# Algebroid Solutions of Second Order Complex Differential Equations

Authors: Lingyun Gao; Yue Wang
Journal: Abstract and Applied Analysis
 (2014)

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## 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.


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## Body

## 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 ν-valued algebroid solution w(z) in the complex plane. We denote by E a subset of [0,∞) for which m(E)<∞ and by K a positive constant, where m(E) denotes the linear measure of E. E or K does not always mean the same one when they appear in the following.Letajk  (j=0,1,…,n;k=0,1,…,qj) be entire functions without common zeroes such that a0q0≠0. We put
(1)Qj(z,w)=∑k=0qj‍ajkwk,qj=degwQj;p=max{qj+j:j=0,1,…,n-1}.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
(2)∑j=0n‍Qj(z,w)(w′)j=0.He obtained the following.Theorem A (see [4]).
Letw(z) be a nonconstant ν-valued algebroid solution of the above differential equation and all ajk are polynomials. If p<n+qn, then w(z) 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:(3)∑j=0n‍Qj(z,w)(w′′)j=0,
where Qj(z,w)=∑k=0qjajkwk, j=0,1,2,…,n.We will prove the following two results.Theorem 1.
Letw(z) be a nonconstant ν-valued algebroid solution of differential equation (3) and all ajk are polynomials. If p≤qn, then w(z) is algebraic, p=max{qj:j=0,1,…,n-1}.Theorem 2.
Letw(z) be a nonconstant ν-valued algebroid solution of differential equation (3) and the orders of all ajk are finite. If q0>max1≤j≤n-1{qj+j}, then the following statements are equivalent:(a)
δ(∞,w)>0;(b)
q0=qn+n;(c)
∞ is a Picard exceptional value of w(z).


## 2. Some Lemmas

Lemma 3 (see [2]).
Suppose thatw(z), ai(z), (i=1,2,…,p) are meromorphic functions, and ap(z)≠0. Then one has
(4)m(r,∑i=1p‍ai(z)wi)≤pm(r,w)+∑i=1p‍m(r,ai(z))+O(1).Examining proof of Lemma  4.5 presented in [2, pp. 192-193], we can verify Lemma 4.Lemma 4.
Letw(z) be a transcendental algebroid function such that w(z) has only finite number of poles, and let w(z), w′(z), and w′′(z) have no poles in |z|>r0. Then, for some constants Ci>0, i=1,2,3, and r≥r1≥r0 it holds:
(5)M(r,w)≤C1+C2r+C3r2M(r,w′′),
where M(r,w)=max|z|=r{|w(z)|}.Lemma 5 (see [11]).
The absolute values of roots of equation(6)zn+a1zn-1+⋯+an=0
are bounded by
(7)max{n|a1|,(n|a2|)1/2,…,(n|an|)1/n}.Lemma 6.
Letw(z) be a nonconstant ν-valued algebroid solution of the differential equation (3) and let ajk be a polynomial. If p<n+qn, then
(8)min{n,qn-p}log+M(r,w)+max{0,qn-p-n}log+M(r,w)≤+O(logr)+O(1),(r∉E),
where M(r,w)=max|z|=r{|w(z)|}, K is a positive constant.Proof.
We first prove that the poles ofw are contained in the zeroes of {anqn(z)}.
Suppose thatz0 is a pole of w of order τ and z0 is not the zeroes of {anqn(z)}. Then
(9)w(z)=(z-z0)-τ/λw1(z),w1(z0)≠0,∞,w′′(z)=(z-z0)-(τ+2λ)/λw2(z),w2(z0)≠0,∞.
We rewrite differential equation (3) as follows:
(10)anqn(w′′)n=∑j=0n-1‍Qj(z,w)(w′′)j.
It follows from (10) that
(11)qnτ+n(τ+2λ)≤pτ+(n-1)(τ+2λ).
Noting that p≤qn, we have
(12)τ+2λ<0.
This is a contradiction.
This shows that the poles ofw are contained in the zeroes of {anqn(z)}.
We rewrite differential equation (3) as follows:
(13)∑j=0n‍Qj(z,w)Qn(z,w)n-j-1(Qnw′′)j=0.(14)M(r,Q1w′′)≥M(r,w)qn+M(r,w′′)-∑k=0qn-1‍M(r,ank)M(r,w)k≥M(r,w)qn+M(r,w)-C1-C2rC3r2-∑k=0qn-1‍M(r,ank)M(r,w)k.
For j=0,1,…,n-1, we have
(15)|Qj(zr,w)Qn(zr,w)n-j-1|≤KM(r,w)qj+qn(n-j-1).
Applying Lemma 5 to (13) at z=zr,
(16)M(r,Q1w′′)≤KM(r,w)max{hj:j=0,1,…,n-1},
where hj=(qj+qn(n-j-1))/(n-j).
From (14) and (15), we have
(17)M(r,w)qn≤K{M(r,w)qn-1+r2M(r,w)max{hj:j=0,1,…,n-1}}≤K{M(r,w)qn-1+r2M(r,w)qn+((p-qn)/n)}.
Note that(18)hj=qj+qn(n-j-1)n-j<p+qn(n-j-1)-jn-j=qn+p-qnn+j(p-qn-n)n(n-j)≤qn+p-qnn,j=0,1,…,n-1.
Dividing the inequality (17) by M(r,w)max{qn-1,qn+((qn-p)/n)}, we obtain, for r∉E,
(19)M(r,w)min{1,(qn-p)/n}≤K4{1+r2M(r,w)max{0,(qn-p-n)/n}},
which reduces to our inequality by calculating log+ of the both sides:
(20)min{n,qn-p}log+M(r,w)≤-max{0,qn-p-n}log+M(r,w)+O(logr)+O(1).
Lemma 6 is complete.


