Fractal Hankel Transform

Abstract

This paper explores the extension of classical transforms to fractal spaces, focusing on the development and application of the Fractal Hankel Transform. We begin with a concise review of fractal calculus to set the theoretical groundwork. The Fractal Hankel Transform is then introduced, along with its formulation and properties. Applications of this transform are presented to demonstrate its utility and effectiveness in solving problems within fractal spaces. Finally, we conclude by summarizing the key findings and discussing potential future research directions in the field of fractal analysis and transformations.

1. Introduction

Fractal geometry has been widely used to study and characterize recurring patterns in natural phenomena, such as blood vessels, mountains, clouds, and coastlines. This article examines the unique properties and measurements of these formations. Common characteristics of fractals include fractional dimensions, self-similarity, and fractal dimensions larger than their topological dimensions [1,2,3,4].

Fractal analysis encompasses techniques from fractional calculus [5], stochastic processes [6], fractional space [7], measure theory [8] and harmonic analysis [9,10,11,12]. Through modifications to the Fundamental Theorem of Calculus, numerous Riemann-type integrals, including s-Riemann, s-HK, and s-first-return integrals, have been developed. Researchers have examined connections between these integrals and the Lebesgue integral related to the Hausdorff measure [13,14,15,16].

Iterated function systems (IFSs) have been introduced for the construction of more general sets that do not necessarily exhibit the strict self-similarity characteristic of traditional IFSs [17]. A class of Hausdorff measures has been examined in a separable metric space, defined through a measure and a premeasure. The discussion has focused on the Hausdorff structure of product sets, emphasizing the importance of weighted Hausdorff measures as a crucial tool for analyzing these product sets [18].

The notion of upper metric mean dimension with potential has been introduced for any subset through Carathéodory–Pesin structures. Several possible versions of upper measure-theoretic mean dimensions with potential have been discussed, and conditions have been found to ensure that these notions coincide [19].

While extensive literature exists on fractal analysis, we focus on select works that introduce novel techniques for Fourier expansions and wavelets tailored to specific classes of fractals [20,21]. Due to their complex geometry and cross-scale self-similarity, fractals often require methods beyond classical mathematics [22].

A new framework for fractal calculus extends classical calculus and provides an algorithmic, geometric, and physically meaningful approach. This generalization enables the study of functions with fractal features, such as Cantor sets and Koch curves [23,24,25,26]. Non-local fractal calculus has been introduced to simulate incompressible visc ous fluids in fractal media and processes with memory, similar to extensions of local calculus by the Riemann–Liouville and Caputo generalizations [27,28,29].

In studying sub- and super-diffusion with fractal local derivatives, researchers have maintained locality and the central limit theorem [26]. Fractal calculus has also contributed to physics [30], enabling models of fractal space [31] and time [32,33,34,35,36] and describing power law and self-similarity solutions in these models. Fractal Laplace, Sumudu, and Fourier transforms applied to Cantor sets and fractal curves have been used to study systems with fractal times [26,37]. Fractal derivatives and integrals have been computed for other functions, including the Weierstrass function [26].

Non-standard analysis using hyperreal and hyperinteger numbers has facilitated defining derivative structures and integrals on fractal curves, as well as specifying fractal integral and differential forms [38]. Fractal calculus has been extended to unbounded functions through gauge function generalizations [26]. Stability analysis of fractal differential equations has also established criteria for unique, stable solutions [26]. Additionally, a novel framework extending mean square calculus has been introduced, including concepts such as the mean square derivative, fractal mean square integral, fractal mean square continuity, and the fractal order of random variables on fractal curves [39].

The Fractal Hankel Transform [40,41,42,43,44] can be applied to systems modeled by the Keller–Segel equations in the context of fractal-like or anomalous behavior, especially when the systems exhibit scale invariance or self-similarity. The use of FHT could reveal deeper insights into the spectral properties of the system or help in analyzing non-integer-dimensional behavior that may arise in certain cases of chemotaxis and pattern formation. Thus, while they stem from different origins (one from the study of fractals, the other from biological diffusion), their connection lies in their ability to describe and analyze complex, non-linear, and potentially fractal-like patterns in systems that exhibit spatial and temporal evolution [45,46,47,48,49].

The main objective of this paper is to develop a generalized Hankel transform on a fractal set.

The paper is organized as follows. Section 2 provides a concise review of fractal calculus. Section 3 introduces the Fractal Hankel Transform, followed by its applications in Section 4. The conclusion is presented in Section 5.

2. Fundamental Definitions in Fractal Calculus

In this section, we provide an overview of fractal calculus as it pertains to the Cantor set $F\subset [a,b]\subset \mathbb{R}$, drawing on foundational insights from [23,26].

Definition 1. 

The indicator function ${⊮}_F\left(I\right)$ for the set F is defined as follows:

$${⊮}_F\left(I\right)=\left\{\begin{array}{cc}1,\hfill & ifF\cap I\ne \varnothing ,\hfill \\ 0,\hfill & otherwise,\hfill \end{array}\right.$$

where $I=[a,b]\subset \mathbb{R}$.

Definition 2. 

The coarse-grained measure ${\mu }_{\delta }^{\alpha }(F,a,b)$ of $F\cap [a,b]$ is defined as

$${\mu }_{\delta }^{\alpha }(F,a,b)=\underset{\left|P\right|\le \delta }{inf}\sum _{j=0}^{n-1}\mathsf{\Gamma }(\alpha +1){(x_{j+1}-x_j)}^{\alpha }{⊮}_F\left([x_j,x_{j+1}]\right),$$

where $\left|P\right|={max}_{0\le j\le n-1}(x_{j+1}-x_j)$, $P_{[a,b]}=\{x_0=a,x_1,\cdots ,x_n=b\}$, $0<\alpha \le 1$, and $\mathsf{\Gamma }(·)$ is the Gamma function.

Definition 3. 

The measure function ${\mu }^{\alpha }(F,a,b)$ for F is given by

$${\mu }^{\alpha }(F,a,b)=\underset{\delta \to 0}{lim}{\mu }_{\delta }^{\alpha }(F,a,b).$$

Definition 4. 

The fractal ν-dimension of $F\cap [a,b]$ is defined by

$${dim}_{\nu }(F\cap [a,b])=inf\{\alpha :{\mu }^{\alpha }(F,a,b)=0\}=sup\{\alpha :{\mu }^{\alpha }(F,a,b)=\infty \}.$$

(1)

Definition 5. 

