Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 17, pp. 1–22.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF MULTIPLE SOLUTIONS FOR QUASILINEAR
ELLIPTIC EQUATIONS IN RN
HONGHUI YIN, ZUODONG YANG
Abstract. In this article, we establish the multiplicity of positive weak solution for the quasilinear elliptic equation
−∆p u + λ|u|p−2 u = f (x)|u|s−2 u + h(x)|u|r−2 u
x ∈ RN ,
x ∈ RN ,
u>0
u∈W
1,p
(RN )
We show how the shape of the graph of f affects the number of positive solutions. Our results extend the corresponding results in [21].
1. Introduction
In this article we consider the existence of solutions for the nonlinear quasilinear
problem
−∆p u + λ|u|p−2 u = f (x)|u|s−2 u + h(x)|u|r−2 u
u>0
x ∈ RN ,
x ∈ RN
(1.1)
u ∈ W 1,p (RN )
where 1 ≤ r < p < s < p∗ , p < N , p∗ =
p
p−r
pN
N −p
denotes the critical Sobolev exponent,
N
λ > 0 is a parameter, h ∈ L
(R )\{0} is nonnegative. For the function f , we
assume the following conditions:
(C1) f ∈ C(RN ) and is nonnegative in RN ;
(C2) f ∞ = lim|x|→∞ f (x) > 0;
(C3) There exist some points x1 , x2 , . . . , xk in RN such that f (xi ) are some strict
maxima and satisfy
f ∞ < f (xi ) = fmax ≡ max{f (x)|x ∈ RN }
for i = 1, 2, . . . , k.
Associated with (1.1), we consider the energy functional
Z
Z
Z
1
1
1
|∇u|p + λ|u|p dx −
f (x)|u|s dx −
h(x)|u|r dx.
Iλ (u) =
p RN
s RN
r RN
2000 Mathematics Subject Classification. 35J62, 35J50 35J92.
Key words and phrases. Nehari manifold; quasilinear; positive solution; (PS)-sequence.
c
2013
Texas State University - San Marcos.
Submitted August 4, 2013. Published January 10, 2014.
1
2
H. YIN, Z. YANG
EJDE-2014/17
It is well known that the functional Iλ ∈ C 1 (W 1,p (RN ), R), and that the solutions
of (1.1) are the critical points of the energy functional Iλ .
When p = 2 and h(x) ≡ 0, Equation (1.1) becomes
−∆u + λu = f (x)|u|s−2 u
u>0
x ∈ RN ,
x ∈ RN ,
1
(1.2)
N
u ∈ H (R ).
It is known that the existence of positive solutions of (1.2) is affected by the shape
of the graph of f (x). This has been the focus of a great deal of research by several
authors [3, 4, 8, 18]. Specially, if f is a positive constant, then (1.2) has a unique
positive solution [15] Adachi and Tanaka [1] showed that there exist at least four
positive solutions of the equation
−∆u + λu = f (x)|u|s−2 u + h(x) x ∈ RN ,
u>0
x ∈ RN ,
1
(1.3)
N
u ∈ H (R )
under the assumptions 0 < f (x) ≤ f ∞ = lim|x|→∞ f (x), h ∈ H −1 (RN )\{0} is
nonnegative and khkH −1 is sufficiently small. Several authors have studied a generalized version of (1.3),
−∆u + λu = f (x, u) + h(x) x ∈ RN ,
u>0
x ∈ RN ,
1
(1.4)
N
u ∈ H (R )
where f (x, u) and h(x) satisfy some suitable conditions. They showed the existence
of at least two positive solutions when khkH −1 is sufficiently small, see [2, 9, 14].
Wu [21] considered the problem (1.1) with p = 2, under some suitable assumptions on f (x), h(x). The author obtained the existence of multiple positive solution
by variational methods. Several publications [5, 6, 10, 22] show results about the
quasilinear elliptic equation
−∆p u + λ|u|p−2 u = f (x, u) x ∈ Ω,
u ∈ W01,p (Ω), u 6= 0
(1.5)
where 1 < p < N , N ≥ 3, Ω is an unbounded domain in RN . Because of the
unboundedness of the domain, the Sobolev compact embedding does not hold.
There are many methods to overcome the difficulty. In [22], the authors used the
concentration-compactness principle posed by Lions and the mountain pass lemma
to solve problem (1.5). In [5, 6], the authors studied the problem in symmetric
Sobolev spaces which posses Sobolev compact embedding.
Especially, when λ = 1, f (x, u) = q(x)uα and Ω is replaced by RN , using a minmax procedure formulated by Bahri and Li [4], Citti and Uguzzoni [10] obtained
1+β
the existence of a solution u ∈ W 1,p (RN ) ∩ Cloc
(RN ) of (1.5) when p ∈ (1, 2), and
β ∈ (0, 1) is constant. In [19], the authors studied the problem
−∆p u + λa(x)up−1 = f (x)up
∗
−1
1+β
u ∈ D01,p (RN ) ∩ Cloc
(RN ),
+ g(x)uq
x ∈ RN ,
lim u(x) = 0,
|x|→∞
(1.6)
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EXISTENCE OF MULTIPLE SOLUTIONS
3
which is a general case of (1.1). The authors proved that there exists a positive
solution of (1.6) for all λ in some interval [0, λ0 ).
In this article, we consider show the existence of multiple positive solutions of
(1.1). Our arguments are based on a combination of the concentration-compactness
principle of Lions [16], and Ekeland’s variational principle [13]. Our main result is
the following theorem.
p
Theorem 1.1. Assume (C1)–(C3) hold, and h ∈ L p−r (RN )\{0} is nonnegative.
Then there exists λ0 > 0 such that for all λ > λ0 , Equation (1.1) has at least k + 1
positive solutions.
The rest of this article is organized as follows. In Section 2, we give some
preliminaries and some properties of Nehri manifold. In Section 3, we prove the
main result, Theorem 1.1.
2. Preliminaries
Throughout the paper, C, c will denote various positive constants, their values
may vary from place to anther. By the change of variables η = λ−1/p , v(x) =
η p/(s−p) u(ηx), Equation (1.1) can be transformed into
−∆p v + |v|p−2 v = fη |v|s−2 v + η
v>0
p(s−r)
s−p
hη |v|r−2 v
x ∈ RN ,
x ∈ RN ,
(2.1)
v ∈ W 1,p (RN )
where fη = f (ηx), hη = h(ηx).
For u ∈ W 1,p (RN ), c ∈ R, a ∈ C(RN ) nonnegative and bounded, and b ∈
p
L p−r (RN ) non-negative, we define
Z
Z
p(s−r) 1
1
1
p
s
s−p
Ia,b (u) = kuk −
a|u| dx − η
b|u|r dx;
p
s RN
r RN
0
Ma,b (c) = {u ∈ W 1,p (RN )\{0}|hIa,b
(u), ui = c};
αa,b (c) = inf{Ia,b (u)|u ∈ Ma,b (c)},
0
where kuk = ( Ω |∇u|p + |u|p dx)1/p is a standard norm in W 1,p (RN ) and Ia,b
denote the Fréchet derivative of Ia,b . We shall write Ma,b (0), αa,b (0) as Ma,b , αa,b
respectively. Then, we have the following results.
R
Lemma 2.1. Suppose a is a continuous bounded and nonnegative function on RN ,
then αa,0 (c) = pc for c > 0 and
αa,0 ≤ αa,0 (c) + αa,0 (−c) −
s−p
|c|
sp
for all c ∈ R.
