Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 8 (2015), 423–433
Research Article
Some new results for power means
Yu-Qin Xua , Kun Chena,∗, Jing-Feng Tianb
a
School of Electrical and Electronic Engineering, North China Electric Power University, Baoding, Hebei Province, 071003, P. R.
China.
b
College of Science and Technology, North China Electric Power University, Baoding, Hebei Province, 071051, P. R. China.
Communicated by Krzysztof Cieplinski
Abstract
In this paper, we establish some new inequalities for power means with n positive numbers. Moreover, some
new properties of p 7→ Mn (a, p) are obtained, where Mn (a, p) denotes the p-th power mean of first n entry
c
of vector a. 2015
All rights reserved.
Keywords: Mapping, power mean, ratio.
2010 MSC: 26D15, 26D07.
1. Introduction and Preliminaries
Given n positive numbers a1 , a2 , · · · , an , denote a = (a1 , a2 , · · · , an ), we recall that the classical arithmetic mean An (a), the geometric mean Gn (a), the harmonic mean Hn (a), and finally Mn (a, p), the p-th
power mean, are respectively defined by
v
u n
Pn
uY
a
n
k=1 k
n
An (a) =
, Gn (a) = t
ak , Hn (a) = Pn 1 ,
n
k=1 a
k=1
and
∗
 n
1

 1X p p
ak , p 6= 0,
Mn (a, p) =
n
k=1

 √
n a a ···a ,
p = 0.
1 2
n
k
(1.1)
Corresponding author
Email addresses: xuyuqin_ncepu@126.com (Yu-Qin Xu), chenk_ncepu@163.com (Kun Chen), tianjfhxm_ncepu@163.com
(Jing-Feng Tian)
Received 2015-02-04
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
424
We can see that Mn (a, −1) = Hn (a), Mn (a, 0) = Gn (a), Mn (a, 1) = An (a). Then, multivariate means
of classical arithmetic, geometric and harmonic are special cases of power mean, and the relations of these
means can be written by next inequalities
Hn (a) = Mn (a, −1) ≤ Gn (a) = Mn (a, 0) ≤ An (a) = Mn (a, 1)
(1.2)
To date, some excellent methods have been proposed to prove and establish inequalities. For example, in
[9], Ibrahim and Dragomir established inequalities by utilizing power series and Young’s inequality. In [10],
Kouba used classical analysis to obtain some new inequalities. In [12], V. Mascioni discover new inequalities
by the differences of some Stolarsky means. Due to the importance of power mean in modern mathematics,
it has also been given considerable attention by mathematicians. Many remarkable results for power mean
have been presented in the literature (see, for example, [2, 3, 4, 5, 6, 11, 14, 15, 16, 17] and the references
cited therein).
The main purpose of this paper is to establish some new inequalities for Mn (a, p) and to give some new
1
properties for p 7→ limn→∞ Mn (a, p)/Mn (a, p + 1) and p 7→ limn→∞ Mn (a, p1 )/Mn (a, p+1
) in the case when
a is an arithmetic sequence.
2. Some known results
In this section we restate some Lemmas and Theorems, which relate to our main results.
Lemma 2.1 (Hölder’s inequality, [8]). Let ak , bk ≥ 0, for k = 1, 2, · · · , n, and let
Then
n
n
n
1
X
1 X
X
p
q
bqk ,
ak bk ≤
apk
k=1
1
p
+
1
q
= 1 with p, q > 1.
(2.1)
k=1
k=1
with equality if and only if apk = τ bqk (k = 1, 2, · · · , n), τ is constant.
The p-th power mean type Hölder’s inequality can be described by the following Theorem.
Theorem 2.2 (Hölder’s inequality [1], p.211). If ak , bk > 0 (k = 1, 2, · · · , n), and
Then
Mn (ab, 1) ≤ Mn (a, p)Mn (b, q),
1
p
+
1
q
= 1 with p > 1.
(2.2)
where a = (a1 , a2 , · · · , an ), b = (b1 , b2 , · · · , bn ), ab = (a1 b1 , a2 b2 , · · · , an bn ).
