\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 262, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/262\hfil  Infinite system of differential equations]
{Existence of solutions to a boundary-value problem for an infinite system of\\
  differential equations}

\author[J. Bana\'s, M. Mursaleen, S. M. H. Rizvi \hfil EJDE-2017/262\hfilneg]
{J\'ozef Bana\'s, Mohammad Mursaleen, Syed M. H. Rizvi}

\address{J\'ozef Bana\'s \newline
Department of Nonlinear Analysis,
Rzesz\'ow University of Technology,
al. Powsta\'nc\'ow Warszawy 8, 35-959 Rzesz\'ow, Poland}
\email{jbanas@prz.edu.pl}

\address{Mohammad Mursaleen \newline
Department of Mathematics,
Aligarh Muslim University,
Aligarh 202002, India}
\email{mursaleenm@gmail.com}

\address{Syed M. H. Rizvi \newline
Department of Mathematics,
Aligarh Muslim University,
Aligarh 202002, India}
\email{syedrizvi022@gmail.com}

\dedicatory{Communicated by Vicentiu D. Radulescu}

\thanks{Submitted August 3, 2017. Published October 17, 2017.}
\subjclass[2010]{34G20, 47H08}
\keywords{Measure of noncompactness; equicontinuous family;
\hfill\break\indent 
boundary value problem; infinite system of differential equations;
Fredholm integral equation}

\begin{abstract}
 Using techniques associated with measures of noncompactness we prove
 an existence of solutions for a boundary-value problem for an infinite
 system of ordinary differential equations of second order.
 Our approach depends on transforming of the original boundary-value problem
 into an infinite system of integral equations of Fredholm type.
 The settings for this article are in the classical Banach sequence
 space $l_p$ with $p\geq 1$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks
 

\section{Introduction and preliminaries concerning measures of noncompactness}

To formulate a fixed point theorem Darbo used the so-called Kuratowski measure
of noncompactness $\alpha$, introduced in 1930 by  Kuratowski \cite{Kurat}. 
Subsequently, in the literature have appeared a lot of functions being 
 measures of noncompactness  (cf.  \cite{Akhmerov,Ayerbe,Goebel,Mursaleen1}).
Nevertheless,  it  turned  out  that   among   all   classical
 realizations of measures of noncompactness the  Hausdorff measure seems 
to be the most convenient and useful in applications.

The Hausdorff measure of noncompactness was introduced in 1965 \cite{Golden1} 
(see also \cite{Golden2}) by the  formula
$$
\chi (X)=\inf \big\{ \varepsilon >0: X \text{ has a finite $\varepsilon$-net in }
 E \big\},
$$
where $X$ is a nonempty and bounded subset of the Banach space $E$.

The function $\chi$ has a lot of useful properties being essential 
in applications. For example, we have that $\chi (X)=0$ if and only if $X$ 
is a relatively compact subset of $E$. To recall other properties of  
 $\chi$ let us introduce first some auxiliary notation.
Namely, by the symbol $B(x,r)$ we denote the closed ball centered at $x$ 
and with radius $r$. The symbol $B_r$ will denote the ball $B(\theta , r)$, 
where $\theta$ stands for the zero element of $E$. If $X$ is a subset of $E$ 
then $\overline{X}$ and $\operatorname{conv} X$ denote the closure and the closed
convex hull of $X$, respectively. Moreover, we use the standard notation
 $X+Y$, $\lambda X$ to denote the algebraic operations on subsets of $E$.

Now, let us recall a few properties of the Hausdorff measure of noncompactness 
$\chi$ defined above. Namely, if $X$, $Y$ are arbitrary nonempty bounded subsets 
of $E$ and $\lambda \in \mathbb{R}$, then:
\begin{itemize}
  \item[(1)] $ X\subset Y$ implies $\chi (X) \leq \chi (Y)$. 
  \item[(2)] $\chi ( \overline{X} ) = \chi (\operatorname{conv} X) =\chi (X)$.
  \item[(3)] $\chi (X+Y) \leq \chi (X) + \chi (Y)$.
  \item[(4)] $\chi (\lambda X) = | \lambda |  \chi (X)$.
\end{itemize}
For other properties of the Hausdorff measure $\chi$ we refer to 
\cite{Goebel,Mursaleen1}, for example.
In what follows we recall the fixed point theorem of Darbo type
 (cf. \cite{Akhmerov,Goebel}) which will be utilized in our considerations.

\begin{theorem}\label{th11}
Let $\Omega $ be a nonempty, bounded, closed and convex subset of the space 
$E$ and let $ F:\Omega \rightarrow \Omega $ be a continuous operator such
 that $\chi (FX)\leq k\chi ( X)$ for any nonempty subset $X$
 of $\Omega $, where $k\in [0,1)$ is a constant. Then $F$
has at least one fixed point in the set $\Omega $.
\end{theorem}

It is worthwhile mentioning that  to apply efficiently Theorem \ref{th11} 
in a concrete Banach space $E$ we have to know a formula expressing the 
Hausdorff measure of noncompactness $\chi$ in $E$ in a convenient way, 
connected with the structure of the underlying Banach space $E$. 
It turns out that such formulas are known only in a few spaces 
\cite{Goebel,Mursaleen1}. For our purposes we recall such a formula for 
the sequence space $l_p$.

