Occupation time of a randomly accelerated particle on the positive half axis: Results for the first five moments
aa r X i v : . [ m a t h - ph ] A ug Occupation time of a randomly accelerated particle on thepositive half axis: Results for the first five moments
Theodore W. BurkhardtDepartment of Physics, Temple UniversityPhiladelphia, PA 19122, USAAugust 2, 2018
Abstract
In the random acceleration process a point particle moving in one dimension is accel-erated by Gaussian white noise with zero mean. Although several fundamental statisticalproperties of the motion have been analyzed in detail, the statistics of occupation times isstill 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. [1].With an alternate approach utilizing basis functions which have proved useful in otherstudies of randomly accelerated motion, results for the first five moments are obtained inthis paper.Keywords: random acceleration, occupation time, random walk, stochastic processPACS 05.10.Gg, 05.40.-a, 05.40.Fb Introduction
In the random acceleration process the variables, x and v , evolve according to dxdt = v , dvdt = η ( t ) , (1)where η ( t ) is Gaussian white noise, with h η ( t ) i = 0 and h η ( t ) η ( t ′ ) i = 2 δ ( t − t ′ ). The quantities x and v may be interpreted as the position and velocity of a particle moving on the x axisand subject to a random force. Another interpretation is to regard v ( t ) as the position of aparticle making a one-dimensional random walk and x ( t ) as the area under the correspondingBrownian curve. Still other interpretations are found in physical applications to semiflexiblepolymers [2, 3, 4, 5], interface growth [6, 7, 8], and the Burgers equation [9].Several of the fundamental statistical properties of randomly accelerated motion havebeen anayzed in detail. The probability density G ( x, v ; x , v ; t ) for propagation from initialvalues ( x , v ) to ( x, v ) in a time t is known both for propagation on the unbounded line −∞ < x < ∞ and for propagation on the half line x > < x < L [17, 18, 19]. For recent reviews of random acceleration statistics, withemphasis on first-passage properties, see [20, 21].Despite this progress, the statistics of occupation times for randomly accelerated motionis still not well understood. For early work on the statistics of occupation times and applica-tions to various stochastic processes, see [20, 22, 23, 24, 25, 26] and references therein. Theparticular occupation or residence time considered in this paper is the time T + spent on thepositive x axis by a particle with initial position x and initial velocity v , which is randomlyaccelerated on the unbounded x axis for a time t . In the case of random walk statistics, both T + and T m , the time at which the walker reaches its maximum excursion from the startingpoint, are distributed according to L´evy’s arcsine law [22, 27]. For random acceleration statis-tics Majumdar et al. [28] have derived the distribution of T m . The distribution of T + is stillunknown, but its first two moments were recently calculated by Ouandji Boutcheng et al. [1],who obtained h T + i ( x , v , t ) = 12 t + 12 Z t dt ′ erf " √ x + v ( t − t ′ )( t − t ′ ) / , (2)2 T + i (0 , , t ) = 12 t , (3) h T i (0 , , t ) = 3 / π t ≈ .
413 497 t , (4)where erf denotes the error function [29]. From the expression for h T i (0 , , t ) in (4) and theresult h T m i (0 , , t ) = t ≈ .
