Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 25, 2000, 73–84

ON THE GROWTH AND FACTORIZATION OF ENTIRE SOLUTIONS OF ALGEBRAIC DIFFERENTIAL EQUATIONS

Liang-Wen Liao and Chung-Chun Yang

The Hong Kong University of Science & Technology, Department of Mathematics Kowloon, Hong Kong, China; mayang@uxmail.ust.hk

Abstract. In this paper we discuss the growth and factorization of entire solutions of some classes of first-order algebraic differential equations and prove that any entire solution of a firstorder algebraic differential equation must be pseudo-prime.

1. Introduction

In the study of the solutions of complex differential equations, the growth of a solution is a very important property. For linear differential equations of the form

(1.1)  f^{(n)} + a_{n−1}(z)f^{(n−1)} + ⋯ + a_{0}(z)f = a(z),

where a(z), a_{0}(z), …, a_{(n−1)}(z) are polynomials, it is known that any entire solution must be of finite and positive order; see Laine [13, pp. 52–73, pp. 144–164], Gundersen, et al. [6]. This can be proved by mainly using the Wiman–Valiron theory. However, there are only a few results concerning the growth of the solutions of a nonlinear algebraic differential equation

(1.2)  P (z, f, f′, …, f^{(n)}) = 0,

where P is a polynomial in all its arguments. Equation (1.2) can be rewritten in the form

(1.3)  ∑_{λ∈I} a_{λ}(z)f^{i_{0}} ⋯ (f^{(n)})^{i_{n}} = 0,

where I is a finite set of multi-indices (i_{0}, …, i_{n}) = λ. We define a differential monomial in f as

M[f] = a_{λ}(z)f^{i_{0}} ⋯ (f^{(n)})^{i_{n}}.

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

The reseach was partially supported by a UGC grant of Hong Kong (Project No: HKUST 712/96p).

The degree γ_{M} and the weight Γ_{M} of M are defined by

γ_{M} = i_{0} + i_{1} + ⋯ + i_{n}, Γ_{M} = i_{0} + 2i_{1} + ⋯ + (n + 1)i_{n}.

Then the left-hand side of equation (1.3) can be expressed as a finite sum of differential monomials. From now on we shall call this a differential poynomial in f, i.e.

P [f] = P (z, f, f′, …, f^{n}) = ∑_{λ∈I} M_{λ}[f].

The degree γ_{P} and the weight Γ_{P} of P are defined by γ_{P} = max_{λ∈I} γ_{M_{λ}}, Γ_{P} = max_{λ∈I} Γ_{M_{λ}}. Some results have been obtained on the growth estimates for solutions of algebraic differential equations; see e.g. [13]. However, in general, a complete growth estimate for nonlinear algebraic differential equations remains to be resolved. The first important result on the growth estimate was due to A. Gol’dberg [4] (or see e.g. [13]). Some variations of Gol’dberg’s result were obtained by Bank–Kaufman [1], W. Bergweiler [2], and others.

Theorem A (Gold’berg). Let P (u_{1}, u_{2}, u_{3}) be a polynomial in all of its arguments u_{1} , u_{2} and u_{3} and consider the first-order algebraic differential equation

(1.4)  P (z, f, f′) = 0.

Then all meromorphic solutions of (1.4) are of finite order of growth.

Note that the order can be zero (see e.g. [13]).

For nonlinear differential equations of second-order Steinmetz [18] proved the following theorem (see also e.g. [13]):

Theorem B. Suppose that in equation (1.2) P is homogeneous in f, f′ and f′′. Then all meromorphic solutions of (1.2) satisfy

T (r, f) = O(exp r^{b})

as r → ∞ for some b > 1 depending only on the degrees of the polynomial coefficients of (1.3).

Recently W.K. Hayman [8] studied the growth of solutions of (1.2) and posed the following

Conjecture. If f(z) is an entire solution of (1.2), then

T (r, f) < a exp_{n−1}(br^{c}), 0 ≤ r < ∞

where a, b and c are positive constants and exp_{l}(x) is the exponential iterated l times.

Hayman also showed that the conjecture is true for some special class of equations.

Set

Λ = {λ = (i_{0}, i_{1}, …, i_{n}) | γ_{M_{λ}} = γ_{P} , Γ_{M_{λ}} = Γ_{P} }.

