# Study of Growth of Certain Second Order Linear Differential Equations

Naveen Mehra, Garima Pant and S. K. Chanyal  Naveen Mehra, Department of
Mathematics, Kumaun University, D.S.B. Campus, Nainital-263001, Uttarakhand,
India<EMAIL_ADDRESS>Garima Pant, Department of Mathematics,
University of Delhi, Delhi-110007, India<EMAIL_ADDRESS>S. K.
Chanyal, Department of Mathematics, Kumaun Univesity, D.S.B. Campus,
Nainital-263001, Uttarakhand, India<EMAIL_ADDRESS>
###### Abstract.

In this article, we study about the solutions of second order linear
differential equations by considering several conditions on the coefficients
of homogenous linear differential equation and its associated non-homogenous
linear differential equation.

###### Key words and phrases:

entire function, order of growth, homogenous linear differential equation and
non-homogenous linear differential equation

###### 2020 Mathematics Subject Classification:

34M10, 30D35

##  1\. Introduction

Consider homogenous linear complex differential equation

$f^{\prime\prime}+A(z)f^{\prime}+B(z)f=0,$ (1)

where $A(z)$ and $B(z)\not\equiv 0$ are entire functions. All the solutions of
equation (1) are of finite order if and only if the coefficient $A_{(}z)$ and
$B(z)$ are polynomials (see [24]). It is a natural question to arise that what
happens when atleast one of the coefficient is transcendental entire function.
M. Frei [2] answered to this question. He proved that atmost all non-trivial
solution of equation (1) are of infinite order.

The main aim of this work is to find conditions on entire coefficients $A(z)$
and $B(z),$ so that all non-trivial solutions of equation (1) are of infinite
order. Many researchers studied this problem earlier. Gundersen [5] proved
that if $\rho(A)<\rho(B)$, then all non-trivial solutions of equation (1) are
of infinite order. It is clear that if $A(z)$ is a polynomial and $B(z)$ is a
transcendental entire function, then all non-trivial solutions are of infinite
order. But, the case $\rho(A)\geq\rho(B)$ was unexplored until the paper by
Ozawa [19]. After Ozawa’s paper, many other researchers studied the same case
partially. The following result is the collection of those results.

######  Theorem A.

All non-trivial solution of equation (1) are of infinite order if the
coefficients $A(z)$ and $B(z)$ satisfy any of the following conditions

  1. (a)

[5] $\rho(A)<\rho(B)$;

  2. (b)

[7] $\rho(B)<\rho(A)\leq\frac{1}{2}$;

  3. (c)

[5] $A(z)$ is a transcendental entire function with $\rho(A)=0$ and $B(z)$ is
a polynomial;

  4. (d)

[5] $A(z)$ is a polynomial and $B(z)$ is a transcendental entire function.

######  Example 1.

  1. (i)

$f^{\prime\prime}+e^{z}f^{\prime}+(e^{z}-1)f=0$ has solution $f(z)=e^{z}.$

  2. (ii)

$f^{\prime\prime}+(\sin^{2}z-2\tan z)f^{\prime}-\tan zf=0$ has solution
$f(z)=\tan z$.

It can be observed from the above examples that differential equations have
finite order solutions when $\rho(A)=\rho(B)$ or $\rho(A)>\rho(B)$ and
$\rho(A)>1/2.$ There arises a question that under what circumstances equation
(1) possesses all non-trivial solutions of infinite order with these
conditions. In the next section we partially answer this question.

The corresponding non-homogenous second order linear differential equation is

$f^{\prime\prime}+A(z)f^{\prime}+B(z)f=H(z),$ (2)

where $A(z)$, $B(z)$ and $H(z)$ are entire functions. A non-homogenous linear
differential equation can always be reduced back to homogenous linear
differential equation. So, the basic results are similar to as in homogenous
case. If all the coefficients and $H(z)$ are entire functions, then all the
solutions of equation (2) are also entire functions (see [21]). If all the
coefficients are polynomials and $H(z)\neq 0$ have finite order of growth,
then all solutions of equation (2) are of finite order of growth (see [3,
Lemma 2]). Therefore, if atleast one of the coefficient is a transcendental
entire function then atmost all solutions are of infinite order. Let $\rho$ be
the minimal order of solutions of equation (1), then it is completely
elementary that there may exist atmost one solution of order less than $\rho$
of equation (2) (see [13]). Thus, if all non-trivial solutions of equation (1)
are of infinite order, there may exist finite order solution of equation (2).
We illustrate this fact by the following examples.

######  Example 2.

The equation

$f^{\prime\prime}+zf^{\prime}+e^{z}f=e^{-z}(1-z)+1$

has a finite order solution, that is, $f(z)=e^{-z},$ whereas the associated
homogenous equation has all non-trivial solutions of infinite order.

######  Example 3.

Let $b(z)$ be a finite order entire function and has multiply connected Fatou
component. Then, the equation

$f^{\prime\prime}-e^{z}f^{\prime}+b(z)f=0,$

has all non-trivial solutions of infinite order (see [18, Theorem B]). But the
associated non-homogenous equation

$f^{\prime\prime}-e^{z}f^{\prime}+b(z)f=e^{-z}(1+b(z))+1$

has finite order solution $f(z)=e^{-z}.$

##  2\. Results

###  2.1. Second Order Homogenous Linear Differential Eqaution

G. Zhang [25] in his paper investigated the solutions of equation (1), when
$A(z)=e^{P(z)}$ such that $P(z)$ and $B(z)$ are poynomials of degree $m$ and
$n$ respectively.

