In what follows, we assume the reader is familiar with the standard notions of Nevanlinna's value distribution theory as the proximity function $m(r,w)$, the integrated counting function $N(r,w)$, the characteristic function $T(r,w)$, see e.g. [1, 2]. Many authors investigated the algebraic differential equations and obtained many results (see [3-10]).

An analytic function $w(z)$ with $\nu$ branches is an algebroid function if the function $w(z)$ satisfies an equation of the form

$ \psi(z,w)=A_{v}(z)w^{v}+A_{v-1}(z)w^{v-1}+\cdots+A_{0}(z), $

where $A_{j}(z)(j=0,1,\cdots,v)$ are regular functions with no commom zeros and $A_{\nu}\neq0$, especially, when $\nu=1$, $w(z)$ be a meromorphic function; when $A_{j}(z)(j=0,1,\cdots,v)$ are polynomials, $w(z)$ is an algebraic function. Some notations

$N_{x}(r,w)=\frac{1}{\nu}\int_{0}^{r}\frac{n_{x}(t,w)-n_{x}(0,w)}{t}dt+\frac{1}{\nu}n_{x}(0,w)\log r,$

and Toda gave the definition of $N_{b}(r,w)$.

Let $w=w(z)$ be a $\nu$-valued algebroid function and $a$ be a pole of $w$. Then, in the neighbourhood of $a$, we have the following expansions of $w$:

$ w(z)=(z-a)^{-\tau_{i}/\lambda_{i}}S((z-a)^{1/\lambda_{i}}), $

where $i=1,2,\cdots, \mu(a)(\leq\nu)$, $1\leq\tau_{i}$, $1\leq\lambda_{i}$, $\sum\lambda_{i}=\nu$ and $S(t)$ is a regular power series of $t$ such that $S(0)\neq0$. Put

$ n_{b}(r,w)=\sum\limits_{|a|\leq r}\sum\limits_{i=1}^{\mu(a)}(\lambda_{i}-1) $

and

$ N_{b}(r,w)=\frac{1}{\nu}\int_{0}^{r}(n_{b}(t,w)-n_{b}(0,w))/tdt+\frac{1}{\nu}n_{b}(0,w)\log r. $

It is trivial that $ N_{b}(r,w)\leq(\nu-1)\overline{N}(r,w). $

About the differential equation

$\begin{equation} \sum\limits_{(i)\in I}a_{(i)}(z)w^{i_{0}}(w')^{i_{1}}\cdots(w^{(n)})^{i_{n}}=\frac{\sum \limits_{i=0}^{p}a_{i}(z)w^{i}}{\sum\limits_{j=0}^{q}b_{j}(z)w^{j}}, \end{equation}$

(1.1)

where ${a_{(i)}(z)}$, ${a_{i}(z)}$ and ${b_{j}(z)}$ are meromorphic coefficients. He and Laine investigated the problem of the growth of solutions of it and obtain the following result.

Theorem A  [2] Let $w(z)$ be an algebroid solution of $(1.1)$ with $\nu$ branches and $p>q+\lambda$. Then for any $\xi>1$, there exist a positive constant $K$ and $r_{0}$ such that for all $r\geq r_{0}$, we have $ T(r,w)\leq KF(\xi r), $ where

$F(r)=\overline{N}(r,w)+\sum\limits_{(i)}T(r,a_{(i)})+\sum _{i=0}^{p}T(r,a_{i})+\sum\limits_{j=0}^{q}T(r,b_{j})+1.$

In this paper, we discuss the problem of the growth of solutions of generalized higher-order algebraic differential equation of the form

$\begin{equation} \frac{\Omega_{1}(z,w)}{\Omega_{2}(z,w)(w-a)^{\lambda}}=\frac{\sum \limits_{i=0}^{p}a_{i}(z)w^{i}}{\sum\limits_{j=0}^{q}b_{j}(z)w^{j}}, \end{equation}$

(1.2)

where $\Omega_{1}(z,w)$ and $\Omega_{2}(z,w)$ be two differential polynomials, $a$ be a nonzero complex constant, and

