Random Processes in Hyperbolic Spaces Hyperbolic Brownian ...

41 Hyperbolic Brownian Motion
and applying the inverse Fourier transform to both members, we get
∞
−∞
e iλx
2 1 + ξ
f
2
′′
(ξ) + ν + µ + 3
ξ f
2
′
(ξ) + µ + 1
f(ξ) dξ = 0.
2
Since, for k = 0, 1, 2, the function xkf (k) (x) is integrable, we have that the function λku (k) (λ) exists,
is continuous in R and vanishes at infinity. In particular u(λ) is twice continuously differentiable
outside 0. We have
∞
e
−∞
iλx f (k) (x)dx = (−iλ) k u(λ) = (−1) k u (k) (λ),
∞
−∞
∞
−∞
e iλx xf ′ (x)dx = −i ∂
∞
e
∂λ −∞
iλx f ′ (x)dx = −i ∂
∂λ [−iλu(λ)] = −u(λ) − λu′ (λ),
e iλx x 2 f ′′ (x)dx = − ∂2
∂λ2 ∞
e
−∞
iλx f ′′ (x)dx = − ∂2
∂λ2 [(iλ)2u(λ)] = 2u(λ) + 4λu ′ (λ) + λ 2 u ′′ (λ).
It follows that u(λ) is in the kernel of the operator N given by
Nu(λ) = λ2
2 u′′
(λ) − µ − 1
λu
2
′
2 λ
(λ) − + iνλ u(λ).
2
Taking into account the third condition in (3.28) we have u(0) = ∞
−∞ pz(ξ)dξ = 1. With the
change of variable u(λ) = e −λ v(2λ) and ω = 2λ we get
ωv ′′ (ω) + (1 − 2µ − ω)v ′
1
(ω) − − µ + iν v(ω) = 0.
2
Since the confluent hypergeometric differential equation ωv ′′ (ω) + (b − ω)v ′ (ω) − av(ω) = 0 has
solution
Φ(a, b; ω) = 21−b Γ(1 − a)e ω
2
π
π
2
0
ω
cos tan θ + (2a − b)θ cos
2 −b θdθ
where Φ(a, b; ω) is the confluent hypergeometric function of the second kind with Re{b} < 1 and a
not a positive integer (see Gradshteyn and Ryzhik (16) formula 9.216.1), the characteristic function
u(λ) we are looking for is given by
u(λ) = e −λ v(2λ) = Ke −λ
1
Φ − µ + iν, 1 − 2µ; 2λ .
2
for some real constant K. In particular we note that Φ(a, b; ω) has a finite non-zero limit for
ω → 0 + ω − and that limω→∞ e 2 Φ(a, b; ω) = 0. To obtain the value of the constant K we observe
that
1
1 = u(0) = KΦ − µ + iν, 1 − 2µ; 0 = K
2 22µ
π Γ
π
1
2
+ µ − iν cos(2iνθ) cos
2 0
2µ−1 θ dθ.