The integral staircase function $S_F^{\alpha }\left(x\right)$ is defined as

$$S_F^{\alpha }\left(x\right)=\left\{\begin{array}{cc}{\mu }^{\alpha }(F,a_0,x),\hfill & ifx\ge a_0,\hfill \\ -{\mu }^{\alpha }(F,x,a_0),\hfill & ifx<a_0,\hfill \end{array}\right.$$

where $a_0\in \mathbb{R}$ is a fixed constant.

Definition 6. 

The characteristic function ${\chi }_F\left(x\right)$ on a fractal set F is defined by

$${\chi }_F\left(x\right)=\left\{\begin{array}{cc}\mathsf{\Gamma }(\alpha +1),\hfill & ifx\in F,\hfill \\ 0,\hfill & ifx\notin F,\hfill \end{array}\right.$$

where α denotes the fractal dimension as defined above.

Definition 7. 

For a function $g:F\to \mathbb{R}$, the $F_{-}lim$ of $g\left(x\right)$ at a point $x\in F$ is the value L such that for any $ϵ>0$, there exists $\delta >0$ satisfying

$$y\in Fand|y-x|<\delta \Rightarrow \left|g\right(y)-L|<ϵ.$$

If such an L exists, it is denoted by

$$L=\underset{y\to x}{F_{-}lim}g\left(y\right).$$

Definition 8. 

A function $g:F\to \mathbb{R}$ is F-continuous at $x\in F$ if

$$g\left(x\right)=\underset{y\to x}{F_{-}lim}g\left(y\right),$$

whenever the $F_{-}lim$ exists.

Definition 9. 

For a function g on an α-perfect fractal set F, the $F^{\alpha }$-derivative of g at x is defined as

$$D_F^{\alpha }g\left(x\right)=\left\{\begin{array}{cc}\underset{y\to x}{F_{-}lim}{\displaystyle \frac{g\left(y\right)-g\left(x\right)}{S_F^{\alpha }\left(y\right)-S_F^{\alpha }\left(x\right)}},\hfill & ifx\in F,\hfill \\ 0,\hfill & ifx\notin F,\hfill \end{array}\right.$$

provided the $F_{-}lim$ exists.

Definition 10. 

The $F^{\alpha }$-integral of a bounded function $g\left(x\right)$, where $g\in B\left(F\right)$ (i.e., g is bounded on F), is defined as

$${\int }_a^{b}g\left(x\right)d_F^{\alpha }x=\underset{P_{[a,b]}}{sup}\sum _{j=0}^{n-1}\underset{x\in F\cap I}{inf}g\left(x\right)(S_F^{\alpha }\left(x_{j+1}\right)-S_F^{\alpha }\left(x_j\right))$$

$$=\underset{P_{[a,b]}}{inf}\sum _{j=0}^{n-1}\underset{x\in F\cap I}{sup}g\left(x\right)(S_F^{\alpha }\left(x_{j+1}\right)-S_F^{\alpha }\left(x_j\right)),$$

where $x\in F$, and the infimum or supremum is taken over all partitions $P_{[a,b]}$ [23,26].

Remark 1. 

We note that fractal calculus generalizes classical calculus by defining operations on fractal sets and curves. A notable distinction is that in fractal calculus, the support function on fractal sets often leads to setting $S_F^{\alpha }\left(x\right)=x$ when $\alpha =1$. In contrast, fractional calculus operates on the real line and is non-local. While both involve non-integer orders, in fractal calculus, the order corresponds to the fractal dimension of the support, providing a geometric interpretation that fractional calculus lacks.

3. Fractal Hankel Transform

The Hankel transform, which is an integral transform akin to the Fourier transform, is predominantly utilized for functions that exhibit cylindrical symmetry. It proves to be especially effective in addressing problems within cylindrical coordinates, which frequently arise in areas such as heat conduction, wave propagation, and signal processing. In this section, we extend the Hankel transform to functions defined on fractal domains, thereby broadening its applicability to challenges involving fractal geometries.

Definition 11. 

The fractal Hankel transform of order n (with $n\ge 0$) for a function $f\left(r\right)$, where $r\in F$ represents the radial coordinate, is defined as follows:

$${\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}=F_n\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r,$$

(2)

where:

$$J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)=\sum _{m=0}^{\infty }{\displaystyle \frac{{(-1)}^m}{m!\mathsf{\Gamma }(m+n+1)}}{\left({\displaystyle \frac{S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)}{2}}\right)}^{2m+n}$$

(3)

is the fractal Bessel function of the first kind of order n. Here, $S_F^{\alpha }\left(k\right)$ denotes the transform variable (the fractal wavenumber) and $S_F^{\alpha }\left(r\right)$ represents the fractal radial distance in fractal space.

Remark 2. 

In the fractal Hankel transform, the domain of functions is illustrated in Figure 1.

Definition 12. 

The inverse fractal Hankel transform, which allows for the reconstruction of the original function $f\left(S_F^{\alpha }\left(r\right)\right)$ from its transform $F_n\left(S_F^{\alpha }\left(k\right)\right)$, is defined as follows:

$$f\left(S_F^{\alpha }\left(r\right)\right)={\int }_0^{\infty }{F}_n\left(S_F^{\alpha }\left(k\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(k\right)d_F^{\alpha }k.$$

(4)

Proposition 1. 

The fractal Hankel transform is a linear operator, which can be expressed as follows:

$${\mathcal{H}}_n\{af\left(S_F^{\alpha }\left(r\right)\right)+bg\left(S_F^{\alpha }\left(r\right)\right)\}=a{\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}+b{\mathcal{H}}_n\left\{g\left(S_F^{\alpha }\left(r\right)\right)\right\},$$

where a and b are constants.

Proof. 

The fractal Hankel transform of a function $f\left(S_F^{\alpha }\left(r\right)\right)$ of order n is defined as follows:

$${\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r,$$

where $J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)$ denotes the fractal Bessel function of the first kind of order n. Now, let us examine the linear combination $af\left(S_F^{\alpha }\left(r\right)\right)+bg\left(S_F^{\alpha }\left(r\right)\right)$. The fractal Hankel transform of this combination can be expressed as:

$$\begin{array}{cc}\hfill & {\mathcal{H}}_n\{af\left(S_F^{\alpha }\left(r\right)\right)+bg\left(S_F^{\alpha }\left(r\right)\right)\}\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & ={\int }_0^{\infty }\left(af\left(S_F^{\alpha }\left(r\right)\right)+bg\left(S_F^{\alpha }\left(r\right)\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.\hfill \end{array}$$

(5)

Utilizing the linearity of fractal integration, we can break down the terms as follows:

$$\begin{array}{cc}\hfill {\mathcal{H}}_n\{af\left(S_F^{\alpha }\left(r\right)\right)& + bg\left(S_F^{\alpha }\left(r\right)\right)\}\left(S_F^{\alpha }\left(k\right)\right)=a{\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\hfill \\ & + b{\int }_0^{\infty }g\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.\hfill \end{array}$$

(6)

This leads to the following result:

$${\mathcal{H}}_n\{af\left(S_F^{\alpha }\left(r\right)\right)+bg\left(S_F^{\alpha }\left(r\right)\right)\}\left(S_F^{\alpha }\left(k\right)\right)=a{\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)+b{\mathcal{H}}_n\left\{g\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right).$$

Therefore, we conclude that the fractal Hankel transform is indeed linear, as demonstrated. □

Theorem 1. 

