We employ some notations of the *q*-fractional derivative and integral in [38]. For 0<q<1, let {T}_{q} be the time scale

{T}_{q}=\{{q}^{m}:m\in \mathrm{Z}\}\cup \{0\},

(43)

where Z is the set of integers.

**Definition 5.1** More generally, if *α* is a nonnegative real number, then we define the time scale as follows:

{T}_{q}^{\alpha}=\{{q}^{m+\alpha}:m\in \mathrm{Z}\}\cup \{0\}.

(44)

The *q*-fractional derivative and integral have been defined in earlier work [8–10].

**Definition 5.2** The *q*-fractional integral of *α* order is defined by

{}_{q}I_{0}^{\alpha}f(t)=\frac{1}{{\mathrm{\Gamma}}_{q}(\alpha )}{\int}_{0}^{t}{(t-q\tau )}^{(\alpha -1)}f(\tau )\phantom{\rule{0.2em}{0ex}}{d}_{q}\tau ,\phantom{\rule{1em}{0ex}}t\in {T}_{q}^{\alpha}

(45)

and the left Caputo *q*-fractional derivative is defined as

{}_{q}{}^{C}D_{0}^{\alpha}u(t)=\frac{1}{\mathrm{\Gamma}(m-\alpha )}{\int}_{0}^{t}\frac{1}{{(t-\tau )}^{(\alpha -m+1)}}\frac{{d}^{m}}{d{\tau}^{m}}u(\tau )\phantom{\rule{0.2em}{0ex}}d\tau ,\phantom{\rule{1em}{0ex}}0<\alpha ,m=[\alpha ]+1,t\in {T}_{q}^{\alpha}.

(46)

When *v* is not a positive integer, the *q*-factorial function is defined by

{(t-\tau )}^{(v)}={t}^{v}\prod _{n=0}^{\mathrm{\infty}}\frac{1-\frac{{q}^{n}\tau}{t}}{1-\frac{{q}^{v+n}\tau}{t}}.

(47)

The fractional *q*-derivative of the *q*-factorial function with respect to *t* is

{}_{q}{}^{C}D_{a}^{\alpha}{(t-a)}^{(v)}=\frac{{\mathrm{\Gamma}}_{q}(v+1)}{{\mathrm{\Gamma}}_{q}(v-\alpha +1)}{(t-a)}^{(v-\alpha )}

(48)

and

{}_{q}I_{a}^{\alpha}{(t-a)}^{(v)}=\frac{{\mathrm{\Gamma}}_{q}(v+1)}{{\mathrm{\Gamma}}_{q}(\alpha +v+1)}{(t-a)}^{(\alpha +v)},\phantom{\rule{1em}{0ex}}a<t,

(49)

where \alpha \in {R}^{+} and v\in (-1,\mathrm{\infty}).

Now, we introduce the *q*-Laplace transform and some properties.

**Definition 5.3** The *q*-Laplace transform was defined by Hahn [39] in 1949 as follows:

{L}_{q}[h(t)]=\frac{1}{1-q}{\int}_{0}^{1/s}h(t){e}_{2,q}(qst)\phantom{\rule{0.2em}{0ex}}{d}_{q}t,

(50)

where {e}_{2,q}(t)={\prod}_{n=0}^{\mathrm{\infty}}(1-{q}^{n}t), {e}_{2,q}(0)=1.

**Lemma 5.4** ([38])

*Let* g(t) *be an analytic function and assume* g(t)={t}^{v-1} *on* {T}_{q}\setminus \{0\}, *where* v\in R\setminus \{\dots ,-2,-1,0\}. *Then the following convolution theorem can hold*:

{L}_{q}[h(t)\ast g(t)]={L}_{q}[h(t)]{L}_{q}[g(t)],

(51a)

*where the convolution is defined as*

(h\ast g)(s)=\frac{1}{1-q}{\int}_{0}^{s}h(\tau )g[s-q\tau ]\phantom{\rule{0.2em}{0ex}}{d}_{q}\tau \phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}g[s-q\tau ]={(s-q\tau )}^{(v-1)}.

