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Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 143, pp. 1–15.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL
EQUATIONS WITH SPECIAL FUNCTIONS OR EXTREMAL
FUNCTIONS AS COEFFICIENTS
XIUBI WU, JIANREN LONG, JANNE HEITTOKANGAS, KE-E QIU
Abstract. The classical problem of finding conditions on the entire coeffi-
cients A(z) and B(z) guaranteeing that all nontrivial solutions of f00+A(z)f0+
B(z)f= 0 are of infinite order is discussed. Two distinct approaches are used.
In the first approach the coefficient A(z) itself is a solution of a differential
equation w00 +P(z)w= 0, where P(z) is a polynomial. This assumption
yields stability on the behavior of A(z) via Hille’s classical method on asymp-
totic integration. In this case A(z) is a special function of which the Airy
integral is one example. The second approach involves extremal functions. It
is assumed that either A(z) is extremal for Yang’s inequality or B(z) is ex-
tremal for Denjoy’s conjecture. A combination of these two approaches is also
discussed.
1. Introduction and main results
It is well known that if A(z) is an entire function, B(z)6≡ 0 is a transcendental
entire function, and f1, f2are two linearly independent solutions of the equation
f00 +A(z)f0+B(z)f= 0,(1.1)
then at least one of f1, f2must have infinite order. Hence, “most” solutions of
(1.1) have infinite order. On the other hand, there are equations of the form (1.1)
that possess a nontrivial solution of finite order; for example, f(z) = ezsatisfies
f00 +e−zf0−(e−z+ 1)f= 0. Thus a natural question is: What conditions on A(z)
and B(z) will guarantee that every nontrivial solution of (1.1) has infinite order?
We denote the order and the lower order of an entire function fby ρ(f) and
µ(f), respectively. The standard notation and basic results in Nevanlinna theory
of meromorphic functions can be found in [10, 14, 24].
From the work by Gundersen [8], Hellerstein, Miles and Rossi [11], and Ozawa
[18], we know that if A(z) and B(z) are entire functions with ρ(A)< ρ(B); or
A(z) is a polynomial and B(z) is transcendental; or ρ(B)< ρ(A)≤1
2, then every
nontrivial solution of (1.1) has infinite order. Therefore, the main problem left
to consider is that whether every nontrivial solution of (1.1) has infinite order if
2010 Mathematics Subject Classification. 34M10, 30D35.
Key words and phrases. Complex differential equation; entire function; infinite order;
Denjoy’s conjecture; Yang’s inequality.
c
2015 Texas State University - San Marcos.
Submitted November 17, 2014. Published May 22, 2015.
1
2 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
ρ(A) = ρ(B) or if ρ(A)>1
2,ρ(B)< ρ(A). In general, the conclusions are false for
these situations. For example, f(z) = exp(P(z)) satisfies the equation
f00 +A(z)f0+ (−P00 −(P0)2−A(z)P0)f= 0,(1.2)
where A(z) is an entire function and P(z) is a nonconstant polynomial. For the
case of ρ(B)< ρ(A), there are also some examples [8] showing that a nontrivial
solution of (1.1) has finite order.
The problem of finding conditions on A(z) and B(z) under which all nontrivial
solutions of (1.1) are of infinite order has raised considerable interest, see, for
example, [14]. Recently, this problem was studied by using a new idea that a
coefficient of (1.1) is a solution of another equation.
Theorem 1.1 ([21]).Let A(z)be a nontrivial solution of w00 +P(z)w= 0, where
P(z) = anzn+· ·· +a0, an6= 0. Let B(z)be a transcendental entire function with
ρ(B)<1
2. Then every nontrivial solution of (1.1) is of infinite order.
From Bank and Laine’s result [2, Theorem 1], we known that ρ(A) = n+2
2, and
hence ρ(A)> ρ(B) in Theorem 1.1. On the other hand, we known that every
nontrivial solution of (1.1) is of infinite order when ρ(A)< ρ(B) by Gundersen’s
result [8, Theorem 2]. The fact that A(z) solves an equation of the form w00 +
P(z)w= 0 makes A(z) a special function. In the particular cases when P(z) = −z
or P(z) = −zn, the solution A(z) is known as the Airy integral or a generalization of
the Airy integral [9]. Another special case is the Weber-Hermite function, which is a
solution in the case P(z) = ν+1
2−z2
4, where νis a constant. In the case when P(z)
is an arbitrary polynomial, Hille’s classical method on asymptotic integration will
become available. The consequences are summarized in Lemma 2.1 below. This
background motivated the second and the fourth author to prove the following
result.
Theorem 1.2 ([16]).Let A(z)be given as in Theorem 1.1, and let B(z)be a tran-
scendental entire function with µ(B)<1
2and ρ(A)6=ρ(B). Then every nontrivial
solution of (1.1) is of infinite order.
Theorem 1.2 is proved by using the cos πρ theorem due to Barry [3], which
does not work for entire functions with lower order (or order) not less than 1/2.
