where χt is the characteristic function of E defined asχ t ⎧ ⎨ ⎩ 1, if t ∈ E, Now let us recall some of the previous results on the linear differential equation where Bz is an entire fun
Trang 1Volume 2010, Article ID 428936, 8 pages
doi:10.1155/2010/428936
Research Article
On the Exponent of Convergence for the Zeros of
Abdullah Alotaibi
Department of Mathematics, King Abdulaziz University, P.O Box 80203, Jeddah 21589, Saudi Arabia
Correspondence should be addressed to Abdullah Alotaibi,mathker11@hotmail.com
Received 1 July 2010; Accepted 12 September 2010
Academic Editor: P J Y Wong
Copyrightq 2010 Abdullah Alotaibi This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
Let B and C be entire functions of order less than 1 with C / ≡ 0 and B transcendental We prove that every solution f / ≡ 0 of the equation y Ay By 0, Az Cze αz , α∈C \ {0} being has zeros with infinite exponent of convergence
1 Introduction
It is assumed that the reader of this paper is familiar with the basic concepts of Nevanlinna theory1,2 such as Tr, f, mr, f, Nr, f, and Sr, f Suppose that f is a meromorphic function, then the order of growth of the function f and the exponent of convergence of the zeros of f are defined, respectively, as
ρ
f
lim sup
r→ ∞
log T
r, f
log r , λ
f
lim sup
r→ ∞
log N
r, 1/f
Let E be a measurable subset of 1, ∞ The lower logarithmic density and the upper logarithmic density of E are defined, respectively, by
log densE lim inf
r→ ∞
r
1
χ tdt/t log r , log densE lim sup
r→ ∞
r
1
χ tdt/t log r , 1.2
Trang 2where χt is the characteristic function of E defined as
χ t
⎧
⎨
⎩
1, if t ∈ E,
Now let us recall some of the previous results on the linear differential equation
where Bz is an entire function of finite order, When Bz is polynomial, many authors 3 6 have studied the properties of the solutions of1.4 If Bz is a transcendental entire function with ρ B / 1, Gundersen 7 proved that every nontrivial solution of 1.4 has infinite order
of growth In8, Wang and Laine considered the nonhomogeneous equation of type
y A1ze az y A0ze bz y Hz, 1.5
where A0z, A1z, Hz are entire functions of order less than one and a, b are complex
numbers In fact, they have proved the following theorem
suppose that a, b ∈ C with ab / 0 and a / b Then every nontrivial solution of 1.5 is of infinite
order.
1 and b ∈ C with b / 0, −1 Then every nontrivial solution of 1.4 is of infinite order.
2 Results
We observe that all the above results concern the order of growth only In this paper, we are going to prove the following theorem which concerns the exponent of convergence
Theorem 2.1 Let B and C be entire functions of order less than 1 with C /≡ 0 and B being
transcendental Then every solution f / ≡ 0 of the equation
y Ay By 0,
has zeros with infinite exponent of convergence.
The hypothesis that B is transcendental is not redundant since Frei4 has shown that
has solutions of finite order, for certain constants K.
Trang 3We notice thatTheorem 2.1fails for ρB ≥ 1 For any entire function D the function
f e Dsolves2.1 with
−B f
f A f
3 Some Lemmas
Throughout this paper we need the following lemmas In 1965, Hayman 9 proved the following lemma
Lemma 3.1 Let the function g be meromorphic of finite order ρ in the plane and let 0 < δ < 1 Then
T
2r, g
≤ Cρ, δ
T
r, g
3.1
for all r outside a set E of upper logarithmic density δ, where the positive constant C ρ, δ depends
only on ρ and δ.
In 1962, Edrei and Fuchs10 proved the following lemma
Lemma 3.2 Let g be a meromorphic function in the complex plane and let I Ir ⊆ 0, 2π have
1
2π
I
log g re iθ dθ ≤ 22μ 1 log1
2r, g
In 2007, Bergweiler and Langley11 proved the following lemma
Lemma 3.3 Let H be a transcendental entire function of order ρ < ∞ For large r > 0 define θr
to be the length of the longest arc of the circle |z| r on which |Hz| > 1, with θr 2π if the
minimum modulus m0r, H min{|Hz| : |z| r} satisfies m0r, H > 1 Then at least one of the
following is true:
i there exists a set F ⊆ 1, ∞ of positive upper logarithmic density such that m0r, H > 1
for r ∈ F;
ii for each τ ∈ 0, 1 the set F r {r : θr > 2π1 − τ} has lower logarithmic density at
least 1 − 2ρ1 − τ/τ.
We deduce the following
density 1 Let F be a transcendental entire function such that |Fz| ≤ |z| N on a path γ tending to infinity and for all z with
that
H z F z − Pz
Trang 4is transcendental entire Then we have|Hz| ≤ 1 for all z ∈ γ and for all z with |z| ∈ G and
Lemma 3.3, we see that m0r, H ≤ 1 for all large r, and
so we must have case
Since G has logarithmic density 1 this gives
2ρ1 − τ ≥ 1, ρ ≥ 21 − τ1 π
Let A, B and C be as in the hypotheses We can assume that α 1 Suppose that f is a solution
of2.1 having zeros with finite exponent of convergence Then we can write
whereΠ and h are entire functions with ρΠ < ∞ We can assume that h/ ≡ 0, since if h is constant we can replace hz by hz z and Πz by Πze −z Using2.1 and 4.1, we get
Π
Π 2
Π
Πh h h
Π
Lemma 4.1 One has ρh ≤ 1.
