On abstract Hilfer fractional integrodifferential equations with boundary conditions

Thabet, Sabri T.M.; Ahmad, Bashir; Agarwal, Ravi P.

doi:10.1016/j.ajmsc.2019.03.001

In this paper, we study a Cauchy-type problem for Hilfer fractional integrodifferential equations with boundary conditions. The existence of solutions for the given problem is proved by applying measure of noncompactness technique in an abstract weighted space. Moreover, we use generalized Gronwall inequality with singularity to establish continuous dependence and uniqueness of $ϵ$ -approximate solutions.

1. Introduction

Fractional calculus has emerged as a powerful tool to study complex phenomena in numerous scientific and engineering disciplines such as viscoelasticity, fluid mechanics, physics and heat conduction in materials with memory. For examples and applications, see [2,14,17–21] and references cited therein. Many authors focused on Riemann–Liouville and Caputo type derivatives in investigating fractional differential equations. In [7], Hilfer introduced a new concept of generalized Riemann–Liouville derivative (Hilfer derivative) of order $α$ and type $β$ ⁠. This definition facilitated dynamic modeling of non-equilibrium processes based on interpolation with respect to parameter of the Riemann–Liouville and Caputo type operators; for instance, see [1,4,6,8,10,11].

Furati et al. [6] established the existence and uniqueness of solutions for the problem:

{\begin{cases} D_{a^{+}}^{α, β} y (t) = f (t, y (t)), t \in J = (a, b], 0 < α < 1,0 \leq β \leq 1, \\ I_{a^{+}}^{1 - γ} y (a^{+}) = w, α \leq γ = α + β - α β, \end{cases}

by applying Banach fixed point theorem in weighted space $C_{1 - γ}^{γ} [J, ℝ]$ ⁠. Abbas et al. [1] discussed the above problem by using Kuratowski measure of noncompactness.

Motivated by the works [1,6], we will study a more general problem of Hilfer fractional integrodifferential equations with boundary conditions given by

{\begin{cases} D_{a^{+}}^{α, β} y (t) = f (t, y (t), (S y) (t)), t \in J = (a, b], 0 < α < 1,0 \leq β \leq 1, \\ I_{a^{+}}^{1 - γ} [u y (a^{+}) + v y (b^{-})] = w, α \leq γ = α + β - α β, \end{cases}

(1.1)

where $D_{a^{+}}^{α, β}$ is the left-sided Hilfer fractional derivative of order $α$ and type $β$ ⁠, $f : J \times X \times X \to X$ ⁠, $X$ is an abstract Banach space, $u, v, w \in ℝ, u + v \neq 0$ ⁠, and $S$ is a linear integral operator defined by $(S y) (t) = \int_{a}^{t} k (t, s) y (s) d s$ with $ζ = \max {\int_{a}^{t} k (t, s) d s : (t, s) \in J \times J}$ ⁠, $k \in (J \times J, ℝ)$ ⁠.

This article is constructed as follows: In Section 2, we recall some preliminaries. Section 3 contains the existence result obtained by using measure of noncompactness and Mönch fixed point theorem. We discuss the $ϵ$ -approximate solution of Hilfer fractional integrodifferential equations in Section 4.

2. Preliminaries

In this section, we present some necessary definitions, notations and preliminaries, which will be used throughout this work.

For $- \infty < a < b < \infty$ ⁠, let $C [J, X]$ denote the space of all continuous functions on $J$ into $X$ endowed with supremum norm ${‖ x ‖}_{C} : = \sup {‖ x (t) ‖ : t \in J}$ ⁠. Define by $C_{1 - γ} [J, X] = {f (x) : (a, b] \to X | {(x - a)}^{1 - γ} f (x) \in C [J, X]}$ the weighted space of the abstract continuous functions. Obviously, $C_{1 - γ} [J, X]$ is a Banach space equipped with the norm ${‖ f ‖}_{C_{1 - γ}} = {‖ {(x - a)}^{1 - γ} f (x) ‖}_{C}$ ⁠, and $C_{1 - γ}^{n} [J, X] = {f \in C^{n - 1} [J, X] : f^{(n)} \in C_{1 - γ} [J, X]}$ is the Banach space endowed with the norm

{‖ f ‖}_{C_{1 - γ}^{n}} = \sum_{i = 0}^{n - 1} {‖ f^{(k)} ‖}_{C} + {‖ f^{(n)} ‖}_{C_{1 - γ}}, n \in ℕ,

where, $C_{1 - γ}^{0} : = C_{1 - γ}$

Definition 2.1

(See [13]). The left-sided Riemann–Liouville fractional integral of order $α > 0$ of function $f : [a, \infty) \to ℝ$ is defined by

(I_{a^{+}}^{α} f) (t) = \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s) d s, t > a,

where $a \in ℝ$ and $Γ$ is the Gamma function.

Definition 2.2

(See [13]). The left-sided Riemann–Liouville fractional derivative of order $α \in (n - 1, n]$ of function $f : [a, \infty) \to ℝ$ ⁠, is defined by

(D_{a^{+}}^{α} f) (t) = \frac{1}{Γ (n - α)} {(\frac{d}{d t})}^{n} \int_{a}^{t} {(t - s)}^{n - α - 1} f (s) d s, t > a,

where $n = [α] + 1, [α]$ denotes the integer part of $α$ ⁠.

Remark 2.1.

If $f$ is an abstract function with values in X, then the integrals appearing in Definitions 2.1 and 2.2 are taken in Bochner’s sense.

