Growth of transcendental entire solution of some q-difference equation

Ilham Eli, Kayoko Gotoh and Niro Yanagihara

Abstract We consider a linear q-difference equation qzf(qz) + (1 − Az)f(z) = 1, with q = e^{2πiβ}, β ∈ (0, 1) \ ℚ and A = e^{2πiα}, α ∈ (0, 1). The equation is known to admit a transcendental entire solution f(z) for suitably chosen β and α. We will show here that f(z) is of positive order for some β, contrary to q-difference equations with |q| ≠ 0, 1.

Keywords and phrases: linear q-difference equation, growth of entire function.

AMS Subject Classification: 39A13, 30D20

1 Introduction

We consider here a q-difference equation

b_{p}(z)f(q^{p}z) + ⋯ + b_{0}(z)f(z) = 𝔟(z), b_{j}(z), 𝔟(z) ∈ ℂ[z], (1.1)

with b_{j}(z) = ∑_{k=0}^{B_{j}} b_{k}^{(j)}z^{k} (b_{B_{j}}^{(j)} ≠ 0), 0 ≤ j ≤ p.

When |q| ≠ 0, 1, a transcendental entire solution f(z) of (1.1) are of order 0. In fact, when 0 < |q| < 1, it satisfies

log M(r, f) = \frac{σ}{−2 log |q|}(log r)^{2}(1 + o(1)), r → ∞,

in which σ is a slope of the Newton diagram for (1.1) [1].

When |q| = 1, that is q = e^{2πiλ}, there is no such regularity. For example, when q = −1, λ = 1/2, the equation f(−z)−f(z) = 0 has solutions of behaviors of several type. We ask here what can be said for the case that

q = e^{2πiβ}, β ∈ (0, 1) \ ℚ. (1.2)

Driver et al. [3] showed that there exist (q, A), with q in (1.2) and A, |A| = 1, such that the equation

qzf(qz) + (1 − Az)f(z) = 1 (1.3)

has a transcendental entire solution. We will show here the following theorem, contrary to the case |q| ≠ 0, 1 :

Theorem 1.1 The solution f(z) of (1.3) is of positive order, supposed β in (1.2) is suitably chosen, as shown at the end of the proof.

2 Proof of Theorem 1.1

We denote by {a} the fractional part of a.

The solution f(z) of (1.3) is written as

(2.1)  f(z) = ∑_{n=0}^{∞} α_{n}z^{n}, α_{n} = ∏_{k=1}^{n}(A − q^{k}),

in which we write A = e^{2πiα}, q = e^{2πiβ}. Then

(2.2)  |A − q^{k}| = 2| sin[π(α − kβ)]|, |α_{n}| = 2^{n} ∏_{k=1}^{n} | sin[π(α − kβ)]|.

Write

β = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ⋯

and define the k-th convergent p_{k}/q_{k} by p_{0} = 0, p_{1} = 1, q_{0} = 1, q_{1} = a_{1}, and

p_{k} = a_{k}p_{k−1} + p_{k−2}, q_{k} = a_{k}q_{k−1} + q_{k−2}, k ≥ 2.

Put

s_{j} = ∑_{k=0}^{j} q_{k}.

By [3] p.476,

{s_{j}β} = β + ∑_{k=1}^{j}(q_{k}β − p_{k}) ∈ (0, β).

By [3] p.474,

(2.3)  (−1)^{k}(q_{k}β − p_{k}) = |q_{k}β − p_{k}| ≤ 1/q_{k+1},

and |q_{k}β − p_{k}| ↓ 0. We put

(2.4)  α = β + ∑_{k=1}^{∞}(q_{k}β − p_{k}) = lim_{j→∞}{s_{j}β}.

We have by [4] p.24 Theorem 13,

(2.5)  |q_{k}β − p_{k}| > 1/(q_{k+1} + q_{k}).

Further, we have by [3] p.475, for 1 ≤ j ≤ q_{k} − 1, with k even,

(2.6)  {jβ} = {j(β − p_{k}/q_{k})} + {jp_{k}/q_{k}}
{{jp_{k}/q_{k}}; 1 ≤ j ≤ q_{k} − 1} = {j/q_{k}; 1 ≤ j ≤ q_{k} − 1},

since (p_{k}, q_{k}) = 1. We assume that, following [3] pp.474, 471,

(2.7) lim_{j→∞}(log q_{j+1})/q_{j} = ∞, hence lim_{j→∞}(log q_{j+1})/s_{j} = ∞.

Take n = s_{j} − 1 with j even. For 1 ≤ k ≤ n,

α − kβ = (α − s_{j}β) + (s_{j} − k)β = ∑_{ℓ=j+1}^{∞}(q_{ℓ}β − p_{ℓ}) − ∑_{ℓ=0}^{j} p_{ℓ} + q_{j}β + (s_{j−1} − k)β.

