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APS/123-QED Tempered Fractional Feynman-Kac Equation Xiaochao Wu1, Weihua Deng1, and Eli Barkai2 1School of Mathematics and Statistics, Gansu Key Laboratory of Applied Mathematics and Complex Systems, Lanzhou University, Lanzhou 730000, P.R. China 2Department of Physics, Advanced Materials and Nanotechnology Institute, Bar Ilan University, Ramat-Gan 52900, Israel Functionals of Brownian/non-Brownian motions have diverse applications and attracted a lot of interestofscientists. ThispaperfocusesonderivingtheforwardandbackwardfractionalFeynman- Kacequationsdescribingthedistributionofthefunctionalsofthespaceandtimetemperedanoma- lous diffusion, belonging to the continuous time random walk class. Several examples of the func- 6 1 tionals are explicitly treated, including the occupation time in half-space, the first passage time, 0 the maximal displacement, the fluctuations of the occupation fraction, and the fluctuations of the 2 time-averaged position. n PACSnumbers: 02.50.-r,05.30.Pr,02.50.Ng,05.40.-a,05.10.Gg a J 0 I. INTRODUCTION length η(x). The exponential tempering offers technical 3 advantagessincethetemperedprocessisstillaninfinitely divisible L´evy process [15]. ] Normal diffusion describes the Brownian dynamics n It is well known that there are many physical quanti- a characterizedbyalargenumberofsmallevents,e.g.,the ties used to describe the motion features of a Brownian - motionofpollengrainsinwater. However,inmanycases, a particle. An example one, Brownian functionals, being t the (rare) large fluctuations result in the non-Brownian defined as A= tU[x(τ)]dτ, where U(x) is a prescribed a motion, anomalous diffusion, being carefully studied in 0 d function and x(t) is a trajectory of a Brownian parti- physics [1, 2], finance [3], hydrology [4], and many other R . cle. Here A is a random variable since x(t) is a stochas- s fields. In particular, based on the continuous time ran- c tic process. Functionals of diffusion motion have diverse dom walk (CTRW) model, the corresponding fractional i applications and have been well studied, including func- s Fokker-Planckordiffusionequationsarederived(seethe y tionals of Brownian motion [16] and non-Brownian mo- review article [1] and numerical methods [5]). h tion [17–19]. In particular, based on the sub-diffusive p Tempered anomalous diffusion describes the very slow CTRW, a widely investigated process being continually [ transitionfromanomaloustonormaldiffusionandithas usedtocharacterizethemotionofparticlesindisordered 1 many applications in physical, biological, and chemical systems [20–22], the fractionalFeynman-Kac equationis v processes [6–9]; and for numerical methods see [10, 11]. derived [17]. Taking the tempered power-law function 1 In some cases,the transition even does not appear at all ψ(t,λ) as the waiting time distribution in the CTRW 7 in the observation time because of the finite lifespan of model,inthispaperwederivetheforwardandbackward 0 the particles or the finite observation time of the exper- Feynman-Kacequationsgoverningthedistributionofthe 0 imentalist. As a generalization of the Brownian walk, functionalsofthetemperedanomalousdiffusion;andthe 0 . the CTRW model allows the incorporation of the wait- tempered fractional substantial derivative [23] is used in 2 ing time distribution ψ(t) and the general jump length theequations. Thederivationsincludeseveralcases: ran- 0 distribution η(x). The CTRW model describes the nor- dom walk on lattice; randomwalk onlattice with forces; 6 mal diffusion if ψ(t) has bounded first moment and η(x) power-law jump distribution; tempered power-law jump 1 : bounded second moment, e.g., ψ(t) is exponential dis- distribution. After deriving the equations, several con- v tribution and η(x) is Gaussian distribution. Anomalous creteexamplesofthefunctionalsofthetemperedanoma- i