If $f\left(S_F^{\alpha }\left(r\right)\right)$ and $g\left(S_F^{\alpha }\left(r\right)\right)$ are two functions, then the fractal convolution $(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)$ in the fractal radial domain is equivalent to the product of their corresponding fractal Hankel transforms.

$${\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}={\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}·{\mathcal{H}}_n\left\{g\left(S_F^{\alpha }\left(r\right)\right)\right\}.$$

Proof. 

The convolution of two functions $f\left(S_F^{\alpha }\left(r\right)\right)$ and $g\left(S_F^{\alpha }\left(r\right)\right)$ in the fractal radial domain is defined by:

$$(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(s\right)\right)g\left(\right|S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right)\left|\right)S_F^{\alpha }\left(s\right)d_F^{\alpha }s.$$

The fractal Hankel transform of order n for a function $f\left(S_F^{\alpha }\left(r\right)\right)$ is given by:

$${\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.$$

Now, consider the fractal Hankel transform of the convolution $(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)$:

$${\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.$$

By substituting the expression for the convolution, we arrive at the following:

$$\begin{array}{cc}\hfill & {\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & ={\int }_0^{\infty }\left({\int }_0^{\infty }f\left(S_F^{\alpha }\left(s\right)\right)g\left(\right|S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right)\left|\right)S_F^{\alpha }\left(s\right)d_F^{\alpha }s\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.\hfill \end{array}$$

(7)

Interchanging the order of fractal integration, we get the following:

$$\begin{array}{cc}\hfill & {\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & ={\int }_0^{\infty }f\left(S_F^{\alpha }\left(s\right)\right)S_F^{\alpha }\left(s\right)\left({\int }_0^{\infty }g\left(\right|S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right)\left|\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\right)d_F^{\alpha }s.\hfill \end{array}$$

(8)

Utilizing the identity for the fractal Hankel transform of a convolution, we obtain the following:

$${\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(k\right)={\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)·{\mathcal{H}}_n\left\{g\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right).$$

Hence, the fractal convolution theorem for the fractal Hankel transform holds as follows:

$${\mathcal{H}}_n\left\{(f\ast g)\left(S_F^{\alpha }\left(r\right)\right)\right\}={\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}·{\mathcal{H}}_n\left\{g\left(S_F^{\alpha }\left(r\right)\right)\right\}.$$

□

Theorem 2. 

For a function $f\left(S_F^{\alpha }\left(r\right)\right)$ and its fractal Hankel transform $F_n\left(S_F^{\alpha }\left(k\right)\right)$, the following relation holds:

$${\int }_0^{\infty }|f\left(S_F^{\alpha }\left(r\right)\right){|}^2{S}_F^{\alpha }\left(r\right)d_F^{\alpha }r={\int }_0^{\infty }{\left|F_n\left(S_F^{\alpha }\left(k\right)\right)\right|}^2{S}_F^{\alpha }\left(k\right)d_F^{\alpha }k$$

where $F_n\left(S_F^{\alpha }\left(k\right)\right)$ is the fractal Hankel transform of $f\left(S_F^{\alpha }\left(r\right)\right)$ of order n and $S_F^{\alpha }\left(k\right)$ is the fractal wavelength variable.

Proof. 

From the definition of the Hankel transform, we have the following:

$$F_n\left(S_F^{\alpha }\left(k\right)\right)={\mathcal{H}}_n\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r.$$

Thus, the magnitude squared of $F_n\left(S_F^{\alpha }\left(k\right)\right)$ is as follows:

$$\begin{array}{cc}\hfill & |F_n\left(S_F^{\alpha }\left(k\right)\right){|}^2=\left({\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)dr\right)\hfill \\ & \ast \left({\int }_0^{\infty }f\left(S_F^{\alpha }\left(s\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(s\right)\right)S_F^{\alpha }\left(s\right)d_F^{\alpha }s\right).\hfill \end{array}$$

(9)

Then, it follows that:

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\left|F_n\left(S_F^{\alpha }\left(k\right)\right)\right|}^2{S}_F^{\alpha }\left(k\right)d_F^{\alpha }k& ={\int }_0^{\infty }\left({\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\right)\hfill \end{array}$$

$$\begin{array}{cc} & \hfill \left({\int }_0^{\infty }f\left(S_F^{\alpha }\left(s\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(s\right)\right)S_F^{\alpha }\left(s\right)d_F^{\alpha }s\right)S_F^{\alpha }\left(k\right)d_F^{\alpha }k\end{array}$$

(10)

The fractal Bessel functions $J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)$ satisfy the following orthogonality condition:

$${\int }_0^{\infty }{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(s\right)\right)S_F^{\alpha }\left(k\right)d_F^{\alpha }k={\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}\delta (S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right)),$$

where $\delta (S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right))$ is the fractal Dirac delta function. Using this orthogonality, we can simplify Equation (10) as follows:

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{\left|F_n\left(S_F^{\alpha }\left(k\right)\right)\right|}^2{S}_F^{\alpha }\left(k\right)d_F^{\alpha }k\hfill \\ & ={\int }_0^{\infty }{\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)f\left(S_F^{\alpha }\left(s\right)\right){\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}\delta (S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right))S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(s\right)d_F^{\alpha }r{d}_F^{\alpha }s.\hfill \end{array}$$

(11)

The fractal Dirac delta function $\delta (S_F^{\alpha }\left(r\right)-S_F^{\alpha }\left(s\right))$ simplifies the fractal double integral to a fractal single integral:

$${\int }_0^{\infty }|F_n\left(S_F^{\alpha }\left(k\right)\right){|}^2{S}_F^{\alpha }\left(k\right)d_F^{\alpha }k={\int }_0^{\infty }{\left|f\left(S_F^{\alpha }\left(r\right)\right)\right|}^2{S}_F^{\alpha }\left(r\right)d_F^{\alpha }r.$$

Thus, we have:

$${\int }_0^{\infty }|f\left(S_F^{\alpha }\left(r\right)\right){|}^2{S}_F^{\alpha }\left(r\right)d_F^{\alpha }r={\int }_0^{\infty }{\left|F_n\left(S_F^{\alpha }\left(k\right)\right)\right|}^2{S}_F^{\alpha }\left(k\right)d_F^{\alpha }k,$$

which completes the proof. □

Example 1. 

