Hindawi Publishing Corporation
Boundary Value Problems
Volume 2009, Article ID 273063, 18 pages
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 alipk@yahoo.com
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 −x t ft, xt, yt, t ∈ 0, 1, −y t gt, xt, yt, t ∈ 0, 1,
x0 y0 0, x1 αxη, y1 αyη, is established. The nonlinearities f, g : 0, 1 ×
0, ∞ × 0, ∞ → 0, ∞ are continuous and may be singular at t 0, t 1, x 0, and/or y 0,
while the parameters η, α satisfy η ∈ 0, 1, 0 < α < 1/η. An example is also included to show the
applicability of our result.
Copyright q 2009 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 problems BVPs arise in different areas of applied mathematics
and physics. For example, the vibration of a guy wire composed of N parts 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 problem 2.
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 Gupta 7. Since then the study of
nonlinear three-point BVPs has attracted much attention of many researchers, see 8–11 and
references therein for boundary value problems with ordinary differential equations and also
12 for boundary value problems on time scales. Recently, the study of singular BVPs has
attracted the attention of many authors, see for example, 13–18 and the recent monograph
by Agarwal et al. 19.
2
Boundary Value Problems
The study of system of BVPs has also fascinated many authors. System of BVPs with
continuous nonlinearity can be seen in 20–22 and 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:
−x t f t, x t , y t ,
−y t g t, x t , y t ,
t ∈ 0, 1 ,
t ∈ 0, 1 ,
x 0 0,
x 1 0,
y 0 0,
y 1 0,
1.1
where f, g ∈ C0, 1 × 0, ∞ × 0, ∞, 0, ∞, and may be singular at t 0, t 1, x 0 and/or
y 0.
By using fixed point theorem in cone, Yuan et al. 26 studied the following coupled
system of nonlinear singular boundary value problem:
x4 t f t, x t , y t ,
−y t g t, x t , y t ,
t ∈ 0, 1 ,
t ∈ 0, 1 ,
x 0 x 1 x 0 x 1 0,
1.2
y 0 y 1 0,
f, g are allowed to be superlinear and are singular at t 0 and/or t 1. Similarly, results are
studied in 8, 21, 23.
In this paper, we generalize the results studied in 25, 26 to the following more general
singular system for three-point nonlocal BVPs:
−x t f t, x t , y t ,
−y t g t, x t , y t ,
x 0 0,
y 0 0,
t ∈ 0, 1 ,
t ∈ 0, 1 ,
x 1 αx η ,
y 1 αy η ,
1.3
where η ∈ 0, 1, 0 < α < 1/η, f, g ∈ C0, 1 × 0, ∞ × 0, ∞, 0, ∞. We allow f and g to be
singular at t 0, t 1, and also x 0 and/or y 0. We study the sufficient conditions for
existence of positive solution for the singular system 1.3 under weaker hypothesis on f and
g as compared to the previously studied results. We do not require the system 1.3 to have
lower and upper solutions. Moreover, the cone we consider is more general than the cones
considered in 20, 21, 26.
By singularity, we mean the functions ft, x, y and gt, x, y are allowed to be
unbounded at t 0, t 1, x 0, and/or y 0. To the best of our knowledge, existence
of positive solutions for a system 1.3 with singularity with respect to dependent variables
has not been studied previously. Moreover, our conditions and results are different from those
Boundary Value Problems
3
studied in 21, 24–26. Throughout this paper, we assume that f, g : 0, 1 × 0, ∞ × 0, ∞ →
0, ∞ are continuous and may be singular at t 0, t 1, x 0, and/or y 0. We also assume
that the following conditions hold:
A1 f·, 1, 1, g·, 1, 1 ∈ C0, 1, 0, ∞ and satisfy
1
a :
1
t 1 − t f t, 1, 1 dt < ∞,
b :
0
t 1 − t g t, 1, 1 dt < ∞.
1.4
0
A2 There exist real constants αi , βi such that αi ≤ 0 ≤ βi < 1, i 1, 2, β1 β2 < 1 and for
all t ∈ 0, 1, x, y ∈ 0, ∞,
c β1 f t, x, y ≤ f t, cx, y ≤ cα1 f t, x, y , if 0 < c ≤ 1,
cα1 f t, x, y ≤ f t, cx, y ≤ c β1 f t, x, y , if c ≥ 1,
c β2 f t, x, y ≤ f t, x, cy ≤ cα2 f t, x, y , if 0 < c ≤ 1,
cα2 f t, x, y ≤ f t, x, cy ≤ c β2 f t, x, y , if c ≥ 1.