Thus we need new ideas when the lower order (or order) of the coefficients is not
less than 1/2. In the present paper, we will prove the following improvement of
Theorem 1.2 by using a modification of the Phragm´en-Lindel¨of principle, as well as
Hille’s classical results on asymptotic integration.
Theorem 1.3. Let A(z)be given as in Theorem 1.1, and let B(z)be a transcenden-
tal entire function with µ(B)<1
2+1
2(n+1) and ρ(A)6=ρ(B). Then every nontrivial
solution of (1.1) is of infinite order.
In 1988, Gundersen proved the following result.
Theorem 1.4 ([8]).Let {φk}and {θk}be two finite collections of real numbers
that satisfy φ1< θ1< φ2< θ2<··· < φn< θn< φn+1 where φn+1 =φ1+ 2π, and
set ν= max1≤k≤n{φk+1 −θk}. Suppose that A(z)and B(z)are entire functions
such that for some constant α > 0,
|A(z)|=O(|z|α) (1.3)
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 3
as z→ ∞ in φk≤arg z≤θkfor k= 1,2, . . . , n, and where B(z)is transcendental
with ρ(B)< π/ν. Then every nontrivial solution of (1.1) is of infinite order.
The usual order ρ(B) in Theorem 1.4 can be replaced with the lower order
µ(B)(≤ρ(B)).
Theorem 1.5. Let {φk},{θk},νand A(z)be given as in Theorem 1.4, and let
B(z)be transcendental with µ(B)< π/ν. Then every nontrivial solution of (1.1)
is of infinite order.
The proof of Theorem 1.5 deviates from that of Theorem 1.4 in the sense that we
require a modification of the Phragm´en-Lindel¨of principle, see Lemma 3.2 below.
In addition, we make use of the cos πρ theorem, which is not needed in proving
Theorem 1.4.
We proceed to consider conditions on the coefficients A(z) and B(z) involving
value distribution instead of just growth. We begin by recalling a conjecture due
to Denjoy [4] from 1907, verified by Ahlfors [1] in 1930.
Denjoy’s Conjecture. Let fbe an entire function of finite order ρ. If fhas k
distinct finite asymptotic values, then k≤2ρ.
An entire function fis called extremal for Denjoy’s conjecture if it is of finite
order ρand has k= 2ρdistinct finite asymptotic values. These functions are
investigated by Ahlfors [1], Drasin [5], Kennedy [13] and Zhang [27], to mention a
few. An example of a function extremal for Denjoy’s conjecture is
f(z) = Zz
0
sin tq
tqdt, (1.4)
where q > 0 is an integer. We know that the order of fequals to q, and fhas 2q
distinct finite asymptotic values
al=elπi
qZ∞
0
sin rq
rqdr,
where l= 1,2,...,2q, see [28, p. 210].
Theorem 1.6. Let A(z)be given as in Theorem 1.1, and let B(z)be a function
extremal for Denjoy’s conjecture and ρ(A)6=ρ(B). Then every nontrivial solution
of (1.1) is of infinite order.
We recall the definition of Borel direction as follows [25].
Definition 1.7. Let fbe a meromorphic function in the finite complex plane C
with 0 < µ(f)<∞. A ray arg z=θ∈[0,2π) from the origin is called a Borel
direction of order ≥µ(f) of f, if for any positive number ε > 0 and for any complex
number a∈CS{∞}, possibly with two exceptions, the following inequality holds
lim sup
r→∞
log n(S(θ−ε, θ +ε, r ), a, f)
log r≥µ(f),(1.5)
where n(S(θ−ε, θ +ε, r), a, f ) denotes the number of zeros, counting the multiplic-
ities, of f−ain the region S(θ−ε, θ +ε, r) = {z:θ−ε < arg z < θ +ε, |z|< r}.
The definition of Borel direction of order ρ(f) of fcan be found in [28, p. 78], it
is defined similarly with the only exception that “≥µ(f)” in (1.5) is to be replaced
with “= ρ(f)”.
In the sequel we will require the following result, known as Yang’s inequality, on
general value distribution theory.
4 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
Theorem 1.8 ([25]).Suppose that fis an entire function of finite lower order
µ > 0. Let q < ∞denote the number of Borel directions of order ≥µ, and let p
denote the number of finite deficient values of f. Then p≤q/2.
An entire function fis called extremal for Yang’s inequality if fsatisfies the
assumptions of Theorem 1.8 with p=q
2. These functions were introduced in [23].
The simplest entire function extremal for Yang’s inequality is ez. A slightly more
complicated example is f(z) = Rz
0e−tndt,n≥2 is teger, which has ndeficient
values
al=ei2πl
nZ∞
0
e−tndt, l = 1,2, . . . , n,
and q= 2nBorel directions arg z=2k−1
2nπ,k= 1,2,...,2n, see [24, pp. 210-211]
for more details.
Theorem 1.9 ([15]).Let A(z)be an entire function extremal for Yang’s inequality,
and let B(z)be a transcendental entire function such that ρ(B)6=ρ(A). Then every
nontrivial solution of (1.1) is of infinite order.
