Volume 2009, Article ID 273063,18pages doi:10.1155/2009/273063
Research Article
Positive Solutions for a Class of Coupled System of
Singular Three-Point Boundary Value Problems
Naseer Ahmad Asif and Rahmat Ali Khan
Centre for Advanced Mathematics and Physics, Campus of College of Electrical and Mechanical
Engineering, National University of Sciences and Technology, Peshawar Road, Rawalpindi 46000, Pakistan
Correspondence should be addressed to Rahmat Ali Khan,rahmat [email protected] Received 27 February 2009; Accepted 15 May 2009
Recommended by Juan J. Nieto
Existence of positive solutions for a coupled system of nonlinear three-point boundary value problems of the type−xt ft, xt, yt, t ∈ 0,1, −yt gt, xt, yt, t ∈ 0,1,
x0 y0 0,x1 αxη,y1 αyη, is established. The nonlinearitiesf,g : 0,1×
0,∞×0,∞ → 0,∞are continuous and may be singular att0, t1, x0, and/ory0, while the parametersη,αsatisfyη∈0,1,0< α <1/η. An example is also included to show the applicability of our result.
Copyrightq2009 N. A. Asif and R. A. Khan. 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.
1. Introduction
Multipoint boundary value problemsBVPsarise in different areas of applied mathematics and physics. For example, the vibration of a guy wire composed ofNparts with a uniform cross-section and different densities in different parts can be modeled as a Multipoint boundary value problem 1. Many problems in the theory of elastic stability can also be modeled as Multipoint boundary value problem2.
The study of Multipoint boundary value problems for linear second order ordinary differential equations was initiated by Il’in and Moiseev, 3, 4, and extended to nonlocal linear elliptic boundary value problems by Bitsadze et al.5,6. Existence theory for nonlinear three-point boundary value problems was initiated by Gupta7. Since then the study of nonlinear three-point BVPs has attracted much attention of many researchers, see8–11and references therein for boundary value problems with ordinary differential equations and also
The study of system of BVPs has also fascinated many authors. System of BVPs with continuous nonlinearity can be seen in20–22and the case of singular nonlinearity can be seen in 8,21, 23–26. Wei 25, developed the upper and lower solutions method for the existence of positive solutions of the following coupled system of BVPs:
−xt ft, xt, yt, t∈0,1,
−yt gt, xt, yt, t∈0,1,
x0 0, x1 0,
y0 0, y1 0,
1.1
wheref, g∈C0,1×0,∞×0,∞,0,∞, and may be singular att0,t1,x0 and/or
y0.
By using fixed point theorem in cone, Yuan et al.26studied the following coupled system of nonlinear singular boundary value problem:
x4t ft, xt, yt, t∈0,1,
−yt gt, xt, yt, t∈0,1,
x0 x1 x0 x1 0,
y0 y1 0,
1.2
f, gare allowed to be superlinear and are singular att0 and/ort1. Similarly, results are studied in8,21,23.
In this paper, we generalize the results studied in25,26to the following more general singular system for three-point nonlocal BVPs:
−xt ft, xt, yt, t∈0,1,
−yt gt, xt, yt, t∈0,1,
x0 0, x1 αxη,
y0 0, y1 αyη,
1.3
whereη∈0,1, 0< α <1/η,f, g∈C0,1×0,∞×0,∞,0,∞. We allowfandgto be singular att 0,t 1, and alsox 0 and/ory 0. We study the sufficient conditions for existence of positive solution for the singular system1.3under weaker hypothesis onfand
gas compared to the previously studied results. We do not require the system1.3to have lower and upper solutions. Moreover, the cone we consider is more general than the cones considered in20,21,26.
