Study of Complex Oscillation of Solutions of a Second Order Linear
Differential Equation With Entire Coefficients of (α, β, γ)-Order
BENHARRAT BELAÏDI∗
Department of Mathematics, Laboratory of Pure and Applied Mathematics
University of Mostaganem (UMAB)
B. P. 227 Mostaganem
ALGERIA
TANMAY BISWAS
Rajbari, Rabindrapally, R. N. Tagore Road, P.O.- Krishnagar,
P.S. Kotwali, Dist-Nadia, PIN-741101, West Bengal
INDIA
Abstract: In this paper, we deal with the complex oscillation of solutions of linear differential equation. We
mainly study the interaction between the growth, zeros of solutions with the coefficients of second order linear
differential equations in terms of (α, β, γ)-order and obtain some results in general form which considerably
extend some results of [5], [18] and [21].
Key-Words: Linear differential equations, (α, β, γ)-order, (α, β, γ)-exponent of convergence of zero sequence.
AMS Subject Classification (2010):30D35, 34M10.
1 Introduction, Definitions and
Notations
Throughout this paper, we assume that the reader
is familiar with the fundamental results and the stan-
dard notations of the Nevanlinna value distribution
theory of entire and meromorphic functions which are
available in [11, 19, 25] and therefore we do not ex-
plain those in details. The theory of complex linear
equations has been developed since 1960s. Many au-
thors have investigated the second order linear differ-
ential equation
f00 +A(z)f= 0,(1)
where A(z)is an entire function or a meromorphic
function of finite order or finite iterated order, and
have obtained many results about the interaction be-
tween the solutions and the coefficient of (1) (see
[1, 2, 3, 18]). Moreover some authors have investi-
gated the exponent of convergence of zero sequence
and pole-sequence of the solutions of second order
differential equations and have obtained some inter-
esting results (see [7, 8, 18, 24]). Mulyava et al. [20]
have investigated the properties of solutions of a het-
erogeneous differential equation of the second order
under some different conditions using the concept of
generalized order. For details one may see [20].
We denote the linear measure and the loga-
rithmic measure of a set E⊂(1,+∞)by mE =
REdt and mlE=RE
dx
x. Now let Lbe a class of con-
tinuous non-negative on (−∞,+∞)function αsuch
that α(x) = α(x0)≥0for x≤x0and α(x)↑+∞
as x0≤x→+∞.
Recently Heittokangas et al. [14] have intro-
duced a new concept of ϕ-order of entire and mero-
morphic function considering ϕas subadditive func-
tion. For details one may see [14]. Now it is interest-
ing to investigate the interaction between the growth,
zeros of solutions with the coefficients of second or-
der linear differential equations using the revised idea
of Heittokangas et al. [14], which is the main aim of
this paper. For this purpose, we introduce the defini-
tion of the (α, β, γ)-order of a meromorphic function
in the following way:
Definitions 1. Let α∈L,β∈Land γ∈L. The
(α, β, γ)-order denoted by σ(α,β,γ)[f]and (α, β, γ)-
lower order denoted by µ(α,β,γ)[f]of a meromorphic
function fare, respectively, defined by
σ(α,β,γ)[f] = lim sup
r→+∞
α(log T(r, f))
β(log γ(r))
5HFHLYHG-XO\5HYLVHG$SULO$FFHSWHG0D\3XEOLVKHG-XQH
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and
µ(α,β,γ)[f] = lim inf
r→+∞
α(log T(r, f))
β(log γ(r)) .
Remark 1. Let fbe a meromorphic function. One
can see that α(r) = log[p]r,(p≥0),β(r) = log[q]r,
(q≥0) and γ(r) = rbelong to the class L, where
log[k]x=log(log[k−1] x) (k≥1), with convention
that log[0] x=x. So, when p= 0 and q= 0, i.e.,
α(r) = β(r) = r, the Definition 1 coincides with the
usual order and lower order, when α(r) = log[p−1] r
(p≥1) and β(r) = r, we obtain the iterated p-order
and iterated lower p-order (see [18, 22]), moreover
when α(r) = log[p−1] rand β(r) = log[q−1] r,(p≥
q≥1), we get the (p, q)-order and lower (p, q)-order
(see [15, 16]). Further, if α(r) = ϕ(er), where ϕ
is an increasing unbounded function on [1,+∞)and
β(r) = r, we obtain the ϕ-order and the lower ϕ-
order (see [4, 9]). Finally if α(r) = β(r) = rand
γ(r) = ϕ(r),where ϕ: (R0,+∞)→(0,+∞)is
a non-decreasing unbounded function satisfying the
condition ϕ(a+b)≤ϕ(a) + ϕ(b)for all a, b ≥R0,
we obtain the new definition of ϕ-order and the lower
ϕ-order introduced by Heittokangas et al. [14].
