Meromorphic solutions of certain nonlinear difference equations

Theorem A

[7] Let n ≥ 2 be an integer, c ∈ ℂ \ { 0 } , and let p ( z ) , q ( z ) ( ≢ 0 ) , Q ( z ) be polynomials such that Q ( z ) is not a constant. Then we identify finite-order entire solutions f of (1.2) as follows:

1. every solution f satisfies σ ( f ) = deg Q and is of mean type;

2. every solution f satisfies λ ( f ) = σ ( f ) if and only if p ( z ) ≢ 0 ;

3. a solution f ∈ Γ 0 if and only if p ( z ) ≡ 0 . In particular, this is the case if n ≥ 3 ;

4. if a solution f ∈ Γ 0 , and if g is any other finite-order entire solution of (1.2), then f = η g , where η n − 1 = 1 ;

5. if f is an exponential polynomial solution of the form (1.3), then f ∈ Γ 1 . Moreover, if f ∈ Γ 1 \ Γ 0 , then σ ( f ) = 1 .

Theorem B

[11] Let n ≥ 2 , k ≥ 1 be an integer, c ∈ ℂ \ { 0 } , and let p ( z ) , q ( z ) ( ≢ 0 ) , Q ( z ) be polynomials such that Q ( z ) is not a constant. Then every finite-order transcendental entire solution f of (1.4) should satisfy results (a), (b), and (d) of Theorem A. Moreover,

1. a solution f ∈ { B ( z ) e α ( z ) : B , α polynomials , α ≢ const . } if and only if p ( z ) ≡ 0 . In particular, this is case if n ≥ 3 ;

2. if f is an exponential polynomial solution of the form (1.3), then f ∈ { B ( z ) e α ( z ) + h ( z ) : B , α , h polynomials , α ≢ const . } .

Theorem 1.1

Let n ≥ 2 be an integer, c ∈ ℂ \ { 0 } , and let p ( z ) , q ( z ) ( ≢ 0 ) , Q ( z ) be polynomials such that Q ( z ) is not a constant. Then every finite-order nonconstant entire solution f of (1.5) has the following properties.

1. σ ( f ) = deg ( Q ( z ) ) and f is of mean type;

2. λ ( f ) = σ ( f ) if and only if p ( z ) ≢ 0 ;

3. λ ( f ) = σ ( f ) − 1 if and only if p ( z ) ≡ 0 . In particular, this is the case if n ≥ 3 .

Theorem 1.2

Let n , p ( z ) , q ( z ) , Q ( z ) , c satisfy the hypotheses of Theorem 1.1. If n ≥ 3 or p ( z ) ≡ 0 , then every finite-order nonconstant entire solution f of (1.5) is of the form

where s = deg Q , ω is a nonzero constant, and A ( z ) ( ≢ 0 ) is an entire function satisfying σ ( A ) = λ ( A ) = deg Q − 1 . In particular, if deg Q = 1 , then A ( z ) reduces to a polynomial.

Remark 1.1

Examples 1.1–1.3 show that the difference equation (1.5) does have such solution defined as Theorem 1.2.

Example 1.1

The equation f 2 ( z ) + e z 2 Δ c f ( z ) = 0 has a solution f ( z ) = ( 1 − e 2 c z ) e z 2 , where c is a nonzero constant such that e c 2 = 1 . Here, λ ( f ) = λ ( 1 − e 2 c z ) = 1 = σ ( f ) − 1 .

Example 1.2

The equation f 2 ( z ) + ( z + 1 ) 2 e β z Δ c f ( z ) = 0 has a solution f ( z ) = − c ( z + 1 ) e β z , where c , β are nonzero constants such that e β c = 1 . Here, λ ( f ) = 0 = σ ( f ) − 1 .

Example 1.3

The equation f 2 ( z ) + 2 e z Δ c f ( z ) = 0 has a solution f ( z ) = e z , where c = − log 2 . Here, λ ( f ) = 0 = σ ( f ) − 1 .

