Multiple positive solutions for a class of Kirchhoff type equations in \(\mathbb{R}^{N}\)

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R E S E A R C H

Open Access

Multiple positive solutions for a class of

Kirchhoff type equations in

R

N

Lian-bing She

1

_{, Xin Sun}

2

_{and Yu Duan}

2*

*_{Correspondence:}

[email protected]

2_{Collect of Science, GuiZhou}

University of Engineering Science, BiJie, Guizhou 551700, People’s Republic of China

Full list of author information is available at the end of the article

Abstract

In this paper, we study the following nonlinear Kirchhoﬀ type equation:

–

a+b

RN|∇u|

2_dx_u₊_Vu₌_f₍_u_{) +}_h₍_x_), _x_∈_RN_,

wherea,b,Vare positive constants,N= 2 or 3. Under appropriate assumptions onf andh, we get that the equation has two positive solutions by using variational methods.

MSC: 35J60; 35B09; 35A15

Keywords: Kirchhoﬀ type equation; variational methods; Pohozaev equality; monotonicity trick

1 Introduction and main results

We consider the following nonlinear Kirchhoﬀ type equation:

–

a+b

RN|∇

u|2dx

u+Vu=f(u) +h(x), x∈RN, (1.1)

wherea,b,Vare positive constants,N= 2 or 3.

In recent years, the existence or multiplicity of solutions for the following Kirchhoﬀ type equation

–

a+b

RN|∇

u|2dx

u+V(x)u=f(x,u), x∈RN,

wherea,b are positive constants,N= 1, 2, 3, has been widely investigated by many au-thors, for example [1–6], etc. But in those papers, the nonlinearityfsatisﬁes 3-superlinear growth at inﬁnity, which assures the boundedness of any Palais-Smale sequence or Cerami sequence.

Very recently, Guo [7], Li and Ye [8], Liu and Guo [9], Tang and Chen [10] studied re-spectively the following equation:

–

a+b

R3|∇u|

2_dx

u+V(x)u=f(u), x∈R3,

(2)

wherea,bare positive constants,f only needs to satisfy superlinear growth at inﬁnity. By using the Pohozaev equality, it is easy to obtain a bounded Palais-Smale sequence. Thus they obtained the existence of positive solution.

Inspired by [7–10], we study equation (1.1); in here, very weak conditions are assumed onf. Exactly,f ∈C(R+_,_R_{) satisﬁes}

(f1) when N = 2, there exists p∈ (2, +∞) such that limt→+∞_tfp–1(t) = 0; when N = 3,

limt→+∞f_t(5t)= 0;

(f2) limt→0+ f(t)

t =m∈(–∞,V);

(f3) limt→+∞f(tt)= +∞.

Onh, we make the following hypotheses:

(h1) h∈L2₍_RN₎_∩_C1₍_RN₎_{is nonnegative and}_h_≡₀_;

(h2) whenN= 2,0≤(∇h(x),x)∈L2₍_R2₎_{; when}_N_{= 3}_,₍_∇h₍_x_),_x₎_∈_L2₍_R3₎_;

(h3) his radially symmetric.

By using Ekeland’s variational principle [11] and Struwe’s monotonicity trick [12], we get the following.

Theorem 1.1 Suppose that(f1)-(f3)and(h1)-(h3)hold.Then there exists m0> 0such that,

when(_RNh2dx) 1

2 _<_m0_,_equation_(1.1)_{has two positive solutions}_.

Whenf(t) < 0, by (f2) and (f3), there existsl> 0 such thatf(t) +lt≥0 for allt≥0. Thus

equation (1.1) is equivalent to the following equation:

–

a+b

RN|∇u|

2_dx

u+Wu=k(u) +h(x), x∈RN, (1.2)

whereW=V+l> 0 andk(t) =f(t) +lt∈C(R+_,_R+_{) satisﬁes}

(k1) when N = 2, there exists p∈ (2, +∞) such that limt→+∞_tkp–1(t) = 0; when N = 3,

limt→+∞k_t(5t)= 0;

(k2) limt→0+ k(t)

t =m+l:=d∈[0,W);

(k3) limt→+∞k(tt) = +∞.

Hence in order to prove Theorem 1.1, we only need to prove the following.

Theorem 1.2 Suppose that(k1)-(k3)and(h1)-(h3)hold.Then there exists m0> 0such that when(_RNh2dx)

1

2 <m₀,equation(1.2)has two positive solutions.

