Coupled fixed points in partially ordered metric spaces and application

Nonlinear Analysis 74 (2011) 983–992

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Coupled fixed points in partially ordered metric spaces and application Nguyen Van Luong ∗ , Nguyen Xuan Thuan Department of Natural Sciences, Hong Duc University, Thanh Hoa, Viet Nam

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Article history: Received 3 June 2010 Accepted 27 September 2010 MSC: 54H25 47H10

abstract In this paper, we prove some coupled fixed point theorems for mappings having a mixed monotone property in partially ordered metric spaces. The main results of this paper are generalizations of the main results of Bhaskar and Lakshmikantham [T. Gnana Bhaskar, V. Lakshmikantham, Fixed point theorems in partially ordered metric spaces and applications, Nonlinear Anal. TMA 65 (2006) 1379–1393]. As an application, we discuss the existence and uniqueness for a solution of a nonlinear integral equation. © 2010 Elsevier Ltd. All rights reserved.

Keywords: Coupled fixed point Mixed monotone mapping Partially ordered set Integral equation

1. Introduction and preliminaries The existence of a fixed point for contraction type mappings in partially ordered metric spaces has been considered recently by Ran and Reurings [1], Bhaskar and Lakshmikantham [2], Nieto and Lopez [3,4], Agarwal et al. [5], Lakshmikantham and Ćirić [6] and Samet [7]. The existence of solutions for matrix equations or ordinary differential equations by applying fixed point theorems are presented in [8,2,9,3,4,1]. In [2], Bhaskar and Lakshmikantham introduced notions of a mixed monotone mapping and a coupled fixed point and proved some coupled fixed point theorems for mixed monotone mapping and discussed the existence and uniqueness of a solution for a periodic boundary value problem. Definition 1.1 ([2]). Let (X , ≤) be a partially ordered set and F : X × X → X . The mapping F is said to have the mixed monotone property if F (x, y) is monotone non-decreasing in x and is monotone non-increasing in y, that is, for any x, y ∈ X , x1 , x2 ∈ X ,

x1 ≤ x2 ⇒ F (x1 , y) ≤ F (x2 , y)

y1 , y2 ∈ X ,

y1 ≤ y2 ⇒ F (x, y1 ) ≥ F (x, y2 ) .

and

Definition 1.2 ([2]). An element (x, y) ∈ X × X is called a coupled fixed point of the mapping F : X × X → X if x = F (x, y)

and y = F (y, x) .

The main theoretical results of Bhaskar and Lakshmikantham in [2] are the following coupled fixed point theorems. Theorem 1.3 ([2]). Let (X , ≤) be a partially ordered set and suppose there exists a metric d on X such that (X , d) is a complete metric space. Let F : X × X → X be a continuous mapping having the mixed monotone property on X . Assume that there exists

∗

Corresponding author. E-mail addresses: [email protected], [email protected] (N.V. Luong), [email protected] (N.X. Thuan).

0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.09.055

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a k ∈ [0, 1) with d (F (x, y) , F (u, v)) ≤

k 2

[d (x, u) + d (y, v)]

for all x ≥ u and y ≤ v.

If there exist two elements x0 , y0 ∈ X with x0 ≤ F (x0 , y0 )

and

y0 ≥ F (y0 , x0 )

then there exist x, y ∈ X such that x = F (x, y)

and

y = F (y, x) .

Theorem 1.4 ([2]). Let (X , ≤) be a partially ordered set and suppose there exists a metric d on X such that (X , d) is a complete metric space. Assume that X has the following property: (i) if a non-decreasing sequence {xn } → x, then xn ≤ x for all n, (ii) if a non-increasing sequence {yn } → y, then y ≤ yn for all n. Let F : X × X → X be a mapping having the mixed monotone property on X . Assume that there exists a k ∈ [0, 1) with d (F (x, y) , F (u, v)) ≤

k 2

[d (x, u) + d (y, v)]

for all x ≥ u and y ≤ v.

If there exist two elements x0 , y0 ∈ X with x0 ≤ F (x0 , y0 )

and

y0 ≥ F (y0 , x0 )

then there exist x, y ∈ X such that x = F (x, y)

and

y = F (y, x) .

