Study of Solutions of Certain Kind of Non-Linear Differential Equations

Abstract

In this paper we study about the existence of solutions of certain kind of non-linear differential and differential-difference equations. We give partial answer to a problem which was asked by chen et al. in [13].

Full text

arXiv:2202.04062v1 [math.CV] 8 Feb 2022
STUDY OF SOLUTIONS OF CERTAIN KIND OF NON-LINEAR
DIFFERENTIAL EQUATIONS
GARIMA PANT AND MANISHA SAINI
Abstract. In this paper we study about the existence of solutions of certain kind
of non-linear differential and differential-difference equations. We give partial an-
swer to a problem which was asked by chen et al. in [13].
We assume that the readers are familiar with the standard notations of Nevanlinna
theory such as characteristic function T(r, f), proximity function m(r, f), integrated
counting function N(r, f), unintegrated counting function n(r, f) and first main the-
orem for a meromorphic function fwhich is defined in the complex plane [2, 3, 18].
To make our paper self contained, we present the elementary definitions of the order
of growth ρ(f) and exponent of convergence of zeros λ(f) of a meromorphic function
f
ρ(f) = lim sup
r→∞
log+T(r, f)
log r
and
λ(f) = lim sup
r→∞
log+n(r, 1/f)
log r,
respectively. If fis a meromorphic function, then S(r, f) denotes such quantities
which are of growth o(T(r, f )) as r→ ∞, outside of a possible exceptional set of
finite linear measure. A meromorphic function h(z) is said to a small function of f(z)
if T(r, h) = S(r, f) and vice versa. Note that finite sum of the quantities which are of
S(r, f) kind is again S(r, f ). A differential polynomial in fis a polynomial in fand
its derivatives with small function of fas its coefficients. For a given meromorphic
function fand a constant c,f(z+c) is known as a shift of f.
To study solubility and existence of solutions of non-linear differential equations or
differential-difference equations are interesting but also difficult. Many researchers
have been studied these equations, one may refer to see [9–11, 15, 16].
In this paper, our aim is to present some studies on the solutions of certain kind
of non-linear differential equations and differential-difference equations which have
been studied even before. In 2021, Chen et al. [13] studied the following non-linear
differential-difference equation
fn(z) + wf n−1(z)f′(z) + q(z)eQ(z)f(z+c) = u(z)eν(z),(1)
2020 Mathematics Subject Classification. 34M05, 30D35, 39B32.
Key words and phrases. Nevanlinna theory, entire function, difference equation, differential-
difference equation.
Research work of the first author is supported by research fellowship from University Grants
Commission (UGC), New Delhi, India.
The second author is Senior Research Fellow (UGC, New Delhi, India).
1

2 G. PANT, M. SAINI
where nis a natural number, wand c6= 0 are constants. q, Q, u and νare non
constant polynomials.
In our first result, we study the same non-linear differential-difference equation (1)
and prove the following result:
Theorem A. Suppose that n≥3for w6= 0 and n≥2for w= 0. Let fbe a finite
order transcendental entire solution of equation (1) satisfying λ(f)< ρ(f). Then f
satisfies one of the following:
(i)ρ(f)<deg ν= deg Qand f=Ce−z /w, where Cis a non-zero constant.
(ii)ρ(f) = deg ν= deg Q
Note that every non-vanishing transcendental entire function fof non-zero finite
order satisfies λ(f)< ρ(f). Therefore we get an immediate corollary from the above
theorem.
Corollary 1. Suppose that n≥3for w6= 0 and n≥2for w= 0. If fis a non-
vanishing transcendental solution of non-zero finite order of equation (1). Then f
satisfies the conclusion of Theorem A.
Prior to Theorem A , Chen et al. [13] proved the following result:
Theorem 1. Assume n≥3for w6= 0 and n≥2for w= 0. Let fbe a non-vanishing
transcendental solution of finite order of equation (1). Then each fsatisfies one of
the following:
(i)ρ(f)<deg ν= deg Qand f=Ce−z /w, where Cis a constant.
(ii)ρ(f) = deg Q≥deg ν
After stating the above theorem, Chen et al. [13] posed a problem namely Problem
1.10, in which they asked “Can the conclusion (ii) in the above theorem be further
improved to ρ(f) = deg Q= deg ν?”
In the Corollary 1, we have given partial answer to that problem.
Chen et al. [13] also studied a non-linear differential-difference equation in which
left hand side of equation (1) is unaltered and right hand side of the same equation
is replaced by P1eλz +P2e−λz. Then they provided the following result:
Theorem 2. Suppose fis a transcendental entire solution with finite order of equa-
tion
fn(z) + wf n−1(z)f′(z) + q(z)eQ(z)f(z+c) = P1eλz +P2e−λz,(2)
where nis a natural number, wis a constant and c, P1, P2, λ are non-zero constants,
q6≡ 0is a polynomial and Qis a non-constant polynomial. Then the following
conclusions hold:
(i) If n≥4for w6= 0 and n≥3for w= 0, then each solution fsatisfies
ρ(f) = deg Q= 1.
(ii) If n≥1and fis a solution which belongs to Γ0={eα(z):α(z)is a non-constant
polynomial}, then
f(z) = e(λz/n)+a, Q(z) = −n+ 1
nλz +b

