Unicity of Meromorphic Solutions of Some Nonlinear Difference Equations

ISSN Online: 2160-0384 ISSN Print: 2160-0368

DOI: 10.4236/apm.2019.97030 Jul. 25, 2019 611 Advances in Pure Mathematics

Unicity of Meromorphic Solutions of Some

Nonlinear Difference Equations

Baoqin Chen

Faculty of Mathematics and Computer Science, Guangdong Ocean University, Zhangjiang, China

Abstract

This paper is to study the unicity of transcendental meromorphic solutions to some nonlinear difference equations. Let m∈ ± ±

{

2, 1,0

}

be a nonzero ra-tional function. Consider the uniqueness of transcendental meromorphic so-lutions to some nonlinear difference equations of the form

(

1

) (

1

)

( ) ( )

m

w z+ w z− =R z w z . For two finite order transcendental mero-morphic solutions of the equation above, it shows that they are almost equal to each other except for a nonconstant factor, if they have the same zeros and poles counting multiplicities, when m∈

{

2, 1,0±

}

. Two relative results are proved, and examples to show sharpness of our results are provided.

Keywords

Unicity, Meromorphic Solution, Difference Equation

1. Introduction

It is well known that a given nonconstant monic polynomial is determined by its zeros. But it is not true for transcendental entire or meromorphic functions. Take ez _{and e}−z _{for example, they are essentially different even have the} same zeros, 1-value points and poles. This indicates that it is complex and inter-esting to determine a transcendental meromorphic function uniquely. Nevan-linna then proves his famous NevanNevan-linna’s 5 CM (4 IM) Theorem (see e.g. [1] [2]):

Theorem A:Let w(z) and u(z) be two nonconstant meromorphic functions. If

w(z) and u(z) share 5 values IM (4 values CM, respectively) in the extended complex plane, then w z

( )

≡u z w z

( ) ( )

(

=T u z

(

( )

)

)

_{, where T is a Möbius}

transformation, respectively).

Here and in the following, for two nonconstant meromorphic functions w(z)

How to cite this paper: Chen, B.Q. (2019) Unicity of Meromorphic Solutions of Some Nonlinear Difference Equations. Advances in Pure Mathematics, 9, 611-618.

https://doi.org/10.4236/apm.2019.97030

Received: June 18, 2019 Accepted: July 22, 2019 Published: July 25, 2019

Copyright © 2019 by author(s) and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

DOI: 10.4236/apm.2019.97030 612 Advances in Pure Mathematics

and u(z), and a complex constant a, we say w(z) and u(z) share a IM (CM), if

w(z)-a and u(z)-a have the same zeros ignoring multiplicities (counting multip-licities); and we say w(z) and u(z) share ∞ IM(CM), if they have the same poles ignoring multiplicities (counting multiplicities).

Our aim is to study the unicity of meromorphic solutions to the nonlinear difference equation of the form

(

1

) (

1

)

( ) ( )

m

w z+ w z− =R z w z _{, (1.1)}

where R(z) is a nonzero rational function and m∈ ± ±

{

2, 1,0

}

The Equation (1.1) comes from the family of Painlevé III equations which are given by Ron-kainen in [3] when he classifies the difference equation

(

1

) (

1

)

(

, ,

)

w z+ w z− =R z w

where R(z, w) is irreducible and rational in w and meromorphic in z. This is a natural idea which comes from the topic on the growth, value distribution and unicity on the meromorphic solutions to difference equations (see e.g. [4] [5] [6] [7] [8]). The first result is as follows.

Theorem 1.1. Let w(z) and u(z) be two finite order transcendental mero-morphic solutions to the Equation (1.1), where m∈

{

2, 1,0±

}

. If w(z) and u(z) share 0, ∞ CM, then w z

( )

≡λu z

( )

, where

λ

_{is a constant such that}

2 m ₁

λ − _{= .}

The following examples show that all cases in Theorem 1.1. can happen, and the “CM” cannot be relaxed to “IM”.

Example 1. In the following examples, w zj

( )

and u zj

( )

share 0, ∞ CM,

while w zj

( )

and v zj

( )

share 0, ∞ IM

(

j=1,2,3,4

)

:

1) u z1

( )

tan ₂z ,w z1

( )

iu z1

( )

π  

= _{ } =

  and

( )

( )

2

1 1

v z =u z satisfy the difference equation

(

₁

) (

₁

)

2

( )

_. w z₊ w z_{− =}w z− here m= −2,λ=i _{such that λ}2− −( )2 _=1.

