Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 1, 1-13; http://www.math.u-szeged.hu/ejqtde/

Growth of meromorphic solutions of higher-order

linear differential equations

Wenjuan Chen^{1} Junfeng Xu^{2}^{∗}

1.Department of Mathematics, Shandong University, Jinan, Shandong 250100, P.R.China 2.Department of Mathematics, Wuyi University, Jiangmen, Guangdong 529020, P.R.China

Abstract. In this paper, we investigate the higher-order linear differential equations with meromorphic coefficients. We improve and extend a result of M.S. Liu and C.L. Yuan, by using the estimates for the logarithmic derivative of a transcendental meromorphic function due to Gundersen, and the extended Winman-Valiron theory which proved by J. Wang and H.X. Yi. In addition, we also consider the nonhomogeneous linear differential equations.

Keywords: Linear differential equation; meromorphic function; growth order; Nevanlinna theorey; iterated order 2000 AMS Subject Classifications: 34M10, 30D35

1 Introduction and main results

In this paper, we shall assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory of meromorphic functions (see [11, 22]). The term “meromorphic function” will mean meromorphic in the whole complex plane ℂ.

For the second order linear differential equation

f′′ + e^{−z}f′ + B(z)f = 0,  (1.1)

where B(z) is an entire function of finite order. It is well known that each solution f of (1.1) is an entire function, and that if f_{1} and f_{2} are any two linearly independent solutions of (1.1), then at least one of f_{1}, f_{2} must have infinitely order([12]). Hence, “most” solutions of (1.1) will have infinite order. However, the equation (1.1) with B(z) = −(1 + e^{−z}) possesses a solution f = e^{z} of finite order.

Thus a natural question is: what condition on B(z) will guarantee that every solution f ≢ 0 of (1.1) will have infinite order? Frei, Ozawa, Amemiya and Langley, and Gunderson studied the question. For the case that B(z) is a transcendental entire function, Gundersen [8] proved that if ρ(B) ≠ 1, then for every solution f ≢ 0 of (1.1) has infinite order.

For the above question, there are many results for second order linear differential equations (see, for example [1, 4, 6, 7, 10, 15]). In 2002, Z. X. Chen considered the problem and obtained the following result in [4].

∗Corresponding author.

E-mail: chenwenjuan@mail.sdu.edu.cn(W.J. Chen); xujunf@gmail.com(J.F. Xu)

Theorem 1.1. Let a, b be nonzero complex numbers and a ≠ b, Q(z) ≢ 0 be a nonconstant polynomial or Q(z) = h(z)e^{bz}, where h(z) is a nonzero polynomial. Then every solution f ≢ 0 of the equation

f′′ + e^{bz}f′ + Q(z)f = 0

has infinite order and σ_{2}(f) = 1.

In 2006, Liu and Yuan generalized Theorem 1.1 and obtained the following result.

Theorem 1.2 (see. [17, Theorem 1]). Suppose that a, b are nonzero complex numbers, h_{j}(j = 0, 1, ⋯, k − 1)(h_{0} ≢ 0) be meromorphic functions that have finite poles and σ = max{σ(h_{j}) : j = 0, 1, ⋯, k − 1} < 1. If arg a ≠ arg b or a = cb(0 < c < 1), then every transcendental meromorphic solution f of the equation

f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + e^{az}f^{(s)} + ⋯ + h_{1}f′ + h_{0}e^{bz}f = 0.  (1.2)

have infinite order and σ_{2}(f) = 1.

It is natural to ask the following question: What can we say if we remove the condition h_{j}(j = 0, 1, ⋯, k − 1) have finite poles in Theorem 1.2. In this paper, we first investigate the problem and obtain the following result.

Theorem 1.3. Let P (z) and Q(z) be a nonconstant polynomials such that

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

for some complex numbers a_{i}, b_{i}(i = 0, 1, 2, ⋯, n) with a_{n} ≠ 0, b_{n} ≠ 0 and let h_{j}(j = 0, 1, ⋯, k − 1)(h_{0} ≢ 0) be meromorphic functions and σ = max{σ(h_{j}) : j = 0, 1, ⋯, k−1} < n. If arg a_{n} ≠ arg b_{n} or a_{n} = cb_{n}(0 < c < 1), suppose that all poles of f are of uniformly bounded multiplicity. Then every transcendental meromorphic solution f of the 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 = 0  (1.3)

have infinite order and σ_{2}(f) = n.

Next, we continue to investigate the problem and extend Theorem 1.2.

Theorem 1.4. Let P (z) and Q(z) be a nonconstant polynomials as the above, for some complex numbers a_{i}, b_{i}(i = 0, 1, 2, ⋯, n) with a_{n} ≠ 0, b_{n} ≠ 0 and let h_{j}(j = 0, 1, ⋯, k − 1)(h_{0} ≢ 0) be meromorphic functions and σ = max{σ(h_{j}) : j = 1, ⋯, k − 1} < n. Suppose all poles of f are of uniformly bounded multiplicity. Then the following three statements hold:

1. If a_{n} = b_{n}, and deg(P − Q) = m ≥ 1, σ < m, then every transcendental meromorphic solution f of the equation (1.3) have infinite order and m ≤ σ_{2}(f) ≤ n.

