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Bull. Cal. Math. Soc., 111, (4) 331–340 (2019)
ON SOLUTIONS OF SECOND ORDER
COMPLEX DIFFERENTIAL EQUATIONS
DINESH KUMAR1, SANJAY KUMAR2AND MANISHA SAINI3
(Received 2 February 2019 and revision received 16 May 2019)
Abstract : In this paper, we establish transcendental entire function 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 order.
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)≡ 0 (1)
where A(z) and B(z) are entire functions. It is well known that all solutions of the
equation (1) consist of entire functions (Hille, 1969 and Laine, 1993). For an entire
function f, we define the order, lower order and exponent of convergence of zeros of f,
respectively, as 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
and λ(f) = lim sup
r→∞
log nr, 1
f
log r
where M(r, f ) = max{|f(z)|:|z|=r}is maximum modulus of fand nr, 1
fis the
number of zeros of fin the disk |z|≤r. Using Wiman -Valiron theory, it is proved
that the equation (1) has all solutions of finite order if and only if A(z) and B(z) are
polynomials (Laine, 1993). Therefore, if either A(z)orB(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 (1988) proved the following result :
331
332 dinesh kumar, sanjay kumar and manisha saini
THEOREM A. A necessary condition for the equation (1) to have a non-trivial solution
fof finite order is
ρ(B)≤ρ(A).(2)
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)=−1equation (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. (Heittokangas, et. al. 2015) 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)<1then ρ(f)is
infinite for all non-trivial solutions of the equation (1).
Frei (1962) 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 (1980) proved that the equation (3)
with B(z)=az+b, a = 0 has all solutions of infinite order. Amemiya and Ozawa (1981)
and Gundersen (1986) studied the equation (3) for B(z) being particular polynomial.
Langley (1986) proved that this is true for any non-constant polynomial. Gundersen
(1988) proved the following result :
THEOREM 1. Let fbe a non-trivial solution of the equation (1) where either
(1) ρ(B)<ρ(A)<1
2or
(2) A(z)is transcendental entire function with ρ(A)=0and B(z)is a polynomial
then ρ(f)is infinite.
Hellerstein, Miles and Rossi (1991) proved Theorem 1 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 deficient value, Borel direction and
function extremal for Yang’s inequality one may refer to (Yang, 1993).
on solutions of second order complex differential equations 333
In 2013, J.R. Long et. al. proved that if A(z) is an entire function extremal for
Yang’s inequality and B(z) a transcendental entire function with ρ(B)=ρ(A), then
all solution of the equation (1) are of infinite order.
In 2018, J.R. Long et. al. replaced the condition ρ(B)=ρ(A) with the condition
that B(z) is an entire function with Fabry gaps.
X.B. Wu et. al., (2015) proved that if A(z) is a non-trivial solution of w +Q(z)w=
0, where Q(z)=bmzm+... +b0,b
m= 0 and B(z) be an entire function with
μ(B)<1
2+1
2(m+1) , then all solutions of equation (1) are of infinite order. J. R. Long
(2018) 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)=ρ(A).
Furthermore, J.R. Long et. al., (2018) gave partial proof of the question by
Gundersen (2017) and proved the following theorem :
THEOREM B. Let A(z)=d(z)eP(z),where d(z)(≡ 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=2kn for all integers k;
(3) m+2=2nand a2
n
bmis not a negative real.
Motivated by Theorem B, we consider entire function A(z) and B(z) such that
ρ(A)>nand B(z) a polynomial. To state and prove our theorem we give some
definitions and notations below :
DEFINITION 1. (Long, Shi, Wu and Zhang, 2018) Let P(z)=anzn+...+a0,a
n=0
and δ(P, θ)=Re(aneιnθ).Arayγ=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 2ndistinct sectors of equal length π
n. Also δ(P, θ)>0in
nsectors and δ(P, θ)<0 in remaining nsectors. 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}
334 dinesh kumar, sanjay kumar and manisha saini
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 (Wu, Long, Heittokangas and Qiu, 2015).
