Hyperbolic and fractional hyperbolic Brownian motion

Stochastics

ISSN: 1744-2508 (Print) 1744-2516 (Online) Journal homepage: http://www.tandfonline.com/loi/gssr20

Hyperbolic and fractional hyperbolic Brownian motion Lanjun Lao & Enzo Orsingher To cite this article: Lanjun Lao & Enzo Orsingher (2007) Hyperbolic and fractional hyperbolic Brownian motion, Stochastics, 79:6, 505-522, DOI: 10.1080/17442500701433509 To link to this article: http://dx.doi.org/10.1080/17442500701433509

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Date: 26 October 2016, At: 07:35

Stochastics: An International Journal of Probability and Stochastics Processes, Vol. 79, No. 6, December 2007, 505–522

Hyperbolic and fractional hyperbolic Brownian motion LANJUN LAO† and ENZO ORSINGHER‡* †Department of Finance, School of Management, Fudan University, Handan Road 220, Shanghai, P. R. China ‡Department of Statistics, Probability and Applied Statistics, University of Rome, “La Sapienza”, pl. A. Moro 5, 00185 Rome, Italy (Received 9 May 2006; in final form 24 April 2007) We examine the hyperbolic, planar Brownian motion and its time-fractional version. The analogy between the hyperbolic Brownian motion and Brownian motion on the sphere is also analysed. We examine in detail the connection between the equations governing the distributions in the Cartesian and hyperbolic coordinates. We discuss the time-fractional generalization of hyperbolic Brownian motion and give a representation of it as composition of classical hyperbolic Brownian motion with a reflecting Brownian motion on the line. Keywords: Poincare´ half-plane; Fractional equations; Mittag-Leffler functions; Legendre equation; Hyperbolic coordinates

1. Introduction The hyperbolic Brownian motion is a diffusion on the half-plane H þ 2 ¼ {x; y : x [ R; y . 0}, with transition function pH ¼ pH ðx; y; tÞ satisfying the following heat-type equation ›pH 1 2 ›2 ›2 ¼ y þ ð1:1Þ pH 2 ›t ›x 2 ›y 2 with initial condition pH ðx; y; 0Þ ¼ dðxÞdð y 2 1Þ:

ð1:2Þ

For the study of hyperbolic Brownian motion a convenient couple of coordinates are the hyperbolic coordinates (h, a) related to the cartesian coordinates by the relationship 8 sinh h cos a < x ¼ cosh h2sinh h sin a ; ð1:3Þ 1 : y ¼ cosh h2sinh h sin a ;

*Corresponding author. Email: [email protected] Stochastics: An International Journal of Probability and Stochastics Processes ISSN 1744-2508 print/ISSN 1744-2516 online q 2007 Taylor & Francis http://www.tandf.co.uk/journals DOI: 10.1080/17442500701433509

506

L. Lao and E. Orsingher

where h is the hyperbolic distance of (x, y) from the origin (0,1) of H þ 2 . The angle a varies in (2 p, p) and corresponds to the angle of the tangent in (0,1) to the circumference passing through (x, y). The geodesic lines of H þ 2 passing through the origin have the form ðx 2 tan aÞ2 þ y 2 ¼

1 ; cos2 a

and for a ¼ p/2 degenerate into the positive y axis. By means of the transformation (1.3) the equation (1.1) is reduced to ›pH 1 1 › › 1 ›2 ¼ sinh h þ pH ; 2 sinh h ›h ›h sinh2 h ›a 2 ›t

ð1:4Þ

ð1:5Þ

subject to the initial condition pH ðh; a; 0Þ ¼ dðhÞ: In the pioneering paper by Gertsenshtein and Vasiliev [6] the exact distribution of the hyperbolic distance h ¼ h(t), t . 0 of Brownian motion is given (without proof) by resolving the Cauchy problem: 8 < ››ptH ¼ sinh1 h ››h sinh h ››h pH ; ð1:6Þ : pH ðh; 0Þ ¼ dðhÞ: The distribution emerging from (1.6) appeared in many papers with different normalizing constants. Our first result is to give an explicit and detailed derivation of ð1 2 e2ðt=4Þ we2ðw =4tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw: pH ðh; tÞ ¼ pffiffiffiffi pffiffiffiffi3 ð1:7Þ cosh w 2 cosh h h p 2t We also examine some properties of (1.7) including some striking relationship with the distribution of three-dimensional hyperbolic Brownian motion in H þ 3 ¼ {x; y; z : ðx; yÞ [ R2 ; z . 0}, first studied by Karpelevich et al. [11]. Our main contribution concerns the fractional hyperbolic Brownian motion whose probability law paH ðx; y; tÞ ¼ paH satisfies the time-fractional equation h i 8 a < ››tpaH ¼ 12 y 2 ›22 þ ›22 pH ; 0 , a # 1; ›x ›y ð1:8Þ : pH ðx; y; 0Þ ¼ dðxÞdðy 2 1Þ: Our result is obtained by solving the fractional equation 8 a < ››tpaH ¼ sinh1 h ››h sinh h ››h pH ; : pH ðh; 0Þ ¼ dðhÞ: and we prove that paH ðh; tÞ coincides with a ð1 ð 2 1 t sin xw dw a 2 a xEa;1 2 2 x t dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : pH ðh; tÞ ¼ p 0 4 2 cosh w 2 2 cosh h h

