Positive solutions for a system of singular second order nonlocal boundary value problems

J. Korean Math. Soc. 47 (2010), No. 5, pp. 985–1000 DOI 10.4134/JKMS.2010.47.5.985

POSITIVE SOLUTIONS FOR A SYSTEM OF SINGULAR SECOND ORDER NONLOCAL BOUNDARY

VALUE PROBLEMS

Naseer Ahmad Asif, Paul W. Eloe, and Rahmat Ali Khan

Abstract. Sufficient conditions for the existence of positive solutions for a coupled system of nonlinear nonlocal boundary value problems of the type

−x

⁰⁰

(t) = f (t, y(t)), t ∈ (0, 1),

−y

⁰⁰

(t) = g(t, x(t)), t ∈ (0, 1),

x(0) = y(0) = 0, x(1) = αx(η), y(1) = αy(η),

are obtained. The nonlinearities f, g : (0, 1) × (0, ∞) → (0, ∞) are con- tinuous and may be singular at t = 0, t = 1, x = 0, or y = 0. The parameters η, α satisfy η ∈ (0, 1), 0 < α < 1/η. An example is provided to illustrate the results.

1. Introduction

Nonlocal 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 nonlocal boundary value problem [18]; problems in the theory of elastic stability can also be modeled as nonlocal boundary value problems [19].

The study of nonlocal BVPs for linear second order ordinary differential equations was initiated by Il’in and Moiseev in [10, 11] and extended to nonlo- cal linear elliptic boundary value problems by Bitsadze and Samarskiˇı, [2, 3, 4].

Existence theory for nonlinear three-point boundary value problems was initi- ated by Gupta [9]. Since then the study of nonlinear regular multi-point BVPs has attracted the attention of many researchers; see for example, [5, 9, 13, 14, 15, 17, 18, 20] for scalar equations, and for systems of ordinary differential equations, see [6, 7, 12].

Received January 1, 2009; Revised June 2, 2009.

2000 Mathematics Subject Classification. 34B16, 34B18.

Key words and phrases. positive solutions, singular system of ordinary differential equa- tions, three-point nonlocal boundary value problem.

c

°2010 The Korean Mathematical Society

985

Recently, the study of singular BVPs has also attracted some attention. An excellent resource with an extensive bibliography was produced by Agarwal and O’Regan [1]. Recently, S. Xie and J. Zhu [21] applied topological degree theory in a cone to study the following two point BVP for a coupled system of nonlinear fourth-order ordinary differential equations

−x ⁽⁴⁾ = f 1 (t, y), t ∈ (0, 1),

−y ⁰⁰ = f 2 (t, x), t ∈ (0, 1), x(0) = x(1) = x ⁰⁰ (0) = x ⁰⁰ (1) = 0, y(0) = y(1) = 0.

(1.1)

In [21], the nonlinearities f i ∈ C((0, 1) × R ⁺ , R ⁺ ) satisfy f i (t, 0) ≡ 0 (i = 1, 2) and may be singular at t = 0 or t = 1 only.

More recently, Y. Zhou and Y. Xu [23] studied the following nonlocal BVP for a system of second order regular ordinary differential equations

−x ⁰⁰ (t) = f (t, y), t ∈ (0, 1),

−y ⁰⁰ (t) = g(t, x), t ∈ (0, 1), x(0) = 0, x(1) = αx(η), y(0) = 0, y(1) = αy(η), (1.2)

where η ∈ (0, 1), 0 < α < 1/η, f, g ∈ C([0, 1] × [0, ∞), [0, ∞)), f (t, 0) ≡ 0, g(t, 0) ≡ 0. The above system was extended to the singular case by B. Liu, L. Liu, and Y. Wu [16], where the nonlinearities f, g were assumed to be singular at t = 0 or t = 1 together with the assumption that f (t, 0) ≡ 0, g(t, 0) ≡ 0, t ∈ (0, 1).

In this paper, we generalize the system (1.2) by allowing f, g to be singular at t = 0, t = 1, x = 0, or y = 0 and obtain sufficient conditions for the existence of a positive solution of the BVP for the system of singular equations, (1.2).

By singularity we mean that the functions f (t, u) or g(t, u) are allowed to be unbounded at t = 0, t = 1, or u = 0. In general, the assumption that there exist singularities with respect to the dependent variable is not new; see [1, 6], for example. However, in the case of nonlocal boundary conditions and coupled systems of ordinary differential equations, we believe this assumption is new.

