# Delayed Gronwall inequality with weakly the singular kernel

Javad A. Asadzade<EMAIL_ADDRESS>Jasarat J. Gasimov
<EMAIL_ADDRESS>Nazim I. Mahmudov<EMAIL_ADDRESS>Department of Mathematics, Eastern Mediterranean University, Mersin 10, 99628,
T.R. North Cyprus, Turkey  Research Center of Econophysics, Azerbaijan State
University of Economics (UNEC), Istiqlaliyyat Str. 6, Baku 1001, Azerbaijan

###### Abstract

Delay Gronwall inequality with a weakly singular kernel has been a subject of
interest in various mathematical studies. In this article, we will delve into
the consideration of this inequality and its application in the study
continuity of the state trajectory for a Volterra integral equation with
delay. Using delay Gronwall inequality with a singular kernel, we investigate
the behavior and properties of the state trajectory if there are delays. This
analysis aims to improve our understanding of the dynamics associated in del
Volterra integral equations with delay. In addition, we present a
comprehensive example demonstrating the practical significance of our results.

###### keywords:

Gronwall inequality, well-posedness, singular kernel

##  1 INTRODUCTION

The delay Gronwall inequality with a singular kernel is a fundamental result
in the field of mathematical analysis and is widely used in various areas of
mathematics, physics, and engineering. The inequality provides a powerful tool
for estimating the growth and behavior of functions and has applications in
diverse fields such as differential equations,delay systems, mathematical
modeling, control theory, and stochastic processes.

The Gronwall inequality, named after American mathematician Thomas H.
Gronwall, establishes an upper bound for the growth of a function based on its
integral and the behavior of a singular kernel. The inequality has proven to
be an indispensable tool for studying the properties of solutions to
differential equations and integral equations.

The delay Gronwall inequality with a singular kernel offers a rigorous
mathematical framework to analyze the growth and stability of functions
subject to delays and singular influences. By considering the integral of a
function multiplied by the singular kernel, the inequality provides an upper
bound on the growth rate of delayed functions, enabling researchers to derive
essential bounds and estimates. This inequality is indispensable for studying
the stability and convergence properties of delayed differential equations,
integral equations, and time-delay systems.

Furthermore, the delay Gronwall inequality with a singular kernel has profound
implications in various scientific fields. In control theory, it plays a vital
role in analyzing the stability and performance of control systems subjected
to delays, ensuring robustness and preventing undesired behaviors. In signal
processing and communication systems, it assists in understanding the impact
of time delays on signal transmission and reception, contributing to the
design of efficient and reliable communication networks. Its versatility and
broad applicability make the delay Gronwall inequality with a singular kernel
a valuable asset in both theoretical and applied mathematics.

In this paper, we aim to provide a comprehensive overview of the delay
Gronwall inequality with a singular kernel. We will explore its mathematical
formulation, discuss its properties, and examine its applications in different
areas of mathematics and science. By understanding the intricacies of this
inequality, researchers and practitioners can employ it effectively to tackle
various problems and gain insights into the behavior of complex systems.

In recent research [1]-[15], different generalizations of the Gronwall
inequalities have been developed. In particular, Ping Lin and Jiongmin Yong
[1] derived a notable inequality under certain conditions.

We will denote the following necessary theorem, which we use it for proof of
the delayed Gronwall’s inequality for singular integral.

######  Lemma 1.1.

([1]) Let $\beta\in(0,1)$ and $q>1/\beta$. Let $L(\cdot)$, $a(\cdot)$,
$y(\cdot)$ be nonnegative functions with $L(\cdot)\in L^{q}(0,T)$ and
$a(\cdot)$, $y(\cdot)\in L^{\frac{q}{q-1}}(0,T)$. Suppose

$\displaystyle y(t)\leq
a(t)+\int_{0}^{t}\frac{L(s)y(s)}{(t-s)^{1-\beta}}ds,\quad a.e.\quad
t\in[0,T].$ (1.1)

Then there exists a constant $K>0$ such that

$\displaystyle y(t)\leq
a(t)+K\int_{0}^{t}\frac{L(s)a(s)}{(t-s)^{1-\beta}}ds,\quad a.e.\quad
t\in[0,T].$

Another interesting scenario, if we replace assumption (1.1) with the
following condition:

$\displaystyle\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{0}^{t}\frac{L(s)\xi(s-h)}{(t-s)^{1-\nu}}ds,\quad\text{a.e.,}\quad
t\in[0,T].$ (1.2)

Then, there exists a constant $K>0$ such that

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta_{n}(t)+K\int_{0}^{t}\frac{L(S)\vartheta(s)}{(t-s)^{1-\nu}}ds,\quad
t\in[0,T].$ (1.3)

Where

$\displaystyle\vartheta_{n}(t)=$
$\displaystyle\vartheta(t)+K\sum_{k=1}^{n}\int_{0}^{t-kh}\frac{L(s)\vartheta(s)}{(t-kh-s)^{1-\nu}}ds+K\sum_{k=0}^{n-1}\int_{h}^{t-kh}\frac{L(s)\vartheta(s-h)}{(t-kh-s)^{1-\nu}}ds.$

The assumption (1.2) is considered natural and plays a crucial role in
investigating the well-posedness problem discussed in Section 4. These results
highlight the significant advancements in understanding the properties and
generalizations of Gronwall inequalities, paving the way for further
exploration and applications in various mathematical analyses.