2. If a_{n} = cb_{n} with c > 1, and deg(P − Q) = m ≥ 1, σ < m, then every solution f ≢ 0 of the equation (1.3) is of infinite order, and σ_{2}(f) = n.

3. If σ < σ(h_{0}) < 1/2, a_{n} = cb_{n} with c ≥ 1 and P (z) − cQ(z) is a constant, then every solution f ≢ 0 of equation (1.3) is of infinite order, and σ(h_{0}) ≤ σ_{2}(f) ≤ n.

Remark 1.1 Setting h_{j}(j = 1, 2, …, k − 1) be entire functions in Theorem 1.3 and Theorem 1.4, we get Theorem 1 in [17].”

Considering nonhomogeneous linear differential equations

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.  (1.4)

Corresponding to (1.3), we obtain the following result:

Theorem 1.5. Let k ≥ 2, s ∈ {1, ⋯, k − 1}, h_{0} ≢ 0, h_{1}, ⋯, h_{k−1}; P (z),Q(z) satisfy the hypothesis of Theorem 1.4; 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 three statements of Theorem 1.4 hold, then all solutions f of non-homogeneous linear differential equation (1.4) 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).

Futhermore, if such an exceptional solution f_{0} of finite order of (1.4) exists, then we have

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

Remark 1.2. Setting h_{j}(j = 1, 2, …, k − 1) and F (z) be entire functions in Theorem 1.5, we get Theorem 2 in [17].

2 Lemmas

The linear measure of a set E ⊂ [0, +∞) is defined as m(E) = ∫_{0}^{+∞} χ_{E}(t) dt. The logarithmic measure of a set E ⊂ [1, +∞) is defined by lm(E) = ∫_{1}^{+∞} χ_{E}(t)/t dt, where χ_{E}(t) is the characteristic function of E. The upper and lower densities of E are

\overline{dens}E = lim sup_{r→+∞} \frac{m(E ∩ [0, r])}{r}, \underline{dens}E = lim inf_{r→+∞} \frac{m(E ∩ [0, r])}{r}.

Lemma 2.1 (see. [4]). Let f(z) be a entire function with σ(f) = ∞, and σ_{2}(f) = α < ∞, let a set E ⊂ [1, ∞) that has finite logarithmic measure. Then there exists {z_{k} = r_{k}e^{iθ_{k}}} such that |f(z_{k})| = M(r_{k}, f), θ_{k} ∈ [0, 2π), lim_{k→∞} θ_{k} = θ_{0} ∈ [0, 2π),r_{k} ∉ E, r_{k} → ∞, and for any given ε > 0, for a sufficiently large r_{k}, we have

lim sup_{r→∞} \frac{log ν_{f}(r_{k})}{log r_{k}} = +∞,  (2.1)
exp{r_{k}^{α−ε}} < ν_{f}(r_{k}) < exp{r_{k}^{α+ε}}  (2.2)

Lemma 2.2 (see.[2, 14]). Let F (r) and G(r) be monotone nondecreasing functions on (0, ∞) such that (i) F (r) ≤ G(r)n.e. or (ii) for r ≠ H ∪ [0, 1] having finite logarithmic measure, then for any constant α > 1, there exists r_{0} > 0 such that F (r) ≤ G(αr) for all r > r_{0}.

Lemma 2.3 (see. [9]). Let f be a transcendental meromorphic function. Let α > 1 be a constant, and k and j be integers satisfying k > j ≥ 0. Then the following two statements hold:

(a) There exists a set E_{1} ⊂ (1, ∞) which has finite logarithmic measure, and a constant C > 0, such that for all z satisfying |z| ∉ E_{1} ⋃[0, 1], we have (with r = |z|)

|\frac{f^{(k)}(z)}{f^{(j)}(z)}| ≤ C[\frac{T (αr, f)}{r}(log r)^{α} log T (αr, f)]^{k−j}.  (2.3)

(b) There exists a set E_{2} ⊂ [0, 2π) which has linear measure zero, such that if θ ∈ [0, 2π) − E_{2}, then there is a constant R = R(θ) > 0 such that (2.3) holds for all z satisfying arg z = θ and R ≤ |z|.

Lemma 2.4 ([18], pp. 253-255). Let n be a positive integer, and let P (z) = a_{n}z^{n} + a_{n−1}z^{n−1} + ⋯ + a_{1}z + a_{0} with a_{n} = α_{n}e^{iθ_{n}}, α_{n} > 0. For given ε, 0 < ε < π/4n, we introduce 2n closed angles

D_{j} : −\frac{θ_{n}}{n} + (2j − 1)\frac{π}{2n} + ε ≤ θ ≤ −\frac{θ_{n}}{n} + (2j + 1)\frac{π}{2n} − ε(j = 0, 1, ⋯, 2n − 1).

Then, there exists a positive number R = R(ε) such that

Re P (z) > α_{n}r^{n}(1 − ε) sin nε

if |z| = r > R and z ∈ D_{j}, where j is even, while

Re P (z) < −α_{n}r^{n}(1 − ε) sin nε

if |z| = r > R and z ∈ D_{j}, where j is odd.