Throughout this paper, we shall assume that f, g : (0, 1) × (0, ∞) → (0, ∞)

are continuous and may be singular at t = 0, t = 1, or u = 0. We also assume that f (t, 0), g(t, 0) are not identically 0. Let N > max{ _η ¹ , _1−η ¹ , _1−αη ^2−α } denote a fixed positive integer. Assume that the following conditions hold:

(A 1 ) there exist K, L ∈ C((0, 1), (0, ∞)) and F, G ∈ C((0, ∞), (0, ∞)) such that

f (t, u) ≤ K(t)F (u), g(t, u) ≤ L(t)G(u), t ∈ (0, 1), u ∈ (0, ∞)

and a :=

Z ₁

0

t(1 − t)K(t)dt < +∞, b :=

Z ₁

0

t(1 − t)L(t)dt < +∞;

(A 2 ) there exist α 1 , α 2 ∈ (0, ∞) with α 1 α 2 ≤ 1 such that

u→∞ lim F (u)

u ^α

¹

→ 0, lim

u→∞

G(u) u ^α

²

→ 0;

(A ₃ ) there exist β ₁ , β ₂ ∈ (0, ∞) with β ₁ β ₂ ≥ 1 such that lim inf

u→0

⁺

min

t∈[η,1]

f (t, u)

u ^β

¹

> 0, lim inf

u→0

⁺

min

t∈[η,1]

g(t, u) u ^β

²

> 0;

(A 4 ) f (t, u), G(u) are non-increasing with respect to u and for each fixed n ∈ {N, N + 1, N + 2, . . .}, there exists a constant M 1 > 0 such that t ∈ [ _n ¹ , 1 − _n ¹ ],

f ¡

t, _n ¹ + b µ n G( _n ¹ ) ¢

≥ M 1

à ν n

Z _1−1/n

η

(s − _n ¹ )(1 − ¹ _n − s)ds

! ₋₁

;

(A 5 ) F (u), g(t, u) are non-increasing with respect to u and for each fixed n ∈ {N, N + 1, N + 2, . . .}, there exists a constant M 2 > 0 such that

F Ã

ν n

Z _1−1/n

η

(s − ¹ _n )(1 − _n ¹ − s)g(s, M 2 )ds

!

≤ M 2 − _n ¹ a µ n . The parameters µ n and ν n in (A 4 ) and (A 5 ) are given by

µ n = max{1, α}

1 − _n ² + ^α _n − αη , ν n = min{1, α} min{η − ¹ _n , 1 − _n ¹ − η}

1 − _n ² + ^α _n − αη . Since N > max{ ¹ _η , _1−η ¹ , _1−αη ^2−α }, µ n , ν n > 0.

We state the main results of this paper here.

Theorem 1.1. Assume that (A 1 ) − (A 3 ) hold. Then the system (1.1) has at least one positive solution.

Theorem 1.2. Assume that (A 1 ), (A 2 ) and (A 4 ) hold. Then the system (1.1) has at least one positive solution.

Theorem 1.3. Assume that (A 1 ), (A 3 ) and (A 5 ) hold. Then the system (1.1) has at least one positive solution.

Theorem 1.4. Assume that (A 1 ), (A 4 ) and (A 5 ) hold. Then the system (1.1)

has at least one positive solution.

2. Preliminaries

For each x ∈ C[0, 1] we write kxk = max{|x(t)| : t ∈ [0, 1]}. Clearly, C[0, 1]

with the norm k · k is a Banach space. For n ≥ N , define a cone P , and a cone K n of C[ _n ¹ , 1 − _n ¹ ] as follows:

P = {x ∈ C[0, 1] : x(t) ≥ 0, t ∈ [0, 1]} , P n = ©

x ∈ P : x is concave on [0, 1], min

t∈ [

n¹

,1−

_n¹

] x(t) ≥ _n ¹ ª ,

K n = ©

x ∈ C £ ₁

n , 1 − _n ¹ ¤

: x is concave on [0, 1] ª .

