Fractals
©World Scientific Publishing Company

A DIFFERENCE METHOD FOR SOLVING THE NONLINEAR $q$-FACTIONAL DIFFERENTIAL
EQUATIONS ON TIME SCALES

TIE ZHANG***Corresponding author and CAN TONG

Department of Mathematics and the State Key Laboratory
of Synthetical Automation for Process Industries
Northeastern University, Shenyang 110004, China
<EMAIL_ADDRESS>

The $q$-fractional differential equation usually describe the physics process
imposed on the time scale set $T_{q}$. In this paper, we first propose a
difference formula for discretizing the fractional $q$-derivative
${}^{c}D_{q}^{\alpha}x(t)$ on the time scale set $T_{q}$ with order
$0<\alpha<1$ and scale index $0<q<1$. We establish a rigours truncation error
boundness and prove that this difference formula is unconditionally stable.
Then, we consider the difference method for solving the initial problem of
$q$-fractional differential equation: ${}^{c}D_{q}^{\alpha}x(t)=f(t,x(t))$ on
the time scale set. We prove the unique existence and stability of the
difference solution and give the convergence analysis. Numerical experiments
show the effectiveness and high accuracy of the proposed difference method.

Keywords: The fractional $q$-derivative; Difference formula; Truncation error;
Unconditional stability; The $q$-fractional differential equation; Convergence
analysis.

## 1\. INTRODUCTION

In recent years, the $q$-calculus (also called quantum calculus) and
$q$-fractional calculus have attracted much attention in mathematical and
physical fields, such as number theory, special functions, basic hyper-
geometric functions, operator theory, quantum dynamics and stochastic analysis
[1, 2, 3, 4, 5, 6, 7]. The $q$-calculus is a class of calculus defined on a
$q$-geometry set $T_{q}$ where $q$ is the scale index of set $T_{q}$. The
$q$-fractional differential equations usually describe some physics processes
imposed on one time scale set $T_{q,b}=\\{q^{n}b,n=0,1,\cdots\\}\cup\\{0\\}$
where $q$ is used to indicate the discrete time path on which the
corresponding physical process varies. The $q$-derivative was originally
presented by Jackson [8] in 1908 and the $q$-integral originated from a
$q$-analogue of the usual integral proposed by Al-Salam [9] in 1966. Following
Al-Salam’s work, Agarwal [10] further gave the definition of $q$-fractional
calculus. We refer the readers to [1, 2, 4, 11, 12, 13] for some basic
introduction on $q$-calculus and $q$-fractional calculus. Compared with the
classical fractional calculus, the study of $q$-fractional calculus is still
immature.

The existence of solutions for the Caputo type $q$-fractional initial value
and boundary value problems has been studied by many researches, see [14, 15,
16, 17, 19]. By using the Lyapunov’s direct method, Jarad et. al. [6] studied
the stability of Caputo type $q$-fractional non-autonomous systems. Wang et.
al. [20] investigated the existence of extremal solutions of the Caputo type
$q$-fractional differential equation. Zhang et. al. [18, 19] gave new
existence theorems for the initial value problem of $q$-fractional
differential equation under different conditions of source term functions. On
the discrete or numerical methods, Abdeljawad and Baleanu [21] first presented
the successive approximation method to obtain an approximate solution of the
Caputo type $q$-fractional differential equation. Then, Salahshour et. al.
[22] investigated the convergence of the successive approximation method
presented by Abdeljawad. Wu and Baleanu [23] used the variational iteration
method and the Lagrange multiplier method to study the Caputo type
$q$-fractional initial value problem. Recently, Zhang and Tang [24]
established a difference method for solving the initial problem of the
$q$-fractional differential equation, but no stability and convergence
analysis are given for this method. At present, only a few discrete or
approximate methods are proposed for the $q$-fractional problems. In
particular, almost no difference method is analyzed theoretically for solving
the $q$-fractional differential equations.

In this article, we first present a difference formula (called the $L_{1,q}$
formula) to discretize the fractional $q$-derivative
${}^{c}D_{q}^{\alpha}x(t)$ of Caputo type on the time scale set
$T_{q,b}=\\{q^{n}b,n=0,1,\cdots\\}\cup\\{0\\}$. Different from the standard
difference formula, we must establish the difference formula on set $T_{q,b}$
, that is, the mesh points $\\{t_{k}\\}$ should be in set $T_{q,b}$. Thus, the
mesh partition with nodes $\\{t_{k}=bq^{k},\,k=0,\cdots,N\\}$ is nonuniform
and the maximal mesh step does not vanish as $N\rightarrow\infty$. This makes
the error analysis and stability estimate much more difficult than that of the
standard difference methods which can be established on any proper meshes, for
example, the uniform mesh. Our difference formula is constructed by using the
piecewise linear interpolation to approximate the integrand function on set
$T_{q,b}$. The coefficients of this formula are analyzed in detail. We derive
the truncation error boundness and prove that this difference formula is
unconditionally stable. Then, using this difference formula, we establish a
difference method for solving the initial problem of $q$-fractional
differential equation: ${}^{c}D_{q}^{\alpha}x(t)=f(t,x(t)),t\in T_{q,b}$. We
prove that this difference method is stale and give an error estimation of
$\triangle t_{n}^{2}$-order. Numerical experiments show the high accuracy and
effectiveness of this difference formula. To the authors’ best knowledge, it
is the first time that an unconditionally stable difference formula is
presented and analyzed for the $q$-fractional problems. Our work provides a
numerical approach for solving the $q$-fractional problems.

This paper is organized as follows. In Section 2, we introduce some basic
notations and operations on $q$-fractional calculus. In Section 3, we present
the $L_{1,q}$ difference formula and derive the truncation error boundness.
Section 4 is contributed to the stability analysis of this difference formula.
In Section 5, this difference formula is used to solve the initial value
problem of $q$-fractional differential equation and the unique existence of
the difference solution, stability and error estimation are given. Numerical
experiments are provided in Section 6. Section 7 gives some conclusions.

## 2\. THE $q$-CALCULUS AND $q$-FRACTIONAL CALCULUS

In this section, we introduce some notations and operations on the
$q$-calculus and the $q$-fractional calculus.

