Let f be a transcendental meromorphic solution of the equation (3). By Lemma 2, we know that the order of f is finite. Since every pole of f must come from the pole of \(R(z)\), \(p_{1}(z)\) or \(p_{2}(z)\), we know that f has at most finitely many poles.
Next we prove that all meromorphic solutions of equation (3) must be of the form \(f(z)=q(z)e^{P(z)}\), where \(q(z)\) is a rational function and \(P(z)\) is a polynomial.
From equation (
3), we have
$$ nf^{n-1}\bigl(f'\bigr)^{2}+f^{n}f''+R'(z)= \bigl(p_{1}'+p_{1}\alpha_{1}' \bigr)e^{\alpha_{1}(z)} +\bigl(p_{2}'+p_{2} \alpha_{2}'\bigr)e^{\alpha_{2}(z)}. $$
(5)
By eliminating
\(e^{\alpha_{2}(z)}\) from equations (
3) and (
5), we have
$$ \bigl(p_{2}'+p_{2} \alpha_{2}'\bigr)f^{n}f'+ \bigl(p_{2}'+p_{2}\alpha _{2}' \bigr)R(z)-np_{2}f^{n-1}\bigl(f' \bigr)^{2} -p_{2}f^{n}f''-p_{2}R'(z)=A(z), $$
(6)
where
\(A(z)=[p_{1}(p_{2}'+p_{2}\alpha_{2}')-p_{2}(p_{1}'+p_{1}\alpha _{1}')]e^{\alpha_{1}(z)}\).
If
\(A(z)\equiv0\), then
$$ \alpha_{2}'-\alpha_{1}'= \frac{p_{1}'}{p_{1}}-\frac{p_{2}'}{p_{2}}. $$
(7)
Thus
\(\alpha_{2}'-\alpha_{1}'\equiv0\). From equation (
6), we have
$$ \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)f^{n}f'-nf^{n-1}\bigl(f' \bigr)^{2}-f^{n}f''= R'(z)- \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)R(z). $$
(8)
Next, we discuss two cases.
Case 1:
\(R'(z)- (\frac{p_{2}'}{p_{2}}+\alpha_{2}' )R(z)\equiv0\). From equation (
8), we have
$$ \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)ff'-n\bigl(f'\bigr)^{2}-ff''= 0. $$
(9)
Suppose that
\(f(z)\) has infinitely many zeros, then we know from (
9) that the multiplicity of each zero of
\(f(z)\) is not less than 2, possibly except finite many zeros of
\(f(z)\). Let
\(z_{0}\) be a zero of
f with multiplicity
k, which is not a zero or pole of
\(\frac{p_{2}'}{p_{2}}+\alpha_{2}'\), then in some small neighborhood of
\(z_{0}\), we have
\(f(z)=a_{k}(z-z_{0})^{k}+a_{k+1}(z-z_{0})^{k+1}+\cdots\), where
\(a_{k}, a_{k+1}, \ldots \) are complex numbers,
\(a_{k}\neq0\) and
\(k\geq2\). By calculating the coefficient of the lowest power of
\(z-z_{0}\) in the left of equation (
9), we have
$$-n(ka_{k})^{2}-k(k-1)a_{k}^{2}=0, $$
that is,
thus
\(k=0\) or
\(k=\frac{1}{n+1}\), this is impossible. This contradiction lead to that
f has at most finitely many zeros. Thus,
\(f(z)=q(z)e^{P(z)}\), where
\(q(z)\) is a rational function and
\(P(z)\) is a polynomial.
Case 2:
\(R'(z)- (\frac{p_{2}'}{p_{2}}+\alpha_{2}' )R(z)\not \equiv0\). Rewriting equation (
8) as follows:
$$ f^{n-1} \biggl( \biggl(\frac{p_{2}'}{p_{2}}+ \alpha_{2}' \biggr)ff'-n\bigl(f' \bigr)^{2}-ff'' \biggr)= R'(z)- \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)R(z). $$
(10)
Let
$$ \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)ff'-n\bigl(f'\bigr)^{2}-ff''=a(z), $$
(11)
then
\(a(z)\) has only finitely many poles. It follows from the Clunie lemma that
\(m(r,a(z))=O(\log r)\), thus
\(a(z)\) is a rational function.
If
\(a(z)\equiv0\), then from equation (
10), we have
$$R'(z)- \biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr)R(z)\equiv0, $$
this is a contradiction.
If \(a(z)\not\equiv0\), then from (10), we know that f is a rational function. This is impossible.
