230
ON THE HYPER-ORDER OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS WITH MEROMORPHIC COEFFICIENTS
Jianren Long
School of Mathematics and Computer Science, Guizhou Normal University, Guiyang, 550001, P.R.China ABSTRACT
In this paper, we investigate the growth of solutions of the linear differential equation
f A f Bf = 0
, where) (z
A
andB ( z )( 0)
are meromorphic functions. More specifically, we estimate the lower bounded of hyper- order of solutions of the equation with respect to the conditions ofA (z )
andB ( z )( 0)
if solutionsf ( 0
) ofthe equation is of infinite order.
keywords: Deficient value, Complex differential equations, Meromorphic function, Hyper-order.
2010 Mathematical Subject Classification: 34M10; 30D35.
1. INTRODUCTION AND MAIN RESULTS
We shall assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna theory of meromorphic functions (see [10], [14] or [17]). In addition, for a meromorphic function
f (z )
in thecomplex plane
, we shall use the notation ( f )
and ( f )
to denote its order and lower order respectively.In order to estimate the rate of growth of meromorphic function of infinite order more precisely, we recall the following definition.
Definition A.([19]) Let
f (z )
be a meromorphic function inC
, then we define the hyper-order
2( f )
off (z )
log .
) , log ( lim log
= )
2
(
r f r f
rT
Consider the second order linear differential equation
0,
= ) ( )
( z f B z f A
f
(1.1)where
A (z )
andB (z )
are meromorphic functions. It is well known that ifA (z )
is entire function andB (z )
is transcendental entire function, andf
1,f
2 are two linearly independent solutions of equation (1.1), then at least one off
1,f
2 must have infinite order. On the other hand, there are some equations of the form (1.1) that possess a solutionf ( 0)
of finite order; for example,f ( z ) = e
z satisfiesf e
zf ( e
z 1) f = 0
. Thus a natural question is: what conditions onA (z )
andB (z )
can guarantee that every solutionf ( 0)
of the equation (1.1) has infinite order? There are many work done on this subject. From the works of Gundersen (see [4]), Hellerstein, Miles and Rossi (see [8]), we know that ifA (z )
andB (z )
are entire functions with ( A ) < ( B )
; or ifA (z )
is a polynomial, and
B (z )
is transcendental; or if2 ) 1 (
<
)
( B A
, then every solutionf ( 0)
of theequation (1.1) has infinite order. More results can be found in [6], [9], [11] and [16]. Then for a great deal of solutions with infinite order, more precise estimates for their rate of growth is a very important aspect. There are many authors investigate the hyper-order
2( f )
of solutions of (1.1) (see, [2-3] and [12-13]). Ki-Ho Kwon investigate the problem and obtained the following result in [12].*The work is supported by the united technology Foundation of Science and Technology Department of Guizhou Province and Guizhou Normal University(Grant No.LKS[2012]12), and National Natural Science Foundation of China (Grant No.11171080).
231
Theorem B.([12]) Let
A (z )
andB (z )
be entire functions such that ( A ) < ( B )
or2
< 1 ) (
<
)
( B A
.Then every solution
f 0
of (1.1) satisfies
2( f ) max { ( A ), ( B )}
.It seems that there are few work done on the equation (1.1), where
A (z )
andB (z )
are meromorphic functions, but the growth of meromorphic solution of the equation (1.1) is a very important aspect (see [1-2]). It would be interesting to get some relations between the equation (1.1) and some deep results in value distribution theory of meromorphic functions. Thus, we shall introduce the deficient value into the studies of the equation (1.1). It is well- know that deficient value plays fundamental role in the theory of value distribution of meromorphic function and many important work done on this aspect (for example [18]). There are some results on the value distribution theory of the solutions of the equation (1.1) having connections with deficient values (see [6] or [11]). Furthermore, we mention that the author consider the equation (1.1) and obtain the result following.Theorem C.([15]) Let
A (z )
be an meromorphic function having a finite deficient value, and letB (z )
be a transcendental meromorphic function with2
< 1 )
(B
and ( , B ) = 1
. Then every solutionf ( 0)
ofequation
(1.1)
is of infinite order.We will consider the equation (1.1) again, and obtain some precise estimates of hyper-order of solutions, when the solutions
f ( 0
) of the equation (1.1) is of infinite order. Here we shall prove some results in which the deficient value is involved.Theorem 1.1. Let
A (z )
be an meromorphic function having a finite deficient value, and letB (z )
be a transcendental meromorphic function with2
< 1 )
(B
and ( , B ) = 1
. Then every solutionf ( 0)
ofequation
(1.1)
satisfies
2( f ) ( B )
.Theorem 1.2. Let
A (z )
be an meromorphic function having a finite deficient value, and letB (z )
be a transcendental meromorphic function with2
< 1 )
(B
and ( , B ) = 1
. Then every solutionf ( 0)
ofequation
(1.1)
satisfies
2( f ) ( B )
.Using the same method of the proof of Theorem 1.1, we can easily obtain the following result.