Similarly to Definition 1, we can also define
the (α, β, γ)-exponent of convergence of the zero-
sequence and (α, β, γ)-exponent of convergence of
the distinct zero sequence of a meromorphic function
fin the following way:
Definitions 2. Let α∈L,β∈Land γ∈L.
The (α, β, γ)-exponent of convergence of the zero-
sequence denoted by λ(α,β,γ)[f]of a meromorphic
function fis defined by
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) .
Similarly, the (α, β, γ)-exponent of convergence of
the distinct zero-sequence denoted by λ(α,β,γ)[f]of f
is defined by
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) .
We say that α∈L1, if α(a+b)≤α(a) +
α(b) + cfor all a, b ≥R0and fixed c∈(0,+∞).
Further we say that α∈L2, if α∈Land α(x+
O(1)) = (1 + o(1))α(x)as x→+∞. Finally, α∈
L3, if α∈Land α(a+b)≤α(a) + α(b)for all
a, b ≥R0,i.e., αis subadditive. Clearly L3⊂L1.
Particularly, when α∈L3, then one can eas-
ily verify that α(mr)≤mα(r), m ≥2is an inte-
ger. Up to a normalization, subadditivity is implied
by concavity. Indeed, if α(r)is concave on [0,+∞)
and satisfies α(0) ≥0, then for t∈[0,1],
α(tx) = α(tx + (1 −t)·0)
≥tα(x) + (1 −t)α(0) ≥tα(x),
so that by choosing t=a
a+bor t=b
a+b,
α(a+b) = a
a+bα(a+b) + b
a+bα(a+b)
≤αa
a+b(a+b)+αb
a+b(a+b)
=α(a) + α(b),a, b ≥0.
As a non-decreasing, subadditive and unbounded
function, α(r)satisfies
α(r)≤α(r+R0)≤α(r) + α(R0)
for any R0≥0. This yields that α(r)∼α(r+R0)
as r→+∞.
Now we add two conditions on α, β and γ:
(i) Always α∈L1, β ∈L2and γ∈L3; and (ii)
α(log[p]x) = o(β(log γ(x))), p ≥2is an integer as
x→+∞.
Throughout this paper, we assume that α, β
and γalways satisfy the above two conditions unless
otherwise specifically stated.
Proposition 1. Let f1, f2be non-constant meromor-
phic functions with σ(α,β,γ)[f1]and σ(α,β,γ)[f2]as
their (α, β, γ)-order. Then
(i) σ(α,β,γ)[f1±f2]≤max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]};
(ii) σ(α,β,γ)[f1·f2]≤max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]};
(iii) If σ(α,β,γ)[f1]6=σ(α,β,γ)[f2], then σ(α,β,γ)[f1±
f2] = max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]};
(iv) If σ(α,β,γ)[f1]6=σ(α,β,γ)[f2], then σ(α,β,γ)[f1·
f2] = max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]}.
Proof. (i) Without loss of generality, we assume that
σ(α,β,γ)[f1]≤σ(α,β,γ)[f2]<+∞. From the defini-
tion of (α, β, γ)-order, for any ε > 0, we obtain for
all sufficiently large values of rthat
T(r, f1)<exp(α−1((σ(α,β,γ)[f1] + ε)β(log γ(r))))
(2)
and
T(r, f2)<exp(α−1((σ(α,β,γ)[f2] + ε)β(log γ(r)))).
(3)
Since T(r, f1±f2)≤T(r, f1) + T(r, f2) + log 2for
all large r, we get from (2) and (3) for all sufficiently
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large values of rthat
T(r, f1±f2)<2exp(α−1((σ(α,β,γ)[f2]
+ε)β(log γ(r)))) + log 2
i.e., T (r, f1±f2)<3exp(α−1((σ(α,β,γ)[f2]
+ε)β(log γ(r))))
i.e., log T(r, f1±f2)< α−1((σ(α,β,γ)[f2]
+ε)β(log γ(r))) + log 3
i.e., α(log T(r, f1±f2)) <(σ(α,β,γ)[f2]
+ε)β(log γ(r)))
+α(log 3) + c, (c > 0) ,
which implies that
lim sup
r→+∞
α(log T(r, f1±f2))
β(log γ(r))) ≤σ(α,β,γ)[f2] + ε
holds for any ε > 0. Hence
σ(α,β,γ)[f1±f2]≤max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]}.