Also here the usual order ρ(B) can be replaced with the lower order µ(B).
Theorem 1.10. Let A(z)be an entire function extremal for Yang’s inequality, and
let B(z)be a transcendental entire function such that µ(B)6=ρ(A). Then every
nontrivial solution of (1.1) is of infinite order.
Let λ(A) be the converge exponent of the zero sequence of A(z). By Lemma 2.1
below and by similar reasoning used in proving Theorems 1.9 and 1.10, we can
easily obtain the following result.
Theorem 1.11. Let A(z)be given as in Theorem 1.1 with λ(A)< ρ(A), and let
B(z)be a transcendental entire function satisfying one of the following conditions.
(1) ρ(B)6=ρ(A),
(2) µ(B)6=ρ(A).
Then every nontrivial solution of (1.1) is of infinite order.
2. Auxiliary results
Let α < β be such that β−α < 2π, and let r > 0. Denote
S(α, β) = {z:α < arg z < β},
S(α, β, r) = {z:α < arg z < β}∩{z:|z|< r}.
Let Fdenote the closure of F. Let fbe an entire function of order ρ(f)∈(0,∞).
For simplicity, set ρ=ρ(f) and S=S(α, β). We say that fblows up exponentially
in Sif for any θ∈(α, β)
lim
r→∞
log log |f(reiθ )|
log r=ρ(2.1)
holds. We also say that fdecays to zero exponentially in Sif for any θ∈(α, β)
lim
r→∞
log log |f(reiθ )|−1
log r=ρ(2.2)
holds.
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 5
The following lemma, originally due to Hille [12, Chapter 7.4], see also [6, 19],
plays an important role in proving our results. The method used in proving the
lemma is typically referred to as the method of asymptotic integration.
Lemma 2.1. Let fbe a nontrivial solution of f00 +P(z)f= 0, where P(z) =
anzn+··· +a0,an6= 0. Set θj=2jπ−arg(an)
n+2 and Sj=S(θj, θj+1), where j=
0,1,2, . . . , n + 1 and θn+2 =θ0+ 2π. Then fhas the following properties.
(1) In each sector Sj,feither blows up or decays to zero exponentially.
(2) If, for some j,fdecays to zero in Sj, then it must blow up in Sj−1and
Sj+1. However, it is possible for fto blow up in many adjacent sectors.
(3) If fdecays to zero in Sj, then fhas at most finitely many zeros in any
closed sub-sector within Sj−1∪Sj∪Sj+1.
(4) If fblows up in Sj−1and Sj, then for each ε > 0,fhas infinitely many
zeros in each sector S(θj−ε, θj+ε), and furthermore, as r→ ∞,
n(S(θj−ε, θj+ε, r),0, f ) = (1 + o(1)) 2p|an|
π(n+ 2)rn+2
2,
where n(S(θj−ε, θj+ε, r),0, f)is the number of zeros of fin the region
S(θj−ε, θj+ε, r).
The Lebesgue linear measure of a set E⊂[0,∞) is m(E) = REdt, and the
logarithmic measure of a set F⊂[1,∞) is ml(F) = RF
dt
t. The upper and lower
logarithmic densities of F⊂[1,∞) are given, respectively, by
log dens(F) = lim sup
r→∞
ml(F∩[1, r])
log r,
log dens(F) = lim inf
r→∞
ml(F∩[1, r])
log r.
A lemma on logarithmic derivatives due to Gundersen [7] plays an important
role in proving our results.
Lemma 2.2. Let fbe a transcendental meromorphic function of finite order ρ(f).
Let ε > 0be a given real constant, and let kand jbe integers such that k >
j≥0. Then there exists a set E⊂[0,2π)of linear measure zero, such that if
ψ0∈[0,2π)−E, then there is a constant R0=R0(ψ0)>0such that for all z
satisfying arg z=ψ0and |z| ≥ R0, we have
f(k)(z)
f(j)(z)
≤ |z|(k−j)(ρ(f)−1+ε).(2.3)
The following result is due to Barry [3].
Lemma 2.3. Let fbe an entire function with 0≤µ(f)<1. Then, for every
α∈(µ(f),1),logdens({r∈[1,∞) : m(r)> M(r) cos πα})≥1−µ(f)
α, where
m(r) = inf|z|=rlog |f(z)|, and M(r) = sup|z|=rlog |f(z)|.
The following result was proved in [8].
Lemma 2.4. Let A(z)and B(z)6≡ 0be two entire functions such that for real
constants α, β, θ1, θ2, where α > 0,β > 0and θ1< θ2, we have
|A(z)| ≥ exp{(1 + o(1))α|z|β},(2.4)
6 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
|B(z)| ≤ exp{o(1)|z|β}(2.5)
as z→ ∞ in S(θ1, θ2) = {z:θ1≤arg z≤θ2}. Let ε > 0be a given small constant,
and let S(θ1+ε, θ2−ε) = {z:θ1+ε≤arg z≤θ2−ε}.