ð1:9Þ

ð1:10Þ

Hyperbolic Brownian motion

507

This generalizes the well-known distribution (1.7) with the role of the exponentials replaced by the Mittag-Leffler function Ea;1 ðxÞ ¼

1 X k¼0

xk ; Gðak þ 1Þ

x [ R:

In the special case where a ¼ 1/2 the distribution emerging from (1.10) coincides with that of the process

h1=2 ðtÞ ¼ hðjBðtÞjÞ;

ð1:11Þ

where h is the hyperbolic Brownian motion and B is the classical Brownian motion. The process defined by (1.11) is a sort of iterated Brownian motion obtained by means of the composition of hyperbolic Brownian motion h with a Brownian time. The hyperbolic Brownian motion is formally similar to Brownian motion on the sphere with imaginary radius. For this reason, some analogies between hyperbolic Brownian motion in H þ 2 and spherical Brownian motion are discussed. Generalizations in higher spaces are also mentioned by discussing the stochastic representation of hyperbolic Brownian motion n21 which easily extends to H þ ; xn . 0}: n ¼ {x1 ; . . . ; xn : ðx1 ; . . . ; xn21 Þ [ R In some papers (see, for example, [3]) the hyperbolic Brownian motion is examined in the unit disk D ¼ {w : w ¼ reiu ; jrj # 1} which can be obtained by means of the conformal transformation of the upper half-plane {z ¼ x þ iy, y . 0} w¼

iz þ 1 : zþi

In this case, the law of hyperbolic Brownian motion can be derived by solving the equation ›p ð1 2 r 2 Þ2 1 › › 1 ›2 ¼ r þ 2 2 p; r ›r ›r r ›u ›t 4

ð1:12Þ

where (r, u) are the usual polar coordinates.

2. The transition function of hyperbolic Brownian motion on the upper half-plane The transition function pH ¼ pH ðh; tÞ; h . 0, t . 0 of hyperbolic Brownian motion has first appeared in the paper by Gertsenshtein and Vasiliev [6], without proof. Many times, these formulas have been reproduced and unfortunately with contradictory constants therein (for example [7 – 9,13,16]). We start our analysis by showing that equation (1.1) is converted into equation ›pH 1 1 › › 1 ›2 ¼ sinh h þ pH ; 2 sinh h ›h ›h sinh2 h ›a 2 ›t

ð2:1Þ

by means of the transformation (1.3). We note that formulas (1.3) can be obtained by suitably combining the form of the equations (1.4) of geodesic lines and the expression of the hyperbolic distance of (x, y) from

508


(0,1) that is cosh h ¼

x2 þ y2 þ 1 : 2y

ð2:2Þ

Theorem 2.1. The hyperbolic Laplacian in cartesian coordinates y

2

›2 ›2 þ ›x 2 ›y 2

ð2:3Þ

is expressed in hyperbolic coordinates (h, a) by means of the differential operator 1 › › 1 ›2 sinh h : þ 2 sinh h ›h ›h sinh h ›a 2

ð2:4Þ

Proof. By successive derivatives of (1.3) we get that

› › cos a › 2sinh h þ cosh h sin a ¼ þ 2 ›h ›x ðcosh h 2 sinh h sin aÞ ›y ðcosh h 2 sinh h sin aÞ2

ð2:5Þ

› › sinh2 h 2 sinh h cosh h sin a › cos a sinh h ¼ þ ›a ›x ðcosh h 2 sinh h sin aÞ2 ›y ðcosh h 2 sinh h sin aÞ2

ð2:6Þ

and

›2 ›2 cos2 a ›2 ð2sinh h þ cosh h sin aÞ2 ¼ 2 þ 2 4 2 ›h ›x ðcosh h 2 sinh h sin aÞ ›y ðcosh h 2 sinh h sin aÞ4 þ2

þ

›2 cos að2sinh h þ cosh h sin aÞ › 22 cos aðsinh h 2 cosh h sin aÞ þ ›x ›x›y ðcosh h 2 sinh h sin aÞ4 ðcosh h 2 sinh h sin aÞ3

› 2cos2 a þ ðsinh h 2 cosh h sin aÞ2 ›y ðcosh h 2 sinh h sin aÞ3

ð2:7Þ

›2 ›2 sinh2 hðsinh h 2 cosh h sin aÞ2 ›2 sinh2 h cos2 a ¼ þ 4 ›a 2 ›x 2 ›y 2 ðcosh h 2 sinh h sin aÞ4 ðcosh h 2 sinh h sin aÞ þ2

›2 sinh2 h cos aðsinh h 2 cosh h sin aÞ ›x›y ðcosh h 2 sinh h sin aÞ4

þ

› sinh h cos aðsinh2 h 2 1 2 sinh h cosh h sin aÞ ›x ðcosh h 2 sinh h sin aÞ3

þ

› sinh hð2sin a cosh h þ sinh h þ sinh h cos2 aÞ : ›y ðcosh h 2 sinh h sin aÞ3

ð2:8Þ


509

By multiplying (2.7) by sinh2h and then summing (2.8) we get that 2 2 ›2 ›2 ›2 2 2 cos a þ ðsinh h 2 cosh h sin aÞ sinh h þ ¼ sinh h ›h 2 ›a 2 ›x 2 ðcosh h 2 sinh h sin aÞ4

