Painlev´e Analysis for Supersymmetric Extensions of the Sawada-Kotera Equation

Painlev´e Analysis for Supersymmetric Extensions of the Sawada-Kotera Equation

Li Li^a, Meng-Xia Zhang^b, and Shu-Qiang Lv^c

aAnhui Agricultural University, School of Economics and Technology, Anhui 230601, China

bChina University of Mining and Technology, Beijing 100083, China

cCollege of Arts and Science of Beijing Union University, Beijing 100191, China Reprint requests to M.-X. Z.; E-mail: zmx@cumtb.edu.cn

Z. Naturforsch.66a,165 – 171 (2011); received February 4, 2010 / revised June 26, 2010

In this paper, Painlev´e analysis of supersymmetric extensions of the Sawada-Kotera (SK) equation is performed. It is shown that only two simple supersymmetric extensions of the Sawada-Kotera equation pass the Painlev´e test. One of them was proposed by Tian and Liu, the other one is a B- extension of the SK equation.

Key words:Supersymmetric Sawada-Kotera Equation; Supersymmetric Extension;

Painlev´e Analysis.

1. Introduction

The following fifth-order evolution equation, u_t+u_xxxxx+5uu_xxx+5u_xu_xx+5u²u_x=0, (1) is a well-known system in soliton theory. It was pro- posed by Sawada and Kotera [1], also by Caudrey, Dodd, and Gibbon independently more than thirty years ago [2]. So it is referred as Sawada-Kotera (SK) equation or Caudrey-Dodd-Gibbon-Sawada-Ko- tera equation in literature. Now there are a lot of pa- pers about it and thus its various properties have been established. For example, its Darboux transformation was given in [3] and [4], its bi-Hamiltonian structure was worked out by Fuchssteiner and Oevel [5], its Painlev´e property was verified by Harada and Oishi [6], its B¨acklund transformation and Lax representa- tion were provided in [7] and [8].

Soliton equations or integrable systems have many interesting extensions and one of them is the su- persymmetric extension. Up to now, many equations such as Korteweg-de Vries (KdV), modified Korteweg- de Vries (mKdV), Kadomtsev-Petviashvili (KP), and nonlinear Schr¨odinger (NLS) equations have been em- bedded into their supersymmetric counterparts. The point is that these supersymmetric systems have also remarkable properties and potential applications.

Our aim is to construct supersymmetric counter- parts for the SK equation. On this regard, the following

0932–0784 / 11 / 0300–0165 $ 06.00 c2011 Verlag der Zeitschrift f¨ur Naturforschung, T ¨ubingen·http://znaturforsch.com

equation, proposed by Carstea [9] based on the Hirota bilinear approach is known as

Φt+Φxxxxx

+

10Φxx(DΦ) +5Φ(DΦxx) +5Φ(DΦ)²

x=0, whereΦ=Φ(x,t,θ)is a fermionic super variable de- pending on the super spatial variables x, θ ^{and the} usual temporal variablet.D=_∂θ^∂ +θ_∂^∂_x is the super derivative. However, there is not much known about its integrabilty apart from the two-soliton solution. Very recently, Tian and Liu [10] obtained another version for the supersymmetric SK (sSK) equation. It reads as

Φt+Φxxxxx+5Φxxx(DΦ) +5Φxx(DΦx) +5Φx(DΦ)²=0. (2) They showed that their system is indeed integrable.

Popowicz [11] studied this system further and showed that it is an remarkable supersymmetric equation with unusal properties.

An interesting question arises: is there any other su- persymmetric system which qualifies as a supersym- metric SK equation? In this paper, we will show that a general supersymmetric extension of the SK equa- tion with free parameters is integrable only for two cases. One is the sSK equation (2) and the other is a B-extended equation [12]. We will adopt the sin- gularity analysis or Painlev´e analysis to reach above conclusion.

The article is organized as follows: In Section 2, we present the supersymmetric extensions of the SK equa- tion containing three free parameters. In Section 3, we proceed with the Painlev´e analysis for the equation ob- tained in Section 2. In Section 4, our conclusion is re- ported.

