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

10.4. Some applications				115
where
			t
(10.42)	F(x, t) =	0	f (x, s)ds.
From (10.33) it is readily verified that G is continuously differentiable on D and
(10.43)	(G (u), v)/2 = a(u, v) − ( f (u), v),			u, v D,
where we write	f (u) in place of f (x, u) (cf., e.g., [112]). We note that G (u) satisfies

(10.5) on E = D : for, if uk → u weakly in D, then a(uk , v) → a(u,2v) for each
v D, and there is a renamed subsequence such that V uk		→ V u in L ( ). Since
V (x) > 0 in , there is a renamed subsequence such that uk		→ u a.e. in . Thus,
(10.44)	f (x, uk )v → f (x, u)v a.e.

Since

| f (x, uk )v| ≤ |V uk V v| + W |V v|

and the right-hand side converges to |V uV v|+ W |V v| in L1( ), we see that ( f (uk ), v)

→( f (u), v). Thus, G (u) satisfies (10.5). I claim that

(10.45)	G(v) → −∞	as v D → ∞,	v N,
(10.46)	G(w) → ∞	as w D → ∞,	w M,
where u D	= | A|1/2u . To prove (10.45), let		{vk } be a sequence such that

ρk = vk D → ∞, and let v˜k = vk /ρk . Then v˜k D = 1, and there is a renamed subsequence that converges weakly in D and a.e. in to a function v˜ D (again using the properties of V ). I claim that

(10.47)				2	F(x, vk )dx/ρk2 → α(v˜)
and
(10.48)	G(vk )/ρk2 → −1 − α(v˜) = ( v˜ 2D − 1) − ( v˜ 2D + α(v˜)) ≤ 0.
To see this, note that
	2F(x, vk )		=	2F(x, vk )		v2	α (v+)2	+	α (v−)2	a.e.
	ρk2
					vk2	˜k −→	+ ˜		− ˜

Moreover, by (10.33),

|F(x, vk )|/ρk2 ≤ C(|V v˜k |2 + W |V v˜k |/ρk ) −→ C|V v˜|2

in L1( ). This implies (10.47) and (10.48). Now, the only way the right-hand side of (10.48) can vanish is if

(10.49)

v˜ D = 1

116			10. Weak Linking
and
(10.50)			( Av˜, v˜) = α(v˜).
Let			(v) = ( Av, v) − α(v).
Then
			(v) ≤ 0, v N.
˜	=	˜		=
			˜		0, i.e.,
If (v)		0, then v	is a maximum point for on N. This means that (v )

that

( Av˜, v) − α(v˜, v) = 0, v N.

Thus, v˜ is a solution of (10.38). However, this contradicts (10.49). Hence, the righthand side of (10.48) must be negative. This proves (10.45). The proof of (10.46) is similar. Note that (10.46) implies

b0 = inf G > −∞,

M

for if G(wk ) → b0, {wk } M, then wk D ≤ C by (10.46). But then (10.33) implies that G is bounded on bounded sets in D. From (10.45) we see that there is an R such that (10.1) holds with A = N ∩ ∂ BR , B = M, Q = N ∩ BR , and F = P, the orthogonal projection of D onto N, p = 0. If S is a finite-dimensional subspace of D such that P S = {0}, let S0 = P S. Clearly, (10.6) is satisfied. We can now apply Theorem 10.2 to conclude that there is a sequence {uk } D such that

(10.51)	G(uk ) → c, b0 ≤ c ≤ a1, G (uk ) → 0.
Assume first that ρk	= uk D → ∞, and write u˜ k = uk /ρk . Then u˜k D = 1.

Consequently, there is a renamed subsequence such that u˜k → u˜ weakly in D and a.e. in (using the properties of V ). Since

| f (x, uk )v|/ρk ≤ |V u˜k V v| + W |V v|/ρk

and the right-hand side converges in L1( ), we see by (10.34) that

(10.52)			( f (uk ), v)/ρk → α(u˜ , v).
Thus,	(u
(G		k	), v)/2ρ	a(u, v)	−	α(u, v), v		D.
			k →	˜		˜

Since this limit vanishes by (10.51), we see that u˜ is a solution of (10.38). Hence, u˜ ≡ 0 by hypothesis. Let u˜k = v˜k + w˜ k , where v˜k N, w˜ k M. Arguments similar to that given in the proof of (10.47) show that

