Uniqueness problem of meromorphic functions is one of the important research directions in Nevanlinna theory. In the 1920s, R. Nevanlinna [1] proved the famous five-value theorem that if two non-constant meromorphic functions $ f $ and $ g $ share five distinct values, then $ f(z)\equiv g(z) $. Later on, the research on the uniqueness problem has been widely promoted, such as references [2-4]. Among them, Fujimoto [5] proposed the definition of a finite range set, and got the following result.

A finite subset S of $ \mathbb{C} $ is said to be a unique range set for meromorphic functions if $ f^*(S) = g^*(S) $ implies $ f = g $ for arbitrary nonconstant meromorphic functions $ f $ and $ g $ on $ \mathbb{C} $, where $ f^*(S) $ and $ g^*(S) $ denote the pull-backs of S considered as a divisor, namely, the inverse images of S counted with multiplicities by $ f $ and $ g $ respectively. A nonconstant monic polynomial $ P(w) $ is called a uniqueness polynomial for meromorphic functions if for any nonconstant meromorphic functions $ f $ and $ g $ on $ \mathbb{C} $, the equation $ P(f) = cP(g) $ implies $ f = g $, where c is a nonzero constant that possibly depends on $ f $ and $ g $.

For $ S=\{a_1, \cdots, a_q\} $, we consider the polynomial

$ \begin{equation} P_S(w)=(w-a_1)\cdots(w-a_q). \end{equation} $

(1.1)

Obviously, if $ S $ is a unique range set of meromorphic functions, then $ P_S(w) $ is a uniqueness polynomial of meromorphic functions.

Definition 1.1 [5, Definition1.2] If for any given nonconstant meromorphic function $ g $, there exist only finitely many nonconstant meromorphic functions $ f $ such that $ f^*(S) = g^*(S) $, then a finite subset $ S $ of $ \mathbb{C} $ is called a finite range set for meromorphic functions.

According to the above definition, Fujimoto obtained the following result.

Theorem 1.2 [5, Theorem1.3] Take a finite set $ S=\{a_1, \cdots, a_q\} $ and assume that for the polynomial $ P_S(w) $ defined by (1.1), $ P'_S(w) $ has exactly $ k $ distinct zeros. If $ P_S(w) $ is a uniqueness polynomial for meromorphic functions and $ q>k+2 $, then $ S $ is a finite range set for meromorphic functions. More precisely, for an arbitrarily given nonconstant meromorphic function $ g $, there exist at most $ \frac{2q-2}{q-k-2} $ meromorphic functions $ f $ such that $ f^*(S) = g^*(S) $.

It is an interesting question whether meromorphic functions on other domains also have corresponding results. Recently, Quang [6] weakened the condition $ f^*(S) = g^*(S) $, and obtained the corresponding results for meromorphic functions on annulus $ \mathbb{A}(R_0) $. In the light of Quang [6], we consider the meromorphic functions on the complex disc with finite growth index and obtain some results.

For $ 0<R\le\infty $, we set a complex disc $ \Delta(R)=\{z \in \mathbb{C};|z|< R\} $. According to Ru and Sibony [7], the growth index of meromorphic function $ f $ on $ \Delta(R) $ is defined by

$ c_f=\inf\{c>0;\int^{R}_0\exp(cT(r, f))dr=+\infty.\} $

Obviously, $ R=+\infty $, then $ c_f=0 $. For convenient, if $ \{c>0;\int^{R}_0\exp(cT(r, f))dr=+\infty.\}=\emptyset, $ we will set $ c_f=+\infty $. All the meromorphic functions on $ \Delta(R) $ we discussed in this paper have finite growth index.

Definition 1.3 Let $ S=\{a_1, \cdots, a_q\} $ be a set of distinct values in $ \mathbb{C} $ and let $ \ell_{i}(i=1, \cdots, q) $ be some positive integers (may be $ +\infty $). Two meromorphic functions $ f $ and $ g $ on $ \Delta(R) $ are said to share $ S $ with truncated multiplicity if

$ \sum^q\limits_{i=1}\min\{\ell_{i}, v_{f-a_i}\}=\sum^q\limits_{i=1}\min\{\ell_{i}, v_{g-a_i}\}. $

If for an arbitrarily given meromorphic function $ g $ on $ \Delta(R) $, there exist finitely many meromorphic functions $ f $ on $ \Delta(R) $ only such that $ f $ and $ g $ share $ S $ with truncated multiplicity, then the set $ S $ is said to be a finite range set with truncated multiplicity for meromorphic functions on $ \Delta(R) $.

We note that $ \ell_{i}=+\infty(i=1, \cdots, q) $, Definition 1.3 implies Definition 1.1. Using Quang's method in [6], we obtain the following second main theorem.

Theorem 1.4 Let $ f $ be a non-constant meromorphic function on $ \Delta(R) $ and let $ a_1, \cdots, a_q $ be $ q(q\ge5) $ distinct meromorphic functions (may be equal to $ \infty $). Let $ \gamma(r) $ be a non-negative measurable function defined on $ (0, R) $ with $ \int^{R}_0\gamma(r)dr=+\infty $. Then, for every $ \varepsilon>0 $,

$ \begin{equation*} \label{eq3} \begin{split} \parallel_E\frac{2q}{5}T(r, f)\le&\sum^q_{i=1}\overline N(r, v^0_{f-a_i})+35\sum^q_{i=1}T(r, a_i)+17((1+\varepsilon)\log\gamma(r)+\varepsilon\log r)+o(T(r, f)). \end{split} \end{equation*} $

Here and later on, we use the notation $ \parallel_EP $ to say that the assertion $ P $ holds for all $ r\in(0;R) $ outside a subset $ E $ of $ (0;R) $ with $ \int^{R}_0\gamma(r)dr<+\infty $.

