Abstract: -Inside this paper, we introduce two new partial b-metric contractions utilizing a rational expression

simulation function. The following conclusions extend, generalize, and integrate the earlier findings in two ways:

in contraction terms and in the abstract environment. We provide an example to establish the main theorem’s

validity.

Key-Words: Contraction, Fixed point, partial b-metric space, Simulation functions

Received: August 25, 2021. Revised: May 17, 2022. Accepted: June 7, 2022. Published: July 1, 2022.

1 Introduction

Banach published his foundational work on fixed

point theory approximately a century ago. Banach’s

basic yet deep theorem has been extended, enhanced,

and generalized by researchers since his first study

(see [1, 2, 3, 4, 5, 6, 7, 8, 9]). This was accomplished

by examining the terms of the contraction inequal-

ity and Banach’s theorem’s abstract structure. We’ll

combine these two tendencies by employing simula-

tion functions that include rational terms to create two

new type contractions in partial b-metric space.

2 Preliminaries

Definition 2.1. [10] Let Abe a non-empty set, a func-

tion p:A × A → R+

0is a partial metric if the follow-

ing conditions:

(p1)α=δif and only if p(α, α) = p(α, δ) = p(δ, δ);

(p2)p(α, α)≤p(α, δ);

(p3)p(α, δ) = p(δ, α);

(p4)p(α, δ)≤p(α, υ) + p(υ, δ)−p(υ, υ),

for all α, υ, δ ∈ A.

The pair (A, p)is called a partial-metric space.

Lemma 2.2. [11] If {αλ},{δλ}are two sequences in

a partial-metric space (A, p)such that

lim

λ→∞

p(αλ, µ) = lim

λ→∞

p(αλ, αλ) = p(µ, µ),

lim

λ→∞

p(δλ, κ) = lim

λ→∞

p(δλ, δλ) = p(κ, κ).

Then limλ→∞ p(αλ, δλ) = p(µ, κ).

Moreover,limλ→∞ p(αλ, σ) = p(µ, σ),for each

σ∈ A.

Denote by L(αλ)the set of limit points (if there

exist any),

L(αλ) = {σ∈ A :lim

λ→∞

p(αλ, σ) = p(σ, σ)}.

Lemma 2.3. [12] Let {αλ}be a Cauchy sequence on

a complete partial-metric space (A, p).If there exists

µ∈ L({αλ})with p(µ, µ) = 0,then µ∈ L({αλ(ι)}),

for every subsequence {αλ(ι)}of {αλ}.

Lemma 2.4. [12] On a complete partial-metric space

(A, p),let Fbe a continuous mapping and {αλ}be a

Cauchy sequence in A. If there exists µ∈ L({αλ})

with p(µ, µ) = 0.Then Fµ∈ L({Fαλ}).

Definition 2.5. [13] Let Abe a non-empty set and

s≥1.A function pb:A × A → R+

0is a partial b-

metric with a coefficient sif the following conditions

hold for all α, δ, υ ∈ A

(pb1)α=δif and only if pb(α, α) = pb(α, δ) =

pb(δ, δ);

(pb2)pb(α, α)≤pb(α, δ);

(pb3)pb(α, δ) = pb(δ, α);

(pb4)pb(α, δ)≤s(pb(α, υ) + pb(υ, δ)) −pb(υ, υ).

Improvements to the fixed point results by the use of a simulation

function employing rational terms

1SUNISA SAIUPARAD, 1KANIKAR MUANGCHOO, 1SUKJIT TANGCHAROEN,

1PHANNIKA MEE-ON, 1SAKULBUTH EKVITTAYANIPHON, 2DUANGKAMON KITKUAN

1Department of Mathematics and Statistics, Faculty of Science and Technology, Rajamangala

University of Technology Phra Nakhon (RMUTP), 1381 Pracharat 1 Road, Wongsawang, Bang Sue, Bangkok

10800, THAILAND

2Department of Mathematics, Faculty of Science and Technology, Rambhai Barni Rajabhat University,

Chanthaburi 22000, THAILAND

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In this case, we say that (A, pb, s)is a partial b-metric

space.