(51b)

**Lemma 5.5** *For the Caputo* *q*-*derivative of* g(t), *the following Laplace transform holds*:

{L}_{q}{[}_{q}^{C}{D}_{0}^{\alpha}g(t)]=\frac{{s}^{\alpha}}{{(1-q)}^{\alpha}}{L}_{q}[g(t)]-\sum _{i=0}^{m-1}\frac{{d}_{q}^{i}g(0)}{{d}_{q}{t}^{i}}\frac{{s}^{\alpha -i-1}}{{(1-q)}^{\alpha -i}},\phantom{\rule{1em}{0ex}}0<\alpha ,m=[\alpha ]+1

(52a)

*and*

{L}_{q}{[}_{q}{I}_{0}^{\alpha}g(t)](s)=\frac{{(1-q)}^{\alpha}}{{s}^{\alpha}}{L}_{q}[g(t)](s).

(52b)

**Lemma 5.6** ([38])

{(t-\tau )}^{(\beta +\gamma )}={(t-\tau )}^{(\beta )}{(t-{q}^{\beta}\tau )}^{(\gamma )}, \beta ,\gamma \in R.

The existence and uniqueness of the solutions of the Caputo *q*-initial value problems have been discussed in [40].

**Lemma 5.7** *Considering the initial value problems of the Caputo* *q*-*fractional equations*,

{}_{q}{}^{C}D_{0}^{\alpha}u(t)+f(t,u)=0,\phantom{\rule{2em}{0ex}}\frac{{d}_{q}}{{d}_{q}t}{u}_{i}(0)={a}_{i},\phantom{\rule{1em}{0ex}}m=[\alpha ]+1,i=0,\dots ,m-1,

(53)

*we construct a* *q*-*fractional correction functional*

{u}_{n+1}={u}_{n}{+}_{q}{I}_{0}^{\alpha}\lambda (t,\tau ){[}_{q}^{C}{D}_{0}^{\alpha}{u}_{n}+f(\tau ,{u}_{n})].

(54)

*One of the Lagrange multipliers can be identified as* \lambda (t,\tau )=-1.

*Proof* Take the Laplace transform of both sides of (54)

\begin{array}{rcl}{L}_{q}[{u}_{n+1}]& =& {L}_{q}[{u}_{n}]+(1-q){L}_{q}[\frac{1}{(1-q)}{\int}_{0}^{t}\frac{{(t-q\tau )}^{(\alpha -1)}}{{\mathrm{\Gamma}}_{q}(\alpha )}\lambda (t,\tau )\\ \times {[}_{q}^{C}{D}_{0}^{\alpha}{u}_{n}+f(\tau ,{u}_{n})\left]\right]\phantom{\rule{0.2em}{0ex}}{d}_{q}\tau .\end{array}

(55)

From Lemma 5.6, we set \lambda (t,\tau )={\sum}_{i=0}^{\mathrm{\infty}}{a}_{i}{[t-{q}^{\alpha}\tau ]}^{({\beta}_{i})}, where {\beta}_{i}\ge 0. Then the Lagrange multiplier is ‘good’ enough so that the product of \frac{{(t-q\tau )}^{(\alpha -1)}}{{\mathrm{\Gamma}}_{q}(\alpha )} and \lambda (t,\tau ) is similar as the function g[s-q\tau ] in (51b) and \frac{1}{(1-q)}{\int}_{0}^{t}\frac{{(t-q\tau )}^{(\alpha -1)}}{{\mathrm{\Gamma}}_{q}(\alpha )}\lambda (t,\tau ){[}_{q}^{C}{D}_{0}^{\alpha}{u}_{n}+f(\tau ,{u}_{n})] becomes a convolution (51b).

