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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 94, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/94\hfil Third-order $q$-difference equations]
{Boundary-value problems for nonlinear third-order $q$-difference
equations}

\author[B. Ahmad\hfil EJDE-2011/94\hfilneg]
{Bashir Ahmad}  % in alphabetical order

\address{Bashir Ahmad \newline
Department of Mathematics,
Faculty of Science, King Abdulaziz University\\
P. O. Box 80203, Jeddah 21589, Saudi Arabia}
\email{bashir\_qau@yahoo.com}

\thanks{Submitted December 2, 2010. Published July 28, 2011.}
\subjclass[2000]{39A05, 39A13}
\keywords{$q$-difference equations; boundary value problems;
\hfill\break\indent Leray-Schauder degree theory; fixed point theorems}

\begin{abstract}
 This article shows existence results for a boundary-value problem
 of nonlinear third-order $q$-difference equations.
 Our results are based on Leray-Schauder degree theory and some
 standard fixed point theorems.
\end{abstract}

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


\section{Introduction}

The subject of $q$-difference equations, initiated in the
beginning of the 19th century
\cite{adam,car,jac,mas}, has
evolved into a multidisciplinary subject; see for example
\cite{ern,fink1,fink2,flo1,flo2,flo3,freu,gas1,han,jaul,kc}
and references therein. For some recent
work on $q$-difference equations, we refer the reader to
\cite{ann,ban,bas,dob,gas2,isma,sha}.
However, the theory of boundary-value
problems for nonlinear $q$-difference equations is still in the
initial stages and many aspects of this theory need to be
explored. To the best of our knowledge, the theory of
boundary-value problems for third-order nonlinear $q$-difference
equations  is yet to be developed.

In this paper, we discuss the existence of solutions for the
 nonlinear boundary-value problem (BVP) of third-order
$q$-difference equation
\begin{equation}\label{e1}
\begin{gathered}
D_q^3u(t)=f(t,u(t)), \quad 0\le  t \le 1,  \\
 u(0)=0,\quad D_qu(0)=0, \quad u(1)=0,
\end{gathered}
\end{equation}
where $f$ is a given continuous function.

\section{Preliminaries}

Let us recall some basic concepts of $q$-calculus \cite{gas1,kc}.

For $0<q<1$, we define the $q$-derivative of a real valued
function $f$ as
$$
D_qf(t) = \frac{f(t)-f(qt)}{(1-q)t}, \quad
D_qf(0)=\lim_{t \to 0}D_qf(t).
$$
Higher order $q$-derivatives are given by
$$
D^0_q f(t) = f(t),  \quad
D^n_q f(t) = D_q D^{n-1}_q f(t), \quad n \in \mathbb{N}.
$$
 The $q$-integral of a function $f$ defined
in the interval $[a, b]$ is given by
$$
\int_{a}^{x}f(t)d_q t:=\sum_{n=0}^{\infty}x(1-q)q^nf(xq^n)-a f(q^n a),
\quad x \in [a, b],
$$
and for $a=0$, we denote
$$
I_qf(x)=\int_{0}^{x}f(t)d_q
t=\sum_{n=0}^{\infty}x(1-q)q^nf(xq^n),
$$
provided the series converges. If $a\in [0, b]$ and $f$
is defined on the interval $[0, b]$, then
$$
\int_{a}^{b}f(t)d_q t=\int_{0}^{b}f(t)d_q t-\int_{0}^{a}f(t)d_q t.
$$
Similarly,  we have
$$
I^0_q f(t) = f(t),  \quad
I^n_q f(t) = I_q I^{n-1}_q f(t), \quad n \in \mathbb{N}.
$$
Observe that
\begin{equation}\label{eFT}
D_qI_q f(x) = f(x),
\end{equation}
 and if $f$ is continuous at $x = 0$, then
 $I_qD_q f(x) = f(x) - f(0)$.
In $q$-calculus, the product rule and integration by parts formula
are
\begin{gather}\label{epr}
 D_q(gh)(t)=D_qg(t)h(t)+g(qt)D_qh(t), \\
\label{eIP}
 \int_0^x f(t)D_q g(t)dqt = \Big[f(t)g(t)\Big]_0^x- \int_0^x D_q f(t)
 g(qt)d_q t.
\end{gather}
In the limit $q \to 1$ the above results correspond to their
counterparts in standard calculus.