The purpose of this paper is to present some coupled fixed point theorems for a mixed monotone mapping in a partially ordered metric space which are generalizations of the results of Bhaskar and Lakshmikantham [2] and give an existence and uniqueness for a solution of a nonlinear integral equation. 2. The main results Let Φ denote all functions ϕ : [0, ∞) → [0, ∞) which satisfy (i) ϕ is continuous and non-decreasing, (ii) ϕ(t ) = 0 if and only if t = 0, (iii) ϕ (t + s) ≤ ϕ(t ) + ϕ (s) , ∀t , s ∈ [0, ∞) and Ψ denote all functions ψ : [0, ∞) → [0, ∞) which satisfy limt →r ψ(t ) > 0 for all r > 0 and limt →0+ ψ(t ) = 0. For example, functions ϕ1 (t ) = kt where k > 0, ϕ2 (t ) = t +t 1 , ϕ3 (t ) = ln (t + 1), and ϕ4 (t ) = min {t , 1} are in Φ ;

ψ1 (t ) = kt where k > 0, ψ2 (t ) = ln(2t2+1) , and 1, t =0    t   , 01 2 are in Ψ . Now, we prove our main results.

Theorem 2.1. Let (X , ≤) be a partially ordered set and suppose there is a metric d on X such that (X , d) is a complete metric space. Let F : X × X → X be a mapping having the mixed monotone property on X such that there exist two elements x0 , y0 ∈ X with x0 ≤ F (x0 , y0 )

and

y0 ≥ F (y0 , x0 ) .

Suppose there exist ϕ ∈ Φ and ψ ∈ Ψ such that

ϕ (d (F (x, y) , F (u, v))) ≤

1 2

ϕ (d (x, u) + d (y, v)) − ψ

for all x, y, u, v ∈ X with x ≥ u and y ≤ v . Suppose either (a) F is continuous or



d (x, u) + d (y, v) 2

 (2.1)

N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992

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(b) X has the following property: (i) if a non-decreasing sequence {xn } → x, then xn ≤ x, for all n, (ii) if a non-increasing sequence {yn } → y, then y ≤ yn , for all n then there exist x, y ∈ X such that x = F (x, y)

and

y = F (y, x)

that is, F has a coupled fixed point in X . Proof. Let x0 , y0 ∈ X be such that x0 ≤ F (x0 , y0 ) and y0 ≥ F (y0 , x0 ). We construct sequences {xn } and {yn } in X as follows xn+1 = F (xn , yn )

and

yn+1 = F (yn , xn )

for all n ≥ 0.

(2.2)

We shall show that x n ≤ x n +1

for all n ≥ 0

(2.3)

yn ≥ yn+1

for all n ≥ 0.

(2.4)

and

We shall use the mathematical induction. Let n = 0. Since x0 ≤ F (x0 , y0 ) and y0 ≥ F (y0 , x0 ) and as x1 = F (x0 , y0 ) and y1 = F (y0 , x0 ), we have x0 ≤ x1 and y0 ≥ y1 Thus (2.3) and (2.4) hold for n = 0. Suppose now that (2.3) and (2.4) hold for some fixed n ≥ 0. Then, since xn ≤ xn+1 and yn ≥ yn+1 , and by mixed monotone property of F , we have xn+2 = F (xn+1 , yn+1 ) ≥ F (xn , yn+1 ) ≥ F (xn , yn ) = xn+1

(2.5)

yn+2 = F (yn+1 , xn+1 ) ≤ F (yn , xn+1 ) ≤ F (yn , xn ) = yn+1 .

(2.6)

and

Now from (2.5) and (2.6), we obtain xn+1 ≤ xn+2

and yn+1 ≥ yn+2 .

Thus by the mathematical induction we conclude that (2.3) and (2.4) hold for all n ≥ 0. Therefore, x0 ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1 ≤ · · ·

(2.7)

y0 ≥ y1 ≥ y2 ≥ · · · ≥ yn ≥ yn+1 ≥ · · · .

(2.8)

and

Since xn ≥ xn−1 and yn ≤ yn−1 , from (2.1) and (2.2), we have

ϕ (d (xn+1 , xn )) = ϕ (d (F (xn , yn ) , F (xn−1 , yn−1 ))) ≤

1 2

ϕ (d (xn , xn−1 ) + d (yn , yn−1 )) − ψ



d (xn , xn−1 ) + d (yn , yn−1 )



2

.

(2.9)

.

(2.10)

Similarly, since yn−1 ≥ yn and xn−1 ≤ xn , from (2.1) and (2.2), we also have

ϕ (d (yn , yn+1 )) = ϕ (d (F (yn−1 , xn−1 ) , F (yn , xn ))) ≤

1 2

ϕ (d (yn−1 , yn ) + d (xn−1 , xn )) − ψ



d (yn−1 , yn ) + d (xn−1 , xn )



2

From (2.9) and (2.10), we have

ϕ (d (xn+1 , xn )) + ϕ (d (yn+1 , yn )) ≤ ϕ (d (xn , xn−1 ) + d (yn , yn−1 )) − 2ψ



d (xn , xn−1 ) + d (yn , yn−1 ) 2



.

