Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 280, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF POSITIVE SOLUTIONS FOR
KIRCHHOFF PROBLEMS
JIA-FENG LIAO, PENG ZHANG, XING-PING WU
Abstract. We study problems for the Kirchhoff equation
Z
“
”
|∇u|2 dx ∆u = νu3 + Q(x)uq , in Ω,
− a+b
Ω
u = 0,
on ∂Ω,
where Ω ⊂ R3 is a bounded domain, a, b ≥ 0 and a + b > 0, ν > 0, 3 < q ≤ 5
and Q(x) > 0 in Ω. By the mountain pass lemma, the existence of positive
solutions is obtained. Particularly, we give a condition of Q to ensure the
existence of solutions for the case of q = 5.
1. Introduction and main results
In this article, we consider the Kirchhoff type problem
Z
− a+b
|∇u|2 dx ∆u = νu3 + Q(x)uq ,
in Ω,
Ω
u = 0,
(1.1)
on ∂Ω,
where Ω ⊂ R3 is a bounded domain, a, b ≥ 0 and a + b > 0, ν > 0, 3 < q ≤ 5
are four parameters. The coefficient function Q is a positive function in Ω. When
a = 0, b > 0, problem (1.1) is called degenerate, and the case of a, b > 0 is called
non-degenerate.
When a ≥ 0 and b > 0, problem (1.1) is called the Kirchhoff type problem.
Kirchhoff type problems
R are often referred to as being nonlocal because of the
presence of the term ( Ω |∇u|2 dx)∆u which implies that the equation in (1.1) is
no longer a pointwise equation. The existence and multiplicity of solutions for the
problem
Z
− a+b
|∇u|2 dx ∆u = f (x, u), in Ω,
(1.2)
Ω
u = 0, on ∂Ω,
on a smooth bounded domain Ω ⊂ R3 and f : Ω × R → R a continuous function,
has been extensively studied (see [1, 3],[7]-[23], [25]-[28],[30, 31]).
2010 Mathematics Subject Classification. 35B09, 35B33, 35J20.
Key words and phrases. Kirchhoff type equation; resonance; positive solution;
mountain pass lemma.
c
2015
Texas State University.
Submitted June 25, 2015. Published November 10, 2015.
1
2
J.-F. LIAO, P. ZHANG, X.-P. WU
EJDE-2015/280
Particularly, Sun and Tang [26] considered the problem
Z
− a+b
|∇u|2 dx ∆u = λu3 + g(u) − h(x),
in Ω,
Ω
u = 0,
(1.3)
on ∂Ω,
where h ∈ L2 (Ω) and g ∈ C(R, R) satisfies
lim
|t|→∞
g(t)
= 0.
t3
(1.4)
Under a Landesman-Lazer type condition, by the minimax methods, they obtained
the existence of solutions for problem (1.3).
When a = 1 and b = 0, problem (1.1) reduces to the semilinear elliptic problem
−∆u = νu3 + λuq ,
u = 0,
in Ω,
(1.5)
on ∂Ω.
Obviously when 3 < q < 5, problem (1.5) has a positive solution for all ν, λ > 0.
While for q = 5, λ = 1, Brézis and Nirenberg [6] studied problem (1.5). By the
variant of the mountain pass theorem of Ambrosetti and Rabinowitz without the
(PS) condition, they obtained that there exists ν0 > 0 such that problem (1.5) has
a positive solution for each ν ≥ ν0 .
For u ∈ H01 (Ω), we define
Z
Z
Z
2 ν Z
b
1
a
2
2
4
|∇u| dx +
|∇u| dx −
|u| dx −
Q(x)|u|q+1 dx,
I(u) =
2 Ω
4 Ω
4 Ω
q+1 Ω
1/2
R
.
where H01 (Ω) is a Sobolev space equipped with the norm kuk = Ω |∇u|2 dx
Note that a function u is called a weak solution of (1.1) if u ∈ H01 (Ω) such that
Z
Z
Z
Z
3
2
(∇u, ∇ϕ)dx − ν
u ϕdx −
Q(x)uq ϕdx = 0,
(1.6)
a+b
|∇u| dx
Ω
Ω
Ω
Ω
for all ϕ ∈ H01 (Ω).
