Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 262, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE GROUND STATE SOLUTION FOR KIRCHHOFF
EQUATIONS WITH SUBCRITICAL GROWTH AND ZERO MASS
YU DUAN, JIU LIU, CHUN-LEI TANG
Abstract. In this article, we study the Kirchhoff equation
Z
“
”
− a+b
|∇u|2 dx ∆u = K(x)f (u), x ∈ RN ,
RN
u ∈ D1,2 (RN ),
where a > 0, b > 0 and N ≥ 3. Under suitable conditions on K and f , we
obtain four existence results and two nonexistence results, using variational
methods.
1. Introduction and statement of main results
We consider the Kirchhoff equation with zero mass
Z
− a+b
|∇u|2 dx ∆u = K(x)f (u),
x ∈ RN ,
RN
u∈D
1,2
(1.1)
N
(R ),
where a > 0, b > 0 and N ≥ 3. The potential function K : RN → R satisfies:
∗
(K1) K be a nonnegative function and K ∈ Lr (RN )\{0}, where r = 2∗2−p ,
∗
2∗ = N2N
−2 and 2 < p < 2 ;
(K2) |(∇K(x), x)| ≤ K(x) for a.e. x ∈ RN , where (·, ·) denotes the scalar product
in RN .
The nonlinear term f ∈ C(R+ , R+ ) satisfies:
(s)
(F1) lim sups→0+ sfp−1
< +∞;
(s)
(F2) lim sups→+∞ sfp−1
< +∞;
f (s)
(F3) lims→+∞ s = +∞.
When Ω ⊂ RN is a smooth bounded domain, the equation
Z
− a+b
|∇u|2 dx ∆u = f (x, u), x ∈ Ω,
Ω
(1.2)
x ∈ ∂Ω.
u = 0,
2010 Mathematics Subject Classification. 35J20, 35J60, 35A01.
Key words and phrases. Kirchhoff equation; subcritical growth; zero mass; Pohozaev identity.
c
2015
Texas State University.
Submitted September 7, 2015. Published October 8, 2015.
1
2
Y. DUAN, J. LIU, C.-L. TANG
is related to the stationary analogue of the Kirchhoff equation
Z
utt − a + b
|∇u|2 dx ∆u = f (x, u)
EJDE-2015/262
(1.3)
Ω
which was proposed by Kirchhoff [11] in 1883 as an extension of the classical
D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model
takes into account the changes in length of the string produced by transverse vibrations. Some early classical studies of Kirchhoff equations were those of Bernstein
[4] and Pohozaev [18]. However, equation (1.3) received great attention only after Lions [16] proposed an abstract framework for the problem. Some interesting
results can be found in [1], [5], [6] and the references therein.
There are many recent articles studying the Kirchhoff equations with subcritical
growth in RN , see for example [7, 8, 9, 12, 13, 15, 17, 20, 21] and so on. But for the
Kirchhoff equations with subcritical growth and zero mass, to our best knowledge,
there are very few results up to now except [2] and [14]. Azzollini [2] obtained the
existence of positive radial solution under K = 1 and f satisfies the BerestyckiLions conditions [3]. In [14], Li et al. obtain a positive solution for b > 0 small
enough when the potential K satisfies
(K3) K ∈ L∞ (RN ),
and other conditions similar to (K1), (K3), and the nonlinear term f satisfies (F3),
and
(F5) lims→+∞ sf2∗(s)
−1 = 0,
f (s)
(F6) lims→0+ s2∗ −1 = 0.
Obviously, condition (F6) is stronger than (F1) and there exist functions which
satisfy (F1), but do not satisfy (F6), such as f (s) = sp−1 . Thus, in the present
paper, we will remove the assumption (F6) to study equation (1.1). Inspired
by
Rs
[14], we will use the monotonicity trick to investigate it. Set F (s) = 0 f (τ )dτ .
Our results read as follows.
Theorem 1.1. Assume that a > 0 and b > 0. Suppose that (F1)–(F3) hold. Then
there exists b0 > 0 such that for any b ∈ (0, b0 ), equation (1.1) has a positive ground
state solution, under one of the following conditions:
(1) N ≥ 4 and (K1) holds,
(2) N = 3, 2 < p < 4 and (K1) holds,
(3) N = 3, 4 ≤ p < 6, (K1) and (K2) hold.
Theorem 1.2. Assume that a > 0, b > 0 and N = 3. Suppose that (K1) with
4 ≤ p < 6, (F1) and (F2) hold. In addition, f satisfies
= +∞ and sf (s) ≥ 4F (s) for all s ≥ 0.
