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

10.4. Some applications					125
In view of (10.116), this implies
(10.119)					2F(x, y, h) ≡ μ0h2 − λ0 y2.
For ζ		C	0
			∞( ) and t > 0 small, we have
(10.120)			2[F(x, y + tζ, h) − F(x, y, h)]/t ≤ −λ0[(y + tζ )2 − y2]/t.
Taking t → 0, we have
(10.121)					f (x, y, h)ζ ≤ −λ0 yζ.
Since this is true for all ζ					0
					C∞( ), we have
(10.122)					f (x, y, h) = −λ0 = − Ay.
Similarly,
(10.123)			2[F(x, y, h + tζ ) − F(x, y, h)]/t ≤ μ0[(h + tζ )2 − h2]/t,
and, consequently,
(10.124)					g(x, y, h)ζ ≤ μ0hζ
and
(10.125)					g(x, y, h) = μ0h = Bh.

We see from (15.89) and (10.125) that (10.76), (10.77) has a nontrivial solution. Thus, we may assume that (10.115) holds. But then we can combine it with (10.106) and (10.107) to conclude from Theorem 10.2 that there is a sequence {uk } D such that

(10.3) holds with c ≥ . Arguing as in the proof of Theorem 10.7, we see that there is a u D such that G(u) = c, G (u) = 0. Since c = 0 and G(0) = 0, we see that

u = 0, and the proof is complete.

Theorem 10.10. Assume that λ0 is simple and the eigenfunctions corresponding to λ0 and μ0 are bounded and = 0 a.e. in . Assume (10.80), (10.105) and

(10.126)	2F(x, 0, t) ≤ μ(x)t2,	x , t R,
where
(10.127)	μ(x) ≤≡ μ0,	x .

Then (10.76), (10.77) has a nontrivial solution.

Proof. As in the proof of Theorem 10.7, (10.80) implies (10.93). As in the proof of Theorem 10.9, (10.105) implies (10.115) for y N0, w M unless (10.76), (10.77) has a nontrivial solution. Thus, we may assume that (10.115) holds. Next we note that there is a ν > 0 such that

(10.128)

G(0, w) ≥ νb(w), w M.

126					10. Weak Linking
Assuming this for the moment, we see that
(10.129)	inf G		≥ ε1 >	0	,
	B
where
(10.130)	B = {w M : b(w) ≥ ρ2} {u = (sϕ0, w) : s ≥ 0, w M},
and 1 = min{ , νρ2}. By (10.93), there is an R > ρ such that
(10.131)		sup G ≤ 0,
		A

where A = N ∩ ∂ BR . We can now apply Theorem 10.2 to conclude that there is a sequence in D satisfying (10.3). As in the proof of Theorem 10.7, this leads to a solution of (10.86) satisfying G(u) = c ≥ 1. Hence, u = 0, and we have a nontrivial solution of (10.76), (10.77). It therefore remains only to prove (10.128). Clearly, ν ≥ 0. If ν = 0, then there is a sequence {wk } M such that

(10.132) G(0, wk ) → 0, b(wk ) = 1.

Thus, there is a renamed subsequence such that wk → w weakly in M, strongly in N, and a.e. in . Consequently,

(10.133)	G(0, wk ) ≥ 1 − μ(x)wk2dx ≥ [μ0 − μ(x)]wk2dx → 0
and
(10.134)	1 =	μ(x)w2dx ≤ μ0 w 2 ≤ b(x) ≤ 1,

which means that we have equality throughout. It follows that we must have w E(μ0), the eigenspace of μ0. Since w ≡ 0, we have w = 0 a.e. But

(10.135) [μ0 − μ(x)]w2dx = 0

implies that the integrand vanishes identically on , and consequently μ(x) ≡ μ0, violating (10.127). This establishes (10.128) and completes the proof of the theorem.