X diffusion is characterized by the CTRW model with the lous diffusion are analytically and explicitly analyzed, r waiting time distribution ψ(t) having divergentfirst mo- covering the occupation time in half-space [24], the first a ment and/or the jump length distribution η(x) having passagetime,themaximaldisplacement,thefluctuations divergentsecondmoment,e.g.,ψ(t) t−α−1(0<α<1) of the occupation fraction, the fluctuations of the time- ≃ and/orη(x) x−β−1 (0<β <2). Sometimesthemore averagedposition, and the ergodicbehavior of the parti- ≃| | reasonable/physical choice for ψ(t) is to make it have cle. finite first moment; similarly, sometimes the bounded Thepaperisorganizedasfollows. InSectionII,wede- physical space implies that η(x) should have finite sec- rive the forward and backward fractional Feynman-Kac ond moment. These can be realized by truncating the equations for the functionals of the tempered anomalous heavy tail of the power-law distribution [12]. The tem- diffusionwithdifferentjumplengthdistributions. InSec- pered anomalous diffusion is described by the CTRW tion III, we present the solutions of the equations for a model with truncated power-law waiting time and/or few functionals of interest of free particles. Then we dis- jumplengthdistribution(s). Inthispaper,weusetheex- cuss the occupation time in half-space, the first passage ponentiallytruncatedstabledistribution(ETSD)[13,14] time, the maximal displacement, and the fluctuations of waiting time ψ(t,λ) and exponentially truncated jump the occupation fraction with the obtained solutions. In 2 α=0.8, λ=0.01 α=0.8, λ=0.1 The processstarts atx=x ,andthe particle waitsat 1.5 0 x fortimetdrawnfromψ(t,λ)andthenjumpstoeither 0.4 0 1 x +a(withprobabilityR(x ))orx a(withprobability 0 0 0 0.2 − L(x )),afterwhichtheprocessisrenewed. Furthermore, x(t)−0.02 x(t) 0.05 wfixeecd0onnusimdebretrh,ethmeosrteepgseinzeersaalrcearsaesn,dio.em.,vinasrtieaabdleos,fsbuebinjegcat −0.4 to power-law or tempered power-law distribution. −0.5 −0.6 −0.8 −1 0 200 400 600 800 0 200 400 600 800 t t FIG. 1: Trajectories of 30 particles, moving on the lattice with thetempered waiting times, where a=0.01, Kα =1/2, and λ=0.01 and 0.1. B. A random walk on a one-dimensional lattice Section IV, we investigate the fluctuations of the time- averaged position. The paper is concluded with some comments in the last section. WenowconsidertheCTRWonalattice. LetG(x,A,t) be the joint probability density function (PDF) of find- II. DERIVATION OF THE EQUATIONS ing the particle at position x and time t with the func- t tional value A. Here the functional A = U[x(t)]dt 0 A. Model as usual. And G(k,p,s) is the Fourier transformation R x k, A p, and Laplace transformation t s of → → → G(x,A,t). Inthissubsection,basedontheCTRWmodel We use the CTRW model with tempered power-law withETSDdescribingthetemperedanomalousdiffusion, waiting time distributionas the underlying processlead- we derivethe forwardandbackwardtemperedfractional ingtotemperedanomalousdiffusion,whichcharacterizes Feynman-Kac equations governing G(x,p,t). the slow transition from the anomalous to normal diffu- Derivation of the forward tempered fractional Feynman- sion which is controlled by the parameter λ (see Fig. 1). Kac equation from the random walk on a lattice. — For First, we consider the CTRW on a lattice, i.e., a parti- a random walk on a lattice with a general given wait- cle is placed on an infinite one-dimensional lattice with ing time PDF ψ(t,λ), the following formal solution was spacingaandis allowedtojumpto itsnearestneighbors obtained by Carmi et al (see Appendix A) [17] only. The probabilities of jumping left L(x) and right R(x) depend on F(x), the force at x (see subsection C for a derivationofthese probabilities). If F(x)=0, then R(x)=L(x)=1/2. Waiting times between jump events are independent