STUDY OF SOLUTIONS... 3
or
f(z) = e(−λz/n)+a, Q(z) = n+ 1
nλz +b,
where aand bare constants.
Now it is natural to ask what can we say about the solutions of equation (2) when
n= 3 and w6= 0? In this sequence, we prove the following result:
Theorem B. Suppose that w, c, P1, P2and λare non-zero constants, q(z)is a non-
zero polynomial, Q(z)is a non constant polynomial. Then there does not exist any
finite order transcendental entire function satisfying equation
f3(z) + wf 2(z)f′(z) + q(z)eQ(z)f(z+c) = P1eλz +P2e−λz.(3)
with a finite non-zero Borel exceptional value.
The following example shows that if fis a transcendental entire function satisfying
equation (3), then fmay have zero as a Borel exceptional value.
Example 1. The function f(z) = 2e3zsatisfies differential-difference equation
f3+f2f′+e12zf(z+ log 2) = 32e9z+ 16e−9z.
Here f(z)has zero as a Borel exceptional value.
In 2020, Xue [7] and Chen et al. [1] studied a new kind of non-linear differential
equations and proved the following results:
Theorem 3. [7] Suppose n≥2and kare integers, Pn−1(z, f )is an algebraic dif-
ferential polynomial in f(z)of degree at most n−1. Let Piand αibe non-zero
constants for i= 1,2,3, and |α1|>|α2|>|α3|. If fis a transcendental entire
function satisfying the following differential equation
fn(z) + Pn−1(z, f ) = P1eα1z+P2eα2z+P3eα3z,(4)
then f(z) = Ceα1z/n, where Cis a non-zero constant such that Cn=P1, and
α1, α2, α3are in one line.
Note that an algebraic differential polynomial P(z, f ) is a polynomial in fwhen-
ever fis a polynomial in f(z) and its derivatives with polynomials as its coefficients.
Theorem 4. [1] Suppose n≥5is an integer and Qd(z, f )is a differential polynomial
in fof degree d≤n−4with rational functions as its coefficients. Also suppose
p1(z), p2(z), p3(z)are non-zero rational functions and α1, α2, α3are non constant
polynomials such that α′
1, α′
2, α′
3are distinct to each other. If fis a meromorphic
solution with finitely many poles of the following differential equation
fn(z) + Qd(z, f ) = p1(z)eα1(z)+p2(z)eα2(z)+p3(z)eα3(z),(5)
then α′
1
α′
2
,α′
2
α′
3
are rational numbers, and f(z)must be of
f(z) = r(z)eP(z),
where r(z)is a non-zero rational function and P(z)is a non constant polynomial.
Furthermore, there must exist positive integers m1, m2, m3with {m1, m2, m3}=
{1,2,3}and distinct integers k1, k2with 1≤k1, k2≤dsuch that α′
m1:α′
m2:α′
m3=
n:k1:k2,nP ′=α′
m1,Qd(z, f )≡pm2(z)eαm2(z)+pm3(z)eαm3(z).

