and thus,
ω
1
= −
n
2
2
√
α
[
3
4n
2
+
3A
2
2n
2
].
Thus, the solution for x
1
is given by
x
1
=
B
1
α(1 − n
2
)
cos(τ) −
B
2
8α
cos(
3τ
n
) −
B
3
9α(n
2
−1)
cos(3τ)
−
B
4
α(n − 1)(n − 3)
cos((1 −
2
n
)τ) −
B
5
α(n + 1)(n + 3)
cos((1 +
2
n
)τ)
−
B
6
4αn(n −1)
cos((2 −
1
n
)τ) −
B
7
4αn(n + 1)
cos((2 +
1
n
)τ).
The total solution can be written as
ω = n
√
α + εω
1
,
x(τ) = x
0
+ εx
1
.
293
10.22. For the combination harmonic response, derive Equation 10.80.
Solution:
Solution: Start with
..
x + ω
2
n
x = −εβω
2
n
x
3
+ F
1
cos(Ω
1
t) + F
2
cos(Ω
2
t)
Expand x as x(t) = x
0
+ εx
1
, and substitute into the equation of motion to obtain
..
x
0
+ ε
..
x
1
+ ω
2
n
(x
0
+ εx
1
) = −εβω
2
n
(x
0
+ εx
1
)
3
+ F
1
cos(Ω
1
t) + F
2
cos(Ω
2
t).
Separating by terms produces
..
x
0
+ ω
2
n
x
0
= F
1
cos(Ω
1
t) + F
2
cos(Ω
2
t)
..
x
1
+ ω
2
n
x
1
= −βω
2
n
x
3
0
.
The solution for x
0
is given by
x
0
(t) = G
1
cos(Ω
1
t) + G
2
cos(Ω
2
t).
Substitution of x
0
into the equation for x
1
produces
..
x
1
+ ω
2
n
x
1
= −βω
2
n
(G
1
cos(Ω
1
t) + G
2
cos(Ω
2
t))
3
.
Expansion of the cubic term produces
..
x
1
+ ω
2
n
x
1
= −βω
2
n
(G
3
1
cos
3
(Ω
1
t) + G
3
2
cos
3
(Ω
2
t) + 3G
1
cos(Ω
1
t)G
2
2
cos
2
(Ω
2
t) + 3G
2
1
cos
2
(Ω
1
t)G
2
cos(Ω
2
t)).
Using the following trigonometric identities given by
(cos(x))
3
=
1
4
(3 cos(x) + cos(3x))
cos
2
(x) cos(y) =
1
4
(cos(y −x) + cos(2x + y) + 2 cos(y)),
and simplifying produces
..
x
1
+ ω
2
n
x
1
= C
1
cos(Ω
1
t) + C
2
cos(Ω
2
t)
+C
3
[cos((2Ω
1
+ Ω
2
)t) + cos((2Ω
1
−Ω
2
)t)]
+C
4
[cos((Ω
1
+ 2Ω
2
)t) + cos((Ω
1
−2Ω
2
)t)]
+C
6
cos(3Ω
1
t) + C
6
cos(3Ω
2
t),
where
C
1
= −
3
4
βω
2
n
G
1
(G
2
1
+ G
2
2
),
C
2
= −
3
4
βω
2
n
G
2
(2G
2
1
+ G
2
2
),
C
3
= −
3
4
βω
2
n
G
2
1
G
2
,
C
4
= −
3
4
βω
2
n
G
1
G
2
2
,
C
5
= −
1
4
βω
2
n
G
3
1
,
C
6
= −
1
4
βω
2
n
G
3
2
.
294
Section 10.4: The Mathieu Equation
10.23. Derive Equation 10.93.
Solution: For n = 2, we have
θ
0
= cos(2t).
Then,
..
θ
1
+ 4θ
1
= −(δ
1
+ 2 cos(2t))θ
0
= −(δ
1
+ 2 cos(2t)) cos(2t)
= −δ
1
cos(2t) + 2 cos
2
(2t)
= −δ
1
cos(2t) + cos(4t) + 1.
For periodicity, δ
1
= 0, and the solution for θ
1
is given by
θ
1
(t) =
1
12
cos(4t) −
1
4
.
Substituting this into the equation for θ
2
yields
..
θ
2
+ 4θ
2
= −(δ
1
+ 2 cos(2t))θ
1
−δ
2
θ
0
= −2 cos(2t)(
1
12
cos(4t) −
1
4
) − δ
2
cos(2t)
= −
1
6
cos(2t) cos(4t) +
1
2
cos(2t) − δ
2
cos(2t)
= −
1
12
(cos(2t) + cos(6t)) +
1
2
cos(2t) − δ
2
cos(2t)
= −
1
12
cos(6t) − (δ
2
−
1
2
+
1
12
) cos(2t).
To satisfy periodicity, then δ
2
=
5
12
. Thus,
δ = 4 +
5
12
ε
2
+ ...
295
10.24. Derive Equation 10.94.
Solution: For n=2,
θ
0
= sin(2t).
Then,
..
θ
1
+ 4θ
1
= −(δ
1
+ 2 cos(2t))θ
0
,
= −(δ
1
+ 2 cos(2t)) sin(2t),
= −δ
1
sin(2t) + 2 cos(2t) sin(2t),
= −δ
1
sin(2t) + sin(4t).
For periodicity, δ
1
= 0, and the solution for θ
1
is given by
θ
1
(t) =
1
12
sin(4t).
Substituting this into the equation for θ
2
yields
..
θ
2
+ 4θ
2
= −(δ
1
+ 2 cos(2t))θ
1
−δ
2
θ
0
,
= −2 cos(2t)(
1
12
sin(4t)) − δ
2
sin(2t),
= −
1
6
cos(2t) sin(4t) −δ
2
sin(2t),
= −
1
12
sin(2t) −
1
12
sin(6t) − δ
2
sin(2t).
To satisfy periodicity, then δ
2
= −
1
12
. Thus,
δ = 4 −
1
12
ε
2
+ ...
296
Section 10.5 The van der Pol Equation
10.25. Consider the van der Pol Equation 10.97. Suppose α is a random variable. How do the statistics
of α affect the trajectories shown in Figure 10.14 and the time traces in Figure 10.15.
Solution: Since the curves in both sets of figures depend on the value of α, uncertainty in that value results
in uncertainties in the curves. One expects that a given variance in α results in a similar variance in the
location of the curves. One can interpret the given curves in Figure 10.11 and the time traces in Figure 10.12
as mean values for a given mean value of α.
297
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