# Multivalued Problems, Orthogonal Mappings, and Fractional Integro-Differential Equation

Authors: R. K. Sharma; Sumit Chandok
Journal: Journal of Mathematics
 (2020)

---

## Abstract

In this manuscript, we propose some sufficient conditions for the existence of solution for the multivalued orthogonalℱ-contraction mappings in the framework of orthogonal metric spaces. As a consequence of results, we obtain some interesting results. Also as application of the results obtained, we investigate Ulam’s stability of fixed point problem and present a solution for the Caputo-type nonlinear fractional integro-differential equation. An example is also provided to illustrate the usability of the obtained results.


---

## Body

## 1. Introduction and Preliminaries

The theory of multivalued mappings has an important role in mathematics and allied sciences because of its many applications, for instance, in real and complex analysis as well as in optimal control problems. Over the years, this theory has increased its significance, and hence in the literature, there are many papers focusing on the discussion of abstract and practical problems involving multivalued mappings. As a matter of fact, amongst the various approaches utilized to develop this theory, one of the most interesting approaches is based on methods of fixed point theory.Acknowledging the work of Nadler [1], Gordji et al. [2], and Wardowski [3–5], the aim of this paper is to introduce the notion of multivalued orthogonal ℱ-contraction mappings in the framework of orthogonal metric space and to establish some sufficient conditions for the existence of fixed points for such class of mappings. Many researchers [6–11] proved the existence of fixed points using the concept of ℱ-contraction introduced by Wardowski [3–5]. In 1974, Reich [12, 13] asked whether we can take into account nonempty closed and bounded set instead of nonempty compact set. Although a lot of fixed point theorists studied this problem, it has not been completely solved. There are some partial affirmative answers to this problem, for instance, Mizoguchi et al. [14] and Olgun et al. [15]. We provide a partial solution to Reich’s original problem using multivalued orthogonal ℱ-contraction mappings in the setting of orthogonal metric spaces. Also, as application of the interesting and new results obtained, we investigate Ulam’s stability of fixed point problem and present a solution for a Caputo-type nonlinear fractional integro-differential equation.Recently, Gordji et al. [2] introduced the concept of an orthogonal set (briefly, O-set) and presented some fixed point theorems in orthogonal metric spaces.Definition 1.
LetX≠∅ and ⊥⊂X×X be a binary relation. If ⊥ satisfies the following condition: there exists x0∈X such that for all y∈X, y⊥x0, or for all y∈X, x0⊥y, then it is called an orthogonal set (briefly O-set). We denote this O-set by X,⊥.Example 1.