Next we consider that
\(A(z)\not\equiv0\). In order to the convenience of calculation, we denote
$$\begin{aligned}& A_{1}=p_{2}'+p_{2} \alpha_{2}', \\& A_{2}=p_{1}\bigl(p_{2}'+p_{2} \alpha_{2}'\bigr)-p_{2}\bigl(p_{1}'+p_{1} \alpha _{1}'\bigr)\not\equiv0. \end{aligned}$$
Thus equation (
6) becomes
$$ A_{1}f^{n}f'-np_{2}f^{n-1} \bigl(f'\bigr)^{2} -p_{2}f^{n}f''+R_{1}(z)=A_{2}e^{\alpha_{1}(z)}, $$
(12)
where
\(R_{1}(z)=(p_{2}'+p_{2}\alpha_{2}')R(z)-p_{2}R'(z)\) is a rational functions. Differentiating both sides of (
12), we have
$$\begin{aligned} &A_{1}'f^{n}f'+n \bigl(A_{1}-p_{2}'\bigr)f^{n-1} \bigl(f'\bigr)^{2}+\bigl(A_{1}-p_{2}' \bigr)f^{n}f''-n(n-1)p_{2}f^{n-2} \bigl(f'\bigr)^{3} \\ &\quad {}-3np_{2}f^{n-1}f'f''-p_{2}f^{n}f'''+R_{1}'(z)= \bigl(A_{2}'+A_{2}\alpha _{1}'(z) \bigr)e^{\alpha_{1}(z)}. \end{aligned}$$
(13)
By eliminating
\(e^{\alpha_{1}(z)}\) from equations (
12) and (
13), we obtain
$$\begin{aligned} &B_{1}f^{n}f'-B_{2}f^{n-1} \bigl(f'\bigr)^{2}-B_{3}f^{n}f''+n(n-1)p_{2}A_{2}f^{n-2} \bigl(f'\bigr)^{3} \\ &\quad {}+3np_{2}A_{2}f^{n-1}f'f''+p_{2}A_{2}f^{n}f'''=Q_{1}(z), \end{aligned}$$
(14)
where
$$\begin{aligned}& B_{1}(z)=A_{1}\bigl(A_{2}'+A_{2} \alpha_{1}'(z)\bigr)-A_{2}A_{1}', \\& B_{2}(z)=np_{2}\bigl(A_{2}'+A_{2} \alpha _{1}'(z)\bigr)+n\bigl(A_{1}-p_{2}' \bigr)A_{2}, \\& B_{3}(z)=p_{2}\bigl(A_{2}'+A_{2} \alpha _{1}'(z)\bigr)+\bigl(A_{1}-p_{2}' \bigr)A_{2}, \\& Q_{1}(z)=-\bigl(A_{2}'+A_{2} \alpha _{1}'(z)\bigr)R_{1}(z)+A_{2}R_{1}'(z), \end{aligned}$$
are rational functions.
From equation (
14), we have
$$\begin{aligned} &f^{n-2} \bigl(B_{1}f^{2}f'-B_{2}f \bigl(f'\bigr)^{2}-B_{3}f^{2}f''+n(n-1)p_{2}A_{2} \bigl(f'\bigr)^{3} \\ &\quad {}+3np_{2}A_{2}ff'f''+p_{2}A_{2}f^{2}f''' \bigr)=Q_{1}(z). \end{aligned}$$
(15)
Let
$$ B_{1}f^{2}f'-B_{2}f \bigl(f'\bigr)^{2}-B_{3}f^{2}f''+n(n-1)p_{2}A_{2} \bigl(f'\bigr)^{3} +3np_{2}A_{2}ff'f''+p_{2}A_{2}f^{2}f'''=Q(z). $$
(16)
It follows from the Clunie lemma that
\(Q(z)\) is a rational function.
First we assume that
\(Q(z)\equiv0\). Suppose that
\(f(z)\) has infinitely many zeros, then we know from (
16) that the multiplicity of each zero of
\(f(z)\) is not less than 2, possibly except finite many zeros of
\(f(z)\). Let
\(z_{0}\) be a zero of
f with multiplicity
k, which is not a zero or pole of
\(B_{1}\),
\(B_{2}\),
\(B_{3}\) and
\(p_{2}A_{2}\), then
\(f(z)=a_{k}(z-z_{0})^{k}+a_{k+1}(z-z_{0})^{k+1}+\cdots\) (
\(a_{k}\neq0\),
\(k\geq2\)) holds in some small neighborhood of
\(z_{0}\). By calculating the coefficient of the lowest power of
\(z-z_{0}\) in the left of equation (
16), we have
$$n(n-1) (ka_{k})^{3}+3nk^{2}(k-1)a_{k}^{3}+k(k-1) (k-2)a_{k}^{3}=0, $$
thus
$$n(n-1)k^{2}+3nk(k-1)+(k-1) (k-2)=0, $$
this is impossible since
\(n\geq2\) and
\(k\geq2\). This contradiction lead to that
f has at most finitely many zeros. Thus,
\(f(z)=q(z)e^{P(z)}\), where
\(q(z)\) is a rational function and
\(P(z)\) is a polynomial.
