# Terminal Value Problem for Implicit Katugampola Fractional Differential Equations inb-Metric Spaces

Authors: Salim Krim; Saïd Abbas; Mouffak Benchohra; Erdal Karapinar
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This manuscript deals with a class of Katugampola implicit fractional differential equations inb-metric spaces. The results are based on the α−φ-Geraghty type contraction and the fixed point theory. We express an illustrative example.


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## Body

## 1. Introduction and Preliminaries

An interesting extension and unification of fractional derivatives of the type Caputo and the type Caputo-Hadamard is called Katugampola fractional derivative that has been introduced by Katugampola [1, 2]. Some fundamental properties of this operator are presented in [3, 4]. Several results of implicit fractional differential equations have been recently provided (see [4–14] and the references therein). A new class of mixed monotone operators with concavity and applications to fractional differential equations has been considered in [15]. In [16], the authors presented some existence and uniqueness results for a class of terminal value problem for differential equations with Hilfer-Katugampola fractional derivative.On the other side, a novel extension ofb-metric was suggested by Czerwik [17, 18]. Although the b-metric standard looks very similar to the metric definition, it has a quite different structure and properties. For example, in the b-metric topology framework, an open (closed) set is not open (closed). Additionally, the b-metric function is not continuous. These weaknesses make this new structure more interesting (see [19–28]).Throughout the paper, any mentioned set is nonempty. We consider the following type of terminal value problems of Katugampola implicit differential equations of noninteger orders:(1)ρD0+r+ϑτ=κτ,ϑτ,ρD0+r+ϑτ,τ∈I≔0,T,ϑT=ϑT∈ℝ,with T>0 and the function κ:I×ℝ×ℝ→ℝ is continuous. Here, ρD0+r is the Katugampola fractional derivative of order r∈0,1.SetCI≔h∣hrealcontinuousfunctionsonI≔0,T. Then, CI forms a Banach space with the norm ϑ∞=supτ∈Iϑτ.SetL1I≔ϑ:I→ℝϑismeasurablefunctionandLebesgueintegrable. Then, L1I becomes a Banach space with the norm ϑL1=∫0Tϑτdt.SetCr,ρI=ϑ:0,T→ℝτρ1−rϑτ∈CI. Then, it forms a Banach space ϑC≔supτ∈Iτρ1−rϑτ. Here, Cr,ρI is called the weighted space of continuous functions.Definition 1 (Katugampola fractional integral) [1].
