1 Complex differential equations from
${2\,\mathbf{sin}\, z\,\mathbf{cos}\, z=\mathbf{sin}\, 2z}$
The existence and growth of transcendental meromorphic solutions of complex differential equations have been widely studied (see Laine’s monograph [Reference Laine10]). Interesting nonlinear complex differential equations can be derived from trigonometric identities. Observe the well-known trigonometric identity
$$ \begin{align} 2\sin z \cos z=\sin 2z \end{align} $$
and the formula
$(\sin z)'=\cos z$
. Zhang and Yi [Reference Zhang and Yi27, Corollary 1.7] studied the complex differential equation corresponding to (1.1) and the existence of meromorphic solutions, obtaining the following result.
Theorem 1.1. The nonlinear complex differential equation
$$ \begin{align} 2f(z)f'(z)=\sin 2z \end{align} $$
has only four meromorphic solutions,
$f(z)=\pm \sin z$
and
$f(z)=\pm i\cos z$
.
Remark 1.2.
$(1)$
The trigonometric identity (1.1) is also connected to functional- differential equations; Beardon [Reference Beardon1] considered the pantograph-type equation
$$ \begin{align*}f(2z)=2f'(z)f(z)\end{align*} $$
and obtained all the formal power series solutions. Ludwig and Tomaschek [Reference Reich and Tomaschek20] perturbed
$2z$
and the constant
$2$
in (1.1) and considered the equation
$$ \begin{align*}f(\varphi(z))=a(z)f'(z)f(z),\end{align*} $$
where
$\varphi (z)=kz+\varphi _{2}z^{2}+\cdots $
and
$a(z)=a_{0}+a_{1}z+\cdots $
, and either
$|k|>1$
and k is a Siegel number, or
$|k|<1$
.
$(2)$
Theorem 1.1 is a corollary of [Reference Zhang and Yi27, Theorem 1.6] dealing with a generalised complex differential equation of the form
$$ \begin{align*}[f(z)^{2}]^{(k)}=\sin 2z+p(z),\end{align*} $$
where
$p(z)$
is a polynomial and k is a positive integer. The main proof idea relies on the integral form, that is,
$$ \begin{align*}f(z)^{2}=\frac{1}{2^{k}}\sin\bigg(2z-\frac{k}{2}\pi\bigg)+P(z),\end{align*} $$
where
$P^{(k)}(z)=p(z)$
.
$(3)$
The periodicity of
$f(z)$
when
$f(z)f'(z)$
is a periodic function was confirmed by Wang and Hu [Reference Wang and Hu23]. Theorems 1.1 and 4.1 below have inspired recent research on the periodicity of complex differential polynomials (see [Reference Liu, Liu, Korhonen and Filipuk18, Reference Zemirni, Laine and Latreuch26]).
We will provide a new method to deal with the meromorphic solutions of (1.2) by differentiating and using the following basic lemmas of Nevanlinna theory. For the basic notations and fundamental results of Nevanlinna theory, see [Reference Hayman8, Reference Laine10].
Lemma 1.3 (Clunie’s lemma; [Reference Laine10, Lemma 2.4.2]).
Let f be a transcendental meromorphic solution of
$$ \begin{align*}f^{n}P(z,f)=Q(z,f),\end{align*} $$
where
$P(z,f)$
and
$Q(z,f)$
are polynomials in f and its derivatives with meromorphic coefficients, say
$\{a_{\mu }(z)$
,
$\mu \in I\}$
, such that
$m(r,a_{\mu }(z))=S(r,f)$
for all
$\mu \in I$
. If the total degree of
$Q(z,f)$
as a polynomial in f and its derivatives does not exceed n, then
$$ \begin{align*} m(r,P(z,f))=S(r,f). \end{align*} $$
Lemma 1.4 (Valiron–Mohon’ko lemma; [Reference Laine10, Theorem 2.2.5]).
Let f be a meromorphic function and
$$ \begin{align*} R(z,f)=\frac{a_0+a_1f+\cdots+a_pf^p} {b_0+b_1f+\cdots+b_qf^q} \end{align*} $$
be irreducible and rational in
$f(z)$
with meromorphic coefficients
$a_j,b_k$
of small growth
$S(r,f)$
. Then
$T(r,R(z,f))=\max (p,q)T(r,f)+S(r,f)$
.
Proof of Theorem 1.1.