identically distributed (i.i.d.) random G(k,p,s)= 1−ψˆ[s+pU(−i∂∂k),λ] 1 , variables with exponentially truncated stable distribu- s+pU(−i∂∂k) · 1−cos(ka)ψˆ[s+pU(−i∂∂k),λ] (2) tion(ETSD)ψ(t,λ),andareindependentoftheexternal where U(x) is a prescribed function. Recently, Cairoli force. This ETSD is useful for rigorous analysis of diffu- andBaule[26]derivedageneralizedFeynman-Kacequa- sion behavior because it is an infinitely divisible distri- tion for CTRW functionals; one of our goals is to now bution, and thus its distribution or characteristic func- show that this equation can be directly derived from tioncanbeexplicitlyderived. TheLaplacetransformfor Eq. (2), thus hopefully clarifying better the meaning ψ(t,λ) is given by [25] ofthe latterequationgivenin[17]. NotethatinEq. (2), +∞ cos(ka)istheFouriertransformofthejumplengthdistri- eφˆ(s,λ) = ψ(t,λ)e−stdt, bution,whichonalatticeisasumoftwodeltafunctions Z0 of step sizes a. ± whereφˆ(s,λ)= B (λ+s)α+B λα. Hence,theLaplace − α α Substituting Eq. (1) into Eq. (2) and using cos(ka) transform (for small s and λ) of ETSD ψ(t,λ) results in 1 a2k2 for the long wavelength k 0 corresponds t≃o − 2 → ψˆ(s,λ)=e−Bα(λ+s)α+Bαλα large x (or the continuous limit a 0) as is well known, (1) then → 1 B (λ+s)α+B λα. α α ≃ − 3 B [λ+s+pU( i ∂ )]α B λα 1 G(k,p,s) α − ∂k − α . (3) ≃ s+pU( i ∂ ) × a2k2 +B [λ+s+pU( i ∂ )]α B λα − ∂k 2 α − ∂k − α After some rearrangements to Eq. (3), we have a2k2 ∂ 1−α ∂ ∂ 1−α λ+s+pU i + λ+s+pU i λα λ+s+pU i G(k,p,s) 1 (2Bα (cid:20) (cid:18)− ∂k(cid:19)(cid:21) (cid:20) (cid:18)− ∂k(cid:19)(cid:21)− (cid:20) (cid:18)− ∂k(cid:19)(cid:21) ) − (4) λ λα λ+s+pU i ∂ 1−α = − − ∂k . s+pU( i ∂ ) (cid:2) − ∂(cid:0)k (cid:1)(cid:3) Inverting to the space-time domain k x and s t , from the well-known Fourier transformation xf(x);k = → → F{ } i ∂ fˆ(k); the operator U( i ∂ ) on the right hand side (rhs) of Eq. (4) is operating on 1, which means that we use − ∂k − ∂k ∞ n ∂ the TaylorexpansionU( i ∂ )= c i andthus we consider the analyticalfunctions U(x). Note that the − ∂k n − ∂k n=0 (cid:18) (cid:19) X order of the terms is important: for instance, k2 does not commute with U( i ∂ ); thus we finally find the tempered − ∂k fractional Feynman-Kac equation: ∂ ∂2 G(x,p,t)= λαD1−α,λ λ G(x,p,t) e−pU(x)tδ(x) pU(x)G(x,p,t)+K D1−α,λG(x,p,t), (5) ∂t t − − − α∂x2 t h ih i with the initial condition G(x,A,t = 0) = δ(x)δ(A) or G(x,p,t = 0) = δ(x), where δ() is the Dirac delta function · and a2 K = , (6) α 2B α with units m2/secα, is finite for a 0,B 0. Eq. (6) is a generalized Einstein relation for tempered motion. In α → → Laplace space, D1−α,λ [λ+s+pU(x)]1−α; and in t space, the tempered fractional substantial derivative t → 1 ∂ t e−(t−τ)·(λ+pU(x)) D1−α,λG(x,p,t)= λ+pU(x)+ G(x,p,τ)dτ. (7) t Γ(α) ∂t (t τ)1−α (cid:20) (cid:21)Z0 − Thus, due to the long waiting times, the evolution of diffusion equation [1]: G(x,p,t) is non-Markovian and depends on the entire history. Throughout the paper, the operator D1−α,λ is ∂ ∂2 t G(x,t)=K D1−αG(x,t), defined as in Eq. (7). ∂t α∂x2 RL,t (a). Withthecasethatλisfinitebutα=1,theequation leads to where D1−α is the Riemann-Liouville fractional deriva- RL,t tive operator. ∂ ∂2 G(x,p,t)=K G(x,p,t) pU(x)G(x,p,t). (d). When p=0 and λ is finite, Eq. (5) becomes: ∂t 1∂x2 − ∂ This is simply the famed Feynman-Kac equation [16]. G(x,t)= λαD1−α,λ λ [G(x,t) δ(x)] Namely, exponentially truncation has no effects on nor- ∂t t − − (9) maldiffusion. AswellknowntheFeynman-Kacequation h ∂2 i +K D1−α,λG(x,t), is the imaginary time Schr¨odinger equation, where U(x) α∂x2 t serves as the potential field. (b). Whenλ=0,Eq. (5)reducestotheimaginarytime where fractional Schr¨odinger equation [17], namely the tradi- 1 ∂ t e−(t−τ)λ tional Feynman-Kac equation. D1−α,λG(x,t)= λ+ G(x,τ)dτ. t Γ(α) ∂t (t τ)1−α ∂ G(x,p,t)=K ∂2 D1−αG(x,p,t) pU(x)G(x,p,t). (cid:20) (cid:21)Z0 − ∂t α∂x2 t − (8) This is a diffusion equation for tempered CTRW pro- (c). Furthermore,ifp=0,Eq. (8)turnstothefractional cesses. 