TEST BANK FOR Chaotic Dynamics_ An Introduction Based on Classical Mechanics By Tamás Tél, Márton

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).