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

8.4. Unbounded domains

73

and

(8.56)

uk 2D − ( f (·, xk ), uk )

≤ K .

Consequently,

(8.57)

H (x, uk) dx

≤ K .

= 1.

If

ρk = uk D → ∞, let u˜ k = uk /ρk . Then u˜k

D

2

Consequently, there is a

renamed subsequence such that u˜k

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

. In view of (8.55), we have

(8.58)

1 − 2

F(x, uk )/ρk2 dx → 0.

But by (8.50), we have

(8.59)

2F(x, u

k

)/ρ2

≤

a

(u

−)2

+

(a

)(u+)2

+

W (x)/ρ2.

k

2

˜k

2

˜k

2

k

In the limit this implies

1

≤

a2

u

− 2

+

(a2) u+ 2.

˜

˜

This shows that u˜ ≡ 0. Let 0 be the subset of on which u˜ = 0. Then

(8.60)

|uk (x)| = ρk |u˜k (x)| → ∞,

x 0.

If 1 = \ 0, then we have

(8.61)

H (x, uk) dx =

+

≥

H (x, uk) dx −

W1(x) dx → ∞.

0

1

0

1

This contradicts (8.57), and we see that ρk = uk D is bounded. Once we know that the ρk are bounded, we can apply Theorem 3.4.1 of [122] to obtain the desired conclusion.

Remark 8.8. It should be noted that the crucial element in the proof of Theorem 8.7 was (8.56). If we had been dealing with an ordinary Palais–Smale sequence, we could

only conclude that

uk 2D − ( f (·, uk ), uk ) = o(ρk ),

which would imply only that

H (x, uk)dx = o(ρk ).

This would not contradict (8.61), and the argument would not go through.

74	8. Semilinear Problems

We also have

Theorem 8.9. The conclusion of Theorem 8.7 holds if in place of (8.47), (8.48) we assume that

(8.62)	H (x, t) ≤ W1(x) L1( ),				x , t R,
and
(8.63)	H (x, t) → −∞ a.e. as				|t| → ∞.
Proof. We use (8.62) and (8.63) to replace (8.61) with
(8.64)	H (x, uk) dx = 0	+ 1	≤ 0	H (x, uk) dx + 1		W1(x) dx → −∞.

We then proceed as before.

As a specific example of an operator A satisfying the hypotheses of Theorems 8.7 and 8.9, let g(x) be a measurable function satisfying

g(x) ≥ c0 > 0, x Rn ,

for some positive constant c0. We consider the problem

(8.65)

− u(x) + g(x)2u(x) = f (x, u(x)),

x Rn .

H 1,2 = H 1,2(Rn ) and

on L

2

=

L2

(

Rn

)

by u

(

A

)

and Au

=

f if u

D

=

We define the operator A

D

(u, v)D = ( u, v) + (gu, gv) = ( f, v),

v H 1,2.

We assume that g(x) L2loc and that multiplication by g−1 is a compact operator from H 1,2 to L2. It follows that A is a self-adjoint operator on L2 that is bijective. Moreover, A−1 is a compact operator on L2. It follows that the spectrum of A consists of isolated eigenvalues of finite multiplicity satisfying (8.49). Thus, A satisfies the hypotheses of Theorems 8.7 and 8.9. Solutions of (8.11) satisfy (8.65). Hence, Theorems 8.7 and 8.9 produce weak solutions of (8.65). We summarize this as

Theorem 8.10. Let g(x) be a function satisfying the conditions described above. Then there exists a sequence of eigenvalues for the equation

(8.66)

− u(x) + g(x)2u(x) = λu(x), x Rn ,

satisfying (8.49). Let f (x, t) be a Caratheodory´ function satisfying hypothesis (A) for= Rn , and assume that (8.47), (8.48), and (8.50) hold for some > 0. Then (8.65) has at least one solution.

