Weak Solution for a Fractional Langevin Inclusion with the Katugampola–Caputo Fractional Derivative

Abstract

In this work, we examine the existence of weak solution for a class of boundary value problems involving fractional Langevin inclusion with the Katugampola–Caputo fractional derivative under specified conditions contain the Pettis integrability assumption. The Mönch fixed point theorem is used with the weak noncompactness measure approach to investigate the existence results. In order to illustrate our results, we present an example.

1. Introduction

Recent years have seen a lot of interest in fractional differential equations [1,2]. Many authors have researched boundary value problems of fractional differential equations and inclusions subject to various boundary conditions, for example, see [3,4,5,6,7] and the references mentioned therein. In the study of dynamical systems and stochastic processes, differential inclusions have been considered to be especially useful. See [8,9] for some recent results on fractional differential inclusion boundary value problems.

There are several approaches of the fractional integrals and derivatives such as Riemann–Liouville, Caputo, Hadamard, Erdelyi–Kober, Conformable, etc. Katugampola fractional operators are considered one of the most important due to they generalize the Riemann–Liouville and the Hadamard fractional operators into a unique form. The Katugampola fractional integral generalizes both the Riemann–Liouville fractional integral and the Hadamard fractional integral into a single form and it is also closely related to the Erdelyi–Kober operator that generalizes the Riemann–Liouville fractional integral. The Katugampola fractional derivative has been defined using the Katugampola fractional integral and as with any other fractional differential operator, it also extends the possibility of taking real number powers or complex number powers of the integral and differential operators [10,11,12,13,14].

When the random fluctuation force is supposed to be white noise, the Langevin equation successfully captures the Brownian motion. Otherwise, the particle motion is portrayed by the generalized Langevin equation [15,16,17]. Langevin’s equation has been popularized in fractal media in order to depict dynamical operations. As a different approach, differential equation models of fractional order are now employed in place of the differential equations of integer order to describe experiential and area measurement data. Mainradi and Pironi [18] reconstructed Brownian motion using the fractional Langevin equation. The two fluctuation-dissipation theorems and fractional calculus methodologies were used to obtain analytical formulas for the correlation functions. Many scientists have been interested in the fractional Langevin equation because of its numerous applications in various disciplines of science and because it has been studied under various conditions [19,20,21].

One of the most popular methods for demonstrating the existence of a certain operator equation is to reformulate the problem as a fixed point problem and determine whether the latter can be solved using a fixed point justification. Measures of noncompactness are important in fixed point theory and have a wide range of applications in nonlinear analysis, differential equations, integral and integro-differential equations, fractional integral equations, optimization and other fields. Kuratowski was the first to establish the notion of measure of noncompactness in 1930. Since then, other noncompactness measures have been proposed. The Hausdorff measure of noncompactness, which was established by [22] in 1957 (and later investigated by [23]), the inner Hausdorff measure of noncompactness, and the Istr$\check{a}$tescu measure, which was introduced by [24] in 1972, are the most important. De Blasi pioneered the theory of weak noncompactness measures in his paper [25,26], where he developed the first weak noncompactness measure. De Blasi’s measure of noncompactness might be thought of as a counterpart to the classical Hausdorff measure. Due to the extreme role of measure on noncompactness techniques for investigation the existence and uniqueness of solution to differential and integral equations and inclusions, several contributors have applied them in their studies, see [27,28,29,30,31] and the references cited therein.

In 2007, Ouahab [32] used a selection theorem and a fixed point theorem to study the existence of solutions for fractional differential inclusions. The research [33] constructed some new existence findings for fractional differential inclusions based on the multivalued map fixed point theorem. The work in [34] examined the problem for the single-valued map of the form

$$\left\{\begin{array}{c}{}^c{D}^{\beta }{(}^c{D}^{\alpha }+\lambda )u\left(t\right)=f(t,u\left(t\right)),0<\alpha ⩽1,1<\beta \le 2,\hfill \\ u\left(0\right)+u\left(1\right)=0,u^{\prime }\left(0\right)=0,{}^c{D}^{\alpha }u\left(1\right)=\frac{\mu }{\Gamma \left(\gamma \right)}{\int }_0^{\eta }{(\eta -s)}^{\gamma -1}u\left(s\right)ds\hfill \end{array}\right.$$

where $0<\eta <1$, $\gamma >0$ and $\mu \in \mathbb{R}$. Motivated by the correct method which is presented in [34], in this article, we investigate the existence of solutions for the following boundary value problem of Langevin fractional inclusion with the generalized Caputo fractional derivative (2)

$$\begin{array}{cc}\hfill & {}^c{D}^{\beta ,\rho }{(}^c{D}^{\alpha ,\rho }+\lambda )x\left(t\right)\in H(t,x\left(t\right)),t\in \mathcal{J}=[0,1],\hfill \end{array}$$

(1)

$$\begin{array}{cc}\hfill & x\left(0\right)=0,\underset{t\to 0}{lim}{t}^{1-\rho }{x}^{\prime }\left(t\right)=0,x\left(1\right)=\mu I^{\zeta ,\rho }x\left(\eta \right)\hfill \end{array}$$

(2)

where $\lambda >0$ is the dissipative parameter, $\mu \in \mathbb{R}$, $\rho >1$, $0<\eta <1$, $I^{\zeta ,\rho }$ is the Katugampola fractional integral of order $\zeta >0$ and ${}^c{D}^{\alpha ,\rho }$ and ${}^c{D}^{\beta ,\rho }$ are the K-C fractional derivatives of orders $0<\alpha \le 1$ and $1<\beta \le 2$, respectively. $H:\mathcal{J}\times \mathbb{E}\to \mathcal{P}\left(\mathbb{E}\right)$ is a multivalued map, $\mathbb{E}$ is a real Banach space with norm $\parallel ·\parallel$ and dual space ${\mathbb{E}}^{*}$ and by $\mathcal{P}\left(\mathbb{E}\right)$, we denote the family of all nonempty subsets of $\mathbb{E}$ and let $(\mathbb{E},w)$ stand for the space $\mathbb{E}$ with its weak topology.

