ρk

2

Hence,

Hμ (t, x (k) (t)) dt/ρk ≥ 0.

2

lim inf

I

Thus, by (17.26),

μ

’ 1 r 2 ≥ 0.

2

If r = 0, this contradicts the fact that μ < 2. If r = 0, then w(k) ’ 0 uniformly

˜

(k) |2 = v (k) 2 ’ 1. Hence, there is a renamed

in I by Lemma 17.8. Moreover, T |v ˜ ˜

(k) ’ v in N with |v|2 = 1/T. Hence, x (k) ’ v uniformly in

subsequence such that v ˜ ˜ ˜ ˜ ˜

(k) (t)| ’ ∞ uniformly in I. Thus, by Hypothesis 4,

I. Consequently, |x

V (t, x (k) (t)) (k) 2

(k)

V (t, x (t)) dt/ρk ≥ lim inf |x (t)| dt > 0.

˜

2

lim inf

|x (k) (t)|2

I e

This contradicts (17.24). Hence, the ρk are bounded, and the proof is complete.

In proving Theorem 17.4, we follow the proof of Theorem 17.1 until (17.26).

Assume ¬rst that r > 0. Note that (17.21) and (17.23) imply that

H (t, x (k)(t)) dt ’ ’c.

(17.28)

I

17.3. Nonconstant solutions 221

On the other hand, by hypothesis 1A, we have

0 ← [x (k) ]· V (t, x (k) (t)) dt/ρk

2 2

˜ ’2

I

’2

≥ [x (k) ]· (|x (k) (t)|2 + ρk ) dt

˜ ’ 2C ˜

2

I

’ r 2 ’ 2C |x(t)|2 dt.

˜

I

Thus, x(t) ≡ 0. Let 0 ‚ I be the set on which x(t) = 0. The measure of

˜ ˜ is

0

(k) (t)| ’ ∞ as k ’ ∞ for t ∈

positive. Moreover, |x 0 . Hence,

H (t, x (k)(t)) dt ¤ H (t, x (k)(t)) dt + W (t) dt ’ ’∞

I\

I 0 0

by hypothesis 3A. But this contradicts (17.28). If r = 0, then w(k) ’ 0 uniformly

˜

(k) |2 = v (k) 2 ’ 1. Thus, there is a renamed

in I by Lemma 17.8. Moreover, T |v ˜ ˜

(k) ’ v in N with |v|2 = 1/T. Hence, x (k) ’ v uniformly in

subsequence such that v ˜ ˜ ˜ ˜ ˜

(k) | ’ ∞ uniformly in I. Thus, by hypothesis 4,

I. Consequently, |x

V (t, x (k) (t)) (k) 2

(k)

V (t, x (t)) dt/ρk ≥ lim inf |x (t)| dt > 0.

˜

2

lim inf

|x (k) (t)|2

I e

This contradicts (17.24). Hence, the ρk are bounded, and the proof is complete.

17.3 Nonconstant solutions

We now turn to the proofs of Theorems 17.5“17.7.

First, we prove Theorem 17.5.

Proof. As before, we de¬ne

G(x) = x

™ ’2 V (t, x(t)) dt, x ∈ X.

2

(17.29)

I

For each x ∈ X, write x = v + w, where v ∈ N, w ∈ M. As before, if x ∈ M and

12 2

2

= ρ2 =

™ m,

x

T

then Lemma 17.8 and hypothesis 2 imply that

≥ ™ ’2 ± dt

2

(17.30) G(x) x

|x|<m

≥ ρ 2 ’ 2±T ≥ 0.

Note that hypothesis 4A is equivalent to

V (t, x) ≥ β|x|2 ’ C, t ∈ I, x ∈ Rn ,

(17.31)

for some constant C.

222 17. Second-Order Periodic Systems

Next, let

y(t) = v + sw0 ,

where v ∈ N, s ≥ 0, and

w0 = (sin(2πt/T ), 0, . . . , 0).

Then w0 ∈ M, and

2 2

= 2π 2 /T.

w0 = T /2, w0

™

Note that

2 2

+ s 2 T /2 = T |v|2 + T s 2 /2.

=v

y

Consequently,

G(y) = s 2 w0

™ ’2 V (t, y(t)) dt

2

I

¤ 2π 2 s 2 /T ’ 2β |y(t)|2 dt + T C

I

¤ 2π 2 s 2 /T ’ 2β( v 2

+ T s 2 /2) + T C

¤ (2π 2 ’ βT 2 )s 2 /T ’ 2Tβ|v|2 + T C

’ ’∞ as s 2 + |v|2 ’ ∞.

We also note that hypothesis 1 implies

G(v) ¤ 0, v ∈ N.