Lemma 2.6. [14] Let (A, pb, s)be a partial b-metric

space. If pb(α, δ)=0then α=δand pb(α, δ)>0

for all α=δ.

Definition 2.7. [15] A sequence {αλ}on a par-

tial b-metric space (A, pb, s)is 0-pb-Cauchy if

limλ,ρ→∞ pb(αλ, αρ) = 0.Moreover, the space

(A, pb, s)is said to be 0-pb-complete if for each 0-

pb-Cauchy sequence in A,there is σ∈ A,such that

lim

λ,ρ→∞

pb(αλ, αρ) = lim

λ→∞

pb(αλ, σ) = pb(σ, σ) = 0.

Lemma 2.8. [15] If the partial b-metric space

(A, pb, s)is pb-complete, then it is 0-pb-complete.

Lemma 2.9. [16] Let (A, pb, s ≥1) be a partial b-

metric space, F:A→Aa mapping and a number

γ∈[0,1).If {αλ}is a sequence in A,where αλ=

Fαλ−1and

pb(αλ, αλ+1)≤γpb(αλ−1, αλ),

for each λ≥1,then the sequence {αλ}is 0-pb-

Cauchy.

Let Υbe the set of all non-decreasing and con-

tinuous functions ψ: [0,∞)→[0,∞)such that

ψ(0) = 0.

Definition 2.10. [17] A function η:R+

0×R+

0→Ris

aψ-simulation function if there exists ψ∈such that

the following conditions hold:

(η1)η(ζ, ξ)< ψ(ξ)−ψ(ζ)for all ζ, ξ ∈R+;

(η2)if {ζλ},{ξλ}are two sequences in [0,∞)such

that limλ→∞ ζλ=limλ→∞ ξλ>0,then

lim sup

λ→∞

η(ζλ, ξλ)<0.

Denote by Zψthe family of all ψ-simulation functions

(see [18, 19, 20, 21, 22]). It is clear, due to the axiom

(η1),that

η(ζ, ζ)<0for all ζ > 0.

3 Main Results

Definition 3.1. Let (A, pb, s ≥1) be a partial b-

metric space. A mapping F:A → A is called η-

rational contraction of type A, if there exists a func-

tion η∈ Zψsuch that

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

η(stpb(Fα, Fδ),DA(α, δ)≥0,

(1)

for every α, δ ∈ A,where DAis defined as

DA(α, δ)

=max pb(α, δ), pb(α, Fα), pb(δ, Fδ),

pb(δ, Fδ)(1 + pb(α, Fα))

1 + pb(α, δ),

pb(δ, Fδ)(1 + pb(α, Fα))

1 + pb(Fα, Fδ),

pb(α, Fδ) + pb(δ, Fα)

2s.

(2)

Theorem 3.2. Let (A, pb, s > 1) be a pb-complete

partial b-metric space and F:A → A be a η-

rational contraction of type A. Then Fadmits exactly

one fixed point.

Proof. Let α0∈ A be an arbitrary but fixed point and

{αλ}be the sequence defined in Aas follows:

αλ=Fαλ−1,∀λ≥1.(3)

Suppose that αλ−1=αλfor every τ≥1.Indeed, if

we assume that there exists λ0∈Nsuch that αλ0−1=

αλ0.Taking (3) into consideration, we get αλ0−1=

Fαλ0−1., that is, αλ0−1is a fixed point of F. Hence,

substituting α=αλ−1and δ=αλin (2), we obtain

DA(αλ−1, αλ)

=max pb(αλ−1, αλ), pb(αλ−1,Fαλ−1), pb(αλ,Fαλ),

pb(αλ,Fαλ)(1 + pb(αλ−1,Fαλ−1))

1 + pb(αλ−1, αλ),

pb(αλ,Fαλ)(1 + pb(αλ−1,Fαλ−1))

1 + pb(Fαλ−1,Fαλ),

pb((αλ−1,Fαλ) + pb(αλ,Fαλ−1)