We denote by ν1 is the first eigenfunction of the eigenvalue problem
Z
|∇u|2 dx ∆u = νu3 , x ∈ Ω,
−
Ω
u = 0,
x ∈ ∂Ω.
From [23], we know that ν1 > 0. Let S be the best Sobolev constant, namely
R
|∇u|2 dx
Ω
S :=
inf
(1.7)
1/3 .
R
u∈H01 (Ω)\{0}
|u|6 dx
Ω
As well known, the function
U (x) =
(3ε2 )1/4
(ε2 + |x|2 )
1/2
,
x ∈ R3 ,
(1.8)
is an extremal function for the minimum problem (1.7); that is, it is a positive
solution of the problem
−∆u = u5 , ∀x ∈ RN .
Problem (1.1) with 3 < q ≤ 5 does not satisfy condition (1.4). It is natural to ask
whether (1.1) has a positive solution. Using the mountain pass theorem, we study
(1.1) and give a positive answer. It is worth pointing out that the result of the case
EJDE-2015/280
KIRCHHOFF PROBLEMS
3
of q = 5 is more meaningful. Because of q = 5 is critical case and the coefficient
of critical term is no longer constant. We give the suitable condition (A1) in the
Theorem 1.3 below to ensure the existence of solutions to problem (1.1).
Our main results are described as follows.
6
Theorem 1.1. Assume a, b > 0, 3 < q < 5 and Q ∈ L 5−q (Ω) is a positive function,
then (1.1) possesses a positive solution u∗ for all ν > 0, and I(u∗ ) > 0.
Remark 1.2. Obviously, Theorem 1.1 does not apply to (1.4). For all ν > 0,
we obtain the existence of positive solutions for problem (1.1). For the degenerate
case, that is a = 0, b > 0, we can also obtain that problem (1.1) possesses a positive
solution for all 0 < ν < bν1 .
Theorem 1.3. Assume a, b > 0, q = 5, Q ∈ C(Ω) is a positive function and
satisfies the assumption
(A1) There exists x0 ∈ Ω such that Q(x0 ) = QM = maxx∈Ω Q(x) and
Q(x) − Q(x0 ) = o(|x − x0 |),
as x → x0 .
∗
Then there exists ν > 0 such that (1.1) possesses a positive solution u∗ for all
ν > ν ∗ , and I(u∗ ) > 0.
Remark 1.4. This case is the critical exponent problem, and Theorem 1.3 does not
apply to (1.4). When Q(x) ≡ 1, the Kirchhoff type problems with critical exponent
have been considered by several papers, such as [1, 9, 12] [20]-[22], [27, 28, 30].
Particularly, problem (1.1) with Q(x) ≡ 1 was been considered in [21]. However,
there exists a flaw in the proof of [21, Theorem 1.3] with the case θ = 4.
To our best knowledge, problem (1.1) with Q(x) not constant has not been considered yet. When Q(x) is not constant, the analysis of the compactness becomes
complicated, which results in much difficulty. It is worth pointing out that (A1) ensures the existence of solutions. Obviously, Theorem 1.3 extends the corresponding
result of [21].
This article is organized as follows. In Section 2, we consider the case of 3 < q < 5
and prove Theorem 1.1 by the variational methods. We study the critical case of
problem (1.1) with q = 5 and give the proof of Theorem 1.3 in Section 3.
2. The case 3 < q < 5
6
In this section, suppose that Q ∈ L 5−q (Ω) is a positive function and 3 < q < 5.
We will prove Theorem 1.1 by the mountain pass theorem. Before proving Theorem
1.1, we give the following lemma.
6
Lemma 2.1. Assume a, b > 0, 3 < q < 5 and Q ∈ L 5−q (Ω) is a positive function,
then the functional I satisfies the (PS)c condition for all ν > 0.
Proof. Suppose that {un } is a (PS)c sequence of I, that is,
I(un ) → c,
I 0 (un ) → 0,
H01 (Ω).
(2.1)
as n → +∞. We claim that {un } is bounded in
In fact, from (2.1) one has
1
1 + c + o(1)kun k ≥ I(un ) − hI 0 (un ), un i
4
Z
1
a
1 2
−
= kun k +
Q(x)|un |q+1 dx
4
4 q+1 Ω
4
J.-F. LIAO, P. ZHANG, X.-P. WU
≥
EJDE-2015/280
a
kun k2 .
