Schechter Minimax Systems and Critical Point Theory (Springer, 2009)

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 − x0 ≤ R0}

and

I0 = {t R : |t − t0| ≤ T0}.

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

(4.1)	g(t, x) − g(t, y) ≤ K0 x − y ,					x, y B0, t I0,
and
(4.2)				g(t, x) ≤ M0,	x B0, t I0.
Let T1 be such that
(4.3)				T1 ≤ min(T0, R0/M0),		K0T1 < 1.
Then there is a unique solution x(t) of
(4.4)		dx(t)		= g(t, x(t)), |t	− t0| ≤ T1, x(t0) = x0.
		dt

Lemma 4.2. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) nonnegative and ρ(t) positive and nondecreasing. Assume that

∞	dτ			T
(4.5)			>	γ (s) ds,
	ρ(τ )
u0				t0

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_4, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

32					4. Ordinary Differential Equations
where t0 < T and u0 ≥ 0 are given numbers. Then there is a unique solution of
(4.6)	u (t) = γ (t)ρ(u(t)),		t [t0, T ), u(t0) = u0
that is positive in [t0, T ) and depends continuously on u0.
Proof. One can separate variables to obtain
	u		dτ		t
	W (u) ≡ u0				= t0	γ (s) ds.
		ρ(τ )

The function W (u) is differentiable and increasing in R, is positive in (u0, ∞), depends continuously on u0, and satisfies

	∞ dτ				T	γ (s) ds as u → ∞.
W (u) → L = u0				>
		ρ(τ )			t0
Thus, for each t [t0, T ), there is a unique u [u0, ∞) such that
				t
	W (u) = t0				γ (s) ds.
Hence,					t
	=	W −1
u					γ (s) ds
				t0

is the unique solution of (4.6) and depends continuously on u0.

Lemma 4.3. Let γ (t) and ρ(t) be continuous functions on [0, ∞), with γ (t) nonnegative and ρ(t) positive and nondecreasing. Assume that

u0	dτ			T
(4.7)			>	γ (s) ds,
	ρ(τ )
m				t0

where t0 < T and m < u0 are given numbers. Then there is a unique solution of

(4.8)

u (t) = −γ (t)ρ(u(t)), t [t0, T ), u(t0) = u0,

which is ≥ m in [t0, T ) and depends continuously on u0.

Proof. One can separate variables to obtain

u0

W (u) ≡

u

	dτ		t
			= t0	γ (s) ds.
	ρ(τ )

The function W (u) is differentiable and decreasing in R, is positive in [m, u0], depends continuously on u0, and satisfies

u0	dτ			T	γ (s) ds as u → m.
W (u) → L = m			>
	ρ(τ )			t0

4.2. Estimating solutions				33
Thus, for each t [t0, T ), there is a unique u [m, u0] such that
			t
	W (u) = t0			γ (s) ds.
Hence,				t
		W −1
u	=			γ (s) ds
			t0

is the unique solution of (4.8) and depends continuously on u0.

4.2 Estimating solutions

Theorem 4.4. Assume, in addition to the hypotheses of Theorem 4.1, that

(4.9) g(t, x) ≤ γ (t)ρ( x ), x B0, t I0,

where γ (t) and ρ(t) satisfy the hypotheses of Lemma 4.2 with T = t0 + T1. Let u(t) be the positive solution of

(4.10)	u (t) = γ (t)ρ(u(t)), t [t0, T ), u(t0) = u0 ≥ x0
provided by that lemma. Then the unique solution of (4.4) satisfies
(4.11)	x(t) ≤ u(t),	t [t0, T ).
Proof. Assume that there is a t1 [t0, T ) such that
	u(t1) < x(t1) .
For ε > 0, let uε (t) be the solution of
(4.12)	u (t) = [γ (t) + ε]ρ(u(t)),	t [t0, T ), u(t0) = u0.

By Lemma 4.2, a solution exists for ε > 0 sufficiently small. Moreover, uε (t) → u(t) uniformly on any compact subset of [t0, T ). Let

w(t) = x(t) − uε (t).

Then we may take ε sufficiently small so that

w(t0) ≤ 0, w(t1) > 0.

Let t2 be the largest number in [t0, t1) such that w(t2) = 0 and w(t) > 0, t (t2, t1].

For h > 0 sufficiently small, we have

w(t2 + h) − w(t2) > 0.

h

34

Consequently,

But (4.13)

4. Ordinary Differential Equations

D+w(t2) ≥ 0.

D+ w(t2) = D+ x(t2) − uε (t2)

≤ x (t2) − uε (t2)

=g(t2, x(t2)) − [γ (t2) + ε]ρ(uε (t2)) ≤ γ (t2)ρ( x(t2) ) − [γ (t2) + ε]ρ(uε (t2))

=− ερ(uε (t2))

< 0.

This contradiction proves the theorem.