Lemma 2.5 ([4], Lemma 1). Let g(z) be a meormorphic function with σ(g) = β < ∞. Then for any ε > 0, there exists a set E ⊂ (1, ∞) with lmE < ∞, such that for all z with |z| = r ∉ ([0, 1] ∪ E), r → ∞, then

|g(z)| ≤ exp{r^{β+ε}}.

Applying Lemma 2.5 to 1/g(z), we can obtain that for any given ε > 0, there exists a set E ⊂ (1, ∞) with lmE < ∞, such that for all z with |z| = r ∉ ([0, 1] ∪ E), r → ∞, then

exp{−r^{β+ε}} ≤ |g(z)| ≤ exp{r^{β+ε}}.  (2.4)

It is well known that the Wiman-Valiron theory (see, [14]) is an indispensable device while considering the growth of entire solution of a complex differential equation. In order to consider the growth of meromorphic function solutions of a complex differential equation, Wang and Yi [19] extended the Wiman-Valiron theory from entire functions to meromorphic functions. Here we give the special form where meromorphic function has infinite order:

Lemma 2.6 ([19, 20]). Let f(z) = g(z)/d(z) be the infinite order meromorphic function and σ_{2}(f) = σ, where g(z) and d(z) are entire function, σ(d) < ∞, there exists a sequence r_{j}(r_{j} → ∞) satisfying z_{j} = r_{j}e^{iθ_{j}}, θ_{j} ∈ [0, 2π), lim_{j→∞} θ_{j} = θ_{0} ∈ [0, 2π), |g(z_{j})| = M(r_{j}, g) and j is sufficient large, we have

\frac{f^{(n)}(z_{j})}{f(z_{j})} = (\frac{ν_{g}(r_{j})}{z_{j}})^{n}(1 + o(1))(n ∈ ℕ),

lim sup_{r→∞} \frac{log log ν_{g}(r)}{log r} = σ_{2}(g).

Lemma 2.7. Let k ≥ 2 and A_{0}, A_{1}, ⋯, A_{k−1} are meromorphic function. Let σ = max{σ(A_{j}), j = 0, 1, ⋯, k−1} and all poles of f are of uniformly bounded multiplicity. Then every transcendental meromorphic solution of the differential equation

f^{(k)} + A_{k−1}f^{(k−1)} + ⋯ + A_{0}f = 0,  (2.5)

satisfies σ_{2}(f) ≤ σ.

Proof. Since f ≢ 0 is a transcendental meromorphic solution of the equation (2.5). If σ(f) < ∞, then σ_{2} = 0 ≤ σ. If σ(f) = ∞. We can rewrite (2.5) to

−\frac{f^{(k)}}{f} = A_{k−1} \frac{f^{(k−1)}}{f} + ⋯ + A_{1} \frac{f′}{f} + A_{0}.  (2.6)

Obviously, the poles of f must be the poles of A_{j}(j = 0, 1, ⋯, k − 1), note that all poles of f are of uniformly bounded multiplicity, then λ(1/f) ≤ σ. By Hadmard factorization theorem, we know f can be written to f(z) = \frac{g(z)}{d(z)}, where g(z) and d(z) are entire function, and λ(d) = σ(d) = λ(1/f) ≤ σ, σ_{2}(f) = σ_{2}(g). By Lemma 2.5 and Lemma 2.6, for any small ε > 0, there exists a sequence r_{j}(r_{j} → ∞) satisfying z_{j} = r_{j}e^{iθ_{j}}, θ_{j} ∈ [0, 2π), lim_{j→∞} θ_{j} = θ_{0} ∈ [0, 2π), |g(z_{j})| = M(r_{j}, g) and j is sufficient large, we have

\frac{f^{(n)}(z_{j})}{f(z_{j})} = (\frac{ν_{g}(r_{j})}{z_{j}})^{n}(1 + o(1)), (n ∈ ℕ),  (2.7)
lim sup_{r→∞} \frac{log log ν_{g}(r)}{log r} = σ_{2}(g),  (2.8)
|A_{j}(z)| ≤ e^{r_{j}^{σ+ε}}, (j = 1, 2, ⋯, k − 1),  (2.9)

Substituting (2.7),(2.9) into (2.6), we obtain

v_{g}(r_{j})(1 + o(1)) ≤ 2r_{j} exp{r_{j}^{σ+ε_{j}}}.  (2.10)

Then by (2.8), (2.10) and for the arbitrary ε, we can obtain σ_{2}(f) ≤ σ. We complete the proof of the lemma.

Remark 3. Here we point out that the condition all poles of f are of uniformly bounded multiplicity in Theorem 1 of [3] and Theorem 1.3 of [20] was missing. Since the growth of the coefficients A_{j} gives only an estimate for the counting function of the distinct poles of f, but not for N(r, f).

Lemma 2.8. (see. [5]) Let A_{0}, A_{1}, …, A_{k−1},F ≢ 0 are finite order meromorphic function. If f(z) is an infinite order meromorphic solution of the equation

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

then f satisfies λ(f) = \overline{λ}(f) = σ(f) = ∞.