(F4) lims→+∞ Fs(s)
4
Then equation (1.1) has a positive ground state solution.
Theorem 1.3. Assume that a > 0 and b > 0. Suppose that (F1), (F3), (F5) hold.
Then there exists b0 > 0 such that for any b ∈ (0, b0 ), equation (1.1) has a positive
ground state solution, under one of the following conditions:
(1) N ≥ 4, (K1) and (K3) hold,
(2) N = 3, (K1)–(K3) hold.
Theorem 1.4. Assume that a > 0, b > 0 and N = 3. Suppose that (K1) with
4 ≤ p < 6, (K3), (F1), (F4), (F5) hold. Then (1.1) has a positive ground state
solution.
EJDE-2015/262
POSITIVE GROUND STATE SOLUTION
3
Remark 1.5. Just like the example which was given by Li et al. in [14], when
∗
1
p ∈ (max{2, 223 }, 2∗ ), one can easily verify that the function K(x) = 1+|x|
α, α ∈
∗
( 32 2−3p
, 1] satisfies (K1 ) − (K3 ).
∗
For equation (1.1) with b > 0 large enough, we have the following nonexistence
results.
Theorem 1.6. Assume that a > 0, b > 0, N ≥ 3 and f satisfies (F1) and (F2).
Suppose that when N = 3, (K1) with 2 < p < 4 holds and when N ≥ 4, (K1) holds.
Then there exists B > 0 such that for any b > B, equation (1.1) has only zero
solution.
Theorem 1.7. Assume that a > 0, b > 0 and N ≥ 4. Suppose that (K1), (K3),
(F1), (F5) hold. Then there exists B > 0 such that for any b > B, equation (1.1)
has only zero solution.
Theorems 1.1–1.4 and 1.6–1.7 can be seen as an extension of the results in [14].
This article is organized as follows. In Section 2 we give some preliminary knowledge. Section 3 is devoted to the proofs of Theorem 1.1 and 1.2. Finally, in Section
4 we complete the proofs of Theorems 1.3, 1.4, 1.6, and 1.7.
2. Preliminaries
In what follows, we use the following notation.
• E := D1,2 (RN ) is the closure of the compactly supported smooth functions with
respect to the norm
Z
1/2
.
kuk =
|∇u|2 dx
RN
• Ls (RN ) is the usual Lebesgue space endowed with the norm
1/s
Z
|u|s dx
, ∀s ∈ [1, +∞), |u|∞ = ess supx∈RN |u(x)|.
|u|s =
RN
∗
• S denotes the best constant of Sobolev embedding D1,2 (RN ) ,→ L2 (RN ), that
is,
S|u|22∗ ≤ kuk2 , for all u ∈ D1,2 (RN ).
(2.1)
• h·, ·i denotes the dual pairing.
• E ∗ is the dual space of E.
• C, Ci denote various positive constants.
Since we are looking for positive solution, we assume that f (s) = 0 for all s ≤ 0.
By (F1) and (F2), there exists C1 > 0 such that
|f (s)| ≤ C1 |s|p−1 , for all s ∈ R,
C1 p
|F (s)| ≤
|s| , for all s ∈ R.
p
(2.2)
(2.3)
By (F1) and (F5), for any ε > 0, there exists Cε > 0 such that
∗
|f (s)| ≤ ε|s|2 −1 + Cε |s|p−1 , for all s ∈ R,
∗
ε
|F (s)| ≤ ∗ |s|2 + Cε |t|p , for all s ∈ R.
2
(2.4)
(2.5)
4
Y. DUAN, J. LIU, C.-L. TANG
EJDE-2015/262
By f ∈ C(R+ , R+ ) and (F3), for any L > 0, there exists CL > 0 such that
F (s) ≥ L|s|2 − CL , for all s ∈ R+ .
(2.6)
F (s) ≥ L|s|4 − CL , for all s ∈ R+ .
(2.7)
By (F4), one has
The energy functional I : E → R defined by
Z
Z
2 Z
a
b
2
2
I(u) =
|∇u| dx +
|∇u| dx −
K(x)F (u)dx.
2 RN
4 RN
RN
Obvious, I is of class C 1 and has the derivative given by
Z
Z
Z
hI 0 (u), vi = a + b
|∇u|2 dx
(∇u, ∇v)dx −
RN
RN
K(x)f (u)vdx,
RN
for all u, v ∈ E. As well known, the critical point of the functional I is solution of
equation (1.1). For proving our theorems, we need the following proposition.
