**Example 1**

Let us consider the following time-fractional initial boundary value problem:

\begin{array}{c}{D}_{t}^{\alpha}u(x,t)=\frac{1}{2}{x}^{2}{u}_{xx}(x,t),\phantom{\rule{1em}{0ex}}0<\alpha \le 1,t>0,\hfill \\ u(x,0)={x}^{2},\phantom{\rule{2em}{0ex}}u(0,t)=0,\phantom{\rule{2em}{0ex}}u(1,t)={e}^{t}.\hfill \end{array}

The exact solution, for the special case \alpha =1, is given by

Now we can apply series (4) to construct the solution u(x,t) for 0<\alpha \le 1, and we have

u(x,t)={x}^{2}[1+\frac{{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}+\frac{{t}^{2\alpha}}{\mathrm{\Gamma}(2\alpha +1)}+\frac{{t}^{3\alpha}}{\mathrm{\Gamma}(3\alpha +1)}+\cdots ],

which is exactly the same solution as in [18].

**Example 2**

Let us consider the following time-fractional initial boundary value problem:

\begin{array}{c}{D}_{t}^{\alpha}u={u}_{xx}(x,t)+x{u}_{x}(x,t)+u(x,t),\phantom{\rule{1em}{0ex}}0<\alpha \le 1,t>0,\hfill \\ u(x,0)=x,\phantom{\rule{2em}{0ex}}{u}_{x}(x,0)=1,\phantom{\rule{2em}{0ex}}u(0,t)=0.\hfill \end{array}

The exact solution, for the special case \alpha =1, is given by

Now we can apply series (4) to u(x,t) for 0<\alpha \le 1, and we have

u(x,t)={x}^{2}[1+\frac{2{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}+\frac{4{t}^{2\alpha}}{\mathrm{\Gamma}(2\alpha +1)}+\frac{8{t}^{3\alpha}}{\mathrm{\Gamma}(3\alpha +1)}+\cdots ],

which is exactly the same solution as in [19].

**Example 3**

Let us consider the nonlinear time-fractional Fisher’s equation

\begin{array}{c}{D}_{t}^{\alpha}u={u}_{xx}(x,t)+6u(x,t)(1-u(x,t)),\phantom{\rule{1em}{0ex}}0<\alpha \le 1,t>0,\hfill \\ u(x,0)=\frac{1}{{(1+{e}^{x})}^{2}}.\hfill \end{array}

The exact solution, for the special case \alpha =1, is given by

u(x,t)=\frac{1}{{(1+{e}^{x-5t})}^{2}}.

As in the previous examples, we can apply series (4) to obtain the solution u(x,t) for 0<\alpha \le 1, and we get

u(x,t)=\frac{1}{{(1+{e}^{x})}^{2}}+\frac{10{e}^{x}}{{(1+{e}^{x})}^{3}}\frac{{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}+\frac{50{e}^{x}(-1+2{e}^{x})}{{(1+{e}^{x})}^{4}}\frac{{t}^{2\alpha}}{\mathrm{\Gamma}(2\alpha +1)}+\cdots ,

which is totaly the same solution as in [20].

**Example 4**

Let us consider the following time-fractional initial boundary value problem:

{D}_{t}^{\alpha}u=\pm {u}_{xx}(x,t),\phantom{\rule{1em}{0ex}}0<\alpha \le 1,t>0,

(17)

u(x,0)={x}^{2},\phantom{\rule{2em}{0ex}}u(0,t)=\frac{2k{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)},\phantom{\rule{2em}{0ex}}u(\ell ,t)={\ell}^{2}+\frac{2k{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)},

(18)

where the boundary conditions are given in fractional terms.

Boundary value problem (17)-(18), for the special case \alpha =1, becomes as follows:

\begin{array}{c}{u}_{t}=\pm {u}_{xx}(x,t),\phantom{\rule{1em}{0ex}}0<\alpha \le 1,t>0,\hfill \\ u(x,0)={x}^{2},\phantom{\rule{2em}{0ex}}u(0,t)=2kt,\phantom{\rule{2em}{0ex}}u(\ell ,t)={\ell}^{2}+2kt,\hfill \end{array}

and its analytic solution is obtained as follows:

Now we can apply series (4) to u(x,t) for 0<\alpha \le 1, and we have

u(x,t)={x}^{2}+2k\frac{{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)},

which is exactly the same solution as in [21].

