Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 31, 2006, 191–211

FIXED POINTS OF MEROMORPHIC SOLUTIONS OF HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS

Liu Ming-Sheng and Zhang Xiao-Mei

South China Normal University, Department of Mathematics Guangzhou 510631, Guangdong, P. R. China; liumsh@scnu.edu.cn

Abstract. In this paper we discuss the problems on the fixed points of meromorphic solutions of higher order linear differential equations with meromorphic coefficients and their derivatives. Because of the restriction of differential equations, we obtain that the properties of fixed points of meromorphic solutions of higher order linear differential equations with meromorphic coefficients and their derivatives are more interesting than those of general transcendental meromorphic functions. Some estimates of the exponent of convergence of fixed points of solutions and their derivatives are obtained.

1. Introduction and main results

Many important results have been obtained on the fixed points of general transcendental meromorphic functions for almost four decades (see [14]). However, there are few studies on the fixed points of solutions of differential equations. In [3], Chen Zong-Xuan at first studied the problems on the fixed points and hyper-order of solutions of second order linear differential equations with entire coefficients. In [11], Wang and Yi studied the problems on the fixed points and hyper order of differential polynomials generated by solutions of second order linear differential equations with meromorphic coefficients. In [9], I. Laine and J. Rieppo had given an extension and improvement of the results in [11]; they studied the problems on the fixed points and iterated order of differential polynomials generated by solutions of second order linear differential equations with meromorphic coefficients. In [10], Wang and Lü studied the problems on the fixed points and hyper-order of solutions of second order linear differential equations with meromorphic coefficients and their derivatives. The main purpose of this paper is to extend some results in [10] to the case of higher order linear differential equations with meromorphic coefficients.

In this paper, we shall assume that the reader is familiar with the fundamental results and the standard notation of R. Nevanlinna’s theory of meromorphic functions (see [4], [13], [12], [8]). In addition, let σ(f) denote the order of growth of the meromorphic function f(z), λ̄(f) denote the exponent of convergence of the sequence of distinct zeros of f, v_{f}(r) denote the central index of f(z). In order to give some estimates of fixed points, we recall the following definitions (see [3], [7], [9]).

2000 Mathematics Subject Classification: Primary 30D35, 34M05.

This work is supported by the National Natural Science Foundation of China (No. 10471048).

Definition 1.1. Let z_{1}, z_{2}, …, |z_{j}| = r_{j}, 0 ≤ r_{1} ≤ r_{2} ≤ ⋯, be the sequence of distinct fixed points of a transcendental meromorphic function f. Then \overline{τ}(f), the exponent of convergence of the sequence of distinct fixed points of f, is defined by

\overline{τ}(f) = inf\{τ > 0 | ∑_{j=1}^{∞} |z_{j}|^{−τ} < +∞\}.

It is evident that

\overline{τ}(f) = \overline{lim}_{r→∞} \frac{log \overline{N} (r, \frac{1}{f − z} )}{log r}.

Definition 1.2. Suppose that f(z) be a meromorphic function of infinite order. Then the hyper-order σ_{2}(f) of f(z) is defined by

σ_{2}(f) = \overline{lim}_{r→∞} \frac{log log T (r, f)}{log r}.

Definition 1.3. Let f be a meromorphic function. Then \overline{λ_{2}}(f), the hyperexponent of convergence of the sequence of distinct zeros of f, is defined by

\overline{λ_{2}}(f) = \overline{lim}_{r→∞} \frac{log log \overline{N} (r, \frac{1}{f} )}{log r},

and \overline{τ}_{2}(f), the hyper-exponent of convergence of the sequence of distinct fixed points of f, is defined by

\overline{τ}_{2}(f) = \overline{lim}_{r→∞} \frac{log log \overline{N} (r, \frac{1}{f − z} )}{log r}.

Definition 1.4 ([2], [6]). Let P (z) be rational and of the form P_{1}(z)/P_{2}(z), where P_{1}(z) and P_{2}(z) are two polynomials. We denote by di(P), the “degree at infinity” of P, defined by

(1.1)  di(P) = deg P_{1}(z) − deg P_{2}(z).

In [10], Wang and Lü obtained the following results.

Theorem A. Suppose that P (z) = P_{1}(z)/P_{2}(z) ≢ 0 be a rational function with n = di(P). Then every transcendental meromorphic solution f of the equation

(1.2)  f′′ + P (z)f = 0

satisfies that f and f′, f′′ all have infinitely many fixed points and

\overline{τ}(f) = \overline{τ}(f′) = \overline{τ}(f′′) = σ(f) = max\{\frac{1}{2}(n + 2), 0\}.

Theorem B. Suppose that A(z) be a transcendental meromorphic function satisfying δ(∞, A) > 0, σ(A) = σ < +∞. Then every meromorphic solution f ≢ 0 of the equation

(1.3)  f′′ + A(z)f = 0

satisfies that f and f′, f′′ all have infinitely many fixed points and

\overline{τ}(f) = \overline{τ}(f′) = \overline{τ}(f′′) = σ(f) = +∞, \overline{τ}_{2}(f) = \overline{τ}_{2}(f′) = \overline{τ}_{2}(f′′) = σ_{2}(f) = σ.

Remark 1.1. In [10], Wang and Lü did not give the detailed proof of Theorem A, they only proved that σ(f) = max\{\frac{1}{2}(n + 2), 0\}, and omit the proof of \overline{τ}(f) = \overline{τ}(f′) = \overline{τ}(f′′) = σ(f). In fact, when n = −2, we cannot prove that \overline{τ}(f′) = \overline{τ}(f′′) = σ(f) by using the similar method as the proof of Theorem B in [10]; please look at Remark 3.1 in Section 3.

In this paper, we shall prove the following two theorems.

Theorem 1.1. Suppose that P (z) = P_{1}(z)/P_{2}(z) ≢ 0 be a rational function with n = di(P), and k be an integer with k ≥ 2. Then:

(1) Every transcendental meromorphic solution f of the equation

(1.4)  f^{(k)} + P (z)f = 0

satisfies σ(f) = max{(n + k)/k, 0}.

(2) If n ≠ −k, then every meromorphic solution f ≢ 0 of the equation (1.4) satisfies that f and f′, f′′, …, f^{(k)} all have infinitely many fixed points and

\overline{τ}(f) = \overline{τ}(f′) = ⋯ = \overline{τ}(f^{(k)}) = max\{\frac{n + k}{k} , 0\}.

(3) If n = −k, then every transcendental meromorphic solution f of the equation (1.4) satisfies that f and f′, f′′, …, f^{(k−2)} all have infinitely many fixed points and

\overline{τ}(f) = \overline{τ}(f′) = ⋯ = \overline{τ}(f^{(k−2)}) = 0.

