Chaos and Time-Series Analysis. Julien Clinton Sprott 13 April 2003

Transcription

1 Chaos and Time-Series Analysis Julien Clinton Sprott 3 April 003 v

2 Preface To be written

3

4 . Change the variable X n to X n = Y n+ A + = A Yn A + Yn Y n A sides by A to get : Y n+ = B Y n with B = A 4 A. A +. Substitution in.4 yields: = A 4 Y n A. Multiply both Fixed point: X = / The first ten iterations are: x = 0.63, x = 0.65, x 3 = 0.635, x 4 = 0.649, x 5 = 0.638, x 6 = x 7 = 0.640, x 8 = 0.645, x 9 = 0.64, x 0 = which indicate an oscillation about X.3 Let fx = AX X and consider X 0 = X + ε 0, with ε 0 very small and positive. Then X = fx 0 = AX AX AX ε 0 + Aε 0 Aε 0. X satisfies X = AX AX which gives X = X AX ε 0 +Aε 0 Aε 0. The Aε 0 term is smaller than the ε 0 terms and may be neglected. Hence X = X + ε 0 A X = X + ε 0 f X. X may be written as X = X + ε thus giving ε = ε 0 f X. After n iterations: ε n = ε 0 f X n Hence the fixed point is unstable when f X >. For f X > all ε s are positive which means X + as n +. This is not the case for negative slopes since the ε s can both be positive and negative depending whether n is even or odd..4 See text on page 7..5 The second iterate map is given by gx = ffx with fx = AX X.Let p and q be the fixed points of the map these two points are given by eqn.7.

5 ix g X X=p = f fxf X X=p = f fpf p = f qf p = A p q = A [ p + q + 4pq] From eqn.6 observe that p + q = [AA + ]/A and pq = A + /A Hence g X X=p = A [ AA + /A + 4 A + /A ] = 4 + A A Eqn.8 is found by setting g X =.In general the -cycle is stable when g X < ie when 3 < A < + 6.6?.7 The map fx = 4X X has pre-images X A,B = ± X. Then f X A,B = 4 X and P X A,B = π. Use these two equations X in eqn. to obtain P X as in eqn..8?.9 The pre-images of X n+ = fx n = A minx n, X n are X A = X/A and X B = X/A. Use these together with the fact that f X A,B = in eqn. to obtain P X = /A..0 Let fx = A X α. Then f X = αasignx X α. Obviously f / is discontinuous for α not defined for α < and having a jump discontinuity for α =. The second derivative is given by: f X = α α AsignX X α = α α A X α. f / is not defined for α <, but for α, f is continuous everywhere and therefore f is smooth Note: signx = for x 0 and for x < 0.. From Exercise.0 the derivative at 0 is given by f 0 = αa α. Since α < / and 0 < A < we may conclude that f 0 < and hence that the point X=0 is stable. The existence of a stable point eliminates the possibility of presence of chaos in the map.

6 3 3. The solution to eqn 3. is of the form x = x 0 e at, where x 0 is the initial condition. A nearby trajectory starting from x 0 + ε with ε small is given by x = x 0 + ε e at. Their separation x = x x = εe at increases exponentially at a rate a. 3. The solution x = has derivative + e x at 0 dx dt = [ + a e at x 0 ]. e x at 0 Use x and its derivative in the expression dx ax x. Verifying that dt this expression equals 0 proves that eqn 3.3 satisfies eqn This equation is a separable st order ODE. It can be written in the form csc x dx = dt, but the solution can be given only in implicit form. Integrating both sides and applying the initial condition x 0 = x 0 we obtain : ln [csc x cot x / csc x 0 cot x 0 ] = t csc x cot x = csc x 0 cot x 0 e t 3.4 dx/dt = ay and dy/dt = bx. Differentiate the first equation with respect to time and substitute dy/dt to get d x/dt + abx = 0. This can abt have a solution of the form x = x 0 sin. Substitution in the second equation yields dy/dt = bx 0 sin. Integrating once we abt get b abt y = x 0 a. cos The equations for x and y generate points that lie on an ellipse with x max y max = a b.

7 3.5 v t = dx dt = v 0 cos ωt+ AΩ ω Ω cos Ωt and d x dt = ωv 0 sin ωt AΩ ω sin Ωt. Ω Using the expressions for x and d x/dt in the left hand side of ean 3.0 we finally obtain A sin Ωt which equals the right hand side. 3.6 d x dt b dx ω = v 4 e ω t b +b 4 e bt 0 b 3 and dx dt t b +b e bt = v 0 ω e ω t b +b e bt b xi. Let = d x dt + dt + ω x = v 0 ω ω e ω b.then comparing with each term 3 from eqn 3.8 we get / d x dt = ω e ω t b b e bt ω e ω t b b e bt =, b e ω b t b e bt ω e ω b t b e bt / b dx dt = ω b 4 and / ω x = ω4 e ω t b +b ω e bt ω4 e ω ω 4 e ω t b +b 4 e bt t b +b 4 e bt = ω 4 e ω b t +b 4 e bt. Since each term is much less than we may neglect, thus verifying that the eqn 3. satisfies eqn x = v 0 te bt/, vt = dx/dt = v 0 e bt v 0tbe bt and d x/dt = v 0 be bt + 4 v 0tb e bt. Substitute these in the right hand side of eqn 3.8 to verify that it equals zero. 3.8 x = v 0 /ω e bt sin ωt, vt = dx/dt = v 0e bt b sin ωt cos ωtω ω and d x/dt = 4 v 0e bt b sin ωt 4bcos ωtω 4sin ωtω ω. Substitution in eqn 3.8 yields = d x/dt + bdx/dt + ω x = 4 v 0e bt sin ωt b ω. Compare with each term from eqn 3.8 to get : / ω x 4 = v 0e bt sin ωt b ω v 0 ωe bt sin ωt = b 4 ω, b dx = 4 v 0e bt b sin ωt ω 4 dt v 0be bt b sin ωt cos ωtω < b sin ωt ω ω b and / d x dt = 4 v 0e bt b sin ωt ω 4 v 0e bt b sin ωt 4bcos ωtω 4sin ωtω ω b 4 ω = 4 b sin ωt ω b 4 ω 4ω sin ωt = < b sin ωt ω. Since is much smaller than each of the terms in eqn 3.8 we may consider the eqn 3.3 roughly satisfying eqn x = A sinωt φ AΩ cosωt φ, vt = dx/dt = ω Ω +b Ω ω and Ω +b Ω d x/dt = AΩ sinωt φ ω. Ω +b Ω Substitution in the right hand side of eqn 3.7 yields: d x dt + bdx dt + ω x = A ω Ω sinωt φ+bω cosωt φ. ω Ω +b Ω Use eqn 3.5 and the trigonometric identities cos φ = / + tan φ =