studied in21,24–26. Throughout this paper, we assume thatf, g:0,1×0,∞×0,∞ →
0,∞are continuous and may be singular att0,t1,x0, and/ory0. We also assume that the following conditions hold:
A1f·,1,1, g·,1,1∈C0,1,0,∞and satisfy
a: 1
0
t1−tft,1,1dt < ∞, b: 1
0
t1−tgt,1,1dt < ∞. 1.4
A2There exist real constantsαi, βisuch thatαi≤0≤βi <1,i1,2,β1 β2<1 and for allt∈0,1,x, y∈0,∞,
cβ1ft, x, y≤ft, cx, y≤cα1ft, x, y, if 0< c≤1,
cα1ft, x, y≤ft, cx, y≤cβ1ft, x, y, if c≥1,
cβ2ft, x, y≤ft, x, cy≤cα2ft, x, y, if 0< c≤1,
cα2ft, x, y≤ft, x, cy≤cβ2ft, x, y, ifc≥1.
1.5
A3There exist real constantsγi, ρisuch thatγi≤0≤ρi<1,i1,2,ρ1 ρ2<1 and for allt∈0,1,x, y∈0,∞,
cρ1gt, x, y≤gt, cx, y≤cγ1gt, x, y, if 0< c≤1,
cγ1gt, x, y≤gt, cx, y≤cρ1gt, x, y, if c≥1,
cρ2gt, x, y≤gt, x, cy≤cγ2gt, x, y, if 0< c≤1,
cγ2gt, x, y≤gt, x, cy≤cρ2gt, x, y, ifc≥1,
1.6
for example, a function that satisfies the assumptionsA2andA3is
ht, x, y
m
i1 n
j1
pijtxriysj, 1.7
wherepij∈C0,1,0,∞,ri, sj<1,i1,2, . . . , m;j1,2, . . . , nsuch that
max
1≤i≤mri max1≤j≤nsj<1. 1.8
The main result of this paper is as follows.
2. Preliminaries
For eachu∈E:C0,1, we writeumax{ut:t∈0,1}. LetP {u∈E:ut≥0, t∈
0,1}. Clearly,E, · is a Banach space andP is a cone. Similarly, for eachx, y ∈E×E, we writex, y1x y. Clearly,E×E, · 1is a Banach space andP×Pis a cone in
E×E. For any real constantr >0, defineΩr {x, y∈E×E:x, y1 < r}. By a positive solution of1.3, we mean a vectorx, y∈C0,1∩C20,1×C0,1∩C20,1such thatx, y satisfies1.3andx >0,y >0 on0,1. The proofs of our main resultTheorem 1.1is based on the Guo’s fixed-point theorem.
Lemma 2.1Guo’s Fixed-Point Theorem27. LetKbe a cone of a real Banach spaceE,Ω1,Ω2
be bounded open subsets ofEandθ∈Ω1 ⊂Ω2. Suppose thatT :K∩Ω2\Ω1 → Kis completely
continuous such that one of the following condition hold:
iTx ≤ xforx∈∂Ω1∩KandTx ≥ xforx∈∂Ω2∩K;
iiTx ≤ xforx∈∂Ω2∩KandTx ≥ xforx∈∂Ω1∩K.
Then,T has a fixed point inK∩Ω2\Ω1.
The following result can be easily verified.
Result 1. Lett1, t2 ∈Rsuch thatt1 < t2. Letx∈Ct1, t2,x≥0 and concave ont1, t2. Then,
xt≥min{t−t1, t2−t}maxs∈t1,t2xsfor allt∈t1, t2.