As its application, we obtain the results as follows.

Theorem 1.5 Take a finite set $ S=\{a_1, \cdots, a_q\} $ of $ q $ distinct values in $ \mathbb{C} $ and assume that for the polynomial $ P_S(w) $ defined by (1.1), $ P'_S(w) $ has exactly $ k $ distinct zeros. Let $ \ell_{i}(1\le i\le q) $ be some positive integers (may be $ \ell_{i}=+\infty $). If $ P_S(w) $ is a uniqueness polynomial for meromorphic functions on $ \Delta(R) $, then any given nonconstant meromorphic function $ g $ on $ \Delta(R) $ with finite growth index, if $ \sum_{f^*(S) = g^*(S)}c_{f}<\frac{1}{372}\{2(q-1)-5(k+1+\sum^q_{i=1}\frac{35}{\ell_i})\} $, then the number of elements in set $ A=\{f|f^*(S) = g^*(S)\} $ does not exceed four, namely, $ S $ is a finite range with truncated multiplicity.

Remark 1.6 If $ R=+\infty $, we have $ c_{f_i}=0(1\le i\le q) $, let $ \ell_{i}=\ell(1\le i\le q) $, then the condition of the above theorem becomes $ q>\frac{(5k+7)\ell}{2\ell-175} $. Furthermore, $ \ell=+\infty $, we get $ q>\frac{(5k+7)}{2} $, which is much greater than the number $ k + 2 $ in Theorem 1.2. Then, the most difficult part comes from the fact that "how to get a better number in our situation".

In this section, we introduce the preliminaries of Nevanlinna theory for meromorphic functions on $ \Delta(R) $, detailed reference [8-10].

Let $ v $ be a divior on $ \Delta(R) $, which is regarded as a function on $ \Delta(R) $ with values in $ \mathbb{Z} $ such that Supp$ (v)=\{z;v(z)\ne0\} $ is a discrete subset of $ \Delta(R) $. We define the counting function of $ v $ to be:

$ n(t)=\sum\limits_{|z|\le t}v(z)(0\le t\le R_0) , N(r, v)=\int^r_0\frac{n(t)-n(0)}{t}dt. $

Let $ f $ be a non-constant meromorphic function on $ \Delta(R) $, we define

(1) $ v^0_f $ (resp.$ v^\infty_f $) is the divisor of zeros (resp. divisor of poles) of $ f $.

(2) $ v^0_{f, \ge k}=\max\{k, v^0_f\}. $

For a meromorphic function $ a $, we then define $ N(r, a, f)=N(r, v^0_{f-a}) $ and the truncated counting function is defined by

$ \overline N(r, a, f)=\overline N(r, v^0_{f-a})=N(r, \min\{1, v^0_{f-a}\}). $

The proximity function of $ f $ is defined by

$ m(r, f)=\int^{2\pi}_0\log^+|f(re^{i\theta})|\frac{d\theta}{2\pi}. $

For a meromorphic function $ a $, we define $ m(r, a, f)=m(r, \frac{1}{f-a}) $, (we regard $ \frac{1}{f-\infty} $ as $ f $), where $ \log^+x=\max\{0, \log x\} $.

The characteristic function of $ f $ is defined by $ T(r, f)=m(r, f)+N(r, v^\infty_{f}). $ The first main theorem states that

$ m(r, \frac{1}{f-a})+N(r, v^0_{f-a})=T(r, f)+O(1) $

for arbitrarily meromorphic function $ a $.

Lemma 2.1 [Lemma on logarithmic derivative, [7, Theorem 5.1]] Let $ f(z) $ be a meromorphic function on $ \Delta(R)(0<R\le\infty) $, and let $ \gamma(r) $ be a non-negative measurable function defined on $ (0, R) $ with $ \int^{R}_0\gamma(r)dr=+\infty $. Then, for $ \varepsilon>0 $, we have

$ \parallel_Em(r, \frac{f'}{f})\le(1+\varepsilon)\log\gamma(r)+\varepsilon\log r+o(\log T(r, f)). $

Theorem 2.2 [Corollary 1.8 in [7], second main theorem, Theorem2.4 in [11]] Let $ f $ be a non-constant meromorphic function on $ \Delta(R)(0<R\le\infty) $, let $ \gamma(r) $ be a non-negative measurable function defined on $ (0, R) $ with $ \int^{R}_0\gamma(r)dr=+\infty $ and let $ a_1, \cdots, a_q $ be $ q $ distinct values in $ \mathbb{C}\cup\{\infty\} $. Then, for $ \varepsilon>0 $, we have

$ \parallel_E(q-2)T(r, f)\le\sum^q\limits_{i=1}\overline N(r, v^0_{f-a_i})+(1+\varepsilon)\log\gamma(r)+\varepsilon\log r+o(\log T(r, f)). $

In order to prove our main theorem, we need the following lemmas.

Lemma 3.1 [6, Lemma 3.1] Let $ f_1 $ and $ f_2 $ be two meromorphic functions on $ \Delta(R) $, and let $ a_1, a_2 $, and $ a_3 $ be three distinct meromorphic functions on $ \Delta(R) $ (being not equal to $ \infty $) such that $ f_2=\frac{(f_1-a_1)}{(f_1-a_2)} $. Then we have

(a) $ T(r, f_2)\ge T(r, f_1)-\sum^2_{i=1}T(r, a_i)+O(1), $

(b) $ \overline N(r, v^0_{f_2})+\overline N(r, v^0_{f_2-1})+\overline N(r, v^\infty_{f_2}) \le\overline N(r, v^\infty_{f_1})+\sum^2_{i=1}(\overline N(r, v^0_{f_1-a_i})+2T(r, a_i))+O(1). $

