Alligood K., Sauer T., Yorke J.A. Chaos: An Introduction to Dynamical Systems
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A NSWERS AND H INTS TO S ELECTED E XERCISES
6.3 The fixed points (⫺1, 0), (0, 0), (1, 0) are forward limit sets; (⫺1, 0) and (1, 0)
are attracting fixed points. The left half-plane is the basin of (⫺1, 0) and the
right half-plane is the basin of (1, 0). The basin boundary is the y-axis. These
points tend to the saddle (0, 0) on iteration.
6.4 F(x
0
, 兵 0其) ⫽
0ifx
0
⫽ 0
1ifx
0
⫽ 0
. F(x
0
,N(r, 兵0其)) ⫽ 1forallx
0
,r⬎ 0. F(x
0
, [0,
⬁
)) ⫽
1 2ifx
0
⫽ 0
1ifx
0
⫽ 0
. The natural measure
f
is the atomic measure located at 0.
6.7 Hint: This map is conjugate to the tent map. See Exercise 3.3 and Challenge 6.
6.8 (e) p(x) ⫽
2 7if0ⱕ x ⱕ 1 4
9 7if1 4 ⱕ x ⱕ 3 4
8 7if3 4 ⱕ x ⱕ 1
T7.2 (a) The characteristic equation of A is f(
) ⫽ (
⫺ 3)
2
. Solving (A ⫺ 3I)v ⫽
0forv yields only v ⫽ (1, 0). (b) The first equation is
˙
x ⫺ 3x ⫽ y ⫽ c
1
e
3t
.
Multiplying through by e
⫺3t
yields (xe
⫺3t
)
⫽ c
1
,orxe
⫺3t
⫽ c
1
t ⫹ c
2
. Therefore
(x(t),y(t)) ⫽ (c
1
te
3t
⫹ c
2
e
3t
,c
1
e
3t
).
T7.4 The ellipses are y
2
⫺ xy ⫹
5
2
x
2
⫽ y
2
0
,wherey
0
is the y-intercept. Solutions are
x(t) ⫽⫺
2
3
y
0
sin 3t, y(t) ⫽ y
0
cos 3t ⫺
y
0
3
sin 3t, using initial condition (0,y
0
).
T7.7 L ⫽
a
2
⫹ b
2
⫹ c
2
⫹ d
2
works. This constant is not optimal. The smallest Lip-
schitz constant for A is the square root of the largest eigenvalue of AA
T
.
T7.9 The existence and uniqueness Theorem 7.14 applies to (7.32). Solutions with
positive x(0) are of form x(t) ⫽ (2t ⫹ 1 x
2
0
)
⫺1 2
; those with negative x(0) are
x(t) ⫽⫺(2t ⫹ 1 x
2
0
)
⫺1 2
. All approach the equilibrium x ⫽ 0 monotonically as
t
→ 0, so x ⫽ 0 is asymptotically stable.
T7.11 The orbit starting at (x,
˙
x) ⫽ (0, 0); the orbit that begins along one wall and has
just enough initial energy to reach (0, 0); etc.
T7.14 Hint: Try the function E ⫽ x
2
⫹ y
2
.
T7.17 (b) First, determine which points on the x-axis belong to the largest possible W.
Then extend W to the plane so that it will be a forward invariant set.
T7.21 The constant K may be set so that E(
c
d
,
a
b
) ⫽ 0. Show that E(
c
d
,
a
b
) is a relative
minimum of E. (A test for finding relative maximum/minimum values that in-
volves the second-order partial derivatives can be found in most standard calculus
texts.)
7.3 The repeller (0, 0) and the saddle (⫺8, ⫺16) are both unstable.
7.4 (b) x ⫽ (b ⫺ cKe
a(b⫺c)t
) (1 ⫹ Ke
a(b⫺c)t
), where K ⫽ (b ⫺ x
0
) (x
0
⫺ c).
(c) T ⫽
ln
|x
0
⫺b|
|x
0
⫺c|
a(c ⫺ b)
.