Theorem 1.3. Let
A (z )
be an meromorphic function having a finite deficient value, and letB ( z )( 0)
be an meromorphic function. Suppose that there exist two constants > 0
and > 0
, for any given > 0
, two finite set of real numbers{
k}
and{
k}
that satisfy
1<
1<
2<
2< <
m<
m<
m1(
m1=
1 2 )
and
,
<
) (
11
=
k km
k
(1.2)such that
}
|
| (1)) {(1
exp
| ) (
| B z o z
(1.3)as
z
in
k arg z
k( k = 1,2, , m )
. Then every solutionf ( 0)
of equation(1.1)
satisfies
2( f )
.The paper is organized as the following. In Section 2, we shall state and prove some lemmas related with our Theorems. In Section 3, we shall prove Theorem 1.1-1.3.
232 2. LEMMAS
For the proofs of our theorems, we need the following lemmas.
Lemma 2.1. (5) Let
( f , )
denote a pair that consists of a transcendental meromorphic functionf (z )
and a finite set.)}
, ( , ), , ( ), , {(
= k
1j
1k
2j
2 k
qj
q
of distinct pairs of integers that satisfies
k
i> j
i 0
fori = 1, , q
. Let > 1
and > 0
be given real constants. Then the following three statements hold.(i)There exists a set
E
1 [0,2 )
that has linear measure zero, and there exists a constantc > 0
that depend only on
and
, such that if
0 [0,2 ) E
1, then there is a constantR
0= R
0(
0) > 0
such that for allz
satisfying
arg z =
0 and| z |= r R
0, and for all( k , j )
, we have. )) , ( log log
) , ( ( ) | (
)
| (
) (
) (
j k j
k
f r T r r
f r c T z f
z
f
(2.1) (ii) There exists a set
E
2 (1, )
that has finite logarithmic measure, and there exists a constantc > 0
that depend only on
and
, such that for allz
satisfying| z |= r ( E
2 [0,1]) and for all ( k , j )
, the
inequality (2.1) holds.
(iii) There exiets a set
E
3 [0, )
that has finite linear measure, and there exists a constantc > 0
that depend only on
and
, such that for allz
satisfying| z |= r E
3 and for all( k , j )
, we have. )) , ( log ) , ( ( ) | (
)
| (
) (
) (
j k j
k
f r T r f r T z c
f z
f
(2.2)To state the following lemma, we define the upper and lower logarithmic measure of
E [1, )
respectively bylog , ]) [1, (
= lim
log r
r E E m
dens
lr
and
log , ]) [1, lim (
=
log r
r E E m
dens
lr
where
t r dt
E
m
l( [1, ]) = E [1,r]
.Lemma 2.2. (7) Suppose that
g (z )
is a transcendental and meromorphic in the complex plane, of lower order1
<
<
, and defineL ( r , g ) = min {| g ( z ) |:| z |= r }
and)}
, ( 1) ) , ( cos
(
>
) , ( log : 1
>
{
= r L r g g T r g
E
,
= sin
. ThenE
has upper logarithmic density at least 1 -
.Remark 1. In Lemma 2.2, if the order of
g (z )
is less than2
1
, then, for every ( ( g ),1)
, there exists a set[1, )
E
such that
( ) 1
log g
E
dens
, where233
)}
, ( 1) ) , ( cos
(
>
) , ( log : 1
>
{
= r L r g g T r g
E
,
= sin
.Lemma 2.3. (15) Let
g (z )
be a transcendental meromorphic function with2
< 1 ) (
<
0 g
and ( , g ) = 1
,and let
A (z )
be a meromorphic function with ( A ) <
. IfA (z )
has a finite deficient valuea
withdeficiency
= ( a , A )
, then for any given constant > 0
, there exists a sequence( R
n)
withR
n< R
n1 and
n
R
(n
), such that the following two inequalities), [0,2
}, {
exp
|>
) (
| g R
ne
iR
n(g)
0
>
)}
, 4 (
| ) (
| log : ) [0,2 {
=:
)
( F mes A R e a T R A d
mes
n
n i
n
hold for all sufficiently large
n
, whered
is a constant depending only on ( A ), ( g )
and
.Remark 2. If
g
is a transcendental meromorphic function with (g ) = 0
and ( , g ) = 1
, according to Lemma 2.2, we only need to give an appropriate modification and Lemma 2.3 still holds.Remark 3. If
g
is a meromorphic function with2
< 1 )
(g
and ( , g ) = 1
, according to Remark 1, we only need to give an appropriate modification for method of proof of Lemma 2.3 and Lemma 2.3 still holds.3. PROOF OF THEOREMS In this section,we prove our Theorems.