The Katugampola fractional integrals of orderr>0 and ρ>0 of a function y∈XcpI are defined by
(2)ρT0+ryτρ1−rΓr∫0tsρ−1ysτρ−sρ1−rds,τ∈I.Definition 2 (Katugampola fractional derivatives) [1, 2].
The generalized fractional derivatives of orderr>0 and ρ>0 corresponding to the Katugampola fractional integrals (2) defined for any τ∈I by
(3)ρD0+ryτ=τ1−ρddtnρT0+n−ryτ=ρr−n+1Γn−rτ1−ρddtn∫0tsρ−1ysτρ−sρr−n+1ds,where n=r+1; if the integrals exist.Remark 1 ([1, 2]).
As a basic example, we quote forr,ρ>0 and θ>−ρ,
(4)ρD0+rτθ=ρr−1Γ1+θ/ρΓ1−r+θ/ρτθ−rρ.
Giving in particular,(5)ρD0+rτρr−i=0,foreachi=1,2,⋯,n.
In fact, forr,ρ>0 and θ>−ρ, we have
(6)ρD0+rτθ=ρr−n+1Γn−rτ1−ρddtn∫0tsρ+θ−1τρ−sρn−r−1ds=ρr−1Γ1+θ/ρΓ1+n−r+θ/ρn−r+θρ⋯1−r+θρτθ−rρ=ρr−1Γ1+θ/ρΓ1−r+θ/ρτθ−rρ.
If we puti=r−θ/ρ, we obtain from (6):
(7)ρD0+rτθr−i=ρr−1Γr−i+1Γn−i+1n−in−i−1⋯1−mτ−ρi.
So,ρD0+rτρr−i=0,∀r,ρ>0.Theorem 1 ([2]).
Letr,ρ,c∈ℝ, be such that r,ρ>0. Then, for any κ,ω∈XcpI, where 1≤p≤∞, we have
(1)
Inverse property:(8)ρD0+rρI0+rκτ=κτ,forallr∈0,1.(2)
Linearity property: for allr∈0,1, we have(9)ρD0+rκ+ωτ=ρD0+rκτ+ρD0+rωτ.ρI0+rκ+ωτ=ρI0+rκτ+ρI0+rωτ.Lemma 1 ([2]).
Letr,ρ>0. Ifϑ∈CI; then the fractional differential equation ρD0+r+ϑτ=0, has a unique solution
(10)ϑτ=C1τρr−1+C2τρr−2+⋯+Cnτρr−n,where Ci∈ℝ with i=1,2,⋯,n.Proof.
Letr,ρ>0. from Remark 1, we have
(11)ρD0+rτρr−i=0,foreachi=1,2,⋯,n.
Then, the fractional equationρD0+rϑτ=0 has a particular solution as follows:
(12)ϑτ=Ciτρr−i,Ci∈ℝ,foreachi=1,2,⋯,n.
Thus, the general solution ofρD0+rϑτ=0 is a sum of particular solutions (12), i.e.
(13)ϑτ=C1τρr−1+C2τρr−2+⋯+Cnτρr−n,Ci∈ℝ;i=1,2,⋯,n.Lemma 2.
Letr,ρ>0. If ϑ,ρD0+rϑ∈CI and 0<r≤1, then
(14)ρI0+rρD0+rϑτ=ϑτ+cτρr−1,for some constant c∈ℝ.Proof.
LetρD0+rϑ∈CI be the fractional derivative (3) of order 0<r≤1. If we apply the operator ρD0+r to ρI0+rρD0+rϑτ−ϑτ and use the properties (8) and (9), we get
(15)ρD0+rρI0+rρD0+rϑτ−ϑτ=ρD0+rρI0+rρD0+rϑτ−ρD0+rϑτ=ρD0+rϑτ−ρD0+rϑτ=0.
From the proof of Lemma1, there exists c∈ℝ, such that
(16)ρI0+rρD0+rϑτ−ϑτ=cτρr−1,which implies (14).Lemma 3.
Leth∈L1I,ℝ and 0<r≤1 and ρ>0. A function ϑ∈CI forms a solution for
(17)ρD0+rϑτ=zτ,τ∈I,ϑT=ϑT,if and only if ϑ fulfills
(18)ϑτ=ϑT−ρI0+rzTτTρr−1+ρr−1Γr∫0τsρ−1τρ−sρr−1zsds.Proof.
Letr,ρ>0. and 0<r≤1. Suppose that ϑ satisfies (17). Employing the operator ρI0+r to the each side of the equation
(19)ρD0+rϑτ=zτ,we find
(20)ρI0+rρD0+rϑτ=ρI0+rzτ.