Consider $f\left(S_F^{\alpha }\left(r\right)\right)=S_F^{\alpha }{\left(r\right)}^{-1}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))$. To obtain the zero-order fractal Hankel transforms of it by using Equation (2), we have:

$${\mathcal{H}}_0\left\{f\left(S_F^{\alpha }\left(r\right)\right)\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r,$$

where $J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)$ is the zeroth-order fractal Bessel function of the first kind. We need to compute the following:

$$\begin{array}{cc}\hfill & {\mathcal{H}}_0\left\{{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))\right\}\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & ={\int }_0^{\infty }{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\hfill \\ & ={\int }_0^{\infty }exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(12)

Using tables of fractal integral transforms, and the fractal Hankel transform, we have the following:

$${\mathcal{H}}_0\left\{{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))\right\}\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{1}{\sqrt{S_F^{\alpha }{\left(a\right)}^2+S_F^{\alpha }{\left(k\right)}^2}}}.$$

Example 2. 

Consider $f\left(S_F^{\alpha }\left(r\right)\right)=\delta \left(S_F^{\alpha }\left(r\right)\right)/S_F^{\alpha }\left(r\right)$; to find its fractal Hankel transform, we need to compute the following:

$$\begin{array}{cc}\hfill {\mathcal{H}}_0\left\{{\displaystyle \frac{\delta \left(S_F^{\alpha }\left(r\right)\right)}{S_F^{\alpha }\left(r\right)}}\right\}\left(S_F^{\alpha }\left(k\right)\right)& ={\int }_0^{\infty }{\displaystyle \frac{\delta \left(S_F^{\alpha }\left(r\right)\right)}{S_F^{\alpha }\left(r\right)}}{J}_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\hfill \\ & ={\int }_0^{\infty }\delta \left(S_F^{\alpha }\left(r\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(13)

Since $\delta \left(S_F^{\alpha }\left(r\right)\right)$ is the fractal Dirac delta function, the fractal integral evaluates to:

$${\mathcal{H}}_0\left\{{\displaystyle \frac{\delta \left(S_F^{\alpha }\left(r\right)\right)}{S_F^{\alpha }\left(r\right)}}\right\}\left(S_F^{\alpha }\left(k\right)\right)=J_0\left(S_F^{\alpha }\left(0\right)\right)={\chi }_F.$$

Example 3. 

Consider the function $f\left(S_F^{\alpha }\left(r\right)\right)=H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))$, where $H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))$ represents the fractal Heaviside step function. To find its fractal Hankel transform, we need to evaluate the following expression:

$$\begin{array}{cc}\hfill & {\mathcal{H}}_0\left\{H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))\right\}\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r\hfill \\ & ={\int }_0^a{J}_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)S_F^{\alpha }\left(r\right)d_F^{\alpha }r={\displaystyle \frac{S_F^{\alpha }\left(a\right)J_1\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(a\right)\right)}{S_F^{\alpha }\left(k\right)}}.\hfill \end{array}$$

(14)

Example 4. 

If $f\left(S_F^{\alpha }\left(r\right)\right)=exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))$, then its first-order fractal Hankel transform is expressed as:

$$F_1\left(S_F^{\alpha }\left(k\right)\right)=H_1\{exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))={\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))J_1\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.$$

Using standard fractal integral results, we get the following:

$$F_1\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{{(S_F^{\alpha }{\left(a\right)}^2+S_F^{\alpha }{\left(k\right)}^2)}^{\frac{3}{2}}}}.$$

Example 5. 

Given the function

$$f(S_F^{\alpha }\left(r\right))={\displaystyle \frac{exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))}{S_F^{\alpha }\left(r\right)}},$$

its first-order fractal Hankel transform can be written as:

$$\begin{array}{cc}\hfill F_1\left(S_F^{\alpha }\left(k\right)\right)& =H_1\left\{{\displaystyle \frac{exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))}{S_F^{\alpha }\left(r\right)}}\right\}\hfill \\ & ={\int }_0^{\infty }exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right))J_1\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(15)

Evaluating the fractal integral gives the following:

$$F_1\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{1}{S_F^{\alpha }\left(k\right)}}\left(1-S_F^{\alpha }\left(a\right){\left(S_F^{\alpha }{\left(k\right)}^2+S_F^{\alpha }{\left(a\right)}^2\right)}^{-{\displaystyle \frac{1}{2}}}\right).$$

Example 6. 

Consider the following fractal function

$$f\left(S_F^{\alpha }\left(r\right)\right)={\displaystyle \frac{sin\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)}{S_F^{\alpha }\left(r\right)}}.$$

Its first-order fractal Hankel transform is as follows:

$$F_1\left(S_F^{\alpha }\left(k\right)\right)={\int }_0^{\infty }{\displaystyle \frac{sin\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)}{S_F^{\alpha }\left(r\right)}}{J}_1\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.$$

Using fractal integral formulas, the result is as follows:

$$F_1\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{S_F^{\alpha }\left(a\right)H(S_F^{\alpha }\left(k\right)-S_F^{\alpha }\left(a\right))}{S_F^{\alpha }\left(k\right)\sqrt{S_F^{\alpha }{\left(k\right)}^2-S_F^{\alpha }{\left(a\right)}^2}}},$$

where $H(S_F^{\alpha }\left(k\right)-S_F^{\alpha }\left(a\right))$ is the fractal Heaviside step function.

Example 7. 

Consider $f\left(S_F^{\alpha }\left(r\right)\right)=S_F^{\alpha }{\left(r\right)}^{n}H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))$. Then, its nth-order fractal Hankel transform is as follows:

$$F_n\left(S_F^{\alpha }\left(k\right)\right)=H_n\left[S_F^{\alpha }{\left(r\right)}^{n}H(S_F^{\alpha }\left(a\right)-S_F^{\alpha }\left(r\right))\right]={\int }_0^a{S}_F^{\alpha }{\left(r\right)}^{n+1}{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.$$

(16)

After evaluating the fractal integral, the result is as follows:

$$F_n\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{S_F^{\alpha }{\left(a\right)}^{n+1}}{S_F^{\alpha }\left(k\right)}}{J}_{n+1}\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(k\right)\right).$$

Example 8. 

