Chaos and Time-Series Analysis

Julien Clinton Sprott 13 April 2003

Preface

To be written

1

2.1

Change the variable X_{n} to X_{n} = (\frac{Y_{n}}{A} + \frac{1}{2}). Substitution in (2.4) yields:

\frac{Y_{n+1}}{A} + \frac{1}{2} = A (\frac{Y_{n}}{A} + \frac{1}{2}) (1 − \frac{Y_{n}}{A} − \frac{1}{2}) = A (\frac{1}{4} − \frac{Y_{n}^{2}}{A^{2}} ). Multiply both

sides by A to get : Y_{n+1} = B − Y_{n}^{2} with B = \frac{A^{2}}{4} − \frac{A}{2}

2.2

Fixed point: X^{∗} = 1 − 1/2.8 ≃ 0.643 The first ten iterations are: x_{1} = 0.631 , x_{2} = 0.652 , x_{3} = 0.635 , x_{4} = 0.649 , x_{5} = 0.638 , x_{6} = 0.647 x_{7} = 0.640 , x_{8} = 0.645 , x_{9} = 0.641 , x_{10} = 0.644 which indicate an oscillation about X^{∗}

2.3

Let f(X) = AX(1 − X) and consider X_{0} = X^{∗} + ε_{0} , with ε_{0} very small and positive. Then X_{1} = f(X_{0}) = AX^{∗} − AX^{∗2} − 2AX^{∗}ε_{0} + Aε_{0} − Aε_{0}^{2}. X^{∗}satisfies X^{∗} = AX^{∗}−AX^{∗2} which gives X_{1} = X^{∗}−2AX^{∗}ε_{0}+Aε_{0}−Aε_{0}^{2}. The −Aε_{0}^{2} term is smaller than the ε_{0} terms and may be neglected. Hence X_{1} = X^{∗} + ε_{0}A (1 − 2X^{∗}) = X^{∗} + ε_{0}f′(X^{∗}). X_{1} may be written as X_{1} = X^{∗} + ε_{1} thus giving ε_{1} = ε_{0}f′(X^{∗}). After n iterations: ε_{n} = ε_{0} (f′(X^{∗}))^{n} Hence the fixed point is unstable when |f′(X^{∗})| > 1. For f′(X^{∗}) > 1 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.

2.4

See text on page 17.

2.5

The second iterate map is given by g(X) = f(f(X)) with f(X) = AX(1 − X).Let p and q be the fixed points of the map (these two points are given by eqn 2.7). g′(X)|_{X=p} = f′(f(X))f′(X)|_{X=p} = f′(f(p))f′(p) = f′(q)f′(p) = A^{2}(1 − 2p)(1 − 2q) = A^{2}[1 − 2(p + q) + 4pq] From eqn 2.6 observe that p + q = [A(A + 1)]/A^{2} and pq = (A + 1)/A^{2} Hence g′(X)|_{X=p} = A^{2}[1 − 2A(A + 1)/A^{2} + 4 (A + 1) /A^{2}] = 4 + 2A − A^{2} Eqn 2.8 is found by setting g′(X) = −1.In general the 2-cycle is stable when |g′(X)| < 1 ie when 3 < A < 1 + \sqrt{6}

2.6

?

2.7

The map f(X) = 4X(1−X) has pre-images X_{A,B} = \frac{1}{2}(1±\sqrt{1 − X}). Then

f′(X_{A,B}) = ∓4\sqrt{1 − X} and P (X_{A,B}) = \frac{2}{π\sqrt{X}} . Use these two equations

in eqn 2.12 to obtain P (X) as in eqn 2.11

2.8

?

2.9

The pre-images of X_{n+1} = f(X_{n}) = A min(X_{n}, 1 − X_{n}) are X_{A} = X/A and X_{B} = 1 − X/A . Use these together with the fact that |f′(X_{A,B})| = 1 in eqn 2.12 to obtain P (X) = 1/A.

2.10

Let f(X) = A (1 − |2X − 1|^{α}) . Then f′(X) = −2αAsign(2X−1) |2X − 1|^{α−1}. Obviously f′ (1/2) is discontinuous for α ≤ 1 (not defined for α < 1 and having a jump discontinuity for α = 1). The second derivative is given by: f′′(X) = −2α (α − 1) A(sign(2X−1))^{2} |2X − 1|^{α−2} = −2α (α − 1) A |2X − 1|^{α−2}. f′′(1/2) is not defined for α < 2 , but for α ≥ 2, f′′ is continuous everywhere and therefore f is smooth (Note: sign(x) = 1 for x ≥ 0 and −1 for x < 0).

2.11

From Exercise 2.10 the derivative at 0 is given by f′(0) = 2αA |−1|^{α−1}. Since α < 1/2 and 0 < A < 1 we may conclude that |f′(0)| < 1 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.

3.1

The solution to eqn 3.1 is of the form x_{1} = 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_{2} = (x_{0} + ε) e^{at}. Their separation Δx = x_{2} − x_{1} = εe^{at} increases exponentially at a rate a.

3.2

The solution x = \frac{1}{1 + (\frac{1}{x_{0}} − 1) e^{−at}} has derivative \frac{dx}{dt} = \frac{a (\frac{1}{x_{0}} − 1) e^{−at}}{[1 + (\frac{1}{x_{0}} − 1) e^{−at}]^{2}} .

Use x and its derivative in the expression \frac{dx}{dt} − ax (1 − x). Verifying that this expression equals 0 proves that eqn 3.3 satisfies eqn 3.2.

3.3

This equation is a separable 1st 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^{2}x/dt^{2} + abx = 0. This can have a solution of the form x = x_{0} sin (\sqrt{ab}t). Substitution in the sec-

ond equation yields dy/dt = −bx_{0} sin (\sqrt{ab}t) . Integrating once we get

y = x_{0}\sqrt{\frac{b}{a}} cos (\sqrt{ab}t). The equations for x and y generate points that lie

on an ellipse with \frac{x_{max}}{y_{max}} = \sqrt{\frac{a}{b}} .

3.5

v (t) = \frac{dx}{dt} = v_{0} cos ωt+\frac{AΩ}{ω^{2} − Ω^{2}} cos Ωt and \frac{d^{2}x}{dt^{2}} = −ωv_{0} sin ωt−\frac{AΩ^{2}}{ω^{2} − Ω^{2}} sin Ωt.

Using the expressions for x and d^{2}x/dt^{2}in the left hand side of ean 3.10 we finally obtain A sin Ωt which equals the right hand side.

3.6

\frac{d^{2}x}{dt^{2}} = −v_{0} \frac{−ω^{4}e^{−ω^{2} \frac{t}{b}} +b^{4}e^{−bt}}{b^{3}} and \frac{dx}{dt} = v_{0} \frac{−ω^{2}e^{−ω^{2} \frac{t}{b}} +b^{2}e^{−bt}}{b^{2}} . Let Δ = \frac{d^{2}x}{dt^{2}} +

b\frac{dx}{dt} + ω^{2}x = −v_{0}ω^{2} \frac{−ω^{2}e^{−ω^{2} \frac{t}{b}} +b^{2}e^{−bt}}{b^{3}} .Then comparing Δ with each term

from eqn 3.18 we get |Δ/\frac{d^{2}x}{dt^{2}} | = \frac{ω^{2}e^{−ω^{2} \frac{t}{b}} −b^{2}e^{−bt}}{b^{2}e^{−ω^{2} \frac{t}{b}} −b^{2}e^{−bt}} ≪ \frac{ω^{2}e^{−ω^{2} \frac{t}{b}} −b^{2}e^{−bt}}{ω^{2}e^{−ω^{2} \frac{t}{b}} −b^{2}e^{−bt}} = 1 ,

|Δ/ (b\frac{dx}{dt})| = \frac{ω^{2}}{b^{2}} ≪ \frac{1}{4} and |Δ/ (ω^{2}x)| = \frac{−ω^{4}e^{−ω^{2} \frac{t}{b}} +b^{2}ω^{2}e^{−bt}}{−ω^{4}e^{−ω^{2} \frac{t}{b}} +b^{4}e^{−bt}} ≪ \frac{−ω^{4}e^{−ω^{2} \frac{t}{b}} +b^{4}e^{−bt}}{−ω^{4}e^{−ω^{2} \frac{t}{b}} +b^{4}e^{−bt}} =

1. Since each term is much less than 1 we may neglect Δ, thus verifying that the eqn 3.21 satisfies eqn 3.18 .

3.7

x = v_{0}te^{−bt/2}, v(t) = dx/dt = v_{0}e^{− \frac{1}{2} bt} − \frac{1}{2}v_{0}tbe^{− \frac{1}{2} bt} and d^{2}x/dt^{2} = −v_{0}be^{− \frac{1}{2} bt} + \frac{1}{4}v_{0}tb^{2}e^{− \frac{1}{2} bt} . Substitute these in the right hand side of eqn 3.18 to verify that it equals zero.

3.8

x = (v_{0}/ω) e^{− \frac{1}{2} bt} sin ωt , v(t) = dx/dt = −\frac{1}{2}v_{0}e^{− \frac{1}{2} bt} \frac{b sin ωt−2(cos ωt)ω}{ω} and d^{2}x/dt^{2} = \frac{1}{4}v_{0}e^{− \frac{1}{2} bt} \frac{b^{2} sin ωt−4b(cos ωt)ω−4(sin ωt)ω^{2}}{ω}. Substitution in eqn 3.18 yields Δ = d^{2}x/dt^{2} + bdx/dt + ω^{2}x = −\frac{1}{4}v_{0}e^{− \frac{1}{2} bt}b^{2} \frac{sin ωt}{ω}. Compare Δ with each

term from eqn 3.18 to get : |Δ/ (ω^{2}x)| = \frac{\frac{1}{4}v_{0}e^{− \frac{1}{2} bt}b^{2} \frac{sin ωt}{ω}}{v_{0}ωe^{− \frac{1}{2} bt} sin ωt} = \frac{1}{4} \frac{b^{2}}{ω^{2}} ≪ 1,

|\frac{Δ}{b\frac{dx}{dt}} | = \frac{\frac{1}{4}v_{0}e^{− \frac{1}{2} bt}b^{2} |\frac{sin ωt}{ω} |}{\frac{1}{2}v_{0}be^{− \frac{1}{2} bt} |\frac{b sin ωt−2(cos ωt)ω}{ω} |} < \frac{\frac{1}{4}b^{2} |\frac{sin ωt}{ω} |}{b} = \frac{1}{4}b |\frac{sin ωt}{ω} | ≤ \frac{1}{4} \frac{b}{ω} ≪

1 and |Δ/\frac{d^{2}x}{dt^{2}} | = \frac{\frac{1}{4}v_{0}e^{− \frac{1}{2} bt}b^{2} |\frac{sin ωt}{ω} |}{\frac{1}{4}v_{0}e^{− \frac{1}{2} bt} |\frac{b^{2} sin ωt−4b(cos ωt)ω−4(sin ωt)ω^{2}}{ω} |} < \frac{b^{2} |\frac{sin ωt}{ω} |}{|4ω sin ωt|} =

\frac{1}{4} \frac{b^{2}}{ω^{2}} ≪ 1. Since Δ is much smaller than each of the terms in eqn 3.18 we may consider the eqn 3.23 roughly satisfying eqn 3.18.