Let $\mathbb{N}=\left\\{1,2,3,...\right\\}$ be the positive integer set.
Define the $q$-shifted factorial operation by

$(t-s)^{(0)}=1,(t-s)^{(k)}=\prod\limits_{i=0}^{k-1}(t-q^{i}s),\,k\in\mathbb{N}.$
(1)

If $\alpha\in R$ and $\pm\alpha\notin\mathbb{N}$, then

$(t-s)^{(\alpha)}=t^{\alpha}\prod\limits_{i=0}^{\infty}\frac{t-q^{i}s}{t-q^{\alpha+i}s},\,0\leq
s\leq t.$ (2)

Let $\mathbb{C}$ be the complex set. For $0<q<1$ and
$\alpha\in\mathbb{C}\setminus\\{-n:n\in\mathbb{N}\cup\\{0\\}\\}$, the
$q$-gamma function $\Gamma_{q}(\alpha)$ is defined by

$\Gamma_{q}(\alpha)=(1-q)^{(\alpha-1)}(1-q)^{1-\alpha}.$ (3)

Introduce the notations

$[\alpha]_{q}=\frac{1-q^{\alpha}}{1-q},\;\;[n]_{q}!=[n]_{q}[n-1]_{q}\cdots[1]_{q}.$

By a straightforward computation, it holds that

$\Gamma_{q}(1)=1,\;\Gamma_{q}(\alpha+1)=[\alpha]_{q}\Gamma_{q}(\alpha),\;\Gamma_{q}(n+1)=[n]_{q}!\,.$

For $\mathrm{Re}(\alpha)>0$, $\mathrm{Re}(\beta)>0$, the $q$-beta function is
defined by

$B_{q}(\alpha,\beta)=\int_{0}^{1}t^{\alpha-1}(1-qt)^{(\beta-1)}d_{q}t,$ (4)

where the $q$-integral $\int\,d_{q}t$ is defined by (5)–(6). The relation
between the $q$-beta function and the $q$-gamma function is given as follows
[25].

$B_{q}(\alpha,\beta)=\frac{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)}{\Gamma_{q}(\alpha+\beta)}.$

For a fixed point $q\in R$, a subset $A\subset R$ is called $q$-geometric if
$qt\in A$ whenever $t\in A$. It is easy to see that $\forall\,t\in A$, $A$
contains all geometric sequences $\\{tq^{n}\\}_{n=0}^{\infty}$. A typical
$q$-geometry set is the time scale set $T_{q}$ defined by
$T_{q}=\left\\{q^{n}:n\in\mathbb{Z}\right\\}\cup\left\\{0\right\\}$, where
$0<q<1$, $\mathbb{Z}=\left\\{0,\pm 1,\pm 2,...\right\\}$.
Definition 1. [8] Let $f(t)$ be a real valued function on set $A$, $0<q<1$.
The $q$-derivative of $f(t)$ is defined by

$\displaystyle D_{q}f(t)$ $\displaystyle=$
$\displaystyle\frac{d_{q}f(t)}{d_{q}t}=\frac{f(qt)-f(t)}{(q-1)t},\,\,t\in
A\backslash\left\\{0\right\\},$ $\displaystyle D_{q}f(0)$ $\displaystyle=$
$\displaystyle\frac{d_{q}f(t)}{d_{q}t}|_{t=0}=\lim_{n\to\infty}\frac{f(tq^{n})-f(0)}{tq^{n}},\,\,0<q<1,\,t\neq
0.$

The $q$-derivative is a discrete analogy of the classical derivative.
Obviously, for a function $f(t)$ defined on $R$, its $q$-derivative
$D_{q}f(t)$ always exists for $t\not=0$ and $D_{q}f(0)$ is well defined if the
classical derivative $f^{\prime}(0)$ exists.

The high order $q$-derivative $D_{q}^{n}f(t)$ is defined by
$D_{q}^{n}f(t)=D_{q}(D^{n-1}_{q}f(t)),\,n\geq 2$.

Let $f(t)$ and $g(t)$ be real valued functions on $R$ and both $f(t)$ and
$g(t)$ are differentiable at $t=0$. Then, the following $q$-derivative
operation rules hold [1]

$\displaystyle D_{q}\left(af(t)\pm bg(t)\right)=aD_{q}f(t)\pm
bD_{q}g(t),\,a,b\in R,$ $\displaystyle
D_{q}\left(f(t)g(t)\right)=g(t)D_{q}f(t)+f(qt)D_{q}g(t),$ $\displaystyle
D_{q}\big{(}\frac{f(t)}{g(t)}\big{)}=\frac{g(t)D_{q}f(t)-f(t)D_{q}g(t)}{g(t)g(qt)},\,g(t)\not=0,\,g(qt)\not=0.$

Definition 2. [26] The $q$-integral on interval $(a,b)$ is defined by

$\int_{a}^{b}f(t)d_{q}t=\int_{0}^{b}f(t)d_{q}t-\int_{0}^{a}f(t)d_{q}t,\,\,a,\,b\in
A,$ (5)

where

$\int_{0}^{x}f(t)d_{q}t=(1-q)\sum_{n=0}^{\infty}xq^{n}f(xq^{n}),\,\,x\in A.$
(6)

By this definition, it is easy to see that

$\displaystyle\Big{|}\int_{0}^{b}f(t)d_{q}t\Big{|}\leq\int_{0}^{b}|f(t)|d_{q}t,\;0<b.$
(7)
$\displaystyle\int_{a}^{b}f(t)d_{q}t=\int_{a}^{c}f(t)d_{q}t+\int_{c}^{b}f(t)d_{q}t,\;a<c<b.$
(8)

The following lemma gives the $q$-integration by parts rule.
Lemma 1. Let $f(t)$ and $g(t)$ are real valued functions on set $A$, $0\leq
a<b,\,a,b\in A$ and $0<q<1$, then

$\displaystyle\int_{a}^{b}g(t)D_{q}f(t)d_{q}t=(fg)(b)-(fg)(a)-\int_{a}^{b}f(qt)D_{q}g(t)d_{q}t,$
(9)
$\displaystyle\int_{a}^{b}g(qt)D_{q}f(t)d_{q}t=(fg)(b)-(fg)(a)-\int_{a}^{b}f(t)D_{q}g(t)d_{q}t.$
(10)

Proof. We only prove conclusion (9), the proof of conclusion (10) is
completely similar. When $a=0$, we have from (6) that

$\displaystyle\int_{0}^{b}g(t)D_{q}f(t)d_{q}t$
$\displaystyle=(1-q)\sum_{n=0}^{\infty}bq^{n}g(bq^{n})D_{q}f(bq^{n})$
$\displaystyle=(1-q)b\sum_{n=0}^{\infty}q^{n}g(bq^{n})\frac{f(bq^{n})-f(bq^{n+1})}{(1-q)bq^{n}}$
$\displaystyle=\sum_{n=0}^{\infty}g(bq^{n})\left(f(bq^{n})-f(bq^{n+1})\right),$

and

$\displaystyle\int_{0}^{b}f(qt)D_{q}g(t)d_{q}t$
$\displaystyle=(1-q)\sum_{n=0}^{\infty}bq^{n}f(bq^{n+1})D_{q}g(bq^{n})$
$\displaystyle=(1-q)b\sum_{n=0}^{\infty}q^{n}f(bq^{n+1})\frac{g(bq^{n})-g(bq^{n+1})}{(1-q)bq^{n}}$
$\displaystyle=\sum_{n=0}^{\infty}f(bq^{n+1})\left(g(bq^{n})-g(bq^{n+1})\right).$

So

$\displaystyle\int_{0}^{b}g(t)D_{q}f(t)d_{q}t+\int_{0}^{b}f(qt)D_{q}g(t)d_{q}t=\sum_{n=0}^{\infty}\Big{(}f(bq^{n})g(bq^{n})-f(bq^{n+1})g(bq^{n+1})\Big{)}$
$\displaystyle=f(b)g(b)-\lim_{n\to\infty}(fg)(bq^{n+1})=(fg)(b)-(fg)(0).$ (11)

When $a>0$, we have from (5) and (11)