######  Theorem B.

[25] Suppose $A(z)=e^{p(z)}$, where $p(z)$ is a polynomial with degree $n\geq
2$ and $B(z)=Q(z)$ is also a nonconstant polynomial with degree $m$. If
$m+2>2n$ and $n\nmid m+2$. Then, every solution $f(\not\equiv 0)$ of equation

$f^{\prime\prime}+e^{p(z)}f^{\prime}+Q(z)f=0$

is of infinite order.

In our first main result, we replace $A(z)=e^{P(z)}$ by $A(z)=h(z)e^{P(z)}$ in
Theorem B, where $\rho(h)<n$.

######  Theorem 1.

Suppose $A(z)=h(z)e^{p(z)}$, where $p(z)$ is a polynomial with degree $n\geq
2$ and $B(z)=Q(z)$ is also a nonconstant polynomial with degree $m$. If
$m+2>2n$ and $n\nmid m+2$. Then, every solution $f(\not\equiv 0)$ of equation

$f^{\prime\prime}+h(z)e^{p(z)}f^{\prime}+Q(z)f=0$ (3)

is of infinite order.

Following Lemma is due to Bank, et al.[1] that gives an estimate for an entire
function with an integral order and the asymptotic properties on most rays of
the function $h(z)e^{P(z)}$.

######  Lemma 1.

[1] Let $A(z)=h(z)e^{P(z)}$ be an entire function with $\lambda(A)<\rho(A)=n$,
where $P(z)$ is a polynomial of degree $n$. Then, for every $\epsilon>0,$
there exists $E\subset[0,2\pi)$ of linear measure zero satisfying

  1. (i)

for $\theta\in[0,2\pi)\setminus E$ with $\delta(P,\theta)>0$, there exists
$R>1$ such that

$exp((1-\epsilon)\delta(P,\theta)r^{n})\leq|A(re^{\iota\theta})|$

for $r>R$;

  2. (ii)

for $\theta\in[0,2\pi)\setminus E$ with $\delta(P,\theta)<0$, there exists
$R>1$ such that

$|A(re^{\iota\theta})|\leq exp((1-\epsilon)\delta(P,\theta)r^{n})$

for $r>R$.

The following Lemma is given by Langley[15].

######  Lemma 2.

[15] Let $S$ be the strip

$z=x+\iota y,\ \ x\geq x_{0},\ \ |y|\leq 4.$

Suppose that in $S$

$Q(z)=a_{n}z^{n}+O(|z|^{n-2}),$

where $n$ is positive integer and $a_{n}>0.$ Then, there exists a path
$\Gamma$ tending to $\infty$ in $S$ such that all solutions of

$y^{\prime\prime}+Q(z)y=0$

tend to zero on $\Gamma.$

The following Lemma gives the logarithmic estimate of a meromorphic function
outside an $R$-set.

######  Lemma 3.

[14] Let $f$ be a meromorphic function of finite order. Then, there exists
$N=N(f)>0$ such that

$\left|\frac{f^{\prime}(z)}{f(z)}\right|=O(r^{N})$

holds outside an $R$-set.

The growth estimate in the following Lemma deduced in [1] from Herold
Comparison Theorem [8].

######  Lemma 4.

[15] Suppose that $A(z)$ is an analytic in a sector containing the ray
$\Gamma:re^{\iota\theta}$ and that as $r\to\infty$,
$A(re^{\iota\theta})=O(r^{n})$ for some $n\geq 0.$ Then, all solutions of
$y^{\prime\prime}+A(z)y=0$ satisfy

$\log^{+}|y(re^{\iota\theta})|=O(r^{\frac{(n+2)}{2}})$

on $\Gamma$.

######  Remark 1.

If $f(z)\to a$ as $z\to\infty$ along a straight line, $f(z)\to b$ as
$z\to\infty$ along another straight line and $f(z)$ is analytic and bounded in
the angle between, then $a=b$ and $f(z)\to a$ uniformly in the angle. The
straight lines may be replaced by the curves approaching $\infty$.

######  Proof of Theorem 1.

We assume that (3) has a solution $f(z)$ with finite order. Set

$f=y\exp\\{-\frac{1}{2}\int_{0}^{z}h(z)e^{p(z)}dz\\}.$ (4)

Equation (3) can be transformed into

$y^{\prime\prime}+\left(Q(z)-\frac{1}{4}(he^{p(z)})^{2}-\frac{1}{2}h^{\prime}(z)e^{p(z)}-\frac{1}{2}h(z)p^{\prime}(z)e^{p(z)}\right)y=0.$
(5)

By a translation, we may assume that

$Q(z)=a_{m}z^{m}+a_{m-2}z^{m-2}+\cdots,\ \ m>2.$

We define the critical ray for $Q(z)$ as those ray $re^{\iota\theta_{j}}$ for
which

$\theta_{j}=\frac{-\arg a_{m}+2j\pi}{m+2},$

where $j=0,1,2,\ldots,m+1$ and note that the substitution
$z=xe^{\iota\theta_{j}}$ transforms equation (5) into

