Dynamics M5
Method of Multiple Scales Tutorial 2 Solutions
Lecturer: Andrea Cammarano

̈ + 2 sin = 0

Expand sin :
̈ + 2 ( −
3
6
) = 0
̈ + 2 − 2
3
6
= 0

Introduce weak cubic nonlinearity:
̈ + 2 − 2
3
6
= 0

Let us say ℎ =
2
6
, so:
̈ + 2 − ℎ3 = 0

Now introduce the following:
= 0 + 1

= 0 + 1
2
2
= 0
2 + 201
Hence:
̈ =
2
2

= (0
2 + 201)(0 + 1)
= 0
20 + ε0
21 + 201ϕ0 + {2
201ϕ1}
= 0
20 + ε0
21 + 201ϕ0

2 =
20 +
21

ℎ3 = ℎ(0 + 1)
3
= ℎ0
3 + {320
21 + 3
21
20 +
31
3}
= ℎ0
3
Hence we have
0
20 + ε0
21 + 201ϕ0 +
20 +
21 − ℎ0
3 = 0
Sorting into terms in 0 and 1 to get zeroth and first order perturbation equations respectively:
0: 0
20 +
20 = 0
1: 0
21 +
21 = ℎ0
3 − 201ϕ0
Zeroth order equation has the solution:
0 = ⅇ
0 + ̅ⅇ−0
Substituting this into the first order equation:
0
21 +
21 = ℎ(ⅇ
0 + ̅ⅇ−0 )3 − 201(ⅇ
0 + ̅ⅇ−0 )
Expanding the cubic and carrying out the 0 derivatives
0
21 +
21 = ℎ
3ⅇ30 + ℎ̅3ⅇ−30 + 3ℎ2̅ⅇ0 + 3ℎ̅2ⅇ−0 − 2 ⅇ0 1 + 2 ⅇ
−0 1̅
Taking a common factor of ⅇ0:
0
21 +
21 = ⅇ
0(ℎ3ⅇ20 + ℎ̅3ⅇ−40 + 3ℎ2̅ + 3ℎ̅2ⅇ−20 − 2 1 + 2 ⅇ
−20 1̅)
Repeating for ⅇ−0 :
0
21 +
21 = ⅇ
−0 (ℎ3ⅇ40 + ℎ̅3ⅇ−20 + 3ℎ2̅ⅇ20 + 3ℎ̅2 − 2 1ⅇ
20 + 2 1̅)

Secular terms:
3ℎ2̅ − 2 1 = 0
3ℎ̅2 + 2 1̅ = 0
Note that
=

2
; ̅ =
ⅇ−
2

3ℎ
3ⅇ
8
− 2 1

2
= 0 →
3
8
ℎ3ⅇ − 1(ⅇ
) = 0
3ℎ
3ⅇ−
8
+ 2 1
ⅇ−
2
= 0 →
3
8
ℎ3ⅇ− + 1(ⅇ
−) = 0

If both and depend on the timescale 1, we get

1(ⅇ
) = ′ⅇ + ′ⅇ
1(ⅇ
−) = ′ⅇ− − ′ⅇ−

Therefore
3
8
ℎ3ⅇ − (′ⅇ + ′ⅇ ) = 0
(
3
8
ℎ3 − ′ + ′) ⅇ = 0
and
3
8
ℎ3ⅇ− + (′ⅇ− − ′ⅇ−) = 0
(
3
8
ℎ3 + ′ + ′) ⅇ− = 0

The exponential ⅇ± can be written as cos( ) ± sin()

(
3
8
ℎ3 − ′ + ′) (cos( ) + sin()) = 0
3
8
ℎ3 cos( ) − ′ cos( ) + ′ cos( ) +
3
8
ℎ3 sin() + ′ sin() + ′ sin() = 0

And

(
3
8
ℎ3 + ′ + ′) (cos( ) − sin()) = 0
3
8
ℎ3 cos( ) + ′ cos( ) + ′ cos( ) −
3
8
ℎ3 sin() + ′ sin() − ′ sin() = 0

Separating the real from the imaginary terms we get

3
8
ℎ3 cos( ) + ′ cos( ) + ′ sin() = 0

3
8
ℎ3 sin() + ′ sin() − ′ cos( ) = 0

Let’s multiply the first equation by sin() and the second by cos( )

3
8
ℎ3 cos( ) sin() + ′ cos( ) sin() + ′ sin2() = 0
3
8
ℎ3 sin() cos( ) + ′ sin() cos( ) − ′ cos2( ) = 0
and let’s subtract the second from the first equation

′ sin2() + ′ cos2( ) = 0
In other terms

′ = 0 → ′ = 0
The amplitude of the oscillation must remain constant if there is no damping and no forcing.

Let’s now multiply the first equation by cos() and the second by sin( )

3
8
ℎ3 cos2( ) + ′ cos2( ) + ′ sin() cos() = 0

3
8
ℎ3 sin2() + ′ sin2() − ′ cos( ) sin() = 0

and let’s sum the second to the first equation

3
8
ℎ3(cos2( ) + sin2()) + ′(cos2( ) + sin2()) = 0
Or
3
8
ℎ3 + ′ = 0
Which gives
′ = −
3
8
2ℎ

= −
3
8
2ℎ1 = −
3
8
2ℎ0
This is obtained using the definition 1 = 0. Remembering that ℎ =
2
6
we have

= −
1
16
2 0

Remembering that 0 = the solution becomes
0 = cos ( 0 −
1
16
2 0) = cos ( (1 −
1
16
2) )
The correction term 1can be found from the first order equation where the secular terms have been
eliminated:
0
21 +
21 = ℎ
3ⅇ30 + ℎ̅3ⅇ−30

The solution to this second order linear differential equation with constant coefficient is given by
1 = 1 + 1
Where 1is the solution of the homogeneous equation and 1 is the particular integral. The solution of
1would be formally identical to 0 and its contribution can be easily assimilated to 0by choosing an
opportune constant .

The solution of the particular integral will be in the form
1 = 1ⅇ
30 + 2ⅇ
−30
01 = 31ⅇ
30 − 32ⅇ
−30
0
21 = −9
21ⅇ
30 − 922ⅇ
−30

Substituting in the previous equation
−921ⅇ
30 − 922ⅇ
−30 + 21ⅇ
30 + 22ⅇ
−30 = ℎ3ⅇ30 + ℎ̅3ⅇ−30
Separating the terms with positive exponents from those with negative exponents we have
−921 +
21 = ℎ
3 → 1 = −
ℎ3
82

−922 +
22 = ℎ̅
3 → 2 = −
ℎ̅3
82

Using the definition of and ̅

1 = −
ℎ3
82
= −
ℎ3ⅇ3
82

2 = −
ℎ̅3
82
= −
ℎ3ⅇ−3
82

So
1 = 1ⅇ
30 + 2ⅇ
−30 = −
ℎ3
82
(ⅇ3(0+) + ⅇ−3(0+)) = −
ℎ3
82
cos(3(0 + ))
But = −
1
16
2 0

1 = 1 = −
ℎ3
82
cos (30 (1 −
1
16
2))
The approximated solution
= 0 + 1 = cos ( (1 −
1
16
2) ) −
ℎ3
82
cos (3 (1 −
1
16
2))
Removing h from the equation

= cos ( (1 −
1
16
2) ) −
3
48
cos (3 (1 −
1
16
2))  