By taking the first derivative of (1.2),
$$ \begin{align*}f^{\prime2}+ff"=\cos 2z\end{align*} $$
and, by taking the derivative of this equation,
$$ \begin{align*}3f'f"+ff"'=-2 \sin 2z=-4ff'.\end{align*} $$
Thus,
$$ \begin{align} 3\frac{f"}{f}+\frac{f"'}{f'}=-4. \end{align} $$
Let
${f"}/{f}=\alpha $
. We will affirm that
$\alpha $
is a constant. Otherwise, if
$\alpha $
is not a constant, from (1.3),
$$ \begin{align*}3\alpha+\frac{\alpha'f+\alpha f'}{f'}=-4,\end{align*} $$
so we conclude that
$$ \begin{align*}\frac{f'}{f}=-\frac{1}{4}\frac{\alpha'}{\alpha+1}.\end{align*} $$
Taking the derivative of this equation,
$$ \begin{align*}-\frac{1}{4}\frac{\alpha"(\alpha+1)-\alpha^{\prime2}}{(\alpha+1)^{2}}=\alpha-\frac{\alpha^{\prime2}}{16(\alpha+1)^{2}},\end{align*} $$
that is,
$$ \begin{align} 4\alpha"(\alpha+1)+16\alpha(\alpha+1)^{2}=5\alpha^{\prime2}, \end{align} $$
which we rewrite as
$$ \begin{align}16\alpha^{3}=5\alpha^{\prime2}-4\alpha"(\alpha+1)-32\alpha^{2}-16\alpha.\end{align} $$
If
$\alpha $
is a nonconstant entire function, then
$m(r,\alpha )=S(r,\alpha )$
by Lemma 1.3, which is impossible.
If
$\alpha $
is a properly meromorphic function and
$z_{0}$
is a pole of
$\alpha $
, then
$z_{0}$
must be the simple pole, since
$z_{0}$
is a simple zero of
$f(z)$
by (1.2). Computing the multiplicity of the pole at
$z_{0}$
from (1.5) gives a contradiction.
Hence,
$\alpha $
is a constant and it follows from (1.4) that
$\alpha =-1$
. Therefore,
${f(z)=c_{1}e^{iz}+c_{2}e^{-iz}}$
. Substituting back into (1.2) gives
$$ \begin{align*}i(c_{1}^{2}e^{2iz}-c_{2}^{2}e^{-2iz})=\frac{e^{2iz}-e^{-2iz}}{4i},\end{align*} $$
that is,
$$ \begin{align*}\bigg(ic_{1}^{2}-\frac{1}{4i}\bigg)e^{4iz}+\bigg(\frac{1}{4i}-ic_{2}^{2}\bigg)=0.\end{align*} $$
To avoid a contradiction with Lemma 1.4, we obtain
$c_{1}^{2}=-\tfrac 14$
and
$c_{2}^{2}=-\tfrac 14$
. So
${f(z)=\pm \sin z}$
or
$f(z)=\pm i\cos z$
. Theorem 1.1 is thus proved.
Next, we consider a polynomial perturbation of (1.2). Mao and Liu [Reference Mao and Liu19, Theorem 1.3] also considered the general perturbed equation
$f(z)f'(z)=p(z)\sin (\alpha (z))$
, where
$p(z)$
is a polynomial and
$\alpha (z)$
is an entire function, using a different method.
Theorem 1.5. If the complex differential equation
$$ \begin{align*}f(z)f'(z)=p(z)\sin 2z\end{align*} $$
admits a transcendental meromorphic solution
$f(z)$
, then
$p(z)$
must be a nonzero constant
$\tfrac 12\lambda ^{2}$
and
$f(z)=\pm \lambda \sin z$
or
$f(z)=\pm \lambda i\cos z$
.
Before giving the proof of Theorem 1.5, we need the following lemma.
Lemma 1.6 ([Reference Mao and Liu19, Lemma 2.2]).
Let k be a positive integer,
$\lambda $
a nonzero constant and let
$Q_{j}\ (j=1,2,3)$
be small functions of
$e^{z^{k}}$
such that
$Q_{1}Q_{2}\not \equiv 0$
. If the equation
$$ \begin{align*}f(z)^{2}=Q_{1}(z)e^{\lambda z^{k}}+Q_{2}(z)e^{-\lambda z^{k}}+Q_{3}(z)\end{align*} $$
admits a meromorphic solution
$f(z)$
, then
$4Q_{1}(z)Q_{2}(z)=Q_{3}(z)$
. Furthermore,
$f(z)=\beta _{1}(z)e^{{\lambda z^{k}}/{2}}+\beta _{2}(z)e^{-{\lambda z^{k}}/{2}}$
, where
$\beta _{1}(z)$
and
$\beta _{2}(z)$
are small functions of f satisfying
$\beta _{j}^{2}=Q_{j}$
and
$2\beta _{1}\beta _{2}=Q_{3}$
.
Proof of Theorem 1.5.