Next we assume that
\(Q(z)\not\equiv0\). By differentiating (
16), we have
$$\begin{aligned} &B_{1}'f^{2}f'+ \bigl(2B_{1}-B_{2}'\bigr)f\bigl(f' \bigr)^{2}+\bigl(B_{1}-B_{3}' \bigr)f^{2}f''+ \bigl(n(n-1) (p_{2}A_{2})'-B_{2} \bigr) \bigl(f'\bigr)^{3} \\ &\quad {}+ \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)ff'f''+ \bigl((p_{2}A_{2})'-B_{3} \bigr)f^{2}f'''+3n^{2}p_{2}A_{2} \bigl(f'\bigr)^{2}f'' \\ &\quad {}+3np_{2}A_{2}f\bigl(f'' \bigr)^{2}+(3n+2)p_{2}A_{2}ff'f'''+p_{2}A_{2}f^{2}f^{(4)}=Q'(z). \end{aligned}$$
(17)
It follows from equations (
16) and (
17) that
$$\begin{aligned} &\bigl(B_{1}Q'-B_{1}'Q \bigr)f^{2}f'- \bigl(B_{2}Q'+ \bigl(2B_{1}-B_{2}'\bigr)Q \bigr)f \bigl(f'\bigr)^{2}- \bigl(B_{3}Q'+ \bigl(B_{1}-B_{3}'\bigr)Q \bigr)f^{2}f'' \\ &\quad {}+ \bigl(n(n-1)p_{2}A_{2}Q'-n(n-1) (p_{2}A_{2})'Q+B_{2}Q \bigr) \bigl(f'\bigr)^{3} \\ &\quad {}+ \bigl(3np_{2}A_{2}Q'- \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)Q \bigr)ff'f'' \\ &\quad {}+ \bigl(p_{2}A_{2}Q'- \bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr)f^{2}f'''-3n^{2}p_{2}A_{2}Q \bigl(f'\bigr)^{2}f''-3np_{2}A_{2}Q f\bigl(f''\bigr)^{2} \\ &\quad {}-(3n+2)p_{2}A_{2}Q ff'f'''-p_{2}A_{2}Qf^{2}f^{(4)}=0. \end{aligned}$$
(18)
If
\(f(z)\) has infinitely many zeros and
\(z_{0}\) is a zero of
\(f(z)\) which is not a zero or pole of the coefficients in (
16) and (
18), then by (
16) we know that a zero
\(z_{0}\) of
\(f(z)\) is simple, and it follows from (
18) that
\(z_{0}\) is a zero of
\(3n^{2}p_{2}A_{2}Qf''- (n(n-1)p_{2}A_{2}Q'-n(n-1)(p_{2}A_{2})'Q+B_{2}Q )f'\). Let
$$ b(z)=\frac{3n^{2}p_{2}A_{2}Qf''- (n(n-1)p_{2}A_{2}Q'-n(n-1)(p_{2}A_{2})'Q+B_{2}Q )f'}{p_{2}A_{2}Qf}. $$
(19)
Then
\(b(z)\) has only finitely many poles and it follows from the lemma of the logarithmic derivative that
\(m(r,b(z))=O(\log r)\). Hence
\(b(z)\) is a rational function.
If
\(b(z)\equiv0\), then we have
$$\begin{aligned} \frac{f''}{f'} &=\frac {n(n-1)p_{2}A_{2}Q'-n(n-1)(p_{2}A_{2})'Q+B_{2}Q}{3n^{2}p_{2}A_{2}Q} \\ &=\frac{n-1}{3n}\frac{Q'}{Q}-\frac{n-1}{3n}\frac {A_{2}'}{A_{2}}- \frac{n-1}{3n}\frac{p_{2}'}{p_{2}}+\frac {1}{3n^{2}}\frac{B_{2}}{p_{2}A_{2}}. \end{aligned}$$
Noticing that
$$B_{2}(z)=np_{2}\bigl(A_{2}'+A_{2} \alpha_{1}'\bigr)+n p_{2}\alpha_{2}'A_{2}, $$
thus we have
$$\frac{f''}{f'}=\frac{n-1}{3n}\frac{Q'}{Q}-\frac{n-2}{3n} \frac {A_{2}'}{A_{2}}-\frac{n-1}{3n}\frac{p_{2}'}{p_{2}}+\frac {1}{3n}\bigl( \alpha_{1}'+\alpha_{2}'\bigr). $$
By solving the above equation, we obtain
$$f'(z)=m(z)e^{\frac{\alpha_{1}+\alpha_{2}}{3n}}, $$
where
\(m(z)= (\frac{CQ^{n-1}}{p_{2}^{n-1}A_{2}^{n-2}} )^{\frac {1}{3n}}\) is a rational function.
From equation (
18) and the fact
\(b(z)\equiv0\), we have
$$\begin{aligned} & \bigl(B_{1}Q'-B_{1}'Q \bigr)ff'- \bigl(B_{2}Q'+ \bigl(2B_{1}-B_{2}'\bigr)Q \bigr) \bigl(f'\bigr)^{2}- \bigl(B_{3}Q'+ \bigl(B_{1}-B_{3}'\bigr)Q \bigr)ff'' \\ &\quad {}+ \bigl(3np_{2}A_{2}Q'- \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)Q \bigr)f'f'' \\ &\quad {}+ \bigl(p_{2}A_{2}Q'- \bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr)ff'''-3np_{2}A_{2}Q \bigl(f''\bigr)^{2} \\ &\quad {}-(3n+2)p_{2}A_{2}Q f'f'''-p_{2}A_{2}Qff^{(4)}=0. \end{aligned}$$
By calculating and substituting
\(f'\),
\(f''\),
\(f'''\),
\(f^{(4)}\) into the above equation, we can obtain
$$ s(z)f+t(z)e^{\frac{\alpha_{1}+\alpha_{2}}{3n}}=0, $$
(20)
where
