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Coupled fixed point theorems for generalized Mizoguchi-Takahashi contractions with applications

Abstract

We derive some new coupled fixed point theorems for nonlinear contractive maps that satisfied a generalized Mizoguchi-Takahashi's condition in the setting of ordered metric spaces. Presented theorems extends and generalize many well-known results in the literature. As an application, we give an existence and uniqueness theorem for the solution to a two-point boundary value problem.

2000 Mathematics Subject Classification: 54H25; 47H10.

1 Introduction

Let (X, d) be a metric space. Denote by P(X) the set of all nonempty subsets of X and CB(X) the family of all nonempty closed and bounded subsets of X. A point x in X is a fixed point of a multivalued map T : XP(X), if x Tx. Nadler [1] extended the Banach contraction principle to multivalued mappings.

Theorem 1.1 (Nadler[1]) Let (X, d) be a complete metric space and let T : XCB(X) be a multivalued map. Assume that there exists r [0,1) such that

H ( T x , T y ) r d ( x , y )

for all x, y X, where H is the Hausdorff metric with respect to d. Then T has a fixed point.

Reich [2] proved the following generalization of Nadler's fixed point theorem.

Theorem 1.2 (Reich[2]) Let (X, d) be a complete metric space and T : XC(X) be a multi-valued map with non empty compact values. Assume that

H ( T x , T y ) φ ( d ( x , y ) ) d ( x , y )

for all x, y X, where φ is a function from [0, ∞) into [0,1) satisfying lim  sup s t + φ ( s ) <1for all t > 0. Then T has a fixed point.

Mizoguchi and Takahashi [3] proved the following generalization of Nadler's fixed point theorem for a weak contraction which is a partial answer of Problem 9 in Reich [4].

Theorem 1.3 (Mizoguchi and Takahashi[3]) Let (X, d) be a complete metric space and T : XCB(X) be a multivalued map. Assume that

H ( T x , T y ) φ ( d ( x , y ) ) d ( x , y )

for all x, y X, where φ is a function from [0, ∞) into [0,1) satisfying lim  sup s t + φ ( s ) <1for all t ≥ 0. Then T has a fixed point.

Suzuki [5] gave a very simple proof of Theorem 1.3.

Very recently, Amini-Harandi and O'Regan [6] obtained a nice generalization of Mizoguchi and Takahashi's fixed point theorem. Throughout the article, let Ψ be the family of all functions ψ : [0, ∞) → [0, ∞) satisfying the following conditions:

  1. (a)

    ψ(s) = 0 s = 0,

  2. (b)

    ψ is nondecreasing,

  3. (c)

    lim  sup s 0 + s ψ ( s ) < .

We denote by Φ the set of all functions φ : [0, ∞) → [0,1) satisfying lim sup r t + φ ( r ) <1 for all t ≥ 0.

Theorem 1.4 (Amini-Harandi and O'Regan[6]) Let (X, d) be a complete metric space and T : XCB(X) be a multivalued map. Assume that

ψ ( H ( T x , T y ) ) φ ( ψ ( d ( x , y ) )) ψ ( d ( x , y ) )

for all x, y X, where ψ Ψ is lower semicontinuous and φ Φ. Then T has a fixed point.

The existence of fixed point in partially ordered sets has been investigated recently in [727] and references therein.

Du [13] proved some coupled fixed point results for weakly contractive single-valued maps that satisfy Mizoguchi-Takahashi's condition in the setting of quasiordered metric spaces. Before recalling the main results in [13], we need some definitions.

Definition 1.1 (Bhaskar and Lakshmikantham[11]) Let X be a nonempty set and A : X × XX be a given map. We call an element (x, y) X × X a coupled fixed point of A if x = A(x, y) and y = A(y, x).

Definition 1.2 (Bhaskar and Lakshmikantham[11]) Let (X, ) be a quasiordered set and A : XXX a map. We say that A has the mixed monotone property on X if A(x, y) is monotone nondecreasing in x X and is monotone nonincreasing in y X, that is, for any x, y X,

x 1 , x 2 X w i t h x 1 x 2 A ( x 1 , y ) A ( x 2 , y ) , y 1 , y 2 X w i t h y 1 y 2 A ( x , y 1 ) A ( x , y 2 ) .

