Random motion with uniformly distributed directions and random velocity

Transcription

1 Noname manuscrip No. (will be insered by he edior) Anaoliy A. Pogorui Ramón M. Rodríguez-Dagnino Random moion wih uniformly disribued direcions and random velociy Received: dae / Acceped: dae Absrac In his paper we deal wih uniformly disribued direcion of moion or isoropic moion a random speed or velociy where he direcion alernaions occur according o he renewal epochs of a general disribuion. We derive he renewal equaion for he characerisic funcion of he ransiion densiy of he mulidimensional moion. Then, by using he renewal equaion, we sudy he behavior of he ransiion densiy near he sphere of is singulariy for one-, wo-, hree-, and four-dimensional cases. To illusrae our soluion mehodology we presen deailed calculaions of many solvable examples. Keywords Random evoluions semi-markov processes general disribuions random velociy Mahemaics Subjec Classificaion () 6K35 6K99 6K15 1 Inroducion Mos of he papers on random moion wih uniformly disribued direcions, or isoropic moion, on he muidimensional space are devoed o he analysis of models in which moions are driven by he homogeneous Poisson process Gran CAT148 from Tecnológico de Monerrey. A. Pogorui Deparmen of Mahemaics, Zhyomyr Sae Universiy, Valyka Berdychivska S., 4, 18, Zhyomyr, Ukraine 18 pogor@zu.edu.ua R. M. Rodríguez-Dagnino Elecrical and Compuer Engineering, Tecnológico de Monerrey, Av. Eugenio Garza Sada 51 Sur, C.P , Monerrey, N.L., México rmrodrig@iesm.mx

2 and consan speed or consan velociy, so heir processes are Markovian, e.g., [1, ], and references herein. One-dimensional non-markovian random evoluions generalizes hese resuls by changing he underlying Poisson process, where moion is driven by an alernaing semi-markov process wih Erlang disribued inerrenewal imes [3 6]. Random flighs in R n wih K Erlang disribued displacemens and uniformly disribued direcions have been analyzed in [7]. A planar random moion performed by a paricle ha changes direcion a even-valued Poisson epochs is sudied in [8]. A random walk wih seps of Dirichle disribued lenghs and uniformly disribued changes of direcion is repored in [9, 1]. Closed-form expressions for he ransiion densiies in he cases of wo- and four-dimensional Markovian random moions have been derived by E. Orsingher and A. de Gregorio in [1], and by A. Kolesnik in []. In his work, we consider mulidimensional random moions wih uniformly disribued direcions and having random velociy. Thus, we are exending some of he resuls found in [1,,7]. Furhermore, we consider his random moion in one-, wo-, hree-, and four-dimensions, and we show ha he ransiion densiies in some of he cases have an explosive growh near he sphere of heir singulariy. Le {ν(), } be a renewal process such ha ν() = max{m : τ m }, where τ m = m k=1 θ k, τ = and θ k, k = 1,,..., are nonnegaive iid random variables denoing he inerarrival imes. We assume ha hese random variables have a cdf G() and ha here exiss he pdf G(). In mos of he worked examples in his paper, we have consid- g() = d d ered exponenially disribued inerrenewal imes, i.e., g() = λe λ I { }. We will sudy he random moion of a paricle ha sars from he coordinae origin = (,,..., ) of he space R n, a ime =, and coninues is moion wih a velociy γ > along he direcion η (n), where γ i, i =, 1,,..., are iid random variables ha have he same cdf Z() and pdf z() = d d Z(), >, and η(n) i = (x i1, x i,, x in ) = (x 1, x,, x n ), i =, 1,,..., are iid random n-dimensional vecors uniformly disribued on he uni sphere Ω1 n 1 = {(x 1, x,..., x n ) : x 1 + x + + x n = 1}. The random variables γ i are independen of he random vecors η (n) i, i =, 1,,.... A insan τ 1 he paricle changes is direcion o η (n) 1 = (x 11, x 1,, x 1n ), and he paricle coninues is moion wih a velociy γ 1 > along he direcion of η (n) 1. Then, a insan τ he paricle changes is direcion o η (n) = (x 1, x,, x n ), and coninues is moion wih a velociy γ along he direcion of η (n), and so on. Denoe by X (n) (),, he paricle posiion a ime. We have ha ν() X (n) () = η (n) j 1 γ j 1 (τ j τ j 1 ) + η (n) ν() γ ν() ( τ ν() ). (1) j=1 Basically, Eq. (1) deermines he random evoluion in he semi-markov (or renewal) medium ν(). Thus, ν() denoes he number of velociy alernaions occurred in he inerval (, ).

