Localization and Ballistic Diffusion for the Tempered Fractional Brownian–Langevin Motion

Abstract

This paper discusses the tempered fractional Brownian motion (tfBm), its ergodicity, and the derivation of the corresponding Fokker–Planck equation. Then we introduce the generalized Langevin equation with the tempered fractional Gaussian noise for a free particle, called tempered fractional Langevin equation (tfLe). While the tfBm displays localization diffusion for the long time limit and for the short time its mean squared displacement (MSD) has the asymptotic form $t^{2H},$ we show that the asymptotic form of the MSD of the tfLe transits from $t^2$ (ballistic diffusion for short time) to $t^{2-2H},$ and then to $t^2$ (again ballistic diffusion for long time). On the other hand, the overdamped tfLe has the transition of the diffusion type from $t^{2-2H}$ to $t^2$ (ballistic diffusion). The tfLe with harmonic potential is also considered.

Appendices

Appendix 1: Derivation of (2.6) and (2.7)

For tfBm, denoting $g(z)=C_z^2|z|^{2H}$ with $C_z^2$ defined by (2.3), the average of $\bar{\delta }^2(x(t))$ is

$$\begin{aligned} \left\langle \bar{\delta }^2(x(t))\right\rangle= & {} \frac{\int _0^{t-\varDelta }\langle [x(t{'}+\varDelta )-x(t{'})]^2\rangle \text {d} t{'}}{t-\varDelta } \\= & {} \frac{\int _0^{t-\varDelta }\langle x^2(t{'}+\varDelta )-2x(t{'}+\varDelta )x(t{'})+x^2(t{'})\rangle \text {d} t{'}}{t-\varDelta } \\= & {} \frac{\int _0^{t-\varDelta }g(t{'}+\varDelta )-[g(t{'}+\varDelta )+g(t{'})-g(\varDelta )]+g(t{'})\text {d} t{'}}{t-\varDelta } \\= & {} C_ {\varDelta }^2|\varDelta |^{2H}, \end{aligned}$$

and the variance of $\bar{\delta }^2(x(t))$ is

$$\begin{aligned} \text {Var}\left[ \bar{\delta }^2(x(t))\right]= & {} \left\langle \left[ \bar{\delta }^2(x(t))\right] ^2\right\rangle -\left\langle \bar{\delta }^2(x(t))\right\rangle ^2 \\= & {} \frac{\int _0^{t-\varDelta }\int _0^{t-\varDelta }\text {d} t_1\text {d} t_2\langle [x(t_1+\varDelta )-x(t_1)]^2[x(t_2+\varDelta )-x(t_2)]^2\rangle }{(t-\varDelta )^2}- C_ {\varDelta }^4|\varDelta |^{4H} \\= & {} \frac{\frac{1}{2}\int _0^{t-\varDelta }\int _0^{t-\varDelta }\text {d} t_1\text {d} t_2(g(t_1-t_2+\varDelta )+g(t_2-t_1+\varDelta )-2g(t_1-t_2))}{(t-\varDelta )^2} \\\simeq & {} \frac{\frac{1}{2}\int _0^{t-\varDelta }\int _0^{t-\varDelta }\text {d} t_1\text {d} t_2(-G(t_1-t_2+\varDelta )-G(t_2-t_1+\varDelta )+2G(t_1-t_2))^2}{(t-\varDelta )^2}\\= & {} \frac{\int _0^{t-\varDelta }\int _{t{'}}^{t-\varDelta }\text {d} t_1\text {d} t{'}(-G(\varDelta +t{'})-G(\varDelta -t{'})+2G(t{'}))^2}{(t-\varDelta )^2}\\= & {} \frac{\int _0^{t-\varDelta }\text {d} t{'}(-G(\varDelta +t{'})-G(\varDelta -t{'})+2G(t{'}))^2(t-t{'}-\varDelta )}{(t-\varDelta )^2}\\= & {} \frac{4\varGamma ^2\left( H+\frac{1}{2}\right) }{(2\lambda )^{2H+1}}\frac{\int _0^{\frac{t}{\varDelta }-1}\text {d}\tau \varDelta ^{2H}(t-\varDelta -\varDelta \tau )Q(\tau )}{(t-\varDelta )^2} \\= & {} \frac{4\varGamma ^2\left( H+\frac{1}{2}\right) \varDelta ^{2H}}{(2\lambda )^{2H+1}(t-\varDelta )}\int _0^{\frac{t}{\varDelta }-1}Q(\tau )\text {d}\tau -\frac{4\varGamma ^2\left( H+\frac{1}{2}\right) \varDelta ^{2H+1}}{(2\lambda )^{2H+1}(t-\varDelta )^2}\int _0^{\frac{t}{\varDelta }-1}\tau Q(\tau )\text {d}\tau \\\simeq & {} D\frac{4\varGamma ^2\left( H+\frac{1}{2}\right) }{(2\lambda )^{2H+1}}\varDelta ^{2H}t^{-1}, \end{aligned}$$

