- Email: [email protected]
On ground state solutions for a Kirchhoff type equation with critical growth
Computers and Mathematics with Applications (
)
–
Contents lists available at ScienceDirect
Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa
On ground state solutions for a Kirchhoff type equation with critical growth✩ Chun-Yu Lei ∗ , Hong-Min Suo, Chang-Mu Chu, Liu-Tao Guo School of Sciences, GuiZhou Minzu University, Guiyang 550025, China
article
info
abstract
Article history: Received 25 November 2015 Received in revised form 11 April 2016 Accepted 22 May 2016 Available online xxxx Keywords: Kirchhoff type equation Critical growth Concentration compactness principle Variational method
In this paper, with the aid of variational method and concentration-compactness principle, a positive ground state solution is obtained for a class of Kirchhoff type equations with critical growth
2 − a+b |∇ u| dx ∆u = u5 + λk(x)uq−1 , x ∈ R3 , R3 u ∈ D1,2 (R3 ), where λ > 0 is a parameter, 2 < q < 6 and k satisfies some conditions. © 2016 Elsevier Ltd. All rights reserved.
1. Introduction and main results We are concerned with the existence of positive solutions for the following Kirchhoff type problems involving the critical growth
− a + b
1 ,2
u∈D
R3 3
|∇ u| dx 1u = u5 + λk(x)uq−1 , 2
x ∈ R3 ,
(1.1)
(R ),
where a, b > 0 and 2 < q < 6, λ > 0 is a real number. It is known that Kirchhoff type equations have a strong physical meaning because they appear in mechanics models. For example, for problem
− a + b |∇ u|2 dx 1u = f (x, u) Ω u = 0,
in Ω ,
(1.2)
on ∂ Ω ,
is related to the stationary analogue of the Kirchhoff equation which was proposed by Kirchhoff in [1] as a generation of the d’Alembert’s equation
∂ 2u ρ 2 − ∂t
L 2 ∂u dx ∂ u = 0 + h 2L 0 ∂ x ∂ x2
P0
E
✩ Supported by Science and Technology Foundation of Guizhou Province (No.LH[2015]7207; No.J[2013]2141).
∗
Corresponding author. E-mail addresses: [email protected] (C.-Y. Lei), [email protected] (H.-M. Suo).
http://dx.doi.org/10.1016/j.camwa.2016.05.027 0898-1221/© 2016 Elsevier Ltd. All rights reserved.
2
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
for free vibrations of elastic strings. In [2], Lions introduced an abstract functional analysis framework to the equation
utt −
a+b Ω
|∇ u|2 1u = f (x, u).
After that there are abundant results about it, for example, in [3], Perera and Zhang obtained nontrivial solutions via the Yang index and critical group; Zhang and Perera in [4] obtained multiple and sign-changing solutions by using the invariant sets of descent flow. In particular, when the nonlinearity f (x, u) is critical growth in Eq. (1.2), some important and interesting results can be found when Ω ⊂ R3 is a smooth bounded domain (see [5–15] and their references therein). Problem (1.2) on unbounded domains involving critical growth has attracted a lot of attention, for example, in [16–20], the existence of positive solutions of (1.2) has been obtained via the variational methods. In recent years, there have been many papers concerned with the existence of positive ground state solutions for Kirchhoff type problems. Results relating to these problems can be found in [15,19,21,22]. Wang et al. in [19] have considered the following Kirchhoff type problem involving critical growth:
− ε2 a + bε |∇ u|2 dx 1u + M (x)u = |u|4 u + λf (u), Ω u ∈ H 1 (R3 ),
in R3 ,
where f ≥ 0 satisfies the following conditions:
(A1 ) f ∈ C 1 (R), f (t ) = o(t 3 ) as t → 0; (A2 ) f t(3t ) is strictly increasing on interval (0, ∞); (A3 ) |f (t )| ≤ c (t + |t |p−1 ) for some c > 0, 4 < p < 6. Under the assumptions of the above, they showed the existence of positive ground state solutions by the variational method. Very recently, in [22], Li and Ye have proved the existence of 2 < p < 5 such that the problem
− a + b R3
|∇ u|2 dx 1u + V (x)u = |u|p−1 u,
x ∈ R3 ,
u ∈ H 1 (R3 )
has a positive ground state solution by using a monotonicity trick and a new version of global compactness lemma. The main purpose of this paper is to establish the existence of a positive ground state solution of (1.1) by means of the variational method and concentration-compactness principle. Before stating our results, we need the following assumptions: 6
(k1 ) k ∈ L 6−q (R3 ) and k(x) ≥ 0 for any x ∈ R3 and k ̸≡ 0; (k2 ) There exist x0 ∈ R3 and δ, ρ1 > 0 such that k(x) ≥ δ|x − x0 |−β for |x − x0 | < ρ1 and 0 < β < 3. Now our main results can be described as follows: Theorem 1.1. Assume the hypotheses (k1 )–(k2 ) and 2 < q < 4 with 3 − 3 < β < 3 hold, then there exists λ∗ > 0 such that problem (1.1) has at least one positive ground state solution for all 0 < λ < λ∗ . 2q
Theorem 1.2. Assume the hypotheses (k1 )–(k2 ) and 4 ≤ q < 6 with 0 < β < 3 hold, then problem (1.1) has at least one positive ground state solution for all λ > 0. Throughout this paper, we make use of the following notations: The norm in D1,2 (R3 ) equipped with the norm ∥u∥2 = R3 |∇ u|2 dx, we shall simply write | · |p as the norm in Lp (R3 ). → (respectively, ⇀) denotes strong (respectively, weak) convergence; C , C1 , C2 , . . . denote various positive constants, which may vary from line to line; We denote by Sr (respectively, Br ) the sphere (respectively, the closed ball) of center zero and radius r, i.e., Sr = {u ∈ D1,2 (R3 ) : ∥u∥ = r }, Br = {u ∈ D1,2 (R3 ) : ∥u∥ ≤ r }; • Denoting by S the best constant of the above inequality, namely
• • • •
S :=
inf
|∇ u|2 dx 1 . 6 dx 3 | u | R3
R3
u∈D1,2 (R3 )\{0}
(1.3)
This work is organized as follows. In the next section we present some preliminary results. In Section 3, we give the proof of Theorems 1.1–1.2.
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
3
2. Preliminaries The existence of solutions of (1.1) understood as critical points of the energy functional a
I λ ( u) =
2
b
1
4
6
∥u∥2 + ∥u∥4 −
|u|6 dx − R3
λ
q
R3
k(x)|u|q dx.
Lemma 2.1 (See [23]). Suppose the hypothesis (k1 ) holds, the function K : D1,2 (R3 ) → R, given by
K (u) =
R3
k(x)|u|q dx,
2
is weakly continuous. Moreover, K is continuously differentiable with derivative K ′ : D1,2 (R3 ) → (D1,2 (R3 ))∗ given by
⟨K ′ (u), v⟩ = q
k(x)|u|q−2 uv dx.
R3
Lemma 2.2. Suppose the hypotheses (k1 )–(k2 ) hold, then: (i) There exist two constants α, ρ > 0 such that Iλ |Sρ ≥ α ;
(ii) There is u0 ∈ D1,2 (R3 ) with ∥u0 ∥ > ρ , satisfies Iλ (u0 ) < 0. Proof. (i) Let u ∈ D1,2 (R3 ) \ {0}, for 2 < q < 6, it follows from (1.3) that a
Iλ (u) =
2 a
≥
2
b
1
4 b
6 1
4
6S 3
∥ u∥ 2 + ∥ u∥ 4 − ∥ u∥ 2 + ∥ u∥ 4 −
|u|6 dx − R3
∥ u∥ 6 −
λ q
λ
q
|k|
R3
k(x)|u|q dx q
6 6−q
S − 2 ∥ u∥ q .
Let ρ = ∥u∥ be small enough such that a 2
b
1
4
6S 3
ρ2 + ρ4 −
ρ6 −
λ q
|k|
q
6 6−q
S− 2 ρq >
a 4
ρ2.
Consequently a
I λ ( u) ≥
4
ρ 2 , α.
(ii) It follows from (k1 ) and (k2 ) that, then Iλ (tu) =
≤
at 2 2 at 2 2
∥u∥2 + ∥u∥2 +
bt 4 4 bt 4 4
∥ u∥ 4 − ∥ u∥ 4 −
t6 6 t6 6
|u|6 dx −
R3
λt q q
R3
k(x)|u|q dx
|u|6 dx
R3
→ −∞ as t → +∞. thus there exists u0 ∈ D1,2 (R3 ) with ∥u0 ∥ > ρ , such that Iλ (u0 ) < 0. Then (ii) follows. This completes the proof of Lemma 2.2. Denote M + as a cone of positive finite Radon measures over R3 , and δx is the Dirac mass at point x. Lemma 2.3 (See [24,25]). Let {un } ⊂ D1,2 (R3 ) be a bounded sequence, due to Hardy–Sobolev inequality; going if necessary to a subsequence, we may assume that un ⇀ u in D1,2 (R3 ), |∇ un |2 ⇀ µ in M + , |un |6 ⇀ ν in M + . Define
µ∞ := lim lim
ν∞ := lim lim
R→∞ n→∞
R→∞ n→∞
R3 ∩|x|>R
R3 ∩|x|>R
|∇ un |2 dx, |un |6 dx.
Then for each j in an at most countable set J, we have 1
3 (1) µ∞ ≥ S ν∞ ; (2) ν = |u|6 + ν0 δ0 + δxj νj > 0, µ ≥ ∥∇ u∥2 + µ0 δ0 + δxj µj ;
4
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
1
(3) µj ≥ S νj3 ; (4) limn→∞
R3
|un |6 dx =
R3
|u|6 dx + ∥ν∥ + ν∞ .
