Rigorous prediction of quadratic hyperchaotic attractors of the plane

Transcription

1 Rigorous predition of quadrati hyperhaoti attrators of the plane Zeraoulia Elhadj 1, J. C. Sprott 2 1 Department of Mathematis, University of Tébéssa, 12000), Algeria. zeraoulia@mail.univ-tebessa.dz and zelhadj12@yahoo.fr. 2 Department of Physis, University of Wisonsin, Madison, WI 53706, USA. sprott@physis.wis.edu. February 13, 2008 Abstrat This paper derives suffiient onditions for the existene of hyperhaoti attrators in the general 2-D quadrati map. PACS numbers: a, Gg. Keywords: 2-D quadrati map, hyperhaos. 1 Introdution The most general 2-D quadrati map is given by f x, y) = a 0 + x + a 2 y + a 3 x 2 + a 4 y 2 + a 5 xy = z a x, y) b 0 + b 1 x + b 2 y + b 3 x 2 + b 4 y 2 + b 5 xy = z b x, y) 1) where a i, b i ) 0 i 5 R 12 are the bifurations parameters. Some speial ases of the map 1) an be used in potential appliations in several different ways and types of studies [ ]. Some important results about the dynamial properties, bifurations, and stability of some speial ases of the 2-D 1

2 map 1) are given in [ ]. However, there are a few papers that fous on the general ase of this map. For example, in [11] some solutions of low-dimensional, low-order polynomial maps were lassified numerially as either fixed point, limit yle, haoti, or unstable using Lyapunov exponent alulations, with the result that a few perent are haoti. For the 2-D quadrati maps, this perentage is about ± 0.36%. Furthermore, in [12] the orrelation dimension was alulated for the strange attrators obtained numerially for some ases of the map 1), and it was found that the average orrelation dimension sales approximately as the square root of the dimension of the system with a small variation. In [10-13] a systemati searh for haoti orbits of the general 2-D quadrati map 1) with randomly hosen oeffiients was desribed using a simple omputer program that gives different attrators. Some simple speial ases of the general 2-D quadrati map 1) were studied in detail in [ ], with analytial results in [1-2-3]. In [16] the number of possible haoti attrators for the map 1) was redued to 30 types, and the existene of unbounded and bounded orbits was investigated analytially with analytial preditions of some system orbits. Furthermore, a lassifiation of the possible haoti orbits was given aording to the number of nonlinearities, showing how to redue all the dynamis of the general ase 1) to a finite number of maps with well known formulas. On the other hand, if two or more Lyapunov exponents of a dynamial system are positive throughout a range in the parameters spae, then the resulting attrators are alled hyperhaoti. The importane of these attrators is that they are less regular and the iteration points seemingly almost fill the spae, whih suggest one of appliations of haos in fluid mixing, for example in [23-24]. This paper offers a rigorous proof of the hyperhaotiity of the general map 1) using the so-alled seond-derivative test defined for real funtions. Indeed, the notions of ritial points and the seond-derivative test are well defined for funtions of two variables. The ritial points of funtion f x, y) are solutions of the equations: fx,y) = 0, and fx,y) = 0, whih must be x y solved simultaneously. Let x, y ) be a ritial point, and define d f x, y ) = 2 f x, y) x x 2, y ) 2 f x, y) 2 2 f x, y) x y 2, y ) x, y )) 2) x y We have the following ases: If d f x, y ) > 0 and 2 fx,y) x x 2, y ) < 0, then f x, y) has a relative maximum at x, y ). If d f x, y ) > 0 and 2

