Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 24, 1999, 215–224

ON THE ZEROS AND HYPER-ORDER OF MEROMORPHIC SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS

Zong-Xuan Chen and Chung-Chun Yang

Jiangxi Normal University, Department of Mathematics Nanchang, 330027, P.R. China The Hong Kong University of Science and Technology, Department of Mathematics Kowloon, Hong Kong; mayang@uxmail.ust.hk

Abstract. In this paper we investigate the exponent of convergence of the zeros and the hyper-order of meromorphic solutions of higher-order homogeneous linear differential equations with transcendental coefficients.

1. Introduction and results

Throughout the presentation, f will denote a transcendental meromorphic function in the finite complex plane, and the standard notations of the Nevanlinna theory will be employed (see e.g. [8]). In addition, we will use notations λ(f) and λ̄(f) to denote the exponent of convergence of the zeros and distinct zeros of f(z), respectively; λ̄(1/f) will be used to denote the exponent of convergence of the distinct poles of meromorphic function f(z), and σ(f) will denote the order of growth of f(z) and μ(f) the lower order of f(z). The hyper-order of f, σ_{2}(f), is defined to be \overline{lim}_{r→∞} log log T (r, f)/ log r. The following two results were proved by S. Bank and I. Laine in [1] and [2], respectively.

Theorem A. Let A(z) be a transcendental entire function of order σ (0 < σ ≤ ∞), with λ(A) < σ. Then any solution f of

(1.1)  f′′ + Af = 0

satisfies λ(f) ≥ σ.

Theorem B. Let A(z) be a transcendental meromorphic function of order σ, where 0 < σ ≤ +∞, and assume that λ̄(A) < σ. Then if f(z) ≢ 0 is a meromorphic solution of (1.1), we have

max{λ̄(f), λ̄(1/f)} ≥ σ.

In this paper, the exponent of convergence of the distinct zeros of an arbitrary solution of higher-order linear equations will be considered.

1991 Mathematics Subject Classification: Primary 34A20, 30D35.

∗ The work of the former author was supported by the National Natural Science Foundation of China and that of the latter by a U.G.C. grant of Hong Kong.

Theorem 1. Let A_{0}, …, A_{k−1} be entire functions such that

(1.2) max{σ(A_{1}), σ(A_{2}), …, σ(A_{k−1}), λ(A_{0})} < σ(A_{0}) = σ  (0 < σ ≤ ∞),

and that A_{0} has at least one zero whose multiplicity is not a multiple of k. Then every solution f of the equation

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

satisfies λ̄(f) ≥ σ. Moreover, if A_{0} is transcendental with σ(A_{0}) = 0 having at least one zero whose multiplicity is not a multiple of k, and A_{1}, …, A_{k−1} are polynomials, every solution f of (1.3) has infinitely many zeros.

We will give some further estimates on the growth of infinite order solutions of the equation (1.3). It is well known that all solutions of (1.3) are entire functions, and when some of the coefficients of (1.3) are transcendental, (1.3) has at least one solution f with σ(f) = ∞.

Recently, Ki-Ho Kwon obtained the following result in [8] for the second-order linear differential equation

(1.4)  f′′ + A(z)f′ + B(z)f = 0.

Theorem C. Let A(z) and B(z) be entire functions such that σ(A) < σ(B) or σ(B) < σ(A) < \frac{1}{2} . Then every solution f ≢ 0 of (1.4) satisfies

\overline{lim}_{r→∞} \frac{log log T (r, f)}{log r} ≥ max\{σ(A), σ(B)\}.

We shall investigate the above types of problems as generalizations for higherorder homogeneous linear differential equations, with one coefficient whose growth rate is the dominant one. The reader is referred to [9] and [11] for some of the results that relate to Theorem 2.4 below.

Theorem 2. Suppose that A_{0}, …, A_{k−1} are entire functions and there exists one A_{s} (0 ≤ s ≤ k − 1) satisfying for j ≠ s such that

σ(A_{j}) < σ(A_{s}).

Then the equation (1.3) has at least one solution f satisfying either λ(f) ≥ σ(A_{s}) or σ_{2}(f) = σ(A_{s}).