In order to identify solutions to differential equations and investigate the existence of solutions to the problems (1) and (2), we utilize Mönch’s fixed point theorem in conjunction with the approach of measures of weak noncompactness. An overview of the genesis of this technique can be found in a study by Banaś and Goebel [35], which has since been developed and applied many studies; for example, [36,37,38,39]. Additionally, in many studies, this method has been used. For instance, in [40], the authors studied the existence results for the fractional differential inclusion with non-separated boundary conditions. In [41], the contributions follow this methodology to discuss the existence of weak solution to the coupled Hadamard fractional system. The work in [42] also deals with the study of existence of weak solutions for the problem of Hilfer fractional differential inclusions.

However, what distinguishes this research is that it is, to the best of our knowledge, the first study that deals with Langevin fractional inclusion with the Katugampola–Caputo fractional derivative with this method in the study. This derivative is considered to be a generalized of classical Caputo fractional derivatives by letting $\rho \to 1$ and Hadamard fractional derivatives by letting $\rho \to 0$. Therefore, our results yield three approaches at the same time.

The remainder of this article is organized as follows. The foundations of fractional calculus and multivalued maps, including definitions and notations, are covered in Section 2. Section 3 presents the main findings for fractional differential inclusions. In the final section, we provide an example to demonstrate our essential point.

2. Preliminaries

Assume that $C(\mathcal{J},\mathbb{E})$ is the Banach space containing all continuous functions u from $\mathcal{J}$ to $\mathbb{E}$ in accordance with the norm

$${\parallel u\parallel }_{\infty }=sup\{\parallel u\left(t\right)\parallel :t\in \mathcal{J}\},$$

and let ${\mathcal{L}}^1(\mathcal{J},\mathbb{E})$ denote the Banach space of functions $u:\mathcal{J}\to \mathbb{E}$ that are the Lebesgue integrable in accordance with the norm

$${\parallel u\parallel }_{{\mathcal{L}}^1}={\int }_0^1\parallel u\left(s\right)\parallel ds.$$

The space of fundamentally bounded measurable functions is represented by ${\mathcal{L}}^{\infty }(\mathcal{J},\mathbb{E})$ in accordance with the norm

$${\parallel u\parallel }_{{\mathcal{L}}^{\infty }}=inf\{l>0:\parallel u\left(t\right)\parallel \le l,a.e.t\in \mathcal{J}\}.$$

Definition 1.

[43] The function $v:\mathcal{J}\to \mathbb{E}$ is said to be the Pettis integrable on $\mathcal{J}$ if and only if there is an element $v_{\mathcal{I}}\in \mathbb{E}$ equivalent to every $\mathcal{I}\subset \mathcal{J}$ such that $\phi \left(v_{\mathcal{I}}\right)={\int }_{\mathcal{I}}\phi \left(v\left(s\right)\right)ds$ for all $\phi \in {\mathbb{E}}^{*}$, where it is claimed that the integral on the right exists in the Lebesgue sense. It follows that $v_{\mathcal{I}}={\int }_{\mathcal{I}}v\left(s\right)ds$. Assume that $P(\mathcal{J},\mathbb{E})$ would be the space of all Pettis integrable functions with E values in the $\mathcal{J}$ interval.

Proposition 1.

If $v\in P(\mathcal{J},E)$ and f is a measurable and essentially bounded E-valued function, then $v·f\in P(\mathcal{J},E)$.

Lemma 1.

Suppose that $\mathbb{E}$ is a normed space with $v_0\ne 0$. Then, there exists $\phi \in {\mathbb{E}}^{*}$ where $\parallel \phi \parallel =1$ and $\phi \left(v_0\right)=\parallel v_0\parallel$.

The previous result is a direct consequence of the Hahn–Banach theorem. For completeness, we review the definitions of the K-C derivative of the fractional and the Pettis integrable. For $a,b,c\in \mathbb{R}$, let the space $X_c^p$ be defined as:

$$X_c^p=\left\{g:[a,b]\to \mathbb{R}:∥g∥={\left({\int }_a^b{\left|t^{c}g\left(t\right)\right|}^p\frac{dt}{t}\right)}^{\frac{1}{p}}<\infty ,1⩽p<\infty \right\}.$$

For $p=\infty ,∥g∥=esssu{p}_{a⩽t⩽b}(t^c\left|g\left(t\right)\right|)$.

We also recall that $AC[a,b]$ is the absolutely continuous functions space on $[a,b]$ and

$$A{C}_{\mathrm{Y}}^m=\left\{g:[a,b]\to \mathbb{C}\mathrm{and}{\mathrm{Y}}^{m-1}g\in AC[a,b],\mathrm{Y}=t^{1-\rho }\frac{d}{dt}\right\}$$

where $∥g∥={\Sigma }_{k=0}^{m-1}∥{\mathrm{Y}}^{k}g∥$ (see [44,45]).

Definition 2 ([44]).

The Katugampola fractional Pettis integrable operator of order ϑ of $g:\mathcal{J}\to \mathbb{E}$ is defined as:

$$I^{\vartheta ,\rho }g\left(t\right)={\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\vartheta -1}g\left(s\right)\frac{ds}{s^{1-\rho }}$$

where the symbol $“\int ”$ indicate the Pettis integral.

Lemma 2 ([44]).

Let $\alpha >0$ and $\rho >1$. Then,

$$I^{\vartheta ,\rho }{t}^s=\frac{\Gamma (\frac{s}{\rho }+1)}{\Gamma (\frac{s}{\rho }+\vartheta +1)}\frac{t^{\rho \vartheta +s}}{{\rho }^{\vartheta }}.$$

Lemma 3 ([45]).

Let $Re\left(\vartheta \right)⩾0$,$m=\left[Re\right(\vartheta \left)\right]+1$ and $g\in A{C}_{\mathrm{Y}}^m$. Then,

1. 

If$\vartheta \notin {\mathbb{N}}_0$,

$${}^c{D}^{\vartheta ,\rho }g\left(t\right)=I^{m-\vartheta ,\rho }\left({\mathrm{Y}}^{m}g\right)\left(t\right),\mathrm{Y}=t^{1-\rho }\frac{d}{dt}.$$

2. 

If$\vartheta \in \mathbb{N}$,

$${}^c{D}^{\vartheta ,\rho }g\left(t\right)={\mathrm{Y}}^{m}g.$$

Lemma 4.

