HHitting spheres on hyperbolic spaces
Valentina Cammarota † Enzo Orsingher ‡‡ Abstract
For a hyperbolic Brownian motion on the Poincar´e half-plane H , starting from a point of hy-perbolic coordinates z = ( η, α ) inside a hyperbolic disc U of radius ¯ η , we obtain the probability ofhitting the boundary ∂U at the point (¯ η, ¯ α ). For ¯ η → ∞ we derive the asymptotic Cauchy hittingdistribution on ∂ H and for small values of η and ¯ η we obtain the classical Euclidean Poisson kernel.The exit probabilities P z { T η < T η } from a hyperbolic annulus in H of radii η and η are derivedand the transient behaviour of hyperbolic Brownian motion is considered. Similar probabilities arecalculated also for a Brownian motion on the surface of the three dimensional sphere.For the hyperbolic half-space H n we obtain the Poisson kernel of a ball in terms of a series involvingGegenbauer polynomials and hypergeometric functions. For small domains in H n we obtain the n -dimensional Euclidean Poisson kernel. The exit probabilities from an annulus are derived also in the n -dimensional case. Keywords:
Hyperbolic spaces, Hyperbolic Brownian motion, Spherical Brownian motion, Poisson kernel, Dirich-let problem, Hypergeometric functions, Gegenbauer polynomials, Cauchy distribution, Hyperbolic and sphericalCarnot formulas
Hyperbolic Brownian motion has been studied over the years by several authors on the half-plane H and on the Poincar´e disc D and more recently in the n -dimensional hyperbolic space (see, for example,Matsumoto and Yor [13], Gruet [10], Byczkowski et al. [2] and Byczkowski and Malecki [3]). Thehyperbolic half-space H n is given by H n = { z = ( x, y ) : x ∈ R n − , y > } with the distance formula cosh η ( z (cid:48) , z ) = 1 + || z (cid:48) − z || yy (cid:48) . The hyperbolic Brownian motion is a diffusion governed by the generator∆ n = y (cid:32) n − (cid:88) i =1 ∂ ∂x i + ∂ ∂y (cid:33) − ( n − y ∂∂y (1.1) † Dipartimento di Statistica, Sapienza Universit`a di Roma, P.le Aldo Moro 5, 00185 Rome, Italy. Tel.: +390649910499,fax: +39064959241. E-mail address: [email protected]. ‡‡ Corresponding author . Dipartimento di Statistica, Sapienza Universit`a di Roma, P.le Aldo Moro 5, 00185 Rome, Italy.Tel.: +390649910585, fax: +39064959241. E-mail address: [email protected]. a r X i v : . [ m a t h . P R ] A p r see, for example, Gruet [10]). Therefore the probability density p ( x , . . . , x n − , y, t ) of hyperbolic Brow-nian motion is solution to the Cauchy problem ∂p∂t = y (cid:32) n − (cid:88) i =1 ∂ p∂x i + ∂ p∂y (cid:33) − ( n − y ∂p∂y subject to the initial condition p ( x , . . . , x n − , y,
0) = n − (cid:89) j =1 δ ( x j ) δ ( y − . For our purposes it is important to express the generator (1.1) in hyperbolic coordinates ( η, α ) =( η, α , . . . , α n − ) as follows ∆ n = ∂ ∂η + n − η ∂∂η + 1sinh η ∆ S n − (1.2)where ∆ S n − is the Laplace operator on the ( n − H n , n >
2, is studied and the paperby Byczkowski and Malecki [3] where the Poisson kernel of a ball in the Poincar´e disc D n , n >
2, isconsidered.The first part of our paper concerns the derivation of the Poisson kernel of a hyperbolic disc in H bysolving the Dirichlet problem (cid:40)(cid:104) ∂ ∂η + η ∂∂η + η ∂ ∂α (cid:105) u ( η, α ; ¯ η, ¯ α ) = 0 , < η < ¯ η < ∞ ,u (¯ η, α ; ¯ η, ¯ α ) = δ ( α − ¯ α ) , α, ¯ α ∈ ( − π, π ] . (1.3)The interplay between Dirichlet problems and hitting probabilities in various contexts is outlined, forexample, in Grigor’yan [8]. The explicit solution of (1.3) is u ( η, α ; ¯ η, ¯ α ) = 12 π cosh ¯ η − cosh η cosh η cosh ¯ η − − sinh η sinh ¯ η cos( α − ¯ α ) (1.4)and represents the hitting distribution on the hyperbolic circumference of radius ¯ η for the hyperbolicBrownian motion starting at ( η, α ).We show that for ¯ η → ∞ the distribution (1.4) tends to the Cauchy distribution as was found bymeans of other arguments in Baldi et al. [1].The solution to the Dirichlet problem (1.3) is carried out by two different approaches. One is basedon the direct solution of the hyperbolic Laplace equation and the second one is based on some integralrepresentation of the associated Legendre polynomials.The derivation of the n -dimensional Poisson kernel for n > η and the angle α − ¯ α between the geodesic lines with ends points ( η, α ) and (¯ η, ¯ α ). Its explicit form reads u ( η, α ; ¯ η, ¯ α ) 2 Ω n − Ω n ∞ (cid:88) k =0 (cid:18) kn − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos( α − ¯ α )) sin n − ( α − ¯ α ) , (1.5)where n >
2, 0 < η < ¯ η < ∞ , α − ¯ α ∈ (0 , π ], Ω n = π n Γ( n ) is the surface area of the n -dimensionalEuclidean unit sphere, F ( α, β ; γ, x ) is the hypergeometric function and C ( n ) k ( x ) are the Gegenbauerpolynomials.The Poisson kernel of a ball in the hyperbolic disc D n , n >
2, is obtained in Byczkowski and Malecki[3], formula (16) and must be compared with (1.5) above.Unfortunately formula (1.5) cannot be reduced to a fine form as (1.4). However, for sufficiently smalldomains, we extract from (1.5) the n -dimensional Euclidean Poisson kernel.Section 3 is devoted to the exit probabilities P z { T η < T η } from a hyperbolic annulus of radii η and η . We examine in detail both the planar and the higher dimensional case discussing also the transientbehaviour of hyperbolic Brownian motion.In the last section the hitting probabilities on a spherical circle for a spherical Brownian motionstarting from p = ( ϑ, ϕ ) are considered. In particular the most interesting result here is that P p { B S ( T ¯ ϑ ) ∈ d ¯ ϕ } = 12 π cos ϑ − cos ¯ ϑ − cos θ cos ¯ ϑ − sin ϑ sin ¯ ϑ cos( ϕ − ¯ ϕ ) d ¯ ϕ, < ¯ ϑ < ϑ < π, ϕ, ¯ ϕ ∈ (0 , π ] . H n We study here the Poisson kernel of the circle in the hyperbolic plane H = { ( x, y ) : x ∈ R , y > } endowed with the Riemannian metric d s = d x + d y y , and the distance formula cosh η ( z (cid:48) , z ) = ( x (cid:48) − x ) + y (cid:48) + y yy (cid:48) . (2.1)We denote with η the hyperbolic distance from the origin O = (0 ,
