Startseite Mathematik Non-trivial solutions for Schrödinger-Poisson systems involving critical nonlocal term and potential vanishing at infinity
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Non-trivial solutions for Schrödinger-Poisson systems involving critical nonlocal term and potential vanishing at infinity

  • Liuyang Shao EMAIL logo
Veröffentlicht/Copyright: 18. Oktober 2019

Abstract

The present study is concerned with the following Schrödinger-Poisson system involving critical nonlocal term

u+V(x)ul(x)ϕ|u|3u=ηK(x)f(u),inR3,ϕ=l(x)|u|5,inR3, (1.1)

where the potential V(x) and K(x) are positive continuous functions that vanish at infinity, and l(x) is bounded, nonnegative continuous function. Under some simple assumptions on V, K, l and f, we prove that the problem (1.1) has a non-trivial solution.

MSC 2010: 35B09; 35J20

1 Introduction and main results

The aim of this paper is to investigate the existence of non-trivial solutions for the following Schrödinger-Poisson system involving critical nonlocal term and potential vanishing at infinity

u+V(x)ul(x)ϕ|u|3u=ηK(x)f(u),inR3,ϕ=l(x)|u|5,inR3, (1.1)

where V, KC(ℝ3, ℝ), fC(ℝ3 × ℝ, ℝ), l(x) is bounded, nonnegative continuous function, and V, K are nonnegative functions which can be vanishing at infinity. η > 0 is a parameter and 2 := 6 is the critical Sobolev exponent. Similar problems have been widely investigated, and it is well known that they have a strong physical meaning, because they appear in quantum mechanics models and in semiconductor theory. In particular, systems like (1.1) have been introduced in [1] as a model describing solitary waves, for nonlinear stationary equations of Schrödinger type interacting with an electrostatic field, and are usually known as Schrödinger-Poisson systems. Indeed, in (1.1) the first equation is a nonlinear stationary Schrödinger equation that is coupled with a Poisson equation, to be satisfied by ϕ, meaning that the potential is determined by the charge of the wave function. For more details, we refer the readers to [2, 3, 4, 5] and the references therein.

In recent years, with the aid of variational methods, there has been increasing attention to problems like (1.1) on the existence and non-existence of positive solutions, positive ground states, multiple solutions, sign-changing solutions and so on. see for instance [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21], and the references therein. In fact, the most of the above results focused on the subcritical nonlocal term. However, the system (1.1) with critical nonlocal term has only been studied in [22, 23, 24, 25].

In [22], A. Azzollini and P. d’Avenia firstly studied the following Schrödinger-Poisson system with critical nonlocal term

u=μu+pϕ|u|3u=f(x,u),xBR,ϕ=p|u|5,xBR,u=ϕ=0,onBR. (1.2)

They proved that the existence and nonexistence results for system (1.1) depend on the value of λ.

In [23], Liu studied the following Schrödinger-Poisson system with critical nonlocal term

u+V(x)uK(x)ϕ|u|3u=f(x,u),inR3,ϕ=K(x)|u|5,inR3. (1.4)

Under the condition V(x), K(x), f are asymptotically periodic, the author proved the system (1.4) has at least a positive solution by the mountain pass theorem and the concentration-compactness principle.

In [25], F. Li, Y. Li and J. Shi proved

u+buϕ|u|3u=f(u),inR3,ϕ=|u|5,inR3, (1.3)

possesses at least one positive radially symmetric solution when b > 0 is a constant. To the best of our knowledge, there seems to be little progress on the existence of nontrivial solution for Schrödinger-Poisson systems involving critical nonlocal term and potential vanishing at infinity.

By the motivation of above work, In our article, we establish the existence of non-trivial solution for problem (1.1) with critical nonlocal term and potential vanishing at infinity. Firstly the critical growth causes a lack of compactness, and it is much more difficult to obtain the existence of non-trivial solutions. Secondly since V(x) is potential vanishing at infinity, which makes our studies more interesting. At last, we obtain a non-trivial solution by using the mountain pass theorem without (PS) condition.