LetX=ℤ. Define m⊥n if there exists k∈ℤ such that m=kn. It is easy to see that 0⊥n for all n∈ℤ. Hence, X,⊥ is an O-set [2].Example 2.
LetX,d be a metric space and T:X⟶X be a Picard operator, that is, T has a unique fixed point x∗∈X and limn⟶+∞Tny=x∗ for all y∈X. We define the binary relation ⊥ on X by x⊥y if(1)limn⟶∞dx,Tny=0.
Then,X,⊥ is an O-set [2].Example 3.
LetX be an inner product space with the inner product .,.. Define the binary relation ⊥ on X by x⊥y if x,y=0. It is easy to see that 0⊥x for all x∈X. Hence, X,⊥ is an O-set [2].
For more interesting examples for an O-set, see [2].Definition 2.
LetX,⊥ be an O-set. A sequence xnn∈ℕ is called an orthogonal sequence (briefly, O-sequence) if for all n,xn⊥xn+1, or for all n,xn+1⊥xn.Definition 3.
LetX,⊥,d be an orthogonal metric space (X,⊥ is an O-set, and X,d is a metric space). Then T:X⟶X is said to be orthogonally continuous (or ⊥-continuous) at a∈X if, for each O-sequence an in X with an⟶a, we have Tan⟶Ta. Also, T is said to be ⊥-continuous on X if T is ⊥-continuous for each a∈X.
It is easy to see that every continuous mapping is⊥-continuous, but the converse is not true [2].Definition 4.
LetX,⊥,d be an orthogonal set with the metric d. Then X is said to be orthogonally complete (briefly, O-complete) if every Cauchy O-sequence is convergent.
It is easy to see that every complete metric space is O-complete, but the converse is not true [2].Definition 5.
LetX,⊥ be an O-set. A mapping T:X⟶X is said to be ⊥-preserving if Tx⊥Ty, whenever x⊥y. Also, T:X⟶X is said to be weakly ⊥-preserving if Tx⊥Ty or Ty⊥Tx, whenever x⊥y.
It is easy to see that every⊥-preserving mapping is weakly ⊥-preserving. But the converse is not true [2].Definition 6.
(see [3, 5]). Let F:0,+∞⟶ℝ be a mapping satisfying the following:(F1) For alla,b>0, a>b implies Fa>Fb(F2) For every sequencean in ℝ+, we have limn⟶+∞an=0 if and only if limn⟶+∞Fan=−∞(F3) There exists a numberk∈0,1 such that lima⟶0+akFa=0Iflimt⟶0+Ft=−∞, then using (F1), we have Ftn⟶−∞⇒tn⟶0 [5, 11].Inspired by the work of Wardowski [3–5], we denote ℱ be the family of all the functions F:0,+∞⟶ℝ satisfying (F1) and (F3)We denoteℱ1 be the family of all the functions F:0,+∞⟶ℝ satisfying (F1), (F3), and (F4) FinfA=infFA for all A⊂0,∞ with infA>0
Here,limc⟶d−Fc=Fd−0=limε⟶0+Fd−ε (left limit at d) and limc⟶d+Fc=Fd+0=limε⟶0+Fd+ε (right limit at d) for all d∈0,+∞. From mathematical analysis, the following is true for all d∈0,+∞:(2)Fd−0≤Fd≤Fd+0.Example 4.
Let functionsF1,F2,F3:0,+∞⟶ℝ defined as follows:(1)
F1a=−1/a, for all a>0.(2)
F2a=lna, for all a>0.(3)
F3a=a+lna, for all a>0.
ThenF1,F2,F3∈ℱ.
LetX,d be a metric space and H be a Hausdorff–Pompeiu metric induced by metric d on a set X. Denote PX the family of all nonempty subsets of X, CℬX the family of all nonempty, and closed and bounded subsets of X and KX the family of all nonempty compact subsets of X. H:CℬX×CℬX⟶ℝ defined by, for every A,B∈CℬX:(3)HA,B=maxsupx∈Adx,B,supy∈Bdy,A,where dx,A=infdx,y:y∈A.


