R E S E A R C H
Open Access
On the meromorphic solutions of certain
class of nonlinear differential equations
Nana Liu
1, Weiran Lü
1*and Chungchun Yang
2Dedicated to Professor George Csordas on the occasion of his retirement.
*Correspondence: [email protected] 1Department of Mathematics, China University of Petroleum, Qingdao, 266580, P.R. China
Full list of author information is available at the end of the article
Abstract
Let
α
be an entire function,an–1,. . .,a1,a0,Rbe small functions off, and letn≥2 bean integer. Then, for any positive integerk, the differential equation fnf(k)+a
n–1fn–1+· · ·+a1f+a0=Reαhas transcendental meromorphic solutions
under appropriate conditions on the coefficients. In addition, forn= 1 andk= 1, we have extended some well-known and relevant results obtained by others, by using different arguments.
MSC: 34M10; 30D35
Keywords: entire function; differential polynomial; Nevanlinna theory; Hayman’s alternative; differential equation
1 Introduction and main results
In this paper, a meromorphic function means meromorphic in the whole complex plane. We shall adopt the standard notations in Nevanlinna’s value distribution theory of mero-morphic functions (see,e.g., [, ]).
Given a meromorphic functionf, recall thatα≡,∞is a small function with respect tof, ifT(r,α) =S(r,f), whereS(r,f) denotes any quantity satisfyingS(r,f) =o{T(r,f)}as
r→ ∞, possibly outside a set ofrof finite linear measure.
Theorem A Let f be a transcendental meromorphic function,n(≥)be an integer.Then F=fnfassumes all finite values,except possibly zero,infinitely many times.
The above theorem was derived by Hayman [] in . Later, he conjectured [] that Theorem A remains valid even ifn= orn= . Mues [] proved the result forn= and the casen= was proved by Bergweiler and Eremenko [] and independently by Chen and Fang []. For entire functions and difference polynomials, similar results have been obtained by others earlier (see,e.g., [–]).
Theorem B([]) If f is a transcendental meromorphic function of finite order and a(≡)
is a polynomial,then ff–a has infinitely many zeros. Wang [] obtained the following result.
Theorem C Let f be a transcendental entire function and n,k be positive integers,and let c(z) (≡)be a small function with respect to f.If T(r,f)= τN)(r, /f) +S(r,f),then
fn(z)f(k)(z) –c(z)has infinitely many zeros,whereτ= if n≥or k= ;τ = otherwise. In this paper, by using methods different from that were used by others (see,e.g., [, ] and []), we shall extend and generalize the above results withfnf(k)being replaced by a differential polynomialPn+(f). Specifically, our main results can be stated as follows. Theorem . Letαbe an entire function,R and ai(i= , , . . . ,n– )be small functions of f
with a≡.If,for n≥,a transcendental meromorphic function f satisfies the differential
equation
fnf(k)+an–fn–+· · ·+af +a=Reα, (.)
then, for any positive integer k, we have f =gexp(α/(n+ )) – (n+ )a
a with g
n+ =
[(n+)a
a ]
n+R a,and(
a
a)
(k)+ n n+(
n+
a
a)
na
≡.
Remark . Letaandabe non-zero constants in Theorem .. Then (.) has no
tran-scendental meromorphic solutions.
A meromorphic solutionf of (.) is called admissible, ifT(r,αj) =S(r,f) holds for all
coefficientsαj(j= , . . . ,n– ) andT(r,R) =S(r,f).
Remark . Ifa≡ andn≥,k≥, then the other coefficientsa, . . . ,an–must be
identically zero. In this case, (.) becomesfnf(k)=Reαandf has the formf=uexp(α/(n+
)) as the only possible admissible solution of (.), whereuis a small function off. We have the following corollary by Theorem ..
Corollary . Let f be a transcendental meromorphic function with N(r,f) =S(r,f),n≥
be an integer.If(a
a)
(k)+ n n+(
n+
a
a)
na
≡,then F=fnf(k)+an–fn–+· · ·+af+ahas
infinitely many zeros,where ai(i= , , . . . ,n– )are small functions of f such that a≡.