$\begin{eqnarray*} &&\Omega_{1}(z,w)=\sum\limits_{(i)\in I}a_{(i)}(z)w^{i_{0}}(w')^{i_{1}}\cdots(w^{(n)})^{i_{n}}(n\geq1), \\ &&\Omega_{2}(z,w)=\sum\limits_{(j)\in J}b_{(j)}(z)w^{j_{0}}(w')^{j_{1}}\cdots(w^{(m)})^{j_{m}}(m\geq1), \end{eqnarray*}$

other notations

$\begin{eqnarray*}&&\lambda_{1}=\max\limits_{(i\in I)}\{\sum\limits_{\alpha=0}^{n}i_{\alpha}\},\lambda_{2}=\max\limits_{(j\in J)}\{\sum\limits_{\beta=0}^{m}j_{\beta}\}, \lambda=\max\{\lambda_{1},\lambda_{2}\}, \\ &&\overline{\mu}_{1}=\max\limits_{(i\in I)}\{\sum\limits_{\alpha=0}^{n}\alpha i_{\alpha}\}, \overline{\mu}_{2}=\max\limits_{(j\in J)}\{\sum\limits_{\beta=0}^{m}\beta j_{\beta}\}, \overline{\mu}=\max\{\overline{\mu}_{1},\overline{\mu}_{2}\}, \\ &&\Delta_{1}=\max\limits_{(i\in I)}\{\sum\limits_{\alpha=0}^{n}(\alpha+1) i_{\alpha}\}, \Delta_{2}=\max\limits_{(j\in J)}\{\sum\limits_{\beta=0}^{m}(\beta+1) j_{\beta}\}, \Delta=\max\{\Delta_{1},\Delta_{2}\}, \\ &&\sigma_{1}=\max\limits_{(i\in I)}\{\sum\limits_{\alpha=0}^{n}(2\alpha-1)i_{\alpha}\}, \sigma_{2}=\max\limits_{(j\in J)}\{\sum\limits_{\beta=0}^{m}(2\beta-1)j_{\beta}\}, \sigma=\max\{\sigma_{1},\sigma_{2}\}, \\ &&l_{1}=\max\limits_{(i\in I)}\{\sum\limits_{\alpha=0}^{n}(\alpha-1)i_{\alpha}\}, l_{2}=\max\limits_{(j\in J)}\{\sum\limits_{\beta=0}^{m}(\beta-1)j_{\beta}\}, l=\max\{l_{1},l_{2}\},\end{eqnarray*}$

and obtain the following result.

Theorem 1 Let $w(z)$ be an algebroid solution of $(1.2)$ with $\nu$ branches and $p>q+2\lambda$. Then for any $\xi>1$, there exist a positive constant $K$ and $r_{0}$ such that for all $r\geq r_{0}$, we get $ T(r,w)\leq KF(\xi r), $ where

$\begin{eqnarray*}F(r)&=&\overline{N}(r,w)+N_{x}(r,w)+N_{b}(r,w)\\ &&+\sum\limits_{(i)}T(r,a_{(i)})+\sum\limits_{(j)}T(r,b_{(j)})+\sum _{i=0}^{p}T(r,a_{i})+\sum\limits_{j=0}^{q}T(r,b_{j})+1,\\ \lambda&=&\max\{\lambda_{1},\lambda_{2}\}.\end{eqnarray*}$

Lemma 1 [1] Let $R(z,w)=\frac{\sum\limits_{i=0}^{p}a_{i}(z)w^{i}}{\sum\limits_{j=0}^{q}b_{j}(z)w^{j}}$ be an irreducible rational function in $w(z)$ with the meromorphic coefficients ${a_{i}(z)}$ and ${b_{j}(z)}$. If $w(z)$ is an algebroid function, then

$ T(r,R(z,w))=\max\{p,q\}T(r,w)+O\{\sum T(r,a_{i})+\sum T(r,b_{j})\}. $

Lemma 2 Let $w(z)$ be an algebroid function, and $\Omega(z,w)$ be as in (1.2), $a$ be a nonzero complex constant. Then

$ T(r,\frac{\Omega(z,w)}{(w-a)^{\lambda}})\leq\lambda T(r,w)+\overline{\mu}[\overline{N}(r,w)+N_{b}(r,w)]+\sum\limits_{(i)}T(r,a_{(i)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})+O(1). $

Proof Let $w$ be an algebroid function with $\nu$ branches, we denote by $L$ a curve connecting all branch points of $w(z)$ and $C'=C\backslash L$, then every branch $w_{j}(z)(j=1,2,\cdots,\nu)$ of $w(z)$ is single-valued in $C'$. Set $E=\{z,|z|=r\}$, $E_{1}^{j}=\{z\epsilon E,|w_{j}(z)-a|\geq1\}$, $E_{2}^{j}=E\backslash E_{1}^{^{j}}$, and $z=re^{i\theta}$, so that

$ \frac{1}{2\pi}\int_{E}\log^{+}|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|d\theta=\frac{1}{2\pi}(\int_{E_{1}^{j}}+\int_{E_{2}^{j}})\log^{+}|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|d\theta. $

