We use the standard notations of the Nevanlinna theory of meromorphic functions(see e.g.[1]). In addition we denote the order of $f(z)$ by $\rho(f)$.

We set up some notations on difference. Let $c$ be a fixed, non-zero complex number, $\Delta_{c}f(z)=f(z+c)-f(z)$, and $\Delta^{n}_{c}f(z)=\Delta_{c}(\Delta^{n-1}_{c}f(z))=\Delta^{n-1}_{c}f(z+c)-\Delta^{n-1}_{c}f(z)$ for each integer $n\geq2$. In order to simplify our notations, we shall use the same notation $\Delta$ for both a general $c$ and when $c=1$. The context will make clear which quantity is under discussion. Equations written with the above difference operators $\Delta^{n}_{c}f(z)$ are difference discussions. Let $E$ be a subset on the positive real axis. We define the logarithmic measure of $E$ to be

$ \log (E)=\int_{E\bigcap(1+\infty)}\frac{dr}{r}. $

A set $E\subseteq (1, +\infty)$ is said to have finite logarithmic measure if $\log (E) < \infty.$

Recently, Laine, Rieppo [5], Heittokangas, Korhonen [3], Halburd and Korhonen [4], Chiang and Feng [2] etc. investigated the existence or growth of solutions of complex difference equations, Gao [9-12] investigated the existence or growth of solutions of systems of complex difference equations, they obtain some main results.

In 2005, I. Laine, J. Rieppo[5] considered the following difference equation

$ \sum\limits_{j = 1}^n {{\alpha _j}} (z)f(z + {c_j}) = \frac{{P(z,f(z))}}{{Q(z,f(z))}}, $

where the coefficients $\alpha_{j}(z)$ are non-vanishing small functions relative to $f$ and where $P, Q$ are relatively prime polynomials in $f$ over the field of small functions relative to $f$. He obtained

Theorem A[5]   Suppose that $c_{1}, c_{2}, \cdots, c_{n}$ are distinct, non-zero complex numbers and that $f$ is a transcendental meromorphic solution of the above equation, $q=\deg^{Q}_{f}>0, n=\max \{p, q\}=\max \{\deg^{P}_{f}, \deg^{Q}_{f}\}$ and that, without restricting generality, $Q$ is a monic polynomial. If there exists $\alpha \in [0, n)$ such that for all $r$ sufficiently large,

$ \bar N(r,\sum\limits_{j = 1}^n {{\alpha _j}} (z)f(z + {c_j})) \le \alpha \bar N(r + C,f(z)) + S(r,f(z)), $

where $C=\max\{|c_{1}|, |c_{2}|, \cdots, |c_{n}|\}, $ then either the order $\rho(f)=\infty$, or

$ Q(z, f(z))\equiv (f(z)+h(z))^{q}, $

where $h(z)$ is a small meromorphic function.

The natural question arises whether or not the assertion of Theorem A remains valid, if we replace the difference equations with systems of complex difference equations. In this paper, our aim is to consider the problem of the growth of solution of systems of complex difference equations of the form

$ \begin{equation} \begin{cases} \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j})=\dfrac{P_{1}(z, f_{2})}{Q_{1}(z, f_{2})}, \\ \sum\limits^{n_{2}}_{j=1}\beta_{j}(z)f_{2}(z+c_{j})=\dfrac{P_{2}(z, f_{1})}{Q_{2}(z, f_{1})}, \end{cases} \end{equation} $

(1.1)

where the coefficients $\alpha_{j}(z), \beta_{j}(z)$ are non-vanishing small functions relative to $f_{1}, f_{2}$ and $P_{1}, Q_{1}$ are relatively prime polynomials in $f_{2}$, $P_{2}, Q_{2}$ are relatively prime polynomials in $f_{1}$, where $q_{k}=\deg^{Q_{k}}_{f}>0, p_{k}=\deg^{P_{k}}_{f}(k=1, 2)$. We assume that $ n_{1}=\max \{p_{1}, q_{1}\}, n_{2}=\max \{p_{2}, q_{2}\}, Q_{1}, Q_{2}$ are monic polynomials.