(2.11)

By property (iii) of ϕ , we have

ϕ (d (xn+1 , xn ) + d (yn+1 , yn )) ≤ ϕ (d (xn+1 , xn )) + ϕ (d (yn+1 , yn )) .

(2.12)

From (2.11) and (2.12), we have

ϕ (d (xn+1 , xn ) + d (yn+1 , yn )) ≤ ϕ (d (xn , xn−1 ) + d (yn , yn−1 )) − 2ψ



d (xn , xn−1 ) + d (yn , yn−1 ) 2

 (2.13)

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which implies

ϕ (d (xn+1 , xn ) + d (yn+1 , yn )) ≤ ϕ (d (xn , xn−1 ) + d (yn , yn−1 )) . Using the fact that ϕ is non-decreasing, we get d (xn+1 , xn ) + d (yn+1 , yn ) ≤ d (xn , xn−1 ) + d (yn , yn−1 ) . Set δn = d (xn+1 , xn ) + d (yn+1 , yn ) then sequence {δn } is decreasing. Therefore, there is some δ ≥ 0 such that lim δn = lim [d (xn+1 , xn ) + d (yn+1 , yn )] = δ.

n→∞

(2.14)

n→∞

We shall show that δ = 0. Suppose, to the contrary, that δ > 0. Then taking the limit as n → ∞ (equivalently, δn → δ ) of both sides of (2.13) and have in mind that we suppose limt →r ψ(t ) > 0 for all r > 0 and ϕ is continuous, we have

]   [  δ n −1 δn−1 = ϕ (δ) − 2 lim ψ < ϕ (δ) ϕ (δ) = lim ϕ (δn ) ≤ lim ϕ (δn−1 ) − 2ψ n→∞

δn−1 →δ

2

n→∞

2

a contradiction. Thus δ = 0, that is, lim δn = lim [d (xn+1 , xn ) + d (yn+1 , yn )] = 0.

n→∞

(2.15)

n→∞

In what follows, we shall prove that {xn } and {yn } are Cauchy sequences. Suppose, to the contrary,  that   at least  of {xn } or {yn } is not Cauchy sequence. Then there exists an ε > 0 for which we can find subsequences xn(k) , xm(k) of {xn } and yn(k) , ym(k) of {yn } with n(k) > m(k) ≥ k such that



 



d xn(k) , xm(k) + d yn(k) , ym(k) ≥ ε.









(2.16)

Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k) > m(k) ≥ k and satisfying (2.16). Then d xn(k)−1 , xm(k) + d yn(k)−1 , ym(k) < ε.









(2.17)

Using (2.16), (2.17) and the triangle inequality, we have

    ε ≤ rk := d xn(k) , xm(k) + d yn(k) , ym(k)         ≤ d xn(k) , xn(k)−1 + d xn(k)−1 , xm(k) + d yn(k) , yn(k)−1 + d yn(k)−1 , ym(k)     ≤ d xn(k) , xn(k)−1 + d yn(k) , yn(k)−1 + ε. Letting k → ∞ and using (2.15) lim rk = lim d xn(k) , xm(k) + d yn(k) , ym(k)

 

k→∞





k→∞



= ε.

(2.18)

By the triangle inequality rk = d xn(k) , xm(k) + d yn(k) , ym(k)









      ≤ d xn(k) , xn(k)+1 + d xn(k)+1 , xm(k)+1 + d xm(k)+1 , xm(k)       + d yn(k) , yn(k)+1 + d yn(k)+1 , ym(k)+1 + d ym(k)+1 , ym(k)     = δn(k) + δm(k) + d xn(k)+1 , xm(k)+1 + d yn(k)+1 , ym(k)+1 . Using the property of ϕ , we have

     ϕ (rk ) = ϕ δn(k) + δm(k) + d xn(k)+1 , xm(k)+1 + d yn(k)+1 , ym(k)+1         ≤ ϕ δn(k) + δm(k) + ϕ d xn(k)+1 , xm(k)+1 + ϕ d yn(k)+1 , ym(k)+1 .