1.5
A3 There exist real constants γi , ρi such that γi ≤ 0 ≤ ρi < 1, i 1, 2, ρ1 ρ2 < 1 and for
all t ∈ 0, 1, x, y ∈ 0, ∞,
c ρ1 g t, x, y ≤ g t, cx, y ≤ cγ1 g t, x, y , if 0 < c ≤ 1,
cγ1 g t, x, y ≤ g t, cx, y ≤ c ρ1 g t, x, y , if c ≥ 1,
c ρ2 g t, x, y ≤ g t, x, cy ≤ cγ2 g t, x, y , if 0 < c ≤ 1,
cγ2 g t, x, y ≤ g t, x, cy ≤ c ρ2 g t, x, y , if c ≥ 1,
1.6
for example, a function that satisfies the assumptions A2 and A3 is
n
m h t, x, y pij t xri ysj ,
1.7
i1 j1
where pij ∈ C0, 1, 0, ∞, ri , sj < 1, i 1, 2, . . . , m; j 1, 2, . . . , n such that
max ri max sj < 1.
1≤i≤m
1≤j≤n
1.8
The main result of this paper is as follows.
Theorem 1.1. Assume that A1 –A3 hold. Then the system 1.3 has at least one positive solution.
4
Boundary Value Problems
2. Preliminaries
For each u ∈ E : C0, 1, we write u max{ut : t ∈ 0, 1}. Let P {u ∈ E : ut ≥ 0, t ∈
0, 1}. Clearly, E, · is a Banach space and P is a cone. Similarly, for each x, y ∈ E × E,
we write x, y1 x y. Clearly, E × E, · 1 is a Banach space and P × P is a cone in
E × E. For any real constant r > 0, define Ωr {x, y ∈ E × E : x, y1 < r}. By a positive
solution of 1.3, we mean a vector x, y ∈ C0, 1 ∩ C2 0, 1 × C0, 1 ∩ C2 0, 1 such that x, y
satisfies 1.3 and x > 0, y > 0 on 0, 1. The proofs of our main result Theorem 1.1 is based
on the Guo’s fixed-point theorem.
Lemma 2.1 Guo’s Fixed-Point Theorem 27. Let K be a cone of a real Banach space E, Ω1 , Ω2
be bounded open subsets of E and θ ∈ Ω1 ⊂ Ω2 . Suppose that T : K ∩ Ω2 \ Ω1 → K is completely
continuous such that one of the following condition hold:
i T x ≤ x for x ∈ ∂Ω1 ∩ K and T x ≥ x for x ∈ ∂Ω2 ∩ K;
ii T x ≤ x for x ∈ ∂Ω2 ∩ K and T x ≥ x for x ∈ ∂Ω1 ∩ K.
Then, T has a fixed point in K ∩ Ω2 \ Ω1 .
The following result can be easily verified.
Result 1. Let t1 , t2 ∈ R such that t1 < t2 . Let x ∈ Ct1 , t2 , x ≥ 0 and concave on t1 , t2 . Then,
xt ≥ min{t − t1 , t2 − t}maxs∈t1 ,t2 xs for all t ∈ t1 , t2 .
Choose n0 ∈ {3, 4, 5, . . .} such that n0 > max{1/η, 1/1 − η, 2 − α/1 − αη}. For fixed
n ∈ {n0 , n0 1, n0 2, . . .} and z ∈ C0, 1, the linear three-point BVP
1
1
,1 −
,
n
n
1
u 1−
αu η ,
n
−u t z t ,
1
0,
u
n
t∈
2.1
has a unique solution
u t 1−1/n
Hn t, s z s ds,
2.2
1/n
where Hn : 1/n, 1 − 1/n × 1/n, 1 − 1/n → 0, ∞ is the Green’s function and is given by
⎧
α t − 1/n η − s
t−1/n 1− 1/n−s
⎪
⎪
⎪
−
− t − s ,
⎪
⎪
1 − 2/n α/n − αη 1 − 2/n α/n − αη
⎪
⎪
⎪
⎪
⎪
⎪
⎪
α t − 1/n η − s
t − 1/n 1 − 1/n − s
⎪
⎪
−
,
⎪
⎨ 1 − 2/n α/n − αη
1 − 2/n α/n − αη
Hn t, s ⎪
⎪
t − 1/n 1 − 1/n − s
⎪
⎪
,
⎪
⎪
⎪ 1 − 2/n α/n − αη
⎪
⎪
⎪
⎪
⎪
⎪
t − 1/n 1 − 1/n − s
⎪
⎪
− t − s ,
⎩
1 − 2/n α/n − αη
1
1
≤ s ≤ t ≤ 1 − , s ≤ η,
n
n
1
1
≤ t ≤ s ≤ 1 − , s ≤ η,
n
n
1
1
≤ t ≤ s ≤ 1 − , s ≥ η,
n
n
1
1
≤ s ≤ t ≤ 1 − , s ≥ η.