By (2.3) and (2.5),

| ∑_{ℓ=j+1}^{∞}(q_{ℓ}β − p_{ℓ})| ≤ |q_{j+1}β − p_{j+1}| ≤ \frac{1}{q_{j+2}} ,

1/(q_{j+1} + q_{j}) ≤ {q_{j}β} = q_{j}β − p_{j} ≤ 1/q_{j+1},

further, by (2.3) and (2.6), for 1 ≤ k ≤ s_{j−1} − 1, noting that s_{j−1} − k < q_{j} − 1 by (2.7), {(s_{j−1} − k)β} ≥ {(s_{j−1} − k)p_{j}/q_{j}} − |(s_{j−1} − k)(β − p_{j}/q_{j})| ≥ 1/q_{j} − 1/q_{j+1}. Thus for k, 1 ≤ k ≤ s_{j−1} − 1,

(2.8)  {|α − kβ|} ≥ 1/q_{j} − 2/q_{j+1} − 1/q_{j+2} > 1/(2q_{j}).

For k = s_{j−1}, by (2.5)

(2.9)  {|α − kβ|} > 1/(q_{j+1} + q_{j}) − 1/q_{j+2} > 1/(2q_{j+1}).

For k, s_{j−1} < k < s_{j}, we get 1 ≤ s_{j} − k < s_{j} − s_{j−1} = q_{j}, and

{(s_{j} − k)β} = {(s_{j} − k)(β − p_{j}/q_{j})} + {(s_{j} − k)p_{j}/q_{j}} ≥ −1/q_{j+1} + 1/q_{j}, hence

(2.10)  {|α − kβ|} ≥ 1/q_{j} − 1/q_{j+1} − 1/q_{j+2} > 1/(2q_{j}).

Noting sin x > (2/π)x, 0 < x < π/2, we obtain by (2.2), (2.8), (2.9), (2.10),

|α_{n}| ≥ 2^{n}(2/π)^{n}(π/(2q_{j}))^{n−1}(π/(2q_{j+1})) = 2^{n}q_{j}^{−n+1}q_{j+1}^{−1}.

Noting that log n = log(s_{j} − 1) = log q_{j} + log(1 + \frac{s_{j−1}−1}{q_{j}}), we get, writing n_{j} = s_{j} − 1,

lim inf_{j→∞} \frac{log(1/|α_{n_{j}}|)}{n_{j} log n_{j}} ≤ lim_{j→∞} \frac{(n_{j} − 1)}{n_{j}} \frac{log q_{j}}{log n_{j}} + lim inf_{j→∞} \frac{log q_{j+1}}{(s_{j} − 1) log(s_{j} − 1)}

= 1 + lim inf_{j→∞} \frac{log q_{j+1}}{(s_{j} − 1) log(s_{j} − 1)}.

Suppose q_{j} are chosen so that, together with (2.7), the following holds:

(2.11)  lim inf_{j→∞} \frac{log q_{j+1}}{(s_{j} − 1) log(s_{j} − 1)} = lim inf_{j→∞} \frac{log q_{j+1}}{q_{j} log q_{j}} = γ < ∞,

then we have, by [2] p.9, that the order ρ(f) of f(z) satisfies

ρ(f) ≥ 1/(1 + γ).

The condition (2.11) can be satisfied by assigning the partial quotients a_{j} in the continued fraction suitably.

References

[1] W. Bergweiler, K. Ishizaki, and N. Yanagihara: Growth of meromorphic solutions of some functional equations I. Aeq. Math. 63 (2002), 140–151.

[2] R.P. Boas: Entire Functions. Academic Press, New York 1954.

[3] K.A. Driver, D.S. Lubinsky, G. Petruska, and P. Sarnak: Irregular distribution of {nβ}, n = 1, 2, 3, …, quadrature of singular integrands, and curious basic hypergeometric series. Indag. Math., N.S., 2 (4) (1991), 469–481.

[4] A.Ya. Khinchin: Continued Fractions. 3rd Ed. Moskwa 1961. (in Russian)

Ilham Eli Department of Mathematics, Kashgar Teacher’s College, Xinjiang, CHINA.

Present address of Professor Ilham Eli: Department of Mathematics, Faculty of Science, Chiba University, 1-33 Yayoicho, Inage-ku, Chiba City, Chiba-ken, 263-8522 JAPAN.

Kayoko Gotoh Part–time Lecturer, Department of Mathematical Science, Faculty of Science, Tokai University, 1117 Kita-kaname, Hiratsuka City, Kanagawa-ken, 259-1292 JAPAN.

Niro Yanagihara Department of Mathematics, Faculty of Science, Chiba University, 1-33 Yayoicho, Inage-ku, Chiba City, Chiba-ken, 263-8522 JAPAN.