2. The Supersymmetric Extensions for the SK Equation

In this section, we are to construct a supersymmetric analogy for the SK equation. To this aim, we extend the ordinary fieldsu(x,t)to the superfieldΦ(x,t,θ):

u(x,t)→Φ(x,t,θ) =θ^u(x,t) +ξ(x,t),

where θ is a Grassmannian variable: θ² =0, and ξ(x,t) is a fermionic field. Then we proceed with a direct extension of the SK equation, multiplying each term byθ and rewriting the results of superfields:

ut→Φt,

u²u_x→k₃Φ(DΦ)(DΦx) + (5−k₃)Φx(DΦ)², u_xu_xx→k₂Φx(DΦxx) + (5−k₂)Φxx(DΦx), uu_xxx→k₁Φ(DΦxxx) + (5−k₁)Φxxx(DΦ), uxxxxx→Φxxxxx,

wherek₁,k₂,k₃are the free parameters. We notice that the nonlinear terms do not have the unique extensions in the term of superfields. Therefore, this direct exten- sion supplies us with a supersymmetric version of the SK equation containing three free parameters:

Φt+Φxxxxx+k₁Φ(DΦxxx) + (5−k₁)Φxxx(DΦ) +k₂Φx(DΦxx) + (5−k₂)Φxx(DΦx)

+ (5−k₃)Φx(DΦ)²+k₃Φ(DΦ)(DΦx) =0. (3)

In components, it reads as:

u_t+u_xxxxx+5uu_xxx+5u_xu_xx+5u²u_x

−k₁ξxxxx+ (5−k₁−k₂)ξxξxxx−k₃uξ ξxx

−k₃u_xξ ξx=0,

(4a)

ξt+ξxxxxx+k₁u_xxxξ+k₂u_xxξx

+k₃ξ^uux+ (5−k₁)uξxxx+ (5−k₂)uxξxx

+ (5−k₃)u²ξx=0.

(4b)

It is clear that this system does reduce to the SK equa- tion when the fermionic variable is absent. This equa- tion is our general sSK equation.

3. Painlev´e Analysis

In this section, we perform the Painlev´e analysis for the system (4a – 4b). To this end, we consider an ex- pansion of the component fields about a movable sin- gular manifoldφ(x,t) =0 :

u(x,t) =

∑

j=0

uj(x,t)φ(x,t)^j−α, ξ(x,t) =

∑

j=0ξj(x,t)φ(x,t)^j−β,

(5)

whereξj(x,t)are fermionic fields. In order to simplify the calculation, we use the Kruskal’s ansatz [13, 14]:

u_j(x,t) =u_j(t), ξj(x,t) =ξj(t),

φ(x,t) =x+f(t), (6) where f(t)is an arbitrary function. Besides, we com- mandu₀=0 andξ0=0.

3.1. Leading Order Analysis

In this step, we start with determining the possible values ofα^andβin the expression (5). By substituting u≈u₀φ^−α,ξ ≈ξ0φ^−β into (4a – 4b), we findα =2 andβ is an arbitrary integer.

3.2. Resonance Structure Analysis

It is known that a ‘resonance’ occurs at jwhen the coefficient u_j(t) of the termφ^j−α or the coefficient ξj(t)of the termφ^j−βin the expression (5) is arbitrary.

Substituting (5), (6), andα=2 into the system (4a – 4b) leaves us with the following recursion formula:

0=u_j−5,t+ (j−6)u_j−4f +5(j−6)

∑

n=0(n−2)(n−3)unu_j−n

−k₃

∑

n+2β−3

m=0

∑

·ξmu_j−n+ (5−k₁−k₂)^j+2

∑

n=0 (j+β−3−n)(n−β)

·(n−β−1)(n−β−2)ξj+2β−3−nξn

+5

∑

_n

m=0

∑

u_nu_n−m

(j−n−2)u_j−n+ (j−2)

·(j−3)(j−4)(j−5)(j−6)u_j−k₁

j+2β−3 n=0

∑

·(n−β−1)(n−β−2)(n−β−3)ξj+2β−3−nξn, (7)

0=ξj−5,t+ (j−β−4)ξj−4f +k₃

∑

_j−n

m=0

∑

ξn

+ (5−k₃)

∑

n=0

_j−n

m=0

∑

u_nu_j−n−m

(n−β)ξn

+k₂

∑

n=0(n−β)(j−n−2)(j−n−3)uj−nξn

+k₁

∑

(j−n−2)(j−n−3)(j−n−4)u_j−nξn

+ (5−k₂)

∑

n=0(n−β)(n−β−1)(j−n−2)u_j−nξn

+ (5−k₁)

∑

n=0(n−β)(n−β−1)(n−β−2)uj−nξn

+ (j−β)(j−β−1)(j−β−2)

·(j−β−3)(j−β−4)ξj.