Theorem C (Hayman [8]). Suppose that equation (1.3) holds with Λ defined as above. Let d be the maximum degree among all the polynomials a_{λ}(z) in (1.3) and suppose that

(1.5)  ∑_{λ∈Λ} a_{λ}(z) ≢ 0.

Then any entire solution of (1.3) has finite order ρ, with ρ ≤ max{2d, 1 + d}.

On the other hand, Sh. Strelitz [20] proved

Theorem D. Every entire transcendental solution of a first-order algebraic differential equation with rational coefficients has an order no less than \frac{1}{2} .

Definition (Gross [5]). A meromorphic function f is called pseudo-prime if whenever f = g(h) with g, h entire or meromorphic, implies that either g is a rational function or h is a polynomial.

As regards the factorization of the solutions of differential equations, the following two results are well known.

Theorem E (Steinmetz [17]). Any meromorphic solution of (1.1) is pseudoprime.

Theorem F (Mues [14]). Let f be a meromorphic solution of the Riccati differential equation

w′ = a(z) + b(z)w + c(z)w^{2},

where a(z), b(z), c(z) are polynomials. Then f is pseudo-prime.

For some other related results on the factorization of solutions of some firstorder algebraic differential equations, see e.g. He–Laine [9] and He–Yang [10]. In this paper we shall mainly discuss the growth and factorization of the transcendental entire solutions of the first-order algebraic differential equation in its most general form. Subsequently, we always assume that f denotes an entire function and write

f(z) = ∑_{n=0}^{∞} a_{n}z^{n}.

We denote the maximum term of f by μ(r, f), the central index by ν(r, f), and the maximum modulus by M(r, f), i.e.

μ(r, f) = max_{|z|=r} |a_{n}z^{n}|, ν(r, f) = sup{n | |a_{n}|r^{n} = μ(r, f)}, M(r, f) = max_{|z|=r} |f(z)|.

As usual, we use T (r, f) to denote the Nevanlinna characteristic function of f and ρ(f) to denote the order of f.

2. Lemmas

In order to obtain our theorems, we need some basic results of the Wiman– Valiron theory .

Lemma 1 (Laine [13]). If f is an entire function of order ρ, then

(2.1)  ρ = lim sup_{r→∞} \frac{log^{+} ν(r, f)}{log r} = lim sup_{r→∞} \frac{log log^{+} μ(r, f)}{log r}.

Lemma 2 (Laine [13]). Let f be a transcendental entire function, and let 0 < δ < \frac{1}{4} . Suppose that at the point z with |z| = r the inequality

(2.2)  |f(z)| > M(r, f)ν(r, f)^{−(1/4)+δ}

holds. Then there exists a set F ⊂ R^{+} of finite logarithmic measure, i.e., ∫ dt/t < +∞ such that

(2.3)  f^{(m)}(z) = (\frac{ν(r, f)}{z})^{m}(1 + o(1))f(z)

holds for all m ≥ 0 and all r ∉ F.

By Lemma 1, we can easily derive

Lemma 3. Let f be a transcendental entire function of order ρ < 1. Then

(2.4)  lim_{r→∞} \frac{ν(r, f)}{r} = 0.

The following lemma due to Polya [15] plays a very important role in our discussions.

Lemma 4. If f and g are entire functions, the composite function f ∘ g is of infinite order unless (a) f is of finite order and g is a polynomial or (b) f is of zero order and g is of finite order.

Lemma 5 (Steinmetz [16]). Let f_{1}, …, f_{n} and g be entire functions and let h_{1}, …, h_{n} be meromorphic functions such that the inequality

∑_{j=1}^{n}T (r, h_{j}) ≤ KT (r, g)

holds, with K a constant. Suppose that f_{j} and h_{j} (j = 1, 2, …, n) satisfy

f_{1}(g(z))h_{1}(z) + ⋯ + f_{n}(g(z))h_{n}(z) ≡ 0.

Then there exist two sets of polynomials {P_{j}} and {Q_{j}} (j = 1, 2, …, n) not all identically zero in either of the two sets such that

(2.5)  P_{1}(g(z))h_{1}(z) + ⋯ + P_{n}(g(z))h_{n}(z) ≡ 0

and

(2.6)  f_{1}(z)Q_{1}(z) + ⋯ + f_{n}(z)Q_{n}(z) ≡ 0.

Lemma 6 (Zimogljad [21]). Every entire transcendental solution of a secondorder algebraic differential equation with rational coefficients has a positive order.