8 xii ω Ω ω and sin φ = cos bω φ =. Eqn can Ω +b Ω ω Ω +b Ω then be written as: d x/dt +bdx/dt+ω x = A cos φ sin Ωt φ + sin φ cos Ωt φ = A sin Ωt. 3.0 A x m = and the derivative dx m ω Ω +b Ω dω = AΩ ω +Ω +b 3. ω 4 ω Ω +Ω 4 +b Ω It becomes zero for Ω = 4ω b and Ω = 4ω b. x m becomes maximum for Ω = 4ω b and the maximum is A b 4ω b. 3. Find the Ω s for which x m A = = ω Ω +b Ω bω b ω = ω Ω + b Ω. Then Ω 4 + b ω Ω + ω 4 b ω = 0 Ω/ω 4 + b/ω Ω/ω + b/ω = 0. The Ω s satisfy Ω ω + Ω ω = Q and Ω Ω ω ω = Q Ω Ω ω ω Q for large Q. Hence Ω ω = Ω ω + Ω ω Ω Ω ω ω Q Q = Q Ω ω = Q 3. Start with x + v = x + dx dt 4 and take the derivative at each side to obtain x dx dt + dx d x dt dt 0 x + d x dt 0 which corresponds to eqn 3.6 in the limit as b Use v= dx dv dt to write eqn3.6 as x + dt + bv v + x = 0 vx + v dv dt + bv v + x = 0 d dt x + v + bv v + x = 0 which is apparently satisfied by v + x =, the limit cycle. 3.4? 3.5 SAME AS Prob f = x y x 3 and g = x x y. At 0, 0 we have dx/dt = dy/dt = 0 and the Jacobian evaluated there is:

9 J0, 0 = 0 0 The eigenvalues can then be found to be λ = 0 and from which we may conclude that the origin is at an unstable equilibrium. For solutions approximately on the unit circle ie satisfying x + y we have dx/dt x x y = xy y. Multiply dx/dt by x and dy/dt by y to get: x dx/dt x y xy and y dy/dt = yx x y. Adding those together we get x dx/dt + y dy/dt = d dt x + y 0. Therefore the points near the unit circle tend to remain close to it since the rate of change of their distance from the origin is close to zero. 3.7 xiii Write eqn 3.8 as a system of st order ODEs ie dy dt = by ω x = y x = gx, y and dx dt = y = fx, y. The equations corresponding to 3.33 and 3.34 with h = are x n+ = x n + y n and y n+ = x n. These two can be written in a single equation x n+ = x n x n. Starting with x 0 = y = β and x = α iterate to get x = α β, x 3 = β, x 4 = α, x 5 = α + β, x 6 = β = x t Total error: eh, t, ɛ = th + ɛ h e h = t ɛ t h. For minimum set 3 e h = 0 to get h = 3 ɛ 4t. For this value of h, e h > 0 which guarantees that h = 3 ɛ 4t gives the minimum. 3.9 The equations become x n+ = x n +h yn++y n and y n+ = y n +h which give x = x 0 +h y +y 0 and y = y 0 h x +x 0. Solving this system we obtain y = 4h 4+h and x = 4 h 4+h. These give y +x = 4 h +6h 4+h =, so the error is zero. 3.0 dy dt = by ω x = gx, y and dx dt = y = fx, y. Then : x n+ = x n +hf xn+ +x n, y n++y n and y n+ = y n +hg xn+ +x n which give x n+ = x n +h y n++y n and y n+ = y n +h 3. xn++xn, y n++y n b y n++y n ω x n+ +x n If we use the leapfrog method to the x variable once the same results are obtained as in the text on page 46, namely x = and y = h which as before gives an error of the order of h since R = h /. However when we.

10 xiv use the leapfrog method to the y variable we get x = h and y = h which give x + y = h + h 4 h. We also get an error of the order of h, but in this case R = h /. 3. Let P n = X n, Y n and P 0 =, 0. This problem is solved in two ways:. Using the leapfrog method in the x variable: The equations become: X n+ = X n + Y n and Y n+ = Y n X n+. Then P =,, P = 0,, P 3 =, 0, P 4 =,, P 5 = 0,, P 6 = P 0 =, 0.. Using the leapfrog method in the y variable: The equations are: Y n+ = Y n X n and X n+ = X n + Y n+ and we get P =, 0, P =,, P 3 = 0,, P 4 =, 0, P 5 =,, P 6 = P 0 =, 0. In either case we still get a solution of period 6 which is comparable to the period π 6.8 of the exact solution. 3.3? 3.4?

11 Expand eqn 4.5 to get λ a + d λ+ad bc = 0 and then use the quadratic formula to obtain the two eigenvalues. 4.3 DONE IN TEXTBOOK ON PAGE The Jacobian is given by: J = is λ + = 0. Hence λ = ±i whose characteristic polynomial The eigenvalues given by eqn 4. can be written as λ, = b ± ω b The term under the square root can never exceed, so the two eigenvalues can either be complex conjugates or real with the same sign. Thus a saddle point cannot occur since this requires λ and λ to be real with opposite signs. For b > ω we get stable nodes λ < λ < 0 if b is positive and.