3 Proofs of main results

3.1 Proof of Theorem 1.3

Proof. Let f ≢ 0 be a transcendental solution of the equation (1.1). We consider two case:

Case 1 : When arg a_{n} ≠ arg b_{n}, by Lemma 2.4, there exist constants c > 0, R_{1} > 0 and θ_{1} < θ_{2} such that for all r ≥ R_{1} and θ ∈ (θ_{1}, θ_{2}), we have

\begin{matrix}Re P (re^{iθ}) < 0, \\ Re Q(re^{iθ}) > br^{n}. \end{matrix}  (3.1)

Note that σ = max{σ(h_{j}), j = 0, 1, ⋯, k−1} < n. Then by Lemma 2.5, for any ε(0 < ε < (n−σ)/2), there exists a set E_{1} ⊂ (1, ∞) that has finite linear measure such that when |z| = r ∉ ([0, 1] ∪ E), r → ∞, we have

|\frac{h_{j}}{h_{0}} | ≤ exp\{r^{\frac{σ+n}{2}} \}, (j = 0, 1, ⋯, k − 1).  (3.2)

Since f is a transcendental meromorphic function, by Lemma 2.3, there exists a set E_{2} ⊂ (1, ∞) that has finite logarithmic measure such that when |z| = r ∉ ([0, 1] ∪ E), r → ∞, we have

|\frac{f^{(j)}(z)}{f(z)}| ≤ Br[T (2r, f)]^{j+1}, (j = 0, 1, ⋯, k − 1).  (3.3)

From the equation (1.1), we obtain

|e^{Q}| ≤ |\frac{1}{h_{0}} | |\frac{f^{(k)}}{f} | + |\frac{h_{k−1}}{h_{0}} | |\frac{f^{(k−1)}}{f} | + ⋯ + |\frac{h_{s}}{h_{0}} | |e^{P} | |\frac{f^{(s)}}{f} | + ⋯ + |\frac{h_{1}}{h_{0}} | |\frac{f′}{f} | .  (3.4)

Therefore, from (3.1)-(3.4), for z = re^{iθ}, θ ∈ (θ_{1}, θ_{2}), r ∉ [0, 1] ∪ E_{1} ∪ E_{2}, we have

exp\{br^{n}\} ≤ kAr exp\{r^{\frac{σ+n}{2}} \}[T (2r, f)]^{k+1}.

Hence by Lemma 2.2, we obtain σ(f) = ∞ and σ_{2}(f) ≥ n.

On the other hand, by Lemma 2.7, we have σ_{2}(f) ≤ n, hence σ_{2}(f) = n.

Case 2 : When a_{n} = cb_{n} with 0 < c < 1. Since deg Q = n > n − 1 = deg(P − cQ), By Lemma 2.4, there exist constant c > 0, R_{2} > 0 and θ_{1} < θ_{2} such that for all r ≥ R_{2} and θ ∈ (θ_{1}, θ_{2}), we have

\begin{matrix}Re Q(re^{iθ}) > br^{n} > 0, \\ Re \{P (re^{iθ}) − cQ(re^{iθ})\} ≤ M. \end{matrix}  (3.5)

From the equation (1.1), we obtain

|e^{(1−c)Q}|

≤ |\frac{1}{h_{0}} | |e^{−cQ}| |\frac{f^{(k)}}{f} | + |\frac{h_{k−1}}{h_{0}} | |e^{−cQ}| |\frac{f^{(k−1)}}{f} |

+ ⋯ + |\frac{h_{s}}{h_{0}} | |e^{P −cQ}| |\frac{f^{(s)}}{f} | + ⋯ + |\frac{h_{1}}{h_{0}} | |e^{−cQ}| |\frac{f′}{f} | .

Therefore, from this and (3.2),(3.3) and (3.5), for z = re^{iθ}, θ ∈ (θ_{1}, θ_{2}), r ∉ [0, 1] ∪ E_{1} ∪ E_{2}, we have

exp\{b(1 − c)r^{n}\} ≤ (k + 1)Br exp\{r^{\frac{σ+n}{2}} \}[T (2r, f)]^{k+1}.

Hence by Lemma 2.2 again, we can obtain σ(f) = ∞ and σ_{2}(f) ≥ n.

On the other hand, by Lemma 2.7, we have σ_{2}(f) ≤ n, hence σ_{2}(f) = n, and the proof of Theorem 1.3 is completed.

3.2 Proof of Theorem 1.4

Proof. We distinguish three cases:

(1) Suppose that a_{n} = cb_{n} with c ≥ 1, and deg(P − cQ) = m ≥ 1, σ < m. We claim that σ(f) = ∞ and m ≤ σ_{2}(f) ≤ n.

Since deg P (z) = n > m = deg(Q − P/c), by Lemma 2.4, there exist a real number b > 0 and a continuous curve Γ tending ∞ such that for all z ∈ Γ with |z| = r, we have

\begin{matrix}Re P (z) = 0, \\ Re [Q(z) − \frac{1}{c}P (z)] ≥ br^{m}. \end{matrix}  (3.6)

From the equation (1.3), we obtain

|e^{Q−P/c}|

≤ |\frac{1}{h_{0}} | |e^{−P/c}| |\frac{f^{(k)}}{f} | + |\frac{h_{k−1}}{h_{0}} | |e^{−P/c}| |\frac{f^{(k−1)}}{f} |

+ ⋯ + |\frac{h_{s}}{h_{0}} | |e^{(1−1/c)P} | |\frac{f^{(s)}}{f} | + ⋯ + |\frac{h_{1}}{h_{0}} | |e^{−P/c}| |\frac{f′}{f} | .

