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arXiv:1810.02704v2 [math.CV] 8 Oct 2018
On Solution of Second Order Complex
Differential Equation
Dinesh Kumar, Sanjay Kumar and Manisha Saini
Abstract. In this paper, we establish transcendental entire func-
tion A(z) and polynomial B(z) such that the differential equation
f′′ +A(z)f′+B(z)f= 0, has all non-trivial solution of infinite or-
der. We use the notion of critical rays of the function eP(z), where
A(z) = d(z)eP(z)with some restrictions.
1. Introduction
Consider a second order linear differential equation of the form
f′′ +A(z)f′+B(z)f= 0; B(z)6≡ 0 (1)
where A(z) and B(z) are entire functions. It is well known that all
solutions of the equation (1) are entire functions [9], [11]. For an
entire function f, we define the order and lower order of fas follows
ρ(f) = lim sup
r→∞
log+log+M(r, f )
log r, µ(f) = lim inf
r→∞
log+log+M(r, f )
log r
where M(r, f) = max|z|=r|f(z)|is maximum modulus of f. Using
Wiman-Valiron theory, it is proved that the equation (1) has all solu-
tions of finite order if and only if A(z) and B(z) are polynomials [11].
Therefore, if either A(z) or B(z) is transcendental entire function then
almost all solutions of equation (1) are of infinite order. Therefore, it
is natural to find conditions on A(z) and B(z) such that all solutions
of the equation (1) are of infinite order. Gundersen [5] proved the
following result:
Theorem A. A necessary condition for equation (1) to have a
non-trivial solution fof finite order is
ρ(B)≤ρ(A).(2)
2010 Mathematics Subject Classification. 34M10, 30D35.
Key words and phrases. entire function, meromorphic function, order of
growth, complex differential equation.
The research work of the last author is supported by research fellowship from
University Grants Commission (UGC), New Delhi.
1
2 D. KUMAR, S. KUMAR AND M. SAINI
We illustrate it with some examples,
Example 1.f(z) = ezsatisfies f′′ +ezf′−(ez+ 1)f= 0,where
ρ(A) = ρ(B) = 1.
Example 2.With A(z) = ezand B(z) = −1 equation (1) has
finite order solution f(z) = 1 −e−z, where ρ(B)< ρ(A).
Thus if ρ(A)< ρ(B) then all solutions of the equation (1) are of
infinite order. However, given necessary condition is not sufficient, for
example
Example 3.[7] If A(z) = P(z)ez+Q(z)e−z+R(z), where P, Q, R
are polynomials and B(z) is an entire function with ρ(B)<1 then
ρ(f) is infinite; for all non-trivial solutions of the equation (1).
Frei [2] showed that the differential equation
f′′ +e−zf′+B(z)f= 0 (3)
with B(z) a polynomial, then the equation (3) has a non-trivial solution
fof finite order if and only if B(z) = −n2, n ∈N. Ozawa [16] proved
that the equation (3) with B(z) = az +b, a 6= 0 has all solution of
infinite order. Amemiya and Ozawa [1], and Gundersen [3] studied
the equation (3) for B(z) being a particular polynomial. Langley [15],
proved that this is true for any non-connstant polynomial. Gundersen
[5] proved the following result:
Theorem 4.Let fbe a non-trivial solution of the equation (1)
where either
(i) ρ(B)< ρ(A)<1
2
or
(ii) A(z) is transcendental entire function with ρ(A) = 0 and B(z) is
a polynomial
then ρ(f) is infinite.
Hellerstein, Miles and Rossi [8] proved Theorem 4 for ρ(B)<
ρ(A) = 1
2.
J.R. Long introduced the notion of the deficient value and Borel
direction into the studies of the equation (1). For the definition of defi-
cient value, Borel direction and function extremal for Yang’s inequality
one may refer to [18].
In [14], J.R .Long proved that if A(z) is an entire function extremal
for Yang’s inequality and B(z) a transcendental entire function with
ρ(B)6=ρ(A), then all solution of the equation (1) are of infinite order.
ON SOLUTION OF 3
In [12], J.R. Long replaced the condition ρ(B)6=ρ(A) with the condi-
tion that B(z) is an entire function with Fabry gaps.
X.B.Wu [17] proved that if A(z) is a non-trivial solution of w′′ +
Q(z)w= 0, where Q(z) = bmzm+... +b0, bm6= 0 and B(z) be
an entire function with µ(B)<1
2+1
2(m+1) , then all solutions of equa-
tion (1) are of infinite order. J.R. Long [12] replaced the condition
µ(B)<1
2+1
2(m+1) with B(z) being an entire function with Fabry gaps
such that ρ(B)6=ρ(A).
