Exam (elaborations) TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton Gruiz(Solutions Manual)

Exam (elaborations) TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton Gruiz(Solutions Manual) 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 small triangles is therefore A0 ∞ n=1 3 × 4n−19−n = A0/3 ∞ n=0(4/9)n . 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 = r 2 1 , the only positive solution of the quadratic equation r D0 1 + r 2D0 1 = 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 corners of the original square, ε = λn πn−1 j=1 (1 − λj )  2−n+1; we therefore cover the set with intervals of this size. In analogy with the solution of Problem 2.13, we find that α = 2 lnλ/ln (λ/2). With λ = 0.6, α = 0.849. Solution 2.16 See Fig. 1. The possible box probabilities are again pm = pm 1 pn−m 2 , m = 0, 1, ..., n. The number of boxes carrying pm at level n is Nm = 2mn m  (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 16:45 2 Solutions to the problems 10 5 15 x P 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 lnm − (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(2p1 ln p1 + p2 ln p2). Since the resolution at level n is ε = 3−n , D1 = −2p1 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∗ = D1n 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 3 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 − ϕ3/6) (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 (s0t), where s0 = √ g/l. Consequently, the linear approximation is valid over time t = 1 s0 ln 15◦ 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 F(x) = −2kx(1 − l0/h). The frequency in the range h l0 is therefore given by ω0 =  2k(h − l0)/h. The derivative of the force at x∗ =  l2 0 − h2 is, from (P.4), F(x∗) = −2k(1 − h2/l2 0 ). Thus, the frequency around x∗ is ω0 =  2k(1 − h2/l2 0 ), 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 sinϕ 2l. From the components perpendicular to the rod, the resulting force per unit mass is given by F(ϕ) = sin ϕ ( 2l cos ϕ − g). CUUK255-Tell Et al June 9, 2006 16:45 4 Solutions to the problems hc= l0 h T Fig. 2. Period T of small oscillations vs. h around the bifurcation point. Ω = Ωc Ω ϕ – * 2 − π 2 − π 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: ϕ ∗ 0 = 0. Above the critical angular velocity, c = √ g/l, the solutions ϕ ∗± = arc cos (g/( 2l)) 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, F(ϕ) ≈ sin ϕ ( 2l − g) − 2l 2 ϕ2 ≈ − 2l 2 ϕ  ϕ2 − 2  1 − g 2l  , (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 F(x) = −g l x + γ a − x (d2 + (a − x)2)3/2 − γ a + x (d2 + (a + x)2)3/2 . CUUK255-Tell Et al June 9, 2006 16:45 Solutions to the problems 5 a a d x y 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 F(x) ≈  −g l + 2γ (2a2 − d2) (a2 + d2)5/2  x. 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 F(x) = −2kx  1 − √ l0 h2 + x2  − gα. Around the bifurcation point, the displacement is small. Expanding the force in powers of x/h and assuming that h ≈ l0 (see (P.5)), F(x) ≈ −2kx  1 − l0 h  − k l2 0 x3 + gα ≈ 2kl0  gα 2kl0 + x l0  1 − h l0  − 1 2  x l0 3  . 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, hc, 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) CUUK255-Tell Et al June 9, 2006 16:45 6 Solutions to the problems x− –4 4 –3 –2 –1 1 2 3 1 –1 3 –3 p 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). x p- –4 4 –3 –2 –1 1 2 3 1 –1 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). (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(x2 1 − 1). For |x1| 1, the phase space volume decreases; α(x2 1 − 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 7 x1 x2 –4 –3 –2 –1 1 2 3 4 –3 –2 –1 1 2 3 Fig. 7. Phase portrait of the van der Pol oscillator