## 2. Multivalued Results

In this section, we establish some results on the existence of fixed point for weak orthogonal multivalued contraction mappings using conditions of Wardowski [3–5].Now, we define the following orthogonal relation between two nonempty subsets of an orthogonal set.Definition 7.
LetA and B be two nonempty subsets of an orthogonal set X,⊥. The set A is orthogonal to set B is denoted by ⊥1 and defined as follows: A⊥1B, if for every a∈A and b∈B, a⊥b.
It is easy to observe the following results.Lemma 1.
LetX,⊥,d be an orthogonal metric space, x∈X and A∈KX. Then there exists a∈A such that dx,A=dx,a.Lemma 2.
LetX,⊥,d be an orthogonal metric space, and A,B∈KX, a∈A. Then there exists b∈B such that da,b≤HA,B.
Now, we are ready to present our first result.Theorem 1.
LetX,⊥,d be an O-complete orthogonal metric space and T:X⟶KX be a multivalued mapping on X. Assume that the following conditions are satisfied:(i)
There existsx0∈X such that x0⊥1Tx0 or Tx0⊥1x0(ii)
For allx,y∈X, x⊥y implies Tx⊥1Ty(iii)
Ifxn is an orthogonal sequence in X such that xn⟶x∗∈X, then xn⊥x∗ or x∗⊥xn for all n∈ℕ(iv)
IfF∈ℱ, there exists τ>0 such that for all x,y∈X with x⊥y satisfying the following:(4)HTx,Ty>0,τ+FHTx,Ty≤Fdx,y.ThenT has at least a fixed point.Proof.
By assumption (i), there existsx1∈Tx0 such that x0⊥x1 or x1⊥x0. By assumption (ii), we get Tx0⊥1Tx1; that is, there exists x2∈Tx1 such that x1⊥x2 or x2⊥x1. If x1∈Tx1, then x1 is a fixed point of T. Suppose that x1∉Tx1. Since Tx1 is compact, dx1,Tx1>0. As dx1,Tx1≤HTx0,Tx1, using (F1), we have Fdx1,Tx1≤FHTx0,Tx1. Therefore, using (iv), we get(5)Fdx1,Tx1≤FHTx0,Tx1≤Fdx0,x1−τ.
Continuing this process inductively, we can construct an orthogonal sequencexn in X such that xn+1∈Txn, for all n∈ℕ∪0. Thus we have xn⊥xn+1 or xn+1⊥xn for all n∈ℕ∪0.
Ifxk∈Txk for some k∈ℕ∪0, then xk is a fixed point of T.
So, we may assume thatxn∉Txn for all n∈ℕ∪0. Since Txn is closed, we have dxn,Txn>0, for all n∈ℕ∪0. Also dxn,Txn≤HTxn−1,Txn. So using (F1), we have Fdxn,Txn≤FHTxn−1,Txn. Further from (iv) and for every n≥1, we have(6)Fdxn,Txn≤FHTxn−1,Txn≤Fdxn−1,xn−τ.
Hence from the strictly increasing property ofF, we get HTxn,Txn−1<dxn,xn−1. We know that xn+1∈Txn, dxn,xn+1=dxn,Txn≤HTxn−1,Txn<dxn−1,xn. Therefore, the sequence dxn+1,xn is strictly decreasing sequence. Suppose that tn=dxn+1,xn⟶t, for some t≥0.
Furthermore, for alln≥n0, we have(7)τ+Fdxn+1,xn≤τ+FHTxn,Txn−1≤Fdxn,xn−1.
Takingn⟶+∞ in (7), we get τ+Ft+0≤Ft+0, which is contradiction, and hence tn=dxn+1,xn⟶0. By (F3), there exists k∈0,1 such that(8)limn⟶+∞tnkFtn=0.
Using (6), we get(9)Ftn≤Ftn−1−τ≤Ftn−2−2τ≤…≤Ft0−nτ.
From (9), the following holds for all n∈ℕ:(10)tnkFtn−tnkFt0≤−tnknτ≤0.
Lettingn⟶∞ in (10), we get limn⟶+∞ntnk=0. Hence there exists n1∈ℕ such that ntnk≤1 for all n≥n1. So, we have all for all n≥n1:(11)tn≤1n1/k.
