Electronic Journal of Qualitative Theory of Differential Equations 2006, No. 9, 1-11; http://www.math.u-szeged.hu/ejqtde/

On the iterated order and the fixed points of entire solutions of some complex linear differential equations

Benharrat BELAI^{..}DI

Department of Mathematics Laboratory of Pure and Applied Mathematics University of Mostaganem B. P 227 Mostaganem-(Algeria) belaidi@univ-mosta.dz belaidi.benharrat@caramail.com

Abstract. In this paper, we investigate the iterated order of entire solutions of homogeneous and non-homogeneous linear differential equations with entire coefficients.

2000 Mathematics Subject Classification: 34M10, 30D35. Key words: Linear differential equation, growth of entire function, iterated order.

1 Introduction and statement of results:

For the definition of the iterated order of an entire function, we use the same definition as in [9] , [2, p. 317] , [10, p. 129] . For all r ∈ R, we define exp_{1} r := e^{r} and exp_{p+1} r := exp (exp_{p} r) , p ∈ N. We also define for all r sufficiently large log_{1} r := log r and log_{p+1} r := log (log_{p} r) , p ∈ N. Moreover, we denote by exp_{0} r := r, log_{0} r := r, log_{−1} r := exp_{1} r and exp_{−1} r := log_{1} r.

Definition 1.1 Let f be an entire function. Then the iterated p−order σ_{p} (f) of f is defined by

σ_{p} (f) = \overline{lim}_{r→+∞} \frac{log_{p}T (r, f)}{log r} = \overline{lim}_{r→+∞} \frac{log_{p+1}M (r, f)}{log r} (p ≥ 1 is an integer) ,

(1.1) where T (r, f) is the Nevanlinna characteristic function of f and M (r, f) = max_{|z|=r} |f (z)| (see [7] , [13]) . For p = 1, this notation is called order and for p = 2 hyper-order.

Definition 1.2 The finiteness degree of the order of an entire function f is defined by

i (f) = \{ \begin{matrix}0,  for f polynomial, \\ min \{j ∈ N : σ_{j} (f) < ∞\} , for f transcendental for which \\ some j ∈ N with σ_{j} (f) < ∞ exists, \\ ∞,  for f with σ_{j} (f) = ∞ for all j ∈ N. \end{matrix}

(1.2)

Definition 1.3 Let f be an entire function. Then the iterated convergence exponent of the sequence of distinct zeros of f (z) is defined by

\overline{λ}_{p} (f) = \overline{lim_{r→+∞}} \frac{log_{p} \overline{N} (r, \frac{1}{f})}{log r},  (1.3)

where \overline{N} (r, \frac{1}{f}) is the counting function of distinct zeros of f (z) in \{|z| < r\}.

Thus \overline{λ}_{p} (f − z) is an indication of oscillation of the fixed points of f (z) .

For k ≥ 2, we consider the linear differential equations

f^{(k)} + A_{k−1} (z) f^{(k−1)} + ... + A_{1} (z) f′ + A_{0} (z) f = 0,  (1.4)
f^{(k)} + A_{k−1} (z) f^{(k−1)} + ... + A_{1} (z) f′ + A_{0} (z) f = F (z) ,  (1.5)

where A_{0} (z) , ..., A_{k−1} (z) and F (z) / ≡ 0 are entire functions. It is well-known that all solutions of equations (1.4) and (1.5) are entire functions.

Extensive work in recent years has been concerned with the growth of solutions of complex linear differential equations. Many results have been obtained. Examples of such results are the following two theorems:

Theorem A [4]. Let A_{0} (z) , ..., A_{k−1} (z) be entire functions such that there exists one transcendental A_{s} (0 ≤ s ≤ k − 1) satisfying σ (A_{j}) ≤ σ (A_{s}) for all j ≠ s. Then equation (1.4) has at least one solution f that satisfies σ_{2} (f) = σ (A_{s}) .

Theorem B [4]. Let A_{0} (z) , ..., A_{k−1} (z) satisfy the hypotheses of Theorem A and F (z) / ≡ 0 be an entire function with σ (F) < +∞. Assume that f_{0} is a solution of (1.5) , and g_{1}, ..., g_{k} are a solution base of the corresponding homogeneous equation (1.4) of (1.5) . Then there exists a g_{j} (1 ≤ j ≤ k) , say g_{1}, such that all the solutions in the solution subspace {cg_{1} + f_{0}, c ∈ C} satisfy σ_{2} (f) = \overline{λ_{2}} (f) = σ (A_{s}) , with at most one exception.