From Lemma2, we get
(21)ϑτ+cτρr−1=ρI0+rzτ,for some c∈ℝ. If we use the terminal condition ϑT=ϑT in (21), we find
(22)ϑT=ϑT=ρI0+rzT−cTρr−1,which shows
(23)c=ρI0+rzT−ϑTTρ1−r.
Henceforth, we deduce (18).
Contrariwise, ifϑ achieves (18), then ρD0+rϑτ=zτ; for τ∈I and ϑτ=ϑT.Lemma 4.
Contemplate the problem (1), and set g∈CI, and ωτ=ϰτ,ϑτ,ωτ.
We presumeϑ achieves
(24)ϑτ=ϑT−ρI0+rωTτTρr−1+ρ1−rΓr∫0τsρ−1τρ−sρr−1ωsds.
Then,ϑ forms a solution of (1).Definition 3 [29, 30].
A functiond:S×S⟶0,∞ is called b-metric if there is c≥1 and d fulfills
(i)
bM1dν,ϑ=0ifandonlyifν=ϑ(ii)
bM2dν,μ=dμ,ν(iii)
bM3dμ,ϑ≤cdμ,ν+dν,ϑ
for allμ,ν,ϑ∈S. We say that the tripled (S,d,c) is b-metric space (in short, b.m.s.).Example 1 [29, 30].
Letd:CI×CI⟶0,∞be described as(25)dν,ϑ=ν−ϑ2∞≔supτ∈Iντ−ϑτ2,forallν,ϑECI.
Ergo,CI,d,2 is b-metric space.Example 2 [29, 30].
SetS=0,1 and d:S×S⟶0,∞ be designated by
(26)dν,ϑ=νr−ϑr,forallν,ϑ∈S.
Henceforth, (S,d,r) with r≥2 is b-metric space.We set the following:ϕ:0,∞→0,∞ϕiscontinuous,increasing,ϕ0=0andϕcμ≤cϕμ≤cμforc>1.For somec≥1, we set F≔λ:0,∞→0,1/c2λisnondecreasing.Definition 4 [29, 30].
A self-operatorT, on a b.m.s. (S,d,c), is called a generalized α−ϕ− Geraghty contraction whenever there exists α:S×S⟶0,∞, and some L≥0 such that for
(27)Dν,ϑ=maxdν,ϑ,dϑ,Tϑ,dν,Tν,dν,Tϑ+dϑ,Tν2s,(28)Nν,ϑ=mindν,ϑ,dϑ,Tϑ,dν,Tν,we have
(29)αμ,νφc3dTμ,Tν≤λφDμ,νφDμ,ν+LψNμ,ν,for all μ,ν,ϑ∈S, where λ∈F,φ,ψ∈Φ.Remark 2.
In the case whenL=0 in Definition 4 and the fact that
(30)dμ,ν≤Dμ,ν,forallμ,ν∈S,the inequality (29) becomes
(31)αμ,νφc3dTμ,Tν≤λφdμ,νφdμ,ν.Definition 5 [29, 30].
Setα:S×S⟶0,∞. An operator T:S⟶S, is α− admissible if
(32)αμ,ν≥1⇒αTμ,Tν≥1,for all μ,ν∈S.Definition 6 [29, 30].
Let (S,d,c) with c≥1 be a b.m.s and α:S×Sℝ+∗.
We say thatS is α− regular if for any sequence νnn∈ℕ in S such that xn⟶x as n⟶∞ and ανn,νn+1≥1 for each n; there exists a subsequence νnκκ∈ℕ of νnn with ανnκ,x≥1 for all k.Theorem 2 [29, 30].
We presume that a self-operatorT over a complete b.m.s.
(S,d,c) with c≥1 forms a generalized α−φ− Geraghty contraction. Furthermore,
(i)
T is α− admissible with initial value αμ0,Tμ0≥1 for some μ0∈M(ii)
eitherT is continuous or M is α− regular
ThenT possesses a fixed point. Furthermore, if
(iii)
for all fixed points μ,νofT, either αμ,ν≥1 or αν,μ≥1, then the found fixed point is uniqueThis manuscript launches the study of Katugampola implicit fractional differential equations on b.m.s.


