Instructor Solution Manual To Accompany Chaotic Dynamics: An Introduction Based on Classical Mechanics (Solutions) [1 ed.] 0521547830, 9780521839129, 0521839122, 9780521547833, 9780511803277, 0511803273

Citation preview

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems1

Solution 2.2 Let the area of the largest regular triangle inscribed into the island be A0 , which is also the area in the zeroth step of the Koch construction. In the first step, three smaller triangles are added, with an area of A0 /9 each. In the nth step, the number of new triangles of area A0 /9n is 3 × 4n−1 . The total area of all the   n−1 −n n 9 = A0 /3 ∞ small triangles is therefore A0 ∞ n=1 3 × 4 n=0 (4/9) . This is a geometrical series of quotient 4/9. Thus, the area increment is finite: A0 /(3(1 − 4/9)). The total area of the Koch island is (8/5)A0 . Solution 2.4 D0 = ln 3/ ln 2 = 1.585. Solution 2.5 (a) D0 = ln 4/ ln (1/r ); (b) D0 = ln 3/ ln 2 = 1.585; (c) D0 = ln 5/ ln 3 = 1.465. Solution 2.7 (a) D0 = ln 20/ ln 3 = 2.727; (b) D0 = 2. Solution 2.8 With r2 = r12 , the only positive solution of the quadratic equation √ r1D0 + r12D0 = 1 is D0 = ln[( 5 − 1)/2]/ ln r1 , which yields, for r1 = 1/2, D0 = 0.694. Solution 2.9 The fractal can be decomposed into five similar parts. Four of these are identical to the entire fractal reduced by a factor of r1 = 2/5, while the reduction factor for the fifth part is r2 = 1/5. The equation for the dimension is therefore 4(2/5) D0 + (1/5) D0 = 1, yielding D0 = 1.601. Solution 2.14 The area (volume) of the preserved rectangles in the nth step is Vn = nj=1 (1 − λ2 j ). (Its limiting value for λ = 0.6 is V = 0.517.) The smallest distance occurring in this step is the size of the holes in the squares situated at the   −n+1 j ; we therefore cover the corners of the original square, ε = λn π n−1 j=1 (1 − λ ) 2 set with intervals of this size. In analogy with the solution of Problem 2.13, we find that α = 2 ln λ/ln (λ/2). With λ = 0.6, α = 0.849. Solution 2.16 See Fig. 1. The possible box probabilities are again pm = p1m p2n−m ,   m = 0, 1, ..., n. The number of boxes carrying pm at level n is Nm = 2m mn (the total number of the boxes is 3n ). The logarithm of the total probability, Nm pm , is,

1 These

are the solutions which do not appear in the book.

1

CUUK255-Tell Et al

June 9, 2006

2

16:45

Solutions to the problems

P(x)

15

10

5

x Fig. 1. Distribution after the seventh step of construction. Similarly to Fig. 2.15 the typical intervals are shown under the graph, and the height of the columns on these typical intervals is marked by a dashed line. (The continuous support is not displayed, and only part of the central peak, of height 61.2, is visible.)

according to Stirling’s formula,

ln (Nm pm ) = n ln n − m ln m − (n − m) ln (n − m) + m ln 2 + m ln p1 + (n − m) ln p2 . The extremum belongs to a value m ∗ = 2np1 . The number of such boxes is N ∗ ≡ Nm ∗ , leading to ln N ∗ = −n(2 p1 ln p1 + p2 ln p2 ). Since the resolution at level n is ε = 3−n , D1 = −

2 p1 ln p1 + p2 ln p2 . ln 3

This is always less than unity for p1 = 1/3, and it is smaller, the smaller p1 is. The same result follows from definition (2.18) by writing it as N ∗ pm ∗ ln pm ∗ = D1 n ln 3, since N ∗ pm ∗ = 1. Solution 3.1 The point mass moves tangentially. Since the tangent forms a slope of inclination ϕ (see Fig. 3.1), the tangential acceleration is given by g sin ϕ, where g is gravitational acceleration. This is, at the same time, the peripheral acceleration, l ϕ; ¨ the Newtonian equation is therefore given by l ϕ¨ = g sin ϕ. For small ϕ, the sine can be approximated by its argument, and the equation becomes ϕ¨ = (g/l)ϕ, which √ is of the same type as equation (3.2), with repulsion parameter s0 = g/l.

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

Solution 3.3 For weak friction, λ± = ±s0 − α/2. The slopes of the asymptotes decrease by the same small quantity (α/2) corresponding to identical (yet small) angular rotation. Solution 3.4 The tangential acceleration is now −g sin ϕ (see Fig. 3.6). The equation of motion is therefore l ϕ¨ = −g sin ϕ. For small deviations, ϕ¨ = −(g/l)ϕ, √ which is an equation of the same type as (3.23), with natural frequency ω0 = g/l. Solution 3.6 The equation of the trajectories is given by (3.17), but both exponents, λ± , are now negative. Solution 3.7 At x ∗ = c, the derivative of the force is F  (x ∗ = c) = ac. The fixed √ point x ∗ = c is stable for ac < 0, and its natural frequency is ω0 = −ac. The √ fixed point is unstable for ac > 0, with repulsion parameter s0 = ac. Solution 3.9 The ‘force law’ is F(ϕ) = (g/l) sin ϕ ≈ (g/l)(ϕ − ϕ 3 /6) (see Problem 3.1). The second term is negligible compared with the first one as long as √ ϕ 6 = 2.450. In order to find the range of validity of the linear approximation, we have to agree in the meaning of ‘a number being much less than another’. A difference of an order of magnitude is a sensible choice. The linear approximation is √ therefore valid as long as the angular deviation is less than one-tenth of 6. This is a quarter of a radian, approximately 15◦ . In the initial phase, the velocity is small, √ friction is negligible and the deviation is ϕ(t) = ϕ0 exp (s0 t), where s0 = g/l. Consequently, the linear approximation is valid over time t=