4 G. PANT, M. SAINI
After stating the above result, Chen et al. [1] mentioned that “in Theorem 4, the
condition n≥5 is necessary” in a remark namely ’Remark 1.2’. By some examples,
they have also shown that if n≤4,then Theorem 4 may not hold. In this sequence
we prove the following results:
Theorem C. Let n≥2, and P(z, f )be a linear differential polynomial in f(z),
its derivatives with coefficients being rational functions. Suppose Pj(z)and αj(z)
(j= 1,2,3) are non-zero rational functions and non constant polynomials, respec-
tively. Also α′
1(z), α′
2(z)and α′
3(z)are distinct to each other. If fis a transcendental
meromorphic solution of the following differential equation
fn(z) + P(z, f ) = P1(z)eα1(z)+P2(z)eα2(z)+P3(z)eα3(z),(6)
such that fhas finitely many poles, then fis of finite order but not in the form
of f(z) = s(z)eq(z), where s(z)is a non-zero rational function and q(z)is a non
constant polynomial.
Note that Theorem 4 is an improvement of Theorem 3 which can be seen by the
following examples.
Example 2. The differential equation
f2+f′′ −2f′−f= 2e2z+ 2e3z+e4z
has a solution f(z) = 1 + ez+e2zwhich is not in the required form as in Theorem 3.
Example 3. The differential equation
f3−2f′−4f=e9z+ 2e3z+ 6ez
has a solution f(z) = e3z+ 2 which is not in the required form as in Theorem 3.
Example 4. The differential equation
f4+3
2f′+f= 16e8z−32e6z+ 23e4z
has a solution f(z) = 2e2z−1which is not in the required form as in Theorem 3.
We observe that the above examples also hold Theorem C. Next, we combine
Theorem 4, Theorem C and we have an immediate result:
Corollary 2. Let n≥5, and P(z, f )be a linear differential polynomial in f(z), its
derivatives with rational functions as its coefficients. Suppose Pj(z)and αj(z) (j=
1,2,3) are non-zero rational functions and non constant polynomials, respectively.
Also α′
1(z), α′
2(z), α′
3(z)are distinct to each other. Then there exist no meromorphic
solution having finitely many poles which satisfies equation (6).
Before stating next result we first define that Q⋆
2(z, f ) is a differential polynomial
of degree 2 with rational functions ai(z) as its coefficients and at least one coefficient
of f(l)(l≥0) must be non-zero.
Theorem D. Suppose P1, P2, P3are non-zero constants, and α1, α2, α3are non-zero
and distinct constants. If fis an entire solution of the following differential equation
fn(z) + Q⋆
2(z, f ) = P1eα1z+P2eα2z+P3eα3z,(7)

STUDY OF SOLUTIONS... 5
where n≥3and satisfying λ(f)< ρ(f). Then ρ(f) = 1 and f(z)must be in the
form of f(z) = Ceaz, where Cand aare non-zero constant. Particularly, we can
say that there exist positive integers k1, k2, k3with {k1, k2, k3}={1,2,3}such that
αk1:αk2:αk3=n: 2 : 1,na =αk1and C=P1/n
k1.
By the following examples, we show that our hypothesis keeps necessary conditions
to conclude the result.
Example 5. The differential equation
f3+ 3f2=e3z−6e2z+ 9ez
has a solution f(z) = ez−3, which is not in the required form, here 3f2is not in
the form of Q⋆
2(z, f ).
Example 6. The differential equation
f3+ 2f2+f′2=e3z−3e2z+ 4ez
has a solution f(z) = ez−2, which is not in the required form, here 2f2+f′2is not
in the form of Q⋆
2(z, f ).
Example 7. The differential equation
f3+ff ′−f=e3z−2e2z+ez
has a solution f(z) = ez−1, which is not in the required form and λ(f) = ρ(f).
Example 8. The differential equation
f4−f2−5f′2=e4z−4e3z−2ez
has a solution f(z) = ez−1, which is not in the required form and given coefficients
of exponent in the right hand side of the above equation don’t preserve required ratio,
here −f2−5f′2is not in the form of Q⋆
2(z, f ).
Note that Theorem D is an improvement of Theorem 3.
1. Preliminary Results
In this section, we collect those lemmas which we use to prove our desired results.
The following lemma is related to the proximity function of logarithmic derivative
of a meromorphic function f.
Lemma 1. [2] Suppose fis a transcendental meromorphic function and k≥1is an
integer. Then
mr, f(k)
f=S(r, f).
The following lemma estimates the characteristic function of a shift of a meromor-
phic function f.
Lemma 2. [14] Suppose fis a meromorphic function of finite order ρand cis a
non-zero complex constant. Then for every ǫ,
T(r, f(z+c)) = T(r, f ) + O(rρ−1+ǫ) + O(log r).

6 G. PANT, M. SAINI
Next lemma is the difference analogue of the lemma on the logarithmic derivative
of a meromorphic function fhaving finite order.
Lemma 3. [14] Suppose fis a meromorphic function with ρ(f)<∞and c1, c2∈C
such that c16=c2, then for each ǫ > 0, we have
mr, f(z+c1)
f(z+c2)=O(rρ−1+ǫ).
The following lemma estimates the characteristic function of an exponential poly-
nomial f. This lemma can be seen in [17].
Lemma 4. Suppose fis an entire function given by
f(z) = A0(z) + A1(z)ew1zs+A2(z)ew2zs+... +Am(z)ewmzs,
where Ai(z); 0 ≤i≤mdenote either exponential polynomial of degree < s or poly-
nomial in z,wi; 1 ≤i≤mdenote the constants and sdenotes a natural number.
Then
T(r, f) = C(Co(W0)) rs
2π+o(rs),
Here C(Co(W0)) is the perimeter of the convex hull of the set W0={0,˜w1,˜w2, ..., ˜wm}.
The following lemma plays a key role to prove all our results in this article.
Lemma 5. [12] Suppose f1, f2, ..., fn(n≥2) are meromorphic functions and h1, h2, ..., hn
are entire functions satisfying
(1) Pn
i=1 fiehi≡0.
(2) For 1 ≤j < k ≤n,hj−hkare not constants .
(3) For 1 ≤i≤n, 1≤m < k ≤n,
T(r, fi) = o(T(r, e(hm−hk))) as r→ ∞, outside a set of finite logarithmic
measure.
Then fi≡0 (i= 1,2, ..., n).
2. Proof of Theorems
Proof of Theorem A.Given that fis a transcendental entire solution of finite
order of equation (1) satisfying λ(f)< ρ(f).
Case 1: First, we consider n≥3 for w6= 0, then using Weierstrass factorisation
theorem, fcan be written as
f(z) = α(z)eβ(z),(8)
where β(z) is a non constant polynomial of degree tand α(z) is an entire function
satisfies ρ(α) = λ(f)< ρ(f) = t.
For simplicity, rewriting (1) as
fn+wf n−1f′+qeQfc=ueν,(9)
where fc=f(z+c). Now, we study the following three sub cases.
Case 1.1: When ρ(f)<deg Q, then we have T(r, f ) = S(r, eQ).