Lemma 2.3 ([1], p.31). Let ψ(x) be twice differentiable and ψ 00 (x) ≥ 0. Then
ψ
n
1 X
n
xk
k=1
n
1X
≤
ψ(xk ).
n
(2.3)
k=1
Lemma 2.4 (Minkowski’s inequality, [13]). Let ak , bk ≥ 0 (k = 1, 2, · · · , n) and p > 1. Then
n
n
n
1 X
1
hX
i1 X
p
p
p
apk
+
bpk .
(ak + bk )p ≤
k=1
k=1
(2.4)
k=1
The inequality is reversed for p < 1 (p 6= 0). In each case, the sign of equality holds if and only if ak = bk
(k = 1, 2, · · · , n).
Lemma 2.5 (Jensen’s inequality[7], p.28). Let ak > 0 (k = 1, 2, · · · , n), 0 < r < s. Then
n
X
k=1
ask
1
s
<
n
X
k=1
ark
1
r
.
(2.5)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
425
Theorem 2.6 ([7], p.26). Let a1 , a2 , · · · , an be a positive sequence, n ∈ N. If p ∈ R, then Mn (a, p) is increasing for fixed a1 , a2 , · · · , an ; if p ∈ (−∞, 0)∪(0, +∞), then Mn (a, p1 ) is decreasing for fixed a1 , a2 , · · · , an .
Lemma 2.7 ([1], p.26). Let Φ be a differentiable function defined on D, if
Ψ(x) =
Φ(x) − Φ(x0 )
,
x − x0
x, x0 ∈ D,
x 6= x0 .
Then Φ(x) is (strictly) convex if and only if Ψ(x) is (strictly) increasing on D.
3. Some new inequalities for power means
In this section, we establish some new inequalities for Mn (a, p) and Mn (a, p1 ).
Theorem 3.1. Let ak > 0, k = 1, 2, · · · , n, p ∈ [1, +∞). Then
n
1 X
n
1
akp
p
≤
k=1
n
1 X
n
apk
1
p
k=1
max {ak } X
n
1 p
1
1≤k≤n
p
≤ p
a
,
Q
k
n
n
n
k=1 ak
(3.1)
k=1
the sign of equality holds if and only if a1 = a2 = · · · = an .
Proof. Let
F (p) =
Mn (a, p)
.
Mn (a, p1 )
(3.2)
By Theorem 2.6, we know that F (p) is an increasing function for p ∈ (0, +∞).
Note that
F (1) = 1,
lim Mn (a, p)
lim F (p) =
p→+∞
p→+∞
1
lim Mn (a, )
p→+∞
p
=
max {ak }
1≤k≤n
p
.
Qn
n
k=1 ak
(3.3)
(3.4)
Then, for p ∈ [1, +∞), we have
max {ak }
1≤k≤n
1 ≤ F (p) ≤ p
.
Qn
n
k=1 ak
(3.5)
Combining equations (3.2), (3.3) and inequality (3.5) lead to inequality (3.1) immediately. The proof of
Theorem 3.1 is completed.
Corollary 3.2. Let a1 , a2 , · · · , an be an increasing arithmetic sequence in inequality (3.1), p ≥ 1. Then we
obtain
n
n
n
1 X
1 X
1
1 X
1 p
1 p
p
akp ≤
apk
<e
akp ,
(3.6)
n
n
n
k=1
k=1
k=1
and the bound e for the right-side of inequality (3.6) is optimal.
Remark 3.3. Let ak = k in inequality (3.6), p ≥ 1. Then we have
n
1 X
n
k=1
1
kp
p
≤
n
1 X
n
kp
1
p
k=1
n
n
1 X
1 p
n 1 X p1 p
p
≤ √
k
<
e
k
.
n
n
n! n k=1
k=1
(3.7)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
426
Using Lemma 2.3, we can obtain the following Theorem easily.
Theorem 3.4. Let ak > 0 (k = 1, 2, · · · , n), p > 1. Then
1
1
1
Mn (a, ) ≤ Mnp (ap , ),
p
p
(3.8)
The inequality is reversed for 0 < p < 1, where ap = (ap1 , ap2 , · · · , apn ).
