999精品在线视频,手机成人午夜在线视频,久久不卡国产精品无码,中日无码在线观看,成人av手机在线观看,日韩精品亚洲一区中文字幕,亚洲av无码人妻,四虎国产在线观看 ?

The growth of difference equations and differential equations

2021-02-23 12:07:14陳宗煊,張然然,藍雙婷

We use σ(f) to denote the order of f(z), λ(f) to denote the convergence exponent of zeros of f(z), and λ(f) to denote the convergence exponent of distinct zeros of f(z).

For n ∈N, we define Over the course of the last 15 years, many authors have paid great attention to complex difference equations and to the difference analogues of Nevanlinna’s theory, and have obtained many interesting results, including [2-6, 8, 10, 11, 15, 18, 19, 21, 23].

Ishizaki and Yanagihara [19] considered the growth of transcendental entire solutions of difference equations

where Q,··· ,Qare polynomials, and obtained the following theorem:

Theorem A Let f(z) be a transcendental entire solution of (1.1), and let its order χ <1/2. Then

where a rational number χ is the slope of a Newton polygon for the equation (1.1), and L >0 is a constant. In particular, we have that χ >0.

Note that the equation (1.1) can be rewritten as

Example 1.1 (see [19]) Suppose that f(z) is a transcendental entire solution of the difference equation

holds. Supposing that f(z) is a meromorphic solution of (1.2), we then have that σ(f)≥1.Chen weakened the condition (1.3) of Theorem B and obtained

Theorem C (see [4,5]) Let F(z),P(z),··· ,P(z)be polynomials such that FPP/≡0 and

Then every finite order transcendental meromorphic solution f(z) of

satisfies σ(f)≥1 and λ(f)=σ(f).

Theorem D (see [4, 5]) Let P(z),··· ,P(z) be polynomials such that PP/≡0 and(1.4) is satisfied. Then every finite order transcendental meromorphic solution f(z)(/≡0) of equation (1.2) satisfies σ(f)≥1, and f(z) assumes every non-zero value a ∈C infinitely often and λ(f -a)=σ(f).

Chen considered difference equations with constant coefficients,and obtained the following two theorems:

Theorem E (see [4]) Let C,··· , Cbe constants such that CC/= 0 and such that they satisfy

Then every finite order transcendental meromorphic solution f(z)(/≡0) of the equation

satisfies σ(f)≥1,f(z)assumes every nonzero value a ∈C infinitely often,and λ(f-a)=σ(f).

Theorem F Let C,··· , Cbe constants and let F(z) be a polynomial such that FCC/≡0 and (1.6) is satisfied. Then every finite order transcendental meromorphic solution f(z) of the equation

satisfies that λ(f)=σ(f)≥1.

Question 1.1 Can the condition (1.6) be deleted from Theorems E and F?

In this paper, we answer this question in the affirmative and delete condition (1.6) from Theorems E and F, and obtain the following theorems:

Theorem 1.1 Let C,··· , Cbe constants such that CC/=0. Then every finite order transcendental meromorphic solution f(z)(/≡0) of the equation (1.7) satisfies that σ(f)≥1.

Theorem 1.2 Let C,··· , Cbe constants, and let F(z) be a polynomial such that FCC/≡0. Then every finite order transcendental meromorphic solution f(z)of the equation(1.8) satisfies that λ(f)=σ(f)≥1.

Remark 1.1 In Theorems 1.1 and 1.2, we have deleted condition (1.6) of Theorems E and F.

In Theorem 1.1, we cannot give the result that every finite order transcendental meromorphic solution f(z) of (1.7) assumes every nonzero value a ∈C infinitely often and that λ(f -a)=σ(f).

In Theorem 1.2, we obtain the same results as for Theorem F.

Remark 1.2 By Theorems 1.1 and 1.2, we see that in Theorems C and D, the condition(1.4) can be weakened as

where “≥1” of (1.4) is deleted. Thus, we can obtain the following corollaries:

Corollary 1.1 Let F(z),P(z),··· ,P(z)be polynomials such that FPP/≡0 and such that they satisfy (1.4). Then every finite order transcendental meromorphic solution f(z) of(1.5) satisfies that σ(f)≥1 and λ(f)=σ(f).

