Estimation of the hyper-order of entire solutions of complex linear ordinary differential equations

whose coefficients are entire functions

Benharrat BELAI^{..}DI Department of Mathematics University of Mostaganem B. P 227 Mostaganem-(Algeria) belaidi.benharrat@caramail.com

Abstract. We investigate the growth of solutions of the differential equation f^{(n)} + A_{n−1} (z) f^{(n−1)} + ... + A_{1} (z) f′ + A_{0} (z) f = 0, where A_{0} (z) , ..., A_{n−1} (z) are entire functions with A_{0} (z) ≢ 0. We estimate the hyper-order with respect to the conditions of A_{0} (z) , ..., A_{n−1} (z) if f ≢ 0 has infinite order.

2000 Mathematics Subject Classification. 30D35, 34M10, 34C10, 34C11.

Key words. Linear differential equations, growth of entire functions, hyper-order.

1 Introduction and statement of results

We assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s value distribution theory of meromorphic functions ( see ([5])). Let σ (f) denote the order of the growth of an entire function f as defined in ([5]) :

σ (f) = \overline{lim}_{r→+∞} \frac{log T (r, f)}{log r} = \overline{lim}_{r→+∞} \frac{log log M (r, f)}{log r},

where T (r, f) is the Nevanlinna characteristic of f ( see [5]), and M (r, f) = max_{|z|=r} |f (z)| .

Definition 1. ([1] , [2] , [8]) Let f be a meromorphic function. Then the hyper-order σ_{2} (f) of f (z) is defined by

σ_{2} (f) = \overline{lim_{r→∞}} \frac{log log T (r, f)}{log r}.  (1.1)

Note. Clearly, if f (z) is entire, then

σ_{2} (f) = \overline{lim_{r→∞}} \frac{log log log M (r, f)}{log r} = \overline{lim_{r→∞}} \frac{log log T (r, f)}{log r}.  (1.2)

We define the linear measure of a set H ⊂ [0, +∞[ by m (H) = ∫_{H} dt and the

logarithmic measure of a set F ⊂ [1, +∞[ by m_{l} (F) = ∫_{F} \frac{dt}{t}. The upper and the lower densities of H are defined by

\overline{dens}H = \overline{lim_{r→∞}} \frac{m (H ∩ [0, r])}{r}, \underline{dens} H = \underline{lim}_{r→∞} \frac{m (H ∩ [0, r])}{r}.

Recently in [1], [2], [3] the concept of hyper-order was used to further investigate the growth of infinite order solutions of complex differential equations.

The following results have been obtained for the second order equation

f′′ + A (z) f′ + B (z) f = 0  (1.3)

where A (z), B (z) ≢ 0 are entire functions.

Theorem A.([1]) Let H be a set of complex numbers satisfying \overline{dens} \{|z| : z ∈ H\} > 0, and let A (z) and B (z) be entire functions such that for some constants α, β > 0,

|A (z)| ≤ exp {o (1) |z|^{β}}  (1.4)

and

|B (z)| ≥ exp {(1 + o (1)) α |z|^{β}}  (1.5)

as z → ∞ for z ∈ H. Then every solution f ≢ 0 of equation (1.3) satisfies σ (f) = +∞ and σ_{2} (f) ≥ β.

Theorem B.([2]) Let H be a set of complex numbers satisfying \overline{dens} \{|z| : z ∈ H\} > 0, and let A (z) and B (z) be entire functions, with σ (A) ≤ σ (B) = σ < +∞ such that for some real constant C (> 0) and for any given ε > 0,

|A (z)| ≤ exp {o (1) |z|^{σ−ε}}  (1.6)

and

|B (z)| ≥ exp {(1 + o (1)) C |z|^{σ−ε}}  (1.7)

as z → ∞ for z ∈ H. Then every solution f ≢ 0 of equation (1.3) satisfies σ (f) = +∞ and σ_{2} (f) = σ (B) .

For n ≥ 2, we consider a linear differential equation of the form

f^{(n)} + A_{n−1} (z) f^{(n−1)} + ... + A_{1} (z) f′ + A_{0} (z) f = 0  (1.8)

where A_{0} (z) , ..., A_{n−1} (z) are entire functions with A_{0} (z) ≢ 0. It is well-known that all solutions of equation (1.8) are entire functions and if some of the coefficients of (1.8) are transcendental, (1.8) has at least one solution with σ (f) = +∞.

The main purpose of this paper is to investigate the growth of infinite order solutions of the linear differential equation (1.8) .