n
n
2.3
Boundary Value Problems
5
We note that Hn t, s → Ht, s as n → ∞, where
⎧
t 1 − s αt η − s
⎪
⎪
⎪
−
− t − s ,
⎪
⎪
1 − αη
1 − αη
⎪
⎪
⎪
⎪
⎪
⎪
⎪
t 1 − s αt η − s
⎪
⎪
⎪
⎨ 1 − αη − 1 − αη ,
H t, s ⎪
⎪
t 1 − s
⎪
⎪
⎪
⎪ 1 − αη ,
⎪
⎪
⎪
⎪
⎪
⎪
⎪
t 1 − s
⎪
⎪
− t − s ,
⎩
1 − αη
0 ≤ s ≤ t ≤ 1, s ≤ η,
0 ≤ t ≤ s ≤ 1, s ≤ η,
2.4
0 ≤ t ≤ s ≤ 1, s ≥ η,
0 ≤ s ≤ t ≤ 1, s ≥ η,
is the Green’s function corresponding the boundary value problem
−u t z t ,
t ∈ 0, 1 ,
u 1 αu η
u 0 0,
2.5
whose integral representation is given by
u t 1
2.6
H t, s z s ds.
0
Lemma 2.2 see 9. Let 0 < α < 1/η. If z ∈ C0, 1 and z ≥ 0, then then unique solution u of the
problem 2.5 satisfies
min u t ≥ γu,
2.7
t∈η,1
where γ min{αη, α1 − η/1 − αη, η}.
We need the following properties of the Green’s function Hn in the sequel.
Lemma 2.3 see 11. The function Hn can be written as
Hn t, s Gn t, s α t − 1/n
Gn η, s ,
1 − 2/n α/n − αη
2.8
where
⎧
1
1
⎪
⎪
1
−
−
t
,
s
−
⎪
n
n
n ⎨
Gn t, s n−2⎪
⎪
1
⎪
⎩ t− 1
1− −s ,
n
n
1
1
≤s≤t≤1− ,
n
n
1
1
≤t≤s≤1− .
n
n
2.9
6
Boundary Value Problems
Following the idea in 10, we calculate upper bound for the Green’s function Hn in
the following lemma.
Lemma 2.4. The function Hn satisfies
1
1
Hn t, s ≤ μn s −
1− −s ,
n
n
t, s ∈
1
1
1
1
,1 −
×
,1 −
,
n
n
n
n
2.10
where μn max{1, α}/1 − 2/n α/n − αη.
Proof. For t, s ∈ 1/n, 1 − 1/n × 1/n, 1 − 1/n, we discuss various cases.
Case 1. s ≤ η, s ≤ t; using 2.3, we obtain
Hn t, s s −
1
t − 1/n s − 1/n
α − 1
.
n
1 − 2/n α/n − αη
2.11
If α > 1, the maximum occurs at t 1 − 1/n, hence
s − 1/n 1 − 1/n − η
1
Hn t, s ≤ Hn 1 − , s α
n
1 − 2/n α/n − αη
1
1
s − 1/n 1 − 1/n − s
≤ μn s −
≤α
1− −s ,
1 − 2/n α/n − αη
n
n
2.12
and if α ≤ 1, the maximum occurs at t s, hence
s − 1/n 1 − 1/n − s α s − η
Hn t, s ≤ Hn s, s 1 − 2/n α/n − αη
1
1
s − 1/n 1 − 1/n − s
≤ μn s −
1− −s .
≤
1 − 2/n α/n − αη
n
n
2.13
Case 2. s ≤ η, s ≥ t; using 2.3, we have
t − 1/n η − s
t − 1/n 1 − 1/n − s
t − 1/n 1 − 1/n − s
−α
≤
1 − 2/n α/n − αη
1 − 2/n α/n − αη
1 − 2/n α/n − αη
1
1
s − 1/n 1 − 1/n − s
≤ μn s −
≤
1− −s .
