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

200							16. Multiple Solutions
Proof. For u = v + w, v Nm , w Mm , we have
G(u) ≤ I (u, a, b) + C	|u|q dx + W1(x) dx ≤ I (u, a, b) + K .
|u|<K
Thus,			(v + w)
J (v) = w Mm		G
inf
							K
inf		I (v		+	w, a, b)	+
≤ w	Mm

= F2m (v, a, b) + K ≤ m(a, b) v 2D + K .

If b ≥ μm (a), then m(a, b) ≤ 0. This proves (16.38). If b > μm (a), then m(a, b) < 0. This proves (16.39).

Lemma 16.15. If l < m and λl < a, b < λm+1, then there are continuous functions ξ : Nm ∩ Ml → Nl , η : Nm ∩ Ml → Mm homogeneous of degree 1 and such that

(16.40)	I (v +w+ y, a, b) = w Mm		(v +w+ y, a, b)
I (ξ(y)+η(y)+ y, a, b) = sup w Mm		sup I
inf	inf
v Nl		v Nl
for y Nm ∩ Ml .

Proof. Let

L y (v, w) = I (v + w + y, a, b).

Then L y is a strictly convex, lower semicontinuous functional in w Mm , and strictly concave and continuous in v Nl . By Theorem 13.6 and Corollary 13.12, for each y0 Nm ∩ Ml , there are unique elements v0 = ξ(y0) Nl , w0 = η(y0) Mm such that (16.40) holds, i.e., that

L y0 (v, w0) ≤ L y0 (v0, w0) ≤ L y0 (v0, w), v Nl , w Mm .

The functions ξ, η are clearly homogeneous of degree 1. To prove continuity, let y j → y0 in Nl ∩ Mm , and let v j = ξ(y j ), w j = η(y j ). We note that the functions v j , w j are bounded in D. Otherwise, it is easy to show that

I (v + w j + y j , a, b) −→ ∞ as j −→ ∞

for any v Nl , and

I (v j + w + y j , a, b) −→ −∞ as j −→ ∞

for any w Mm . This would contradict (16.40). Thus there are renamed subsequences such that v j → v1, w j w1 in D. Since

I (v + w j + y j , a, b)

≤ I (v j + w j + y j , a, b)

≤ I (v j + w + y j , a, b), v Nl , w Mm ,

16.4. Some lemmas

201

we have in the limit

I (v + w1 + y0, a, b)

	≤ I (v1 + w1 + y0, a, b)
	≤ I (v1 + w + y0, a, b), v Nl , w Mm ,
showing that v1	= v0, w1 = w0. Since this is true for any subsequence, the result
follows.
Lemma 16.16. If
(16.41)	2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| ≤ δ,

for some δ > 0, with a1, b1 > λl , b1 < νl (a1), l < m, then there are ε > 0, r > 0 such that

(16.42)

J (y + ξ(y)) ≥ ε y 2D ,

y Nm ∩ Ml ∩ Br .

Proof. By Lemma 16.15

1) = w Mm v Nl

(16.43)

w Mm

I

(ξ(

y

) +

y

+ w,

a

1,

b

(v +

y

+ w,

a

1,

b

1)

inf

inf

sup I

for y Nm ∩ Ml . Then, for y (Nm ∩ Ml ∩ Br )\{0},

J (ξ(y) + y) = G(ξ(y) + y + ϕ(ξ(y) + y))

≥ I (ξ(y) + y + ϕ(ξ(y) + y), a1, b1) − o( y 2D)

≥ w Mm

I (ξ(y) + y + w, a1

, b1) − o( y D)

inf

2

sup I (v + y + w, a1, b1) − o( y D)

= w Mm

inf

v Nl

2

inf

M a

b

)

y

2

−

o

y 2

≥ w

Mm

l ( ,

+ w D

( D)

= Ml (a, b) y 2D − o( y 2D) ≥ ε y 2D.

Lemma 16.17. Assume
(16.44)	t[ f (x, t1) − f (x, t0)] ≥ a(t−)2 + b(t+)2,	t j R, t = t1 − t0.
Then
(16.45)	(G (v1 + w) − G (v0 + w), v) ≤ 2I (v, a, b),	v j , w D, v = v1 − v0.

