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

94

9. Superlinear Problems

in view of the fact that y 2D = λ0 y 2 and (9.28) holds. Thus, by (8.3),

(9.29) G

(u)

(λ

1)

y 2

λ

λ0

C w q−2

w 2 ,

u

D

ρ.

≥

−

+

− λ1

−

≤

λ

D

D

D

We take ρ > 0 to satisfy

1− λ0 > C ρq−2. λ1

This gives

	Gλ(u) ≥ (λ − 1)ρ2 +	λ −	λ0	− C	ρq−2 − λ + 1				w 2D
			λ1
	≥ (λ − 1)ρ2, u D = ρ.
	Hence, (9.23) holds.					=	˜		=
	To prove (9.24) under hypothesis (C ), let σ							and		(σ, K ). Under
							λ/λ0
	hypothesis (C ), we have in place of (9.29)
	(9.30) Gλ(u) ≥ (λ − σ ) y D +	λ −	λ	− C	w D−			w D , u D ≤ ρ.
			λ1
	2		˜				q 2		2

We take ρ > 0 to satisfy

˜

σ − λ > C ρq−2. λ1

Consequently,

Gλ(u) ≥ (λ − σ )ρ	2	+	λ −	λ	− C ρ	q	−	2	− λ + σ	w D
				λ1
				˜						2

≥ (λ − σ )ρ2, u D = ρ.

This gives (9.24), and the proof is complete.

We now turn to the proofs of Theorems 9.6 and 9.7. We prove the latter first. We shall prove Theorem 9.7 by applying Theorems 9.15 and 9.16.

Proof. We take E

=

D,

= (σ,

K

),

where

σ

=

˜

0

, K

> 1 is a finite number,

λ/λ

and

I (u) = u 2D ,

J (u) = 2

F(x, u) dx.

For the purpose of this application, it is sufficient to know that the sets

A± = [0, ± Rϕ0], B = {x D : x D = ρ}

link each other if R > ρ (cf., e.g., [120]). In our case hypothesis (H1) is satisfied. We now check that (H3) holds. We observe that

Gλ(u) = 0

9.5. The parameter problem		95
is equivalent to (9.10) with β = 1/λ. Now, at least one of the expressions
J (± Rϕ0)/ R2 = 2	F(x, ± Rϕ0) dx/ R2 → ∞ as R → ∞
by hypothesis (D ). Hence, for R sufficiently large, one of the inequalities
Gλ(± Rϕ0)/ R2 ≤ K ϕ0 2D − 2		{F(x, ± Rϕ0)/ R2 ≤ 0
holds. Thus,
	a0(λ) ≤ 0,	λ .

Moreover, it follows from Theorem 9.16 that (9.24) holds. Hence, b0(λ) ≥ (λ − σ )ρ2, λ .

This shows that hypothesis (H3) holds. We can now apply Theorem 9.15 to con-

clude that for almost all λ , there exists uk (λ) D such that supk uk (λ) <

∞, Gλ(uk (λ)) → 0, and

Gλ(uk (λ)) → a(λ) ≥ b0(λ).

Once it is known that the sequence {uk } is bounded, we can apply the usual theory to conclude that there is a solution of

G		(u)	=	0, G	(u)	=	a(λ)
	λ			λ

(cf., e.g., [122, p. 64]). Moreover, from the definition, we see that a(λ) ≥ (λ − σ )ρ2.

λ	=	0 has a nontrivial solution for almost every λ		. This
Hence, the equation G (u)

is equivalent to (9.10) having a nontrivial solution for almost every β (K −1, σ −1). Since K was arbitrary, the result follows.

						˜	=	0	and show that hypothesis (D)
To prove Theorem 9.6, it suffices to take λ							λ
implies hypothesis (D ). To see this, we note that
F(x,	±	Rϕ	0	) dx/ R2	=	F(x, ± Rϕ0)				ϕ2dx	→ ∞
							R2ϕ2			0
							0

by hypothesis (D) and the fact that ϕ0(x) = 0 a.e.

To prove Corollary 9.8, we let ε be any positive number. By hypothesis (C ), there is a δ > 0 such that

F(x, t)/t2 ≤ ε, |t| ≤ δ, x .

By Theorem 9.7, (9.10) has a nontrivial solution for a.e. β (0, λ0/ε). Since ε was arbitrary, the result follows.

We now give the proof of Theorem 9.4.

96	9. Superlinear Problems

Proof. By Theorem 2.4, for each arbitrary K > 1, and a.e. λ (1, K ), there exists uλ such that Gλ(uλ) = 0, Gλ(uλ) = a(λ) ≥ (λ − 1)ρ2. Choose λn → 1, λn > 1. Then there exists un such that

G

(un )

=

0,

G

λn

(un )

=

a(λn )

≥

a(1)

≥

b0(1).

