Citation: Samir Kumar Bhandari, Dhananjay Gopal, Pulak Konar. Probabilistic α-min Ciric type contraction results using a control function[J]. AIMS Mathematics, 2020, 5(2): 1186-1198. doi: 10.3934/math.2020082
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In this current paper, the probabilistic outcomes of Ciric contraction of
Being a control function, “altering distance function”, alters the distance between two points in a metric space and Khan, Swaleh and Sessa in 1984 showed us the property in their paper [17]. Some generalized works in this line may be referred as [16,18,19,21,22,24,25,28].
In recent time, the concept of altering distance function is extended to the context of Menger spaces in [6]. This control function is known as
Main features of this paper are following:
(1) A new probabilistic
(2) For such contraction, unique fixed point is obtained.
(3) The use of control function to prove the theorems.
(4) A corollary.
(5) Two illustrative examples validating our theorems.
(6) An application of our results on integral calculus.
(7) An important conclusion which may incur new problems.
Some important definitions and mathematical preliminaries are discussed before we want to prove our main results.
Definition 2.1. [15,26] A distribution function is a mapping
Definition 2.2.
(i)
(ii)
(iii)
(iv)
The examples of
(ⅰ)
(ⅱ)
Definition 2.3. Menger space [15,26] A triplet
(i)
(ii)
(iii)
(iv)
Definition 2.4. [15,26] A sequence
Fxn,x(ϵ)≥1−λ | (2.2) |
Definition 2.5. [15,26] A sequence
Fxn,xm(ϵ)≥1−λfor all m,n>Nϵ,λ. | (2.3) |
The equivalent of Definition 2.4 and 2.5 is to replace
Definition 2.6. [15,26] A Menger space
We use the following control function
Definition 2.7.
(i)
(ii)
(iii)
(iv)
In numerous research works, many authors [4,8,9,10,11] use this function.
We begin this section by introducing the concept of
Recent documents, such as [13,14] motivated us.
Definition 3.1. Let
α(x,y,t)(1Ffx,fy(ϕ(t))−1)≤min(1Fx,y(ϕ(tc))−1,1Fx,fx(ϕ(tc))−1,1Fy,fy(ϕ(tc))−1) | (3.1) |
for all
Definition 3.2. ([14]) Let
α(x,y,t)≥1⇒α(fx,fy,t)≥1 |
Theorem 3.1. Let
(i)
(ii) there exists
(iii) if
Then
Proof. Let
Then by using the fact
α(x0,fx0,t)=α(x0,x1,t)≥1⇒α(fx0,fx1,t)=α(x1,x2,t)≥1, |
and, by induction, we get
α(xn,xn+1,t)≥1,for all n∈N and for all t>0. |
From the properties of function
Now, we have from (3.1) for
1Fxn+1,xn(ϕ(t))−1=1Ffxn,fxn−1(ϕ(t))−1≤α(xn,xn−1,t)1Ffxn,fxn−1(ϕ(t))−1≤min(1Fxn,xn−1(ϕ(tc))−1,1Fxn,fxn(ϕ(tc))−1,1Fxn−1,fxn−1(ϕ(tc))−1)=min(1Fxn,xn−1(ϕ(tc))−1,1Fxn,xn+1(ϕ(tc))−1,1Fxn−1,xn(ϕ(tc))−1)=min(1Fxn+1,xn(ϕ(tc))−1,1Fxn,xn−1(ϕ(tc))−1). | (3.2) |
We now claim that for all
min(1Fxn+1,xn(ϕ(tc))−1,1Fxn,xn−1(ϕ(tc))−1)=1Fxn,xn−1(ϕ(tc))−1. | (3.3) |
If possible, let for some
min(1Fxn+1,xn(ϕ(sc))−1,1Fxn,xn−1(ϕ(sc))−1)=1Fxn+1,xn(ϕ(sc))−1, |
then we have from (3.2),
