2011
DOI: 10.1002/mana.201010038
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Zero distribution of solutions of complex linear differential equations determines growth of coefficients

Abstract: It is shown that the exponent of convergence λ(f) of any solution f of with entire coefficients A0(z), …, Ak−2(z), satisfies λ(f) ⩽ λ ∈ [1, ∞) if and only if the coefficients A0(z), …, Ak−2(z) are polynomials such that $ \deg (A_j)\le (k-j)(\lambda -1) $ for j = 0, …, k − 2. In the unit disc analogue of this result certain intersections of weighted Bergman spaces take the role of polynomials. The key idea in the proofs is W. J. Kim’s 1969 representation of coefficients in terms of ratios of linearly indepe… Show more

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Cited by 10 publications
(23 citation statements)
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“…has only one dominant term (highest-degree term), then (1.1) has no admissible transcendental meromorphic solutions with a few poles. There are also many other papers concerning the structure of solutions to various differential equations (for example [1,4,[10][11][12][13][14]). In this paper, we consider the algebraic differential equation…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…has only one dominant term (highest-degree term), then (1.1) has no admissible transcendental meromorphic solutions with a few poles. There are also many other papers concerning the structure of solutions to various differential equations (for example [1,4,[10][11][12][13][14]). In this paper, we consider the algebraic differential equation…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…For example, the choice E(z) = , j = 1, 2, then reduce to constant multiples of the functions f 1 , f 2 in (2.9), respectively. We have seen that it is possible to construct zero-free solutions bases {f 1 , f 2 } for (1.1) in the cases of C and D. It has recently been proved [14] in both cases that arbitrary linear combinations f = C 1 f 1 + C 2 f 2 typically have the maximal quantity of zeros when compared to the growth of f 1 , f 2 .…”
Section: Concluding Remarks On Zero-free Solution Basesmentioning
confidence: 99%
“…. Therefore, to avoid unnecessary repetition, we merely sketch a proof of (a), and refer to [10] for a further discussion on the topic. This is not a surprise, because the growth of g is determined via g = h −k by a solution h of (1·3) with B entire.…”
Section: Functions Of Maximal Growth In αmentioning
confidence: 99%
“…, k −1, see [10] for details. Therefore one implication in Theorem B(a) follows by the growth estimates for the solutions of (1·8), see Lemma D(a) below.…”
Section: It Is Well Known That λ(H) σ (H) For All H ∈ H(d)mentioning
confidence: 99%
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