Proof of Theorem 1.1. From the equation (1.1), we have the following inequalities
. ) | (
)
|| ( ) (
| ) | (
)
| (
| ) (
|
' '
z f
z z f
z A f
z z f
B
(3.1)By using Lemma 2.1 that there exists a set
E
1 [0,2 )
that has linear measure zero, if
0 [0,2 ) E
1, then there is a constantR
0= R
0(
0) > 0
such that for allz
satisfyingarg z =
0 and| z |= r R
0, we have3 )
(
) , (2 ) |
( )
| ( T r f
z f
z
f
k
(3.2)holds for
k = 1
and 2.In the following, we treat two cases.
Case 1.
2
< 1 ) (
<
0 B
. Suppose thatA (z )
has a finite deficient valuea
with deficiency = ( a , A )
. Byusing Lemma 2.3, we have that for any given constant
> 0
, there exists a sequence( r
n)
withr
n< r
n1 and
n
r
(n
), such that for everyn
, we have0,
>
)}
, 4 (
| ) (
| log : ) [0,2 {
=:
)
( F mes A r e a T r A d
mes
n
n i
n
(3.3)
), [0,2
}, {
exp
|>
) (
| B r
ne
ir
n(B)
(3.4)where
d
is a constant depending only on ( A ), ( B )
and
.For every
n n
0, there exists a constantR
1> 0
, such thatr
n> R
1 (n n
0), we have (3.3) and (3.4) hold. Thus234
there must exist real numbers
0 ([0,2 ) F
n) E
1, andR
2> max { R
0, R
1}
, such that for allz
satisfying=
0arg z
and| z |= r R
2, we have (3.2), (3.3) and (3.4) hold.Hence, calculating at the points
z
n= r
ne
i0 withr
n R
2(n > n
0), we get from (3.1), (3.2), (3.3) and (3.4) that|).
| )) , 4 ( ( exp (1 ) , (2 ) (
exp r
n(B) T r
nf
3 T r
nA a
Thus
).
log (
) , ( log
lim log B
r f r T
r
(3.5)Therefore, from (3.5), we have
2( f ) ( B )
.Case 2:
(B ) = 0
. Obviously, we have
2( f ) ( B )
. The proof of Theorem is completed.Proof of Theorem 1.2. It follows from the Remark 3 and same method of proof of Theorem 1.1, we can easily obtain
2( f ) ( B )
, so, we omit the detail.Proof of Theorem 1.3. Suppose that
A (z )
has a finite deficient valuea
with deficiency = ( a , A )
. By using Lemma 2.1 that there exists a setE
2 [0,2 )
that has linear measure zero, if
0 [0,2 ) E
2, then there is a constantR
3= R
3(
0) > 0
such that for allz
satisfyingarg z =
0 and| z |= r R
3, we have (3.2) holds for1
=
k
and 2. From the proof of Lemma 2.3, there still exists a sequence( r
n)
that satisfies (3.3).Let
0 < < d 2
. Then for every positive integern
, we choose( [ , ])
1
= k k
m n k
n
F
. So there exists a
integer
k {1,2, , m }
such that for every integern
satisfies
n F
n [ k,
k]
(otherwise we use the
subsequence
nj
instead of
n). Thus form our hypothesis, we have}, (1)) {(1
exp
| ) (
| B r
ne
in o r
n (3.6)as
n
.Hence, calculating at the points
z
n= r
ne
in withr
n> R
3( n n
0)
, we get from (3.1), (3.2), (3.3) and (3.6) that|),
| )) , 4 ( ( exp (1 ) , (2 } (1)) {(1
exp o r
n T r
nf
3 T r
nA a
(3.7)holds for
n n
0. Thuslog .
) , ( log
lim
log
r f r T
r (3.8)
Therefore, from (3.8), we have
2( f )
. The proof of Theorem is completed.References
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