Now, we have to show thatxn is a Cauchy orthogonal sequence. Consider m,n∈ℕ such that m>n≥n1. Using the triangle inequality and (11), we have(12)dxn,xm≤dxn,xn+1+dxn+1,xn+2+⋯+dxm−1,xm,=tn+tn+1+⋯+tm−1=∑i=nm−1ti≤∑i=n+∞ti≤∑i=n+∞1i1/k.
By the convergence of series,∑i=n+∞1/i1/k, passing to limit n⟶+∞, we get dxn,xm⟶0.
This shows thatxn is a Cauchy orthogonal sequence. Since X is O-complete, there exists x∗∈X such that limn⟶+∞xn=x∗.
Now, we claim thatx∗∈Tx∗. Assume the contrary that x∗∉Tx∗. Hence there exists n1∈ℕ such that x∗∉xnn≥n1, HTxn,Tx∗>0. Therefore, further by our assumption, xn⊥x∗ or x∗⊥xn, and using (iv), we get(13)Fdxn+1,Tx∗≤τ+FHTxn,Tx∗≤Fdxn,x∗.
Now using strict increasing property ofF and τ>0, we get dxn+1,Tx∗<dxn,x∗. Taking n⟶+∞, we get x∗∈Tx∗¯=Tx∗. Hence, the result is obtained.
Here it should be noted that in Theorem1, Tx is compact for all x∈X. Now, we have the following result in which we give a partial answer to Reich’s problem for a closed and bounded set.Theorem 2.
LetX,⊥,d be an O-complete orthogonal metric space and T:X⟶CℬX be a multivalued mapping on X. Assume that the following conditions are satisfied:(i)
There existsx0∈X such that x0⊥1Tx0 or Tx0⊥1x0(ii)
For allx,y∈X, x⊥y implies Tx⊥1Ty(iii)
Ifxn is an orthogonal sequence in X such that xn⟶x∗∈X, then xn⊥x∗ or x∗⊥xn for all n∈ℕ.(iv)
IfF∈ℱ1, there exists τ>0 such that for all x,y∈X with x⊥y satisfying the following:(14)HTx,Ty>0,τ+FHTx,Ty≤Fdx,y.ThenT has at least a fixed point.Proof.
Letx0∈X. Since Tx is nonempty for all x∈X, by assumption (i), we can choose x1∈Tx0 such that x0⊥x1 or x1⊥x0. If x1∈Tx1, then x1 is a fixed point of T. Let x1∉Tx1. Then dx1,Tx1>0 since Tx1 is closed. Since dx1,Tx1≤HTx0,Tx1, then from (F1), we get(15)Fdx1,Tx1≤FHTx0,Tx1.
Using (iv), we get(16)Fdx1,Tx1≤FHTx0,Tx1≤Fdx0,x1−τ.
From (F4), we getFdx1,Tx1=infy∈Tx1Fdx1,y. So from (16), we have(17)Fdx1,Tx1=infy∈Tx1Fdx1,y≤FHTx0,Tx1,≤Fdx0,x1−τ,<Fdx0,x1−τ2.
By assumption (ii), we getTx0⊥1Tx1. Continuing this process, we construct an orthogonal sequence xn in X such that xn+1∈Txn for all n∈ℕ∪0. Thus we have xn⊥xn+1 or xn+1⊥xn for all n∈ℕ∪0.
Ifxk∈Txk for some k∈ℕ∪0, then xk is a fixed point of T, and so the proof is completed.
So, we may assume thatxn∉Txn for all n∈ℕ∪0. Since Txn is closed, we have dxn,Txn>0, for all n∈ℕ∪0. Also dxn,Txn≤HTxn−1,Txn, and from (F1), we get Fdxn,Txn≤FHTxn−1,Txn.
Furthermore, using (iv), we have(18)Fdxn,Txn≤FHTxn,Txn+1≤Fdxn,xn+1−τ<Fdxn,xn+1−τ2.
SinceFdxn,Txn=infy∈TxnFdxn,y. Therefore, using (18), we get(19)Fdxn,Txn=infy∈TxnFdxn,y≤FHTxn−1,Txn<Fdxn−1,xn−τ2.
So from (19), we can get a sequence xn in X such that there exists xn+1∈Txn and Fdxn,xn+1<Fdxn−1,xn for all n∈ℕ. Now, proceeding on the same lines of Theorem 1, we get the result.


