0(t)=0,t∈J.Proof.
We only prove the case thatp≥2. Another case can be proved in a similar way. By Lemma 6, we know that A:P→P is completely continuous. From the definition of A, (A1), and (A3), we can easily get that Ax1≤Ax2 for any x1,x2∈P with x1≤x2, x1′≤x2′. Denote that
(48)P¯d={x∈P∣∥x∥D≤d}.
In what follows, we first prove that A:P¯d→P¯d. If x∈P¯d, then ∥x∥D≤d. By (6), (40), (42), (44), (H3), (A2), and (A3), we get that
(49)∥Ax∥F=supt∈J{∑tk<t‍11+t×|11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))]dt+∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(x(tk))|}≤11-∫0+∞‍g(t)dt×∫0+∞‍tg(t)dt·φp-1×(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+φp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+11-∫0+∞‍g(t)dt(a*+b*d)≤21/(p-1)[11-∫0+∞‍g(t)dt∫0+∞‍tg(t)dt+1]×[φp-1(φp(d3m))·φp-1(∫0+∞‍q(τ)dτ)+x∞]+11-∫0+∞‍g(t)dt(a*+b*d)≤d3+d3+d3=d,(50)∥(Ax)′∥B=supt∈J{|φp-1(∫t+∞‍q(s)f(s,x(s),x′(s))ds+φp(x∞))|}≤21/(p-1)[φp-1(∫0+∞‍q(s)ds)φp-1(φp(d3m))+x∞]≤d.Thus, we get that ∥Ax∥D≤d. Hence, we have proved that A:P¯d→P¯d.
Letw0(t)=d+dt, 0≤t<+∞, then w0(t)∈P¯d. Let w1=Aw0, w2=A2w0, then by Lemma 6, we have that w1∈P¯d and w2∈P¯d. Denote that
(51)wn+1=Awn=An+1w0,n=0,1,2,….
Since A:P¯d→P¯d, we have that
(52)wn∈A(P¯d)⊂P¯d,n=1,2,3,….
It follows from the complete continuity of A that {wn}n=1∞ is a sequentially compact set. We assert that {wn}n=1∞ has a convergent subsequence {wnk}k=1∞, and there exists w*∈P¯d such that wnk→w*.
By (51), (A1)–(A3), we get that
(53)w1(t)=11-∫0+∞‍g(t)dt×∫0+∞‍g(t)×[∑tk<t‍∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,w0(τ),w0′(τ))dτ+φp(x∞)∫s+∞‍)ds+∑tk<t‍Ik(w0(tk))]dt+∫0t‍φp-1×(∫s+∞‍q(τ)f(τ,w0(τ),w0′(τ))dτ+φp(x∞))ds+∑tk<t‍Ik(w0(tk))≤11-∫0+∞‍g(t)dt×∫0+∞‍tg(t)dt·φp-1×(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+tφp-1(∫0+∞‍q(τ)f(τ,x(τ),x′(τ))dτ+φp(x∞))+11-∫0+∞‍g(t)dt(a*+b*d)≤21/(p-1)(11-∫0+∞‍g(t)dt·∫0+∞‍tg(t)dt)×[φp-1(∫0+∞‍q(τ)dτ)φp-1(φp(d3m))+x∞]+21/(p-1)t[φp-1(∫0+∞‍q(τ)dτ)φp-1(φp(d3m))+x∞]+11-∫0+∞‍g(t)dt(a*+b*d)≤d+dt=w0(t),w1′(t)=(Aw0)′(t)=φp-1(∫t+∞‍q(s)f(s,w0(s),w0′(s))ds+φp(x∞))≤21/(p-1)[φp-1(∫0+∞‍q(τ)dτ)φp-1(φp(d3m))+x∞]≤d=w0′(t),0≤t<+∞.So, by (53) (A1)–(A3) we have
(54)w2(t)=(Aw1)(t)≤(Aw0)(t)=w1(t),0≤t<+∞,w2′(t)=(Aw1)′(t)≤(Aw0)′(t)=(w1)′(t),0≤t<+∞.
By induction, we get that
(55)wn+1(t)≤wn(t),wn+1′(t)≤wn′(t),0≤t<+∞,n=0,1,2,….
Hence, we claim that wn→w* as n→∞. Applying the continuity of A and wn+1=Awn, we get that Aw*=w*.
Letv0(t)=0, 0≤t<+∞, then v0(t)∈P¯d. Let v1=Av0,v2=A2v0. By Lemma 6, we have that v1∈P¯d and v2∈P¯d. Denote
(56)vn+1=Avn=An+1v0,n=0,1,2,….
Since A:P¯d→P¯d, we have that vn∈A(P¯d)⊂P¯d,n=1,2,3,…. It follows from the complete continuity of A that {vn}n=1∞ is a sequentially compact set. And, we assert that {vn}n=1∞ has a convergent subsequence {vnk}k=1∞ and there exists v*∈P¯d such that vnk→v*.
Sincev1=Av0∈P¯d, we have
(57)v1(t)=(Av0)(t)=(A0)(t)≥0,0≤t<+∞,v1′(t)=(Av0)′(t)=(A0)′(t)≥0=v0′(t),0≤t<+∞.
By (A1)–(A3), we have
(58)v2(t)=(Av1)(t)≥(A0)(t)=v1(t),0≤t<+∞,v2′(t)=(Av1)′(t)≥(A0)′(t)=v1′(t),0≤t<+∞.
By induction, we get that
(59)vn+1(t)≥vn(t),vn+1′(t)≥vn′(t),0≤t<+∞,n=0,1,2,….
Hence, we claim that vn→v* as n→∞. Applying the continuity of A and vn+1=Avn, we get that Av*=v*.
Sincef(t,0,0)≢0,0≤t<∞, then the zero function is not the solution of BVP (4). Thus, v* is a positive solution of BVP (4). By Lemma 4 we know that w* and v* are positive, nondecreasing on [0,+∞) and concave solutions of the BVP (4).
We can easily get that Theorem7 holds for 1<p<2 in a similar manner.Remark 8.
The iterative schemes in Theorem7 are w0(t)=d+dt, wn+1=Awn=An+1w0, n=0,1,2,… and v0(t)=0, vn+1=Avn=An+1v0, n=0,1,2,…. They start off with a known simple linear function and the zero function respectively. This is convenient in application.Theorem 9.
Assume that(H1)–(H3) hold, and there exist
(60)dn>dn-1>⋯>d1>{3Λ,asp≥2,3Λ′,as1<p<2
such that (A1′)
f(t,x1,y1)≤f(t,x2,y2) for any 0≤t<+∞, 0≤x1≤x2, 0≤y1≤y2.(A2′)
(61)f(t,(1+t)u,v)≤{φp(dk3m),asp≥2,φp(dk3m′),as1<p<2(t,u,v)∈[0,+∞)×[0,dk]×[0,dk],k=1,2,…,n.(A3′)
Ik(x1)≤Ik(x2)(k=1,2,…,), for any 0≤x1≤x2.
Then, the boundary value problem (4) admits positive nondecreasing on [0,+∞) and concave solutions wk* and vk*, such that 0<∥wk*∥D≤dk, and limn→∞wkn=limn→∞Anwk0=wk*, where
(62)w0(t)=dk+dkt,t∈J,
and 0<∥vk*∥D≤dk, limn→∞vkn=limn→∞Anvk0=vk*, where v0(t)=0, t∈J.Remark 10.
It is easy to see thatw* and v* in Theorem 7 may coincide, and then the boundary value problem (4) has only one solution in P. Similarly, positive solutions wk* and vk* may also coincide.


