\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 $-\infty0$ 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 $-\infty0$, $\alpha+\beta\not=0$, 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 $\delta$. 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