202							16.	Multiple Solutions
Proof. We have
Hence,	( f (x, v1 + w) − f (x, v0 + w), v) ≥ a v− 2 + b v+ 2.
(G (v1 + w) − G (v0+w), v)/2
				= v 2D − ( f (x, v1 + w) − f (x, v0 + w), v)
				≤ I (v, a, b).
Lemma 16.18. If		f (x, t) satisfies (16.44), and b > γm (a), then there is a continuous
map ψ from Mm → Nm such that						1			), w Mm ,
(16.46) J	(w) ≡		G	(w + ψ(w)) = v Nm G(v + w) C			(Mm ,
				max				R
and
(16.47)				J (w) = G (w + ψ(w)), w Mm .
Proof. In view of Lemmas 16.7 and 16.17, we have
	(G (v1 + w) − G (v0 + w), v) ≤ − v 2D , v Nm .
We can now apply Lemma 16.9 to obtain the conclusion.
Lemma 16.19. If, in addition,
(16.48)			a0(t−)2 + b0(t+)2 ≤ 2F(x, t), |t| < δ,

for some δ > 0, with a0, b0 < λl+1, b0 > μl (a0), l > m, then there are ε > 0, r > 0 such that

(16.49) J (y + η(y)) ≤ −ε y 2D , y Nl ∩ Mm ∩ Br .

Proof. For y Mm ∩ Nl , let u = y + η(y) Mm . By (16.2),

J (u) = G(u + ψ(u)) ≤ I (u + ψ(u), a0, b0) + o( u 2D )

≤sup I (u + v, a0, b0) + o( u 2D )

v Nm

= I (y + η(y) + ξ(y), a0, b0) + o( u 2D )

= v Nm	w Ml	I (y + v + w, a0	, b0) + o( u D )
sup	inf		2

=sup F2l (y + v, a0, b0) + o( u 2D )

v Nm

≤sup ml (a0, b0) y + v 2D + o( u 2D )

v Nm

≤−ε y 2D

for r sufficiently small.

16.4. Some lemmas	203
Lemma 16.20. If
(16.50)	2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| ≤ δ,

for some δ > 0, with a1, b1 > λl , b1 < νl (a1), l > m, then there are ε > 0, r > 0 such that

(16.51) J (w) ≥ ε w 2D , w Ml ∩ Br .

Proof. We recall from Theorem 13.6 that there is a continuous map θ : Ml → Nl such that

(16.52)	θ (s w) = s θ (w), s ≥ 0,
(16.53)	I (θ (w) + w, a1, b1) = sup I (v + w, a1, b1), w Ml .
		v Nl
Thus,
	J (w) ≥ G(w + θ (w), a1, b1)
	≥ I (w + θ (w), a1, b1) − o( w 2D)
	= sup I (v + w, a1, b1) − o( w 2D )
	v Nl
	= F1l (w, a1, b1) − o( w 2D )
	≥ Ml (a1, b1) w 2D − o( w 2D )
	≥ ε w 2D
for r sufficiently small.
Lemma 16.21. Assume that
(16.54)	2F(x, t) ≤ a(t−)2 + b(t+)2 + W1(x), |t| > K ,
for some K ≥ 0, where a, b > λm ,		b ≤ νm (a), l ≥ m, and W1 L1( ). Then there
is a K1 < ∞ such that
(16.55)	J (w) ≥ −K1, w Mm .
If b < νm (a), then
(16.56)	J (w) −→ ∞ as w D −→ ∞.
Proof. For u = v + w, v Nm , w Mm , we have
G(u) ≥ I (u, a, b) − C		|u|q dx − W1(x)dx ≥ I (u, a, b) − K .
	|u|<K

204	16. Multiple Solutions

Thus,

J (w) = sup G(v + w)

v Nm

≥ sup I (v + w, a, b) − K

vNm

=F1m (w, a, b) − K

≥ Mm (a, b) w 2D − K .

If b ≤ νm (a), then Mm (a, b) ≥ 0. This proves (16.55). If b < νm (a), then Mm (a, b) > 0. This proves (16.56).

Lemma 16.22. If
(16.57)	a0(t−)2 + b0(t+)2 ≤ 2F(x, t), |t| < δ,

for some δ > 0, with b0 > γl (a0), l ≤ m, then there are ε > 0, r > 0 such that

(16.58)	J (v) ≤ −ε v 2D , v Nl ∩ Br ,
where	Br = {u D : u D ≤ r }.
Proof. Let q be any number satisfying
	2 < q ≤ 2n/(n − 2T ), 2T < n
	2 < q < ∞, n ≤ 2T.
By (16.2),
J (v) ≤ G(v) ≤ I (v, a0, b0) +		[a0(v−)2 + b0(v+)2 − 2F(x, v)] dx
		|v |>δ
	≤ − v 2D + C	|v|q dx
		|v |>δ

≤ − v 2D + o( v 2D ) ≤ −ε v 2D

for r sufficiently small.

