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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# The existence of solutions to certain type of nonlinear difference-differential equations

Weiran Lü
/ Linlin Wu
/ Dandan Wang
/ Chungchun Yang
Published Online: 2018-07-17 | DOI: https://doi.org/10.1515/math-2018-0071

## Abstract

In this paper we study the entire solutions to a certain type of difference-differential equations. We also give an affirmative answer to the conjecture of Zhang et al. In addition, our results improve and complement earlier ones due to Yang-Laine, Latreuch, Liu-Lü et al. and references therein.

MSC 2010: 34M05; 30D35; 39A10; 39B32

## 1 Introduction and main results

In studying difference-differential equations in the complex plane ℂ, it is always an interesting and quite difficult problem to prove the existence or uniqueness of the entire or meromorphic solutions to a given difference-differential equation. There have been many studies and results obtained lately that relate to the existence or growth of the entire or meromorphic solutions of various types of difference or differential equations, see, e.g., [1, 2, 3, 4, 5, 6, 7, 8, 9] and references therein.

Herein let f denote a non-constant meromorphic function and we assume that the reader is familiar with the standard terminology and results of Nevanlinna theory such as the characteristic function T(r, f), the proximity function m(r, f) and the counting function N(r, f) ( see, e.g., [10, 11, 12] ). However, for the convenience of the reader, we shall repeat some notations needed below.

We call a meromorphic function α ≢ 0, ∞ a small function with respect to f, if T(r, α) = S(r, f), where S(r, f) denotes any quantity satisfying S(r, f) = o{T(r, f)} as r → ∞, possibly outside a set of r of finite linear measure. The order of f is

$ρ(f)=lim supr→∞log⁡T(r,f)log⁡r,$

and the hyper-order ρ2(f) is defined as

$ρ2(f)=lim supr→∞loglog⁡T(r,f)log⁡r.$

#### Definition 1.1

A difference polynomial, respectively, a difference-differential polynomial, in f is a finite sum of difference products of f and its shifts, respectively, of products of f, derivatives of f and of their shifts, with all the coefficients of these monomials being small functions of f.

#### Definition 1.2

Given a nonzero constant c, we define the difference operators by

$Δcf(z)=f(z+c)−f(z)andΔcnf(z)=Δc(Δcn−1f(z))(n≥2).$

For the sake of simplicity, we let Δf(z) = f(z + 1) – f(z) and Δnf(z) = Δ(Δn–1f(z)) (n ≥ 2) for the case c = 1 (see, e. g., [2, 13] and [14]).

For the benefit of the readers, we shall give some related results. Yang and Laine considered the following difference equation and proved:

#### Theorem A

([7]). A nonlinear difference equation

$f3(z)+q(z)f(z+1)=csin⁡bz,$

where q is a non-constant polynomial and b, c ∈ ℂ ∖ {0}, does not admit entire solutions of finite order. If q is a nonzero constant, then the above equation possesses three distinct entire solutions of finite order, provided that b = 3nπ and q3 = (–1)n+1c227/4 for a nonzero integer n.

In 2014, Liu and Lü et al. proved the following result.

#### Theorem B

([15]). Let n ≥ 4 be an integer, q be a polynomial, and p1, p2, α1, α2 be nonzero constants such that α1α2. If there exists some entire solution f of finite order to the following equation

$fn(z)+q(z)Δf(z)=p1eα1z+p2eα2z,$

then q is a constant, and one of the following relations holds:

1. $\begin{array}{}f\left(z\right)={c}_{1}{\mathrm{e}}^{\frac{{\alpha }_{1}}{n}z},\end{array}$ and c1(eα1/n – 1)q = p2, α1 = 2,

2. $\begin{array}{}f\left(z\right)={c}_{2}{\mathrm{e}}^{\frac{{\alpha }_{2}}{n}z},\end{array}$ and c2(eα2/n – 1)q = p1, α2 = 1,

where c1, c2 are constants satisfying $\begin{array}{}{c}_{1}^{n}={p}_{1},{c}_{2}^{n}={p}_{2}.\end{array}$

Recently, Zhang et al. obtained the following result.