1 − 2/n α/n − αη
n
n
Hn t, s 2.14
Case 3. s ≥ η, t ≤ s; using 2.3, we have
Hn t, s 1
1
t − 1/n 1 − 1/n − s s − 1/n 1 − 1/n − s
≤
≤ μn s −
1− −s .
1 − 2/n α/n − αη
1 − 2/n α/n − αη
n
n
2.15
Boundary Value Problems
7
Case 4. s ≥ η, t ≥ s; using 2.3, we have
1 α η − 1/n − s − 1/n
1
.
Hn t, s s − t −
n
n
1 − 2/n α/n − αη
2.16
For αη − 1/n > s − 1/n, the maximum occurs at t 1 − 1/n, hence
η − 1/n 1 − 1/n − s
1
s − 1/n 1 − 1/n − s
Hn t, s ≤ Hn 1 − , s α
≤α
n
1 − 2/n α/n − αη
1 − 2/n α/n − αη
1
1
1− −s .
≤ μn s −
n
n
2.17
For αη − 1/n ≤ s − 1/n, the maximum occurs at t s, so
Hn t, s ≤ Hn s, s 1
1
s − 1/n 1 − 1/n − s
≤ μn s −
1− −s .
1 − 2/n α/n − αη
n
n
2.18
Now, we consider the nonlinear nonsingular system of BVPs
1 1
1 1
, max y t ,
,
−x t f t, max x t ,
n n
n n
1 1
1 1
−y t g t, max x t ,
, max y t ,
,
n n
n n
1
1
x
0,
x 1−
αx η ,
n
n
1
1
y
0,
y 1−
αy η .
n
n
1
1
t∈
,1 −
,
n
n
1
1
,1 −
t∈
,
n
n
2.19
We write 2.19 as an equivalent system of integral equations
x t 1−1/n
1/n
y t 1−1/n
1/n
Hn t, s f
1 1
1 1
s, max x s ,
, max y s ,
ds,
n n
n n
1 1
1 1
Hn t, s g s, max x s ,
, max y s ,
ds.
n n
n n
2.20
By a solution of the system 2.19, we mean a solution of the corresponding system of integral
equations 2.20. Define a retraction σn : 0, 1 → 1/n, 1−1/n by σn t max{1/n, min{t, 1−
1/n}} and an operator Tn : E × E → P × P by
Tn x, y An x, y , Bn x, y ,
where operators An , Bn : E × E → P are defined by
2.21
8
Boundary Value Problems
1−1/n
1 1
An x, y t , max y s Hn σn t , s f s, max x s ,
n n
1/n
1−1/n
1 1
, max y s Hn σn t , s g s, max x s ,
Bn x, y t n n
1/n
1 1
,
n n
1 1
,
n n
ds,
ds.
2.22
Clearly, if xn , yn ∈ E × E is a fixed point of Tn , then xn , yn is a solution of the system 2.19.
Lemma 2.5. Assume that A1 –A3 holds. Then Tn : P × P → P × P is completely continuous.
Proof. Clearly, for any x, y ∈ P × P , An x, y, Bn x, y ∈ P . We show that the operator An :
P × P → P is uniformly bounded. Let d > 0 be fixed and consider
D
x, y ∈ P × P : x, y1 ≤ d .