Proof. The case p = 2 was proved by Cao-Noussair [8, Lemma 2.2]. By a modification of the method given in [8], we obtain our result. For the readers convenience,
we give a sketch here. For any c > 0, let u ∈ Ma,0 (c). Then
Z
p
kuk =
a|u|s dx + c ≥ c.
RN
Thus
1
1
Ia,0 (u) = kukp −
p
s
Z
RN
1 1
c
c
a|u|s dx = ( − )kukp + ≥ .
p s
s
p
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H. YIN, Z. YANG
EJDE-2014/17
To show that the equality holds, choose v ∈ W 1,p (RN ) with
any σ > 0, define
uσ (x) = σ
N −p
p
v(σx),
R
RN
|∇v|p dx = c, for
wσ (x) = (1 + θ)uσ
where θ > 0 being selected so that wσ ∈ Ma,0 (c). It is easy to see that
Z
|∇uσ |p dx = c,
RN
Z
Z
(N −p)q
|v|q dx → 0 as σ → ∞
|uσ |q dx = σ p −N
RN
RN
∗
for q < p . Obviously, such a θ = θ(σ) exists when σ large enough and θ → 0 as
σ → +∞. Therefore,
Z
1
1
c
Ia,0 (wσ ) = kwσ kp −
a|wσ |s dx →
as σ → +∞.
p
s RN
p
Hence
c
.
p
To complete the proof of Lemma 2.1, let c > 0 and u ∈ Ma,0 (−c). Then
Z
Z
p
s
kuk =
a|u| dx − c <
a|u|s dx.
αa,0 (c) =
RN
RN
It is easy to see that there exist unique t ∈ (0, 1) such that v = tu ∈ Ma,0 . Then
we have
1 1
Ia,0 (v) = ( − )kvkp
p s
1 1
= ( − )tp kukp
p s
1 1
c c
< ( − )kukp + −
p s
s s
c
1 1
= Ia,0 (u) + + ( − )c
p
s p
s−p
c.
≤ Ia,0 (u) + αa,0 (c) −
sp
The required inequality then follows by taking the infimum over Ma,0 (−c).
Define
ψ(u) = hIf0 η ,hη (u), ui = kukp −
Z
fη |u|s dx − η
p(s−r)
s−p
RN
Z
hη |u|r dx.
RN
Then for u ∈ Mfη ,hη , we have
0
p
Z
p(s−r)
s−p
Z
fη |u| dx − rη
hη |u|r dx
RN
Z
= (p − r)kukp − (s − r)
fη |u|s dx.
hψ (u), ui = pkuk − s
s
RN
RN
Using the same methods as [20], we split Mfη ,hη into three parts:
Z
+
p
Mfη ,hη = {u ∈ Mfη ,hη |(p − r)kuk − (s − r)
fη |u|s dx > 0};
RN
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EXISTENCE OF MULTIPLE SOLUTIONS
Mf0η ,hη = {u ∈ Mfη ,hη |(p − r)kukp − (s − r)
Mf−η ,hη
Z
fη |u|s dx = 0};
RN
Z
p
5
fη |u|s dx < 0}.
= {u ∈ Mfη ,hη |(p − r)kuk − (s − r)
RN
Then we have the following result.
Lemma 2.2. There exists η1 > 0 such that for all η ∈ (0, η1 ), we have Mf0η ,hη = ∅.
Proof. Assume the contrary, that is Mf0η ,hη 6= ∅ for all η > 0. Then for u ∈ Mf0η ,hη ,
we have
Z
s−r
fη |u|s dx
(2.2)
kukp =
p − r RN
Z
Z
Z
p(s−r)
s−p
η s−p
hη |u|r dx = kukp −
fη |u|s dx =
fη |u|s dx.
(2.3)
p − r RN
RN
RN
Moreover,
s−p
kukp = kukp −
s−r
Z
= η β khk
fη |u|s dx ≤ η
p
L p−r
where β =
p(s−r)
s−p
−
p−r
p N.
p(s−r)
s−p
khη k
p
L p−r
RN
kukr
kukr ,
Also we have
kuk ≤ [
1
s−r β
p ] p−r .
η khk p−r
L
s−p
(2.4)
Let K : Mfη ,hη → R be given by
s−1
K(u) = c(s, r)( R
p(s−r)
p−1
kukp p−1
) s−p − η s−p
s
f |u| dx
RN η
Z
hη |u|r dx,
RN
1−s
s−r s−p s−p
0
where c(s, r) = ( p−r
)
p−r . Then K(u) = 0 for all η > 0 and u ∈ Mfη ,hη . From
0
(2.2) and (2.3), it follows that for u ∈ Mfη ,hη , and
K(u) = c(s, r)[
s−r
( p−r
s−1
RN
fη |u|s dx) p−1
RN
fη |u|s dx
R
R
p−1
] s−p −
s−p
p−r
Z
fη |u|s dx = 0.
(2.5)
RN
However, by (2.4), the Hölder and Sobolev inequalities and
(R
RN
p−1
S s p−1
kuks
s−p > (
)
) s−p
fmax
fmax |u|s dx
where S = inf u∈W 1,p (RN )\{0}
kuk
kukLs
s−1
for all u ∈ Mfη ,hη ,
is the best Sobolev constant. Also we have
p−1
kukp p−1
r
p kuk
) s−p − η β khk p−r
s
L
f
|u|
dx
η
N
R
S s p−1
r
p ]
≥ kuk [c(s, r)(
) s−p kuk1−r − η β khk p−r
L
fmax
1−r
S s p−1
s−r
β
p ) p−r − η khk
p ]
≥ kukr [c(s, r)(
) s−p (η β
khk p−r
L
L p−r
fmax
s−p
K(u) ≥ c(s, r)( R
6
H. YIN, Z. YANG
EJDE-2014/17
p−r
1−r
for all u ∈ Mf0η ,hη , where β = p(s−r)
s−p − p N > 0 (see Lemma 3.11). Since p−r ≤ 0,
there exists η1 > 0 such that for each η ∈ (0, η1 ) and u ∈ Mf0η ,hη , we have K(u) > 0,
this contradicts to (2.5). We can conclude that Mf0η ,hη = ∅ for all η ∈ (0, η1 ).
By Lemma 2.2 for η ∈ (0, η1 ) we write Mfη ,hη = Mf+η ,hη ∪ Mf−η ,hη and define
αf+η ,hη =
inf
+
u∈Mfη ,hη
Ifη ,hη ,
αf−η ,hη =
inf
u∈Mf−η ,hη
Ifη ,hη .
The following Lemma shows that the minimizers on Mfη ,hη are “usually” critical
points for Ifη ,hη .
Lemma 2.3. For η ∈ (0, η1 ), if u0 is a local minimizer for Ifη ,hη on Mfη ,hη , then
If0 η ,hη (u0 ) = 0 in W −1 (RN ), where W −1 (RN ) is the dual space of W 1,p (RN ).
Proof. If u0 is a local minimizer for Ifη ,hη on Mfη ,hη , then u0 is a solution of the
optimization problem
minimize Ifη ,hη (u) subject to ψ(u) = 0.
Hence, by the theory of Lagrange multipliers, there exists θ ∈ R such that
If0 η ,hη (u0 ) = θψ 0 (u0 )
in W −1 (RN ).
This implies
hIf0 η ,hη (u0 ), u0 i = θhψ 0 (u0 ), u0 i.
Since u0 ∈ Mfη ,hη and by Lemma 2.2, Mf0η ,hη = ∅ when η ∈ (0, η1 ), we have
hIf0 η ,hη (u0 ), u0 i = 0 and hψ 0 (u0 ), u0 i =
6 0.