Lemma 16.23. If

(16.59) 2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| ≤ δ,

for some δ > 0, with b1 < l (a1), l < m, then there are ε > 0, r > 0 such that

(16.60) J (v) ≥ ε v 2D , v Nm ∩ Ml ∩ Br .

16.4. Some lemmas	205
Proof. Let u = v + ϕ(v) Ml . Then
J (v) = G(u) ≥ I (u, a1, b1) +	[a0(u−)2 + b0(u+)2 − 2F(x, u)] dx
	|u|>δ
≥ u 2D − C	|u|q dx

|u|>δ

≥u 2D − o( u 2D)

≥v 2D − o( v 2D )

≥ε v 2D

for r sufficiently small, since

	v D ≤ u D ≤ C v D .
Lemma 16.24. If
(16.61)	a0(t−)2 + b0(t+)2 ≤ 2F(x, t), |t| < δ,

for some δ > 0, with b0 > γl (a0), l ≥ m, then there are ε > 0, r > 0 such that

(16.62)	J (w) ≤ −ε w 2D , w Nl ∩ Mm ∩ Br .
Proof. For w Mm ∩ Nl , let u = w + ψ(w) Nl . By (16.2),
	J (w) = G(w + ψ(w)) = G(u)
	≤ I (u, a0, b0) +	[a0(v−)2 + b0(u+)2 − 2F(x, u)] dx
		|u|>δ
	≤ − u 2D + C	|u|q dx
		|u|>δ

≤− u 2D + o( u 2D)

≤−ε u 2D

for r sufficiently small. Since

	w D ≤ u D ≤ C w D ,
the result follows.
Lemma 16.25. If
(16.63)	2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| ≤ δ

for some δ > 0, with b1 < l (a1), l > m, then there are ε > 0, r > 0 such that

(16.64) J (w) ≥ ε w 2D , w Ml ∩ Br .

206		16. Multiple Solutions
Proof. We have
G(w) ≥ I (w, a1, b1) +		[a0(u−)2 + b0(w+)2 − 2F(x, w)] dx
		|w|>δ
≥ w 2D	− C	|w|q dx
	|w|>δ

≥ w 2D − o( w 2D ) ≥ w 2D − o( w 2D ) ≥ ε w 2D

for r sufficiently small. Since

J (w) = sup G(v + w) ≥ G(w),

v Nl

the result follows.

16.5 Local linking

The following theorem will also be used in the proofs of the theorems of Section 16.3. It is also of interest in its own right.

Theorem 16.26. Let M, N be closed subspaces of a Hilbert space E such that
0 < dim N < ∞ and M = N . Let G C1(E, R) satisfy the PS condition and
G(v) ≤ 0, v N ∩ BR ,
G(w) ≥ 0, w M ∩ BR ,
for some R > 0. Assume that
−∞ < α = E	<	0	.
inf G
Then G has at least three critical points.
Proof. Since G satisfies the PS condition, it	has a minimum point satisfying

G(u0) = α. Clearly, 0 is also a critical point. Assume that there are no others. Then

ˆ

= {

u

E : G (u)

=

0

}

contains all points except u0

and 0. If θ < 1, then

the set E

there is a mapping Y (u) from

E to E that is locally Lipschitz continuous and satisfies

ˆ

≤

ˆ

≤

1,

G (u)

(G (u), Y (u)), u

Y (u)

θ

E.

For v N ∩ BR \{0}, let σ (t)v be the solution of

σ (t) = −Y (σ (t)),

t ≥ 0, σ (0) = v.

16.5. Local linking

207

Then

dG(σ (t)v)/dt = (G (σ ), σ ) = −(G (σ ), Y (σ )) ≤ −θ G (σ ) < 0

as long as

σ (

)v

is

in

ˆ

σ (

)v

is continuous in t and

v

for

v = 0. For each

t

E. Note that

t

v N ∩ BR\{0}, there is a maximal interval 0 < t < Tv

in which σ (t)v exists and

satisfies G (σ (t)v) = 0 and G(σ (t)v) < 0. I claim that

(16.65)

σ (t)v → u0 as t → Tv .

To see this, suppose that tk → Tv . Then

tk

σ (tk )v − σ (t j )v ≤ |

t j

Y (σ (t)v) dt| ≤ |tk − t j | → 0.