8.5. Further applications		75
Remark 8.11. We could have assumed
a1 ≤ limt	inf 2F(x, t)/t2	≤ lim sup 2F(x, t)/t2 ≤ a2
→−∞		t→−∞
and	inf 2F(x, t)/t2	≤ lim sup 2F(x, t)/t2 ≤ (a2)
γ (a1) ≤ limt
→∞		t→∞

in place of (8.50).

Remark 8.12. This theorem generalizes results of several authors, including [10], [46], [71], [98], and [104], with various conditions on the function g(x) to ensure that the spectrum of (8.66) is discrete. We guarantee it by assuming that multiplication by g−1 is a compact operator from H 1,2 to L2. A sufficient condition for this is given in [106]. Since g−1 is bounded, a simple sufficient condition is that for each constant b > 0,

(8.67) m{x Rn : |x − y| < 1, g(x) < b} → 0 as |y| → ∞.

Remark 8.13. If we choose a1 = λ and a2 = λ +1, inequality (8.50) reduces to

(8.68) λ t2 − W1(x) ≤ 2F(x, t) ≤ λ +1t2 + W2(x), x Rn, t R.

By choosing a1, a2 to be different values, we allow a wider range of possibilities for F(x, t).

8.5 Further applications

Now we look for solutions of (8.13) under different conditions. Let A be a self-adjoint operator on L2( ). We assume that A ≥ λ0 > 0 and that

C0∞( ) D := D( A1/2) H m,2( )

for some m > 0. Let q be given by

q =	2n/(n − 2m),	2m < n
=	∞,	n ≤ 2m,

and let f (x, t) be a Carath´eodory function on × R. We assume the following.

I. There are positive functions V (x), V0(x), V1(x), W (x), W1(x), r0(t), r1(t) and constants q, q1 such that the following hold.

(a)2 < q < q , 1 < q1 < q .

(b)Multiplication by V or W is a bounded operator from D to L2( ), multiplica-

tion by V is compact from D to Lq−1(K ) for each compact subset K of , and W is in Lq ( ). Multiplication by V0 is compact from D to Lq ( ), V0 is locally bounded, and

(8.69) | f (x, st)| ≤ V0(x)(|V (x)s|q−1 + W (x)q−1)tr0(t), s R, t ≥ 1.

76 8. Semilinear Problems

(c) Multiplication by V1 is a compact operator from D to Lq1 ( ), W1 is in L1( ),

and
(8.70)	|H (x, st)| ≤ (|V1(x)s|q1 + W1(x))r1(t),			s R, t ≥ 1.
(d) There is a ψ such that
(8.71)	ψ(t) ≤ tr0(t),	tψ(t) ≤ r1(t),		t ≥ 1,
and
(8.72)	ri (t) → ri	as t → ∞,	i = 0, 1,
with
(8.73)		r0 < ∞
and
(8.74)		r1 ≤ ∞.
II. There are positive constants α, β such that
(8.75)	2F(x, t) ≤ λ0(1 − α)t2,		t2 < β.

III. (a) If r0 = 0, it is assumed that there is a measurable function γ (x, a) on × R

such that
(8.76)	f (x, st)/tr0(t) → γ (x, a) a.e. as s → a, t → ∞.

(b) If r0 = 0 and r1 = 0, it is assumed that there is a measurable function (x, a)

on × R such that
(8.77)	H (x, st)/r1(t) → (x, a) a.e. as s → a, t → ∞.
If r1 = ∞, then (8.77) is required to hold only for a = 0.
IV. If r0	= 0 and r1 = 0, then we assume that there does not exist a function
u D\{0} and b > 0 such that
(8.78)		a(u) :=		1		( Au, u) = 1,
					2
(8.79)	r −1 Au		=	γ (x, u) a.e.,
	0
(8.80)	(x, u) dx = −b/r1.

V. If r0 = 0 and r1 = ∞, we assume that all solutions u ≡ 0 of (8.79) are nonzero a.e. We have the following.

8.5. Further applications	77
Theorem 8.14. Under hypotheses I to V, there is a solution of
(8.81)	Au = f (x, u)
in D. If there is a u0 D\{0} such that
(8.82)	G(u0) ≤ 0,
where
(8.83)	G(u) = a(u) − F(x, u) dx,

then (8.81) has a nonzero solution.

