In this section the existence of a unique solution is introduced in Theorem 6.2. Furthermore, the convergence of VIM solution (22) is proved in Theorem 6.3. Finally, the maximum absolute truncation error of VIM solution (22) is given in Theorem 6.4. In this section we prove theorems for the space-time fractional Burgers’ equation. This theorem covers the time-fractional Burgers’ equation on setting \(\beta=1\). We define \((C(I), \Vert \cdot \Vert )\) as a Banach space, the space of all continuous functions on \(I=R \times R^{+} \) with the norm \(\Vert v(x,t) \Vert =\max_{(x,t)\in I} \vert {v(x,t)} \vert\).

### Lemma 6.1

*Suppose that*
\(v(x,t)\)
*and their partial derivatives are continuous*. *Then the derivatives*
\(D_{t}^{\alpha}v(x,t)\), \(D_{x}^{\beta}v(x,t)\)
*and*
\(D_{x}^{2\beta }v(x,t)\)
*are bounded*.

### Proof

We prove that \(D_{t}^{\alpha}v(x,t)\) is bounded. From the fractional derivative definition (3) we have

$$\begin{aligned} \bigl\Vert D_{t}^{\alpha}v(x,t) \bigr\Vert &= \biggl\Vert \frac{1}{\Gamma(m-\alpha)} \int_{a}^{b} (t-\tau)^{m-\alpha-1}v^{(m)}(x, \tau) \,d\tau \biggr\Vert \\ &\leq \frac{ \vert b-a \vert }{ \vert (m-\alpha)\Gamma(m-\alpha ) \vert } \bigl\Vert v(x,t) \bigr\Vert =L_{1} \bigl\Vert v(x,t) \bigr\Vert , \end{aligned}$$

(44)

where \(L_{1}=\frac{ \vert b-a \vert }{ \vert (m-\alpha)\Gamma (m-\alpha) \vert }\). In the same manner \(\Vert D_{x}^{\beta}v(x,t) \Vert \leq L_{2} \Vert v(x,t) \Vert \) and \(\Vert D_{x}^{2\beta}v(x,t) \Vert \leq L_{3} \Vert v(x,t) \Vert \). □

### Theorem 6.2

*Let*
\(F(v)=-a v v_{x}^{\beta}\)
*satisfy the Lipschitz condition with the Lipschitz constant *
\(L_{4}\). *Then problem* (21) *has the unique solution*
\(v(x,t)\)
*whenever*
\(0 < \gamma<1\).

### Proof

Let *u* and *v* be two different solutions of the space-time fractional Burgers’ equation (21). For all \(t\in[0,T]\) and \(\tau\in[0,t]\), these solutions are bounded on using Lemma 6.1. Let \(M=\max_{0\leq\tau\leq t, 0\leq t\leq T} \vert (t-\tau)^{\alpha-1} \vert \). Then

$$\begin{aligned} & u-v = J^{\alpha}_{t} \bigl(-ahD_{x}^{\beta}u(x, \tau) + cD_{x}^{2\beta}u(x,\tau) + F(u) \bigr) \\ &\phantom{u-v =}{} -J^{\alpha}_{t} \bigl(-ahD_{x}^{\beta}v(x, \tau)+cD_{x}^{2\beta}v(x,\tau) + F(v) \bigr) \\ &\phantom{u-v }= \frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-\tau)^{\alpha-1} \bigl[-ahD_{x}^{\beta}u(x,\tau) + cD_{x}^{2\beta}u(x, \tau) + F(u) \bigr]\,d\tau \\ &\phantom{u-v =} {}-\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-\tau)^{\alpha-1} \bigl[ -ahD_{x}^{\beta}v(x,\tau) + cD_{x}^{2\beta}v(x, \tau) + F(v) \bigr]\,d\tau, \\ \begin{aligned} &\max\vert u-v \vert= \max \biggl\vert \frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-\tau)^{\alpha-1} \bigl[-ahD_{x}^{\beta}u(x,\tau) + cD_{x}^{2\beta}u(x, \tau) + F(u) \bigr]\,d\tau \\ &\phantom{\max\vert u-v \vert=}{} -\frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-\tau)^{\alpha-1} \bigl[- ahD_{x}^{\beta}v(x,\tau) +cD_{x}^{2\beta}v(x, \tau) + F(v) \bigr]\,d\tau\biggr\vert \\ &\phantom{\max\vert u-v \vert}\leq \frac{[ -ahL_{2} + cL_{3} + L_{4}]}{\Gamma(\alpha)}\max \int_{0}^{t}|t - \tau|^{\alpha-1} \vert u - v \vert \,d\tau, \end{aligned} \\ &\Vert u-v \Vert \leq \frac{[ -a h L_{2}+c L_{3} + L_{4}]}{\Gamma(\alpha)}\max \int_{0}^{t}|t - \tau|^{\alpha-1} \vert u - v \vert \,d\tau, \\ &\Vert u-v \Vert \leq \gamma \Vert u-v \Vert , \\ &(1-\gamma) \Vert u-v \Vert \leq 0, \end{aligned}$$