(8)

Solving these equations we must rememember that u_j(t)andξj(t)are even and odd elements of the Grass- man algebra, respectively, i. e.

u_j=d_j+U_j, d_j∈C, U_jis nilpotent, ξ²j =0. Forj=0, we get

0=u³₀+18u²₀+72u₀+σ1u₀−σ2−σ3, (9) 0=ξ0

(5−k₃)β+2k₃ u²₀

+β(β+1)(β+2)(β+3)(β+4) +

(5−k₁)β(β+1)(β+2)

+2(5−k₂)β(β+1) +6k₂β+24k₁ u₀

. (10)

Forβ≤1, we have

σ1=σ2=σ3=0, u₀=−12, or u₀=−6. Forβ≥2, we obtain

σ1= k₃ 10

2β−3

m=0

∑

10

2β−3

∑

n=0(n−β)(n−β−2)(n−β−3)

·(n−β−4)ξ2β−3−nξn, σ3= 1

10(5−k₁−k₂)^2β

∑

n=0

(β−3−n)(n−β)

·(n−β−1)(n−β−2)ξ2β−3−nξn,

σi^β=0, i=1,2,3, and

u₀=d₀+U₀=−6+1 2(1+√

3i)(X+Y) or

u₀=d₀+U₀=−12+1 2(1−√

3i)(X+Y), where

X=−

√3 3 (√

3+i)KL, Y=−

√3 3 (√

3−i)KLV+

√3(√ 3+i) 36 σ1V, V=1+1

3KL+ (1

3KL)²+ (1

3KL)³+..., K=

1 5184i√

3(2592i√

3PH−648σ1−108σ2−108σ3), L=1+1

3K+5

9K²+10

81K³+..., H=1+1

4P+1 4P+1

8P³+..., P=− 1

559872

46656σ1+1620σ1²+12σ1³

+972σ1σ2+972σ3σ1+81σ2²

+162σ3σ2+81σ3²

with i²=−1. (10) implies thatξ0is an arbitrary odd Grassmania function oft.

Now we determine resonances and recursion rela- tions foru_j andξj. We can rewrite (7) and (8) in the form:

F_j=u_jP(j) +X_jξ0ξj+2β−3, G_j=ξjQ(j) +Y_jξ0u_j, where

P(j) = (j−2)(j−3)(j−4)(j−5)(j−6) +5(j−6)(j²−5j+12)u₀+5(j−6)u²₀

−k₃

2β−3 m

∑

(m−β)(j+m−β−3)ξ2β−3−mξm, Q(j) = (5−k₃)(j−β)−2k₃

u²₀ + (5−k₁)(j−β)(j−β−1)(j−β−2)

−2(5−k₂)(j−β)(j−β−1) +6k₂(j−β)−24k₁

u₀ + (j−β)(j−β−1)(j−β−2)(j−β−3)(j−β−4),

Xj= (5−k₁−k₂)β(j+β−3) (β+1)(β+2)

−(j+β−4)(j+β−5)

+k₁ (j+β−3)(j+β−4)

·(j+β−5)(j+β−6)−β(β+1)(β+2)(β+3) , Y_j= (5−k₂)β(β+1)(j−2) +k₃(j−4)u₀

−2β(5−k₃)u₀+k₁(j−2)(j−3)(j−4)

−(5−k₁)β(β+1)(β+2)−k₂β(j−2)(j−3),

F_j=−5

j−1 n=1

∑

_n

m=1

∑

u_mu_n−m

(j−n−2)u_j−n−u_j−5,t

−(j−6)u_j−4f−5(j−6)

∑

n=1(n−2)(n−3)u_nu_j−n +k₃

j−1

∑

n=1

_n+2β−3

m=1

∑

·ξn+2β−3−mξm

u_j−n+k₁

j+2β−4 n=1

∑

(n−β)(n−β−1)

·(n−β−2)(n−β−3)ξj+2β−3−nξn−(5−k₁−k₂)

·^j+2

∑

n=1 (j+β−3−n)(n−β)(n−β−1)

