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 : X → P(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 : X → CB(X) be a multivalued map. Assume that there exists r ∈ [0,1) such that

$$H\left(Tx,Ty\right)\le rd\left(x,y\right)$$

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 : X → C(X) be a multi-valued map with non empty compact values. Assume that

$$H\left(Tx,Ty\right)\le \phi \left(d\left(x,y\right)\right)d\left(x,y\right)$$

for all x, y ∈ X, where φ is a function from [0, ∞) into [0,1) satisfying$\mathrm{lim }{\sup }_{s\to t^{+}}\phi \left(s\right)<1$for 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 : X → CB(X) be a multivalued map. Assume that

$$H\left(Tx,Ty\right)\le \phi \left(d\left(x,y\right)\right)d\left(x,y\right)$$

for all x, y ∈ X, where φ is a function from [0, ∞) into [0,1) satisfying$\mathrm{lim }{\sup }_{s\to t^{+}}\phi \left(s\right)<1$for 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:

• (a) ψ(s) = 0 ⇔ s = 0,

• (b) ψ is nondecreasing,

• (c) $\mathrm{lim }{\sup }_{s\to 0^{+}}\frac{s}{\psi \left(s\right)}<\infty$.

We denote by Φ the set of all functions φ : [0, ∞) → [0,1) satisfying $\lim {\sup }_{r\to t^{+}}\phi \left(r\right)<1$ for all t ≥ 0.

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

$$\psi \mathsf{\text{(}}H\left(Tx,Ty\right)\mathsf{\text{)}}\le \phi \mathsf{\text{(}}\psi \mathsf{\text{(}}d\left(x,y\right)\mathsf{\text{))}}\psi \mathsf{\text{(}}d\left(x,y\right)\mathsf{\text{)}}$$

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 [7–27] 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 × X → X 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 : XX → X 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,

$$\begin{array}{c}{x}_1,x_2\in Xwith{x}_1\preccurlyeq x_2\Rightarrow A\left(x_1,y\right)\preccurlyeq A\left(x_2,y\right),\\ y_1,y_2\in Xwith{y}_1\preccurlyeq y_2\Rightarrow A\left(x,y_1\right)\succcurlyeq A\left(x,y_2\right).\end{array}$$

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 × X → X be a continuous map having the mixed monotone property on X. Assume that there exists a function φ ∈ Φ such that

$$d\left(A\left(x,y\right),A\left(u,v\right)\right)\le \frac{1}{2}\phi \left(d\left(x,u\right)+d\left(y,v\right)\right)\left(d\left(x,u\right)+d\left(y,v\right)\right)$$

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 × X → X 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

$$d\left(A\left(x,y\right),A\left(u,v\right)\right)\le \frac{1}{2}\phi \left(d\left(x,u\right)+d\left(y,v\right)\right)\left(d\left(x,u\right)+d\left(y,v\right)\right)$$

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 : X → X 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 : X → X 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

$$\psi \left(d\left(Tx,Ty\right)\right)\le \phi (\psi \left(d\left(x,y\right)\right)\psi \left(d\left(x,y\right)\right)$$

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

$$\left(x,y\right),\left(u,v\right)\in X\times X,\left(x,y\right)\preccurlyeq \left(u,v\right)\iff x\preccurlyeq u,y\succcurlyeq v.$$

Our first result is the following.

Theorem 2.1 Let (X, d, ≼) be a sequentially ≼↕-complete metric space and A : X × X → X 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),

$$\psi \left(d\left(A\left(x,y\right),A\left(u,v\right)\right)\right)\le \phi \left(\psi \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}\right)\right)\psi \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}\right).$$

(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

$$x_{n+1}=A\left(x_n,y_n\right),y_{n+1}=A\left(y_n,x_n\right)\mathsf{\text{for}}\mathsf{\text{all}}n\ge 0.$$

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

• Step 1. x _n ≼ x_n+1and y _n ≽ y_n+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 ≼ x_n+1and y _n ≽ y_n+1, and as A has the mixed monotone property, we get

$$x_{n+1}=A\left(x_n,y_n\right)\preccurlyeq A\left(x_{n+1},y_n\right)\preccurlyeq A\left(x_{n+1},y_{n+1}\right)=x_{n+2}$$

and

$$y_{n+1}=A\left(y_n,x_n\right)\succcurlyeq A\left(y_{n+1},x_n\right)\succcurlyeq A\left(y_{n+1},x_{n+1}\right)=y_{n+2}.$$

This proves our claim.