$$\begin{aligned} &{s(z)=\bigl(B_{1}Q'-B_{1}'R \bigr)m(z)- \bigl(B_{3}Q'+\bigl(B_{1}-B_{3}' \bigr)Q \bigr) \biggl(m'(z)+\frac{1}{3n}m(z) \bigl( \alpha_{1}'+\alpha_{2}'\bigr) \biggr)} \\ &{\hphantom{s(z)=}{}+ \bigl(p_{2}A_{2}Q'- \bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr) \biggl(m''(z)+\frac{1}{3n}m(z) \bigl(\alpha_{1}''+\alpha_{2}'' \bigr)} \\ &{\hphantom{s(z)=}{}+\frac{2}{3n}m'(z) \bigl(\alpha_{1}'+ \alpha_{2}'\bigr) +\frac {1}{9n^{2}}m(z) \bigl( \alpha_{1}'+\alpha_{2}' \bigr)^{2} \biggr)} \\ &{\hphantom{s(z)=}{}-p_{2}A_{2}Q \biggl(m'''(z)+ \frac{1}{n}m'(z) \bigl(\alpha_{1}''+ \alpha _{2}''\bigr)+\frac{1}{3n}m(z) \bigl(\alpha_{1}'''+ \alpha_{2}'''\bigr)+ \frac{1}{n}m''(z) \bigl(\alpha_{1}'+ \alpha_{2}'\bigr)} \\ &{\hphantom{s(z)=}{}+\frac{1}{3n^{2}}m'(z) \bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2}+\frac {1}{3n^{2}}m(z) \bigl(\alpha_{1}'+\alpha_{2}' \bigr) \bigl(\alpha_{1}''+ \alpha_{2}''\bigr)+ \frac{1}{27n^{3}}m(z) \bigl(\alpha_{1}'+\alpha_{2}' \bigr)^{3} \biggr),}\\ &{t(z)=- \bigl(B_{2}Q'+\bigl(2B_{1}-B_{2}' \bigr)Q \bigr)m^{2}(z)-3np_{2}A_{2}Q \biggl(m'(z)+\frac{1}{3n}m(z) \bigl(\alpha_{1}'+ \alpha_{2}'\bigr) \biggr)^{2}} \\ &{\hphantom{t(z)=}{}+ \bigl(3np_{2}A_{2}Q'- \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)Q \bigr)m(z) \biggl(m'(z)+\frac{1}{3n}m(z) \bigl( \alpha_{1}'+\alpha _{2}'\bigr) \biggr)} \\ &{\hphantom{t(z)=}{}-(3n+2)p_{2}A_{2}Q m(z) \biggl(m''(z)+ \frac{1}{3n}m(z) \bigl(\alpha _{1}''+ \alpha_{2}''\bigr)} \\ &{\hphantom{t(z)=}{}+ \frac{2}{3n}m'(z) \bigl(\alpha_{1}'+\alpha_{2}' \bigr)+\frac{1}{9n^{2}}m(z) \bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2} \biggr).} \end{aligned}$$
Next we show that
\(s(z)\not\equiv0\) and
\(t(z)\not\equiv0\). Suppose that the assertion is not correct. Obviously, If
\(s(z)\equiv 0\), then from (
20), we must have
\(t(z)\equiv0\). By
\(s(z)\equiv0\), we have
$$\begin{aligned} \frac{s(z)}{m(z)p_{2}A_{2}Q} =&\frac{B_{1}}{p_{2}A_{2}}\frac {Q'}{Q}- \frac{B_{1}'}{p_{2}A_{2}} \\ &{}- \biggl(\frac{B_{3}}{p_{2}A_{2}} \frac{Q'}{Q}+ \frac{B_{1}}{p_{2}A_{2}}-\frac{B_{3}'}{p_{2}A_{2}} \biggr)\hspace{-1pt} \biggl(\frac{m'(z)}{m(z)}+ \frac{1}{3n}\bigl(\alpha_{1}'+\alpha _{2}'\bigr) \biggr) \\ &{}+ \biggl(\frac{Q'}{Q}-\frac{(p_{2}A_{2})'}{p_{2}A_{2}}+\frac {B_{3}}{p_{2}A_{2}} \biggr)\hspace{-1pt} \biggl(\frac{m''(z)}{m(z)}+\frac{1}{3n}\bigl(\alpha_{1}''+ \alpha _{2}''\bigr)+\frac{2}{3n} \frac{m'(z)}{m(z)}\bigl(\alpha_{1}'+\alpha _{2}'\bigr) \\ & {}+\frac{1}{9n^{2}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2} \biggr) \\ &{}- \biggl( \frac{m'''(z)}{m(z)}+ \frac{1}{n}\frac{m'(z)}{m(z)}\bigl( \alpha_{1}''+\alpha_{2}'' \bigr)+\frac {1}{3n}\bigl(\alpha_{1}'''+ \alpha_{2}'''\bigr)+ \frac{1}{n}\frac{m''(z)}{m(z)}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr) \\ &{}+\frac {1}{3n^{2}}\frac{m'(z)}{m(z)} \bigl(\alpha_{1}'+\alpha_{2}' \bigr)^{2} +\frac{1}{3n^{2}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr) \bigl(\alpha_{1}''+ \alpha _{2}''\bigr) \\ &{}+ \frac{1}{27n^{3}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{3} \biggr) \\ \equiv&0. \end{aligned}$$
(21)
Similarly, by
\(t(z)\equiv0\), we can get
$$\begin{aligned} \frac{t(z)}{m^{2}(z)p_{2}A_{2}Q} =&-\frac{B_{2}}{p_{2}A_{2}}\frac {Q'}{Q}- \frac{2B_{1}}{p_{2}A_{2}}+\frac{B_{2}'}{p_{2}A_{2}}-3n \biggl(\frac{m'(z)}{m(z)}+ \frac{1}{3n}\bigl(\alpha_{1}'+\alpha_{2}' \bigr) \biggr)^{2} \\ &{}+ \biggl(3n\frac{Q'}{Q}-3n\frac{(p_{2}A_{2})'}{p_{2}A_{2}}+\frac {2B_{2}}{p_{2}A_{2}}+ \frac{2B_{3}}{p_{2}A_{2}} \biggr) \biggl(\frac{m'(z)}{m(z)}+\frac{1}{3n}\bigl( \alpha_{1}'+\alpha_{2}'\bigr) \biggr) \\ &{}-(3n+2) \biggl(\frac{m''(z)}{m(z)}+\frac{1}{3n}\bigl( \alpha_{1}''+\alpha_{2}'' \bigr)+ \frac{2}{3n}\frac{m'(z)}{m(z)}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr) \\ &{}+\frac{1}{9n^{2}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2} \biggr) \\ \equiv&0, \end{aligned}$$