Note that in Theorem ., it is assumed thatn≥ andk≥. However, forn= and
k= , we can derive the following result.
Theorem . Let p,q,and R be non-zero polynomials,αbe an entire function.Then the differential equation pff–q=Reαhas no transcendental meromorphic solutions,where p,
q,and R are small functions of f with pq≡.
Remark . From the proof of Theorem ., we see that the restriction in Theorem . top,q, andRmay extend to small functions. In fact, it is easy to find that the conclusion is valid provided thatp,q, andRare non-vanishing small functions off. The following corollary arises directly from an immediate consequence of Theorem ..
2 Some lemmas and proofs of theorems
In order to prove our conclusions, we need some lemmas. The following lemma is funda-mental to Clunie’s theorem [].
Lemma .([, ]) Let f be a transcendental meromorphic solution of
fnP(z,f) =Q(z,f),
where P(z,f)and Q(z,f)are polynomials in f and its derivatives with meromorphic coef-ficients{aλ|λ∈I}such that m(r,aλ) =S(r,f)for all r∈I.If the total degree of Q(z,f)as a
polynomial in f and its derivatives is less than or equal to n,then m(r,P(r,f)) =S(r,f). The following lemma is crucial to the proof of our theorems.
Lemma .([, ]) Let f be a meromorphic solution of an algebraic equation
Pz,f,f, . . . ,f(n)= , (.)
where P is a polynomial in f,f, . . . ,f(n) with meromorphic coefficients small with respect
to f.If a complex constant c does not satisfy(.),then
m
r,
f –c
=S(r,f).
Proof of Theorem. Letf be a transcendental meromorphic function that satisfies (.). Then two cases are to be treated, namely case :N(r,f)=S(r,f), and case :N(r,f) =S(r,f). For case , it is impossible asαis an entire function andR,a, . . . ,anare small functions
off.
To prove Theorem ., we now suppose thatN(r,f) =S(r,f). Denotingφ:=fnf(k)+a
n–fn–+· · ·+af, and assuming thatT(r,φ) =S(r,f), then by
Lemma ., we getm(r,f(k)) =S(r,f) and thenT(r,f(k)) =S(r,f), sinceN(r,f) =S(r,f) by
the assumption. The contradictionT(r,f) =S(r,f) now follows by the theorem in [] and combining it with the proof of Proposition E in []. Thus, for any transcendental meromorphic functionf under the condition:N(r,f) =S(r,f),
Tr,fnf(k)+an–fn–+· · ·+af
=S(r,f). (.)
From (.) and the result of Milloux (see,e.g., [], Theorem .), one can easily show that
Tr, eα≤(n+ )T(r,f) +S(r,f),
which leads toT(r,α) +T(r,α) =S(r,f).
By taking the logarithmic derivative on both sides of (.), we have
nfn–ff(k)+fnf(k+)+a
n–fn–+· · ·+af+af+a
fnf(k)+a
n–fn–+· · ·+af +a
=R
R +α
It follows by (.) that
–
R R +α
fnf(k)+nfn–ff(k)+fnf(k+)+
an––
R R +α
a n–
fn–
+ (n– )an–fn–f+· · ·+
a–
R R +α
a
f+af=
R R +α
a
–a. (.)
If (RR +α)a–a≡, thenAa=Reα, whereAis a non-zero constant. From (.), we get
fnf(k)+an–fn–+· · ·+af = (A– )a. (.)
IfA= , then from (.), we obtain
fnf(k)+an–fn–+· · ·+af ≡,
which contradicts (.). However, ifA= , then again from (.), we would derive
Tr,fnf(k)+an–fn–+· · ·+af
=S(r,f), a contradiction.
Thus
R R +α
a
–a:=ϕ≡.