Theorem 1.3

Let p ( z ) ( ≢ 0 ) , q ( z ) ( ≢ 0 ) , Q ( z ) be polynomials such that Q ( z ) is not a constant. If f is an exponential polynomial solution of ( 1.5 ) of the form ( 1.3 ), then f is of the form

where H 1 , H 0 are nonconstant polynomials, ω 1 is a nonzero constant satisfying e ω 1 c = 1 .

Remark 1.2

Example 1.4 shows that entire solutions of the form (1.6) do exist.

Example 1.4

The equation f 2 ( z ) + ( z + 1 ) 2 e 2 π i z ( f ( z + 1 ) − f ( z ) ) = ( z 2 + 3 z + 2 ) 2 has a solution f ( z ) = z 2 + 3 z + 2 − ( z + 1 ) e 2 π i z .

Lemma 2.1

[12] Let f ( z ) be a finite-order meromorphic function in the complex plane, and c be a fixed nonzero constant. Then, for each ε > 0 , we have

Lemma 2.2

[13] Let f ( z ) be a transcendental meromorphic function of finite order. Then

where S ( r , f ( z ) ) denotes any quantity satisfying S ( r , f ) = o ( T ( r , f ) ) as r → ∞ , possibly outside of a set of finite logarithmic measure.

Lemma 2.3

[14] Let f j ( z ) , g j ( z )   ( j = 1 , … , n )   ( n ≥ 2 ) be entire functions satisfying the following conditions.

1. ∑ j = 1 n f j ( z ) e g j ( z ) ≡ 0 ;

2. The order of f j is less than that of e g t ( z ) − g k ( z ) for 1 ≤ j ≤ n , 1 ≤ t < k ≤ n .

   Then f j ( z ) ≡ 0 ,   ( j = 1 , … , n ) .

Lemma 2.4

[14] Let f ( z ) be an entire function of finite order. Then f ( z ) = u ( z ) e v ( z ) , where u ( z ) is the canonical product of f ( z ) formed with the zeros of f, and v ( z ) is a polynomial of degree at most σ ( f ) .

Proof of Theorem 1.1

Let f be a finite-order nonconstant entire solution of (1.5). If Δ c f ≡ 0 , then we get f is transcendental, which contradicts that f n = p is a polynomial by (1.5). So we have Δ c f ≢ 0 .

Proof of (i). From Lemma 2.1 and (1.5), it follows that for each ε > 0 ,

Hence, by (2.1) and (2.2), we get σ ( f ) = deg ( Q ( z ) ) , and f is of mean type, that is, its type τ ( f ) = lim ¯ r → ∞ T ( r , f ) r σ ( f ) ∈ ( 0 , + ∞ ) .

Proof of (ii) “ ⇒ .” Suppose that P ( z ) ≡ 0 , then (1.5) can be represented as

By (2.3), we get

Then, combining Lemma 2.1, we get for each ε > 0 ,

Hence, from (2.3) and (2.4), it follows that for each ε > 0 ,

By (2.5), we get λ ( f ) < σ ( f ) , which contradicts the condition λ ( f ) = σ ( f ) . So P ( z ) ≢ 0 .

Proof of (ii) “ ⇐ .” Since P ( z ) ≢ 0 , from Lemma 2.1, (1.5), and the second main theorem related to three small functions [9, Theorem 2.5], we get for each ε > 0 ,

(2.6) implies that σ ( f ) ≤ λ ( f ) . Then from the fact σ ( f ) ≥ λ ( f ) , we get σ ( f ) = λ ( f ) .

Proof of (iii). First, we prove that if n ≥ 3 , then p ( z ) ≡ 0 . In fact, if p ( z ) ≢ 0 , then by (2.6), we get T ( r , f ) = O ( r σ ( f ) − 1 + ε ) + S ( r , f ) , which yields σ ( f ) ≤ σ ( f ) − 1 . This is a contradiction.

Second, from result (ii) of Theorem 1.1, it is easy to obtain that λ ( f ) < σ ( f ) if and only if P ( z ) ≡ 0 . So, we only prove λ ( f ) = σ ( f ) − 1 provided P ( z ) ≡ 0 .