Corollary 3.5. Let ak , bk > 0 (k = 1, 2, · · · , n), p, q > 1. Then
1
1
1
1
1
1
Mn (a, )Mn (b, ) ≤ Mnp (ap , )Mnq (bq , ),
p
q
p
q
1
p
Theorem 3.6. Let ak , bk > 0 (k = 1, 2, · · · , n), and let
1
+
1
q
(3.9)
= 1 with p, q > 1. Then
1
max{Mn (ab, 1), Mnp (a p , p)Mnq (b q , q)} ≤ Mn (a, p)Mn (b, q).
(3.10)
Proof. By Theorem 2.2, we can obtain
Mn (ab, 1) ≤ Mn (a, p)Mn (b, q).
(3.11)
And by Theorem 3.4, we have
Mn (a,
Let p0 =
1
p
1
1
0 1
p0
(ap , 0 ),
)
≥
M
n
0
p
p
0 < p0 < 1.
(3.12)
in inequality (3.12). Then
1
Mn (a, p) ≥ Mnp (a p , p),
p > 1.
(3.13)
q > 1.
(3.14)
Similarly,
1
Mn (b, q) ≥ Mnq (b q , q),
Combining equations (3.11), (3.13) and (3.14) lead to equation (3.10) easily.
Remark 3.7. Let 0 < a1 < a2 < · · · < an , 0 < b1 < b2 < · · · < bn in Theorem 3.6. Then we have
1
1
Mnp (a p , p)Mnq (b q , q) ≤ Mn (ab, 1) ≤ Mn (a, p)Mn (b, q).
(3.15)
Theorem 3.8. Let ak , bk ≥ 0 (k = 1, 2, · · · , n), λ, µ > 0, and let p ≥ 1. Then
Mn (λa + µb, p) ≤ λMn (a, p) + µMn (b, p).
(3.16)
The inequality is reversed for p < 1.
Proof. Using Minkowski inequality in Lemma 2.4, we have
Mn (λa + µb, p) =
≤
n
1 X
n
(λa + µb)p
1
p
k=1
n
1 h X
n1/p
k=1
(λa)p
1
p
+
n
X
k=1
= λMn (a, p) + µMn (b, p).
Therefore, we obtain the desired result (3.16).
(µb)p
1 i
p
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
427
Remark 3.9 (Minkowski’s inequality [7], p.30). Let λ = µ = 1 in Theorem 3.8. Then we can easily obtain
the p-th power mean type Minkowski’s inequality
Mn (a + b, p) ≤ Mn (a, p) + Mn (b, p).
(3.17)
The inequality is reversed for p < 1.
Theorem 3.10 (Young’s inequality). Let ak , bk ≥ 0 (k = 1, 2, · · · , n), λ + µ = 1, λ > 0, and let p ≥ 1.
Then
Mn (aλ bµ , p) ≤ λMn (a, p) + µMn (b, p).
(3.18)
Proof. By using inequality
Aλ B µ ≤ λA + µB,
λ + µ = 1, A, B, λ > 0,
we get
Mn (aλ bµ , p) ≤ Mn (λa + µb, p).
(3.19)
Then, combining inequalities (3.16) and (3.19), we obtain the desired result (3.18).
Theorem 3.11. If ak ≥ 0 (k = 1, 2, · · · , n), 0 < ρ ≤ ν. Then
ρ
Mn (a, ν1 )
1 h Mn (a, ν1 ) i ν
1
≥ ν−ρ .
1 ≥ nν−ρ
1
n
Mn (a, − ρ )
Mn (a, − ν )
(3.20)
Proof. Using Jensen inequality in Lemma 2.5, and simple computations lead to
X
ν X
X
n
ν n
n
n
1
1
1
1 ρ
1
1 X ν1 ρ ρ
ν
ak ·
= ν+ρ
ak
1
1
n
n
n
ρ
ρ
k=1
k=1 a
k=1 a
k=1
k
k
X
ν ρ
X
n n
n n
X
1
ρ
1
1 X ν1 ρ
1
1
ν
ak
ak
= ν+ρ
≥ ν+ρ
1
1
n
n
ν
ν
k=1 ak k=1
k=1 ak k=1
≥
(3.21)
n2ρ
1
= ν−ρ .
