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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