Choosen0∈ {3,4,5, . . .}such thatn0>max{1/η,1/1−η,2−α/1−αη}. For fixed
n∈ {n0, n0 1, n0 2, . . .}andz∈C0,1, the linear three-point BVP
−ut zt, t∈
1
n,1−
1
n
,
u
1
n 0, u
1− 1
n αu
η,
2.1
has a unique solution
ut
1−1/n
1/n
Hnt, szsds, 2.2
whereHn :1/n,1−1/n×1/n,1−1/n → 0,∞is the Green’s function and is given by
Hnt, s
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
t−1/n 1−1/n−s
1− 2/n α/n−αη −
αt−1/nη−s
1− 2/n α/n−αη −t−s,
1
n ≤s≤t≤1−
1
n, s≤η,
t−1/n 1− 1/n−s
1−2/n α/n−αη −
αt−1/nη−s
1− 2/n α/n−αη,
1
n ≤t≤s≤1−
1
n, s≤η,
t−1/n 1− 1/n−s
1− 2/n α/n−αη ,
1
n ≤t≤s≤1−
1
n, s≥η,
t−1/n 1− 1/n−s
1− 2/n α/n−αη −t−s,
1
n ≤s≤t≤1−
1
n, s≥η.
We note thatHnt, s → Ht, sasn → ∞, where
Ht, s
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
t1−s
1−αη −
αtη−s
1−αη −t−s, 0≤s≤t≤1, s≤η, t1−s
1−αη −
αtη−s
1−αη , 0≤t≤s≤1, s≤η, t1−s
1−αη , 0≤t≤s≤1, s≥η,
t1−s
1−αη −t−s, 0≤s≤t≤1, s≥η,
2.4
is the Green’s function corresponding the boundary value problem
−ut zt, t∈0,1,
u0 0, u1 αuη 2.5
whose integral representation is given by
ut
1
0
Ht, szsds. 2.6
Lemma 2.2see9. Let0< α <1/η. Ifz∈C0,1andz≥0, then then unique solutionuof the problem2.5satisfies
min
t∈η,1ut≥γu, 2.7
whereγmin{αη, α1−η/1−αη, η}.
We need the following properties of the Green’s functionHnin the sequel.
Lemma 2.3see11. The functionHncan be written as
Hnt, s Gnt, s
αt−1/n
1− 2/n α/n−αηGn
η, s, 2.8
where
Gnt, s
n n−2
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
s− 1 n
1− 1
n−t ,
1
n ≤s≤t≤1−
1
n,
t− 1 n
1−1
n−s ,
1
n ≤t≤s≤1−
1
n.
Following the idea in10, we calculate upper bound for the Green’s functionHn in
the following lemma.
Lemma 2.4. The functionHnsatisfies
Hnt, s≤μn
s− 1 n
1− 1
n−s , t, s∈
1
n,1−
1
n
×
1
n,1−
1
n
, 2.10
whereμnmax{1, α}/1−2/n α/n−αη.
Proof. Fort, s∈1/n,1−1/n×1/n,1−1/n, we discuss various cases.
Case 1. s≤η,s≤t; using2.3, we obtain
Hnt, s s− 1
n α−1
t−1/n s−1/n
1−2/n α/n−αη. 2.11
Ifα >1, the maximum occurs att1−1/n, hence
Hnt, s≤Hn
1− 1
n, s α
s−1/n1−1/n−η
1−2/n α/n−αη
≤αs−1/n 1−1/n−s
1−2/n α/n−αη ≤μn
s− 1 n
1− 1
n−s ,
2.12
and ifα≤1, the maximum occurs atts, hence
Hnt, s≤Hns, s
s−1/n1−1/n−s αs−η
1−2/n α/n−αη
≤ s−1/n 1−1/n−s
1−2/n α/n−αη ≤μn
s− 1 n
1− 1
n−s .
2.13
Case 2. s≤η,s≥t; using2.3, we have
Hnt, s
t−1/n 1−1/n−s
1−2/n α/n−αη −α
t−1/nη−s
1−2/n α/n−αη ≤
t−1/n 1−1/n−s
1−2/n α/n−αη
≤ s−1/n 1−1/n−s
1−2/n α/n−αη ≤μn
s−1 n
1− 1
n−s .