Definition 2.3

(See [7]). The left-sided Hilfer fractional derivative of order $0 < α < 1$ and type $0 \leq β \leq 1$ ⁠, of function $f (t)$ is defined by

(D_{a^{+}}^{α, β} f) (t) = (I_{a^{+}}^{β (1 - α)} D (I_{a^{+}}^{(1 - β) (1 - α)})) (t),

where $D : = \frac{d}{d t} .$

Remark 2.2

(See [7]). From Definition 2.3, we observe that:

(i)
the operator $D_{a^{+}}^{α, β}$ can be written as
$D_{a^{+}}^{α, β} = I_{a^{+}}^{β (1 - α)} D I_{a^{+}}^{(1 - γ)} = I_{a^{+}}^{β (1 - α)} D^{γ}, γ = α + β - α β;$
(ii)
The Hilfer fractional derivative can be regarded as an interpolator between the Riemann–Liouville derivative (⁠ $β = 0$ ⁠) and Caputo derivative (⁠ $β = 1$ ⁠) as
$D_{a^{+}}^{α, β} = {\begin{cases} D I_{a^{+}}^{(1 - α)} = D_{a^{+}}^{α}, & if β = 0; \\ I_{a^{+}}^{(1 - α)} D =^{C} D_{a^{+}}^{α}, & if β = 1. \end{cases}$

In the forthcoming analysis, we need the spaces:

C_{1 - γ}^{α, β} [J, X] = {f \in C_{1 - γ} [J, X], D_{a^{+}}^{α, β} f \in C_{1 - γ} [J, X]},

and

C_{1 - γ}^{γ} [J, X] = {f \in C_{1 - γ} [J, X], D_{a^{+}}^{γ} f \in C_{1 - γ} [J, X]} .

Since $D_{a^{+}}^{α, β} f = I_{a^{+}}^{β (1 - α)} D^{γ} f$ ⁠, it is obvious that $C_{1 - γ}^{γ} [J, X] \subset C_{1 - γ}^{α, β} [J, X]$ ⁠.

Now, we state some known results related to our work.

Lemma 2.1

(See [5]). Let $β > 0$ and $α > 0$ . Then

[I_{a^{+}}^{α} {(t - a)}^{β - 1}] (x) = \frac{Γ (β)}{Γ (β + α)} {(x - a)}^{β + α - 1}

and

[D_{a^{+}}^{α} {(t - a)}^{α - 1}] (x) = 0, 0 < α < 1.

Lemma 2.2

(See [5]). If $α > 0$ and $β > 0$ , and $f \in L^{1} (J)$ for $t \in [a, b]$ , then the following properties hold:

(I_{a^{+}}^{α} I_{a^{+}}^{β} f) (t) = (I_{a^{+}}^{α + β} f) (t) and (D_{a^{+}}^{α} I_{a^{+}}^{β} f) (t) = f (t) .

In particular, if $f \in C_{γ} [J, X]$ or $f \in C [J, X]$ , then the above properties hold for each $t \in (a, b]$ or $t \in [a, b]$ respectively.

Lemma 2.3

(See [5]). If $0 < α < 1$ , $0 \leq γ < 1$ and that $f \in C_{γ} [J, X]$ , $I_{a^{+}}^{1 - α} f \in C_{γ}^{1} [J, X]$ , then

I_{a^{+}}^{α} D_{a^{+}}^{α} f (t) = f (t) - \frac{(I_{a^{+}}^{1 - α} f) (a)}{Γ (α)} {(t - a)}^{α - 1}, \forall t \in J .

Lemma 2.4

(See [6]). If $0 \leq γ < 1$ and $f \in C_{γ} [J, X]$ , then

(I_{a^{+}}^{α} f) (a) = \lim_{t \to a^{+}} I_{a^{+}}^{α} f (t) = 0, 0 \leq γ < α .

Lemma 2.5

(See [6]). Let $α > 0$ , $β > 0$ and $γ = α + β - α β$ . If $f \in C_{1 - γ}^{γ} [J, X]$ , then

I_{a^{+}}^{γ} D_{a^{+}}^{γ} f = I_{a^{+}}^{α} D_{a^{+}}^{α, β} f, D_{a^{+}}^{γ} I_{a^{+}}^{α} f = D_{a^{+}}^{β (1 - α)} f .

Lemma 2.6

(See [6]). Let $f \in L^{1} (J)$ and $D_{a^{+}}^{β (1 - α)} f \in L^{1} (J)$ exists, then

D_{a^{+}}^{α, β} I_{a^{+}}^{α} f = I_{a^{+}}^{β (1 - α)} D_{a^{+}}^{β (1 - α)} f .

Lemma 2.7

(Theorem 23, [6]). Let $f : J \times ℝ \to ℝ$ be a function such that $f \in C_{1 - γ} [J, ℝ]$ for any $y \in C_{1 - γ} [J, ℝ]$ . Then $y \in C_{1 - γ}^{γ} [J, ℝ]$ is a solution of the initial value problem:

{\begin{cases} D_{a^{+}}^{α, β} y (t) = f (t, y (t)), t \in J = (a, b], 0 < α < 1,0 \leq β \leq 1, \\ I_{a^{+}}^{1 - γ} y (a^{+}) = y_{a}, α \leq γ = α + β - α β, \end{cases}

if and only if $y$ satisfies the following Volterra integral equation:

y (t) = \frac{y_{a}}{Γ (γ)} {(t - a)}^{γ - 1} + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s)) d s .

Next we obtain the integral solution of the problem (1.1) by using Lemma 2.7.

Lemma 2.8.

Let $f : J \times X \times X \to X$ be a function such that $f \in C_{1 - γ} [J, X]$ for any $y \in C_{1 - γ} [J, X]$ . Then $y \in C_{1 - γ}^{γ} [J, X]$ is a solution of the problem (1.1) if and only if $y$ satisfies the following integral equation

\begin{matrix} y (t) = \frac{w}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} - \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \\ \times \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s \\ + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s . \end{matrix}

(2.1)

Proof. In view of Lemma 2.7, the solution of (1.1) can be written as

y (t) = \frac{I_{a^{+}}^{1 - γ} y (a^{+})}{Γ (γ)} {(t - a)}^{γ - 1} + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s .

(2.2)

Applying $I_{a^{+}}^{1 - γ}$ on both sides of (2.2) and taking the limit $t \to b^{-}$ ⁠, we obtain

I_{a^{+}}^{1 - γ} y (b^{-}) = I_{a^{+}}^{1 - γ} y (a^{+}) + \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s .

(2.3)

In a similar manner, we find that

\begin{array}{l} I_{a^{+}}^{1 - γ} y (a^{+}) = \frac{1}{1 + \frac{v}{u}} {\frac{w}{u} - \frac{v}{u} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s} \\ = \frac{1}{u + v} {w - v \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s} . \end{array}

(2.4)

Submitting (2.4) into (2.2), we obtain

\begin{array}{l} y (t) = \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{u + v} {w - v \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s} \\ + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s, \\ = \frac{w}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} - \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \\ \times \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s \\ + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s . \end{array}

Conversely, applying $I_{a^{+}}^{1 - γ}$ on both sides of (2.1) and using Lemmas 2.1 and 2.2, we get

I_{a^{+}}^{1 - γ} y (t) = \frac{w}{u + v} - \frac{v}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s + I_{a^{+}}^{1 - β (1 - α)} f (t, y (t), (S y) (t)) .