## 4. An Example

Example 11.
Consider the following impulsive integral boundary value problem:(63)(|x′|x′)′+e-6tf(t,x(t),x′(t))=0,t∈J+,Δx|t=k=19[12k+2x(k)+12k+1(1+x(k))1/6],x(0)=∫0+∞‍1(1+t)3x(t)dt,x′(∞)=23,
where
(64)f(t,u,v)={164|sin(101t+20)|+172(u1+t)3+110(v20),u≤2,164|sin(101t+20)|+172(21+t)3+110(v20),u≥2.
It is clear that conditions (H1), (A1), and (A3) hold for p=3, q(t)=e-6t,  g(t)=1/(1+t)3. By direct computation, we obtain that
(65)∫0+∞q(t)dt=16,∫0+∞‍φp-1(∫s+∞‍q(τ)dτ)ds=318,
which implies that (H2) holds.
Obviously,Ik∈C(J,J). Using a simple inequality
(66)(1+u)α≤1+αu,∀u≥0,0<α<1,
we get that
(67)Ik(x(k))≤19[12k+2x(k)+12k+1(1+16x(k))]≤19·12k+1+227·12k+1x(k).
Thus, (H3) holds for ak=(1/9)·(1/2k+1), bk=(2/27)·(1/2k+1). Considering that
(68)∫0+∞‍g(t)dt=∫0+∞‍1(1+t)3dt=12,∫0+∞‍tg(t)dt=∫0+∞‍t(1+t)3dt=12,φp-1(∫0+∞‍q(τ)dτ)=φp-1(16)=66,
we can obtain that
(69)a*=∑k=1∞‍ak=118,b*=227∑k=1∞‍1+k2k+1=227(∑k=1∞‍12k+1+∑k=1∞‍k2k+1)=19,m=3,n=2,Λ=2.
Take d=8. In this case, we have
(70)φp(d3m)=φp(833)=6427.
On the other hand, nonlinear term f satisfies
(71)f(t,(1+t)u,v)≤164+19+125=240114400,t∈[0,+∞),u,v∈[0,8],
which means that (A2) holds. Thus, we have checked that all the conditions of Theorem 7 are satisfied. Therefore, the conclusion of Theorem 7 holds.



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*Source: 708281-2013-02-25.xml*