4 Another formulation for Eq. (5). — Rearranging Eq. (3) leads to ∂ a2k2 ∂ G(k,p,s) s+pU i G(k,p,s) 1= s+pU i , (10) (cid:20) (cid:18)− ∂k(cid:19)(cid:21) − −2Bα (cid:20) (cid:18)− ∂k(cid:19)(cid:21)Φˆ[s+pU(−i∂∂k),λ] where Φˆ(s,λ) = (λ+s)α λα. Inverting to the time-space domain s t and k x, from the well-known Fourier transformation g (x)g −(x);k =gˆ( i ∂ )gˆ(k), the tempered fractio→nal Feynma→n-Kac equation is obtained as F{ 1 2 } 1 − ∂k 2 ∂ ∂2 ∂ t G(x,p,t)+pU(x)G(x,p,t)=K +pU(x) K(t τ,λ)e−pU(x)(t−τ)G(x,p,τ)dτ, (11) ∂t α∂x2 ∂t − (cid:20) (cid:21)Z0 with the initial condition G(x,A,t = 0) = δ(x)δ(A) or distribution of A, so integrating G(x,A,t) over all x is G(x,p,t=0)=δ(x),wherethememorykernelisrelated necessary. Therefore, it would be convenient to obtain to Φ by Kˆ(s,λ) = Φˆ(s,λ)−1 and given by K(t,λ) = an equation for Gx0(A,t), which is the PDF of the func- e−λttα−1Eα,α((λt)α). Here Eα,α() is the Mittag-Leffler tionalAattimetforaprocessstartingatx0. According function. It can be noted that E·q. (11) is exactly the to the CTRW model, the particle starts at x=x0; after same as Eq. (15) of [26]. its first jump at time τ, it is at either x0+a or x0 a. − Derivation of the backward tempered Feynman-Kac Alternatively,the particledoesn’t moveatallduring the equation from the random walk on a lattice. — Now we measurement time (0,t). Translating this process to an derive a backward equation which turns out to be very equation, there exists [17] useful. In some cases we may be just interested in the t 1 G (A,t)= dτψ(τ,λ) G [A τU(x ),t τ]+G [A τU(x ),t τ] +W(t,λ)δ[A tU(x )], (12) x0 2 { x0+a − 0 − x0−a − 0 − } − 0 Z0 where τU(x ) is the contribution to A from the pausing follows from the form for the Laplace transform of an 0 timeonx0 inthetimeinterval(0,τ);thelasttermonthe integral [27] and reads Wˆ(s,λ) = 1−ψˆ(s,λ), here ψˆ(s,λ) right hand side of Eq. (12) shows motionless particles, is also given by Eq. (1). Taking the Lsaplace transform t for which A(t)=tU(x0); and W(t,λ)=1− 0 ψ(τ,λ)dτ t → s and Fourier transforms x0 → k and A → p, we is the probability that particle remained motionless on have R its initial location. The Laplace transform of W(t,λ) ∂ ∂ G (p,s)=ψˆ pU i +s,λ cos(ka)G (p,s)+Wˆ pU i +s,λ δ(k) k k − ∂k − ∂k (cid:20) (cid:18) (cid:19) (cid:21) (cid:20) (cid:18) (cid:19) (cid:21) ∂ α B [λ+pU i ∂ +s]α B λα = 1 B λ+pU i +s +B λα cos(ka)G (p,s)+ α − ∂k − α δ(k). − α − ∂k α k pU i ∂ +s (cid:26) (cid:20) (cid:18) (cid:19) (cid:21) (cid:27) (cid:0) − ∂(cid:1)k (13) (cid:0) (cid:1) Rearranging the expressions and taking approximation k 0, cos(ka) 1 a2k2 in the last equation we find → ≃ − 2 ∂ 1−α a2k2 ∂ ∂ 1−α λ+pU i +s G (p,s)+ λ+pU i +s λα λ+pU i +s G (p,s) k k (cid:20) (cid:18)− ∂k(cid:19) (cid:21) 2Bα ((cid:20) (cid:18)− ∂k(cid:19) (cid:21)− (cid:20) (cid:18)− ∂k(cid:19) (cid:21) ) (14) λ λα[λ+pU( i ∂ )+s]1−α δ(k)= − − ∂k δ(k). − pU( i ∂ )+s − ∂k Inverting to the space-time domain s t and k x similartothatusedinthederivationoftheforwardequa- 0 → → 5 tion, in the continuum limit, we get the backward tem- pered fractional Feynman-Kac equation ∂ ∂2 G (p,t)= λαD1−α,λ λ G (p,t) e−pU(x0)t pU(x )G (p,t)+K D1−α,λ G (p,t). (15) ∂t x0 t − x0 − − 0 x0 α t ∂x2 x0 0 h ih i The initial condition is G (A,t = 0) = δ(A), or in ing left (L(x)) and right(R(x)) are no longer equal. As- x0 Laplace space G (p,t = 0) = 1, where δ() is also the sume the system is coupled to a heat bath at temper- x0 · Diracdeltafunction. ThesymbolD1−α,λisthetempered ature T and detailed balance, i.e., L(x)exp V(x) = fractional substantial derivative, detfined as Eq. (7). No- −kBT tice that here, this operator appears to the left of the R(x−a)exp −Vk(xB−Ta) , where a is the spahcing ofithe Laplacian ∂2 in Eq. (15), in contrast