(2.19)

Since n(k) > m(k), hence xn(k) ≥ xm(k) and yn(k) ≤ ym(k) , from (2.1) and (2.2)

         ϕ d xn(k)+1 , xm(k)+1 = ϕ d F xn(k) , yn(k) , F xm(k) , ym(k) ≤ =

1 2 1 2

  ϕ d xn(k) , xm(k) + d yn(k) , ym(k)  

ϕ (rk ) − ψ



r  k

2



.



−ψ

d xn(k) , xm(k) + d yn(k) , ym(k)







2 (2.20)

N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992

987

Similarly, we also have

         ϕ d ym(k)+1 , yn(k)+1 = ϕ d F ym(k) , xm(k) , F yn(k) , xn(k) ≤ =

1 2 1 2

  ϕ d ym(k) , yn(k) + d xm(k) , xn(k) 

 

r  k

ϕ (rk ) − ψ

2





d ym(k) , yn(k) + d xm(k) , xn(k)

−ψ







2

.

(2.21)

From (2.19)–(2.21), we have

r    k . ϕ (rk ) ≤ ϕ δn(k) + δm(k) + ϕ (rk ) − 2ψ 2

Letting k → ∞ and using (2.15) and (2.18), we have

ϕ (ε) ≤ ϕ(0) + ϕ (ε) − 2 lim ψ k→∞

r  k

2

= ϕ (ε) − 2 lim ψ

r  k

2

rk →ε

< ϕ (ε)

a contradiction. This shows that {xn } and {yn } are Cauchy sequences. Since X is a complete metric space, there exist x, y ∈ X such that lim xn = x and

n→∞

lim yn = y.

(2.22)

n→∞

Now, suppose that assumption (a) holds. Taking the limit as n → ∞ in (2.2) and by (2.22), we get x = lim xn = lim F (xn−1 , yn−1 ) = F



y = lim yn = lim F (yn−1 , xn−1 ) = F



n→∞

n→∞

lim xn−1 , lim yn−1

n→∞



= F (x, y)



= F (y, x) .

n→∞

and n→∞

n→∞

lim yn−1 , lim xn−1

n→∞

n→∞

Therefore x = F (x, y) and y = F (y, x). Finally, suppose that (b) holds. Since {xn } is non-decreasing sequence and xn → x and as {yn } is a non-increasing sequence and yn → y, by assumption (b), we have xn ≥ x and yn ≤ y for all n. Since d (x, F (x, y)) ≤ d (x, xn+1 ) + d (xn+1 , F (x, y)) = d (x, xn+1 ) + d (F (xn , yn ) , F (x, y)) therefore

ϕ (d (x, F (x, y))) ≤ ϕ (d (x, xn+1 )) + ϕ (d (F (xn , yn ) , F (x, y))) 1

≤ ϕ (d (x, xn+1 )) + ϕ (d (xn , x) + d (yn , y)) − ψ



2

d (xn , x) + d (yn , y) 2



.

Taking the limit as n → ∞ in the above inequality, using (2.22) and the property of ψ , we get ϕ (d (x, F (x, y))) = 0, thus d (x, F (x, y)) = 0. Hence x = F (x, y). Similarly, one can show that y = F (y, x). Thus we proved that F has a coupled fixed point.  Corollary 2.2. Let (X , ≤) be a partially ordered set and suppose there is a metric d on X such that (X , d) is a complete metric space. Let F : X × X → X be a mapping having the mixed monotone property on X such that there exist two elements x0 , y0 ∈ X with x0 ≤ F (x0 , y0 )

and

y0 ≥ F (y0 , x0 ) .

Suppose there exists ψ ∈ Ψ such that d (F (x, y) , F (u, v)) ≤

d (x, u) + d (y, v) 2

−ψ



d (x, u) + d (y, v) 2

for all x, y, u, v ∈ X with x ≥ u and y ≤ v . Suppose either (a) F is continuous or (b) X has the following property: (i) if a non-decreasing sequence {xn } → x, then xn ≤ x, for all n, (ii) if a non-increasing sequence {yn } → y, then y ≤ yn , for all n then F has a coupled fixed point in X . Proof. In Theorem 2.1, taking ϕ(t ) = t, we get Corollary 2.2.





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N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992

Corollary 2.3. Let (X , ≤) be a partially ordered set and suppose there is a metric d on X such that (X , d) is a complete metric space. Let F : X × X → X be a mapping having the mixed monotone property on X such that there exist two elements x0 , y0 ∈ X with x0 ≤ F (x0 , y0 )

and

y0 ≥ F (y0 , x0 ) .