$\displaystyle\int_{a}^{b}g(t)D_{q}f(t)d_{q}t$ $\displaystyle=$
$\displaystyle\int_{0}^{b}g(t)D_{q}f(t)d_{q}t-\int_{0}^{a}g(t)D_{q}f(t)d_{q}t$
$\displaystyle=$
$\displaystyle(fg)(b)-(fg)(0)-\int_{0}^{b}f(qt)D_{q}g(t)d_{q}t$
$\displaystyle-(fg)(a)+(fg)(0)+\int_{0}^{a}f(qt)D_{q}g(t)d_{q}t$
$\displaystyle=$
$\displaystyle(fg)(b)-(fg)(a)-\int_{a}^{b}f(qt)D_{q}g(t)d_{q}t.$

The proof is completed. $\hfill\Box$

Now we introduce the $q$-fractional calculus.
Definition 3. [3] Let $t\in A$, $\alpha\neq-1,-2,...$ and $a\geq 0$. The
$\alpha$-order $q$-fractional integral of the Riemann-Liouville type with the
lower limit point $a$ is defined by $I_{q,a}^{0}f(t)=f(t)$ and

$I_{q,a}^{\alpha}f(t)=\frac{1}{\Gamma_{q}(\alpha)}\int_{a}^{t}(t-qs)^{(\alpha-1)}f(s)d_{q}s.$
(12)

Definition 4. [27] Let $a\geq 0,a\in A,n=\lceil\alpha\rceil$. The
$\alpha$-order Riemann-Liouville type fractional $q$-derivative of function
$f(t):(a,\infty)\rightarrow\mathrm{R}$ is defined by

$\displaystyle
D_{q,a}^{\alpha}f(t)=\left\\{\begin{array}[]{lll}&I_{q,a}^{-\alpha}f(t),&\alpha\leq
0,\\\ &D_{q}^{n}I_{q,a}^{n-\alpha}f(t),&\alpha>0,\end{array}\right.$ (15)

where $\lceil\alpha\rceil$ denotes the smallest integer that is greater or
equal to $\alpha$.
Definition 5. [27] Let $a\geq 0,\,a\in A,\,n=\lceil\alpha\rceil$. The
$\alpha$-order Caputo type fractional $q$-derivative of function
$f(t):(a,\infty)\rightarrow\mathrm{R}$ is defined by

${}^{c}D_{q,a}^{\alpha}f(t)=\left\\{\begin{array}[]{lll}&I_{q,a}^{-\alpha}f(t),&\alpha\leq
0,\\\ &I_{q,a}^{n-\alpha}D_{q}^{n}f(t),&\alpha>0.\end{array}\right.$ (18)

For simplicity, we often use the notations $I_{q}^{\alpha}f(t)$ instead of
$I_{q,0}^{\alpha}f(t)$, $D_{q}^{\alpha}f(t)$ instead of $D_{q,0}^{\alpha}f(t)$
and ${}^{c}D_{q}^{\alpha}f(t)$ instead of ${}^{c}D_{q,0}^{\alpha}f(t)$,
respectively.

Under a certain condition, the following relations between the Caputo type
fractional $q$-derivative and the Riemann-Liouville type fractional
$q$-derivative hold [1, 21]

${}^{c}D_{q}^{\alpha}f(t)=D_{q}^{\alpha}\Big{(}f(t)-\sum_{j=0}^{n-1}\frac{D_{q}^{j}f(0)}{\Gamma_{q}(j+1)}t^{j}\Big{)},\;n=\lceil\alpha\rceil,\,\alpha>0.$
(19)

In particular, for $0<\alpha<1$,

${}^{c}D_{q}^{\alpha}f(t)=D_{q}^{\alpha}f(t)-\frac{t^{-\alpha}}{\Gamma_{q}(1-\alpha)}f(0).$
(20)

## 3\. THE $L_{1,q}$ DIFFERENCE FORMULA

In this section, we establish a difference formula (called the $L_{1,q}$
formula) for discretizing the Caputo type fractional $q$-derivative
${}^{c}D_{q}^{\alpha}x(t)$ with $0<\alpha,q<1$ and give the truncation error
boundness.

Let us consider the Caputo type fractional $q$-derivative:

${}^{c}D_{q}^{\alpha}x(t)=\frac{1}{\Gamma_{q}(1-\alpha)}\int_{0}^{t}(t-qs)^{(-\alpha)}D_{q}x(s)d_{q}s,\,0<\alpha,q<1,\;0<t\leq
b,\,t\in T_{q,b}.$ (21)

We need to construct a difference formula to approximate
${}^{c}D_{q}^{\alpha}x(t)$ on the discrete point set $\\{t_{k}\\}\subset
T_{q,b}$ where the time scale set
$T_{q,b}=\\{bq^{n}:n=0,1,\dots\\}\bigcup\\{0\\}$.

Let $0=t_{0}<t_{1}<\cdots<t_{N}=b$ be a nonuniform partition of $[0,b]$ with
mesh point $t_{k}=bq^{N-k}\in T_{q,b}$ and step sizes $\Delta
t_{k}=t_{k}-t_{k-1},\,1\leq k\leq N$ where $N\geq 1$ is a positive integer.
Introduce the piecewise linear interpolation of function $x(s)$:

$L_{1,k}(s)=\frac{s-t_{k-1}}{\Delta t_{k}}x(t_{k})+\frac{t_{k}-s}{\Delta
t_{k}}x(t_{k-1}),\,s\in[t_{k-1},t_{k}],\,k=1,2,...,N,$ (22)

with the interpolation error

$R_{k}(s)=x(s)-L_{1,k}(s),s\in[t_{k-1},t_{k}],\;k=1,2,...,N.$ (23)

Since

${}^{c}D_{q}^{\alpha}x(t_{n})=\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}\int_{t_{k-1}}^{t_{k}}(t_{n}-qs)^{(-\alpha)}D_{q}x(s)d_{q}s,$
(24)

replacing $D_{q}x(s)$ by $D_{q}L_{1,k}x(s)$ in (24) and noting that
$D_{q}L_{1,k}x(s)=(x(t_{k})-x(t_{k-1}))/\triangle t_{k}$, we obtain

${}^{c}D_{q}^{\alpha}x(t_{n})=\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}b_{k}\big{(}x(t_{k})-x(t_{k-1})\big{)}+R^{n}_{q},$
(25) $\displaystyle b_{k}=\frac{1}{\triangle
t_{k}}\int_{t_{k-1}}^{t_{k}}(t_{n}-qs)^{(-\alpha)}d_{q}s,\;\,k=1,2,\dots,n\,,$
(26)

where the truncation error

$R^{n}_{q}=\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}\int_{t_{k-1}}^{t_{k}}(t_{n}-qs)^{(-\alpha)}D_{q}R_{k}(s)d_{q}s.$
(27)

Thus, we derive the $L_{1,q}$ difference formula:

$\triangle_{q}^{\alpha}x^{n}\doteq\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}b_{k}\big{(}x^{k}-x^{k-1}),\;\;^{c}D_{q}^{\alpha}x(t_{n})=\triangle^{\alpha}_{q}x(t_{n})+R^{n}_{q}\,,$
(28)

where $x^{k}$ is any one mesh function defined on $\\{t_{k}\\}$ and when
$x(t)$ is continuous on $[0,t_{N}]$, we set $x^{k}=x(t_{k})$.
Remark 1. In constructing the difference scheme, we used $D_{q}L_{1,k}(t)$ to
approximate the integrand function $D_{q}x(t)$ on interval $[t_{k-1},t_{k}]$
in (20). Since (noting that the $q$-mesh point $t_{k-1}=qt_{k}$)

$D_{q}L_{1,k}(t)=\frac{x(t_{k})-x(t_{k-1})}{t_{k}-t_{k-1}}=\frac{x(t_{k})-x(qt_{k})}{(1-q)t_{k}}=D_{q}x(t_{k}),\;t\in[t_{k-1},t_{k}],$

so actually, we used the $q$-rectangle quadrature formula to discretize the
integral on $[t_{k-1},t_{k}]$.