Theorem 3. Suppose that A_{0}, …, A_{k−1} are entire functions such that max{σ(A_{j}) : j = 1, …, k − 1} < σ(A_{0}).

Then every solution f ≢ 0 of (1.3) satisfies σ_{2}(f) ≥ σ(A_{0}). Furthermore, if λ(f) < +∞, then σ_{2}(f) = σ(A_{0}).

Theorem 4. Suppose that A_{0}, …, A_{k−1} are entire functions such that A_{0} ≢ 0 and

max\{σ(A_{j}) : j = 0, 2, …, k − 1\} < σ(A_{1}) < \frac{1}{2}.

Then every solution f ≢ 0 of (1.3) satisfies σ_{2}(f) ≥ σ(A_{1}). Moreover, if λ(f) < +∞, then σ_{2}(f) = σ(A_{1}).

On the zeros and hyper-order of meromorphic solutions differential equations 217

2. Proof of Theorem 1

Assume that f ≢ 0 is a solution of (1.3). It is easy to see that f is entire with σ(f) = ∞. Now we assume that the assertion of Theorem 1 is false. That is,

(2.1)  λ̄(f) < σ.

Set f′/f = g. Then, by an elementary computation, it follows that

(2.2)  f^{(j)}/f = g^{j} + \frac{1}{2}j(j − 1)g^{j−2}g′ + H_{j−2}(g),

where H_{j−2}(g) is a differential polynomial in g and its derivatives with constant coefficients, and the degree of H_{j−2}(g) is no greater than j −2. Substituting (2.2) in (1.3), we get

(2.3)  −A_{0} = g^{k} + \frac{1}{2}k(k − 1)g^{k−2}g′ + A_{k−1}g^{k−1} + P_{k−2}(g),

where P_{k−2}(g) is a differential polynomial in g and its derivatives with coefficients that are linear combinations of A_{1}, …, A_{k−2} with constant coefficients, and the degree of P_{k−2}(g) is no greater than k − 2. Since the poles of g = f′/f can only occur at the zeros of f, and all poles of g are simple, it follows from (2.1) that

(2.4)  σ(N(r, g)) = λ̄(f) < σ.

By (2.3), we have

(2.5)  T (r, A_{0}) ≤ M(T (r, g) + ∑_{j=1}^{k−1} T (r, A_{j})) + S(r, g)  (r ∉ E),

where E ⊂ (0, +∞) with a finite linear measure and M (> 0) is a constant. By (1.2) and hence because μ(A_{0}) = σ(A_{0}) (see e.g. [5, Corollary 6.1]), we have σ(A_{j}) < μ(A_{0}) = σ(A_{0}) = σ for j = 1, …, k − 1. It follows that the inequality

(2.6)  M ∑_{j=1}^{k−1} T (r, A_{j}) < \frac{1}{2}T (r, A_{0})

holds for r ∉ E and sufficiently large r. By (2.5) and (2.6), we have

(2.7)  T (r, A_{0}) ≤ (2M + o(1))T (r, g)

for r ∉ E and sufficiently large r. Hence, again by [5] as well as by (2.4), (1.2), and (2.7), we have μ(g) = σ(g) ≥ σ,

T (r, A_{j}) = S(r, g) (j = 1, …, k),

and, for r ∉ E,

N(r, g) = o(1)T (r, g) = S(r, g).

Thus by (1.2) and μ(g) ≥ σ, N(r, 1/A_{0}) = S(r, g) holds. Therefore (2.3) satisfies all the conditions of the Tumura–Clunie theorem [8, p. 69], and it follows that

−A_{0} = h^{k}(z), h(z) = g(z) + \frac{1}{k} a(z),

where h satisfies

\begin{aligned}h^{k−1}a & = \frac{1}{2}k(k − 1)h^{k−2}h′ + A_{k−1}h^{k−1} \\ a & = \frac{1}{2}k(k − 1)\frac{h′}{h} + A_{k−1}, \frac{h′}{h} = \frac{1}{k} \frac{A′_{0}}{A_{0}} . \end{aligned}

Hence

−A_{0} = (\frac{f′}{f} + \frac{k − 1}{2k} \frac{A′_{0}}{A_{0}} + \frac{A_{k−1}}{k} )^{k}.