##  2 Preliminaries

In the forthcoming section, we will provide initial findings that will prove
beneficial in subsequent analysis. To begin, consider a predetermined time
horizon denoted as $T>0$. We will now define the subsequent spaces:

$\displaystyle L^{p}(0,T;R^{n})=\bigg{\\{}\phi:[0,T]\to
R^{n}\quad|\quad\phi(\cdot)\quad is\quad measurable,$
$\displaystyle\|\phi(\cdot)\|_{p}\equiv\bigg{(}\int_{0}^{T}|\phi(t)|^{p}dt\bigg{)}^{1/p}<\infty\bigg{\\}},1\leq
p<\infty,$ $\displaystyle L^{\infty}(0,T;R^{n})=\bigg{\\{}\phi:[0,T]\to
R^{n}\quad|\quad\phi(\cdot)\quad is\quad
measurable,\quad\|\phi(\cdot)\|_{\infty}\equiv\operatorname*{ess\,sup}_{t\in[0,T]}|\phi(t)|<\infty\bigg{\\}}.$

Also, we define

$\displaystyle L^{p+}(0,T;R^{n})=\bigcup_{r>p}L^{r}(0,T;R^{n}),\quad 1\leq
p<\infty,$ $\displaystyle
L^{p-}(0,T;R^{n})=\bigcap_{r<p}L^{r}(0,T;R^{n}),\quad 1<p<\infty.$

In the subsequent analysis, we utilize the notation
$\Delta=\\{(t,s)\in[0,T]^{2}|0\leq s<t\leq T\\}$. It is important to note that
the ”diagonal line” represented by $\\{(t,t)|t\in[0,T]\\}$ does not belong to
$\Delta$. Consequently, if we consider a continuous mapping $\phi:\Delta\to
R^{n}$ where $(t,s)\mapsto\phi(t,s)$, the function $\phi(\cdot,\cdot)$ may
become unbounded as the difference $|t-s|\to 0$.

In the article, we adopt the notation $t_{1}\vee t_{2}=\max\\{t_{1},t_{2}\\}$
and $t_{1}\wedge t_{2}=\min\\{t_{1},t_{2}\\}$, for any $t_{1},t_{2}\in R$.
Notably, $t^{+}=t\vee 0$.

######  Lemma 2.1.

(Lemma 2.1, page 138, [1]) Let $p,q,r\geq 1$ satisfy
$\frac{1}{p}+1=\frac{1}{q}+\frac{1}{r}$. Then for any $f(\cdot)\in
L^{q}(R^{n}),g(\cdot)\in L^{r}(R^{n})$,

$\displaystyle\|f(\cdot)*g(\cdot)\|_{L^{p}(R^{n})}\leq\|f(\cdot)\|_{L^{q}(R^{n})}\|g(\cdot)\|_{L^{r}(R^{n})}.$
(2.1)

######  Corollary 2.1.

(Corollary 2.2, page 138, [1]) Let $\beta\in(0,1)$, $1\leq
r<\frac{1}{1-\beta}$, and $\frac{1}{p}+1=\frac{1}{q}+\frac{1}{r},\quad p,q\geq
1$. Then for any $a<b,\quad 0<\delta\leq b-a$, and $\varphi(\cdot)\in
L^{q}(a,b)$,

$\displaystyle\bigg{(}\int_{a}^{a+\delta}\bigg{|}\int_{a}^{t}\frac{\varphi(s)ds}{(t-s)^{1-\beta}}\bigg{|}^{p}dt\bigg{)}^{1/p}\leq\bigg{(}\frac{\delta^{1-r(1-\beta)}}{1-r(1-\beta)}\bigg{)}^{1/r}\|\varphi(\cdot)\|_{L^{q}(a,b)}.$
(2.2)

##  3 Main result

In this section, we will present the following theorem that demonstrates the
delay Gronwall inequality with a singular kernel.

######  Theorem 3.1.

Let $\nu\in(0,1)$, $h>0$, and $q>\frac{1}{\nu}$. Suppose that
$L(\cdot),\vartheta(\cdot),\xi(\cdot)$ be nonnegative functions with
$\vartheta(t)=0$, $L(t)=0\quad t<0$, where $L(\cdot)\in L^{q}(0,T)$ and
$\vartheta(\cdot),\xi(\cdot)\in L^{\frac{q}{q-1}}[-h,T)$. Assume

$\begin{cases}\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{0}^{t}\frac{L(s)\xi(s-h)}{(t-s)^{1-\nu}}ds,&a.e.\quad
t\in[0,T],\\\ \xi(t)=0,&t<0.\end{cases}$ (3.1)

Then there is a constant $K>0$ such that

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta_{n}(t)+K\int_{0}^{t}\frac{L(s)\vartheta(s)}{(t-s)^{1-\nu}}ds,\quad
a.e.\quad t\in[0,T],$ (3.2)

where

$\displaystyle\vartheta_{n}(t)=$
$\displaystyle\vartheta(t)+K\sum_{k=1}^{n}\int_{0}^{t-kh}\frac{L(s)\vartheta(s)}{(t-kh-s)^{1-\nu}}ds+K\sum_{k=0}^{n-1}\int_{h}^{t-kh}\frac{L(s)\vartheta(s-h)}{(t-kh-s)^{1-\nu}}ds.$
(3.3)

###### Proof.

Let’s contemplate the partition of $[0,T]$ by intervals of length $h$, where
there exists an integer $n$ such that $0<h<2h<\dots<nh\leq T<(n+1)h$.

$\bullet$ Step 1. Consider (3.1) on the $0<t\leq h$,

$\displaystyle\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{0}^{t}\frac{L(s)\xi(s-h)}{(t-s)^{1-\nu}}ds.$

Since $\xi(t)=0,\quad-h\leq t\leq 0$, it follows

$\displaystyle\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds.$

By using Lemma 1.1, we obtain

$\displaystyle\xi(t)\leq\vartheta_{0}(t)+K_{0}\int_{0}^{t}\frac{L(s)\vartheta(s)}{(t-s)^{1-\nu}}ds,\quad
t\in[0,h],$ (3.4)

where $\vartheta_{0}(t)=\vartheta(t)$.

$\bullet$ Step 2. Now, employing the same principle, we take $h<t\leq 2h$,

$\displaystyle\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)\xi(s-h)}{(t-s)^{1-\nu}}ds$

Substituting (3.4) into the above

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)}{(t-s)^{1-\nu}}\bigg{[}\vartheta(s-h)+K_{0}\int_{0}^{s-h}\frac{L(\tau)\vartheta(\tau)}{(s-h-\tau)^{1-\nu}}d\tau\bigg{]}ds.$

By applying the Fubini’s theorem,

$\displaystyle\int_{h}^{t}\frac{L(s)}{(t-s)^{1-\nu}}\int_{0}^{s-h}\frac{L(\tau)\vartheta(\tau)}{(s-h-\tau)^{1-\nu}}d\tau
ds=\int_{0}^{t-h}L(\tau)\bigg{[}\int_{\tau+h}^{t}\frac{L(s)ds}{(t-s)^{1-\nu}(s-h-\tau)^{1-\nu}}\bigg{]}\vartheta(\tau)d\tau$

