On certain non-linear differential monomial sharing non-zero polynomial
UDC 517.5 With the idea of normal family we study the uniqueness of meromorphic functions $f$ and $g$ when $f^{n}(\mathcal{L}(f))^{m}-p$ and $g^{n}(\mathcal{L}(g))^{m}-p$ share two values, where $\mathcal{L}(f)= a_{k}f^{(k)}+a_{k-1} f^{(k-1)}+\ldots+a_{1} f'+a_{0}f...
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| author | Majumder, S. Dam, A. Majumder, S. Dam, A. |
| author_facet | Majumder, S. Dam, A. Majumder, S. Dam, A. |
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With the idea of normal family we study the uniqueness of meromorphic functions $f$ and $g$ when $f^{n}(\mathcal{L}(f))^{m}-p$ and $g^{n}(\mathcal{L}(g))^{m}-p$ share two values, where $\mathcal{L}(f)= a_{k}f^{(k)}+a_{k-1} f^{(k-1)}+\ldots+a_{1} f'+a_{0}f,$ $a_{k}(\ne 0),a_{k-1},\ldots,a_{1},a_{0}\in\mathbb{C}$ and $p(z)(\not\equiv 0)$ is a polynomial. The obtained result significantly improves and generalizes the result in [A. Banerjee, S. Majumder, On certain non-linear differential polynomial sharing a non-zero polynomial, Bol. Soc. Mat. Mex. (2016),https://doi.org/10.1007/s40590-016-0156-0]. |
| doi_str_mv | 10.37863/umzh.v73i2.99 |
| first_indexed | 2026-03-24T02:04:18Z |
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DOI: 10.37863/umzh.v73i2.99
UDC 517.5
S. Majumder (Raiganj Univ., West Bengal, India),
A. Dam (North Bengal St. Xavier’s College, West Bengal, India)
ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL
SHARING NON-ZERO POLYNOMIAL
ПРО НЕЛIНIЙНИЙ ДИФЕРЕНЦIАЛЬНИЙ ОДНОЧЛЕН
ЗI СПIЛЬНИМ НЕНУЛЬОВИМ МНОГОЧЛЕНОМ
With the idea of normal family we study the uniqueness of meromorphic functions f and g when fn(\scrL (f))m - p and
gn(\scrL (g))m - p share two values, where \scrL (f) = akf
(k)+ak - 1f
(k - 1)+ . . .+a1f
\prime +a0f, ak( \not = 0), ak - 1, . . . , a1, a0 \in \BbbC
and p(z)(\not \equiv 0) is a polynomial. The obtained result significantly improves and generalizes the result in [A. Banerjee,
S. Majumder, On certain non-linear differential polynomial sharing a non-zero polynomial, Bol. Soc. Mat. Mex. (2016),
https://doi.org/10.1007/s40590-016-0156-0].
На базi iдеї про нормальнi сiм’ї функцiй вивчається єдинiсть мероморфних функцiй f i g у випадку, коли
fn(\scrL (f))m - p i gn(\scrL (g))m - p мають спiльнi значення, де \scrL (f) = akf
(k) + ak - 1f
(k - 1) + . . . + a1f
\prime + a0f ,
ak( \not = 0), ak - 1, . . . , a1, a0 \in \BbbC , а p(z)( \not \equiv 0) — полiном. Отриманий результат є iстотним узагальненням результату з
[A. Banerjee, S. Majumder, On certain non-linear differential polynomial sharing a non-zero polynomial, Bol. Soc. Mat.
Mex. (2016), https://doi.org/10.1007/s40590-016-0156-0].
1. Introduction definitions and results. In this paper, by meromorphic functions we mean that
meromorphic functions in the whole complex plane \BbbC . We adopt the standard notations of value
distribution theory (see [9]). We denote by T (r) the maximum of T (r, f) and T (r, g). The notation
S(r) denotes any quantity satisfying S(r) = o(T (r)) as r - \rightarrow \infty , outside of a possible exceptional
set of finite linear measure. A meromorphic function a is said to be a small function of f if
T (r, a) = S(r, f). We denote by S(f) the set of all small functions of f. We use the symbol \rho (f)
to denote the order of f.
Let f(z) and g(z) be two nonconstant meromorphic functions. Let a(z) \in S(f) \cap S(g). We
say that f(z) and g(z) share a(z) counting multiplicities (CM) if the zeros of f(z) - a(z) and
g(z) - a(z) have the same locations and same multiplicities, and we say that f(z) and g(z) share
a(z) ignoring multiplicities (IM) if the zeros of f(z) - a(z) and g(z) - a(z) have the same locations
but different multiplicities.
We say that a finite value z0 is called a fixed point of f if f(z0) = z0. For the sake of simplicity,
we use the notion (m)\ast defined by (m)\ast = m - 1, if m is a positive integer; (m)\ast = [m], if m is
positive rational, where [m] denotes the greatest integer not exceeding m.
Let h be a meromorphic function in \BbbC . Then h is called a normal function if there exists a
positive real number M such that h\#(z) \leq M \forall z \in \BbbC , where
h\#(z) =
| h\prime
(z)|
1 + | h(z)| 2
denotes the spherical derivative of h.
c\bigcirc S. MAJUMDER, A. DAM, 2021
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2 201
202 S. MAJUMDER, A. DAM
Let \scrF be a family of meromorphic functions in a domain D \subset \BbbC . We say that \scrF is normal in
D if every sequence \{ fn\} n \subseteq \scrF contains a subsequence which converges spherically and uniformly
on the compact subsets of D (see [16]).
The following well-known theorem in value distribution theory was posed by Hayman and settled
by several authors almost at the same time [3, 5].
Theorem A. Let f be a transcendental meromorphic function and n \in \BbbN . Then fnf \prime = 1 has
infinitely many solutions.
To investigate the uniqueness result corresponding to Theorem A, both Fang and Hua [7], Yang
and Hua [20] obtained the following result.
Theorem B. Let f and g be two non-constant entire (meromorphic) functions, n \in \BbbN with
n \geq 6 (n \geq 11). If fnf \prime and gng\prime share 1 CM, then either f(z) = c1e
cz and g(z) = c2e
- cz, where
c, c1, c2 \in \BbbC \setminus \{ 0\} satisfying 4(c1c2)
n+1c2 = - 1 or f \equiv tg, t \in \BbbC \setminus \{ 0\} such that tn+1 = 1.
Considering the uniqueness question of entire or meromorphic functions having fixed points,
Fang and Qiu [8] obtained the following theorem.
Theorem C. Let f and g be two non-constant meromorphic (entire) functions, n \in \BbbN with
n \geq 11 (n \geq 6). If fn(z)f \prime (z) - z and gn(z)g\prime (z) - z share 0 CM, then either f(z) = c1e
cz2 and
g(z) = c2e
- cz2 , where c, c1, c2 \in \BbbC \setminus \{ 0\} satisfying 4(c1c2)
n+1c2 = - 1 or f \equiv tg, t \in \BbbC \setminus \{ 0\}
such that tn+1 = 1.
It is instinctive to ask what happens if the first derivative f \prime in Theorem A is replaced by the
general derivative f (k). By considering this problem, Xu et al. [17] and Li [24], respectively, proved
the following result.
Theorem D. Let f be a transcendental meromorphic function and k, n \in \BbbN with n \geq 2. Then
fnf (k) takes every finite non-zero value infinitely many times or has infinitely many fixed points.
Recently, Cao and Zhang [6] proved the following theorem.
Theorem E. Let f, g be two non-constant meromorphic functions, whose zeros are of multipli-
cities at least k + 1, k \in \BbbN with 1 \leq k \leq 5 and let n \in \BbbN with n \geq 10. If fnf (k) and gng(k) share
1 CM, f (k) and g(k) share 0 CM, f and g share \infty IM, then one of the following two conclusions
hold:
(i) f \equiv tg, t \in \BbbC \setminus \{ 0\} such that tn+1 = 1;
(ii) f(z) = c1e
az and g(z) = c2e
- az, where a, c1, c2 \in \BbbC \setminus \{ 0\} such that ( - 1)k(c1c2)
n+1a2k = 1.
Regarding Theorem E, the following questions are inevitable.
Question 1. Can the lower bound of n be further reduced in Theorem E?
Question 2. Can the condition “Let f and g be two non-constant meromorphic functions, whose
zeros are of multiplicities at least k + 1, k \in \BbbN ” in Theorem E be further weakened?
Question 3. Does Theorem E hold for k \geq 6?
We now explain the notation of weighted sharing as introduced in [11].