Definition 1.3 (Du [13]) Let (X, d) be a metric space with a quasi-order . A nonempty subset M of X is said to be

(i) sequentially ↑-complete if every -nondecreasing Cauchy sequence in M converges;

(ii) sequentially ↓-complete if every -nonincreasing Cauchy sequence in M converges;

(iii) sequentially ↕-complete if it is both ↑-complete and ↓-complete.

Theorem 1.5 (Du[13]) Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a continuous map having the mixed monotone property on X. Assume that there exists a function φ Φ such that

d ( A ( x , y ) , A ( u , v ) ) 1 2 φ ( d ( x , u ) + d ( y , v ) ) ( d ( x , u ) + d ( y , v ) )

for all x u and y v. If there exist x0, y0 X such that x 0 A (x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then A has a coupled fixed point.

Theorem 1.6 (Du[13]) Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a map having the mixed monotone property on X. Assume that

(i) any -nondecreasing sequence (x n ) with x n x implies x n x for all n,

(ii) any -nonincreasing sequence (y n ) with y n y implies for all n.

Assume also that there exists a function φ Φ such that

d ( A ( x , y ) , A ( u , v ) ) 1 2 φ ( d ( x , u ) + d ( y , v ) ) ( d ( x , u ) + d ( y , v ) )

for all x u and y v. If there exist x0,y0 X such that x 0 A(x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then A has a coupled fixed point.

Very recently, Gordji and Ramezani [14] established a new fixed point theorem for a self-map T : XX satisfying a generalized Mizoguchi-Takahashi's condition in the setting of ordered metric spaces. The main result in [14] is the following.

Theorem 1.7 (Gordji and Ramezani[14]) Let (X, d, ) be a complete ordered metric space and T : XX an increasing mapping such that there exists an element x0 X with x 0 Tx 0 . Suppose that there exists a lower semicontinuous function ψ Ψ and φ Φ such that

ψ ( d ( T x , T y ) ) φ ( ψ ( d ( x , y ) ) ψ ( d ( x , y ) )

for all x, y X such that x and y are comparable. Assume that either T is continuous or X is such that the following holds: any -nondecreasing sequence (x n ) with x n x implies x n x for all n. Then T has a fixed point.

In this article, we present new coupled fixed point theorems for mixed monotone mappings satisfying a generalized Mizoguchi-Takahashi's condition in the setting of ordered metric spaces. Presented theorems extend and generalize Du [[13], Theorems 2.8 and 2.10], Bhaskar and Lakshmikantham [[11], Theorems 2.1 and 2.2], Harjani et al. [[15], Theorems 2 and 3], and other existing results in the literature. Moreover, some applications to ordinary differential equations are presented.

2 Main results

Through this article, we will use the following notation: if (X, ) is an ordered set, we endow the product set X × X with the order given by

( x , y ) , ( u , v ) X × X , ( x , y ) ( u , v ) x u , y v .

Our first result is the following.

Theorem 2.1 Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a map having the mixed monotone property on X. Suppose that there exist ψ Ψ and φ Φ such that for any (x, y),(u, v) X×X with (u, v) (x, y),

ψ ( d ( A ( x , y ) , A ( u , v ) ) ) φ ( ψ ( max { d ( x , u ) , d ( y , v ) } ) ) ψ ( max { d ( x , u ) , d ( y , v ) } ) .
(1)

Suppose also that either A is continuous or (X, d, ) has the following properties:

(i) any -nondecreasing sequence (x n ) with x n x implies x n x for each n,

(ii) any -nonincreasing sequence (y n ) with y n y implies y n y for each n.

If there exist x0,y0 X such that x 0 A(x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then there exist a, b X such that a = A(a, b) and b = A(b, a).

Proof. Define the sequences (x n ) and (y n ) in X by

x n + 1 = A ( x n , y n ) , y n + 1 = A ( y n , x n ) for all n 0 .

In order to make the proof more comprehensive we will divide it into several steps.

Step 1. x n xn+1and y n yn+1for all n ≥ 0.

We use mathematical induction.

As x 0 A(x 0 , y 0 ) = x1 and y 0 A(y 0 , x 0 ) = y1, our claim is satisfied for n = 0.

Suppose that our claim holds for some fixed n ≥ 0. Then, since x n xn+1and y n yn+1, and as A has the mixed monotone property, we get

x n + 1 = A ( x n , y n ) A ( x n + 1 , y n ) A ( x n + 1 , y n + 1 ) = x n + 2

and

y n + 1 = A ( y n , x n ) A ( y n + 1 , x n ) A ( y n + 1 , x n + 1 ) = y n + 2 .