(2.5)

Next, taking the limit $t \to a^{+}$ of (2.5) and using Lemma 2.4, with $1 - γ < 1 - β (1 - α)$ ⁠, we obtain

I_{a^{+}}^{1 - γ} y (a^{+}) = \frac{w}{u + v} - \frac{v}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s .

(2.6)

Now, taking the limit $t \to b^{-}$ of (2.5), we get

I_{a^{+}}^{1 - γ} y (b^{-}) = \frac{w}{u + v} - \frac{v}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s + I_{a^{+}}^{1 - β (1 - α)} f (b, y (b), (S y) (b)) .

(2.7)

From (2.6) and (2.7), we find that

\begin{array}{l} u I_{a^{+}}^{1 - γ} y (a^{+}) + v I_{a^{+}}^{1 - γ} y (b^{-}) \\ = \frac{u w}{u + v} - \frac{u v}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s \\ + \frac{v w}{u + v} - \frac{v^{2}}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s . \\ + v I_{a^{+}}^{1 - β (1 - α)} f (b, y (b), (S y) (b)) \\ = \frac{w (u + v)}{u + v} - \frac{v (u + v)}{u + v} I_{a^{+}}^{1 - γ + α} f (b, y (b), (S y) (b)) \\ + v I_{a^{+}}^{1 - β (1 - α)} f (b, y (b), (S y) (b)) \\ = w, \end{array}

which shows that the boundary condition $I_{a^{+}}^{1 - γ} [u y (a^{+}) + v y (b^{-})] = w$ is satisfied.

Next, applying $D_{a^{+}}^{γ}$ on both sides of (2.1) and using Lemmas 2.1 and 2.5, we have

D_{a^{+}}^{γ} y (t) = D_{a^{+}}^{β (1 - α)} f (t, y (t), (S y) (t)) .

(2.8)

Since $y \in C_{1 - γ}^{γ} [J, X]$ and by definition of $C_{1 - γ}^{γ} [J, X]$ ⁠, we have $D_{a^{+}}^{γ} y \in C_{1 - γ} [J, X]$ ⁠, therefore, $D_{a^{+}}^{β (1 - α)} f = D I_{a^{+}}^{1 - β (1 - α)} f \in C_{1 - γ} [J, X]$ ⁠. For $f \in C_{1 - γ} [J, X]$ ⁠, it is clear that $I_{a^{+}}^{1 - β (1 - α)} f \in C_{1 - γ} [J, X]$ ⁠. Hence $f$ and $I_{a^{+}}^{1 - β (1 - α)} f$ satisfy the hypothesis of Lemma 2.3.

Now, applying $I_{a^{+}}^{β (1 - α)}$ on both sides of (2.8), and using Lemma 2.3, we get

D_{a^{+}}^{α, β} y (t) = f (t, y (t), (S y) (t)) - \frac{I_{a^{+}}^{1 - β (1 - α)} f (a, y (a), (S y) (a))}{Γ (β (1 - α))} {(t - a)}^{β (1 - α) - 1} .

By Lemma 2.4, we have $I_{a^{+}}^{1 - β (1 - α)} f (a, y (a), (S y) (a)) = 0$ ⁠. Therefore, we have $D_{a^{+}}^{α, β} y (t) = f (t, y (t), (S y) (t))$ ⁠. This completes the proof. □

Next, we recall definition of noncompactness measure of Hausdorff $Ψ (\cdot)$ on each bounded subset $Ω$ of Banach space $X$ defined by

Ψ (Ω) = \inf {r > 0, Ω can be covered by finite number of balls with radii r} .

Lemma 2.9

([3]). For all nonempty subsets $A, B \subset X$ , the Hausdorff measure of noncompactness $Ψ (\cdot)$ satisfies the following properties:

(1) $A$ is precompact if and only if $Ψ (A) = 0;$
(2) $Ψ (A) = Ψ (\bar{A}) = Ψ (c o n v A)$ , where $\bar{A}$ and $c o n v A$ denote the closure and convex hull of $A$ respectively;
(3) $Ψ (A) \leq Ψ (B)$ when $A \subseteq B;$
(4) $Ψ (A + B) \leq Ψ (A) + Ψ (B)$ ⁠, where $A + B = {a + b; a \in A, b \in B};$
(5) $Ψ (A \cup B) \leq \max {Ψ (A), Ψ (B)};$
(6) $Ψ (λ A) = | λ | Ψ (A)$ for any $λ \in ℝ$ ;
(7) $Ψ ({x} \cup A) \leq Ψ (A)$ for any $x \in X$ .

Lemma 2.10

([3]). If $B \subseteq C ([a, b], X)$ is bounded and equicontinuous, then $Ψ (B (t))$ is continuous for $t \in [a, b]$ and $Ψ (B) = \sup {Ψ (B (t)), t \in [a, b]}$ , where $B (t) = {x (t); x \in B} \subseteq X$ .

Lemma 2.11

([16]). If ${u_{n}}_{n = 1}^{\infty}$ is a sequence of Bochner integrable functions from $J$ into $X$ with $‖ u_{n} (t) ‖ \leq μ (t)$ for almost all $t \in J$ and every $n \geq 1$ , where $μ \in L^{1} (J, R)$ , then the function $Ψ (t) = Ψ ({u_{n} (t) : n \geq 1})$ belongs to $L^{1} (J, R)$ with

Ψ ({\int_{0}^{t} u_{n} (s) d s : n \geq 1}) \leq 2 \int_{0}^{t} Ψ (s) d s .

In order to prove the existence of solutions for our problem with lesser number of constraints, we will introduce another type of measure of noncompactness as follows.