If fis a nontrivial solution of (1.1) with ρ(f)<∞, then the following conclu-
sions hold:
(1) There exists a constant b(6= 0) such that f(z)→bas z→ ∞ in S(θ1+
ε, θ2−ε). Furthermore,
|f(z)−b| ≤ exp{−(1 + o(1))α|z|β}(2.6)
as z→ ∞ in S(θ1+ε, θ2−ε).
(2) For each integer k > 1,
|f(k)(z)| ≤ exp{−(1 + o(1))α|z|β}
as z→ ∞ in S(θ1+ε, θ2−ε).
3. Modified Phragm´
en-Lindel¨
of principle
We recall a result due to Phragm´en and Lindel¨of [17, Theorem 7.5].
Lemma 3.1. Let fbe an analytic function in Dand continuous in D, where
D=S(α, β)∩ {z:|z|> r0}, and α, β , r0are constants such that 0< β −α≤2π
and r0>0. Suppose that there exists a constant M > 0such that |f(z)| ≤ Mfor
z∈∂D. If
lim inf
r→∞
log log M(r, D, f )
log r<π
β−α,
where M(r, D, f ) = max|z|=r
z∈D|f(z)|, then |f(z)| ≤ Mfor all z∈D.
Next we introduce a key lemma in which the Phragm´en-Lindel¨of principle is
tailored to suit for our purposes.
Lemma 3.2. Let fbe an entire function of lower order µ(f)∈[1
2,∞). Then there
exists a sector S(α, β) = {z:α < arg z < β}with β−α≥π
µ(f), such that
lim sup
r→∞
log log |f(reiθ )|
log r≥µ(f)
holds for all the rays arg z=θ∈(α, β), where 0≤α < β ≤2π.
Proof. Suppose on the contrary to the assertion that any sector S(α, β) with β−α≥
π
µ(f)there exists at least one ray arg z=ψ1∈(α, β ) such that
lim sup
r→∞
log log |f(reiψ1)|
log r=µ1< µ(f),
where µ1is a constant.
Let ψ0
1=ψ1+π
µ(f). From our assumption, there exists at least one ray arg z=
ψ2∈(ψ1, ψ0
1), such that
lim sup
r→∞
log log |f(reiψ2)|
log r=µ2< µ(f),(3.1)
where µ2is a constant. Let η1= max{µ1, µ2}, and let λ1∈(η1, µ(f)) ∩Qbe a
constant, where Qdenotes the set of rational numbers. Suppose that ω0=zλ1is
the principal branch of ω=zλ1. Then S(ψ1, ψ2) is mapped onto a subsector of
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 7
the right half-plane by transformation ζ=eiθ1zλ1, where θ1∈(0,2π) is a constant
depending on λ1, ψ1and ψ2.
Let H(z) = f(z)
exp(eiθ1zλ1). Then we obtain
lim
r→∞ |H(reiψ1)|= lim
r→∞ |H(reiψ2)|= 0 (3.2)
and
lim inf
r→∞
log log M(r, S(ψ1, ψ2), H )
log r≤lim inf
r→∞
log log M(r, S(ψ1, ψ2), f )
log r
≤lim inf
r→∞
log log M(r, f)
log r
=µ(f)<π
ψ2−ψ1
.
(3.3)
By Lemma 3.1, there exists a constant M1>0 such that
|H(z)| ≤ M1
holds for all z∈S(ψ1, ψ2); that is,
|f(z)| ≤ M1exp(|z|λ1) (3.4)
holds for all z∈S(ψ1, ψ2).
Let ψ0
2=ψ2+π
µ(f). From our assumption, there exists at least one ray arg z=
ψ3∈(ψ2, ψ0
2) such that
lim sup
r→∞
log log |f(reiψ3)|
log r=µ3< µ(f),
where µ3is a constant. Similarly as above, there exist constants M2>0, λ2< µ(f)
and θ2∈(0,2π) such that
|f(z)| ≤ M2exp(|z|λ2) (3.5)
holds for all z∈S(ψ2, ψ3). We proceed in this way until there exists a ray arg z=
ψmsuch that
lim sup
r→∞
log log |f(reiψm)|
log r=µm< µ(f) (3.6)
and ψ1+ 2π−ψm<π
µ(f), where µmis a constant. By the discussion above, there
exist constants Mm−1>0, λm−1< µ(f) and θm−1∈(0,2π) such that
|f(z)| ≤ Mm−1exp(|z|λm−1) (3.7)
holds for all z∈S(ψm−1, ψm). By (3.4), (3.5), (3.7) and Lemma 3.1, we have
|f(z)| ≤ Mexp(|z|λ) for all z∈C,(3.8)
where M= max{M1, M2, . . . , Mm−1}and λ= max{λ1, λ2, . . . , λm−1}< µ(f). By
(3.8), we obtain
µ(f) = lim inf
r→∞
log log M(r, f)
log r≤lim
r→∞
log log M+λlog r
log r=λ,
which is a contradiction with the fact λ<µ(f). This completes the proof.