(22)
where
$$\begin{aligned}& \frac{B_{1}}{p_{2}A_{2}}=\biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr) \biggl(\frac {A_{2}'}{A_{2}}-\frac{A_{1}'}{A_{1}}+\alpha_{1}' \biggr), \\& \frac{B_{2}}{p_{2}A_{2}}=n\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)+n\frac {A_{2}'}{A_{2}}, \\& \frac{B_{3}}{p_{2}A_{2}}=\alpha_{1}'+\alpha_{2}'+ \frac {A_{2}'}{A_{2}},\\& \frac{B_{1}'}{p_{2}A_{2}}=\biggl(\frac{p_{2}'}{p_{2}}+\alpha_{2}' \biggr) \biggl(\frac {A_{2}''}{A_{2}} +\frac{A_{2}'}{A_{2}}\alpha_{1}'+ \frac{A_{1}'}{A_{1}}\alpha _{1}'-\frac{A_{1}''}{A_{1}}+ \alpha_{1}''\biggr), \\& \frac{B_{2}'}{p_{2}A_{2}}=n \biggl(\frac{p_{2}'}{p_{2}}\frac {A_{2}'}{A_{2}}+ \frac{A_{2}''}{A_{2}}+ \biggl(\frac{A_{2}'}{A_{2}}+\frac{p_{2}'}{p_{2}}\biggr) \bigl( \alpha_{1}'+\alpha _{2}'\bigr)+ \alpha_{1}''+\alpha_{2}'' \biggr), \\& \frac{B_{3}'}{p_{2}A_{2}}=\frac{p_{2}'}{p_{2}}\frac {A_{2}'}{A_{2}}+\frac{A_{2}''}{A_{2}}+ \biggl(\frac{A_{2}'}{A_{2}}+\frac{p_{2}'}{p_{2}}\biggr) \bigl(\alpha_{1}'+ \alpha _{2}'\bigr)+\alpha_{1}''+ \alpha_{2}''. \end{aligned}$$
From (
21) and (
22), we see that the highest degree terms may appear in
$$ -\frac{1}{3n}\alpha_{1}'\alpha_{2}' \bigl(\alpha_{1}'+\alpha_{2}' \bigr)+\frac {1}{9n^{2}}\bigl(\alpha_{1}' + \alpha_{2}'\bigr)^{3}-\frac{1}{27n^{3}}\bigl( \alpha_{1}'+\alpha _{2}' \bigr)^{3} $$
and
$$ -2\alpha_{1}'\alpha_{2}'- \frac{1}{3n}\bigl(\alpha_{1}'+\alpha _{2}'\bigr)^{2}+\frac{2(n+1)}{3n}\bigl( \alpha_{1}'+\alpha_{2}' \bigr)^{2} -\frac{3n+2}{9n^{2}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2}. $$
Namely,
$$ -\frac{1}{3n}\alpha_{1}' \alpha_{2}'\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)+\frac {3n-1}{27n^{3}}\bigl( \alpha_{1}'+\alpha_{2}' \bigr)^{3} $$
(23)
and
$$ -2\alpha_{1}'\alpha_{2}'+ \frac{6n^{2}-2}{9n^{2}}\bigl(\alpha_{1}'+\alpha _{2}'\bigr)^{2}. $$
(24)
Let \(\alpha_{1}(z)=a_{p}z^{p}+a_{p-1}z^{p-1}+\cdots+a_{0}\), \(\alpha _{2}(z)=b_{q}z^{q}+b_{q-1}z^{q-1}+\cdots+b_{0}\), where \(a_{i}\) (\({i=0,1,\ldots,p}\)), \(b_{j}\) (\(i=0,1,\ldots,q\)) are complex numbers, \(a_{p}b_{q}\neq0\) and \(p\geq1\), \(q\geq1\).
If \(p\neq q\), without loss of generality, we assume that \(p>q\). Since \(t(z)\equiv0\), by (24) we have \(\frac{6n^{2}-2}{9n^{2}}p^{2}a_{p}^{2}=0\), so \(n^{2}=\frac{1}{3}\), this is impossible. Next we suppose that \(p=q\).
If \(a_{p}+b_{p}=0\), then by \(t(z)\equiv0\) and (24), we have \(a_{p}b_{p}=0\), this is a contradiction.
If \(a_{p}+b_{p}\neq0\), then from \(s(z)\equiv0\), \(t(z)\equiv0\), (23) and (24), we can obtain \(-\frac{1}{3n}a_{p}b_{p}+\frac{3n-1}{27n^{3}}(a_{p}+b_{p})^{2}=0\) and \(-2a_{p}b_{p}+\frac{6n^{2}-2}{9n^{2}}(a_{p}+b_{p})^{2}=0\), thus we have \(\frac{2(3n-1)}{27n^{3}}=\frac{1}{3n}\frac {6n^{2}-2}{9n^{2}}\). So \(n=0\) or \(n=1\), this is a contradiction with the assumption \(n\geq2\).
Therefore, \(s(z)\not\equiv0\) and \(t(z)\not\equiv0\). Hence \(f(z)=-\frac{t(z)}{s(z)}e^{\frac{\alpha_{1}+\alpha_{2}}{3n}}\). Obviously, \(\alpha_{1}'(z)+\alpha_{2}'(z)\not\equiv0\), otherwise, \(f(z)\) is a rational function, this is impossible. Thus, \(f(z)\) is of the form \(q(z)e^{P(z)}\), where \(q(z)=-\frac{t(z)}{s(z)}\) is a rational function, \(P(z)=\frac {\alpha_{1}+\alpha_{2}}{3n}\) is a nonconstant polynomial.
