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

190	15. Weak Sandwich Pairs

holds with c ≥ ε1. Arguing as in the proof of Theorem 15.8, we see that there is a u D such that G(u) = c ≥ ε1 > 0, G (u) = 0. Since c = 0 and G(0) = 0, we see

that u = 0, and we have a nontrivial solution of the system (15.48), (15.49).

It therefore remains only to prove (15.93). Clearly, ν ≥ 0. If ν = 0, then there is a sequence {wk } M such that

(15.97)		G(0, wk ) → 0,	b(wk ) = 1.
Thus, there is a renamed subsequence such that				wk → w weakly in M, strongly in
L2( ), and a.e. in . Consequently,
(15.98)	[μ0 − μ(x)]wk2 dx ≤ 1 − μ(x)wk2 dx ≤ G(0, wk ) → 0
and
(15.99)	1 =	μ(x)w2dx ≤ μ0 w 2 ≤ b(w) ≤ 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

(15.100) [μ0 − μ(x)]w2dx = 0

implies that the integrand vanishes identically on , and consequently μ(x) ≡ μ0, violating (15.73). This establishes (15.93) and completes the proof of the theorem.

15.4 Notes and remarks

Weak sandwich pairs were introduced in [133].

Chapter 16

Multiple Solutions

16.1 Introduction

Nonlinear problems are characterized by the fact that very often solutions are not unique. However, it is sometimes quite difficult to show that more than one solution exists. In this respect, variational methods can be very helpful. We have chosen some examples that illustrate this point.

16.2 Two examples

Let be a smooth, bounded domain in Rn , and let A be a self-adjoint operator on L2( ). We assume that

(16.1)	C	∞( )		D :	=	D(	A	1/2)		H T ,2( )
		0				|	|

holds for some T > 0 (T need not be an integer), and the eigenvalues of A satisfy

0 < λ0 < λ1 < · · · < λk < · · · .

Let f (x, t) be a Carath´eodory function on × R. We assume that the function f (x, t) satisfies

(16.2) | f (x, t)| ≤ C(|t| + 1), x , t R.

We shall prove

Theorem 16.1. Assume that for some integers l < m, the following inequalities hold:

(16.3)	t[ f (x, t1) − f (x, t0)] ≤ at2, t j R, t = t1 − t0,
where a < λm+1,
(16.4)	a0t2 ≤ 2F(x, t) ≤ a1t2, |t| < δ,

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_16, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

192	16. Multiple Solutions
for some δ > 0, with λl < a0 ≤ a1 < λl+1, where
t
F(x, t) := f (x, s)ds,
0
and

(16.5) a2t2 − W1(x) ≤ 2F(x, t), |t| > K ,

for some K ≥ 0, where a2 > λm and W1 L1( ). Then the equation

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

has at least two nontrivial solutions.

Theorem 16.2. Equation (16.6) will have at least two nontrivial solutions if we assume that for some integers l > m, the following inequalities hold:

(16.7)	t[ f (x, t1) − f (x, t0)] ≥ at2, t j	R, t = t1 − t0,
where a > λm ,
(16.8)	a0t2 ≤ 2F(x, t) ≤ a1t2,	|t| < δ,
for some δ > 0, with λl < a0 ≤ a1 < λl+1,
(16.9)	2F(x, t) ≤ a2t2 + W2(x),	|t| > K ,

for some K ≥ 0, where a2 < λm+1 and W2 L1( ).

More comprehensive theorems will be presented in the next section. Proofs will be given in Section 16.6.

16.3 Statement of the theorems

We use the notation
	a(u, v) = ( Au, v), a(u) = a(u, u), u, v D.
We define
(16.10)	u D := A1/2u
and
	G(u) := u 2D − 2	F(x, u)dx.

It is known that G is a continuously differentiable functional on the whole of D (cf.,

e.g., [120]) and

(G (u), v)D = 2(u, v)D − 2( f (u), v),

16.3. Statement of the theorems	193
where we write f (u) in place of	f (x, u(x)). In connection with the operator A, the

following quantities are very useful. For each fixed positive integer , we let N denote

the subspace of D spanned by the eigenfunctions corresponding to λ0, . . . , λ , and let M = N ∩ D. Then D = M N . For real a, b, we define

I (u, a, b) = ( Au, u) − a u− 2 − b u+ 2,

γ (a) = sup{I (v, a, 0) : v N , v+ = 1},(a) = inf{I (w, a, 0) : w M , w+ = 1},

F1 (w, a, b) = sup{I (v + w, a, b) : v N }, F2 (v, a, b) = inf{I (v + w, a, b) : w M },

