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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 619296, 6 pages
On Growth of Meromorphic Solutions for Linear Difference Equations
1School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China
2Department of Mathematics, College of Natural Sciences, Pusan National University, Pusan 609-735, Republic of Korea
Received 9 June 2013; Accepted 19 August 2013
Academic Editor: Norio Yoshida
Copyright © 2013 Zong-Xuan Chen and Kwang Ho Shon. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We mainly study growth of linear difference equations and where are polynomials such that and give the most weak condition to guarantee that orders of all transcendental meromorphic solutions of the above equations are greater than or equal to 1.
1. Introduction and Results
Consider growth of meromorphic solutions of the following linear difference equations:
where are polynomials such that .
Theorem A (see ). Let be a transcendental entire solution of where are polynomials, , , and of order . Then one has where a rational number is a slope of the Newton polygon for (3) and is a constant. In particular, one has .
Remark 1. In , Ishizaki and Yanagihara give an example. The difference equation
admits an entire solution of order .
In , Ishizaki and Yanagihara do not give a concrete solution of order . In fact, we assert that (5) has no entire solution of order . Contrary to the assertion, we assume that is an entire solution of order of . Set . Then is an entire function of order . Substituting in to , we obtain
Using the same method as in the proof of Case 1, in proof of Theorem 4, we obtain the order of which is greater than or equal to 1. It is a contradiction.
Chiang and Feng proved the following theorem.
In this paper, we use the basic notions of Nevanlinna’s theory (see [9, 10]). In addition, we use the notations to denote the order of growth of a meromorphic function and to denote the exponent of convergence of zeros of .
Remark 2. Comparing Theorem A with Theorem B, we see that since (3) can be rewritten as (1), Theorem A shows that, under general case, (1) may have transcendental meromorphic solution with . In Theorem B, the condition (7) guarantees that all meromorphic solutions of (1) satisfy .
The author who weakened the condition (7) of Theorem B proved the following results.
From we see that the sum of coefficients of , which is equal to , does not satisfy the condition (8), but all transcendental entire solutions of have order .
Thus, a natural question to ask is whether the condition (8) can be weakened.
In this note, we consider this question, again weaken the condition (8) and prove the following results.
Theorem 3. Let be polynomials such that and satisfy Then every-finite order transcendental meromorphic solution of (1) satisfies , assumes every nonzero value infinitely often, and .
Theorem 4. Let be polynomials such that . Then every finite-order transcendental meromorphic solution of (2) satisfies .
Remark 5. For the homogeneous equation (1) by Theorems B, C and 3, we see that the condition (9) is weaker than (7) and (8). For the nonhomogeneous equation (2) by Theorem 4, we see that the condition (9) is omitted. But, under the condition (8), (1) has no nonzero rational solution, and under the condition (9), (1) may have nonzero rational solution. For example, has a rational solution . This shows that Theorem C can not be replaced by Theorem 3 completely.
Example 6. The equation has a solution ; here satisfies for any nonzero finite value , and has no zero. This shows that in Theorem 3, the condition cannot be omitted.
By Theorem 3, we can obtain the following corollary.
Corollary 8. Let be polynomials such that . If (1) has a transcendental meromorphic solution with , then
Consider the growth of the second order linear difference equation
with is a meromorphic function. Since (14) is closely related with the difference Riccati equation
Ishizaki  proved the following result.
Theorem F (see ). Suppose that is a rational function in (14) and has no transcendental meromorphic solutions of order less than . Further, one assumes that (14) possesses a rational solution. Then, every transcendental meromorphic solution of (14) has order of at least .
In this note, we improve this result, omit the condition of Theorem F “(14) possesses a rational solution”, and prove the same result.
Theorem 9. Let be a rational function. Then every transcendental meromorphic solution of (14) has order of at least .
Further, If , where and are nonconstant polynomials such that , then (14) has no nonzero rational solution.
For the linear difference equation with transcendental coefficients, one has
Chiang and Feng proved the following result.
Laine and Yang  prove that if are entire functions of finite order so that among those having the maximal order , exactly one has its type strictly greater than the others, then every meromorphic solution of (16) satisfies .
Remark 10. If are meromorphic functions satisfying (17), then Theorem G does not hold. For example,
has a solution , in which .
Thus, a natural question to ask is what conditions will guarantee that every transcendental meromorphic solution of (16) satisfies .
We answer this question and prove the following result.
Theorem 11. Let be meromorphic functions such that there exists an integer , such that If is a meromorphic solution of (16), then one has .
2. Proofs of Theorems
Remark 12. Following Hayman [11, p. 75-76], we define an -set to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If is an -set, then the set of for which the circle meets has finite logarithmic measure, and for almost all real the intersection of with the ray is bounded.
Lemma 13 (see ). Let be a function transcendental and meromorphic in the plane of order less than 1. Let . Then there exists an -set such that uniformly in for . Further, may be chosen so that for large not in the function has no zeros or poles in .
Proof of Theorem 4. Suppose that is a transcendental meromorphic solution of (2) with . We divide this proof into the following two cases.
Case 1. Suppose that has only finitely many poles. Now we suppose that . By Lemma 13, there exists an -set such that where satisfy Set . By Remark 12, is of finite logarithmic measure. Substituting (24) into (2), we obtain, as in , that is, Thus, since has only finitely many poles, we deduce that when ,
This contradicts with the fact that is transcendental. Hence .
Case 2. Suppose that has infinitely many poles. Thus, by Theorem D, we see that .
Finally, we prove that . By (2), and we set Thus, By Lemma 14, we have so that
Thus, Theorem 4 is proved.
Proof of Theorem 3. Suppose that is a transcendental meromorphic solution of (1) with and that is a constant. Set . Then, .
Substituting into (1), we obtain
Since , we see that (33) satisfies the conditions of Theorem 4. Thus, we deduce that .
Finally, we prove that assumes every nonzero value infinitely often and that . Set Thus, since and (9), we have By Lemma 14 and (35), we have so that Hence . Theorem 3 is thus proved.
Proof of Theorem 9. Suppose that is a transcendental meromorphic solution of (14). We rewrite (14) as
If , then by (38), we obtain
We affirm that . In fact, if , then has infinitely many zeros, or infinitely many poles. If has infinitely many zeros, then by (39), we see that if is a zero of , then , are also zeros of . Thus, . If has infinitely many poles, then by using the same method, we can obtain .
Now we suppose that . Set , where and are nonzero polynomials. By (38), we have Since
by Theorem 3, we see that .
Further, if and are nonconstant polynomials such that , then (41) satisfies the condition of Theorem C. Thus, we see that (14) has no rational solution. Thus, Theorem 9 is proved.
Proof of Theorem 11. Clearly, (16) has no nonzero rational solution.
Now suppose that is a transcendental meromorphic solution of (16) with . By (16), we obtain Set Thus, we have By Lemma 15, we see that for given , Thus, by (42), (45), and (46), we have By (43), we see that for given above, Since , we see that there is a sequence satisfying Thus, by (47)–(49), we obtain If we combine this with , it follows that So that, it follows that . Thus, Theorem 11 is proved.
Zong-Xuan Chen was supported by the National Natural Science Foundation of China (no. 11171119). Kwang Ho Shon was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and Future Planning (2013R1A1A2008978).
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