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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 20, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/20\hfil Nonexistence of solutions to BVP]
{Nonexistence of solutions to some boundary-value problems for
second-order ordinary differential equations}

\author[G. L. Karakostas \hfil EJDE-2012/20\hfilneg]
{George L. Karakostas}

\address{George L. Karakostas \newline
 Department of Mathematics, University of Ioannina,
 451 10 Ioannina, Greece}
\email{gkarako@uoi.gr}

\thanks{Submitted December 9, 2011. Published February 2, 2012.}
\subjclass[2000]{34B15, 34B99, 34K10}
\keywords{Boundary value problems; non-existence of solutions}

\begin{abstract}
 We present a method to obtain  concrete sufficient conditions
 which guarantee non-existence of solutions lying into a prescribed
 domain of six two- or three-point boundary value problems for
 second-order ordinary differential equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

 \section{Introduction}

  Let $I$ be the interval [0,1] of the real line $\mathbb{R}$ and
let $C(I)$ be the Banach space of all continuous  functions
$x: I\to\mathbb{R}$, endowed with the sup-norm $\|\cdot\|$.
In this article, we investigate the  non-existence of a solution $x$
of the equation
  \begin{equation}\label{e7}
 x''(t)+(Fx)(t)=0,\quad \text{a. a. }t\in{I},
 \end{equation}
satisfying one of the following conditions:
\begin{gather}
\label{e1} x(0)=0,\quad  x'(1)=\alpha x'(0),\\
\label{e2} x(0)=0,\quad  x(1)=\alpha x'(0),\\
\label{e3} x'(0)=0,\quad  x(1)=\alpha x(\eta),\\
\label{e4} x'(0)=0,\quad  x'(1)=\alpha x(0),\\
\label{e5} x(0)=0,\quad  x(1)=\alpha x(\eta), \\
\label{e6} x(0)=\alpha x'(0),\quad  x(1)=0,
\end{gather}
and lying into a prescribed domain of the space $C(I)$.
Here assume that  $\alpha\geq 0$ and $\eta\in[0,1]$.
The dependence of the response $F$ on the function $x$ might be in a moment
or a functional way.

Some publications which deal with the existence of positive solutions of
equations of the form \eqref{e7} lying in a certain domain, associated
with the conditions (or the multi-point version of them) respectively, are,
for instance,  \cite{bf, ma, kt1, kt2, kt3} for \eqref{e1};
\cite{LG1} for \eqref{e2}, \cite{aoy1, JGZ} for \eqref{e3};
\cite{cz1}  for   \eqref{e4};
\cite{bf1, chen, chen1, CR1, DG1, db1, HG1, KT1,  LP1, MW, YR, SW1,
 xx, YS, w1, WL1, wi1, ZS1} for  \eqref{e5};
\cite{mt} for \eqref{e6}. (See, also, the references therein)

In most of these cases the response factor $(Fx)(t)$ has the
Hammerstein form $q(t)f(x(t))$ and the quotient ${f(u)}/{u}$ plays
a central role. In particular, its least and upper limiting values
at $0+$ and at $+\infty$ are combined with some arguments related
to the Green's function, in order to ensure the applicability of
a method leading to the existence of a fixed point of an appropriate
integral operator. But, as the sufficient conditions are
important for the existence of  solutions, the necessary conditions
are  (more or less)  equally important.

 Working in this direction, in this paper
we provide some rather simple sufficient conditions for the nonexistence
of (positive) solutions, which lie in an angular domain of the origin.
Our discussion refers to the existence of a real number $\rho>0$
such that no solution of the problem exists satisfying the inequality
$$
|(Fx)(t)|<\rho\|x\|,
$$
for almost all $t\in I$. This fact implies that, if we have  the
response $q(t)f(x(t))$, then  we can proceed  further to obtain a
more concrete domain not containing solutions. Indeed, assume that
$(Fx)(t)$ is of the form $q(t)f(x(t))$, where, in the simplest case,
$q$ is measurable and essentially bounded and $f$ is nondecreasing.
Then, the general result is formulated as follows:
\emph{There is no solution $x$ of equation \eqref{e7} satisfying the
conditions $(1,j)$  such that}
$$
f(\|x\|)<\frac{\rho_j}{\|q\|_{\infty}}\|x\|,
$$
where $j=2,\dots,7$.

\section {Problem \eqref{e1}-\eqref{e7}}

 We start with the following theorem.

\begin{theorem}\label{th1}
 Assume that $F$ is a function defined in a domain ${\mathcal{D}}(F)$
of the space $C(I)$ such that  for each $x\in{\mathcal{D}}(F)$
the value $Fx:I\to{\mathbb{R}}$ is a  Lebesgue measurable function.
\begin{itemize}
\item[(i)] If $\alpha>1$, then  there is no positive solution $x$ of
the problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and
such that $(Fx)(t)\geq 0$, a.e. on $I$.

\item[(ii)] If $\alpha\in[0,1]$, then there is no solution  $x$ of
 problem \eqref{e1}-\eqref{e7} lying in ${\mathcal{D}}(F)$ and such
that
\begin{equation}\label{e01}
\operatorname{ess\,sup}\frac{(Fx)(t)}{x^2(t)}<+\infty.
\end{equation}
\end{itemize}
\end{theorem}

\begin{proof}
(i) Assume that $\alpha>1$ and let $x$ be a positive solution of the
problem in ${\mathcal{D}}(F)$.  Consider a real number
 $\lambda\neq 0$ and write equation \eqref{e7} in the form
$$
x''(t)+\lambda x'(t)=\lambda x'(t)-(Fx)(t).
$$
Multiply both sides with $e^{\lambda t}$ and take
$$
\big(x'(t)e^{\lambda t}\big)'=[x''(t)+\lambda x'(t)]e^{\lambda t}
=[\lambda x'(t)-(Fx)(t)]e^{\lambda t}.
$$
 Integrating from 0 to $t$, we obtain
\begin{align*}
x'(t)e^{\lambda t}&=x'(0)+{\lambda}\int_0^tx'(s)e^{\lambda s}ds
-\int_0^t(Fx)(s)e^{\lambda s}ds.\\
&=x'(0)+\lambda x(t)e^{\lambda t}-\lambda x(0)
-\int_0^tz(s)e^{\lambda s}ds,
\end{align*}
where
$z(s):=\lambda^2x(s)+(Fx)(s)$.
Thus we have
\begin{equation}\label{e8}
x'(t)-\lambda x(t)=x'(0)e^{-\lambda t}
-\int_0^tz(s)e^{-\lambda (t-s)}ds.
\end{equation}
Multiplying both sides by $e^{-\lambda t}$, we obtain
\[
\big(x(t)e^{-\lambda t}\big)'
=[x'(t)-\lambda x(t)]e^{-\lambda t}
=x'(0)e^{-2\lambda t}-\int_0^tz(s)e^{-\lambda (2t-s)}ds.
\]
Integrate both sides from 0 to $t$ and take
\begin{align*}
x(t)e^{-\lambda t}
&=x'(0)\int_0^te^{-2\lambda s}ds-\int_0^t\int_0^rz(s)e^{-\lambda (2r-s)}dsdr\\
&=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t})
 -\int_0^tz(s)e^{\lambda s}\int_s^te^{-2\lambda r}drds\\
&=\frac{x'(0)}{2\lambda}(1-e^{-2\lambda t})
 -\frac{1}{2\lambda}\int_0^tz(s)e^{\lambda s}(e^{-2\lambda s}-e^{-2\lambda t}),
\end{align*}
because of \eqref{e1}.
Finally we obtain
\begin{equation}\label{e9}
x(t)=\frac{x'(0)}{\lambda}\sinh({\lambda t})
-\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.
\end{equation}
By using \eqref{e1} and \eqref{e8} it follows that
\begin{align*}
\alpha x'(0)=x'(1)
&=\lambda x(1)+x'(0)e^{-\lambda}-\int_0^1z(s)e^{-\lambda (1-s)}ds\\
&=x'(0)\cosh({\lambda})-\int_0^1z(s)\cosh({\lambda(1-s)})ds
\end{align*}
and therefore the solution $x$ must satisfy
\begin{equation}\label{e10}
x'(0)(\cosh(\lambda)-\alpha)=\int_0^1z(s)\cosh({\lambda(1-s)})ds,
\end{equation}
for all $\lambda\neq 0$.
Observe that the right side is a positive quantity, while the left
side depends on $\lambda$. Hence, if it holds $x'(0)>0$, choose
$\lambda$ such that  $\cosh(\lambda)<\alpha$ and if $x'(0)<0$,
choose $\lambda$ such that  $\cosh(\lambda)>\alpha$, to get a contradiction.
\smallskip

(ii) Next assume that $\alpha\in[0,1]$. Let $x$ be a solution satisfying
relation \eqref{e01}. Choose $\lambda$ negative large enough such that
$$
z(t):=(Fx)(t)+\lambda x^2(t)<0,
$$
for a.a. $t\in I$. This and \eqref{e10} imply that
$x'(0)<0$ and so, due to the fact that $x(0)=0$, the solution $x$
can not be positive. The proof is complete.
\end{proof}

In the sequel we shall assume  that $0<\alpha<1$ and moreover the
function $F$ satisfies the following condition:
\begin{itemize}
\item[(C)]  $F$ is a function defined in a domain ${\mathcal{D}}(F)$
of the space $C(I)$ and for each $x\in{\mathcal{D}}(F)$ the value
$Fx:I\to{\mathbb{R}}$ is an element of ${\mathcal{L}}_{\infty}$.
\end{itemize}

For each $\rho>0$ we shall denote by ${\mathcal{A}}_{\rho}$ the set
of all  functions $x\in{\mathcal{D}}(F)$ satisfying the inequality
$$
\|Fx\|_{\infty}<\rho\|x\|.
$$

\begin{theorem}  \label{thm2.2}
Assume that $F$ satisfies condition $(C)$.
Then  there is none $x\in {\mathcal{A}}_{\rho_1}$ which solves
 problem \eqref{e1}-\eqref{e7}, where 
$$
\rho_1:=1-\alpha.
$$
\end{theorem}

\begin{proof}
Assume that a solution exists satisfying the requirements of the theorem
and take $\rho'$ such that  $\rho'<\rho_1$ and
$\|Fx\|_{\infty}< \rho'\|x\|$.  Since $1-\alpha$ is the maximum of
the function
$$
\psi(\lambda):=\frac{\cosh(\lambda)-\sinh^2(\lambda)-\alpha}{\sinh^2(\lambda)}
\lambda^2,\quad \lambda\geq 0,
$$
we can choose $\lambda\geq 0$ such that
\begin{equation}\label{e11}
\rho'<\psi(\lambda).
\end{equation}

Next,  as in Theorem \ref{th1}, we obtain relation \eqref{e9}.
By using relation \eqref{e8} and the boundary condition \eqref{e1}
we obtain
$$
\alpha x'(0)=x'(1)=\lambda x(1)+x'(0)e^{-\lambda}-\int_0^1z(s)e^{-\lambda(1-s)}ds,
$$
and due to \eqref{e9} it follows that
$$
x'(0)=\frac{1}{\cosh(\lambda)-\alpha}\int_0^1z(s)\cosh(\lambda(1-s))ds.
$$
Hence we have
\begin{equation}\label{e12}
x(t)
=\frac{\sinh({\lambda t})}{\lambda[\cosh(\lambda)
 -\lambda\alpha]}\int_0^1z(s)\cosh(\lambda(1-s))ds
 -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.
\end{equation}

Assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides
of \eqref{e12} by $x(t_0)$, we obtain
\[
\lambda[\cosh(\lambda)-\alpha]
\leq\sinh(\lambda)\int_0^1\Big[\lambda^2\frac{x(s)}{\|x\|}
+\frac{|(Fx)(s)|}{\|x\|}\Big]\cosh({\lambda(1-s)})ds.
\]
From this relation we obtain
$$
\lambda(\cosh(\lambda)-\alpha)< \sinh({\lambda})\int_0^1[\lambda^2+\rho']
\cosh({\lambda(1-s)})ds
$$
and so
$$
\lambda^2(\cosh(\lambda)-\alpha)< \sinh^2({\lambda})[\lambda^2+\rho'].
$$
The latter contradicts to \eqref{e11}  and so there is no solution
of the problem.
\end{proof}

\section {Problem \eqref{e2}-\eqref{e7}}

Before we will discuss the problem, we need some
information about the function defined by
$$
\phi(\lambda):=\frac{\sinh(\lambda)}{\lambda}(2-\cosh(\lambda)), \quad
\lambda\in[0,1].
$$
Observe that $2-\cosh(\lambda)>0$ for all $\lambda\in[0,1]$.
Also, since $\phi(0)=1$  we  conclude  that for each $\alpha\in(0,1)$
there is a $\lambda\in (0,1)$ such that
$\alpha<\phi(\lambda)$.
Thus the set
$$
D_{\alpha}:=\{\lambda\in[0,1): \phi(\lambda)>\alpha \}
$$
is nonempty and it contains a right neighborhood of 0.

\begin{theorem} \label{thm3.1}
Assume that $F$ satisfies condition {\rm (C)}.
Then  there is none $x\in {\mathcal{A}}_{\rho_2}$ which solves
 problem  \eqref{e2}-\eqref{e7}, where
 $$
 \rho_2:=2(1-\alpha).
 $$
\end{theorem}

\begin{proof}
Assume that a solution exists satisfying the requirements of the theorem
and take $\rho'$ such that  $\rho'<\rho_2$ and $\|Fx\|_{\infty}< \rho'\|x\|$.
Since the number $2(1-\alpha)$ is the maximum of the quantity
$$
\psi_1(\lambda):= \frac{\phi(\lambda)-\alpha}{[\cosh(\lambda)-1]
\sinh(\lambda)} \lambda^3,\quad  \lambda\geq 0,
$$
we can choose $\lambda>0$ such that
\begin{equation}\label{e13}
\rho'<\psi_1(\lambda).
\end{equation}
Next, following the same method as in Theorem \ref{th1}, we obtain
relation \eqref{e9}. By using the boundary condition \eqref{e2} we obtain
$$
\alpha x'(0)=x(1)=\frac{x'(0)}{\lambda}\sinh({\lambda})
 -\frac{1}{\lambda}\int_0^1z(s)\sinh(\lambda(1-s))ds,
$$
from which it follows that
$$
x'(0)=\frac{1}{\sinh(\lambda)-\lambda\alpha}\int_0^1z(s)\sinh(\lambda(1-s))ds.
$$
Hence we have
\begin{equation}\label{e14}
x(t)
=\frac{\sinh({\lambda t})}{\lambda[\sinh(\lambda)
 -\lambda\alpha]}\int_0^1z(s)\sinh(\lambda(1-s))ds
 -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.
\end{equation}
From here and our assumptions we conclude that
$\sinh(\lambda)>\lambda\alpha$.

Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$.
Dividing both sides of \eqref{e14} by $x(t_0)$, we obtain
\[
\lambda[\sinh(\lambda)-\lambda\alpha]
\leq\sinh(\lambda)\int_0^1\Big[\lambda^2\frac{x(s)}{\|x\|}
+\frac{|(Fx)(s)|}{\|x\|}\Big]\sinh({\lambda(1-s)})ds.
\]
From this relation we obtain
$$
\lambda(\sinh(\lambda)-\lambda\alpha)
< \sinh({\lambda})\int_0^1[\lambda^2+\rho']\sinh({\lambda(1-s)})ds
$$
and so
$$
\lambda^2(\sinh(\lambda)-\lambda\alpha)
< \sinh({\lambda})[\lambda^2+\rho'](\cosh({\lambda)-1)}.
$$
The latter contradicts to \eqref{e13}  and so there is no solution
of the problem.
\end{proof}

\section {Problem \eqref{e3}-\eqref{e7}}

\begin{theorem}\label{th2}
Assume that $F$ satisfies condition {\rm (C)}.
Then  there is no $x\in {\mathcal{A}}_{\rho_3}$ that solves
 problem  \eqref{e22}-\eqref{e1}, where
 $$
 \rho_3:=\sup_{\lambda>0}\frac{2-\cosh(\lambda)-\alpha
 e^{\lambda(\eta-1)}}{\cosh(\lambda)-1} \lambda^2.
 $$
\end{theorem}

\begin{proof}
Assume that a solution exists satisfying the requirements of the
theorem and take $\rho'$ such that  $\rho'<\rho_3$ and
$\|Fx\|_{\infty}< \rho'\|x\|$.  Choose $\lambda>0$ such that
\begin{equation}\label{e15}
\rho'<\frac{2-\cosh(\lambda)-\alpha e^{\lambda(\eta-1)}}{\cosh(\lambda)-1}
\lambda^2.
\end{equation}
Following the same method as in Theorem \ref{th1}, we obtain
relation \eqref{e8}, which due to \eqref{e3} becomes
\begin{equation}\label{e16}
x'(t)-\lambda x(t)=-\int_0^tz(s)e^{-\lambda (t-s)}ds.
\end{equation}
Multiplying both sides with $e^{\lambda t}$ we obtain
\begin{equation}\label{e17}
x(t)=x(0)e^{\lambda t}-\int_0^tz(s)\sinh(\lambda(t-s))ds.
\end{equation}
From this relation and \eqref{e3} we derive
$$
x(0)=\frac{1}{\lambda(e^{\lambda}-\alpha e^{\lambda\eta})}
\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds
-\alpha\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big],
$$
and therefore we have
\begin{equation}\label{e18}\begin{aligned}
x(t)&=\frac{e^{\lambda t}}{\lambda(e^{\lambda}
 -\alpha e^{\lambda\eta})}\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds\\
&\quad-\alpha\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big]
 -\frac{1}{\lambda}\int_0^tz(s)\sinh(\lambda(t-s))ds.
 \end{aligned}
\end{equation}

Assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$.
Dividing both sides of \eqref{e18} by $x(t_0)$, we obtain
$$
\lambda^2[e^{\lambda}-\alpha e^{\lambda\eta}]\leq e^{\lambda}
[\lambda^2+\rho'](\cosh({\lambda)-1)}).
$$
The latter contradicts to \eqref{e15}  and so there is no solution of
the problem.
\end{proof}

The following table shows the values of the bound $\rho_3$ for
some values of the coefficient $\alpha$ and the argument $\eta$.
\begin{center}
\begin{tabular}{|l||l|l|l|l|l|l|l|l|l|l|l|}
\hline
$\displaystyle{}_\eta\quad ^{\alpha}_{\rho_3}$ &0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\
\hline\hline
$0$& 1.808&1.628&1.464&1.307&1.160&1.020&0.888&0.761&0.639\\
\hline
$0.1$& 1.806&1.625&1.453&1.291&1.137&0.990&0.849&0.714&0.585 \\
\hline
$0.2$& 1.805&1.620&1.444&1.275&1.114&0.959&0.811&0.667&0.529 \\
\hline
$0.3$& 1.804&1.615&1.434&1.260&1.092&0.929&0.772&0.620&0.472 \\
\hline
$0.4$& 1.803&1.611&1.426&1.246&1.071&0.900&0.735&0.573&0.416 \\
\hline
$0.5$& 1.802&1.608&1.418&1.233&1.051&0.873&0.699&0.528&0.361 \\
\hline
$0.6$& 1.801&1.605&1.412&1.221&1.034&0.849&0.667&0.487&0.310 \\
\hline
$0.7$& 1.800&1.603&1.407&1.212&1.019&0.828&0.639&0.451&0.265 \\
\hline
$0.8$& 1.800&1.599&1.403&1.205&1.009&0.813&0.618&0.423&0.230 \\
\hline
$0.9$& 1.800&1.600&1.400&1.201&1.002&0.803&0.604&0.379&0.207 \\
\hline
\end{tabular}
\end{center}


\section {Problem \eqref{e4}-\eqref{e7}}

\begin{theorem} \label{thm5.1}
 Assume that $F$ satisfies condition {\rm (C)}.
 Then  there is none $x\in {\mathcal{A}}_{\rho_4}$ which solves
problem  \eqref{e4}-\eqref{e7},  where
$$
\rho_4:=\sup_{\lambda>0}\frac{\lambda[e^{\lambda}-\alpha]}{\lambda\cosh(\lambda)
-\lambda+1-e^{-\lambda}}-\lambda^2.
$$
\end{theorem}

\begin{proof}
Assume that a solution $x$ exists satisfying the requirements of the theorem and 
take $\rho'$ such that  $\rho'<\rho_4$ and $\|Fx\|_{\infty}< \rho'\|x\|$.  
Choose $\lambda>0$ such that
\begin{equation}\label{e19}
\rho'<\frac{\lambda[e^{\lambda}-\alpha]}{\lambda\cosh(\lambda)-\lambda+1-e^{-\lambda}}
-\lambda^2.
\end{equation}
Next, following the same method as in Theorem \ref{th2}, we obtain relation \eqref{e17}, 
which because of \eqref{e4} gives
$$
\alpha x(0)=\frac{1}{\lambda e^{\lambda}-\alpha}\int_0^1z(s)\Big[\sinh(\lambda(1-s))
+e^{-\lambda(1-s)}\Big]ds.
$$ 
Therefore,
\begin{equation}\label{e20}
x(t)=\frac{e^{\lambda t}}{\lambda e^{\lambda}-\alpha)}\int_0^1z(s)
 \Big[\sinh(\lambda(1-s))+e^{-\lambda(1-s)}\Big]ds
 -\int_0^tz(s)e^{\lambda(t-s)}ds.
\end{equation}
Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides 
of \eqref{e20} by $x(t_0)$, we, finally, obtain
$$
\lambda[\lambda- e^{-\lambda}\alpha]\leq[\lambda^2+\rho']
(\lambda\cosh(\lambda)-\lambda+1-e^{-\lambda}).
$$
The latter contradicts \eqref{e19}  and so there is no solution of the problem.
\end{proof}

The parameter $\rho_4$ for various values of the coefficient $\alpha$ 
is given in the following table.

\begin{center}
\begin{tabular}{|l||l|l|l|l|l|l|l|l|l|}
\hline
$\alpha$&0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\
\hline
$\rho_4$& 1.480&1.375&1.271&1.167&1.063&0.961&0.859&0.757&0.657 \\
\hline
\end{tabular}
\end{center}

\section {Problem \eqref{e5}-\eqref{e7}}

\begin{theorem} 
Assume that $F$ satisfies condition {\rm (C)}  and, moreover, assume
 that $0<\alpha<1$  and $0<\eta<1$. Then  there is none 
$x\in {\mathcal{A}}_{\rho_5}$ which solves  the problem  \eqref{e5}-\eqref{e7}, 
where 
$$
\rho_5:=\sup_{\lambda>0}\frac{\lambda^2[\sinh(\lambda)
-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda)\big(\cosh(\lambda)-1\big)}-\lambda^2.
$$
\end{theorem}

\begin{proof} 
Assume that a solution $x$ exists satisfying the requirements of the theorem and 
take $\rho'$ such that  $\rho'<\rho_5$ and $\|Fx\|_{\infty}< \rho'\|x\|$.  
Choose $\lambda>0$ such that
\begin{equation}\label{e21}
\rho'<\frac{\lambda^2[\sinh(\lambda)-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda)
\big(\cosh(\lambda)-1\big)}-\lambda^2.
\end{equation}
Next, following the same method as in Theorem \ref{th1}, we obtain
 relation \eqref{e9}, which because of \eqref{e5} gives
\[
x'(0)=\frac{1}{\sinh(\lambda)-\alpha\sinh(\lambda\eta)}
\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds
-\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big].
\]
Therefore,
\begin{equation}\label{e22}
\begin{aligned}
x(t)&=\frac{\sinh(\lambda t)}{\lambda\big(\sinh(\lambda)
 -\alpha\sinh(\lambda\eta)\big)}\Big[\int_0^1z(s)\sinh(\lambda(1-s))ds\\
&\quad -\int_0^{\eta}z(s)\sinh(\lambda(\eta-s))ds\Big]-\frac{1}{\lambda}
 \int_0^tz(s)\sinh({\lambda(t-s)})ds.
\end{aligned}
\end{equation}
Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides
 of \eqref{e22} by $x(t_0)$, we, finally, obtain
$$
\lambda^2[\sinh(\lambda)-\alpha\sinh(\lambda\eta)]
\leq[\sinh(\lambda)[\lambda^2+\rho'](\cosh(\lambda)-1).
$$
The latter contradicts \eqref{e21}  and so there is no solution of the problem.
\end{proof}

The following table shows the values of the bound $\rho_5$  for  some values 
of the coefficient $\alpha$ and the argument $\eta$. 

\begin{center}
\begin{tabular}{|l||l|l|l|l|l|l|l|l|l|l|}
\hline
$\displaystyle{_\alpha\quad ^{\eta}_{\rho_5}}$ &0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\
\hline\hline
$0.1$& 1.978&1.959&1.939&1.919&1.899&0.878&0.858&0.839&0.818 \\
\hline
$0.2$& 1.958&1.919&1.878&1.839&1.798&1.759&1.718&1.679&1.637 \\
\hline
$0.3$& 1.938&1.919&1.878&1.839&1.798&1.759&1.718&1.518&1.458 \\
\hline
$0.4$& 1.917&1.838&1.758&1.679&1.598&1.519&1.439&1.355&1.267 \\
\hline
$0.5$& 1.897&1.797&1.697&1.599&1.498&1.398&1.298&1.195&1.097 \\
\hline
$0.6$& 1.879&1.759&1.638&1.519&1.399&1.279&1.159&1.037&0.918 \\
\hline
$0.7$& 1.858&1.719&1.578&1.439&1.296&1.159&1.017&0.879&0.736 \\
\hline
$0.8$& 1.839&1.678&1.518&1.359&1.195&1.039&0.879&0.719&0.559 \\
\hline
$0.9$& 1.819&1.637&1.458&1.278&1.099&0.917&0.738&0.559&0.379 \\
\hline
\end{tabular}
\end{center}

\section {Problem \eqref{e6}-\eqref{e7}}

\begin{theorem} 
Assume that $F$ satisfies condition {\rm (C)}. Then  there is no
 $x\in {\mathcal{A}}_{\rho_6}$ which solves problem  \eqref{e6}-\eqref{e7}, 
 where 
$$
\rho_6:=\sup_{\lambda>0}\frac{\lambda^2[\alpha\lambda
+\cosh(\lambda)]}{\sinh(\lambda)\big(\lambda\alpha e^{\lambda}
+\cosh(\lambda)\big)}-\lambda^2.
$$
\end{theorem}

\begin{proof} 
Assume that a solution $x$ exists satisfying the requirements of the theorem 
and take $\rho'$ such that  $\rho'<\rho_5$ and $\|Fx\|_{\infty}< \rho'\|x\|$. 
 Choose $\lambda>0$ such that
\begin{equation}\label{e23}
\rho'<\frac{\lambda^2[\sinh(\lambda)-\alpha\sinh(\eta\lambda)]}{\sinh(\lambda)
\big(\cosh(\lambda)-1\big)}-\lambda^2.
\end{equation}
Next, following the same method as previously, we obtain 
\[
x'(0)=\frac{1}{\alpha\lambda+\cosh(\lambda)}\int_0^1z(s)\sinh(\lambda(1-s))ds.
\]
Therefore,
\begin{equation}\label{e24}
x(t)=\frac{\lambda\alpha e^{\lambda t}+\sinh(\lambda t)}{\lambda
 \big(\alpha\lambda+\cosh(\lambda)\big)}\int_0^1z(s)\cosh(\lambda(1-s))ds
-\frac{1}{\lambda}\int_0^tz(s)\cosh({\lambda(t-s)})ds.
\end{equation}
Next assume that $\|x\|=x(t_0)$, for some $t_0\in[0,1]$. Dividing both sides 
of \eqref{e24} by $x(t_0)$, we, finally, obtain
$$
\lambda^2[\alpha\lambda+\cosh(\lambda)]\leq\sinh(\lambda)
[\lambda^2+\rho'](\lambda\alpha e^{\lambda}+\sinh(\lambda)).
$$
The latter contradicts \eqref{e23}  and so there is no solution of the problem.
\end{proof}

The following tableshows the values of the bound $\rho_6$, for  some values
 of the coefficient $\alpha$. 

\begin{center}
\begin{tabular}{|l||l|l|l|l|l|l|l|l|l|}
\hline
$\alpha$&0.1&0.2&0.3&0.4&0.5&0.6&0.7&0.8&0.9 \\
\hline
$\rho_6$& 0.223&0.216&0.211&0.206&0.202&0.198&0.195&0.192&0.190 \\
\hline
\end{tabular}
\end{center}

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\end{document}