Similar, we can get (3.2) and (3.3). Therefore, from this and (3.2),(3.3) and (3.6), for z = re^{iθ}, θ ∈ (θ_{1}, θ_{2}), r ∉ [0, 1] ∪ E_{1} ∪ E_{2}, we have

exp\{br^{m}\} ≤ (k + 1)Br exp\{r^{\frac{σ+n}{2}} \}[T (2r, f)]^{k+1}.

Hence by Lemma 2.2, from this we obtain σ(f) = ∞ and σ_{2}(f) ≥ m. On the other hand, by Lemma 2.7, we have σ_{2}(f) ≤ n, hence m ≤ σ_{2}(f) ≤ n.

(2) We shall verify that σ_{2}(f) = n. If it is not true, then it follows from the proof of Part (1) that σ_{2}(f) = α(m ≤ α < n), we shall arrive at a contradiction in the sequel.

Since σ = max{σ(h_{j}) : j = 0, 1, ⋯, k − 1} < m, then by Lemma 2.5, for any given ε(0 < ε < min\{\frac{m−σ}{3} , \frac{n−σ}{3} , \frac{π}{4n}\}), there is a set E_{3} ⊂ [1, ∞) having finite logarithmic measure such that for all z satisfying |z| = r ∉ E_{3} ∪ [0, 1], we have

exp{−r^{σ+ε}} ≤ |h_{j}(z)| ≤ exp{r^{σ+ε}}, (j = 0, 1, ⋯, k − 1).  (3.7)
exp{−r^{m+ε}} ≤ | exp{(P (z) − cQ(z))}| ≤ exp{r^{m+ε}}.  (3.8)

Let f(z) = g(z)/d(z) be the infinite order meromorphic function and σ_{2}(f) = σ, where g(z) and d(z) are entire function, σ(d) < ∞, there exists a sequence r_{k}(r_{k} → ∞) satisfying z_{k} = r_{k}e^{iθ_{k}}, θ_{k} ∈ [0, 2π), lim_{k→∞} θ_{k} = θ_{0} ∈ [0, 2π), |g(z_{k})| = M(r_{k}, g) and k is sufficient large, we have

\frac{f^{(j)}(z_{k})}{f(z_{k})} = (\frac{ν_{g}(r_{k})}{z_{k}})^{j}(1 + o(1)), (j = 0, 1, ⋯, k − 1)  (3.9)

and

exp{r_{k}^{σ−ε}} ≤ ν_{g}(r_{k}) ≤ exp{r_{k}^{σ+ε}}.  (3.10)

Let Q(z) = b_{n}z^{n} + b_{n−1}z^{n−1} + ⋯ + b_{1}z + b_{0}, where b_{n} = |b_{n}|e^{iθ}, |b_{n}| > 0, θ_{n} ∈ [0, 2π). By Lemma 2.4, for the above ε, there are 2n opened angles

G_{j} : −\frac{θ}{n} + (2j − 1)\frac{π}{2n} + ε < θ < −\frac{θ_{n}}{n} + (2j + 1)\frac{π}{2n} − ε(j = 0, 1, ⋯, 2n − 1).  (3.11)

and a positive number R = R(ε) such that

Re Q(z) > |b_{n}|r^{n}(1 − ε) sin nε

if |z| = r > R and z ∈ D_{j}, where j is even, while

Re Q(z) < −|b_{n}|r^{n}(1 − ε) sin nε

if |z| = r > R and z ∈ D_{j}, where j is odd.

For the above θ, if θ_{0} ≠ −\frac{θ}{n} + (2j − 1)\frac{π}{2n}(j = 0, 1, ⋯, 2n − 1), then we may take ε sufficiently small, and there is some G_{j}, j ∈ {0, 1, ⋯, 2n − 1} such that θ_{0} ∈ G_{j}. Hence there are three cases:

(i) θ_{0} ∈ G_{j} for some odd number j; (ii) θ_{0} ∈ G_{j} for some even number j; (iii) θ_{0} = −\frac{θ_{n}}{n} + (2j − 1)\frac{π}{2n} for some j ∈ {0, 1, ⋯, 2n − 1}.

