Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 63, 1-18; http://www.math.u-szeged.hu/ejqtde/

Oscillation of solutions of some higher order linear differential equations^{∗}

Hong-Yan XU^{a}^{†} and Ting-Bin CAO^{b}

aDepartment of Informatics and Engineering, Jingdezhen Ceramic Institute (Xianghu Xiaoqu),

Jingdezhen, Jiangxi, 333403, China <e-mail:xhyhhh@126.com> bDepartment of Mathematics, Nanchang University Nanchang, Jiangxi 330031, China <e-mail: tbcao@ncu.edu.cn or ctb97@163.com>

Abstract

In this paper, we deal with the order of growth and the hyper order of solutions of higher order linear differential equations

f^{(k)} + B_{k−1}f^{(k−1)} + ⋯ + B_{1}f′ + B_{0}f = F

where B_{j}(z) (j = 0, 1, …, k − 1) and F are entire functions or polynomials. Some results are obtained which improve and extend previous results given by Z.-X. Chen, J. Wang, T.-B. Cao and C.-H. Li. Key words: linear differential equation; growth order; entire function. Mathematical Subject Classification (2000): 34M10, 30D35.

1 Introduction and Main Results

We shall assume that reader is familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions(see [11,14]). In addition, we will use the notation σ(f) to denote the order of growth of entire function f(z), σ_{2}(f) to denote the hyper-order of f(z), λ(f)(λ_{2}(f)) to denote the exponent(hyper-exponent) of convergence of the zero-sequence of f(z) and \overline{λ}(f)(\overline{λ}_{2}(f)) to denote exponent(hyper-exponent) of convergence of distinct zero sequence of meromorphic function f(z). We also define

\overline{λ}(f − φ) = lim sup_{r→∞} \frac{log \overline{N}(r, \frac{1}{f−φ})}{log r}, and \overline{λ}_{2}(f − φ) = lim sup_{r→∞} \frac{log log \overline{N}(r, \frac{1}{f−φ})}{log r},

for any meromorphic function φ(z).

∗This research was supported by the NSF of Jiangxi of China (2008GQS0075).

†Corresponding author

For a set E ⊂ R^{+}, let m(E), respectively m_{l}(E), denote the linear measure, respectively the logarithmic measure of E. By χ_{E}(t), we denote the characteristic function of E. Moreover, the upper logarithmic density and the lower logarithmic density of E are defined by

\overline{log dens}(E) = lim sup_{r→∞} \frac{m_{l}(E ∩ [1, r])}{log r}, \underline{log dens}(E) = lim inf_{r→∞} \frac{m_{l}(E ∩ [1, r])}{log r}.

Observe that E may have a different meaning at different occurrences in what follows. We now recall some previous results concerning linear differential equations

f′′ + e^{−z}f′ + Q(z)f = 0, (1)

where Q(z) is an entire function of finite order. It is well known that each solution f of (1) is an entire function and that if f_{1} and f_{2} are any two linearly independent solutions of (1), then at least one of f_{1}, f_{2} must have infinitely order(see [13, P167-168]). Hence, ”most” solutions of (1) will have infinite order. But the equation (1) with Q(z) = −(1 + e^{−z}) possesses a solution f = e^{z} of finite order.

Thus a natural question is: what condition on Q(z) will guarantee that every solution f ≢ 0 of (1) has infinite order? Many authors, such as Amemiya and Ozawa [1], Gundersen [10] and Langley [15], Frei [6], Ozawa [20] have studied the problem. They proved that when Q(z) is a nonconstant polynomial or Q(z) is a transcendental entire function with order σ(Q) ≠ 1, then every solution f ≢ 0 of (1) has infinite order.

For the above question, some mathematicians investigated the second order linear differential equations and obtained many results (see REF.[2,3,5,6,10,15,16,20,24]). In 2002, Chen [3] considered the question: what condition on Q(z) when σ(Q) = 1 will guarantee every nontrivial solution of (1) has infinite order? He proved the following result, which greatly extended and improved results of Frei, Ozawa, Langley and Gundersen.

Theorem A(see. [3]) Let A_{j}(z)(≢ 0)(j = 0, 1) be an entire function with σ(A_{j}) < 1. Suppose a, b are complex constants such that ab ≠ 0 and a = cb(c > 1). Then every nontrivial solution f of

f′′ + A_{1}(z)e^{az}f′ + A_{0}(z)e^{bz}f = 0 (2)

has infinite order.

Recently, some mathematicians investigate the non-homogeneous equations of second order and higher order linear equations such as Li and Wang [18], Cao [5], Wang and Laine [22] and proved that every solution of these equation has infinite order.

In 2008, Li and Wang [18] investigated the non-homogeneous equation related to (1) in the case when Q(z) = h(z)e^{bz}, where h(z) is a transcendental entire function of order σ(h) < \frac{1}{2}, and b is a real constant and obtained the following results.

Theorem B(see. [18]) If Q(z) = h(z)e^{bz}, where h(z) is a transcendental entire function of order σ(h) < \frac{1}{2}, and b is a real constant. Then all nontrivial solutions f of equation

f′′ + e^{−z}f′ + Q(z)f = H(z)

satisfies σ(f) = \overline{λ}(f − z) = ∞, provided that σ(H) < 1.

In 2008, Wang and Laine [22] investigated the non-homogeneous equation related to (2) and obtained the following result.

Theorem C(see. [22, Theorem 1.1]) Suppose that A_{j} ≢ 0(j = 0, 1), H are entire functions of order less than one, and the complex constants a, b satisfy ab ≠ 0 and a ≠ b. Then every nontrivial solution f of equation

f′′ + A_{1}(z)e^{az}f′ + A_{0}(z)e^{bz}f = H(z)

is of infinite order.

For equation (2), Li and Huang [17], Tu and Yi [21], Chen and Shon [4] and Gan and Sun [7] investigated the higher order homogeneous and non-homogeneous linear differential equations and obtained many results. In 2009, Chen and Xu [23] investigated the higher order non-homogeneous linear differential equations and obtained the following result.

Theorem D(see. [23, Theorem 1.5]) Let k ≥ 2, s ∈ {1, …, k−1}, h_{0} ≢ 0, h_{1}, …, h_{k−1} be meromorphic functions and σ = max{σ(h_{j}) : j = 1, …, k − 1} < n; P (z) = a_{n}z^{n} + a_{n−1}z^{n−1} + ⋯ + a_{1}z + a_{0} and Q(z) = b_{n}z^{n} + b_{n−1}z^{n−1} + ⋯ + b_{1}z + b_{0} be two nonconstant polynomials, where a_{i}, b_{i}(i = 0, 1, 2, …, n) with a_{n} ≠ 0, b_{n} ≠ 0; F ≢ 0 be an meromorphic function of finite order. Suppose all poles of f are of uniformly bounded multiplicity and if at least one of the following statements hold

1. If a_{n} = b_{n}, and deg(P − Q) = m ≥ 1, σ < m;

2. If a_{n} = cb_{n} with c > 1, and deg(P − Q) = m > 1, σ < m;

3. If σ < σ(h_{0}) < 1/2, a_{n} = cb_{n} with c ≥ 1 and P (z) − cQ(z) is a constant,

then all solutions f of non-homogeneous linear differential equation

f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{s}e^{P (z)}f^{(s)} + ⋯ + h_{1}f′ + h_{0}e^{Q(z)}f = F, (3)

with at most one exceptional solution f_{0} of finite order, satisfy

λ(f) = \overline{λ}(f) = σ(f) = ∞, λ_{2}(f) = \overline{λ}_{2}(f) = σ_{2}(f).

