Multiple solutions for a quasilinear
Schrödinger equation
Xiang-Dong Fang∗
School of Mathematical Sciences, Dalian University of Technology
116024 Dalian, PR China
E-mail: [email protected]
Andrzej Szulkin∗∗
Department of Mathematics, Stockholm University
106 91 Stockholm, Sweden
E-mail: [email protected]
Abstract
In this paper we consider the quasilinear Schrödinger equation
−∆u + V (x)u − ∆(u2 )u = g(x, u),
x ∈ RN ,
where g and V are periodic in x1 , . . . , xN and g is odd in u, subcritical and satisfies a monotonicity condition. We employ the approach
developed in [14, 15] and obtain infinitely many geometrically distinct
solutions.
Keywords: Quasilinear Schrödinger equation; Multiplicity of solutions; Nehari manifold.
∗
∗∗
Supported by the China Scholarship Council.
Supported in part by the Swedish Research Council.
1
1
Introduction
The quasilinear Schrödinger equation
−∆u + V (x)u − ∆(u2 )u = g(x, u),
x ∈ RN
(1.1)
has been studied recently by several authors, see [1]-[4], [6]-[10], [12], [16],
[18] and the references there. This is a special case of a more general equation which models certain phenomena in physics, see e.g. [6, 8, 10] for an
explanation. In most of the papers mentioned above the existence of positive
solutions has been considered, under different assumptions on V and g.
A major difficulty here is that the natural functional corresponding to
(1.1) is not well defined for all u ∈ H 1 (RN ) if N ≥ 2, see Section 2. Despite
this problem the existence of a positive solution has been proved in [8] by
using a constrained minimization argument. In [6], by a change of variables,
the quasilinear problem (1.1) was transformed to a semilinear one and an
Orlicz space framework was used. A slightly different change of variables
was used in [1], and a simpler and shorter proof of the results in [6], which
does not use Orlicz spaces, was given (see also [12]). In [4, 16] equation (1.1),
with ε2 in front of ∆u and ∆(u2 )u has been studied, with g respectively of
subcritical and critical growth. It was shown that there exists a positive
solution uε which concentrates at a local minimum of V as ε → 0. In a
recent paper [9] existence of multibump solutions was shown for a quasilinear Schrödinger equation which is more general than (1.1). Instead of
transforming the problem to a semilinear one (which does not seem to be
possible in this case), the authors have worked with the natural functional
which, after a truncation, is defined and continuous on H 1 (RN ). Since this
truncated functional is not of class C 1 , nonsmooth critical point theory was
used.
To the best of our knowledge, except for [9], there are no results about
the existence of infinitely many solutions for the equation (1.1) (however, see
[10] where infinitely many solutions were found in the ODE case). Motivated
by the arguments in [1, 12, 16], we also make a change of variables, and then
we use the method developed in [14, 15] in order to obtain such result. We
consider the problem
−∆u + V (x)u − ∆(u2 )u = g(x, u), u ∈ H 1 (RN ),
(1.2)
Ru
Setting G(x, u) := 0 g(x, s)ds, we suppose that V and g satisfy the
following assumptions:
2
(V ) V is continuous, 1-periodic in xi , 1 ≤ i ≤ N , and there exists a
constant a0 > 0 such that V (x) ≥ a0 for all x ∈ RN .
(g1 ) g is continuous, 1-periodic in xi , 1 ≤ i ≤ N , and |g(x, u)| ≤ a(1 +
|u|p−1 ) for some a > 0 and 4 < p < 2 · 2∗ , where 2∗ := 2N/(N − 2) if
N ≥ 3, 2∗ := ∞ if N = 1 or 2.
(g2 ) g(x, u) = o(u) uniformly in x as u → 0.
(g3 ) G(x, u)/u4 → ∞ uniformly in x as |u| → ∞.
(g4 ) u 7→ g(x, u)/u3 is positive for u 6= 0, nonincreasing on (−∞, 0) and
nondecreasing on (0, ∞).
Note that (g2 ), (g4 ) imply g(x, u) = O(u3 ) as u → 0. Note also that
Let ∗ denote the action of ZN on H 1 (RN ) given by
(k ∗ u)(x) := u(x − k),
k ∈ ZN .
(1.3)
It follows from (V ) and (g1 ) that if u0 is a solution of (1.2), then so is k ∗ u0
for all k ∈ ZN . Set
O(u0 ) := {k ∗ u0 : k ∈ ZN }.
O(u0 ) is called the orbit of u0 with respect to the action of ZN , and it is
called a critical orbit for a functional F if u0 is a critical point of F and F is
ZN - invariant, i.e., F (k ∗ u) = F (u) for all k ∈ ZN and all u (then of course
all points of O(u0 ) are critical). Two solutions u1 , u2 of (1.2) are said to be
geometrically distinct if O(u1 ) 6= O(u2 ).
The first main result of this paper, which we prove in Section 3, is the
following
Theorem 1.1. Suppose that (V ), (g1 )-(g4 ) are satisfied and g is odd in u.
Then (1.2) admits infinitely many pairs ±u of geometrically distinct solutions.
In Section 4 we let g(x, u) = q(x)u3 , that is, we consider the problem
−∆u + V (x)u − ∆(u2 )u = q(x)u3 ,
u ∈ H 1 (RN ),
(1.4)
and we assume that
(Q) q is continuous, 1-periodic in xi , 1 ≤ i ≤ N , and there exists a constant
b0 > 0 such that q(x) ≥ b0 for all x ∈ RN .
Our second main result is
3
Theorem 1.2. Suppose that (V ) and (Q) are satisfied. Then (1.4) admits
infinitely many pairs ±u of geometrically distinct solutions.