Let $C=\max \{|c_{1}|, |c_{2}|, \cdots, |c_{n}|\}.$ We obtain the following result

Theorem 1  Let $(f_{1}(z), f_{2}(z))$ be transcendental meromorphic solution of (1.1). If there exist $\alpha, \beta \in[0, n), n=\min \{n_{1}, n_{2}\}$ such that for all $r$ sufficiently large,

$ \begin{equation*} \begin{cases} \overline{N}(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))\leq \alpha \overline{N}(r+C, f_{1}(z))+S(r, f_{1}), \\ \overline{N}(r, \sum\limits^{n_{2}}_{j=1}\beta_{j}(z)f_{2}(z+c_{j}))\leq \beta \overline{N}(r+C, f_{2}(z))+S(r, f_{2}), \end{cases} \end{equation*} $

(1.2)

then either the order $\rho (f_{1})=\infty, \rho (f_{2})=\infty$ at least one of them will be true, or

$ Q_{1}(z, f_{2})\equiv (f_{2}+h_{1}(z))^{q_{1}}, Q_{2}(z, f_{1})\equiv (f_{1}+h_{2}(z))^{q_{2}} $

at least one of them will be true where $h_{1}(z), h_{2}(z)$ are small meromorphic functions.

Lemma 1[6]  Let $f(z)$ be a meromorphic function and let $\phi$ be given by

$ \begin{eqnarray*}&& \phi=f^{n}+a_{n-1}f^{n-1}+\cdots+a_{0}, \\ && T(r, a_{j})=S(r, f), j=0, 1, \cdots, n-1.\end{eqnarray*} $

Then either

$ \phi\equiv (f+\frac{a_{n-1}}{a_{n}})^{n} $

or

$ T(r, f)\leq \overline{N}(r, \frac{1}{\phi})+\overline{N}(r, f)+S(r, f). $

Lemma 2[5]   Let $f(z)$ be a non-constant meromorphic function and let $P(z, f), Q(z, f)$ be two polynomials in $f$ with meromorphic coefficients small relative to $f$. If $P$ and $Q$ have no common factors of positive degree in $f$ over the field of small functions relative to $f$, then

$ \overline{N}(r, \frac{1}{Q(z, f)})\leq \overline{N}(r, \frac{P(z, f)}{Q(z, f)})+S(r, f). $

Lemma 3[7]   Let $T : [0, +\infty)\rightarrow [0, +\infty)$ be a non-decreasing continuous function, $\delta\in(0, 1)$ and $s\in (0, +\infty].$ If $T$ is of finite order, i.e.,

$ \lim\limits_{r\rightarrow\infty}\frac{\log T(r)}{\log r} < \infty, $

then

$ T(r+s)=T(r)+o(\frac{T(r)}{r^{\delta}}), $

where $r$ runs to infinity outside of a set of finite logarithmic measure.

Lemma 4[8]   Let $f(z)$ be a non-constant meromorphic function. Then for all irreducible rational functions in $f(z)$,

$ R(z, f(z))=\dfrac{\sum\limits^{p}_{j=0}a_{j}(z)f(z)^{j}}{\sum\limits^{q}_{j=0}b_{j}(z)f(z)^{j}} $

with meromorphic coefficients $a_{j}(z), b_{j}(z)$, the characteristic function of $R(z, f(z))$ satisfies

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

When the second alternative of assertion is not hand. Then by Lemma 1 and Lemma 2, we obtain

$ T(r, f_{2})\leq \overline{N} (r, \frac{1}{Q_{1}})+\overline{N}(r, f_{2})+S(r, f_{2})\leq \overline{N}(r, \frac{P_{1}}{Q_{1}}) +\overline{N}(r, f_{2})+S(r, f_{2}). $