Utilizing the Holder inequality, and letting $r=\frac{s-h-\tau}{t-h-\tau}$
implies

$\displaystyle\int_{\tau+h}^{t}\frac{L(s)ds}{(t-s)^{1-\nu}(s-h-\tau)^{1-\nu}}\leq$
$\displaystyle\bigg{(}\int_{\tau+h}^{t}L(s)^{q}ds\bigg{)}^{\frac{1}{q}}\bigg{(}\int_{\tau+h}^{t}\frac{ds}{(t-s)^{(1-\nu)\frac{q}{q-1}}(s-h-\tau)^{(1-\nu)\frac{q}{q-1}}}\bigg{)}^{\frac{q-1}{q}}$
$\displaystyle\leq$
$\displaystyle\|L(\cdot)\|_{q}\frac{1}{(t-h-\tau)^{2(1-\nu)-\frac{q-1}{q}}}\bigg{(}\int_{0}^{1}\frac{dr}{(1-r)^{(1-\nu)\frac{q}{q-1}}r^{(1-\nu)\frac{q}{q-1}}}\bigg{)}^{\frac{q-1}{q}}.$

Since $q>\frac{1}{\nu}$ that is equivalent to

$\displaystyle 0<1-\frac{(1-\nu)q}{q-1}=\frac{\nu q-1}{q-1},$ (3.5)

we get

$\displaystyle\int_{\tau+h}^{t}\frac{L(s)ds}{(t-s)^{1-\nu}(s-h-\tau)^{1-\nu}}\leq\frac{\|L(\cdot)\|_{q}}{(t-h-\tau)^{2(1-\nu)-\frac{q-1}{q}}}B\bigg{(}\frac{\nu
q-1}{q-1},\frac{\nu
q-1}{q-1}\bigg{)}^{\frac{q-1}{q}}\equiv\frac{K_{1}}{(t-h-\tau)^{1-\nu_{1}}},$

where, $B(\cdot,\cdot)$ is the well-known Beta function.

Therefore,

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)\vartheta(s-h)}{(t-s)^{1-\nu}}ds+K_{1}\int_{0}^{t-h}\frac{L(s)\vartheta(s)}{(t-h-s)^{1-\nu_{1}}}ds,$
(3.6)

such as

$\displaystyle K_{1}=\|L(\cdot)\|_{q}B\bigg{(}\frac{\nu q-1}{q-1},\frac{\nu
q-1}{q-1}\bigg{)}^{\frac{q-1}{q}},$
$\displaystyle\nu_{1}=1-\Big{(}2(1-\nu)-\frac{q-1}{q}\Big{)}=\nu+(\nu-\frac{1}{q})>\nu.$

By using the following inequality in (3.6)

$\displaystyle\frac{1}{(t-s)^{1-\nu_{1}}}\leq\frac{C}{(t-s)^{1-\nu}},\quad
0\leq s<t\leq T,$ (3.7)

we receive

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)\vartheta(s-h)}{(t-s)^{1-\nu}}ds+K_{1}\int_{0}^{t-h}\frac{L(s)\vartheta(s)}{(t-h-s)^{1-\nu}}ds.$

By using Lemma 1.1, we achieve

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta_{1}(t)+K_{1}\int_{0}^{t}\frac{L(s)\vartheta(s)}{(t-s)^{1-\nu}}ds,$
(3.8)

where

$\displaystyle\vartheta_{1}(t)=\vartheta(t)+K_{1}\int_{h}^{t}\frac{L(s)\vartheta(s-h)}{(t-s)^{1-\nu}}ds+K_{2}\int_{0}^{t-h}\frac{L(s)\vartheta(s)}{(t-h-s)^{1-\nu}}ds.$

$\bullet$ Step 3. Subsequently, we examine the range where $2h\leq t\leq 3h$,

$\displaystyle\xi(t)\leq\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)\xi(s-h)}{(t-s)^{1-\nu}}ds$

Again substituting (3.8) into the last expression

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta(t)+\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+\int_{h}^{t}\frac{L(s)}{(t-s)^{1-\nu}}\bigg{[}\vartheta_{1}(s-h)+K_{1}\int_{0}^{s-h}\frac{L(\tau)\vartheta(s)}{(s-h-\tau)^{1-\nu}}d\tau\bigg{]}ds$
By using the Fubini’s theorem and inequality (3.7) $\displaystyle=$
$\displaystyle\vartheta(t)+K_{3}\int_{0}^{t}\frac{L(s)\xi(s)}{(t-s)^{1-\nu}}ds+K_{3}\int_{h}^{t}\frac{L(s)\vartheta(s-h)}{(t-s)^{1-\nu}}ds+K_{3}\int_{0}^{t-h}\frac{L(s)\vartheta(s)}{(t-h-s)^{1-\nu}}ds$
$\displaystyle+$ $\displaystyle
K_{3}\int_{h}^{t-h}\frac{L(s)\vartheta(s-h)}{(t-h-s)^{1-\nu}}ds+K_{3}\int_{0}^{t-2h}\frac{L(s)\vartheta(s)}{(t-2h-s)^{1-\nu}}ds$

Applying Lemma 1.1, we get

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta_{3}(t)+K_{4}\int_{0}^{t}\frac{L(s)\vartheta(s)}{(t-s)^{1-\nu}}ds,$
(3.9)

where

$\displaystyle\vartheta_{3}(t)=$
$\displaystyle\vartheta(t)+K_{4}\int_{0}^{t-h}\frac{L(s)\vartheta(s)}{(t-h-s)^{1-\nu}}ds+K_{4}\int_{0}^{t-2h}\frac{L(s)\vartheta(s)}{(t-2h-s)^{1-\nu}}ds$
$\displaystyle+$ $\displaystyle
K_{4}\int_{h}^{t}\frac{L(s)\vartheta(s-h)}{(t-s)^{1-\nu}}ds+K_{4}\int_{h}^{t-h}\frac{L(s)\vartheta(s-h)}{(t-h-s)^{1-\nu}}ds.$

As $T$ is finite, there exists $n\in N$ such that $T<(n+1)h$. Continuing the
given recursive relationship in the same manner for $nh\leq t\leq(n+1)h$, we
obtain the following inequality:

$\displaystyle\xi(t)\leq$
$\displaystyle\vartheta_{n}(t)+K\int_{0}^{t}\frac{L(s)\vartheta(s)}{(t-s)^{1-\nu}}ds,\quad
a.e.\quad t\in[0,T],$

where, $K$ is the maximum value of $K_{i}$ for $i\in N$, and

$\displaystyle\vartheta_{n}(t)=\vartheta(t)+K\sum_{k=1}^{n}\int_{0}^{t-kh}\frac{L(s)\vartheta(s)}{(t-kh-s)^{1-\nu}}ds+K\sum_{k=0}^{n-1}\int_{h}^{t-kh}\frac{L(s)\vartheta(s-h)}{(t-kh-s)^{1-\nu}}ds.$

∎

##  4 Well-posedness in $L^{p}$ space and continuity of the state trajectory.

In this section, we consider the continuity of the state trajectory for the
delayed Volterra integral equation with initial condition, and we will show
the application of the delayed Gronwall inequality.

