Av = ’ f (x, v, w)

(15.48)

Bw = g(x, v, w).

(15.49)

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

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

(15.50)

Bw = γ+ v + ’ γ’ v ’ + δ+ w+ ’ δ’ w’

(15.51)

is v = w = 0. Our ¬rst result is

Theorem 15.8. Assume

2F(x, s, 0) ≥ ’»2 ’ W1 (x), x∈ , t ∈ R,

(15.52) 0

and

2F(x, 0, t) ¤ μ0 t 2 + W2 (x), x∈ , t ∈ R,

(15.53)

where Wi (x) ∈ L 1 ( ). Then the system (15.48), (15.49) 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

(15.54) u D

We de¬ne

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

(15.55)

where

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

(15.56)

15.3. Applications 185

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 ),

(15.57)

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 (15.48), (15.49) is equivalent to

G (u) = 0.

(15.58)

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.

(15.59)

If we de¬ne

Lu = 2(’v, w), u = (v, w) ∈ D,

(15.60)

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

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

(15.61)

where

c0 (u) = ’(A’1 f (u), B ’1 g(u))

(15.62)

is compact on D. This follows from (15.45) and the fact that A and B have compact

resolvents. It also follows that G has weak-to-weak continuity, for if u k ’ u weakly,

then Lu k ’ Lu weakly and c0 (u k ) has a convergent subsequence. Now, by (15.53),

G(0, w) ≥ b(w) ’ μ0 w ’ W2 (x) d x, (0, w) ∈ M.

2

(15.63)

Thus,

inf G ≥ ’ W (x) d x ≡ b0 .

(15.64)

M

On the other hand, (15.52) implies

G(v, 0) ¤ ’a(v) + »0 v + W1 (x) d x, (v, 0) ∈ N.

2

(15.65)

Thus,

sup G ¤ W1 (x) d x ≡ a0 .

(15.66)

N

We can now apply Theorem 15.2 to conclude that there is a sequence {u k } ‚ D such

that (15.8) holds. Let u k = (v k , wk ). I claim that

ρk = a(v k ) + b(wk ) ¤ C,

2

(15.67)

186 15. Weak Sandwich Pairs

for 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

(15.68)

(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 (15.45)“(15.47), we see that u = (v, w) is a solution

˜ ˜˜

of (15.50), (15.51). 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 (15.46), (15.47), and (15.57),

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

˜ ˜ ˜

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

’ ’a ’

˜ ˜ ˜ ˜ ˜˜

and

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

˜ ˜ ˜

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

’b’ ˜ ˜ ˜ ˜˜

Thus, by (15.8),

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

a=’

˜ ˜ ˜ ˜ ˜˜

(15.69)

and