þ

2 2 ›2 2 cos a þ ðsinh h 2 cosh h sin aÞ sinh h ›y 2 ðcosh h 2 sinh h sin aÞ4

þ

› sinh h cos að2cosh2 h þ sinh h cosh h sin aÞ ›x ðcosh h 2 sinh h sin aÞ3

› sinh h cosh hðsinh h 2 cosh h sin aÞ ›y ðcosh h 2 sinh h sin aÞ2 2 › ›2 › ¼ sinh2 h y 2 þ 2 sinh h cosh h cos a y 2 2 2 ›x ›x ›y þ

þ sinh h cosh hðsinh h 2 cosh h sin aÞy 2 2

¼ ½in view ofð2:5Þsinh h y

2

› ›y

›2 ›2 › þ 2 sinh h cosh h : ›h ›x 2 ›y 2

and this lets (2.4) emerge.

A

Remark 2.1. The transformation (1.3) relating cartesian and hyperbolic coordinates can be found in Ref. [15], page 213 without proof. However, an easy derivation can be obtained by solving (2.2) with respect to x and then substituting into (1.4). After some calculations this leads to the second relationship of (1.3) while the first one can be derived by exploiting (2.2) once more. The hyperbolic coordinates are much more convenient in deriving the distribution of the hyperbolic distance of Brownian motion in H þ 2. The factor 1/2 appearing in (1.5) can be easily eliminated by means of the time change t 0 ¼ (1/2)t. This different time scale originates many of the inconsistencies appearing in the literature on this point. Remark 2.2. The conformal mapping f ðzÞ ¼

z2i ; 2iz þ 1

ð2:9Þ

2 2 converts the Poincare´ half-space H þ 2 into the disk D ¼ {x; y : x þ y , 1}: þ A point ðx; yÞ [ H 2 is mapped into the point (u, v) with coordinates

u¼

2x ; 2 x þ ð y þ 1Þ2

v¼

x2 þ y2 2 1 : x 2 þ ð y þ 1Þ2

ð2:10Þ

A point (u, v) [ D is mapped into the point ðx; yÞ [ H þ 2 with coordinates x¼

2u ; 2 u þ ð1 2 vÞ2

y¼

1 2 ðu 2 þ v 2 Þ : u 2 þ ð1 2 vÞ2

ð2:11Þ

510


In polar coordinates u ¼ r cos u; v ¼ r sin u we can represent (x, y) as x¼

r2

2r cos u ; 2 2r sin u þ 1

y¼

r2

1 2 r2 : 2 2r sin u þ 1

ð2:12Þ

By means of (2.12) the hyperbolic laplacian appearing in (1.1) is converted into ð1 2 r 2 Þ2 1 › › 1 ›2 r þ 2 2 r ›r ›r r ›u 22

ð2:13Þ

and this explains the form of the heat-equation (1.12). The calculations leading to (2.13) are similar to those of Theorem 2.1. Up to the factor ð1 2 r 2 Þ2 =22 the laplacian (2.13) coincides with the well-known corresponding operator in polar coordinates. Remark 2.3. We now present a detailed derivation of the distribution (1.7). While the distribution of h ¼ hðtÞ has appeared in some papers (without proof) the joint distribution of (a(t), h(t)) is not known (as well as the distribution of hyperbolic Brownian motion in cartesian coordinates). Theorem 2.2. The explicit form of the distribution of h ¼ h(t), t . 0 with respect to the hyperbolic area element sinh hdh is given by e2ðt=4Þ pH ðh; tÞ ¼ pffiffiffiffi pffiffiffiffi3 p 2t

ð1 h

2

we2ðw =4tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw; cosh w 2 cosh h

h . 0:

Proof. In order to obtain (2.14) we must resolve the initial value problem 8 < ››ut ¼ sinh1 h ››h sinh h ››h u; : uðh; 0Þ ¼ dðhÞ;

ð2:14Þ

ð2:15Þ

because we disregard the dependence of the distribution from a. It is now convenient to perform the change of variable cosh h ¼ y

ð2:16Þ

which yields

›u ¼ ›t

›2 › ðy 2 1Þ 2 þ 2y u: ›y ›y 2

ð2:17Þ

In Ref. [6], unfortunately, the change of variable is written as tanh h ¼ y. The method of separation of variables produces two ordinary equations, one of which is strictly related to Legendre equation. We now put uðt; yÞ ¼ TðtÞFð yÞ

ð2:18Þ


511

and easily get the following equations 8 0 ðtÞ < TTðtÞ ¼ 2v; : ðy 2 2 1ÞF 00 þ 2yF 0 þ vF ¼ 0;

ð2:19Þ

where v is an arbitrary positive constant, the choice of which is an essential part of the derivation of (2.14). The bounded solution of equation (2.15), because of its homogeneity and linearity has thus the form uðh; tÞ ¼

ð1

Tðt; vÞFðh; vÞGðvÞdv;