( f (uk ), v˜k )/ρk → α(u˜ , v˜)

and

( f (uk ), w˜ k )/ρk → α(u˜, w),˜

10.4. Some applications

117

where u˜ = v˜ + w˜ . Since v˜k 2D + w˜ k 2D = 1, we have for a renamed subsequence

(G	(u	), v	)/2ρ	→	γ	1	−	α(u, v ),
	k	˜k	k					˜ ˜
(G (uk ), wk )/2ρk				→	γ2		−	α(u, w),
		˜						˜ ˜

where γ2 + γ1 = 1. By (10.51), both of these expressions must vanish. But this cannot happen if u˜ ≡ 0. Thus, the ρk must be bounded, and consequently there is a renamed subsequence of {uk } such that uk → u weakly in D and a.e. in . As before, this implies ( f (uk ), v) → ( f (u), v) for v D. Hence,

(G (uk ), v) → (G (u), v), v D.

By (10.51), this limit must vanish for each v in D. We conclude that G (u) = 0 and that u is a solution of (10.39).

	2				Rn and let A		≥ λ0	>			2
As another application, let									0 be a self-adjoint operator
on L		( ) with compact resolvent. Let F(x, s, t) be a function on × R such that
(10.53)			f (x, s, t) = ∂ F/∂ t,			g(x, s, t) = ∂ F/∂ s
are Carath´eodory functions satisfying
(10.54)			| f (x, s, t)| + |g(x, s, t)| ≤ C(|s| + |t| + 1),							s, t R,
and
(10.55)			f (x, s, t)/ρ → α(x)s˜ + β(x)t˜,
(10.56)			g(x, s, t)/ρ → γ (x)s˜ + δ(x)t˜
as ρ2 = s2 + t2 → ∞, s/ρ → s˜, t/ρ → g˜. We wish to solve the system
(10.57)			Av = f (x, v, w),		Aw = g(x, v, w),			v, w D( A).
We assume that the only solution of
(10.58)			Av = αv + βw,			Aw = γ v + δw
is v	≡ w ≡ 0 in . If λ0 is an eigenvalue of A, we assume that corresponding
eigenfunctions are = 0 a.e. on . Finally, we assume that
(10.59)			β + γ − α − δ ≤≡ 2λ0,			β + γ + α + δ ≤≡ 2λ0.

118 10. Weak Linking

We have

Theorem 10.6. Under the above assumptions, system (10.57) has a solution.

Proof. Let D = D( A1/2), E = D × D. Then D becomes a Hilbert space with norm given by

(10.60)	u 2E = ( Av, v) + ( Aw, w), u = (v, w) D.
We define
(10.61)	G(u) = ( Av, w) −	F(x, v, w) dx, u D.
In view of (10.54), G C1(D, R) and
(10.62)	(G (u), h) = ( Av, h2) + ( Aw, h1) − ( f (u), h2) − (g(u), h1)
for u = (v, w) E, h = (h1, h2)		E, and we write f (u), g(u) in place of
f (x, v, w),	g(x, v, w), respectively. It follows from (10.62) that u E is a solu-
tion of (10.57) if u satisfies
(10.63)	G (u) = 0.

Let N denote the subspace of E consisting of those u = (v, w) for which w = −v and let M consist of those for which w = v. Then M = N . We note that

(10.64)	G(v, −v) → −∞ as v D → ∞
and
(10.65)	G(w, w) → +∞ as w D → ∞.

To prove (10.64), let {vk } D be such that ρk = vk D → ∞, and take v˜k = vk /ρk .

Since v˜k D 2= 1, there is a renamed subsequence such that v˜k → v˜ weakly in N,
strongly in L ( ), and a.e. in such that
(10.66) 2G(vk , −vk )/ρk2 = −2 v˜k 2D − 2	F(x, vk , −vk )dx/ρk2

→ −2 − [γ − δ − α + β]v˜2dx

= − [2λ0 + β + γ − α − δ]v˜2dx + 2λ0 v˜ 2 − 2.

In view of (10.59), this is negative unless λ0 v˜ 2 = 1. But this would mean that v˜ E(λ0), the eigenspace of λ0. If λ0 is not an eigenvalue, then we conclude immediately that

(10.67)	lim sup G(v, −v)/ v 2D < 0
	v D →∞

10.4. Some applications

119

and (10.64) holds. If λ0 is an eigenvalue, then v˜ = 0 a.e. by hypothesis. But then the last integral in (10.66) must be positive by (10.59). Again, this implies (10.67) and (10.64). A similar argument implies (10.65). Now that we have (10.64) and (10.65), we know that (10.1) holds for A = N ∩ ∂ BR and B = M when R is sufficiently large. If we let F be the projection of E onto N, we can take Q = N ∩ BR and p = 0.