For any real constant r > 0, define

Ω r = {x ∈ C[0, 1] : kxk < r}

as an open neighborhood of 0 ∈ C[0, 1] of radius r. (x(t), y(t)) is called a positive solution of (1.1) if

(x, y) ∈ ¡

C[0, 1] ∩ C ² (0, 1) ¢

× ¡

C[0, 1] ∩ C ² (0, 1) ¢ ,

x(t) > 0, y(t) > 0 on (0, 1) and (x, y) satisfies (1.1).

The proofs of our main results (Theorems 1.1-1.4) are based on the Guo- Krasnosel’skii fixed-point theorem.

Lemma 2.1 ([8, Guo Krasnosel’skii Fixed-Point Theorem]). Let K be a cone of a real Banach space E, and let Ω 1 , Ω 2 be bounded open neighborhoods of 0 ∈ E, and assume Ω 1 ⊂ Ω 2 . Suppose that T : K ∩ (Ω 2 \Ω 1 ) → K is completely continuous such that one of the following conditions holds:

(i) kT xk ≤ kxk for x ∈ ∂Ω 1 ∩ K; kT xk ≥ kxk for x ∈ ∂Ω 2 ∩ K;

(ii) kT xk ≤ kxk for x ∈ ∂Ω 2 ∩ K; kT xk ≥ kxk for x ∈ ∂Ω 1 ∩ K.

Then, T has a fixed point in K ∩ (Ω 2 \Ω 1 ).

For fixed n ≥ N and z ∈ C[0, 1], the linear boundary value problem

−u ⁰⁰ (t) = z(t), t ∈ [ ¹ _n , 1 − _n ¹ ],

u( _n ¹ ) = _n ¹ , u(1 − _n ¹ ) = αu(η) + ^1−α _n , (2.1)

has a unique solution

(2.2) u(t) = 1

n +

Z _1−1/n

1/n

H n (t, s)z(s)ds,

where H n : £ ₁

n , 1 − _n ¹ ¤

× £ ₁

n , 1 − _n ¹ ¤

→ [0, ∞) is an associated Green’s function and is defined by

(2.3)

H n (t, s) =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



( ^t−

n¹

)(( ¹⁻

n¹

−s ) ^{−α(η−s)} )

1−

_n²

+

^α_n

−αη − (t − s), _n ¹ ≤ s ≤ t ≤ 1 − _n ¹ , s ≤ η, ( ^t−

n¹

)(( ¹⁻

n¹

−s ) ^{−α(η−s)} )

1−

_n²

+

^α_n

−αη , _n ¹ ≤ t ≤ s ≤ 1 − _n ¹ , s ≤ η, ( ^t−

n¹

)( ¹⁻

n¹

−s )

1−

_n²

+

^α_n

−αη , _n ¹ ≤ t ≤ s ≤ 1 − _n ¹ , s ≥ η, ( ^t−

n¹

)( ¹⁻

n¹

−s )

1−

_n²

+

^α_n

−αη − (t − s), _n ¹ ≤ s ≤ t ≤ 1 − _n ¹ , s ≥ η.

We note that H n (t, s) → H(t, s) as n → ∞, where

H(t, s) =

 

 

 

 

 

 



 

 

 

 

 

 

t(1−s)

1−αη − ^αt(η−s) _1−αη − (t − s), 0 ≤ s ≤ t ≤ 1, s ≤ η,

t(1−s)

1−αη − ^αt(η−s) _1−αη , 0 ≤ t ≤ s ≤ 1, s ≤ η,

t(1−s)

1−αη , 0 ≤ t ≤ s ≤ 1, s ≥ η,

t(1−s)

1−αη − (t − s), 0 ≤ s ≤ t ≤ 1−, s ≥ η, is the Green’s function corresponding the boundary value problem

−u ⁰⁰ (t) = z(t), t ∈ [0, 1], u(0) = 0, u(1) = αu(η) with

u(t) = Z ₁

0

H(t, s)z(s)ds,

as its integral representation. We need the following properties of the Green’s function H _n in the sequel. For the proof, see [22].

Lemma 2.2. The function H n can be written as (2.4) H n (t, s) = G n (t, s) + α ¡

t − _n ¹ ¢

1 − _n ² + ^α _n − αη G n (η, s), where

(2.5) G n (t, s) = n n − 2

(¡ s − _n ¹ ¢ ¡

1 − _n ¹ − t ¢

, _n ¹ ≤ s ≤ t ≤ 1 − _n ¹ ,

¡ t − ¹ _n ¢ ¡

1 − _n ¹ − s ¢

, _n ¹ ≤ t ≤ s ≤ 1 − _n ¹ .

