On deformable fractional impulsive implicit boundary value problems with delay

Abstract

This paper deals with some existence and uniqueness results for a class of deformable fractional differential equations. These problems encompassed nonlinear implicit fractional differential equations involving boundary conditions and various types of delays, including finite, infinite, and state-dependent delays. Our approach to proving the existence and uniqueness of solutions relied on the application of the Banach contraction principle and Schauder’s fixed-point theorem. In the last section, we provide different examples to illustrate our obtained results.

1 Introduction

Recently, fractional differential equations have been used in engineering, mathematics, physics, and other applied disciplines. The existence of solutions to the ordinary and fractional differential equations with various conditions has received much attention; see the monographs [1, 2, 4, 5, 17, 33, 34, 38] and the papers [3, 6, 8, 13, 23, 27, 29,30,31,32]. Several results of implicit fractional differential equations have been recently provided, see [18,19,20,21, 28], and the references therein.

As a model of equations, functional differential equations with delay are commonly used. Several authors studied differential equations with delay [10,11,12, 14, 15]. For more details, see the papers which are concerned with finite delay [29, 30], infinite delay [1, 3, 10, 13], and state-dependent delay [1, 12].

In their work [39], Zulfeqarr et al. introduced a novel concept known as the deformable fractional derivative. This derivative is constructed using a limit technique, similar to the standard derivative, and is named “deformable” because of its unique ability to continuously transform a function into its derivative. Deformable derivatives can be considered as fractional order derivatives, which is supported by their linear relationship with functions and their corresponding derivatives. The study of various properties of deformable derivatives includes investigations into forms of Rolle’s theorem, Mean Value theorem, and Taylor’s theorem. It is worth noting that most definitions of fractional derivatives typically rely on integral forms. In contrast, R. Khalil introduced a limit-based definition of the fractional derivative, referred to as the conformable fractional derivative, in 2014 [16]. However, Khalil’s definition has limitations, particularly in its inability to handle zero and negative numbers. Zulfeqarr et al. claim that the deformable derivative offers a simpler and more versatile approach compared to Khalil’s definition, overcoming not only this limitation but also encompassing a broader class of functions.

The author of Ref. [24] investigated further properties of the new concept of deformable derivative and used the results to study the following Cauchy problem with non-local condition:

$$\begin{aligned} \begin{gathered} {\mathfrak {D}}_{0}^{\alpha } x(t)=f(t, x(t)), \quad t \in (0, T], \\ x(0)+g(x)=x_{0}, \end{gathered} \end{aligned}$$

where ${\mathfrak {D}}_{0}^{\alpha }$ is the deformable derivative of order $\alpha \in (0,1)$, and $g: {\mathcal {C}} \rightarrow {\mathbb {R}}$ is a continuous function. Their arguments are based on Krasnoselskii’s fixed-point theorem.

In Ref. [26], Meraj and Pandey studied the existence and uniqueness of mild solution for the following initial value problem:

$$\begin{aligned} {\mathfrak {D}}_{0}^{\alpha } x(t)&=A x(t)+f(t, x(t)), \quad t \in J, \\ x(0)&=x_{0}, \end{aligned}$$

where $A: D(A) \subset X \rightarrow X$ is an infinitesimal generator of a $C_{0}$-semigroup $T(t)(t \geqslant 0)$ on a suitable space $X, x_{0} \in X$, and $J=[0, b], b>0$ is a constant. The results are obtained with the help of semigroup theory, Banach fixed-point theorem, and Schauder fixed-point theorem.

Given the scarcity of existing papers on deformable differential equations and with the goal of extending prior results in the field to contribute to the advancement of fractional calculus, this paper aims to investigate a specific class of deformable fractional differential equations with boundary condition, instantaneous impulses and finite delay:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= {\Psi }\left( {{t}},{{x}}_{{t}}, {\mathfrak {D}}_{0}^{\alpha } {x}({{t}}) \right) , \quad {t}\in {\overline{\Theta }}:=\Theta \backslash \{{t}_1,\ldots ,{t}_{{\mathfrak {m}}}\},\ {\Theta }:=(0,\varpi ],\\ \Delta {x}|_{{t}={{t}_\ell }}=\hbar _\ell ({x}({{t}_\ell ^-})),\quad \ell =1,2,\ldots ,{\mathfrak {m}},\\ {{x}}({{t}}) =\zeta ({{t}}),\ \ {{t}}\in (-{\kappa },0],\\ \imath {x}(0)+\jmath {x}(\varpi )=\varrho , \end{array}\right. } \end{aligned}$$

(1)

where ${\mathfrak {D}}_{0}^{\alpha } {{x}}({{t}})$ is the deformable fractional derivative starting from the initial time a of the function ${\Psi }$ of order $\alpha \in (0,1)$, ${\Psi }: {\Theta }\times C([-{\kappa },0],{{\mathbb {R}}})\times {{\mathbb {R}}}$, $\zeta \in C((-{\kappa },0],{{\mathbb {R}}})$, $0<\varpi <+\infty$, ${x}_\varpi \in {{\mathbb {R}}}$, $\hbar _\ell : C((-{\kappa },0],{{\mathbb {R}}})\rightarrow {\mathbb {R}}$ are given continuous functions, and ${\kappa }>0$ is the time delay and $0={t}_0<{t}_1<\cdots<{t}_{\mathfrak {m}}<{t}_{{\mathfrak {m}}+1}={\varpi }<\infty$, ${x}_{{t}_\ell ^+}=\lim \limits _{\epsilon \rightarrow 0^+} {x}({{t}_\ell +\epsilon })$ and ${x}_{{t}_\ell ^-}=\lim \limits _{\epsilon \rightarrow 0^-} {x}({{t}_\ell -\epsilon })$ represent the right and left hand limits of x(t) at ${t}={t}_\ell$, $\Delta {x}\left| _{{t}={t}_\ell }={x}({t}_\ell ^+)-{x}({t}_\ell ^-) \right.$.

For any ${{t}}\in {\Theta },$ we give ${{x}}_{{t}}$ by

$$\begin{aligned} {{x}}_{{t}}({\vartheta })={{x}}({{t}}+{\vartheta });\ \hbox {for}\ {\vartheta }\in [-{\kappa },0]. \end{aligned}$$

Next, we consider the following deformable fractional differential boundary problem with instantaneous impulses and infinite delay:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= {\Psi }\left( {{t}},{{x}}_{{t}}, {\mathfrak {D}}_{0}^{\alpha } {x}({{t}}) \right) , \quad {t}\in {\overline{\Theta }},\\ \Delta {x}|_{{t}={{t}_\ell }}=\hbar _\ell ({x}({{t}_\ell ^-})),\quad \ell =1,2,\ldots ,{\mathfrak {m}},\\ {{x}}({{t}}) =\zeta ({{t}}),\ \ {{t}}\in (-\infty ,0],\\ \imath {x}(0)+\jmath {x}(\varpi )=\varrho , \end{array}\right. } \end{aligned}$$

(2)

where ${\Psi }:{\Theta }\times {{{\mathcal {G}}}}\times {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$, $\zeta :(-\infty ,0]\rightarrow {{\mathbb {R}}}$, $\hbar _\ell :{{{\mathcal {G}}}}\rightarrow {\mathbb {R}}$ are given continuous functions, and ${{{\mathcal {G}}}}$ is a phase space. For ${{t}}\in {\Theta },$ we define ${{x}}_{{t}}\in {{{\mathcal {G}}}}$ by

$$\begin{aligned} {{x}}_{{t}}({\vartheta })={{x}}({{t}}+{\vartheta });\ \hbox {for}\ {\vartheta }\in (-\infty ,0]. \end{aligned}$$

In the next segment, we look into the following deformable fractional differential boundary problem with instantaneous impulses and state-dependent finite delay:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= {\Psi }\left( {{t}},{{x}}_{\rho ({{t}},{{x}}_{{t}})}, {\mathfrak {D}}_{0}^{\alpha } {x}({t}) \right) , \quad {t}\in {\overline{\Theta }},\\ \Delta {x}|_{{t}={{t}_\ell }}=\hbar _\ell ({x}({{t}_\ell ^-})),\quad \ell =1,2,\ldots ,{\mathfrak {m}},\\ {{x}}({{t}}) =\zeta ({{t}}),\ \ {{t}}\in (-{\kappa },0],\\ \imath {x}(0)+\jmath {x}(\varpi )=\varrho , \end{array}\right. } \end{aligned}$$

(3)

where ${\Psi }\in C((0,{\varpi }],{{\mathbb {R}}})$, $\rho :{\Theta }\times C([-{\kappa },{\varpi }],{{\mathbb {R}}})\rightarrow {{\mathbb {R}}}$, $\zeta \in C((-{\kappa },{\varpi }],{{\mathbb {R}}})$ are given continuous functions.

Finally, we study the following deformable fractional differential boundary problem with instantaneous impulses and state-dependent infinite delay:

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= {\Psi }\left( {{t}},{{x}}_{\rho ({{t}},{{x}}_{{t}})}, {\mathfrak {D}}_{0}^{\alpha } {x}({t}) \right) , \quad {t}\in {\overline{\Theta }},\\ \Delta {x}|_{{t}={{t}_\ell }}=\hbar _\ell ({x}({{t}_\ell ^-})),\quad \ell =1,2,\ldots ,{\mathfrak {m}},\\ {{x}}({{t}}) =\zeta ({{t}}),\ \ {{t}}\in (-\infty ,0],\\ \imath {x}(0)+\jmath {x}(\varpi )=\varrho , \end{array}\right. } \end{aligned}$$

(4)

where ${\Psi }:{\Theta }\times {{{\mathcal {G}}}}\times {{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$, $\zeta :(-\infty ,0]\rightarrow {{\mathbb {R}}}$ are given continuous functions.

The paper is organized as follows: Section 2 introduces some preliminaries, definitions, lemmas and auxiliary results about the deformable fractional derivative. In Sect. 3, we give some existence and uniqueness results for the first problem (1) that are based on the Banach contraction principle and Schauder’s fixed-point theorem. Section 4 deals with the deformable fractional boundary problem with instantaneous impulses and infinite delay (2). The same methods are used with the inclusion of the properties of the phase space ${{{\mathcal {G}}}}$. In Sections 5 and 7, we mention the problems (3) and (4), the results can be obtain using the same steps of the previous sections. Finally, we present several examples to show the validity of our results.

2 Preliminaries

First, we give the definitions and the notations that we will use throughout this paper. We denote by $C(\Theta ,{{\mathbb {R}}})$ and $C([-{\kappa },0],{{\mathbb {R}}})$ the Banach spaces of all continuous functions from $\Theta$ and $[-{\kappa },0]$ into ${{\mathbb {R}}}$, respectively, with the following norms:

$$\begin{aligned} \Vert {x}\Vert _{\infty }=\displaystyle \sup _{ {t}\in \Theta }\{|{x}({t})|\} \end{aligned}$$

and

$$\begin{aligned} \Vert {x}\Vert _{[-{\kappa },0]}=\displaystyle \sup _{ {t}\in [-{\kappa },0]}\{|{x}({t})|\}. \end{aligned}$$

Let ${{{\mathcal {C}}}}:=C([-{\kappa },{\varpi }])$ be a Banach space with the norm

$$\begin{aligned} \Vert {{x}}\Vert _{C}:=\displaystyle \sup _{{{t}}\in [-{\kappa },{\varpi }]}|{{x}}({{t}})|. \end{aligned}$$

Let $L^{1}(\Theta )$ be the Banach space of measurable and Lebesgue integrable functions ${x}:\Theta \rightarrow {\mathbb {R}}$ with

$$\begin{aligned} \Vert {x}\Vert _{L^{1}}=\displaystyle \int _{0}^{\beta }|{x}({t})|\textrm{d}t. \end{aligned}$$

$C^n(\Theta )$ denotes the set of mappings having n times continuously differentiable on $\Theta$, where $n\in {\mathbb {N}}$.