582
A NSWERS AND H INTS TO S ELECTED E XERCISES
7.7 (a) Hint: Setting y ⫽
˙
x, try E(x, y) ⫽ y
2
2 ⫹ x
2
2 ⫹ x
4
4.
7.8 Hint: Use the LaSalle Corollary.
7.9 (a) (x
1
, 0) and (x
2
, 0) are stable; there is an unstable equilibrium (x
3
, 0) in
between. (b) (x
1
, 0) stable, (x
2
, 0) unstable.
7.14 x ⬅ 0andx(t) ⫽
0ifx ⬍ 0
2t
3 2
3ifx ⱖ 0
T8.4 Let z be in the orbit of u. Then there is a number t
1
such that z ⫽ F(t
1
, y). Since
u is in
(v
0
), there is a sequence 兵F(t
n
, v
0
)其, 兵t
n
其 →
⬁
as n
→
⬁
, converging to
u. By Theorem 7.16, given
⑀
⬎ 0, there is an N ⬎ 0, such that F(t
1
,F(t
n
, v
0
))
is in N
⑀
(z), for n ⬎ N. Now use the composition property to get these points on
the orbit of v
0
.
T8.6 Assume that there are distinct points v
1
and v
2
in
(v
0
)suchthatV(v
1
) ⫽ a
and V(v
2
) ⫽ b, where a ⬍ b. Reach a contradiction by showing that the value
of V along F(t, v
0
) must increase (with t) from values near a to values near b.
T8.7 If v(t
1
) ⫽ v(t
2
),t
1
⬍ t
2
, and if v(t) is not an equilibrium, then it can be shown (by
the composition property of autonomous equations) that v(t ⫹ (t
2
⫺ t
1
)) ⫽ v(t),
for any t. In this case, t
2
⫺ t
1
must be an integer multiple of the period of the
orbit.
8.2 The
-limit sets of (0, 0) and (1 2, 0) are both the origin. The
-limit set of
(1, 0) is the circle r ⫽ 1andof(2, 0) is the circle r ⫽ 3.
8.3 (a) Refer to the similarity types given in Appendix A.
8.6 D.
8.7 A, C, D, P, Y.
8.9 Hint: Set E(x, y) ⫽ y
2
2 ⫹
g(x) dx where y ⫽
˙
x. Show that
˙
E(x, y) ⱕ 0, with
equality possible only when y ⫽ 0.
8.10 Yes.
8.12 (a) Hint: Use the Monotonicity Lemma 8.12.
8.16 If
(t) is the direction of travel of x(t), measured in radians, show that |
˙
| ⱕ L.
T9.3 Hints: Find some r
0
and
⑀
⬎ 0 depending only on the constants a
i
,b
i
, and c such
that (9.2) implies
˙
V(u) ⬍⫺
⑀
whenever |u| ⱖ r
0
.Definem to be the maximum
of the function V(u) on the ball of radius r
0
.ThesetS ⫽ 兵u : V(u) ⱕ m其 is
forced to be a bounded set by the hypotheses. Trajectories can get into S from
the outside but no trajectory can leave S, because all points of the boundary are
outside radius r
0
. Choose B large enough so that a ball of radius B contains S.
583
A NSWERS AND H INTS TO S ELECTED E XERCISES
9.1 Hint: Set x ⫽ y ⫽ 0 in (9.1) and solve.
9.3 Area contraction rate is e
⫺2
c
per iterate; the system is dissipative if c ⬎ 0.
9.6 Both are 0.
9.7 Note that all orbits are identical except for a rescaling of time. The Lyapunov
exponents of ˙v ⫽ cf(v) are obtained by multiplying c by each of the Lyapunov
exponents of ˙v ⫽ f(v).