3 3 The probabilisic properies of he random vecor X (n) () are compleely deermined by hose of is projecion X (n) () = m j=1 η(n) j 1 γ j 1(τ j τ j 1 ) + η (n) ν() γ ν()( τ ν() ) on a fixed line, where η (n) j is he projecion of η (n) j on he line. Indeed, le us consider he cdf F X (y) = P ( X (n) () y ). Then, he characerisic funcion (Fourier ransform) H(, α) = H() of X (n) (), where α = α = α1 + α + + α n, is given by [ { ( )}] { ( )} H() = E exp i α, X (n) () = E exp i α e, X (n) () { } = E exp i αx (n) () = exp {i αy} df X (y), where X (n) () is he projecion of X (n) () ono he uni vecor e and i has a cdf F X (y). Le us denoe by f η (n)(x) he pdf of he projecion η (n) j of he vecor η (n) j ono a fixed line. I is shown in [5] ha f η (n)(x) is of he following form Γ ( ) n ( f η (n)(x) = π Γ n 1 ) (1 x ) (n 3)/, if x [ 1, 1], (), if x / [ 1, 1]. Hence, i is no hard o verify ha he cdf F η γ(x) = P [η (n) γ x] is of (n) he following form ( 1 + Γ n ) 1 ( x ) ( π Γ n 1 ) Z (1 ) (n 3)/ d, if x, F η γ(x) = ( (n) 1 Γ n ) 1 ( ( π Γ n 1 ) Z x ) (1 ) (n 3)/ d, if x <. Le us denoe he characerisic funcion of η (n) j γ j as { } ϕ() = E exp iαγ j η (n) j = e iαx df η γ(x). (n) We should noice ha he funcion ϕ( ) is also used in [,7], and we have ( ϕ() = n n ) J n (αv) Γ (αv) n when γ j = v > is a deerminisic consan. However, when γ j is a random variable wih pdf z(v), v, hen { ϕ() = Ee iαγ j η (n) j } ( = n n ) Γ J n (αv) z(v) dv. (αv) n (3)

4 4 A renewal equaion for he characerisic funcion As we know, he characerisic funcion H() of he random moion X (n) () is given by [ { ( )}] H() = E exp i α, X (n) (), and i fulfills he following heorem Theorem 1 H(),, is a soluion of he following Volerra inegral equaion H() = (1 G()) ϕ() + or in he convoluion form Proof I follows from Eq. (1) ha H() = (1 G()) ϕ() + (g ϕ) H(). [ { ( )}] H() = E exp i α, X (n) () g(u) ϕ(u) H( u) du, (4) ν() = E i α, η (n) j 1 γ j 1 θ j + η (n) ν() γ ν() ( τ ν() ) j=1 = (1 G()) Ee iγ (α,η (n) ) + g(u)ee iuγ (α,η (n) ) H( u) du. We should observe ha ϕ() = Ee iγ (α,η (n) ), and passing o he Laplace ransform Ĥ(s) = L(H()) = H()e s d we obain Ĥ(s) = and i complees he proof. (1 G())ϕ()e s d 1 g()ϕ()e s d, Denoe by f n (, X) he pdf of he paricle posiion a ime. I is easily seen ha f n (, X) = F 1 (H()), where F 1 ( ) is he n dimensional inverse Fourier ransform wih respec o α. We will proceed o find f n (, X) = F 1 (H()) for some ineresing cases from one o four dimensions.

