
Citation: Mohammed A. Almalahi, Mohammed S. Abdo, Satish K. Panchal. On the theory of fractional terminal value problem with ψ-Hilfer fractional derivative[J]. AIMS Mathematics, 2020, 5(5): 4889-4908. doi: 10.3934/math.2020312
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In the recent years, scientific community renders more attention on fractional differential equations, since their are effective tools in modeling many phenomena in applied sciences and engineering application such as acoustic control, rheology, polymer physics, porous media, medicine, electrochemistry, proteins, electromagnetics, economics, astrophysics, chemical engineering, signal processing, optics, chaotic dynamics, statistical physics and so on for more details, see [1,2,3,4]. Since boundary value problems of fractional differential equations represent an important class of applied analysis, therefore the said area was given more importance, see [5,6,7,8,9,10] and references therein.
Terminal value problems for differential equation nowadays play an essential part in the modeling of numerous phenomena in physical science, engineering, and so forth. Also, it arise naturally in the simulation of techniques that are watched at a later point, eventually after the methodology has started.
Existence theory for classical terminal value problems have been investigated by several researchers [11,12,13,14,15,16,17,18,19,20,21,22]. It is well known [23] that the comparison principle for initial value problems of ordinary differential equations is a very useful tool in the study of qualitative and quantitative theory. Recently, attempts have been made to study the corresponding comparison principle for terminal value problems (TVP) [24]. Benchohra et. al. [14], studied the existence results and uniqueness of solutions for a class of boundary value problems of terminal type for fractional differential equations with the Hilfer–Katugampola fractional derivative by using different types of classical fixed point theory such as the Banach contraction principle and Krasnoselskii's fixed point theorem.
Motivated by the above-mentioned works, the objective of this work is to study conditions for the existence and uniqueness of the solutions for terminal value problem for fractional differential equations of the type
Dα,β;ψa+y(t)=f(t,y(t),Dα,β;ψa+y(t)), t∈(a,T],a>0 | (1.1) |
y(T)=w∈R, | (1.2) |
where Dα,β;ψa+(⋅) is the ψ-Hilfer fractional derivative of order α∈(0,1), type β∈[0,1] and f:(a,T]×R×R→R is a given function. Moreover, we study the uniqueness and continuous dependence of the δ-approximate solution by generalized Gronwall inequality. To our knowledge, no papers on terminal value problem for implicit fractional differential equations exist in the literature, in particular for those involving the ψ-Hilfer fractional derivative.
The rest of the paper is organized as follows. In section 2, we present some necessary definitions and results which are used throughout this paper. In section 3, we study the existence and uniqueness results of ψ-Hilfer fractional differential equation with the terminal condition by using some fixed point theorems such as Banach and Krasnoselskii. In section 4, we study the δ-approximate solution of the problem (1.1), (1.2). Also, four examples are included to illustrate the applicability of the results obtained.
In this section, we recall some notations, definitions of the fractional differential equation which are using throughout this paper. Let [a,T]⊂R+ with (0<a<T<∞), and let C[a,T] be the Banach space of continuous function y:[a,T]→R with the norm ‖y‖C[a,T]=max{|y(t)|:a≤t≤T}. The weighted space C1−γ;ψ[a,T] of continuous function y is defined by [25]
C1−γ;ψ[a,T]={y:(a,T]→R;[ψ(t)−ψ(a)]1−γy(t)∈C[a,T]}, 0≤γ<1 |
Obviously, C1−γ;ψ[a,T] is a Banach space endowed with the norm
‖y‖c1−γ;ψ=maxt∈[a,T]|[ψ(t)−ψ(a)]1−γy(t)|. |
Definition 2.1. [26] Let α>0, y∈L1[a,b] and ψ∈C1[a,b] be an increasing function with ψ′(t)≠0, for all t∈[a,b]. Then, the left-sided ψ-Riemann-Liouville fractional integral of a function y is defined by
Iα,ψa+y(t)=1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1y(s)ds, |
where Γ(⋅) is the Euler gamma function defined by Γ(α)=∫∞0sα−1e−sds, α>0.
Definition 2.2. [27] Let n−1<α<n, ( n=[α]+1), and y,ψ∈Cn[a,b] be two functions with an increasing ψ and ψ′(t)≠0, for all t∈[a,b]. Then, the left-sided ψ-Riemann-Liouville fractional (ψ-Caputo) derivative of a function y of order α is defined by
Dα,ψa+y(t)=(1ψ′(t)ddt)nIn−α,ψa+y(t), |
and
CDα,ψa+y(t)=In−α,ψa+(1ψ′(t)ddt)ny(t), |
respectively.
Definition 2.3. [25] Let n−1<α<n, ( n∈N), and y,ψ∈Cn[a,T] be two functions such that ψ is increasing and ψ′(t)≠0, for all t∈[a,T]. Then, the left-sided ψ-Hilfer fractional derivative of a function y of order α and type 0≤β≤1 is defined by
Dα,β,ψa+y(t)=Iβ(n−α);ψa+(1ψ′(t)ddt)nI(1−β)(n−α);ψa+y(t)=Iβ(n−α);ψa+Dγ;ψa+y(t), (γ=α+nβ−αβ). | (2.1) |
In this paper we consider the case n=1, because 0<α<1.
Lemma 2.4. [2] Let α>0 and 0≤γ<1. Then Iα,ψa+ is bounded from C1−γ;ψ[a,b] into C1−γ;ψ[a,b].