Furthermore, if such an exceptional solution f_{0} of finite order of (1.3) exists, then we have

σ(f_{0}) ≤ max\{n, σ(F), \overline{λ}(f_{0})\}.

We find that there is an exceptional possible solution with finite order for equation (3). It is natural to ask the following question: what condition on the coefficients of equation f^{(k)} + B_{k−1}(z)f^{(k−1)} + ⋯ + B_{1}(z)f′ + B_{0}(z)f = F (4)

when F ≢ 0 will guarantee every nontrivial solution has infinite order?

The main purpose of this paper is to study the above problem and the relation between small functions and solutions of higher order linear differential equation related to (4). We will prove the following results.

Theorem 1.1 Let P (z) and Q(z) be a nonconstant polynomials as above, for some complex numbers a_{i}, b_{i}, (i = 0, 1, …, n) with a_{n}b_{n} ≠ 0 and a_{n} ≠ b_{n}. Suppose that h_{i−1}(2 ≤ i ≤ k − 1) are polynomials of degree no more n − 1 in z, A_{j}(z) ≢ 0 (j = 0, 1) and H(z) are entire functions satisfying σ := max{σ(A_{j}), j = 0, 1} < n and σ(H) < n, and φ(z) is an entire function of finite order. Then every nontrivial solution f of equation

f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′ + A_{1}e^{P (z)}f′ + A_{0}e^{Q(z)}f = H (5)

satisfies σ(f) = ∞ , σ(f) = λ(f) = \overline{λ}(f) = \overline{λ}(f − φ) = ∞ and σ_{2}(f) = λ_{2}(f) = \overline{λ}_{2}(f) = \overline{λ}_{2}(f − φ) ≤ n.

Remark 1.1 We can see that the conclusions of Theorem 1.1 improve Theorem D and extend Theorem B and Theorem C.

Theorem 1.2 Suppose that A_{j}(z) ≢ 0, D_{j}(z)(j = 0, 1), and H(z) are entire functions satisfying σ(A_{j}) < n, σ(D_{j}) < n(j = 0, 1), and σ(H) < n, and P (z), Q(z), h_{i−1}(2 ≤ i ≤ k − 1) are as in Theorem 1.1 satisfying a_{n}b_{n} ≠ 0 and a_{n}b_{n} < 0. Then every nontrivial solution f of equation

f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′ + (A_{1}e^{P (z)} + D_{1})f′ + (A_{0}e^{Q(z)} + D_{0})f = H (6)

is of infinite order.

Remark 1.2 From Theorem 1.1 and Theorem 1.2, we give an answer to the above question.

2 Some Lemmas

To prove the theorems, we need the following lemmas:

Lemma 2.1 (see. [24, Lemma 1.10]) Let f_{1}(z) and f_{2}(z) be nonconstant meromorphic functions in the complex plane and c_{1}, c_{2}, c_{3} be nonzero constants. If c_{1}f_{1} + c_{2}f_{2} ≡ c_{3}, then

T (r, f_{1}) < \overline{N} (r, \frac{1}{f_{1}} ) + \overline{N} (r, \frac{1}{f_{2}} ) + \overline{N}(r, f_{1}) + S(r, f_{1}).

Lemma 2.2 (see. [3,19]) Suppose that P (z) = (α + βi)z^{n} + ⋯ (α, β are real numbers, |α| + |β| ≠ 0) is a polynomial with degree n ≥ 1, that A(z)(≢ 0) is an entire function with σ(A) < n. Set g(z) = A(z)e^{P (z)}, z = re^{iθ}, δ(P, θ) = α cos nθ − β sin nθ. Then for any given ε > 0, there exists a set H_{1} ⊂ [0, 2π) that has the linear measure zero, such that for any θ ∈ [0, 2π)\(H_{1} ∪ H_{2}), there is R > 0 such that for |z| = r > R, we have:

(i) If δ(P, θ) > 0, then

exp{(1 − ε)δ(P, θ)r^{n}} < |g(re^{iθ})| < exp{(1 + ε)δ(P, θ)r^{n}};

(ii) If δ(P, θ) < 0, then

exp{(1 + ε)δ(P, θ)r^{n}} < |g(re^{iθ})| < exp{(1 − ε)δ(P, θ)r^{n}},

where H_{2} = {θ ∈ [0, 2π); δ(P, θ) = 0} is a finite set.

Lemma 2.3 (see. [9]) Let f(z) be a transcendental meromorphic function of finite order σ(f) = σ < ∞, and let ε > 0 be a given constant. Then there exists a set H ⊂ (1, ∞) that has finite logarithmic measure, such that for all z satisfying |z| ∉ H ∪ [0, 1] and for all k, j, 0 ≤ j < k, we have

|\frac{f^{(k)}(z)}{f^{(j)}(z)}| ≤ |z|^{(k−j)(σ−1+ε)}.

Similarly, there exists a set E ⊂ [0, 2π) of linear measure zero such that for all z = re^{iθ}with |z| sufficiently large and θ ∈ [0, 2π) \ E, and for all k, j, 0 ≤ j < k, we have

|\frac{f^{(k)}(z)}{f^{(j)}(z)}| ≤ |z|^{(k−j)(σ−1+ε)}.

Lemma 2.4 (see. [22, Lemma 2.4]) Let f(z) be an entire function of finite order σ, and M(r, f) = f(re^{iθ_{r}}) for every r. Given ζ > 0 and 0 < C(σ, ζ) < 1, there exists a constant 0 < l_{0} < \frac{1}{2} and a set E_{ζ} of lower logarithmic density greater than 1 − ζ such that

e^{−5π}M(r, f)^{1−C(σ,ζ)} ≤ |f(re^{iθ})|

for all r ∈ E_{ζ} large enough and all θ such that |θ − θ_{r}| ≤ l_{0}.

Lemma 2.5 (see. [8,12]) Let f(z) be a transcendental entire function, ν_{f}(r) be the central index of f(z) and δ be a constant satisfying 0 < δ < \frac{1}{8}. Suppose z lying in the circle |z| = r satisfies |f(z)| > M(r, f)ν_{f}(r)^{− \frac{1}{8} +δ}. Then except a set of r with finite logarithmic measure, we have

\frac{f^{(j)}(z)}{f(z)} = \{\frac{ν_{f}(r)}{z}\}^{j} (1 + η_{j}(z)),

where η_{j}(z) = O(ν_{f}(r)^{− \frac{1}{8} +δ}), j ∈ N.