3 2 fx,y) x x 2, y ) > 0, then f x, y) has a relative minimum at x, y ). If d f x, y ) < 0, then f x, y) has a saddle point at x, y ). If d f x, y ) = 0, the seond-derivative test is inonlusive. The Jaobian matrix of the map 1) is given by ) a1 x + 2a J x, y) = 3 x + a 5 y a 2 y + 2a 4 y + a 5 x. 3) b 1 x + 2b 3 x + b 5 y b 2 y + 2b 4 y + b 5 x For the map 1) assume that a 3 > 0, 4a 3 a 4 > a 2 5, a 4 > 0 C 1 : b 3 < 0, 4b 3 b 4 > b 2 5, b 4 < 0, 4) where C 1 defines a subset of the elements a i, b i ) 0 i 5 R 12. If the seondderivative test for both z a x, y) and z b x, y) is used separately, then one has for all x, y) R 2 that { za x, y) a 0a 2 5 a 2 a 5 4a 0 a 3 a 4 + a 4+a 2 2 a 3 a 2 5 4a 3a 4 = L a z b x, y) b 0b 2 5 b 1b 2 b 5 4b 0 b 3 b 4 +b 2 1 b 4+b 2 2 b 3 b 2 5 4b 3b 4 = L b, 5) i.e., for all iterations x, y) R 2 of the map 1), one has x L a and y L b. 6) It is shown in [17] that a system x k+1 = f x k ), x k Ω R n, suh that fx) = J = λ max J T J) N < +, 7) with a smallest eigenvalue of J T J that satisfies λ min J T J ) θ > 0, 8) where N 2 θ, then, for any x 0 Ω, all the Lyapunov exponents at x 0 are loated inside [ ln θ 2, ln N]. That is, ln θ 2 l i x 0 ) ln N, i = 1, 2,..., n, 9) where l i x 0 ) are the Lyapunov exponents for the map f. 3

4 For the map 1) one has J T J11 J J = 12 J 12 J 22 ) 10) where J 12 = J 21 beause J T J is symmetri and J 11 = + 2a 3 ) x + a 5 y + b 1 + 2b 3 ) x + b 5 y J 12 = + 2a 3 ) x + a 5 y) a 5 x + a 2 + 2a 4 ) y) + b 1 + 2b 3 ) x + b 5 y) b 5 x + b 2 + 2b 4 ) y) J 22 = a 5 x + a 2 + 2a 4 ) y + b 5 x + b 2 + 2b 4 ) y. 11) Beause J T J is at least a positive semi-definite matrix, then all its eigenvalue are real and positive, i.e., λ max J T J ) λ min J T J ) 0. Hene the eigenvalues of J T J are given by λ max J T J ) = J 11+J 22 + J11 2 2J 11J 22 +4J12 2 +J λ min J T J ) 12) = J 11+J 22 J11 2 2J 11J 22 +4J12 2 +J where We have J 11 = C 1 x 2 + C 2 y 2 + C 3 xy J 12 = 1 2 C 3x 2 + C 4 y 2 + C 5 xy J 22 = C 2 x 2 + C 6 y 2 + 2C 4 xy 13) 4

5 C 1 = 2a b 3 + b 1 0 C 2 = a b C 3 = 2 + 2a 3 ) a 5 + b 1 + 2b 3 ) b 5 ) C 4 = a 2 + 2a 4 ) a 5 + b 2 + 2b 4 ) b 5 14) C 5 = + 2a 3 ) a 2 + 2a 4 ) + b 1 + 2b 3 ) b 2 + 2b 4 ) + a b 2 5 C 6 = 2a 4 + a 2 + 2b 4 + b 2 0. The 2-D quadrati map 1) has hyperhaoti attrators if λ min J T J ) > 1, i.e., J 11 + J 22 > 2 J 11 + J 22 < J 11 J 22 J ) The first ondition of 15) gives J 11 J 22 J 2 12 > 1. where ξ 1 x 2 + ξ 3 y) x + ξ 2 y 2 2 > 0 16) ξ 1 = C 1 + C 2 0 ξ 2 = C 2 + C 6 0 ξ 3 = C 3 + 2C 4. 17) The seond-derivative test is inonlusive for the funtion ξ 1 x 2 + ξ 3 y) x + ξ 2 y 2 2 beause it minimum is 2 and does not depend on bifuration parameters. Hene it is not possible to hoose a lower bound for it. Thus inequality 16) has a solution if ξ 2 3 4ξ 1 ξ 2 > 0 and x > ξ 3 y+ q ξ 2 3 4ξ 1 ξ 2)y 2 +8ξ 1 2 ξ 1 for all y L b. We hoose y L b suh that x L a ξ 3 y+ q ξ 2 3 4ξ 1 ξ 2)y 2 +8ξ 1 2 ξ 1, i.e., 5

