Coupled fixed point theorems in complete metric spaces endowed with a directed graph and application

Definition 1.1

([15]). Let (X,≼) be a partially ordered set and F : X²⟶X be a mapping. Then a map F is said to have the mixed monotone property if F (x, y) is monotone nondecreasing in x and is monotone non-increasing in y; that is, for any x, y ∈ X,

and

Definition 1.2

([15]). An element (x, y) ∈ X² is said to be a coupled fixed point of the mapping F : X²⟶X if F (x, y) = x and F (y, x) = y.

Lakshmikantham and Ćirić [16] extended the concept of mixed monotone property to mixed g-monotone property as follows.

Definition 1.3

([16]). Let (X,≼) be a partially ordered set and F : X²⟶X and g : X ⟶ X.We say F is called the mixed g-monotone property if for any x, y ∈ X,

and

Definition 1.4

([16]). An element (x, y) ∈ X² is said to be a coupled coincidence point of the mappings F : X²⟶X and g : X⟶X if F (x, y)=gx and F (y, x) = gy.

Definition 1.5

([16]). An element (x, y) ∈ X² is said to be a coupled common fixed point of the mappings F : X²⟶X and g : X⟶X if F (x, y) = gx = x and F (y, x) = gy = y.

Definition 1.6

([16]). Let X be a nonempty set and F : X²⟶X and g : X⟶X. We say F and g are commutative if gF (x, y) = F (gx, gy)for all x,y ∈ X.

In 2010, Choudhury and Kundu [17] introduced the notion of compatibility in the context of coupled coincidence point problems as follows.

Definition 1.7

([17]). The mappings F : X²⟶X and g : X⟶X are said to be compatible if

whenever {x_n} and {y_n} are sequences in X such that lim_{n ⟶ ∞} F (x_n, y_n)= lim_{n ⟶ ∞} gx_n = x and lim_{n ⟶ ∞}F (y_n,x_n) = lim_{n ⟶ ∞} gy_n = y with x,y ∈ X.

Definition 1.8

([18]). Let Φ denote the class of functions ϕ : [0,∞) ⟶ [0, ∞) which satisfies the following conditions:

• (ϕ₁) ϕ is lower semi-continuous and (strictly) increasing;

• (ϕ₂)ϕ(t) < t for all t > 0;

• (ϕ₃) ϕ (t + s) ≤ ϕ(t) + ϕ (s) for all t,s ∈ [0,∞).

• Note that lim_{n ⟶ ∞} ϕ(t_n) = 0 ⇔ lim_{n ⟶ ∞} t_n = 0for t_n ∈ [0,∞).

• Also, for ϕ ∈ Φ, Ψ_ϕdenote all functions ψ : [0,∞) ⟶ [0,∞)which satisfy the following conditions:

• (Ψ₁)lim sup_{n ⟶ ∞} Ψ (t_n) < ϕ (r) if lim_{n ⟶ ∞}t_n = r > 0;

• (Ψ₂) lim_{n ⟶ ∞} Ψ (t_n) = 0 if lim_{n ⟶ ∞} t_n = 0 for t_n ∈ [0,∞).

Now, we have the following coupled fixed point theorems as the main result of Jain et al. in [18].

Theorem 1.9

([18]). Let (X, ≤ ) be a partially ordered set and there is a metric d on X such that (X, d ) is a complete metric space. Suppose that F : X² ⟶ X is a mapping having the mixed monotone property on X. Assume there exists ϕ ∈ Φ and Ψ∈ Ψ_ϕ such that

for all x, y, u, v ∈ X with x ≥u and y ≤ v.

Suppose that either

1. F is continuous or;

2. X has the following properties:

1. if a non-decreasing sequence {x_n} → x, then x_n≤ x for all n,

2. if a non-increasing sequence {y_n} → y, then y ≤ y_n for all n.

The fixed point theorem using the context of metric spaces endowed with a graph was initiated by Jachymski [19]. Other results for single valued and multivalued operators in such metric spaces were given by Beg et al. [20], Bajor [21], Alfuraid [22, 23], Chifu and Petrusel [24] and Suantai et al. [25].