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∈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)we have
lim
r→∞
log log
1
A(reιθ)
log r=ρ(A).
We illustrate these notions with an example :
EXAMPLE 4. 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
2and decays to zero exponentially
in Sπ
2,3π
2.
We are now able to state our main theorem :
THEOREM 2. 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 exponentially blows up in E+and E−respectively and B(z)be a
polynomial. Then all non-trivial solutions of the equation (1) are of infinite order.
We present here some examples to show that the hypothesis of the Theorem 2 can
not be relaxed :
EXAMPLE 5. The function f(z)=e−z−1satisfies f +ezf−f=0. The coefficient
A(z)satisfies λ(A)<ρ(A)and B(z)is a constant polynomial.
on solutions of second order complex differential equations 335
EXAMPLE 6. The differential equation f +(1+z+zez)f+zf =0possesses
finite order solution f(z)=e−z+1, here A(z)=1+z+zez=d(z)eP(z)satisfies
ρ(A)=λ(A)=1and λ(A)=ρ(d)≥n, where nis degree of P(z)and n=0or 1. The
fact that λ(A)=1follows from λ(z+zez)=1and nr, 1
1+z+zez=nr, 1
z+zez.
EXAMPLE 7. Suppose f +((z+ 1)(ez+1)+1)f+(z+1)f=0with coefficient
A(z)=(z+1)(ez+1)+1 = d(z)eP(z)satisfying λ(A)=ρ(A)=1and λ(A)=ρ(d)≥n,
where nis degree of P(z)which is a polynomial of at most degree 1. The finite order
entire function f(z)=e−z+1 satisfies given differential equation.
EXAMPLE 8. Suppose A(z)=d(z)eP(z)where d(z) = expz2+1 and eP(z)=ezwith
λ(A)=ρ(d)=2. The function A(z)satisfies that ρ(d)>nbut it is not exponentially
blows up in E−of ez. The differential equation
f +(ez2+1)ezf+(ez2+z+ez−1)f=0
possesses finite order solution f(z)=e−z.The coefficients A(z)and B(z)both fails to
satisfies the hypothesis of the theorem.
The paper is organised as follows: in Section 2, we have stated preliminary lemmas
and proved some required results. In Section 3, we have proved Theorem 2.
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. (Gundersen, 1988) Let fbe a transcendental meromorphic function with
finite order and (k, j)be a pair of integers that satisfies k>j≥0. Then for >0
there exists a set E⊂[0,2π]with linear measure zero such that for θ∈[0,2π)\Ethere
exists R(θ)>1such that
f(k)(z)
f(j)(z)
≤|z|(k−j)(ρ(f)−1+)(4)
for all |z|>R(θ)and arg z=θ.
LEMMA 2. (Gundersen, 1988) Let fbe an analytic function on a ray γ=reιθ and
suppose that for some constant α>1we have
f(z)
f(z)
=O(|z|−α)
336 dinesh kumar, sanjay kumar and manisha saini
as z→∞along arg z=θ. Then there exists a constant c=0 such that
f(z)→c
as z→∞along arg z=θ.
We now prove a result which would be required for proving Theorem 2.
LEMMA 3. Let A(z)=d(z)eP(z)be an entire function, where P(z)is a non-constant
polynomial of degree nand d(z)satisfies the condition of Theorem 2. Then there exists
a set E⊂[0,2π]of linear measure zero such that for >0the following holds :
(i) for θ∈E+\Ethere exists R(θ)>1such that
|A(reιθ)|≥exp {(1 −)δ(P, θ)rn}
for r>R(θ).
