Generalized Space–Time Fractional Dynamics in Networks and Lattices

Abstract

We analyze generalized space–time fractional motions on undirected networks and lattices. The continuous-time random walk (CTRW) approach of Montroll and Weiss is employed to subordinate a space fractional walk to a generalization of the time fractional Poisson renewal process. This process introduces a non-Markovian walk with long-time memory effects and fat-tailed characteristics in the waiting time density. We analyze ‘generalized space–time fractional diffusion’ in the infinite d-dimensional integer lattice $\mathbb {Z}^d$. We obtain in the diffusion limit a ‘macroscopic’ space–time fractional diffusion equation. Classical CTRW models such as with Laskin’s fractional Poisson process and standard Poisson process which occur as special cases are also analyzed. The developed generalized space–time fractional CTRW model contains a four-dimensional parameter space and offers therefore a great flexibility to describe real-world situations in complex systems.

Appendix: Laplace Transforms and Fractional Operators

Here, we derive briefly some basic mathematical apparatus used in the paper in the context of causal functions and distributions involving fractional operators and Heaviside calculus. All functions and distributions considered are to be conceived as generalized functions and distributions in the Gelfand–Shilov sense [23]. Let us first introduce the Heaviside step function

$$\begin{aligned} \varTheta (t) = \left\{ \begin{array}{l} 1 ,\qquad t \ge 0 \\ 0 ,\qquad t < 0. \end{array}\right. \end{aligned}$$

(14.80)

A function is causal if it has the form $\varTheta (t) f(t)$, i.e., is null for $t<0$ and nonvanishing only for nonnegative times t. We introduce the Laplace transform of $\varTheta (t) f(t)$ by

$$\begin{aligned} {\tilde{f}}(s) = \mathcal{L}(f(t)) = \int _{-\infty }^{\infty } e^{-st} \varTheta (t)f(t)\mathrm{d}t = \int _{0}^{\infty }f(t)e^{-st}\mathrm{d}t ,\quad s=\sigma +i\omega \end{aligned}$$

(14.81)

with suitably chosen $\sigma >\sigma _0$ in order to guarantee convergence of (14.81). In view of the fact that (14.81) can be read as Fourier transform of the causal function $e^{-\sigma t}f(t)\varTheta (t)$, it is straightforward to see that the Laplace inversion corresponds to the representation of this function as Fourier integral, namely

$$\begin{aligned} e^{-\sigma t}f(t)\varTheta (t) = \frac{1}{2\pi } \int _{-\infty }^{\infty } e^{i\omega t} {\tilde{f}}(\sigma +i\omega )\mathrm{d}\omega \end{aligned}$$

(14.82)

which can be rewritten as

$$\begin{aligned} f(t)\varTheta (t) = \frac{ e^{\sigma t} }{2\pi } \int _{-\infty }^{\infty } e^{i\omega t} {\tilde{f}}(\sigma +i\omega )\mathrm{d}\omega = \frac{1}{2\pi i} \int _{-i\infty }^{+i\infty } e^{st} {\tilde{f}}(s)\mathrm{d}s . \end{aligned}$$

(14.83)

Sometimes when there is no time derivative involved, we skip the Heaviside $\varTheta (t)$-function implying that all expressions are written for $t\ge 0$. Then, we mention that

$$\begin{aligned} \varTheta (t)f(t) = \int _{-\infty }^{\infty } \delta (t-\tau )\varTheta (\tau )f(\tau )\mathrm{d}\tau \end{aligned}$$

(14.84)

and introduce the shift operator $e^{-\tau \frac{d}{dt}}$ acting on a function g(t) as $e^{-\tau \frac{d}{dt}}g(t)=g(t-\tau )$ thus

$$\begin{aligned} e^{-\tau \frac{d}{dt}} \delta (t) = \delta (t-\tau ) . \end{aligned}$$

(14.85)

Substituting this relation into (14.84) yields

$$\begin{aligned} \varTheta (t)f(t) = \left\{ \int _{-\infty }^{\infty } e^{-\tau \frac{d}{dt}} \varTheta (\tau )f(\tau )\mathrm{d}\tau \right\} \,\, \delta (t) =\mathcal{L}^{-1}\{ {\tilde{f}}(s) \} = {\tilde{f}}\left( \frac{d}{dt}\right) \,\, \delta (t) \end{aligned}$$

