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

8.8. Notes and remarks				83
and
(8.123)	|H (x, t)| ≤	1	V (x)t2 +	3	W (x)|t|,
		2		2
Moreover, by (8.114),
(8.124)	2F(x, t)/t2 → b±(x) a.e. as t → ±∞.
Consequently, by (8.2),
	H (x, t)/t2 → 0 a.e. as t → ±∞.

Thus r (x, a) → 0 a.e. and (8.80) is always satisfied. Thus, we must assume that (8.79) has no solutions in D\{0}.

8.8 Notes and remarks

The material in this chapter is from [120], [110], [107], and [132].

Chapter 9

Superlinear Problems

9.1 Introduction

Consider the problem

(9.1)

− u = f (x, u), x ; u = 0 on ∂ ,

where Rn is a bounded domain whose boundary is a smooth manifold, and f (x, t)

is a continuous function on ¯ × R This semilinear Dirichlet problem has been studied

.

by many authors. It is called sublinear if there is a constant C such that

| f (x, t)| ≤ C(|t| + 1), x , t R.

Otherwise, it is called superlinear. Beginning in [7], almost all researchers studying the superlinear problem assumed

(a1) There are constants c1, c2 ≥ 0 such that

| f (x, t)| ≤ c1 + c2|t|s ,

where 0 ≤ s < (n + 2)/(n − 2) if n > 2. (a2) f (x, t) = o(|t|) as t → 0.

(a3) There are constants μ > 2, r ≥ 0 such that

(9.2) 0 < μF(x, t) ≤ t f (x, t), |t| ≥ r,

where

t

F(x, t) = f (x, s) ds.

0

They proved

Theorem 9.1. Under hypotheses (a1)–(a3), problem (9.1) has a nontrivial weak solution.

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_9, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

86	9. Superlinear Problems

The condition (a3) is convenient, but it is very restrictive. In particular, it implies that there exist positive constants c3, c4 such that

(9.3)

F(x, t) ≥ c3|t|μ − c4, x , t R.

Although this condition is weaker, it still eliminates many superlinear problems. A much weaker condition that implies superlinearity is

(a3 ) Either

F(x, t)/t2 → ∞ as t → ∞

or

F(x, t)/t2 → ∞ as t → −∞.

The purpose of the present chapter is to explore what happens when (a3) is replaced with (a3 ). Surprisingly, we find the following to be true.

Theorem 9.2. Under hypotheses (a1), (a2), (a3 ), the boundary-value problem

(9.4) − u = β f (x, u), x ; u = 0 on ∂ ,

has a nontrivial solution for almost every positive β.

We generalize this theorem and present some variations below.

9.2 The main theorems

We consider the boundary-value problems described in Sections 8.2 and 8.4 under assumption (A) stipulated there. We allow the domain to be unbounded. We also assume

(B)The point λ0 is an isolated simple eigenvalue with a bounded eigenfunction

ϕ0(x) = 0 a.e. in .

(C) There is a δ > 0 such that

	2F(x, t) ≤ λ0t2,		|t| ≤ δ, x ,
where
			t
(9.5)	F(x, t) :=	0	f (x, s)ds.
(D)	There is a function W (x) L1( ) such that either
	W (x) ≤ F(x, t)/t2 → ∞ as t → ∞,			x
or	W (x) ≤ F(x, t)/t2 → ∞ as t → −∞,			x .

(The function W (x) need not be positive.)

(E)There are constants μ > 2, C ≥ 0 such that

[μF(x, t) − t f (x, t)]/(t2 + 1) ≤ C, t R, x .

9.2. The main theorems

87

We shall prove

Theorem 9.3. Under the above hypotheses, the problem

(9.6) Au = f (x, u), u D

has at least one nontrivial solution.

We also have

Theorem 9.4. If we replace hypothesis (E) with (E ) The function

(9.7)

H (x, t) := t f (x, t) − 2F(x, t)

is convex in t.

then problem (9.6) has at least one nontrivial solution.

As we noted, problem (9.6) is called sublinear if f (x, t) satisfies

(9.8)

| f (x, t)| ≤ C(|t| + 1), x , t R.

Otherwise, it is called superlinear. Hypothesis (D) requires (9.6) to be superlinear.

If we drop hypothesis (E) completely, then we are able to prove the following theorems.