ð2:20Þ

0

where G(v), v . 0, is an arbitrary function which is determined by applying the initial condition. The equation of F for v ¼ 2 n(n þ 1) becomes the classical Legendre equation

ð1 2 z 2 Þ

d2 u du 2 2z þ nðn þ 1Þu ¼ 0 dz 2 dz

ð2:21Þ

which is a special case of the hypergeometric equation tð1 2 tÞ

d2 u du þ ½g 2 ða þ b þ 1Þt 2 abu ¼ 0; dt 2 dt

ð2:22Þ

for z ¼ 1 2 2t, a ¼ 2 n, b ¼ n þ 1, g ¼ 1. A solution to (2.22) is the hypergeometric function Fða; b; g ; tÞ ¼

1 X ðaÞk ðbÞk k t k!ðgÞk k¼0

¼

1 X aða þ 1Þ . . . ða þ k 2 1Þbðb þ 1Þ . . . ðb þ k 2 1Þ k t k!g ðg þ 1Þ . . . ðg þ k 2 1Þ k¼0

¼

1 X Gða þ kÞGðb þ kÞ GðgÞ t k; k!Gð Gð g þ kÞ a ÞGð b Þ k¼0

jtj , 1:

ð2:23Þ

This means that a solution to Legendre equation (2.21), known as Legendre polynomial, is thus X 1 12z ð2nÞk ðn þ 1Þk 1 2 z k Pn ðzÞ ¼ F 2n; n þ 1; 1; ¼ 2 2 k!ð1Þk k¼0 k 1 X ð2nÞk ðn þ 1Þk 1 2 z ¼ ; j1 2 zj , 2: 2 ðk!Þ2 k¼0

ð2:24Þ

512


By following Ref. [12], page 165, it is possible to represent the Legendre polynomials in a more suitable form by considering that 2 p

ð p=2

1 2k

sin w dw ¼

0

2 k

Gðk þ 1Þ

:

ð2:25Þ

By substituting (2.25) in (2.24) we get that ð 1 X ð2nÞk ðn þ 1Þk 1 2 z k 2 p=2 2k Pn ðzÞ ¼ sin w dw 2 ð1=2Þk k! p 0 k¼0 ¼

2 p

ð p=2 X 1 0

k¼0

ð2nÞk ðn þ 1Þk 12z k sin2 w dw 2 ð1=2Þk k!

ð 2 p=2 1 2 12z ¼ F 2n; n þ 1; ; sin w dw : 2 2 p 0

ð2:26Þ

For us, the following step is essential in order to obtain (2.14). The function in the integrand of (2.26) can be written as pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi2nþ1 pffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffi22n21 þ 1þvþ v 1þvþ v 1 pffiffiffiffiffiffiffiffiffiffiffiffi f n ðvÞ ¼ F 2n; n þ 1; ; 2v ¼ : 2 2 1þv

ð2:27Þ

This can be shown by proving that fn(v) solves the hypergeometric equation (2.22) for a ¼ 2 n, b ¼ n þ 1, g ¼ 1/2, t ¼ 2 v, that is

vð1 þ vÞ

d2 f n 1 df n þ 2 þ v 2 n ðn þ 1Þf n ¼ 0; 2 dv 2 dv

ð2:28Þ

which can also be rewritten in the more convenient form pffiffiffiffihpffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi 0 i0 1 2 v v 1 þ v 1 þ vfn 2 n þ f n ¼ 0: 2

ð2:29Þ

It is now a relatively simple matter to show that (2.27) is a solution to (2.29) and also to check that f n ðvÞ ¼ f 2n21 ðvÞ: By some calculations it can be seen that fn

sin2 w ðcosh h 2 1Þ 2

¼

cosh uðn þ ð1=2ÞÞ coshðu=2Þ

ð2:30Þ

where

u h sinh ¼ sinh sin w: 2 2

ð2:31Þ


513

Therefore, the Legendre polynomial (2.26) ð p=2 1 sin2 w ð1 2 cosh hÞ dw F 2n; n þ 1; ; 2 2 0 2 ð 2 p=2 sin w ðcosh h 2 1Þ dw; ¼ fn p 0 2

Pn ðcosh hÞ ¼

2 p

in view of (2.31), takes the following form Pn ðcosh hÞ ¼

1 p

ðh

eðnþð1=2ÞÞu pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi du: 2h 2ðcosh h 2 cosh uÞ

ð2:32Þ

We skip the next step which consists in applying a contour integration to the complexvalued function eðnþð1=2ÞÞt f ðtÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; p 2ðcosh h 2 cosh tÞ

t [ C;

and permits us to express the Legendre function (2.32) by means of an integral on [h,1) as follows ð 2 1 1 sinh{ðn þ ð1=2ÞÞw} pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw: Pn ðcosh hÞ ¼ cot p n þ ð2:33Þ 2 h 2 cosh w 2 2 cosh h p The details of this transformation can be found in Ref. [12], page 172. If we now choose v ¼ (1/4) þ x 2 in the second equation of (2.19) we note that it coincides with (2.21) when n ¼ 2 (1/2) ^ ix. From (2.20), by considering (2.33) we get that ð1 2 uðh; tÞ ¼ e2ðt=4Þ2x t P2ð1=2Þþix ðcosh hÞxGðxÞdx 0