Suppose there exists a real number k ∈ [0, 1) such that d (F (x, y) , F (u, v)) ≤

k 2

[d (x, u) + d (y, v)]

for all x, y, u, v ∈ X with x ≥ u and y ≤ v . Suppose either (a) F is continuous or (b) X has the following property: (i) if a non-decreasing sequence {xn } → x, then xn ≤ x, for all n, (ii) if a non-increasing sequence {yn } → y, then y ≤ yn , for all n then F has a coupled fixed point in X . k Proof. In Corollary 2.2, taking ψ(t ) = 1− t, we get Corollary 2.3. 2



Now we shall prove the uniqueness of the coupled fixed point. Note that if (X , ≤) is a partially ordered set, then we endow the product X × X with the following partial order relation: for all (x, y) , (u, v) ∈ X × X ,

(x, y) ≤ (u, v) ⇔ x ≤ u, y ≥ v.

Theorem 2.4. In addition to hypotheses of Theorem 2.1, suppose that for every (x, y), (z , t ) in X × X , there exists a (u, v) in X × X that is comparable to (x, y) and (z , t ), then F has a unique coupled fixed point. Proof. From Theorem 2.1, the set of coupled fixed points of F is non-empty. Suppose (x, y) and (z , t ) are coupled fixed points of F , that is, x = F (x, y) , y = F (y, x) , z = F (z , t ) and t = F (t , z ). We shall show that x = z and y = t. By assumption, there exists (u, v) ∈ X × X that is comparable to (x, y) and (z , t ). We define sequences {un } , {vn } as follows u0 = u,

v0 = v,

un+1 = F (un , vn )

and

vn+1 = F (vn , un )

for all n.

Since (u, v) is comparable with (x, y), we may assume that (x, y) ≥ (u, v) = (u0 , v0 ). By using the mathematical induction, it is easy to prove that

(x, y) ≥ (un , vn ) ,

for all n.

(2.23)

From (2.1) and (2.23), we have

ϕ (d (x, un+1 )) = ϕ (d (F (x, y) , F (un , vn ))) ≤

1 2

ϕ (d (x, un ) + d (y, vn )) − ψ



d (x, un ) + d (y, vn )



2

.

(2.24)

.

(2.25)

Similarly,

ϕ (d (vn+1 , y)) = ϕ (d (F (vn , un ) , F (y, x))) ≤

1 2

ϕ (d (vn , y) + d (un , x)) − ψ



d (vn , y) + d (un , x) 2



From (2.24), (2.25) and the property of ϕ , we have

ϕ (d (x, un+1 ) + d (y, vn+1 )) ≤ ϕ (d (x, un+1 )) + ϕ (d (y, vn+1 ))   d (x, un ) + d (y, vn ) ≤ ϕ (d (x, un ) + d (y, vn )) − 2ψ 2

(2.26)

which implies

ϕ (d (x, un+1 ) + d (y, vn+1 )) ≤ ϕ (d (x, un ) + d (y, vn )) . Thus, d (x, un+1 ) + d (y, vn+1 ) ≤ d (x, un ) + d (y, vn ) . That is, the sequence {d (x, un ) + d (y, vn )} is decreasing. Therefore, there exists α ≥ 0 such that lim [d (x, un ) + d (y, vn )] = α.

n→∞

(2.27)

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We shall show that α = 0. Suppose, to the contrary, that α > 0. Taking the limit as n → ∞ in (2.26), we have

ϕ (α) ≤ ϕ (α) − 2 lim ψ



d (x, un ) + d (y, vn )



2

n→∞

< ϕ (α)

a contradiction. Thus, α = 0, that is, lim [d (x, un ) + d (y, vn )] = 0.

n→∞

It implies lim d (x, un ) = lim d (y, vn ) = 0.

n→∞

(2.28)

n→∞

Similarly, we show that lim d (z , un ) = lim d (t , vn ) = 0.

(2.29)

n→∞

n→∞

From (2.28) and (2.29), we have x = z and y = t.



Corollary 2.5. In addition to hypotheses of Corollary 2.2, suppose that for every (x, y), (z , t ) in X × X , there exists a (u, v) in X × X that is comparable to (x, y) and (z , t ), then F has a unique coupled fixed point. Theorem 2.6. In addition to hypotheses of Theorem 2.1, if x0 and y0 are comparable then F has a fixed point. Proof. Following the proof of Theorem 2.1, F has a coupled fixed point (x, y). We only have to show that x = y. Since x0 and y0 are comparable, we may assume that x0 ≥ y0 . By using the mathematical induction, one can show that xn ≥ y n

for all n ≥ 0

(2.30)

where xn+1 = F (xn , yn ) and yn+1 = F (yn , xn ) , n = 0, 1, 2, . . . . We have d (x, y) ≤ d (x, xn+1 ) + d (xn+1 , yn+1 ) + d (yn+1 , y)

