On buried points and phase transition points in the Julia sets concerning renormalization transformation
© Gao; licensee Springer. 2014
Received: 4 April 2014
Accepted: 13 August 2014
Published: 11 September 2014
Considering a family of rational maps concerning renormalization transformation, we give a perfect description of buried points and phase transition points in the Julia set . Furthermore, we prove that contains an open interval where all points are buried points for some parameters n and λ, which is according to the problem that Curry and Mayer proposed.
with two parameters () and . This physical model can be derived from the limit distribution of zeroes of the partition function in the famous Yang-Lee theory [8–11]. In fact, some interesting relationships among the phase transitions, the critical exponents, the critical amplitudes and the shape of the Julia sets were found in . After this, many works have been devoted to the Julia sets of the renormalization transformation [13–17].
In , Qiao proved that when n is an even number and when n is an odd number. Moreover, he also proved that the equation has one unique real solution when n is an even number. Furthermore, in this paper we prove the following.
if n is an even number, then contains buried points if and only if ;
if , then contains buried points if and only if ;
if , then contains buried points if and only if , where is an absolute constant.
In , Curry and Mayer proposed some questions about buried points. One of these questions is the following:
Is there the set of buried points whose components are not either points, circles, or the irrational points of the Sierpinski curve?
In this paper, considering the Julia sets , we prove that the set of buried points may contain an open interval of the real axis ℝ. In fact, we have the following.
For this physical model, it is well known that there is a specific physical significance about the intersection points between the Julia set and the positive real axis when λ is a positive integer. In fact, these intersection points are the locations of phase transitions of this model. A natural problem is how to distribute on the locations of phase transitions? In the following we will show that the number of this kind of intersection points may be 1, 2, 3, or ∞. Denote by the number of the intersection points of and . Then we have the following.
2 Some notations and preliminary results
Let f be a rational map with degree from a complex sphere to itself. The notation means the k th iteration of f. A point z is called a critical point if . A point z is called a periodic point if for some , the minimal of such k is called the period of z. For a periodic point with periodic k, the multiplier of is defined as . The periodic point is either attracting, indifferent, or repelling according to , , or . The Julia set, denoted by , is the closure of repelling periodic points. Its complement is called the Fatou set, denoted by . A component D of is called completely invariant if . Moreover, if D is a completely invariant component, then . Let be the post-critical set of f, i.e., the closure of the forward orbits of critical points. It plays a crucial rule in the study of complex dynamics. For the classical results in complex dynamics, see [18, 19] and .
In order to prove our theorems, we need the following lemmas.
Let f be a rational map f with and connected and locally connected, then contains buried points if and only if has no completely invariant Fatou component.
Lemma 2 ()
If is connected, then it is locally connected.
Lemma 3 ()
If is an odd integer, then has only four real fixed points (, ) for .
Lemma 4 ()
is an attracting fixed point for ;
is a parabolic fixed point and for or ;
is a repelling fixed point for .
If is an even integer, then contains a completely invariant Fatou component if and only if . Furthermore, is connected for .
If is an odd integer, then contains a completely invariant Fatou component for . Furthermore, contains at most three Fatou periodic (not completely invariant) cycles for .
3 Proof of Theorem 1
So has only four critical values 1, ∞, 0 and .
By Lemma 1, Lemma 2 and Lemma 5, it is easy to see that contains buried points if and only if for even integer n. Similarly, if is an odd number, contains no buried point for , contains buried points for . Hence we need only to investigate the case that is an odd integer and .
- (I)If and . It is easy to verify that since for . Hence . Next we define
Note that , then . Since , then is completely invariant. This implies that and thus contains no buried point.
- (II)If . Let and be two numbers stated in Lemma 4. Below we prove(7)
By the proof of Proposition 2 in , we know that is monotone decreasing from to on . So we conclude that since is a parabolic fixed point of by Lemma 4.
So and thus (7) is obvious.
Below we prove (8).
We get (8).
If . By (5) and (7), we know . By the same discussion as used in the case , we can deduce that is completely invariant for . This implies that and thus contains no buried point.
If , it is easy to see that contains at most another k-periodic cycle except for and . By (3) and (4), we can deduce that since is a repelling fixed point by Lemma 4 (in fact, the periodic Fatou components lie on both sides of ). Then cannot contain a completely invariant Fatou component. This shows that contains buried points.
If . By (5), . By (3), . It shows that is not completely invariant since it contains only one critical point and , and thus contains buried points. It completes the proof of Theorem 1.
4 Proof of Theorem 2
In fact, let Γ be the circle centered at with radius . Then Γ passes through critical points 1, and , and . We conclude that and . For , there exists a positive integer k such that . This implies . Hence .
Let be the component of containing 1. Then and . Therefore and .
Obviously, has only two components and . By and , we know that for every component W of . Similarly, by (3) and (4), we can deduce that for every component of except for . Hence the closures of all pull-backs of W and are disjoint with ℝ since maps ℝ into . By , we know that there exist at least two real numbers such that and are Feigenbaum-like maps. Obviously, and contain only two Fatou periodic domains and . By the above analysis and (4), we conclude that lies in the Julia sets and , and each point of this open interval does not belong to the boundaries of any Fatou components. This implies that is a buried interval of and (in fact, by a calculation, there exists such that is strictly eventually periodic, then is also a buried interval of ). It completes the proof of Theorem 2.
5 Proof of Theorem 3
If n is an even number. We consider the real fixed points of , i.e., the set . It is easy to see that has no fixed point on since for . By (2), is monotone increasing on from 0 to +∞. Note that and is an attracting fixed point, then there exists at least one fixed point such that for and .
Next we claim that α is the unique fixed point in .
Otherwise, if there exists another fixed point γ in . Without loss of generality, we assume , then or on . Note that is monotone increasing on , then one of them (α or γ) is an attracting fixed point or a parabolic fixed point. Since , , and , then the immediate basin contains no critical point. But the Sullivan theorem  says that each immediate basin contains at least one critical point. It is a contradiction. Hence α is the unique fixed point on . Similarly, we can deduce that there also exists a unique fixed point β in . Hence has only four real fixed points 0, 1, α, β. α and β are two repelling fixed points.
- (2)If is an odd number. By Lemma 3, we know that has only four fixed points , and . By a similar discussion as used in (3) in Theorem 1, we can deduce that there exist only two points and such that and . Furthermore, we can deduce that
We know that has only one fixed point . Obviously, q is a repelling fixed point and . In what follows we distinguish two cases to discuss.
If and . , has only six critical points ±1, ±i, 0, ∞ and , , , . Considering the real fixed points of , has no fixed point on . Note that is monotone decreasing on from +∞ to 1, then has no fixed point on .
If and . By a similar discussion as (6) in Theorem 1, we can easily deduce that . Then . We conclude since for . By (2), is monotone decreasing on . Then there exists a unique point such that . Note that , then , i.e., .
We also distinguish two cases to discuss.
If and . We know that has a unique repelling point on . It is easy to see that for and for . Note that and is monotone decreasing for , then for . Then and . We get , i.e., .
If and . By a similar discussion as (6) in Theorem 1, we also get . Hence since . Note that is monotone decreasing for , then for . This means . We obtain , i.e., . The proof of Theorem 3 is completed.
The author is entirely responsible for this research. The author read and approved the final manuscript.
The author would like to thank the referees for their valuable suggestions for improving this paper. This work was supported by the National Natural Science Foundation of China (No. 11371363, 11231009, 11261002, 11201474) and the Special Basic Scientific Research Funds of Central Universities in China.
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