$\frac{d^{2}y}{dx^{x}}+(Q_{1}(x)+P_{1}(x))y=0,$ (6)

where

$Q_{1}(x)=\alpha_{1}x^{m}+O(x^{m-2}),\alpha_{1}>0$

and

$P_{1}(x)=-\frac{1}{4}(he^{p(xe^{\iota\theta_{j}})})^{2}-\frac{1}{2}h^{\prime}(xe^{\iota\theta_{j}})e^{p(xe^{\iota\theta_{j}})}-\frac{1}{2}h(xe^{\iota\theta_{j}})p^{\prime}(xe^{\iota\theta_{j}})e^{p(xe^{\iota\theta_{j}})}.$

For the polynomial $p(z)$ with degree $n$, set
$p(z)=(\alpha+\iota\beta)z^{n}+p_{n-1}(z)$ with $\alpha,\ \beta$ real, and
denote $\delta(p,\theta)=\alpha\cos n\theta-\beta\sin n\theta.$ The rays

$\arg z=\theta_{k}=\frac{arc\tan\frac{\alpha}{\beta}+k\pi}{n},\\\
k=0,1,2,\ldots,2n-1$

satisfying $\delta(p,\theta_{k})=0$ can split the complex domain into $2n$
equal angular domains. Without loss of generality, denote these angle domains
as

$\omega^{+}=\\{re^{\iota\theta}:0<r<+\infty,\frac{2i\pi}{n}<\theta<\frac{(2i+1)}{n}\\},$
$\omega^{-}=\\{re^{\iota\theta}:0<r<+\infty,\frac{(2i+1)}{n}<\theta<\frac{2(i+1)}{n}\\},$

$\iota=0,1,\ldots,n-1,$ where $\delta(p,\theta)>0$ on $\omega^{+}$ and
$\delta(p,\theta)<0$ on $\omega^{-}$. By Lemma 1, we obtain

$\displaystyle|P_{1}(x)|\leq$
$\displaystyle|(h(xe^{\iota\theta_{j}})e^{p(xe^{\iota\theta_{j}})})^{2}|+|h^{\prime}(xe^{\iota\theta_{j}})e^{p(xe^{\iota\theta_{j}})}|+|h(xe^{\iota\theta_{j}})e^{p(xe^{\iota\theta_{j}})}p^{\prime}(xe^{\iota\theta_{j}})|$
$\displaystyle\leq\exp\\{\delta(P,\theta)x^{n}\\}+\exp\\{\frac{1}{2}\delta(P,\theta)x^{n}\\}+\exp\\{\frac{1}{2}\delta(P,\theta)x^{n}\\}O(x^{n-1})\to
0$

for $xe^{\iota\theta_{j}}\in\omega^{-}$ as $x\to\infty$, then by Lemma 2 and
(6), for any critical line $\arg z=\theta_{j}$ lying in $\omega^{-}$ there
exists a path $\Gamma_{\theta_{j}}$ tending to $\infty$, such that $\arg
z\to\theta_{j}$ on $\Gamma_{\theta_{j}}$ while $y(z)\to 0$ there. Moreover, by

$\displaystyle|\exp\\{-\frac{1}{2}\int_{0}^{z}h(z)e^{p(z)}\\}|$
$\displaystyle\leq\exp\\{\frac{1}{2}\left|\int_{0}^{z}h(z)e^{p(z)}\right|\\}$
(7) $\displaystyle\leq\exp\\{\frac{1}{2}r\exp\\{\delta(p,\theta)r^{n}\\}\\}\to
1$

for $z>\omega^{-}$ as $r\to\infty$, together with (4) we have $f(z)\to 0$
along $\Gamma_{\theta_{j}}$ tending to $\infty$. Setting
$V=\frac{f^{\prime}}{f}$ , equation (3) can be written as

$V^{\prime}+V^{2}+h(z)e^{p(z)}V+Q(z)=0.$

By Lemma 3, we have

$|V^{\prime}|+|V|^{2}=O(|z|^{N})$

outside an $R$-set $U$, where $N$ is a positive constant. Moreover, if
$z=re^{\iota\phi}\in\omega^{+}$ is such that the ray $\arg z=\phi$ meets only
finitely many discs of $U$ we see that $V=o(|z|^{-2})$ as $z$ tends to
$\infty$ on this ray and hence $f$ tends to a finite, nonzero limit. Applying
this reasoning to a set of $\phi$ outside a set of $0$ measure we deduce by
the Phragmén-Lindelöf principle that without loss of generality, for any small
enough given positive $\epsilon$,

$f(re^{\iota\theta})\to 1,$ (8)

as $r\to\infty$ with

$z=re^{\iota\theta}\in\omega^{+}_{\epsilon}=\\{z=re^{\iota\theta}:0<r<\infty,\frac{2i\pi}{n}+\epsilon<\theta<\frac{(2i+1)\pi}{n}-\epsilon\\}.$

For any $z=re^{\iota\theta}\in\omega^{-}$, we have that $\delta(p,\theta)<0$,
and by Lemma 1, we have