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2 Preliminaries

From now on, we will use the following notations.

• E:={u∈H1₍_RN_{) :}_u₍_x_{) =}_u₍_|x|₎_}_{is the usual Sobolev space endowed with the norm}

u=

RN|∇

u|2+u2dx

1 2

.

• D1,2₍_RN₎_{is completion of}_C∞

0 (RN)with respect to the norm uD1,2₍_RN₎=

RN|∇u|

2_dx

1 2

.

• For any1≤p<∞,Lp₍_RN₎_{denotes the Lebesgue space and its norm is denoted by}

|u|p=

RN|u|

p_dx

1 p

.

• ·,·denotes the action of dual,(·,·)denotes the inner product inRN_. • C,Cidenote various positive constants.

Since we are looking for positive solution, we may assume thatk(t) = 0 for allt< 0. Under the assumptions onkandh, it is obvious that the functionalI:E→Rdeﬁned by

I(u) =a 2

RN|∇

u|2dx+b 4

RN|∇

u|2dx

2

+W 2

RN

u2dx–

RN

K(u)dx–

RN

hu dx

is of classC1_{, where}_K₍_t_{) =}t

0k(s)dsand

I(u),v =

a+b

RN|∇u|

2_dx

RN(∇u,∇v)dx+W

RNuv dx–

RNk(u)v dx

–

RN

hv dx,

for allu,v∈E. As is well known, the weak solution of equation (1.2) is the critical point of

IinE.

3 Proof of the main results

Next lemma can be viewed as a generalization of Struwe’s monotonicity trick [12] and is the main tool for obtaining a bounded Palais-Smale sequence.

Lemma 3.1(see [13] or [14]) Let X be a Banach space equipped with a norm · X,and let J⊂R+be an interval.We consider a family{μ}μ∈Jof C1-functionals on X of the form

μ(u) =A(u) –μB(u), ∀μ∈J,

where B(u)≥0for all u∈X and such that either A(u)→+∞or B(u)→+∞asuX→

+∞.We assume that there are two points v1,v2in X such that cμ=inf

γ∈tmax∈[0,1]μ

(4)

where

=γ ∈C[0, 1],X:γ(0) =v1,γ(1) =v2

.

Then,for almost everyμ∈J,there is a bounded(PS)cμsequence forμ,that is,there exists

a sequence{un} ⊂X such that (1) {un}is bounded inX, (2) μ(un)→cμ,

(3) _μ(un)→0inX∗,whereX∗is the dual ofX.

Remark 3.2 In [13], it is also proved that, under the assumptions of Lemma 3.1, the map

μ→cμis left-continuous.

In the paper, we set X:=E, · X:= · andJ:= [1₂, 1]. Let us deﬁneIμ:E→Rby

Iμ(u) =A(u) –μB(u), where

A(u) =a 2

RN|∇

u|2dx+b 4

RN|∇

u|2dx

2

+W 2

RN

u2dx–

RN

hu dx,

B(u) =

RNK(u)dx.

Then I1(u) =I(u). By (k1)-(k3) and (h1), it is obvious thatIμ∈C1(E,R),B(u)≥0 for all

u∈EandA(u)≥min{₂a,W}u2–C|h|2u →+∞asu →+∞.

Lemma 3.3 Assume that(k1)-(k3)and(h1)hold.Then there existρ> 0,α> 0and m0> 0

such that Iμ(u)|u=ρ≥αfor all h satisfying|h|2<m0and for allμ∈J.

Proof First, we considerN= 2. It follows from (k1) and (k2) that, for allt∈R, we have

K(t)≤W+d 4 |t|

2₊_C_|_t_|p_. _(3.1)

By (3.1), the Hölder inequality and the Sobolev inequality, for allμ∈Jandu∈E, one has

Iμ(u)≥

a

2

R2|∇u|

2_dx₊W

2

R2u

2_dx_–

R2K(u)dx–

R2hu dx

≥ a

2

R2|∇u|

2_dx₊W

2

R2u

2_dx_–W+d

4

R2u

2_dx_–_C

R2|u|

p_dx_–_|_h_|2|_u_|2

≥ min{2a,W–d}

4 u

2_–_C1_up_–_C2|_h_|2_u

=u

min{2a,W–d}

4 u–C1u

p–1_–_C_2|_h_|2

.