Then by (1.1) and (1.2), we have

$ \begin{equation*} \begin{split} T(r, f_{2})-\overline{N}(r, f_{2})&\leq \overline{N}(r, \frac{P_{1}}{Q_{1}})+S(r, f_{2})\\ & =\overline{N}(r, \sum^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{2})\\ &\leq\alpha\overline{N}(r+C, f_{1}(z))+S(r, f_{1})+S(r, f_{2}). \end{split} \end{equation*} $

Namely,

$ T(r, f_{2})-\overline{N}(r, f_{2})\leq\alpha\overline{N}(r+C, f_{1}(z))+S(r, f_{1})+S(r, f_{2}), $

where $\alpha \in[0, n), n=\min \{n_{1}, n_{2}\}$. Similarly, we obtain

$ T(r, f_{1})-\overline{N}(r, f_{1})\leq \beta\overline{N}(r+C, f_{2}(z))+S(r, f_{1})+S(r, f_{2}), $

where $\beta \in[0, n), n=\min \{n_{1}, n_{2}\}.$

Assuming, contrary to the assertion, that $\rho(f_{i}) < \infty, i=1, 2.$ Then it implies that

$ S(r, f_{1}(z+c_{j}))=S(r, f_{1}), \ \ \ S(r, f_{1}(z+c_{j}))=S(r, f_{2}). $

Hence

$ T(r, f_{2}(z+c_{j}))-\overline{N}(r, f_{2}(z+c_{j}))\leq \alpha\overline{N}(r+C, f_{1}(z+c_{j}))+S(r, f_{1}(z))+S(r, f_{2}(z)), $

(3.1)

$ T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))\leq \beta\overline{N}(r+C, f_{2}(z+c_{j}))+S(r, f_{1}(z))+S(r, f_{2}(z)). $

(3.2)

By (1.1), (1.2), (3.2) and Lemma 4, we have

$ \begin{equation*} \begin{split} n_{1}T(r, f_{2})&\leq T(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ & \leq \sum\limits^{n_{1}}_{j=1}[T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))]+\overline{N}(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ &\leq \sum\limits^{n_{1}}_{j=1}\beta\overline{N}(r+C, f_{2}(z+c_{j}))+\\ &\alpha \overline{N}(r+C, f_{1})+S(r, f_{1})\\ &\leq n_{1}\beta\overline{N}(r+2C, f_{2}(z))+\alpha \overline{N}(r+C, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{split} \end{equation*} $

Namely,

$ T(r, f_{2})\leq \beta\overline{N}(r+2C, f_{2}(z))+\frac{\alpha}{n_{1}} \overline{N}(r+C, f_{1})+S(r, f_{1})+S(r, f_{2}). $

(3.3)

Similarly, we have

$ T(r, f_{1})\leq \alpha\overline{N}(r+2C, f_{1}(z))+\frac{\beta}{n_{2}} \overline{N}(r+C, f_{2})+S(r, f_{1})+S(r, f_{2}). $

(3.4)

Hence

$ \begin{eqnarray}&& T(r, f_{2})-\overline{N}(r, f_{2})\nonumber\\ &\leq& \beta\overline{N}(r+2C, f_{2}(z))+\frac{\alpha}{n_{1}}\overline{N}(r+C, f_{1})-\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}), \end{eqnarray} $

(3.5)

$ \begin{eqnarray} && T(r, f_{1})-\overline{N}(r, f_{1})\nonumber\\ &\leq& \alpha\overline{N}(r+2C, f_{1}(z))+\frac{\beta}{n_{2}}\overline{N}(r+C, f_{2})-\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}).\end{eqnarray} $

(3.6)

By (3.5) and Lemma 3, we have

$ \begin{eqnarray*} &&T(r, f_{2}(z+c_{j}))-\overline{N}(r, f_{2}(z+c_{j}))\\ &\leq& \beta\overline{N}(r+2C, f_{2}(z+c_{j}))+\frac{\alpha}{n_{1}}\overline{N}(r+C, f_{1}(z+c_{j}))-\overline{N}(r, f_{2}(z+c_{j}))\\ &&+ S(r, f_{1}(z+c_{j}))+S(r, f_{2}(z+c_{j})) \\ &\leq& \beta\overline{N}(r+3C, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r+2C, f_{1})-\overline{N}(r-C, f_{2})+S(r, f_{1})+S(r, f_{2})\\ &\leq& \beta\overline{N}(r+3C, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})-\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}). \end{eqnarray*} $