$\begin{cases}\xi(t)=\zeta(t)+\int_{0}^{t}\frac{\kappa(t,s,\xi(s),\xi(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds,\quad
a.e.\quad t\in[0,T]\\\ \xi(t)=0,\quad-h\leq t\leq 0,\quad h\geq 0.\end{cases}$
(4.1)

We present a set of the following conditions that apply to the generator
function $\kappa(\cdot,\cdot,\cdot,\cdot,\cdot)$ used in our state equation.
Let $U$ be a separable metric space with the metric $d$ ,which could be a
nonempty bounded or unbounded set in $R^{n}$ with the metric induced by the
usual Euclidean norm. With the Borel $\sigma$-field, $U$ is regarded as a
measurable space. Let $u_{0}\in U$ be fixed. For any $p\geq 1$, we define

$\displaystyle\mathcal{U}_{p}[0,T]=\bigl{\\{}u:[0,T]\rightarrow
U\bigm{|}u(\cdot)\quad is\quad measurable,\quad d(u(\cdot),u_{0})\in
L^{p}(0,T)\\}.$

(H): Let $\kappa:[0,T]\times[0,T]\times R^{n}\times R^{n}\times U\rightarrow
R^{n}$ be a transformation with $(t,s)\mapsto\kappa(t,s,\xi,\xi_{h},\upsilon)$
being measurable, $(\xi,\xi_{h})\mapsto\kappa(t,s,\xi,\xi_{h},\upsilon)$ being
continuously differentiable,
$(\xi,\xi_{h},\upsilon)\mapsto\kappa(t,s,\xi,\xi_{h},\upsilon)$ ,
$(\xi,\xi_{h},\upsilon)\mapsto\kappa_{\xi}(t,s,\xi,\xi_{h},\upsilon)$ and
$(\xi,\xi_{h},\upsilon)\mapsto\kappa_{\xi_{h}}(t,s,\xi,\xi_{h},\upsilon)$
being continuous. There exist non-negative functions. $L_{0}(\cdot),L(\cdot)$
with

$\displaystyle L_{0}(\cdot)\in L^{\frac{1}{\nu}{+}}(0,T),\quad L(\cdot)\in
L^{\frac{p}{p\nu-1}{+}}(0,T)$ (4.2)

for some $p>\frac{1}{\nu}$ and $\nu\in(0,1),\upsilon_{0}\in U$.

$\displaystyle|\kappa(t,s,0,0,\upsilon_{0})|\leq L_{0}(s),\quad t\in[0,T],$
(4.3)
$\displaystyle|\kappa(t,s,\xi,\xi_{h},\upsilon)-\kappa(t,s,\xi^{\prime},\xi^{\prime}_{h},\upsilon^{\prime})|\leq
L(s)[|\xi-\xi^{\prime}|+|\xi_{h}-\xi^{\prime}_{h}|+d(\upsilon,\upsilon^{\prime})],$
(4.4) $\displaystyle
for\quad(t,s)\in\Delta,\quad\xi,\xi^{\prime},\xi_{h},\xi^{\prime}_{h}\in
R^{n},\quad\upsilon,\upsilon^{\prime}\in U.$

We point out (4.3)-(4.4) declare

$\displaystyle|\kappa(t,s,\xi,\xi_{h},\upsilon)|\leq
L_{0}(s)+L(s)[|\xi|+|\xi_{h}|+d(\upsilon,\upsilon_{0})],\quad(t,\xi,\xi_{h},\upsilon)\in[0,T]\times
R^{n}\times R^{n}\times U.$ (4.5)
$\displaystyle|\kappa(t,s,\xi,\xi_{h},\upsilon)-\kappa(t^{\prime},s,\xi,\xi_{h},\upsilon)|\leq
K\omega(|t-t^{\prime}|)(1+|\xi|+|\xi_{h}|),\quad
t,t^{\prime}\in[0,T],\quad\xi,\xi_{h}\in R^{n},\quad\upsilon\in U,$ (4.6)

for some modulus of continuity $\omega(\cdot)$. Moreover, it is evident that
$L$ is included in a smaller space, compared to the space to which $L_{0}$
belongs.

We will now demonstrate the well-posedness of the state equation (4.1) when
considering $L^{p}$ spaces.

######  Theorem 4.1.

Assume that (H) satisfies with some $p\geq 1$ and $\nu\in(0,1)$. Hence for
each $\zeta(\cdot)\in L^{p}(-h,T;R^{n})$ and
$\upsilon\in\mathscr{U}^{p}[0,T]$, (4.1) admits a unique solution
$\xi(\cdot)\equiv\xi(\cdot,\zeta(\cdot),\upsilon(\cdot))\in
L^{p}(-h,T;R^{n})$, and the following estimate satisfies:

$\displaystyle\|\xi(\cdot)\|_{p}\leq\|\zeta(\cdot)\|_{p}+K\Big{(}1+\|d(\upsilon(\cdot),\upsilon_{0})\|_{L^{p}(0,T)}\Big{)}.$
(4.7)

If
$(\zeta_{1}(\cdot),\upsilon_{1}(\cdot)),(\zeta_{2}(\cdot),\upsilon_{2}(\cdot))\in
L^{p}(-h,T;R^{n})\times\mathscr{U}^{p}[0,T]$ and
$\xi_{1}(\cdot),\xi_{2}(\cdot)$ are the solutions of (4.1) corresponding to
$(\zeta_{1}(\cdot),,\upsilon_{1}(\cdot))$, and
$(\zeta_{2}(\cdot),\upsilon_{2}(\cdot))$, accordingly, then