The purpose of this paper is to extend the above two results by considering the iterated order. We will prove the following theorems:

Theorem 1.1 Let A_{0} (z) , ..., A_{k−1} (z) be entire functions such that there exists one transcendental A_{s} (0 ≤ s ≤ k − 1) satisfying σ_{p} (A_{j}) ≤ σ_{p} (A_{s}) < +∞ for all j ≠ s. Then equation (1.4) has at least one solution f that satisfies i (f) = p + 1 and σ_{p+1} (f) = σ_{p} (A_{s}) .

Theorem 1.2 Let A_{0} (z) , ..., A_{k−1} (z) satisfy the hypotheses of Theorem 1.1 and F (z) / ≡ 0 be an entire function with i (F) = q. Assume that f_{0} is a solution of (1.5) , and g_{1}, ..., g_{k} are a solution base of the corresponding homogeneous equation (1.4) of (1.5) . If either i (F) = q < p+1 or q = p+1 and σ_{p+1} (F) < σ_{p} (A_{s}) < +∞, then there exist a g_{j} (1 ≤ j ≤ k) , say g_{1}, such that all the solutions in the solution subspace {cg_{1} + f_{0}, c ∈ C} satisfy i (f) = p + 1 and σ_{p+1} (f) = \overline{λ}_{p+1} (f) = σ_{p} (A_{s}) , with at most one exception.

Set g (z) = f (z) − z. Then clearly \overline{λ}_{p+1} (f − z) = \overline{λ}_{p+1} (g) and σ_{p+1} (g) = σ_{p+1} (f) . By Theorem 1.1 and Theorem 1.2, we can get the following corollaries.

Corollary 1 Under the hypotheses of Theorem 1.1, if A_{1} + zA_{0} / ≡ 0, then equation (1.4) has at least one solution f that satisfies i (f) = p + 1 and \overline{λ}_{p+1} (f − z) = σ_{p+1} (f) = σ_{p} (A_{s}) .

Corollary 2 Under the hypotheses of Theorem 1.2, if F −A_{1}−zA_{0} / ≡ 0, then every solution f of (1.5) with i (f) = p+1 and σ_{p+1} (f) = \overline{λ}_{p+1} (f) = σ_{p} (A_{s}) satisfies \overline{λ}_{p+1} (f − z) = σ_{p} (A_{s}) .

2 Preliminary Lemmas

Our proofs depend mainly upon the following lemmas.

Lemma 2.1 ([3] , [11]) . Let f (z) = ∞ ; ∑_{n=0}a_{n} z^{n} be an entire with σ_{p+1} (f) = σ,

let μ (r) be the maximum term, i.e., μ (r) = max{|a_{n}| r^{n}; n = 0, 1, ...} and let ν_{f} (r) be the central index of f, i.e., ν_{f} (r) = max{m, μ (r) = |a_{m}| r^{m}}. Then

\overline{lim}_{r→∞} \frac{log_{p+1} ν_{f} (r)}{log r} = σ.  (2.1)

Lemma 2.2 (Wiman-Valiron, [8] , [12]) . Let f (z) be a transcendental entire function, and let z be a point with |z| = r at which |f (z)| = M (r, f). Then the estimation

\frac{f^{(k)} (z)}{f (z)} = (\frac{ν_{f} (r)}{z})^{k} (1 + o (1)) (k is an integer),  (2.2)

holds for all |z| outside a set E_{2} of r of finite logarithmic measure lm (E_{2}) =

∫_{1}^{+∞} \frac{χ_{E_{2}} (t)}{t}dt, where χ_{E_{2}} is the characteristic function of E_{2}.

Lemma 2.3 (See Remark 1.3 of [9]) . If f is a meromorphic function with i (f) = p ≥ 1, then σ_{p} (f) = σ_{p} (f′).

Lemma 2.4 ([5]) . Let f_{1}, ..., f_{k} be linearly independent meromorphic solutions of the differential equation

f^{(k)} + A_{k−1} (z) f^{(k−1)} + ... + A_{1} (z) f′ + A_{0} (z) f = 0,  (2.3)

with meromorphic coefficients A_{0} (z) , ...A_{k−1} (z). Then

m (r, A_{j}) = O {log ( max_{1≤n≤k}T (r, f_{n}))} (j = 0, ..., k − 1) .  (2.4)

Lemma 2.5 ([9]) . Let f be a meromorphic function for which i (f) = p ≥ 1 and σ_{p} (f) = σ, and let k ≥ 1 be an integer. Then for any ε > 0,

m (r, \frac{f^{(k)}}{f} ) = O (exp_{p−2} \{r^{σ+ε}\}) ,  (2.5)

outside of a possible exceptional set E_{3} of finite linear measure.