$\displaystyle|Q(z)-\frac{1}{4}(he^{p(z)})^{2}-\frac{1}{2}h^{\prime}(z)e^{p(z)}$
$\displaystyle-\frac{1}{2}h(z)p^{\prime}(z)e^{p(z)}|\leq|Q(z)|+|(he^{p(z)})^{2}|$
$\displaystyle+|h^{\prime}(z)e^{p(z)}|+|h(z)p^{\prime}(z)e^{p(z)}|$
$\displaystyle\leq
O(r^{m})+\exp\\{\delta(P,\theta)x^{n}\\}+\exp\\{\frac{1}{2}\delta(P,\theta)x^{n}\\}$
$\displaystyle+\exp\\{\frac{1}{2}\delta(P,\theta)x^{n}\\}O(x^{n-1})$
$\displaystyle\leq O(r^{m})$ (9)

for sufficiently large $r$. Applying, Lemma 4 to (5) and together with (2.1),
$y(z)$ satisfies

$\log^{+}|y(re^{\iota\theta}|=O(r^{\frac{m+2}{2}})$

as $r\to\infty$ for any $z=re^{\iota\theta}\in\omega^{-}.$ From (5) and (7),
we have

$\log^{+}|f(re^{\iota\theta}|=O(r^{\frac{m+2}{2}})$ (10)

as $r\to\infty$ for any $z=re^{\iota\theta}\in\omega^{-}.$ On the rays $\arg
z=\theta_{k}$ such that $\delta(p,\theta_{k})=0,$ we have
$|e^{p(z)}|=|e^{p_{n-1}(z)}|.$ Consider the two cases
$\delta(p_{n-1},\theta_{k})>0$ or $\delta(p_{n-1},\theta_{k})<0$, by the same
method above, we get $f(z)\to 1$ or $\log^{+}|f(z)|=O(r^{\frac{m+2}{2}}),$
respectively, on the ray $\arg z=\theta_{k}.$ If
$\delta(p_{n-1},\theta_{k})=0$ also, repeating these arguments again. Finally,
we deduce that either $f(z)\to 1$ or $\log^{+}|f(z)|=O(r^{\frac{m+2}{2}})$ on
the rays $\arg z=\theta_{k},\ \ k=0,1,\ldots,2n-1.$ Thus, (4), (8), (10) and
the fact $\epsilon$ is arbitrary imply that, by the Phragmén-Lindelöf
principle,

$\rho(f)\leq\frac{m+2}{2}.$ (11)

We claim that $\frac{(2i+1)}{n},\ \ (i=0,1,\ldots,n-1)$ are critical rays for
$Q(z)$. Otherwise, there exists a critical $\theta_{j}$ for $Q(z)$ in

$\frac{2i+1\pi}{n}<\theta_{j}<\frac{2(i+1)\pi}{n}+\frac{2\pi}{m+2}\ \
(i=0,1,\ldots,n-1)$

because $m+2>2n.$ This implies the existence of an unbounded domain of angular
measure at most $\frac{2\pi}{m+2}+\epsilon$, bounded by a path on which
$f(z)\to 0$ and a ray on which $f(z)\to 1.$ By Remark 1 implies that
$\rho(f)>\frac{m+2}{2},$ contradicting (11). Then, there exists a positive
integer $k$ satisfying $\frac{2\pi}{n}=k\frac{2}{m+2},$ that is, $m+2=kn$,
which contradicts $n\nmid m+2.$ Thus, we complete the proof. ∎

Theorem 2 is motivated by Theorem C given by Kumar and Saini[12]. They
considered, $A(z)$ has Fabry gaps and $\rho(B)<\rho(A)$. We changed the
conditions on $A(z)$ to have multiply connected Fatou component.

######  Theorem C.

[12] Let $A(z)$ and $B(z)$ be an entire functions such that $\rho(B)<\rho(A)$
and $A(z)$ has Fabry gaps. Then, $\rho(f)=\infty$ and $\rho_{2}(f)=\rho(A),$
where $f$ is a non-trivial solution of equation (1).

######  Theorem 2.

Let $A(z)$ be a transcendental entire function with a multiply-connected Fatou
component and $B(z)$ be an entire function satisfying $\rho(B)<\rho(A)$. Then,
every non-trivial solution of equation (1) is of infinite order. Moreover,

$\rho_{2}(f)=\rho(A).$

Lemma 5 is given by Gundersen[4]. He generalized the estimates of logarithmic
derivatives of transcendental meromorphic function of finite order.

######  Lemma 5.

[4] Let $f$ be a transcendental meromorphic function with finite order and
$(k,j)$ be a finite pair of integers that satisfies $k>j\geq 0$ and let
$\epsilon>0$ be a given constant. Then following statements holds:

  1. (a)

there exists a set $E_{1}\subset[0,2\pi]$ with linear measure zero such that
for $\theta\in[0,2\pi)\setminus E_{1}$ there exists $R(\theta)>0$ such that

$\left|\frac{f^{(k)}(z)}{f^{(j)}(z)}\right|\leq|z|^{(k-j)(\rho(f)-1+\epsilon)}$

for all $k,j$; $|z|>R(\theta)$ and $argz=\theta$

  2. (b)

there exists a set $E_{2}\subset(1,\infty)$ with finite logarithmic measure
such that for all $|z|\not\in E_{2}\cup[0,1]$ such that inequality in (a)
holds for all $k,j$ and $|z|\geq R(\theta)$.

  3. (c)

there exists a set $E_{3}\subset[0,\infty)$ with finite linear measure such
that for all $|z|\not\in E_{3}$ such that

$\left|\frac{f^{(k)}(z)}{f^{(j)}(z)}\right|\leq|z|^{(k-j)(\rho(f)+\epsilon)}$

holds for all $k,j$.