2.23
Choose a constant c ∈ 0, 1 such that cx 1/3 ≤ 1, cy 1/3 ≤ 1, x, y ∈ D. Then, for
every x, y ∈ D, using 2.22, Lemma 2.4, A1 and A2 , we have
An x, y t 1−1/n
s, x s Hn σn t , s f
1/n
1−1/n
Hn σn t , s f
1/n
1
1
, y s n
n
ds
x s 1/n y s 1/n
,c
s, c
c
c
ds
β1 1−1/n
y s 1/n
1
1
≤
,c
ds
Hn σn t , s f s, c x s c
n
c
1/n
β1 β2 1−1/n
1
1
1
1
≤
, c y s ds
Hn σn t , s f s, c x s c
c
n
n
1/n
1−1/n
1
1 α1
≤ cα1 −β1 −β2
Hn σn t , s xs f s, 1, c y s ds
n
n
1/n
1−1/n
1 α2
1 α1
Hn σn t , s xs f s, 1, 1 ds
ys ≤ cα1 −β1 α2 −β2
n
n
1/n
1−1/n
α1 α2
1
1
α1 −β1 α2 −β2
≤c
Hn σn t , s
f s, 1, 1 ds
n
n
1/n
1−1/n 1
1
α1 −β1 α2 −β2 −α1 −α2
≤ μn c
n
s−
1 − − s f s, 1, 1 ds
n
n
1/n
1−1/n
≤ μn cα1 −β1 α2 −β2 n−α1 −α2
s 1 − s f s, 1, 1 ds
1/n
≤ μn cα1 −β1 α2 −β2 n−α1 −α2
1
0
s 1 − s f s, 1, 1 ds aμn cα1 −β1 α2 −β2 n−α1 −α2 ,
2.24
Boundary Value Problems
9
which implies that
An x, y ≤ aμn cα1 −β1 α2 −β2 n−α1 −α2 ,
2.25
that is, An D is uniformly bounded. Similarly, using 2.22, Lemma 2.4, A1 and A3 , we
can show that Bn D is also uniformly bounded. Thus, Tn D is uniformly bounded. Now we
show that An D is equicontinuous. Define
ω max
max
f
t,x,y∈1/n,1−1/n×0,d×0,d
t, x 1
1
,y n
n
,
2.26
1
1
.
g t, x , y max
n
n
t,x,y∈1/n,1−1/n×0,d×0,d
Let t1 , t2 ∈ 0, 1 such that t1 ≤ t2 . Since Hn t, s is uniformly continuous on 1/n, 1 − 1/n ×
1/n, 1 − 1/n, for any ε > 0, there exist δ δε > 0 such that |t1 − t2 | < δ implies
|Hn σn t1 , s − Hn σn t2 , s| <
ε
ω 1 − 2/n
for s ∈
1
1
,1 −
.
n
n
2.27
For x, y ∈ D, using 2.22–2.27, we have
An x, y t1 − An x, y t2 1−1/n 1
1
ds
Hn σn t1 , s − Hn σn t2 , s f s, x s , y s 1/n
n
n
≤
1−1/n
|Hn σn t1 , s − Hn σn t2 , s| f
1/n
≤ω
1−1/n
1
1
s, x s , y s n
n
ds
2.28
∀ x, y ∈ D, |t1 − t2 | < δ,
2.29
|Hn σn t1 , s − Hn σn t2 , s| ds
1/n
<ω
ε
ω 1 − 2/n
1−1/n
1/n
ds ε
1 − 2/n
1−
2
n
ε.
Hence,
An x, y t1 − An x, y t2 < ε,
which implies that An D is equicontinuous. Similarly, using 2.22–2.27, we can show
that Bn D is also equicontinuous. Thus, Tn D is equicontinuous. By Arzelà-Ascoli theorem,
Tn D is relatively compact. Hence, Tn is a compact operator.
10
Boundary Value Problems
Now we show that Tn is continuous. Let xm , ym , x, y ∈ P × P such that xm , ym −
x, y1 → 0 as m → ∞. Then by using 2.22 and Lemma 2.4, we have
An xm , ym t − An x, y t
1−1/n
1
1
1
1
− f s, x s , y s ds
Hn σn t , s f s, xm s , ym s 1/n
n
n
n
n
1−1/n
1
1 1
1
ds
− f s, x s , y s Hn σn t , s f s, xm s , ym s n
n
n
n 1/n
1−1/n 1 1
1
1
1
1
≤ μn
ds.
s−
1− −s f s, xm s , ym s
−f s, x s , y s
n
n
n
n
n
n 1/n
2.30
≤
Consequently,
An xm , ym − An x, y 1−1/n 1
1
1− −s
≤ μn
s−
n
n
1/n
1 1
1
1
× f s, xm s , ym s ds.
− f s, x s , y s n
n
n
n 2.31
By Lebesgue dominated convergence theorem, it follows that
An xm , ym − An x, y −→ 0
as m −→ ∞.
2.32
as m −→ ∞.
2.33
as m −→ ∞,
2.34
Similarly, by using 2.22 and Lemma 2.4, we have
Bn xm , ym − Bn x, y −→ 0
From 2.32 and 2.33, it follows that
Tn xm , ym − Tn x, y −→ 0
1
that is, Tn : P × P → P × P is continuous. Hence, Tn : P × P → P × P is completely continuous.