Definition 1 [11]. Let k \in \BbbN \cup \{ 0\} \cup \{ \infty \} . For a \in \BbbC \cup \{ \infty \} we denote by Ek(a; f) the set of
all a-points of f, where an a-point of multiplicity m is counted m times if m \leq k and k + 1 times
if m > k. If Ek(a; f) = Ek(a; g), we say that f and g share the value a with weight k. We write f
and g share (a, k) to mean that f and g share the value a with weight k.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 203
Keeping in mind the above questions, Banerjee and Majumder [2] obtained the following result
in 2016.
Theorem F. Let f, g be two transcendental meromorphic functions, whose zeros are of multi-
plicities at least k \in \BbbN and n \in \BbbN such that n >
\biggl(
k2 + 4k + 4
k
\biggr) \ast
. Let p(z)(\not \equiv 0) be a polynomial
such that either \mathrm{d}\mathrm{e}\mathrm{g}(p) \leq n - 1 or zeros of p(z) be of multiplicities at most n - 1. If fnf (k) - p
and gng(k) - p share (0, k1), where k1 =
\biggl[
k + 2
n - k
\biggr]
+ 3 and f, g share \infty IM and f (k), g(k) share
0 CM, then f \equiv tg, t \in \BbbC \setminus \{ 0\} such that tn+1 = 1.
Throughout this paper, we always use \scrL (f) to denote a differential polynomial as follows:
\scrL (f) = akf
(k) + ak - 1f
(k - 1) + . . .+ a1f
\prime + a0f, ak( \not = 0), ak - 1, . . . , a1, a0 \in \BbbC . (1.1)
Now we observe Theorem F. Then it is natural to ask the following questions which are the motive
of the present paper.
Question 4. Can one remove the condition “\mathrm{d}\mathrm{e}\mathrm{g}(p) \leq n - 1 or zeros of p(z) be of multiplicities
at most n - 1” in Theorem F?
Question 5. What happens when “fn(\scrL (f))m - p and gn(\scrL (g))m - p” share the value 0 CM,
where p(z)(\not \equiv 0) is a polynomial in Theorem F?
Question 6. Can the lower bound of n be further reduced in Theorem F?
2. Main result. In this paper, taking the possible answers of the above questions into background
we obtain the following result which significantly improves and generalizes Theorem F.
Theorem 1. Let f and g be two transcendental meromorphic functions having zeros of multi-
plicities at least k \in \BbbN . Let m,n \in \BbbN such that n \geq k2 + 2mk + 6
k
and p(z)(\not \equiv 0) be a polynomial.
If fn(\scrL (f))m - p and gn(\scrL (g))m - p share (0, k1), where k1 =
\biggl[
3 + (k - 1)m
n+m+ (m - 2)k - 1
\biggr]
+3 and
f, g share \infty IM and \scrL (f), \scrL (g) share 0 CM, then f \equiv tg, where t \in \BbbC \setminus \{ 0\} with tn+m = 1.
Remark 1. It is easy to see that the condition “Let f and g be two transcendental meromorphic
functions having zeros of multiplicities at least k \in \BbbN ” in Theorem 1 is sharp by the following
example.
Example 1. Let
f(z) = c1e
az and g(z) = c2e
- az,
where a, c1, c2 \in \BbbC \setminus \{ 0\} . Note that
\scrL (f(z)) = a2f
\prime \prime (z) + a1f
\prime (z) + a0f(z) = c1
\bigl(
a2a
2 + a1a+ a0
\bigr)
eaz
and
\scrL (g(z)) = a2g
\prime \prime (z) + a1g
\prime (z) + a0g(z) = c2
\bigl(
a2a
2 - a1a+ a0
\bigr)
e - az,
where a2(\not = 0), a1, a0 \in \BbbC such that
cn+m
1
\bigl(
a2a
2 + a1a+ a0
\bigr) m
= cn+m
2
\bigl(
a2a
2 - a1a+ a0
\bigr) m
, m, n \in \BbbN .
Since f and g have no zeros, it follows that the condition “Let f and g be two transcendental
meromorphic functions having zeros of multiplicities at least k \in \BbbN ” does not hold. Here we see
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
204 S. MAJUMDER, A. DAM
that f, g share \infty CM and \scrL (f), \scrL (g) share 0 CM. On the other hand, we see that
fn(z)(\scrL (f(z)))m - p(z) = cn+m
1
\bigl(
a2a
2 + a1a+ a0
\bigr) m \Bigl(
ea(n+m)z - 1
\Bigr)
and
gn(z)(\scrL (g(z)))m - p(z) = cn+m
2
\bigl(
a2a
2 - a1a+ a0
\bigr) m \Bigl(
e - a(n+m)z - 1
\Bigr)
,
where p(z) = cn+m
1
\bigl(
a2a
2 + a1a+ a0
\bigr) m
. Clearly fn(\scrL (f))m - p and gn(\scrL (g))m - p share (0,\infty ),
but f \not \equiv tg, where t \in \BbbC \setminus \{ 0\} with tn+m = 1.
We now explain some definitions and notations which are used in the paper.
Definition 2 [14]. Let p \in \BbbN and a \in \BbbC \cup \{ \infty \} .
(i) N(r, a; f | \geq p) (N(r, a; f | \geq p)) denotes the counting function (reduced counting function)
of those a-points of f whose multiplicities are not less than p.
(ii) N(r, a; f | \leq p) (N(r, a; f | \leq p)) denotes the counting function (reduced counting function)
of those a-points of f whose multiplicities are not greater than p.
Definition 3 [22]. For a \in \BbbC \cup \{ \infty \} and p \in \BbbN we denote by Np(r, a; f) the sum N(r, a; f) +
+N(r, a; f | \geq 2) + . . .+N(r, a; f | \geq p). Clearly N1(r, a; f) = N(r, a; f).
Definition 4. We denote by N(r, a; f | = k) the reduced counting function of those a-points of
f whose multiplicities exactly k \in \BbbN . Clearly N(r, a; f | = 1) = N(r, a; f | = 1).
Definition 5 [1]. Let f and g be two non-constant meromorphic functions such that f and g
share 1 IM. Let z0 be a 1-point of f with multiplicity p and a 1-point of g with multiplicity q.
We denote by NL(r, 1; f), the counting function of those 1-points of f and g where p > q and by
N
(l
E(r, 1; f), the counting function of those 1-points of f and g where p = q \geq l, each point in
these counting functions is counted only once, where l \in \BbbN \setminus \{ 1\} . In the same way we can define
NL(r, 1; g) and N
(l
E(r, 1; g).
Definition 6 [11]. Let f, g share a value a IM. We denote by N\ast (r, a; f, g) the reduced counting
function of those a-points of f whose multiplicities differ from the multiplicities of the corresponding
a-points of g. Clearly N\ast (r, a; f, g) = NL(r, a; f) +NL(r, a; g).
3. Lemmas. In this section, we present some lemmas which will be needed in the sequel. Now
we define the following two auxiliary functions H and G, respectively:
H =
\biggl(
F \prime \prime
F \prime - 2F \prime
F - 1
\biggr)
-
\biggl(
G\prime \prime
G\prime - 2G\prime
G - 1
\biggr)
(3.1)
and
V =
\biggl(
F \prime
F - 1
- F \prime
F
\biggr)
-
\biggl(
G\prime
G - 1
- G\prime
G
\biggr)
=
F \prime
F (F - 1)
- G\prime
G(G - 1)
, (3.2)
where F and G are two non-constant meromorphic functions.
Lemma 1 [23]. Let f be a non-constant meromorphic function and L(f) be a differential poly-
nomial defined as follows:
L(f) = f (k) + ak - 1f
(k - 1) + ak - 2f
(k - 2) + . . .+ a1f
\prime + a0f,
where k \in \BbbN , aj \in S(f), j = 0, 1, . . . , k - 1. If L(f) \not \equiv 0 and p \in \BbbN , we have
Np(r, 0;L(f)) \leq kN(r,\infty ; f) +Np+k(r, 0; f) + S(r, f).
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 205
Lemma 2 [12]. If N(r, 0; f (k) | f \not = 0) denotes the counting function of those zeros of f (k)
which are not the zeros of f, where a zero of f (k) is counted according to its multiplicity. Then
N
\Bigl(
r, 0; f (k) | f \not = 0
\Bigr)
\leq kN(r,\infty ; f) +N(r, 0; f | < k) + kN(r, 0; f | \geq k) + S(r, f).
Lemma 3 [19]. Let f be a non-constant meromorphic function and P (f) = a0+a1f+a2f
2+. . .
. . .+ anf
n, where a0, a1, a2 . . . , an( \not = 0) \in \BbbC . Then T (r, P (f)) = nT (r, f) +O(1).
Lemma 4 [13]. Let f be a transcendental meromorphic function and \alpha (\not \equiv 0,\infty ) \in S(f), then
\psi = \alpha (f)n(f (k))p \not \in \BbbC , where n \in \BbbN \cup \{ 0\} and p, k \in \BbbN .