Theorem 1. Let H be a set of complex numbers satisfying \overline{dens} \{|z| : z ∈ H\} > 0, and let A_{0} (z) , ..., A_{n−1} (z) be entire functions such that for some constants 0 ≤ β < α and μ > 0, we have

|A_{0} (z)| ≥ e^{α|z|^{μ}}  (1.9)

and

|A_{k} (z)| ≤ e^{β|z|^{μ}}, k = 1, ..., n − 1  (1.10)

as z → ∞ for z ∈ H. Then every solution f ≢ 0 of equation (1.8) satisfies σ (f) = +∞ and σ_{2} (f) ≥ μ.

Theorem 2. Let H be a set of complex numbers satisfying \overline{dens} \{|z| : z ∈ H\} > 0, and let A_{0} (z) , ..., A_{n−1} (z) be entire functions with max {σ (A_{k}) : k = 1, ..., n − 1} ≤ σ (A_{0}) = σ < +∞ such that for some real constants 0 ≤ β < α, we have

|A_{0} (z)| ≥ e^{α|z|^{σ−ε}}  (1.11)

and

|A_{k} (z)| ≤ e^{β|z|^{σ−ε}}, k = 1, ..., n − 1  (1.12)

as z → ∞ for z ∈ H. Then every solution f ≢ 0 of equation (1.8) satisfies σ (f) = +∞ and σ_{2} (f) = σ (A_{0}) .

2 Preliminary Lemmas

Our proofs depend mainly upon the following Lemmas.

Lemma 1. ([4] , p. 90) Let f be a transcendental entire function of finite order σ. Let Γ = {(k_{1}, j_{1}) , (k_{2}, j_{2}) , ..., (k_{m}, j_{m})} denote a finite set of distinct pairs of integers satisfying k_{i} > j_{i} ≥ 0 for i = 1, ..., m and let ε > 0 be a given constant. Then there exists a set E ⊂ [0, ∞) with finite linear measure, such that for all z satisfying |z| ∉ E and for all (k, j) ∈ Γ, we have

|\frac{f^{(k)} (z)}{f^{(j)} (z)}| ≤ |z|^{(k−j)(σ−1+ε)} .  (2.1)

Lemma 2. ([4]) Let f (z) be a nontrivial entire function, and let α > 1 and ε > 0 be given constants. Then there exist a constant c > 0 and a set E ⊂ [0, ∞) having finite linear measure such that for all z satisfying |z| = r ∉ E, we have

|\frac{f^{(k)} (z)}{f (z)}| ≤ c [T (αr, f) r^{ε} log T (αr, f)]^{k} , k ∈ N.  (2.2)

Let f (z) = ∞ ; ∑_{n=0}a_{n} z^{n} be an entire function, 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}} .

Lemma 3.([2]) Let f (z) be an entire function of infinite order with the hyperorder σ_{2} (f) = σ, and let ν_{f} (r) be the central index of f. Then

\overline{lim}_{r→∞} \frac{log log ν_{f} (r)}{log r} = σ.  (2.3)

Lemma 4.(W iman − V aliron, [6] , [7]) 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 for all |z| outside a set E of r of finite logarithmic measure, we have

\frac{f^{(k)} (z)}{f (z)} = (\frac{ν_{f} (r)}{z})^{k} (1 + o (1)) , (k is an integer, r ∉ E)  (2.4)

where ν_{f} (r) is the central index of f.

3 Proof of Theorem 1

Suppose that f ≢ 0 is a solution of equation (1.8) with σ (f) < ∞. By (1.8) we can write

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

or

\frac{1}{A_{0} (z)} \frac{f^{(n)}}{f} + n−1 ; ∑_{k=1} \frac{A_{k} (z)}{A_{0} (z)} \frac{f^{(k)}}{f} = −1.  (3.2)

Then, by Lemma 1, there exists a set E_{1} ⊂ [0, ∞) with finite linear measure, such that for all z satisfying |z| ∉ E_{1} and for all k = 1, 2, ...n, we have

|\frac{f^{(k)} (z)}{f (z)}| ≤ |z|^{k c} , k = 1, ..., n ; c = σ − 1 + ε.  (3.3)

Also, by the hypothesis of Theorem1, there exists a set E_{2} with \overline{dens} \{|z| : z ∈ E_{2}\} > 0 such that for all z satisfying z ∈ E_{2}, we have

|A_{0} (z)| ≥ e^{α|z|^{μ}}  (3.4)

and

|A_{k} (z)| ≤ e^{β|z|^{μ}}, k = 1, ..., n − 1  (3.5)

as z → ∞. Hence from (3.3) , (3.4) and (3.5) it follows that for all z satisfying z ∈ E_{2} and |z| ∉ E_{1}, we have

|\frac{A_{k} (z)}{A_{0} (z)}| |\frac{f^{(k)} (z)}{f (z)}| ≤ \frac{1}{e^{(^{α−β})}^{|z|}^{μ}} |z|^{k c} , k = 1, ..., n − 1 ; c = σ − 1 + ε (3.6)