So we obtain θ = 0. This completes the proof.
For each u ∈ W 1,p (RN )\{0}, we define
tmax = (
1
kukp
p−r
R
) s−p > 0.
s
s − r RN fη |u| dx
Then we have the following Lemma.
Lemma 2.4. There exists η2 > 0 such that for each u ∈ W 1,p (RN )\{0} and
η ∈ (0, η2 ), we have
(i) there is a unique t− = t− (u) > tmax > 0 such that t− u ∈ Mf−η ,hη and
Ifη ,hη (t− u)
R = maxt≥tmax Ifη ,hη (tu);
(ii) if RN hη |u|r dx > 0, then there is a unique 0 < t+ = t+ (u) < tmax < t− =
t− (u) such that t+ u ∈ Mf+η ,hη , t− u ∈ Mf−η ,hη and
Ifη ,hη (t+ u) = min Ifη ,hη (tu),
t− ≥t≥0
Ifη ,hη (t− u) = max Ifη ,hη (tu).
t≥tmax
hη |u|r dx ≥ 0. Let
Z
m(t) = tp−r kukp − ts−r
fη |u|s dx,
Proof. (i) Since h(x) is nonnegative, then
R
RN
RN
clearly, m(t) is increasing in (0, tmax ) and is decreasing in (tmax , +∞), also, we have
m(0) = 0,
lim m(t) = −∞,
t→+∞
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EXISTENCE OF MULTIPLE SOLUTIONS
7
i.e. m(t) is concave and achieve its maximum at tmax . Moreover,
m(tmax ) = (
p−r
s−r
p−r
kukp
kukp
p−r
R
R
) s−p kukp − (
) s−p
s
s
s − r RN fη |u| dx
s − r RN fη |u| dx
Z
fη |u|s dx
RN
s−r
s−r
kukp s−p
p − r p−r
p − r s−p
= R
) s−p − (
)
]
p−r [(
s−r
( RN fη |u|s dx) s−p s − r
p−r
s − p p − r p−r
kuks
(
) s−p ( R
) s−p kukr
s
s−r s−r
f |u| dx
RN η
s
p−r
s − p p − r p−r
S
≥
(
) s−p (
) s−p kukr = Ckukr .
s−r s−r
fmax
=
Since C > 0,
0≤η
p(s−r)
s−p
Z
hη |u|r dx ≤ η β khk
p
L p−r
RN
kukr
and β > 0, there exists η2 > 0, such that for any η ∈ (0, η2 ), we have
Z
p(s−r)
m(tmax ) > η s−p
hη |u|r dx.
RN
Case (a): RN hη |u| dx = 0. Then there is unique t− > tmax such that m(t− ) = 0
and m0 (t− ) < 0. Now
Z
hψ 0 (t− u), t− ui = (p − r)kt− ukp − (s − r)
fη |t− u|s dx = (t− )r+1 m0 (t− ) < 0
R
r
RN
and
Z
p(s−r)
hη |t− u|r dx
fη |t− u|s dx − η s−p
N
N
R
R
Z
− r
− p−r
p
− s−r
= (t ) [(t ) kuk − (t )
fη |u|s dx]
hIf0 η ,hη (t− u), t− ui = kt− ukp −
Z
RN
= (t− )r m(t− ) = 0.
Thus, t− u ∈ Mf−η ,hη . Moreover, we have
d
If ,h (tu) = 0,
dt η η
d2
If ,h (tu) < 0,
dt2 η η
for t = t− .
Then we haveR Ifη ,hη (t− u) = maxt≥tmax Ifη ,hη (tu).
Case (b): RN hη |u|r dx > 0. There are unique t+ and t− such that 0 < t+ <
tmax < t− such that
Z
p(s−r)
hη |u|r dx = m(t− )
m(t+ ) = η s−p
RN
0
and m (t ) > 0 > m (t ). Similar to the argument in Case a, we have t± u ∈
Mf±η ,hη , and Ifη ,hη (t− u) ≥ Ifη ,hη (tu) ≥ Ifη ,hη (t+ u) for each t ∈ [t+ , t− ], and
Ifη ,hη (tu) ≥ Ifη ,hη (t+ u) for each t ∈ [0, t+ ].
(ii) By case (b) it follows part (i)
+
0
−
To establish the existence of a local minimum for Ifη ,hη on Mfη ,hη , we need the
following results.
8
H. YIN, Z. YANG
Lemma 2.5. (i) For each u ∈ Mf+η ,hη , we have
In particular αfη ,hη ≤ αf+η ,hη < 0.
EJDE-2014/17
R
RN
hη |u|r dx > 0 and Ifη ,hη (u) < 0.
1
β
(ii) Ifη ,hη is coercive and bounded below on Mfη ,hη for all η ∈ (0, ( s−p
s−r ) ).
Moreover, αfη ,hη → 0 as η → 0.
Proof. (i) For each u ∈ Mf+η ,hη , we have
Z
(p − r)kukp − (s − r)
fη |u|s dx > 0,
N
R
Z
Z
p(s−r)
p
s
s−p
hη |u|r dx.
kuk =
fη |u| dx + η
RN
RN
By (C1), we have
Z
Z
p(s−r)
r
p
s−p
hη |u| dx = kuk −
η
RN
s−p
fη |u| dx >
p−r
s
RN
Z
fη |u|s dx ≥ 0
RN
and
Z
Z
1 1 p(s−r)
1 1
s
s−p
fη |u| dx + ( − )η
hη |u|r dx
Ifη ,hη (u) = ( − )
p s RN
p r
RN
Z
Z
1 1
1 1 s−p
<( − )
fη |u|s dx + ( − )
fη |u|s dx
p s RN
p r p − r RN
Z
1
1
fη |u|s dx ≤ 0
= (s − p)( − )
ps pr RN
R
p(s−r) R
(ii) For each u ∈ Mfη ,hη , we have kukp = RN fη |u|s dx + η s−p RN hη |u|r dx.
Then by the Hölder and Young inequalities,
s−p
s−r β
r
p kuk
kukp −
η khk p−r
L
ps
rs
p
s−p s−r β
(p − r)(s − r)
≥(
−
η )kukp − η β
khk p−rp .
ps
ps
prs
L p−r
Ifη ,hη (u) ≥
1
β
Thus, Ifη ,hη is coercive and bounded below on Mfη ,hη for all η ∈ (0, ( s−p
s−r ) ) and
αfη ,hη → 0 as η → 0, where β =
p(s−r)
s−p
−
p−r
p N
> 0 as above.
3. Proofs of main results
Now, we use the graph of the coefficient f to find some Palais-Smale sequences
which are used to prove Theorem 1.1. For a > 0, let Ca (xi ) denote the hypercube
i
i
i
i
i
i
i
ΠN
j=1 (xj − a, xj + a) centered at x = (x1 , x2 , . . . , xN ) for i = 1, 2, . . . , k. Let Ca (x )
i
i
and ∂Ca (x ) denote the closure and the boundary of Ca (x ) respectively. By the
conditions (C1) and (C3), we can choose numbers K, l > 0 such that Cl (xi ) are
disjoint, f (x) < f (xi ) for x ∈ ∂Cl (xi ) for all i = 1, 2, . . . , k and ∪ki=1 Cl (xi ) ⊂
ΠN
i=1 (−K, K).
Define φη ∈ C(R, R), gη ∈ (W 1,p (RN ), RN ) by
 2K
t > 2K

 η
η ,
2K
φη (t) = t
− η ≤ t ≤ 2K
η ,

 2K
2K
− η t<− η .