Lemma 5 [21]. Let fj , j = 1, 2, 3, be a meromorphic and f1 be non-constant. Suppose that\sum 3
j=1
fj \equiv 1 and
\sum 3
j=1
N(r, 0; fj) + 2
\sum 3
j=1
N(r,\infty ; fj) < (\lambda + o(1))T1(r) as r \rightarrow +\infty ,
r \in I, \lambda < 1 and T1(r) = \mathrm{m}\mathrm{a}\mathrm{x}1\leq j\leq 3 T (r, fj), where I is a set of infinite linear measure. Then
either f2 \equiv 1 or f3 \equiv 1.
Lemma 6 ([21], Theorem 1.24). Let f be a non-constant meromorphic function and k \in \BbbN .
Suppose that f (k) \not \equiv 0, then N
\bigl(
r, 0; f (k)
\bigr)
\leq N(r, 0; f) + kN(r,\infty ; f) + S(r, f).
Lemma 7. Let f, g be two non-constant meromorphic functions, whose zeros are of multipli-
cities at least k, where k \in \BbbN and F = fn(\scrL (f))m/p, G = gn(\scrL (g))m/p, where p(z)(\not \equiv 0) is a
polynomial and m,n \in \BbbN such that n+m+ (m - 2)k > 1. Suppose H \not \equiv 0. If F, G share (1, k1)
except for the zeros of p and f, g share (\infty , 0), where 0 \leq k1 \leq \infty , then
N(r,\infty ; f) \leq k + 1
k(n+m+ (m - 2)k - 1)
(T (r, f) + T (r, g))+
+
1
n+m+ (m - 2)k - 1
N\ast (r, 1;F,G) + S(r, f) + S(r, g).
Proof. First, we suppose \infty is a Picard exceptional value of both f and g. Then the lemma
follows immediately. Next we suppose \infty is not a Picard exceptional value of both f and g. We
claim that V \not \equiv 0. If possible suppose V \equiv 0. Then by integration we obtain 1 - 1
F
\equiv A
\biggl(
1 - 1
G
\biggr)
,
A \in \BbbC \setminus \{ 0\} . It is that if z0 is a pole of f, then it is a pole of g. Hence from the definition of F and
G we have
1
F (z0)
= 0 and
1
G(z0)
= 0. So, A = 1 and hence F \equiv G. Since H \not \equiv 0, it follows that
F \not \equiv G. Therefore we arrive at a contradiction. Hence V \not \equiv 0. Also m(r, V ) = S(r, f) + S(r, g).
Let z0 be a pole of f with multiplicity q and a pole of g with multiplicity r such that p(z0) \not = 0.
Clearly z0 is a pole of F with multiplicity (n + m)q + mk and a pole of G with multiplicity
(n+m)r +mk. Clearly
F \prime (z)
F (z)(F (z) - 1)
= O
\Bigl(
(z - z0)
(n+m)q+mk - 1
\Bigr)
and
G\prime (z)
G(z)(G(z) - 1)
= O
\Bigl(
(z - z0)
(n+m)r+mk - 1
\Bigr)
.
Consequently V (z) = O
\bigl(
(z - z0)(n+m)t+mk - 1
\bigr)
, where t = \mathrm{m}\mathrm{i}\mathrm{n}\{ q, r\} . Since f and g share (\infty , 0),
from the definition of V it is clear that z0 is a zero of V with multiplicity at least n+m+mk - 1.
So from the definition of V and using Lemma 2 we have
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
206 S. MAJUMDER, A. DAM
(n+m+mk - 1)N(r,\infty ; f) \leq
\leq N(r, 0;V ) +O(\mathrm{l}\mathrm{o}\mathrm{g} r) \leq T (r, V ) + S(r, f) + S(r, g) \leq
\leq N(r,\infty ;V ) + S(r, f) + S(r, g) \leq
\leq N(r, 0;F ) +N(r, 0;G) +N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq N(r, 0; f) +N
\Bigl(
r, 0; f (k) | f \not = 0
\Bigr)
+N(r, 0; g) +N
\Bigl(
r, 0; g(k) | g \not = 0
\Bigr)
+
+N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq N(r, 0; f) + kN(r,\infty ; f) +Nk(r, 0; f) +N(r, 0; g) + kN(r,\infty ; g)+
+Nk(r, 0; g) +N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq k + 1
k
N(r, 0; f) +
k + 1
k
N(r, 0; g) + 2kN(r,\infty ; f) +N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq k + 1
k
(T (r, f) + T (r, g)) + 2kN(r,\infty ; f) +N\ast (r, 1;F,G) + S(r, f) + S(r, g).
Lemma 7 is proved.
Lemma 8. Let f be a non-constant meromorphic function and let F = fn(\scrL (f))m, where
m,n, k \in \BbbN satisfying n > m. Then
(n - m)T (r, f) \leq T (r, F ) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f).
Proof. Note that
N(r,\infty ;F ) = N (r,\infty ; fn) +N (r,\infty ; (\scrL (f))m) =
= N (r,\infty ; fn) +mN(r,\infty ; f) +mkN(r,\infty ; f) + S(r, f),
i.e.,
N (r,\infty ; fn) = N(r,\infty , F ) - mN(r,\infty ; f) - mkN(r,\infty , f) + S(r, f).
Also
m (r, fn) = m
\biggl(
r,
F
(\scrL (f))m
\biggr)
\leq
\leq m(r, F ) +m
\biggl(
r,
1
(\scrL (f))m
\biggr)
+ S(r, f) =
= m(r, F ) + T (r, (\scrL (f))m) - N (r, 0; (\scrL (f))m) + S(r, f) =
= m(r, F ) +N (r,\infty ; (\scrL (f))m) +m (r, (\scrL (f))m) - N (r, 0; (\scrL (f))m) + S(r, f) \leq
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 207
\leq m(r, F ) +mN(r,\infty ; f) +mkN(r,\infty ; f) +m
\biggl(
r,
(\scrL (f))m
fm
\biggr)
+
+m (r, fm) - N (r, 0; (\scrL (f))m) + S(r, f) =
= m(r, F ) +mT (r, f) +mkN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f).
Now
nT (r, f) = N (r,\infty ; fn) +m (r, fn) \leq
\leq T (r, F ) +mT (r, f) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f),
i.e.,
(n - m)T (r, f) \leq T (r, F ) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f).
Lemma 8 is proved.
Lemma 9. Let f be a transcendental meromorphic function and let a(z)(\not \equiv 0,\infty ) \in S(f). If
n > m+ 1, then fn(\scrL (f))m - a has infinitely many zeros, where n,m, k \in \BbbN .
Proof. Let F = fn(\scrL (f))m. Note that
T (r, F ) = N(r,\infty ;F ) +m(r, F ) \leq
\leq N (r,\infty ; fn) +N (r,\infty ; (\scrL (f))m) +m
\bigl(
r, fn+m
\bigr)
+m
\biggl(
r,
\biggl(
\scrL (f)
f
\biggr) m\biggr)
\leq
\leq nN(r,\infty ; f) +mN(r,\infty ;\scrL (f)) + (n+m)m(r, f) +mm
\biggl(
r,
\scrL (f)
f
\biggr)
\leq
\leq nN(r,\infty ; f) +m(N(r,\infty ; f) + kN(r,\infty ; f)) + (n+m)m(r, f) + S(r, f) \leq
\leq (n+ (k + 1)m)N(r,\infty ; f) + (n+m)m(r, f) + S(r, f) \leq
\leq (n+ (k + 1)m)T (r, f) + S(r, f). (3.3)
Also by Lemma 8 we have
(n - m)T (r, f) \leq T (r, F ) + S(r, f). (3.4)
Since n > m + 1, from (3.3) and (3.4) we conclude that S(r, F ) = S(r, f). Now we prove
that F - a has infinitely many zeros. If possible suppose F - a has finitely many zeros. Then
N(r, a;F ) = O(\mathrm{l}\mathrm{o}\mathrm{g} r) = S(r, f) = o(T (r, f)). Now in view of Lemma 8, (3.3) and the second
fundamental theorem for small functions (see [18]) we get
(n - m)T (r, f) \leq T (r, F ) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f) \leq
\leq N(r, 0;F ) +N(r,\infty ;F ) +N(r, a;F ) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m)+
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208 S. MAJUMDER, A. DAM
+(\varepsilon + o(1))T (r, F ) + S(r, f) \leq
\leq N(r, 0; f) +N (r, 0; (\scrL (f))m) +N(r,\infty ; f) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m)+
+\varepsilon T (r, F ) + o(T (r, F )) + S(r, f) \leq
\leq N(r, 0; f) + \varepsilon T (r, F ) + S(r, F ) + S(r, f) \leq
\leq T (r, f) + (n+ (k + 1)m)\varepsilon T (r, f) + \varepsilon S(r, f) + S(r, f)
for all \varepsilon > 0. Therefore,
(n - m - 1)T (r, f) \leq (n+ (k + 1)m)\varepsilon T (r, f) + S(r, f). (3.5)
If we take 0 < \varepsilon <
n - m - 1
n+ (k + 1)m
, then from (3.5) we arrive at a contradiction. Hence F - a has
infinitely many zeros.