And that, if we set $ b=\frac{a_3-a_1}{a_3-a_2} $, then

(c) $ \overline N(r, v^0_{f_2-b})\le\overline N(r, v^0_{f_1-a_3})+T(r, a_1)+2T(r, a_2)+T(r, a_3)+O(1). $

Lemma 3.2 [6, Lemma 3.2] Let $ f_1 $ and $ f_2 $ be two meromorphic functions on $ \Delta(R) $, and let $ a_1 $, $ a_2 $, $ a_3 $ and $ a_4 $ be four distinct meromorphic functions on $ \Delta(R) $ such that

$ f_2=\frac{f_1-a_1}{f_1-a_2}\cdot\frac{a_3-a_2}{a_3-a_1}. $

Then we have

(a) $ T(r, f_2)\ge T(r, f_1)-\sum^3_{i=1}T(r, a_i)+O(1), $

(b) $ \overline N(r, v^0_{f_2})+\overline N(r, v^0_{f_2-1})+\overline N(r, v^\infty_{f_2})\le\sum^3_{i=1}(\overline N(r, v^0_{f_1-a_i})+3T(r, a_i))+O(1). $

If we set $ b=\frac{a_4-a_1}{a_4-a_2}\cdot\frac{a_3-a_2}{a_3-a_1} $, then

(c) $ \overline N(r, v^0_{f_2-b})\le\overline N(r, v^0_{f_1-a_4})+2T(r, a_1)+3T(r, a_2)+2T(r, a_3)+T(r, a_4). $

In fact, Quang proved the above two lemmas for meromorphic function on $ \mathbb{A}(R_0) $. But their proof process still holds for meromorphic functions on $ \Delta(R) $. To prove Theorem 1.4, we just only prove the following lemma.

Lemma 3.3 Let $ g $ be a nonconstant meromorphic function on $ \Delta(R) $. Let $ a_1, a_2, a_3, a_4 $ and $ a_5 $ be five distinct meromorphic functions on $ \Delta(R) $ (may be equal to $ \infty $). We have

$ \parallel_E2T(r, g)\le\sum^5\limits_{i=1}\overline N(r, v^0_{g-a_i})+35\sum^5\limits_{i=1}T(r, a_i)+17S(r)+o(T(r, g)), $

where $ S(r)=(1+\varepsilon)\log\gamma(r)+\varepsilon\log r. $

Proof The proof of the lemma 3.3 is divided into two parts.

Part 1. We first consider the case where $ a_i\not\equiv\infty $ for all $ i=1, \cdots, 5. $ Put

$ \begin{equation*} \begin{split} f=&\frac{g-a_1}{g-a_2}\cdot\frac{a_3-a_2}{a_3-a_1}, b_1=\frac{a_4-a_1}{a_4-a_2}\cdot\frac{a_3-a_2}{a_3-a_1}, \\ b_2=&\frac{a_5-a_1}{a_5-a_2}\cdot\frac{a_3-a_2}{a_3-a_1}, b_3=0, b_4=1. \end{split} \end{equation*} $

By Lemma 3.2, we have

$ \begin{equation} \begin{split} T(r, g) &\le T(r, f)+\sum^3_{i=1}T(r, a_i)+O(1), \\ T(r, b_1)&\le \sum^4_{i=1}T(r, a_i)+O(1), \\ T(r, b_2)&\le \sum^3_{i=1}T(r, a_i)+T(r, a_5)+O(1), \end{split} \end{equation} $

(3.1)

and

$ \begin{equation} \begin{split} \overline N(r, v^\infty_{f})+\sum^4_{i=1}\overline N(r, v^0_{f-b_i})\le&\sum^5_{i=1}\overline N(r, v^0_{g-a_i})+7T(r, a_1)+9T(r, a_2)\\ &+7T(r, a_3)+T(r, a_4)+T(r, a_5)+O(1). \end{split} \end{equation} $

(3.2)

We need to prove the following proposition.

proposition 3.4

$ \begin{equation*} \begin{split} \parallel_E2T(r, f)\le&\overline N(r, v^\infty_f)+\sum^4_{i=1}\overline N(r, v^0_{f-b_i})+\sum^2_{i=1}18T(r, b_i)+17S(r)+o(T(r, f)). \end{split} \end{equation*} $

Actually, if $ b_1 $ or $ b_2 $ is constant, then the proposition directly follows from Theorem 2.2. Therefore, we may assume that both $ b_1 $ and $ b_2 $ are not constant. We define

$ \begin{equation} \begin{split} F=\left |\begin{array}{cccc} ff' & f'& f^2-f \\ b_1b'_1 &b'_1 & b^2_1-b_1 \\ b_2b'_2 & b'_2 &b^2_2-b_2 \\ \end{array}\right|. \end{split} \end{equation} $

(3.3)

Consider the following two cases.

Case 1: $ F(z)\equiv0 $. For (3.3), by elementary transformation of determinants, we have

$ F=f(f-1)b_1(b_1-1)b_2(b_2-1)\left |\begin{array}{cccc} \frac{f'}{f-1}-\frac{b'_2}{b_2-1} & \frac{f'}{f}-\frac{b'_2}{b_2}& 0 \\ \frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1} &\frac{b'_1}{b_1}-\frac{b'_2}{b_2} & 0 \\ \frac{b'_2}{b_2-1} & \frac{b'_2}{b_2} &1 \\ \end{array}\right|. $

Therefore

$ \begin{equation} \begin{split} (\frac{b'_1}{b_1}-\frac{b'_2}{b_2})(\frac{f'}{f-1}-\frac{b'_2}{b_2-1})\equiv(\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1})(\frac{f'}{f}-\frac{b'_2}{b_2}) \end{split}. \end{equation} $

(3.4)

We discuss the following four subcases.