To avoid some problems caused by the exceptional set we recall the following Lemma.

Lemma 2.6 ([1, p. 68] , [9]) . Let g : [0, +∞) → R and h : [0, +∞) → R be monotone non-decreasing functions such that g (r) ≤ h (r) outside of an exceptional set E of finite linear measure. Then for any α > 1, there exists r_{0} > 0 such that g (r) ≤ h (αr) for all r > r_{0}.

3 Proof of Theorem 1.1

Suppose that f is a solution of (1.4) . We can rewrite (1.4) as

\frac{f^{(k)}}{f} + A_{k−1} (z) \frac{f^{(k−1)}}{f} + ... + A_{s+1} (z) \frac{f^{(s+1)}}{f} + A_{s} (z) \frac{f^{(s)}}{f}

+A_{s−1} (z) \frac{f^{(s−1)}}{f} + ... + A_{1} (z) \frac{f′}{f} + A_{0} (z) = 0.  (3.1)

By Lemma 2.2, there exists a set E_{2} ⊂ (1, +∞) with logarithmic measure lm (E_{2}) < +∞ and we can choose z satisfying |z| = r ∉ [0, 1] ∪ E_{2} and |f (z)| = M (r, f), such that (2.2) holds. For given small ε > 0 and sufficiently large r, we have

|A_{j} (z)| ≤ exp_{p} {r^{σ_{p}(A_{s})+ε}} (j = 0, 1, ..., k − 1) .  (3.2)

Substituting (2.2) into (3.1), we obtain by using (3.2)

(\frac{ν_{f} (r)}{|z|})^{k} |1 + o (1)| ≤ k (\frac{ν_{f} (r)}{|z|})^{k−1} |1 + o (1)| exp_{p} \{r^{σ_{p}(A_{s})+ε}\}, (3.3)

(r ∉ [0, 1] ∪ E_{2}) . By Lemma 2.1, Lemma 2.6 and (3.3) , we obtain that i (f) ≤ p + 1 and

σ_{p+1} (f) = \overline{lim}_{r→∞} \frac{log_{p+1} ν_{f} (r)}{log r} ≤ σ_{p} (A_{s}) + ε.  (3.4)

Since ε > 0 is arbitrary, then σ_{p+1} (f) ≤ σ_{p} (A_{s}) .

Assume that {f_{1}, ..., f_{k}} is a solution base of (1.4). Then by Lemma 2.4

m (r, A_{s}) ≤ M log ( max_{1≤n≤k}T (r, f_{n})) .  (3.5)

We assert that there exists a set E ⊂ (0, +∞) of infinite linear measure such that

lim_{r→∞_{r∈E}} \frac{log_{p} m (r, A_{s})}{log r} = σ_{p} (A_{s}) .  (3.6)

In fact, there exists a sequence {r_{n}} (r_{n} → ∞) such that

lim_{r_{n}→∞} \frac{log_{p} m (r_{n}, A_{s})}{log r_{n}} = σ_{p} (A_{s}) .  (3.7)

We take E = ∪_{n=1}^{∞} [r_{n}, 2r_{n}]. Then on E, (3.6) holds obviously. Now by setting E_{n} = {r : r ∈ E and m (r, A_{s}) ≤ M log T (r, f_{n}) (n = 1, ..., k)}, we

have ∪_{n=1}^{k}E_{n} =E. It is easy to see that there exists at least one E_{n}, say E_{1}, which has an infinite linear measure and on which

lim_{r→∞_{r∈E_{1}}} \frac{log_{p} m (r, A_{s})}{log r} = σ_{p} (A_{s}) ,  (3.8)

and

m (r, A_{s}) ≤ M log T (r, f_{1}) (r ∈ E_{1}) .  (3.9)

From (3.8) and (3.9) we have i (f_{1}) ≥ p + 1 and σ_{p+1} (f_{1}) ≥ σ_{p} (A_{s}) . This and the fact that i (f_{1}) ≤ p + 1 and σ_{p+1} (f_{1}) ≤ σ_{p} (A_{s}) yield i (f_{1}) = p + 1 and σ_{p+1} (f_{1}) = σ_{p} (A_{s}). The proof of Theorem 1.1 is complete.