We define
Λ=
abS 3 4
+
3
b3 S 6
+
24
(b2 S 4 + 4aS ) 2 24
. 6
Lemma 2.4. Assume 2 < q < 4, let {un } be a (PS )c sequence for Iλ at level c < Λ − Dλ 6−q (where the real positive number D is independent of λ), then there exists u ∈ D1,2 (R3 ) such that |un |66 → |u|66 as n → ∞. Proof. For 2 < q < 4, let {un } ⊂ D1,2 (R3 ) be a (PS )c sequence for Iλ at level c < Λ, namely Iλ (un ) → c ,
Iλ′ (un ) → 0
as n → ∞.
We firstly shall claim that {un } is bounded in D enough, it follows from (2.1) that 1 + o(∥un ∥) + c ≥ Iλ (un ) −
1,2
(2.1)
(R ). Otherwise, we can assume that ∥un ∥ → ∞ as n → ∞. For n large 3
1 ′ ⟨Iλ (un ), un ⟩
4
b 1 λ ∥ un ∥ 2 + ∥ un ∥ 4 − |un |6 dx − k(x)|un |q dx 2 4 6 R3 q R3 1 − a∥un ∥2 + b∥un ∥4 − |un |6 dx − λ k(x)|un |q dx 4 R3 R3 1 1 1 a |un |6 dx − λ − k(x)|un |q dx = ∥ un ∥ 2 + a
=
4 a
≥
4
12
q
R3
4
R3
6
∥un ∥2 − Dλ 6−q ,
where the last inequality comes from the fact, i.e., by Hölder and Young inequalities, then
λ
1 q
−
1 4
R3
k(x)|un |q dx ≤ λ
≤
1 12
1 q
−
R3
1 4
6−6 q
6
R3
k(x) 6−q dx
|un |6 dx
6q
R3
6
|un |6 dx + Dλ 6−q ,
where the real positive number D is independent of λ. Therefore {un } is bounded in D1,2 (R3 ). Clearly Iλ (|un |) = Iλ (un ), for this reason we assume straight away that un (x) ≥ 0 a.e. in R3 for all n. Then there exist u ∈ D1,2 (R3 ) and a subsequence (still denoted by {un }), such that
u ⇀ u, weakly in D1,2 (R3 ), n un (x) → u(x), a.e. in R3 , |∇ u |2 ⇀ µ, in M+ , 6n |un | ⇀ ν, in M+ . Then u(x) ≥ 0 a.e. in R3 . Let xj be a singular point of the measures µ and ν . For any ε > 0 small, we define a function φε,j ∈ C0∞ (R3 , [0, 1]) such that
φε,j (x) = 1, in B(xj , ε), φε,j (x) = 0, in R3 \ B(xj , 2ε), |∇φε,j (x)| ≤ 4/ε, in R3 . It is easy to see that {φε,j un } is bounded in D1,2 (R3 ). Then by (2.1) we also obtain ⟨Iλ′ (un ), φε,j un ⟩ → 0, i.e.,
R3
u6n φε,j dx + λ
R3
= ( a + b ∥ un ∥ 2 )
= ( a + b ∥ un ∥ 2 )
k(x)uqn φε,j dx
R3
R3
(∇ un , ∇(φε,j un ))dx + o(1) (∇ un , ∇φε,j )un dx + (a + b∥un ∥2 )
R3
|∇ un |2 φε,j dx + o(1).
(2.2)
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
5
We claim that lim lim (a + b∥un ∥2 )
ε→0 n→∞
R3
(∇ un , ∇φε,j )un dx = 0.
Indeed, since {un } is bounded, then for any ε > 0, using Hölder’s inequality and the definition of φε,j , one has
lim sup (∇ un , ∇φε,j )un dx ≤ lim sup (∇ un , ∇φε,j )un dx n→∞ n→∞ B(xj ,ε) R3 4 |∇ un | · |un | · 1dx ≤ lim sup n→∞ ε B(xj ,ε) 31 16
4
≤
ε 4
=
ε
C2 ε
13 dx
|u|6 dx
C1
B(xj ,ε)
B(xj ,ε)
61
6
|u| dx B(xj ,ε)
16
= 4C2
,
6
|u| dx B(xj ,ε)
where C1 , C2 > 0, the above inequality suggests that
lim lim
ε→0 n→∞ Ω
(∇ un , ∇φε,j )un dx = 0.
Moreover, Lemma 2.3(2)–(3) suggests that
lim lim
ε→0 n→∞ R3
u6n φε,j (x)dx = lim
φε,j (x)dν = νj ,
ε→0 R3
and
lim sup n→∞
R3
|∇ un |2 φε,j (x)dx =
R
3
≥ R3
φε,j (x)dµ φε,j (x)|∇ u|2 dx +
δj φε,j (x)µj
≥ µj , so that lim sup(a + b∥un ∥2 ) n→∞
R3
|∇ un |2 φε,j (x)dx ≥ lim sup a + b n→∞
≥ (a + bµj )µj .
R3
|∇ un |2 φε,j (x)dx
Since the weak continuity of K , one has
lim lim sup
ε→0
n→∞
R3
φε,j (x)k(x)uqn dx = lim
ε→0 B(x ,ε) j
≤S
− 2q
φε,j (x)k(x)uq dx
|k|
6 6−q
Hence, by (2.2) one has
νj ≥ aµj + bµ2j . Combining with Lemma 2.3(3) we have that either
(i) µj = 0 or (ii) µj ≥
√
b2 S 6 + 4aS 3 2
2
|∇ u| dx
lim
ε→0
= 0.
bS 3 +
2q
.
B(xj ,ε)
R3
|∇ un |2 φε,j (x)dx
6
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Now, define φR ∈ C0∞ (R3 , [0, 1]) such that
φR (x) = 0, on |x| < R, φR (x) = 1, on |x| > 2R, |∇φR (x)| ≤ 4 , in R3 . R
Since {φR un } is bounded in D1,2 (R3 ), we have limn→∞ ⟨Iλ′ (un ), φR un ⟩ = 0, i.e.,
R3
φR dν∞ + λ
R3
φR k(x)|u|q dx = lim (a + b∥un ∥2 )
n→∞
R3
(∇ un , ∇φR )un dx +
R3
|∇ un |2 φR dx ,
(2.3)
by Hölder’s inequality, the boundedness of {un } in D1,2 (R3 ), and the weak continuity of K , then passing the limit as R → ∞ in (2.3), we have
ν∞ ≥ aµ∞ + bµ2∞ . Hence it follows from Lemma 2.3(1) that
(iii) µ∞ = 0 or (iv) µ∞ ≥
bS 3 +
√
b2 S 6 + 4aS 3 2
, A.
Now we claim that (ii) and (iv) cannot occur. Indeed, from the weak lower semicontinuity of the norm and the weak continuity of K , and (2.1) and Young inequality, one gets
c = lim
n→∞
I λ ( un ) −
1 ′ ⟨Iλ (un ), un ⟩ 4
b 1 λ ∥ un ∥ 2 + ∥ un ∥ 4 − |un |6 dx − k(x)|un |q dx n→∞ 2 4 6 R3 q R3 1 2 4 6 q − a∥ un ∥ + b ∥ un ∥ − |un | dx − λ k(x)|un | dx 4 R3 R3 1 1 1 1 1 1 1 1 = lim − a∥ un ∥ 2 + − b∥un ∥4 + − |un |6 dx − λ − k(x)|un |q dx n→∞ 2 4 4 4 4 6 q 4 R3 R3 1 1 1 1 1 1 1 1 1 − aµ ∞ + − bµ2∞ + − ν∞ + |u|6 dx − λ − k(x)|u|q dx ≥ 2 4 4 4 4 6 12 R3 q 4 R3 6 1 1 1 1 1 1 A3 ≥ − aA + − bA2 + − − Dλ 6−q . 3
= lim
2
a
4
4
4
4
6
S
In the following we claim that
aA
b
+ A2 −
A3 S −3
= Λ,
2 4 6 aA + bA2 − A3 S −3 = 0.
Indeed, 2
aA + bA −
A3 S3
= A a+b
= A a+
=A = 0,
√
b2 S 6 + 4aS 3 2
−
1 2b2 S 6 + 4aS 3 + 2bS 3 S3
√
bS 3 +
b2 S 3 + b b2 S 6 + 4aS 3 2
√
b2 S 6 + 4aS 3
4
√
−
√
2S 3 (b2 S 3 + 2a + b b2 S 6 + 4aS 3 ) 4S 3
√
2a + b2 S 3 + b b2 S 6 + 4aS 3 − b2 S 3 − 2a − b b2 S 6 + 4aS 3 2
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
7
and aA 2
b
2
+ A − 4
A3 6S 3
=A
a 2
=A
=A
+
4
·
bS 3 +
√
b2 S 6 + 4aS 3 2
−
√
a 2
b
b2 S 3 + b b2 S 6 + 4aS 3
+
8
√
8a + b2 S 3 + b b2 S 6 + 4aS 3
2b2 S 6 + 4aS 3 + 2bS 3
b2 S 6 + 4aS 3
24S 3
√
−
√
b2 S 3 + 2 + b b2 S 6 + 4aS 3
12
24
√
12abS 3 + 2b3 S 6 + (2b2 S 3 + 8a) b2 S 6 + 4aS 3
=
48
= Λ. 6
Consequently Λ ≤ c < Λ − Dλ 6−q , this is a contradiction. Similarly, we can also prove that (ii) cannot occur for each j. Up to now, we have shown that
lim
n→∞
R3
u6n dx =
R3
u6 dx.
(2.4)
This completes the proof of Lemma 2.4.
