Existence of a common solution to systems of integral equations via fixed point results

Definition 1.1

[7] An element (u, v) ∈ X ² is said to be a coupled fixed point of a mapping A: X ² → X if u = A(u, v) and v = A(v, u).

Definition 1.2

[17] An element (u, v) ∈ X ² is called a coupled common fixed point of mappings A, B: X ² → X if A(u, v) = B(u, v) = u and A(u, v) = B(u, v) = v. The set of all coupled common fixed points of A and B is indicated by ℱ ( A , B ) .

Definition 1.3

[17] Let ( X , ≺ ̲ ) be an ordered set. Two mappings A, B: X ² → X are said to be weakly increasing with respect to ≺ ̲ if

hold for all (u,v) ∈ X ².

Following Su [22], we define the set Ψ = {ψ:[0, +∞) → [0, +∞):ψ] that satisfies the conditions (i)–(iii)}, where

1. is nondecreasing,

2. ψ(t) = 0 if and only if t = 0,

3. is subadditive, that is, ψ(t + s) ≤ ψ(t) + ψ(s) for all t, s ∈ [0, +∞).

The set Φ = {ϕ:[0, +∞) → [0, +∞):ϕ is a nondecreasing and right upper semi-continuous function with ψ(t) > ϕ(t) for all t > 0, where ψ ∈ Ψ}.

Throughout the article, ( X , d , ≺ ̲ ) states an ordered metric space where d is a metric on X and ≺ ̲ is a partial order on the set X. In addition, we say that (x,y) ∈ X ² is comparable to (u,v) ∈ X ² if x ≺ ̲ u and y ≺ ̲ v , or u ≺ ̲ x and v ≺ ̲ y . For brevity, we denote by ( x , y ) ≺ ̲ ( u , v ) or ( x , y ) ≻ ̲ ( u , v ) .

If d is a metric on X, then δ: X ² × X ² → [0, +∞), defined by δ((x,y),(u,v)) = d(x,u) + d(y,v) for all (x,y),(u,v) ∈ X ², is also a metric on X ².

Now, we define Su type contractive pairs, which will be utilized in our main results.

Definition 1.4

Let ( X , d , ≺ ̲ ) be an ordered metric space and A, B: X ² → X be given mappings. We say that (A, B) is a Su type contractive pair if, for all comparable pairs (x, y),(u, v) ∈ X ²,

holds, where

Remark 1.5

By the definition of Q(x, y, u, v), it is obvious that

Theorem 2.1

Let ( X , d , ≺ ̲ ) be an ordered complete metric space, A, B: X ² → X weakly increasing mappings with respect to ≺ ̲ and (A, B) be a Su type contractive pair. If A (or B) is continuous, then ℱ ( A , B ) ≠ ∅ .

Proof

Let u ₀, v ₀, ∈ X Define sequences {u _n } and {v _n } in X by

and

for all n ≥ 0. Since A and B are weakly increasing, we have

Suppose that u _n ≠ u _n+1 and v _n ≠ v _n+1 for all n ≥ 0 Then, for n = 2m + 1 using (1) and (2), we have

where

Since

and

we obtain

Similarly, by (1) and (2), we obtain

Summing the inequalities (3) and (4) and using the subadditivity property of ψ, we obtain

If Q(u _2m , v _2m , u _2m+1, v _2m+1) = δ((u _2m+1, v _2m+1),(u _2m+2, v _2m+2) for some m, then by (5), we obtain

Since ψ is nondecreasing,

which is a contradiction. Then,

and so, by (5),

Set δ _n := {δ((u _n ,v _n ),(u _n+1,v _n+1))}. Then, the sequence {δ _n } is decreasing. Thus, there exists r ≥ 0 such that lim _n→∞ δ _n = r. Suppose that r > 0. Letting n → ∞ in (6), we deduce

a contradiction, and hence, r = 0, that is,

To prove that {u _n } and {v _n } are Cauchy sequences, it is sufficient to show that {u _2n } and {v _2n } are Cauchy sequences in (X, d). Suppose, to the contrary, that at least one of {u _2n } or {v _2n } is not Cauchy sequence. Then, there exists an ε > 0 for which we can find subsequences { u 2 m k } , { u 2 n k } of {u _2n } and { v 2 m k } , { v 2 n k } of {v _2n }, such that n _k is the smallest index for which n _k > m _k > k and

