We can now follow the usual procedures to obtain a weak solution of (9.6) satisfying

G(u) = c ≥ µ (cf., e.g., [122, p. 64]). Since G(0) = 0, we see that u = 0. This

completes the proof.

We postpone the proof of Theorem 9.4 until the next section.

In proving Theorem 9.5, we shall make use of

Lemma 9.13. Under hypothesis (C ), there is an ± = 0 such that G(±•0 ) < 0.

Proof. We can assume that

•0 = 1.

(9.21) D

Thus,

G(±•0 ) = ± 2 ’ 2 F(x, ±•0 ) d x

˜

¤ ± 2 ’ »± 2 •0 (x)2 d x

|±•0 (x)|<δ

+ V q (|±•0 |q + |±•0 |)

|±•0 (x)|>δ

˜ q

¤ ± 2 ’ »± 2 • 0 + C|±|q V •0

2

q

˜

¤ ± 2 [1 ’ (»/»0 ) + C |±|q’2 ].

This can be made negative by taking ± suf¬ciently small.

Lemma 9.14. Under hypothesis (D ),

G(u) ’ ∞ ’ ∞.

(9.22) as u D

9.4. Proofs 91

Proof. Suppose there is a sequence {u k } ‚ D such that ρk = u k ’ ∞ and

G(u k ) ¤ K .

Write

u k = wk + ±k •0 , u k = u k /ρk ,

˜ wk = wk /ρk ,

˜ ±k = ±k /ρk ,

˜

where wk ⊥•0 . If »1 > »0 is the next point in the spectrum of A, then

»1 w ¤w D, w⊥•0 .

2 2

Thus,

G(u k ) = u k ’ »0 u k ’2 P(x, u k ) d x

2 2

D

»0

≥ 1’ wk ’2 P(x, u k ) d x

2

D

»1

»0 2

≥ 1’ wk ’2 W (x) d x.

D

»1

The only way this would not converge to ∞ is if wk D is bounded. But then wk D ’

˜

0 and |±k | ’ 1. Since u k D = 1, there is a renamed subsequence such that u k ’ u

˜ ˜ ˜ ˜

2 ( ), and a.e. in . Since w = 0 and |±| = 1, we have

˜ ˜

weakly in D, strongly in L

u(x) = ±•0 (x) = 0 a.e. Hence, |u k (x)| = ρk |u k (x)| ’ ∞ a.e. Consequently,

˜ ˜ ˜

P(x, u k ) d x ’ ’∞,

showing that G(u k ) ’ ∞. This completes the proof.

We can now give the proof of Theorem 9.5.

Proof. Let

m = inf G.

D

Then there is a sequence {u k } ‚ D such that G(u k ) ’ m. In view of Lemma 9.14,

we must have u k D ¤ C. Thus, there is a renamed subsequence such that u k ’ u

weakly in D, strongly in L 2 ( ), and a.e. in . Now,

loc

2

G(u) = u ’2 F(x, u) d x

D

= uk ’ 2([u k ’ u], u) D ’ u k ’ u

2 2

D D

’2 F(x, u k ) d x + 2 [F(x, u k ) ’ F(x, u)]d x

¤ G(u k ) ’ 2([u k ’ u], u) D + 2 [F(x, u k ) ’ F(x, u)]d x.

92 9. Superlinear Problems

From our hypotheses, it follows that

F(x, u k ) d x ’ F(x, u) d x

(cf., e.g., [122, p. 64]). We therefore have in the limit G(u) ¤ m, from which we

conclude that G(u) = m and G (u) = 0. Hence, u is a weak solution of (9.6). We see

from Lemma 9.13 that m < 0. Since G(0) = 0, we see that u = 0. This completes

the proof.

9.5 The parameter problem

In this section we shall give the proofs of Theorems 9.4, 9.6, and 9.7. They will be

based on the following result proved in the next section. Let E be a re¬‚exive Banach

space with norm · , and let A, B be two closed subsets of E. Suppose that G ∈

C 1 (E, R) is of the form G(u) := I (u) ’ J (u), u ∈ E, where I, J ∈ C 1 (E, R) map

bounded sets to bounded sets. De¬ne

G » (u) = »I (u) ’ J (u), »∈ ,

where is an open interval contained in (0, +∞). Assume one of the following alter-

natives holds.

(H1 ) I (u) ≥ 0 for all u ∈ E and either I (u) ’ ∞ or |J (u)| ’ ∞ as u ’ ∞.

(H2 ) I (u) ¤ 0 for all u ∈ E and either I (u) ’ ’∞ or |J (u)| ’ ∞ as u ’ ∞.

Furthermore, we suppose that

(H3 ) a0 (») := sup A G » ¤ b0 (») := inf B G » , for any » ∈ .

We let be the set of mappings (t) ∈ C(E — [0, 1], E) described in Chapter 3.

We have

Theorem 9.15. Assume that (H1 ) or (H2 ) holds together with (H3 ).

(1) If A links B [hm] and A is bounded, then, for almost all » ∈ , there exists

u k (») ∈ E such that supk u k (») < ∞, G » (u k (»)) ’ 0, and

G » (u k (»)) ’ a(») := inf G » ( (s)u), k ’ ∞.

sup

∈ s∈[0,1],u∈A

Furthermore, if a(») = b0 (»), then dist(u k (»), B) ’ 0, k ’ ∞.

(2) If B links A[hm] and B is bounded, then, for almost all » ∈ , there exists

v k (») ∈ E such that supk v k (») < ∞, G » (v k (»)) ’ 0, and

G » (v k (»)) ’ b(») := sup G » ( (s, v)), k ’ ∞.

inf

∈ s∈[0,1],v∈B

Furthermore, if a0 (») = b(»), then dist(v k (»), A) ’ 0, k ’ ∞.

9.5. The parameter problem 93