Lemma 2.5. Assume 4 ≤ q < 6, let {un } be a (PS )c sequence for Iλ at level c < Λ, then there exists u ∈ D1,2 (R3 ) such that |un |66 → |u|66 as n → ∞. Proof. For 4 ≤ q < 6, set f (x, u) = u5 + λk(x)uq−1 , then f (x, u) satisfies the (AR) condition, i.e.,
∀ u ∈ D1,2 (R3 ),
4F (x, u) ≤ uf (x, u),
t
where F (x, t ) = 0 f (x, s)ds. Therefore, by the same argument as Lemma 2.4, we can also obtain (2.4). This completes the proof of Lemma 2.5. Moreover, it is well known that S is attained by the functions 1
vε,x0 (x) = C
ε4
1
(ε + |x − x0 |2 ) 2
,
where C is a normalizing constant and x0 is given in (k2 ). Let ϕ ∈ C0∞ (R3 ) be such that 0 ≤ ϕ ≤ 1, ϕ|Br ≡ 1 and suppϕ ⊂ B2r for some r > 0. Set uε (x) = ϕvε,x0 (x), then uε ∈ D1,2 (R3 ) and uε ≥ 0 for any x ∈ R3 . Let ε be small enough, it follows from the paper [26] that
1 2 |∇ uε |2 = K1 + O ε 2 , 1 u2ε dx = O ε 2 ,
|uε |26 = K2 + O(ε), (2.5)
R3
where K1 , K2 are positive constants. Moreover K1 K2
= S.
(2.6)
Lemma 2.6. Suppose the hypotheses (k1 )–(k2 ) hold, then: If 2 < q < 4, there exists u1 ∈ D1,2 (R3 ) such that 6
sup Iλ (tu1 ) < Λ − Dλ 6−q . t ≥0
If 4 ≤ q < 6, there exists u2 ∈ D1,2 (R3 ) such that sup Iλ (tu2 ) < Λ. t ≥0
(2.7)
8
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Proof. We shall prove that there exist tε > 0 and positive constants t0 , T1 independent of ε, λ, such that supt ≥0 Iλ (tuε ) = Iλ (tε uε ) and 0 < t0 ≤ tε ≤ T1 < ∞.
(2.8)
In fact, since limt →+∞ Iλ (tuε ) = −∞, there exists tε > 0, such that Iλ (tε uε ) = sup Iλ (tuε ),
dIλ (tuε )
and
dt
t ≥0
= 0.
(2.9)
t = tε
It follows from (2.9) that 2
3
4
5
tε a∥uε ∥ + tε b∥uε ∥ − tε
uε dx − λtε
q −1
6
R3
R3
k(x)uqε dx = 0,
and a∥uε ∥2 + 3tε2 b∥uε ∥4 − 5tε4
R3
u6ε dx − λ(q − 1)tεq−2
R3
k(x)uqε dx ≤ 0.
(2.10)
On one hand, as 2 < q < 6, we can obtain easily from (2.10) that tε is bounded below, that is, there exists a positive constant t0 > 0 (independent of ε, λ), such that 0 < t0 ≤ tε . On the other hand, it follows from (2.9) that a∥ uε ∥ 2 tε2
+ b∥uε ∥2 = tε2
R3
u6ε dx + λtεq−4
R3
k(x)uqε dx,
we see that tε is bounded above for all ε > 0 small enough. Thus (2.8) holds. Set h(tε uε ) =
a 2
tε2 ∥uε ∥2 +
t6 tε4 ∥uε ∥4 − ε 4 6 b
R3
u6ε dx.
Firstly, we claim that there exists a positive constant C3 (independent of ε, λ), such that 1
h(tε uε ) ≤ Λ + C3 ε 2 .
(2.11)
Indeed, let g¯ (t ) =
a 2
2
b
2
t ∥ uε ∥ +
4
4
t6
4
t ∥ uε ∥ −
6
u6ε dx.
R3
As limt →∞ g¯ (t ) = −∞, g¯ (0) = 0, and limt →0+ g¯ (t ) > 0, it follows that supt ≥0 g¯ (t ) attained at Tε > 0, that is g¯ ′ (t )|Tε = aTε ∥uε ∥2 + bTε3 ∥uε ∥4 − Tε5
R3
u6ε dx = 0,
that is Tε4
R3
u6ε dx − a∥uε ∥2 − bTε2 ∥uε ∥4 = 0,
so that
Tε =
b ∥ uε ∥ 4 +
b2 ∥uε ∥8 + 4a∥uε ∥2 2
R3
R3
u6ε dx
u6ε dx
12 .
Note that g¯ (t ) is increasing in the interval [0, Tε ], then by (2.5) and (2.6) one gets h(tε uε ) ≤ g¯ (Tε )
=
=
ab∥uε ∥6 4
ab 4
R3
u6ε dx
∥ uε ∥ 2 |uε |26
b3 ∥uε ∥12
+ 24
3 +
R3
b3 24
2 +
u6ε dx
∥ uε ∥ 2 |uε |26
3 (b2 ∥uε ∥8 + 4a∥uε ∥2 R3 u6ε dx) 2 2 24
6 +
1 24
R3
u6ε dx
b∥uε ∥8 + 4a∥u∥2 |u|66
|u|86
23
C.-Y. Lei et al. / Computers and Mathematics with Applications (
=
abS 3
)
–
9
3
(b2 S 4 + 4aS ) 2
b3 S 6
+ + 4 24 1 = Λ + O ε2 .
1 + O ε2
24
Therefore, there exists C3 > 0 (independent of ε, λ) such that (2.11) holds. √ Since 2 < q < 6, by (k2 ), let ε > 0 be small enough such that 0 < ε < ρ1 , then
R3
k(x)uqε dx ≥ C
q
|x − x0 |−β ε 4
+
q
(ε + |x − x0 |2 ) 2
|x−x0 |<ρ1 ρ1
q
≥ Cε 4
r2 ρ √1 ε
β
q
3
= Cε 2 − 4 − 2
t2 q
t β (1 + t 2 ) 2
0
≥ Cε
1
3 − 4q − β2 2
dt
dt
2q t β
0 β
q
3
t2
k(x)uqε dx
dr
q
r β (ε + r 2 ) 2
0
|x−x0 |>ρ1
= C4 ε 2 − 4 − 2 ,
(2.12)
where C4 (independent of ε, λ). Case: 2 < q < 4. since 3 − 1
C3 ε 2 −
λ q
3
< β < 3, then
2q 3
β
q
6
T12 C4 ε 2 − 4 − 2 = C3 λ 6−q −
=λ
6
λ q
C3 −
6−q
18−4q−6β 6−q 12
T12 C4 λ 6−q 1 q
T12 C4
λ
< 0. Let ε = λ
3 − 4q − β2 2
18−4q−6β 6−q
12 6−q
, 0 < λ < λ0 =
T12 C4
q(C3 +D)
q 4q+66−β− 18
, then
6
< −Dλ 6−q . Therefore, from (2.11) and (2.12), one has tq Iλ (tε uε ) = h(tε uε ) − λ ε q 1
≤ Λ + C3 ε 2 −
λ q
Ω
k(x)uqε dx 3
q
β
T12 C4 ε 2 − 4 − 2
6
< Λ − Dλ 6−q . Hence, there is u1 ∈ D1,2 (R2 ) such that (2.7) holds. β q Case: 4 ≤ q < 6. since 0 < β < 3, thus 32 − 4 − 2 < 1
C3 ε 2 −
λ q
3
q
1 , 2
for every λ > 0, when ε > 0 is small enough, then
β
T12 C4 ε 2 − 4 − 2 < 0.
Using (2.11) and (2.12) again, we obtain Iλ (tε uε ) < Λ, therefore, there exists u2 ∈ D1,2 (R2 ) such that supt ≥0 Iλ (tu2 ) < Λ. The proof is complete. Theorem 2.7. Under the assumptions of Theorem 1.1, problem (1.1) has at least a positive solution. 6
Proof. For 2 < q < 4, there exists δ ′ > 0 such that Λ − Dλ 6−q > 0 for every λ ∈ (0, δ ′ ). Set λ∗ = min{δ ′ , λ0 }, then Lemmas 2.1–2.4 and 2.6 hold when λ ∈ (0, λ∗ ). By Lemma 2.2, Iλ satisfies the geometry of the mountain-pass lemma [27]. Applying the mountain-pass lemma, there exists a sequence {un } ⊂ D1,2 (R3 ), such that Iλ (un ) → c > α,
and Iλ′ (un ) → 0,
where c = inf max Iλ (γ (t )), γ ∈Γ t ∈[0,1]
and
Γ = γ ∈ C ([0, 1], D1,2 (R3 )) : γ (0) = 0, γ (1) = u0 .
(2.13)
10
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Moreover, by Lemmas 2.5 and 2.6, we can get 6
0 < α < c ≤ max Iλ (t u¯ ) ≤ sup Iλ (t u¯ ) < Λ − Dλ 6−q . t ∈[0,1]
(2.14)
t ≥0
Note that Iλ (|un |) = Iλ (un ), we know that {un } ⊂ D1,2 (R3 ) has a convergent subsequence, still denoted by {un } (un ≥ 0), and there exists u∗ ∈ D1,2 (R3 ) such that weakly in D1,2 (R3 ),
un ⇀ u∗ ,
un → u∗ ,
strongly inLp (R3 )(1 ≤ p < 6).
Set limn→∞ ∥un ∥ = l, for every φ ∈ D1,2 (R3 ), since ⟨Iλ′ (un ), φ⟩ → 0 as n → ∞, there holds
(a + bl ) 2
(∇ u∗ , ∇φ)dx −
R3
u∗ φ dx − λ 5
R3
R3
k(x)u∗q−1 φ dx = 0.
