R E S E A R C H
Open Access
On hyper-order of solutions of higher
order linear differential equations with
meromorphic coefficients
Jianren Long
*and Jun Zhu
*Correspondence:[email protected] School of Mathematics and Computer Science, Guizhou Normal University, Guiyang, 550001, P.R. China
Abstract
In this paper, we investigate the growth of meromorphic solutions of the differential equations
f(k)+Ak–1(z)f(k–1)+· · ·+A1(z)f+A0(z)f= 0
and
f(k)+Ak–1(z)f(k–1)+· · ·+A1(z)f+A0(z)f=F(z),
whereA0(z)≡0,A1(z),. . .,Ak–1(z) andF(z)≡0 are meromorphic functions. A precise
estimation of the hyper-order of meromorphic solutions of the above equations is given provided that there exists one dominant coefficient, which improves and extends previous results given by Belaïdi, Chen,etc.
MSC: Primary 34M10; secondary 30D35
Keywords: complex differential equation; meromorphic function; hyper-order
1 Introduction and main results
For a meromorphic functionf in the complex planeC, the order of growth and the lower order of growth are defined as
ρ(f) =lim sup
r→∞
log+T(r,f)
logr , μ(f) =lim infr→∞
log+T(r,f) logr ,
respectively. The exponent of convergence of the poles sequence off is defined as
λ
f
=lim sup
r→∞
log+N(r,f) logr .
For fast-growing meromorphic functions the growth is typically measured in terms of the hyper-order, defined as
ρ(f) =lim sup
r→∞
log+log+T(r,f) logr .
Nevanlinna theory of meromorphic functions is a powerful tool in the field of complex differential equations. For an introduction to the theory of differential equations in the complex plane by using Nevanlinna theory; see, for example [–]. Active research in this field was started by Wittich [] and his students in the s and s. The order of growth of solutions of the equation
f(k)+Ak–(z)f(k–)+· · ·+A(z)f+A(z)f = (.)
is one of the aims in studying complex differential equations, whereA(z)≡,A(z), . . . ,
Ak(z) are entire functions. For the casek= , from [–], we know that every nontrivial solution of the equation
f+A(z)f+B(z)f= (.)
is of infinite order provided that (i) ρ(A) <ρ(B); or (ii)ρ(B) <ρ(A)≤ ; or (iii)A(z) is polynomial andB(z) is transcendental withρ(B) = . Motivated by the result above, it is obvious that every nontrivial solution of (.) is of infinite order provided that the coeffi-cientA(z) is dominant over others. For example, ifmax{ρ(Aj),j= , , . . . ,k– }<ρ(A), then every nontrivial solution of (.) is of infinite order. Another generalized condition is found by Hellersteinet al.in [].
Theorem .([]) Let A(z),A(z), . . . ,Ak–(z),F(z)be entire functions.Suppose there
ex-ists an integer s, ≤s≤k– ,such that
maxρ(F), max ≤j≤k–
j=s
ρ(Aj)
<ρ(As)≤
.
Then every solution of
f(k)+Ak–(z)f(k–)+· · ·+A(z)f+A(z)f =F(z) (.)
is either a polynomial or an entire function of infinite order.
In , Chen and Yang studied the hyper-order of solutions of (.).
Theorem .([]) Let A(z),A(z), . . . ,Ak–(z)be entire function satisfying
maxρ(Aj),j= , , . . . ,k–
<ρ(A) <∞.
Then every nontrivial solution f of (.)satisfiesρ(f) =ρ(A).
Theorem .([], Theorem ) Let A(z),A(z), . . . ,Ak–(z)be meromorphic functions.
Suppose there exists an integer s, ≤s≤k– ,satisfying
max
ρ(Aj),j=s,λ
As
<μ(As)≤ρ(As) < .
If equation(.)has a meromorphic solution,then every transcendental meromorphic so-lution f of (.)satisfiesρ(f) =ρ(As).
In , Xiao and Chen considered the non-homogeneous equation (.), the following result is proved.
Theorem .([]) Let A(z),A(z), . . . ,Ak–(z),F(z)be meromorphic functions.Suppose
there exists an integer s, ≤s≤k– ,satisfying
max
ρ(Aj),j=s,λ
As
,ρ(F) <μ(As)≤ρ(As) < .
If equation(.)has a meromorphic solution,then every transcendental meromorphic so-lution f of (.)satisfiesρ(f) =ρ(As).
Belaïdi studied equation (.), a precise estimation of hyper-order of solutions of (.) is also obtained by using different conditions from those mentioned above, in which the growths of the coefficients are limited in a set having positive densities.