Motivated by the solution of a classical third-order ordinary
differential equation (see Remark \ref{rmk1}), we can write the solution
of the third-order $q$-difference equation $D_q^3u(t)=v(t)$ in
the  form
\begin{equation}\label{e222}
u=\int_{0}^{t}\Big(\alpha_1(q)t^2+\alpha_2(q)ts
 +\alpha_3(q)s^2\Big)v(s)d_qs+a_0 +a_1t+a_2t^2,
\end{equation}
where $a_0, a_1, a_2$ are arbitrary constants and $\alpha_1(q),
\alpha_2(q), \alpha_3(q)$ can be fixed appropriately.

Choosing $\alpha_1(q)=1/(1+q)$, $\alpha_2(q)=-q$,
$\alpha_3(q)=q^3/(1+q)$ and using \eqref{eFT} and \eqref{epr}, we
find that
$$
D_qu(t)=\int_{0}^{t}tv(s)d_qs-\int_{0}^{t}qsv(s)d_qs,\quad
D^2_qu(t)=\int_{0}^{t}v(s)d_qs, \quad
D_q^3u(t)=v(t).
$$
Thus, the solution \eqref{e222} of
$D_q^3u(t)=v(t)$ takes the form
\begin{equation}\label{e2222}
u=\int_{0}^{t}\Big(\frac{t^2+q^3s^2}{1+q}-qts\Big)v(s)d_qs+a_0
+a_1t+a_2t^2.
 \end{equation}

\begin{lemma} \label{lem1}
The BVP \eqref{e1} is equivalent to the integral equation
\begin{equation}\label{e23}
u=\Gamma u,
\end{equation}
where
$$
\Gamma u=\int_{0}^{1}G(t,s;q)f(s, u(s))d_qs,
$$
and $G(t,s;q)$ is the Green's function given by
\begin{equation}\label{e24}
G(t,s;q)= \frac{1}{(1+q)}
\begin{cases}
qs(1-t)[q^2s(1+t)-(1+q)t],  &0 \le s < t\le 1, \\
t^2(1-qs)(q^2s-1),   &0\le t\le s\le 1.
\end{cases}
\end{equation}
\end{lemma}

\begin{proof}  In view of \eqref{e2222},  the solution of
$D_q^3u=f(t,u)$ can be written as
\begin{equation}\label{e25}
u=\int_{0}^{t}\Big(\frac{t^2+q^3s^2}{1+q}-qts\Big)f(s,
u(s))d_qs+a_0 +a_1t+a_2t^2,
 \end{equation}
where $a_1, a_2, a_2$ are arbitrary constants. Using the boundary
conditions of \eqref{e1} in \eqref{e25}, we find that $a_0=0,
a_1=0$ and
\[
 a_2=-\int_{0}^{1}\Big(\frac{1+q^3s^2}{1+q}-qs\Big)f(s,
u(s))d_qs.
\]
Substituting the values of  $a_0, a_1$ and $a_2$ in \eqref{e25},
we obtain
\begin{align*}
u&=  \int_{0}^{t}\Big(\frac{t^2+q^3s^2}{1+q}-qts\Big)f(s,
u(s))d_qs-t^2\int_{0}^{1}\Big(\frac{1+q^3s^2}{1+q}-qs\Big)f(s,
u(s))d_qs\\
&=   \int_{0}^{1}G(t,s;q)f(s, u(s))d_qs,
\end{align*}
where $G(t,s;q)$ is  given by \eqref{e24}.
\end{proof}

We define
\begin{equation}\label{e230}
G_1=\max_{t \in [0,1]}\big|\int_0^1 G(t,s;q) d_qs\big|
=\frac{(1+q)q^2}{(1+q+q^2)^4}.
\end{equation}

\begin{remark}\label{rmk1} \rm
For $q \to 1$, equation \eqref{e25} takes the form
$$
u=\frac{1}{2}\int_{0}^{t}(t-s)^2f(s, u(s))ds+a_0 +a_1t+a_2t^2,
$$
which is the solution of a classical third-order ordinary
differential equation $u'''(t)=f(t,u(t))$ and the associated form
of Green's function for the classical case is
\[
G(t,s)=\frac{1}{2} \begin{cases}
s(1-t)[s(1+t)-2t], & \text{if }0\le s<t\le 1, \\
-t^2(1-s)^2, & \text{if } 0\le t\le s\le 1.
\end{cases}
\]
\end{remark}