10.5 Notes and remarks

The concept of weak linking was begun in [19] and [78]. Further work in this direction was carried out in [106]. The main thrust was to consider the case when G can be written in the form

(10.136)	G(u) =	1	(Lu, u) + b(u)
		2

10.5. Notes and remarks

127

where L is a self-adjoint operator and b (u) is weakly continuous and E is a Hilbert space. With this method, very few sets have been found to link. The results of [19], [78], and [122] do not require E to be separable. However, they require G to be of the form (10.136) with b (u) compact. In [78] the compactness of b (u) is replaced by the assumption that it be weak-to-weak continuous, and b(u) itself is required to be bounded from below and be weakly lower semicontinuous. A theorem is given in [78] that requires only (10.5) but also requires G to be τ -upper semi-continuous (the τ -topology is specially constructed to accommodate the splitting of E into subspaces). In all of the results mentioned only two examples of linking sets A, B are given. The presentation given here is from [117], [118], [140]. It has several advantages. It does not require G to be of the form (10.136) [which indeed satisfies (10.5)]. It does not require hypotheses on each of an exhausting sequence of finite-dimensional subspaces. Moreover, all sets A, B known to link when one of the subspaces M, N is finite-dimensional will link now as well.

Chapter 11

Fucˇ´ık Spectrum: Resonance

11.1				Introduction
Let			be a bounded domain in		Rn	and let A ≥ λ0	>	0 be a self-adjoint operator on
L	2
		( ) with compact resolvent and eigenvalues
(11.1)				0 < λ0 < λ1 < · · · < λk			< · · · .
If f (x, t) is a Carath´eodory function on × R, then the semilinear problem
(11.2)				Au = f (x, u), u D( A)
is said to have asymptotic resonance at infinity if
(11.3)				f (x, t)/t → λk as |t| → ∞,

where λk is one of the eigenvalues of A. Interest in resonant problems began with the pioneering work of Landesman and Lazer [79] and has continued until the present because such problems are more difficult to solve than nonresonant problems. The reason is that for |u(x)| large, (11.2) approximates the eigenvalue problem

(11.4) Au = λk u

with its inherent instabilities. Even if (11.3) does not occur, but

(11.5)	f (x, t)/t → a a.e. as t → −∞
	→ b a.e. as t → +∞,
one encounters the same difficulties if
(11.6)	Au = bu+ − au−, u± = max{±u, 0}

has a nontrivial solution. In fact, the difficulties are compounded because there is no eigenspace with which to work in this case. We call the set of those (a, b) R2 for

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_11, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

130 11. Fucˇ´ık Spectrum: Resonance

which (11.6) has nontrivial solutions the Fuˇc´ık spectrum of A. For each , it was shown in [111, 122] that in the square [λ −1, λ +1]2 there are decreasing curves C,1, C,2 (which may coincide) passing through the point (λ , λ ) such that all points above or below both curves in the square are not in . We denote the lower curve by C,1 and the upper curve by C,2. Further discussions concerning the Fuˇc´ık spectrum can be found in Chapter 14.

In the present chapter we shall study resonance problems with respect to the Fuˇc´ık

spectrum.

We shall allow

(

a

,

b

)

to be on either of the curves C

, j

in

Q

=

2

(λ −1, λ +1)

. We do not presently consider the case when (a, b) is a point between

the curves (when there are points in in that region). Such points will be discussed in Chapter 14. We have

Theorem 11.1. Let f (x, t) be a Caratheodory´ function such that

(11.7)		| f (x, t)| ≤ C(|t| + 1),					x , t R,
and (11.6) has a nontrivial solution with (a, b) C,1. Assume that
(11.8)		λ −1t2 ≤ 2F(x, t) + W1(x),					x , t R,
and
(11.9)		H (x, t) ≤ W0(x),				x ,		t R,
where Wi (x) L1( ),						t
(11.10)			F(x, t) :=		0	f (x, s)ds,
and
(11.11)			H (x, t) := 2F(x, t) − t f (x, t).
Assume further that
(11.12)		H (x, t) → H±(x) a.e. as t → ±∞
and
(11.13)	H+(x)dx +		H−(x)dx +			u=0	W0(x)dx < −B1 = − W1(x)dx
	u>0		u<0
whenever u is a nontrivial solution of (11.6). Then (11.2) has a solution.
Theorem 11.2. If		f (x, t) satisfies (11.7) and (11.6) has a nontrivial solution with
(a, b) C,2, assume that
(11.14)		2F(x, t) ≤ λ +1t2 + W1(x),					x , t R,
(11.15)		H (x, t) ≥ −W0(x),			x , t R,
and
(11.16)		H+(x)dx +		H−(x)dx −				W0(x)dx > B1
	u>0			u<0			u=0

for all nontrivial solutions u of (11.6). Then (11.2) has a solution.