Proposition 2.1. Let X be a Banach space equipped with a norm k · kX and let
J ⊂ R+ be an interval. We consider a family {Φλ }λ∈J of C 1 -functionals on X of
the form
Φλ (u) = A(u) − λB(u),
∀λ ∈ J,
where B(u) ≥ 0 for all u ∈ X and such that either A(u) → +∞ or B(u) → +∞,
as kukX → +∞. We assume that there are two points v1 , v2 in X such that
cλ = inf max Φλ (γ(t)) > max{Φλ (v1 ), Φλ (v2 )}
γ∈Γ t∈[0,1]
where
Γ = {γ ∈ C([0, 1], X) : γ(0) = v1 , γ(1) = v2 }.
Then for almost every λ ∈ J, there is a bounded (P S)cλ sequence for Φλ , that is,
there exists a sequence {un } ⊂ X such that
(i) {un } is bounded in X,
(ii) Φλ (un ) → cλ ,
(iii) Φ0λ (un ) → 0 in X ∗ , where X ∗ is the dual of X.
Remark 2.2. The above result corresponds to [10, Theorem 1.1] which is reminiscent of Struwe’s monotonicity trick (see [19]) and can be viewed as its generalization.
In [10, Lemma 2.3] it is also proved that under the assumptions of Proposition 2.1,
the map λ → cλ is continuous from the left.
Let X := E, J := [1/2, 1] and Φλ (u) := Iλ (u) = A(u) − λB(u), where
Z
Z
2
b
a
|∇u|2 dx +
|∇u|2 dx ,
A(u) =
2 RN
4 RN
Z
B(u) =
K(x)F (u)dx.
RN
Then I1 (u) = I(u). It is obvious that B(u) ≥ 0 for all u ∈ E and A(u) → +∞ as
kuk → +∞.
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POSITIVE GROUND STATE SOLUTION
5
3. Proofs of Theorem 1.1 and Theorem 1.2
First we give some lemmas.
Lemma 3.1. Assume that a > 0, b > 0 and N ≥ 3. Suppose that (K1), (F1),
(F2) hold. Then there exist ρ > 0 and α > 0 such that Iλ (u)|kuk=ρ ≥ α for all
λ ∈ [1/2, 1].
Proof. By (2.3), the Hölder and Sobolev inequalities, for all u ∈ E and all λ ∈
[1/2, 1], we have
Z
Z
a
Iλ (u) ≥
|∇u|2 dx −
K(x)F (u)dx
2 RN
RN
Z
a
C1
≥ kuk2 −
K(x)|u|p dx
2
p RN
Z
p/2∗
∗
a
C1
2
≥ kuk −
|K|r
|u|2 dx
2
p
RN
≥ C2 kuk2 − C3 kukp
= kuk2 (C2 − C3 kukp−2 ).
1
C2 p−2
Since p > 2, we can choose ρ = 2C
. Then Iλ (u) ≥
3
kuk = ρ. The proof is complete.
C2
2
C2
2C3
2
p−2
:= α for all
For cλ in Proposition 2.1, we have the following lemma.
Lemma 3.2. Assume that a > 0, b > 0 and N ≥ 3. Suppose that (K1) and (F1)–
(F3) hold. Then there exists b0 > 0 such that for any b ∈ (0, b0 ), there are two
points v1 , v2 in E such that cλ > max{Iλ (v1 ), Iλ (v2 )} for all λ ∈ [1/2, 1].
Proof. Choose a nonnegative function ϕ ∈ C0∞ (RN ) such that kϕk = 1, anf
Z
Z
K(x)ϕ2 dx 6= 0,
K(x)dx 6= 0.
supp ϕ
supp ϕ
Let
L= R
2a
2 dx
K(x)ϕ
supp ϕ
in (2.6) and take
t0 =
√
n C Z
1/2 o
L
2 max ρ,
K(x)dx
,
a supp ϕ
where ρ is given by Lemma 3.1. Then kt0 ϕk > ρ and for all λ ∈ [1/2, 1], one has
Z
at2
bt4
Iλ (t0 ϕ) = 0 + 0 − λ
K(x)F (t0 ϕ)dx
2
4
supp ϕ
Z
at20
bt40
1
+
−
K(x)(Lt20 ϕ2 − CL )dx
≤
2
4
2 supp ϕ
Z
Z
at2
bt4
CL
Lt2
= 0+ 0+
K(x)dx − 0
K(x)ϕ2 dx
2
4
2 supp ϕ
2 supp ϕ
Z
bt4
CL
a
= 0+
K(x)dx − t20
4
2 supp ϕ
2
6
Y. DUAN, J. LIU, C.-L. TANG
≤
bt40
CL
−
4
2
EJDE-2015/262
Z
K(x)dx,
supp ϕ
which implies that there exists b0 > 0 such that for any b ∈ (0, b0 ), one has Iλ (t0 ϕ) <
0. Thus letting v1 = 0 and v2 = t0 ϕ, by Lemma 3.1 and definition of cλ , we have
0 < α ≤ c1 ≤ cλ ≤ c 21 < +∞.
and then cλ > Iλ (v1 ) > Iλ (v2 ).