Then we get the following equation:

\begin{array}{rcl}{L}_{q}[{u}_{n+1}]& =& {L}_{q}[{u}_{n}]+(1-q)\overline{a}(s)(\frac{{s}^{\alpha}}{{(1-q)}^{\alpha}}{L}_{q}[{u}_{n}]\\ -\sum _{i=0}^{m-1}\frac{{d}_{q}^{i}{u}_{n}(0)}{{d}_{q}{t}^{i}}\frac{{s}^{\alpha -i-1}}{{(1-q)}^{\alpha -i}}+{L}_{q}[f(t,{u}_{n})]),\end{array}

(56)

where \overline{a}(s) is the Laplace transform of some function.

Considering {L}_{q}[f(t,{u}_{n})] as a restricted variation so that after taking the classical variational derivative to both sides of (56), we can obtain

\delta {L}_{q}[{u}_{n+1}]=\delta {L}_{q}[{u}_{n}]+\overline{a}(s)\frac{{s}^{\alpha}}{{(1-q)}^{\alpha -1}}\delta {L}_{q}[{u}_{n}],

(57)

from which we can derive

\overline{a}(s)=-\frac{{(1-q)}^{\alpha -1}}{{s}^{\alpha}}.

(58)

The inverse Laplace transform of \overline{a}(s) is

a(t)=-\frac{{t}^{\alpha -1}}{{\mathrm{\Gamma}}_{q}(\alpha )}.

(59)

As a result, we can get

\lambda (t,\tau )=-1.

(60)

Substituting \lambda (t,\tau )=-1 into (54), the variational iteration formula is determined as

{u}_{n+1}={u}_{n}{-}_{q}{I}_{0}^{\alpha}{[}_{q}^{C}{D}_{0}^{\alpha}{u}_{n}+f(\tau ,{u}_{n})],\phantom{\rule{1em}{0ex}}0<\alpha .

(61)

□

**Example 5.8** Now consider the application in the initial value problems of the Caputo *q*-fractional difference equations [12],

{}_{q}{}^{C}D_{a}^{\alpha}u=\omega u+f(t),\phantom{\rule{2em}{0ex}}u(a)=c,\phantom{\rule{1em}{0ex}}t\in {T}_{q}^{\alpha},0<\alpha <1.

(62)

We have the following variational iteration formula:

{u}_{n+1}={u}_{n}-{\int}_{a}^{t}\frac{{(t-q\tau )}^{(\alpha -1)}}{{\mathrm{\Gamma}}_{q}(\alpha )}{(}_{q}^{C}{D}_{a}^{\alpha}{u}_{n}-\omega {u}_{n}-f(\tau ))\phantom{\rule{0.2em}{0ex}}{d}_{q}\tau .

(63)

Starting from the initial iteration

the successive solutions can be given as

For n\to \mathrm{\infty}, {u}_{n} tends to the exact solution

u={c}_{q}{E}_{\alpha}(\omega ,(t-a))+{\int}_{a}^{t}(t-q\tau )^{(\alpha -1)}{}_{q}{E}_{\alpha ,\alpha}(\omega ,t-{q}^{\alpha}\tau )f(\tau )\phantom{\rule{0.2em}{0ex}}{d}_{q}\tau ,

where {}_{q}E_{\alpha ,\beta}(\omega ,(t-{q}^{\alpha}\tau )) is the discrete Mittag-Leffler function defined by [12]

{}_{q}E_{\alpha ,\beta}(\omega ,t-{q}^{\alpha}\tau )=\sum _{k=0}^{\mathrm{\infty}}\frac{{\omega}^{k}{(t-{q}^{\alpha}\tau )}^{(k\alpha )}}{{\mathrm{\Gamma}}_{q}(k\alpha +\beta )}

(64)

and

{}_{q}E_{\alpha ,1}(\omega ,(t-a)){=}_{q}{E}_{\alpha}(\omega ,t-a).

(65)

Readers are referred to the recent development in the application of the VIM for solving fuzzy equations [41–43] and the calculus of variations on time scales [44–48]. Since this study only concentrates on the applications of the VIM, other numerical methods in FC can be found in [49–51].