15◦ 1 ln , s0 ϕ0

where ϕ0 should also be measured in degrees. The typical length of the pencil is √ l = 10 cm, thus s0 = g/l ≈ 10 s−1 . Accordingly, the motion of a pencil tipping over from initial angular deviations ϕ0 = 1◦ , 0.01◦ , and 0.0004◦ fulfils the linear equation (3.1) up to times t ≈ 0.3, 0.7 and 1.1 s, respectively. Solution 3.11 Point P−4 provides the initial condition for a point which departs from the origin to the left, then turns back, crosses the origin with a finite velocity and, finally, after turning back again, halts exactly at the origin. Solution 3.14 Around the origin, equation (P.4) is given by l0 is therefore given by F(x) = −2kx(1 − l0 / h). The frequency in the range h >   ω0 = 2k(h − l0 )/ h. The derivative of the force at x ∗ = l02 − h 2 is, from (P.4),  F  (x ∗ ) = −2k(1 − h 2 /l02 ). Thus, the frequency around x ∗ is ω0 = 2k(1 − h 2 /l02 ),  which, in the neighbourhood of the bifurcation point, is ω0 ≈ 4k(l0 − h)/ h. The period of small oscillations around the stable state therefore becomes longer and longer, whichever side the bifurcation point is approached from (Fig. 2). Solution 3.15 In a co-rotating frame the body is subject to a gravitational force and a centrifugal force. At a deflection angle, ϕ, from the vertical direction, the centrifugal acceleration pointing horizontally outwards is l sin ϕ 2 l. From the components perpendicular to the rod, the resulting force per unit mass is given by F(ϕ) = sin ϕ ( 2 l cos ϕ − g).

3

CUUK255-Tell Et al

June 9, 2006

4

16:45

Solutions to the problems

T

hc= l0

h

Fig. 2. Period T of small oscillations vs. h around the bifurcation point.

ϕ* π − 2

–π − 2

Ω Ω = Ωc

Fig. 3. Pitchfork bifurcation of a body moving along a rotating ring (Fig. 3.14(b)) in terms of the angular velocity, .

For small rotational angular velocities (when the expression in the brackets cannot change sign), there exists only one equilibrium state: ϕ0∗ = 0. Above the critical √ angular velocity, c = g/l, the solutions ϕ±∗ = arc cos (g/( 2 l)) also appear (see Fig. 3). In the vicinity of the bifurcation point, the angular deviation is small, and for ϕ 1 the Taylor expansion of the force yields, in leading order,



2 l 2 g  2 l  2 ϕ ≈− ϕ ϕ −2 1− 2 , F(ϕ) ≈ sin ϕ ( 2 l − g) − 2 2 l

(1)

which is of the same type as (3.41). Solution 3.16 The small amplitude swinging of a pendulum has a natural √ frequency ω0 = g/l (see the solution to Problem 3.4). Approximating the force between the ball and the small magnets by a Coulomb-type expression, its magnitude is γ /r 2 , where r is the distance between the ball and the corresponding magnet and γ denotes a positive constant. The horizontal force components can be read off from Fig. 4. The resulting force is given by a+x a−x g −γ 2 . F(x) = − x + γ 2 l (d + (a − x)2 )3/2 (d + (a + x)2 )3/2

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

a

y

a

d

x

Fig. 4. Magnetic forces acting on the pendulum. The magnetic interaction is assumed to be attracting.

For small deviations (compared with a and d) we can perform a Taylor expansion in x, from which we obtain   2γ (2a 2 − d 2 ) g x. F(x) ≈ − + l (a 2 + d 2 )5/2 √ The origin can be unstable only if the magnets are not too far away, i.e. if d < 2a. Furthermore, for instability it is also necessary that the parameter γ is sufficiently large or that g/l is sufficiently small. Solution 3.18 The spring force along the rod is given by (P.4), where x is now the distance measured from the intersection of the rod and the straight line connecting the fixed end-points of the springs. There is an additional force, the weight, that causes an acceleration gα along the rod for small angles of inclination. The force law is therefore given by   l0 − gα. F(x) = −2kx 1 − √ h2 + x 2 Around the bifurcation point, the displacement is small. Expanding the force in powers of x/ h and assuming that h ≈ l0 (see (P.5)),

  l0 k F(x) ≈ −2kx 1 − − 2 x 3 + gα h l0      x h 1 x 3 gα ≈ 2kl0 + 1− − . 2kl0 l0 l0 2 l0 This expression is of the same form as (P.6): around the bifurcation point the system behaves like the merry-go-round model. The quantity analogous to the square of the angular velocity is now 1/ h. The physical reason for the bifurcation is that, for very weak compressions, the force exerted by the springs is not sufficient to ensure an upper equilibrium state. When, however, h is shorter than a critical value, h c , there exist two equilibrium states above the line between the fixed end-points of the springs (the lower one is unstable). Solution 3.20 The phase space trajectories are of the form

 v(x) = ± 2(E + A cos x)

(2)

5

CUUK255-Tell Et al

June 9, 2006

6

16:45

Solutions to the problems

4 3 1 –3

–2

–1

–1

1

2

3

x

− p

–3 –4

Fig. 5. Phase portrait of a frictionless motion on a bumpy road (A = 1). The unstable and stable manifolds of the hyperbolic points constitute a separatrix (thick line), which separates trajectories passing over the local potential hills from those bounded to a given well (or, in the example of the pendulum, turning-overs from swinging). Fig. 6. Phase portrait of a frictionless motion on a bumpy slope. The thick lines are the manifolds of the hyperbolic points (A = 1, F 0 = 0.25).