4.3 Extending solutions

Theorem 4.5. Let g(t, x) be a continuous map from space. Assume that for each point (t0, x0) R × H, that

R × H to H , where H is a Banach there are constants K , b > 0 such

(4.14)	g(t, x) − g(t, y) ≤ K x − y , |t − t0| < b, x − x0 < b, y − x0 < b.
Assume also that
(4.15)	g(t, x) ≤ γ (t)ρ( x ), x H, t [t0, TM ),

where TM ≤ ∞, and γ (t), ρ(t) satisfy the hypotheses of Lemma 4.2. Then, for each x0 H and t0 > 0, there is a unique solution x(t) of the equation

(4.16)	dx(t)	= g(t, x(t)), t	[t0, TM ), x(t0) = x0.
	dt
Moreover, x(t) depends continuously on x0 and satisfies
(4.17)		x(t) ≤ u(t),	t [t0, TM ),

where u(t) is the solution of (4.6) in that interval satisfying u(t0) = u0 ≥ x0 .

Before proving Theorem 4.5, we note that the following is an immediate consequence.

Corollary 4.6. Let V (y) be a locally Lipschitz continuous map from H to itself satisfying

(4.18)	V (y) ≤ C(1 + y ),	y H.
Then, for each y0 H, there is a unique solution of
(4.19)	y (t) = V (y(t)), t R+,	y(0) = y0.

4.4. The proofs

35

4.4 The proofs

We now give the proof of Theorem 4.5.

Proof. By Theorems 4.1 and 4.4, there is an interval [t0, t0 + m], m > 0, in which a unique solution of

(4.20)

dx(t)

= g(t, x(t)),

t

[t0, t0 + m],

x(t0) = x0

dt

exists and satisfies

(4.21)

x(t) ≤ u(t),

t [t0, t0 + m],

where u(t) is the unique solution of

(4.22)

u

(t)

=

γ (t)ρ(u(t)),

t

[t0, TM ), u(t0)

=

u

0

=

x0 .

Let T ≤ TM be the supremum of all numbers t0 + m for which this holds. If t1 < t2 < T, then the solution in [t0, t2] coincides with that in [t0, t1], since such solutions are unique. Thus, a unique solution of (4.20) satisfying (4.21) exists for each t0 < t < T. Moreover, we have

	t2
x(t2) − x(t1) = t1		g(t, x(t)) dt.
Consequently,
x(t2) − x(t1) ≤	t2	g(t, x(t)) dt
	t1

t2

≤γ (t)ρ( x(t) ) dt

t1 t2

≤γ (t)ρ(u(t)) dt

t1

= u(t2) − u(t1).

Assume that T < TM . Let tk be a sequence such that t0 < tk < T and tk → T . Then,

x(tk ) − x(t j ) ≤ u(tk ) − u(t j ) → 0.

Thus, {x(tk )} is a Cauchy sequence in H. Since H is complete, x(tk ) converges to an element x1 H. Since x(tk ) ≤ u(tk ), we see that x1 ≤ u(T ). Moreover, we note that

x(t) → x1 as t → T.

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

x(tk ) − x1 < ε, u(T ) − u(tk ) < ε.

36					4. Ordinary Differential Equations
Then, for tk ≤ t < T,
	x(t) − x1 ≤ x(t) − x(tk ) + x(tk ) − x1
			≤ u(t) − u(tk ) + x(tk ) − x1 < 2ε.
We define x(T )	= x1. Then we have a solution of (4.20) satisfying (4.21) in [0, T ].
By Theorem 4.1, there is a unique solution of
(4.23)		dy(t)			= g(t, y(t)), y(T ) = x1
		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 ]. Define

z(t) = x(t), t0 ≤ t < T, z(T ) = x1,

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 definition 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, x0) 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 − x0 < b, y − x0 < b.

Assume also that g(t, x) satisfies (4.15), where TM ≤ ∞, and γ (t), ρ(t) satisfy the hypotheses of Lemma 4.2 with ρ(t) nondecreasing. Then, for each x0 H and t0 R, there is a unique solution x(t) of (4.16) depending continuously on x0 and satisfying

(4.25) x(t) ≥ u(t), t [t0, TM ),

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

4.5 An important estimate

Lemma 4.8. Let ρ, γ continuous. Let u(t) satisfying

satisfy the hypotheses of Lemma 4.3, with ρ locally Lipschitz be the solution of (4.8), and let h(t) be a continuous function

		t
(4.26)	h(t) ≥ h(s) − s	γ (r )ρ(h(r )) dr, t0 ≤ s < t < T, h(t0) ≥ u0.
Then
(4.27)		u(t) ≤ h(t), t [t0, T ).

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 τ be the largest point < t1 such that y(τ ) = 0. Then

(4.28)		y(t) > 0, t (τ, t1].
Moreover, by (4.8) and (4.26), we have
	t			t
(4.29)	y(t) ≤ − τ	γ (s)[ρ(u(s)) − ρ(h(s))] ds ≤ L	τ	y(s) ds,

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

w(t) = τ	t
	y(s) ds.