To this end let us fix a number $p$, $p \geq 1$, and denote by $l_p$ the 
classical Banach sequence space with the norm
$$
\| x \| _{l_p} = \| ( x_n ) \| _{l_p} = \Big( \sum_{n=1}^{\infty} | x_n | ^p \Big)
 ^{1/p}
$$
for $x=( x_n ) \in l_p$. In some considerations we will drop the index $l_p$ 
if it does not lead to misunderstanding.

If $X$ is a nonempty and bounded subset of $l_p$ then
\begin{equation}\label{eq11}
\chi (X)= \lim _{n \to \infty} \Big\{ \sup _{( x_k ) \in X} 
\Big( \sum_{k=n}^{\infty} | x_k | ^p \Big) ^{1/p} \Big\}
\end{equation}
(cf. \cite{Goebel}). The above formula will be utilized in the sequel of the paper.

Finally, let us mention that results of the present paper generalize those
 obtained in \cite{Mursaleen3}, where the considerations were conducted in 
the classical Banach  sequence spaces $c_0$ and $l_1$.

\section{Main result}

Infinite systems of ordinary differential equations are closely related to 
several important problems appearing naturally in applications 
(cf. \cite{Mursaleen1,Deimling1,Deimling2}). For example, let us mention 
that some methods of solving of partial differential equations based on 
the application of the so-called semidiscretization and numerical analysis 
lead to infinite systems of differential equations. Apart from this one 
can encounter other significant problems in engineering, mechanics, 
in the theory of branching processes and so on, which are associated with the 
theory of infinite systems of differential equations 
(see \cite{Lecko,Mursaleen1,Deimling1,Mursaleen2}).
On the other hand infinite systems of differential equations can be treated 
as ordinary differential equations in some Banach sequence spaces 
\cite{Lecko,Mursaleen1,Deimling1}.
The above mentioned facts justify the interest in the theory of infinite systems
 of differential equations.

In this paper we  study the infinite systems of differential equations of 
 second order having the form
\begin{equation}\label{eq21}
u''_i (t)= - f_i ( t, u_1, u_2, \ldots ),
\end{equation}
where $t \in I=[0,T]$ and $i=1,2, \ldots$. The above system will be studied 
together with the boundary problem
\begin{equation}\label{eq22}
u_i (0)=u_i (T) = 0,
\end{equation}
for each $i=1,2, \ldots$.
In our study of problem \eqref{eq21}-\eqref{eq22} we will apply a 
technique associated with the Hausdorff measure of noncompactness $\chi$,
 and the fixed point theorem of Darbo type presented in Theorem \ref{th11}.

An essential tool applied in our approach to the study of problem 
\eqref{eq21}-\eqref{eq22} depends on converting  \eqref{eq21}-\eqref{eq22}
 into the infinite system of integral equations of Fredholm type of the form
\begin{equation}\label{eq23}
u_i (t) = \int_0^T G(t,s) f_i ( s, u(s) )ds,
\end{equation}
where $G(t,s)$ is the Green function corresponding to problem
 \eqref{eq21}-\eqref{eq22} on the interval $I=[0,T]$ and defined on 
the square $I^2$ in the following way (cf. \cite{Duffy})
\begin{equation}\label{eq24}
G(t,s) =\begin{cases}
\frac tT (T-s) & \text{for }0 \leq t \leq s \leq T,  \\
\frac sT (T-t) & \text{for }0 \leq s \leq t \leq T\,.
\end{cases}
\end{equation}
Using standard methods it is easy to show the estimate
\begin{equation}\label{eq25}
G(t,s) \leq \frac T4
\end{equation}
for all $(t,s) \in I^2$.

In what follows we  write $f_i (t,u)$ instead of 
$f_i ( t, u_1 , u_2 , \ldots )$ ($i=1,2, \ldots$).
To introduce further auxiliary facts let us assume that $E$ is a given
 Banach space with the norm $\| \cdot \| _E$. Denote by $C(I,E)$ the 
Banach space consisting of all functions $u=u(t)$ acting continuously 
from the interval $I$ into the space $E$ and endowed by the classical supremum norm
$$
\| u \| _C =\sup \{ \| u(t) \| _E : t \in I \}.
$$

\begin{remark}\label{rm21} \rm
Assume that $\chi _E$ is the Hausdorff measure of noncompactness in the Banach 
space $E$. Next, let us take an arbitrary bounded subset $X$ of the space 
$C(I,E)$ which is equicontinuous on the interval $I$. Then, it can be shown 
\cite{Goebel} that we have the following formula expressing the Hausdorff 
measure of noncompactness of the set $X$:
\begin{equation}\label{eq26}
\chi (X) = \sup \{ \chi _E ( X(t) ): t \in I \}.
\end{equation}
\end{remark}