4

1 –3

–2

–1

–1

1

2

x p-

3

–4

(see Fig. 5). The separatrices corresponding to energy Ec = A connect the neighbouring hyperbolic points. Solution 3.21 The left branches of the unstable and stable manifolds coincide and separate the periodic motion in the well from non-periodic motion (Fig. 6). The upper branches of the unstable manifolds rise, reflecting the fact that particles passing over the local maximum run downwards faster and faster. Solution 3.23 The phase space contraction rate is given by σ = α(x12 − 1). For |x1 | > 1, the phase space volume decreases; α(x12 − 1) can also be considered as a position-dependent friction coefficient, and energy dissipation takes place. In the range |x1 | < 1, the phase space volume expands since energy is supplied to the system via a negative friction. Solution 3.24 The stability matrix of the mechanical problem, equation (3.52), linearised around the fixed point (x ∗ , 0) is given by



A=

0 1 F  (x ∗ ) −α



.

(3)

Its trace and determinant are −α and −F  (x ∗ ), respectively. Thus, the eigenvalue equation (3.64) becomes λ2 + λα − F  (x ∗ ) = 0, which leads to either (3.14) or (3.29), depending on whether F  (x ∗ ) is positive or negative.

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

x2 4 3 2 1 –3

–2

–1

–1

1

2

3

x1

–2 –3 –4

Fig. 7. Phase portrait of the van der Pol oscillator obtained numerically (α = 0.2).

√ Solution 3.25 The eigenvalues are λ± = −3/4 ± 25/16 + a. The fixed point is hyperbolic for a > −1. In the range −25/16 < a < −1 it is a node, otherwise it is a spiral attractor. Solution 3.28 The origin is a spiral repellor surrounded by a limit cycle attractor (Fig. 7). Solution 4.1 Fg (x, t) = u I (x(t))

∞ 0

δ (t − nT − T ϕ0 /(2π)).

Solution 4.4 From (4.24) and (4.25), xn s0 ± vn = 2s0 c± n± . Using n± = e(−α/2±s0 )T n and taking suitable powers, the right-hand sides of both equations become proportional to the same power of en ; therefore, ( xn s0 − vn )α/2−s0 = constant. ( xn s0 + vn )α/2+s0  Solution 4.5 ± = exp{(−α/2 ± [s02 + α 2 /4]) T }. Solution 4.6 The velocity vn before the nth kick changes suddenly to the value v˜ n = vn + u I (xn ), due to the kick. The motion is that of a damped harmonic oscillation until the next kick; the new co-ordinates therefore fulfil xn+1 = v˜ n E/ω0 and vn+1 = −Eω0 xn . The map is given by xn+1 =

E (vn + u I (xn )) , ω0

vn+1 = −Eω0 xn .

Solution 4.8 The fixed point is x ∗ = I0 E/(1 + E 2 ), v ∗ = I0 /(1 + E 2 ). The eigenvalues are ± = ±i E; the limit cycle is represented as a spiral attractor on the map. Solution 4.9 The images of the vertices of the rectangle P0 P1 P2 P3 with side lengths x1 and x2 are the points P0 = (M1 (x1 , x2 ), M2 (x1 , x2 )), P1 = (M1 (x1 + x1 , x2 ), M2 (x1 + x1 , x2 )), P2 = (M1 (x1 + x1 , x2 + x2 ), M2 (x1 + x1 , x2 + x2 )) and

7

CUUK255-Tell Et al

June 9, 2006

8

16:45

Solutions to the problems

P2' P3'

x2 P3

P2 P1'

Δ x2 P0' P0

Δ x1

P1

x1 Fig. 8. A rectangle of side lengths x1 and x2 and its image.

P3 = (M1 (x1 , x2 + x2 ), M2 (x1 , x2 + x2 )) (see Fig. 8). In a reference frame whose origin is P0 , for small side lengths, the image points are given by



 ∂ M1 ∂ M2

x1 ,

x1 , ∂ x1 ∂ x1   ∂ M ∂ M1 ∂ M2 ∂ M2 1 P2 =

x1 +

x2 ,

x1 +

x2 ∂ x1 ∂ x2 ∂ x1 ∂ x2

P1 =

and

 P3 =

 ∂ M1 ∂ M2

x2 ,

x2 . ∂ x2 ∂ x2

From Fig. 8, the area of the parallelogram is given by   ∂ M1 ∂ M2 ∂ M1 ∂ M2

x1 x2 , − ∂ x1 ∂ x2 ∂ x2 ∂ x1 which is the area x1 x2 of the initial rectangle multiplied by J . Solution 4.10 The Jacobian is J = −1 − a, which is between 0 and 1 for √ −2 < a < −1. The eigenvalues are ± = −3/4 ± 25/16 + a. The fixed point is thus hyperbolic for −3/2 < a < −1, while it is a node or a spiral attractor for −25/16 < a < −3/2 and −2 < a < −25/16, respectively. The stability matrix is formally identical to that of Problem 25. This result illustrates how different the meaning of the same matrix is in maps and flows. Solution 5.2 The two-fold iterated map is given by ⎧ for vn ≤ 1/4, (c2 x , 4vn ), ⎪ ⎪  ⎪ n ⎨ 2 x , 4(v − 1/4) , for 1/4 < vn ≤ 1/2, 1 − c + c n n  B 2 (xn , vn ) =  2 ⎪ (x − 1), 4(v − 1/2) , for 1/2 < vn ≤ 3/4, c + c n n ⎪ ⎪  ⎩ 1 + c2 (xn − 1), 4(vn − 3/4) , for vn > 3/4.     This shows that P1 = 1/(1 + c), 1/3 and P2 = c/(1 + c), 2/3 are the fixed points of the twice iterated map in the ranges 1/4 < vn ≤ 1/2 and 1/2 < vn ≤ 3/4, respectively. Since they are not fixed points of map B, they can only be the points of a two-cycle.

