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BY-NC-ND 3.0 license Open Access Published by De Gruyter October 11, 2016

Ground state solutions for Kirchhoff-type equations with a general nonlinearity in the critical growth

  • Li-Ping Xu EMAIL logo and Haibo Chen

Abstract

In this paper, we concern ourselves with the following Kirchhoff-type equations:

{-(a+b3|u|2𝑑x)u+Vu=f(u)in 3,uH1(3),

where a, b and V are positive constants and f has critical growth. We use variational methods to prove the existence of ground state solutions. In particular, we do not use the classical Ambrosetti–Rabinowitz condition. Some recent results are extended.

MSC 2010: 35J20; 35J65; 35J60

1 Introduction and main results

Consider the following Kirchhoff-type problem:

(1.1){-(a+b3|u|2𝑑x)u+Vu=f(u)in 3,uH1(3),

where a, b and V are positive constants and f(u) satisfies the following hypotheses:

  1. f(u)C(,) is odd.

  2. limu0+f(u)/u=0.

  3. limu+f(u)/u5=μ>0.

  4. There exist M>0 and 2<q<6 such that f(u)μu5+Muq-1 for u0.

Kirchhoff-type problems are related to the stationary analogue of the equation

utt-(a+bΩ|u|2𝑑x)u=f(x,u)in Ω,

where u denotes the displacement, f(x,u) the external force, and b the initial tension while a is related to the intrinsic properties of the string (such as Young’s modulus). Equations of this type arise in the study of string or membrane vibration and were first proposed by Kirchhoff in 1883 (see [15]) to describe the transversal oscillations of a stretched string, particularly, taking into account the subsequent change in string length caused by oscillations. Kirchhoff-type problems are often referred to as being nonlocal because of the presence of the integral over the entire domain Ω, which provokes some mathematical difficulties. Similar nonlocal problems also model several physical and biological systems, where u describes a process which depends on the average of itself, for example, the population density; see [10, 11], and the references therein.

If we set a=1, b=0, then (1.2) reduces to the following Schrödinger equation:

-u+V(x)u=f(u)inN.

There exist many studies on the existence and multiplicity of solutions for this equation. We refer to [5, 4, 3, 2, 21, 20] and the references therein.

There has been a lot of research on the existence of nontrivial solutions to (1.1) with subcritical nonlinearities by variational methods; see, e.g., [24, 19, 18, 17] and the references therein. Recently, some researchers have considered the existence of ground state solutions for Kirchhoff-type problems with critical Sobolev exponent. By using the Nehari manifold, Wang and Tian [23] proved the existence and multiplicity of positive ground state solutions for the following semilinear Kirchhoff problem with critical growth:

(1.2){-(ε2a+bε3|u|2𝑑x)u+M(x)u=f(u)+u5,x3,u>0,uH1(3),

where ε is a positive parameter and f is a C1 and subcritical function such that the following hold:

  1. f(t)=o(t3)ast0andf(t)0for all t0.

  2. f(t)t-3is strictly increasing fort>0.

  3. f(t)=o(t5)as|t|+.

We pull the energy level down below the following critical level:13(aS)32+112b3S6. He and Zou in [13] also considered (1.2), where f(t) satisfies (F1), (F2) and the following:

  1. νF(t)=ν0tf(s)𝑑stf(t) holds for some ν>4.

  2. f(t)=o(tq) as t, 3<q<5.

By the use of variational methods, the authors showed that there exist ε*>0, λ*>0 such that for any ε>ε*, λ>λ*, problem (1.2) has at least one positive ground state solution in H1(3).

Under conditions (F1)–(F4), by variational methods, Li and Ye [16] proved the existence of positive ground state solutions for the following Kirchhoff-type problem:

{-(a+b3|u|2𝑑x)u+u=f(u)+u5,x3,uH1(3),u>0,x3.

Particularly, Alves and Figueiredo [1] obtained the existence of positive solutions for a periodic Kirchhoff equation with critical or subcritical nonlinearity.

Motivated by the above works described, we borrow an idea from [25] to prove the existence of ground state solutions for problem (1.1) with a general nonlinearity in the critical growth. Our main result is the following.

Theorem 1.1.

Assume that (f1)(f4) hold, then for 2<q4 with M>0 sufficiently large or 4<q<6, problem (1.1) possesses a radial ground state solution.

Remark 1.2.