Next we assume that
\(b(z)\not\equiv0\). From (
18) and (
19), we get
$$\begin{aligned} & \bigl(B_{1}Q'-B_{1}'Q \bigr)ff'- \bigl(B_{2}Q'+ \bigl(2B_{1}-B_{2}'\bigr)Q \bigr) \bigl(f'\bigr)^{2}- \bigl(B_{3}Q'+ \bigl(B_{1}-B_{3}'\bigr)Q \bigr)ff'' \\ &\quad {}+ \bigl(3np_{2}A_{2}Q'- \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)Q \bigr)f'f'' \\ &\quad {}+ \bigl(p_{2}A_{2}Q'- \bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr)ff'''-3np_{2}A_{2}Q \bigl(f''\bigr)^{2} \\ &\quad {}-(3n+2)p_{2}A_{2}Q f'f'''-p_{2}A_{2}Qff^{(4)}=b(z)p_{2}A_{2}Q \bigl(f'\bigr)^{2} \end{aligned}$$
(25)
and
$$\begin{aligned} f'' =&\frac {n(n-1)p_{2}A_{2}Q'-n(n-1)(p_{2}A_{2})'Q+B_{2}Q}{3n^{2}p_{2}A_{2}Q}f'+ \frac {p_{2}A_{2}Q b(z)}{3n^{2}p_{2}A_{2}Q}f \\ =& \biggl(\frac{n-1}{3n}\frac{Q'}{Q}-\frac{n-1}{3n} \frac {(p_{2}A_{2})'}{p_{2}A_{2}}+\frac{1}{3n^{2}}\frac {B_{2}}{p_{2}A_{2}} \biggr)f' + \frac{1}{3n^{2}}b(z)f \\ =& \biggl(\frac{n-1}{3n}\frac{Q'}{Q}-\frac{n-1}{3n} \frac {(p_{2}A_{2})'}{p_{2}A_{2}} +\frac{1}{3n}\frac{A_{2}'}{A_{2}}+\frac{1}{3n}\bigl( \alpha_{1}'+\alpha _{2}'\bigr) \biggr)f' +\frac{1}{3n^{2}}b(z)f. \end{aligned}$$
(26)
Let
\(\varphi(z)=\frac{n-1}{3n}\frac{Q'}{Q}-\frac{n-1}{3n}\frac {(p_{2}A_{2})'}{p_{2}A_{2}} +\frac{1}{3n}\frac{A_{2}'}{A_{2}}+\frac{1}{3n}(\alpha_{1}'+\alpha _{2}')\), then by calculating and substituting
\(f''\),
\(f'''\),
\(f^{(4)}\) into (
25), we can get
$$ \mu_{1}(z)f^{2}+\mu_{2}(z)ff'+ \mu_{3}(z) \bigl(f'\bigr)^{2}=0, $$
(27)
where
$$\begin{aligned} &{\mu_{1}(z)=\frac{1}{3n^{2}}\bigl(\varphi b+b'\bigr) \bigl(p_{2}A_{2}Q'-\bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr)-\frac{1}{3n^{2}}b \bigl(B_{3}Q'+ \bigl(B_{1}-B_{3}'\bigr)Q \bigr)} \\ &{\hphantom{\mu_{1}(z)=}{} -p_{2}A_{2}Q \biggl(\frac{1}{3n^{2}} \varphi^{2}b+\frac {2}{3n^{2}}\varphi'b+ \frac{1}{3n^{2}}\varphi b'+\frac {1}{3n^{2}}b''+ \frac{3n+1}{9n^{4}}b^{2} \biggr),}\\ &{\mu_{2}(z)=B_{1}Q'-B_{1}'Q -\varphi B_{3}Q'-\varphi \bigl(B_{1}-B_{3}' \bigr)Q+\frac{1}{n}p_{2}A_{2}Q'b- \frac {1}{n}(p_{2}A_{2})'Qb} \\ &{\hphantom{\mu_{2}(z)=}{} +\frac{2}{3n^{2}}(B_{2}+B_{3})Q b+ \bigl(p_{2}A_{2}Q'-\bigl((p_{2}A_{2})'-B_{3} \bigr)Q \bigr) \biggl(\varphi^{2}+\varphi'+ \frac {1}{3n^{2}}b\biggr)} \\ &{\hphantom{\mu_{2}(z)=}{} -p_{2}A_{2}Q \biggl(\varphi^{3}+3\varphi \varphi'+\varphi ''+\frac{9n+4}{3n^{2}} \varphi b+\frac{3n+4}{3n^{2}}b' \biggr),}\\ &{\mu_{3}(z)=3np_{2}A_{2}Q' \varphi- \bigl(3n(p_{2}A_{2})'-2B_{2}-2B_{3} \bigr)Q\varphi- \bigl(B_{2}Q'+\bigl(2B_{1}-B_{2}' \bigr)Q \bigr)} \\ &{\hphantom{\mu_{3}(z)=}{} -p_{2}A_{2}Q b-(3n+2)p_{2}A_{2}Q \biggl(\varphi^{2}+\varphi'+\frac {1}{3n^{2}}b \biggr)-3np_{2}A_{2}R\varphi^{2},} \end{aligned}$$
are rational functions.