Below, we assume that the pair (V, K) of continuous functions V, K : ℝ3 → ℝ belongs to 𝓚. Throughout the paper, (V, K) ∈ 𝓚 means that

  1. V(x), K(x) > 0 for all x ∈ ℝ3 and KL(ℝ3).

  2. If {An}n ⊂ ℝ3 is a sequence of Borel sets such that the Lebesgue measure meas(An) ≤ R, for all n ∈ ℕ and some R > 0, then

    limR+AnBRc(0)K(x)dx=0,uniformlyinnN.

Furthermore, one of the below conditions occurs:

  1. K/VL(ℝ3) or

  2. there exists p0 ∈ (2, 6) such that

    K(x)V(x)6p040 as|x|+.

The hypotheses (VK1)–(VK4) on functions V and K were first introduced in [26] and characterized problem (1.1) as zero mass. Problems of zero mass have been studied by many authors, see for example, [27, 28, 29, 30, 31, 32] and references therein.

Finally, we assume the following growth conditions at the origin and at the infinity for the continuous function f : ℝ → ℝ:

  1. limt0f(t)t=0, if (VK3) holds; or limt0f(t)|t|p01<, if (VK4) holds.

  2. f has a “quasicritical growth”, namely, limt+f(t)t5=0.

  3. There exists a θ ∈ (2, 2) such that

    0θF(t)tf(t)foralltR,

    where F(u) = 0u f(s)ds.

Furthermore, we make the following hypotheses on the function l(x).

  1. There exists x0, such that

    l(x0)=supxR3l(x).
  2. For x close to x0 we have

    l(x)=l(x0)+O(|xx0|),asxx0.

Now we state our main results as follows.

Theorem 1.1

Suppose that (V, K) ∈ 𝓚, and f satisfies (f1), (f2), (f3), l(x) satisfies (l1)–(l2).

  1. If θ ∈ (1, 3], for sufficiently lange η > 0, then system (1.1) has at least one non-trivial solution.

  2. If θ ∈ (3, 5), for any η > 0, then system (1.1) has at least one non-trivial solution.

Notation. In this paper we make use of the following notations: C will denote various positive constants; the strong (respectively weak) convergence is denoted by → (respectively ⇀); o(1) denotes o(1) → 0 as n → ∞, Bρ(0) denotes a ball centered at the origin with radius ρ > 0.

The remainder of this paper is organized as follows. In Section 2, some preliminary results are presented. In Section 3, we give the proof of our main results.

2 Variational setting and preliminaries

Let us consider the space

E=uD1,2(R3):R3V(x)|u|2dx<+

endowed with the norm

uE2=R3|u|2dx+R3V(x)|u|2dx.

Recall that a weak solution of problem (1.1) is a function uE such that

R3uφdx+R3V(x)uφdxR3l(x)ϕ|u|4φdxηR3K(x)f(u)φdx=0, (2.1)

for all φE.

Then, the weak solutions of (1.1) are the critical points of the energy functional defined on E by

J(u):=12R3(|u|2+V(x)|u|2)dx110R3l(x)ϕu|u|5dxηR3K(x)f(u)dx, (2.2)

where F(u) = 0u f(s)ds. More precisely, JC1(E, ℝ) and its differential J′ : EE′ is defined as

J(u),v=R3(|u||v|+V(x)uv)dxR3l(x)ϕu|u|4vdxηR3K(x)f(u)vdx, (2.3)

for all vE, where E′ is the dual space of E.

We define the Lebesgue space LKp (ℝN) composed by all measurable functions u : ℝN → ℝ such that

LKp(R3)=u:R3R|uismeasurableandR3K(x)|u|pdx<+,

endowed with norm

uLKp(R3):=R3K(x)|u|pdx1p,

and we will state, without proof, two important results of Alves and Souto (see [[26], Lemmas 2.1 and 2.2]).

Proposition 2.1

Assume (V, K) ∈ 𝓚 holds, E is compactly embedded in LKp (ℝN) for every p ∈ (2, 6). If (VK4) holds, E is compactly embedded in LKp (ℝN).