(14.86)

where the operator ${\tilde{f}}(\frac{d}{dt})$ is related with the Laplace transform (14.81) by replacing $s \rightarrow \frac{d}{dt}$. Equation (14.86) is the operator representation of the causal function $\varTheta (t)f(t)$. A convolution of two causal functions $\varTheta (t)f(t), g(t)\varTheta (t)$ then can be represented by

$$\begin{aligned} \begin{array}{l} \displaystyle \int _{0}^t g(t-\tau )f(\tau )\mathrm{d}\tau \\ = \int _{-\infty }^{\infty } \int _{-\infty }^{\infty } \delta (t-\tau _1-\tau _2) g(\tau _1)\varTheta (\tau _1)f(\tau _2)\varTheta (\tau _2) \mathrm{d}\tau _1\mathrm{d}\tau _2 ,\quad t>0 \\[1em] \displaystyle \left( \int _{-\infty }^{\infty } e^{-\tau _1\frac{d}{dt}} \varTheta (\tau _1)f(\tau )\mathrm{d}\tau _1\right) \left( \int _{-\infty }^{\infty } e^{-\tau _2\frac{d}{dt}} \varTheta (\tau _2)f(\tau )\mathrm{d}\tau _2\right) \delta (t) \\[1em] \displaystyle =\mathcal{L}^{-1}\{{\tilde{g}}(s){\tilde{f}}(s)\} = {\tilde{f}}\left( \frac{d}{dt}\right) {\tilde{g}}\left( \frac{d}{dt}\right) \,\, \delta (t) = {\tilde{g}}\left( \frac{d}{dt}\right) {\tilde{f}}\left( \frac{d}{dt}\right) \,\ \delta (t) \end{array} \end{aligned}$$

(14.87)

where it has been used $\delta (t-\tau _1-\tau _2) = e^{-(\tau _1+\tau _2)\frac{d}{dt}}\delta (t)$. We observe that in (14.86) and (14.87), the Laplace variable is replaced $s \rightarrow \frac{d}{dt}$ in the causal time domain. By considering ${\bar{f}}(t)\varTheta (t)= f(t)\varTheta (t) e^{-\lambda t}$, we observe that

$$\begin{aligned} f(t)\varTheta (t) e^{-\lambda t} = \left\{ \int _{-\infty }^{\infty } e^{-\tau (\lambda +\frac{d}{dt})} f(\tau )\varTheta (\tau ) \mathrm{d}\tau \right\} \delta (t) = {\tilde{f}}\left( \lambda +\frac{d}{dt}\right) \delta (t) \end{aligned}$$

(14.88)

where ${\tilde{f}}(s)= \mathcal{L}\{f(t)\}$. We are especially dealing with normalized (probability) distributions

$$\begin{aligned} {\tilde{f}}(s=0)= 1 = \int _{-\infty }^{\infty }\varTheta (t) f(t)\mathrm{d}t. \end{aligned}$$

(14.89)

A very important consequence of relations (14.86) and (14.87) is that they can be used to solve differential equations and to determine causal Green’s functions. As a simple example, consider the trivial algebraic equation in the Laplace domain

$$\begin{aligned} (s+\xi ) \,\, \frac{\xi }{(s+\xi )} = \xi , \qquad \xi >0 \end{aligned}$$

(14.90)

takes with $\mathcal{L}^{-1}\left\{ s+\xi \right\} = \frac{d}{dt}+\xi$ and $\mathcal{L}^{-1}\left\{ \frac{\xi }{s+\xi }\right\} = \varTheta (t) \xi e^{-\xi t}$ where on the right-hand side is used that $\mathcal{L}^{-1}\{1\} = \delta (t)$. In the causal time domain (14.90) then gives the representation

$$\begin{aligned} \left( \frac{d}{dt}+\xi \right) \, \left( \xi \varTheta (t) e^{-\xi t} \right) = \xi \delta (t) , \end{aligned}$$

(14.91)

result which is straightforwardly confirmed, i.e., the normalized causal Green’s function of $\frac{d}{dt}+\xi$ is directly obtained as $(\frac{d}{dt}+\xi )^{-1} \xi \delta (t) = \varTheta (t) \xi e^{-\xi t}$ where it is important that the $\varTheta (t)$-function is taken into account in the Laplace inversion.