(45)

where \(\gamma= (\frac{[ -a h L_{2} +c L_{3} + L_{4}]M T}{\Gamma (\alpha)} )\). Since \(1-\gamma\neq0\), \(\Vert u-v \Vert = 0\). Therefore, \(u=v\), which completes the proof. □

### Theorem 6.3

*The sequence*
\(v_{n}(x,t)\)
*obtained from VIM iteration* (22) *converges to the exact solution of problem* (21) *for*
\(0<\sigma <1\)
*and*
\(0<\gamma_{1}<1\).

### Proof

The approximate solution using VIM is given by

$$\begin{aligned} v_{n+1}(x,t) = v_{n}(x,t)-\frac{1}{\Gamma(\alpha)} \int_{0}^{t} (t-\tau)^{(\alpha-1)} \bigl(D_{\tau}^{\alpha}v_{n}+ahD_{x}^{\beta}v_{n} -c D_{x}^{2\beta}v_{n}-F(v_{n}) \bigr)\,d \tau. \end{aligned}$$

(46)

Since *v* is the exact solution of equation (21), it satisfies VIM formula, i.e.,

$$\begin{aligned} v(x,t) = v(x,t)-\frac{1}{\Gamma(\alpha)} \int_{0}^{t} (t-\tau)^{(\alpha-1)} \bigl(D_{\tau}^{\alpha}v+ahD_{x}^{\beta}v -c D_{x}^{2\beta}v-F(v) \bigr)\,d\tau. \end{aligned}$$

(47)

On subtracting *v* from \(v_{n+1}\) and recalling that \(E_{n}(x,t)=v_{n}(x,t)-v(x,t)\), we obtain

$$\begin{aligned} & E_{n+1}(x,t) = E_{n}(x,t)- \frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t - \tau)^{\alpha-1} \\ &\phantom{E_{n+1}(x,t) =}{}\times\bigl[D_{t}^{\alpha}(v_{n}-v) + ahD_{x}^{\beta}(v_{n}-v) - cD_{x}^{2\beta}(v_{n}-v) - F(v_{n}-v) \bigr]\,d\tau \end{aligned}$$

(48)

$$\begin{aligned} &\max \bigl\vert E_{n+1}(x,t) \bigr\vert \leq \max \bigl\vert E_{n}(x,t) \bigr\vert -\frac{[L_{1}+ ah L_{2} - c L_{3} - L_{4}] M }{ \Gamma(\alpha)}\max \int_{0}^{t} \bigl\vert E_{n}(x,\tau) \bigr\vert \,d \tau, \end{aligned}$$

(49)

$$\begin{aligned} &\bigl\Vert E_{n+1}(x,t) \bigr\Vert \leq \bigl\Vert E_{n}(x,t) \bigr\Vert -\frac{[L_{1}+ ah L_{2} - c L_{3} - L_{4}] M T}{\Gamma(\alpha)} \bigl\Vert E_{n}(x,t) \bigr\Vert . \end{aligned}$$