Lemma 9 is proved.
Lemma 10 [10]. Let f and g be two non-constant meromorphic functions. Suppose that f and
g share 0 and \infty CM, f (k) and g(k) share 0 CM for k = 1, 2, . . . , 6. Then f and g satisfy one of
the following cases:
(i) f \equiv tg, where t \in \BbbC \setminus \{ 0\} ;
(ii) f(z) = eaz+b and g(z) = ecz+d, where a(\not = 0), b, c(\not = 0), d \in \BbbC ;
(iii) f(z) =
a
1 - be\alpha (z)
and g(z) =
a
e - \alpha (z) - b
, where a, b \in \BbbC \setminus \{ 0\} and \alpha is a non-constant
entire function;
(iv) f(z) = a
\bigl(
1 - becz
\bigr)
and g(z) = d
\bigl(
e - cz - b
\bigr)
, where a, b, c, d \in \BbbC \setminus \{ 0\} .
Lemma 11. Let f and g be two transcendental meromorphic functions having zeros of mul-
tiplicities at least k \in \BbbN , m, n \in \BbbN . Let \scrL (f), \scrL (g) share 0 CM and f, g share \infty IM. If
fn(\scrL (f))m \equiv gn(\scrL (g))m. Then f \equiv tg, where t \in \BbbC \setminus \{ 0\} such that tn+m = 1.
Proof. Suppose
fn(\scrL (f))m \equiv gn(\scrL (g))m, (3.6)
i.e.,
fn
gn
\equiv (\scrL (f))m
(\scrL (g))m
. (3.7)
Since f and g share \infty IM, it follows from (3.6) that f and g share \infty CM and so \scrL (f) and
\scrL (g) share \infty CM. Again since \scrL (f) and \scrL (g) share 0 CM, it follows that f and g share 0 CM
also. Let h1 =
f
g
and h2 =
\scrL (f)
\scrL (g)
. Then h1 \not = 0,\infty and h2 \not = 0,\infty . From (3.7) we see that
hn1h
m
2 \equiv 1. (3.8)
First we suppose h1 is a non-constant entire function. Clearly h2 is also a non-constant entire
function. Let F1 = hn1 and G1 = hm2 . Also from (3.8) we get
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 209
F1G1 \equiv 1. (3.9)
Clearly F1 \not \equiv d0G1, where d0 \in \BbbC \setminus \{ 0\} , otherwise F1 \in \BbbC and so h1 will be a constant. Since
F1 \not = 0,\infty and G1 \not = 0,\infty then there exist two non-constant entire functions \alpha and \beta such that
F1 = e\alpha and G1 = e\beta . Now from (3.9) we see that \alpha + \beta = C, where C \in \BbbC . Therefore \alpha \prime = - \beta \prime .
Note that F \prime
1 = \alpha \prime e\alpha and G\prime
1 = \beta \prime e\beta . This shows that F \prime
1 and G\prime
1 share 0 CM. Note that F1 \not = 0,\infty ,
G1 \not = 0,\infty and F1 \not \equiv d0G1, where d0 \in \BbbC \setminus \{ 0\} . Now in view of Lemma 10 we have to consider
the cases F1(z) = c1e
az and G1(z) = c2e
- az, where a, c1, c2 \in \BbbC \setminus \{ 0\} such that c1c2 = 1. Since\biggl(
f(z)
g(z)
\biggr) n
= c1e
az and
\biggl(
\scrL (f(z))
\scrL (g(z))
\biggr) m
= c2e
- az,
it follows that
f(z)
g(z)
= t1e
a
n
z = t1e
cz and
\scrL (f(z))
\scrL (g(z))
= t2e
- a
m
z = t2e
dz, (3.10)
where c, d, t1, t2 \in \BbbC \setminus \{ 0\} such that tn1 = c1, t
m
2 = c2, c =
a
n
and d = - a
m
. Let
\Phi 1 =
\scrL \prime (f)
\scrL (f)
- \scrL \prime (g)
\scrL (g)
. (3.11)
From (3.10), we see that
\Phi 1(z) = d. (3.12)
Again from (3.10) we see that f (j)(z) = t1
\sum j
i=0
Cj
i (e
cz)(i)g(j - i)(z), i.e.,
f (j)(z) = t1e
cz
\biggl(
g(j)(z) + jcg(j - 1)(z) +
j(j - 1)
2
c2g(j - 2)(z) + . . .+ cjg(z)
\biggr)
.
Therefore
\scrL (f(z)) = t1e
cz
\Biggl(
akg
(k)(z) + (kcak + ak - 1)g
(k - 1)(z)+
+
\biggl(
k(k - 1)
2
c2ak + (k - 1)cak - 1 + ak - 2
\biggr)
g(k - 2)(z) + . . .
\Biggr)
(3.13)
and
\scrL \prime (f(z)) = t1e
cz
\Biggl(
akg
(k+1)(z) + ((k + 1)cak + ak - 1) g
(k)(z)+
+
\biggl(
k(k + 1)
2
c2ak + kcak - 1 + ak - 2
\biggr)
g(k - 1)(z) + . . .
\Biggr)
. (3.14)
Now from (3.11), (3.13) and (3.14), we have
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210 S. MAJUMDER, A. DAM
\Phi 1 =
G2 + (k + 1)cg(k)g(k) - kcg(k - 1)g(k+1)
G3 + g(k)g(k)
, (3.15)
where
G2(z) =
\sum
0\leq i\leq k+1
0\leq j\leq k
0\leq i+j\leq 2k - 1
Ai,jg
(i)(z)g(j)(z) and G3(z) =
\sum
0\leq i,j\leq k
0\leq i+j\leq 2k - 1
Bi,jg
(i)(z)g(j)(z),
Ai,j , Bi,j \in \BbbC . Let zp be a zero of g(z) with multiplicity p(\geq k). Then the Taylor expansion of g
about zp is
g(z) = bp(z - zp)
p + bp+1(z - zp)
p+1 + bp+2(z - zp)
p+2 + . . . , bp \not = 0. (3.16)
We now consider following two cases.
Case 1. Suppose p = k. Then
g(k)(z) = k!bk + (k + 1)!bk+1(z - zk) + . . . (3.17)
and
g(k+1)(z) = (k + 1)!bk+1 + (k + 2)!bk+2(z - zk) + . . . . (3.18)
Now from (3.15), (3.17) and (3.18), we have
\Phi 1(zk) = c
(k + 1)(k!)2b2k
(k!)2b2k
= c(k + 1). (3.19)
Therefore, we arrive at a contradiction from (3.12) and (3.19).
Case 2. Suppose p \geq k + 1. Then
g(k - 1)(z) = p(p - 1) . . . (p - k + 2)bp(z - zp)
(p - k+1) + . . . ,
g(k)(z) = p(p - 1) . . . (p - k + 1)bp(z - zp)
(p - k) + . . . ,
and
g(k+1)(z) = p(p - 1) . . . (p - k)bp(z - zp)
(p - k - 1) + . . . .
Therefore
g(k)(z)g(k)(z) = Kb2p(z - zp)
2p - 2k + . . . , (3.20)
g(k - 1)(z)g(k+1)(z) =
p - k
p - k + 1
Kb2p(z - zp)
2p - 2k + . . . , (3.21)
where K = [p(p - 1) . . . (p - k + 1)]2. Also
G2(z) = O
\bigl(
(z - zp)
2p - i - j
\bigr)
and G3(z) = O
\bigl(
(z - zp)
2p - i - j
\bigr)
,
where 2p - 2k + 1 \leq 2p - i - j \leq 2p. Now from (3.15), (3.20) and (3.21), we have
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 211
\Phi 1(zp) =
(k + 1)cKb2p - kc
p - k
p - k + 1
Kb2p
Kb2p
= c
p+ 1
p - k + 1
. (3.22)
Therefore we arrive at a contradiction from (3.12) and (3.22). Thus in either cases one can easily say
that g has no zeros. Since f and g share 0 CM, it follows that f and g have no zeros. But this is
impossible because zeros of f and g are of multiplicities at least k \in \BbbN . Hence h1 \in \BbbC \setminus \{ 0\} . Then
from (3.6) we get hn+m
1 = 1. Therefore, we have f \equiv tg, where t \in \BbbC \setminus \{ 0\} such that tn+m = 1.
