Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 43, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLICITY OF SOLUTIONS TO THE SUM OF
POLYHARMONIC EQUATIONS WITH
CRITICAL SOBOLEV EXPONENTS
WEI LIU, GAO JIA, LU-QIAN GUO
Abstract. In this article, we prove multiplicity of solutions for the sum of
polyharmonic equation with critical Sobolev exponent. The proof is based
upon the methods of weakly lower semi-continuous of the functionals and the
Mountain Pass Lemma without (PS) conditions.
1. Introduction
In this article, we discuss the multiplicity of solutions for the sum polyharmonic
equation
k
X
i
(−∆) u = λ|u|q−2 u + |u|N −2 u + µf (x), in Ω,
(1.1)
i=0
u ∈ H0k (Ω),
where Ω ⊂ R is a bounded smooth domain, k is positive integer, q is a real number
with 2 < q < N , N = 2n/(n−2k) is the critical Sobolev exponent in the embedding
H0k (Ω) ,→ LN (Ω), λ, µ are both positive real parameters and f (x) is continuous
with not identical to 0 in Ω. Our main result is the following theorem.
n
Theorem 1.1. Let Ω ⊂ Rn be a bounded smooth domain, n > 2k and f (x) be
continuous and not identical to 0 in Ω. Then there exist λ0 > 0 and µ0 > 0, such
that for any λ > λ0 and 0 < µ < µ0 , problem (1.1) admits at least two distinct
weak solutions u1 with positive energy and u2 with negative energy.
Remark 1.2. For the highest order term (−∆)k u of problem (1.1), we need to
discuss that k is odd or even. In fact, no matter k is odd or even, we obtain the
similar result of Theorem 1.1. For the sake of simplicity, in the following discussion,
we let k be an even, that is k = 2m and m is positive integer.
Higher-order elliptic boundary problems have abundant applications in physics
and engineering [16] and have also been studied in many areas of mathematics,
including conformal geometry [12], some geometry invariants [5] and non-linear
elasticity [13].
2000 Mathematics Subject Classification. 35J30, 35J60.
Key words and phrases. Polyharmonic equation; multiple solutions; critical Sobolev exponent.
c
2015
Texas State University - San Marcos.
Submitted November 21, 2014. Published February 17, 2015.
1
2
W. LIU, G. JIA, L.-Q. GUO
EJDE-2015/43
The existence of the solutions of the Brezis-Nirenberg problem [9] for the higherorder equations has been studied in many papers [1, 3, 7, 11, 14, 18]. Grunau [15]
considered the existence of positive solution for semilinear polyharmonic Dirichlet
problem with critical Sobolev exponent
s−2
(−∆)k u = λu + |u|
α
D u = 0, |α| ≤ k − 1
u in B,
on ∂B,
(1.2)
where k ∈ N, B is the unit ball centered at the origin, λ ∈ R, n > 2k, s =
2n/(n − 2k) is the critical Sobolev exponent. He proved the existence of a positive
radial solution for: λ ∈ (0, λ1 ), if n ≥ 4k; λ ∈ (λ, λ1 ) for some λ = λ(n, k) ∈ (0, λ1 ),
if 2k + 1 ≤ n ≤ 4k − 1, where λ1 is the first eigenvalue of (−∆)k with homogeneous
Dirichlet boundary conditions.
Recently, Benalili and Tahri [6] considered the multiplicity of solutions considered
for the equation
∆2 u − ∇i (aρ−σ ∆i u) + bρ−µ u = λ|u|q−2 u + f (x)|u|s−2 u
(1.3)
2n
is the
where the function a(x) and b(x) are smooth on M and 1 < q < 2. s = n−4
critical Sobolev exponent. They proved that when 0 < σ < 2 and 0 < µ < 4, there
is λ∗ > 0 such that if λ ∈ (0, λ∗ ), the equation (1.3) possesses at least two distinct
nontrivial solutions in the distribution sense.
The multiplicity of solutions for higher-order equations can be founded in [4] and
the references therein.
Here, our motivation comes from the recent papers [6, 15]. We consider the situation of the multiplicity of the higher-order equation with critical Sobolev exponent
when k ≥ 1 and q > 2.