This contradicts the assumption that A_{0} has at least one zero whose multiplicity is not a multiple of k. Thus we must have λ̄(f) ≥ σ, and Theorem 1 is proved.

3. Lemmas needed for the proof of Theorem 2

Lemma 1 ([6, Lemma 3]). Let the differential equation

(3.1)  w^{(k)} + a_{k−1}w^{(k−1)} + ⋯ + a_{0}w = 0

be satisfied in the complex plane by linearly independent meromorphic functions f_{1}, …, f_{k}. Then the coefficients a_{j} (j = 0, …, k − 1) are meromorphic in the plane satisfying the properties

(3.2)  m(r, a_{j}) = O{log[max(T (r, f_{s}) : s = 1, …, k)]}.

Using the same reasoning as in the proof of the Tumura–Clunie theorem, we can obtain the following lemma.

Lemma 2. Let g^{n+1} = P_{n}(g), and let P_{n}(g) be a differential polynomial in the transcendental entire function g(z) of total degree at most n, with the coefficients in w′/w, …, w^{(n)}/w, and A_{0}, …, A_{k−1} (where w is an entire function with σ(w) < ∞, A_{0}, …, A_{k−1} , satisfying the additional hypotheses of Theorem 2). Then σ(g) ≤ σ(A_{s}).

Proof. Since

(3.3)

m(r, g) = \frac{1}{2π} ∫_{0}^{2π} log^{+} |g(re^{iθ})| dθ

= \frac{1}{2π} ∫_{ε_{1}} log^{+} |g(re^{iθ})| dθ + \frac{1}{2π} ∫_{ε_{2}} log^{+} |g(re^{iθ})| dθ,

where ε_{1} = {θ : |g(re^{iθ})| < 1} ∩ [0, 2π], ε_{2} = [0, 2π] − ε_{1} , we have

(3.4)  ∫_{ε_{1}} log^{+} |g(re^{iθ})| dθ = 0.

On ε_{2} , we have |g(re^{iθ})| ≥ 1. By g = P_{n}(g)/g^{n} and the hypothesis that P_{n}(g) is the sum of a finite number of terms of the type

a(z)g^{l_{0}} · (g′)^{l_{1}} ⋯ (g^{(ν)})^{l_{ν}} ,

where l_{0}, …, l_{ν} are non-negative integers and ∑_{t=0}^{ν} l_{t} = n, a(z) is a combination of the addition, subtraction and multiplication operations of w′/w, ⋯, w^{(n)}/w, A_{0}, …, A_{k−1} . We have

(3.5) \begin{aligned}\frac{1}{2π} ∫_{ε_{2}} log^{+} |g(re^{iθ})| dθ & ≤ M(∑ m(r, a) + ∑_{t=1}^{ν} m(\frac{r, g^{(t)}}{g} )) \\ & ≤ M(∑_{j=1}^{n} m(r, \frac{w^{(j)}}{w} ) + ∑_{j=0}^{k−1} m(r, A_{j}) + S(r, g)) \end{aligned}

for |z| = r outside a set E of finite linear measure, where M is some positive constant. By the fact that m(r, w^{(j)}/w) = O(log r) (j = 1, …, n) and (3.3)–(3.5), we have

(3.6)  m(r, g) ≤ M(∑_{j=0}^{k−2} m(r, A_{j}) + S(r, g)),  r ∉ E.

By S(r, g)/m(r, g) → 0 (r → ∞) and the additional hypotheses, we get σ(g) ≤ σ(A_{s}) from (3.6).

Lemma 3 [12, Theorem 1.45]. Let h(z) be a nonconstant entire function with σ(h) = σ. If f(z) = exp{h(z)}, then σ_{2}(f), the hyper-order of f(z), satisfies σ_{2}(f) = σ.