Lemma 11 is proved.
Lemma 12 [4]. Let f be a meromorphic function on \BbbC with finitely many poles. If f has
bounded spherical derivative on \BbbC , then f is of order at most 1.
Lemma 13 (Zalcman’s [15, 23]). Let F be a family of meromorphic functions in the unit disc
\Delta and \alpha be a real number satisfying - 1 < \alpha < 1. Then if F is not normal at a point z0 \in \Delta there
exist for each \alpha with - 1 < \alpha < 1,
(i) points zn \in \Delta , zn \rightarrow z0,
(ii) positive numbers \rho n, \rho n \rightarrow 0+,
(iii) functions fn \in F,
such that \rho - \alpha
n fn(zn + \rho n\zeta ) \rightarrow g(\zeta ) spherically uniformly on compact subset of \BbbC , where g is
a non-constant meromorphic function. The function g may be taken to satisfy the normalisation
g\#(\zeta ) \leq g\#(0) = 1, \zeta \in \BbbC .
Lemma 14. Let f, g be two transcendental meromorphic functions having zeros of multiplicities
at least k \in \BbbN , and let fn(\scrL (f))m - p, gn(\scrL (g))m - p share 0 CM and f, g share \infty IM, where
p(z)(\not \equiv 0) is a polynomial and m,n \in \BbbN . Then
fn(\scrL (f))mgn(\scrL (g))m \not \equiv p2.
Proof. Suppose
fn(\scrL (f))mgn(\scrL (g))m \equiv p2. (3.23)
Since f and g share \infty IM, from (3.23) one can easily say that f and g are transcendental entire
functions. We consider the following cases.
Case 1. Let \mathrm{d}\mathrm{e}\mathrm{g}(p) \in \BbbN . Now from (3.23) it follows that N(r, 0; f) = O(\mathrm{l}\mathrm{o}\mathrm{g} r) and N(r, 0; g) =
= O(\mathrm{l}\mathrm{o}\mathrm{g} r). Let
F =
fn(\scrL (f))m
p
and G =
gn(\scrL (g))m
p
. (3.24)
From (3.23) we get
FG \equiv 1. (3.25)
If F \equiv d1G, d1 \in \BbbC \setminus \{ 0\} , then F \in \BbbC \setminus \{ 0\} , which is impossible by Lemma 4. Hence F \not \equiv d1G.
Let
\Phi =
fn(\scrL (f))m - p
gn(\scrL (g))m - p
. (3.26)
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212 S. MAJUMDER, A. DAM
Since f and g are transcendental entire functions, it follows that fn(\scrL (f))m - p \not = \infty and
gn(\scrL (g))m - p \not = \infty . Also since fn(\scrL (f))m - p and gn(\scrL (g))m - p share 0 CM, we deduce
from (3.26) that
\Phi \equiv e\beta , (3.27)
where \beta is an entire function. Let f1 = F, f2 = - e\beta G and f3 = e\beta . Here f1 is transcendental.
Now from (3.27), we have f1 + f2 + f3 \equiv 1. Hence, by Lemma 6, we get
3\sum
j=1
N(r, 0; fj) + 2
3\sum
j=1
N(r,\infty ; fj) \leq
\leq N(r, 0;F ) +N(r, 0; e\beta G) +O(\mathrm{l}\mathrm{o}\mathrm{g} r) \leq (\lambda + o(1))T1(r)
as r \rightarrow +\infty , r \in I, \lambda < 1. So, by Lemma 5, we get either e\beta G \equiv - 1 or e\beta \equiv 1. But here the only
possibility is that e\beta G \equiv - 1, i.e., gn(\scrL (g))m \equiv - e - \beta p and so from (3.23) we obtain F \equiv e\gamma 1G,
i.e., fn(\scrL (f))m \equiv e\gamma 1gn(\scrL (g))m, where \gamma 1 is a non-constant entire function. Then, from (3.23), we
get
fn(\scrL (f))m \equiv d2e
1
2
\gamma 1p and gn(\scrL (g))m \equiv d2e
- 1
2
\gamma 1p, (3.28)
where d2 = \pm 1. This shows that fn(\scrL (f))m and gn(\scrL (f))m share 0 CM. Clearly, from (3.28), we
see F and G are entire functions having no zeros.
Let zp be a zero of f(z) of multiplicity p(\geq k) and zq be a zero of g(z) of multiplicity q(\geq k).
Clearly zp will be a zero of fn(\scrL (f))m of multiplicity (n + 1)p - k and zq will be a zero of
gn(\scrL (g))m of multiplicity (n + 1)q - k. Since fn(\scrL (f))m and gn(\scrL (g))m share 0 CM, it follows
that zp = zq and p = q. Consequently f(z) and g(z) share 0 CM. Since N(r, 0; f) = O(\mathrm{l}\mathrm{o}\mathrm{g} r) and
N(r, 0; g) = O(\mathrm{l}\mathrm{o}\mathrm{g} r), so we can take
f(z) = h(z)e\alpha (z) and g(z) = h(z)e\beta (z), (3.29)
where h(z) is a non-constant polynomial and \alpha , \beta are two non-constant entire functions.
We deduce from (3.29) that
fn(\scrL (f))m \equiv P1
\Bigl(
h, h\prime , . . . , h(k), \alpha \prime , \alpha \prime \prime , . . . , \alpha (k)
\Bigr)
e(n+m)\alpha , (3.30)
where
P1
\Bigl(
h, h\prime , . . . , h(k), \alpha \prime , \alpha \prime \prime , . . . , \alpha (k)
\Bigr)
= hn
\Biggl(
k\sum
i=0
ai P1i
\Bigl(
h, h\prime , . . . , h(i), \alpha \prime , \alpha \prime \prime , . . . , \alpha (i)
\Bigr) \Biggr) m
,
P1i
\bigl(
h, h\prime , . . . , h(i), \alpha \prime , \alpha \prime \prime , . . . , \alpha (i)
\bigr)
is a differential polynomial in h, h\prime , . . . , h(i), \alpha \prime , \alpha \prime \prime , . . . , \alpha (i),
i = 1, . . . , k, P10 = a0h and
gn(\scrL (g))m \equiv P2
\Bigl(
h, h\prime , . . . , h(k), \beta \prime , \beta \prime \prime , . . . , \beta (k)
\Bigr)
e(n+m)\beta , (3.31)
where
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 213
P2
\Bigl(
h, h\prime , . . . , h(k), \beta \prime , \beta \prime \prime , . . . , \beta (k)
\Bigr)
= hn
\Biggl(
k\sum
i=0
aiP2i
\Bigl(
h, h\prime , . . . , h(i), \beta \prime , \beta \prime \prime , . . . , \beta (i)
\Bigr) \Biggr) m
,
P2i
\bigl(
h, h\prime , . . . , h(i), \beta \prime , \beta \prime \prime , . . . , \beta (i)
\bigr)
is a differential polynomial in h, h\prime , . . . , h(i), \beta \prime , \beta \prime \prime , . . . , \beta (i),
i = 1, . . . , k, P20 = a0h. Let \scrF = \{ F\omega \} and \scrG = \{ G\omega \} , where F\omega (z) = F (z + \omega ) and G\omega (z) =
= G(z + \omega ), z \in \BbbC . Clearly \scrF and \scrG are two families of entire functions defined on \BbbC . We now
consider following two subcases.
Subcase 1.1. Suppose that one of the families \scrF and \scrG , say \scrF , is normal on \BbbC . Then by Marty’s
theorem F\#(\omega ) = F\#
\omega (0) \leq M for some M > 0 and for all \omega \in \BbbC . Hence, by Lemma 12, we have
F is of order at most 1. Now from (3.25), we obtain
\rho (fn(\scrL (f))m) = \rho (F ) = \rho (G) = \rho (gn(\scrL (g))m) \leq 1. (3.32)
Consequently we get
fn(z)(\scrL (f(z)))m = d3pe
az and gn(z)(\scrL (g(z)))m = d4pe
bz, (3.33)
where a, b, d3, d4 \in \BbbC \setminus \{ 0\} . From (3.23) we see that a+b = 0. We claim that (n+m)\alpha (z) - az \in \BbbC
and (n+m)\beta (z) - bz \in \BbbC . If possible suppose (n+m)\alpha (z) - az \not \in \BbbC and (n+m)\beta (z) - bz \not \in \BbbC .
Let \alpha 1(z) = (n+m)\alpha (z) - az and \beta 1(z) = (n+m)\beta (z) - bz. Note that
T
\bigl(
r, \alpha \prime \bigr) = m
\bigl(
r, \alpha \prime \bigr) \leq m
\bigl(
r, (n+m)\alpha \prime \bigr) +O(1) = m
\bigl(
r, \alpha \prime
1 + a
\bigr)
+O(1) \leq
\leq m
\bigl(
r, \alpha \prime
1
\bigr)
+O(1) = m
\biggl(
(e\alpha 1)\prime
e\alpha 1
\biggr)
+O(1) = S (r, e\alpha 1) .