Subcase 1: $ \frac{b'_1}{b_1}\equiv\frac{b'_2}{b_2} $. Then we have $ \frac{b'_1}{b_1-1}\equiv\frac{b'_2}{b_2-1} $ or $ \frac{f'}{f}\equiv\frac{b'_2}{b_2} $, if $ \frac{b'_1}{b_1-1}\equiv\frac{b'_2}{b_2-1} $, then $ b_1 $ and $ b_2 $ are constant, and get a contradiction. If $ \frac{f'}{f}\equiv\frac{b'_2}{b_2} $, then there is $ f=cb_2 $ with a constant c. This implies that $ T(r, f)=T(r, b_2) $, and the proposition is proved in this subcase.

Subcase 2: $ \frac{b'_1}{b_1-1}\equiv\frac{b'_2}{b_2-1} $. Using the same arguments as in subcase 1, we obtain the inequality of the proposition in this subcase.

Subcase 3: $ \frac{b'_1}{b_1}\not\equiv\frac{b'_2}{b_2}, \frac{b'_1}{b_1-1}\not\equiv\frac{b'_2}{b_2-1}, \frac{b'_1}{b_1}-\frac{b'_2}{b_2}\equiv\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1} $. The identity (3.4) implies that

$ \frac{f'}{f-1}-\frac{f'}{f}\equiv\frac{b'_2}{b_2-1}-\frac{b'_2}{b_2}. $

Then we get

$ \frac{f-1}{f}\equiv c\cdot\frac{b_2-1}{b_2}, $

where c is a constant. Therefore $ f=\frac{b_2}{(c-1)b_2-c}, $ and $ T(r, f)=T(r, b_2). $ Again, we obtain the inequality of the proposition in this subcase.

Subcase 4: $ \frac{b'_1}{b_1}\not\equiv\frac{b'_2}{b_2}, \frac{b'_1}{b_1-1}\not\equiv\frac{b'_2}{b_2-1}, \frac{b'_1}{b_1}-\frac{b'_2}{b_2}\not\equiv\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1} $. The identity (3.4) may be rewritten as

$ \begin{equation} \begin{split} (\frac{b'_1}{b_1}-\frac{b'_2}{b_2})\frac{f'}{f-1}-(\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1})\frac{f'}{f}\equiv\frac{b'_1b'_2}{b_1b_2-1}-\frac{b'_2b'_1}{b_2b_1-1}. \end{split} \end{equation} $

(3.5)

We see that each zero of $ f $ must be a zero or a 1-point or a pole of $ b_j(j=1, 2) $ or a zero of $ \frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1}. $ Therefore,

$ \begin{equation} \begin{split} \min\{1, v^0_{f}\}\le\sum\limits_{i=1, 2}\sum\limits_{a=0, 1, \infty}\min\{1, v^0_{b_i-a}\}+\min\{1, v^0_{\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1}}\}. \end{split} \end{equation} $

(3.6)

Similarly, we have

$ \begin{equation} \begin{split} \min\{1, v^0_{f-1}\}\le\sum\limits_{i=1, 2}\sum\limits_{a=0, 1, \infty}\min\{1, v^0_{b_i-a}\}+\min\{1, v^0_{\frac{b'_1}{b_1}-\frac{b'_2}{b_2}}\}. \end{split} \end{equation} $

(3.7)

From (3.5), we also see that each pole of $ f $ must be a zero or a 1-point or a pole of $ b_j(j=1, 2) $ or a zero of $ (\frac{b'_1}{b_1}-\frac{b'_2}{b_2})-(\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1}). $ By the same arguments, we have

$ \begin{equation} \begin{split} \min\{1, v^\infty_{f}\}\le\sum\limits_{i=1, 2}\sum\limits_{a=0, 1, \infty}\min\{1, v^0_{b_i-a}\}+\min\{1, v^0_{(\frac{b'_1}{b_1}-\frac{b'_2}{b_2})-(\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1})}\}. \end{split} \end{equation} $

(3.8)

Combining (3.6)-(3.8), we have

$ \begin{equation} \begin{split} \sum\limits_{a=0, 1, \infty}\min\{1, v^0_{f-a}\}\le&\sum\limits_{i=1, 2}\sum\limits_{a=0, 1, \infty}\min\{1, v^0_{b_i-a}\}+\min\{1, v^0_{\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1}}\}\\ &+\min\{1, v^0_{\frac{b'_1}{b_1}-\frac{b'_2}{b_2}}\}+\min\{1, v^0_{(\frac{b'_1}{b_1}-\frac{b'_2}{b_2})-(\frac{b'_1}{b_1-1}-\frac{b'_2}{b_2-1})}\}. \end{split} \end{equation} $

(3.9)

By Theorem 2.2, we get

$ \begin{equation*} \begin{split} \parallel_ET(r, f)\le&\overline{N}(r, v^0_{f})+\overline{N}(r, v^0_{f-1})+\overline{N}(r, v^\infty_{f})+S(r)\\ \le&2\sum\limits_{i=1, 2}(T(r, \frac{b'_i}{b_i})+T(r, \frac{b'_i}{b_i-1}))+S(r)\\ \le&2\sum\limits_{i=1, 2}(N(r, v^\infty_{\frac{b'_i}{b_i}})+N(r, v^\infty_{\frac{b'_i}{b_i-1}})) +2\sum\limits_{i=1, 2}(m(r, \frac{b'_i}{b_i})+m(r, \frac{b'_i}{b_i-1}))+S(r)\\ \le&2\sum\limits_{i=1, 2}(N(r, v^0_{b_i})+2N(r, v^\infty_{b_i})+N(r, v^0_{b_i-1}))+9S(r)\\ \le&8T(r, b_1)+8T(r, b_2)+9S(r). \end{split} \end{equation*} $

Then we have the desired inequality of the proposition in this subcase.

