and

(u k , v) D ’ ( f (u k ), v) ’ 0, v ∈ D.

(14.90)

If

ρk = u k ’ ∞,

(14.91) D

let u k = u k /ρk . Then u k D = 1, and there is a renamed subsequence such that

˜ ˜

u k ’ u weakly in D, strongly in L 2 ( ), and a.e. in . By (14.56),

˜ ˜

|F(x, u k )|/ρk ¤ C(u 2 + |u k |/ρk )

˜k ˜

2

(14.92)

and the right-hand side converges to C u 2 in L 1 ( ). Moreover,

˜

2F(x, u k )/ρk ’ b(u + )2 + a(u ’ )2 as k ’ ∞ a.e. in

˜ ˜ .

2

(14.93)

Hence,

F(x, u k )d x/ρk ’ b u + + a u’ 2.

˜ ˜

2 2

(14.94) 2

But the left-hand side of (14.94) converges to 1 by (14.89). Thus, u ≡ 0. Now, by

˜

(14.10),

’1

| f (x, u k )|/ρk ¤ C(|u k | + ρk )

˜

(14.95)

174 14. Type (II) Regions

and the right-hand side converges to C|u| in L 2 ( ). Since

˜

f (x, u k )/ρk ’ b u + ’ a u ’ a.e. in

˜ ˜ as k ’ ∞,

(14.96)

we have

(u, v) D = (b u + ’ a u ’ , v),

˜ ˜ ˜ v ∈ D,

(14.97)

by (14.90). This says that u satis¬es (14.3). Since (a, b) ∈ , we must have u ≡

˜ ˜

0, contradicting the conclusion reached earlier. Thus, (14.91) cannot hold. But then

standard arguments imply that (14.1) has a solution u satisfying G(u) = c ¤ ’µ. This

shows that u ≡ 0, and the proof is complete.

Proof of Theorem 14.3. In this case we take

¯

A R = [N © B R ] ∪ {u = v + sy1 : v ∈ N ’1 , s ≥ 0, u = R},

(14.98) ’1

where y1 is the element of E(» )\{0} given by Lemma 14.5. By (14.18), there is a

continuous map „ from N to M such that (14.41) holds (for „ ) and

ˆ ˆ

G(v + „ (v)) = inf G(v + w).

ˆ

(14.99)

w∈M

If

H (v + w) = v + w + „ (v),

ˆ v∈N , w∈M ,

(14.100)

then H is a homeomorphism of D onto itself. Thus, H A R links H [N © ‚ Bρ ] for

0 < ρ < R (Example 7 of Section 3.5 and Proposition 3.11). Let „ be the map

described in Section 14.2 satisfying (14.41)“(14.45). Then we have

G(v + sy1 + „ (v + sy1 ))

ˆ ¤ G(v + sy1 + „ (v + sy1 ))

= I (v + sy1 + „ (v + sy1 ), a, b)

’2 P(x, v + sy1 + „ (v + sy1 ))d x

¤ B1

for v ∈ N ’1 , s ≥ 0. This implies

G(u) ¤ B1 , u ∈ H A R.

(14.101)

Moreover, we have

2

G(v + „ (v))

ˆ ¤ G(v) = v ’2 F(x, v) d x

D

¤ v ’» v ¤ 0, v∈N ’1 .

2 2

’1

D

Also, Lemma 14.9 implies that

G(w) ≥ µ > 0, w ∈ H [M © ‚ Bρ ],

(14.102) ’1

14.5. Notes and remarks 175

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

latter option implies the conclusion of the theorem. We may therefore assume that

(14.102) holds. We can now apply Theorem 3.15 to conclude that there is a sequence

{u k } ‚ D such that

G(u k ) ’ c, µ ¤ c < ∞, G (u k ) ’ 0.

(14.103)

We now follow the proof of Theorem 14.2 to conclude that there is a solution of (14.1)

satisfying G(u) = c ≥ µ. This implies that u ≡ 0, and the proof is complete.

14.5 Notes and remarks

Since the work of Fuˇ´k , many authors have studied other problems of the form (14.1)

c±

when (14.2) holds (cf., e.g.,[29]“[30], [52], [54], [55], [58], [56], [63], [64], [72], [73],

[80], [81], [82], [90], [105], [122]“[116] and the references quoted in them. In [72],

[73], [33], [30] the problem (14.1) was considered for (a, b) ∈ Q when » is a simple

eigenvalue. If • is a corresponding eigenfunction, then it was shown in [72] that for

each s ∈ R, there are a unique function u s and a unique constant Cs (a, b) such that

Au s = bu + ’ au ’ + Cs (a, b)•, (u s , •) = s,

(14.104) s s

with

Cs = Cs (a, b) = sC1 , s≥0

= sC’1 , s < 0.

Clearly, (a, b) ∈ iff C1 C’1 = 0. The region where C1 > 0, C’1 > 0 is below both

curves, the region C1 < 0, C’1 < 0 is above the curves, and the region C1 C’1 < 0