to the forward lattice. For smhall a, exipanding R(x), L(x), and the ex- ∂x2 0 ponential function leads to equation (5). When λ = 0, Eq. (15) turns to the back- ward fractional Feynman-Kac equation [17]: 1 aF(x) 1 aF(x) ∂ ∂2 R(x) 1+ , L(x) 1 , ∂tGx0(p,t)=KαDt1−α∂x2Gx0(p,t)−pU(x0)Gx0(p,t). ≃ 2(cid:20) 2kBT (cid:21) ≃ 2(cid:20) − 2kBT (cid:21) 0 where F(x)= V′(x) [19]. C. A one-dimensional lattice random walk with − forces Derivation of the forward tempered fractional Feynman-Kac equation with forces. — Using This subsection still considers the CTRW on lattice cos(ka) 1 a2k2 and sin(ka) ka for the long ≃ − 2 ≃ but with forces, which means the probabilities of jump- wavelength k 0 and following Eq. (18) in [19], → 1 ψˆ[s+pU( i ∂ ),λ] 1 G(k,p,s) − − ∂k . (16) ≃ s+pU(−i∂∂k) · 1− 1− a22k2 +i(ka)aF2(−kbiT∂∂K) ψˆ[s+pU(−i∂∂k),λ] h i Substituting ψˆ(s,λ) Eq. (1) into Eq. (16) and rearranging the equation, we obtain a2 F i ∂ ∂ 1−α ∂ k2 ik − ∂k λ+s+pU i G(k,p,s)+ λ+s+pU i G(k,p,s) 2Bα ( − (cid:0)kbT (cid:1))(cid:20) (cid:18)− ∂k(cid:19)(cid:21) (cid:20) (cid:18)− ∂k(cid:19)(cid:21) (17) ∂ 1−α λ λα[λ+s+pU( i ∂ )]1−α λα λ+s+pU i G(k,p,s) 1= − − ∂k . − − ∂k − s+pU( i ∂ ) (cid:20) (cid:18) (cid:19)(cid:21) − ∂k Inverting k x, s t, then → → ∂ G(x,p,t)= λαD1−α,λ λ G(x,p,t) e−pU(x)tδ(x) pU(x)G(x,p,t)+K ∂2 ∂ F(x) D1−α,λG(x,p,t). ∂t t − − − α ∂x2 − ∂x kbT t h i(cid:2) (cid:3) h i (18) Similarly, we can obtain another equation as follows, ∂ ∂2 ∂ F(x) ∂ t G(x,p,t)+pU(x)G(x,p,t)=K +pU(x) K(t τ,λ)e−pU(x)(t−τ)G(x,p,τ)dτ. (19) ∂t α ∂x2 − ∂x k T ∂t − (cid:20) b (cid:21)(cid:20) (cid:21)Z0 If F(x)=0, then Eq. (18) is the same as Eq. (5). If λ=0, Eq. (18) becomes the same as Eq. (22) given in [19], ∂ ∂2 ∂ F(x) G(x,p,t)=K D1−αG(x,p,t) pU(x)G(x,p,t). ∂t α ∂x2 − ∂x k T t − (cid:20) b (cid:21) Derivation of the backward tempered fractional in the above subsection, if we are just interested in the Feynman-Kac equation with forces. — As mentioned 6 distribution of the functional A, the backward equation lowing formal equation holds [19], should be useful and convenient. For G (p,s), the fol- k ∂ ∂ aF( i ∂ ) G (p,s) Wˆ pU i +s,λ δ(k)+ψˆ pU i +s,λ cos(ka) − ∂k isin(ka) G (p,s). (20) k k ≃ (cid:20) (cid:18)− ∂k(cid:19) (cid:21) (cid:20) (cid:18)− ∂k(cid:19) (cid:21)·" − 2kbT # Substituting Wˆ(s) = (1 ψˆ(s,λ))/s and ψˆ(s,λ) (given in Eq. (1)) into Eq. (20), and using cos(ka) 1 k2a2 and − ≃ − 2 sin(ka) ka as ak 0 and small s (B 0) approximation, after some rearrangements,we have α ≃ → → a2 ∂ 1−α F( i ∂ ) ∂ λ+pU i +s k2+ − ∂k (ik) G (p,s)+ λ+pU i +s G (p,s) k k 2Bα (cid:20) (cid:18)− ∂k(cid:19) (cid:21) " kbT # (cid:20) (cid:18)− ∂k(cid:19) (cid:21) (21) ∂ 1−α λ λα[λ+pU( i ∂ )+s]1−α λα λ+pU i +s G (p,s) δ(k)= − − ∂k δ(k). − − ∂k k − pU( i ∂ )+s (cid:20) (cid:18) (cid:19) (cid:21) − ∂k Taking inversion in the above equation, k x and s t, we get 0 → → ∂∂tGx0(p,t)= λαDt1−α,λ−λ Gx0(p,t)−e−pU(x0)t −pU(x0)Gx0(p,t)+KαDt1−α,λ ∂∂x220 + Fk(bxT0)∂∂x0 Gx0(p,t). h i(cid:2) (cid:3) h i (22) If F(x)=0, then Eq. (22) is exactly the same as Eq. (15). For λ=0, Eq. (22) reduces to the one given in [19]: ∂ ∂2 F(x ) ∂ G (p,t)=K D1−α + 0 G (p,t) pU(x )G (p,t). ∂t x0 α t ∂x2 k T ∂x x0 − 0 x0 (cid:20) 0 b 0(cid:21) D. Tempered CTRW with power-law jump length Tempering of jump length will be considered in the next distribution subsection. InsteadofdiscussingthetemperedCTRWonalattice, we further analyze the tempered CTRW with a power- lawjumplengthdistribution,η(x) x−1−β,0<β <2, Derivation of the forward tempered fractional and the Fourier transform of η(x) i≃s [|1]| Feynman-Kac equation with power-law jump length distribution. — From the CTRW model, the main η(k)=exp( C k β) 1 C k β. (23) Eq.