We can conclude from Theorem 10.2 that there is a sequence {uk }

E satisfying

(10.2), (10.3). I claim that

(10.68)

ρk = uk E ≤ C.

To see this, assume that ρk → ∞, and let u˜k

= uk /ρk .2

Then there is a renamed

subsequence such that u˜k → u˜ weakly in E, strongly in L

( ), and a.e. in . If h =

(h1, h2) E, then, by (10.62),

(10.69) (G (u

k

), h)/ρ

( Av

, h

2

)

+

( Aw

, h

1

)

−

( f (u

k

), h

)/ρ

k −

(h(u

k

), h

)/ρ ,

k =

˜k

˜ k

2

1 k

where uk = (vk , wk ). Taking the limit and applying (10.54) – (10.56) we see that u˜ = (v˜, w)˜ is a solution of (10.58). By hypothesis, u˜ ≡ 0. On the other hand, there is a renamed subsequence such that v˜k D → a, w˜ k D → b with a2 + b2 = 1. Moreover,

(10.70)

(G

k

˜k

,

− ˜k

k

= −

˜k D +

k

˜k

k

(u

), (v

v

))/ρ

2 v 2

( f (u

), v

)/ρ

− (g(uk ), v˜k )/ρk

→ − 2a2 + [α + β − γ − δ]v˜2dx

and

˜

˜

= ˜

D −

˜

(10.71)

(G (uk ), (wk , wk ))/ρk

2

wk

2

( f (uk ), wk )/ρk

− (g(uk ), wk )/ρk

→2b2 − [α + β + γ + δ]w˜ 2dx.

By (10.2), both these limits are 0. Since a and b cannot both vanish, the same is true of v˜ and w˜ . Hence, u˜ ≡ 0, and this contradiction shows that (10.68) does not hold. We can now use the usual procedures to show that {uk } has a renamed subsequence

converging weakly in E to a function u, strongly in L2( ), and a.e. in . Then this limit is a solution of (10.63) and consequently of (10.57).

For another application, let A, B be positive, self-adjoint operators on L2( ) with compact resolvents, where Rn. Let F(x, v, w) be a function on × R2 such that

(10.72)	f (x, v, w) = ∂ F/∂ v, g(x, v, w) = ∂ F/∂ w
are Carath´eodory functions satisfying
(10.73)	| f (x, v, w)| + |g(x, v, w)| ≤ C0(|v| + |w| + 1), v, w R,

120	10. Weak Linking
and
(10.74)	f (x, t y, tz)/t → α+(x)v+ − α−(x)v− + β+(x)w+ − β−(x)w−,
(10.75)	g(x, t y, tz)/t → γ+(x)v+ − γ−(x)v− + δ+(x)w+ − δ−(x)w−

as t → +∞, y → v, z → w, where a± = max(±a, 0). We wish to solve the system

(10.76)	Av = − f (x, v, w),
(10.77)	Bw = g(x, v, w).

Let λ0(μ0) be the lowest eigenvalue of A(B). We assume that the only solution of

(10.78)	− Av = α+v+ − α−v− + β+w+ − β−w−,
(10.79)	Bw = γ+v+ − γ−v− + δ+w+ − δ−w−
is v = w = 0. We have

Theorem 10.7. Assume that eigenfunctions of λ0 are = 0 a.e. on ,

(10.80)	α±(x) ≥≡ −λ0,	x ,
and
(10.81)	2F(x, 0, t) ≤ μ0t2 + W (x),	x , t R,

where W (x) L1( ). Then the system (10.76), (10.77) has a solution.

Proof. Let D = D( A1/2) × D(B1/2). Then D becomes a Hilbert space with norm given by

(10.82)	u 2D = ( Av, v) + (Bw, w), u = (v, w) D.
We define
(10.83)	G(u) = b(w) − a(v) − 2	F(x, v, w) dx, u D,
where
(10.84)	a(v) = ( Av, v),	b(w) = (Bw, w).
Then G C1(D, R) and
(10.85)	(G (u), h)/2 = b(w, h2) − a(v, h1) − ( f (u), h1) − (g(u), h2),

where we write f (u), g(u) in place of f (x, v, w), g(x, v, w), respectively. It is readily seen that the system (10.76), (10.77) is equivalent to

(10.86) G (u) = 0.