Now, we introduce the following spaces
Cα,β1−γ;ψ[a,T]={y∈C1−γ;ψ[a,T],Dα,β;ψa+y∈C1−γ;ψ[a,T]}, 0≤γ<1 |
and
Cγ1−γ;ψ[a,T]={y∈C1−γ;ψ[a,T],Dγ;ψa+y∈C1−γ;ψ[a,T]}, 0≤γ<1. | (2.2) |
Lemma 2.5. [25] Let γ=α+β−αβ where α∈(0,1), β∈[0,1], and y∈Cγ1−γ;ψ[a,T]. Then
Iγ;ψa+Dγ;ψa+y=Iα;ψa+Dα,β;ψa+y, |
and
Dγ;ψa+Iα;ψa+y=Dβ(1−α);ψa+y. |
Lemma 2.6. [25] Let α>0, 0≤γ<1 and y∈C1−γ[a,T], β∈[0,1]. Then
Dα,β,ψa+Iα,ψa+y(t)=y(t). |
Lemma 2.7. [2] Let t>a. Then for α≥0 and γ>0, we have
Iα,ψa+[ψ(t)−ψ(a)]γ−1=Γ(γ)Γ(α+γ)(ψ(t)−ψ(a))α+γ−1, t>a. |
and
Dα,ψa+[ψ(t)−ψ(a)]α−1=0, for α∈(0,1). |
Lemma 2.8. [25] Let γ=α+β−αβ where α∈(0,1), β∈[0,1], y∈Cγ1−γ;ψ[a,T] and I1−γ;ψa+y∈C11−γ,ψ[a,T]. Then, we have
Iγ;ψa+Dγ,ψa+y(t)=y(t)−I1−γ;ψa+y(a)Γ(γ)(ψ(t)−ψ(a))γ−1. |
Lemma 2.9. [25] Let α>0,0≤γ<α and y∈C1−γ,ψ[a,T] (0<a<T<∞). If γ<α, then Iα;ψa+:C1−γ,ψ[a,T]→C1−γ,ψ[a,T] is continuous on [a,T] and satisfies
Iα;ψa+ y(a)=limt→a+Iα;ψa+y(t)=0. |
Theorem 2.10. [28] (Krasnoselskii fixed point theorem). Let M be closed, convex, bounded and nonempty subset of a Banach space X and A,B be two operators such that
(1) Au+Bv∈M for all u,v∈M.
(2) A is compact and continuous.
(3) B is contraction mapping.
Then there exists z∈M such that z=Az+Bz.
Theorem 2.11. [29] (Banach's fixed point theorem). Let X be a Banach space and M be a nonempty closed subset of X, then any contraction mapping T:M→M has a unique fixed point.
Lemma 2.12. [30] (Generalized Gronwall's Inequality Lemma) Let α>0 and x,y be two nonnegative function locally integrable on [a,b]. Assume that g is nonnegative and nondecreasing, and let ψ∈C1[a,b] be an increasing function such that ψ′(t)≠0 for all t∈[a,b]. If
x(t)≤y(t)+g(t)∫taψ′(s)(ψ(t)−ψ(s))α−1x(s)ds, t∈[a,b], |
then
x(t)≤y(t)+∫ta∞∑n=1[g(t)Γ(α)]nΓ(nα)ψ′(s)(ψ(t)−ψ(s))nα−1y(s)ds, t∈[a,b]. |
If y be a nondecreasing function on [a,b], then
x(t)≤y(t)Eα{g(t)Γ(α)[ψ(t)−ψ(a)]α}, t∈[a,b], |
where Eα(⋅) is the Mittag-Leffler function defined by
Eα(z)=∞∑n=0znΓ(αn+1), z∈C. |
Theorem 3.1. Let γ=α+β−αβ, where α∈(0,1) and β∈[0,1]. If f:(a,T]→ R is a function such that f(⋅)∈C1−γ,ψ[a,T], then y∈Cγ1−γ,ψ(a,T] satisfies the following problem
HDα,β;ψa+y(t)=f(t),t∈(a,T],a>0 | (3.1) |
y(T)=w∈R | (3.2) |
if and only if y satisfies the following integral equation
y(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1f(s)ds]+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1f(s)ds. | (3.3) |
Proof. First, let y∈Cγ1−γ,ψ(a,T] be a solution of the problem (3.1), (3.2). We prove that y is also a solution of Eq (3.3). From the definition of Cγ1−γ,ψ(a,T], Lemma 2.4, and using the definition 2.3, we have
I1−γ,ψa+y(t)∈C1−γ,ψ[a,T] and Dγ;ψa+y(t)=D1,ψI1−γ,ψa+y(t). |
By the definition of the space Cn1−γ,ψ[a,T], it follows that
I1−γ,ψa+y(t)∈C11−γ,ψ[a,T]. | (3.4) |
Using Lemma 2.8, with α=γ, we obtain
Iγ;ψa+Dγ,ψa+y(t)=y(t)−I1−γ;ψa+y(a)Γ(γ)(ψ(t)−ψ(a))γ−1, t∈(a,T]. | (3.5) |
Since y∈Cγ1−γ,ψ[a,T], and using Lemma 2.5 with Eq (3.1), we have
Iγ;ψa+Dγ;ψa+y(t)=Iα;ψa+ Dα,β;ψa+y(t)=Iα;ψa+f(t). | (3.6) |
Comparing Eqs (3.5) and (3.6), we see that
y(t)=I1−γ;ψa+y(a)Γ(γ)(ψ(t)−ψ(a))γ−1+Iα;ψa+f(t) | (3.7) |
Using Eq (3.2), we get
y(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1f(s)ds]+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1f(s)ds. |
Hence y(t) satisfies the problem (3.1), (3.2).
Conversely, Let y∈Cγ1−γ,ψ[a,T] be a function satisfying Eq (3.3). We prove that y is also a solution of the problem (3.1), (3.2). Apply the operator Dγ;ψa+ on both sides of Eq (3.3). Then, from Lemmas 2.7 and 2.5, we have
Dγ;ψa+y(t)=Dγ;ψa+Iα;ψa+f(t)=Dβ(1−α);ψa+f(t) | (3.8) |
From Eq (3.4), we have Dγ;ψa+y∈C1−γ;ψ[a,T], and hence, Eq (3.8) implies
Dγ;ψa+y(t)=D1,ψI1−γ,ψa+f(t)=Dβ(1−α);ψa+f(t)∈C1−γ;ψ[a,T]. | (3.9) |
As f(t)∈C1−γ;ψ[a,T], and from Lemma 2.4, it follows that
I1−β(1−α);ψa+f∈C11−γ;ψ[a,T] | (3.10) |
From Eqs (3.9) and (3.10) and the definition of the space Cn1−γ;ψ(a,T], we get
I1−β(1−α);ψa+f∈C11−γ;ψ[a,T]. |
Now, by applying operator Iβ(1−α);ψa+ on both sides of Eq (3.9) and using Lemmas 2.9, 2.8, we have
Iβ(1−α);ψa+Dγ;ψa+y(t)=f(t)−I1−β(1−α);ψa+f(a)Γ(β(1−α))(ψ(t)−ψ(a))β(1−α)−1=f(t). | (3.11) |
From Eq (2.1), the Eq (3.11) reduces to
HDα,β;ψa+y(t)=f(t). |
that is, Eq (3.1) holds.
