From this we conclude that

» y0 (x) ≡ f (x, y0 (x)), x∈ .

(14.71)

Since y0 ∈ E(» ), this implies that y0 satis¬es (14.61).

Lemma 14.9. Let „ be a continuous map from N to M such that (14.41) holds and

de¬ne

H (v + w) = v + w + „ (v), v∈N , w∈M .

(14.72)

If F(x, t) satis¬es (14.20), then the following alternative holds: Either (a) for each

ρ > 0 suf¬ciently small, there is an µ > 0 such that

G(H w) ≥ µ, w∈M © ‚ Bρ ,

(14.73) ’1

or (b) there is a y ∈ E(» )\{0} such that (14.61) holds.

Proof. The proof is similar to that of Lemma 14.8. If w = w +y, w ∈ M , y ∈ E(» ),

then H w = w + „ (y), H w D ¤ C w D and

G(H w) ≥ µ w + „ (y) 2 , w∈M © Bρ ,

(14.74) ’1

for ρ suf¬ciently small by Lemma 14.7. If (a) does not hold and wk = wk + yk , wk ∈

M , yk ∈ E(» ), wk ∈ ‚ Bρ , and G(H wk ) ’ 0, then wk + „ (yk ) ’ 0, yk ’

y1 , „ (yk ) ’ „ (y1 ), and wk ’ ’„ (y1 ) for a renamed subsequence. If y1 = 0, then

wk ’ 0, contradicting the fact that wk D = ρ. Now H wk ’ y1 and

G(y1 ) = 0.

(14.75)

If ρ is suf¬ciently small, (14.51) holds and

2F(x, y1(x)) ¤ » y1 (x)2 , x∈ ,

(14.76)

by (14.20). This together with (14.75) implies that

2F(x, y1 (x)) ≡ » y1 (x)2 , x∈ .

(14.77)

This and (14.20) imply that

t ’1 2[F(x, y1 + tζ ) ’ F(x, y1 )] ¤ t ’1 » [(y1 + tζ )2 ’ y1 ]

2

(14.78)

∞

for t small and ζ ∈ C0 ( ). This yields

f (x, y1 ) ≡ » y1 (x), x∈

(14.79)

and y1 is a solution of (14.61).

172 14. Type (II) Regions

14.4 The solutions

We can now present the proof of Theorems 14.2 and 14.3. We begin with the proof of

Theorem 14.2.

Proof. We shall study G given by (14.48) and look for solutions of (14.49). By (14.13)

we have for w ∈ M ’1 , v 0 , v 1 ∈ N ’1 , v = v 1 ’ v 0 ,

1

(G (w + v 1 ) ’ G (w + v 0 ), v)

(14.80)

2

2

=v ’ ( f (w + v 1 ) ’ f (w + v 0 ), v)

D

2

¤ {» ’1 v ’ [ f (x, w + v 1 ) ’ f (x, w + v 0 )]v}d x.

The right-hand side will be negative unless v ≡ 0 in view of (14.13). Thus, there is a

ˆ

unique solution θ (w) of

ˆ

G(w + θ (w)) = sup G(w + v)

(14.81)

v∈N ’1

ˆ

(Lemma 13.5). Moreover, θ is continuous and satis¬es (14.32). De¬ne

ˆ

H (v + w) = v + w + θ (w), v∈N , w∈M ’1 ,

(14.82)

which is continuous as well. Let

¯

A R = [M © B R ] ∪ {u = w + sy0 : w ∈ M , s ≥ 0, u = R}

(14.83)

where y0 is the element of E(» )\{0} given in Lemma 14.4. By Example 7 of Sec-

tion 3.5, A R links N © ‚ Bρ [hm] whenever 0 < ρ < R. In view of Proposition 3.11,

H A R links H [N © ‚ Bρ ] [hm]. Let θ be the map described in Section 14.2 satisfying

(14.32)“(14.36). Then, by (14.81),

ˆ

(14.84) G(w + sy0 + θ (w + sy0 )) ≥ G(w + sy0 + θ (w + sy0 ))

= I (w + sy0 + θ (w + sy0 ), a, b)

’2 P(x, w + sy0 + θ (w + sy0 ))d x

≥ ’ W1 (x) d x

≡ ’B1

for w ∈ M , s ≥ 0. Thus,

G(u) ≥ ’B1 , u ∈ H A R.

(14.85)

14.4. The solutions 173

We can do a bit better on that portion of A R contained in M . In fact, (14.16) implies

ˆ

G(w + sy0 + θ (w)) ≥ G(w) = w ’2 F(x, w) d x

2

(14.86) D

≥ w ’» w

2 2

+1

D

≥ 0

for w ∈ M . On the other hand, we see from Lemma 14.8 that

G(u) ¤ ’µ < 0, u ∈ H [N © ‚ Bρ ],

(14.87)

for ρ > 0 suf¬ciently small unless there is a y ∈ E(» )\{0} satisfying (14.61). But in

the latter case the theorem is proved. Thus, we may assume that (14.87) holds. We can

now apply Theorem 3.15 to conclude that there is a sequence {u k } ‚ D such that

G(u k ) ’ c, ’∞ < c ¤ ’µ, G (u k ) ’ 0.

(14.88)

Thus

2

’2 F(x, u k )d x ’ c

(14.89) uk D