Consider $f\left(S_F^{\alpha }\left(r\right)\right)=S_F^{\alpha }{\left(r\right)}^{n}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }{\left(r\right)}^2)$. Then, its nth-order fractal Hankel transform is calculated as:

$$\begin{array}{cc}\hfill & F_n\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & =H_n\left[S_F^{\alpha }{\left(r\right)}^{n}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }{\left(r\right)}^2)\right]\hfill \\ & ={\int }_0^{\infty }{S}_F^{\alpha }{\left(r\right)}^{n+1}{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }{\left(r\right)}^2)d_F^{\alpha }r.\hfill \end{array}$$

(17)

After solving the fractal integral, the result is as follows:

$$F_n\left(S_F^{\alpha }\left(k\right)\right)={\displaystyle \frac{S_F^{\alpha }{\left(k\right)}^n}{{\left(2S_F^{\alpha }\left(a\right)\right)}^{n+1}}}exp\left(-{\displaystyle \frac{S_F^{\alpha }{\left(k\right)}^2}{4S_F^{\alpha }\left(a\right)}}\right).$$

Theorem 3. 

The fractal Hankel transform of $f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)$ (where $S_F^{\alpha }\left(a\right)>0$) follows the scaling property:

$$\begin{array}{c}\hfill H_n\left\{f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{1}{S_F^{\alpha }{\left(a\right)}^2}}{F}_n\left({\displaystyle \frac{S_F^{\alpha }\left(k\right)}{S_F^{\alpha }\left(a\right)}}\right).\end{array}$$

(18)

Proof. 

By the definition of the nth-order fractal Hankel transform, we get the following:

$$H_n\left\{f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)\right\}={\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.$$

Using the substitution $S_F^{\alpha }\left(s\right)=S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)$, so that

$$S_F^{\alpha }\left(r\right)={\displaystyle \frac{S_F^{\alpha }\left(s\right)}{S_F^{\alpha }\left(a\right)}},$$

and

$$d_F^{\alpha }r={\displaystyle \frac{d_F^{\alpha }s}{S_F^{\alpha }\left(a\right)}},$$

we rewrite the fractal integral:

$$H_n\left\{f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{1}{S_F^{\alpha }\left(a\right)}}{\int }_0^{\infty }{\displaystyle \frac{S_F^{\alpha }\left(s\right)}{S_F^{\alpha }\left(a\right)}}{J}_n\left({\displaystyle \frac{S_F^{\alpha }\left(k\right)}{S_F^{\alpha }\left(a\right)}}{S}_F^{\alpha }\left(s\right)\right)f\left(S_F^{\alpha }\left(s\right)\right)d_F^{\alpha }s.$$

By simplifying, we obtain the following:

$$H_n\left\{f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{1}{S_F^{\alpha }{\left(a\right)}^2}}{\int }_0^{\infty }{S}_F^{\alpha }\left(s\right)J_n\left({\displaystyle \frac{S_F^{\alpha }\left(k\right)}{S_F^{\alpha }\left(a\right)}}{S}_F^{\alpha }\left(s\right)\right)f\left(S_F^{\alpha }\left(s\right)\right)d_F^{\alpha }s.$$

This fractal integral is the fractal Hankel transform of $f\left(S_F^{\alpha }\left(s\right)\right)$, so we have:

$$H_n\left\{f\left(S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{1}{S_F^{\alpha }{\left(a\right)}^2}}{F}_n\left({\displaystyle \frac{S_F^{\alpha }\left(k\right)}{S_F^{\alpha }\left(a\right)}}\right).$$

This completes the proof of the fractal scaling property. □

Theorem 4. 

The fractal Hankel transform of the first $F^{\alpha }$-derivative is given by:

$$\begin{array}{c}\hfill H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left[(n-1)F_{n+1}\left(S_F^{\alpha }\left(k\right)\right)-(n+1)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)\right],n\ge 1.\end{array}$$

(19)

For $n=1$, the result simplifies to:

$$\begin{array}{c}\hfill H_1\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}=-S_F^{\alpha }\left(k\right)F_0\left(S_F^{\alpha }\left(k\right)\right).\end{array}$$

(20)

This result holds provided that $\left[S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)\right]$ vanishes as $S_F^{\alpha }\left(r\right)\to 0$ and $S_F^{\alpha }\left(r\right)\to \infty$.

Proof. 

We have, by definition,

$$\begin{array}{c}\hfill H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}={\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r,\end{array}$$

(21)

which means, fractal integrating by parts, the following:

$$\begin{array}{cc}\hfill & H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}\hfill \\ & ={\left[S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]}_0^{\infty }\hfill \\ & -{\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)D_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(22)

Utilizing the properties of the fractal Bessel function, we get the following:

$$\begin{array}{cc}\hfill & D_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)\hfill \\ & =J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)+S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(k\right)D_{F,r}^{\alpha }{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\hfill \\ & =J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)+S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(k\right)J_{n-1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)-n{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\hfill \\ & =(1-n)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)+S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(k\right)J_{n-1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right).\hfill \end{array}$$

(23)

In view of the given condition, the first term of Equation (21) vanishes as $S_F^{\alpha }\left(r\right)\to 0$ and $S_F^{\alpha }\left(r\right)\to \infty$, and the derivative within the integral in Equation (21) can be replaced by Equation (23), so that Equation (21) becomes:

$$\begin{array}{c}\hfill H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}=(n-1){\int }_0^{\infty }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)dr-S_F^{\alpha }\left(k\right)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right).\end{array}$$

(24)

We next use the standard recurrence relation for the fractal Bessel function:

$$J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)={\displaystyle \frac{S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)}{2n}}\left[J_{n-1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)+J_{n+1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right].$$

Thus, Equation (24) can be rewritten as:

$$\begin{array}{cc}\hfill & H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}=-S_F^{\alpha }\left(k\right)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)\hfill \\ & +{\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left(n-1\right){\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)\left[J_{n-1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)+J_{n+1}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]d_F^{\alpha }r\hfill \\ & =-S_F^{\alpha }\left(k\right)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)+{\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left((n-1)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)+F_{n+1}\left(S_F^{\alpha }\left(k\right)\right)\right)\hfill \\ & ={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left[(n-1)F_{n+1}\left(S_F^{\alpha }\left(k\right)\right)-(n+1)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)\right].\hfill \end{array}$$

(25)

In particular, when $n=1$, Equation (25) follows immediately:

$$H_1\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}=-S_F^{\alpha }\left(k\right)F_0\left(S_F^{\alpha }\left(k\right)\right).$$

□

Theorem 5. 