M (a, b) = inf{F1 (w, a, b) : w M , w D = 1}, m (a, b) = sup{F2 (v, a, b) : v N , v D = 1},

ν (a) = sup{b : M (a, b) ≥ 0}, μ (a) = inf{b : m (a, b) ≤ 0},

where u±(x) = max{±u(x), 0}. Our first result is

Theorem 16.3. Assume that for some integers l < m, the following inequalities hold:

(16.11)	t[ f (x, t1) − f (x, t0)] ≤ a(t−)2 + b(t+)2, t j R, t = t1 − t0,
where b < m (a),
(16.12)	a0(t−)2 + b0(t+)2 ≤ 2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| < δ,

for some δ > 0, with a0, b0 < λl+1, a1, b1 > λl , b0 > μl (a0), and b1 < νl (a1),

(16.13) a2(t−)2 + b2(t+)2 − W1(x) ≤ 2F(x, t), |t| > K ,

for some K ≥ 0, where a2, b2 < λm+1, b2 > μm (a2), and W1 L1( ). Then (16.6) has at least two nontrivial solutions.

In contrast to this, we have

Theorem 16.4. Equation (16.6) will have at least two nontrivial solutions if we assume that for some integers l > m, the following inequalities hold:

(16.14)	t[ f (x, t1) − f (x, t0)] ≥ a(t−)2 + b(t+)2, t j R, t = t1 − t0,
where b > γm (a),
(16.15)	a0(t−)2 + b0(t+)2 ≤ 2F(x, t) ≤ a1(t−)2 + b1(t+)2, |t| < δ

for some δ > 0, with a0, b0 < λl+1, b0 > μl (a0) and a1, b1 > λl , b1 < νl (a1),

(16.16) 2F(x, t) ≤ a2(t−)2 + b2(t+)2 + W2(x), |t| > K ,

for some K ≥ 0, where a2, b2 > λm , b2 < νm (a2), and W2 L1( ).

194	16. Multiple Solutions

It was shown in [114] that the functions γl , μl , νl−1, l−1 all emanate from the point (λl , λl ) and satisfy

l−1(a) ≤ νl−1(a) ≤ μl (a) ≤ γl (a)

on their common domains. It would therefore give a weaker result if we assumed in Theorems 16.3 and 16.4 that b0 > γl (a0) and b1 < l (a1). However, the functions γl , l are defined on the whole of R, while the others are not. For cases in which the other functions are not defined, we state the following.

Theorem 16.5. Theorems 16.3 and 16.4 remain true if we assume that (16.12) holds with b0 > γl (a0), and b1 < l (a1) for some a0, a1 R.

16.4 Some lemmas

The proofs of the theorems of the previous section will be based on a series of lemmas.

Lemma 16.6. If b < l (a), then there is an > 0 such that

(16.17)

I (w, a, b) ≥ w 2D ,

w Ml .

Proof. By the continuity of l , there is a t < 1 such that b/t < l (a/t). Then

I (w, a/t, b/t)

w 2

a

w− 2

b

w+ 2

0,

w

M .

= D − t

− t

≥

l

Therefore,

I (w, a, b) = t

w 2D −

a

w− 2

−

b

w+ 2

+ (1 − t) w 2D ≥ (1 − t) w 2D .

t

t

Lemma 16.7. If b > γl (a), then there is an > 0 such that

(16.18)

I (v, a, b) ≤ − v 2D ,

v Nl .

Proof. By the continuity of γl , there is a t > 1 such that b/t > γl (a/t). Hence,

I (v, a/t, b/t) = v 2D −

a

v− 2 −

b

v+ 2 ≤ 0, v Nl ,

t

t

and

a

b

I (v, a, b) = t v 2D −

v− 2

−

v+ 2

+ (1 − t) v 2D ≤ (1 − t) v 2D .

t

t

16.4. Some lemmas		195
Lemma 16.8. If
(16.19)	t[ f (x, t1) − f (x, t0)] ≤ a(t−)2 + b(t+)2,	t j R, t = t1 − t0,
then
(16.20)	(G (v + w1) − G (v + w0), w) ≥ 2I (w, a, b),	v, w j D, w = w1 − w0.
Proof.	We have

( f (x, v + w1) − f (x, v + w0), w) ≤ a w− 2 + b w+ 2.