2 ¼ p

ð1 xGðxÞe

2ðt=4Þ2x 2 t

ð1 coth pxdx h

0

sin xw pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw: 2 cosh w 2 2 cosh h

ð2:34Þ

We show now that if we choose G(x) ¼ tanh px then the initial condition u(h,0) ¼ d(h) is satisfied. In order to check that (2.14) satisfies the initial condition it suffices to write that limþ pH ðh; tÞ ¼ limþ

t!0

t!0

ð1

2

we2ðw =2tÞ dw qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pffiffi pffiffiffiffiffiffiffiffiffiffi p ffiffi ffi 3 2pt h= 2 coshð 2wÞ 2 cosh h

8 0; for h . 0; < d ð wÞ Ð1 d ð w Þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi q ¼ w ¼ d ffi dw ¼ 1; for h ¼ 0: pffiffi pffiffi pffiffiffi : 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h= 2 coshð 2wÞ21 coshð 2wÞ 2 cosh h ð1

In the steps above, we clearly took into account the behaviour of the distribution of the firstpassage time of a standard Brownian motion. A

514


Remark 2.4. We now show that the density (2.14) integrates to one with respect to the area element sinh hdh: ð1

1 e2ðt=4Þ pH ðh; tÞsinh hdh ¼ pffiffiffiffi pffiffiffiffiffiffiffiffiffi p ð2tÞ3 0 1 e2ðt=4Þ ¼ pffiffiffiffi pffiffiffiffiffiffiffiffiffi p ð2tÞ3 e2ðt=4Þ ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi p ð2tÞ3 2e2ðt=4Þ ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi p ð2tÞ3

ð1

ð1 sinh hdh h

0

ð1

we2ðw

2

=4tÞ

ðw dw

0

ð1

we2ðw

2

we2ðw

2

=4tÞ

2

we2ðw =4tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw cosh w 2 cosh h sinh h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dh 0 cosh w 2 cosh h

h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiih¼w dw 22 cosh w 2 cosh h h¼0

0

ð1

=4tÞ

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh w 2 1dw

0

ð1

2e2ðt=4Þ eðw=2Þ 2 e2ðw=2Þ 2 pffiffiffi ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi we2ðw =4tÞ dw 2 p ð2tÞ3 0 pffiffiffi ð 1 ð1 2 2ððw2tÞ2 =4tÞ 2ððwþtÞ2 =4tÞ ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi we d w 2 we dw 0 p ð2tÞ3 0 ( ) ð 1 pffiffiffiffi pffiffiffiffið 1 pffiffiffiffi p ffiffiffiffi 1 2 2 2t pffi ð 2ty þ tÞe2ð y =2Þ dy 2 2t pffi ð 2ty 2 tÞe2ð y =2Þ dy ¼ pffiffiffiffiffiffiffiffi t t 4pt 2 2t 2 1 ¼ pffiffiffiffiffiffi 2p

ð1

e2ð y

2

=2Þ

dy ¼ 1:

21

The normalizing constant we found coincides with that appearing in Ref. [13], formula (78) for D ¼ 1, a ¼ 1. In the case where we considered equation of (2.15) with the factor 1/2 the distribution (2.14) must be replaced by e2ðt=8Þ pH ðh; tÞ ¼ pffiffiffiffi pffi 3 p t

ð1 h

2

we2ðw =2tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw; cosh w 2 cosh h

ð2:35Þ

and this almost coincides with the formula appearing in Ref. [8], at page 57 and formula (3.4) in [10]. Remark 2.5. We note that the joint density 2

we2ðw =4tÞ sinh h f ðh; w; tÞ ¼ e2ðt=4Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; for w . h . 0; 3 w 2 cosh h cosh ð2tÞ p

ð2:36Þ

has a marginal which coincides with the distribution of the hyperbolic distance h ¼ h(t),


(formula (2.14)) while the other marginal has the interesting form ðw o 2 2 w n f ðh; w; tÞdh ¼ pffiffiffiffiffiffiffiffiffiffi e2ððw2tÞ =4tÞ 2 e2ððwþtÞ =4tÞ 4pt 3 0 2ðw 2 =4tÞ2ðt=4Þ e w pffiffiffiffiffiffiffiffi ¼ sinh ; w . 0: 3 2 pt

515

ð2:37Þ

From (2.36) also the distribution function can be obtained ð1 ð1 2 e2ðt=4Þ we2ðw =4tÞ sinh h Pr {hðtÞ . h} ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi dh pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw cosh w 2 cosh h h p ð2tÞ3 h ðw ð1 e2ðt=4Þ sinh h 2 ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi we2ðw =4tÞ dw pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dh 3 w 2 cosh h cosh h p ð2tÞ h ð 2e2ðt=4Þ 1 2ðw 2 =4tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh w 2 cosh hdw ¼ pffiffiffiffipffiffiffiffiffiffiffiffiffi we p ð2tÞ3 h pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 2ðt=4Þ ð 1 ¼ 2 2e f t ðwÞ cosh w 2 cosh hdw;

ð2:38Þ

h

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 where f t ðwÞ ¼ ððwe2ðw =4tÞ Þ=ð 2pð2tÞ3 ÞÞ; t . 0; w . 0; as a function of t, is the distribution of the first-passage time of Brownian motion (with infinitesimal variance 2) through level w.