Letg1(t) =min{2a₄,W–d}t–C1tp–1_for_t_≥_{0. Since}_p_{> 2, we know that there exists a constant}

ρ> 0 such thatmaxt≥0g1(t) =g1(ρ) > 0. Choosem0=2C12g1(ρ), then there existsα> 0 such

thatIμ(u)|u=ρ≥αfor allhsatisfying|h|2<m0.

Next whenN= 3, it follows from (k1) and (k2) that, for allt∈R, we have

K(t)≤W+d 4 |t|

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By (3.2), the Hölder inequality and the Sobolev inequality, for allμ∈Jandu∈E, one has

Iμ(u)≥

a

2

R3|∇u|

2_dx₊W

2

R3u

2_dx_–

R3K(u)dx–

R3hu dx

≥ a

2

R3|∇u|

2_dx₊W

2

R3u

2_dx_–W+d

4

R3u

2_dx_–_C

R3|u|

6_dx_–_|h|2|u|2

≥ min{2a,W–d}

4 u

2_–_C3u6_–_C4|h|2u

=u

min{2a,W–d}

4 u–C3u

5_–_C4|h|2

.

Letg2(t) = min{2a₄,W–d}t–C3t5_for_t_≥_{0, we know that there exists a constant}_ρ_{> 0 such}

that maxt≥0g2(t) =g2(ρ) > 0. Choose m0= 21C4g2(ρ), then there existsα > 0 such that

Iμ(u)|u=ρ≥αfor allhsatisfying|h|2<m0.

Lemma 3.4 Assume that(k1)-(k3)and(h1)hold.Then–∞<c:=inf{I(u) :u ≤ρ}< 0,

whereρis given by Lemma3.3.

Proof Sinceh∈L2₍_RN_{) and}_h_≡_{0, then for}_ε₌ |h|2

2 , there existsφ∈C0∞(RN) such that |h–φ|2<ε. Thus

RN

h2–hφdx≤

RN

h2–hφdx≤ |h–φ|2|h|2<ε|h|2,

and then

RNh

φdx≥ |h|2₂–ε|h|2=|h|

2 2

2 > 0. Hence

I(tφ)≤at

2

RN|∇

φ|2dx+bt

4

RN|∇

φ|2dx

2

+Wt

2

RN

φ2dx–t

RNh

φdx< 0

fort> 0 small enough. Then we getc=inf{I(u) :u ≤ρ}< 0.c> –∞is obvious. In order to prove the compactness, we deﬁneg(t) =k(t) –dt,∀t∈R. Then, by (k1) and (k2), we get that

lim t→0+

g(t)

t = 0, (3.3)

and whenN= 2,

lim t→+∞

g(t)

tp–1 = 0, (3.4)

whenN= 3,

lim t→+∞

g(t)

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Lemma 3.5 Suppose that(k1)-(k3), (h1)and(h3)hold.Assume that{un} ⊂E is a bounded Palais-Smale sequence of Iμfor eachμ∈J.Then{un}has a convergent subsequence in E.

Proof Since{un}is bounded inEandE →Ls(R3),∀s∈(2, 6),E →Ls(R2),∀s∈(2, +∞) are

compact (see [15]), up to a subsequence, we can assume that there existsu∈Esuch that

unuinE,un→uinLs(R3),∀s∈(2, 6),un→uinLs(R2),∀s∈(2, +∞),un(x)→u(x)

a.e. inRN.

By (3.3) and (3.4), for anyε> 0, we have

g(t)≤ε|t|+Cε|t|p–1, ∀t≥0. (3.6)

Then, by (3.6) and the Hölder inequality, one has

_R2g

(un)(un–u)dx

≤ε

R2|un||un–u|dx+Cε

R2|un|

p–1_|u n–u|dx

≤ε|un|2|un–u|2+Cε

R2|un|

p_dx

p–1 p

|un–u|p ≤Cε+on(1).

Similarly, we can obtain that

_R2g(u)(un–u)dx =on(1).