2.14
Case 3. s≥η,t≤s; using2.3, we have
Hnt, s t−1/n 1−1/n−s
1−2/n α/n−αη ≤
s−1/n 1−1/n−s
1−2/n α/n−αη ≤μn
s− 1 n
1− 1
n−s .
Case 4. s≥η,t≥s; using2.3, we have
Hnt, s s− 1
n
t− 1 n
αη−1/n−s−1/n
1−2/n α/n−αη . 2.16
Forαη−1/n> s−1/n, the maximum occurs att1−1/n, hence
Hnt, s≤Hn
1− 1
n, s α
η−1/n1−1/n−s
1−2/n α/n−αη ≤α
s−1/n 1−1/n−s
1−2/n α/n−αη
≤μn
s− 1 n
1− 1
n−s .
2.17
Forαη−1/n≤s−1/n, the maximum occurs atts, so
Hnt, s≤Hns, s
s−1/n 1−1/n−s
1−2/n α/n−αη ≤μn
s− 1 n
1− 1
n−s . 2.18
Now, we consider the nonlinear nonsingular system of BVPs
−xt f
t,max
xt 1 n, 1 n ,max
yt 1 n,
1
n , t∈
1
n,1−
1
n
,
−yt g
t,max
xt 1 n, 1 n ,max
yt 1 n,
1
n , t∈
1
n,1−
1 n , x 1
n 0, x
1− 1
n αx
η,
y
1
n 0, y
1− 1
n αy
η.
2.19
We write2.19as an equivalent system of integral equations
xt
1−1/n
1/n
Hnt, sf
s,max
xs 1 n, 1 n ,max
ys 1 n,
1
n ds,
yt
1−1/n
1/n
Hnt, sg
s,max
xs 1 n, 1 n ,max
ys 1 n,
1
n ds.
2.20
By a solution of the system2.19, we mean a solution of the corresponding system of integral equations2.20. Define a retractionσn:0,1 → 1/n,1−1/nbyσnt max{1/n,min{t,1−
1/n}}and an operatorTn:E×E → P×Pby
Tn
x, yAn
x, y, Bn
x, y, 2.21
An
x, yt
1−1/n
1/n
Hnσnt, sf
s,max
xs 1 n, 1 n ,max
ys 1 n,
1
n ds,
Bn
x, yt
1−1/n
1/n
Hnσnt, sg
s,max
xs 1 n, 1 n ,max
ys 1 n,
1
n ds.
2.22
Clearly, ifxn, yn∈E×Eis a fixed point ofTn, thenxn, ynis a solution of the system2.19.
Lemma 2.5. Assume thatA1–A3holds. ThenTn:P×P → P×Pis completely continuous.
Proof. Clearly, for anyx, y∈P×P,Anx, y, Bnx, y∈P. We show that the operatorAn :
P×P → Pis uniformly bounded. Letd >0 be fixed and consider
Dx, y∈P×P :x, y1≤d. 2.23
Choose a constantc ∈ 0,1such thatcx 1/3 ≤ 1,cy 1/3≤ 1,x, y ∈D. Then, for everyx, y∈D, using2.22,Lemma 2.4,A1andA2, we have
An
x, yt
1−1/n
1/n
Hnσnt, sf
s, xs 1
n, ys
1
n ds
1−1/n
1/n
Hnσnt, sf
s, cxs 1/n c , c
ys 1/n c ds
≤
1
c
β11−1/n
1/n
Hnσnt, sf
s, c
xs 1 n , c
ys 1/n c ds
≤
1
c
β11
c
β21−1/n
1/n
Hnσnt, sf
s, c
xs 1 n , c
ys 1 n
ds
≤cα1−β1−β2
1−1/n
1/n
Hnσnt, s
xs 1 n
α1
f
s,1, c
ys 1 n
ds
≤cα1−β1 α2−β2
1−1/n
1/n
Hnσnt, s
xs 1 n
α1
ys 1 n
α2
fs,1,1ds
≤cα1−β1 α2−β2
1−1/n
1/n
Hnσnt, s
1
n
α11
n
α2
fs,1,1ds
≤μncα1−β1 α2−β2n−α1−α2
1−1/n
1/n
s− 1 n
1− 1
n−s fs,1,1ds
≤μncα1−β1 α2−β2n−α1−α2
1−1/n
1/n
s1−sfs,1,1ds
≤μncα1−β1 α2−β2n−α1−α2
1
0
s1−sfs,1,1dsaμncα1−β1 α2−β2n−α1−α2,
which implies that
An
x, y ≤aμncα1−β1 α2−β2n−α1−α2, 2.25
that is,AnDis uniformly bounded. Similarly, using2.22,Lemma 2.4,A1andA3, we can show thatBnDis also uniformly bounded. Thus,TnDis uniformly bounded. Now we
show thatAnDis equicontinuous. Define
ωmax
max
t,x,y∈1/n,1−1/n×0,d×0,df
t, x 1 n, y
1
n ,
max
t,x,y∈1/n,1−1/n×0,d×0,dg
t, x 1 n, y
1
n .