Furthermore, J.R. Long [13] proved the following Theorem:
Theorem B. Let A(z) = d(z)eP(z), where d(z)(6≡ 0) is an entire
function and P(z) = anzn+...+a0is a polynomial of degree nsuch that
ρ(d)< n. Let B(z) = bmzm+...+b0be a non-constant polynomial of
degree m, then all non-trivial solutions of the equation (1) have infinite
order if one of the following condition holds:
(1) m+ 2 <2n;
(2) m+ 2 >2nand m+ 2 6= 2kn for all integers k;
(3) m+ 2 = 2nand a2
n
bmis not a negative real.
Motivated by Theorem B, we consider entire functions A(z) and B(z)
such that ρ(A)> n and B(z) a polynomial. To state and prove our
theorem we give some definitions and notations below:
Definition 1.[13] Let P(z) = anzn+an−1zn−1+...+a0,an6= 0
and δ(P, θ) = Re(aneιnθ ). A ray γ=reιθ is called critical ray of eP(z)
if δ(P, θ) = 0.
It can be easily seen that there are 2ndifferent critical rays of eP(z)
which divides the whole complex plane into 2ndistinict sectors of equal
length π
n.Also δ(P, θ)>0 in nsectors and δ(P, θ)<0 in remaining n
sectors. We note that δ(P, θ) is alternatively positive and negative in
the 2nsectors.
We now fix some notations,
E+={θ∈[0,2π] : δ(P, θ)≥0}and E−={θ∈[0,2π] : δ(P, θ)≤
0}.
Let α > 0 and β > 0 be such that 0 ≤α < β ≤2πthen
S(α, β) = {z∈C:α < arg z < β}
We now recall the notion of an entire function to blow up and decay
to zero exponentially [17].
Definition 2.Let A(z) be an entire function with order ρ(A)∈
(0,∞). Then A(z)blows up exponentially in ¯
S(α, β) if for any θ∈
4 D. KUMAR, S. KUMAR AND M. SAINI
S(α, β) we get,
lim
r→∞
log log |A(reιθ )|
log r=ρ(A).
We say A(z)decays to zero exponentially in ¯
S(α, β) if for any θ∈
S(α, β)
lim
r→∞
log log
1
A(reιθ )
log r=ρ(A)
We illustrate these notions with an example
Example 5.The function f(z) = ezhas two critical rays namely
−π
2and π
2. It is easy to show that f(z)blows up exponentially in
¯
S(−π
2,π
2) and decays to zero exponential ly in ¯
S(π
2,3π
2).
We are now able to state our main theorem:
Theorem 6.Consider a transcendental entire function A(z) =
d(z)eP(z), where P(z) is a non-constant polynomial of degree nand
ρ(d)> n. Assume that d(z) is bounded away from zero and exponen-
tially blows up in E+and E−respectively and let B(z) be a polynomial.
Then all non-trivial solutions of the equation (1) are of infinite order.
The paper is organised as follows: in Section 2, we have stated pre-
liminary lemmas and proved some required results. In Section 3, we
have proved Theorem 6.
2. Auxiliary Result
In this section, we present some known results. Next two lemmas
are due to Gundersen which has been used extensively over the years.
Lemma 1.[4] Let fbe a trancendental meromorphic function with
finite order and (k, j) be a pair of integers that satisfies k > j ≥0. Then
for ǫ > 0 there exists a set E⊂[0,2π] with linear measure zero such
that for θ∈[0,2π)\Ethere exist R(θ)>1 such that
f(k)(z)
f(j)(z)
≤| z|(k−j)(ρ(f)−1+ǫ)(4)
for|z|> R(θ) and arg z=θ.
Lemma 2.[5] Let fbe analytic on a ray γ=reιθ and suppose that
for some constant α > 1 we have
f′(z)
f(z)
=O(|z|−α)
as z→ ∞ along arg z=θ. Then there exists a constant c6= 0 such
that f(z)→cas z→ ∞ along arg z=θ.
ON SOLUTION OF 5
We now prove a result which would be required for proving Theorem
6.
Lemma 3.Let A(z) = d(z)eP(z)be an entire function, where P(z)
is a polynomial of degree nand d(z) satisfies the condition of Theorem
6. Then there exists a set E⊂[0,2π] of linear measure zero such that
for ǫ > 0 the following holds:
(i) for θ∈E+\Ethere exists R(θ)>1 such that
|A(reιθ )|≥ exp (1 −ǫ)δ(P, θ)rn
for r > R(θ)
(ii) for θ∈E−\Ethere exists R(θ)>1 such that
|A(reιθ )|≥ exp (1 −ǫ)δ(P, θ)rn
for r > R(θ).