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

Solution 5.4 The two three-cycles are given by         1/C c/C c2 /C 1/C B− B− B+ B− −→ −→ −→ ··· −→ 1/7 2/7 1/7 4/7 and



(c + 1)/C 3/7





   c(c + 1)/C (c2 + 1)/C B+ −→ −→ 6/7 5/7   (c + 1)/C B+ B− −→ ··· , −→ 3/7 B−

where C = c2 + c + 1. Solution 5.7 b = c1 /(c1 + c2 ), and the Jacobian is J = c1 + c2 . Solution 5.8 The area contraction rate in the lower (upper) half-square is a1 c1 (a2 c2 ). The second expression is the reciprocal of the first if a1 c1 = c1 /b = (1 − b)/c2 . There exists a solution only for c1 c2 < 1/4, and two b-values are then possible. The area contraction rate on one of the half-squares is less than unity, while that on the other is greater than unity. In spite of this, the total phase space volume changes by a factor of (c1 + c2 ) < 1 in each step; therefore the map is globally area contracting. Solution 5.10 The map is given by xn+1 = vn ,

vn+1 = −E 2 xn + 1 − avn .

There exists a single fixed point at x ∗ = v∗ =

1 . 1 + E2 + a

Note that f ∗ = −a, and therefore the fixed point is hyperbolic for a > 1 + E 2 . In the ranges 2E < a < 1 + E 2 and a < 2E, the fixed point is a node attractor and a spiral attractor, respectively. Since f is linear, so is the map; consequently, chaos does not exist. The eigenvalues are given by √ −a ± a 2 − 4E 2 ± = . 2 Thus, for a > 1 + E 2 , the attracting and repelling directions are the same on the entire phase space as around the fixed point: the manifolds are perfect straight lines with equation (5.33). Solution 5.11 The kicking strength is constant: f ∗ ≡ 0. The eigenvalues are complex ( f ∗ = 0 < 2E), and the fixed point is a spiral attractor with co-ordinates x ∗ = v∗ =

1 . 1 + E2

The system behaves in a way similar to the continuously driven harmonic oscillator; i.e., the motion converges towards a periodic dynamics governed by a limit cycle whose period is identical to that of the driving.

9

CUUK255-Tell Et al

June 9, 2006

10

16:45

Solutions to the problems

–0.1

(a)

(b) –0.270

–0.2 –0.275

–0.280

–0.3

–0.285 –0.4 –0.290 0.1

0.2

0.3

0.4

0.265

0.5

0.270 0.275

0.280 0.285

0.290 0.295

Fig. 9. The sawtooth attractor around P1 . (a) Rectangle in Fig. 5.22 magnified ×30. (b) Rectangle in part (a) magnified by a further factor of 10.

Solution 5.12 A two-cycle point (xi∗ , vi∗ ) is invariant under the two-fold iterated map: K 2 (xi∗ , vi∗ ) ≡ K (K (xi∗ , vi∗ )) = (xi∗ , vi∗ ). The twice iterated map (5.19) is given by

xn+2 = vn+1 = −E 2 xn + f (vn ), vn+2 = −E 2 xn+1 + f (vn+1 ) = −E 2 vn + f (−E 2 xn + f (vn )). The two-cycle points are obtained with xi∗ ≡ xn = xn+2 , vi∗ ≡ vn = vn+2 as follows:

xi∗ = −E 2 xi∗ + f (vi∗ ), vi∗ = −E 2 vi∗ + f (−E 2 xi∗ + f (vi∗ )) = −E 2 vi∗ + f (xi∗ ). This leads to two independent equations:

⎫ ,⎪ ⎬ ∗ f (v ) ⎪ vi∗ (1 + E 2 ) = f (xi ) = f 1+Ei 2 . ⎭ xi∗ (1 + E 2 ) = f (vi ) = f



f (xi∗ ) 1+E 2

(4)

The equations are of similar form; this does not imply, however, that the x- and v-co-ordinates of a two-cycle are identical. Solution 5.13 See Fig. 9. Solution 5.16 In the general form (4) there are two functions f (x) embedded in each other. Since the sign function can take on two values, there are altogether four different ways of taking the expressions f = axi∗ + 1 and/or f = axi∗ − 1. Taking f = axi∗ − 1 as the ‘inner’ and f = axi∗ + 1 as the ‘outer’ function, we obtain from (4) x1∗ (1 + E 2 ) = a

ax1∗ − 1 +1 1 + E2

with x1∗ =

1 a + 1 + E2

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

as the solution. Equation (4) also indicates that the equation for vi∗ contains the functions f in reversed order; hence, v1∗ (1 + E 2 ) = a