4 Proof of Theorem 1.2

Assume that f is a solution of (1.5) and g_{1}, ..., g_{k} are k entire solutions of the corresponding homogeneous equation (1.4) . Then by the proof of Theorem 1.1, we know that i (g_{j}) ≤ p + 1, σ_{p+1} (g_{j}) ≤ σ_{p} (A_{s}) (j = 1, 2, 3..., k) and there exists a g_{j}, say g_{1}, satisfying i (g_{1}) = p + 1, σ_{p+1} (g_{1}) = σ_{p} (A_{s}) . Thus by variation of parameters, f can be expressed in the form

f (z) = B_{1} (z) g_{1} (z) + ... + B_{k} (z) g_{k} (z) ,  (4.1)

where B_{1} (z) , ..., B_{k} (z) are determined by

B1^{′} (z) g_{1} (z) + ... + Bk^{′} (z) g_{k} (z) = 0
B1^{′} (z) g1^{′} (z) + ... + Bk^{′} (z) gk^{′} (z) = 0

.....................................................

B1^{′} (z) g1^{(k−1)} (z) + ... + Bk^{′} (z) gk^{(k−1)} (z) = F.  (4.2)

Noting that the Wronskian W (g_{1}, g_{2}, ..., g_{k}) is a differential polynomial in g_{1}, g_{2}, ..., g_{k} with constant coefficients, it follows that

σ_{p+1} (W) ≤ max {σ_{p+1} (g_{j}) : j = 1, ..., k} ≤ σ_{p} (A_{s}) .

Set

W_{j} = | \begin{matrix}g_{1}, ..., 0^{(j)}, ..., g_{k} \\ ... \\ ... \\ g_{1}^{(k−1)}, ..., F, ..., g_{k}^{(k−1)} \end{matrix} | = F.G_{j} (j = 1, ..., k) ,  (4.3)

where G_{j} (g_{1}, g_{2}, ..., g_{k}) are differential polynomials in g_{1}, g_{2}, ..., g_{k} and of their derivatives with constant coefficients. So

σ_{p+1} (G_{j}) ≤ max {σ_{p+1} (g_{j}) : j = 1, ..., k} ≤ σ_{p} (A_{s}) (j = 1, ..., k) ,
Bj^{′} = \frac{W_{j}}{W} = \frac{F.G_{j}}{W} (j = 1, ..., k) .  (4.4)

Since i (F) = q < p + 1 or i (F) = p + 1, σ_{p+1} (F) < σ_{p} (A_{s}), then by Lemma 2.3, we obtain

σ_{p+1} (B_{j}) = σ_{p+1} (Bj^{′}) ≤ max (σ_{p+1} (F) , σ_{p} (A_{s})) = σ_{p} (A_{s}) (j = 1, ..., k) .

(4.5)

Then from (4.1) and (4.5) , we get i (f) ≤ p + 1 and

σ_{p+1} (f) ≤ max {σ_{p+1} (g_{j}) , σ_{p+1} (B_{j}) : j = 1, ..., k} ≤ σ_{p} (A_{s}) .  (4.6)

Now we set

H = {f_{c} = cg_{1} + f_{0}, c ∈ C} ,  (4.7)

where f_{0} is a solution of (1.5) . Obviously, every f_{c} in H is a solution of (1.5) . Now we prove that for any two solutions f_{a} and f_{b} (a ≠ b) in H, there is at least one solution, say f_{a}, among f_{a} and f_{b} satisfying i (f_{a}) = p + 1 and σ_{p+1} (f_{a}) = \overline{λ}_{p+1} (f_{a}) = σ_{p} (A_{s}) . Since f_{a} = (a − b) g_{1}+ f_{b}, then

T (r, g_{1}) ≤ T (r, f_{a}) + T (r, f_{b}) + O (1) .  (4.8)

Assume that the set E_{1} satisfies the condition as required in proof of Theorem 1.1. Then there exists at least one of f_{a} and f_{b}, say f_{a}, such that there is a subset E_{4} of E_{1} with infinite linear measure and

T (r, f_{b}) ≤ T (r, f_{a}) , for r ∈ E_{4}.  (4.9)

We get from (4.8) and (4.9)

T (r, g_{1}) ≤ 2T (r, f_{a}) + O (1) , for r ∈ E_{4}.  (4.10)

Thus, i (f_{a}) ≥ p + 1 and σ_{p+1} (f_{a}) ≥ σ_{p+1} (g_{1}) = σ_{p} (A_{s}) and hence i (f_{a}) = p + 1, σ_{p+1} (f_{a}) = σ_{p} (A_{s}) = σ.