Recently, Pant and Saini [20] proved the following result for an entire
function.

######  Lemma 6.

[20] Suppose $f$ is a transcendental entire funtion. Then, there exists a set
$F\subset(0,\infty)$ with finite logarithmic measure such that for all $z$
satisfying $|z|=r\in F$ and $|f(z)|=M(r,f)$ we have

$\left|\frac{f(z)}{f^{(m)}(z)}\right|\leq 2r^{m},$

for all $m\in N.$

######  Lemma 7.

[26] Suppose $f$ is a transcendental meromorphic function having atmost finite
poles. If $J(f)$ has only bounded components, then for any complex number,
there exists a constant $0<\beta<1$ and two sequences of positive numbers
$\\{r_{n}\\}$ and $\\{R_{n}\\}$ with $r_{n}\to\infty$ and
$R_{n}/r_{n}\to\infty(n\to\infty)$ such that

$M(r,f)^{\beta}\leq L(r,f)\quad\mbox{for}\quad r\in H,$

where $H=\cup_{n=1}^{\infty}\\{r:r_{n}<r<R_{n}\\}.$

######  Proof of Theorem 2.

We prove this Theorem by contradiction. Suppose $f$ is a finite order non-
trivial solution of equation (1). Applying Lemma 5, there is a set
$E\subset(1,\infty)$ with finite logarithmic measure such that

$\left|\frac{f^{{}^{\prime\prime}}(z)}{f^{{}^{\prime}}(z)}\right|\leq|z|^{2\rho(f)},$
(12)

holds for all $z$ satisfying $|z|\notin E\cup[0,1]$.
Suppose that $z_{r}=re^{\iota\theta_{r}}$ be the points such that
$|f(z_{r})|=M(r,f)$. Then, applying Lemma 6, there exists a set
$F\subset(0,\infty)$ with $m_{l}(F)<\infty$ such that

$\frac{f(re^{\iota\theta_{r}})}{f^{m}(re^{\iota\theta_{r}})}\leq 2r^{m},$ (13)

holds for all sufficiently large $r\notin F$ and for all $m\in\mathbb{N}$.
Applying Lemma 7, we have

$M(r,A)^{\gamma}\leq|A(re^{\iota\theta})|,$ (14)

for $0<\gamma<1$ and $r\in F_{1}=\cup_{n=1}^{\infty}\\{r:r_{n}<r<R_{n}\\}$.
Let $\rho(B)<\beta<\rho(A)$, then the definition of order of growth of $B(z)$
implies that

$|B(re^{\iota\theta})|\leq\exp r^{\beta},$ (15)

for all sufficiently large $r$. From equations (1), (12), (13), (14) and (15),
there exists a sequence $z=re^{\iota\theta}$ such that for all $r\in
F_{1}\setminus(F\cup E\cup[0,1])$, we have

$\displaystyle|A(re^{\iota\theta})|$
$\displaystyle\leq\left|\frac{f^{{}^{\prime\prime}}(re^{\iota\theta})}{f^{{}^{\prime}}(re^{\iota\theta})}\right|+|B(re^{\iota\theta})|\left|\frac{f(re^{\iota\theta})}{f^{{}^{\prime}}(re^{\iota\theta})}\right|$
$\displaystyle\implies M(r,A)^{\gamma}$ $\displaystyle\leq r^{2\rho(f)}+2r\exp
r^{\beta}$ $\displaystyle\leq 2r\exp r^{\beta}(1+o(1)).$

This gives $\rho(A)\leq\beta$, which is a contradiction. Hence, every non-
trivial solution of equation (1) is of infinite order. ∎

In 2017, Gundersen[6] asked a question, “Does every non-trivial solution $f$
of equation (1) is of infinite order, when $A(z)$ satisfies
$\lambda(A)<\rho(A)$ and $B(z)$ is a non-constant polynomial?” Long, et
al.[16] partially answered the question.

######  Theorem D.

[16] Let $A(z)=h(z)e^{P(z)}$ satisfy $\lambda(A)<\rho(A)$ and
$B(z)=b_{m}z^{m}+b_{m-1}z^{m-1}+\cdots+b_{0}$ is a polynomial of degree $m$
such that:

  1. (a)

$m+2<2n$, or

  2. (b)

$m+2>2n$ and $m+2\neq 2kn$ for all integers $k$, or

  3. (c)

$m+2=2n$ and $\frac{a_{n}^{2}}{b_{m}}$ is not real and negative.

Then, all non-trivial solution of equation (1) have infinite order.

Kumar, et al.[11] motivated by their result, considered $\rho(A)>n$ and $B(z)$
to be a polynomial in Theorem E.

######  Theorem E.

[11] Consider a transcendental entire function $A(z)=h(z)e^{P(z)}$, where
$P(z)$ is a non-constant polynomial of degree $n$ and $\rho(h)>n$. Assume that
$h(z)$ is bounded away from zero and exponentially 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.

We are motivated by Theorem E and replace the condition of $B(z)$ to satisfy
some conditions given in Theorem 3.

######  Theorem 3.

Let $A(z)$ satisfy the conditions of Theorem E and $B(z)$ be a transcendental
entire function satisfying

  1. (i)

$\rho(B)<\rho(A)$ or

  2. (ii)

$\mu(B)<\rho(A)$

Then, all non-trivial solutions of the equation (1) are of infinite order.