Before given our main results, the following conditions must be satisfied
H1 f:(a,T]×R×R→R is continuous function such that f(⋅,x(⋅),y(⋅))∈Cβ(1−α)1−γ;ψ for all x,y∈C1−γ;ψ[a,T].
H2 There exist two constants L>0 and M∈(0,1) such that
|f(t,x1,y1)−f(t,x2,y2)|≤L|x1−x2|+M|y1−y2|, |
for all x1,y1,x2,y2∈R and t∈(a,T].
In the forthcoming theorem, by using the Banach fixed point theorem, we prove the unique solution of the problem (1.1), (1.2)
Theorem 3.2. Assume that (H1) and (H2) hold. If
[2LΓ(γ)(1−M)Γ(α+γ)(ψ(T)−ψ(a))α]<1, | (3.12) |
then the problem (1.1), (1.2) has a unique solution in Cγ1−γ;ψ[a,T]⊂Cα,β1−γ;ψ[a,T].
Proof. In view of Theorem 3.1, the solution of the problem (1.1), (1.2) is given by
y(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1Ky(s)ds]+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1Ky(s)ds, | (3.13) |
where Ky(t)=f(t,y(t),Ky(t)). Consider the operator F:C1−γ;ψ[a,T]→C1−γ;ψ[a,T] defined by
Fy(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1Ky(s)ds]+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1Ky(s)ds, | (3.14) |
by Lemma 2.4, we deduce that Fy∈C1−γ;ψ[a,T]. The proof will be given in two steps
Step(1): We show that the operator F has a unique fixed point ˆy in C1−γ;ψ[a,T]. Let y,y∗∈C1−γ;ψ[a,T] and t∈(a,T]. Then, we have
|Fy(t)−Fy∗(t)|≤1Γ(α)(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ∫Taψ′(s)(ψ(T)−ψ(s))α−1|Ky(s)−Ky∗(s)|ds+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1|Ky(s)−Ky∗(s)|ds, |
where Ky(s),Ky∗(s)∈C1−γ;ψ[a,T] such that
Ky(s)=f(s,y(s),Ky(s)) |
Ky∗(s)=f(s,y∗(s),Ky∗(s)). |
By (H2), we have
|Ky(s)−Ky∗(s)|=|f(s,y(s),Ky(s))−f(s,y∗(s),Ky∗(s))|≤L|y(s)−y∗(s)|+M|Ky(s)−Ky∗(s)|, |
which implies
|Ky(s)−Ky∗(s)|≤L1−M|y(s)−y∗(s)|. | (3.15) |
Then for any t∈(a,T], we have
|Fy(t)−Fy∗(t)|≤L(1−M)Γ(α)(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ∫Taψ′(s)(ψ(T)−ψ(s))α−1|y(s)−y∗(s)|ds+L(1−M)Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1|y(s)−y∗(s)|ds≤L‖y−y∗‖C1−γ,ψ[a,T](1−M)Γ(α)(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ∫Taψ′(s)(ψ(T)−ψ(s))α−1(ψ(s)−ψ(0))γ−1ds+L‖y−y∗‖C1−γ,ψ[a,T](1−M)Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1(ψ(s)−ψ(0))γ−1ds. |
In view of Lemma 2.7, we obtain
|Fy(t)−Fy∗(t)|≤[LΓ(γ)‖y−y∗‖C1−γ,ψ[a,T](1−M)Γ(α+γ)(ψ(T)−ψ(a))α(ψ(t)−ψ(a))1−γ+LΓ(γ)‖y−y∗‖C1−γ,ψ[a,T](1−M)Γ(α+γ)(ψ(t)−ψ(a))α+γ−1]≤[2LΓ(γ)(1−M)Γ(α+γ)(ψ(T)−ψ(a))α](ψ(t)−ψ(a))γ−1‖y−y∗‖C1−γ,ψ[a,T]. |
Hence
|(ψ(t)−ψ(a))1−γ[Fy(t)−Fy∗(t)]|≤[2LΓ(γ)(1−M)Γ(α+γ)(ψ(T)−ψ(a))α]‖y−y∗‖C1−γ,ψ[a,T], |
which implies that
‖Fy−Fy∗‖C1−γ,ψ≤[2LΓ(γ)(1−M)Γ(α+γ)(ψ(T)−ψ(a))α]‖y−y∗‖C1−γ,ψ[a,T]. |
Due to Eq (3.12), we deduce that the operator F is a contraction mapping. According to Banach's contraction principle, we conclude that F has a unique fixed point ˆy∈C1−γ;ψ[a,T].
Step(2): We show that such a fixed point ˆy∈C1−γ;ψ[a,T] is actually in Cγ1−γ;ψ(a,T]. Since ˆy is the unique fixed point of F in C1−γ;ψ[a,T], then, for each t∈(a,T], we have
ˆy(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1Kˆy(s)ds]+1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1Kˆy(s)ds. |
Multiplying both sides of the last equation by Dγ,ψa+, using Lemmas 2.7 and 2.5, we have
Dγ,ψa+ˆy(t)=Dγ,ψa+Iα;ψa+Kˆy(s)(t)=Dβ(1−α);ψa+Kˆy(t), |
Since γ≥α, by (H1), we have Dβ(1−α);ψa+Kˆy(t)∈C1−γ;ψ[a,T], and hence Dγ,ψa+ˆy∈C1−γ;ψ[a,T]. It follows from definition of Cγ1−γ;ψ[a,T] that ˆy∈Cγ1−γ;ψ[a,T]. As a consequence of the above steps, we conclude that the problem (1.1), (1.2) has a unique solution in Cγ1−γ;ψ[a,T].
We present now the second result, which is based on Krasnoselskii fixed point theorem.
Theorem 3.3. Assume that (H1) and (H2) hold. Then the problem (1.1), (1.2) has at least one solution in Cγ1−γ;ψ[a,T].