Proposition 2.2

Suppose that f satisfies (f1) and (f2) and (V, K) ∈ 𝓚, Let {vn} be such that vnv in E. Then,

R3K(x)F(vn)dxR3K(x)F(v)dx (2.4)

and

R3K(x)f(vn)vndxR3K(x)f(v)vdx. (2.5)

Lemma 2.3

[25, Lemma 2.1] For every uL6(ℝ3), there exists a unique ϕuD1,2(ℝ3) which is the solution of

ϕ=|u|5,inR3,

here ϕu can be expressed by the from

ϕu(x)=R3|u(y)|5|xy|dy

Moreover,

  1. ϕuD1,2(R3)2=R3ϕu|u|5;

  2. ϕu(x) > 0 for x ∈ ℝ3;

  3. for any θ > 0, ϕuθ = θ2(ϕu)θ, where uθ(⋅) = u(⋅/θ);

  4. for any t > 0, ϕtu = t5ϕu;

  5. for any uL6(ℝ3),

    ϕuD1,2(R3)S12|u|65,R3ϕu|u|5S1|u|610,

    where S is defined in (1.6);

  6. if unu in L6(ℝ3) and unu a.e in3 as n → ∞, then ϕunϕu in D1,2(ℝ3).

Lemma 2.4

[25, Lemma 2.3] If unu in L6(ℝ3), a, e in3, then as n → ∞,

|un|5|unu|5|u|50inL65(R3),ϕunϕunuϕu0,inD1,2(R3),R3ϕun|un|5dxR3ϕunu|unu|5dxR3ϕu|u|5dx0,

and

|un|3un|unu|(unu)|u|3u0,inD1,2(R3).

3 Proof of Theorem 1.1

To prove Lemma 3.2, we need the following results.

Lemma 3.1

Suppose that (V, K) ∈ 𝓚 hold. Then for p ∈ [2, 6], there is C > 0 such that

uLkp(R3)CuE,uE.

Proof

First we suppose that (VK2) holds. The proof is trivial if p = 2 or 6. Now we prove that the embedding is true for p ∈ (2, 6) under the assumption (VK3). For fixed p ∈ (2, 6), define m=(6p)4, and hence p = 2m + (1 – m)6. so we have that

R3K(x)|u|pdx=R3|u|2m|u|(1m)6dxR3|K(x)|1m|u|2dxmR3|u|6dx1msupxR3|K(x)||V(x)|mR3V(x)u2dxmR3|u|6dx1mCsupxR3|K(x)||V(x)|mR3V(x)u2dxmR3|u|2dx3(1m)CsupxR3|K(x)||V(x)|mR3(|u|2+V(x)u2)dxm+3(1m)=CsupxR3|K(x)||V(x)|msupxR3R3|u|2+V(x)|u|2dxp2.

Since K(x) ∈ L(ℝ3) and K/VL(ℝ3), we have that

uLKp(R3)CuE,forp(2,6).

Next, we suppose that (VK4) holds. Using the same argument as above, we define m0=(6p0)4, and hence p0 = 2m0 + (1 – m0)6 so that we have

R3K(x)|u|p0dx=R3K(x)|u|2m0|u|(1m0)6dxR3|K(x)|1m0|u|2dxm0R3|u|6dx1m0supxR3|K(x)||V(x)|m0R3V(x)|u|2dxm0R3|u|6dx1m0CsupxR3|K(x)||V(x)|m0R3(|u|2+V(x)|u|2)dxp02.

From (VK3) we deduce that |K(x)||V(x)|m0L(R3). It follows from the above inequality that

uLKp0(R3)CuE.

we complete the proof.□

The functional J satisfies the mountain pass geometry.

Lemma 3.2

The functional J satisfies the following conditions:

  1. There exist ρ and α > 0 such that J(u) ≥ α with |u|E = ρ.

  2. There exists eBρ(0) with J(e) < 0.

Proof

  1. Now, we distinguish two case.

    1. We suppose that (VK3) is true. For any ε > 0, it follows from (f1) and (f2) that there exists Cε > 0 such that

      F(u)ε2|u|2+Cε|u|6,foralluE. (3.1)

      Thus, by (3.1) and Lemma 3.1, we get that

      R3K(x)F(u)dxε2R3K(x)|u|2dx+CεR3K(x)|u|6dxε2uE2+CεuE6.

      Hence, in view of Lemma 2.3, we obtain

      J(u)=12R3|u|2+V(x)|u|2dx110R3l(x)ϕu|u|5dxηR3K(x)F(u)dx12uE2CuE10ε2uE2CεuE6=(1ε2)uE2CuE10CεuE6.

      So, taking ε=12, there exists enough small |u|E = ρ, such that J(u) ≥ α.