• Step 2. lim_n→∞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

$$\begin{array}{c}\psi \left(d\left(A\left(x_{n+1},y_{n+1}\right),A\left(x_n,y_n\right)\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\right)\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\\ \le \psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right).\end{array}$$

(2)

Similarly, since (y _{n+ 1},x _{n+ 1}) ≼ (y _n , x _n ), by (1), we have

$$\begin{array}{c}\psi \left(d\left(A\left(y_n,x_n\right),A\left(y_{n+1},x_{n+1}\right)\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right)\right)\psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right)\\ \le \psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right).\end{array}$$

(3)

From (2) and (3), we get

$$\max \{\psi \left(d\left(x_{n+2},x_{n+1}\right),\psi \left(d\left(y_{n+2},y_{n+1}\right)\right)\right\}\le \psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right).$$

Since ψ is nondecreasing, this implies that

$$\psi \left(\max \left\{d\left(x_{n+2},x_{n+1}\right)d\left(y_{n+2},y_{n+1}\right)\right\}\right)\le \psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)$$

(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

$$\underset{n\to \infty }{\lim }\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)=\mu .$$

(5)

Since φ ∈ Φ, we have $\mathrm{lim }{\sup }_{r\to {\mu }^{+}}\phi \left(r\right)<1$ 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 μ ≤ ψ(max{d(x _{n+ 1},x _n ),d(y _{n+ 1},y _n )}) ≤ μ + ε for all n ≥ n₀. Then, from (2), for all n ≥ n₀, we have

$$\begin{array}{c}\psi \left(d\left(x_{n+1},x_{n+2}\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\right)\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\\ \le \alpha \psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right).\end{array}$$

(6)

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

$$\begin{array}{c}\psi \left(d\left(y_{n+1},y_{n+2}\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right)\right)\psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right)\\ \le \alpha \psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right).\end{array}$$

(7)

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

$$\psi \left(\max \left\{d\left(x_{n+1},x_{n+2}\right)d\left(y_{n+1},y_{n+2}\right)\right\}\right)\le \alpha \psi \left(\max \left\{d\left(y_n,y_{n+1}\right),d\left(x_n,x_{n+1}\right)\right\}\right)$$

(8)

for all n ≥ n₀. Letting n → ∞ in the above inequality and using (5), we obtain that

$$\mu \le \alpha \mu .$$

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

$$\underset{n\to \infty }{\lim }\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)=0.$$

(9)

• Step 3. lim_n→∞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

$$\underset{n\to \infty }{\lim }\max \{d(x_{n+1},x_n),d(y_{n+1},y_n)\})={\theta }^{+}.$$

Since ψ is nondecreasing, we have

$$\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\ge \psi \left(\theta \right).$$

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

$$\underset{n\to \infty }{\lim }\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}=0.$$

(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

$$a_n=\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)\mathsf{\text{for}}\mathsf{\text{all}}n\ge 0.$$

From (8), we have

$$a_{n+1}\le \alpha a_n\mathsf{\text{for}}\mathsf{\text{all}}n\ge n_0.$$

Then, we have

$$\sum _{n=0}^{\infty }{a}_n\le \sum _{n=0}^{n_0}{a}_n+\sum _{n=n_0+1}^{\infty }{\alpha }^{n-n_0}{a}_{n0}<\infty .$$

On the other hand, we have

$$\underset{n\to \infty }{\mathrm{lim }\sup }\frac{\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}}{\psi \left(\max \left\{d\left(x_{n+1},x_n\right),d\left(y_{n+1},y_n\right)\right\}\right)}\le \underset{s\to 0^{+}}{\mathrm{lim }\sup }\frac{s}{\psi \left(s\right)}<\infty .$$

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

$$\underset{n\to \infty }{\lim }{x}_n=a.$$

(11)

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

$$\underset{n\to \infty }{\lim }{y}_n=b.$$

(12)

Case 1. A is continuous.