Lemma 2.6 (see. [22, Lemma 2.5]) Let f(z) and g(z) be two nonconstant entire functions with σ(g) < σ(f) < +∞. Given ε with 0 < 4ε < σ(f) − σ(g) and 0 < δ < \frac{1}{8}, there exists a set E with \overline{log dens}(E) > 0 and a positive constant r_{0} such that

|\frac{g(z)}{f(z)}| ≤ exp\{−r^{σ(f)−2ε}\}

for all z such that r ∈ E is sufficiently large and that |f(z)| ≥ M(r, f)ν_{f}(r)^{− \frac{1}{8} +δ}.

Lemma 2.7 (see. [12]) Let f(z) be an entire function of finite order σ(f) = σ < ∞, and let ν_{f}(r) be the central index of f. Then for any ε(> 0), we have

lim sup_{r→∞} \frac{log ν_{f}(r)}{log r} = σ.

Lemma 2.8 (see. [16]) Let f(z) be an entire function of infinite order. Denote M(r, f) = max{|f(z)| : |z| = r}, then for any sufficiently large number λ > 0, and any r ∈ E ⊂ (1, ∞)

M(r, f) > c_{1} exp{c_{2}r^{λ}},

where m_{l}E = ∞ and c_{1}, c_{2} are positive constants.

Lemma 2.9 Suppose B_{0}, B_{1}, …, B_{k−1} and F (≢ 0) are all entire functions of finite order and let ϱ := max{σ(B_{j}), σ(F), j = 0, 1, …, k − 1}, k ≥ 2. Then every solution f of infinite order of equation

f^{(k)} + B_{k−1}f^{(k−1)} + ⋯ + B_{0}f = F

satisfies σ_{2}(f) ≤ ϱ.

Proof: We rewrite the equation as

\frac{f^{(k)}}{f} = \frac{F}{f} − ∑_{j=1}^{k−1} B_{j} \frac{f^{(j)}}{f} − B_{0}.

Since ϱ := max{σ(B_{j}), σ(F), j = 0, 1, …, k −1}, by virtue of [2], for any positive number ε(0 < ε < σ(F) + 1) and r ∉ [0, 1] ∪ E_{1}, we have

|B_{j}(z)| ≤ exp{r^{ϱ+ε}}, |F (z)| ≤ exp{r^{σ(F)+ε}}, j = 0, 1, …, k − 1.

By Lemma 2.5, there exists a set E_{2} ⊂ (1, +∞) satisfying m_{l}E_{2} < ∞, taking z satisfying

\frac{f^{(j)}(z)}{f(z)} = (\frac{ν_{f}(r)}{z})^{j} (1 + o(1)) (j = 0, 1, …, k).

Since σ(f) = ∞ , from Lemma 2.8 there exists |z| = r ∈ H_{1} \ ([0, 1] ∪ E_{1} ∪ E_{2}) satisfying |f(z)| = M(r, f), for λ > 2σ(F) + 1, we have

(\frac{ν_{f}(r)}{|z|})^{k} (1+o(1)) ≤ \frac{1}{c_{1}} exp\{r^{σ(F)+ε}−c_{2}r^{λ}\}+exp\{r^{ϱ+ε}\} (∑_{j=1}^{k−1} (\frac{ν_{f}(r)}{|z|})^{j} (1 + o(1)) + 1) .

Thus, we have

lim sup_{r→∞,r∈H_{1}\([0,1]∪E_{1}∪E_{2})} \frac{log log ν_{f}(r)}{log r} ≤ ϱ + ε.

By the definition of hyper-order, we can get σ_{2}(f) ≤ ϱ.

Therefore, we complete the proof of this lemma. □

3 The Proof of Theorem 1.1

Proof: The growth of solutions We first point out that σ(f) ≥ n.

We rewrite (5) as

A_{1}e^{P (z)}f′ + A_{0}e^{Q(z)}f = H − (f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′) . (7)

If H − (f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′) ≡ 0, by a_{n} ≠ b_{n}, we have

f = K exp \{∫ \frac{A_{0}}{A_{1}} e^{Q(z)−P (z)}dz\} ,

where K is a nonzero constant. If H − (f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′) ≢ 0, rewrite (7) as

\frac{A_{1}e^{P (z)}f′ + A_{0}e^{Q(z)}f}{H − (f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′)} ≡ 1.

Suppose σ(f) < n, then by Lemma 2.1, we can get T (r, e^{P (z)}) = S(r, e^{P (z)}), which is a contradiction.

By Lemma 2.5, for any given 0 < δ < \frac{1}{8}, there exists a set E_{1} of finite logarithmic measure such that

\frac{f^{(j)}(z)}{f(z)} = (\frac{ν_{f}(r)}{z})^{j} (1 + o(1)), j = 1, 2, …, k, (8)

where |f(z)| ≥ M(r, f)ν_{f}(r)^{− \frac{1}{8} +δ}, r ∉ E_{1}. Furthermore, from the definition of the central index, we know that ν_{f}(r) → ∞ as r → ∞. By Lemma 2.7, we have

ν_{f}(r) ≤ r^{σ(f)+1}, (9)

for all r sufficiently large. By Lemma 2.3, we have

|\frac{f^{(j)}(z)}{f(z)}| ≤ |z|^{j(σ(f)−1+ε)}, j = 1, 2, …, k, (10)

for all z satisfying |z| = r ∉ E_{2} where m_{l}(E_{2}) < ∞, and ε is any given constant with 0 < 4ε < min{1, n − σ(H), n − σ, n − t}, where t = max{t_{j} = deg(h_{i}(z)), 2 ≤ i ≤ k − 1}. By Lemma 2.6, there is a set E_{3} with ζ = \overline{log dens}E_{3} > 0 such that

\frac{ν_{f}(r)^{\frac{1}{8} −δ}|H(z)|}{M(r, f)} ≤ exp\{−r^{n−2ε}\}, (11)

when r ∈ E_{3} is large enough. We may take θ_{p} such that M(r, f) = |f(re^{iθ_{p}})| for every p. By Lemma 2.4, given a constant 0 < C < 1, there exists a constant l_{0} and a set E_{4} with 1 − \frac{ζ}{2} ≤ \underline{log dens}(E_{4}) such that

e^{−5π}M(r, f)^{1−C} ≤ |f(re^{iθ})| (12) for all r ∈ E_{4} and |θ − θ_{p}| ≤ l_{0}. Since the characteristic functions of E_{3} and E_{4} satisfy the relation

χ_{E_{3}∩E_{4}}(t) = χ_{E_{3}}(t) + χ_{E_{4}}(t) − χ_{E_{3}∪E_{4}}(t).

Then \overline{log dens}(E_{3} ∪ E_{4}) ≤ 1. Thus, we can get

\frac{ζ}{2} ≤ \overline{log dens}E_{3} + \underline{log dens}(E_{4}) − \overline{log dens}(E_{3} ∪ E_{4}) ≤ \overline{log dens}(E_{3} ∩ E_{4}).

