N©‚ B R

To see this, let (v k , 0) be any sequence in N such that ρk = a(v k ) ’ ∞. Then

2

G(v k , 0)/ρk = ’a(v k ) ’ 2

˜ F(x, v k , 0)d x/ρk ,

2 2

(10.94)

where v k = v k /ρk . Note that a(v k ) = 1. Thus, there is a renamed subsequence v k ’ v

˜ ˜ ˜ ˜

2 ( ), and a.e. in

weakly in N, strongly in L such that

{±+ (v + )2 + ±’ (v ’ )2 }d x

G(v k , 0)/ρk ’ ’ 1 ’ ˜ ˜

2

{(»0 + ±+ )(v + )2 + (»0 + ±’ )(v ’ )2 }d x

=’ ˜ ˜

+ »0 v

˜ ’ 1.

2

This is negative unless »0 v 2 = 1. Since a(v) ¤ 1, this would mean that v ∈ E(»0 ),

˜ ˜ ˜

the eigenspace of »0 . Thus, v = 0 a.e. by hypothesis. But then the integral cannot

˜

vanish by (10.80). Hence,

lim sup G(0, v)/a(v) < 0,

(10.95)

a(v)’∞

122 10. Weak Linking

and (10.93) holds. Since N is an invariant subspace of L, we can apply Theorem 10.2

to conclude that there is a sequence {u k } ‚ D such that (10.2) and (10.3) hold. Let

u k = (v k , wk ). I claim that

ρk = a(v k ) + b(wk ) ¤ C.

2

(10.96)

To see this, assume that ρk ’ ∞, and let u k = u k /ρk . Then there is a renamed

˜

subsequence such that u k ’ u weakly in D, strongly in L 2 ( ), and a.e. in . If h =

˜ ˜

(h 1 , h 2 ) ∈ D, then

(10.97)

(G (u k ), h)/ρk = 2b(wk , h 2 ) ’ 2a(v k , h 1 ) ’ 2( f (u k ), h 1 )/ρk ’ 2(g(u k ), h 2 )/ρk .

˜ ˜

Taking the limit and applying (10.73) “ (10.75), we see that u = (v, w) is a solution of

˜ ˜˜

(10.78) (10.79). Hence, u = 0 by hypothesis. On the other hand, since

˜

˜

a(v k ) + b(wk ) = 1, there is a renamed subsequence such that a(v k ) ’ a, b(wk ) ’ b

˜ ˜ ˜ ˜ ˜

˜˜

with a + b = 1. Thus, by (10.74), (10.75), and (10.85),

(10.98) (G (u k ), (v k , 0))/2ρk = ’a(v k ) ’ ( f (u k ), v k )/ρk

˜ ˜ ˜

(±+ v + ’ ±’ v ’ + β+ w+ ’ β’ w’ )vd x

’ ’a ’

˜ ˜ ˜ ˜ ˜˜

and

(10.99) (G (u k ), (0, wk ))/2ρk = b(wk ) ’ (g(u k ), wk )/ρk

˜ ˜ ˜

˜ (γ+ v + ’ γ’ v ’ + δ+ w+ ’ δ’ w’ )wd x.

’b’ ˜ ˜ ˜ ˜˜

Thus, by (10.42),

(±+ v + ’ ±’ v ’ + β+ w+ ’ β’ w’ )vd x

a=’

˜ ˜ ˜ ˜ ˜˜

(10.100)

and

˜ (γ+ v + ’ γ’ v ’ + δ+ w+ ’ δ’ w’ )wd x.

b= ˜ ˜ ˜ ˜˜

(10.101)

˜˜

Since one of the two numbers a, b is not zero, we see that we cannot have u ≡ 0. This

˜

contradiction proves (10.96). Once this is known, we can use the usual procedures

to show that there is a renamed subsequence such that u k ’ u in D, and u satis¬es

(10.86).

Corollary 10.8. In Theorem 10.7 we can replace (10.80), (10.81) with

2F(x, v, 0) ≥ ’»0 v 2 ’ W (x), x∈ , v ∈ R,

(10.102)

provided eigenfunctions of μ0 are = 0 a.e. in and

δ± (x) ¤≡ μ0 .

(10.103)

10.4. Some applications 123

Proof. We interchange the roles of M and N in the proof of Theorem 10.7.

Theorem 10.9. If »0 is simple and the eigenfunctions of »0 and μ0 are bounded and

= 0 a.e. in , we can replace (10.81) in Theorem 10.7 with

2F(x, t, 0) ≥ ’»1 t 2 , t ∈ R,

(10.104)

2F(x, s, t) ¤ μ0 t 2 ’ »0 s 2 , |t| + |s| ¤ δ,

(10.105)

where »1 is the next eigenvalue of A and δ > 0. Moreover, system (10.76), (10.77) has

a nontrivial solution.

Proof. Let N be the orthogonal complement of N0 = {•0 } in N. Then N = N • N0

and

(10.106)

G(v , 0) = ’a(v ) ’ 2 F(x, v , 0) d x ¤ ’a(v ) + »1 v ¤ 0, v ∈N,

2

and

G(v, 0) ’ ’∞ as a(v) ’ ∞

(10.107)

by (10.80) [cf. (10.93)]. Let M0 be the subspace of M spanned by the eigenfunctions

of B corresponding to μ0 , and let M be its orthogonal complement in M. Since N0

and M0 are contained in L ∞ ( ), there is a positive constant ρ such that

a(y) ¤ ρ 2 ’ y ¤ δ/4, y ∈ N0 ,

(10.108) ∞

b(h) ¤ ρ 2 ’ h ¤ δ/4, h ∈ M0 ,

(10.109) ∞