1) of H . The Laplace operator on H in cartesian coordinates reads ∆ = y (cid:18) ∂ ∂x + ∂ ∂y (cid:19) (2.2)(for a proof see, for example, Chavel [6] page 265). It is convenient to write the Laplace operator inhyperbolic coordinates ( η, α ) ∆ = ∂ ∂η + 1tanh η ∂∂η + 1sinh η ∂ ∂α (2.3)(for information on hyperbolic coordinates see Cammarota and Orsingher [4]). The relationship betweenhyperbolic coordinates ( η, α ) and the cartesian coordinates ( x, y ) is given by (cid:40) x = sinh η cos α cosh η − sinh η sin α ,y = η − sinh η sin α . (2.4)By exploiting (2.4), in the paper by Lao and Orsingher [12], the Laplace operator (2.3) is obtained from(2.2).We have now our first theorem. 3 heorem 2.1. Let U = { ( η, α ) : η < ¯ η } be a hyperbolic disc in H with radius ¯ η and center in O , thesolution to the Dirichlet problem (cid:40)(cid:104) ∂ ∂η + η ∂∂η + η ∂ ∂α (cid:105) u ( η, α ; ¯ η, ¯ α ) = 0 , < η < ¯ η < ∞ ,u (¯ η, α ; ¯ η, ¯ α ) = δ ( α − ¯ α ) , α, ¯ α ∈ ( − π, π ] , (2.5) is given by u ( η, α ; ¯ η, ¯ α ) = 12 π cosh ¯ η − cosh η cosh η cosh ¯ η − − sinh η sinh ¯ η cos( α − ¯ α ) . (2.6) Proof
Our proof is based on the classical method of separation of variables. We assume that u ( η, α ; ¯ η, ¯ α ) = E ( η )Θ( α ) (2.7)and we arrive at the following ordinary equations (cid:40) Θ (cid:48)(cid:48) ( α ) + µ Θ( α ) = 0 , sinh η E (cid:48)(cid:48) ( η ) + cosh η sinh η E (cid:48) ( η ) − µ E ( η ) = 0 , (2.8)where µ is an arbitrary constant. The first equation has general solutionΘ( α ) = A cos( µα ) + B sin( µα ) (2.9)and becomes periodic with period 2 π for µ = m ∈ N . The second equation necessitates some furthertreatment. We start with the change of variable w = cosh η which transforms the second equation of(2.8) into (1 − w ) G (cid:48)(cid:48) ( w ) − wG (cid:48) ( w ) − m − w G ( w ) = 0 . (2.10)The general solution to (2.10) can be conveniently written as G ( w ) = C (cid:12)(cid:12)(cid:12)(cid:12) w + 1 w − (cid:12)(cid:12)(cid:12)(cid:12) m/ + C (cid:12)(cid:12)(cid:12)(cid:12) w − w + 1 (cid:12)(cid:12)(cid:12)(cid:12) m/ , m (cid:54) = 0 , (2.11)(see, for example, Polyanin and Zaitsev [14] Section 2.1.2, formula 233 for a = 1, b = − λ = 0 and µ = − m ). From (2.11) we have that E ( η ) = C (cid:18) cosh η + 1cosh η − (cid:19) m/ + C (cid:18) cosh η − η + 1 (cid:19) m/ = C (cid:18) cosh η + 1sinh η (cid:19) m + C (cid:18) cosh η − η (cid:19) m . (2.12)We disregard the first term of (2.12) since our aim is to extract finite-valued and increasing solutions to(2.5), so that we have E ( η ) = C (cid:18) cosh η − η (cid:19) m = C tanh m η . (2.13)In light of (2.7), (2.9) and (2.13) we can write u ( η, α ; ¯ η, ¯ α ) = ∞ (cid:88) m =0 Θ m ( α ) E m ( η ) = A + ∞ (cid:88) m =1 [ A m cos( mα ) + B m sin( mα )] (cid:18) cosh η − η (cid:19) m . (2.14)4f we take the Fourier expansion of the Dirac delta function δ ( α − ¯ α ) = 12 π + 1 π ∞ (cid:88) m =1 cos[ m ( α − ¯ α )]= 12 π + 1 π ∞ (cid:88) m =1 [cos( mα ) cos( m ¯ α ) + sin( mα ) sin( m ¯ α )] , (2.15)by comparing (2.14) with (2.15) we obtain the Fourier coefficients A m and B m so that we can write u ( η, α ; ¯ η, ¯ α ) = 12 π + 1 π ∞ (cid:88) m =1 [cos( mα ) cos( m ¯ α ) + sin( mα ) sin( m ¯ α )] (cid:18) cosh ¯ η − η (cid:19) − m (cid:18) cosh η − η (cid:19) m = 12 π + 1 π ∞ (cid:88) m =1 cos( m ( α − ¯ α )) (cid:18) cosh ¯ η − η (cid:19) − m (cid:18) cosh η − η (cid:19) m = 12 π (cid:34) ∞ (cid:88) m =1 (cid:20)(cid:18) e i ( α − ¯ α ) sinh ¯ η cosh ¯ η − η − η (cid:19) m + (cid:18) e − i ( α − ¯ α ) sinh ¯ η cosh ¯ η − η − η (cid:19) m (cid:21)(cid:35) = 12 π (cid:16) cosh ¯ η − η (cid:17) − (cid:16) cosh η − η (cid:17) (cid:16) cosh ¯ η − η (cid:17) + (cid:16) cosh η − η (cid:17) − cosh ¯ η − η cosh η − η cos( α − ¯ α ) (2.16)= 12 π tanh η − tanh η tanh η + tanh η − ¯ η tanh η cos( α − ¯ α ) . The expression in (2.16) can be substantially simplified by observing that:(cosh ¯ η − sinh η − (cosh η − sinh ¯ η = (cosh ¯ η − η − η − η ]and (cosh ¯ η − sinh η + (cosh η − sinh ¯ η − η − η −
1) sinh ¯ η sinh η cos( α − ¯ α )= (cosh ¯ η − η − η cosh ¯ η − − η sinh ¯ η cos( α − ¯ α )] . In view of all these calculations we have that the hyperbolic Poisson kernel takes the form u ( η, α ; ¯ η, ¯ α ) = 12 π cosh ¯ η − cosh η cosh η cosh ¯ η − − sinh η sinh ¯ η cos( α − ¯ α ) . (cid:4) Remark 2.1.
It is possible to obtain the expression (2.14) by means of an alternative approach as follows.We start from the associated Legendre equation(1 − z ) y (cid:48)(cid:48) ( z ) − zy (cid:48) ( z ) + (cid:20) ν ( ν + 1) − m − z (cid:21) y ( z ) = 0 (2.17)which coincides with (2.10) for ν = 0 or ν = −
1. In view of Gradshteyn and Ryzhik [9] formula 8.711.2,the solution to (2.17) can be written as P mν ( z ) = ( − m π Γ( ν + 1)Γ( ν − m + 1) (cid:90) π cos( mϕ )( z + √ z − ϕ ) ν +1 d ϕ, | arg z | < π . ν = − P m − (cosh η ) = ( − m π − m ) (cid:90) π cos( mφ )d φ = 0 , for m ∈ Z . If ν = 0 we have P m (cosh η ) = ( − m π − m ) (cid:90) π cos( mφ )cosh η + sinh η cos φ d φ (cid:40) = 0 , for m = 1 , , . . . (cid:54) = 0 , for m = 0 , − , − , . . . It follows that u ( η, α ; ¯ η, ¯ α ) = (cid:88) m = −∞ [ A m cos( mα ) + B m sin( mα )] P m (cosh η )= 1 π (cid:88) m = −∞ ( − m Γ(1 − m ) [ A m cos( mα ) + B m sin( mα )] (cid:90) π cos( mφ )cosh η + sinh η cos φ d φ = 1 π ∞ (cid:88) m =0 ( − m [ A m cos( mα ) + B m sin( mα )] (cid:90) π cos( mφ )cosh η + sinh η cos φ d φ, (2.18)in the last step A n and B n include the multiplicative constant − m ) . Since we have1 + 2 ∞ (cid:88) n =1 (cid:18) − cosh η sinh η (cid:19) n cos( nφ ) = 1cosh η + sinh η cos φ , (2.19)by inserting (2.19) into (2.18) we get u ( η, α ; ¯ η, ¯ α )= 1 π ∞ (cid:88) m =0 ( − m [ A m cos( mα ) + B m sin( mα )] (cid:90) π cos( mφ ) (cid:32) ∞ (cid:88) n =1 (cid:18) − cosh η sinh η (cid:19) n cos( nφ ) (cid:33) d φ = 2 π ∞ (cid:88) m =0 ( − m [ A m cos( mα ) + B m sin( mα )] ∞ (cid:88) n =1 (cid:18) − cosh η sinh η (cid:19) n (cid:90) π cos( mφ ) cos( nφ )d φ = A + ∞ (cid:88) m =1 [ A m cos( mα ) + B m sin( mα )] (cid:18) cosh η − η (cid:19) m and thus we retrieve (2.14). Remark 2.2.
