+ H’ (x) d x

’

+ W0 (x) d x

0

< ’ B1 ,

contradicting (11.66). Thus, (11.70) does not hold. Once we know that the sequence

{u k } is bounded in D, we can use standard techniques [122] to obtain a solution of

G(u) = c, ’B0 ¤ c ¤ B1 , G (u) = 0,

(11.75)

which is a solution of (11.2).

It thus remains to prove (11.64). Let {v k } ‚ N ’1 be a sequence such that ρk =

v k D ’ ∞. Let v k = v k /ρk . Then v k D = 1 and there is a renamed subsequence

˜ ˜

such that v k ’ v in D and a.e. in . Since

˜ ˜

|F(x, t)| ¤ C(t 2 + |t|)

(11.76)

by (11.7), we have

2

˜2

|F(x, v k )/ρk | ¤ C(v k + |v k |/ρk )

˜

(11.77)

and, consequently,

F(x, v k )d x/ρk ’ a v ’ + b v+

˜ ˜

2 2 2

(11.78) 2

by (11.6). This means that

G(v k )/ρk = vk

˜ ’2 F(x, v k )d x/ρk

2 2 2

D

’ a v’ ’ b v+

’ v

˜ ˜ ˜

2 2 2

D

’ a) v ’ ’ b) v + 2 .

¤ (» ˜ + (» ˜

2

’1 ’1

Since both a and b are greater than » ’1 , this will always be positive since v ≡ 0.

˜

Thus, (11.64) holds, and the proof of Theorem 11.1 is complete.

138 11. Fuˇ ´k Spectrum: Resonance

c±

Proof of Theorem 11.2. By (11.15) and (11.61) we have

’2P(x, t) ¤ W0 (x), x∈ , t ∈ R.

(11.79)

Consequently,

G(u) ¤ ’2 P(x, u) d x ¤ B0 , u∈S ,2 .

(11.80)

Moreover, I claim that

G(w) ’ ∞ as w ’ ∞, w∈M .

(11.81)

Assuming this, we note that there is an R > 0 suf¬ciently large that

sup G ¤ inf G,

(11.82)

A

B

where A = M © ‚ B R and B = S ,2 . This is not quite (3.4). To correct the situation,

we let G 1 = ’G. Then (11.82) becomes

sup G 1 ¤ inf G 1 .

(11.83)

B

A

We know from Lemma 11.11 that A links B [hm]. Consequently, we may apply Theo-

rem 3.4 to conclude that there is a sequence {u k } ‚ D such that (11.65) holds with G

replaced by G 1 . Now c satis¬es the estimates

inf G 1 ¤ c ¤ sup G 1

(11.84)

B M©B R

or

inf G ¤ ’c ¤ sup G.

(11.85)

M©B R B

By (11.14), (11.56), and (11.80), we see that c satis¬es (11.66) as well. Thus, for the

given sequence, (11.65), (11.67)“(11.69) hold with c replaced by ’c, while c satis¬es

(11.66). If we assume that (11.70) holds, we can reason as in the proof of Theorem

11.1 that there is a renamed subsequence of {u k } converging in D to a function u ∈ D

˜ ˜

˜

and a.e. in . Moreover, the function u satis¬es (11.72) and (11.73). Also, (11.74)

holds with c replaced by ’c. But then (11.15) and (11.16) imply

(11.86)

c = lim H (x, u k )d x ≥ H+ (x)d x + H’(x)d x ’ W0 (x) d x > B1 .

+ ’ 0

This contradiction shows that the ρk are bounded, and we can now employ standard

techniques to obtain a solution of

G(u) = c1 , ’B1 ¤ c1 ¤ B0 , G (u) = 0.

(11.87)

This produces the desired result.

11.4. Notes and remarks 139

It therefore remains only to prove (11.81). Let {wk } ‚ M be a sequence such that

ρk = wk D ’ ∞. Let wk = wk /ρk . Then we have wk D = 1, and there is a

˜ ˜

renamed subsequence such that wk ’ w weakly in D, strongly in L 2 ( ), and a.e. in

˜ ˜

. Thus,

F(x, wk )d x/ρk ’ a w’ + b w+ 2 .

˜ ˜

2 2

(11.88) 2

Consequently,

G(wk )/ρk ’ 1 ’ a w’ ’ b w+

˜ ˜

2 2 2

’ a) w’ ’ b) w+ 2 ,

2 2

≥1’ w

˜ + (» ˜ + (» ˜

+1 +1

D