2s

=max pb(αλ−1, αλ), pb(αλ−1, αλ), pb(αλ, αλ+1),

pb(αλ, αλ+1)(1 + pb(αλ−1, αλ))

1 + pb(αλ−1, αλ),

pb(αλ, αλ+1)(1 + pb(αλ−1, αλ))

1 + pb(αλ, αλ+1),

pb(αλ−1, αλ+1) + pb(αλ, αλ)

2s

≤max pb(αλ−1, αλ), pb(αλ−1, αλ), pb(αλ, αλ+1),

pb(αλ, αλ+1)(1 + pb(αλ−1, αλ))

1 + pb(αλ−1, αλ),

pb(αλ, αλ+1)(1 + pb(αλ−1, αλ))

1 + pb(αλ, αλ+1),

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s(pb(αλ−1, αλ) + pb(αλ, αλ+1))

−pb(αλ, αλ) + pb(αλ, αλ)

2s

=max {pb(αλ−1, αλ), pb(αλ, αλ+1)}

(4)

Furthermore, by (1), we get

2smin {pb(αλ−1,Fαλ−1), pb(αλ,Fαλ)}

2smin {pb(αλ−1, αλ), pb(αλ, αλ+1)}

≤pb(αλ−1, αλ)

(5)

implies

η(stpb(Fαλ−1,Fαλ),DA(αλ−1, αλ)) ≥0,(6)

Taking (η1)into account, the preceding inequality

provides

0< ψ(DA(αλ−1, αλ)) −ψ(stpb(Fαλ−1,Fαλ)),

(7)

or, equivalently,

ψ(stpb(αλ, αλ+1))

< ψ(DA(αλ−1, αλ))

=ψ(max {pb(αλ−1, αλ), pb(αλ, αλ+1)}).

(8)

As a result of the monotony of the function ψ, we ob-

tain

stpb(αλ, αλ+1)<max {pb(αλ−1, αλ), pb(αλ, αλ+1)}.

(9)

If there exists λ1∈Nsuch that

max{pb(αλ1−1, αλ1), pb(αλ1, αλ1+1)}=pb(αλ1, αλ1+1),

(9) becomes stpb(αλ1, αλ1+1)< pb(αλ1, αλ1+1),

which is a contradiction (because s > 1). Hence, for

any λ∈Nwe obtain

stpb(αλ, αλ+1)< pb(αλ−1, αλ),

pb(αλ, αλ+1)<1

stpb(αλ−1, αλ).(10)

Denoting 1

stby µ, we have

pb(αλ, αλ + 1) < µpb(αλ−1, αλ),

with 0≤µ < 1.Using Lemma 2.9, we see that the

sequence {αρ}is a 0-pb-Cauchy sequence on the pb-

complete partial b-metric space. Since by Lemma 2.8,

the space is also 0-pb-complete, it follows that there

exists σ∈ A such that

lim

λ,ρ→∞

pb(αλ, αρ) = lim

λ→∞

pb(αλ, σ) = pb(σ, σ) = 0.

(11)

We now assert that

2spb(αλ, αλ+1)≤pb(αλ, σ)or

2spb(αλ+1, αλ+2)≤pb(αλ+1, σ).

On the other hand, assuming the opposite, we can ob-

tain λ0∈Nsuch that

pb(αλ0, αλ0+1)

≤s(pb(αλ0, σ) + pb(σ, αλ0+1)) −pb(σ, σ)

< s 1

2spb(αλ0, αλ0+1) + 1

2spb(αλ0+1, αλ0+2)

2(pb(αλ0, αλ0+1) + pb(αλ0+1, αλ0+2))

< pb(αλ0, αλ0+1),

which is a contradiction. Hence, there exists a subse-

quence {αλ(ι)}of {αλ}such that

2smin pb(αλ(ι),Fαλ(ι)), pb(σ, Fσ)

2spb(αλ(ι), αλ(ι)+1)

≤pb(αλ(ι), σ)

implies

η(stpb(Fαλ(ι),Fσ),DA(αλ(ι), σ)) ≥0,

where

pb(σ, Fσ)