Then

[e−Lt w(t)] = e−Lt [y(t) − Lw(t)] ≤ 0, t [τ, t1].

Consequently,

e−Lt w(t) ≤ e−Lτ w(τ ) = 0, t [τ, t1].

Hence,

y(t) ≤ Lw(t) ≤ 0, t [τ, 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

(5.1)	||	G	(u)	≥	ψ(	u )
when
(5.2)	u Q = {u E : |G(u) − a| ≤ 3δ}.

Reduce δ so that 3δ < a − a0. Since G locally Lipschitz continuous mapping Y (u) that

C1(E, R), for any θ					(0, 1), there is a
of E =	{	u		E : G (u)		=	0	}	into E such
ˆ

(5.3)		Y (u)	≤	1, θ		G (u)	≤	(G (u), Y (u)),			ˆ
									u		E
(cf., e.g., [120]). Let
		Q0		= {u E : |G(u) − a| ≤ 2δ},
		Q1		= {u E : |G(u) − a| ≤ δ},
		Q2		= E\Q0,
		η(u) = d(u, Q2)/[d(u, Q1) + d(u, Q2)].

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_5, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

40		5. The Method Using Flows
It is easily checked that η(u) is locally Lipschitz continuous on E and satisfies
	η(u) = 1,	u Q1,
(5.4)	η(u) = 0,	u ¯ 2,
		Q
	η(u) (0, 1),	otherwise.

Let ρ(t) = 1/ψ(t). Then ρ is a positive, nondecreasing, locally Lipschitz continuous function on [0, ∞) such that

				∞ dτ
(5.5)									= ∞
				0		ρ(τ )
by (2.8). Let			W (u) = −η(u)Y (u)ρ( u ).
Then		W (u) ≤ ρ( u ),								u E.
By (5.5), for each u E, there is a unique solution of
(5.6)	σ (t) = W (σ (t)),							t R+, σ (0) = u
(cf. Theorem 4.5). We have
(5.7)	dG(σ (t)u)/dt	=	−	η(σ (t)u)(G (σ (t)u), Y (σ (t)u))ρ( σ (t)u							)
							G
		≤	−	θ η(σ )				(σ ) ρ( σ )
		≤ −θ η(σ ).
Thus,
		G(σ (t)u) ≤ G(u),								t ≥ 0,
		G(σ (t)u) ≤ a0,							u A, t ≥ 0,
and
(5.8)			σ (t)u = u,					u A, t ≥ 0.
This follows from the fact that
	G(σ (t) u) ≤ a0 < a − 3δ, u A, t ≥ 0.
Hence, η(σ (t) u) = 0 for u A, t ≥ 0. This means that
		σ (t) u = 0,						u A, t ≥ 0,
and	σ (t) u = σ (0) u = u, u A, t ≥ 0.

5.3. Theorem 2.12	41
Let
(5.9)	Eα = {u E : G(u) ≤ α}.
There is a T > 0 such that
(5.10)	σ (T )Ea+δ Ea−δ .

In fact, we can take T > 2δ/θ . To see this, let u be any element in Ea+δ . If there is a t1 [0, T ] such that σ (t1)u / Q1, then

G(σ (T )u) ≤ G(σ (t1)u) < a − δ

by (5.7). Hence, σ (T )u Ea−δ . On the other hand, if σ (t)u Q1 for all t [0, T ],

then η(σ (t)u) = 1 for all such t, and (5.7) yields

(5.11)

G(σ (T )u) ≤ G(u) − θ T < a − δ.

Hence, (5.10) holds. Now, by (2.6), there is a K K such that

(5.12)

σ (T )

( A). Let ˜

K Ea+δ .

K

As we saw,

= σ (

T

)(

K

).

˜

by definition. But

K

Then K

sup G = sup G(σ (T )u) < a − δ,

˜

u

K

K

which contradicts (2.6), proving the theorem.

Next, we prove Theorem 2.8.

Proof. By (2.6) and (2.17), we see that b0 ≤ a. This implies (2.7) in view of (2.18). The result now follows from Theorem 2.4.

Theorem 2.11 follows obviously from Theorem 2.8.

5.3 Theorem 2.12

Now we give the proof of Theorem 2.12.

Proof. If a0 < a, the conclusion follows from Theorem 2.4. Assume that a0 = a. If there did not exist a sequence satisfying (2.9), then there would be positive numbers δ, θ , T such that 2δ < θ T and (5.1) holds whenever u Q, where Q is given by (5.2). Since a = a0, we see by (2.6), (2.17), and (2.20) that b0 = a. Define Q0, Q1, Q2, and η(u) as before and let σ (t) be the flow generated by the mapping (5.6). Let u be any element in Ea+δ . If there is a t1 ≤ T such that σ (t1)u / Q1, then

(5.13) G(σ (t1)u) < a − δ.