Since m_{l}(E_{1} ∪ E_{2}) < ∞, we have \overline{log dens}((E_{3} ∩ E_{4}) \ (E_{1} ∪ E_{2})) > 0. Thus, there exists a sequence of points z_{q} = r_{q}e^{iθ_{q}} with r_{q} ↑ ∞ and

|f(z_{q})| = M(r_{q}, f), r_{q} ∈ (E_{3} ∩ E_{4}) \ (E_{1} ∪ E_{2}).

Passing to a sequence of {θ_{q}}, we may assume that lim_{q→∞} θ_{q} = θ_{0} in this paper. We now take the three cases as follows into consideration. Case 1. δ(P, θ_{0}) > 0. From the continuity of δ(P, θ), we have

\frac{1}{3}δ(P, θ_{0}) < δ(P, θ_{q}) < \frac{4}{3}δ(P, θ_{0}) (13)

for sufficiently large q. By Lemma 2.2, we can get

exp \{\frac{1 − ε}{3}δ(P, θ_{0})r_{q}^{n}\} < |A_{1}(z_{q})e^{P (z_{q})}| < exp \{\frac{4(1 + ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (14)

for all q sufficiently large. From (7) we can get

\begin{aligned}|\frac{f′(z_{q})}{f(z_{q})} + \frac{A_{0}(z_{q})}{A_{1}(z_{q})}e^{Q(z_{q})−P (z_{q})}| ≤ & |\frac{e^{−P (zq)}}{A_{1}(z_{q})} | (|\frac{f^{(k)}(z_{q})}{f(z_{q})} | + ∑_{j=2}^{k−1} |h_{j} \frac{f^{(j)}(z_{q})}{f(z_{q})} | \\ & +\frac{|H(z_{q})|}{M(r_{q},f)}) \end{aligned} (15)

We divide the proof in Case 1 in three subcases in the following.

Subcase 1.1. We first assume that θ_{0} satisfies ξ := δ(Q − P, θ_{0}) > 0. From the continuity of δ(Q − P, θ_{0}) and Lemma 2.2, we have

\frac{1}{3}δ(Q − P, θ_{0}) ≤ δ(Q − P, θ_{q}) ≤ \frac{4}{3}δ(Q − P, θ_{0}) (16)

for sufficiently large q. Similar to (14), we have

exp \{\frac{1 − ε}{3}ξr_{q}^{n}\} < |\frac{A_{0}(z_{q})}{A_{1}(z_{q})}e^{Q(z_{q})−P (z_{q})}| < exp \{\frac{4(1 + ε)}{3}ξr_{q}^{n}\} , (17)

for sufficiently large q. Substituting (8)-(11) to (15), for sufficiently large q, we can get

\begin{aligned}& |\frac{ν_{f} (r_{q})}{z_{q}}(1 + o(1)) + \frac{A_{0}(z_{q})}{A_{1}(z_{q})}e^{Q(z_{q})−P (z_{q})}| \\ ≤ & |\frac{e^{−P (zq)}}{A_{1}(z_{q})} | (2r_{q}^{kσ(f)} + Σ_{j=2}^{k}|z_{q}|^{j(σ(f)−1+ε)}r_{q}^{t_{j}+ε} + exp\{−r_{q}^{n−2ε}\}) \\ ≤ & |\frac{e^{−P (zq)}}{A_{1}(z_{q})} | r_{q}^{kσ(f)+t+ε}. \end{aligned} (18)

By (14) and 0 < 4ε < min{1, n − σ, n − σ(H), n − t}, we have

|\frac{e^{−P (z_{q})}}{A_{1}(z_{q})} | r_{q}^{kσ+t+ε} ≤ \frac{r_{q}^{kσ+t+ε}}{exp\{\frac{(1−ε)}{3}δ(P, θ_{0})r_{q}^{n}\}} ≤ exp\{−\frac{(1 − 2ε)}{3}δ(P, θ_{0})r_{q}^{n}\}. (19)

From (17)-(19) and (9), we can obtain

\begin{aligned}exp \{\frac{(1−ε)}{3}ξr_{q}^{n}\} & ≤ |\frac{ν_{f} (r_{q})}{z_{q}}(1 + o(1)) + \frac{A_{0}(z_{q})}{A_{1}(z_{q})}e^{Q(z_{q})−P (z_{q})} − \frac{ν_{f} (r_{q})}{z_{q}}(1 + o(1))| \\ & ≤ exp\{−\frac{(1−2ε)}{3}δ(P, θ_{0})r_{q}^{n}\} + 2r_{q}^{σ(f)} ≤ 3r_{q}^{σ(f)}. \end{aligned} (20)

Thus, we can get a contradiction.

Subcase 1.2. ξ := δ(Q − P, θ_{0}) < 0. Then from Lemma 2.2, for sufficiently large q, we have

exp \{\frac{4(1 + ε)}{3}ξr_{q}^{n}\} ≤ |\frac{A_{0}(z_{q})}{A_{1}(z_{q})}e^{Q(z_{q})−P (z_{q})}| ≤ exp \{\frac{(1 − ε)}{3}ξr_{q}^{n}\} . (21)

From (21) and similar to (20), we can get

\frac{ν_{f}(r_{q})}{r_{q}}(1 + o(1)) ≤ exp \{\frac{(1 − ε)}{3}ξr_{q}^{n}\} + exp \{−\frac{(1 − 2ε)}{3}δ(P, θ_{0})r_{q}^{n}\} ,

when q is large enough. Thus, we can get that ν_{f}(r_{q}) → 0 as q → ∞, which is impossible.

Subcase 1.3. ξ := δ(Q − P, θ_{0}) = 0. From (12), we may construct another sequence of points z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} with lim_{q→∞} θ_{q}^{∗} = θ_{0}^{∗} such that ξ_{1} := δ(Q − P, θ_{0}^{∗}) > 0. Without loss of generality, we may suppose that

δ(Q − P, θ) > 0, θ ∈ (θ_{0} + 2kπ, θ_{0} + (2k + 1)π),
δ(Q − P, θ) < 0, θ ∈ (θ_{0} + (2k − 1)π, θ_{0} + 2kπ),

which k ∈ ℤ. When q is large enough, we have |θ_{0} − θ_{q}| ≤ l_{0}. Choose θ_{q}^{∗} such that \frac{l_{0}}{2} ≤ θ_{q}^{∗} − θ_{q} ≤ l_{0}, i.e., θ_{q} + \frac{l_{0}}{2} ≤ θ_{q}^{∗} ≤ θ_{q} + l_{0}, then

θ_{0} + \frac{l_{0}}{2} ≤ θ_{0}^{∗} ≤ θ_{0} + l_{0}. (22)