STUDY OF SOLUTIONS... 7
Also, it is well known that a meromorphic function and its derivative have same
order, thus we have T(r, f′) = S(r, eQ) = T(r, f). Using Lemma 2, we obtain
T(r, fc) = S(r, eQ)
.
From equation (9), we get
T(r, ueν) = T(r, fn+wf n−1f′+qeQfc)
≤T(r, fn−1) + T(r, f +wf ′) + T(r, q) + T(r, eQ) + T(r, fc) + log 2
≤nT (r, f) + T(r, f′) + T(r, eQ) + T(r, fc) + T(r, q) + O(1)
=T(r, eQ) + S(r, eQ) (Since T(r, f) = T(r, f′) = T(r, fc) = S(r, eQ))
and
T(r, eQ) = Tr, qfceQ
qfc
≤T(r, qfceQ+wf n−1f′+fn) + T(r, wf n−1f′+fn) + Tr, 1
qfc+O(1)
≤T(r, ueν) + nT (r, f) + T(r, f′) + T(r, fc) + T(r, q) + O(1)
=T(r, ueν) + S(r, eQ).
From the last two inequalities,
T(r, eQ) = T(r, ueν) + S(r, eQ).
This implies deg Q= deg ν. Now Differentiation of equation (9) gives
nfn−1f′+w(n−1)fn−2(f′)2+wf n−1f′′ +AeQ= (u′+uν′)eν,(10)
where, A=q′fc+qQ′fc+qf ′
c. Eliminating eνfrom equation (9) and (10), then we
obtain
D1eQ+D2= 0,
where
D1= (u′+uν′)qfc−uA,
D2= (u′+uν′)(fn+wf n−1f′)−u(nfn−1f′+w(n−1)fn−2(f′)2+wf n−1f”).
Using Lemma 5, we have D1=D2≡0.
From D1= 0, we get
u′
u+ν′=q′
q+f′
c
fc
+Q′.
On doing integration of the above equation, we have
qfceQ=C1ueν.
where C1is a non-zero constant. If C1= 1, then qfceQ=ueνand by equation (9),
we obtain
fn+wf n−1f′= 0.
This yields that f=Ce−z/w , where Cis a non-zero constant. This is the required
conclusion (i).
If C16= 1, then qfceQ=C1ueν. This implies that f=C1u−ceν−c−Q−c
q−c,where u−c=