3.9

x = \frac{A sin(Ωt−ϕ)}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} , v(t) = dx/dt = \frac{AΩ cos(Ωt−ϕ)}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} and d^{2}x/dt^{2} = −\frac{AΩ^{2} sin(Ωt−ϕ)}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} .

Substitution in the right hand side of eqn 3.17 yields:

\frac{d^{2}x}{dt^{2}} + b\frac{dx}{dt} + ω^{2}x = A\frac{(ω^{2}−Ω^{2}) sin(Ωt−ϕ)+bΩ cos(Ωt−ϕ)}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} (∗).

Use eqn 3.25 and the trigonometric identities cos ϕ = 1/\sqrt{1 + tan^{2} ϕ} = \frac{ω^{2}−Ω^{2}}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} and sin ϕ = \sqrt{1 − cos^{2} ϕ} = \frac{bΩ}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} . Eqn (∗) can then be written as: d^{2}x/dt^{2}+bdx/dt+ω^{2}x = A (cos ϕ sin (Ωt − ϕ) + sin ϕ cos (Ωt − ϕ)) = A sin Ωt.

3.10

x_{m} = \frac{A}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} and the derivative \frac{dx_{m}}{dω} = \frac{−AΩ(−2ω^{2}+2Ω^{2}+b^{2})}{(\sqrt{(ω^{4}−2ω^{2}Ω^{2}+Ω^{4}+b^{2}Ω^{2})})^{3}} .

It becomes zero for Ω = \frac{1}{2} \sqrt{(4ω^{2} − 2b^{2})} and Ω = −\frac{1}{2} \sqrt{(4ω^{2} − 2b^{2})}. x_{m} becomes maximum for Ω = \frac{1}{2} \sqrt{(4ω^{2} − 2b^{2})} and the maximum is \frac{2A}{b\sqrt{(4ω^{2}−b^{2})}}.

3.11

Find the Ω’s for which \frac{x_{m}}{A} = \frac{1}{\sqrt{(ω^{2}−Ω^{2})^{2}+b^{2}Ω^{2}}} = \frac{1}{\sqrt{2}bω} ⇔

2b^{2}ω^{2} = (ω^{2} − Ω^{2})^{2} + b^{2}Ω^{2}. Then Ω^{4} + (b^{2} − 2ω^{2}) Ω^{2} + ω^{4} − 2b^{2}ω^{2} = 0 ⇔ (Ω/ω)^{4} + ((b/ω)^{2} − 2) (Ω/ω)^{2} + 1 − 2 (b/ω)^{2} = 0. The Ω’s satisfy (\frac{Ω_{1}}{ω} )^{2} + (\frac{Ω_{2}}{ω} )^{2} = 2 − \frac{1}{Q^{2}} and (\frac{Ω_{1}}{ω} )^{2} (\frac{Ω_{2}}{ω} )^{2} = 1 − \frac{2}{Q^{2}} ⇒ \frac{Ω_{1}}{ω} \frac{Ω_{2}}{ω} ≃ 1 − \frac{1}{Q^{2}} for large Q. Hence (\frac{ΔΩ}{ω})^{2} = (\frac{Ω_{1}}{ω} )^{2} + (\frac{Ω_{2}}{ω} )^{2} − 2\frac{Ω_{1}}{ω} \frac{Ω_{2}}{ω} ≃ 2 − \frac{1}{Q^{2}} − 2 (1 − \frac{1}{Q^{2}} ) = \frac{1}{Q^{2}} ⇒ \frac{ΔΩ}{ω} = \frac{1}{Q}

3.12

Start with x^{2} + v^{2} = x^{2} + (\frac{dx}{dt})^{2} ≃ 4 and take the derivative at each side to obtain 2x\frac{dx}{dt} + 2\frac{dx}{dt} \frac{d^{2}x}{dt^{2}} ≃ 0 ⇔ x + \frac{d^{2}x}{dt^{2}} ≃ 0 which corresponds to eqn 3.26 in the limit as b → 0.

3.13

Use v=\frac{dx}{dt} to write eqn3.26 as x + \frac{dv}{dt} + bv (v^{2} + x^{2} − 1) = 0 ⇒ vx + v \frac{dv}{dt} + bv^{2} (v^{2} + x^{2} − 1) = 0 ⇒ \frac{1}{2} \frac{d}{dt} (x^{2} + v^{2}) + bv^{2} (v^{2} + x^{2} − 1) = 0 which is apparently satisfied by v^{2} + x^{2} = 1, the limit cycle.

3.14

?

3.15

SAME AS Prob 3.13

3.16

f = x − y − x^{3} and g = x − x^{2}y . At (0, 0) we have dx/dt = dy/dt = 0 and the Jacobian evaluated there is:

J(0, 0) = ( \begin{matrix}1 & 0 \\ 1 & 0 \end{matrix} )

The eigenvalues can then be found to be λ = 0 and 1 from which we may conclude that the origin is at an unstable equilibrium. For solutions approximately on the unit circle ie satisfying x^{2} + y^{2} ≃ 1 we have dx/dt ≃ x (1 − x^{2}) − y = xy^{2} − y. Multiply dx/dt by x and dy/dt by y to get: x (dx/dt) ≃ x^{2}y^{2} − xy and y (dy/dt) = yx − x^{2}y^{2}. Adding those together we get x (dx/dt) + y (dy/dt) = \frac{1}{2} \frac{d}{dt} (x^{2} + y^{2}) ≃ 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.

3.17

Write eqn 3.18 as a system of 1st order ODEs ie \frac{dy}{dt} = −by−ω^{2}x = −y−x = g(x, y) and \frac{dx}{dt} = y = f(x, y). The equations corresponding to 3.33 and 3.34 with h = 1 are x_{n+1} = x_{n} + y_{n} and y_{n+1} = −x_{n}. These two can be written in a single equation x_{n+1} = x_{n} − x_{n−1} (∗). Starting with x_{0} = y_{1} = β and x_{1} = α iterate (∗) to get x_{2} = α − β , x_{3} = −β , x_{4} = −α , x_{5} = −α + β , x_{6} = β = x_{0}.

3.18

Total error: e(h, t, ϵ) = th + ϵ\sqrt{\frac{t}{h}} ⇒ \frac{∂e}{∂h} = t − \frac{1}{2}ϵ\sqrt{\frac{t}{h^{3}}} . For minimum set

\frac{∂e}{∂h} = 0 to get h = \sqrt[3]{\frac{ϵ^{2}}{4t}}. For this value of h , \frac{∂^{2}e}{∂h^{2}} > 0 which guarantees

that h = \sqrt[3]{\frac{ϵ^{2}}{4t}} gives the minimum.

3.19

The equations become x_{n+1} = x_{n}+h (\frac{y_{n+1}+y_{n}}{2} ) and y_{n+1} = y_{n}+h (−\frac{x_{n+1}+x_{n}}{2} )

which give x_{1} = x_{0}+h (\frac{y_{1}+y_{0}}{2} ) and y_{1} = y_{0}−h (\frac{x_{1}+x_{0}}{2} ). Solving this system

we obtain y_{1} = \frac{−4h}{4+h^{2}} and x_{1} = \frac{4−h^{2}}{4+h^{2}} . These give y_{1}^{2}+x_{1}^{2} = ^{\frac{(4−h^{2})^{2}+16h^{2}}{4+h^{2}}} = 1 , so the error is zero.

3.20

\frac{dy}{dt} = −by − ω^{2}x = g(x, y) and \frac{dx}{dt} = y = f(x, y). Then :

x_{n+1} = x_{n}+hf (\frac{x_{n+1}+x_{n}}{2} , \frac{y_{n+1}+y_{n}}{2} ) and y_{n+1} = y_{n}+hg (\frac{x_{n+1}+x_{n}}{2} , \frac{y_{n+1}+y_{n}}{2} )

which give x_{n+1} = x_{n}+h\frac{y_{n+1}+y_{n}}{2} and y_{n+1} = y_{n}+h (−b\frac{y_{n+1}+y_{n}}{2} − ω^{2} \frac{x_{n+1}+x_{n}}{2} ) .

3.21

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_{1} = 1 and y_{1} = −h which as before gives an error of the order of h^{2} since ΔR = h^{2}/2. However when we use the leapfrog method to the y variable we get x_{1} = 1 − h^{2} and y_{1} = −h which give x_{1}^{2} + y_{1}^{2} = 1 − h^{2} + h^{4} ≃ 1 − h^{2}. We also get an error of the order of h^{2}, but in this case ΔR = −h^{2}/2.

3.22

Let P_{n} = (X_{n}, Y_{n}) and P_{0} = (1, 0) . This problem is solved in two ways:

1. Using the leapfrog method in the x variable: The equations become: X_{n+1} = X_{n} + Y_{n} and Y_{n+1} = Y_{n} − X_{n+1}. Then P_{1} = (1, −1) , P_{2} = (0, −1) , P_{3} = (−1, 0) , P_{4} = (−1, 1) , P_{5} = (0, 1) , P_{6} = P_{0} = (1, 0).

2. Using the leapfrog method in the y variable: The equations are: Y_{n+1} = Y_{n} − X_{n} and X_{n+1} = X_{n} + Y_{n+1} and we get P_{1} = (−1, 0) , P_{2} = (−1, −1) , P_{3} = (0, −1) , P_{4} = (1, 0) , P_{5} = (1, 1) , P_{6} = P_{0} = (1, 0) . In either case we still get a solution of period 6 which is comparable to the period 2π ≃ 6.28 of the exact solution.