Below we estimate the truncation error $R^{n}_{q}$. We first give several
lemmas.
Lemma 2.($q$-Rolle lemma [28]) Let $x(t)$ be a continuous function on $[a,b]$
and $x(a)=x(b)$. Then, for $q\in(0,1)$, there exists point $\xi\in(a,b)$ such
that $D_{q}x(\xi)=0$.

By means of Lemma 2, we can establish the following interpolation error
formula.
Lemma 3. Assume that $x(s)$ and $D_{q}x(s)$ is continuous on
$[t_{k-1},t_{k}]$. Then, the linear interpolation error function
$R_{k}(s)=x(s)-L_{1,k}(s)$ has the following expression

$R_{k}(s)=\frac{1}{1+q}D_{q}^{2}x(\xi_{k})(s-t_{k-1})(s-t_{k}),s\in[t_{k-1},t_{k}],\,\xi_{k}\in(t_{k-1},t_{k}),\,1\leq
k\leq N.$

Proof. Since $R_{k}(t_{k-1})=R_{k}(t_{k})=0$, we may assume that

$R_{k}(s)=K_{q}(s)(s-t_{k-1})(s-t_{k}),\,s\in[t_{k-1},t_{k}],$ (29)

where $K_{q}(s)$ is an undetermined function. For any fixed point
$s\in(t_{k-1},t_{k})$, let
$\varphi_{q}(t)=x(t)-L_{1,k}(t)-K_{q}(s)(t-t_{k-1})(t-t_{k})$. Since
$\varphi_{q}(t_{k-1})=\varphi_{q}(t_{k})=\varphi_{q}(s)=0$, by using the
$q$-Rolle lemma twice, we know that $D_{q}^{2}\varphi_{q}(t)$ has at least one
zero point $\xi_{k}\in(t_{k-1},t_{k})$. So

$\displaystyle D_{q}^{2}\varphi_{q}(\xi_{k})$ $\displaystyle=$ $\displaystyle
D^{2}_{q}\Big{(}x(t)-L_{1,k}(t)-K_{q}(s)(t-t_{k-1})(t-t_{k})\Big{)}\Big{|}_{t=\xi_{k}}$
$\displaystyle=$ $\displaystyle D_{q}^{2}x(\xi_{k})-(1+q)K_{q}(s)=0.$

Hence, we obtain $K_{q}(s)=D_{q}^{2}x(\xi_{k})/(1+q)$. Substituting $K_{q}(s)$
into (29), the proof is completed.$\hfill\Box$
Lemma 4. Let $0<\alpha,q<1$ and $D_{q}$ be the $q$-derivative operator with
respect to variable $s$. Then

$\displaystyle
D_{q}(t-s)^{(-\alpha)}=-[-\alpha]_{q}(t-qs)^{(-\alpha-1)},\,0\leq s\leq t,$
(30) $\displaystyle|(t-qs)^{(-\alpha-1)}|\leq
t^{-\alpha-1}\frac{1}{1-q^{\alpha}}\frac{1}{1-q^{1-\alpha}},\;0\leq s\leq t.$
(31)

Proof. We first prove conclusion (30). By the definition of operator $D_{q}$
and (2), we have

$\displaystyle
D_{q}(t-s)^{(-\alpha)}=\frac{(t-qs)^{(-\alpha)}-(t-s)^{(-\alpha)}}{(q-1)s}$
(32) $\displaystyle=$
$\displaystyle\frac{t^{-\alpha}}{(q-1)s}\lim_{m\rightarrow\infty}S_{m},\;S_{m}=\prod_{i=0}^{m}\frac{(t-q^{i+1}s)}{(t-q^{i+1-\alpha}s)}-\prod_{i=0}^{m}\frac{(t-q^{i}s)}{(t-q^{i-\alpha}s)}.$

By a straightforward computation, it yields

$\displaystyle S_{m}$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{m}\frac{(t-q^{i}s)}{(t-q^{i-\alpha}s)}\Big{[}\frac{(t-q^{m+1}s)}{(t-q^{m+1-\alpha}s)}-\frac{(t-s)}{(t-q^{-\alpha})s}\Big{]}$
$\displaystyle=$
$\displaystyle\prod_{i=1}^{m}\frac{(t-q^{i}s)}{(t-q^{i-\alpha}s)}\Big{[}\frac{st(1-q^{-\alpha})(1-q^{m+1})}{(t-q^{m+1-\alpha}s)(t-q^{-\alpha}s)}\Big{]}$
$\displaystyle=$
$\displaystyle\prod_{i=0}^{m}\frac{(t-q^{i+1}s)}{(t-q^{i-\alpha}s)}\Big{[}\frac{st(1-q^{-\alpha})(1-q^{m+1})}{(t-q^{m+1-\alpha}s)(t-q^{m+1}s)}\Big{]}$
$\displaystyle=$
$\displaystyle\prod_{i=0}^{\infty}\frac{(t-q^{i+1}s)}{(t-q^{i-\alpha}s)}\Big{[}\frac{s(1-q^{-\alpha})}{t}\Big{]},\;m\rightarrow\infty.$

Substituting this into (32), we obtain

$D_{q}(t-s)^{(-\alpha)}=\frac{(1-q^{-\alpha})}{(q-1)}t^{-\alpha-1}\prod_{i=0}^{\infty}\frac{(t-q^{i+1}s)}{(t-q^{i-\alpha}s)}=-[-\alpha]_{q}(t-qs)^{(-\alpha-1)}.$