Now we prove that σ_{p+1} (f_{a}) = \overline{λ}_{p+1} (f_{a}) = σ. By (1.5) , it is easy to see that if f_{a} has a zero at z_{0} of order α (> k), then F must have a zero at z_{0} of order α − k. Hence,

n (r, \frac{1}{f_{a}} ) ≤ k \overline{n} (r, \frac{1}{f_{a}} ) + n (r, \frac{1}{F} )  (4.11)

and

N (r, \frac{1}{f_{a}} ) ≤ k \overline{N} (r, \frac{1}{f_{a}} ) + N (r, \frac{1}{F} ) .  (4.12)

Now (1.5) can be rewritten as

\frac{1}{f_{a}} = \frac{1}{F} (\frac{f_{a}^{(k)}}{f_{a}} + A_{k−1} \frac{f_{a}^{(k−1)}}{f_{a}} + ... + A_{1} \frac{fa^{′}}{f_{a}} + A_{0}) .  (4.13)

By (4.13), we have

m (r, \frac{1}{f_{a}} ) ≤ k ; ∑_{j=1}m (r, \frac{f_{a}^{(j)}}{f_{a}} ) + k ; ∑_{j=1}m (r, A_{k−j}) + m (r, \frac{1}{F} ) + O (1) . (4.14)

Applying the Lemma 2.5, we have

m (r, \frac{f_{a}^{(j)}}{f_{a}} ) = O (exp_{p−1} \{r^{σ+ε}\}) (j = 1, ..., k − 1) , (σ_{p+1} (f_{a}) = σ) ,

(4.15)

holds for all r outside a set E_{3} ⊂ (0, +∞) with a linear measure m (E_{3}) = δ < +∞. By (4.12) , (4.14) and (4.15), we get

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

≤ k\overline{N} (r, \frac{1}{f_{a}} )+ k ; ∑_{j=1}T (r, A_{k−j})+T (r, F)+O (exp_{p−1} \{r^{σ+ε}\}) (|z| = r ∉ E_{3}) .

(4.16)

For sufficiently large r, we have

T (r, A_{0}) + ... + T (r, A_{k−1}) ≤ k exp_{p−1} {r^{σ+ε}} .  (4.17)

If i (F) = q < p + 1, then q − 1 ≤ p − 1 and

T (r, F) ≤ exp_{q−1} {r^{σ_{q}(F)+ε}} ≤ exp_{p−1} {r^{σ_{q}(F)+ε}} (σ_{q} (F) < ∞) . (4.18)

Thus, by (4.16) − (4.18), we have

T (r, f_{a}) ≤ k \overline{N} (r, \frac{1}{f_{a}} ) + k exp_{p−1} \{r^{σ+ε}\}

+ exp_{p−1} {r^{σ_{q}(F)+ε}} + O (exp_{p−1} {r^{σ+ε}})  (|z| = r ∉ E_{3}) . (4.19)

Hence for any f_{a} with σ_{p+1} (f_{a}) = σ, by (4.19) and Lemma 2.6, we have σ_{p+1} (f_{a}) ≤ \overline{λ}_{p+1} (f_{a}). Therefore, \overline{λ}_{p+1} (f_{a}) = σ_{p+1} (f_{a}) = σ. If i (F) = p + 1 and σ_{p+1} (F) < σ_{p} (A_{s}) = σ, then

T (r, F) ≤ exp_{p} {r^{σ_{p+1}(F)+ε}} ≤ exp_{p−1} {r^{σ+ε}} .  (4.20)

Thus, by (4.16) − (4.17) and (4.20), we have

T (r, f_{a}) ≤ k \overline{N} (r, \frac{1}{f_{a}} ) + k exp_{p−1} \{r^{σ+ε}\}

+ exp_{p−1} {r^{σ+ε}} + O (exp_{p−1} {r^{σ+ε}})  (|z| = r ∉ E_{3}) .  (4.21)

By using similar reasoning as above, we obtain from (4.21) and Lemma 2.6 that \overline{λ}_{p+1} (f_{a}) = σ_{p+1} (f_{a}) = σ. The proof of Theorem 1.2 is complete.

Acknowledgement. The author would like to thank the referee for his/her helpful remarks and suggestions.

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(Received April 2, 2006)