\section{Some existence results}

\begin{theorem}\label{thm1}
 Assume that there exist constants $M_1\ge 0$ and  $M_2>0$
such that  $M_1 G_1 < 1$ and $|f(t,u)| \le M_1|u| +M_2$ for all
$t \in [0,1], u \in  C([0,1])$, where $G_1$
is given by  \eqref{e230}. Then the BVP \eqref{e1} has at least
one solution.
\end{theorem}

\begin{proof}
In view of Lemma \ref{lem1}, we just need to prove the
existence of at least one solution $u \in C([0,1])$ such that
$u =\Gamma u$. Thus, it is sufficient to show that $\Gamma :
\overline{B}_R \to C([0,1])$ satisfies
\begin{equation}\label{e31}
u \ne \lambda \Gamma u, \quad \forall u \in \partial B_R
\quad  \forall \lambda \in [0,1],
 \end{equation}
where $B_R \subset C([0,1])$ is a suitable ball with radius $R>0$.
Let us define
$$
H(\lambda, u)=\lambda \Gamma u, \quad
u \in C([0,1]),\;\lambda \in [0,1].
$$
Then, by Arzela-Ascoli theorem,
 $h_\lambda(u)=u-H(\lambda, u)=u-\lambda \Gamma u$
is completely continuous. If \eqref{e31}  is
true, then the following Leray-Schauder degrees are well defined
and by the homotopy invariance of topological degree, it follows
that
\begin{align*}
\deg(h_\lambda, B_R, 0)
&=\deg(I-\lambda \Gamma, B_R,0)
 =\deg(h_1, B_R, 0)\\
&=\deg(h_0, B_R, 0) =\deg(I, B_R, 0)=1\ne 0,
\quad 0 \in B_r,
\end{align*}
where  $I$ denotes the unit operator. By the nonzero
property of Leray-Schauder degree,
$h_1(t)=u- \lambda \Gamma u=0$ for at least one $u \in B_R$.
Let us set
$$
B_R=\{u \in C([0,1]) : \max_{t \in [0,1]}|u(t)|<R\},
$$
where $R$ will be fixed later. In order to prove \eqref{e31},  we
assume that $u = \lambda \Gamma u$ for some $\lambda \in [0,1]$
and for all $t \in [0,1]$ so that
\begin{align*}
|u(t)|
&= | \lambda \Gamma u(t)|\leq
 \big| \int_0^1|G(t,s;q)f(s, u(s))d_qs\big|\\
&\leq  \big| \int_0^1G(t,s;q)(M_1|u(s)| +M_2)d_qs\big|\\
&\leq  (M_1\|u\|+M_2)\max_{t \in [0,1]}\big|\int_0^1G(t,s;q)d_qs\big|\\
&\leq  (M_1\|u\|+M_2)G_1,
\end{align*}
which implies
$$
\|u\| \le  \frac{M_2G_1}{1-M_1G_1}.
$$
Letting $R=\frac{M_2G_1}{1-M_1G_1}+1$, \eqref{e31}  holds. This
completes the proof.
\end{proof}

\begin{example} \label{examp3.1} \rm
Consider the following problem
\begin{equation}\label{examp1}
\begin{gathered}
D_{1/2}^3u(t)=\frac{M_1}{(2 \pi)}\sin(2 \pi u)+\frac{|u|}{1+|u|},
\quad  0\le  t \le 1,  \\
 u(0)=0,\quad D_{1/2}u(0)=0, \quad u(1)=0.
\end{gathered}
\end{equation}
Here $q=1/2$ and  $M_1$ will be fixed later. Observe that
$$
|f(t,u)|=\big|\frac{M_1}{(2 \pi)}\sin(2 \pi u)+\frac{|u|}{1+|u|}\big|
 \le M_1|u|+1,
$$
and
$$
G_1=\frac{q^2(1+q)}{(1+q+q^2)^4}\Big|_{q=1/2}=\frac{96}{2401}.
$$
Clearly $M_2=1$ and and we can choose $M_1 <
\frac{1}{G_1}=\frac{2401}{96}$; that is, $M_1\le 25$. Thus,
Theorem \ref{thm1} applies to the problem \eqref{examp1}.
\end{example}


To prove the next existence result, we need the following known
fixed point theorem \cite{alt}.