6 ξ 2 3 4ξ 1 ξ 2 > 0 C 2 : The seond ondition of 16) gives L b 2 ξ 1 L a ξ 3 ξ 3 < 0 8ξ 2 4ξ 1 ξ 2 L 2 a + L 2 aξ 2 3 < 0. 18) ξ 4 x 4 + ξ 5 y 4 + ξ 6 x 3 y + ξ 7 xy 3 + ξ 8 x 2 y 2 + ξ 1 x 2 + ξ 2 y 2 + ξ 3 xy 1 < 0 19) where ξ 4 = 1 4 C2 3 C 1 C 2 ξ 5 = C 2 4 C 2 C 6 ξ 6 = C 3 C 5 C 2 C 3 2C 1 C 4 20) ξ 7 = 2C 4 C 5 C 3 C 6 2C 2 C 4 ξ 8 = C 2 5 C 1 C 6 C 3 C 4 C 2 2. Now onsider the funtion h x, y) = ξ 4 x 4 + ξ 5 y 4 +ξ 6 x 3 y+ξ 7 xy 3 +ξ 8 x 2 y 2 + ξ 1 x 2 +ξ 2 y 2 +ξ 3 xy 1. The ritial points of h are the solutions of the system 4ξ 4 ) x 3 + 3ξ 6 y) x 2 + 2ξ 1 + 2ξ 8 y 2 ) x + ξ 7 y 3 + ξ 3 y = 0 4ξ 5 ) y 3 + 3xξ 7 ) y 2 + 2ξ 8 x 2 + 2ξ 2 ) y + ξ 6 x 3 + xξ 3 = 0. 21) Assume first that C 3 : ξ 4 0, ξ ) Then both equations in 21) are ubi, and the first equation of 21) has at least ) one real solution z 1) for all values of y, and at most three roots z i) for all values of y. The seond equation of 21) has at least one 1 i 3 ) real solution u 1) for all values of x, and at most three roots u i) for 1 i 3 6

7 all values of x. Thus there is still solutions of equation 21) that are ritial points of the funtion h. On the other hand, one has d 2 h dx 2 x, y) = 12x 2 ξ 4 + 2ξ 8 y 2 + 6ξ 6 xy + 2ξ 1 d h x, y) = ξ 9 x 4 + ξ 10 y 4 + ξ 11 xy 3 + ξ 12 x 3 y + ξ 14 x 2 y 2 + ξ 15 x 2 + ξ 16 y 2 + ξ 17 xy + ξ 18 23) where ξ 9 = 24ξ 4 ξ 8 9ξ 2 6 ξ 10 = 24ξ 5 ξ 8 9ξ 2 7 ξ 11 = 72ξ 5 ξ 6 12ξ 7 ξ 8 ξ 12 = 72ξ 4 ξ 7 12ξ 6 ξ 8 ξ 13 = 144ξ 4 ξ ξ 6 ξ 7 12ξ 2 8 ξ 14 = 144ξ 4 ξ ξ 6 ξ 7 12ξ ) ξ 15 = 24ξ 2 ξ 4 + 4ξ 1 ξ 8 6ξ 3 ξ 6 ξ 16 = 24ξ 1 ξ 5 + 4ξ 2 ξ 8 6ξ 3 ξ 7 ξ 17 = 12ξ 1 ξ ξ 2 ξ 6 8ξ 3 ξ 8 ξ 18 = 4ξ 1 ξ 2 ξ 2 3. ) ) If one root z 1), u 1) exists for equation 21), then assume that d2 h z 1) dx 2, u 1) < ) 0 and d h z 1), u 1) > 0, i.e., C 4 : 12 z 1) ξ4 + 2ξ 8 u 1) + 6ξ6 z 1) u 1) + 2ξ 1 < 0 ξ 19 + ξ 20 > 0 25) 7