(4)
(iii) Further without loss of any generality, let
σ(α,β,γ)[f1]< σ(α,β,γ)[f2]<+∞and f=f1±
f2.Then in view of (4) we get that σ(α,β,γ)[f]≤
σ(α,β,γ)[f2]. As, f2=±(f−f1)and in this
case we obtain that σ(α,β,γ)[f2]≤max {σ(α,β,γ)[f],
σ(α,β,γ)[f1]}. As we assume that σ(α,β,γ)[f1]
< σ(α,β,γ)[f2], therefore we have σ(α,β,γ)[f2]≤
σ(α,β,γ)[f]and hence σ(α,β,γ)[f] = σ(α,β,γ)[f2] =
max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]}.
(ii) and (iv) Similarly, from T(r, f1·f2)≤
T(r, f1) + T(r, f2)for all large r, we can also get
that
σ(α,β,γ)[f1·f2]≤max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]}
and if σ(α,β,γ)[f1]6=σ(α,β,γ)[f2], then
σ(α,β,γ)[f1·f2] = max{σ(α,β,γ)[f1], σ(α,β,γ)[f2]},
which completes the proof of Proposition 1.
Proposition 2. Let f1, f2be non-constant
meromorphic functions with σ(α(log),β,γ)[f1]and
σ(α(log),β,γ)[f2]as their (α(log), β, γ)-order. Then
(i) σ(α(log),β,γ)[f1±f2]≤max{σ(α(log),β,γ)[f1],
σ(α(log),β,γ)[f2]};
(ii) σ(α(log),β,γ)[f1·f2]≤max{σ(α(log),β,γ)[f1],
σ(α(log),β,γ)[f2]};
(iii) If σ(α(log),β,γ)[f1]6=σ(α(log),β,γ)[f2], then
σ(α(log),β,γ)[f1±f2] = max{σ(α(log),β,γ)[f1],
σ(α(log),β,γ)[f2]};
(iv) If σ(α(log),β,γ)[f1]6=σ(α(log),β,γ)[f2], then
σ(α(log),β,γ)[f1·f2] = max {σ(α(log),β,γ)[f1],
σ(α(log),β,γ)[f2]}.
Since α(a+b)≤α(a)+α(b)+cfor all a, b ≥
R0and fixed c∈(0,+∞), the proof of Proposition 2
would run parallel to that of Proposition 1. We omit
the details.
Proposition 3. (i) If fis an entire function, then
σ(α,β,γ)[f] = lim sup
r→+∞
α(log T(r, f))
β(log γ(r))
=lim sup
r→+∞
α(log[2] M(r, f))
β(log γ(r))
and
µ(α,β,γ)[f] = lim inf
r→+∞
α(log T(r, f))
β(log γ(r))
=lim inf
r→+∞
α(log[2] M(r, f))
β(log γ(r)) .
(ii) If fis a meromorphic function, then
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r))
and
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r)) .
Proof. (i) By the inequality T(r, f)≤
log+M(r, f)≤R+r
R−rT(R, f)(0< r < R) (cf.
[11]) for an entire function f, set R=ηr (η > 1),
we have
T(r, f)≤log+M(r, f)≤η+ 1
η−1T(ηr, f). (5)
By (5), α(a+b)≤α(a) + α(b) + cfor all a, b ≥
R0and fixed c∈(0,+∞),β((1 + o(1))x) = (1 +
o(1))β(x)as x→+∞and γ(a+b)≤γ(a) + γ(b)
for all a, b ≥R0, it is easy to see that conclusion (i)
holds.
(ii) Without loss of generality, assume that f(0) 6=
0, then N(r, 1/f) = Rr
0
n(t,1/f)
tdt. We get for 0<
r0< r
N(r, 1/f)−N(r0,1/f) = Zr
r0
n(t, 1/f)
tdt
≤n(r, 1/f)log r
r0
,
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that is
N(r, 1/f)≤N(r0,1/f) + n(r, 1/f)
×log r
r0
(0< r0< r),
i.e., N(r, 1/f)≤1 + N(r0,1/f)
n(r, 1/f)log r
r0
×n(r, 1/f)log r
r0
(0< r0< r),
which implies that
log N(r, 1/f)≤log n(r, 1/f) + log log r
+log 1−log r0
log r+log 1
+N(r0,1/f)
n(r, 1/f)log r
r0(0< r0< r),
(6)
then by the condition on αand (6), we obtain that
lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r))
≤lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) +lim sup
r→+∞
α(log[2] r)
β(log γ(r))
+lim sup
r→+∞
αlog 1−log r0
log r
β(log γ(r))
+lim sup
r→+∞
αlog 1 + N(r0,1/f)
n(r,1/f)log r
r0
β(log γ(r))
+lim sup
r→+∞
c
β(log γ(r))
=lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) ,(c > 0) ,(7)
since α(log[2] x) = o(β(log γ(x))) as x→+∞we
have α(log[2] r)
β(log γ(r)) →0as r→+∞.