≤ DA(αλ(ι), σ)

=max pb(αλ(ι), σ), pb(αλ(ι),Fαλ(ι)), pb(σ, Fσ),

pb(σ, Fσ)(1 + pb(αλ(ι),Fαλ(ι)))

1 + pb(αλ(ι), σ),

pb(σ, Fσ)(1 + pb(αλ(ι),Fαλ(ι)))

1 + pb(Fαλ(ι),Fσ),

pb(αλ(ι),Fσ) + pb(σ, Fαλ(ι))

2s

=max pb(αλ(ι), σ), pb(αλ(ι), αλ(ι)+1), pb(σ, Fσ),

pb(σ, Fσ)(1 + pb(αλ(ι), αλ(ι)+1))

1 + pb(αλ(ι), σ),

pb(σ, Fσ)(1 + pb(αλ(ι), αλ(ι)+1))

1 + pb(αλ(ι)+1,Fσ),

pb(αλ(ι),Fσ) + pb(σ, αλ(ι)+1)

2s.

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Taking ι→ ∞ and using (11) in mind, we arrive at

lim

ι→∞ DA(αλ(ι), σ) = pb(σ, Fσ).(12)

On the one hand, we assume that αλ=σfor an infi-

nite number of λ∈Nwithout sacrificing generality.

So,

η(stpb(Fαλ,Fσ),DA(αλ, σ)) ≥0.

Thus, as a result of η1, leads us to

ψ(stpb(Fαλ,Fσ)) < ψ(DA(αλ, σ)).

Taking into account the fact that has a non-decreasing

property of ψ,

stpb(Fαλ,Fσ)<DA(αλ, σ).

On the alternative,

pb(σ, Fσ)

≤s(pb(σ, Fαλ) + pb(Fαλ,Fσ)) −pb(Fαλ,Fαλ)

≤spb(σ, Fαλ) + stpb(Fαλ,Fσ)−pb(αλ+1, αλ+1)

< spb(σ, Fαλ) + DA(αλ, σ).

Taking λ→ ∞ in the above inequality and using (11)

and (12), we get

pb(σ, Fσ)≤stlim

λ→∞

pb(Fαλ,Fσ)

<lim

λ→∞

DA(αλ, σ)

=pb(σ, Fσ).

Hence, stlimλ→∞ pb(Fαλ,Fσ) = pb(σ, Fσ).

Therefore, putting ζλ=pb(Fαλ,Fσ)

and ξλ=DA(αλ, σ),using η2it follows

lim supλ→∞ η(stζλ, ξλ)<0,which is a contra-

diction. Then pb(σ, Fσ)=0=pb(σ, σ),that is, σis

a fixed point of F.

Finally, we establish uniqueness of the fixed point.

Indeed, if we can find another point, υ∈ A, υ =σ

such that υ=Fυ,

0 = 1

2smin{pb(υ, Fυ), pb(σ, Fσ)} ≤ pb(υ, σ),

implies

0≤η(stpb(Fυ, Fσ),DA(υ, σ))

< ψ(DA(υ, σ)) −ψ(stpb(Fυ, Fσ))

=ψ(pb(υ, σ)) −ψ(stpb(υ, σ)),

which is a contradiction. Hence, σ=υ.

Corollary 3.3. Let F:A → A be a mapping on

apb-complete partial b-metric space (A, pb, s > 1).

Suppose that ψ∈Υand ϕ: [0,∞)→[0,∞)is a

function such that lim infξ→ξ0ϕ(ξ)>0,for ξ0>0

and ϕ(ξ) = 0 ⇔ξ= 0.If for every ζ, ξ ∈ A

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

ψ(stpb(Fα, Fδ)) ≤ψ(DA(α, δ)) −ϕ(DA(α, δ)).

Then Fadmits a unique fixed point.

Proof. Taking η(ζ, ξ) = ψ(ξ)−ϕ(ξ)−ψ(ζ)in The-

orem 3.2.

Corollary 3.4. Let F:A → A be a mapping on

apb-complete partial b-metric space (A, pb, s > 1).