3. Fractional hyperbolic Brownian motion A generalization of the classical hyperbolic Brownian motion can be obtained by considering the time-fractional equation ›a u 1 2 ›2 ›2 y ¼ þ ð3:1Þ u; 0 , a # 1; ›t a 2 ›x 2 ›y 2 subject to the initial condition uðx; y; 0Þ ¼ dðxÞdð y 2 1Þ: We assume that the fractional derivative appearing in (3.1) is understood in the sense of Dzerbashyan-Caputo that is as, for f [ C m ðt da f 1 f ðmÞ ðsÞ ¼ ds; m 2 1 , a # m: a dt Gðm 2 aÞ 0 ðt 2 sÞaþ12m As in Section 2, we are mainly interested in studying the distribution of the hyperbolic distance h ¼ h(t), t . 0 and therefore we are obliged to study the initial value problem (we get rid once again of the factor 1/2 by means of the position t0 ¼ (1/2)t) 8 a < ››t au ¼ sinh1 h sinh h ››h u; ð3:2Þ : uðh; 0Þ ¼ dðhÞ:

516


By applying the transformation cosh h ¼ y and the separation of variables (2.18) we get ( da T

dt a ¼ 2

2vT; ð3:3Þ

ðy 2 1ÞF 00 þ 2yF 0 þ vF ¼ 0:

This can be done because we are considering fractional derivatives in the DzerbashyanCaputo sense which permit us to assume initial conditions with integer derivatives and also with fractional derivatives of constants equal to zero. The solution of the second equation coincides with that appearing in the Theorem 2.2, while the solution of the ordinary fractional equation can be obtained by applying the Laplace transform to both members and then by considering that ð1

2 lt

e 0

da T dt ¼ l a dt a

ð1

2 lt

e

TðtÞdt 2

0

m21 X

l

k¼0

aþ12k

dk T dt k

: t¼0

Therefore ð1

e2lt Tðt; vÞdt ¼

0

l a21 l a21 Tð0; v Þ ¼ ; la þ v la þ v

ð3:4Þ

with T(0,v) ¼ 1. The inverse of the Laplace transform (3.4) is given by Tðt; vÞ ¼ Ea;1 ðvt a Þ

ð3:5Þ

where Ea;1 ðxÞ ¼

1 X k¼0

xk Gðak þ 1Þ

is the Mittag-Leffler function. By making use of the representation (2.20) of the solution to (3.2) with T replaced by (3.5) we have that ua ðh; tÞ ¼

a t Ea;1 2 2 x 2 t a P2ð1=2Þþix ðcosh hÞxGðxÞdx: 4 0

ð1

ð3:6Þ

By choosing again GðxÞ ¼ tanh px we get the solution to (3.2) as 2 ua ðh; tÞ ¼ p

ð1 xEa;1 0

ð1 ta sin xw dw 2 a 2 2 x t dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 4 2 cosh w 2 2 cosh h h

for 0 , a # 1. The classical result (2.14) can be derived from (3.7) as a special case for a ¼ 1.

ð3:7Þ


517

Since for a ¼ 1; E1;1 ðxÞ ¼ ex we get from (3.7) that ð ð1 2 1 2ðt=4Þ2x 2 t sin xw dw xe dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u1 ðh; tÞ ¼ p 0 w 2 2 cosh h 2 cosh h pffiffiffi 2ðt=4Þ ð 1 ð1 2e dw 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xe2x t sin xw dx ¼ p w 2 cosh h cosh 0 h pffiffiffi 2ðt=4Þ ð 1 ð1 2e wd w 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cos xwe2x t dx ¼ 2p t w 2 cosh h cosh 0 h pffiffiffi 2ðt=4Þ pffiffiffiffi ð 1 2 2e p we2ðw =4tÞ p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi dw ¼ p 22 t 3=2 h cosh w 2 cosh h e2ðt=4Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi pð2tÞ3

ð1 h

2

we2ðw =4tÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dw ¼ pH ðh; tÞ cosh w 2 cosh h

In the third step, we used the following formula rffiffiffiffi ð1 1 p 2ðb 2 =4aÞ 2 e2ax cos bx dx ¼ e 2 a 0