Let (X, d) be a metric space, Δ be a diagonal of X², and G be a directed graph with no parallel edges such that the set V (G) of its vertices coincides with X and Δ ⊆ E (G), where E (G) is the set of the edges of the graph. That is, G is determined by (V (G), E (G)). We will use this notation of G throughout this work.

In 2014, Chifu and Petrusel [24] introduced the notion of G—continuity for a mapping F : X² ⟶ X and the property A as follows.

Definition 1.10

([24]). Let (X, d) be a complete metric space, G be a directed graph, and F : X² ⟶ X be a mapping. Then

1. F is called G—continuous if for all (x_*,y_*) ∈ X² and for any sequence {n_i}_i ∈ ℕof positive integers such that F {x_ni, y_ni) ⟶ x_*, F (y_ni,x_ni) ⟶ y_* as i ⟶ ∞ and (F(x_ni, y_ni), F (x_{ni +1}, y_{ni +1})), (F(y_ni,x_ni), F (y_{ni +1}, x_{ni +1})) ∈ E(G), we have that

   F(F(xni,yni),F(yni,xni))→F(x∗,y∗)asi→∞

   and

   F(F(yni,xni),F(xni,yni))→F(x∗,y∗)asi→∞;

2. (X, d, G) has property A if for any sequence {x_n}_{n∊ ℕ} ⊆ X with x_n ⟶ x as n ⟶ ∞ and (x_n, x_n+1)∈ E (G) for n ∈ ℕ, then (x_n, x) ∈ E (G).

Consider the set CcFix (F) of all coupled coincidence points of mappings F : X² ⟶ X, g : X ⟶ X and the set (X2)gF as follows:

and

In 2015, Suantai et al. [25] introduced the concept of G–edge preserving and the transitivity property as follows.

Definition 1.11

([25]). F : X² ⟶ X, g : X ⟶ X are G–edge preserving if

Definition 1.12

([25]). Let (X,d) be a complete metric space, and E (G) be the set of the edges of the graph. E (G) satisfies the transitivity property if and only if, for all x, y, t ∈ X,

The purpose of this paper is to present some existence results for coupled fixed points of (ϕ, Ψ) –contractive mappings in metric spaces endowed with a directed graph. Our results generalize the results obtained by Jain et al. in (International Journal of Analysis, Volume 2014, Article ID 586096, 9 pages). Moreover, we have an application to some integral system to support the results.

Definition 2.1

Let (X, d) be a complete metric space endowed with a directed graph G. The mappings F : X² ⟶ X, g : X ⟶ X are called a (ϕ,Ψ) –contractive if:

1. F and g is G–edge preserving;

2. there exists ϕ ∈ Φ and Ψ ∈ Ψ_ϕ such that for all x, y, u, v ∈ X satisfying (gx, gu), (gy,gv) ∈ E (G),

   φ({d(F(x,y),F(u,v))+d(F(y,x),F(v,u))}×2−1≤ψ({d(gx,gu)+d(gy,gv)}×2−1). (2)

Theorem 2.2

Let (X, d) be a complete metric space endowed with a directed graph G, and let F : X² ⟶ X, g : X ⟶ X be a (ϕ,Ψ) –contractive mapping. Suppose that:

1. g is continuous and g (X) is closed;

2. F (X²) ⊆ g (X), and (F, g) is compatible;

3. 1. F is G–continuous, or

   2. the tripled (X, d, G) has a property A;

4. E (G) satisfies the transitivity property.

   Under these conditions, CcFix (F) ≠ ∅ iff(x2)gF≠∅.

Proof

Consider x₀, y₀, t₀, s₀ ∈ X followed by assumptions. Since F (X²) ⊆ g (X), then there exists x₁, y₁, t₁, s₁ ∈ X such that F (x₀,y₀) = gx₁ and F (y₀, x₀) = gy₁, F (t₀, s₀) = gt₁ and F (s₀, t₀) = gs₁ continuing the procedure above we have the sequence {x_n}, {y_n}, {t_n}, {s_n} in X for which

CcFix (F)≠ ∅. Let (u,v) ∈ CcFix (F) such that

Hence, (gu, F (u,v)), (gv, F (v,u)) ∈ E (G). Then we have (u,v)∈X2gF, thereby X2gF≠∅.