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EXISTENCE OF MULTIPLE SOLUTIONS
9
φ (x )|u|s dx
RNR η j
for j = 1, 2, . . . , N
|u|s dx
RN
= (gη1 (u), gη2 (u), . . . , gηN (u)) ∈ RN .
R
gηj (u) =
gη (u)
i
Let Cl/η
≡ Cl/η (xi /η),
i
Nηi = {u ∈ Mf−η ,hη ”u ≥ 0 and gη (u) ∈ Cl/η
},
i
∂Nηi = {u ∈ Mf−η ,hη : u ≥ 0 and gη (u) ∈ ∂Cl/η
}
for i = 1, 2, . . . , k. It is easy to verify that Nηi and ∂Nηi are nonempty sets for all
i = 1, 2, . . . , k. Consider the minimization problems in Nηi and ∂Nηi for Ifη ,hη ,
γηi = inf i Ifη ,hη (u),
u∈Nη
γ iη = inf i Ifη ,hη (u).
u∈∂Nη
Using the results in [19], we can assume w be a unique positive radial solution of
−∆p u + |u|p−2 u = fmax |u|s−2 u
x ∈ RN ,
u > 0 x ∈ RN ,
u ∈ W 1,p (RN )
and that Ifmax ,0 (w) = αfmax ,0 . By (C3) and the routine computations, we have
αfmax ,0 < αf ∞ ,0 .
√
For small η > 0 satisfying 2 η < 1, we define a function ψη ∈ C 1 (RN , [0, 1]) such
that
(
1
1 |x| < 2√
η − 1,
ψη (x) =
1
0 |x| > 2√η ,
and |∇ψη | ≤ 2 in RN . Let xη =
1
√
2 η (1, 1, . . . , 1)
wη (x) = t−
η w(x −
∈ RN and
xi
xi
+ xη )ψη (x −
+ xη ),
η
η
−
where t−
η > 0 are selected such that wη ∈ Mfη ,hη . Then we have the following
results.
Lemma 3.1. As η → 0, we have
p(s−r) R
i
(i) η s−p RN hη wr (x − xη + xη )ψηr (x −
(ii) t−
η → 1.
Proof. (i) Since β =
p(s−r)
s−p
−
Z
p−r
p N
xi
η
+ xη )dx → 0;
> 0 and hη (x) ≥ 0, we have
xi
xi
+ xη )ψηr (x −
+ xη )dx
η
η
RN
xi
xi
p kw(x −
≤ η β khk p−r
+ xη )ψη (x −
+ xη )kr
L
η
η
0 ≤ eta
p(s−r)
s−p
and
kw(x −
Thus (i) holds.
hη wr (x −
xi
xi
sp
+ xη )ψη (x −
+ xη )kp →
αf ,0 .
η
η
s − p max
10
H. YIN, Z. YANG
EJDE-2014/17
(ii) Since wη ∈ Mf−η ,hη , we have
Z
xi
xi
− p
(tη ) [
+ xη )ψη (x −
+ xη ))|p
|∇(w(x −
η
η
RN
xi
xi
+ (w(x −
+ xη )ψη (x −
+ xη ))p ]
η
η
Z
xi
xi
− s
= (tη )
+ xη )ψηs (x −
+ xη )dx
fη ws (x −
η
η
RN
Z
p(s−r)
xi
xi
− r
s−p
(tη )
hη wr (x −
+ xη )ψηr (x −
+ xη )dx.
+η
η
η
RN
When η → 0, from part (i) it follows that
xi
xi
η
p
p
− p
+
x
)ψ
(x
−
+ xη )kp + o(η)
(t−
)
(kwk
+
o(η))
=
(t
)
kw(x
−
η
η
η
η
η
Z
xi
xi
− s
= (tη )
fη ws (x −
+ xη )ψηs (x −
+ xη )dx + o(η)
η
η
RN
Z
s
= (t−
)
f (ηx + xi − ηxη )ws dx + o(η).
η
RN
Moreover, ηxη → 0 as η → 0, and from kwkp =
R
RN
i
t−
η
> tmax
→(
fmax ws dx, we have
i
kw(x − xη + xη )ψη (x − xη + xη )kp
1
p−r
) s−p
=(
R
i
i
x
x
η
η
s
s − r RN fη |w(x − η + x )ψη (x − η + x )| dx
1
p − r s−p
)
> 0.
s−r
Thus, t−
η → 1 as η → 0 and (ii) holds.
1
β
Let η∗ = min{η1 , η2 , ( s−p
s−r ) }, then we have the following result.
Lemma 3.2. For each ε > 0, there exists ηε ∈ (0, η∗ ] such that
αf−η ,hη ≤ γηi < min{αfmax ,0 + ε, αfη ,hη + αf ∞ ,0 },
i = 1, 2, . . . , k, η ∈ (0, ηε ).
Proof. For i = 1, 2, . . . , k, obviously we have αf−η ,hη ≤ γηi .
i
Now we show the second inequality hold. First, we prove that gη (wη ) ∈ Cl/η
.
For j = 1, 2, . . . , N , since
R
i
i
φ (x )ws (x − xη + xη )ψηs (x − xη + xη )dx
RN η j
j
gη (wη ) =
R
i
i
ws (x − xη + xη )ψηs (x − xη + xη )dx
RN
and
ψη (x −
xi
+ xη ) = 0
η
if |xj −
xij
1
|> √ .
η
η
By the definition of ψη , we have
R
i
i
φη (xj )ws (x − xη + xη )ψηs (x − xη + xη )dx
i
Cl/η
gηj (wη ) =
R
i
i
ws (x − xη + xη )ψηs (x − xη + xη )dx
Ci
l/η
EJDE-2014/17
provided
√1
η
EXISTENCE OF MULTIPLE SOLUTIONS
11
i
< ηl . From the definition of φη and gη we conclude that gη (wη ) ∈ Cl/η
.
Thus, wη ∈ Nηi . Moreover, by Lemma 3.1, we obtain
p Z
(t−
xi
xi
η)
[
+ xη )ψη (x −
+ xη ))|p dx
Ifη ,hη (wη ) =
|∇(w(x −
p
η
η
N
R
Z
xi
xi
+ xη )ψη (x −
+ xη )|p dx]
+
|w(x −
η
η
N
R
s Z
(t−
xi
xi
η)
−
fη ws (x −
+ xη )ψηs (x −
+ xη )dx
s
η
η
N
R
− r Z
p(s−r) (t )
xi
xi
η
s−p
−η
hη wr (x −
+ xη )ψηr (x −
+ xη )dx
r
η
η
RN
Z
Z
1
1
|∇w|p + |w|p dx −
f (ηx + xi − ηxη )ws dx + o(η).
=
p RN
s RN
Since ηxη → 0 as η → 0 and from the above, we have
Ifη ,hη (wη ) = Ifmax ,0 (w) + o(η) = αfmax ,0 + o(η).
Therefore, for any ε > 0 there exists η3 > 0 such that
γηi < αfmax ,0 + ε, i = 1, 2, . . . , k, η ∈ (0, η3 ).
Moreover, αfmax ,0 < αf ∞ ,0 and αfη ,hη → 0 as η → 0, then there exists η4 > 0 such
that
γηi < αfη ,hη + αf ∞ ,0 , i = 1, 2, . . . , k, η ∈ (0, η4 ).