A less trivial example is obtained when considering fractional powers of operators. For instance, let us consider in the Laplace domain the equation

$$\begin{aligned} (s^{\beta }+\xi ) \, \frac{\xi }{(s^{\beta }+\xi )} = \xi , \qquad \xi >0, \qquad 0< \beta \le 1 \end{aligned}$$

(14.92)

which writes in the time domain

$$\begin{aligned} \left\{ \left( \frac{d}{dt}\right) ^{\beta }+\xi \right\} \, \varTheta (t) g_{\beta ,\xi }(t) = \xi \delta (t) \end{aligned}$$

(14.93)

where the fractional derivative $\left( \frac{d}{dt}\right) ^{\beta }$ is determined subsequently. The causal Green’s function $\varTheta (t) g_{\beta ,\xi }(t)$ is obtained from the Laplace inversion

$$\begin{aligned} g_{\beta ,\xi }(t)=\mathcal{L}^{-1}\left\{ \frac{\xi }{s^{\beta }+\xi }\right\} . \end{aligned}$$

(14.94)

The inversion is performed directly when taking into account

$$\begin{aligned} \frac{\xi }{(s^{\beta }+\xi )} = s^{-\beta } \frac{\xi }{(1+\xi s^{-\beta })} = \xi \sum _{n=0}^{\infty } (-1)^n \xi ^n s^{-\beta (n+1) }, \quad \sigma =\mathfrak {R}\{s\} > \xi ^{\frac{1}{\beta }} \end{aligned}$$

(14.95)

where $\sigma =\mathfrak {R}\{s\} > \xi ^{\frac{1}{\beta }}$ guarantees convergence of this geometric series $\forall \omega = \mathfrak {I}\{s\}$, i.e., for the entire interval of integration of the corresponding Laplace inversion integral (14.88). On the other hand, we have

$$\begin{aligned} s^{-\mu } = \mathcal{L}\left\{ \varTheta (t)\frac{t^{\mu -1}}{\varGamma (\mu )} \right\} ,\qquad \mu>0 , \qquad \sigma > 0 , \end{aligned}$$

(14.96)

where we use the notation $\varGamma (\xi +1)= \xi !$ for the gamma function. We then arrive at

$$\begin{aligned} \begin{array}{lll} \displaystyle \varTheta (t) g_{\beta ,\xi }(t)&{}=&{}\displaystyle \mathcal{L}^{-1}\left\{ \frac{\xi }{\xi +s^{\beta }}\right\} \\ \displaystyle \quad &{}=&{} \displaystyle \sum _{n=0}^{\infty } (-1)^n \xi ^{n+1} \mathcal{L} ^{-1}\left\{ s^{-\beta (n+1) }\right\} = \varTheta (t) \sum _{n=0}^{\infty } (-1)^n \xi ^{n+1} \frac{t^{n\beta +\beta -1}}{\varGamma (n\beta +\beta )} \end{array} \end{aligned}$$

(14.97)

with

$$\begin{aligned} g_{\beta ,\xi }(t) = \xi t^{\beta -1} \sum _{n=0}^{\infty }\frac{(-\xi t^{\beta })^n}{\varGamma (n\beta +\beta )} = \xi t^{\beta -1} E_{\beta ,\beta }(-\xi t^{\beta }) . \end{aligned}$$

(14.98)

Here, we have introduced the generalized Mittag-Leffler function, e.g., [4, 30, 44]

$$\begin{aligned} E_{\beta ,\gamma }(z) = \sum _{n=0}^{\infty } \frac{z^n}{\varGamma (\beta n+\gamma )} ,\qquad \beta , \gamma >0 ,\quad z \in \mathbb {C} \end{aligned}$$

(14.99)

It follows that $\xi$ is a dimensional constant having units $\sec ^{-\beta }$ so that (14.98) has physical dimension of $\sec ^{-1}$ of a density. The result (14.98) also is referred to as Mittag-Leffler density and represents the waiting time density of Eq. (14.26) of the fractional Poisson renewal process introduced by Laskin [14]. Generally, Mittag-Leffler type functions play a major role in time fractional dynamics. We further often use the Mittag-Leffler function which is defined as, e.g., [4, 30, 44]

$$\begin{aligned} E_{\beta }(z) = \sum _{n=0}^{\infty } \frac{z^n}{\varGamma (\beta n+1)} ,\qquad \beta >0 ,\quad z \in \mathbb {C} \end{aligned}$$

(14.100)

where with (14.99) we have $E_{\beta }(z)=E_{\beta ,1}(z)$. The Mittag-Leffler function has the important property that for $\beta =1$ it recovers the exponential $E_{1}(z)= e^{z}$.