2.26
Lett1, t2 ∈0,1such thatt1 ≤ t2. SinceHnt, sis uniformly continuous on1/n,1−1/n×
1/n,1−1/n, for anyε >0, there existδδε>0 such that|t1−t2|< δimplies
|Hnσnt1, s−Hnσnt2, s|< ε
ω1−2/n fors∈
1
n,1−
1
n
. 2.27
Forx, y∈D, using2.22–2.27, we have
An
x, yt1−An
x, yt2
1−1/n
1/n
Hnσnt1, s−Hnσnt2, sf
s, xs 1
n, ys
1
n
ds
≤
1−1/n
1/n
|Hnσnt1, s−Hnσnt2, s|f
s, xs 1
n, ys
1
n ds
≤ω
1−1/n
1/n
|Hnσnt1, s−Hnσnt2, s|ds
< ω ε ω1−2/n
1−1/n
1/n
ds ε
1−2/n
1− 2
n ε.
2.28
Hence,
An
x, yt1−An
x, yt2< ε, ∀
x, y∈D, |t1−t2|< δ, 2.29
which implies that AnD is equicontinuous. Similarly, using 2.22–2.27, we can show
thatBnDis also equicontinuous. Thus,TnDis equicontinuous. By Arzel`a-Ascoli theorem,
Now we show thatTnis continuous. Letxm, ym,x, y∈P×P such thatxm, ym−
x, y1 → 0 asm → ∞.Then by using2.22andLemma 2.4, we have
An
xm, ym
t−An
x, yt
1−1/n
1/n
Hnσnt, s
f
s, xms
1
n, yms
1
n −f
s, xs 1
n, ys
1
n
ds
≤
1−1/n
1/n
Hnσnt, s
f
s, xms 1
n, yms
1
n −f
s, xs 1
n, ys
1
n ds
≤μn
1−1/n
1/n
s−1 n
1−1
n−s
f
s, xms 1
n, yms
1
n −f
s, xs 1 n, ys
1
n ds.
2.30
Consequently, An
xm, ym
−An
x, y
≤μn
1−1/n
1/n
s−1 n
1− 1
n−s
×f
s, xms
1
n, yms
1
n −f
s, xs 1
n, ys
1
n ds.
2.31
By Lebesgue dominated convergence theorem, it follows that An
xm, ym
−An
x, y−→0 asm−→ ∞. 2.32
Similarly, by using2.22andLemma 2.4, we have Bn
xm, ym
−Bn
x, y−→0 asm−→ ∞. 2.33
From2.32and2.33, it follows that
Tnxm, ym−Tnx, y
1−→0 asm−→ ∞, 2.34
that is,Tn:P×P → P×Pis continuous. Hence,Tn:P×P → P×Pis completely continuous.