Consider the Banach space

$$\begin{aligned} \mathcal{P}\mathcal{C}({\Theta },{{\mathbb {R}}}) =\Bigg \{{x}:{\Theta }\rightarrow {{\mathbb {R}}}:&\ {x} \in C\left( \Theta _{\jmath }, {\mathbb {R}}\right) ; \jmath =0, \ldots , {\mathfrak {m}},\ \text {and there exist}\\&{x}\left( {t}_{\jmath }^{-}\right) \ \text {and} \ {x}\left( {t}_{\jmath }^{+}\right) ;\jmath =1, \ldots , {\mathfrak {m}},\ \text{ with } \ {x}\left( {t}_{\jmath }^{-}\right) ={x}\left( {t}_{\jmath }\right) \Bigg \}, \end{aligned}$$

with the norm

$$\begin{aligned} \Vert {x}\Vert _\mathcal{P}\mathcal{C}=\displaystyle \sup _{{t} \in {\Theta }}|{x}({t})|. \end{aligned}$$

Definition 2.1

(The deformable fractional derivative [24, 39]). Let ${\Psi }:[0,+\infty ) \longrightarrow {{\mathbb {R}}}$ be a given function, the deformable fractional derivative of ${\Psi }$ of order $\alpha$ is defined by

$$\begin{aligned} \left( {\mathfrak {D}}_{0}^{\alpha }{\Psi }\right) ({t})=\lim _{\varepsilon \rightarrow 0} \frac{(1+\varepsilon \beta )\Psi \left( {t}+\varepsilon \alpha \right) -{\Psi }({t})}{\varepsilon }, \end{aligned}$$

where $\alpha +\beta =1$ and $\alpha \in (0,1].$ If the deformable fractional derivative of ${\Psi }$ of order $\alpha$ exists, then we simply say that ${\Psi }$ is $\alpha$-differentiable.

Definition 2.2

(The $\alpha$-fractional integral [24, 25]). For $\alpha \in (0,1]$ and a continuous function ${\Psi }$, let

$$\begin{aligned} \left( {}{{\mathcal {J}}}_{0^+}^{\alpha }{\Psi }\right) ({t})= \dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }{\Psi }(\tau ) \textrm{d}\tau . \end{aligned}$$

Lemma 2.3

([24, 25]) If $\alpha ,\alpha _1 \in (0,1]$ such that $\alpha +\beta =1$, $\Psi$ and $\Phi$ are two $\alpha$-differentiable functions at a point t and m, n are two given numbers, then the deformable fractional derivative satisfies the following properties:

• 1. ${\mathfrak {D}}_{0}^{\alpha }(\lambda )=\beta \lambda$, for any constant $\lambda$;

• 2. ${\mathfrak {D}}_{0}^{\alpha }(m \Psi +n \Phi )=m {\mathfrak {D}}_{0}^{\alpha }(\Psi )+n {\mathfrak {D}}_{0}^{\alpha }(\Phi )$;

• 3. ${\mathfrak {D}}_{0}^{\alpha }(\Psi \Phi )= \Phi {\mathfrak {D}}_{0}^{\alpha }(\Psi )+\alpha \Psi \Phi ^\prime$;

• 4. ${}{{\mathcal {J}}}_{0^+}^{\alpha }\ {}{{\mathcal {J}}}_{0^+}^{\alpha _1}\Psi ={}{{\mathcal {J}}}_{0^+}^{\alpha +\alpha _1}\Psi$.

Lemma 2.4

([24, 25]) If $\alpha \in (0,1]$ and f is continuous function, then we have:

• 1. $\left( {}{{\mathcal {J}}}_{0^+}^{\alpha }\ {\mathfrak {D}}_{0}^{\alpha }(\Psi )\right) ({t})=\Psi ({t})-e^{\frac{-\beta }{\alpha }{t}}\Psi (0);$

• 2. ${\mathfrak {D}}_{0}^{\alpha }\left( {}{{\mathcal {J}}}_{0^+}^{\alpha }\Psi \right) ({t})=\Psi ({t}).$

Lemma 2.5

Let $\Phi \in L^1(\Theta )$ and $0<\alpha \le 1$. Then, the boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} \left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= \Phi ({t}); \ {t}\in {\overline{\Theta }},\\ \Delta {x}|_{{t}={{t}_\ell }}=\hbar _\ell ({x}({{t}_\ell ^-})),\quad \ell =1,2,\ldots ,{\mathfrak {m}},\\ \imath {x}(0)+\jmath {x}(\varpi )=\varrho , \end{array}\right. } \end{aligned}$$

(5)

has a unique solution defined by

$$\begin{aligned} {x}({t})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\quad if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}_{{t}_k^-}) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\quad if \ {t}\in ({t}_\ell ,{t}_\ell +1]. \end{array} \right. \end{aligned}$$

(6)

Proof

Applying the $\alpha$-fractional integral of order $\alpha$ to both sides the equation $\left( {\mathfrak {D}}_{0}^{\alpha } {x}\right) ({t})= \Phi ({t})$, and using the first property from Lemma 2.4 and if ${t}\in [0,{t}_1]$, we get

$$\begin{aligned} {x}({t})-{x}(0)e^{\frac{-\beta }{\alpha }{t}}=\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau . \end{aligned}$$

(7)

Hence, we get

$$\begin{aligned} {x}({t})={x}(0)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau . \end{aligned}$$

(8)

If ${t}\in ({t}_1,{t}_2]$, then using the first property from Lemma  2.4,

$$\begin{aligned} {x}({t})&= {x}({t}_1^+)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_1}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&= \Delta {x}|_{{t}={{t}_1}}e^{\frac{-\beta }{\alpha }{t}}+ {x}({t}_1^-)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_1}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&={x}(0)e^{\frac{-\beta }{\alpha }({t}+{t}_1)}+\hbar _1({x}({{t}_1^-}))e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}+{t}_1)}\displaystyle \int _{0}^{{t}_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&\quad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_1}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau . \end{aligned}$$

If ${t}\in ({t}_2,{t}_3]$, then using the first property from Lemma  2.4,

$$\begin{aligned} {x}({t})&= {x}({t}_2^+)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_2}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&= \Delta {x}|_{{t}={{t}_2}}e^{\frac{-\beta }{\alpha }{t}}+ {x}({t}_2^-)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_2}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&=\hbar _2({x}({{t}_2^-}))e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_2}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +e^{\frac{-\beta }{\alpha }{t}}\left[ {x}(0)e^{\frac{-\beta }{\alpha }({t}_2+{t}_1)}+\hbar _1({x}({{t}_1^-}))e^{\frac{-\beta }{\alpha }{t}_2}\right. \\&\left. \quad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}_2+{t}_1)}\displaystyle \int _{0}^{{t}_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}_2}\displaystyle \int _{{t}_1}^{{t}_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right] \\&={x}(0)e^{\frac{-\beta }{\alpha }({t}+{t}_2+{t}_1)}+\left[ \hbar _1({x}({{t}_1^-}))e^{\frac{-\beta }{\alpha }({t}+{t}_2)}+\hbar _2({x}({{t}_2^-}))e^{\frac{-\beta }{\alpha }{t}}\right] +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_2}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&\quad +\left[ \dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}+{t}_2+{t}_1)}\displaystyle \int _{0}^{{t}_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}+{t}_2)}\displaystyle \int _{{t}_1}^{{t}_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right] . \end{aligned}$$

If ${t}\in ({t}_3,{t}_4]$, then using the first property from Lemma  2.4,

$$\begin{aligned} {x}({t})&= {x}({t}_3^+)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_3}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&= \Delta {x}|_{{t}={{t}_3}}e^{\frac{-\beta }{\alpha }{t}}+ {x}({t}_3^-)e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_3}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&= \hbar _3({x}({{t}_3^-}))e^{\frac{-\beta }{\alpha }{t}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{{t}_3}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +e^{\frac{-\beta }{\alpha }{t}}\Bigg [{x}(0)e^{\frac{-\beta }{\alpha }({t}_3+{t}_2+{t}_1)}\\&\quad +\hbar _1({x}({{t}_1^-}))e^{\frac{-\beta }{\alpha }({t}_3+{t}_2)}+\hbar _2({x}({{t}_2^-})) e^{\frac{-\beta }{\alpha }{t}_3}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}_3+{t}_2+{t}_1)}\displaystyle \int _{0}^{{t}_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&\quad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}_3+{t}_2)}\displaystyle \int _{{t}_1}^{{t}_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}_3}\displaystyle \int _{{t}_2}^{{t}_3} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\\&=e^{\frac{-\beta }{\alpha }{t}}\Bigg [{x}(0)e^{\frac{-\beta }{\alpha }({t}_3+{t}_2+{t}_1)} +\hbar _1({x}({{t}_1^-}))e^{\frac{-\beta }{\alpha }({t}_3+{t}_2)}\\&\quad +\hbar _2({x}({{t}_2^-}))e^{\frac{-\beta }{\alpha }{t}_3}+\hbar _3({x}({{t}_3^-})) +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}_3+{t}_2+{t}_1)}\displaystyle \int _{0}^{{t}_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\&\quad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }({t}_3+{t}_2)}\displaystyle \int _{{t}_1}^{{t}_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}_3}\displaystyle \int _{{t}_2}^{{t}_3} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_3}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]. \end{aligned}$$

Hence, if ${t}\in ({t}_\ell ,{t}_\ell +1]$, we get

$$\begin{aligned} {x}({t})&=e^{\frac{-\beta }{\alpha }{t}}\Bigg [{x}(0)e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k=\ell }{t}_k} +\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}_{{t}_k^-})e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\quad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]. \end{aligned}$$

From the mixed boundary conditions $\imath {x}(0)+\jmath {x}(\varpi )=\varrho$, we get

$$\begin{aligned} \begin{array}{rrl} \imath {x}(0) + \jmath e^{\frac{-\beta }{\alpha }\varpi }\Bigg [{x}(0)e^{\frac{-\beta }{\alpha } \sum _{k=1}^{k={\mathfrak {m}}}{t}_k}+\displaystyle \sum _{k=1}^{k={\mathfrak {m}}} \hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}&{}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]&{}=\varrho . \end{array} \end{aligned}$$

Thus,

$$\begin{aligned} \begin{array}{rrl} {x}(0) &{} = &{} \frac{-1}{\imath e^{\frac{\beta }{\alpha }\varpi } +\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-})) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} \\ &{} &{} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k} ^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]. \end{array} \end{aligned}$$

Hence, we obtain (6).

Conversely, using the first property from Lemma 2.3 and the second property from Lemma 2.4, we can easily show that if x verifies equation (6) then it satisfied the problem (5). $\square$

Theorem 2.6

(Banach’s fixed-point theorem [9]). Let D be a non-empty closed subset of a Banach space X, then any contraction mapping T of D into itself has a unique fixed point.

Theorem 2.7

(Schauder’s fixed-point theorem [9]) . Let X be a Banach space, D be a bounded closed convex subset of X and $T:D\rightarrow D$ be a compact and continuous map. Then T has at least one fixed point in D.

3 Existence results for the first problem

Definition 3.1

A solution of problem (1) is a function ${{x}}\in C([-{\kappa },{\varpi }],{{\mathbb {R}}})$ where

$$\begin{aligned} {{x}}({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\quad if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}} \Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\quad if \ {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}),\quad if\ {{t}}\in [-{\kappa },0], \end{array} \right. \end{aligned}$$

where $\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}\ne 0$ and ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))$.