Proof. Under hypothesis I(b), it is easily checked that G(u) given by (8.83) is continuously Fr´echet differentiable with

(8.84)	(G (u), v)	=	2a(u, v)	−	f (x, u)vdx,	u, v		D.
Also by hypothesis I(b),
(8.85)	V u q2 + V0u q2 + W u q2 ≤ Ca(u),					u D.

Thus, by (8.75),

(8.86) G(u) ≥ αa(u) + (1 − α)a(u) − 1 λ0(1 − α) u 2

2

− C |V0u|(|V u|q−1 + β(1−q)/2|W u|q−1) dx

u2>β

≥ a(u)(αC1a(u)(q/2)−1).

Hence, there are positive constants ρ, δ such that

(8.87)

G(u) ≥ ρ, a(u) = δ2.

We consider two cases.

Case 1. G(u) ≥ 0 for all u D. In this case 0 is a minimum point and we must have G (0) = 0. By (8.84), we see that 0 is a solution of (8.81).

Case 2. There is a u0 D\{0} such that (8.82) holds. In this case the hypotheses of Theorem 2.18 are satisfied. In particular, there is a sequence {uk } D such that

(8.88)	G(uk ) → b
and
(8.89)	G (uk )/ψ(rk ) → 0,

78	8. Semilinear Problems

where b is given by (8.80), tk2 = a(uk ), and ψ(t) is the function satisfying (8.71). We shall show that this sequence satisfies

(8.90) a(uk ) ≤ C.

If so, we can find a renamed subsequence that converges weakly in D to a function u and such that V0uk → V0u in Lq ( ) and a.e. in while V uk → V u in Lq−1(K ) for each compact subset K of . By (8.84) and (8.89),

(8.91)

2a(uk, v) − ( f (x, uk ), v) → 0,

v D.

If v

C

0

∞( ), then, by hypothesis I(b),

| f (x, uk )v| ≤ C(|V (x)uk | + W (x)q−1)|V0v|,

and the right-hand side converges in L1( ) to

C(|V u|q−1 + W q−1)|V0u|.

Thus,

(8.92)

( f (x, uk ), v)

→

( f (x, u), v), v

C

∞( ).

0

Since uk → u weakly in D, we have

C∞( ),

(8.93)

2a(u, v)

=

( f ( , u), v),

v

·

0

which shows that u is a solution of (8.81).

I claim that u = 0. To see this, note that b > 0 by (8.87). Moreover, by (8.70),

|H (x, uk)| ≤ C(|V1uk |q1 + W1),

and the right-hand side converges in L1( ) to

C(|V1uk |q1 + W1).

Since

H (x, uk) → H (x, u) a.e.,

we have

H (x, uk) dx → H (x, u) dx.

But for any u D,

(8.94)

1

(G (u), u)

=

G(u)

+

H (x, u) dx.

2

Hence, (8.88) and (8.89) imply that

H (x, u) dx = −b = 0.

Thus, u = 0 since H (x, 0) ≡ 0.

8.5. Further applications

It therefore remains only to prove (8.90). Assume that tk Then

79

→ ∞, and let u˜ k = uk /tk .

(8.95) a(u˜k ) = 1

and there is a renamed subsequence {u˜k } that converges weakly in D to a function u˜

˜

→

˜

( ) and a.e. in while V uk

→

˜

and such that V0uk

V0u in Lq

V u in Lq−1(K ) for

each compact subset K of . By (8.69),

|

f (x, u

k

)

/t r

0

(t )

≤

(

V u

q−1

+

W q−1) V u

.

|

k

k

|

˜ k |

|

0 ˜ k |

Thus,

(8.96)

f (x, uk ) dx /tkr0(tk ) ≤ ( V u˜ k qq−1 + W qq−1) V0u˜ k q .

Moreover, by (8.71) and (8.89),

(8.97)

2a(u˜k )/r0(tk ) − ( f (x, uk )/tkr0(tk ), u˜ k ) → 0.