9.9 (a)
˙
x ⫽ y,
˙
y ⫽⫺x.(b)
˙
x ⫽ y,
˙
y ⫽ x.(c)J
1
(x
0
,y
0
) ⫽ [(cos 1)x
0
⫹ (sin 1)y
0
,
(⫺ sin 1)x
0
⫹ (cos 1)y
0
], Lyapunov exponents are 兵0, 0其.(d)J
1
(x
0
,y
0
) ⫽ [
1
2
(e ⫹
e
⫺1
)x
0
⫹
1
2
(e ⫺ e
⫺1
)y
0
,
1
2
(e ⫺ e
⫺1
)x
0
⫹
1
2
(e ⫹ e
⫺1
)y
0
], Lyapunov exponents are
兵1, ⫺1其. (e) The stable equilibria (2n
, 0) for integer n have Lyapunov expo-
nents 兵0, 0其. The saddle equilbria 兵(2n ⫹ 1)
, 0) for integer n have Lyapunov
exponents 兵1, ⫺1其, as do the connecting arcs converging to them. All periodic
orbits have Lyapunov exponents 兵0, 0其.
T10.2 Any point on a stable (resp., unstable) manifold maps to another point on the
stable (resp., unstable) manifold under both f and f
⫺1
. Therefore a homoclinic
point, which lies on both, remains homoclinic under mapping by either f or f
⫺1
.
T10.3 When c
2
⫽ 0, the inverse of the Ik
´
eda map f(z) ⫽ c
1
⫹ c
2
z exp[i(c
3
⫺
a
1 ⫹ |z|
2
)]
is f
⫺1
(w) ⫽
w ⫺ c
1
c
2
exp[i(a (1 ⫹ |
w ⫺ c
1
c
2
|
2
) ⫺ c
3
)].
T10.5 Let Q
a
and f(Q
a
) denote the two pieces of the attractor for a ⬍ a
c
. Each piece
attracts a branch of
U(p). As a crosses a
c
, Q
a
crosses one branch of the stable
manifold of p, while f(Q
a
) simultaneously crosses the other. For each a ⬎ a
c
,
each point in
U(p) is a limit point of forward iterates of Q
a
.NoticethatQ
a
is
invariant under f
2
. There is a crisis at a ⫽ a
c
for the map f
2
in which the attractor
Q
a
suddenly jumps in size to contain f(Q
a
).
10.3 (a) Hint: Show that curves g
1
and g
2
intersect at a nonzero angle at x if and only
if f(g
1
)andf(g
2
) intersect at a nonzero angle at f(x).
10.6 Hint: Sufficiently large forward iterates of an endpoint or a crossing point must
be on the local stable manifold.
10.7 The unstable manifold of the cat map is the line y ⫽ [(1 ⫹
5) 2]x, treated
modulo one. In the unit square, it consists of infinitely many parallel line segments
dense in the square; on the torus, it is a curve which wraps around infinitely many
times without intersecting itself. The stable manifold y ⫽ [(1 ⫺
5) 2]x does
the same, crossing the unstable manifold at right angles because the vectors
(1, (1 ⫹
5) 2) and (1, (1 ⫺
5) 2) are perpendicular.
10.9 The part of the stable manifold lying within the unit square consists of the
horizontal length-one segments whose y-coordinates are middle-third Cantor set
584
A NSWERS AND H INTS TO S ELECTED E XERCISES
endpoints—in other words, 0 ⱕ x ⱕ 1,y⫽ p 3
k
for integers k ⱖ 0 and integers
0 ⱕ p ⱕ 3
k
. These segments are connected via the part of the stable manifold
lying outside the unit square in the order y ⫽ 0,y ⫽ 1,y ⫽ 2 3,y ⫽ 1 3,y ⫽
2 9,y ⫽ 7 9,y ⫽ 8 9,y ⫽ 1 9,.... The part of the unstable manifold lying
within the unit square consists of the vertical length-one segments whose x-
coordinates are middle-third Cantor set endpoints. The second fact follows from
the first by noticing that the unstable manifold of f is the stable manifold of the
inverse map
f
⫺1
(x, y) ⫽
(3x, y 3) if 0 ⱕ x ⱕ 1 3
(3 ⫺ 3x, 1 ⫺ y 3) if 2 3 ⱕ x ⱕ 1,
which is f with the roles of x and y reversed. (In fact, f
⫺1
is conjugate to f via the
conjugacy C(x, y) ⫽ (y, x)).