Let $g\in A{C}_{\mathrm{Y}}^m$ or $C_{\mathrm{Y}}^m$ and $\vartheta \in \mathbb{C}$. Then,

$${\displaystyle I^{\vartheta ,\rho }{}^c{D}^{\vartheta ,\rho }g\left(t\right)=g\left(t\right)-\sum _{k=0}^{m-1}\frac{\left({\mathrm{Y}}^{k}g\right)\left(0\right)}{k!}{\left(\frac{t^{\rho }}{\rho }\right)}^k.}$$

Lemma 5.

Let $0<\nu \le \vartheta$. Then,

$${}^c{D}^{\nu ,\rho }{I}^{\vartheta ,\rho }g\left(t\right)=I^{\vartheta -\nu ,\rho }g\left(t\right).$$

The following equation may be obtained from Lemma 4 and Lemma 5

$${\displaystyle {\mathrm{Y}}^m\left(I^{\vartheta ,\rho }g\left(t\right)\right)=I^{\vartheta ,\rho }{\mathrm{Y}}^{m}g\left(t\right)+\sum _{k=1}^m\frac{\left({\mathrm{Y}}^{m-k}g\right)\left(0\right)}{\Gamma \left(\vartheta -k+1\right)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\vartheta -k}.}$$

By taking $m=1$, the previous formula becomes

$$\mathrm{Y}\left(I^{\vartheta ,\rho }g\left(t\right)\right)=I^{\vartheta ,\rho }\mathrm{Y}g\left(t\right)+\frac{g\left(0\right)}{\Gamma \left(\vartheta \right)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\vartheta -1}.$$

Assume that $(\mathbb{E},\parallel ·\parallel )$ is a Banach space and consider

• $({\mathbb{K}}_1) {\mathcal{P}}_{cl}\left(\mathbb{E}\right)=\{\mathfrak{V}\in \mathcal{P}\left(\mathbb{E}\right):\mathfrak{V}\mathrm{is}\mathrm{closed}\},$
• $({\mathbb{K}}_2) {\mathcal{P}}_b\left(\mathbb{E}\right)=\{\mathfrak{V}\in \mathcal{P}\left(\mathbb{E}\right):\mathfrak{V}\mathrm{is}\mathrm{bounded}\},$
• $({\mathbb{K}}_3) {\mathcal{P}}_{cp}\left(\mathbb{E}\right)=\{\mathfrak{V}\in \mathcal{P}\left(\mathbb{E}\right):\mathfrak{V}\mathrm{is}\mathrm{compact}\},$
• $({\mathbb{K}}_4) {\mathcal{P}}_{cp,cv}\left(\mathbb{E}\right)=\{\mathfrak{V}\in \mathcal{P}\left(\mathbb{E}\right):\mathfrak{V}\mathrm{is}\mathrm{compact}\mathrm{and}\mathrm{convex}\}.$
• $({\mathbb{K}}_5) {\mathcal{P}}_{cl,cv}\left(\mathbb{E}\right)=\{\mathfrak{V}\in \mathcal{P}\left(\mathbb{E}\right):\mathfrak{V}\mathrm{is}\mathrm{closed}\mathrm{and}\mathrm{convex}\}.$

Definition 3.

A multivalued map $\mathfrak{G}:\mathbb{E}\to {\mathcal{P}}_{cl}\left(\mathbb{E}\right)$ is considered to be measurable if for every $\mathfrak{w}\in \mathbb{E}$, the function

$$t\to d(\mathfrak{w},\mathfrak{G}(t\left)\right)=inf\{\parallel \mathfrak{w}-v\parallel :v\in \mathfrak{G}(t\left)\right\}$$

is measurable.

Definition 4.

A multivalued map’s selection set $\mathfrak{G}:\mathbb{E}\to {\mathcal{P}}_{cl}\left(\mathbb{E}\right)$ is defined as

$$S_{\mathfrak{G}}=\{x\in {\mathcal{L}}^1(\mathcal{J},\mathbb{E}):x\left(t\right)\in \mathfrak{G}\left(t\right),a.e.t\in \mathcal{J}\}.$$

For every $x\in C(\mathcal{J},\mathbb{E})$, the set $S_{H\circ x}$ of selections of H is given by

$$S_{H\circ x}=\{\sigma \in {\mathcal{L}}^1(\mathcal{J},\mathbb{E}):\sigma \left(t\right)\in H\circ x,a.e.t\in \mathcal{J}\}.$$

For further information on multivalued maps, see [46,47].

Definition 5 ([48]).

Assume that $\mathbb{E}$ is a Banach space, ${\mathsf{\Phi }}_{\mathbb{E}}$ denotes the set of all bounded subsets of $\mathbb{E}$ and $B_1$ is the unit ball in $\mathbb{E}$. Then, the map $\mathcal{B}:{\mathsf{\Phi }}_{\mathbb{E}}\to [0,\infty )$ which defined by

$$\mathcal{B}\left(X\right)=inf\left\{ϵ>0:\mathit{there}\mathit{exists}\mathit{compact}\mathit{subset}\mathsf{\Phi }\mathit{of}\mathbb{E}\mathit{such}\mathit{that}X\subset ϵB_1+\mathsf{\Phi }\right\}$$

is known as The De Blasi measure of weak noncompactness.

Lemma 6.

Assume that $F\subset C$ is a bounded and equicontinuous subset. After that, the function $t\to \mathcal{B}\left(F\right(t\left)\right)$ is continuous on $\mathcal{J}$ and

$${\mathcal{B}}_C\left(F\right)=\underset{t\in \mathcal{J}}{max}{\mathcal{B}}_C\left(F\left(t\right)\right)$$

and

$$\mathcal{B}\left({\int }_{\mathcal{J}}x\left(s\right)ds\right)\le {\int }_{\mathcal{J}}\mathcal{B}\left(F\left(s\right)\right)ds$$

where $F\left(s\right)=\left\{x\right(s):x\in F,s\in \mathcal{J}\}$ and ${\mathcal{B}}_C$ is the De Blasi measure of weak noncompactness defined on the bounded sets of C.

Lemma 7.

Assume that $\mathbb{E}$ is a Banach space and Q is a nonempty closed bounded convex equicontinuous subset of $C(\mathcal{J},\mathbb{E})$. Let T have a graph that is weakly sequentially closed. The operator T has a solution in Q if the implication

$$\begin{array}{c}\hfill \overline{V}=\overline{conv}(\left\{0\right\}\cup T\left(V\right))⟹V\mathit{is}\mathit{relatively}\mathit{weakly}\mathit{compact}\end{array}$$

(3)

holds for each subset V of Q.