·(n−β−2)ξj+2β−3−nξn,

Gj=−k₂

j−1

∑

n=1(n−β)(j−n−2)(j−n−3)u_j−nξn

−(5−k₁)^j−1

∑

n=1

(n−β)(n−β−1)(n−β−2)u_j−nξn

−(5−k₂)^j−1

∑

n=1(n−β)(n−β−1)(j−n−2)u_j−nξn

−k₃

j−1

∑

n=1

_j−n

m=1

∑

ξn

−(5−k₃)^j−1

∑

n=1

_j−n

m=1

∑

u_mu_j−n−m

(n−β)ξn−ξj−5,t

−(j−β−4)ξj−4f

−k₁

j−1

∑

n=1(j−n−2)(j−n−3)(j−n−4)u_j−nξn. Thus, the resonances of the system (4a – 4b) are the roots of the equationsP(j) =0 andQ(j) =0, respec- tively. We will discuss the resonances and the corre- sponding values of the parametersk₁,k₂,k₃for the case β≤1 and the caseβ≥2 separately.

Forβ≤1,P(j) =0, we have

(j−2)(j−3)(j−4)(j−5)(j−6)

+5(j−6)(j²−5j+12)u₀+5(j−6)u²₀=0. (11) In this case, we have u₀ =−12 or u₀ =−6. In the case u₀=−12, the set of the roots of (11) is {−2,−1,5,6,12}, which contains two negative inte- gers, so (4a – 4b) could not possess the generalized Painlev´e property. Therefore, we takeu₀=−6.

Substituting u₀=−6 into the equationsP(j) =0 andQ(j) =0, respectively, gives

P(j) = (j+1)(j−2)(j−3)(j−6)(j−10), (12) Q(j) = (j+2−β) β⁴+ (12−4j)β³

+

6j²−36j+29+6k₁ β²+

−4j³+36j²−58j

−12k₁j+30k₁+12k₂+42 β+

j⁴−12j³+29j² +6k₁j²−12k₂j+42j−30k₁j−36k₃−72k₁

. (13) Then the resonances of recursion relations (7) and (8) are roots ofP(j) =0 (i. e. j=−1,2,3,6,10) and roots j₁,j₂,j₃,j₄,j₅ofQ(j) =0, where the resonance j=

−1 corresponds to the arbitrary functionφdefining in the singular manifold.

Forβ=1, substitutingβ=1 andu₀=−6 into (10), we have

108k₁+12k₂−36k₃=0, (14) and (13) gives

Q(j) = j(j+1) j³−16j²+ (6k₁+71)j

−(42k₁+12k₂+56)

. (15)

Suppose that three roots of the cubic equation in the square brackets in (15) are j₁,j₂,j₃. Then we obtain three equations from (15):

j₁+j₂+j₃=16, (16)

j₁j₂+j₁j₃+j₂j₃=6k₁+71, (17) j₁j₂j₃=42k₁+12k₂+56. (18) Combining the last two equations with (14), we have the values of k₁,k₂,k₃ in terms of the three undeter- mined roots j₁,j₂,j₃:

k₁=1

6(j₁j₂+j₁j₃+j₂j₃−71), (19) k₂= 1

12[j₁j₂j₃−7(j₁j₂+j₁j₃+j₂j₃)+441], (20)

Table 1.j4=−1,j₅=0.

Case (j1,j2,j3,j4,j5) (k1,k2,k3) (a.1) (1,2,13,−1,0) (−5,15,−10) (a.2) (1,3,12,−1,0)

−¹⁰3,10,−²⁰3

(a.3) (1,4,11,−1,0) (−2,6,−4) (a.4) (1,5,10,−1,0) (−1,3,−2)

(a.5) (1,6,9,−1,0)

−3¹,1,−²3

(a.6) (1,7,8,−1,0) (0,0,0) (a.7) (2,3,11,−1,0)

−3⁵,²⁰3,−²⁵9

(a.8) (2,4,10,−1,0)

−2¹,¹⁵4,−¹4

(a.9) (2,5,9,−1,0) ₁

3,⁵3,¹⁴9

(a.10) (2,6,8,−1,0) ₅

6,12⁵,⁹⁵36

(a.11) (3,4,9,−1,0) ₂

3,2,⁸3

(a.12) (3,5,8,−1,0) ₄

3,²₃,³⁸₉ (a.13) (3,6,7,−1,0) ₅

3,0,5 (a.14) (4,5,7,−1,0) (2,0,6)

Table 2.j4=−2,j5=0.