This proves our claim.

Step 2. limn→∞ψ(max{d(x n+ 1,x n ),d(y n+ 1,y n )}) = 0.

Since x n x n+1 and y n y n+1 (Step 1), we have (x n , y n ) (x n+ 1,y n+ 1), and by (1), we have

ψ ( d ( A ( x n + 1 , y n + 1 ) , A ( x n , y n ) ) ) φ ( ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) ) ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) .
(2)

Similarly, since (y n+ 1,x n+ 1) (y n , x n ), by (1), we have

ψ ( d ( A ( y n , x n ) , A ( y n + 1 , x n + 1 ) ) ) φ ( ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) ) ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) .
(3)

From (2) and (3), we get

max { ψ ( d ( x n + 2 , x n + 1 ) , ψ ( d ( y n + 2 , y n + 1 ) ) } ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) .

Since ψ is nondecreasing, this implies that

ψ max d ( x n + 2 , x n + 1 ) d ( y n + 2 , y n + 1 ) ψ max d ( x n + 1 , x n ) , d ( y n + 1 , y n )
(4)

for all n ≥ 0. Now, (4) means that (ψ(max{d(x n+ 1,x n ),d(y n+ 1,y n )})) is a non increasing sequence. On the other hand, this sequence is bounded below; thus there exists μ ≥ 0 such that

lim n ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) = μ .
(5)

Since φ Φ, we have lim  sup r μ + φ ( r ) <1 and φ (μ) < 1. Then, there exist α [0,1) and ε > 0 such that φ(r) ≤ α for all r [μ,μ + ε). From (5), we can take n0 ≥ 0 such that μψ(max{d(x n+ 1,x n ),d(y n+ 1,y n )}) ≤ μ + ε for all nn0. Then, from (2), for all nn0, we have

ψ ( d ( x n + 1 , x n + 2 ) ) φ ( ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) ) ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) α ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) .
(6)

Similarly, from (3), for all n > n0, we have

ψ ( d ( y n + 1 , y n + 2 ) ) φ ( ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) ) ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) α ψ ( max { d ( y n , y n + 1 ) , d ( x n , x n + 1 ) } ) .
(7)

Now, from (6) and (7), we get

ψ max d ( x n + 1 , x n + 2 ) d ( y n + 1 , y n + 2 ) α ψ max d ( y n , y n + 1 ) , d ( x n , x n + 1 )
(8)

for all nn0. Letting n → ∞ in the above inequality and using (5), we obtain that

μ α μ .

Since α [0,1), this implies that μ = 0. Thus, we proved that

lim n ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) = 0 .
(9)
  • Step 3. limn→∞max{d(x n+ 1,x n ),d(y n+ 1,y n )} = 0.

Since (ψ(max{d(x n+ 1,x n ),d(y n+ 1,y n )})) is a decreasing sequence and ψ is nondecreasing, then (max{d(x n+ 1,x n ),d(y n+ 1,y n )}) is a decreasing sequence of positive numbers. This implies that there exists θ ≥ 0 such that

lim n max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) = θ + .

Since ψ is nondecreasing, we have

ψ ( max d ( x n + 1 , x n ) , d ( y n + 1 , y n ) ) ψ ( θ ) .

Letting n → ∞ in the above inequality, from (9), we obtain that 0 ≥ ψ(θ), which implies that θ = 0. Thus, we proved that

lim n max d ( x n + 1 , x n ) , d ( y n + 1 , y n ) = 0 .
(10)

Step 4. (x n ) and (y n ) are Cauchy sequences in (X, d).

Suppose that max{d(x m+ 1,x m ), d(y m+ 1,y m )} = 0 for some m ≥ 0. Then, we have d(x m +1,x m ) = d(y m+ 1,y m ) = 0, which implies that (x m , y m ) = (x m+ 1,y m+ 1), that is, x m = A(x m , y m ) and y m = A(y m , x m ). Then, (x m , y m ) is a coupled fixed point of A.

Now, suppose that max{d(x n +1,x n ), d(y n+ 1,y n )} 0 for all n ≥ 0.

Denote

a n = ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) for all n 0 .

From (8), we have

a n + 1 α a n for all n n 0 .

Then, we have

n = 0 a n n = 0 n 0 a n + n = n 0 + 1 α n - n 0 a n 0 < .

