Effects of the Tempered Aging and the Corresponding Fokker–Planck Equation

Abstract

In the renewal processes, if the waiting time probability density function is a tempered power-law distribution, then the process displays a transition dynamics; and the transition time depends on the parameter $\lambda$ of the exponential cutoff. In this paper, we discuss the aging effects of the renewal process with the tempered power-law waiting time distribution. By using the aging renewal theory, the p-th moment of the number of renewal events $n_a(t_a, t)$ in the interval $(t_a, t_a+t)$ is obtained for both the weakly and strongly aged systems; and the corresponding surviving probabilities are also investigated. We then further analyze the tempered aging continuous time random walk and its Einstein relation, and the mean square displacement is attained. Moreover, the tempered aging diffusion equation is derived.

Appendices

Appendix 1: Generation of Random Variables (Fig. 1)

When generating the random variables with the PDF Eq. (1) to plot Fig. 1, the Monte Carlo statistical methods [42] is used. We first rewrite $\varphi (t)$ as $\varphi (t)=H(t)f_1(t)$, where $f_1(t)=\alpha t_{0}^\alpha t^{-\alpha -1}$ with $t_0$ being a small number. Denote the maximum of H(x) as M. Then the algorithm can be described as:

• 1. Generate a r.v. $x_{f_{1}}$ with PDF $f_{1}$ and a r.v. $\xi$ being uniformly distributed in the interval [0, 1].

• 2. Accept $x_{f_{1}}$, if $M\xi \le H(x_{f_{1}})$; otherwise, reject.

• 3. Return to Step 1.

Fig. 14

Time evolution of g(t)

Appendix 2: Mittag-Leffler Function

The two-parameter function of the Mittag-Leffler type plays a very important role in the fractional calculus, being introduced by G.M. Mittag-Leffler and studied by A. Wiman [39]. The two-parameter Mittag-Leffler function is defined by the series expansion

$$\begin{aligned} E_{\alpha ,\beta }=\sum _{k=0}^{\infty } \frac{z^{k}}{\Gamma (\alpha k+\beta )} \end{aligned}$$

(67)

with $\alpha >0$ and $\beta >0$. Its one-parameter form ($\beta =1$) is given as

$$\begin{aligned} E_{\alpha }=\sum _{k=0}^{\infty } \frac{z^{k}}{\Gamma (\alpha k+1)}. \end{aligned}$$

(68)

The asymptotic expansions of Mittag-Leffler are important for obtaining the various useful estimates of the long time or short time fractional dynamics. For small z, there exists

$$\begin{aligned} E_{\alpha ,\beta }(z) \sim \frac{1}{\Gamma (\beta )}+\frac{z}{\Gamma (\alpha +\beta )}, \end{aligned}$$

(69)

in the special case, $E_{\alpha }(z) \sim 1+\frac{z}{\Gamma (\alpha +1)}$. Another important and useful formula is the asymptotic expansion of large scale for the Mittag-Leffler function. For $0<\alpha <2$, $\beta$ is an arbitrary complex number and $\mu$ is an arbitrary real number such that $\pi \alpha /2< \mu < \min \{ \pi , \pi \alpha \}$, then for an arbitrary integer $p\ge 1$, the following expansion holds,

$$\begin{aligned} E_{\alpha ,\beta }(z) \sim \frac{1}{\alpha }z^{(1-\beta )/\alpha }\exp (z^{1/\alpha })-\sum _{k=1}^{p}\frac{z^{-k}}{\Gamma (\beta -\alpha k)} \end{aligned}$$

(70)

with $z\rightarrow \infty$ and $\arg (z) \le \mu$. For $z\rightarrow +\infty$, from Eq. (70), we have

$$\begin{aligned} E_{\alpha ,\beta }(z) \sim \frac{1}{\alpha }z^{(1-\beta )/\alpha }\exp (z^{1/\alpha }). \end{aligned}$$

(71)

And if $z\rightarrow -\infty$,

$$\begin{aligned} E_{\alpha ,\beta }(z)\sim -\frac{1}{z \Gamma (\beta -\alpha )}-\frac{1}{z^2 \Gamma (\beta -2\alpha )}; \end{aligned}$$

(72)

when $\alpha =\beta$, we have $|\Gamma (0)|\rightarrow \infty$. Therefore $E_{\alpha ,\alpha }(z)\sim -\frac{1}{z^2 \Gamma (-\alpha )}$.

