Constructions of harmonic maps between Hadamard manifolds

4.2. A COUNTER EXAMPLE 79
where (y, η, ξ, x) ∈ R + × R 3 and (Y,X) ∈ R + × R. Note that the vector field ∂/∂x is
mapped to the vector field ∂/∂y via the complex structure of CH 2 . Let Φ : B 2 → CH 2 and
Ψ:D 2 → RH 2 be the Cayley transforms of CH 2 and RH 2 , respectively. Assume that a
map ũ : CH 2 → RH 2 takes the following form:
ũ(y, η, ξ, x) =(f(y),g(x)).
Then the harmonic map equation for ũ is reduced to the following system of ordinary differential
equations:
⎧
⎪⎨ 4y 2 f yy (y) − 4yf y (y) − y 2 f(y) −1 {4f y (y) 2 − g x (x) 2 } =0,
⎪⎩ g xx (x) =0,
where f y = df/dy, f yy = d 2 f/dy 2 ,g x = dg/dx, g xx = d 2 g/dx 2 .
For any a, b ∈ R, we can easily verify that f(y) =(|a|/2 √ 2)y and g(x) =ax + b are
solutions to this system. Therefore we obtain a family of harmonic maps
Let ⎧⎪ ⎨
⎪ ⎩
˜σ (a,b) (y, x) =
ũ (a,b) : CH 2 ∋ (y, η, ξ, x) ↦→
( |a|
2 √ 2 y, ax + b )
∈ RH 2 .
( |a|
2 √ 2 y, a
2 √ 2 x + b )
: RH 2 → RH 2 ,
σ (a,b) =Ψ −1 ◦ ˜σ (a,b) ◦ Ψ:D 2 → D 2 ,
ũ(y, η, ξ, x) =ũ (2
√
2,0) (y, η, ξ, x) =(y, 2 √ 2x) :CH 2 → RH 2 ,
u (a,b) =Ψ −1 ◦ ũ (a,b) ◦ Φ:B 2 → D 2 .
Note that, if a ≠0, each u (a,b) is smooth up to the boundary except for (0, 1) ∈ ∂B 2 . Thus
we need to investigate the boundary behavior of u (a,b) near the point (0, 1). To this end, since
any σ (a,b) is an isometry of D 2 which fixes the point (0, 1) commonly, it suffices to consider
the case where a =2 √ 2 and b =0. Set u =Ψ −1 ◦ ũ
√
(2 2,0)
◦ Φ.