By using the triangle inequality and (8), we obtain

Taking k → ∞ in the above inequality and using (7), we deduce

Again, from the triangle inequality, we have

Letting k → ∞ in the above inequality and using (7) and (9), we have

Since ( u 2 m k , v 2 m k ) ≺ ̲ ( u 2 n k + 1 , v 2 n k + 1 ) for n _k > m _k , using (1), we obtain

where

Again, since ( v 2 m k , u 2 m k ) ≺ ̲ ( v 2 n k + 1 , u 2 n k + 1 ) , by (1), we also have

Summing the inequalities (11) and (12) and using the subadditivity property of ψ, we obtain

Now, by using (7), (9) and (10) and letting k → ∞ in the above inequality, we deduce

which implies ε = 0 a contradiction with ε > 0. Therefore, {u _n } and {v _n } are Cauchy sequences in X.

Now, we prove the existence of coupled common fixed point of A and B.

Owing to the completeness of (X, d), there exist u, v ∈ X such that

Without loss of generality, we assume that A is continuous. Now we have

and

We now assert that d(u, B(u, v)) = d(v, B(v, u)) = 0. To establish the claim, assume that d(u, B(u, v)) > 0 and d(v, B(v, u)) > 0. Since (u, v) ∈ X ² is comparable to its own, from (1), we obtain

where

Again, since ( v , u ) ≺ ̲ ( v , u ) , by (1), we have

Thus, it follows from (14) and (15) that

which implies d(u, B(u, v)) = d(v, B(v, u)) = 0

Therefore, u = A(u, v) = B(u, v) and v = A(v, u) = B(v, u)□

Example 2.2

Let X = [0,1] be equipped with the usual metric and the partial order defined by

Define mappings A, B: X ² → X by A ( u , v ) = u + v 4 and B ( u , v ) = u + v 3 . Then, A and B are weakly increasing with respect to ≺ ̲ and continuous.

Also, (A, B) is a Su type contractive pair. Indeed, for all comparable (x, y), (u, v) ∈ X ²,

where ψ ( t ) = t and ϕ ( t ) = t 2 . Thus, all the hypotheses of Theorem 2.1 are fulfilled. Therefore, A and B have a coupled common fixed point, which is (0,0).

Definition 2.3

Let ( X , d , ≺ ̲ ) be an ordered metric space. We say that ( X , d , ≺ ̲ ) is regular if each nondecreasing sequence {x _n } with d(x _n ,x) → 0 implies that x n ≺ ̲ x for all n.

We replace the continuity of A (or B) with the regularity of ( X , d , ≺ ̲ ) in the following theorem.

Theorem 2.4

Let ( X , d , ≺ ̲ ) be an ordered complete metric space, A, B: X ² → X weakly increasing mappings with respect to ≺ ̲ and (A, B) be a Su type contractive pair. If ( X , d , ≺ ̲ ) is regular, then A and B have a coupled common fixed point.

Proof

Let u ₀, v ₀ ∈ X. Define sequences {u _n } and {v _n } in X by

and

for all n ≥ 0. Following the proof of Theorem 2.1, we can show that the sequences {u _n } and {v _n } are nondecreasing, lim _n→∞ u _n = u and lim _n→∞ v _n = v. Since ( X , d , ≺ ̲ ) is regular, we deduce that (u _n , v _n ) is comparable to (u, v) for all n. From (1), we obtain

where

Again, by (1), we obtain

Thus, it follows from (16), (17) and the subadditivity property of ψ that

Taking n → ∞ in the above inequality, we obtain

which implies that d(u, B(u, v)) + d(v,B(v, u)) = 0, that is, u = B(u, v) and v = B(v, u).

Since ( u , v ) ≺ ̲ ( u , v ) , by (1), we deduce

where

Again, by (1),

Hence, it follows from (18), (19) and the subadditivity of ψ that

which implies d(u, A(u, v)) = d(v, A(v, u)) = 0. This completes the proof.□