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

$$a=\underset{n\to \infty }{\lim }{x}_{n+1}=\underset{n\to \infty }{\lim }A\left(x_n,y_n\right)=A\left(\underset{n\to \infty }{\lim }{x}_n,\underset{n\to \infty }{\lim }{y}_n\right)=A\left(a,b\right)$$

and

$$b=\underset{n\to \infty }{\lim }{y}_{n+1}=\underset{n\to \infty }{\lim }A\left(y_n,x_n\right)=A\left(\underset{n\to \infty }{\lim }{y}_n,\underset{n\to \infty }{\lim }{x}_n\right)=A\left(b,a\right).$$

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

Since (x _n ) is ≼-nondecreasing and lim_n→∞x _n = a, then from (i), we have x _n ≼ a for all n. Similarly, from (ii), since (y _n ) is ≼-nonincreasing and lim_n→∞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

$$\begin{array}{c}\psi \left(d\left(x_{n+1},A\left(a,b\right)\right)\right)=\psi \left(d\left(A\left(x_n,y_n\right),A\left(a,b\right)\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(x_n,a\right),d\left(y_n,b\right)\right\}\right)\right)\psi \left(\max \left\{d\left(x_n,a\right),d\left(y_n,b\right)\right\}\right)\\ \le \psi \left(\max \left\{d\left(x_n,a\right),d\left(y_n,b\right)\right\}\right).\end{array}$$

Since ψ is nondecreasing, this implies that

$$d\left(x_{n+1},A\left(a,b\right)\right)\le \max \left\{d\left(x_n,a\right),d\left(y_n,b\right)\right\}.$$

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

$$f_{n+1}=A\left(f_n,g_n\right),g_{n+1}=A\left(g_n,f_n\right)\mathsf{\text{for}}\mathsf{\text{all}}n\ge 0.$$

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

$$f_{n+1}=A\left(f_n,g_n\right)\succcurlyeq A\left(a,g_n\right)\succcurlyeq A\left(a,b\right)=a$$

and

$$g_{n+1}=A\left(g_n,f_n\right)\preccurlyeq A\left(b,f_n\right)\preccurlyeq A\left(b,a\right)=b.$$

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

$$\begin{array}{c}\psi \left(d\left(f_{n+1},a\right)\right)=\psi \left(d\left(A\left(f_n,g_n\right),A\left(a,b\right)\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}\right)\right)\psi \left(\max \left\{d\left(f_n,a\right)d\left(g_n,b\right)\right\}\right).\end{array}$$

(13)

Similarly, we have

$$\begin{array}{c}\psi \left(d\left(g_{n+1},b\right)\right)=\psi \left(d\left(A\left(g_n,f_n\right),A\left(b,a\right)\right)\right)\\ \le \phi \left(\psi \left(\max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}\right)\right)\psi \left(\max \left\{d\left(f_n,a\right)d\left(g_n,b\right)\right\}\right).\end{array}$$

(14)

Combining (13) with (14), we obtain

$$\begin{array}{c}\psi \left(\max \left\{d\left(f_{n+1},a\right),d\left(g_{n+1},b\right)\right\}\right)\le \phi \left(\psi \left(\max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}\right)\right)\psi \left(\max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}\right)\\ \le \psi \left(\max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}\right).\end{array}$$

(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 lim_n→∞max{d(f _n , a),d(g _n , b)} = 0.

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

Now, we have

$$d\left(a,c\right)\le d\left(a,f_n\right)+d\left(f_n,c\right)\le \max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}+\max \left\{d\left(f_n,c\right),d\left(g_n,e\right)\right\}$$

and

$$d\left(b,e\right)\le d\left(b,g_n\right)+d\left(g_n,e\right)\le \max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}+\max \left\{d\left(f_n,c\right),d\left(g_n,e\right)\right\}.$$

Then, we have

$$\max \left\{d\left(a,c\right),d\left(b,e\right)\right\}\le \max \left\{d\left(f_n,a\right),d\left(g_n,b\right)\right\}+\max \left\{d\left(f_n,c\right),d\left(g_n,e\right)\right\}.$$