416 667 t of [28], and from Monte Carlo calculations of thedistributions of T + and T m reported in [1], it is clear that the distributions of T + and T m arevery similar, but not identical.Section 2 of this paper begins with the same generating-function formalism for the distri-bution of T + and its moments as in [1]. The generating function is then expanded in a setof basis functions which have proved useful in studying random acceleration on the half line[2, 11, 28]. To calculate the generating function, one needs to determine the expansion coeffi-cients. In Section 3 this crucial step is reduced to the problem of solving an integral equation.In Section 4 the integral equation is solved to low orders in a perturbation expansion, and inSection 5 results for the first five moments of T + are presented. Section 6 contains concludingremarks. Q p ( x , v , t ) A quantity of central importance in our study is the generating function Q p ( x , v , t ) = (cid:28) exp (cid:20) − p Z t dt ′ θ (cid:0) x ( t ′ ) (cid:1)(cid:21)(cid:29) , (5)also considered in [1]. Here θ ( x ) is the usual unit step function, and the angular bracketsdenote an average over the possible paths of a randomly accelerated particle with initialposition x and initial velocity v . The moments of T + are related by h T n + i ( x , v , t ) = ( − n ∂ n ∂p n Q p ( x , v , t ) (cid:12)(cid:12)(cid:12)(cid:12) p =0 (6)to derivatives of the generating function, and the distribution of T + , P ( T + | x , v , t ) ≡ (cid:28) δ (cid:18) T + − Z t dt ′ θ (cid:0) x ( t ′ ) (cid:1)(cid:19)(cid:29) = L − p → T + Q p ( x , v , t ) , (7)is the inverse Laplace transform of the generating function.Below, we will make use of the symmetric property Q p ( x , v , t ) = e − pt Q − p ( − x , − v , t ) , (8)3hich follows from the facts that (i) the residence times T + and T − on the positive andnegative x axes satisfy T + + T − = t , and (ii) the distribution of T − for a particle with initialconditions x , v is the same as the distribution of T + for a particle with initial conditions − x , − v .The generating function defined by (5) satisfies the Fokker-Planck type differential equa-tion [1, 2, 11, 23, 30] ∂∂t − v ∂∂x − ∂ ∂v + pθ ( x ) ! Q p ( x, v, t ) = 0 , (9)as follows, for example, from path integral considerations. To simplify the notation, thesubscripts of x and v have been dropped in (9). Forming the Laplace transform t → s of(9) and using the initial condition Q p ( x, v,
0) = 1 leads to s + pθ ( x ) − v ∂∂x − ∂ ∂v ! ˜ Q p ( x, v, s ) = 1 , (10)where ˜ Q p ( x, v, s ) = R ∞ dt e − st Q p ( x, v, t ). The Laplace transform of the symmetry property(8) has the form ˜ Q p ( x , v , s ) = ˜ Q − p ( − x , − v , s + p ) . (11)Let us look for a solution to the differential equation (10) in terms of the basis functions ψ s,F ( ± v ) = F − / Ai (cid:16) ± F / v + F − / s (cid:17) , (12)where Ai( z ) denotes the standard Airy function [29]. The combined set of functions ψ s,F ( v ), ψ s,F ( − v ), with 0 < F < ∞ , is complete on the interval −∞ < v < ∞ , and this basis has beenused to good advantage in studying randomly accelerated motion on the half line [2, 11, 28].The functions (12) satisfy the Airy differential