The hypotheses:

$(H_1)$:

There exist constants $\omega _1>0,\ 0<\omega _2<1$ such that

$$\begin{aligned} |{\Psi } ({{t}},{{x}}_1, \Im _1)-{\Psi } ({{t}},{{x}}_2, \Im _2)|\le \omega _1\Vert {{x}}_1- {{x}}_2\Vert _{[-{\kappa },0]}+ \omega _2|\Im _1 -\Im _2|, \end{aligned}$$

for any ${{x}}_1,{{x}}_2\in C([-{\kappa },0],{{\mathbb {R}}}),\ \Im _1,\Im _2\in {\mathbb {R}},$ and each ${{t}}\in {\Theta }.$

$(H_{01})$:

There exists a constant $\omega '_1>0$ such that

$$\begin{aligned} |\hbar _\ell ({x})-\hbar _\ell (\Im )|\le \omega '_1\Vert {{x}}- {\Im }\Vert _{[-{\kappa },0]}, \end{aligned}$$

for any ${{x}},{\Im }\in C([-{\kappa },0],{{\mathbb {R}}})$ and each ${{t}}\in {\Theta }.$

Remark 3.2

We note that by taking

$$\begin{aligned} {\varpi _1}=\omega _1, \quad {\varpi _2}=\omega _2,\ \varpi '_1= \omega '_1,\ \varpi '_3=\hbar _\ell (0)\ \text {and} \ {\varpi _3}={\Psi }^*, \end{aligned}$$

where ${\Psi }^*=\displaystyle \sup _{{{t}}\in [0,{\varpi }]}{\Psi }({{t}},0,0)$. Then, hypothesis $(H_1)$ implies that

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )|\le {\varpi _1}\Vert {{x}}\Vert _{[-{\kappa },0]}+ {\varpi _2}|\Im |+{\varpi _3}, \end{aligned}$$

and hypothesis $(H_{01})$ implies that

$$\begin{aligned} |\hbar _\ell ({x})|\le {\varpi '_1}\Vert {{x}}\Vert _{[-{\kappa },0]}+{\varpi '_3}, \end{aligned}$$

for any ${{x}}\in C([-{\kappa },0],{{\mathbb {R}}}),\ \Im \in {\mathbb {R}},$ and each ${{t}}\in {\Theta }.$

Now, we will give our first existence and uniqueness result that is based on Banach’s fixed-point theorem [9].

Theorem 3.3

Assume that the hypotheses $(H_1)$ and $(H_{01})$ hold. If

$$\begin{aligned} \begin{array}{rll} \ell := &{}\max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_1+\frac{({{\mathfrak {m}}}+1)\omega _1}{\beta (1-\omega _2)}\right) +\frac{\omega _1}{\beta (1-\omega _2)}\right] ,\right. \\ &{}\qquad \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_1 +\frac{({{\mathfrak {m}}}+1)\omega _1}{\beta (1-\omega _2)}\right) +{\mathfrak {m}}\omega '_1 +\frac{(\omega +1)\omega _1}{\beta (1-\omega _2)}\right] \right\} <1, \end{array} \end{aligned}$$

(9)

then problem (1) has a unique solution on $[-{\kappa },{\varpi }].$

Proof

Consider the operator ${{\mathcal {H}}}:{{{\mathcal {C}}}}\rightarrow {{{\mathcal {C}}}}$ such that,

$$\begin{aligned}&({{\mathcal {H}}}{{x}})({{t}})\nonumber \\&=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\quad if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}} e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}({{t}_k^-})) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\quad if \ {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}),\quad if\ {{t}}\in [-{\kappa },0], \end{array} \right. \end{aligned}$$

(10)

where ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))$.

Let ${{x}},\Im \in C({\Theta }).$ Then, for each ${{t}} \in [-{\kappa },0],$ we get

$$\begin{aligned} |({{\mathcal {H}}}{{x}})({{t}})-({{\mathcal {H}}}\Im )({{t}})|=0, \end{aligned}$$

and for each ${{t}} \in [0,{t}_1],$ we obtain

$$\begin{aligned}&|({{\mathcal {H}}}{{x}})({{t}})-({{\mathcal {H}}}\Im )({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}({{t}_k^-}))-\hbar _k(\Im ({{t}_k^-}))|e^{\frac{-\beta }{\alpha } \sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i} \displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau , \end{aligned}$$

where ${\Phi },\ {\Upsilon }\in C({\Theta })$ such that

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))\ \ \ and\ \ \ {\Upsilon }({{t}})={\Psi }({{t}},\Im _{{t}},{\Upsilon }({{t}})). \end{aligned}$$

From $(H_1),$ we have

$$\begin{aligned} |{\Phi }({{t}})-{\Upsilon }({{t}})|&= |{\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))-{\Psi }({{t}},\Im _{{t}},{\Upsilon }({{t}}))|\\&\le \omega _1\Vert {{x}}_{{t}}-\Im _{{t}}\Vert _{[-{\kappa },0]}+\omega _2|{\Phi }({{t}})-{\Upsilon }({{t}})|. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert {\Phi }-{\Upsilon }\Vert _\infty \le \frac{\omega _1}{1-\omega _2}\Vert {{x}}-\Im \Vert _{C}. \end{aligned}$$

Then, we get

$$\begin{aligned} |({{\mathcal {H}}}{{x}})({{t}})-({{\mathcal {H}}}\Im )({{t}})|&\le \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_1 +\frac{({{\mathfrak {m}}}+1)\omega _1}{\beta (1-\omega _2)}\right) +\frac{\omega _1}{\beta (1-\omega _2)}\right] \Vert {{x}}-\Im \Vert _{C}\\&\le \ell \Vert {{x}}-\Im \Vert _{C}, \end{aligned}$$

and for each ${{t}} \in ({t}_\ell ,{t}_\ell +1],$ we obtain

$$\begin{aligned}&|({{\mathcal {H}}}{{x}})({{t}})-({{\mathcal {H}}}\Im )({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}({{t}_k^-})) -\hbar _k(\Im ({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({x}({{t}_k^-})) -\hbar _k(\Im ({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ], \end{aligned}$$

where ${\Phi },\ {\Upsilon }\in C({\Theta })$ such that

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))\ \ \ and\ \ \ {\Upsilon }({{t}})={\Psi }({{t}},\Im _{{t}},{\Upsilon }({{t}})). \end{aligned}$$

Then, we get

$$\begin{aligned}&|({{\mathcal {H}}}{{x}})({{t}})-({{\mathcal {H}}}\Im )({{t}})|\\&\quad \le \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_1 +\frac{({{\mathfrak {m}}}+1)\omega _1}{\beta (1-\omega _2)}\right) +{\mathfrak {m}}\omega '_1 +\frac{({\mathfrak {m}}+1)\omega _1}{\beta (1-\omega _2)}\right] \Vert {{x}}-\Im \Vert _{C}\\&\quad \le \ell \Vert {{x}}-\Im \Vert _{C}. \end{aligned}$$

Hence, we get

$$\begin{aligned} \Vert {{\mathcal {H}}}({{x}})-{{\mathcal {H}}}(\Im )\Vert _{C}\le \ell \Vert {{x}}-\Im \Vert _{C}. \end{aligned}$$

Consequently, by Banach’s fixed-point theorem, the operator ${{\mathcal {H}}}$ has a unique fixed point, which is the unique solution of our problem (1) on $[-{\kappa },{\varpi }].$ $\square$

Theorem 3.4

If $(H_1)$ and $(H_{01})$ hold, and

$$\begin{aligned} \begin{array}{rll} &{}\max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_1 +\frac{({{\mathfrak {m}}}+1){\varpi _1}}{\beta (1-\varpi _2)}\right) +\frac{{\varpi _1}}{\beta (1-\varpi _2)}\right] ,\right. \\ &{}\qquad \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_1 +\frac{({{\mathfrak {m}}}+1){\varpi _1}}{\beta (1-\varpi _2)}+ \right) +{\mathfrak {m}}\varpi '_1+\frac{({\mathfrak {m}}+1)\varpi _1+}{\beta (1-\varpi _2)}\right] \right\} <1, \end{array} \end{aligned}$$

then problem (1) has at least one solution on $[-{\kappa },{\varpi }].$

Proof

Consider the operator ${{\mathcal {H}}}$ defined in (10). Let ${\delta }>0$ such that

$$\begin{aligned} \begin{array}{rrl} {\delta }\ge &{} \max &{} \left\{ \Vert \zeta \Vert _{[-{\kappa },0]},\frac{ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_3 +\frac{({{\mathfrak {m}}}+1){\varpi _3}}{\beta (1-\varpi _2)}+\varrho e^{\frac{\beta }{\alpha }\varpi }\right) +\frac{{\varpi _3}}{\beta (1-\varpi _2)}}{1- \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_1 +\frac{({{\mathfrak {m}}}+1){\varpi _1}}{\beta (1-\varpi _2)}\right) +\frac{{\varpi _1}}{\beta (1-\varpi _2)}\right] },\right. \\ &{} &{} \ \ \ \ \ \left. \frac{ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_3 +\frac{({{\mathfrak {m}}}+1){\varpi _3}}{\beta (1-\varpi _2)}+ +\varrho e^{\frac{\beta }{\alpha }\varpi }\right) +{\mathfrak {m}}\varpi '_3 +\frac{({\mathfrak {m}}+1)\varpi _3}{\beta (1-\varpi _2)}}{1-\left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\varpi '_1 +\frac{({{\mathfrak {m}}}+1){\varpi _1}}{\beta (1-\varpi _2)}+ \right) +{\mathfrak {m}}\varpi '_1+\frac{({\mathfrak {m}}+1)\varpi _1+}{\beta (1-\varpi _2)}\right] } \right\} . \end{array} \end{aligned}$$

(11)

Consider the ball

$$\begin{aligned} {\Xi }_{\delta }=\{\xi \in C([-{\kappa },\varpi ],{\mathbb {R}}),\ \Vert \xi \Vert _{C} \le {\delta }\}. \end{aligned}$$

Claim 1. ${{\mathcal {H}}}$ is continuous.

Let $\{{{x}}_n\}_n$ be a sequence such that ${{x}}_n\rightarrow {{x}}$ on ${\Xi }_{\delta }.$ For each ${{t}} \in [-{\kappa },0],$ we have

$$\begin{aligned} |({{\mathcal {H}}}{{x}}_n)({{t}})-({{\mathcal {H}}}{{x}})({{t}})|=0, \end{aligned}$$

and for each ${{t}}\in [0,{t}_1],$ we have

$$\begin{aligned}&|({{\mathcal {H}}}{{x_n}})({{t}})-({{\mathcal {H}}}x)({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}_n({{t}_k^-})) -\hbar _k({x}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi _n}({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau , \end{aligned}$$

and for each ${{t}}\in ({t}_\ell ,{t}_\ell +1],$ we have

$$\begin{aligned}&|({{\mathcal {H}}}{{x_n}})({{t}})-({{\mathcal {H}}}x)({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}_n({{t}_k^-})) -\hbar _k({x}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({x}_n({{t}_k^-})) -\hbar _k({x}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ], \end{aligned}$$

where ${\Phi }_n,\ {\Phi }\in C({\Theta },{\mathbb {R}})$ such that

$$\begin{aligned} {\Phi }_n({{t}})={\Psi }({{t}},{{x}}_{n{t}},{\Phi }_n({{t}}))\ \ \ and\ \ \ {\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}})). \end{aligned}$$

Since

$$\begin{aligned} \Vert {{x}}_n-{{x}}\Vert _{C}\rightarrow 0\text { as } n\rightarrow \infty \end{aligned}$$

and ${\Psi },{\Phi }$ and ${\Phi }_n$ are continuous, we deduce that

$$\begin{aligned} \Vert {{\mathcal {H}}}({{x}}_{n})-{{\mathcal {H}}}({{x}})\Vert _{C} \rightarrow 0 \quad \text {as } n\rightarrow \infty . \end{aligned}$$

Hence, ${{\mathcal {H}}}$ is continuous.