The fractal Hankel transform of the second $F^{\alpha }$-derivative is given by:

$$\begin{array}{cc}\hfill & H_n\left\{D_F^{2\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}\hfill \\ & ={\displaystyle \frac{S_F^{\alpha }{\left(k\right)}^2}{4}}\left[{\displaystyle \frac{n+1}{n-1}}{F}_{n-2}\left(S_F^{\alpha }\left(k\right)\right)-2{\displaystyle \frac{n^2-3}{n^2-1}}{F}_n\left(S_F^{\alpha }\left(k\right)\right)+{\displaystyle \frac{n-1}{n+1}}{F}_{n+2}\left(S_F^{\alpha }\left(k\right)\right)\right].\hfill \end{array}$$

(26)

Proof. 

To prove Theorem 5, we start with the result from Theorem 4 as:

$$H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left[(n-1)F_{n+1}\left(S_F^{\alpha }\left(k\right)\right)-(n+1)F_{n-1}\left(S_F^{\alpha }\left(k\right)\right)\right].$$

The fractal Hankel transform of a second $F^{\alpha }$-derivative is obtained by applying the first $F^{\alpha }$-derivative result twice. Therefore, taking the $F^{\alpha }$-derivative of $H_n\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}$, we get the following:

$$\begin{array}{c}\hfill H_n\left\{D_F^{2\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2n}}\left[(n-1)H_{n+1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}-(n+1)H_{n-1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}\right].\end{array}$$

(27)

Using the result of Theorem 5 for $H_{n+1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}$ and $H_{n-1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}$, we have:

$$\begin{array}{cc}\hfill H_{n+1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}& ={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2(n+1)}}\left[n{F}_{n+2}\left(S_F^{\alpha }\left(k\right)\right)-(n+2)F_n\left(S_F^{\alpha }\left(k\right)\right)\right],\hfill \\ \hfill H_{n-1}\left\{D_F^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}& ={\displaystyle \frac{S_F^{\alpha }\left(k\right)}{2(n-1)}}\left[(n-2)F_n\left(S_F^{\alpha }\left(k\right)\right)-n{F}_{n-2}\left(S_F^{\alpha }\left(k\right)\right)\right].\hfill \end{array}$$

(28)

Substituting Equation (28) into Equation (27), we get the following:

$$\begin{array}{cc}\hfill H_n\left\{D_F^{2\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}& ={\displaystyle \frac{S_F^{\alpha }{\left(k\right)}^2}{4n}}\left[(n-1)\left({\displaystyle \frac{n{F}_{n+2}\left(S_F^{\alpha }\left(k\right)\right)-(n+2)F_n\left(S_F^{\alpha }\left(k\right)\right)}{n+1}}\right)\right.\hfill \\ & -\left.(n+1)\left({\displaystyle \frac{(n-2)F_n\left(S_F^{\alpha }\left(k\right)\right)-n{F}_{n-2}\left(S_F^{\alpha }\left(k\right)\right)}{n-1}}\right)\right].\hfill \end{array}$$

(29)

Finally, it simplifies to:

$$\begin{array}{cc}\hfill & H_n\left\{D_F^{2\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right\}\hfill \\ & ={\displaystyle \frac{S_F^{\alpha }{\left(k\right)}^2}{4}}\left[{\displaystyle \frac{n+1}{n-1}}{F}_{n-2}\left(S_F^{\alpha }\left(k\right)\right)-2{\displaystyle \frac{n^2-3}{n^2-1}}{F}_n\left(S_F^{\alpha }\left(k\right)\right)+{\displaystyle \frac{n-1}{n+1}}{F}_{n+2}\left(S_F^{\alpha }\left(k\right)\right)\right].\hfill \end{array}$$

(30)

This completes the proof. □

Theorem 6. 

The fractal Hankel transform of the operator

$${\nabla }^{2\alpha }-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}},$$

(31)

applied to $f\left(S_F^{\alpha }\left(r\right)\right)$ is given by:

$$\begin{array}{cc}\hfill & H_n\left\{\left({\nabla }^{2\alpha }-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\right)f\left(S_F^{\alpha }\left(r\right)\right)\right\}\hfill \\ & =H_n\left\{{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}f\left(S_F^{\alpha }\left(r\right)\right)\right\}\hfill \\ & =-S_F^{\alpha }{\left(k\right)}^2{F}_n\left(S_F^{\alpha }\left(k\right)\right).\hfill \end{array}$$

(32)

This result holds provided that both $S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)$ and $S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)$ vanish as $S_F^{\alpha }\left(r\right)\to 0$ and $S_F^{\alpha }\left(r\right)\to \infty$.

Proof. 

To prove this theorem, we begin with by Definition 11 as follows:

$$\begin{array}{cc}\hfill H_n& \{{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}f\left(S_F^{\alpha }\left(r\right)\right)\}\hfill \\ & ={\int }_0^{\infty }{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)D_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)d_F^{\alpha }r\hfill \\ & -{\int }_0^{\infty }{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\left(S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(33)

Using fractal integration by parts on the first fractal integral, we have the following:

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{J}_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)D_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)d_F^{\alpha }r\hfill \\ & ={\left[S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]}_0^{\infty }\hfill \\ & -S_F^{\alpha }\left(k\right){\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(34)

The boundary term ${\left[S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]}_0^{\infty }$ vanishes due to the assumption given in the problem (likely smooth behavior at $S_F^{\alpha }\left(r\right)=0$ and $S_F^{\alpha }\left(r\right)\to \infty$). Thus, this term simplifies to:

$$\begin{array}{cc}\hfill & -S_F^{\alpha }\left(k\right){\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r\hfill \\ & -{\int }_0^{\infty }{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\left(S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(35)

We now fractal integrate by parts on the second fractal integral:

$$\begin{array}{cc}\hfill & -S_F^{\alpha }\left(k\right){\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r\hfill \\ & =-{\left[S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]}_0^{\infty }\hfill \\ & +{\int }_0^{\infty }{\displaystyle \frac{d}{dr}}\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(36)

The boundary term ${\left[S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right]}_0^{\infty }$ also vanishes under the given assumptions, so we are left with:

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r\hfill \\ & -{\int }_0^{\infty }{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\left(S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.\hfill \end{array}$$

(37)

Next, we use fractal Bessel’s fractal differential equation:

$$D_{F,r}^{\alpha }\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)\right)+S_F^{\alpha }\left(r\right)\left(S_F^{\alpha }{\left(k\right)}^2-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)=0,$$

to simplify the integral:

$${\int }_0^{\infty }\left(S_F^{\alpha }{\left(k\right)}^2-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\right)S_F^{\alpha }\left(r\right)f\left(S_F^{\alpha }\left(r\right)\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r=0.$$

This reduces the expression to:

$$H_n\left({\nabla }^{2\alpha }-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}\right)f\left(S_F^{\alpha }\left(r\right)\right)=-S_F^{\alpha }{\left(k\right)}^2{\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)J_n\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)f\left(S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r.$$

Thus, we have:

$$\begin{array}{cc}\hfill & H_n\left({\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)-{\displaystyle \frac{n^2}{S_F^{\alpha }{\left(r\right)}^2}}f\left(S_F^{\alpha }\left(r\right)\right)\right)\hfill \\ & =-S_F^{\alpha }{\left(k\right)}^2{H}_n\left[f\left(S_F^{\alpha }\left(r\right)\right)\right]=-S_F^{\alpha }{\left(k\right)}^2{F}_n\left(S_F^{\alpha }\left(k\right)\right).\hfill \end{array}$$

(38)

This completes the proof. □

Remark 3. 