8 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
4. Proofs of Theorems 1.3 and 1.5
We rely heavily on Phragm´en-Lindel¨of principle and modified Phragm´en-Lindel¨of
principle.
Proof of Theorem 1.3. Since the case ρ(A)< ρ(B) is proved in [8], we may assume
ρ(A)> ρ(B). Suppose on the contrary to the assertion that there exists a nontrivial
solution fof (1.1) with ρ(f)<∞. We aim for a contradiction. Set θj=2jπ−arg(an)
n+2
and Sj={z:θj<arg z < θj+1 }, where j= 0,1,2, . . . , n + 1 and θn+2 =θ0+ 2π.
We consider two cases appearing in Lemma 2.1.
Case 1: Suppose that A(z) blows up exponentially in each sector Sj, where j=
0,1, . . . , n + 1; that is, for any θ∈(θj, θj+1), we have
lim
r→∞
log log |A(reiθ )|
log r=ρ(A) = n+ 2
2.(4.1)
Then for any given constant ε∈(0,π
4ρ(A)) and η∈(0,ρ(A)−ρ(B)
4), we have
|A(z)| ≥ exp{(1 + o(1))α|z|n+2
2−η},(4.2)
|B(z)| ≤ exp(|z|ρ(B)+η)≤exp(|z|ρ(A)−2η)≤exp{o(1)|z|n+2
2−η}(4.3)
as z→ ∞ in Sj(ε) = {z:θj+ε < arg z < θj+1 −ε},j= 0,1, . . . , n + 1, where αis
a positive constant depending on ε. Combining (4.2), (4.3), and Lemma 2.4, there
exist corresponding constants bj6= 0 such that
|f(z)−bj| ≤ exp{−(1 + o(1))α|z|n+2
2−η}(4.4)
as z→ ∞ in Sj(2ε), j= 0,1, . . . , n + 1. Therefore, fis bounded in the whole
complex plane by the Phragm´en-Lindel¨of principle. So fis a nonzero constant in
the whole complex plane by Liouville’s theorem. This contradicts with the fact
that equation (1.1) doesn’t have nonzero constant solutions.
Case 2: There exists at least one sector of the n+ 2 sectors, such that A(z) decays
to zero exponentially, say Sj0={z:θj0<arg z < θj0+1}, 0 ≤j0≤n+ 1. That is,
for any θ∈(θj0, θj0+1), we have
lim
r→∞
log log 1
|A(reiθ )|
log r=n+ 2
2.(4.5)
If µ(B)<1
2, the assertion follows by Theorem 1.2.
If 1
2≤µ(B)<1
2+1
2(n+1) , then by Lemma 3.2, there exists a sector S(α, β) =
{z:α < arg z < β}with β−α≥π
µ(B)>π
1
2+1
2(n+1)
= 2π−2π
n+2 , such that
lim sup
r→∞
log log |B(reiθ )|
log r≥µ(B) (4.6)
holds for any θ∈(α, β). Thus, there exists a subsector S(α0, β0), such that for any
θ∈(α0, β0) we have (4.5) and (4.6).
By Lemma 2.2, there exists a set E1⊂[0,2π) of linear measure zero, such that
if ψ0∈[0,2π)−E1, then there is a constant R0=R0(ψ0)>1 such that for all z
satisfying arg z=ψ0and |z| ≥ R0, we have
f(k)(z)
f(z)
≤ |z|2ρ(f), k = 1,2.(4.7)
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 9
Thus, there exists a sequence of points zn=rneiθ with rn→ ∞ as n→ ∞ and
θ∈(α0, β0)−E1, such that (4.5), (4.6) and (4.7) hold.
Combining (4.5), (4.6), (4.7) and (1.1), for every n>n0, we have
exp(rµ(B)−ε
n)≤ |B(rneiθ)|
≤
f00(rneiθ )
f(rneiθ)
+|A(rneiθ)|
f0(rneiθ)
f(rneiθ)
≤r2ρ(f)
n(1 + o(1)).
(4.8)
Obviously, this is a contradiction for sufficiently large nand arbitrary small ε.
Therefore we have ρ(f) = ∞for every nontrivial solution fof (1.1). This completes
the proof.
Proof of Theorem 1.5. Assume on the contrary to the assertion that there is a non-
trivial solution fof (1.1) with ρ(f) = ρ < ∞.
Case 1: Suppose first that µ(B)>0. By Lemma 2.2, there exists a set E2⊂[0,2π)
of linear measure zero, such that if ϕ∈[φk, θk]−E2for some k, 1 ≤k≤n, we have
f0(reiϕ)
f(reiϕ)
=O(rρ),
f00(reiϕ)
f(reiϕ)
=O(r2ρ),(4.9)
as r→ ∞ along arg z=ϕ. Combining (4.9), (1.1), and our assumption, we have
|B(reiϕ)| ≤
f00(reiϕ)
f(reiϕ)
+|A(reiϕ)|
f0(reiϕ)
f(reiϕ)
=O(rσ) (4.10)
in each [φk, θk]−E2, 1 ≤k≤n, as r→ ∞, where σ=α+ 2ρ.
