F(x, R•0 ) d x/R 2 ’ ∞ as R ’ ∞

or

F(x, ’R•0 ) d x/R 2 ’ ∞ as R ’ ∞.

˜

then (9.10) has a nontrivial solution for almost every β ∈ (0, »0 /»).

Corollary 9.8. If we replace hypothesis (C ) in Theorem 9.7 with

(C ) F(x, t)/t 2 ’ 0 uniformly as t ’ 0.

then (9.10) has a nontrivial solution for almost every β ∈ (0, ∞).

9.3 Preliminaries

De¬ne

’2

2

(9.11) G(u) := u F(x, u)d x.

D

Under hypothesis (A), it is known that G is a continuously differentiable functional

on the whole of D. In fact, the following were proved in [122, pp. 56“58].

Proposition 9.9. Under hypothesis (A), F(x, u(x)) and v(x) f (x, u(x)) are in L 1 ( )

whenever u, v ∈ D.

Proposition 9.10. G(u) has a Fr´ chet derivative G (u) on D given by

e

(G (u), v) D = 2(u, v) D ’ 2( f (·, u), v).

(9.12)

Proposition 9.11. The derivative G (u) given by (9.12) is continuous in u.

Theorem 9.12. Under hypotheses (A)“(C), the following alternative holds:

Either

(a) there is an in¬nite number of y(x) ∈ D(A) \ {0} such that

Ay = f (x, y) = »0 y

(9.13)

or

(b) for each ρ > 0 suf¬ciently small, there is an µ > 0 such that

G(u) ≥ µ, = ρ.

(9.14) u D

9.4. Proofs 89

9.4 Proofs

We now give the proof of Theorem 9.3.

Proof. We take

G(u) = u ’2

2

(9.15) F(x, u)d x.

D

Under our hypotheses, Propositions 9.9“9.11 apply, and

(G (u), v) = 2(u, v) D ’ 2( f (·, u), v), u, v ∈ D.

(9.16)

By Theorem 9.12, we see that there are positive constants µ, ρ such that

G(u) ≥ µ, = ρ,

(9.17) u D

unless

Au = »0 u = f (x, u), u ∈ D \ {0},

(9.18)

has a solution. This would give a nontrivial solution of (9.6). We may therefore assume

that (9.17) holds. Next, we note that

G(±R•0 )/R 2 = •0 2

{F(x, ±R•0 )/R 2 •0 }•0 d x ’ ’∞ as R ’ ∞

22

’2

D

by hypothesis (D), depending on which part of (D) is assumed, since •0 = 0 a.e. Since

G(0) = 0 and (9.17) holds, we can now apply Theorem 3.13 to conclude that there is

a sequence {u k } ‚ D such that

G(u k ) ’ c ≥ µ, G (u k ) ’ 0.

Then

G(u k ) = ρk ’ 2 F(x, u k )d x ’ c

2

(9.19)

and

2

(G (u k ), u k ) = 2ρk ’ 2( f (·, u k ), u k ) = o(ρk ),

(9.20)

where ρk = u k D . Assume that ρk ’ ∞, and let u k = u k /ρk . Since u k D = 1,

˜ ˜

there is a renamed subsequence such that u k ’ u weakly in D, strongly in L 2 ( ), and

˜ ˜

a.e. in . By (9.19),

2F(x, u k ) 2

u k d x ’ 1.

˜

u2

k

Let

= {x ∈ : u(x) = 0},

˜ = \ 1.

1 2

90 9. Superlinear Problems

Then

2F(x, u k ) 2

u k ’ ∞,

˜ x∈ 1,

u2

k

by hypothesis (D). If has positive measure, then

1

2F(x, u k ) 2 2F(x, u k ) 2

uk d x ≥

˜ uk d x +

˜ [’W (x)] d x ’ ∞.

u2 u2

k k

1 2

must be 0, i.e., we must have u ≡ 0 a.e. Moreover,

˜

Thus, the measure of 1

2F(x, u k ) ’ μu k f (x, u k ) 2

u k d x ’ 1 ’ μ.

˜

u2

k

But by hypothesis (E),

u2 + 1 2

2F(x, u k ) ’ μu k f (x, u k ) 2

u k ¤ lim sup C k 2 u k = 0,

˜ ˜

lim sup 2

uk uk