λn

By Theorem 3.4, we may assume that b0(1) ≥ ε > 0. Therefore,

2F(x, un )

dx

≤ c.

{

n } is

un 2D

=

n /

n D

Now we prove that

u

2

n D

→ ∞

, let

wn

u

; then

bounded. If

u

u

wn → w weakly in D, strongly in L ( ), and a.e. in .

We now have two cases:

Case 1: w = 0 in D. We get a contradiction as follows:

c ≥

2F(x, un)

2F(x, un )

|wn |2dx

dx =

2

u

2

un D

n

≥

2F(x, un )

|wn |2dx −

W1(x) dx → ∞.

w=0

u2

n

w=0

Case 2: w = 0 in D. We define tn [0, 1] by

G

λn

(tn un ) = t

max G

λn

(tun ).

[0,1]

For any c > 0 and w¯ n = cwn , we have

F(x, w¯ n ) dx → 0

(cf., e.g., [122, p. 64]). Thus,

Gλn (tn un ) ≥ Gλn (cwn ) = c2λn − 2 F(x, w¯ n )

for n large enough. That is, limn→∞ Gλn (tn un ) = ∞ and (Gλn fore,

dx ≥ c2/2

(tn un ), un ) = 0. There-

Gλn (tn un ) = f (x, tn un )tn un − 2F(x, tn un ) dx

=H (x, tnun ) dx → ∞.

By hypothesis (E ),

Gλn (un ) = H (x, un) dx ≥ H (x, tnun ) dx → ∞.

9.6. The monotonicity trick		97
But
Gλn (un) = a(λn) ≤	sup	Gλn ((1 − s)u)
≤	s [0,1] ,u A	G K ((1 − s)u)
	sup
	s [0,1] ,u A
<	c,

a contradiction. Thus, un D ≤ C. It now follows that

G (un ) → 0, G(un) → a(1) ≥ b0(1).

We can now apply Theorem 3.4.1 in [120, p. 64] to obtain the desired solution.

9.6 The monotonicity trick

We now give the proof of Theorem 9.15.

Proof. First, we prove conclusion (1) with the first alternative (H1).

Evidently, a(λ) ≥ b0(λ) since A links B. By (H1), the map λ → a(λ) is non-

decreasing. Hence, a (λ)

=

da(λ)/dλ exists for almost every λ

. From this point

:

on, we consider those λ where a (λ) exists.

For fixed λ

, let λn (λ, 2λ) ∩

, λn → λ as n → ∞. Then there exists n¯(λ) such that

(9.31)

a (λ)

−

1

≤

a(λn ) − a(λ)

≤

a (λ)

+

1

forn

n(λ).

λn − λ

≥ ¯

Next, we note that there exist n , k0 := k0(λ) > 0 such that

(9.32)

n (s)u ≤ k0 whenever

Gλ( n (s)u) ≥ a(λ) − (λn − λ).

In fact, by the definition of a(λn), there exists n such that

(9.33)

sup

Gλ( n (s)u) ≤

sup

Gλn ( n (s)u) ≤ a(λn) + (λn − λ).

s [0,1],u A

s [0,1],u A

If Gλ( n (s)u) ≥ a(λ) − (λn − λ) for some u A, s [0, 1], then, by (9.31) and (9.33), we have that

(9.34)	I (	(s)u)	=		Gλn ( n (s)u) − Gλ( n (s)u)
	n				λn − λ
			≤		a(λn) + (λn − λ) − a(λ) + (λn − λ)
					λn	−	λ
			≤	a (λ) + 3,

and it follows that

(9.35) J ( n (s)u) = λn I ( n (s)u) − Gλn ( n(s)u)

≤λn (a (λ) + 3) − Gλ( n (s)u)

≤λn (a (λ) + 3) − a(λ) + (λn − λ)

≤2λ(a (λ) + 3) − a(λ) + λ.

98		9. Superlinear Problems
On the other hand, by (H1), (9.31), and (9.33),
(9.36)	J ( n (s)u) =	λn I ( n (s)u) − Gλn ( n (s)u)
	≥ −Gλn ( n (s)u)
	≥	−(a(λn ) + (λn − λ))
	≥	−(a(λ) + (λn − λ)(a (λ) + 2))

≥ −a(λ) − λ|a (λ) + 2|.

Combining (9.34)–(9.37) and (H1), we see that there exists k0(λ) := k0 (depending only on λ) such that (9.32) holds.