Special Cases of Equation (38) are:

1. 

For $n=0$:

$$H_0\left({\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)\right)=-S_F^{\alpha }{\left(k\right)}^2{F}_0\left(f\left(S_F^{\alpha }\left(k\right)\right)\right),$$

2. 

For $n=1$:

$$H_1\left({\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }\left(S_F^{\alpha }\left(r\right)D_{F,r}^{\alpha }f\left(S_F^{\alpha }\left(r\right)\right)\right)-{\displaystyle \frac{1}{S_F^{\alpha }{\left(r\right)}^2}}f\left(S_F^{\alpha }\left(r\right)\right)\right)=-S_F^{\alpha }{\left(k\right)}^2{F}_1\left(S_F^{\alpha }\left(k\right)\right).$$

These results can be used to solve fractal partial differential equations in axisymmetric fractal cylindrical coordinates.

4. Application of Fractal Hankel Transform

In this section, we present an application of the fractal Hankel transform by solving a fractal partial differential equation.

Example 9. 

The problem is to solve the free vibration of a large circular elastic membrane governed by the fractal partial differential equation (FPDE):

$$c^2\left(D_{F,r}^{2\alpha }u+{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }u\right)=D_{F,t}^{2\alpha }u,0<r<\infty ,t>0,$$

where $c^2=T/\rho$, T is the tension, and ρ is the surface density of the membrane. The initial conditions are:

$$u(r,0)=f\left(S_F^{\alpha }\left(r\right)\right),D_{F,t}^{\alpha }u(r,0)=g\left(S_F^{\alpha }\left(r\right)\right),0\le r<\infty .$$

By applying the zero-order fractal Hankel transform, we have:

$$U_0(k,t)={\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)u(r,t)d_F^{\alpha }r.$$

Applying this transform to the FPDE results in the fractal differential equation (FDE) for $U_0(k,t)$:

$$D_{F,t}^{2\alpha }{U}_0+c^2{S}_F^{\alpha }{\left(k\right)}^2{U}_0=0,$$

with the transformed initial conditions:

$$U_0(k,0)=\tilde{f}\left(S_F^{\alpha }\left(k\right)\right),D_{F,t}^{\alpha }{U}_0(k,0)=\tilde{g}\left(S_F^{\alpha }\left(k\right)\right).$$

The general solution to the FDE is:

$$U_0(k,t)=\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)cos\left(c{S}_F^{\alpha }\left(k\right)S_F^{\alpha }\left(t\right)\right)+{\displaystyle \frac{\tilde{g}\left(S_F^{\alpha }\left(k\right)\right)}{c{S}_F^{\alpha }\left(k\right)}}sin\left(c{S}_F^{\alpha }\left(k\right)S_F^{\alpha }\left(t\right)\right).$$

Letting initial conditions $u(r,0)=f\left(S_F^{\alpha }\left(r\right)\right)=A{(S_F^{\alpha }{\left(a\right)}^2+S_F^{\alpha }{\left(r\right)}^2)}^{-{\displaystyle \frac{1}{2}}}$ and $D_{F,t}^{\alpha }u(r,0)=g\left(S_F^{\alpha }\left(r\right)\right)=0$, we have $\tilde{g}\left(S_F^{\alpha }\left(k\right)\right)=0$, so the solution simplifies to:

$$U_0(k,t)=\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)cos\left(c{S}_F^{\alpha }\left(k\right)S_F^{\alpha }\left(t\right)\right).$$

The inverse fractal Hankel transform gives the solution in terms of $u(r,t)$:

$$\begin{array}{c}\hfill u(r,t)={\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)cos\left(c{S}_F^{\alpha }\left(k\right)S_F^{\alpha }\left(t\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }k.\end{array}$$

(39)

The fractal Hankel transform of $f\left(r\right)=A{(S_F^{\alpha }{\left(a\right)}^2+S_F^{\alpha }{\left(r\right)}^2)}^{-{\displaystyle \frac{1}{2}}}$ is given by:

$$\begin{array}{cc}\hfill & \tilde{f}\left(\kappa \right)=A{S}_F^{\alpha }\left(a\right){\int }_0^{\infty }{\displaystyle \frac{S_F^{\alpha }\left(r\right)}{{(S_F^{\alpha }{\left(a\right)}^2+S_F^{\alpha }{\left(r\right)}^2)}^{1/2}}}{J}_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }r\hfill \\ & ={\displaystyle \frac{A{S}_F^{\alpha }\left(a\right)}{\kappa }}exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(k\right)).\hfill \end{array}$$

(40)

Substituting Equation (40) into Equation (39), the solution becomes:

$$u(r,t)=A{S}_F^{\alpha }\left(a\right){\int }_0^{\infty }exp(-S_F^{\alpha }\left(a\right)S_F^{\alpha }\left(k\right))cos\left(c{S}_F^{\alpha }\left(k\right)S_F^{\alpha }\left(t\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)d_F^{\alpha }k.$$

Thus, the solution is:

$$\begin{array}{cc}\hfill u(r,t)& =A{S}_F^{\alpha }\left(a\right){\displaystyle \frac{S_F^{\alpha }{\left(a\right)}^2-c^2{S}_F^{\alpha }{\left(t\right)}^2+S_F^{\alpha }{\left(r\right)}^2}{{\left[{(S_F^{\alpha }{\left(a\right)}^2-c^2{S}_F^{\alpha }{\left(t\right)}^2+S_F^{\alpha }{\left(r\right)}^2)}^2+{\left(2S_F^{\alpha }\left(a\right)c{S}_F^{\alpha }\left(t\right)\right)}^2\right]}^{1/2}}}\hfill \end{array}$$

(41)

$$\approx A{a}^{\alpha }{\displaystyle \frac{a^{2\alpha }-c^2{t}^{2\alpha }+r^{2\alpha }}{{\left[{(a^{2\alpha }-c^2{t}^{2\alpha }+r^{2\alpha })}^2+{\left(2c{\left(at\right)}^{\alpha }\right)}^2\right]}^{1/2}}}.$$

(42)

This fractal integral represents the solution to the free vibration problem for the fractal circular membrane.