The paper is organized as follows. In Section 2, we will introduce the Sobolev
spaces and the embedding theorem which is applicable to problem (1.1). In Section
3, since a lack of compactness, we use analytic techniques and variational arguments
to overcome the difficulty and establish some basic lemmas. In Section 4, we give
the proof of two distinct weak solutions of Theorem 1.1. Our methods are mainly
based on the weakly lower semi-continuous of the functional and the Mountain Pass
Lemma without (PS) condition.
2. Preliminaries
Suppose Ω ⊂ Rn is a bounded smooth open domain. We let H02m (Ω) be the
Sobolev space which is the completion of the space C0∞ (Ω) with respect to the
norm
kukH2m = (k∆m uk22 + k∇∆m−1 uk22 + · · · + k∇uk22 + kuk22 )1/2 .
(2.1)
It is well known that a weak solution of the equation (1.1) is a critical point of
the following functional
Z
1
2
2
2
2
Iλ,µ (u) =
((∆m u) + |∇∆m−1 u| + · · · + (∆u) + |∇u| + u2 )
2 Ω
Z
Z
Z
(2.2)
1
1
q
N
|u| −
|u| − µ
f (x)u.
− λ
q Ω
N Ω
Ω
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MULTIPLICITY OF SOLUTIONS
3
Under the above assumptions, it is easy to know that Iλ,µ (u) ∈ C 1 (H02m (Ω), R)
and with the Gâteaux derivative
Z
h∇Iλ,µ (u), vi =
(∆m u∆m v) + (∇∆m−1 u · ∇∆m−1 v) + · · · + ∇u · ∇v + uv)
Ω
Z
Z
Z
q−2
N −2
−λ
|u| uv −
|u|
uv − µ
f (x)v
Ω
Ω
Ω
(2.3)
for every v ∈
H02m (Ω)
(see [15]).
Lemma 2.1 (Mountain Pass Theorem [2]). Let E be a real Banach space and let
I(u) ∈ C 1 (E, R). Suppose I(0) = 0 and
(I1) there is a constant ρ > 0 such that I|∂Bρ (0) > 0,
(I2) there is an e ∈ E\Bρ (0) such that I(e) ≤ 0.
Set
C = inf sup I(γ(t)) > 0
γ∈Γ t∈[0,1]
(2.4)
where Γ denotes the class of paths joining 0 to e. Conclusion: there is a sequence
{uk } in E, such that
I(uk ) → C
and ∇I(uk ) → 0
in a dual space E 0 .
Lemma 2.2 (Sobolev-Rellich-Knodrakov Theorem [19]). Assume that Ω ⊂ Rn is
a bounded domain with Lipschitz boundary, k is positive integer and 1 ≤ p < ∞.
Then the following hold:
np
• if n > kp, then W k,p (Ω) ,→ Ls (Ω), for 1 ≤ s ≤ p∗ = n−kp
;
np
• the embedding is compact, for s < n−kp .
3. Basic Lemmas
To complete the proof of Theorem 1.1, the following lemmas are our main tools.
Lemma 3.1. For each fixed λ > 0, there exist δ > 0, µ0 > 0 and η > 0, such that
for all u ∈ H02m (Ω) with kukH2m = δ and any 0 < µ < µ0 , it holds Iλ,µ (u) > η > 0.
Proof. From (2.1), (2.2) and the Hölder inequality, we deduce that
Z
1
2
Iλ,µ (u) =
((∆m u) + (∇∆m−1 u)2 + · · · + (∆u)2 + |∇u|2 + u2 )
2 Ω
Z
Z
Z
λ
1
q
N
−
|u| −
|u| − µ
f (x)u
(3.1)
q Ω
N Ω
Ω
q
1
1
1
λ
1− N
≥ kuk2H2m − |Ω|1− N kukqN − kukN
kukN .
N − µ max f (x)|Ω|
x∈Ω
2
q
N
By (3.1) and Lemma 2.2, we infer that
q
1
λ
1
Iλ,µ (u) ≥ kuk2H2m − (C)q |Ω|1− N kukqH2m − (C)N kukN
H2m
2
q
N
1
− µ max f (x)|Ω|1− N CkukH2m
x∈Ω
1
N −2
=
− λC1 kukq−2
−
C
kuk
·
kuk
−
µC
2
3 kukH2m ,
H2m
H2m
H2m
2
with some positive constants C1 , C2 , C3 and 2 < q < N .