This shows that u˜(x) ≡ 0. Otherwise, we would have V0u˜ k → 0 in Lq ( ) and consequently the left-hand side of (8.96), which is equal to the second term of (8.97), will converge to 0. But this would mean that a(u˜k ) → 0, contradicting (8.95).

If r0 = 0, we obtain another contradiction, for the second term in (8.97) is bounded

by (8.96) while the first becomes infinite by (8.95) if tk

→ ∞. Hence, r0 = 0 implies

(8.90), and the proof is complete in this case.

It remains to consider the case when r0 = 0. By (8.69),

|

f (x, uk )v

|

/tk r0(tk )

≤

(

V uk

|

q−1

+

W q−1)

V0v

.

|

˜

|

|

When v

C

0

∞( ), the right-hand side converges in L1( ) to

(

V u

q−1

+

W q−1)

V0v

|

|

˜|

|

by hypothesis II(b). By (8.76), the left-hand side converges a.e. to γ (x, u˜ ). Thus,

(8.98)

f (x, uk )v(x)dx/tkr0(tk ) →

γ (x, u˜ )v(x) dx

for each v

0

C∞( ). By (8.71) and (8.89),

(8.99)

2a(u˜k/t0(tk ), v) − ( f (x, uk )/tkr0(tk ), v) → 0

for each v D. Thus, by (8.72) and (8.98),

2a(u/r0

, v)

=

(γ (x, u), v),

v

C∞( ).

˜

0

This shows that u˜ is a solution of (8.79). By (8.70),

(8.100)

|H (x, uk)|/r1(tk ) ≤ |V1u˜k |q1 + W1,

80	8. Semilinear Problems

and the right-hand side converges in L1( ) to the function

|V1u˜|q1 + W1

by hypothesis II(c). If r1 = 0, this produces a contradiction, for by (8.88), (8.89), and (8.94),

(8.101)	H (x, uk) dx/r1(tk )

converges to −b/r1. Since b = 0, this means that expression (8.101) becomes infinite. Thus, the proof is complete for the case when r1 = 0.

It therefore remains to consider the case when r0 = 0, r1 = 0. If r1 = ∞, the integrand of (8.101) converges a.e. to (x, u˜) by hypothesis III(b). If r1 = ∞, this convergence takes place only for those points x where u˜(x) = 0. But by hypothesis V, u˜(x) = 0 a.e. since it is a solution of (8.79). Hence, the convergence is a.e. in all cases, and expression (8.101) converges to

(8.102) (x, u˜) dx.

Since it also converges to −b/r1 by (8.88), (8.89) and (8.94), we see that u˜ is a solution of (8.80) as well as (8.79). As before, (8.97) holds by (8.71) and (8.89). In view of (8.76), this implies that

2/r0 = (γ (x, u˜), u˜ ).

If we combine this with (8.79), we see that u˜ satisfies (8.78) as well as (8.79) and (8.80). A contradiction is now provided by hypothesis IV. Hence, (8.90) holds, and the proof is complete.

8.6 Special cases

In this section we present some consequences of Theorem 8.14.

Theorem 8.15. Assume that
(8.103)	|H (x, t)| ≤ W (x) L1( ),
(8.104)	H (x, t) → H±(x) as t → ±∞ a.e.,
(8.105)	2t−2 F(x, t) → b±(x) as t → ±∞ a.e.,
(8.106)	| f (x, t)| ≤ W1(x) L1( ), |t| ≤ 1,
and that
(8.107)	u>0	H+(x) dx + u<0	H−(x) dx ≥ 0
for all u D\{0} satisfying		Au = b+(x)u+ − b−(x)u−,
(8.108)

8.7. The proofs				81
where u±(x)	=	max[	±	q
				u(x), 0]. Assume that there is a q > 2 such that b1/q , W 1/q are

compact operators from D to L ( ), where b(x) = max |b±(x)|, and that solutions of

(8.108) that are not identically zero a.e. are never zero a.e. Then (8.81) has a solution u D\{0}.