Lemma 2.3. Let

µ n = max{1, α}

1 − ² _n + ^α _n − αη , ν n = min{1, α} min{η − ¹ _n , 1 − _n ¹ − η}

1 − _n ² + ^α _n − αη . Then

(i) H n (t, s) ≤ µ n

¡ s − _n ¹ ¢ ¡

1 − _n ¹ − s ¢

, (t, s) ∈ £ ₁

n , 1 − _n ¹ ¤

× £ ₁

n , 1 − _n ¹ ¤ , (ii) H n (t, s) ≥ ν n

¡ s − ¹ _n ¢ ¡

1 − _n ¹ − s ¢

, (t, s) ∈ £

η, 1 − ¹ _n ¤

× £ ₁

n , 1 − ¹ _n ¤ . Now consider the system of nonlinear non-singular BVPs

−x ⁰⁰ (t) = f (t, max{ _n ¹ , y(t)}), t ∈ [ _n ¹ , 1 − _n ¹ ],

−y ⁰⁰ (t) = g(t, max{ ¹ _n , x(t)}), t ∈ [ _n ¹ , 1 − ¹ _n ], x( _n ¹ ) = _n ¹ , x(1 − _n ¹ ) = αx(η) + ^1−α _n , y( _n ¹ ) = _n ¹ , y(1 − _n ¹ ) = αy(η) + ^1−α _n , (2.6)

where n > N . Write (2.6) as an equivalent system of integral equations x(t) = 1

n +

Z _1−1/n

1/n

H n (t, s)f (s, max{ _n ¹ , y(s)})ds,

y(t) = 1 n +

Z _1−1/n

1/n

H n (t, s)g(s, max{ _n ¹ , x(s)})ds.

(2.7)

Thus, (x, y) is a solution of (2.6) if and only if (x, y) ∈ C[ _n ¹ , 1 − ¹ _n ] × C[ _n ¹ , 1 − _n ¹ ] and (x, y) is a solution of (2.7).

Define operators A n , B n , T n : K n → K n by (A n y)(t) = 1

n +

Z _1−1/n

1/n

H n (t, s)f (s, max{ _n ¹ , y(s)})ds,

(B n x)(t) = 1 n +

Z _1−1/n

1/n

H n (t, s)g(s, max{ _n ¹ , x(s)})ds, (T n x)(t) = (A n (B n x))(t).

(2.8)

If u n ∈ K n is a fixed point of T n , then the system of BVPs (2.6) has a solution (x n , y n ) given by

(

x n (t) = u n (t), y n (t) = (B n u n )(t).

By construction, the system of BVPs (2.6) is regular and so the following lemma is standard.

Lemma 2.4. Assume f, g : (0, 1) × (0, ∞) → [0, ∞) are continuous. Then

T n : K n → K n is completely continuous.

3. Main results

Proof of Theorem 1.1. By (A 2 ), there exist constants C 1 , C 2 , N 1 , N 2 > 0 such that

(3.1) 4 ^α

¹

ab ^α

¹

µ ^α _n

¹

⁺¹ C 1 C ₂ ^α

¹

< 1, and

(3.2) F (x) ≤ C 1 x ^α

¹

+ N 1 , G(x) ≤ C 2 x ^α

²

+ N 2 for x ≥ 1 n . Choose a constant R > 0 such that

(3.3) R ≥

1

n + ²

^α1

_n ^aµ

_α1ⁿ

^C

¹

+ aµ n N 1 + 4 ^α

¹

ab ^α

¹

µ ^α _n

¹

⁺¹ C 1 N ₂ ^α

¹

1 − 4 ^α

¹

ab ^α

¹

µ ^α n

¹

⁺¹ C 1 C ₂ ^α

¹

. For any u ∈ ∂Ω R ∩ K n , using (2.8) and (A 1 ), we have

(T n u)(t) = (A n (B n u))(t) = 1 n +

Z _1−1/n

1/n

H n (t, s)f (s, (B n u)(s))ds

= 1 n +

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F ( 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds.