On the other hand, we have

lim  sup n max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ψ ( max { d ( x n + 1 , x n ) , d ( y n + 1 , y n ) } ) lim  sup s 0 + s ψ ( s ) < .

Then ∑ n max{d(x n , x n +1),d(y n , y n +1)} < ∞. Hence, (x n ) and (y n ) are Cauchy sequences in X.

Step 5. Existence of a coupled fixed point.

Since (X, d, ) is sequentially ↕-complete metric space and (x n ) is -nondecreasing Cauchy sequence, there exists a X such that

lim n x n = a .
(11)

Similarly, since (X, d, ) is sequentially ↕-complete metric space and (y n ) is -noincreasing Cauchy sequence, there exists b X such that

lim n y n = b .
(12)

Case 1. A is continuous.

From the continuity of A and using (11) and (12), we get

a = lim n x n + 1 = lim n A x n , y n = A ( lim n x n , lim n y n ) = A ( a , b )

and

b = lim n y n + 1 = lim n A y n , x n = A ( lim n y n , lim n x n ) = A ( b , a ) .

Case 2. (X, d, ) satisfies (i) and (ii).

Since (x n ) is -nondecreasing and limn→∞x n = a, then from (i), we have x n a for all n. Similarly, from (ii), since (y n ) is -nonincreasing and limn→∞y n = b, we have y n b for all n. Then, we have (x n , y n ) (a, b) for all n. Now, applying our contractive condition (1), we get

ψ ( d ( x n + 1 , A ( a , b ) ) ) = ψ ( d ( A ( x n , y n ) , A ( a , b ) ) ) φ ( ψ ( max { d ( x n , a ) , d ( y n , b ) } ) ) ψ ( max { d ( x n , a ) , d ( y n , b ) } ) ψ ( max { d ( x n , a ) , d ( y n , b ) } ) .

Since ψ is nondecreasing, this implies that

d ( x n + 1 , A ( a , b ) ) max { d ( x n , a ) , d ( y n , b ) } .

letting n → ∞ in the above inequality, we obtain that d(a, A(a, b)) ≤ 0, that is, a = A(a, b). Similarly, we can show that b = A(b, a). □

Remark 2.1 In our presented theorems we don't need the hypothesis: ψ is lower semicontinuous. Such hypothesis is considered in Theorem 1.7 of Gordji and Ramezani[14].

In what follows, we give a sufficient condition for the uniqueness of the coupled fixed point in Theorem 2.1. We consider the following hypothesis:

(H): For all (x, y), (u, v) X × X, there exists (w, z) X × X such that (x, y) (w, z) and (u, v) (w, z).

Theorem 2.2 Adding condition (H) to the hypotheses of Theorem 2.1, we obtain uniqueness of the coupled fixed point of A.

Proof. Suppose that (a, b) and (c, e) are coupled fixed points of A, that is, a = A(a, b), b = A(b, a), c = A(c, e) and e = A(e, c). From (H), there exists (f0, g0) X × X such that (a, b) (f0, g0) and (c, e) (f0, g0).

We construct the sequences (f n ) and (g n ) in X defined by

f n + 1 = A ( f n , g n ) , g n + 1 = A ( g n , f n ) for all n 0 .

We claim that (a, b) (f n , g n ) for all n ≥ 0.

In fact, we will use mathematical induction.

Since (a, b) (f0, g0), then our claim is satisfied for n = 0.

Suppose that our claim holds for some fixed n ≥ 0. Then, we have (a, b) (f n , g n ), that is, a f n and b g n . Using the mixed monotone property of A, we get

f n + 1 = A ( f n , g n ) A ( a , g n ) A ( a , b ) = a

and

g n + 1 = A ( g n , f n ) A ( b , f n ) A ( b , a ) = b .

This proves that (a, b) (f n+1 , g n+1 ). Then, our claim holds.