3. Growth of solutions of certain types of first-order algebraic differential equations

In this section, we shall provide a more precise estimation of the growth of entire solutions of algebraic differentional equations of the form

(3.1)  C(z, w)(w′)^{2} + B(z, w)w′ + A(z, w) = 0,

where C(z, w) ≢ 0, B(z, w), and A(z, w) are polynomials in z and w. Here we refer the reader to Ishizaki’s works [11], [12] for the cases where the cofficients of the powers of w in A(z, w), B(z, w) and C(z, w) are transcendental functions. Steinmetz [19] showed that if (3.1) has a transcendental meromorphic solution, then

deg_{w} C(z, w) = 0,  deg_{w} B(z, w) ≤ 2,  deg_{w} A(z, w) ≤ 4.

Thus we can assume that C(z, w) ≡ a(z) is a polynomial in z only and that (3.1) can be rewritten in the following form:

(3.2) \begin{aligned}a(z)w′^{2} + (b_{2}(z)w^{2} + b_{1}(z)w + b_{0}(z))w′ & = d_{4}(z)w^{4} + d_{3}(z)w^{3} \\ & + d_{2}(z)w^{2} + d_{1}(z)w + d_{0}(z), \end{aligned}

where a(z), b_{i}(z) (i = 0, 1, 2) and d_{j}(z) (j = 0, …, 4) are polynomials. If (3.2) has a transcendental entire solution, one can derive d_{4}(z) ≡ 0 by comparing the characteristic functions on both sides of (3.2). Finally equation (3.2) can be reduced to the following form:

(3.3) a(z)f′^{2} +(b_{2}(z)f^{2} +b_{1}(z)f +b_{0}(z))f′ = d_{3}(z)f^{3} +d_{2}(z)f^{2} +d_{1}(z)f +d_{0}(z).

Theorem 1. If deg d_{2}(z) ≠ deg a(z)−1 in (3.3) and f(z) is a transcendental entire solution of equation (3.2), then ρ(f) ≥ 1.

Proof. We assume that f has an order ρ(f) < 1 and satisfies equation (3.3). Now we can rewrite (3.3) as follows:

(3.4) \begin{aligned}(d_{3}(z) − b_{2}(z)\frac{f′(z)}{f(z)}) & = [a(z)\frac{(f′(z))^{2}}{(f(z))^{2}} + b_{1}(z)\frac{f′(z)}{f(z)} + b_{0}(z)\frac{f′(z)}{(f(z))^{2}} \\ & + d_{2}(z) + \frac{d_{1}(z)}{f(z)} + \frac{d_{0}(z)}{(f(z))^{2}} ]\frac{1}{f(z)}. \end{aligned}

We choose r_{n} ∉ F and z_{n} such that r_{n} → ∞, n → ∞, |z_{n}| = r_{n}, |f(z_{n})| = M(r_{n}, f). From Lemmas 2 and 3 we have

(3.5)  \frac{f′(z_{n})}{f(z_{n})} = \frac{ν(r_{n}, f)}{z_{n}}(1 + o(1)) → 0.

Thus

(3.6) |a(z)\frac{(f′(z))^{2}}{(f(z))^{2}} +b_{1}(z)\frac{f′(z)}{f(z)} +b_{0}(z)\frac{f′(z)}{(f(z))^{2}} +d_{2}(z)+ \frac{d_{1}(z)}{f(z)} + \frac{d_{0}(z)}{(f(z))^{2}} | < cr^{m},

where c is a constant and m is the degree of d_{2}(z). Hence

(3.7)  lim_{n→∞}[d_{3}(z_{n}) − b_{2}(z_{n})\frac{ν(r_{n}, f)}{r_{n}}(1 + o(1))] = 0.

If b_{2}(z) ≢ 0, then

(3.8)  lim_{n→∞} \frac{d_{3}(z_{n}) − b_{2}(z_{n})(ν(r_{n}, f)/r_{n})(1 + o(1))}{b_{2}(z_{n})} = 0.

Equation (3.7) and Lemma 3 yield

(3.9)  lim_{n→∞} \frac{d_{3}(z_{n})}{b_{2}(z_{n})} = 0.

It follows that deg d_{3}(z) < deg b_{2}(z), and hence

(3.10)  lim_{n→∞} \frac{z_{n}d_{3}(z_{n})}{b_{2}(z_{n})} = t,

where t is a finite constant. If d_{3}(z) ≢ 0, then deg b_{2}(z) ≥ 1. Then, again from (3.7), and noting that |z/b_{2}(z)| is bounded for sufficiently large r = |z| and f is a transcendental function, we have

(3.11)  lim_{n→∞} \frac{z_{n}d_{3}(z_{n})}{b_{2}(z_{n})} = lim_{n→∞} ν(r_{n}, f)(1 + o(1)) = ∞.