Set $\lambda_{i}=i_{0}+i_{1}+\cdots+i_{n}$, when $z\epsilon E_{1}^{j}$, it is easy to show

$\begin{eqnarray*} |\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|&=&|\frac{\sum\limits_{(i)}a_{(i)}(z)(\frac{(w_{j}-a)'}{w_{j}-a})^{i_{1}}\cdots(\frac{(w_{j}-a)^{(n)}}{w_{j}-a})^{i_{n}}}{(w_{j}-a)^{\lambda-\lambda_{i}}}| \\&\leq&\sum\limits_{(i)}|a_{(i)}(z)||(\frac{(w_{j}-a)'}{w_{j}-a})^{i_{1}}|\cdots|(\frac{(w_{j}-a)^{(n)}}{w_{j}-a})^{i_{n}}|, \end{eqnarray*}$

by the lemma of Logarithmic Derivate, we have

$\begin{eqnarray*} &&\frac{1}{2\pi}\int_{E_{1}^{j}}\log^{+}|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|d\theta\leq\frac{1}{2\pi}\int_{E_{1}^{j}}\log^{+}\sum\limits_{(i)}|a_{(i)}(z)|d\theta +\sum\limits_{\alpha=1}^{n}m(r,\frac{(w_{j}-a)^{(\alpha)}}{w_{j}-a})\\ &=&\sum\limits_{(i)}m(r,a_{(i)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w_{j}^{(\alpha)}}{w_{j}-a}); \end{eqnarray*}$

when $z\epsilon E_{2}^{j}$, we have

$\begin{eqnarray*} &&|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|=|\frac{\sum\limits_{(i)}a_{(i)}(z)(\frac{(w_{j}-a)'}{w_{j}-a})^{i_{1}}\cdots(\frac{(w_{j}-a)^{(n)}}{w_{j}-a})^{i_{n}}}{(w_{j}-a)^{\lambda-\lambda_{i}}}|\\ &\leq&|\frac{1}{(w_{j}-a)^{\lambda-\lambda_{i}}}|\sum\limits_{(i)}|a_{(i)}(z)||(\frac{(w_{j}-a)'}{w_{j}-a})^{i_{1}}|\cdots|(\frac{(w_{j}-a)^{(n)}}{w_{j}-a})^{i_{n}}|, \end{eqnarray*}$

then

$\begin{eqnarray*} &&\frac{1}{2\pi}\int_{E_{2}^{j}}\log^{+}|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|d\theta\\ &\leq&\frac{1}{2\pi}\int_{E_{2}^{j}}\log^{+}\sum\limits_{(i)}|a_{(i)}(z)|d\theta +\frac{\lambda}{2\pi}\int_{E_{2}^{j}}\log^{+}|\frac{1}{w_{j}-a}|d\theta+\sum\limits_{\alpha=1}^{n}m(r,\frac{(w_{j}-a)^{(\alpha)}}{w_{j}-a})\\ &=&\sum\limits_{(i)}m(r,a_{(i)}(z))+\lambda m(r,\frac{1}{w_{j}-a})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w_{j}^{(\alpha)}}{w_{j}-a}). \end{eqnarray*}$

Hence we obtain

$\begin{eqnarray} &&m(r,\frac{\Omega(z,w)}{(w-a)^{\lambda}})=\frac{1}{\nu}\sum\limits_{j=1}^{\nu}\frac{1}{2\pi}(\int_{E_{1}^{j}}+\int_{E_{2}^{j}})\log^{+}|\frac{\Omega(z,w_{j})}{(w_{j}-a)^{\lambda}}|d\theta \nonumber\\ &\leq&\lambda m(r,\frac{1}{w-a})+\sum\limits_{(i)}m(r,a_{(i)}(z))+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a}). \end{eqnarray}$

(2.1)

Next, we estimate the poles of $\frac{\Omega(z,w)}{(w-a)^{\lambda}}$, we denote by $\tau(z_{0},f)$ the order of pole of $f$ at $z=z_{0}$. Now we discuss the following two cases.