av1∗ + 1 − 1. 1 + E2

Consequently, x1∗ = −v1∗ . By exchanging the sequence of the functions, we obtain the points of the other two-cycle as follows: x2∗ = −x1∗ (= −v2∗ ). (Taking expression f i = axi∗ + 1 or f i = axi∗ − 1 twice, the already known fixed points, (5.35), are obtained, which are, of course, two-cycles as well.) Solution 5.18 According to equation (5.33), the equation of the unstable basic branches is given by vn = v+ (xn ) = + (xn − x±∗ ) + x±∗ , where + is given by (5.29) with f ∗ = a. The basic branches of H− and H+ are straight lines of the half-planes xn < 0 and xn > 0, respectively. The end-points falling on the vn -axis are (0, x±∗ (1 − + )). Solution 5.20 The equation of the unstable basic branch of H+ (see (5.33)) is given by

vn = v+ (xn ) = + (xn − x+∗ ) + x+∗ ,

(5)

where x+∗ is given by (5.38) and + is obtained from (5.29) with f ∗ = −a. Since + is, by definition, the solution with the greater absolute value, the square root in (5.29) has to be taken with the negative sign; thus + is negative. The intersection of the basic branch with the vn -axis is (0, x+∗ (1 − + )), which is mapped into (x+∗ (1 − + ), 1 − ax+∗ (1 − + )). Substituting + and x1∗ , one can check that the image point satisfies (5). The end-points of the basic branch of point H+ are therefore given by   0, x+∗ (1 − + ) ,



 x+∗ (1 − + ), 1 − ax+∗ (1 − + ) .

Solution 5.22 The calculation is similar to that of Problem 16 and yields



 1 + E2 + a 1 + E2 + a , , (1 + E 2 )2 + a 2 (1 + E 2 )2 − a 2   1 + E2 + a 1 + E2 + a P2 = (x2∗ , v2∗ ) = , . (1 + E 2 )2 + a 2 (1 + E 2 )2 + a 2

P1 = (x1∗ , v1∗ ) =

At points P1 and P2 the stability matrix (5.28) has to be taken with f ∗ = −a and f ∗ = a, respectively. The stability matrix of the twice iterated map is the product of these matrices. At point P1 this matrix is given by



−E 2 −a −a E 2 −a 2 − E 2



.

11

CUUK255-Tell Et al

June 9, 2006

12

16:45

Solutions to the problems

At point P2 the stability matrix of the twice iterated map is the same, except that −a has to be replaced by a. The Jacobian of both matrices is E 4 , the square of the Jacobian per iteration. The eigenvalues are given by (2) ±

√ −a 2 − 2E 2 ± a 4 + 4a 2 E 2 . = 2

The eigenvalue does not depend on the sign of a, inaccordance with the requirement that the stability property of a cycle must be independent of the cycle (2) point considered. The eigenvectors are given by u± = 1, −(± + E 2 )/a , and (2)

they differ from the eigenvectors (5.32) of both fixed points. The eigenvalues, ± , (2) characterise two iteration steps. The square root of |± | has to be compared with the eigenvalues ± , (5.29), of the fixed points; they are found to be different. Solution 5.27 In the asymmetric baker map, the segments that cross the entire unit square along the v-direction are stretched by a factor of two; therefore h = ln 2. In map (5.34), segments along the unstable direction are stretched by a factor of the eigenvalue + , given by (5.29) with f ∗ = a. Consequently, h = ln + . Solution 5.29 Distribution P0 ranges, for v < 1/2, between 1 − γ /2 and 1. Its support becomes stretched by a factor of 2 along the v-direction and is mapped onto the column 0 < x < c. The probability that any point in the column takes on velocity v is (1 − γ /2)/2 + vγ /4. Similarly, in the right column, the probability is 1/2 + vγ /4. The probability that the velocity takes on a value v in any of the two columns after one step is the sum of the distributions in the two columns: 1 − γ /4 + vγ /2. Thus, the slope of the distribution along the v-direction is halved. After many steps, therefore, the distribution becomes independent of the velocity. The unevenness along the x-direction becomes negligible for n → ∞ according to the mechanisms presented in the solution to Problem 5.28. Solution 5.30 For 0 < x < c, P1 = g (x/c, v/2) /(2c) and, for 1 − c < x < 1, P1 = g (1 + (x − 1)/c, (1 + v)/2) /(2c). Solution 5.33 The interval lengths in the nth step are of the form lm = c1m c2n−m . The dimension must therefore be calculated using the most probable value lm ∗ that belongs to m ∗ = bn (cf. the solution to Problem 2.15), and this is found to be given by (2)

D1 =

b ln b + (1 − b) ln (1 − b) . b ln c1 + (1 − b) ln c2

Solution 5.34 In the interval of probability pm = bm (1 − b)n−m , the stretching factor along the y-direction over n steps is b−m (1 − b)m−n . The positive local Lyapunov exponents are therefore of the form −(m/n ln b + (n − m)/n ln (1 − b)). Their average is given by  n    n m n−m ln b + ln (1 − b) bm (1 − b)n−m . λ¯ = − m n n m=0 The coefficient of − ln b is

n   n 1 mbm (1 − b)n−m , n m=0 m

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

which turns out to be b, independently of n. To see this, we write the coefficient as

 n   n  1 n n−1 1 m n−m = mb (1 − b) nbbm−1 (1 − b)n−1−(m−1) n m=1 m n m=1 m − 1  n−1   n − 1 m  =b b (1 − b)n−1−m = b (b + 1 − b)n−1 = b.  m  m =0

Analogously, the coefficient of − ln (1 − b) is 1 − b, and thus equation (5.71) is recovered. Solution 5.35 The local Jacobians are c1 /b and c2 /(1 − b) in the lower and upper rectangles that are mapped on the left and right phase space columns of width c1 and c2 , respectively. Their average is ln J = b ln (c1 /b) + (1 − b) ln (c2 /(1 − b)), which is, of course, the sum of the Lyapunov exponents given in equations (5.71) and (5.72). Solution 5.36 Map (5.34) is piecewise linear, and stability matrix (5.28) is the same at any point. The local Lyapunov exponents are the logarithm of the eigenvalues, ± , obtained from (5.29) with f ∗ = a. Consequently, they are also the average values: λ¯ = ln + ,

λ¯ = ln − .