The following Lemma yield us a lower bound for modulus of an entire function
in the neighbourhood of $\theta$, where $\theta\in[0,2\pi).$

######  Lemma 8.

[23] Suppose $f(z)$ is an entire function of finite order $\rho$ and
$M(r,f)=|f(re^{\iota\theta_{r}})|$ for every $r$. Given $\zeta>0$ and
$0<C(\rho,\zeta)<1,$ there exists $0<l_{0}<\frac{1}{2}$ and a set
$S\subset(1,\infty)$ with $\underline{\log dens}(S)\geq 1-\zeta$ such that

$e^{-5\pi}M(r,f)^{1-C}\leq|f(re^{\iota\theta})|,$

for all sufficiently large $r\in S$ and for all $\theta$ satisfying
$|\theta-\theta_{r}|\leq l_{0}$.

The following Lemma is a Proposition in the research paper of Kumar, et
al.[9].

######  Lemma 9.

[9] Suppose $f(z)$ and $g(z)$ be two entire functions satisfying
$\rho(g)<\rho(f)$. Then, for $0<\epsilon\leq
min\\{\frac{3\rho(f)}{4},\frac{\rho(f)-\rho(g)}{2}\\}$, there exists
$S\subset(1,\infty)$ with $\overline{\log dens}(S)=1$ satisfying

$|g(z)|=o(M(|z|,f))$

for sufficiently large $|z|\in S$.

######  Remark 2.

If we replace $\rho(g)$ with $\mu(g)$, then Lemma 9 would be true.

In Lemma 1, consider $A(z)=v(z)e^{P(z)}$, where $v(z)$ is an entire function
and $P(z)$ is a polynomial of degree $n$ satisfying $\rho(v)<\deg P$. But, in
Lemma 10, the authors considered $\rho(v)>\deg P$ and obtained that
$|A(re^{\iota\theta})|\geq exp((1-\epsilon)\delta(P,\theta)r^{n})$ for
$\theta\in E^{+}/E$ and also for $\theta\in E^{-}/E,$ where $E$ is a set of
linear measure $0.$

######  Lemma 10.

[10] Let $A(z)=v(z)e^{P(z)}$ be an entire function, where $P(z)$ is a
polynomial of degree $n$ and $v(z)$ satisfies the condition of Theorem E.
Then, there exists a set $E\subset[0,2\pi]$ of linear measure zero such that
for $\epsilon>0$ the following holds:

  1. (i)

for $\theta\in E^{+}\setminus E,$ there exists $R(\theta)>1$ such that

$|A(re^{\iota\theta})|\geq exp((1-\epsilon)\delta(P,\theta)r^{n})$ (16)

for $r>R(\theta),$

  2. (ii)

for $\theta\in E^{-}\setminus E,$ there exists $R(\theta)>1$ such that

$|A(re^{\iota\theta})|\geq exp((1-\epsilon)\delta(P,\theta)r^{n})$ (17)

for $r>R(\theta)$.

The following Lemma is proved by Gundersen[5], it gives the logarithmic
estimate of the analytic function $f(z).$

######  Lemma 11.

[5] Let $f$ be an analytic on a ray $\gamma=re^{\iota\theta}$ and suppose that
for some constant $\alpha>1,$ we have

$\left|\frac{f^{\prime}(z)}{f(z)}\right|=O(|z|^{-\alpha})$ (18)

as $z\to\infty$ along $argz=\theta$. Then, there exists a constant $c\neq 0$
such that $f(z)\to c$ as $z\to\infty$ along $argz=\theta$.

The proof of Theorem 3 is inspired by the proof of Theorem E. We have slightly
changed the proof according to the conditions of the Theorem.

######  Proof of Theorem 3.

If $\rho(A)=\infty,$ then it is obvious that $\rho(f)=\infty$, for all non-
trivial solution $f$ of the equation (1). Therefore, let us suppose that
$\rho(A)<\infty$ and there exists a non-trivial solution $f$ of the equation
(1) such that $\rho(f)<\infty$. From Lemma 5, there exists
$E_{1}\subset[0,2\pi]$ of linear measure zero and $m>0$ such that,

$\left|\frac{f^{\prime\prime}(re^{\iota\theta})}{f(re^{\iota\theta})}\right|\leq
r^{m},$ (19)

for $\theta\in[0,2\pi]\setminus E_{1}$ and $r>R(\theta)$. Since $A(z)$ is an
entire function of finite order, suppose that
$M(r,A)=|A(re^{\iota\theta_{r}})|$ for every $r$. Then, from Lemma 8, for
$0<\zeta<1$ and $0<C<1$, there exists $0<l_{0}<\frac{1}{2}$ and
$S_{1}\subset(0,\infty)$ with $\underline{\log dens}(S_{1})\geq 1-\zeta$ such
that

$e^{-5\pi}M(r,A)^{1-C}\leq|A(re^{\iota\theta})|,$

for all sufficiently large $r\in S_{1}$ and for all $\theta$ satisfying
$|\theta-\theta_{r}|\leq l_{0}$.