In view of (3.2) and (A 2 ), it follows that (T n u)(t)

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)(C 1 ( 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ ) ^α

¹

+ N 1 )ds

= 1 n + C 1

Z _1−1/n

1/n

H n (t, s)K(s)( 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ ) ^α

¹

ds

+ N 1

Z _1−1/n

1/n

H n (t, s)K(s)ds

≤ 1 n + C 1

Z _1−1/n

1/n

H n (t, s)K(s)( 1 n +

Z _1−1/n

1/n

H n (s, τ )L(τ )G(u(τ ))dτ ) ^α

¹

ds

+ N 1

Z _1−1/n

1/n

H n (t, s)K(s)ds

≤ 1 n + C 1

Z _1−1/n

1/n

H n (t, s)K(s)

· Ã 1

n +

Z _1−1/n

1/n

H n (s, τ )L(τ )(C 2 (u(τ )) ^α

²

+ N 2 )dτ

! α

1

ds

+ N 1

Z _1−1/n

1/n

H n (t, s)K(s)ds.

Employing (i) of Lemma 2.3, we obtain (T n u)(t)

≤ 1

n + C 1 µ n

Z _1−1/n

1/n

(s − 1 n )(1 − 1

n − s)K(s)ds

· Ã

1 n + µ n

Z _1−1/n

1/n

(τ − 1 n )(1 − 1

n − τ )L(τ )(C 2 (u(τ )) ^α

²

+ N 2 )dτ

! _α

₁

+ N 1 µ n

Z _1−1/n

1/n

(s − 1 n )(1 − 1

n − s)K(s)ds

≤ 1

n + C 1 µ n

Z _1−1/n

1/n

s(1 − s)K(s)ds

· Ã 1

n + µ n

Z _1−1/n

1/n

τ (1 − τ )L(τ )(C 2 (u(τ )) ^α

²

+ N 2 )dτ

! α

1

+ N 1 µ n

Z _1−1/n

1/n

s(1 − s)K(s)ds.

Hence,

(T n u)(t)

≤ 1

n + C 1 µ n

Z _1−1/n

1/n

s(1 − s)K(s)ds

· Ã 1

n + µ n

Z _1−1/n

1/n

τ (1 − τ )L(τ )(C 2 kuk ^α

²

+ N 2 )dτ

! _α

₁

+ N 1 µ n

Z _1−1/n

1/n

s(1 − s)K(s)ds

≤ 1

n + µ n C 1

Z ₁

0

s(1 − s)K(s)ds

· µ 1

n + µ n

Z ₁

0

τ (1 − τ )L(τ )dτ (C 2 kuk ^α

²

+ N 2 )

¶ α

1

+ µ n N 1

Z ₁

0

s(1 − s)K(s)ds

≤ 1

n + aµ n N 1 + 2 ^α

¹

aµ n C 1 ( 1

n ^α

¹

+ b ^α

¹

µ ^α _n

¹

(C 2 kuk ^α

²

+ N 2 ) ^α

¹

)

≤ 1

n + 2 ^α

¹

aµ _n C ₁

n ^α

¹

+ aµ n N 1 + 2 ^2α

¹

ab ^α

¹

µ ^α _n

¹

⁺¹ C 1 (C ₂ ^α

¹

kuk ^α

¹

^α

²

+ N ₂ ^α

¹

).

Using (3.3), we obtain

(3.4) kT n uk ≤ kuk for all u ∈ ∂Ω R ∩ K n .

Now, by (A 3 ), there exist constants C 3 , C 4 > 0 and ρ ∈ (0, R) such that (3.5) f (t, x) ≥ C 3 x ^β

¹

, g(t, x) ≥ C 4 x ^β

²

for x ∈ [0, ρ] and t ∈ [η, 1].