In particular, we have
(a + bl2 )∥u∗ ∥ −
R3
u6∗ dx − λ
R3
k(x)uq∗ dx = 0.
(2.15)
As ⟨Iλ′ (un ), un ⟩ → 0 holds true as n → ∞, consequently
(a + b∥un ∥2 )∥un ∥2 −
R3
u6n dx − λ
R3
k(x)uqn dx = o(1).
By Lemma 2.4, there holds
R3
u6n dx →
R3
u6∗ dx,
as n → ∞,
so that
(a + bl )l − 2
2
u∗ dx − λ 6
R3
R3
k(x)uq∗ dx = 0.
(2.16)
Now, from (2.15) and (2.16), one has ∥u∗ ∥ = l, so we obtain that un → u∗ in D1,2 (R3 ) as n → ∞. Hence, it follows from (2.13) and (2.14) that Iλ (u∗ ) = lim Iλ (un ) = c > α > 0.
(2.17)
n→∞
From (2.17), it follows that u∗ ̸≡ 0, note that u∗ ≥ 0 in D1,2 (R3 ). Therefore, by using the strong maximum principle we obtained that u∗ is a positive solution of (1.1). The proof is complete. Remark 2.8. Under the assumptions of Theorem 1.2, problem (1.1) has at least a positive solution. 3. Proof of theorems Proof of Theorems 1.1 and 1.2. By Theorem 2.7 and Remark 2.8, we obtain the existence of positive solutions of (1.1). In the following we shall prove that problem (1.1) admits a positive ground state solution. Let m = inf Iλ (u) : u ∈ D1,2 (R3 ), u ̸= 0, Iλ′ (u) = 0 .
By the definition of m, there exists {vn } ⊂ D1,2 (R3 ) such that vn ̸= 0, and Iλ (vn ) → m,
Iλ′ (vn ) = 0,
as n → ∞.
(3.1)
Obviously, from (3.1), we can easily deduce that {vn } is bounded in D1,2 (R3 ). There then exist a subsequence of {vn } (still denoted by {vn }) and v∗ ∈ D1,2 (R3 ) such that vn ⇀ v∗ weakly in D1,2 (R3 ). We claim that v∗ ̸= 0. Otherwise, vn ⇀ 0 weakly in D1,2 (R3 ), then by Lemma 2.1, we have
R3
k(x)|vn |q = o(1),
as n → ∞.
By (3.1), we have a∥vn ∥2 + b∥vn ∥4 −
R3
|vn |6 dx − λ
R3
k(x)|vn |q dx = 0.
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
11
Consequently a∥vn ∥2 + b∥vn ∥4 −
R3
|vn |6 dx = 0.
(3.2)
Set limn→∞ ∥vn ∥ = l. By (3.2) and (1.3), we can deduce that 2
l ≥
√
bS 3 +
b2 S 6 + 4aS 3 2
.
Therefore, using (3.1) and (3.2) again, one gets
b
a
1
∥vn ∥2 + ∥vn ∥4 − n→∞ 2 4 6 a b = lim ∥vn ∥2 + ∥vn ∥4
m = lim
3
n→∞
= ≥
al2 3
+
abS 3 4
|vn |6 dx − R3
λ q
R3
k(x)|vn |q dx
12
b 12
+
l4
b3 S 6 24
= Λ.
3
+
(b2 S 4 + 4aS ) 2 24
6
By Lemma 2.6, we have Λ ≤ m < Λ − Dλ 6−q (or Λ ≤ m < Λ), this is a contradiction. Then vn ⇀ v∗ ̸= 0 in D1,2 (R3 ) and Iλ′ (v∗ ) = 0 (moreover, vn → v∗ in D1,2 (R3 ), and we also prove that v∗ is a positive solution of (1.1)), it suggests that Iλ (v∗ ) ≥ m. We claim that m ≥ Iλ (v∗ ). Indeed, since Iλ′ (v∗ ) = 0, one gets
λ b 1 ∥v∗ ∥2 + ∥v∗ ∥4 − v∗6 dx − k(x)v∗q dx 2 4 6 R3 q R3 1 − a∥v∗ ∥2 + b∥v∗ ∥4 − v∗6 dx − λ k(x)v∗q dx 6 R3 R3 b 1 1 a − k(x)v∗q dx. = ∥v∗ ∥2 + ∥v∗ ∥4 − λ
Iλ (v∗ ) =
a
3
12
q
6
R3
By Iλ′ (vn ) = 0, we have m + o(1) = Iλ (vn )
=
a 3
∥vn ∥2 +
b 12
∥vn ∥4 − λ
1 q
−
1 6
R3
k(x)vnq dx
and the Fatou’s Lemma, one obtains m≥
a 3
∥v∗ ∥2 +
b 12
∥v∗ ∥4 − λ
1 q
−
1 6
R3
k(x)v∗q dx = Iλ (v∗ ).
Then v∗ ̸= 0 satisfies Iλ′ (v∗ ) = 0 and Iλ (v∗ ) = m. Consequently, v∗ is a positive ground state solution of (1.1). Therefore, we prove the existence of positive ground state solutions of (1.1). The proof is complete. Acknowledgment The authors thank the referee for his or her careful reading of the paper and for the helpful comments, which led to an improvement of this paper. References [1] G. Kirchhoff, Mechanik, Teubner, Leipzig, 1883. [2] J.L. Lions, On some questions in boundary value problems of mathematical physics, in: Contemporary Development in Continuum Mechanics and Partial Differential Equations, in: North-Holland Math. Stud., vol. 30, North-Holland, Amsterdam, New York, 1978, pp. 284–346. [3] K. Perera, Z.T. Zhang, Nontrivial solutions of Kirchhoff-type problems via the Yang index, J. Differential Equations 221 (2006) 246–255. [4] Z.T. Zhang, K. Perera, Sign changing solutions of Kirchhoff type problems via invariant sets of descent flow, J. Math. Anal. Appl. 317 (2006) 456–463. [5] A. Ourraoui, On a p-Kirchhoff problem involving a critical nonlinearity, C. R. Acad. Sci., Paris I 352 (2014) 295–298. [6] A. Fiscella, E. Valdinoci, A critical Kirchhoff type problem involving a nonlocal operator, Nonlinear Anal. 94 (2014) 156–170. [7] C.O. Alves, F.J.S.A. Corrêa, G.M. Figueiredo, On a class of nonlocal elliptic problems with critical growth, Differ. Equ. Appl. 23 (2010) 409–417.
12
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
[8] G.M. Figueiredo, Existence of a positive for a Kirchhoff problem type with critical growth via truncation argument, J. Math. Anal. Appl. 401 (2013) 706–713. [9] A. Hamydy, M. Massar, N. Tsouli, Existence of solution for p-Kirchhoff type problems with critical exponents, Electron. J. Differential Equations 105 (2011) 1–8. [10] T.F. Ma, Remarks on an elliptic equation of Kirchhoff type, Nonlinear Anal. 63 (2005) 1967–1977. [11] F. Cammaroto, L. Vilasi, Multiple solutions for a Kirchhoff-type problem involving the p(x)-Laplacian operator, Nonlinear Anal. 74 (2011) 1841–1852. [12] G.M. Figueiredo, J.R.S. Junior, Multiplicity of solutions for a Kirchhoff equation with subcritical or critical growth, Differential Integral Equations 25 (2012) 853–868. [13] D. Naimen, The critical problem of Kirchhoff type elliptic equations in dimension four, J. Differential Equations 257 (2014) 1168–1193. [14] C.Y. Lei, J.F. Liao, C.L. Tang, Multiple positive solutions for Kirchhoff type of problems with singularity and critical exponents, J. Math. Anal. Appl. 421 (2015) 521–538. [15] C.Y. Lei, C.M. Chu, H.M. Suo, C.L. Tang, On Kirchhoff type problems involving critical and singular nonlinearities, Ann. Polon. Math. 114 (2015) 269–291. [16] X.M. He, W.M. Zou, Existence and concentration behavior of positive solutions for a Kirchhoff equation in R3 , J. Differential Equations 252 (2012) 1813–1834. [17] S. Liang, S.Y. Shi, Soliton solutions to Kirchhoff type problems involving the critical growth in RN , Nonlinear Anal. 81 (2013) 31–41. [18] S. Liang, J. Zhang, Existence of solutions for Kirchhoff type problems with critical nonlinearity in R3 , Nonlinear Anal. Real World Appl. 17 (2014) 126–136. [19] J. Wang, L. Tian, J. Xu, F. Zhang, Multiplicity and concentration of positive solutions for a Kirchhoff type problem with critical growth, J. Differential Equations 253 (2012) 2314–2351. [20] J. Liu, J.F. Liao, C.L. Tang, Positive solutions for Kirchhoff-type equations with critical exponent in RN , J. Math. Anal. Appl. 429 (2015) 1153–1172. [21] J. Sun, T. Wu, Ground state solutions for an indefinite Kirchhoff type problem with steep potential well, J. Differential Equations 256 (2014) 1771–1792. [22] G.B. Li, H.Y. Ye, Existence of positive ground state solutions for the nonlinear Kirchhoff type equations in R3 , J. Differential Equations 257 (2014) 566–600. [23] M. Willem, Minimax Theorems, Birthäuser, Boston, 1996. [24] P.L. Lions, The concentration-compactness principle in the calculus of variations, the limit case, Part I, Rev. Mat. Iberoam. 1 (1985) 145–201. [25] P.L. Lions, The concentration-compactness principle in the calculus of variations, the limit case, Part II, Rev. Mat. Iberoam. 2 (1985) 45–121. [26] H. Brézis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (1983) 437–477. [27] A. Ambrosetti, P.H. Rabinowitz, Dual variational methods in critical point theory and applications, J. Funct. Anal. 14 (1973) 349–381.