Conditions (f1)–(f4) were introduced in [25] to obtain the existence of ground state solutions for a class of Schrödinger–Poisson equations with critical growth. An interesting question now is whether the same existence results occur to the nonlocal problem (1.1) with critical growth. In this paper, we study problem (1.1) and give some positive answers. Theorem 1.1 extends the main result in [25] to the Kirchhoff-type equation.

Remark 1.3.

Set V=1 in (1.1). Compared to [16, Theorem 1.3], we need not consider the usual Ambrosetti–Rabinowitz (AR) condition (F4), which is restrictive. The lack of the (AR)-condition gives rise to two obstacles to the standard mountain pass arguments both in checking the geometrical assumptions in the functional and in proving the boundedness of its (PS) sequences. In the present paper, we use another method to obtain our results. Moreover, there exist some functions which satisfy our conditions (f1)–(f4), but do not satisfy the conditions of [16]. For example, the function f(t)=μt5+Mt3, μ,M>0, satisfies our conditions (f1)–(f4), but does not satisfy conditions (F3) and (F4) in [16]. For this reason, Our main results can be viewed as a partial extension of [16].

The outline of the paper is as follows: In Section 2, we present some preliminary results. In Section 3, we give the proof of Theorem 1.1.

Notations.

We use the following notations:

  1. H1(3) is the Sobolev space equipped with the norm

    uH1(3)=3(|u|2+u2)𝑑x.
  2. Define

    u2:=3(a|u|2+Vu2)𝑑xfor uH1(3).

    Note that is an equivalent norm on H1(3).

  3. For any 1s, we denote by

    uLs:=(3|u|s𝑑x)1s

    the usual norm of the Lebesgue space Ls(3).

  4. For any x3, we set

    Br(x){y3:|x-y|<r}.
  5. Let D1,2(3):={uL6(3):uL2(3)} be the Sobolev space equipped with the norm

    uD1,2(3)2:=3|u|2𝑑x.
  6. S denotes the best Sobolev constant

    S:=infuD1,2(3){0}3|u|2𝑑x(3u6𝑑x)13.
  7. C denotes various positive constants.

2 Preliminaries

It is clear that problem (1.1) are the Euler–Lagrange equations of the functional I:H1(3) defined by

(2.1)I(u)=12u2+b4(3|u|2𝑑x)2-3F(u)𝑑x,

where F(u)=0uf(t)𝑑t. Obviously, by (f1)–(f3), we can obtain that I is a well-defined C1 functional and satisfies

I(u),v=3(auv+Vuv)𝑑x+b3|u|2𝑑x3uvdx-3f(u)v𝑑x

for vH1(3). For simplicity, by (f4), we may assume that μ=1. Let g(t)=f(t)-t5. Then

I(u)=12u2+b4(3|u|2𝑑x)2-3G(u)𝑑x-163u6𝑑x,

where G(u)=0ug(t)𝑑t. It is well known that uH1(3) is a critical point of the functional I if and only if u is a weak solution of (1.1).

Let E:=Hr1(3)={uH1(3):uis radial}. Then, by the principle of symmetric criticality, a critical point of I on E is a critical point of I on H1(3). We refer the readers to [12, 6]. Thus, we only need to look for critical points of I on E. To complete the proof of our theorem, the following result will be needed in our argument.

Theorem 2.1 (See [14]).

Let (X,) be a Banach space and hR+ an interval. Consider the following family of C1 functionals on X:

Iλ(u)=A(u)-λB(u),λh,

with B nonnegative and either A(u)+ or B(u)+ as u. We assume there are two points v1, v2 in X such that

cλ=infγΓλmaxt[0,1]Iλ(γ(t))>max{Iλ(v1),Iλ(v2)}for all λh,

where

Γλ={γC([0,1],X):γ(0)=v1,γ(1)=v2}.

Then for almost every λh there is a sequence {un}X such that

  1. {un} is bounded,

  2. Iλ(un)cλ,

  3. Iλ(un)0 in the dual X-1 of X.

Moreover, the map λcλ is continuous from the left.

3 Proof of the main result

This section is devoted to the proof of Theorem 1.1. According to Theorem 2.1, we need the following lemmas.

Lemma 3.1.

If uλ, for λ[12,1], is a critical point of Iλ, then uλ satisfies the following Pohožaev-type identity:

123a|u|2𝑑x+b2(3|u|2𝑑x)2+323Vu2𝑑x-3λ3F(u)𝑑x=0.

