Bounds for Products of Zeros of Solutions to Nonhomogeneous ODE with Polynomial Coefficients

We consider the equation 𝑢 󸀠󸀠 = 𝑃(𝑧)𝑢 + 𝐹(𝑧) (𝑧 ∈ C ) , where 𝑃(𝑧) is a polynomial and 𝐹(𝑧) is an entire function. Let 𝑧 𝑘 (𝑢) (𝑘 = 1,2,. ..) be the zeros of a solution 𝑢(𝑧) to that equation. Lower estimates for the products ∏ 𝑗𝑘=1 |𝑧 𝑘 (𝑢)| (𝑗 = 1, 2,. ..) are derived. In particular, they give us a bound for the zero free domain. Applications of the obtained estimates to the counting function of the zeros of solutions are also discussed.


Introduction and Statement of the Main Result
In the present paper, we investigate the zeros of solutions to the initial problem = ( ) + ( ) with the initial conditions (0) = 1, where is a polynomial with complex in general coefficients and ( ) is an entire function. It is assumed that there are nonnegative constants , and an integer , such that The literature devoted to the zeros of solutions of homogeneous equations is very rich. Here the main tool is the Nevanlinna theory. An excellent exposition of the Nevanlinna theory and its applications to differential equations is given in book [1]. In connection with the recent results see interesting papers [2][3][4][5][6][7][8][9][10][11][12][13] (see also [14,15]). At the same time the zeros of solutions to nonhomogeneous ODE were not enough investigated in the available literature. Here we can point out [16], only, in which the estimates for the sums of the zeros of solutions to (1) have been derived. In the present paper, lower estimates for the products of the zeros are obtained. In addition, we refine the main result from [16]. Enumerate the zeros ( ) of with their multiplicities in order of increasing absolute values: Now we are in a position to formulate the main result of the paper.

Theorem 1. Let condition (3) hold. Then the zeros of the solution to problem (1) satisfy the inequality
2 International Journal of Differential Equations where = 2 ( 0 ( + )) 1/ 0 , The proof of this theorem is presented in the next two sections. Below we also suggest the sharper but more complicated bound for the products of the zeros.

Lemma 2.
A solution ( ) of (7) satisfies the inequality Proof. For a fixed ∈ [0, 2 ) and = we have Integrating twice this equation in , we obtain Hence, where Due to the comparison lemma [17, Lemma III.2.1], we have ( ) ≤V( ), whereV( ) is a solution of the equation Here is the Volterra operator defined by and, therefore,V But for any positive nondecreasing ℎ( ) we have Similarly, Thus from (15) it follows This implies the required result.
Note that in our reasoning and 0 can be arbitrary piecewise continuous functions.

Corollary 3. A solution ( ) of problem (1) satisfies the inequality
International Journal of Differential Equations 3 This corollary is sharp: as it is well known a solution of the homogeneous equation is an entire function of order no more than ( + 2)/2; see, for example, [1, Proposition 5.1]. Besides, our proof is absolutely different.
Employing the usual method for finding extrema it is easy to see that the function − ( ≥ 1) takes its smallest value in the range > 0 for 0 = 0 ( ) defined by Since =̂− , the lemma is proved.

Corollary 4 implies
Since ≤ 1 + 0 , we obtain Introduce the notations Denote by V , , and the Taylor coefficients of V( ), ( ), and ( ), respectively. Then Lemma 5 yields Let be the Taylor coefficients of . Since | | ≤ |V | + | | + | |, we have Let us consider the entire function Enumerate the zeros ( ) ( = 1, 2, . . .) of with their multiplicities in order of nondecreasing absolute values and assume that This lemma is a particular case of Theorem 2.1 proved in [18].
Proof. Taking into account that we obtain This and Lemma 7 prove the theorem.
International Journal of Differential Equations 5

Sums of Zeros and the Counting Function
In this section we derive a bound for sums of the zeros of solutions. To this end we need the following.