Proof. Defined the closed, bounded, convex and nonempty set
kξ={y∈C1−γ;ψ[a,T]:‖y‖1−γ;ψ≤ξ}, |
with
ξ≥(ψ(T)−ψ(a))1−γ[|w|+RΓ(γ)Γ(α+γ)(ψ(T)−ψ(a))α+γ−1]. |
Set N=supt∈(a,T]|f(t,0,0)|. We split the operator F which defined by Eq (3.14) into two operators F1,F2 in kξ as following
F1y(t)=(ψ(T)−ψ(a))1−γ(ψ(t)−ψ(a))1−γ[w−1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1Ky(s)ds], |
and
F2y(t)=1Γ(α)∫taψ′(s)(ψ(t)−ψ(s))α−1Ky(s)ds. |
Note that Fy(t)=F1y(t)+F2y(t). The proof will be divided into several steps as follows:
Step(1):We show that F1y(t)+F2v(t)∈kξ for any y,v∈kξ.
(ⅰ) For t∈(a,T] and y∈kξ, we have
|(ψ(t)−ψ(a))1−γF1y(t)|≤(ψ(T)−ψ(a))1−γ[|w|+1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1|Ky(s)|ds]≤(ψ(T)−ψ(a))1−γ[|w|+1Γ(α)∫Taψ′(s)(ψ(T)−ψ(s))α−1(ψ(t)−ψ(a))γ−1 |(ψ(t)−ψ(a))1−γKy(s)|ds]. | (3.16) |
From (H2), we have
|Ky(t)|=|f(t,y(t),Ky(t))|=|f(t,y(t),Ky(t))−f(t,0,0)+f(t,0,0)|≤|f(t,y(t),Ky(t))−f(t,0,0)|+|f(t,0,0)|≤L|y(t)|+M|Ky(t)|+N. |
Multiplying both sides of the last inequality by (ψ(t)−ψ(a))1−γ, we get
|(ψ(t)−ψ(a))1−γKy(t)|≤L|(ψ(t)−ψ(a))1−γy(t)|+(ψ(t)−ψ(a))1−γN+M|(ψ(t)−ψ(a))1−γKy(t)|≤Lξ+(ψ(T)−ψ(a))1−γN+M|(ψ(t)−ψ(a))1−γKy(t)|. |
Then, for each t∈(a,T], we have
|(ψ(t)−ψ(a))1−γKy(t)|≤Lξ+(ψ(T)−ψ(a))1−γN1−M:=R |
Thus, the Eq (3.16) and Lemma 2.7, imply that
‖F1y‖1−γ;ψ≤(ψ(T)−ψ(a))1−γ[|w|+RΓ(γ)Γ(α+γ)(ψ(T)−ψ(a))α+γ−1]. | (3.17) |
(ⅱ) In a similar manner, for t∈(a,T], v∈kξ, we get
‖F2v‖1−γ;ψ≤RΓ(γ)Γ(α+γ)(ψ(T)−ψ(a))α+1−γ. | (3.18) |
Linking Eqs (3.17) and (3.18), for any y,v∈kξ, we obtain
‖F1y+F2v‖1−γ;ψ≤max{‖F1y‖1−γ;ψ,‖F2v‖1−γ;ψ}≤(ψ(T)−ψ(a))1−γ[|w|+RΓ(γ)Γ(α+γ)(ψ(T)−ψ(a))α+γ−1]≤ξ, |
which implies that F1y(t)+F2v(t)∈kξ.
Step(2): We show that F1 is a contraction mapping. From Theorem 3.2, we have already proved that F is a contraction mapping and hence F1 is a contraction mapping too in kξ.
Step(3): We show that F2 is a compact and continuous in kξ.
The continuity of F2 follows from the continuity of f. Now, we need only to prove that F2 is compact (i.e F2 uniformly bounded and equicontinuous). From Eq (3.18), for any v∈kξ, we have
‖F2v‖1−γ;ψ≤RΓ(γ)Γ(α+γ)(ψ(T)−ψ(a))α+1−γ. |
This means that F2 is uniformly bounded in kξ. Next, we show that F2 is equicontinuous in kξ. Let y∈kξ and t1,t2∈(a,T] such that t1<t2. Then, we have
|(ψ(t2)−ψ(a))1−γF2y(t2)−(ψ(t1)−ψ(a))1−γF2y(t1)|=|(ψ(t2)−ψ(a))1−γΓ(α)∫t2aψ′(s)(ψ(t2)−ψ(s))α−1Ky(s)ds−(ψ(t1)−ψ(a))1−γΓ(α)∫t1aψ′(s)(ψ(t1)−ψ(s))α−1Ky(s)ds|≤(ψ(t2)−ψ(a))1−γΓ(α)∫t2t1ψ′(s)(ψ(t2)−ψ(s))α−1|Ky(s)|ds+1Γ(α)∫t1a|[ψ′(s)(ψ(t2)−ψ(a))1−γ(ψ(t2)−ψ(s))α−1−ψ′(s)(ψ(t1)−ψ(a))1−γ(ψ(t1)−ψ(s))α−1]||Ky(s)|ds≤‖Ky(⋅)‖C1−γ,ψ(ψ(t2)−ψ(a))1−γΓ(γ)Γ(α+γ)(ψ(t2)−ψ(t1))α+γ−1+‖Ky(⋅)‖C1−γ,ψΓ(α)∫t1a|[ψ′(s)(ψ(t2)−ψ(s))α−1(ψ(t2)−ψ(a))1−γ−ψ′(s)(ψ(t1)−ψ(s))α−1(ψ(t1)−ψ(a))1−γ]|(ψ(s)−ψ(a))γ−1ds→0 as t2→t1. |
This means that \mathcal{F}_{2} is equicontinuous in \mathcal{\Bbbk } _{\xi }. Hence \mathcal{F}_{2} is relatively compact on \mathcal{\Bbbk } _{\xi }. By Arzelá-Ascoli Theorem, we deduce that \mathcal{F}_{2} is compact on \mathcal{\Bbbk }_{\xi }. According to Theorem (2.10), we conclude that \mathcal{F} has at least a fixed
point \widehat{y}\in C_{1-\gamma; \psi }\left[a, T\right] and by the same way of the proof of Theorem 3.2, we can easily show that \widehat{y} \in C_{1-\gamma; \psi }^{\gamma }\left[a, T\right]. Thus the problem (1.1), (1.2) has at least one solution in C_{1-\gamma; \psi }^{\gamma } \left[a, T\right] .