As illustrated in Figure 2, panel (a) shows how Equation (41) varies with r for different values of t, highlighting the Cantor set structure’s influence on $u(r,t)$. Panel (b) displays an approximation of Equation (41) for different α values, showing how changes in α impact the function’s behavior.

Example 10. 

The axisymmetric fractal diffusion equation is given by:

$$D_{F,t}^{\alpha }u=\kappa \left(D_{F,r}^{2\alpha }u+{\displaystyle \frac{1}{S_F^{\alpha }\left(r\right)}}{D}_{F,r}^{\alpha }u\right),0<r<\infty ,t>0,$$

where $\kappa >0$ is the diffusivity constant, and the initial condition is:

$$u(r,0)=f\left(S_F^{\alpha }\left(r\right)\right),0<r<\infty .$$

Applying the zero-order fractal Hankel transform, we have:

$$U_0(k,t)={\int }_0^{\infty }{S}_F^{\alpha }\left(r\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)u(r,t)d_F^{\alpha }r,$$

to obtain the equation:

$$\begin{array}{c}\hfill D_{F,t}^{\alpha }{U}_0+\kappa S_F^{\alpha }{\left(k\right)}^2{U}_0=0,U_0(k,0)=\tilde{f}\left(S_F^{\alpha }\left(k\right)\right),\end{array}$$

(43)

where k is the fractal Hankel transform variable. The solution to Equation (43) is:

$$U_0(k,t)=\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right)).$$

Applying the inverse fractal Hankel transform gives:

$$u(r,t)={\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right))d_F^{\alpha }k.$$

Substituting the expression for $\tilde{f}\left(S_F^{\alpha }\left(k\right)\right)$, we have:

$$\begin{array}{cc}\hfill u(r,t)& ={\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)\left[{\int }_0^{\infty }{S}_F^{\alpha }\left(l\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(l\right)\right)f\left(S_F^{\alpha }\left(l\right)\right)d_F^{\alpha }l\right]\hfill \\ & J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right))d_F^{\alpha }k.\hfill \end{array}$$

(44)

Interchanging the order of fractal integration, we get:

$$\begin{array}{cc}\hfill u(r,t)& ={\int }_0^{\infty }{S}_F^{\alpha }\left(l\right)f\left(S_F^{\alpha }\left(l\right)\right)\hfill \\ & \left[{\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(l\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right))d_F^{\alpha }k\right]d_F^{\alpha }l.\hfill \end{array}$$

(45)

Using the known fractal integral involving fractal Bessel functions, we obtain:

$$\begin{array}{cc}\hfill & {\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(l\right)\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right))d_F^{\alpha }k\hfill \\ & ={\displaystyle \frac{1}{2\kappa S_F^{\alpha }\left(t\right)}}exp\left(-{\displaystyle \frac{S_F^{\alpha }{\left(r\right)}^2+S_F^{\alpha }{\left(l\right)}^2}{4\kappa S_F^{\alpha }\left(t\right)}}\right)I_0\left({\displaystyle \frac{S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(l\right)}{2\kappa S_F^{\alpha }\left(t\right)}}\right),\hfill \end{array}$$

(46)

where $I_0\left(x\right)$ is

$$I_0\left(x\right)=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(k!)}^2}}{\left({\displaystyle \frac{S_F^{\alpha }\left(x\right)}{2}}\right)}^{2k},$$

which might be called the modified fractal Bessel function of the first kind. The solution to the axisymmetric fractal diffusion equation is:

$$u(r,t)={\displaystyle \frac{1}{2\kappa S_F^{\alpha }\left(t\right)}}{\int }_0^{\infty }{S}_F^{\alpha }\left(l\right)f\left(S_F^{\alpha }\left(l\right)\right)I_0\left({\displaystyle \frac{S_F^{\alpha }\left(r\right)S_F^{\alpha }\left(l\right)}{2\kappa S_F^{\alpha }\left(t\right)}}\right)exp\left(-{\displaystyle \frac{S_F^{\alpha }{\left(r\right)}^2+S_F^{\alpha }{\left(l\right)}^2}{4\kappa S_F^{\alpha }\left(t\right)}}\right)d_F^{\alpha }l.$$

In particular, when $S_F^{\alpha }\left(l\right)=0$, the integral simplifies to:

$${\int }_0^{\infty }{S}_F^{\alpha }\left(k\right)J_0\left(S_F^{\alpha }\left(k\right)S_F^{\alpha }\left(r\right)\right)exp(-\kappa S_F^{\alpha }{\left(k\right)}^2{S}_F^{\alpha }\left(t\right))d_F^{\alpha }k={\displaystyle \frac{1}{2\kappa S_F^{\alpha }\left(t\right)}}exp\left(-{\displaystyle \frac{S_F^{\alpha }{\left(r\right)}^2}{4\kappa S_F^{\alpha }\left(t\right)}}\right).$$

Thus, we have:

$$\begin{array}{cc}\hfill u(r,t)& ={\displaystyle \frac{1}{4{\kappa }^2{\left(S_F^{\alpha }\left(t\right)\right)}^2}}exp\left(-{\displaystyle \frac{S_F^{\alpha }{\left(r\right)}^2}{4\kappa S_F^{\alpha }\left(t\right)}}\right)\hfill \end{array}$$

(47)

$$\begin{array}{cc}\hfill & \approx {\displaystyle \frac{1}{4{\kappa }^2{\left(t\right)}^{2\alpha }}}exp\left(-{\displaystyle \frac{r^{2\alpha }}{4\kappa t^{\alpha }}}\right).\hfill \end{array}$$

(48)

As illustrated in Figure 3a,b, the plot of Equation (47) with respect to r at various t values demonstrates how the spatial decay is influenced by the structure of the Cantor set. Additionally, the effect of varying α on the spatial decay can be seen in Figure 3b, where a fixed $t=0.1$ results in sharper decay for higher values of α.

Figure 3c depicts the relationship of Equation (47) with respect to t for a fixed r, further showcasing the Cantor set’s influence on temporal decay. Lastly, Figure 3d provides an approximation of Equation (47) over time for different values of α, indicating that higher values of α lead to faster decay.

5. Conclusions

In this paper, we have extended classical transforms to fractal spaces by developing and applying the Fractal Hankel Transform. We provided a concise review of fractal calculus as a foundation, introduced the Fractal Hankel Transform along with its formulation and properties, and explored its applications to solve complex problems within fractal spaces. This study demonstrates the potential and versatility of the Fractal Hankel Transform in fractal analysis. Our findings suggest that further investigation into fractal transformations could yield new insights and tools for analysis within fractal and non-Euclidean frameworks.