Theorem 9.5. If we replace hypotheses (C), (D) with

˜

(C ) There are a δ > 0 and a λ > λ0 such that

2F

x

t

λt2

t

| ≤

x

(

,

) ≥ ˜

,

|

δ,

and

there is a function W (x) L1( ) such that

(D )

W (x) ≥ P(x, t) → −∞ as |t| → ∞, x ,

where

(9.9)

P(x, t) := F(x, t) −

1

λ0t2.

2

and drop hypothesis (E), then problem (9.6) has at least one nontrivial solution.

We also have

Theorem 9.6. Assume that (A)–(D) hold. Then, for almost every β (0, 1), the equation

(9.10) Au = β f (x, u)

has a nontrivial solution. In particular, the eigenvalue problem (9.10) has infinitely many solutions.

88

9.

Superlinear Problems

Theorem 9.7. If we replace hypothesis (C) in Theorem 9.6 with

(C )

˜

≤

There are a δ > 0 and a λ

λ0

such that

2F

x

t

λt2

t

| ≤ δ,

x

.

(

,

) ≤ ˜

,

|

and (D) with

(D )

Either

or

F(x, Rϕ0) dx/ R2 → ∞ as R → ∞

F(x, − Rϕ0) dx/ R2 → ∞ as R → ∞.

then (9.10) has a nontrivial solution for almost every β

0

˜

(0, λ

/λ).

Corollary 9.8. If we replace hypothesis (C ) in Theorem 9.7 with

(C )

F(x, t)/t2 → 0 uniformly as

t → 0.

then (9.10) has a nontrivial solution for almost every β (0, ∞).

9.3	Preliminaries
Define
(9.11)	G(u) := u 2D − 2	F(x, u)dx.

Under hypothesis (A), it is known that G is a continuously differentiable functional on the whole of D. In fact, the following were proved in [122, pp. 56–58].

Proposition 9.9. Under hypothesis (A), F(x, u(x)) and v(x) f (x, u(x)) are in L1( ) whenever u, v D.

Proposition 9.10. G(u) has a Frechet´ derivative G (u) on D given by

(9.12)	(G	(u), v)D	=	2(u, v)D	−	2( f ( , u), v).
						·

Proposition 9.11. The derivative G (u) given by (9.12) is continuous in u.

Theorem 9.12. Under hypotheses (A)–(C), the following alternative holds:

	Either
(a)	there is an infinite number of y(x) D( A) \ {0} such that
(9.13)		Ay = f (x, y) = λ0 y
or
(b)	for each ρ > 0 sufficiently small, there is an ε > 0 such that

(9.14)

G(u) ≥ ε, u D = ρ.

9.4. Proofs

89

9.4 Proofs

We now give the proof of Theorem 9.3.

Proof.	We take
(9.15)		G(u) = u 2D − 2					F(x, u)dx.
Under our hypotheses, Propositions 9.9–9.11 apply, and
(9.16)	(G	(u), v)	=	2(u, v)D	−	2( f ( , u), v), u, v			D.
							·

By Theorem 9.12, we see that there are positive constants ε, ρ such that

(9.17)	G(u) ≥ ε, u D = ρ,
unless
(9.18)	Au = λ0u = f (x, u), u D \ {0},

has a solution. This would give a nontrivial solution of (9.6). We may therefore assume that (9.17) holds. Next, we note that

G(± Rϕ0)/ R2 = ϕ0 2D − 2 {F(x, ± Rϕ0)/ R2ϕ02}ϕ02dx → −∞ as R → ∞

by hypothesis (D), depending on which part of (D) is assumed, since ϕ0 = 0 a.e. Since G(0) = 0 and (9.17) holds, we can now apply Theorem 3.13 to conclude that there is a sequence {uk } D such that

G(uk ) → c ≥ ε,

G (uk ) → 0.

Then

(9.19)

G(uk ) = ρk2 − 2

F(x, uk )dx → c

and

(9.20)

=

(G (uk ), uk ) = 2ρk2 − 2( f (·, uk ), uk ) = o(ρk ),

˜2

=

k

D. Assume that ρk

→ ∞

˜

=

uk /ρk . Since

1,

where ρ

uk

, and let uk

uk

D

there is a renamed subsequence such that u˜ k

→ u˜ weakly in D, strongly in L ( ), and

a.e. in . By (9.19),

2F(x, uk ) 2

u˜k dx → 1.

u2

k

Let

1 = {x : u˜ (x) = 0}, 2 = \ 1.