4
Hence, we conclude that {un } is bounded in H01 (Ω). Going if necessary to a subsequence, still denoted by {un }, there exists u ∈ H01 (Ω) such that
weakly in H01 (Ω),
un * u,
un → u,
strongly in Ls (Ω), 1 ≤ s < 6,
un (x) → u(x),
(2.2)
a.e. in Ω,
as n → ∞. Now, we only need to prove that un → u as n → ∞ in H01 (Ω). As
usually, letting wn = un − u, we need prove that kwn k → 0 as n → ∞. By the
Vitali theorem (see [24, p.133]), we claim that
Z
Z
lim
Q(x)|un |q+1 dx =
Q(x)|u|q+1 dx.
(2.3)
n→∞
Ω
Ω
Indeed, we only need to prove that { Ω Q(x)|un |p+1 dx, n ∈ N } is equi-absolutelycontinuous. Note that {un } is bounded in H01 (Ω), by the Sobolev embedding theorem, then exists a constant C > 0 such that |un |6 ≤ C < ∞. From the Hölder
inequality, for every ε > 0, setting δ > 0, when E ⊂ Ω with meas E < δ, we have
Z
5−q
Z
6
6
q+1
q+1
< ε,
Q(x)|un | dx ≤ |un |6
Q 5−q (x)dx
R
E
E
R
6
where the last inequality is from the absolutely-continuity of Ω Q 5−q (x)dx. Thus,
our claim is proved. Moreover, one also has
Z
Z
Z
2
2
|∇un | dx =
|∇wn | dx +
|∇u|2 dx + o(1),
(2.4)
Ω
Ω
Ω
Z
2
|∇un |2 dx = kwn k4 + kuk4 + 2kwn k2 kuk2 + o(1).
(2.5)
Ω
0
Since I (un ) → 0, one obtains
akun k2 + bkun k4 − ν
Z
|un |4 dx −
Z
Ω
Q(x)|un |q+1 dx = o(1),
Ω
consequently, from (2.2)-(2.5), we deduce that
2
2
4
2
2
akwn k + akuk + bkwn k + 2bkwn k kuk + bkuk
4
− ν|u|44 −
Z
Q(x)|u|q+1 dx = o(1).
Ω
(2.6)
From (2.1) it follows that
lim hI 0 (un ), ui = akuk2 + bl2 kuk2 + bkuk4 − ν|u|44 −
n→∞
Z
Q(x)|u|q+1 dx = 0, (2.7)
Ω
where l = limn→∞ kwn k. According to (2.6) and (2.7), we have
akwn k2 + bkwn k4 + bkwn k2 kuk2 = o(1),
consequently, one has al2 + bl4 + bl2 kuk2 = 0. Thus l = 0; that is, un → u as
n → ∞ in H01 (Ω). This completes the proof.
EJDE-2015/280
KIRCHHOFF PROBLEMS
5
Proof of Theorem 1.1. The main idea is to construct a suitable geometry of mountain pass lemma (see[2]). Then obtain a critical point of I in H01 (Ω). We claim
that I has the geometry of mountain pass lemma in H01 (Ω). Indeed, since
Z
Z
a
b
ν
1
I(u) = kuk2 + kuk4 −
|u|4 dx −
Q(x)|u|q+1 dx,
2
4
4 Ω
q+1 Ω
then I(0) = 0, and for every u ∈ H01 (Ω)\{0} one has
Z
a
1
I(tu)
I(tu)
2
= kuk ,
=−
lim
Q(x)|u|q+1 dx.
lim
t→+∞ tq+1
t→0+ t2
2
q+1 Ω
R
Since a > 0 and Ω Q(x)|u|q+1 dx > 0, then there exist R, α > 0 and e ∈ H01 (Ω) with
kek > R such that I|∂BR ≥ α and I(e) < 0, where ∂BR = {u ∈ H01 (Ω) | kuk = R}.
Thus, I satisfies the geometry of the mountain-pass lemma.