(2) is modified according β β − | | ≃ − | | 1 ψˆ[s+pU( i ∂ ),λ] 1 G(k,p,s)= − − ∂k . (24) s+pU( i ∂ ) 1 η(k)ψˆ[s+pU( i ∂ ),λ] − ∂k − − ∂k Comparedwith Eq. (2)where randomwalkis onalattice, hence Fouriertransformofjump lengthPDF wascos(ka), now we replace it with the more general form η(k). Substituting the approximation of η(k) (given in Eq. (23)) and ψˆ(s,λ) (given in Eq. (1)) into Eq. (24) leads to B [λ+s+pU( i ∂ )]α B λα 1 G(k,p,s) α − ∂k − α . (25) ≃ s+pU( i ∂ ) · 1 (1 C k β)[1 B (λ+s+pU( i ∂ ))α+B λα] − ∂k − − β| | − α − ∂k α Rearranging Eq. (25) and taking k 0, we obtain the following equation → 1−α C ∂ ∂ β k β λ+s+pU i G(k,p,s)+ λ+s+pU i G(k,p,s) 1 B | | − ∂k − ∂k − α (cid:20) (cid:18) (cid:19)(cid:21) (cid:20) (cid:18) (cid:19)(cid:21) (26) ∂ 1−α λ λα[λ+s+pU( i ∂ )]1−α λα λ+s+pU i G(k,p,s)= − − ∂k . − − ∂k s+pU( i ∂ ) (cid:20) (cid:18) (cid:19)(cid:21) − ∂k 7 Taking k x,s t in the above equation results in the forward Feynman-Kac equation → → ∂ C G(x,p,t)= λαD1−α,λ λ G(x,p,t) e−pU(x)tδ(x) pU(x)G(x,p,t)+ β βD1−α,λG(x,p,t), (27) ∂t t − − − B ∇x t α h ih i wheretheRieszspatialfractionalderivativeoperator β And the symbol D1−α,λ is defined as before. If taking ∇x t and the fractional Laplacian operator ( ∆ )β/2 are β = 2, Eq. (27) reduces to Eq. (5); letting λ = 0 leads x e[1q8u]i;vaanledntin[2x9]s.paIcne,Fourier x → k spac−e ∇−βx → −|k|β to ∂∂tG(x,p,t)= BCαβ∇βxDt1−αG(x,p,t)−pU(x)G(x,p,t), which is the same as the one obtained in [18]. 1 βf(x)= Dβf(x)+ Dβ f(x) , ∇x −2cosβπ −∞ x x +∞ 2 h i where (n 1<β <n) − 1 dn x f(ξ) Dβf(x)= dξ, (28) −∞ x Γ(n β)dxn (x ξ)β+1−n − Z−∞ − Derivation of the backward tempered Feynman-Kac Dβ f(x)= (−1)n dn +∞ f(ξ) dξ. equationwithpower-law jumplengthdistribution. —Fol- x +∞ Γ(n β)dxn (ξ x)β+1−n lowing [28] and replacing cos(ka) with η(k) in Eq. (13) − Zx − (29) corresponded to the general jump lengths, there exists ∂ ∂ G (p,s)=ψˆ pU i +s,λ η(k)G (p,s)+Wˆ pU i +s,λ δ(k). (30) k k − ∂k − ∂k (cid:20) (cid:18) (cid:19) (cid:21) (cid:20) (cid:18) (cid:19) (cid:21) Substituting the approximationof η(k) (given in Eq. (23)), Wˆ(s,λ)=(1 ψˆ(s,λ))/s, and ψˆ(s,λ) (given in Eq. (1)) − into Eq. (30), taking k 0 and rearrangingthe terms, we find → 1−α 1−α ∂ C ∂ ∂ λ+pU i +s β k βG (p,s)+ λ+pU i +s λα λ+pU i +s G (p,s) k k (cid:20) (cid:18)− ∂k(cid:19) (cid:21) Bα| | ((cid:20) (cid:18)− ∂k(cid:19) (cid:21)− (cid:20) (cid:18)− ∂k(cid:19) (cid:21) ) λ λα[λ+pU( i ∂ )+s]1−α δ(k)= − − ∂k δ(k). − pU( i ∂ )+s − ∂k (31) Taking the inverse transforms k x ,s t, we get 0 → → ∂ C G (p,t)= λαD1−α,λ λ G (p,t) e−pU(x0)t pU(x )G (p,t)+ βD1−α,λ β G (p,t). (32) ∂t x0 t − x0 − − 0 x0 B t ∇x0 x0 α h ih i If taking β =2, Eq. (32) reduces to Eq. (15); while λ=0, it reduces to Eq. (21) in [28] as expected. E. Tempered CTRW with tempered power-law where θ =arg(γ+ik), Aθβ =2Aβcos(βθ) . jump length distribution Now we are going to discuss the tempered power-law Derivation of the forward tempered fractional jump length distribution η(x) Aβ e−γ|x| x−β−1, Feynman-Kac equation with tempered power-law jump ≃ |Γ(−β)| | | where 0< γ, 0 < β < 2; and the asymptotic form (with length distribution. — Similar to the above analysis, small k) of the Fourier transform of η(x) is [14] substituting the approximation of η(k) (given in Eq. (33)) and ψˆ(s,λ) (given in Eq. (1)) into Eq. (24), we η(k) 1 Aθ(γ2+k2)β/2+2A γβ, (33) get ≃ − β β 8 B [λ+s+pU( i ∂ )]α B λα 1 G(k,p,s) α − ∂k − α . ≃ s+pU( i ∂ ) ·1 [1 Aθ(γ2+k2)β/2+2A γβ] 1 B [λ+s+pU( i ∂ )]α+B λα − ∂k − − β β { − α − ∂k α } (34) Taking k 0 makes Aθ 2A . Rearranging Eq. (34), we have → β → β 1−α 1−α ∂ ∂ K (γ2+k2)β/2 λ+s+pU i K γβ λ+s+pU i G(k,p,s) α,β α,β ( (cid:20) (cid:18)− ∂k(cid:19)(cid:21) − (cid:20) (cid:18)− ∂k(cid:19)(cid:21) ) 1−α ∂ ∂ + λ+s+pU i G(k,p,s) λα λ+s+pU i G(k,p,s) 1 (35) − ∂k − − ∂k − (cid:20) (cid:18) (cid:19)(cid:21) (cid:20) (cid:18) (cid:19)(cid:21) λ λα[λ+s+pU( i ∂ )]1−α = − ∂k , s+pU( i ∂ ) − s+pU( i ∂ ) − ∂k − ∂k where K = 2Aβ. Taking the inversion transforms k x and s t results in α,β Bα → → ∂ G(x,p,t)= λαD1−α,λ λ G(x,p,t) e−pU(x)tδ(x) pU(x)G(x,p,t)+K