We assume that
\(\mu_{1}(z)\equiv0\),
\(\mu_{2}(z)\equiv0\),
\(\mu _{3}(z)\equiv0\) hold simultaneously. By
\(\mu_{1}(z)\equiv0\), we have
$$\begin{aligned} &\frac{1}{3n^{2}}\biggl(\varphi+\frac{b'}{b}\biggr) \biggl(\frac{Q'}{Q}-\frac {(p_{2}A_{2})'}{p_{2}A_{2}} +\frac{B_{3}}{p_{2}A_{2}} \biggr)- \frac{1}{3n^{2}} \biggl(\frac {B_{3}}{p_{2}A_{2}}\frac{Q'}{Q}+\frac{B_{1}}{p_{2}A_{2}}- \frac {B_{3}'}{p_{2}A_{2}} \biggr) \\ &\quad {}- \biggl(\frac{1}{3n^{2}}\varphi^{2}+\frac{2}{3n^{2}} \varphi'+\frac {1}{3n^{2}}\varphi\frac{b'}{b}+ \frac{1}{3n^{2}}\frac{b''}{b}+\frac {3n+1}{9n^{4}}b \biggr)\equiv0. \end{aligned}$$
(28)
By
\(\mu_{2}(z)\equiv0\), we have
$$\begin{aligned} &\frac{B_{1}}{p_{2}A_{2}}\frac{Q'}{Q}-\frac{B_{1}'}{p_{2}A_{2}} -\varphi \frac{B_{3}}{p_{2}A_{2}}\frac{Q'}{Q} -\varphi\frac{B_{1}-B_{3}'}{p_{2}A_{2}}+\frac{1}{n} \frac{Q'}{Q}b- \frac{1}{n}\frac{(p_{2}A_{2})'}{p_{2}A_{2}}b \\ &\quad {}+\frac{2}{3n^{2}}\frac{B_{2}+B_{3}}{p_{2}A_{2}} b+ \biggl(\frac {Q'}{Q}- \frac{(p_{2}A_{2})'}{p_{2}A_{2}}+\frac{B_{3}}{p_{2}A_{2}} \biggr) \biggl(\varphi^{2}+ \varphi'+\frac{1}{3n^{2}}b\biggr) \\ &\quad {}- \biggl(\varphi^{3}+3\varphi\varphi'+ \varphi''+\frac {9n+4}{3n^{2}}\varphi b+ \frac{3n+4}{3n^{2}}b' \biggr)\equiv0. \end{aligned}$$
(29)
By
\(\mu_{3}(z)\equiv0\), we have
$$\begin{aligned} &3n\frac{Q'}{Q}\varphi- \biggl(3n\frac {(p_{2}A_{2})'}{p_{2}A_{2}}- \frac{2B_{2}}{p_{2}A_{2}}-\frac {2B_{3}}{p_{2}A_{2}} \biggr)\varphi - \biggl(\frac{B_{2}}{p_{2}A_{2}} \frac{R'}{R}+\frac {2B_{1}}{p_{2}A_{2}}-\frac{B_{2}'}{p_{2}A_{2}} \biggr) \\ &\quad {}-b-(3n+2) \biggl(\varphi^{2}+\varphi'+ \frac{1}{3n^{2}}b\biggr)-3n\varphi ^{2}\equiv0. \end{aligned}$$
(30)
Noticing the expressions of
φ,
\(\frac{B_{1}}{p_{2}A_{2}}\),
\(\frac{B_{2}}{p_{2}A_{2}}\),
\(\frac{B_{3}}{p_{2}A_{2}}\),
\(\frac{B_{1}'}{p_{2}A_{2}}\),
\(\frac {B_{2}'}{p_{2}A_{2}}\),
\(\frac{B_{3}'}{p_{2}A_{2}}\), we know that the highest powers of
z in the left hand side of (
28), (
29), (
30) may, respectively, appear in
$$\begin{aligned}& \frac{3n-1}{27n^{4}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2}-\frac {1}{3n^{2}} \alpha_{1}'\alpha_{2}'- \frac{3n+1}{9n^{4}}b, \end{aligned}$$
(31)
$$\begin{aligned}& \frac{3n-1}{27n^{3}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{3}-\frac {1}{3n} \alpha_{1}'\alpha_{2}'\bigl( \alpha_{1}'+\alpha_{2}'\bigr)+ \frac {6n^{2}-4}{9n^{3}}\bigl(\alpha_{1}'+\alpha_{2}' \bigr)b, \end{aligned}$$
(32)
$$\begin{aligned}& \frac{6n^{2}-2}{9n^{2}}\bigl(\alpha_{1}'+ \alpha_{2}'\bigr)^{2}-2\alpha _{1}'\alpha_{2}'- \frac{3n^{2}+3n+2}{3n^{2}}b. \end{aligned}$$
(33)
Let
\(\alpha_{1}(z)=a_{p}z^{p}+a_{p-1}z^{p-1}+\cdots+a_{0}\),
\(\alpha _{2}(z)=b_{q}z^{q}+b_{q-1}z^{q-1}+\cdots+b_{0}\), where
\(a_{i}\) (
\(i=0,1,\ldots,p\)),
\(b_{j}\) (
\(i=0,1,\ldots,q\)) are complex numbers,
\(a_{p}b_{q}\neq0\) and
\(p\geq1\),
\(q\geq1\).
If
\(p\neq q\), without loss of generality, we assume that
\(p>q\). Obviously,
\(\deg_{\infty} b \leq2p-2\), otherwise we can get a contradiction from (
31) immediately. If
\(\deg_{\infty } b < 2p-2\), then from (
31), we have
\(\frac {3n-1}{27n^{4}}=0\). So
\(n=\frac{1}{3}\), this is impossible. If
\(\deg_{\infty} b = 2p-2\), then from (
31) and (
33), we obtain
$$\frac{\frac{3n-1}{27n^{4}}}{\frac{6n^{2}-2}{9n^{2}}}=\frac{\frac {3n+1}{9n^{4}}}{\frac{3n^{2}+3n+2}{3n^{2}}}, $$
by solving above equation, we have
\(n=\pm1\), this is impossible. Next we suppose that
\(p=q\).