For sufficiently large q, we can get (12) for z_{q}^{∗}, and ξ_{1} := δ(Q − P, θ_{0}^{∗}) > 0. Hence we can get

|\frac{H(z_{q}^{∗})}{f(z_{q}^{∗})}| ≤ \frac{ν_{f}(r)^{\frac{1}{8} −δ}M(r_{q}, H)}{e^{−5π}M(r_{q}, f)^{1−C}}, (23)

and

exp \{\frac{(1 − ε)}{3}ξ_{1}r_{q}^{n}\} ≤ |\frac{A_{0}(z_{q}^{∗})}{A_{1}(z_{q}^{∗})}e^{Q(z_{q}^{∗})−P (z_{q}^{∗})}| ≤ exp \{\frac{4(1 + ε)}{3}ξ_{1}r_{q}^{n}\} . (24)

By virtue of [22], we may assume that M(r_{q}, f) ≥ exp{r_{q}^{σ(f)−ε}}. From the above argument, we can know that z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} satisfies (9). Then from (23) and for all large enough q, we have

(25)  |\frac{H(z_{q}^{∗})}{f(z_{q}^{∗})}| ≤ r_{q}^{(σ(f)+1)( \frac{1}{8} −δ)} \frac{exp\{r_{q}^{σ(H)+ε}\}}{exp\{r_{q}^{σ(f)− \frac{3}{2} ε}\}} ≤ exp\{−r_{q}^{n−2ε}\}.

Taking now l_{0} small enough, we have δ(P, θ_{0}^{∗}) > 0 by the continuity of δ(P, θ). Thus, we have

exp \{\frac{1 − ε}{3}δ(P, θ_{0}^{∗})r_{q}^{n}\} < |A_{1}(z_{q}^{∗})e^{P (z_{q}^{∗})}| < exp \{\frac{4(1 + ε)}{3}δ(P, θ_{0}^{∗})r_{q}^{n}\} . (26)

Substituting (10) and (25) into (15) and by (26), we have

|\frac{A_{0}(z_{q}^{∗})}{A_{1}(z_{q}^{∗})}e^{Q(z_{q}^{∗})−P (z_{q}^{∗})}| ≤ exp \{−\frac{(1 − 3ε)}{3}δ(P, θ_{0}^{∗})r_{q}^{n}\} + 2r_{q}^{σ(f)} ≤ 3r_{q}^{σ(f)}. (27)

Combining (27) with (24), for large enough q, we can get a contradiction easily.

Case 2. Suppose that δ(P, θ_{0}) < 0. Then from the continuity of δ(P, θ) and Lemma 2.2, we have

exp \{\frac{4(1 + ε)}{3}δ(P, θ_{0})r_{q}^{n}\} ≤ |A_{1}(z_{q})e^{P (z_{q})}| ≤ exp \{\frac{(1 − ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (28)

for all sufficiently large q. From (5), we can get

\begin{aligned}|\frac{f^{(k)}(z_{q})}{f(z_{q})} + ∑_{j=2}^{k−1} h_{j}(z_{q})\frac{f^{(j)}(z_{q})}{f(z_{q})} & + & A_{0}(z_{q})e^{Q(z_{q})}| \\ & ≤ & |A_{1}(z_{q})e^{P (z_{q})}| |\frac{f′(z_{q})}{f(z_{q})} | + \frac{|H(z_{q})|}{M(r_{q},f)} \end{aligned} (29)

as q → ∞. Again, we divide the proof in Case 2 in three subcases in the following.

Subcase 2.1. δ(Q, θ_{0}) > 0. From the continuity of δ(Q, θ) and Lemma 2.2, for large enough q, we have

exp \{\frac{(1 − ε)}{3}δ(q, θ_{0})r_{q}^{n}\} ≤ |A_{0}(z_{q})e^{Q(z_{q})}| ≤ exp \{\frac{4(1 + ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} . (30)

Substituting (8)-(11) and (28) into (29), we get

|\frac{f^{(k)}(z_{q})}{f(z_{q})} + ∑_{j=2}^{k−1} h_{j}(z_{q})\frac{f^{(j)}(z_{q})}{f(z_{q})} + A_{0}(z_{q})e^{Q(z_{q})}| ≤ exp \{−r_{q}^{n−3ε}\} . (31)

From (8)-(11),(30),(31) and enough large q, we have

exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ |\frac{f^{(k)}(z_{q})}{f(z_{q})} + ∑_{j=2}^{k−1} h_{j}(z_{q})\frac{f^{(j)}(z_{q})}{f(z_{q})} + A_{0}(z_{q})e^{Q(z_{q})}

− (\frac{f^{(k)}(z_{q})}{f(z_{q})} + ∑_{j=2}^{k−1} h_{j}(z_{q})\frac{f^{(j)}(z_{q})}{f(z_{q})})|

i.e.,

exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ exp \{−r_{q}^{n−3ε}\} + r_{q}^{kσ(f)+t+ε}.

Thus, we can get a contradiction.

Subcase 2.2. δ(Q, θ_{0}) < 0. By the continuity of δ(Q, θ) and Lemma 2.2, for all sufficiently large q, we have

exp \{\frac{4(1 + ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ |A_{0}(z_{q})e^{Q(z_{q})}| ≤ exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} . (32)

From (28), (29) and (32), we can get

2 (\frac{ν_{f}(r_{q})}{z_{q}})^{k} ≤ exp\{−r_{q}^{n−2ε}\} + |(\frac{ν_{f}(r_{q})}{z_{q}})^{k−1}| r_{q}^{t+ε} + exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} .

Since 0 < 4ε < min{1, n − σ, n − σ(H), n − t}, we can get a contradiction as q → ∞.

Subcase 2.3. δ(Q, θ_{0}) = 0. Using the same argument as in Subcase 1.3, we can construct another sequence of points z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} satisfying \frac{l_{0}}{2} ≤ |θ_{q}^{∗} − θ_{q}| ≤ l_{0} such that δ(P, θ_{0}^{∗}) < 0 < δ(Q, θ_{0}^{∗}) where θ_{0}^{∗} = lim_{q→∞} θ_{q}^{∗}. Then, we have (28) for δ(P, θ_{0}^{∗}) and (30) for δ(Q, θ_{0}^{∗}). Using the same argument as in Subcase 1.3, we also have (25) for the sequence of points z_{q}^{∗}. From (29) and sufficiently large q, we have

|A_{0}(z_{q}^{∗})e^{Q(z_{q}^{∗})}| ≤ |A_{1}(z_{q}^{∗})e^{P (z_{q}^{∗})}|r_{q}^{σ(f)+ε} + exp{−r_{q}^{n−2ε}} + r_{q}^{kσ(f)+t+ε}.

Thus, we can also get a contradiction.

Case 3. Suppose that δ(P, θ_{0}) = 0. We discuss three subcases according to δ(Q, θ_{0}) as follows.

Subcase 3.1. δ(Q, θ_{0}) > 0. By the same argument as in Subcase 1.3, we can also construct another sequence of points z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} with θ_{0}^{∗} = lim_{q→∞} θ_{q}^{∗} and \frac{l_{0}}{2} ≤ |θ_{q}^{∗} − θ_{q}| ≤ l_{0} such that z_{q}^{∗} satisfies (25) and δ(P, θ_{0}^{∗}) < 0 < δ(Q, θ_{0}^{∗}). Using the same argument as in Subcase 2.3, we can get a contradiction easily as n → ∞.