By applying the hyperbolic Carnot formula we note that it is possible to write the hyper-bolic Poisson kernel (2.6) in a new form. We construct a hyperbolic triangle with sides of length η , ¯ η andˆ η , and angle between the two sides of length η and ¯ η equal to θ = α − ¯ α , see Figure 1. The hyperbolicCarnot formula cosh ˆ η = cosh η cosh ¯ η − sinh η sinh ¯ η cos( α − ¯ α ) , permits us to write (2.6) as u ( η, α ; ¯ η, ¯ α ) = 12 π cosh ¯ η − cosh η cosh ˆ η − , (2.20)where the dependence of u from α and ¯ α is hidden in ˆ η .6 O (η,α) (η,α)α-α η Figure 1: Hyperbolic triangle in H with sides of length η , ¯ η and ˆ η . Remark 2.3.
We observe that the hyperbolic Poisson kernel (2.6) is a proper probability law. In fact: • It is non-negative because, for η > ¯ η , we have cosh ¯ η − cosh η > η cosh ¯ η − − sinh η sinh ¯ η cos( α − ¯ α ) = cosh ˆ η − > . • It integrates to one since it is well-known that (cid:90) π d θa + b cos θ = 2 π √ a − b (2.21)where, in this case, a = cosh η cosh ¯ η − b = − sinh η sinh ¯ η . Remark 2.4.
The kernel appearing in formulas (2.6) and (2.20) represents the law of the positionoccupied by the hyperbolic Brownian motion { B H ( t ) : t ≥ } on H starting from z = ( η, α ) ∈ H whenit hits for the first time the boundary ∂U of the hyperbolic disc U . In other words P z { B H ( T ¯ η ) ∈ d¯ α } = 12 π cosh ¯ η − cosh η cosh η cosh ¯ η − − sinh η sinh ¯ η cos( α − ¯ α ) d¯ α, ¯ α ∈ [0 , π ) , where T ¯ η = inf { t > B H ( t ) ∈ ∂U } , see Figure 2. Remark 2.5.
For small values of η and ¯ η the hyperbolic Poisson kernel (2.6) is approximated by theEuclidean Poisson kernel u ( η, α ; ¯ η, ¯ α ) ∼ π ¯ η − (1 + η )(1 + ¯ η )(1 + η ) − − η ¯ η cos( α − ¯ α ) = 12 π ¯ η − η ¯ η + η − η ¯ η cos( α − ¯ α )that represents the law of the position occupied by the Euclidean Brownian motion { B ( t ) , t ≥ } on R starting from a point z = ( η, α ) when it hits for the first time the boundary ∂U of the Euclidean disc U = { ( η, α ) , η < ¯ η } with Euclidean radius ¯ η . This is a consequence of the fact that in sufficiently smalldomains of the Lobatchevskian space, the Euclidean geometry is in force.7 O (η,α) (η,α)α-α Figure 2: Brownian motion on H starting at ( η, α )and hitting the boundary of the hyperbolic disc U . ˆ O (η,α) η1 η2 Figure 3: Hyperbolic Brownian motion starting in-side the hyperbolic annulus A with radii η and η . Remark 2.6.
We also note that: • For η = 0 formula (2.6) becomes the uniform distribution as expected. • For ¯ η → ∞ we have that˜ u ( η, α ; ¯ α ) := lim ¯ η →∞ u ( η, α ; ¯ η, ¯ α ) = 12 π η − sinh η cos( α − ¯ α ) . (2.22)In view of (2.19), the limiting distribution (2.22) can also be written as˜ u ( η, α ; ¯ α ) = 12 π η − sinh η cos( α − ¯ α ) = 12 π (cid:34) ∞ (cid:88) n =1 (cid:18) cosh η − η (cid:19) n cos n ( α − ¯ α ) (cid:35) . We note that ˜ u represents the hitting distribution of the hyperbolic Brownian motion, starting at z =( η, α ), on the horizontal axis ∂ H = { (¯ η, ¯ α ) : ¯ η = ∞} = { (¯ x, ¯ y ) : ¯ y = 0 } , see Figure 4. We observe thatthe boundary ∂ H represents the point at infinity of H . We can write the ‘hitting’ probability on ∂ H in the following form P z { B H ( T ∞ ) ∈ d¯ α } = 12 π η − sinh η cos α cos ¯ α − sinh η sin α sin ¯ α d¯ α. (2.23)We write now the distribution (2.23) in cartesian coordinates. In view of (2.4) we have that xy = sinh η cos α, tan α = x + y − x . (2.24)The first relation is an immediate consequence of (2.4) and for a proof of the second equality, see Cam-marota and Orsingher [5]. From (2.1) and (2.24) it follows thatsinh η sin α = (cid:113) cosh η − α √ α = x + y − y . (2.25)8etting ¯ η → ∞ we note, in view of (2.4), that for a point (¯ x, ¯ y ) ∈ ∂ H it holds that (cid:40) ¯ x = cos ¯ α − sin ¯ α , ¯ y = 0 . (2.26)Formula (2.26) implies that ¯ x − cos ¯ α = ¯ x √ − cos ¯ α and this leads to the following relationscos ¯ α = 2¯ x x , sin ¯ α = 1 − ¯ x x . (2.27)In view of (2.24), (2.25) and (2.27) and since d¯ α = x d¯ x , we can write ˜ u ( η, α ; ¯ α )d¯ α in cartesiancoordinates as follows˜ u ( x, y ; ¯ x )d¯ x = 12 π x + y +12 y − xy x x − x + y − y − ¯ x x
21 + ¯ x d¯ x = 1 π y ( x + y + 1)(1 + ¯ x ) − x ¯ x − ( x + y − − ¯ x ) d¯ x = 1 π y ¯ x ( x + y ) − x ¯ x + 1 d¯ x = 1 π y (cid:20) ¯ x (cid:112) x + y − x √ x + y (cid:21) − x x + y + 1 d¯ x = 1 π yx + y (cid:104) ¯ x − xx + y (cid:105) + (cid:104) yx + y (cid:105) d¯ x. (2.28)Formula (2.28) says that the probability that the hyperbolic Brownian motion starting at ( x, y ) ∈ H hits the boundary of H at (¯ x,
0) is Cauchy distributed with scale parameter y (cid:48) = yx + y and positionparameter x (cid:48) = xx + y depending on the starting point. In particular, if the hyperbolic Brownian motionstarts at the origin O of H , we obtain a standard Cauchy. We note that (2.22) can be viewed as a Cauchydensity in hyperbolic coordinates. Remark 2.7.
In view of formula (2.28), we also note that the probability that the hyperbolic Brownianmotion starting at z = ( x, y ) = ( η, α ) ∈ H hits ∂ H at (¯ x,
0) is equal to the probability that a EuclideanBrownian motion starting at z (cid:48) = ( x (cid:48) , y (cid:48) ) = ( η, α (cid:48) ) hits the x -axis at (¯ x, z and z (cid:48) have the samehyperbolic distance η from the origin but α (cid:48) = − α , see Figure 5. In factcosh η (cid:48) = x ( x + y ) + y ( x + y ) + 1 yx + y = x + y + 12 y = cosh η, tan α (cid:48) = x ( x + y ) + y ( x + y ) − xx + y = 1 − x − y x = − tan α. Formula (2.28) is in accordance with formula (1.2) in Baldi et al. [1]. In this paper the hitting distributionon the horizontal axis, for the hyperbolic Brownian with horizontal and vertical drift, is obtained fromthe hitting distribution on the horizontal lines H a = { ( x, y ) ∈ H : y = a > } when a → Remark 2.8.
The Poisson kernel (2.6) can be conveniently written also in cartesian coordinates byexploiting the relations (2.24), (2.25) and the hyperbolic distance formulacosh η = x + y + 12 y .