$\displaystyle\|\xi_{1}(\cdot)-\xi_{2}(\cdot)\|_{p}\leq
K\Big{\\{}\|\zeta_{1}(t)-\zeta_{2}(t)\|_{p}$ $\displaystyle+$
$\displaystyle\bigg{[}\int_{0}^{T}\bigg{(}\int_{0}^{t}\frac{|\kappa(t,s,\xi_{1}(s),\xi_{1}(s-h),\upsilon_{1}(s))-\kappa(t,s,\xi_{2}(s),\xi_{2}(s-h),\upsilon_{2}(s))|}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\bigg{]}^{1/p}\Big{\\}}.$
(4.8)

###### Proof.

Assuming a fixed function $\zeta(\cdot)$ in the space $L^{p}(-h,S;R^{n})$, and
a control function $\upsilon$ belonging to the set $\mathscr{U}^{p}[0,T]$. For
each $z(\cdot)\in L^{p}(-h,S;R^{n})$ with $0<S\leq T$, such that $z(t)=0,$
$-h\leq t\leq 0.$ Define

$\displaystyle\mathscr{T}[z(\cdot)](t)=\zeta(t)+\int_{0}^{t}\frac{\kappa(t,s,z(s),z(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds,\quad
a.e.\quad t\in[0,S].$

Hence by Corollary 2.1, for each $q\geq 1,0\leq\varepsilon<\frac{\nu}{1-\nu}$,
with $\frac{1}{p}+1=\frac{1}{q}+\frac{1}{1+\varepsilon}$,

$\displaystyle\|\mathscr{T}[z(\cdot)]\|_{L^{p}(-h,S;R^{n})}\leq\|\zeta(\cdot)\|_{L^{p}(-h,S;R^{n})}+\bigg{[}\int_{0}^{S}\bigg{|}\int_{0}^{t}\frac{\kappa(t,s,z(s),z(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds\bigg{|}^{p}dt\bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\bigg{[}\int_{0}^{S}\Big{(}\int_{0}^{t}\frac{|\kappa(t,s,z(s),z(s-h),\upsilon(s))|}{(t-s)^{1-\nu}}ds\Big{)}^{p}dt\bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\bigg{[}\int_{0}^{S}\Big{(}\int_{0}^{t}\frac{L_{0}(s)+L(s)[|z(s)|+|z(s-h)|+d(\upsilon(s),\upsilon_{0})]}{(t-s)^{1-\nu}}ds\Big{)}^{p}dt\bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\bigg{(}\frac{S^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L_{0}(\cdot)+L(\cdot)[|z(\cdot)|+|z(\cdot-h)|+d(\upsilon(\cdot),\upsilon_{0})]\|_{L^{q}(0,S)}.$
(4.9)

We will examine three cases.

Case 1. $p>\frac{1}{1-\nu}$. In this part, $\frac{1}{\nu}>\frac{p}{p-1}$ and
$\frac{p}{1+\nu p}>1$. For every $\varepsilon\in(0,\frac{\nu}{1-\nu})$, that
is equivalent to $(1-\nu)(1+\varepsilon)<1$, we obtain (without any constraint
on $p>\frac{1}{1-\nu}$ for the current case)

$\displaystyle\frac{1}{q}=\frac{1}{p}+1-\frac{1}{1+\varepsilon}<\frac{1}{p}+1-\frac{1}{1+\frac{\nu}{1-\nu}}=\frac{1}{p}+\nu<1,\quad\frac{1}{q}-\frac{1}{p}=1-\frac{1}{1+\varepsilon}<\nu.$

Therefore,

$\displaystyle q\searrow\frac{p}{1+\nu
p}>1,\quad\frac{pq}{p-q}\searrow\frac{1}{\nu},\quad
as\quad\varepsilon\nearrow\frac{\nu}{1-\nu}.$

As $L_{0}(\cdot)$ belongs to $L^{\frac{p}{1+\nu p}+}(0,T)$ and $L(\cdot)$
belongs to $L^{\frac{1}{\nu}+}(0,T)$, we was able to find $\varepsilon$ close
enough to $\frac{\nu}{1-\nu}$ such that $L_{0}(\cdot)$ belongs to $L^{q}(0,T)$
and $L(\cdot)$ belongs to $L^{\frac{pq}{p-q}}(0,T)$.Hence

$\displaystyle\|L_{0}(\cdot)+L(\cdot)[|z(\cdot)|+|z(\cdot-h)|+d(\upsilon(\cdot),\upsilon_{0})]\|_{L^{q}(0,S)}\leq\|L_{0}(\cdot)\|_{L^{q}(0,T)}$
$\displaystyle+$
$\displaystyle\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\||z(\cdot)|+|z(\cdot-h)|+d(\upsilon(\cdot),\upsilon_{0})\|_{L^{p}(0,S)}.$
(4.10)

which yields

$\displaystyle\|\mathscr{T}[z(\cdot)]\|_{L^{p}(-h,S;R^{n})}\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\bigg{(}\frac{S^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\Big{\\{}\|L_{0}(\cdot)\|_{L^{q}(0,T)}$
$\displaystyle+$
$\displaystyle\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\||z(\cdot)|+|z(\cdot-h)|+d(\upsilon(\cdot),\upsilon_{0})\|_{L^{p}(0,S)}\Big{\\}}$
(4.11) $\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\bigg{(}\frac{S^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\Big{\\{}\|L_{0}(\cdot)\|_{L^{q}(0,T)}$
$\displaystyle+$
$\displaystyle\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}(2\|z(\cdot)\|_{p}+\|d(\upsilon(\cdot),\upsilon_{0})\|_{L^{p}(0,T)})\Big{\\}}$

Hence, we can conclude that the operator $\mathscr{T}$ maps functions in
$L^{p}(-h,S;R^{n})$ to functions in $L^{p}(-h,S;R^{n})$ for all $S$ belonging
to the interval $(0,T]$. Moving forward, suppose $\delta\in(0,T]$ be
undetermined, and consider two functions $z_{1}(\cdot)$ and $z_{2}(\cdot)$
belonging to $L^{p}(-h,\delta;R^{n})$. We will investigate the following
expression (using Corollary 2.1):

