Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.3, No.7, 2013
The Dynamics of the Fixed Points to Duffing Map
Iftichar M.T.AL-Shara'a 1 May Alaa Abdul-khaleq AL-Yaseen 2* Mathematics Department ,University of Babylon ,Hilla , Babylon, Iraq * E-mail of the corresponding author : [email protected]
Abstract We study the fixed points of Duffing map and the general properties of them , we find the contracting and expanding area of this map .So, we divide the parameter space for many regions to determinate the fixed points attracting , repelling or saddle . Also we determinate the bifurcation points of the parameter space . Keywords: Duffing map , fixed point , bifurcation point , attracting- expanding area.
1. Introduction Chaos Theory is a synonym for dynamical systems theory, a branch of mathematics. Dynamical systems come in three flavors: flows (continuous dynamical systems), cascades (discrete, reversible, dynamical systems), and semi-cascades (discrete, irreversible, dynamical systems).The concept of a dynamical system has its origins in Newtonian mechanics. There, as in other natural sciences and engineering disciplines, the evolution rule of dynamical systems is given implicitly by a relation that gives the state of the system only a short time into the future.In a dynamical system, a bifurcation is a period doubling, quadrupling, etc., that accompanies the onset of chaos. It represents the sudden appearance of a qualitatively different solution for a nonlinear system as some parameter is varied. Most commonly applied to the mathematical study of dynamical systems, a bifurcation occurs when a small smooth change made to the parameter values (the bifurcation parameters) of a system causes a sudden 'qualitative' or topological change in its behaviour. Bifurcations occur in both continuous systems (described by ordinary differential equations , delay differential equations, or partial differential equations), and discrete systems (described by maps). The name "bifurcation" was first introduced by Henri Poincaré in 1885 in the first paper in mathematics showing such a behavior. Henri Poincaré also later named various types of stationary points and classified them. The Duffing map is a 2 – D discrete dynamical system , its form is . It has fractal y F , = attractor studied by Paul Bourke 1998 .Some literatures studied its chaotic properties−bx + , ayMishra − y and Mankar [3] constructed encryption method using Duffing map, this method is desined using message embedded scheme and is analyzed for its validity, for plaintext sensitivity , key sensitivity, known plaintext and brute-force attacks. Tarai and Khan[2] applied backstepping method for projective synchronization of Duffing map ,in this case the error goes to zero through oscillation , the numerical simulation results indicate that their approach works very well .In this research we show that the Duffing map is diffeomorphism map and it has three fixed points , in our research we divide the parameter space into six regions to determinate types of the fixed points, and to find the bifurcation parameter points .