10.4. Some applications

121

We let N be the set of those (v, 0) D and M the set of those (0, w) D. Then M, N are orthogonal closed subspaces such that

(10.87)	D = M N.
If we define
(10.88)	Lu = 2(−v, w), u = (v, w) D
then L is a self-adjoint, bounded operator on D. Also,
(10.89)	G (u) = Lu + c0(u),
where
(10.90)	c0(u) = −( A−1 f (u), B−1g(u))

is compact on D. This follows from (10.73) and the fact that A and B have compact resolvents. Now, by (10.81),

(10.91)	G(0, w) ≥ b(w) − μ0 w 2 −		W (x) dx, (0, w) M.
Thus,
(10.92)	M G ≥ −	W (x)dx.
	inf
On the other hand, (10.80) implies
(10.93)	sup G → −∞ as R → ∞.
	N∩∂ BR

To see this, let (vk , 0) be any sequence in N such that ρk2 = a(vk ) → ∞. Then

(10.94) G(vk , 0)/ρk2 = −a(v˜k ) − 2 F(x, vk , 0)dx/ρk2,

where v˜k = vk /ρk . Note that a(v˜k ) = 1. Thus, there is a renamed subsequence v˜k → v˜ weakly in N, strongly in L2( ), and a.e. in such that

G(vk , 0)/ρ2

→ −

1

−

α

(v

+ )2

+

α (v

−)2

}

dx

k

{

+

˜

− ˜

= −

{

(λ0

+

α

+

)(v+)2

+

(λ0

+

α

)(v−)2

}

dx

˜

−

˜

+

This is negative unless λ0 v˜ 2 the eigenspace of λ0. Thus, v˜ vanish by (10.80). Hence,

λ0 v˜ 2 − 1.

= 1. Since a(v˜) ≤ 1, this would mean that v˜ E(λ0),= 0 a.e. by hypothesis. But then the integral cannot

(10.95)

lim sup G(0, v)/a(v) < 0,

a(v )→∞

122	10. Weak Linking

and (10.93) holds. Since N is an invariant subspace of L, we can apply Theorem 10.2 to conclude that there is a sequence {uk } D such that (10.2) and (10.3) hold. Let uk = (vk , wk ). I claim that

(10.96)			ρk2 = a(vk ) + b(wk ) ≤ C.
To see this, assume that ρk			→ ∞, and let u˜k				= uk /ρk .2	Then there is a renamed
subsequence such that u˜ k → u˜ weakly in D, strongly in L								( ), and a.e. in . If h =
(h1, h2) D, then
(10.97)		2b(wk, h2)			2a(vk , h1)		2( f (uk ), h1)/ρk			2(g(uk ), h2)/ρk .
(G (uk ), h)/ρk	=			−		−			−
		˜			˜

Taking the limit and applying (10.73) – (10.75), we see that u˜ = (v˜, w)˜ is a solution of

(10.78)

(10.79).

Hence,

u˜

=

0

by

hypothesis.

On

the

other hand,

since

a

˜

+

˜

=

˜

→ ˜

˜

→

b

(vk )

b(wk ) 1, there is a renamed subsequence such that a(vk )

a, b(wk )

˜

˜ +

b

=

1. Thus, by (10.74), (10.75), and (10.85),

with a

˜

˜k

k = −

˜ k

−

˜k

(10.98)

(G

(u

k

)

( f (u

k

k

), (v

, 0))/2ρ

a (v

), v

)/ρ

a

(α v+

−

α v−

+

β

w+

−

β

w−)vdx

→ − ˜ −

+

˜

−

˜

+

˜

− ˜ ˜

and

(10.99)

(G

(u

k

), (0, w

))/2ρ

b (w

)

−

(g(u

k

), w

)/ρ

˜ k

k =

˜ k

k

b

(γ

v+

−

γ

v

−

+

δ

w+

−

δ

w−)wdx.

→ ˜ −

+ ˜

−

˜

+

˜

−

˜ ˜

Thus, by (10.42),

(10.100)

a

(α

v+

−

v

−

+

β

+

w+

−

β

−

w−)v dx

˜ = −

+ ˜

− ˜

˜

˜ ˜

and

(10.101)

b

(γ

+

v+

−

γ

−

+

δ

w+

−

δ

−

w−)wdx.