for long times and moderate $\lambda ,$ where

$$\begin{aligned} G(t)=\frac{2\varGamma \left( H+\frac{1}{2}\right) }{(2\lambda )^{H+\frac{1}{2}}}|t|^{H-\frac{1}{2}}\text {e}^{-\lambda |t|}, \end{aligned}$$

$$\begin{aligned} Q(t)=\left[ (1+t)^{H-\frac{1}{2}}\text {e}^{-\lambda \varDelta (1+t)}+ |t-1|^{H-\frac{1}{2}}\text {e}^{-\lambda |\varDelta (t-1)|}-2t^{H-\frac{1}{2}}\text {e}^{-\lambda \varDelta t}\right] ^2, \end{aligned}$$

$$\begin{aligned} D=\int _0^{\infty }\text {d}\tau \left[ (1+\tau )^{H-\frac{1}{2}}\text {e}^{-\lambda \varDelta (1+\tau )}+ |\tau -1|^{H-\frac{1}{2}}\text {e}^{-\lambda |\varDelta (\tau -1)|}-2\tau ^{H-\frac{1}{2}}\text {e}^{-\lambda \varDelta \tau }\right] ^2. \end{aligned}$$

Appendix 2: Asymptotic Behavior of the Covariance Function of tfGn

Fixing the value of $\lambda$ and considering sufficiently small t,  we have

$$\begin{aligned} K(t)= & {} 2\langle \gamma (0)\gamma (t)\rangle \nonumber \\= & {} \frac{1}{h^2}\left( C_{t+h}^2|t+h|^{2H}+C_{t-h}^2|t-h|^{2H}-2C_t^2|t|^{2H}\right) \nonumber \\\simeq & {} C_t^2\frac{1}{h^2}\left( |t+h|^{2H}+|t-h|^{2H}-2|t|^{2H}\right) \nonumber \\\simeq & {} 2C_t^2H(2H-1)t^{2H-2}, \end{aligned}$$

(B.1)

where $C_t^2$ is a constant. Figure 9 is plotting the asymptotic second integration of K(t),  i.e., $C_t^2|t|^{2H}.$ The convexity (concavity) of $C_t^2|t|^{2H}$ implies the positivity (negativity) of K(t). It is clear that $K(t)>0$ (see Fig. 9a) for $\frac{1}{2}<H<1,$ and $K(t)<0$ (see Fig. 9b) for $0<H<\frac{1}{2}.$

Fig. 9

Evolution of the asymptotic second integration of K(t) defined in Eqs. (B.1) and (B.3). a $H=0.7$ and $\lambda =0.1$. It is clear that $C_t^2t^{2H}$ is a convex function when t is small and then along with t’s increasing, $C_t^2t^{2H}$ becomes a concave function. So the second derivation of $C_t^2t^{2H}$ goes from positive to negative, and then it approaches to zero. b $H=0.3$ and $\lambda =0.1.$ It is clear that $C_t^2t^{2H}$ is always a concave function. So the second derivation of $C_t^2t^{2H}$ is always negative, and in the end it tends to zero

