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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 150, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/150\hfil Second order iterative BVP]
{Existence and uniqueness of solutions for a second-order
iterative boundary-value problem}

\author[E. R. Kaufmann \hfil EJDE-2018/150\hfilneg]
{Eric R. Kaufmann}

\address{Eric R. Kaufmann \newline
Department of Mathematics \& Statistics\\
University of Arkansas at Little Rock,
Little Rock, AR 72204, USA}
\email{erkaufmann@ualr.edu}

\thanks{Submitted September 20 2017. Published August 8, 2018.}
\subjclass[2010]{34B15, 34K10, 39B05}
\keywords{Iterative differential equation; Schauder fixed point theorem;
\hfill\break\indent  contraction mapping principle}

\begin{abstract}
 We consider the existence and uniqueness of solutions to the second-order
 iterative boundary-value problem
 \[
 x''(t) = f(t, x(t), x^{[2]}(t)), \quad a \leq t \leq b,
 \]
 where $x^{[2]}(t) = x(x(t))$, with solutions satisfying one of the
 boundary conditions $x(a) = a$, $x(b) = b$ or $x(a) = b$, $x(b) = a$.
 The main tool employed to establish our results is the Schauder fixed
 point theorem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction}

The study of iterative differential equations can be traced back to papers 
by Petuhov \cite{vrp} and Eder \cite{ee}. In 1965 Petuhov \cite{vrp} considered 
the existence of solutions to the functional differential equation
 $x'' = \lambda x(x(t))$ under the condition that $x(t)$ maps the interval 
$[-T, T]$ into itself and that $x(0) = x(T) = \alpha$. He obtained conditions
 on $\lambda$ and $\alpha$ for the existence and uniqueness of solutions. 
In 1984, Eder \cite{ee} studied solutions of the first order equation 
$x'(t) = x(x(t))$. The author proved that every solution either vanishes
 identically or is strictly monotonic. The author established conditions 
for the existence, uniqueness, analyticity, and analytic dependence of 
solutions on initial data. In 1990, using Schauder's fixed point theorem 
Wang \cite{kw} obtained a solution of $x' = f(x(x(t))), x(a) = a$, where 
$a$ is one endpoint of the interval of existence. In 1993, Fe\v{c}kan showed 
the existence of local solutions via the Contraction Mapping Principle for 
the initial value problem for the iterative differential equation 
$x'(t) = f(x(x(t))), x(0) = 0$. For more on iterative differential equations 
see the papers \cite{pa,vb} \cite{mf}-\cite{xpl}, \cite{pz1}-\cite{hz} 
and references therein.

In this paper we consider the existence and uniqueness of solutions to the
 second-order iterative boundary-value problem
\begin{equation}\label{eq01}
 x''(t) = f(t, x(t), x^{[2]}(t)), \quad a < t < b,
\end{equation}
where $x^{[2]}(t) = x(x(t))$, with solutions satisfying one of the following 
boundary conditions:
\begin{gather}\label{eq02}
 x(a) = a, \quad x(b) = b; \\
\label{eq03}
 x(a) = b, \quad x(b) = a.
\end{gather}
We assume throughout that $f\colon [a, b] \times \mathbb{R} \times \mathbb{R} 
\to \mathbb{R}$ is continuous. Due to the iterative term $x^{[2]}(t)$, 
in order for solutions to be well-defined, we require that the image 
of $x$ be in the interval $[a, b]$; that is, we need 
$a \leq x(t) \leq b$ for all $t \in [a, b]$.

In Section 2, we first rewrite \eqref{eq01}, \eqref{eq02} as an integral 
equation and then state a condition under which solutions of the 
integral equation will be solutions of the boundary value problem.
 We also state properties of the kernel that will be needed in the sequel. 
In Section 3, we state and prove theorems on the existence and uniqueness
 of solutions for the boundary value problems \eqref{eq01}, \eqref{eq02} 
and \eqref{eq01}, \eqref{eq03}. We provide an example to demonstrate our results.