Thus,

σ (tk )v → h

in E. By continuity,

G (σ (tk )v) → G (h).

If G (h) = 0, the solution can be continued beyond Tv , contrary to the way it was chosen. Thus, G (h) = 0, showing that h = u0. Consequently, σ (tk )v → u0. Since

this is true for any such sequence, (16.65) holds. Note that Tv is continuous in v for v = 0.

Define
(16.66)			σˆ (t)v = σ (t)v, t < Tv ,
					= u0,	t ≥ Tv .
Let w0 be an element of M with unit norm, and take
	K0 = {sw0 + v : v N, s ≥ 0, sw0 + v = R}.
Let	Q = {sw0 + v : v N, s ≥ 0, sw0 + v ≤ R}.
Let ε > 0 be given, and let T > 0 satisfy
		T 2		ε2
(16.67)		v	+		≤ 1,	v N, v = ε.
		T 2		R2
Let ξ(u) be the continuous map from ∂ Q to E such that
			ξ(v) = v, v N ∩ BR ,
and for u = sw0 + v K0,
(16.68)	ξ(sw0 + v) = σˆ (T s/ R)v, v ≥ ε,
					= u0, v < ε.

208				16. Multiple Solutions
By (16.67) and (16.68),
	σˆ (T s/ R)v = u0, v = ε.
Hence, ξ is continuous on ∂ Q. Moreover,
		G(ξ(u)) ≤ 0,	u ∂ Q.
In addition,
	ξ(u) ≥ r > 0, u K0.
Let
	B = ∂ Bδ ∩ M, 0 < δ < r < R.
By Corollary 16.11, A = ξ(∂ Q) links B [mm]. Since
(16.69)	a	0 := sup G ≤ b0	inf G	,
			:= B
		A

we can apply Theorem 2.12 to conclude that (1.4) holds. If a > 0, this provides a third critical point by the PS condition. If a = 0, then there is a sequence satisfying (1.4) and

(16.70) d(uk , B) → 0, k → ∞.

Since G satisfies the PS condition, there is a subsequence converging to a critical point on B. Again, this provides a third critical point.

16.6 The proofs

We prove the theorems of Section 16.3. First, we prove Theorem 16.3.

Proof. By Lemma 16.12, it suffices to show that J (v) has two nontrivial solutions. Now J is bounded from above by Lemma 16.14 and satisfies (PS) by (16.39). Moreover,

(16.71)	J (v) < 0,	v Nl ∩ Br \{0},
by Lemma 16.13, and
(16.72)	J (ξ(y) + y) > 0,	y Nm ∩ Ml ∩ Br \{0},

by Lemma 16.16. Thus, J has a positive maximum on Nm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion.

16.7. Notes and remarks

209

Similarly, we prove Theorem 16.4.

Proof. By Lemma 16.18, it suffices to show that J (w) given by (16.46) has two nontrivial solutions. Now J is bounded from below by Lemma 16.21 and satisfies (PS) by (16.56). Moreover,

(16.73)	J (w + η(w)) < 0,	w Nl ∩ Mm ∩ Br \{0},
by Lemma 16.19, and
(16.74)	J (w) > 0,	w Ml ∩ Br \{0}

by Lemma 16.20. Thus, J has a negative minimum on Mm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion.

Next, we prove Theorem 16.5.

Proof. With reference to Theorem 16.3, we note that, by Lemma 16.12, it suffices to show that J (v) has two nontrivial solutions. Now J is bounded from above by Lemma 16.14 and satisfies (PS) by (16.39). Moreover,

(16.75)	J (v) < 0, v Nl ∩ Br \{0}
by Lemma 16.22, and
(16.76)	J (v) > 0, v Nm ∩ Ml ∩ Br \{0}

by Lemma 16.23. Thus J has a positive maximum on Nm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion. With respect to Theorem 16.4, we note that by Lemma 16.18, it suffices to show that J (w) given by (16.46) has two nontrivial solutions. Now J is bounded from below by Lemma 16.21 and satisfies (PS) by (16.56). Moreover,

(16.77)	J (w) < 0, w Nl ∩ Mm ∩ Br \{0},
by Lemma 16.24, and
(16.78)	J (w) > 0, w Ml ∩ Br \{0},

by Lemma 16.25. Thus, J has a negative minimum on Mm . We can now apply Theorem 16.26 and Lemma 16.9 to obtain the desired conclusion.

16.7 Notes and remarks

In his studies of semilinear elliptic problems with jumping nonlinearities, C´ac [29]– [34] proved the following.