12 xvi unstable nodes if b is negative λ > λ > 0. When b = ω we either get a degenerate stable node for b > 0 or a degenerate unstable node for b < 0. When b < ω the term in the square root becomes negative thus giving complex solutions. Under this condition we get an unstable focus for b < 0 and a stable focus for b > If the arrows are in the direction towards the spiral and radial points, each of these points are stable, otherwise they are unstable. Doing the same on the saddle point has no effect on its overall stability. 4.7 In order to find the eigenvectors v one must solve the equation J λi v = λ 0 0 ie ω v = The solution is v b λ 0 i =, i =,. λi For b =.5ω > 0, λ = 4b/5 and λ = b/5 we obtain the figure shown below which is similar to the middle plot of Fig The Jacobian J = gives the characteristic equation λ 4 λ+ 4 = 0 with eigenvalues λ = 3 and λ = and eigenvectors v = and v =. 4.9 Add the two solutions given by 4.6 to obtain λ + λ = a + d =tracej and multiply them to get λ λ = ad bc =detj.

13 4.0 xvii The Jacobian of the damped harmonic oscillator has tracej = b and detj = ω. For the critically damped case b = ω. Hence tracej = ± det J. 4. Denote by τ the trace of J and by δ its determinant. The eigenvalues given by eqn 4.6 may be written as λ, = τ± τ 4δ. The ultimate fate for each romantic style can be determined by the values of τ and δ. Therefore we can have the following cases: CASE : τ > 0 and τ > 4δ > 0. In this case both λ and λ are positive which means that the origin is an unstable node and therefore the couple may end up extremely loving or hating each other or each of them having opposite feelings for each other. What decides their final state is a matter of the initial conditions. CASE : τ > 0 and τ < 4δ. In such a case λ and λ are complex conjugates with positive real part which means that the origin is an unstable spiral. No matter what the initial conditions are the couple will never be indifferent for each other. They will go through cycles of growing love and growing hate forever ie oscillatory feelings of growing intensity, never reaching some final feeling for his/her mate. CASE 3. τ = 0 and δ > 0. The origin becomes a center. In this case the couple will pass through never- ending periods of love and hate, but these feelings are oscillatory with fixed maximum intensity and do not lead to such an explosion of feelings as in the cases and. Moreover the couple will never be indifferent for each other like in cases and. CASE 4: δ < 0 The origin is a saddle.depending on the values of the parameters a, b, c and d we can have two cases: 4a. The couple will either end up extremely loving each other or extremely hating each other. 4b. The couple will have growing opposite feelings for each other eg Romeo extremely loving her and Juliet extremely hating him or vice versa. b Since the eigenvectors may be written as v i = a λ i with i =,, we b end up in case 4a if for the positive eigenvalue we have a λ < 0 and in case 4b if this expression is positive for the positive eigenvalue. The outcomes of 4a and 4b are dependent upon the initial feelings for each other. In both cases 4a and 4b there is a possibility of that Romeo and Juliet become completely indifferent for each other. This may happen if and only if initially the couple has feelings that lie on the eigendirection whose eigenvalue is negative. CASE 5: τ < 0. In this case the couple will end up being indifferent for one another. The way this happens depends on whether we have τ > 4δ > 0, where we get a stable node or τ < 4δ, where we get a stable spiral.

14 xviii 4. IMPOSSIBLE TO MAKE SUCH PLOTS. The trace and determinant are not sufficient to give the nature of the eigenvalues. For the case this was possible since knowing the sum and product of the eigenvalues suffice to fully determine the eigenvalues. 4.3 The equation λ 3 +.4λ + = 0 is solved using a computer and we obtain λ =.33 and λ,3 = ± 0.649i which verifies that the origin is a spiral saddle of index. 4.4 The equilibria can be found by solving dx/dt = dy/dt = dz/dt = 0 using eqns It can be shown that the solutions are p = 0, 0, 0 and p,3 = ± br, ± br, r. Obviously the first bifurcation occurs when r = at which the equilibria p and p 3 are created. The Jacobian is Jx, y, z = r z x which gives the characteristic equations σ σ 0 y x b b + λλ + σ + λ + σr = 0 for p and λ 3 + σ + b + λ + bσ + rλ + bσr = 0 for p,3. For < r < r the equilibria are spiral nodes and for r > r they become unstable spiral saddle points with index see text on page 69. Since this happens, at the point where r = r the two of the three eigenvalues should be complex conjugates with no real part. Assuming solutions of the form λ = iω substitute in the characteristic equation for p,3. The real and imaginary parts of the left hand side should both equal to zero which give ω = b σ + r and ω σ + b + = bσ r respectively. Substituting ω in the second equation and solving for r we obtain r σb + σ + 3 = σ b 4.5 Solving the system of equations dx/dt = dy/dt = dz/dt = 0 we obtain two equilibria, namely p, =, ±, 0. The Jacobian is given by Jx, y, z = 0 z 0 x y x 0. Hence Jp, = ± ± 0. For both p and p the Jacobian has the same characteristic polynomial λ 3 + λ + λ + = 0. Solving this equation using a computer we get λ =.35 and λ,3 = 0.77 ±.0i. 4.6 Use eqns to solve dx/dt = dy/dt = dz/dt = 0. These give x = az, y = z and z = c± c 4ab a. The Jacobian is given by Jx, y, z =

15 xix 0 a 0 z 0 x c λ 3 λ b z, whose characteristic equation is: + a + λ z + ba z + az b z = 0.

16 5 5. Working as in section 5.. on page 8 the result is λ =ln 0 5. Since we have f x n = A > 0 we get λ = ln A which is negative for 0 < A < so there exists a stable fixed point for 0 < A <. For A > we always have f x > 0 at x = x so it is impossible to have stable orbits. 5.3 NOT SURE ABOUT REASONING. After many iterations the map settles to either a fixed point or a cycle. At bifurcation points, all points which are fixed or belong to a cycle have their derivatives equal to and hence the logarithm of their derivatives equal to zero. Since the Lyapunov exponent equals to the average of those logarithms after many iterations see eqn 5.3, we may conclude that the Lypunov exponent equals to zero. 5.4 Since there is only one fixed point X = /A the probability distribution may take the form P X = δx X, where δ is the Dirac delta function. So using λ = 0 P X ln A X dx we obtain λ = ln A X = ln A. 5.5 Start from gx = ffx = A X A A + X + A 3 X 3 A 3 X 4 which gives g X = A A A + X + 6A [ 3 X 4A 3 X 3. Then the probability distribution is given by P X = δx X + + δx X ] with X± being given in eqn.7. Use λ = 0 P X ln g X dx to obtain λ = ln g X+g X. A lot of algebra would be required to evaluate this expression, but there is an easier way for doing so if we consider g X± = f fx±f X± = f X f X± = A X X± = A [ X + X± + 4X X ±]. From equation.6 X+ + X = [AA + ]/A and X X + = A + /A. Hence g X± = A [ AA + /A + 4 A + /A ] = 4 + A A and therefore λ = ln 4 + A A.