4. Proof of Theorem 2

Proof. By Lemma 1 and σ(A_{j}) < σ(A_{s}) (j ≠ s), it follows that the equation (1.3) has at least one solution f with σ(f) = ∞. Now assume that λ(f) < σ(A_{s}). Then f can be expressed in the form f = we^{h} such that σ(w) < σ(A_{s}) and h is a transcendental entire function. By mathematical induction, we can prove for n = 1, …, k

(4.1)  f^{(n)}(z) = we^{h}(h′)^{n} + we^{h}P_{n−1}(h′),

where P_{n−1}(h′) is a differential polynomial in h′, …, h^{(n)} of total degree n − 1 with its coefficients being polynomials in w′/w, ⋯, w^{(n)}/w. Substituting (4.1) in (1.3), we obtain

(4.2)  (h′)^{k} + P_{k−1}(h′) + A_{k−1}[(h′)^{k−1} + P_{k−2}(h′)] + ⋯ + A_{0} = 0,

i.e.,

(4.3)  (h′)^{k} = P_{k−1}^{∗}(h′),

where P_{k−1}^{∗}(h′) is a differential polynomial in h′, …, h^{(k)} of total degree k − 1 with its coefficients being polynomials in w′/w, …, w^{(k)}/w, A_{0}, …, A_{s}, …, A_{k−1} . By (4.3) and Lemma 2, we have σ(h) = σ(h′) ≤ σ(A_{s}). Since (h′)^{s} + P_{s−1}(h′) = f^{(s)}/f ≢ 0 and (4.2), we get σ(h′) ≥ σ(A_{s}). Hence σ(h) = σ(A_{s}).

By Lemma 3 and f = we^{h} with σ(w) < ∞, it follows that σ_{2}(f) = σ(h) = σ(A_{s}).

5. Proof of Theorem 3

Set max{σ(A_{j}) : j = 1, …, k − 1} = ρ, < σ(A_{0}) = α. We can rewrite (1.3) as

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

Hence the inequality

(5.2)  m(r, A_{0}) ≤ ∑_{j=1}^{k−1} m(r, A_{j}) + O{log T (r, f) + log r}

holds for all r outside a set E ⊂ (0, +∞) with a linear measure mE = δ < +∞. Since σ(A_{0}) = α, there exists {r′_{n}} (r′_{n} → ∞) such that

(5.3)  lim_{r′_{n}→∞} \frac{log m(r′_{n}, A_{0})}{log r′_{n}} = α.

By mE = δ < ∞, there exists a point r_{n} ∈ [r′_{n}, r′_{n} + δ + 1] − E. From

\frac{log m(r_{n}, A_{0})}{log r_{n}} ≥ \frac{log m(r′_{n}, A_{0})}{log(r′_{n} + δ + 1)} = \frac{log m(r′_{n}, A_{0})}{log r′_{n} + log(1 + (δ + 1)/r′_{n})}

we get

(5.4)  \underline{lim}_{r_{n}→∞} \frac{log m(r_{n}, A_{0})}{log r_{n}} ≥ α.

So, for any given ε (0 < 2ε < α − ρ), and for j = 1, …, k − 1,

m(r_{n}, A_{j}) ≤ r_{n}^{ρ+ε}  and  m(r_{n}, A_{0}) > r_{n}^{α−ε}

hold for any sufficiently large r_{n}. Therefore,

(5.5)  ∑_{j=1}^{k−1} m(r_{n}, A_{j}) ≤ \frac{1}{2}m(r_{n}, A_{0})

holds for a sufficiently large r_{n}. By (5.2) and (5.5), we get for a sufficiently large r_{n}

(5.6)  m(r_{n}, A_{0}) ≤ M{log T (r_{n}, f) + log r_{n}}.

By (5.4) and (5.6), we have σ_{2}(f) ≥ σ(A_{0}). By this and Theorem 2, we easily conclude that σ_{2}(f) = σ(A_{0}) if λ(f) < +∞.

6. Proof of Theorem 4

Our proof depends mainly on the following four known results.

Lemma 4 ([6]). Let f be a transcendental meromorphic function, and let α > 1 be a given constant. Then there exists a set E_{1} ⊂ (1, ∞) that has finite logarithmic measure and a constant B > 0 that depends only on α and j = 1, …, k, such that for all z satisfying |z| = r ∉ [0, 1] ∪ E_{1} and for j = 2, …, k, we have

|\frac{f^{(j)}(z)}{f′(z)}| ≤ B(\frac{T (αr, f)}{r}(log^{α} r) log T (αr, f))^{k−j}.