Remark 1.2. Setting k = 2 in Theorem 1.1, we get Theorem 1 of [10] or Theorem A, which contains the related results in [1]. From Theorem 1.1, we know that every transcendental meromorphic solution f of equation (1.4) has infinitely many fixed points. In addition, if n = −k, there exists some equations of the form (1.4) that may have solutions with finite fixed points. For example, the equation

f′′′ − \frac{6}{z(z − 2)(z − 5)}f = 0

has a family of polynomial solutions {f_{c} = cz(z −2)(z −5); c is a constant}, which have at most three fixed points.

Theorem 1.2. Suppose that k ≥ 2 and A(z) be a transcendental meromorphic function satisfying δ(∞, A) = δ > 0, σ(A) = σ < +∞. Then every meromorphic solution f ≢ 0 of the equation

(1.5)  f^{(k)} + A(z)f = 0

satisfies that f and f′, f′′, …, f^{(k)} all have infinitely many fixed points and

(1.6)  \overline{τ}(f) = \overline{τ}(f^{(j)}) = σ(f) = +∞, \overline{τ}_{2}(f) = \overline{τ}_{2}(f^{(j)}) = σ_{2}(f) = σ,

for j = 1, …, k.

Remark 1.3. Setting k = 2 in Theorem 1.2, we get Theorem 2 of [10] or Theorem B.

2. Some lemmas

Lemma 2.1 ([11]). Suppose that f(z) = g(z)/d(z) is a meromorphic function with σ(f) = ϱ, where g(z) is an entire function and d(z) is a polynomial. Then there exists a sequence {r_{j}}, r_{j} → ∞, such that for all z satisfying |z| = r_{j}, |g(z)| = M(r_{j}, g), when j sufficiently large, we have

\frac{f^{(n)}(z)}{f(z)} = (\frac{v_{g}(r_{j})}{z})^{n}(1 + o(1)), n ≥ 1,
σ(f) = lim_{j→∞} \frac{log v_{g}(r_{j})}{log r_{j}}.

Lemma 2.2. Suppose that k ≥ 2 and A(z) is a transcendental meromorphic function with δ(∞, A) = δ > 0 and σ(A) = σ < +∞. Then every meromorphic solution f ≢ 0 of equation (1.5) satisfies σ(f) = +∞ and σ_{2}(f) = σ.

Proof. Suppose that f ≢ 0 is a meromorphic solution of (1.5). Rewrite (1.5) as

(2.1)  A = −\frac{f^{(k)}}{f} .

By the lemma of logarithmic derivatives, there exists a set E with finite linear measure such that m(r, A) ≤ c log(rT (r, f)) for r ∉ E, where c is a positive constant.

It follows from the definition of deficiency that for sufficiently large r, we have m(r, A) ≥ \frac{1}{2}δT (r, A). So when r ∉ E sufficiently large, we have

(2.2)  T (r, A) ≤ \frac{2c}{δ} log(rT (r, f)).

Hence by the definition of hyper-order, we obtain that σ(f) = +∞, and

(2.3)  σ_{2}(f) = \overline{lim}_{r→+∞} \frac{log log T (r, f)}{log r} ≥ σ(A) = σ.

On the other hand, we know that the poles of f can only occur at the poles of A, so λ(1/f) ≤ σ(A). According to the Hadamard factorization theorem, we can write f as f(z) = g(z)/d(z), where d(z) and g(z) are entire functions satisfying λ(d) = σ(d) = λ(1/f) ≤ σ ≤ σ_{2}(g) = σ_{2}(f). Substituting f(z) = g(z)/d(z) into (1.5), by Theorem 2.1 in [7], we get

(2.4)  σ_{2}(f) = σ_{2}(g) ≤ σ.

Hence σ_{2}(f) = σ. This completes the proof.

Lemma 2.3. Suppose that k be a positive integer and f(z) be a nonzero solution of equation (1.5). Let w_{0} = f − z, w_{1} = f′ − z, w_{2} = f′′ − z. Then w_{0} , w_{1} and w_{2} satisfy the following equations:

− w_{0}^{(k)} − Aw_{0} = zA, if k ≥ 2; (2.5)

− Aw_{1}^{(k)} + A′w_{1}^{(k−1)} − A^{2}w_{1} = zA^{2}, if k ≥ 3; (2.6)

− A^{2}w_{2}^{(k)} + 2AA′w_{2}^{(k−1)} + (AA′′ − 2(A′)^{2})w_{2}^{(k−2)} − A^{3}w_{2} (2.7)

= zA^{3}, if k ≥ 4;

− Aw′′_{1} + A′w′_{1} − A^{2}w_{1} = zA^{2} − A′, if k = 2; (2.8) − A^{2}w′′_{2} + 2AA′w′_{2} + (AA′′ − 2(A′)^{2} − A^{3})w_{2} (2.9)

= zA^{3} − 2AA′ + 2z(A′)^{2} − zAA′′, if k = 2;

− A^{2}w′′′_{2} + 2AA′w′′_{2} + (AA′′ − 2A^{′2})w′_{2} − A^{3}w_{2} (2.10)

= zA^{3} − AA′′ + 2A^{′2}, if k = 3;

− A^{3}w′′′_{3} + 3A^{2}A′w′′_{3} + 3A(AA′′ − 2(A′)^{2})w′_{3} (2.11)

+ (A^{2}A′′′ − 6AA′A′′ + 6A^{′3} − A^{4})w_{3}
= zA^{4} − 3A(AA′′ − 2A^{′2})
− z(A^{2}A′′′ − 6AA′A′′ + 6A^{′3}), if k = 3.

Proof. From f = w_{0} + z, we have that f^{(k)} = w_{0}^{(k)} for k ≥ 2. Substituting it into (1.5), we get f = −w_{0}^{(k)}/A, that is, w_{0} + z = −w_{0}^{(k)}/A. From this, we obtain (2.5).

According to the equality f′ = w_{1} + z, we obtain that f^{(k)} = w_{1}^{(k−1)} for k ≥ 3. Substituting it into (1.5), we get f = −w_{1}^{(k−1)}/A, so that f′ = (−w_{1}^{(k)}A + w_{1}^{(k−1)}A′)/A^{2} , that is, w_{1} + z = (−w_{1}^{(k)}A + w_{1}^{(k−1)}A′)/A^{2} . From this, we obtain (2.6).