Fixing $\lambda$ and letting $t \rightarrow +\infty ,$ i.e., $\lambda t$ is large, we have

$$\begin{aligned} K(t)= & {} 2\langle \gamma (0)\gamma (t)\rangle \nonumber \\= & {} \frac{1}{h^2}\left( C_{t+h}^2|t+h|^{2H}+C_{t-h}^2|t-h|^{2H}-2C_t^2|t|^{2H}\right) \nonumber \\= & {} \frac{1}{h^2}\left[ \left( C_{t+h}^2+C_t^2-C_t^2\right) |t+h|^{2H}+\left( C_{t-h}^2+C_t^2-C_t^2\right) |t-h|^{2H} \right. \nonumber \\&\left. -\,2C_t^2|t|^{2H}\right] \nonumber \\= & {} \frac{C_t^2}{h^2}|t|^{2H}\left( \left| 1+\frac{h}{t}\right| ^{2H}+\left| 1-\frac{h}{t}\right| ^{2H}-2\right) \nonumber \\&+\,\frac{1}{h^2}\left[ \left( C_{t+h}^2-C_t^2\right) |t+h|^{2H}+\left( C_{t-h}^2-C_t^2\right) |t-h|^{2H}\right] . \end{aligned}$$

(B.2)

Using Taylor’s series expansion, along with $C_t^2=\frac{2\varGamma (2H)}{(2\lambda |t|)^{2H}}-\frac{2\varGamma \left( H+\frac{1}{2}\right) K_H(\lambda |t|)}{\sqrt{\pi }(2\lambda |t|)^H}=\bar{A}|t|^{-2H}-\bar{B}|t|^{-H}K_H(\lambda |t|)$ and $K_H(\lambda t)\simeq \sqrt{\pi }(2\lambda t)^{-\frac{1}{2}}\text {e}^{-\lambda t}$ as $\lambda t$ is large, the first term in (B.2) is

$$\begin{aligned} 2C_t^2H(2H-1)t^{2H-2}\simeq 2\bar{A}H(2H-1)t^{-2}-2\bar{B}H(2H-1)\sqrt{\pi }(2\lambda )^{-\frac{1}{2}}t^{H-\frac{5}{2}}\text {e}^{-\lambda t}, \end{aligned}$$

and the second term in (B.2) is

$$\begin{aligned}&\frac{1}{h^2}\left[ \left( C_{t+h}^2-C_t^2\right) |t+h|^{2H}+\left( C_{t-h}^2-C_t^2\right) |t-h|^{2H}\right] \\&\quad ={-}\frac{\bar{A}}{h^2}\left\{ \left| 1+\frac{h}{t}\right| ^{2H}+\left| 1-\frac{h}{t}\right| ^{2H}-2\right\} \\&\qquad +\,\frac{\bar{B}}{h^2}\left\{ -|t+h|^HK_H(\lambda |t+h|)+\frac{|t+h|^{2H}}{t^H}K_H(\lambda t)-|t-h|^HK_H(\lambda |t-h|) \right. \\&\qquad \left. +\frac{|t-h|^{2H}}{t^H}K_H(\lambda t)\right\} \\&\quad \simeq {-}2\bar{A}H(2H-1)t^{-2} \\&\qquad -\,\sqrt{\pi }(2\lambda )^{-\frac{1}{2}}\frac{\bar{B}}{h^2}\left\{ g(t+h)+g(t-h)-\left[ \left( 1+\frac{h}{t}\right) ^{2H}+\left( 1-\frac{h}{t}\right) ^{2H}\right] g(t)\right\} \\&\quad \simeq {-}2\bar{A}H(2H-1)t^{-2}\\&\qquad -\,\sqrt{\pi }(2\lambda )^{-\frac{1}{2}}\frac{\bar{B}}{h^2}\{g(t+h)+g(t-h)-2g(t)\}\\&\quad \simeq -\,2\bar{A}H(2H-1)t^{-2}-\sqrt{\pi }(2\lambda )^{-\frac{1}{2}}\bar{B}\frac{\text {d}^2g(t)}{\text {d} t^2}\\&\quad \simeq -\,2\bar{A}H(2H-1)t^{-2}-\sqrt{\pi }(2\lambda )^{-\frac{1}{2}}\bar{B}\lambda ^2t^{H-\frac{1}{2}}\text {e}^{-\lambda t}, \end{aligned}$$