On the other hand, we have
N(er, 1/f) = Zer
0
n(t, 1/f)
tdt
≥Zer
r
n(t, 1/f)
tdt
≥n(r, 1/f)log e=n(r, 1/f).
(8)
By (8), we have
lim sup
r→+∞
α(log N(er, 1/f))
β(log γ(r)) ≥lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) .
By the conditions β((1 + o(1))x) = (1 + o(1))β(x)
as x→+∞and γ(er)≤γ(3r)≤3γ(r),we can
write
lim sup
r→+∞
α(log N(er, 1/f))
β(log γ(r)) ≤lim sup
r→+∞
α(log N(er, 1/f))
β(log 1
3γ(er))
=lim sup
r→+∞
α(log N(er, 1/f))
β(log 1
3+log γ(er))
=lim sup
r→+∞
α(log N(er, 1/f))
β((1 + o(1)) log γ(er))
=lim sup
r→+∞
α(log N(er, 1/f))
(1 + o(1))β(log γ(er))
=lim sup
r→+∞
α(log N(er, 1/f))
β(log γ(er))
=lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r)) ,
it follows that
lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r)) ≥lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r)) .
(9)
By (7) and (9), it is easy to see that
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r)) .
By the same proof above, we can obtain the conclu-
sion
λ(α,β,γ)[f] = lim sup
r→+∞
α(log n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log N(r, 1/f))
β(log γ(r)) .
Proposition 4. (i) If fis an entire function, then
σ(α(log),β,γ)[f] = lim sup
r→+∞
α(log[2] T(r, f))
β(log γ(r))
=lim sup
r→+∞
α(log[3] M(r, f))
β(log γ(r))
and
µ(α(log),β,γ)[f] = lim inf
r→+∞
α(log[2] T(r, f))
β(log γ(r))
=lim inf
r→+∞
α(log[3] M(r, f))
β(log γ(r)) .
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(ii) If fis a meromorphic function, then
λ(α(log),β,γ)[f] = lim sup
r→+∞
α(log[2] n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log[2] N(r, 1/f))
β(log γ(r))
and
λ(α(log),β,γ)[f] = lim sup
r→+∞
α(log[2] n(r, 1/f))
β(log γ(r))
=lim sup
r→+∞
α(log[2] N(r, 1/f))
β(log γ(r)) .
Since α(a+b)≤α(a)+α(b)+cfor all a, b ≥
R0and fixed c∈(0,+∞),β((1 + o(1))x) = (1 +
o(1))β(x)as x→+∞and γ(a+b)≤γ(a) + γ(b)
for all a, b ≥R0, the proof of Proposition 4 would run
parallel to that of Proposition 3. We omit the details.
2 Main Results
In this paper, our aim is to make use of the con-
cept of (α, β, γ)-order of entire functions to investi-
gate the growth, zeros of the solutions of equation (1)
which considerably extend some results of [21].
Theorem 1. Let A(z)be an entire function satisfying
σ(α,β,γ)[A]>0. Then σ(α(log),β,γ)[f] = σ(α,β,γ)[A]
holds for all non-trivial solutions of (1).
Theorem 2. Let A(z)be an entire function satisfy-
ing σ(α,β,γ)[A]>0, let f1and f2be two linearly
independent solutions of (1) and denote F=f1·
f2. Then max {λ(α(log),β,γ)[f1], λ(α(log),β,γ)[f2]}=
λ(α(log),β,γ)[F] = σ(α(log),β,γ)[F]≤σ(α,β,γ)[A]. If
σ(α(log),β,γ)[F]< σ(α,β,γ)[A], then λ(α(log),β,γ)[f] =
σ(α,β,γ)[A]holds for all solutions of type f=c1f1+
c2f2, where c1·c26= 0.
Theorem 3. Let A(z)be an entire function satisfy-
ing λ(α,β,γ)[A]< σ(α,β,γ)[A]. Then λ(α(log),β,γ)[f]≤
σ(α,β,γ)[A]≤λ(α,β,γ)[f]holds for all non-trivial so-
lutions of (1).
Remark 2. This article may be understood as an ex-
tension and an improvement of [5], [18] and [21].
3 Some Lemmas
In this section we present some lemmas which
will be needed in the sequel.