ν+ρ
n
n
The proof of Theorem 3.11 is completed.
4. Some known results
In this section, we give some new properties for the ratio of power means.
Theorem 4.1. Let {an } be an increasing arithmetic sequence with a1 > 0, n ∈ N, denote Cn (a, p) =
Mn (a,p)
Mn (a,p+1) , and C(p) = lim Cn (a, p). Then
n→∞
√
p+2
√
, p ∈ (−1, 0) ∪ (0, ∞),
p
p+1
2
C(p) =

,
p = 0,


 e
0,
p ≤ −1,





p+1
and C(p) is strictly increasing on (−1, +∞).
(4.1)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
Proof. Denote a =
d
a1
428
(d is tolerance). Then simple computation leads to
n
1 X
Cn (a, p) =
apk
1
p
n
Mn (a, p)
= nk=1 1
Mn (a, p + 1)
1 X p+1 p+1
ak
n
k=1
n
1 X
d p1
[1 + (k − 1) ]p
n
a1
= nk=1
1
1X
d
p+1
[1 + (k − 1) ]p+1
n
a1
k=1
(4.2)
n
1 X
1
p
[1 + (k − 1)a]p
n
= nk=1
1
1X
p+1
[1 + (k − 1)a]p+1
n
k=1
n h
1 X
1 + (k − 1)a ip p1
n
n
= nk=1
.
1 X h 1 + (k − 1)a ip+1 1
p+1
n
n
k=1
Let ψ(x) = xp . Then applying Lagrange mean value theorem for ψ(x) on interval
have
h 1 + (k − 1)a ip h (k − 1)a ip
1
−
= pξkp−1 · ,
n
n
n
where
(k − 1)a
1 + (k − 1)a
< ξk <
.
n
n
Denote
n 1 X h 1 + (k − 1)a ip h (k − 1)a ip
rn =
−
.
n
n
n
h
(k−1)a 1+(k−1)a
n ,
n
i
, we
(4.3)
k=1
Now, we prove that
lim rn = 0.
(4.4)
n→∞
We divide three cases to prove equation (4.4).
Case I. When p ≥ 1, we get
rn =
<
1
n1+p
1
n1+p
+
n
n−1
1 X p−1
1
p X 1 + ka p−1
pξ
+
≤
k
n2
n1+p n2
n
k=2
n−1
X
α 1
+ ·
n n
k=1
k=1
k p−1
,
n
α = p(a + 1)
(4.5)
p−1
.
Moreover,
n−1
1 X k p−1
=
n→∞ n
n
Z
lim
k=1
0
1
1
xp−1 dx = ,
p
Thus, inequality (4.5) and equation (4.6) lead to equation (4.4) immediately.
(4.6)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
429
Case II. When 0 < p < 1, simple computation leads to
rn =
<
<
1
n1+p
1
n1+p
+
+
n
n
p X p−1
p X 1 1−p
1
+
ξ
=
k
n2
n1+p n2
ξk
p
n2
k=2
n−1
X
k=1
k=2
n−1
β X n 1−p
n 1−p
1
= 1+p + 2
ka
n
n
k
(4.7)
k=1
β
1
+
→0
np+1 np
(n → ∞),
where β = pap−1 .
Obviously, inequality (4.7) is equivalent to inequality (4.4)
Case III. When −1 < p < 0, denote t = −p (0 < t < 1). Then
n
1
t X −t−1 ξk
rn = 1−t − 2
n
n
<
<
<
Note that
1
n1−t
1
n1−t
1
n1−t
+
t
n2
t
+ 2
n
+
γ
k=2
n−1
X
k=1
n−1
X
1 t+1
ξk
(4.8)
n t+1
ka
k=1
n
X
n1−t
k=1
1
,
k
n
1 X1
= 0,
n→∞ nε
k
lim
γ=
t
at+1
.
ε > 0.
(4.9)
k=1
Thus, inequality (4.8) and equation (4.9) lead to equation (4.4) immediately.