= d (x, xn+1 ) + d (yn+1 , y) + d (F (xn , yn ) , F (yn , xn )) . Hence

ϕ (d (x, y)) ≤ ϕ (d (x, xn+1 ) + d (yn+1 , y)) + ϕ (d (F (xn , yn ) , F (yn , xn ))) 1

≤ ϕ (d (x, xn+1 ) + d (yn+1 , y)) + ϕ (2d (xn , yn )) − ψ (d (xn , yn )) 2

≤ ϕ (d (x, xn+1 ) + d (yn+1 , y)) + ϕ (d (xn , yn )) − ψ (d (xn , yn )) .

(2.31)

Suppose x ̸= y, that is, d (x, y) > 0, taking the limit as n → ∞ in (2.31) we get

ϕ (d (x, y)) ≤ ϕ(0) + ϕ (d (x, y)) − lim ψ (d (xn , yn )) n→∞

or lim

d(xn ,yn )→d(x,y)

ψ (d (xn , yn )) ≤ 0

a contradiction. Thus, x = y.



Corollary 2.7. In addition to hypotheses of Corollary 2.2, if x0 and y0 are comparable then F has a fixed point. 3. Application In this section, we study the existence of a unique solution to a nonlinear integral equation, as an application to the fixed point theorem proved in Section 2. Let Θ denote the class of those functions θ : [0, ∞) → [0, ∞) which satisfies the following conditions: (i) θ is increasing. (ii) There exists ψ ∈ Ψ such that θ (x) = For example, θ (x) = kx, where 0 ≤ k < Consider the integral equation x(t ) =

1 2

x 2

−ψ

x 2

, θ (x) =

for all x ∈ [0, ∞).

x2 2(x+1)

, θ (x) =

x 2

−

ln(x+1) 2

are in Θ .

b

∫

(K1 (t , s) + K2 (t , s)) (f (s, x(s)) + g (s, x(s))) ds + h(t ) a

t ∈ I = [a, b] .

(3.1)

990

N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992

We assume that K1 , K2 , f , g satisfy the following conditions Assumption 3.1. (i) K1 (t , s) ≥ 0 and K2 (t , s) ≤ 0 for all t , s ∈ [a, b]. (ii) There exist λ, µ > 0 and θ ∈ Θ such that for all x, y ∈ R, x ≥ y, 0 ≤ f (t , x) − f (t , y) ≤ λθ(x − y) and

−µθ (x − y) ≤ g (t , x) − g (t , y) ≤ 0 b (iii) max{λ, µ} supt ∈I a (K1 (t , s) − K2 (t , s)) ds ≤ 12 . Definition 3.2. An element (α, β) ∈ C (I , R) × C (I , R) is called a coupled lower and upper solution of the integral equation (3.1) if α(t ) ≤ β(t ) and

α(t ) ≤

b

∫

K1 (t , s) (f (s, α(s)) + g (s, β(s))) ds +

b

∫

K2 (t , s) (f (s, β(s)) + g (s, α(s))) ds + h(t ) a

a

and

β(t ) ≥

b

∫

K1 (t , s) (f (s, β(s)) + g (s, α(s))) ds + a

b

∫

K2 (t , s) (f (s, α(s)) + g (s, β(s))) ds + h(t ) a

for all t ∈ [a, b]. Theorem 3.3. Consider the integral equation (3.1) with K1 , K2 ∈ C (I × I , R) , f , g ∈ C (I × R, R) and h ∈ C (I , R) and suppose that Assumption 3.1 is satisfied. Then the existence of a coupled lower and upper solution for (3.1) provides the existence of a unique solution of (3.1) in C (I , R). Proof. Let X := C (I , R). X is a partially ordered set if we define the following order relation in X: x, y ∈ C (I , R) ,

x ≤ y ⇔ x(t ) ≤ y(t ),

for all t ∈ [a, b] .