#### Theorem C

([16]). Let q be a polynomial, and p1, p2, α1, α2 be nonzero constants such that α1α2. If f is an entire solution of finite order to the following equation:

$f3(z)+q(z)Δf(z)=p1eα1z+p2eα2z,$(1)

then q is a constant, and one of the following relations holds:

1. T(r, f) = N1)(r, $\begin{array}{}\frac{1}{f}\end{array}$) + S(r, f),

2. f(z) = c1exp( $\begin{array}{}\frac{{\alpha }_{1}z}{3}\end{array}$), and c1(exp( $\begin{array}{}\frac{{\alpha }_{1}}{3}\end{array}$) – 1)q = p2, α1 = 3α2,

3. f(z) = c2 exp( $\begin{array}{}\frac{{\alpha }_{2}z}{3}\end{array}$), and c2(exp( $\begin{array}{}\frac{{\alpha }_{2}}{3}\end{array}$) – 1)q = p1, α2 = 3α1,

where N1)(r, $\begin{array}{}\frac{1}{f}\end{array}$) denotes the counting function corresponding to simple zeros of f, and c1, c2 are constants satisfying $\begin{array}{}{c}_{1}^{3}={p}_{1},\text{\hspace{0.17em}}{c}_{2}^{3}={p}_{2}.\end{array}$

#### Remark 1.3

In [16], the authors also gave an example to show that the case (1) occurs indeed.

#### Example 1.4

Let f(z) = eπiz + eπiz = 2i sin(πiz). Then f is a solution of the following equation:

$f3(z)+3Δf(z)=e3πiz+e−3πiz.$

Obviously, T(r, f) = N1)(r, $\begin{array}{}\frac{1}{f}\end{array}$) + S(r, f). Thus, the case (1) occurs indeed.

Since in the above example α1 + α2 = 3πi + (–3πi) = 0, consequently, Zhang et al. posed the following conjecture.

#### Conjecture 1.5

([16]). If α1α2, α1 + α2 ≠ 0, then the conclusion (1) of Theorem A is impossible. In fact, any entire solution f of (1) must have 0 as its Picard exceptional value.

In 2017, Latreuch gave an affirmative answer to Conjecture 1.5. In fact, he obtained the following result.

#### Theorem D

([17]). Let q be a polynomial, and p1, p2, α1, α2 be nonzero constants such that α1α2 and α1 + α2 ≠ 0. If f is an entire solution of finite order of (1), then q is a constant, and one of the following relations holds:

1. f(z) = c1 exp( $\begin{array}{}\frac{{\alpha }_{1}z}{3}\end{array}$), and c1(exp( $\begin{array}{}\frac{{\alpha }_{1}}{3}\end{array}$) – 1)q = p2, α1 = 3α2;

2. f(z) = c2 exp( $\begin{array}{}\frac{{\alpha }_{2}z}{3}\end{array}$), and c2(exp( $\begin{array}{}\frac{{\alpha }_{2}}{3}\end{array}$) – 1)q = p1, α2 = 3α1,

where c1, c2 are constants satisfying $\begin{array}{}{c}_{1}^{3}={p}_{1},{c}_{2}^{3}={p}_{2}.\end{array}$ Furthermore, (1) does not have any entire solution of infinite order satisfies any one of the following conditions:

3. ρ2(f) < 1;

4. λ(f) < ρ(f) = ∞ and ρ2(f) < ∞,

here λ(f) denotes the exponent of convergence of zeros sequence of f.

In the present paper we continue discussing Conjecture 1.5. Moreover, our result will include several known results for difference or differential equations obtained earlier as its special case. In fact, we consider a slightly more general form of (1) and obtain the following result.

#### Theorem 1.6

Let L(z, f) denote a difference-differential polynomial in f of degree one with small functions as its coefficients such that L(z, 0) ≡ 0, and let p1, p2, α1, α2 be nonzero constants such that α1α2. If f is an entire solution with ρ2(f) < 1 to the following equation:

$f3+L(z,f)=p1eα1z+p2eα2z,$(2)

then one of the following relations holds:

1. f(z) = c1exp( $\begin{array}{}\frac{{\alpha }_{1}z}{3}\end{array}$) + c2exp( $\begin{array}{}\frac{{\alpha }_{2}z}{3}\end{array}$), where c1 and c2 are two nonzero constants satisfying $\begin{array}{}{c}_{1}^{3}={p}_{1},{c}_{2}^{3}={p}_{2}\end{array}$ and α1 + α2 = 0;

2. f3(z) = (p1c1) exp(α1z), and L(z, f) = c1 exp(α1z) + p2 exp(α2z), where c1 is a constant;

3. f3(z) = (p2c2) exp(α2z), and L(z, f) = p1 exp(α1z) + c2 exp(α2z), where c2 is a constant.