Case 2: $ F(z)\not\equiv0. $ we set

$ \begin{equation*} \begin{split} \delta(z)&=min\{1, \mid b_1(z)\mid, \mid b_2(z)\mid, \mid b_1(z)-1\mid, \mid b_2(z)-1\mid, \mid b_1(z)-b_2(z)\mid\}, \\ \theta_j(r)&=\{\theta:\mid f(re^{i\theta})-b_j(re^{i\theta})\mid\le\delta(re^{i\theta})\}(j=1, 2), \\ \theta_3(r)&=\{\theta:\mid f(re^{i\theta})\mid\le\delta(re^{i\theta})\}, \\ \theta_4(r)&=\{\theta:\mid f(re^{i\theta})-1\mid\le\delta(re^{i\theta})\}. \end{split} \end{equation*} $

It is clear that the sets $ \theta_i(r)\cap\theta_j(r)(i\not=j, i, j=1, 2, 3, 4) $ have at most many finite points. Therefore we easily get that

$ \begin{equation} \begin{split} \frac{1}{2\pi}\int_0^{2\pi}\log\frac{1}{\delta(re^{i\theta})}d\theta\le& m(r, \frac{1}{b_1})+m(r, \frac{1}{b_2})+m(r, \frac{1}{b_1-1})\\ &+m(r, \frac{1}{b_2-1})+m(r, \frac{1}{b_1-b_2})\\ \le& T(r)+O(1), \end{split} \end{equation} $

(3.10)

where $ T(r)=3T(r, b_1)+3T(r, b_2) $. We also get that

$ \begin{equation*} \begin{split} ff'&=(f-b_1)(f'-b'_1)+b'_1(f-b_1)+b_1(f'-b'_1)+b_1b'_1, \\ f'&=(f'-b'_1)+b'_1, \\ f^2-f&=(f-b_1)^2+(2b_1-1)(f-b_1)+b^2_1-b_1. \end{split} \end{equation*} $

Substituting these functions (on the right-hand side) into (3.3), we have

$ \begin{equation} \begin{split} F=\left |\begin{array}{cccc} \varphi & f'-b'_1 & \psi \\ b_1b'_1 &b'_1 & b^2_1-b_1 \\ b_2b'_2 & b'_2 &b^2_2-b_2 \\ \end{array}\right|, \end{split} \end{equation} $

(3.11)

where

$ \begin{equation*} \begin{split} \varphi&=(f-b_1)(f'-b'_1)+b'_1(f-b_1)+b_1(f'-b'_1), \\ \psi&=(f-b_1)^2+(2b_1-1)(f-b_1). \end{split} \end{equation*} $

From (3.11), we see that each zero with multiplicity $ p(p>1) $ of $ (f-b_1) $ is neither a pole of $ b_1 $ nor a pole of $ b_2 $, which must be a zero of $ F $ with multiplicity at least $ p-1 $. Similarly, each zero with multiplicity $ p(p>1) $ of $ (f-b_2) $ is neither a pole of $ b_1 $ nor a pole of $ b_2 $, which must be a zero of $ F $ with multiplicity at least $ p-1 $. Furthermore, from (3.3), we see that each zero of multiplicity $ p(p>1) $ of $ f $ or $ f-1 $ is neither a pole of $ b_1 $ nor a pole of $ b_2 $, which must be a zero of $ F $ with multiplicity at least $ p-1 $. This implies that

$ \begin{equation} \begin{split} \sum^4_{i=1}(N(r, v^0_{f-b_i})-\overline{N}(r, v^0_{f-b_i}))\le N(r, v^0_{F}). \end{split} \end{equation} $

(3.12)

We now estimate this quantities $ m(r, \frac{1}{f-b_j})(1\le j\le4) $.

We first analyze $ j=3 $, since $ \mid f(re^{i\theta})-b_3\mid=\mid f(re^{i\theta})\mid\le1 $ for each $ \theta\in\theta_3(r) $, we have

$ \begin{equation} \begin{split} F=fb_1b_2\left |\begin{array}{cccc} {f'} & \frac{f'}{f}& f-1 \\ b'_1 &\frac{b'_1}{b_1} & b_1-1 \\ b'_2 & \frac{b'_2}{b_2} &b_2-1 \\ \end{array}\right|=fb_1b_2G \end{split}. \end{equation} $

(3.13)

For $ G $,

$ \begin{equation} \begin{split} \log^+G\le(\log^+f+\log^+\frac{f'}{f})+\sum^2_{i=1}(\log^+b_i+\log^+\frac{b'_i}{b_i}). \end{split} \end{equation} $

(3.14)

Thus

$ \begin{equation} \begin{split} \log^+F\le\log^+f+\sum^2_{i=1}\log^+b_i+\log^+G. \end{split} \end{equation} $

(3.15)

Hence

$ \begin{equation} \begin{split} \parallel_E\frac{1}{2\pi}&\int_{\theta_3(r)}\log^+|\frac{F}{f-b_3}|d\theta\le\frac{1}{2\pi}\int_{\theta_3(r)}\log^+\frac{\mid f'\mid}{\mid f\mid}d\theta\\ &+\sum^2_{i=1}\frac{1}{2\pi}\int_{\theta_3(r)}(2\log^+\mid b_i\mid+\log^+\frac{\mid b'_i\mid}{\mid b'_i\mid})+O(1)\\ \le& m(r, \frac{f'}{f})+m(r, \frac{b'_1}{b_1})+m(r, \frac{b'_2}{b_2})\\ &+\frac{1}{2\pi}\int_{\theta_3(r)}2\log^+|b_1|d\theta+\frac{1}{2\pi}\int_{\theta_3(r)}2\log^+|b_2|d\theta+O(1)\\ \le&\frac{1}{2\pi}\int_{\theta_3(r)}2\log^+|b_1|d\theta+\frac{1}{2\pi}\int_{\theta_3(r)}2\log^+|b_2|d\theta+3S(r)+O(1). \end{split} \end{equation} $

(3.16)

Here the above inequality comes from the fact that

$ \begin{equation} \begin{split} \log^+|\det(x_{ij};1\le i, j\le3)|\le\sum^3_{i=1}\log^+\max\{|x_{ij}|;1\le j\le3\}+O(1) \end{split} \end{equation} $

(3.17)

for every $ 3\ast3 $ matrix of complex numbers $ (x_{ij})_{1\le i, j\le3} $.