3. Main Results
Proof of Theorem 1.1. Let M max{μn0 , max{1, α}/1 − αη}. Choose a constant R > 0 such
that
R ≥ max 2aM1/1−α1 −α2 , 2bM1/1−γ1 −γ2 .
3.1
Boundary Value Problems
11
Choose a constant c1 ∈ 0, 1 such that c1 xt 1/n0 ≤ 1, c1 yt 1/n0 ≤ 1, x, y ∈
∂ΩR ∩ P × P , t ∈ 0, 1. For any x, y ∈ ∂ΩR ∩ P × P , using 2.22, 3.1, A1 , and A2 , we
have
An x, y t 1−1/n
Hn σn t , s f
1
1
ds
s, x s , y s n
n
Hn σn t , s f
y s 1/n
x s 1/n
, c1
ds
s, c1
c1
c1
1/n
1−1/n
1/n
≤
≤
≤
1
c1
1
c1
β1 1−1/n
1/n
β1 α −β −β
c1 1 1 2
1
c1
β2 1−1/n
1/n
α −β1 α2 −β2
α −β1 α2 −β2
Hn σn t , s f
s, c1
1
x s n
1
, c1 y s ds
n
1
1 α1
ds
Hn σn t , s xs f s, 1, c1 y s n
n
1−1/n
1/n
≤ c1 1
y s 1/n
1
ds
s, c1 x s , c1
n
c1
1/n
1−1/n
≤ c1 1
≤
Hn σn t , s f
1−1/n
1 α2
1 α1
Hn σn t , s xs f s, 1, 1 ds
ys n
n
α
Hn σn t , s xsα1 ys 2 f s, 1, 1 ds
1/n
α −β α −β
μn c 1 1 1 2 2
α −β1 α2 −β2
≤ μn c 1 1
1−1/n 1/n
1−1/n
s−
1
n
α
1
1 − − s xsα1 ys 2 f s, 1, 1 ds
n
α
s 1 − s xsα1 ys 2 f s, 1, 1 ds
1/n
α −β1 α2 −β2
≤ μn c 1 1
1
α
s 1 − s xsα1 ys 2 f s, 1, 1 ds
0
α −β1 α2 −β2
≤ Mc1 1
1
α
s 1 − s xsα1 ys 2 f s, 1, 1 ds.
0
3.2
Since,
α −β α −β
Mc1 1 1 2 2
≤M
1
1
α
s 1 − s xsα1 ys 2 f s, 1, 1 ds
0
α
s 1 − s xsα1 ys 2 f s, 1, 1 ds
0
≤ aMR α1 α2 ≤
R
,
2
3.3
12
Boundary Value Problems
it follows that
x, y
1
An x, y ≤ R ,
2
2
∀ x, y ∈ ∂ΩR ∩ P × P .
3.4
Similarly, using 2.22, 3.1, A1 , and A3 , we have
x, y1
Bn x, y ≤
,
2
∀ x, y ∈ ∂ΩR ∩ P × P .
3.5
∀ x, y ∈ ∂ΩR ∩ P × P .
3.6
From 3.4, and 3.5, it follows that
Tn x, y ≤ x, y ,
1
1
Choose a real constant r ∈ 0, R such that
r ≤ min 2aM1/1−β1 −β2 , 2bM1/1−ρ1 −ρ2 .
3.7
Choose a constant c2 ∈ 0, 1 such that c2 xt 1/n0 ≤ 1, c2 yt 1/n0 ≤ 1, x, y ∈
∂Ωr ∩ P × P , t ∈ 0, 1. For any x, y ∈ ∂Ωr ∩ P × P , using 2.22, 3.7, A1 , and A2 , we
have
An x, y t 1−1/n
s, x s Hn σn t , s f
1/n
1−1/n
Hn σn t , s f
1/n
≥
≥
1
c2
1
c2
α1 1−1/n
β −α1 −α2
≥ c2 1
≥
1
c2
1/n
β −α1 β2 −α2
ds
ds
y s 1/n
1
ds
s, c2 x s , c2
n
c2
Hn σn t , s f
1/n
1−1/n
β −α β −α
c2 1 1 2 2
≥ c2 1
Hn σn t , s f
α2 1−1/n
y s 1/n
x s 1/n
, c2
c2
c2
1/n
α1 s, c2
1
1
, y s n
n
1
1
, c2 y s ds
s, c2 x s n
n
1
1 β1
ds
Hn σn t , s xs f s, 1, c2 y s n
n
1−1/n
1/n
1−1/n
1/n
1
Hn σn t , s xs n
β1 1 β2
f s, 1, 1 ds
ys n
β
r
Hn σn t , s xsβ1 ys 2 f s, 1, 1 ds ≥ .