3.23

?

3.24

?

4.1

4.2

Expand eqn 4.5 to get λ^{2}−(a + d) λ+ad−bc = 0 and then use the quadratic formula to obtain the two eigenvalues.

4.3

DONE IN TEXTBOOK ON PAGE 58

4.4

The Jacobian is given by: J = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) whose characteristic polynomial

is λ^{2} + 1 = 0 . Hence λ = ±i

4.5

The eigenvalues given by eqn 4.12 can be written as λ_{1,2} = −\frac{b}{2} (1 ± \sqrt{1 − (\frac{2ω}{b} )^{2}}).

The term under the square root can never exceed 1, 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_{1} and λ_{2} to be real with opposite signs. For b > 2ω we get stable nodes (λ_{1} < λ_{2} < 0) if b is positive and unstable nodes if b is negative (λ_{1} > λ_{2} > 0) . When b = 2ω we either get a degenerate stable node (for b > 0 ) or a degenerate unstable node (for b < 0 ). When b < 2ω 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 > 0.

4.6

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 =

0 ie ( \begin{matrix}−λ & 1 \\ −ω^{2} & b − λ \end{matrix} ) v = ( \begin{matrix}0 \\ 0 \end{matrix} ) The solution is v_{i} = ( \begin{matrix}1 \\ λ_{i} \end{matrix} ) , i = 1, 2.

For b = 2.5ω > 0, λ_{1} = −4b/5 and λ_{2} = −b/5 we obtain the figure shown below which is similar to the middle plot of Fig 4.1.

4.8

The Jacobian J = ( \begin{matrix}1 & 1 \\ −2 & 4 \end{matrix} ) gives the characteristic equation (1 − λ) (4 − λ)+

2 = 0 with eigenvalues λ_{1} = 3 and λ_{2} = 2 and eigenvectors v_{1} = ( \begin{matrix}1 \\ 2 \end{matrix} )

and v_{2} = ( \begin{matrix}1 \\ 1 \end{matrix} ).

4.9

Add the two solutions given by 4.6 to obtain λ_{1} + λ_{2} = a + d =trace(J) and multiply them to get λ_{1}λ_{2} = ad − bc =det(J).

4.10

The Jacobian of the damped harmonic oscillator has trace(J) = −b and det(J) = ω^{2}. For the critically damped case b = 2ω. Hence trace(J) = ±2\sqrt{det (J)}.

4.11

Denote by τ the trace of J and by δ its determinant. The eigenvalues given by eqn 4.6 may be written as λ_{1,2} = \frac{τ±\sqrt{τ^{2}−4δ}}{2} . 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 1: τ > 0 and τ^{2} > 4δ > 0. In this case both λ_{1} and λ_{2} 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 2: τ > 0 and τ^{2} < 4δ. In such a case λ_{1} and λ_{2} 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 1 and 2. Moreover the couple will never be indifferent for each other like in cases 1 and 2. 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).

Since the eigenvectors may be written as v_{i} = ( −1^{\frac{b}{a−λ_{i}}} ) with i = 1, 2, we

end up in case 4a if for the positive eigenvalue we have \frac{b}{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^{2} > 4δ > 0 , where we get a stable node or τ^{2} < 4δ, where we get a stable spiral.

4.12

IMPOSSIBLE TO MAKE SUCH PLOTS. The trace and determinant are not sufficient to give the nature of the eigenvalues. For the 2 × 2 case this was possible since knowing the sum and product of the eigenvalues suffice to fully determine the eigenvalues.

4.13

The equation λ^{3} + 2.14λ^{2} + 1 = 0 is solved using a computer and we obtain λ_{1} = −2.33 and λ_{2,3} = 0.0925 ± 0.649i which verifies that the origin is a spiral saddle of index 2.

4.14

The equilibria can be found by solving dx/dt = dy/dt = dz/dt = 0 using eqns 4.33-4.35. It can be shown that the solutions are p_{1}^{∗} = (0, 0, 0) and p_{2,3}^{∗} = (±\sqrt{b(r − 1)}, ±\sqrt{b(r − 1)}, r − 1). Obviously the first bifurcation occurs when r = 1 at which the equilibria p_{2}^{∗} and p_{3}^{∗} are created. The Jacobian is

J(x, y, z) = ( \begin{matrix}−σ & σ & 0 \\ r − z & −1 & −x \\ y & x & −b \end{matrix} ) which gives the characteristic equations

(b + λ)(λ^{2} + (σ + 1)λ + σ(r − 1) = 0 for p_{1}^{∗} and λ^{3} + (σ + b + 1)λ^{2} + b(σ + r)λ + 2bσ(r − 1) = 0 for p_{2,3}^{∗}. For 1 < r < r^{∗} the equilibria are spiral nodes and for r > r^{∗} they become unstable spiral saddle points with index 2 (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_{2,3}^{∗}. The real and imaginary parts of the left hand side should both equal to zero which give ω^{2} = b (σ + r^{∗}) and ω^{2} (σ + b + 1) = 2bσ (r^{∗} − 1) respectively. Substituting ω^{2} in the second

equation and solving for r^{∗} we obtain r^{∗} = \frac{σ(b + σ + 3)}{σ − b − 1}

4.15

Solving the system of equations dx/dt = dy/dt = dz/dt = 0 we obtain two equilibria, namely p_{1,2}^{∗} = (∓1, ±1, 0) . The Jacobian is given by

J(x, y, z) = ( \begin{matrix}−1 & −1 & 0 \\ −z & 0 & −x \\ y & x & 0 \end{matrix} ). Hence J(p_{1,2}^{∗}) = ( \begin{matrix}−1 & −1 & 0 \\ 0 & 0 & ±1 \\ ±1 & ∓1 & 0 \end{matrix} ). For

both p_{1}^{∗} and p_{2}^{∗} the Jacobian has the same characteristic polynomial λ^{3} + λ + λ^{2} + 2 = 0. Solving this equation using a computer we get λ_{1} = −1.35 and λ_{2,3} = 0.177 ± 1.20i.

4.16

Use eqns 4.36-4.38 to solve dx/dt = dy/dt = dz/dt = 0. These give x^{∗} = az^{∗} , y^{∗} = −z^{∗} and z^{∗} = \frac{c±\sqrt{c^{2}−4ab}}{2a} . The Jacobian is given by J(x, y, z) = ( \begin{matrix}0 & −1 & −1 \\ 1 & a & 0 \\ z & 0 & x − c \end{matrix} ), whose characteristic equation is:

λ^{3} − λ^{2} (\frac{b}{z^{∗}} + a) + λ (z^{∗} + \frac{ba}{z^{∗}} + 1) − az − \frac{b}{z^{∗}} = 0.

5.1

Working as in section 5.1.1 on page 81 the result is λ =ln 10

5.2

Since we have |f′(x_{n})| = A > 0 we get λ = ln |A| which is negative for 0 < A < 1 so there exists a stable fixed point for 0 < A < 1. For A > 1 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 1 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^{∗} = 1−1/A the probability distribution may take the form P (X) = δ(X −X^{∗}) , where δ is the Dirac delta function. So using λ = ∫_{0}^{1} P (X) ln |A(1 − 2X)| dX we obtain λ = ln |A(1 − 2X^{∗})| = ln |2 − A| .

5.5

Start from g(X) = f(f(X)) = A^{2}X − A^{2} (A^{2} + 1) X^{2} + 2A^{3}X^{3} − A^{3}X^{4} which gives g′(X) = A^{2} − 2A^{2} (A^{2} + 1) X + 6A^{3}X^{2} − 4A^{3}X^{3} . Then the probability distribution is given by P (X) = \frac{1}{2} [δ(X − X_{+}^{∗}) + δ(X − X_{−}^{∗})] with X_{±}^{∗} being given in eqn 2.7. Use λ = ∫_{0}^{1} P (X) ln |g′(X^{∗})| dX to obtain λ = \frac{1}{2} 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′(f(X_{±}^{∗}))f′(X_{±}^{∗}) = f′(X_{∓}^{∗})f′(X_{±}^{∗}) = A^{2}(1 − 2X_{∓}^{∗})(1 − 2X_{±}^{∗}) = A^{2}[1 − 2(X_{∓}^{∗} + X_{±}^{∗}) + 4X_{∓}^{∗}X_{±}^{∗}] . From equation (2.6) X_{+}^{∗} + X_{−}^{∗} = [A(A + 1)]/A^{2} and X_{−}^{∗}X_{+}^{∗} = (A + 1)/A^{2}. Hence g′(X_{±}^{∗}) = A^{2}[1 − 2A(A + 1)/A^{2} + 4 (A + 1) /A^{2}] = 4 + 2A − A^{2} and therefore λ = ln |4 + 2A − A^{2}|.

5.6

The fixed point for A = 2 is X^{∗} = 1−1/2 = 1/2 . 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} = 0.1 we get X_{1} = 0.2952 , X_{2} = 0.41611392 , X_{3} = 0.4859262512 , X_{4} = 0.4996038592 , X_{5} = 0.4999996862 and X_{6} = 0.5000000000 .

5.7

Substitution of A = 1+\sqrt{5} in eqn 2.7 we get X_{+}^{∗} = \frac{3+\sqrt{5}}{2(1+\sqrt{5})}and X_{−}^{∗} = \frac{1}{2}. 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_{±}^{∗}) = 0.

5.8

???????

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 (2α) .

5.10

\frac{ΔR_{n+1}^{2}}{ΔR_{n}^{2}} = \frac{(aΔX_{n} + bΔY_{n})^{2} + (cΔX_{n} + dΔY_{n})^{2}}{ΔX_{n}^{2} + ΔY_{n}^{2}} = \frac{(a + bYn^{′})^{2} + (c + dYn^{′})^{2}}{1 + Y_{n}^{′2}}

with Yn^{′} = \frac{ΔY_{n}}{ΔX_{n}} = \frac{cΔX_{n−1} + dΔY_{n−1}}{aΔX_{n−1} + bΔY_{n−1}} = \frac{c + dYn−1^{′}}{a + bYn−1^{′}} which is the same

as eqn 5.12. Substitution of \frac{ΔR_{n+1}^{2}}{ΔR_{n}^{2}} in eqn 5.10 yields eqn 5.11.