)
–
Contents lists available at ScienceDirect
Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa
On ground state solutions for a Kirchhoff type equation with critical growth✩ Chun-Yu Lei ∗ , Hong-Min Suo, Chang-Mu Chu, Liu-Tao Guo School of Sciences, GuiZhou Minzu University, Guiyang 550025, China
article
info
abstract
Article history: Received 25 November 2015 Received in revised form 11 April 2016 Accepted 22 May 2016 Available online xxxx Keywords: Kirchhoff type equation Critical growth Concentration compactness principle Variational method
In this paper, with the aid of variational method and concentration-compactness principle, a positive ground state solution is obtained for a class of Kirchhoff type equations with critical growth
2 − a+b |∇ u| dx ∆u = u5 + λk(x)uq−1 , x ∈ R3 , R3 u ∈ D1,2 (R3 ), where λ > 0 is a parameter, 2 < q < 6 and k satisfies some conditions. © 2016 Elsevier Ltd. All rights reserved.
1. Introduction and main results We are concerned with the existence of positive solutions for the following Kirchhoff type problems involving the critical growth
− a + b
1 ,2
u∈D
R3 3
|∇ u| dx 1u = u5 + λk(x)uq−1 , 2
x ∈ R3 ,
(1.1)
(R ),
where a, b > 0 and 2 < q < 6, λ > 0 is a real number. It is known that Kirchhoff type equations have a strong physical meaning because they appear in mechanics models. For example, for problem
− a + b |∇ u|2 dx 1u = f (x, u) Ω u = 0,
in Ω ,
(1.2)
on ∂ Ω ,
is related to the stationary analogue of the Kirchhoff equation which was proposed by Kirchhoff in [1] as a generation of the d’Alembert’s equation
∂ 2u ρ 2 − ∂t
L 2 ∂u dx ∂ u = 0 + h 2L 0 ∂ x ∂ x2
P0
E
✩ Supported by Science and Technology Foundation of Guizhou Province (No.LH[2015]7207; No.J[2013]2141).
∗
Corresponding author. E-mail addresses: [email protected] (C.-Y. Lei), [email protected] (H.-M. Suo).
http://dx.doi.org/10.1016/j.camwa.2016.05.027 0898-1221/© 2016 Elsevier Ltd. All rights reserved.
2
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
for free vibrations of elastic strings. In [2], Lions introduced an abstract functional analysis framework to the equation
utt −
a+b Ω
|∇ u|2 1u = f (x, u).
After that there are abundant results about it, for example, in [3], Perera and Zhang obtained nontrivial solutions via the Yang index and critical group; Zhang and Perera in [4] obtained multiple and sign-changing solutions by using the invariant sets of descent flow. In particular, when the nonlinearity f (x, u) is critical growth in Eq. (1.2), some important and interesting results can be found when Ω ⊂ R3 is a smooth bounded domain (see [5–15] and their references therein). Problem (1.2) on unbounded domains involving critical growth has attracted a lot of attention, for example, in [16–20], the existence of positive solutions of (1.2) has been obtained via the variational methods. In recent years, there have been many papers concerned with the existence of positive ground state solutions for Kirchhoff type problems. Results relating to these problems can be found in [15,19,21,22]. Wang et al. in [19] have considered the following Kirchhoff type problem involving critical growth:
− ε2 a + bε |∇ u|2 dx 1u + M (x)u = |u|4 u + λf (u), Ω u ∈ H 1 (R3 ),
in R3 ,
where f ≥ 0 satisfies the following conditions:
(A1 ) f ∈ C 1 (R), f (t ) = o(t 3 ) as t → 0; (A2 ) f t(3t ) is strictly increasing on interval (0, ∞); (A3 ) |f (t )| ≤ c (t + |t |p−1 ) for some c > 0, 4 < p < 6. Under the assumptions of the above, they showed the existence of positive ground state solutions by the variational method. Very recently, in [22], Li and Ye have proved the existence of 2 < p < 5 such that the problem
− a + b R3
|∇ u|2 dx 1u + V (x)u = |u|p−1 u,
x ∈ R3 ,
u ∈ H 1 (R3 )
has a positive ground state solution by using a monotonicity trick and a new version of global compactness lemma. The main purpose of this paper is to establish the existence of a positive ground state solution of (1.1) by means of the variational method and concentration-compactness principle. Before stating our results, we need the following assumptions: 6
(k1 ) k ∈ L 6−q (R3 ) and k(x) ≥ 0 for any x ∈ R3 and k ̸≡ 0; (k2 ) There exist x0 ∈ R3 and δ, ρ1 > 0 such that k(x) ≥ δ|x − x0 |−β for |x − x0 | < ρ1 and 0 < β < 3. Now our main results can be described as follows: Theorem 1.1. Assume the hypotheses (k1 )–(k2 ) and 2 < q < 4 with 3 − 3 < β < 3 hold, then there exists λ∗ > 0 such that problem (1.1) has at least one positive ground state solution for all 0 < λ < λ∗ . 2q
Theorem 1.2. Assume the hypotheses (k1 )–(k2 ) and 4 ≤ q < 6 with 0 < β < 3 hold, then problem (1.1) has at least one positive ground state solution for all λ > 0. Throughout this paper, we make use of the following notations: The norm in D1,2 (R3 ) equipped with the norm ∥u∥2 = R3 |∇ u|2 dx, we shall simply write | · |p as the norm in Lp (R3 ). → (respectively, ⇀) denotes strong (respectively, weak) convergence; C , C1 , C2 , . . . denote various positive constants, which may vary from line to line; We denote by Sr (respectively, Br ) the sphere (respectively, the closed ball) of center zero and radius r, i.e., Sr = {u ∈ D1,2 (R3 ) : ∥u∥ = r }, Br = {u ∈ D1,2 (R3 ) : ∥u∥ ≤ r }; • Denoting by S the best constant of the above inequality, namely
• • • •
S :=
inf
|∇ u|2 dx 1 . 6 dx 3 | u | R3
R3
u∈D1,2 (R3 )\{0}
(1.3)
This work is organized as follows. In the next section we present some preliminary results. In Section 3, we give the proof of Theorems 1.1–1.2.
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
3
2. Preliminaries The existence of solutions of (1.1) understood as critical points of the energy functional a
I λ ( u) =
2
b
1
4
6
∥u∥2 + ∥u∥4 −
|u|6 dx − R3
λ
q
R3
k(x)|u|q dx.
Lemma 2.1 (See [23]). Suppose the hypothesis (k1 ) holds, the function K : D1,2 (R3 ) → R, given by
K (u) =
R3
k(x)|u|q dx,
2
is weakly continuous. Moreover, K is continuously differentiable with derivative K ′ : D1,2 (R3 ) → (D1,2 (R3 ))∗ given by
⟨K ′ (u), v⟩ = q
k(x)|u|q−2 uv dx.
R3
Lemma 2.2. Suppose the hypotheses (k1 )–(k2 ) hold, then: (i) There exist two constants α, ρ > 0 such that Iλ |Sρ ≥ α ;
(ii) There is u0 ∈ D1,2 (R3 ) with ∥u0 ∥ > ρ , satisfies Iλ (u0 ) < 0. Proof. (i) Let u ∈ D1,2 (R3 ) \ {0}, for 2 < q < 6, it follows from (1.3) that a
Iλ (u) =
2 a
≥
2
b
1
4 b
6 1
4
6S 3
∥ u∥ 2 + ∥ u∥ 4 − ∥ u∥ 2 + ∥ u∥ 4 −
|u|6 dx − R3
∥ u∥ 6 −
λ q
λ
q
|k|
R3
k(x)|u|q dx q
6 6−q
S − 2 ∥ u∥ q .
Let ρ = ∥u∥ be small enough such that a 2
b
1
4
6S 3
ρ2 + ρ4 −
ρ6 −
λ q
|k|
q
6 6−q
S− 2 ρq >
a 4
ρ2.
Consequently a
I λ ( u) ≥
4
ρ 2 , α.
(ii) It follows from (k1 ) and (k2 ) that, then Iλ (tu) =
≤
at 2 2 at 2 2
∥u∥2 + ∥u∥2 +
bt 4 4 bt 4 4
∥ u∥ 4 − ∥ u∥ 4 −
t6 6 t6 6
|u|6 dx −
R3
λt q q
R3
k(x)|u|q dx
|u|6 dx
R3
→ −∞ as t → +∞. thus there exists u0 ∈ D1,2 (R3 ) with ∥u0 ∥ > ρ , such that Iλ (u0 ) < 0. Then (ii) follows. This completes the proof of Lemma 2.2. Denote M + as a cone of positive finite Radon measures over R3 , and δx is the Dirac mass at point x. Lemma 2.3 (See [24,25]). Let {un } ⊂ D1,2 (R3 ) be a bounded sequence, due to Hardy–Sobolev inequality; going if necessary to a subsequence, we may assume that un ⇀ u in D1,2 (R3 ), |∇ un |2 ⇀ µ in M + , |un |6 ⇀ ν in M + . Define
µ∞ := lim lim
ν∞ := lim lim
R→∞ n→∞
R→∞ n→∞
R3 ∩|x|>R
R3 ∩|x|>R
|∇ un |2 dx, |un |6 dx.
Then for each j in an at most countable set J, we have 1
3 (1) µ∞ ≥ S ν∞ ; (2) ν = |u|6 + ν0 δ0 + δxj νj > 0, µ ≥ ∥∇ u∥2 + µ0 δ0 + δxj µj ;
4
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
1
(3) µj ≥ S νj3 ; (4) limn→∞
R3
|un |6 dx =
R3
|u|6 dx + ∥ν∥ + ν∞ .