2. Preliminaries ∞ Let be a map .We say F is C , if its mixed k-th partial derivatives exist and are continuous for all ∞ ∞ F ∶ ℝ, and→ it ℝ is called a diffeomorphism if it is one-to-one, onto, C and its inverse is C . Let V be a subset of , and be any element in .Consider be a map. Furthermore assume that the first partials k ∈ ℤ of theℝ coordinate maps f and g ofℝ F exist at .The ∶ differential → ℝ of F at is the linear map defined v v DF v on by , for all in .The determinant of is called the Jacobian v v ℝ DF v = v ℝ DF v of F at and it is denoted v by v , so F is said to be area-contracting at if , F is v said to be area-expandingJF v at = if det . Let be an matrix .The real|det number |
41 Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.3, No.7, 2013
′ parameterized family has a bifurcation of parameter if the number or nature (attracting vs. repelling ) ′ , ′ ′ of periodic point of the family changes as the parameter a or b passes through . In this case, is said to ′ ′ be a bifurcation point .We denote to the inverse of A. 3.The General Properties of Duffing Map In this section , we find the fixed point and we study the general properties of Duffing map , for example : is ∞ , one to one , onto , C and invertible so it is diffeomorphism, and we find the parameter which hasF area contracting or expanding . , Proposition 3-1 : Let be a Duffing map , if then has three fixed points : and 0 − − 1 , :ℝ → ℝ a > + 1 , , √ 0 . √ − − 1 − − − 1 √ Proof−√ : − By −definition 1 of fixed point , then and , = = = = – + − , so , therefore − + − then is the first fixed point , and 0 = − + − − +,then − − 1 = 0 and= therefore 0 – + − − 1 = 0 that…….( is 3.1) = and − − 1 = ±√ −……( − 3.2) 1 are the = second ±√ − and − third 1 fixed points . − − 1 − − − 1 = √ = √ ¶ √ − − 1 −√ − − 1 Proposition 3-2 : Let be the Duffing map then the Jacobian of is b . , Proof : : ℝ → ℝ = , so . ¶ 0 1 0 1 , , = = Proposition 3-3 : − − 3 − − 3
The eigenvalues of are λ ; . x ± x DF , , = ∀ ∈ ℝ Proof : If is eigen value ofy then must be satisfied the characteristicy equation −1 , = − + 3 then and the solutions of this equation are where λ . ± 0¶ − + 3 + = 0 , , = Remark : 1. It is clear that if , then the eigenvalues of are real . x , |3y − a| > 2√b b > 0 DF y 2. The eigenvalues of at fixed points are for fixed point and ± , , = P = for fixed point and λ for fixed points . ± ± Proposition 3-4 : P , = P Let be a Duffing map , If then is diffeomorphism . ProofF , : To:ℝ prove→ ℝ is diffeomorphism ,b we ≠ 0must proveF , : 1) is one-to-oneF , map : Let such that then x , , ∈ ℝ F , = F , y so and = hence = and so , that is , is − + − − + − one-to-one.−bx + ay − y = −bx + ay − y −bx = −bx b ≠ 0 x = x F , 2) is ∞ : then all first partial derivatives exist and continuous. Note that , ℂ = − + − , , and , , , , , and , , .We get that all its = ixed 0 ∀ k-th ∈ ℕpartial derivatives = exist 0 and continuou= 0 s∀ for ≥ all 2 k. From definition = 0 of∀ diffeomorfism ≥ 4 , is F∞ , . F , ℂ
42 Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.3, No.7, 2013
3) is onto : Let any element in such that and . Then there exist , ℝ = = ∈ .Then is onto . ℝ ∋ , = F , 4) has an inverse : Let such that . Note , , = , ∘ , = , ∘ , = that , , and , ∘ , = , = = , ∘ , = − + − . Then has an inverse and it is invertible . ¶ , , = − + − = F It is clear that if or then is a hyperbolic matrix , the question what about , , these proposition answer= ≠ this 1 question−1 . = Remark1 3-5 : If then , so then is not one to one, that y = 0 , = , = : ∈ ℝ , is , is not diffeomorphism . ay − y Proposition , 3-6 : 1. is non hyperbolic matrix if and . Also , it is non hyperbolic matrix if , and . y = ± b = 1 y = ± 2. If b = −1 and then one of eigenvalues for every fixed point of is . b = −1 = 2 + 2 F , −1 Proof : 1. let then and hence = ± 3y = a − 2 4 + 4 3y − a + 3y − a = 3y − a − 4 so λ . By follow the same steps we get is non hyperbolic matrix if ± x , , = 1 = 1 DF y and . y = ± b = −1 2. since and then λ hence λ , ± = a = 2b + 2 2 = − b + 1 ± b − 1 = −1 also λ , and we have λ then ± λ = −1 ,since= so λ .