For any given ε∈(0,min{π
2µ(B)−ν
2,µ(B)
4}), for any ϕ0∈(θk−ε, θk)−E2,
ϕ00 ∈(φk+1, φk+1 +ε)−E2,k= 1,2, . . . , n, we obtain
|B(reiϕ0)|=O(rσ),|B(reiϕ00 )|=O(rσ),
as r→ ∞.
Similarly as in the proof of Lemma 3.2, let H(z) = B(z)
exp(azλ), where λ∈(0, µ(B)−
4ε)∩Q,Qdenotes the set of rational numbers, a=eiτ ,τ∈(0,2π) is a constant
depending on λ,ϕ0and ϕ00, and zλdenotes the principal branch. Note that
ϕ00 −ϕ0< φk+1 −θk+ 2ε<ν+ 2ε < π
µ(B)<π
λ.
So, S(ϕ0, ϕ00) is mapped onto a subsector of the right half-plane by ζ=azλ. Thus,
we obtain
lim
r→∞ |H(reiϕ0)|= lim
r→∞ |H(reiϕ00 )|= 0
and
lim inf
r→∞
log log M(r, S(ϕ0, ϕ00), H)
log r≤lim inf
r→∞
log log M(r, S(ϕ0, ϕ00), B)
log r
≤lim inf
r→∞
log log M(r, B)
log r
=µ(B)<π
ϕ00 −ϕ0.
By Lemma 3.1, there exists a constant M > 0 such that
|H(z)| ≤ M
10 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
for all z∈S(ϕ0, ϕ00); that is,
|B(z)| ≤ Mexp(|z|λ) (4.11)
for all z∈S(ϕ0, ϕ00). So we have (4.11) in each (θk−ε, φk+1 +ε)−E2, 1 ≤k≤n.
By (4.10), (4.11) and Phragm´en-Lindel¨of principle, we obtain
|B(z)| ≤ Mexp(|z|λ)
for all z∈C. Thus, we have
µ(B) = lim inf
r→∞
log log M(r, B)
log r≤lim
r→∞
log log M+λlog r
log r=λ,
which contradicts with the fact that λ<µ(B).
Case 2: Suppose that µ(B) = 0. By using Lemma 2.3, there exists a set E3⊂
[1,∞) with log dens(E3) = 1 such that for all zsatisfying |z|=r∈E3, we have
log |B(z)|>√2
2log M(r, B),(4.12)
where M(r, B) = max|z|=r|B(z)|.
It follows from (1.1), (1.3), (4.9) and (4.12) that there exists a sequence (Rn)
with Rn→ ∞ as n→ ∞, such that
M(Rn, B)√2/2<|B(Rneiϕ )| ≤ R2ρ(f)
n(1 + Rα
n),(4.13)
as n→ ∞,ϕ∈ ∪n
k=1[φk, θk]−E2. However, B(z) is a transcendental entire function,
so that
lim inf
r→∞
log M(r, B)
log r=∞,
which contradicts with (4.13). This completes the proof.
5. Proof of Theorem 1.6
We begin by recalling some properties satisfied by entire functions that are ex-
tremal for Denjoy’s conjecture.
Lemma 5.1 ([28, Theorem 4.11]).Let fbe an entire function extremal for Denjoy’s
conjecture. Then, for any θ∈(0,2π), either ∆(θ)is a Borel direction of order ρ(f)
of for there exists a constant σ(0 < σ < π
4), such that
lim
|z|→∞
z∈(S(θ−σ,θ+σ)−E)
log log |f(z)|
log r=ρ(f),
where ∆(θ)is a half-line from the origin, Edenotes a subset of S(θ−σ, θ +σ), and
satisfies
lim
r→∞ m(S(θ−σ, θ +σ;r, ∞)∩E)=0,
where S(θ−σ, θ +σ;r, ∞) = {z:θ−σ < arg z < θ +σ, r < |z|<∞}.
Lemma 5.2. Let fbe an entire function of order ρ∈(0,∞), and let S(φ1, φ2) =
{z:φ1<arg z < φ2}be a sector with φ2−φ1<π
ρ. If there exists a Borel direction
of order ρof fin S(φ1, φ2), then for at least one of the two rays Lj: arg z=φj(j=
1,2), say L2, we have
lim sup
r→∞
log log |f(reiφ2)|
log r=ρ.
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 11
Lemma 5.2 is [26, Lemma 1], which can be proved by using a result in [20, pp.
119-120].
We may assume ρ(A)> ρ(B) due to Gundersen’s result [8, Theorem 2]. If A(z)
blows up exponentially in each sector Sj,j= 0,1, . . . , n + 1, then the assertion
follows by the proof of Theorem 1.3. Suppose there exists at least one sector of the
n+ 2 sectors, such that A(z) decays to zero exponentially, say Sj0={z:θj0<
arg z < θj0+1 }, 0 ≤j0≤n+ 1. That is, for any θ∈(θj0, θj0+1 ), we have
lim
r→∞
log log 1
|A(reiθ )|
log r=n+ 2
2.