Lemma 1. ([12, 13, 19]) Let fbe a transcendental
entire function, and let zbe a point with |z|=rat
which |f(z)|=M(r, f). Then, for all |z|outside a
set E1of rof finite logarithmic measure, we have
f(j)(z)
f(z)=ν(r, f)
zj(1 + o(1)) (j∈N),(10)
where ν(r, f)is the central index of f.
Lemma 2. ([10, 19]) Let g: [0,+∞)→Rand h:
[0,+∞)→Rbe monotone nondecreasing functions
such that g(r)≤h(r)outside of an exceptional set E2
of finite linear measure or finite logarithmic measure.
Then, for any d > 1, there exists r0>0such that
g(r)≤h(dr)for all r > r0.
Lemma 3. ([[13], Theorems 1.9 and 1.10, or [17],
Satz 4.3 and 4.4]) Let f(z) = P+∞
n=0 anznbe any
entire function, µ(r, f)be the maximum term, i.e.,
µ(r, f) = max {|an|rn;n= 0,1, ...}, and ν(r, f)be
the central index of f.
(i) If |a0| 6= 0, then
log µ(r, f) = log |a0|+
r
Z
0
ν(t, f)
tdt.(11)
(ii) For r < R, we have
M(r, f)< µ(r, f)ν(R, f) + R
R−r.(12)
Lemma 4. Let fbe an entire function satisfying
σ(α,β,γ)[f] = σ1and µ(α,β,γ)[f] = µ1, and let ν(r, f)
be the central index of f. Then
lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) =σ1
and
lim inf
r→+∞
α(log ν(r, f))
β(log γ(r)) =µ1.
Proof. In view of the first part of Lemma 3, one may
obtain that
log µ(2r, f) = log |a0|+
2r
Z
0
ν(t, f)
tdt
≥log |a0|+
2r
Z
r
ν(t, f)
tdt ≥log |a0|+ν(r, f)log 2.
(13)
Also by Cauchy’s inequality, it is well known that (cf.
[23])
µ(r, f)≤M(r, f). (14)
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Therefore, one may obtain from (13) and (14) that
ν(r, f)log 2≤log M(2r, f)−log |a0|.
Thus from above, we get that
log ν(r, f) + log[2] 2≤log[2] M(2r, f)
+log 1−log |a0|
log M(2r, f).
By using condition on α, we obtain that
lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) ≤lim sup
r→+∞
α(log[2] M(2r, f))
β(log γ(r))
+lim sup
r→+∞
αlog 1−log |a0|
log M(2r,f)
β(log γ(r))
+lim sup
r→+∞
α(−log[2] 2)
β(log γ(r)) +lim sup
r→+∞
c
β(log γ(r))
=lim sup
r→+∞
α(log[2] M(2r, f))
β(log γ(r)) .
By using γ(2r)≤2γ(r),it follows that
lim sup
r→+∞
α(log ν(r, f))
β(log γ(r))
≤lim sup
r→+∞
α(log[2] M(2r, f))
β(log 1
2γ(2r))
=lim sup
r→+∞
α(log[2] M(2r, f))
β((1 + o(1)) log γ(2r))
=lim sup
r→+∞
α(log[2] M(2r, f))
(1 + o(1)) β(log γ(2r))
=lim sup
r→+∞
α(log[2] M(r, f))
β(log γ(r)) =σ1,
i.e., σ1≥lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) (15)
and consequently
µ1≥lim inf
r→+∞
α(log ν(r, f))
β(log γ(r)) . (16)
Further for any constant K1, one may get from the
second part of Lemma 3, that (cf. [6])
log M(r, f)< ν(r, f)log r+log ν(2r, f) + K1.
Therefore from above we obtain that
log M(r, f)< ν(2r, f)log r+ν(2r, f) + K1,
i.e., log M(r, f)< ν(2r, f)(1 + log r) + K1,
i.e., log M(r, f)< ν(2r, f)log(e·r) + K1,
i.e., log[2] M(r, f)<log ν(2r, f) + log[2](e·r)
+log 1 + K1
ν(2r, f)log(e·r),
i.e., lim sup
r→+∞
α(log[2] M(r, f))
β(log γ(r))
≤lim sup
r→+∞
α(log ν(2r, f))
β(log γ(r)) +lim sup
r→+∞
α(log[2](e·r))
β(log γ(r))
+lim sup
r→+∞
αlog 1 + K1
ν(2r,f)log(e·r)
β(log γ(r))
+lim sup
r→+∞
c
β(log γ(r)) =lim sup
r→+∞
α(log ν(2r, f))
β(log γ(r)) ,
where c > 0. Since γ(2r)≤2γ(r), so from above
we have
σ1=lim sup
r→+∞
α(log[2] M(r, f))
β(log γ(r))
≤lim sup
r→+∞
α(log ν(2r, f))
β(log γ(r))
≤lim sup
r→+∞
α(log ν(2r, f))
β(log 1
2γ(2r))
=lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) ,
i.e., σ1≤lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) (17)
and accordingly
µ1≤lim inf
r→+∞
α(log ν(r, f))
β(log γ(r)) .(18)
Combining (15), (17) and (16), (18) we obtain that
lim sup
r→+∞
α(log ν(r, f))
β(log γ(r)) =σ1
and
lim inf
r→+∞
α(log ν(r, f))
β(log γ(r)) =µ1.