Equations (4.3) and (4.4) imply that
X
1
n h
1 + (k − 1)a ip p
1
n
n
lim Cn (a, p) = lim nk=1
1
n→∞
n→∞
1 X h 1 + (k − 1)a ip+1 p+1
n
n
k=1
1
n
1 X h (k − 1)a ip p
lim
n→∞ n
n
k=1
=
1
n
1 X h (k − 1)a ip+1 p+1
lim
n→∞ n
n
k=1
1
R1
√
p
p+1
p
p+2
0 (ax) dx
= R
.
1 = √
p
1
p+1
(ax)p+1 dx p+1
(4.10)
0
Hence,
√
p+1
p+2
C(p) = √
,
p
p+1
p ∈ (−1, 0) ∪ (0, +∞).
(4.11)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
Note that
v
u n
uY
n
t
ak
430
n
1 X ak
ln
n
an
k=1
Mn (a, 0)
e k=1
= lim
=
lim
n
n
n→∞ Mn (a, 1)
n→∞ 1 X
n→∞
1 X ak
ak
n
n
an
C(0) = lim
k=1
2
= ,
e
(4.12)
k=1
and
i
h 1
1
√
ln(p + 2) − ln(p + 1)
lim
2
p+2
p
lim C(p) = lim √
= .
= ep→0 p + 1
p
p→0
p→0
e
p+1
Thus, C(p) is continuous at p = 0, and we obtain
 p+1
√
p+2

√
, p ∈ (−1, 0) ∪ (0, +∞),
p
C(p) =
p+1
 2
p = 0.
e,
p+1
(4.13)
(4.14)
Similarly, when p ≤ −1, we can obtain
C(p) = 0.
(4.15)
Combining equations (4.14) and (4.15) lead to equation (4.1) immediately.
Next, we prove that C(p) is strictly increasing on (−1, +∞).
Denote
g(p + 1) − g(p)
f (p) = lnC(p) =
, (p > −1),
(4.16)
p+1−p
where
1
g(p) = ln(p + 1), g(0) = 1.
(4.17)
p
By the Lemma 2.7, we know that f (p) is strictly increasing on (−1, +∞) if and only if g(p) is strictly
convex on (−1, +∞).
Simple computations lead to
p
− ln(p + 1)
(p + 1)
0
g (p) =
, p ∈ (−1, 0) ∪ (0, +∞),
(4.18)
p2
1
g 0 (0) = lim g 0 (p) = − ,
p→0
2
h(p)
g 00 (p) = 3 , p ∈ (−1, 0) ∪ (0, +∞),
p
where
(4.19)
(4.20)
h(p) = 2ln(1 + p) −
3p2 + 2p
.
(p + 1)2
(4.21)
lim h(p) = −∞,
h(0) = 0,
(4.22)
Simple computations give
p→−1+
2
g 00 (0) = lim g 00 (p) = ,
p→0
3
(4.23)
2p2
≥ 0, p ∈ (−1, +∞).
(4.24)
(p + 1)3
Hence, h(p) is increasing on (−1, +∞). It follows from (4.21) and (4.22) together with the monotonicity of
h(p) that h(p) < 0 when p ∈ (−1, 0), and h(p) > 0 when p ∈ (0, +∞). Combining equations (4.20) and
(4.23) lead to g 00 (p) > 0 immediately. Hence g(p) is strictly convex on (−1, +∞).
Therefore, C(p) is strictly increasing on (−1, +∞). The proof of Theorem 4.1 is completed.
h0 (p) =
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
431
Theorem 4.2. Let {an } be an increasing arithmetic sequence with a1 > 0, n ∈ N, denote En (a, p) =
Mn (a, p1 )
1
Mn (a, p+1
)
, and E(p) = lim En (a, p). Then
n→∞
 p
p (p + 2)p+1


,
p > 0,

(p + 1)2p+1
E(p) =
(−p)p (−p − 2)p+1



, p < −2.
(−p − 1)2p+1
(4.25)
Moreover, E(p) is decreasing on (0, +∞) and increasing on (−∞, −2).
