for some positive unless there is a solution of

1

Ay = » y = f (x, y), y ∈ E(» )\{0}

(cf. [122]). Since such a solution would solve (8.13), we may assume that (8.16) holds.

Since

v 2 ¤ » v 2 , v ∈ N,

D

and

2 2

» w ¤w D, w ∈ M,

+1

we have, by (8.6),

G(w) ¤ » w + B1 ’ w ¤ B1 , w ∈ A,

2 2

D

8.2. Bounded domains 67

where B1 = W1 (x) d x. Moreover, (8.8) implies

G(v ) ≥ (ν ’ » ’1 ) v , v ∈N.

2

Hence, there is an µ > 0 such that

G(v) ≥ µ, v ∈ B.

In view of these inequalities, we can now apply Proposition 7.3 to conclude that there

is a sequence {u k } ‚ D such that

G(u k ) ’ c, µ ¤ c ¤ B1 , G (u k ) ’ 0.

(8.17)

Let ρk = u k D. If ρk ’ ∞, then

G(u k ) = 2 F(x, u k ) d x ’ ρk ’ c

2

(8.18)

and

2

(G (u k ), u k )/2 = f (x, u k )u k d x ’ ρk = o(ρk ).

Hence,

H (x, u k ) d x = o(ρk ).

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 . By (8.9) and (8.10),

˜ ˜

H (x, u k ) d x/ρk ¤ lim sup[H (x, u k )/|u k |]|u k | d x

˜

lim sup

= σ (x)|u| d x.

˜

Since σ (x) < 0 a.e. in , the last two statements imply that u ≡ 0. However, we see

˜

from (8.18) that

2

F(x, u k ) d x/ρk ’ 1,

2

while (8.6) implies

2

u 2 d x,

F(x, u k ) d x/ρk ¤ » ˜

lim sup 2

showing that u ≡ 0. This contradiction tells us that the ρk must be bounded. We can

˜

now apply Theorem 3.4.1 of [122] to conclude that there is a u ∈ D satisfying

G(u) = c, G (u) = 0.

(8.19)

Since c ≥ µ > 0, we see that u = 0, and the proof is complete.

68 8. Semilinear Problems

The proof of Theorem 8.1 implies

Corollary 8.2. If » is a simple eigenvalue, then hypothesis (8.6) in Theorem 8.1 can

be weakened to

2

2F(x, t) ¤ » + W1 (x), x∈ , t ∈ R,

(8.20) +1 t

for some W1 (x) ∈ L 1 (R).

Remark 8.3. The proof of Theorem 8.1 is much simpler if = 0. In this case N = {0}

and (8.14) immediately implies (8.16). The rest of the proof is unchanged.

We now show that we can essentially reverse the inequalities (8.6)“(8.10) and

obtain the same results. In fact, we have

Theorem 8.4. Equation (8.13) has at least one nontrivial solution if we assume > 0

and

» t 2 ¤ 2F(x, t) + W1 (x), x∈ , t ∈ R,

(8.21)

for some W1 (x) ∈ L 1 (R),

2F(x, t) ¤ » t 2 , |t| ¤ δ,

(8.22)

for some δ > 0,

2F(x, t) ¤ νt 2 , x∈ , t ∈ R,

(8.23)

for some ν < » +1 ,

H (x, t) ≥ ’C(|t| + 1), x∈ , t ∈ R,

(8.24)

and

lim inf H (x, t)/|t| > 0

(8.25) a.e.

|t |’∞

Proof. In this case we take G to be the functional (8.5). We take A = N, N=

N {v 0 }, and consider the mapping

F(v + w + sv 0 ) = v + [s + δ ’ δ•( w 2 /δ 2 )]v 0 , v ∈ N , w ∈ M, s ∈ R,

where • satis¬es the hypotheses of Proposition 7.3. By (8.21), we have

G(v) ¤ B1 , v ∈ N.

For w ∈ M1 , we write w = w + y, where w ∈ M and y ∈ E(» ). Then (8.22) implies

G(w) ≥ µ1 , w = ρ, w ∈ M1 ,

D

unless (8.13) has a nontrivial solution. Hence, by the argument given in the proof of

˜ ˜

Theorem 8.1 we have a sequence satisfying (8.17). If u k and u are as in the proof of

8.3. Some useful quantities 69

Theorem 8.1, then (8.24), (8.25) imply that u ≡ 0 as in that proof. However, (8.18)

˜

implies

F(x, u k ) d x/ρk ’ 1,

2