2
3.8
Boundary Value Problems
13
We used the fact that
β −α1 β2 −α2
1−1/n
c2 1
β
Hn σn t , s xsβ1 ys 2 f s, 1, 1 ds
1/n
≥
1−1/n
β
Hn σn t , s xsβ1 ys 2 f s, 1, 1 ds
3.9
1/n
≥ aMr β1 β2 .
Thus,
x, y1
An x, y ≥
,
2
∀ x, y ∈ ∂Ωr ∩ P × P .
3.10
Similarly, using 2.22, 3.7, A1 and A3 , we have,
x, y1
Bn x, y ≥
,
2
∀ x, y ∈ ∂Ωr ∩ P × P .
3.11
∀ x, y ∈ ∂Ωr ∩ P × P .
3.12
From 3.10 and 3.11, it follows that
Tn x, y ≥ x, y ,
1
1
Hence by Lemma 2.1, Tn has a fixed point xn , yn ∈ P × P ∩ ΩR \ Ωr , that is,
xn An xn , yn ,
yn Bn xn , yn .
3.13
Moreover, by 3.4, 3.5, 3.10 and 3.11, we have
r
R
≤ xn ≤ ,
2
2
R
r ≤ yn ≤ .
2
2
3.14
Since xn , yn is a solution of the system 2.19, hence xn and yn are concave on 1/n, 1 − 1/n.
Moreover, maxt∈1/n,1−1/n xn t xn and maxt∈1/n,1−1/n yn t yn . For h ∈ 1/n, 1/2,
using result 2.2 and 3.14, we have
rh
R
≤xn t ≤ ,
2
2
∀t ∈ h, 1 − h ,
rh
R
≤yn t ≤ ,
2
2
∀t ∈ h, 1 − h ,
3.15
14
Boundary Value Problems
which implies that {xn , yn } is uniformly bounded on h, 1−h. Now we show that {xn , yn }
is equicontinuous on h, 1 − h. Choose η ∈ h, 1 − h and 0 < α < 1 − 2h/η − h and consider
the integral equation
xn 1 − h − α xn η − 1 − α xn h
xn t t − h xn h
1 − 2h αh − αη
1−h
1
1
ds, t ∈ h, 1 − h .
Hh−1 t, s f s, xn s , yn s n
n
h
3.16
Using Lemma 2.3, we have
xn 1 − h − α xn η − 1 − α xn h
xn t t − h xn h
1 − 2h αh − αη
1−h−t t
1
1
ds
s − h f s, xn s , yn s 1 − 2h h
n
n
1
1
t − h 1−h
ds
1 − h − s f s, xn s , yn s 1 − 2h t
n
n
1−h
α t − h
1
1
ds,
Gh−1 η, s f s, xn s , yn s 1 − 2h αh − αη h
n
n
t ∈ h, 1 − h .
3.17
Differentiating with respect to t, we obtain
xn
xn 1 − h − α xn η − 1 − α xn h
t 1 − 2h αh − αη
t
1
1
1
ds
−
s − h f s, xn s , yn s 1 − 2h h
n
n
1−h
1
1
1
ds
1 − h − s f s, xn s , yn s 1 − 2h t
n
n
1−h
α
1
1
ds,
Gh−1 η, s f s, xn s , yn s 1 − 2h αh − αη h
n
n
t ∈ h, 1 − h ,
3.18
which implies that
xn t ≤
1−h 1
1
f s, xn s , yn s ds
n
n
h
1−h
1
1
α
Gh−1 η, s f s, xn s , yn s ds,
1 − 2h αh − αη h
n
n
1 α R
1 − 2h αh − αη
t ∈ h, 1 − h ,
3.19
Boundary Value Problems
15
In view of A2 and 3.15, we have
xn t ≤
α1 α2 1−h
hr
1 α R
α −β α −β
c1 1 1 2 2
f s, 1, 1 ds
1 − 2h αh − αη
2
h
α1 α2 1−h
α
α1 −β1 α2 −β2 hr
c
Gh−1 η, s f s, 1, 1 ds,
1 − 2h αh − αη 1
2
h
t ∈ h, 1 − h ,
3.20
which implies that
xn α1 α2 1−h
1 α R
α1 −β1 α2 −β2 hr
c1
≤
f s, 1, 1 ds
1 − 2h αh − αη
2
h
α1 α2 1−h
α
α1 −β1 α2 −β2 hr
c
Gh−1 η, s f s, 1, 1 ds,
1 − 2h αh − αη 1
2
h
t ∈ h, 1 − h .