5.11

Using the notation of the text we get |a| = A , b = c = 0 and |d| =

B. Substitution in eqn 5.11 yields λ_{1} = lim_{N→∞} \frac{1}{2N} ∑_{n=0}^{N−1} ln [\frac{A^{2}+B^{2}Y_{n}^{′2}}{1+Y_{n}^{′2}} ] =

\frac{1}{2} lim_{n→∞} ln [B^{2} + \frac{A^{2}−B^{2}}{1+Y_{n}^{′2}} ] with |Yn^{′}| = |^{\frac{BY′_{n−1}}{A}} | = |Y_{0}| |\frac{B}{A} |^{n} . Then we have :

lim_{n→∞} |Yn^{′}| = \{ \begin{matrix}+∞ & for & A < B \\ 0 & for & A > B \end{matrix} Substitution of the limit in the equa-

tion for λ_{1} gives λ_{1} = ln[max (A, B)]. Also use λ_{1} + λ_{2} = det (J) = |AB| to determine λ_{2}.

5.12

A.1.1 Logistic Map : J(X) = A(1 − 2X) A.1.2 Sine Map : J(X) = Aπ cos πX A.1.3 Tent Map : J(X) = A sign(1/2 − X) A.1.4 Linear Congruential Generator : J(X) = A(mod C) A.1.5 Gauss Map : J(X) = −1/X^{2}(mod 1) A.2.1 Hènon Map : J(X, Y) = ( \begin{matrix}−2aX & b \\ 1 & 0 \end{matrix} ) A.2.2 Lozi Map : J(X, Y) = ( \begin{matrix}−a sign(X) & b \\ 1 & 0 \end{matrix} )

A.2.3 Tinkerbell Map : J(X, Y) = ( \begin{matrix}2X + a & −2Y + b \\ 2Y + c & 2X + d \end{matrix} )

A.2.4 Burgers Map : J(X, Y) = ( \begin{matrix}a & −2Y \\ Y & X + b \end{matrix} ) A.2.5 Kaplan-Yorke Map : J(X, Y) = ( \begin{matrix}a & 0 \\ −4π sin(4πX) & b \end{matrix} ) (mod 1) A.2.6 Dissipative Standard Map: J(X, Y) = ( \begin{matrix}1 & 1 \\ k cos X(mod 2π) & b \end{matrix} ) A.3.1 Chirikov Map : J(X, Y) = ( \begin{matrix}1 & 1 \\ k cos(X) & 1 \end{matrix} ) (mod 2π) A.3.2 Hènon Area-preserving Quadratic Map : J(X, Y) = ( \begin{matrix}cos α + 2X sin α & − sin α \\ sin α − 2X cos α & cos α \end{matrix} ) A.3.3 Arnold Cat Map: J(X, Y) = ( \begin{matrix}1 & 1 \\ 1 & k \end{matrix} ) (mod 1) A.3.4 Gingerbreadman Map : J(X, Y) = ( \begin{matrix}sign(X) & −1 \\ 1 & 0 \end{matrix} ) A.3.5 Stochastic Web Map : J(X, Y) = ( \begin{matrix}cos α − k sin α cos X & − sin α \\ sin α + k cos α cos X & cos α \end{matrix} ) A.3.6 Lorenz Simplest 3-D Map : J(x, y, z) = ( \begin{matrix}Y & X & −1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} )

5.13

Using eqn 5.20 we get λ = a (1 − 2x) for the local Lyapunov exponent. Since at x^{∗} = 1 we have f (x^{∗}) = 0 the global Lyapunov exponent is λ = −a.

5.14 5.15

Since J(x, y) = ( \begin{matrix}0 & 1 \\ −1 − 2bxy & b(1 − x^{2}) \end{matrix} ), we have det(J) = 2bxy + 1.

For area expansion we require det(J) > 0, ie xy > −1/2b and for area contraction we must have xy < −1/2b.

5.16

INCOMPLETE We have a = d = 0 , b = c = 1 . From eqn 5.23 we get λ_{1} + λ_{2} = 0 and

from eqn 5.21 λ_{1} = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} \frac{2y′}{1 + y^{′2}} dt with \frac{dy′}{dt} = 1 − y^{′2} for Δt → 0

∫ \frac{1}{1 − y^{2}} dy = arctanh y

∫ \frac{2 tanh t}{1 + tanh^{2} t}dt = −\frac{1}{2} ln (−1 + tanh t)−\frac{1}{2} ln (1 + tanh t)+\frac{1}{2} ln (1 + tanh^{2} t)

lim_{t→∞} (−\frac{1}{2} ln (−1 + tanh t) − \frac{1}{2} ln (1 + tanh t) + \frac{1}{2} ln (1 + tanh^{2} t)) = ∞

5.17 5.18

Since \frac{∂}{∂x} [σ (y − x)] + \frac{∂}{∂y} [rx − y − xz] + \frac{∂}{∂z} [xy − bz] = −b − 1 − σ < 0 for

all σ, b > 0 we have volume contraction with V (t) ∼ e^{−(b+1+σ)t} .

5.19

A.4.1 Damped Driven Pendulum :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ − cos x & −b & A cos Ωz \\ 0 & 0 & 0 \end{matrix} )

A.4.2 Driven Van Der Pol Oscillator :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ −1 − 2bxy & b(1 − x^{2}) & A cos Ωz \\ 0 & 0 & 0 \end{matrix} )

A.4.3 Ueda Attractor :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ −3x^{2} & −b & A cos Ωz \\ 0 & 0 & 0 \end{matrix} )

A.5.1 Lorenz Attractor :

J(x, y, z) = ( \begin{matrix}−σ & σ & 0 \\ r − z & −1 & −x \\ y & x & −b \end{matrix} )

A.5.2 Rössler Attractor :

J(x, y, z) = ( \begin{matrix}0 & −1 & −1 \\ 1 & a & 0 \\ z & 0 & x − c \end{matrix} )

A.5.3 Diffusionless Lorenz Attractor :

J(x, y, z) = ( \begin{matrix}−1 & −1 & 0 \\ −z & 0 & −x \\ y & x & 0 \end{matrix} )

A.5.4 Chua’s circuit :

J(x, y, z) = ( \begin{matrix}α(b − 1 + 1/2(a − b)(sign(x + 1) − sign(x − 1)) & α & 0 \\ 1 & −1 & 1 \\ 0 & −β & 0 \end{matrix} )

A.5.5 Moore-Spiegel Oscillator :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ −T − 2Rxy & −(T − R + Rx^{2}) & −1 \end{matrix} )

A.5.6 Thomas’ Cyclically Symmetric Attractor :

J(x, y, z) = ( \begin{matrix}−b & cos y & 0 \\ 0 & −b & cos z \\ cos x & 0 & −β \end{matrix} )

A.5.7 Simplest Quadratic Chaotic Flow :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ −1 & 2y & −a \end{matrix} )

A.5.8 Simplest Cubic Chaotic Flow :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ y^{2} − 1 & 2yx & −a \end{matrix} )

A.5.9 Simplest Piecewise Linear Chaotic Flow :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ sign(x) & −1 & −a \end{matrix} )

A.6.1 Simplest Driven Chaotic Flow :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 3x^{2} & 0 & cos Ωz \\ 0 & 0 & 0 \end{matrix} )

A.6.2 Nosé-Hoover Oscillator :

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ −1 & z & y \\ 0 & −2y & 0 \end{matrix} )

A.6.3 Labyrinth Chaos :

J(x, y, z) = ( \begin{matrix}0 & cos y & 0 \\ 0 & 0 & cos z \\ cos x & 0 & 0 \end{matrix} )

A.6.4 Hènon-Heilles System :

J(x, y, v, w) = ( \begin{matrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ −1 − 2y & −2x & 0 & 0 \\ −2x & 2y − 1 & 0 & 0 \end{matrix} )

5.20

A. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} zdt = ⟨z⟩ Numerical calculation is needed. B,C,E,I,L,M,R,S. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} −1dt = −1 D,O. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} xdt = ⟨x⟩ Numerical calculation is needed.

F,H,Q. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} (0.5 − 1) dt = −0.5

G. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} (0.4 − 1) dt = −0.6

J,N. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} −2dt = −2

K. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} (y − 1 + 0.3) dt = ⟨y⟩ − 0.7 Numerical calculation is needed. P. ∑ λ = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} (2y) dt = 2 ⟨y⟩Numerical calculation is needed.

5.21

Since one of the eigenvalues is positive we have a strange attractor of dimension greater than 2.

5.22

D_{KY} = 3 + \frac{1}{27} (0.11 + 0.02) = 3. 004 8

5.23

D_{KY} = 4 + \frac{1}{0.6} (0.5 + 0.1 + 0 − 0.3) = 4. 5

5.24

We first have to find the equation of a parabola y = ax^{2} + bx + c which passes through the points (0, p_{1}) , (1, p_{2}) and (2, p_{3}) . Solving the resulting system of three equations in three unknowns we get

y = (\frac{1}{2}p_{1} − p_{2} + \frac{1}{2}p_{3}) x^{2} + (4p_{2} − \frac{3}{2}p_{3} − \frac{5}{2}p_{1}) x + p_{3} − 3p_{2} + 3p_{1}.

Hence we get : A.5.1 p_{1} = p_{2} = 0.9056, p_{3} = −13.6667 and dimension ≃ 2.111 A.5.2 p_{1} = p_{2} = 0.0714, p_{3} = −5.3229 and dimension ≃ 2.025 A.5.3 p_{1} = p_{2} = 0.2101, p_{3} = −1 and dimension ≃ 2.273 A.5.4 p_{1} = p_{2} = 0.3271, p_{3} = −2.1926 and dimension ≃ 2.213 A.5.5 p_{1} = p_{2} = 0.1109, p_{3} = −1 and dimension ≃ 2.171 A.5.6 p_{1} = p_{2} = 0.0349, p_{3} = −0.54 and dimension ≃ 2.109 A.5.7 p_{1} = p_{2} = 0.0551, p_{3} = −2.017 and dimension ≃ 2.051 A.5.8 p_{1} = p_{2} = 0.0837, p_{3} = −2.028 and dimension ≃ 2.074 A.5.9 p_{1} = p_{2} = 0.0362, p_{3} = −0.6 and dimension ≃ 2.103 A.6.2 p_{1} = p_{2} = 0.0138, p_{3} = 0 and dimension = 3 A.6.3 p_{1} = p_{2} = 0.1402, p_{3} = 0 and dimension = 3 .