Suppose on the contrary to the assertion that there is a nontrivial solution f
of (1.1) with ρ(f)<∞. By Lemma 2.2, there exists a set E1⊂[0,2π) of linear
measure zero, such that if ψ0∈[0,2π)−E1, then there is a constant R0=R0(ψ0)>
1 such that for all zsatisfying arg z=ψ0and |z| ≥ R0, we have (4.7). Next we
consider the two cases appearing in Lemma 5.1.
Case 1: Suppose that the ray arg z=θis a Borel direction of order ρ(B) of B(z),
where θj0< θ < θj0+1. Choose φ1∈(θj0, θ)−E1and φ2∈(θ, θj0+1)−E1. Then
φ2−φ1<π
ρ(A)<π
ρ(B). By Lemma 5.2, at least one of two rays L1: arg z=φ1and
L2: arg z=φ2, say L1, satisfies
lim sup
r→∞
log log |B(reiφ1)|
log r=ρ(B).
Thus, there exists a sequence of points zn=rneiφ1with rn→ ∞ as n→ ∞, such
that
lim
n→∞
log log |B(rneiφ1)|
log rn
=ρ(B),(5.1)
lim
n→∞
log log 1
|A(rneiφ1)|
log rn
=n+ 2
2,(5.2)
f(k)(rneiφ1)
f(rneiφ1)
≤r2ρ(f)
n, k = 1,2.(5.3)
Combining (5.1)-(5.3) and (1.1), we arrive at a contradiction as in the proof of
Theorem 1.3. Thus, we have that every nontrivial solution fof (1.1) satisfies
ρ(f) = ∞.
Case 2: Suppose that the ray arg z=θis not a Borel direction of order ρ(B)
of B(z), where θj0< θ < θj0+1 . By Lemma 5.1, there exists a constant σ∈
0,min{θ−θj0
2,θj0+1−θ
2,π
4}, such that
lim
|z|→∞
z∈(S(θ−σ,θ+σ)−E2)
log log |B(z)|
log r=ρ(B),(5.4)
where E2denotes a subset of S(θ−σ, θ +σ), and satisfies
lim
r→∞ m(S(θ−σ, θ +σ;r, ∞)∩E2) = 0.
Let ∆ = {z: arg z=ψ, ψ ∈E1}. We can easily see that there exists a sequence of
points znwith zn→ ∞ as n→ ∞,{zn} ⊂ (S(θ−σ, θ +σ)−E2)∩(Sj0−∆), such
12 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
that
lim
n→∞
log log |B(zn)|
log |zn|=ρ(B),(5.5)
lim
n→∞
log log 1
|A(zn)|
log |zn|=n+ 2
2,(5.6)
f(k)(zn)
f(zn)
≤ |zn|2ρ(f), k = 1,2.(5.7)
Combining (5.5)-(5.7) and (1.1), we arrive at a contradiction as in the proof of
Theorem 1.3. Thus, we have ρ(f) = ∞for every nontrivial solution fof (1.1).
This completes the proof.
6. Proof of Theorem 1.10
We begin by recalling some basic properties satisfied by entire functions that are
extremal for Yang’s inequality. To this end, if Ais a function extremal for Yang’s
inequality, then the rays arg z=θk, denote the qdistinct Borel directions of order
≥µ(A) of A, where k= 1,2, . . . , q and 0 ≤θ1< θ2<· ·· < θq< θq+1 =θ1+ 2π.
Lemma 6.1 ([23]).Suppose that Ais a function extremal for Yang’s inequality.
Then µ(A) = ρ(A). Moreover, for every deficient value ai,i= 1,2, . . . , p, there
exists a corresponding sector domain S(θki, θki+1) = {z:θki<arg z < θki+1}such
that for every ε > 0the inequality
log 1
|A(z)−ai|> C(θki, θki+1 , ε, δ(ai, A))T(|z|, A) (6.1)
holds for z∈S(θki+ε, θki+1 −ε;r, +∞) = {z:θki+ε < arg z < θki+1 −ε, r <
|z|<∞}, where C(θki, θki+1, ε, δ(ai, A)) is a positive constant depending only on
θki, θki+1, ε and δ(ai, A).
In the sequel, we shall say that Adecays to aiexponentially in S(θki, θki+1),
if (6.1) holds in S(θki, θki+1 ). Note that if Ais a function extremal for Yang’s
inequality, then µ(A) = ρ(A). Thus, for these functions, we need only to consider
the Borel directions of order ρ(A).
Lemma 6.2 ([15]).Let Abe an entire function extremal for Yang’s inequality.
Suppose that there exists arg z=θwith θj< θ < θj+1,1≤j≤q, such that
lim sup
r→∞
log log |A(reiθ )|
log r=ρ(A).(6.2)
Then θj+1 −θj=π
ρ(A).