17 5.6 xxi The fixed point for A = is X = / = /. Moreover since f X = 0 we may conclude that there exists a supercycle since these results would force the Lyapunov exponent to tend to. Iterating the map starting from X 0 = 0. we get X = 0.95, X = , X 3 = , X 4 = , X 5 = and X 6 = Substitution of A = + 5 in eqn.7 we get X + = and X =. In order to verify that we have supercycle we have to show that g X = 0 with g being given in the solution of Problem 5.5. With a bit of algebra one can verify that g X ± = ??????? 5.9 Since the map for those range of values of α has a single fixed point the Lyapunov exponent equals to λ = ln f X. Since X = 0 the fixed point we may write λ = ln f 0 = ln α. 5.0 R n+ R n = a X n + b Y n + c X n + d Y n X n + Y n with Y n = Y n X n = c X n + d Y n a X n + b Y n as eqn 5.. Substitution of R n+ R n 5. = = c + dy n a + by n in eqn 5.0 yields eqn 5.. a + by n + c + dy n + Y n which is the same Using the notation of the text we get a = A, b = c = [ 0 and d ] = N A B. Substitution in eqn 5. yields λ = lim ln +B Y n N N +Y = n=0 n [ ] lim ln B + A B n +Y with Y n = BY n A = Y 0 B n A. Then we have : n { lim Y + for A < B n = Substitution of the limit in the equation for λ gives λ = ln[max A, B]. Also use λ + λ = det J = AB n 0 for A > B to determine λ.

18 xxii 5. A.. Logistic Map : JX = A X A.. Sine Map : JX = Aπ cos πx A..3 Tent Map : JX = A sign/ X A..4 Linear Congruential Generator : JX = Amod C A..5 Gauss Map : JX = /X mod ax b A.. Hènon Map : JX, Y = 0 a signx b A.. Lozi Map : JX, Y = 0 X + a Y + b A..3 Tinkerbell Map : JX, Y = Y + c X + d a Y A..4 Burgers Map : JX, Y = Y X + b a 0 A..5 Kaplan-Yorke Map : JX, Y = mod 4π sin4πx b A..6 Dissipative Standard Map: JX, Y = k cos Xmod π b A.3. Chirikov Map : JX, Y = k cosx mod π cos α + X sin α sin α sin α X cos α cos α A.3. Hènon Area-preserving Quadratic Map : JX, Y = A.3.3 Arnold Cat Map: JX, Y = mod k signx A.3.4 Gingerbreadman Map : JX, Y = 0 cos α k sin α cos X sin α A.3.5 Stochastic Web Map : JX, Y = sin α + k cos α cos X cos α A.3.6 Lorenz Simplest 3-D Map : Jx, y, z = Y X Using eqn 5.0 we get λ = a x for the local Lyapunov exponent. Since at x = we have f x = 0 the global Lyapunov exponent is λ = a Since Jx, y =, we have detj = bxy +. 0 bxy b x For area expansion we require detj > 0, ie xy > /b and for area contraction we must have xy < /b.

19 5.6 xxiii INCOMPLETE We have a = d = 0, b = c =. From eqn 5.3 we get λ + λ = 0 and T y dy from eqn 5. λ = lim dt with T T 0 + y dt = y for t 0 dy = arctanh y y tanh t + tanh t dt = ln + tanh t ln + tanh t+ ln + tanh t ln + tanh t ln + tanh t + ln + tanh t = lim t Since x [σ y x] + y [rx y xz] + [xy bz] = b σ < 0 for z all σ, b > 0 we have volume contraction with V t e b++σt. 5.9 A.4. Damped Driven Pendulum : 0 0 Jx, y, z = cos x b A cos Ωz A.4. Driven Van Der Pol Oscillator : 0 0 Jx, y, z = bxy b x A cos Ωz A.4.3 Ueda Attractor : Jx, y, z = A.5. Lorenz Attractor : A.5. Rössler Attractor : Jx, y, z = 3x b A cos Ωz r z x σ σ 0 y x b

20 xxiv Jx, y, z = 0 a 0 z 0 x c A.5.3 Diffusionless Lorenz Attractor : 0 Jx, y, z = z 0 x y x 0 A.5.4 Chua s circuit : αb + /a bsignx + signx α 0 Jx, y, z = 0 β 0 A.5.5 Moore-Spiegel Oscillator : 0 0 Jx, y, z = 0 0 T Rxy T R + Rx A.5.6 Thomas Cyclically Symmetric Attractor : b cos y 0 Jx, y, z = 0 b cos z cos x 0 β A.5.7 Simplest Quadratic Chaotic Flow : Jx, y, z = y a A.5.8 Simplest Cubic Chaotic Flow : Jx, y, z = y yx a A.5.9 Simplest Piecewise Linear Chaotic Flow : 0 0 Jx, y, z = 0 0 signx a A.6. Simplest Driven Chaotic Flow :

21 xxv Jx, y, z = A.6. Nosé-Hoover Oscillator : A.6.3 Labyrinth Chaos : Jx, y, z = Jx, y, z = A.6.4 Hènon-Heilles System : 5.0 A. T λ = lim T T 0 Jx, y, v, w = B,C,E,I,L,M,R,S. λ = lim D,O. λ = lim T T F,H,Q. λ = lim G. λ = lim T J,N. λ = lim K. λ = lim 0 0 3x 0 cos Ωz z y y cos z 0 cos y 0 cos x y x 0 0 x y 0 0 zdt = z Numerical calculation is needed. T 0 T T T 0 T T T 0 T T 0 T needed. P. T λ = lim T T 0 5. T T 0 T T T 0 dt = xdt = x Numerical calculation is needed. 0.5 dt = dt = 0.6 dt = y dt = y 0.7 Numerical calculation is y dt = y Numerical calculation is needed. Since one of the eigenvalues is positive we have a strange attractor of dimension greater than. 5. D KY = =