We let C,1

(11.24)

11.2. The curves	131
Note that (11.13) is automatically satisfied if
(11.17)	H (x, t) → −∞ a.e. as |t| → ∞
and (11.16) is automatically satisfied if
(11.18)	H (x, t) → +∞ a.e. as |t| → ∞.

11.2 The curves

In this section we shall describe the curves C,1 and C,2 and discuss some of their properties. For each fixed positive integer , we let N denote the subspace of E spanned by the eigenfunctions corresponding to λ0, . . . , λ , and we let M = N ∩ D, where D = D( A1/2). Then D = M N . For a, b R, we define

(11.19)	I (u, a, b) = ( Au, u) − a u− 2 − b u+ 2, u D,
(11.20)	M	(a, b) = w M	v N	(v + w,	a	,	b	),
		inf	sup I
		w=1
(11.21)	m (a, b) = v N		w M I (v + w, a, b),
		sup	inf
		v=1
(11.22)		ν (a) = sup{b : M (a, b) ≥ 0},
(11.23)		μ (a) = inf{b : m (a, b) ≤ 0}.

We have

Lemma 11.3. [111, 122] In the square Q the functions μ (a), ν −1(a) have the following properties:

(a)μ (λ ) = ν −1(λ ) = λ ;

(b)μ (a), ν −1(a) are continuous and strictly decreasing;

(c)(a, μ (a)) and (a, ν −1(a)) are in ;

(d)if (a, b) Q and b > μ (a), then (a, b) / ;

(e)if (a, b) Q and b < ν −1(a), then (a, b) / .

It follows from Lemma 11.3 that ν −1(a) ≤ μ (a) in Q . If ν −1(a) < μ (a), this lemma does not cover those points in Q for which

ν −1(a) < b < μ (a).

be the lower curve b = ν −1(a) and we take C,2 to be the upper curve b = μ (a). We shall need

132	11. Fucˇ´ık Spectrum: Resonance
Lemma 11.4. [111, 122] If (a, b) Q , w M −1, v1, v2 N −1, and
(11.25)	I (w + v j , a, b) N −1, j = 1, 2,

then v1 = v2.

Lemma 11.5. [111, 122] I f (a, b) Q , v N , w1, w2 M , and

(11.26) I (v + w j , a, b) M , j = 1, 2,

then w1 = w2.

Lemma 11.6. [111, 122] If (a, b) Q , then, for each w M −1, there is a v0 N −1 such that

(11.27)	I (w + v0, a, b) = sup I (w + v, a, b).
	v N −1

Lemma 11.7. [111, 122] If (a, b) Q , then, for each v N , there is a w0 M such that

(11.28)	inf			I (v	+ w, a, b).
	I (v + w0, a, b) = w		M
Lemma 11.8. If (a, b) C,1, then
(11.29)	I (u, a, b) ≥ 0,			u S,1,
where
(11.30)	S,1 = {u E : I (u, a, b) N −1}.
Proof. If (a, b) C,1, then b = ν −1(a). Thus, by (11.22),
(11.31)	M −1(a, b) = 0.
This implies by (11.20) that
(11.32)	sup I (v + w, a, b) ≥ 0,				w M −1.
	v N −1

Let u0 be any element in S,1. Then u0 = w0 + v0, w0 M −1, v0 N −1, and

(11.33) I (w0 + v0, a, b) N −1

by (11.30). In view of Lemma 11.6, there is a v1 N −1 such that

(11.34)	I (w0 + v1, a, b) = sup I (w0 + v, a, b).
	v N −1
From this and (11.32), it follows that
(11.35)	I (w0 + v1, a, b) N −1

11.2. The curves	133
and
(11.36)	I (w0 + v1, a, b) ≥ 0.

But (11.33) and (11.35) imply via Lemma 11.4 that v1 = v0. Thus,

I (u0, a, b) ≥ 0,

and the lemma is proved.

Lemma 11.9. If (a, b) C,2, then

(11.37)	I (u, a, b) ≤ 0, u S,2,
where
(11.38)	S,2 = {u E : I (u, a, b) M }.

Proof. If (a, b) C,2, then b = μ (a). Consequently, by (11.23),

(11.39)	m (a, b) = 0.
By (11.21), this implies
(11.40)	w M I (v + w, a, b) ≤ 0, v N .
	inf

Let u0 be any element in

(11.41)

By Lemma 11.7, there is

S,2 and write u0 = v0 + w0 with v0 N , w M . Then

I (v0 + w0, a, b) M . a w1 M such that

(11.42)	I	(v0	+ w1, a, b) = w M I (v0 + w, a, b).
			inf
Consequently,
(11.43)			I (v0 + w1, a, b) M
and
(11.44)			I (v0 + w1, a, b) ≤ 0.

