Existence of a common solution for a system of nonlinear integral equations via fixed point methods in b-metric spaces

The function φ:[0, ∞) → [0, ∞) is called an altering distance function if the following properties hold:

1. φis continuous and non-decreasing;

2. φ(t) = 0 if and only if t = 0.

([11]).Let X be a nonempty set and s > 1. Suppose that the mapping d : X × X → ℝ₊satisfies the following conditions: for all x, y, z, ∈ X,

(BM1) d(x, y) = 0 if and only if x = y;

(BM2) d(x, y) = d(y, x);

(BM3) d(x, y) ≤ s[d(x, z) + d(z, y)].

Then (X, d) is called a b-metric space with coefficient s.

Let X = ℝ and define a mapping d : X × X → ℝ₊by

for all x, y ∈ X. Then(X, d) is a b-metric space with coefficient s = 2.

The set l_p(ℝ) with 0 < p < 1,where

together with the mapping d : l_p (ℝ) × l_p (ℝ) → ℝ₊defined by

for all x = {x_n}, y = {y_n ∈ l_p(ℝ)is a b-metric space with coefficients=21p>1. The above result also holds for the general case l_p (X) with 0 < p < 1, where X is a Banach space.

Let p be a given real number in(0, 1).The space

together with the mapping d : L_p [0,1] × L_p [0,1] → ℝ₊defined by

for all x, y ∈ L_p [0, 1] is a b-metric space with constants=21p>1.

([12]).Let(X, d)be a b-metric space. Then a sequence{x_n} in X is called:

1. b-convergent if there exists x ∈ X such thatd(x_n, x) → 0 as n → ∞. In this case, we writelimn→∞ xn = x;

2. a b-Cauchy sequence if d(x_n, x_m) → 0 as n, m → ∞.

([12]).In a b-metric space(X, d), the following assertions hold:

1. a b-convergent sequence has a unique limit;

2. each b-convergent sequence is a b-Cauchy sequence;

3. in general, a b-metric is not continuous.

([13]).Let(X, d) be a b-metric space with coefficient s ≥ 1 and let{x_n}, {y_n} be b-convergent to the points x, y ∈ X, respectively. Then we have

In particular, if x = y, then we havelimn→∞ d(xn, yn) = 0.Moreover, for all z ∈ X, we have

([12]).A b-metric space(X, d)is said to be b-complete if every b-Cauchy sequence in X b-converges.

([12]).Let(X, d)and(X′, d′)be two b-metric spaces. A function f: X → X′ is said to be b-continuous at a point x ∈ X if it is b-sequentially continuous at x, that is, whenever {x_n} is b-convergent to x, {fx_n}is b-convergent to fx.

([12]).Let Y be a nonempty subset of a b-metric space(X, d).The closure of Y is denoted by Y and it is the set of limits of all b-convergent sequences of points in Y, that is,

([12]).Let(X, d)be a b-metric space. Then a subset Y ⊆ X is said to be closed if, for each sequence{x_n} inY which b-converges to an element x, we have x ∈ Y(i.e., Y = Y).

Let (X, d) be a b-metric space with coefficient s ≥ 1 and f, g : X → X be two mappings. The pair (f, g)is said to be compatible if

whenever {x_n} is a sequence in X such that

for some t ∈ X. If s = 1, then it becomes compatible in the sense of Junck [14].

([15]).Let f and g be two self mappings on a nonempty set X. The pair (f, g) is said to be weakly compatible if f and g commute at their coincidence point (i.e., fgx = gfx whenever fx = gx).

Let (X, d) be a b-metric space with coefficient s ≥ 1 and let α : X × X → [0, ∞) be a given mapping. The b-metric space X is said to be α-complete if every Cauchy sequence {x_n} ⊆ X with

for all n ∈ ℕ converges in X.

Let X be a nonempty set. The mapping α : X × X →[0, ∞)is said to be transitive if, for x, y, z∈X, we have

Let X be a nonempty set, α : X × X → [0, ∞) and f, g : X → X be three mappings. The ordered pair (f, g) is said to be:

1. α-weakly increasing if α(fx, gfx) ≥ 1 and α(gx, fgx) ≥ 1 for all x ∈ X;

2. partially α-weakly increasing if α(fx, gfx)≥1 for all x ∈ X.