  1. (i)

Let $\rho(B)<\rho(A)$, from Lemma 9 for $0<\epsilon\leq
min\\{\frac{3\rho(A)}{4},\frac{\rho(A)-\rho(B)}{2}\\}$, there exists
$S_{2}\subset(1,\infty)$ with $\overline{\log dens}(S_{2})=1$ satisfying

$\frac{|B(z)|}{M(|z|,A)}\to 0$ (20)

for sufficiently large $|z|\in S_{2}$. Using properties of logarithmic density
and the fact that $\overline{\log dens}(S_{1}\cup S_{2})\leq 1$, we get

$\displaystyle\overline{\log dens}(S_{1}\cap S_{2})$
$\displaystyle\geq\underline{\log dens}(S_{1})+\underline{\log
dens}(S_{2})-\overline{\log dens}(S_{1}\cup S_{2})$ $\displaystyle\geq
1-\zeta+1-1=1-\zeta.$

Thus, we can choose $z_{r}=re^{\iota\theta_{r}}$ with $r\to\infty$ such that
$r\in(S_{1}\cap S_{2})$ and $|A(re^{\iota\theta_{r}})|=M(r,A)$. We may
consider $<\theta_{r}>$ as a sequence, where $r\in(S_{1}\cap S_{2})$ such that
$\theta_{r}\to\theta_{0}$ and $r\in(S_{1}\cap S_{2})$.
We may consider following three cases:-

    1. (a)

$\delta(P,\theta_{0})>0.$
From Lemma 10(i), we have

$|A(re^{\iota\theta_{0}})|\geq\exp(\frac{1}{2}\delta(P,\theta_{0})r),$ (21)

for sufficiently large $r$, where $r\in(S_{1}\cap S_{2})$ and $\theta_{0}\in
E^{+}/E_{2}$ and $E_{2}$ is a set of critical rays of $e^{P(z)}$ of linear
measure $0$.
From equation (1) we get,

$\left|\frac{f^{\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|\leq\left|\frac{f^{\prime\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|\frac{1}{|A(re^{\iota\theta_{0}})|}+\frac{\left|B(re^{\iota\theta_{0}})\right|}{M(r,A))},$
(22)

for $r\in(S_{1}\cap S_{2})$ and $\theta_{0}\in E^{+}/(E_{1}\cup E_{2})$. Using
equations (19), (20), (21) and (22), we get,

$\left|\frac{f^{\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|\to
0$

for $r\in(S_{1}\cap S_{2})$, $r\to\infty$ and $\theta_{0}\in E^{+}/(E_{1}\cup
E_{2}).$ This implies that

$\left|\frac{f^{\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|=O\left(\frac{1}{r^{2}}\right),$
(23)

as $r\to\infty$ and $r\in S_{1}\cap S_{2}$. From Lemma 11,

$f(re^{\iota\theta_{0}})\to a$ (24)

as $r\to\infty$ and $r\in(S_{1}\cap S_{2})$ for $\theta_{0}\in
E^{+}\setminus(E_{1}\cup E_{2})$, where $a$ is a non-zero finite constant.
Since $f(re^{\iota\theta_{r}})\to f(re^{\iota\theta_{0}})$ and using (24), we
get

$f(re^{\iota\theta_{r}})\to a$

for $r\to\infty$ and $r\in(S_{1}\cap S_{2})$. Thus, entire function $f$ is
bounded over domain. But since function $f$ is entire and non-constant,
$f(re^{\iota\theta})$ is unbounded for all $\theta\in[0,2\pi]$. Thus, for
$\theta_{r}\in[0,2\pi]$, function $f(re^{\iota\theta_{r}})$ is also unbounded,
which is a contradiction.

    2. (b)

$\delta(P,\theta)<0.$
From Lemma 10(ii), we have

$|A(re^{\iota\theta_{0}})|\geq\exp(\frac{1}{2}\delta(P,\theta_{0})r^{n}),$
(25)

for $\theta_{0}\in E^{-}/E_{1}$ for large $r$. Using equation (19), (20) and
(25), we have

$\left|\frac{f^{\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|\to
0,$ (26)

as $r\to\infty$ and $\theta_{0}\in E^{-}/(E_{1}\cup E_{2})$. From Lemma 11,

$f(re^{\iota\theta_{0}})\to b,$ (27)

as $r\to\infty$ and $r\in(S_{1}\cap S_{2})$ for $\theta_{0}\in
E^{-}\setminus(E_{1}\cup E_{2})$, where $b$ is a non-zero finite constant.
Since $f(re^{\iota\theta_{r}})\to f(re^{\iota\theta_{0}})$ and using (27), we
get

$f(re^{\iota\theta_{r}})\to b,$

for $r\to\infty$ and $r\in(S_{1}\cap S_{2})$. Thus, entire function $f$ is
bounded over whole domain. Since function $f$ is entire and non-constant, then
$f(re^{\iota\theta})$ is unbounded for all $\theta\in[0,2\pi]$. Thus, for
$\theta_{r}\in[0,2\pi]$, function $f(re^{\iota\theta_{r}})$ is also unbounded,
which is a contradiction.

    3. (c)

$\delta(P,\theta_{0})=0.$
Suppose $\theta^{*}_{0}\in[0,2\pi]$ in the neighbourhood of $\theta_{0}$ such
that $\delta(P,\theta^{*}_{0})>0$. Letting $r\to\infty$, we get
$|\theta_{0}-\theta^{*}_{0}|\leq l_{0}$. Choosing $C$ and $\zeta$ such that
$l_{0}\to 0$.