5 5 3 One-dimensional case Le us consider he 1-D case, i.e., n = 1, wih consan velociy v >, and ϕ() = cos(αv). We have, for he exponenial pdf g() = λe λ I { }, he renewal equaion H() = e λ cos(αv) + λ e λu cos(αuv)h( u) du. (5) We should noice ha he process X (1) () is no he elegrapher process because a renewal epochs he paricle may no change is direcion. Denoe by h() = H()e λ, hen i follows from Eq. (5) ha h() = cos(αv) + λ cos(αuv)h( u) du. Since L(cos(αv)) = e su s cos(αuv)du = s + v, hen afer applying he Laplace ransform o Eq. (5) (or is equivalen in h()) we α obain L(h()) = e s h()d = Thus, he inverse Laplace ransform gives H() = e λ h() s s λs + v α. ( ) = e (cosh λ/ λ 4v α + λ sinh ( λ 4v α ) ). (6) λ 4v α We should recall he formula [1], p. 57, no. 47, sinh( 1 α ) 1 α = 1 which can be ailored o our case as λ sinh( λ 4v α ) λ 4v α = λ v v v I ( x ) e ixα dx, I ( λ v (v) x ) e ixα dx. Therefore, by applying he inverse Fourier ransform (wih respec o α) o Eq. (6), we obain f 1 (, x) = e (δ λ/ ( (v) x ) + λ ( ) λ v I (v) x v + λ I 1 ( ) (v) x ). 4 (v) x

6 6 However, we should noice ha in some siuaions he dependence wih λ disappears in f 1 (, x), as i is elaboraed in he following random velociy case. Suppose ha γ j is a random variable wih pdf Then, we obain ϕ() = z(v) = π(1 + v ), v. π(1 + v ) cos(αv) dv = e α. Hence, we have he following renewal equaion for he characerisic funcion H() = e λ e α + λ The soluion of Eq. (7) is given by H() = e α. e λu e u α H( u) du. (7) Now, by obaining he inverse Fourier ransform wih respec o α we have f 1 (, x) = π( + x ). As we can see, in his case he pdf of X (1) () does no depend on λ. Now, le us presen a hird example where he corresponding pdf f 1 (, x) has an explosive behavior. Suppose ha he velociy γ j is a random variable wih pdf z(v) = π 1 v, v 1. Then, ϕ() = cos(αv) π 1 v dv = J (α). Now, le us assume an exponenial pdf for he inerrenewal imes, i.e., g() = λe λ I { }, hen we have he renewal equaion H() = e λ J (α) + λ or is equivalen form h() = J (α) + λ e λ( u) J (α( u))h(u) du. (8) J (α( u))h(u) du, (9)

7 7 where h() = e λ H() as above. As i is well-known he Laplace ransform of he Bessel funcion J (α) wih respec o is given by L(J (α)) = 1 s + α. We obain, afer applying he Laplace ransform o he renewal equaion (9), L(h()) = 1 s + α + λ s + α L(h()), or equivalenly, L(h()) = 1 s + α λ = 1 λ ( ) m λ. s + α On he oher hand, we have he following inverse Laplace ransform Therefore, ( ( ) m ) 1 L 1 λ = λ s + α H() = e λ π Γ (m/) π Γ (m/) ( λ ) m 1 ( λ ) m 1 J m 1 J m 1 α m 1 α m 1 (α), (α). Thus, we can calculae he corresponding pdf, afer using he Hankel ransform in [11], p. 75, for he one-dimensional case, f 1 (, x) = F 1 (H()) = e λ λ m 1 ( x ) (Γ (m/)) m 1 () m 1 m I {> x }, (1) where he inverse Fourier ransform is aken wih respec o α. The inverse Fourier ransform of he firs erm, i.e. m = 1 in Eq. (1), is given by F 1 (e λ J (α)) = e λ 1 π x I {>x} Hence, we should noice ha f 1 (, x) + as x. So, we have an explosive effec close o he line x =. I seems ha his explosive effec is absen in he cases of consan velociy for he one-dimensional random moion. However, i has been observed in wo-dimensional cases [14] and in hree-dimensional cases [] for consan velociy for he absolue coninuous par of he disribuion of he paricle posiion. In his paper we also include a four-dimensional case where his phenomenon is presen as well.