From Theorem 1.6, we have

#### Corollary 1.7

Equation (2) does not have any entire solution f with ρ(f) = ∞ and ρ2(f) < 1.

#### Remark 1.8

It is obvious from Theorem 1.6 that the above conjecture is true. We also point out that if an entire function solution f in Theorem 1.6 is replaced by a meromorphic solution with N(r, f) = S(r, f), the conclusion of Theorem 1.6 still holds.

#### Remark 1.9

Lemma 2.2 (in section 2) is crucial to the proofs of our main results. However, it may be false if the conditionρ2(f) < 1 ” is violated. There is no difficulty in showing that f(z) = exp(exp(z)) is a counterexample. Now one may raise the questions: what will happen if we delete the condition ρ2(f) < 1 in Theorem 1.6, Corollary 1.7 and so on?

## 2 Some lemmas

In order to prove Theorem 1.6, we need the following results.

#### Lemma 2.1

([18]). Let m, n be positive integers satisfying $\begin{array}{}\frac{1}{m}+\frac{1}{n}<1.\end{array}$ Then there are no transcendental entire solutions of f and g satisfy the following equation

$a(z)f(z)n+b(z)g(z)m=1,$(3)

with a, b being small functions of f, and g, respectively.

#### Lemma 2.2

([19]). Let f be a transcendental meromorphic function of hyper-order ρ2(f) < 1. Then for c ∈ ℂ, we have

$m(r,f(z+c)f(z))=S(r,f),$(4)

outside of a possible exceptional set with finite logarithmic measure.

#### Remark 2.3

The following result is the analogue of the logarithmic derivatives lemma [10, 11] for the difference-differential polynomials of a meromorphic function f. It can be proved by applying Lemma 2.2 and the logarithmic derivatives lemma with a similar reasoning as in [19, 20, 21, 22] and stated as follows.

#### Lemma 2.4

Let f be a transcendental meromorphic function with ρ2(f) < 1. Given L(z, f) as to Theorem 1.6, then for any positive integer k, we have

$m(r,L(z,f)f(z))+m(r,L(k)(z,f)f(z))=S(r,f),$(5)

outside of a possible exceptional set with finite logarithmic measure.

#### Lemma 2.5

([12], Theorem 1.55). Let g1, g2,⋯,gp be transcendental meromorphic functions satisfying Θ(∞, gj) = 1 (j = 1, 2, ⋯, p). If $\begin{array}{}\sum _{j=1}^{p}{a}_{j}{g}_{j}=1,\end{array}$ then for aj ∈ ℂ ∖ {0} (j = 1, 2, ⋯, p), we have $\begin{array}{}\sum _{j=1}^{p}\end{array}$ δ(0, gj) ≤ p – 1.

#### Remark 2.6

By the same methods as in the proof of Theorem 1.55 used in [12], we also point out that if nonzero constants a1, a2, ⋯, ap are replaced by small functions of g1, g2, ⋯, gp, the conclusion of Lemma 2.5 still holds.

#### Lemma 2.7

([22]). Suppose that f is a transcendental meromorphic function, a, b, c and d are small functions of f such that acd ≢ 0. If

$af2+bff′+c(f′)2=d,$

then

$c(b2−4ac)d′d+b(b2−4ac)−c(b2−4ac)′+(b2−4ac)c′=0.$

#### Lemma 2.8

([23]). Assume that c ∈ ℂ is a nonzero constant, α is a non-constant meromorphic function. Then the differential equation f2 + (cf(n))2 = α has no transcendental meromorphic solutions satisfying T(r, α) = S(r, f).

#### Lemma 2.9

([12]). Assume that f is a meromorphic function. Then for all irreducible rational functions in f,

$R(z,f)=Σi=0paifiΣj=0qbjfj,$

with meromorphic coefficients ai, bj satisfying

$T(r,ai)=S(r,f),i=0,⋯,p,T(r,bj)=S(r,f),j=0,⋯,q,$

the characteristic function of R(z, f) satisfies

$T(r,R(z,f))=dT(r,f)+S(r,f),$

where d = max(p, q).