˜ =

˜

− ˜

+

˜

˜ ˜

˜ ≡

˜

0. This

b is not zero, we see that we cannot have u

Since one of the two numbers a,

˜

contradiction proves (10.96). Once this is known, we can use the usual procedures to show that there is a renamed subsequence such that uk → u in D, and u satisfies (10.86).

Corollary 10.8. In Theorem 10.7 we can replace (10.80), (10.81) with

(10.102)	2F(x, v, 0) ≥ −λ0v2 − W (x), x , v R,
provided eigenfunctions of μ0 are = 0 a.e. in and
(10.103)	δ±(x) ≤≡ μ0.

10.4. Some applications

123

Proof. We interchange the roles of M and N in the proof of Theorem 10.7.

Theorem 10.9. If λ0 is simple and the eigenfunctions of λ0 and μ0 are bounded and= 0 a.e. in , we can replace (10.81) in Theorem 10.7 with

(10.104)	2F(x, t, 0) ≥ −λ1t2, t R,
(10.105)	2F(x, s, t) ≤ μ0t2 − λ0s2, |t| + |s| ≤ δ,

where λ1 is the next eigenvalue of A and δ > 0. Moreover, system (10.76), (10.77) has a nontrivial solution.

Proof. Let N be the orthogonal complement of N0 = {ϕ0} in N. Then N = N N0 and

(10.106)
G(v , 0) = −a(v ) − 2	F(x, v , 0) dx ≤ −a(v ) + λ1 v 2 ≤ 0, v N ,
and
(10.107)	G(v, 0) → −∞ as a(v) → ∞

by (10.80) [cf. (10.93)]. Let M0 be the subspace of M spanned by the eigenfunctions of B corresponding to μ0, and let M be its orthogonal complement in M. Since N0 and M0 are contained in L∞( ), there is a positive constant ρ such that

(10.108)	a(y) ≤ ρ2	y ∞ ≤ δ/4,	y N0,
(10.109)	b(h) ≤ ρ2	h ∞ ≤ δ/4,	h M0,
where δ is the number given in (10.105). If
(10.110)	a(y) ≤ ρ2, b(w) ≤ ρ2, |y(x)| + |w(x)| ≥ δ,
we write w = h + w , h M0, w M , and

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

Thus,
(10.112)	|y(x)| + |h(x)| ≤ δ/2 ≤ |w (x)|
and
(10.113)	|y(x)| + |w(x)| ≤ 2|w (x)|.

124					10. Weak Linking
Now, by (10.105) and (10.113),
G(y, w) =b(w) − a(y) − 2			F(x, y, w)dx
≥b(w) − a(y) −				{μ0w2 − λ0 y2}dx
		|y|+|w|<δ
− c0		(|y| + |w| + 1)dx
|y|+|w|>δ
≥b(w) − a(y) − μ0 w 2 + λ0 y 2 − c1					|w |4dx
				2|w |>δ
≥b(w ) − μ0 w 2 − c2b(w )2
	μ0
≥ 1 −		− c2b(w ) b(w )
	μ1
when	a(y) ≤ ρ2,			b(w) ≤ ρ2,

where μ1 is the next eigenvalue of B after μ0. If we reduce ρ accordingly, we can find a positive constant ν such that

(10.114)	G(y, w) ≥ νb(w ), a(y) ≤ ρ2, b(w) ≤ ρ2.
I claim that either	(a) (15.48), (15.49) has a nontrivial solution or (b) there is an
> 0 such that
(10.115)	G(y, w) ≥ , a(y) + b(w) = ρ2.

To see this, suppose (10.115) did not hold. Then there would be a sequence {yk , wk } such that a(yk ) + b(wk ) = ρ2 and G(yk , wk ) → 0. If we write wk = wk + hk , wk

M , hk M0, then (10.114) tells us that b(wk ) → 0. Thus, a(yk ) + b(hk ) → ρ2.

Since N0, M0 are finite-dimensional, there is a renamed subsequence such that yk → y

in N0 and hk → h in M0. By (10.108) and (10.109), y ∞ ≤ δ/4 and h ∞ ≤ δ/4. Consequently, (10.105) implies

(10.116)	2F(x, y, h) ≤ μ0h2 − λ0 y2.
Since
(10.117)	G(y, h) = b(h) − a(y) − 2	F(x, y, h) dx = 0,
we have
(10.118)	{2F(x, y, h) + λ0 y2 − μ0h2}dx = 0.