Choose

(3.6) r n = min

 

 ρ, C 3 C ₄ ^β

¹

ν _n ^β

¹

⁺¹ n ^β

¹

^β

²

ÃZ _1−1/n

η

(s − _n ¹ )(1 − ¹ _n − s)ds

! _β

₁

₊₁ 



 . For any u ∈ ∂Ω r

n

∩ K n , using (2.8), (3.5) and (ii) of Lemma 2.3, we have

(T n u)(t) = (A n (B n u))(t)

= 1 n +

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≥

Z _1−1/n

η

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≥ C 3

Z _1−1/n

η

H n (t, s)

ÃZ _1−1/n

η

H n (s, τ )g(τ, u(τ ))dτ

! _β

₁

ds

≥ C 3 ν n

Z _1−1/n

η

(s − 1 n )(1 − 1

n − s)ds

· Ã

ν n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )g(τ, u(τ ))dτ

! _β

₁

≥ C 3 ν _n ^β

¹

⁺¹

Z _1−1/n

η

(s − 1 n )(1 − 1

n − s)ds

· Ã

C 4

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )(u(τ )) ^β

²

dτ

! _β

₁

≥ C 3 C ₄ ^β

¹

ν _n ^β

¹

⁺¹ n ^β

¹

^β

²

ÃZ _1−1/n

η

(s − 1 n )(1 − 1

n − s)ds

! _β

₁

₊₁ .

Thus, in view of (3.6), it follows that

(3.7) kT n uk ≥ kuk for u ∈ ∂Ω r

n

∩ K n . By Lemma 2.1, T n has a fixed point u n ∈ K n ∩ (Ω R \Ω r

n

).

Note that

(3.8) r n ≤ u n (t) ≤ R for all t ∈ [ _n ¹ , 1 − _n ¹ ]

and r n → 0 as n → ∞. Thus, we have exhibited a uniform bound for each u n ∈ [ _n ¹ , 1 − ¹ _n ] and for m ≥ n, {u m } is uniformly bounded on [ _n ¹ , 1 − _n ¹ ].

To show that {u m } for m ≥ n, is equicontinuous on [ _n ¹ , 1 − _n ¹ ], consider for t ∈ [ _n ¹ , 1 − _n ¹ ], the integral equation

u m (t) = u m ( _m ¹ ) +

Z _1−1/m

1/m

H m (t, s)f (s, (B m u m )(s))ds.

Employ Lemma 2.2 to obtain u _m (t)

= u m ( _m ¹ )+

Z _1−1/m

1/m

£ G m (t, s) + α(t − _m ¹ )

1 − _m ² + _m ^α − αη G m (η, s) ¤

f (s, (B m u m )(s))ds

= u m ( _m ¹ )+ m m − 2

Z _t

1/m

(s − 1

m )(1 − 1

m − t)f (s, (B m u m )(s))ds

+ m

m − 2

Z _1−1/m

t

(t − 1

m )(1 − 1

m − s)f (s, (B m u m )(s))ds + α(t − _m ¹ )

1 − _m ² + _m ^α − αη

Z _1−1/m

1/m

G m (η, s)f (s, (B m u m )(s))ds.

Differentiate with respect to t to obtain u ⁰ _m (t) = − m

m − 2 Z _t

1/m

(s − 1

m )f (s, (B m u m )(s))ds

+ m

m − 2

Z _1−1/m

t

(1 − 1

m − s)f (s, (B m u m )(s))ds

+ α

1 − _m ² + _m ^α − αη

Z _1−1/m

1/m

G m (η, s)f (s, (B m u m )(s))ds, which implies that for t ∈ [ _n ¹ , 1 − _n ¹ ]

|u ⁰ _m (t)| ≤

Z _1−1/m

1/m

f (s, (B m u m )(s))ds

+ α

1 − _m ² + _m ^α − αη

Z _1−1/m

1/m

G m (η, s)f (s, (B m u m )(s))ds.

(3.9)

Hence, for m ≥ n, {u m } is equicontinuous on [ _n ¹ , 1 − ¹ _n ].

For m ≥ n, define

v m =

 



 

u m ( ¹ _n ), if 0 ≤ t ≤ _n ¹ ,

u m (t), if _n ¹ ≤ t ≤ 1 − _n ¹ ,

αu m (η), if 1 − _n ¹ ≤ t ≤ 1.

Since v m is a constant extension of u m to [0, 1], the sequence {v m } is uniformly bounded and equicontinuous on [0, 1]. Thus, there exists a subsequence {v n

k

} of {v m } converging uniformly on [0, 1] to v ∈ P ∩ (Ω R \Ω r ).

We introduce the notation x n

k

(t) = v n

k

(t), y n

k

(t) = 1

n k +

Z _1−1/n

_k

1/n

_k

H n

k

(t, s)g(s, v n

k

(s))ds, x(t) = lim

n

k

→∞ x n

_k

(t), y(t) = lim

n

k

→∞ y n

_k

(t), and for t ∈ [0, 1] consider the integral equation

x n

k

(t) = x n

k

( _n ¹

k

) +

Z _1−1/n

_k

1/n

k

H n

k

(t, s)f (t, y n

k

(s))ds.

