820 CHAPTER 24. CRITICAL POINTS

where r→ h(r) is a decreasing function which has values in (0,1] and equals 1 for r ∈ [0,1]and equals 1/r for r > 1.

1h(r) = 1

r

h(r) = 1

Here u is given and the η ′ is the time derivative is with respect to t. By Corollary24.1.11, there exists a solution to this for t ∈ [0,1].

Then for this solution, η (0,u) = u because of the above initial condition. If u ∈I−1 (X \ [c− ε,c+ ε]) , then u ∈ A and so g(u) = 0 so η (t,u) = u for all t ∈ [0,1]. Thisgives the first two conditions. Consider the third.

ddt

(I (η (t,u))) =⟨I′ (η) ,η ′

⟩=−

⟨I′ (η) ,g(u)h(∥V (η (t,u))∥)V (η (t,u))

⟩= −g(u)h(∥V (η (t,u))∥)

⟨I′ (η) ,V (η (t,u))

⟩≤ −g(u)h(∥V (η (t,u))∥)

∥∥I′ (η)∥∥2

X ′ ≤ 0

this last inequality from the inequalities of 24.1.5, and so this implies the third conditionsince it says that the function t→ I (η (t,u)) is decreasing.

It remains to consider the last condition. This involves an appropriate choice of smallδ . It was chosen small and now it will be seen how small. It is desired to verify that

η(1, I−1 ((−∞,c+δ ])

)⊆ I−1 ((−∞,c−δ ])

Suppose it is not so. Then there exists u such that I (u) ∈ (c− δ ,c+ δ ] but I (η (1,u)) >c−δ . We can assume that I (u)∈ (c−δ ,c+δ ] because if I (u)≤ c−δ , then so is I (η (1,u))from what was just shown. Hence g(u) = 1. Then using the fact that g(u) = 1,

c−δ < I (η (1,u)) = I (u)+∫ 1

0

ddt

(I (η))dt

= I (u)−∫ 1

0

⟨I′ (η) ,h(∥V (η)∥)V (η)

⟩dt

= I (u)+∫ 1

0−h(∥V (η)∥)

⟨I′ (η) ,V (η)

⟩dt

≤≤c+δ

I (u) +∫ 1

0

(−h(∥V (η)∥)

∥∥I′ (η)∥∥2)

dt

Then

c−δ +∫ 1

0h(∥V (η)∥)

∥∥I′ (η)∥∥2 dt < I (u)≤ c+δ

Thus

c−2δ +∫ 1

0h(∥V (η)∥)

∥∥I′ (η)∥∥2 dt < c