Let X be a nonempty set, α : X × X → [0, ∞) and f, g, h : X → X be four mappings such that f(X) ⊆ h(X) and g(X) ⊆ h(X). The ordered pair (f, g) is said to be:

1. α-weakly increasing with respect to h if, for all x ∈ X, we have α(fx, gy) ≥ 1 for all y ∈ h^−l(fx) and α(gx, fy) ≥ 1 for all y ∈ h^−l(gx);

2. partially a-weakly increasing with respect to h if α(fx, gy) ≥ 1 for all y ∈ h⁻¹(fx).

From 3.2, we have the following assertions:

1. If h = I_X (: the identity mapping on X), then 3.2 reduces to 3.1;

2. If g = f, then we say that f is α-weakly increasing with respect to h(partially α-weakly increasing with respect to h). Also, if h = I_X, then we say that f is α-weakly increasing (partially α-weakly increasing).

Let (X, d) be a b-metric space, α : X × X → [0, ∞) and f, g : X → X be three mappings. The pair (f, g) is said to be α-compatible if

whenever {x_n} is a sequence in X such that

for all n ϵ ℕ and

for some t ϵ X.

Let (X, d) be a b-metric space, α : X × X → [0, ∞) and f : X → X be two mappings. We say that f is a-continuous at a point x ϵ X if, for each sequence {x_n} in X with x_n → x as n → ∞ and α(x_n, x_n+1) ≥ 1 for all n ϵ ℕ, we have

Let (X, d) be a b-metric space with coefficient s ≥ 1 and f, g, R, S : X → X be four mappings. Throughout this paper, unless otherwise stated, for all x, y ϵ X, let

Now, we give a coincidence point result in this paper.

Let (X, d) be a b-metric space with coefficient s ≥ 1, α : X × X → [0, ∞) and f, g, R, S : X → Xbe five mappings such that f (X) ⊆ R(X) and g(X) ⊆ S(X). Suppose that, for all x, y ϵ X, we have

where ψ φ : [0, ∞) → [0, ∞) are altering distance functions. If the following conditions hold:

1. (X, d) is α-complete;

2. f, g, R and S are α-continuous;

3. the pairs (f, S) and (g, R) are α-compatible;

4. the pairs (f, g) and (g, f) are partially α-weakly increasing with respect to R and S, respectively;

5. α is a transitive mapping,

then the pair (f, S) and (g, R) have a coincidence pointz ϵ X. Moreover, if α(Rz, Sz) ≥ 1 or α(Sz, Rz) ≥ 1, then z is a coincidence point of f, g, R and S.

Let x₀ be an arbitrary point of X. Choose x₁ ϵ X such that fx₀ = Rx₁ and x₂ ϵ X such that gx₁ = Sx₂. Now, we can construct a sequence {z_n} defined by

for all n ϵ ℕ ∪ {0}. Since x₁ ϵ R⁻¹(fx₀), x₂ ϵ S⁻¹ (gx₁) and the pair (f, g) and (g, f) are partially α-weakly increasing with respect to R and S, respectively, we have

Repeating this process, we obtain

for all n ϵ ℕ ∪ {0}. Furthermore, by the transitive property of α, we have

and

for all n ϵ ℕ.

Now, we will complete the proof in three steps:

Step I. We prove that limk→∞ d(zk, zk+1) = 0. For all k ϵ ℕ ∪ {0}, we define d_k: = d(z_k, Z_k+1). We assume that d_k0 = 0 for some k₀ ϵ ℕ ∪ {0}, which implies that Z_k0 = Z_k0+1. If k₀ = 2n such that n ϵ ℕ ∪ {0}, then Z_2n = Z_2n+1. Next, we show that Z_2n+1 = Z_2n+2. Since α(Sx_2n, Rx_2n+1) = α(z_2n, z_2n+1) ≥ 1, we have

where

Therefore, from (6), we have

This implies that

and so

Therefore, we have Z_2n+1 = Z_2n+2. Similarly, if k₀ = 2n+1 such that n ϵ ℕ ∪ {0}, then Z_2n+1 = Z_2n+2 gives Z_2n+2 = Z_2n+3. Consequently, the sequence {Z_n} becomes constant for k ≥ k₀ and hence lim_k→œd(Z_k, Z_k+1) = 0. This completes this step.