9e have that u ( x, y ; ¯ x, ¯ y ) = 12 π ¯ x +¯ y +12¯ y − x + y +12 yx + y +12 y ¯ x +¯ y +12¯ y − − xy ¯ x ¯ y − x + y − y ¯ x +¯ y − y = 1 π (¯ x + ¯ y + 1) y − ( x + y + 1)¯ yx + y + ¯ x + ¯ y − y ¯ y − x ¯ x = 1 π (¯ x + ¯ y ) y − ( x + y )¯ y + y − ¯ y ( x − ¯ x ) + ( y − ¯ y ) . In the special case where ¯ y = 0 the previous expression becomes u ( x, y ; ¯ x,
0) = 1 π (1 + ¯ x ) y ( x − ¯ x ) + y and thus multiplying by x we get the Cauchy density as expected. ˆ O (η,α) (∞,α)α-α Figure 4: Brownian motion on H starting at ( η, α )and hitting the boundary of the hyperbolic plane. ˆ O (x,0)α z=(η,α)-α z’=(η,-α) Figure 5: Hyperbolic Brownian motion starting at z and Euclidean Brownian motion starting at z (cid:48) . Let H n = { z = ( x, y ) : x ∈ R n − , y > } be the n -dimensional hyperbolic plane, n >
2, with origin O = (0 , . . . , ,
1) endowed with the Riemannian metricd s = d x + · · · + d x n − + d y y and the distance formula cosh η ( z (cid:48) , z ) = 1 + || z (cid:48) − z || yy (cid:48) with η := η ( O, z ). The Laplace operator on H n in cartesian coordinates is given by∆ n = y (cid:32) n − (cid:88) i =1 ∂ ∂x i + ∂ ∂y (cid:33) − ( n − y ∂∂y n in hyperbolic coordinates ( η, α ) =( η, α , . . . , α n − ), reads ∆ n = ∂ ∂η + n − η ∂∂η + 1sinh η ∆ S n − (2.29)where ∆ S n − is the Laplace operator on the ( n − η in H n . This result permits us,in Theorem 2.2, to determine the hyperbolic Laplacian of a smooth function f ( η ). The statement of thisresult is given, for example, in Davies [7] page 117 without proof where cosh ρ must be replaced by coth ρ . Lemma 2.1.
For z = ( x, y ) and z (cid:48) = ( x (cid:48) , y (cid:48) ) in H n we have that the hyperbolic distance η ( z, z (cid:48) ) is asolution of ∆ n η ( z, z (cid:48) ) = n − η ( z, z (cid:48) ) . Proof
Since coth(arcosh( x )) = x √ x − and η ( z, z (cid:48) ) = arcosh || x − x (cid:48) || + y + y (cid:48) yy (cid:48) , we have to prove that∆ n η ( z, z (cid:48) ) = ( n − || x − x (cid:48) || + y + y (cid:48) (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] . In fact we have ∂∂y η ( z, z (cid:48) ) = − || x − x (cid:48) || + y (cid:48) − y y (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] , (2.30) ∂ ∂y η ( z, z (cid:48) ) = || x − x (cid:48) || + y (cid:48) − y y (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ]+ 4 || x − x (cid:48) || [ || x − x (cid:48) || + y (cid:48) + y ] (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] . On the other side, for i = 1 , . . . , n −
1, we have ∂∂x i η ( z, z (cid:48) ) = 2( x i − x (cid:48) i ) (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] , (2.31) ∂ ∂x i η ( z, z (cid:48) ) = 2 (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] − x i − x (cid:48) i ) [ || x − x (cid:48) || + y + y (cid:48) ] (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] , n − (cid:88) i =1 ∂ ∂x i η ( z, z (cid:48) ) = 2( n − (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] − || x − x (cid:48) || [ || x − x (cid:48) || + y + y (cid:48) ] (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] . So, finally, we obtain that∆ n η ( z, z (cid:48) ) = y (cid:34) n − (cid:88) i =1 ∂ ∂x i η ( z, z (cid:48) ) + ∂ ∂y η ( z, z (cid:48) ) (cid:35) − ( n − y ∂∂y η ( z, z (cid:48) )11 2( n − y (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ]+ || x − x (cid:48) || + y (cid:48) − y (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ]+ ( n − || x − x (cid:48) || + y (cid:48) − y ] (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ]= ( n − || x − x (cid:48) || + y (cid:48) + y (cid:112) [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] . (cid:4) In view of Lemma 2.1 the following theorem holds:
Theorem 2.2. If f is a smooth function on R , it holds that ∆ n f ( η ) = f (cid:48)(cid:48) ( η ) + n − η f (cid:48) ( η ) . Proof
We have ∆ n f ( η ( z, z (cid:48) ))= y (cid:34) n − (cid:88) i =1 ∂ ∂x i f ( η ( z, z (cid:48) )) + ∂ ∂y f ( η ( z, z (cid:48) )) (cid:35) − ( n − y ∂∂y f ( η ( z, z (cid:48) ))= y (cid:34) n − (cid:88) i =1 (cid:32) ∂ f∂η (cid:18) ∂η∂x i (cid:19) + ∂f∂η ∂ η∂x i (cid:33) + ∂ f∂η (cid:18) ∂η∂y (cid:19) + ∂f∂η ∂ η∂y (cid:35) − ( n − y ∂f∂η ∂η∂y = ∂ f∂η y (cid:34) n − (cid:88) i =1 (cid:18) ∂η∂x i (cid:19) + (cid:18) ∂η∂y (cid:19) (cid:35) + ∂f∂η (cid:34) y (cid:32) n − (cid:88) i =1 ∂ η∂x i + ∂ η∂y (cid:33) − ( n − y ∂η∂y (cid:35) = ∂ f∂η y (cid:34) n − (cid:88) i =1 (cid:18) ∂η∂x i (cid:19) + (cid:18) ∂η∂y (cid:19) (cid:35) + ∂f∂η ∆ n η = ∂ f∂η + ∂f∂η ∆ n η, since, in view of formula (2.30) and (2.31), it holds that n − (cid:88) i =1 (cid:18) ∂η∂x i (cid:19) + (cid:18) ∂η∂y (cid:19) = 4 || x − x (cid:48) || y + [ || x − x (cid:48) || + y (cid:48) − y ] y [ || x − x (cid:48) || + ( y + y (cid:48) ) ] [ || x − x (cid:48) || + ( y − y (cid:48) ) ] = 1 y . From Lemma 2.1 we obtain the final result∆ n f ( η ( z, z (cid:48) )) = ∂ f∂η + ∂f∂η ∆ n η = ∂ f∂η + n − η ∂f∂η . (cid:4) We denote with { B H n ( t ) , t ≥ } the hyperbolic Brownian motion on H n with starting point z =( η, α ) ∈ H n where α = ( α , . . . , α n − ) ∈ [0 , π ] n − × [0 , π ), and we assume that z is inside the n -dimensional hyperbolic ball U with hyperbolic radius ¯ η (see Figure 6).12e are interested in obtaining the law of the position occupied by the hyperbolic Brownian motionon H n when it hits the boundary ∂U for the first time.Since the Laplace operator is invariant under rotations (see, for example, Helgason [11] Proposition2.4), without loss of generality we can assume that the starting point is z = ( η, α , , . . . ,
0) and theprocess hits the boundary of the ball U at some point ¯ z = (¯ η, ¯ α , , . . . , α − ¯ α is the anglebetween the vectors z and ¯ z . For a function on the ( n − S n − depending onlyon one angle θ we have ∆ S n − = 1sin n − θ ∂∂θ (cid:18) sin n − θ ∂∂θ (cid:19) = ∂ ∂θ + n − θ ∂∂θ . (2.32)In view of (2.29) and (2.32) we have that the hitting distribution on ∂U is obtained from the solution ofthe following Dirichlet problem (cid:40)(cid:104) ∂ ∂η + n − η ∂∂η + η (cid:16) ∂ ∂α + n − α ∂∂α (cid:17)(cid:105) u ( η, α ; ¯ η, ¯ α ) = 0 , < η < ¯ η < ∞ ,u (¯ η, α ; ¯ η, ¯ α ) = δ ( α − ¯ α ) , α − ¯ α ∈ (0 , π ] . (2.33) Theorem 2.3.