Remark 1.1. In [9] it was assumed that g and ∂g/∂u are continuous, g
satisfies (g1 ), (g2 ) and the Ambrosetti-Rabinowitz type condition ug(x, u) ≥
pG(x, u) > 0 for some p > 4 and all u 6= 0. Here our (g3 ) is weaker and
g need not be differentiable, but instead we need a monotonicity condition
(g4 ) which does not follow from the hypotheses in [9]. A simple example
of a function satisfying our assumptions but not conditions in [9] is g(u) =
u3 ln(1 + |u|). Another example is the function g in our Theorem 1.2. Also,
our proofs are different and much simpler. On the other hand, g in [9] need
not be odd and the equation is more general than (1.1) (and our solutions
are not of multibump type).
We would also like to point out that in one of the main results of a
recent paper [18] the existence of a positive solution has been shown under
assumptions which are somewhat stronger than in our Theorem 1.1. It will
be easily seen from our arguments that we may also obtain such a solution
under the assumptions of Theorem 1.1 as well as under the assumptions of
Theorem 1.2. More precisely, the functional I introduced in the next section
attains a minimum on the Nehari manifold M (cf. [14, 15]) and if u is a
minimizer, then so is |u|. Hence the conclusion.
As we have mentioned, we do not assume the Ambrosetti-Rabinowitz
condition. Different ways of weakening or replacing this condition have been
studied extensively in recent years, starting with [5].
Notation. C, C1 , C2 , . . . will denote different positive constants whose exact
value is inessential. |A| is the Lebesgue measure of a measurable set A ⊂ RN .
Bρ (y) := {x ∈ RN : |x − y| < ρ}. The usual norm in the Lebesgue space
Lp (Ω) is denoted by kukp,Ω , and by kukp if Ω = RN . E denotes the Sobolev
space H 1 (RN ) and S is the unit sphere in E. It follows from (V ) that
Z
kuk :=
1/2
(|∇u| + V (x)u )
2
2
RN
is an equivalent norm in E. It is more convenient for our purposes than the
standard one and will be used henceforth. For a functional F we put
F d := {u : F (u) ≤ d},
Fc := {u : F (u) ≥ c},
Fcd := {u : c ≤ F (u) ≤ d}.
Acknowledgements. The second author would like to thank Louis Jeanjean for helpful comments on an earlier version of this paper and for pointing
out the reference [4].
4
2
Preliminary results
In this section we introduce a variational framework associated with Problem (1.2). We observe that (1.2) is formally the Euler-Lagrange equation
associated with the energy functional
Z
Z
Z
1
1
2
2
2
G(x, u).
(2.1)
J(u) :=
(1 + 2u )|∇u| +
V (x)u −
2 RN
2 RN
RN
It is difficult to apply variational methods in order to study the functional
J because, unless N = 1, J is not defined for all u in the space E ≡ H 1 (RN )
which is natural for the problem. To overcome this difficulty, we employ
an argument developed in [1] (see also [12]). As we have mentioned in
the introduction, a somewhat earlier and slightly different variant of this
argument may be found in [6]. We make a change of variables v := f −1 (u),
where f is defined by
f 0 (t) =
1
on [0, +∞)
(1 + 2f 2 (t))1/2
and f (t) = −f (−t) on (−∞, 0].
Below we summarize the properties of f . Proofs may be found in [1, 2, 3, 16].
Lemma 2.1. The function f satisfies the following properties:
(1) f is uniquely defined, C ∞ and invertible;
(2) |f 0 (t)| ≤ 1 for all t ∈ R;
(3) |f (t)| ≤ |t| for all t ∈ R;
(4) f (t)/t → 1 as t → 0;
√
(5) f (t)/ t → 21/4 as t → +∞;
(6) f (t)/2 ≤ tf 0 (t) ≤ f (t) for all t > 0;
(7) |f (t)| ≤ 21/4 |t|1/2 for all t ∈ R;
(8) f 2 (t) − f (t)f 0 (t)t ≥ 0 for all t ∈ R;
(9) there exists a positive constant C such that |f (t)| ≥ C|t| for |t| ≤ 1
and |f (t)| ≥ C|t|1/2 for |t| ≥ 1;
√
(10) |f (t)f 0 (t)| < 1/ 2 for all t ∈ R;
(11) the function f (t)f 0 (t)t−1 is strictly decreasing for t > 0;
5
(12) the function f p (t)f 0 (t)t−1 is strictly increasing for p ≥ 3 and t > 0.
In [3, 16] it is stated that the functions in (11) and (12) are respectively
decreasing and increasing but it is easy to see from the proofs there that
they are strictly decreasing and strictly increasing.
Consider the functional
Z
Z
Z
1
1
2
2
I(v) :=
|∇v| +
V (x)f (v) −
G(x, f (v)).
(2.2)
2 RN
2 RN
RN
Then I is well defined on E and I ∈ C 1 (E, R) under the hypotheses (V ),
(g1 ) and (g2 ). Note also that (V ), (g1 ) imply I is invariant with respect to
the action of ZN given by (1.3). It is easy to see that
Z
Z
0
hI (v), wi =
∇v∇w +
V (x)f (v)f 0 (v)w
N
N
R
ZR
−
g(x, f (v))f 0 (v)w
(2.3)
RN
for all v, w ∈ E and the critical points of I are weak solutions of the problem
−∆v + V (x)f (v)f 0 (v) = g(x, f (v))f 0 (v),
v ∈ E.
It has been shown in [1] that if v ∈ E is a critical point of the functional I,
then u = f (v) ∈ E and u is a solution of (1.2).
Let
M := {v ∈ E \ {0} : hI 0 (v), vi = 0}.