This contradicts (3.10). Hence d_{3}(z) ≡ 0. It follows that equation (3.3) becomes

(3.12) a(z)f′^{2} + (b_{2}(z)f^{2} + b_{1}(z)f + b_{0}(z))f′ = d_{2}(z)f^{2} + d_{1}(z)f + d_{0}(z).

Assume that a(z) ≢ 0. If b_{2}(z) ≢ 0, then

(3.13)  \begin{aligned}f′(z) & = \frac{d_{2}(z)}{b_{2}(z)} + \frac{d_{1}(z)}{b_{2}(z)} \frac{1}{f(z)} + \frac{d_{0}(z)}{b_{2}(z)} \frac{1}{(f(z))^{2}} \\ & − \frac{b_{1}(z)}{b_{2}(z)} \frac{f′(z)}{f(z)} − \frac{b_{0}(z)}{b_{2}(z)} \frac{f′(z)}{(f(z))^{2}} − \frac{a(z)}{b_{2}(z)} \frac{(f′(z))^{2}}{(f(z))^{2}} . \end{aligned}

Applying Lemmas 2 and 3 to the above equation, the following result holds for a sequence of r_{n} → ∞:

(3.14)  (1 + o(1))ν(r_{n}, f)M(r_{n}, f) ≤ Ar_{n}^{B},

where A and B are constants. This is impossible. Thus b_{2}(z) ≡ 0. Similarly, if a(z) ≡ 0, we can also conclude b_{2}(z) ≡ 0. Therefore (3.12) reduces to

(3.15)  a(z)f′^{2} + (b_{1}(z)f + b_{0}(z))f′ = d_{2}(z)f^{2} + d_{1}(z)f + d_{0}(z).

Now if a(z) ≢ 0, by (3.15) and Lemma 2 we have for a sequence of r_{n} → ∞ (as in (3.5))

(3.16)  \begin{matrix}a(z_{n}) & (\frac{ν(r_{n}, f)}{r_{n}})^{2}(1 + o(1)) + b_{1}(z_{n})\frac{ν(r_{n}, f)}{r_{n}}(1 + o(1)) − d_{2}(z_{n}) \\ = [b_{0}(z_{n})\frac{ν(r_{n}, f)}{r_{n}}(1 + o(1)) + d_{1}(z_{n}) + \frac{d_{0}(z_{n})}{f(z_{n})}]\frac{1}{f(z_{n})}. \end{matrix}

By Lemma 3, (3.16) and noting lim_{r→∞}(r^{k}/M(r, f)) = 0 for any k, we have

(3.17) lim_{n→∞}\{a(z_{n})(\frac{ν(r_{n}, f)}{r_{n}})^{2}(1 + o(1)) + b_{1}(z_{n})\frac{ν(r_{n}, f)}{r_{n}} − d_{2}(z_{n})\} = 0.

Now we will discuss three cases separately.

Case 1: deg b_{1}(z) > deg a(z). From (3.17) we have

(3.18)  lim_{n→∞} \frac{d_{2}(z_{n})}{b_{1}(z_{n})} = 0.

Thus deg d_{2}(z) < deg b_{1}(z) and hence

(3.19)  lim_{n→∞} \frac{z_{n}d_{2}(z_{n})}{b_{1}(z_{n})} = c,

where c is a finite constant. But, on the other hand, from (3.17) we have

(3.20)  lim_{n→∞}[\frac{b_{1}(z_{n})}{a(z_{n})} \frac{ν(r_{n}, f)}{r_{n}} − \frac{d_{2}(z_{n})}{a(z_{n})}] = 0.

By (20), and noting that |za(z)/b_{1}(z)| is bounded for sufficiently large r = |z| and f is a transcendental function, we can get a conclusion which contradicts (3.19) by using the same argument as in the derivation of (3.11).

Case 2: deg b_{1}(z) = deg a(z). Then
(3.21)  lim_{n→∞} \frac{d_{2}(z_{n})}{a(z_{n})} = c ≠ 0.

From (3.17) and Lemma 3 we have

(3.22)  lim_{n→∞} \frac{d_{2}(z_{n})}{a(z_{n})} = 0.