Claim 2. ${{\mathcal {H}}}({\Xi }_{\delta })\subset {\Xi }_{\delta }.$

Let ${{x}}\in {\Xi }_{\delta },$ If ${{t}}\in [-{\kappa },0]$ then $\Vert ({{\mathcal {H}}}{{x}})({{t}})\Vert \le \Vert \zeta \Vert _{C}\le {\delta }.$ From Remark 3.2, for each ${{t}}\in [0,{t}_1],$ we have

$$\begin{aligned} |{\Phi }({{t}})|&\le |{\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))| \\&\le {\varpi _1}\Vert {{x}}_{{t}}\Vert _{[-{\kappa },0]}+{\varpi _2}|{\Phi }({{t}})|+{\varpi _3} \\&\le {\varpi _1}\Vert {{x}}\Vert _C+{\varpi _2}\Vert {\Phi }\Vert _{\infty }+{\varpi _3}\\&\le {\varpi _1}{\delta }+{\varpi _2}\Vert {\Phi }\Vert _{\infty }+{\varpi _3}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert {\Phi }\Vert _{\infty }\le \frac{{\delta }{\varpi _1}+{\varpi _3}}{1-{\varpi _2}}. \end{aligned}$$

Thus,

$$\begin{aligned} |({{\mathcal {H}}}{{x}})({{t}})| \le&\left| \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right. \\&\left. +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right| \\ \le&\Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&+\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau +\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\ \le&\frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}(\varpi '_1\delta +\varpi '_3) +\frac{({{\mathfrak {m}}}+1)({\delta }{\varpi _1}+{\varpi _3})}{\beta (1-\varpi _2)}+\varrho e^{\frac{\beta }{\alpha }\varpi }\right) +\frac{({\delta }{\varpi _1}+{\varpi _3})}{\beta (1-\varpi _2)}\\ \le&{\delta }, \end{aligned}$$

and for each ${{t}} \in ({t}_\ell ,{t}_\ell +1],$ we obtain

$$\begin{aligned}&|({{\mathcal {H}}}{{x}})({{t}})|\\&\quad \le \left| \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+e^{\frac{-\beta }{\alpha }{t}} \Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}_{{t}_k^-})e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\right| \\&\quad \le \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({x}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\qquad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau +\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }| \hbar _k({x}_{{t}_k^-})|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\\&\quad \le \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} (\varpi '_1\delta +\varpi '_3)+\frac{({{\mathfrak {m}}}+1)({\delta }{\varpi _1}+{\varpi _3})}{\beta (1-\varpi _2)} +\varrho e^{\frac{\beta }{\alpha }\varpi }\right) \\&\qquad +{\mathfrak {m}}(\varpi '_1\delta +\varpi '_3)+\frac{({\mathfrak {m}}+1)(\delta \varpi _1+\varpi _3)}{\beta (1-\varpi _2)} \\&\quad \le {\delta }. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert {{\mathcal {H}}}({{x}})\Vert _{C}\le {\delta }. \end{aligned}$$

Consequently, ${{\mathcal {H}}}({\Xi }_{\delta })\subset {\Xi }_{\delta }.$

Claim 3. ${{\mathcal {H}}}({\Xi }_{\delta })$ is equicontinuous.

For $0\le {{\varrho }}_1\le {{\varrho }}_2\le {\varpi },$ and ${x}\in {\Xi }_{\delta },$ we get

$$\begin{aligned}&|{{\mathcal {H}}}({{x}})({{\varrho }}_1)-{{\mathcal {H}}}({{x}})({{\varrho }}_2)|\\&\quad \le \left| \frac{e^{\frac{-\beta }{\alpha }({\varrho }_2+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\\&\qquad -\frac{e^{\frac{-\beta }{\alpha }({\varrho }_1+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha } \sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]\right| \\&\qquad +\left| e^{\frac{-\beta }{\alpha }{\varrho }_2}\Bigg [\displaystyle \sum _{0\le {t}_\ell \le {t}} \hbar _\ell ({x}({{t}_\ell ^-}))e^{\frac{-\beta }{\alpha }\sum _{{t}_{\ell +1}\le {t}_i\le {t}}{t}_i} \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{0\le {t}_\ell \le {t}}e^{\frac{-\beta }{\alpha } \sum _{{t}_{\ell }\le {t}_i\le {t}}{t}_i}\displaystyle \int _{{t}_{\ell -1}}^{{t}_{\ell }} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{\varrho }_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\\&\qquad -e^{\frac{-\beta }{\alpha }{\varrho }_1}\Bigg [\displaystyle \sum _{0\le {t}_\ell \le {t}} \hbar _\ell ({x}({{t}_\ell ^-}))e^{\frac{-\beta }{\alpha }\sum _{{t}_{\ell +1}\le {t}_i\le {t}}{t}_i}\\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{0\le {t}_\ell \le {t}}e^{\frac{-\beta }{\alpha } \sum _{{t}_{\ell }\le {t}_i\le {t}}{t}_i}\displaystyle \int _{{t}_{\ell -1}}^{{t}_{\ell }} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{\varrho }_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\right| \\&\quad \le \frac{\left| e^{\frac{-\beta }{\alpha }({\varrho }_2+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )} -e^{\frac{-\beta }{\alpha }({\varrho }_1+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}\right| }{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}(\varpi '_1\delta +\varpi '_3) +\frac{({{\mathfrak {m}}}+1)({\delta }{\varpi _1}+{\varpi _3})}{\beta (1-\varpi _2)}\right. \\&\qquad \left. + \varrho e^{\frac{\beta }{\alpha }\varpi }\right) +\left| e^{\frac{-\beta }{\alpha }{\varrho }_2} -e^{\frac{-\beta }{\alpha }{\varrho }_1}\right| \left[ {\mathfrak {m}}(\varpi '_1\delta +\varpi '_3) +\frac{{\mathfrak {m}}(\delta \varpi _1+\varpi _3)}{\beta (1-\varpi _2)} \right] \\&\qquad + \frac{{\delta }{\varpi _1}+{\varpi _3}}{\beta (1-{\varpi _2})} \left| e^{\frac{-\beta }{\alpha }({\varrho }_2-{t}_\ell )}-e^{\frac{-\beta }{\alpha }({\varrho }_1-{t}_\ell )}\right| . \end{aligned}$$

As ${{\varrho }}_2\rightarrow {{\varrho }}_1$ then $|{{\mathcal {H}}}({{x}})({{\varrho }}_1) -{{\mathcal {H}}}({{x}})({{\varrho }}_2)|\rightarrow 0$. We deduce that ${{\mathcal {H}}}({\Xi }_{\delta })$ is equicontinuous.

Consequently, Arzelá–Ascoli theorem implies that ${{\mathcal {H}}}$ is continuous and compact. Thus, by Schauder’s fixed-point theorem [9], we deduce that ${{\mathcal {H}}}$ has at least a fixed point which is a solution of (1). $\square$

4 Existence results for the second problem

In this section, we are concerned with the existence results of (2). Let the space $({{{\mathcal {G}}}},\Vert \cdot \Vert _{{{\mathcal {G}}}})$ be a seminormed linear space of functions mapping $(-\infty ,0]$ into ${\mathbb {R}}$ given by

$$\begin{aligned} {{{\mathcal {G}}}} =\Bigg \{&\ {x}:{(-\infty ,0]}\rightarrow {{\mathbb {R}}}:\ \tau \rightarrow {x}(\tau ) \in C\left( (\tau _{\ell }, \tau _{\ell +1}], {\mathbb {R}}\right) ;\ell =0, \ldots , {\mathfrak {m}},\\&\text {and there exist}\ {x}\left( \tau _{\jmath }^{-}\right) \ \text {and} \ {x}\left( \tau _{\ell }^{+}\right) ;\ell =1, \ldots , {\mathfrak {m}},\ \text{ with } \\ {}&\ {x}\left( \tau _{\ell }^{-}\right) ={x}\left( \tau _{\ell }\right) \ \text {and}\ \tau _\ell ={t}_\jmath -{t}\ \text {for each}\ {t}\in ({t}_{\ell }, {t}_{\ell +1}] \Bigg \}, \end{aligned}$$

and verifying the following axioms which were derived from Hale and Kato’s originals [10]:

• $(Ax_1)$ If $y:(-\infty ,0]\rightarrow {\mathbb {R}},$ and ${{x}}_0 \in {{{\mathcal {G}}}},$ then there exist constants ${\xi _1},{\xi _2},{\xi _3}>0$, such that for each ${{t}}\in {\Theta };$ we have:

  (a):

  ${{x}}_{{t}}\ is\ in\ {{{\mathcal {G}}}},$

  (b):

  $\Vert {{x}}_{{t}}\Vert _{{{\mathcal {G}}}}\le {\xi _1} \Vert {{x}}_1\Vert _\mathcal{G}+ {\xi _2} \sup _{{\vartheta }\in [0,{{t}}]}|{{x}}({\vartheta })|,$

  (c):

  $\Vert {{x}}({{t}})\Vert \le {\xi _3}\Vert {{x}}_{{t}}\Vert _{{{\mathcal {G}}}}.$

• $(Ax_2)$ For the function ${{x}}(\cdot )\ in\ (Ax_1)$, $y_{t}$ is a ${{{\mathcal {G}}}}-$ valued continuous function on ${\Theta }.$

• $(Ax_3)$ The space ${{{\mathcal {G}}}}$ is complete.

Consider the space

$$\begin{aligned} \Xi =\{{{x}}:(-\infty ,{\varpi }]\rightarrow {\mathbb {R}},\ {{x}}|_{(-\infty ,0]}\in {{{\mathcal {G}}}},\ {{x}}|_{{\Theta }}\in \mathcal{P}\mathcal{C}({\Theta },{{\mathbb {R}}})\}. \end{aligned}$$

Definition 4.1

By a solution of problem (2), we mean a function ${{x}}\in \Xi$ such that

$$\begin{aligned} {{x}}({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\quad if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell } \hbar _k({x}_{{t}_k^-})e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\quad if \ {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}),\quad if\ {{t}}\in (-{\infty },0], \end{array} \right. \end{aligned}$$

where $\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}\ne 0$ and ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))$.

The hypothesis:

$(H_2)$:

The function ${\Psi }$ verifies:

$$\begin{aligned} |{\Psi } ({{t}},{{x}}_1,\Im _1)-{\Psi } ({{t}},{{x}}_2,\Im _2)|\le b_1\Vert {{x}}_1- {{x}}_2\Vert _{{{\mathcal {G}}}} + b_2|\Im _1 -\Im _2|, \end{aligned}$$

for any ${{x}}_1,x_2\in {{{\mathcal {G}}}}$, $\Im _1,\Im _2\in {\mathbb {R}},$ and each ${{t}}\in {\Theta },$ where $b_1>0$ and $0<b_2<1.$

$(H_{02})$:

There exists a constant $b'_1>0$ such that:

$$\begin{aligned} |\hbar _\ell ({x})-\hbar _\ell (\Im )|\le b'_1\Vert {{x}}- {\Im }\Vert _\mathcal{G}, \end{aligned}$$

for any ${{x}},{\Im }\in {{{\mathcal {G}}}}$.