## 3. Consequences

In this section, we give some interesting consequences of the results proved in the previous section.The following result is an immediate consequence of Theorem1.Corollary 1.
LetX,⊥,d be an O-complete orthogonal metric space and T:X⟶KX. Assume that the following conditions are satisfied:(i)
There existsx0∈X such that x0⊥1Tx0 or Tx0⊥1x0(ii)
For allx,y∈X, x⊥y implies Tx⊥1Ty(iii)
Ifxn is an orthogonal sequence in X such that xn⟶x∗∈X, then xn⊥x∗ or x∗⊥xn for all n∈ℕ.(iv)
There exists someτi>0, i=1,2,3 such that for all x,y∈X with x⊥y, HTx,Ty>0, either of the following contractive conditions hold:(20)τ1+HTx,Ty≤dx,y;τ2−1HTx,Ty≤−1dx,y;τ3+11−eHTx,Ty≤11−edx,y.ThenT has at least a fixed point in each of these cases.Proof.
As each functionsF1r=r, F2r=−1/r, and F3r=1/1−er, where r=dx,y>0, is strictly increasing on 0,+∞, so the proof immediately follows from Theorem 1.
As a consequence of Theorem1, we have the following result for single-valued mapping by replacing condition (iii) with T is ⊥-continuous.Corollary 2.
LetX,⊥,d be an O-complete orthogonal metric space and T:X⟶X. Assume that the following conditions are satisfied:(i)
There exists someτ>0, such that for all x,y∈X with x⊥y, dTx,Ty>0:(21)τ+FdTx,Ty≤Fdx,y,whereF∈ℱ.(ii)
There existsx0∈X such that x0⊥Tx0 or Tx0⊥x0.(iii)
For allx,y∈X, x⊥y implies Tx⊥Ty(iv)
T is ⊥-continuous
Then,T has a fixed point.Proof.
Here, we can chooseT as a multivalued mapping by considering Tx is a singleton set for every x∈X. Arguing on the same lines of Theorem 1, we consider xn is a Cauchy orthogonal sequence and limn⟶∞xn=x∗. As T is ⊥-continuous, we have(22)dx∗,Tx∗=limn⟶+∞dTxn,Tx∗=0,i.e., x∗ is a fixed point of T.
As a consequence of Corollary2, we have the following result by taking Fr=lnr, r>0.Corollary 3.
LetX,⊥,d be an O-complete orthogonal metric space and T:X⟶X. Assume that the following conditions are satisfied:(i)
There exists someτ>0, such that for all x,y∈X with x⊥y, dTx,Ty>0:(23)dTx,Ty≤e−τdx,y,whereF∈ℱ.(ii)
There existsx0∈X such that x0⊥Tx0 or Tx0⊥x0.(iii)
For allx,y∈X, x⊥y implies Tx⊥Ty.(iv)
T is ⊥-continuous.
ThenT has a fixed point.


## 4. Illustration

In this section, we illustrate an example which shows thatT is a multivalued orthogonal mapping and satisfies the condition (iv) of Theorem 1, but it is not multivalued orthogonal contraction (HTx,Ty≤kdx,y, for k∈0,1 with x⊥y).Example 5.
LetX=Sn=nn+1/2:n∈ℕ and d:X×X⟶0,∞ be a mapping defined by dx,y=x−y for all x,y∈X.
Define a relation⊥ on X by x⊥y if and only if xy∈x,y⊆X=Sn.
ThusX,⊥,d is an O-complete orthogonal metric space. Now, we define a mapping T:X⟶KX by(24)Tx=x1,x=x1,x1,…,xn−1,x=xn,n≥1.
We claim thatT is a multivalued orthogonal mapping satisfying condition (iv) of Theorem 1 with respect to Fα=α+lnα,α>0 and τ=1. To see this, we have the following cases.
First, we observe that for allm,n∈ℕ, HTx,Ty>0 if and only if m>2 and n=1 or m>n>1.Case 1. Form>2 and n=1, we have(25)HTxm,Tx1dxm,x1eHTxm,Tx1−dxm,x1=xm−1−x1xm−x1exm−1−x1=m2−m−2m2+m−2e−m<e−m<e−1.Case 2. Form>n>1, we get(26)HTxm,Txndxm,xneHTxm,Txn−dxm,xn=xm−1−xn−1xm−xnexm−1−xn−1−xm+xn=m+n−1m+n+1en−m<en−m≤e−1.
This shows thatT satisfies (iv) of Theorem 1. Hence, T has a fixed point.
On the contrary,T is not multivalued orthogonal contraction (HTx,Ty≤kdx,y, k∈0,1), as(27)limn⟶+∞HTxn,Tx1dxn,x1=limn⟶+∞xn−1−1xn−1=1.