Case (j1,j2,j3,j4,j5) (k1,k2,k3) (b.1) (1,2,9,−2,0) (0,5,0) (b.2) (1,3,8,−2,0) (1,3,2)

(b.3) (1,4,7,−2,0) ₅

3,⁵3,¹⁰3

(b.4) (1,5,6,−2,0) (2,1,4) (b.5) (2,3,7,−2,0) (2,2,4)

(b.6) (2,4,6,−2,0) ₅

2,⁵4,5 (b.7) (3,4,5,−2,0) (3,1,6)

k₃= 1

36[j₁j₂j₃+2(j₁j₂+j₁j₃+j₂j₃)−198]. (21) Now we need to find the values ofj₁,j₂,j₃. Since they are positive distinct integers, we can list up all the pos- sible combinations of them satisfying the constraint (16). In this way, we obtain Table 1 with j₄=−1 and

j₅=0.

Forβ=0, substitutingβ=0 andu₀=−6 into (10), we get

144k₁−72k₃=0, and (13) in this case yields

Q(j) = (j+2)j

j³−12j²+ (29+6k₁)j +(−12k₂−30k₁+42) . Similarly, we get Table 2 withj₄=−2 and j₅=0.

Forβ≥2, as a result ofu₀=d₀+U₀, we can write P(j) =0 andQ(j) =0 in the following form:

P(j) = (j−2)(j−3)(j−4)(j−5)(j−6)

+5(j−6)(j²−5j+12)d₀+5(j−6)d²₀+p,

Q(j) = (5−k₃)(j−β)−2k₃

d₀² + (5−k₁)(j−β)(j−β−1)(j−β−2)

−2(5−k₂)(j−β)(j−β−1) +6k₂(j−β)−24k₁

d₀ + (j−β)(j−β−1)(j−β−2)(j−β−3)(j−β−4)+q, where

p=10(j−6)U₀²+5(j−6)(j²−5j+12)U₀

−k₃

2β−3 m=0

∑

(m−β)(j+m−β−3)ξ2β−3−mξm, q=2 (5−k₃)(j−β)−2k₃

U₀²

+ (5−k₁)(j−β)(j−β−1)(j−β−2)

−2(5−k₂)(j−β)(j−β−1)+6k₂(j−β)−24k₁

U₀. The contribution to the resonances is the non-fermioic part ofP(j) =0 (also withQ(j) =0), namely

(j−2)(j−3)(j−4)(j−5)(j−6) +5(j−6)(j²−5j+12)d₀+5(j−6)d₀². Since we have knownd₀=−12 ord₀=−6, similarly to (11), we taked₀=−6. So the resonances of the re- cursion relations (7) and (8) are also roots ofP(j) =0 and roots ofQ(j) =0.

Forβ=2, substitutingβ=2 andu₀=−6 into (10), we get

0≡0.

Substitutingβ=2 andu₀=−6 into (13), we get j

j⁴−20j³+ (6k₁+125)j²−(54k₁+250+12k₂)j + (144+24k₂+156k₁−36k₃)

.

Similarly to Table 1, we get Table 3 with j₅=0.

3.3. Resonance Criteria Analysis

As the last step, we have to check the resonance cri- teria for β =0,1,2. We take advantage of the sym- bolic software Maple 11. The final result is that only the three cases (a.6), (b.1), and (c.15), see Tables 1 – 3, have satisfied all criteria.

Therefore, the system (4a – 4b) has the Painlev´e property only for

case (i): k₁=k₂=k₃=0 and case (ii): k₁=k₃=0,k₂=5.

Case (j1,j2,j3,j4,j5) (k1,k2,k3) Case (j1,j2,j3,j4,j5) (k1,k2,k3) (c.1) (1,2,3,14,0) (−5,15,−10) (c.2) (1,2,4,13,0)

−¹⁰₃,10,−²⁰₃ (c.3) (1,2,5,12,0) (−2,6,−4) (c.4) (1,2,6,11,0) (−1,3,−2) (c.5) (1,2,7,10,0)

−¹₃,1,−²₃

(c.6) (1,2,8,9,0) (0,0,0) (c.7) (1,3,4,12,0)

−⁵3,²⁰3,−²⁵9

(c.8) (1,3,5,11,0)