Since P ( z ) ≡ 0 , we get λ ( f ) < σ ( f ) . From this equality and Lemma 2.4, it yields

where α ( z ) is a polynomial, A ( z ) ( ≢ 0 ) is the canonical product of f formed with the zeros of f, such that

Substituting (2.7) into (1.5), we get

where α ( z ) = a s z s + α 0 ( z ) , Q ( z ) = b s z s + Q 0 ( z ) . Here, a s , b s are nonzero constants, α 0 , Q 0 are polynomials of degree ≤ s − 1 . By (2.8), (2.9), and Lemma 2.3, we get n a s = a s + b s , and

Next, we divide three cases to complete the proof.

Case 1. s − 1 > 0 and σ ( A ) < s − 1 . If deg ( Q 0 ( z ) + ( 1 − n ) α 0 ( z ) ) < s − 1 , then by deg ( α ( z + c ) − α ( z ) ) = s − 1 , Lemmas 2.2, 2.3, and (2.10), we get q ( z ) A ( z + c ) ≡ 0 , which contradicts the fact that q ( z ) ≢ 0 and A ( z ) ≢ 0 . Similarly, if deg ( Q 0 ( z ) + ( 1 − n ) α 0 ( z ) ) = s − 1 , then by Lemmas 2.2, 2.3, and (2.10), it yields q ( z ) A ( z ) ≡ 0 , and we again have a contradiction.

Case 2. 0 < s − 1 ≤ σ ( A ) . It follows from (2.10) that

Then by Lemma 2.1, we get for each ε > 0 ,

Combining (2.10) and (2.11), we get for each ε > 0 ,

Hence, from deg ( α ( z + c ) − α ( z ) ) = s − 1 , (2.8), and (2.12), it follows λ ( A ) ≤ s − 1 . Since λ ( A ) = σ ( A ) ≥ s − 1 , we get λ ( f ) = λ ( A ) = s − 1 = σ ( f ) − 1 .

Case 3. s − 1 = 0 . By (2.8), we get σ ( A ) = λ ( A ) < 1 . If A ( z ) is transcendental, then A ( z ) has infinitely many zeros. But (2.12) implies N r , 1 A ( z ) = O ( log r ) . This is a contradiction. So A ( z ) is a polynomial and λ ( f ) = λ ( A ) = 0 = σ ( f ) − 1 .□

Theorem C

[15] Let f be given by (3.1). Then

Theorem D

[15] Let f be given by (3.1). If H 0 ( z ) ≢ 0 , then

while if H 0 ( z ) ≡ 0 , then

Lemma 3.1

Let f be given by (3.1), and be a solution of (1.5), where n = 2 . If the points 0 , ω 1 , … , ω m are collinear, then m = 1 .

Proof

Suppose contrary to the assertion that m ≥ 2 . Without loss of generality, we assume that ω i = k i ω ( i = 1 , … , m ) , where ω is a nonzero complex number, k i are distinct nonzero real numbers such that k i > k j for i > j . Set Q ( z ) = b s z s + Q 0 ( z ) , where Q 0 ( z ) is a polynomial of degree ≤ s − 1 . Substituting the expression (3.1) of f into (1.5), we get

where k 0 = 0 , l 0 ( z ) = H 0 ( z + c ) − H 0 ( z ) , l t ( z ) = H t ( z + c ) e ω t ( z + c ) s − ω t z s − H t ( z ) ( t = 1 , … , m ).

Now we divide two cases k m > 0 and k m < 0 to derive a contradiction. Considering the similarity of proofs in these two cases, we only discuss the case k m > 0 in detail.

Let A = { k i + k j : i , j = 0 , 1 , … , m } , we have

From the proof of Theorem 1.1, we get Δ c f ≢ 0 , which implies that at least one of l m , … , l 0 is not vanishing.

Case 1. l m ( z ) = ⋯ = l 1 ( z ) ≡ 0 . Then we have l 0 ( z ) ≢ 0 , which implies H 0 ( z ) ≢ 0 . If 2 k m ω ≠ b s , then by (3.2), (3.3), and Lemma 2.3, we get H m 2 ( z ) ≡ 0 , which contradicts H m ( z ) ≢ 0 . If 2 k m ω = b s , then by (3.2), (3.3), and Lemma 2.3, we get H 1 ( z ) ≡ 0 , which also contradicts H 1 ( z ) ≢ 0 .