Let $Φ$ denote the measure of noncompactness in the Banach space $C [J, X]$ defined by

Φ (Ω) = \max_{E \in Δ (Ω)} (δ (E), m o d_{c} (E)),

(2.9)

for all bounded subsets $Ω$ of $C [J, X]$ ⁠, where $Δ (Ω)$ is the set of countable subsets of $Ω$ ⁠, $δ$ is the real measure of noncompactness given by

δ (E) = \sup_{t \in [0, b]} e^{- L t} Ψ (E (t)),

with $E (t) = {x (t) : x \in E}, t \in J$ ⁠, $L$ is a suitably chosen constant and $m o d_{c} (E)$ is the modulus of equicontinuity of the function set $E$ defined as

\mod_{c} (E) = \lim_{δ \to 0} \sup_{x \in E} \max_{| t_{2} - t_{1} | \leq δ} ‖ x (t_{2}) - x (t_{1}) ‖ .

Observe that $Φ$ is well defined [9] (i.e., $E_{0} \in Δ (Ω)$ which attends the maximum in (2.9)) and is nonsingular, monotone and regular measure of noncompactness.

Lemma 2.12

(Mönch fixed point theorem, [15]). Let $D$ be a closed convex subset of a Banach space $X$ with $0 \in D$ . Suppose that $F : D \to X$ is a continuous map satisfying the Mönch’s condition (if $M \subseteq D$ is countable and $M \subseteq c o n v ({0} \cup F (M))$ , then $\bar{M}$ is compact), then $F$ has a fixed point in $D$ .

3. Existence of solutions

Let us begin this section by introducing the hypotheses needed to prove the existence of solutions for the problem at hand.

(H1)

The function $f : J \times X \times X \to X$ satisfies (i) $f (\cdot, x, y) : J \to X$ is measurable for all $x, y \in X$ and (ii) $f (t, \cdot, \cdot) : X \times X \to X$ is continuous for a.e $t \in J$ ⁠.

(H2)

There exists a constant $N > 0$ such that

‖ f (t, y, S y) ‖ \leq N (1 + ζ ‖ y ‖),

for each $t \in J$ and all $y \in X$ ⁠.

(H3)

There exist constants $m_{1}, m_{2} > 0$ such that

Ψ (f (t, x, y)) \leq m_{1} Ψ (x) + m_{2} Ψ (y),

for bounded sets $x, y \subset X$ ⁠, a.e $t \in J$ ⁠.

Now, we are ready to present the existence result for the problem (1.1), which is based on Mönch fixed point theorem.

Theorem 3.1.

Suppose that $f : J \times X \times X \to X$ is such that $f (\cdot, y (\cdot), S y (\cdot)) \in C_{1 - γ}^{β (1 - α)} [J, X]$ for any $y \in C_{1 - γ} [J, X]$ and satisfies the hypotheses (H1)-(H3). Then the Hilfer problem (1.1) has at least one solution in $C_{1 - γ}^{γ} [J, X] \subset C_{1 - γ}^{α, β} [J, X]$ , provided that

Q ≔ \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N ζ}{Γ (1 - γ + α)} {(b - a)}^{α} B (γ, α - γ + 1) + \frac{N ζ}{Γ (α)} {(b - a)}^{α} B (γ, α) < 1 .

Proof. Introduce the operator $Q : C_{1 - γ} [J, X] \to C_{1 - γ} [J, X]$ defined by

\begin{matrix} (Q y) (t) = \frac{w}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} - \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \\ \times \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s \\ + \frac{1}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s . \end{matrix}

(3.1)

Notice that the solutions of problem (1.1) are the fixed points of the operator $Q$ ⁠. Define a bounded closed convex set $B_{r} : = {y \in C_{1 - γ} [J, X] : {‖ y ‖}^{C}_{1 - γ} \leq r, t \in J}$ with $r \geq \frac{ω}{1 - ϱ} (ϱ < 1)$ and

ω : = \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{N {(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N {(b - a)}^{α - γ + 1}}{Γ (2 - γ + α)} .

In order to satisfy the hypotheses of the Mönch fixed point theorem, we split the proof into four steps.

Step 1. The operator $Q$ maps the set $B_{r}$ into itself.

By the assumption (H2), we have

‖ (Q y) (t) {(t - a)}^{1 - γ} ‖ = ‖ \frac{1}{Γ (γ)} \frac{w}{u + v} - \frac{1}{Γ (γ)} \frac{v}{u + v} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} f (s, y (s), (S y) (s)) d s + \frac{{(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} f (s, y (s), (S y) (s)) d s ‖ \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{1}{Γ (1 - γ + α)} \times \int_{a}^{b} {(b - s)}^{α - γ} ‖ f (s, y (s), (S y) (s)) ‖ d s + \frac{{(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} ‖ f (s, y (s), (S y) (s)) ‖ d s \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} N (1 + ζ ‖ y (s) ‖) d s + \frac{{(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} N (1 + ζ ‖ y (s) ‖) d s \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} d s + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} ζ ‖ y (s) ‖ d s + \frac{N {(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} d s + \frac{N {(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} ζ ‖ y (s) ‖ d s \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N}{Γ (1 - γ + α)} \frac{{(b - a)}^{α - γ + 1}}{(α - γ + 1)} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N ζ}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} {(s - a)}^{γ - 1} {‖ y ‖}_{c_{1 - γ}} d s + \frac{N {(t - a)}^{1 - γ}}{Γ (α)} \frac{{(t - a)}^{α}}{α} + \frac{N ζ {(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} {(s - a)}^{γ - 1} {‖ y ‖}_{C_{1 - γ}} d s \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N {(b - a)}^{α - γ + 1}}{Γ (2 - γ + α)} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N ζ r}{Γ (1 - γ + α)} {(b - a)}^{α} B (γ, α - γ + 1) + \frac{N {(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \frac{N ζ r}{Γ (α)} {(t - a)}^{α} B (γ, α) \leq \frac{1}{Γ (γ)} \frac{| w |}{| u + v |} + \frac{N {(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N {(b - a)}^{α - γ + 1}}{Γ (2 - γ + α)} + [\frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{N ζ}{Γ (1 - γ + α)} {(b - a)}^{α} B (γ, α - γ + 1) + \frac{N ζ}{Γ (α)} {(b - a)}^{α} B (γ, α)] r,

where we used the fact

\begin{matrix} \int_{a}^{t} {(t - s)}^{α - 1} ‖ y (s) ‖ d s \leq (\int_{a}^{t} {(t - s)}^{α - 1} {(s - a)}^{γ - 1} d s) {‖ y ‖}_{C_{1 - γ}} \\ = {(t - a)}^{α + γ - 1} B (γ, α) {‖ y ‖}_{C_{1 - γ}} \end{matrix}

In consequence, we get ${‖ Q y ‖}_{C_{1 - γ}} \leq ω + ϱ r \leq r$ ⁠, that is, $Q B_{r} \subset B_{r}$ ⁠. Thus $Q : B_{r} \to B_{r}$ ⁠.