Subcase 3.2. δ(Q, θ_{0}) < 0. By Lemma 2.2, we first define

δ′(P, θ) := −nα sin(nθ) − nβ cos(nθ)

where a_{n} = α + iβ. Since a_{n} ≠ 0, we have δ′(P, θ_{0}) ≠ 0. Take z′_{q} = r_{q}e^{iθ′_{q}} satisfying 0 < |θ′_{q} − θ_{0}| ≤ l_{0}, we have (25) for z′_{q} and δ(P, θ′_{q}) ≠ 0. By the continuity of δ(Q, θ), we may assume that δ(Q, θ′_{q}) < 0 < δ(P, θ′_{q}) for a suitable l_{0}, 0 < θ′_{q} − θ_{0} ≤ l_{0}. For a suitable l_{0}, we have δ′(P, θ_{0}) > 0 and

\frac{1}{3}δ′(P, θ_{0}) ≤ δ′(P, θ) ≤ \frac{4}{3}δ′(P, θ_{0}), θ ∈ (θ_{0}, θ_{0} + l_{0}). (33)

Since we have |f(z_{q})| = M(r_{q}, f) and θ_{q} → ∞ as q → ∞ for the sequence of points z_{q}, we have |f(r_{q}e^{iθ_{0}})| ≥ M(r_{q}, f)ν_{f}(r_{q})^{− \frac{1}{8} +δ} for sufficiently large q. From (7), we have

\begin{aligned}|\frac{f′(z′_{q})}{f(z′_{q})} | ≤ & |\frac{e^{−P (z′_{q})}}{A_{1}(z′_{q})} | (|\frac{f^{(k)}(z_{q})}{f(z_{q})} | + ∑_{j=2}^{k−1} |h_{j}(z_{q})| |\frac{f^{(j)}(z_{q})}{f(z_{q})} | \\ & + |\frac{H(z′_{q})}{M(r_{q},f)} | + |A_{0}(z′_{q})e^{Q(z′_{q})}|) . \end{aligned} (34)

By Lemma 2.2, we have

exp \{−(1 + ε)δ(P, θ′_{q})r_{q}^{n}\} ≤ |\frac{e^{−P (z′_{q})}}{A_{1}(z′_{q})} | ≤ exp \{−(1 − ε)δ(P, θ′_{q})r_{q}^{n}\} (35)

and exp{(1 + ε)δ(Q, θ′_{q})r_{q}^{n}} ≤ |A_{0}(z′_{q})e^{Q(z′_{q})}| ≤ exp{(1 − ε)δ(Q, θ′_{q})r_{q}^{n}} (36)

for sufficiently large q. From (9),(25),(35),(36) and (34), we can get

|\frac{f′(z′_{q})}{f(z′_{q})}| ≤ exp\{−(1 − 2ε)δ(P, θ′_{q})r_{q}^{n}\}.

Since θ′_{q} is arbitrary in (θ_{0}, θ_{0} + l_{0}), for sufficiently large r_{q}, we can obtain

|\frac{f′(r_{q}e^{iθ})}{f(r_{q}e^{iθ})}| ≤ exp\{−(1 − 2ε)δ(P, θ)r_{q}^{n}\}, θ ∈ (θ_{0}, θ_{0} + l_{0}). (37)

Therefore, for θ ∈ (θ_{0}, θ_{0} + l_{0}), we have

γ(r_{q}, θ) = r_{q} ∫_{θ_{0}}^{θ} |\frac{f′(r_{q}e^{iθ})}{f(r_{q}e^{iθ})}| dθ ≤ r_{q} ∫_{θ_{0}}^{θ} e^{κ_{1}(θ)r_{q}^{n}}dθ = ∫_{θ_{0}}^{θ} \frac{−1}{κ_{2}(θ)r_{q}^{n−1}} e^{κ_{1}(θ)r_{q}^{n}}d(κ_{1}(θ)r_{q}^{n}),

where κ_{1}(θ) = −(1 − 2ε)δ(P, θ), κ_{2}(θ) = (1 − 2ε)δ′(P, θ).

Since δ(P, θ) > 0 for all θ ∈ (θ_{0}, θ_{0} + l_{0}), we can get

0 ≤ γ(r_{q}, θ) ≤ \frac{2}{(1 − 2ε)δ′(P, θ_{0})r_{q}^{n−1}} (e^{κ_{1}(θ_{0})r_{q}^{n}} − e^{κ_{1}(θ)r_{q}^{n}}).

For sufficiently large q, we can get

0 ≤ γ(r_{q}, θ) ≤ \frac{2}{κ_{2}(θ_{0})}. (38)

By the proof of Lemma 2.4 in REF.[22], we have

ν_{f}(r_{q})^{− \frac{1}{8} +δ′}M(r_{q}, f) = exp\{−2π − 2/κ_{2}(θ_{0})\}ν_{f}(r_{q})^{− \frac{1}{8} +δ}M(r_{q}, f) ≤ |f(r_{q}e^{iθ})| (39)

for θ ∈ (θ_{0}, θ_{0} + l_{0}), where 0 < δ′ < δ < \frac{1}{8}. Therefore, we can take the sequence of points z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} satisfying θ_{q}^{∗} = \frac{l_{0}}{2} + θ_{0} and (25) for z_{q}^{∗}. Furthermore, from (39), we have (8) for z_{q}^{∗} when q is sufficiently large. Thus, from (8) and (37), we can deduce that ν_{f}(r_{q}) → ∞ as q → ∞, which is impossible.

When δ(Q, θ′_{q}) < 0 < δ(P, θ′_{q}) for −l_{0} < θ′_{q}−θ_{0} < 0. Then, we deduce that γ(r_{q}, θ) ≤ 0 for all θ ∈ (θ_{0} − l_{0}, θ_{0}). Similarly, we can get

ν_{f}(r_{q})^{− \frac{1}{8} +δ′}M(r_{q}, f) = exp\{−2π\}ν_{f}(r_{q})^{− \frac{1}{8} +δ}M(r_{q}, f) ≤ |f(r_{q}e^{iθ})| (40)

for θ ∈ (θ_{0} − l_{0}, θ_{0}), where 0 < δ′ < δ < \frac{1}{8}. Thus, we can also get a contradiction.

Subcase 3.3. δ(Q, θ_{0}) = 0. We have a_{n} = cb_{n} and c ∈ ℝ \ {0, 1}. Then we have

P (z) = cb_{n}z^{n} + ⋯ + a_{1}z + a_{0}, Q(z) − P (z) = (1 − c)b_{n}z^{n} + R_{n−1}(z),

where R_{n−1}(z) is a polynomial of degree at most n − 1.