8 G. PANT, M. SAINI
u(z−c), ν−c=ν(z−c) and Q−c=Q(z−c).
Substituting the above value of finto the equation (9), we have
C1u−c
q−cn−1"C1u−c
q−c
+C1w u−c
q−c′
+u−c
q−c
(ν−c−Q−c)′!#en(ν−c−Q−c)=
(1 −C1)ueν.
Since deg Q= deg ν > ρ(f) = deg β≥1 and ρ(f) = deg(ν−c−Q−c). Using Lemma
5, we have (1 −C1)u≡0. This implies either C1= 1 or u= 0, which is not possible.
Case 1.2: When ρ(f) = deg Q, we substitute the value of ffrom equation (8) into
equation (9)
(αn+wαn−1(α′+β′α))enβ +qαceQ+βc=ueν,(11)
where βc=β(z+c). Now, we consider β(z) = atzt+at−1zt−1+... +a0,(at6= 0)
and Q(z) = btzt+bt−1zt−1+... +b0,(bt6= 0) are polynomials of degree tprovided
ρ(f) = deg β=t.
Let deg ν < deg Q=ρ(f) = deg β=t, then from equation (11), we have
[αn+wαn−1(α′+αβ′)]eα1enatzt+qαceα2e(bt+at)zt=ueν,(12)
where α1=n(at−1zt−1+...+a0) and α2= (bt−1+ctat+at−1)zt−1+...+ (b0+cat+a0)
are polynomials of degree at most t−1.
If at6=±bt, then using Lemma 5, we get u≡0, which is not possible.
If at=bt, then again using Lemma 5, we get the same contradiction.
If at=−bt, then
[αn+wαn−1(α′+αβ′)]eα1enatzt=ueν−qαceα2.
Using Lemma 5, we get
αn−1(α+w(α′+αβ′)) ≡0.
This gives α=C2e−z/w−β, where C2is a non-zero constant, which implies ρ(α) =
deg β=ρ(f). This is a contradiction to the fact that ρ(α)< ρ(f). Thus deg ν <
deg Qis not possible.
Let deg ν > deg Q=ρ(f) = tand we consider ν(z) = νszs+νs−1zs−1+... +ν0such
that s > t, then from equation (11), we have
[αn+wαn−1(α′+αβ′)]eα1enatzt+qαceα2e(bt+at)zt=ueα3eνszs,
where α1,α2are same polynomials as in equation (12) and α3=νs−1zs−1+... +ν0
is a polynomial of degree at most s−1.
If at6=±bt, then using Lemma 5, we get u≡0, which is not possible.
If at=bt, then again using Lemma 5, we get u≡0, which is not possible.
If at=−bt, then we also get same contradiction as for the above cases. Therefore
deg ν > deg Qis not possible. Hence deg ν= deg Q=ρ(f), which is the required
conclusion (ii).
Case 1.3: If ρ(f)>deg Q, then from equation (9)
T(r, ueν) = T(r, fn+wf n−1f′+qeQfc)
≤T(r, fn−1) + T(r, f +wf ′) + T(r, q) + T(r, eQ) + T(r, fc) + O(1)

STUDY OF SOLUTIONS... 9
≤(n+ 1)T(r, f) + T(r, f+wf ′
f) + S(r, f)
= (n+ 1)T(r, f) + S(r, f ).
This implies deg ν≤ρ(f) = deg β. Now, we discuss the following sub cases:
(i) If deg ν < deg β, we consider β(z) = atzt+at−1zt−1+... +a0,(at6= 0) and
Q(z) = bdzd+bd−1zd−1+... +b0,(bd6= 0) such that d < ρ(f) = t, then from
equation (11), we have
[αn+wαn−1(α′+αβ′)]eα1enatzt+qαce˜α2eatzt=ueν,
where ˜α2=Q(z) + at−1zt−1+... +a0is a polynomial of degree at most t−1.
Then using Lemma 5 into the above equation, we get u≡0, which is not
possible.
(ii) If deg ν= deg β, then we consider ν(z) = νtzt+νt−1zt−1+... +ν0, νt6= 0. From
equation (11)
[αn+wαn−1(α′+αβ′)]eα1enatzt+qαce˜α2eatzt=ue ˜α3eνtzt,
where ˜α3=νt−1zt−1+... +ν0is a polynomial of degree at most t−1.
If νt6=natand νt6=at, then using Lemma 5, qαc≡0, which is not possible.
If νt=nat, then again using Lemma 5, we get the same contradiction as above.
If νt=at, then
[αn+wαn−1(α′+αβ′)]eα1enatzt+ (qαce˜α2−ue ˜α3)eatzt≡0,
Using Lemma 5,
αn−1(α+w(α′+αβ′)) ≡0.
=⇒α+w(α′+αβ′) = 0.
This gives that α=C3e−z/w−β. This implies deg α= deg β=ρ(f), which is
not possible.
Case 2: If we consider n≥2 with w= 0, then similarly we study 3 sub cases as for
Case 1.
Case 2.1: When ρ(f)<deg Q, then proceeding similar lines as in Case 1.1, we get
deg Q= deg ν. On differentiating equation (9), we have
nfn−1f′+ (q′fc+qQ′fc+qf ′
c)eQ= (u′+uν′)eν,(13)
Eliminating eνfrom equation (9) and (13), then we obtain
D1eQ+D0= 0,
where
D1= (u′+uν′)qfc−u(q′fc+qQ′fc+qf ′
c),
D0= (u′+uν′)fn−nufn−1f′.
Using Lemma 5, we have D1=D0≡0.
From D1≡0, we get
u′
u+ν′=q′
q+f′
c
fc
+Q′.
On doing integration of the above equation, we have
qfceQ=C1ueν.