Clearly \alpha (i) \in S(\alpha 1) for i \in \BbbN . Therefore P1 \in S(\alpha 1) and so
p
P1
\in S(\alpha 1). Similarly we have
p
P2
\in S(\beta 1). Now from (3.30), (3.31) and (3.33), we conclude that e\alpha 1 \in S
\bigl(
e\alpha 1
\bigr)
and e\beta 1 \in S
\bigl(
e\beta 1
\bigr)
,
which is a contradiction. Hence \alpha 1, \beta 1 \in \BbbC and so both \alpha and \beta are polynomials of degree 1.
Finally, we take
f(z) = d5h(z)e
az and g(z) = d6h(z)e
- az, (3.34)
where d5, d6 \in \BbbC \setminus \{ 0\} . Now from (3.34), we get
fn(z)(\scrL (f(z)))m = dn+m
5 hn(z)
\left( a0h(z) + k\sum
j=1
aj
\Biggl(
j\sum
i=0
Cj
i a
j - i h(i)(z)
\Biggr) \right) m
e(n+m)az,
where we define h(0)(z) = h(z). Similarly we obtain
gn(z)(\scrL (g(z)))m =
= dn+m
6 hn(z)
\left( a0h(z) + k\sum
j=1
aj
\Biggl(
j\sum
i=0
Cj
i ( - 1)j - i aj - i h(i)(z)
\Biggr) \right) m
e - (n+m)az.
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214 S. MAJUMDER, A. DAM
Since fn(\scrL (f))m and gn(\scrL (g))m share 0 CM, it follows that
a0h(z) +
k\sum
j=1
aj
\Biggl(
j\sum
i=0
Cj
i a
j - i h(i)(z)
\Biggr)
\equiv
\equiv d7
\left( a0h(z) + k\sum
j=1
aj
\Biggl(
j\sum
i=0
Cj
i ( - 1)j - iaj - i h(i)(z)
\Biggr) \right) , (3.35)
where d7 \in \BbbC \setminus \{ 0\} . But the relation (3.35) does not hold.
Subcase 1.2. Suppose that one of the families \scrF and \scrG , say \scrF is not normal on \BbbC . Now by
Marty’s theorem there exists a sequence of meromorphic functions \{ F (z+\omega j)\} \subset \scrF , where z \in \{ z :
| z| < 1\} and \{ \omega j\} \subset \BbbC is some sequence such that F\#(\omega j) \rightarrow \infty , as | \omega j | \rightarrow \infty . Then by Lemma 13
there exist:
(i) points zj , | zj | < 1,
(ii) positive numbers \rho j , \rho j \rightarrow 0+,
(iii) a subsequence \{ F (\omega j + zj + \rho j\zeta )\} of \{ F (\omega j + z)\}
such that
\^hj(\zeta ) = \rho
- 1
2
j F (\omega j + zj + \rho j\zeta ) \rightarrow \^h(\zeta ) (3.36)
spherically uniformly on compact subset of \BbbC , where \^h(\zeta ) is non-constant holomorphic function
such that \^h\#(\zeta ) \leq \^h\#(0) = 1. Now from Lemma 12 we see that \rho (\^h) \leq 1. By Hurwitz’s theorem
we can see that \^h(\zeta ) \not = 0. In the proof of Zalcman’s lemma (see [15, 23]) we see that
\rho j =
1
F\#(bj)
(3.37)
and
F\#(bj) \geq F\#(\omega j), (3.38)
where bj = \omega j + zj . Let
\v hj(\zeta ) = \rho
1
2
j G(\omega j + zj + \rho j\zeta ). (3.39)
(3.25) yields F (\omega j + zj + \rho j\zeta )G(\omega j + zj + \rho j\zeta ) \equiv 1 and so, from (3.36) and (3.39), we get
\^hj(\zeta )\v hj(\zeta ) \equiv 1. (3.40)
Now, from (3.36) and (3.40), we can deduce that
\v hj(\zeta ) \rightarrow \v h(\zeta ) (3.41)
spherically uniformly on compact subset of \BbbC , where \v h(\zeta ) is some non-constant holomorphic func-
tion in the complex plane. By Hurwitz’s theorem we can see that \v h(\zeta ) \not = 0. From (3.36), (3.40) and
(3.41), we get
\^h(\zeta )\v h(\zeta ) \equiv 1. (3.42)
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 215
Now, from (3.42) and \rho (\^h) \leq 1, we see that
\rho (\^h) = \rho (\v h) \leq 1. (3.43)
Noting that \^h and \v h are transcendental entire functions having no zeros, we observe from (3.43) that
\^h(z) = d8e
cz and \v h(z) = d9e
- cz, (3.44)
where c, d8, d9 \in \BbbC \setminus \{ 0\} such that d8d9 = 1. Also from (3.44), we have
\^h\prime j(\zeta )
\^hj(\zeta )
= \rho j
F \prime (wj + zj + \rho j\zeta )
F (wj + zj + \rho j\zeta )
\rightarrow
\^h\prime (\zeta )
\^h(\zeta )
= c, (3.45)
spherically uniformly on compact subset of \BbbC . Now from (3.37) and (3.45), we obtain\bigm| \bigm| \bigm| \bigm| \bigm| \^h\prime j(0)\^hj(0)
\bigm| \bigm| \bigm| \bigm| \bigm| = \rho j
\bigm| \bigm| \bigm| \bigm| F \prime (\omega j + zj)
F (\omega j + zj)
\bigm| \bigm| \bigm| \bigm| = 1 + | F (\omega j + zj)| 2
| F \prime (\omega j + zj)|
| F \prime (\omega j + zj)|
| F (\omega j + zj)|
=
=
1 + | F (\omega j + zj)| 2
| F (\omega j + zj)|
\rightarrow
\bigm| \bigm| \bigm| \bigm| \bigm| \^h\prime (0)\^h(0)
\bigm| \bigm| \bigm| \bigm| \bigm| = | c| , (3.46)
which implies that
\mathrm{l}\mathrm{i}\mathrm{m}
j\rightarrow \infty
F (\omega j + zj) \not = 0,\infty . (3.47)
From (3.36) and (3.47) we see that
\^hj(0) = \rho
- 1
2
j F (\omega j + zj) \rightarrow \infty . (3.48)
Again from (3.36) and (3.44), we have
\^hj(0) \rightarrow \^h(0) = c1. (3.49)
Now from (3.48) and (3.49) we arrive at a contradiction.
Case 2. Let p \in \BbbC \setminus \{ 0\} . Then from (3.23) we get fn(\scrL (f))mgn(\scrL (g))m \equiv b2, where f and g
are transcendental entire functions. Clearly f and g have no zeros. But this is impossible because
zeros of f and g are of multiplicities at least k \in \BbbN .
Lemma 14 is proved.
Lemma 15. Let f, g be two transcendental meromorphic functions having zeros of multiplicities
at least k \in \BbbN and let F =
fn(\scrL (f))m
p
, G =
gn(\scrL (g))m
p
, where p(z)(\not \equiv 0) is a polynomial and
m,n \in \BbbN such that n >
mk + k2 + k + 2
k
. Suppose fn(\scrL (f))m - p, gn(\scrL (g))m - p share (0, k1)
where k1 \in \BbbN \cup \{ 0\} \cup \{ \infty \} and f, g share (\infty , 0). If H \equiv 0, then either fn(\scrL (f))mgn(\scrL (g))m \equiv p2,
where fn(\scrL (f))m - p, gn(\scrL (g))m - p share 0 CM or fn(\scrL (f))m \equiv gn(\scrL (g))m.
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216 S. MAJUMDER, A. DAM
Proof. Since H \equiv 0, by integration, we get
F \prime
(F - 1)2
= d10
G\prime
(G - 1)2
, where d10 \in \BbbC \setminus \{ 0\} ,
i.e., \biggl(
F1 - p
p
\biggr) \prime
\biggl(
F1 - p
p
\biggr) 2 = d10
\biggl(
G1 - p
p
\biggr) \prime
\biggl(
G1 - p
p
\biggr) 2 ,
where F1 = fn(\scrL (f))m and G1 = fn(\scrL (g))m. This shows that
F1 - p
p
and
G1 - p
p
share 0 CM.
Since F1 - p and G1 - p share (0, k1), it follows that F1 - p and G1 - p share 0 CM. Finally by
integration we get
1
F - 1
\equiv d12G+ d11 - d12
G - 1
, (3.50)
where d11(\not = 0), d12 \in \BbbC . We now consider the following cases.