Lemma 3.3. Assume that a > 0, b > 0 and N = 3. Suppose that (K1) with
4 ≤ p < 6, (F1), (F2), (F4) hold. Then there are two points v1 , v2 in E such that
cλ > max{Iλ (v1 ), Iλ (v2 )} for all λ ∈ [1/2, 1].
Proof. Choose a nonnegative function ϕ ∈ C0∞ (R3 ) such that kϕk = 1, and
Z
Z
4
K(x)ϕ dx 6= 0,
K(x)dx 6= 0.
supp ϕ
Let L =
b
R
K(x)ϕ4 dx
supp ϕ
supp ϕ
in (2.7). Then for any t > 0 and for all λ ∈ [1/2, 1], one
has
Iλ (tϕ) =
≤
bt4
at2
+
−λ
2
4
Z
at2
bt4
1
+
−
2
4
2
Z
K(x)F (tϕ)dx
supp ϕ
K(x)(Lt4 ϕ4 − CL )dx
supp ϕ
Z
Z
at2
bt4
CL
Lt4
=
+
+
K(x)dx −
K(x)ϕ4 dx
2
4
2 supp ϕ
2 supp ϕ
Z
at2
CL
bt4
=
+
K(x)dx −
,
2
2 supp ϕ
4
which indicates that there exists t0 > 0 such that kt0 ϕk > ρ and Iλ (t0 ϕ) < 0. Thus
letting v1 = 0 and v2 = t0 ϕ, by Lemma 3.1 and definition of cλ , we have
0 < α ≤ c1 ≤ cλ ≤ c 21 < +∞.
and then cλ > Iλ (v1 ) > Iλ (v2 ).
Lemma 3.4. Assume that {un } is bounded in Ls (RN ), where 1 < s < +∞. Suppose that un (x) → u(x) a.e. in RN . Then up to a subsequence, un * u in Ls (RN ).
s
Proof. Suppose that |un |s ≤ M and v ∈ L s−1 (RN ) is fixed. Then for any ε > 0,
there exists r > 0 such that
Z
s−1
Z
s
s
≤ M ε.
un vdx ≤ |un |s
|v| s−1 dx
|x|≥r
|x|≥r
Similarly, combining the Fatou’s lemma, we have
Z
s
Z
s−1
s
uvdx ≤ |u|ss
|v| s−1 dx
|x|≥r
|x|≥r
Z
Z
s−1
s
=
lim inf |un |s dx
|v| s−1 dx
RN n→∞
|x|≥r
Z
Z
s−1
s
≤ lim inf
|un |s dx
|v| s−1 dx
n→∞
RN
|x|≥r
EJDE-2015/262
POSITIVE GROUND STATE SOLUTION
7
≤ M s εs .
s
Since v ∈ L s−1 (Br (0)) with Br (0) := {x ∈ RN : |x| < r}, there exists δ > 0 such
s−1
R
s
that for any A ⊂ Br (0), when measA < δ, one has A |v| s−1 dx s < ε. Thus for
all n, we have
Z
Z
s−1
s
s
s−1
un vdx ≤ |un |s
|v|
< M ε.
dx
A
A
By Vitali’s theorem, one gets
Z
Z
un vdx =
Br (0)
Hence we have
Z
Z
(un − u)vdx ≤ RN
|x|≥r
uvdx + o(1).
Br (0)
Z
(un − u)vdx + Br (0)
(un − u)vdx ≤ 2M ε + o(1).
By the arbitrariness of ε, we complete the proof.
Lemma 3.5. For any λ ∈ [1/2, 1], if {un } ⊂ E is a bounded and nonnegative
Palais-Smale sequence of the function Iλ , there exists a nonnegative function u ∈ E
such that, up to a subsequence, un → u in E.
Proof. Since {un } is bounded and nonnegative in E, up to a subsequence, there
exists a nonnegative function u ∈ E such that
R un * u in E, un (x) → u(x) a.e. in
RN and there exists d ≥ 0 such that d = RN |∇un |2 dx + o(1). By (2.2) and the
Hölder inequality, one has
Z
Z
2∗ (p−1)
2∗
2∗
|un | p |un − u| p dx
|f (un )(un − u)| p dx ≤ C1
RN
RN
≤ C1
Z
p−1
Z
p
|un | dx
2∗
RN
1/p
∗
|un − u|2 dx
RN
≤ C.