$\displaystyle\Big{\|}\mathscr{T}[z_{1}(\cdot)](t)-\mathscr{T}[z_{2}(\cdot)](t)\Big{\|}_{L^{p}(-h,S;R^{n})}\equiv\Bigg{[}\int_{0}^{\delta}\bigg{|}\mathscr{T}[z_{1}(\cdot)](t)-\mathscr{T}[z_{2}(\cdot)](t)\bigg{|}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle=$
$\displaystyle\Bigg{[}\int_{0}^{\delta}\bigg{|}\int_{0}^{t}\frac{\kappa(t,s,z_{1}(s),z_{1}(s-h),\upsilon(s))-\kappa(t,s,z_{2}(s),z_{2}(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds\bigg{|}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\Bigg{[}\int_{0}^{\delta}\bigg{(}\int_{0}^{t}\frac{|\kappa(t,s,z_{1}(s),z_{1}(s-h),\upsilon(s))-\kappa(t,s,z_{2}(s),z_{2}(s-h),\upsilon(s))|}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\bigg{(}\frac{\delta^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\Big{(}\|z_{1}(\cdot)-z_{2}(\cdot)\|_{L^{p}(-h,\delta;R^{n})}$
$\displaystyle+$
$\displaystyle\|z_{1}(\cdot-h)-z_{2}(\cdot-h)\|_{L^{p}(-h,\delta;R^{n})}\Big{)}$
$\displaystyle\leq$ $\displaystyle
2\bigg{(}\frac{\delta^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\|z_{1}(\cdot)-z_{2}(\cdot)\|_{L^{p}(-h,\delta;R^{n})}.$

Consider choosing a value of $\delta$ from the interval $(0,T]$ such that
$2\bigg{(}\frac{\delta^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}<1$.
This choice of $\delta$ ensures that the operator
$\mathscr{T}:L^{p}(-h,\delta;R^{n})\to L^{p}(-h,\delta;R^{n})$ is a
contraction, and therefore has a unique fixed point $\xi(\cdot)$ in
$L^{p}(-h,\delta;R^{n})$. This fixed point is the unique solution of the state
equation (4.1) on the interval $[0,\delta]$, with $\xi(t)=0$, $-h\leq t\leq
0$.

Next, we focus on the state equation (4.1) over the interval $[0,2\delta]$.
Let $z(\cdot)$ be any function belonging to $L^{p}(\delta,2\delta;R^{n})$, and
define the following expression:

$\displaystyle\mathscr{T}[z(\cdot)](t)=\zeta(t)+\int_{0}^{\delta}\frac{\kappa(t,s,\xi(s),\xi(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds+\int_{\delta}^{t}\frac{\kappa(t,s,z(s),z(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds,\quad
a.e\quad t\in[\delta,2\delta].$

Denote
$\bar{z}(\cdot)=\xi(\cdot)\chi_{[0,\delta)}(\cdot)+z(\cdot)\chi_{[\delta,2\delta]}(\cdot)\in
L^{p}(-h,2\delta;R^{n}),\quad\bar{z}(\cdot-h)=\xi(\cdot-h)\chi_{[0,\delta)}(\cdot)+z(\cdot-h)\chi_{[\delta,2\delta]}(\cdot)\in
L^{p}(-h,2\delta;R^{n})$. Hence similarly to (4.9) - (4.11)(with $S=2\delta$)

$\displaystyle\bigg{\|}\zeta(\cdot)+\int_{0}^{\delta}\frac{\kappa(t,s,\xi(s),\xi(s-h),\upsilon(s))}{(\cdot-s)^{1-\nu}}ds\bigg{\|}_{L^{p}(\delta,2\delta;R^{n})}$
$\displaystyle=$
$\displaystyle\Bigg{[}\int_{\delta}^{2\delta}\bigg{|}\zeta(t)+\int_{0}^{\delta}\frac{\kappa(t,s,\xi(s),\xi(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds\bigg{|}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\Big{(}\int_{\delta}^{2\delta}|\zeta(t)|^{p}dt\Big{)}^{1/p}+\Bigg{[}\int_{\delta}^{2\delta}\bigg{(}\int_{0}^{\delta}\frac{|\kappa(t,s,\xi(s),\xi(s-h),\upsilon(s))|}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\Bigg{[}\int_{\delta}^{2\delta}\bigg{(}\int_{0}^{\delta}\frac{L_{0}(s)+L(s)\big{[}|\xi(s)|+|\xi(s-h)|+d(\upsilon(s),\upsilon_{0})\big{]}}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+\Bigg{[}\int_{0}^{2\delta}\bigg{(}\int_{0}^{\delta}\frac{L_{0}(s)}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle+$
$\displaystyle\Bigg{[}\int_{0}^{2\delta}\bigg{(}\int_{0}^{\delta}\frac{L(s)\big{[}|\bar{z}(s)|+|\bar{z}(s-h)|+d(\upsilon(s),\upsilon_{0})\big{]}}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\|\zeta(\cdot)\|_{p}+K\bigg{[}1+2\|\xi(\cdot)\|_{L^{p}(-h,\delta;R^{n})}+2\|z(\cdot)\|_{L^{p}(\delta,2\delta;R^{n})}+\|d(\upsilon(\cdot),\upsilon_{0})\|_{p}\bigg{]}$

Following a similar argument to the proof of (4.11), we can conclude that the
operator $\mathscr{T}$ maps functions from $L^{p}(\delta,2\delta;R^{n})$ to
$L^{p}(\delta,2\delta;R^{n})$.Suppose we have two functions, $z_{1}(\cdot)$
and $z_{2}(\cdot)$, both belonging to $L^{p}(\delta,2\delta;R^{n})$. By using
Corollary 2.1, we obtain