Suppose that ψ∈Υand φ: [0,∞)→[0,1) is a

function such that lim supξ→ξ0φ(ξ)<1,for ξ0>0

and φ(ξ) = 0 ⇔ξ= 0.If for every ζ, ξ ∈ A

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

ψ(stpb(Fα, Fδ)) ≤φ(DA(α, δ))ψ(DA(α, δ)).

Then Fadmits a unique fixed point.

Proof. Taking η(ζ, ξ) = φ(ξ)ψ(ξ)−ψ(ζ)in Theo-

rem 3.2.

Definition 3.5. Let (A, pb, s > 1) be a partial b-

metric space. A mapping F:A → A is called η-

rational contraction of type B, if there exists a func-

tion η∈ Zψsuch that

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

η(stpb(Fα, Fδ),DB(α, δ)≥0,

(13)

for every α, δ ∈ A,where DBis defined as

DB(α, δ)

=max pb(α, δ), pb(α, Fα), pb(δ, Fδ),

pb(δ, Fδ)pb(α, Fα)

1 + pb(α, δ),pb(δ, Fδ)pb(α, Fα)

1 + pb(Fα, Fδ),

pb(α, Fα) + pb(δ, Fδ)

2s.

(14)

Theorem 3.6. Let (A, pb, s > 1) be a pb-complete

partial b-metric space and F:A → A be a η-

rational contraction of type B. Then Fadmits exactly

one fixed point.

Proof. Let the sequence {αλ}be defined by (3). Be-

cause αλ−1=αλ,for each λ∈N, using logic similar

to that used to prove Theorem 3.2, we have

2smin {pb(αλ,Fαλ), pb(αλ+1,Fαλ+1)}

2smin {pb(αλ, αλ+1), pb(αλ+1, αλ+2)}

≤pb(αλ, αλ+1)

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implies

0≤η(stpb(Fαλ,Fαλ+1),DB(αλ, αλ+1))

< ψ(DB(αλ, αλ+1)) −ψ(stpb(Fαλ,Fαλ+1)),

(15)

where

DB(αλ, αλ+1)

=max pb(αλ, αλ+1), pb(αλ,Fαλ), pb(αλ+1,Fαλ+1),

pb(αλ+1,Fαλ+1)pb(αλ,Fαλ)

1 + pb(αλ, αλ+1),

pb(αλ+1,Fαλ+1)pb(αλ,Fαλ)

1 + pb(Fαλ,Fαλ+1),

pb(αλ,Fαλ) + pb(αλ+1,Fαλ+1)

2s

=max pb(αλ, αλ+1), pb(αλ, αλ+1), pb(αλ+1, αλ+2),

pb(αλ+1, αλ+2)pb(αλ, αλ+1)

1 + pb(αλ, αλ+1),

pb(αλ+1, αλ+2)pb(αλ, αλ+1)

1 + pb(αλ+1, αλ+2),

pb(αλ, αλ+1) + pb(αλ+1, αλ+2)

2s

=max{pb(αλ, αλ+1), pb(αλ+1, αλ+2)}.

Hence,

ψ(stpb(αλ+1, αλ+2))

< ψ(DB(αλ, αλ+1))

≤ψ(max{pb(αλ, αλ+1), pb(αλ+1, αλ+2)}.

Taking into account the fact that has a non-decreasing

property of ψ,

stpb(αλ+1, αλ+2)

<max{pb(αλ, αλ+1), pb(αλ+1, αλ+2)}.

If max{pb(αλ, αλ+1), pb(αλ+1, αλ+2)}=

pb(αλ+1, αλ+2), we get a contradiction, and then it

follows that

pb(αλ+1, αλ+2)<1

stpb(αλ, αλ+1).

Using Lemma 2.9, we conclude that {αλ}is a 0-pb-

Cauchy on a pb-complete b-partialmetric space, and

there exists σ∈ A such that limλ→∞ αλ=σ.