Assume that
$p(z)$
is a nonconstant polynomial of degree m. From the hypothesis of the theorem,
$f(z)f'(z) = \tfrac 12(f(z)^{2})'=p(z)\sin 2z$
. Integrating,
$$ \begin{align} \tfrac{1}{2}f(z)^{2}=Q(z)\cos 2z+R(z)\sin 2z +c, \end{align} $$
where
$$ \begin{align*}Q(z)=-\dfrac{1}{2}p(z)+\dfrac{1}{2^{3}}p"(z)-\dfrac{1}{2^{5}}p^{(4)}(z)+\cdots+(-1)^{[{m+1}/{2}]}\frac{1}{2^{2m+1}}p^{(2m)}(z)\end{align*} $$
and
$$ \begin{align*}R(z)=\dfrac{1}{2^{2}}p'(z)-\dfrac{1}{2^{4}}p"'(z)+\dfrac{1}{2^{6}}p^{(5)}(z)+\cdots+(-1)^{[{m}/{2}]}\frac{1}{2^{2m}}p^{(2m-1)}(z).\end{align*} $$
Now,
$$ \begin{align*}Q(z)\cos 2z+R(z)\sin 2z +c=\bigg(\frac{Q(z)}{2}+\frac{R(z)}{2i}\bigg)e^{2iz}+\bigg(\frac{Q(z)}{2}-\frac{R(z)}{2i}\bigg)e^{-2iz}+c.\end{align*} $$
Since
$p(z)$
is a nonconstant polynomial,
${Q(z)}/{2}+{R(z)}/{2i} = \tfrac 12(Q(z)-iR(z))$
and
${Q(z)}/{2}-{R(z)}/{2i} = \tfrac 12({}Q(z)+iR(z))$
are nonconstant polynomials. If
$c=0$
, then
$$ \begin{align*}Q(z)\cos 2z+R(z)\sin 2z=\frac{\frac{1}{2}(Q(z)-iR(z))e^{4iz}+\frac{1}{2}(Q(z)+iR(z))}{e^{2iz}} = \frac{G(z)}{e^{2iz}} \quad \text{(say)}.\end{align*} $$
From the second main theorem of Nevanlinna theory for three small functions [Reference Hayman8, Theorem 2.5] and (1.6),
$$ \begin{align*} T(r,e^{4iz})&\leq \overline{N}(r,e^{4iz})+\overline{N}\bigg(r,\frac{1}{e^{4iz}}\bigg)+\overline{N}\bigg(r,\frac{1}{G(z)}\bigg)+S(r,e^{4iz})\\ &\leq \overline{N}\bigg(r,\frac{1}{G(z)}\bigg)+S(r,e^{4iz}) \leq \frac{1}{2} N\bigg(r,\frac{1}{G(z)}\bigg)+S(r,e^{4iz})\\ &\leq \frac{1}{2}T(r,e^{4iz})+S(r,e^{4iz}), \end{align*} $$
which is impossible.
If
$c\not =0$
, by Lemma 1.6,
$$ \begin{align*}\bigg(\frac{Q(z)}{2}+\frac{R(z)}{2i}\bigg)\bigg(\frac{Q(z)}{2}-\frac{R(z)}{2i}\bigg)=\frac{c}{8},\end{align*} $$
which is impossible. In both cases, (1.6) leads to a contradiction. Hence,
$p(z)$
must be a constant and
$p(z)={\lambda ^{2}}/{2}$
. The expressions for
$f(z)$
are given by Theorem 1.1.
We next consider a generalisation of the complex differential equation (1.2):
$$ \begin{align} f(z)f^{(4k+1)}(z)=\tfrac{1}{2}\sin 2z, \end{align} $$
where k is a positive integer. Obviously, (1.7) also has four meromorphic solutions
$f(z)=\pm \sin z$
and
$f(z)=\pm i\cos z$
and all meromorphic solutions of (1.7) are entire functions. However, the uniqueness of the meromorphic solutions to this equation remains an open question. So far, neither the integral method in Zhang and Yi [Reference Zhang and Yi27] nor the differential method in the paper has yielded an answer. The following question is a natural one to ask.
Question 1.7. Let
$f(z)$
be a transcendental meromorphic solution of
$$ \begin{align} f(z)f^{(k)}(z)=p(z)\sin 2z. \end{align} $$
Does it follow that
$k=4k_{1}+1$
,
$p(z)=\tfrac 12\lambda ^{2}$
is a nonzero constant and
$f(z)=\pm \lambda \sin z$
or
$f(z)=\pm \lambda i\cos z$
, where
$k_{1}$
is a nonnegative integer and
$\lambda $
is a nonzero constant?