$\displaystyle\Big{\|}\mathscr{T}[z_{1}(\cdot)](t)-\mathscr{T}[z_{2}(\cdot)](t)\Big{\|}_{L^{p}(\delta,2\delta;R^{n})}\equiv\bigg{[}\int_{\delta}^{2\delta}\bigg{|}\mathscr{T}[z_{1}(\cdot)](t)-\mathscr{T}[z_{2}(\cdot)](t)\bigg{|}^{p}dt\bigg{]}^{1/p}$
$\displaystyle=$
$\displaystyle\bigg{[}\int_{\delta}^{2\delta}\bigg{|}\int_{\delta}^{t}\frac{\kappa(t,s,z_{1}(s),z_{1}(s-h),\upsilon(s))-\kappa(t,s,z_{2}(s),z_{2}(s-h),\upsilon(s))}{(t-s)^{1-\nu}}ds\bigg{|}^{p}dt\bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\bigg{[}\int_{\delta}^{2\delta}\bigg{(}\int_{\delta}^{t}\frac{L(s)[|z_{1}(s)-z_{2}(s)|+|z_{1}(s-h)-z_{2}(s-h)|]}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\bigg{]}^{1/p}$
$\displaystyle\leq$
$\displaystyle\bigg{(}\frac{\delta^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\Big{[}\|z_{1}(\cdot)-z_{2}(\cdot)\|_{L^{p}(\delta,2\delta;R^{n})}$
$\displaystyle+$
$\displaystyle\|z_{1}(\cdot-h)-z_{2}(\cdot-h)\|_{L^{p}(\delta,2\delta;R^{n})}\Big{]}$
$\displaystyle\leq$ $\displaystyle
2\bigg{(}\frac{\delta^{1-(1+\varepsilon)(1-\nu)}}{1-(1+\varepsilon)(1-\nu)}\bigg{)}^{\frac{1}{1+\varepsilon}}\|L(\cdot)\|_{L^{\frac{pq}{p-q}}(0,T)}\|z_{1}(\cdot)-z_{2}(\cdot)\|_{L^{p}(\delta,2\delta;R^{n})}$

Hence we get the existence and uniqueness of the solution to the state
equation on $[\delta,2\delta]$. Using the induction, we can obtain the
solution
$\xi(\cdot)\in[-h,\delta],[\delta,2\delta],\cdots,[(\frac{T}{\delta})\delta,T]$.

Now, suppose
$\zeta_{1}(\cdot),\upsilon_{1}(\cdot)),\quad(\zeta_{2}(\cdot),\upsilon_{2}(\cdot))\in
L^{p}(-h,T;R^{n})\times\mathscr{U}^{p}[0,T]$ and
$\xi_{1}(\cdot),\xi_{2}(\cdot)$ become the corresponding solutions. Hence

$\displaystyle|\xi_{1}(t)-\xi_{2}(t)|\leq$
$\displaystyle|\zeta_{1}(t)-\zeta_{2}(t)|+\int_{0}^{t}\frac{|\kappa(t,s,\xi_{1}(s),\xi_{1}(s-h),\upsilon_{1}(s))-\kappa(t,s,\xi_{2}(s),\xi_{2}(s-h),\upsilon_{2}(s))|}{(t-s)^{1-\nu}}ds$
$\displaystyle+$
$\displaystyle\int_{0}^{t}\frac{|\kappa(t,s,\xi_{1}(s),\xi_{1}(s-h),\upsilon_{2}(s))-\kappa(t,s,\xi_{2}(s),\xi_{2}(s-h,\upsilon_{2}(s))|}{(t-s)^{1-\nu}}ds$
$\displaystyle\leq$
$\displaystyle|\zeta_{1}(t)-\zeta_{2}(t)|+\int_{0}^{t}\frac{|\kappa(t,s,\xi_{1}(s),\xi_{1}(s-h),\upsilon_{1}(s))-\kappa(t,s,\xi_{2}(s),\xi_{2}(s-h),\upsilon_{2}(s))|}{(t-s)^{1-\nu}}ds$
$\displaystyle+$
$\displaystyle\int_{0}^{t}\frac{L(s)\Big{[}|\xi_{1}(s)-\xi_{2}(s)|+|\xi_{1}(s-h)-\xi_{2}(s-h)|\Big{]}}{(t-s)^{1-\nu}}ds$
$\displaystyle\equiv$
$\displaystyle\vartheta(t)+\int_{0}^{t}\frac{L(s)|\xi_{1}(s)-\xi_{2}(s)|}{(t-s)^{1-\nu}}ds+\int_{0}^{t}\frac{L(s)|\xi_{1}(s-h)-\xi_{2}(s-h)|}{(t-s)^{1-\nu}}ds$

Then, by using Theorem 3.1, we will get

$\displaystyle|\xi_{1}(t)-\xi_{2}(t)|\leq$
$\displaystyle\vartheta(t)+K\sum_{k=0}^{n}\int_{0}^{t-kh}\frac{L(s)\vartheta(s)}{(t-kh-s)^{1-\nu}}ds+K\sum_{k=0}^{n-1}\int_{h}^{t-kh}\frac{L(s)\vartheta(s-h)}{(t-kh-s)^{1-\nu}}ds,\quad
a.e.\quad t\in[0,T],$ (4.12)

for some constant $K>0$. As a result,

$\displaystyle\|\xi_{1}(\cdot)-\xi_{2}(\cdot)\|_{p}\leq$ $\displaystyle
K\Bigg{[}\int_{0}^{T}\Big{(}\vartheta(t)+K\sum_{k=0}^{n}\int_{0}^{t-kh}\frac{L(s)\vartheta(s)}{(t-kh-s)^{1-\nu}}ds+K\sum_{k=0}^{n-1}\int_{h}^{t-kh}\frac{L(s)\vartheta(s-h)}{(t-kh-s)^{1-\nu}}ds\Big{)}^{p}dt\Bigg{]}^{1/p}$
$\displaystyle\leq$ $\displaystyle
K\Big{(}\int_{0}^{T}\big{(}\vartheta(t)\big{)}^{p}dt\Big{)}^{1/p}\leq
K\Bigg{\\{}\|\zeta_{1}(t)-\zeta_{2}(t)\|_{p}$ $\displaystyle+$
$\displaystyle\bigg{[}\int_{0}^{T}\bigg{(}\int_{0}^{t}\frac{|\kappa(s,\xi_{1}(s),\xi_{1}(s-h),\upsilon_{1}(s))-\kappa(s,\xi_{2}(s),\xi_{2}(s-h,\upsilon_{2}(s))|}{(t-s)^{1-\nu}}ds\bigg{)}^{p}dt\bigg{]}^{1/p}\Bigg{\\}}.$

proving the stability estimate. We are able to the similar argument to prove
(4.7).

