Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 47, pp. 1–10.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

POSITIVE SOLUTIONS OF SCHRÖDINGER-POISSON SYSTEMS

WITH HARDY POTENTIAL AND INDEFINITE

NONLINEARITY

YONGYI LAN, BIYUN TANG, XIAN HU

Abstract. In this article, we study the nonlinear Schrödinger-Poisson system

−∆u+ u− µ
u

|x|2
+ l(x)φu = k(x)|u|p−2u x ∈ R3,

−∆φ = l(x)u2 x ∈ R3,

where k ∈ C(R3) and 4 < p < 6, k changes sign in R3 and lim sup|x|→∞ k(x) =

k∞ < 0. We prove that Schrödinger-Poisson systems with Hardy potential and
indefinite nonlinearity have at least one positive solution, using variational

methods.

1. Introduction and statement of the main result

In recent years, the Schrödinger-Poisson system

−∆u+ V (x)u+ l(x)φu = f(x, u) x ∈ R3,

−∆φ = l(x)u2 x ∈ R3,
(1.1)

have been widely investigated, because of its importance in quantum mechanics
models and in semiconductor theory. For more details about its physical aspects,
see [4, 7] and the references therein. There have been several results about nontriv-
ial solutions, radial and nonradial solutions, ground states, multiplicity of solutions,
and concentration of solutions, depending on assumptions of the potential V . Most
of the literature focuses on the study of (1.1) with V ≡ 1, see e.g. [3, 5, 8, 11, 15, 22].
Azzollini and Pomponio [3] proved the existence of a ground state solution when
f(x, u) = |u|p−1u with p ∈ (3, 5), V ≡ 1, and l(x) ≡ 1. Ruiz [15] proved the
existence and nonexistence of nontrivial solutions by using a constrained mini-
mization method. Cerami et al. [5] obtained the existence of ground states and
bound states, under suitable assumptions. Huang et al. [11] considered the case
when f(x, u) is a combination of a superlinear and linear terms. More precisely,
f(x, u) = k(x)|u|p−1u + µh(x)u, where 4 < p < 6 and µ > 0, k(x) ∈ C(R3), k
changes sign in R3, and lim|x|→∞ k(x) = k∞ < 0. They proved the existence of at
least two positive solutions. In [22] the authors obtained the existence of ground
state and multiple solutions using critical growth by variational methods. Existence

2010 Mathematics Subject Classification. 35J20, 35J70.
Key words and phrases. Hardy potential; variational methods; indefinite nonlinearity;

positive solution.
c©2020 Texas State University.

Submitted April 6, 2020. Published May 21, 2020.

1



2 Y. LAN, B. TANG, X. HU EJDE-2020/47

of multiple positive solutions of Schrödinger-Poisson type equations with indefinite
nonlinearity was proved in [8] using the mountain pass theorem. For V 6≡ 1 with
infx∈R3 V (x) > 0, there are also many results (see [6, 10, 16, 17, 18, 21]). For
other interesting results on the Schrödinger-Poisson system, we refer readers to
[12, 13, 14, 19, 20] and references therein.

In this article, we consider the system

−∆u+ u− µ u

|x|2
+ l(x)φu = k(x)|u|p−2u x ∈ R3,

−∆φ = l(x)u2 x ∈ R3,
(1.2)

where 0 < µ < µ := (N−2)2

4 = 1
4 , 4 < p < 6, k(x) ∈ C(R3), k changes sign in R3,

and lim sup|x|→∞ k(x) = k∞ < 0.
To the best of our knowledge, the literature does not have results on the existence

of positive solutions to (1.2) with Hardy potential. The aim of this article is to
show the existence of positive solutions of problem (1.2). Our approach combines
variational techniques based on critical point theory and some analysis techniques.

Hereafter we use the following notation: For 1 ≤ s < +∞, Ls(R3) is the Lebesgue
space endowed with norm

‖u‖ss :=

∫
R3

|u|s dx.

H1(R3) is the Sobolev space endowed with the scalar product and norm

(u, v) :=

∫
R3

(
∇u · ∇v + uv − µ uv

|x|2
)

dx; ‖u‖2 :=

∫
R3

(
|∇u|2 + u2 − µ u2

|x|2
)

dx.

By Hardy inequality [9], we easily derive that this norm is equivalent to the usual
norm, in H1(R3),

‖u‖0 =
(∫

R3

|∇u|2 + u2 dx
)1/2

.