(2.4)
Recall that M is called the Nehari manifold. We do not know whether M
is of class C 1 under our assumptions and therefore we cannot use minimax
theory directly on M. To overcome this difficulty, we employ an argument
developed in [14, 15].
3
Proof of Theorem 1.1
We assume that (V ) and (g1 )-(g4 ) are satisfied from now on. First, (g1 ) and
(g2 ) imply that for each ε > 0 there is Cε > 0 such that
|g(x, u)| ≤ ε|u| + Cε |u|p−1
for all u ∈ R.
Lemma 3.1. G(x, u) ≥ 0 and 41 g(x, u)u ≥ G(x, u).
6
(3.1)
This follows immediately from (g2 ) and (g4 ).
For t > 0, let
Z
Z
Z
t2
1
2
2
h(t) := I(tu) =
|∇u| +
V (x)f (tu) −
G(x, f (tu)).
2 RN
2 RN
RN
Lemma 3.2. For each u 6= 0 there is a unique tu > 0 such that h0 (t) > 0
for 0 < t < tu and h0 (t) < 0 for t > tu . Moreover, tu ∈ M if and only if
t = tu .
Proof. By (3.1) and Lemma 2.1-(7), for ε sufficiently small we obtain
Z
Z
Z
Z
t2
1
ε
Cε
h(t) ≥
|∇u|2 +
V (x)f 2 (tu) −
f 2 (tu) −
|f (tu)|p
2 RN
2 RN
2 RN
p RN
Z
Z
t2
2
p/2
|u|p/2 .
≥
|∇u| − C1 t
2 RN
RN
Since p > 4 and u is not constant, h(t) > 0 whenever t > 0 is small enough.
Using Lemma 2.1-(3), we have
Z
Z
Z
t2
t2
2
2
h(t) ≤
|∇u| +
V (x)u −
G(x, f (tu))
2 RN
2 RN
RN
Z
Z
Z
t2
G(x, f (tu)) f 4 (tu) 2
t2
≤
|∇u|2 +
V (x)u2 − t2
·
·u .
2 RN
2 RN
f 4 (tu)
(tu)2
u6=0
It follows from (g3 ), Lemma 2.1-(5) and Fatou’s lemma that the last integral
on the right-hand side above tends to infinity with t. Hence h(t) → −∞ as
t → ∞ and h has a positive maximum.
The condition h0 (t) = 0 is equivalent to
Z
Z
g(x, f (tu))f 0 (tu) V (x)f 0 (tu)f (tu) 2
|∇u|2 =
−
u .
tu
tu
RN
u6=0
Let
g(x, f (s))f 0 (s) V (x)f 0 (s)f (s)
−
.
s
s
By (g4 ) and Lemma 2.1-(12) the function
Z(s) :=
g(x, f (s)) f 0 (s)f 3 (s)
g(x, f (s))f 0 (s)
=
·
s
f 3 (s)
s
is strictly increasing for s > 0. Hence also s 7→ Z(s) is strictly increasing
according to Lemma 2.1-(11). So there exists a unique tu > 0 such that
h0 (tu ) = 0 and the first conclusion follows. The second conclusion is an
immediate consequence of the fact that h0 (t) = t−1 hI 0 (tu), tui.
7
Lemma 3.3. (1) There exists ρ > 0 such that c := inf M I ≥ inf Sρ I > 0,
where Sρ := {u ∈ E : kuk = ρ}.
(2) kuk2 ≥ 2c for all u ∈ M.
Proof. (1) We claim that there are C1 , ρ > 0 such that
Z
(|∇u|2 + V (x)f 2 (u)) ≥ C1 kuk2 whenever kuk ≤ ρ.
(3.2)
RN
If the claim
is false, then for all n there exists un 6= 0 such that un → 0 in
R
E and RN (|∇un |2 + V (x)f 2 (un )) ≤ n1 kun k2 . Let vn := kuunn k . Then
2
Z
Z
f (un )
1
2
2
(|∇vn | + V (x)vn ) +
V (x)
− 1 vn2 ≤ .
2
un
n
RN
RN
Going if necessary to a subsequence,
un → 0 a.e. Since
un → 0 in L2 (RN ),
N
for every ε > 0 the measure {x ∈ R : |un (x)| > ε} → 0 as n → ∞. Hence
by the Hölder inequality,
Z
(r−2)/r
kvn k2r → 0,
(3.3)
vn2 ≤ {x ∈ RN : |un (x)| > ε}
|un |>ε
where r = 2∗ if N ≥ 3 and r > 2 if N = 1 or 2. Now it follows from Lemma
2.1-(4) that the second integral above goes to 0. So kvn k = 1 and vn → 0 in
E, a contradiction.
Using (3.1), Lemma 2.1-(3),(7) and the Sobolev inequality, it is easy to
see that
Z
Z
Z
ε
Cε
2
G(x, f (u)) ≤
f (u) +
|f (u)|p ≤ C2 εkuk2 + C3 Cε kukp/2 ,
2 RN
p RN
RN
so choosing ε small enough and using the claim we have, for sufficiently
small ρ, I(u) ≥ C4 kuk2 −C5 kukp/2 and inf Sρ I > 0. The inequality inf M I ≥
inf Sρ I is a consequence of Lemma 3.2 since for every u ∈ M there is s > 0
such that su ∈ Sρ (and I(tu u) ≥ I(su)).
(2) For u ∈ M, by Lemma 2.1-(3) we have
Z
Z
Z
1
1
2
2
c≤
|∇u| +
V (x)f (u) −
G(x, f (u))
2 RN
2 RN
RN
Z
Z
1
1
≤
|∇u|2 +
V (x)f 2 (u)
2 RN
2 RN
Z
Z
1
1
1
2
≤
|∇u| +
V (x)u2 = kuk2 .