If c < 0, we may take l_{0} small enough such that δ(Q, θ) < 0 < δ(P, θ), provided that either θ ∈ (θ_{0}, θ_{0} +l_{0}) or θ ∈ (θ_{0} −l_{0}, θ_{0}). Using the same argument as in Subcase 3.2, we can get (37) and (39). Therefore, by a standard Wiman-Valiron theory, we can deduce that ν_{f}(r_{q}) → ∞ as q → ∞. Thus, we can get a contradiction.

If 0 < c < 1, for small enough l_{0}, we also obtain δ(Q − P, θ) > 0 and δ(P, θ) > 0, provided that either θ ∈ (θ_{0}, θ_{0} + l_{0}) or θ ∈ (θ_{0} − l_{0}, θ_{0}). Using the same argument as in Subcase 1.3, we can get a contradiction easily.

If c > 1, from the above argument, we can obtain δ(Q − P, θ) < 0 < δ(P, θ) provided that either θ ∈ (θ_{0}, θ_{0} + l_{0}) or θ ∈ (θ_{0} − l_{0}, θ_{0}). Furthermore, we can take the sequence of points z′_{q} = r_{q}e^{iθ′_{q}} satisfying (25), provided that either θ′_{q} ∈ (θ_{0}, θ_{0}+l_{0}) or θ′_{q} ∈ (θ_{0}−l_{0}, θ_{0}). Therefore, from (7), we have

|\frac{f′(z′_{q})}{f(z′_{q})}| ≤ |\frac{A_{0}(z′_{q})}{A_{1}(z′_{q})}e^{(1−c)b_{n}(z′_{q})^{n}+R_{n−1}(z′_{q})}| + |\frac{e^{−P (z′_{q})}}{A_{1}(z′_{q})} | + |\frac{H(z′_{q})}{f(z′_{q})}|

+ (|\frac{f^{(k)}(z_{q})}{f(z_{q})}| + ∑_{j=2}^{k−1} |h_{j}(z_{q})| |\frac{f^{(j)}(z_{q})}{f(z_{q})}|) .

Similarly as in Subcase 3.2, we get (37) and (39). By the Wiman-Valiron theory, we can also get a contradiction.

Thus,from the above argument, we can prove that every solution f of equation (4) satisfies σ(f) = ∞.

The exponent of convergence of the zero points

Rewrite (4) as

\frac{1}{f} = \frac{1}{H} (\frac{f^{(k)}}{f} + ∑_{j=2}^{k−1} h_{j} \frac{f^{(j)}}{f} + A_{1}e^{P (z)} \frac{f′}{f} + A_{0}e^{Q(z)}) . (41)

If f has z_{0} as its zeros with multiplicity of s(> k), then z_{0} is the zeros of H with order s − k. Therefore, we have

N (r, \frac{1}{f} ) ≤ k\overline{N} (r, \frac{1}{f} ) + N (r, \frac{1}{H} ) . (42)

On the other hand, from (41), we have

m (r, \frac{1}{f} ) ≤ m (r, \frac{1}{H} ) + ∑_{j=2}^{k−1} m(r, h_{j}) + m(r, A_{1}e^{P}) + m(r, A_{0}e^{Q}) + S(r, f). (43)

Since σ(f) = ∞, σ := max{σ(A_{j}), j = 0, 1} < n and σ(H) < n, and from (42) and (43), we have

T (r, f) = T (r, \frac{1}{f} ) + O(1) ≤ (k + 4)k\overline{N} (r, \frac{1}{f} ) + S(r, f).

Thus, by Lemma 2.9, we can get σ(f) = λ(f) = \overline{λ}(f) = ∞ and σ_{2}(f) = λ_{2}(f) = \overline{λ}_{2}(f) ≤ n.

Next, we will prove that σ(f) = \overline{λ}(f − φ) = ∞ and σ_{2}(f) = \overline{λ}_{2}(f − φ) ≤ n.

First, setting ω_{0} = f − φ. Since σ(φ) < ∞, then we have σ(ω_{0}) = σ(f). From (4), we have

ω_{0}^{(k)} + ∑_{j=2}^{k−1} h_{j}ω′′_{0} + A_{1}e^{P} ω′_{0} + A_{0}e^{Q}ω_{0} = H − (A_{0}e^{Q}φ + A_{1}e^{P} φ′ + ∑_{j=2}^{k−1} h_{j}φ^{(j)} + φ^{(k)}).

Since σ(H) < n, σ < n, σ(φ) < ∞ and a_{n} ≠ b_{n}, we have H − (A_{0}e^{Q}φ + A_{1}e^{P} φ′ + ∑_{j=2}^{k−1} h_{j}φ^{(j)} + φ^{(k)}) ≢ 0 whether H ≢ 0 or H ≡ 0. Thus,

\begin{matrix}\frac{1}{ω_{0}} = \\ _{\frac{1}{H−(A_{0}e^{Q}φ+A_{1}e^{P} φ′+∑_{j=2}^{k−1} h_{j}φ^{(j)}+φ^{(k)})}} (\frac{ω_{0}^{(k)}}{ω_{0}} + ∑_{j=2}^{k−1} h_{j} \frac{ω_{0}^{(j)}}{ω_{0}} + A_{1}e^{P} \frac{ω′_{0}}{ω_{0}} + A_{0}e^{Q}) . \end{matrix} (44)

If ω_{0} has z_{1} as its zero with multiplicity of l(> k), then z_{1} is the zeros of H − (A_{0}e^{Q}φ + A_{1}e^{P} φ′ + ∑_{j=2}^{k−1} h_{j}φ^{(j)} + φ^{(k)}) with multiplicity l − k. Then, we have

N (r, \frac{1}{ω_{0}} ) ≤ k\overline{N} (r, \frac{1}{ω_{0}} ) + N (r, \frac{1}{H − (A_{0}e^{Q}φ + A_{1}e^{P} φ′ + ∑_{j=2}^{k−1} h_{j}φ^{(j)} + φ^{(k)})}) .

On the other hand, from (44), we have

m (r, \frac{1}{ω_{0}} ) ≤ m (r, \frac{1}{H − (A_{0}e^{Q}φ + A_{1}e^{P} φ′ + ∑_{j=2}^{k−1} h_{j}φ^{(j)} + φ^{(k)})}) + m(r, A_{1}e^{P})

+m(r, A_{0}e^{Q}) + ∑_{j=2}^{k−1} m(r, h_{j}) + S(r, f).

Using the above argument, we obtain

T (r, ω_{0}) = T (r, f) + S(r, f) ≤ K_{1}\overline{N} (r, \frac{1}{ω_{0}} ) + S(r, f) = K_{1}\overline{N} (r, \frac{1}{f − φ}) + S(r, f),

where K_{1} is a constant.

Thus, by Lemma 2.9, we can get σ(f) = σ(ω_{0}) = \overline{λ}(f − φ) = ∞ and σ_{2}(f) = \overline{λ}_{2}(f − φ) ≤ n.

Hence, we can get σ(f) = λ(f) = \overline{λ}(f) = \overline{λ}(f −φ) = ∞ and σ_{2}(f) = λ_{2}(f) = \overline{λ}_{2}(f) = \overline{λ}_{2}(f − φ) ≤ n.