Now, we can apply (1) with (u, v) = (a, b) and (x, y) = (f n , g n ). We get

ψ ( d ( f n + 1 , a ) ) = ψ ( d ( A ( f n , g n ) , A ( a , b ) ) ) φ ( ψ ( max { d ( f n , a ) , d ( g n , b ) } ) ) ψ ( max { d ( f n , a ) d ( g n , b ) } ) .
(13)

Similarly, we have

ψ ( d ( g n + 1 , b ) ) = ψ ( d ( A ( g n , f n ) , A ( b , a ) ) ) φ ( ψ ( max { d ( f n , a ) , d ( g n , b ) } ) ) ψ ( max { d ( f n , a ) d ( g n , b ) } ) .
(14)

Combining (13) with (14), we obtain

ψ ( max { d ( f n + 1 , a ) , d ( g n + 1 , b ) } ) φ ( ψ ( max { d ( f n , a ) , d ( g n , b ) } ) ) ψ ( max { d ( f n , a ) , d ( g n , b ) } ) ψ ( max { d ( f n , a ) , d ( g n , b ) } ) .
(15)

Consequently, (ψ(max{d(f n , a), d(g n , b)})) is a nonnegative decreasing sequence and hence possesses a limit γ ≥ 0. Following the same strategy used in the proof of Theorem 2.1, one can show that γ = 0 and limn→∞max{d(f n , a),d(g n , b)} = 0.

Analogously, it can be proved that limn→∞max{d(f n , c),d(g n , e)} = 0.

Now, we have

d ( a , c ) d ( a , f n ) + d ( f n , c ) max { d ( f n , a ) , d ( g n , b ) } + max { d ( f n , c ) , d ( g n , e ) }

and

d ( b , e ) d ( b , g n ) + d ( g n , e ) max { d ( f n , a ) , d ( g n , b ) } + max { d ( f n , c ) , d ( g n , e ) } .

Then, we have

max { d ( a , c ) , d ( b , e ) } max { d ( f n , a ) , d ( g n , b ) } + max { d ( f n , c ) , d ( g n , e ) } .

Letting n → ∞ in the above inequality, we get

max { d ( a , c ) , d ( b , e ) } = 0 ,

which implies that d(a, c) = d(b, e) = 0. Then, (a, b) = (c, e). □

Theorem 2.3 Under the assumptions of Theorem 2.1, suppose that x0and y 0 are comparable, then the coupled fixed point (a, b) X × X satisfies a = b.

Proof. Assume that x 0 y 0 (the same strategy can be used if y 0 x 0 ). Using the mixed monotone property of A, it is easy to show that x n y n for all n ≥ 0.

Now, using the contractive condition, as x n y n , we have

ψ ( d ( y n + 1 , x n + 1 ) ) = ψ ( d ( A ( y n , x n ) , A ( x n , y n ) ) ) φ ( ψ ( d ( x n , y n ) ) ψ ( d ( x n , y n ) ) ψ ( d ( x n , y n ) ) .
(16)

Thus limn→∞ψ(d(x n , y n )) = θ for certain θ ≥ 0. Since φ Φ, we have lim  sup r θ + φ ( r ) <1 and φ(θ) < 1. Then, there exist α [0,1) and ε > 0 such that φ(r) <α for all r [θ,θ + ε).

Now, we take n0 ≥ 0 such that θψ(d(x n , y n )) ≤ θ + ε for all n ≥ n0. Then, from (16), for all nn0, we have

ψ ( d ( y n + 1 , x n + 1 ) ) α ψ ( d ( x n , y n ) ) .

Letting n → ∞ in the above inequality, we obtain that

θ α θ .

Since α [0,1), this implies that θ = 0. Thus, we proved that

lim n ψ ( d ( x n , y n ) ) = 0 ,

which implies that limn→ ∞d(x n , y n ) = 0. Now, we have

0 = lim n d ( x n , y n ) = d ( lim n x n , lim n y n ) = d ( a , b )

and thus a = b. This finishes the proof. □

Now, we present some consequences of our theorems.

Corollary 2.1 Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a map having the mixed monotone property on X. Suppose that there exist ψ Ψ and φ ̃ : [ 0 , ) [ 0 , ) with lim inf s t + ( φ ̃ ( s ) / ψ ( s ) ) >0for all t ≥ 0 such that for any (x, y),(u, v) X × X with(u, v) (x, y),

ψ ( d ( A ( x , y ) , A ( u , v ) ) ) ψ ( max { d ( x , u ) , d ( y , v ) } ) - φ ̃ ( ψ ( max { d ( x , u ) , d ( y , v ) } ) ) .

Suppose also that either A is continuous or (X, d, ) has the following properties:

(i) any -nondecreasing sequence (x n ) with x n x implies x n x for each n,

(ii) any -nonincreasing sequence (y n ) with y n y implies y n y for each n.