Namely,

$ T(r, f_{2}(z+c_{j}))-\overline{N}(r, f_{2}(z+c_{j}))\leq \beta\overline{N}(r+3C, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})-\\\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}). $

(3.7)

Similarly, we have

$ T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))\leq \alpha\overline{N}(r+3C, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})-\\\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). $

(3.8)

By (1.1), (1.2), (3.8) and Lemma 4, we have

$ \begin{align*} n_{1}T(r, f_{2})\leq& T(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq& \sum\limits^{n_{1}}_{j=1}[T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))]+\\ &\overline{N}(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq& \sum\limits^{n_{1}}_{j=1}[\alpha\overline{N}(r+3C, f_{1}(z))+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})-\overline{N}(r, f_{1})]+\alpha \overline{N}(r+C, f_{1})\\ &+S(r, f_{1})+S(r, f_{2})\\ \leq& n_{1}\alpha\overline{N}(r+3C, f_{1}(z))+\frac{n_{1}\beta}{n_{2}}\overline{N}(r, f_{2})-n_{1}\overline{N}(r, f_{1})\\ &+\alpha \overline{N}(r+C, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

(3.9)

Then by Lemma 3 and (3.9), we have

$ \begin{align*} T(r, f_{2})-\overline{N}(r, f_{2})\leq & \alpha\overline{N}(r+3C, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})-\overline{N}(r, f_{1})\\ &-\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

Similarly, we have

$ \begin{align*} T(r, f_{1})-\overline{N}(r, f_{1})\leq & \beta\overline{N}(r+3C, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})-\overline{N}(r, f_{2})\\ &-\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

Namely,

$ \begin{equation*} \begin{cases} T(r, f_{2})-\overline{N}(r, f_{2})\leq &\alpha\overline{N}(r+3C, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})-\overline{N}(r, f_{1})\\ &-\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}), \\ T(r, f_{1})-\overline{N}(r, f_{1})\leq &\beta\overline{N}(r+3C, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})-\overline{N}(r, f_{2})\\ &-\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{cases} \end{equation*} $

(3.10)

We now proceed, inductively, to prove

$ \begin{equation*} \begin{cases} T(r, f_{2})-\overline{N}(r, f_{2})\leq &\alpha\overline{N}(r+(2m+1)C, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})-\\&m\overline{N}(r, f_{1})\\ &-m\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}), \\ T(r, f_{1})-\overline{N}(r, f_{1})\leq &\beta\overline{N}(r+(2m+1)C, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2})-\\&m\overline{N}(r, f_{2})\\ &-m\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{cases} \end{equation*} $

Having already proved the case $m=1$ in (3.10), we continue to the inductive step. To this end, by the above inequalities, we obtain

$ \begin{align*} &T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))\\ \leq &\beta\overline{N}(r+(2m+1)C, f_{2}(z+c_{j}))+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1}(z+c_{j}))+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2}(z+c_{j}))\\ &-m\overline{N}(r, f_{2}(z+c_{j}))-m\overline{N}(r, f_{1}(z+c_{j}))+S(r, f_{1}(z+c_{j}))+S(r, f_{2}(z+c_{j}))\\ \leq & \beta\overline{N}(r+(2m+2)C, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r+C, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r+C, f_{2})-m\overline{N}(r-C, f_{2})\\ &-m\overline{N}(r-C, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

(3.11)