2 RN
2 RN
2
8
Lemma 3.4. I is coercive on M, i.e., I(u) → ∞ as kuk → ∞, u ∈ M.
Proof. We employ a similar argument as in [14], Proposition 2.7. Arguing
by contradiction, let (un ) ⊂ M be a sequence such that kun k → ∞ and
I(un ) ≤ d for some d. Let vn := un /kun k. Then vn * v in E and vn (x) →
v(x) a.e. in RN up to a subsequence. Choose yn ∈ RN so that
Z
Z
2
vn = max
vn2 .
(3.4)
B1 (yn )
y∈RN
B1 (y)
Since I and M are invariant with respect to the action of ZN given by (1.3),
we may assume that (yn ) is bounded in RN . If
Z
vn2 → 0 as n → ∞,
(3.5)
B1 (yn )
then it follows that vn → 0 in Lr (RN ) for 2 < r < 2∗ by P.L. Lions’ lemma
(cf.
[17], Lemma 1.21). Using (3.1) and Lemma 2.1-(3),(7) we see that
R
G(x,
f (svn )) → 0 for all s ∈ R. We have by Lemma 2.1-(9)
RN
d ≥ I(un ) ≥ I(svn )
Z
Z
Z
s2
1
2
2
=
|∇vn | +
V (x)f (svn ) −
G(x, f (svn ))
2 RN
2 RN
RN
Z
Z
Z
s2
C 2 s2
≥
|∇vn |2 +
V (x)vn2 −
G(x, f (svn ))
2 RN
2
|svn |≤1
RN
Z
Z
C 2 s2
s2
=
|∇vn |2 +
V (x)vn2
2 RN
2
N
R
Z
Z
C2
2
−
V (x)(svn ) −
G(x, f (svn ))
2 |svn |≥1
RN
Z
Z
s2
s2
2
p/2
≥
min{1, C } − C1
(svn ) −
G(x, f (svn )) →
min{1, C 2 }.
2
2
N
N
R
R
Taking sufficiently large s we obtain a contradiction. Hence (3.5) cannot
hold and since vn → v in L2loc (RN ), v 6= 0. Since |un (x)| → ∞ if v(x) 6= 0, it
follows from Lemma 2.1-(5), (g3 ) and Fatou’s lemma that
Z
Z
G(x, f (un ))
G(x, f (un )) f 4 (un ) 2
=
·
· vn → ∞.
kun k2
f 4 (un )
u2n
RN
RN
Thus
Z
I(un )
1
G(x, f (un ))
0≤
≤ −
→ −∞,
2
kun k
2
kun k2
RN
a contradiction. This completes the proof.
9
Lemma 3.5. If V is a compact subset of E \ {0}, then there exists R > 0
such that I ≤ 0 on (R+ V) \ BR (0).
Proof. We may assume without loss of generality that V ⊂ S. Arguing
by contradiction, suppose there exist un ∈ V and wn = tn un such that
I(wn ) ≥ 0 and tn → ∞ as n → ∞. Passing to a subsequence, we may
assume that un → u ∈ S. Since |wn (x)| → ∞ if u(x) 6= 0, it follows from
(g3 ), Lemma 2.1-(5) and Fatou’s lemma that
Z
Z
G(x, f (wn ))
G(x, f (wn ))u2n
=
t2n
wn2
RN
RN
Z
G(x, f (wn )) f 4 (wn ) 2
=
·
· un → ∞.
f 4 (wn )
wn2
RN
By Lemma 2.1-(3),
I(wn )
1
0≤
≤ −
2
tn
2
Z
RN
G(x, f (wn ))
→ −∞,
t2n
a contradiction.
Recall that S is the unit sphere in E and define the mapping m : S → M
by setting
m(w) := tw w,
where tw is as in Lemma 3.2. Note that km(w)k = tw . Lemmas 3.6 and 3.7
below are taken from [15] (see Proposition 8 and Corollary 10 there). That
the hypotheses in [15] are satisfied is a consequence of Lemmas 3.2, 3.3 and
3.5 above. Indeed, if h(t) = I(tw) and w ∈ S, then h0 (t) > 0 for 0 < t < tw
and h0 (t) < 0 for t > tw by Lemma 3.2, tw ≥ δ > 0 by Lemma 3.3 and
tw ≤ R for w ∈ V ⊂ S by Lemma 3.5.
Lemma 3.6. The mapping m is a homeomorphism between S and M, and
u
the inverse of m is given by m−1 (u) = kuk
.
We shall consider the functional Ψ : S → R given by
Ψ(w) := I(m(w)).
Lemma 3.7. (1) Ψ ∈ C 1 (S, R) and
hΨ0 (w), zi = km(w)khI 0 (m(w)), zi
for all z ∈ Tw (S).
(2) If (wn ) is a Palais-Smale sequence for Ψ, then (m(wn )) is a Palais-Smale
sequence for I. If (un ) ⊂ M is a bounded Palais-Smale sequence for I, then
10
(m−1 (un )) is a Palais-Smale sequence for Ψ.
(3) w is a critical point of Ψ if and only if m(w) is a nontrivial critical
point of I. Moreover, the corresponding values of Ψ and I coincide and
inf S Ψ = inf M I.
(4) If I is even, then so is Ψ.
From now on we assume that the nonlinearity g = g(x, u) is odd in u.
As a consequence of Lemma 3.7 we see as in Remark 2.12 of [14] that since
m, m−1 are equivariant and I, Ψ are invariant with respect to the action of
ZN given by (1.3), there is a one-to-one correspondence between the critical
orbits of I|M and Ψ. Hence the proof of Theorem 1.1 will be completed
once we show that Ψ has infinitely many critical orbits.