## 3 Proof of Theorem 1.6

Suppose that f is an entire solution with ρ2(f) < 1 to (2). Obviously, f is a transcendental function. For the simplicity, we replace f(z), f′(z) and L(z, f) by f, f′ and L, respectively.

By differentiating both sides of (2), we obtain

$3f2f′+L′=α1p1eα1z+α2p2eα2z.$(6)

Combining (2) and (6) yields

$α2f3+α2L−3f2f′−L′=(α2−α1)p1eα1z.$(7)

By differentiating (7) again, we derive that

$3α2f2f′+α2L′−6f(f′)2−3f2f″−L″=α1(α2−α1)p1eα1z.$(8)

It follows from (7) and (8) that

$fφ=T(z,f),$(9)

where

$φ=α1α2f2−3(α1+α2)ff′+6(f′)2+3ff″,$(10)

$T(z,f)=−α1α2L+(α1+α2)L′−L″.$

Two cases will now be considered below, depending on whether or not φ vanishes identically.

If φ ≡ 0, then (9) shows that T(z, f) ≡ 0, namely

$L″−(α1+α2)L′+α1α2L=0.$(11)

Further, the general solution of (11) is given by

$L=c1eα1z+c2eα2z,$(12)

where c1 and c2 are constants. Thus, (2) and (12) would give

$f3=(p1−c1)eα1z+(p2−c2)eα2z.$(13)

We claim that p1 = c1 or p2 = c2. Assume now, contrary on the assertion, that p1c1 and p2c2. We rewrite (13) as

$(e−α2z/3fp2−c23)3−(p1−c1p2−c23e(α1−α2)z/3)3=1,$

which contradicts Lemma 2.1. Hence, p1 = c1 or p2 = c2. In this case, we can derive the conclusions (2) and (3).

In the following, we will consider the case φ ≢ 0. In this case, (9) gives

$φ(z)=T(z,f)f(z).$(14)

Since T(z, f) is a difference-differential polynomial in f of degree 1, and L(z, 0) ≡ 0, it follows from (14), Lemma 2.4 and the lemma on the logarithmic derivatives that m(r, φ) = S(r, f). Note that φ is an entire function, so T(r,φ) = m(r, φ) = S(r, f), which means that φ is a small function of f.

Now, we rewrite (10) as

$φf2=α1α2−3(α1+α2)f′f+6(f′f)2+3f″f.$(15)

Applying the lemma on the logarithmic derivatives to (15), we find m(r, $\begin{array}{}\frac{\phi }{{f}^{2}}\end{array}$) = S(r, f). Since φ is a small function of f, one can get m(r, $\begin{array}{}\frac{1}{f}\end{array}$) = S(r, f). Thus, the first fundamental theorem implies T(r, f) = N(r, $\begin{array}{}\frac{1}{f}\end{array}$) + S(r, f).

On the other hand, by (10) again, we have N(2(r, 1/f) = S(r, f), and

$T(r,f)=N1)(r,1f)+S(r,f),$(16)

where N1)(r, $\begin{array}{}\frac{1}{f}\end{array}$) denotes the counting function corresponding to simple zeros of f. Differentiating (10) yields

$φ′=2α1α2ff′−3(α1+α2)[ff″+(f′)2]+15f′f″+3ff‴.$(17)

For brevity, in the following, we assume that z0 is a simple zero of f, and we can assume, without loss of generality, by (16) that ψ(z0) ≠ 0, ∞, where ψ is any non-vanishing small function of f. Thus, (10) enables us to deduce the following fact

$φ(z0)=6(f′(z0))2.$(18)

Now, we are ready to present φ′ ≢ 0. Suppose, contrary to our assertion, that φ′ ≡ 0, namely, φ is a constant, say A.

If z0 is a zero of f′(z) – $\begin{array}{}\sqrt{A/6},\end{array}$ then we set

$h(z)=f′(z)−A/6f(z).$(19)

Trivially, h ≢ 0. Then by the lemma on the logarithmic derivatives, the facts m(r,1/f) = S(r, f), N(2(r,1/f) = S(r, f) and (18), we have T(r, h) = S(r, f).