Corollary 1.2 Let P(z),··· ,P(z) be polynomials such that PP/≡0 and such that they satisfy (1.4). Then every finite order transcendental meromorphic solution f(z)(/≡0) of(1.2) satisfies that σ(f)≥1.

Since 1982,many authors have studied the growth of solutions of linear differential equations and obtained many good results, see [7, 9, 12, 14].

Now, we consider the growth of solutions of homogeneous and non-homogeneous linear differential equations with constant coefficients, and obtain the following results:

Theorem 1.3 Let C,··· , Cbe constants such that CC/=0. Then every meromorphic solution f(z)(/≡0) of the homogeneous differential equation

Remark 1.3 From Theorems 1.1 and 1.3, we see that for homogeneous equations (1.7)or (1.10), we can only obtain that σ(f)≥1 or σ(f)=1.

From Theorems 1.2 and 1.4,we see that for non-homogeneous equations(1.8)or(1.11),we can obtain that λ(f)=σ(f)≥1 or λ(f)=σ(f)=1.

Remark 1.4 From Theorem 1.3,we see that homogeneous equation(1.10)does not have a polynomial solution. From Theorem 1.4, we see that non-homogeneous equation (1.11) may have a polynomial solution.

Example 1.4 The differential equation

has solutions f(z)=cos z and σ(f)=1.

Example 1.5 The differential equation has a polynomial solution f(z)=z+z+1.

2 The Asymptotic Method

Theorem G (see [17(p.30), 20]; the Wiman-Valiron Theory) Suppose that f is a transcendental entire function, and that for any given 0 <δ <1/8, there exists a set H of finite logarithmic measure such that

and the corresponding homogeneous linear differential equation

Since the solution νof (2.9) is an algebraic function of z, setting the principal part of νas a(ρ)z(a, ρ are nonzero real numbers) in the neighborhood of z =∞, we get that,

By (2.9) and (2.10), it is easy to see that the degrees of all of the terms of the left of (2.9) are

Since ν(r) is the solution of (2.9), we see that in (2.11), at least, there are two terms that are both the largest numbers and equal, and that the sum of coefficients of their corresponding terms in (2.9) is zero. Hence, ρ satisfies that we have i and j such that

Thus, we see that ρ is a rational number, and we have at most n such rational numbers that are not less than 1/n.

3 Proofs of Theorems 1.1 and 1.2

We need the following lemmas to prove Theorems 1.1 and 1.2:

Lemma 3.1 (see [2, 4]) Let n ∈N, and let f be a transcendental meromorphic function of an order less than 1. Then there exists an ε-set Esuch that

Since Eare ε-set, we see that a set His of finite logarithmic measure. By (3.1)and (3.2), we obtain that By (3.7) and (3.8), we see that (3.6) is a contradiction.

Hence, every finite order transcendental meromorphic solution f(z)(/≡0) of the equation(1.7) satisfies that σ(f)≥1.□

Proof of Theorem 1.2 Using the same method as in the proof of Theorem 1.1,we can prove that every finite order transcendental meromorphic solution f(z) of the equation (1.8)satisfies σ(f)≥1.

Now, we prove that every finite order transcendental meromorphic solution f(z) of the equation (1.8) satisfies λ(f)=σ(f).

By (1.8), we set

Hence, λ(f)=σ(f).□

4 Proofs of Theorems 1.3 and 1.4

Proof of Theorem 1.3 It is well known that all meromorphic solutions of equations(1.10) are entire functions.

Suppose that f(z) is a solution of (1.10).

First, we prove that f(z) cannot be a polynomial. If f(z) is a nonzero constant, then f(z) = ··· = f(z) = 0, and this is not possible. If f(z) is a polynomial with deg f(z) ≥1,then deg f(z)<deg f(z) (j =1,··· ,n) which is also not possible.

Now, we suppose that f(z) is a transcendental entire function with σ(f)=σ.

