where δ is the number given in (10.105). If

a(y) ¤ ρ 2 , b(w) ¤ ρ 2 , |y(x)| + |w(x)| ≥ δ,

(10.110)

we write w = h + w , h ∈ M0 , w ∈ M , and

δ ¤ |y(x)| + |w(x)| ¤ |y(x)| + |h(x)| + |w (x)| ¤ (δ/2) + |w (x)|.

(10.111)

Thus,

|y(x)| + |h(x)| ¤ δ/2 ¤ |w (x)|

(10.112)

and

|y(x)| + |w(x)| ¤ 2|w (x)|.

(10.113)

124 10. Weak Linking

Now, by (10.105) and (10.113),

G(y, w) =b(w) ’ a(y) ’ 2 F(x, y, w)d x

{μ0 w2 ’ »0 y 2 }d x

≥b(w) ’ a(y) ’

|y|+|w|<δ

’ c0 (|y| + |w| + 1)d x

|y|+|w|>δ

≥b(w) ’ a(y) ’ μ0 w + »0 y ’ c1 |w |4 d x

2 2

2|w |>δ

≥b(w ) ’ μ0 w ’ c2 b(w )2

2

μ0

≥ 1’ ’ c2 b(w ) b(w )

μ1

when

a(y) ¤ ρ 2 , b(w) ¤ ρ 2 ,

where μ1 is the next eigenvalue of B after μ0 . If we reduce ρ accordingly, we can ¬nd

a positive constant ν such that

a(y) ¤ ρ 2 , b(w) ¤ ρ 2 .

G(y, w) ≥ νb(w ),

(10.114)

I claim that either (a) (15.48), (15.49) has a nontrivial solution or (b) there is an

> 0 such that

G(y, w) ≥ , a(y) + b(w) = ρ 2 .

(10.115)

To see this, suppose (10.115) did not hold. Then there would be a sequence {yk , wk }

such that a(yk ) + b(wk ) = ρ 2 and G(yk , wk ) ’ 0. If we write wk = wk + h k , wk ∈

M , h k ∈ M0 , then (10.114) tells us that b(wk ) ’ 0. Thus, a(yk ) + b(h k ) ’ ρ 2 .

Since N0 , M0 are ¬nite-dimensional, there is a renamed subsequence such that yk ’ y

in N0 and h k ’ h in M0 . By (10.108) and (10.109), y ∞ ¤ δ/4 and h ∞ ¤ δ/4.

Consequently, (10.105) implies

2F(x, y, h) ¤ μ0 h 2 ’ »0 y 2 .

(10.116)

Since

G(y, h) = b(h) ’ a(y) ’ 2 F(x, y, h) d x = 0,

(10.117)

we have

{2F(x, y, h) + »0 y 2 ’ μ0 h 2 }d x = 0.

(10.118)

10.4. Some applications 125

In view of (10.116), this implies

2F(x, y, h) ≡ μ0 h 2 ’ »0 y 2 .

(10.119)

∞

For ζ ∈ C0 ( ) and t > 0 small, we have

2[F(x, y + tζ, h) ’ F(x, y, h)]/t ¤ ’»0 [(y + tζ )2 ’ y 2 ]/t.

(10.120)

Taking t ’ 0, we have

f (x, y, h)ζ ¤ ’»0 yζ.

(10.121)

∞

Since this is true for all ζ ∈ C0 ( ), we have

f (x, y, h) = ’»0 = ’Ay.

(10.122)

Similarly,

2[F(x, y, h + tζ ) ’ F(x, y, h)]/t ¤ μ0 [(h + tζ )2 ’ h 2 ]/t,

(10.123)

and, consequently,

g(x, y, h)ζ ¤ μ0 hζ

(10.124)

and

g(x, y, h) = μ0 h = Bh.

(10.125)

We see from (15.89) and (10.125) that (10.76), (10.77) has a nontrivial solution. Thus,

we may assume that (10.115) holds. But then we can combine it with (10.106) and

(10.107) to conclude from Theorem 10.2 that there is a sequence {u k } ‚ D such that

(10.3) holds with c ≥ . Arguing as in the proof of Theorem 10.7, we see that there

is a u ∈ D such that G(u) = c, G (u) = 0. Since c = 0 and G(0) = 0, we see that

u = 0, and the proof is complete.

Theorem 10.10. Assume that »0 is simple and the eigenfunctions corresponding to »0

and μ0 are bounded and = 0 a.e. in . Assume (10.80), (10.105) and

2F(x, 0, t) ¤ μ(x)t 2 , x∈ , t ∈ R,

(10.126)

where

μ(x) ¤≡ μ0 , x∈ .

(10.127)

Then (10.76), (10.77) has a nontrivial solution.

Proof. As in the proof of Theorem 10.7, (10.80) implies (10.93). As in the proof of

Theorem 10.9, (10.105) implies (10.115) for y ∈ N0 , w ∈ M unless (10.76), (10.77)

has a nontrivial solution. Thus, we may assume that (10.115) holds. Next we note that

there is a ν > 0 such that

G(0, w) ≥ νb(w), w ∈ M.

(10.128)

126 10. Weak Linking

Assuming this for the moment, we see that

inf G ≥ µ1 > 0,

(10.129)

B

where