First, we consider the case of a(λ) > b0(λ). For each n, we define

(9.37)	Qn (λ) := {u E : u ≤ k0 + 1, |Gλ(u) − a(λ)| ≤ λn − λ}.
We claim that Qn (λ) = φ and that									} →
(9.38)	{	λ		: u		Q	n	(λ)		0
	inf	G (u)

as n → ∞.

By the definition of a(λ), there exists (s0, u0) [0, 1] × A such that

Gλ( n (s0)u0) ≥ a(λ) − (λn − λ).

Then, by (9.32), n (s0)u0 ≤ k0(λ). It follows that n (s0)u0 Qn (λ). Therefore,

Qn (λ) = φ.

Next, we want to prove (9.38). If this were not so, there would exist a positive

ε < [a(λ) − b0(λ)]/3 such that Gλ(u) ≥ 3ε for all u Qn (λ). We take n so large that (a (λ) + 2)(λn − λ) ≤ ε, λn − λ ≤ ε. By Lemma 2.10.1 of [122], we may construct

a locally Lipschitz continuous map Y											E such that
	Yλ (u) ≤ 1,	u		ˆ						λ of	ˆ
1.					,
		≥		E
	λ									Qn (λ),
2.	(G (u), Yλ(u))	≥	2ε,			u
	λ									ˆ
3.	(G (u), Yλ(u))		0,		u				E.

Consider the initial boundary-value problem:

dσ (t)u	= −Yλ(σ (t)u), σ (0, u) = u.
dt

By Theorem 4.5, there exists a unique continuous solution σ (t)u such that Gλ(σ (t)u) is nonincreasing in t. Define

(s)u :

σ (2s)u,

0 ≤ s ≤ 1/2,

˜

=

σ (1) n (2s − 1)u,

1/2 ≤ s ≤ 1.

˜

. We want to prove that

Then it is easy to check that

×

(9.39)

G

λ( ˜

≤

a(λ)

−

(λ

n −

λ),

[0, 1]

A,

(s)u)

(s, u)

9.6. The monotonicity trick

99

which provides the desired contradiction. Choose any u A. If 0 ≤ s ≤ 1/2, by (H3), we get

(9.40)	G	λ( ˜
		(s)u)

= Gλ(σ (2s)u)

≤Gλ(u)

≤a0(λ)

≤b0(λ)

< a(λ) − (λn − λ).

If 1

/

2

≤

s

≤

˜

=

n(

2s

−

1

)

u. If G

λ( n

(

2s

−

1

)

u

) <

a

(λ) − (λn − λ)

1, then (s)u

σ (1)

for s [1/2, 1], then

(9.41)

G

λ( ˜

=

λ

n

−

1)u)

(s)u)

G

(σ (1) (2s

≤ Gλ(σ (0) n(2s − 1)u)

<a(λ) − (λn − λ).

If there exists an s0 [1/2, 1] such that Gλ( n (2s0 − 1)u) ≥ a(λ) − (λn − λ), then, by (9.32), n (2s0 − 1)u ≤ k0. Since

σ (t)u − u ≤ t,

it follows that

(9.42)

σ (t) n (2s0 − 1)u ≤ n (2s0 − 1)u + t ≤ k0 + 1

for all t [0, 1].

If there is a t0 [0, 1] such that

Gλ(σ (t0) n (2s0 − 1)u) < a(λ) − (λn − λ)

for some t0 [0, 1],

we obtain

(9.43)

G

λ( ˜ 0

=

G

λ

n

0 −

1)u) < a(λ)

−

(λ

n −

λ).

(s )u)

(σ (1) (2s

If there is no such t0 [0, 1], then in view of (9.32), we see that

(9.44)

a(λ) − (λn − λ) ≤

Gλ(σ (1) n(2s0 − 1)u)

≤Gλ(σ (t) n (2s0 − 1)u)

≤Gλ( n(2s0 − 1)u)

≤a(λ) − (λn − λ)

100 9. Superlinear Problems

for all t [0, 1]. Thus, (9.42) and (9.44) imply that σ (t) n (2s0 − 1)u Qn (λ) for all

λ

≥

2ε on Qn (λ), we have

t

[0, 1]. Since (G (u), Yλ(u))

Gλ(σ (t) n (2s0 − 1)u) − Gλ( n (2s0 − 1)u)

=

t

dGλ(σ (θ ) n (2s0 − 1)u)

dθ

0

dθ

t

≤

−

G

(σ (θ ) (2s

1)u), Y (σ (θ ) (2s

−

1)u) dθ

0

λ

n

0 −

λ

n 0

≤ −2tε.