It follows that from this, (3.16), and (3.21) we have

(3.23)  |(c + o(1))\frac{ν(r_{n}, f)}{r_{n}} |M(r_{n}, f) ≤ Ar_{n}^{k},

where A, k are constants. This is impossible.

Case 3: deg b_{1}(z) < deg a(z). In this case, we also have (3.22). This means deg a(z) > deg d_{2}(z). However, by the assumption of the theorem, we have deg a(z) > deg d_{2}(z) + 1. If d_{2}(z) ≢ 0, then deg a(z) > 1. Thus

(3.24)  lim_{n→∞} \frac{r_{n}b_{1}(z_{n})}{a(z_{n})} = c_{1},  lim_{n→∞} \frac{r_{n}^{2}d_{2}(z_{n})}{a(z_{n})} = c_{2}.

From (3.17) we get

(3.25)  lim_{n→∞}\{(ν(r_{n}, f))^{2} + \frac{r_{n}b_{1}(z_{n})}{a(z_{n})}ν(r_{n}, f) − \frac{r_{n}^{2}d_{2}(z_{n})}{a(z_{n})}\} = 0.

This is also impossible. Now we assume that d_{2}(z) ≡ 0. Then by (3.16) and noting that |zb_{1}(z)/a(z)| is bound for sufficiently large |z|, we have

(3.26)  |(1 + o(1))a(z_{n})(ν(r_{n}, f))^{2}|M(r_{n}, f) < Ar_{n}^{k},

where A, k are constants. This is again impossible. From the above discussions we can conclude that a(z) ≡ 0 if f satisfies (3.15), i.e., f must satisfy

(3.27)  (b_{1}(z)f + b_{0}(z))f′ = d_{2}(z)f^{2} + d_{1}(z)f + d_{0}(z).

By Lemmas 2 and 3, and noting lim_{r→∞}(r^{k}/M(r, f)) = 0 for any k, we have

(3.28)  lim_{n→∞}\{b_{1}(z_{n})\frac{ν(r_{n}, f)}{r_{n}}(1 + o(1)) − d_{2}(z_{n})\} = 0.

By the same argument as in the case of (3.7), we have b_{1}(z) ≡ 0 and d_{2}(z) ≡ 0. Thus (3.15) becomes

(3.29)  b_{0}(z)f′(z) = d_{1}(z)f(z) + d_{0}(z).

From this and Lemma 2, we have for a sequence of z_{n}, (|z_{n}| = r_{n} → ∞),

(3.30)  b_{0}(z_{n})\frac{ν(r_{n}, f)}{r_{n}}(1 + o(1)) = d_{1}(z_{n}) + \frac{d_{0}(z_{n})}{f(z_{n})},

where f(z_{n}) = M(r_{n}, f). From this it is easily seen that either lim_{n→∞} ν(r_{n}, f) = c, where c is a constant, or ν(r_{n}, f) ≥ Ar_{n}^{k} + o(1), where A is a constant, and k is a positive integer. However, both cases are impossible. The proof is thus completed.

Remark 1. The condition deg d_{2}(z) ≠ deg a(z) − 1 in Theorem 1 cannot be omitted. For example, f(z) = cos \sqrt{z} , which has an order ρ(f) = \frac{1}{2} . However, it satisfies the following first-order algebraic differential equation:

4z(w′)^{2} + w^{2} − 1 = 0.

Remark 2. The conclusion is sharp in Theorem 1. There exists the function f(z) = z sin z, which is of order one and satisfies the following first-order algebraic differential equation:

z^{3}(1 − z^{2})(w′)^{2} − 2z^{2}ww′ + (z^{2} + z)w^{2} − z^{5} = 0.

4. Factorization of the solutions of first-order algebraic differential equations

In general, a solution of a higher-order algebraic differential equation may not be pseudo-prime. For example f_{2}(z) = e^{e^{z}} satisfies the homogeneous second-order algebraic differential equation

ww′′ − w′^{2} − ww′ = 0.

Furthermore, according to Hayman [8], f_{n}(z) = exp_{n}(z) satisfies a homogeneous nth-order algebraic differential equation, where exp_{n}(z) is the nth iterate of exponential function. In this section, we consider the factorization of the entire solutions of the most general first-order algebraic differential equation,

(4.1)  ∑_{λ∈I} a_{λ}(z)f^{i_{0}}(f′)^{i_{1}} = 0,

where a_{λ}(z) denotes a rational function. First we prove Theorem 2. All transcendental entire solutions of (4.1) are pseudo-prime.