1Fxn+1,xn(ϕ(s))−1≤1Fxn+1,xn(ϕ(sc))−1, |
that is,
Fxn+1,xn(ϕ(s))≥Fxn+1,xn(ϕ(sc)), | (3.4) |
which is impossible as for
Then, for all
that is,
Fxn+1,xn(ϕ(t))≥Fxn,xn−1(ϕ(tc))≥Fxn−1,xn−2(ϕ(tc2))≥..............................≥Fx1,x0(ϕ(tcn)), |
Therefore,
Fxn+1,xn(ϕ(t))≥Fx1,x0(ϕ(tcn)). | (3.5) |
Now, taking limit as
limn→∞Fxn+1,xn(ϕ(t))=1. | (3.6) |
Now, we prove that
On the contrary, there exist
Fxm(k),xn(k)(ϵ)<1−λ. | (3.7) |
We take
Fxm(k)−1,xn(k)(ϵ)≥1−λ. | (3.8) |
If
Fxm(k),xn(k)(ϵ1)≤Fxm(k),xn(k)(ϵ). |
So, it is feasible to construct
Then, by the above logic, it is possible to get an increasing sequence of integers
Fxm(k),xn(k)(ϕ(ϵ2))<1−λ, | (3.9) |
and
Fxm(k)−1,xn(k)(ϕ(ϵ2))≥1−λ. | (3.10) |
Now, from (3.9), we get
1−λ>Fxm(k),xn(k)(ϕ(ϵ2)), |
that is,
11−λ<1Fxm(k),xn(k)(ϕ(ϵ2)), |
that is,
11−λ−1<1Fxm(k),xn(k)(ϕ(ϵ2))−1, |
which implies,
λ1−λ<1Fxm(k),xn(k)(ϕ(ϵ2))−1, ≤α(xm(k)−1,xn(k)−1,t)(1Ffxm(k)−1,fxn(k)−1(Φ(ϵ2))−1), ≤min(1Fxm(k)−1,xn(k)−1(ϕ(ϵ2c))−1,1Fxm(k)−1,xm(k)(ϕ(ϵ2c))−1,1Fxn(k)−1,xn(k)(ϕ(ϵ2c))−1) (using the inequality (3.1)) | (3.11) |
Now, using the property of (ⅳ) of the Menger space, we have
Fxm(k)−1,xn(k)−1(ϕ(ϵ2c))≥Δ(Fxm(k)−1,xn(k)(ϕ(ϵ2)),Fxn(k),xn(k)−1(ϕ(ϵ2c))−ϕ(ϵ2))≥Δ(1−λ,1−λ)(using (3.6) and (3.10))=1−λ, |
that is,
1Fxm(k)−1,xn(k)−1(ϕ(ϵ2c))−1≤11−λ−1=λ1−λ. | (3.12) |
Now, using (3.6), for sufficiently large
Fxm(k)−1,xm(k)(ϕ(ϵ2c))≥1−λ, |
1Fxm(k)−1,xm(k)(ϕ(ϵ2c))−1≤11−λ−1=λ1−λ. | (3.13) |
Fxn(k)−1,xn(k)(ϕ(ϵ2c))≥1−λ, |
that is,
1Fxn(k)−1,xn(k)(ϕ(ϵ2c))−1≤11−λ−1=λ1−λ. | (3.14) |
Now using (3.12), (3.13) and (3.14) in (3.11), we have
λ1−λ<min(1Fxm(k)−1,xn(k)−1(ϕ(ϵ2c))−1,1Fxm(k)−1,xm(k)(ϕ(ϵ2c))−1,1Fxn(k)−1,xn(k)(ϕ(ϵ2c))−1)≤min(λ1−λ,λ1−λ,λ1−λ)=λ1−λ, |
that is,
λ1−λ<λ1−λ, |
which is a contradiction.
Hence
Since
Ffu,u(ϵ)≥Δ(Ffu,xn+1(ϵ2),Fxn+1,u(ϵ2)). | (3.15) |
Next, using the properties of function
1Fxn+1,fu(ϵ2)−1≤1Ffxn,fu(ϕ(t2))−1≤α(xn,u,t)(1Ffxn,fu(ϕ(t2))−1)≤min(1Fxn,u(ϕ(t2c))−1,1Fxn,fxn(ϕ(t2c))−1,1Fu,fu(ϕ(t2c))−1)=min(1Fxn,u(ϕ(t2c))−1,1Fxn,xn+1(ϕ(t2c))−1,1Fu,fu(ϕ(t2c))−1). |
Taking limit
1Fu,fu(ϕ(t2))−1≤min(0,0,1Fu,fu(ϕ(t2c))−1)=0⇒1Fu,fu(ϕ(t2))≤1⇒Fu,fu(ϕ(t2))≥1. |
Thus,
The uniqueness of the fixed point is established next. Let
1Ffx,fy(ϕ(s))−1≤α(x,y,t)(1Ffx,fy(ϕ(s))−1)≤min(1Fx,y(ϕ(sc))−1,1Fx,fx(ϕ(sc))−1,1Fy,fy(ϕ(sc))−1)=min(1Fx,y(ϕ(sc))−1,1Fx,x(ϕ(sc))−1,1Fy,y(ϕ(sc))−1)=min(1Fx,y(ϕ(sc))−1,0,0)=0, |
which implies,
1Ffx,fy(ϕ(s))−1≤0,⇒Ffx,fy(ϕ(s))≥1, |
that is,
Fx,y(ϕ(s))=1. |
Hence
If we replace
Corollary 3.1. Let
1Ffx,fy(t)−1≤min(1Fx,y(tc)−1,1Fx,fx(tc)−1,1Fy,fy(tc)−1) | (3.16) |
where
Example 4.1. Let
Fx,y(t)=tt+|x−y|. |
Then
fx=x6 for all x∈[0,1], |
then the mapping
Example 4.2. Let
Fβ,γ(t)=Fγ,α(t)={0,ift≤0,0.75,if0<t≤2,1,ift>2, |
and Fα,β(t)={0,ift≤0,1,ift>0, |
Then
Some recent references [5,14,23] help us to establish the following application.