    2. We suppose that (VK4) holds. By (f1) and (f2), there exist Cε>0 and Cε>0 such that

      F(u)Cε|u|p0+Cε|u|6,foralluE.

      Therefore

      J(u)=12R3|u|E2+V(x)u2dx110R3l(x)ϕu|u|5dxηR3K(x)F(u)dx12uE2CuE6CuE10CuEp0.

      The same as Case 1, we can take |u|E = ρ such that J(u) ≥ α.

  2. For every t > 0 we obtain

    J(tu)=t22uE2t1010R3K(x)ϕu|u|5dxηR3K(x)F(tu)dx.

From (f3), we obtain J(tu) → –∞ as t → +∞, so it is satisfies (ii). We complete the proof.□

As a consequence of Lemma 3.2, we can find a (PS) sequence of the functional J(u) at the level

c:=infγΓmaxt[0,1]J(γ(t))>0, (3.2)

where the set of paths is defined as

J:={γC([0,1],H1(R3)):γ(0)=0,J(γ(1))<0}.

Lemma 3.3

Let {un} be a (PS)c sequence for J. Then {un} is bounded in E.

Proof

Let {un} ⊂ E be a (PS)c sequence for J, that is

J(un)candJ(un)0asn+.

Therefore, From (f3), we have

c+1+unEJ(un)1θJ(un),un=121θunE2+ηθR3K(x)(f(un)unθF(un))dx+1θ110R3l(x)ϕu|u|5dx121θunE2,

which implies that {un} is bounded in E.

Because of the appearance of the critical nonlocal term, we have to estimate the Mountain-pass value given by (3.2) carefully, To do it, we choose the extremal function Uε(x)=(3ε2)14(ε2+|xx0|2)12 to solve –△u = u5 in ℝ3. Let φ C0 (ℝ3) be a cut-off function verifying that 0 ≤ φ(x) ≤ 1 for all x ∈ ℝ3, suppφB2(x0), and φ(x) ≡ 1 on B1(x0). Set Vε = φUε, then thanks to the asymptotic estimates from [8], we have

vε|22=S32+O(ε),|vε|62=S12+O(ε)

and for s ∈ [2, 6)

|vε|ss=O(εs2),ifs[2,3),O(ε32|logε|),ifs=3,O(ε6s2),ifs(3,6),

where S denotes the best constant for the embedding D1,2(ℝ3) ↪ L6(ℝ3), namely,

S:=infuD1,2(R3)R3|u|2dx,R3|u|6dx=1.

We define

Vmax:=maxxB2R(x0)V(x)

and

Kmin:=minxB2R(x0)K(x).

By the assumption (l2) we also have

l(x0)B2R(x0)ϕ|u|5dxB2R(x0)l(x)ϕ|u|5dx.

Lemma 3.4

Suppose that (V, K) ∈ 𝓚, (f1)–(f3) hold, and l(x) satisfies (l1)–((l2). Then, there exists a u0E ∖ {0} such that

0<supt0J(tu0)<25S32|l(x)|L(R3)12.

Proof

We now consider

I(tvε)=t22R3(|vε|2+V(x)|vε|2)dxt1010R3l(x)ϕvε|vε|5dxηR3K(x)F(tvε)dx.

By Lemma 3.1, we know that, there exists tε > 0 such that supt0 J(tvε) > 0 is attained and limt J(tvε) = –∞ for any ε > 0.

We suppose that there exists ϱ1, ϱ2 such that ϱ1 < tε < ϱ2 for small enough ε > 0. In fact, J(tε vε) = supt0 J(tvε), and hence dJ(tvε)/dt|t=tε = 0, we obtain that

tεB2R(x0)(|vε|2+V(x)|vε|2)dxηB2R(x0)K(x)f(tεvε)vεdxtε10B2R(x0)l(x)ϕvε|vε|5dx=0. (3.3)

Now, we prove that tε → +∞ as εn → 0+ does not hold. By (3.3)

tεnB2R(x0)(|vεn|2+V(x)|vεn|2)dxtεn10B2R(x0)l(x)ϕvε|vε|5dx,

which is a contradiction when tε → +∞.