Proof. If f is a transcendental entire solution of (4.1), then ρ(f) < ∞ by Theorem A. Now we assume that f is not pseudo-prime. i.e.,

(4.2)  f = g(h),

where g is a transcendental meromorphic function and h is a transcendental entire function. First we assume that g is entire. From Lemma 4 we have ρ(g) = 0. By substituting (4.2) into (4.1), we have

(4.3)  ∑_{λ∈I} a_{λ}(z)[g(h(z))]^{i_{0}}[g′(h(z))]^{i_{1}}[h′(z)]^{i_{1}} = 0.

We denote F_{λ}(w) = (g(w))^{i_{0}}(g′(w))^{i_{1}} . Then we can rewrite (4.3) as

(4.4)  ∑_{λ∈I} a_{λ}(z)F_{λ}(h(z))[h′(z)]^{i_{1}} = 0.

Noting that T (r, h′) = m(r, h′) ≤ m(r, h) + m(r, h′/h) = T (r, h) + S(r, h) and using Lemma 5, we see that there exist some polynomials Q_{λ} which are not all identically zero such that

∑_{λ∈I} Q_{λ}F_{λ} ≡ 0,

i.e.

(4.5)  ∑_{λ∈I} Q_{λ}(z)(g(z))^{i_{0}}(g′(z))^{i_{1}} ≡ 0.

From this and Theorem D we have ρ(g) > 0. This contradicts the fact ρ(g) = 0. Now if g is not entire, it is easily shown that g has one and only one pole w_{1} . Thus g(w) = g_{1}(w)/(w − w_{1})^{n}, where g_{1}(z) is a transcendental entire function. Then

(4.6)  f(z) = \frac{g_{1}(h(z))}{h(z) − w_{1}} .

By substituting (4.6) into (4.1), we get

(4.7)  ∑_{λ∈J} b_{λ}(z)Q_{λ}(h(z), h′(z))[g_{1}(h(z))]^{i_{0}}[g′_{1}(h(z))]^{i_{1}} = 0,

where b_{λ}(z) denotes a rational function, Q_{λ}(η, ζ) being a rational function of η and ζ. From this and (2.6) it follows that g_{1} satisfies a first-order algebraic differential equation. This will lead to ρ(g_{1}) > 0, which contradicts the fact that ρ(g_{1}) = ρ(g) = 0. Thus f must be pseudo-prime, which also completes the proof of the theorem.

By using Lemmas 5 and 6, and the argument similar to that used in the proof of Theorem 2, we can prove

Theorem 3. Every finite-order transcendental entire solution of a secondorder algebraic differential equation with rational coefficients must be pseudoprime.

Remark 3. The proof of Theorem 2 cannot be used to show that every transcendental meromorphic solution of (4.1) must be pseudo-prime, since it is known that there exists a transcendental meromorphic (non-entire) function f which satisfies a first-order algebraic differential equation and is of zero order. For example, there exists a meromorphic function H (see [13]), which satisfies the differential equation

(4.8)  (z^{2} − 4)H′(z)^{2} = 4(H(z) − e_{1})(H(z) − e_{2})(H(z) − e_{3})

with the growth condition

(4.9)  T (r, H) = O(log r)^{2}, r → ∞.

Question. Is every transcendental meromorphic solution of a first-order algebraic differential equation pseudo-prime?

The answer to the question is negative. As indicated by W. Bergweiler, the meromorphic function f(z) = H(g(z)) satisfies the differential equation

f′(z)^{2} = 4(f(z) − e_{1})(f(z) − e_{2})(f(z) − e_{3}),

where H satisfies (4.8) and g satisfies (g^{2} − 4) = g′^{2} . Apparently f is not pseudo-prime.

Remark 4. For some higher-order algebraic differential equations with rational coefficients, periodic entire solutions of finite order have been presented which are not pseudo-prime (see e.g. [3, pp. 164, Theorem 4.13]). Also all known nonpseudo-prime entire solutions of some higher-order algebraic differential equations are periodic functions. However, by using the same method as in [3], one can easily construct non-periodic entire functions which are not pseudo-prime and satisfy some higher-order algebraic differential equations.

Acknowledgement. The authors want to express their thanks to the anonymous referee for his valuable comments and suggestions.

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84 Liang-Wen Liao and Chung-Chun Yang

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Received 23 March 1998