Thus, {x(tk )} is a Cauchy sequence in H. Since H is complete, x(tk ) converges to an

element x 1 ∈ H. Since x(tk ) ¤ u(tk ), we see that x 1 ¤ u(T ). Moreover, we note

that

x(t) ’ x 1 as t ’ T.

To see this, let µ > 0 be given. Then there is a k such that

x(tk ) ’ x 1 < µ, u(T ) ’ u(tk ) < µ.

36 4. Ordinary Differential Equations

Then, for tk ¤ t < T,

x(t) ’ x 1 ¤ x(t) ’ x(tk ) + x(tk ) ’ x 1

¤ u(t) ’ u(tk ) + x(tk ) ’ x 1 < 2µ.

We de¬ne x(T ) = x 1 . Then we have a solution of (4.20) satisfying (4.21) in [0, T ].

By Theorem 4.1, there is a unique solution of

d y(t)

= g(t, y(t)), y(T ) = x 1

(4.23)

dt

satisfying y(t) ¤ u(t) in some interval |t ’ T | < δ. By uniqueness, the solution of

(4.23) coincides with the solution of (4.20) in the interval (T ’ δ, T ]. De¬ne

z(t) = x(t), t0 ¤ t < T,

z(T ) = x 1 ,

z(t) = y(t), T < t ¤ T + δ.

This gives a solution of (4.20) satisfying (4.21) in the interval [t0 , T + δ), contradicting

the de¬nition of T. Hence, T = TM .

We also have the following.

Theorem 4.7. Let g(t, x) be a continuous map from R— H to H , where H is a Banach

space. Assume that for each point (t0 , x 0 ) ∈ R — H, there are constants K , b > 0 such

that

(4.24) g(t, x)’ g(t, y) ¤ K x ’ y , |t ’t0 | < b, x ’ x 0 < b, y ’ x 0 < b.

Assume also that g(t, x) satis¬es (4.15), where TM ¤ ∞, and γ (t), ρ(t) satisfy the

hypotheses of Lemma 4.2 with ρ(t) nondecreasing. Then, for each x 0 ∈ H and t0 ∈ R,

there is a unique solution x(t) of (4.16) depending continuously on x 0 and satisfying

x(t) ≥ u(t), t ∈ [t0 , TM ),

(4.25)

where u(t) is the solution of (4.8) in that interval.

4.5 An important estimate

Lemma 4.8. Let ρ, γ satisfy the hypotheses of Lemma 4.3, with ρ locally Lipschitz

continuous. Let u(t) be the solution of (4.8), and let h(t) be a continuous function

satisfying

t

h(t) ≥ h(s) ’ γ (r )ρ(h(r )) dr, t0 ¤ s < t < T, h(t0 ) ≥ u 0 .

(4.26)

s

Then

u(t) ¤ h(t), t ∈ [t0 , T ).

(4.27)

4.5. An important estimate 37

Proof. Assume that there is a point t1 in the interval such that

h(t1 ) < u(t1 ).

Let

y(t) = u(t) ’ h(t), t ∈ [t0 , T ).

Then y(t0 ) ¤ 0 and y(t1 ) > 0. Let „ be the largest point < t1 such that y(„ ) = 0. Then

y(t) > 0, t ∈ („, t1 ].

(4.28)

Moreover, by (4.8) and (4.26), we have

t t

y(t) ¤ ’ γ (s)[ρ(u(s)) ’ ρ(h(s))] ds ¤ L

(4.29) y(s) ds,

„ „

where L is the Lipschitz constant for ρ at u(„ ) times the maximum of γ in the interval.

Let

t

w(t) = y(s) ds.

„

Then

[e’Lt w(t)] = e’Lt [y(t) ’ Lw(t)] ¤ 0, t ∈ [„, t1 ].

Consequently,

e’Lt w(t) ¤ e’L„ w(„ ) = 0, t ∈ [„, t1 ].

Hence,

y(t) ¤ Lw(t) ¤ 0, t ∈ [„, t1 ],

contradicting (4.28). This completes the proof.

Chapter 5

The Method Using Flows

5.1 Introduction

In this chapter we give the proofs of the theorems of Chapter 2. They rely on the

theorems for ordinary differential equations in abstract spaces developed in Chapter 4.

5.2 Theorem 2.4

We begin with the proof of Theorem 2.4.

Proof of Theorem 2.4. First, we note that if the theorem were false, there would be a

δ > 0 and a ψ satisfying the hypotheses of Theorem 2.4 such that

||G (u) ≥ ψ( u )

(5.1)

when

u ∈ Q = {u ∈ E : |G(u) ’ a| ¤ 3δ}.

(5.2)

Reduce δ so that 3δ < a ’ a0 . Since G ∈ C 1 (E, R), for any θ ∈ (0, 1), there is a

ˆ

locally Lipschitz continuous mapping Y (u) of E = {u ∈ E : G (u) = 0} into E such

that

ˆ

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

(5.3)

(cf., e.g., [120]). Let

= {u ∈ E : |G(u) ’ a| ¤ 2δ},

Q0

= {u ∈ E : |G(u) ’ a| ¤ δ},

Q1

= E\Q 0 ,

Q2

·(u) = d(u, Q 2 )/[d(u, Q 1 ) + d(u, Q 2 )].

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_5,

© Birkh¤user Boston, a part of Springer Science+Business Media, LLC 2009

40 5. The Method Using Flows

It is easily checked that ·(u) is locally Lipschitz continuous on E and satis¬es

§

⎪·(u) = 1, u ∈ Q1,

⎪

⎨

¯

·(u) = 0, u ∈ Q2,

(5.4)

⎪

⎪

©·(u) ∈ (0, 1), otherwise.