Remark 1.8. Question 1.7 has a connection with Hayman’s conjecture [Reference Hayman7]. Equation (1.8) can be written as
$$ \begin{align*} g(z)g^{(k)}(z)&= -p\bigg(z-\frac{\pi}{4}\bigg)\cos 2z=-p\bigg(z-\frac{\pi}{4}\bigg)+2p\bigg(z-\frac{\pi}{4}\bigg)\sin^{2}z\\&= p\bigg(z-\frac{\pi}{4}\bigg)-p\bigg(z-\frac{\pi}{4}\bigg)\cos^{2}z \end{align*} $$
by letting
$g(z)=f(z-{\pi }/{4})$
. The two complex differential polynomials
$$ \begin{align*}g(z)g^{(k)}(z)\pm p\bigg(z-\frac{\pi}{4}\bigg)\end{align*} $$
have infinitely many zeros which are double except for the finitely many zeros of
${p(z-{\pi }/{4})}$
. Hayman’s conjecture states that if
$f(z)$
is a transcendental meromorphic function, then
$f(z)^{n}f'(z)-a$
has infinitely many zeros, where a is a nonzero constant and n is a positive integer. Hayman’s conjecture has been confirmed, but we do not know if
$g(z)g^{(k)}(z)-p(z)$
has infinitely many zeros, where
$p(z)$
is a nonzero polynomial. The case in which
$k=2$
and
$p(z)$
is a nonzero constant can be seen in Langley [Reference Langley11].
2 Complex differential systems from
$2\,\mathbf{sin}\ z\,\mathbf{cos}\, z=\mathbf{sin}\, 2z$
Complex differential systems have been studied recently by Gao [Reference Gao3, Reference Gao, Liu and Gao5, Reference Liu, Gao and Wang17]. Results on systems of complex delay-differential equations [Reference Liu, Laine and Yang16, Ch. 11], Malmquist- or Riccati-type differential systems [Reference Gao, Liu and Qi4, Reference Tu and Xiao21, Reference Wang and Liu22] and Fermat-type differential-difference systems [Reference Gao3, Reference Li and Liu14] have also been derived from the corresponding differential equations. We will use the results in Section 1 to investigate the meromorphic solutions of the corresponding complex differential systems, where the coefficients are not necessarily small functions.
Theorem 2.1. Let
$p_{1}(z)$
and
$p_{2}(z)$
be nonzero polynomials and n be a positive integer, and consider the nonlinear differential system
$$ \begin{align} \begin{cases} f(z)^{n}g'(z)=p_{1}(z)\sin 2z, \\ g(z)^{n}f'(z)=p_{2}(z)\sin 2z. \end{cases} \end{align} $$
(1) If
$n\geq 2$
, then (2.1) has no meromorphic solutions.
(2) Assume that
$n=1$
,
$f(z)=\alpha (z)\sin z$
and
$g(z)=\beta (z)\sin z$
, where
$\alpha (z)$
and
$\beta (z)$
are small functions with respect to
$\sin z$
. Then
$p_{1}(z)=p_{2}(z)=p$
where p is a nonzero constant. Thus,
$f(z)=\gamma g(z)$
, and
$f(z)=\tau \sin z$
or
$f(z)=i\tau \cos z$
where
$2p={\tau ^{2}}/{\gamma }$
.
Proof. (1) Assume that
$n\geq 2$
and
$f(z)$
and
$g(z)$
are properly transcendental meromorphic functions with infinitely many poles. Assume that
$z_{0}$
is a pole of
$f(z)$
with multiplicity p which is not a zero of
$p_{1}(z)$
and
$p_{2}(z)$
. Then
$z_{0}$
is a zero of
$g'(z)$
with multiplicity at least
$np$
and at most
$np+1$
by the first equation of (2.1), and
$z_{0}$
is a zero of
$g(z)$
with multiplicity at least
${(p+1)}/{n}$
and at most
${(p+2)}/{n}$
by the second equation of (2.1). Therefore,
$z_{0}$
is a zero of
$g(z)$
and
$g'(z)$
. This gives the four possibilities listed below, all of which are impossible because they imply that
$p\leq 0$
when
$n\geq 2$
.
Case 1. If
${(p+2)}/{n}-1=np$
, then
$p={(n-2)}/{(1-n^{2})}$
.
Case 2. If
${(p+2)}/{n}-1=np+1$
, then
$p={(2n-2}){(1-n^{2})}$
.
Case 3. If
${(p+1)}/{n}-1=np$
, then
$p={(n-1)}/{(1-n^{2})}$
.
Case 4. If
${(p+1)}/{n}-1=np+1$
, then
$p={2n-1}/{(1-n^{2})}$
.