The solution to the Dirichlet problem (2.33) is given by u ( η, α ; ¯ η, ¯ α )= Ω n − Ω n ∞ (cid:88) k =0 (cid:18) kn − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos( α − ¯ α )) sin n − ( α − ¯ α ) , (2.34) where n > , < η < ¯ η < ∞ and α − ¯ α ∈ (0 , π ] . Proof
As in Theorem 2.1 our proof is based on the method of separation of variables. We assume that u ( η, α ; ¯ η, ¯ α ) = Θ( α ) E ( η ) . Since we have thatΘ( α ) E (cid:48)(cid:48) ( η ) + Θ( α ) n − η E (cid:48) ( η ) + E ( η )sinh η (cid:20) Θ (cid:48)(cid:48) ( α ) + n − α Θ (cid:48) ( α ) (cid:21) = 0 , there exists a constant µ such that (cid:40) Θ (cid:48)(cid:48) ( α ) + ( n −
2) cot α Θ (cid:48) ( α ) + µ Θ( α ) = 0 , sinh η E (cid:48)(cid:48) ( η ) + ( n −
1) cosh η sinh η E (cid:48) ( η ) − µ E ( η ) = 0 . (2.35)The first equation in (2.35) can be reduced to the Gegenbauer equation. With the change of variable ω = cos α and for µ = k ( k + n − − ω ) G (cid:48)(cid:48) ( ω ) − ( n − ωG (cid:48) ( ω ) + k ( k + n − G ( ω ) = 0 . (2.36)The Gegenbauer polynomials C ( n − ) k ( ω ) satisfy (2.36) (see, for example, Polyanin and Zaitsev [14] S.2.11-4or Helgason [11] page 16) and this implies thatΘ( α ) = AC ( n − ) k (cos α ) . (2.37)13e transform the second equation of (2.35)sinh ηE (cid:48)(cid:48) ( η ) + ( n −
1) cosh η sinh ηE (cid:48) ( η ) − µ E ( η ) = 0 (2.38)into a hypergeometric equation. The first step is based on the change of variable ζ = tanh η . We havethat dd η = 12 cosh η dd ζ , d d η = 14 cosh η d d ζ − sinh η η dd ζ . By taking into account that sinh η = 2 sinh η cosh η and that cosh η = 2 cosh η −
1, equation (2.38)becomes4 sinh η η (cid:34)
14 cosh η d d ζ − sinh η η dd ζ (cid:35) E ( ζ )+ ( n − η η (cid:16) η − (cid:17)
12 cosh η dd ζ E ( ζ ) − µ E ( ζ ) = 0 . And since − η cosh η + ( n −
1) sinh η cosh η (cid:16) η − (cid:17) = tanh η (cid:104) − η n − (cid:16) η − (cid:17)(cid:105) = tanh η (cid:104) n − (cid:16) η − (cid:17)(cid:105) = tanh η (cid:34) n −
2) 1 + tanh η − tanh η (cid:35) , we can write equation (2.38) astanh η E (cid:48)(cid:48) ( ζ ) + tanh η (cid:34) n −
2) 1 + tanh η − tanh η (cid:35) E (cid:48) ( ζ ) − µ E ( ζ ) = 0 , that is ζ E (cid:48)(cid:48) ( ζ ) + ζ (cid:20) n −
2) 1 + ζ − ζ (cid:21) E (cid:48) ( ζ ) − µ E ( ζ ) = 0 . (2.39)We now assume that E ( ζ ) = ζ k f ( ζ ) . Since E (cid:48) ( ζ ) = kζ k − f ( ζ ) + 2 ζ k +1 f (cid:48) ( ζ ) , (2.40) E (cid:48)(cid:48) ( ζ ) = k ( k − ζ k − f ( ζ ) + 2 kζ k f (cid:48) ( ζ ) + 2( k + 1) ζ k f (cid:48) ( ζ ) + 4 ζ k +2 f (cid:48)(cid:48) ( ζ )= k ( k − ζ k − f ( ζ ) + 2(2 k + 1) ζ k f (cid:48) ( ζ ) + 4 ζ k +2 f (cid:48)(cid:48) ( ζ ) , (2.41)by replacing (2.40) and (2.41) into (2.39) (with µ = k ( k + n − k ( k − ζ k f ( ζ ) + 2(2 k + 1) ζ k +2 f (cid:48) ( ζ ) + 4 ζ k +4 f (cid:48)(cid:48) ( ζ )14 (cid:20) n −
2) 1 + ζ − ζ (cid:21) (cid:2) kζ k f ( ζ ) + 2 ζ k +2 f (cid:48) ( ζ ) (cid:3) − k ( k + n − ζ k f ( ζ ) = 0and with obvious simplifications we have that k ( k − f ( ζ ) + 2(2 k + 1) ζ f (cid:48) ( ζ ) + 4 ζ f (cid:48)(cid:48) ( ζ )+ (cid:20) n −
2) 1 + ζ − ζ (cid:21) (cid:2) kf ( ζ ) + 2 ζ f (cid:48) ( ζ ) (cid:3) − k ( k + n − f ( ζ )= 4 ζ f (cid:48)(cid:48) ( ζ ) + 2 ζ (cid:20) k + 1) + ( n −
2) 1 + ζ − ζ (cid:21) f (cid:48) ( ζ ) + 2( n − k ζ − ζ f ( ζ ) = 0 . After some additional manipulations we arrive at the following equation ζ (1 − ζ ) f (cid:48)(cid:48) ( ζ ) + (cid:104) k + n − (cid:16) k + 2 − n (cid:17) ζ (cid:105) f (cid:48) ( ζ ) + k (cid:16) n − (cid:17) f ( ζ ) = 0 . (2.42)Equation (2.42) coincides with the hypergeometric equation t (1 − t ) f (cid:48)(cid:48) ( t ) + [ γ − ( α + β + 1) t ] f (cid:48) ( t ) − αβf ( t ) = 0for t = ζ , α = k , β = 1 − n and γ = k + n . In view of the position E ( ζ ) = ζ k f ( ζ ) and ζ = tanh η , weconclude that a solution to (2.38) is given by E ( η ) = tanh k η F (cid:16) k, − n k + n η (cid:17) . (2.43)Equations (2.37) and (2.43) imply that u ( η, α ; ¯ η, ¯ α ) = ∞ (cid:88) k =0 E k ( η )Θ k ( α )= ∞ (cid:88) k =0 A k tanh k η F (cid:16) k, − n k + n η (cid:17) C ( n − ) k (cos α ) . In order to determine the coefficients A k by applying the boundary conditions we have that u (¯ η, α ; ¯ η, ¯ α ) = δ ( α − ¯ α ) = ∞ (cid:88) k =0 A k tanh k ¯ η F (cid:16) k, − n k + n ¯ η (cid:17) C ( n − ) k (cos α ) . By multiplying both members by C ( n − ) m (cos α ) sin n − α and then integrating we have that (cid:90) π δ ( α − ¯ α ) C ( n − ) m (cos α ) sin n − α d α = ∞ (cid:88) k =0 A k tanh k ¯ η F (cid:16) k, − n k + n ¯ η (cid:17) (cid:90) π C ( n − ) k (cos α ) C ( n − ) m (cos α ) sin n − α d α = A m tanh m ¯ η F (cid:16) m, − n m + n ¯ η (cid:17) π − n Γ( m + n − m ! (cid:0) m + n − (cid:1) Γ (cid:0) n − (cid:1) , because the functions C ( n ) k ( x ) form an orthogonal system on the interval x ∈ ( − ,
1) (see Gradshteynand Ryzhik [9] formula 7.313). This implies that A m = C ( n − ) m (cos ¯ α ) sin n − ¯ α tanh m ¯ η F (cid:0) m, − n ; m + n ; tanh η (cid:1) m ! (cid:0) m + n − (cid:1) Γ (cid:0) n − (cid:1) π − n Γ( m + n − .
15e finally obtain u ( η, α ; ¯ η, ¯ α )= Γ (cid:0) n − (cid:1) sin n − ¯ α − n π ∞ (cid:88) k =0 k ! (cid:0) k + n − (cid:1) Γ( k + n −
2) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos ¯ α ) C ( n − ) k (cos α ) . By rotational invariance and since C ( n − ) k (1) = (cid:0) n + k − k (cid:1) , the last expression reduces to u ( η, α ; ¯ η, ¯ α )= Γ (cid:0) n − (cid:1) sin n − ( α − ¯ α )2 − n ( n − π ∞ (cid:88) k =0 (cid:18) k + n − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos( α − ¯ α )) . We arrive at formula (2.34) by observing thatΓ (cid:0) n − (cid:1) − n ( n − π = 2 n − n − Ω n where Ω n = π n/ Γ( n/ is the surface area of the n -dimensional Euclidean unit sphere. In fact, since2 − n √ π Γ( n −
1) = Γ( n )Γ( n − ), we haveΓ (cid:0) n − (cid:1) − n ( n − π = 12 − n ( n − π (cid:0) n − (cid:1) (cid:0) n − (cid:1) Γ (cid:18) n − (cid:19) = 12 − n ( n − π Γ (cid:0) n (cid:1) (cid:0) n − (cid:1) = 2 n − (cid:0) n (cid:1) √ π Γ (cid:0) n (cid:1) − n √ π Γ( n −
1) = 2 n − (cid:0) n (cid:1) √ π (cid:0) n − (cid:1) = 2 n − n − Ω n and this concludes the proof of the theorem. (cid:4) Remark 2.9.