¶ = b + 1 − √b + b + 1 b = −1 = −1 Proposition 3-7 : For all , and , and , is a hyperbolic matrix if and , only if ∈ ℝ . ≠ ± ≠ ± |3 − | > 2√ | | = 1 Proof : since and , then or . , , ⇒ ∈ 2, ℤ = | | = 1 −1 Let . Then since we have λ λ , that is λ λ . x ⇐ b = 1 det D F , = b = b = 1 Then λ . If λ λ y, or λ λ . λ = > 1 therefore < 1 < 1 so > 1 So is a hyperbolic matrix . ¶ x , PropositionD F y 3-8 : If then is area-contracting and its area-expanding if . | | < 1 , | | > 1 Proof : If then , by proposition 3-2 , that is is area-contracting map , , | | < 1 < 1 and if then this implies that is an area-expanding map. ¶ | | > 1 , > 1 , 4.The Fixed Point’s Properties of Duffing Map we divide the parameter space into six regions to determinate types of the fixed points such that where and two of these fixed point are attracting and the other one is repelling ,so all of|b | fixed< 1 b + 1 < < 2 + 2
43 Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.3, No.7, 2013 points are repelling if and or , and all of them are saddle where . Finally , if the fixed|b| > points 1 area < repelling 2 + 2 werea > 2 + 2 and two are repelling and one|b is| = saddle 1 were a > 2 + 2 . 0 ≤ < 1 Proposition−1 < < 0 4-1 : The Duffing map has only one fixed point if . This fixed point is 0 a ≤ b + 1 ∀b ∈ ℝ attracting if and saddle if . 0 Proof : by (3.1)|b| ≤and 1 (3.2 ) if |b| > 1then , that mean the Duffing map has one fixed point . 0 ≤ + 1 , ∉ ℝ So , if the equation , then Duffing map has one fixed point . To show type of0 this 0 = + 1 √ − − 1 = 0 point , since we have and then so and0 . Then is ± ± 0 , = a ≤ b + 1 , < < < 1 attracting if and saddle if ¶ 0 From Proposition|b| ≤ 1 4-1 we conclude|b| > that 1 if then is bifurcation point , since the map has attracting fixed point and saddle if ≤ . + 1 = 1 Proposition 4-2 : | | ≤ 1 | | > 1 If then the region get has two attracting fixed points 3 and− − , 4 and ≥ has 0 one repelling fixed| |point<1 , +1< <2 +2 . , Proof : since so ,therefore , then we get and b + 1 < – a + b + 1 < then 0 – a + b + 1 < + 1 < , so – a + b < < − 1 –2a+2b<2 <2 −2 −2a−b<− <2 −2−3 − 2 − 3 − 3 + 1 − 4 − 4 < Therefore− < 2 − 3 − 3 + 1 and , by adding for both sides we get − 2 − 3 − 3 + 1 < − < 2 − 3 − 3 + 1 −4 2 − 3 − 3 + 4 < −4 < , we have 4 2 − 3 − 3 + 4 2 − 3 − 3 2 − 3 − 3 − 2 < 2 − 3 − 3 − 4 < 2 − 3 − 3 + 2 then −2 < 2 − 3 − 3 − 2 < ± 2 − 3 − 3 − 4 < 2 − 3 − 3 + 2 < 2 , from proposition 2 , then and are attracting fixed ± ± points−1 < . To show that is a repelling< 1 fixed point , since| | < 1 we have so therefore that is + 1 < , by adding− < for − both + sides 1 we– get + 2 + 1 < + 1 − < + 1 < 4 +4<−4 <4−4 by adding for both sides we have − − 2 < − 4 < 2 − ⇒ − − 2 < ±√ − 4 < 2 − . That is , so . ¶ ± −2From < Propositions ± √ − 4 4-1< and 2 4-2 , we−1 conclude < that