Remark 4.2

We note that by taking:

$$\begin{aligned} B_1=b_1, \quad B_2=b_2,\ b'_1=B'_1,\ B'_3=\hbar _\ell (0)\ \text {and} \ B_3={\Psi }^*, \end{aligned}$$

where ${\Psi }^*=\displaystyle \sup _{{{t}}\in [0,{\varpi }]}{\Psi }({{t}},0,0)$. Then, hypothesis $(H_2)$ implies that

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )|\le B_1\Vert {{x}}\Vert _{{{\mathcal {G}}}}+ B_2|\Im |+B_3, \end{aligned}$$

and hypothesis $(H_{02})$ implies that

$$\begin{aligned} |\hbar _\ell ({x})|\le {B'_1}\Vert {{x}}\Vert _{{{\mathcal {G}}}}+{B'_3}, \end{aligned}$$

for any ${{x}}\in {{{\mathcal {G}}}},\ \Im \in {\mathbb {R}},$ and each ${{t}}\in {\Theta }.$

Theorem 4.3

If $(H_2)$ and $(H_{02})$ hold, and

$$\begin{aligned} \begin{array}{rll} \lambda := &{}\max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +\frac{b_1}{\beta (1-b_2)}\right] ,\right. \\ &{}\qquad \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +{\mathfrak {m}} b'_1+\frac{({\mathfrak {m}}+1)b_1}{\beta (1-b_2)}\right] \right\} <1, \end{array} \end{aligned}$$

(12)

then (2) admit a unique solution on $(-\infty ,{\varpi }].$

Proof

Define the operator $N_1:\Xi \rightarrow \Xi$ where

$$\begin{aligned} (N_1{{x}})({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau , \ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}} \Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ], {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}), \ {{t}}\in (-{\infty },0], \end{array} \right. \end{aligned}$$

where ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{{t}},{\Phi }({{t}}))$.

Let ${\varkappa _1}:(-\infty ,{\varpi }]\rightarrow {\mathbb {R}}$ be a function given by

$$\begin{aligned} {\varkappa _1}({{t}})=\left\{ \begin{array}{ll} \zeta ({{t}});&{} {{t}}\in (-\infty ,0], \\ {x}(\varpi )&{} {{t}}\in {\Theta }. \end{array} \right. \end{aligned}$$

Then ${\varkappa _1}_0=\zeta$. For each ${\varkappa _2}\in C({\Theta }),$ with ${\varkappa _2}(0)=0,$ we denote by $\overline{{\varkappa _2}}$ the function defined by

$$\begin{aligned} \overline{{\varkappa _2}}=\left\{ \begin{array}{ll} 0, &{} {{t}}\in (-\infty ,0], \\ {\varkappa _2}({{t}}), &{} {{t}}\in {\Theta }. \end{array} \right. \end{aligned}$$

If ${{x}}(\cdot )$ satisfies the integral equation

$$\begin{aligned} {{x}}({{t}}) = \left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\ if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell } \hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\ if \ {t}\in ({t}_\ell ,{t}_\ell +1]. \end{array} \right. \end{aligned}$$

We can decompose ${{x}}(\cdot )$ as ${{x}}({{t}}) = \overline{{\varkappa _2}}({{t}}) + {\varkappa _1}({{t}});$ for ${{t}}\in {\Theta },$ which implies that ${{x}}_{{t}} =\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}}$ for every ${{t}}\in {\Theta },$ and the function ${\varkappa _2}(\cdot )$ satisfies

$$\begin{aligned} {\varkappa _2}({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-})) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell } \hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ] \ \ \ \ if \ {t}\in ({t}_\ell ,{t}_\ell +1], \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},\overline{{\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}));\ {{t}}\in {\Theta }. \end{aligned}$$

Set

$$\begin{aligned} {{\mathcal {D}}_0}=\{{\varkappa _2}\in C({\Theta });\ {\varkappa _2}_0=0\}, \end{aligned}$$

and let $\Vert \cdot \Vert _\varpi$ be the norm in ${{\mathcal {D}}_0}$ defined by

$$\begin{aligned} \Vert {\varkappa _2}\Vert _\varpi =\Vert {\varkappa _2}_0\Vert _{{{\mathcal {G}}}}+\sup _{{{t}}\in {\Theta }}|{\varkappa _2}({{t}})|=\sup _{{{t}}\in {\Theta }}|{\varkappa _2}({{t}})|;\ {\varkappa _2}\in {{\mathcal {D}}_0}, \end{aligned}$$

where ${{\mathcal {D}}_0}$ is a Banach space with norm $\Vert \cdot \Vert _T.$ Define the operator ${{\mathcal {K}}}: {{\mathcal {D}}_0} \rightarrow {{\mathcal {D}}_0}$ by

$$\begin{aligned} ({{\mathcal {K}}}{\varkappa _2})({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau ,\ if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-})) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell } \hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i} \displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ],\ if \ {t}\in ({t}_\ell ,{t}_\ell +1], \end{array} \right. \end{aligned}$$

(13)

where

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}));\ {{t}}\in {\Theta }. \end{aligned}$$

We shall show that ${{\mathcal {K}}}:{{\mathcal {D}}_0}\rightarrow {{\mathcal {D}}_0}$ is a contraction map. Let ${\varkappa _2},{\varkappa _2}'\in {{\mathcal {D}}_0},$ then we have for each ${{t}}\in [0,{t}_1]$

$$\begin{aligned}&|{{\mathcal {K}}}({\varkappa _2})({{t}})-{{\mathcal {K}}}({\varkappa _2}')({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}({{t}_k^-}))-\hbar _k({\varkappa _2}'({{t}_k^-}))| e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau , \end{aligned}$$

where ${\Phi },\ {\Upsilon }\in C({\Theta })$ such that

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}))\ \ \ and\ \ \ {\Upsilon }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}'}_{{t}} + {\varkappa _1}_{{t}},{\Upsilon }({{t}})). \end{aligned}$$

For each ${{t}} \in {\Theta }$, we have

$$\begin{aligned} |{\Phi }({{t}})-{\Upsilon }({{t}})|\le \frac{b_1}{1-b_2}\Vert \overline{ {\varkappa _2}}_{{t}}-\overline{ {\varkappa _2}'}_{{t}}\Vert _{{{\mathcal {G}}}}. \end{aligned}$$

Then, for each ${{t}} \in {\Theta },$ we get

$$\begin{aligned} \begin{array}{ccl} |{{\mathcal {K}}}({\varkappa _2})({{t}})-{{\mathcal {K}}}({\varkappa _2}')({{t}})| &{}\le &{}\left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +\frac{b_1}{\beta (1-b_2)}\right] \Vert \overline{ {\varkappa _2}}_{{t}}-\overline{ {\varkappa _2}'}_{{t}}\Vert _{{{\mathcal {G}}}}\\ &{}\le &{}\left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +\frac{b_1}{\beta (1-b_2)}\right] \Vert \overline{ {\varkappa _2}}-\overline{ {\varkappa _2}'}\Vert _{{\varpi }}\\ &{}=&{}\lambda \Vert \overline{ {\varkappa _2}}-\overline{ {\varkappa _2}'}\Vert _{{\varpi }}, \end{array} \end{aligned}$$

and for each ${{t}} \in ({t}_\ell ,{t}_\ell +1],$ we obtain

$$\begin{aligned}&|{{\mathcal {K}}}({\varkappa _2})({{t}})-{{\mathcal {K}}}({\varkappa _2}')({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}({{t}_k^-})) -\hbar _k({\varkappa _2}'({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({\varkappa _2}({{t}_k^-})) -\hbar _k({\varkappa _2}'({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i} \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }({\tau })-{\Upsilon }({\tau })| \textrm{d}\tau \Bigg ], \end{aligned}$$

where ${\Phi },\ {\Upsilon }\in C({\Theta })$ such that

$$\begin{aligned} {\Phi }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}))\ \ \ and\ \ \ {\Upsilon }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}'}_{{t}} + {\varkappa _1}_{{t}},{\Upsilon }({{t}})). \end{aligned}$$

Then, we get

$$\begin{aligned}&|{{\mathcal {K}}}({\varkappa _2})({{t}})-{{\mathcal {K}}}({\varkappa _2}')({{t}})|\\&\quad \le \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +{\mathfrak {m}}\omega '_1 +\frac{({\mathfrak {m}}+1)b_1}{\beta (1-b_2)}\right] \Vert \overline{ {\varkappa _2}}_{{t}}-\overline{ {\varkappa _2}'}_{{t}}\Vert _{{{\mathcal {G}}}}\\&\quad \le \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_1+\frac{({{\mathfrak {m}}}+1)b_1}{\beta (1-b_2)}\right) +{\mathfrak {m}}\omega '_1 +\frac{({\mathfrak {m}}+1)b_1}{\beta (1-b_2)}\right] \Vert \overline{ {\varkappa _2}}-\overline{ {\varkappa _2}'}\Vert _{{\varpi }}\\&\quad =\lambda \Vert \overline{ {\varkappa _2}}-\overline{ {\varkappa _2}'}\Vert _{{\varpi }}. \end{aligned}$$

Thus, we get

$$\begin{aligned} \Vert {{\mathcal {K}}}({\varkappa _2})-{{\mathcal {K}}}({\varkappa _2}')\Vert _\varpi \le \lambda \Vert \overline{ {\varkappa _2}}-\overline{ {\varkappa _2}'}\Vert _{{\varpi }}. \end{aligned}$$

Hence, from the Banach contraction principle, ${{\mathcal {K}}}$ has a unique fixed point. Thus, the existence of the unique solution of problem (2). $\square$

Theorem 4.4

If $(H_2)$ and $(H_{02})$ hold. Then (2) admit at least one solution on $(-\infty ,{\varpi }].$

Proof

Consider ${{\mathcal {K}}}:{{\mathcal {D}}_0}\rightarrow {{\mathcal {D}}_0}$ given in (13), For ${\delta }>0,$ let

$$\begin{aligned} {\Xi }_{\delta }=\{{\varkappa _1}\in {{\mathcal {D}}_0},\ \Vert {\varkappa _1}\Vert _{{\varpi }} \le {\delta }\}. \end{aligned}$$

Claim 1. ${{\mathcal {H}}}$ is continuous.

Let ${{\varkappa _2}_n}$ be a sequence where ${\varkappa _2}_n\rightarrow {\varkappa _2}$ in ${{\mathcal {D}}_0}.$ For ${{t}}\in [0,{t}_1],$ we have

$$\begin{aligned}&|({{\mathcal {K}}}{\varkappa _2}_n)({{t}})-({{\mathcal {K}}}{\varkappa _2})({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}_n({{t}_k^-})) -\hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau , \end{aligned}$$

and for each ${{t}}\in ({t}_\ell ,{t}_\ell +1],$ we have

$$\begin{aligned}&|({{\mathcal {K}}}{\varkappa _2}_n)({{t}})-({{\mathcal {K}}}{\varkappa _2})({{t}})|\\&\quad \le \frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}_n({{t}_k^-})) -\hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau } |{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ]\\&\qquad +e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({\varkappa _2}_n({{t}_k^-})) -\hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i} \displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|{\Phi }_n({\tau })-{\Phi }({\tau })| \textrm{d}\tau \Bigg ], \end{aligned}$$

where ${\Phi }_n,\ {\Phi }\in C({\Theta })$ such that

$$\begin{aligned} {\Phi }_n({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}}_{n{t}} + {\varkappa _1}_{{t}},{\Phi }_n({{t}}))\ \ \ and\ \ \ {\Phi }({{t}})={\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}})). \end{aligned}$$

Since $\Vert {\varkappa _2}_n-{\varkappa _2}\Vert _{{\varpi }}\rightarrow 0\text { as } n\rightarrow \infty$ and ${\Psi },{\Phi }$ and ${\Phi }_n$ are continuous, then

$$\begin{aligned} \Vert {{\mathcal {K}}}({{x}}_{n})-{{\mathcal {K}}}({{x}})\Vert _{{\varpi }} \rightarrow 0 \quad \text {as } n\rightarrow \infty . \end{aligned}$$

Hence, ${{\mathcal {K}}}$ is continuous.

Claim 2. ${{\mathcal {K}}}({\Xi }_{\delta })$ is bounded.