By the equality f′′ = w_{2} + z, we get that f^{(k)} = w_{2}^{(k−2)} for k ≥ 4. Substituting it into (1.5), we get f = −w_{2}^{(k−2)}/A, so that

\begin{aligned}f′′ & = (f′)′ = (\frac{−w_{2}^{(k−1)}A + w_{2}^{(k−2)}A′}{A^{2}} )′ \\ & = \frac{−A^{2}w_{2}^{(k)} + 2AA′w_{2}^{(k−1)} + (AA′′ − 2(A′)^{2})w_{2}^{(k−2)}}{A^{3}} , \end{aligned}

that is,

w_{2} + z = \frac{−A^{2}w_{2}^{(k)} + 2AA′w_{2}^{(k−1)} + (AA′′ − 2(A′)^{2})w_{2}^{(k−2)}}{A^{3}} .

From this, we obtain (2.7).

We may prove equations (2.9)–(2.11) similarly. Hence the proof of Lemma 2.3 is complete.

Let i, j are two non-negative integers. Now we define the notation H_{ij}(A) as follows.

(1) For any non-negative integer i, we define H_{i0}(A) = −A^{i};

(2) For any non-negative integer i, H_{ii}(A) and H_{i(i−1)}(A) are defined by the following recurrence formula:

(2.12)  \begin{aligned}H_{(i+1)(i+1)}(A) & = A · [H_{ii}(A)]′ − (i + 1)A′ · H_{ii}(A), \\ H_{(i+2)(i+1)}(A) & = A · [H_{(i+1)i}(A)]′ − (i + 2)A′ · H_{(i+1)i}(A) \\ & + A · H_{(i+1)(i+1)}(A), \\ H_{00}(A) & = −1,  H_{10}(A) = −A; \end{aligned}

(3) For i ≥ 3 and 1 ≤ j ≤ i − 2, H_{ij}(A) are defined as a sum of a finite number of terms of the type

(2.13)  B = bA^{l_{0}}(A′)^{l_{1}} ⋯ (A^{(i+1)})^{l_{i+1}},

where b is a constant, l_{0}, l_{1}, …, l_{i+1} are non-negative integers such that l_{0} + l_{1} + ⋯ + l_{i+1} = i and l_{1} + 2l_{2} + ⋯ + (i + 1)l_{i+1} = j. It is obvious that H_{ij}(A) is a differential polynomial of A. Moreover, this notation H_{ij}(A), 1 ≤ j ≤ i − 2, may represent a different differential polynomial of A in different occurrences, even within one single formula.

By the definition of H_{ij}(A), simple computation yields

(2.14)

H_{00}(A) = −1,
H_{10}(A) = −A,
H_{11}(A) = A′;
H_{20}(A) = −A^{2},
H_{21}(A) = 2AA′,
H_{22}(A) = AA′′ − 2A^{′2},
H_{32}(A) = 3A^{2}A′′ − 6AA^{′2},
H_{33}(A) = A^{2}A′′′ − 6AA′A′′ + 6A^{′3};
A · (H_{ij}(A))′ = H_{(i+1)(j+1)}(A) for 0 < j ≤ i − 2;
A′ · H_{ij}(A) = H_{(i+1)(j+1)}(A) for 0 < j ≤ i − 2;
A · H_{ij}(A) = H_{(i+1)j}(A) for 0 ≤ j ≤ i − 1.

Lemma 2.4. Suppose that f(z) is a nonzero solution of equation (1.5) and k ≥ 2. Let w_{i} = f^{(i)} − z, i = 0, 1, …, k − 2. Then w_{i} satisfy the following equations:

(2.15)  ∑_{j=0}^{i} H_{ij}(A)w_{i}^{(k−j)} − A^{i+1}w_{i} = zA^{i+1}, i = 0, 1, …, k − 2,

where H_{ij}(A) are defined by (2.12)–(2.14).

Proof. When k = 2 or 3, 4, by Lemma 2.3 and (2.12)–(2.14), it is evident that (2.15) hold. In the following, we suppose that k ≥ 5. We shall use an inductive method to prove it.

At first, by Lemma 2.3 and (2.14), we get that (2.15) hold for i = 0, 1, 2. Next, suppose that w_{i}, 2 ≤ i ≤ k − 3, satisfy (2.15). Now we verify that

w_{i+1} also satisfies (2.15).

198 Liu Ming-Sheng and Zhang Xiao-Mei

From w_{i} + z = f^{(i)} and w_{i+1} + z = f^{(i+1)} , we know that

w_{i}^{(k−j)} = w_{i+1}^{(k−j−1)}, j = 0, 1, …, i, i ≤ k − 3.

Since w_{i} satisfies (2.15), we have

f^{(i)} = w_{i} + z = \frac{∑_{j=0}^{i} H_{ij}(A)w_{i}^{(k−j)}}{A^{i+1}} ,

so by (2.14) and (2.12), we obtain

f^{(i+1)} = (f^{(i)})′ = (\frac{∑_{j=0}^{i} H_{ij}(A)w_{i}^{(k−j)}}{A^{i+1}} )′

= (\frac{∑_{j=0}^{i} H_{ij}(A)w_{i+1}^{(k−j−1)}}{A^{i+1}} )′

= \frac{1}{A^{2(i+1)}} \{[∑_{j=0}^{i}(H_{ij}(A))′w_{i+1}^{(k−j−1)} + ∑_{j=0}^{i} H_{ij}(A)w_{i+1}^{(k−j)}]A^{i+1}

− (i + 1)A^{i}A′ ∑_{j=0}^{i} H_{ij}(A)w_{i+1}^{(k−j−1)}}

= \frac{1}{A^{i+2}} \{∑_{j=0}^{i−2} H_{(i+1)(j+1)}(A)w_{i+1}^{(k−j−1)} + A(H_{i(i−1)}(A))′w_{i+1}^{(k−i)}

+ A(H_{ii}(A))′w_{i+1}^{(k−i−1)} + ∑_{j=0}^{i} AH_{ij}(A)w_{i+1}^{(k−j)}

− (i + 1)A′ ∑_{j=0}^{i} H_{ij}(A)w_{i+1}^{(k−j−1)}}

= \frac{1}{A^{i+1+1}} \{∑_{j=1}^{i−1} H_{(i+1)j}(A)w_{i+1}^{(k−j)} + H_{(i+1)i}(A)w_{i+1}^{(k−i)}

+ H_{(i+1)(i+1)}(A)w_{i+1}^{(k−i−1)}

+ ∑_{j=1}^{i−1} H_{(i+1)j}(A)w_{i+1}^{(k−j)} + H_{i0}(A) · Aw_{i+1}^{(k)}

− (i + 1) ∑_{j=1}^{i−1} H_{(i+1)j}(A)w_{i+1}^{(k−j)}}

\begin{matrix}= \frac{1}{A^{i+1+1}} \{−A^{i+1}w_{i+1}^{(k)} + ∑_{j=1}^{i+1} H_{(i+1)j}(A)w_{i+1}^{(k−j)}\} \\ = \frac{1}{A^{i+1+1}} ∑_{j=0}^{i+1} H_{(i+1)j}(A)w_{i+1}^{(k−j)} = w_{i+1} + z. \end{matrix}

From the above equality, we obtain that w_{i+1} also satisfies (2.15). This completes the proof.

