G(u 0 ) = ±. Clearly, 0 is also a critical point. Assume that there are no others. Then

ˆ

the set E = {u ∈ E : G (u) = 0} contains all points except u 0 and 0. If θ < 1, then

ˆ

there is a mapping Y (u) from E to E that is locally Lipschitz continuous and satis¬es

ˆ

Y (u) ¤ 1, θ G (u) ¤ (G (u), Y (u)), u ∈ E.

For v ∈ N © B R \{0}, let σ (t)v be the solution of

σ (t) = ’Y (σ (t)), t ≥ 0, σ (0) = v.

16.5. Local linking 207

Then

d G(σ (t)v)/dt = (G (σ ), σ ) = ’(G (σ ), Y (σ )) ¤ ’θ G (σ ) < 0

ˆ

as long as σ (t)v is in E. Note that σ (t)v is continuous in t and v for v = 0. For each

v ∈ N © B R \{0}, there is a maximal interval 0 < t < Tv in which σ (t)v exists and

satis¬es G (σ (t)v) = 0 and G(σ (t)v) < 0. I claim that

σ (t)v ’ u 0 as t ’ Tv .

(16.65)

To see this, suppose that tk ’ Tv . Then

tk

σ (tk )v ’ σ (t j )v ¤ | Y (σ (t)v) dt| ¤ |tk ’ t j | ’ 0.

tj

Thus,

σ (tk )v ’ h

in E. By continuity,

G (σ (tk )v) ’ G (h).

If G (h) = 0, the solution can be continued beyond Tv , contrary to the way it was

chosen. Thus, G (h) = 0, showing that h = u 0 . Consequently, σ (tk )v ’ u 0 . Since

this is true for any such sequence, (16.65) holds. Note that Tv is continuous in v for

v = 0.

De¬ne

σ (t)v = σ (t)v,

ˆ t < Tv ,

(16.66)

= u0, t ≥ Tv .

Let w0 be an element of M with unit norm, and take

K 0 = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v = R}.

Let

Q = {sw0 + v : v ∈ N, s ≥ 0, sw0 + v ¤ R}.

Let µ > 0 be given, and let T > 0 satisfy

µ2

Tv2

+ 2 ¤ 1, v ∈ N, v = µ.

(16.67)

T2 R

Let ξ(u) be the continuous map from ‚ Q to E such that

ξ(v) = v, v ∈ N © BR ,

and for u = sw0 + v ∈ K 0 ,

ξ(sw0 + v) = σ (T s/R)v,

ˆ v ≥ µ,

(16.68)

= u0, v < µ.

208 16. Multiple Solutions

By (16.67) and (16.68),

σ (T s/R)v = u 0 ,

ˆ v = µ.

Hence, ξ is continuous on ‚ Q. Moreover,

G(ξ(u)) ¤ 0, u ∈ ‚ Q.

In addition,

ξ(u) ≥ r > 0, u ∈ K0.

Let

B = ‚ Bδ © M, 0 < δ < r < R.

By Corollary 16.11, A = ξ(‚ Q) links B [mm]. Since

a0 := sup G ¤ b0 := inf G,

(16.69)

B

A

we can apply Theorem 2.12 to conclude that (1.4) holds. If a > 0, this provides a third

critical point by the PS condition. If a = 0, then there is a sequence satisfying (1.4)

and

d(u k , B) ’ 0, k ’ ∞.

(16.70)

Since G satis¬es the PS condition, there is a subsequence converging to a critical point

on B. Again, this provides a third critical point.

16.6 The proofs

We prove the theorems of Section 16.3. First, we prove Theorem 16.3.

Proof. By Lemma 16.12, it suf¬ces to show that J (v) has two nontrivial solutions.

Now J is bounded from above by Lemma 16.14 and satis¬es (PS) by (16.39). More-

over,

J (v) < 0, v ∈ Nl © Br \{0},

(16.71)

by Lemma 16.13, and

J (ξ(y) + y) > 0, y ∈ Nm © Ml © Br \{0},

(16.72)

by Lemma 16.16. Thus, J has a positive maximum on Nm . We can now apply Theo-

rem 16.26 and Lemma 16.9 to obtain the desired conclusion.

16.7. Notes and remarks 209

Similarly, we prove Theorem 16.4.

Proof. By Lemma 16.18, it suf¬ces to show that J (w) given by (16.46) has two non-

trivial solutions. Now J is bounded from below by Lemma 16.21 and satis¬es (PS) by

(16.56). Moreover,

J (w + ·(w)) < 0, w ∈ Nl © Mm © Br \{0},

(16.73)

by Lemma 16.19, and

J (w) > 0, w ∈ Ml © Br \{0}

(16.74)

by Lemma 16.20. Thus, J has a negative minimum on Mm . We can now apply Theo-

rem 16.26 and Lemma 16.9 to obtain the desired conclusion.

Next, we prove Theorem 16.5.

Proof. With reference to Theorem 16.3, we note that, by Lemma 16.12, it suf¬ces to

show that J (v) has two nontrivial solutions. Now J is bounded from above by Lemma

16.14 and satis¬es (PS) by (16.39). Moreover,

J (v) < 0, v ∈ Nl © Br \{0}

(16.75)

by Lemma 16.22, and