Now we split this into three cases to prove:

Case (i): θ_{0} ∈ G_{j} for some odd number j. Since G_{j} is an open set and lim_{k→∞} θ_{k} = θ_{0}, there is a K > 0 such that θ_{k} ∈ G_{j} for k > K. By Lemma 2.4, we have

Re {Q(r_{k}e^{iθ_{k}})} < −σr_{k}^{n}(σ > 0), i.e., Re {−Q(r_{k}e^{iθ_{k}})} > σr_{k}^{n}(σ > 0).  (3.12)

Since deg(P − cQ) = m ≥ 1, from (3.12), we obtain that for a sufficiently large k,

Re {P (z_{k}) − Q(z_{k})} = Re {(c − 1)Q + (P − cQ} < −(c − 1)σr_{k}^{n} + dr_{k}^{n} < 0,  (3.13)

where Re{P (z_{k}) − Q(z_{k})} < dr_{k}^{n} for a sufficiently large k. Substituting (3.10) into (1.3), we get for {z_{k} = r_{k}e^{iθ_{k}}},

\begin{matrix}− e^{−Q(z_{k})}[ν_{g}^{k}(r_{k})(1 + o(1)) + z_{k}h_{k−1}ν_{g}^{k−1}(r_{k})(1 + o(1)) + ⋯ + \\ z_{k}^{k−s−1}h_{s+1}(z_{k})ν_{g}^{s+1}(r_{k})(1 + o(1)) + z_{k}^{k−s+1}h_{s−1}(z_{k})ν_{g}^{s−1}(r_{k})(1 + o(1)) \\ + ⋯ + z_{k}^{k−1}h_{1}(z_{k})ν_{g}(r_{k})(1 + o(1))] \\ = z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−Q(z_{k})}ν_{g}^{s}(r_{k})(1 + o(1)) + z_{k}^{k}h_{0}(z_{k}). \end{matrix}  (3.14)

Thus from (3.10) and (3.12), we obtain, for a sufficiently large k,

\begin{matrix}| − e^{−Q(z_{k})}[ν_{g}^{k}(r_{k})(1 + o(1)) + z_{k}h_{k−1}ν_{g}^{k−1}(r_{k})(1 + o(1)) + ⋯ + \\ z_{k}^{k−s−1}h_{s+1}(z_{k})ν_{g}^{s+1}(r_{k})(1 + o(1)) + z_{k}^{k−s+1}h_{s−1}(z_{k})ν_{g}^{s−1}(r_{k})(1 + o(1)) \\ + ⋯ + z_{k}^{k1}h_{1}(z_{k})ν_{g}(r_{k})(1 + o(1))]| \\ > e^{σr_{k}^{n}}e^{kr_{k}^{σ−ε}}[\frac{1}{2} − 2r_{k}|h_{k}(z_{k})|/ν_{g}(r_{k}) − ⋯ − 2r_{k}^{k−1}|h_{1}(z_{k})|/ν_{g}^{k−1}(r_{k})] > \frac{1}{4}e^{σr_{k}^{n}}. \end{matrix}  (3.15)

And from (3.7), (3.10) and (3.13), we have

\begin{matrix}|z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−Q(z_{k})}ν_{g}^{s+1}(r_{k})(1 + o(1)) & z_{k}^{k}h_{0}(z_{k})| \\ ≤ 2r_{k}^{k−s}e^{r_{k}^{σ+ε}} & e^{sr_{k}^{β+ε}} + r_{k}^{k}e^{r_{k}^{σ+ε}} ≤ e^{r_{k}^{σ+2ε}}. \end{matrix}  (3.16)

From (3.14) we see that (3.16) is in contradiction to (3.15).

Case (ii): θ_{0} ∈ G_{j} where j is even. Since G_{j} is an open set and lim_{k→∞} θ_{k} = θ_{0}, there is K > 0 such that θ_{k} ∈ G_{j} for k > K. By Lemma 2.4, we have

\begin{matrix}Re \{Q(r_{k}e^{iθ_{k}})\} & > σr_{k}^{n},  Re \{−cQ(r_{k}e^{iθ_{k}})\} < −cσr_{k}^{n}, \\ Re \{(1 − c)Q(r_{k}e^{iθ_{k}})\} < (1 − c)σr_{k}^{n}. \end{matrix}  (3.17)

We may rewrite (3.14) to

\begin{matrix}− z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−cQ(z_{k})}ν_{g}^{s}(r_{k})(1 + o(1)) = e^{−cQ(z_{k})}[ν_{g}^{k}(r_{k})(1 + o(1)) \\ + z_{k}h_{k−1}ν_{g}^{k−1}(r_{k})(1 + o(1)) + ⋯ + z_{k}^{k−s−1}h_{s+1}(z_{k})ν_{g}^{s+1}(r_{k})(1 + o(1)) \\ + z_{k}^{k−s+1}h_{s−1}(z_{k})ν_{g}^{s−1}(r_{k})(1 + o(1)) + ⋯ + z_{k}^{k−1}h_{1}(z_{k})ν_{g}(r_{k})(1 + o(1))] \\ + z_{k}^{k}h_{0}(z_{k})e^{(1−c)Q(z_{k})}. \end{matrix}  (3.18)

Thus from (3.7),(3.8),(3.10),(3.17) and (3.18), we have

\begin{aligned}e^{−r_{k}^{m+ε}} & < \frac{1}{2}r_{k}^{k−s}e^{−r_{k}^{σ+ε}}e^{−r_{k}^{m+ε}}e^{sr_{k}^{σ−ε}} \\ & | − z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−cQ(z_{k})}ν_{g}^{s}(r_{k})(1 + o(1))| < e^{\frac{1−c}{2} σr_{k}^{n}} \end{aligned}  (3.19)

This is in contradiction to n > m + ε and c > 1.