\section{Preliminaries}

Our goals in this section are to convert the boundary value \eqref{eq01}, 
\eqref{eq02} to a fixed point problem and to state theorems we will need 
to prove the existence and uniqueness. To this end, let $x \in C^{2}[a, b]$ 
be a solution of
\begin{gather*}
 x''(t) = f (t, x(t), x^{[2]}(t) ), \quad a < t < b,\\
 x(a) = a, \quad x(b) = b.
\end{gather*}
We begin by integrating the equation $x''(t) = f (t, x(t), x^{[2]}(t) )$ twice.
\begin{equation}\label{eq06}
 x(t) = a + x'(a)(t-a) + \int_a^t  (t-s) f (s, x(s), x^{[2]}(s) ) \, ds.
\end{equation}
After applying the boundary condition $x(b) = b$, we can solve for $x'(a)$ to obtain,
\[
 x'(a) = 1 - \frac{1}{b-a}\int_a^b  (b-s) f (s, x(s), x^{[2]}(s) ) \, ds.
\]
Now substitute this expression for $x'(a)$ into \eqref{eq06}.
\[
 x(t) = t - \frac{(t-a)}{b-a}\int_a^b  (b-s) f (s, x(s), x^{[2]}(s) ) \, ds 
+ \int_a^t  (t-s) f (s, x(s), x^{[2]}(s) ) \, ds.
\]
We can rewrite this equation in the form
\begin{align*}
 x(t) &=  t - \frac{1}{b-a} \int_t^b  (t-a)(b-s) f(s, x(s), x^{[2]}(x)) \, ds\\
 &\quad -  \frac{1}{b-a} \int_a^t  (t-a)(b-s) f(s, x(s), x^{[2]}(s)) \, ds\\
 & \quad +  \frac{1}{b-a} (t-s) f(s, x(s), x^{[2]}(s)) \, ds.
\end{align*}
Finally, we combine the last two integrals and simplify the integrand.
\begin{align*}
 x(t) & =  t + \frac{1}{b-a} \int_t^b  (t-a)(s-b) f(s, x(s), x^{[2]}(x)) \, ds\\
 & \quad +  \frac{1}{b-a} \int_a^t  (t-b)(s-a) f(s, x(s), x^{[2]}(s)) \, ds.
\end{align*}
Thus, if $x \in C^{2}[a, b]$ is a solution of
\begin{gather*}
 x''(t) = f (t, x(t), x^{[2]}(t) ), \quad a < t < b,\\
 x(a) = a, \quad x(b) = b,
\end{gather*}
then $x \in C[a, b]$ must satisfy the integral equation
\begin{equation}\label{eq07}
 x(t) = t + \int_a^b  G(t,s) f(s, x(s), x^{[2]}(s) ) \, ds, \quad a \leq t \leq b,
\end{equation}
where
\[
 G(t,s) = \frac{1}{b-a} 
\begin{cases} (t-b)(s-a), &  a \leq s \leq t \leq b,\\ 
(t-a)(s-b), &  a \leq t \leq s \leq b.
 \end{cases} 
\]

Define the operator $T_1: C[a, b] \to C[a, b]$ by
\[
 (T_1x)(t) = t + \int_a^b  G(t,s) f(s, x(s), x^{[2]}(s) ) \, ds.
\]
Note that $(T_1x)(a) = a$ and $(T_1x)(b) = b$. Also,
\begin{align*}
 (T_1x)'(t) 
& =  1 + \frac{1}{b-a} \int_a^t  (s-a) f(s, x(s), x^{[2]}(s)) \, ds\\
&  \quad - \frac{1}{b-a} \int_t^b  (b-s) f(s, x(s), x^{[2]}(s)) \, ds,
\end{align*}
and
\[
 (T_1x)''(t) = f(s, x(s), x^{[2]}(s)).
\]
Recall that in order for the solution of \eqref{eq01}, \eqref{eq02} to be 
well-defined we need $a \leq x(t) \leq b$, for all $a \leq t \leq b$. 
As such, if $x \in C[a,b]$ is a fixed point of $T_1$ such that
 $a \leq (T_1x)(t) \leq b$ for all $t \in [a,b]$, then $x$ is a solution 
of \eqref{eq01}, \eqref{eq02}. We have the following lemma.

\begin{lemma}\label{lemma01}
 The function $x$ is a solution of \eqref{eq01}, \eqref{eq02} if and only 
if $a \leq (T_1x)(t) \leq b$ and $x$ is a fixed point of $T_1$.
\end{lemma}

To establish our uniqueness results we will need the following results concerning
 the kernel of \eqref{eq07}. The proof of this lemma is straight forward and 
hence omitted.