Riemann–Liouville fractional integral and derivative

Now let us derive the kernel of the fractional power of time-derivative operator of Eq. (14.93) where we consider now exponents $\gamma >0$. This kernel is then obtained with above-introduced methods in the following short way

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{L}^{-1}\{s^{\gamma }\}= \mathcal{L}^{-1} \{s^{{\lceil }\gamma {\rceil }} s^{\gamma -{\lceil }\gamma {\rceil }}\} = e^{\sigma t}\left( \sigma +\frac{d}{dt}\right) ^{{\lceil }\gamma {\rceil }}\left( \sigma +\frac{d}{dt}\right) ^{\gamma -{\lceil }\gamma {\rceil }} \delta (t)\\ \displaystyle \quad = e^{\sigma t}\left( \sigma +\frac{d}{dt}\right) ^{{\lceil }\gamma {\rceil }} \int _{-\infty }^{\infty }\frac{\mathrm{d}\omega }{(2\pi )}e^{i\omega t} (\sigma +i\omega )^{\gamma -{\lceil }\gamma {\rceil }} \\ \displaystyle \quad = e^{\sigma t}\left( \sigma +\frac{d}{dt}\right) ^{{\lceil }\gamma {\rceil }} \left\{ e^{-\sigma t}\varTheta (t) \frac{t^{{\lceil }\gamma {\rceil }-\gamma -1}}{({\lceil }\gamma {\rceil }-\gamma -1)!}\right\} = \frac{d^{{\lceil }\gamma {\rceil }}}{dt^{{\lceil }\gamma {\rceil }}} \left( \varTheta (t) \frac{t^{{\lceil }\gamma {\rceil }-\gamma -1}}{\varGamma ({\lceil }\gamma {\rceil }-\gamma )} \right) . \end{array} \end{aligned}$$

(14.101)

with $\gamma >0 , \gamma \notin \mathbb {N}$. Here, we introduced the ceiling function ${\lceil }\gamma {\rceil }$ indicating the smallest integer greater or equal to $\gamma$. In this way, the Fourier integral in the second line is integrable around $\omega =0$ $\forall \sigma \ge 0$ since $\gamma -{\lceil }\gamma {\rceil } >-1$. We then obtain for the Laplace inversion

$$\begin{aligned} \mathcal{L}^{-1}\{s^{-\gamma }\}= \varTheta (t) \frac{t^{\gamma -1}}{\varGamma (\gamma )} ,\qquad \gamma >0 \end{aligned}$$

(14.102)

as a fractional generalization of integration operator. This kernel indeed can be identified with the kernel of the Riemann–Liouville fractional integral operator of order $\gamma$ [45,46,47] which recovers for $\gamma \in \mathbb {N}$ the multiple integer-order integrations.

On the other hand, the kernel (14.101) with explicit representation in (14.101)$_3$ can be conceived as the ‘fractional derivative’ operator $(\frac{d}{dt})^{\gamma }$. The fractional derivative acts on causal functions $\varTheta (t)f(t)$ as

$$\begin{aligned} \begin{array}{l} \displaystyle \mathcal{L}^{-1}\{s^{\gamma }\} \cdot f(t)\varTheta (t) =: _0\!D_t^{\gamma } f(t) = \frac{d^{{\lceil }\gamma {\rceil }}}{dt^{{\lceil }\gamma {\rceil }}}\int _{-\infty }^{\infty } \left\{ \varTheta (t-\tau ) \frac{(t-\tau )^{{\lceil }\gamma {\rceil }-\gamma -1}}{{\varGamma ({\lceil }\gamma {\rceil }-\gamma )}} \right\} f(\tau )\varTheta (t) \mathrm{d}\tau ,\\ \displaystyle _0\!D_t^{\gamma } f(t) = \frac{1}{{\varGamma ({\lceil }\gamma {\rceil }-\gamma )}} \frac{d^{{\lceil }\gamma {\rceil }}}{d\tau ^{{\lceil }\gamma {\rceil }}} \int _0^t (t-\tau )^{{\lceil }\gamma {\rceil }-\gamma -1} f(\tau )\mathrm{d}\tau . \end{array} \end{aligned}$$

(14.103)

with $\gamma >0$. We identify in the last line this operator with the Riemann–Liouville fractional derivative [45,46,47] which recovers for $\gamma \in \mathbb {N}$ integer-order standard derivatives. We emphasize that (14.101) requires causality, i.e., a distribution of the form $f(t)\varTheta (t)$; thus, the Laplace transform captures the entire nonzero contributions of the causal distribution. In the diffusion equation (14.72), the Riemann–Liouville fractional derivative is of order $0<\beta <1$ with ${\lceil }\beta {\rceil }=1$. In this case (14.103) then has representation

$$\begin{aligned} \displaystyle _0\!D_t^{\beta } f(t) =\frac{1}{\varGamma (1-\beta )}\frac{d}{dt}\int _0^t (t-\tau )^{-\beta }f(\tau )\mathrm{d}\tau ,\qquad 0<\beta <1 , \quad t>0 . \end{aligned}$$

(14.104)