Lemma 2.5. Suppose that f(z) ≢ 0 is a solution of equation (1.5) and k ≥ 2. Let w_{k−1} = f^{(k−1)} − z. Then w_{k−1} satisfies

(2.16)  ∑_{j=0}^{k−1} H_{(k−1)j}(A)w_{k−1}^{(k−j)} − A^{k}w_{k−1} = zA^{k} − H_{(k−1)(k−1)}(A),

where H_{ij} are defined by (2.12)–(2.14).

Proof. When k = 2 or 3, by (2.8), (2.10) and (2.14), it is evident that (2.16) holds. When k ≥ 4, from w_{k−2} = f^{(k−2)} − z and w_{k−1} = f^{(k−1)} − z, we obtain

w_{k−2}^{(k−j)} = w_{k−1}^{(k−j−1)}, j = 0, 1, 2, …, k − 3, w′′_{k−2} = w′_{k−1} + 1.

Since w_{k−2} satisfies (2.15), rewrite it as

f^{(k−2)} = w_{k−2} + z = \frac{∑_{j=0}^{k−2} H_{(k−2)j}(A)w_{k−2}^{(k−j)}}{A^{k−1}} ,

so by (2.14) and (2.12), we have

f^{(k−1)} = (f^{(k−2)})′

= (\frac{∑_{j=0}^{k−2} H_{(k−2)j}(A)w_{k−2}^{(k−j)}}{A^{k−1}} )′

= (\frac{∑_{j=0}^{k−3} H_{(k−2)j}(A)w_{k−1}^{(k−j−1)} + H_{(k−2)(k−2)}(A)(w′_{k−1} + 1)}{A^{k−1}})′

= \frac{1}{A^{2(k−1)}} \{[∑_{j=0}^{k−3}(H_{(k−2)j}(A))′w_{k−1}^{(k−j−1)} + ∑_{j=0}^{k−3} H_{(k−2)j}(A)w_{k−1}^{(k−j)}

+ (H_{(k−2)(k−2)}(A))′(w′_{k−1} + 1) + H_{(k−2)(k−2)}(A)w′′_{k−1}]A^{k−1}

− [∑_{j=0}^{k−3} H_{(k−2)j}(A)w_{k−1}^{(k−j−1)}

+ H_{(k−2)(k−2)}(A)(w′_{k−1} + 1)](k − 1)A^{k−2}A′}

= \frac{1}{A^{k}} \{∑_{j=0}^{k−4} A(H_{(k−2)j}(A))′w_{k−1}^{(k−j−1)} + ∑_{j=0}^{k−3} AH_{(k−2)j}(A)w_{k−1}^{(k−j)}

+ A(H_{(k−2)(k−3)}(A))′w′′_{k−1} − (k − 1)A′H_{(k−2)(k−3)}(A)w′′_{k−1} + AH_{(k−2)(k−2)}(A)w′′_{k−1} + A(H_{(k−2)(k−2)}(A))′(w′_{k−1} + 1) − (k − 1)A′H_{(k−2)(k−2)}(A)(w′_{k−1} + 1)

− ∑_{j=0}^{k−4}(k − 1)A′H_{(k−2)j}(A)w_{k−1}^{(k−j−1)}}

= \frac{1}{A^{k}} \{∑_{j=0}^{k−4} H_{(k−1)(j+1)}(A)w_{k−1}^{(k−j−1)} + ∑_{j=1}^{k−3} H_{(k−1)j}(A)w_{k−1}^{(k−j)}

+ AH_{(k−2)0}(A)w_{k−1}^{(k)} + H_{(k−1)(k−2)}(A)w′′_{k−1} + H_{(k−1)(k−1)}(A)(w′_{k−1} + 1)

− ∑_{j=0}^{k−4}(k − 1)H_{(k−1)(j+1)}(A)w_{k−1}^{(k−j−1)}}

= \frac{1}{A^{k}} \{−A^{k−1}w_{k−1}^{(k)} + ∑_{j=1}^{k−3} H_{(k−1)j}(A)w_{k−1}^{(k−j)}

+ H_{(k−1)(k−2)}(A)w′′_{k−1} + H_{(k−1)(k−1)}(A)(w′_{k−1} + 1)}

= \frac{1}{A^{k}} \{∑_{j=0}^{k−1} H_{(k−1)j}(A)w_{k−1}^{(k−j)} + H_{(k−1)(k−1)}(A)\} = w_{k−1} + z.

From the above equality we get (2.16). This completes the proof. Lemma 2.6. Suppose that f(z) ≢ 0 is a solution of equation (1.5) and

k ≥ 2. Let w_{k} = f^{(k)} − z. Then w_{k} satisfies

(2.17) ∑_{j=0}^{k−1} H_{kj}(A)w_{k}^{(k−j)}+(H_{kk}(A)−A^{k+1})w_{k} = zA^{k+1}−H_{k(k−1)}(A)−zH_{kk}(A),

where H_{ij} are defined by (2.12)–(2.14).

Proof. When k = 2, by (2.9) and (2.14), it is evident that (2.17) holds. When k ≥ 3, we obtain from w_{k−1} = f^{(k−1)} − z and w_{k} = f^{(k)} − z,

\begin{aligned}w_{k−1}^{(k−j)} & = w_{k}^{(k−j−1)}, j = 0, 1, …, k − 3, \\ w′′_{k−1} & = w′_{k} + 1, \\ w′_{k−1} & = w_{k} + z − 1. \end{aligned}

Fixed points of meromorphic solutions 201

Since w_{k−1} satisfies (2.16), rewrite it as

f^{(k−1)} = w_{k−1} + z = \frac{∑_{j=0}^{k−1} H_{(k−1)j}(A)w_{k−1}^{(k−j)} + H_{(k−1)(k−1)}(A)}{A^{k}},

so by (2.14) and (2.12), we have

f^{(k)} = (f^{(k−1)})′

= (\frac{∑_{j=0}^{k−1} H_{(k−1)j}(A)w_{k−1}^{(k−j)} + H_{(k−1)(k−1)}(A)}{A^{k}})′

= (\frac{∑_{j=0}^{k−3} H_{(k−1)j}(A)w_{k}^{(k−j−1)} + H_{(k−1)(k−2)}(A)(w′_{k} + 1)}{A^{k}}

+ \frac{H_{(k−1)(k−1)}(A)(w_{k} + z − 1) + H_{(k−1)(k−1)}(A)}{A^{k}})′

= (\frac{∑_{j=0}^{k−3} H_{(k−1)j}(A)w_{k}^{(k−j−1)} + H_{(k−1)(k−2)}(A)(w′_{k} + 1)}{A^{k}}

