Take

= {v ∈ N : v ¤ R} ∪ {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R},

A

= ‚ Bρ © M, 0 < ρ < R,

B

where

Bσ = {x ∈ X : x < σ }.

X

By Example 3 of Section 3.4, A links B. Moreover, if R is suf¬ciently large,

sup G ¤ 0 ¤ inf G.

(17.33)

B

A

Hence, we may apply Theorem 3.4 to conclude that there is a sequence {x (k) } ‚ X

such that

G(x (k) ) = x (k) V (t, x (k) (t)) dt ’ c ≥ 0,

™ ’2

2

(17.34)

I

(G (x (k) ), z)/2 = (x (k) , z ) ’ ∇x V (t, x (k) ) · z(t) dt ’ 0,

™ ™ z ∈ X,

(17.35)

I

17.3. Nonconstant solutions 223

and

(G (x (k) ), x (k) )/2 = x (k) ∇x V (t, x (k) ) · x (k) dt ’ 0.

™ ’

2

(17.36)

I

If

ρk = x (k) ¤ C,

X

then there is a renamed subsequence such that x (k) converges to a limit x ∈ X weakly

in X and uniformly on I. From (17.35), we see that

(G (x), z)/2 = (x, z ) ’

™™ ∇x V (t, x(t)) · z(t) dt = 0, z ∈ X,

I

from which we conclude easily that x is a solution of (17.1). By (17.34), we see that

G(x) ≥ c ≥ 0,

showing that x(t) is not a constant, for if c > 0 and x ∈ N, then

G(x) = ’2 V (t, x(t)) dt ¤ 0.

I

If c = 0, we know that d(x (k) , B) ’ 0 by Theorem 2.12. Hence, there is a sequence

{y (k) } ‚ B such that x (k) ’ y (k) ’ 0 in X. If v ∈ N, then

(x, v) = (x ’ x (k) , v) + (x (k) ’ y (k) , v) ’ 0

since y (k) ∈ M. Thus, x ∈ M.

If

ρk = x (k) ’ ∞,

X

let x (k) = x (k) /ρk . Then x (k) X = 1. Let x (k) = w(k) + v (k) , where w(k) ∈ M and

˜ ˜ ˜ ˜ ˜ ˜

(k) ∈ N. There is a renamed subsequence such that [w (k) ]· ’ r and v (k) ’ „,

v

˜ ˜ ˜

2 + „ 2 = 1. From (17.21) and (17.23) we obtain

where r

[x (k) ]· V (t, x (k) (t)) dt/ρk ’ 0

2 2

˜ ’2

I

and

[x (k) ]· ∇x V (t, x (k) ) · x (k) dt/ρk ’ 0.

˜ ’

2 2

I

Thus,

V (t, x (k) (t)) dt/ρk ’ r 2

2

(17.37) 2

I

and

∇x V (t, x (k) ) · x (k) dt/ρk ’ r 2 .

2

(17.38)

I

224 17. Second-Order Periodic Systems

Hence,

μ

Hμ (t, x (k) (t)) dt/ρk ’

2

’ 1 r 2.

(17.39)

2

I

By hypothesis 3, the left-hand side of (17.37) is

≥ (2β x (k) ’ 4πC)/ρk ’ 2β„ 2 .

2 2

Hence,

r 2 ≥ 2β„ 2 = 2β(1 ’ r 2 ),

showing that

2β

r2 ≥ > 0.

1 + 2β

Note that

|x (k) (t)| ¤ C x (k)

˜ ˜ = C.

X

If

|x (k) (t)| ’ ∞,

then

Hμ (t, x (k) (t)) Hμ (t, x (k) (t)) (k) 2

¤ lim sup |x (t)| ¤ 0.

˜

lim sup

|x (k) (t)|2

ρk2

If

|x (k) (t)| ¤ C,

then