If there exist x0, y0 X such that x 0 A (x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then there exist a, b X such that a = A(a, b) and b = A(b, a).

Proof. It follows immediately from Theorem 2.1 by considering φ ( s ) =1- φ ̃ ( s ) /ψ ( s ) . □

Remark 2.2 Corollary 2.1 is an extension of Harjani et al. [[15], Theorems 2 and 3].

Corollary 2.2 Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a map having the mixed monotone property on X. Suppose that there exists a nondecreasing function φ : [0, ∞) → [0, 1) such that for any (x, y), (u, v) X × X with (u, v) (x, y),

d ( A ( x , y ) , A ( u , v ) ) φ ( 2 max { d ( x , u ) , d ( y , v ) } ) max { d ( x , u ) , d ( y , v ) } .

Suppose also that either A is continuous or (X, d, ) has the following properties:

(i) any -nondecreasing sequence (x n ) with x n x implies xn x for each n,

(ii) any -nonincreasing sequence (y n ) with y n y implies yn y for each n.

If there exist x0, y0 X such that x 0 A(x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then there exist a, b X such that a = A(a, b) and b = A(b, a).

Proof. It follows from Theorem 2.1 by considering ψ(s) = 2s

Remark 2.3 If φ is nondecreasing, Corollary 2.2 generalizes Du [[13], Theorems 1.5 and 1.6].

Corollary 2.3 Let (X, d, ) be a sequentially ↕-complete metric space and A : X × XX be a map having the mixed monotone property on X. Suppose that there exists k [0, 1) such that for any (x, y),(u, v) X × X with (u, v) (x, y),

d ( A ( x , y ) , A ( u , v ) ) k max { d ( x , u ) , d ( y , v ) } .

Suppose also that either A is continuous or (X, d, ) has the following properties:

(i) any -nondecreasing sequence (x n ) with x n x implies x n x for each n,

(ii) any -nonincreasing sequence (y n ) with y n y implies y n y for each n.

If there exist x0, y0 X such that x 0 A(x 0 , y 0 ) and y 0 A(y 0 , x 0 ), then there exist a, b X such that a = A(a, b) and b = A(b, a).

Proof. It follows immediately from Corollary 2.2 by considering φ(s) = k

Remark 2.4 Corollary 2.3 is a generalization of Bhaskar and Lakshmikantham [[11], Theorems 2.1 and 2.2].

3 An application

In this section, we apply our main results to study the existence and uniqueness of solution to the two-point boundary value problem

- d 2 x d t 2 ( t ) = f ( t , x ( t ) , x ( t ) ) , t [ 0 , 1 ] x ( 0 ) = x ( 1 ) = 0 ,
(17)

where f : [0, 1] × × is a continuous function.

Previously we considered the space X = C(I, )(I = [0, 1]) of continuous functions defined on I. Obviously, this space with the metric given by

d ( x , y ) = max { | x ( t ) - y ( t ) | : t I } for x , y I

is a complete metric space. The space X can also be equipped with a partial order given by

x , y I , x y x ( t ) y ( t ) for all t I .

Obviously, (X, ) satisfies condition (H) since for x, y X the functions max{x, y} and min{x, y} are least upper and greatest lower bounds of x and y, respectively. Moreover, in [21] it is proved that (X, d, ) satisfies conditions (i) and (ii) of Theorem 2.1.

Now, we consider the following assumptions:

  1. (a)

    f : [0, 1] × × is continuous.

  2. (b)

    For all t I, z ≥ h, w ≤ r,

    0 f ( t , z , w ) - f ( t , h , r ) 4 [ ln ( z - h + 1 ) + ln ( r - w + 1 ) ] .
  3. (c)

    There exists (α, β) C 2 (I, ) × C 2 (I, ) solution to

    - d 2 α d t 2 ( t ) f ( t , α ( t ) , β ( t ) ) , t [ 0 , 1 ] - d 2 β d t 2 ( t ) f ( t , β ( t ) , α ( t ) ) , t [ 0 , 1 ] α ( 0 ) = α ( 1 ) = β ( 0 ) = β ( 1 ) = 0 .
    (18)
  4. (d)

    α β or β α.

Theorem 3.1 Under the assumptions (a)-(d), problem (17) has one and only one solution x* C2(I, ).