4
W. LIU, G. JIA, L.-Q. GUO
EJDE-2015/43
Thus for any λ > 0, there exist δ = δ(λ) > 0, sufficiently small µ0 = µ0 (δ) > 0,
and η0 = η0 (µ0 ) > 0, such that for all u ∈ H02m (Ω) with kukH2m = δ and for any
0 < µ < µ0 , it holds Iλ,µ (u) > η0
Lemma 3.2. Suppose f (x) is continuous and not identical to 0 in Ω. For any
µ0 > 0, there exist λ0 > 0 and v0 ∈ H02m (Ω), such that for any λ ≥ λ0 , we have
2m ∗ −n/(2m)
0 < sup Iλ,µ (tv0 ) <
(C )
,
(3.2)
n
t≥0
where C ∗ is the best Sobolev constant of H02m (Ω) ,→ LN (Ω), N = 2n/(n − 4m).
Proof. By the conditions of f (x), we can choose v0 ∈ H02m (Ω), such that
Z
Z
N
f (x)v0 > 0 and
|v0 | = 1.
Ω
Ω
Thus from (2.2), we obtain
t2
λ
Iλ,µ (tv0 ) = kv0 k2H2m − tq
2
q
Z
1
|v0 | − tN − tµ
N
Ω
q
Z
f (x)v0 .
(3.3)
Ω
For any λ, µ > 0, we have
lim Iλ,µ (tv0 ) = −∞.
(3.4)
t→+∞
Using Lemma 3.1 and (3.4), there exists tλ,µ > 0, such that
Iλ,µ (tλ,µ v0 ) = sup Iλ,µ (tv0 ) > 0.
(3.5)
t≥0
By (3.3) and (3.5), one gets
1 2
λ
1
t kv0 k2H2m − ( tqλ,µ kv0 kqq + tN
) − tλ,µ µ
2 λ,µ
q
N λ,µ
Z
f (x)v0 > 0.
(3.6)
Ω
That is,
Z
λ
1 N −q
1
q
2
kv0 kq + tλ,µ
< kv0 kH2m − µ
f (x)v0 .
q
N
2
Ω
By simple analysis, we obtain
1 −q
λ
lim ( kv0 kqq + tN
) = +∞,
λ→+∞ q
N λ,µ
and
lim tλ,µ = 0.
tq−1
λ,µ
λ→+∞
(3.7)
From (3.3) and (3.7), we obtain
q−1
lim λtλ,µ
≤ 0.
λ→+∞
(3.8)
Taking into account of (3.5), (3.7) and (3.8), we obtain
lim sup Iλ,µ (tλ,µ v0 ) = 0.
λ→+∞ t≥0
(3.9)
Then there exist λ0 such that for any λ > λ0 , we have
2m ∗ −n/(2m)
0 < sup Iλ,µ (tv0 ) <
(C )
.
n
t≥0
The proof is complete.
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MULTIPLICITY OF SOLUTIONS
5
Lemma 3.3. For any λ > 0, there exists sufficiently small µ0 > 0, such that for
any 0 < µ < µ0 , the Iλ,µ (u) satisfies the (P S)Cλ,µ -condition for all Cλ,µ in the
interval
2m ∗ −n/(2m)
0 < Cλ,µ <
(C )
.
(3.10)
n
Proof. First we prove that each (P S)Cλ,µ sequence is bounded in H02m (Ω). Let
{uk } ⊂ H02m (Ω) be a (P S)Cλ,µ sequence for Iλ,µ (u), defined by (2.2), i.e.,
Iλ,µ (uk ) → Cλ,µ ,
0
in H02m (Ω) , as k → ∞.