5.25 5.26

λ = ln(a(modc))

6.1

In order to determine the inverse map interchange X_{n} and X_{n+1}, Y_{n} and Y_{n+1} to get Y_{n} = bX_{n+1} and X_{n} = 1+Y_{n+1} −aX_{n+1}^{2}. Solving for Y_{n+1} and X_{n+1} gives the inverse map. Answer: X_{n+1} = Y_{n}/b, Y_{n+1} = X_{n}+aY_{n}^{2}/b−1.

6.2

The fixed points satisfy X_{n+1} = X_{n} and Y_{n+1} = Y_{n} . By eqn 6.2, this means that X_{n+1} = X_{n} = Y_{n+1} = Y_{n} = X^{∗} = Y^{∗}. Eqn 6.1 becomes (a_{4} + a_{5} + a_{6}) X^{∗2} + (a_{2} + a_{3} − 1) X^{∗} + a_{1} = 0 . This equation has real solutions when Δ = (a_{2} + a_{3} − 1)^{2} −4a_{1} (a_{4} + a_{5} + a_{6}) ≥ 0. Under this condition the fixed points are given by X^{∗} = Y^{∗} = (1 − a_{2} − a_{3} ± \sqrt{Δ}) / (2a_{4} + 2a_{5} + 2a_{6}) .

6.3

The fixed points of the Hénon Map are given by the equation 1.4X^{2} + 0.7X − 1 = 0, whose solution is X = −1. 131 4 and X = 0. 631 35. Their position relative to the attractor is shown in the figure below:

6.4

Interchange X_{n} and X_{n+1}, Y_{n} and Y_{n+1} to get : X_{n} = a_{1} + a_{2}X_{n+1} + a_{3}Y_{n+1} + a_{4}X_{n+1}^{2} + a_{5}X_{n+1}Y_{n+1} and X_{n+1} = Y_{n} . Solving for Y_{n+1} and

X_{n+1} gives the inverse map. Answer: X_{n+1} = Y_{n} , Y_{n+1} = \frac{X_{n}−a_{1}−a_{2}Y_{n}−a_{4}Y_{n}^{2}}{a_{3}−a_{5}Y_{n}} .

6.5

As in the previous problem use X_{n+1} = Y_{n} and X_{n} = a_{1} + a_{2}X_{n+1} + a_{3}Y_{n+1} + a_{4}X_{n+1}^{2} + a_{5}X_{n+1}Y_{n+1} + a_{6}Y_{n+1}^{2}. The second equation becomes a_{6}Y_{n+1}^{2} + (a_{5}Y_{n} + a_{3}) Y_{n+1} + X_{n} − a_{1} − a_{2}Y_{n} − a_{4}Y_{n}^{2} = 0, whose solution is:

Y_{n+1} = \frac{1}{2a_{6}} [−a_{5}Y_{n} − a_{3} ± \sqrt{(a_{5}Y_{n} + a_{3})^{2} − 4a_{6} (X_{n} − a_{1} − a_{2}Y_{n} − a_{4}Y_{n}^{2})}]

6.6

Antisymmetry about r = 0.5 can be verified if we notice that f_{i} (0.5 − r) = −f_{i} (0.5 + r) for i = 1, 2, 3. It is obvious that f_{1} is infinite for r = 0 and r = 1 since tan (±π/2) = ±∞. 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 −1 < f < 1 since we have f(0.5±0.25) = ±1.

6.7 6.8

Eqn 6.6 may be also written as a system of differential equations of the form: ẋ = y , ẏ = z and ż = −\frac{3}{5}z − y + |x| − 1. Equilibrium points satisfy ẋ = ẏ = 0 . This happens only when y = z = 0 and x = ±1. The Jacobian takes the form:

J(x, y, z) = ( \begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ sign(x) & −1 & −a \end{matrix} )

When x = −1 the eigenvalues are −0.58846, 0.59423 ± 1.1603i and when x = 1 the eigenvalues are 0.83555, −0.11778 ± 1.0876i.

6.9

The following calculations are valid as long as the a_{4} and a_{5} are posi-

tive numbers. Eqn 6.17 becomes \frac{α}{β^{2}} \frac{dx^{′2}}{dt^{′2}} + a_{1} (α^{2}x^{2} − a_{4}) \frac{α}{β} \frac{dx′}{dt′} + a_{5}αx′ =

a_{2} sin (a_{3}t′) ⇒

\frac{dx^{′2}}{dt^{′2}} +a_{1}βα^{2} (x^{2} − \frac{a_{4}}{α^{2}} ) \frac{dx′}{dt′} +a_{5}β^{2}x′ = \frac{β^{2}}{α} a_{2} sin (a_{3}βt′) . Choose β so that

a_{5}β^{2} = 1 ie β = ±\sqrt{a_{5}^{−1}} and α = ±\sqrt{a_{4}}. In such a case the equation be-

comes \frac{dx^{′2}}{dt^{′2}} + a′_{1} (x^{2} − 1) \frac{dx′}{dt′} + x′ = a′_{2} sin (a′_{3}t′) with a′_{1} = ±a_{1}a_{4}\sqrt{a_{5}^{−1}} ,

a′_{2} = ±\frac{a_{2}}{a_{5}\sqrt{a_{4}}} and a′_{3} = ±a_{3}\sqrt{a_{5}^{−1}}.

6.10

In this problem we have to find the a ’s and b’s for which X_{n} = b_{1} tanh(a_{10}+ a_{11}X_{n−1}) + b_{2} tanh(a_{20} + a_{21}X_{n−1}). The information we have in order to solve the problem is that when X_{n−1} is 0 or 1 then X_{n} = 0 and that the maximum should be at X_{n} = 1 when X_{n−1} = 1/2. Since we have six unknowns and 4 equations we may have two free variables. The equations are: b_{1} tanh(a_{10})+b_{2} tanh(a_{20}) = b_{1} tanh(a_{10} +a_{11})+b_{2} tanh(a_{20} +a_{21}) = 0 , b_{1}a_{11} − b_{1}a_{11} tanh^{2}(a_{10} + a_{11}/2) + b_{2}a_{21} − b_{2}a_{21} tanh^{2}(a_{20} + a_{21}/2) = 0 and b_{1} tanh(a_{10} + a_{11}/2) + b_{2} tanh(a_{20} + a_{21}/2) = 1

6.11

Cannot determine the LE accurately enough from the diagram in order to compare them with those in the appendix.

6.12

Since we are interested only for the maximum, fit the parabola x = at^{2} + bt+c through the points (−1, x_{1}) , (0, x_{2}) and (1, x_{3}) . In this case we would get a = \frac{1}{2} (x_{1} + x_{3}) , b = \frac{1}{2} (x_{3} − x_{1}) and c = x_{2}. The maximum of the parabola occurs at t = −b/ (2a) and equals to x_{max} = c − b^{2}/4a.

6.13

The intersection with the Poincaré section occurs only when z_{1}z_{2} < 0 . Each point (x, y, z) on the line joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is given by x = x_{1} + λ (x_{2} − x_{1}) , y = y_{1} + λ (y_{2} − y_{1}) and z = z_{1} + λ (z_{2} − z_{1}) with 0 ≤ λ ≤ 1. For λ = \frac{z_{1}}{z_{1} − z_{2}} we have z = 0, ie an intersection

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_{1} + \frac{z_{1} (x_{2} − x_{1})}{z_{1} − z_{2}} and

y = y_{1} + \frac{z_{1} (y_{2} − y_{1})}{z_{1} − z_{2}}.

6.14 6.15 6.16 6.17 6.18

The fixed points of the Lozi map satisfy X^{∗} = Y^{∗} and 0.5X^{∗} = 1−1.7 |X^{∗}|. This gives either 0.5X^{∗} = 1 − 1.7X^{∗} ⇒ X^{∗} = Y^{∗} = 0. 454 55 or 0.5X^{∗} = 1 + 1.7X^{∗} ⇒ X^{∗} = Y^{∗} = −0. 833 33.

6.19

The fixed points satisfy X^{∗} = 10X^{∗2} − 10Y^{∗2} − 6Y^{∗} and X^{∗} = \frac{Y^{∗}}{4 (Y^{∗} + 1)}.

Solving these equations we finally obtain the pairs (0, 0) and (1. 027 6, −1. 321 5) .

6.20

The fixed points satisfy X = X^{2} − Y^{2} − 1 and Y (1 − 2X) = 0 ⇒ (i) Y = 0 and (ii) X = 1/2 . Then for Y = 0 (case (i))we get the equation X^{2} −X −1 = 0 whose solutions are X_{1}^{∗} = \frac{1}{2} \sqrt{5}+ \frac{1}{2} and X_{2}^{∗} = \frac{1}{2} − \frac{1}{2} \sqrt{5}. The Jacobian of the map has characteristic polynomial (2X − λ)^{2} + 4Y^{2} = 0 and eigenvalues λ = 2X ± 2Y i. Therefore X_{1}^{∗} is unstable since λ > 0, whereas X_{2}^{∗} is stable. For X = 1/2 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^{2} − Y^{2} − 1) = Y ⇒ Y [4X (X^{2} − Y^{2} − 1) − 1] = 0. The equation 4X (X^{2} − Y^{2} − 1) − 1 = 0 together with the X component of the second iterate map do not have real solutions. However when Y = 0, using (X^{2} − Y^{2} − 1)^{2} − 4X^{2}Y^{2} − 1 = X we get X(X^{3} − 2X − 1) = 0. This gives X = 0 and X = −1. ,ie in this case we have a 2-period cycle with oscillations between the points (0, 0) and (−1, 0) .

6.21

By definition, the derivative of f with respect to Z̄ is \frac{∂f}{∂Z̄} = \frac{1}{2} (\frac{∂}{∂X} + i\frac{∂}{∂Y} ) f =

\frac{1}{2} (\frac{∂F}{∂X} − \frac{∂G}{∂Y})+ \frac{i}{2} (\frac{∂G}{∂X} + \frac{∂f}{∂Y} ) . Since f is a function of Z only we have

\frac{∂f}{∂Z̄} = 0. This condition requires \frac{∂F}{∂X} = \frac{∂G}{∂Y} and \frac{∂G}{∂X} = −\frac{∂F}{∂Y} .