Step 2. The operator $Q$ is continuous.

Suppose that ${y_{n}}$ is a sequence such that $y_{n} \to y$ in $B_{r}$ as $n \to \infty$ ⁠. Since $f$ satisfies (H1), for each $t \in J$ ⁠, we get

\begin{array}{l} ‖ ((Q y_{n}) (t) - (Q y) (t)) {(t - a)}^{1 - γ} ‖ \\ \leq \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{1}{Γ (1 - γ + α)} \times \int_{a}^{b} {(b - s)}^{α - γ} ‖ f (s, y_{n} (s), (S y_{n}) (s)) - f (s, y (s), (S y) (s)) ‖ d s \\ + \frac{{(t - a)}^{1 - γ}}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} ‖ f (s, y_{n} (s), (S y_{n}) (s)) - f (s, y (s), (S y) (s)) ‖ d s \\ \leq \frac{1}{Γ (γ)} \frac{| v |}{| u + v |} \frac{{(b - a)}^{α} B (γ, α - γ + 1)}{Γ (1 - γ + α)} \\ \times {‖ f (\cdot, y_{n} (\cdot), (S y_{n}) (\cdot)) - f (\cdot, y (\cdot), (S y) (\cdot)) ‖}_{C_{1 - γ}} \\ + \frac{{(t - a)}^{α}}{Γ (α)} B (γ, α) {‖ f (\cdot, y_{n} (\cdot), (S y_{n}) (\cdot)) - f (\cdot, y (\cdot), (S y) (\cdot)) ‖}_{C_{1 - γ}} . \end{array}

By (H1) and using the Lebesgue dominated convergence theorem, we have

{‖ (Q y_{n} - Q y) ‖}_{C_{1 - γ}} \to 0 a s n \to \infty,

which implies that the operator $Q$ is continuous on $B_{r}$ ⁠.

Step 3. The operator $Q$ is equicontinuous.

For any $a < t_{1} < t_{2} < b$ and $y \in B_{r}$ ⁠, we get

\begin{array}{l} ‖ {(t_{2} - a)}^{1 - γ} (Q y) (t_{2}) - {(t_{1} - a)}^{1 - γ} (Q y) (t_{1}) ‖ \\ \leq \frac{1}{Γ (α)} | | {(t_{2} - a)}^{1 - γ} {\int_{a}^{t_{2}} (t_{2} - s)}^{α - 1} f (s, y (s), (S y) (s)) d s \\ - {(t_{1} - a)}^{1 - γ} \int_{a}^{t_{1}} {(t_{1} - s)}^{α - 1} f (s, y (s), (S y) (s)) d s | | \\ \leq \frac{{‖ f ‖}_{C_{1 - γ}}}{Γ (α)} | | {(t_{2} - a)}^{1 - γ} \int_{a}^{t_{2}} {(t_{2} - s)}^{α - 1} {(s - a)}^{γ - 1} d s \\ - {(t_{1} - a)}^{1 - γ} \int_{a}^{t_{1}} {(t_{1} - s)}^{α - 1} {(s - a)}^{γ - 1} d s | | \\ \leq \frac{{‖ f ‖}_{C_{1 - γ}}}{Γ (α)} B (γ, α) ‖ {(t_{2} - a)}^{1 - γ} {(t_{2} - a)}^{α + γ - 1} - {(t_{1} - a)}^{1 - γ} {(t_{1} - a)}^{α + γ - 1} ‖ \\ \leq \frac{{‖ f ‖}_{C_{1 - γ}}}{Γ (α)} B (γ, α) ‖ {(t_{2} - a)}^{α} - {(t_{1} - a)}^{α} ‖, \end{array}

which tends to zero as $t_{2} \to t_{1}$ ⁠, independent of $y \in B_{r}$ ⁠. Thus we conclude that $Q (B_{r})$ is equicontinuous, that is, $m o d_{c} (Q (B_{r})) = 0$ ⁠.

Step 4. The Mönch condition is satisfied.

Suppose that $D \subset B_{r}$ is a countable set and $D \subseteq c o n v ({0} \cup Q (D))$ ⁠. In order to show that $D$ is precompact, it is enough to obtain that $Φ (D) = (0, 0)$ ⁠. Since $Φ (Q (D))$ is maximum, let ${x_{n}}_{n = 1}^{\infty} \subseteq Q (D)$ be a countable set attaining its maximum. Then, there exists a set ${y_{n}}_{n = 1}^{\infty} \subseteq D$ such that $x_{n} = (Q y_{n}) (t)$ for all $t \in J, n \geq 1$ ⁠.