8 8 4 Two-dimensional case Now, le us suppose ha he velociy γ j is a random variable wih he following heavy-ail pdf Then, z(v) = ϕ() = v ; v. (1 + v ) 3/ J (αv)z(v)dv = e α. Now, le us assume an exponenial pdf g() = λe λ I { }, hence we have he renewal equaion H() = e λ e α + By solving Eq. (11) we obain e λ( u) e α( u) H(u)du. (11) H() = e α. Now, we can obain he wo-dimensional inverse Fourier ransform of H() wih respec o α, and we can apply he Hankel ransform [11] o obain he pdf of X () () f (, x) = 1 π e α αj (αx)dα = We should noice ha his pdf does no depend on λ. π( + x ) 3/. Now, we will consider ha he random velociy γ j has he pdf z(v) = v, v 1. Then, ϕ() = J (αv)z(v)dv = 1 J (αv) v dv = J 1(α). α Le us assume an exponenial pdf for g(), i.e., g() = λe λ I { }, hen we have he renewal equaion or equivalenly, H() = e λ J 1(α) α h() = J 1(α) α + λ + λ e λ( u) J 1(α( u)) H(u)du, α( u) J 1 (α( u)) h(u)du, (1) α( u) where h() = e λ H(). We should recall he following Laplace ransform (wih respec o ) ( ) J1 (α) L = s s + α α α.

9 9 Thus, we can apply he Laplace ransform o Eq. (1), and i gives L(h()) = s s + α α + λ s s + α α L(h()), or L(h()) = s s + α α +4λ (s s + α ) α 4 +8λ (s s + α ) 3 α 6 +. We should recall he inverse Laplace ransform ( L 1 (λ) m (s ) s + α ) m λα m = m(λ) m J m(α) λα m. and Therefore, h() = H() = e λ m(λ) m J m(α) λα m, m(λ) m J m(α) λα m. We need o calculae he wo-dimensional inverse Fourier ransform of H() (wih respec o α) by using he Hankel ransform and Formula of [13]. Namely, Hence, F 1 ( Jm (α) λα m = e λ π ) = 1 J m (α) π λα m αj (αx)dα = ( x ) m 1 π m λ m+1 m!. f (, x) = F 1 (H()) = e λ λ m 1 ( x ) m 1 m 1 (m 1)! m(λ) m ( x ) m 1 π m λ m+1 m! = e λ π e λ ( x ) = 1 π exp ( λ x ). 5 Three-dimensional case We need o find he pdf of X (3) (), i.e., f 3 (, x). Le us remember he case of consan velociy, v =consan >. In his case we have π J 1/ (αv) ϕ() = = sin(αv). αv αv Then, he corresponding renewal equaion is given by H() = (1 G()) sin(αv) αv + g() sin(αuv) αuv H( u) du.

10 1 Now, suppose ha he random velociy γ j is disribued according o he Maxwell disribuion wih pdf z(v) = π v e 1 v, v >. Then, i follows ha f η γ(x) = df η γ(x) (n) = 1 e 1 (n) x. dx π Hence, [ ( )] ϕ() = E exp iαγ j η (n) j = 1 π e 1 α. For he random velociy we can calculae ϕ() in an equivalen manner as follows ϕ() = π v e 1 v sin(αv) dv = e 1 α. αv Therefore, he renewal equaion for he characerisic funcion H() is given by H() = e λ e 1 α + λ e λu e 1 u α H( u) du, (13) where we have assumed an exponenial pdf for g(). In our opinion Eq. (13) is he equaion for he characerisic funcion of an alernaive o he Wiener process model of he Brownian moion. Unforunaely, we were no able o solve Eq. (13) in a closed-form. We sae his opinion since ha he subsiuion of e 1 α for e 1 α ino Eq. (13) yields a solvable renewal equaion H() = e λ e 1 α + λ e λu e 1 uα H( u) du, (14) wih soluion H() = e 1 α which corresponds o he characerisic funcion of a Brownian moion. Now, we can apply he Hankel inversion formula [11] o obain he inverse Fourier ransform of H() wih respec o α f 3 (, X) = 1 (π) 3/ e 1 α α J 1/(αx) 1 dα = e x /(), (15) αx (π) 3/ where x = X. We should observe ha he pdf f 3 (, X) does no depend on λ and his fac reflecs ha he Wiener process has zero lengh of free pah. In our view, his is a drawback of he Wiener process as a model of Brownian moion. In addiion, we should remark ha here is no a pdf such ha z(v) sin(αv) αv dv = e 1 α.