Let
c = inf max I(γ(t)),
γ∈Γ t∈[0,1]
where Γ = {γ ∈ C([0, 1], H01 (Ω)) : γ(0) = 0, γ(1) = e}. Then c ≥ α. According to
Lemma 2.1, I satisfies the conditions of the mountain pass lemma. Applying the
mountain-pass lema, there exists a sequence {un } ⊂ H01 (Ω), such that I(un ) → c
and I 0 (un ) → 0 as n → ∞. Then c is a critical value of I and c > α > 0. Moreover,
{un } ⊂ H01 (Ω) has a convergent subsequence, still denoted by {un }, we may assume
that un → u∗ in H01 (Ω) as n → ∞. Thus I(u∗ ) = c > 0 and u∗ is a nonzero solution
of (1.1). Since I(|u|) = I(u), by a result due to Brézis and Nirenberg [4, Theorem
10], we conclude that u∗ ≥ 0. By the strong maximum principle, one has u∗ > 0
in Ω. Therefore, u∗ is a positive solution of problem (1.1) with I(u∗ ) > 0. This
completes the proof.
3. The case q=5
In this part, assume that Q ∈ C(Ω) is a positive function and satisfies (A1).
We study the case of q = 5. This case is more delicate, because of the Sobolev
embedding H01 (Ω) ,→ L6 (Ω) is not compact. Thus the functional I does not satisfy
the (P S)c condition. When Q(x) is not constant, the analysis of (P S) sequences
becomes complicated, which results in much difficulty. We will complete the proof
of Theorem 1.3 by the mountain pass lemma. Now, we prove that I satisfies the
local (P S)c condition.
Lemma 3.1. Assume a, b > 0 and the positive function Q ∈ C(Ω) satisfies (A1),
then I satisfies the (P S)c condition, where c ∈ (0, Λ) with
p
p
abS 3
b3 S 6
aS b2 S 4 + 4aSQM
b2 S 4 b2 S 4 + 4aSQM
Λ=
+
+
+
.
4QM
24Q2M
6QM
24Q2M
Proof. Suppose that {un } is a (P S)c sequence for c ∈ (0, Λ); that is,
I(un ) → c,
I 0 (un ) → 0,
(3.1)
as n → +∞. According to Lemma 2.1, we can easy obtain that {un } is bounded
in H01 (Ω). Going if necessary to a subsequence, there exists u ∈ H01 (Ω) such that
(2.2) holds. As usually, letting wn = un − u, we need prove that kwn k → 0 as
6
J.-F. LIAO, P. ZHANG, X.-P. WU
EJDE-2015/280
n → ∞. We denote limn→∞ kwn k = l. As in Lemma 2.1, we have (2.4) and (2.5).
By Brézis-Lieb’s Lemma [5], one has
Z
Z
Z
Q(x)|un |6 dx =
Q(x)|wn |6 dx +
Q(x)|u|6 dx + o(1).
(3.2)
Ω
Ω
Ω
From (3.1) and (2.2), one obtains
akun k2 + bkun k4 − ν
Z
Z
|u|4 dx −
Ω
Q(x)|un |6 dx = o(1),
Ω
consequently, from (2.4)-(2.5) and (3.2) it follows that
akuk2 + akwn k2 + bkuk4 + bkwn k4 + 2bkwn k2 kuk2
Z
Z
Z
6
6
−
Q(x)|wn | dx −
Q(x)|u| dx − ν
|u|4 dx = o(1).
Ω
Ω
(3.3)
Ω
From (3.1) it follows that
0
2
4
2
2
Z
lim hI (un ), ui = akuk +bkuk +bl kuk −
n→∞
6
Z
Q(x)|u| dx−ν
Ω
|u|4 dx = 0. (3.4)
Ω
On the one hand, from (3.4), we have
Z
Z
b
ν
1
a
2
4
4
|u| dx −
Q(x)|u|6 dx
I(u) = kuk + kuk −
2
4
4 Ω
6 Ω
Z
a
1
bl2
2
= kuk +
Q(x)|u|6 dx −
kuk2
4
12 Ω
4
bl2
≥ − kuk2 .
