Dynamics of the functions \( f_\mu (z)=z\exp (z+\mu ) \) with the real parameter

Deng et al. SpringerPlus (2016) 5:850 DOI 10.1186/s40064-016-2411-2 Open Access RESEARCH Dynamics of the functions fµ(z) = z exp(z + µ) with the real parameter Xiaocheng Deng1, Fanning Meng1*, Jianming Lin2 and Wenjun Yuan1 *Correspondence: 1 School of Mathematics and Information Science, Key Laboratory of Mathematics and Interdisciplinary Sciences of Guangdong Higher Education Institutes, Guangzhou University, Guangzhou 510006, People’s Republic of China Full list of author information is available at the end of the article Abstract In this paper, the dynamics of the functions fµ (z) = z exp(z + µ) with the real parameter is studied. We say that a real parameter µ belongs to the set Bn for a positive integer n if fµ has an attracting cycle of n-order. We prove that the Fatou set F(fµ ) is a completely invariant attracting basin for every parameter µ < 0. Further, regarding the set Bn for n > 1, we prove the following results: (1) There exists µ∗ �= +∞ such that B2 = (2, µ∗ ). (2) For every positive integer n > 2, the set Bn is non-empty. (3) For every prime number p > 3, the set Bp has at least two components. Keywords: Julia set, Fatou set, Periodic point, Critical value Mathematics Subject Classification: Primary 37F10, Secondary 30D05 Introduction and main results Let f n be the n-th iterate of a transcendental entire function f. The maximal open set F(f) where the family {f n }∞ n=0 is normal in the sense of Montel is called the Fatou set, and its complement J (f ) := C\F (f ) is called the Julia set. The dynamics given by the iteration of transcendental entire maps has been widely studied (cf. Eremenko and Lyubich 1992). Baker (1970) first obtained an entire function f with the property J (f ) = C. He proved the following Theorem. Theorem 1 For a certain real positive value k , the function f (z) = kzez has the whole plane for its set J(f). After that, many authors (cf. Fagella 1995; Jang 1992; Kuroda and Jang 1997; Morosawa 1998) studied the dynamics of the functions fµ (z) := z exp(z + µ). Jang (1992) proved that the set B0 := {µ ∈ R|J (fµ ) = C} is an infinite set. Further, Morosawa (1998) proved that the one-dimensional Lebesgue measure of B0 is positive. The function fµ has only two singular values: an asymptotic value 0 and a critical value fµ (−1), hence the Fatou set F (fµ ) has no wandering components. The asymptotic value is fixed, hence there is only one free singular orbit. It follows that there is at most one © 2016 The Author(s). This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Deng et al. SpringerPlus (2016) 5:850 Page 2 of 15 cycle of periodic Fatou components, either attracting, parabolic or Siegel. Since for real parameters the orbit of the free critical value is entirely real, there is no possibility of Siegel discs. Hence only attracting or parabolic cycles are possible and attracting or parabolic periodic points (if they exist) are real. In this paper, our main goal is to study the structure of Bn, where Bn := {µ ∈ R|fµ has a cycle of attracting periodic points of n-order}, for every positive integer n. For every real parameter µ, fµ has two real fixed points 0 and −µ. The multiplier of 0 is eµ, and the multiplier of −µ is 1 − µ. Hence µ ∈ B1 if and only if µ satisfies the following condition: −1 < eµ < 1 or − 1 < 1 − µ < 1. This immediately implies that B1 = (−∞, 0) ∪ (0, 2). Since a completely invariant domain contains all singular values, it is easy to see that if µ ∈ (0, 2), then the Fatou set F (fµ ) is not a completely invariant attracting basin. However, for µ ∈ (−∞, 0), we have the following result. Theorem 2 For every parameter µ < 0, the Fatou set F (fµ ) is a completely invariant attracting basin. Regarding the set Bn for n > 1, we prove the following Theorems. Theorem 3 There exists µ∗ � = +∞ such that B2 = (2, µ∗ ). Theorem 4 For every positive integer n > 2, Bn � = ∅. Theorem 5 For every prime number p > 3, the set Bp has at least two components. Remark 6 We believe that B3 is also an interval and Theorem 5 holds also for every integer n > 3. An interesting problem is how many components contained in Bp. The Proof of Theorem 2 In order to prove Theorem 2, we need the hr (x) := r 2 exp(−2x) − x2 and �r := {z ∈ C| |z| < r}. following Lemmas. Set Lemma 7 Let r ∈ (0, e−1 ), then hr has 3 distinct zeros x1 < −1, x2 ∈ (−1, 0) and x3 > 0. Moreover, the solving set of inequality hr (x) ≥ 0 is the union of I1 = (−∞, x1 ] and I2 = [x2 , x3 ]. Proof Noting f0 (x) = xex and hr (x) = e−2x (r 2 − x2 e2x ), we have hr (x) = 0 ⇔ |f0 (x)| = |r|, (1) Deng et al. SpringerPlus (2016) 5:850 Page 3 of 15 and (2) hr (x) > 0 ⇔ |f0 (x)| < |r|. From f0′ (x) = (x + 1)ex, we see that f0 (x) is decreasing in (−∞, −1] and increasing in [−1, +∞), and f0 (−1) = −e−1 is the minimum value of f0 (x). Note that f0 (0) = 0, lim f0 (x) = 0 and x→−∞ lim f0 (x) = +∞, x→+∞ if r ∈ (0, e−1 ), then we infer that f0 (x) = r has the only one root x3 > 0, and f0 (x) = −r has two roots x1 < −1 and x2 ∈ (−1, 0). Moreover, the solving set of inequality |f0 (x)| < |r| is the union of (−∞, x1 ) and (x2 , x3 ). Hence from (1) and (2), we obtain the  assertion.  Lemma 8 Let r ∈ (0, e−1 ), then f0−1 (�r ) has two connected components D1 and D2, and the set D1 ∪ D2 ∪ (−∞, 0) is connected. Proof For every z = x + iy ∈ f0−1 (�r ), we have |f0 (z)| = |z exp(z)| < r, which implies  x2 + y2 exp(x) < r. It follows that  |y| < hr (x). From Lemma 7, we know that the graph of |y| =  L1 : |y| = hr (x), x ∈ I1 √ hr (x) consists of two curves and L2 : |y| =  hr (x), x ∈ I2 . Therefore, f0−1 (�r ) has two connected components D1 and D2, where ∂D1 = L1 and ∂D2 = L2. Obviously the set D1 ∪ D2 ∪ (−∞, 0) is connected. Hence we obtain the asser tion.  Lemma 9 Let I = (a, b) be an open interval, and f : I → I be a continuous mapping. (1) If f (x) > x for every x ∈ I , then we have lim f n (x) = b. n→+∞ (2) If f (x) < x for every x ∈ I , then we have lim f n (x) = a. n→+∞ Deng et al. SpringerPlus (2016) 5:850 Proof (1) Suppose f (x) > x for every x ∈ I . Then it follows that the sequence {f n (x)}∞ n=1 +∞ x < +∞ either tends to or tends to . is increasing. Hence the sequence {f n (x)}∞ 0 n=1 If the first case happens, then we have b = +∞. If the second case happens, then we infer x0 = b. Otherwise, x0 < b, and then x0 is a fixed point of f, which contradicts that f (x) > x for every x ∈ I . Thus, we obtain that the sequence {f n (x)}∞ n=1 tends to b.  (2) Similar as the proof of (1), we can obtain (2) easily.  Proof of Theorem 2 Proof Let µ < 0. Then singular value 0 of fµ is an attracting fixed (...truncated)