Solution 5.37 According to the Kaplan–Yorke relation (5.74), D0 = D 1 = 1 +

ln + , | ln − |

where ± are the eigenvalues, (5.29), with f ∗ = a. The two dimensions are identical because the local Lyapunov exponents are position-independent. Using the parameters of Fig. 5.22, + = 1.532, − = 0.418, and thus D0 = D1 = 1.489. Solution 5.39 With the given data, qc = κ 2 α/(g Rπ ) ≈ 10−3 kg s−1 , which implies 3.6 litres of rain per hour. At q = 2qc , frequency (5.100) is ω = κ, and the period of the rotation is therefore 2π/κ = 62.8 s, i.e. about one minute. The amplitudes are A∗ = B ∗ = κα/(g Rπ ) ≈ 10−2 kg. Solution 5.41 The equations linearised around the non-trivial fixed point are given by

x˙ = σ ( y − x),

y˙ = r x − y − z ∗ x − x ∗ z,

z˙ = − z + y ∗ x + x ∗ y. The stability matrix is therefore given by ⎛ ⎞ −σ σ 0 ⎜ ⎟ A = ⎝ r − z ∗ −1 −x ∗ ⎠ . x∗ x ∗ −1 The eigenvalue equation for r > 1 is thus λ3 + λ2 (σ + 2) + λ(σ + r ) + 2σ (r − 1) = 0. Right above the bifurcation point rc0 = 1, all three eigenvalues are real and negative, i.e. the non-trivial fixed points are stable. Increasing parameter r , two eigenvalues suddenly become complex. At this point, rc , of stability loss, the

13

CUUK255-Tell Et al

June 9, 2006

14

Fig. 10. The map zn+1 vs. zn for σ = 10, r = 28. An approximately one-dimensional map is obtained in accordance with the strong phase space contraction.

16:45

Solutions to the problems

50

zn+1

30 30

zn

50

Fig. 11. The R¨ ossler attractor with parameters a = b = 0.2, c = 5.7.

real parts of the eigenvalues vanish and, for r > rc they become positive. At rc , the complex roots are therefore of the form λ = ±iωc . Substituting these into the eigenvalue equation and grouping the real and the imaginary parts, one can see that the equation can only be fulfilled if σ > 2 and parameter r takes on the value  = rc0

σ (σ + 4) . σ −2

(In this case the imaginary part of the eigenvalues fulfils ωc2 = 2σ (σ + 1)/  . (σ − 2).) The non-trivial fixed points are therefore stable for rc0 = 1 < r < rc0  With σ = 10 , stability is lost at rc0 = 17.5. Solution 5.42 See Fig. 10. Solution 5.43 See Fig. 11.

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

Solution 6.1 The points P1 = (1/(1 + c), 1/(1 + a)) and P2 = (c/(1 + c), a/(1 + a)) form a two-cycle. The two three-cycles are given by         c/C c2 /C 1/C 1/C B− B− B+ B− −→ −→ ··· −→ −→ a/A 1/A a 2 /A 1/A and



(c + 1)/C (a + 1)/A





   c(c + 1)/C (c2 + 1)/C B+ −→ a(a + 1)/A (a 2 + 1)/A   (c + 1)/C B+ B− −→ ··· , −→ (a + 1)/A B

− −→

where A = a 2 + a + 1, C = c2 + c + 1. Solution 6.2 Each velocity interval of length a −m is stretched by a factor of a m (see Figs. 6.6 (a) and (e)), and is mapped onto the interval (0, 1). Thus, in the m-fold iterated map, the graph of function vn+m (vn ) is a set of straight line segments of slope a m , defined over 2n intervals. The m-cycles are the solutions of the equation vn+m (vn ) = vn , i.e. the intersection points of the graph and the diagonal. There are 2m such intersection points. Solution 6.7 Using x±∗ defined by (5.26) and introducing the new variables x  = (x − x−∗ )/(x+∗ − x−∗ ), v  = (v − x−∗ )/(x+∗ − x−∗ ), we obtain H− = (0, 0),  = vn , and H+ = (1, 1). Now the map is xn+1  vn+1 = −E 2 xn +

f (vn (x+∗ − x−∗ ) + x−∗ ) − (1 + E 2 )x−∗ . x+∗ − x−∗

(6)

Assuming that for a 1 the points of the saddle have co-ordinates close to those of one of the fixed points only (numerical solutions support this assumption), function f in (6) can be expanded around x−∗ and x+∗ up to linear terms. We thus obtain f (x−∗ ) + f  (x−∗ )(x+∗ − x−∗ )vn and f (x+∗ ) + f  (x+∗ )(x+∗ − x−∗ )(vn − 1), respectively. Substituting these into (6) and taking into account equation (5.27): ∗

 = −E 2 xn + f  − vn , vn+1

for |vn | 1,

∗

 vn+1 − 1 = −E 2 (xn − 1) + f  + (vn − 1), for |vn − 1| 1,

where f  ∗± denotes the derivative of f at x±∗ (see Section 5.2.2). The eigenvalues of this piecewise linear map in the lower  and upper half-planes are given by  ( f  ∗− ± f  ∗− 2 − 4E 2 )/2 and ( f  ∗+ ± f  ∗+ 2 − 4E 2 )/2, respectively. In the limit a 1, the derivatives are large, and the eigenvalues with absolute value larger (smaller) than unity are therefore f  ∗± (E 2 / f  ∗± ) in the two half-planes. Since these are the respective slopes of the unstable and stable manifolds, the manifolds are almost vertical and horizontal lines. Accordingly, the map is an asymmetric baker map with parameters a1,2 = | f  ∗± | and c1,2 = |E 2 / f  ∗± |. If the signs of f  ∗± are identical, then the dynamics is given by (6.13); otherwise, it is given by (6.14). Solution 6.9 With a vertical initial line segment of length 1, L n = 2n , and thus h = ln 2.