Definition 4.1 A function y\in C_{1-\gamma, \psi }^{\gamma }\left[a, T \right] satisfying the \psi -Hilfer implicit fractional differential inequality
\begin{equation} \left\Vert ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z(t)-f\left( t, z(t), ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z(t)\right) \right\Vert \leq \delta , t\in \left( a, T\right] , \end{equation} | (4.1) |
and
\begin{equation*} z(T) = w^{\ast }, \end{equation*} |
is called \delta -approximate solutions of \psi -Hilfer implicit fractional differential (1.1), (1.2)
Theorem 4.2. Let f:\left(a, T\right] \times \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} be a continuous function satisfies the condition (H _{2} ) for each t\in J and. Let z_{i}\in C_{1-\gamma, \psi }^{\gamma }\left(a, T\right], i = 1, 2, be a \delta -approximation solutions of the following \psi -Hilfer implicit fractional differential equation
\begin{equation} \begin{array}{c} ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z_{i}(t) = f\left( t, z_{i}(t), ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z_{i}(t)\right) , ~~t\in \left( a, T\right] \\ z_{i}(T) = w_{i}^{\ast }. \end{array} , \end{equation} | (4.2) |
Then
\begin{eqnarray*} &&\left\Vert z_{1}-z_{2}\right\Vert _{C_{1-\gamma , \psi }} \\ &\leq &\Upsilon ^{-1}\left\{ \left( \delta _{1}+\delta _{2}\right) \left[ \frac{\left( \psi (t)-\psi (a)\right) ^{\alpha -\gamma +1}}{\Gamma (\alpha +1)}+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\left( \psi (t)-\psi (a)\right) ^{(n+1)\alpha -\gamma +1}}{\Gamma ((n+1)\alpha +1)} \right] \right. \\ &&\left. +\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\right\vert \left[ 1+\sum\limits_{n = 1}^{ \infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \right\} , \end{eqnarray*} |
where
\begin{equation*} \Upsilon = \left\{ 1-\left( \psi (T)-\psi (a)\right) ^{\alpha }\frac{L}{1-M} \frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left[ 1+\sum \limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma ) }{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \right\} \end{equation*} |
Proof. Let z_{i}\in C_{1-\gamma, \psi }^{\gamma }\left(a, T\right], i = 1, 2, be an \delta -approximation solutions of the problem (4.2). Then, we have
\begin{equation} \left\Vert ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z_{i}(t)-f\left( t, z_{i}(t), ^{H}D_{a^{+}}^{\alpha , \beta ;\psi }z_{i}(t)\right) \right\Vert \leq \delta _{i}, \ t\in \left( a, T\right] , i = 1, 2 \end{equation} | (4.3) |
and
\begin{equation*} z_{i}(T) = w_{i}^{\ast }. \end{equation*} |
Applying I_{a^{+}}^{\alpha, \psi } on both sides of the above inequality, and using lemma 2.8, we get
\begin{eqnarray} &&\left. \left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{i}}{\Gamma (\alpha +1)}\geq \right. \\ &&\left\vert z_{i}(t)-w_{i}^{\ast }\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\text{ } I_{a^{+}}^{\alpha , \psi }K_{z_{i}}(T)-\text{ }I_{a^{+}}^{\alpha , \psi }K_{z_{i}}(t)\right\vert \end{eqnarray} |
Using the fact \left\vert x\right\vert -\left\vert y\right\vert \leq \left\vert x-y\right\vert \leq \left\vert x\right\vert +\left\vert y\right\vert in the above inequality, we have
\begin{eqnarray*} &&\left. \left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}\geq \right. \\ &&\left\vert z_{1}(t)-w_{1}^{\ast }\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\text{ } I_{a^{+}}^{\alpha , \psi }K_{z_{1}}(T)-\text{ }I_{0^{+}}^{\alpha , \psi }K_{z_{1}}(t)\right\vert \\ &&+\left\vert z_{2}(t)-w_{2}^{\ast }\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\text{ } I_{a^{+}}^{\alpha , \psi }K_{z_{2}}(T)-\text{ }I_{0^{+}}^{\alpha , \psi }K_{z_{2}}(t)\right\vert \\ &\geq &\left\vert \left[ z_{1}(t)-w_{1}^{\ast }\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\text{ } I_{a^{+}}^{\alpha , \psi }K_{z_{1}}(T)-\text{ }I_{0^{+}}^{\alpha , \psi }K_{z_{1}}(t)\right] \right. \\ &&\left. -\left[ z_{2}(t)-w_{2}^{\ast }\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\text{ } I_{a^{+}}^{\alpha , \psi }K_{z_{2}}(T)-\text{ }I_{a^{+}}^{\alpha , \psi }K_{z_{2}}(t)\right] \right\vert \\ &\geq &\left\vert \left( z_{1}(t)-z_{2}(t)\right) -\left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{ (\psi (t)-\psi (a))^{1-\gamma }}+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(T)-K_{z_{2}}(T)\right] \right. \\ &&-\left. I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(t)-K_{z_{2}}(t)\right] \right\vert \\ &\geq &\left\vert \left( z_{1}(t)-z_{2}(t)\right) \right\vert -\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\right\vert +\left\vert \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(T)-K_{z_{2}}(T)\right] \right\vert \\ &&-\left\vert I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(t)-K_{z_{2}}(t) \right] \right\vert \end{eqnarray*} |
In consequence, we have
\begin{eqnarray*} &&\left\vert \left( z_{1}(t)-z_{2}(t)\right) \right\vert \\ &\leq &\left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\right\vert \\ &&-\left\vert \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(T)-K_{z_{2}}(T)\right] \right\vert +\left\vert I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(t)-K_{z_{2}}(t)\right] \right\vert \\ &\leq &\left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\right\vert \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\left\vert I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(T)-K_{z_{2}}(T)\right] \right\vert +\left\vert I_{a^{+}}^{\alpha , \psi }\left[ K_{z_{1}}(t)-K_{z_{2}}(t)\right] \right\vert \\ &\leq &\left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\right\vert \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\frac{L}{1-M}\frac{1}{\Gamma (\alpha )}\int_{a}^{T}\psi ^{\prime }(s)\left( \psi (T)-\psi (s)\right) ^{\alpha -1}\left\vert z_{1}(s)-z_{2}(s)\right\vert ds \\ &&+\frac{1}{\Gamma (\alpha )}\int_{a}^{t}\psi ^{\prime }(s)\left( \psi (t)-\psi (s)\right) ^{\alpha -1}\left\vert y_{1}(s)-y_{2}(s)\right\vert ds \\ &\leq &\left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }}\right\vert \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{\alpha }}{(\psi (t)-\psi (a))^{1-\gamma }}\frac{L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert z_{1}-z_{2}\right\Vert _{1-\gamma ;\psi } \\ &&+\frac{L}{1-M}\frac{1}{\Gamma (\alpha )}\int_{a}^{t}\psi ^{\prime }(s)\left( \psi (t)-\psi (s)\right) ^{\alpha -1}\left\vert z_{1}(s)-z_{2}(s)\right\vert ds \\ &\leq &\Lambda (t)+\frac{L}{1-M}\frac{1}{\Gamma (\alpha )}\int_{a}^{t}\psi ^{\prime }(s)\left( \psi (t)-\psi (s)\right) ^{\alpha -1}\left\vert z_{1}(s)-z_{2}(s)\right\vert ds, \end{eqnarray*} |