Case (iii). θ_{0} = −\frac{θ_{n}}{n} + (2j − 1)\frac{π}{2n} for some j ∈ \{0, 1, ⋯, 2n − 1\}. Since Re \{Q(r_{k}e^{iθ_{k}})\} = 0 when r_{k} is sufficiently large and a ray arg z = θ_{0} is an asymptotic line of {r_{k}e^{θ_{k}}}, where is a K > 0 such that when k > K, we have

−1 < Re {Q(r_{k}e^{iθ_{k}})} < 1.  (3.20)

Since a_{n} = cb_{n}, so the head terms of P (z) and Q(z) have the same argument, therefore by Lemma 2.4, Re {P (z)/c} and Re {Q(z)} possesses the same property in the above G_{j}(j = 0, 1, ⋯, 2n − 1), i.e., when k > K, we have

−1 < Re {P (r_{k}e^{iθ_{k}})/c} < 1.  (3.21)

Hence when k > K, we have

−2c < Re {P (r_{k}e^{iθ_{k}}) − cQ(r_{k}e^{iθ_{k}})} < 2c.  (3.22)

We may rewrite (3.14) to

\begin{matrix}− e^{−cQ(z_{k})}[ν_{g}^{k}(r_{k})(1 + o(1)) + z_{k}h_{k−1}ν_{g}^{k−1}(r_{k})(1 + o(1)) + ⋯ + \\ z_{k}^{k−s−1}h_{s+1}(z_{k})ν_{g}^{s+1}(r_{k})(1 + o(1)) + z_{k}^{k−s+1}h_{s−1}(z_{k})ν_{g}^{s−1}(r_{k})(1 + o(1)) \\ + ⋯ + z_{k}^{k−1}h_{1}(z_{k})ν_{g}(r_{k})(1 + o(1))] \\ = z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−cQ(z_{k})}ν_{g}^{s}(r_{k})(1 + o(1)) + z_{k}^{k}h_{0}(z_{k})e^{(1−c)Q(z_{k})}. \end{matrix}  (3.23)

Thus from (3.7), (3.10) and (3.21)-(3.23), we obtain, for a sufficiently large k,

\begin{matrix}\frac{1}{4}e^{−c}ν_{g}^{k}(r_{k}) < & | − e^{−cQ(z_{k})}[ν_{g}^{k}(r_{k})(1 + o(1)) + z_{k}h_{k−1}ν_{g}^{k−1}(r_{k})(1 + o(1)) + ⋯ + \\ z_{k}^{k−s−1}h_{s+1}(z_{k})ν_{g}^{s+1}(r_{k})(1 + o(1)) + z_{k}^{k−s+1}h_{s−1}(z_{k})ν_{g}^{s−1}(r_{k})(1 + o(1)) \\ + ⋯ + z_{k}^{k−1}h_{1}(z_{k})ν_{g}(r_{k})(1 + o(1))]| \\ = |z_{k}^{k−s}h_{s}(z_{k})e^{P (z_{k})−cQ(z_{k})}ν_{g}^{s}(r_{k})(1 + o(1)) + z_{k}^{k}h_{0}(z_{k})e^{(1−c)Q(z_{k})}| \\ ≤ 2r_{k}^{k−s}e^{r_{k}^{σ+ε}}ν_{g}^{k}(r_{k}) + r_{k}^{k}e^{r_{k}^{σ+ε}}e^{r_{k}^{σ+ε}}e^{c−1} ≤ ν_{g}^{k}(r_{k})e^{r_{k}^{σ+2ε}}. \end{matrix}  (3.24)

This is in contradiction to ν_{g}(r_{k}) ≥ exp{r_{k}^{σ−ε}}. Thus we complete the proof of Part (2) of Theorem 1.4.

(3). By using the same argument as in Theorem 2 (iv) of [13], we can prove part (3). Here we omit the detail.

3.3 Proof of Theorem 1.5

Proof. Assume f_{0} is a solution of finite order of (1.4). If there exists another solution f_{1}(≢ f_{0}) of finite order of (1.4), then σ(f_{1} − f_{0}) < ∞, and f_{1} − f_{0} is a solution of the corresponding homogeneous differential equation (1.3). However, by Theorem 1, we get that σ(f_{1} − f_{0}) = ∞, which is in contradiction to σ(f_{1} − f_{0}) < ∞. Hence all solutions f of non-homogeneous linear differential equation (1.4), with at most one exceptional solution f_{0} of finite order, satisfy σ(f) = ∞.

Now suppose that f is a solution of infinite order of (1.4), then by Lemma 2.8, we obtain

λ(f) = \overline{λ}(f) = σ(f) = ∞.

In the following, we shall verify that every solution f of infinite order of (1.4) satisfy \overline{λ}_{2}(f) = σ_{2}(f). In fact, by (1.4), it is easy to see that the zeros of f occurs at the poles of h_{j}(z)(j = 1, …, k − 1) or the zeros of F (z). If f has a zero at z_{0} of order n, n > k, then F (z) must have a zero at z_{0} of order n − k. Therefore we get by F ≢ 0 that

N(r, \frac{1}{f}) ≤ k\overline{N}(r, \frac{1}{f}) + N(r, \frac{1}{F} ) + ∑_{j=0}^{k−1} N(r, h_{j}).

