completes the proof of the lemma.

The following can be proved in the same way.

Lemma 11.11. M © ‚ B R links S [hm] for each R > 0.

,2

11.3 Existence

In this section we give the proofs of Theorems 11.1 and 11.2. Let

p(x, t) = f (x, t) + at ’ ’ bt + , t ± = max{±t, 0},

(11.54)

and

t

P(x, t) =

(11.55) p(x, s) ds.

0

Under hypothesis (11.7), it is readily checked [122] that the functional

G(u) = u ’2 F(x, u) d x = I (u, a, b) ’ 2

2

(11.56) P(x, u) d x

D

is in C 1 on D with

(G (u), v) = 2(u, v) D ’ 2( f (·, u), v) = (I (u, a, b), v) ’ 2( p(·, u), v)

(11.57)

and

(I (u, a, b), v)/2 = (u, v) D + a(u ’ , v) ’ b(u + , v).

(11.58)

From (11.5) and (11.54) we see that

p(x, t)/t ’ 0, 2P(x, t)/t 2 ’ 0 as |t| ’ ∞.

(11.59)

This and the fact that

‚(F/t 2 )/‚t = ‚(P/t 2 )/‚t = ’t ’3 H

(11.60)

imply that

∞

s ’3 H (x, s)ds,

= t >0

t2

(11.61) P(x, t)

t

t

s ’3 H (x, s)ds,

2

= ’t t < 0.

’∞

136 11. Fuˇ ´k Spectrum: Resonance

c±

Proof of Theorem 11.1. By (11.9) and (11.61), we have

2P(x, t) ¤ W0 (x), x∈ , t ∈ R.

(11.62)

In view of Lemma 11.8, this implies

G(u) ≥ ’2 P(x, u)d x ≥ ’B0 = ’ W0 (x)d x, u∈S ,1 .

(11.63)

I claim that

G(v) ’ ’∞ as v ’ ∞, v∈N ’1 .

(11.64)

Assume this for the moment. Then there is an R > 0 suf¬ciently large that (3.4) holds

with A = N ’1 © ‚ B R and B = S ,1 . Moreover, for this choice of A and B, A links

B [hm] by Lemma 11.10. We can now apply Theorem 3.4 to conclude that there is a

sequence {u k } ‚ D such that

G(u k ) ’ c, (1 + u k D )G (u k ) ’ 0,

(11.65)

and from (11.56) we readily estimate c by

’B0 ¤ c ¤ B1 .

(11.66)

From (11.65) we ¬nd

I (u k , a, b) ’ 2 P(x, u k )d x ’c,

(11.67)

I (u k , a, b) ’ ( p(·, u k ), u k ) ’0,

(11.68)

(I (u k , a, b), v) ’ 2( p(·, u k ), v) ’0, v ∈ D.

(11.69)

Assume that

ρk = u k ’ ∞,

(11.70) D

and let u k = u k /ρk . Then u k D = 1. Thus, there is a renamed subsequence such that

˜ ˜

u k ’ u weakly in D, strongly in L 2 ( ), and a.e. in . As a consequence, (11.59) and

˜ ˜

(11.68) imply

1 = a u’ + b u+ 2.

˜ ˜

2

(11.71)

This shows that u ≡ 0. Moreover, (11.69) implies

˜

I (u, a, b) = 0.

˜

(11.72)

˜

Thus, u is a solution of (11.5) and satis¬es

I (u, a, b) = 0.

˜

(11.73)

11.3. Existence 137

If we combine this with (11.71), we see that u D = 1. Consequently, u k ’ u

˜ ˜ ˜

strongly in D. Combining (11.67) and (11.68), we obtain

H (x, u k ) d x ’ ’c.

(11.74)

= {x ∈ : ±u(x) > 0},

˜ = {x ∈ : u(x) = 0}. Then u k (x) ’ ±∞ for

˜

Let ± 0

x∈ ± . By (11.9) and (11.13),

’c = lim H (x, u k ) d x

¤ H+ (x) d x