Taking into account the continuity of the mapping F,

we have

σ=lim

λ→∞

αλ+1 =lim

λ→∞

Flim

λ→∞

αλ=Fσ,

that is, σis a fixed point of the mapping F.

Finally, we establish uniqueness of the fixed point. In-

deed, if we can find another point, υ∈ A, υ =σsuch

that υ=Fυ,

0 = 1

2smin{pb(υ, Fυ), pb(σ, Fσ)} ≤ pb(υ, σ),

implies

0≤η(stpb(Fυ, Fσ),DB(υ, σ))

< ψ(DB(υ, σ)) −ψ(stpb(Fυ, Fσ))

=ψ(pb(υ, σ)) −ψ(stpb(υ, σ)),

which is a contradiction. Hence, pb(σ, υ)=0,that is

σ=υ.

Corollary 3.7. Let F:A → A be a continu-

ous mapping on a pb-complete partial b-metric space

(A, pb, s > 1).Suppose that ψ∈Υand ϕ: [0,∞)→

[0,∞)is a function such that lim infξ→ξ0ϕ(ξ)>0,

for ξ0>0and ϕ(ξ) = 0 ⇔ξ= 0.If for every

ζ, ξ ∈ A

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

ψ(stpb(Fα, Fδ)) ≤ψ(DB(α, δ)) −ϕ(DB(α, δ)).

Then Fadmits a unique fixed point.

Proof. Taking η(ζ, ξ) = ψ(ξ)−ϕ(ξ)−ψ(ζ)in The-

orem 3.6.

Corollary 3.8. Let F:A → A be a continu-

ous mapping on a pb-complete partial b-metric space

(A, pb, s > 1).Suppose that ψ∈Υand φ: [0,∞)→

[0,1) is a function such that lim supξ→ξ0φ(ξ)<1,

for ξ0>0and φ(ξ)=0⇔ξ= 0.If for every

ζ, ξ ∈ A

2smin {pb(α, Fα), pb(δ, Fδ)} ≤ pb(α, δ)implies

ψ(stpb(Fα, Fδ)) ≤φ(DB(α, δ))ψ(DB(α, δ)).

Then Fadmits a unique fixed point.

Proof. Taking η(ζ, ξ) = φ(ξ)ψ(ξ)−ψ(ζ)in Theo-

rem 3.6.

4 Illustrative example

Example 4.1. Let the set A={7,8,9,10}and ρb be

the partial b-metric on A(s= 2),where

pb(α, δ) = 0.000002,for α=δ= 10

|α−δ|2,otherwise.

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Define the mapping F:A → A,

Fα=7,for α∈ {7,8,9},

8,for α= 10.

We choose ϕ∈Υ, ϕ(ξ) = ξ

2and η(ζ, ξ) =

8ξ−ζ

It is easy to see that η∈ Zψ. We have

αFα pb(α, Fα)

7 7 0

8 7 1

9 7 4

10 8 4

and will take into account the following scenarios:

1. For α, δ ∈ {7,8,9},we have pb(Fα, Fδ) = 0,

and then

4min {pb(α, Fα), pb(δ, Fδ)} ≤ 1≤pb(α, δ)

implies

2pb(Fα, Fδ) = 0 ≤12

13DA(α, δ)≥0,

2. For α= 7, δ = 10 we have

pb(α, δ) = 9, pb(7,F7) = 0,

pb(10,F10) = pb(10,8) = 4,

pb(F7,F10) = pb(7,8) = 1

and then

4min {pb(7,F7), pb(10,F10)}= 0 ≤9 = pb(α, δ)

implies

2pb(F7,F10) = 2 ≤63

8=7

8pb(7,10).

3. For α= 8, δ = 10 we have

pb(α, δ) = 4,

pb(8,F8) = 1,

pb(10,F10) = pb(10,8) = 4,

pb(F8,F10) = pb(7,8) = 1

and then

4min {pb(8,F8), pb(10,F10)}=1

4<4 = pb(α, δ)

implies

2pb(F8,F10) = 2 ≤7

2=7

8pb(8,10).