4
On the other hand, from (3.3) and (3.4) it follows that
Z
akwn k2 + bkwn k4 + bkwn k2 kuk2 −
Q(x)|wn |6 dx = o(1),
(3.5)
(3.6)
Ω
and
b
b
1
a
I(un ) = I(u) + kwn k2 + kwn k4 + kwn k2 kuk2 −
2
4
2
6
Z
Q(x)|wn |6 dx + o(1). (3.7)
Ω
From (A1) and (1.7), one has
Z
Z
kwn k6
6
Q(x)|wn | dx ≤ QM
|wn |6 dx ≤ QM
,
S3
Ω
Ω
consequently, it follows from (3.6) that
al2 + bl4 + bl2 kuk2 ≤ QM
l6
,
S3
which implies that
1 h bS 3
l ≥
+
2 QM
2
p
b2 S 6 + 4S 3 QM (a + bkuk2 ) i
.
QM
(3.8)
Thus, from (3.6)-(3.8), we obtain
Z
h
i
a
b
b
1
2
4
2
2
I(u) = lim I(un ) − kwn k − kwn k − kwn k kuk +
Q(x)|wn |6 dx
n→∞
2
4
2
6 Ω
a
b
b
=c−
l2 + l4 + l2 kuk2
3
12
3
EJDE-2015/280
KIRCHHOFF PROBLEMS
h a bS 3
7
p
b2 S 6 + 4S 3 QM (a + bkuk2 ) +
6 QM
QM
p
3
2
6
3
b S + 4S QM (a + bkuk2 ) 2
b bS
+
+
48 QM
QM
p
b2 S 6 + 4S 3 QM (a + bkuk2 ) i bl2
bkuk2 bS 3
−
+
+
kuk2
24
QM
QM
4
p
p
abS 3
aS b2 S 4 + 4aSQM
b3 S 6
b2 S 4 b2 S 4 + 4aSQM +
≤c−
+
+
4QM
24Q2M
6QM
24Q2M
≤c−
bl2
kuk2
4
bl2
< − kuk2 ,
4
which contradicts (3.5). Hence, l ≡ 0; that is, un → u in H01 (Ω) as n → ∞.
Therefore, I satisfies the (P S)c condition for all c < Λ. This completes the proof.
−
Next, we estimate the level value of functional I and obtain the following lemma.
Lemma 3.2. Assume that a, b > 0 and the positive function Q ∈ C(Ω) satisfies
(A1). Then there exists u0 ∈ H01 (Ω), such that supt≥0 I(tu0 ) < Λ for all ν > ν ∗ ,
where Λ is defined by Lemma 3.1 and ν ∗ independent of u0 is a positive constant.
Proof. Define a cut-off function η ∈ C0∞ (Ω) such that 0 ≤ η ≤ 1, |∇η| ≤ C1 . For
some δ̃ > 0, we define
(
1, |x − x0 | ≤ 2δ̃ ,
η(x) =
0, |x − x0 | ≥ δ̃,
where x0 is defined by (A1). Set uε = η(x)U (x − x0 ). As well known(see [6, 29]),
one has
kuε k2 = kUε k2 + O(ε) = S 3/2 + O(ε),
|uε |66
=
|Uε |66
3
+ O(ε ) = S
3/2
(3.9)
3
+ O(ε ),
(3.10)
and
s
2
Z
s
usε dx ≤ C3 ε 2 , 1 ≤ s < 3,
Ω
Z
s
s
C4 ε 2 | ln ε| ≤
usε dx ≤ C5 ε 2 | ln ε|, s = 3,
Z Ω
6−s
6−s
2
C6 ε
≤
usε dx ≤ C7 ε 2 , 3 < s < 6.
C2 ε ≤
(3.11)
Ω
Moreover, from [30], we have
9
kuε k4 = S 3 + O(ε),
kuε k6 = S 2 + O(ε),
kuε k8 = S 6 + O(ε),
kuε k12 = S 9 + O(ε).
(3.12)
For all t ≥ 0, we define I(tuε ) by
a
b
ν
I(tuε ) = t2 kuε k2 + t4 kuε k4 − t4
2
4
4
t6
|uε | dx −
6
Ω
Z
4
Z
Ω
Q(x)|uε |6 dx,
8
J.-F. LIAO, P. ZHANG, X.-P. WU
EJDE-2015/280
then we have
lim I(tuε ) = 0,
uniformly for all 0 < ε < ε0 ,
t→+0
lim I(tuε ) = −∞,
t→+∞
uniformly for all 0 < ε < ε0 ,
where ε0 > 0 is a small constant. Thus supt≥0 I(tuε ) attains for some tε > 0.