Next, assume that X2gF≠∅. Let x₀, y₀ ∈ X such that (x0,y0)∈X2Fg. In that case, (gx₀, F (x₀,y₀)), (gy₀,F (y₀,x₀)) ∈ E (G). By (3), we obtain

Since F and g are G–edge preserving and by (4), we get

Continuing this process, we can construct (gx_n, gx_n+1) and (gy_n, gy_n+1) ∈ E (G) for each n ∈ ℕ.

• Denote κ_n := (d (gx_n,gx_n+1) + d (gy_n,gy_n+1)) × 2^-1 for all n ∈ ℕ .

• Using the (ϕ,Ψ) –contractive type operator (2) and (3), we get that

for all n ∈ ℕ , then we obtain

From (6) and monotonicity of ϕ, we get that {κ_n} is a nonnegative decreasing. Then κ_n ⟶κ as n ⟶ ∞ for some κ ≥ 0. If possible, let κ > 0. Taking the limit as n ⟶ ∞ in (6) and using the properties of ϕ, Ψ, we obtain

which is a contradiction. Thus κ = 0 and then we have

Now we shall prove that {gx_n} and {gy_n} are Cauchy sequences. Let at least one of {gx_n} and {gy_n} be not Cauchy sequences. Then there exists ɛ > 0 for which we can find subsequences {gx_n(k)}, {gx_m(k)} of {gx_n} and {gy_n(k)}, {gy_m(k)} of {gy_n} with n(k) > m (k) ≥ k such that

Further, corresponding to m (k), we can choose n (k) in such a manner that it is the smallest integer for which (9) holds. Then,

From (9), (10), and triangular inequality, we get

By using (8) in (11), we obtain

Again, by the triangle inequality

From monotonicity of ϕ and property (ϕ₃), we obtain

As ϕ is lower semi-continuous by taking limit as k ⟶ ∞, we obtain

a contradiction. Therefore, {gx_n} and {gy_n} are Cauchy sequences in X. From assumption (i) there exist u, v ∈ g (X) such that lim_{n ⟶ ∞} F (x_n,y_n) = lim_{n ⟶ ∞} gx_n = u and lim_{n ⟶ ∞} F (y_n,x_n) = lim_{n ⟶ ∞} gy_n = v. Since F and g are compatible mappings, we have

Let the assumption (1) hold. For all n ≥ 0, we have

Letting n ⟶ ∞, using (15), by assumption (i) and (iii), we have d (F (u,v), gu) = 0, that is, F (u,v) = gu.

Similarly, we also have F (u, v) = gv. Then CcFix (F) ≠ ∅ .

Next, we suppose that the assumption (2) holds. Let u = gx and v = gy for some x, y ∈ X. In this way, (gx_n, gx), (gy_n, gy) ∈ E (G) for each n ∈ ℕ. Then

and

Combining (16) and (17), we have

by the monotonicity of ϕ

As ϕ is lower semi-continuous, letting k ⟶ ∞ and by (Ψ₂), we obtain gx = F (x, y) and gy = F (y, x). □

Denote by CmFix (F) the set of all common fixed points of mappings F : X² ⟶ X, g : X ⟶ X, that is,

Theorem 2.3

In addition to Theorem 2.2, suppose that

(v) for any two elements (x, y) , (u, v) ∈ X², there exists (t, s) ∈ X² such that (gx, gt), (gu, gt), (gy, gs), (gv, gs) ∈ E(G).

Then, CmFix (F)≠ ∅ iff (X2)gF≠∅.

Proof

Theorem 2.2 implies that there exists (x, y) ∈ X² such that F (x, y) = gx and F (y, x) = gy. Assume that there exists another (u, v) ∈ X² such that F (u, v) = gu and F (v, u) = gv. Now, we shall prove that gu = gx and gv = gy.