Proof. It is well known that the solution (in C2 (I, )) of problem (17) is equivalent to the solution (in C(I, )) of the following Hammerstein integral equation:

x ( t ) = 0 1 G ( t , s ) f ( x ( s ) ) d s , t I ,

where G(t, s) is the Green function of differential operator -d2/dt2 with Dirichlet boundary condition x(0) = x(1) = 0, i.e.,

G ( t , s ) = s ( 1 - t ) , 0 s t 1 , t ( 1 - s ) , 0 t s 1 .

Define A : X × XX by

A ( x , y ) ( t ) = 0 1 G ( t , s ) f ( s , x ( s ) , y ( s ) ) d s , t I ,

for all x, y X.

From (b), it is clear that A has the mixed monotone property with respect to the partial order in X.

Let x, y, u, v X such that x u and y υ. From (b), we have

d ( A ( x , y ) , A ( u , v ) ) = sup t I | A ( x , y ) ( t ) - A ( u , v ) ( t ) | = sup t I 0 1 G ( t , s ) f ( s , x ( s ) , y ( s ) ) - f ( s , u ( s ) , v ( s ) ) d s sup t I 0 1 4 G ( t , s ) [ ln ( x ( s ) - u ( s ) + 1 ) + ln ( v ( s ) - y ( s ) + 1 ) ] d s ( ln ( d ( x , u ) + 1 ) + ln ( d ( y , v ) + 1 ) ) sup t I 0 1 4 G ( t , s ) d s sup t I 0 1 8 G ( t , s ) d s ln ( max { d ( x , u ) , d ( y , v ) } + 1 )

On the other hand, for all t I, we have

0 1 G ( t , s ) d s = 1 2 t ( 1 - t ) ,

which implies that

sup t I 0 1 G ( t , s ) d s = 1 8 .

Then, we get

d ( A ( x , y ) , A ( u , v ) ) ln ( max { d ( x , u ) , d ( y , v ) } + 1 ) .

This implies that

ln d ( A ( x , y ) , A ( u , v ) ) + 1 ln ln max { d ( x , u ) , d ( y , v ) } + 1 + 1 = ln ln max { d ( x , u ) , d ( y , v ) } + 1 + 1 ln max { d ( x , u ) , d ( y , v ) } + 1 ln max { d ( x , u ) , d ( y , v ) } + 1 .

Thus, the contractive condition (1) of Theorem (2.1) is satisfied with ψ(t) = ln(t + 1) and φ(t) = ψ(t)/t.

Now, let (α, β) C2 (I, ) × C2 (I, ) be a solution to (18). We will show that α A(α, β) and β A(β, α). Indeed,

- α ( s ) f ( s , α ( s ) , β ( s ) ) , s [ 0 , 1 ] .

Multiplying by G(t, s), we get

0 1 - α ( s ) G ( t , s ) d s A ( α , β ) ( t ) , t [ 0 , 1 ] .

Then, for all t [0, 1], we have

- ( 1 - t ) 0 1 s α ( s ) d s - t t 1 ( 1 - s ) α ( s ) d s A ( α , β ) ( t ) .

Using an integration by parts, and since α(0) = α(1) = 0, for all t [0, 1], we get

- ( 1 - t ) ( t α ( t ) - α ( t ) ) - t ( - ( 1 - t ) α ( t ) - α ( t ) ) A ( α , β ) ( t ) .

Thus, we have

α ( t ) A ( α , β ) ( t ) , t [ 0 , 1 ] .

This implies that α A(α, β). Similarly, one can show that β A(β, α).

Now, applying our Theorems 2.1 and 2.2, we deduce the existence of a unique (x, y) X2 solution to x = A(x, y) and y = A(y, x). Moreover, from (d), and using Theorem 2.3, we get x = y. Thus, we proved that x* = x = y C2([0, 1], ) is the unique solution to (17). □

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Acknowledgements

L. Ćirić and B. Damjanović were supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.

M. Jleli extends his appreciation to the Deanship of Scientific Research at King Saud University for funding the work through the research group project No RGP-VPP-087.

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Ćirić, L., Damjanović, B., Jleli, M. et al. Coupled fixed point theorems for generalized Mizoguchi-Takahashi contractions with applications. Fixed Point Theory Appl 2012, 51 (2012). https://doi.org/10.1186/1687-1812-2012-51

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Keywords

  • coupled fixed point
  • ordered metric space
  • mixed monotone mapping
  • generalized Mizoguchi-Takahashi's contraction
  • two-point boundary value problem