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: 欧美 亚洲 日韩 国产| 98超碰在线观看| 在线观看视频一区二区| 国产成人免费高清AⅤ| 国产手机在线观看| 欧美日韩国产一级| 国产精品无码AⅤ在线观看播放| 亚洲综合专区| 国产第一页亚洲| 2020极品精品国产| 精品欧美一区二区三区久久久| 国产成人一区二区| 国产香蕉国产精品偷在线观看| 国内精品91| 国产麻豆福利av在线播放 | 欧美中文字幕一区| 在线国产资源| 91无码人妻精品一区二区蜜桃| 91美女视频在线| 刘亦菲一区二区在线观看| 黄色在线不卡| 91美女视频在线观看| 少妇精品久久久一区二区三区| 久久免费视频播放| 亚洲欧美不卡视频| 国产日韩欧美在线播放| 毛片免费观看视频| 六月婷婷综合| 国产福利拍拍拍| 国产超薄肉色丝袜网站| 超碰色了色| 久热中文字幕在线| 中文一级毛片| 波多野衣结在线精品二区| 国产人免费人成免费视频| 国模在线视频一区二区三区| 国产丝袜91| 亚洲成aⅴ人片在线影院八| 亚洲福利片无码最新在线播放| 91综合色区亚洲熟妇p| 久久99热66这里只有精品一| 亚洲中文字幕97久久精品少妇| 2020久久国产综合精品swag| 国产麻豆91网在线看| 国产亚洲一区二区三区在线| 日韩黄色大片免费看| 亚洲精品福利网站| 亚洲色图欧美激情| 操国产美女| 亚洲婷婷六月| 亚洲 欧美 日韩综合一区| 久久精品中文字幕免费| 91精品久久久无码中文字幕vr| 国产制服丝袜无码视频| 狠狠色综合久久狠狠色综合| 久久semm亚洲国产| 国产一区在线视频观看| 综合五月天网| 久久91精品牛牛| 露脸一二三区国语对白| 99免费在线观看视频| 亚洲欧洲日产无码AV| 国产亚洲精品自在线| 全午夜免费一级毛片| 青青草久久伊人| 免费又爽又刺激高潮网址 | 国产自产视频一区二区三区| 日韩精品亚洲一区中文字幕| 亚洲日韩AV无码精品| 欧美一级一级做性视频| 91国内外精品自在线播放| 狠狠久久综合伊人不卡| 天堂久久久久久中文字幕| 国产精品亚洲а∨天堂免下载| 在线视频精品一区| 日韩欧美国产综合| 久久精品国产精品青草app| 国产毛片片精品天天看视频| 91色国产在线| 男人天堂亚洲天堂| 国产福利2021最新在线观看| 久久精品国产免费观看频道|