We take ηε = min{η3 , η4 }, this implies
γηi < min{αfmax ,0 + ε, αfη ,hη + αf ∞ ,0 },
for i = 1, 2, . . . , k and η ∈ (0, ηε ). This completes the proof.
Since W 1,p (RN ) is not a Hilbert space in general, even if the (P S) sequence {un }
of Iλ (u) is bounded, hence there exists u ∈ W 1,p (RN ) such that
un * u in W 1,p (RN ),
we can not ensure
p
|∇unk |p−2 ∇unk * |∇u|p−2 ∇u in L p−1 (RN )
for some subsequence {unk } of {un }, so we can not use Brezis-Lieb lemma [20]
directly. We use the following results.
Lemma 3.3. If {un } ⊂ W 1,p (RN ) is a (P S)c sequence of Ifη ,hη , then there exists
a subsequence {uk } such that uk * u0 in W 1,p (RN ) for some u0 ∈ W 1,p (RN ), and
I 0 (u0 ) = 0, ∇uk → ∇u0 a.e. in RN .
The proof of the above lemma was given in [12, Lemma 2.1], also in [17]. We
omit it here.
Lemma 3.4. There are positive numbers δ and ηδ ∈ (0, η∗ ] such that for i =
1, 2, . . . , k, we have
fi > αf ,0 + δ
γ
η
max
for all η ∈ (0, ηδ ).
12
H. YIN, Z. YANG
EJDE-2014/17
Proof. Fix i ∈ {1, 2, . . . , k}. Suppose the contrary that there exists a sequence {ηn }
g
i
with ηn → 0 as n → ∞ such that γ
ηn → c ≤ αfmax ,0 . Consequently, there exists a
i
sequence {un } ⊂ ∂Nηn such that gηn (un ) ∈ ∂C i l and
ηn
hIf0 ηn ,hηn (un ), un i
= 0,
Ifηn ,hηn (un ) → c ≤ αfmax ,0 .
By Lemma 2.5, {un } is uniformly bounded in W 1,p (RN ). For un ∈ Mf−η ,hη , we
n
n
deduce from the Sobolev imbedding theorem that there exists a constant ν > 0
such that kun k > ν for all n. Applying the concentration-compactness principle of
Lions [16] to |un |s , there are positive constants R, µ and {yn } ⊂ RN such that
Z
|un |s dx ≥ µ for all n,
B N (yn ,R)
N
where B (yn , R) = {x ∈ RN ||x − yn | < R}. Let u
en = un (x + yn ), and define
feηn (x) = f (ηn x + ηn yn ), e
hηn (x) = h(ηn x + ηn yn ).
Then we have
hIf0e
ηn ,hηn
e
Ifeη
(e
un ), u
en i = 0,
(3.1)
(e
un ) → c.
n ,hηn
e
By Lemma 3.3, Sobolev imbedding theorem and Riesz’s theorem, there is a u0 ∈
W 1,p (RN ) and a subsequence of {e
un }, still denoted by {e
un } such that
u
en * u0
in W 1,p (RN ),
a.e. in RN ,
Z
|e
un |s dx →
|u0 |s dx ≥ µ,
u
en → u0
Z
B N (0,R)
B N (0,R)
and
∇e
un → ∇u0
a.e. in RN ,
|∇e
un |p−2 ∇e
un * |∇u0 |p−2 ∇u0
p
in L p−1 (RN ).
Set wn = u
en − u0 . By the Brezis-Lieb lemma [20], we have
Z
Z
Z
feηn |e
un |s dx =
feηn |u0 |s dx +
feηn |wn |s dx + o(1).
RN
RN
Since {un } is uniformly bounded and u
en * u0 , we obtain
Z
Z
p(s−r)
p(s−r)
e
hηn |un |r dx = ηn s−p
hηn |e
un |r dx → 0
ηn s−p
RN
(3.2)
RN
as n → ∞
(3.3)
RN
and
ke
un kp = ku0 kp + kwn kp + o(1).
Combining (3.1)-(3.4), we have
Z
Z
p
s
p
e
kwn k −
fηn |wn | dx = −(ku0 k −
feηn |u0 |s dx) + o(1).
RN
(3.4)
(3.5)
RN
We distinguish the two cases: (A) kwn k → 0 and (B) kwn k → c > 0.
Case (A): From condition (C3) we can choose a positive constant δ such that
i
i
f (x) < fmax for x ∈ C l+δ \Cl−δ
.
EJDE-2014/17
EXISTENCE OF MULTIPLE SOLUTIONS
13
We complete the proof by establishing the contradiction
lim Ifηn ,hηn (un ) > αfmax ,0 .
n→∞
Consider the sequence {ηn yn }. By passing to a subsequence if necessary, we may
assume that one of the following cases occur:
i
i
(A1) {ηn yn } ⊂ C l+δ \Cl−δ
,
i
(A2) {ηn yn } ⊂ C l−δ ,
i
(A3) {ηn yn } ⊂ RN \Cl+δ
and {ηn yn } is bounded;
(A4) {ηn yn } is unbounded.
Let > 0 and R > 0 be such that
R
|e
un |s dx
|x|≥R
R
≤ .
|e
un |s dx
RN
(3.6)
i
i
and f (e
y ) < fmax . Consequently
In case (A1), we assume ηn yn → ye ∈ C l+δ \Cl−δ
by (3.3) and (3.4), we have
Z
Z
p(s−r)
1
1
s−p
p
s
e
e
hηn |e
un |r dx]
lim Ifηn ,hηn (un ) = lim [ ke
un k −
fη (x)|e
un | dx − ηn
n→∞ p
n→∞
s RN n
RN
Z
1
1
p
= ku0 k −
f (e
y )|u0 |s dx
p
s RN
≥ αf (ey),0 > αfmax ,0 ,
we also have
p
Z
ku0 k −
f (e
y )|u0 |s dx = 0,
RN
which is a contradiction.
In case (A2),
R
un |s dx
N φηn (xj + (yn )j )|e
j
R
gηn (un ) = R
|e
un |s dx
RN
R
R
un |s dx
φ (xj + (yn )j )|e
un |s dx + |x|≥R φηn (xj + (yn )j )|e
|x|≤R ηn
R
=
.
|e
un |s dx
RN
In the region |x| ≤ R , when n is sufficiently large, we have
xj + (yn )j ∈ (
xij − (l − δ)
xij + (l − δ)
2K 2K
− R ,
+ R ) ⊂ (−
,
).
ηn
ηn
ηn ηn
It follows from (3.6) and the definition of φηn that
xij − (l − δ)
2K
− R )(1 − ) −
,
ηn
ηn
xij + (l − δ)
2K
gηj n (un ) < (
+ R )(1 − ) +
.
ηn
ηn
It is clear from the above inequalities that we can choose > 0, δ > sufficiently
small such that
xij − l xij + l
gηj n (un ) ∈ (
,
)
ηn
ηn
i
for n large enough, which contradicts gηn (un ) ∈ ∂Cl/η
.
n
gηj n (un ) > (
14
H. YIN, Z. YANG
EJDE-2014/17
i
In case (A3), we assume that ηn yn → ye 6∈ Cl+δ
as n → ∞, then for some
i
j ∈ {1, 2, . . . , N }, we have yej ≥ xj + (l + δ) or yej ≤ xij − (l + δ).
xi +(l+ δ )
First, we assume yej ≥ xij + (l + δ) occurs, then (yn )j > j ηn 2 for all n. When
|xj | ≤ R , we have
xij + (l + 2δ )
xj + (yn )j >
− R
ηn
and
xij + l
xij + (l + 2δ )
2K
gηj n (un ) > (
− R )(1 − ) −
>
,
ηn
ηn
ηn
for sufficiently small > 0, δ > and n large enough. This contradicts to gηn (un ) ∈
∂C i l . When yej ≤ xij − (l + δ), the argument is similar.