## 2. Main Results

Observe thatCr,ρI,d,2 is a complete b.m.s. with d:Cr,ρI×Cr,ρI⟶0,∞ described as
(33)dν,ϑ=ν−ϑ2C≔supτ∈Iτρ1−rντ−ϑτ2.A functionϑ∈Cr,ρI is called a solution of (1) if it archives
(34)ϑτ=ϑT−ρI0+rωTτTρr−1+ρ1−rΓr∫0τsρ−1τρ−sρr−1ωsds,with ωτ=κτ,ϑτ,ωτ∈CI.In the sequel, we shall need the following hypotheses:(H1) There exist φ∈Φ,p:CI×CI⟶0,∞ and q:I⟶0,1 so that for each ϑ,v,ϑ1,v1∈Cr,ρI, and τ∈I(35)κτ,ϑ,υ−κτ,ϑ1,υ1≤τρ/21−rpϑ,υϑ−ϑ1+qτυ−υ1,with
(36)ρ1−rΓr∫0Tsρ−1Tρ−sρr−1pϑ,υ1−q∗dsC2+ρ1−rΓr∫0τsρ−1τρ−sρr−1pϑ,υ1−q∗dsC2≤φϑ−υ2C(H2) There are μ0∈Cr,ρ(I) and θ:Cr,ρI×Cr,ρI⟶ℝ, so that
(37)θμ0τ,ϑT−ρI0+rωTτTρr−1+ρ1−rΓr∫0τsρ−1τρ−sρr−1ωsds≥0,with g∈CI and ωτ=κτ,μ0τ,ωτ(H3) For any τ∈I, and ϑ,v∈Cr,ρI, θϑτ,vτ≥0 implies
(38)θρ1−rΓr∫0τsρ−1τρ−sρr−1ωsds,ρ1−rΓr∫0τsρ−1τρ−sρr−1zsds≥0,with ω,z∈CI so that
(39)zτ=κτ,υτ,zτ,ωτ=κτ,ϑτ,ωτ.(H4) If ϑnn∈N⊂CI with ϑn⟶ϑ and θϑn,ϑn+1≥, then
(40)θϑn,ϑ≥1.Theorem 3.
We presume (H1)–(H4). Then, the problem (1) possesses at least a solution on I.Proof.
Take the operatorN:Cr,ρI⟶Cr,ρI into account that is described as
(41)Nϑτ=ϑT−ρI0+rωTτTρr−1+ρ1−rΓr∫0τsρ−1τρsρr−1ωsds,where ω∈CI, with ωτ=κτ,ϑτ,ωτ.On account of Lemma4, we deduce that solutions of (1) are the fixed points of N.LetCr,ρI×Cr,ρI⟶0,∞ be the function defined by
(42)αϑ,υ=1,ifθϑτυτ≥0,τ∈I,αϑ,υ=0,otherwise.First, we demonstrate thatN form a generalized α−φ-Geraghty operator. For any τ∈I and each ϑ,v∈CI, we derive that
(43)τρ1−rNϑτ−τρ1−rNυτ≤τρ1−rρI0+rg−hTτTρr−1+ρ1−rτρ1−rΓr∫0τsρ−1τρ−sρr−1ωs−zsds,where ω, z∈CI, with
(44)ωτ=κτ,ϑτ,ωτ,(45)zτ=κτ,vτ,zτ.From (H1), we have
(46)ωτ−zτ=κτ,ϑτ,ωτ−κτ,υτ,zτ≤pϑ,υτρ/21−rϑτ−υτ+qτωτ−zτ≤pϑ,υτρ1−rϑτ−υτ21/2+qτωτ−zτ.Thus,(47)ωτ−zτpϑ,υ1−q∗ϑ−υ2C1/2,where q∗=supτ∈Iqτ.Next, we have(48)τρ1−rNϑτ−τρ1−rNυτ≤τρ1−rρI0+rg−hTτTρr−1+ρ1−rτρ1−rΓr∫0τsρ−1τρ−sρr−1pϑ,υ1−q∗ϑ−υ2C1/2ds≤ρ1−rτρ1−rΓr∫0Τsρ−1Τρ−sρr−1pϑ,υ1−q∗ϑ−υ2C1/2ds+ρ1−rτρ1−rΓr∫0τsρ−1τρ−sρr−1pϑ,υ1−q∗ϑ−υ2C1/2ds.Thus,(49)αϑ,υτρ1−rNϑτ−τρ1−rNυτ2≤ϑ−υ2Cαϑ,υρ1−rΓr∫0Tsρ−1Tρ−sρr−1pϑ,υ1−q∗dsC2+ϑ−υ2Cαϑ,υρ1−rΓr∫0Tsρ−1τρ−sρr−1pϑ,υ1−q∗dsC2≤ϑ−υ2Cϕϑ−υ2C.Hence,(50)αϑ,vφ23dNϑ,Nv≤λφdϑ,vφdϑ,v,where λ∈ϝ,φ∈Φ, with λτ=1/8t, and φτ=τ.So,N is generalized α−φ− Geraghty operator.Letϑ,v∈Cr,ρI such that
(51)αϑ,v≥1.Accordingly, for anyt∈I, we find
(52)θϑτ,vτ≥0.This implies from (H3) that
(53)θNuτ,Nvτ≥0,which gives αNϑ,Nv≥1.Ergo,N is a α-admissible.Now, from (H2), there exists μ0∈Cr,ρI such that
(54)αμ0,Nμ0≥1.Finally, from (H4), if μnn∈N⊂M with μn⟶μ and αμn,μn+1≥1, then,
(55)αμn,μ≥1.Theorem2 implies that fixed point ϑ of N forms a solution for (1).