\begin{lemma}\label{lemma02}
 The function
 \[
 G(t,s) = \frac{1}{b-a} \begin{cases} 
(t-b)(s-a), &  a \leq s \leq t \leq b,\\ 
(t-a)(s-b), &  a \leq t \leq s \leq b
 \end{cases}
 \]
 satisfies
 \begin{gather*}
 |G(t, s)| \leq |G(s, s)|, \quad t, s \in [a, b] \times [a, b], \\
 \int_a^b  |G(s, s)| \, ds = \frac{1}{6}(b-a)^2.
 \end{gather*}
\end{lemma}

We conclude this section with Schauder's fixed point theorem \cite{tab}.

\begin{theorem}[Schauder]
Let $A$ be a nonempty compact convex subset of a Banach space and let 
$T: A \to A$ be continuous. Then $T$ has a fixed point in $A$.
\end{theorem}


\section{Existence and uniqueness of solutions}

We present our main results in this section. 
From Lemma \ref{lemma01} we note that we need $a \leq (T_1x)(t) \leq b$ 
for all $t \in [a, b]$. The following condition will be used to control 
the range of $T_1x$.
\begin{itemize}
 \item[(H1)] There exists constants $K, L > 0$ such that $-K \leq f(t, u, v) \leq L$ 
for all $t \in [a, b]$, $ u, v \in \mathbb{R}$ and $1 - \frac{b-a}{2}(K+L) > 0$.
\end{itemize}
We are now ready to state our first result.

\begin{theorem}\label{thm01}
 Suppose that condition {\rm (H1)} holds. The there exists a solution of the 
boundary-value problem \eqref{eq01}, \eqref{eq02}.
\end{theorem}

\begin{proof}
 Consider the Banach space $\Phi = ( C[a, b], \| \cdot\| )$ with the norm 
defined by $\|x \| = \max_{t \in [a,b]} |x(t)|$. 
Let $m = \max \{ |a|, |b|\}$ and let 
$\Phi_m =  \{ x \in \Phi : \|x \| \leq m \}$. Since (H1) holds,
 \begin{align*}
 (T_1x)'(t) 
& =  1 + \frac{1}{b-a} \int_a^t  (s-a) f(s, x(s), x^{[2]}(s)) \, ds\\
&  \quad  -  \frac{1}{b-a} \int_t^b (b-s)f(s, x(s), x^{[2]}(s))\, ds\\
& \geq  1 - \frac{K}{b-a} \int_a^t (s-a) \, ds 
 - \frac{L}{b-a} \int_t^b  (b-s) \, ds\\
& \geq  1 - \frac{b-a}{2}(K+L) > 0.
 \end{align*}
Consequently $T_1x$ is increasing. Since $(T_1x)(a) = a$ and $(T_1x)(b) = b$, 
then $a \leq (T_1x)(t) \leq b$ for all $t \in [a, b]$. An application 
of Schauder's theorem yields a fixed point $x$ of $T_1$ and the proof is complete.
\end{proof}

 By Lemma \ref{lemma01} the function $x$ is a solution of 
\eqref{eq01}, \eqref{eq02}.

Using the same technique as in Section 2, we can show that the 
boundary-value problem \eqref{eq01}, \eqref{eq03} is equivalent to the 
integral equation
\[
 (T_2x)(t) = (b+a) - t + \int_a^b  G(t,s) f(s, x(s), x^{[2]}(s)) \, ds
\]
provided $a \leq (T_2x)(t) \leq b$.

\begin{theorem}\label{thm02}
 Suppose that condition {\rm (H1)} holds. The there exists a solution of 
the boundary-value problem \eqref{eq01}, \eqref{eq03}.
\end{theorem}

\begin{proof}
 As in the proof of Theorem \ref{thm01}, we first show that $T_2x$ is monotone.
 From condition (H1) we have
 \begin{align*}
 (T_2x)'(t) & =  -1 + \frac{1}{b-a} \int_a^t  (s-a)f(s, x(s), x^{[2]}(s)) \, ds\\
 & \quad -  \frac{1}{b-a} \int_t^b  (b-s) f(s, x(s), x^{[2]}(s)) \, ds\\
 & \leq  -1 + \frac{b-a}{2}(K+L) < 0.
 \end{align*}
 The rest of the proof is the same as in Theorem \ref{thm01}.
\end{proof}

\begin{example}\label{ex01} \rm
Consider the following boundary-value problem with parameter $k$.
 \begin{gather}
 x''(t) = k\cos(x^{[2]}(t)) \label{eq04}\\
 x(0) = 0, \, \, x (\pi ) = \pi.\label{eq05}
 \end{gather}
Here, $f(t, u, v) = k \cos v$. Since $-|k| \leq k \cos v \leq |k|$, then 
$1 - \frac{b-a}{2}(K+L) = 1 - \frac{\pi}{2} |k|$. By Theorem \ref{thm01} 
there exists a solution of \eqref{eq04}, \eqref{eq05} for all values of $k$ 
such that $|k| < \frac{2}{\pi}$.
\end{example}