22 xxvi 5.3 D KY = = We first have to find the equation of a parabola y = ax + bx + c which passes through the points 0, p,, p and, p 3. Solving the resulting system of three equations in three unknowns we get y = p p + p 3 x + 4p 3 p 3 5 p x + p 3 3p + 3p. Hence we get : A.5. p = p = , p 3 = and dimension. A.5. p = p = 0.074, p 3 = 5.39 and dimension.05 A.5.3 p = p = 0.0, p 3 = and dimension.73 A.5.4 p = p = 0.37, p 3 =.96 and dimension.3 A.5.5 p = p = 0.09, p 3 = and dimension.7 A.5.6 p = p = , p 3 = 0.54 and dimension.09 A.5.7 p = p = 0.055, p 3 =.07 and dimension.05 A.5.8 p = p = , p 3 =.08 and dimension.074 A.5.9 p = p = 0.036, p 3 = 0.6 and dimension.03 A.6. p = p = 0.038, p 3 = 0 and dimension = 3 A.6.3 p = p = 0.40, p 3 = 0 and dimension = λ = lnamodc

23 6 6. In order to determine the inverse map interchange X n and X n+, Y n and Y n+ to get Y n = bx n+ and X n = +Y n+ ax n+. Solving for Y n+ and X n+ gives the inverse map. Answer: X n+ = Y n /b, Y n+ = X n +ay n /b. 6. The fixed points satisfy X n+ = X n and Y n+ = Y n. By eqn 6., this means that X n+ = X n = Y n+ = Y n = X = Y. Eqn 6. becomes a 4 + a 5 + a 6 X + a + a 3 X + a = 0. This equation has real solutions when = a + a 3 4a a 4 + a 5 + a 6 0. Under this condition the fixed points are given by X = Y = a a 3 ± / a 4 + a 5 + a The fixed points of the Hénon Map are given by the equation.4x + 0.7X = 0, whose solution is X =. 3 4 and X = Their position relative to the attractor is shown in the figure below: 6.4 Interchange X n and X n+, Y n and Y n+ to get : X n = a + a X n+ + a 3 Y n+ + a 4 X n+ + a 5 X n+ Y n+ and X n+ = Y n. Solving for Y n+ and X n+ gives the inverse map. Answer: X n+ = Y n, Y n+ = X n a a Y n a 4 Y n a 3 a 5Y n.

24 xxviii 6.5 As in the previous problem use X n+ = Y n and X n = a + a X n+ + a 3 Y n+ + a 4 Xn+ + a 5 X n+ Y n+ + a 6 Yn+. The second equation becomes a 6 Yn+ + a 5 Y n + a 3 Y n+ + X n a a Y n a 4 Yn = 0, whose solution is: Y n+ = a 6 [ a 5 Y n a 3 ± a 5 Y n + a 3 4a 6 X n a a Y n a 4 Yn 6.6 Antisymmetry about r = 0.5 can be verified if we notice that f i 0.5 r = f i r for i =,, 3. It is obvious that f is infinite for r = 0 and r = since tan ±π/ = ±. Morever, one can verify that eqns 6.4 and 6.5 are also infinite using the Del Hôpital s rule. All three functions have half their values in the range < f < since we have f0.5±0.5 = ± Eqn 6.6 may be also written as a system of differential equations of the form: ẋ = y, ẏ = z and ż = 3 5z y + x. Equilibrium points satisfy ẋ = ẏ = 0. This happens only when y = z = 0 and x = ±. The Jacobian takes the form: Jx, y, z = signx a When x = the eigenvalues are , ±.603i and when x = the eigenvalues are , ±.0876i. 6.9 The following calculations are valid as long as the a 4 and a 5 are positive numbers. Eqn 6.7 becomes α dx β dt + a α x α dx a 4 β dt + a 5αx = a sin a 3 t dx dt +a βα x a 4 dx α dt +a 5β x = β α a sin a 3 βt. Choose β so that a 5 β = ie β = ± a 5 and α = ± a 4. In such a case the equation becomes dt + a x dx dt + x = a sin a 3t with a = ±a a 4 a 5, a = ± a and a 3 = ±a 3 a 5 a 5 a4 6.0 In this problem we have to find the a s and b s for which X n = b tanha 0 + a X n + b tanha 0 + a X n. The information we have in order to ]

25 xxix solve the problem is that when X n is 0 or then X n = 0 and that the maximum should be at X n = when X n = /. Since we have six unknowns and 4 equations we may have two free variables. The equations are: b tanha 0 +b tanha 0 = b tanha 0 +a +b tanha 0 +a = 0, b a b a tanh a 0 + a / + b a b a tanh a 0 + a / = 0 and b tanha 0 + a / + b tanha 0 + a / = 6. Cannot determine the LE accurately enough from the diagram in order to compare them with those in the appendix. 6. Since we are interested only for the maximum, fit the parabola x = at + bt+c through the points, x, 0, x and, x 3. In this case we would get a = x + x 3, b = x 3 x and c = x. The maximum of the parabola occurs at t = b/ a and equals to x max = c b /4a. 6.3 The intersection with the Poincaré section occurs only when z z < 0. Each point x, y, z on the line joining x, y, z and x, y, z is given by x = x + λ x x, y = y + λ y y and z = z + λ z z z with 0 λ. For λ = we have z = 0, ie an intersection z z of the line with the plane z = 0. Therefore the coordinates of the point x, y on the Poincaré section are given by x = x + z x x and z z y = y + z y y. z z The fixed points of the Lozi map satisfy X = Y and 0.5X =.7 X. This gives either 0.5X =.7X X = Y = or 0.5X = +.7X X = Y = The fixed points satisfy X = 0X 0Y 6Y and X Y = 4 Y +. Solving these equations we finally obtain the pairs 0, 0 and. 07 6,. 3 5.