Therefore, for $ j=3 $, we get

$ \begin{equation} \begin{split} \parallel_Em(r, \frac{1}{f-b_3})\le&\frac{1}{2\pi}\int_{\theta_3(r)}\log^+|\frac{1}{f-b_3}|d\theta+\frac{1}{2\pi}\int_0^{2\pi}\log\frac{1}{\delta(re^{i\theta})}d\theta\\ \le&\frac{1}{2\pi}\int_{\theta_3(r)}\log^+|\frac{F}{f-b_3}|d\theta\\ &+\frac{1}{2\pi}\int_{\theta_3(r)}\log^+|\frac{1}{F}|d\theta+T(r)+O(1)\\ \le&\frac{1}{2\pi}\int_{\theta_3(r)}(\log^+\frac{1}{\mid F \mid}+2\log^+\mid b_1\mid+2\log^+\mid b_2\mid)d\theta\\ &+3S(r)+T(r)+O(1). \end{split} \end{equation} $

(3.18)

Similarly, we get

$ \begin{equation} \begin{split} \parallel_Em(r, \frac{1}{f-b_4})\le&\frac{1}{2\pi}\int_{\theta_4(r)}(\log^+\frac{1}{\mid F \mid}+2\log^+\mid b_1\mid+2\log^+\mid b_2\mid)d\theta\\ &+3S(r)+T(r)+O(1). \end{split} \end{equation} $

(3.19)

On the other hand, Since $ |f(re^{i\theta})-b_1(re^{i\theta})|\le\delta(re^{i\theta})\le1 $ for every $ \theta\in\theta_1(r) $, by (3.11),

$ \begin{equation*} \begin{split} \log^+&|\frac{F(re^{i\theta})}{f(re^{i\theta})-b_1}|\le\log^+\frac{\mid f'(re^{i\theta})-b'_1(re^{i\theta})\mid}{\mid f(re^{i\theta})-b_1(re^{i\theta})\mid}d\theta+2\log^+\mid b_1(re^{i\theta})\mid\\ &+\log^+\frac{\mid b'_1(re^{i\theta})\mid}{\mid b_1(re^{i\theta})\mid} +\sum^2_{i=1}(2\log^+\mid b_i(re^{i\theta})\mid+\log^+\frac{\mid b'_i(re^{i\theta})\mid}{\mid b_i(re^{i\theta})\mid})+O(1). \end{split} \end{equation*} $

Similarly, for $ j=1 $, we get

$ \begin{equation} \begin{split} \parallel_Em(r, \frac{1}{f-b_1})\le&\frac{1}{2\pi}\int_{\theta_1(r)}(\log^+\frac{1}{\mid F \mid}+4\log^+\mid b_1\mid+2\log^+\mid b_2\mid)d\theta\\ &+4S(r)+T(r)+O(1). \end{split} \end{equation} $

(3.20)

$ \begin{equation} \begin{split} \parallel_Em(r, \frac{1}{f-b_2})\le&\frac{1}{2\pi}\int_{\theta_2(r)}(\log^+\frac{1}{\mid F \mid}+2\log^+\mid b_1\mid+4\log^+\mid b_2\mid)d\theta\\ &+4S(r)+T(r)+O(1). \end{split} \end{equation} $

(3.21)

Combining (3.18)–(3.21), we obtain

$ \begin{equation*} \begin{split} \parallel_E\sum^4_{i=1}m(r, \frac{1}{f-b_i})&\le m(r, \frac{1}{F})+4m(r, b_1)+4m(r, b_2)+4T(r)+14S(r). \end{split} \end{equation*} $

Therefore,

$ \begin{equation*} \begin{split} \parallel_E&4T(r, f)\le N(r, v^0_f)+N(r, v^0_{f-1})+N(r, v^0_{f-b_1})+N(r, v^0_{f-b_2})-N(r, v^0_{F})\\ &+T(r, F)+4m(r, b_1)+4m(r, b_2)+4T(r)+14S(r)+o(T(r, f)). \end{split} \end{equation*} $

Combining with (3.12), we have

$ \begin{equation} \begin{split} \parallel_E4T(r, f)\le&\overline N(r, v^0_f)+\overline N(r, v^0_{f-1})+\overline N(r, v^0_{f-b_1})+\overline N(r, v^0_{f-b_2})+T(r, F)\\ &+4m(r, b_1)+4m(r, b_2)+4T(r)+14S(r)+o(T(r, f)). \end{split} \end{equation} $

(3.22)

Moreover, from (3.3) and (3.17),

$ \begin{equation*} \begin{split} m(r, F)&\le2m(r, f)+m(r, \frac{f'}{f})+\sum^2_{i=1}(2m(r, b_i)+m(r, \frac{b'_i}{b_i}))\\ &\le2m(r, f)+2m(r, b_1)+2m(r, b_2)+3S(r) \end{split} \end{equation*} $

and

$ N(r, v^\infty_F)\le2N(r, v^\infty_f)+\overline N(r, v^\infty_f)+3\sum^2\limits_{i=1}N(r, v^\infty_{b_i}). $

These inequalities imply that

$ \begin{equation} \begin{split} \parallel_ET(r, F)\le2T(r, f)+\overline N(r, v^\infty_f)+2\sum^2_{i=1}N(r, v^\infty_{b_i})+\frac{2}{3}T(r)+3S(r). \end{split} \end{equation} $

(3.23)

From (3.22) and (3.23), we get

$ \begin{equation*} \begin{split} \parallel_E2T(r, f)&\le\overline N(r, v^0_f)+\overline N(r, v^0_{f-1})+\overline N(r, v^0_{f-b_1})+\overline N(r, v^0_{f-b_2})\\ &+\overline N(r, v^\infty_f)+18T(r, b_1)+18T(r, b_2)+17S(r)+o(T(r, f)). \end{split} \end{equation*} $

Therefore, this proposition is proved.