By (3.3) and (3.5), for anyε> 0, we have

g(t)≤ε|t|+|t|5+Cε|t|3, ∀t≥0. (3.7)

Hence, by (3.7) and the Hölder inequality, one has

_R3g(un)(un–u)dx

≤ε

R3|

un||un–u|dx+ε

R3|

un|5|un–u|dx+Cε

R3|

un|3|un–u|dx

≤ε|un|2|un–u|2+ε

R3|un|

6_dx

5 6

|un–u|6+Cε

R3|un| 9 2_dx

2 3

|un–u|3 ≤Cε+on(1).

Similarly, we can obtain that

_R3

g(u)(un–u)dx

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Hence whenN= 2 or 3, one has

_RN

g(un) –g(u)

(un–u)dx

=on(1).

It is clear that

Iμ(un) –Iμ(u),un–u =on(1)

and

b

RN

|∇u|2_–_|∇u n|2

dx

RN

∇u,∇(un–u)

dx=on(1).

Note that

Iμ(un) –Iμ(u),un–u =

a+b

RN|∇un|

2_dx

RN

_∇(un–u) 2

dx

+ (W–μd)

RN|un–u|

2_dx

–b

RN

|∇u|2_–_|∇u n|2

dx

RN

∇u,∇(un–u)

dx

–μ

RN

g(un) –g(u)

(un–u)dx ≥min{a,W–μd}un–u2

–b

RN

|∇u|2_–_|∇u n|2

dx

RN

∇u,∇(un–u)

dx

–μ

RN

g(un) –g(u)

(un–u)dx.

Therefore we get thatun–u →0 asn→ ∞. Proof of the ﬁrst solution of Theorem1.2 By Lemma 3.4 and Ekeland’s variational principle [11], there exists a sequence{un} ⊂Esuch thatun ≤ρ,I(un)→candI(un)→0 as n→ ∞. From Lemma 3.5 withμ= 1, there existsu0∈Esuch thatun→u0inEand then I(u0) = 0 andI(u0) =c< 0. Putu–0:=max{–u0, 0}, one has

0 =I(u0),u–₀

= –a

RN _∇u–

0 2

dx–b

RN|∇u0|

2_dx

RN _∇u–

0 2

dx–W

RN u–₀2dx

–

RN

hu–₀dx, (3.8)

which implies that u–₀ = 0 and then u0≥0. By the strong maximum principle, we get

u0> 0.

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Lemma 3.6 Assume that(k1)-(k3)and(h1)hold.Then

(∗) ∃v2∈Ewithv2>ρsuch thatIμ(v2) < 0,∀μ∈J.

(∗∗) cμ=infγ∈maxt∈[0,1]Iμ(γ(t)) >max{Iμ(0),Iμ(v2)},∀μ∈J,where

=γ ∈C[0, 1],E:γ(0) = 0,γ(1) =v2.

Proof It follows from (k3) that, for anyL> 0, there existsCL> 0 such that, for allt≥0, one

has

K(t)≥Lt2–CL. (3.9)

Fix 0≤w∈C₀∞(RN) withsuppw⊂B1:={x∈RN:|x|< 1}andw≡0. Deﬁnewt(x) =tw(_tx2)

fort> 0, then

suppwt=

t2y:y∈suppw.

By direct computation, we have

RN|∇wt|

2_dx₌_t2N–2

RN|∇w|

2_dx_,

RNw

2

tdx=t2N+2

RNw

2_dx

and, by (3.9),

RNK(wt)dx=

suppwt

K(wt)dx ≥L

suppwt

w2_tdx–CL

suppwt

dx

≥Lt2N+2

suppw

w2dx–CL

{t2_y_:_y_∈_B 1}

dx

=Lt2N+2

RNw

2_dx_–_C LCt2N.

Therefore

Iμ(wt)

=a 2

RN|∇wt|

2_dx₊b

4

RN|∇wt|

2_dx

2

+W 2

RNw

2 tdx

–μ

RN

K(wt)dx–

RN

hwtdx

≤at2N–2

2

RN|∇w|

2_dx₊bt4N–4

4

RN|∇w|

2_dx

2

+Wt

2N+2

2

RNw

2_dx

–Lt

2N+2

2

RNw

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for all μ∈J. When N = 2, we chooseL= 2W. When N = 3, we choose L= 2W +

b(

RN|∇w|2dx)2

RNw2dx . ThenIμ(wt)→–∞ast→+∞. Hence there existst

_{> 0 such that}_v2_:=_w t

withv2>ρandIμ(v2) < 0,∀μ∈J. This completes the proof of (∗).