Since the proof can be done as in [16], we omit it here.

Lemma 3.2.

Assume that conditions (f1), (f2)(f4) are satisfied. Then the conclusions of Theorem 2.1 hold.

Proof.

By Theorem 2.1, we set

X=E,h=[12,1],A(u)=12u2+b4(3|u|2𝑑x)2,B(u)=3F(u)𝑑x.

By (f3)–(f4), we obtain that f(u) is odd and by the definition of A(u), we can see that B(u)0 for uE and A(u)+ as u. By (f1)–(f3), for any ε>0, there exists C(ε)>0 such that

(3.1)|F(u)|εu2+C(ε)u6.

Then, by the Sobolev embedding theorem, there holds

Iλ(u)=12u2+b4(3|u|2𝑑x)2-λ3F(u)𝑑x12u2-3εu2𝑑x-3C(ε)u6𝑑xCu2-Cu6,

which implies that there exists ρ>0 small enough and α>0 such that

(3.2)Iλ(u)α>0for allu=ρ.

By (f4) and (2.1), for ϕE, ϕ0 and ϕ0, we have

Iλ(tϕ)t22ϕ2+bt44(3|ϕ|2𝑑x)2-Mtq2q3|ϕ|q𝑑x-t6123|ϕ|6𝑑x-ast+.

Taking v2=t1ϕ with t1>0 large enough, we have v2>ρ and

(3.3)Iλ(v2)<0for all λ[12,1].

On the other hand, Iλ(0)=0. Set v1=0; then inequalities (3.2) and (3.3) imply that the conclusions of Theorem 2.1 hold. ∎

Lemma 3.3.

Assume that (f1)(f4) hold. Then

0<cλ<cλ*:=abS34λ+b3S624λ2+(b2S4+4aλS)3224λ2

for q(2,4] with M sufficiently large or q(4,6).

Proof.

For ε,r>0, define

uε(x)=φ(x)ε14(ε+|x|2)12,

where φC0(B2r(0)), 0φ1 and φ|Br(0)1. Using the method of [9], we obtain

(3.4)3|uε|2𝑑x=K1+O(ε12),3|uε|6𝑑x=K2+O(ε32)

and

(3.5)3|uε|t𝑑x={Kεt4,t[2,3),Kε34|lnε|,t=3,Kε6-t4,t(3,6),

where K1, K2, K are positive constants. Moreover, the best Sobolev constant is S=K1K2-1/3. By (3.4), we have

3|uε|2𝑑x(3uε6𝑑x)13=S+O(ε12).

By Lemma 3.2 and the definition of cλ, we can deduce that cλsupt0Iλ(tuε). Let

h(t)=t22uε2+bt44(3|uε|2𝑑x)2-λt663uε6𝑑xfor all t0.

Note that h(t) attains its maximum at

t0=(b(3|uε|2𝑑x)2+b2(3|uε|2𝑑x)4+4λuε23uε6𝑑x2λ3uε6𝑑x)12.

Then

maxt0h(t)=buε2(3|uε|2𝑑x)24λ3uε6𝑑x+b3(3|uε|2𝑑x)624λ2(3uε6𝑑x)2+[b2(3|uε|2𝑑x)4+4λ3uε6𝑑xuε2]3224λ2(3uε6𝑑x)2
=abS34λ+b3S624λ2+(b2S4+4aλS)3224λ2+O(ε12):=cλ*+O(ε12)

for ε>0 small enough.

Obviously, we see that there exists 0<t1<1 such that for ε<1, we have

(3.6)sup0tt1Iλ(tuε)sup0tt1[12t2uε2+b4t4(3|uε|2𝑑x)2]sup0tt1(12t2uε2+Ct4uε4)cλ*.

By (f4) , we have

(3.7)Iλ(tuε)=h(t)-λ3G(tuε)𝑑xh(t)-λMqtq3|uε|q𝑑xh(t)-CMtq3|uε|q𝑑x.

It follows from (3.4), (3.5) and (3.7) that there exists 0<ε0<1 such that for ε<ε0, we have

limnIλ(tuε)=-.

Thus, there exists t2>0 such that

(3.8)suptt2Iλ(tuε)cλ*.