Now, we start proving the auxiliary lemma, which will be essential in the following section.

Lemma 8.

The unique representation of solution to the following linear fractional differential equation

$$\begin{array}{cc}\hfill & {}^c{D}^{\beta ,\rho }{(}^c{D}^{\alpha ,\rho }+\lambda )x\left(t\right)=h\left(t\right),t\in [0,1]\hfill \\ \hfill & x\left(0\right)=0,\underset{t\to 0}{lim}{t}^{1-\rho }{x}^{\prime }\left(t\right)=0,x\left(1\right)=\mu I^{\zeta ,\rho }x\left(\eta \right),\hfill \end{array}$$

for $\rho >1$, $0<\beta \le 2$ and $0<\alpha <1$ is given by

$$\begin{array}{cc}\hfill x\left(t\right)& =\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{h\left(s\right)}{s^{1-\rho }}ds-\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & +t^{\rho (\alpha +1)}[\frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}\frac{x\left(s\right)}{s^{1-\rho }}ds-\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{h\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & +\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds].\hfill \end{array}$$

(4)

and when $\alpha =1$, it can be represented in the following form:

$$\begin{array}{cc}\hfill x\left(t\right)& =\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(v\right)}{v^{1-\rho }}dv\right)ds\hfill \\ \hfill & +N\left(t\right)[\frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}x\left(s\right)ds\hfill \\ \hfill & -\frac{1}{\Gamma \left(\beta \right)}{\int }_0^1{e}^{-\lambda \left(\frac{1-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(v\right)}{v^{1-\rho }}dv\right)ds]\hfill \end{array}$$

(5)

where

$$N\left(t\right)=\frac{\lambda t^{\rho }-\rho (1-e^{\frac{-\lambda t^{\rho }}{\rho }})}{\lambda -\rho (1-e^{\frac{-\lambda }{\rho }})}.$$

Proof. 

From Lemmas 4 and 5, we can obtain

$$\begin{array}{cc}\hfill x\left(t\right)& =\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{h\left(s\right)}{s^{1-\rho }}ds-\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & +\frac{c_0}{\Gamma (\alpha +1)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\alpha }+\frac{c_1}{\Gamma (\alpha +2)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\alpha +1}+c_2,\hfill \\ \hfill \mathrm{Y}x\left(t\right)& =\frac{1}{\Gamma (\alpha +\beta -1)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -2}\frac{h\left(s\right)}{s^{1-\rho }}ds-\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}{x}^{\prime }\left(s\right)ds\hfill \\ \hfill & +\frac{1}{\Gamma \left(\alpha \right)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\alpha -1}[c_0-\lambda x\left(0\right)]+\frac{c_1}{\Gamma (\alpha +1)}{\left(\frac{t^{\rho }}{\rho }\right)}^{\alpha }.\hfill \end{array}$$

(6)

Using the boundary conditions, we find $c_2=c_0=0$ and

$$\begin{array}{cc}\hfill c_1& ={\rho }^{\alpha +1}\Gamma (\alpha +2)[\frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}\frac{x\left(s\right)}{s^{1-\rho }}ds-{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{h\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & +\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds].\hfill \end{array}$$

By substituting into (6), we obtain (4). When $\alpha =1$, we obtain

$$\begin{array}{cc}\hfill x\left(t\right)=& \frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(v\right)}{v^{1-\rho }}dv\right)ds,\hfill \\ \hfill & +\frac{c_0}{\lambda }\left[1-e^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}\right]+\frac{c_1}{{\lambda }^2\rho }\left[\lambda t^{\rho }-\rho -e^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}\right]+c_2{e}^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}\hfill \\ \hfill \mathrm{Y}x\left(t\right)=& \frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(s\right)}{s^{1-\rho }}ds-\frac{\lambda }{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(v\right)}{v^{1-\rho }}dv\right)ds\hfill \\ \hfill & +c_0{e}^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}+\frac{c_1}{\lambda \rho }\left[\rho -\rho e^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}\right]-\lambda c_2{e}^{-\lambda \left(\frac{t^{\rho }}{\rho }\right)}.\hfill \end{array}$$

(7)

Thus,

$$\mathrm{Y}x\left(t\right)=I^{\beta ,\rho }h\left(t\right)-\lambda x\left(t\right)+c_0+c_1\left(\frac{t^{\rho }}{\rho }\right).$$

From the boundary condition, we find that $c_0=c_2=0$ and

$$\begin{array}{cc}\hfill c_1=& \frac{{\lambda }^2\rho }{\lambda -\rho (1+e^{\frac{-\lambda }{\rho }})}[\frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}x\left(s\right)ds\hfill \\ \hfill -& \frac{1}{\Gamma \left(\beta \right)}{\int }_0^1{e}^{-\lambda \left(\frac{1-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{h\left(v\right)}{v^{1-\rho }}dv\right)ds].\hfill \end{array}$$

Substituting the values of $c_0,c_1,c_2$ in (7), we obtain (5). This completes the proof. □

3. Main Results

The following assumptions are required to prove the major findings:

$\left({\mathcal{H}}_1\right)$

$H:\mathcal{J}\times \mathbb{E}\to P_{cp,cv}\left(\mathbb{E}\right)$ has a weakly sequentially closed graph;

$\left({\mathcal{H}}_2\right)$

For every $x:\mathcal{J}\to \mathbb{E}$, there is a measurable scalar function $\sigma :\mathcal{J}\to \mathbb{E}$ which has $\sigma \left(t\right)\in H(t,x(t\left)\right)$ a.e. on $\mathcal{J}$ and $\sigma$ is Pettis integrable on $\mathcal{J}$;

$\left({\mathcal{H}}_3\right)$

There is a $\delta \in L^{\infty }(\mathcal{J},{\mathbb{R}}^{+})$ and a non-decreasing continuous function $\omega :[0,\infty )\to [0,\infty )$ as well as

$$\Vert H(t,x)\Vert =sup\left\{\right|\sigma |:\sigma \in H(t,x\left)\right\}⩽\delta \left(t\right)\omega (\Vert x\Vert );$$

$\left({\mathcal{H}}_4\right)$

The following inequality holds for every bounded set $\Delta \subset \mathbb{E}$ and each $t\in \mathcal{J}$:

$$\mathcal{B}\left(H\right(t,\Delta \left)\right)⩽\delta \left(t\right)\mathcal{B}(\Delta ).$$

Theorem 1.