From assumption (v), there exists (t, s) ∈ X² such that (gx, gt), (gu, gt), (gy, gs), (gv, gs) ∈ E (G). By using (3), we define the sequences {x_n}, {y_n}, {u_n}, {v_n}, {t_n} and {s_n} in X as follows:

for all n ∈ ℕ. From the properties of coincidence points, x = x_n, y = y_n and u = u_n, v = v_n, namely,

for all n ∈ ℕ. As (gx, gt), (gy, gs) ∈ E (G), we get (gx, gt₀), (gy, gs₀) ∈ E (G). Since F and g are G–edge preserving, we obtain (F (x, y) , F (t₀, s₀)) = (gx, gt₁) and (F (y, x) , F (s₀, t₀)) = (gy, gs₁) ∈ E(G). Similarly, (gx, gt_n) and (gy, gs_n) ∈ E (G). By (2), we have

Then, we get ϕ(ρ_n+1) ≤ Ψ (ρ_n), where ρ_n := {d(gt_n, gx) + d(gs_n, gy)} x 2^-1. The sequence {ρ_n} is a monotonically decreasing sequence of nonnegative real numbers, thus there exists some ρ ≥ 0 such that ρ_n → ρ as n → ∞. Now, we shall show that ρ = 0. Suppose, to the contrary, that ρ > 0. Letting n → ∞ in (19), we have

a contradiction. Hence, ρ = 0; that is; lim_n→∞ [{d (gt_n, gx) + d (gs_n, gy)} x 2^-1] = 0, which implies

Similarly, (gt, gu), (gs, gv) ∈ E (G), we have

From the triangular inequality we obtain

for all n ∈ ℕ. Letting n → ∞ in (20), we obtain that d (gu, gx) = 0 = d (gv, gy). Hence, we get gu = gx and gv = gy.

The proof of the existence and uniqueness of the coupled common fixed point for our main results can be obtained by using a similar assertion as in Theorem 9 in [18]. □

Remark 2.4

In this case where (X, ≼) is partially ordered complete metric space, taking E (G) ={(x, y) ∈ X² : x ≼ y}, we obtain Theorem 5 in [18]. By using Remark 6, Remark 7 in [18], we get that our results generalize the results obtained by Bhaskar and Lakshmikantham in Theorem 2.1 of [15], Luong and Thuan in Theorem 2.1 of [26], Berinde in Theorem 3 of [27] and Berinde in Theorem 2 of [28].

Theorem 3.1

Suppose that conditions (1) — (4) are satisfied. Then (21) has a unique solution in C (I, ℝⁿ).

Proof

Let F : X² → X, (x, y) → F (x, y), where

and define g : X → X by gx(t) = 2x (t) . (21) can be stated as

Let x, y, u, v ∈ X be such that gx ≤ gu and gy ≤ gv. We get x ≤ u and y ≤ v and

Then, F and g are G—edge preserving.

From (22) and (23) for all t ∈ I, we have

Then, there exists ϕ (t) = t and ψ(t) = αt for α ∈ [0, 1) such that

Thus, F and g are a (ϕ, Ψ) —contractive.

By assumption (4) shows that there exists (x₀, y₀) ∈ X² such that gx₀ ≤ F (x₀, y₀) and gy₀ ≤ F (y₀, x₀), which implies that (X2)gF≠∅.

Hence, there exists a coupled common fixed point (x_*, y_*) ∈ X² of the mapping F and g, which is the solution of the integral system (21). □

The following illustrative example, considered as X := C (I, ℝ), h : I × ℝ × ℝ → ℝ, q : I → ℝ, x, y, u, v ∈ ℝ and g : X → X by gx = x. Inspired and motivated by Example 3.1 of [29], we present an example of a functional integral equation.

Example 3.2

Consider the following functional integral equation:

for t ∈ I. Observe that this equation is a special case of (21) with

It also easily seen that these functions are continuous.

For arbitrary x, y, u, v ∈ ℝ and for all s ∈ I, we have

Therefore, the function h satisfies (22) with α=19.

For all t, s ∈ I, there exists a continuous A : I × ℝ → ℝ such that

We put x0(t)=5t27(1+t4), we obtain