Combining f (un (x))(un (x) − u(x)) → 0 a.e. in RN with Lemma 3.4, up to a
2∗
subsequence, we get f (un )(un − u) * 0 in L p (RN ). Since K ∈ Lr (RN ), we have
Z
K(x)f (un )(un − u)dx = o(1).
RN
Thus by Iλ0 (un ) → 0 in E ∗ , one has
0 = hIλ0 (un ), un − ui + o(1)
Z
Z
Z
2
=a
(∇un , ∇(un − u))dx + b
|∇un | dx
(∇un , ∇(un − u))dx
RN
RN
RN
Z
−λ
K(x)f (un )(un − u)dx + o(1)
=
RN
a(kun k2
− kuk2 ) + bd(kun k2 − kuk2 ) + o(1),
which implies kun k → kuk. Combining un * u in E, we get un → u in E. The
proof is complete.
8
Y. DUAN, J. LIU, C.-L. TANG
EJDE-2015/262
Remark 3.6. According to Proposition 2.1 and Lemma 3.2, for almost every λ ∈
[1/2, 1], there exists a bounded sequence {un } ⊂ E such that Iλ (un ) → cλ and
Iλ0 (un ) → 0 in E ∗ . Define u± = max{±u, 0}, then
o(1) = hIλ0 (un ), u−
ni
Z
Z
Z
2
−
= a+b
|∇un | dx
(∇un , ∇un )dx − λ
K(x)f (un )u−
n dx
RN
RN
RN
Z
Z
2
=− a+b
|∇un |2 dx
|∇u−
n | dx,
RN
RN
which implies u−
n → 0 in E. Thus one has
cλ = Iλ (un ) + o(1)
Z
2
Z
Z
b
a
2
2
|∇un | dx +
|∇un | dx − λ
K(x)F (un )dx + o(1)
=
2 RN
4
RN
RN
Z
Z
a
2
2
|∇u+
|∇u−
=
n | dx +
n | dx
2 RN
RN
Z
Z
Z
2
b
2
− 2
+
|∇u+
|
dx
+
|∇u
|
dx
−
λ
K(x)F (u+
n
n
n )dx + o(1)
4 RN
N
N
R
R
Z
2
Z
Z
b
a
+ 2
+ 2
|∇un | dx +
=
|∇un | dx − λ
K(x)F (u+
n )dx + o(1)
2 RN
4
RN
RN
= Iλ (u+
n ) + o(1)
and
0 = hIλ0 (un ), ϕi + o(1)
Z
Z
Z
(∇un , ∇ϕ)dx − λ
K(x)f (un )ϕdx + o(1)
= a+b
|∇un |2 dx
RN
RN
RN
Z
Z
Z
2
− 2
= a+b
|∇u+
|
dx
+
b
|∇u
|
dx
(∇u+
n
n
n , ∇ϕ)dx
RN
RN
RN
Z
Z
−
(∇u−
K(x)f (u+
n , ∇ϕ)dx − λ
n )ϕdx + o(1)
RN
RN
Z
Z
Z
2
+
= a+b
|∇u+
|
dx
(∇u
,
∇ϕ)dx
−
λ
K(x)f (u+
n
n
n )ϕdx + o(1)
=
RN
0
+
hIλ (un ), ϕi
RN
RN
+ o(1),
uniformly for all ϕ ∈ E and kϕk = 1. That is, {u+
n } is a bounded Palais-Smale
sequence of Iλ . By Lemma 3.5, there exists a nonnegative function u ∈ E such
0
∗
that, up to a subsequence, u+
n → u in E. Thus Iλ (u) = cλ and Iλ (u) = 0 in E .
Proof of Theorem 1.1. Set λj ∈ [1/2, 1] and λj → 1− . Then there exists a sequence
nonnegative functions uj ∈ E such that Iλj (uj ) = cλj and Iλ0 j (uj ) = 0. If N ≥ 4,
EJDE-2015/262
POSITIVE GROUND STATE SOLUTION
2 < p < 2∗ ≤ 4. By (K1), (2.2) and hIλ0 j (uj ), uj i = 0, we have
Z
akuj k2 + bkuj k4 = λj
K(x)f (uj )uj dx
N
ZR
≤ C1
K(x)|uj |p dx
RN
9
(3.1)
≤ C1 |K|r |uj |p2∗
≤ Ckuj kp ,
which implies kuj k ≤ C. If N = 3 and 2 < p < 4, by (3.1) with N = 3, we have
kuj k ≤ C. If N = 3 and 4 ≤ p < 6, by (K1), (K2), Iλ0 j (uj ) = 0, one has the
following Pohozaev equality (see [14, Lemma 2.2])
Z
Z
a
b
2
4
kuj k + kuj k = 3λj
K(x)F (uj )dx + λj
(∇K(x), x)F (uj )dx.