8 where ξ 19 = ξ 9 z 1) ) 4 + ξ10 u 1) ) 4 + ξ11z 1) u 1) ) 3 + ξ12 z 1) ) 3 u 1) ξ 20 = ξ 14 z 1) u 1) + ξ15 z 1) + ξ16 u 1) + ξ17z 1) u 1) + ξ 18. 6) Hene the funtion h has a relative maximum at z 1), u 1), i.e., h x, y) ) h z 1), u 1) for all x, y) R 2, espeially for x L a and y L b. In this ) ase we hoose h z 1), u 1) < 0, i.e., C 5 : ξ 21 + ξ 22 < 0 27) where ξ 21 = ξ 4 ξ 22 = ξ 8 z 1) z 1) ) 4 + ξ5 u 1) u 1) + ξ1 ) 4 + ξ6 z 1) z 1) + ξ2 ) 3 1) u + ξ 7 z 1) u 1) + ξ3 z 1) u 1) u 1) ) ) Therefore, inequality 19) holds for all x L a and y L b. ) If equation 21) has more than one root, then one alulates d2 h z i) dx 2, u i), ) ) d h z i), u i), and h z i), u i) and determines the type of eah point by imposing some onditions as above, and aording to the values of h z i), u i), ) one an determine the global ) maximum of the funtion h, and finally make the quantity h, u i) stritely negative. z i) The third ondition of 16) gives ξ 4 x 4 + ξ 5 y 4 + ξ 6 x 3 y + ξ 7 xy 3 + ξ 8 x 2 y < 0. 29) Now onsider the funtion g x, y) = ξ 4 x 4 +ξ 6 y) x 3 +ξ 8 y 2 ) x 2 +ξ 7 y 3 ) x+ ξ 5 y The ritial points of the funtion g are the solutions of the system 4ξ 4 ) x 3 + 3yξ 6 ) x 2 + 2y 2 ξ 8 ) x + ξ 7 y 3 = 0 30) 4ξ 5 ) y 3 + 3xξ 7 ) y 2 + 2x 2 ξ 8 ) y + ξ 6 x 3 = 0. 8

9 Using the same analysis as above, one has that the first equation of 30) has at ) least one real solution x 1) for all values of y, and at most three roots x i) for all values of y, and the seond equation of 30) has at least 1 i 3 ) one real solution y 1) for all values of x, and at most three roots y i) 1 i 3 for all values of x. Thus there are still solutions of equation 30) that are ritial points for the funtion g. On the other hand, one has d 2 g dx 2 x, y) = 12ξ 4 x 2 + 2ξ 8 y 2 + 6ξ 6 xy d g x, y) = ξ 9 x 4 + ξ 10 y 4 + ξ 11 xy 3 + ξ 12 x 3 y + ξ 13 x 2 y 2. ) If one root x 1), y 1) ) 0 and d g x 1), y 1) > 0, i.e., C 6 : ξ 9 x 1) ) 4 + ξ10 31) exists for equation 30), then assume that d2 g x 1) dx 2, y 1) y 1) 12ξ 4 x 1) + 2ξ8 ) 4 + ξ11x 1) y 1) y 1) ) 3 + ξ12 + 6ξ6 x 1) y 1) < 0 x 1) ) 3 y 1) + ξ 13 x 1) ) 32) Hene the funtion g has a relative maximum at x 1), y 1), i.e., g x, y) ) ) g x 1), y 1) for all x, y) R 2, and in this ase we hoose g x 1), y 1) < 0, i.e., C 7 : ξ 4 x 1) ) 4+ ξ5 y 1) ) 4+ξ6 y 1) ) x 1) ) 3+ξ7 y 1) ) 3 x 1) ) +ξ8 y 1) ) < y 1) ) x 1+1 < 0. 33) Therefore, inequality 29) holds for all x L a and y ) L b. If equation ) 30) has more than one root, then one alulates d2 g x i) dx 2, y i), d g x i), y i) ) and g x i), y i), and determines the type of eah point by imposing some ) onditions as above, and aording to the values g x i), y i) one an determine the global ) maximum of the funtion g and finally make the quantity g, y i) stritly negative. x i) Condition 7) gives > 0. 9

