Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

39 Hyperbolic Brownian Motion
2 φ
2 cos 2 g(φ), we obtain
2
d d 1 + ξ
f(ξ) + νf(ξ) + µ −
dξ dξ 2
1
ξf(ξ)
2
2 φ d 2 φ d
= 2 cos 2 cos g(φ) + ν + µ −
2 dφ 2 dφ 1
tan
2
φ
2 φ
2 cos
2 2 g(φ)
2 φ d d
= 4 cos g(φ) + ν + µ −
2 dφ dφ 1
tan
2
φ
2 φ
g(φ) cos .
2
2
since d
dξ
= 2 cos2 φ
2
d
dφ
2 φ d
4 cos
2 dφ
and is proportional to g1(φ) that satisfies
. The function g(φ) is solution to the differential equation
d
g(φ) + ν + µ −
dφ 1
tan
2
φ
g(φ) cos
2
= 0
2
2 φ
d
dφ g1(φ)
+ ν + µ − 1
tan
2
φ
−2 φ
g1(φ) = k cos
2
2 .
The function g(φ) has the following form
−F (φ)
g(φ) = c1g1(φ) = c1e
where c, c1 are two real constants and
Then
F (φ) =
c
2 +
φ
0
F (ψ) k
e
cos2 ψ
2
ν + µ − 1
tan
2
φ
dφ = νφ − 2 µ −
2
1
log cos
2
φ
.
2
g(φ) = c1e −νφ
2µ−1 φ c
cos + k
2 2
φ
0
dψ
e νψ −2µ−1 ψ
cos
2 dψ
c, c1, k ∈ R. Since µ > 0 and φ ∈ (−π, π) the integral takes values of either sign. Taking into
account the second condition in (3.28) we have that k = 0 and
g(φ) = c2
2 e−νφ 2µ−1 φ
cos
2
for some real constant c2. Recalling that φ = 2 arctan ξ and f(ξ) = 2
1+ξ 2 g(2 arctan ξ) we have
and finally
g(φ) =
1 + ξ2
2
= c2 arctan ξ
e−2ν
2
f(ξ) = c2
2 e−νφ cos
1
1 + ξ2 f(ξ) = c2
2µ−1 φ
2
2µ−1
−2ν arctan ξ e
(1 + ξ2 1
) µ+ 2
= c2
2 e−2ν arctan ξ cos 2µ−1 (arctan ξ)
By imposing the third condition in (3.28) that is π
g(φ)dφ = 1 we obtain the value of the constant
−π
.