Case 1. Let d12 \not = 0 and d11 \not = d12. If d12 = - 1, then from (3.50) we have
F \equiv - d11
G - d11 - 1
.
Therefore N(r, a+ 1;G) = N(r,\infty ;F ) = N(r,\infty ; f) +N(r, 0; p). Now in view of Lemma 8 and
the second fundamental theorem we get
(n - m)T (r, g) \leq T (r,G) - mN(r,\infty ; g) - N (r, 0; (\scrL (g))m) + S(r, g) \leq
\leq N(r,\infty ;G) +N(r, 0;G) +N(r, a+ 1;G) - mN(r,\infty ; g) - N (r, 0; (\scrL (g))m) + S(r, g) \leq
\leq N(r, 0; g) +N (r, 0; (\scrL (g))m) +N(r,\infty ; f) - N (r, 0; (\scrL (g))m) + S(r, g) \leq
\leq N(r, 0; g) +N(r,\infty ; g) + S(r, g) \leq
\leq 1
k
N(r, 0; g) +N(r,\infty ; g) + S(r, g) \leq k + 1
k
T (r, g) + S(r, g),
which is contradiction since n >
mk + k + 1
k
.
If d12 \not = - 1, from (3.50) we obtain
F -
\biggl(
1 +
1
d12
\biggr)
\equiv - d11
d212
\biggl(
G+
d11 - d12
d12
\biggr) .
So, N
\biggl(
r,
d12 - d11
d12
;G
\biggr)
= N(r,\infty ;F ) = N(r,\infty ; f) + N(r, 0; p). By using Lemma 8 and the
same argument as used in the case when d12 = - 1, we can get a contradiction.
Case 2. Let d12 \not = 0 and d11 = d12. If d12 = - 1, then from (3.50) we have FG \equiv 1, i.e.,
fn(\scrL (f))mgn(\scrL (g))m \equiv p2, where fn(\scrL (f))m - p and gn(\scrL (g))m - p share 0 CM.
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 217
If d12 \not = - 1, from (3.50) we have
1
F
\equiv d12G
(1 + d12)G - 1
.
Therefore N
\biggl(
r,
1
1 + d12
;G
\biggr)
= N(r, 0;F ). So, in view of Lemmas 1 and 8 and the second funda-
mental theorem we get
(n - m)T (r, g) \leq T (r,G) - mN(r,\infty ; g) - N (r, 0; (\scrL (g))m) + S(r, g) \leq
\leq N(r,\infty ;G) +N(r, 0;G) +N
\biggl(
r,
1
1 + d12
;G
\biggr)
- mN(r,\infty ; g) -
- N (r, 0; (\scrL (g))m) + S(r, g) \leq N(r, 0; g)+
+N (r, 0; (\scrL (g))m) +N(r, 0;F ) - N (r, 0; (\scrL (g))m) + S(r, g) \leq
\leq N(r, 0; g) +N(r, 0; f) +N(r, 0;\scrL (f)) + S(r, g) \leq
\leq N(r, 0; g) +N(r, 0; f) +Nk+1(r, 0; f) + kN(r,\infty ; f) + S(r, g) \leq
\leq 1
k
T (r, g) +
1
k
T (r, f) + T (r, f) + k T (r, f) + S(r, f) + S(r, g).
We suppose that there exists a set I with infinite measure such that T (r, f) \leq T (r, g) for r \in I and
so for r \in I we have (n - m) T (r, g) \leq k2 + k + 2
k
T (r, g) + S(r, g), which is a contradiction
since n >
mk + k2 + k + 2
k
.
Case 3. Let d12 = 0. From (3.50) we obtain
F \equiv G+ d11 - 1
d11
.
If d11 \not = 1 then we obtain N(r, 1 - d11;G) = N(r, 0;F ). We can similarly deduce a contradiction
as in Case 2. Therefore d11 = 1 and so we obtain F \equiv G, i.e., fn(\scrL (f))m \equiv gn(\scrL (g))m.
Lemma 15 is proved.
Lemma 16 [1]. Let f and g be non-constant meromorphic functions sharing (1, k1), where
2 \leq k1 \leq \infty . Then
N(r, 1; f | = 2) + 2N(r, 1; f | = 3) + . . .+ (k1 - 1)N(r, 1; f | = k1) + k1 NL(r, 1; f)+
+(k1 + 1)NL(r, 1; g) + k1N
(k1+1
E (r, 1; g) \leq N(r, 1; g) - N(r, 1; g).
4. Proof of Theorem 1. Let F =
fn(\scrL (f))m
p
and G =
gn(\scrL (g))m
p
. Clearly F, G share (1, k1)
except for the zeros of p and f, g share (\infty , 0).
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218 S. MAJUMDER, A. DAM
Case 1. Let H \not \equiv 0.
From (3.1) it can be easily calculated that the possible poles of H occur at (i) multiple zeros of
F and G, (ii) those 1 points of F and G whose multiplicities are different, (iii) those poles of F and
G whose multiplicities are different, (iv) zeros of F \prime which are not the zeros of F (F - 1), (v) zeros
of G\prime which are not the zeros of G(G - 1). Since H has only simple poles we get
N(r,\infty ;H) \leq N\ast (r,\infty ; f, g) +N\ast (r, 1;F,G) +N(r, 0;F | \geq 2) +N(r, 0;G | \geq 2)+
+N0(r, 0;F
\prime ) +N0(r, 0;G
\prime ), (4.1)
where N0(r, 0;F
\prime ) is the reduced counting function of those zeros of F \prime which are not the zeros of
F (F - 1) and N0(r, 0;G
\prime ) is similarly defined. Now from Nevanlinna’s fundamental estimate of the
logarithmic derivative we obtain m(r,H) = S(r, F ) + S(r,G). Since
T (r, F ) \leq [n+ (k + 1)m]T (r, f) + S(r, f), T (r,G) \leq [n+ (k + 1)m]T (r, g) + S(r, g),
it follows that
m(r,H) = S(r, f) + S(r, g).
Let z0 be a simple zero of F - 1 but p(z0) \not = 0. Clearly z0 is a simple zero of G - 1. Then an
elementary calculation gives that H(z) = O(z - z0), which proves that z0 is a zero of H. By the
first fundamental theorem of Nevanlinna we get
N(r, 1;F | = 1) \leq N(r, 0;H) \leq T (r,H) +O(1) =
= N(r,\infty ;H) +m(r,H) +O(1) \leq N(r,\infty ;H) + S(r, f) + S(r, g). (4.2)
By using (4.1) and (4.2), we obtain
N(r, 1;F ) \leq N(r, 1;F | = 1) +N(r, 1;F | \geq 2) \leq
\leq N\ast (r,\infty ; f, g) +N(r, 0;F | \geq 2) +N(r, 0;G | \geq 2) +N\ast (r, 1;F,G)+
+N(r, 1;F | \geq 2) +N0(r, 0;F
\prime ) +N0(r, 0;G
\prime ) + S(r, f) + S(r, g) \leq
\leq N(r,\infty ; f) +N(r, 0;F | \geq 2) +N(r, 0;G | \geq 2) +N\ast (r, 1;F,G)+
+N(r, 1;F | \geq 2) +N0(r, 0;F
\prime ) +N0(r, 0;G
\prime ) + S(r, f) + S(r, g). (4.3)
Now in view of Lemmas 2 and 16 we have
N0(r, 0;G
\prime ) +N(r, 1;F | \geq 2) +N\ast (r, 1;F,G) \leq
\leq N0(r, 0;G
\prime ) +N(r, 1;F | = 2) +N(r, 1;F | = 3) + . . .+N(r, 1;F | = k1)+
+N
(k1+1
E (r, 1;F ) +NL(r, 1;F ) +NL(r, 1;G) +N\ast (r, 1;F,G) \leq
\leq N0(r, 0;G
\prime ) - N(r, 1;F | = 3) - . . . - (k1 - 2)N(r, 1;F | = k1) -
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ON CERTAIN NONLINEAR DIFFERENTIAL MONOMIAL SHARING NON-ZERO POLYNOMIAL 219
- (k1 - 1)NL(r, 1;F ) - k1NL(r, 1;G) - (k1 - 1)N
(k1+1
E (r, 1;F )+
+N(r, 1;G) - N(r, 1;G) +N\ast (r, 1;F,G) \leq
\leq N0(r, 0;G
\prime ) +N(r, 1;G) - N(r, 1;G) - (k1 - 2)NL(r, 1;F ) - (k1 - 1)NL(r, 1;G) \leq
\leq N(r, 0;G\prime | G \not = 0) - (k1 - 2)NL(r, 1;F ) - (k1 - 1)NL(r, 1;G) \leq
\leq N(r, 0;G) +N(r,\infty ; g) - (k1 - 2)NL(r, 1;F ) - (k1 - 1)NL(r, 1;G) =
= N(r, 0;G) +N(r,\infty ; g) - (k1 - 2)N\ast (r, 1;F,G) - NL(r, 1;G). (4.4)
Hence, by using (4.3), (4.4) and Lemma 1, we get from second fundamental theorem that
T (r, F ) \leq N(r, 0;F ) +N(r,\infty ;F ) +N(r, 1;F ) - N0(r, 0;F
\prime ) \leq
\leq 2N(r,\infty , f) +N2(r, 0;F ) +N(r, 0;G | \geq 2) +N(r, 1;F | \geq 2)+
+N\ast (r, 1;F,G) +N0(r, 0;G