This proves the lemma.
Lemma 5. Let fbe an entire function satisfying
σ(α(log),β,γ)[f] = σ2and µ(α(log),β,γ)[f] = µ2, and
let ν(r, f)be the central index of f. Then
lim sup
r→+∞
α(log[2] ν(r, f))
β(log γ(r)) =σ2
and
lim inf
r→+∞
α(log[2] ν(r, f))
β(log γ(r)) =µ2.
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In the line of Lemma 4 one can easily deduce
the conclusion of Lemma 5 and so its proof is omitted.
Lemma 6. Let f1and f2be entire functions
of (α, β, γ)-exponent of convergence of the zero-
sequence and denote F=f1·f2. Then
λ(α,β,γ)[F] = max{λ(α,β,γ)[f1], λ(α,β,γ)[f2]}.
Proof. Let n(r, 0, F ),n(r, 0, f1)and n(r, 0, f2)be
unintegrated counting functions for the number of ze-
ros of F,f1and f2. For any r > 0, it is easy to see
that
n(r, 0, F )≥max{n(r, 0, f1), n(r, 0, f2)}. (19)
By Definition 2 and (19), we have
λ(α,β,γ)[F]≥max{λ(α,β,γ)[f1], λ(α,β,γ)[f2]}. (20)
On the other hand, since the zeros of Fmust be the
zeros of f1and the zeros of f2, for any r > 0, we have
n(r, 0, F ) = n(r, 0, f1) + n(r, 0, f2)
≤2max{n(r, 0, f1), n(r, 0, f2)}.(21)
By Definition 2 and (21), we get that
λ(α,β,γ)[F]≤max{λ(α,β,γ)[f1], λ(α,β,γ)[f2]}. (22)
Therefore, by (20) and (22), we have
λ(α,β,γ)[F] = max{λ(α,β,γ)[f1], λ(α,β,γ)[f2]}.
This complete the proof.
Lemma 7. Let f1and f2be entire functions of
(α(log), β, γ)-exponent of convergence of the zero-
sequence and denote F=f1·f2. Then
λ(α(log),β,γ)[F] = max{λ(α(log),β,γ)[f1], λ(α(log),β,γ)[f2]}.
In the line of Lemma 6 one can easily deduce
the conclusion of Lemma 7 and so its proof is omitted.
Lemma 8. Let fbe a transcendental meromorphic
function satisfying σ(α,β,γ)[f] = σ3and let k≥1be
an integer. Then, for any ε > 0, there exists a set E3
having finite linear measure such that for all r/∈E3,
we have
mr, f(k)
f=O(α−1((σ3+ε)β(log γ(r)))).
Proof. Set k= 1. Since σ(α,β,γ)[f] = σ3<+∞, for
sufficiently large rand for any given ε > 0, we have
T(r, f)<exp(α−1((σ3+ε)β(log γ(r)))).(23)
By the lemma of logarithmic derivative, we have
mr, f0
f=O(log r+log T(r, f)) (r/∈E3),
(24)
where E3⊂[0,+∞)is a set of finite linear mea-
sure, not necessarily the same at each occurrence.
By (23) and (24) and the condition α(log[2] x) =
o(β(log γ(x))) as x→+∞, we have
mr, f0
f=O(α−1((σ3+ε)β(log γ(r)))) (r/∈E3).
We assume that
mr, f(k)
f=O(α−1((σ3+ε)β(log γ(r)))) (r/∈E3)
(25)
holds for a certain integer k≥1. By N(r, f(k))≤
(k+ 1)N(r, f), for all r/∈E3, we have
T(r, f(k)) = m(r, f(k)) + N(r, f(k))
≤mr, f(k)
f+m(r, f)
+ (k+ 1)N(r, f)
≤(k+ 1)T(r, f)
+O(α−1((σ3+ε)β(log γ(r)))).