By (19), we therefore have

$f′=hf+A/6,f″=(h′+h2)f+hA6.$(20)

Substituting (20) into (10) yields

$[α1α2−3(α1+α2)h+3h′+9h2]f=3[(α1+α2)−5h]A6,$

which implies

$α1+α2≡5h,α1α2−3(α1+α2)h+9h2≡0.$

Thereby we have

$h=α1+α25=α13,orh=α1+α25=α23.$(21)

Thus, (19) and (21) would give

$f(z)=Bhehz−1hA6,$(22)

where B is a nonzero constant.

On the other hand, substituting (22) into (2), it follows by Lemma 2.5 that $\begin{array}{}\frac{1}{h}\sqrt{\frac{A}{6}}=0.\end{array}$ This, however, contradicts (16), and thus φ′ ≢ 0.

Using the same way as above, φ′ ≢ 0 is also obtained by setting

$ℏ(z)=f′(z)+A/6f(z)$

assuming that f′(z0) + $\begin{array}{}\sqrt{A/6}=0.\end{array}$

Moreover, (17) gives

$φ′(z0)=[−3(α1+α2)(f′)2+15f′f″](z0)=0.$(23)

In order to prove Theorem 1.6, we discuss two cases below:

• Case 1

[2φ′ + (α1 + α2)φ]f′ – 5φf″ ≡ 0.

In this case, let us write it in the following form

$f″=[25φ′φ+15(α1+α2)]f′:=sf′,$(24)

and consequently

$f‴=(s′+s2)f′.$(25)

Substituting (24) and (25) into (17), we then immediately derive

$α1α2φ′f=[2α1α2φ−3(α1+α2)sφ+3(s′+s2)φ+3(α1+α2)φ′−3sφ′]f′,$

which is impossible by (16) and the facts that the coefficients φ′(≢ 0), 2α1α2φ – 3(α1 + α2) + 3(s′ + s2)φ + 3(α1 + α2)φ′ – 3′ are small functions of f. So, this case can not occur.

• Case 2

[2φ′ + (α1 + α2)φ]f′ – 5φf″ ≢ 0.

Obviously, in this case, by (18), (23) and f′(z0) ≠ 0, we then see that z0 is a zero of the function [2φ′ + (α1 + α2)φ]f′ – 5φf″.

Accordingly, we set

$ϕ(z)=[2φ′(z)+(α1+α2)φ(z)]f′(z)−5φf″(z)f(z).$(26)

Then by the lemma on the logarithmic derivatives, the facts N(2(r, 1/f) = S(r, f), m(r, 1/f) = S(r, f), and (17), we have T(r, ϕ) = S(r, f). Thereby, from (26), we obtain

$f″=[25φ′φ+15(α1+α2)]f′−ϕ5φf:=sf′+tf.$(27)

Trivially, s, t are small functions of f.

By (10) and (27), we have

$af2+bff′+6(f′)2=φ,$(28)

where a = α1α2 + 3t, b = 3[s – (α1 + α2)].

In the following, we consider two subcases.

• Subcase 2.1

Suppose that a ≡ 0. In this case, (28) becomes

$(bf+6f′)f′=φ,$

which gives

$f′=φ1eβ,bf+6f′=φ2e−β,$(29)

where β, φ1 and φ2 are entire functions such that φ1φ2 = φ.

Trivially, in this case, b ≢ 0, and it follows by (29) that

$f=φ2e−β−6φ1eβb.$(30)

Thus, by (29) and (30), we have

$[φ1+6(φ1b)′+6φ1bβ′]e2β=(φ2b)′−φ2bβ′,$

which shows that

$(φ2b)′−φ2bβ′≡0.$(31)

Obviously, (31) gives log $\begin{array}{}\frac{{\phi }_{2}}{b}=\beta +C,\end{array}$ where C is a constant. Therefore, eβ is a small function of f, this shows that f′ is also a small function of f. The contradiction T(r, f) = S(r, f) now follows by (30).

• Subcase 2.2

a ≢ 0. In this case, applying Lemma 2.7 to (28), we immediately get the following equation

$6(b2−24a)φ′φ+b(b2−24a)−6(b2−24a)′≡0.$(32)

Now, we consider two cases.

Firstly, assume that b2 – 24a ≢ 0.