Case 2. At least one of l m , … , l 1 is not vanishing. Set Λ = { t ∈ { 0 , 1 , … , m } : l t ( z ) ≢ 0 } , then Λ is a nonempty set. Denote t 0 = max Λ .

Subcase 2.1. l 0 ( z ) ≢ 0 . If there exists some j 0 ∈ Λ \ { t 0 } such that 2 k m ω = b s + k j 0 ω , then by (3.2), (3.3), Lemma 2.3, and the fact

we get q ( z ) e Q 0 ( z ) l t 0 ( z ) ≡ 0 , which contradicts q ( z ) ≢ 0 . If 2 k m ω = b s + k t 0 ω , then by (3.2), (3.3), Lemma 2.3, and the fact

we get H 1 ( z ) ≡ 0 , which contradicts H 1 ( z ) ≢ 0 .

Subcase 2.2. l 0 ( z ) ≡ 0 and H 0 ( z ) ≢ 0 . Using the similar proof to subcase 2.1, we get a contradiction.

Subcase 2.3. l 0 ( z ) ≡ 0 and H 0 ( z ) ≡ 0 . Then by (3.2), (3.3), and Lemma 2.3, we get there exists some j 0 ∈ Λ \ { 0 } such that 2 k m ω = b s + k j 0 ω . Otherwise, we have H m 2 ( z ) ≡ 0 , which contradicts H m ( z ) ≢ 0 . While if 2 k m ω = b s + k j 0 ω , then by (3.2), (3.3) and the fact

we get H 1 2 ( z ) ≡ 0 , which contradicts H 1 ( z ) ≢ 0 .□

Lemma 3.2

Let f be given by (3.1). If f is a solution of (1.5), where n = 2 , then m = 1 .

Proof

Suppose contrary to the assertion that m ≥ 2 . Substituting the expression (3.1) of f into (1.5), we get

where M ( z ) is either an exponential polynomial of degree ≤ s − 1 or ordinary polynomial in z, l j ( z ) = H j ( z + c ) e ω j ( z + c ) s − ω j z s − H j ( z ) ( j = 0 , 1 , … , m ), ω 0 = 0 . From the proof of Theorem 1.1, we get Δ c f ≢ 0 , which implies that at least one of l m , … , l 0 is not vanishing.

Set

It is obvious by convexity that

If l m ( z ) = ⋯ = l 1 ( z ) ≡ 0 , then we have l 0 ( z ) ≢ 0 , which implies H 0 ( z ) ≢ 0 . From the fact F ( z ) = − q ( z ) l 0 ( z ) e Q ( z ) and Theorem C, we get

On the other hand, by (3.4), Theorems C and D, we get

This yields a contradiction. So, there exists some t 0 ∈ { 1 , … , m } such that l t 0 ( z ) ≢ 0 . Set Λ 1 = { t ∈ { 1 , … , m } : l t ( z ) ≢ 0 } . Denote by T the set consisted of nonzero elements ω ¯ t ( t ∈ Λ 1 ) and T 0 = T ∪ { 0 } . It is obvious that

By (3.4), Theorems C and D, we get

Case 1. Suppose that M ( z ) ≢ 0 , then by (3.4) and Theorem D, we get

So, by Theorem C, (3.4), (3.6), and (3.7), we get

which contradicts C ( c o ( W 0 ) ) ≥ C ( c o ( T 0 ) ) ≥ C ( c o ( T ) ) .

Case 2. Suppose that M ( z ) ≡ 0 , then by (3.4) and Theorem D, we get

Subcase 2.1. If H 0 ( z ) ≡ 0 , then we must have l 0 ( z ) ≡ 0 . So, by (3.6) and (3.8), we get C ( c o ( T ) ) = C ( c o ( X 2 ) ) . From this equality and (3.5), it follows that