Similarly, we suppose that there is a sequence εn → 0 as εn → 0+. Firstly, if (VK3) holds, from (f1) and (f2), for all δ > 0 there exists Cδ > 0 such that

R3K(x)f(t~εnvεn)vεndxδt~εnR3K(x)|vεn|2dx+Cδ(t~εn)5R3K(x)|vεn|6dxδCt~εnR3(|vεn|2+V(x)|vεn|2)dx+Cδ(t~εn)5R3K(x)|vεn|6dx.

Choosing δ=12C, it follows from (3.3) that

t~εn2R3(|vεn|2+V(x)|vεn|2)dxCδ(t~εn)5R3K(x)|vεn|6dx+(t~εn)9R3l(x)ϕvn|vεn|5dx.

Next, we suppose that (VK4) holds. By (f1), (f2), there is a constant C > 0, such that

R3K(x)f(t~εnvεn)dx(t~εn)p01R3K(x)|vεn|p0dx+C¯(t~εn)5R3K(x)|vεn|6dx.

It again follows from (3.3) that

t~εnR3(|vεn|2+V(x)|vεn|2)dx(t~εn)p01R3K(x)|vεn|p0dx+C¯(t~εn)5R3K(x)|vεn|6dx+(t~εn)9R3l(x)ϕvεn|vεn|5dx,

We arrive at a contradiction because p0 > 2. So we complete the proof.

Since 0 < ϱ1 < tε < ϱ2 < ∞, together with the definitions of Vmax and Kmin, we have

J(tvε)=t22R3(|vε|2+V(x)|vε|2)dxt1010R3l(x)ϕvε|vε|5dxηR3K(x)F(tvε)dxtε22R3(|vε|2+V(x)|vε|2)dxtε1010R3l(x)ϕvε|vε|5dxηR3K(x)F(tεvε)dxtε22R3|vε|2dx+tε22Vmax(x)R3|vε|2dxtε1010R3l(x)ϕvε|vε|5dxηKmin(x)R3F(tεvε)dx.

We define

h(t):=t22R3|vε|2dxt1010R3l(x)ϕvε|vε|5dx.

By some elementary calculations, we obtain

maxt0h(t)=4(12R3|vε|2dx)545(12R3l(x)ϕvε|vε|5dx)14=25(R3|vε|2dx)54(R3l(x)ϕvε|vε|5dx)14.

The Poisson equation –△ϕvε = |vε|5 and Cauchy’s inequality give

R3|vε|6dx=R3ϕvε|vε|dx12|l(x)|R3|ϕvε|2dx+|l(x)|2R3|vε|2dx=12|l(x)|R3ϕvε|vε|5dx+|l(x)|2R3|vε|2dx.

This implies that

R3l(x)ϕvε|vε|5dx2|l(x)|R3l(x)|vε|6dx|l(x)|2R3|vε|2dx=|l(x)|2S32+O(ε).

As a consequence of the above fact, one has

maxt0h(t)25(S25+O(ε))54(|l(x)|2S32+O(ε))14=25|l(x)|S32+O(ε).

On the other hand, from (f3), we obtain that F(s) ≥ Csθ, for s > 0. Hence,

B2R(x0)F(tεvε)dxCB2R(x0)(tεvε)θdxCϱ1θBR(x0)(vε)θdx=Cϱ1θO(εθ+12),ifθ[1,2),Cϱ1θO(ε32|logε|),ifθ=2,Cϱ1θO(ε5θ2),ifθ(2,5). (3.4)

We have maxt0J(tvε)=J(tεvε) at the beginning, that is,

maxt0J(tvε)=tε22R3|vε|2+V(x)|vε|2dxtε1010R3l(x)ϕvε|vε|5dxηR3K(x)F(tvε)dxh(tε)+Vmax2R3|vε|2dxηR3K(x)F(tvε)dx25|l(x)|S32+CO(ε)ηR3K(x)F(tvε)dx.

Using (3.4), we have

CO(ε)ηR3K(x)F(tvε)dxCO(ε)Cηϱ1θO(εθ+12),ifθ[1,2),Cηϱ1θO(ε32|logε|),ifθ=2,Cηϱ1θO(ε5θ2),ifθ(2,5).

If (1, 2) and η=ε12, then 12<θ+1212<1 and hence for small enough ε > 0

CO(ε)ηR3K(x)F(tvε)dxCO(ε)Cϱ1θε12O(εθ+12)<0,

when η > 0 is enough large.