## 3. An Example

The tripledCr,ρ0,1,d,2 is a complete b.m.s. with d:Cr,ρ0,1×Cr,ρ0,1⟶0,∞ such that
(56)dμ,ϑ=μ−ϑ2C.We take the following fractional differential problem into consideration(57)ρD0+rμτ=κτ,μτ,ρD0+rμτ,τ∈0,1,μ1=2,with
(58)κτ,μτ,ϑτ=τρ/21−r1+sinμτ41+μτ+e−τ21+ϑτ;τ∈0,1.Letτ∈0,1, and μ,ϑ∈Cr,ρ0,1. If |μτ∣≤∣ϑτ∣, then
(59)κτ,μτ,μ1τ−κτ,ϑτ,ϑ1τ=τρ/21−r1+sinμτ41+μτ−1+sinϑτ41+ϑτ+e−τ21+μ1τ−e−τ21+ϑ1τ≤τρ/21−r4μτ−ϑτ+τρ/21−r4sinμτ−sinϑτ+τρ/21−r4μτsinϑτ−ϑτsinμτ∣+e−τ2μ1τ−ϑ1τ≤τρ/21−r4μτ−ϑτ+τρ/21−r4sinμτ−sinϑτ+τρ/21−r4ϑτsinϑτ−ϑτsinμτ∣+e−τ2μ1τ−ϑ1τ=τρ/21−r4μτ−ϑτ+τρ/21−r41+υτsinμτ−sinϑτ+e−τ2u1τ−ϑ1τ≤τρ/21−r4μτ−ϑτ+τρ/21−r21+ϑτ×sinμτ−ϑτ2cosμτ+ϑτ2+e−τ2μ1τ−ϑ1τ≤τρ/21−r42+υτμτ−ϑτ+e−τ2μ1τ−ϑ1τ.In the case when∣ϑτ∣≤∣μτ∣, we get
(60)κτ,μτ−κτ,ϑτ≤τp/21−r42+μτμτ−ϑτ+e−τ2μ1τ−ϑ1τ.Hence,(61)κτ,μτ−κτ,ϑτ≤Τp/21−r4minτ∈I2+μτ,2+ϑτμτ−ϑτ+e−τ2μ1τ−ϑ1τ.Thus, hypothesis (H1) is achieved with
(62)pμ,ϑ=Tρ/21−r4minr∈I2+μτ,2+ϑτ,(63)qτ=12e−τ.Define the functionsλτ=1/8t,ϕτ=τ,α:Cr,ρ0,1×Cr,ρ0,1→ℝ+∗ with
(64)αμ,ϑ=1,ifδμτ,ϑτ≥0,τ∈I,αμ,ϑ=0,elseand δ:Cr,ρ0,1×Cr,ρ0,1⟶R with δμ,ϑ=kμ−ϑkC.Hypothesis (H2) is satisfied with μ0τ=μ0. Also, (H3) holds the definition of the function δ. So, Theorem 3 yields that problem (57) admits a solution.



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