D1,2(R3) is the completion of C∞0 (R3) with respect to the norm

‖u‖D1,2 :=
(∫

R3

|∇u|2 dx
)1/2

.

on(1) is a quantity that approaches zero as n→∞.
Recall that by the Lax-Milgram theorem, for each u ∈ H1(R3), there exists a

unique solution φu ∈ D1,2(R3) of

−∆φ = l(x)u2, x ∈ R3. (1.3)

Using this in (1.2) gives

−∆u+ u− µ u

|x|2
+ l(x)φuu = k(x)|u|p−2u, x ∈ R3. (1.4)

Moreover one has

φu(x) =
1

4π

∫
R3

l(y)u2(y)

|x− y|
dy,

‖φu‖2D1,2(R3) =

∫
R3

|∇φu|2 dx =

∫
R3

l(x)φuu
2 dx.



EJDE-2020/47 SCHRÖDINGER-POISSON SYSTEMS WITH HARDY POTENTIAL 3

There is a one-to-one correspondence between the solution of (1.3) and the critical
points of the functional defined in H1(R3) by

I(u) =
1

2

∫
R3

(
|∇u|2 + |u|2 − µ u2

|x|2
)

dx+
1

4

∫
R3

l(x)φu(x)u2(x) dx

− 1

p

∫
R3

k(x)|u|p dx .

(1.5)

Then I ′(u) is defined by

〈I ′(u), v〉 =

∫
R3

(
∇u · ∇v + uv − µ uv

|x|2
)

dx+

∫
R3

l(x)φuuv dx

−
∫
R3

k(x)|u|p−2uv dx.

(1.6)

A pair of functions (u, φ) is called a positive solution of problem (1.2) if it satisfies
(1.2) and u > 0, φ > 0 for a.a. x ∈ R3.

Let us introduce some hypotheses on k(x) and l(x):

(H1) k ∈ C(R3) and k changes sign in R3;
(H2) lim sup|x|→∞ k(x) = k∞ < 0;

(H3) l ∈ L2(R3), l(x) ≥ 0 for all x ∈ R3 and l 6≡ 0;
(H4) l = 0 a.e. in Ω0 := {x ∈ R3 : k(x) = 0} and Ω0 coincides with the closure

of its interior.

The main result of this article is the following theorem.

Theorem 1.1. Assume that (H1)–(H4) hold, and 4 < p < 6. Then problem (1.2)
has at least one positive solution in H1(R3)×D1,2(R3).

In the following discussions, c or ci (i = 0, 1, . . . ) we denote positive constants.

2. Proof of Theorem 1.1

The proof is presented in three steps.

Step 1: The (PS) condition. Let {un} be any sequence in H1(R3) such that I(un)
is bounded and I ′(un) converges to zero; that is,

I(un) =
1

2
‖un‖2 +

1

4

∫
R3

l(x)φun(x)u2
n(x) dx− 1

p

∫
R3

k(x)|un|p dx→ c, (2.1)

and

〈I ′(un), ϕ〉 =

∫
R3

(
∇un · ∇ϕ+ unϕ− µ

unϕ

|x|2
)

dx+

∫
R3

l(x)φununϕdx

−
∫
R3

k(x)|un|p−2unϕdx→ 0,

(2.2)

for any ϕ ∈ H1(R3) as n→∞.
We now prove that {un} is bounded in H1(R3). By contradiction, we assume

‖un‖ → ∞. Let vn = un/‖un‖, then ‖vn‖ = 1 for each n ∈ N. Then there exists a



4 Y. LAN, B. TANG, X. HU EJDE-2020/47

v ∈ H1(R3) such that for each bounded domain Ω ⊂ R3,

vn ⇀ v in H1(R3),

vn → v a.e. in R3,

vn → v in Lsloc(R3), where 2 ≤ s < 2∗ = 6,

‖vn‖ ≤ wΩ(x), for some wΩ(x) ∈ Ls(Ω).

