Proof. We refer to Theorem 2.24. We take a sequence of points such that G(v k ) >

a0 ’ (1/k) and a sequence {Rk } such that Rk > 2 v k and

’1

Rk +2βk

ck = 2 k ψ(t) dt ’ 0, k ’ ∞,

2βk

where βk = v k . We then take

ψk (t) = ck ψ(t + βk )

in Theorem 2.24. Then (2.53) holds. We used the fact that ‚ B Rk +βk links {v k } for each

k (Theorem 3.12). The conclusion follows since

u ¤ d(u, v k ) + βk .

In the second case, we replace G with ’G.

Corollary 3.22. If (3.36) holds, then there is a sequence satisfying

G(u k ) ’ a0 , (1 + u k )G (u k ) ’ 0.

(3.40)

If (3.38) holds, there is a sequence satisfying

G(u k ) ’ b0 , (1 + u k )G (u k ) ’ 0.

(3.41)

3.8 Notes and remarks

The results of Section 3.2 are from [136], [114], and [120] (cf. also [122]). Sections 3.5

and 3.6 are from [122]. The results of Section 3.7 come from [143], [109], [108], [129],

and [132].

Chapter 4

Ordinary Differential Equations

4.1 Extensions of Picard™s theorem

In proving the theorems of Chapter 2, we shall make use of various extensions of

Picard™s theorem in a Banach space. Some are well known, and some are of interest in

their own right. All of them will be used in proving the theorems of Chapter 2.

Theorem 4.1. Let X be a Banach space, and let

B0 = {x ∈ X : x ’ x 0 ¤ R0 }

and

I0 = {t ∈ R : |t ’ t0 | ¤ T0 }.

Assume that g(t, x) is a continuous map of I0 — B0 into X such that

g(t, x) ’ g(t, y) ¤ K 0 x ’ y , x, y ∈ B0 , t ∈ I0 ,

(4.1)

and

g(t, x) ¤ M0 , x ∈ B0 , t ∈ I 0 .

(4.2)

Let T1 be such that

T1 ¤ min(T0 , R0 /M0 ), K 0 T1 < 1.

(4.3)

Then there is a unique solution x(t) of

d x(t)

= g(t, x(t)), |t ’ t0 | ¤ T1 , x(t0 ) = x 0 .

(4.4)

dt

Lemma 4.2. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) non-

negative and ρ(t) positive and nondecreasing. Assume that

∞ T

d„

> γ (s) ds,

(4.5)

ρ(„ )

u0 t0

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_4,

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

32 4. Ordinary Differential Equations

where t0 < T and u 0 ≥ 0 are given numbers. Then there is a unique solution of

u (t) = γ (t)ρ(u(t)), t ∈ [t0 , T ), u(t0 ) = u 0

(4.6)

that is positive in [t0 , T ) and depends continuously on u 0 .

Proof. One can separate variables to obtain

u t

d„

W (u) ≡ = γ (s) ds.

ρ(„ )

u0 t0

The function W (u) is differentiable and increasing in R, is positive in (u 0 , ∞), depends

continuously on u 0 , and satis¬es

∞ T

d„

W (u) ’ L = > γ (s) ds as u ’ ∞.

ρ(„ )

u0 t0

Thus, for each t ∈ [t0 , T ), there is a unique u ∈ [u 0 , ∞) such that

t

W (u) = γ (s) ds.

t0

Hence,

t

’1

u=W γ (s) ds

t0

is the unique solution of (4.6) and depends continuously on u 0 .

Lemma 4.3. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) non-

negative and ρ(t) positive and nondecreasing. Assume that

u0 T

d„

> γ (s) ds,

(4.7)

ρ(„ )

m t0

where t0 < T and m < u 0 are given numbers. Then there is a unique solution of

u (t) = ’γ (t)ρ(u(t)), t ∈ [t0 , T ), u(t0 ) = u 0 ,

(4.8)

which is ≥ m in [t0 , T ) and depends continuously on u 0 .

Proof. One can separate variables to obtain

u0 t

d„

W (u) ≡ = γ (s) ds.

ρ(„ )

u t0

The function W (u) is differentiable and decreasing in R, is positive in [m, u 0 ], depends

continuously on u 0 , and satis¬es

u0 T

d„

W (u) ’ L = > γ (s) ds as u ’ m.

ρ(„ )

m t0

4.2. Estimating solutions 33

Thus, for each t ∈ [t0 , T ), there is a unique u ∈ [m, u 0 ] such that

t

W (u) = γ (s) ds.

t0