26 xxx 6.0 The fixed points satisfy X = X Y and Y X = 0 i Y = 0 and ii X = /. Then for Y = 0 case iwe get the equation X X = 0 whose solutions are X = 5+ and X = 5. The Jacobian of the map has characteristic polynomial X λ + 4Y = 0 and eigenvalues λ = X ± Y i. Therefore X is unstable since λ > 0, whereas X is stable. For X = / we get non-real Y case ii. In order to determine the periodic cycle consider the fixed points of the second iterate map. For the Y component we get 4XY X Y = Y Y [ 4X X Y ] = 0. The equation 4X X Y = 0 together with the X component of the second iterate map do not have real solutions. However when Y = 0, using X Y 4X Y = X we get XX 3 X = 0. This gives X = 0 and X =.,ie in this case we have a -period cycle with oscillations between the points 0, 0 and, By definition, the derivative of f with respect to Z is f Z = F X G + i G Y X + f Y f F = 0. This condition requires Z X = G Y 6. Hénon Map: F X X + i f = Y. Since f is a function of Z only we have and G X = F Y. G F = ax 0 =. Lozi map: Y Y Julia set: F X = G G = X and Y X = F Y = Y. 6.3 = 0.5 = G X. The Cauchy-Riemann equations require F X = a +a 4 X +a 5 Y = 0 = G Y. Hence we must have a = a 4 = a 5 = 0. Also F Y = a 3 + a 5 X + a 6 Y = = G X which gives a 3 =, a 5 = a 6 = 0. Therefore in this case we must have X n+ = a Y n and Y n+ = X n which is obviously a non-chaotic system. 6.4 Let F = b + b X + b 3 Y + b 4 X + b 5 XY + b 6 Y and G = c + c X + c 3 Y + c 4 X + c 5 XY + c 6 Y. The Cauchy-Riemann equations require F X = b + b 4 X + b 5 Y c 3 + c 5 X + c 6 Y = G Y and F Y = b 3 + b 5 X + b 6 Y

27 xxxi c c 4 X c 5 Y = G. Equating the coefficients with the same powers X of X and Y we get : b = c 3 = a, b 4 = c 5 / = a 4, b 5 / = c 6 = a 5, b 3 = c = a 3, b 5 / = c 4 = a 5 and b 6 = c 5 / = a 4. If we also let b = a and c = a 6 we obtain eqns 6.8 and 6.9.

28 7 7. This is done in the textbook on page 3. Eigenvalues? 7. Done in textbook p.33. Eigenvalues? 7.3 When y =, ẏ = 0 for all values of the parameter a. Now, expand about y = by making a change of variable from y to v = y. In this case v is small and we have ẏ = v = a lnv + + v a[v v +...] + v = a + v av + Ov 3. Hence a transcritical bifurcation occurs when a =. In order to write the last equation in normal form, let v = cξ which transforms it to ξ = a + ξ caξ + Ov 3. Letting c = /a and µ = a + we finally obtain the normal form with ξ = a y. 7.4 The fixed points satisfy f = µx x 3 = 0. Solving for x we obtain x = 0 and x = ± µ. Then df dx = µ 3x and for x = 0 we get df dx = µ. Thus for µ > 0, x is unstable and for µ < 0 it is stable. However for the points x = ± µ, which exist only for µ > 0 we get df dx = µ > 0 which are stable. 7.5 The fixed points satisfy f = µx +x 3 = 0. Solving for x we obtain x = 0 and x = ± µ. Then df dx = µ + 3x and for x df = 0 we get dx = µ. Thus for µ > 0, x is unstable and for µ < 0 it is stable. However for the points x = ± df µ, which exist only for µ < 0 we get dx = µ > 0 which are unstable. 7.6 For a > the only equilibrium point we have is x = 0 which is unstable since df dx = a cos x > 0. For 0 < a <, the equilibrium points satisfy sin x = a. INCOMPLETE-NEEDS FIGURE. x

29 7.7 xxxiii As shown in the text eqns 7.8 and 7.9 may be written in polar coordinates as dr dt = r µ r = fr and dθ dt =. The equation fr = 0 is satisfied for the points r = ± µ which exist for µ > 0 and for r = 0. For r = 0 we have df dr = µ which means that it is stable for µ < 0 and unstable for µ > 0. The points satisfying r = ± µ give df dr = µ which is always stable given that µ > Multiply eqn 7.8 by x and eqn 7.9 by y and add them together. Then x dx dt + y dy dt = dr dt = r dr dt = r µ r. 7.9 The equation d x dt + b x dx dt + x = a can be written as ẋ = y and ẏ = a b x y x. The Jacobian of this system is thus: 0 Jx, y = bxy bx The fixed points satisfy ẋ = y = 0 and ẏ = 0 which gives x = a. In such a case the characteristic polynomial is λ [ λ + b a ] + = 0 which has solutions λ = ba ±i 4 b a = Aa, b ± iba, b. We have Hopf db = a bifurcations when b = 0 since Aa, 0 = 0, Ba, 0 0 and da which is greater than zero provided that a <. We may also have Hopf bifurcations when a = ± since A±, 0 = 0, B±, b 0 and da da = b which is greater than zero when b is positive for a = and when b is negative for a =. 7.0 The equation dr dt = rµ + r r 4 = fr gives the fixed points r = 0 and r± = ± +4µ r,3 = ± +4µ. To classify their stability compute df dr = µ+3r 5r 4 = 4µ r +5µ+r r 4. Obviously for r df dr = µ which is unstable for µ > 0 and stable for µ < 0. We have three cases : i µ < /4 in which the only fixed point is r which is stable. ii /4 < µ < 0 in which we find r+ to be stable and r to be unstable.iiiµ > 0. The cycle r vanishes, r+ is still stable whereas r becomes unstable.hysteresis 7. HARD 7. Hilborn pages 57 and 58. Too hard to be included. At least a number of hints should be provided.