Suppose that $\mathbb{E}$ is a Banach space and let hypotheses $\left({\mathcal{H}}_1\right)-\left({\mathcal{H}}_4\right)$ be verified. If

$$\begin{array}{c}\hfill \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}<1,0<\alpha <1\end{array}$$

(8)

or if

$$\begin{array}{c}\hfill \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\beta }\Gamma \left(\beta \right)}<1,\alpha =1.\end{array}$$

(9)

then there is at least one weak solution to the $\mathcal{J}$ issue (1) and (2).

Proof. 

Consider the multivalued map $\Xi :C(\mathcal{J},\mathbb{E})\to P_{cl,cv}$ which is defined as follows

• If $0<\alpha <1$

  $$\begin{array}{cc}\hfill \Xi \left(x\right)=& \{q\in C(\mathcal{J},E):q\left(t\right)=k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds,\sigma \in S_{H\circ x}\}\hfill \end{array}$$

  (10)

  where

  $$\begin{array}{cc}\hfill k=& \frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}\frac{q\left(s\right)}{s^{1-\rho }}ds-\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill +& \frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^1{\left(\frac{1-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds.\hfill \end{array}$$

  (11)
• If $\alpha =1$

  $$\begin{array}{cc}\hfill \Xi \left(x\right)=& \{q\in C(\mathcal{J},E):q\left(t\right)=k_{1}N\left(t\right)+\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\hfill \\ \hfill & \times \left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds,\sigma \in S_{H\circ x}\},\hfill \end{array}$$

  (12)

  where

  $$\begin{array}{cc}\hfill k_1=& [\frac{\mu }{\Gamma \left(\zeta \right)}{\int }_0^{\eta }{\left(\frac{{\eta }^{\rho }-s^{\rho }}{\rho }\right)}^{\zeta -1}q\left(s\right)ds\hfill \\ \hfill & -\frac{1}{\Gamma \left(\beta \right)}{\int }_0^1{e}^{-\lambda \left(\frac{1-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds].\hfill \end{array}$$

  (13)

To begin, Proposition 1 indicates that the function

$$\frac{1}{\Gamma (\alpha +\beta )}{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }},t\in \mathcal{J}$$

is Pettis integrable. As a result, $\Xi$ is a well-defined operator. Assume $r>0$ and that

$$\begin{array}{c}\hfill r>\frac{{\Lambda }_1}{{\Lambda }_2}\end{array}$$

(14)

where

$${\Lambda }_1=\frac{2\delta \omega \left(r\right)}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)},{\Lambda }_2=1-\left(\frac{2\lambda }{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}+\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}\right),\mathrm{if}0<\alpha <1$$

and

$${\Lambda }_1=\left(1+\frac{\lambda +\rho }{|\lambda -\rho |}\right)\frac{\delta \omega \left(r\right)}{\lambda {\rho }^{\beta }\Gamma (\beta +1)},{\Lambda }_2=1-\frac{\lambda +\rho }{|\lambda -\rho |}\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)},\mathrm{when}\alpha =1.$$

Consider the set

$$\begin{array}{cc}\hfill {\Theta }_{\alpha }=& \{x\in {C(\mathcal{J},E):\parallel x\parallel }_{\infty }\le r,\parallel x\left(t_2\right)-x\left(t_1\right)\parallel \le k^{*}\left(t_2^{\rho (\alpha +1)}-t_1^{\rho (\alpha +1)}\right)\hfill \\ \hfill & +\frac{\delta \omega \left(r\right)}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}\left(t_2^{\rho (\alpha +\beta )}-t_1^{\rho (\alpha +\beta )}\right)+\frac{2\lambda r}{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}{(t_2^{\rho }-t_1^{\rho })}^{\alpha },\mathrm{for}{t}_1,t_2\in \mathcal{J}\}\hfill \end{array}$$

(15)

when $0<\alpha <1$ where

$$k^{*}=\left[\frac{\delta \omega \left(r\right)}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}+\left(\frac{\lambda }{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}+\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}\right)r\right].$$

and

$$\begin{array}{cc}\hfill {\Theta }_1=& \{x\in {C(\mathcal{J},E):\parallel x\parallel }_{\infty }\le r,\parallel x\left(t_2\right)-x\left(t_1\right)\parallel \le k_1^{*}(t_2^{\rho }-t_1^{\rho })\hfill \\ \hfill & +\frac{2\delta \omega \left(r\right)}{{\rho }^{\beta }\Gamma (\beta +1)}\left[1-e^{-\lambda (\frac{t_2^{\rho }-t_1^{\rho }}{\rho })}\right],\mathrm{for}{t}_1,t_2\in \mathcal{J}\};\hfill \end{array}$$

(16)

when $\alpha =1$ where

$$k_1^{*}=\frac{\lambda }{|\lambda -\rho |}\left[\frac{\delta \omega \left(r\right)}{\lambda {\rho }^{\beta }\Gamma \left(\beta +1\right)}+\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}r\right].$$

The subsets ${\Theta }_{\alpha }$ and ${\Theta }_1$ are convex, closed and equicontinuous. We shall demonstrate that the operator $\Xi$ meets every requirement in our theorem. The next stages will provide the evidence

step1:

Ξ maps ${\Theta }_{\alpha },0<\alpha \le 1$ into itself. Consider $q\in \Xi \left(x\right)$. Then, there is $x\in {\Theta }_{\alpha }$ with $q\in \Xi \left(x\right)$ and there exists a Pettis integrable $\sigma :\mathcal{J}\to \mathbb{E}$ with $\sigma \left(t\right)\in H(t,x(t\left)\right)$. Assume that $q\left(t\right)\ne 0$ for all $t\in \mathcal{J}$, then there exist $\phi \in {\mathbb{E}}^{*}$ with $\parallel \phi \parallel =1$ as well as $\parallel q\parallel =\phi \left(q\right(t\left)\right)$. Then,

(i)