15

CUUK255-Tell Et al

June 9, 2006

16

x

16:45

Solutions to the problems

y

x

y

x

(a)

(b)

Fig. 12. Natural distribution on the chaotic saddle of (a) map (6.13) and (b) map (6.14) for a1 = 3, a2 = 2.5, c1 = 0.25, c2 = 0.45 (b = 1/2). The distributions are different, although their information dimension is the same.

Solution 6.10 Points surviving one iteration start (in both maps) in two horizontal bands (cf. Fig. 6.4) of heights 1/a1 and 1/a2 . The probability of falling, after one step, in the left (right) rectangle of width c1 (c2 ), both of height unity, is given by (a1 /(a1 + a2 )) a2 /(a1 + a2 ). Such points have been subjected to an expansion (contraction) of rate a1 (c1 ) or a2 (c2 ), respectively. The average Lyapunov exponents are thus given by a2 ln a1 + a1 ln a2 > 0, λ¯ = a1 + a2

a2 ln c1 + a1 ln c2 λ¯ = < 0. a1 + a2

Because of the simplicity of the map, they follow from observing one iteration step ¯ only. The escape rate is given by κ = ln (a1 a2 /(a1 + a2 )) < λ. (2)

Solution 6.11 See Fig. 12. The partial information dimension, D1 , along the stable manifold is the result of a construction similar to that of Problem 5.33. The probability contents after one step are now (for both maps) 1/a1 /(1/a1 + 1/a2 ) and 1/a2 /(1/a1 + 1/a2 ), replacing b and 1 − b, respectively. We thus obtain: (2)

D1 =

1 a1

1 ln 1/a1/a + 1 +1/a2

1 a1

ln c1 +

1 a2 1 a2

ln

1/a2 1/a2 +1/a2

ln c2

a2 ln a2 + a1 ln a1 − (a1 + a2 ) ln (a1 + a2 ) = . a2 ln c1 + a1 ln c2 (1)

When determining D1 , c1 and c2 are replaced by 1/a1 and 1/a2 , i.e. the contraction rates of the inverse map, and the probability contents remain unchanged (as the distribution is invariant under both the forward and the inverse map). Thus, (1)

D1 = =

1 a1

1 ln 1/a1/a + 1 +1/a2

1 2 ln 1/a1/a a2 2 +1/a2 − a11 ln a1 − a12 ln a2

a2 ln a2 + a1 ln a1 − (a1 + a2 ) ln (a1 + a2 ) . −a2 ln a1 − a1 ln a2

Since the information dimension is determined solely by the stretching and contraction factors (a1 , a2 , c1 , c2 ), and not by the geometrical arrangements, the

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

Fig. 13. The small-size roof attractor and the saddle around it for a = 1.62, E = 0.7.

n

1

–2

–1

17

attractor

1

2

xn

–1

–2

partial information dimensions (just like the Lyapunov exponents) of both distributions are the same. Solution 6.12 The branches of the stable manifolds emanating from H± are horizontal lines. All the branches of the unstable manifolds lie over the interval [0, 1] in the vn -direction, and therefore just touch the stable manifolds of H± . (The solution of Problem 5.15 shows a baker attractor before crisis.) Solution 6.15 The equation of the unstable manifold emanating from H− is √ v+ (xn ) = x−∗ + + (xn − x−∗ ), ± = (a ± a 2 − 4E 2 )/2, and the co-ordinates of the fixed points are given by (5.38). The upper end-point of this branch is on the v-axis (see Fig. 5.32): (0, x−∗ (1 − + )). Its image under (5.37) is the point (x−∗ (1 − + ), 1 − ax−∗ (1 − + )). The crisis value is the parameter, ac , for which this point falls onto the stable manifold v− (xn ) = x−∗ + − (xn − x−∗ ) of H− , i.e. for which 1 − ac x−∗ (1 − + ) = x−∗ + − (x−∗ (1 − + ) − x−∗ ). This leads to + = 2 − 2E 2 /ac , and after rearrangement we obtain ac3 − ac2 (2 + 3E 2 /2) + ac 4E 2 − 2E 4 = 0. For E = 0.7, the only real solution is ac = 1.7898. Solution 6.16 The condition for crisis is that the top of the function touches the upper edge of the square of size |2x−∗ |, centred at the origin. (According to the notation of Fig. P.28, f (0) = −x−∗ .) Solution 6.17 See Fig. 13. Solution 6.18 See Fig. 14. Solution 7.1 The eigenvalues of any 2 × 2 matrix are determined by the trace and the determinant of the matrix. The determinant is unity due to area preservation. Stability is therefore uniquely determined by TrL (see the solution to Problem 4.11). The eigenvalues are ± = {Tr L ± [(Tr L)2 − 4]1/2 }/2. For |Tr L| < 2, they

CUUK255-Tell Et al

June 9, 2006

18

Fig. 14. Basins of attraction (black and white) of the two attractors of the two-fold iterated roof map for a = 1.55, E = 0.7.