(26)
By (24) and (26), for r/∈E3, we obtain that
mr, f(k+1)
f(k)=mr, f(k)0
f(k)
=O(log r+log T(r, f(k)))
=O(α−1((σ3+ε)β(log γ(r)))).
(27)
Therefore, by (25) and (27), for r/∈E3, we have that
mr, f(k+1)
f≤mr, f(k+1)
f(k)+mr, f(k)
f
=O(α−1((σ3+ε)β(log γ(r)))).
Hence the lemma follows.
4 Proof of the Main Results
Proof of Theorem 1. Set σ(α,β,γ)[A] = σ4>0.
First, we prove that every solution of (1) satisfies
σ(α(log),β,γ)[f]≤σ4. If fis a polynomial solution
of (1), it is easy to show that σ(α(log),β,γ)[f] = 0 ≤σ4
holds. Let fbe a transcendental solution of (1). By
(1), we can write that
f00(z)
f(z)
=|A(z)|,
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so, by Lemma 1, there exists a set E1⊂(1,+∞)
having finite logarithmic measure such that for all z
satisfying |z|=r/∈[0,1]∪E1and |f(z)|=M(r, f),
we have
ν(r, f)
r2(1 + o(1))
≤exp[2] α−1σ4+ε
2β(log γ(r)),
and hence, we obtain for r/∈E1that
ν(r, f)≤rexp[2](α−1((σ4+ε)β(log γ(r)))). (28)
Therefore by (28) and Lemma 2, there exists some
η1>1such that for all r > r1, we have
ν(r, f)≤η1rexp[2](α−1((σ4+ε)β(log γ(η1r)))).
(29)
By (29), Lemma 5, and the conditions on α, β and γ,
we obtain that
σ(α(log),β,γ)[f] = lim sup
r→+∞
α(log[2] ν(r, f))
β(log γ(r)) ≤σ4.
(30)
On the other hand, since fis a transcendental, so by
(1), we get that
m(r, A) = mr, −f00
f=O(log rT (r, f))
=O(log r+log T(r, f)),(r/∈E3),
where E3⊂[0,+∞)is a set of finite linear measure.
By using Lemma 2, for any η2>1such that for all
r > r2,we get that
m(r, A) = mr, −f00
f≤K2(log η2r+log T(η2r, f)),
(31)
where K2>0is some constant. Since A(z)is an
entire function, so by (31) and using the inequality
log(x+y)≤log x+log y+log 2 (x, y ≥1), we
have
σ(α,β,γ)[A] = lim sup
r→+∞
α(log m(r, A))
β(log γ(r))
≤lim sup
r→+∞
α(log 2K2)
β(log γ(r))
+lim sup
r→+∞
α(log log η2r)
β(log γ(r))
+lim sup
r→+∞
α(log log T(η2r, f))
β(log γ(r))
+lim sup
r→+∞
c
β(log γ(r))
=lim sup
r→+∞
α(log log η2r)
β(log γ(r))
+lim sup
r→+∞
α(log log T(η2r, f))
β(log γ(r)) (c > 0) .
Since γ(η2r)≤γ(([η2] + 1) r)≤([η2] + 1) γ(r),
where [η2]is the integer part of the number η2,so
from the inequality above and (30), we get that
σ(α(log),β,γ)[f] = σ(α,β,γ)[A]holds for all non-trivial
solutions of (1).
Thus Theorem 1 follows.
Proof of Theorem 2. Set σ(α,β,γ)[A] = σ5>
0, by Theorem 1, we have σ(α(log),β,γ)[f1] =
σ(α(log),β,γ)[f2] = σ(α,β,γ)[A] = σ5. Hence, we get
λ(α(log),β,γ)[F]≤σ(α(log),β,γ)[F]
≤max{σ(α(log),β,γ)[f1], σ(α(log),β,γ)[f2]}
=σ(α,β,γ)[A].
(32)
By (32) and Lemma 7, we have
max{λ(α(log),β,γ)[f1], λ(α(log),β,γ)[f2]}=λ(α(log),β,γ)[F]
≤σ(α(log),β,γ)[F]
≤σ(α,β,γ)[A].
(33)
It remains to show that λ(α(log),β,γ)[F] =
σ(α(log),β,γ)[F].By (1), we have (see [18, pp.
76-77]) that all zeros of Fare simple and that
F2=C2F0
F2−2F00
F−4A−1, (34)
where C6= 0 is a constant. Hence,
2T(r, F ) = Tr, F0
F2−2F00
F−4A+O(1)
≤ONr, 1
F+mr, F0
F+mr, F00
F+m(r, A).