equation s ± F v − ∂ ∂v ! ψ s,F ( ± v ) = 0 , (13)the orthonormality conditions Z ∞−∞ dv v ψ s,F ( − v ) ψ s,G ( − v ) = δ ( F − G ) , (14) Z ∞−∞ dv v ψ s,F ( − v ) ψ s,G ( v ) = 0 , (15)and the closure relation Z ∞ dF (cid:2) ψ s,F ( − v ) ψ s,F ( − v ′ ) − ψ s,F ( v ) ψ s,F ( v ′ ) (cid:3) = v − δ ( v − v ′ ) . (16)4n terms of these basis functions the most general solution of (10) which does not divergefor x → ±∞ is given by˜ Q p ( x, v, s ) = s + Z ∞ dF a ( F ) e F x ψ s,F ( − v ) , x < , s + p + Z ∞ dF b ( F ) e − F x ψ s + p,F ( v ) , x > . (17)To avoid a term proportional to δ ( x ) when the general solution (17) is substituted into dif-ferential equation (10), ˜ Q p ( x, v, s ) must be continuous at x = 0. This leads to the matchingcondition ps ( s + p ) = Z ∞ dF [ − a ( F ) ψ s,F ( − v ) + b ( F ) ψ s + p,F ( v )] . (18) a ( F ) and b ( F ) To calculate the generating function Q ( x, v, t ) using (17), one must first determine the un-known functions a ( F ) and b ( F ). In this section separate integral equations for a ( F ) and b ( F )are derived from the matching condition (18).With the help of the expansion1 s = Z ∞ dF F − / [ ψ s,F ( − v ) + ψ s,F ( v )] (19)in terms of the basis functions (12), which is derived in Appendix A, equation (18) may berewritten as ps + p Z ∞ dF F − / [ ψ s,F ( − v ) + ψ s,F ( v )]= ps Z ∞ dF F − / [ ψ s + p,F ( − v ) + ψ s + p,F ( v )]= Z ∞ dF [ − a ( F ) ψ s,F ( − v ) + b ( F ) ψ s + p,F ( v )] . (20)Solving (20) for a in terms of b and b in terms of a , using the orthonormality properties (14),(15), we obtain the coupled integral equations a ( F ) = − ps + p F − / + Z ∞ dG k ( F, G ) b ( G ) , (21) b ( F ) = ps F − / − Z ∞ dG k ( G, F ) a ( G ) , (22)where k ( F, G ) ≡ Z ∞−∞ dv v ψ s,F ( − v ) ψ s + p,G ( v )= − p ( F G ) − / ( F + G ) − / Ai (cid:18) ( s + p ) F + sG ( F + G ) / ( F G ) / (cid:19) . (23)5 derivation of the result (23) is given in Appendix B. To decouple (21) and (22), we substitute(22) on the right side of (21). This leads to an inhomogeneous Fredholm integral equation ofthe second kind, with a symmetric kernel, a ( F ) = a ( F ) + Z ∞ dG K ( F, G ) a ( G ) , (24) a ( F ) = − ps + p F − / + ps Z ∞ dG k ( F, G ) G − / , (25) K ( F, G ) = − Z ∞ dH k ( F, H ) k ( G, H ) , (26)for the unknown function a ( F ). The corresponding integral equation for b ( F ) is the sameas for a ( F ), except that the variables s and s + p are switched. (Under this switching, theprefactor p = ( s + p ) − s in the last line of (23) transforms to s − ( s + p ) = − p .) Q p (0 , , t ) to 4th order in p Together with equations (17), (23), (25), and (26), the solution to the integral equation(24) completely determines the generating function. However, solving the integral equationexactly appears extremely difficult, due to the complicated kernel (23), (26). In this sectionwe calculate a ( F ) and ˜ Q p (0 , , s ) perturbatively to fourth order in powers of p , confirmingand extending the second order results in [1]. With little extra work the fourth order resultsare extended to fifth order in the following section.Expanding the quantity a ( F ), defined in (23) and (25), to fourth order in p , one finds a ( F ) = − ps − ps + p