2)

( )

2 2

(

)

( )

2₅

( )

2 2 2

2 1

tan tan , e

3 6

i

z z

u z = _π _ _ − π_ w z = πu z

  _{ } and

( )

( )

2

2 2

v z =u z

satisfy the difference equation

(

₁

) (

₁

)

1

( )

_. w z₊ w z_{− =}w z−

here _{1, e}23

i

m= − λ= π _{such that λ}2− −( )1 ₌₁

. 3) u z3

( )

tan ₄z ,w z3

( )

u z3

( )

π  

= _{ } = −

  and

( )

( )

2

3 3

v z =iu z _{satisfy the}

differ-ence equation

(

1

) (

1

)

1.

w z+ w z− = −

here m=0,λ= −1 _{such that λ}2 0− _=1.

4) 4

( )

(

)

4

( )

4

( )

1

tan tan ,

6 6

z z

u z = _π _ _π − _ w z =u z

  _{ } and

( )

( )

3

4 4

DOI: 10.4236/apm.2019.97030 613 Advances in Pure Mathematics

the difference equation

(

1

) (

1

)

( )

.

w z+ w z− = −w z

here m=1,λ=1 such that _λ2 1− _=1.

Theorem 1.2. Let w(z) and u(z) be two finite order transcendental mero-morphic solutions to the Equation (1.1), where m∈

{

2, 1,0±

}

. If w(z) and u(z) share 0, ∞ CM, then

( )

2

( )

2 1 0

ea z a z a ,

w z _≡ + + u z _(1.2)

where a a a0, ,1 2 are constants such that e2a2 =1. What is more, w z

( )

≡u z

( )

ifw z u z

( ) ( )

− has a zero z1 of multiplicity ≥3 such that

( )

1

( )

1 0

w z =u z = ≠c .

The following example shows that all conclusions in Theorem 1.2 can happen, and the “CM” cannot be relaxed to “IM”.

Example 2. Let _{u z}

( )

_=tan

( ) ( )

_{πz v z, =u z}2

( )

_and

_{( )}

2

_{( )}

1 eiz

w z ₌ π u z _,

( )

( )

2 ez

w z = u z , w z3

( )

=u

( )

z . Then w zj

( )

and u z

( )

share 0, ∞ CM, while

( )

j

w z and v z

( )

share 0, ∞ IM (j = 1, 2, 3), and they solve the equation

(

₁

) (

₁

)

2

( )

_. w z+ w z− =w z

Theorem 1.3. Let w(z) and u(z) be two finite order transcendental mero-morphic solutions to the Equation (1.1), where m∈ ±

{

1,0

}

. If w(z) and u(z) share 1, ∞ CM, then

( )

1 ea z a1 0

(

( )

1 ,

)

w z _{− ≡} + u z _{− (1.3)}

where a a0, 1 are constants such that:

1) 1

1 k₂

a = πi_{, when} m=0_{; 2)} 2

1 2₃k

a = πi_{, when} m= −1_{; (3)} 3 1 k₃

a = πi_,

when

1

m= , where k k k1, ,2 3 are some integers. What is more, w z

( )

≡u z

( )

if

one of the following additional condition holds:

a) w z u z

( ) ( )

− has a zero z1 of multiplicity ≥2 such that

( )

1

( )

1 0

w z =u z = ;

b) there exist two constants z z2, 3 such that w z

( ) ( )

j =u zj ≠1

(

j=2,3

)

and z2− ∈/z3 .

Remark 1. We have tried hard but failed to provide some similar results as Theorem 1.3 for the cases m= ±2 _{so far.}

2. Proof of Theorem 1.1

Since w(z) and u(z) are finite order transcendental meromorphic functions and share 0, ∞ CM, we see that

( )

( )

ep z( ),

w z u z =

where p z

( )

is a polynomial such that it is of degree

DOI: 10.4236/apm.2019.97030 614 Advances in Pure Mathematics

Next, we discuss case by case.