−2¹,¹⁵4,−¹4

(c.9) (1,3,6,10,0) ₁

3,⁵3,¹⁴9

(c.10) (1,3,7,9,0) ₅

6,12⁵,⁹⁵36

(c.11) (1,4,5,10,0) ₂

3,2,⁸3

(c.12) (1,4,6,9,0) ₄

3,²3,³⁸9

(c.13) (1,4,7,8,0) ₅

3,0,5

(c.14) (1,5,6,8,0) (2,0,6) (c.15) (2,3,4,11,0) (0,5,0) (c.16) (2,3,5,10,0) (1,3,2) (c.17) (2,3,6,9,0) ₅

3,⁵3,¹⁰3

(c.18) (2,3,7,8,0) (2,1,4) (c.19) (2,4,5,9,0) (2,2,4) (c.20) (2,4,6,8,0) ₅

2,⁵4,5 (c.21) (2,5,6,7,0) (3,1,6) (c.22) (3,4,5,8,0) (3,2,5) (c.23) (3,4,6,7,0) ₁₀

3,⁵3,⁵⁰9

Table 3.j4=0,j5=0.

For case (i), we get the sSK equation

Φt+Φxxxxx+5Φxxx(DΦ) +5Φxx(DΦx) +5Φx(DΦ)²=0. (22) For case (ii), we obtain the sSK equation

Φt+Φxxxxx+5Φxxx(DΦ) +5Φx(DΦxx) +5Φx(DΦ)²=0. (23) Equation (22) is identical to (2), which was proposed by Kai and Liu while (23) is the B-extension of the SK equation.

To complete the Painlev´e analysis of the two cases, we must check the resonance of their other branches.

For case (i), from (10) and (13), the resonances are given by

1. β=−6, j₁=−8, j₂=−7, j₃=−6, j₄=0, j₅=1;

2. β=−7, j₁=−9, j₂=−8, j₃=−7, j₄=−1, j₅=0;

and for case (ii), from (10) and (13), the resonances are 1. β=−1, j₁=−3, j₂=−1,

j₃=0, j₄=1, j₅=8;

2. β=−2, j₁=−4, j₂=−2, j₃=−1, j₄=0, j₅=7;

3. β=−9, j₁=−11, j₂=−9, j₃=−8, j₄=−7, j₅=0.

One can convince oneself that the compatibility condi- tions for these branches are satisfied.

However, there are still other possibilities that must be considered, namely those withβ≥3. Here we only

prove that the compatibility condition is contradictory forβ =3. Substitutingu₀=−6 andβ =3 into (10), we get

k₃=15+6k₁+k₂. (24)

From (7),j=−2β+4, one obtains the condition 2ξ0ξ1(β+1)(k₂β²−5β²+3β^2k1

+3k₁β−k₂β+5β+k₃u₀) =0, (25) and then it is easy to see that

ξ0ξ1=0

which is in contradiction with the arbitrariness ofξ0

andξ1, because of j=1 and j=0 are two roots of Q(j) =0 withβ=3.

So, the result of the Painlev´e analysis is that the sSK equations possesses Painlev´e property if and only if k₁=k₂=k₃=0 andk₁=k₃=0,k₂=5. That is we only obtain two integrable supersymmetric extensions of the SK equations, and one is the sSK equation (2) and the other is the B-extension of the SK equation.

4. Discussions

We have shown that the sSK equation containing three parameters possesses the Painlev´e property for k₁=k₂=k₃=0 andk₁=k₃=0,k₂=5, which corre- spond to the sSK equation (2) and a B-extension of the SK equation, respectively. So no new integrable sSK equation could arise by the means of the Painlev´e analysis and the one proposed by Castera does not appear in our study in particular. As a further prob- lem, it would be interesting to study systematically the

Painlev´e property of the supersymmetric extensions of the fifth-order KdV type equations in the form

u_xxxxx+a₁uu_xxx+a₂u_xu_xx+a₃u²u_x+a₄u_t=0, wherea₁,a₂,a₃are parameters. We may return to this problem later and hope some new supersymmetric equations of fifth order will appear.

Acknowledgement

We would like to thank Q. P. Liu for interesting dis- cussions. This work is supported by the National Na- ture Science Foundation of China with grant numbers 10671206 and 10971222. Also, it is supported by the Fundamental Research Funds for the Central Universi- ties.

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