s − p s ! F − / − p s Z ∞ dG G − / ( F G ) − / ( F + G ) − / × (cid:20) Ai (cid:18) s (cid:16) F − + G − (cid:17) / (cid:19) + p G − (cid:16) F − + G − (cid:17) − / Ai ′ (cid:18) s (cid:16) F − + G − (cid:17) / (cid:19) + 12 p G − (cid:16) F − + G − (cid:17) − / Ai ′′ (cid:18) s (cid:16) F − + G − (cid:17) / (cid:19)(cid:21) + O (cid:16) p (cid:17) . (27)On changing the integration variable from G to z = 1 + F/G , integrating the term involvingAi ′ ( ξ ) by parts, and utilizing the Airy differential equation Ai ′′ ( ξ ) = ξ Ai( ξ ), equation (27)6akes the form a ( F ) = − ps − ps + p s − p s ! F − / − p s F − / Z ∞ dz Ai (cid:16) sF − / z / (cid:17) ( ( z − z − / − p s ddz h ( z − z − / i + 12 p sF − ( z − z − / ) + O (cid:16) p (cid:17) . (28)With the help of Mathematica , all of the integrals in (28) can be evaluated explictly in termsof generalized hypergeometric functions.The integral equation a = a + R Ka in (24) has the iterative solution a = a + R Ka + R R
KKa + ... Since a ( F ) in equations (23)-(26) is O( p ) and K ( F, G ) is of order O (cid:0) p (cid:1) ,we only need to consider the first two terms, a + R Ka , to obtain a ( F ) to 4th order in p .The first term, a ( F ), is shown to 4th order in p in (28). The second term, R Ka , has theexpansion Z Ka ≡ Z ∞ dG K ( F, G ) a ( G ) = a ( F ) + a ( F ) + a ( F ) + O (cid:16) p (cid:17) , (29) a ( F ) = p s (cid:18) − ps (cid:19) × Z ∞ dG Z ∞ dH G − / ( F H G ) − / ( F + H ) − / ( G + H ) − / × Ai (cid:18) s (cid:16) F − + H − (cid:17) / (cid:19) Ai (cid:18) s (cid:16) G − + H − (cid:17) / (cid:19) , (30) a ( F ) = p s Z ∞ dG Z ∞ dH G − / ( F H G ) − / ( F + H ) − / ( G + H ) − / × H − (cid:20)(cid:16) F − + H − (cid:17) − / Ai ′ (cid:18) s (cid:16) F − + H − (cid:17) / (cid:19) Ai (cid:18) s (cid:16) G − + H − (cid:17) / (cid:19) + (cid:16) G − + H − (cid:17) − / Ai (cid:18) s (cid:16) F − + H − (cid:17) / (cid:19) Ai ′ (cid:18) s (cid:16) G − + H − (cid:17) / (cid:19)(cid:21) ,a ( F ) = p s Z ∞ dG Z ∞ dH ( F H G ) − / ( F + H ) − / ( G + H ) − / × Ai (cid:18) s (cid:16) F − + H − (cid:17) / (cid:19) Ai (cid:18) s (cid:16) G − + H − (cid:17) / (cid:19) × G − / Z ∞ dz ( z − z − / Ai (cid:16) sG − / z / (cid:17) . (31)In analyzing the integrals in (30)-(31), we make the substitutions H = F/h and G = H/g = F/ ( gh ), which lead to a ( F ) = p s (cid:18) − ps (cid:19) F − / Z ∞ dg Z ∞ dh g h / (1 + g ) − / (1 + h ) − / Ai (cid:16) sF − / (1 + h ) / (cid:17) Ai (cid:16) sF − / h / (1 + g ) / (cid:17) , (32) a ( F ) = p s F − / Z ∞ dg Z ∞ dh g h / (1 + g ) − / (1 + h ) − / × h h (1 + h ) − / Ai ′ (cid:16) sF − / (1 + h ) / (cid:17) Ai (cid:16) sF − / h / (1 + g ) / (cid:17) + h / (1 + g ) − / Ai (cid:16) sF − / (1 + h ) / (cid:17) Ai ′ (cid:16) sF − / h / (1 + g ) / (cid:17)i ,a ( F ) = p s F − / Z ∞ dg Z ∞ dh g / h / (1 + g ) − / (1 + h ) − / × Ai (cid:16) sF − / (1 + h ) / (cid:17) Ai (cid:16) sF − / h / (1 + g ) / (cid:17) × Z ∞ dz ( z − z − / Ai (cid:16) sF − / ( hg ) / z / (cid:17) . (33)We now specialize to the case of a particle which begins at the origin with velocity zero.According to equations (12) and (17), the generating function ˜ Q p ( x , v , s ) for these initialconditions is given by˜ Q p (0 , , s ) − s = Z ∞ dF a ( F ) ψ s,F (0)= Z ∞ dF a ( F ) F − / Ai (cid:16) sF − / (cid:17) = 32 s / Z ∞ dx a (cid:18) sx (cid:19) / ! x − / Ai( x ) , x = sF − / . (34)Let us denote the contributions to the right side of (34) from a ( F ),..., a ( F ) in equations(28), (32), (33), and (33) by ˜ Q (0) p (0 , , s ),..., ˜ Q (3) p (0 , , s ). Substituting expression (28) for a ( F ) in (34) and making use of the integrals Z ∞ dx Ai( x ) = 13 , (35) Z ∞ dx x Ai( x )Ai( ax ) = 12 π √ a − a − , (36) Z ∞ dx x Ai( x )Ai( ax ) = √ π ( a + 1) (cid:18) a − a − (cid:19) , (37)one obtains ˜ Q (0) p (0 , , s ) = − p s + 3 / π p s + / π − ! p s + / π − ! p s + O (cid:16) p (cid:17) . (38)Substituting expression (32) for a ( F ) in (34) leads to the triple integral˜ Q (1) p (0 , , s ) = 32 p s (cid:18) − ps (cid:19) Z ∞ dx x Ai( x )8 Z ∞ dg Z ∞ dh g h / (1 + g ) − / (1 + h ) − / × Ai (cid:16) x (1 + h ) / (cid:17) Ai (cid:16) xh / (1 + g ) / (cid:17) . (39)Integrating over g analytically and then over h and x numerically, using Mathematica , gave aresult consistent with ˜ Q (1) p (0 , , s ) = − / π ! p s (cid:18) − ps (cid:19) (40)with an uncertainty in the numerical prefactor less than 10 − . That (40) is not just anexcellent approximation but is exact is shown in the next section.Substituting expression (33) for a ( F ) in (34) leads to the triple integral˜ Q (2) p (0 , , s ) = 32 p s Z ∞ dx x Ai( x ) × Z ∞ dg Z ∞ dh g h / (1 + g ) − / (1 + h ) − / × h h (1 + h ) − / Ai ′ (cid:16) x (1 + h ) / (cid:17) Ai (cid:16) xh / (1 + g ) / (cid:17) + h / (1 + g ) − / Ai (cid:16) x (1 + h ) / (cid:17) Ai ′ (cid:16) xh / (1 + g ) / (cid:17)i . (41)This integral can be written more compactly, without derivatives of Airy functions, by rewrit-ing the square bracket as[ ... ] = x − (cid:18) h ∂∂h − g ∂∂g (cid:19) Ai (cid:16) x (1 + h ) / (cid:17) Ai (cid:16) xh / (1 + g ) / (cid:17) (42)and then integrating by parts. This yields˜ Q (2) p (0 , , s ) = − p s Z ∞ dx x Ai( x ) × Z ∞ dg Z ∞ dh g h / (1 + g ) − / (1 + h ) − / × (cid:18) h −
21 + g (cid:19) Ai (cid:16) x (1 + h ) / (cid:17) Ai (cid:16) xh / (1 + g ) / (cid:17) . (43)Performing the integral in (43) over g analytically and then over h and x numerically,using Mathematica , one finds that the contributions of the terms 2 / (1 + h ) and − / (1 + g )in the integrand cancel on integration with an uncertainty less than 10 − . Presumably thecancellation is exact, although we have not succeeded in showing this analytically. Thus,within this minute uncertainty, ˜ Q (2) p (0 , , s ) is entirely determined by the integral of the first9f the sum of three terms. Since the integral of that term is the same as in (39) and (40),apart from the prefactor, we conclude that˜ Q (2) p (0 , , s ) = / π − ! p s . (44)Substituting expression (33) for a ( F ) in (34) leads to the quadruple integral˜ Q (3) p (0 , , s ) = 32 p s Z ∞ dx x Ai( x ) × Z ∞ dg Z ∞ dh g / h / (1 + g ) − / (1 + h ) − / × Ai (cid:16) x (1 + h ) / (cid:17) Ai (cid:16) xh / (1 + g ) / (cid:17) × Z ∞ dz ( z − z − / Ai (cid:16) xg / h / z / (cid:17) . (45)Integrating over z analytically and then over g , h , and x numerically, using Mathematica , gave˜ Q (3) p (0 , , s ) ≡ ǫ p s , ǫ = 0 .