+ \frac{H_{(k−1)(k−1)}(A)(w_{k} + z)}{A^{k}})′

= \frac{1}{A^{2k}} \{[∑_{j=0}^{k−3}(H_{(k−1)j}(A))′w_{k}^{(k−j−1)} + ∑_{j=0}^{k−3} H_{(k−1)j}(A)w_{k}^{(k−j)}

+ (H_{(k−1)(k−2)}(A))′(w′_{k} + 1) + H_{(k−1)(k−2)}(A)w′′_{k} + (H_{(k−1)(k−1)}(A))′(w_{k} + z) + H_{(k−1)(k−1)}(A)(w′_{k} + 1)]A^{k}

− [∑_{j=0}^{k−3} H_{(k−1)j}(A)w_{k}^{(k−j−1)} + H_{(k−1)(k−2)}(A)(w′_{k} + 1)

+ H_{(k−1)(k−1)}(A)(w_{k} + z)]kA^{k−1}A′}

= \frac{1}{A^{k+1}} \{∑_{j=0}^{k−3} H_{k(j+1)}(A)w_{k}^{(k−j−1)}

+ AH_{(k−1)0}(A)w_{k}^{(k)} + ∑_{j=1}^{k−3} H_{kj}(A)w_{k}^{(k−j)}

+ A(H_{(k−1)(k−2)}(A))′(w′_{k} + 1) + AH_{(k−1)(k−2)}(A)w′′_{k} + A(H_{(k−1)(k−1)}(A))′(w_{k} + z)

\begin{matrix}+ AH_{(k−1)(k−1)}(A)(w′_{k} + 1) − kA′H_{(k−1)(k−2)}(A)(w′_{k} + 1) \\ − kA′H_{(k−1)(k−1)}(A)(w_{k} + z) − ∑_{j=0}^{k−3} kH_{k(j+1)}(A)w_{k}^{(k−j−1)}\} \end{matrix}

= \frac{1}{A^{k+1}} \{∑_{j=1}^{k−2} H_{kj}(A)w_{k}^{(k−j)} − A^{k}w_{k}^{(k)} + H_{k(k−1)}(A)(w′_{k} + 1)

+ H_{k(k−2)}(A)w′′_{k} + H_{kk}(A)(w_{k} + z)}

= \frac{1}{A^{k+1}} \{∑_{j=0}^{k} H_{kj}(A)w_{k}^{(k−j)} + H_{k(k−1)}(A) + zH_{kk}(A)\} = w_{k} + z.

From the above equality we obtain (2.17). This completes the proof.

Lemma 2.7. Suppose that k ≥ 2 and P (z) is a rational function with n = di(P) < −k ≤ −2. Then there exist two nonzero constants c_{1} and c_{2} such that

(2.18)  H_{(k−1)(k−1)}(P) = \frac{c_{1} + o(1)}{z^{−nk−1+k+n}} as z → ∞,

and

(2.19)  H_{k(k−1)}(P) + zH_{kk}(P) = \frac{c_{2} + o(1)}{z^{−n(k+1)−1+k+n}} as z → ∞.

Proof. At first, we prove (2.18) by means of induction.

Since P (z) is a rational function with n = di(P) < −k ≤ −2, we have

(2.20)  P (z) = \frac{c_{0} + o(1)}{z^{−n}} as z → ∞,

where c_{0} ≠ 0 is a constant. From this and (2.14), simple computation yields

H_{11}(P) = P′ = \frac{nc_{0} + o(1)}{z^{−n+1}} as z → ∞,

where nc_{0} ≠ 0. Hence (2.18) holds, when k = 2.

Furthermore, if (2.18) is assumed to be valid when k = m (> 2), that is, there exists a constant c′′_{1} ≠ 0 such that

H_{(m−1)(m−1)}(P) = \frac{c′′_{1} + o(1)}{z^{−nm−1+m+n}} as z → ∞,

it must also hold when k = m + 1, since, by (2.12),

H_{mm}(P) = P · [H_{(m−1)(m−1)}(P)]′ − mP′ · H_{(m−1)(m−1)}(P)

= \frac{c_{0} + o(1)}{z^{−n}} · [\frac{c′′_{1} + o(1)}{z^{−nm−1+m+n}} ]′ − m · \frac{nc_{0} + o(1)}{z^{−n+1}} · \frac{c′′_{1} + o(1)}{z^{−nm−1+m+n}}

= \frac{(1 − m − n)c_{0}c′′_{1} + o(1)}{z^{−n(m+1)−1+(m+1)+n}}

= \frac{c_{1} + o(1)}{z^{−n(m+1)−1+(m+1)+n}} as z → ∞,

where c_{1} = (1 − m − n)c_{0}c′′_{1} ≠ 0. Hence (2.18) holds.

Next, we prove (2.19) by means of induction. By (2.20) and (2.14), simple computation yields

\begin{aligned}H_{21}(P) + zH_{22}(P) & = 2P P′ + z(P P′′ − 2P^{′2}) = \frac{n(1 − n)c_{0}^{2} + o(1)}{z^{−2n+1}} \\ & = \frac{c′_{2} + o(1)}{z^{−4n+3}} as z → ∞, \end{aligned}

where c′_{2} = n(1 − n)c_{0}^{2} ≠ 0. Hence (2.19) holds, when k = 2.

Furthermore, if (2.19) is assumed to be valid when k = m (> 2), that is, there exists a constant c′′_{2} ≠ 0 such that

H_{m(m−1)}(P) + zH_{mm}(P) = \frac{c′′_{2} + o(1)}{z^{−n(m+1)−1+m+n}} as z → ∞,

it must also hold when k = m + 1, since, by (2.12),

H_{(m+1)m}(P) + zH_{(m+1)(m+1)}(P) = P [H_{m(m−1)}(P)]′ − (m + 1)P′H_{m(m−1)}(P)

+ P H_{mm}(P) + zP · [H_{mm}(P)]′

− (m + 1)zP′ · H_{mm}(P) = P [H_{m(m−1)}(P) + zH_{mm}(P)]′

− (m + 1)[H_{m(m−1)}(P) + zH_{mm}(P)]

= \frac{c_{0} + o(1)}{z^{−n}} · [\frac{c′′_{2} + o(1)}{z^{−n(m+1)−1+m+n}} ]′

− (m + 1) · \frac{nc_{0} + o(1)}{z^{−n+1}} · \frac{c′′_{2} + o(1)}{z^{−n(m+1)−1+m+n}}

= \frac{(1 − m − n)c_{0}c′′_{2} + o(1)}{z^{−n(m+2)−1+(m+1)+n}}

= \frac{c_{2} + o(1)}{z^{−n(m+2)−1+(m+1)+n}} as z → ∞,

where c_{2} = (1 − m − n)c_{0}c′′_{2} ≠ 0. Hence (2.19) holds, and the proof is complete.