Next, we consider estimation (31). Since

$\displaystyle(t-qs)^{(-\alpha-1)}=t^{-\alpha-1}\lim_{m\rightarrow\infty}S_{m}^{\prime},\;S^{\prime}_{m}=\prod_{i=0}^{m}\frac{(t-q^{i+1}s)}{(t-q^{i-\alpha}s)},$
(33)

and

$\displaystyle\max_{0\leq s\leq
t}\big{|}\frac{(t-qs)}{(t-q^{-\alpha}s)}\big{|}=\max\big{\\{}1,\big{|}\frac{1-q}{1-q^{-\alpha}}\big{|}\big{\\}}\leq\frac{1-q}{1-q^{\alpha}},$
$\displaystyle\max_{0\leq s\leq
t}\frac{(t-q^{i+1}s)}{(t-q^{i-\alpha}s)}=\frac{(1-q^{i+1})}{(1-q^{i-\alpha})},\;i\geq
1,$

so

$\displaystyle|S^{\prime}_{m}|$ $\displaystyle\leq$
$\displaystyle\frac{1-q}{1-q^{\alpha}}\prod_{i=1}^{m}\frac{(1-q^{i+1})}{(1-q^{i-\alpha})}$
$\displaystyle=$
$\displaystyle\frac{1}{1-q^{\alpha}}\frac{1}{1-q^{1-\alpha}}\frac{1-q}{1-q^{2-\alpha}}\frac{1-q^{2}}{1-q^{3-\alpha}}\cdots\frac{1-q^{m-1}}{1-q^{m-\alpha}}\frac{1-q^{m}}{1}\frac{1-q^{m+1}}{1}$
$\displaystyle\leq$
$\displaystyle\frac{1}{1-q^{\alpha}}\frac{1}{1-q^{1-\alpha}}.$

Substituting this into (33), the proof is completed.$\hfill\Box$

Now we can give the truncation error estimation.
Theorem 1. Let $x(t)$ and $D_{q}x(t)$ are continuous on $[0,b]$. Then, the
truncation error of the $L_{1,q}$ difference formula (28) satisfies the
following estimate

$|R_{q}^{n}|\leq\frac{1}{4\Gamma_{q}(1-\alpha)}\frac{1}{1-q^{2}}\frac{1}{q^{\alpha}-q}\,t_{n}^{-\alpha}\triangle
t_{n}^{2}\max_{0\leq t\leq t_{n}}|D_{q}^{2}x(t)|,\;\,n\geq 1.$ (34)

Proof. Let $\bar{R}(s)=R_{k}(s),\,s\in[t_{k-1},t_{k}],\,1\leq k\leq N$. From
(27), the $q$-integration by parts formula (10) and Lemma 4, we obtain

$\displaystyle R^{n}_{q}$ $\displaystyle=$
$\displaystyle\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}\int_{t_{k-1}}^{t_{k}}(t_{n}-qs)^{(-\alpha)}D_{q}R_{k}(s)d_{q}s$
$\displaystyle=$
$\displaystyle\frac{(t_{n}-qs)^{(-\alpha)}}{\Gamma_{q}(1-\alpha)}R_{k}(s)\big{|}_{t_{k-1}}^{t^{k}}-\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}\int_{t_{k-1}}^{t_{k}}D_{q}(t_{n}-s)^{(-\alpha)}R_{k}(s)d_{q}s$
$\displaystyle=$
$\displaystyle\frac{[-\alpha]_{q}}{\Gamma_{q}(1-\alpha)}\int_{0}^{t_{n}}(t_{n}-qs)^{(-\alpha-1)}\bar{R}(s)d_{q}s.$

Hence, using (7)–(8), Lemma 3 and (31) it yields

$\displaystyle|R_{q}^{n}|$ $\displaystyle\leq$
$\displaystyle\frac{|[-\alpha]_{q}|}{\Gamma_{q}(1-\alpha)}\int_{0}^{t_{n}}|(t_{n}-qs)^{(-\alpha-1)}\bar{R}(s)|d_{q}s$
$\displaystyle=$
$\displaystyle\frac{|[-\alpha]_{q}|}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}\int_{t_{k-1}}^{t_{k}}|(t_{n}-qs)^{(-\alpha-1)}R_{k}(s)|d_{q}s$
$\displaystyle\leq$
$\displaystyle\frac{|[-\alpha]_{q}|}{\Gamma_{q}(1-\alpha)}\frac{1}{1+q}\frac{1}{4}\max_{1\leq
k\leq n}|\triangle t_{k}|^{2}\max_{0\leq t\leq
t_{n}}|D_{q}^{2}x(t)|\int_{0}^{t_{n}}|(t_{n}-qs)^{(-\alpha-1)}|d_{q}s$
$\displaystyle\leq$
$\displaystyle\frac{|[-\alpha]_{q}|}{\Gamma_{q}(1-\alpha)}\frac{1}{1+q}\frac{1}{4}\max_{1\leq
k\leq n}|\triangle t_{k}|^{2}\max_{0\leq t\leq
t_{n}}|D_{q}^{2}x(t)|t_{n}^{-\alpha}\frac{1}{1-q^{\alpha}}\frac{1}{1-q^{1-\alpha}}$
$\displaystyle=$
$\displaystyle\frac{1}{\Gamma_{q}(1-\alpha)}\frac{q^{-\alpha}}{1-q^{2}}\frac{1}{4}\frac{1}{1-q^{1-\alpha}}t_{n}^{-\alpha}\triangle
t_{n}^{2}\max_{0\leq t\leq t_{n}}|D_{q}^{2}x(t)|\,.$

This gives the desired estimate.$\hfill\Box$

## 4\. STABILITY OF THE $L_{1,q}$ DIFFERENCE FORMULA

We first give two useful lemmas.
Lemma 5. For $0<q,\alpha<1$, the following inequality holds

$t_{n}^{-\alpha}<(t_{n}-q^{i+1}s)^{(-\alpha)}\leq(t_{n}-qs)^{(-\alpha)},\;i\geq
0,\;0\leq s\leq t_{n}.$ (35)

Proof. For the left side inequality, since
$(t_{n}-q^{i+j+1}s)/(t_{n}-q^{i+j+1-\alpha}s)>1$ for any $i,j\geq 0$, we have

$(t_{n}-q^{i+1}s)^{(-\alpha)}=t_{n}^{-\alpha}\prod_{j=0}^{\infty}\frac{(t_{n}-q^{i+j+1}s)}{(t_{n}-q^{i+j+1-\alpha}s)}>t_{n}^{-\alpha},\,i\geq
0,\,0\leq s\leq t_{n}.$

Next, by a straightforward computation, it yields

$\displaystyle(t_{n}-q^{i+1}s)^{(-\alpha)}-(t_{n}-qs)^{(-\alpha)}$
$\displaystyle=$ $\displaystyle
t_{n}^{-\alpha}\prod_{j=0}^{\infty}\frac{(t_{n}-q^{i+j+1}s)}{(t_{n}-q^{i+j+1-\alpha}s)}-t_{n}^{-\alpha}\prod_{j=0}^{\infty}\frac{(t_{n}-q^{j+1}s)}{(t_{n}-q^{j+1-\alpha}s)}$
$\displaystyle=$ $\displaystyle
t_{n}^{-\alpha}\prod_{j=i}^{\infty}\frac{(t_{n}-q^{j+1}s)}{(t_{n}-q^{j+1-\alpha}s)}\Big{(}1-\prod_{j=0}^{i-1}\frac{(t_{n}-q^{j+1}s)}{(t_{n}-q^{j+1-\alpha}s)}\Big{)}<0,\,i\geq
1.$

The proof is completed.$\hfill\Box$
Lemma 6. Let coefficient series $\\{b_{k}\\}$ be defined by (26). Then, it
holds