where
\begin{eqnarray*} \Lambda (t) & = &\left( \psi (t)-\psi (a)\right) ^{\alpha }\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)}+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{ (\psi (t)-\psi (a))^{1-\gamma }}\right\vert \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{\alpha }}{(\psi (t)-\psi (a))^{1-\gamma }}\frac{L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert z_{1}-z_{2}\right\Vert _{1-\gamma ;\psi }. \end{eqnarray*} |
Using Lemma 2.12, we obtain
\begin{eqnarray*} &&\left\vert \left( z_{1}(t)-z_{2}(t)\right) \right\vert \\ &\leq &\Lambda (t)+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}I_{0^{+}}^{n\alpha , \psi }\Lambda (s)ds \\ &\leq &\Lambda (t)+\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)} \sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}I_{a^{+}}^{n\alpha , \psi }\left( \psi (t)-\psi (a)\right) ^{\alpha } \\ &&+\left\vert \left( w_{1}-w_{2}\right) \right\vert \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M} \right) ^{n}I_{a^{+}}^{n\alpha , \psi }(\psi (t)-\psi (a))^{\gamma -1} \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{\alpha }L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert z_{1}-z_{2}\right\Vert _{1-\gamma ;\psi }\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}I_{0^{+}}^{n\alpha , \psi }(\psi (t)-\psi (a))^{\gamma -1} \\ &\leq &\Lambda (t)+\frac{\delta _{1}+\delta _{2}}{\Gamma (\alpha +1)} \sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\alpha +1)}{\Gamma ((n+1)\alpha +1)}\left( \psi (t)-\psi (a)\right) ^{(n+1)\alpha } \\ &&+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \right\vert \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\sum\limits_{n = 1}^{\infty }\left( \frac{ L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha +\gamma -1} \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{\alpha }L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert z_{1}-z_{2}\right\Vert _{1-\gamma ;\psi }\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n} \frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha +\gamma -1} \\ & = &\left( \delta _{1}+\delta _{2}\right) \left[ \frac{\left( \psi (t)-\psi (a)\right) ^{\alpha }}{\Gamma (\alpha +1)}+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\left( \psi (t)-\psi (a)\right) ^{(n+1)\alpha }}{\Gamma ((n+1)\alpha +1)}\right] \\ &&+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \frac{\left( \psi (T)-\psi (a)\right) ^{1-\gamma }}{(\psi (t)-\psi (a))^{1-\gamma }} \right\vert \left[ 1+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \\ &&+\frac{\left( \psi (T)-\psi (a)\right) ^{\alpha }}{(\psi (t)-\psi (a))^{1-\gamma }}\frac{L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert z_{1}-z_{2}\right\Vert _{1-\gamma ;\psi }\left[ 1+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \end{eqnarray*} |
Hence for each t\in \left[a, b\right], we have
\begin{eqnarray*} &&\left\Vert z_{1}-z_{2}\right\Vert _{C_{1-\gamma , \psi }} \\ &\leq &\left( \delta _{1}+\delta _{2}\right) \left[ \frac{\left( \psi (t)-\psi (a)\right) ^{\alpha -\gamma +1}}{\Gamma (\alpha +1)} +\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\left( \psi (t)-\psi (a)\right) ^{(n+1)\alpha -\gamma +1}}{\Gamma ((n+1)\alpha +1)} \right] \\ &&+\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\right\vert \left[ 1+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \\ &&+\left( \psi (T)-\psi (a)\right) ^{\alpha }\frac{L}{1-M}\frac{\Gamma (\gamma )}{\Gamma (\alpha +\gamma )}\left\Vert y_{1}-y_{2}\right\Vert _{1-\gamma ;\psi }\left[ 1+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M} \right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] . \end{eqnarray*} |
Thus
\begin{eqnarray} &&\left\Vert z_{1}-z_{2}\right\Vert _{C_{1-\gamma , \psi }} \\ &\leq &\Upsilon ^{-1}\left\{ \left( \delta _{1}+\delta _{2}\right) \left[ \frac{\left( \psi (t)-\psi (a)\right) ^{\alpha -\gamma +1}}{\Gamma (\alpha +1)}+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\left( \psi (t)-\psi (a)\right) ^{(n+1)\alpha -\gamma +1}}{\Gamma ((n+1)\alpha +1)} \right] \right. \\ &&\left. +\left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\right\vert \left[ 1+\sum\limits_{n = 1}^{ \infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \right\} . \end{eqnarray} | (4.4) |
Remark 4.3 If \delta _{1} = \delta _{2} = 0 in the inequality (4.3), then z_{1}, z_{2} are solutions of the problem (1.1) and the inequality (4.4) reduces to
\begin{eqnarray*} &&\left\Vert z_{1}-z_{2}\right\Vert _{C_{1-\gamma , \psi }} \\ &\leq &\Upsilon ^{-1}\left\{ \left\vert \left( w_{1}^{\ast }-w_{2}^{\ast }\right) \left( \psi (T)-\psi (a)\right) ^{1-\gamma }\right\vert \left[ 1+\sum\limits_{n = 1}^{\infty }\left( \frac{L}{1-M}\right) ^{n}\frac{\Gamma (\gamma )}{\Gamma (n\alpha +\gamma )}(\psi (t)-\psi (a))^{n\alpha }\right] \right\} , \end{eqnarray*} |
which provides the continuous dependence of the problem (1.1). Also if w_{1}^{\ast } = w_{2}^{\ast }, we have \left\Vert z_{1}-z_{2}\right\Vert _{C_{1-\gamma, \psi }} = 0, which provides the uniqueness of a solution of problem (1.1).