We now prove the lemma of part 1. From (3.1), (3.2) and proposition 3.4, we get

$ \begin{equation*} \begin{split} \parallel_E2T(r, g)\le&\sum^5_{i=1}\overline N(r, v^0_{g-a_i})+45\sum\limits_{j=1, 3}T(r, a_{i_j})+47T(r, a_{i_2})\\ &+19\sum\limits_{j=4, 5}T(r, a_{i_j})+17S(r)+o(T(r, g)), \end{split} \end{equation*} $

for any permutation$ (i_1, \cdots, i_5) $ of $ \{1, \cdots, 5\} $.Summing up both sides of the above inequalities over all such permutations, we have

$ \begin{equation} \begin{split} \parallel_E2T(r, g)\le\sum^5_{i=1}\overline N(r, v^0_{g-a_i})+35\sum^5_{i=1}T(r, a_i)+17S(r)+o(T(r, g)). \end{split} \end{equation} $

(3.24)

Therefore, the part 1 is proved.

Part 2. We now prove the part 2, where there is a function among $ \{a_1, \cdots, a_5\} $ equal to $ \infty $; for example, $ a_1\equiv\infty $. We set $ h=\frac{g-a_2}{g-a_3}, c_1=\frac{a_4-a_2}{a_4-a_3}, c_2=\frac{a_5-a_2}{a_5-a_3}, c_3=0, c_4=1. $ By Lemma 3.1, we have

$ \begin{equation} \begin{split} T(r, g) &\le T(r, h)+\sum^3_{i=2}T(r, a_i)+O(1), \\ T(r, c_1)&\le \sum^4_{i=2}T(r, a_i)+O(1), \\ T(r, c_2)&\le \sum^3_{i=2}T(r, a_i)+T(r, a_5)+O(1), \end{split} \end{equation} $

(3.25)

and

$ \begin{equation} \begin{split} &\overline N(r, v^0_h)+\overline N(r, v^\infty_h)+\overline N(r, v^0_{h-1})+\overline N(r, v^0_{h-c_1})+\overline N(r, v^0_{h-c_2})\\ \le&\sum^5_{i=1}\overline N(r, v^0_{g-a_i})+4T(r, a_2)+8T(r, a_3)+T(r, a_4)+T(r, a_5)+O(1). \end{split} \end{equation} $

(3.26)

Applying Proposition 3.4 for functions $ h, c_1, c_2, c_3, $ and $ c_4 $, we get

$ \parallel_E2T(r, h)\le\overline N(r, v^\infty_h)+\sum^4\limits_{i=1}\overline N(r, v^0_{h-c_i})+\sum^2\limits_{i=1}18T(r, c_i)+17S(r)+O(1). $

Combining (3.25), (3.26) and the above inequality, we get

$ \begin{equation*} \begin{split} \parallel_E2T(r, g)\le&\sum^5_{i=1}\overline N(r, v^0_{g-a_i})+42T(r, a_{i_2})+46T(r, a_{i_3})\\ &+19\sum\limits_{j=4, 5}T(r, a_{i_j})+17S(r)+o(T(r, g)), \end{split} \end{equation*} $

for any permutation $ (i_2, \cdots, i_5) $ of $ \{2, \cdots, 5\} $. Summing up both sides of the above inequalities over all such permutations, we get

$ \begin{equation} \parallel_E2T(r, g)\le\sum^5\limits_{i=1}\overline N(r, v^0_{g-a_i})+\frac{63}{2}\sum^5\limits_{i=2}T(r, a_i)+17S(r)+o(T(r, g)). \end{equation} $

(3.27)

Therefore, the part 2 is proved. We complete the proof of the lemma.

Assuming there exists a meromorphic function $ g $ on $ \Delta(R) $ and mutually distinct meromorphic functions $ f_i(1<i<5) $ on $ \Delta(R) $ such that $ f_i $ and $ g $ share the set $ S $ with truncated multiplicity, where $ g=f_1 $.

Put $ P_S(w)=(w-a_1)\cdots(w-a_q), \psi_i=P_S(f_i)/P_S(g)(1<i<5), \ \varphi_j=P'_S(f_j)f'_j/P_S(f_j) $ and $ \varphi=\varphi_1. $ We have

$ \varphi=\alpha_j+\varphi_j, T(r, P_S(f_j))=qT(r, f_j)+O(1) $

where $ \alpha_j=-\frac{\psi'_j}{\psi_j}(1<j<5). $ Let $ T(r)=\sum^5_{i=1}T(r, f_i) $, by lemma 2.1, we have

$ m(r, \varphi_j)=S_{P_S(f_j)}(r)\le S(r)+o(T(r))\ \text{for all}\ 1<j<5. $

Since each pole of $ \varphi_j $ is a simple pole, which is either zero or a pole of $ P_S(f_j) $, we get

$ N(r, v^\infty_{\varphi_j})=\overline N(r, v^0_{P_S(f_j)})+\overline N(r, v^\infty_{P_S(f_j)})\le2qT(r, f_j). $

This implies

$ \parallel_ET(r, \varphi_j)=m(r, \varphi_j)+N(r, v^\infty_{\varphi_j})\le2qT_0(r, f_j)+S(r)+o(T(r)). $