Now, using (H3) together with Lemmas 2.9–2.11, we obtain

Ψ ({x_{n}}_{n = 1}^{\infty}) = Ψ ({(Q y_{n}) (t)}_{n = 1}^{\infty}) \leq \frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \times \int_{a}^{b} {(b - s)}^{α - γ} Ψ (f (s, {y_{n} (s)}_{n = 1}^{\infty}, (S {y_{n} (s)}_{n = 1}^{\infty}))) d s + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} Ψ (f (s, {y_{n} (s)}_{n = 1}^{\infty}, (S {y_{n} (s)}_{n = 1}^{\infty}))) d s \leq \frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \times \int_{a}^{b} {(b - s)}^{α - γ} (m_{1} Ψ ({y_{n} (s)}_{n = 1}^{\infty}) + m_{2} Ψ ((S {y_{n} (s)}_{n = 1}^{\infty}))) d s + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} (m_{1} Ψ ({y_{n} (s)}_{n = 1}^{\infty}) + m_{2} Ψ ((S {y_{n} (s)}_{n = 1}^{\infty}))) d s \leq \frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \times \int_{a}^{b} {(b - s)}^{α - γ} (m_{1} sup_{t \in [a, b]} Ψ ({y_{n} (t)}_{n = 1}^{\infty}) + 2 m_{2} ζ \sup_{t \in [a, b]} Ψ ({y_{n} (t)}_{n = 1}^{\infty})) d s + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} (m_{1} \sup_{t \in [a, b]} Ψ ({y_{n} (t)}_{n = 1}^{\infty}) + 2 m_{2} ζ \sup_{t \in [a, b]} Ψ ({y_{n} (t)}_{n = 1}^{\infty})) d s \leq \frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} \times e^{L s} (m_{1} \sup_{t \in [a, b]} e^{- L t} Ψ ({y_{n} (t)}_{n = 1}^{\infty}) + 2 m_{2} ζ \sup_{t \in [a, b]} e^{- L t} Ψ ({y_{n} (t)}_{n = 1}^{\infty})) d s + \frac{2}{Γ (α)} \int_{a}^{b} {(t - s)}^{α - 1} \times e^{L s} (m_{1} \sup_{t \in [a, b]} e^{- L t} Ψ ({y_{n} (t)}_{n = 1}^{\infty}) + 2 m_{2} ζ \sup_{t \in [a, b]} e^{- L t} Ψ ({y_{n} (t)}_{n = 1}^{\infty})) d s \leq \frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{δ ({y_{n}}_{n = 1}^{\infty})}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} e^{L s} (m_{1} + 2 m_{2} ζ) d s + \frac{2 δ ({y_{n}}_{n = 1}^{\infty})}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} e^{L s} (m_{1} + 2 m_{2} ζ) d s \leq [\frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} e^{L s} (m_{1} + 2 m_{2} ζ) d s + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} e^{L s} (m_{1} + 2 m_{2} ζ) d s] δ ({y_{n}}_{n = 1}^{\infty}) .

Hence

\begin{array}{l} δ ({x_{n}}_{n = 1}^{\infty}) \\ \leq \sup_{t \in [a, b]} e^{- L t} [\frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \int_{a}^{b} {(b - s)}^{α - γ} e^{L s} (m_{1} + 2 m_{2} ζ) d s \\ + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} e^{L s} (m_{1} + 2 m_{2} ζ) d s] δ ({y_{n}}_{n = 1}^{\infty}) . \end{array}

Fixing a suitable constant $0 < L^{'} < 1$ given by

\begin{array}{l} L^{'} = \sup_{t \in [a, b]} e^{- L t} [\frac{2 | v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{1}{Γ (1 - γ + α)} \\ \times \int_{a}^{b} {(b - s)}^{α - γ} e^{L s} (m_{1} + 2 m_{2} ζ) d s \\ + \frac{2}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} e^{L s} (m_{1} + 2 m_{2} ζ) d s], \end{array}

we get $δ ({x_{n}}_{n = 1}^{\infty}) \leq L^{'} δ ({y_{n}}_{n = 1}^{\infty})$ ⁠. Thus

δ ({y_{n}}_{n = 1}^{\infty}) \leq δ (D) \leq δ (c o n v ({0} \cup Q (D))) = δ ({x_{n}}_{n = 1}^{\infty}) \leq L^{'} δ ({y_{n}}_{n = 1}^{\infty}),

which implies that $δ ({y_{n}}_{n = 1}^{\infty}) = 0$ and hence $δ ({x_{n}}_{n = 1}^{\infty}) = 0$ ⁠.

Now, according to the Step 3, we have found an equicontinuous set ${x_{n}}_{n = 1}^{\infty}$ on $J$ ⁠. Hence $Φ (D) \leq Φ (c o n v ({0} \cup Q (D))) \leq Φ (Q (D))$ ⁠, where $Φ (Q (D)) = Φ ({x_{n}}_{n = 1}^{\infty}) = (0, 0)$ ⁠. Therefore, $D$ is precompact. Hence, by Lemma 2.12, there is a fixed point $y$ of operator $Q$ ⁠, which is a solution of the problem (1.1) in $C_{1 - γ} [J, X]$ ⁠.

Next, we show that such a solution is indeed in $C_{1 - γ}^{γ} [J, X]$ ⁠. By applying $D_{a^{+}}^{γ}$ on both sides of (2.1), we get

D_{a^{+}}^{γ} y (t) = D_{a^{+}}^{β (1 - α)} f (t, y (t), (S y) (t)) .

Since $f (t, y (t), (S y) (t)) \in C_{1 - γ}^{β (1 - α)} [J, X]$ ⁠, it follows by definition of the space $C_{1 - γ}^{β (1 - α)} [J, X]$ that $D_{a^{+}}^{γ} y (t) \in C_{1 - γ} [J, X]$ ⁠, which implies that $y \in C_{1 - γ}^{γ} [J, X]$ ⁠. □

4. $ϵ -$ Approximate solution

Definition 4.1.

A function $z \in C_{1 - γ}^{γ} [J, X]$ satisfying the Hilfer fractional integrodifferential inequality

‖ D_{a^{+}}^{α, β} z (t) - f (t, z (t), (S z) (t)) ‖ \leq ϵ, t \in J,

and

I_{a^{+}}^{1 - γ} [u z (a^{+}) + v z (b^{-})] = \bar{w},

is called an $ϵ -$ approximate solutions of Hilfer fractional integrodifferential equation (1.1).

Lemma 4.1

(See [22]). For $β > 0$ , let $v (t)$ be a nonnegative function locally integrable on $0 < t < T$ (some $T \leq + \infty$ ) and $g (t)$ be a nonnegative, nondecreasing continuous function defined on $0 < t < T$ with $g (t) \leq M$ (constant) and $u (t)$ be a nonnegative and locally integrable function on $0 < t < T$ such that

u (t) \leq v (t) + g (t) \int_{0}^{t} {(t - s)}^{β - 1} u (s) d s, 0 < t < T .

Then

u (t) \leq v (t) + \int_{0}^{t} [\sum_{n = 1}^{\infty} \frac{{(g (t) Γ (β))}^{n}}{Γ (n β)} {(t - s)}^{n β - 1} v (s)] d s, 0 < t < T .

Theorem 4.1.