If $0<\alpha <1$

$$\begin{array}{cc}\hfill \parallel q\parallel =\phi \left(q\right(t\left)\right)& =\phi (k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds)\hfill \\ \hfill & \le \phi (k{t}^{\rho (\alpha +1)})+\phi \left(\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\right)\hfill \\ & +\phi \left(\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds\right)\hfill \\ \hfill & \le \frac{2\delta \omega \left(r\right)}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}+\frac{\mu {\eta }^{\zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}r+\frac{2\lambda }{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}r.\hfill \end{array}$$

(ii)

If $\alpha =1$

$$\begin{array}{cc}\hfill \parallel q\parallel =\phi \left(q\right(t\left)\right)& =\phi \left(k_{1}N\left(t\right)+{\int }_0^t\frac{e^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}}{\Gamma \left(\beta \right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds\right)\hfill \\ \hfill & \le \left(1+\frac{\lambda +\rho }{|\lambda -\rho |}\right)\frac{\delta \omega \left(r\right)}{\lambda {\rho }^{\beta }\Gamma (\beta +1)}+\frac{\lambda +\rho }{|\lambda -\rho |}\left(\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}\right)r.\hfill \end{array}$$

Therefore, from (14), $\parallel q\parallel <r$. Next, suppose $q\in \Xi \left(x\right)$ and $t_1,t_2\in \mathcal{J}$, with $t_1<t_2$ to ensure $q\left(t_2\right)-q\left(t_1\right)\ne 0$. Eventually, there exists $\phi \in {\mathbb{E}}^{*}$ as well as $\parallel q\left(t_2\right)-q\left(t_1\right)\parallel =\phi (q\left(t_2\right)-q\left(t_1\right))$. Hence,

(i)

If $0<\alpha <1$

$$\begin{array}{cc}\hfill & \parallel q\left(t_2\right)-q\left(t_1\right)\parallel =\phi (k{t_2}^{\rho (\alpha +1)}-k{t}_1^{\rho (\alpha +1)}+{\int }_0^{t_2}{\left(\frac{{t_2}^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }\Gamma (\alpha +\beta )}ds\hfill \\ \hfill & -{\int }_0^{t_1}{\left(\frac{t_1^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }\Gamma (\alpha +\beta )}ds-\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^{t_2}{\left(\frac{{t_2}^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & +\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^{t_1}{\left(\frac{t_1^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds)\hfill \\ \hfill & \le ∥k{t_2}^{\rho (\alpha +1)}-k{t}_1^{\rho (\alpha +1)}∥+∥{\int }_{t_1}^{t_2}{\left(\frac{t_2^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }\Gamma (\alpha +\beta )}ds∥\hfill \\ \hfill & +∥{\int }_0^{t_1}\left[{\left(\frac{t_2^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}-{\left(\frac{t_1^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\right]\frac{\sigma \left(s\right)}{s^{1-\rho }\Gamma (\alpha +\beta )}ds∥\hfill \\ \hfill & +∥\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^{t_1}[{\left(\frac{t_1^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}-{\left(\frac{t_2^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}]\frac{q\left(s\right)}{s^{1-\rho }}ds∥\hfill \\ \hfill & +∥\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_{t_1}^{t_2}{\left(\frac{t_2^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q\left(s\right)}{s^{1-\rho }}ds∥\hfill \\ \hfill & \le \left[\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}+\frac{\delta }{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}+\frac{\lambda }{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}r\right]\left(t_2^{\rho (\alpha +1)}-t_1^{\rho (\alpha +1)}\right)\hfill \\ \hfill & +\frac{\delta \omega \left(r\right)}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}(t_2^{\rho (\alpha +\beta )}-t_1^{\rho (\alpha +\beta )})+\frac{2\lambda r}{{\rho }^{\alpha }\Gamma \left(\alpha +1\right)}{(t_2^{\alpha }-t_1^{\alpha })}^{\rho }.\hfill \end{array}$$

(ii)

If $\alpha =1$. Likewise, we find

$$\begin{array}{cc}\hfill & \parallel q\left(t_2\right)-q\left(t_1\right)\parallel \le \frac{\lambda }{|\lambda -\rho |}\left[\frac{\delta \omega \left(r\right)}{\lambda {\rho }^{\beta }\Gamma \left(\beta +1\right)}+\frac{\mu {\eta }^{\rho \zeta }}{{\rho }^{\zeta }\Gamma \left(\zeta +1\right)}r\right](t_2^{\rho }-t_1^{\rho })\hfill \\ \hfill & +\frac{2\delta \omega \left(r\right)}{{\rho }^{\beta }\Gamma (\beta +1)}\left[1-e^{-\lambda \left(\frac{t_2^{\rho }-t_1^{\rho }}{\rho }\right)}\right].\hfill \end{array}$$

step2:

For any $x\in \Theta \alpha$, the operator $\Xi \left(x\right)$ is convex. In fact, if $q_1$ and $q_2$ are components of $\Xi \left(x\right)$, then Pettis’s integrable functions ${\sigma }_1,{\sigma }_2\in H(t,x\left(t\right))$ exist such that we have for any $\mathcal{J}$

$$\begin{array}{cc}\hfill q_i\left(t\right)& =k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{{\sigma }_i\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{q_i\left(s\right)}{s^{1-\rho }}ds,i=1,2.\hfill \end{array}$$

(17)

Assume that $0\le z\le 1$. Afterward, for each $t\in \mathcal{J}$, we have

(i)

If $0<\alpha <1$

$$\begin{array}{cc}\hfill [z{q}_1+(1-z)q_2]\left(t\right)& =k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{[z{\sigma }_1\left(s\right)+(1-z){\sigma }_2\left(s\right)]}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{[z{q}_1\left(s\right)+(1-z)q_2\left(s\right)]}{s^{1-\rho }}ds.\hfill \end{array}$$

(ii)

If $\alpha =1$

$$\begin{array}{cc}\hfill [z{q}_1+(1-z)q_2]\left(t\right)& =k_{1}N\left(t\right)+{\int }_0^t\frac{e^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}}{\Gamma \left(\beta \right)}{s}^{\rho -1}\hfill \\ \hfill & \times \left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{[z{\sigma }_1\left(v\right)+(1-z){\sigma }_2\left(v\right)]}{v^{1-\rho }}dv\right)ds.\hfill \end{array}$$

Since $S_{H\circ x}$ is convex (relying on that H has convex values), it follows that

$$z{q}_1+(1-z)q_2\in \Xi \left(x\right).$$

step3:

Ξ has a weakly sequentially closed graph. Let $(x_m,y_m)$ be a sequence in ${\Theta }_{\alpha }\times {\Theta }_{\alpha }$, which has $x_m\left(t\right)\to x\left(t\right)$ in $(\mathbb{E},\omega )$ for every $t\in \mathcal{J}$ and $y_m\left(t\right)\to y\left(t\right)$ in $(\mathbb{E},\omega )$ for every $t\in \mathcal{J}$ and $y_m\in \Xi \left(x_m\right)$ for $m\in \mathbb{N}$. We demonstrate $y\in \Xi \left(x\right)$. Since $y_m\in \Xi \left(x_m\right)$, there exist ${\sigma }_m\in S_{H\circ x_m}$ as well as

(i)

If $0<\alpha <1$

$$\begin{array}{cc}\hfill y_m\left(t\right)& =k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{{\sigma }_m\left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{y_m\left(s\right)}{s^{1-\rho }}ds.\hfill \end{array}$$

We demonstrate there exists ${\sigma }_m\in S_{H\circ x_m}$ as well as, for every $t\in \mathcal{J}$,

$$\begin{array}{cc}\hfill y\left(t\right)& =k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{y\left(s\right)}{s^{1-\rho }}ds.\hfill \end{array}$$

Therefore, there exists a subsequence ${\sigma }_{m_l}$ as well as ${\sigma }_{m_l}$ being Pettis integrable because $H(·,·)$ has compact values.

$$\begin{array}{c}\hfill {\sigma }_{m_l}\in H(t,x_m\left(t\right))\mathrm{a}.\mathrm{e}.t\in \mathcal{J}\mathrm{and}{\sigma }_{m_l}(·)\to \sigma (·)\mathrm{in}(\mathbb{E},\omega )\mathrm{as}l\to \infty .\end{array}$$

We find that $\sigma$ belongs to H since $H(t,x(t\left)\right)$ has a weakly sequentially closed graph. The Lebesgue Dominated Convergence Theorem for the Pettis integrable gives us the following

$$\begin{array}{cc}\hfill \phi \left(y_m\left(t\right)\right)& \to \phi (k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hfill & -\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{y\left(s\right)}{s^{1-\rho }}ds);\hfill \end{array}$$

which means that $y_m\left(t\right)\to (\Xi x)\left(t\right)$ in $(\mathbb{E},\omega )$. Since this convergence holds, for every values of $t\in \mathcal{J}$, we obtain $y\in \Xi$.

(ii)

If $\alpha =1$, the same approach can be taken to obtain the result directly.

step4:

The implication (3) holds. Suppose that V is a subset of ${\Theta }_{\alpha }$ as well as $\overline{V}=\overline{\mathrm{conv}}(\left\{0\right\}\cup \Xi \left(V\right))$. Obviously, $V\left(t\right)\subset \overline{\mathrm{conv}}(\left\{0\right\}\cup (\Xi \left(V\left(t\right)\right))$ for every $t\in \mathcal{J}$. Further, if the function is continuous on $\mathcal{J}$, the closure of it is compact, so the value of measure $\mathcal{B}$ of continuous function is 0. By (${\mathcal{H}}_4$) and Lemma 6 of the measure $\mathcal{B}$, for any $t\in \mathcal{J}$, we have

(i)

If $0<\alpha <1$,

$$\begin{array}{c}\mathcal{B}(\Xi \left(V\right)\left(t\right))=\mathcal{B}\{k{t}^{\rho (\alpha +1)}+\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hspace{1em}-\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds:\sigma \in S_{H\circ x},x\in V,t\in \mathcal{J}\}\hfill \\ \hspace{1em}\le \mathcal{B}\left\{k{t}^{\rho (\alpha +1)}\right\}+\mathcal{B}\left\{\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds:\sigma \in S_{H\circ x},t\in \mathcal{J}\right\}\hfill \\ \hspace{1em}+\mathcal{B}\left\{\frac{\lambda }{\Gamma \left(\alpha \right)}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha -1}\frac{x\left(s\right)}{s^{1-\rho }}ds:x\in V,t\in \mathcal{J}\right\}\hfill \\ \hspace{1em}=\mathcal{B}\left\{\frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\sigma \left(s\right)}{s^{1-\rho }}ds:\sigma \in S_{H\circ x},t\in \mathcal{J}\right\}\hfill \\ \hspace{1em}\le \frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\mathcal{B}H(s,V(s\left)\right)}{s^{1-\rho }}ds\hfill \\ \hspace{1em}\le \frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\delta \left(s\right)\mathcal{B}\left(V\right(s\left)\right)}{s^{1-\rho }}ds\hfill \\ \hspace{1em}\le \frac{1}{\Gamma (\alpha +\beta )}{\int }_0^t{\left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}^{\alpha +\beta -1}\frac{\delta \left(s\right)\sigma \left(s\right)}{s^{1-\rho }}ds\hfill \\ \hspace{1em}\le \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}{\parallel \sigma \parallel }_{\infty }.\hfill \end{array}$$

which leads to

$${\parallel \sigma \parallel }_{\infty }\le \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}{\parallel \sigma \parallel }_{\infty }.$$

This implies that

$${\parallel \sigma \parallel }_{\infty }·\left[1-\frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}\right]\le 0.$$

(ii)

If $\alpha =1$

$$\begin{array}{c}\beta (\Xi \left(V\right)\left(t\right))=\beta \{k_{1}N\left(t\right)+\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\hfill \\ \hspace{1em}\times \left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds:\sigma \in S_{H\circ x},x\in V,t\in \mathcal{J}\}\hfill \\ \hspace{1em}\le \beta \{k_{1}N\left(t\right)\}+\beta \{\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\hfill \\ \hspace{1em}\times \left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds:\sigma \in S_{H\circ x},t\in \mathcal{J}\}\hfill \\ \hspace{1em}=\beta \left\{\frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\sigma \left(v\right)}{v^{1-\rho }}dv\right)ds:\sigma \in S_{H\circ x},t\in \mathcal{J}\right\}\hfill \\ \hspace{1em}\le \frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\beta H(v,V(v\left)\right)}{v^{1-\rho }}dv\right)ds\hfill \\ \hspace{1em}\le \frac{1}{\Gamma \left(\beta \right)}{\int }_0^t{e}^{-\lambda \left(\frac{t^{\rho }-s^{\rho }}{\rho }\right)}{s}^{\rho -1}\left({\int }_0^s{\left(\frac{s^{\rho }-v^{\rho }}{\rho }\right)}^{\beta -1}\frac{\delta \left(s\right)\beta \left(V\right(s\left)\right)}{v^{1-\rho }}dv\right)ds\hfill \\ \hspace{1em}\le \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\beta }\Gamma (\beta +1)}{\parallel \sigma \parallel }_{\infty }\hfill \end{array}$$

which leads to

$${\parallel \sigma \parallel }_{\infty }\le \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\beta }\Gamma (\beta +1)}{\parallel \sigma \parallel }_{\infty }.$$