11 11 6 Four-dimensional case We have ha ϕ() = J 1(αv) when v = consan > in he four-dimensional αv case. Now, le us assume ha he velociy γ j is a random variable wih he pdf z() = π, v 1. Then, for he random velociy we have ha 1 v ϕ() is given by ϕ() = J 1 (αv) αv π 1 v dv = 1 π ( ) sin(α), α wih he corresponding renewal equaion ( ) H() = e λ sin(α) + λ ( ) sin(α( u)) e λ( u) H(u) du. (16) π α π α( u) for an exponenial pdf g(). We canno solve Eq. (16) bu i is easily seen ha ( (sin(α) ) ) f 4 (, x) = F 1 (H()) e λ π F 1. α We have, by using he Hankel ransform, ha ( (sin(α) ) ) F 1 = 1 ( ) J 1 (αv) sin(α) α 4π α 3 dα αv α Therefore, = 1 1 4π x, x <. x f 4 (, x) as x. This explosive effec phenomenon was no observed in he four-dimensional case for consan velociy [], for he absolue coninuous par of he disribuion of he paricle posiion. References 1. Orsingher, E., De Gregorio, A.: Random flighs in higher spaces. J. Theore. Prob., (7). Kolesnik, A.D.: Random moions a finie speed in higher dimensions. J. Sa. Phys. 131, (8) 3. Di Crescenzo, A.: On random moions wih velociies alernaing a Erlangdisribued random imes. Adv. Appl. Prob. 61, (1) 4. Pogorui, A.A., Rodriguez-Dagnino, R.M.: One-dimensional semi-markov evoluions wih general Erlang sojourn imes. Random Operaors and Soch. Equa. 13, (5) 5. Pogorui, A.A.: Fading evoluion in mulidimensional spaces. Ukrainian Mahemaical Journal 6, No. 11, (1)

12 1 6. Pogorui, A.A.: The disribuion of random evoluion in Erlang semi-markov media. Theory of Sochasic Processes 17(1), 9-99 (11) 7. Pogorui, A.A., Rodriguez-Dagnino, R.M.: Isoropic random moion a finie speed wih K-Erlang disribued direcion alernaions. J. Sa. Phys. 145, (11) 8. Beghin, L., Orsingher, E.: Moving randomly amid scaered obsacles. Sochasics 8, 1 9 (1) 9. Le Caer, G.: A Pearson-Dirichle random walk. J. Sa. Phys. 14, (1) 1. Le Caer, G.: A new family of solvable Pearson-Dirichle random walks. J. Sa. Phys. 144, 3 45 (11) 11. Bochner, S., Chandrasekharan, K.: Fourier Transforms, Annals of Mahemaics Sudies, No. 19, Princeon Universiy Press (1949) 1. Erdélyi, A. e al.: Tables of Inegral Transforms, The Baeman Projec, vol. I, McGraw Hill, New York, (1953) 13. Gradsheyn, I. S., Ryzhik, I. M. Table of Inegrals, Series, and Producs, 4h edn. Translaed from he Russian by Scripa Technica, Academic Press, Kolesnik, A.D., Orsingher, E.: A planar random moion wih infinie number of direcions conrolled by he damped wave equaion. J. Appl.. Prob. 4, (5)