In this section, we present illustrative examples to validate our results.
Example 5.1. Consider the following terminal value problem
\begin{equation} \left\{ \begin{array}{c} D_{1^{+}}^{\frac{1}{2}, 0;e^{t}}y(t) = \frac{1}{10e^{-t+2}}\left( 1+\left\vert y(t)\right\vert +\left\vert D_{1^{+}}^{\frac{1}{2}, 0;e^{t}}y(t)\right\vert \right) , ~~ t\in \left( 1, 2\right] , \\ y(2) = w\in \mathbb{R} . \end{array} \right. \end{equation} | (5.1) |
Set f(t, u, v) = \frac{t}{10}\left(1+u+v\right), for each u, v\in \mathbb{R} , t\in \left(1, 2\right],
\begin{equation*} C_{1-\gamma ;\psi }^{\beta (1-\alpha )}\left[ 1, 2\right] = C_{\frac{1}{2} ;e^{t}}^{0}\left[ 1, 2\right] = \left\{ f:\left( 1, 2\right] \times \mathbb{R} ^{2}\rightarrow \mathbb{R} ;\left( e^{t}-e\right) ^{\frac{1}{2}}f\in C\left[ 1, 2\right] \right\} , \text{ } \end{equation*} |
with \alpha = \frac{1}{2}, \beta = 0, \gamma = \frac{1}{2}, \psi (t) = e^{t}, \left(a, T\right] = \left(1, 2\right], K_{y}(t) = f(t, y(t), K_{y}(t)). Clearly, the function f\in C_{\frac{1}{2};e^{t}}\left[1, 2\right] . Hence condition ( H_{1} ) is satisfied. For u, v, u^{\ast }, v^{\ast }\in \mathbb{R} , t\in \left(1, 2\right], we have
\begin{eqnarray*} \left\vert f(t, u, v)-f(t, u^{\ast }, v^{\ast })\right\vert &\leq &\frac{1}{ 10e^{-t+2}}\left[ \left\vert u-u^{\ast }\right\vert +\left\vert v-v^{\ast }\right\vert \right] \\ &\leq &\frac{1}{10e}\left[ \left\vert u-u^{\ast }\right\vert +\left\vert v-v^{\ast }\right\vert \right] . \end{eqnarray*} |
Hence the hypothesis ( H_{2} ) is satisfied with M = L = \frac{1}{10e}. By some simple calculations, the condition:
\begin{equation*} \left[ \frac{2L\Gamma (\gamma )}{\left( 1-M\right) \Gamma (\alpha +\gamma )} \left( e^{T}-e^{a}\right) ^{\alpha }\right] \approx 0.3 \lt 1 \end{equation*} |
is satisfied with T = 2 and a = 1 . Thus all assumptions in Theorem 3.2 are satisfied. It follows from Theorem 3.2 that the problem (5.1) has a unique solution in C_{\frac{1}{3};e^{t}}^{\frac{1}{2}}[1,2].
Example 5.2 Consider the following terminal value problem
\begin{equation} \left\{ \begin{array}{c} D_{1^{+}}^{\frac{1}{2}, 0;\ln t}y(t) = \frac{1}{20e^{e+1-t}}\left[ \ln t^{\frac{ 1}{2}}\left\vert \cos y(t)\right\vert +\left\vert D_{1^{+}}^{\frac{1}{2} , 0;\ln t}y(t)\right\vert \right] , ~~ t\in \left( 1, e\right] \\ y(e) = w\in \mathbb{R} . \end{array} \right. \end{equation} | (5.2) |
Set f(t, u, v) = \frac{1}{20e^{e+1-t}}\left(\ln t^{\frac{1}{2}}\cos u+v\right), for each u, v\in \mathbb{R} , t\in \left(1, e\right],
\begin{equation*} C_{1-\gamma ;\psi }^{\beta (1-\alpha )}\left[ 1, e\right] = C_{\frac{1}{2};\ln t}^{0}\left[ 1, e\right] = \left\{ f:\left( \ln t\right) ^{\frac{1}{2}}f\in C \left[ 1, e\right] \right\} , \text{ } \end{equation*} |
with \alpha = \frac{1}{2}, \beta = 0, \gamma = \frac{1}{2}, \psi (t) = \ln t, \left(a, T\right] = \left(1, e\right]. Clearly, the function f\in C_{\frac{ 1}{2};\ln t}\left[1, e\right] . Hence condition ( H_{1} ) is satisfied. For u, v, u^{\ast }, v^{\ast }\in \mathbb{R} , t\in \left(1, e\right], we have
\begin{eqnarray*} \left\vert f(t, u, v)-f(t, u^{\ast }, v^{\ast })\right\vert &\leq &\frac{1}{ 20e^{e+1-t}}\left[ \left\vert u-u^{\ast }\right\vert +\left\vert v-v^{\ast }\right\vert \right] \\ &\leq &\frac{1}{20e}\left[ \left\vert u-u^{\ast }\right\vert +\left\vert v-v^{\ast }\right\vert \right] . \end{eqnarray*} |
Hence the hypothesis ( H_{2} ) is satisfied with M = L = \frac{1}{20e}. By some simple calculations, the condition:
\begin{equation*} \left[ \frac{2L\Gamma (\gamma )}{\left( 1-M\right) \Gamma (\alpha +\gamma )} \left( \ln \left( \frac{T}{a}\right) \right) ^{\alpha }\right] = 6.642\, 7\times 10^{-2} \lt 1 \end{equation*} |
is satisfied with with T = e and a = 1 . Thus all assumptions in Theorem 3.2 are satisfied. It follows from Theorem 3.2 that the problem (5.2) has a unique solution in C_{\frac{1}{2};\ln t}^{\frac{1}{2}}\left[1, e\right].