Theorem 8.16. If we replace (8.103), (8.104) in Theorem 8.15 by

(8.109)	|H (x, t)| ≤ V (x)|t|γ + W (x), 0 < γ < 2,
and
(8.110)	H (x, t)/|t|γ → H±(x) as t → ±∞ a.e.,
the theorem will hold if
(8.111)	u>0	H±(x)|u|γ dx +	u<0	H−(x)|u|γ dx > 0

for all u D\{0} satisfying (8.108), and V 1/q is a compact operator from D to Lq ( ). In this case the requirement that solutions of (8.108) that are not identically zero a.e. are never zero a.e. can be removed.

Theorem 8.17. Assume that there is a number γ such that 0 ≤ γ < 1 and

(8.112)	| f (x, t)| ≤ V (x)|t|γ + W (x),
with V 1/q ,	W 1/q compact operators from D to Lq ( ), where q > 2. Then (8.81) has
a solution u D\{0}.
Theorem 8.18. Assume that
(8.113)	| f (x, t)| ≤ V (x)|t| + W (x)
and
(8.114)	f (x, t)/t → b±(x) as t → ±∞ a.e.,
with b1/q ,	V 1/q , and W 1/q compact operators from D to Lq ( ), where b(x) =

max |b ± (x)|, and q > 2. Assume in addition that (8.108) has no nontrivial solutions. Then (8.81) has a solution u D\{0}.

8.7 The proofs

In this section we give the proofs of Theorems 8.15 to 8.18. First, we give the

Proof of Theorem 8.15. We apply Theorem 8.14.		We take r0(t) = r1(t)	= 1,
ψ(t) = t−1. Inequality (8.70) is satisfied with q1		= q, V1 arbitrary, and W1	= W.
Since
(8.115)	∂ (Ft−2)/∂ t = −2t−3 H (x, t),

82

8.

Semilinear Problems

we have

(8.116)

F(x, t) =

21 b+(x)t2 + F0(x, t), t > 0,

21 b−(x)t2 + F0(x, t),

t < 0,

where

2t2

∞ s

−

3 H

(

x

,

s

)

ds

,

t

>

0

,

(8.117)

F0

(x, t) =

t

t

−2t

2

3

t < 0.

−∞ s−

H (x, s) ds,

Thus,

b+(x)t + 4t ( t∞) − 2H (x, t)/t,

t > 0,

(8.118)

f (x, t) = b−(x)t − 4t (

−t ∞) − 2H (x, t)/t

t < 0.

Consequently,

(8.119)

| f (x, t)| ≤ b(x)|t| + 4W |t|−1,

|t| ≥ 1.

This combined with (8.106) gives (8.69) with V0 = V = b1/q . Moreover, by (8.119),

(8.120) f (x, st)/t → ±b± as s → a, t → ∞.

Thus, (8.79) becomes (8.108). Moreover, by (8.104), the left-hand side of (8.80) is the left-hand side of (8.107). Hence, (8.107) assures us that (8.80) cannot hold for any solution u D of (8.79). Thus, all of the hypotheses of Theorem 8.14 are satisfied.

Proof of Theorem 8.16. In this case we take r0(t) = 1, r1(t) = tγ . Since r1 = ∞, the right-hand side of (8.80) vanishes. Thus, strict inequality in (8.111) is needed to guarantee that no solutions of (8.78)–(8.80) exist. Moreover, we do not require hypothesis V for this case.

Proof of Theorem 8.17. Here we take r0(t) = tγ −1 and r1(t) = tγ +1. Note that by (8.112),

(8.121)	|F(x, t)| ≤ p−1V (x)|t| p + W (x)|t|,
where p = 1 + γ . Thus, by (8.2),
(8.122)	|H (x, t)| ≤	1	+	1	V (x)|t| p +	3	W (x)|t|.
		2		p		2

Since r0 = 0 and r1 = ∞, all of the hypotheses of Theorem 8.14 hold.

Proof of Theorem 8.18. Now we take r0(t) = 1, r1(t) = t2. By (8.113), |F(x, t)| ≤ 1 V (x)t2 + W (x)|t|

2