Lemma 3.8. The mapping m−1 defined in Lemma 3.6 is Lipschitz continuous.
Proof. We use an argument from [14]. By Lemma 3.3-(2) we have, for all
u, v ∈ M,
u
u − v (kvk − kuk)v v −1
−1
km (u) − m (v)k = =
−
+
kuk kvk kuk
kukkvk r
2
2
≤
ku − vk ≤
ku − vk.
kuk
c
Below we shall use the notation
K := {w ∈ S : Ψ0 (w) = 0},
Kd := {w ∈ K : Ψ(w) = d}.
Choose a subset F of K such that F = −F and each orbit O(w) ⊂ K has a
unique representative in F. According to the remark preceding Lemma 3.8
we must show that the set F is infinite. Arguing indirectly, assume
F is a finite set.
(3.6)
In Lemma 3.13 below we shall show that Palais-Smale sequences have a certain discreteness property. First we need some preparations. The following
result has been proved in Lemma 2.13 of [14]:
Lemma 3.9. κ := inf{kv − wk : v, w ∈ K, v 6= w} > 0.
Lemma 3.10. For every C > 0 there exists δ > 0 such that
δ > 0 as |t| ≤ C.
11
d
0
dt (f (t)f (t))
≥
Proof. Since
d 0
f 0 (t)
1
f (t)
d
=
=
(f (t)f (t)) =
1/2
3/2
2
2
dt
dt (1 + 2f (t))
(1 + 2f 2 (t))2
(1 + 2f (t))
and f is bounded for |t| ≤ C, the conclusion follows.
Lemma 3.11. If {u1n }, {u2n } are bounded in E, then there exists C > 0,
depending only on the bound on ku1n k and ku2n k, such that
Z
Z
|∇(u1n − u2n )|2 +
V (x)[f 0 (u1n )f (u1n ) − f 0 (u2n )f (u2n )](u1n − u2n )
N
N
R
R
Z
≥C
|∇(u1n − u2n )|2 + V (x)(u1n − u2n )2 ≡ Cku1n − u2n k2 .
RN
Proof. We may assume u1n 6= u2n (otherwise the conclusion is trivial). Set
wn :=
u1n − u2n
ku1n − u2n k
and gn (x) :=
f 0 (u1n )f (u1n ) − f 0 (u2n )f (u2n )
.
u1n − u2n
We argue by contradiction and assume that u1n , u2n may be found so that
Z
Z
2
|∇wn | +
V (x)gn (x)wn2 → 0.
(3.7)
RN
RN
Since f 0 (t)f (t) is strictly increasing by Lemma 3.10, gn (x) is positive if
wn (x) 6= 0. Hence
Z
Z
|∇wn |2 → 0 and
V (x)wn2 → 1.
(3.8)
RN
RN
For a given C1 > 0, let An := {x ∈ RN : |u1n (x)| ≥ C1 or |u2n (x)| ≥ C1 },
Bn := RN \ An . Then for each ε > 0, C1 may be chosen so that |An | ≤ ε. It
follows from Lemma 3.10, (3.7) and the Mean Value Theorem that
Z
Z
δ
V (x)wn2 ≤
V (x)gn (x)wn2 → 0.
(3.9)
Bn
Bn
Choosing ε small enough and arguing as in (3.3) (with the same r), we
obtain
Z
1
V (x)wn2 ≤ C2 ε(r−2)/r ≤
2
An
because C2 is independent of ε. This and (3.9) contradict (3.8).
12
Lemma 3.12. If {u1n }, {u2n } are bounded in E, then for each ε > 0 there
exists Cε > 0, depending only on the bound on ku1n k and ku2n k, such that
Z
1
2 1
2
1
2
h
(x)(u
−
u
)
n
n
n
≤ εkun − un k + Cε kun − un kp/2 ,
RN
where
hn (x) := g(x, f (u1n ))f 0 (u1n ) − g(x, f (u2n ))f 0 (u2n ) = h1n (x) − h2n (x).
Proof. By (3.1), Lemma 2.1-(6),(3),(7), the Hölder and the Sobolev inequalities, we have
Z
Z
(ε|f (u1n )| + Cε |f (u1n )|p−1 )f 0 (u1n )u1n
1
1
2
· |u1n − u2n |
N hn (x)(un − un ) ≤ N
1
u
R
R
n
Z 2 1
f (un ) 1
|f (u1n )|p 1
2
2
≤
ε
|un − un | + Cε
|un − un |
|u1n |
|u1n |
RN
Z ≤
ε|u1n ||u1n − u2n | + 2p/4 Cε |u1n |(p−2)/2 |u1n − u2n |
RN
(p−2)/2
≤ C1 εku1n − u2n k + 2p/4 Cε ku1n kp/2
ku1n − u2n kp/2
≤ C1 εku1n − u2n k + C2 Cε ku1n − u2n kp/2 .
Here it should be understood that the integrands in the first two lines are
zero if u1n (x) = 0. Similarly we have
Z
2
1
2 hn (x)(un − un ) ≤ C1 εku1n − u2n k + C2 Cε ku1n − u2n kp/2 .
RN
Since C1 , C2 depend only on the bound on ku1n k and ku2n k but are independent of ε and the particular choice of u1n , u2n , we may replace C1 ε by ε/2 and
C2 Cε by Cε /2. The conclusion follows.
Lemma 3.13. Let d ≥ c. If (vn1 ), (vn2 ) ⊂ Ψd are two Palais-Smale sequences
for Ψ, then either kvn1 − vn2 k → 0 as n → ∞ or lim supn→∞ kvn1 − vn2 k ≥
ρ(d) > 0, where ρ(d) depends on d but not on the particular choice of PalaisSmale sequences.