Case 2. $1<p\leq\frac{1}{1-\nu}$. In this case,

$\displaystyle\frac{1}{\nu}\leq\frac{p}{p-1},\quad\frac{p}{1+\nu p}\leq 1.$

Also, since $1-\nu\leq\frac{1}{p}<1$ for each $\varepsilon\in(0,p-1)$, the
following satisfies:

$\displaystyle 1-\nu\leq\frac{1}{p}<\frac{1}{1+\varepsilon}.$

This imply $(1-\nu)(1+\varepsilon)<1$. Hence

$\displaystyle\frac{1}{p}<\frac{1}{q}=\frac{1}{p}+1-\frac{1}{1+\varepsilon}\nearrow
1,\quad\frac{p-q}{pq}=\frac{1}{p}-\frac{1}{q}=1-\frac{1}{1+\varepsilon}\nearrow\frac{p-1}{p},\quad
as\quad\varepsilon\nearrow p-1.$

Therefore, we can choose a value for $\varepsilon$ close to $p-1$, given that
$L_{0}(\cdot)\in L^{1+}(0,T)$ and $L(\cdot)\in L^{\frac{p}{p-1}+}(0,T)$. This
will ensure that $L_{0}(\cdot)\in L^{q}(0,T)$ and $L(\cdot)\in
L^{\frac{pq}{p-q}}(0,T)$. The remaining steps of the proof will be identical
to Case 1.

Case 3. $p=1$. In this part, the condition reads $L_{0}(\cdot)\in L^{1+}(0,T)$
and $L(\cdot)\in L^{\infty}(0,T)$. Hence we take $\varepsilon=0$, and (4.11)
reads

$\displaystyle\|\mathscr{T}[z(\cdot)]\|_{L^{1}(-h,S;R^{n})}\leq\|\zeta(\cdot)\|_{1}+\frac{S^{\nu}}{\nu}\\{\|L_{0}(\cdot)\|_{L^{1}(0,T)}+\|L(\cdot)\|_{L^{\infty}(0,T)}(2\|z(\cdot)\|+\|d(\upsilon(\cdot),\upsilon_{0})\|_{L^{1}(0,S)})\\}.$
(4.13)

The remainder of the proof is comparable to that of Case 1.

∎

######  Example 4.1.

Let

$\displaystyle\kappa(t,s,\xi(s),\xi(s-h),\upsilon)=\frac{\sqrt{|s-1|^{2\delta-2}|t+1|^{2-2\gamma}+|\xi(s)|+|\xi(s-h)|}}{|s-1|^{1-\nu}|t+1|^{1-\gamma}},$
$\displaystyle\forall(t,s,\xi,\xi_{s-h},\upsilon)\in\Delta\times R^{2}\times
U,\quad\gamma<1,\quad s\not=1.$

Consider the following Volterra integral equation

$\xi(t)=\frac{1}{|t-1|^{1-\sigma}}+\int_{0}^{t}\frac{\sqrt{|s-1|^{2\delta-2}|t+1|^{2-2\gamma}+|\xi(s)|+|\xi(s-h)|}}{|s-1|^{1-\nu}|t+1|^{1-\gamma}(t-s)^{1-\beta}}ds,\quad
a.e,\quad t\in[0,T]$ (4.14)

for some $\nu,\beta\in(0,1)$, $\sigma,\delta\in(0,1]$ , and $\gamma>1$. We can
take

$\displaystyle\zeta(t)=\frac{1}{|t-1|^{1-\sigma}},\quad
L_{0}(s)=\frac{1}{|s-1|^{2-\nu-\delta}},\quad
L(s)=\frac{1}{|s-1|^{1-\nu}},\quad t\not=1,\quad s\not=1.$

We see that

$\displaystyle\zeta(\cdot)\in L^{p}(0,1)\leftarrow p(1-\sigma)<1\iff
p<\frac{1}{1-\sigma}\quad(\frac{1}{0}=\infty),$
$\displaystyle\begin{cases}(2-\nu-\delta)\frac{p}{1+\beta p}<1\iff
2-\nu-\beta-\delta<\frac{1}{p}\\\ \\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\iff
p<\frac{1}{(2-\nu-\beta-\delta)_{+}},\\\ \\\
2-\nu-\delta<1\quad\quad\quad\iff\nu+\beta>1.\end{cases}$
$\displaystyle\implies L(\cdot)\in L^{(\frac{1}{\beta}\vee 1)+}(0,T),$
$\displaystyle\begin{cases}\frac{1-\nu}{\beta}<1\iff\nu+\beta>1,\\\ \\\
(1-\nu)\frac{p}{p-1}<1\iff 1-\nu<1-\frac{1}{p}\iff p>\frac{1}{\nu}\end{cases}$
$\displaystyle L(\cdot)\in L^{(\frac{1}{\beta}\vee\frac{p}{p-1})+}(0,T).$

Then, (4.14) has a unique solution $\xi(\cdot)\in L^{p}(0,T)$ for any

$\displaystyle
p\in\Big{(}\frac{1}{\nu},\frac{1}{(1-\sigma)\vee(2-\nu-\beta-\delta)+}\Big{)},$

supplied

$\displaystyle\nu+\beta>1,\quad\nu+\delta>1.$ (4.15)

Let us highlight that the solution $\xi(\cdot)$ of equation (4.10) does not
have to be continuous in general, even if the free term $\zeta(\cdot)$ is
continuous. Specifically, when we set $\sigma$ to 1 and consider $\zeta(t)$ as
a constant function equal to 1, which is continuous, we observe that the
solution $\xi(\cdot)$ is positive. This positivity can be determined by
applying the Picard iteration method. Thus,

$\displaystyle\lim_{t\to 1}\xi(t)\geq 1+\lim_{t\to
1}\int_{0}^{t}\frac{ds}{|s-1|^{2-\nu-\delta}(t-s)^{1-\beta}}=\int_{0}^{1}\frac{ds}{(1-s)^{3-\nu-\beta-\delta}}=\infty,$
(4.16)

supplied

$\displaystyle 3-\nu-\beta-\delta>1\iff\nu+\beta+\delta<2.$ (4.17)

This will be the case if we take

$\displaystyle\nu=\frac{2}{3},\quad\beta=\delta=\frac{1}{2}.$

In this scenario, there is a solution $\xi(\cdot)\in L^{p}(0,T)$ where
$p\in(\frac{3}{2},3)$. However, it should be noted that the solution is not
continuous at $t=1$.

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