From (3.7) and the definition of h(t), we have

(3.9)supt1tt2Iλ(tuε)supt0h(t)-CM3|uε|q𝑑xabS34λ+b3S624λ2+(b2S4+4aλS)3224λ2+O(ε12)-CM3|uε|q𝑑x.

For q(2,4], fix ε(0,ε0), it follows from (3.9) that

(3.10)supt1tt2Iλ(tuε)<cλ*forMsufficiently large.

For q(4,6), by (3.5) and (3.9), we obtain

(3.11)supt1tt2Iλ(tuε)cλ*+O(ε12)-CO(ε6-q4).

Since 6-q4<12, then there exists ε1(0,ε0) small enough such that for ε(0,ε1), we have

(3.12)supt1tt2Iλ(tuε)cλ*.

By (3.6), (3.8) and (3.10)–(3.12), the proof of Lemma 3.3 is complete. ∎

Lemma 3.4.

Set

Λ=limn3|un|2𝑑x𝑎𝑛𝑑Jλ(u)=12u2+bΛ223|u|2𝑑x-λ3F(u)𝑑x.

If Jλ(u)=0, where λ[12,1], then Jλ(u)0.

Proof.

Since Jλ(u),u=0, by Lemma 3.1, we get the following the Pohožaev-type identity:

123a|u|2𝑑x+bΛ223|u|2𝑑x+323Vu2𝑑x-3λ3F(u)𝑑x=0.

Then we get that

(3.13)163a|u|2𝑑x+bΛ263|u|2𝑑x+123Vu2𝑑x-λ3F(u)𝑑x=0.

Combining (3.13) with the definition of Jλ(u), we obtain that

Jλ(u)=133a|u|2𝑑x+bΛ233|u|2𝑑x0.

Using a notion similar to [16, Lemma 3.6], we can obtain the following result.

Lemma 3.5.

For s,t>0 , the system

{Γ(t,s)=t-aS(s+tλ)13=0,Υ(t,s)=s-bS2(s+tλ)23=0

has a unique solution (t0,s0). Moreover, if

{Γ(t,s)0,Υ(t,s)0,

then tt0 and ss0, where

t0=abS3+ab2S6+4λaS32λ,s0=bS6+2λabS3+b2S3b3S6+4λaS32λ2.

Lemma 3.6.

Assume that (f1)(f3) hold. If {un}E, for λ[12,1], is a sequence such that un<C, Iλ(un)cλ, Iλ(un)0, and, moreover, cλ<cλ*, then {un} has a strong convergent subsequence in E.

Proof.

Since un<C in E, there exists a uE such that

unuweakly inH,unustrongly inLs(3) for all s(2,6).

Set

Λ=limn3|un|2𝑑x.

By Lemma 3.4 and the fact that cλ<cλ*, we have

(3.14)cλ-Jλ(u)cλ*.

Using an argument similar to [7, Radial Lemma A.II.], by the boundedness of {un}, we have lim|x|un(x)=0. Since

lim|t|G(t)t2+t6=0andlimt0G(t)t2+t6=0,

we also get

3(un2+un6)𝑑x<.

By the compactness lemma of Strass [22], one has

(3.15)limn3(G(un)-G(u))𝑑x=0.

Similarly,

(3.16)limn3(g(un)un-g(u)u)𝑑x=0.

Setting ωn=un-u, due to Brezis–Lieb (see [8]), we have

3|ωn|2𝑑x=3|un|2𝑑x-3|uλ|2𝑑x+o(1),
3|ωn|6𝑑x=3|un|6𝑑x-3|uλ|6𝑑x+o(1),

and

Λ2+o(1)=3|un|2𝑑x=3|ωn|2𝑑x+3|u|2𝑑x+o(1).

Then, from (3.15) and (3.16), we have

Iλ(un)-b4(3|un|2𝑑x)2+bΛ223|un|2𝑑x-Jλ(u)
=Jλ(un)-Jλ(u)
=12ωn2+bΛ243|ωn|2𝑑x
(3.17)+b4[(3|ωn|2𝑑x)2+3|ωn|2𝑑x3|u|2𝑑x]-λ63|ωn|6𝑑x+o(1),

and

Jλ(u)=0.

Then, by (3.16) and the fact that Iλ(un)0, we have

o(1)=Jλ(un),un-Jλ(u),u
(3.18)=ωn2+b(3|ωn|2𝑑x)2+b3|ωn|2𝑑x3|u|2𝑑x-λ3ωn6𝑑x+o(1).