## 5. Applications

In this section, we present the Ulam stability and solve a nonlinear fractional differential-type equation using Corollary3.


### 5.1. Ulam Stability

The Ulam [16, 17] stability has attracted attention of several authors in fixed point theory [18]. On orthogonal metric space X,⊥,d, T:X⟶X, we investigate the fixed point equation:(28)Tv=v,and the inequality (for ε>0):(29)dTx,x≤ε.Equation (28) is called the Ulam stable if it satisfies the following condition:(A) There is a constantδ>0, for each ε>0, and for every solution x∗ of the inequality (29), there is a solution v∗∈X for equation (28) such that(30)dv∗,x∗≤δε.Theorem 3.
Under the hypothesis of Corollary3, the fixed point equation (28) is Ulam stable.Proof.
On account of Corollary3, we guarantee a unique v∗∈X such that v∗=Tv∗, that is, v∗∈X forms a solution of (28). Let ε>0 and x∗∈X be an ε-solution, that is,(31)dTx∗,x∗≤ε.
We have(32)dv∗,x∗=dTv∗,x∗≤dTv∗,Tx∗+dTx∗,x∗≤e−τdv∗,x∗+ε.
Hence,dv∗,x∗≤1/1−e−τε=kε, where k=1/1−e−τ>0. Therefore, equation (28) is Ulam stable.