And (X , d) is a complete metric space with metric d (x, y) = sup |x(t ) − y(t )| ,

x, y ∈ C (I , R) .

t ∈I

Suppose {un } is a monotone non-decreasing in X that converges to u ∈ X . Then for every t ∈ I, the sequence of real numbers u1 (t ) ≤ u2 (t ) ≤ · · · ≤ un (t ) ≤ · · · converges to u(t ). Therefore, for all t ∈ I , n ∈ N, un (t ) ≤ u(t ). Hence un ≤ u, for all n. Similarly, we can verify that limit v(t ) of a monotone non-increasing sequence vn (t ) in X is a lower bound for all the elements in the sequence. That is, v ≤ vn for all n. Therefore, condition (b) of Corollary 2.2 holds. Also, X × X = C (I , R) × C (I , R) is a partially ordered set if we define the following order relation in X × X

(x, y) , (u, v) ∈ X × X , (x, y) ≤ (u, v) ⇔ x(t ) ≤ u(t ) and y(t ) ≥ v(t ),

∀t ∈ I .

For any x, y ∈ X , max {x(t ), y(t )} and min {x(t ), y(t )}, for each t ∈ I, are in X and are the upper and lower bounds of x, y, respectively. Therefore, for every (x, y) , (u, v) ∈ X × X , there exists a (max {x, u} , min {y, v}) ∈ X × X that is comparable to (x, y) and (u, v). Define F : X × X → X by F (x, y) (t ) =

b

∫

K1 (t , s) (f (s, x(s)) + g (s, y(s))) ds + a

∫

b

K2 (t , s) (f (s, y(s)) + g (s, x(s))) ds + h(t )

a

for all t ∈ [a, b]. Now we shall show that F has the mixed monotone property. Indeed, for x1 ≤ x2 , that is, x1 (t ) ≤ x2 (t ), for all t ∈ [a, b], we have F (x1 , y) (t ) − F (x2 , y) (t ) =

b

∫

K1 (t , s) (f (s, x1 (s)) + g (s, y(s))) ds a b

∫

K2 (t , s) (f (s, y(s)) + g (s, x1 (s))) ds + h(t )

+ a b

∫

K1 (t , s) (f (s, x2 (s)) + g (s, y(s))) ds

− a

N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992 b

∫

K2 (t , s) (f (s, y(s)) + g (s, x2 (s))) ds − h(t )

− a b

∫

K1 (t , s) (f (s, x1 (s)) − f (s, x2 (s))) ds

= a

b

∫

K2 (t , s) (g (s, x1 (s)) − g (s, x2 (s))) ds ≤ 0,

+ a

by Assumption 3.1. Hence F (x1 , y) (t ) ≤ F (x2 , y) (t ), ∀t ∈ I, that is, F (x1 , y) ≤ F (x2 , y). Similarly, if y1 ≥ y2 , that is, y1 (t ) ≥ y2 (t ), for all t ∈ [a, b], we have F (x, y1 ) (t ) − F (x, y2 ) (t ) =

b

∫

K1 (t , s) (f (s, x(s)) + g (s, y1 (s))) ds a b

∫

K2 (t , s) (f (s, y1 (s)) + g (s, x(s))) ds + h(t )

+ a b

∫

K1 (t , s) (f (s, x(s)) + g (s, y2 (s))) ds

− a b

∫

K2 (t , s) (f (s, y2 (s)) + g (s, x(s))) ds − h(t )

− a b

∫

K1 (t , s) (g (s, y1 (s)) − g (s, y2 (s))) ds

= a

b

∫

K2 (t , s) (f (s, y1 (s)) − f (s, y2 (s))) ds ≤ 0,

+ a

by Assumption 3.1. Hence F (x, y1 ) (t ) ≤ F (x, y2 ) (t ), ∀t ∈ I, that is, F (x, y1 ) ≤ F (x, y2 ). Thus, F (x, y) is monotone non-decreasing in x and monotone non-increasing in y. Now, for x ≥ u, y ≤ v , that is, x(t ) ≥ u(t ), y(t ) ≤ v(t ) for all t ∈ I, we have d (F (x, y), F (u, v)) = sup |F (x, y)(t ) − F (u, v)(t )| t ∈I

∫  b  K1 (t , s) (f (s, x(s)) + g (s, y(s))) ds = sup t ∈I  a ∫ b + K2 (t , s) (f (s, y(s)) + g (s, x(s))) ds + h(t ) a ∫ b

K1 (t , s) (f (s, u(s)) + g (s, v(s))) ds

− a b

∫ + a

   K2 (t , s) (f (s, v(s)) + g (s, u(s))) ds + h(t )  