We note that for small values of η and ¯ η we obtain the Euclidean Poisson kernel. In fact,since tanh η ∼ η , C ( n )1 ( t ) = 2 nt , C ( n )0 ( t ) = 1 and kC ( n ) k ( t ) = 2 n [ tC ( n +1) k − ( t ) − C ( n +1) k − ( t )], we have that ∞ (cid:88) k =0 (cid:18) kn − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos( α − ¯ α )) ∼ ∞ (cid:88) k =0 (cid:18) kn − (cid:19) (cid:18) η ¯ η (cid:19) k C ( n − ) k (cos( α − ¯ α ))= 2 n − (cid:34) ∞ (cid:88) k =2 k (cid:18) η ¯ η (cid:19) k C ( n − ) k (cos( α − ¯ α )) + ( n − η ¯ η cos( α − ¯ α )+ n − ∞ (cid:88) k =0 (cid:18) η ¯ η (cid:19) k C ( n − ) k (cos( α − ¯ α )) (cid:35) = 2 (cid:34) cos( α − ¯ α ) ∞ (cid:88) k =2 (cid:18) η ¯ η (cid:19) k C ( n ) k − (cos( α − ¯ α )) 16 ∞ (cid:88) k =2 (cid:18) η ¯ η (cid:19) k C ( n ) k − (cos( α − ¯ α )) + η ¯ η cos( α − ¯ α ) + 12 ∞ (cid:88) k =0 (cid:18) η ¯ η (cid:19) k C ( n − ) k (cos( α − ¯ α )) (cid:35) = 2 (cid:34) η ¯ η cos( α − ¯ α ) ∞ (cid:88) k =0 (cid:18) η ¯ η (cid:19) k C ( n ) k (cos( α − ¯ α )) − (cid:18) η ¯ η (cid:19) ∞ (cid:88) k =0 (cid:18) η ¯ η (cid:19) k C ( n ) k (cos( α − ¯ α )) + 12 ∞ (cid:88) k =0 (cid:18) η ¯ η (cid:19) k C ( n − ) k (cos( α − ¯ α )) (cid:35) = 2 (cid:34)(cid:18) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:19) − n (cid:18) η ¯ η cos( α − ¯ α ) − η ¯ η (cid:19) + 12 (cid:18) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:19) − n − (cid:35) = 2 (cid:18) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:19) − n (cid:20) η ¯ η cos( α − ¯ α ) − η ¯ η + 12 (cid:18) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:19)(cid:21) = (cid:18) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:19) − n (cid:18) − η ¯ η (cid:19) = 1 − η ¯ η (cid:16) − η ¯ η cos( α − ¯ α ) + η ¯ η (cid:17) n . Remark 2.10.
The kernel (2.34) represents the marginal, with respect to ¯ α , . . . , ¯ α n − , of the distributionof the position occupied by the hyperbolic Brownian motion { B H n ( t ) , t ≥ } starting from z = ( η, α ) ∈ H n when it hits for the first time the boundary ∂U of the n -dimensional hyperbolic hypersphere of radius ¯ η .For z = ( η, ), such distribution is given by P z { B H n ( T ¯ η ) ∈ d ¯ α } = ∞ (cid:88) k =0 (cid:18) kn − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) C ( n − ) k (cos ¯ α ) f ( ¯ α )d ¯ α , (2.44)where n > η < ¯ η , ¯ α ∈ [0 , π ) is the angle between z and ¯ z , and f ( ¯ α ) = 1Ω n sin n − ¯ α sin n − ¯ α . . . sin ¯ α n − is the uniform density on S n − . Remark 2.11.
We observe that (2.44) is a proper probability law. In fact: • The non negativity is due to the non negativity of solutions of Dirichlet problems with non-negativeboundary conditions. • It integrates to one, in fact (cid:90) π . . . (cid:90) π (cid:90) π P z { B H n ( T ¯ η ) ∈ d ¯ α } = Ω n − Ω n ∞ (cid:88) k =0 (cid:18) kn − (cid:19) tanh k η F (cid:0) k, − n ; k + n ; tanh η (cid:1) tanh k ¯ η F (cid:0) k, − n ; k + n ; tanh η (cid:1) (cid:90) π C ( n − ) k (cos ¯ α ) sin n − ¯ α d¯ α = Ω n − Ω n F (cid:0) , − n ; n ; tanh η (cid:1) F (cid:0) , − n ; n ; tanh η (cid:1) (cid:90) π C ( n − )0 (cos ¯ α ) sin n − ¯ α d¯ α
17 Ω n − Ω n (cid:90) π sin n − ¯ α d¯ α = 1 , since, if k >
0, we have (cid:90) π C ( n ) k (cos θ ) sin n − θ d θ = 0(see Gradshteyn and Ryzhik [9] formula 7.311.1) and F (0 , β ; γ ; z ) = 1, C n ( x ) = 1, (cid:90) π sin n − θ d θ = B (cid:18) , n − (cid:19) . Remark 2.12.
We also note that: • For η → S n − as expected. • For ¯ η → ∞ , since tanh ¯ η → F ( α, β ; γ ; 1) = Γ( γ )Γ( γ − α − β )Γ( γ − α )Γ( γ − β ) if γ > α + β (see Gradshteyn andRyzhik [9] formula 9.122.1), we have thatlim ¯ η →∞ P z { B H n ( T ¯ η ) ∈ d ¯ α } = ∞ (cid:88) k =0 (cid:18) kn − (cid:19) Γ( k + n − k + n ) tanh k η F (cid:16) k, − n k + n η (cid:17) C ( n − ) k (cos ¯ α ) f ( ¯ α )d ¯ α . Remark 2.13.
Byczkowski et al. in [2] provide an integral formula for the hyperbolic Poisson kernel ofthe half-space H a = { ( x, y ) ∈ H n : y > a } for n > a >
0, and show that for a →
0, it converges to theCauchy-type distribution Γ( n − π n − Γ( n − ) (cid:18) yy + | x | (cid:19) n − . H n Suppose the hyperbolic Brownian motion { B H ( t ) , t ≥ } starts at z = ( η, α ) ∈ H inside the hyperbolicannulus A with radii 0 < η < η < ∞ A = { ( η, α ) : η < η < η } (see Figure 3). We define the hitting times T η i = inf { t > η ( O, B H ( t )) = η i } , i = 1 , , and T = T η ∧ T η . In the next theorem we evaluate the exit probabilities P z { T η < T η } . Since these aregiven in terms of harmonic functions on the annulus A , they are closely related to the Dirichlet problem. Theorem 3.1.
Let { B H ( t ) : t ≥ } be a hyperbolic Brownian motion starting at z = ( η, α ) ∈ A . Thefollowing result holds true P z { T η < T η } = log tanh η − log tanh η log tanh η − log tanh η , η < η < η . (3.1)18 roof Since the probability in (3.1) is spherically symmetric we are lead to study the solution v : ( η , η ) → R to the Laplace equation involving only the radial part: (cid:20) ∂ ∂η + 1tanh η ∂∂η (cid:21) v ( η ) = 0subjected to the boundary conditions v ( η ) = 1 and v ( η ) = 0. With the change of variable w = cosh η we immediately get (1 − w ) K (cid:48)(cid:48) ( w ) − wK (cid:48) ( w ) = 0 , (3.2)whose general solution is K ( w ) = C + C log (cid:12)(cid:12)(cid:12)(cid:12) w − w + 1 (cid:12)(cid:12)(cid:12)(cid:12) (see, for example, Polyanin and Zaitsev [14], Section 2.1.2 Formula 233 for a = 1, b = − λ = 0 and µ = 0). It follows that v ( η ) = C + C log (cid:18) cosh η − η + 1 (cid:19) = C + C log tanh η . (3.3)By imposing the boundary conditions we get P z { T η < T η } = v ( η ) − v ( η ) v ( η ) − v ( η ) = log tanh η − log tanh η log tanh η − log tanh η . (cid:4) Starting from (3.1) and letting η go to infinity we have that Theorem 3.1 leads to the followingcorollary. Corollary 3.1.
