(10.68) E

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 E, strongly in L 2 ( ), and a.e. in . If h =

˜ ˜

(h 1 , h 2 ) ∈ E, then, by (10.62),

(10.69) (G (u k ), h)/ρk = (Av k , h 2 )+(Awk , h 1 )’( f (u k ), h 2 )/ρk ’(h(u k ), h 1 )/ρk ,

˜ ˜

where u k = (v k , wk ). Taking the limit and applying (10.54) “ (10.56) we see that

u = (v, w) is a solution of (10.58). By hypothesis, u ≡ 0. On the other hand, there

˜ ˜˜ ˜

is a renamed subsequence such that v k D ’ a, wk D ’ b with a 2 + b2 = 1.

˜ ˜

Moreover,

2

(G (u k ), (v k , ’v k ))/ρk = ’ 2 v k

˜ ˜ ˜ + ( f (u k ), v k )/ρk

˜

(10.70) D

’ (g(u k ), v k )/ρk

˜

’ ’ 2a 2 + [± + β ’ γ ’ δ]v 2 d x

˜

and

(G (u k ), (wk , wk ))/ρk =2 wk

˜˜ ˜ ’ ( f (u k ), wk )/ρk

˜

2

(10.71) D

’ (g(u k ), wk )/ρk

’2b2 ’ [± + β + γ + δ]w2 d x.

˜

By (10.2), both these limits are 0. Since a and b cannot both vanish, the same is true

of v and w. Hence, u ≡ 0, and this contradiction shows that (10.68) does not hold.

˜ ˜ ˜

We can now use the usual procedures to show that {u k } has a renamed subsequence

converging weakly in E to a function u, strongly in L 2 ( ), and a.e. in . Then this

limit is a solution of (10.63) and consequently of (10.57).

For another application, let A, B be positive, self-adjoint operators on L 2 ( ) with

compact resolvents, where ‚ Rn . Let F(x, v, w) be a function on — R2 such that

f (x, v, w) = ‚ F/‚v, g(x, v, w) = ‚ F/‚w

(10.72)

are Carath´ odory functions satisfying

e

| f (x, v, w)| + |g(x, v, w)| ¤ C0 (|v| + |w| + 1), v, w ∈ R,

(10.73)

120 10. Weak Linking

and

f (x, t y, tz)/t ’ ±+ (x)v + ’ ±’ (x)v ’ + β+ (x)w+ ’ β’ (x)w’ ,

(10.74)

g(x, t y, tz)/t ’ γ+ (x)v + ’ γ’ (x)v ’ + δ+ (x)w+ ’ δ’ (x)w’

(10.75)

as t ’ +∞, y ’ v, z ’ w, where a ± = max(±a, 0). We wish to solve the system

Av = ’ f (x, v, w),

(10.76)

Bw = g(x, v, w).

(10.77)

Let »0 (μ0 ) be the lowest eigenvalue of A(B). We assume that the only solution of

’Av = ±+ v + ’ ±’ v ’ + β+ w+ ’ β’ w’ ,

(10.78)

Bw = γ+ v + ’ γ’ v ’ + δ+ w+ ’ δ’ w’

(10.79)

is v = w = 0. We have

Theorem 10.7. Assume that eigenfunctions of »0 are = 0 a.e. on ,

±± (x) ≥≡ ’»0 , x∈ ,

(10.80)

and

2F(x, 0, t) ¤ μ0 t 2 + W (x), x∈ , t ∈ R,

(10.81)

where W (x) ∈ L 1 ( ). Then the system (10.76), (10.77) has a solution.

Proof. Let D = D(A1/2 ) — D(B 1/2 ). Then D becomes a Hilbert space with norm

given by

= (Av, v) + (Bw, w), u = (v, w) ∈ D.

2

(10.82) u D

We de¬ne

G(u) = b(w) ’ a(v) ’ 2 F(x, v, w) d x, u ∈ D,

(10.83)

where

a(v) = (Av, v), b(w) = (Bw, w).

(10.84)

Then G ∈ C 1 (D, R) and

(G (u), h)/2 = b(w, h 2 ) ’ a(v, h 1 ) ’ ( f (u), h 1 ) ’ (g(u), h 2 ),

(10.85)

where we write f (u), g(u) in place of f (x, v, w), g(x, v, w), respectively. It is readily

seen that the system (10.76), (10.77) is equivalent to

G (u) = 0.

(10.86)

10.4. Some applications 121

We let N be the set of those (v, 0) ∈ D and M the set of those (0, w) ∈ D. Then M, N

are orthogonal closed subspaces such that

D = M • N.

(10.87)

If we de¬ne

Lu = 2(’v, w), u = (v, w) ∈ D

(10.88)

then L is a self-adjoint, bounded operator on D. Also,

G (u) = Lu + c0 (u),

(10.89)

where

c0 (u) = ’(A’1 f (u), B ’1 g(u))

(10.90)

is compact on D. This follows from (10.73) and the fact that A and B have compact

resolvents. Now, by (10.81),

2

G(0, w) ≥ b(w) ’ μ0 w ’ W (x) d x, (0, w) ∈ M.

(10.91)

Thus,

inf G ≥ ’ W (x)d x.

(10.92)

M

On the other hand, (10.80) implies

sup G ’ ’∞ as R ’ ∞.