Proof Suppose that |hz| ≥ 1 Dividing 4.2 by h, we get
hz ≤ Πz
Πz
2 ΠΠzz h hz z
|Az| ΠΠzz 1 |Bz|. 4.3
Hence, provided r lies outside a set of finite measure,
T
r, h
mr, h
≤ Olog r
Tr, A Tr, B oT
r, h
and so, using the fact that B and C have order less than 1, we obtain
T
r, h
This holds outside a set E0 of finite measure and so for all large r, since we may take s / ∈ E0
Trang 5with r ≤ s ≤ 2r so that
T
r, h
≤ Ts, h
Lemma 4.1is proved
Let M1, M2, denote large positive constants Choose σ with
max
ρ B, ρC< σ < 1. 4.7
There exists an R-set U2, page 84 such that for all large z outside U, we have
ΠΠzz ΠΠzz h hz z
and using the Poisson-Jensen formula,
Moreover, there exists a set G ⊆ 1, ∞ of logarithmic density 1 such that for r ∈ G the circle
|z| r does not meet the R-set U.
or h A is a polynomial Let z be large
B z O|z| M2
by arg z θ is bounded ApplyingLemma 3.4to the function B−z, with γ a subpath of L,
The next step is to estimate hin the right half-plane
arg z ≤ π
one has, either
or
hz Az ≤ |z| N 4.12
Trang 6Proof Let z be large and satisfy 4.10, and assume that 4.11 does not hold Then 4.8 implies that
ΠΠzz hz
Also,4.7, and 4.9 give
log|Bz| ≤ |z|σ , log|Az| ≥ Rez − |z|σ≥ |z|
2 cos 2
c1|z|. 4.14
4.12 and 4.14 we get, from 4.2,
log hz ≥ c2|z|. 4.15 Now divide4.2 by hz We obtain, using 4.15,
hz Az
⎛
⎜
⎝1 O |z|
M1
hz
⎞
⎟
⎠ O |z| M1
which gives|hz| ∼ |Az| and 4.12 This provesLemma 4.3
Lemma 4.3 Choose θ0∈ −π/4, π/4 such that the ray arg z θ0
has bounded intersection with the R-set U Let V be the union of the ray arg z θ0 and the arcs
4.9 Then one of
the following holds:
i one has 4.11 for all large z in V ;
ii one has 4.12 for all large z in V
Proof This follows simply from continuity For each large z in V we have4.11 or 4.12, but
we cannot have both because of4.14 This provesLemma 4.4
log hz O|z| σ
, log hz Az O|z| σ
Proof Let z be as in the hypotheses Since A z o1 we only need to prove 4.17 for |hz|.
Assume that|hz| ≥ 1 Then dividing 4.2 by hgives
4.18
by4.8, and so 4.17 follows using 4.7 This provesLemma 4.5
Trang 7Lemma 4.6 If conclusion (i) of Lemma 4.4 holds then ρ h < 1, while if conclusion (ii) of Lemma 4.4 holds then ρ h A < 1.
and let δ > 0 be small compared to δ1 Lemma 4.4is small compared to δ,
in particular so small that
1 log 1 C
ρ h, δ≤ 1
where Cρh, δ is the positive constant fromLemma 3.1 Let
I r π
2
π
2
∪
3π
2
3π
2
and let E be the exceptional set ofLemma 3.1, with g h Then for large r ∈ G \ E we have,
using4.20 and Lemmas3.1,3.2, and4.5,
T
r, h
mr, h
≤ Or σ Olog r
1
2π
I rlog h re iθ dθ
≤ Or σ
1 log 1 T
2r, h
≤ Or σ
1 log 1 C
ρ h, δT
r, h
≤ Or σ 1
2T
r, h
.
4.22
We then have
T
r, h
for large r ∈ G\E Now take any large r Since G has logarithmic density 1, while E has upper logarithmic density at most δ, and since δ/δ1is small, there exists s with
r ≤ s ≤ r1δ 1, s ∈ G \ E, Tr, h
≤ Ts, h
Os σ O r σ 1δ1
, 4.24
which provesLemma 4.6in this case The alternative case, in which we have conclusionii
inLemma 4.4, is proved the same way, using h A in place of h
Trang 8To finish the proof suppose first that conclusion ii of Lemma 4.4 holds Then
Lemma 3.4implies that hhas order at least π/
small, this contradictsLemma 4.6 The same contradiction, with hreplaced by h A, arises
if conclusioni ofLemma 4.4holds, and the proof of the theorem is complete
Acknowledgment
The author thanks Professor J K Langley for the invaluable discussions on the results of this paper during his visit in summer 2008 and summer 2010 to the University of Nottingham in the U.K
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Proceedings of the London Mathematical... this contradictsLemma 4.6 The same contradiction, with hreplaced by h A, arises
if conclusioni ofLemma 4.4holds, and the proof of the theorem... the ray arg z θ0
has bounded intersection with the R-set U Let V be the union of the ray arg z θ0 and the arcs
4.9 Then one