000 872 073 2 , (46)believed to be correct to the number of digits shown. This completes the evaluation of all thecontributions to ˜ Q p (0 ,
0) through order p . T + Continuing to specialize to initial conditions x = 0, v = 0, we note that the generatingfunctions Q p (0 , , t ) and e pt/ Q p (0 , , t ) defined by (5) have the expansions Q p (0 , , t ) = ∞ X n =0 ( − p ) n n ! (cid:10) T n + (cid:11) , (47) e pt/ Q p (0 , , t ) = ∞ X n =0 ( − p ) n n ! D(cid:16) T + − t (cid:17) n E , (48)in terms of moments of T + . Here and throughout the remainder of the paper we use theshorthand h T n + i ≡ h T n + i (0 , , t ). According to the symmetry property (8), e pt/ Q p (0 , , t ) = e − pt/ Q − p (0 , , t ) is an even function of p . Together with (48), this implies D(cid:16) T + − t (cid:17) n E = 0 for n odd . (49)This result can be understood very simply. For a particle which begins at rest at the origin,the residence times T + and T − = t − T + on the positive and negative axes are identically10istributed. Thus, T + − t and T − − t = − ( T + − t ) also have identical distributions, whichleads directly to (49).For n =1, 3, 5, and 7, equation (49) implies h T + i = t , (50) D T E = (cid:10) T (cid:11) t − t , (51) D T E = (cid:10) T (cid:11) t − (cid:10) T (cid:11) t + t . (52)In general (cid:10) T n + (cid:11) for n odd is determined by the 0 th , 2 nd , ..., ( n − st moments of T + .Combining the contributions (38), (40), (44), and (46) to ˜ Q p (0 , , s ) through order p ,performing the inverse Laplace transform s → t , comparing with (6) or (47) to identify themoments, and making use of (52), we obtain Q P (0 , , t ) to order p and the corresponding 5moments of T + . In terms of the quantity ǫ = 0 .
000 872 073 2 , (53)defined in (45) and (46) and obtained by numerical integration,˜ Q p (0 , , s ) = 1 s − ps + 3 / π p s − / π − ! p s + · / π − ǫ ! p s − · / π −
192 + 52 ǫ ! p s + O (cid:16) p (cid:17) , (54) Q p (0 , , t ) = 1 − pt
1! + 3 / π ( pt ) − / π − ! ( pt ) · / π − ǫ ! ( pt ) − · / π −
192 + 52 ǫ ! ( pt )
5! + O (cid:16) p (cid:17) , (55) h T + i = t , (56) D T E = 3 / π t = 0 .
413 496 672 t , (57) D T E = / π − ! t = 0 .
370 245 007 t , (58) D T E = · / π − ǫ ! t = 0 .
342 587 125 t , (59) D T E = · / π −
192 + 52 ǫ ! t = 0 .
322 726 133 t . (60)11hese are our main findings. The results (56) and (57) for the first two moments areconsistent with [1], and the third moment (58) follows immediately from the second momentand (51). That the O (cid:0) p (cid:1) calculation in the preceding section leads to the same result for thethird moment proves our conjecture that the quantity ˜ Q (1) p (0 , , s ) defined in (39) is exactlygiven by (40). The results (59) and (60) for the fourth and fifth moments depend on the value(53) of ǫ obtained by numerical integration and are believed to be exact to the number ofdigits shown.These results for the moments can also be summarized as D(cid:16) T + − t (cid:17) n E = , n odd , / π − ! t = 0 .
163 496 672 t , n = 2 , / π − ǫ ! t = 0 .