For any z = ( η, α ) outside the hyperbolic disc of radius η and center O , we have P z { T η < ∞} = log (cid:16) cosh η − η +1 (cid:17) log (cid:16) cosh η − η +1 (cid:17) = log tanh η log tanh η , η < η. (3.4)It is possible to show with simple computations that the functions in (3.1) and (3.4) are genuineprobabilities since they vary in (0 , It is possible to generalize the exit probabilities from a hyperbolic annulus to the case of the n -th dimen-sional hyperbolic Brownian motion.In order to evaluate the exit probabilities from the hyperbolic annulus A in H n , with hyperbolic radii η and η with η < η , we are interested in obtaining a solution v n : ( η , η ) → R to the radial part of thehyperbolic Laplace equation in H n . We have proved in Lemma 2.1 and Theorem 2.2 that it is equivalentto solve (cid:20) d d η + n − η dd η (cid:21) v n ( η ) = 0 . (3.5)In what follows we will assume that c ( n,
0) = 1 , c ( n, k ) = ( n − n − · · · ( n − k − n − n − · · · ( n − k − , k = 1 , . . . n − . heorem 3.2. For a hyperbolic Brownian motion { B H n ( t ) : t ≥ } started at z = ( η, α ) ∈ A , we have thatFor n = 3 , , , . . . P z { T η < T η } = (cid:80) n − k =0 ( − k − c ( n, k ) (cid:104) cosh η sinh n − k − η − cosh η sinh n − k − η (cid:105)(cid:80) n − k =0 ( − k − c ( n, k ) (cid:104) cosh η sinh n − k − η − cosh η sinh n − k − η (cid:105) . (3.6) For n = 4 , , , . . . P z { T η < T η } = (cid:80) n − k =0 ( − k − c ( n, k ) (cid:104) cosh η sinh n − k − η − cosh η sinh n − k − η (cid:105) + ( − n − ( n − n − log tanh η tanh η (cid:80) n − k =0 ( − k − c ( n, k ) (cid:104) cosh η sinh n − k − η − cosh η sinh n − k − η (cid:105) + ( − n − ( n − n − log tanh η tanh η . Proof
The general solution to equation (3.5) is given by v n ( η ) = C + C (cid:90) n − η d η. For n = 2 m + 1, m = 1 , , . . . we have v n ( η ) = C + C (cid:90) m η d η = C + C cosh η m − (cid:34) − m − η + m − (cid:88) k =1 ( − k − k ( m − m − · · · ( m − k )(2 m − m − · · · (2 m − k −
1) 1sinh m − k − η (cid:35) = C + C n − (cid:88) k =0 ( − k − C ( n, k ) cosh η sinh n − k − η (3.7)(see Gradshteyn and Ryzhik [9] formula 2.416.2).For n = 2 m + 2, m = 1 , , . . . we have v n ( η ) = C + C (cid:90) m +1 η d η = C + C cosh η m (cid:34) − m η + m − (cid:88) k =1 ( − k − (2 m − m − · · · (2 m − k + 1)2 k ( m − m − · · · ( m − k ) 1sinh m − k η (cid:35) + C ( − m (2 m − m )!! log tanh η C + C n − (cid:88) k =0 ( − k − C ( n, k ) cosh η sinh n − k − η + ( − n − ( n − n − η (3.8)(see Gradshteyn and Ryzhik [9] formula 2.416.3). With computations analogous to those performed inthe two dimensional case, we obtain the statement. (cid:4) From this it follows immediately that: 20 orollary 3.2.
For z = ( η, α ) outside the hyperbolic ball in H n with radius η and center in O , we havethatFor n = 3 , , , . . . P z { T η < ∞} = (cid:80) n − k =0 ( − k c ( n, k ) cosh η sinh n − k − η + ( − n − ( n − n − (cid:104) − cosh η sinh η (cid:105)(cid:80) n − k =0 ( − k c ( n, k ) cosh η sinh n − k − η + ( − n − ( n − n − (cid:104) − cosh η sinh η (cid:105) , η < η. For n = 4 , , , . . . P z { T η < ∞} = (cid:80) n − k =0 ( − k c ( n, k ) cosh η sinh n − k − η + ( − n ( n − n − log tanh η (cid:80) n − k =0 ( − k c ( n, k ) cosh η sinh n − k − η + ( − n ( n − n − log tanh η , η < η. Remark 3.1.
For the space H formula (3.6) takes the simple form P z { T η < T η } = coth η − coth η coth η − coth η , η < η < η , and for η → ∞ yields P z { T η < ∞} = 1 − coth η − coth η < . This shows that there is a positive probability that the hyperbolic Brownian motion never hits the ballof radius η . Remark 3.2.
We note that for small values of η we have cosh η sinh p η ∼ η p and log tanh η ∼ log η . From (3.3),(3.7) and (3.8) it follows that v n ( η ) ∼ (cid:40) C + C log η, if n = 2 ,C + C η − n , if n = 3 , , . . . This means that, for sufficiently small domains, we obtain the exit probabilities of Euclidean Brownianmotion from an annulus: P z { T η < T η } ∼ log η − log η log η − log η , if n = 2 , η − n − η − n η − n − η − n , if n = 3 , , . . . (3.9) Remark 3.3.
It is important to note that for a planar hyperbolic Bownian motion the probability thatthe process goes to infinity before hitting the hyperbolic circle of radius η is strictly less then one P z { T η < ∞} = log tanh η log tanh η < P z { T η < ∞} = 1 . Hyperbolic Brownian motion is, in fact, transient for every dimension n ≥ Hitting distribution on a hyperbolic circle in D D instead of the half-plane model H .The half-plane H can be mapped onto the disc D = { ( r, θ ) : r ∈ [0 , , θ ∈ ( − π, π ] } by means of theconformal mapping f : H → D such that f ( z ) = iz + 1 z + i . (4.1)The x -axis of H is mapped onto ∂ D while the origin O = (0 ,
1) of H is mapped into the origin O = (0 , D . An arbitrary point z = ( x, y ) ∈ H is mapped into a point Q = ( r, θ ) ∈ D such that (cid:40) x = r cos θ r − r sin θ ,y = − r r − r sin θ (4.2)(for details see Lao and Orsingher [12]). In view of (2.4) and (4.2) we have xy = sinh η cos α = 2 r cos θ − r . Since we have that cos α = cos θ and sinh η = 2 r − r , for θ, α ∈ ( − π, π ], we easily arrive at (cid:40) r = cosh η − η = (cid:113) cosh η − η +1 = tanh η ,θ = α. (4.3)The hyperbolic metric and the distance formula in D becomed s = 4(1 − r ) d r , d( O, Q ) = log 1 + r − r . By means of (4.2) the hyperbolic Laplacian in (2.2) is converted into(1 − r ) (cid:20) r ∂∂r (cid:18) r ∂∂r (cid:19) + 1 r ∂ ∂θ (cid:21) , and the Dirichlet problem for the hyperbolic disc U = { ( r, θ ) : r < ¯ r } in D reads (cid:40) (1 − r ) (cid:104) r ∂∂r (cid:0) r ∂∂r (cid:1) + r ∂ ∂θ (cid:105) u ( r, θ ; ¯ r, ¯ θ ) = 0 , < r < ¯ r < ,u ( r, θ ; ¯ r, ¯ θ ) = δ ( θ − ¯ θ ) , θ, ¯ θ ∈ ( − π, π ] . (4.4)Since (1 − r ) >