Therefore,

(9.45) λ ˜ s0 u G ( ( ) )

= Gλ(σ (1) n (2s0 − 1)u)

≤Gλ( n (2s0 − 1)u) − 2ε

≤a(λ) − (λn − λ).

Combining (9.40), (9.41), (9.43), and (9.45), we get										×
G	λ( ˜	≤	a(λ)	−	n −	λ),			[0, 1]		A,
	(s)u)			(λ			(s, u)

which contradicts the definition of a(λ). This implies that (9.38) holds in the case of a(λ) > b0(λ). Clearly, (9.38) yields conclusion (1) of this theorem.

We prove that conclusion (1) of Theorem 9.15 is still true in case a(λ) = b0(λ). Since A is bounded, dA := max{ u : u A} < ∞. For ε > 0, T > 0, we define

¯ (ε,

T

, λ)

={

u

E :

≤

k

0(λ) +

4

+

d

A,

Q

:

u

¯

|Gλ(u) − a(λ)| ≤ 3ε, d(u, B) ≤ 4T }.

We claim that

(ε,

T

, λ) = φ.

By (9.33), we may choose n so large that

Q

(9.46)

sup

Gλ( n (s)u) ≤

sup

Gλn ( n (s)u) ≤ a(λ) + 3ε.

s [0,1],u A

s [0,1],u A

Since A links B [hm], there exists (s0, u0) [0, 1] × A such that n (s0)u0 B. Hence dist( n (s0)u0, B) = 0 and

(9.47)

G

λ( n (s0)u0) ≥ b0(λ) = B

λ

= a(λ) > a(λ) − (λn − λ) ≥

a

(λ) −

3

ε.

inf G

By (9.32), n ( 0)

u

0 ≤

k

0.

Hence,

n (

0)

u

0

¯ (ε,

T

, λ).

s

s

Q

Next, we prove that

{

¯

} =

(9.48)

inf

λ

(u)

: u

0

G

Q(ε, T, λ)

for ε, T sufficiently small. If not, there would exist δ > 0, ε1 > 0, and T1

(0, 1)

such that

≥

¯

λ

, T1, λ).

(9.49)

G (u)

3δ

for u

Q(ε1

9.6. The monotonicity trick

101

Define

(9.50)

Q

(ε

, T , λ) :

= {

u

E : u

k

0 +

4

+

d

A

,

dist(u, B)

≤

4T ,

¯

1

1

≤

1

a(λ) − (λn − λ) ≤ Gλ(u) ≤ a(λ) + 3ε1}.

¯

=

¯

¯

, T1

By (9.46) and (9.47), Q (ε1, T1, λ)

φ and Q (ε1, T1

, λ)

Q(ε1

, λ). Let n be

so large that λn − λ ≤ ε1, (a (λ) + 2)(λn − λ) < ε1 and λn − λ < δT1. Similarly, we

Y of E such that

may construct a locally Lipschitz continuous map ¯λ

ˆ

1. ¯λ(

u

) ≤

1

,

ˆ

,

Y

¯

≥

u

E

¯

λ

2. (G (u), Yλ(u))

≥

2δ,

u

Q(ε1

, T1

, λ),

λ

¯

ˆ

3. (G (u), Yλ(u))

0,

u

E.

Define

(9.51) Q1 := {u E : u ≤ k0 + 2 + dA, |Gλ(u) − a(λ)| ≤ 2ε1, dist(u, B) ≤ 3T1}.

As in the proof of (9.38), Q1

= φ

and Q

1

Q

T

Moreover, the distance

¯ (ε1,

1, λ).

between Q

1 and E\ ¯

(ε1,

1, λ)

is positive. Thus, we may choose a Lipschitz continu-

Q

T

ous map

γ

from E into [0

,

1] that equals 1 on Q

Q

,

T

, λ)

.

1 and vanishes outside ¯ (ε1

1

Consider the following initial boundary-value problem:

d(σ1(t)u)

Y

(0)u = u.

dt

= −γ (σ1) ¯λ(σ1), σ1

Let σ1(t)u be the unique continuous solution. Then we have that

(9.52)

dGλ(σ1(t)u)

≤ −2δγ (σ1(t)u) ≤ 0.

dt

First, we note that

(9.53)

σ1(t)u B, s [0, T1], u A.

For u A, by (9.52), we have that

(9.54)

Gλ(σ1(t)u) ≤ Gλ(u) ≤ a0(λ) ≤ b0(λ) = a(λ),

t [0, T1],

and

(9.55)

=

Gλ(u) +

t dGλ(σ1(θ )u)

Gλ(σ1(t)u)

dθ

0

dθ

t

≤Gλ(u) − 2δγ (σ1(θ )u)dθ

0

for all t [0, T1].