Theorem .([]) Let E be a set of complex numbers satisfyingdens({|z|:z∈E}) > ,
and let Aj(z),j= , , . . . ,k– ,be entire functions such that
maxρ(Aj),j= , . . . ,k– ≤ρ(A) =ρ<∞;
and for some constants≤β<α,we have,for allε> sufficiently small,
A(z)≥exp
α|z|ρ–ε, Aj(z)≤expβ|z|ρ–ε, j= , , . . . ,k– ,
as z→ ∞for z∈E.Then every nontrivial solution f of (.)satisfiesρ(f) =ρ(A).
Theorem . and the remaining theorems involve the logarithmic measure and densities of set, which will be recalled in Section . In this paper, we study the growth of solutions of (.) and (.), and one of the goals is to extend Theorems . and . in which the conditionρ(As) <is deleted. On the other hand, we consider the case of a meromorphic coefficient in Theorem .. The following results are proved by combining the methods of Theorems ., ., and ..
Theorem . Let E be a set of complex numbers satisfyingml({|z|:z∈E}) =∞,and let
Aj(z),j= , , . . . ,k– ,be meromorphic functions.Suppose there exists an integer s, ≤s≤ k– ,satisfying
max
ρ(Aj),j=s,λ
and for some constants≤β<α,we have,for allε> sufficiently small,
as z→ ∞for z∈E.Then every nontrivial meromorphic solution f whose poles are of uni-formly bounded multiplicities of equation(.)satisfiesρ(f) =ρ(As).
For the case of non-homogeneous equation, we get the following result.
Theorem . Let E and Aj(z),j= , , . . . ,k– be defined as Theorem.,and let F(z)≡
be meromorphic function.Suppose there exists an integer s, ≤s≤k– ,satisfying
max
and(.)hold as z→ ∞for z∈E.Then every nontrivial meromorphic solution f whose poles are of uniformly bounded multiplicities of equation(.)satisfiesρ(f) =ρ(As).
Remark . From [], Remark ., we know that the condition that the multiplicity of poles of the meromorphic solutionf is uniformly bounded is necessary. Hence the condi-tion was missing in Theorems . and .. Of course, the condicondi-tion could also be changed byδ(∞,f) > .
A lemma on logarithmic derivatives due to Gundersen [] plays an important role in proving our results.
The following result was proved originally in []; see also [], Lemma .
Lemma . Let f be a meromorphic function of orderρ(f) =β<∞.Then,for any given
ε> ,there exists a set E⊂(,∞)withml(E) <∞andm(E) <∞,such that,for all z
satis-fying|z|=r∈/([, ]∪E),
f(z)≤exprβ+ε.
The next lemma is related to the central index.
Lemma .([], Lemma ) Let f(z) = dg((zz)) be a meromorphic function,where g(z)and d(z)are entire functions satisfying
μ(g) =μ(f) =μ≤ρ(g) =ρ(f)≤ ∞, λ(d) =ρ(d) =λ
f
=β<μ.
Then there exists a set E⊂(,∞)withml(E) <∞,such that,for all z satisfying|z|=r∈/ ([, ]∪E)and|g(z)|=M(r,g),M(r,g) =max|z|=r|g(z)|,we have
f(n)(z)
f(z) =
νg(r) z
n
+o(), n≥,
whereνg(r)denotes the central index of g(z).
Lemma . Let f(z) = dg((zz)) be a meromorphic function, where g(z)and d(z)are entire functions. If≤ρ(d) <μ(f), thenμ(g) =μ(f), ρ(g) =ρ(f). Moreover,if ρ(f) =∞,then
ρ(f) =ρ(g).
Proof We divide the proof into the following three cases.
Case .ρ(f) <∞. SinceT(r,f)≤T(r,g) +T(r,d) +O(), for any givenε∈(,ρ(f)–ρ(d)), there exists an increasing sequence (rn) withrn→ ∞asn→ ∞, such that
rρn(f)–ε≤T(rn,f), T(rn,d)≤rnρ(d)+ε
hold for all sufficiently largen. This implies that, for all sufficiently largen,
rρn(f)–ε≤T(rn,g) +rnρ(d)+ε+O().
Thus, for all sufficiently largen,
rρn(f)–ε –o()≤T(rn,g) +O().
Hence,ρ(f)≤ρ(g). On the other hand, sinceT(r,g)≤T(r,f) +T(r,d), andρ(d) <ρ(f), we get
ρ(g)≤ρ(f).
In a similar way to above and the definition of the lower order of growth, we can prove
μ(g) =μ(f).