Case 1: m = −2. From (1.1) and (2.1) we get

(

) (

) ( )

( ) ( ) ( )

(

) (

) ( )

( )

(

) (

) ( )

1 1 2 2

2 2

1 1 e

1 1 1 1 ,

p z p z p z

u z u z u z

w z w z w z R z u z u z u z

+ + − +

+ −

= + − = = + −

which gives

( ) ( ) ( )

(

_ep z+ +1 p z− +1 2p z ₋₁

)

_{u z}

(

₊₁

) (

_{u z−1}

) ( )

_{u z}2 _≡0.

Thus, we have

( 1) ( 1 2) ( ) ep z+ +p z− + p z _≡1.

(2.2) Since

(

)

(

)

( )

(

)

( )

deg p z+ +1 p z− +1 2p z =degp z =p,

from (2.2), it is easy to find that p = 0. Therefore, there exists some constant p0,

such that p z

( )

≡ p0 and

( ) ( ) ( )

0 1 1 2

4

e p ₌ep z+ +p z− + p z _≡1.

That is, for λ =ep0, we have w z

( )

≡λu z

( )

and λ =4 1.

Case 2: m = −1. Now, we obtain from (1.1) and (2.1) that

(

) (

) ( )

( ) ( ) ( )

(

) (

) ( )

( )

(

) (

) ( )

1 1 2

1 1 e

1 1 1 1 .

p z p z p z

u z u z u z

w z w z w z R z u z u z u z

+ + − +

+ −

= + − = = + −

With this equation and similar reasoning as in Case 1, we can deduce that

( )

( )

w z ≡λu z holds for some

λ

_{such that λ =}3 _1.

Case 3: m = 0. From (1.1) and (2.1), we have

(

₁

) (

_{1 e}

)

p z( 1) ( )p z1

₍

₁

_{) (}

₁

₎

_{( )}

₍

₁

_{) (}

_{1 .}

₎

u z₊ u z₋ + + − ₌w z₊ w z_{− =}R z ₌u z₊ u z₋

Similarly, we can prove that w z

( )

≡λu z

( )

holds for some

λ

_{such that}

2 ₁

λ = _.

Case 4: m = 1. Now (1.1) is of the form

(

1

) (

1

)

( ) ( )

.

w z+ w z− =R z w z (2.3)

Thus,

(

2

) ( )

(

1

) (

1 .

)

w z+ w z =R z+ w z+

It follows from these two equations above and (2.1) that

(

) (

)

( ) ( )

(

) (

)

(

) ( )

(

) (

)

2 1

2 1 e

2 1 1 2 1 ,

p z p z

u z u z

w z w z R z R z u z u z

+ + −

+ −

= + − = + = + −

with which we can show that w z

( )

≡λu z

( )

holds for some

λ

such that

2 ₁

λ = _{. However, if} w z

( )

≡ −u z

( )

, we find that

(

)

(

−w z+1

)

(

−w z

(

−1

)

)

=u z

(

+1

) (

u z− =1

)

R z u z

( ) ( )

= −R z w z

( ) ( )

. _(2.4)

Combining (2.3) and (2.4), we get R z w z

( ) ( )

≡0, which is impossible. Thus,

1

DOI: 10.4236/apm.2019.97030 615 Advances in Pure Mathematics

3. Proof of Theorem 1.2

Notice that (2.1) still holds for this case. We can get from (1.1) and (2.1) that

(

) (

)

( ) ( )

( )

( )

(

) (

)

( )

( )

(

) (

( )

)

1 1 2 2 2 2

1 1 e 1 1 1 1

. e

p z p z

p z

u z u z w z w z u z u z

R z

w z u z

u z

+ + −

+ − + − + −

= = =

Thus, we have

( 1) ( 1 2) ( )

ep z+ +p z− + p z _≡1. _(3.1)

If p≤1, then our conclusion holds for a2=0. If p≥2, set

( )

1

1 1 0,

p p

p p

p z a z a z − a z a

−

= + + +_ + _(3.2)

where ap≠0,ap−1, , , a a1 0 are constants.

From (3.2), we see that

(

₁

)

(

_{1 2}

)

( )

(

₁

)

p 2

( )

_,

p

p z_{+ +}p z_{− −} p z ₌ p p₋ a z − ₊q z

(3.3) where q(z) is a polynomial such that q z

( )

≡0 when p=2_{, or}

( )

degq z < −p 2 when p≥3.