and

|\frac{1}{A_{0} (z)}| |\frac{f^{(n)} (z)}{f (z)}| ≤ \frac{1}{e^{α|z|^{μ}}} |z|^{n c} , c = σ − 1 + ε  (3.7)

as z → ∞. Thus there exists a set H ⊂ [0, ∞) with a positive upper density such that (3.6) , (3.7) hold. Since

lim_{z→∞_{z∈H}} \frac{1}{e^{(^{α−β})}^{|z|}^{μ}} |z|^{k c} = 0, k = 1, ..., n − 1

and

lim_{z→∞_{z∈H}} \frac{1}{e^{α|z|^{μ}}} |z|^{n c} = 0,

it follows that

lim_{z→∞_{z∈H}} |\frac{A_{k} (z)}{A_{0} (z)}| |\frac{f^{(k)} (z)}{f (z)}| = 0, k = 1, ..., n − 1

and

lim_{z→∞_{z∈H}} |\frac{1}{A_{0} (z)}| |\frac{f^{(n)} (z)}{f (z)}| = 0.

By making z → ∞ for z ∈ H in the relation (3.2) , we get a contradiction. Then every solution f/ ≡0 of equation (1.8) has infinite order.

Now from (1.8) , it follows that

|A_{0} (z)| ≤ |\frac{f^{(n)}}{f} | + |A_{n−1} (z)| |\frac{f^{(n−1)}}{f} | + ... + |A_{1} (z)| |\frac{f^{/}}{f} | .  (3.8)

Then, by Lemma 2, there exists a set E_{3} ⊂ [0, +∞) with a finite linear measure such that for all z satysfying |z| = r ∉ E_{3}, we have

|\frac{f^{(k)} (z)}{f (z)}| ≤ r [T (2r, f)]^{k+1} , k = 1, ..., n.  (3.9)

Also, by the hypothesis of the Theorem1, there exists a set E_{4} with \overline{dens} \{|z| : z ∈ E_{4}\} > 0 such that for all z satisfying z ∈ E_{4}, we have

|A_{0} (z)| ≥ e^{α|z|^{μ}}  (3.10)

and

|A_{k} (z)| ≤ e^{β|z|^{μ}}, k = 1, ..., n − 1  (3.11)

as z → ∞. Hence from (3.8) , (3.9) , (3.10) and (3.11) it follows that for all z satisfying z ∈ E_{4} and |z| ∉ E_{3}, we have

e^{α|z|^{μ}} ≤ |z| [T (2 |z| , f)]^{n+1} [1 + (n − 1) e^{β|z|^{μ}}]  (3.12)

as z → ∞. Thus there exists a set H ⊂ [0, +∞) with positive upper density such that

e^{(α−β)r^{μ}(1−o(1))} ≤ [T (2r, f)]^{n+1}

as r → ∞ in H. Therefore

\overline{lim_{r→∞}} \frac{log log T (r, f)}{log r} ≥ μ.

This proves Theorem 1.

4 Proof of Theorem 2

Assume that f ≢ 0 is a solution of equation (1.8) . Using the same arguments as in Theorem 1, we get σ (f) = +∞.

Now we prove that σ_{2} (f) = σ (A_{0}) = σ. By Theorem 1, we have σ_{2} (f) ≥ σ −ε, and since ε is arbitrary, we get σ_{2} (f) ≥ σ (A_{0}) = σ.

On the other hand, by Wiman-Valiron theory, there is a set E ⊂ [1, +∞) with logarithmic measure m_{l} (E) < ∞ and we can choose z satisfying |z| = r ∉ [0, 1] ∪ E and |f (z)| = M (r, f), such that (2.4) holds. For any given ε > 0, if r is sufficiently large, we have

|A_{k} (z)| ≤ e^{r^{σ+ε}}, k = 0, 1, ..., n − 1.  (4.1)

Substituting (2.4) and (4.1) into(1.8) , we obtain

(\frac{ν_{f} (r)}{|z|})^{n} |1 + o (1)| ≤ e^{r^{σ+ε}} (\frac{ν_{f} (r)}{|z|})^{n−1} |1 + o (1)| +

+e^{r^{σ+ε}} (\frac{ν_{f} (r)}{|z|})^{n−2} |1 + o (1)| + ... + e^{r^{σ+ε}} (\frac{ν_{f} (r)}{|z|}) |1 + o (1)| + e^{r^{σ+ε}}  (4.2)

where z satisfies |z| = r ∉ [0, 1] ∪ E and |f (z)| = M (r, f). By (4.2), we get

\overline{lim}_{r→∞} \frac{log log ν_{f} (r)}{log r} ≤ σ + ε.  (4.3)

Since ε is arbitrary, by (4.3) and Lemma 3 we have σ_{2} (f) ≤ σ. This and the fact that σ_{2} (f) ≥ σ yield σ_{2} (f) = σ.

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

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