3. Proofs of Theorems 1.1 and 1.2

Proof of Theorem 1.2. Suppose that f ≢ 0 is a meromorphic solution of (1.5). Then, by Lemma 2.2, we obtain that

σ(f) = ∞, σ_{2}(f) = σ.

Set w_{i} = f^{(i)} − z, i = 0, 1, …, k. Then for every i, a point z_{0} is a fixed points of f^{(i)} if and only if z_{0} is a zero of w_{i}, and

(3.1)  σ(w_{i}) = σ(f^{(i)}) = σ(f) = +∞, \overline{τ}(f^{(i)}) = λ̄(w_{i}),

and

(3.2)  σ_{2}(w_{i}) = σ_{2}(f^{(i)}) = σ_{2}(f) = σ, \overline{τ}_{2}(f^{(i)}) = \overline{λ_{2}}(w_{i}).

By Lemmas 2.4–2.6, we know that w_{i}, i = 0, 1, …, k, satisfy the following equations:

∑_{j=0}^{i} H_{ij}(A)w_{i}^{(k−j)} − A^{i+1}w_{i} = zA^{i+1}, i = 0, 1, …, k − 2; (3.3)

∑_{j=0}^{k−1} H_{(k−1)j}(A)w_{k−1}^{(k−j)} − A^{k}w_{k−1} = zA^{k} − H_{(k−1)(k−1)}(A); (3.4)

∑_{j=0}^{k−1} H_{kj}(A)w_{k}^{(k−j)} + (H_{kk}(A) − A^{k+1})w_{k} (3.5)

= zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A).

Since A is a transcendental meromorphic function, A^{i+1}z ≢ 0, i = 0, 1, …, k − 2. We claim that

zA^{k} − H_{(k−1)(k−1)}(A) ≢ 0 and zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A) ≢ 0.
In fact, if zA^{k} − H_{(k−1)(k−1)}(A) ≡ 0, rewrite it as

A = \frac{H_{(k−1)(k−1)}(A)}{zA^{k−1}}.

From this, we obtain that m(r, A) = S(r, A), which contradicts δ(∞, A) > 0. If zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A) ≡ 0, rewrite it as

A = \frac{H_{k(k−1)}(A)}{zA^{k}} + \frac{H_{kk}(A)}{A^{k}}.

Similarly, we may obtain that m(r, A) = S(r, A), which contradicts δ(∞, A) > 0; hence the claims hold.

Rewrite (3.3), (3.4) and (3.5) as

\frac{1}{zA^{i+1}} (∑_{j=0}^{i} H_{ij}(A)\frac{w_{i}^{(k−j)}}{w_{i}} − A^{i+1}) = \frac{1}{w_{i}} , i = 0, 1, …, k − 2, (3.6)

\frac{1}{zA^{k} − H_{(k−1)(k−1)}(A)}(∑_{j=0}^{k−1} H_{(k−1)j}(A)\frac{w_{k−1}^{(k−j)}}{w_{k−1}} − A^{k}) = \frac{1}{w_{k−1}} , (3.7)

\frac{1}{zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A)} (3.8)

× (∑_{j=0}^{k−1} H_{kj}(A)\frac{w_{k}^{(k−j)}}{w_{k}} + H_{kk}(A) − A^{k+1}) = \frac{1}{w_{k}} .

Hence there exists a set E with finite linear measure such that for r ∉ E, we have

(3.9)  m(r, \frac{1}{w_{i}} ) ≤ O(m(r, \frac{1}{A})) + C(log rT (r, w_{i})), i = 0, 1, …, k.

On the other hand, since H_{ij}(A) are differential polynomials of A, according to their definition, it is easy to get that if z_{0} is a pole of order l (≥ 1) of A, then z_{0} is a pole of order li + j of H_{ij}(A). In the following, we split it into three cases to discuss the poles of w_{i}:

Case (1): i = 0, 1, …, k − 2. Since the coefficients of w_{i}^{(k−j)} are H_{ij}(A), j = 0, 1, …, i, and the right-hand sides of the equations (3.3) are A^{i+1}z, by (3.3), all zeros (except for z = 0) of w_{i}, whose order is larger than k, are zeros of A(z) (each is not the pole of A, if not, it will lead to a contradiction as follows). Hence

(3.10)  N(r, \frac{1}{w_{i}} ) ≤ k\overline{N} (r, \frac{1}{w_{i}} ) + N(r, \frac{1}{zA^{i+1}} ).

Case (2): i = k −1. Suppose that w_{k−1} has a zero of order k_{1} (≥ k^{2} −k +2) at point z_{1} , and z_{1} is also the pole of order l of A(z). We claim that 0 ≤ l ≤ k−1. In fact, if l ≥ k, then z_{1} is a pole of order kl of zA^{k} − H_{(k−1)(k−1)}(A), and z_{1} is also a pole of order l_{1} , l_{1} ≤ kl − k_{1} < kl, of the left-hand side in equality (3.4); this is a contradiction; hence 0 ≤ l ≤ k − 1. At this time, z_{1} is a pole of order l(k −1)+j of H_{(k−1)j}(A)) for j = 0, 1, …, k −1. Therefore, by (3.4), direct computation yields that z_{1} is a zero of order k_{1}−k(l+1)+l of the left-hand side of equality (3.4), so z_{1} is also a zero of order k_{1}−k(l+1)+l of zA^{k} −H_{(k−1)(k−1)}(A). Especially, k_{1} − k(l + 1) + l = k_{1} − (k^{2} − k + 1) for l = k − 1. Hence

(3.11) N(r, \frac{1}{w_{k−1}} ) ≤ (k^{2} −k +2)\overline{N} (r, \frac{1}{w_{k−1}} )+N(r, \frac{1}{zA^{k} − H_{(k−1)(k−1)}(A)}).

Case (3): i = k. Suppose that w_{k} has a zero of order k_{2} (≥ k^{2}+k+1) at point z_{2} , and z_{2} is also the pole of order t of A(z). We claim that 0 ≤ t ≤ k. In fact, if t ≥ k + 1, then z_{2} is a pole of order (k + 1)t of zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A), and z_{2} is also a pole of order (k + 1)t − k_{2} of the left-hand side in equality (3.5); this is a contradiction. Hence 0 ≤ t ≤ k. At this time, z_{2} is a pole of order tk + j of H_{kj}(A) for j = 0, 1, …, k. Therefore, by (3.5), direct computation yields that z_{2} is a zero of order k_{2} − k(t + 1) of the left-hand side of equality (3.5), so z_{2} is also a zero of order k_{2} − k(t + 1) of zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A). Especially, k_{1} − k(t + 1) = k_{1} − (k^{2} + k) for t = k. Hence (3.12)

N(r, \frac{1}{w_{k}} ) ≤ (k^{2} + k + 1)\overline{N} (r, \frac{1}{w_{k}} ) + N(r, \frac{1}{zA^{k+1} − H_{k(k−1)}(A) − zH_{kk}(A)}).

