˜ ˜˜

and

( f (u k ), wk )/ρk ’ ±(u, w),

˜ ˜˜

10.4. Some applications 117

where u = v + w. Since v k

˜ ˜ ˜ ˜ + wk

˜ = 1, we have for a renamed subsequence

2 2

D D

(G (u k ), v k )/2ρk

˜ ’ γ1 ’ ±(u, v),

˜˜

(G (u k ), wk )/2ρk

˜ ’ γ2 ’ ±(u, w),

˜˜

where γ2 + γ1 = 1. By (10.51), both of these expressions must vanish. But this cannot

happen if u ≡ 0. Thus, the ρk must be bounded, and consequently there is a renamed

˜

subsequence of {u k } such that u k ’ u weakly in D and a.e. in . As before, this

implies ( f (u k ), v) ’ ( f (u), v) for v ∈ D. Hence,

(G (u k ), v) ’ (G (u), v), v ∈ D.

By (10.51), this limit must vanish for each v in D. We conclude that G (u) = 0 and

that u is a solution of (10.39).

As another application, let ‚ Rn and let A ≥ »0 > 0 be a self-adjoint operator

on L 2 ( ) with compact resolvent. Let F(x, s, t) be a function on — R2 such that

f (x, s, t) = ‚ F/‚t, g(x, s, t) = ‚ F/‚s

(10.53)

are Carath´ odory functions satisfying

e

| f (x, s, t)| + |g(x, s, t)| ¤ C(|s| + |t| + 1), s, t ∈ R,

(10.54)

and

˜

f (x, s, t)/ρ ’ ±(x)˜ + β(x)t ,

(10.55) s

˜

g(x, s, t)/ρ ’ γ (x)˜ + δ(x)t

(10.56) s

as ρ 2 = s 2 + t 2 ’ ∞, s/ρ ’ s , t/ρ ’ g. We wish to solve the system

˜ ˜

Av = f (x, v, w), Aw = g(x, v, w), v, w ∈ D(A).

(10.57)

We assume that the only solution of

Av = ±v + βw, Aw = γ v + δw

(10.58)

is v ≡ w ≡ 0 in . If »0 is an eigenvalue of A, we assume that corresponding

eigenfunctions are = 0 a.e. on . Finally, we assume that

β + γ ’ ± ’ δ ¤≡ 2»0 , β + γ + ± + δ ¤≡ 2»0 .

(10.59)

118 10. Weak Linking

We have

Theorem 10.6. Under the above assumptions, system (10.57) has a solution.

Proof. Let D = D(A1/2 ), E = D — D. Then D becomes a Hilbert space with norm

given by

= (Av, v) + (Aw, w), u = (v, w) ∈ D.

2

(10.60) u E

We de¬ne

G(u) = (Av, w) ’ F(x, v, w) d x, u ∈ D.

(10.61)

In view of (10.54), G ∈ C 1 (D, R) and

(G (u), h) = (Av, h 2 ) + (Aw, h 1 ) ’ ( f (u), h 2 ) ’ (g(u), h 1 )

(10.62)

for u = (v, w) ∈ E, h = (h 1 , h 2 ) ∈ E, and we write f (u), g(u) in place of

f (x, v, w), g(x, v, w), respectively. It follows from (10.62) that u ∈ E is a solu-

tion of (10.57) if u satis¬es

G (u) = 0.

(10.63)

Let N denote the subspace of E consisting of those u = (v, w) for which w = ’v and

let M consist of those for which w = v. Then M = N ⊥ . We note that

G(v, ’v) ’ ’∞ as v ’∞

(10.64) D

and

G(w, w) ’ +∞ as w ’ ∞.

(10.65) D

To prove (10.64), let {v k } ‚ D be such that ρk = v k D ’ ∞, and take v k = v k /ρk .

˜

Since v k D = 1, there is a renamed subsequence such that v k ’ v weakly in N,

˜ ˜ ˜

2 ( ), and a.e. in

strongly in L such that

2G(v k , ’v k )/ρk = ’2 v k

˜ ’2 F(x, v k , ’v k )d x/ρk

2 2 2

(10.66) D

’ ’2 ’ [γ ’ δ ’ ± + β]v 2 d x

˜

[2»0 + β + γ ’ ± ’ δ]v 2 d x + 2»0 v 2

=’ ˜ ˜ ’ 2.

In view of (10.59), this is negative unless »0 v 2 = 1. But this would mean that

˜

v ∈ E(»0 ), the eigenspace of »0 . If »0 is not an eigenvalue, then we conclude immedi-

˜

ately that

lim sup G(v, ’v)/ v <0

2

(10.67) D

v D ’∞

10.4. Some applications 119

and (10.64) holds. If »0 is an eigenvalue, then v = 0 a.e. by hypothesis. But then the

˜

last integral in (10.66) must be positive by (10.59). Again, this implies (10.67) and

(10.64). A similar argument implies (10.65). Now that we have (10.64) and (10.65),

we know that (10.1) holds for A = N © ‚ B R and B = M when R is suf¬ciently large.

If we let F be the projection of E onto N, we can take Q = N © B R and p = 0.

We can conclude from Theorem 10.2 that there is a sequence {u k } ‚ E satisfying

(10.2), (10.3). I claim that