\begin{theorem}\label{thm2}
 Let $\Omega$ be an open bounded subset
of a Banach space $E$ with $0 \in \Omega$ and $B:
\overline{\Omega} \to E$ be a compact operator. Then $B$ has a
fixed point in $\overline{\Omega}$ provided $\|Bu-u\|^2 \ge
\|Bu\|^2-\|u\|^2, ~ u \in \partial \Omega$.
\end{theorem}

\begin{theorem}\label{thm3}
If there exists  a constant $M_3$ such that
$$
|f(t,u)| \le  \frac{M_3}{G_1},  \quad\forall t \in [0,1],\;
u \in [-M_3, M_3],
$$
where $G_1$ is given by  \eqref{e230}. Then  \eqref{e1}
has at least one solution.
\end{theorem}

\begin{proof}
Let us define $B_{M_3} =\{u \in C([0,1])
: \max_{t\in [0,1]}|u(t)| <M_3\}$. In view of Theorem \ref{thm2},
we just need to show that
\begin{equation}\label{e32}
\|\Gamma u\| \le \|u\|, \quad \forall u \in \partial B_{M_{3}}.
 \end{equation}
For all $t \in [0,1]$, $u \in \partial B_{M_{3}}$, we have
\begin{align*}
|\Gamma u(t)|= \big|\int_0^1G(t,s;q)f(s, u(s))d_qs\big| \le
 \frac{M_3}{G_1}\big|\int_0^1 G(t,s;q)d_qs \big|\le
M_3.
\end{align*}
Thus \eqref{e32} holds, which completes the proof.
\end{proof}

In view of the assumption $|f(t,u)| \le
M_1|u| +M_2$ of Theorem \ref{thm1}, we find that
$M_3=M_2G_1(1-M_1G_1)^{-1}$.

\begin{theorem}\label{thm4}
 Suppose that $f$ is of class $C^1$
in the second variable and  there exists a constant
 $0 \le M_4 < \frac{1}{G_1}$ ($G_1$ is given by  \eqref{e230})
such that $|f_u(t,u)| \le M_4$ for all $t \in [0,1], u \in C([0,1])$,
then \eqref{e1} has at least one solution.
\end{theorem}

\begin{proof} For all $t \in [0,1]$, we find that
\begin{align*}
|\Gamma u(t)|
&=   \Big|\int_0^1 G(t,s;q) f(s, u(s))d_qs\Big|
\le  \Big|\int_0^1 G(t,s;q)(f_u(s,u(s))u(s)+\nu)d_qs\Big|\\
&\leq   \Big|\int_0^1 G(t,s;q)d_qs\Big|(M_4 \|u\|+\nu) \le M_4 G_1
\|u\|+\nu_1,
\end{align*}
where $\nu_1=G_1 \nu$ ($\nu$ is a positive constant). For $R>0$,
we define
 $$
B_R=\{u \in C([0,1]): \max_{t \in [0,1]}|u(t)|<R \},
$$
so that
 $$
\|\Gamma u\| \le M_4 G_1 R + \nu_1
=R \Big(M_4 G_1+\frac{\nu_1}{R}\Big) \le R,
$$
for sufficiently large $R$. Therefore, by Schauder fixed point
theorem,  $\Gamma$ has a fixed point. This completes the proof.
 \end{proof}


\begin{example} \label{examp3.2}\rm
 Consider the problem
\begin{equation}\label{examp2}
\begin{gathered}
D_{\frac{1}{4}}^3u(t)=\frac{1}{12
}\big(\frac{1-u^2}{1+u^2})\sin(2 \pi t), \quad 0\le  t \le 1,  \\
 u(0)=0, \quad D_{\frac{1}{4}}u(0)=0, \quad u(1)=0.
\end{gathered}
\end{equation}
Clearly $f(t,u)= \frac{1}{12 }\big(\frac{1-u^2}{1+u^2})\sin(2 \pi
t)$ and
$$
G_1=\frac{q^2(1+q)}{(1+q+q^2)^4}\Big|_{q=1/4}=\frac{5120}{194481}.
$$
Furthermore,
$$
|f_u(t,u)| \le \frac{1}{3}\Big(\frac{|u|}{(1+u^2)^2}\Big)
 < \frac{1}{G_1}=\frac{194481}{5120}.
$$
Thus, by Theorem \ref{thm4}, there exists one solution for
problem \eqref{examp2}.
\end{example}