30 xxxiv 7.3 Write the second iterate map keeping the terms up to 3rd order : F F X = + µ [ + µ X + X 3] + [ + µ X + X 3] 3 X +Xµ+Xµ X 3 4X 3 µ 3X 3 µ X 3 µ 3 = + µ X X 3 +4µ+ 3µ +µ F F X + µ X + µ + µ + µ X 3, which is identical to eqn 7.9. Then taking F F ± µ = ± + µ µ µ + µ + µ + µ [ µ = µ 5 7 ] 9 ± 3 µ 3 µ µ. Since µ is small then we may neglect the 5th, 7th and 9th powers of µ to get F F ± µ ± µ which shows that ± µ is approximately a fixed point. 7.4 Substitute X = ln A in F X = AXe X to get F X = X, thus showing that X is a fixed point. In order to show the existence of the flip bifurcation make the change of variable Y = X ln A and define GY = F X X to get GY = A Y + ln A e Y +ln A ln A = Y + ln A e Y ln A. The map Y n+ = GY n has a fixed point at Y = 0. Expand GY about 0 up to order 3 to get : GY Y + ln A Y + Y 6 Y 3 ln A = ln A Y + ln A Y + 6 ln A Y 3 INCOMPLETE F X ln A+ ln A X ln A+ + ln A X ln A ln A X ln A3 = + ln A ln3 A + ln A X + ln A ln A X + 6 ln A X ln4 A

31 8 8. dh dt = i H dp i p i dt + H dq i q i dt = i H H p i q i + H H q i p i = 0 8. H = p ml + mgl cos x. In this case we have q x. Eqn 8.5 gives : dp dt = H θ ω = g L. = mgl sin x = ml d x dt dv dt = ω sin x with v = dx dt and The Jacobian Matrix is given by: Jx, v = ω which, evaluated at x = π, v = 0 gives Jπ, 0 = cos x 0 0 ω. The characteristic 0 polynomial is λ ω = 0 which gives λ = ±ω = at x = π, v = 0 we have a saddle point. 8.4 dx mgl Start from E mgl = ml dt cos x. Let x0 be the maximum value of x, which satisfies cos x 0 = E mgl mgl. Hence L dx g dt = cos x cos x0. L Solving for t : dt = dx g cos x cos x0. Integrate from 0 to x 0 to get /4 L x0 of the period T which can be written as T = 4 dx g 0 cos x cos x0 = L x0 dx g. Make the substitution sin y = sin 0 sin x sin x x/ sin x 0 0 L to convert it to T = 4 g Ksin x 0 where Ks = π/ ds 0 k is sin s the complete elliptic integral of the first kind. Since sin x cos x 0 = 0 = E mgl, use these formulas to get T E/mgL : L T 0. = g, T L.0 = 7.46 g, T.0 =.

32 xxxvi 8.5 In this case we have q x, q y, p v, p w. Using eqns dp 8.4 and 8.5 we get : = dv dt dt = H = H q x = ω sin x + Ay, dp dt = dw dt = H = H q y = Ω y + Ax, dq dt = dx dt = H = H p v = v and dq dt = dy dt = H = H p w = w. 8.6 We have q x, q y, p v, p w. Use eqns 8.4 and 8.5 to get : dq dt = dx dt = H = H p v = v, dq dt = dy dt = H = H p w = w, dp dt = dv dt = H = H q x = xy x and dp dt = dw dt = H = H q y = y x +y. 8.7 LOOK UP FORMULA IN FORD S REFERENCE 8.8 First note that x = χ + ψ dv dt = x d dt = d χ dt, y = χ ψ. Eqn 8.8 may be written as + d ψ dt = χ ψ χ + ψ and eqn 8.9 with y replaced by y as dw dt = y d dt = d χ dt d ψ dt = χ + ψ χ ψ. Solve for d χ dt and d ψ to obtain eqns 8. and 8.. dt SOLUTION?? 8.9 Use eqn 5.4 with f = y, g = x + yz and h = y, from which it follows that dv V dt = lim T zdt = z T T If we make the change of variables t t, z z and y y we still get the same equations. Therefore the system is time reversible. 8. Since performing the change of variables t t, z z and x x we still get the same equations, the system is time reversible. Also, using eqn 5.4 together with f = y, g = x + z and h = xz + 3y we get for the time-reversed system : V dv dt = lim T T 0 T xdt = x.

33 xxxvii 8. Use dx dt = y, dy dz = x+yz and dt dt = y.since dy dt = d x dt = x+dx dt z d x z = dt + x / dx. Taking the derivative of z with respect to t we dt get dz d 3 dt = x dt 3 + dx / dx dt dt d x d x dx dx dt dt + x / =. dt dt Simplifying the last result yields eqn The differential equation for the simple pendulum can be written as dx dt = y = f x, y and dy = k sin x = g x, y. For the case of the Leapfrog dt method see solutions to problems 8.0 and 8.. For the Runge-Kutta method use k xn = Y n, k yn = k sin X n, k xn = Y n + k sin X n and k yn = k sinx n + Y n. Then X n+ = X n + Y n + k sin X n and Y n+ = Y n + k sinx n + Y n. The Jacobian of this map has determinant det J = + k cos X n + k cosx n + Y n k sinx n + Y n. Hence the map in this case is not simplectic area-preserving. 8.4 Use the Jacobian of the map is given at problem 5.. whose determinant is det J = cos α + sin α =. 8.5 The Hénon area-preserving quadratic map is not invertible. 8.6 a + a The Jacobian is given by JX, Y = 4 X + a 5 Y a 3 + a 5 X + a 6 Y 0 The map is area preserving if det J = a 3 a 5 X a 6 Y = for all X and Y. This is the case only if a 3 = and a 5 = a 6 = The Jacobian is JX, Y = whose eigenvalues are λ ± = 3± 5 with eigenvectors v ± = = λ ± ± 5 Since v + v = 0 we may conclude that the two manifolds are perpendicular to each other. The unstable manifold corresponds to λ +. The tangent of the angle of v + with the x-axis is given by the ratio of the y over the x component of the vector which is obviously equal to the golden mean.