10 G. PANT, M. SAINI
where C1is a non-zero constant.
If C1= 1, then qfceQ=ueνand by equation (9), we obtain
fn≡0.
Which is not possible. Hence conclution (i) does not hold.
If C16= 1, then again proceeding same lines to Case 1.1, we get same contradiction.
Case 2.2: When ρ(f) = deg Q, Using same argument as for Case 1.2, we get required
conclusion (ii).
Case 2.3: When ρ(f)>deg Q, Using same argument as for Case 1.3,we get same
conradiction.
Hence we complete the proof. 
Proof of Theorem B.We prove our result by contradiction. Suppose that fis
a finite order transcendental entire function satisfying equation (3) with a non-zero
finite Borel exceptional value. Then we study following three cases:
Case 1: If ρ(f)<1, then using similar technique as in [ [13],Theorem 1.11], we get
contradiction.
Case 2: If ρ(f)>1 and let a(6= 0) be a finite Borel exceptional value of f. Then
using Weierstrass factorisation theorem, f(z) can be represented in the form
f(z) = α(z)eβ(z)+a, (14)
where β(z) = atzt+arzt+... +a0;at6= 0 is a polynomial of degree t > 1 and α(z)
is an entire function such that ρ(α)<deg β.
From equation (3), we have
m(r, eQ(z)) = mr, q(z)eQ(z)f(z+c)
q(z)f(z+c)
=mr, P1eλz +P2e−λz −f2(z)(f(z) + wf′(z))
q(z)f(z+c)
≤mr, P1eλz +P2e−λz
q(z)f(z+c)+mr, f2(z)(f(z) + wf′(z))
q(z)f(z+c)+O(1)
≤mr, 1
q(z)f(z+c)+m(r, P1eλz +P2e−λz) + m(r, f 2)+
mr, 1
q(z)+mr, f(z)
q(z)f(z+c)+mr, f′(z)
q(z)f(z+c)+O(1)
Applying Lemma 1 and 3 into the above inequality, we have
T(r, eQ(z))≤mr, 1
q(z)f(z+c)+ 2T(r, f) + mr, f′(z)
q(z)f(z+c)+S(r, f)
≤mr, f(z)
q(z)f(z+c)+mr, 1
f(z)+ 2T(r, f) + mr, f′(z)f(z)
q(z)f(z)f(z+c)
+S(r, f)
≤3T(r, f) + S(r, f).

STUDY OF SOLUTIONS... 11
This implies that deg Q≤ρ(f) = tand we also know that deg Q≥1.
First suppose 1 ≤deg Q < t, then substituting the value of f(z) from equation (14)
into equation (3), we have
A1(z)e3β(z)+A2(z)e2β(z)+A3(z)eβ(z)+aq(z)eQ(z)+
q(z)α(z+c)eQ(z)+β(z+c)−P1eλz −P2e−λz +a3= 0,(15)
where
A1(z) = α3(z) + wα3(z)β′(z) + wα2(z)α′(z)
A2(z) = 3aα2(z) + 2waβ′(z)α2(z) + 2waα(z)α′(z)
A3(z) = 3a2α(z) + wa2α(z)β′(z) + wa2α′(z).
On simplifying equation (15), we get
A1(z)e3β(z)+A2(z)e2β(z)+ (A3(z)eβ1(z)+q(z)α(z+c)eδ(z))eatzt+
(aq(z)eQ(z)−P1eλz −P2e−λz +a3) = 0,(16)
where β1(z) = at−1zt−1+at−2zt−2+...+a0and δ(z) = Q(z)+ at[t
1czt−1+t
2c2zt−2+
... +ct] + at−1(z+c)t−1+... +a0are polynomials of degree at most t−1. Applying
Lemma 5 into equation (16), we have
aq(z)eQ(z)−P1eλz −P2e−λz +a3≡0 (17)
Again applying Lemma 5 into equation (17), we have the following sub cases:
(i) If Q(z)6=±λz, then we have P1≡0≡P2=a3, which is not possible.
(ii) If Q(z) = λz, then we have P2≡0≡a3, which is not possible
(iii) If Q(z) = −λz, then we have P1≡0≡a3, which is not possible.
Now suppose deg Q=t, and let Q(z) = btzt+bt−1zt−1+...+b0;bt6= 0 is a polynomial.
Then from equation (15), we have
A1(z)e3β1(z)e3atzt+A2(z)e2β1(z)e2atzt+A3(z)eβ1(z)eatzt+aq(z)eQ1(z)ebtzt+
q(z)α(z+c)eQ1(z)+β1(z+c)ebtzt+at(z+c)t−(P1eλz +P2e−λz −a3) = 0,
(18)
where β1(z) = at−1zt−1+at−2zt−2+... +a0and Q1(z) = bt−1zt−1+bt−2zt−2+... +b0
are polynomials of degree at most t−1. Next we discuss the following sub cases:
(i) If bt6=katfor any k= 1,2,3, and bt6=−atthen applying Lemma 5, we have
aq(z)eQ1(z)≡0, this implies either a≡0 or q(z)≡0, which is not possible.
(ii) If bt=katfor some k= 1,2,3, say bt=at, then equation (18) becomes
A1(z)e3β1(z)e3atzt+ (A2(z)e2β1(z)+q(z)α(z+c)e˜
δ(z))e2atzt+
(A3(z)eβ1(z)+aq(z)eQ1(z))eatzt−(P1eλz +P2e−λz −a3) = 0,
where ˜
δ(z) = Q1(z) +β1(z+c) +at[t
1czt−1+t
2c2zt−2+...+ct] is a polynomial
of degree at most t−1. Applying Lemma 5 into the above equation, we have
P1eλz +P2e−λz −a3≡0, which is not possible.