Hence,

(G (v + w1) − G (v+w0), w)/2

=w 2D − ( f (x, v + w1) − f (x, v + w0), w)

≥I (w, a, b).

Lemma 16.9. Suppose there are closed subspaces X,Y of E such that E = X Y, and

(16.21) (G

(x

+

y1)

−

G

(x

+

y2), y1

−

y2)

≥

m(

y1

−

y2

),

x

X, y1, y2

Y,

for some function m(t) satisfying

(a)

m(t) > 0, t > 0,

(b) there is a t0 > 0 such that

(c)

and

(d)

1

M

		t0 m(t)				dt <	∞,
		0			t
m(t)		→ 0			i mpli es		t	→ 0,
t
M m(t)
			dt → ∞ as					M → ∞.
0		t

Then there exists a continuous function ϕ : X → Y satisfying

(i) G(x + ϕ(x)) = miny Y G(x + y).

: X

R is of class C1 and satisfies

G

x

) =

G

x

+ ϕ(

x

))

→

(ii) The function ˜ (

(

(16.22)

(G

(x), h)

=

(G

(x

+

ϕ(x)), h), h

X.

˜

196 16. Multiple Solutions

Proof. For each x X, we define Gx : Y → R by Gx (y) = G(x + y). From (16.21),

we have

(Gx (y1) − Gx (y2), y1 − y2) ≥ m( y1 − y2 ).

In particular, Gx (y) is strictly convex (Theorem 13.7). Thus, for each x X, Gx (y) has a unique critical point ϕ(x). To see this, note that

(G

(y)

−

G

(0), y)

≥

m(

y ),

y

Y.

x

x

Thus,

(G

(y), y)

≥

(G

(0), y)

+

m(

y ).

Consequently,

x

x

1

d

Gx (y) − Gx (0) = 0

Gx (t y) dt

dt

=

1 1

(G

(t y), t y) dt

0

t

x

≥1 1 [(Gx (0), t y) + m( t y )] dt

	0		t
=	(G		(0), y)	+		1	1	m( t y ) dt
		x			0		t
=	(G		(0), y)	+		y			m(s)	ds
		x			0				s

→ ∞ as y → ∞.

Hence, Gx (y) has a unique minimum ϕ(x) (Lemma 13.5). Moreover,

Gx

(ϕ(

x

)) = y Y

x

(

y

) = y Y

(

x

+

y

).

min G

min G

Therefore, ϕ(x) is the only element of Y such that

(16.23)

0

=

(G

x

=

(G

(x

+

ϕ(x)), y),

y

Y.

(ϕ(x)), y)

Note that ϕ is continuous. Suppose, on the contrary, that {xn } is a sequence in X such that xn → x X. Let tn = ϕ(xn ) − ϕ(x) . Since G is continuous, for each ε > 0 and n sufficiently large, we have

(16.24)	|(G (xn + ϕ(x)), y)| < ε y ,	y Y.
Because of (16.21), we obtain
(16.25)	(G (xn + ϕ(x)) − G (xn + ϕ(xn )), ϕ(x) − ϕ(xn )) ≥ m(tn ).
Since, by (16.23),		y Y,
	(G (xn + ϕ(xn )), y) = 0,

16.4. Some lemmas

197

inequalities (16.24) and (16.25) imply

m(tn )/tn < ε.

Thus,

m(tn)/tn → 0,

and consequently, tn → 0. Hence, ϕ is continuous. This proves part (i).

For t > 0 and h X, we have

(16.26)

˜ (

x

+

th

) −

˜

(

x

)

G

(

x

+

th

+ ϕ(

x

+

th

)) −

(

x

+ ϕ(

x

))

G

G

=

G

t

t

≤

G(x + th + ϕ(x)) − G(x + ϕ(x))

t

1

=

0

(G (x

+

ϕ(x)

+

sth), h) ds.

In a similar manner, we see that

˜ (

x

+

th

) −

˜ (

x

)

G

(

x

+

th

+ ϕ(

x

+

th

)) −

(

x

+ ϕ(

x

))

G

G

=

G

t

t

≥

G(x + th + ϕ(x + th)) − G(x + ϕ(x + th))

t

1

(x

≥

0

(G

+

ϕ(x

+

th)

+

sth), h) ds.

Therefore, since G and ϕ are continuous, we have

G

x

+

th

) −

G

x

)

t

lim

˜

(

˜ (

=

(G (x

+

ϕ(x)), h).

0

t

→ +

This shows that

G has a continuous Gateaux derivative and hence is of class C1

.