β,γ +γβ D1−α,λG(x,p,t). ∂t t − − − α,β ∇x t h ih i (36) (cid:0) (cid:1) The tempered fractional Riesz derivative (TFRD) operator β,γ is defined in Fourier x k space as β,γ (γ2+k2)β/2; and in x space, the operator is defined as (for m∇oxre details, see Appendix B):→ ∇x → − 1 β,γf(x)= Dβ,γf(x)+ Dβ,γ f(x) . (37) ∇x −2cos(βπ) −∞ x x +∞ 2 h i When γ =0, Eq. (36) becomes Eq. (27) as expected. Derivation of the backward tempered fractional Feynman-Kac equation with tempered power-law jump length distri- bution. —Againfollowing[28]andinsertingtheapproximationofη(k)(giveninEq. (33)), Wˆ(s,λ)=(1 ψˆ(s,λ))/s, − and ψˆ(s,λ) (given in Eq. (1)) into Eq. (30), we have ∂ ∂ G (p,s)=ψˆ pU i +s,λ η(k)G (p,s)+Wˆ pU i +s ,λ δ(k) k k − ∂k − ∂k (cid:20) (cid:18) (cid:19) (cid:21) (cid:20) (cid:18) (cid:19) (cid:21) α ∂ ≃ 1−Bα λ+pU −i∂k +s +Bαλα 1−Aθβ(γ2+k2)β/2+2Aβγβ Gk(p,s) (38) (cid:26) (cid:20) (cid:18) (cid:19) (cid:21) (cid:27)h i B [λ+pU( i ∂ )+s]α B λβ + α − ∂k − α δ(k). pU( i ∂ )+s − ∂k Letting k 0 makes Aθ 2A . Rearranging the last equation leads to → β → β 1−α ∂ λ+pU i +s [Kθ (γ2+k2)β/2 K γβ]G (p,s) − ∂k α,β − α,β k (cid:20) (cid:18) (cid:19) (cid:21) 1−α ∂ ∂ + λ+pU i +s λα λ+pU i +s Gk(p,s) (39) ((cid:20) (cid:18)− ∂k(cid:19) (cid:21)− (cid:20) (cid:18)− ∂k(cid:19) (cid:21) ) λδ(k) λα[λ+pU( i ∂ )+s]1−α = − ∂k δ(k)+δ(k), pU( i ∂ )+s − pU( i ∂ )+s − ∂k − ∂k where Kθ = Aθβ , K = 2Aβ. Taking the inverse Laplace and Fourier transformations, k x and s t, there α,β Bα α,β Bα → → exists ∂ G (p,t)= λαD1−α,λ λ G (p,t) e−pU(x0)t pU(x )G (p,t)+K D1−α,λ β,γ +γβ G (p,t). (40) ∂x x0 t − x0 − − 0 x0 α,β t ∇x0 x0 Notice that D1−hα,λ is on the lefithof β,γ in Eq. (40), iin contrastto the forwardequation(cid:0)Eq. (36). W(cid:1)hen γ =0, Eq. t ∇x0 (40) reduces to Eq. (32) as expected. III. SOLUTIONS TO THE DERIVED tempered anomalous dynamics. EQUATIONS In this section, we present the distributions of four concrete functionals of the paths of particles performing 9 A. Occupation time in half-space zero otherwise. For example, for Brownian motion the PDF of T+ is the famous Arcsine distribution. In order The occupation time of a particle in half-space is tofindthePDFoftheoccupationtime,hereweconsider widely used in physics [24,31, 32] andmathematics [33]. thebackwardfractionalFeynman-Kacequation(15)with Define the occupation time in x > 0 as T+ = A = regular jump length in Laplace s space: t Θ[x(τ)]dτ, i.e., U(x) = Θ(x) = 1 for x 0 and is 0 ≥ R ∂2 λ λα(λ+s)1−α K (λ+s)1−α G (p,s)+(λ+s)G (p,s) 1=λα(λ+s)1−αG (p,s)+ − ,x <0. (41) − α ∂x2 x0 x0 − x0 s 0 0 ∂2 λ λα(λ+s+p)1−α K (λ+s+p)1−α G (p,s)+(λ+s+p)G (p,s) 1=λα(λ+s+p)1−αG (p,s)+ − ,x >0. − α ∂x2 x0 x0 − x0 s+p 0 0 (42) HenceherethetemperingisintimeonlyandK isgiven i.e., α in Eq. (6). Rewriting the above equations leads to s (λ+s+p)α λα+(s+p) (λ+s)α λα Kα(λ+s)1α−λα ∂∂x220Gx0(p,s)+ 1s, x0 <0; G0(p,s)= s(sp+p)( (λ+s−+p)α λα+p(λ+s)α−λα), Gx0(p,s)= − (4−7)  Kα(λ+s+p1)α−λα∂∂x22Gx0(p,s)+ s+1p, x0 >0. whichdescribesthepPDFofT+ andisvalidpforalltimes. 0 (43) However, it seems difficult to invert Eq. (47) analyti- They both are second order, ordinary differential equa- cally. We’ll soon analyse the moments of this equation tions in x . Solving the equations in each half-space in- in the following discussion. Specially, if α = 1, then 0 dividually,requiringthatGx0(p,s)isfinitefor|x0|→∞, Gth0e(pe,qsu)il=ibsri−u1m/2(PsD+Fp)o−f1/ε2, thTis+c/atn, boeritnhveerotecdcutpoagtiiovne ≡ C exp x (λ+s)α−λα + 1, x <0; fraction, 0 0 Kα s 0  (cid:18) q (cid:19) 1 Gx0(p,s)= C1exp −x0 (λ+sK)αα−λα + s+1p, G(ε)= π ε(1−ε), (cid:18) q (cid:19) x >0. which is the arcsine law of Lp´evy [32].  