34 xxxviii 8.8 As indicated in the text on page 69 if we start from, 3, 3, 5, 3 3, 5, 3 which is a period 5 cycle. 8.9 Use the Jacobian of the map is given at problem 5.. whose determinant is det J = cos α + sin α =. 8.0 he differential equation for the simple pendulum d x dt + k sin x = 0 can be transformed to a system of differential equations : dx dt = y = f x, y and dy dt = k sin x = g x, y. The Leapfrog method in the y-variable can be written as Y n+ = Y n + hgx n, Y n and X n+ = X n + hfx n, Y n+. Since we are dealing with angles we may take the equations modulo π and if we use h = we finally obtain eqns 8.40 and Use the Jacobian from the solution to problem 5. and calculate its determinant which equals det J = + k cos X k cos X =. Therefore the map is area-preserving. 8. The fixed points satisfy Y n modπ = 0 and k sin X n modπ = 0. The fixed points are Y = 0 and X = 0 or π. The Jacobian of the map is given in the solution of problem 5. with characteristic polynomial λ λ + k cos X + = 0. When X = 0 we get λ + k λ + = 0 whose solution is λ = + k ± k 4 k +. So for all k we have a saddle point.when X = π we have the equation λ k λ + = 0 whose solution is λ = k + k 4 k. In this case we have an unstable spiral point for 0 < k <, a stable spiral for < k < 4 and a stable fixed point for k > Start with X 0 = 0 and Y 0 = π. The first iteration will give Y = πmodπ = π and X = πmodπ = π. The second iteration will give Y = πmodπ = π and X = πmodπ = 0 which correspond to the initial point X 0, Y One solution is 0, π 3 π 3, π 3 4π 3, π 3 0, π 3

35 8.5 xxxix The sum of the eigenvalues equals to the logarithm of the determinant which in this case equals to log det J = log b + k cos X bk cos X = log b. 8.6 In order to show that Q is conserved we have to show that A = X n + X n +X n X n X n X n X n +X n +X n 3 X n X n X n 3 = 0. A simplifies to A = X n X n 3 X n X n X n X n 3 = X n X n X n X n X n X n X n X n X n = 0. Hence Q is conserved.

36 9

37 0 0.? 0. Assume the polynomial to be of the form y = az + bz + c and passing through the points, X n, 0, X n and, X n. Solving the resulting system for a, b and c we get a = X n + X n X n, b = X n X n and c = X n. Using these coefficients set z equal to to find X n Assume the polynomial to be of the form y = az 3 +bz +cz +d and passing trough the points, X n 3,, X n, 0, X n and, X n.solving the resulting system for a, b, c and d we get a = X n + 6 X n + X n 6 X n 3, b = X n + X n + X n, c = X n + 3 X n X n + 6 X n 3 and d = X n. Using these coefficients set z equal to to find X n Assume the polynomial to be of the form y = az 4 + bz 3 + cz + dz + e and passing through the points, X n 4,, X n 3, 0, X n,, X n and, X n. Solving the resulting system for a, b, c, d and e we get a = 4 X n + 4 X n + 4 X n 4 6 X n 6 X n 3, b = X n X n 4 6 X n + 6 X n 3, c = 5 4 X n 4 X n 4 X n X n + 3 X n 3, d = X n + X n X n 3 X n 3 and e = X n. Using these coefficients set z equal to 3 to find X n Assume the polynomial to be of the form y = az 5 + bz 4 + cz 3 + dz + ez+f and passing through the points 3, X n 5,, X n 4,, X n 3, 0, X n,, X n and, X n. Solving the resulting system for a, b, c, d, e and f we get: a = X n X n 3 4 X n + 0 X n 0 X n 5+ 4 X n 4, b = 4 X n + 4 X n + 4 X n 4 6 X n 6 X n 3, c = 5 X n + 7 X n X n + 4 X n + 4 X n X n 4, d = 5 4 X n + 3 X n X n 4 X n 4 X n 4, e = 0 X n 30 X n X n 4 X n X n + X n and f = X n. Using these coefficients set z equal to 3 to find X n+.

38 xlii 0.6 The values to be predicted are: X = sin = , X = sin = and X 3 = sin 3 = Using eqn 0. we obtain: X =. 879, X = and X 3 = Using eqn 0.: X =.500, X =.4564 and X 3 = 3.46 Using eqn 0.3: X =.3483, X =.47 and X 3 = Using eqn 0.4: X = , X =. 73 and X 3 =.66 Using eqn 0.5: X = 0.3, X = and X 3 = Using the original map, the first 6 points are: X = 0., X = 0.36, X 3 = 0.96, X 4 = 0.890, X 5 = and X 6 = Knowing the first 3 points we may use eqn 0. to get X 4 =.483, X 5 =.0448 and X 6 = Applying eqn 0. to the same set of 3 points we get X 4 =.7848, X 5 =.9496 and X 6 = Set j = m = in eqns 0.7 and 0.8 and the result follows immediately. 0.9 a = X 4X 3 + X 3 X + X X X 3 + X + X = =. Hence X n+ = X n. 0.0 a = X X X = 5. Hence X n+ = 5X n so X 3 = 5, X 4 = 5 and X 5 = 65 0.

39

40 . D = log 3/ log 5 = D = log 3/ log 7 = D = log 4/ log 3 =.6.4 Answers given in section.4.5 Answer given in section.5.6 At the n-th stage of the construction we need a minimum of m boxes with L = /3 m log m, so D 0 = log/3 = log m log 3 = By the definition of Cr we have C0 = 0. As r all Heaviside functions evaluate to since a circle with infinite radius contains all X n s. In that case the double sum evaluates to N N j = N = NN. This gives C =..0 Cr = j= r frdr, where fr is the distribution of points on the real line. 0

41 . xlv Referring to problem.0,let fr =. In this case the integral is performed on a surface so Cr = rdrdθ = πr. Hence D = d logπr π r d log r = d[ log r+log π] d log r = The answers for D =,,..., 0 are 5, 63, 585, 398, 0000, 59, 63096, , , 0 6 respectively D = log log 5 3 = D = log log 5 3 = D = log 5a+b 3a+b log 5 3