We state one more auxiliary result that covers one particular case of the proof
of Theorem 1.10.
Lemma 6.3 ([22]).Let A(z)be a finite order entire function having a finite defi-
cient value, and let B(z)be a transcendental entire function with µ(B)<1
2. Then
every nontrivial solution of (1.1) is of infinite order.
By [8, Theorem 2] and Lemma 6.3, we just need prove the case 1/2≤µ(B)<
ρ(A). Suppose on the contrary to the assertion that there is a nontrivial solution
fof (1.1) with ρ(f) = ρ < ∞. We aim for a contradiction.
EJDE-2015/143 SECOND-ORDER COMPLEX LINEAR DIFFERENTIAL EQUATIONS 13
Suppose that ai,i= 1,2, . . . , p, are all the finite deficient values of A(z). Thus
we have 2psectors Sj={z|θj<arg z < θj+1 },j= 1,2,...,2p, such that A(z) has
the following properties. In each sector Sj, either there exists some aisuch that
log 1
|A(z)−ai|> C(θj, θj+1 , ε, δ(ai, A))T(|z|, A) (6.3)
holds for z∈S(θj+ε, θj+1 −ε;r, +∞), where C(θj, θj+1, ε, δ(ai, A)) is a positive
constant depending only on θj, θj+1, ε and δ(ai, A), or there exists arg z=θ∈
(θj, θj+1) such that
lim sup
r→∞
log log |A(reiθ )|
log r=ρ(A).(6.4)
For the sake of simplicity, in the sequel we use Cto represent C(θj, θj+1, ε, δ(ai, A)).
Note that if there exists some aisuch that (6.3) holds in Sj, then there exists
arg z=θsuch that (6.4) holds in Sj−1and Sj+1. If there exists θ∈(θj, θj+1) such
that (6.4) holds, then there are ai(ai0) such that (6.3) holds in Sj−1and Sj+1,
respectively.
Without loss of generality, we assume that there is a ray argz=θin S1such
that (6.4) holds. Therefore, there exists a ray in each sector S3, S5, . . . , S2p−1, such
that (6.4) holds. By using Lemma 6.2, we know that all the sectors have the same
magnitude π
ρ(A).
Note that B(z) is an entire function of lower order 1/2≤µ(B)< ρ(A). By
Lemma 3.2 we see that there exists a sector S(α, β) with β−α≥π
µ(B), 0 ≤α <
β≤2π, such that for all the rays arg z=θ∈(α, β ) we have
lim sup
r→∞
log log |B(reiθ )|
log r≥µ(B).(6.5)
Note that ρ(A)> µ(B). It is not hard to see that there exists a sector S(α0, β0),
where α < α0< β0< β, such that there is an aj0such that
log 1
|A(reiθ )−aj0|> C T (r, A) (6.6)
holds for all θ∈[α0, β0]. By using Lemma 2.2, there exists a θ0∈[α0, β0] and R > 1
such that for k= 1,2,
f(k)(reiθ0)
f(reiθ0)
≤r2ρ(f)(6.7)
holds for all r > R. Note that (6.5) holds for θ=θ0. Thus there is a sequence (rn)
with rn→ ∞ as n→ ∞, such that
|B(rneiθ0)| ≥ exp(rµ(B)−ε
n) (6.8)
holds for every ε∈(0, µ(B)). Therefore, we deduce from (6.6)-(6.8) that
exp(rµ(B)−ε
n)≤ |B(rneiθ0)|
≤
f00 (rneiθ0)
f(rneiθ0)
+
f0(rneiθ0)
f(rneiθ0)
(|A(rneiθ0)−aj0|+|aj0|)
≤r2ρ(f)
n(1 + |aj0|+ exp(−CT (rn, A)))
holds for all sufficiently large n. Obviously, this is a contradiction. Hence Theo-
rem 1.10 holds.
14 X. WU, J. LONG, J. HEITTOKANGAS, K. QIU EJDE-2015/143
Acknowledgements. This research work is supported in part by the Academy
of Finland, project ]268009, and the Centre for International Mobility of Finland
(CIMO) (Grant TM-13-8775).
The authors want to thank the anonymous reviewer for his/her valuable sugges-
tions.
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Xiubi Wu
School of Science, Guizhou University, Guiyang 550025, China
E-mail address:basicmath@163.com
Jianren Long (corresponding author)
School of Mathematics and Computer Science, Guizhou Normal University, Guiyang
550001, China.
Department of Physics and Mathematics, University of Eastern Finland, P.O. Box 111,
80101 Joensuu, Finland
E-mail address:longjianren2004@163.com, jianren.long@uef.fi
Janne Heittokangas
Department of Physics and Mathematics, University of Eastern Finland, P.O. Box 111,
80101 Joensuu, Finland
E-mail address:janne.heittokangas@uef.fi
Ke-e Qiu
School of Mathematics and Computer Science, Guizhou Normal Colleage, Guiyang
550018, China
E-mail address:qke456@sina.com