## 3. Proof of Theorem1

First, we considerN(r,w).Letz0 be a pole of w of τ. Let t be the order of zero of anqn(z) at z0.(i) When the order of the pole ofQn(w)(w′′)n is not equal to that of other terms of the left-hand side of (10) at z0, we get
(21)qnτ+n(τ+2λ)-tλ≤pτ;
that is,
(22)τ≤(t-2n)λqn+n-p.(ii) When the order of pole ofQn(w)(w′′)n is equal to that of some term Qk(w)(w′′)n of the left-hand side of (10) at z0, we get
(23)qnτ+n(τ+2λ)-tλ≤pτ+n(τ+2λ);
that is,
(24)τ≤tλqn-p.Combining cases (i) and (ii), we obtain(25)N(r,w)≤K8N(r,1anqn),
where K8 is a positive constant.Secondly, by Lemma6, we obtain
(26)min{n,qn-p}m(r,w)≤K[∑i=0p‍m(r,ai)+∑k=0qn-1‍m(r,bk)]+O(logr),(r∉E).Combining the inequalities (25) and (26), we have
(27)T(r,w)=O(logr),(r∉E),
which shows that w is an algebraic solution of (3).This completes the proof of Theorem1.


## 4. Proof of Theorem2

(i) (
a)
⇒
(
b).
Suppose that δ(∞,w)>0. If q0>qn+n, then we have by (3)
(28)wq0=-1a0q0{∑j=1n‍Qj(w)wj(w′′w)j-∑k=0q0-1‍a0kwk}.
Applying Lemma 3 to (28),
(29)q0m(r,w)≤(q0-1)m(r,w)+∑j,k‍m(r,ajk)+Km(r,w′′w)+m(r,1a0q0)+O(1).
Since w(z) is admissible solution, we have
(30)m(r,w)=S(r,w),
so that
(31)δ(w,∞)=0.
This is a contradiction. Thus, q0≤qn+n.Ifq0<qn+n, by Theorem 1, w(z) is nonadmissible. Thus,
(32)q0=qn+n.(ii) (
b)
⇒
(
c).
Let q0=qn+n. Then, similar to the proof of Lemma 6, we obtain that the poles of w(z) are contained in the set of anqn and ∞ is a Picard exceptional value of w(z).(iii) (
c)⇒ (
a).
Let ∞ be a Picard exceptional value of w(z). Then δ(w,∞)=1.


## 5. Some Examples

Example 1.
The differential equation(33)ν2w2ν-1w′′-[((2ν+1)w4ν+2w2ν-ν+1)]=0
has a transcendental algebroid solution w(z)=(tanz)1/ν. In this case
(34)p=q0=4ν>1+q1=2ν-1+1=2ν.Remark 7.
Example1 shows that the condition in Theorem 1 is sharp.Example 2.
Transcendental algebroid functionw(z)=(sinz)1/2 is a 2-valued solution of the following differential equation:
(35)16w6(w′′)2+2w5w′′-(w8-12w6+2w4-12w2+1)=0.
In this case
(36)q0=8,n=2,q1=5,q2=6.
By Theorem 2, for transcendental algebroid function w(z)=(sinz)1/2, ∞ is a Picard exceptional value.Remark 8.
Example2 shows that the result in Theorem 2 holds.



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*Source: 123049-2014-01-02.xml*