12 G. PANT, M. SAINI
(iii) If bt=−at, then equation (18) becomes
A1(z)e3β1(z)e3atzt+A2(z)e2β1(z)e2atzt+A3(z)eβ1(z)eatzt+aq(z)eQ1(z)ebtzt+
(q(z)α(z+c)e˜
δ(z)−P1eλz +P2e−λz −a3) = 0,
Applying Lemma 5 into the above equation, we have aq(z)eQ1(z)≡0, this
implies either a≡0 or q(z)≡0, which is not possible.
Case 3: If ρ(f) = 1, then we can represent f(z) as
f(z) = γ(z)ebz+d+a, (19)
where b6= 0, d ∈Cand γ(z) is an entire such that ρ(γ)<1.
Now proceeding to similar lines as in case of ρ(f)>1, we have deg Q= 1. Thus we
can put Q(z) = pz +s, (0 6=p, s ∈C) in the equation (3), we have
B1(z)e3bz +B2(z)e2bz +B3(z)ebz +B4(z)e(p+b)z+aq(z)esepz +a3−P1eλz −P2e−λz = 0,
(20)
where
B1(z) = (γ3(z) + wγ2(z)γ′(z) + wbγ3(z))e3d
B2(z) = (3aγ2(z) + 2awγ(z)γ′(z) + 2abwγ2(z))e2d
B3(z) = (3a2γ(z) + a2wγ′(z) + a2bwγ(z))ed
B4(z) = q(z)γ(z+c)ebc+s+d.
Here we discuss the following sub cases:
(i) If kb 6=pand kb 6=±λfor any k= 1,2,3, then applying Lemma 5 to equation
(20), we have P1≡0≡P2, which is not possible.
(ii) If kb =pfor some k= 1,2,3 and kb 6=±λfor any k= 1,2,3, then applying
Lemma 5 to equation (20), we obtain same contradiction as for previous case.
(iii) If kb 6=pfor any k= 1,2,3 and kb =λfor some k= 1,2,3, then applying
Lemma 5 to equation (20), we have P2≡0, which is not possible.
(iv) If kb 6=pfor any k= 1,2,3 and kb =−λfor some k= 1,2,3, then applying
Lemma 5 to equation (20), we have P1≡0, which is not possible.
(v) If kb =pand kb =λfor some k= 1,2,3, then applying Lemma 5 to equation
(20), we have P2≡0≡a3, which is not possible.
(vi) If kb =pand kb =−λfor some k= 1,2,3, then again applying Lemma 5 to
equation (20), we have P1≡0≡a3, which is not possible.
Thus we complete the proof.

Proof of Theorem C.Suppose that fis a transcendental meromorphic function
satisfying equation (6) and given that fhas finitely many poles, then from equation
(6), and Lemma 1 and 4,
nT (r, f) = T(r, fn) = T(r, P1eα1(z)+P2eα2(z)+P3eα3(z)−P(z, f ))
≤T(r, P1eα1(z)+P2eα2(z)+P3eα3(z)) + T(r, P (z, f )) + log 2
≤T(r, P1eα1(z)+P2eα2(z)+P3eα3(z)) + mr, P(z, f )
ff+