Letting n → ∞ in the above inequality, we get

$$\max \left\{d\left(a,c\right),d\left(b,e\right)\right\}=0,$$

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

$$\begin{array}{c}\psi \left(d\left(y_{n+1},x_{n+1}\right)\right)=\psi \left(d\left(A\left(y_n,x_n\right),A\left(x_n,y_n\right)\right)\right)\\ \le \phi (\psi \left(d\left(x_n,y_n\right)\right)\psi \left(d\left(x_n,y_n\right)\right)\\ \le \psi \left(d\left(x_n,y_n\right)\right).\end{array}$$

(16)

Thus lim_n→∞ψ(d(x _n , y _n )) = θ for certain θ ≥ 0. Since φ ∈ Φ, we have $\mathrm{lim }{\sup }_{r\to {\theta }^{+}}\phi \left(r\right)<1$ 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 θ ≤ ψ(d(x _n , y _n )) ≤ θ + ε for all n ≥ n₀. Then, from (16), for all n ≥ n₀, we have

$$\psi \left(d\left(y_{n+1},x_{n+1}\right)\right)\le \alpha \psi \left(d\left(x_n,y_n\right)\right).$$

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

$$\theta \le \alpha \theta .$$

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

$$\underset{n\to \infty }{\lim }\psi \left(d\left(x_n,y_n\right)\right)=0,$$

which implies that lim_{n→ ∞}d(x _n , y _n ) = 0. Now, we have

$$0=\underset{n\to \infty }{\lim }d\left(x_n,y_n\right)=d\left(\underset{n\to \infty }{\lim }{x}_n,\underset{n\to \infty }{\lim }{y}_n\right)=d\left(a,b\right)$$

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 × X → X be a map having the mixed monotone property on X. Suppose that there exist ψ ∈ Ψ and$\tilde{\phi }:\left[0,\infty \right)\to \left[0,\infty \right)$with$\lim {\inf }_{s\to t^{+}}\left(\tilde{\phi }\left(s\right)/\psi \left(s\right)\right)>0$for all t ≥ 0 such that for any (x, y),(u, v) ∈ X × X with(u, v)≼ (x, y),

$$\psi \left(d\left(A\left(x,y\right),A\left(u,v\right)\right)\right)\le \psi \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}\right)-\tilde{\phi }\left(\psi \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}\right)\right).$$

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 $\phi \left(s\right)=1-\tilde{\phi }\left(s\right)/\psi \left(s\right)$. □

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 × X → X 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),

$$d\left(A\left(x,y\right),A\left(u,v\right)\right)\le \phi \left(2\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}\right)\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}.$$

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 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 × X → X 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),

$$d\left(A\left(x,y\right),A\left(u,v\right)\right)\le k\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}.$$

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

$$\left\{\begin{array}{c}-\frac{d^{2}x}{d{t}^2}\left(t\right)=f\left(t,x\left(t\right),x\left(t\right)\right),t\in \left[0,1\right]\\ x\left(0\right)=x\left(1\right)=0,\end{array}\right.$$

(17)

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

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

$$d\left(x,y\right)=\max \left\{|x\left(t\right)-y\left(t\right)|:t\in I\right\}\mathsf{\text{for}}x,y\in I$$

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

$$x,y\in I,x\preccurlyeq y\iff x\left(t\right)\le y\left(t\right)\mathsf{\text{for}}\mathsf{\text{all}}t\in I.$$

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:

• (a) f : [0, 1] × ℝ × ℝ → ℝ is continuous.

• (b) For all t ∈ I, z ≥ h, w ≤ r,

  $$0\le f\left(t,z,w\right)-f\left(t,h,r\right)\le 4\left[\ln \left(z-h+1\right)+\ln \left(r-w+1\right)\right].$$

• (c) There exists (α, β) ∈ C ² (I, ℝ) × C ² (I, ℝ) solution to

  $$\left\{\begin{array}{c}-\frac{d^2\alpha }{d{t}^2}\left(t\right)\le f\left(t,\alpha \left(t\right),\beta \left(t\right)\right),t\in \left[0,1\right]\\ -\frac{d^2\beta }{d{t}^2}\left(t\right)\ge f\left(t,\beta \left(t\right),\alpha \left(t\right)\right),t\in \left[0,1\right]\\ \alpha \left(0\right)=\alpha \left(1\right)=\beta \left(0\right)=\beta \left(1\right)=0.\end{array}\right.$$

  (18)

• (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, ℝ).