Suppose that the function $f : J \times X \times X \to X$ satisfies the condition:

‖ f (t, y_{1}, x_{1}) - f (t, y_{2}, x_{2}) ‖ \leq n_{1} ‖ y_{1} - y_{2} ‖ + n_{2} ‖ x_{1} - x_{2} ‖,

for each $t \in J$ and all $y_{1}, y_{2}, x_{1}, x_{2} \in X$ , where $n_{1}, n_{2} > 0$ are constants. Let $z_{i} \in C_{1 - γ}^{γ} [J, X], i = 1,2$ , be an $ϵ -$ approximate solution of the following Hilfer fractional integrodifferentialequation

{\begin{array}{l} D_{a^{+}}^{α, β} z_{i} (t) = f (t, z_{i} (t), (S z_{i}) (t)), t \in J, 0 < α < 1,0 \leq β \leq 1, \\ I_{a^{+}}^{1 - γ} [u z_{i} (a^{+}) + v z_{i} (b^{-})] = \bar{w_{i}}, α \leq γ = α + β - α β, i = 1,2. \end{array}

(4.1)

Then

{‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \leq Z^{- 1} \times [(ϵ_{1} + ϵ_{2}) (\frac{{(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ ((n + 1) α + 1)} {(b - a)}^{(n + 1) α - γ + 1}) + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α})],

(4.2)

where

Z = (1 - \frac{| v |}{| u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) \times {\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α}}) \neq 0 .

(4.3)

Proof. Let $z_{i} \in C_{1 - γ}^{γ} [J, X], (i = 1,2)$ be an $ϵ -$ approximate solution of problem (4.1). Then $I_{a^{+}}^{1 - γ} [u z_{i} (a^{+}) + v z_{i} (b^{-})] = \bar{w_{i}}$ and

‖ D_{a^{+}}^{α, β} z_{i} (t) - f (t, z_{i} (t), (S z_{i}) (t)) ‖ \leq ϵ_{i}, i = 1,2, t \in J .

(4.4)

Applying $I_{a^{+}}^{α}$ on both sides of the above inequality and using Lemma 2.3, we get

I_{a^{+}}^{α} ϵ_{i} \geq I_{a^{+}}^{α} ‖ D_{a^{+}}^{α, β} z_{i} (t) - f (t, z_{i} (t), (S z_{i}) (t)) ‖ \geq ‖ z_{i} (t) - \frac{\bar{w_{i}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{i} (b), (S z_{i}) (b)) - I_{a^{+}}^{α} f (t, z_{i} (t), (S z_{i}) (t)) ‖,

which implies that

\frac{ϵ_{i}}{Γ (α + 1)} {(t - a)}^{α} \geq ‖ z_{i} (t) - \frac{\bar{w_{i}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{i} (b), (S z_{i}) (b)) - I_{a^{+}}^{α} f (t, z_{i} (t), (S z_{i}) (t)) ‖, i = 1,2.

Using $| x | - | y | \leq | x - y | \leq | x | + | y |$ in the above inequality yields

\begin{array}{l} \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} \\ \geq ‖ z_{1} (t) - \frac{\bar{w_{1}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{1} (b), (S z_{1}) (b)) - I_{a^{+}}^{α} f (t, z_{1} (t), (S z_{1}) (t)) ‖ \\ + ‖ z_{2} (t) - \frac{\bar{w_{2}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{2} (b), (S z_{2}) (b)) - I_{a^{+}}^{α} f (t, z_{2} (t), (S z_{2}) (t)) ‖ \\ \geq ‖ [z_{1} (t) - \frac{\bar{w_{1}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{1} (b), (S z_{1}) (b)) \\ - I_{a^{+}}^{α} f (t, z_{1} (t), (S z_{1}) (t))] - [z_{2} (t) - \frac{\bar{w_{2}}}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \\ + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} f (b, z_{2} (b), (S z_{2}) (b)) - I_{a^{+}}^{α} f (t, z_{2} (t), (S z_{2}) (t))] ‖ \\ \geq | | (z_{1} (t) - z_{2} (t)) - \frac{(\bar{w_{1}} - \bar{w_{2}})}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \\ + \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} [f (b, z_{1} (b), (S z_{1}) (b)) - f (b, z_{2} (b), (S z_{2}) (b))] \\ - I_{a^{+}}^{α} [f (t, z_{1} (t), (S z_{1}) (t)) - f (t, z_{2} (t), (S z_{2}) (t))] | | \\ \geq ‖ (z_{1} (t) - z_{2} (t)) ‖ - | \frac{(\bar{w_{1}} - \bar{w_{2}})}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} | \\ + ‖ \frac{v}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} [f (b, z_{1} (b), (S z_{1}) (b)) - f (b, z_{2} (b), (S z_{2}) (b))] ‖ \\ - ‖ I_{a^{+}}^{α} [f (t, z_{1} (t), (S z_{1}) (t)) - f (t, z_{2} (t), (S z_{2}) (t))] ‖ . \end{array}

In consequence, we have

\begin{array}{l} ‖ (z_{1} (t) - z_{2} (t)) ‖ \\ \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + | \frac{(\bar{w_{1}} - \bar{w_{2}})}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} | \\ - ‖ \frac{| u |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} I_{a^{+}}^{α - γ + 1} [f (b, z_{1} (b), (S z_{1}) (b)) - f (b, z_{2} (b), (S z_{2}) (b))] ‖ \\ + ‖ I_{a^{+}}^{α} [f (t, z_{1} (t), (S z_{1}) (t)) - f (t, z_{2} (t), (S z_{2}) (t))] ‖ \\ \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{u + v} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \\ + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} ‖ I_{a^{+}}^{α - γ + 1} [f (b, z_{1} (b), (S z_{1}) (b)) - f (b, z_{2} (b), (S z_{2}) (b))] ‖ \\ + ‖ I_{a^{+}}^{α} [f (t, z_{1} (t), (S z_{1}) (t)) - f (t, z_{2} (t), (S z_{2}) (t))] ‖ \\ \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \\ + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} \int_{a}^{b} {(b - s)}^{α - γ} ‖ z_{1} (s) - z_{2} (s) ‖ d s \\ + \frac{(n_{1} + ζ n_{2})}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - γ} ‖ z_{1} (s) - z_{2} (s) ‖ d s \\ \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \\ + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \\ + \frac{(n_{1} + ζ n_{2})}{Γ (α)} \int_{a}^{t} {(t - s)}^{α - 1} ‖ z_{1} (s) - z_{2} (s) ‖ d s . \end{array}