and ∇Iλ,µ (uk ) → 0,
That is,
1
2
Z
2
((∆m uk ) + (∇∆m−1 uk )2 + · · · + (∆uk )2 + |∇uk |2 + u2k )
Ω
Z
Z
Z
1
λ
(3.11)
q
N
|uk | −
|uk | − µ
f (x)uk
−
q Ω
N Ω
Ω
= Cλ,µ + o(1)
Iλ,µ (uk ) =
and
Z
h∇Iλ,µ (uk ), uk i =
2
((∆m uk ) + (∇∆m−1 uk )2 + · · · + (∆uk )2 + |∇uk |2 + u2k )
Z
Z
Z
q
N
−λ
|uk | −
|uk | − µ
f (x)uk
Ω
Ω
Ω
Ω
= o(1)kuk kH2m ,
(3.12)
as k → ∞. By (3.11), (3.12), the Hölder inequality and Lemma 2.2, we obtain
1
Iλ,µ (uk ) − h∇Iλ,µ (uk ), uk i
q
Z
Z
1 1
1
1
1
N
= ( − )kuk k2H2m + ( − )
|uk | − (1 − )µ
f (x)uk
2 q
q
N Ω
q
Ω
Z
(3.13)
1
1 1
f (x)uk
≥ ( − )kuk k2H2m − (1 − )µ
2 q
q
Ω
1
1 1
1
2
≥ ( − )kuk kH2m − (1 − )µ max f (x)|Ω|1− N (C ∗ )kuk kH2m .
2 q
q x∈Ω
It follows from (3.11), (3.12) and (3.13) that
o(1) + Cλ,µ + o(1)kuk kH2m
1
1 1
1
≥ ( − )kuk k2H2m − (1 − )µ max f (x)|Ω|1− N (C ∗ )kuk kH2m ,
2 q
q x∈Ω
i.e.
1 1
( − )kuk k2H2m
2 q
(3.14)
1
1− 1
≤ o(1) + Cλ,µ + (1 − )µ max f (x)|Ω| N (C ∗ ) + o(1) kuk kH2m ,
q x∈Ω
where q > 2. Hence, for each λ , µ > 0, fixed Cλ,µ ∈ R, we conclude that the
sequence {uk } is bounded in H02m (Ω).
Now, we show that the (P S)Cλ,µ sequence contains a strongly convergent subsequence.
6
W. LIU, G. JIA, L.-Q. GUO
EJDE-2015/43
Since the sequence {uk } is bounded in H02m (Ω) and the well-known Sobolev’s
embedding, there exists a subsequence, still denoted by {uk }, and u ∈ H02m (Ω),
such that
uk * u weakly in H02m (Ω),
uk → u strongly in Li (Ω), for 1 < i < N =
2n
,
n − 4m
∇uk → ∇u strongly in L2 (Ω),
∆uk → ∆u strongly in L2 (Ω),
...
∇∆
m−1
uk → ∇∆
m−1
strongly in L2 (Ω),
u,
uk → u a.e. in Ω.
Thus
Z
u2k →
Z
Ω
Z
u2 ,
Ω
Z
2
|∇uk | →
2
|∇u| ,
Ω
Ω
...
Z
Z
2
2
|∇∆m−1 uk | →
|∇∆m−1 u| ,
Ω
Ω
Z
Z
q
q
|uk | →
|u| , q < N,
Ω
Ω
Z
Z
f (x)uk →
f (x)u,
Ω
Ω
as k → ∞. By Brezis-Lieb Lemma [8], we have
k∆m uk k22 − k∆m uk22 = k∆m (uk − u)k22 + o(1),
Z
Z
N
N
(|uk | − |u|N ) =
|uk − u| + o(1).
Ω
Ω
Now, by doing some calculations, we obtain
1
1
Iλ,µ (uk ) − Iλ,µ (u) = k∆m (uk − u)k22 −
2
N
Z
N
|uk − u| + o(1).
(3.15)
Ω
By (3.12) and {uk } being bounded, we have
o(1) = h∇Iλ,µ (uk ), uk − ui
Z
Z
2
N
N
=
(∆m (uk − u)) −
(|uk | − |u| ) + o(1).
Ω
(3.16)
Ω
From (3.15) and (3.16), we have
1
1
(3.17)
Iλ,µ (uk ) − Iλ,µ (u) = ( − )k∆m (uk − u)k22 + o(1).
2 N
On the other hand, the Vitali convergence theorem see [17, chap. III.2] yields
Z
Z
Z
N
N −2
2
N
|uk | −
(|uk |
|uk − u| ) →
|u| , as k → ∞.