Thus, we can complete the proof of Theorem 1.1. □

4 The proof of Theorem 1.2

Proof: Let f be a nontrivial solution of (5) with finite order. By [19], similar to Theorem 1.1, we can get σ(f) ≥ n. Now rewrite (5) as

\frac{f^{(k)}}{f} + ∑_{j=2}^{k−1} h_{j} \frac{f^{(j)}}{f} + (A_{1}(z)e^{P (z)} + D_{1}(z)) \frac{f′}{f} + (A_{0}(z)e^{Q(z)} + D_{0}(z)) = \frac{H(z)}{f}. (45)

Since D = max{σ(D_{j}), j = 0, 1} < n, then for any ε(0 < 4ε < min{1, n − σ, n − σ(H), n − t, n − D}), we have

|D_{j}(z)| ≤ exp{r^{D+ε}}, j = 0, 1. (46)

Similarly as in the proof of Theorem 1.1, we can take a sequence of points z_{q} = r_{q}e^{iθ_{q}}, r_{q} → ∞, such that lim_{q→∞} θ_{q} = θ_{0} and

|f(z_{q})| = M(r_{q}, f), r_{q} ∈ (E_{3} ∩ E_{4}) \ (E_{1} ∪ E_{2}),

and the sequence of points satisfies (8)-(12).

Suppose that a_{n}/b_{n} = c < 0, we will discuss three cases according to the signs of δ(P, θ_{0}) and δ(Q, θ_{0}) as follows.

Case 1. Suppose that δ(P, θ_{0}) < 0 < δ(Q, θ_{0}). By Lemma 2.2 and the continuity of δ(P, θ), δ(Q, θ), we have

exp \{\frac{4(1 + ε)}{3}δ(P, θ_{0})r_{q}^{n}\} ≤ |A_{1}(z_{q})e^{P (z_{q})}| ≤ exp \{\frac{(1 − ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (47)

and

exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ |A_{0}(z_{q})e^{Q(z_{q})}| ≤ exp \{\frac{4(1 + ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} (48)

for all sufficiently large q. From (45), we have

\begin{aligned}|A_{0}(z_{q})e^{Q(z_{q})} + D_{0}(z_{q})| ≤ & |\frac{f^{(k)}(z_{q})}{f(z_{q})} | + ∑_{j=2}^{k−1} |h_{j}(z_{q})| |\frac{f^{(j)}(z_{q})}{f(z_{q})} | + \frac{|H(z_{q})|}{M(r_{q},f)} \\ & + |(|A_{1}(z_{q})e^{P (z_{q})} + D_{1}(z_{q})|)\frac{f′(z_{q})}{f(z_{q})} | . \end{aligned} (49)

From (46),(47) and (48), we have

|A_{1}(z_{q})e^{P (z_{q})} + D_{1}(z_{q})| ≤ exp{r^{D+2ε}} (50)

and

|A_{0}(z_{q})e^{Q(z_{q})} + D_{0}(z_{q})| ≥ exp \{\frac{(1 − 2ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} (51)

for large enough q.

Substituting (10),(11),(47) and (48) into (49), we can obtain

exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ r_{q}^{kσ(f)+ε} + exp\{r_{q}^{D+2ε}\} + (k − 2)r_{q}^{(k−1)σ(f)+t+ε} + exp\{−r_{q}^{n−2ε}\}
≤ exp{r_{q}^{D+3ε}}.

Since D < n, we can obtain a contradiction.

Case 2. Suppose that δ(Q, θ_{0}) < 0 < δ(P, θ_{0}). By Lemma 2.2 and the continuity of δ(P, θ), δ(Q, θ), we have

exp \{\frac{(1 − ε)}{3}δ(P, θ_{0})r_{q}^{n}\} ≤ |A_{1}(z_{q})e^{P (z_{q})}| ≤ exp \{\frac{4(1 + ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (52)

and

exp \{\frac{4(1 + ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} ≤ |A_{0}(z_{q})e^{Q(z_{q})}| ≤ exp \{\frac{(1 − ε)}{3}δ(Q, θ_{0})r_{q}^{n}\} (53)

for all sufficiently large q. From (45), we have

\begin{aligned}|(A_{1}(z_{q})e^{P (z_{q})} + D_{1}(z_{q}))\frac{f′(z_{q})}{f(z_{q})} | ≤ & |\frac{f^{(k)}(z_{q})}{f(z_{q})} | + ∑_{j=2}^{k−1} |h_{j}(z_{q})| |\frac{f^{(j)}(z_{q})}{f(z_{q})} | + \frac{|H(z_{q})|}{M(r_{q},f)} \\ & +|A_{0}(z_{q})e^{Q(z_{q})} + D_{0}(z_{q})|. \end{aligned} (54)

From (46), (52) and (53), we have

|A_{0}(z_{q})e^{Q(z_{q})} + D_{0}(z_{q})| ≤ exp{r^{D+2ε}} (55)

and

|A_{1}(z_{q})e^{P (z_{q})} + D_{1}(z_{q})| ≥ exp \{\frac{(1 − 2ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (56)

for large enough q.

Substituting (10),(11),(55) and (56) into (54), we obtain

ν_{f}(r_{q}) ≤ 2r_{q} exp \{−\frac{(1 − 2ε)}{3}δ(P, θ_{0})r_{q}^{n}\} (2Kr_{q}^{kσ(f)+t+ε} + exp\{r_{q}^{D+2ε}\}) (57)

for sufficiently large q, where K is a constant. From (9), (57) and D < n, we can deduce that ν_{f}(r_{q}) → 0 as q → ∞, which is a contradiction.

Case 3. Suppose that δ(Q, θ_{0}) = 0 = δ(P, θ_{0}). Similarly as in Subcase 1.3 of the proof of Theorem 1.1, from (12), we can construct a sequence of points z_{q}^{∗} = r_{q}e^{iθ_{q}^{∗}} with lim_{q→∞} θ_{q}^{∗} = θ_{0} such that δ(P, θ_{0}^{∗}) < 0 and (25) holds for z_{q}^{∗}.

Without loss of generality, we can assume that

δ(P, θ) > 0, θ ∈ (θ_{0} + 2mπ, θ_{0} + (2m + 1)π)

and

δ(P, θ) < 0, θ ∈ (θ_{0} + (2m − 1)π, θ_{0} + 2mπ)

for all m ∈ ℤ.

For sufficiently large q, we can have |θ_{0} − θ_{q}| ≤ l_{0}. Taking θ_{0}^{∗} such that \frac{l_{0}}{2} ≤ θ_{q} − θ_{q}^{∗} ≤ l_{0}, then θ_{0} − l_{0} ≤ θ_{0}^{∗} ≤ θ_{0} − \frac{l_{0}}{2} and δ(P, θ_{0}^{∗}) < 0. Since δ(Q, θ_{0}^{∗}) > 0, by using the same argument as in Case 2, we can get a contradiction easily.

□

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(Received June 8, 2009) EJQTDE, 2009 No. 63, p. 18