Proof. Our argument is an adaptation of the proof of Lemma 2.14 in [14].
Let u1n := m(vn1 ) and u2n := m(vn2 ). Then (u1n ), (u2n ) are Palais-Smale sequences for I and since (u1n ), (u2n ) ⊂ I d , these sequences are bounded. We
consider two cases.
13
Case 1: ku1n − u2n kp/2 → 0 as n → ∞. It follows from Lemmas 3.11 and 3.12
that for each ε > 0 and n large enough,
Z
Z
1
2 2
1
2 2
V (x)gn (x)(u1n − u2n )2
|∇(un − un )| +
Ckun − un k ≤
N
N
R
R
Z
0 1
1
2
0 2
1
2
hn (x)(u1n − u2n )
= hI (un ), un − un i − hI (un ), un − un i +
RN
≤
2εku1n
−
u2n k
+
Cε ku1n
−
u2n kp/2 ,
where gn , hn are as in Lemmas 3.11 and 3.12, and C does not depend on the
choice of ε. Therefore ku1n − u2n k → 0 and Lemma 3.8 implies kvn1 − vn2 k =
km−1 (u1n ) − m−1 (u2n )k → 0.
Case 2: ku1n − u2n kp/2 does not go to 0 as n → ∞. It follows from Lemma
1.21 in [17] that there exist ε > 0 and yn ∈ RN such that
Z
Z
1
2 2
(un − un ) = max
(u1n − u2n )2 ≥ ε for all n
(3.10)
y∈RN
B1 (yn )
B1 (y)
after passing to a subsequence. Since m, m−1 and I 0 , Ψ0 are equivariant
with respect to the action of ZN given by (1.3), we may assume that the
sequence (yn ) is bounded in RN . Passing to a subsequence once more, there
exist u1 , u2 and α1 , α2 such that
u1n * u1 ,
u2n * u2 ,
ku1n k → α1 ,
ku2n k → α2
and I 0 (u1 ) = I 0 (u2 ) = 0. According to (3.10), u1 6= u2 and by Lemma
3.3-(2),
√
2c ≤ αi ≤ ν(d) < ∞, where ν(d) := sup{kuk : u ∈ I d ∩ M} and i = 1, 2
(that ν(d) < ∞ is a consequence of Lemma 3.4).
Suppose u1 , u2 6= 0. Then u1 , u2 ∈ M and v 1 := m−1 (u1 ) ∈ K, v 2 :=
m−1 (u2 ) ∈ K, v 1 6= v 2 . Hence
1
1
un
u
u2n u2 1
2
lim inf kvn − vn k = lim inf 1 − 2 ≥ 1 − 2 = kβ1 v 1 − β2 v 2 k,
n→∞
n→∞
kun k kun k
α
α where
√
ku1 k
2c
β1 := 1 ≥
α
ν(d)
√
ku2 k
2c
and β2 := 2 ≥
.
α
ν(d)
Since kv 1 k = kv 2 k = 1, it is easy to see from the inequalities above that
√
κ 2c
1
2
1
2
1
2
lim inf kvn − vn k ≥ kβ1 v − β2 v k ≥ min{β1 , β2 }kv − v k ≥
(3.11)
n→∞
ν(d)
14
(κ is the constant in Lemma 3.9). Hence (3.11) implies lim inf n→∞ kvn1 −
vn2 k ≥ ρ(d) > 0, where ρ(d) depends only on d (via ν(d)).
If u2 = 0, then u1 6= 0 and
√
1
un
u2n ku1 k
2c
1
2
lim inf kvn − vn k = lim inf 1 − 2 ≥ 1 ≥
.
n→∞
n→∞
kun k kun k
α
ν(d)
The case u1 = 0 is similar.
It is well known that Ψ admits a pseudo-gradient vector field H : S \K →
T S (see e.g. [11], Remark A.17-(iv) or [13], p. 86). Moreover, since Ψ is
even, we may assume H is odd. Let η : G → S \ K be the flow defined by
(
d
dt η(t, w) = −H(η(t, w)),
(3.12)
η(0, w) = w,
where
G := {(t, w) : w ∈ S \ K, T − (w) < t < T + (w)}
and (T − (w), T + (w)) is the maximal existence time for the trajectory t 7→
η(t, w). Note that η is odd in w because H is and t 7→ Ψ(η(t, w)) is strictly
decreasing by the properties of a pseudogradient.
Let P ⊂ S, δ > 0 and define
Uδ (P ) := {w ∈ S : dist(w, P ) < δ}.
Below we summarize the properties of Ψ and η which will be needed in the
proof of Theorem 1.1.
Lemma 3.14. Let d ≥ c. Then for every δ > 0 there exists ε = ε(δ) > 0
such that
(a) Ψd+ε
d−ε ∩ K = Kd and
(b) limt→T + (w) Ψ(η(t, w)) < d − ε for w ∈ Ψd+ε \ Uδ (Kd ).
Part (a) is an immediate consequence of (3.6) and (b) has been proved
in [14], see Lemmas 2.15 and 2.16 there. The argument is exactly the same
except that S + should be replaced by S. We point out that an important
role in the proof of Lemma 3.14 is played by the discreteness property of the
Palais-Smale sequences expressed in Lemma 3.13.
Proof of Theorem 1.1. Our argument is borrowed from [14] where it
appears in the proof of Theorem 1.2. Let
Σ := {A ⊂ S : A = A, A = −A}.
15
Recall [11, 13] that for A ⊂ Σ, the Krasnoselskii genus γ(A) is by definition
the smallest integer k for which there exists an odd mapping A → Rk \{0}. If
there is no such mapping for any k, then γ(A) := +∞. Moreover, γ(∅) := 0.