4. For α= 9, δ = 10 we have

pb(α, δ) = 1,

pb(9,F9) = 4,

pb(10,F10) = pb(10,8) = 4,

pb(F9,F10) = pb(7,8) = 1

and then

4min {pb(9,F9), pb(10,F10)}= 1 = pb(α, δ)

implies

2pb(F9,F10) = 2

≤70

pb(9,F9)(1 + pb(10,F10)

1 + pb(F9,F10)

≤7

8DA(9,10).

Hence, the hypothesis of Theorem 3.2 are satisfied

and α= 10 is the fixed point of the mapping F.

Example 4.2. Let the set A= [0,1],and pb:A ×

A → [0,∞), pb(α, δ) = (max{α, δ})2be a partial b-

metric on A.Let the continuous mapping F:A → A

be defined by

Fα=





α2,for α∈0,2

5,

25 ,for α∈2

5,1,

and the functions ψ∈Υ, η ∈ Zψ,where ψ(ξ) = ξ

and η(ζ, ξ) =

25 ξ−ζ

We verify that Fis a η-rational contraction of type

1. For α, δ ∈[0,2

5],if α > δ, (the case α≤δis

similar), we have

pb(α, δ) = (max{α, δ})2=α2,

pb(α, Fα) = (max{α, α2)2=α2,

pb(δ, Fδ) = δ2,

pb(Fα, Fδ) = (max{α2, δ2})2=α4,

and then

4min {pb(α, Fα), pb(δ, Fδ)} ≤ 1

4≤α2≤pb(α, δ)

implies

2pb(Fα, Fδ) = 2α4≤8

25α2≤8

25DB(α, δ).

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DOI: 10.37394/23206.2022.21.54

Sunisa Saiuparad, Kanikar Muangchoo,

Sukjit Tangcharoen, Phannika Mee-On,

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2. For α, δ ∈(2/5,1],if α > δ, (the case α≤δis

similar), we have

pb(α, δ) = (max{α, δ})2=α2,

pb(α, Fα) = (max{α, 4

25)2=α2,

pb(δ, Fδ) = δ2,

pb(Fα, Fδ) = 16

625,

and then

4min {pb(α, Fα), pb(δ, Fδ)} ≤ 1

4α2≤α2≤pb(α, δ)

implies

2pb(Fα, Fδ) = 32

625 ≤8

25α2≤8

25DB(α, δ).

3. For α∈[0,2/5], δ ∈(2/5,1],we have

pb(α, δ) = (max{α, δ})2=δ2,

pb(α, Fα) = α2,

pb(δ, Fδ) = δ2,

pb(Fα, Fδ) = 16

625,

and then

4min {pb(α, Fα), pb(δ, Fδ)} ≤ 1

4δ2≤δ2≤pb(α, δ)

implies

2pb(Fα, Fδ) = 32

625

≤8

25δ2

25pb(δ, Fδ)

≤8

25DB(α, δ).

Hence, all the hypotheses of Theorem (3.6) are satis-

fied and α= 0 is the unique fixed point of F.

5 Conclusion

The search for fixed point theorems involving con-

tractive type conditions has received much interest in

recent decades. In this context, we analyzed conver-

gence point results for such mappings and illustrative

for support theorem based on the new idea of fixed

point results by the use of a simulation function em-

ploying rational terms in partial b-metric space metric

spaces. The fresh ideas inspire more research and ap-

plications. It will be fascinating to apply these princi-

ples, for example, in metric spaces that do not involve

the entire form of triangle inequality, such as partial

order b-metric spaces.

6 Acknowledgments

The authors were financially supported by Raja-

mangala University of Technology Phra Nakhon

(RMUTP) Research Scholarship.

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Sukjit Tangcharoen, Phannika Mee-On,

Sakulbuth Ekvittayaniphon, Duangkamon Kitkuan

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DOI: 10.37394/23206.2022.21.54

Sunisa Saiuparad, Kanikar Muangchoo,

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Sakulbuth Ekvittayaniphon, Duangkamon Kitkuan

E-ISSN: 2224-2880

475

Volume 21, 2022