By Lemma 3.3 and the deﬁnition ofcμ, for allμ∈J, we have

0 <α≤c1≤cμ≤c1

2 ≤_tmax_∈[0,1]I12(tv2) < +∞.

Therefore, byIμ(0) = 0 andIμ(v2) < 0, we obtain the proof of (∗∗).

So far we have veriﬁed all the conditions of Lemma 3.1. Then there exists{μj} ⊂Jsuch

that

(i) μj→1–asj→ ∞,{ujn}is bounded inE; (ii) Iμj(u

j

n)→cμj asn→ ∞;

(iii) I_μ_j(ujn)→0asn→ ∞.

Using (i)-(iii) and Lemma 3.5, there existsuj∈E such thatunj →uj inE asn→ ∞

and thenIμj(uj) =cμj andIμj(uj) = 0. Hence, fromIμj(uj) =cμj andIμj(uj),uj= 0, we get

respectively

a

2

RN|∇uj|

2_dx₊b

4

RN|∇uj|

2_dx

2

+W 2

RNu

2 jdx

–μj

RN

K(uj)dx–

RN

hujdx=cμj, (3.10)

a

RN|∇uj|

2_dx₊_b

RN|∇uj|

2_dx

2

+W

RNu

2 jdx

–μj

RNk(uj)ujdx–

RNhujdx= 0. (3.11)

Next, for obtaining{uj}is bounded inE, we need the following lemma (Pohozaev type

identity). The proof is similar to Lemma 2.6 in [16], and we omit its proof in here.

Lemma 3.7 Suppose that(h1)and(h2)hold.If Iμ(u) = 0,we have

a(N– 2) 2

RN|∇

u|2dx+b(N– 2) 2

RN|∇

u|2dx

2

+NW 2

RN

u2dx

–Nμ

RNK(u)dx–N

RNhu dx–

RN

∇h(x),xu dx= 0.

SinceI_μ_j(uj) = 0, by Lemma 3.7, we get that

a(N– 2) 2

RN|∇uj|

2_dx₊b(N– 2)

2

RN|∇uj|

2_dx

2

+NW 2

RNu

2 jdx

–Nμj

RN

K(uj)dx–N

RN

hujdx–

RN

∇h(x),xujdx= 0. (3.12)

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Proof It follows from (3.10) and (3.12) that

a

RN|∇uj|

2_dx₊b(4 –N)

4

RN|∇uj|

2_dx

2

+

RN

∇h(x),xujdx=Ncμj. (3.13)

Be similar to (3.8), byIμj(uj) = 0, we obtainuj≥0.

Firstly, we considerN= 2. From (3.13) andcμj≤c1₂, we get

a

R2|∇

uj|2dx≤a

R2|∇

uj|2dx+ b

2

R2|∇

uj|2dx

2

– 2cμj+ 2cμj

= –

R2

∇h(x),xujdx+ 2cμj. (3.14)

Since (∇h(x),x)≥0, by (3.14) anduj≥0, one has{

R2|∇uj|2dx} is bounded. Next we

prove{_R2u2_jdx} is bounded. Inspired by [14], we suppose by contradiction that λj:= |uj|2→+∞. Deﬁnewj:=uj(λjx), then

R2|∇wj|

2_dx₌

R2|∇uj|

2_dx_≤_C

and

R2|wj|

2_dx₌ 1

λ2_j

R2|uj|

2_dx_{= 1.} _(3.15)

Hence{wj}is bounded inE. Up to a subsequence, we may assume thatwjwinE,wj→ winLs₍_R2_),_∀s_∈_{(2, +}_∞_),_w

j→winLs_loc(R2),∀s∈[1, +∞),wj(x)→w(x) a.e. inR2. By Iμj(uj) = 0, one has

–

a+b

R2|∇wj|

2_dx

1

λ2_jwj+ (W–dμj)wj=μjg(wj) +h(λjx). (3.16)

For anyv∈C₀∞(R2_{), one has}

_R2h(

λjx)v dx

≤ |v|2

R2

h(λjx)2dx

1 2

= 1

λj

|v|2|h|2→0 (3.17)

and by the Lebesgue dominated convergence theorem, we have

_R2g(wj)v dx–

R2g(w)v dx

≤C

suppv

g(wj) –g(w)dx→0. (3.18)