where $g(x)=x^{H-\frac{1}{2}}\text {e}^{-\lambda x}.$ Combining the above two estimations leads to

$$\begin{aligned} \begin{array}{ll}&K(t)\simeq {-}\frac{2\varGamma \left( H+\frac{1}{2}\right) \lambda ^2}{(2\lambda )^{H+\frac{1}{2}}}t^{H-\frac{1}{2}}\text {e}^{-\lambda t}<0, \end{array} \end{aligned}$$

(B.3)

for large t;  see Fig. 9. The K(t) can also be obtained by making second derivative on the asymptotic expression of $C_t^2t^{2H}$ for large t. So the asymptotic behavior of tfGn’s covariance function is $\langle \gamma (0)\gamma (t)\rangle \simeq {-}\frac{\varGamma \left( H+\frac{1}{2}\right) \lambda ^2}{(2\lambda )^{H+\frac{1}{2}}}t^{H-\frac{1}{2}}\text {e}^{-\lambda t}$ for large t and $\langle \gamma (0)\gamma (t)\rangle \simeq C_t^2H(2H-1)t^{2H-2}$ with a constant $C_t^2$ for small t.

Appendix 3: Algorithm for Numerical Simulations

To generate tfGn $X_0,\,X_1,\ldots ,$ we adopt the Hosking method [7], which works for the general stationary Gaussian process. The key observation of this algorithm is to generate $X_{n+1}$ by the conditional distribution of $X_{n+1}$ given $X_n,\ldots ,X_0$ recursively. The covariance function of tfGn is

$$\begin{aligned} \rho (k):=\mathbb {E}X_nX_{n+k}, \end{aligned}$$

for $n,\,k=0,\,1,\,2,\ldots$ Note that $\langle B_{\alpha ,\lambda }^2(t)\rangle =C_t^2|t|^{2H}$ from (2.2) and $B_{\alpha ,\lambda }(t)$ has the stationary increments, $X_n:=B_{\alpha ,\lambda }(n+1)-B_{\alpha ,\lambda }(n)\sim N(0,\,C_1^2),$ which means $\rho (0)=\mathbb {E}(X_n^2)=C_1^2.$ Furthermore, let $D(n)=(\rho (i-j))_{i,j=0,\ldots ,n}$ be the covariance matrix and c(n) be the $(n+1)$-column vector with elements $c(n)_k=\rho (k+1),\,k=0,\ldots ,n.$ Define the $(n+1)\times (n+1)$ flipping matrix $F(n)=(\mathbf {1}(i=n-j))_{i,j=0,\ldots ,n}$, where $\mathbf {1}$ denotes the indicator function.

We claim that the conditional distribution of $X_{n+1}$ is Gaussian with expectation $\mu _n$ and variance $\sigma _n^2$ given by

$$\begin{aligned} \begin{array}{ll} \mu _n:=c(n){'}D(n)^{-1} \left( \begin{array}{c} X_n \\ \vdots \\ X_1 \\ X_0 \end{array}\right) , \quad \sigma _n^2:=C_1^2-c(n){'}D(n)^{-1}c(n). \end{array} \end{aligned}$$

(C.1)

To avoid matrix inversion in (C.1) in each step, define $d(n):=D(n)^{-1}c(n),$ and $\tau _n:=d(n){'}F(n)c(n)=c(n){'}F(n)d(n).$ Split the matrix $D(n+1)$ as follows:

$$\begin{aligned} D(n+1)= & {} \left( \begin{array}{cc} C_1^2 &{} c(n){'} \\ c(n) &{} D(n) \end{array} \right) \\= & {} \left( \begin{array}{cc} D(n) &{} F(n)c(n) \\ c(n){'}F(n) &{} C_1^2 \end{array} \right) . \end{aligned}$$