**Example 5**

Let us consider the following space-fractional initial boundary value problem:

{u}_{t}(x,t)={D}_{x}^{\beta}u(x,t)-(1+tant)u(x,t),\phantom{\rule{1em}{0ex}}1<\beta \le 2,

(19)

u(x,0)={e}^{x},\phantom{\rule{2em}{0ex}}u(0,t)=cost,\phantom{\rule{2em}{0ex}}{u}_{x}(0,t)=cost.

(20)

The exact solution, for the special case \beta =2, is given by

Now we can apply series (9) to u(x,t) for 1<\beta \le 2, then we have

u(x,t)=cost(1+x+\frac{{x}^{\beta}}{\mathrm{\Gamma}(\beta +1)}+\frac{{x}^{\beta +1}}{\mathrm{\Gamma}(\beta +2)}+\frac{{x}^{2\beta}}{\mathrm{\Gamma}(2\beta +1)}+\frac{{x}^{2\beta +1}}{\mathrm{\Gamma}(2\beta +2)}+\cdots ).

**Example 6**

Let us consider the following space and time-fractional initial boundary value problem:

{D}_{t}^{\alpha}u(x,t)=2{D}_{x}^{\beta}u(x,t)+cost,\phantom{\rule{1em}{0ex}}0<\alpha \le 1,1<\beta \le 2,

(21)

u(x,0)={e}^{x},\phantom{\rule{2em}{0ex}}u(0,t)={e}^{2t}+sint,\phantom{\rule{2em}{0ex}}{u}_{x}(0,t)={e}^{2t}+sint.

(22)

The exact solution, for the special case \alpha =1 and \beta =2, is given by

u(x,t)={e}^{2t}{e}^{x}+sint.

Now we can apply the series (4)-(9) to u(x,t) for 0<\alpha \le 1 and 1<\beta \le 2, then we have

\begin{array}{rl}u(x,t)=& (1+\frac{2{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}+\frac{4{t}^{2\alpha}}{\mathrm{\Gamma}(2\alpha +1)}+\cdots )(1+x+\frac{{x}^{\beta}}{\mathrm{\Gamma}(\beta +1)}+\frac{{x}^{\beta +1}}{\mathrm{\Gamma}(\beta +2)}+\cdots )\\ +(\frac{{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}-\frac{{t}^{3\alpha}}{\mathrm{\Gamma}(3\alpha +1)}+\frac{{t}^{5\alpha}}{\mathrm{\Gamma}(5\alpha +1)}-\cdots ).\end{array}

**Example 7**

Let us consider the following space and time-fractional initial boundary value problem:

{D}_{t}^{\alpha}u(x,t)=2{D}_{x}^{\beta}u(x,t)+2sinx,\phantom{\rule{1em}{0ex}}0<\alpha \le 1,1<\beta \le 2,

(23)

u(x,0)={e}^{x}+sinx,\phantom{\rule{2em}{0ex}}u(0,t)={e}^{2t},\phantom{\rule{2em}{0ex}}{u}_{x}(0,t)={e}^{2t}+1.

(24)

The exact solution, for the special case \alpha =1 and \beta =2, is given by

u(x,t)={e}^{2t}{e}^{x}+sinx.

Now we can apply the series (4)-(9) to u(x,t) for 0<\alpha \le 1 and 1<\beta \le 2, then we have

\begin{array}{rl}u(x,t)=& (1+\frac{2{t}^{\alpha}}{\mathrm{\Gamma}(\alpha +1)}+\frac{4{t}^{2\alpha}}{\mathrm{\Gamma}(2\alpha +1)}+\cdots )(1+x+\frac{{x}^{\beta}}{\mathrm{\Gamma}(\beta +1)}+\frac{{x}^{\beta +1}}{\mathrm{\Gamma}(\beta +2)}+\cdots )\\ +(x-\frac{{x}^{\beta +1}}{\mathrm{\Gamma}(\beta +2)}+\frac{{x}^{2\beta +1}}{\mathrm{\Gamma}(2\beta +2)}-\frac{{x}^{3\beta +1}}{\mathrm{\Gamma}(3\beta +2)}-\cdots ).\end{array}

Figures 7-9 show the evolution results for the approximate solutions of problem (23)-(24) obtained for different values of *α* and *β*.