Therefore, we will suppose that

for all k ϵ ℕ ∪ {0}. Next, we show that

for all k ϵ ℕ ∪ {0}. Assume that

for some k ϵ ℕ ∪ {0}. If k = 2n such that n ϵ ℕ ∪ {0}, then we have

Since

using (2), we obtain

where

Since ψ is nondecreasing and

we have

Now, the inequality (9) implies that

which is possible only M_s (x_2n, x_2n+1) = 0, that is, d(z_2n+1, z_2n+2) = 0, which contradicts (7). Hence we have

Therefore, (8) is proved for k = 2n. Also, we have

Similarly, we can shown that

for all n ϵ ℕ ∪ {0}. Hence the inequality (8) holds and then {d(z_k, z_k+1)} is a nonincreasing sequence of nonnegative real numbers. Since {d(z_k, z_k+1)} is bounded below, there exists r ≥ 0 such that

and then

Furthermore, we get

for all k ϵ ℕ ∪ {0}. Letting k → ∞ in (13), using (11), (12) and the continuity of ψ, φ, we have

This implies that φ(r) = 0. From the property of φ, we have r = 0 and so

Step II. We now claim that {Z_n} is a b-Cauchy sequence in X. That is, for any 𝜖 ≥ 0, there exists k ϵ ℕ such that d(Z_m, Z_n) < 𝜖, for all m, n ≥ k. Assume that there exists 𝜖 ≥ 0 for which we can find subsequences {Z_m(k)} and {Z_n(k)} of {Z_n} such that n(k) ≥ m(k) ≥ k and

1. m(k) = 2t and n(k) = 2t′ + 1, where t, t′ ϵ ℕ;

2. (15)d (zm  (k) , zn (k)) ≥ ∈ ;

3. n(k) is the smallest number such that the condition (b) holds, i.e.,

   (16)d (zm  (k) , zn (k) − 1) < ∈ .

By the triangle inequality, (15) and (16), we obtain

Taking limit supremum as k → ∞ in (17) and using (14), we have

From the triangle inequality, we have

and

Taking limit supremum as k → ∞ in (19) and (20), from (14) and (18), it follows that

and

This implies that

Again, using above process, we get

Finally, we obtain that

Taking limit supremum as k → ∞ in (23), from (14) and (21), we obtain that

By similar method, we have

From (24) and (25) implies that

Since α is transitive, we have

From (2), it follows that

where

Taking limit supremum as k → ∞ in the above equation and using (14), (18), (21), (23) and (22), we have

Also, we can show that

Taking limit supremum as k → ∞ in (27), we have

This implies that φ(ε) ≤ 0 and so ϵ = 0, which is a contradiction. Therefore, {z_n} is a b-Cauchy sequence.

Step III. We will show that f, g, R and S have a coincidence point. From Step II, we show that {z_n} is a b-Cauchy sequence in X. Since the inequality (3) holds, by the a-completeness of b-metric space X, there exists z ∈ X such that

and so

and

From (29) and (30), we have fx_2n → z and Sx_2n → z as n → ∞. Since (f, S) is a-compatible, by (4), we have

By (4), the α-continuity of S, f and 2.7, we obtain

By the triangle inequality, we have

for all n ∈ ℕ ∪ {0}. Taking limit as n → ∞ in the above inequality and using (23) and (31), we obtain

This implies that d(SZ, fz) = 0 and so fz = Sz, that is, Z is a coincidence point of f and S. Similarly, we can prove that z is also a coincidence point of g and R.

Finally, we prove that z is a coincidence point of f, g, R and S provide that

From (2) and (34), we have

where

Therefore, (35) implies

Now we obtain that

This implies that d(fZ, gZ) = 0 and so

Hence z is a coincidence point of f, g, R and S. This completes the proof. □

Next, we give an example to illustrate Theorem 3.6.

Let X = ℝ and b-metric d : X χ X → [0, ∞) be given by d(x, y) = |x — y|²for all x, y ∈ X. Define mapping f, g, R, S : X → X and α : X χ X → [0, ∞)by

and

It is easy to see that (X, d) is an a-complete b-metric space with coefficient s = 2. Also, we can see that f, g, R and S are α-continuous.

To prove that (f, g) is partially a-weakly increasing with respect to ℝ. Let x, y ∈ X besuchthat y ∈ R⁻¹ (fx), that is, Ry = fx. By the definition of f and R, we divide two cases, that is,

If x, y ∈ [0,1), then we have

and so

Since sinh x ≥ sinh⁻¹x for all x ≥ 0, we have 12x ≥ sinh⁻¹ (sinh⁻¹x)or