3.21
Similarly, consider the integral equation
yn 1 − h − α yn η − 1 − α yn h
yn t t − h yn h
1 − 2h αh − αη
1−h
1
1
ds, t ∈ h, 1 − h ,
Hh−1 t, s g s, xn s , yn s n
n
h
3.22
using A3 and 3.15, we can show that
yn γ1 γ2 1−h
1 α R
γ1 −ρ1 γ2 −ρ2 hr
c1
≤
g s, 1, 1 ds
1 − 2h αh − αη
2
h
γ1 γ2 1−h
α
hr
γ −ρ γ −ρ
c11 1 2 2
Gh−1 η, s g s, 1, 1 ds,
1 − 2h αh − αη
2
h
t ∈ h, 1 − h .
3.23
In view of 3.21 and 3.23, {xn , yn } is equicontinuous on h, 1 − h. Hence by ArzelàAscoli theorem, the sequence {xn , yn } has a subsequence {xnk , ynk } converging uniformly
on h, 1 − h to x, y ∈ P × P ∩ ΩR \ Ωr . Let us consider the integral equation
xnk 1 − h − α xnk η − 1 − α xnk h
xnk t t − h xnk h
1 − 2h αh − αη
1−h
1
1
Hh−1 t, s f s, xnk s , ynk s ds, t ∈ h, 1 − h .
nk
nk
h
3.24
16
Boundary Value Problems
Letting nk → ∞, we have
x 1 − h − α x η − 1 − α x h
x t t − h x h
1 − 2h αh − αη
1−h
Hh−1 t, s f s, x s , y s ds, t ∈ h, 1 − h .
3.25
h
Differentiating twice with respect to t, we have
−x t f t, x t , y t ,
t ∈ h, 1 − h .
3.26
t ∈ 0, 1 .
3.27
Letting h → 0, we have
−x t f t, x t , y t ,
Similarly, consider the integral equation
ynk t ynk 1 − h − α ynk η − 1 − α ynk h
t − h ynk h
1 − 2h αh − αη
1−h
1
1
Hh−1 t, s g s, xnk s , ynk s ds, t ∈ h, 1 − h ,
nk
nk
h
3.28
we can show that
−y t g t, x t , y t ,
t ∈ 0, 1 .
3.29
Now, we show that x, y also satisfies the boundary conditions. Since,
1
x 0 lim x
lim xnk
0,
nk → ∞
nk → ∞
nk
1
1
x 1 lim x 1 −
lim xnk 1 −
lim αxnk η αx η .
nk → ∞
nk → ∞
nk → ∞
nk
nk
1
nk
3.30
Similarly, we can show that
y t 0,
y 1 αy η .
3.31
Equations 3.27–3.31 imply that x, y is a solution of the system 1.3. Moreover, x, y is
positive. In fact, by 3.27 x is concave and by Lemma 2.2
x 1 ≥ min x t ≥ γ x > 0,
t∈η,1
3.32
Boundary Value Problems
17
implies that xt > 0 for all t ∈ 0, 1. Similarly, yt > 0 for all t ∈ 0, 1. The proof of
Theorem 1.1 is complete.
Example 3.1. Let
m n
f t, x, y tpi 1 − tqj xri ysj ,
i1 j1
m n
g t, x, y tpk 1 − tql xrk ysl ,
3.33
k1 l1
where the real constants pi , qj , ri , sj satisfy pi , qj > −2, ri , sj < 1, i 1, 2, . . . , m; j 1, 2, . . . , n,
with max1≤i≤m ri max1≤j≤n sj < 1 and the real constants pk , ql , rk , sl satisfy pk , ql > −2, rk , sl <
1, k 1, 2, . . . , m ; l 1, 2, . . . , n , with max1≤k≤m rk max1≤l≤n sl < 1. Clearly, f and g satisfy the
assumptions A1 –A3 . Hence, by Theorem 1.1, the system 1.3 has a positive solution.
Acknowledgment
Research of R. A. Khan is supported by HEC, Pakistan, Project 2- 350/PDFP/HEC/2008/1.
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