(50)

Hence,

$$ \bigl\Vert E_{n+1}(x,t) \bigr\Vert \leq\sigma \bigl\Vert E_{n}(x,t) \bigr\Vert , $$

(51)

where \(\sigma=(1-\gamma_{1})\), \(\gamma_{1}=\frac{[L_{1}+ ah L_{2} - c L_{3} - L_{4}] M T}{\Gamma(\alpha)}\) for all \(n = 0,1,2,\ldots\) Now, if \(n=0\),

$$\begin{aligned} \bigl\Vert E_{1}(x,t) \bigr\Vert \leq \sigma \bigl\Vert E_{0}(x,t) \bigr\Vert , \end{aligned}$$

(52)

while if \(n=1\),

$$\begin{aligned} \bigl\Vert E_{2}(x,t) \bigr\Vert \leq \sigma^{2} \bigl\Vert E_{0}(x,t) \bigr\Vert . \end{aligned}$$

(53)

If \(n=2\), then

$$\begin{aligned} &\bigl\Vert E_{3}(x,t) \bigr\Vert \leq \sigma^{3} \bigl\Vert E_{0}(x,t) \bigr\Vert , \ldots \end{aligned}$$

(54)

$$\begin{aligned} &\bigl\Vert E_{n}(x,t) \bigr\Vert \leq \sigma^{n} \bigl\Vert E_{0}(x,t) \bigr\Vert . \end{aligned}$$

(55)

Since \(0<\sigma<1\), then \(\Vert E_{n}(x,t) \Vert \rightarrow0\) as \(n\rightarrow\infty\), i.e., \(v_{n}\rightarrow v\) and the sequence \(\{v_{n}(x,t) \}_{n=1}^{\infty}\) converges to \(v(x,t)\). □

### Theorem 6.4

*The maximum absolute truncation error of the approximate solution*
\({v_{n}(x,t)}\)
*of the time*-*space fractional Burgers’ equation* (21) *can be estimated as*
\(\Vert E_{n}(x,t) \Vert \leq\frac{\sigma^{n}}{1-\sigma} \Vert v_{1}(x,t) \Vert \).

### Proof

$$\begin{aligned} v_{n}(x,t)-v_{n+1}(x,t) &= \bigl(v_{n}(x,t)-v(x,t) \bigr)+ \bigl(v(x,t)-v_{n+1}(x,t) \bigr) \\ &= E_{n}(x,t)-E_{n+1}(x,t). \end{aligned}$$

(56)

Hence,

$$\begin{aligned} \begin{aligned} &E_{n}(x,t) = E_{n+1}(x,t)- \bigl(v_{n+1}(x,t)-v_{n}(x,t) \bigr), \\ &\bigl\Vert E_{n}(x,t) \bigr\Vert = \bigl\Vert E_{n+1}(x,t)- \bigl(v_{n+1}(x,t)-v_{n}(x,t) \bigr) \bigr\Vert \\ &\bigl\Vert E_{n}(x,t) \bigr\Vert \leq \bigl\Vert E_{n+1}(x,t) \bigr\Vert + \bigl\Vert \bigl(v_{n+1}(x,t)-v_{n}(x,t) \bigr) \bigr\Vert \\ &\phantom{\bigl\Vert E_{n}(x,t) \bigr\Vert }\leq \sigma \bigl\Vert E_{n}(x,t) \bigr\Vert + \bigl\Vert \bigl(v_{n+1}(x,t)-v_{n}(x,t) \bigr) \bigr\Vert . \end{aligned} \end{aligned}$$

(57)

Therefore,

$$\begin{aligned} \bigl\Vert E_{n}(x,t) \bigr\Vert &\leq \frac{ \Vert (v_{n+1}(x,t)-v_{n}(x,t)) \Vert }{1-\sigma} \end{aligned}$$

(58)

$$\begin{aligned} &\leq \frac{\sigma^{n}}{1-\sigma} \bigl\Vert v_{1}(x,t) \bigr\Vert . \end{aligned}$$

(59)

□