Let ${\varkappa _2}\in {\Xi }_{\delta },$ for ${{t}}\in [0,{t}_1],$ we obtain

$$\begin{aligned} |{\Phi }({{t}})|&\le |{\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}))| \\&\le B_1\Vert \overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}}\Vert _{{{\mathcal {G}}}}+B_2|{\Phi }({{t}})|+B_3 \\&\le B_1 \left[ \Vert \overline{ {\varkappa _2}}_{{t}} \Vert _\mathcal{G}+\Vert {\varkappa _1}_{{t}}\Vert _\mathcal{G}\right] +B_2\Vert {\Phi }\Vert _{\infty }+B_3\\&\le B_1{\xi _2}\delta +B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}} +B_2\Vert {\Phi }\Vert _{\infty }+B_3. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert {\Phi }\Vert _{\infty }\le \frac{B_1{\xi _2}\delta +B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3}{1-B_2}. \end{aligned}$$

Thus,

$$\begin{aligned} |({{\mathcal {K}}}{\varkappa _2})({{t}})|&\le \left| \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right. \\&\quad \left. +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right| \\&\le \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\quad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\le \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}(B'_1\delta +B'_3) +\frac{({{\mathfrak {m}}}+1)(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)}\right) \\&\quad +\frac{(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)}:={\tilde{\ell }}, \end{aligned}$$

and for each ${{t}} \in ({t}_\ell ,{t}_\ell +1],$ we obtain

$$\begin{aligned}&|({{\mathcal {K}}}{\varkappa _2})({{t}})|\\&\quad \le \left| \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1} ^{k=\ell }\hbar _k({x}_{{t}_k^-})e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\right| \\&\quad \le \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}({{t}_k^-})) |e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\qquad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+e^{\frac{-\beta }{\alpha }{t}} \Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({\varkappa _2}_{{t}_k^-})|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\\&\quad \le \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}(B'_1\delta +B'_3) +\frac{({{\mathfrak {m}}}+1)(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)} \right) \\&\qquad +{\mathfrak {m}}(B'_1\delta +B'_3)+\frac{({\mathfrak {m}}+1)(B_1{\xi _2}{\delta } +B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)} \\&\quad :={\tilde{\ell }}'. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert {{\mathcal {K}}}({\varkappa _2})\Vert _{{\varpi }}\le \max \left\{ {\tilde{\ell }},{\tilde{\ell }}'\right\} . \end{aligned}$$

Claim 3. ${{\mathcal {K}}}({\Xi }_{\delta })$ is equicontinuous.

For $0\le {\varrho }_1\le {\varrho }_2\le {\varpi },$ and ${\varkappa _2}\in {\Xi }_{\delta },$ we have

$$\begin{aligned}&|{{\mathcal {K}}}({{\varkappa _2}})({\varrho }_1)-{{\mathcal {K}}}({{\varkappa _2}})({\varrho }_2)|\\&\quad \le \left| \frac{e^{\frac{-\beta }{\alpha }({\varrho }_2+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\\&\qquad -\frac{e^{\frac{-\beta }{\alpha }({\varrho }_1+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i}\\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\right| \\&\qquad +\left| e^{\frac{-\beta }{\alpha }{\varrho }_2}\Bigg [\displaystyle \sum _{0\le {t}_\ell \le {t}} \hbar _\ell ({\varkappa _2}({{t}_\ell ^-}))e^{\frac{-\beta }{\alpha }\sum _{{t}_{\ell +1}\le {t}_i\le {t}}{t}_i} \right. \\&\qquad +\dfrac{1}{\alpha }\displaystyle \sum _{0\le {t}_\ell \le {t}}e^{\frac{-\beta }{\alpha }\sum _{{t}_{\ell } \le {t}_i\le {t}}{t}_i}\displaystyle \int _{{t}_{\ell -1}}^{{t}_{\ell }} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{\varrho }_2} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ] \\ {}&\qquad -e^{\frac{-\beta }{\alpha }{\varrho }_1}\Bigg [\displaystyle \sum _{0\le {t}_\ell \le {t}} \hbar _\ell ({\varkappa _2}({{t}_\ell ^-}))e^{\frac{-\beta }{\alpha }\sum _{{t}_{\ell +1}\le {t}_i\le {t}}{t}_i} \\&\qquad \left. +\dfrac{1}{\alpha }\displaystyle \sum _{0\le {t}_\ell \le {t}}e^{\frac{-\beta }{\alpha } \sum _{{t}_{\ell }\le {t}_i\le {t}}{t}_i}\displaystyle \int _{{t}_{\ell -1}}^{{t}_{\ell }} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha } \displaystyle \int _{{t}_\ell }^{{\varrho }_1} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]\right| \\&\quad \le \frac{\left| e^{\frac{-\beta }{\alpha }({\varrho }_2+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )} -e^{\frac{-\beta }{\alpha }({\varrho }_1+\sum _{0\le {t}_\ell \le {t}}{t}_\ell )}\right| }{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}} \left( {\mathfrak {m}}(B'_1\delta +B'_3)\right. \\&\qquad \left. +\frac{({{\mathfrak {m}}}+1)(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)}\right) \\&\qquad +\left| e^{\frac{-\beta }{\alpha }{\varrho }_2}-e^{\frac{-\beta }{\alpha } {\varrho }_1}\right| \left[ {\mathfrak {m}}(B'_1\delta +B'_3)+\frac{{\mathfrak {m}}(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-B_2)} \right] \\&\qquad + \frac{(B_1{\xi _2}{\delta }+B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3)}{\beta (1-{B_2})} \left| e^{\frac{-\beta }{\alpha }({\varrho }_2-{t}_\ell )}-e^{\frac{-\beta }{\alpha }({\varrho }_1-{t}_\ell )}\right| . \end{aligned}$$

As ${\varrho }_2\rightarrow {\varrho }_1$ then $|{{\mathcal {H}}}({{x}})({\varrho }_1)-{{\mathcal {H}}}({{x}})({\varrho }_2)|\rightarrow 0$. Thus, ${{\mathcal {K}}}:{{\mathcal {D}}_0}\rightarrow {{\mathcal {D}}_0}$ is completely continuous.

Claim 4. The priori bounds.

We prove that the set

$$\begin{aligned} {{{\mathcal {E}}}}=\left\{ {{x}}\in {{\mathcal {D}}_0}:{x}=\lambda {{\mathcal {K}}}({{x}});\ \text {for some} \ \lambda \in (0,1)\right\} \end{aligned}$$

is bounded. Let ${\varkappa _2}\in {{\mathcal {D}}_0}.$ Let $u\in {{\mathcal {D}}_0},$ such that ${\varkappa _2}=\lambda {{\mathcal {K}}}({\varkappa _2})$; for some $\lambda \in (0,1).$ Then for each ${{t}}\in {\Theta },$ we have

$$\begin{aligned} {\varkappa _2}({{t}})&=\lambda ({{\mathcal {K}}}{\varkappa _2})({{t}})\\&=\left\{ \begin{array}{ll} \frac{-\lambda e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-}))e^{\frac{-\beta }{\alpha } \sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} &{}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{\lambda }{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]&{}\\ +\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau , \ if\ {t}\in [0,{t}_1],\\ \frac{-\lambda e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({\varkappa _2}({{t}_k^-})) e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} &{}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ]&{}\\ +\lambda e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({\varkappa _2} ({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ], \ if \ {t}\in ({t}_\ell ,{t}_\ell +1]. \end{array} \right. \end{aligned}$$

By Remark 4.2 we have

$$\begin{aligned} |{\Phi }({{t}})|&\le |{\Psi }({{t}},\overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}},{\Phi }({{t}}))| \\&\le B_1\Vert \overline{ {\varkappa _2}}_{{t}} + {\varkappa _1}_{{t}}\Vert _{{{\mathcal {G}}}}+B_2|{\Phi }({{t}})|+B_3 \\&\le B_1 \left[ \Vert \overline{ {\varkappa _2}}_{{t}} \Vert _{{{\mathcal {G}}}}+\Vert {\varkappa _1}_{{t}}\Vert _{{{\mathcal {G}}}}\right] +B_2\Vert {\Phi }\Vert _{\infty }+B_3\\&\le B_1{\xi _2}\Vert {\varkappa _2}\Vert _\varpi +B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}} +B_2\Vert {\Phi }\Vert _{\infty }+B_3. \end{aligned}$$

This gives

$$\begin{aligned} \Vert {\Phi }\Vert _{\infty }\le \frac{B_1{\xi _2}\Vert {\varkappa _2}\Vert _\varpi +B_1{\xi _1} \Vert \varphi \Vert _{{{\mathcal {G}}}}+B_3}{1-B_2}:=\eta . \end{aligned}$$

Thus, for each ${{t}}\in [0,{t}_1],$ we obtain

$$\begin{aligned} |{\varkappa _2}({{t}})|&\le \, \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}|\hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\quad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\frac{e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\&\le \, \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}(B'_1\delta +B'_3) +\frac{({{\mathfrak {m}}}+1)\eta }{\beta }\right) +\frac{\eta }{\beta } \\&:= \,\eta ', \end{aligned}$$

and for each ${{t}} \in ({t}_\ell ,{t}_\ell +1],$ we obtain

$$\begin{aligned} |{\varkappa _2}({{t}})|&\le \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}| \hbar _k({\varkappa _2}({{t}_k^-}))|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \\ {}&\quad +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\frac{e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}}+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }|\hbar _k({\varkappa _2}_{{t}_k^-})|e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i} \\&\quad +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i} \displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }|\Phi (\tau )| \textrm{d}\tau \Bigg ]\\&\le \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}} \left( {\mathfrak {m}}(B'_1\delta +B'_3)+\frac{({{\mathfrak {m}}}+1)\eta }{\beta } \right) +{\mathfrak {m}}(B'_1\delta +B'_3)+\frac{({\mathfrak {m}}+1)\eta }{\beta } \\&:= \eta ''. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert {\varkappa _2}\Vert _{{\varpi }}\le \max \left\{ \eta ',\eta ''\right\} . \end{aligned}$$

This shows that the set ${{{\mathcal {E}}}}$ is bounded. Thus, by Schaefer’s fixed-point theorem [9], ${{\mathcal {H}}}$ has a fixed point which is a solution of problem (2). $\square$

5 Existence results for the third problem

Definition 5.1

A solution of (3) is a function ${{x}}\in C([-{\kappa },{\varpi }],{{\mathbb {R}}})$ where

$$\begin{aligned} {{x}}({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau , \ if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}}\Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x} ({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i} \\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ], \ if \ {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}), \ if\ {{t}}\in [-{\kappa },0], \end{array} \right. \end{aligned}$$

where $\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}\ne 0$ and ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{\rho ({{t}},{{x}}_{{t}})},{\Phi }({{t}}))$.

The hypotheses:

$(H_3)$:

The function ${\Psi }$ verifies

$$\begin{aligned} |{\Psi } ({{t}},{{x}}_1, \Im _1)-{\Psi } ({{t}},{{x}}_2, \Im _2)|\le \omega _4\Vert {{x}}_1- {{x}}_2\Vert _{[-{\kappa },0]}+ \omega _5|\Im _1 -\Im _2|, \end{aligned}$$

for any ${{x}}_1,x_2\in C([-{\kappa },0],{\mathbb {R}}),\ \Im _1,\Im _2\in {\mathbb {R}},$ and each ${{t}}\in {\Theta },$ where $\omega _4>0,\ 0<\omega _5<1.$

$(H_{03})$:

There exist a constant $\omega '_4>0$ such that

$$\begin{aligned} |\hbar _\ell ({x})-\hbar _\ell (\Im )|\le \omega '_4\Vert {{x}}- {\Im }\Vert _{[-{\kappa },0]}, \end{aligned}$$

for any ${{x}},{\Im }\in C([-{\kappa },0],{{\mathbb {R}}})$.

