Constructions of harmonic maps between Hadamard manifolds

1.1. LOCAL EXISTENCE OF SOLUTIONS TO (1.1.1) WITH (1.1.2) 27
we present here a proof in our context.
Take τ 1 ∈ (−∞,τ 0 ) arbitrarily, and let A be a positive constant such that P (τ) ≤ A for
any τ ∈ (−∞,τ 1 ]. Fix τ 2 ∈ (−∞,τ 1 − 1] arbitrarily, and let ρ be a unique solution to the
following ordinary differential equation with the initial values:
⎧
d
⎪⎨
2 ρ
dτ (τ)+Adρ (τ) =0,
2 dτ
⎪⎩ ρ(τ 2 )=r(τ 2 )
Then it is easy to see that ρ is given by
and
dρ
dτ (τ 2)= dr
dτ (τ 2).
ρ(τ) = 1 A (1 − e−A(τ−τ 2) ) dr
dτ (τ 2)+r(τ 2 ).
Let
R(τ) =ρ(τ) dr (τ) − r(τ)dρ
dτ dτ (τ).
Then R(τ 2 ) = 0 and (dR/dτ)(τ 2 ) > 0. Moreover, R ≥ 0on(τ 2 ,τ 1 ]. Indeed, if this is not the
case, then there exists a point τ ∗ ∈ (τ 2 ,τ 1 ] such that
R(τ ∗ )=0,
dR
dτ (τ ∗) < 0, and R>0 on (τ 2 ,τ ∗ ).
On the other hand, we have
dR
dτ (τ ∗)=ρ(τ ∗ ) d2 r
dτ (τ ∗) − r(τ 2 ∗ ) d2 ρ
dτ (τ ∗)
2
= ρ(τ ∗ ){−P (τ ∗ ) dr
dτ (τ ∗)+Q(τ ∗ ,r(τ ∗ ))} + Ar(τ ∗ ) dρ
dτ (τ ∗)
≥ A{r(τ ∗ ) dρ
dτ (τ ∗) − ρ(τ ∗ ) dr
dτ (τ ∗)} + Q(τ ∗ ,r(τ ∗ ))ρ(τ ∗ )
= Q(τ ∗ ,r(τ ∗ ))ρ(τ ∗ ) > 0,
which leads to a contradiction. Thus R ≥ 0on[τ 2 ,τ 1 ]. Since r>0 and ρ>0, it holds that
r ≥ ρ on [τ 2 ,τ 1 ].
As a consequence, we obtain
(1.1.7) r(τ 2 +1)− r(τ 2 ) ≥ ρ(τ 2 +1)− ρ(τ 2 )= 1 dr
A dτ (τ 2)(1 − e −A )