Applying (3.11) and Lemma 3, and using (1.1) and (1.2) we conclude that

$ \begin{align*} n_{1}T(r, f_{2})\leq& T(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq& \sum\limits^{n_{1}}_{j=1}[T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))]+\overline{N}(r, \sum^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq& \sum\limits^{n_{1}}_{j=1}[\beta\overline{N}(r+(2m+2)C, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r+C, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r+C, f_{2})\\ &-m\overline{N}(r-C, f_{2})-m\overline{N}(r-C, f_{1})]+\alpha \overline{N}(r+C, f_{1})+S(r, f_{1})\\ \leq& n_{1}\beta\overline{N}(r+(2m+2)C, f_{2})+\frac{n_{1}m\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{n_{1}m\beta}{n_{2}}\overline{N}(r, f_{2})\\ &- mn_{1}\overline{N}(r, f_{2})-mn_{1}\overline{N}(r, f_{1})+\alpha \overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

Therefore, we obtain

$ \begin{equation*} \begin{split} T(r, f_{2})-\overline{N}(r, f_{2})\leq & \beta\overline{N}(r+(2m+2)C, f_{2})+\frac{(m+1)\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2})\\ &-(m+1)\overline{N}(r, f_{2})-m\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{split} \end{equation*} $

Similarly, we have

$ \begin{equation*} \begin{split} T(r, f_{1})-\overline{N}(r, f_{1})\leq & \alpha\overline{N}(r+(2m+2)C, f_{1})+\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &-(m+1)\overline{N}(r, f_{1})-m\overline{N}(r, f_{2})+S(r, f_{2})+S(r, f_{1}). \end{split} \end{equation*} $

The above inequality applies to the functions $f(z+c_{j})$ as well instead of $f(z)$. Therefore, by Lemma 3, we obtain

$ \begin{align*} &T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))\\ \leq & \alpha\overline{N}(r+(2m+2)C, f_{1}(z+c_{j}))+\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2}(z+c_{j}))+\\ &\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1}(z+c_{j}))\\ &-(m+1)\overline{N}(r, f_{1}(z+c_{j}))-m\overline{N}(r, f_{2}(z+c_{j}))+S(r, f_{2})+S(r, f_{1}) \\ \leq &\alpha\overline{N}(r+(2m+3)C, f_{1})+\frac{(m+1)\beta}{n_{2}}\overline{N}(r+C, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r+C, f_{1})\\ &-(m+1)\overline{N}(r-C, f_{1})-m\overline{N}(r-C, f_{2})+S(r, f_{2})+S(r, f_{1})\\ \leq &\alpha\overline{N}(r+(2m+3)C, f_{1})+\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &-(m+1)\overline{N}(r, f_{1})-m\overline{N}(r, f_{2})+S(r, f_{2})+S(r, f_{1}). \end{align*} $

(3.12)

Applying (3.12) and Lemma 3, and using (1.1) and (1.2) we conclude that

$ \begin{align*} n_{1}T(r, f_{2})\leq& T(r, \sum\limits^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq & \sum\limits^{n_{1}}_{j=1}[T(r, f_{1}(z+c_{j}))-\overline{N}(r, f_{1}(z+c_{j}))]+\overline{N}(r, \sum^{n_{1}}_{j=1}\alpha_{j}(z)f_{1}(z+c_{j}))+S(r, f_{1})\\ \leq &\sum\limits^{n_{1}}_{j=1}[\alpha\overline{N}(r+(2m+3)C, f_{1})+\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &-(m+1)\overline{N}(r, f_{1})-m\overline{N}(r, f_{2})]+\alpha \overline{N}(r+C, f_{1})+S(r, f_{1})+S(r, f_{2})\\ \leq& n_{1}\alpha\overline{N}(r+(2m+3)C, f_{1})+\frac{n_{1}(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{n_{1}m\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &- n_{1}(m+1)\overline{N}(r, f_{1})-n_{1}m\overline{N}(r, f_{2})+\alpha\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

This implies that

$ \begin{align*} T(r, f_{2})-\overline{N}(r, f_{2})\leq & \alpha\overline{N}(r+[2(m+1)+1]C, f_{2})+\\ &\frac{(m+1)\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})\\ &-(m+1)\overline{N}(r, f_{2})-(m+1)\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

Similarly, we obtain

$ \begin{align*} T(r, f_{1})-\overline{N}(r, f_{1})\leq & \beta\overline{N}(r+[2(m+1)+1]C, f_{1})+\\ &\frac{(m+1)\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{(m+1)\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &-(m+1)\overline{N}(r, f_{1})-(m+1)\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}). \end{align*} $