∫  b  = sup K1 (t , s) [(f (s, x(s)) − f (s, u(s))) + (g (s, y(s)) − g (s, v(s)))] ds t ∈I  a  ∫ b   + K2 (t , s) [(f (s, y(s)) − f (s, v(s))) + (g (s, x(s)) − g (s, u(s)))] ds  a ∫  b  = sup K1 (t , s) [(f (s, x(s)) − f (s, u(s))) − (g (s, v(s)) − g (s, y(s)))] ds t ∈I  a  ∫ b   K2 (t , s) [(f (s, v(s)) − f (s, y(s))) − (g (s, x(s)) − g (s, u(s)))] ds −  a ∫  b  ≤ sup K1 (t , s) [λθ (x(s) − u(s)) + µθ (v(s) − y(s))] ds t ∈I  a

991

992

N.V. Luong, N.X. Thuan / Nonlinear Analysis 74 (2011) 983–992

   K2 (t , s) [λθ (v(s) − y(s)) + µθ (x(s) − u(s))] ds −  a ∫ b ≤ max {λ, µ} sup (K1 (t , s) − K2 (t , s)) [θ (x(s) − u(s)) + θ (v(s) − y(s))] ds. ∫

b

t ∈I

a

As the function θ is increasing and x ≥ u, y ≤ v , then θ (x(s) − u(s)) ≤ θ (d (x, u)) , θ (v(s) − y(s)) ≤ θ (d (v, y)), for all s ∈ [a, b], we obtain d (F (x, y), F (u, v)) ≤ max {λ, µ} sup t ∈I

b

∫

(K1 (t , s) − K2 (t , s)) a

× [θ (x(s) − u(s)) + θ (v(s) − y(s))] ds ≤ max {λ, µ} . [θ (d (x, u)) + θ (d (v, y))] . sup t ∈I

≤

1 2

b

∫

(K1 (t , s) − K2 (t , s)) ds a

[θ (d (x, u)) + θ (d (v, y))]

≤ θ (d (x, u) + d (v, y)) =

d (x, u) + d (v, y) 2

−ψ



d (x, u) + d (v, y)



2

.

Therefore, for x ≥ u, y ≤ v , we have d (F (x, y), F (u, v)) ≤

d (x, u) + d (v, y) 2

−ψ



d (x, u) + d (v, y) 2



.

Now, let (α, β) be a coupled lower and upper solution of the integral equation (3.1) then we have α(t ) ≤ F (α, β) (t ) and β(t ) ≥ F (β, α) (t ) for all t ∈ [a, b], that is, α ≤ F (α, β) and β ≥ F (β, α). Finally, Corollaries 2.2 and 2.5 give that F has a unique coupled fixed point (x, y). Since α ≤ β , then the hypothesis of Corollary 2.7 is satisfied. Therefore x = y, that is, x(t ) = y(t ) for all t ∈ [a, b], implying x = F (x, x) and x(t ) is the unique solution of Eq. (3.1).  Acknowledgements The authors would like to express their sincere appreciation to the referees for their very helpful suggestions and many kind comments. References [1] A.C.M. Ran, M.C.B. Reurings, A fixed point theorem in partially ordered sets and some applications to matrix equations, Proc. Amer. Math. Soc. 132 (2004) 1435–1443. [2] T. Gnana Bhaskar, V. Lakshmikantham, Fixed point theorems in partially ordered metric spaces and applications, Nonlinear Anal. TMA 65 (2006) 1379–1393. [3] J.J. Nieto, R. Rodriguez-Lopez, Contractive mapping theorems in partially ordered sets and applications to ordinary differential equation, Order 22 (2005) 223–239. [4] J.J. Nieto, R. Rodriguez-Lopez, Existence and uniqueness of fixed point in partially ordered sets and applications to ordinary differential equations, Acta Math. Sin. (Engl. Ser.) 23 (12) (2007) 2205–2212. [5] R.P. Agarwal, M.A. El-Gebeily, D. O’Regan, Generalized contractions in partially ordered metric spaces, Appl. Anal. 87 (2008) 1–8. [6] V. Lakshmikantham, L. Ćirić, Coupled fixed point theorems for nonlinear contractions in partially ordered metric spaces, Nonlinear Anal. TMA 70 (2009) 4341–4349. [7] B. Samet, Coupled fixed point theorems for a generalized Meir–Keeler contraction in partially ordered metric spaces, Nonlinear Anal. TMA (2010) doi:10.1016/j.na.2010.02.026. [8] I. Altun, H. Simsek, Some fixed point theorems on ordered metric spaces and application, Fixed Point Theory Appl. 2010 (2010) 17 pages. Article ID 621469. [9] J. Harjani, K. Sadarangani, Generalized contractions in partially ordered metric spaces and applications to ordinary differential equations, Nonlinear Anal. TMA 72 (2010) 1188–1197.