We consider the following boundary value problem of second order differential equation :
−d2xdt2=g(t, x(t)),t∈[0,1] |
x(0)=x(1)=0, |
where
x′′=0⇒D2x=0 | (4.1) |
and boundary values are
The auxiliary equation is
D2=0. |
Therefore, the general solution is
x(t)=At+B. |
Now, The Green's function
G(t,s)= {a1t+a2,0≤t<sb1t+b2,s<t≤1 |
The Green's function must satisfy the following three properties:
ⅰ)
i.e.,
b1s+b2=a1s+a2⇒s(b1−a1)+b2−a2=0 | (4.2) |
ⅱ) The determination of
i.e.,
(∂G∂t)t=s+0−(∂G∂t)t=s−0=−1⇒b1−a1=−1 | (4.3) |
ⅲ)
G(0,s)=0⇒a2=0 | (4.4) |
and
G(1,s)=0⇒b1+b2=0. | (4.5) |
Therefore,G(t,s)= {t(1−s),0≤t≤s≤1−st+s,0≤s≤t≤1 |
Let
d(x,y)=‖x−y‖∞=maxt∈I|x(t)−y(t)| |
is a complete metric space.
We have to show that the above mentioned differential equation satisfies the following inequality,
α(x,y,t)(1Ffx,fy(ϕ(t))−1)≤min(1Fx,y(ϕ(tc))−1,1Fx,fx(ϕ(tc))−1,1Fy,fy(ϕ(tc))−1) |
taking
we have
1Ffx,fy(ϕ(t))−1≤min(1Fx,y(ϕ(tc))−1,1Fx,fx(ϕ(tc))−1,1Fy,fy(ϕ(tc))−1). |
Taking
that is,
1tt+d(fx,fy)−1≤min(1tctc+d(x,y)−1,1tctc+d(x,fx)−1,1tctc+d(y,fy)−1), |
that is,
t+d(fx,fy)t−1≤min(tc+d(x,y)tc−1,tc+d(x,fx)tc−1,tc+d(y,fy)tc−1), |
that is,
d(fx,fy)t≤min(tc+d(x,y)−tctc,tc+d(x,fx)−tctc,tc+d(y,fy)−tctc), |
that is,
d(fx,fy)t≤min(cd(x,y)t,cd(x,fx)t,cd(y,fy)t),fort≠0 |
that is,
d(fx,fy)≤minc(d(x,y),d(x,fx),d(y,fy)). |
We have
|g(t,a)−g(t,b)|≤cmin{|x(s)−y(s)|,|x(s)−fx(s)|,|y(s)−fy(s)|}. |
Now, It is well known that
x(t)=∫10G(t,s)g(s,x(s))ds,for allt∈I. | (4.6) |
Define the operator
f(x(t))=∫10G(t,s)g(s,x(s))ds,for allt∈I. |
To find
So,
|f(x(t))−f(y(t))|=|∫10G(t,s)[g(s,x(s))−g(s,y(s))]ds|≤∫10G(t,s)|g(s,x(s))−g(s,y(s))|ds≤∫10G(t,s)c⋅min{d(x,y),d(x,fx),d(y,fy))}ds=c⋅min{d(x, y),d(x, fx), d(y, fy)}∫10G(t,s)ds≤c⋅min{d(x,y),d(x,fx),d(y,fy))}×18=0. |
Note that for all
∫10G(t,s)ds=−t22+t2, |
which implies that,
supt∈I∫10G(t,s)ds=18. |
Also,
min{d(x,y),d(x,fx),d(y,fy)}=min{d(x,y),0,0}=0. |
implies
d(fx,fy)=min{d(x,y),d(x,fx),d(y,fy)},for allx,y∈C([0,1],R). |
Therefore by Theorem 3.1 with
In the course of mathematical analysis and allied stream related to it, probabilistic metric spaces has an important role. The structural theory was created primarily after 1960. Many researchers have taken their interest in this area of research. Some authors have recently demonstrated that PM spaces are also applicable in nuclear fusion. One of the references may be noted as [29]. This paper [29] outlines the application to identify regimes of containment and disruption of plasma.
The authors declare no conflict of interest.
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