10 ξ 1 x 2 + ξ 2 y 2 + ξ 3 xy 2N 0 ξ 4 x 4 + ξ 5 y 4 + ξ 6 x 3 y + ξ 7 xy 3 + ξ 8 x 2 y 2 N 2 ξ 1 x 2 N 2 ξ 2 y 2 N 2 ξ 3 xy N ) The aim of the following investigation is to determine an interval for the quantity N > 1. For this purpose, begin with the first ondition of 34) and onsider the funtion m x, y) = ξ 1 x 2 + ξ 2 y 2 + ξ 3 xy 2N, assuming that Then from 5) and 6) one has Thus we an hoose L a x a 3x 2 a 4y 2 C 8 : < 0. 35) x 1 L a x a 3x 2 a 4y 2 a 5xy a 5xy a 2y a 2y + L a a 0. 36) + L a a 0 x 2 37) where x 1 and x 2 are the roots of the equation m x, y) = 0 with respet to x, i.e., its disriminant is 8Nξ 1 + ) ξ 2 3 4ξ 1 ξ 2 y 2 > 0 for all y R 1 > 0. Then one has q x 1 = ξ 3 y 8Nξ 1 +ξ 2 3 4ξ 1 ξ 2)y 2 2ξ 1 q 38) x 2 = ξ 3 y+ 8Nξ 1 +ξ 2 3 4ξ 1 ξ 2)y 2. 2ξ 1 The inequality x 1 L a holds for all y L b if C 9 : L b 2ξ 1L a ξ 3, 39) and the inequality a 3x 2 + La a 0 x 2 holds for all y L b if w 1 x, y) + w 2 x, y) + ξ ) a 4y 2 a 5xy a 2y where { w1 x, y) = ξ 23 x 4 + ξ 24 y 4 + ξ 25 x 2 y 2 + ξ 26 x 3 y + ξ 27 xy 3 + ξ 28 y 3 w 2 x, y) = ξ 29 x 2 y + ξ 30 xy 2 + ξ 31 x 2 + ξ 32 y 2 + ξ 33 xy + ξ 34 y 10 41)

11 and ξ 23 = 4a2 3 ξ2 1 ξ 24 = 4a2 4 ξ2 1 ξ 25 = 42a 3a 4 +a 2 5)ξ 2 1 ξ 26 = 8a 3a 5 ξ 2 1 ξ 27 = 8a 4a 5 ξ 2 1 ξ 28 = 4ξ 3 2a 2 ξ 1 )a 4 ξ 1 ξ 29 = 4ξ 3 2a 2 ξ 1 )a 3 ξ 1 42) ξ 30 = 4ξ 3 2a 2 ξ 1 )a 5 ξ 1 ξ 31 = 8La a 0)a 3 ξ 2 1 ξ 32 = 4 a 2 ξ 3 2a 0 a 4 ξ 1 +2a 4 ξ 1 L a ξ 2 a2 2 ξ 1)ξ 1 ξ 33 = 8L a a 0 )a 5 ξ 2 1 ξ 34 = 4L a a 0 ) ξ 3 2a 2 ξ 1 )ξ 1 ξ 35 = 4 a 2 0 ξ 1 2Na2 1 2a 0ξ 1 L a +ξ 1 L 2 a)ξ 1. Now onsider the funtion w x, y) = w 1 x, y)+w 2 x, y)+ξ 35. The ritial points of w are the solutions of the system 11

12 4ξ 23 x 3 + 3ξ 26 yx 2 + 2ξ yξ y 2 ξ 25 ) x + ξ 27 y 3 + ξ 30 y 2 + ξ 33 y = 0 4ξ 24 y 3 + 3ξ ξ 27 x) y 2 + 2ξ ξ 30 x + 2ξ 25 x 2 ) y + ξ 26 x 3 + ξ 29 x 2 + ξ 33 x + ξ 34 = 0. )43) With the same analysis as above, there are still solutions, q i) of equation 43) that are ritial points for the funtion w. On the other hand, one has d 2 w dx 2 x, y) = 12ξ 23 x 2 + 2ξ 25 y 2 + 6ξ 26 xy + 2ξ 29 y + 2ξ 31 d w x, y) = d 1 x, y) + d 2 x, y) where d 1 x, y) = ξ 36 x 4 + ξ 37 y 4 + ξ 38 x 2 y 2 + ξ 39 x 3 y + ξ 40 xy 3 + ξ 41 y 3 + ξ 42 x 2 y s i) 44) d 2 x, y) = ξ 43 xy 2 + ξ 44 y 3 + ξ 45 x 2 + ξ 46 y 2 + ξ 47 xy + ξ 48 x + ξ 49 y + ξ 5045) and ξ 36 = 24ξ 23 ξ 25 ξ 37 = 24ξ 24 ξ 25 ξ 38 = 144ξ 23 ξ ξ 26 ξ ξ 2 25 ξ 39 = 72ξ 23 ξ ξ 25 ξ 26 ξ 40 = 12ξ 25 ξ ξ 24 ξ 26 46) ξ 41 = 24ξ 24 ξ ξ 25 ξ 28 ξ 42 = 12ξ 30 ξ ξ 25 ξ ξ 23 ξ 28 ξ 43 = 4ξ 30 ξ ξ 26 ξ ξ 27 ξ 29 12