Note that for sufficiently large r, C log(rT (r, w_{i})) ≤ \frac{1}{2}T (r, w_{i}), i = 0, 1, …, k. Because w_{i}, i = 0, 1, …, k, are transcendental meromorphic functions and σ(A) = σ < +∞, so for any given ε > 0, it follows from (3.10)–(3.12) and (3.9) that

(3.13)  T (r, w_{i}) ≤ 2(k^{2} + k + 1)\overline{N} (r, \frac{1}{w_{i}} ) + O(r^{σ+ε}), i = 0, 1, …, k,

for r ∉ E_{2} sufficiently large, where E_{2} has finite linear measure.

By Definition 1.2, (3.1), (3.2) and (3.13), we obtain that for i = 0, 1, …, k, we have

(3.14)  \overline{τ}(f^{(i)}) = λ̄(w_{i}) = σ(w_{i}) = +∞, \overline{τ}_{2}(f^{(i)}) = \overline{λ_{2}}(w_{i}) = σ_{2}(w_{i}) = σ.

That is, each meromorphic solution f ≢ 0 of (1.5) and its f′, …, f^{(k)} all have infinitely many fixed points and satisfy (1.6). This completes the proof.

Proof of Theorem 1.1. (1) Suppose that f(z) is a transcendental meromorphic solution of (1.4). Then we know that the poles of f can only occur at the poles of P (z), so f has only finite poles. According to the Hadamard factorization theorem, we can write f as f(z) = g(z)/d(z), where g(z) is an entire function satisfying σ(g) = σ(f) and d(z) is a polynomial. By Lemma 2.1, for any ε > 0, there exists a sequence {r_{j}}, r_{j} → ∞, such that for all z satisfying |z| = r_{j}, |g(z)| = M(r_{j}, g), when j is large enough, we have

(3.15)  \frac{f^{(k)}(z)}{f(z)} = (\frac{v_{g}(r_{j})}{z})^{k}(1 + o(1)),
(3.16)  σ(f) = lim_{j→∞} \frac{log v_{g}(r_{j})}{log r_{j}}.

Substituting (3.15) into (1.4), we have

(3.17)  (\frac{v_{g}(r_{j})}{r_{j}})^{k}(1 + o(1)) = r_{j}^{n}(a + o(1)),

where a is a nonzero constant. From (3.16) and (3.17), we obtain that σ(f) = max{(n + k)/k, 0}.

(2): di(P) = n ≠ −k. Suppose that f ≢ 0 is a meromorphic solution of (1.4). Then f must be a transcendental meromorphic function. In fact, if not, then f is a rational function. We conclude that f^{(k)}(z)/f(z) = (b + o(1))/z^{k} as z → ∞ (refer to p. 62 in [4]). On the other hand, since di(P) = n ≠ −k and P (z) ≢ 0, we have that −P (z) = (b_{0} + o(1))/z^{−n} as z → ∞, where b_{0} ≠ 0; this contradicts f^{(k)}(z)/f(z) = −P (z) and −n ≠ k. Hence f(z) is a transcendental meromorphic function. From part (1), we have

(3.18)  σ(f) = max\{\frac{n + k}{k} , D\}.

Set w_{i} = f^{(i)} −z, i = 0, 1, …, k. Then for every i, a point z_{0} is a fixed point of f^{(i)} if and only if z_{0} is a zero of w_{i}, and

(3.19)  σ(w_{i}) = σ(f^{(i)}) = σ(f) = max\{\frac{n + k}{k} , D\}, \overline{τ}(f^{(i)}) = λ̄(w_{i}).

According to Lemmas 2.4–2.6, we know that w_{i}, i = 0, 1, …, k, satisfy the following equations:

(3.20)  ∑_{j=0}^{i} H_{ij}(P)w_{i}^{(k−j)} − P^{i+1}w_{i} = zP^{i+1}, i = 0, 1, …, k − 2;

(3.21) ∑_{j=0}^{k−1} H_{(k−1)j}(P)w_{k−1}^{(k−j)} − P^{k}w_{k−1} = zP^{k} − H_{(k−1)(k−1)}(P), if k ≥ 3;

(3.22) ∑_{j=0}^{k−1} H_{kj}(P)w_{k}^{(k−j)}+(H_{kk}(P)−P^{k+1})w_{k} = zP^{k+1}−H_{k(k−1)}(P)−zH_{kk}(P).

Since P ≢ 0, we conclude that P^{i+1}z ≢ 0 for i = 0, 1, …, k − 2. We claim that

zP^{k} − H_{(k−1)(k−1)}(P) ≢ 0 and zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≢ 0.

We discuss three subcases below.

Subcase I: di(P) = n ≥ 0. If zP^{k} − H_{(k−1)(k−1)}(P) ≡ 0, rewrite it as P = H_{(k−1)(k−1)}(P)/P^{k−1}z. Notice that P (z) is a rational function. We have that P^{(j)}(z)/P (z) → 0 as z → ∞. From this we obtain that H_{(k−1)(k−1)}(P)/zP^{k−1} → 0 as z → ∞, which contradicts that P (z) → ∞ or a nonzero constant. If zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≡ 0, rewrite it as

P = \frac{H_{k(k−1)}(P)}{zP^{k}} + \frac{H_{kk}(P)}{P^{k}}.

Similarly, we may obtain that

\frac{H_{k(k−1)}(P)}{zP^{k}} + \frac{H_{kk}(P)}{P^{k}} → 0 as z → ∞,

which contradicts that P (z) → ∞ or a nonzero constant; hence the claims hold.

Subcase II: When −k < di(P) = n ≤ −1, we have P (z) = (c_{0} + o(1))/z^{−n} as z → ∞, where c_{0} is a nonzero constant. From this, we have

(3.23)  zP^{k} = \frac{c_{0}^{k} + o(1)}{z^{−nk−1}}, zP^{k+1} = \frac{c_{0}^{k+1} + o(1)}{z^{−n(k+1)−1}} as z → ∞.

On the other hand, by (2.12)–(2.14), using a similar argument as in the proof of Lemma 2.7, we have

(3.24)  H_{(k−1)(k−1)}(P) = \frac{d_{1} + o(1)}{z^{−nk−1+k+n}} as z → ∞,

and

(3.25)  H_{k(k−1)}(P) + zH_{kk}(P) = \frac{d_{2} + o(1)}{z^{−n(k+1)−1+k+n}} as z → ∞,

where d_{1} , d_{2} are two constants (note that d_{1} or d_{2} may be zero). Hence it follows from (3.23)–(3.25) and k + n > 0 that

zP^{k} − H_{(k−1)(k−1)}(P) ≢ 0 and zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≢ 0,

that is, the claims hold.