Letting n k → ∞, we have

x(t) = x(0) + Z ₁

0

H(t, s)f (t, y(s))ds,

and

y(t) = Z ₁

0

H(t, s)g(s, x(s))ds, t ∈ [0, 1].

Moreover,

x(0) = 0, x(1) = αx(η), y(0) = 0, y(1) = αy(η).

Hence, (x(t), y(t)) is a solution of the system (1.2).

Since

f, g : (0, 1) × (0, ∞) → (0, ∞),

f (t, 0), g(t, 0) are not identically 0, and H is of fixed sign on (0, 1) × (0, 1), it

follows that x, y > 0 on (0, 1). ¤

Example 3.1. Let f (t, y) = 1

t(1 − t) µ 1

y + 3y ^1/3

¶

, g(t, x) = 1 t(1 − t)

µ 1 x + 4x

¶

and α = 2, η = ¹ ₃ . Choose K(t) = L(t) = 1

t(1 − t) , F (y) = 1

y + 3y ^1/3 , G(x) = 1

x + 4x,

and α 1 = ¹ ₂ , α 2 = 2, β 1 = β 2 = 1. Then (A 1 ) − (A 3 ) are satisfied. Hence, by

Theorem 1.1, system (1.2) has a positive solution.

Proof of Theorem 1.2. For u ∈ ∂Ω M

1

∩ K n , using (2.8), we obtain for t ∈ [ _n ¹ , 1 − ¹ _n ]

(T _n u)(t) = (A _n (B _n u))(t) = 1 n +

Z _1−1/n

1/n

H _n (t, s)f (s, (B _n u)(s))ds

= 1 n +

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds.

Using (A 1 ), (A 4 ) and Lemma 2.3, we have (T n u)(t)

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1 n + µ n

Z _1−1/n

1/n

(τ − 1 n )(1 − 1

n − τ )g(τ, u(τ ))dτ )ds

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1 n + µ n

Z _1−1/n

1/n

(τ − 1 n )(1 − 1

n − τ )L(τ )G(u(τ ))dτ )ds

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1

n + µ n G( 1 n )

Z _1−1/n

1/n

(τ − 1 n )(1 − 1

n − τ )L(τ )dτ )ds

≥

Z _1−1/n

1/n

H n (t, s)f (s, 1

n + b µ n G( 1 n ))ds

≥ M 1

Z _1−1/n

1/n

H n (t, s)ds(ν n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )dτ ) ⁻¹ ≥ M 1 , which implies that

(3.10) kT n uk ≥ kuk for all u ∈ ∂Ω M

1

∩ K n .

In view of (A 2 ), we can choose R > M 1 such that (3.4) holds. Hence, by Lemma 2.1, T n has a fixed point u n ∈ K n ∩ (Ω R \Ω M

1

). By the same process as done in Theorem 1.1, the system (1.2) has a positive solution. ¤ Example 3.2. Let

f (t, y) = e

^1y

t(1 − t) , g(t, x) = e

^x1

t(1 − t) and α = 2, η = ¹ ₃ . Choose

K(t) = L(t) = 1

t(1 − t) , F (y) = e

^1y

, G(x) = e

^1x

. Choose constant M 1 such that M 1 ≤ ⁴⁽ⁿ⁻³⁾ _n e

^1+6nenn

R _1−1/n

1/3 (s − _n ¹ )(1 − ¹ _n − s)ds.

Then (A 1 ), (A 2 ) and (A 4 ) are satisfied. Hence, by Theorem 1.2, system (1.2)

has a positive solution.

Proof of Theorem 1.3. For u ∈ ∂Ω M

2

∩ K n , using (2.8), we have (T n u)(t) = 1

n +

Z _1−1/n

1/n

H n (t, s)f (s, (B n u)(s))ds

= 1 n +

Z _1−1/n

1/n

H n (t, s)f (s, 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds.