Note that $\begin{array}{}b=\frac{6}{5}\frac{{\phi }^{\prime }}{\phi }-\frac{12}{5}\left({\alpha }_{1}+{\alpha }_{2}\right),\end{array}$ and we then rewrite (32) as

$115φ′φ−125(α1+α2)=(b2−24a)′b2−24a.$

By integrating the above equation, we have

$11log⁡φ−12(α1+α2)z=5log⁡(b2−24a)+log⁡D,$(33)

where D is a constant. Obviously, (33) gives φ11 = D(b2 – 24a)5e12(α1+α2)z. If α1 + α2 ≠ 0, then eα1z, eα2z are also two small functions of f because φ and b2 – 24a are small functions of f. Rewrite (2) as

$f2=−Lf+p1eα1z+p2eα2zf.$

Then

$2m(r,f)=m(r,f2)=m(r,−Lf+p1eα1z+p2eα2zf)≤m(r,1f)+S(r,f)=S(r,f),$

which is a contradiction, since f is an entire function. Therefore, α1 + α2 = 0.

Combining (2) and (6) yields

$α1f3+α1L−3f2f′−L′=(α1−α2)p2eα2z.$(34)

It follows by (7), (34) and α1 + α2 = 0 that

$f4[−α12f2+9(f′)2]=2α12Lf3−6L′f2f′+(α1L)2−(L′)2−4α12p1p2.$(35)

Obviously, P4(f) := 2 $\begin{array}{}{\alpha }_{1}^{2}\end{array}$Lf3 – 6Lf2f′ + (α1L)2 – (L′)2 – 4 $\begin{array}{}{\alpha }_{1}^{2}\end{array}$p1p2 is a difference-differential polynomial of f, and its total degree at most 4.

If P4(f) ≡ 0, it follows from (35) that 9(f′)2$\begin{array}{}{\alpha }_{1}^{2}\end{array}$f2 ≡ 0, and then $\begin{array}{}{f}^{\prime }=±\frac{{\alpha }_{1}}{3}f.\end{array}$ Substituting the above expression into (10), we arrive at φ ≡ 0, a contradiction. Therefore, P4(f) ≢ 0. Set β = 9(f′)2$\begin{array}{}{\alpha }_{1}^{2}\end{array}$f2. In this case, we rewrite (35) as

$β=P4(f)f4,$

which, Lemma 2.4 and the fact that we have proved m(r, $\begin{array}{}\frac{1}{f}\end{array}$) = S(r, f) must show that m(r, β) = S(r, f), i.e. β is a small function of f. Moreover, from Lemma 2.8, it is easy to see that β is a constant. By differentiating both sides of β = 9(f′)2$\begin{array}{}{\alpha }_{1}^{2}\end{array}$f2, we get

$f″−(α13)2f=0.$(36)

It follows from (36) that

$f(z)=c1eα13z+c2e−α13z,$(37)

where c1, c2 are constants. By (37) and (2), we have $\begin{array}{}{c}_{1}^{3}={p}_{1},\phantom{\rule{thinmathspace}{0ex}}{c}_{2}^{3}={p}_{2}.\end{array}$ Conclusion (1) has consequently been proved.

Now, we assume that b2 – 24a ≡ 0.

By making use of (28), we have $\begin{array}{}6\left({f}^{\prime }+\frac{b}{12}f{\right)}^{2}=\phi ,\end{array}$ which shows that $\begin{array}{}\gamma :={f}^{\prime }+\frac{b}{12}f\end{array}$ is a small function of f. Thus, $\begin{array}{}{f}^{″}=\left(\frac{{b}^{2}}{144}-\frac{{b}^{\prime }}{12}\right)f+{\gamma }^{\prime }-\frac{b}{12}\gamma .\end{array}$

Substituting two above expressions into (10), we obtain

$[α1α2+b4(α1+α2)+b2144−b′4]f2+[−3(α1+α2)γ−bγ+3(γ′−b′12γ)]f=φ.$(38)

Using (38) and Lemma 2.9, we therefore have

$α1α2+b4(α1+α2)+b2144−b′4≡0,−3(α1+α2)γ−bγ+3(γ′−b′12γ)≡0andφ≡0.$

Thus, we finish the proof of Theorem 1.6.

## Acknowledgement

The authors would like to thank the referees for their several important suggestions and for pointing out some errors in our original manuscript. These comments greatly improved the readability of the paper.

The research was supported by NNSF of China Project No. 11601521, and the Fundamental Research Fund for Central Universities in China Project Nos. 15CX05061A, 18CX02048A and 18CX02045A.

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Accepted: 2018-03-22

Published Online: 2018-07-17

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 806–815, ISSN (Online) 2391-5455,

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