2
2
R3
R3
Combining (K2) with Iλj (uj ) = cλj , we have
Z
Z
b
a
kuj k2 + kuj k4 = 3λj
K(x)F (uj )dx + λj
(∇K(x), x)F (uj )dx
2
2
3
R3
ZR
≥ 2λj
K(x)F (uj )dx
R3
b
= akuj k2 + kuj k4 − 2cλj ,
2
which implies that
a
kuj k2 .
2
Hence kuj k ≤ C. Since I(uj ) = Iλj (uj ) + o(1) = cλj + o(1) = c + o(1) and
I 0 (uj ) = Iλ0 j (uj ) + o(1) = o(1) in E ∗ , according to Lemma 3.5, there exists a
nonnegative function u ∈ E such that uj → u in E. Thereby I(uj ) → I(u) = c
and I 0 (uj ) → I 0 (u) = 0 in E ∗ . That is, u is a nonnegative solution of equation
(1.1). To obtain the ground state solution, we set π = inf u∈Π I(u), where Π =
{u ∈ E\{0}|I 0 (u) = 0, u ≥ 0} and then π ≤ c. Obviously, π > −∞. Since
Π 6= ∅, there exist a nonnegative sequence un ∈ E such that I 0 (un ) = 0 and
I(un ) → π. With the same method, we can obtain {un } is bounded in E and
then there exists a nonnegative function u ∈ E such that un → u in E. Hence we
have I(un ) → I(u) = π, I 0 (un ) → I 0 (u) = 0. Because of the strongly maximum
principle, we know u > 0. So we complete the proof of Theorem 1.1.
2cλj ≥
Proof of Theorem 1.2. By Lemma 3.1, Lemma 3.3 and the mountain pass theorem,
there exists a sequence {un } ⊂ E such that I(un ) → c and I 0 (un ) → 0 in E ∗ . By
(F4), for n large enough, we have
1
c + 1 + kun k = I(un ) − hI 0 (un ), un i
4Z
a
1
= kun k2 +
K(x)[ f (un )un − F (un )]dx
4
4
R3
a
2
≥ kun k .
4
Thus {un } is bounded in E. By Remark 3.6, {u+
n } is a bounded sequence satisfying
0 +
∗
I(u+
)
→
c
and
I
(u
)
→
0
in
E
.
By
Lemma
3.5, there exists a nonnegative
n
n
10
Y. DUAN, J. LIU, C.-L. TANG
EJDE-2015/262
function u ∈ E such that, up to a subsequence, u+
n → u in E. Thus I(u) = c
and I 0 (u) = 0 in E ∗ . The rest of proof is the same as Theorem 1.1. The proof is
complete.
4. Proofs of Theorem 1.3, 1.4, 1.6 and 1.7
We establish parallel steps as Lemmas 3.1, 3.2, 3.3 and 3.5.
Lemma 4.1. Assume that a > 0, b > 0 and N ≥ 3. Suppose that (K1), (K3),
(F1), (F5) hold. Then there exist ρ > 0 and α > 0 such that Iλ (u)|kuk=ρ ≥ α for
all λ ∈ [1/2, 1].
Proof. By (2.5) with ε = 1, the Hölder and Sobolev inequalities, for all u ∈ E and
all λ ∈ [1/2, 1], we have
Z
Z
a
|∇u|2 dx −
K(x)F (u)dx
Iλ (u) ≥
2 RN
RN
Z
Z
∗
a
1
≥ kuk2 − ∗
K(x)|u|2 dx − C
K(x)|u|p dx
2
2 RN
RN
Z
∗
1
a
2
≥ kuk − ∗ |K|∞
|u|2 dx − C|K|r |u|p2∗
2
2
RN
∗
≥ C2 kuk2 − C3 kuk2 − C4 kukp
∗
= kuk2 (C2 − C3 kuk2
−2
− C4 kukp−2 ).
∗
Since p > 2, we can choose ρ > 0 small enough such that C2 −C3 ρ2 −2 −C4 ρp−2 > 0.