With some simple calculations, one gets

$$\begin{aligned} D(n+1)^{-1}= & {} \frac{1}{\sigma _n^2}\left( \begin{array}{cc} 1 &{} -d(n){'} \\ -d(n) &{} \sigma _n^2D(n)^{-1}+d(n)d(n){'} \end{array} \right) \nonumber \\= & {} \frac{1}{\sigma _n^2}\left( \begin{array}{cc} \sigma _n^2D(n)^{-1}+F(n)d(n)d(n){'}F(n) &{} -F(n)d(n) \\ -d(n){'}F(n) &{} 1 \end{array} \right) . \end{aligned}$$

(C.2)

From (C.2), for each $x\in \mathbb {R}^{n+1}$ and $y\in \mathbb {R},$ we have

$$\begin{aligned} (\begin{array}{cc} y&x{'} \end{array}) D(n+1)^{-1} \left( \begin{array}{c} y \\ x \end{array}\right) =\frac{(y-d(n){'}x)^2}{\sigma _n^2}+x{'}D(n)^{-1}x. \end{aligned}$$

This implies that the conditional distribution of $X_{n+1}$ is indeed Gaussian with expectation $\mu _n$ and variance $\sigma _n^2.$ On the other hand, by (C.2), some recursions are as follows:

$$\begin{aligned} \sigma _{n+1}^2=\sigma _n^2-\frac{(\rho (n+2)-\tau _n)^2}{\sigma _n^2}, \end{aligned}$$

and

$$\begin{aligned} d(n+1)=\left( \begin{array}{c} d(n)-\phi _nF(n)d(n) \\ \phi _n \end{array}\right) , \end{aligned}$$

with

$$\begin{aligned} \phi _n=\frac{\rho (n+2)-\tau _n}{\sigma _n^2}. \end{aligned}$$

We start the recursion with $\mu _0=\rho (1)X_0,\,\sigma _0^2=C_1^2-\rho (1)^2$ and $\tau _0=\frac{1}{C_1^2}\rho (1)^2.$ Taking cumulative sums on the generated tfGn samples $X_0,\ldots ,X_n,$ one obtains the tfBm sample $B_{\alpha ,\lambda }(k),\,k=0,\ldots ,n.$

Next, we consider to generate a series tfGn samples of number N for obtaining $B_{\alpha ,\lambda }(T).$ The scaling property of tfBm shows that $B_{\alpha ,\lambda }(T)=\left( \frac{T}{N}\right) ^H B_{\alpha ,\lambda T/N}(N)$ [14]. Denoting $\gamma (T):=\gamma _{\alpha ,\lambda }(T)=\left( \frac{T}{N}\right) ^{H-1}\gamma _{\alpha ,\lambda T/N}(N),$ i.e., $h=\frac{T}{N}$ in (2.8), for simulating the second moment of position $\langle x^2(t)\rangle ,$ we solve (3.1) with the scheme:

$$\begin{aligned}&m\frac{v(t_{n+1})-v(t_n)}{h}\\&\quad ={-}\xi \int _0^{t_{n+1}}K\left( t_{n+1}-\tau \right) v(\tau )\text {d}\tau +\sqrt{2k_BT\xi }\gamma \left( t_{n+1}\right) \\&\quad ={-}\xi \left( \frac{h}{2}\left( K\left( t_{n+1}\right) v(0)+K(0)v\left( t_{n+1}\right) \right) +h\sum \limits _{i=1}^nK\left( t_i\right) v\left( t_{n+1-i}\right) \right) \\&\qquad +\,\sqrt{2k_BT\xi }\gamma \left( t_{n+1}\right) , \end{aligned}$$

where h is the step length, and $K(t)=\frac{1}{h^2}(C_{t+h}^2|t+h|^{2H}+C_{t-h}^2|t-h|^{2H}-2C_t^2|t|^{2H}).$ In simulating the normalized displacement correlation function $C_x(t),$ we take the initial conditions as $v_0\sim N\left( 0,\,\frac{k_BT}{m}\right) ,$ and $x_0\sim U[-r,\,r],$ where $r=\sqrt{\frac{3k_BT}{m}}\frac{1}{\omega }.$