In particular, $ T(r, \varphi_j)=O(T(r, g)) $. In addition, by Theorem 2.2, we get

$ \begin{equation*} \begin{split} \parallel_EN(r, v^\infty_{\varphi})&=\overline N(r, v^0_{P_S(g)})+\overline N(r, v^\infty_{P_S(g)})\\ &=\overline N(r, v^0_{P_S(f_j)})+\overline N(r, v^\infty_{P_S(f_j)})\\ &=\sum^q_{i=1}\overline N(r, v^0_{f_j-a_i})+q\overline N(r, v^\infty_{f_j})\\ &\ge(q-1)(T(r, f_j)+\overline N(r, v^\infty_{f_j}))-S_{f_j}(r). \end{split} \end{equation*} $

It yields that

$ \begin{equation} \begin{split} \parallel_E(q-1)T(r, f_j)\le T(r, \varphi)-(q-1)\overline N(r, v^\infty_{f_j})+o(T(r)). \end{split} \end{equation} $

(4.1)

Hence

$ \begin{equation*} \begin{split} T(r, \alpha_j)&=m(r, \alpha_j)+N(r, v^\infty_{\alpha_j})\\ &\le m(r, \varphi)+m(r, \varphi_j)+\overline N(r, v^0_{\psi_j})+\overline N_0(r, v^\infty_{\psi_j})+O(1)\\ &\le\overline N(r, v^0_{P_S(g)/P_S(f_j)})+\overline N(r, v^\infty_{P_S(g)/P_S(f_j)})+2S(r)+o(T(r))\\ &\le\sum^q_{i=1}\overline N(r, v^0_{f_j-a_i, \ge\ell_i})+2S(r)+o(T(r))\\&\le\sum^q_{i=1}\frac{1}{\ell_i}T(r, f_j)+2S(r)+o(T(r)). \end{split} \end{equation*} $

This also implies $ T(r, \alpha_j)=O(T(r, g)). $

We now show that $ \alpha_1, \cdots, \alpha_5 $ are mutually distinct. Indeed, supposing that $ \alpha_i=\alpha_j $ for some $ i \ne j $. Then there exists a nonzero constant $ c_0 $ with $ \psi_i=c_0\psi_j $, and hence $ c_0P_S(f_i)=P_S(f_j). $ Since $ P_S(w) $ is a uniqueness polynomial for meromorphic functions on $ \Delta(R_0) $, we have $ f_i = f_j $, it is a contradiction.

Applying Theorem 1.4, we have

$ \parallel_E2T(r, \varphi)\le\sum^5\limits_{j=1}\overline N(r, v^0_{\varphi-\alpha_j})+35\sum^5\limits_{j=1}T(r, \alpha_j)+17S(r)+o(T(r, \varphi)). $

On the other hand,

$ \overline N(r, v^0_{\varphi-\alpha_j})=\overline N(r, v^0_{\varphi_j})\le\overline N(r, v^0_{f'_j})+\sum^k\limits_{t=1}\overline N(r, v^0_{f_j-e_t}), $

where $ e_1, \cdots, e_k $ are all of distinct zeros of $ P'_S(w) $. Since $ \overline N(r, v^0_{f_j-e_t})\le T(r, f_j)+O(1) $ and

$ \begin{equation*} \begin{split} \overline N(r, v^0_{f'_j})&\le T(r, f'_j)+O(1)=m(r, f'_j)+N(r, v^\infty_{f'_j})+O(1)\\ &\le m(r, f_j)+m(r, \frac{f'_j}{f_j})+N(r, v^\infty_{f_j})+\overline N(r, v^\infty_{f_j})+O(1)\\ &\le T(r, f_j)+\overline N(r, v^\infty_{f_j})+S(r)+o(T(r)), \end{split} \end{equation*} $

we have

$ \begin{equation} \begin{split} \sum^5_{j=1}\overline N(r, v^0_{\varphi-\alpha_j})\le(k+1)\sum^5_{j=1}T(r, f_j)+\sum^5_{j=1}\overline N(r, v^\infty_{f_j})+5S(r)+o(T(r)). \end{split} \end{equation} $

(4.2)

Combining (4.1) and (4.2), we obtain

$ \begin{equation*} \begin{split} \parallel_E2(q-1)T(r, f_i)\le&2T(r, \varphi)-2(q-1)\overline N(r, v^\infty_{f_i})\\ \le&(k+1+\sum^q_{i=1}\frac{35}{\ell_i})\sum^5_{j=1}T(r, f_j)+\sum^5_{j=1}\overline N(r, v^\infty_{f_j})\\ &-2(q-1)\overline N(r, v^\infty_{f_i})+372S(r)+o(T(r)), \end{split} \end{equation*} $

for all $ i=1, \cdots, 5. $

Summing up this inequality over all $ i=1, \cdots, 5, $

$ \parallel_E2(q-1)\sum^5\limits_{i=1}T(r, f_i)\le5(k+1+\sum^q\limits_{i=1}\frac{35}{\ell_i})\sum^5\limits_{j=1}T(r, f_j)+372\times 5S(r)+o(T(r)). $

Set $ \gamma(r)=\exp\{\min_{1\le j\le5}\{c_{f_j}\}+\varepsilon\}T(r). $ Then $ S(r)=(1+\varepsilon)(\min_{1\le j\le5}\{c_{f_j}\}+\varepsilon)T(r)+\varepsilon\log r. $ Letting $ \varepsilon\to0 $, and $ r\to R_0 $ ($ r\notin E $), we get

$ 372\sum^5\limits_{i=1}c_{f_i}\ge372\times5\min\limits_{1\le j\le5}\{c_{f_j}\}\ge\{2(q-1)-5(k+1+\sum^q\limits_{i=1}\frac{35}{\ell_i})\}. $

It is a contradiction. We complete the proof of the theorem.