13 and ξ 44 = 24ξ 24 ξ ξ 25 ξ 28 ξ 45 = 24ξ 23 ξ ξ 31 ξ 25 16ξ 2 25 ξ 46 = 24ξ 31 ξ ξ 32 ξ ξ 28 ξ 29 36ξ 2 27 ξ 47 = 12ξ 31 ξ ξ 32 ξ ξ 30 ξ 29 48ξ 25 ξ 27 47) ξ 48 = 4ξ 30 ξ 31 16ξ 30 ξ 25 If one root s 1) 0 and d w s 1), q 1) 12ξ 23 C 10 : ξ 49 = 12ξ 31 ξ 28 24ξ 30 ξ ξ 32 ξ 29 ξ 50 = 4ξ 31 ξ 32 4ξ x 3 ξ 30 ξ 23. ), q 1) exists for equation 43), then assume that d2 w s 1) dx 2, q 1) ) > 0, i.e., s 1) + 2ξ25 q 1) + 6ξ26s 1) q 1) + 2ξ 29 q 1) + 2ξ 31 < 0 ) ) d 1 s 1), q 1) + d 2 s 1), q 1) > 0. ) Hene the funtion w has a relative maximum at s 1), q 1) ) w s 1), q 1) for all x, y) R 2, and in this ase we hoose w i.e., 48), i.e., w x, y) ) s 1), q 1) < 0, ) ) w 1 s 1), q 1) + w2 s 1), q 1) + ξ35 < 0 49) or ) )) w 1 s 1), q 1) + w 2 s 1), q 1) + 4a L 2 a 8a 0 L a ) ξ 2 1 C 11 : N > = N 8ξ ) beause only the oeffiient ξ 35 depends on N and beause we want N > 1, then we must assume N 1 > 1, i.e., C 12 : 4a L 2 a 8a 0 L a ) ξ w 1 + w 2 8ξ 1 ) > 0. 51) 13 ) <

14 Therefore, the first inequality of 34) holds for all x L a and y L b. For the seond ondition of 34), onsider the funtion v x, y) = v 1 x, y)+ v 2 x, y), where v 1 x, y) = ξ 4 x 4 + ξ 5 y 4 + ξ 6 x 3 y + ξ 7 xy 3 + ξ 8 x 2 y v 2 x, y) = N 2 N 2 ξ 1 x 2 + ξ 2 y 2 + ξ 3 xy)). 52) The ritial points of the funtion v are the solutions of the system 4ξ 4 x 3 + 3ξ 6 y) x 2 + 2ξ 8 y 2 2N 2 ξ 1 ) x + ξ 7 y 3 N 2 ξ 3 y = 0 4ξ 5 y 3 + 3ξ 7 x) y 2 + 2ξ 8 x 2 2N 2 ξ 2 ) y + ξ 6 x 3 N 2 ξ 3 x = 0. With the same analysis as above, there are still solutions k i) 53) ), l i) of equation 53) that are ritial points for the funtion v. On the other hand, one has where and and d 2 v dx 2 x, y) = 12ξ 4 x 2 + 2ξ 8 y 2 + 6ξ 6 xy 2N 2 ξ 1 d v x, y) = p 1 x, y) + p 2 x, y) p 1 x, y) = ξ 51 x 4 + ξ 52 y 4 + ξ 53 x 3 y + ξ 54 xy 3 + ξ 55 x 2 y 2 54) p 2 x, y) = N 2 h 1 x, y) + h 2 x, y) + 4ξ 1 ξ 2 N 4 55) h 1 x, y) = 12 ξ 2 ξ 6 ξ 1 ξ 7 ) xy 4 6ξ 2 ξ 4 + ξ 1 ξ 8 ) x 2 4 6ξ 1 ξ 5 + ξ 2 ξ 8 ) y 2 h 2 x, y) = 16ξ 2 8x ξ 2 7y ξ 7 ξ 8 xy ) 56) 14