STUDY OF SOLUTIONS... 13
N(r, P (z, f )) + log 2
≤Arm+T(r, f ) + S(r, f ),
where
A=sum of the leading coefficients ofα1(z), α2(z)andα3(z)
π
and
m= max{deg α1(z),deg α2(z),deg α3(z)}.
Thus we have (n−1)T(r, f)≤Arm+S(r, f ) which implies that fhas finite order.
Now we prove rest part of the theorem by contradiction. Suppose f(z) = s(z)eq(z)
is a solution of equation (6), where s(z) is a non-zero rational function and q(z) is a
non constant polynomial, then from equation (6), we get
P1(z)eα1(z)+P2(z)eα2(z)+P3(z)eα3(z)=fn(z) + P(z, f )
=fn(z) +
i=l
X
i=0
ai(z)f(i)(z) + ˜a0(z)
= (s(z)eq(z))n+
i=l
X
i=0
ai(z)(s(z)eq(z))(i)+ ˜a0(z)
This implies
P1(z)eα1(z)+P2(z)eα2(z)+P3(z)eα3(z)−sn(z)enq(z)−Q(z)eq(z)−˜a0(z) = 0,(21)
where Q(z) = a0(z)s(z) + a1(z)[s′(z) + s(z)q′(z)] + ... +al(z)[sl(z) + ... +s(z)(q′(z))l]
is a rational function. Now we discuss the following cases:
(i) If q(z)−αj(z)6= constant, and nq(z)−αj(z)6= constant, for any j= 1,2,3.
Then applying Lemma 5 into equation (21), we have P1(z)≡0≡P2(z) = P3(z),
which is not possible.
(ii) If q(z)−αj(z) = constant, for some j= 1,2,3, say q(z)−α1(z) = constant, and
nq(z)−αj(z)6= constant, for any j= 1,2,3. Let q(z) = brzr+br−1zr−1+...+b0
and α1(z) = brzr+br−1zr−1+... +c0. Then q(z)−α1(z) = b0−c0and equation
(21) becomes
(P1(z)ec0−Q(z)eb0)e˜α1(z)+P2(z)eα2(z)+P3(z)eα3(z)−sn(z)enq(z)−˜a0(z) = 0,
where ˜α1(z) = brzr+br−1zr−1+... +b1z. Applying Lemma 5 to the above
equation, we have P2(z)≡0≡P3(z), which is not possible.
(iii) If q(z)−αj(z)6= constant, for any j= 1,2,3 and nq(z)−αj(z) = constant, for
some j= 1,2,3, say nq(z)−α1(z) = constant. Then applying Lemma 5 into
equation (21), we get the same contradiction as for just previous case.
(iv) If q(z)−αj(z) = constant, for some j= 1,2,3, say q(z)−α1(z) = constant
and nq(z)−αk(z) = constant, for some k, say nq(z)−α2(z) = constant. Then
applying Lemma 5 into equation (21), we have P3(z)≡0, which is not possible.
Hence we complete the proof. 
Proof of Theorem D.Since fis an entire function, then from equation (7) and
Lemma 1, we have
T(r, P1eα1z+P2eα2z+P3eα3z) = T(r, f n+Q∗
2(z, f ))

14 G. PANT, M. SAINI
≤nT (r, f) + Tr, Q∗
2(z, f )
f2f2
≤nT (r, f) + Tr, Q∗
2(z, f )
f2+ 2T(r, f)
≤(n+ 3)T(r, f) + S(r, f)
As we know that ρ(P1eα1z+P2eα2z+P3eα3z) = 1 by using Lemma 4, thus we get
ρ(f)≥1.
First suppose ρ(f)>1 and given that fis an entire satisfying equation (7) with
λ(f)< ρ(f). Then using Weierstrass factorization theorem, fmust be a transcen-
dental entire function in the form of
f(z) = H(z)eβ(z),
where β(z) is a polynomial of degree r > 1 and H(z) is an entire function such that
ρ(H)< r. Now from equation (7), we have
Hn(z)enβ(z)+
i=2
X
i=1
bi(z)eiβ(z)+a0(z) = P1eα1z+P2eα2z+P3eα3z,
where b1(z), b2(z) are meromorphic functions of order < r and a0(z) is a rational
function. This implies
Hn(z)enβ(z)+b2(z)e2β(z)+b1(z)eβ(z)+ (a0(z)−P1eα1z−P2eα2z−P3eα3z) = 0
Applying Lemma 5 into the above equation, we have
a0(z)−P1eα1z−P2eα2z−P3eα3z≡0,(22)
which is not possible. Otherwise, again applying Lemma 5 into equation (22), we
get P1≡0≡P2=P3, which is contradiction. Thus ρ(f) = 1.
Now considering
f(z) = H(z)eaz+b,(23)
where 0 6=a, b ∈Cand H(z) is an entire function of order <1. From equation (7),
we have
Hn(z)en(az+b)+b2(z)e2(az+b)+b1(z)eaz+b+a0(z)−P1eα1z−P2eα2z−P3eα3z= 0
Applying Lemma 5 into the above equation, there must exist positive integers k1, k2
and k3with {k1, k2, k3}={1,2,3}such that na =αk1,2a=αk2, a =αk3which gives
αk1:αk2:αk3=n: 2 : 1. We also have Hn(z)enb =Pk1, this gives H(z)≡P1/n
k1e−b
is a constant, b2(z)e2b≡Pk2,b1(z)eb≡Pk3, and a0(z)≡0. Substituting the value
of H(z) into equation (23), we have f(z) = P1/n
k1eaz =Ceaz, where Cis a non-zero
constant. 
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Garima Pant; department of mathematics, university of delhi, delhi-110007, india.
Email address :garimapant.m@gmail.com
department of mathematics, university of delhi, delhi-110007, india.
Email address :msaini@maths.du.ac.in, sainimanisha210@gmail.com