Example 5.3 Consider the following terminal value problem
\begin{equation} \left\{ \begin{array}{c} D_{1^{+}}^{\frac{1}{2}, 0;\sqrt{t}}y(t) = \frac{1}{10}\left[ t^{2}\left\vert \cos y(t)\right\vert +\left\vert D_{1^{+}}^{\frac{1}{2}, 0;\sqrt{t} }y(t)\right\vert \right] , ~~ t\in \left( 1, 2\right] \\ y(2) = w\in \mathbb{R} . \end{array} \right. . \end{equation} | (5.3) |
Set f(t, u, v) = \frac{1}{10}\left(t^{2}\cos u+v\right), for each u, v\in \mathbb{R} , t\in \left(1, 2\right],
\begin{equation*} C_{1-\gamma ;\psi }^{\beta (1-\alpha )}\left[ 1, 2\right] = C_{\frac{1}{2}; \sqrt{t}}^{0}\left[ 1, 2\right] = \left\{ f:\sqrt{2}\sqrt{\sqrt{t}-1}f\in C \left[ 1, 2\right] \right\} , \text{ } \end{equation*} |
with \alpha = \frac{1}{2}, \beta = 0, \gamma = \frac{1}{2}, \psi (t) = t^{\rho } (\rho = \frac{1}{2}), \left(a, T\right] = \left(1, 2\right]. Clearly, the function f\in C_{\frac{1}{2};t^{\rho }}\left[1, 2\right] . Hence condition ( H_{1} ) is satisfied. For u, v, u^{\ast }, v^{\ast }\in \mathbb{R} , t\in \left(1, 2\right], we have
\begin{equation*} \left\vert f(t, u, v)-f(t, u^{\ast }, v^{\ast })\right\vert \leq \frac{1}{10} \left[ \left\vert u-u^{\ast }\right\vert +\left\vert v-v^{\ast }\right\vert \right] . \end{equation*} |
Hence the hypothesis ( H_{2} ) is satisfied with M = L = \frac{1}{10}. By some simple calculations, the condition:
\begin{equation*} \left[ \frac{2L\Gamma (\gamma )}{\left( 1-M\right) \Gamma (\alpha +\gamma )} \left( \frac{T^{\rho }-a^{\rho }}{\rho }\right) ^{\alpha }\right] \approx 0.4 \lt 1 \end{equation*} |
is satisfied with \rho = \frac{1}{2} T = 2 and a = 1 . Thus all assumptions in Theorem 3.2 are satisfied. It follows from Theorem 3.2 that the problem (5.3) has a unique solution in C_{\frac{1}{2};t^{\rho }}^{ \frac{1}{2}}\left[1, 2\right].
Example 5.4 Consider the following terminal value problem
\begin{equation} \left\{ \begin{array}{c} D_{a^{+}}^{\alpha , \beta ;\psi }y(t) = K_{y}(t), ~~t\in \left( a, T \right] , \\ y(T) = w\in \mathbb{R} . \end{array} \right. \end{equation} | (5.4) |
By Theorem 3.1, the implicit solution of problem (5.4) is given by
\begin{eqnarray*} y(t) & = &\frac{\left[ \psi (T)-\psi (a)\right] ^{1-\gamma }}{\left[ \psi (t)-\psi (a)\right] ^{1-\gamma }}\left[ w-\frac{1}{\Gamma (\alpha )} \int_{a}^{T}\psi ^{\prime }(s)\left( \psi (T)-\psi (s)\right) ^{\alpha -1}K_{y}(s)ds\right] \\ &&+\frac{1}{\Gamma (\alpha )}\int_{a}^{t}\psi ^{\prime }(s)\left( \psi (t)-\psi (s)\right) ^{\alpha -1}K_{y}(s)ds, ~~ t\in \left( a, T\right] . \end{eqnarray*} |
Here, we consider K_{y}(t) = f(t, y(t), K_{y}(t)) = 1, w = 1, a = 1\ and T = 2 .
Case (ⅰ) If \psi (t) = t, the exact solution of problem (5.4) is defined by
\begin{equation*} y(t) = \left( t-1\right) ^{_{^{\gamma -1}}}-\frac{\left( t-1\right) ^{_{^{\gamma -1}}}}{\Gamma (\alpha +1)}+\frac{\left( t-1\right) ^{_{\alpha }} }{\Gamma (\alpha +1)}, \text{ }t\in (1, 2]. \end{equation*} |
Case (ⅱ) If \psi (t) = \log t, the exact solution of problem (5.4) is defined by
\begin{equation*} y(t) = \left( \log t\right) ^{_{^{\gamma -1}}}-\frac{\left( \log t\right) ^{_{^{\gamma -1}}}}{\Gamma (\alpha +1)}+\frac{\left( \log t\right) ^{_{\alpha }}}{\Gamma (\alpha +1)}, \text{ }t\in (1, e]. \end{equation*} |
Case (ⅲ) If \psi (t) = t^{\rho }, \rho > 0, the exact solution of problem (5.4) is defined by
\begin{equation*} y(t) = \left( \sqrt{2}-1\right) \left( t^{\rho }-1\right) ^{_{^{\gamma -1}}}\left( 1-\frac{\sqrt{2}\left( \sqrt{2}-1\right) ^{\alpha }}{\Gamma (\alpha +1)}\right) +\frac{\sqrt{2}\left( t^{\rho }-1\right) ^{_{\alpha }}}{ \Gamma (\alpha +1)}, \text{ }t\in (1, 2] \end{equation*} |
Figure 1, presents the solution curves with some values of \alpha and \gamma , when \psi (t) = t. Figure 2, presents the solution curves with some values of \alpha and \gamma , when \psi (t) = log(t). Figure 3, presents the solution curves with some values of \alpha and \gamma , when \psi (t) = t^{\rho}.
We have provided sufficient conditions ensuring the existence and uniqueness of solutions to a class of terminal value problem for differential equations with the \psi -Hilfer type fractional derivative. The arguments are based on the classical Banach contraction principle, and the Krasnoselskii's fixed point theorem. Moreover, we used generalized Gronwall inequality with singularity to established uniqueness and continuous dependence of the \delta -approximate solution. Four examples are included to show the applicability of our results.
All authors declare no conflicts of interest.
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