We can assume that there exists li0 (i=1,2,3) such that

ωn2l1,b(3|ωn|2𝑑x)2+b3|ωn|2𝑑x3|u|2𝑑xl2,λ3ωn6𝑑xl3.

Then by (3.18) and (3.17), we have

{l1+l2-l3=0,12l1+14l2-16l3+bΛ24limn3|ωn|2𝑑x=cλ+bΛ24-Jλ(u).

By the definition of S, we see that

3|ωn|2𝑑xSλ1/3(λ3|ωn|6𝑑x)13,b(3|ωn|𝑑x)2bS2λ2/3(λ3|ωn|6𝑑x)23.

Then

l1aS(l1+l2λ)13andl2bS2(l1+l2λ)23.

Obviously, if l1>0, then l2,l3>0. By Lemma 3.5, we have that

cλ+bΛ24-Jλ(u)=13l1+112l2+bΛ24limn3|ωn|2𝑑x
13abS3+ab2S6+4λaS32λ+112bS6+2λabS3+b2S3b3S6+4λaS32λ2+bΛ24
=abS34λ+b3S624λ2+(b2S4+4aλS)3224λ2+bΛ24
=cλ*+bΛ24,

which is contrary to (3.14). Therefore, ωn0 and Lemma 3.6 is complete. ∎

Lemma 3.7.

Under the assumptions of Theorem 1.1, for almost every λ[12,1] there exists uλE, uλ0 such that Iλ(uλ)=cλ and Iλ(uλ)=0.

Proof.

By Lemma 3.2, there is a sequence{un}E satisfying un<C, Iλ(un)cλ and Iλ(un)0. Moreover, 0<cλ<cλ*. Then, by Lemma 3.6, the sequence {un} has a strong convergent subsequence, still denoted by{un}. In other words, there exists uλE such that Iλ(uλ)=cλ and Iλ(uλ)=0. ∎

Lemma 3.8.

Under the assumptions of Theorem 1.1, the functional I admits nontrivial critical points.

Proof.

From Lemma 3.7, there exist cλn(0,cλn*), λn[12,1] and uλnE with uλn0 such that λn1, Iλn(uλn)=0 and Iλn(uλn)=cλn. Then, by Lemma 3.1, we have

3cλn=3a|uλn|2𝑑x+b4(3|uλn|2𝑑x)2.

The Sobolev embedding theorem implies the boundedness of 3|uλn|6𝑑x. From (3.1) and Lemma 3.1 we get

123a|uλn|2𝑑x+323Vuλn2𝑑x+b2(3|uλn|2𝑑x)2=3λλn3F(uλn)𝑑xε3|uλn|2𝑑x+C(ε)3|uλn|6𝑑x

for any ε>0 and some C(ε)>0. Hence, uλn is bounded. Since

I(uλn)=Iλn(uλn)+(λn-1)3F(uλn)𝑑x,

we have

I(uλn),uλn=Iλn(uλn),uλn+(λn-1)3f(uλn)uλn𝑑x

and

limncλn=c1(0,c*),wherec*=abS34+b3S624+(b2S4+4aS)3224.

By a standard argument, we obtain that limnI(uλn)=c1 and limnI(uλn)=0. By the boundedness of uλn, similarly to the proof of Lemma 3.6, we can prove that there exists u0E such that I(u0)=0. We claim that u00. Otherwise, if u0=0, then uλn0 weakly in E and

(3.19)uλn0strongly inLs(3),s(2,6).

From (3.19) and (f1)–(f3), we have

3G(uλn)=o(1)and3g(uλn)uλn=o(1).

From limnI(uλn)=c1 and limnI(uλn)=0, we have that

(3.20)c1+o(1)=12uλn2+b4(3|uλn|2𝑑x)2-163|uλn|6𝑑x

and

(3.21)o(1)=uλn2+b(3|uλn|2𝑑x)2-3|uλn|6𝑑x.

Assume that

uλn2h10,b(3|uλn|2𝑑x)2h20,3uλn6𝑑xh30.

Then by (3.21) and (3.20), we have

{h1+h2-h3=0,c1=12h1+14h2-16h3,

where c1>0 implies that h1,h2,h3>0. By the definition of S, we see that

h1aS(h1+h2)13andh2bS2(h1+h2)23.