ηn
In case (A4), we assume ηn yn → ∞ as n → ∞, using a similar argument to case
(A1) and condition (C3), we can also obtain a contradiction.
Case (B): Set
Z
ku0 kp −
feη |u0 |s dx = A + o(1),
n
RN
then by (3.5),
kwn kp −
Z
feηn |wn |s dx = −A + o(1).
RN
Without loss of generality, we may assume that A > 0(A < 0 can be considered
similarly). We can choose a sequence {tn } with tn → 1 as n → ∞ such that
vn = tn wn satisfies
Z
kvn kp −
feη |vn |s dx = −A.
n
RN
Since u0 ∈ Mfeη
(A + o(1)), by (3.2)-(3.4) and Lemma 2.1 we have
Z
1
1
feη (x)|u0 |s dx
Ifηn ,hηn (un ) = ku0 kp −
p
s RN n
Z
1
1
+ kwn kp −
feη (x)|wn |s dx + o(1)
p
s RN n
Z
A + o(1) 1
1
p
≥
+ kvn k −
feη (x)|vn |s dx + o(1)
p
p
s RN n
≥ αfeη ,0 (A) + αfeη ,0 (−A) + o(1)
n ,0
n
n
s−p
> αfeη ,0 +
A + o(1)
n
2sp
s−p
≥ αfmax ,0 +
A + o(1),
2sp
which is a contradiction. If A = 0, we can find two sequences {tn } and {sn } with
tn → 1, sn → 1 as n → ∞ such that wn = tn wn , v n = sn u0 satisfy
Z
kwn kp −
feηn |wn |s dx = 0,
RN
Z
p
kv n k −
feηn |v n |s dx = 0.
RN
Thus
lim Ifηn ,hηn (un )
n→∞
EJDE-2014/17
EXISTENCE OF MULTIPLE SOLUTIONS
= lim
h1
p
> αfmax ,0 ,
n→∞
kwn kp −
1
s
Z
RN
1
1
feηn |wn |s dx + kv n kp −
p
s
Z
feηn |v n |s dx
15
i
RN
which is a contradiction. This completes the proof.
From now on, taking δ > 0 as in Lemma 3.4, and fixing ε > 0 such that ε ≤ δ,
consider ηε as in Lemma 3.2, ηδ as in Lemma 3.4, and denote η0 = min{ηε , ηδ }.
Lemma 3.5. If η ∈ (0, η0 ), then for each u ∈ Mfη ,hη , there exist εu > 0 and
a differentiable function ξu : B(0, εu ) ⊂ W 1,p (RN ) → R+ such that ξu (0) =
1, ξu (v)(u − v) ∈ Mfη ,hη , and
h Z
0
hξu (0), vi = p
|∇u|p−2 ∇u∇v + |u|p−2 uvdx
RN
Z
Z
i
p(s−r)
hη |u|r−2 uvdx
−s
fη |u|s−2 uvdx − η s−p r
RN
RN
Z
h
i
÷ (p − r)kukp − (s − r)
fη |u|s dx
RN
for all v ∈ W
1,p
N
(R ).
Proof. For u ∈ Mfη ,hη , define a function F : R × W 1,p (RN ) → R by
Fu (ξu , w) = hIf0 η ,hη (ξu (u − w)), ξu (u − w)i
Z
Z
p
p
p
s
= ξu
|∇(u − w)| + |u − w| dx − ξu
fη |u − w|s dx
RN
RN
Z
p(s−r)
hη |u − w|r dx.
− η s−p ξur
RN
Then Fu (1, 0) = hIf0 η ,hη (u), ui = 0 and
Z
Z
p(s−r)
d
hη |u|r dx
Fu (1, 0) = pkukp − s
fη |u|s dx − η s−p r
dξu
RN
RN
Z
p
= (p − r)kuk − (s − r)
fη |u|s dx 6= 0.
RN
According to the implicit function theorem, there exist εu > 0 and a differentiable
function ξu : B(0, εu ) ⊂ W 1,p (RN ) → R+ such that ξu (0) = 1, and
h Z
0
hξu (0), vi = p
|∇u|p−2 ∇u∇v + |u|p−2 uvdx
RN
Z
Z
i
p(s−r)
−s
fη |u|s−2 uvdx − η s−p r
hη |u|r−2 uvdx
RN
RN
Z
h
i
÷ (p − r)kukp − (s − r)
fη |u|s dx ,
RN
and Fu (ξu (v), v) = 0 for all v ∈ B(0, εu ), which is equivalent to
hIf0 η ,hη (ξu (v)(u − w)), ξu (v)(u − w)i = 0,
That is, ξu (v)(u − v) ∈ Mfη ,hη .
∀v ∈ B(0, εu ).
16
H. YIN, Z. YANG
EJDE-2014/17
Lemma 3.6. If η ∈ (0, η0 ), then for each u ∈ Nηi , there exist εu > 0 and
a differentiable function ξu− : B(0, εu ) ⊂ W 1,p (RN ) → R+ such that ξu− (0) =
1, ξu− (v)(u − v) ∈ Nηi for all v ∈ B(0, εu ), and
h Z
h(ξu− )0 (0), vi = p
|∇u|p−2 ∇u∇v + |u|p−2 uvdx
RN
Z
Z
i
p(s−r)
s−2
s−p
r
hη |u|r−2 uvdx
−s
fη |u| uvdx − η
RN
RN
Z
h
i
p
÷ (p − r)kuk − (s − r)
fη |u|s dx
RN
for all v ∈ W
1,p
N
(R ).
Proof. Similar to the argument in Lemma 3.5, there exist εu > 0 and a differentiable
function ξu− : B(0, εu ) ⊂ W 1,p (RN ) → R+ such that ξu− (0) = 1, ξu− (v)(u − v) ∈
Mfη ,hη for all v ∈ B(0, εu ), Since
Z
p
(p − r)kuk − (s − r)
fη |u|s dx < 0,
RN
thus, if εu small enough, by the continuity of the functions ξu− and gη , we have
hψ 0 (ξu− (v)(u − v))), ξu− (v)(u − v)i
= (p − r)kξu− (v)(u − v)kp − (s − r)
Z
RN
fη |ξu− (v)(u − v)|s dx < 0.
i
and gη (ξu− (v)(u − v)) ∈ Cl/η
.
Proposition 3.7. (i) If η ∈ (0, η0 ), then there exists a (P S)αfη ,hη sequence {un } ⊂
Mfη ,hη in W 1,p (RN ) for Ifη ,hη .
(ii) If η ∈ (0, η0 ), then there exists a (P S)γηi sequence {un } ⊂ Nηi in W 1,p (RN )
for Ifη ,hη , i = 1, 2, . . . , k.
Proof. Since the proof of (i) is similar to that of (ii), but simpler, we only prove
(ii) here. We denote by Nηi the closure of Nηi , then we note that
Nηi = Nηi ∪ ∂Nηi ,
for each i = 1, 2, . . . , k.
From Lemma 3.2 and Lemma 3.4, we obtain
fi },
γηi < min{αfη ,hη + αf ∞ ,0 , γ
η
i = 1, 2, . . . , k, η ∈ (0, η0 ).
(3.7)
Hence
γηi = inf{Ifη ,hη (u) : u ∈ Nηi }
for i = 1, 2, . . . , k.