Thus, if
$n\geq 2$
, both
$f(z)$
and
$g(z)$
are transcendental entire functions or properly transcendental meromorphic functions with finitely many poles. From the first and the second equation of (2.1),
$$ \begin{align*}nT(r,f)\leq T(r,g')+T(r,\sin 2z)+O(\log r)\leq T(r,g)+T(r,\sin 2z)+O(\log r)\end{align*} $$
and
$$ \begin{align*}nT(r,g)\leq T(r,f')+T(r,\sin 2z)+O(\log r)\leq T(r,f)+T(r,\sin 2z)+O(\log r).\end{align*} $$
Hence,
$$ \begin{align*}(n-1) (T(r,f)+T(r,g))\leq 2T(r,\sin 2z)+O(\log r).\end{align*} $$
Thus,
$f(z)$
and
$g(z)$
must be transcendental entire functions or properly transcendental meromorphic functions having finitely many poles with order at most
$1$
. Since
$n\geq 2$
,
$f(z)$
and
$g(z)$
have finitely many zeros. Therefore, we can assume that
$f(z)=t_{1}(z)e^{s_{1}z}$
and
$g(z)=t_{2}(z)e^{s_{2}z}$
, where
$t_{1}(z)$
and
$t_{2}(z)$
are rational functions. Substituting
$f(z)$
and
$g(z)$
into (2.1),
$$ \begin{align*} t_{1}(z)^{n}(t_{2}'(z)+t_{2}(z)s_{2})e^{ns_{1}z+s_{2}z}=p_{1}(z)\sin 2z, \end{align*} $$
and checking the numbers of zeros on the two sides of this equation gives a contradiction.
Thus, we have proved that
$n=1$
. Consider
$$ \begin{align} \begin{cases} f(z)g'(z)=p_{1}(z)\sin 2z, \\ g(z)f'(z)=p_{2}(z)\sin 2z, \end{cases} \end{align} $$
and let
$f(z)=\alpha (z)\sin z$
and
$g(z)=\beta (z)\sin z$
.
Case 1. If
$\alpha (z)$
and
$\beta (z)$
are nonzero constants, the conclusion of the theorem is obviously true.
Case 2. If one of
$\alpha (z)$
and
$\beta (z)$
is a nonconstant small function with respect to
$\sin z$
, then (2.2) reduces to
$$ \begin{align*} \begin{cases} \alpha(z)(\beta'(z)\sin z+\beta(z)\cos z)=2p_{1}(z)\cos z, \\ \beta(z)(\alpha'(z)\sin z+\alpha(z)\cos z)=2p_{2}(z)\cos z. \end{cases} \end{align*} $$
Hence,
$$ \begin{align} \sin z(\alpha(z)\beta'(z)-\beta(z)\alpha'(z))=2\cos z(p_{1}(z)-p_{2}(z)). \end{align} $$
If
$\alpha (z)\beta '(z)-\beta (z)\alpha '(z)\not \equiv 0$
, since
$\alpha (z)$
and
$\beta (z)$
are small functions with respect to
$\sin z$
and
$\cos z$
, (2.3) contradicts Lemma 1.4.
If
$\alpha (z)\beta '(z)-\beta (z)\alpha '(z)\equiv 0$
, then
$p_{1}(z)-p_{2}(z)\equiv 0$
. We have
$f(z)=\gamma g(z)$
, where
$\gamma $
is a nonzero constant. By Theorem 1.5,
$p_{1}(z)$
and
$p_{2}(z)$
must be constants and
$f(z)=\tau \sin z$
or
$f(z)=i\tau \cos z$
, where
$2p={\tau ^{2}}/{\gamma }$
.
Remark 2.2. We have assumed that
$f(z)=\alpha (z)\sin z$
and
$g(z)=\beta (z)\sin z$
in Theorem 2.1, where
$\alpha (z)$
and
$\beta (z)$
are small functions with respect to
$\sin z$
. We believe that the condition on
$\alpha (z)$
and
$\beta (z)$
can be removed, but we have not yet been able to show this.
3 Complex differential equations from
$ {\mathbf{cos}\ 2z=\mathbf{cos} ^{2}\,z-\mathbf{sin}^{2}\,z}$
Gundersen et al. [Reference Gundersen, Lü, Ng and Yang6] derived another nonlinear complex differential equation from the trigonometric identity
$\cos 2z=\cos ^{2}z-\sin ^{2}z$
.
Theorem 3.1 [Reference Gundersen, Lü, Ng and Yang6].
The complex differential equation
$$ \begin{align} f(z)^{2}-f'(z)^{2}=\cos 2z \end{align} $$
has only four meromorphic solutions
$f(z)=\pm \cos z$
and
$f(z)=\pm i\sin z$
.
Gao and Gao [Reference Gao and Gao2] extended (3.1) by proving the following result.
Theorem 3.2 [Reference Gao and Gao2, Theorem 7].
The complex differential equation
$$ \begin{align*} f(z)^{4}-f'(z)^{4}=\cos 2z \end{align*} $$
has only eight meromorphic solutions
$$ \begin{align*}f_{j}(z)=e^{{2j\pi i}/{4}}\cos z, \quad \tilde{f}_{k}(z)=e^{{(2k+1)\pi i}/{4}}\sin z,\end{align*} $$
for
$j=0,1,2,3$
and
$k=0,1,2,3$
.