We now consider uniqueness of solutions of \eqref{eq01}, \eqref{eq02} 
and \eqref{eq01}, \eqref{eq03}. To this end, we need the following condition.
\begin{itemize}
 \item[(H2)] There exists $M, N > 0$ such that 
$|f(t, u_1, v_1) - f(t, u_2, v_2)| \leq M|u_1 - u_2| + N|v_1 - v_2|$ 
for all $t \in [a, b], \, u_1, u_2, v_1, v_2 \in \mathbb{R}$.
\end{itemize}

\begin{theorem}\label{thm03}
 Suppose that {\rm (H1)} and {\rm (H2)} hold. Assume that
 \[
 \frac{1}{6}(M+N)(b - a)^2 < 1.
 \]
 Then there exists a unique solution of \eqref{eq01}, \eqref{eq02}.
\end{theorem}

\begin{proof}
 Since  (H1) holds, then there exists a fixed point $x$ of $T_1$. 
Suppose that $x_1$ and $x_2$ are two distinct fixed points of $T_1$. Then for all $t \in [a, b]$ we have,
 \begin{align*}
 |x_1(t) - x_2(t) | 
& =  |(T_1x_1)(t) - (T_1x_2)(t)|\\
& =  \Big| \int_a^b  G(t, s) \big ( f(s, x_1(s), x_1^{[2]}(s)) 
 - f(s, x_2(s), x_2^{[2]}(s)) \big) \, ds \Big|\\
& \leq  \int_a^b  |G(t,s)| \big| f(s, x_1(s), x_1^{[2]}(s)) 
 - f(s, x_2(s), x_2^{[2]}(s)) \big| \, ds\\
& \leq  \int_a^b  |G(s,s)| \Big(M |x_1(s) - x_2(s)| + N|x_1^{[2]}(s) 
 - x_2^{[2]}(s)| \Big) \, \\
& \leq  \frac{1}{6} (M+N)(b-a)^2 \|x_1 - x_2\|\\
& <  \|x_1 - x_2\|.
 \end{align*}
 Thus, $\|x_1 - x_2\| < \|x_1 - x_2\|$ and we have a contradiction. 
Consequently, the fixed point $x$ of $T_1$ is unique. 
By Lemma \ref{lemma01} $x$ is the unique solution of \eqref{eq01}, \eqref{eq02} 
and the proof is complete.
\end{proof}

In a similar manner we can prove the following theorem.

\begin{theorem}
 Suppose that {\rm (H1)} and {\rm (H2)} hold. Assume that
 \[
 \frac{1}{6}(M+N)(b - a)^2 < 1.
 \]
 Then there exists a unique solution of \eqref{eq01}, \eqref{eq03}.
\end{theorem}

\begin{example} \rm
We again consider the boundary-value problem \eqref{eq04}, \eqref{eq05},
 \begin{gather*}
 x''(t) = k\cos(x^{[2]}(t))\\
 x(0) = 0, \quad x (\pi ) = \pi.
 \end{gather*}
By the Mean Value Theorem we know there exists a $\xi \in [0, \pi]$ such that
 \[
 |k \cos v_1 - k \cos v_2| = |k| |\sin \xi| |v_1 - v_2| \leq |k| |v_1 - v_2|.
 \]
 We have $\frac{1}{6} (M + N) (b-a)^2 = |k| \pi^2/6$.
 By Theorem \ref{thm03} there exists a unique solution of \eqref{eq04}, \eqref{eq05} 
for all values of $k$ such that $|k| < 6/\pi^2$.
\end{example}

Note that the results in this paper can be extended to boundary-value problems 
of the form
\begin{gather*}
 x'' = f \big (t, x(t), x^{[2]}(t), \dots , x^{[n]}(t) \big ),\\
 x(a) = a, \quad x(b) = b,
\end{gather*}
as well as boundary-value problems of the form
\begin{gather*}
 x'' = f \big (t, x(t), x^{[2]}(t), \dots , x^{[n]}(t) \big ),\\
 x(a) = b, \quad x(b) = a.
\end{gather*}

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\end{document}