On the other hand, (1.4) may be rewritten as follows

\frac{1}{f} = \frac{1}{F} [\frac{f^{(k)}}{f} + h_{k−1} \frac{f^{(k−1)}}{f} + ⋯ + h_{s}e^{P} \frac{f^{(s)}}{f} ⋯ + h_{1} \frac{f′}{f} + h_{0}e^{Q}].

So

m(r, \frac{1}{f}) ≤ m(r, \frac{1}{F} ) + ∑_{j=0}^{k−1} m(r, h_{j}) + m(r, e^{P}) + m(r, e^{Q}) + ∑_{j=0}^{k−1} m(r, \frac{f^{(j)}}{f}) + O(1).

Hence by the logarithmic derivative lemma, there exists a set E having finite linear measure such that for all r ∉ E, we have

T (r, f) = T (r, \frac{1}{f}) + O(1)

≤T (r, \frac{1}{F} ) + k\overline{N}(r, \frac{1}{f}) + ∑_{j=0}^{k−1} T (r, h_{j}) + m(r, e^{P}) + m(r, e^{Q}) + ∑_{j=0}^{k−1} m(r, \frac{f^{(j)}}{f}) + O(1)
≤T (r, F) + ∑_{j=0}^{k−1} T (r, h_{j}) + C log(rT (r, f)) + T (r, e^{P}) + T (r, e^{Q}) + k\overline{N}(r, \frac{1}{f}) + O(log r),

where C is a positive constant. Since for any ε > 0 and sufficiently large r, we have

C log(rT (r, f)) ≤ \frac{1}{2}T (r, f), T (r, F) ≤ r^{σ(F)+ε}, T (r, e^{P}) ≤ r^{n+ε};

T (r, e^{Q}) ≤ r^{n+ε}, T (r, h_{j}) ≤ r^{σ+ε}, j = 0, 1, ⋯, k;

so that for r ∉ E and sufficiently large r, we have

T (r, f) ≤ 2k\overline{N}(r, \frac{1}{f}) + (4k + 5)r^{σ+ε} + 4r^{n+ε} + 2r^{σ(F)+ε}.

Hence by Lemma 2.2, we get that σ_{2}(f) ≤ \overline{λ_{2}}(f). It is obvious that λ_{2}(f) ≥ \overline{λ}_{2}(f) ≥ σ_{2}(f), hence λ_{2}(f) = \overline{λ}_{2}(f) = σ_{2}(f).

Finally, let f_{0} be a solution of finite order of (1.4), then f_{0} ≢ 0. Substitute it into (1.4), and rewrite it as follows

\frac{1}{f_{0}} = \frac{1}{F} [\frac{f_{0}^{(k)}}{f_{0}} + h_{k−1} \frac{f_{0}^{(k−1)}}{f_{0}} + ⋯ + h_{s}e^{P} \frac{f_{0}^{(s)}}{f_{0}} + ⋯ + h_{1} \frac{f′_{0}}{f_{0}} + h_{0}e^{Q}].

Thus

m(r, \frac{1}{f_{0}} ) ≤ m(r, \frac{1}{F} ) + ∑_{j=0}^{k−1} m(r, h_{j}) + m(r, e^{P}) + m(r, e^{Q}) + ∑_{j=0}^{k−1} m(r, \frac{f_{0}^{(j)}}{f_{0}} ) + O(1).

It is easy to see that f_{0} occurs at the poles of h_{j}(z)(j = 1, …, k − 1) or the zeros of F (z). If f_{0} has a zero at z_{0} of order n, n > k, then F (z) must have a zero at z_{0} of order n − k. Therefore we get by F ≢ 0 that

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

So by the logarithmic derivative lemma, and noting that σ(f_{0}) < +∞, we can obtain that

\begin{matrix}T (r, f) = T (r, \frac{1}{f}) + O(1) \\ ≤ & T (r, F) + ∑_{j=0}^{k−1} T (r, h_{j}) + T (r, e^{P}) + T (r, e^{Q}) + k\overline{N}(r, \frac{1}{f}) + O(log r). \end{matrix}

Hence σ(f_{0}) ≤ max\{n, σ(F), \overline{λ}(f_{0})\}, and this completes the proof of the theorem.

Example 1. Consider the non-homogeneous linear differential equation

f^{(k)} + h_{k−1}f^{(k−1)} + ⋯ + h_{2}f′′ + \frac{1}{z}e^{iz}f′ − \frac{1}{z^{2}} e^{−iz}f = \frac{1}{z}2i sin z,

where h_{2}, ⋯, h_{k−1} are meromorphic functions. It has a solution f_{0}(z) = z of finite order.

Example 2.(see. [16]) Consider the non-homogeneous linear differential

f′′′ + e^{z^{2}}f′′ − f′ + ze^{z^{2}−z}f = ze^{z^{2}} + e^{z^{2}+z}.

It has a solution f_{0}(z) = e^{z} of finite order. σ(f_{0}) = 1 < 2 = max\{2, σ(F), \overline{λ}(f_{0})\}.

Acknowledgements

Project supported by the NSFC-RFBR and the NSFC (No. 10771121).

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(Received August 6, 2008)