3. Main Results
Proof ofTheorem 1.1. LetM max{μn0,max{1, α}/1−αη}. Choose a constantR > 0 such that
Choose a constant c1 ∈ 0,1such that c1 xt 1/n0 ≤ 1,c1yt 1/n0 ≤ 1,x, y ∈
∂ΩR∩P×P,t∈0,1. For anyx, y∈∂ΩR∩P×P, using2.22,3.1,A1, andA2, we have
An
x, yt
1−1/n
1/n
Hnσnt, sf
s, xs 1
n, ys
1
n ds
1−1/n
1/n
Hnσnt, sf
s, c1
xs 1/n c1
, c1
ys 1/n c1 ds ≤ 1 c1
β11−1/n
1/n
Hnσnt, sf
s, c1
xs 1 n , c1
ys 1/n c1 ds ≤ 1 c1
β11
c1
β21−1/n
1/n
Hnσnt, sf
s, c1
xs 1 n , c1
ys 1 n
ds
≤cα1−β1−β2 1
1−1/n
1/n
Hnσnt, s
xs 1 n
α1
f
s,1, c1
ys 1 n
ds
≤cα1−β1 α2−β2 1
1−1/n
1/n
Hnσnt, s
xs 1 n
α1
ys 1 n
α2
fs,1,1ds
≤cα1−β1 α2−β2 1
1−1/n
1/n
Hnσnt, s xsα1
ysα2fs,1,1ds
≤μncα11−β1 α2−β2
1−1/n
1/n
s−1 n
1− 1
n−s xs
α1ysα2fs,1,1ds
≤μncα11−β1 α2−β2
1−1/n
1/n
s1−s xsα1ysα2
fs,1,1ds
≤μncα11−β1 α2−β2
1
0
s1−s xsα1ysα2fs,1,1ds
≤Mcα1−β1 α2−β2 1
1
0
s1−s xsα1ysα2fs,1,1ds.
3.2
Since,
Mcα1−β1 α2−β2 1
1
0
s1−s xsα1ysα2
fs,1,1ds
≤M
1
0
s1−s xsα1ysα2fs,1,1ds
≤aMRα1 α2 ≤ R 2,
it follows that
An
x, y≤ R
2
x, y1
2 , ∀
x, y∈∂ΩR∩P×P. 3.4
Similarly, using2.22,3.1,A1, andA3, we have
Bn
x, y≤ x, y1
2 , ∀
x, y∈∂ΩR∩P×P. 3.5
From3.4, and3.5, it follows that
Tnx, y1 ≤x, y1, ∀
x, y∈∂ΩR∩P×P. 3.6
Choose a real constantr∈0, Rsuch that
r≤min2aM1/1−β1−β2,2bM1/1−ρ1−ρ2. 3.7
Choose a constantc2 ∈ 0,1 such that c2xt 1/n0 ≤ 1, c2yt 1/n0 ≤ 1, x, y ∈
∂Ωr∩P×P,t∈0,1. For anyx, y∈∂Ωr∩P×P, using2.22,3.7,A1, andA2, we have
An
x, yt
1−1/n
1/n
Hnσnt, sf
s, xs 1
n, ys
1
n ds
1−1/n
1/n
Hnσnt, sf
s, c2xs 1/n
c2
, c2
ys 1/n c2
ds
≥
1
c2
α11−1/n
1/n
Hnσnt, sf
s, c2
xs 1 n , c2
ys 1/n c2 ds
≥
1
c2
α11
c2
α21−1/n
1/n
Hnσnt, sf
s, c2
xs 1 n , c2
ys 1 n
ds
≥cβ1−α1−α2 2
1−1/n
1/n
Hnσnt, s
xs 1 n
β1
f
s,1, c2
ys 1 n
ds
≥cβ1−α1 β2−α2 2
1−1/n
1/n
Hnσnt, s
xs 1 n
β1
ys 1 n
β2
fs,1,1ds
≥cβ1−α1 β2−α2 2
1−1/n
1/n
Hnσnt, s xsβ1
ysβ2fs,1,1ds≥ r 2.