We define
Λ=
abS 3 4
+
3
b3 S 6
+
24
(b2 S 4 + 4aS ) 2 24
. 6
Lemma 2.4. Assume 2 < q < 4, let {un } be a (PS )c sequence for Iλ at level c < Λ − Dλ 6−q (where the real positive number D is independent of λ), then there exists u ∈ D1,2 (R3 ) such that |un |66 → |u|66 as n → ∞. Proof. For 2 < q < 4, let {un } ⊂ D1,2 (R3 ) be a (PS )c sequence for Iλ at level c < Λ, namely Iλ (un ) → c ,
Iλ′ (un ) → 0
as n → ∞.
We firstly shall claim that {un } is bounded in D enough, it follows from (2.1) that 1 + o(∥un ∥) + c ≥ Iλ (un ) −
1,2
(2.1)
(R ). Otherwise, we can assume that ∥un ∥ → ∞ as n → ∞. For n large 3
1 ′ ⟨Iλ (un ), un ⟩
4
b 1 λ ∥ un ∥ 2 + ∥ un ∥ 4 − |un |6 dx − k(x)|un |q dx 2 4 6 R3 q R3 1 − a∥un ∥2 + b∥un ∥4 − |un |6 dx − λ k(x)|un |q dx 4 R3 R3 1 1 1 a |un |6 dx − λ − k(x)|un |q dx = ∥ un ∥ 2 + a
=
4 a
≥
4
12
q
R3
4
R3
6
∥un ∥2 − Dλ 6−q ,
where the last inequality comes from the fact, i.e., by Hölder and Young inequalities, then
λ
1 q
−
1 4
R3
k(x)|un |q dx ≤ λ
≤
1 12
1 q
−
R3
1 4
6−6 q
6
R3
k(x) 6−q dx
|un |6 dx
6q
R3
6
|un |6 dx + Dλ 6−q ,
where the real positive number D is independent of λ. Therefore {un } is bounded in D1,2 (R3 ). Clearly Iλ (|un |) = Iλ (un ), for this reason we assume straight away that un (x) ≥ 0 a.e. in R3 for all n. Then there exist u ∈ D1,2 (R3 ) and a subsequence (still denoted by {un }), such that
u ⇀ u, weakly in D1,2 (R3 ), n un (x) → u(x), a.e. in R3 , |∇ u |2 ⇀ µ, in M+ , 6n |un | ⇀ ν, in M+ . Then u(x) ≥ 0 a.e. in R3 . Let xj be a singular point of the measures µ and ν . For any ε > 0 small, we define a function φε,j ∈ C0∞ (R3 , [0, 1]) such that
φε,j (x) = 1, in B(xj , ε), φε,j (x) = 0, in R3 \ B(xj , 2ε), |∇φε,j (x)| ≤ 4/ε, in R3 . It is easy to see that {φε,j un } is bounded in D1,2 (R3 ). Then by (2.1) we also obtain ⟨Iλ′ (un ), φε,j un ⟩ → 0, i.e.,
R3
u6n φε,j dx + λ
R3
= ( a + b ∥ un ∥ 2 )
= ( a + b ∥ un ∥ 2 )
k(x)uqn φε,j dx
R3
R3
(∇ un , ∇(φε,j un ))dx + o(1) (∇ un , ∇φε,j )un dx + (a + b∥un ∥2 )
R3
|∇ un |2 φε,j dx + o(1).
(2.2)
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
5
We claim that lim lim (a + b∥un ∥2 )
ε→0 n→∞
R3
(∇ un , ∇φε,j )un dx = 0.
Indeed, since {un } is bounded, then for any ε > 0, using Hölder’s inequality and the definition of φε,j , one has
lim sup (∇ un , ∇φε,j )un dx ≤ lim sup (∇ un , ∇φε,j )un dx n→∞ n→∞ B(xj ,ε) R3 4 |∇ un | · |un | · 1dx ≤ lim sup n→∞ ε B(xj ,ε) 31 16
4
≤
ε 4
=
ε
C2 ε
13 dx
|u|6 dx
C1
B(xj ,ε)
B(xj ,ε)
61
6
|u| dx B(xj ,ε)
16
= 4C2
,
6
|u| dx B(xj ,ε)
where C1 , C2 > 0, the above inequality suggests that
lim lim
ε→0 n→∞ Ω
(∇ un , ∇φε,j )un dx = 0.
Moreover, Lemma 2.3(2)–(3) suggests that
lim lim
ε→0 n→∞ R3
u6n φε,j (x)dx = lim
φε,j (x)dν = νj ,
ε→0 R3
and
lim sup n→∞
R3
|∇ un |2 φε,j (x)dx =
R
3
≥ R3
φε,j (x)dµ φε,j (x)|∇ u|2 dx +
δj φε,j (x)µj
≥ µj , so that lim sup(a + b∥un ∥2 ) n→∞
R3
|∇ un |2 φε,j (x)dx ≥ lim sup a + b n→∞
≥ (a + bµj )µj .
R3
|∇ un |2 φε,j (x)dx
Since the weak continuity of K , one has
lim lim sup
ε→0
n→∞
R3
φε,j (x)k(x)uqn dx = lim
ε→0 B(x ,ε) j
≤S
− 2q
φε,j (x)k(x)uq dx
|k|
6 6−q
Hence, by (2.2) one has
νj ≥ aµj + bµ2j . Combining with Lemma 2.3(3) we have that either
(i) µj = 0 or (ii) µj ≥
√
b2 S 6 + 4aS 3 2
2
|∇ u| dx
lim
ε→0
= 0.
bS 3 +
2q
.
B(xj ,ε)
R3
|∇ un |2 φε,j (x)dx
6
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Now, define φR ∈ C0∞ (R3 , [0, 1]) such that
φR (x) = 0, on |x| < R, φR (x) = 1, on |x| > 2R, |∇φR (x)| ≤ 4 , in R3 . R
Since {φR un } is bounded in D1,2 (R3 ), we have limn→∞ ⟨Iλ′ (un ), φR un ⟩ = 0, i.e.,
R3
φR dν∞ + λ
R3
φR k(x)|u|q dx = lim (a + b∥un ∥2 )
n→∞
R3
(∇ un , ∇φR )un dx +
R3
|∇ un |2 φR dx ,
(2.3)
by Hölder’s inequality, the boundedness of {un } in D1,2 (R3 ), and the weak continuity of K , then passing the limit as R → ∞ in (2.3), we have
ν∞ ≥ aµ∞ + bµ2∞ . Hence it follows from Lemma 2.3(1) that
(iii) µ∞ = 0 or (iv) µ∞ ≥
bS 3 +
√
b2 S 6 + 4aS 3 2
, A.
Now we claim that (ii) and (iv) cannot occur. Indeed, from the weak lower semicontinuity of the norm and the weak continuity of K , and (2.1) and Young inequality, one gets
c = lim
n→∞
I λ ( un ) −
1 ′ ⟨Iλ (un ), un ⟩ 4
b 1 λ ∥ un ∥ 2 + ∥ un ∥ 4 − |un |6 dx − k(x)|un |q dx n→∞ 2 4 6 R3 q R3 1 2 4 6 q − a∥ un ∥ + b ∥ un ∥ − |un | dx − λ k(x)|un | dx 4 R3 R3 1 1 1 1 1 1 1 1 = lim − a∥ un ∥ 2 + − b∥un ∥4 + − |un |6 dx − λ − k(x)|un |q dx n→∞ 2 4 4 4 4 6 q 4 R3 R3 1 1 1 1 1 1 1 1 1 − aµ ∞ + − bµ2∞ + − ν∞ + |u|6 dx − λ − k(x)|u|q dx ≥ 2 4 4 4 4 6 12 R3 q 4 R3 6 1 1 1 1 1 1 A3 ≥ − aA + − bA2 + − − Dλ 6−q . 3
= lim
2
a
4
4
4
4
6
S
In the following we claim that
aA
b
+ A2 −
A3 S −3
= Λ,
2 4 6 aA + bA2 − A3 S −3 = 0.
Indeed, 2
aA + bA −
A3 S3
= A a+b
= A a+
=A = 0,
√
b2 S 6 + 4aS 3 2
−
1 2b2 S 6 + 4aS 3 + 2bS 3 S3
√
bS 3 +
b2 S 3 + b b2 S 6 + 4aS 3 2
√
b2 S 6 + 4aS 3
4
√
−
√
2S 3 (b2 S 3 + 2a + b b2 S 6 + 4aS 3 ) 4S 3
√
2a + b2 S 3 + b b2 S 6 + 4aS 3 − b2 S 3 − 2a − b b2 S 6 + 4aS 3 2
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
7
and aA 2
b
2
+ A − 4
A3 6S 3
=A
a 2
=A
=A
+
4
·
bS 3 +
√
b2 S 6 + 4aS 3 2
−
√
a 2
b
b2 S 3 + b b2 S 6 + 4aS 3
+
8
√
8a + b2 S 3 + b b2 S 6 + 4aS 3
2b2 S 6 + 4aS 3 + 2bS 3
b2 S 6 + 4aS 3
24S 3
√
−
√
b2 S 3 + 2 + b b2 S 6 + 4aS 3
12
24
√
12abS 3 + 2b3 S 6 + (2b2 S 3 + 8a) b2 S 6 + 4aS 3
=
48
= Λ. 6
Consequently Λ ≤ c < Λ − Dλ 6−q , this is a contradiction. Similarly, we can also prove that (ii) cannot occur for each j. Up to now, we have shown that
lim
n→∞
R3
u6n dx =
R3
u6 dx.
(2.4)
This completes the proof of Lemma 2.4.