Using Lemma 4.1 with $u (t) = ‖ (z_{1} (t) - z_{2} (t)) ‖$ ⁠, $g (t) = \frac{(n_{1} + ζ n_{2})}{Γ (α)}$ and $v (t) = \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}}$ ⁠, we get

‖ (z_{1} (t) - z_{2} (t)) ‖ \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} + \int_{a}^{t} \sum_{n = 1}^{\infty} \frac{{(n_{1} + ζ n_{2})}^{n}}{Γ (n α)} {(t - s)}^{n α - 1} (\frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(s - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(s - a)}^{γ - 1}}{Γ (γ)} + \frac{| v |}{| u + v |} \frac{{(s - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}}) d s \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} + \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} I_{a^{+}}^{n α} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{Γ (γ) | u + v |} \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} I_{a^{+}}^{n α} {(t - a)}^{γ - 1} + \frac{| v | {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}}}{Γ (γ) | u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) \times \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} I_{a^{+}}^{n α} {(t - a)}^{γ - 1} \leq \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} {(t - a)}^{α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \frac{| v |}{| u + v |} \frac{{(t - a)}^{γ - 1}}{Γ (γ)} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} + \frac{(ϵ_{1} + ϵ_{2})}{Γ (α + 1)} \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{Γ (α + 1)}{Γ ((n + 1) α + 1)} {(t - a)}^{(n + 1) α} + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{Γ (γ) | u + v |} \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{Γ (γ)}{Γ (n α + γ)} {(t - a)}^{n α + γ - 1} + \frac{| v | {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}}}{Γ (γ) | u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) \times \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{Γ (γ)}{Γ (n α + γ)} {(t - a)}^{n α + γ - 1} = (ϵ_{1} + ϵ_{2}) (\frac{{(t - a)}^{α}}{Γ (α + 1)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ ((n + 1) α + 1)} {(t - a)}^{(n + 1) α}) + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(t - a)}^{n α + γ - 1}) + \frac{| v |}{| u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \times (\frac{{(t - a)}^{γ - 1}}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(t - a)}^{n α + γ - 1}) .

Hence, for each $t \in J$ ⁠, we have

\begin{array}{l} {(t - a)}^{1 - γ} ‖ (z_{1} (t) - z_{2} (t)) ‖ \\ \leq (ϵ_{1} + ϵ_{2}) (\frac{{(t - a)}^{α - γ + 1}}{Γ (α + 1)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ ((n + 1) α + 1)} {(t - a)}^{(n + 1) α - γ + 1}) \\ + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(t - a)}^{n α}) \\ + \frac{| v |}{| u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \\ \times (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(t - a)}^{n α}) . \end{array}

Thus

\begin{array}{l} {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \\ \leq (ϵ_{1} + ϵ_{2}) (\frac{{(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ ((n + 1) α + 1)} {(b - a)}^{(n + 1) α - γ + 1}) \\ + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α}) \\ + \frac{| v |}{| u + v |} \frac{(n_{1} + ζ n_{2})}{Γ (α - γ + 1)} {(b - a)}^{α} B (γ, α - γ + 1) {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \\ \times (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α}), \end{array}

which, together with (4.3), yields

\begin{array}{l} {‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \leq Z^{- 1} [(ϵ_{1} + ϵ_{2}) \\ \times (\frac{{(b - a)}^{α - γ + 1}}{Γ (α + 1)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ ((n + 1) α + 1)} {(b - a)}^{(n + 1) α - γ + 1}) \\ + \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α})] . □ \end{array}

(4.5)

Remark 4.1.

If $ϵ_{1} = ϵ_{2} = 0$ in the inequality (4.4), then $z_{1}, z_{2}$ are solutions of the problem (1.1) in the space $C_{1 - γ}^{γ} [J, X]$ and the inequality (4.5) takes the form

{‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} \leq Z^{- 1} \frac{| \bar{w_{1}} - \bar{w_{2}} |}{| u + v |} (\frac{1}{Γ (γ)} + \sum_{n = 1}^{\infty} {(n_{1} + ζ n_{2})}^{n} \frac{1}{Γ (n α + γ)} {(b - a)}^{n α}),

which provides the information with respect to continuous dependence on the solution of the problem (1.1). In addition, if $\bar{w_{1}} = \bar{w_{2}}$ we get ${‖ z_{1} - z_{2} ‖}_{C_{1 - γ}} = 0,$ which proves the uniqueness of solutions of the system (1.1).

Remark 4.2.

One can note that our results for the Hilfer fractional integrodifferential equation (1.1) correspond to initial boundary value problem for $u = 1, v = 0$ ⁠, terminal boundary value problem for $u = 0, v = 1$ and anti-periodic problem for $u = 1, v = 1, w = 0$ ⁠.

Remark 4.3.

If $β = 1$ ⁠, then Eq. (1.1) reduces to the Caputo fractional integrodifferential equation with boundary conditions as in [12].

The publisher wishes to inform readers that the article “On abstract Hilfer fractional integrodifferential equations with boundary conditions” was originally published by the previous publisher of the Arab Journal of Mathematical Sciences and the pagination of this article has been subsequently changed. There has been no change to the content of the article. This change was necessary for the journal to transition from the previous publisher to the new one. The publisher sincerely apologises for any inconvenience caused. To access and cite this article, please use Thabet, S.T.M., Ahmad, B. and Agarwal, R.P. (2019), “On abstract Hilfer fractional integrodifferential equations with boundary conditions”, Arab Journal of Mathematical Sciences, Vol. 26 No. 1/2, pp. 107-125. The original publication date for this paper was 14/03/2019.

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Sabri T.M. Thabet, Bashir Ahmad and Ravi P. Agarwal

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) license. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this license may be seen at http://creativecommons.org/licences/by/4.0/legalcode

On abstract Hilfer fractional integrodifferential equations with boundary conditions

1. Introduction

2. Preliminaries

3. Existence of solutions

4. $ϵ -$ Approximate solution

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On abstract Hilfer fractional integrodifferential equations with boundary conditions

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4. $ϵ -$ Approximate solution