We complete the induction. Moreover, we immediately see from (3.10) that

$ \begin{equation*} \begin{cases} T(r, f_{2})-\overline{N}(r, f_{2})\leq &\alpha\overline{N}(r+(2m+1)C, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})\\ &-m\overline{N}(r, f_{1})-m\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}), \\ T(r, f_{1})-\overline{N}(r, f_{1})\leq &\beta\overline{N}(r+(2m+1)C, f_{2})+\frac{m\alpha}{n_{1}}\overline{N}(r, f_{1})+\frac{m\beta}{n_{2}}\overline{N}(r, f_{2})\\ &-m\overline{N}(r, f_{2})-m\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{cases} \end{equation*} $

(3.13)

By (3.13) and Lemma 3, we have

$ \begin{equation*} \begin{cases} \overline{N}(r, f_{1})+\overline{N}(r, f_{2})\leq &(\frac{\alpha}{m}+\frac{\alpha}{n_{1}})\overline{N}(r, f_{1})+\frac{\beta}{n_{2}}\overline{N}(r, f_{2})+S(r, f_{1})+S(r, f_{2}), \\ \overline{N}(r, f_{2})+\overline{N}(r, f_{1})\leq &(\frac{\beta}{m}+\frac{\beta}{n_{2}})\overline{N}(r, f_{2})+\frac{\alpha}{n_{1}}\overline{N}(r, f_{1})+S(r, f_{1})+S(r, f_{2}). \end{cases} \end{equation*} $

(3.14)

For sufficiently large $m$ we see that

$ \begin{eqnarray*}&& \frac{\alpha}{m}+\frac{\alpha}{n_{1}}=\alpha(\frac{1}{m}+\frac{1}{n_{1}}) < 1, \frac{\beta}{n_{2}} < 1, \\ && \frac{\beta}{m}+\frac{\beta}{n_{2}}=\beta(\frac{1}{m}+\frac{1}{n_{2}}) < 1, \frac{\alpha}{n_{1}} < 1.\end{eqnarray*} $

From (3.14), we have

$ \begin{equation*} \begin{cases} \left[1-(\frac{\alpha}{m}+\frac{\alpha}{n_{1}})\right]\overline{N}(r, f_{1})+ \left[1-\frac{\beta}{n_{2}}\right]\overline{N}(r, f_{2})\leq &S(r, f_{1})+S(r, f_{2}), \\ \left[1-(\frac{\beta}{m}+\frac{\beta}{n_{2}})\right]\overline{N}(r, f_{2})+ \left[1-\frac{\alpha}{n_{1}}\right]\overline{N}(r, f_{1})\leq &S(r, f_{1})+S(r, f_{2}). \end{cases} \end{equation*} $

Thus, for sufficiently large $m$, we have

$ \overline{N}(r, f_{1})=S(r, f_{1})+S(r, f_{2}), \overline{N}(r, f_{2})=S(r, f_{1})+S(r, f_{2}). $

(3.15)

By (3.3), (3.4), (3.15) and Lemma 3, we have

$ T(r, f_{1})=S(r, f_{1})+S(r, f_{2}), T(r, f_{2})=S(r, f_{1})+S(r, f_{2}). $

Namely,

$ [1+o(1)]T(r, f_{1})=S(r, f_{2}), [1+o(1)]T(r, f_{2})=S(r, f_{1}). $

Hence, we have

$ [1+o(1)][1+o(1)]T(r, f_{1})T(r.f_{2})=S(r, f_{1})S(r, f_{2})=o(T(r, f_{1}))o(T(r, f_{2})). $

Namely,

$ [1+o(1)]T(r, f_{1})T(r, f_{2})=o(T(r, f_{1}))o(T(r, f_{2})). $

Then, we have $1=0$, which is a contradiction. Thus, the order $\rho (f_{1})=\infty, \rho (f_{2})=\infty$ at least one of them will be true.