This implies that

$${\parallel \sigma \parallel }_{\infty }·\left[1-\frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\beta }\Gamma (\beta +1)}\right]\le 0.$$

By (8) and (9), we find ${\parallel \sigma \parallel }_{\infty }=0$ which means that $\sigma =0$ for all $t\in \mathcal{J}$. The proof is completed. □

4. An Example

Let the space

$$\mathbb{E}={\ell }_1=\left\{x=\left(x_1,x_2,\cdots ,x_n,\cdots \right),\sum _{n=1}^{\infty }\left|x_n\right|<\infty ,x_n\in C(\mathcal{J},\mathbb{R})\right\}$$

be the real sequence space which is a Banach space with the norm

$${\parallel x\parallel }_{\mathbb{E}}=\sum _{n=1}^{\infty }\left|x_n\right|.$$

We consider the following boundary value problem of Langevin fractional inclusion with the K-C fractional derivative type to illustrate our results

$$\left\{\begin{array}{cc}& {}^c{D}^{\frac{9}{8},\frac{5}{2}}\left({}^c{D}^{\frac{7}{8},\frac{5}{2}}+\frac{6}{5}\right)x\left(t\right)\in H(t,x\left(t\right)),t\in \mathcal{J},\hfill \\ & x\left(0\right)=0,\underset{t\to 0}{lim}{t}^{-\frac{3}{2}}{x}^{\prime }\left(t\right)=0,x\left(1\right)=I^{\frac{1}{4},\frac{5}{2}}x\left(\frac{1}{3}\right).\hfill \end{array}\right.$$

Here, we take $\rho =\frac{5}{2}$, $\beta =\frac{9}{8}$, $\alpha =\frac{7}{8}$, $\lambda =\frac{6}{5}$, $\eta =\frac{1}{3}$, $\zeta =\frac{1}{4}$ and $\mu =1$. Set the multivalued function $H:\mathcal{J}\times {\ell }_1\to \mathcal{P}\left({\ell }_1\right)$ as

$$H=\left(H_1,H_2,\cdots ,H_n,\cdots \right)$$

where

$$H_n(t,x_n)=\frac{s(1-t^2)}{6e^{t+8}}(1+x_n),t\in \mathcal{J}.$$

Let $\left(x^m\right)=(x_1^m,x_2^m,\cdots )\in {\ell }_1$ be a convergent sequence of the space ${\ell }_1$ and let there exist $x^0=(x_1^0,x_2^0,\cdots )\in {\ell }_1$ such that $x_n^m\to x_n^0$ as $m\to \infty$ for all $n\in \mathbb{N}$. This means that $x^m\to x^0$ as $m\to \infty$. Now, let the sequence of multivalued functions $\left(H^m\right)=(H_1^m,H_2^m,\cdots )$. Then,

$$H_n^m(t,x_n^m)=\frac{s(1-t^2)}{6e^{t+8}}(1+x_n^m),t\in \mathcal{J}$$

and so

$$\underset{m\to \infty }{lim}{H}_n^m(t,x_n^m)=\underset{m\to \infty }{lim}\frac{s(1-t^2)}{6e^{t+8}}(1+x_n^m)=\frac{s(1-t^2)}{6e^{t+8}}(1+x_n^0)\in {\ell }_1\triangleq H_n^0(t,x_n^0),t\in \mathcal{J}$$

which implies that there exists $\left(H_n^0\right)=(H_1^0,H_2^0,\cdots )\in {\ell }_1$ such that $H_n^m\to H_n^0$. This verifies conditions (${\mathcal{H}}_1$).

For each $x_n\in C(\mathcal{J},\mathbb{R})$ and $t\in \mathcal{J}$, we have

$$|H_n(t,x_n)|\le \frac{s}{6e^{t+8}}(1+|x_n\left|\right),s<75e^8.$$

Hence, conditions (${\mathcal{H}}_2$) and (${\mathcal{H}}_3$) hold with $\delta \left(t\right)=\frac{s}{6e^{t+8}}$, $t\in \mathcal{J}$ and $\omega \left(x\right)=1+x$ which is increasing for all x.

For every bounded set $\Delta \subset \mathbb{E}$ and each $t\in \mathcal{J}$,

$$\mathcal{B}\left(H(t,\Delta )\right)⩽\frac{s}{6e^{t+8}}·\mathcal{B}(\Delta ).$$

Hence, the fourth condition is also fulfilled. We demonstrate that condition (8) holds. Indeed,

$$\begin{array}{c}\hfill \frac{{\parallel \delta \parallel }_{\infty }}{{\rho }^{\alpha +\beta }\Gamma (\alpha +\beta +1)}=\frac{4s}{300e^8}<1.\end{array}$$

(18)

Thus, as stated in Theorem 1, problems (1) and (2) has at least one solution on $\mathcal{J}$.

5. Conclusions

In this study, we discussed the existence of weak solution to a class of boundary problem for fractional Langevin equations. We chose the Katugampola–Caputo fractional derivatives to be a basic derivative to construct our model. This choice was made as they are a generalization of Caputo and Hadamard fractional derivatives and that this is also closely related to the Erdelyi–Kober approch.

It is worth noting that this derivative tends to Caputo approach as $\rho \to 1$ and tends to Hadamard approach as $\rho \to 0$ and this is considered a great contribution of this derivative to the generalization of other derivatives, which prompted us to choose it as a subject of study in our paper.

The Mönch fixed point theorem is used with the weak noncompactness measure approach to investigate the existence results under specified conditions that contain the Pettis integrability assumption. In order to illustrate our results, we presented a numerical example.