ð3:8Þ

ð3:9Þ

(see Gradshteyn-Ryzhik (1981), page 480, formula 3.896.4). For arbitrary values of a the transformations converting (3.7) into (2.14) are not possible. For a ¼ 1/2, we can obtain an explicit and interesting representation of the distribution (3.7) because ð 2 1 2 Eð1=2Þ;1 ðxÞ ¼ pffiffiffiffi e2y þ2xy dy; ð3:10Þ p 0 as can be checked by observing that pffiffiffiffi k kþ1 1 Gðk þ 1Þ þ G þ1 ¼G ¼ p22k kþ1 : 2 2 2 G 2 Therefore by plugging (3.10) into (3.7) for a ¼ 1/2 we get that ð ð ð1 2 1 2 1 2y 2 2yðpffit=2Þ22x 2 ypffit sin xw dw dx x pffiffiffiffi e dy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u1=2 ðh; tÞ ¼ p 0 p 0 h 2 cosh w 2 2 cosh h pffi pffi ð ð ð1 2 2 2 2 1 1 e2y 2yð t=2Þ22x y t w cos xw dw pffi ¼ pffiffiffiffiffiffi dxdy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ðby ð3:9ÞÞ p 2p 0 0 w 2 cosh h cosh 4 ty h ð 1 2y 2 2yðpffit=2Þ pffiffiffiffi ð 1 2ðw 2 =8pffityÞ 22 e p e w dw pffi pffiffiffiffiffi p4 ffi dy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffi w 2 cosh h cosh 4 t y 2y 2 t p 2p 0 h pffi ð ð 1 2ðw 2 =ð4pffit2yÞÞ 2 22 1 e2y e2ð2y t=4Þ e wd w ¼ pffiffiffi pffiffiffiffiffiffi pffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffi 3 dy pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 cosh w 2 cosh h 2 0 2p p 2 ð2y tÞ h 22 ¼ pffiffiffi 2

ð1 0

2

pffi e2y pffiffiffiffiffiffi pH ðh;2 tyÞdy ¼ 2 2p

ð1 0

2

e2ðw =4tÞ pffiffiffiffiffiffiffiffiffiffi pH ðh;wÞdw: 2p2t

ð3:11Þ

518


This result represents the distribution of a hyperbolic Brownian motion at a reflected Brownian time. In other words, the result u1/2(h, t) is the distribution of the process

h1=2 ¼ hðjBðtÞjÞ

ð3:12Þ

where h is the hyperbolic Brownian motion and B is a Brownian motion independent of h. We can reformulate the result (3.11) as follows. The distribution (3.12) is the solution to the initial value problem for the fractional equation 8 1=2 < › 1=2u ¼ sinh1 h ››h sinh h ››h u; ›t ð3:13Þ : uðh; 0Þ ¼ dðhÞ: It is possible to obtain similar, but more complicated representations of fractional hyperbolic Brownian motion for the case a ¼ 1/n, in the spirit of the papers [1,14]. Remark 3.1. For 1 , a # 2, when the initial condition ut(h,0) ¼ 0 is assumed the analysis follows in the same way. This is because the extra term in the Laplace transform vanishes and therefore we get (3.5) as in the previous case. Therefore the result (3.7) remains valid also for 1 , a # 2 although in this case representations of solutions of the form (3.12) seem not possible.

4. Analogies between Brownian motion on the sphere and hyperbolic Brownian motion On the sphere, the probability law pS ¼ pS ðu; w; tÞ of the spherical Brownian motion is a solution to ›pS 1 1 › › 1 ›2 ¼ sin u ð4:1Þ þ 2 pS 2 sin u ›u ›u sin u ›w 2 ›t and, in particular, the latitude u ¼ u(t), t . 0 has a transition law pL ¼ pL ðu; tÞ which satisfies ›pL 1 1 › › ¼ sin u ð4:2Þ pL : 2 sin u ›u ›u ›t For Brownian motion on the sphere see [2] where a review of this topic is presented. If we put u ¼ ^ ih and thus sin u ¼ î sinh h; ð›=›hÞ ¼ îð›=›uÞ, equation (4.2) becomes (the sign is irrelevant) ›pL 1 1 › › 2 ¼ sin h pL 2 sin h ›h ›h ›t and with the further time change t ¼ 2 t 0 we reobtain equation (1.6). This suggests to imagine hyperbolic Brownian motion as a Brownian motion on a sphere with imaginary radius. The same type of formal analogy between spherical trigonometry and hyperbolic trigonometry has been established since the beginning of non-Euclidean hyperbolic geometry ([4], page 241). The analogy between spherical Brownian motion and its hyperbolic version can be made more transparent by resorting to the so-called stochastic representation of hyperbolic Brownian motion.


519

From the form of the generator of hyperbolic Brownian motion in cartesian coordinates it can be inferred that the hyperbolic Brownian particle satisfies the following stochastic differential system (

dXðtÞ ¼ YðtÞdB1 ; dYðtÞ ¼ YðtÞdB2 ;

ð4:3Þ

where B1 and B2 are standard, independent Brownian motions. The processes X(t), Y(t), t . 0 are subject to the initial conditions Xð0Þ ¼ 0

Yð0Þ ¼ 1:

If we consider the generator in hyperbolic coordinates (see (2.1)) we have the alternative representation 8 dt < dh ¼ 2 tanh h þ dB1 ; : da ¼ sinh1 h dB2 ;

ð4:4Þ

with h(0) ¼ 0. This system of stochastic differential equations for hyperbolic Brownian motion is the counterpart of the well-known equations of Brownian motion on the sphere 8 dt < du ¼ 2 tan u þ dB1 ; : dw ¼ sin1 u dB2 :