034 842 117 t , n = 4 (61)For comparison we show the generating function and moments of the time T m at whicha particle, initially at rest at the origin and randomly accelerted for a time t , reaches itsmaximum displacement [28]. In terms of the hypergeometric function F and the modifiedBessel function I ν [29],˜ Q ( m ) p (0 , , s ) = s − F ( , ; − p/s ) . (62) Q ( m ) p (0 , , t ) = 2 − / Γ (cid:16) (cid:17) ( pt ) / e − pt/ I − ( pt/ , (63) h T nm i = Γ ( ) Γ ( + n ) Γ ( + n ) Γ ( ) t n , (64) h T m i = t , (65) D T m E = t = 0 .
416 666 667 t , (66) D T m E = t = 0 .
375 000 000 t , (67) D T m E = t = 0 .
348 214 286 t , (68) D T m E = t = 0 .
328 869 048 t , (69) D(cid:16) T m − t (cid:17) n E = , n odd , / − n Γ ( ) Γ ( + n ) Γ ( ) Γ ( + n ) t n , n even , t = 0 .
166 666 667 t , n = 2 , t = 0 .
035 714 286 t , n = 4 . (70)12part from the first moment, all of the moments of T + are slightly smaller than the corre-sponding moments of T m , in accordance with the conclusion of [1] that the distributions of T + and T m are very similar but not identical. We close by comparing the approach of this paper with that of Ouandji Boutcheng et al. [1].In both papers moments of T + are calculated by solving an integral equation to low orders in p , but the integral equations considered in the two papers are different.The results of [1] are based on the integral equation Q p ( x, v, t ) = 1 − p Z t dt ′ Z ∞ dx ′ Z ∞−∞ dv ′ Q p ( x ′ , v ′ , t ′ ) G ( x ′ , v ′ ; x, v ; t − t ′ ) . (71)It is obtained by integrating differential equation (9) with the Green’s function [2] G ( x ′ , v ′ ; x, v ; t ) = (cid:16) / / πt (cid:17) × exp h − x ′ − x − v ′ t )( x ′ − x − vt ) /t − ( v ′ − v ) /t i , (72)which is the probability density for propagation on the unbounded x axis from ( x, v ) to ( x ′ , v ′ )in a time t . The expressions (2)-(4) for the first two moments of T + follow from iterating (71)to order p and using the relation (6) between the moments and the generating function. Dueto the complicated kernel and the triple integral on the right-hand side of (71), obtaininganalytic results beyond order p with this approach does not seem feasible. Since the kernelin (71) is proportional to p , a fourth order calculation requires four iterations of the integralequation, i.e. evaluation of a 12-fold integral.The results of this paper, on the other hand, were derived from integral equation (24)for the expansion coefficient a ( F ) in the general solution (17) for the generating function. Incontrast to the triple integral in (71), there is only a single integral on the right-hand sideof (24). However, the complicated kernel K ( F, G ), shown in (23) and (26), involves specialfunctions and an additional integral over the variable H . Since a ( F ) and K ( F, G ) are of order p and p , respectively, only one iteration of the integral equation (24) is required to obtain The integration over t ′ in (71) can be eliminated by forming the Laplace transform t → s of (71) andusing the Laplace transform of G given in equation (8) of [2]. Even with these steps it still appears extremelydifficult to solve the integral equation analytically beyond order p . ( F ) to fourth order. Even so, we needed numerical integration to evaluate the fourth-ordercontribution (45).Each of the integral equations (24) and (71) determines not only the first few moments of T + but the entire distribution (7) of T + . The determination of this distribution is a challengingunsolved problem. Appendix A: Derivation of the identity (19)