We used the fact that
cβ1−α1 β2−α2 2
1−1/n
1/n
Hnσnt, s xsβ1
ysβ2fs,1,1ds
≥
1−1/n
1/n
Hnσnt, s xsβ1
ysβ2
fs,1,1ds
≥aMrβ1 β2.
3.9
Thus,
An
x, y≥ x, y1
2 , ∀
x, y∈∂Ωr∩P×P. 3.10
Similarly, using2.22,3.7,A1andA3, we have,
Bn
x, y≥ x, y1
2 , ∀
x, y∈∂Ωr ∩P×P. 3.11
From3.10and3.11, it follows that
Tnx, y1≥x, y1, ∀
x, y∈∂Ωr∩P×P. 3.12
Hence byLemma 2.1,Tnhas a fixed pointxn, yn∈P×P∩ΩR\Ωr, that is,
xnAn
xn, yn
, ynBn
xn, yn
. 3.13
Moreover, by3.4,3.5,3.10and3.11, we have
r
2 ≤ xn ≤
R
2,
r
2 ≤yn≤
R
2.
3.14
Sincexn, ynis a solution of the system2.19, hencexnandynare concave on1/n,1−1/n.
Moreover, maxt∈1/n,1−1/nxnt xnand maxt∈1/n,1−1/nynt yn. Forh ∈ 1/n,1/2,
using result2.2and3.14, we have
rh
2 ≤xnt≤
R
2, ∀t∈h,1−h,
rh
2 ≤ynt≤
R
2, ∀t∈h,1−h,
which implies that{xn, yn}is uniformly bounded onh,1−h. Now we show that{xn, yn}
is equicontinuous onh,1−h. Chooseη∈h,1−hand 0< α <1−2h/η−hand consider the integral equation
xnt
xn1−h−α xn
η−1−αxnh
1−2h αh−αη t−h xnh
1−h
h
Hh−1t, sf
s, xns
1
n, yns
1
n ds, t∈h,1−h.
3.16
UsingLemma 2.3, we have
xnt
xn1−h−α xn
η−1−αxnh
1−2h αh−αη t−h xnh
1−h−t
1−2h
t
h
s−hf
s, xns 1
n, yns
1
n ds
t−h
1−2h
1−h
t
1−h−sf
s, xns 1
n, yns
1
n ds
αt−h
1−2h αh−αη
1−h
h
Gh−1η, sf
s, xns
1
n, yns
1
n ds, t∈h,1−h.
3.17
Differentiating with respect tot, we obtain
xnt xn1−h−α xn
η−1−αxnh
1−2h αh−αη
− 1
1−2h
t
h
s−hf
s, xns 1
n, yns
1
n ds
1 1−2h
1−h
t
1−h−sf
s, xns
1
n, yns
1
n ds
α
1−2h αh−αη
1−h
h
Gh−1η, sf
s, xns
1
n, yns
1
n ds, t∈h,1−h,
3.18
which implies that
x nt≤
1 αR
1−2h αh−αη
1−h
h
f
s, xns
1
n, yns
1
n ds
α
1−2h αh−αη
1−h
h
Gh−1
η, sf
s, xns 1
n, yns
1
n ds, t∈h,1−h,
In view ofA2and3.15, we have
xnt≤ 1 αR
1−2h αh−αη c
α1−β1 α2−β2 1
hr
2
α1 α21−h
h
fs,1,1ds
α
1−2h αh−αηc
α1−β1 α2−β2 1
hr
2
α1 α21−h
h
Gh−1
η, sfs,1,1ds, t∈h,1−h,
3.20
which implies that
xn ≤ 1 αR
1−2h αh−αη c
α1−β1 α2−β2 1
hr
2
α1 α21−h
h
fs,1,1ds
α
1−2h αh−αηc
α1−β1 α2−β2 1
hr
2
α1 α21−h
h
Gh−1η, sfs,1,1ds, t∈h,1−h.