Moreover, we can claim that there exist two constants t0 , T0 > 0, which independent
of ε, such that t0 < tε < T0 . In fact, from limt→+0 I(tuε ) = 0 uniformly for all
ε, we choose = I(tε4uε ) > 0, then there exists t0 > 0 such that |I(t0 uε )| =
|I(t0 uε ) − I(0)| < . Then according to the monotonicity of I(tuε ) near t = 0, we
have tε > t0 . Similarly, we can obtain that tε < T0 . Therefore, our claim is proved.
Set I(tuε ) = Iε,1 (t) − νIε,2 (t), where
Z
a 2
b 4
t6
2
4
Q(x)u6ε dx,
Iε,1 (t) = t kuε k + t kuε k −
2
4
6 Ω
and
Z
t4
u4 dx.
4 Ω ε
R
0
First, we estimate the value Iε,1 . Since Iε,1
(t) = atkuε k2 +bt3 kuε k4 −t5 Ω Q(x)u6ε dx,
0
letting Iε,1
(t) = 0; that is,
Z
akuε k2 + bt2 kuε k4 − t4
Q(x)u6ε dx = 0,
(3.13)
Iε,2 (t) =
Ω
one obtains
q
R
b2 kuε k8 + 4akuε k2 Ω Q(x)u6ε dx
R
Tε2 =
.
2 Ω Q(x)u6ε dx
0
0
Then Iε,1
(t) > 0 for all 0 < t < Tε and Iε,1
(t) < 0 for all t > Tε , so Iε,1 (t) attains
+
its maximum at Tε . From (A1), let ε → 0 , we claim that
Z
1/3
1/3
6
Q(x)uε dx
= QM |uε |26 + o(ε).
(3.14)
bkuε k4 +
Ω
In fact, for all ε > 0, it follows that
Z
Z
Z
6
≤
Q(x)u6ε dx −
Q
u
dx
|Q(x) − Q(x0 )|u6ε dx
M ε
Ω
Ω
Ω
Z
≤
|Q(x) − Q(x0 )|u6ε dx.
{x∈Ω: |x−x0 |≤δ̃}
From (A1), for all η > 0, there exists δ > 0 such that
|Q(x) − Q(x0 )| < η|x − x0 |,
for all 0 < |x − x0 | < δ.
When ε > 0 small enough, for δ > ε1/2 , it follows from (3.15) and (A1) that
Z
Z
Q(x)u6ε dx −
QM u6ε dx
Ω
Ω
Z
≤
|Q(x) − Q(x0 )|u6ε dx
{x∈Ω:|x−x0 |≤δ̃}
Z
η|x − x0 |
<
{x∈Ω:|x−x0 |≤δ}
[ε2
(3ε2 )3/2
dx
+ |x − x0 |2 ]3
(3.15)
EJDE-2015/280
KIRCHHOFF PROBLEMS
9
(3ε2 )3/2
dx
+ |x − x0 |2 ]3
{x∈Ω: δ<|x−x0 |≤δ̃}
Z δ
√
√ Z δ̃
ε3
ε3 r 2
3
= 27η
dr
+
dr
r 2
27
2
2 3
(ε + r2 )3
0
δ (ε + r )
δ̃
Z δε
√
√ Z ε
r2
r3
= 27ηε
dr
+
27
dr
2
3
δ
(1 + r2 )3
0 (1 + r )
ε
Z
+
[ε2
≤ C8 ηε + C9 ε3 .
Consequently, one has
R
R
Q(x)u6ε dx − QM u6ε dx
Ω
Ω
≤ C8 η + C9 ε2 ,
ε
which implies
R
R
Q(x)u6ε dx − QM u6ε dx
Ω
Ω
≤ C8 η.
lim sup
ε
ε→0+
Then from the arbitrariness of η, we obtain (3.14). Thus, from (3.10) and (3.14),
one gets
Z
Q(x)u6ε dx = QM |uε |66 + o(ε) = QM S 3/2 + o(ε).