Subcase III: n = di(P) < −k. By Lemma 2.7, there exist two nonzero constants c_{1} , c_{2} such that

(3.26)  H_{(k−1)(k−1)}(P) = \frac{c_{1} + o(1)}{z^{−nk−1+k+n}} as z → ∞,

Fixed points of meromorphic solutions 209

and

(3.27)  H_{k(k−1)}(P) + zH_{kk}(P) = \frac{c_{2} + o(1)}{z^{−n(k+1)−1+k+n}} as z → ∞.

Hence it follows from (3.23), (3.26), (3.27) and k + n < 0 that

zP^{k} − H_{(k−1)(k−1)}(P) ≢ 0 and zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≢ 0, that is, the claims hold.

Rewrite (3.20), (3.21) and (3.22) as

(3.28)  \frac{1}{zP^{i+1}} (∑_{j=0}^{i} H_{ij}(P)\frac{w_{i}^{(k−j)}}{w_{i}} − P^{i+1}) = \frac{1}{w_{i}} , i = 0, 1, …, k − 2,
(3.29)  \frac{1}{zP^{k} − H_{(k−1)(k−1)}(P)}(∑_{j=0}^{k−1} H_{(k−1)j}(P)\frac{w_{k−1}^{(k−j)}}{w_{k−1}} − P^{k}) = \frac{1}{w_{k−1}} ,

(3.30)

\frac{1}{zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P)}(∑_{j=0}^{k−1} H_{kj}(P)\frac{w_{k}^{(k−j)}}{w_{k}} + H_{kk}(P) − P^{k+1}) = \frac{1}{w_{k}} .

Hence there exists a set E with finite linear measure such that for r ∉ E, and we have

(3.31)  m(r, \frac{1}{w_{i}} ) ≤ O(m(r, \frac{1}{P} )) + C(log rT (r, w_{i})), i = 0, 1, …, k.

Applying similar arguments as in the proof of Theorem 1.2, we obtain that

(3.32)  N(r, \frac{1}{w_{i}} ) ≤ k\overline{N} (r, \frac{1}{w_{i}} ) + N(r, \frac{1}{zP^{i+1}} ), i = 0, 1, …, k − 2;
(3.33) N(r, \frac{1}{w_{k−1}} ) ≤ (k^{2} −k+2)\overline{N} (r, \frac{1}{w_{k−1}} )+N(r, \frac{1}{zP^{k} − H_{(k−1)(k−1)}(P)});

(3.34)

N(r, \frac{1}{w_{k}} ) ≤ (k^{2} + k + 1)\overline{N} (r, \frac{1}{w_{k}} ) + N(r, \frac{1}{zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P)}).

Note that for sufficiently large r, C log(rT (r, w_{i})) ≤ \frac{1}{2}T (r, w_{i}), i = 0, 1, …, k. Because w_{i}, i = 0, 1, …, k, are transcendental meromorphic functions and P is a rational function, so for any given ε > 0, from (3.31)–(3.34), we have

T (r, w_{i}) ≤ 2(k^{2} + k + 1)\overline{N} (r, \frac{1}{w_{i}} ) + O(log r), i = 0, 1, …, k,

for r ∉ E_{3} sufficiently large, where E_{3} has finite linear measure.

By (3.19), we obtain that for i = 0, 1, …, k, f^{(i)}(z) all have infinitely many fixed points and

(3.35)  \overline{τ}(f^{(i)}) = λ̄(w_{i}) = σ(w_{i}) = max\{\frac{n + k}{k} , 0\}.

That is, each meromorphic solution f ≢ 0 of (1.4) and its f′, …, f^{(k)} all have infinitely many fixed points and satisfy (3.35).

Case (3): di(P) = n = −k. Suppose that f(z) is a transcendental meromorphic solution of (1.4). Set w_{i}(z) = f^{(i)}(z) − z, i = 0, 1, …, k − 2, since P^{i+1}z ≢ 0 for i = 0, 1, …, k − 2. Using similar arguments as in part (2), we may prove that for any given ε > 0,

(3.36)  T (r, w_{i}) ≤ 2k\overline{N} (r, \frac{1}{w_{i}} ) + O(log r), i = 0, 1, …, k − 2,

for r ∉ E_{4} sufficiently large, where E_{4} has finite linear measure.

Note that w_{i}, i = 0, 1, …, k − 2, are transcendental meromorphic functions. It follows from (3.36) that f and f′, f′′, …, f^{(k−2)} all have infinitely many fixed points and

\overline{τ}(f) = \overline{τ}(f′) = ⋯ = \overline{τ}(f^{(k−2)}) = σ(w_{i}) = max\{\frac{n + k}{k} , 0\} = 0.

Hence the proof of Theorem 1.1 is complete.

Remark 3.1. When n = −k in Theorem 1.1, we cannot prove that \overline{τ}(f^{k−1}) = \overline{τ}(f^{(k)}) = σ(f) by using similar arguments as in part (2) of the proof of Theorem 1.1. Since we cannot prove that zP^{k} − H_{(k−1)(k−1)}(P) ≢ 0 and zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≢ 0 in (3.21) and (3.22), respectively. In fact, if n = −k, there exist some equations of the form (1.4) such that zP^{k}−H_{(k−1)(k−1)}(P) ≡ 0 or zP^{k+1} − H_{k(k−1)}(P) − zH_{kk}(P) ≡ 0. For example, when k = 2, by (2.12)–(2.14), simple computation yields

zP^{2} − H_{11}(P) = zP^{2} − P′ ≡ 0 for P (z) = \frac{−2}{z^{2}},

and

Fixed points of meromorphic solutions 211

zP^{3} − H_{21}(P) − zH_{22}(P) = zP^{3} − 2P P′ + 2z(P′)^{2} − zP P′′ ≡ 0 for P (z) = \frac{−6}{z^{2}}.

When k = 3, simple computation yields zP^{3}−H_{22}(P) = zP^{3}−P P′′+2P^{′2} ≡ 0 for P (z) = −6/z^{3} , and for P (z) = −24/z^{3} , we have

\begin{matrix}zP^{4} − H_{32}(P) − zH_{33}(P) = zP^{4} − 3P (P P′′ − 2P^{′2}) \\ + 6zP P′P′′ − zP^{2}P′′′ − 6zP^{′3} ≡ 0. \end{matrix}

Acknowledgements. The authors thank the referees for their helpful comments and suggestions to improve our manuscript.

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Received 30 December 2004