If θ = 2 and η=ε12, then for small enough ε > 0

CO(ε)ηR3K(x)F(tvε)dxCO(ε)Cϱ1θε12O(ε32)|logε|<0,

when η > 0 is enough large.

If θ ∈ (2, 3] and η=ε12, then 125θ212<1 and hence for small enough ε > 0

CO(ε)ηR3K(x)F(tvε)dxCO(ε)Cϱ1θε12O(ε5θ2)<0,

when η > 0 is enough large.

If θ ∈ (3, 5) then 0<5θ2<1 and for small enough ε > 0

CO(ε)ηR3K(x)F(tvε)dxCO(ε)Cηϱ1θO(ε5θ2)<0,

for any η > 0.

Consequently, we show that for θ ∈ (3, 5) with any η > 0, or θ ∈ (1, 3] with enough large η > 0

CO(ε)ηR3K(x)F(tvε)dx<0.

So, we can obtain that supt0J(tvε)<25S32|l(x)|L(R3)12.

Proof of Theorem 1.1

Proof. From Lemma 3.3, we know that {un}n is bounded in E. Up to a subsequence, we have

unuinE,unuinLlocr(R3),forr[2,6),unua.einR3. (3.5)

By (2.1)(2.3), we get

J(un)=12R3(|un|2+V(x)|un|2)dxηR3K(x)F(un)dx110R3l(x)ϕun|un|5dx=c+o(1)

and

J(un),un=R3(|un|2+V(x)|un|2)dxηR3K(x)F(un)dxR3l(x)ϕun|un|5dx=o(1).

Assuming vn = unu, in view of (3.5), Proposition 2.2, Lemma 2.4 and the Brézis-Lieb Lemma [33, 34],

J(un)=J(u)+12R3(|vn|2+V(x)|vn|2)dx110R3l(x)ϕvn|vn|5dx (3.6)

and

J(un),un=R3(|un|2+V(x)|un|2)dxηR3K(x)f(un)undxR3l(x)ϕun|un|5dx=R3|u|2+V(x)|u|2)dxR3l(x)ϕu|u|5dxηR3K(x)f(u)udx+R3(|vn|2+V(x)|vn|2)dxR3l(x)ϕvn|vn|5dx+on(1). (3.7)

Since J′(un) → 0 as n → +∞ and by (3.5) again, we get

J(un),u=R3(|u|2+V(x)u2)dxR3l(x)ϕu|u|5dxηR3K(x)f(u)udx. (3.8)

From (3.7) and (3.8), we obtain

R3|vn|2dx+R3V(x)|vn|2dxR3l(x)ϕvn|vn|5dx0asn+ (3.9)

and

J(u)=12R3(|u|2+V(x)|u|2)dx110R3R3l(x)ϕu|u|5dxηR3K(x)F(u)dx=12ηR3K(x)f(u)udxηR3K(x)F(u)dx+25R3l(x)ϕu|u|5dx0. (3.10)

Without loss of generality we can suppose that

R3(|vn|2+V(x)|vn|2)dxlasn (3.11)

and from (3.9)

R3l(x)ϕvn|vn|5dxlasn. (3.12)

By estimate

R3l(x)ϕvn|vn|5dx|l(x)|2|vn|610S|l(x)|2vn10S6. (3.13)

Combining (3.10)(3.13), which implies l|l(x)|2l5S6. Therefore, either l = 0 or l|l(x)|12S32.

If l > 0, we have l|l(x)|12S32. Taking the limit in (3.6) as n → +∞, and using (3.10), we obtain c0 25|l(x)|12S32. On the other hand, from (3.2) and Lemma 3.4, we obtain c025|l(x)|12S32. We get a contradiction. This shows that l = 0. Thus

J(u)=c>0and J(u)=0,

i.e. u is a non-trivial solution of (1.1). We complete the proof.□

Acknowledgements

The authors thank the referees for valuable comments and suggestions which improved the presentation of this manuscript. This work is partially supported by the Fundamental Research Funds for the National Natural Science Foundation of China 11671403 and Guizhou University of Finance and Economics of 2018YJ20.

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Received: 2019-04-04
Accepted: 2019-08-16
Published Online: 2019-10-18

© 2019 Shao, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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