Theorem 3.3. Let
$p(z)\not \equiv 0$
be a polynomial. If the complex differential equation
$$ \begin{align} f(z)^{4}-f'(z)^{4}=p(z)\cos 2z \end{align} $$
admits meromorphic solutions
$f(z)$
, then
$p(z)$
must be a nonzero constant p. Furthermore, all meromorphic solutions of (3.2) are
$$ \begin{align*}f_{j}(z)=p^{{1}/{4}}e^{{2j\pi i}/{4}}\cos z, \quad \tilde{f}_{k}(z)=p^{{1}/{4}}e^{{(2k+1)\pi i}/{4}}\sin z,\end{align*} $$
for
$j=0,1,2,3$
and
$k=0,1,2,3$
.
For the proof of Theorem 3.3, we need the following lemma.
Lemma 3.4. If
$p(z)$
is a nonzero polynomial and
$f(z)$
is a meromorphic solution of (3.2), then
$N(r,{1}/{f})\not =S(r,f)$
and
$N(r,{1}/{f'})\neq S(r,f)$
.
Proof. Obviously, the meromorphic solution
$f(z)$
must be a transcendental entire function. Define
$\phi (z)=1-({f'}/{f})^{4}$
. If
$N(r,{1}/{f})=S(r,f)$
, then
$$ \begin{align*}N\bigg(r,1-\bigg(\frac{f'}{f}\bigg)^{4}\bigg)=4\overline{N}\bigg(r,\frac{1}{f}\bigg)=S(r,f).\end{align*} $$
We also have
$$ \begin{align*}m\bigg(r,1-\bigg(\frac{f'}{f}\bigg)^{4}\bigg)=S(r,f)\end{align*} $$
by the lemma of the logarithmic derivative. Thus,
$T(r,\phi (z))=S(r,f)$
. Rewrite (3.2) as
$$ \begin{align*}\phi(z)f(z)^{4}+p(z)=2p(z)\cos^{2}(z).\end{align*} $$
By the second main theorem of Nevanlinna theory for three small functions [Reference Hayman8, Theorem 2.5],
$$ \begin{align*} T(r,f^{4})&\leq \overline{N}(r,f^{4})+\overline{N}\bigg(r,\frac{1}{f^{4}}\bigg)+\overline{N}\bigg(r,\frac{1}{f^{4}+{p(z)}/{\phi}}\bigg)+S(r,f^{4})\\ &\leq \overline{N}\bigg(r,\frac{1}{f}\bigg)+\frac{1}{2}N\bigg(r,\frac{1}{f^{4}+{p(z)}/{\phi}}\bigg)+S(r,f^{4})\\ &\leq 3 T(r,f)+S(r,f) \end{align*} $$
which is impossible. The same reasoning can be used to show
$N(r,{1}/{f'})\neq S(r,f)$
.
Proof of Theorem 3.3.
Taking the first derivative of (3.2),
$$ \begin{align} 4f^{3}f'-4(f')^{3}f"=p'\cos 2z-2p\sin 2z. \end{align} $$
Computing
$(p'-2ip)$
times (3.2) minus p times (3.3) gives
$$ \begin{align} (p'-2ip)f^{4}-(p'-2ip)(f')^{4}-4pf^{3}f'+4p(f')^{3}f"=-2ip^{2}e^{2iz}. \end{align} $$
Taking the derivative of (3.4) and eliminating
$e^{2iz}$
,
$$ \begin{align} A_{1}f^{4} & +A_{2}f^{3}f'+A_{3}f^{3}f"+A_{4}f^{2}(f')^{2}\nonumber\\ &-A_{1}(f')^{4}-A_{2}(f')^{3}f"-A_{3}(f')^{3}f"'-A_{4}(f')^{2}(f")^{2}=0, \end{align} $$
where
$A_{i}~(i=1,2,3,4)$
are the polynomials
$$ \begin{align*} \begin{cases} A_{1}=(2p'+2ip)(p'-2ip)-p(p"-2ip'),\\ A_{2}=-8pp',\\ A_{3}=4p^{2},\\ A_{4}=12p^{2}.\\ \end{cases} \end{align*} $$
Dividing the two sides of (3.5) by
$f^{3}f'$
gives
$$ \begin{align*}m\bigg(r,\frac{f}{f'}\bigg)=S(r,f)\end{align*} $$
by the lemma of the logarithmic derivative. Let
$z_{0}$
be a zero of
$f'$
but not a zero of
$p(z)$
. From (3.2),
$z_{0}$
is not a zero of f because all zeros of
$\cos 2z$
are simple. By (3.5),
$z_{0}$
is a zero of
$A_{1}f+A_{3}f"$
. Define
$\varphi (z):={(A_{1}f+A_{3}f")}/{f'}$
. Then
$T(r,\varphi (z))=S(r,f)$
. From the definition of
$\varphi (z)$
,
$$ \begin{align} A_{3}f"=\varphi f'-A_{1}f. \end{align} $$