ð4:5Þ

By means of the transformation u ¼ ih and dt ¼ 2 dt0 the first equation of (4.5) can be reduced to the first equation of (4.4). This shows that the hyperbolic distance of Brownian motion behaves as the latitude of Brownian motion on a sphere with imaginary radius. The hyperbolic distance h ¼ h(t), t . 0 is a stochastic process with drift equal to ðcoth hÞ=2 and therefore is big near the origin of H þ 2 . This means that hyperbolic Brownian motion is pushed away from the origin because the hyperbolic Brownian particle experiences an infinite drift there. The same happens for spherical Brownian motion near the poles and thus the Brownian particle wanders mostly on equatorial latitudes. The solution to system (4.3) is straightforward and is given by the couple 8 < YðtÞ ¼ eB2 ðtÞ2ðt=2Þ ; Ðt : XðtÞ ¼ 0 eB2 ðsÞ2ðs=2Þ dB1 ðsÞ:

ð4:6Þ

In view of (2.2), this also gives a stochastic representation of the hyperbolic distance as ð 2 n t o 1 2B2 ðtÞþðt=2Þ t B2 ðsÞ2ðs=2Þ e dB1 ðsÞ : þ e cosh hðtÞ ¼ cosh B2 ðtÞ 2 2 2 0

ð4:7Þ

520


From this, we can give the mean value of the hyperbolic distance as follows 1 E cosh hðtÞ ¼ E{eB2 ðtÞ2ðt=2Þ þ e2B2 ðtÞþðt=2Þ } 2 ðt ðt 1 eB2 ðsÞ2ðs=2Þ eB2 ðwÞ2ðw=2Þ dB1 ðsÞdB1 ðwÞ þ E e2ðB2 ðtÞ2ðt=2ÞÞ 2 0 0 ðt 1 1 ¼ ð1 þ et Þ þ E{e2B2 ðsÞ2s e2ðB2 ðtÞ2ðt=2ÞÞ }ds 2 2 0 ð t 1 t 2sþðt=2Þ t=2 ¼ et=2 cosh þ e e ds ¼ et : 2 2 0

ð4:8Þ

In the calculations above, we considered that E{e2B2 ðsÞ2B2 ðtÞ } ¼ E{E{eB2 ðsÞ2ðB2 ðtÞ2B2 ðsÞÞ jF s }} ¼ E{eB2 ðsÞ E{e2ðB2 ðtÞ2B2 ðsÞÞ }} ¼ eð1=2Þðt2sÞ es=2 ¼ et=2 ; where Ft, t . 0 is the filtration to which the Brownian motion B2 is adapted. In analogy with the case of planar hyperbolic Brownian motion, it is possible to give a stochastic representation to hyperbolic Brownian motion in H þ n ¼ {x1 ; . . . ; xn : ðx1 ; . . . ; xn21 Þ [ Rn21 ; xn . 0} first introduced by Gruet [8]. The transition law pH of this n-dimensional hyperbolic Brownian motion satisfies the n-th order heat-equation " # n ›pH 1 2 X ›2 › x ¼ þ ð2 2 nÞxn pH 2 n j¼1 ›x2j ›x n ›t

ð4:9Þ

subject to the initial condition pH ðx1 ; . . . ; xn ; 0Þ ¼ dðx1 Þ . . . dðxn21 Þdðxn 2 1Þ:

ð4:10Þ

The stochastic representation can be obtained by observing that the form of the generator shows that the coordinates Xj(t), j ¼ 1, . . . , n satisfy the stochastic differential system 8 dX 1 ¼ X n ðtÞdB1 ; > > > > > < dX 2 ¼ X n ðtÞdB2 ; ... > > > > > : dX n ¼ X n ðtÞdBn þ 12 ð2 2 nÞX n ðtÞdt;

ð4:11Þ

with initial conditions X 1 ð0Þ ¼ · · · ¼ X n21 ð0Þ ¼ 0; X n ð0Þ ¼ 1: By applying the Ito formula, we get that 8 < X n ðtÞ ¼ eBn ðtÞþðð12nÞ=2Þt ; Ðt : X j ðtÞ ¼ 0 eBn ðsÞþðð12nÞ=2Þs dBj ðsÞ;

j ¼ 1; . . . ; n 2 1:

ð4:12Þ


521

In order to study the distribution of the hyperbolic distance in H þ n , we must examine the following initial value problem 8 n21 › 1 › < ››ut ¼ 12 ðsinh h Þ ›h u; ðsinh hÞn21 ›h ð4:13Þ : uðh; 0Þ ¼ dðhÞ: In particular, for n ¼ 3 this has been done in the paper by Karpelevich et al. and Monthus and Texier [11,13]. Since equation (4.13) can be written as › u 1 ›2 n21 › ¼ þ u; ð4:14Þ ›t 2 ›h 2 tanh h ›h the process h(t), t . 0 in H þ n satisfies the stochastic equation dh ¼ dB þ

n21 dt: 2 tanh h

ð4:15Þ

The analogy of equation (4.15) with the stochastic equation governing the Bessel process vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi uX u n 2 RðtÞ ¼ t Bj ðtÞ; j¼1

(B1, . . . , Bn are independent Brownian motions), namely dR ¼ dB þ

n21 dt; 2R

shows that the hyperbolic distance h ¼ h(t), t . 0 plays the same role in H þ n of the Bessel n process in the Euclidean space R .

Acknowledgement We greatfully acknowledge the support of the NSFC (No. 70371010) and the Ministry of Education of China for the visit of the second author at the Fudan University.

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