Nonlinear Equations - UFRJ

[SEC. 10.2: PROOF OF THEOREM 10.5 145
Lemma 10.12. Assume the conditions of Lemma 10.8, and assume
furthermore that ‖F ti − F ti+1 ‖ = ɛ 1 /µ F (f ti , x i ). Then,
Proof.
L(f t , z t ; t i , t i+1 ) ≥
1
CD 3/2√ n
L(f t , z t ; t i , t i+1 ) =
≥
≥
∫ ti+1
t i
∫ ti+1
t i
µ(f s , z s )‖( f ˙ s , z˙
s )‖ fs,z s
ds
µ(f s , z s )‖ f ˙ s ‖ fs ds
µ
1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ D
∫ ti+1
t i
‖ f ˙ s ‖ fs ds
The rightmost integral evaluates to d Riem (f ti , f ti+1 ). Assume that
tan θ 1 = ‖F ti − F 0 ‖ and tan θ 2 = ‖F ti+1 − F 0 ‖
We know from elementary calculus that
tan θ 2 − tan θ 1
θ 2 − θ 1
≤
1
cos 2 θ 2
= 1 + tan 2 θ 2
Therefore, using tan θ 2 ≤ ‖F 1 − F 0 ‖, we obtain that
Using that bound,
θ 2 − θ 1 ≥ 1 2 ‖F t i+1
− F ti ‖
L(f t , z t ; t i , t i+1 ) ≥ 1 µ
2 1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ D ‖F t i
− F ti+1 ‖
√
2
ɛ 1
≥
D 3/2√ n 1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ D
Numerically, we obtain
√
2
ɛ 1
1 + ɛ 1 + π(1 − ɛ 1 )αr 0 (α)/ √ 2 ≥ C−1 . (10.17)