Remark 5.2

We note that by taking:

$$\begin{aligned} A_1=\omega _3, \quad A_2=\omega _4,\ A_4=\omega '_4,\ A'_3=\hbar _\ell (0)\ \text {and} \ A_3={\Psi }^*, \end{aligned}$$

where ${\Psi }^*=\displaystyle \sup _{{{t}}\in [0,{\varpi }]}{\Psi }({{t}},0,0)$. Then, hypothesis $(H_3)$ implies that

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )|\le A_1\Vert {{x}}\Vert _{[-{\kappa },0]}+ A_2|\Im |+A_3, \end{aligned}$$

and hypothesis $(H_{03})$ implies that

$$\begin{aligned} |\hbar _\ell ({x})|\le A_4\Vert {{x}}\Vert _{[-{\kappa },0]}+{A'_3}. \end{aligned}$$

for any ${{x}}\in C([-{\kappa },0],{\mathbb {R}}),\ v\in {\mathbb {\Im }},$ and each ${{t}}\in {\Theta }.$

As in Theorems 3.3 and 3.4, we obtain the following.

Theorem 5.3

If $(H_3)$ and $(H_{03})$ hold and

$$\begin{aligned}{} & {} \max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_4+\frac{({{\mathfrak {m}}}+1) \omega _4}{\beta (1-\omega _5)}\right) +\frac{\omega _4}{\beta (1-\omega _5)}\right] ,\right. \nonumber \\ {}{} & {} \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}}\omega '_4+\frac{({{\mathfrak {m}}}+1) \omega _4}{\beta (1-\omega _5)}\right) +{\mathfrak {m}}\omega '_4+\frac{(\omega +1)\omega _4}{\beta (1-\omega _5)}\right] \right\} <1, \end{aligned}$$

(14)

then (3) admit a unique solution on $[-{\kappa },{\varpi }].$

Theorem 5.4

Assume that the hypotheses $(H_3)$ and $(H_{03})$ hold. If

$$\begin{aligned}{} & {} \max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} A_4+\frac{({{\mathfrak {m}}}+1)A_1}{\beta (1-A_2)}\right) +\frac{A_1}{\beta (1-A_2)}\right] ,\right. \nonumber \\ {}{} & {} \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} A_4+\frac{({{\mathfrak {m}}}+1)A_1}{\beta (1-A_2)}\right) +{\mathfrak {m}} A_4+\frac{(\varpi +1)A_1}{\beta (1-A_2)}\right] \right\} <1, \end{aligned}$$

(15)

then (3) admit at least one solution on $[-{\kappa },{\varpi }].$

6 The existence results for the fourth problem

Definition 6.1

A solution of (4) is a function ${{x}}\in \Xi$ where

$$\begin{aligned} {{x}}({{t}})=\left\{ \begin{array}{ll} \frac{-e^{\frac{-\beta }{\alpha }{t}}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+\dfrac{1}{\alpha }e^{\frac{-\beta }{\alpha }{t}}\displaystyle \int _{0}^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau , \ if\ {t}\in [0,{t}_1],\\ \frac{-e^{\frac{-\beta }{\alpha }({t}+\sum _{k=1}^{k=\ell }{t}_k)}}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}} \Bigg [\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha } \sum _{i=k+1}^{{\mathfrak {m}}}{t}_i} +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k={\mathfrak {m}}}e^{\frac{-\beta }{\alpha } \sum _{i=k}^{i={\mathfrak {m}}}{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau &{}\\ +\dfrac{1}{\alpha }\displaystyle \int _{{t}_{\mathfrak {m}}}^{\varpi } e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau -\varrho e^{\frac{\beta }{\alpha }\varpi }\Bigg ]+e^{\frac{-\beta }{\alpha }{t}} \Bigg [\displaystyle \sum _{k=1}^{k=\ell }\hbar _k({x}({{t}_k^-}))e^{\frac{-\beta }{\alpha }\sum _{i=k+1}^{\ell }{t}_i}\\ +\dfrac{1}{\alpha }\displaystyle \sum _{k=1}^{k=\ell }e^{\frac{-\beta }{\alpha }\sum _{i=k}^{i=\ell }{t}_i}\displaystyle \int _{{t}_{k-1}}^{{t}_{k}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau +\dfrac{1}{\alpha }\displaystyle \int _{{t}_\ell }^{{t}} e^{\frac{\beta }{\alpha }\tau }\Phi (\tau ) \textrm{d}\tau \Bigg ], \ if \ {t}\in ({t}_\ell ,{t}_\ell +1],\\ \zeta ({{t}}),\ if\ {{t}}\in (-{\infty },0], \end{array} \right. \end{aligned}$$

where $\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta }{\alpha }\sum _{k=1}^{k={\mathfrak {m}}}{t}_k}\ne 0$ and ${\Phi }\in C({\Theta })$, with ${\Phi }({{t}})={\Psi }({{t}},{{x}}_{\rho ({{t}},{{x}}_{{t}})},{\Phi }({{t}}))$.

Set

$$\begin{aligned} {{\delta }}':={{\delta }}'_{\rho ^-}=\{\rho ({{t}},{{x}}):{{t}}\in {\Theta },\ {{x}}\in {{{\mathcal {G}}}} \ \rho ({{t}},{{x}})<0 \}. \end{aligned}$$

Suppose that $\rho :{\Theta }\times {{{\mathcal {G}}}}\rightarrow {\mathbb {R}}$ is continuous and ${{t}}\rightarrow {{x}}_{{t}}$ is continuous from ${{\delta }}'$ into ${{{\mathcal {G}}}}.$ $(H_\eta )$ There exists a continuous bounded function ${\varrho }:{{\delta }}'_{\rho ^-}\rightarrow (0,\infty )$ where

$$\begin{aligned} \Vert \eta _{{t}}\Vert _{{{\mathcal {G}}}}\le {\varrho }({{t}})\Vert \eta \Vert _{{{\mathcal {G}}}},\ \text { for any} \ {{t}}\in {{\delta }}'. \end{aligned}$$

Lemma 6.2

If ${{x}}\in \Xi$, then

$$\begin{aligned} \Vert {{x}}_{{t}}\Vert _{{{\mathcal {G}}}}= ({\xi _2}+{\varpi }')\Vert \eta \Vert _\mathcal{G}+{\xi _1}\sup _{{\tau }\in [0,\max \{0,{{t}}\}]}\Vert {{x}}({\tau })\Vert , \end{aligned}$$

where

$$\begin{aligned} {\varpi }'=\sup _{{{t}}\in {{\delta }}'}{\varpi }({{t}}). \end{aligned}$$

The hypotheses:

$(H_4)$:

The function ${\Psi }$ verifies

$$\begin{aligned} |{\Psi } ({{t}},{{x}}_1, \Im _1) - {\Psi } ({{t}},{{x}}_2, \Im _2)|\le b_3\Vert {{x}}_1- {{x}}_2\Vert _{{{\mathcal {G}}}} + b_4|\Im _1 -\Im _2|, \end{aligned}$$

for any ${{x}}_1,x_2\in {{{\mathcal {G}}}}$, $\Im _1,\Im _2\in {\mathbb {R}},$ and each ${{t}}\in {\Theta },$ where $b_3>0$ and $0<b_4<1.$

$(H_{04})$:

There exists a constant $b'_4>0$ such that

$$\begin{aligned} |\hbar _\ell ({x})-\hbar _\ell (\Im )|\le b'_4\Vert {{x}}- {\Im }\Vert _\mathcal{G}, \end{aligned}$$

for any ${{x}},{\Im }\in {{{\mathcal {G}}}}$.

Remark 6.3

We note that by taking

$$\begin{aligned} B_4=b_3, \quad B_5=b_4,\ B'_4=b'_4,\ B'_6=\hbar _\ell (0),\ \text {and} \ B_6={\Psi }^*, \end{aligned}$$

where ${\Psi }^*=\displaystyle \sup _{{{t}}\in [0,{\varpi }]}{\Psi }({{t}},0,0)$. Then, hypothesis $(H_4)$ implies that

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )|\le B_4\Vert {{x}}\Vert _{{{\mathcal {G}}}}+ B_5|\Im |+B_6, \end{aligned}$$

and hypothesis $(H_{04})$ implies that

$$\begin{aligned} |\hbar _\ell ({x})|\le B'_4\Vert {{x}}\Vert _{{{\mathcal {G}}}}+{B'_6}, \end{aligned}$$

for any ${{x}}\in {{{\mathcal {G}}}},\ \Im \in {\mathbb {R}},$ and each ${{t}}\in {\Theta }.$

As in Theorems 4.3 and 4.4, we have the following.

Theorem 6.4

If $(H_4)$ and $(H_{04})$ hold and

$$\begin{aligned}{} & {} \max \left\{ \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_4+\frac{({{\mathfrak {m}}}+1)b_3}{\beta (1-b_4)}\right) +\frac{b_3}{\beta (1-b_4)}\right] ,\right. \nonumber \\ {}{} & {} \quad \left. \left[ \frac{1}{\imath e^{\frac{\beta }{\alpha }\varpi }+\jmath e^{\frac{-\beta {\mathfrak {m}}\varpi }{\alpha }}}\left( {\mathfrak {m}} b'_4+\frac{({{\mathfrak {m}}}+1)b_3}{\beta (1-b_4)}\right) +{\mathfrak {m}} b'_4+\frac{(\varpi +1)b_4}{\beta (1-b_4)}\right] \right\} <1, \end{aligned}$$

(16)

then (4) admit a unique solution on $(-\infty ,{\varpi }].$

Theorem 6.5

If $(H_\zeta )$, $(H_4)$ and $(H_{04})$ are met, then (4) admit at least one solution on $(-\infty ,{\varpi }].$

7 Some examples

We give now some examples that illustrate our obtained results throughout the paper.

Example 7.1

Consider the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} ({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})=\dfrac{1}{90\left( 1+\Vert {{x}}_{{{t}}}\Vert \right) } +\dfrac{1}{30\left( 1+|({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})|\right) };\ {{t}}\in \Theta _0\cup \Theta _1,\\ \Delta {x}|_{{t}=\frac{1}{2}}=\frac{{x}_{{t}_\jmath ^-}}{30+{x}_{{t}_\jmath ^-}},\\ {{x}}(0)+{{x}}(1)=0,\\ {{x}}({{t}})=1+{{t}}^{2};\ {{t}}\in [-1,0], \end{array}\right. } \end{aligned}$$

(17)

where $\Theta _0=[0,\frac{1}{2}]$,$\Theta _1=(\frac{1}{2},1]$, ${t}_0=0$ and ${t}_1=1$.

Set

$$\begin{aligned} {\Psi }({{t}},{{x}},\Im )=\frac{1}{90\left( 1+\Vert {{x}}\Vert \right) }+\frac{1}{30\left( 1+|\Im |\right) };\ {{t}}\in [0,1],\ {{x}}\in {{{\mathcal {C}}}},\ \Im \in {\mathbb {R}}. \end{aligned}$$

For any ${{x}},\widetilde{{{x}}}\in {{{\mathcal {C}}}},\ {{x}},\widetilde{{{x}}}\in {\mathbb {R}},$ and ${{t}}\in [0,1],$ we have

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )-{\Psi }({{t}},\widetilde{{{x}}},{\widetilde{\Im }})|\le \frac{1}{90}\Vert {{x}} -\widetilde{{{x}}}\Vert _{[-1,0]}+\frac{1}{30}|\Im -{\widetilde{\Im }}|, \end{aligned}$$

and

$$\begin{aligned} \Vert \hbar _\ell ({x})-\hbar _\ell (\widetilde{{{x}}})\Vert \le \frac{1}{30} \Vert ({x}-\widetilde{{{x}}})\Vert _{[-1,0]}. \end{aligned}$$

Hence, hypothesis $(H_1)$ is satisfied with

$$\begin{aligned} \omega _1=\frac{1}{90}\ \ \ and\ \ \ \omega _2=\frac{1}{30}, \end{aligned}$$

and hypothesis $(H_{01})$ is satisfied with

$$\begin{aligned} \omega '_1=\frac{1}{30}. \end{aligned}$$

Some calculations indicate that all of the requirements of Theorem 3.3 are verified. Thus, (17) has a unique solution.