Hence by (3.16)-(3.18), we have (W–d)w=g(w) inR2_{, from which we get that}_w_{= 0.}

(11)

and (3.16), one has

W–d= (W–d)

R2|wj|

2_dx

≤

a+b

R2|∇wj|

2_dx

1

λ2_j

R2|∇wj|

2_dx_{+ (}_W_–_d_μ j)

R2|wj|

2_dx

=μj

R2g(wj)wjdx+

R2h(

λjx)wjdx ≤ε

R2|wj|

2_dx₊_C

ε

R2|wj|

p_dx₊ 1

λj

|h|2|wj|2

≤Cε+on(1),

which implies a contradiction. Hence{_R2|uj|2dx}is bounded and then{uj}is bounded

inE.

Secondly, forN= 3, we have a simple proof. From (3.13), (h2) andcμj≤c1₂, we get

a

R3|∇uj|

2_dx_≤_a

R3|∇uj|

2_dx₊b

4

R3|∇uj|

2_dx

2

– 3cμj+ 3cμj

= –

R3

∇h(x),xujdx+ 3cμj

≤∇h(x),x₂|uj|2+ 3c1 2

≤C|uj|2+ 3c1

2. (3.19)

We prove directly{_R3u2_jdx}is bounded. Similar to (3.19), we obtain

b

4

R3|∇uj|

2_dx

2 ≤a

R3|∇uj|

2_dx₊b

4

R3|∇uj|

2_dx

2

– 3cμj+ 3cμj

≤C|uj|2+ 3c1

2. (3.20)

By the Hölder inequality, we have

R3|∇uj|

2_dx

3 ≤C

R3u

2 jdx

3 4

+C. (3.21)

By (3.3) and (3.5), for allt∈R, one has

g(t)t≤W–d

2 |t|

2₊_C_|_t_|6_. _(3.22)

From (3.11), (3.21), (3.22),μj≤1 andD1,2(R3)→L6(R3), it follows that

(W–d)

R3u

2 jdx≤a

R3|∇uj|

2_dx₊_b

R3|∇uj|

2_dx

2

+ (W–μjd)

R3u

2 jdx

=μj

R3g(uj)ujdx+

(12)

≤W–d

2

R3u

2 jdx+C

R3u

6

jdx+|h|2

R3u

2 jdx

1 2

≤W–d

2

R3

u2_jdx+C

R3|∇

uj|2dx

3

+|h|2

R3

u2_jdx

1 2

≤W–d

2

R3u

2 jdx+C

R3u

2 jdx

3 4

+C+|h|2

R3u

2 jdx

1 2

,

which implies that {_R3u2_jdx} is bounded. Combining with (3.19), we get that {uj} is

bounded inE.

Proof of the second solution of Theorem1.2 ByIμj(uj) =cμj,Iμj(uj) = 0,μj→1

–_and

Re-mark 3.2, we getI(uj)→c1andI(uj)→0 asn→+∞. By Lemmas 3.5 and 3.8, there exists v0∈Esuch thatuj→v0inEasn→+∞and thenI(v0) =c1> 0,I(v0) = 0. Be similar to

(3.8), we getv0≥0. By the strong maximum principle, one hasv0> 0.

4 Conclusions

The goal of this paper is to study the multiplicity of positive solutions for the following nonlinear Kirchhoﬀ type equation:

–

a+b

RN|∇u|

2_dx

u+Vu=f(u) +h(x), x∈RN,

wherea,b,Vare positive constants,N= 2 or 3. Under very weak conditions onf, we get that the equation has two positive solutions by using variational methods.

Acknowledgements

The authors thank the referees for valuable comments and suggestions which improved the presentation of this manuscript.

Funding

This work was supported by the Natural Science Foundation of Education of Guizhou Province (No. KY[2016]103, KY[2016]281, KY[2017]297); the Science and Technology Foundation of Guizhou Province (No. LH[2015]7595, LH[2016]7054).

Abbreviations Not applicable.

Availability of data and materials Not applicable.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Each of the authors contributed to each part of this study equally. All authors read and approved the ﬁnal vision of the manuscript.

Author details

1_{School of Mathematics and Information Engineering, Liupanshui Normal College, LiuPanshui, Guizhou 553004, People’s}

Republic of China.2_{Collect of Science, GuiZhou University of Engineering Science, BiJie, Guizhou 551700, People’s}

Republic of China.

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