Definition 1 (see [8].)A function S_b : Z³⟶[0, ∞) is said to be S_b-metric on a nonempty set Z, with b ≥ 1, if it satisfies the following properties: for each ϱ, ς, ϑ, τ ∈ Z

• i.

  S_b(ϱ, ς, ϑ) = 0 if and only if ϱ = ς = ϑ;

• ii.

  S_b(ϱ, ς, ϑ) ≤ b[S_b(ϱ, ϱ, τ) + S_b(ς, ς, τ) + S_b(ϑ, ϑ, τ)].

Definition 2 (see [10].)Let Z be a nonempty set, and B : Z³⟶[1, ∞) be a positive real valued function. Then a function $$ is said to be EPSb-metric on Z if for each ϱ, ς, ϑ, τ ∈ Z and λ > 0, it satisfies

•  

  $$$$ if and only if ϱ = ς = ϑ;

•  

  $$$$.

Example 1. Let $$. Define function B : Z³⟶[1, ∞) by

and a function $$ by for each $$ and λ > 0. Then $$ is an EPSb-metric space.

Definition 3 (see [10].)Suppose $$ is an EPSb-metric space. The pair $$ is said to be symmetric EPSb-metric space, if for all ϱ, ς ∈ Z, λ > 0, it satisfies condition

Definition 4. [10] A sequence 〈ϱ_n〉 in an EPSb-metric space $$ is said to be:

• i.

  Convergent, and converges to ϱ if and only if $$ for all n ≥ n₀ and λ > 0.

• ii.

  Cauchy sequence if $$ for all n, m, λ > 0 with n > m.

Definition 5 (see [10].)The space “$$ is said to be complete if every Cauchy sequence is convergent in Z.”

Lemma 1 (see [10].)Let $$ be a symmetric EPSb-metric space. Suppose there exist two sequences such that

Lemma 2. Let $$ be an EPSb-metric space and suppose there exist a sequence {ϱ_n} ∈ Z and a q ∈ (0, 1) such that the following inequality holds:

Definition 6 (see [10].)Let Φ denote the class of all functions ϕ : [0, ∞)⟶[0, ∞) which is increasing, continuous, and such that $$, for all t > 0. Then ϕⁿ(t)⟶0 as n⟶∞.

Lemma 3. Suppose that $$ be an EPSb-metric spaces, and further suppose that there exists a sequence {ϱ_n} ∈ Z such that following inequality holds:

where ϕ ∈ Φ. Then {ϱ_n} is a Cauchy sequence in EPSb-metric spaces.

Proof 1. By iterating equation (7), we derive the following inequality:

From Definition 2, for all m > n we have

Since for all t > 0, $$,

In what follows, {ϱ_n} is a Cauchy sequence.

Theorem 1. Suppose that two self-maps $$ and $$ defined on a complete symmetric EPSb-metric spaces $$ are such that for each ϱ, ς, ϑ ∈ Z and λ > 0 with 0 < q < 1, it satisfies

where

Proof 2. Let us begin the proof by claiming that if ϱ is any arbitrary fixed point of the map $$, then it is also the fixed point of the map $$ and vice versa.

First assume that $$, and claim that: $$.

Let us substitute ϱ = ς = ϱ and $$ in equation (11), and we have

where

Thus, from equation (14), we have

This is only possible if $$, and hence we have our claim that $$. Similarly, if we assume that $$, one can get that $$. This proves that ϱ is a unique common fixed point of maps $$ and $$.

In order to prove our theorem, it is sufficient to deduce that either $$ or $$ has a fixed point. Let ϱ₀ ∈ Z be any arbitrary point. Since $$ and $$ are two self-maps defined on Z, there exists ϱ₁, ϱ₂ ∈ Z such that $$ and $$.

Proceeding in this manner we can, in a general sense, choose two sequences ϱ_2n, ϱ_2n+1 ∈ Z such that

Assume that ϱ_2n ≠ ϱ_2n+1 for any n ∈ ℕ.

Substitute ϱ = ϱ_2n, ς = ϱ_2n, and ϑ = ϱ_2n−1 in equation (11), and we obtain

where

Two alternatives emerge here.