Example 7.2

Consider the following example:

$$\begin{aligned} {\left\{ \begin{array}{ll} ({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})=\dfrac{{{x}}_{{{t}}}e^{-\gamma {{t}}+{{t}}}}{180\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+\Vert {{x}}_{{{t}}}\Vert \right) } +\dfrac{{{x}}({{t}})e^{-\gamma {{t}}+{{t}}}}{60\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+|({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})|\right) };\ {{t}}\in \Theta _0\cup \Theta _1,\\ \Delta {x}|_{{t}=\frac{1}{2}}=\frac{{x}_{{t}_\jmath ^-}}{30+{x}_{{t}_\jmath ^-}},\\ {{x}}(0)+{{x}}(1)=0,\\ {{x}}({{t}})={{t}}+1;\ {{t}}\in (-\infty ,0]. \end{array}\right. } \end{aligned}$$

(18)

Let $\gamma$ be a positive real constant and

$$\begin{aligned} B_\gamma =\{{{x}}\in \mathcal{P}\mathcal{C}((-\infty ,1],{{\mathbb {R}}}):\ \lim _{{\tau }\rightarrow -\infty }e^{\gamma {\tau }}{{x}}({\tau })\ exists\ in\ {\mathbb {R}} \}. \end{aligned}$$

(19)

The norm of $B_\gamma$ is given by

$$\begin{aligned} \Vert {{x}}\Vert _\gamma =\sup _{{\tau }\in (-\infty ,1]}e^{\gamma {\tau }}|{{x}}({\tau })|. \end{aligned}$$

Let ${{x}}:(-\infty ,0]\rightarrow {\mathbb {R}}$ be such that ${{x}}_0\in B_\gamma .$ Then,

$$\begin{aligned} \begin{array}{rl} \lim _{{\tau }\rightarrow -\infty }e^{\gamma {\tau }}{{x}}_{{t}}({\tau })&{}= \lim _{{\tau }\rightarrow -\infty }e^{\gamma {\tau }}{{x}}({{t}}+{\tau }-1)= \lim _{{\tau }\rightarrow -\infty }e^{\gamma ({\tau }-{{t}}+1)}{{x}}({\tau })\\ \\ &{}=e^{\gamma (-{{t}}+1)} \lim _{{\tau }\rightarrow -\infty }e^{\gamma ({\tau })}{{x}}_1({\tau })<\infty . \end{array} \end{aligned}$$

Hence, ${{x}}_{{t}}\in B_\gamma .$ Finally, we prove that

$$\begin{aligned} \Vert {{x}}_{{t}}\Vert _{\gamma }\le {\xi _1} \Vert {{x}}_1\Vert _{\gamma }+ {\xi _2} \sup _{{\vartheta }\in [0,{{t}}]}|{{x}}({\vartheta })|, \end{aligned}$$

where ${\xi _1}={\xi _2}=1$ and ${\xi _3}=1.$ We have

$$\begin{aligned} \Vert {{x}}_{{t}}({\tau })\Vert =|{{x}}({{t}}+{\tau })|. \end{aligned}$$

If ${{t}}+{\tau }\le 1,$ we get

$$\begin{aligned} \Vert {{x}}_{{t}}(\xi )\Vert \le \sup _{{\vartheta }\in (-\infty ,0]}|{{x}}({\vartheta })|. \end{aligned}$$

For ${{t}}+{\tau }\ge 0,$ then we have

$$\begin{aligned} \Vert {{x}}_{{t}}(\xi )\Vert \le \sup _{{\vartheta }\in [0,{{t}}]}|{{x}}({\vartheta })|. \end{aligned}$$

Thus, for all ${{t}}+{\tau }\in {\Theta },$ we get

$$\begin{aligned} \Vert {{x}}_{{t}}(\xi )\Vert \le \sup _{{\vartheta }\in (-\infty ,0]}|{{x}}({\vartheta })|+\sup _{{\vartheta }\in [0,{{t}}]}|{{x}}({\vartheta })|. \end{aligned}$$

Then,

$$\begin{aligned} \Vert {{x}}_{{t}}\Vert _{\gamma }\le \Vert {{x}}_0\Vert _{\gamma }+\sup _{{\vartheta }\in [0,{{t}}]}|{{x}}({\vartheta })|. \end{aligned}$$

It is clear that $(B_\gamma ,\Vert \cdot \Vert )$ is a Banach space. We can conclude that $B_\gamma$ is a phase space.

Set

$$\begin{aligned} {\Psi }({{t}},{{x}},\Im )=\frac{e^{-\gamma {{t}}+{{t}}}}{180\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+\Vert {{x}} \Vert _{B_\gamma }\right) } +\frac{e^{-\gamma {{t}}+{{t}}}}{60\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+|\Im |\right) };\ {{t}}\in [0,1],\ {{x}}\in {B_\gamma },\Im \in {\mathbb {R}}. \end{aligned}$$

For any ${{x}},\tilde{{{x}}}\in {B_\gamma },\ \Im ,{\tilde{\Im }}\in {\mathbb {R}}$ and ${{t}}\in [0,1],$ we have

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )-{\Psi }({{t}},\tilde{{{x}}},{\tilde{\Im }})|\le \frac{1}{180}\Vert {{x}}-\tilde{{{x}}} \Vert _{B_\gamma }+\frac{1}{60}|\Im -{\tilde{\Im }}|, \end{aligned}$$

and

$$\begin{aligned} \Vert \hbar _\ell ({x})-\hbar _\ell (\widetilde{{{x}}})\Vert \le \frac{1}{30} \Vert ({x}-\widetilde{{{x}}})\Vert _{B_\gamma }. \end{aligned}$$

Hence, hypothesis $(H_2)$ is satisfied with

$$\begin{aligned} b_1=\frac{1}{180}\ \ \ and\ \ \ b_2=\frac{1}{60}, \end{aligned}$$

and hypothesis $(H_{02})$ is satisfied with

$$\begin{aligned} b'_1=\frac{1}{30}. \end{aligned}$$

All requirements of Theorem 4.4 are met. Then, problem (18) has at least one solution defined on $(-\infty ,1].$

Example 7.3

We consider the following problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} ({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})=\dfrac{1}{90(1+|{{x}}({{t}}-\sigma ({{x}}({{t}})))|)} +\dfrac{1}{30\left( 1+|({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})|\right) };\ {{t}}\in \Theta _0\cup \Theta _1,\\ \Delta {x}|_{{t}=\frac{1}{2}}=\frac{{x}_{{t}_\jmath ^-}}{30+{x}_{{t}_\jmath ^-}},\\ {{x}}(0)+{{x}}(1)=0,\\ {{x}}({{t}})=1+{{t}}^{2};\ {{t}}\in [-1,0], \end{array}\right. } \end{aligned}$$

(20)

where $\sigma \in C({\mathbb {R}},[0,1]).$ Set

$$\begin{aligned} \rho ({{t}},\zeta )= & {} {{t}}-\sigma (\zeta (0)), \ \ ({{t}},\zeta ) \in [0,1]\times C([-1,0],{\mathbb {R}}),\\ {\Psi }({{t}},{{x}},\Im )= & {} \frac{1}{90(1+|{{x}}({{t}}-\sigma ({{x}}({{t}})))|)} +\frac{1}{30\left( 1+|\Im ({{t}})|\right) };\ {{t}}\in [0,1],\ {{x}}\in {{{\mathcal {C}}}},\ \Im \in {\mathbb {R}}. \end{aligned}$$

For ${{x}},\widetilde{{{x}}}\in {{{\mathcal {C}}}},\ \Im ,{\widetilde{\Im }}\in {\mathbb {R}}$ and ${{t}}\in [0,1],$ we have

$$\begin{aligned} |{\Psi }({{t}},{{x}},\Im )-{\Psi }({{t}},\widetilde{{{x}}},{\widetilde{\Im }})|\le \frac{1}{90}\Vert {{x}}-\widetilde{{{x}}} \Vert _{[-1,0]} +\frac{1}{30}|\Im -{\widetilde{\Im }}|, \end{aligned}$$

and

$$\begin{aligned} \Vert \hbar _\ell ({x})-\hbar _\ell (\widetilde{{{x}}})\Vert \le \frac{1}{30} \Vert ({x}-\widetilde{{{x}}})\Vert _{[-1,0]}. \end{aligned}$$

Hence, hypothesis $(H_3)$ is satisfied with

$$\begin{aligned} \omega _4=\frac{1}{90}\ \ \ and\ \ \ \omega _5=\frac{1}{30}, \end{aligned}$$

and hypothesis $(H_{03})$ is satisfied with

$$\begin{aligned} b'_4=\frac{1}{30}. \end{aligned}$$

All requirements of Theorem 5.3 are verified. Thus, (20) has a unique solution defined on $[-1,1].$

Example 7.4

Consider now the problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} ({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})=\frac{{{x}}({{t}}-\lambda ({{x}}({{t}}))) e^{-\gamma {{t}}+{{t}}}}{180\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+|{{x}}({{t}}-\sigma ({{x}}({{t}}))|\right) } +\frac{{{x}}({{t}})e^{-\gamma {{t}}+{{t}}}}{60\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+|({\mathfrak {D}}_{0}^{\frac{1}{2}}{{x}})({{t}})|\right) };\ {{t}}\in \Theta _0\cup \Theta _1,\\ \Delta {x}|_{{t}=\frac{1}{2}}=\frac{{x}_{{t}_\jmath ^-}}{30+{x}_{{t}_\jmath ^-}},\\ {{x}}(0)+{{x}}(1)=0,\\ {{x}}({{t}})={{t}};\ {{t}}\in (-\infty ,0]. \end{array}\right. } \end{aligned}$$

(21)

Define

$$\begin{aligned} \rho ({{t}},\zeta )={{t}}-\lambda (\zeta (0)), \ \ ({{t}},\zeta ) \in [0,2]\times B_{\gamma }, \end{aligned}$$

and set

$$\begin{aligned} {\Psi }({{t}},{{x}},\Im )=\frac{e^{-\gamma {{t}}+{{t}}}}{180\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+\Vert {{x}} \Vert _{B_\gamma }\right) } +\frac{e^{-\gamma {{t}}+{{t}}}}{60\left( e^{{t}}-e^{-{{t}}}\right) \left( 1+|\Im |\right) }, \end{aligned}$$

where ${{t}}\in [0,2],\ {{x}}\in {B_\gamma },\ \Im \in {\mathbb {R}}$. We can demonstrate that all conditions of Theorem 4.4 are verified. Then, (21) has at least one solution defined on $(-\infty ,2].$

8 Conclusion

In this research, we have established the existence and uniqueness of solutions for a class of deformable fractional differential problems. These problems encompassed nonlinear implicit fractional differential equations that were accompanied by boundary conditions and various types of delays, including finite, infinite, and state-dependent delays. Our approach to proving the existence and uniqueness of solutions relied on the application of the Banach contraction principle and Schauder’s fixed-point theorem. To illustrate the practicality of our key findings and to demonstrate that the prerequisites of our theorems can be satisfied, we provided several specific examples. Our results in this particular configuration represent a novel contribution to the literature in this emerging field of study. Given the limited number of publications on deformable differential equations, we believe there are numerous potential research avenues to explore. These may include investigating coupled systems, problems involving non-local conditions, hybrid problems, problems with non-instantaneous impulses, and various qualitative studies such as Ulam stability. We aspire for this article to serve as a point for further exploration in these directions.