(3.18)
Ω
Ω
Ω
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MULTIPLICITY OF SOLUTIONS
7
From (3.15) and (3.18), one gets
Iλ,µ (uk ) − Iλ,µ (u) =
1 m
1
k∆ (uk − u)k22 −
2
N
Z
N −2
|uk |
2
|uk − u| + o(1).
(3.19)
Ω
Using (3.19), Lemma 2.2 and Hölder inequality, we obtain
1
1
2
N −2
Iλ,µ (uk ) − Iλ,µ (u) ≥
− kuk kN
(C ∗ ) k∆m (uk − u)k22 + o(1).
(3.20)
2 N
Taking account of (3.17) and (3.20), we obtain
1
1
1
1
−2
∗ 2
− kuk kN
( − )k∆m (uk − u)k22 ≥
(C
)
k∆m (uk − u)k22 + o(1),
N
2 N
2 N
i.e.
2
N −2
1 − kuk kN
(C ∗ ) k∆m (uk − u)k22 ≤ o(1).
Hence
2
lim sup kuk kN < (C ∗ )− N −2 ,
(3.21)
k→+∞
which implies
k∆m (uk − u)k22 = o(1), k → ∞.
Thus uk → u strongly in H02m (Ω).
Now, we verify (3.21). Using (3.11), (3.12) and that the sequence {uk } is bounded
in H02m (Ω), we have
Z
1
1 1
µ
1
q
+
(
−
)λku
k
−
f uk = Cλ,µ + o(1).
(3.22)
( − )kuk kN
k q
N
2 N
2 q
2 Ω
By (3.14), for any ε > 0 and µ is sufficiently small, we have
Z
µ
| f uk | < ε.
2 Ω
Thus for any λ > 0 and µ is sufficiently small, we obtain
1
1
( − )kuk kN
N ≤ Cλ,µ .
2 N
∗ −n/(2m)
By the assumption 0 < Cλ,µ < 2m
, we have thus (3.21).
n (C )
(3.23)
Lemma 3.4. For all λ > 0 and µ > 0, the function Iλ,µ (u) is weak lower semicontinuous on the set
{u ∈ H02m (Ω) : kukH2m ≤ r0 },
1/(N −2)
where r0 = 22N −2N(C ∗ )N
.
Proof. Let {uk } be a sequence in H02m (Ω), and 0 < r < 22N −3N(C ∗ )N
that
uk * u, in H02m (Ω) and kuk kH2m ≤ r.
1/(N −2)
, such
Then we have kukH2m ≤ r. Up to a subsequence, we obtain
uk → u,
strongly in Lp (Ω), for all p < N,
uk → u a.e. in Ω.
Thus
Z
q
Z
|uk | →
Ω
q
|u| ,
Ω
2 < q < N,
(3.24)
8
W. LIU, G. JIA, L.-Q. GUO
Z
EJDE-2015/43
Z
f (x)uk →
f (x)u.
Ω
(3.25)
Ω
By the Brezis-Lieb Lemma [8], we have
k∆i uk k22 − k∆i uk22 = k∆i (uk − u)k22 + o(1),
k∇∆
i
uk k22
− k∇∆
i
uk22
i
= k∇∆ (uk −
Z
N
u)k22
N
Z
(|uk | − |u| ) =
Ω
i = 0, 1, 2, . . . , m,
(3.26)
i = 0, 1, 2, . . . , m − 1, (3.27)
+ o(1),
N
|uk − u| + o(1).
(3.28)
Ω
From (3.26) and (3.27), we obtain
kuk k2H2m − kuk2H2m = kuk − uk2H2m + o(1).
(3.29)
Using (3.28) and Lemma 2.2, we have
Z
|uk − u|N ≤ (C ∗ )N kuk − ukN
H2m
Ω
≤ (C ∗ )N kuk − uk2H2m 2N −2 (kuk kH2m + kukH2m )N −2
(3.30)
≤ 22N −4 (C ∗ )N rN −2 kuk − uk2H2m .
To sum up (3.24), (3.25), (3.29) and (3.30), we obtain
Z
1
1
N
|uk − u| + o(1)
Iλ,µ (uk ) − Iλ,µ (u) = kuk − uk2H2m −
2
N Ω
1
1
N
≥
− 22N −4 (C ∗ ) rN −2 kuk − uk2H2m + o(1).