In this case, from (.), we have
N(
r,
f
≤N
r,
ϕ
+S(r,f)≤T(r,ϕ) +S(r,f) =S(r,f),
whereN((r,f), as usually, denotes the counting function of zeros offwhose multiplicities
are not less than , which implies that the zeros off are mainly simple zeros. Again, from (.), the fact that αis a small function off and Lemma . (wherec= is used), we concludem(r,f) =S(r,f). This together with Nevanlinna’s first theorem will result in
T(r,f) =N
r,
f
+S(r,f) =N)
r,
f
+S(r,f), (.)
where inN)(r, /f) only the simple zeros off are to be considered.
Assume thata≡. It follows by (.) andn≥ thatN)(r, /f) =S(r,f), which
contra-dicts (.). Thusa≡. Letzbe a simple zero off, andzbe not a pole of one of the
coefficientsai, (R
R +α)ai–a
i(i= , , . . . ,n– ). From (.), we see thatzis a zero of
af+a– (R
R +α)a. Set
h=af
+a
– (R
R +α)a
f . (.)
Then (.) givesT(r,h) =S(r,f). We have
f=
a
hf–a+
R R +α
a
Clearly, it follows from (.) and T(r,μ) +T(r,ν) =S(r,f) thatμν≡. By (.), we
obtain
fn–ψ=P
n–(f), (.)
whereψ = –(RR +α)ff(k)+nff(k)+ff(k+),P
n–(f) = (R
R +α)(an–fn–+· · ·+af +a) –
(an–fn–+· · ·+af+a). It follows by (.) thatPn–(f)≡. Thusψ≡. Moreover, by
applying Lemma . to (.), we getm(r,ψ) =S(r,f). It is easy to see byN(r,f) =S(r,f) thatT(r,ψ) =S(r,f).
From (.) and induction, we havef= (μ+μ)f +μν+ν:=μf+ν, and
f(k)=μkf +νk, (.)
whereμk,νkare small functions off. By the expression ofψ and (.), we getνk≡. If
μk≡, then (.) givesT(r,f(k)) =S(r,f), which is impossible. Therefore,μk≡.
By (.), (.) becomes
μkfn++νkfn+an–fn–+· · ·+af +a=Reα. (.)
By applying the Tumura-Clunie lemma (see,e.g., [], Theorem .) to the left-hand side of (.), we haveμk[f+
νk
(n+)μk]
n+=Reαandf =geα/(n+)– νk
(n+)μk withg
n+= R μk.
In view of (.), we have
μkfn++νkfn+an–fn–+· · ·+af +a=μk
f + νk (n+ )μk
n+
.
Thus, we have
n+
νk
μk
= (n+ )a
a
and μk=
n+
a
a
n+
a. (.)
By (.), we obtainνk= (n+ )(n+ aa)naandgn+= [(n+)aa]n+aR.
Set (n+ )γ =α. It follows by (.) andf =geγ – (n+ )a
a that
f(k)=
n+
a
a
n+
a
geγ– (n+ )a
a
+ (n+ )
n+
a
a
n
a. (.)
In addition, byf=geγ– (n+ )a
a we get
f(k)=Qg,g, . . . ,g(k)eγ– (n+ )
a
a
(k)
, (.)
whereQ(g,g, . . . ,g(k)) is a differential polynomial ofg.
Thus, (.) and (.) imply
Qg,g, . . . ,g(k)=
n+
a
a
n+
and
(n+ )
a
a
(k)
=
n+
a
a
n+
a(n+ )
a
a
– (n+ )
n+
a
a
n
a. (.)
It follows by (.) that
a
a
(k)
+ n
n+
n+
a
a
n
a= .
This completes the proof of Theorem ..
Proof of Remark. Letf be a transcendental meromorphic solution of (.). Sincea≡,
we have N(r, /f)≤N(r, /R) +S(r,f) =S(r,f). Obviously, N(r,f) =S(r,f). In this case, there exist a meromorphic function uand an entire functionvsuch thatf =uev, and
N(r, /u) +N(r,u) =S(r,f). Clearly, from the expressions offand the Borel lemma (see,e.g., [], Theorem .), all theaj(j= , , . . . ,n– ) must be identically zero. Thus, Remark .
follows.