(ⅰ) When $z_{0}$ is not a pole of $w(z)$, we have

$\begin{eqnarray} \tau(z_{0},\frac{\Omega(z,w)}{(w-a)^{\lambda}}) &\leq&\tau(z_{o},\frac{1}{(w-a)^{\lambda}})+\sum\tau(z_{0},a_{(i)}(z))\nonumber\\ &=&\lambda\tau(z_{0},\frac{1}{w-a})+\sum\tau(z_{0},a_{(i)}(z)). \end{eqnarray}$

(2.2)

(ⅱ) When $z_{0}$ is a pole of $w(z)$, in a neighbourhood of $z_{0}$,

$\begin{eqnarray*} &&w-a=(z-z_{0})^{-\tau/\beta}S((z-z_{0})^{1/\beta})(\beta\geq1,\tau\geq1), \\&&w^{(\alpha)}=(w-a)^{(\alpha)}=(z-z_{0})^{(-\tau+\beta \alpha)/\beta}S_{\alpha}(z),S_{\alpha}(z_{0})\neq0, \infty, \end{eqnarray*}$

then

$ \tau(z_{0},(\frac{w^{(\alpha)}}{w-a})^{i_{\alpha}})=\tau(z_{0},(\frac{(w-a)^{(\alpha)}}{w-a})^{i_{\alpha}})=\beta \alpha i_{\alpha}. $

Set $a_{(i)}(z)w^{i_{0}}(w')^{i_{1}}\cdots(w^{n})^{i_{n}}$ is a general term of $\Omega(z,w(z))$, we obtain

$ \tau(z_{0},a_{(i)}(z)w^{i_{0}}(w')^{i_{1}}\cdots(w^{(n)})^{i_{n}}/(w-a)^{\lambda})\leq\beta\sum\limits_{a=1}^{n}\alpha i_{\alpha}+\tau(z_{0},a_{(i)}(z)), $

then

$\begin{eqnarray} \tau(z_{0},\frac{\Omega(z,w)}{(w-a)^{\lambda}})&\leq&\beta \max\{\sum\limits_{a=1}^{n}\alpha i_{\alpha}\}+\tau(z_{0},a_{(i)}(z))=\beta \overline{\mu}+\sum\tau(z_{0},a_{(i)}(z))\nonumber\\ &=&\overline{\mu}(\beta-1)+\overline{\mu}+\tau(z_{0},a_{(i)}(z)). \end{eqnarray}$

(2.3)

Combining (2.2) with (2.3) we deduce

$\begin{eqnarray} N(r,\frac{\Omega(z,w)}{(w-a)^{\lambda}})\leq\lambda N(r,\frac{1}{w-a})+\overline{\mu}N_{b}(r,w)+\overline{\mu}\overline{N}(r,w)+\sum\limits_{(i)}N(r,a_{(i)}(z)). \end{eqnarray}$

(2.4)

Combining (2.1) with (2.4) we have

$\begin{eqnarray*} &&T(r,\frac{\Omega(z,w)}{(w-a)^{\lambda}})\\ &\leq&\lambda T(r,\frac{1}{w-a})+\overline{\mu}N_{b}(r,w)+\overline{\mu}\overline{N}(r,w)+\sum\limits_{(i)}T(r,a_{(i)(z)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})\\ &=&\lambda T(r,w-a)+\overline{\mu}[N_{b}(r,w)+\overline{N}(r,w)]+\sum\limits_{(i)}T(r,a_{(i)(z)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})+O(1)\\ &=&\lambda T(r,w)+\overline{\mu}[N_{b}(r,w)+\overline{N}(r,w)]+\sum\limits_{(i)}T(r,a_{(i)(z)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})+O(1). \end{eqnarray*}$

Lemma 3 Let $w(z)$ be an algebroid function, then we have

$ T(r,\Omega(z,w))\leq\lambda T(r,w)+\overline{\mu}\overline{N}(r,w)+\sigma N_{x}(r,w)-lN_{b}(r,w)+\sum\limits_{(i)}T(r,a_{(i)(z)})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w}). $

Proof Proceeding similary as in the proof the Lemma $2$, we can verify the assertion.

Lemma 4 [2] Let $U(r)$, $H(r)(r\in[0,\infty))$ be two nonnegative and nondecreasing functions, $H(r)\rightarrow\infty$ as $r\rightarrow\infty$, $a$ and $b$ be two positive numbers,

$H(r_{0})\geq\max\{(a+b)\log2, 2^{2+\frac{b}{a}}a(a+b)\},$

if for all $r$ and $t$, when $0<r_{0}\leq r<t$, satisfies

$ U(r)<a\log^{+}U(t)+b\log\frac{t}{t-r}+H(r), $

then for $0<r_{0}\leq r<t$, we have

$ U(r)<(a+b)\log\frac{t}{t-r}+2H(r). $

We discuss the following two cases.