T11.1 Yes. The chain rule shows that f
2
⫽ f
(f(x))f
(x), and f
2
⫽ f
(f(x))f
(x)
2
⫹
f
(f(x))f
(x) ⫽ f
(x)[f
(x)
2
⫹ f
(x)] ⫽ f
(x)[1 ⫺ 1] ⫽ 0.
T11.3 (a) See, for fixed d ⫽ 0, how the family of graphs y ⫽ f
a,d
(x) varies with a.
T11.5 Hint: See the proof of Theorem 11.9.
T11.8 (a) Show that e
1
⫹ e
2
⫽ m
11
⫹ m
22
and e
1
e
2
⫽ det A ⫽ m
11
m
22
⫺ m
21
m
12
.
T11.14 (b) 0 ⬍ a ⬍ 2.
T11.15 (a) First show that for r ⬍ 1, the sequence is increasing and bounded and,
therefore, must have a limit. Conclude that the limit is the fixed point r ⫽ 1. A
similar argument works for r ⬎ 1. (b) e
⫺2
b
.(c)r ⱖ 0.
T11.16 Linked. The outside edge of a M
¨
obius strip, regarded as a closed curve, is linked
with the equator.
T11.17 Hint: Let g be a periodic orbit and let L be a line segment which crosses g
transversally. A period-doubling bifurcation from g would produce a periodic
orbit which crosses L on both sides of g, violating uniqueness of solutions.
11.2 There is a period-doubling bifurcation of the origin for a ⫽ 1. The period-two
orbit created is a sink for 1 ⬍ a ⬍
6 ⫺ 1, when it loses stability in a period-
doubling bifurcation to a period-four sink.
11.3 Period-doubling bifurcation of the origin for a ⫽⫺1.
11.5 (a) Saddle-node at a ⫽ 1, no period-doubling. (b) No saddle-node, period-
doubling at a ⫽ 1. (c) Saddle-node at a ⫽ e, period-doubling at a ⫽⫺1 e.
(d) Period-doubling at a ⫽ 2.
11.6 The origin is a fixed point for all a. The other path of fixed points (⫺(1 ⫺ a)
3
,
⫺(1 ⫺ a)
2
) crosses the origin at a ⫽ 1, at which point the origin changes from
attracting to repelling.
585
A NSWERS AND H INTS TO S ELECTED E XERCISES
11.10 (b) Hint: 0 is attracting for g if and only if 0 is attracting for g ◦ g. Find the power
series expansion for g ◦ g, and apply part(a) of this problem.
11.12 The minus sign can be moved to the denominator of the left side of the equation.
Reverse the direction of time.
11.14 (c) In Chapter 7, H(x, y) ⫽⫺c ln x ⫹ dx ⫺ a ln y ⫹ by was shown to be a
Lyapunov function for the equilibrium (
c
d
,
a
b
)whenr ⫽ 0. Calculate
dH
dt
⫽
H
x
dx
dt
⫹
H
y
dy
dt
for r ⫽ 0.
T12.1 Hint: If 兵x
1
,x
2
,x
3
其 is a periodic orbit of f
a
,thenf
3
a
(x
1
) ⫽ f
3
a
(x
2
) ⫽ f
3
a
(x
3
)bythe
Chain Rule.
T12.5 Assume that there are infinitely many zeros of g in the interval. First show that
g
can be zero at only a finite subset of these zeros. Is this fact compatible with
the hypotheses?
T12.8 Hint: Check the direction of the snake that returns to a ⫽ a
c
to obtain a contra-
diction.
T12.9 The snake cannot have an endpoint.
T12.11 First argue that the snake must pass through period–doubling bifurcation orbits
of these periods.
12.1 Use the fact that f
2
is conjugate to the logistic map g(x) ⫽ 4x(1 ⫺ x), and hence
to the tent map T
2
, to show that all periodic orbits of f
2
are (hyperbolic) repellers.
12.3 Hint: Use symbol sequences to classify regular and flip unstable orbits.
586
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