Suppose that p≥3_{, we obtain from (3.1) and (3.3) that}

( ) ( )_{1 1 2} ( ) ( ₁) 2 ( )

1 e_≡ p z+ +p z− + p z ₌ep p− a zp p− +q z,

which is impossible. Thus, p=2_{, then from (3.1) and (3.3), we get} e2a2 =1

immediately. To sum up, we prove that (1.2) holds.

Next, we use

( )

2

2 1 0

p z =a z +a z a+ and prove our additional conclusion. From (1.2), we see that _ep z( )1 _=1.

Differentiating both sides of (1.2), we can deduce that

( )

_ep z( )

_{( )}

_{( )}

_ep z( )

_{( )}

p z′ u z =w z′ − u z′

and

( )

( )

_{( )}

_{( )}

( )

_{( )}

₍

_{( )}

₎

2 ( )

_{( )}

_{( )}

( )

_{( )}

ep z ep z ep z 2 ep z .

p z′′ u z =w z′′ u z = p z′ u z − p z′ u z′

By our assumption, (1.2}), (3.4) and the fact that ep z( )1 =1, we have

( )

( ) ( )

( )

( )

_{( )}

( )

( )

_{( )}

( )

( )

1

1

1 1 1 1 1

1 1 1 1 e e 0. p z p z

p z p z u z p z u z

w z u z

w z u z

′ = ′ = ′

′ ′

= −

′ ′

= − =

Therefore, similarly, it follows from (3.5) that

( )

( )

( )

_{( )}

( )

( )

_{( )}

₍

_{( )}

₎

( )

_{( )}

_{( )}

( )

_{( )}

( )

( )

1

1 1 1

1 1 1

2

1 1 1 1 1 1

1 1

e

e e 2 e

0.

p z

p z p z p z

p z p z u z

w z u z p z u z p z u z

w z u z

′′ = ′′

′′ ′′ ′ ′ ′

= − − −

′′ ′′ = − =

As a result, we obtain

( )

( )

2 ( )₁

2 1 0 1 0

2

2 1 2 1 1 1

2_{a = p z′′ =}0,2_{a z a+ = p z′ =}0,e a z a z a+ + ₌ep z ₌1,

that is, 0

2 1 0,ea 1

DOI: 10.4236/apm.2019.97030 616 Advances in Pure Mathematics

4. Proof of Theorem 1.3

Here, we need the lemma below, where the case that R(z) is a nonzero constant has been proved by Zhang and Yang [7] and the case that R(z) is a nonconstant rational function by Lan and Chen [8].

Lemma 4.1. [7] [8] Let w(z) be a finite order transcendental meromorphic solution to

the Equation (1.1), where m∈ − ±

{

2, 1,0

}

and a be a constant. Then

(

w a

)

( )

1w

( )

w 1.

λ − =λ =ρ ≥

Proof of Theorem 1.3. Since w z

( )

and u z

( )

are finite order transcen-dental meromorphic functions and share 1, ∞ CM, we see that

( )

( )

11 e ( ),

p z w z

u z −

=

− (4.1)

where p z

( )

is a polynomial such that

( )

1

1 0,

p p

p p

p z a z a z − a

−

= + + +_{ (4.2)}

where ap ≠0, , a0 are constants and p=degp z

( )

≤max

{

ρ

( ) ( )

w ,

ρ

u

}

.

Case 1: m = 0. From (1.1) and (4.1), we obtain

(

)

( )

(

(

)

)

1

( )

4 3

: 1

u z R z

R z

u z R z

+ +

= =

+ (4.3)

and

( )

₍

₍

₎

₎

( )

₍

_{( )}

₎

(

)

( )

(

(

)

)

( )

4

1

e 4 1 1 4 3

, 1

e 1 1

p z

p z

u z w z R z

R z

w z R z

u z

+ _{+ − +}

+ +

= = =

+

− + (4.4)

where R z1

( )

is a rational function. Combining (4.1}), (4.3) and (4.4), we have

( ) ( )

(

4

)

( ) ( )

(

)

(

( )

)

(

( 4)

)

1 1

ep z+ ₋ep z _{R z u z − = −}1 1 _{R z} ep z+ ₋1 . _(4.5)

Now, if _ep z(+4) _ep z( )

≡/ , then p≥1 _{and it follows from (4.5) that}

( )

_{( )}

1

( )

_{( ) (}( 4) ₎

4 1

1 1 e _1.

