ZHANG Shao-hua,WU Zhao-jun,SUN Dao-chun
(1.School of Mathematics and Statistics,Hubei University of Science and Technology,Xianning 437100, China;2.School of Mathematics,South China Normal University,Guangzhou 510631,China)
The Lower Order and Type of Solutions of Second Order Diff erential Equation
ZHANG Shao-hua1,WU Zhao-jun1,SUN Dao-chun2
(1.School of Mathematics and Statistics,Hubei University of Science and Technology,Xianning 437100, China;2.School of Mathematics,South China Normal University,Guangzhou 510631,China)
The main purpose of this paper is to study the lower order and type of second order diff erential equation w′′(z)−A(z)w=0,where A(z)is a polynomial.In the case ofthe authors prove that the lower order and the type of all non-trivial solutions w of w′′(z)−A(z)w=0 are equal toandrespectively.In the case of A(z)=adzd+ad−1zd−1+···+a1z+a0,ad>0,ad−1≥0,···,a1≥0,a0≥0,the authors prove that the lower order of all non-trivial solutions w of w′′(z)−A(z)w=0 is
type;order;lower order;diff erential equation
Since 1982 when the article by Bank and Laine[1]appeared in Trans Amer Math Soc,there have appeared many papers on the global theory of complex diff erential equations which were studied from the point of view of Nevanlinna theory.We refer the reader to the books by Laine[4]and by Gao et al[2].In[1],Bank and Laine have proved.
Theorem A Suppose that A(z)=adzd+ad−1zd−1+···+a1z+a0,ad/=0 is a polynomial. Then allnon-trivialsolutions f of
have the order of growthσ(f)=
Theorem A study the order of solutions of(1.1).It’s naturalto ask what are the lower order and the type of any one of non-trivial solutions.Upon to now,there isn’t a paper to study this problem.This paper is devoted to consider it.More precisely,we shallprove the following theorems.
Theorem 1 Assume that A(z)=adzd,ad/=0.Then
(i)Allnon-trivialsolutions f of(1.1)have the lower order of growthµ(f)=
(ii)All non-trivial solutions f of(1.1)have the typeTheorem 2 Assume that A(z)=adzd+ad−1zd−1+···+a1z+a0,ad>0,ad−1≥0,···,a1≥0,a0≥0.Then allnon-trivialsolutions f of(1.1)have the lower order ofgrowth µ(f)=
From Theorem 1,Theorem 2 and Theorem A,we can get
Corollary 1 Assume that A(z)=adzd,ad/=0 or A(z)=adzd+ad−1zd−1+···+a1z+ a0,ad>0,ad−1≥0,···,a1≥0,a0≥0.Then all non-trivial solutions f of(1.1)are regular growth.
In Theorem 2,we add that condition ad>0,ad−1≥0,···,a1≥0,a0≥0.Here,we pose the following question.
Question 1 Does Theorem 2 holds if we remove the additional condition of ad>0,ad−1≥0,···,a1≥0,a0≥0?
As is well-known,allsolution of(1.1)are entire functions.Let
Suppose that
is a non-trivial solution of(1.1).Then
where xv=0 for any v<0.We can get from(1.1),(2.1)and(2.2)that
Hence,for any k∈N−{0,1},
Replace k by k−j1in(2.3),we can get
Subsisting the above equality to(2.3),for any k/=0,1,j1,j1+1,we have
Replace k by k−j1−j2in(2.3),we can get
Subsisting the above equality to(2.3),for any k/=0,1,j1,j1+1,j1+j2,j1+j2+1,we have
Put x0=c0,x1=c1.By induction,we can get the following coeffi cients formula ofnon-trivial solution of(1.1),
where the Sum for any combination jusuch thatFor any positive integer u,ju≥2.So n≤k/2.Noting that k−0,1,2,···,n−1).Hence,(2.5)is well defined.
Especially,if
then
Where the sum for any combination jpsuch that
If A(z)=adzd,ad/=0.Then
for k=n(d+2);
for k=n(d+2)+1 and xk=0 for others.
In order to prove Theorems 1 and 2,we need it following results related to the growth of Taylor series.
Suppose Taylor series
satisfies
Then(3.1)is convergent in the whole complex plane C.We denote it by f(z)and f(z)is an entire function.The maximum module M(r,f),maximum term m(r,f)and central index ν(r,f)of f(z)are defined as the following
The orderσ(f)and the lower orderµ(f)of f(z)is defined by
Ifσ(f)=µ(f),then we say f(z)has regular growth orderσ(f).Suppose that the order of f(z) isρ∈(0,∞),we define the type of f(z)by
Lemma 1[3,7]Suppose the Taylor series(3.1)satisfies(3.2).Then
If{nk}is the sequence of centralindex,then
Lemma 2[3,7]Suppose the Taylor series(3.1)satisfies(3.2).Then
where the Max for any increasing sequence{nt}.
Lemma 3[3,7]Suppose that the order of f(z)isρ∈(0,∞).Then
Theorem 3 Suppose the Taylor series(3.1)satisfies(3.2).Then f(z)has regular growth orderρif and only if the series(3.1)satisfies the following conditions
i)
ii)There is an increasing sequence{nt}such that
Proof Suppose that the series(3.1)satisfies the following conditions
i)
ii)There is an increasing sequence{nt}such that
Then
By Lemma 2,we haveµ(f)≥ρ.By Lemma 1,we haveσ(f)≤ρ.So f(z)has regular growth orderρ.
Suppose that f(z)has regular growth orderρ.In virtue Lemmas 1 and 2,we have
i)
and ii)
for the sequence of central index{nk},whereis a subsequence of{nk}.
We are now in the position to prove theorems 1 and 2.
Proof of Theorem 1 Since
So
Suppose that
is a non-trivial solution of(1.1).By Theorem A and Lemma 1,we have
Take c0/=0 in Section 2.By(2.7)and(3.3),for n=k(d+2),we have
Take c1/=0 in Section 2.By(2.8)and(3.3),for n=k(d+2)+1,we also obtain(3.4).Hence
Noting that
By Theorem 3,we derive that f(z)is regular growth and the lower order of f(z)is
Take c0/=0 in Section 2.By Lemma 3,(2.7)and(3.3),for n=k(d+2),we have
Take c1/=0 in Section 2.By(2.8)and(3.3),for n=k(d+2)+1,we also obtain(3.5).
Proof of Theorem 2 Take c1/=0 in Section 2.By Lemma 3,(2.6)and(3.3),for n(t):=t(d+2)+1,we have
If c1=0,then c0/=0 in Section 2.By(2.8)and(3.3),for n=k(d+2),we also obtain (3.6).By Theorem 3,we derive that f(z)is regular growth and the lower order of f(z)is
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tion:30D30
1002–0462(2014)01–0014–08
Chin.Quart.J.of Math. 2014,29(1):14—21
date:2013-04-08
Supported by the NNSF of China(11201395);Supported by the Science Foundation of Educational Commission of Hubei Province(Q20132801)
Biographies:ZHANG Shao-hua(1961-),male,native of Xianning,Hubei,an associate professor of Hubei University of Science and Technology,engages in complex analysis;WU Zhao-jun(1978-),male,native of Xianning,Hubei,an associate professor of Hubei University of Science and Technology,Ph.D.,engages in complex analysis.
CLC number:O174.32 Document code:A
Chinese Quarterly Journal of Mathematics2014年1期