If \(\deg_{\infty} b > 2p -2\), then by \(\mu_{1}(z)\equiv0\) and (31), we have \(-\frac{3n+1}{9n^{4}}=0\), so \(n=-\frac{1}{3}\), this is impossible. Thus we may assume that \(\deg_{\infty} b \leq2p -2\).
If \(a_{p}+b_{p}=0\) and \(\deg_{\infty} b < 2p -2\), then by \(\mu _{1}(z)\equiv0\) and (31), we have \(-\frac {1}{3n^{2}}p^{2}a_{p}b_{p}=0\), this is impossible. If \(a_{p}+b_{p}=0\) and \(\deg_{\infty} b = 2p -2\), then by \(\mu _{1}(z)\equiv0\), \(\mu_{3}(z)\equiv0\) and (31), (33) we have \(\frac{2(3n+1)}{9n^{4}}=\frac{3n^{2}+3n+2}{9n^{4}}\), hence \(n=1\), this is also impossible.
If \(a_{p}+b_{p}\neq0\) and \(\deg_{\infty} b < 2p -2\), then by \(\mu _{1}(z)\equiv0\), \(\mu_{3}(z)\equiv0\) and (31), (33) we have \(\frac{2(3n-1)}{27n^{4}}=\frac{6n^{2}-2}{27n^{4}}\), hence \(n=1\), this is a contradiction with the assumption \(n\geq2\). If \(a_{p}+b_{p}\neq 0\) and \(\deg_{\infty} b = 2p -2\), then from (31), (32) and the assumption \(\mu_{1}(z)\equiv0\), \(\mu_{2}(z)\equiv0\), we have \(-\frac{3n+1}{9n^{4}}=\frac {6n^{2}-4}{9n^{4}}\), thus \(n=\frac{1}{2}\) or \(n=-1\), this is impossible. Therefore, \(\mu_{1}(z)\equiv0\), \(\mu_{2}(z)\equiv0\), \(\mu _{3}(z)\equiv0\) cannot hold simultaneously.
If
\(\mu_{3}(z)\not\equiv0\). Then equation (
27) can be rewritten as
$$ \mu_{3}(z) \bigl(f'\bigr)^{2}=- \mu_{1}(z)f^{2}-\mu_{2}(z)ff'. $$
(34)
Suppose that
\(f(z)\) has infinitely many zeros, then we know that the multiplicity of each zero of
\(f(z)\) is not less than 2, possibly except finitely many zeros of
\(f(z)\). Let
\(z_{0}\) be a zero of
f with multiplicity
k, but that is not the zero and pole of
\(\mu_{1}(z)\),
\(\mu_{2}(z)\),
\(\mu_{3}(z)\), then
\(z_{0}\) is a zero with multiplicity
\(2k-2\) in the left side and a zero with multiplicity at least
\(2k-1\) in the right side of equation (
34). This contradiction lead to that
f has at most finitely many zeros.
If
\(\mu_{3}(z)\equiv0\), then
\(\mu_{1}(z)\not\equiv0\) and
\(\mu _{2}(z)\not\equiv0\), equation (
27) is simplified to the following form:
$$\mu_{1}(z)f=-\mu_{2}(z)f', $$
by a similar discussion to above, we see that
f has at most finitely many zeros. Thus,
\(f(z)=q(z)e^{P(z)}\), where
\(q(z)\) is a rational function and
\(P(z)\) is a polynomial.
Substituting
\(f(z)=q(z)e^{P(z)}\) into the equation (
3) yields
$$q(z)^{n}\bigl(q'(z)+q(z)p'(z) \bigr)e^{(n+1)P(z)}+R(z)=p_{1}(z)e^{\alpha _{1}(z)}+p_{2}(z)e^{\alpha_{2}(z)}. $$
If
\(\alpha_{1}'(z)\not\equiv\alpha_{2}'(z)\), it follows from Lemma
1 that
\(R(z)\equiv0\). Furthermore, either
\((n+1)P(z)=\alpha_{1}(z)+C_{1}\),
\(q(z)^{n}(q'(z)+q(z)P'(z))=D_{1}p_{1}(z)\),
\(p_{2}(z)\equiv0\) or
\((n+1)P(z)=\alpha_{2}(z)+C_{2}\),
\(q(z)^{n}(q'(z)+q(z)P'(z))=D_{2}p_{2}(z)\),
\(p_{1}(z)\equiv0\), where
\(C_{1}\),
\(C_{2}\),
\(D_{1}\),
\(D_{2}\) are constants and
\(D_{1}e^{C_{1}}=D_{2}e^{C_{2}}=1\). This is a contradiction with the assumption
\(p_{1}\),
\(p_{2}\) are nonzero rational functions. If
\(\alpha_{1}'(z)\equiv\alpha_{2}'(z)\), then
\(\alpha_{2}(z)=\alpha _{1}(z)+C\), where
C is a constant, and it follows from Lemma
1 that
\(R(z)\equiv0\) and the equation (
3) is reduced to the following form:
$$f^{n}f'=\bigl(p_{1}+p_{2}e^{C} \bigr)e^{\alpha_{1}(z)}. $$
By Theorem
A, we have
$$f(z)=q(z)e^{\frac{\alpha_{1}(z)}{n+1}}, $$
where
\(q(z)\) is a rational function with
\(q^{n}((n+1)q'+q\alpha_{1}')=(n+1)(p_{1}+p_{2} e^{\alpha_{2} (z)-\alpha_{1} (z)})\).