Differentiating (3.6),
$$ \begin{align} A_{3}f"'=(\varphi-A^{\prime}_{3})f"+(\varphi'-A_{1})\varphi f'-A^{\prime}_{1}f. \end{align} $$
$$ \begin{align} A_{3}f"'=\bigg(\frac{\varphi(\varphi-A^{\prime}_{3})}{A_{3}}+(\varphi'-A_{1})\varphi\bigg) f'-\bigg(\frac{A_{1}(\varphi-A^{\prime}_{3})}{A_{3}}+A^{\prime}_{1}\bigg)f. \end{align} $$
Substituting (3.6) and (3.8) into (3.5),
$$ \begin{align} H_{1}f^{3}f'+H_{2}f^{2}(f')^{2}+H_{3}f(f')^{3}+H_{4}(f')^{4}=0, \end{align} $$
where
$$ \begin{align*} \begin{cases} H_{1}=A_{2}+\varphi,\\ H_{2}=A_{4}-\dfrac{A_{4}(A_{1})^{2}}{(A_{3})^{2}},\\ H_{3}=\dfrac{A_{1}A_{2}}{A_{3}}+\dfrac{A_{1}}{A_{3}}(\varphi-A^{\prime}_{3})+A^{\prime}_{1}+\dfrac{2A_{1}A_{4}\varphi}{(A_{3})^{2}},\\ H_{4}=-A_{1}-\dfrac{A_{2}\varphi}{A_{3}}-\dfrac{\varphi(\varphi-A^{\prime}_{3})}{A_{3}}-(\varphi'-A_{1})\varphi-\dfrac{A_{4}(\varphi)^{2}}{(A_{3})^{2}}.\\ \end{cases} \end{align*} $$
By Lemma 3.4, noting that the zeros of
$f(z)$
and
$f'(z)$
are different except for the zeros of
$p(z)$
, we have
$N(r,{1}/{f})=S(r,f)$
unless
$H_{i}\not \equiv 0$
. But
$H_{2}\equiv 0$
gives
$A_{3}=\pm A_{1}$
, so that
$$ \begin{align*}(2p'+2ip)(p'-2ip)-p(p"-2ip')=\pm4p^{2},\end{align*} $$
that is,
$$ \begin{align*}4p^{2}+2p^{\prime2}-pp"=\pm4p^{2}.\end{align*} $$
It follows that
$p(z)$
is a constant and
$A_{3}=A_{1}$
. We now rewrite (3.2) as
$$ \begin{align*} (F(z))^{4}-(F'(z))^{4}=\cos 2z, \end{align*} $$
where
$F(z)=p^{{-1}/{4}}f(z)$
. From Theorem 3.2, all solutions of (3.2) are
$$ \begin{align*}f_{j}(z)=p^{{1}/{4}}e^{{2j\pi i}/{4}}\cos z, \quad \tilde{f}_{k}(z)=p^{{1}/{4}}e^{{(2k+1)\pi i}/{4}}\sin z,\end{align*} $$
for
$j=0,1,2,3$
and
$k=0,1,2,3$
.
There are many other ways to perturb (3.1). For example, Gundersen et al. [Reference Gundersen, Lü, Ng and Yang6, Theorem 7] asked for all entire solutions of
$$ \begin{align} (z+1)f(z)^{2}+(z-1)f'(z)^{2}=z+\cos 2z. \end{align} $$
They stated that
$f(z)=\pm \cos z$
and
$f(z)=\pm i\sin z$
are the entire solutions of the above equation. However,
$f(z)=\pm i\sin z$
are not solutions. The remaining question is to prove the uniqueness of
$f(z)=\pm \cos z$
for (3.10). We make some observations which may contribute to the uniqueness problem. Let
$f(z)=\cos z+F(z)$
so that
$$ \begin{align*}(z+1)(\cos z+F(z))^{2}+(z-1)(F'(z)-\sin z)^{2}=z+\cos 2z.\end{align*} $$
By elementary calculation,
$$ \begin{align*}(z+1)(2\cos z+F(z))F(z)=(z-1)(2\sin z-F'(z))F'(z).\end{align*} $$
Let
$G(z)=2\cos z+F(z)$
. Then
$$ \begin{align*}(z+1)G(z)F(z)=(1-z)G'(z)F'(z),\end{align*} $$
where
$G(z)$
and
$F(z)$
are entire functions. If
$G(z)\equiv 0$
, then
$f(z)=-\cos z$
; if
$F(z)\equiv 0$
, then
$f(z)=\cos z$
. If
$G(z)\not \equiv 0$
and
$F(z)\not \equiv 0$
,
$$ \begin{align*}\frac{G'(z)}{G(z)}\frac{F'(z)}{F(z)}=\frac{z+1}{1-z}.\end{align*} $$
We conjecture that this equation has no entire solutions satisfying
$G(z)=2\cos z+F(z)$
.
To close this section, we pose the following question for further study.