(3.16)
Ω
Thus from (3.9),(3.12),(3.13) and (3.16), we have
Iε,1 (t) ≤ Iε,1 (Tε )
a
= Tε2 kuε k2 +
2
a
= Tε2 kuε k2 +
3
b 2
T4
Tε kuε k4 − ε
4
6
b 2
Tε kuε k4
12 q
Z
Q(x)u6ε dx
Ω
abkuε k6 + akuε k2
R
b2 kuε k8 + 4akuε k2 Ω Q(x)u6ε dx
R
=
6 Ω Q(x)u6ε dx
R
b3 kuε k12 + 2abkuε k6 Ω Q(x)u6ε dx
+
2
R
24 Ω Q(x)u6ε dx
q
R
b2 kuε k4 b2 kuε k8 + 4akuε k2 Ω Q(x)u6ε dx
+
2
R
24 Ω Q(x)u6ε dx
=
abkuε k6
b3 kuε k12
R
+
4 Ω Q(x)u6ε dx 24( Ω Q(x)u6ε dx)2
q
R
akuε k2 b2 kuε k8 + 4akuε k2 Ω Q(x)u6ε dx
R
+
6 Ω Q(x)u6ε dx
q
R
b2 kuε k4 b2 kuε k8 + 4akuε k2 Ω Q(x)u6ε dx
+
2
R
24 Ω Q(x)u6ε dx
R
9
ab(S 2 + O(ε))
b3 (S 9 + O(ε))
=
+
4(QM S 3/2 + o(ε)) 24(QM S 3/2 + o(ε))2
10
J.-F. LIAO, P. ZHANG, X.-P. WU
p
a(S 3/2 + O(ε)) b2 S 6 + 4aS 3 + O(ε)
+
6(QM S 3/2 + o(ε))
p
2
6
b (S + O(ε)) b2 S 6 + 4aS 3 + O(ε)
+
24(QM S 3/2 + o(ε))2
p
abS 3
b3 S 6
aS b2 S 4 + 4aSQM
=
+
+
4QM
24Q2M
6QM
p
2 4
b S
b2 S 4 + 4aSQM
+
+ O(ε)
24Q2M
= Λ + O(ε).
EJDE-2015/280
(3.17)
Second, we estimate the value of Iε,2 . From (3.11), since 0 < t0 < tε < T0 , one
has
Z
Z
t4
t4
u4ε dx ≥ 0
u4 dx ≥ C10 ε.
(3.18)
Iε,2 (tε ) = ε
4 Ω
4 Ω ε
Thus, from (3.17) and (3.18), one gets
I(tuε ) = Iε,1 (t) − νIε,2 (t)
≤ Iε,1 (tε ) − νIε,2 (tε )
Z
t40
≤ Iε,1 (Tε ) − ν
u4ε dx
4
Ω
≤ Λ + O(ε) − C10 νε < Λ,
provided that ν is large enough. Thus there exists ν ∗ > 0 such that I(tuε ) < Λ for
all ν > ν ∗ . This completes the proof.
Proof of Theorem 1.3. As in the proof of Theorem 1.1, we can obtain that I has
the geometry of mountain pass lemma in H01 (Ω). According to Lemmas 3.1 and
3.2, it follows that I satisfies the conditions of the mountain pass lemma. Then as
in the proof of Theorem 1.1, we obtain that (1.1) has a positive solution u∗ with
I(u∗ ) > 0 as long as ν > ν ∗ . The proof is complete.
Acknowledgments. This research was supported by the National Science Foundation of China (No. 11471267), by the Science and Technology Foundation of
Guizhou Province (No. LKZS[2014]30, No. LH[2015]7001, No. LH[2015]7049).
The authors would like to thank the anonymous referees and Professor G. Molica
Bisci for their valuable suggestions.
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Jia-Feng Liao
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China.
School of Mathematics and Computational Science, Zunyi Normal College, Zunyi, China
E-mail address: liaojiafeng@163.com
12
J.-F. LIAO, P. ZHANG, X.-P. WU
EJDE-2015/280
Peng Zhang
School of Mathematics and Computational Science, Zunyi Normal College, Zunyi, China
E-mail address: gzzypd@sina.com
Xing-Ping Wu (corresponding author)
School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
E-mail address: wuxp@swu.edu.cn