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

$$x\left(t\right)=\underset{0}{\overset{1}{\int }}G\left(t,s\right)f\left(x\left(s\right)\right)ds,t\in I,$$

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

$$G\left(t,s\right)=\left\{\begin{array}{c}s\left(1-t\right),0\le s\le t\le 1,\\ t\left(1-s\right),0\le t\le s\le 1.\end{array}\right.$$

Define A : X × X → X by

$$A\left(x,y\right)\left(t\right)=\underset{0}{\overset{1}{\int }}G\left(t,s\right)f\left(s,x\left(s\right),y\left(s\right)\right)ds,t\in I,$$

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

$$\begin{array}{c}d\left(A\left(x,y\right),A\left(u,v\right)\right)=\underset{t\in I}{\sup }|A\left(x,y\right)\left(t\right)-A\left(u,v\right)\left(t\right)|\\ =\underset{t\in I}{\sup }\underset{0}{\overset{1}{\int }}G\left(t,s\right)\left[f\left(s,x\left(s\right),y\left(s\right)\right)-f\left(s,u\left(s\right),v\left(s\right)\right)\right]ds\\ \le \underset{t\in I}{\sup }\underset{0}{\overset{1}{\int }}4G\left(t,s\right)\left[\ln \left(x\left(s\right)-u\left(s\right)+1\right)+\ln \left(v\left(s\right)-y\left(s\right)+1\right)\right]ds\\ \le \left(\ln \left(d\left(x,u\right)+1\right)+\ln \left(d\left(y,v\right)+1\right)\right)\underset{t\in I}{\sup }\underset{0}{\overset{1}{\int }}4G\left(t,s\right)ds\\ \le \left(\underset{t\in I}{\sup }\underset{0}{\overset{1}{\int }}8G\left(t,s\right)ds\right)\ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right)\end{array}$$

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

$$\underset{0}{\overset{1}{\int }}G\left(t,s\right)ds=\frac{1}{2}t\left(1-t\right),$$

which implies that

$$\underset{t\in I}{\sup }\underset{0}{\overset{1}{\int }}G\left(t,s\right)ds=\frac{1}{8}.$$

Then, we get

$$d\left(A\left(x,y\right),A\left(u,v\right)\right)\le \ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right).$$

This implies that

$$\begin{array}{c}\ln \left(d\left(A\left(x,y\right),A\left(u,v\right)\right)+1\right)\le \ln \left(\ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right)+1\right)\\ =\frac{\ln \left(\ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right)+1\right)}{\ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right)}\ln \left(\max \left\{d\left(x,u\right),d\left(y,v\right)\right\}+1\right).\end{array}$$

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

$$-{\alpha }^{″}\left(s\right)\le f\left(s,\alpha \left(s\right),\beta \left(s\right)\right),s\in \left[0,1\right].$$

Multiplying by G(t, s), we get

$$\underset{0}{\overset{1}{\int }}-{\alpha }^{″}\left(s\right)G\left(t,s\right)ds\le A\left(\alpha ,\beta \right)\left(t\right),t\in \left[0,1\right].$$

Then, for all t ∈ [0, 1], we have

$$-\left(1-t\right)\underset{0}{\overset{1}{\int }}s{\alpha }^{″}\left(s\right)ds-t\underset{t}{\overset{1}{\int }}\left(1-s\right){\alpha }^{″}\left(s\right)ds\le A\left(\alpha ,\beta \right)\left(t\right).$$

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

$$-\left(1-t\right)\left(t{\alpha }^{\prime }\left(t\right)-\alpha \left(t\right)\right)-t\left(-\left(1-t\right){\alpha }^{\prime }\left(t\right)-\alpha \left(t\right)\right)\le A\left(\alpha ,\beta \right)\left(t\right).$$

Thus, we have

$$\alpha \left(t\right)\le A\left(\alpha ,\beta \right)\left(t\right),t\in \left[0,1\right].$$

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], ℝ) is the unique solution to (17). □