Define
ck := inf{d ∈ R : γ(Ψd ) ≥ k}, k ≥ 1.
Thus the ck ’s are those numbers at which the sets Ψd change genus and it
is easy to see that ck ≤ ck+1 . Let k ≥ 1 and set d := ck . By Lemma 3.9, Kd
is either empty or a discrete set, hence γ(Kd ) = 0 or 1. By the continuity
property of the genus there exists δ > 0 such that γ(U ) = γ(Kd ), where
U := Uδ (Kd ) and δ < κ2 . For such δ, choose ε > 0 so that the conclusions of
Lemma 3.14 hold. Then for each w ∈ Ψd+ε \ U there exists t ∈ [0, T + (w))
such that Ψ(η(t, w)) < d − ε. Let e = e(w) be the infimum of the time for
which Ψ(η(t, w)) ≤ d − ε. Since d − ε is not a critical value of Ψ, it is easy
to see by the implicit function theorem that e is a continuous mapping and
since Ψ is even, e(−w) = e(w). Define a mapping h : Ψd+ε \ U → Ψd−ε
by setting h(w) := η(e(w), w). Then h is odd and continuous, so it follows
from the properties of the genus and the definition of ck that
γ(Ψd+ε ) ≤ γ(U ) + γ(Ψd−ε ) ≤ γ(U ) + k − 1 = γ(Kd ) + k − 1.
If γ(Kd ) = 0, then γ(Ψd+ε ) ≤ k − 1, contrary to the definition of ck . So
γ(Kd ) = 1 and Kd 6= ∅. If ck+1 = ck = d, then γ(Kd ) > 1 (see e.g. [11],
Proposition 8.5 or [13], Lemma II.5.6). Since this is impossible, we must
have ck+1 > ck and Kck 6= ∅ for all k ≥ 1, a contradiction to (3.6). The
proof is complete.
4
Proof of Theorem 1.2
In this section we look for critical points of the functional I : E → R given
by
Z
Z
Z
1
1
1
2
2
I(u) :=
|∇u| +
V (x)f (u) −
q(x)f 4 (u).
2 RN
2 RN
4 RN
Note that it follows from Lemma 2.1-(3),(7) that
|f (t)| ≤ 2|t|s
Let
E :=
1
≤ s ≤ 1.
2
Z
|∇u|2 <
q(x)u2 .
for all t ∈ R,
Z
u∈E:
RN
RN
Since q ≥ b0 > 0, it is easy to see that E 6= ∅.
Recall that h(t) := I(tu), t > 0.
16
(4.1)
Lemma 4.1. (1) For each u ∈ E there is a unique tu > 0 such that h0 (t) > 0
for 0 < t < tu and h0 (t) < 0 for t > tu . Moreover, tu ∈ M if and only if
t = tu .
(2) If u 6∈ E, then tu 6∈ M for any t > 0.
Proof. (1) For each u ∈ E, we have, by the Lebesgue dominated convergence
theorem and Lemma 2.1-(5),(4)
Z
Z
I(tu)
f 2 (tu) q(x) f 4 (tu)
1
2
lim
V
(x)
|∇u|
+
u2
=
lim
−
2
2
t→∞ t2
t→∞ 2
(tu)
2
(tu)
N
u6=0
R
Z
Z
1
2
2
=
|∇u| −
q(x)u < 0
2 RN
RN
and
Z
Z
I(tu)
1
f 2 (tu) q(x) f 4 (tu)
2
lim
= lim
|∇u| +
V (x)
−
u2
2
2
t→0 t2
t→0 2
(tu)
2
(tu)
N
R
u6=0
Z
Z
1
2
2
=
|∇u| +
V (x)u > 0.
(4.2)
2 RN
RN
Hence h has a positive maximum. The condition h0 (t) = 0 is equivalent to
Z
Z
q(x)f 3 (tu)f 0 (tu) V (x)f (tu)f 0 (tu) 2
2
−
u .
|∇u| =
tu
tu
RN
u6=0
Let
q(x)f 3 (s)f 0 (s) V (x)f (s)f 0 (s)
−
.
s
s
As in the proof of Lemma 3.2 we see that s 7→ Z(s) is strictly increasing, hence the first conclusion. The second conclusion follows because as
previously we have h0 (t) = t−1 hI 0 (tu), tui.
(2) If tu ∈ M for some t > 0, then hI 0 (tu), ui = 0 and therefore
Z
Z
f 3 (tu)f 0 (tu)
f (tu)f 0 (tu) 2
2
|∇u| =
q(x)
− V (x)
u
tu
tu
RN
u6=0
Z
Z
f 3 (tu)f 0 (tu) 2
q(x)
<
u ≤
q(x)u2 ,
tu
u6=0
RN
Z(s) :=
where the first inequality follows from the fact that f (t)f 0 (t)/t is positive
for t 6= 0 and the second one from Lemma 2.1-(7),(10). Hence u ∈ E.
17
Lemma 4.2. (1) There exists ρ > 0 such that c := inf M I ≥ inf Sρ I > 0.
(2) kuk2 ≥ 2c for all u ∈ M.
Proof. (1) Using (4.1) with r = 4s ∈ (2, min{2∗ , 4}) and the Sobolev inequality we obtain
Z
Z
1
4
q(x)|u|r ≤ C2 kukr .
q(x)f (u) ≤ 4
4 RN
N
R
Hence it follows from (3.2) that inf Sρ I > 0 if ρ is small enough. That
inf M I ≥ inf Sρ I is seen as in Lemma 3.3.
(2) The argument is the same as in Lemma 3.3-(2).
We do not know whether I is coercive on M. However, we prove
Lemma 4.3. All Palais-Smale sequences (un ) ⊂ M are bounded.