102 9. Superlinear Problems

If (9.53) were not true, then there would be a t0 [0, T1] such that σ1(t0)u B. Then Gλ(σ1(t0)u) ≥ a(λ) = b0(λ) = infB Gλ. By (9.52)–(9.56), we see that

t0

2δγ (σ1(θ )u)dθ = 0.

0

Hence, γ (σ1(θ )u) = 0 for θ [0, t0]; i.e., σ1(θ )u Q1 θ [0, t0]. Therefore, one of the following three cases occurs:

(9.56)	σ1(θ )u > k0 + 2 + dA;
or
(9.57)	|Gλ(σ1(θ )u) − a(λ)| > 2ε1;
or
(9.58)	dist(σ1(θ )u, B) > 3T1.
Inequality (9.56) cannot hold, since
(9.59)	σ1(θ , u) − σ1(θ )u ≤ |θ − θ |
and

σ1(θ )u ≤ σ1(0)u + T1 ≤ dA + 1, θ [0, t0].

If (9.57) holds, then Gλ(σ1(θ )u) < a(λ) − 2ε1. Hence, σ1(θ )u B. Evidently, (9.58) implies that σ1(θ )u B. Therefore, (9.53) is true.

Next, we note that

(9.60) σ1(T1) n (2s − 1)u B, u A, s [1/2, 1].

For any fixed u A and s [1/2, 1], we divide the proof into two cases.

Case 1: If σ1(θ ) n (2s − 1)u Q1 for all θ [0, T1], by (9.52) and (9.32), we

have that
Gλ(σ1(T1) n (2s − 1)u)							T1
=	G	( (2s	−	1)u)	+				dGλ(σ1(θ ) n (2s − 1)u)	dθ
	λ	n				0			dθ
≤ Gλ( n (2s − 1)u) − 0							T1	2δγ (σ1(θ ) n (2s − 1)u)dθ

= Gλ( n (2s − 1)u) − 2δT1

≤ a(λ) − 2δT1 + (a (λ) + 2)(λn − λ),

which implies that σ1(T1) n (2s − 1)u B since a(λ) = b0(λ).

9.6. The monotonicity trick

103

Case 2: If there exists t0 [0, T1] such that σ1(t0) n (2s − 1)u Q1, then one of the following alternatives holds:

either
(9.61)	σ1(t0) n(2s − 1)u > k0 + 2 + dA;
or
(9.62)	|Gλ(σ1(t0) n (2s − 1)u) − a(λ)| > 2ε1;
or
(9.63)	dist(σ1(t0) n(2s − 1)u, B) > 3T1.

Assume that (9.61) holds. If σ1(T1) n (2s − 1)u B, then

b0(λ) = a(λ) ≤ Gλ(σ1(T1) n (2s − 1)u) ≤ Gλ( n (2s − 1)u).

By (9.32), n (2s − 1)u ≤ k0. Furthermore, since

σ1(t0) n (2s − 1)u − σ1(0) n(2s − 1)u ≤ t0,

it follows that

σ1(t0) n (2s − 1)u ≤ k0 + t0 ≤ k0 + 1,

which contradicts (9.61). Hence, σ1(T1) n (2s − 1)u B. Assume that (9.62) holds. Note, by (9.32), that

Gλ(σ1(t0) n (2s − 1)u) ≤ Gλ(σ1(0) n (2s − 1)u) = Gλ( n (2s − 1)u)

≤ a(λ) + ε1.

Therefore, (9.62) implies that

Gλ(σ1(T1) n (2s − 1)u) ≤ Gλ(σ1(t0) n (2s − 1)u) ≤ a(λ) − 2ε1.

It follows that σ1(T1) n(2s − 1)u B since a(λ) = b0(λ).

Assume (9.63) holds. Note that σ1(t)u − σ1(t )u ≤ |t − t |. It therefore follows that

σ1(t) n (2s − 1)u − w ≥ σ1(t0) n (2s − 1)u − w − |t − t0|

for all w B, t [0, T1]. Hence, dist(σ1(t) n (2s − 1)u, B) ≥ 2T1 for all t [0, T1]. In particular,

		σ1(T1) n (2s − 1)u B.
Hence, (9.60) holds in this case as well.
Define		σ1(2sT1)u,	0 ≤ s ≤ 1/2,
	(s)u :
1	=	σ1(T1) n(2s − 1)u,	1/2 ≤ s ≤ 1.