As a special case of the above discussed space $C(I,E)$ we will consider 
the space $C(I, \mathbb{R})$, which will be denoted by $C(I)$. 
The supremum norm in the space $C(I)$ will be denoted by $\| \cdot \| _{\infty}$.
Moreover, the symbol $C^2(I)$ denotes the space of real functions defined 
and twice conti\-nuously differentiable on $I$ with the standard norm 
$\| \cdot \| _{C^2}$ defined as follows
$$
\| u \| _{C^2}=\sum_{i=0}^2 \| u^{(i)} \| _{\infty}.
$$

Now, let us observe that problem \eqref{eq21}-\eqref{eq22} has a solution 
$u=( u_i )$ belonging to the space $C^2 (I)$ if and only if the infinite 
system of integral equations \eqref{eq23} has a solution $u \in C(I)$.
To prove this fact, for arbitrarily fixed $i$ and $t \in I$, let us write 
(cf. \eqref{eq24}):
$$
u_i (t) = \int_0^t \frac sT (T-t) f_i ( s, u(s) )ds 
+ \int_t^T \frac tT (T-s) f_i ( s, u(s) )ds.
$$
Hence, differentiating the above equality we subsequently obtain
\begin{gather*}
u_i' (t) = - \frac 1T \int_0^t s f_i ( s, u(s) )ds 
 + \frac 1T \int_t^T (T-s) f_i ( s, u(s) )ds,\\
u_i'' (t)= - f_i ( t, u(t)),
\end{gather*}
which proves our assertion.

We will investigate the infinite system of integral equations \eqref{eq23} 
imposing the following assumptions.
\begin{itemize}
\item[(i)] The function $f_i$ is defined on the set $I \times \mathbb{R}^{\infty}$
 and takes real values for $i=1,2, \ldots$.

\item[(ii)] The operator $f$ defined on the space $I \times l_p$ by the formula
$$
(fu)(t)= ( f_i (t,u) )= ( f_1(t,u), f_2 (t,u), \ldots )
$$
transforms the space $I \times l_p$ into $l_p$ and is such that the family
 of functions $\{ (fu)(t) \}_{t \in I}$ is equicontinuous
at each point of the space $l_p$ i.e., for each arbitrarily fixed 
$u \in l_p$ and for a given $\varepsilon >0$ there exists 
$\delta >0$ such that
$$
\| (fv)(t) - (fu)(t) \| _{l_p} < \varepsilon
$$
for each $t \in I$ and for any $v \in l_p$ such that $\| v-u \|_{l_p} <\delta$.

\item[(iii)] For any natural $i$ there exist functions $g_i$, 
$h_i : I \to \mathbb{R}_{+}$ such that the inequality
$$
| f_i (t,u) | ^p \leq g_i (t) + h_i (t) | u_i |^p
$$
is satisfied for $t \in I$, $u=( u_i ) \in l_p$ and $i=1,2, \ldots$.\\
Moreover, we assume that the function $g_i$ is continuous on $I$ 
($i=1,2, \ldots $) and the function series $\sum _{i=1} ^{\infty} g_i (t)$
 is uniformly convergent, while the function sequence $( h_i (t) )$ 
is equibounded on $I$.
\end{itemize}

Observe that in view of assumption (iii) the constant 
$$
H:=\sup \{ h_i (t): t \in I, \, i=1,2, \ldots\}
$$
is finite. Moreover, we can define the function $g=g(t)$ on the interval 
$I$ by putting
$$
g(t)= \sum _{i=1} ^{\infty} g_i (t) .
$$
Obviously the function $g(t)$ is continuous on $I$. Therefore, we can 
define the finite constant
$$
G= \max \{ g(t): t \in I \}.
$$

%\begin{remark}\label{rm22} \rm
The concept of the equicontinuity utilized in assumption (iii) was 
introduced in the book \cite{Sikorski}.
Now, we can formulate our main result.

\begin{theorem}\label{th23}
Under assumptions {\rm (i)-(iii)}, if additionally $( H/T )^{1/p} \, T^2 <4$ and 
$T \leq 1$, the infinite system of integral equations  \eqref{eq23} has at least 
one solution $u(t)=( u_i (t) )$ in the space $l_p$ i.e., 
$( u_i (t) ) \in l_p$ for each $t \in I$.
\end{theorem}

\begin{proof}
 At the beginning let us consider the space $C ( I, l_p )$ of all functions 
continuous on the interval $I=[0,T]$ with values in the space 
$l_p$ and furnished with the classical supremum norm
$$
\| u \| =\sup \{ \| u(t) \| _{l_p} :t \in I \}.
$$
This space is a particular case of the space $C(I,E)$ introduced previously.
Next, let us denote by $F$ the operator defined on the space $C ( I, l_p )$ 
by the formula
\begin{align*}
(Fu)(t)
&= ( (Fu)_i (t) )= \Big( \int_0^T G(t,s) f_i ( s, u(s) )ds \Big)\\
&= \Big(\int_0^T G(t,s) f_1 ( s, u(s) )ds, \, 
\int_0^T G(t,s) f_2 ( s, u(s) )ds, \ldots \Big).
\end{align*}
At first, let us notice that the operator $F$ maps the space 
$C ( I, l_p )$ into itself. Indeed, for a fixed 
$u=u(t)=( u_i (t) ) \in C ( I, l_p )$ and for an arbitrary $t \in I$, 
using the imposed assumptions and the H\"older inequality, we obtain
\begin{align*}
\| (Fu)(t)\| ^p _{l_p} 
&= \sum _{i=1} ^{\infty} \Big| \int_0^T G(t,s) f_i ( s, u(s) )ds \Big| ^p \\
& \leq \sum _{i=1} ^{\infty} \Big( \int_0^T | G(t,s) | ^p | f_i ( s, u(s) )| ^p ds \Big)
  \Big( \int_0^T ds \Big)^{p/q} \\
& \leq T^{p/q} \sum _{i=1} ^{\infty} \int_0^T | G(t,s) | ^p 
 [ g_i (s) + h_i (s) | u_i (s) | ^p ] ds \\
& \leq T^{p/q} \sum _{i=1} ^{\infty} \Big[ \int_0^T | G(t,s) | ^p g_i (s) ds 
+  \int_0^T | G(t,s) | ^p h_i (s) | u_i (s) | ^p  ds \Big],
\end{align*}
where $q>1$ is a number such that $1/p+1/q=1$.

Further, applying the Lebesgue dominated convergence theorem, from the above 
estimate we obtain
\begin{align*}
\| (Fu)(t)\| ^p _{l_p} 
&\leq T^{p/q} \int_0^T | G(t,s) | ^p g(s) ds \\
&\quad + T^{p/q} H \int_0^T | G(t,s) | ^p \{ | u_1 (s) | ^p + | u_2 (s) | ^p
+ \ldots \} ds.
\end{align*}
Hence, in view of \eqref{eq25}, we derive the  estimate
\begin{equation} \label{eq27}
\begin{aligned}
\| (Fu)(t)\| ^p _{l_p} 
& \leq T^{p/q} \int_0^T ( T/4 )^p g(s) ds + T^{p/q} H \int_0^T ( T/4 )^p
\Big( \sum _{i=1} ^{\infty} | u_i (s) | ^p \Big) ds \\
& \leq ( T^{2p-1} / 4^p ) \int_0^T g(s)ds + ( H T^{2p-1} / 4^p )
\int_0^T \Big(\sum _{i=1} ^{\infty} | u_i (s) | ^p \Big) ds \\
& \leq ( T^{2p} / 4^p )G + H ( T^{2p-1} / 4^p ) \int_0^T \| u \| ^p ds   \\
& = ( T^{2p} / 4^p ) [ G + H \| u \| ^p ].
\end{aligned}
\end{equation}
Hence we deduce that $Fu$ is bounded on the interval $I$. This implies that
the operator $F$ transforms the space $C ( I, l_p )$ into itself.
Apart from this, from estimate \eqref{eq27} we derive the following inequality
\begin{equation}\label{eq28}
\| Fu \| \leq \frac {T^2}{4} ( G + H \| u \| ^p  ) ^{1/p} .
\end{equation}
Further, taking into account assumption (iii) we deduce that the positive number
$$
r_0 = \frac {T^2 \, G^{1/p}}{( 4^p - H T^{2p} ) ^{1/p}}
$$
is the optimal solution of the inequality
$$
\frac {T^2}{4} ( G + H r^p  ) ^{1/p} \leq r.
$$
Thus, in view of \eqref{eq28} we infer that the operator $F$ transforms
the ball $B_{r_0}$ in the space $C ( I, l_p )$ into itself.

Next we show that the operator $F$ is continuous on the ball $B_{r_0}$.
To this end let us fix arbitrarily a number $\varepsilon >0$ and a function 
$u \in B_{r_0}$. Then, for an arbitrary $v \in B_{r_0}$ such that 
$\| v-u \| \leq \varepsilon$ and for a fixed number $t \in I=[0,T]$, 
in view of the imposed assumptions we obtain
\begin{align*}
&\| (Fv)(t) - (Fu)(t) \| ^p _{l_p} \\
& = \sum _{i=1} ^{\infty} \Big| \int_0^T G(t,s) [ f_i ( s, v(s) ) - f_i ( s, u(s) )
 ] ds \Big| ^p \\
& \leq \sum _{i=1} ^{\infty} \Big( \int_0^T | G(t,s) | ^p | f_i ( s, v(s) ) 
 - f_i ( s, u(s) ) | ^p ds \Big) \Big( \int_0^T ds \Big)^{p/q} \\
& \leq \sum _{i=1} ^{\infty} T^{p/q} \int_0^T | G(t,s) | ^p 
| f_i ( s, v(s) ) - f_i ( s, u(s) ) | ^p ds.
\end{align*}
Hence, using estimate \eqref{eq25} and applying assumption (ii) concerning 
the equicontinuity of the family of functions $\{ (fu)(t)\}_{t \in I}$,
 we obtain
\begin{equation} \label{eq29}
\begin{aligned}
&\| (Fv)(t) - (Fu)(t) \| ^p _{l_p} \\ 
& \leq T^{p/q} ( T/4 )^p
 \sum _{i=1} ^{\infty} \int_0^T | f_i ( s, v(s) ) - f_i ( s, u(s) ) | ^p ds  \\
& = ( T^{2p-1} / 4^p ) \lim_{k \to \infty} \sum _{i=1} ^{k}
  \int_0^T | f_i ( s, v(s) ) - f_i ( s, u(s) ) | ^p ds   \\
& = ( T^{2p-1} / 4^p ) \lim_{k \to \infty} \int_0^T
\Big( \sum _{i=1} ^{k} | f_i ( s, v(s) ) - f_i ( s, u(s) ) | ^p \Big) ds.
\end{aligned}
\end{equation}
Further on, let us observe that keeping in mind assumption (ii) on the
equicontinuity of the family $\{ (fu)(t) \}_{t \in I}$ at every point
$u \in l_p$, we conclude that $\delta (\varepsilon ) \to 0$ as $\varepsilon \to 0$,
where $\delta = \delta (\varepsilon)$ is the function defined by the equality
$$
\delta (\varepsilon) = \sup \big\{ | f_i (t,v) - f_i (t, u) | :
 u,v \in l_p , \, \| v-u \| _{l_p} \leq \varepsilon , \,
 t \in I, \, i=1,2, \ldots \big\}.
$$
Combining the above fact with \eqref{eq29} and applying the Lebesgue dominated
convergence theorem we arrive at the  estimate
$$
\| (Fv)(t) - (Fu)(t) \| ^p _{l_p}
\leq \frac {T^{2p-1}}{4^p} \int_0^T ( \delta (\varepsilon) )^p ds
= (T/2)^{2p} ( \delta (\varepsilon) )^p .
$$
The above  estimate allows us to infer that the operator $F$ is continuous on
 the ball $B_{r_0}$.

Further, let us notice that the function $G(t,s)$ defined by \eqref{eq24} 
is uniformly continuous on the square $I^2$. Combining this fact with the 
definition of the operator $F$, it is easily seen that the set $FB_{r_0}$ 
is equicontinuous on the interval $I$.
Now, let us consider the set $B_{r_0}^1 = \operatorname{conv} FB_{r_0}$. 
Obviously $B_{r_0}^1 \subset B_{r_0}$ and the functions from the set $B_{r_0}^1$ 
are equicontinuous on $I$.

Next, let us take a nonempty subset $X$ of the set $B_{r_0}^1$. 
From the above  facts we conclude that $X$ is equicontinuous on the interval $I$. 
Choose a function $u \in X$ and fix an arbitrary natural number $n$. 
Then, for arbitrarily fixed $t \in I$, in view of assumption (iii) we obtain
\begin{align*}
\sum _{i=n} ^{\infty} | (Fu)_i (t) | ^p 
&= \sum _{i=n} ^{\infty} | \int_0^T G(t,s) f_i ( s, u(s) )ds | ^p \\
&\leq \sum _{i=n} ^{\infty} ( \int_0^T | G(t,s) | | f_i ( s, u(s) ) | ds ) ^p .
\end{align*}
Hence, using the H\"older inequality, we obtain
\begin{align*}
\sum _{i=n} ^{\infty} | (Fu)_i (t) | ^p  
& \leq \sum _{i=n} ^{\infty} \Big[ \Big( \int_0^T | G(t,s) | ^p 
 | f_i ( s, u(s) ) |^p ds \Big)^{1/p} \Big( \int_0^T ds \Big)^{1/q} \Big] ^p \\
& \leq T^{p/q} \sum _{i=n} ^{\infty} \int_0^T | G(t,s) | ^p | f_i ( s, u(s) ) |^p ds \\
& \leq T^{p/q} (T/4)^p \sum _{i=n} ^{\infty} \int_0^T | f_i ( s, u(s) ) |^p ds.
\end{align*}
Now, applying the earlier conducted reasoning depending on the use of the 
Lebesgue dominated convergence theorem, from the above estimate we derive
 the inequality
\begin{align*}
&\sum _{i=n} ^{\infty} | (Fu)_i (t) | ^p  \\
& \leq ( T^{2p-1} / 4^p ) \int_0^T 
 \Big( \sum _{i=n} ^{\infty} | f_i ( s, u(s) ) |^p \Big)ds \\
& \leq ( T^{2p-1} / 4^p ) \int_0^T
 \Big( \sum _{i=n} ^{\infty} [ g_i (s) + h_i(s) | u_i (s) |^p  ] \Big)ds \\
& \leq ( T^{2p-1} / 4^p ) \Big\{ \int_0^T \Big( \sum _{i=n} ^{\infty} g_i (s) \Big)ds 
 + \int_0^T ( \sum _{i=n} ^{\infty} h_i(s) | u_i (s) |^p )ds \Big\} \\
& \leq ( T^{2p-1} / 4^p ) \Big\{ \int_0^T \Big( \sum _{i=n} ^{\infty} g_i (s) \Big)ds
 +H \int_0^T \Big( \sum _{i=n} ^{\infty} | u_i (s) |^p \Big)ds  \Big\} .
\end{align*}
Hence, we obtain
\begin{equation} \label{eq210}
\begin{aligned}
\sup _{u \in X} \sum _{i=n} ^{\infty} | (Fu)_i (t) | ^p
& \leq ( T^{2p-1} / 4^p ) \int_0^T \Big( \sum _{i=n} ^{\infty} g_i (s) \Big)ds  \\
&\quad + ( H T^{2p-1} / 4^p ) \sup _{u \in X} \int_0^T
 \Big(\sum _{i=n} ^{\infty} | u_i (s) |^p \Big)ds .
\end{aligned}
\end{equation}
Now, keeping in mind assumption (iii) and formula \eqref{eq11} expressing
the Hausdorff measure of noncompactness $\chi$ in the space $l_p$ and taking
into account the fact that the set $X$ consists of functions equicontinuous
 on the interval $I$ on the base of Remark \ref{rm21} and estimate \eqref{eq210}
 we infer the following inequality
$$
\big( \chi (FX) \big)^p \leq ( H T^{2p-1} / 4^p ) ( \chi (X) )^p .
$$
Consequently, we obtain
$$
\chi (FX) \leq [ (H/ T)^{1/p} \, T^2 / 4 ] \chi (X) .
$$
The above obtained inequality in conjunction with Theorem \ref{th11}
applied to the operator $F$ on the set $B_{r_0} ^1$ completes the proof.
\end{proof}

Further, let us notice that in view of the equivalence of the infinite system 
of integral equations \eqref{eq23} and the boundary-value
 problem \eqref{eq21}-\eqref{eq22}, we obtain the following reformulation 
of Theorem \ref{th23}.

\begin{theorem}\label{th24}
Under assumptions of Theorem  \ref{th23} the infinite system of 
differential equations of the second order {\rm \eqref{eq21}} satisfying 
boundary conditions {\rm \eqref{eq22}}, has at least one solution
 $u(t)= ( u_1(t), u_2(t), \ldots )$ such that $u_i \in C^2(I)$ for 
$i=1,2, \ldots$ and $u(t) \in l_p$ for any $t \in I$.
\end{theorem}

\begin{remark}\label{rm25} \rm
Observe that reasoning conducted in the proof of Theorem \ref{th23} 
require to impose the additional assumption that $p>1$. 
The case $p=1$ was considered in \cite{Mursaleen3}.
\end{remark}

Next we provide an example illustrating our considerations and results 
covered by Theorems \ref{th23} and \ref{th24}.

\begin{example}\label{ex26} \rm
Let us consider the infinite system of differential equations of the second order 
of the form
\begin{equation}\label{eq211}
u_i'' = - \frac {\sqrt{t} \, e^{-it}}{i}
 - \sum _{k=i} ^{\infty} \frac {\ln (1+t)}{k(1+i)}\cdot 
\frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)},
\end{equation}
for $t \in I=[0,T]$ and $i=1,2, \ldots$. The system \eqref{eq211} is 
considered together with the  boundary conditions
\begin{equation}\label{eq212}
u_i (0) = u_i (T) =0
\end{equation}
for $i=1,2, \ldots$.

In our further considerations we will apply the fact that for any fixed positive 
$\beta$ we have the equality
\begin{equation}\label{eq213}
\max _{a>0} \frac {a}{1 + \beta ^2 a^2} = \frac {1}{2 \beta} .
\end{equation}

Further, let us observe that problem \eqref{eq211}-\eqref{eq212} is a particular 
case of problem \eqref{eq21}-\eqref{eq22} if we put
$$
f_i ( t, u_1 , u_2 , \ldots ) = \frac {\sqrt{t} \, e^{-it}}{i} 
+ \sum _{k=i} ^{\infty} \frac {\ln (1+t)}{k(1+i)}\cdot
 \frac {u_k}{1+ (k-i)^2 \, u_k^2} 
$$
for an arbitrary $i=1,2, \ldots$.

Next fixing arbitrarily a natural number $i$, we obtain the  estimates
\begin{align*}
&| f_i ( t, u_1 , u_2 , \ldots ) | ^2 \\
& = \Big| \frac {\sqrt{t} \, e^{-it}}{i} 
 + \sum _{k=i} ^{\infty} \frac {\ln (1+t)}{k(1+i)}\cdot
  \frac {u_k}{1+ (k-i)^2 \, u_k^2} \Big| ^2 \\
& \leq 2 \Big\{ \frac {t \, e^{-2it}}{i^2} 
+ \Big[ \sum _{k=i} ^{\infty} \frac {\ln (1+t)}{k(1+i)}\cdot 
 \frac {u_k}{1+ (k-i)^2 \, u_k^2} \Big]^2 \Big\} \\
& \leq 2 \Big\{ \frac {t \, e^{-2it}}{i^2} 
 + \Big[ \Big( \sum _{k=i} ^{\infty} 
 \frac {\ln ^2(1+t)}{k^2(1+i)^2} \Big)^{1/2}
\Big( \sum _{k=i} ^{\infty} ( \frac {u_k}{1+ (k-i)^2 \, u_k^2} )^2 \Big)^{1/2} \Big]^2 
 \Big\} \\
& \leq \frac {2t e^{-2it}}{i^2} + 2 \Big( \sum _{k=i} ^{\infty} 
\frac {\ln ^2(1+t)}{k^2(1+i)^2} \Big) \cdot \sum _{k=i} ^{\infty} 
\Big( \frac {u_k}{1+ (k-i)^2 \, u_k^2} \Big)^2 .
\end{align*}
Hence, applying \eqref{eq213}, we obtain
\begin{equation} \label{eq214}
\begin{aligned}
&| f_i ( t, u_1 , u_2 , \ldots ) | ^2 \\
& \leq \frac {2t e^{-2it}}{i^2} + \frac {2 \ln ^2(1+t)}{(1+i)^2}
 \cdot \frac {\pi ^2}{6} \big( u_k^2 + \frac 14 + \frac {1}{4 \cdot 2^2}
 + \frac {1}{4 \cdot 3^2} + \cdots \big)  \\
& \leq \frac {2t e^{-2it}}{i^2} + \frac {2 \ln ^2(1+t)}{(1+i)^2} \cdot
\frac {\pi ^2}{24} \big( \frac {1}{1^2} + \frac {1}{2^2} + \cdots \big) +
\frac {2 \ln ^2(1+t)}{(1+i)^2} \cdot \frac {\pi ^2}{6} \cdot u_k^2  \\
& \leq \frac {2t e^{-2it}}{i^2} + \frac {\pi ^4}{72} \cdot
\frac {\ln ^2(1+t)}{(1+i)^2} + \frac {\pi ^2}{3} \cdot
\frac {\ln ^2(1+t)}{(1+i)^2} \cdot u_k^2 .
\end{aligned}
\end{equation}
Now, if we put
\[
g_i (t)  = \frac {2t e^{-2it}}{i^2} + \frac {\pi ^4}{72} \cdot
  \frac {\ln ^2(1+t)}{(1+i)^2} , \quad
h_i (t)  = \frac {\pi ^2}{3} \cdot \frac {\ln ^2(1+t)}{(1+i)^2},
\]
then the functions $g_i$, $h_i$ ($i=1,2, \ldots$) are continuous
on the interval $I$. Moreover, let us notice that the series
$\sum  _{i=1} ^{\infty} g_i (t)$ is uniformly convergent on $I$ which is a
simple consequence of the estimate
$$
| g_i (t) | = g_i (t) \leq \frac {2T}{i^2} + \frac {\pi ^4}{72} \ln ^2 (1+T)
\cdot \frac {1}{(i+1)^2} \leq [ 2T + \frac {\pi ^4}{72} \ln ^2 (1+T) ] \frac {1}{i^2}
$$
($t \in I$) and the classical Weierstrass test for uniform convergence of a
function series. On the other hand we have the obvious inequality
$$
| h_i (t) | = h_i (t) \leq \frac {\pi ^2  \ln ^2 (1+T)}{12}
$$
for $t \in I$ and for all $i=1,2, \ldots$. This means that the function sequence
$( h_i (t) )$ is equibounded on $I$.

Combining the above established facts with \eqref{eq214} we see that
$$
| f_i ( t, u_1 , u_2 , \ldots ) | ^2 \leq g_i (t) + h_i (t) | u_i | ^2
$$
for $t \in I$ and $i=1,2, \ldots$. This shows that assumption (iii) is satisfied. 
Moreover, we can accept the following constants $G$ and $H$ appearing in our 
considerations:
\begin{gather*}
G  = \frac {\pi ^2}{6} [ 2T + \frac {\pi ^4}{72} \ln ^2(1+T) ] , \\
H  = \frac {\pi ^2}{12} \ln ^2(1+T).
\end{gather*}
It is easily seen that the functions $f_i$ satisfy assumption (i) for each 
$i=1,2, \ldots$.

To prove that assumption (ii) is satisfied let us fix arbitrarily $t \in I$ 
and $u= ( u_i ) = ( u_1 , u_2 , \ldots ) \in l_2$. Then, in view of the above 
estimates, we obtain
$$
\sum _{i=1} ^{\infty} | f_i ( t, u_1 , u_2 , \ldots ) | ^2 
\leq \sum _{i=1} ^{\infty} g_i (t) + \sum _{i=1} ^{\infty} h_i (t) | u_i | ^2
\leq G+H \sum _{i=1} ^{\infty} | u_i | ^2 .
$$
Hence we deduce that the operator $f=( f_i )$ transforms the space
 $I \times l_2$ into $l_2$.

To show the remainder part of assumption (ii) let us fix arbitrarily a positive 
number $\varepsilon >0$ and an arbitrary point $u= ( u_i ) \in l_2$. 
Then, taking a point $v= ( v_i ) \in l_2$ such that 
$\| v-u \| _{l_2} < \varepsilon$, we have
\begin{align*}
&\| (fv)(t) - (fu)(t) \| _{l_2} \\
&= \sum _{i=1} ^{\infty} | f_i ( t, v_1 , v_2 , \ldots ) - f_i ( t, u_1 , u_2 , 
\ldots ) | ^2  \\
& = \sum _{i=1} ^{\infty} \Big| \sum _{k=i} ^{\infty} \frac {\ln (1+t)}{k(1+i)} \cdot 
\frac {v_k (t)}{1+ (k-i)^2 \, v_k^2 (t)} 
- \sum _{k=i} ^{\infty}  \frac {\ln (1+t)}{k(1+i)} \cdot 
\frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)} \Big| ^2 \\
& \leq  \sum _{i=1} ^{\infty} \Big| \sum _{k=i} ^{\infty} 
\frac {\ln (1+t)}{k(1+i)} \big[ \frac {v_k (t)}{1+ (k-i)^2 \, v_k^2 (t)} 
- \frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)}  \big] \Big| ^2 \\
& \leq \sum _{i=1} ^{\infty} 
\Big[ \frac {\ln (1+t)}{1+i} \sum _{k=i} ^{\infty} \frac 1k | 
\frac {v_k (t)}{1+ (k-i)^2 \, v_k^2 (t)} 
- \frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)} | \Big] ^2 \\
& \leq \sum _{i=1} ^{\infty} \frac {\ln ^2 (1+t)}{(1+i)^2} 
\Big[ \sum _{k=i} ^{\infty} \frac 1k | \frac {v_k (t)}{1+ (k-i)^2 \, v_k^2 (t)}
 - \frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)} | \Big] ^2 .
\end{align*}
Hence, applying the classical Cauchy-Schwarz inequality, we derive the  estimates
\begin{align*}
&\| (fv)(t) - (fu)(t) \| _{l_2} ^2 \\
& \leq \sum _{i=1} ^{\infty} \frac {\ln ^2 (1+t)}{(1+i)^2}
\Big[ \Big( \sum _{k=i} ^{\infty} \frac {1}{k^2} \Big) ^{1/2}
\Big( \sum _{k=i} ^{\infty} | \frac {v_k (t)}{1+ (k-i)^2 v_k^2 (t)} 
- \frac {u_k (t)}{1+ (k-i)^2  u_k^2 (t)} |^2 \Big) ^{1/2} \Big] ^2 \\
& \leq \frac {\pi ^2}{6} \ln ^2 (1+t) \sum _{i=1} ^{\infty} \frac {1}{ (1+i)^2} 
\Big(  \sum _{k=i+1} ^{\infty} \Big[ \Big| \frac {v_k (t)}{1+ (k-i)^2 \, v_k^2 (t)} 
- \frac {u_k (t)}{1+ (k-i)^2 \, u_k^2 (t)} \Big|^2  \\
&  + | v_i (t) - u_i (t) | ^2 \Big]  \Big) .
\end{align*}
Hence, we obtain
$$
\| (fv)(t) - (fu)(t) \| _{l_2} ^2 
\leq \frac {\pi ^2}{6} \ln ^2 (1+t) \sum _{i=1} ^{\infty} \frac {1}{ (1+i)^2}
\Big(  \sum _{k=i} ^{\infty} | v_k - u_k | ^2 \Big).
$$
Consequently, from the above inequality we derive the following one
$$
\| (fv)(t) - (fu)(t) \| _{l_2} ^2
 \leq \frac {\pi ^2}{6} \ln ^2 (1+t) \sum _{i=1} ^{\infty} 
\frac {1}{ (1+i)^2} \varepsilon ^2 
\leq \varepsilon ^2 \frac {\pi ^4}{36} \ln ^2 (1+t).
$$
Finally, we obtain
$$
\| (fv)(t) - (fu)(t) \| _{l_2} 
\leq \varepsilon \frac {\pi ^2}{6} \ln (1+t) 
\leq \varepsilon \frac {\pi ^2}{6} \ln (1+T) 
$$
for any $t \in I$.
This   estimate   proves   the   desired   equicontinuity   of   the   family   
of   functions 
$\{ (fu)(t) \}_{t \in I}$ and simultaneously shows that problem 
\eqref{eq211}-\eqref{eq212} has at least one solution $u(t)= ( u_i (t) )$ 
in the space $C(I, l_2)$ provided $(H/T)^{1/2} \, T^2 <4$ and $T \leq 1$.
Keeping in mind our earlier obtained evaluation concerning the constant $H$,
 we conclude that in our considerations connected with problem 
\eqref{eq211}-\eqref{eq212} we can take any $T$ such that 
$T \sqrt{T} \ln (1+T) < 8 \sqrt{3} / \pi$. For example, the number $T=1$ 
satisfies this inequality.
\end{example}


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\end{document}