Proof. Suppose kun k → ∞ and set vn := un /kun k. As in the proof of Lemma
3.4 we may assume that (3.4) holds with (yn ) bounded and we see that (3.5)
implies vn → 0 in Lr (RN ) for all 2 < r < 2∗ . By (4.1),
1
G(x, f (svn )) := q(x)f 4 (svn ) ≤ 4sr q(x)|vn |r ,
4
where we can choose r ∈ (2, min{2∗ , 4}). So
Z
Z
1
4
r
|vn |r → 0.
q(x)f (svn ) ≤ Cs
4
N
N
R
R
Now the same argument as in the proof of Lemma 3.4 shows that, up to a
subsequence, vn → v in L2loc (RN ) and a.e., and v 6= 0.
Let ϕ ∈ C0∞ (RN ). Then hI 0 (un ), ϕi → 0 and hence
Z
Z f (un )f 0 (un )
f 3 (un )f 0 (un )
∇vn ∇ϕ + V (x)
vn ϕ −
q(x)
vn ϕ → 0.
un
un
RN
RN
Since |un (x)| → ∞ if v(x) 6= 0 and f (t)f 0 (t) → 2−1/2 as t → ∞, we get
using Lemma 2.1-(5) that
f (un )f 0 (un )
→0
un
and
f 3 (un )f 0 (un )
→1
un
for such x. By the Lebesgue dominated convergence theorem we therefore
have
Z
Z
∇v∇ϕ =
q(x)vϕ.
RN
RN
So v 6= 0 and −∆v = q(x)v. This is impossible because −∆−q has absolutely
continuous spectrum. The proof is complete.
18
Lemma 4.4. If V is a compact subset of E, then there exists R > 0 such
that I ≤ 0 on (R+ V) \ BR (0).
Proof. We argue by contradiction as at the beginning of the proof of Lemma
3.5 and we obtain un ∈ V ⊂ S, wn = tn un , where un → u, tn → ∞ and
I(wn ) ≥ 0. We have
Z
I(tn un )
1
0≤
|∇un |2
=
t2n
2 RN
Z
f 2 (tn un ) q(x) f 4 (tn un )
+
V (x)
u2n .
−
(tn un )2
2 (tn un )2
un 6=0
It follows from Lemma 2.1-(5) and the Lebesgue dominated convergence
theorem that
Z
Z
f 2 (tn un ) q(x) f 4 (tn un )
2
0≤
|∇un | +
V (x)
−
u2n
2
2
(t
u
)
2
(t
u
)
N
n
n
n
n
R
u 6=0
Zn
Z
2
→
|∇u| −
q(x)u2 < 0,
RN
RN
a contradiction.
Let U := E ∩ S and define the mapping m : U → M by setting
m(w) := tw w,
where tw is as in Lemma 4.1.
Lemma 4.5. U is an open subset of S.
Proof. Obvious because E is open in E.
Lemma 4.6. Assume un ∈ U, un → u0 ∈ ∂U. If tn un ∈ M, then tn → ∞.
Proof. If (tn ) is bounded, we may assume tn → t0 and since M is bounded
away from 0 by Lemma 4.2-(2), t0 6= 0. So
0 = hI 0 (tn un ), un i → hI 0 (t0 u0 ), u0 i
and t0 u0 ∈ M. Hence u0 ∈ U .
The following two lemmas correspond to Lemma 3.6 and Lemma 3.7 in
the preceding section.
19
Lemma 4.7. The mapping m is a homeomorphism between U and M, and
u
the inverse of m is given by m−1 (u) = kuk
.
We consider the functional Ψ : U → R given by
Ψ(w) := I(m(w)).
Lemma 4.8. (1) Ψ ∈ C 1 (U, R) and
hΨ0 (w), zi = km(w)khI 0 (m(w)), zi
for all z ∈ Tw (U ).
(2) If (wn ) is a Palais-Smale sequence for Ψ, then (m(wn )) is a Palais-Smale
sequence for I. If (un ) ⊂ M is a bounded Palais-Smale sequence for I, then
(m−1 (un )) is a Palais-Smale sequence for Ψ.
(3) w is a critical point of Ψ if and only if m(w) is a nontrivial critical
point of I. Moreover, the corresponding values of Ψ and I coincide and
inf U Ψ = inf M I.
(4) Ψ is even (because I is).
A remark is in order here. Lemmas 4.7 and 4.8 are taken from [15] again.
Although in [15] they are formulated for U = S, it is easy to see that they
also hold if U is a proper open subset of S provided the other hypotheses
are satisfied. That this is the case follows from Lemmas 4.1 and 4.4.
Let p ∈ (4, 2 · 2∗ ). Then for each ε > 0 there exists Cε such that
|q(x)u3 | ≤ ε|u| + Cε |u|p−1 , hence Lemma 3.12 holds for g(x, u) = q(x)u3 and
so does Lemma 3.13 (we use Lemma 4.3 to assure boundedness of PalaisSmale sequences).
Proof of Theorem 1.2. The argument is the same as in Theorem 1.1.
However, there are two details which need to be clarified.
Let η be the flow given by (3.12). If T + (w) < ∞, then limt→T + (w) η(t, w)
exists (cf. [14], Lemma 2.15, Case 1) but unlike the situation in [14], this
limit may be a point w0 ∈ ∂U . This possibility is ruled out by Lemma 4.6.
Now we see as before that the conclusions of Lemma 3.14 hold.
Finally, we need to show that U contains sets of arbitrarily large genus.
Since the spectrum of −∆ − q in L2 (RN ) is absolutely continuous and q ≥
b0 > 0, E ∪{0} contains an infinite-dimensional subspace E0 . Hence E0 ∩S ⊂
U and γ(E0 ∩ S) = ∞.
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