2
N
1/(N −2)
N
in above equation,
Taking r = r0 = 22N −2 (C ∗ )N
1
kuk − uk2H2m + o(1).
4
→ 0, as k → ∞, by the (3.15) and (3.16), we have
Iλ,µ (uk ) − Iλ,µ (u) ≥
If kuk − ukH2m
lim inf Iλ,µ (uk ) = Iλ,µ (u).
k→∞
If kuk − ukH2m → 0, as k → ∞, thus
lim inf Iλ,µ (uk ) ≥ Iλ,µ (u).
k→∞
In brief, we obtain
lim inf Iλ,µ (uk ) ≥ Iλ,µ (u).
k→∞
This completes the proof.
4. Proof of main results
Proposition 4.1. Suppose that f (x) is continuous with not identical to 0 in Ω and
λ > 0. For µ0 > 0 small enough such that for any 0 < µ < µ0 , then (1.1) has a
solution with negative energy.
Proof. Since f (x) isR continuous and not identical to 0 in Ω, then there exists φ ∈
H02m (Ω), such that Ω f (x)φ > 0. For any t > 0, we have
Z
1 N
1 2
1 q
2
N
q
f (x)φ.
(4.1)
Iλ,µ (tφ) = t kφkH2m − t kφkN − λ t kφkq − µt
2
N
q
Ω
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MULTIPLICITY OF SOLUTIONS
9
Hence, there exists t0 (λ, µ) > 0, such that 0 < t ≤ t0 (λ, µ) and
Iλ,µ (tφ) < 0.
By Lemma 3.4, there exist r0 > 0 and v ∈ H02m (Ω) with kvkH2m ≤ r0 , such that
Iλ,u (v) =
inf
kukH
2m
≤r0
Iλ,µ (u) < 0.
(4.2)
Thus v is a weak solution of (1.1) with negative energy.
Proposition 4.2. Suppose f (x) is continuous and not identical to 0 in Ω. If λ > 0
is sufficiently large and µ > 0 is enough small, then (1.1) has a weak solution with
positive energy.
Proof. Lemma 3.1 implies that Iλ,µ (u) satisfies the condition (I1) in Lemma 2.1.
On the other hand, from (4.1), we obtain
lim Iλ,µ (tφ) = −∞.
t→+∞
There exists a constant T > 0, taking e = T φ with kekH2m > δ such that
Iλ,µ (e) < 0,
where δ > 0 is the constant in Lemma 3.1. Thus the condition (I2) of Lemma 2.1
holds. Denote
Cλ,µ = inf sup Iλ,µ (γ(t)),
(4.3)
γ∈Γ t∈[0,1]
where
Γ = {γ ∈ C([0, 1], H02m (Ω)) : γ(0) = 0,
γ(1) = e}.
From Lemma 3.2 and (4.3), it follows that
0 < Cλ,µ <
2m ∗ −n/(2m)
(C )
.
n
Applying Lemma 2.1, there exists a sequence {uk } ⊂ H02m (Ω), such that
Iλ,µ (uk ) → Cλ,µ ,
and ∇Iλ,µ (uk ) → 0,
as k → ∞.
By Lemma 3.3, there exists a subsequence of {uk } which strongly converges to u
in H02m (Ω). Thus Iλ,µ (u) has a critical point u with Iλ,µ (u) = Cλ,µ > 0. Hence we
obtain a weak solution of equation (1.1) with positive energy.
Proof of Theorem 1.1. From Propositions 4.1 and 4.2, problem (1.1) has two distinctic solutions u1 , u2 with Iλ,µ (u1 ) < 0 < Iλ,µ (u2 ).
Acknowledgments. This research was supported by the National Natural Science
Foundation of China (11171220), by the Shanghai Leading Academic Discipline
Project (XTKX2012), and by Hujiang Foundation of China (B14005).
10
W. LIU, G. JIA, L.-Q. GUO
EJDE-2015/43
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Wei Liu
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: liuweimaths@hotmail.com
Gao Jia (corresponding author)
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: gaojia89@163.com
Lu-Quian Guo
College of Science, University of Shanghai for Science and Technology, Shanghai
200093, China
E-mail address: xiaoshudian.2008@163.com