(ii) for θ∈E−\Ethere exists R(θ)>1such that
|A(reιθ)|≥exp {(1 −)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 Eis set of critical rays of eP(z),
which implies that Ehas linear measure zero. Let >0 then
(i) for δ(P, θ)>0,exp (−δ(P, θ)rn)→0asr→∞. Thus there exists R(θ)>1
such that
exp (−δ(P, θ)rn)≤|h(reιθ)|(5)
for r>R(θ)>1.Now
|exp (an(reιθ)n)|= exp (δ(P, θ)rn) (6)
Thus using equations (5) and (6) we have
|A(reιθ)|+|h(reιθ)|| exp (an(reιθ)n)|≥exp ((1 −)δ(P, θ)rn)
for θ∈E+\Eand r>R(θ).
on solutions of second order complex differential equations 337
(ii) Since d(z) blows up exponentially in E−therefore, h(z) also blows up exponen-
tially in E−. Let >0 and δ(P, θ)<0. Then ρ(exp (−anzn)) = n<ρ(h).
Thus, using definition 2, for any θ∈E+\Ethere exists R(θ)>1 such that
|h(reιθ)|≥exp (−δ(P, θ)rn) (7)
for r>R(θ). Using equations (6) and (7) we obtain that
|A(reιθ)|≥exp ((1 −)δ(P, θ)rn)
for r>R(θ) and θ∈E−\E.
3. Proof of Main Theorem. In this section we will establish Theorem 2 which
is the main result of this paper.
Proof: Let us suppose that 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 part (i) of Lemma 3 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 equations (1), (8) and (9), for θ∈E+\(E1∪E2) we 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)
338 dinesh kumar, sanjay kumar and manisha saini
as r→∞, for θ∈E+\(E1∪E2), where ais a non-zero finite constant. Applying
Maximum Modulus principle for the function fover the domain E+it can be concluded
that fis bounded over E+. Now using Phragm´en-Lindel¨of principle,
f(reιθ)→a(12)
as r→∞, for θ∈E+. From part (ii) of Lemma 3 implies that
|A(reιθ)|≥exp −1
2δ(P, θ)rn(13)
for θ∈E−\E2and for r>R
(θ). Using equations (1), (8) and (13) we have,
f(reιθ)
f(reιθ)
→0 (14)
as r→∞,for θ∈E−\(E1∪E2). From here we can obtain equation (10) for
θ∈E−\(E1∪E2). Again using Maximum Modulus principle for the 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.
In our discussion with Professor Gundersen, the following and crucial problem was
suggested by him to strengthen the result of the paper. Although we have shown that
hypothesis of the theorem are necessary but interestingly we could not find example of
function A(z) which satisfies the conditions of the theorem. So Gundersen’s question
is still valid.
PROBLEM 1. Construct the function A(z)=d(z)eP(z), where d(z)is an entire
function and P(z)is a non-constant polynomial, such that A(z)satisfies the conditions
of the Theorem 2.
All our work is directed to prove the following problem :
PROBLEM 2. If A(z)is a transcendental entire function such that λ(A)<ρ(A)and
if B(z)is a non-constant polynomial, then ρ(f)=∞for all non-trivial solutions fof
the equation (1)
on solutions of second order complex differential equations 339
Acknowledgment. We profusely thank the referee for asking pointed questions
whose resolution have enhanced the value of the paper. We are grateful to Professor
Gundersen for reading the paper and suggesting many things which have improved the
paper considerably . The third author is thankful to the department of mathematics,
Deen Dayal Upadhyaya College for providing the proper research facility.
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1DEPARTMENT OF MATHEMATICS
DEEN DAYAL UPADHYAYA COLLEGE
UNIVERSITY OF DELHI
NEW DELHI 110078
INDIA
E-MAIL : dinukumar680@gmail.com
2DEPARTMENT OF MATHEMATICS
DEEN DAYAL UPADHYAYA COLLEGE
UNIVERSITY OF DELHI
NEW DELHI 110078
INDIA
E-MAIL : sanjpant@gmail.com;skpant@ddu.du.ac.in
3DEPARTMENT OF MATHEMATICS
UNIVERSITY OF DELHI
NEW DELHI 110007
INDIA
E-MAIL : sainimanisha210@gmail.com