•  

  Case 1 Either $$,

•  

  Case 2 or $$.

Suppose Case 1 is true; then from equation (18) and using the fact that 0 < q < 1, we obtain that

This gives that ϱ_2n = ϱ_2n+1. This is a contradiction to our assumption that ϱ_2n ≠ ϱ_2n+1. Thus, Case 2 must be true.

Hence, from equation (18), we have

Again, on substituting ϱ = ς = ϱ_2n and ϑ = ϱ_2n+1, in equation (11), we obtain

where

By following the similar argument as above, we can get

On combining equations (21) and (24), we obtain

Thus, for all n ≥ 2, we can conclude that

On repeating the procedure n times, we obtain

On applying Lemma 2, we obtain our claim that the sequence {ϱ_n} is a Cauchy sequence. Since Z is a complete EPSb-metric space, there exist some ϱ ∈ Z such that

Next we will now prove that ϱ is the fixed point of $$, that is, we have to show that $$. On the contrary, assume that $$.

From equation (11), on substituting ϱ = ϱ_2n, ς = ϱ_2n and ϑ = ϱ, we obtain

where

By taking upper limit as n⟶∞ in above equality and utilizing equation (28), we get

By substituting equation (31) and utilizing the fact that 0 < q < 1 in equation (29), we obtain $$. Consequently, it follows that $$. Thus, the maps $$ and $$ have a fixed point in Z.

In order to guarantee the uniqueness of fixed point of maps, assume on contrary that there exists another common fixed point κ, different from ϱ, of maps $$ and $$ in Z such that $$.

Consider

where

Thus, from above equality, we have

This is possible only if $$ and so κ = ϱ. Thus, ϱ is a unique common fixed point of $$ and $$. This finally completes the theorem proof.

Theorem 2. Suppose $$ is an EPSb-metric space, which is symmetric and complete. Let $$ be two self-mappings such that for all ϱ, ς, ϑ ∈ Z and λ > 0, it satisfies

where ϕ ∈ Φ and

Proof 3. Let us begin the proof by claiming that if ϱ is any arbitrary fixed point of the map $$, then it is also the fixed point of the map $$ and vice versa.

First assume that $$, and claim that: $$.

Let us substitute ϱ = ς = ϱ and $$ in equation (35), and we have

where

Thus, from equation (37), we have

and we arrive at a contradiction as ϕ ∈ Φ. This is only possible if $$, and hence we have our claim that $$. Similarly, if we assume that $$, one can get that $$. This proves that ϱ is a unique common fixed point of maps $$ and $$.

In order to prove our theorem, it is sufficient to deduce that either $$ or $$ has a fixed point. Let ϱ₀ ∈ Z be any arbitrary point. Since $$ and $$ are two self-maps defined on Z, there exists ϱ₁, ϱ₂ ∈ Z such that $$ and $$.

Proceeding in this manner, we can, in a general sense, choose two sequences ϱ_2n, ϱ_2n+1 ∈ Z such that

Assume that ϱ_2n ≠ ϱ_2n+1 for any n ∈ ℕ.

Let us substitute ϱ = ϱ_2n, ς = ϱ_2n, and ϑ = ϱ_2n−1 in equation (35), and we obtain

where

Two alternatives emerge here.

•  

  Case 1 Either $$,

•  

  Case 2 or $$.

Suppose Case 1 is true; then from equation (41),

and we arrive at a contradiction as ϕ ∈ Φ. Thus, Case 2 must be true.

Hence, from equation (41), we have

Again, on substituting ϱ = ϱ_2n, ς = ϱ_2n, and ϑ = ϱ_2n+1, in equation (35), we obtain

where

By following the similar argument as above, we can get

From equations (44) and (47), we obtain

Thus, for all n ≥ 2, we can conclude that

On repeating the procedure n times, we obtain

Thus, on using Lemma 3, {ϱ_n} is a Cauchy sequence in Z. Since Z is a complete EPSb-metric space; therefore, there exists ϱ ∈ X such that

Next we will now prove that ϱ is the fixed point of $$, that is, we have to show that $$. On the contrary, assume that $$.