(2.3)

So, for any ϕ ∈ H1(R3), we have∫
R3

(
∇vn∇ϕ+ vnϕ− µ

vnϕ

|x|2
)

dx→
∫
R3

(
∇v∇ϕ+ vϕ− µ vϕ

|x|2
)

dx. (2.4)

We claim that v(x) = 0 a.e. in R3. In fact, since un = ‖un‖vn, (2.2) becomes∫
R3

(
∇vn∇ϕ+ vnϕ− µ

vnϕ

|x|2
)

dx+ ‖un‖2
∫
R3

l(x)φvnvnϕdx

− ‖un‖p−2

∫
R3

k(x)|vn|p−2vnϕdx→ 0, as n→∞.
(2.5)

Next we prove the claim for x in Ω+, Ω− and Ω0. From (H1), we see that Ω+ 6= ∅
and Ω− 6= ∅.

First, let x ∈ Ω+. Since k ∈ C(R3), there exists δ > 0 such that

k(y) > 0, ∀y ∈ Bδ(x). (2.6)

We define a continuous function ζm ∈ C(R3) (m > 2) such that ζm(y) ≥ 0 for any
y ∈ R3 and

ζm(y) =

{
1, if y ∈ B( 1

2−
1
m2 )δ(x),

0, if y ∈ R3 \Bδ/2(x).
(2.7)

Taking ϕ = vζm in (2.5), we know that suppϕ ⊂ Bδ/2(x) for any m ∈ N and m > 2.
In view of (2.3), we have

k(y)|vn(y)|p−2vn(y)ϕ(y)→ k(y)|v(y)|p−2v(y)ϕ(y) for y ∈ Bδ/2(x),

and
|k(y)vn(y)p−1ϕ(y)| ≤ C|wΩ(y)|p−1|ϕ(y)| ∈ L1(Bδ/2(x)).

Therefore, by the Lebesgue dominated convergent theorem, we have∫
Bδ/2(x)

k(y)|vn(y)|p−2vn(y)ϕ(y) dy →
∫
Bδ/2(x)

k(y)|v(y)|p−2v(y)ϕ(y) dy. (2.8)

Dividing (2.5) by ‖un‖p−2 and passing to the limit as n → ∞, in view of the
boundedness of vn, we obtain

0 = lim
n→∞

∫
R3

k(y)|vn(y)|p−2vn(y)ϕ(y) dy

=

∫
Bδ/2(x)

k(y)|v(y)|p−2v(y)ϕ(y) dy

=

∫
B

( 1
2
− 1
m2 )δ

(x)

k(y)|v(y)|p dy +

∫
Bδ/2(x)\B

( 1
2
− 1
m2 )δ

(x)

k(y)|v(y)|pζm dy,

(2.9)

for any m ∈ N and m > 2. Passing to the limit in (2.9) as m→∞, we obtain∫
Bδ/2(x)

k(y)|v(y)|p dy = 0.



EJDE-2020/47 SCHRÖDINGER-POISSON SYSTEMS WITH HARDY POTENTIAL 5

It follows from (2.6) that v = 0 a.e. in Bδ/2(x). Since x ∈ Ω+ is arbitrarily, we can

obtain that v = 0 a.e. in Ω+.
A similar argument shows that v = 0 a.e. in Ω−. Next, we prove that v = 0

a.e. in Ω0. If |Ω0| = 0, the claim is true. If |Ω0| 6= 0, we take ϕ ∈ C(R3) with
suppϕ ⊂ Ω0 in (2.5). From the definition of Ω0 and the assumption that l = 0 a.e.
in Ω0, it follows that∫

R3

k(y)|vn(y)|p−2vn(y)ϕ(y) dy =

∫
suppϕ

k(y)|vn(y)|p−2vn(y)ϕ(y) dy = 0, (2.10)∫
R3

l(y)φvnvnϕdy =

∫
suppϕ

l(y)φvnvnϕdy = 0, (2.11)

for any n ∈ N. By (2.4), (2.10), (2.11), passing to the limit in (2.5) as n → ∞ we
obtain ∫

R3

(
∇v∇ϕ+ vϕ− µ vϕ

|x|2
)

dx = 0. (2.12)

From (2.12) and v = 0 a.e. in Ω+ ∪ Ω−, we obtain∫
Ω0

(
∇v∇ϕ+ vϕ− µ vϕ

|x|2
)

dx = 0. (2.13)

Therefore, v = 0 a.e. in Ω0. Hence, vn ⇀ 0 in H1(R3).
In the second place, choosing ϕ = vn in (2.2), dividing (2.1) by ‖un‖2 and

dividing (2.2) by ‖un‖, we obtain

1

2
+

1

4
‖un‖2

∫
R3

l(x)φvn(x)v2
n(x) dx− 1

p

∫
R3

k(x)|un|p−2v2
n dx→ 0, (2.14)

1 + ‖un‖2
∫
R3

l(x)φvn(x)v2
n(x) dx−

∫
R3

k(x)|un|p−2v2
n dx→ 0, (2.15)

as n→∞.
From (2.14), (2.15) and the assumption of 4 < p < 6, it follows that

lim
n→∞

∫
R3

k(y)|un(y)|p−2v2
n(y) dy =

p

4− p
< 0.

