Non-Chaotic Systems 151
with Hamiltonian:
x
4
+ y
4
4
− b
2
x
2
+ y
2
2
There are five stable equilibrium points, one at the origin (0, 0), and the other four
located at the symmetric points (b, b), (b, −b), (−b, b) and (−b, −b). There are also
four unstable equilibrium points located at (0, b), (0, −b), (b, 0) and (−b, 0). The
level curve passing through those equilibrium points has level −
b
4
4
. In Figure 7.8(a),
a heavy line marks this critical level curve.
Introducing a new parameter a into (7.26), the following system arises:
˙x = −y(y
2
− b
2
)
˙y = x(x
2
− a
2
)
(7.27)
Its Hamiltonian is
x
4
+ y
4
4
−
a
2
x
2
+ b
2
y
2
2
This system has a richer structure, as illustrated in Figure 7.8(b) (a = 0.7, b = 0.5).
Two distinct level curves appear, for values H = −
1
4
a
4
and H = −
1
4
b
4
respectively.
The first curve is associated with two 8-shape forms of Figure 7.8(b). The second
curve is associated with the other more complicated form, also marked with a heavy
line. The four points located on the x axis (y = 0) are given by:
x = ±
q
a
2
±
√
a
4
− b
4
If the parameter a is introduced in both equations, then an interesting system
arises:
˙x = −y(y − a)(y − b)
˙y = x(x − a)(x − b)
(7.28)
Its Hamiltonian is:
H(x, y) =
x
4
+ y
4
4
− (a + b)
x
3
+ y
3
3
+ ab
x
2
+ y
2
2
The model, with parameters equal to a = 0.8 and b = 0.3, is illustrated in Fig-
ure 7.8(c).
There is an outer characteristic level curve, with level h
b
= −
1
12
b
4
+
1
6
ab
3
, and
an inner characteristic level curve, with level h
a
= −
1
12
a
4
+
1
6
a
3
b. There are five
stable points located at (0, 0), (a, a), (b, b), (a, 0) and (0, a), and four unstable points
at (b, 0), (0, b), (a, b) and (b, a).
When b =
1
2
a, then h
b
= 0, and the system leads to a perfect symmetry, illustrated
in Figure 7.8(d) when the parameters are a = 0.8 and b =
a
2
= 0.4.