(35)
By σ(α(log),β,γ)[f] = σ(α,β,γ)[A] = σ5<+∞
and Lemma 8, for all r/∈E3, we have m(r, A) =
mr, f00
f=O(exp(α−1((σ5+ε)β(log γ(r))))),
mr, F0
F=O(exp(α−1((σ5+ε)β(log γ(r))))) and
mr, F00
F=O(exp(α−1((σ5+ε)β(log γ(r))))).
Therefore, by (35), for all r/∈E3, we obtain
T(r, F )
=ONr, 1
F+exp(α−1((σ5+ε)β(log γ(r)))).
(36)
Now let us assume that λ(α(log),β,γ)[F]<κ<
σ(α(log),β,γ)[F]. Since all zeros of Fare simple, we
have
Nr, 1
F=Nr, 1
F
=O(exp[2](α−1(κβ(log γ(r))))).
(37)
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Hence by (36) and (37), for all r/∈E3, we get that
T(r, F ) = O(exp[2](α−1(κβ(log γ(r))))).
By Definition 1 and Lemma 2, we have
σ(α(log),β,γ)[F]≤κ < σ(α(log),β,γ)[F], this is a
contradiction. Therefore, the first assertion is proved.
If σ(α(log),β,γ)[F]< σ(α,β,γ)[A], let us assume that
λ(α(log),β,γ)[f]< σ(α,β,γ)[A]holds for any solu-
tion of type f=c1f1+c2f2(c1c26= 0). We
denote F=f1·f2and F1=f·f1, then we have
λ(α(log),β,γ)[F]< σ(α,β,γ)[A]and λ(α(log),β,γ)[F1]<
σ(α,β,γ)[A]. Since (36) holds for Fand F1,where
F1=f·f1= (c1f1+c2f2)f1=c1f2
1+c2F, then
we get that
T(r, f1) = O(T(r, F1) + T(r, F ))
=ONr, 1
F1+Nr, 1
F
+exp(α−1((σ5+ε)β(log γ(r)))).
(38)
By λ(α(log),β,γ)[F]< σ(α,β,γ)[A],λ(α(log),β,γ)[F1]<
σ(α,β,γ)[A]and (37), for some κ < σ(α,β,γ)[A], we
obtain
T(r, f1) = O(exp[2](α−1(κβ(log γ(r))))). (39)
By Definition 1 and (39), we have σ(α(log),β,γ)[f1]≤
κ < σ(α,β,γ)[A], this is a contradiction
with Theorem 1. Therefore, we have that
λ(α(log),β,γ)[f] = σ(α,β,γ)[A]holds for all solu-
tions of type f=c1f1+c2f2, where c1c26= 0.
Hence the theorem follows.
Proof of Theorem 3. By Theorem 1 and
λ(α(log),β,γ)[f]≤σ(α(log),β,γ)[f], it is easy to show
that λ(α(log),β,γ)[f]≤σ(α,β,γ)[A]holds. It remains to
show that σ(α,β,γ)[A]≤λ(α,β,γ)[f]. Let us assume
that σ(α,β,γ)[A]> λ(α,β,γ)[f]. By (1) and a similar
proof of Theorem 5.6 in [18, pp. 82], we obtain
Tr, f
f0=ONr, 1
f+Nr, 1
A (r/∈E3).
(40)
By (40), the assumption σ(α,β,γ)[A]> λ(α,β,γ)[f]and
λ(α,β,γ)[A]< σ(α,β,γ)[A], for some κ < σ(α,β,γ)[A],
we obtain that
Tr, f
f0=O(exp(α−1(κβ(log γ(r))))). (41)
Further by Definition 1 and (41), we have
σ(α,β,γ)hf
f0i=σ(α,β,γ)hf0
fi≤κ < σ(α,β,γ)[A].
Therefore by
−A(z) = f0
f0+f0
f2,
we get that σ(α,β,γ)[A]≤σ(α,β,γ)hf0
fi< σ(α,β,γ)[A],
this is a contradiction. Hence, we have that
λ(α(log),β,γ)[f]≤σ(α,β,γ)[A]≤λ(α,β,γ)[f]holds for
all non-trivial solutions of (1).
The proof is complete.
5 Conclusion
Throughout this article, we have generalized some
previous results to general (α, β, γ)-order. Defining
new order of growth in the complex plane is discussed
and is applied to complex differential equations with
entire coefficients to solve some problems related to
growth of solutions. It is interesting now to study the
growth of solutions of complex differential equations
with meromorphic coefficients.
6 Acknowledgements
The authors are grateful to the referees for their many
valuable remarks and suggestions which lead to the
improvement of the original version of this paper.
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