0 (44) Theparticlecanneverarriveatx>0forx ;thus 0 G (T+,t)=δ(T+) and G (p,s)= 1, in co→nf−or∞mity to B. First passage time x0 x0 s Eq. (44). Likewise,forx + ,theparticleisneverat 0 x1<,0aasnedxptheuctseGdxi0n(TE+q,.t)→(=44δ)∞(.T+Th−ent)daenmdaGndx0in(pg,st)ha=t asAthsewteimlleknTowitn,tatkheesfiarsptaprtaiscsleagsetatritmineg(aFtPxT)=is debfi(nbe>d s+p f 0 − G (p,s)anditsfirstderivativearecontinuousatx =0, 0)tohitx=0forthefirsttime[35]andiswidelyapplied x0 0 yields a pair of equations about C ,C : in physics and other disciplines. A relationship between 0 1 the distribution of first passage time and the occupation C + 1 =C + 1 0 s 1 s+p time functional was put forward by Kac [36]:   C0 (λ+s)α−λα =−C1 (λ+s+p)α−λα. Pr{Tf >t}=Pr{0m≤τa≤xtx(τ)<b}=pl→im∞Gx0(p,t), (45) Bysolvinpg these equations, we gpet where G (p,t) describes the Laplace transform of the x0 p√(λ+s+p)α−λα PDF of functional T+ = tΘ[x(τ)]dτ. The last equa- C = 0 0 −s(s+p)(√(λ+s+p)α−λα+√(λ+s)α−λα) tionistrueforG (p,t)= ∞e−pT+G (T+,t)dT+,and  (46) x0 R0 x0 thus,iftheparticlehasneverpassagedthroughx=0,we C = p√(λ+s)α−λα . have T+ = 0 and e−pT+ =R 1, while otherwise, T+ > 0, 1 s(s+p)(√(λ+s+p)α−λα+√(λ+s)α−λα) andforp ,e−pT+ =0. Forx = bandp ,ac- Assume that the particle starts at x0 = 0. Substituting cording to→E∞q. (44) and (46) of th0e p−revious su→bse∞ction, x =0 in Eq. (44), then G (p,s)=C + 1 = C + 1 , we have 0 0 0 s 1 s+p 10 1 p (λ+s+p)α λα (λ+s)α λα lim G (p,s)= lim − exp b − −b p→∞ s −p→∞s(s+p)( (λ+ps+p)α λα+ (λ+s)α λα) − s Kα  − − (48) p p   1 1 (λ+s)α λα = exp b − . s − s − s Kα    Inaccordancewiththedefinitionofthefirstpassagetime, PDF of FPT with λ=0.01 its PDF satisfies 100 α=0.6 ∂ ∂ f(t)= (1 P T >t )= lim G (p,t). Asymptotic estimation r f −b ∂t − { } −∂tp→∞ ) −1 Hence, in Laplace space, we have (t10 f α=0.9 α=0.8 f(s)= s lim G (p,s)+1 −b − p→∞ (λ+s)α λα (49) 10−2 =exp−bs Kα− . 103 t   The inversionof Eq. (49) is done numerically [34] seeing FIG. 2: Behavior of f(t) (i.e., Eq. (49)) with different values Fig. 2, Fig. 3 and Fig. 4. of the parameter α, and lattice spacing a = 0.01, starting Expanding Eq. (49) in small s, point b = 0.05 and diffusion constant Kα = 1/2. The solid dotted(green)lineistheasymptoticestimationwiththeslope of -3/2 for long time, confirming that the standard Sparre αλα−1s f(s) 1 b . Andersen scaling also holds for thetempered sub-diffusion. ≃ − s Kα Taking inverse Laplace transform for long times, s t, → in Fig. 5 and Fig. 6, this is expected the particle cannot we have reachtheoriginwhent 0andshowsasL´evybehaviour → for short but not too short times. b αλα−1 f(t) t−32, (50) When λ = 0, then waiting times show as power-law ≃ |Γ(−21)|s Kα distributed. Eq. (49) becomes for all α, whichcoincideswith the famoust−23 decaylaw b α of a one-dimensionalrandomwalk[35,37]and decreases f(s)=exp s2 . (52) −√K with increasing λ, being confirmed in Fig. 2 and Fig. 3. (cid:18) α (cid:19) Hence, In t space, Eq. (52) is the one-sided L´evy laws L (t). α/2 And then f(t) decays very fast to zero when t 0. For ∞ b αλα−1 t , f(t) behaves as t−(1+α/2), which is in a→greement Pr{Tf >t}=Zt f(Tf)dTf ≃ √πs Kα t−1/2, w→ith∞the results given in [18, 39], indicating that hti is (51) infinite for all α. the lastequationisexactly the resultgivenin[38]andis illustrated in Fig. 5 and Fig. 6. However,ifs ,correspondingtosmallt,fromEq. C. Maximal displacement →∞ (49), we have Now we develop another application of Eq. (44). The b α f(s) exp s2 . maximaldisplacementofadiffusing particle is arandom ≃ −√K (cid:18) α (cid:19) variable, which has been studied in recent years [40, 41]. In order to get the distribution of this variable, we have In t space, the above equation tends to be the one-sided t L´evy laws Lα/2(t). Hence f(t) decays very fast to zero Gx0(p,t) describes the functional A= 0 U[x(τ)]dτ with U(x) = 1 for x > 0, otherwise, U(x) = 0. Let x when t 0 and behaviors as t−1−α/2 for short but not R m ≡ → max x(τ); andthenP x <b = lim G (p,t). From tooshorttimes,correspondedprobabilityPrisillustrated 0≤τ≤t r{ m } p→∞ x0

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