$\displaystyle b_{k}=(t_{n}-qt_{k})^{(-\alpha)},\;k\geq 2,$ (36)
$\displaystyle t_{n}^{-\alpha}<b_{1}<b_{2}<\cdots<b_{k-1}<b_{k},\;2\leq k\leq
n\,.$ (37)

Proof. Since $\triangle
t_{k}=t_{k}-t_{k-1}=t_{k}(1-q),\,t_{k-1}=qt_{k},\,k\geq 2$, then by (26) we
have

$\displaystyle b_{k}$ $\displaystyle=$ $\displaystyle\frac{1}{\triangle
t_{k}}\int_{t_{k-1}}^{t_{k}}(t_{n}-qs)^{(-\alpha)}d_{q}s$ $\displaystyle=$
$\displaystyle\frac{1}{\triangle
t_{k}}\int_{0}^{t_{k}}(t_{n}-qs)^{(-\alpha)}d_{q}s-\frac{1}{\triangle
t_{k}}\int^{t_{k-1}}_{0}(t_{n}-qs)^{(-\alpha)}d_{q}s$ $\displaystyle=$
$\displaystyle(1-q)\sum_{i=0}^{\infty}\frac{t_{k}}{\triangle
t_{k}}q^{i}(t_{n}-q^{i+1}t_{k})^{(-\alpha)}-(1-q)\sum_{i=0}^{\infty}\frac{t_{k-1}}{\triangle
t_{k}}q^{i}(t_{n}-q^{i+1}t_{k-1})^{(-\alpha)}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}q^{i}(t_{n}-q^{i+1}t_{k})^{(-\alpha)}-\sum_{i=0}^{\infty}q^{i+1}(t_{n}-q^{i+2}t_{k})^{(-\alpha)}=(t_{n}-qt_{k})^{(-\alpha)}.$

This gives equality (36). For (37), we first obtain from Lemma 5

$\displaystyle b_{1}$ $\displaystyle=$ $\displaystyle\frac{1}{\triangle
t_{1}}\int_{0}^{t_{1}}(t_{n}-qs)^{(-\alpha)}d_{q}s=(1-q)\sum_{i=0}^{\infty}\frac{t_{1}}{\triangle
t_{1}}q^{i}(t_{n}-q^{i+1}t_{1})^{(-\alpha)}$ (38) $\displaystyle>$
$\displaystyle t_{n}^{-\alpha}(1-q)\sum_{i=0}^{\infty}q^{i}=t_{n}^{-\alpha},$

and

$\displaystyle b_{1}$ $\displaystyle=$
$\displaystyle(1-q)\sum_{i=0}^{\infty}\frac{t_{1}}{\triangle
t_{1}}q^{i}(t_{n}-q^{i+1}t_{1})^{(-\alpha)}$ (39) $\displaystyle\leq$
$\displaystyle(1-q)\sum_{i=0}^{\infty}q^{i}(t_{n}-qt_{1})^{(-\alpha)}=(t_{n}-qt_{1})^{(-\alpha)}.$

Next, it follows from (36), (39) and Lemma 5

$b_{2}-b_{1}\geq(t_{n}-qt_{2})^{(-\alpha)}-(t_{n}-qt_{1})^{(-\alpha)}=(t_{n}-qt_{2})^{(-\alpha)}-(t_{n}-q^{2}t_{2})^{(-\alpha)}>0.$
(40)

Moreover, from (36) and Lemma 5, it yields

$\displaystyle b_{k}-b_{k-1}$ $\displaystyle=$
$\displaystyle(t_{n}-qt_{k})^{(-\alpha)}-(t_{n}-qt_{k-1})^{(-\alpha)}$ (41)
$\displaystyle=$
$\displaystyle(t_{n}-qt_{k})^{(-\alpha)}-(t_{n}-q^{2}t_{k})^{(-\alpha)}>0,\;k\geq
3.$

Combining (38), (40) and (41), estimation (37) is derived.$\hfill\Box$

Now let us consider the stability of the $L_{1,q}$ difference formula (28).
For this end, we consider the following difference equation:

$\triangle_{q}^{\alpha}x^{n}=f^{n},\;n=1,2,\dots,N,\;\;\;\triangle_{q}^{\alpha}x^{n}\doteq\frac{1}{\Gamma_{q}(1-\alpha)}\sum_{k=1}^{n}b_{k}(x^{k}-x^{k-1}),$
(42)

where the initial value $x^{0}$ and the source term $f^{n}$ are given. Using
the identity

$\sum_{k=1}^{n}b_{k}(x^{k}-x^{k-1})=b_{n}x^{n}+\sum_{k=1}^{n-1}(b_{k}-b_{k+1})x^{k}-b_{1}x^{0},$

difference equation (42) can be rewritten as

$b_{n}x^{n}=b_{1}x^{0}+\sum_{k=1}^{n-1}(b_{k+1}-b_{k})x^{k}+\Gamma_{q}(1-\alpha)f^{n},\;n\geq
1\,.$ (43)

Theorem 2. The $L_{1,q}$ difference formula is unconditionally stable such
that the solution of difference equation (43) satisfies

$|x^{n}|\leq|x^{0}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|f^{k}|,\;n\geq 1.$

Proof. It follows from (43) and Lemma 6 that

$\displaystyle b_{n}|x^{n}|$ $\displaystyle\leq$ $\displaystyle
b_{1}|x^{0}|+\sum_{k=1}^{n-1}(b_{k+1}-b_{k})\max_{1\leq k\leq
n}|x^{k}|+\Gamma_{q}(1-\alpha)|f^{n}|$ (44) $\displaystyle\leq$ $\displaystyle
b_{1}|x^{0}|+(b_{n}-b_{1})\max_{1\leq k\leq
n}|x^{k}|+\Gamma_{q}(1-\alpha)|f^{n}|\,.$

Let $1\leq n_{0}\leq n$ be such that $|x^{n_{0}}|=\displaystyle{\max_{1\leq
k\leq n}}|x^{k}|$. Taking $n=n_{0}$ in (44), it yields

$b_{1}\max_{1\leq k\leq n}|x^{n}|\leq
b_{1}|x^{0}|+\Gamma_{q}(1-\alpha)\max_{1\leq k\leq n}|f^{k}|.$

Hence, using (37), we complete the proof.$\hfill\Box$

## 5\. DIFFERENCE METHOD FOR THE $q$-FRACTIONAL
DIFFERENTIAL EQUATION

Consider the initial value problem of $q$-fractional differential equation:

$\displaystyle\left\\{\begin{array}[]{lll}&{}^{c}D_{q}^{\alpha}x(t)=f(t,x(t)),\;0<t\leq
b,\;0<q,\,\alpha<1,\;t\in T_{q,b},\\\ &x(0)=x^{0},\end{array}\right.$ (47)

where $f:[0,b]\times R^{d}\rightarrow R^{d}$ is continuous, $d\geq 1$. Let the
partition of interval $[0,b]$ be given as that in Section 3. By means of the
$L_{1,q}$ difference formula, we establish the difference method for solving
problem (47) as follows (see (43))