Moreover, in view of (2.14), we deduce that

lim
n→∞

∫
R3

k(y)|un(y)|p−2v2
n(y) dy > 0.

which yields a contradiction. Hence {un} is bounded in H1(R3).
Now we prove that {un} has a convergent subsequence. Since {un} is bounded

in H1(R3). Going if necessary to a subsequence (still denoted by {un}), we may
assume that

un ⇀ u in H1(R3), un → u a.e. R3,

∇un ⇀ ∇u in L2(R3), un ⇀ u in L2(R3),

un → u in Lsloc(R3), where 2 ≤ s < 2∗ = 6.

We define wn = k(x)|un|p−2un and w = k(x)|u|p−2u. Then wn → w a.e. in R3.
Since {un} is bounded in Lp(R3) for 4 < p < 6 and k is bounded in R3, it follows

that {wn} is bounded in L
p
p−1 (R3), so there exists M > 0 such that(∫

E

|wn(x)− w(x)|
p
p−1 dx

) p−1
p ≤M, (2.16)



6 Y. LAN, B. TANG, X. HU EJDE-2020/47

and there exists w̃ ∈ L
p
p−1 (R3) such that wn ⇀ w̃ in L

p
p−1 (R3) with 4 < p < 6.

Moreover w = w̃ a.e. in R3; indeed, let f ∈ (L
p
p−1 (R3))∗ = Lp(R3), for any ε > 0

there exists r > 0 such that∫
{x∈R3:|x|≥r}

|f(x)|p dx ≤ εp

2Mp
. (2.17)

Moreover, for any ε > 0 there exists δ > 0 such that for every E ⊆ R3 and
measE < δ, one has ∫

E

|f(x)|p dx ≤ εp

2Mp
. (2.18)

From Hölder’s inequality and (2.16)–(2.18), for every E ⊆ R3 and measE < δ, we
have ∣∣ ∫

E

f(x)(wn(x)− w(x)) dx
∣∣

≤
(∫

E

|f(x)|p dx
)1/p(∫

E

|wn(x)− w(x)|
p
p−1 dx

) p−1
p

≤M
(∫

E∩{x∈R3: |x|≥r}
|f(x)|p dx+

∫
E∩{x∈R3: |x|≤r}

|f(x)|p dx
)1/p

≤M
( εp

2Mp
+

∫
E∩{x∈R3: |x|≤r}

|f(x)|p dx
)1/p

≤M
( εp

2Mp
+

εp

2Mp

)1/p

≤ ε,

hence {
∫
R3 f(x)(wn(x)−w(x)) dx, n ∈ N} is equi-absolutely-continuous. It follows

easily from Vitali Convergence Theorem that∫
R3

f(x)(wn(x)− w(x)) dx→
∫
R3

f(x)(w(x)− w(x)) dx = 0. (2.19)

therefore w = w̃ a.e. in R3.
Note that, for any ψ ∈ H1(R3), one has∫

R3

k(x)|un|p−2up−2
n ψ dx→

∫
R3

k(x)|u|p−2up−2ψ dx, (2.20)

and ∫
R3

(
∇un∇ψ + unψ − µ

unψ

|x|2
)

dx→
∫
R3

(
∇u∇ψ + uψ − µ uψ

|x|2
)

dx, (2.21)

as n→∞.
Now we prove that∫

R3

l(x)φun(x)u2
n dx→

∫
R3

l(x)φu(x)u2 dx, (2.22)

as n→∞ and that for all ψ ∈ H1(R3),∫
R3

l(x)φun(x)u2
nψ dx→

∫
R3

l(x)φu(x)u2ψ dx. (2.23)



EJDE-2020/47 SCHRÖDINGER-POISSON SYSTEMS WITH HARDY POTENTIAL 7

We borrow the strategy from [5]. In fact, by the continuity of embedding, we have
un ⇀ u in H1(R3) that

un ⇀ u in L6(R3),

u2
n → u2 in L3

loc(R3),

φun ⇀ φu in D1,2(R3),

φun ⇀ φu in L3
loc(R3).