Lemma 5 ([10]). Let f(z) be a nonconstant entire function. Then there exists a real number R such that for all r ≥ R there is some corresponding z_{r} with |z_{r}| = r satisfying

|f(z_{r})/f′(z_{r})| ≤ r.

Lemma 6 ([3]). Let f(z) be entire of order σ(f) = ρ < \frac{1}{2} . Denote A(r) = inf_{|z|=r} log |f(z)|, B(r) = sup_{|z|=r} log |f(z)|. If ρ < α < 1, then

\underline{log dens}\{r : A(r) > (cos πα)B(r)\} ≥ 1 − ρ/α,

where

\underline{log dens}(H) = \underline{lim}_{r→∞}\{lm(H ∩ [1, r])/ log r\}

and

\overline{log dens}(H) = \overline{lim}_{r→∞}\{lm(H ∩ [1, r])/ log r\}.

Lemma 7 ([4]). Let f(z) be entire with μ(f) = μ < \frac{1}{2} and μ < ρ = σ(f). If μ ≤ δ < min(ρ, \frac{1}{2}) and δ < α < \frac{1}{2} , then

\overline{log dens}\{r : A(r) > (cos πα)B(r) > r^{δ}\} > C(ρ, δ, α),

where C(ρ, δ, α) is a positive constant depending only on ρ, δ, and α.

Proof of Theorem 4. Suppose that ρ and α are real numbers satisfying

max{σ(A_{j}) : j = 0, 2, …, k − 1} < ρ < α < σ(A_{1}).

It is easy to see that every solution f ≢ 0 of (1.3) is transcendental and one can have from (1.3)

(6.1)  |A_{1}| ≤ |\frac{f^{(k)}}{f′} | + |A_{k−1}||\frac{f^{(k−1)}}{f′} | + ⋯ + |A_{2} \frac{f′′}{f′} | + |A_{0}| · |\frac{f}{f′} |.

By Lemma 4, there exists a set E_{1} ⊂ (1, ∞) with the logarithmic measure lmE_{1} < ∞, such that for j = 2, …, k, and for all z satisfying |z| = r ∉ [0, 1] ∪ E_{1} , we have

(6.2)  |\frac{f^{(j)}(z)}{f′(z)}| ≤ log^{2k} r · [T (2r, f′)]^{2k}.

By Lemma 6 (if μ(A_{1}) = σ(A_{1})) or Lemma 7 (if μ(A_{1}) < σ(A_{1})), there exists a set H ⊂ (1, +∞) with lmH = ∞, such that for all z satisfying |z| = r ∈ H we have

(6.3)  |A_{1}(z)| ≥ exp{r^{α}}

and

(6.4)  |A_{j}(z)| ≤ exp{r^{ρ}}  (j = 0, 2, …, k − 1).

By Lemma 5, there exists a number R > 0 such that for all r ≥ R there is a corresponding z_{r} with |z_{r}| = r satisfying

(6.5)  |f(z_{r})/f′(z_{r})| ≤ r.

By (6.1)–(6.5), it follows that there is a sequence {z_{r}} with |z_{r}| = r ∈ H −([0, R)∪ E_{1}) (z_{r} → ∞) such that

exp{r^{α}} ≤ M exp{r^{ρ}}r log^{2k} r[T (2r, f′)]^{2k},

where M (> 0) is a suitable constant. Thus

\overline{lim}_{r→∞} \frac{log^{+} log^{+} T (r, f′)}{log r} ≥ α.

Since α is arbitrary and

T (r, f′) = m(r, f′) + N(r, f′) ≤ m(r, f) + m(r, \frac{f′}{f} ) + 2N(r, f)
≤ 2T (r, f) + M{log rT (r, f)}
≤ (2 + M)T (r, f) + M log r (r ∉ E),

where mE < +∞, we get σ_{2}(f) ≥ σ(A_{1}). By Theorem 2, we easily conclude that σ_{2}(f) = σ(A_{1}) if λ(f) < +∞.

Acknowledgement. The authors are indebted to the referee for his/her helpful remarks and suggestions.

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224 Zong-Xuan Chen and Chung-Chun Yang

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Received 6 August 1997