6.22

Hénon Map: \frac{∂F}{∂X} = −2aX ≠ 0 = \frac{∂G}{∂Y}. Lozi map: \frac{∂F}{∂Y} = 0.5 ≠ 1 = \frac{∂G}{∂X} .

Julia set: \frac{∂F}{∂X} = \frac{∂G}{∂Y} = 2X and \frac{∂G}{∂X} = −\frac{∂F}{∂Y} = 2Y.

6.23

The Cauchy-Riemann equations require\frac{∂F}{∂X} = a_{2} +2a_{4}X +a_{5}Y = 0 = \frac{∂G}{∂Y}.

Hence we must have a_{2} = a_{4} = a_{5} = 0. Also \frac{∂F}{∂Y} = a_{3} + a_{5}X + 2a_{6}Y =

−1 = −\frac{∂G}{∂X} which gives a_{3} = −1, a_{5} = a_{6} = 0. Therefore in this case we must have X_{n+1} = a_{1}−Y_{n} and Y_{n+1} = X_{n} which is obviously a non-chaotic system.

6.24

Let F = b_{1} + b_{2}X + b_{3}Y + b_{4}X^{2} + b_{5}XY + b_{6}Y^{2} and G = c_{1} + c_{2}X +

c_{3}Y + c_{4}X^{2} + c_{5}XY + c_{6}Y^{2}. The Cauchy-Riemann equations require\frac{∂F}{∂X} =

b_{2} + 2b_{4}X + b_{5}Y ≡ c_{3} + c_{5}X + 2c_{6}Y = \frac{∂G}{∂Y} and \frac{∂F}{∂Y} = b_{3} + b_{5}X + 2b_{6}Y ≡

−c_{2} −2c_{4}X −c_{5}Y = −\frac{∂G}{∂X} . Equating the coefficients with the same powers of X and Y we get : b_{2} = c_{3} = a_{2} , b_{4} = c_{5}/2 = a_{4} , b_{5}/2 = c_{6} = a_{5} , b_{3} = −c_{2} = a_{3} , b_{5}/2 = −c_{4} = a_{5} and b_{6} = −c_{5}/2 = −a_{4}. If we also let b_{1} = a_{1} and c_{1} = a_{6} we obtain eqns 6.18 and 6.19.

7.1

This is done in the textbook on page 132. Eigenvalues?

7.2

Done in textbook (p.133). Eigenvalues?

7.3

When y = 1, ẏ = 0 for all values of the parameter a. Now, expand about y = 1 by making a change of variable from y to v = y − 1. In this case v is small and we have ẏ = v̇ = a ln(v + 1) + v ≈ a[v − \frac{1}{2}v^{2} + ...] + v = (a + 1)v − \frac{1}{2}av^{2} + O(v^{3}). Hence a transcritical bifurcation occurs when a = −1. In order to write the last equation in normal form, let v = cξ which transforms it to ξ̇ = (a + 1)ξ − \frac{1}{2}caξ^{2} + O(v^{3}). Letting c = 2/a and μ = (a + 1) we finally obtain the normal form with ξ = \frac{1}{2}a (y − 1) .

7.4

The fixed points satisfy f = μx^{∗} −x^{∗3} = 0. Solving for x^{∗} we obtain x^{∗} = 0 and x^{∗} = ±\sqrt{μ}. Then \frac{df}{dx^{∗}} = μ − 3x^{∗2} and for x^{∗} = 0 we get \frac{df}{dx^{∗}} = μ. Thus for μ > 0, x^{∗} is unstable and for μ < 0 it is stable. However for the points x^{∗} = ±\sqrt{μ}, which exist only for μ > 0 we get \frac{df}{dx^{∗}} = −2μ > 0 which are stable.

7.5

The fixed points satisfy f = μx^{∗} +x^{∗3} = 0. Solving for x^{∗} we obtain x^{∗} = 0 and x^{∗} = ±\sqrt{−μ}. Then \frac{df}{dx^{∗}} = μ + 3x^{∗2} and for x^{∗} = 0 we get \frac{df}{dx^{∗}} = μ. Thus for μ > 0, x^{∗} is unstable and for μ < 0 it is stable. However for the points x^{∗} = ±\sqrt{−μ}, which exist only for μ < 0 we get \frac{df}{dx^{∗}} = −2μ > 0 which are unstable.

7.6

For a > 1 the only equilibrium point we have is x^{∗} = 0 which is unstable since \frac{df}{dx^{∗}} = a − cos x^{∗} > 0. For 0 < a < 1, the equilibrium points satisfy

\frac{sin x^{∗}}{x^{∗}} = a. INCOMPLETE-NEEDS FIGURE.

7.7

As shown in the text eqns 7.8 and 7.9 may be written in polar coordinates as \frac{dr}{dt} = r (μ − r^{2}) = f(r) and \frac{dθ}{dt} = 1. The equation f(r) = 0 is satisfied for the points r^{∗} = ±\sqrt{μ} which exist for μ > 0 and for r^{∗} = 0. For r^{∗} = 0 we have \frac{df}{dr} = μ which means that it is stable for μ < 0 and unstable for μ > 0. The points satisfying r^{∗} = ±\sqrt{μ} give \frac{df}{dr} = −μ which is always stable (given that μ > 0 ).

7.8

Multiply eqn 7.8 by x and eqn 7.9 by y and add them together. Then x\frac{dx}{dt} + y \frac{dy}{dt} = \frac{1}{2} \frac{dr^{2}}{dt} = r \frac{dr}{dt} = r^{2} (μ − r^{2}) .

7.9

The equation \frac{d^{2}x}{dt^{2}} + b (x^{2} − 1) \frac{dx}{dt} + x = a can be written as ẋ = y and ẏ = a − b (x^{2} − 1) y − x. The Jacobian of this system is thus:

J(x, y) = ( \begin{matrix}0 & 1 \\ −2bxy − 1 & −b(x^{2} − 1) \end{matrix} )

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^{2} − 1)] + 1 = 0 which has

solutions λ = ^{\frac{−b(a^{2}−1)±i\sqrt{4−b^{2}(a^{2}−1)^{2}}}{2}} = A(a, b) ± iB(a, b). We have Hopf

bifurcations when b = 0 since A(a, 0) = 0, B(a, 0) ≠ 0 and \frac{dA}{db} = −\frac{(a^{2}−1)}{2} which is greater than zero provided that a < 1. We may also have Hopf bifurcations when a = ±1 since A(±1, 0) = 0, B(±1, b) ≠ 0 and \frac{dA}{da} = ∓b which is greater than zero when b is positive for a = −1 and when b is negative for a = 1.

7.10

The equation \frac{dr}{dt} = r(μ + r^{2} − r^{4}) = f(r) gives the fixed points r_{1}^{∗} = 0 and r_{±}^{∗2} = \frac{1±\sqrt{1+4μ}}{2}r_{2,3}^{∗2} = \frac{1±\sqrt{1+4μ}}{2}. To classify their stability compute \frac{df}{dr} = μ+3r^{2} −5r^{4} = −4μ−2r^{2} +5(μ+r^{2} −r^{4}). Obviously for r_{1}^{∗} \frac{df}{dr} = μ which is unstable for μ > 0 and stable for μ < 0. We have three cases : (i) μ < −1/4 in which the only fixed point is r_{1}^{∗} which is stable. (ii) −1/4 < μ < 0 in which we find r_{+}^{∗} to be stable and r_{−}^{∗} to be unstable.(iii)μ > 0. The cycle r_{−}^{∗} vanishes, r_{+}^{∗} is still stable whereas r_{1}^{∗} becomes unstable.HYSTERESIS

7.11

HARD

7.12

Hilborn pages 257 and 258. Too hard to be included. At least a number of hints should be provided.

7.13

Write the second iterate map keeping the terms up to 3rd order : F (F (X)) = − (1 + μ) [− (1 + μ) X + X^{3}] + [− (1 + μ) X + X^{3}]^{3} ≃ X +2Xμ+Xμ^{2} −2X^{3} −4X^{3}μ−3X^{3}μ^{2} −X^{3}μ^{3} = (1 + μ)^{2} X −X^{3}2+4μ+ 3μ^{2}+μ ⇒ F (F (X)) ≃ (1 + μ)^{2} X−(1 + μ) (2 + 2μ + μ^{2}) X^{3}, which is identical to eqn 7.19. Then taking F (F (±\sqrt{μ})) = ± (1 + μ)^{2} \sqrt{μ}∓μ (1 + μ) (2 + 2μ + μ^{2}) \sqrt{μ} = ± [\sqrt{μ} − 3 (\sqrt{μ})^{5} − 3 (\sqrt{μ})^{7} − (\sqrt{μ})^{9}] . Since μ is small then we may neglect the 5th, 7th and 9th powers of \sqrt{μ} to get F (F (±\sqrt{μ})) ≃ ±\sqrt{μ} which shows that ±\sqrt{μ} is approximately a fixed point.

7.14

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 G(Y) = F (X)−X^{∗}to get G(Y) = A (Y + ln A) e^{−(Y +ln A)} − ln A = (Y + ln A) e^{−Y} − ln A. The map Y_{n+1} = G(Y_{n}) has a fixed point at Y = 0. Expand G(Y) about 0 up to order 3 to get : G(Y) ≃ (Y + ln A) (1 − Y + \frac{1}{2}Y^{2} − \frac{1}{6}Y^{3})−ln A = (1 − ln A) Y +(\frac{1}{2} ln A − 1) Y^{2}+ (\frac{1}{2} − \frac{1}{6} ln A) Y^{3} INCOMPLETE F (X) ≃ ln A+(1 − ln A) (X − ln A)+\frac{1}{2} (−2 + ln A) (X − ln A)^{2}+\frac{1}{6} (3 − ln A) (X − ln A)^{3} = (1 + ln A − \frac{1}{2} ln^{3} A + \frac{1}{2} ln^{2} A) X−(1 + ln A − \frac{1}{2} (ln^{2} A)) X^{2}+(\frac{1}{2} − \frac{1}{6} ln A) X^{3}+ \frac{1}{6} ln^{4} A

7.15 7.16 7.17 7.18 7.19 7.20 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 8.1

\frac{dH}{dt} = ∑_{i} (\frac{∂H}{∂p_{i}} \frac{dp_{i}}{dt} + \frac{∂H}{∂q_{i}} \frac{dq_{i}}{dt} ) = ∑_{i} (−\frac{∂H}{∂p_{i}} \frac{∂H}{∂q_{i}} + \frac{∂H}{∂q_{i}} \frac{∂H}{∂p_{i}} ) = 0

8.2

H = \frac{p^{2}}{2mL^{2}} + mgL (1 − cos x) . In this case we have q ≡ x. Eqn 8.5 gives :

\frac{dp}{dt} = −\frac{∂H}{∂θ} = −mgL sin x = mL^{2} \frac{d^{2}x}{dt^{2}} ⇒ \frac{dv}{dt} = −ω^{2} sin x with v = \frac{dx}{dt} and

ω = \sqrt{\frac{g}{L}}.

