Abstract
Using value distribution theory and maximum modulus principle, the problem of the algebroid solutions of second order algebraic differential equation is investigated. Examples show that our results are sharp.
1. Introduction and Main Results
We use the standard notations and results of the Nevanlinna theory of meromorphic or algebroid functions; see, for example, [1, 2].
In this paper we suppose that second order algebraic differential equation (3) admit at least one nonconstant -valued algebroid solution in the complex plane. We denote by a subset of for which and by a positive constant, where denotes the linear measure of . or does not always mean the same one when they appear in the following.
Let be entire functions without common zeroes such that . We put
Some authors had investigated the problem of the existence of algebroid solutions of complex differential equations, and they obtained many results ([2–10], etc.).
In 1989, Toda [4] considered the existence of algebroid solutions of algebraic differential equation of the form
He obtained the following.
Theorem A (see [4]). Let be a nonconstant -valued algebroid solution of the above differential equation and all are polynomials. If , then is algebraic.
The purpose of this paper is to investigate algebroid solutions of the following second order differential equation in the complex plane with the aid of the Nevanlinna theory and maximum modulus principle of meromorphic or algebroid functions: where , .
We will prove the following two results.
Theorem 1. Let be a nonconstant -valued algebroid solution of differential equation (3) and all are polynomials. If , then is algebraic, .
Theorem 2. Let be a nonconstant -valued algebroid solution of differential equation (3) and the orders of all are finite. If , then the following statements are equivalent:(a);(b);(c) is a Picard exceptional value of .
2. Some Lemmas
Lemma 3 (see [2]). Suppose that , , are meromorphic functions, and . Then one has
Examining proof of Lemma 4.5 presented in [2, pp. 192-193], we can verify Lemma 4.
Lemma 4. Let be a transcendental algebroid function such that has only finite number of poles, and let , , and have no poles in . Then, for some constants , , and it holds: where .
Lemma 5 (see [11]). The absolute values of roots of equation are bounded by
Lemma 6. Let be a nonconstant -valued algebroid solution of the differential equation (3) and let be a polynomial. If , then where , is a positive constant.
Proof. We first prove that the poles of are contained in the zeroes of .
Suppose that is a pole of of order and is not the zeroes of . Then
We rewrite differential equation (3) as follows:
It follows from (10) that
Noting that , we have
This is a contradiction.
This shows that the poles of are contained in the zeroes of .
We rewrite differential equation (3) as follows:
For , we have
Applying Lemma 5 to (13) at ,
where .
From (14) and (15), we have
Note that
Dividing the inequality (17) by , we obtain, for ,
which reduces to our inequality by calculating of the both sides:
Lemma 6 is complete.
3. Proof of Theorem 1
First, we consider .
Let be a pole of of . Let be the order of zero of at .
(i) When the order of the pole of is not equal to that of other terms of the left-hand side of (10) at , we get that is,
(ii) When the order of pole of is equal to that of some term of the left-hand side of (10) at , we get that is,
Combining cases (i) and (ii), we obtain where is a positive constant.
Secondly, by Lemma 6, we obtain
Combining the inequalities (25) and (26), we have which shows that is an algebraic solution of (3).
This completes the proof of Theorem 1.
4. Proof of Theorem 2
(i) (a) ⇒ (b). Suppose that . If , then we have by (3) Applying Lemma 3 to (28), Since is admissible solution, we have so that This is a contradiction. Thus, .
If , by Theorem 1, is nonadmissible. Thus,
(ii) (b) ⇒ (c). Let . Then, similar to the proof of Lemma 6, we obtain that the poles of are contained in the set of and is a Picard exceptional value of .
(iii) (c) ⇒ (a). Let be a Picard exceptional value of . Then .
5. Some Examples
Example 1. The differential equation has a transcendental algebroid solution . In this case
Remark 7. Example 1 shows that the condition in Theorem 1 is sharp.
Example 2. Transcendental algebroid function is a 2-valued solution of the following differential equation: In this case By Theorem 2, for transcendental algebroid function , is a Picard exceptional value.
Remark 8. Example 2 shows that the result in Theorem 2 holds.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This project is project Supported by National Natural Science Foundation (10471065) of China and NSF of Guangdong Province (04010474).