Proof. Denote a =
d
a1
(d is tolerance), p = −t (t > 2). When p ∈ (−∞, −2), simple computation leads to
Mn (a, − 1t )
1
Mn (a, 1−t
)
n
1 X
− 1 −t
ak t
n
= nk=1
1 1−t
1 X − t−1
ak
n
k=1
X
n h
i 1 t−1
n
1
t−1
d
n
1 + (k − 1) a1
= k=1n h
i 1 t
1X
n
t
d
n
1 + (k − 1) a1
k=1
X
n h
i 1 t−1
n
1
En (a, p) = En (a, −t) =
(4.26)
t−1
n
1 + (k − 1)a
k=1
X
n h
i 1 t
1
n
t
n
1 + (k − 1)a
=
.
k=1
− 1t
h
Let ϕ(x) = xi , then ϕ0 (x) = − 1t x
(k−1)a 1+(k−1)a
, we have
n ,
n
− 1t −1
. Applying Lagrange mean value theorem for ϕ(x) on interval
h 1 + (k − 1)a i− 1
t
n
(k−1)a
n
where
Denote
< ξk <
−
h (k − 1)a i− 1
t
n
1 − 1 −1 1
= − ξk t · ,
t
n
1+(k−1)a
.
n
n 1
1 X h 1 + (k − 1)a i− 1t h (k − 1)a i− 1t δn = −
.
1 +
n
n
n
n1− t
k=2
(4.27)
lim δn = 0.
(4.28)
Now, we prove that
n→∞
Simple computations lead to
δn ≤
1
1
n1− t
1
<
n
1− 1t
1
=
n
1− 1t
+
+
+
n
1 X 1 1+ 1t
tn2
ξk
1
tn2
1
k=2
n−1
X
k=1
n−1
X
1− 1t
cn
n 1+ 1t
ka
k=1
1 1+ 1t
,
k
(4.29)
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
where c = ta1+1/t .
432
n−1
X
1 1+ 1t
is convergent for t > 2.
n→∞
k
k=1
Combining equation (4.27) and inequality (4.29) lead to equation (4.28). Thus
By the property of p-series, we know that lim
n−1
1 t−1
1 X ka − t−1
lim
n→∞ n
n
k=1
lim En (a, −t) =
n−1
n→∞
1 X ka − 1t t
lim
n→∞ n
n
k=1
R
t−1
1
1
− t−1
dx
0 (ax)
= R
t
1
− 1t
dx
(ax)
0
=
(t − 1)2t−1
,
tt (t − 2)t−1
(4.30)
(t > 2).
Let t = −p in the last equation of (4.30), we obtain
E(p) =
(−p − 1)−2p−1
(−p)p (−p − 2)p+1
=
.
(−p)−p (−p − 2)−p−1
(−p − 1)2p+1
(4.31)
Next, we prove E(p) is increasing on (−∞, −2).
When p < −2, denote
φ(p) = lnE(p) = pln(−p) + (p + 1)ln(−p − 2) − (2p + 1)ln(−p − 1).
(4.32)
Simple computations lead to
φ0 (p) = ln(−p) + ln(−p − 2) − 2ln(−p − 1) +
φ00 (p) =
3p + 4
> 0.
p(p + 1)2 (p + 2)2
Then, φ0 (p) is increasing on (−∞, −2).
Note that
lim φ0 (p) = lim
p→−∞
p→−∞
1
1
−
,
p+1 p+2
h
ln
(4.33)
(4.34)
i
p(p + 2)
1
+
= 0.
(p + 1)2
(p + 1)(p + 2)
Thus
φ0 (p) > 0,
(4.35)
Therefore, φ(p) is increasing on (−∞, −2).
When p ∈ (0, +∞). By the same method, we can obtain
E(p) =
pp (p + 2)p+1
.
(p + 1)2p+1
And E(p) is decreasing on (0, +∞). The proof of Theorem 4.2 is completed.
Acknowledgements
The authors would like to express their sincere thanks to the anonymous referees for making great
efforts to improve this paper. This work was supported by the Fundamental Research Funds for the Central
Universities (Grant no. 13ZD19).
Y.-Q. Xu, K. Chen, J.-F. Tian, J. Nonlinear Sci. Appl. 8 (2015), 423–433
433
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