1 e

p z p z p z

R z u z

R z

− +

− +

− −

= +

− (4.6)

Notice that deg

(

p z

( )

−p z

(

+4

)

)

≤ −p 1. From (4.6), we can find that

(

_{u 1}

)

_{p p 1}

(

_{1 e}p z p z( ) ( 4)

)

1 _.

u

λ _{− = > − ≥}ρ ₋ − + λ  ≥  _{ }

This is a contradiction to the conclusion of Lemma 4.1. Thus, _ep z( +4) _≡ep z( )_.

From (4.2) there exists some integer k1 such that

(

) ( )

1

1

2 4 4 p ,

p

k i p z_{π = + −}p z ₌ pa z − _+ which yields obviously that p=1_{. Therefore, we see that}

1 1 ₂

p k

a =a = πi _{and hence}

( )

1

0

2 k

p z = π +iz a _{for some constant} a₀. Case 2: m = −1. Now (1.1) is of the form

(

1

) (

1

) ( )

( )

,

DOI: 10.4236/apm.2019.97030 617 Advances in Pure Mathematics

which gives

(

)

( )

(

(

)

)

2

( )

3 2

: .

1

u z R z

R z

u z R z

+ +

= =

+

With this equation and a similar arguing as in Case 1, we can prove that

( )

2

0

2 3

k

p z = π +iz a _{for some integer} k₂ and some constant a0.

Case 3: m = 1. Now (1.1) is of the form

(

1

) (

1

)

( ) ( )

,

u z+ u z− =R z u z

which gives

(

3

) ( )

(

2

) (

1 .

)

u z+ u z =R z+ R z+

And hence we have

(

)

( )

(

(

) (

) (

)

)

3

( )

6 5 4

: .

2 1

u z R z R z

R z

u z R z R z

+ + +

= =

+ +

It follows this equation that

( )

3 0

3

k

p z = π +iz a _{for some integer} k₃ and some constant a0, and (1.3) holds.

Now, if w z u z

( ) ( )

− has a zero z1 of multiplicity ≥2 such that w z

( )

1 =0,

then from (4.1), we see that _ep z( )1 _=1.

Rewrite (4.1) as the form

( )

_{1 e}p z( )

₍

_{( )}

_{1 .}

₎

w z − = u z −

Differentiating both sides of the equation above, we have

( )

_ep z( )

₍

₁

_{( )}

₎

_ep z( )

_{( )}

_{( )}

_.

p z′ −u z = u z w z′ − ′

Since z1 is a zero of w z u z

( ) ( )

− with multiplicity ≥2 such that

( )

1

( )

1 0

w z =u z = _{, from the fact that e}p z( )1 _{=1 and (4.7), we find that}

( )

( )

( )1

₍

_{( )}

₎

( )1

_{( )}

_{( )}

1 1 ep z 1 1 ep z 1 1 0.

p z′ = p z′ −u z = u z′ −w z′ =

Thus, a1= p z′

( )

1 =0, and hence ep z( ) ≡ep z( )1 =1. This implies that

( )

( )

w z ≡u z _.

Finally, we discuss the Case 2). Since w z

( ) ( )

j =u zj ≠1 and z2− ∈/z3 ,

then from (4.1), we can deduce that ep z( )2 = =1 ep z( )3 . Therefore, there exists an

integer k0 such that

(

)

( )

( )

1 2 3 2 3 2 0 .

a z −z = p z −p z = k iπ

If a1≠0, from the equation above, considering each form of a1 for 1,0,1

m= − _{, we can find that} z₂−z_{3 must be a nonzero rational number. This}

contradicts our assumption that z2− ∈/z3 . Thus a1=0, and hence ep z( ) ≡1.

This gives w z

( )

≡u z

( )

_again.

5. Conclusion

DOI: 10.4236/apm.2019.97030 618 Advances in Pure Mathematics

Equation (1.1) is mainly determined by its zeros (or 1-value points) and poles. Examples are provided to show sharpness of our results.

Acknowledgements

The author is very appreciated for the editors and reviewers for their construc-tive suggestions and comments for the readability of this paper.

Funding

This work was supported by the Natural Science Foundation of Guangdong Province (2018A030307062).

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this pa-per.

References

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