Expression (19) may be checked quickly and non-rigorously by integrating over F numericallyfor a variety of numerical values of v . It follows analytically from integrating the closurerelation (16) over all v ′ to obtain1 = v Z ∞ dF F − / [ ψ s,F ( − v ) − ψ s,F ( v )] (73)and rewriting this as1 = s − ∂ ∂v ! I ( v ) , I ( v ) ≡ Z ∞ dF F − / [ ψ s,F ( − v ) + ψ s,F ( v )] , (74)with the help of the Airy differential equation (13). Differential equation (74) for I ( v ) has thesolution I ( v ) = s − + A exp ( √ sv ) + B exp ( −√ sv ), where, however, the constants A and B both vanish, since I ( v ), as defined by the integral in (74), remains finite for v → ±∞ . Thus, I ( v ) = s − , which, together with the definition of I ( v ) in (74), establishes (19).We note that (19) is also consistent with the exact result Z ∞ dF F − / ψ s,F ( − v ) = (2 s ) − (cid:16) − e −√ s | v | (cid:17) , (2 s ) − e −√ s | v | , v > ,v < . (75)It may be derived by first calculating ˜ Q p (0 , v, s ) to first order in p , using the upper line of(17) and equation (21), which imply˜ Q p (0 , v, s ) = 1 s − ps Z ∞ dF F − / ψ s,F ( − v ) + O (cid:16) p (cid:17) . (76)Differentiating this expression with respect to p , as in (6), we then obtain Z ∞ dt e − st h T + i (0 , v, t ) = 1 s Z ∞ dF F − / ψ s,F ( − v ) (77)for the Laplace transform of h T + i (0 , v, t ). Equating this result to the Laplace transform ofexpression (2) for h T + i (0 , v, t ) and explicitly evaluating the latter leads directly to (75). By14ombining (75) with the Airy differential equation (13), the integrals R ∞ dF F − n ψ s,F ( − v ),where n = , , , ... , can all be evaluated analytically. Appendix B: Evaluation of the integral in equation (23)
Combining the relation Z ∞−∞ dv " ∂ ∂v ψ s,F ( − v ) ψ s + p,G ( v ) = Z ∞−∞ dv ψ s,F ( − v ) ∂ ∂v ψ s + p,G ( v ) , (78)which follows from integration by parts, with the Airy differential equation (13) leads to k ( F, G ) ≡ Z ∞−∞ dv v ψ s,F ( − v ) ψ s + p,G ( v )= − p ( F + G ) − Z ∞−∞ dv ψ s,F ( − v ) ψ s + p,G ( v ) . (79)With the help of definition (12) and the integral representation [29]Ai( z ) = 12 π Z ∞−∞ dq e i ( q + qz ) , (80)the integral on the right side of (79) can be written as Z ∞−∞ dv ψ s,F ( − v ) ψ s + p,G ( v ) = (2 π ) − ( F G ) − / Z ∞−∞ dv Z ∞−∞ dk Z ∞−∞ dℓ × exp (cid:26) i (cid:20) k + k (cid:16) − F / v + sF − / (cid:17) + 13 ℓ + ℓ (cid:16) G / v + ( s + p ) G − / (cid:17)(cid:21)(cid:27) . (81)First integrating over v in (81), which leads to a factor 2 πδ (cid:16) kF / − ℓG / (cid:17) , then integratingover ℓ , and finally making the substitution k = [ G/ ( F + G )] / q , we obtain Z ∞−∞ dv ψ s,F ( − v ) ψ s + p,G ( v ) = ( F G ) − / ( F + G ) − / × π Z ∞−∞ dq exp (cid:26) i (cid:20) q + q (cid:18) ( s + p ) F + sG ( F + G ) / ( F G ) / (cid:19)(cid:21)(cid:27) (82)= ( F G ) − / ( F + G ) − / Ai (cid:18) ( s + p ) F + sG ( F + G ) / ( F G ) / (cid:19) . (83)In rewriting (82) in the form (83), we have again utilized the integral representation (80) ofthe Airy function.The final expression for k ( F, G ) in (23) follows from substituting (83) in (79). Note that k ( F, G ) vanishes in the limit p →
0, in accordance with the orthonormality property (15).15 cknowledgements
I thank Hermann Jo¨el Ouandji Boutcheng, Alberto Rosso, and Andrea Zoia for helpful cor-respondence.