If we now apply Lemma 11.5, we see that w1 = w0. Thus, I (u0, a, b) ≤ 0, and the proof is complete.

Lemma 11.10. N −1 ∩ ∂ BR links S,1 [hm] for each R > 0.

134	11. Fucˇ´ık Spectrum: Resonance

Proof. We suppress the subscript − 1. Let P be the orthogonal projection of D onto N, and define

(11.45) F(u) = P I (u), u D.

I claim that the restriction F0 of F to N is a homeomorphism of N onto N. If this is so,

∩ ¯

then the image of N BR under F0 is the closure of a bounded, open set. Moreover,

this open set contains the point 0 since F0(0) = 0 and 0 is an interior point of N ∩ BR . Proposition 3.10 can then be used to show that N ∩ ∂ BR links F−1(0) = S,1. Clearly, F0 is a continuous map of N into itself. It is surjective. To see this, let h be any element of N and take

(11.46)	G0(v) = I (v, a, b) − (h, v).
I claim that
(11.47)	G0(v) → −∞ as v → ∞, v N.

Assuming this for the moment, we see that G0(v) has a maximum on N. At a point of

	0		N, producing a solution of
maximum we have G (v)
(11.48)			P I (v, a, b) = h.
Moreover, if
(11.49)	P I (v0, a, b) = h0, P I (v1, a, b) = h1,
then
(11.50)	(h, v) = ( Av, v) + a(v1− − v0−, v) − b(v1+ − v0+, v)
where h = h1 − h0, v = v1 − v0. Thus,
(11.51)	(b − λ −1)(v1+ − v0+, v) − (a − λ −1)(v1− − v0−, v),
			+ [λ −1 v 2 − ( Av, v)] = −(h, v).

Since a, b > λ −1, the left-hand side of (11.51) is positive for all v ≡ 0. Consequently, there is a δ > 0 such that

(11.52) δ v 2 ≤ h v

showing not only that F0 is injective but that F0−1 is continuous. To prove (11.47), let {vk } N be a sequence such that ρk2 = ( Avk , vk ) → ∞, and let v˜k = vk /ρk . Thenv˜k 2D = ( Av˜k , v˜k ) = 1 and there is a subsequence such that v˜k → v˜ in D. We have

(11.53) G0(vk )/ρk2 = I (v˜k , a, b) − (h, v˜k )/ρk

→I (v˜, a, b)

=[ v˜ 2D − λ −1 v˜ 2] +(λ −1 − a) v˜− 2 +(λ −1 − b) v˜+ 2.

11.3. Existence

135

Since v˜ D = 1, the right-hand side of (11.53) is negative. This gives (11.47) and completes the proof of the lemma.

The following can be proved in the same way.

Lemma 11.11. M ∩ ∂ BR links S,2 [hm] for each R > 0.

11.3 Existence

In this section we give the proofs of Theorems 11.1 and 11.2. Let

(11.54)	p(x, t) = f (x, t) + at− − bt+, t± = max{±t, 0},
and
					t
(11.55)		P(x, t) =		0	p(x, s) ds.
Under hypothesis (11.7), it is readily checked [122] that the functional
(11.56)	G(u) = u 2D − 2		F(x, u) dx = I (u, a, b) − 2			P(x, u) dx
is in C1 on D with
(11.57)	(G (u), v) = 2(u, v)D − 2( f (·, u), v) = (I (u, a, b), v) − 2( p(·, u), v)
and
(11.58)	(I (u, a, b), v)/2 = (u, v)D + a(u−, v) − b(u+, v).
From (11.5) and (11.54) we see that
(11.59)	p(x, t)/t → 0, 2P(x, t)/t2 → 0 as |t| → ∞.
This and the fact that
(11.60)	∂ (F/t2)/∂ t = ∂ ( P/t2)/∂ t = −t−3 H
imply that
			∞	s−3 H (x, s)ds, t > 0
(11.61)	P(x, t)	=	t2
			t
			t	s−3 H (x, s)ds, t < 0.
		= −t2

−∞