Lemma 2.5. Assume 4 ≤ q < 6, let {un } be a (PS )c sequence for Iλ at level c < Λ, then there exists u ∈ D1,2 (R3 ) such that |un |66 → |u|66 as n → ∞. Proof. For 4 ≤ q < 6, set f (x, u) = u5 + λk(x)uq−1 , then f (x, u) satisfies the (AR) condition, i.e.,
∀ u ∈ D1,2 (R3 ),
4F (x, u) ≤ uf (x, u),
t
where F (x, t ) = 0 f (x, s)ds. Therefore, by the same argument as Lemma 2.4, we can also obtain (2.4). This completes the proof of Lemma 2.5. Moreover, it is well known that S is attained by the functions 1
vε,x0 (x) = C
ε4
1
(ε + |x − x0 |2 ) 2
,
where C is a normalizing constant and x0 is given in (k2 ). Let ϕ ∈ C0∞ (R3 ) be such that 0 ≤ ϕ ≤ 1, ϕ|Br ≡ 1 and suppϕ ⊂ B2r for some r > 0. Set uε (x) = ϕvε,x0 (x), then uε ∈ D1,2 (R3 ) and uε ≥ 0 for any x ∈ R3 . Let ε be small enough, it follows from the paper [26] that
1 2 |∇ uε |2 = K1 + O ε 2 , 1 u2ε dx = O ε 2 ,
|uε |26 = K2 + O(ε), (2.5)
R3
where K1 , K2 are positive constants. Moreover K1 K2
= S.
(2.6)
Lemma 2.6. Suppose the hypotheses (k1 )–(k2 ) hold, then: If 2 < q < 4, there exists u1 ∈ D1,2 (R3 ) such that 6
sup Iλ (tu1 ) < Λ − Dλ 6−q . t ≥0
If 4 ≤ q < 6, there exists u2 ∈ D1,2 (R3 ) such that sup Iλ (tu2 ) < Λ. t ≥0
(2.7)
8
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Proof. We shall prove that there exist tε > 0 and positive constants t0 , T1 independent of ε, λ, such that supt ≥0 Iλ (tuε ) = Iλ (tε uε ) and 0 < t0 ≤ tε ≤ T1 < ∞.
(2.8)
In fact, since limt →+∞ Iλ (tuε ) = −∞, there exists tε > 0, such that Iλ (tε uε ) = sup Iλ (tuε ),
dIλ (tuε )
and
dt
t ≥0
= 0.
(2.9)
t = tε
It follows from (2.9) that 2
3
4
5
tε a∥uε ∥ + tε b∥uε ∥ − tε
uε dx − λtε
q −1
6
R3
R3
k(x)uqε dx = 0,
and a∥uε ∥2 + 3tε2 b∥uε ∥4 − 5tε4
R3
u6ε dx − λ(q − 1)tεq−2
R3
k(x)uqε dx ≤ 0.
(2.10)
On one hand, as 2 < q < 6, we can obtain easily from (2.10) that tε is bounded below, that is, there exists a positive constant t0 > 0 (independent of ε, λ), such that 0 < t0 ≤ tε . On the other hand, it follows from (2.9) that a∥ uε ∥ 2 tε2
+ b∥uε ∥2 = tε2
R3
u6ε dx + λtεq−4
R3
k(x)uqε dx,
we see that tε is bounded above for all ε > 0 small enough. Thus (2.8) holds. Set h(tε uε ) =
a 2
tε2 ∥uε ∥2 +
t6 tε4 ∥uε ∥4 − ε 4 6 b
R3
u6ε dx.
Firstly, we claim that there exists a positive constant C3 (independent of ε, λ), such that 1
h(tε uε ) ≤ Λ + C3 ε 2 .
(2.11)
Indeed, let g¯ (t ) =
a 2
2
b
2
t ∥ uε ∥ +
4
4
t6
4
t ∥ uε ∥ −
6
u6ε dx.
R3
As limt →∞ g¯ (t ) = −∞, g¯ (0) = 0, and limt →0+ g¯ (t ) > 0, it follows that supt ≥0 g¯ (t ) attained at Tε > 0, that is g¯ ′ (t )|Tε = aTε ∥uε ∥2 + bTε3 ∥uε ∥4 − Tε5
R3
u6ε dx = 0,
that is Tε4
R3
u6ε dx − a∥uε ∥2 − bTε2 ∥uε ∥4 = 0,
so that
Tε =
b ∥ uε ∥ 4 +
b2 ∥uε ∥8 + 4a∥uε ∥2 2
R3
R3
u6ε dx
u6ε dx
12 .
Note that g¯ (t ) is increasing in the interval [0, Tε ], then by (2.5) and (2.6) one gets h(tε uε ) ≤ g¯ (Tε )
=
=
ab∥uε ∥6 4
ab 4
R3
u6ε dx
∥ uε ∥ 2 |uε |26
b3 ∥uε ∥12
+ 24
3 +
R3
b3 24
2 +
u6ε dx
∥ uε ∥ 2 |uε |26
3 (b2 ∥uε ∥8 + 4a∥uε ∥2 R3 u6ε dx) 2 2 24
6 +
1 24
R3
u6ε dx
b∥uε ∥8 + 4a∥u∥2 |u|66
|u|86
23
C.-Y. Lei et al. / Computers and Mathematics with Applications (
=
abS 3
)
–
9
3
(b2 S 4 + 4aS ) 2
b3 S 6
+ + 4 24 1 = Λ + O ε2 .
1 + O ε2
24
Therefore, there exists C3 > 0 (independent of ε, λ) such that (2.11) holds. √ Since 2 < q < 6, by (k2 ), let ε > 0 be small enough such that 0 < ε < ρ1 , then
R3
k(x)uqε dx ≥ C
q
|x − x0 |−β ε 4
+
q
(ε + |x − x0 |2 ) 2
|x−x0 |<ρ1 ρ1
q
≥ Cε 4
r2 ρ √1 ε
β
q
3
= Cε 2 − 4 − 2
t2 q
t β (1 + t 2 ) 2
0
≥ Cε
1
3 − 4q − β2 2
dt
dt
2q t β
0 β
q
3
t2
k(x)uqε dx
dr
q
r β (ε + r 2 ) 2
0
|x−x0 |>ρ1
= C4 ε 2 − 4 − 2 ,
(2.12)
where C4 (independent of ε, λ). Case: 2 < q < 4. since 3 − 1
C3 ε 2 −
λ q
3
< β < 3, then
2q 3
β
q
6
T12 C4 ε 2 − 4 − 2 = C3 λ 6−q −
=λ
6
λ q
C3 −
6−q
18−4q−6β 6−q 12
T12 C4 λ 6−q 1 q
T12 C4
λ
< 0. Let ε = λ
3 − 4q − β2 2
18−4q−6β 6−q
12 6−q
, 0 < λ < λ0 =
T12 C4
q(C3 +D)
q 4q+66−β− 18
, then
6
< −Dλ 6−q . Therefore, from (2.11) and (2.12), one has tq Iλ (tε uε ) = h(tε uε ) − λ ε q 1
≤ Λ + C3 ε 2 −
λ q
Ω
k(x)uqε dx 3
q
β
T12 C4 ε 2 − 4 − 2
6
< Λ − Dλ 6−q . Hence, there is u1 ∈ D1,2 (R2 ) such that (2.7) holds. β q Case: 4 ≤ q < 6. since 0 < β < 3, thus 32 − 4 − 2 < 1
C3 ε 2 −
λ q
3
q
1 , 2
for every λ > 0, when ε > 0 is small enough, then
β
T12 C4 ε 2 − 4 − 2 < 0.
Using (2.11) and (2.12) again, we obtain Iλ (tε uε ) < Λ, therefore, there exists u2 ∈ D1,2 (R2 ) such that supt ≥0 Iλ (tu2 ) < Λ. The proof is complete. Theorem 2.7. Under the assumptions of Theorem 1.1, problem (1.1) has at least a positive solution. 6
Proof. For 2 < q < 4, there exists δ ′ > 0 such that Λ − Dλ 6−q > 0 for every λ ∈ (0, δ ′ ). Set λ∗ = min{δ ′ , λ0 }, then Lemmas 2.1–2.4 and 2.6 hold when λ ∈ (0, λ∗ ). By Lemma 2.2, Iλ satisfies the geometry of the mountain-pass lemma [27]. Applying the mountain-pass lemma, there exists a sequence {un } ⊂ D1,2 (R3 ), such that Iλ (un ) → c > α,
and Iλ′ (un ) → 0,
where c = inf max Iλ (γ (t )), γ ∈Γ t ∈[0,1]
and
Γ = γ ∈ C ([0, 1], D1,2 (R3 )) : γ (0) = 0, γ (1) = u0 .
(2.13)
10
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
Moreover, by Lemmas 2.5 and 2.6, we can get 6
0 < α < c ≤ max Iλ (t u¯ ) ≤ sup Iλ (t u¯ ) < Λ − Dλ 6−q . t ∈[0,1]
(2.14)
t ≥0
Note that Iλ (|un |) = Iλ (un ), we know that {un } ⊂ D1,2 (R3 ) has a convergent subsequence, still denoted by {un } (un ≥ 0), and there exists u∗ ∈ D1,2 (R3 ) such that weakly in D1,2 (R3 ),
un ⇀ u∗ ,
un → u∗ ,
strongly inLp (R3 )(1 ≤ p < 6).
Set limn→∞ ∥un ∥ = l, for every φ ∈ D1,2 (R3 ), since ⟨Iλ′ (un ), φ⟩ → 0 as n → ∞, there holds
(a + bl ) 2
(∇ u∗ , ∇φ)dx −
R3
u∗ φ dx − λ 5
R3
R3
k(x)u∗q−1 φ dx = 0.
In particular, we have
(a + bl2 )∥u∗ ∥ −
R3
u6∗ dx − λ
R3
k(x)uq∗ dx = 0.