In view of (A 1 ), (A 5 ) and Lemma 2.3, we obtain

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F ( 1 n +

Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F ( Z _1−1/n

1/n

H n (s, τ )g(τ, u(τ ))dτ )ds

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F ( Z _1−1/n

1/n

H n (s, τ )g(τ, M 2 )dτ )ds

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F ( Z _1−1/n

η

H n (s, τ )g(τ, M 2 )dτ )ds

≤ 1 n +

Z _1−1/n

1/n

H n (t, s)K(s)F (ν n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )g(τ, M 2 )dτ )ds

= 1

n + F (ν n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )g(τ, M 2 )dτ )

Z _1−1/n

1/n

H n (t, s)K(s)ds

≤ 1

n + µ n F (ν n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )g(τ, M 2 )dτ )

·

Z _1−1/n

1/n

(s − 1 n )(1 − 1

n − s)K(s)ds

≤ 1

n + aµ _n F (ν _n

Z _1−1/n

η

(τ − 1 n )(1 − 1

n − τ )g(τ, M ₂ )dτ ) ≤ M ₂ , which implies that

(3.11) kT n uk ≤ kuk for all u ∈ ∂Ω M

2

∩ K n .

By (A 3 ), we can choose ρ ∈ (0, M 2 ) such that (3.7) holds. Hence, T n has a fixed point u _n ∈ K _n ∩ (Ω _M

₂

\Ω _ρ ). By the same process as done in Theorem 1.1,

the system (1.2) has a positive solution. ¤

Example 3.3. Let

f (t, y) =

 



ye

^y1

t(1−t) , y ≤ 1,

e

t(1−t) , y > 1, g(t, x) = ( xe

^1x

t(1−t) , x ≤ 1,

e

t(1−t) , x > 1,

and α = 2, η = ¹ ₃ . Choose K(t) = L(t) = 1

t(1 − t) , F (y) = (

ye

^1y

, y ≤ 1,

e, y > 1, G(x) =

( xe

^1x

, x ≤ 1, e, x > 1, and β 1 = β 2 = 1. Choose constant M 2 such that

M 2 ≥ max (

1, 1

n + 6F (e(1 − 3/n)

Z _1−1/n

1/3

(s − 1/n)(1 − 1/n − s) s(1 − s) ds)

) .

Then (A 1 ), (A 3 ) and (A 5 ) are satisfied. Hence, by Theorem 1.3, system (1.2) has a positive solution.

Proof of Theorem 1.4. By (A 1 ) and (A 4 ), we obtain (3.10). By (A 5 ) we can choose a constant M 2 > M 1 such that (3.11) holds. Then T n has a fixed point u _n ∈ K _n ∩ (Ω _M

₂

\Ω _M

₁

). By the same process as done in Theorem 1.1, the

system (1.2) has a positive solution. ¤

Example 3.4. Let

f (t, y) = 1 t(1 − t)

√ 1

y , g(t, x) = 1 t(1 − t)

1 x ² and α = 2, η = ¹ ₃ . Choose

K(t) = L(t) = 1

t(1 − t) , F (y) = 1

√ y , G(x) = 1 x ² . Choose constants M 1 and M 2 such that M 1 ≤ √ ⁴⁽ⁿ⁻³⁾

n(6n

³

+1)

R ₁₋

¹

n

1/3 (s − _n ¹ )(1 − _n ¹ − s)ds and M 2 ≥ _6n ¹ ( ¹ ₆ − q

n

n−3 ( R _1−1/n

1/3

(s−1/n)(1−1/n−s)

s(1−s) ds) ^−1/2 ) ⁻¹ . Then (A 1 ), (A 4 ) and (A 5 ) are satisfied. Hence, by Theorem 1.4, system (1.2) has a positive solution.

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Naseer Ahmad Asif

Centre for Advanced Mathematics and Physics National University of Sciences and Technology

Campus of College of Electrical and Mechanical Engineering Peshawar Road, Rawalpindi, Pakistan

E-mail address: [email protected]

Paul W. Eloe

Department of Mathematics University of Dayton Dayton, Ohio 454-2316, USA

E-mail address: [email protected]

Rahmat Ali Khan

Centre for Advanced Mathematics and Physics National University of Sciences and Technology

Campus of College of Electrical and Mechanical Engineering

Peshawar Road, Rawalpindi, Pakistan

and

Department of Mathematics University of Dayton

Dayton, Ohio 45469-2316, USA

E-mail address: rahmat [email protected]