(2.24)

Thus, given ε > 0, we have∣∣ ∫
R3

l(x)(φun(x)− φu(x))u2
n dx

∣∣ ≤ ε, (2.25)

for n large enough. And for any fixed ψ,∣∣∣ ∫
R3

l(x)φu(x)(un − u)(x) dx
∣∣∣ ≤ ε, (2.26)

for n large enough. Moreover, by (2.24)2 and (2.24)4, we can assert that for any
choice of ε and ρ > 0, the relations(∫

Bρ(0)

|u2
n − u2|3 dx

)1/3

≤ ε, (2.27)(∫
Bρ(0)

|φun − φu|6 dx
)1/6

≤ ε, (2.28)

hold for n large enough.
Noting that {un} is bounded in H1(R3). It is deduced from this and the con-

tinuity of the Sobolev embedding of D1,2(R3) in L6(R3) that {φun} is bounded in
D1,2(R3) and in L6(R3). Since l ∈ L2(R3), then lu2

n and lu2 belong to L6/5(R3)
and that for any ε > 0 there exists ρ = ρ(ε) such that(∫

R3\Bρ(0)

|l(x)|2 dx
)1/2

≤ ε for ρ ≥ ρ. (2.29)

Thus, from (2.22), (2.25), (2.27) and (2.29), we obtain∣∣∣ ∫
R3

l(x)φun(x)u2
n dx−

∫
R3

l(x)φu(x)u2 dx
∣∣∣

≤
∣∣∣ ∫

R3

l(x)φun(x)(u2
n − u2) dx

∣∣∣+
∣∣∣ ∫

R3

l(x)(φun(x)− φu(x))u2 dx
∣∣∣

≤ ‖φun‖6
(∫

R3

|l(x)(u2
n − u2)|6/5 dx

)5/6

+ ε

≤ C
(∫

R3\Bρ(0)

|l(x)(u2
n − u2)|6/5 dx+

∫
Bρ(0)

|l(x)(u2
n − u2)|6/5 dx

)5/6

+ ε

≤ C
[( ∫

R3\Bρ(0)

|l(x)|2 dx
)3/5

|u2
n − u2|6/53

+ |l|6/52

(∫
Bρ(0)

|u2
n − u2|3 dx

)2/5]5/6
+ ε ≤ Cε,

for n large enough.



8 Y. LAN, B. TANG, X. HU EJDE-2020/47

By a similar argument, we conclude from (2.23) (2.26), (2.28) and (2.29) that∣∣∣ ∫
R3

l(x)φun(x)unψ(x) dx−
∫
R3

l(x)φu(x)uψ(x) dx
∣∣∣

≤
∣∣∣ ∫

R3

l(x)φu(x)(un − u)ψ(x) dx
∣∣∣+
∣∣∣ ∫

R3

l(x)(φun(x)− φu(x))unψ(x) dx
∣∣∣

≤ ‖un‖6‖ψ‖6
(∫

R3

|l(x)(φun(x)− φu(x))| 32 dx
)2/3

+ ε

≤ Cε,

for n large enough.
From (2.20) and (2.21) with (2.23), one has

〈I ′(un), ψ〉

=

∫
R3

(
∇un∇ψ + unψ − µ

unψ

|x|2
)

dx+

∫
R3

l(x)φun(x)unψ(x) dx

−
∫
R3

k(x)|un|p−2unψ dx

→
∫
R3

(
∇u∇ψ + uψ − µ uψ

|x|2
)

dx+

∫
R3

l(x)φu(x)uψ(x) dx

−
∫
R3

k(x)|u|p−2uψ dx

= 〈I ′µ(u), ψ〉.