Then there exists α > 0 such that Iλ (u) ≥ α for all kuk = ρ. The proof is
complete.
The proofs of the following two lemmas are the same as Lemma 3.2 and 3.3.
Lemma 4.2. Assume that a > 0, b > 0 and N ≥ 3. Suppose that (K1), (K3),
(F1), (F3), (F5) hold. Then there exists b0 > 0 such that for any b ∈ (0, b0 ), there
are two points v1 , v2 in E such that cλ > max{Iλ (v1 ), Iλ (v2 )} for all λ ∈ [1/2, 1].
Lemma 4.3. Assume that a > 0, b > 0 and N = 3. Suppose that (K1) with
4 ≤ p < 6, (K3), (F1), (F4), (F5) hold. Then there are two points v1 , v2 in E such
that cλ > max{Iλ (v1 ), Iλ (v2 )} for all λ ∈ [1/2, 1].
Lemma 4.4. For any λ ∈ [1/2, 1], if {un } ⊂ E is a bounded and nonnegative
Palais-Smale sequence of the function Iλ , there exists a nonnegative function u ∈ E
such that, up to a subsequence, un → u in E.
Proof. Since {un } is bounded and nonnegative in E, up to a subsequence, there
exists a nonnegative function u ∈ E suchRthat un * u in E, un (x) → u(x) a.e. in
RN and there exists d ≥ 0 such that d = RN |∇un |2 dx + o(1). Since {|un |p−1 |un −
2∗
u|} is bounded in L p (RN ) and |un (x)|p−1 |un (x) − u(x)| → 0 a.e. in RN , by
2∗
Lemma 3.4, up to a subsequence, we have |un |p−1 |un − u| * 0 in L p (RN ). In view
of K ∈ Lr (RN ), one has
Z
K(x)|un |p−1 |un − u|dx = o(1).
RN
EJDE-2015/262
POSITIVE GROUND STATE SOLUTION
Combining this with (2.4), we have
Z
K(x)f (un )(un − u)dx
N
R
Z
Z
∗
≤ε
K(x)|un |2 −1 |un − u|dx + Cε
RN
11
K(x)|un |p−1 |un − u|dx
RN
∗
≤ ε|K|∞ |un |22∗ −1 |un − u|2∗ + o(1) = Cε + o(1).
Hence by Iλ0 (un ) → 0 in E ∗ , we deduce that
0 = hIλ0 (un ), un − ui + o(1)
Z
Z
Z
2
=a
(∇un , ∇(un − u))dx + b
|∇un | dx
(∇un , ∇(un − u))dx
RN
RN
RN
Z
−λ
K(x)f (un )(un − u)dx + o(1)
=
RN
a(kun k2
− kuk2 ) + bd(kun k2 − kuk2 ) + o(1),
which implies kun k → kuk. Combining un * u in E, we get un → u in E. The
proof is complete.
According to Proposition 2.1 and Lemma 4.2, for almost every λ ∈ [1/2, 1], there
exists a bounded sequence {un } ⊂ E such that Iλ (un ) → cλ and Iλ0 (un ) → 0 in E ∗ .
By Remark 3.6, we can assume that un is nonnegative. Thus from Lemma 4.4, we
know that there exists a nonnegative function u ∈ E such that un → u in E and
then for almost every λ ∈ [1/2, 1], Iλ (u) = cλ and Iλ0 (u) = 0.
Proof of Theorem 1.3. Set λj ∈ [1/2, 1] and λj → 1− . T hen there exists a nonnegative sequence {uj } ⊂ E such that Iλj (uj ) = cλj and Iλ0 j (uj ) = 0. If N ≥ 4,
2 < p < 2∗ ≤ 4. By (K1), (K3), (2.4) and hIλ0 j (uj ), uj i = 0, we have
Z
2
4
akuj k + bkuj k = λj
K(x)f (uj )uj dx
N
Z R
Z
∗
≤ε
K(x)|uj |2 dx + Cε
K(x)|uj |p dx
N
N
R
R
Z
2∗
≤ ε|K|∞
|uj | dx + Cε |K|r |uj |p2∗
≤ Cεkuj k
RN
2∗
+ Cε kuj kp ,
b
. If N = 3, by (K1), (K2), Iλ0 j (uj ) = 0, one has
which implies kuj k ≤ C, for ε = 2C
the Pohozaev equality
Z
Z
a
b
2
4
kuj k + kuj k = 3λj
K(x)F (uj )dx + λj
(∇K(x), x)F (uj )dx.