Fix some i ∈ {1, 2, . . . , k}. Applying the Ekeland variational principle [17] there
exists a minimizing sequence {un } ⊂ Nηi such that
Ifη ,hη (un ) < γηi +
1
,
n
(3.8)
1
kw − un k for all w ∈ Nηi .
(3.9)
n
From (3.7) we may assume that un ∈ Nηi for n sufficiently large. Applying Lemma
3.6 with u = un we obtain the functional ξu−n : B(0, εun ) → R for some εun > 0
such that ξu−n (w)(un − w) ∈ Nηi . Choose 0 < ρ < εun and u ∈ W 1,p (RN ) with
Ifη ,hη (un ) < Ifη ,hη (w) +
EJDE-2014/17
u 6≡ 0. Set wρ =
(3.9) that
EXISTENCE OF MULTIPLE SOLUTIONS
ρu
kuk
17
and zρn = ξu−n (wρ )(un − wρ ). Since zρn ∈ Nηi , we deduce from
1
Ifη ,hη (zρn ) − Ifη ,hη (un ) ≥ − kzρn − un k.
n
By the mean value theorem, we have
1
hIf0 η ,hη (un ), zρn − un i + o(kzρn − un k) ≥ − kzρn − un k.
n
Thus,
1
hIf0 η ,hη (un ), −wρ i+(ξu−n (wρ )−1)hIf0 η ,hη (un ), un −wρ i ≥ − kzρn −un k+o(kzρn −un k).
n
(3.10)
Since ξu−n (wρ )(un − wρ ) ∈ Nηi and consequently from (3.10) we obtain
u
i + (ξu−n (wρ ) − 1)hIf0 η ,hη (un ) − If0 η ,hη (zρn ), un − wρ i
− ρhIf0 η ,hη (un ),
kuk
1
≥ − kzρn − un k + o(kzρn − un k).
n
Thus,
hIf0 η ,hη (un ),
(ξ − (wρ ) − 1) 0
u
i ≤ un
hIfη ,hη (un ) − If0 η ,hη (zρn ), un − wρ i
kuk
ρ
kzρn − un k o(kzρn − un k)
+
+
.
nρ
ρ
(3.11)
Since
kzρn − un k ≤ ρ|ξu−n (wρ )| + |ξu−n (wρ ) − 1|kun k
and
|ξu−n (wρ ) − 1|
≤ k(ξu−n )0 (0)k,
ρ→0
ρ
if we let ρ → 0 in (3.11) for a fixed n, and by Lemma 2.5 (ii) we can find a constant
C > 0, independent of ρ, such that
u
C
hIf0 η ,hη (un ),
i ≤ (1 + k(ξu−n )0 (0)k).
kuk
n
lim
We are done once we show that k(ξu−n )0 (0)k is uniformly bounded in n. By Lemma
2.5 (ii), Lemma 3.6 and the Hölder inequality, we have
0
h(ξu−n ) (0), vi ≤
bkvk
R
|(p − r)kun kp − (s − r) RN fη |un |s dx|
for some b > 0.
We only need to show that
|(p − r)kun kp − (s − r)
Z
fη |un |s dx| > C
RN
for some C > 0 and n large. We argue by way of contradiction. Assume that there
exists a subsequence {un } satisfy
Z
(p − r)kun kp − (s − r)
fη |un |s dx = o(1).
(3.12)
RN
By the fact that un ∈
Mf−η ,hη (un )
and (3.12), we obtain that
Z
RN
fη |un |s dx > 0.
18
H. YIN, Z. YANG
EJDE-2014/17
So we have
1
s−r β
p ] p−r + o(1)
η khk p−r
L
s−p
1
p − r S s s−p
kun k > (
)
+ o(1).
s − r fmax
kun k ≤ [
(3.13)
(3.14)
Then
K(un ) = c(s, r)
h ( s−r
p−r
s−1
fη |un |s dx + o(1)) p−1 i p−1
s−p
R
RN
R
RN
fη |un |s dx
−
s−p
p−r
Z
fη |un |s dx
RN
= o(1).
(3.15)
However, by (3.13)-(3.14), the Hölder and Sobolev inequalities, combining with
β > 0 and η ∈ (0, η0 ), we have
s−1
p−1
kun kp p−1
r
p ku k + o(1)
) s−p − η β khk p−r
n
s
L
f
|u
|
dx
η
n
N
R
S s p−1
r
p ] + o(1)
≥ kun k [c(s, r)(
) s−p kun k1−r − η β khk p−r
L
fmax
1−r
S s p−1
s−r
β
p ) p−r − η khk
p ] + o(1)
≥ kun kr [c(s, r)(
) s−p (η β
khk p−r
L
L p−r
fmax
s−p
≥d
K(un ) ≥ c(s, r)( R
for some d > 0 and n large enough. This is a contradiction to (3.15). So we have
Ifη ,hη (un ) = γηi + o(1)
and If0 η ,hη (un ) = 0 in W −1 (RN ). Thus we complete the proof of (ii).
Theorem 3.8. For each η ∈ (0, η0 ), Equation (2.1) has a positive solution uη ∈
Mf+η ,hη such that Ifη ,hη (uη ) = αfη ,hη = αf+η ,hη .
Proof. By Proposition 3.7 (i), there exists a (P S)αfη ,hη sequence {un } ⊂ Mfη ,hη , by
Lemma 2.5 (ii) and Lemma 3.3, there exist a subsequence {un } and uη in W 1,p (RN )
such that
un * uη
weakly in W 1,p (RN ),
un → uη
un → uη
a.e. in RN ,
in Lq (RN ) for1 ≤ q ≤ p∗ ,
∇un → ∇uη
a.e. in RN ,
|∇un |p−2 ∇un * |∇uη |p−2 ∇uη
p
in L p−1 (RN ),
It is easy to see that uη is a solution of (2.1)
Moreover, by the Egorov theorem and the Hölder inequality and condition h ∈
p
L p−r (RN ), we obtain
Z
Z
r
hη |un | dx →
hη |uη |r dx.
RN
RN
R
We claim that RN hη |uη |r dx 6= 0. If not,
Z
p
kun k =
fη |un |s dx + o(1),
RN
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EXISTENCE OF MULTIPLE SOLUTIONS
19
and
Z
1 1
( − )
fη |un |s dx
p s RN
Z
Z
p(s−r) 1
1
1
p
s
s−p
= kun k −
fη |un | dx − η
hη |un |r dx + o(1)
p
s RN
r RN
= αfη ,hη + o(1),
this contradicts αfη ,hη < 0. Thus, uη is a nontrivial solution of (2.1). Now we show
that un → uη strongly in W 1,p (RN ). If not, kuη k < lim inf n→∞ kun k, so we have
Z
1 1 p(s−r)
1 1
p
s−p
hη |uη |r dx
αfη ,hη ≤ Ifη ,hη (uη ) = ( − )kuη k − ( − )η
p s
r
s
RN
< lim Ifη ,hη (un ) = αfη ,hη ,
n→∞
this is a contradiction. Thus Ifη ,hη (uη ) = αfη ,hη . At last, we show uη ∈ Mf+η ,hη . If
not, by Lemma 2.2, we know that uη ∈ Mf−η ,hη , by Lemma 2.4, there exist unique
+
−
−
−
−
+
+
t+
0 and t0 such that t0 uη ∈ Mfη ,hη and t0 uη ∈ Mfη ,hη , and t0 < t0 = 1. Since
d2
If ,h (t+ uη ) > 0,
dt2 η η 0
−
+
e
there exists e
t ∈ (t+
0 , t0 ] such that Ifη ,hη (t0 uη ) < Ifη ,hη (tuη ). By Lemma 2.4,
d
If ,h (t+ uη ) = 0,
dt η η 0
−
e
Ifη ,hη (t+
0 uη ) < Ifη ,hη (tuη ) ≤ Ifη ,hη (t0 uη ) = Ifη ,hη (uη ),
which is a contradiction. Thus, Ifη ,hη (uη ) = αfη ,hη = αf+η ,hη . Since Ifη ,hη (uη ) =
Ifη ,hη (|uη |) and |uη | ∈ Mf+η ,hη , by Lemma 2.3 and the maximum principle, we may
assume that uη is a positive solution of (2.1).