Since I ′(un) → 0 in H−1(R3), we have 〈I ′(un), ψ〉 →0 for any ψ ∈ H1(R3). So
〈I ′(u), ψ〉=0 for any ψ ∈ H1(R3), and

〈I ′(u), u〉 = 0. (2.30)

We denote vn = un − u. Then vn ⇀ 0 in H1(R3). By using this and (2.22), we
deduce that

lim
n→∞

∫
R3

l(x)φvn(x)v2
n(x) dx = 0. (2.31)

From the Brézis-Lieb lemma, we derive

‖un‖2 = ‖vn‖2 + ‖u‖2 + o(1),∫
R3

k(x)|un|p dx =

∫
R3

k(x)|vn|p dx+

∫
R3

k(x)|u|p dx+ o(1),

for n large enough.
It follows from (2.22) that

〈I ′(un), un〉 = 〈I ′(u), u〉+ ‖vn‖2 +

∫
R3

l(x)φvn(x)v2
n(x) dx−

∫
R3

k(x)|vn|p dx+ o(1).

By using this, (2.30), and (2.31), we deduce that

lim
n→∞

(
‖un − u‖2 −

∫
R3

k(x)|un − u|p dx
)

= 0. (2.32)

Next, without loss of generality we can assume that k∞ < −1. By (H2), there
is R0 > 0 such that

k(x) < −1 if |x| > R0. (2.33)



EJDE-2020/47 SCHRÖDINGER-POISSON SYSTEMS WITH HARDY POTENTIAL 9

Moreover, since k ∈ C(R3) and 4 < p < 6, we obtain∫
|x|≤R0

k(x)|un − u|p dx→ 0, (2.34)

as n→∞. It follows from (2.32)–(2.34) that

0 ≤ lim
n→∞

‖un − u‖2

= lim inf
n→∞

∫
R3

k(x)|un − u|p dx

≤ lim
n→∞

∫
|x|≤R0

k(x)|un − u|p dx = 0.

(2.35)

Thus we have un → u in H1(R3), which means that I satisfies (PS) condition.

Step 2: Mountain-pass geometric structure. It follows from (H3) that∫
R3

l(x)φu(x)u2(x) dx ≥ 0.

From (H1) and (H2), we have k is bounded in R3. It follows from the continuity of
the Sobolev embedding of H1(R3) in Lp(R3) that

I(u) =
1

2
‖u‖2 +

1

4

∫
R3

l(x)φu(x)u2(x) dx− 1

p

∫
R3

k(x)|u|p dx

≥ 1

2
‖u‖2 − C‖u‖p.

Choosing ρ = ‖u‖ small enough such that 1
2‖u‖

2−C‖u‖p > 0, we obtain I(u) > 0.

Choose ϕ ∈ H1(R3) with suppϕ ⊆ Ω+ such that ϕ(x) ≥ 0 with strict inequality
holding on a subset of positive measure, for all x ∈ Ω+. Then we have

I(sϕ) =
s2

2
‖ϕ‖2 +

s4

4

∫
R3

l(x)φϕ(x)ϕ2(x) dx− sp

p

∫
R3

k(x)|ϕ|p dx→ −∞,

as s→ +∞. Thus there is u1 := sϕ ∈ H1(R3) with ‖u1‖ > ρ such that I(u1) < 0.
Therefore I has a mountain pass geometry.

Step 3: Critical value of I. For u1 in step 2, we define

Γ := {γ : C[0, 1]→ H1(R3)|γ(0) = 0, γ(1) = u1},
c := inf

γ∈Γ
max

0≤t≤1
I(γ(t)).

It turns out that the Mountain Pass Theorem holds. Then c > 0 is critical value of
I.

Since I(u) = I(|u|) in H1(R3), we conclude that u ≥ 0 a.e. in R3 with I(u) > 0
and it is a solution of (1.3). The strong Maximum Principle implies that u > 0 in
R3.

Acknowledgments. This research was supported by the National Natural Sci-
ence Foundation of China (11671331), and by the National Foundation Training
Program of Jimei University (ZP2020057). The authors want to thank the anony-
mous referees for their valuable suggestions.



10 Y. LAN, B. TANG, X. HU EJDE-2020/47

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Yongyi Lan (corresponding author)

School of Science, Jimei University, Xiamen 61021, China
Email address: lanyongyi@jmu.edu.cn

Biyun Tang

School of Science, Jimei University, Xiamen 61021, China
Email address: 1520840642@qq.com

Xian Hu
School of Science, Jimei University, Xiamen 61021, China

Email address: 2321894958@qq.com


	1. Introduction and statement of the main result
	2. Proof of Theorem ??
	Acknowledgments

	References