90

9. Superlinear Problems

Then

2F(x, uk ) 2

u˜k → ∞,

x 1,

u2

k

by hypothesis (D). If 1 has positive measure, then

2F(x, uk )

2

2F(x, uk ) 2

u˜ k dx ≥

u˜k

dx + [−W (x)] dx → ∞.

u2

u2

k

1

k

2

Thus, the measure of 1 must be 0, i.e., we must have u˜ ≡ 0 a.e. Moreover,

2F(x, uk ) − μuk f (x, uk )

u2 dx

→

1

−

μ.

u2

˜ k

k

But by hypothesis (E),

lim sup

2F(x, uk ) − μuk f (x, uk )

u2

≤

lim sup C

uk2 + 1

u2

=

0,

uk2

˜k

uk2

˜ k

which implies that 1 − μ ≤ 0, contrary to our assumption. Hence, the ρk are bounded. We can now follow the usual procedures to obtain a weak solution of (9.6) satisfying

G(u) = c ≥ ε (cf., e.g., [122, p. 64]). Since G(0)						= 0, we see that u = 0. This
completes the proof.
We postpone the proof of Theorem 9.4 until the next section.
In proving Theorem 9.5, we shall make use of
Lemma 9.13. Under hypothesis (C ), there is an α = 0 such that G(αϕ0) < 0.
Proof. We can assume that
(9.21)		ϕ0 D = 1.
Thus,
G(αϕ0) = α2 − 2			F(x, αϕ0) dx
≤	α2	− ˜			0	(x)2dx
		λα2		ϕ

|αϕ0(x)|<δ

+V q (|αϕ0|q + |αϕ0|)

α2

|αϕ0(x)|>δ

2

C

α

q

V ϕ

q

≤

−

λα2 ϕ

+

˜

0

|

|

0

≤

α2

˜

+

C

|

|

q−2].

q

[1

−

0

(λ/λ )

α

This can be made negative by taking α sufficiently small.

Lemma 9.14. Under hypothesis (D ),

(9.22)

G(u) → ∞

as

u D → ∞.

9.4. Proofs	91
Proof. Suppose there is a sequence {uk } D such that ρk = uk → ∞ and
G(uk ) ≤ K .
Write

uk = wk + αk ϕ0, u˜k = uk /ρk , w˜ k = wk /ρk , α˜ k = αk /ρk ,

where wk ϕ0. If λ1 > λ0 is the next point in the spectrum of A, then

	λ1 w 2 ≤ w 2D , w ϕ0.
Thus,
G(uk ) = uk 2D − λ0 uk 2 − 2				P(x, uk ) dx
≥	1 −	λ0	wk 2D − 2	P(x, uk ) dx
		λ1
≥	1 −	λ0	wk 2D − 2	W (x) dx.
		λ1

The only way this would not converge to ∞ is if wk D is bounded. But then w˜ k D → 0 and |α˜ k | → 1. Since u˜k D = 1, there is a renamed subsequence such that u˜k → u˜ weakly in D, strongly in L2( ), and a.e. in . Since w˜ = 0 and |α˜ | = 1, we have u˜(x) = αϕ˜ 0(x) = 0 a.e. Hence, |uk (x)| = ρk |u˜k (x)| → ∞ a.e. Consequently,

P(x, uk ) dx → −∞,

showing that G(uk ) → ∞. This completes the proof.

We can now give the proof of Theorem 9.5.

Proof. Let

m = inf G.

D

Then there is a sequence {uk } D such that G(uk ) → m. In view of Lemma 9.14,

	u	k D ≤	2	Thus, there is a renamed subsequence such that u		k →	u
we must have			C.
weakly in D, strongly in Lloc( ), and a.e. in . Now,
G(u) = u 2D − 2				F(x, u) dx
= uk 2D − 2([uk − u], u)D − uk − u 2D
		− 2	F(x, uk ) dx + 2		[F(x, uk ) − F(x, u)]dx
≤ G(uk ) − 2([uk − u], u)D + 2 [F(x, uk ) − F(x, u)]dx.

