**Example 3.1** Consider the following equation:

\{\begin{array}{c}\frac{\rho {t}^{1/2}}{1.128}\frac{{\partial}^{1/2}u(t,z)}{\partial {t}^{1/2}}+{D}_{z}^{\beta}u(t,z)=0,\phantom{\rule{1em}{0ex}}t\in J=[0,1]\hfill \\ u(0,z)=0,\phantom{\rule{1em}{0ex}}\text{in a neighborhood of}z=0,\hfill \end{array}

(14)

where u(t,z) is the unknown function \rho \in (0,1) and \beta \in (0,1].

We propose to show that (14) has a unique analytic solution by using the Banach fixed point theorem. By assuming

u(t,z)=\mu (z)t+v(t,z)\phantom{\rule{1em}{0ex}}(v(t,z)=O\left({t}^{2}\right))

as a formal solution, where \mu (z)=O({z}^{\beta}), calculations imply

{t}^{1/2}\frac{{\partial}^{1/2}u(t,z)}{\partial {t}^{1/2}}=1.128\mu (z)t+{t}^{1/2}{v}_{\alpha}(t,z),\phantom{\rule{1em}{0ex}}\alpha =1/2,

and

{D}_{z}^{\beta}u(t,z)={D}_{z}^{\beta}(\mu (z)t+v(t,z))=t{D}_{z}^{\beta}\mu (z)+{v}_{\beta}(t,z).

Therefore, \mu (z) satisfies

\rho \mu (z)+{D}_{z}^{\beta}\mu (z)=0,

which is equivalent to

{D}_{z}^{\beta}\mu (z)=g(z,\mu (z)),

(15)

where

g(z,\mu (z))=-\rho \mu (z).

Now g(z,\mu (z)) is a contraction mapping whenever \rho \in (0,1); therefore, in view of the Banach fixed point theorem, Eq. (15) has a unique analytic solution in the unit disk and consequently the problem (14).

To calculate the fractal index for the equation

{D}_{z}^{\beta}\mu (z)+\rho \mu (z)=0,\phantom{\rule{1em}{0ex}}\mu (0)=1,

(16)

we assume the transform Z={z}^{\beta} and the solution can be expressed in a series in the form

\mu (Z)=\sum _{m=0}^{\mathrm{\infty}}{\mu}_{m}{Z}^{m},

(17)

where {\mu}_{m} are constants. Substituting (17) into Eq. (16) yields

\frac{\partial}{\partial Z}\sum _{m=0}^{\mathrm{\infty}}{\theta}_{\beta m}{\mu}_{m}{Z}^{m}+\rho \sum _{m=0}^{\mathrm{\infty}}{\mu}_{m}{Z}^{m}=0.

(18)

Since

{\theta}_{\beta m}=\frac{\mathrm{\Gamma}(1+m\beta )}{m\mathrm{\Gamma}(1+m\beta -\beta )},

then the computation imposes the relation

\frac{\mathrm{\Gamma}(1+m\beta )}{\mathrm{\Gamma}(1+m\beta -\beta )}{\mu}_{m}+\rho {\mu}_{m-1}=0

with \mu (0)=1, and consequently we obtain

{\mu}_{m}=\frac{{(-\rho )}^{m}}{\mathrm{\Gamma}(1+m\beta )}.

Thus, we have the following solution:

\mu (Z)=\sum _{m=0}^{\mathrm{\infty}}\frac{{(-\rho )}^{m}}{\mathrm{\Gamma}(1+m\beta )}{Z}^{m}

which is equivalent to

\mu (z)=\sum _{m=0}^{\mathrm{\infty}}\frac{{(-\rho )}^{m}}{\mathrm{\Gamma}(1+m\beta )}{z}^{m\beta}={E}_{\beta}(-\rho {z}^{\beta}),

where {E}_{\beta} is a Mittag-Leffler function. The last assertion is the exact solution for the problem (16) and consequently for (14).

**Example 3.2** Consider the following equation:

\{\begin{array}{c}\frac{{t}^{1/2}/2}{1.128}\frac{{\partial}^{1/2}u(t,z)}{\partial {t}^{1/2}}+4z\frac{\partial u(t,z)}{\partial z}+{D}_{z}^{\beta}u(t,z)={z}^{\beta}t,\phantom{\rule{1em}{0ex}}t\in J=[0,1],\hfill \\ u(0,z)=0,\phantom{\rule{1em}{0ex}}\text{in a neighborhood of}z=0,\hfill \end{array}

(19)

where u(t,z) is the unknown function and \beta \in (0,1]. In the same manner of Example 3.1, we let

u(t,z)=\mu (z)t+v(t,z)\phantom{\rule{1em}{0ex}}(v(t,z)=O\left({t}^{2}\right))

as a formal solution, where \mu (z)=O({z}^{\beta}) and

|{\mu}^{\prime}(z)-{\nu}^{\prime}(z)|<\lambda |\mu (z)-\nu (z)|,\phantom{\rule{1em}{0ex}}\lambda <\frac{1}{8}.

Estimations imply

and

{D}_{z}^{\beta}u(t,z)={D}_{z}^{\beta}(\mu (z)t+v(t,z))=t{D}_{z}^{\beta}\mu (z)+{v}_{\beta}(t,z).

Therefore, \mu (z) satisfies

\frac{\mu (z)}{2}+4z{\mu}^{\prime}(z)+{D}_{z}^{\beta}\mu (z)-{z}^{\beta}=0,

which is equivalent to

{D}_{z}^{\beta}\mu (z)=G(z,\mu (z),z{\mu}^{\prime}(z)),

(20)

where

G(z,\mu (z),z{\mu}^{\prime}(z))={z}^{\beta}-1/2\mu (z)-4z{\mu}^{\prime}(z).