˜

From the above, we see that

(16.27)

(G (x), h)

=

(G

(x

+

ϕ(x)), h),

h

X.

˜

Lemma 16.10. Let M, N be closed subspaces of

a Banach space E such that

dim N < ∞ and E = M N. Let w0 = 0 be an element of M, and take

K0 = {v N : v ≤ R} {sw0 + v : v N, s ≥ 0, sw0 + v = R},

B = ∂ Bδ ∩ M, 0 < δ < R.

Let

Q = {sw0 + v : v N, s ≥ 0, sw0 + v ≤ R},

198 16. Multiple Solutions

Then K0 = ∂ Q. Let ξ(u) be a continuous map from Q to E such that

ξ(v) = v, v N ∩ BR

and

ξ(u) ≥ r > δ > 0, u K0 ∩ ∂ BR .

Then

ξ(Q) ∩ B = φ.

Proof. Let P be the projection of E onto N via M. Then the conclusion of the lemma is true iff there is a u Q such that

S(u) := Pξ(u) + ξ(u) w0 = δw0.

Let

ξt (u) = (1 − t)ξ(u) + tu

and

St (u) = Pξt (u) + [(1 − t) ξ(u) + t u ]w0.

Then S0 = S and

S1(u) = Pu + u w0.

Clearly, there is a unique u Q such that S1(u) = δw0. The St map Q into E, but there is no point on ∂ Q that gets mapped into δw0. If v N ∩ BR , then St (v) = v + v w0, which cannot equal δw0. If u ∂ Q\N, then

[(1 − t) ξ(u) + t u ] ≥ r > δ.

Thus, no such point can be mapped into δw0. Hence, the Brouwer degree d(St , Q, δw0) is defined for t [0, 1], and d(S0, Q, δw0) = d(S1, Q, δw0) = 1. This completes the proof.

Corollary 16.11. A = ξ(∂ Q) links B [mm].

Proof. Let K be the collection of all sets

K = {η(Q) : η C(Q, E), η(u) = ξ(u), u ∂ Q}.

It is easily checked that K is a minimax system. If ϕ ( A), then ϕ(K ) K for K K. Since A ∩ B = φ and K ∩ B = φ for K K, we see that A links B [mm].

Lemma 16.12. If f (x, t) satisfies (16.19) and b < m (a), then there is a continuous map ϕ from Nm → Mm such that

(16.28)	J	(v) ≡	G	(v + ϕ(v)) = w Mm G(v + w) C		(Nm ,		), v Nm ,
				min	1		R
and
(16.29)				J (v) = G (v + ϕ(v)),	v Nm .

16.4. Some lemmas

199

Proof. In view of Lemmas 16.6 and 16.8, we have

(G (v + w1) − G (v + w0), w) ≥ w 2D , w Mm .

We can now apply Lemma 16.9 to arrive at the conclusion.

Lemma 16.13. If, in addition,

(16.30) 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

(16.31)	J (v) ≤ −ε v 2D , v Nl ∩ Br ,
where		Br = {u D : u D ≤ r }.
Proof. Let q be any number satisfying
(16.32)	2	< q	≤ 2n/(n − 2T ), 2T < n
(16.33)	2	< q	< ∞, n ≤ 2T.

It follows from Theorem 13.6 that there is a continuous map τ : Nl → Ml such that

(16.34)	τ (s v) = s τ (v),				s ≥ 0,
(16.35)	I (v + τ (v), a0, b0) = w Ml			I (v + w, a0, b0), v Nl ,
		inf
(16.36)	τ (v) D ≤ C v D ,				v Nl .
Then, for u = v + τ (v), we have, by (16.2),
J (v) ≤ G(u) ≤ I (u, a0, b0) +		[a0(u−)2 + b0(u+)2 − 2F(x, u)] dx
		|u|>δ
≤ F2l (v, a0, b0) + C		|u|q dx ≤ ml (a0, b0) v 2D + o( v 2D ) ≤ −ε v 2D
	|u|>δ
for r sufficiently small.
Lemma 16.14. Assume that
(16.37)	a(t−)2 + b(t+)2 − W1(x) ≤ 2F(x, t), |t| > K ,

for some K ≥ 0, where a, b < λm+1, b ≥ μm (a), l ≤ m, and W1 L1( ). Then there is a K1 < ∞ such that

(16.38)	J (v) ≤ K1.
If b > μm (a), then
(16.39)	J (v) −→ −∞ as v D −→ ∞.