Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

33 Hyperbolic Brownian Motion
equation (3.12), because of its homogeneity and linearity, has thus the form
u(η, t) =
∞
0
T (t, ω)F (η, ω)Q(ω)dω, (3.14)
where Q(ω), ω > 0, is an arbitrary function which is determined by applying the initial condition.
The equation satisfied by F , for ω = −ν(ν + 1), becomes the classical Legendre equation
which is a special case of the hypergeometric equation
(1 − z 2 ) d2u du
− 2z + ν(ν + 1)u = 0 (3.15)
dz2 dz
t(1 − t) d2u du
+ [γ − (α + β + 1)t] − αβu = 0, (3.16)
dt2 dt
for z = 1 − 2t, α = −ν, β = ν + 1 and γ = 1. A solution to (3.16) is the hypergeometric function
2F1(α, β, γ; t) =
=
=
∞ (α)k(β)k
t
k!(γ)k
k
∞ α(α + 1)...(α + k − 1)β(β + 1)...(β + k − 1)
t
k!γ(γ + 1)...(γ + k − 1)
k
∞ Γ(α + k)Γ(β + k) Γ(γ)
k!Γ(γ + k) Γ(α)Γ(β) tk , |t| < 1.
k=0
k=0
k=0
This means that a solution to the Legendre equation (3.15) is the Legendre polynomial
Pν(z) = 2F1
=
∞
k=0
∞
k 1 − z (−ν)k(ν + 1)k 1 − z
−ν, ν + 1, 1; =
2
k!(1)k 2
k=0
(−ν)k(ν + 1)k
(k!) 2
k 1 − z
, |1 − z| < 2. (3.17)
2
By following Lebedev (27), page 165, it is possible to represent the Legendre polynomials in a more
suitable form by considering that
By substituting (3.18) in (3.17) we get that
Pν(z) =
π
2 2
sin
π 0
2k
1
2 k
ϕ dϕ = . (3.18)
Γ(k + 1)
∞ (−ν)k(ν + 1)k
1
k=0 2 k k!
k 1 − z 2
2 π
= 2
π ∞ 2 (−ν)k(ν + 1)k
π
1
0
2 k k!
sin 2 ϕ
k=0
= 2
π
2
2F1
π 0
−ν, ν + 1, 1
2 ; sin2 ϕ
π
2
0
1 − z
2
sin 2k ϕ dϕ
k 1 − z
dϕ
2
dϕ. (3.19)