Now, to show that G(z,\mu (z),z{\mu}^{\prime}(z)) is a contraction mapping,

Thus, in view of the Banach fixed point theorem, Eq. (20) has a unique analytic solution in the unit disk and consequently the problem (19).

To evaluate the fractal index for the equation

{D}_{z}^{\beta}\mu (z)+\frac{\mu (z)}{2}+4z{\mu}^{\prime}(z)-{z}^{\beta}=0,\phantom{\rule{1em}{0ex}}\mu (0)=1,

(21)

we assume the transform Z={z}^{\beta} and the solution can be articulated as in (17). Substituting (17) into Eq. (21), we have

\frac{\partial}{\partial Z}\sum _{m=0}^{\mathrm{\infty}}{\theta}_{\beta m}{\mu}_{m}{Z}^{m}+\frac{1}{2}\sum _{m=0}^{\mathrm{\infty}}{\mu}_{m}{Z}^{m}+4\sum _{m=1}^{\mathrm{\infty}}m{\mu}_{m}{Z}^{m}-Z=0,

(22)

where

{\theta}_{\beta m}=\frac{\mathrm{\Gamma}(1+m\beta )}{m\mathrm{\Gamma}(1+m\beta -\beta )}.

Hence, the computation imposes the relation

(\frac{\mathrm{\Gamma}(1+m\beta )}{\mathrm{\Gamma}(1+m\beta -\beta )}+4m){\mu}_{m}+\frac{1}{2}{\mu}_{m-1}=0

with \mu (0)=1, and consequently we obtain

{\mu}_{m}:=\frac{{({B}_{m})}^{m}}{\mathrm{\Gamma}(1+m\beta )},

where {B}_{m} in terms of a gamma function. If we let B:={max}_{m}\{{B}_{m}\}, then the solution approximates to

\mu (Z)\simeq \sum _{m=0}^{\mathrm{\infty}}\frac{{(B)}^{m}}{\mathrm{\Gamma}(1+m\beta )}{Z}^{m},

which is equivalent to

\mu (z)=\sum _{m=0}^{\mathrm{\infty}}\frac{{(B)}^{m}}{\mathrm{\Gamma}(1+m\beta )}{z}^{m\beta}={E}_{\beta}\left(B{z}^{\beta}\right).

The last assertion is the exact solution for the problem (21) and consequently for (19).

Next, we consider the Cauchy problem by employing the generalized fractional differential operator (4). We shall show that the solution of such a problem can be determined in terms of the Fox-Wright function [37]:

where {A}_{j}>0 for all j=1,\dots ,p, {B}_{j}>0 for all j=1,\dots ,q, and 1+{\sum}_{j=1}^{q}{B}_{j}-{\sum}_{j=1}^{p}{A}_{j}\ge 0 for suitable values |w|<1 and {a}_{i}, {b}_{j} are complex parameters.

**Example 3.3** Consider the Cauchy problem in terms of the differential operator (4)

{D}_{z}^{\alpha ,\mu}u(z)=F(z,u(z)),

(23)

where F(z,u(z)) is analytic in *u* and u(z) is analytic in the unit disk. Thus, *F* can be expressed by

Let Z={z}^{\alpha (\mu +1)}. Then the solution can be formulated as follows:

u(Z)=\sum _{m=0}^{\mathrm{\infty}}{u}_{m}{Z}^{m},

(24)

where {u}_{m} are constants. Substituting (24) into Eq. (23) implies

\frac{\partial}{\partial Z}\sum _{m=0}^{\mathrm{\infty}}{\mathrm{\Theta}}_{\alpha ,\mu ,m}{u}_{m}{Z}^{m}-\varphi \sum _{m=0}^{\mathrm{\infty}}{u}_{m}{Z}^{m}=0.

(25)

Since

{\mathrm{\Theta}}_{\alpha ,\mu ,m}=\frac{{(\mu +1)}^{\alpha}\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1)}{m\mathrm{\Gamma}(\frac{n\alpha}{\mu +1}+1-\alpha )},

then the calculation yields the relation

\frac{{(\mu +1)}^{\alpha}\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1)}{\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1-\alpha )}{u}_{m}-\varphi {u}_{m-1}=0;

consequently, we obtain

{u}_{m}={\left[\frac{\varphi}{{(\mu +1)}^{\alpha}}\right]}^{m}\frac{\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1-\alpha )\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1-\alpha )}{\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1)\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1)}.

Thus, we have the following solution:

u(Z)=\sum _{m=0}^{\mathrm{\infty}}{\left[\frac{\varphi}{{(\mu +1)}^{\alpha}}\right]}^{m}\frac{\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1-\alpha )\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1-\alpha )}{\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1)\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1)}{Z}^{m}

which is equivalent to

u(Z)=\sum _{m=0}^{\mathrm{\infty}}{\left[\frac{\varphi}{{(\mu +1)}^{\alpha}}\right]}^{m}\frac{\mathrm{\Gamma}(m+1)\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1-\alpha )\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1-\alpha )}{\mathrm{\Gamma}(\frac{(m-1)\alpha}{\mu +1}+1)\mathrm{\Gamma}(\frac{m\alpha}{\mu +1}+1)}\frac{{Z}^{m}}{m!}.

Since *ϕ* is an arbitrary constant, we assume that

\varphi :={(\mu +1)}^{\alpha}.

Thus, for a suitable *α*, we present

or

where |z|<1.