Proof. Here A(z) = d(z)eP(z)=h(z)eanzn, where h(z) is also an
entire function. Let E={θ∈[0,2π] : δ(P, θ) = 0}. This means that E
is set of critical rays of eP(z), which implies that Ehas linear measure
zero. Let ǫ > 0. Then
(i) for δ(P, θ)>0, exp(−ǫδ(P, θ)rn)→0 as r→ ∞. Thus there
exists R(θ)>1 such that
exp(−ǫδ(P, θ)rn)≤| h(reιθ )|(5)
for r > R(θ). Now
|exp(an(reιθ )n)|= exp(δ(P, θ)rn) (6)
Thus using (5) and (6) we have
|A(reιθ )|=|h(reιθ )|| exp(an(reιθ )n)|≥ exp (1 −ǫ)δ(P, θ)rn
for θ∈E+\Eand r > R(θ).
(ii) Since d(z) blows up exponentially in E−therefore, h(z) also blows
up exponentially in E−. Let ǫ > 0 and δ(P, θ)<0. Then
ρ(exp(−ǫanzn)) = n < ρ(h). Thus, using definition 2, for any
θ∈E−\Ethere exists R(θ)>1 such that
|h(reιθ )|≥ exp(−ǫδ(P, θ)rn) (7)
for r > R(θ). Using equation (6) and (7) we obtain that
|A(reιθ )|≥ exp (1 −ǫ)δ(P, θ)rn
for r > R(θ) and θ∈E−\E.
6 D. KUMAR, S. KUMAR AND M. SAINI
3. Proof of Main Theorem
In this section we will establish Theorem 6 which is the main result
of this paper.
Proof. If ρ(A) = ∞then it is obvious that ρ(f) = ∞, for all non-
trivial solution fof the equation (1).Therefore, let us suppose that
ρ(A)<∞and there exists a non-trivial solution fof the equation (1)
such that ρ(f)<∞. Then from Lemma 1, we have that there exist
E1⊂[0,2π] of linear measure zero and m > 0 such that,
f′′(reιθ)
f(reιθ )
≤rm(8)
for θ∈[0,2π]\E1and r > R(θ). From Lemma 3, part (i) we have,
|A(reιθ )|≥ exp 1
2δ(P, θ)rn(9)
for θ∈E+\E2and r > R′(θ) where E2is set of critical rays of
eP(z)of linear measure zero. Using equation (1), (8) and (9), for θ∈
E+\E1∪E2we get,
f′(reιθ )
f(reιθ )
→0
as r→ ∞. This implies that for θ∈E+\E1∪E2
f′(reιθ )
f(reιθ )
=O1
r2(10)
as r→ ∞. From Lemma 2,
f(reιθ )→a(11)
as r→ ∞, for θ∈E+\E1∪E2, where ais a non-zero finite con-
stant. Applying Maximum Modulus principle for the function fover
the domain E+this can be concluded that fis bounded over E+. Now
using Phragm´en- Lindel¨of principle,
f(reιθ )→a(12)
as r→ ∞, for θ∈E+.
Lemma 3, part (ii) implies that,
|A(reιθ )|≥ exp −1
2δ(P, θ)rn(13)
for θ∈E−\E2and for r > R′′ (θ). Using equation (1), (8) and (13)
we have,
f′(reιθ )
f(reιθ )
→0 (14)
as r→ ∞, for θ∈E−\E1∪E2. From here we can obtain equations
(10) for θ∈E−\E1∪E2. Again using Maximum Modulus principle
ON SOLUTION OF 7
for function fover the domain E−we get fis bounded over E−. Which
further using Phragm´en-Lindel¨of principle implies,
f(reιθ )→b(15)
as r→ ∞, for θ∈E−, where bis a non-zero finite constant. Again
using Phragm´en-Lindel¨of principle we get,
f(reιθ )→a(16)
as r→ ∞, for all θ∈E+∪E−, which is a contradiction to the
Liouville’s theorem.
Acknowlegement: We are thankful to Professor Gundersen for
reading the paper and suggesting many things. In fact, he has asked
to construct an example in support of the main theorem. We are still
working on construction of such type of example. We invite readers for
their comments. Paper is not for publication in any academic journal
as of now.
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Department of Mathematics, Deen Dayal Upadhyaya College, Uni-
versity of Delhi, New Delhi–110 078, India
E-mail address:dinukumar680@gmail.com
Department of Mathematics, Deen Dayal Upadhyaya College, Uni-
versity of Delhi, New Delhi–110 078, India
E-mail address:sanjpant@gmail.com
Department of Mathematics, University of Delhi, Delhi–110 007,
India
E-mail address:sainimanisha210@gmail.com