(2.15)
As ⟨Iλ′ (un ), un ⟩ → 0 holds true as n → ∞, consequently
(a + b∥un ∥2 )∥un ∥2 −
R3
u6n dx − λ
R3
k(x)uqn dx = o(1).
By Lemma 2.4, there holds
R3
u6n dx →
R3
u6∗ dx,
as n → ∞,
so that
(a + bl )l − 2
2
u∗ dx − λ 6
R3
R3
k(x)uq∗ dx = 0.
(2.16)
Now, from (2.15) and (2.16), one has ∥u∗ ∥ = l, so we obtain that un → u∗ in D1,2 (R3 ) as n → ∞. Hence, it follows from (2.13) and (2.14) that Iλ (u∗ ) = lim Iλ (un ) = c > α > 0.
(2.17)
n→∞
From (2.17), it follows that u∗ ̸≡ 0, note that u∗ ≥ 0 in D1,2 (R3 ). Therefore, by using the strong maximum principle we obtained that u∗ is a positive solution of (1.1). The proof is complete. Remark 2.8. Under the assumptions of Theorem 1.2, problem (1.1) has at least a positive solution. 3. Proof of theorems Proof of Theorems 1.1 and 1.2. By Theorem 2.7 and Remark 2.8, we obtain the existence of positive solutions of (1.1). In the following we shall prove that problem (1.1) admits a positive ground state solution. Let m = inf Iλ (u) : u ∈ D1,2 (R3 ), u ̸= 0, Iλ′ (u) = 0 .
By the definition of m, there exists {vn } ⊂ D1,2 (R3 ) such that vn ̸= 0, and Iλ (vn ) → m,
Iλ′ (vn ) = 0,
as n → ∞.
(3.1)
Obviously, from (3.1), we can easily deduce that {vn } is bounded in D1,2 (R3 ). There then exist a subsequence of {vn } (still denoted by {vn }) and v∗ ∈ D1,2 (R3 ) such that vn ⇀ v∗ weakly in D1,2 (R3 ). We claim that v∗ ̸= 0. Otherwise, vn ⇀ 0 weakly in D1,2 (R3 ), then by Lemma 2.1, we have
R3
k(x)|vn |q = o(1),
as n → ∞.
By (3.1), we have a∥vn ∥2 + b∥vn ∥4 −
R3
|vn |6 dx − λ
R3
k(x)|vn |q dx = 0.
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
11
Consequently a∥vn ∥2 + b∥vn ∥4 −
R3
|vn |6 dx = 0.
(3.2)
Set limn→∞ ∥vn ∥ = l. By (3.2) and (1.3), we can deduce that 2
l ≥
√
bS 3 +
b2 S 6 + 4aS 3 2
.
Therefore, using (3.1) and (3.2) again, one gets
b
a
1
∥vn ∥2 + ∥vn ∥4 − n→∞ 2 4 6 a b = lim ∥vn ∥2 + ∥vn ∥4
m = lim
3
n→∞
= ≥
al2 3
+
abS 3 4
|vn |6 dx − R3
λ q
R3
k(x)|vn |q dx
12
b 12
+
l4
b3 S 6 24
= Λ.
3
+
(b2 S 4 + 4aS ) 2 24
6
By Lemma 2.6, we have Λ ≤ m < Λ − Dλ 6−q (or Λ ≤ m < Λ), this is a contradiction. Then vn ⇀ v∗ ̸= 0 in D1,2 (R3 ) and Iλ′ (v∗ ) = 0 (moreover, vn → v∗ in D1,2 (R3 ), and we also prove that v∗ is a positive solution of (1.1)), it suggests that Iλ (v∗ ) ≥ m. We claim that m ≥ Iλ (v∗ ). Indeed, since Iλ′ (v∗ ) = 0, one gets
λ b 1 ∥v∗ ∥2 + ∥v∗ ∥4 − v∗6 dx − k(x)v∗q dx 2 4 6 R3 q R3 1 − a∥v∗ ∥2 + b∥v∗ ∥4 − v∗6 dx − λ k(x)v∗q dx 6 R3 R3 b 1 1 a − k(x)v∗q dx. = ∥v∗ ∥2 + ∥v∗ ∥4 − λ
Iλ (v∗ ) =
a
3
12
q
6
R3
By Iλ′ (vn ) = 0, we have m + o(1) = Iλ (vn )
=
a 3
∥vn ∥2 +
b 12
∥vn ∥4 − λ
1 q
−
1 6
R3
k(x)vnq dx
and the Fatou’s Lemma, one obtains m≥
a 3
∥v∗ ∥2 +
b 12
∥v∗ ∥4 − λ
1 q
−
1 6
R3
k(x)v∗q dx = Iλ (v∗ ).
Then v∗ ̸= 0 satisfies Iλ′ (v∗ ) = 0 and Iλ (v∗ ) = m. Consequently, v∗ is a positive ground state solution of (1.1). Therefore, we prove the existence of positive ground state solutions of (1.1). The proof is complete. Acknowledgment The authors thank the referee for his or her careful reading of the paper and for the helpful comments, which led to an improvement of this paper. References [1] G. Kirchhoff, Mechanik, Teubner, Leipzig, 1883. [2] J.L. Lions, On some questions in boundary value problems of mathematical physics, in: Contemporary Development in Continuum Mechanics and Partial Differential Equations, in: North-Holland Math. Stud., vol. 30, North-Holland, Amsterdam, New York, 1978, pp. 284–346. [3] K. Perera, Z.T. Zhang, Nontrivial solutions of Kirchhoff-type problems via the Yang index, J. Differential Equations 221 (2006) 246–255. [4] Z.T. Zhang, K. Perera, Sign changing solutions of Kirchhoff type problems via invariant sets of descent flow, J. Math. Anal. Appl. 317 (2006) 456–463. [5] A. Ourraoui, On a p-Kirchhoff problem involving a critical nonlinearity, C. R. Acad. Sci., Paris I 352 (2014) 295–298. [6] A. Fiscella, E. Valdinoci, A critical Kirchhoff type problem involving a nonlocal operator, Nonlinear Anal. 94 (2014) 156–170. [7] C.O. Alves, F.J.S.A. Corrêa, G.M. Figueiredo, On a class of nonlocal elliptic problems with critical growth, Differ. Equ. Appl. 23 (2010) 409–417.
12
C.-Y. Lei et al. / Computers and Mathematics with Applications (
)
–
[8] G.M. Figueiredo, Existence of a positive for a Kirchhoff problem type with critical growth via truncation argument, J. Math. Anal. Appl. 401 (2013) 706–713. [9] A. Hamydy, M. Massar, N. Tsouli, Existence of solution for p-Kirchhoff type problems with critical exponents, Electron. J. Differential Equations 105 (2011) 1–8. [10] T.F. Ma, Remarks on an elliptic equation of Kirchhoff type, Nonlinear Anal. 63 (2005) 1967–1977. [11] F. Cammaroto, L. Vilasi, Multiple solutions for a Kirchhoff-type problem involving the p(x)-Laplacian operator, Nonlinear Anal. 74 (2011) 1841–1852. [12] G.M. Figueiredo, J.R.S. Junior, Multiplicity of solutions for a Kirchhoff equation with subcritical or critical growth, Differential Integral Equations 25 (2012) 853–868. [13] D. Naimen, The critical problem of Kirchhoff type elliptic equations in dimension four, J. Differential Equations 257 (2014) 1168–1193. [14] C.Y. Lei, J.F. Liao, C.L. Tang, Multiple positive solutions for Kirchhoff type of problems with singularity and critical exponents, J. Math. Anal. Appl. 421 (2015) 521–538. [15] C.Y. Lei, C.M. Chu, H.M. Suo, C.L. Tang, On Kirchhoff type problems involving critical and singular nonlinearities, Ann. Polon. Math. 114 (2015) 269–291. [16] X.M. He, W.M. Zou, Existence and concentration behavior of positive solutions for a Kirchhoff equation in R3 , J. Differential Equations 252 (2012) 1813–1834. [17] S. Liang, S.Y. Shi, Soliton solutions to Kirchhoff type problems involving the critical growth in RN , Nonlinear Anal. 81 (2013) 31–41. [18] S. Liang, J. Zhang, Existence of solutions for Kirchhoff type problems with critical nonlinearity in R3 , Nonlinear Anal. Real World Appl. 17 (2014) 126–136. [19] J. Wang, L. Tian, J. Xu, F. Zhang, Multiplicity and concentration of positive solutions for a Kirchhoff type problem with critical growth, J. Differential Equations 253 (2012) 2314–2351. [20] J. Liu, J.F. Liao, C.L. Tang, Positive solutions for Kirchhoff-type equations with critical exponent in RN , J. Math. Anal. Appl. 429 (2015) 1153–1172. [21] J. Sun, T. Wu, Ground state solutions for an indefinite Kirchhoff type problem with steep potential well, J. Differential Equations 256 (2014) 1771–1792. [22] G.B. Li, H.Y. Ye, Existence of positive ground state solutions for the nonlinear Kirchhoff type equations in R3 , J. Differential Equations 257 (2014) 566–600. [23] M. Willem, Minimax Theorems, Birthäuser, Boston, 1996. [24] P.L. Lions, The concentration-compactness principle in the calculus of variations, the limit case, Part I, Rev. Mat. Iberoam. 1 (1985) 145–201. [25] P.L. Lions, The concentration-compactness principle in the calculus of variations, the limit case, Part II, Rev. Mat. Iberoam. 2 (1985) 45–121. [26] H. Brézis, L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (1983) 437–477. [27] A. Ambrosetti, P.H. Rabinowitz, Dual variational methods in critical point theory and applications, J. Funct. Anal. 14 (1973) 349–381.