8.3

The Jacobian Matrix is given by: J(x, v) = ( \begin{matrix}0 & 1 \\ −ω^{2} cos x & 0 \end{matrix} ) which, evaluated at x = π , v = 0 gives J(π, 0) = ( \begin{matrix}0 & 1 \\ ω^{2} & 0 \end{matrix} ). The characteristic polynomial is λ^{2} − ω^{2} = 0 which gives λ = ±ω ⟹ at x = π , v = 0 we have a saddle point.

8.4

Start from E−mgL = \frac{mL^{2}}{2} (\frac{dx}{dt})^{2}−mgL cos x. Let x_{0} be the maximum value of x, which satisfies cos x_{0} = −\frac{E−mgL}{mgL} . Hence \frac{L}{2g} (\frac{dx}{dt})^{2} = cos x − cos x_{0}. Solving for t : dt = \sqrt{\frac{L}{2g}} \frac{dx}{\sqrt{cos x−cos x_{0}}} . Integrate from 0 to x_{0} to get 1/4 of the period T which can be written as T = 4\sqrt{\frac{L}{2g}} ∫_{0}^{x_{0}} \frac{dx}{\sqrt{cos x−cos x_{0}}} = 2\sqrt{\frac{L}{g}} ∫_{0}^{x_{0}} _{\frac{dx}{\sqrt{sin^{2} \frac{1}{2} x−sin^{2} \frac{1}{2} x_{0}}}} . Make the substitution sin y = sin^{2} \frac{1}{2}x/ sin^{2} \frac{1}{2}x_{0} to convert it to T = 4\sqrt{\frac{L}{g}}K(sin \frac{1}{2}x_{0}) where K(s) = ∫_{0}^{π/2} \frac{ds}{\sqrt{1−k^{2} sin^{2} s}} is the complete elliptic integral of the first kind. Since sin \frac{1}{2}x_{0} = \sqrt{\frac{1−cos x_{0}}{2}} = \sqrt{\frac{E}{2mgL}}, use these formulas to get T_{E/mgL} : T_{0.1} = 6.364\sqrt{\frac{L}{g}} , T_{1.0} = 7.416\sqrt{\frac{L}{g}} , T_{2.0} = ∞.

8.5

In this case we have q_{1} ≡ x , q_{2} ≡ y , p_{1} ≡ v , p_{2} ≡ w . Using eqns

8.4 and 8.5 we get : \frac{dp_{1}}{dt} = \frac{dv}{dt} = −\frac{∂H}{∂q_{1}} = −\frac{∂H}{∂x} = −ω^{2} sin x + Ay ,

\frac{dp_{2}}{dt} = \frac{dw}{dt} = −\frac{∂H}{∂q_{2}} = −\frac{∂H}{∂y} = −Ω^{2}y + Ax , \frac{dq_{1}}{dt} = \frac{dx}{dt} = \frac{∂H}{∂p_{1}} = \frac{∂H}{∂v} = v

and \frac{dq_{2}}{dt} = \frac{dy}{dt} = \frac{∂H}{∂p_{2}} = \frac{∂H}{∂w} = w .

8.6

We have q_{1} ≡ x , q_{2} ≡ y , p_{1} ≡ v , p_{2} ≡ w . Use eqns 8.4 and 8.5 to get :

\frac{dq_{1}}{dt} = \frac{dx}{dt} = \frac{∂H}{∂p_{1}} = \frac{∂H}{∂v} = v , \frac{dq_{2}}{dt} = \frac{dy}{dt} = \frac{∂H}{∂p_{2}} = \frac{∂H}{∂w} = w , \frac{dp_{1}}{dt} = \frac{dv}{dt} =

−\frac{∂H}{∂q_{1}} = −\frac{∂H}{∂x} = −2xy−x and \frac{dp_{2}}{dt} = \frac{dw}{dt} = −\frac{∂H}{∂q_{2}} = −\frac{∂H}{∂y} = −y−x^{2}+y^{2}

.

8.7

LOOK UP FORMULA IN FORD’S REFERENCE

8.8

First note that x = \frac{χ + ψ}{2} , y = \frac{χ − ψ}{2} . Eqn 8.18 may be written as

2\frac{dv}{dt} = 2\frac{d^{2}x}{dt^{2}} = \frac{d^{2}χ}{dt^{2}} + \frac{d^{2}ψ}{dt^{2}} = −χ − ψ − χ^{2} + ψ^{2} and eqn 8.19 (with y^{2}

replaced by −y^{2}) as 2\frac{dw}{dt} = 2\frac{d^{2}y}{dt^{2}} = \frac{d^{2}χ}{dt^{2}} − \frac{d^{2}ψ}{dt^{2}} = −χ + ψ − χ^{2} − ψ^{2}. Solve

for \frac{d^{2}χ}{dt^{2}} and \frac{d^{2}ψ}{dt^{2}} to obtain eqns 8.21 and 8.22. SOLUTION??

8.9

Use eqn 5.24 with f = y , g = −x + yz and h = 1 − y^{2}, from which it

follows that \frac{1}{V} \frac{dV}{dt} = lim_{T →∞} \frac{1}{T} ∫_{0}^{T} zdt = ⟨z⟩

8.10

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.

8.11

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 5.24 together with f = −y, g = x + z and h = xz + 3y^{2} we get for the time-reversed system : \frac{1}{V} \frac{dV}{dt} = lim_{T →∞} \frac{1}{T} ∫_{T}^{0} xdt = − ⟨x⟩ .

8.12

Use \frac{dx}{dt} = y , \frac{dy}{dt} = −x+yz and \frac{dz}{dt} = 1−y^{2} .Since \frac{dy}{dt} = \frac{d^{2}x}{dt^{2}} = −x+\frac{dx}{dt}z ⇒

z = (\frac{d^{2}x}{dt^{2}} + x) /\frac{dx}{dt} . Taking the derivative of z with respect to t we

get \frac{dz}{dt} = (\frac{d^{3}x}{dt^{3}} + \frac{dx}{dt}) /\frac{dx}{dt} − \frac{d^{2}x}{dt^{2}} (\frac{d^{2}x}{dt^{2}} + x) / (\frac{dx}{dt})^{2} = 1 − (\frac{dx}{dt})^{2} .

Simplifying the last result yields eqn 8.31.

8.13

The differential equation for the simple pendulum can be written as \frac{dx}{dt} =

y = f (x, y) and \frac{dy}{dt} = k sin x = g (x, y) . For the case of the Leapfrog method see solutions to problems 8.20 and 8.21. For the Runge-Kutta method use k_{1xn} = Y_{n} , k_{1yn} = k sin X_{n} , k_{2xn} = Y_{n} + \frac{1}{2}k sin X_{n} and k_{2yn} = k sin(X_{n} + \frac{1}{2}Y_{n}). Then X_{n+1} = X_{n} + Y_{n} + \frac{1}{2}k sin X_{n} and Y_{n+1} = Y_{n} + k sin(X_{n} + \frac{1}{2}Y_{n}). The Jacobian of this map has determinant det J = (1 + \frac{1}{2}k cos X_{n}) (1 + \frac{1}{2}k cos(X_{n} + \frac{1}{2}Y_{n})) − k sin(X_{n} + \frac{1}{2}Y_{n}) ≠ 1. Hence the map in this case is not simplectic (area-preserving).

8.14

Use the Jacobian of the map is given at problem 5.12. whose determinant is det J = cos^{2} α + sin^{2} α = 1.

8.15

The Hénon area-preserving quadratic map is not invertible.

8.16

The Jacobian is given by J(X, Y) = ( \begin{matrix}a_{2} + 2a_{4}X + a_{5}Y & a_{3} + a_{5}X + 2a_{6}Y \\ 1 & 0 \end{matrix} )

The map is area preserving if | det J| = | − a_{3} − a_{5}X − 2a_{6}Y | = 1 for all X and Y. This is the case only if |a_{3}| = 1 and a_{5} = a_{6} = 0.

8.17

The Jacobian is J(X, Y) = ( \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} ) whose eigenvalues are λ_{±} = \frac{3±\sqrt{5}}{2}

with eigenvectors v_{±} = ( \begin{matrix}1 \\ λ_{±} − 1 \end{matrix} ) = ( 1_{\frac{1±\sqrt{5}}{2}} ) 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.

8.18

As indicated in the text on page 169 if we start from (−1, 3) → (−1, −1) → (3, −1) → (5, 3) → (3, 5) → (−1, 3) which is a period 5 cycle.

8.19

Use the Jacobian of the map is given at problem 5.12. whose determinant is det J = cos^{2} α + sin^{2} α = 1.

8.20

he differential equation for the simple pendulum \frac{d^{2}x}{dt^{2}} + k sin x = 0 can be transformed to a system of differential equations :\frac{dx}{dt} = y = f (x, y) and \frac{dy}{dt} = k sin x = g (x, y). The Leapfrog method in the y-variable can be written as Y_{n+1} = Y_{n} + hg(X_{n}, Y_{n}) and X_{n+1} = X_{n} + hf(X_{n}, Y_{n+1}). Since we are dealing with angles we may take the equations modulo 2π and if we use h = 1 we finally obtain eqns 8.40 and 8.41.

8.21

Use the Jacobian from the solution to problem 5.12 and calculate its determinant which equals det J = 1 + k cos X − k cos X = 1. Therefore the map is area-preserving.