3.21
Similarly, consider the integral equation
ynt
yn1−h−α yn
η−1−αynh
1−2h αh−αη t−h ynh
1−h
h
Hh−1t, sg
s, xns
1
n, yns
1
n ds, t∈h,1−h,
3.22
usingA3and3.15, we can show that
yn ≤ 1 αR
1−2h αh−αη c
γ1−ρ1 γ2−ρ2 1
hr
2
γ1 γ21−h
h
gs,1,1ds
α
1−2h αh−αηc
γ1−ρ1 γ2−ρ2 1
hr
2
γ1 γ21−h
h
Gh−1η, sgs,1,1ds, t∈h,1−h.
3.23
In view of 3.21 and 3.23,{xn, yn} is equicontinuous on h,1 −h. Hence by
Arzel`a-Ascoli theorem, the sequence{xn, yn}has a subsequence{xnk, ynk}converging uniformly
onh,1−htox, y∈P×P∩ΩR\Ωr. Let us consider the integral equation
xnkt
xnk1−h−α xnk
η−1−αxnkh
1−2h αh−αη t−h xnkh
1−h
h
Hh−1t, sf
s, xnks
1
nk
, ynks
1
nk
ds, t∈h,1−h.
Lettingnk → ∞, we have
xt x1−h−α x
η−1−αxh
1−2h αh−αη t−h xh
1−h
h
Hh−1t, sf
s, xs, ysds, t∈h,1−h.
3.25
Differentiating twice with respect tot, we have
−xt ft, xt, yt, t∈h,1−h. 3.26
Lettingh → 0, we have
−xt ft, xt, yt, t∈0,1. 3.27
Similarly, consider the integral equation
ynkt
ynk1−h−α ynk
η−1−αynkh
1−2h αh−αη t−h ynkh
1−h
h
Hh−1t, sg
s, xnks
1
nk
, ynks
1
nk
ds, t∈h,1−h,
3.28
we can show that
−yt gt, xt, yt, t∈0,1. 3.29
Now, we show thatx, yalso satisfies the boundary conditions. Since,
x0 lim
nk→ ∞
x
1
nk
lim
nk→ ∞
xnk
1
nk
0,
x1 lim
nk→ ∞
x
1− 1
nk
lim
nk→ ∞
xnk
1− 1
nk
lim
nk→ ∞
αxnk
ηαxη.
3.30
Similarly, we can show that
yt 0, y1 αyη. 3.31
Equations3.27–3.31imply thatx, yis a solution of the system1.3. Moreover,x, yis positive. In fact, by3.27xis concave and byLemma 2.2
x1≥ min
implies thatxt > 0 for all t ∈ 0,1. Similarly, yt > 0 for all t ∈ 0,1. The proof of Theorem 1.1is complete.
Example 3.1. Let
ft, x, y
m
i1
n
j1
tpi1−tqjxriysj,
gt, x, y
m
k1 n
l1
tpk1−tqlxrk ysl,
3.33
where the real constantspi, qj, ri, sj satisfypi, qj > −2,ri, sj < 1, i 1,2, . . . , m;j 1,2, . . . , n,
with max1≤i≤mri max1≤j≤nsj <1 and the real constantspk, ql, rk, slsatisfypk, ql>−2,rk, sl <
1, k 1,2, . . . , m;l1,2, . . . , n,with max1≤k≤mrk max1≤l≤nsl<1. Clearly,fandgsatisfy the
assumptionsA1–A3. Hence, byTheorem 1.1, the system1.3has a positive solution.
Acknowledgement
Research of R. A. Khan is supported by HEC, Pakistan, Project 2- 350/PDFP/HEC/2008/1.
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