In the analysis of this paper, we use the following function several times.

$$\begin{aligned} g(z)=z^{\alpha -1}\exp (-\lambda z)E_{\alpha , \alpha }(\lambda ^{\alpha }z^{\alpha }). \end{aligned}$$

(73)

When $z \rightarrow +\infty$, from (71), there exists

$$\begin{aligned} g(z) \sim \frac{\lambda ^{1-\alpha }}{\alpha }; \end{aligned}$$

(74)

see Fig. 14.

While for $z \rightarrow 0$, from (69), we have $g(z) \sim 1/\Gamma (\alpha )z^{\alpha -1}$.

Appendix 3: Laplace Transform of $\varphi (t)$

Here we present the Laplace transform of Eq. (1). From the definition of the survival probability on a site, i.e., the probability that the waiting time on a site exceeds t,

$$\begin{aligned} \Psi (t) = \int _{t}^{\infty } \varphi (\tau )d\tau =1-\int _{0}^{t} \varphi (\tau )d\tau . \end{aligned}$$

(75)

Using the definition of incomplete Gamma function, for Eq. (1) we have

$$\begin{aligned} \begin{array}{ll} \Psi (t) &{}\sim \lambda ^\alpha \int _{\lambda t}^{\infty } \exp (-z)z^{-\alpha -1} dz \\ &{} =\lambda ^\alpha \Gamma (-\alpha , \lambda t). \end{array} \end{aligned}$$

(76)

According to the Laplace transform of the incomplete Gamma function $\mathcal {L}[\Gamma (-\alpha ,\lambda t)]= \Gamma (-\alpha )[1-(\frac{u+\lambda }{\lambda })^\alpha ]/u$ with $\mathfrak {R}e(-\alpha )>-1$ and $\Psi (t)$, we have

$$\begin{aligned} \hat{\varphi }(u) \sim 1+\lambda ^\alpha -(u+\lambda )^{\alpha }, \end{aligned}$$

(77)

which can also be obtained by using the Laplace transform of the one side stable law with a shift. From Eq. (77), we have two useful asymptotics. For the long time scale, $t \gg 1/\lambda$, (i.e., $u \ll \lambda$) by the Taylor expansion, we have

$$\begin{aligned} \begin{array}{ll} \hat{\varphi }(u)\sim &{}1-\alpha \lambda ^{\alpha -1}u+\alpha (1-\alpha )\lambda ^{\alpha -2}u^2+\cdots \\ &{}+(-1)^{n+1}(-\alpha )(1-\alpha )\ldots ((n-1)-\alpha )\lambda ^{\alpha -n}u^n. \end{array} \end{aligned}$$

(78)

Notice that $\hat{\varphi }(0)=1$, so the PDF is normalized. From the definition of $\langle \tau ^{n} \rangle =\int _{0}^{\infty } \tau ^{n} \psi (\tau )d\tau$, we can get $\langle \tau \rangle =\alpha \lambda ^{\alpha -1}$, and $\langle \tau ^2 \rangle =(1-\alpha )(-\alpha )\lambda ^{\alpha -2}$. For the general cases, $\langle \tau ^n \rangle =-\Gamma (n-\alpha )/\Gamma (-\alpha )\lambda ^{\alpha -n}$, i.e., $\hat{\varphi }(u) \sim 1-\langle \tau \rangle u+\langle \tau ^2 \rangle u^2+\cdots +(-1)^n\langle \tau ^n \rangle u^n$. For short time scale, $t_{0}\ll t \ll 1/\lambda$ (i.e. $u \gg \lambda$), we have

$$\begin{aligned} \hat{\varphi }(u) \sim 1+u^\alpha \left[ \left( \frac{\lambda }{u}\right) ^\alpha -\left( 1+\frac{\lambda }{u}\right) ^\alpha \right] \sim 1-u^{\alpha }. \end{aligned}$$

(79)