Our final result deals with  the uniqueness of solutions to
\eqref{e1}.

\begin{theorem}\label{thm5}
Let $f : [0,1]\times \mathbb{R} \to \mathbb{R}$ be a
jointly continuous function satisfying the condition
$$
|f(t,u)-f(t,v)| \le L |u-v|, \quad \forall t \in [0,1], \;
u, v\in \mathbb{R},
$$
where $L$ is a Lipschitz constant.  Then \eqref{e1} has a unique
solution provided that $L < 1/G_1$, where $G_1$ is given
by \eqref{e230}.
\end{theorem}

\begin{proof}
 For $t \in [0,1]$, we define
$\Gamma : C([0,1], \mathbb{R}) \to C([0,1], \mathbb{R})$
by
$$
\Gamma u=\int_{0}^{1}G(t,s;q)f(s, u(s))d_qs,
$$
where
$G(t,s;q)$ is the Green's function given by \eqref{e24}.

  Let us set $M=\max_{t \in [0,1]}|f(t,0)|$ and choose
\begin{equation}\label{e33}
r \ge \frac{MG_1}{1-L G_1}.
 \end{equation}
 Now we show that
$\Gamma B_r \subset B_r$, where $B_r=\{u \in C([0,1],
\mathbb{R}): \|u\|\le r \}$. For $u \in B_r$, we have
\begin{align*}
\|\Gamma u\|
&=  \max_{t \in [0,1]}\big|
\int_{0}^{1}G(t,s;q)f(s, u(s))d_qs\big|\\
&=  \max_{t \in [0,1]}\big| \int_{0}^{1}G(t,s;q)[(f(s,u(s))
 -f(s,0))+f(s,0)]d_qs\big|\\
&\leq  \max_{t \in [0,1]}\big |\int_{0}^{1}G(t,s;q)d_qs\big|
 (L\|u\|+M|)\\
&\leq  G_1(Lr+M) \le  r.
 \end{align*}
where we have used \eqref{e33}.
Now, for $u, v\in \mathbb{R}$ and for each $t \in [0,1]$, we obtain
\begin{align*}
\|(\Gamma u)-(\Gamma v)\|
&= \max_{t \in [0, 1]} |(\Gamma u)(t)-(\Gamma v)(t)|\\
&\leq  \max_{t \in [0, 1]}\big|
\int_{0}^{1}G(t,s;q)[f(s,u(s))-f(s,v(s))]d_qs \big|\\
&\leq  L\max_{t \in [0, 1]}\big|
\int_{0}^{1}G(t,s;q)d_qs \big|\|u-v\|
\\
&\leq  L G_1 \|u-v\|.
\end{align*}
As $L<1/G_1$, therefore
$\Gamma$ is a contraction. Thus, the conclusion of the theorem
follows by the contraction mapping principle. This completes the
proof.
\end{proof}


\begin{example} \label{examp3.3}\rm  Consider
\begin{equation}\label{examp3}
\begin{gathered}
 D_{\frac{3}{4}}^3u(t)=L \big(\cos t+ \tan^{-1}u\big), \quad
 0\le  t \le 1,  \\
 u(0)=0,\quad D_{\frac{3}{4}}u(0)=0, \quad u(1)=0.
\end{gathered}
\end{equation}
With  $  f(t,u)=L (\cos t+ \tan^{-1}u)$, we find that
$$
|f(t,u)-f(t,v)| \le L|\tan^{-1}u-\tan^{-1}v| \le L|u-v|
$$
and
$$
G_1=\frac{q^2(1+q)}{(1+q+q^2)^4}\Big|_{q=3/4}
=\frac{64512}{1874161}.
$$
Fixing $L < \frac{1}{G_1}=\frac{1874161}{64512}$,
it follows by Theorem \ref{thm5} that  \eqref{examp3}
has a unique solution.
\end{example}


 \begin{remark} \label{rmk2} \rm
In the limit as $q \to 1$, our results reduce to the
ones for the classical third-order boundary-value problem
\begin{gather*}
u'''(t)=f(t,u(t)) \quad t \in [0,1]\\
u(0)=0, \quad u'(0)=0, \quad u(1)=0.
\end{gather*}
\end{remark}

\subsection*{Acknowledgments}
The author wants to thank anonymous  referees
for their useful comments.

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