2
2
R3
R3
Combining Iλj (uj ) = cλj , we have
Z
Z
a
b
kuj k2 + kuj k4 = 3λj
K(x)F (uj )dx + λj
(∇K(x), x)F (uj )dx
2
2
3
R3
ZR
≥ 2λj
K(x)F (uj )dx
R3
b
= akuj k2 + kuj k4 − 2cλj
2
12
Y. DUAN, J. LIU, C.-L. TANG
EJDE-2015/262
which implies
a
kuj k2 .
2
Hence kuj k ≤ C. The rest of the proof is similar with Theorem 1.1.
2cλj ≥
The proof of Theorem 1.4 is same as that of Theorem 1.2 and it is omitted.
Proof of Theorem 1.6. Suppose that u ∈ E is a nonzero solution of (1.1). Then
combining (2.1), (2.2), the Hölder and Young inequalities, we have
Z
akuk2 + bkuk4 =
K(x)f (u)udx
RN
Z
≤ C1
K(x)|u|p dx
RN
≤ C1 |K|r |u|p2∗
p
≤ C1 S − 2 |K|r kukp
p
4 − p 4−p
2a 4−p
2
2
=
kuk4−p C1 S − 2 |K|r
kuk2p−4
4−p
2a
2
p−2
p
p − 2
4 − p 4−p
2
C1 S − 2 |K|r
≤ akuk2 +
kuk4
2
2a
< akuk2 + bkuk4 ,
for any b > B :=
is complete.
2
4−p 4−p p−2
p−2
−p
2 |K| (
,
r 2a ) 2 ]
2 [C1 S
which is a contradiction. The proof
Proof of Theorem 1.7. Suppose that u ∈ E is a nonzero solution of (1.1). Then
combining (2.1), (2.4) and Hölder’s inequality, we have
Z
2
4
akuk + bkuk =
K(x)f (u)udx
RN
Z
Z
∗
(4.1)
≤ε
K(x)|u|2 dx + Cε
K(x)|u|p dx
RN
RN
∗
≤ ε|K|∞ S
− 22
When N = 4, 2∗ = 4. Choose ε =
2∗
kuk
p
+ Cε |K|r S − 2 kukp .
b
2|K|∞ S −
2∗
2
. Using (4.1) and the Young
inequality, one has
p
b
akuk2 + kuk4 ≤ C|K|r S − 2 kukp
2
p
2a 4−p
4 − p 4−p
2
2
=
kuk4−p CS − 2 |K|r
kuk2p−4
4−p
2a
2
p−2
p − 2 −p
4 − p 4−p
2
CS 2 |K|r
kuk4
≤ akuk2 +
2
2a
b
< akuk2 + kuk4 ,
2
2
4−p
p−2
p
2
for any b > B := (p − 2) CS − 2 |K|r 4−p
, which is a contradiction. When
2a
N > 4, 2 < 2∗ < 4. Choose ε = 1. Using (4.1) and the Young inequality, one has
akuk2 + bkuk4 ≤ |K|∞ S −
2∗
2
∗
p
kuk2 + C|K|r S − 2 kukp
EJDE-2015/262
POSITIVE GROUND STATE SOLUTION
13
∗
∗ 4−2∗
∗
∗
a 4−2
4−2∗
− 22 4 − 2
2
2
kuk
|K|
S
kuk22 −4
∞
∗
4−2
a
a 4−p
4 − p 4−p
4−p
−p
2
2
+
kuk CS 2 |K|r
kuk2p−4
4−p
a
∗
2∗2−2
2∗
2∗ − 2 4 − 2∗ 4−2
2
≤ akuk2 +
kuk4
|K|∞ S − 2
2
a
2
p−2
p − 2 −p
4 − p 4−p
2
+
CS 2 |K|r
kuk4
2
a
< akuk2 + bkuk4 ,
=
∗
−p
∗
∗ 4−2 ∗2
2∗
2
2 −2
for any b > B := 2 2−2 |K|∞ S − 2 4−2
+ p−2
CS 2 |K|r
a
2
which is a contradiction. The proof is complete.
4−p
a
2
p−2
4−p
2
,
Acknowledgments. This research was supported by the National Natural Science
Foundation of China (No. 11471267), and by the Fundamental Research Funds for
the Central Universities (No. XDJK2013C006).
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Yu Duan
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China.
College of Science, Guizhou University of Engineering Science, Bijie, Guizhou 551700,
China
E-mail address: duanyu3612@163.com
Jiu Liu
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: jiuliu2011@163.com
Chun-Lei Tang (corresponding author)
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: tangcl@swu.edu.cn, Pone +86 23 68253135, fax +86 23 68253135