92	9. Superlinear Problems
From our hypotheses, it follows that
F(x, uk ) dx → F(x, u) dx

(cf., e.g., [122, p. 64]). We therefore have in the limit G(u) ≤ m, from which we conclude that G(u) = m and G (u) = 0. Hence, u is a weak solution of (9.6). We see from Lemma 9.13 that m < 0. Since G(0) = 0, we see that u = 0. This completes the proof.

9.5 The parameter problem

In this section we shall give the proofs of Theorems 9.4, 9.6, and 9.7. They will be based on the following result proved in the next section. Let E be a reflexive Banach

space with norm · , and let A, B be two closed subsets of E. Suppose that G C1(E, R) is of the form G(u) := I (u) − J (u), u E, where I, J C1(E, R) map

bounded sets to bounded sets. Define

Gλ(u) = λI (u) − J (u), λ ,

where

is an open interval contained in (0, +∞). Assume one of the following alter-

natives holds.

(H1)

I (u) ≥ 0 for all u E and either I (u) → ∞ or | J (u)| → ∞ as u → ∞.

(H2)

I (u) ≤ 0 for all u E and either I (u) → −∞ or | J (u)| → ∞ as u → ∞.

Furthermore, we suppose that

(H3)

a0(λ) := supA Gλ ≤ b0(λ) := infB Gλ, for any λ .

We let be the set of mappings (t) C(E × [0, 1], E) described in Chapter 3.

We have

Theorem 9.15. Assume that (H1) or (H2) holds together with (H3).

(1) If A links B [hm] and A is bounded, then, for almost all λ

, there exists

uk (λ)

k

uk (λ)

∞

λ

→

E such that sup

<

, G

(uk (λ))

0, and

→ ∞.

Gλ(uk (λ)) → a(λ) := s [0,1],u A Gλ( (s)u),

k

inf

sup

Furthermore, if a(λ) = b0(λ), then dist(uk (λ), B) → 0, k → ∞.

(2) If B links A[hm] and B is bounded, then, for almost all λ

, there exists

k

∞

λ

→

vk (λ)

E such that sup

vk (λ)

<

, G (vk (λ))

0, and

→ ∞.

G

λ(vk (λ)) → b(λ) := s [0,1],v B Gλ( (s, v)),

k

sup

inf

Furthermore, if a0(λ) = b(λ), then dist(vk (λ), A) → 0, k → ∞.

9.5. The parameter problem

93

We shall also need the following extension of Theorem 9.12.

Theorem 9.16. Let λ be a parameter satisfying 1 < λ ≤ K < ∞. Under hypotheses

(A)–(D), for each ρ > 0 sufficiently small (not depending on λ), we have

(9.23)

Gλ(u) := λ u 2D − 2

F(x, u)dx ≥ (λ − 1)ρ2, u D = ρ.

∞

˜

≤

K

If we replace hypothesis (C) with hypothesis (C ), assuming 1 < λ/λ0

< λ

<

,

then we have

˜

2

(9.24)

Gλ(u) ≥

λ −

λ

ρ , u D = ρ.

λ0

Proof. Let λ1 > λ0 be the next point in the spectrum of A, and let N0

denote the

eigenspace of λ0. We take M = N0 ∩ D. By hypothesis (B), there is a ρ > 0 such that

	y D ≤ ρ |y(x)| ≤ δ/2,		y N0.
Now suppose u D satisfies
(9.25)	u D ≤ ρ	and |u(x)| ≥ δ
for some x . We write
(9.26)	u = w + y,	w M,	y N0.

Then for those x satisfying (9.25) we have

δ ≤ |u(x)| ≤ |w(x)| + |y(x)| ≤ |w(x)| + (δ/2).

Hence,
(9.27)	|y(x)| ≤ δ/2 ≤ |w(x)|,
and consequently,
(9.28)		|u(x)| ≤ 2|w(x)|
for all such x. Now we have by (9.26) and (9.28),
Gλ(u) ≥ λ u 2D	− λ0	u2dx − C	(|V u|q + V q |u|)dx
		|u|<δ	|u|>δ
≥ λ u 2D	− λ0 u 2 − C		|V u|q dx
		|u|>δ
≥ (λ − 1) y 2D + λ w 2D − λ0 w 2 − C				|V w|q dx
				2|w|>δ