Proposition 3.9. Assume that {un } ⊂ Mf−η ,hη is a (P S)c sequence, where c <
αfη ,hη + αf ∞ ,0 . Then there exists a subsequence, still denoted by {un }, and u0 in
W 1,p (RN ) such that un → u0 strongly in W 1,p (RN ) and Ifη ,hη (u0 ) = c.
Proof. By Lemma 2.5 (ii), there exists a subsequence {un } and u0 in W 1,p (RN )
such that
un * u0 weakly in W 1,p (RN ).
p
First, we claim that u0 ≡ 0 is impossible. If not, by h ∈ L p−r (RN ), the Egorov
theorem and the Hölder inequality, we have
kun kp = o(1).
(3.16)
Moreover, {un } ⊂ Mf−η ,hη , we deduce from the Sobolev imbedding theorem that
kun k > C
for some C > 0, n = 1, 2, . . . .
which contradicts to (3.16). Thus, by Lemma 3.3, u0 is a nontrivial solution of (2.1)
and Ifη ,hη (u0 ) ≥ αfη ,hη . We write un = u0 + vn with vn * 0 weakly in W 1,p (RN ).
By the Brezis-Lieb lemma [16], we have
Z
Z
Z
fη |un |p dx =
fη |u0 |p dx +
fη |vn |p dx + o(1)
RN
RN
RN
Z
Z
p
=
fη |u0 | dx +
f ∞ |vn |p dx + o(1).
RN
RN
20
H. YIN, Z. YANG
EJDE-2014/17
Since {un } is a bounded sequence in W 1,p (RN ), we have {vn } is also a bounded
p
sequence in W 1,p (RN ). Moreover, by h ∈ L p−r (RN ), the Egorov theorem and the
Hölder inequality, we have
Z
Z
Z
r
r
hη |vn | dx =
hη |un | dx −
hη |u0 |r dx + o(1) = o(1).
RN
RN
RN
Hence, for n large enough, we can conclude that
αfη ,hη + αf ∞ ,0 > Ifη ,hη (u0 + vn )
Z
1
1
f ∞ |vn |s dx + o(1)
≥ Ifη ,hη (u0 ) + kvn kp −
p
s RN
Z
1
1
≥ αfη ,hη + kvn kp −
f ∞ |vn |s dx + o(1),
p
s RN
we obtain
Z
1
1
kvn kp −
f ∞ |vn |s dx < αf ∞ ,0 + o(1).
(3.17)
p
s RN
Also from If0 η ,hη (un ) = o(1) in W −1 (RN ), {un } is uniformly bounded and u0 is a
solution of (2.1), we obtain
Z
hIf0 η ,hη (un ), un i = kvn kp −
f ∞ |vn |s dx + o(1) = o(1).
(3.18)
RN
We claim that (3.17) and (3.18) can be hold simultaneously only if {vn } admits a
subsequence which converges strongly to zero. If not, then kvn k is bounded away
from zero; that is,
kvn k ≥ C for some C > 0.
From (3.18), it follows that
Z
sp
f ∞ |vn |s dx ≥
αf ∞ ,0 + o(1).
s
−p
N
R
By (3.17) and (3.18), for n large enough
Z
1 1
αf ∞ ,0 ≤ ( − )
f ∞ |vn |s dx + o(1)
p s RN
Z
1
1
= kvn kp −
f ∞ |vn |s dx + o(1) < αf ∞ ,0 ,
p
s RN
which is a contradiction. Therefore, un → u0 strongly in W 1,p (RN ) and Ifη ,hη (u0 ) =
c.
Proof of Theorem 1.1. By Lemma 3.2, Proposition 3.7 and Proposition 3.9, for
each η ∈ (0, η0 ) and i ∈ {1, 2, . . . , k}, there exist a sequence {uin } ⊂ Nηi and
ui0 ∈ W 1,p (RN )\{0} such that
Ifη ,hη (uin ) = γηi + o(1),
If0 η ,hη (uin ) = o(1)
and uin → ui0 strongly in W 1,p (RN ). Obviously, the function ui0 is a solution of the
equation (2.1) and Ifη ,hη (ui0 ) = γηi . Similar to the argument in Theorem 3.8, we
have ui0 is positive. Since gηi (ui0 ) ∈ Cl/η (xi ), uη ∈ Mf+η ,hη and ui0 ∈ Mf−η ,hη , where
uη is a positive solution of Eq.(2.1) as in Theorem 3.8. This implies uη , ui0 and uj0
are different for i 6= j.
EJDE-2014/17
EXISTENCE OF MULTIPLE SOLUTIONS
1
21
1
Letting λ0 = η0−p , Uλ (x) = λ s−p uη (λ1/p x) and Ui (x) = λ s−p ui0 (λ1/p x). We
obtain Uλ and Ui are positive solutions of the (1.1) with i = 1, 2, . . . , k. This
completes the proof.
Remark 3.10. It is easy to see from the proof of Theorem 1.1 that the solutions
Uλ , Ui (i = 1, 2, . . . , k) satisfy
(1) kUλ kL∞ (RN ) , kUi kL∞ (RN ) → ∞ as λ → ∞;
(2) kUλ kLp (RN ) , kUi kLp (RN ) → ∞ as λ → ∞ if p < s <
(3) kUλ kLp (RN ) , kUi kLp (RN ) → 0 as λ → ∞ if
p2
N
p2
N
+ p;
+ p < s < p∗ .
Lemma 3.11. When 1 ≤ r < p < s < p∗ and N ≥ 1, we have
p(s−r)
s−p
− (p−r)N
> 0.
p
Proof. When N ≤ p and 1 ≤ r < p < s < p∗ , obviously, we have
p(s − r) (p − r)N
−
> 0.
s−p
p
We consider only the case N > p. Set
L(s) = p2 (s − r) − (p − r)N (s − p),
s ∈ (p, p∗ ).
Then it is easy to see that
L(s) ≥ min{L(p), L(p∗ )} = min{p2 (p − r),
This completes the proof.
p3 r
} > 0.
N −p
Acknowledgements. The authors wish to thank the anonymous reviewers and
the editor for their helpful comments. We would like to thank Professot D. M. Cao
and T. F. Wu for their help and advice for completing this article. This research was
supported by the National Natural Science Foundation of China (No. 11171092);
the Natural Science Foundation of the Jiangsu Higher Education Institutions of
China (No. 08KJB110005); the Natural Science Foundation of Jiangsu Education
Office (No. 12KJB110002)
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Honghui Yin
Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China.
School of Mathematical Sciences, Huaiyin Normal University, Jiangsu Huaian 223001,
China
E-mail address: yinhh771109@163.com
Zuodong Yang
Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China.
School of Teacher Education, Nanjing Normal University, Jiangsu Nanjing 210097,
China
E-mail address: zdyang jin@263.net