Proof : let , since then that is ± ± ± ± so = . Then and are repelling > fixed + points. 1 2 > + 1 ± − 1 > ± ± > 1 To show that is a repelling fixed point , we have and then ± , by follows the above proof then we get = .¶ > + 1 2 > + 1 ± From + Propositions 1 − 4 4-2 and 4-3 , we conclude that if > 1 then is bifurcation point , since the map has two attracting fixed points and one repelling +1< <2 +2 if and all of them = are 1 repelling if . Proposition 4-4 : | | < 1 | | > 1 If and then the fixed points of are repelling . | | >1 , >2 +2 3 − − 4 ≥ 0 , Proof : since and then we get ± ± = so > + , since 1 then hence is repelling ± ± ± ± 2 > − − 1 ± − + 1 − 4 > | | > 1 > 1 fixed point . by follows the above proof then we get is repelling fixed point . Now , and ± then , hence since= , then > 2 + 2 2 > 2 so + 1 ± 4 then + we +have 1 . also> + 1 + − 1 + 3 | | > then 1 > + 1 ± that − 1 is > 1 so , |then | >we 1 get is >a repelling + 1 − fixed + point 2 −. ¶ 5 + 1 > + 1 − + 2 > −1 | | > 1
44 Mathematical Theory and Modeling www.iiste.org ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online) Vol.3, No.7, 2013
Proposition 4-5 : If and then the fixed points of are saddle . Proof = : −1 > 0 , By hypothesis and so . It is clear , and − ± = √ = √ = ± + 1 > 0 + 1 > 0 hence are distinct real number . √ ± √ −√ Let so , since then so ± | | ≥ that 1 is ± √ , this+ 1contradiction ≥ 1 with√ + 1 > ,√ hence ∀ > 0 and√ we+ have 1 − ≥ 1 then + 1 ≥ 1 + 2 so + 2 but ≤ 0 then that mean > 0 is saddle| | < fixed 1 point , so as =. To show| that | = 1 | | = | | < 1 | | > 1 is a saddle fixed point , since then and so that is . Also we have > then 0 + 4, hence > 4 − 4 + is a saddle√ fixed + point 4 > 2 . ¶ − > 1 Proposition > 1 4-6 : = | | < 1 If and then the fixed points of are saddle . = 1 > 2 , Proof : By hypothesis and so and √ − 2 −√ − 2 ± = = = − − 3 ± √ − 6 + 8 = √ − 2 −√ − 2 . since we get and so that is ± > 2 then2 > 4 . Since2 − > − 6 so+ 8 2 − − then√ − 6 + that 8 > mean 0 is− saddle − 3 fixed− √ point− 6 , so + as 8 > 1 . Now by >follows 1 the same above= 1 steps| we| =get and| | < 1 then is a saddle fixed point . ¶ > 1 | | < 1 Proposition 4-7 : If and then : 1. The fixed points | of| < 1 are repelling > 2 + if 2 . 2. Two fixed points , , of are saddle0 ≤ and < 1 is repelling if , −1 < < 0 Proof : since and then ± = ,so > hence 2 + 2 for all b , and 2 > − + 1 ,since± + 1 − 4 > then| | > 1 . Now if 2 > − +we 1 get− + 1 and− 4 is repelling– + fixed 1 < point − , so + 1 as − 4 . If > − we + 1 get 0 ≤ < 1 so | | > 1 −1 < < 0 0<− −1− +1<2 then is saddle fixed point , so as .To show is repelling fixed point0 < if for all b , since< 1 we have| | < 1 then so that is 2 > ± √ − 4 hence > + 1, by− √ hypothesis + + 1 then> + 1 − . ¶ + 2 − 5 + 1 > + 1 − + 2 | | > 1 > | | > 1From Proposition 4-7 , we conclude that if then is bifurcation point , since the map has two saddle fixed points and one repelling if >and 2 all + of 2 them are = repelling 0 if . < 0 > 0 References [1] Gulick D., "Encounters with Chaos", McGraw Hill, Inc, USA, 1992 . [2] Tarai A. and Khan M. A. , “Projective Synchronization of Chaotic Systems Via Backstepping Design “, International Journal of Applied Mathematical Research , 4(2012),531-540. [3] Misra M. and Mankar V. H. ,“Message Embedded Cipher Using 2-D Chaotic Map “,International Journal of Chaos ,Control, Modelling and simulation (IJCCMS), 1(2012),13-24 .
45