$\displaystyle
b_{n}x^{n}=b_{1}x^{0}+\sum_{k=1}^{n-1}(b_{k+1}-b_{k})x^{k}+\Gamma_{q}(1-\alpha)f(t_{n},x^{n}),\;n=1,2,\dots,N.$
(48)

We first consider the unique existence of the difference solution.
Theorem 3. Assume that $f:[0,b]\times R^{d}\rightarrow R^{d}$ is continuous
and satisfies the Lipschitz condition:

$|f(t,x)-f(t,x^{\prime})|\leq L|x-x^{\prime}|,\;x,x^{\prime}\in
R^{d},\;L_{1}=L\Gamma_{q}(1-\alpha)b^{\alpha}<1.$ (49)

Then, difference equation (48) has a unique solution.
Proof. We only need to prove that for any fixed $n\geq 1$, when
$\\{x^{k}\\}_{k=0}^{n-1}$ are known, there exists a unique $x^{n}$ satisfying
Eq. (48). Consider the iteration scheme:

$b_{n}x^{n,l}=b_{1}x^{0}+\sum_{k=1}^{n-1}(b_{k+1}-b_{k})x^{k}+\Gamma_{q}(1-\alpha)f(t_{n},x^{n,l-1}),\,l=1,2,\dots,\;x^{n,0}=x^{n-1}.$
(50)

Noting that $b_{n}\geq t_{n}^{-\alpha}\geq b^{-\alpha}$, we obtain from
(49)–(50) that

$\displaystyle|x^{n,l+1}-x^{n,l}|$ $\displaystyle=$ $\displaystyle
b_{n}^{-1}\Gamma_{q}(1-\alpha)|f(t_{n},x^{n,l})-f(t_{n},x^{n,l-1})|$
$\displaystyle\leq$ $\displaystyle L_{1}|x^{n,l}-x^{n,l-1}|\leq\cdots\leq
L_{1}^{l}|x^{n,1}-x^{n,0}|,\;l\geq 1.$

Since $L_{1}<1$, this implies that the series $\\{x^{n,l}\\}$ is convergent.
Letting $l\rightarrow\infty$ in Eq. (50), we see that $x^{n}\doteq
x^{n,\infty}$ is the solution of difference equation (48). Now for fixed $n$,
assume that Eq. (48) has two solutions $x^{n}$ and $\bar{x}^{n}$. Then,
$x^{n}$ and $\bar{x}^{n}$ satisfy

$|x^{n}-\bar{x}^{n}|=b_{n}^{-1}\Gamma_{q}(1-\alpha)|f(t_{n},x^{n})-f(t_{n},\bar{x}^{n})|\leq
L_{1}|x^{n}-\bar{x}^{n}|<|x^{n}-\bar{x}^{n}|\,.$

This implies $x^{n}=\bar{x}^{n}$, the uniqueness is proved. $\hfill\Box$

Next, we consider the stability of the difference solution.
Theorem 4. Assume that condition (49) holds. Then, the solution of difference
equation (48) satisfies the stability estimation

$|x^{n}|\leq\frac{1}{1-L_{1}}\big{[}\,|x^{0}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq
k\leq n}|f(t_{k},0)|\,\big{]},\;n\geq 1\,.$

Proof. Let $x^{n}$ be the solution of Eq. (48). By a similar argument to that
of Theorem 2, we have

$|x^{n}|\leq|x^{0}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|f(t_{k},x^{k})|,\;n\geq 1.$ (51)

Using condition (49), it yields

$|f(t_{k},x^{k})|=|f(t_{k},x^{k})-f(t_{k},0)|+|f(t_{k},0)|\leq
L|x_{k}|+|f(t_{k},0)|\,.$

Combining this with (51), we arrive at

$|x^{n}|\leq|x^{0}|+L_{1}\max_{1\leq k\leq
n}|x^{k}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|f(t_{k},0)|,\,n\geq 1\,.$ (52)

Now, let $1\leq n_{0}\leq n$ be such that
$|x^{n_{0}}|=\displaystyle{\max_{1\leq k\leq n}}|x^{k}|$. Then, taking
$n=n_{0}$ in (52), we can derive

$(1-L_{1})\max_{1\leq k\leq
n}|x^{k}|\leq|x^{0}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|f(t_{k},0)|,\;n\geq 1\,.$

This gives the desired estimation.$\hfill\Box$

Finally, we can give the error estimation.
Theorem 5. Let $x(t)$ and $x^{n}$ be the solution of Eqs. (47) and (48),
respectively. Assume that $x(t)$ and $D_{q}x(t)$ are continuous on $[0,b]$ and
condition (49) holds. Then, we have the following error estimation.

$|x(t_{n})-x^{n}|\leq\frac{1}{1-L_{1}}\frac{1}{4}\frac{1}{1-q^{2}}\frac{1}{q^{\alpha}-q}\triangle
t_{n}^{2}\max_{0\leq t\leq t_{n}}|D_{q}^{2}x(t)|,\;n\geq 1.$ (53)

Proof. From Eq. (47), we see that the solution $x(t)$ satisfies the discrete
equation: $\triangle_{q}^{\alpha}x(t_{n})=f(t_{n},x(t_{n}))-R_{q}^{n}$, or

$b_{n}x(t_{n})=b_{1}x^{0}+\sum_{k=1}^{n-1}(b_{k+1}-b_{k})x(t_{k})+\Gamma_{q}(1-\alpha)f(t_{n},x(t_{n}))-\Gamma_{q}(1-\alpha)R_{q}^{n},$
(54)

where the truncation error
$R_{q}^{n}=^{c}\\!D_{q}^{\alpha}x(t_{n})-\triangle_{q}^{\alpha}x(t_{n})$ is
given by (27). Let the error function $e_{n}=x^{n}-x(t_{n})$. From Eqs. (48)
and (54), we obtain

$\displaystyle
b_{n}e_{n}=\sum_{k=1}^{n-1}(b_{k+1}-b_{k})e_{k}+\Gamma_{q}(1-\alpha)\big{(}f(t_{n},x^{n})-f(t_{n},x(t_{n}))\big{)}+\Gamma_{q}(1-\alpha)R_{q}^{n}\,.$

Then, it follows a similar argument to that of Theorem 2

$|e_{n}|\leq\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|f(t_{k},x^{k})-f(t_{k},x(t_{k}))|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq
k\leq n}|R_{q}^{k}|,\;n\geq 1.$ (55)

Hence, using condition (49), it yields

$|e_{n}|\leq L_{1}\max_{1\leq k\leq
n}|e_{k}|+\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq k\leq
n}|R_{q}^{k}|,\;n\geq 1.$

Which implies

$|e_{n}|\leq\frac{1}{1-L_{1}}\Gamma_{q}(1-\alpha)t_{n}^{\alpha}\max_{1\leq
k\leq n}|R_{q}^{k}|,\;n\geq 1.$ (56)

Using Theorem 1 and noting that $t_{k}=\triangle t_{k}/(1-q)$, we obtain

$|R_{q}^{k}|\leq\frac{1}{4\Gamma_{q}(1-\alpha)}\frac{1}{1-q^{2}}\frac{(1-q)^{\alpha}}{q^{\alpha}-q}\triangle
t_{k}^{2-\alpha}\max_{0\leq t\leq t_{k}}|D_{q}^{2}x(t)|\,.$ (57)