By Lemma 3.5, we have that

c1=13h1+112h2
13abS3+ab2S6+4aS32+112bS6+2abS3+b2S3b3S6+4aS32
=abS34+b3S624+(b2S4+4aS)3224=c*,

which is contrary to c1<c*. ∎

In the following, we prove c1I(u0). From Iλn(uλn)=0, I(u0)=0 and by Lemma 3.1, we obtain that

cλn=Iλn(uλn)=a3R3|uλn|2𝑑x+b12(R3|uλn|2𝑑x)2

and

I(u0)=a33|u0|2𝑑x+b12(3|u0|2𝑑x)2.

Thus, by Fatou’s Lemma,

c1=limncλna33|u0|2𝑑x+b12(3|u0|2𝑑x)2=I(u0).

Proof of Theorem 1.1.

Let

m=inf{I(u):uE,u0,I(u)=0inE-1}.

By Lemma 3.8 and Lemma 3.4, we have 0mI(u0)c1<c*. We choose a minimizing sequence {un} for m, i.e. un0, I(un)m and I(un)=0. Now, we prove that {un} is bounded. The proof is divided into two steps.

Step 1: {unL2} is bounded. By contradiction, we assume that unL2 as n. Set

vn=ununL2,Xn=3|un|2𝑑xunL2-2,Yn=(3|un|2𝑑x)2unL2-2,Zn=3F(un)𝑑xunL2-2.

Since I(un)m and I(un)=0, using (2.1) and Lemma 3.1, we have

(3.22){123a|un|2𝑑x+123V|un|2𝑑x+b4(3|un|2𝑑x)2-3F(un)𝑑x=m+o(1),123a|un|2𝑑x+323V|un|2𝑑x+b2(3|un|2𝑑x)2-33F(un)𝑑x=0,

and m is bounded. Multiplying (3.22) by 1unL2, we get

(3.23){12aXn+12V+b4Yn-Zn=o(1),12aXn+32V+b2Yn-3Zn=0,

where o(1) denotes that the quantity tends to zero as n. Solving (3.23), we have

(3.24)Xn=-b4aYn+o(1).

Since Yn0, Xn0 and a,b>0 for all nN, equation (3.24) is a contradiction for n large enough. Thus, {unL2} is bounded.

Step 2: unL2 is bounded. Similarly, by contradiction, we can assume that unL2 as n. Set

vn=ununL2,Mn=3un2𝑑xunL2-2,Nn=(3|un|2𝑑x)2unL2-2,Sn=3F(un)𝑑xunL2-2.

Then, multiplying (3.22) by 1unL2, we get

(3.25){12a+12VMn+b4Nn-Sn=o(1),12a+32VMn+b2Nn-3Sn=0.

Solving (3.25), we have

(3.26)Nn=-4ab+o(1).

Since Nn0 and a,b>0, equation (3.26) is a contradiction for n large enough. Thus, {unL2} is bounded. Thus, we prove the boundedness of {un}. Similarly to the proofs of Lemma 3.6 and Lemma 3.8, we can prove that there exists u0E such that I(u)=0.

Next, we will give the proof of mI(u). In fact, by I(u)=0, I(un)=0 and Lemma 3.1, we have that

I(u)=a33|u|2𝑑x+b12(3|u|2𝑑x)2

and

I(un)=a33|un|2𝑑x+b12(3|un|2𝑑x)2.

Then

m+o(1)=I(un)=a33|un|2𝑑x+b12(3|un|2𝑑x)2.

Thus, by Fatou’s Lemma,

ma33|u|2𝑑x+b12(3|u|2𝑑x)2=I(u).

By combination with the definition of m, there exists u0 satisfying m=I(u) and I(u)=0, which completes the proof of Theorem 1.1. ∎

Award Identifier / Grant number: 11271372

Award Identifier / Grant number: 11671236

Award Identifier / Grant number: 11671403

Funding statement: Research supported by the National Natural Science Foundation of China (grants 11271372, 11671236, 11671403), by the Natural Science Foundation of Hunan Province (grant 12JJ2004), and by the Major State Basic Research of Higher Education of Henan Provincial of China (grant 17A110019).

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Received: 2016-03-29
Revised: 2016-06-03
Accepted: 2016-06-28
Published Online: 2016-10-11
Published in Print: 2018-11-01

© 2018 Walter de Gruyter GmbH, Berlin/Boston

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