### 5.2. Application to Nonlinear Fractional Integro-Differential Equation

Here, we give a solution for a Caputo-type nonlinear fractional integro-differential equation. For more details on fractional calculus, see [19–25] and references cited therein.The Caputo derivative of a continuous mappingg:0,∞⟶ℝ (order δ>0) is given by(33)DCδgt≔1Γn−δ∫0tgnsdst−sδ−n+1,n−1≤δ<n,n=δ+1,where Γ represents the gamma function and δ refers to the integer part of the positive real number δ.In this section, we examine the nonlinear fractional integro-differential equation of the Caputo type:(34)DδCut=Gt,ut,t∈I=0,1,1<δ≤2,u0=0,u1=∫0θusds,where u∈C0,1,ℝ, θ∈0,1, and G:I×ℝ⟶ℝ is a continuous function (for more details, see [20]).We considerX=u:u∈C0,1,ℝ with supremum norm u=supt∈0,1ut. So X,. is a Banach space.The spaceX:=C0,1,ℝ endowed with the metric d:X×X⟶0,∞ defined as du,v=u−v=supt∈0,1tut−vt and define an orthogonal relation u⊥v if and only if uv≥0, for all u,v∈X. Then X,⊥,d is an orthogonal metric space.Clearly, a solution of equation (34) is a fixed point of the integral equation:(35)Tut=1Γδ∫0tt−sδ−1Gs,usds,−2t2−θ2Γδ∫011−sδ−1Gs,usds,+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds.Theorem 4.
Assume thatG:I×ℝ⟶ℝ is a continuous function satisfying(36)Gs,us−Gs,vs≤Γδ+15e−τus−vs,for each s∈0,1, for some τ>0 and for all u,v∈C0,1,ℝ. Then the fractional differential equation (34) with given boundary conditions has a solution.Proof. The space X:=C0,1,ℝ endowed with the metric d:X×X⟶0,∞ defined as du,v=supt∈0,1ut−vt, for all u,v∈X. Define an orthogonal relation u⊥v if and only if uv≥0, for all u,v∈X. Then X,⊥,d is an orthogonal metric space. Define T:X⟶X as in (35). So T is ⊥-continuous. First, we show that T is ⊥-preserving, let ut⊥vt for all t∈0,1. Now, from (35). we have(37)Tut=1Γδ∫0tt−sδ−1Gs,usds,−2t2−θ2Γδ∫011−sδ−1Gs,usds,+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds>0,which implies that Tu⊥Tv.Now, we have to show thatT satisfies (i) of Corollary 2 for Fr=lnr,r>0. For all t∈0,1, ut⊥vt, we have(38)Tut−Tvt=|1Γδ∫0tt−sδ−1Gs,usds−2t2−θ2Γδ∫011−sδ−1Gs,usds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds,−1Γδ∫0tt−sδ−1Gs,vsds−2t2−θ2Γδ∫011−sδ−1Gs,vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,vmdmds|,≤1Γδ∫0tt−sδ−1Gs,us−Gs,vsds−2t2−θ2Γδ∫011−sδ−1Gs,us−Gs,vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,um−Gs,vmdmds,≤1Γδ∫0tt−sδ−1Γδ+15e−τsups∈0,1us−vsds−2t2−θ2Γδ∫011−sδ−1Γδ+15e−τsups∈0,1us−vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Γδ+15e−τsups∈0,1us−vsdmds,≤Γδ+15e−τsups∈0,1us−vs×supt∈0,11Γδ∫0tt−sδ−1ds−2t2−θ2Γδ∫011−sδ−1ds+2t2−θ2Γδ∫0θ∫0ss−mδ−1dmds≤e−τsups∈0,1us−vs=e−τdu,v,for all u,v∈X. Therefore, the condition (i) of Corollary 2 holds. Accordingly, all axioms of Corollary 2 are verified, and T has a fixed point. The Caputo-type nonlinear fractional differential equation (34) possesses a solution is yielded.


## 5.1. Ulam Stability

The Ulam [16, 17] stability has attracted attention of several authors in fixed point theory [18]. On orthogonal metric space X,⊥,d, T:X⟶X, we investigate the fixed point equation:(28)Tv=v,and the inequality (for ε>0):(29)dTx,x≤ε.Equation (28) is called the Ulam stable if it satisfies the following condition:(A) There is a constantδ>0, for each ε>0, and for every solution x∗ of the inequality (29), there is a solution v∗∈X for equation (28) such that(30)dv∗,x∗≤δε.Theorem 3.
Under the hypothesis of Corollary3, the fixed point equation (28) is Ulam stable.Proof.
On account of Corollary3, we guarantee a unique v∗∈X such that v∗=Tv∗, that is, v∗∈X forms a solution of (28). Let ε>0 and x∗∈X be an ε-solution, that is,(31)dTx∗,x∗≤ε.
We have(32)dv∗,x∗=dTv∗,x∗≤dTv∗,Tx∗+dTx∗,x∗≤e−τdv∗,x∗+ε.
Hence,dv∗,x∗≤1/1−e−τε=kε, where k=1/1−e−τ>0. Therefore, equation (28) is Ulam stable.