Substituting this into (56), it yields

$|e_{n}|\leq\frac{1}{1-L_{1}}\frac{1}{4}\frac{1}{1-q^{2}}\frac{(1-q)^{\alpha}}{q^{\alpha}-q}t_{n}^{\alpha}\max_{1\leq
k\leq n}\triangle t_{k}^{2-\alpha}\max_{0\leq t\leq
t_{n}}|D_{q}^{2}x(t)|,\;n\geq 1.$

The proof is completed, noting that $\displaystyle{\max_{1\leq k\leq
n}}\triangle t_{k}=\triangle t_{n},\,t_{n}=\triangle t_{n}/(1-q)$.$\hfill\Box$

Since $\triangle t_{n}=t_{n}-t_{n-1}=bq^{N-n}(1-q)$, then from (53) we also
obtain

$|x(t_{n})-x^{n}|\leq\frac{1}{1-L_{1}}\frac{1}{4}\frac{1-q}{q^{\alpha}-q}b^{2}q^{2(N-n)}\max_{0\leq
t\leq t_{n}}|D_{q}^{2}x(t)|,\;\,n\geq 1.$ (58)

This implies that for any fixed mesh point $t_{n}\in(0,b)$ with
$n\leq(1-\delta)N$ where $0<\delta<1$, the difference solution
$x^{n}\rightarrow x(t_{n})$ as $N\rightarrow\infty$ and the convergence rate
is of $O(q^{2\delta N})$-order.
Remark 2. Since the $q$-mesh step sizes $\triangle t_{1}<\triangle
t_{2}<\cdots<\triangle t_{N}$ and $\triangle t_{N}=(1-q)b$, therefore, in
general speaking, the difference solution $x^{n}$ defined on the time scale
$T_{q,b}$ can not admits a global convergence rate that is valid for all
$q$-mesh points $\\{t_{n}\\}$.

## 6\. NUMERICAL EXPERIMENT

In this section, we use numerical examples to verify the high accuracy and
effectiveness of the $L_{1,q}$ difference formula. In the numerical
experiment, all computations are carried out by using Mathematica 11.3.
Example 1. In this example, we use the difference scheme (48) to solve the
linear $q$-fractional differential equation:

${}^{c}D_{q}^{\frac{1}{2}}x(t)=\frac{1+q}{\Gamma_{q}(3/2)}t^{\frac{3}{2}}+\frac{1}{\Gamma_{q}(3/2)}\sqrt{t},\quad
0<t\leq 1,\,t\in T_{q,1},\,\quad x(0)=1.$ (59)

The exact solution is $x(t)=t^{2}+t+1$. The numerical results are given in
Table 1.

Table 1. Numerical results for problem (59), $q=1/4,N=10$.

$t_{n}=q^{N-n}$ | $x^{n}$ | $x(t_{n})$ | $|x(t_{n})-x^{n}|$
---|---|---|---
$(1/4)^{9}$ | $1.0000$ | $1.0000$ | $3.2911\mathrm{E}{-7}$
$(1/4)^{8}$ | $1.0000$ | $1.0000$ | $1.5456\mathrm{E}{-7}$
$(1/4)^{7}$ | $1.0001$ | $1.0001$ | $7.5569\mathrm{E}{-8}$
$(1/4)^{6}$ | $1.0002$ | $1.0002$ | $2.8026\mathrm{E}{-8}$
$(1/4)^{5}$ | $1.0010$ | $1.0010$ | $1.3998\mathrm{E}{-7}$
$(1/4)^{4}$ | $1.0039$ | $1.0039$ | $2.5337\mathrm{E}{-6}$
$(1/4)^{3}$ | $1.0159$ | $1.0159$ | $4.0685\mathrm{E}{-5}$
$(1/4)^{2}$ | $1.0664$ | $1.0671$ | $6.5104\mathrm{E}{-4}$
$(1/4)^{1}$ | $1.3125$ | $1.3129$ | $1.4035\mathrm{E}{-4}$
$(1/4)^{0}$ | $3.0000$ | $3.0002$ | $2.1218\mathrm{E}{-4}$

Example 2. In this example, we use the difference scheme (48) to solve the
nonlinear $q$-fractional differential equation:

${}^{c}D_{q}^{\frac{2}{3}}x(t)=\frac{1+q}{\Gamma_{q}(7/3)}(x-1)^{\frac{2}{3}},\quad
0<t\leq 1,\;t\in T_{q,1},\quad x(0)=1.$ (60)

By a straightforward computation, using formulas (4) and (21), we can verify
that $x(t)=t^{2}+1$ is a solution of problem (60). The numerical results are
given in Table 2.

From the numerical results in Table 1 and Table 2, we see that our difference
method is stable, high accurate and effective.

Table 2. Numerical results for problem (60), $q=2/3,N=10$.  $t_{n}=q^{N-n}$
$x(t_{n})$ $x^{n}$ $|x(t_{n})-x^{n}|$ $(2/3)^{9}$ $1.0007$ $1.0005$
$2.0209\mathrm{E}{-4}$ $(2/3)^{8}$ $1.0015$ $1.0014$ $1.6013\mathrm{E}{-4}$
$(2/3)^{7}$ $1.0034$ $1.0033$ $1.3277\mathrm{E}{-4}$ $(2/3)^{6}$ $1.0077$
$1.0076$ $1.1208\mathrm{E}{-4}$ $(2/3)^{5}$ $1.0173$ $1.0172$
$9.4942\mathrm{E}{-5}$ $(2/3)^{4}$ $1.0390$ $1.0389$ $7.9273\mathrm{E}{-5}$
$(2/3)^{3}$ $1.0878$ $1.0877$ $6.2639\mathrm{E}{-5}$ $(2/3)^{2}$ $1.1975$
$1.1975$ $4.0754\mathrm{E}{-5}$ $(2/3)^{1}$ $1.4444$ $1.4444$
$4.6923\mathrm{E}{-6}$ $(2/3)^{0}$ $2.0000$ $2.0001$ $6.5151\mathrm{E}{-5}$

## 7\. CONCLUSION

An unconditionally stable difference formula is first presented to discretize
the fractional $q$-derivative ${}^{c}D^{\alpha}_{q}x(t)$ of Caputo type with
$0<q,\,\alpha<1$ on the time scale set $T_{q,b}$. The rigours truncation error
boundness is derived. This difference formula can be used to solve the
$q$-fractional differential equation: ${}^{c}D^{\alpha}_{q}x(t)=f(t,x(t))$ on
set $T_{q,b}$. We prove the unique existence and stability of the difference
solution and derive an error estimate of $\triangle t_{n}^{2}$-order.
Numerical experiments show the effectiveness and high accuracy of this
difference method. This difference formula provides a useful tool for solving
numerically the $q$-fractional problems on the time scale set.

## ACKNOWLEDGMENTS

The authors would like to thank the anonymous referees for many helpful
suggestions which improved the presentation of this paper.

This work was supported by the State Key Laboratory of Synthetical Automation
for Process Industries Fundamental Research Funds (Grant No. 2013ZCX02).

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