0, we can derive the Poisson kernel related to the Dirichlet problem (4.4) from theEuclidean case: u ( r, θ ; ¯ r, ¯ θ ) = 12 π ¯ r − r ¯ r + r − r ¯ r cos( θ − ¯ θ ) . (4.5)Alternatively it is possible to obtain formula (4.5) from the Poisson kernel in H with a change ofcoordinates. In fact, in view of (2.16) and (4.3), formula (4.5) immediately follows.22he Poisson kernel in (4.5) represents the law of the position occupied by the hyperbolic Brownianmotion { B D ( t ) : t ≥ } on D starting from Q = ( r, θ ) ∈ D when it hits for the first time the boundary ∂U . We have P Q { B D ( T ¯ r ) ∈ d¯ θ } = 12 π ¯ r − r ¯ r + r − r ¯ r cos( θ − ¯ θ ) d¯ θ. We note that for ¯ r → u ( r, θ ; ¯ θ ) := lim ¯ r → u ( r, θ ; ¯ r, ¯ θ ) = 12 π − r r − r cos( θ − ¯ θ ) , r < . (4.6)Again ˜ u ( r, θ ; ¯ θ ) represents the law of the position occupied by the hyperbolic Brownian motion in D when it hits for the first time the boundary of the hyperbolic disc U with hyperbolic radius that goes toinfinity. Result (4.6) is stated in Helgason [11] page 34. For the n -dimensional case see Byczkowski andMalecki [3] formula (16).In view of (4.3), we can write (4.6) in hyperbolic coordinates as follows12 π − r r − r cos( θ − ¯ θ ) = 12 π + 1 π ∞ (cid:88) n =1 r n cos n ( θ − ¯ θ ) = 12 π + 1 π ∞ (cid:88) n =1 (cid:18) cosh η − η (cid:19) n cos n ( θ − ¯ θ )= 12 π η − sinh η cos( θ − ¯ θ ) , which coincides with (2.22). On the other side it is well-known that under the conformal mapping (4.1)the Poisson kernel (4.6) takes the form of the Cauchy distribution as it is shown in formula (2.28).Since (1 − r ) >
0, the exit probabilities from the hyperbolic annulus A = { ( r, θ ) : r < r < r } areeasily derived from the Euclidean case. If the hyperbolic Brownian motion starts at Q = ( r, θ ) ∈ A , wehave P Q { T r < T r } = log r − log r log r − log r , < r < r < r < . (4.7)Letting r → P Q { T r < ∞} = log r log r < , < r < r < . The surface S of the unit-radius three dimensional sphere is a model of the elliptic geometry if geodesiclines are represented by great circles. We specify the position of an arbitrary point p ∈ S with the couple( ϑ, ϕ ) of spherical coordinates where ϑ ∈ [0 , π ] and ϕ ∈ [0 , π ).If U = { ( ϑ, ϕ ) : ϑ > ¯ ϑ } is the surface of a spherical cap on S with center in the south pole, theDirichlet problem on the surface of the sphere S reads: (cid:40)(cid:104) ∂ ∂ϑ + ϑ ∂∂ϑ + ϑ ∂ ∂ϕ (cid:105) u ( ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = 0 , < ¯ ϑ < ϑ < π,u ( ¯ ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = δ ( ϕ − ¯ ϕ ) , ϕ, ¯ ϕ ∈ [0 , π ) . Assuming that u ( ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = T ( ϑ ) F ( ϕ ) we immediately arrive at the following ordinary equations (cid:40) F (cid:48)(cid:48) ( ϕ ) + µ F ( ϕ ) = 0 , sin ϑ T (cid:48)(cid:48) ( ϑ ) + cos ϑ sin ϑ T (cid:48) ( ϑ ) − µ T ( ϑ ) = 0 , (5.1)23ith µ ∈ R . With the change of variable w = cos θ , in the second equation of (5.1), we arrive at equation(2.10) with general solution (2.11). Therefore, for µ = m ∈ N , the general solution to the second equationof (5.1) can be written as T ( ϑ ) = C (cid:32)(cid:114) ϑ − cos ϑ (cid:33) m + C (cid:32)(cid:114) − cos ϑ ϑ (cid:33) m . (5.2)We restrict ourselves to the increasing component of (5.2) so that we have u ( ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = ∞ (cid:88) m =0 [ A m cos( mϕ ) + B m sin( mϕ )] (cid:32)(cid:114) − cos ϑ ϑ (cid:33) m . By imposing the boundary condition u ( ¯ ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = δ ( ϕ − ¯ ϕ ) and in view of (2.15), we finally obtain that u ( ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = 12 π + 1 π ∞ (cid:88) m =1 cos( m ( ϕ − ¯ ϕ )) (cid:114) − cos ϑ ϑ (cid:115) ϑ − cos ¯ ϑ m = 12 π − − cos ϑ ϑ ϑ − cos ¯ ϑ − cos ϑ ϑ ϑ − cos ¯ ϑ − (cid:113) − cos ϑ ϑ ϑ − cos ¯ ϑ cos( ϕ − ¯ ϕ )= 12 π cos ϑ − cos ¯ ϑ − cos ϑ cos ¯ ϑ − sin ϑ sin ¯ ϑ cos( ϕ − ¯ ϕ ) . (5.3) Remark 5.1.
Since for spherical triangles the following Carnot formula holdscos ˆ ϑ = cos ϑ cos ¯ ϑ + sin ϑ sin ¯ ϑ cos( ϕ − ¯ ϕ ) , we can rewrite (5.3) as follows u ( ϑ, ϕ ; ¯ ϑ, ¯ ϕ ) = 12 π cos ϑ − cos ¯ ϑ − cos ˆ ϑ . (5.4) Remark 5.2.
We note that the Poisson kernel (5.3) is a proper probability law. In fact • In view of (5.4) and observing that ¯ ϑ < ϑ we have that (5.3) is positive. • Applying (2.21) with a = 1 − cos ϑ cos ¯ ϑ and b = − sin ϑ sin ¯ ϑ we obtain that (5.3) integrates to one.For ϑ = 0 we obtain from (5.3) the uniform law, while for ¯ ϑ = π we get u ( ϑ, ϕ ; π , ¯ ϕ ) = 12 π cos ϑ − sin ϑ cos( ϕ − ¯ ϕ ) . Remark 5.3.
Let { B S ( t ) : t ≥ } be a Brownian motion on the surface of the three dimensional sphere S with starting point p = ( ϑ, ϕ ) ∈ S (see Figure 7). The kernel in (5.3) represents the law of the positionoccupied by the spherical Brownian motion when it hits for the first time the boundary of the sphericalcap U .In order to obtain the exit probabilities of { B S ( t ) : t ≥ } from a spherical annulus A = { ( ϑ, ϕ ) : ϑ < ϑ < ϑ } with center in the south pole of S , we consider the solution v : ( ϑ , ϑ ) → R to the Laplaceequation involving only the radial part (cid:20) ∂ ∂ϑ + 1tan ϑ ∂∂ϑ (cid:21) v ( ϑ ) = 0 . O =(0,0,1) α(η,α1,α2) -α(η,α1,α2)(0,0,cosh η) α1-α1 Figure 6: Brownian motion on H starting at ( η, α )and hitting the boundary of the hyperbolic ball ˆ (1,θ,φ) (1,θ,φ) O =(0,0,0) Figure 7: Spherical Brownian motion starting at( ϑ, ϕ ) and hitting the boundary of the spherical discWith the change of variable w = cos ϑ we arrive at equation (3.2). With calculations analog to thoseperformed in the proof of Theorem 3.1 we get P p { T ϑ < T ϑ } = log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) − log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) − log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) , ϑ < ϑ < ϑ . (5.5)In particular for ϑ → π formula (5.5) reads P p { T ϑ < T π } = log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) log (cid:12)(cid:12)(cid:12) cos ϑ − ϑ +1 (cid:12)(cid:12)(cid:12) = log (cid:12)(cid:12) tan ϑ (cid:12)(cid:12) log (cid:12)(cid:12) tan ϑ (cid:12)(cid:12) , π < ϑ < ϑ . Remark 5.4.
We note that replacing formally ϑ with iϑ it is possible to extract from (5.3) and (5.5) thePoisson kernel and the exit probabilities obtained for the hyperbolic plane, namely (2.6) and (3.1). Thisis because H can be viewed formally as a sphere with imaginary radius. For small values of θ we obtaininstead results analogous to these obtained for the Euclidean Brownian motion. In fact, for sufficientlysmall domains, Euclidean geometry is in force. References [1] Baldi, P., Casadio Tarabusi, E., Fig`a-Talamanca, A., Yor, M.: Non-symmetric hitting distributionson the hyperbolic half-plane and subordinated perpetuities.
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