24.1. MOUNTAIN PASS THEOREM IN HILBERT SPACE 809

Taking inner products with η (t,u) , and integrating∫ t

0 for this local solution,

12∥η (t,u)∥2− 1

2∥u∥2 +

∫ t

0g(u)h

(∥∥I′ (Pη (s,u))∥∥)(I′ (Pη (s,u)) ,η (s,u)

)ds = 0

12∥η (t,u)∥2 ≤ 1

2∥u∥2 +

∫ t

0g(u)h

(∥∥I′ (Pη (t,u))∥∥)∥∥I′ (Pη (t,u))

∥∥∥η (s,u)∥ds

It follows that for t ≤ 1,

∥η (t,u)∥2 ≤ ∥u∥2 +2∫ t

0∥η (s,u)∥ds

≤ ∥u∥2 +1+∫ t

0∥η (s,u)∥2 ds

and so from Gronwall’s inequality, for t ≤ 1,

∥η (t,u)∥2 ≤(∥u∥2 +1

)e1

Thus we pick M > e(∥u∥2 +1

)and then we obtain that for t ∈ [0,1] , the projection map

does not change anything. Hence there exists a solution to 24.1.1, 24.1.2 on [0,1] as desired.Then for this solution, η (0,u) = u because of the above initial condition. If u ∈

[I (u) /∈ [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

(∥∥I′ (η)∥∥) I′ (η)

)= −g(u)h

(∥∥I′ (η)∥∥)∥∥I′ (η)

∥∥2

and so this implies the third condition since it says that the function t → I (η (t,u)) isdecreasing.

It remains to consider the last condition. This involves choosing δ still smaller if nec-essary. It is desired to verify that

η (1, [I (u)≤ c+δ ])⊆ [I (u)≤ c−δ ]

Suppose it is not so. Then there exists u ∈ [I (u)≤ c+δ ] but I (η (1,u))> c−δ .

c−δ < I (η (1,u)) = I (u)−∫ 1

0

(I′ (η) ,g(u)h

(∥∥I′ (η)∥∥) I′ (η)

)dt

= I (u)−g(u)∫ 1

0h(∥∥I′ (η)

∥∥)∥∥I′ (η (t,u))∥∥2 dt

< c+δ −g(u)∫ 1

0h(∥∥I′ (η)

∥∥)∥∥I′ (η (t,u))∥∥2 dt

Then

c−2δ +g(u)∫ 1

0h(∥∥I′ (η (t,u))

∥∥)∥∥I′ (η (t,u))∥∥2 dt < c

24.1. MOUNTAIN PASS THEOREM IN HILBERT SPACE 809Taking inner products with 1 (t,w) , and integrating {j for this local solution,= 5 In (e.u) u = 5 ll + [elu h(||Z’ (Pn (s,u))||) (Pn (s,4)), 1 (s,u)) ds =0slin(e, uw)" <5 [lel os h (|| (Pn (t,u))|]) [0 (Pn @,x))|| {In (s,u) ll dsIt follows that fort < 1,tIn (@wIP a? +2 [ In (s,u)|| dst< |ul?+1+ [In (s,u) Pasand so from Gronwall’s inequality, for t < 1,2 2In (cu) |? < (lll? +1) e!Thus we pick M > e (ill? + 1) and then we obtain that for ¢ € [0,1], the projection mapdoes not change anything. Hence there exists a solution to 24.1.1, 24.1.2 on [0, 1] as desired.Then for this solution, 7 (0,u) = u because of the above initial condition. If u €(I (u) ¢ |c—e€,c+e]], then uw € A and so g(u) =0 so n(t,u) =u for all t € [0,1]. Thisgives the first two conditions. Consider the third.dqlntw)) = mon == ('(n),.8 (wh (ll (a) ())2= ~8 (h(i m)|) [||and so this implies the third condition since it says that the function t + I(7 (t,u)) isdecreasing.It remains to consider the last condition. This involves choosing 6 still smaller if nec-essary. It is desired to verify thatn (1, [Z(u) <c4+6]) C [I (u) <c—4]Suppose it is not so. Then there exists u € [I (u) <c+ 6] but /(n (1,u)) >c—6.c-8 < Tin) =10)— [Cn .gn(r |) nyaHuw) eu) [a(\\r |) ie ney |a[eduh(|I'(n)Ac+5—g(u) II (n (t,u)) || atThenc—26-+8(u) | h(|\1' (n (t,u))|]) [IZ (n (t,u)) |? dt <e