## 5.2. Application to Nonlinear Fractional Integro-Differential Equation

Here, we give a solution for a Caputo-type nonlinear fractional integro-differential equation. For more details on fractional calculus, see [19–25] and references cited therein.The Caputo derivative of a continuous mappingg:0,∞⟶ℝ (order δ>0) is given by(33)DCδgt≔1Γn−δ∫0tgnsdst−sδ−n+1,n−1≤δ<n,n=δ+1,where Γ represents the gamma function and δ refers to the integer part of the positive real number δ.In this section, we examine the nonlinear fractional integro-differential equation of the Caputo type:(34)DδCut=Gt,ut,t∈I=0,1,1<δ≤2,u0=0,u1=∫0θusds,where u∈C0,1,ℝ, θ∈0,1, and G:I×ℝ⟶ℝ is a continuous function (for more details, see [20]).We considerX=u:u∈C0,1,ℝ with supremum norm u=supt∈0,1ut. So X,. is a Banach space.The spaceX:=C0,1,ℝ endowed with the metric d:X×X⟶0,∞ defined as du,v=u−v=supt∈0,1tut−vt and define an orthogonal relation u⊥v if and only if uv≥0, for all u,v∈X. Then X,⊥,d is an orthogonal metric space.Clearly, a solution of equation (34) is a fixed point of the integral equation:(35)Tut=1Γδ∫0tt−sδ−1Gs,usds,−2t2−θ2Γδ∫011−sδ−1Gs,usds,+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds.Theorem 4.
Assume thatG:I×ℝ⟶ℝ is a continuous function satisfying(36)Gs,us−Gs,vs≤Γδ+15e−τus−vs,for each s∈0,1, for some τ>0 and for all u,v∈C0,1,ℝ. Then the fractional differential equation (34) with given boundary conditions has a solution.Proof. The space X:=C0,1,ℝ endowed with the metric d:X×X⟶0,∞ defined as du,v=supt∈0,1ut−vt, for all u,v∈X. Define an orthogonal relation u⊥v if and only if uv≥0, for all u,v∈X. Then X,⊥,d is an orthogonal metric space. Define T:X⟶X as in (35). So T is ⊥-continuous. First, we show that T is ⊥-preserving, let ut⊥vt for all t∈0,1. Now, from (35). we have(37)Tut=1Γδ∫0tt−sδ−1Gs,usds,−2t2−θ2Γδ∫011−sδ−1Gs,usds,+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds>0,which implies that Tu⊥Tv.Now, we have to show thatT satisfies (i) of Corollary 2 for Fr=lnr,r>0. For all t∈0,1, ut⊥vt, we have(38)Tut−Tvt=|1Γδ∫0tt−sδ−1Gs,usds−2t2−θ2Γδ∫011−sδ−1Gs,usds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,umdmds,−1Γδ∫0tt−sδ−1Gs,vsds−2t2−θ2Γδ∫011−sδ−1Gs,vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,vmdmds|,≤1Γδ∫0tt−sδ−1Gs,us−Gs,vsds−2t2−θ2Γδ∫011−sδ−1Gs,us−Gs,vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Gs,um−Gs,vmdmds,≤1Γδ∫0tt−sδ−1Γδ+15e−τsups∈0,1us−vsds−2t2−θ2Γδ∫011−sδ−1Γδ+15e−τsups∈0,1us−vsds+2t2−θ2Γδ∫0θ∫0ss−mδ−1Γδ+15e−τsups∈0,1us−vsdmds,≤Γδ+15e−τsups∈0,1us−vs×supt∈0,11Γδ∫0tt−sδ−1ds−2t2−θ2Γδ∫011−sδ−1ds+2t2−θ2Γδ∫0θ∫0ss−mδ−1dmds≤e−τsups∈0,1us−vs=e−τdu,v,for all u,v∈X. Therefore, the condition (i) of Corollary 2 holds. Accordingly, all axioms of Corollary 2 are verified, and T has a fixed point. The Caputo-type nonlinear fractional differential equation (34) possesses a solution is yielded.


## 6. Conclusions

In this manuscript, we prove some existence results for the multivalued orthogonal mappings using the conditions (F1) and (F2) of Wardowski’s and obtain the stability of a fixed point problem and a solution for the Caputo-type nonlinear fractional differential equation.Now, we have an open question, whether we can obtain Theorems1 and 2 with condition (F1) only of Wardowski in the setting of orthogonal metric space?



---

*Source: 6615478-2020-11-21.xml*