Case .μ(f) =∞. Suppose on the contrary to the assertion thatμ(g) <μ(f). We aim for a contradiction. By the definition of the lower order of growth, there exist a constant
M> and an increasing sequence (rn) withrn→ ∞asn→ ∞, such that
T(rn,g) <rMn, T(rn,d)≤rMn
hold for all sufficiently largen. Then, for all sufficiently largen,
T(rn,f)≤rMn +O().
Therefore,μ(f)≤M. This is a contradiction with our assumption.
Case .μ(f) <∞andρ(f) =∞. In a similar way to proving cases and , we can prove case .
Finally, we will proveρ(f) =ρ(g). Suppose thatρ(f) =∞. Then there exists an increas-ing sequence (rn) withrn→ ∞asn→ ∞, such that
ρ(f) = lim
n→∞
log logT(rn,f) logrn
.
Combiningρ(d) <μ(f) and the definitions of the order and the lower order, we get
lim
n→∞ T(rn,d)
T(rn,f)
= .
Then there exists a positive integerN, such that, forn>N,
T(rn,f)≤T(rn,g) +O().
Hence,ρ(f)≤ρ(g). In a similar way to proving case , we getρ(f)≥ρ(g). Therefore,
ρ(f) =ρ(g).
Lemma .([], Lemma ) Let f be an entire function of infinite order,with the hyper-orderρ(f) <∞,and letν(r)be the central index of f.Then
lim sup
r→∞
log+log+ν(r) logr =ρ(f).
Lemma .([], Lemma ) Let g: [,∞)→Rand h: [,∞)→Rbe monotonically non-decreasing functions such that g(r)≤h(r)outside of an exceptional set E withml(E) <∞.
Then,for anyα> ,there exists an r> such that g(r)≤h(αr)for all r>r.
Lemma . Let f be a meromorphic solution of equation (.), where A(z),A(z), . . . ,
that satisfies
Combining the formula above and the first main theory in Nevanlinna theory, we get
T(r,As)≤
Combining the two inequalities above and (.), we get
. . . ,k– }that satisfies
maxρ(F),ρ(Aj),j=s
<μ(As),
thenρ(f)≥ρ(As),μ(f)≥μ(As).
Proof In a similar way to proving Lemma ., we can get the proof of Lemma ., here we
omit the details.
Then
νgk(r) +o()≤kexprρ(As)+ε|z|kνk–
g (r) +o().
By Lemmas . and ., we get
ρ(g)≤ρ(As).
Combining Lemma . and the inequality above, we have
ρ(f)≤ρ(As).
It follows from (.) and (.) that
Combining (.) and the inequality above, we get
Combining Lemmas ., ., and the inequality above, we get
ρ(g)≤ρ(As).
Combining Lemma . and the inequality above, we have
ρ(f)≤ρ(As).
This completes the proof.
3 Proof of Theorems 1.6 and 1.7
Proof of Theorem. By Lemma ., it is easy to see that equation (.) cannot have any nonzero rational solution. By (.),
Considering the equation above, there exists a constantR(>R), such that, for allz sat-isfying|z|=r>R,νg(r) > ,| +o()|>, and|g(z)|=M(r,g),M(r,g) > ,
ff(s()z(z))
SetH={|z|:z∈E}\([,R]∪E∪E). Then ml(H) =∞. It follows from (.), (.), (.), (.) and conditions (.), (.) that, for allzsatisfying|z|=r∈Hand|g(z)|=M(r,g),
expα|z|ρ–ε≤kBrs+T(r,f)k+expβ|z|ρ–ε.
Therefore, we get
ρ(f)≥ρ=ρ(As).
It follows from the inequality above and Lemma . that
ρ(f) =ρ(As).
This completes the proof.
Proof of Theorem. By Lemma ., we know that if equation (.) has solutions, then the solution must be transcendental. By equation (.),
It follows from (.), (.), (.), (.), (.), and conditions (.), (.) that, for allz satis-fying|z|=r∈Hand|g(z)|=M(r,g),
expαrρ–ε≤(k+ )Brs+T(r,f)k+
expβrρ–εexprη.
Byη<μ(As)≤ρand for any givenε∈(,μ(As)–η), we get
ρ(f)≥ρ=ρ(As).
It follows from the inequality above and Lemma . that
ρ(f) =ρ(As).
This completes the proof.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the manuscript. All authors read and approved the final manuscript.
Acknowledgements
The authors thank the referee for his/her valuable suggestions to improve the paper. This research work is supported by the Foundation of Science and Technology of Guizhou Province of China (Grant No. [2015]2112), the National Natural Science Foundation of China (Grant No. 11501142).
Received: 25 October 2015 Accepted: 8 March 2016
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