8.22

The fixed points satisfy Y_{n}(mod2π) = 0 and k sin X_{n}(mod2π) = 0. The fixed points are Y^{∗} = 0 and X^{∗} = 0 or π. The Jacobian of the map is given in the solution of problem 5.12 with characteristic polynomial λ^{2} − λ (2 + k cos X) + 1 = 0. When X^{∗} = 0 we get λ^{2} − (2 + k) λ + 1 = 0

whose solution is λ = 1 + \frac{1}{2}k ± \frac{1}{2}k\sqrt{(\frac{4}{k} + 1)}. So for all k we have a saddle point.When X^{∗} = π we have the equation λ^{2} − (2 − k) λ + 1 = 0 whose

solution is λ = 1 − \frac{1}{2}k + \frac{1}{2}k\sqrt{(1 − \frac{4}{k} )}. In this case we have an unstable spiral point for 0 < k < 2, a stable spiral for 2 < k < 4 and a stable fixed point for k > 4.

8.23

Start with X_{0} = 0 and Y_{0} = π. The first iteration will give Y_{1} = π(mod2π) = π and X_{1} = π(mod2π) = π . The second iteration will give Y_{2} = π(mod2π) = π and X_{2} = 2π(mod2π) = 0 which correspond to the initial point (X_{0}, Y_{0}) .

8.24

One solution is (0, \frac{2π}{3} ) → (\frac{2π}{3} , \frac{2π}{3} ) → (\frac{4π}{3} , \frac{2π}{3} ) → (0, \frac{2π}{3} )

8.25

The sum of the eigenvalues equals to the logarithm of the determinant which in this case equals to log | det J| = log |b(1 + k cos X) − bk cos X| = log |b|.

8.26

In order to show that Q is conserved we have to show that A = (X_{n}^{2} + X_{n−1}^{2}+X_{n−2}^{2}−X_{n}X_{n−1}X_{n−2})−(X_{n−1}^{2}+X_{n−2}^{2}+X_{n−3}^{2}−X_{n−1}X_{n−2}X_{n−3}) = 0. A simplifies to A = X_{n}^{2} − X_{n−3}^{2} − X_{n−1}X_{n−2} (X_{n} − X_{n−3}) = X_{n}^{2} − (X_{n−1}X_{n−2} − X_{n})^{2} −X_{n−1}X_{n−2} (2X_{n} − X_{n−1}X_{n−2}) = 0. Hence Q is conserved.

9

10.1

?

10.2

Assume the polynomial to be of the form y = az^{2} + bz + c and passing through the points (−1, X_{n−2}) , (0, X_{n−1}) and (1, X_{n}). Solving the resulting system for a, b and c we get a = \frac{1}{2}X_{n} + \frac{1}{2}X_{n−2} − X_{n−1} , b = \frac{1}{2}X_{n} − \frac{1}{2}X_{n−2} and c = X_{n−1}. Using these coefficients set z equal to 2 to find X_{n+1}.

10.3

Assume the polynomial to be of the form y = az^{3} +bz^{2} +cz +d and passing trough the points (−2, X_{n−3}), (−1, X_{n−2}) , (0, X_{n−1}) and (1, X_{n}).Solving the resulting system for a, b, c and d we get a = −\frac{1}{2}X_{n−1} + \frac{1}{6}X_{n} + \frac{1}{2}X_{n−2} − \frac{1}{6}X_{n−3} , b = −X_{n−1} + \frac{1}{2}X_{n} + \frac{1}{2}X_{n−2} , c = \frac{1}{2}X_{n−1} + \frac{1}{3}X_{n} − X_{n−2} + \frac{1}{6}X_{n−3} and d = X_{n−1}. Using these coefficients set z equal to 2 to find X_{n+1}.

10.4

Assume the polynomial to be of the form y = az^{4} + bz^{3} + cz^{2} + dz + e and passing through the points (−2, X_{n−4}) , (−1, X_{n−3}) , (0, X_{n−2}) , (1, X_{n−1}) and (2, X_{n}). Solving the resulting system for a, b, c, d and e we get a = \frac{1}{4}X_{n−2} + \frac{1}{24}X_{n} + \frac{1}{24}X_{n−4} − \frac{1}{6}X_{n−1} − \frac{1}{6}X_{n−3} , b = \frac{1}{12}X_{n} − \frac{1}{12}X_{n−4} − \frac{1}{6}X_{n−1} + \frac{1}{6}X_{n−3} , c = −\frac{5}{4}X_{n−2} − \frac{1}{24}X_{n} − \frac{1}{24}X_{n−4} + \frac{2}{3}X_{n−1} + \frac{2}{3}X_{n−3} , d = −\frac{1}{12}X_{n} + \frac{1}{12}X_{n−4} + \frac{2}{3}X_{n−1} − \frac{2}{3}X_{n−3} and e = X_{n−2}. Using these coefficients set z equal to 3 to find X_{n+1}.

10.5

Assume the polynomial to be of the form y = az^{5} + bz^{4} + cz^{3} + dz^{2} + ez+f and passing through the points (−3, X_{n−5}), (−2, X_{n−4}), (−1, X_{n−3}), (0, X_{n−2}), (1, X_{n−1}) and (2, X_{n}). Solving the resulting system for a, b, c, d, e and f we get: a = \frac{1}{12}X_{n−2}− \frac{1}{12}X_{n−3}− \frac{1}{24}X_{n−1}+ \frac{1}{120}X_{n}− \frac{1}{120}X_{n−5}+ \frac{1}{24}X_{n−4} , b = \frac{1}{4}X_{n−2} + \frac{1}{24}X_{n} + \frac{1}{24}X_{n−4} − \frac{1}{6}X_{n−1} − \frac{1}{6}X_{n−3} , c = −\frac{5}{12}X_{n−2} + \frac{7}{12}X_{n−3} + \frac{1}{24}X_{n−1} + \frac{1}{24}X_{n} + \frac{1}{24}X_{n−5} − \frac{7}{24}X_{n−4} , d = −\frac{5}{4}X_{n−2} + \frac{2}{3}X_{n−3} + \frac{2}{3}X_{n−1} − \frac{1}{24}X_{n} − \frac{1}{24}X_{n−4} , e = −\frac{1}{20}X_{n} − \frac{1}{30}X_{n−5} + \frac{1}{4}X_{n−4} − X_{n−3} + \frac{1}{3}X_{n−2} + \frac{1}{2}X_{n−1} and f = X_{n−2}. Using these coefficients set z equal to 3 to find X_{n+1}.

10.6

The values to be predicted are: X_{11} = sin 11 = −0. 99999, X_{12} = sin 12 = −0. 53657 and X_{13} = sin 13 = 0. 42017 Using eqn 10.1 we obtain: X_{11} = −1. 8791, X_{12} = −3. 593 1 and X_{13} = −5. 686 Using eqn 10.2: X_{11} = −1.5002, X_{12} = −2.4564 and X_{13} = −3.4126 Using eqn 10.3: X_{11} = −1.3483, X_{12} = −1.47 and X_{13} = −0.3784 Using eqn 10.4: X_{11} = −0. 51206, X_{12} = 2. 7113 and X_{13} = 12.166 Using eqn 10.5: X_{11} = −0.23111, X_{12} = 4.397 and X_{13} = 18.035

10.7

Using the original map, the first 6 points are: X_{1} = 0.1, X_{2} = 0.36, X_{3} = 0 .9216, X_{4} = 0.28901, X_{5} = 0.821 93 and X_{6} = 0.58544. Knowing the first 3 points we may use eqn 10.1 to get X_{4} = 1.4832, X_{5} = 2.0448 and X_{6} = 2.6064. Applying eqn 10.2 to the same set of 3 points we get X_{4} = 1.7848, X_{5} = 2.9496 and X_{6} = 4.416.

10.8

Set j = m = 1 in eqns 10.7 and 10.8 and the result follows immediately.

10.9

a_{1} = \frac{X_{4}X_{3} + X_{3}X_{2} + X_{2}X_{1}}{X_{3}^{2} + X_{2}^{2} + X_{1}^{2}} = \frac{32 + 8 + 2}{16 + 4 + 1} = 2. Hence X_{n+1} = 2X_{n}.

10.10

a_{1} = \frac{X_{2}X_{1}}{X_{1}^{2}} = 5. Hence X_{n+1} = 5X_{n} so X_{3} = 25, X_{4} = 125 and X_{5} = 625

10.11

11

12.1

D = log 3/ log 5 = 0.6826

12.2

D = log 3/ log 7 = 0.6826

12.3

D = log 4/ log 3 = 1.262

12.4

Answers given in section 11.4

12.5

Answer given in section 11.5

12.6

At the n-th stage of the construction we need a minimum of 2^{m} boxes with L = (1/3)^{m}, so D_{0} = −\frac{− log 2^{m}}{log(1/3)^{m}} = \frac{log 2}{log 3} = 0.6309

12.7 12.8 12.9

By the definition of C(r) we have C(0) = 0. As r → ∞ all Heaviside functions evaluate to 1 since a circle with infinite radius contains all X_{n}’s. In that case the double sum evaluates to ∑_{j=1}^{N}(N − j) = 1 + 2 + ... + N − 1 =

\frac{N(N−1)}{2}. This gives C(∞) = 1.

12.10

C(r) = ∫_{0}^{r} f(r)dr, where f(r) is the distribution of points on the real line.

12.11

Referring to problem 12.10,let f(r) = 1. In this case the integral is performed on a surface so C(r) = ∫_{0}^{2π} ∫_{0}^{r} rdrdθ = πr^{2}. Hence D_{2} = \frac{d log(πr^{2})}{d log r} =

\frac{d[2 log r+log π]}{d log r} = 2

12.12 12.13 12.14 12.15

The answers for D = 1, 2, ..., 10 are 252, 631, 1585, 3982, 10000, 25119, 63096, 1.58× 10^{5}, 3.98 × 10^{5}, 10^{6} respectively.

12.16 12.17 12.18

D = log \frac{494}{298} log \frac{5}{3} = 0.989

12.19

D = log \frac{244036}{88804} log \frac{5}{3} = 1.979

12.20

D = \begin{matrix}log & \frac{5(a+b)−2}{3(a+b)−2} \\ log & \frac{5}{3} \end{matrix}