Occupation Time of a Randomly Accelerated Particle on the Positive Half Axis: Results for the First Five Moments

Abstract

In the random acceleration process a point particle is accelerated by Gaussian white noise with zero mean. Although several fundamental statistical properties of the motion have been analyzed in detail, the statistics of occupation times is still not well understood. We consider the occupation or residence time $T_+$ on the positive x axis of a particle which is randomly accelerated on the unbounded x axis for a time t. The first two moments of $T_+$ were recently derived by Ouandji Boutcheng et al. (J Stat Mech 053213:1–10, 2016). With an alternate approach utilizing basis functions which have proved useful in other studies of randomly accelerated motion, results for the first five moments are obtained in this paper.

Appendices

Appendix A: Derivation of the Identity (19)

Expression (19) may be checked quickly and non-rigorously by integrating over F numerically for a variety of numerical values of v. It follows analytically from integrating the closure relation (16) over all $v'$ to obtain

$$\begin{aligned} 1= v\int _0^\infty dF F^{-1/2}\left[ \psi _{s,F}(-v)-\psi _{s,F}(v)\right] \end{aligned}$$

(75)

and rewriting this as

$$\begin{aligned} 1= \left( s-{\partial ^2\over \partial v^2}\right) I(v),\quad I(v)\equiv \int _0^\infty dF F^{-3/2}\left[ \psi _{s,F}(-v)+\psi _{s,F}(v)\right] , \end{aligned}$$

(76)

with the help of the Airy differential equation (13). Differential equation (76) for I(v) has the solution $I(v)=s^{-1}+A\exp \left( \sqrt{s}v\right) +B\exp \left( -\sqrt{s}v\right)$, where, however, the constants A and B both vanish, since I(v), as defined by the integral in (76), remains finite for $v\rightarrow \pm \infty$. Thus, $I(v)=s^{-1}$, which, together with the definition of I(v) in (76), establishes (19).

We note that (19) is also consistent with the exact result

$$\begin{aligned} \int _0^\infty dF F^{-3/2}\psi _{s,F}(-v)=\left\{ \begin{array}{l}(2s)^{-1}\left( 2-e^{-\sqrt{3s} |v|}\right) ,\\ (2s)^{-1}e^{-\sqrt{3s} |v|}, \end{array}\right. \begin{array}{l}v>0,\\ v<0.\end{array} \end{aligned}$$

(77)

It may be derived by first calculating $\tilde{Q}_p(0,v,s)$ to first order in p, using the upper line of (17) and equation (21), which imply

$$\begin{aligned} \tilde{Q}_p(0,v,s)={1\over s}-{p\over s} \int _0^\infty dF F^{-3/2}\psi _{s,F}(-v)+\mathrm{O}\left( p^2\right) . \end{aligned}$$

(78)

Differentiating this expression with respect to p, as in (6), we then obtain

$$\begin{aligned} \int _0^\infty dt e^{-st}\langle T_+\rangle (0,v,t)={1\over s} \int _0^\infty dF F^{-3/2}\psi _{s,F}(-v) \end{aligned}$$

(79)

for the Laplace transform of $\langle T_+\rangle (0,v,t)$. Equating this result to the Laplace transform of expression (2) for $\langle T_+\rangle (0,v,t)$ and explicitly evaluating the latter leads directly to (77). By combining (77) with the Airy differential equation (13), the integrals $\int _0^\infty dF F^{-n}\psi _{s,F}(-v)$, where $n={1\over 2}$, ${3\over 2}$, ${5\over 2}$,..., can all be evaluated analytically.

Appendix B: Evaluation of the Integral in Eq. (23)

Combining the relation

$$\begin{aligned} \int _{-\infty }^\infty dv \left[ {\partial ^2\over \partial v^2} \psi _{s,F}(-v)\right] \psi _{s+p,G}(v)=\int _{-\infty }^\infty dv \psi _{s,F}(-v){\partial ^2\over \partial v^2} \psi _{s+p,G}(v), \end{aligned}$$

(80)

which follows from integration by parts, with the Airy differential equation (13) leads to

$$\begin{aligned} k(F,G)\equiv & {} \int _{-\infty }^\infty dv v \psi _{s,F}(-v)\psi _{s+p,G}(v)\nonumber \\ {}= & {} -p(F+G)^{-1}\int _{-\infty }^\infty dv \psi _{s,F}(-v)\psi _{s+p,G}(v). \end{aligned}$$

(81)

With the help of definition (12) and the integral representation [29]

$$\begin{aligned} \mathrm{Ai}(z)={1\over 2\pi }\int _{-\infty }^\infty dq e^{i\left( {1\over 3}q^3+qz\right) }, \end{aligned}$$

(82)

the integral on the right side of (81) can be written as

$$\begin{aligned} \int _{-\infty }^\infty dv \psi _{s,F}(-v) \psi _{s+p,G}(v)= & {} (2\pi )^{-2}(FG)^{-1/6}\int _{-\infty }^\infty dv\int _{-\infty }^\infty dk\int _{-\infty }^\infty d\ell \nonumber \\&\times \exp \left\{ i\left[ {1\over 3}k^3+k\left( -F^{1/3}v+sF^{-2/3}\right) \right. \right. \nonumber \\&\left. \left. +{1\over 3}\ell ^3+\ell \left( G^{1/3}v+(s+p)G^{-2/3}\right) \right] \right\} . \end{aligned}$$

(83)

First integrating over v in (83), which leads to a factor $2\pi \delta \left( kF^{1/3}-\ell G^{1/3}\right)$, then integrating over $\ell$, and finally making the substitution $k=[G/(F+G)]^{1/3}q$, we obtain

$$\begin{aligned}&\int _{-\infty }^\infty dv \psi _{s,F}(-v) \psi _{s+p,G}(v)=(FG)^{-1/6}(F+G)^{-1/3}\nonumber \\&\qquad \times {1\over 2\pi }\int _{-\infty }^\infty dq \exp \left\{ i\left[ {1\over 3}q^3+q\left( {(s+p)F+sG\over (F+G)^{1/3}(FG)^{2/3}}\right) \right] \right\} \end{aligned}$$

(84)

$$\begin{aligned}&\quad =(FG)^{-1/6}(F+G)^{-1/3}\mathrm{Ai}\left( {(s+p)F+sG\over (F+G)^{1/3}(FG)^{2/3}}\right) . \end{aligned}$$

(85)

In rewriting (84) in the form (85), we have again utilized the integral representation (82) of the Airy function.

The final expression for k(F, G) in (23) follows from substituting (85) in (81). Note that k(F, G) vanishes in the limit $p\rightarrow 0$, in accordance with the orthonormality property (15).