$\left|\frac{f^{\prime}(re^{\iota\theta_{0}})}{f(re^{\iota\theta_{0}})}\right|\sim\left|\frac{f^{\prime}(re^{\iota\theta^{*}_{0}})}{f(re^{\iota\theta^{*}_{0}})}\right|\leq\left|\frac{f^{\prime\prime}(re^{\iota\theta^{*}_{0}})}{f(re^{\iota\theta^{*}_{0}})}\right|\frac{1}{|A(re^{\iota\theta^{*}_{0}})|}+\frac{\left|B(re^{\iota\theta^{*}_{0}})\right|}{M(r,A))},$
(28)

Remaining proof is similar to part (i).

  2. (ii)

Let $\mu(B)<\rho(A)$, from Remark 2 for $0<\epsilon\leq
min\\{\frac{3\rho(A)}{4},\frac{\rho(A)-\mu(B)}{2}\\}$, there exist
$S_{2}\subset(1,\infty)$ with $\overline{\log dens}(S_{2})=1$ satisfying

$\frac{|B(z)|}{M(|z|,A)}\to 0.$ (29)

Remaining proof is similar to part (i).

∎

###  2.2. Second Order Non-Homogenous Linear Differential Equation

Kumar and Saini[12] gave severeal results for equation (2). In one of their
result, they considered $A(z)$ to have Fabry gaps,
$\max(\rho(H),\rho(B))<\rho(A)$ and proved the following result. We change the
condition on $A(z)$ and consider $A(z)$ to be transcendental entire function
having a multiply-connected Fatou component and prove Theorem 4.

######  Theorem F.

[12] Let the coefficients and H(z) of equation (2) are entire functions such
that $\max(\rho(H),\rho(B))<\rho(A)$ and $A(z)$ has Fabry gaps. Then, any non-
trivial solution of equation (2) are of infinite order.

######  Theorem 4.

Let $A(z)$ be a transcendental entire function having a multiply-connected
Fatou component and $B(z)$, $H(z)$ be entire functions such that
$\max(\rho(H),$$\rho(B))<\rho(A)$. Then, any non-trivial solution of (2) is of
infinite order.

######  Proof of Theorem 4.

Suppose $f$ is a finite order solution of equation (2). Then, applying Lemma
5, there is a set $E\subset(1,\infty)$ with finite logarithmic measure such
that

$\left|\frac{f^{{}^{\prime\prime}}(z)}{f^{{}^{\prime}}(z)}\right|\leq|z|^{2\rho(f)},$
(30)

holds for all $z$ satisfying $|z|\notin E\cup[0,1]$.
Given that $\max(\rho(H),\rho(B))<\rho(A)$, so let $\beta$ be such that
$\max(\rho(H),\rho(B))<\beta<\rho(A)$, then applying the definition of order
of growth on $B(z)$ and $H(z)$ gives

$|B(re^{\iota\theta})|\leq\exp
r^{\beta}\qquad\text{and}\qquad|H(re^{\iota\theta})|\leq\exp r^{\beta},$ (31)

holds for all sufficiently large $r$. Suppose that
$z_{r}=re^{\iota\theta_{r}}$ be the points such that $|f(z_{r})|=M(r,f)$.
Then, applying Lemma 6, there exists a set $F\subset(0,\infty)$ with
$m_{l}(F)<\infty$ such that

$\frac{f(re^{\iota\theta_{r}})}{f^{m}(re^{\iota\theta_{r}})}\leq 2r^{m},$ (32)

holds for all sufficiently large $r\notin F$ and for all $m\in\mathbb{N}$.
Applying Lemma 7, we have

$M(r,A)^{\gamma}\leq|A(re^{\iota\theta})|,$ (33)

for $0<\gamma<1$ and $r\in F_{1}=\cup_{n=1}^{\infty}\\{r:r_{n}<r<R_{n}\\}$.
From equations (2), (30), (31), (32), and (33), there exists a sequence
$z=re^{\iota\theta}$ such that for all $r\in F_{1}\setminus(E\cup
F\cup[0,1])$, we have

$\displaystyle|A(re^{\iota\theta})|$
$\displaystyle\leq\left|\frac{f^{{}^{\prime\prime}}(re^{\iota\theta})}{f^{{}^{\prime}}(re^{\iota\theta})}\right|+|B(re^{\iota\theta})|\left|\frac{f(re^{\iota\theta})}{f^{{}^{\prime}}(re^{\iota\theta})}\right|+\left|\frac{H(re^{\iota\theta})}{f(re^{\iota\theta})}\right|\left|\frac{f(re^{\iota\theta})}{f^{{}^{\prime}}(re^{\iota\theta})}\right|$
$\displaystyle M(r,A)^{\gamma}$ $\displaystyle\leq r^{2\rho(f)}+2r\exp
r^{\beta}+2r\left|\frac{H(re^{\iota\theta})}{M(r,f)}\right|$
$\displaystyle\leq r^{2\rho(f)}+4r\exp r^{\beta}$ $\displaystyle\leq 4r\exp
r^{\beta}(1+o(1)).$

This gives $\rho(A)\leq\beta$, which is a contradiction. Hence, every non-
trivial solution of equation (2) is of infinite order. ∎

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