\prime ) + S(r, f) + S(r, g) \leq
\leq 3N(r,\infty ; f) +N2(r, 0;F ) +N2(r, 0;G) - (k1 - 2)N\ast (r, 1;F,G)+
+S(r, f) + S(r, g) \leq 3N(r,\infty ; f) + 2N(r, 0; f) +N2 (r, 0; (\scrL (f))m)+
+2N(r, 0; g) +mN2(r, 0;\scrL (g)) - (k1 - 2)N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq 3N(r,\infty ; f) + 2N(r, 0; f) +N (r, 0; (\scrL (f))m) + 2N(r, 0; g)+
+m Nk+2(r, 0; g) +mkN(r,\infty ; g) - (k1 - 2)N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq (3 +mk)N(r,\infty ; f) + 2N(r, 0; f) + 2N(r, 0; g) +mN(r, 0; g)+
+N (r, 0; (\scrL (f))m) - (k1 - 2)N\ast (r, 1;F,G) + S(r, f) + S(r, g). (4.5)
Now, by using Lemmas 7 and 8, we get from (4.5)
(n - m)T (r, f) \leq T (r, F ) - mN(r,\infty ; f) - N (r, 0; (\scrL (f))m) + S(r, f) \leq
\leq (3 + (k - 1)m)N(r,\infty ; f) + 2N(r, 0; f) + 2N(r, 0; g) +mN(r, 0; g) -
- (k1 - 2)N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq (k + 1)(3 + (k - 1)m)
k(n+m+ (m - 2)k - 1)
(T (r, f) + T (r, g))+
+
2
k
(T (r, f) + T (r, g)) +
3 + (k - 1)m
n+m+ (m - 2)k - 1
N\ast (r, 1;F,G)+
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
220 S. MAJUMDER, A. DAM
+mT (r, g) - (k1 - 2)N\ast (r, 1;F,G) + S(r, f) + S(r, g) \leq
\leq
\Biggl(
(mk + 4)n+m2k2 +
\bigl(
m2 + 3m - 2
\bigr)
k + 2(m+ 1)
k(n+m+ (m - 2)k - 1)
\Biggr)
T (r) + S(r). (4.6)
In a similar way we can obtain
(n - m)T (r, g) \leq
\Biggl(
(mk + 4)n+m2k2 +
\bigl(
m2 + 3m - 2
\bigr)
k + 2(m+ 1)
k(n+m+ (m - 2)k - 1)
\Biggr)
T (r) + S(r). (4.7)
Combining (4.6) and (4.7) we see that
(n - m)T (r) \leq
\Biggl(
(mk + 4)n+m2k2 +
\bigl(
m2 + 3m - 2
\bigr)
k + 2(m+ 1)
k(n+m+ (m - 2)k - 1)
\Biggr)
T (r) + S(r),
i.e.,
(k(n - K1)(n - K2))T (r) \leq S(r), (4.8)
where
K1 =
(2 - m)k2 + (m+ 1)k + 4 +
\surd
L1
2k
, K2 =
(2 - m)k2 + (m+ 1)k + 4 -
\surd
L1
2k
,
L1 =
\bigl[
(2 - m)k2 + (m+ 1)k + 4
\bigr] 2
+ 8k
\bigl\{ \bigl(
m2 - m
\bigr)
k2 +
\bigl(
m2 +m - 1
\bigr)
k + (m+ 1)
\bigr\}
=
= m2k4 + 9m2k2 + 2mk2 + 6m2k3 - 6mk3+
+4k4(1 - m) + 16k(m+ 1) + 9k2 + 4k3 + 16 <
< m2k4 + 9m2k2 + 6m2k3 + 10mk2 - 2mk3 + 16(3m - 1)k+
+k2 + 64 + 8k2(1 - m) + 4k3(1 - m) + 32k(1 - m) \leq
\bigl[
mk2 + (3m - 1)k + 8
\bigr] 2
.
Therefore,
K1 <
(2 - m)k2 + (m+ 1)k + 4 +mk2 + (3m - 1)k + 8
2k
=
k2 + 2mk + 6
k
.
Since n \geq k2 + 2mk + 6
k
, (4.8) leads to a contradiction.
Case 2. Let H \equiv 0. Then theorem follows from Lemmas 15, 11 and 14.
Theorem 1 is proved.
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(1994).
Received 18.05.18,
after revision — 29.01.19
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 2
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| id | umjimathkievua-article-99 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:04:18Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/ed/a3a1cb945b6212a3865b68a6beb35ded.pdf |
| spelling | umjimathkievua-article-992025-03-31T08:48:28Z On certain non-linear differential monomial sharing non-zero polynomial On certain non-linear differential monomial sharing non-zero polynomial Majumder, S. Dam, A. Majumder, S. Dam, A. Uniqueness Meromorphic function small functions Non-linear differential polynomials and normal family Uniqueness Meromorphic function small functions Non-linear differential polynomials and normal family UDC 517.5 With the idea of normal family we study the uniqueness of meromorphic functions $f$ and $g$ when $f^{n}(\mathcal{L}(f))^{m}-p$ and&nbsp;$g^{n}(\mathcal{L}(g))^{m}-p$ share two values, where&nbsp;$\mathcal{L}(f)= a_{k}f^{(k)}+a_{k-1} f^{(k-1)}+\ldots+a_{1} f'+a_{0}f,$&nbsp;$a_{k}(\ne 0),a_{k-1},\ldots,a_{1},a_{0}\in\mathbb{C}$ and $p(z)(\not\equiv 0)$ is a polynomial.&nbsp;The obtained result significantly improves and generalizes the result in [A. Banerjee, S. Majumder, On certain non-linear differential polynomial sharing a non-zero polynomial, Bol. Soc. Mat. Mex. (2016),https://doi.org/10.1007/s40590-016-0156-0]. УДК 517.5 Про нелінійний диференціальний одночлен зі спільним ненульовим многочленом На базі ідеї про нормальні сім'ї функцій вивчається єдиність мероморфних функцій $f$ і $g$ у випадку, коли $f^{n}(\mathcal{L}(f))^{m}-p$ і&nbsp;$g^{n}(\mathcal{L}(g))^{m}-p$ мають спільні значення, де&nbsp;$\mathcal{L}(f)= a_{k}f^{(k)}+a_{k-1} f^{(k-1)}+\ldots +a_{1} f'+a_{0}f$, $a_{k}(\ne 0),a_{k-1},\ldots,a_{1},a_{0}\in\mathbb{C}$, а $p(z)(\not\equiv 0)$ — поліном.&nbsp;&nbsp;Отриманий результат є істотним узагальненням результату з&nbsp;[A. Banerjee, S. Majumder, &nbsp;On certain non-linear differential polynomial sharing a non-zero polynomial, Bol. Soc. Mat. Mex. (2016),&nbsp;https://doi.org/10.1007/s40590-016-0156-0]. Institute of Mathematics, NAS of Ukraine 2021-02-22 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/99 10.37863/umzh.v73i2.99 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 2 (2021); 201 - 221 Український математичний журнал; Том 73 № 2 (2021); 201 - 221 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/99/8939 Copyright (c) 2021 Sujoy Majumder, Arup Dam |
| spellingShingle | Majumder, S. Dam, A. Majumder, S. Dam, A. On certain non-linear differential monomial sharing non-zero polynomial |
| title | On certain non-linear differential monomial sharing non-zero polynomial |
| title_alt | On certain non-linear differential monomial sharing non-zero polynomial |
| title_full | On certain non-linear differential monomial sharing non-zero polynomial |
| title_fullStr | On certain non-linear differential monomial sharing non-zero polynomial |
| title_full_unstemmed | On certain non-linear differential monomial sharing non-zero polynomial |
| title_short | On certain non-linear differential monomial sharing non-zero polynomial |
| title_sort | on certain non-linear differential monomial sharing non-zero polynomial |
| topic_facet | Uniqueness Meromorphic function small functions Non-linear differential polynomials and normal family Uniqueness Meromorphic function small functions Non-linear differential polynomials and normal family |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/99 |
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