Question 3.5. Let
$p_{1}(z)$
and
$p_{2}(z)$
be nonzero polynomials and n be a positive integer. How can we describe the transcendental meromorphic solutions of the system
$$ \begin{align*} \begin{cases} f(z)^{n}-g'(z)^{n}=p_{1}(z)\cos 2z,\ \ \\ g(z)^{n}-f'(z)^{n}=p_{2}(z)\cos 2z? \end{cases} \end{align*} $$
4 Discussion
Observe the trigonometric identity
$$ \begin{align*} 4\sin^{3}z-3\sin z=-\sin 3z. \end{align*} $$
The corresponding nonlinear complex differential equation
$$ \begin{align} 4f(z)^{3}+3f"(z)=-\sin 3z=\frac{e^{-3iz}-e^{3iz}}{2i}, \end{align} $$
was studied by Yang [Reference Yang24], Heittokangas et al. [Reference Heittokangas, Korhonen and Laine9] and Yang and Li [Reference Yang and Li25]. They pointed out that
$f_{1}(z)=\sin z$
,
$f_{2}(z)=-\tfrac 12\sqrt {3}\cos z-\tfrac 12\sin z$
and
$f_{3}(z)=\tfrac 12\sqrt {3}\cos z-\tfrac 12\sin z$
are the only entire solutions of (4.1). Equation (4.1) can also be viewed as a differential equation of Tumura–Clunie type,
$$ \begin{align*} f(z)^{n}+P(z,f(z))=g(z), \end{align*} $$
where
$P(z,f(z))$
is a differential polynomial in
$f(z)$
and
$g(z)$
is a meromorphic function (see the related results in Liao et al. [Reference Liao, Yang and Zhang15], and Mao and Liu [Reference Mao and Liu19]).
From the trigonometric identity
$$ \begin{align*}\cos^{3}z-3\sin^{2}z\cos z=\cos 3z,\end{align*} $$
Gao et al. [Reference Gao, Liu and Gao5] showed that the only entire solutions of the complex differential equations
$$ \begin{align} f(z)^{3}-3f(z)f'(z)^{2}=\cos 3z \end{align} $$
and
$$ \begin{align} f(z)^{3}+3f"(z)f'(z)^{2}=\cos 3z \end{align} $$
are
$f(z)=c_{1}e^{z}+c_{2}e^{-z}$
, where
$c_{1}^{3}=\tfrac 18$
and
$c_{2}^{3}=\tfrac 18$
. Furthermore, they considered
$$ \begin{align*} p(z)f(z)^{3}+q(z)f(z)f'(z)^{2}=\cos (az+b) \end{align*} $$
and showed that if this equation has entire solutions, then
$p(z)$
and
$q(z)$
are constants. Very recently, Li et al. [Reference Li, Wu and Xu12] found corresponding results when
$\cos 3z$
is replaced by
$p_{1}e^{\lambda _{1}(z)}+p_{2}e^{\lambda _{2}(z)}$
in (4.2) and (4.3).
We can consider the related complex differential system
$$ \begin{align} \begin{cases} 4f(z)^{3}-3g'(z)=\cos 3z, \\ 4g(z)^{3}+3f'(z)=-\sin 3z. \end{cases} \end{align} $$
Obviously, system (4.4) has the entire solution
$f(z)=\cos z$
and
$g(z)=\sin z$
, but we do not know if this entire solution is unique.
Finally, consider the trigonometric identity
$$ \begin{align*}\sin z \sin"z=-\sin^{2}z.\end{align*} $$
Li et al. [Reference Li, Lü and Yang13, Theorem 1] obtained the following result by Laguerre’s theorem on the separation of zeros.
Theorem 4.1. If
$f(z)$
is an entire solution of
$$ \begin{align*} f(z)f"(z)=p(z)\sin^{2}z, \end{align*} $$
where
$p(z)$
is a nonzero polynomial with real coefficients and real zeros, then
$p(z)$
must be a nonzero constant
$p=-a^{2}$
and
$f(z)=a\sin z$
.
We pose the following question on a related nonlinear complex differential system for further investigation.
Question 4.2. Let
$n, m, k_{1}, k_{2}, t_{1}, t_{2}$
be positive integers, and
$p_{1}(z)$
and
$p_{2}(z)$
be nonzero small functions with respect to
$\sin z$
. Suppose that
$f(z)$
and
$g(z)$
solve the complex differential system
$$ \begin{align} \begin{cases} f(z)^{n}g^{(k_{1})}(z)=p_{1}(z)\sin^{t_{1}}z, \ \ \\ g(z)^{m}f^{(k_{2})}(z)=p_{2}(z)\sin^{t_{2}}z. \end{cases} \end{align} $$
Can we describe the properties of transcendental meromorphic solutions of (4.5)?
Acknowledgement
The authors are very grateful to the reviewer and Professor Feng Lü for useful suggestions and valuable comments on the present version of the paper.