By the Wiman-Valiron theory,we see that there exists a set Hof finite logarithmic measure such that (3.4) holds, where |f(z)| = M(r,f),|z| = r /∈[0,1]∪H∪H, ν(r,f) is the central index of f(z). By (3.4) and (1.10), we obtain

Thus, by (4.4) and (4.5), we see that (4.1) is a contradiction.

Hence, by (4.4) and (4.5), we see that σ = 1, that is, every solution f(z)(/≡0) of the equation (1.10) satisfies σ(f)=1.

Theorem 1.3 is thus proved.□

Proof of Theorem 1.4

We need the following lemma:

Lemma 4.1 (see [13, pp. 168]) Suppose that a, a, ··· , a, F /≡0 are entire functions, that f satisfies the differential equation

Proof of Theorem 1.4 Using the same method as in the proof of Theorem 1.3, we see that all meromorphic solutions of equations (1.11) are entire functions, and if f(z) is a transcendental entire solution of (1.11), then σ(f)=1.

Since C(j =0,··· ,n)are constants,F(z)/≡0 is a polynomial,and thus C,F(z)and f(z)satisfy condition (4.6) of Lemma 4.1. By Lemma 4.1, we obtain that λ(f)=λ(f)=σ(f)=1.

Theorem 1.4 is thus proved.□

主站蜘蛛池模板: 狠狠色婷婷丁香综合久久韩国| 2021精品国产自在现线看| 97视频在线观看免费视频| 亚洲国产理论片在线播放| 日本人又色又爽的视频| 久久综合丝袜长腿丝袜| 亚洲精品麻豆| 国产美女主播一级成人毛片| 狠狠色成人综合首页| 91网站国产| 亚洲欧美日韩高清综合678| 国产精品亚洲专区一区| 欧美日韩一区二区在线免费观看 | 国产91无码福利在线| 麻豆精选在线| 色婷婷成人网| 色综合久久无码网| 91免费在线看| 国产91在线|日本| 伊人久久大香线蕉成人综合网| 成人永久免费A∨一级在线播放| 国产产在线精品亚洲aavv| 2021国产精品自产拍在线观看| 尤物精品视频一区二区三区| 免费一级毛片完整版在线看| 欧美成一级| 91成人在线观看| 国产门事件在线| 亚洲床戏一区| 欧美www在线观看| 五月天久久综合| 亚洲 欧美 偷自乱 图片 | 欧美亚洲另类在线观看| 欧洲日本亚洲中文字幕| 国产精品专区第一页在线观看| 少妇精品在线| A级毛片高清免费视频就| 亚洲AV免费一区二区三区| 国产一级妓女av网站| 久久精品最新免费国产成人| 免费在线不卡视频| 欧美国产成人在线| 男女性色大片免费网站| 欧美激情福利| 色香蕉影院| 99一级毛片| 四虎国产精品永久在线网址| 九月婷婷亚洲综合在线| 国产成人综合亚洲欧美在| 婷婷五月在线| 日韩精品亚洲一区中文字幕| 久爱午夜精品免费视频| 黄色成年视频| 国产精品视频观看裸模| 丁香婷婷综合激情| 色首页AV在线| 黄色污网站在线观看| 无码日韩人妻精品久久蜜桃| 在线欧美日韩| 91系列在线观看| 怡红院美国分院一区二区| 国产乱人激情H在线观看| 欧美激情第一欧美在线| 精品国产香蕉伊思人在线| 尤物特级无码毛片免费| 色综合天天综合中文网| av大片在线无码免费| 国产精品无码翘臀在线看纯欲| 久久影院一区二区h| 亚洲天堂视频在线播放| 女人天堂av免费| 精品视频在线一区| 欧美一级99在线观看国产| 免费高清自慰一区二区三区| 国产一在线观看| 亚洲无码高清一区二区| 欧美日本激情| 日韩精品无码免费专网站| 男人的天堂久久精品激情| 99精品热视频这里只有精品7| 亚洲一区网站| 久久青草免费91观看|