16:45

Solutions to the problems

2

n

0

–2 –2.7

xn

0

2.7

are complex with unit absolute value. The periodic motion is then stable, and the fixed point is elliptic. For |Tr L| > 2, the fixed point is hyperbolic. Solution 7.2 The number of unstable cycles of length m is 2m , just as in the dissipative case (see (5.13)). The same growth rate governs the stretching of vertical line segments (see (5.51)). The topological entropy is therefore h = ln 2. √ Solution 7.4 The origin is hyperbolic with eigenvalues ± = (3 ± 5)/2. Since the map is linear, these characterise the neighbourhood of any point. The natural distribution is uniform (see Fig. 15); the average Lyapunov exponents are therefore √ the logarithm of the eigenvalues: λ¯ = ln ((3 + 5)/2) = 0.962 = −λ¯ . The √ unstable (stable) manifold consists of straight lines of slope ( 5 − 1)/2, √ (−( 5 + 1)/2). An arbitrary line segment becomes stretched along this direction, and its length is multiplied by a factor of + in each step. The topological entropy ¯ is therefore h = λ. Solution 7.5 xn+1 = xn + vn+1 , vn+1 = vn + f (xn ). Solution 7.6 With a linear damping, the velocity decreases exponentially in time; therefore, after the nth kick, v(t) = vn exp (−αt), where α is the friction coefficient and t < T . The displacement is, accordingly, x(t) = xn + vn (1 − exp (−αt))/α. With the notation E ≡ exp (−αT /2), the maps connecting the states after and before the kicks are given by xn+1 = xn + vn

(1 − E 2 ) , α

vn+1 = vn E 2 + f (xn+1 ),

and xn+1 = xn + vn+1 respectively.

(E −2 − 1) , α

vn+1 = (vn + f (xn ))E 2 ,

CUUK255-Tell Et al

June 9, 2006

16:45

Solutions to the problems

Fig. 15. Points of one trajectory of the cat map. The points visit the unit square in a visibly uniform manner. The initial condition and the number of iterates are the same as in Fig. 7.2 for the baker map.

Solution 7.7 If the electric field points horizontally in the plane of Fig. 7.6, the tangential component of the momentum transfer due to the field is proportional to sin xn+1 and we obtain the map (7.10). √ Solution 7.8 The eigenvectors are given by (1, (a ± 4a + a 2 )/2), with the positive sign corresponding to the unstable direction. Solution 7.9 The two-fold iterated standard map is given by

xn+2 = xn + 2vn + a sin (xn + vn ),

⎫ ⎬ (7)

vn+2 = vn + a sin (xn + vn ) + a sin xn+2 . ⎭

The condition for the existence of a two-cycle is that there exists a point xn ≡ x ∗ , vn ≡ v ∗ , for which xn+2 = x ∗ + 2π and vn+2 = v ∗ = π , since, due to periodicity, this implies that (x ∗ , v ∗ ) is a fixed point of the twice iterated map. Point (0, π ) and point (π, π ) both satisfy this condition. The matrix of the map linearised around the former is given by

 L=

1−a −a 2

2−a 1 + a − a2

 .

(8)

The two-cycle is unstable for a > 2. Solution 7.13 The stability matrix of the fixed point is given by

⎛

tan α −1 − 12 sin (4α) tan α √ L=⎝ z∗ √ z ∗ sin (4α)

1 + 14 sin (4α) tan α −1 − 12 sin (4α) tan α

⎞ ⎠.

(9)

Solution 7.14 Map (7.25) yields in this case z n+1 = 1 − z n . The velocity component perpendicular to the slope is therefore decoupled from the parallel component and follows a simple linear rule; consequently, the motion cannot be chaotic. Solution 7.16 The eigenvalues, ± , of an elliptic fixed point are complex numbers of unit absolute value (see Table (4.2) ± = exp (±iφ), where φ is the phase of the

19

CUUK255-Tell Et al

June 9, 2006

20

Fig. 16. Manifolds of the chaotic saddle of map (8.6) with f (x) = −7(x − x 2 )e−x . Points of trajectories of lifetime at least n = 8 in the depicted region are plotted at n = 0 and n = 8.

16:45

Solutions to the problems

n

1 stable manifold

–1

1

2

xn

unstable manifold

–1

complex number. The angle of deviation of the iterated point is φ in one step. The winding number is thus v = φ/(2π), proportional to the phase of the eigenvalue. Solution 7.17 The elements are given by π − 3 = [7, 15, 1, 292, 1], e − 2 = √ √ [1, 2, 1, 1, 4], 2 − 1 = [2, 2, 2, 2, 2], 3 − 1 = [1, 2, 1, 2, 1] and √ ( 5 − 1)/2 = [1, 1, 1, 1, 1]. The respective deviations from the exact numbers are 3 × 10−10 , 5 × 10−4 , 7 × 10−5 , 1 × 10−3 and 7 × 10−3 . Solution 7.18 Their general form is gk = 1/(k + g); the ‘silver mean’ is thus given by g2 = 0.382, and the next noble number is g3 = 0.276. Solution 8.4 From the solutions of Problems 8.2 and 8.3 the expansion and contraction rates of both periodic orbits are, for a/R 1, proportional to a/R and R/a, respectively. Accordingly, in the notation of (6.3), a ∼ a/R and c ∼ R/a. (1) (2) Consequently (see Section 6.1), κ ∼ λ¯ ∼ ln (a/R), D0 = D0 ∼ ln 2/ ln (a/R) and h = ln 2. The chaotic saddle is, in this limit, rather unstable and its fractal dimension is small. Solution 8.5 See Fig. 16.