Case 1 If $w(z)$ satisfies $ \sum\limits_{i=0}^{p}a_{i}(z)w^{i}\equiv0, $ then $ a_{p}(z)w^{p}=-a_{p-1}(z)w^{p-1}-\cdots-a_{0}(z). $ From Lemma 1, there exists a positive constant $K$ such that

$\begin{eqnarray*} &&pT(r,w)+T(r,a_{p})\leq(p-1)T(r,w)+\sum\limits_{i=0}^{p-1}T(r,a_{i}(z)), \\ &&T(r,w)\leq K\sum\limits_{i=0}^{p}T(r,a_{i}(z))\leq KF(r).\end{eqnarray*}$

Case 2 If $\sum\limits_{i=0}^{p}a_{i}(z)w^{i}\neq0$, we rewrite equation (1.2) as follows

$ Q(z,w)\frac{\Omega_{1}(z,w)}{\Omega_{2}(z,w)(w-a)^{\lambda}}=P(z,w). $

Using Lemma 1, Lemma 2 and Lemma 3, we get

$\begin{eqnarray} &&T(r,Q(z,w)\frac{\Omega_{1}(z,w)}{\Omega_{2}(z,w)(w-a)^{\lambda}}) \leq T(r,Q(z,w))+T(r,\frac{\Omega_{1}(z,w)}{(w-a)^{\lambda}})+T(r,\frac{1}{\Omega_{2}(z,w)})\nonumber\\ &\leq& T(r,Q(z,w)+T(r,\frac{\Omega_{1}(z,w)}{(w-a)^{\lambda}})+T(r,\Omega_{2}(z,w))+O(1)\nonumber\\ &\leq&(q+2\lambda)T(r,w)+2\overline{\mu}\overline{N}(r,w)+(\overline{\mu}-l)N_{b}(r,w)+\sigma N_{x}(r,w)+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})\nonumber\\ &&+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w})+\sum T(r,a_{(i)})+\sum T(r,b_{(j)})+\sum\limits_{j=0}^{q}T(r,b_{j})+O(1). \end{eqnarray}$

(3.1)

By means of Lemma 1, we obtain

$\begin{eqnarray} T(r,P(z,w))=p T(r,w)+O\{\sum\limits_{i=0}^{p}T(r,a_{i})\}. \end{eqnarray}$

(3.2)

It follows from (3.1) and (3.2) that

$\begin{eqnarray*} &&pT(r,w)\\ &<&(q+2\lambda)T(r,w)+2\overline{\mu}\overline{N}(r,w)+(\overline{\mu}-l)N_{b}(r,w)+\sigma N_{x}(r,w)+\sum\limits_{i=0}^{p}T(r,a_{i})\\ &&+\sum T(r,a_{(i)})+\sum T(r,b_{(j)})+\sum\limits_{j=0}^{q}T(r,b_{j})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w}). \end{eqnarray*}$

We note that $p>q+2\lambda$. Thus

$\begin{eqnarray} T(r,w)&<&\frac{2\overline{\mu}}{p-(q+2\lambda)}\overline{N}(r,w)\nonumber\\ &&+\frac{\overline{\mu}-l}{p-(q+2\lambda)}N_{b}(r,w)+\frac{\sigma}{p-(q+2\lambda)} N_{x}(r,w)+F_{1}(r)+D(r), \end{eqnarray}$

(3.3)

where

$\begin{eqnarray*} F_{1}(r)&=&\frac{1}{p-(q+2\lambda)}\{\sum\limits_{i=0}^{p}T(r,a_{i})+\sum T(r,a_{(i)})+\sum T(r,b_{(j)})+\sum\limits_{j=0}^{q}T(r,b_{j})\}, \\ D(r)&=&\frac{1}{p-(q+2\lambda)}\{\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w-a})+\sum\limits_{\alpha=1}^{n}m(r,\frac{w^{(\alpha)}}{w})\}. \end{eqnarray*}$

We apply the generalized Lemma of Logarithmic derivate to $D(r)$. Then inequality (3.3) becomes

$\begin{eqnarray} T(r,w)<a\log T(t,w)+b \log\frac{t}{t-r}+H(r), \end{eqnarray}$

(3.4)

where $a$ and $b$ are constants, and

$ H(r)=\frac{2\overline{\mu}}{p-(q+2\lambda)}\overline{N}(r,w)+\frac{\overline{\mu}-l}{p-(q+2\lambda)}N_{b}(r,w)+\frac{\sigma}{p-(q+2\lambda)} N_{x}(r,w)+F_{1}(r).\\ $

Applying Lemma 4 to (3.4) and we get

$ T(r,w)<(a+b)\log\frac{t}{t-r}+2H(t). $

Set $t=\xi r,\xi>1$. Then $T(r,w)\leq KF(\xi r)$.

Combining Case 1 and Case 2 we complete the proof of Theorem 1.