BOUNDS FOR THE SUMS OF ZEROS OF SOLUTIONS OF

The main purpose of this paper is to consider the differential equation u (m) = P (z)u (m ≥ 2) where P is a polynomial with complex, in general, coefficients. Let z k (u), k = 1, 2,. .. be the zeros of a nonzero solution u to that equation. We obtain bounds for the sums j X k=1 1 |z k (u)| (j ∈ N) which extend some recent results proved by Gil'.


Introduction and main results
It is well known that Nevanlinna theory has appeared to be a powerful tool in the theory of ordinary differential equations in the complex plane C. For the linear differential equation ( 1) whose coefficients A 0 (z), ..., A k−1 (z) are entire functions, and A 0 (z) is not equal to zero identically, it is well known that all solutions of (1) are entire functions, and that if some coefficients of (1) are transcendental then (1) has at least one solution with order ρ(f ) = ∞.The active research of the asymptotic distribution of the zeros of linear differential equations in the complex plane was started by Bank and Laine [1].They investigated the equation f ′′ + A(z)f = 0 with an entire function A(z).We refer to the book [11] and some recent works [2,3,4,5,6,12,13,14,16] for the literature on asymptotic distribution and counting functions of zeros, and the growth of solutions of complex differential equations.
At the same time, bounds for the zeros of solutions are very important in various applications.Recently, Gil' [10] obtained some results on the bounds of the sums of the zeros of solutions for the second order differential equation u ′′ = P (z)u with polynomial coefficients.
In this paper, we consider the differential equation ( 2) where is a polynomial of degree n with in general complex coefficients.Denote by 2) with multiplicities taken into account.Without loss of generality we assume that the set of the zeros of u is infinite.If u has a finite number l of zeros, then we put Rearrange the zeros of u in order of increasing modulus: We obtain the following theorem which is an extension of Theorem 1.1 from [10].
holds for any j ∈ N.
Here and below in this section, we take γ := 1 n+m .Denote by ν(u, r) (r > 0) the counting function of the zeros of u in |z| ≤ r.We can get the following corollary from Theorem 1.1 above and Corollary 2.2 from [10] (see also [7]).This is an extension of Corollary 4.1 in [10].
In particular, for any p ≥ 1 and j = 2, 3, . . ., Finally, in the light of Theorem 1.1 above and Corollary 2.4 from [10] (see also [7]) we obtain the following result which is an extension of Corollary 4.3 in [10].
Corollary 1.3.Let Φ(t 1 , t 2 , . . ., t j ) be a scalar-valued function with an integer j defined on the domain and satisfying the condition Then under the hypothesis of Theorem 1.1, EJQTDE, 2011 No. 60, p. 3 In particular, let {d k } ∞ k=1 be a decreasing sequence of positive numbers with holds for j = 2, 3, . . . .

Preliminaries and some lemmas
Consider the entire function with in general complex coefficients and finite order ρ(f ).Denote by z k (f ), k = 1, 2, . . . the zeros of f with multiplicities taken into account.Similar discussion as in the first section, without loss of generality we assume that the set of the zeros of f is infinite.Enumerate the zeros of f in order of increasing modulus: The entire function f can be rewritten in the form ( 4) The following result is proved by Gil' in [7] (see also in Section 5.1 from [9]).
where the numbers ν k , k = 0, 1, . . .satisfy the condition EJQTDE, 2011 No. 60, p. 4 Proof.It follows from the Wiman-Valiron theory (see page 281 from [15]) that any nonzero solution u of ( 2) is of order Here and below we put c j = 0 for j > n.It follows from the above equality that Take γ := 1 n+m and ν k := (k!) γ u k .Then we have We now take into account two cases as follows.
In the case k > n, it follows from ( 6) that Using the inequality between the arithmetic and geometric means and similar discussion as in [10], For the other case where k ≤ n, it follows from ( 6) that ( 8) Again by the inequality between the arithmetic and geometric means, Thus (8) gives that If m − 1 ≤ k ≤ n, then we also have inequality Similar discussion as in [10], by (7), ( 9)-( 10) and the comparison theorem (see section 1.6 in [8]), we have |ν j | ≤ w j , where w j is a solution of the equation  Then Note that c j = 0 for j > n, and in consideration of ( 11) and ( 12), where Let z = re iθ for a fixed θ ∈ [0, 2π) and f (r) = F (re iθ ), thus ( 13) yields (m + 1) 2 4(m − 1) 2 µ 2 (P ).