From equation (35), on substituting ϱ = ϱ_2n, ς = ϱ_2n, and ϑ = ϱ, we obtain

where

By taking upper limit as n⟶∞ in above equality altogether and utilizing equation (51), we get

This is a contradiction. Hence, $$. Consequently, it follows that $$. Thus, the maps $$ and $$ have a fixed point in Z.

In order to guarantee the uniqueness of fixed point of maps, assume on contrary that there exists another common fixed point κ, different from ϱ, of maps $$ and $$ in Z such that $$.

Consider

where

Thus, from above equality, we have

which is a contradiction. Thus, our assumption is wrong, and hence ϱ = κ. That is, ϱ is a unique common fixed point of $$ and $$. This completes the proof.

Theorem 3. Suppose that $$ and $$ are self-maps defined on a complete symmetric EPSb-metric spaces $$, such that for each ϱ, ς, ϑ ∈ Z and λ > 0, it satisfies

for all ϱ, ς, ϑ ∈ Z, where a₁, a₂, a₃, a₄, a₅ > 0 with $$. Then $$ and $$ have a unique common fixed point in Z.

Proof 4. Given that maps $$ and $$ are self-maps such that for each ϱ, ς, ϑ ∈ Z and λ > 0, it satisfies

Since each a_i > 0,

where $$ and

This expression reduces to the form of Theorem 1, and as a result, the remainder of the proof can be directly inferred from the conclusions of Theorem 1. Thus, the maps $$ and $$ have a unique common fixed point in Z.

Corollary 1. Suppose that $$ is a self-map defined on a complete symmetric EPSb-metric space $$, such that for each ϱ, ς, ϑ ∈ Z and λ > 0 with 0 < q < 1, it satisfies

where

Corollary 2. Suppose that $$ is a self-map defined on a complete symmetric EPSb-metric space $$, such that for each ϱ, ς, ϑ ∈ Z and λ > 0, there exists a function ϕ ∈ Φ such that

where

Corollary 3. Suppose that $$ is a self-map defined on a complete symmetric EPSb-metric space $$, such that for each ϱ, ς, ϑ ∈ Z and λ > 0, it satisfies

for all ϱ, ς, ϑ ∈ Z, where a₁, a₂, a₃, a₄, a₅ > 0 with $$. Then the map $$ has a unique fixed point in Z.

Example 2. Let Z = [1, ∞) ∪ {0}. For each λ > 0 and ϱ, ς, ϑ ∈ Z, define function B : Z³⟶[1, ∞) by

and function $$ by

Example 3. Let Z = (0, 1]. For each λ > 0 and ϱ, ς, ϑ ∈ Z, define function B : Z³⟶[1, ∞) by

and function $$ by

Example 4. Let Z = [0, 1). Define function B : Z³⟶[1, ∞) by

and a function $$ by for each λ > 0 and ϱ, ς, ϑ ∈ Z. Then, $$ is an EPSb space.

Theorem 4. Suppose the set $$ is closed and bounded on an interval [a, b]. Then, the system of linear integral equations

under the following assumptions:

• i.

  $$ are continuous functions,

• ii.

  for every ϱ, ς ∈ C[a, b],

  $$ ()

•  

  have a unique common solution ϱ(g₂) ∈ C[a, b].

Proof 5. Define maps $$ by:

and

It is evident to note that existence of the solution of system of integral equations is equivalent to find the common fixed point of the maps $$ and $$.

For all ϱ, ς ∈ C[a, b], consider

On using Cauchy–Schwar inequality, we obtain

Assume that

Clearly, for every a > 0, b − a > 0, and so ((b − a)²/9) > 0. Thus, ϕ is positive, increasing, and continuous mapping, and so ϕ ∈ Φ.

As a result (from equation (87)), we have

Hence, all the conditions of Theorem 2 are satisfied. Therefore, there exists a unique common fixed point, say ϱ ∈ C[a, b] of the maps $$ and $$. Consequently, ϱ(g₂) is the unique common solution of the system of linear equations.

Example 5. Consider the following system of integral equations: