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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 257, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/257\hfil BVPs with an integral constraint]
{Boundary value problems with an integral constraint}

\author[E. M. Mangino, E. Pascali \hfil EJDE-2015/257\hfilneg]
{Elisabetta M. Mangino, Eduardo Pascali}

\address{Elisabetta M. Mangino \newline
Dipartimento di Matematica e Fisica ``E. De Giorgi''\\
Universit\`a del Salento\\
I-73100 Lecce, Italy}
\email{elisabetta.mangino@unisalento.it}

\address{Eduardo Pascali\\
Dipartimento di Matematica e Fisica ``E.De Giorgi''\\
Universit\`a del Salento\\
I-73100 Lecce, Italy}
\email{eduardo.pascali@unisalento.it}

\thanks{Submitted  April 13, 2015. Published October 2, 2015.}
\subjclass[2010]{34B15}
\keywords{Second order ODE; boundary condition; integral condition}


\begin{abstract}
 We show the existence of solutions for  a second-order ordinary differential
 equation coupled with a boundary-value condition and an integral condition.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks

\section{Introduction and preliminaries}

Ordinary differential equations are usually associated  with further conditions,
such as prescribed initial or boundary values, periodicity  etc.
(see e.g. \cite{A, PVS}). Apart from the previous ones, other classes of
problems where ordinary differential equations are coupled with more
elaborate conditions  have been  studied (see e.g. \cite{C, N,S, W, Z, Z1}
and the references quoted in the survey \cite{M}).
The aim  in these cases is usually to get existence and uniqueness results
from the assigned conditions.

In  the present note we investigate  a problem in which a general
ordinary differential equation of the second order is coupled with a boundary
quasi-linearity condition and an integral condition.  More precisely
we consider the  problem
\begin{gather}
 y''(x)=f(x, y(x), y'(x))\quad a\leq x\leq b \label{1a}\\
\alpha y(a) + \beta y(b)=\gamma, \label{1b}\\
\int_a^b y'(t)^2 dt=\delta,\label{1c}
\end{gather}
where $-\infty<a<b<+\infty$, $\alpha, \beta,\gamma, \delta\in \mathbb{R}$ with
$\delta>0$ and  $f:[a,b]\times\mathbb{R}^2\to \mathbb{R}$  is a continuous
function such that
 \begin{itemize}
\item[(H1)] There exist $\sigma_1, \sigma_2\in\mathbb{R}$ such that
$0<\sigma_1\leq f(x,u,v)\leq \sigma_2$
for all  $(x,u,v)\in [a,b]\times\mathbb{R}^2$;

\item[(H2)]  there exists  $L_f>0$ such that for  all
$x\in [a,b]$ and all $(u_1,v_1), (u_2,v_2) \in \mathbb{R}^2$, we have
\[
|f(x,u_1,v_1)-f(x,u_2,v_2)| \leq L_f(|u_1-u_2| + |v_1-v_2|).
\]
\end{itemize}

We will prove that, under some additional condition on $f$,
problem \eqref{1a}--\eqref{1c}  has at least two solutions if
 $\alpha+\beta\not=0$, while it has at least one solution if $\alpha+\beta=0$.
The main tool will be the classical Shaefer's Fixed Point Theorem
(see e.g. \cite[Chapter 9]{E}):

\begin{theorem}[Schaefer's theorem] \label{thm1.1}
Let $T$ be a continuous and compact mapping of a Banach space $X$ into itself,
such that the set
\[
\{ x \in X: x = \lambda T x \text{ for some } 0 \leq \lambda \leq 1 \}
\]
is bounded. Then $T$ has a fixed point.
\end{theorem}

We start with some preliminary observations.


\begin{lemma} \label{lem1.2}
 If $y\in C^2[a,b]$ satisfies \eqref{1a} and \eqref{1c}, then
\begin{equation} 
\begin{aligned}
\Delta(y)&:=\Big[ \int_a^b  \int_a^t f(t,y(s),y'(s))ds dt \Big]^2 \\
&- (b-a) \Big[\int_a^b\Big[ \int_a^t f(t,y(s),y'(s))ds\Big]^2dt-\delta\Big] 
\geq 0
\end{aligned} \label{Delta}
\end{equation}
and either for every $x\in [a,b]$,
\begin{align*} 
y(x)&=y(a) - \frac{\int_a^b\int_a^t f(s, y(t), y'(s))\,ds\,dt
+ \sqrt{\Delta(y)}}{b-a}(x-a) \\
&\quad + \int_a^x\int_a^t f(s, y(s), y'(s))\,ds\,dt\,,
\end{align*}
or for every $x\in [a,b]$,
\begin{align*} 
y(x)&=y(a) - \frac{\int_a^b\int_a^t f(s, y(s), y'(s))\,ds\,dt
- \sqrt{\Delta(y)}}{b-a}(x-a) \\
&\quad + \int_a^x\int_a^t f(s, y(s), y'(s))\,ds\,dt\,.
\end{align*}
\end{lemma}


\begin{proof}
 By integrating \eqref{1a}, we obtain
\begin{equation*}
y'(x)^2=y'(a)^2 + 2y'(a)\int_a^x f(t,y(t), y'(t))dt 
+ \Big[ \int_a^x f(t,y(t),y'(t))dt\Big]^2.
\end{equation*}
Hence, by \eqref{1c}, $y'(a)$ is a solution of the equation
\begin{equation}\label{eq3}
\begin{aligned}
&z^2(b-a) + 2z\int_a^b\int_a^t f(s,y(s), y'(s))\,ds\,dt \\
&+ \int_a^b \Big[ \int_a^t f(s,y(s),y'(s))ds\Big]^2 dt-\delta=0.
\end{aligned}
\end{equation}
Therefore,
\begin{align*} 
\Delta(y)&:= \Big[ \int_a^b  \int_a^t f(s,y(s),y'(s))ds dt\Big]^2 \\
&- (b-a) \Big[\int_a^b\Big[ \int_a^t f(s,y(s),y'(s))ds\Big]^2dt-\delta\Big] 
\geq 0
\end{align*}
and
\begin{equation}
y'(a)=\frac{-\int_a^b\int_a^t f(s, y(s), y'(s))\,ds\,dt
\pm \sqrt{\Delta(y)}}{b-a}. \label{y'}
\end{equation}
We obtain the assertion by observing that, if $y\in C^2[a,b]$ is a solution 
of \eqref{1a}, then
\begin{equation}\label{equat} 
y(x)=y(a)+y'(a)(x-a) + \int_a^x \int_a^t f(s, y(s),y'(s))\,ds\,dt.
 \end{equation}
\end{proof}


\section{The case $\alpha+\beta=0$}

If $\alpha+\beta =0$, $\alpha\neq 0$  and
$y\in C^2[a,b]$ is a solution of \eqref{1a}--\eqref{1c}, then by \eqref{equat},
\[ 
y(b)=y(a) + y'(a)(b-a) + \int_a^b\int_a^t f(s, y(s), y'(s))ds
\]
and therefore, by \eqref{1b},
\begin{equation} \label{7}
 y'(a)= -\frac{\gamma}{\alpha(b-a)} -\frac{1}{b-a}
\int_a^b\int_a^t f(s, y(s), y'(s))ds.
\end{equation}
But $y'(a)$ solves  \eqref{eq3},  therefore comparing \eqref{y'} and \eqref{7},  
we get  that $\Delta(y)=\gamma=0$ and
\begin{equation*} 
y'(a)= -\frac{1}{b-a}\int_a^b\int_a^t f(s, y(s), y'(s))ds.
\end{equation*}
Thus problem \eqref{1a}--\eqref{1c} turns into
\begin{gather} %{(P_0)\quad}
 y''(x)=f(x, y(x), y'(x)),\quad a\leq x\leq b, \label{1a0}\\
y(a)= y(b), \label{1b0}\\
\int_a^b y'(t)^2 dt=\delta.\label{1c0} 
 \end{gather}
Integrating by parts \eqref{1c0}, we find that
\begin{align*}
 \delta
&=\int_a^b y'(x)^2 dx\\
&= y(b)y'(b)-y(a)y'(a)-\int_a^b y(s)f(s, y(s), y'(s))ds \\
&=  y(a)(y'(b)-y'(a))- \int_a^b y(s)f(s, y(s), y'(s)ds \\
&=y(a)\cdot  \int_a^b f(s,y(s),y'(s))ds  - \int_a^b y(s)f(s, y(s), y'(s))ds.
\end{align*}
Hence
\begin{equation*} 
y(a)\cdot \int_a^b f(s, y(s), y'(s))ds=
\delta + \int_a^by(s) f(s, y(s), y'(s))ds
\end{equation*}
Assuming (H1), we obtain
\begin{equation*} 
\int_a^b f(s, y(s), y'(s))ds\geq \sigma_1(b-a) > 0.
 \end{equation*}
Therefore
\begin{equation*}
y(a)=\frac{\delta+ \int_a^by(s) f(s, y(s), y'(s))ds}{ \int_a^b f(s, y(s), y'(s))ds}
\end{equation*}
 and
\begin{align*} 
y(x)&=\frac{\delta + \int_a^by(s) f(s, y(s), y'(s))ds}{ \int_a^b f(s, y(s), y'(s))ds}
 -\frac{x-a}{b-a}\int_a^b\int_a^t f(s, y(s), y'(s))ds\\
&\quad + \int_a^x\int_a^t f(s, y(s), y'(s))\,ds\,dt.
\end{align*}
As a consequence, we easily obtain the following characterization of 
the solutions of \eqref{1a0}--\eqref{1c0}.

\begin{lemma}\label{le3} 
Assume that {\rm (H1)} holds.
Then   $y\in C^2[a,b]$ is a solution of  \eqref{1a0}--\eqref{1c0}  if and only if
$y\in C^1[a,b]$ and for every $x\in [a,b]$
\begin{align*}
 y(x)&=\frac{\delta +\int_a^by(s) f(s, y(s), y'(s))ds}
 { \int_a^b f(s, y(s), y'(s))ds} 
 - \frac{x-a}{b-a}\int_a^b\int_a^t f(s, y(s), y'(s))ds \\
&\quad + \int_a^x\int_a^t f(s, y(s), y'(s))\,ds\,dt.
\end{align*}
\end{lemma}

We consider now the following assumptions:
\begin{itemize}
 \item[(H1')] There exists $\sigma_3>0$ such that $|uf(x,u,v)|\leq \sigma_3$
for all $(x,u,v)\in [a,b]\times\mathbb{R}^2$.
 
\item[(H2)']  There exists $L>0$ such that fore all $x\in [a,b]$ and all 
$(u_1,v_1), (u_2,v_2) \in \mathbb{R}^2$,
\[  
|u_1f(x,u_1,v_1)-u_2f(x,u_2,v_2)| \leq L(|u_1-u_2| + |v_1-v_2|).
\]
\end{itemize}


\begin{theorem}\label{th2} 
If {\rm (H1), (H2), (H1'), (H2')}  hold,   then there exists  at least 
one solution of \eqref{1a0}--\eqref{1c0}.
\end{theorem}


\begin{proof}
Consider the map $T:C^1[a,b]\to C^1[a,b]$ defined by
\begin{align*}
T y(x)&= \frac{\delta +\int_a^by(s) f(s, y(s), y'(s))ds}{ \int_a^b f(s, y(s), 
y'(s))ds} + \frac{x-a}{b-a}\int_a^b\int_a^t f(s, y(s), y'(s))ds \\
&\quad+ \int_a^x\int_a^t f(s, y(s), y'(s))\,ds\,dt.
\end{align*}

By Lemma \ref{le3}, a function $y\in C^2[a,b]$ is a solution of 
\eqref{1a0}--\eqref{1c0}  if and only if $y\in C^1[a,b]$ is a fixed point of $T$.

Observe that for every $y\in C^1[a,b]$ and every $x\in [a,b]$:
\begin{gather*}
(Ty)'(x)= \frac{1}{b-a}\int_a^b\int_a^t f(s, y(s), y'(s))ds + \int_a^x f(s,y(s), y'(s))ds\\
(Ty)''(x)=f(x, y(x), y'(x)),
\end{gather*}
hence for every $x\in [a,b]$,
\begin{gather}
 |Ty(x)| \leq \frac{\delta + \sigma_3(b-a)}{\sigma_1(b-a)} 
 + \sigma_2 (b-a)^2\label{T1} \\
|(Ty)'(x)| \leq \frac{3}{2}(b-a)\sigma_2\label{T2}\\
 |(Ty)''(x)|\leq \sigma_2.
\end{gather}
Moreover, for every $y, z\in C^1[a,b]$, $x\in [a,b]$:
\begin{align*}
& |Ty(x)-Tz(x)| \\
&\leq   \Big| \bigg(\Big(\delta+ \int_a^by(s) f(s, y(s), y'(s))ds\Big)
 \int_a^b f(s, z(s), z'(s))ds \\
&\quad -\Big(\delta +\int_a^bz(s) f(s, z(s), z'(s))ds\Big)
 \int_a^b f(s, y(s), y'(s))ds \bigg)\\
&\quad \div \Big(\int_a^b f(s, y(s), y'(s))ds
\int_a^b f(s, z(s), z'(s))ds\Big) \Big| +L_f (b-a)^2\|y-z\|_{C^1} \\
&\leq \frac{\delta L_f}{(b-a)\sigma_1^2} \|y-z\|_{C^1}
+ \frac{ L_f\sigma_3 + L\sigma_2}{\sigma_1^2} \|y-z\|_{C^1}
+ L_f (b-a)^2\|y-z\|_{C^1}
\end{align*}
and
\begin{equation*}
|(Ty)'(x)-(Tz)'(x)|\leq \frac 3 2 L_f (b-a) \cdot \|y-z\|_{C^1},
\end{equation*}
 hence
 \[ 
\|Ty-Tz\|_{C^1} \leq \Big[\frac{ L_f\sigma_3 + L\sigma_2}{\sigma_1^2} 
+ L_f\Big( \frac 3 2 (b-a) + (b-a)^2+ \frac{\delta}{(b-a)\sigma_1^2}\Big)\Big] 
\|y-z\|_{C_1}.
\]
Thus $T$ is continuous on $C^1[a,b]$.

 We  prove that $T$ is also compact.
Let $(y_n)$ be a sequence in $C^1[a,b]$.
Then $((Ty_n)')_n$ is a bounded sequence of continuous functions such that 
$((Ty_n)'')_n$ is also bounded.
By Ascoli-Arzel\`a's theorem there exists a subsequence $(y_{k_n})_n$ 
such that $((Ty_{k_n})')_n$ is uniformly convergent on $[a,b]$.
 On the other hand, $(Ty_n)_n$ is a bounded sequence in $C^1[a,b]$, hence, 
passing to a subsequence if necessary, we can assume that $(Ty_{k_n}(a))_n$ 
is convergent. It follows that $(Ty_{k_n})_n$ converges in $C^1[a,b]$.

Thus, observing that the set
\[ 
\{y\in C^1[a,b]: y=\lambda Ty  \text{ for some } \lambda\in [0,1]\} 
\]
is clearly bounded by \eqref{T1} and \eqref{T2},  by Schaefer's theorem, 
$T$ has a fixed point. 
\end{proof}

An immediate application of the Schauder's Fixed Point Theorem  gives
the following result.

\begin{corollary} If
\[ 
\frac{ L_f\sigma_3 + L\sigma_2}{\sigma_1^2} + L_f\Big( \frac 3 2 (b-a) 
+ (b-a)^2+ \frac{\delta}{(b-a)\sigma_1^2}\Big)<, 1
\]
 then  problem \eqref{1a0}--\eqref{1c0} has a unique  solution in $C^2[a,b]$.
\end{corollary}


\section{The case $\alpha+\beta\not=0$}
	
 The main result of this section is the following Theorem.

\begin{theorem}\label{th1} 
If $\alpha+\beta\not=0$, {\rm (H1)} and {\rm (H2)} hold and 
\begin{itemize}
\item[(H3)] 
$(3\sigma_1^2-4\sigma_2^2)(b-a)^3 + 12\delta> 0$,
\end{itemize}
then  there exist at least two solutions to \eqref{1a}--\eqref{1c}.
\end{theorem}

We need first some lemmas about $\Delta(y)$,  defined for every 
$y\in C^1[a,b]$ as in \eqref{Delta}.

\begin{lemma}\label{Deltay} 
If {\rm (H1), (H2)}  hold, then for every $y\in C^1[a,b]$,
\begin{equation}\label{est1}
\begin{aligned}
M_1&:=\frac{ (3\sigma_1^2-4\sigma_2^2)(b-a)^4 + 12\delta(b-a)}{12}\\
&\leq \Delta(y)
\leq  M_2:=\frac{(3\sigma_2^2-4\sigma_1^2)(b-a)^4+ 12\delta(b-a)}{12}
\end{aligned}
 \end{equation}
and for every $y,z\in C^1[a,b]$,
\begin{equation}\label{est1a}
 |\Delta(y)-\Delta(z)| \leq  \frac 7 6 \sigma_2L (b-a)^4 \|y-z\|_{C^1}.
\end{equation}
\end{lemma}

\begin{proof}
By (H1), it holds that
\begin{align*}\Delta(y) &
\geq \sigma_1^2\Big[ \int_a^b (t-a)dt\Big]^2 + \delta (b-a) 
 - (b-a) \sigma_2^2 \int_a^b(t-a)^2dt \\
&= \sigma_1^2 \frac{(b-a)^4}{4} + \delta(b-a) -\sigma_2^2\frac{(b-a)^4}{3}\\
&=  \frac{ (3\sigma_1^2-4\sigma_2^2)(b-a)^4 + 12\delta(b-a)}{12}.
\end{align*}
On the other hand,
\begin{equation*} 
\Delta(y) \leq \sigma_2^2 \frac{(b-a)^4}{2} + \delta (b-a) 
-  \sigma_1^2 \frac{(b-a)^4}{3}
= \frac{(3\sigma_2^2-4\sigma_1^2)(b-a)^4+ 12\delta(b-a)}{12}.
\end{equation*}
Moreover,
 \begin{align*}
 &|\Delta(y)-\Delta(z)|  \\
&\leq \Big| \Big[ \int_a^b  \int_a^t f(s,y(s),y'(s))ds dt\Big]^2
 -  \Big[\int_a^b  \int_a^t f(s,z(s),z'(s))ds dt\Big]^2\Big|  \\
&\quad +(b-a) \Big| \int_a^b\Big[ \int_a^t f(s,z(s),z'(s))ds\Big]^2dt
 - \int_a^b\Big[ \int_a^t f(s,y(s),y'(s))ds\Big]^2dt\Big|  \\
&= \Big| \int_a^b\int_a^t[f(s,y(s),y'(s))- f(s, z(s), z'(s))]\,ds\,dt
\Big|\\
&\quad\times \Big| \int_a^b\int_a^x[f(s,y(s),y'(s))+ f(s, z(s), z'(s))]\,ds\,dt\Big|
\\
&\quad +(b-a) \Big| \int_a^b\Big[ \int_a^t[f(s,y(s),y'(s))- f(s, z(s), z'(s))]ds\Big]\\
&\quad\times \Big[\int_a^t[f(s,y(s),y'(s))+ f(s, z(s), z'(s))]ds\Big]  
dt\Big| \\
&\leq \Big(\sigma_2L_f \frac{(b-a)^4}{2} + \frac{2}{3}\sigma_2 L_f (b-a)^4g\Big) 
\|y-z\|_{C^1}\\
&= \frac 7 6 \sigma_2L_f (b-a)^4 \|y-z\|_{C^1}.  
\end{align*}
\end{proof}

As immediate consequence we have the lemma.

\begin{lemma}\label{Ai} 
Assume that {\rm (H1), (H2), (H3)} hold, and
 for every $y\in C^1[a,b]$  define
\begin{gather}
A_1(y)= \frac{-\int_a^b\int_a^x f(t, y(t), y'(t))\,dt\,dx+\sqrt{\Delta(y)}}{b-a}\\
A_2(y) = \frac{-\int_a^b\int_a^x f(t, y(t), y'(t))\,dt\,dx-\sqrt{\Delta(y)}}{b-a}
\end{gather}
Then, for $i=1,2$,  for every $y,z\in C^1[a,b]$,
\begin{gather}\label{est2}
|A_i(y)| \leq \frac{1}{b-a}\Big[ \sigma_2\frac{(b-a)^2}{2} + \sqrt{M_2} \Big] \\
 \label{est2a}
|A_i(y)-A_i(z)| \leq  L_f(b-a)\Big[\frac{1}{2}+\frac{7\sigma_2}{12\sqrt{M_1} } 
(b-a)^2\Big] \|y-z\|_{C^1} ,
\end{gather}
 where $M_1$ and $M_2$ are defined in Lemma \ref{Deltay}.
\end{lemma}

\begin{proof} 
First observe that $A_1$ and $A_2$ are well-defined on $C^1[a,b]$, 
since $\Delta(y)\geq 0$ for every $y\in C^1[a,b]$ by Lemma \ref{Deltay} and (H3).
For every $y\in C^1[a,b]$ and  $i=1,2$, by \eqref{est1},  we have 
\begin{align*}
|A_i(y)|
&\leq \frac{1}{b-a}\Big( \Big|\int_a^b\int_a^t f(s, y(s), y'(s))
\,ds\,dt\Big| +|\sqrt{ \Delta_i(y)}  | \Big)  \\
&\leq
 \frac{1}{b-a}\Big( \sigma_2\frac{(b-a)^2}{2} + \sqrt{M_2}\Big).
\end{align*}
Moreover, by \eqref{est1a}, for every $y,z\in C^1[a,b]$,
\begin{align*} 
&|A_i(y)-A_i(z)|\\
&= \Big| \frac{-\int_a^b\int_a^t \left[f(s, y(s), y'(s))- f(s, z(s),z'(s))\right]
\,ds\,dt\pm \left( \sqrt{\Delta(y)}- \sqrt{\Delta(z)} \right) } {b-a} 
\Big| \\
&\leq \frac{1}{b-a}\Big[ L_f\|y-z\|_{C^1} \frac{(b-a)^2}{2} 
+ \frac{\vert \Delta(y)-\Delta(z)\vert}{|\sqrt{\Delta(y)}
+\sqrt{\Delta(z)}|} \Big] \\
&\leq  \frac{1}{b-a}\Big[ L_f\|y-z\|_{C^1} \frac{(b-a)^2}{2} 
+ \frac{\vert \Delta(y)-\Delta(z)\vert}{2\sqrt{M_1}}\Big] \\
& \leq \frac{1}{b-a}\Big[ L_f\|y-z\|_{C^1} \frac{(b-a)^2}{2}
+ \frac{7}{12\sqrt{M_1} } \sigma_2L_f(b-a)^4 \|y-z\|_{C^1}\Big].
\end{align*}
\end{proof}

\begin{proof}[Proof of Theorem \ref{th1}]
 Observe first  that  if $\alpha+\beta\not=0$ and  $y \in C^2[a,b]$ 
is a solution of \eqref{1a}, \eqref{1b}, then
\begin{equation}
  \begin{aligned}
 y(x)
&=y'(a)(x-a) + \int_a^x\int_a^t f(s, y(s), y'(s)\,ds\,dt \\
&\quad +\frac{1}{\alpha+\beta}
\Big[ \gamma - \beta y'(a)(b-a) - \beta  \int_a^b\int_a^t f(s, y(s), y'(s))\,ds\,dt
\Big]
\end{aligned}\label{eqa0}
\end{equation}

By comparing \eqref{eqa0} with \eqref{y'} and by Lemma \ref{Ai}, we obtain
 that for $i=1$ or $i=2$,
\begin{equation}
\begin{aligned}
 y(x)&=A_i(y)(x-a) + \int_a^x\int_a^t f(s, y(s), y'(s)\,ds\,dt \\
&\quad +\frac{1}{\alpha+\beta}\Big[ \gamma - \beta A_i(y)(b-a)
- \beta  \int_a^b\int_a^t f(s, y(s), y'(s))\,ds\,dt\Big]
\end{aligned} \label{eqa1}
\end{equation}
It is immediate to prove that, conversely, if $y\in C^1[a,b]$
satisfies \eqref{eqa1}, then $y\in C^2[a,b]$ and $y$ is a solution
of \eqref{1a}--\eqref{1c}.

Thus $y\in C^2[a,b]$ is a solution of \eqref{1a}--\eqref{1c} if and 
only if $y$ is a fixed point of one of the operators 
$T_i:C^1[a,b]\to C^1[a,b]$, $i=1,2$
defined by
\begin{align*}
T_iy&=A_i(y)(x-a) + \int_a^x\int_a^t f(s, y(s), y'(s)\,ds\,dt \\
&\quad +\frac{1}{\alpha+\beta}\Big[ \gamma - \beta A_i(y)(b-a) 
- \beta  \int_a^b\int_a^t f(s, y(s), y'(s)\,ds\,dt\Big]
\end{align*}
Observe that for every $y\in C^1[a,b]$ and every $x\in [a,b]$,
\begin{gather*}
 (T_iy)'(x)= A_i(y) + \int_a^xf(s,y(s), y'(s))ds\\
 (T_iy)''(x)=f(x,y(x), y'(x)).
\end{gather*}
By Lemma \ref{Ai} we get that $T_i:C^1[a,b]\to C^1[a,b]$ is bounded.  
Moreover, if $y,z\in C^1[a,b]$ and $x\in [a,b]$,
\begin{align*}
|T_iy(x)-T_iz(x)| 
&\leq \Big( 1 +| \frac{\beta}{\alpha+\beta}|\Big)
\Big( |A_i(y)-A_i(z)|(b-a) + \frac{L_f}{2}(b-a)^2\|y-z\|_{C^1}\Big) \\
&\leq \Big( 1 +| \frac{\beta}{\alpha+\beta}|\Big)  L_f(b-a)^2
\Big[1+\frac{7\sigma_2}{12\sqrt{M_1} }(b-a)^2 \Big] \|y-z\|_{C^1}, 
\end{align*}
while
\begin{align*}
|(T_iy)'(x)-(T_iz)'(x)| 
&\leq |A_i(y)-A_i(z)| + L_f(b-a)\|y-z\|_{C^1} \\
&\leq  L_f(b-a)\Big[\frac{3}{2}+\frac{7\sigma_2}{12\sqrt{M_1} } (b-a)^2\Big] 
\|y-z\|_{C^1} .
\end{align*}
Thus $T_i$ is Lipschitz continuous with Lipschitz constant:
\begin{equation} \label{eq:5} 
\begin{aligned}
L_{a,b}&= \Big( 1 +| \frac{\beta}{\alpha+\beta}|\Big) 
 L_f(b-a)^2\big[1+\frac{7\sigma_2}{12\sqrt{M_1} }(b-a)^2 \big] \\
&\quad +  L_f(b-a)\big[\frac{3}{2}+\frac{7\sigma_2}{12\sqrt{M_1} } (b-a)^2\big] .
\end{aligned}
\end{equation}
With the same argument as in the proof of Theorem \ref{th2}, 
we can prove that  $T_i$  is compact and that, by Schaefer's fixed point theorem,  
$T_i$ has  a fixed point.  
\end{proof}

\begin{corollary} 
For every $a\in\mathbb{R}$, $\alpha, \beta,\gamma, \delta\in\mathbb{R}$ such that
$\alpha+\beta\not=0$,  for every $\sigma_1, \sigma_2>0, L_f>0$ such that 
(H2), (H1) hold, there exists $b>a$ such that the problem 
\eqref{1a}--\eqref{1c} has two solutions in $C^2[a,b]$.
\end{corollary}

\begin{proof}
  First observe that
\[
\lim_{b\to a^+} (3\sigma_1^2-4\sigma_2^2)(b-a)^3 + 12\delta =\delta> 0,
\]
thus $b$  can be chosen in such a way that (H3) is satisfied.  Moreover,
considering  the Lipschitz constant $L_{a,b}$ in  \eqref{eq:5} and observing that
$$
M_1=\frac{ (3\sigma_1^2-4\sigma_2^2)(b-a)^4 + 12\delta(b-a)}{12}\sim \delta (b-a) 
\quad \text{as } b\to a^+,
$$
we obtain $\lim_{b\to a^+} L_{a,b}= 0$.
Thus we can choose $b$ in such a way that $T_1$ and $T_2$ are  
contractions from $C^1[a,b]$ into itself and therefore have a unique 
fixed point.
\end{proof}


We conclude this article pointing out some problems that can be approached 
with similar considerations.


\begin{remark} \rm
Consider the problem
\begin{gather*}
 y''(x)=f(x, y(x), y'(x))\quad a\leq x\leq b \\
\alpha y(a) + \beta y(b)=\gamma \\
\int_a^b \exp(y'(t))dt=\delta,  
\end{gather*}
where $-\infty<a<b+\infty$, $\alpha, \beta,\gamma, \delta\in \mathbb{R}$ with
$\delta>0$, $\alpha+\beta\not=0$,  and  $f:[a,b]\times\mathbb{R}^2\to \mathbb{R}$ continuous.
Then
\[ 
y'(x)=y'(a)+\int_a^x f(s, y(s), y'(s))ds
\]
hence
\[ 
\delta=\exp(y'(a)) \int_a^b \exp\Big(\int_a^x f(s, y(s), y'(s))ds\Big)dx.
\]
Thus
\[
y'(a)=\log\Big(\frac{\delta}{ \int_a^b 
\exp\left(\int_a^t f(s, y(s), y'(s))ds\right)dt}\Big).
\]
It is immediate to prove that $y\in C^2[a,b]$ is a solution of the problem 
if and only if $y\in C^1[a,b]$ and $y$ is a fixed point of the operator
 $T:C^1[a,b]\to C^1[a,b]$ defined by
\begin{align*}
Ty(x)
&=- \frac{\beta}{\alpha+\beta}(b-a) 
\log\Big(\frac{\delta}{ \int_a^b \exp\left(\int_a^x f(s, y(s), y'(s))ds\right)dx}
\Big) \\
&\quad - \frac{\beta}{\alpha+\beta}  \int_a^b \int_a^x f(s, y(s), y'(s))ds\\
&\quad + (x-a)\log\Big(\frac{\delta}{ \int_a^b 
\exp\left(\int_a^x f(s, y(s), y'(s))ds\right)dx}\Big) \\
&\quad +  \int_a^x\int_a^tf(s, y(s), y'(s))ds + \frac{\gamma}{\alpha+\beta}.
\end{align*}
\end{remark}


\begin{remark} \rm
The approach we have used for the case $\alpha + \beta=0$ can be used also to 
study the case $\alpha+\beta\not=0$.  Indeed
if we integrate by parts  condition \eqref{1c},
we find that
\begin{align*}
\delta&=\int_a^b y'(x)^2 dx\\
&= y(b)y'(b)-y(a)y'(a)-\int_a^b y(s)f(s, y(s), y'(s)ds \\
&= \frac{\gamma-\alpha y(a)}{\beta}
y'(b) - y(a)y'(a)- \int_a^b y(s)f(s, y(s), y'(s)ds \\
&=\frac{\gamma-\alpha y(a)}{\beta} \Big(y'(a)+ \int_a^b f(s,y(s),y'(s))ds\Big)  
- y(a)y'(a)\\
&\quad - \int_a^b y(s)f(s, y(s), y'(s)ds.
\end{align*}
Hence
\begin{align*} 
&y(a)\Big[ (\alpha+\beta)y'(a) + \alpha\int_a^b f(s, y(s), y'(s))ds\Big] \\
&= -\beta\delta +\gamma y'(a) -\beta \int_a^by(s) f(s, y(s), y'(s))ds
\end{align*}
If
\begin{equation*} (\alpha+\beta)y'(a) + \alpha\int_a^b f(s, y(s), y'(s))ds\not=0 
\end{equation*}
then
\begin{equation*}
y(a)=\frac{-\beta\delta +\gamma y'(a) 
-\beta \int_a^by(s) f(s, y(s), y'(s))ds}{ (\alpha+\beta)y'(a) 
+ \alpha\int_a^b f(s, y(s), y'(s))ds}
\end{equation*}
and therefore
\begin{align*} 
y(x)&=\frac{-\beta\delta +\gamma y'(a) -\beta \int_a^by(s) f(s, y(s), y'(s))ds}
{ (\alpha+\beta)y'(a) + \alpha\int_a^b f(s, y(s), y'(s))ds}
+y'(a)(x-a) \\
&\quad + \int_a^x\int_a^t f(s, y(s), y'(s))ds
\end{align*}
As a consequence,  we easily obtain that
for  $y\in C^1[a,b]$ such that
\begin{equation*} 
(\alpha+\beta)y'(a) + \alpha\int_a^b f(s, y(s), y'(s))ds\not=0,
\end{equation*}
the following conditions are equivalent:
\begin{itemize}
\item[(i)] $y\in C^2[a,b]$ is a solution of  \eqref{1a}--\eqref{1c}

\item[(ii)]  $y\in C^1[a,b]$ and either for $i=1$ or $i=2$,
\begin{align*} 
y(x)&=\frac{-\beta\delta +\gamma A_i(y) 
 -\beta \int_a^by(s) f(s, y(s), y'(s))ds}{ (\alpha+\beta)A_i(y) 
 + \alpha\int_a^b f(s, y(s), y'(s))ds} \\
&\quad+ A_i(y)(x-a) + \int_a^x\int_a^t f(s, y(s), y'(s))ds.
\end{align*}
\end{itemize}
Anyway, with this approach one has to require additional conditions on $f$ 
such as (H1') and (H2').
\end{remark}

\begin{remark} \rm
Similar results can be obtained if  \eqref{1c} is replaced with
\[ 
\int_a^b [ \alpha(x)[y'(x)]^2 + \beta(x) y'(x)+ \gamma(x)]dx=\delta\in\mathbb{R}
\]
with suitable conditions on the functions $\alpha, \beta, \gamma$ and on $\delta$.
Moreover it could be of interest the study of systems such as
\begin{gather*} 
y^{(n)}(x) =f(x, y(x), y'(x), \dots,y^{(n-1)}(x)), \quad a\leq x\leq b,\\
\alpha y(a)+\beta y(b)=\gamma\\
\int_a^b y'(x)^2 dx=\delta_1\\
\int_a^b y''(x)^2 dx=\delta_2\\
\dots\\
\int_a^b y^{(n-1)}(x) dx=\delta_n
\end{gather*}
or analogous problems.
\end{remark}

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\end{document}