Proof of Theorem. Now we proceed to prove the theorem by contradiction. Letf be a transcendental meromorphic function that satisfiespff–q=Reα. Then two cases are
to be retreated, namely N(r,f)=S(r,f) andN(r,f) =S(r,f). ForN(r,f)=S(r,f), this is impossible asαis an entire function andR,p,qare non-zero polynomials.
To prove Theorem ., we now suppose thatN(r,f) =S(r,f). We differentiatepff–q=
Reαand eliminate eα,
tff+p
f+pff=t, (.)
wheret=p– (R
R +α)p,t=q– ( R
R +α)q.
If t≡, then, by integrating the definition of t,α must be a constant, hence ff is
rational, and then, by Lemma .,m(r,f) =S(r,f). HenceT(r,f) =S(r,f). This is a con-tradiction by Proposition E in []. Thus,t≡, and then by (.), we get (.). By
dif-ferentiating both sides of (.), we have
tff+t+p
f+t+p
ff+ pff+pff=t. (.)
Lettingzbe a simple zero off, (.) and (.) imply
pf–t
(z) = (.)
and
t+p
f+ pff–t(z) = . (.)
Let
g=ptf
+ [t
(t+p) –tp]f
From (.), (.), and (.), we get
T(r,g) =S(r,f).
By (.), we obtain
f=αf +βf, (.)
where
α=
g
pt
, β=
tp–t(t+p)
pt
and
T(r,α) =S(r,f), T(r,β) =S(r,f).
Substituting (.) into (.) yields
(t+pβ)ff+p
f+αpf=t. (.)
On the other hand, from (.), we have
f=αf +βf, (.)
whereα=α+αβ,β=α+β+β, and
T(r,α) =S(r,f), T(r,β) =S(r,f).
Substituting (.) into (.), we have
t+β
t+p
+ pα+pβ
ff+t+p+ pβ
f+α
t+p
+αp
f=t. (.)
It follows by (.) and (.) that
pt+β
t+p
+ pα+pβ
–t+p+ pβ
(t+pβ)
ff
+pα
t+p
+αp
–αp
t+p+ pβ
f=tp–t
t+p+ pβ
. (.)
From the definition ofβ, we now claimtp–t(t+p+ pβ)≡. To show this, we
assume the contrary, that is, tp–t(t+p+ pβ)≡. Then from the fact thattp–
t(t+p+ pβ) is a small function off and (.), we get
N)
r,
f
≤N
r,
tp–t(t+p+ pβ)
≤Tr,tp–t
t+p+ pβ
and from this and (.) we deduceT(r,f) =S(r,f), a contradiction. Thus, we have
tp–t
t+p+ pβ
≡. (.)
Now, (.) and (.) lead to
pα
t+p
+αp
–αp
t+p+ pβ
≡. (.)
From the definition ofαand (.), we deduce
α≡αβ. (.)
It follows from (.) and the definitions oft,βthat
αp≡teα.
In the beginning of the proof it was already shown thatt≡. Hence, the contradiction
here is immediate.
This also completes the proof of Theorem ..
3 Remarks and a conjecture
Remark . Corollary . or Corollary . can be strengthened to
N
r,
F
=S(r,f).
Remark . What can be said if ‘pff–q’ is replaced by ‘pff(k)–q’, for any integerk≥,
in Theorem .?
Remark . Takingf(z) = ez, we have
N
r,
ff(k)–a
∼T(r,f) +S(r,f),
wherekis a positive integer, andais a non-zero constant.
Finally, we present the following more general and quantitative conjecture.
Conjecture . Let f be a transcendental entire function.Then for any integer k≥,and any small function a(≡),
N
r,
ff(k)–a
∼T(r,f) +S(r,f).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1Department of Mathematics, China University of Petroleum, Qingdao, 266580, P.R. China.2Department of Mathematics,
Nanjing University, Nanjing, 210093, P.R. China.
Acknowledgements
The authors would like to thank the referee for his/her several important suggestions and valuable comments to our original manuscript which enabled us to improve greatly the quality and readability of the paper.
Received: 22 September 2014 Accepted: 21 April 2015 References
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