15 ξ 51 = 24ξ 4 ξ 8 ξ 52 = 24ξ 5 ξ 8 ξ 53 = 72ξ 4 ξ ξ 6 ξ 8 ξ 54 = 12ξ 7 ξ ξ 5 ξ 6 ξ 55 = 144ξ 4 ξ ξ 6 ξ 7 + 4ξ 2 8 ξ 56 = 16ξ 2 8 4N 2 6ξ 2 ξ 4 + ξ 1 ξ 8 ) ξ 57 = 36ξ 2 7 4N 2 6ξ 1 ξ 5 + ξ 2 ξ 8 ) 57) ξ 58 = 12N 2 ξ 2 ξ 6 ξ 1 ξ 7 ) 48ξ 7 ξ 8 ξ 59 = 4N 4 ξ 1 ξ 2. ) ) If one root k 1), l 1) exists for equation 53), then assume that d2 v k 1) dx 2, l 1) > ) 0 and d v k 1), l 1) > 0, i.e., C 13 : N < 12ξ 4 k 1) 2+2ξ8 l 1) 2+6ξ6 k 1) l 1) 2ξ 1 = N 2 ) ) )] 4ξ 1 ξ 2 N 4 + h 1 k 1), l 1) N 2 + [p 1 k 1), l 1) + h 2 k 1), l 1) > 0. 58) The first ondition of 58) is possible if 12ξ 4 k 1) + 2ξ8 l 1) + 6ξ6 k 1) l 1) C 14 : > 0, 59) 2ξ 1 and the seond ondition of 58) is possible for all N R if ) ) )) C 15 : h 2 1 k 1), l 1) 44ξ1 ξ 2 p1 k 1), l 1) + h2 k 1), l 1) < 0 60) beause ξ 1 ξ 2 > 0, and from 50) one has that N i, i = 1, 2 must satisfy the inequalities 1 < N 1 < N 2, i.e., 15

16 C 16 : 4a L 2 a 8a 0 L a ) ξ w 1 + w 2 8ξ 1 ) > 0 w 1 s 1),q +w 1) 2 s 1),q 1) +4a2 0 +4L2 a 8a 0 L a)ξ 2 1 8ξ 1 < 12ξ 4 k 1) 2+2ξ8 l 1) 2+6ξ6 k 1) l 1) 2ξ 1. 61) Therefore, there is an N suh that 1 < N 1 < N < N 2 in whih inequality 34) holds for all x L a and y L b. Finally, the general map 1) has hyperhaoti attrators if all the above inequalities hold. Hene we have proved the following theorem: Theorem 1 If i=16 i=1 C i, then the general quadrati map of the plane given by equation 1) has hyperhaoti attrators x, y) with the ondition x L a and y L b, where L a and L b are given by 6). We onlude with the following remarks: 1) The above inequalities do not guarantee the boundedness of the attrators. 2) Not all hyperhaoti attrators satisfy the above inequalities. 3) Finding a speifi example is not simple beause at eah step the of third-degree equations and very ompliated inequalities with 12 unknown variables are required. 4) It may be possible to onvert the proof to a numerial algorithm. 5) Some of the above hyperhaoti attrators an be infinitely or very large. 2 Conlusion We have given a rigorous proof of hyperhaos in the general quadrati map of the plane. The proof shows how to loate speifi type of orbits in some ases. Referenes [1] M. Hénon, Numerial study of quadrati area preserving mappings, Q. Appl. Math ,

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