h 1 ¼
1
8
cA
3 e
i3T 0 À
1
2
qx
1
x þ 2
Ae
iðx þ 1ÞT 0 þ
1
x À 2
Ae
iðxÀ1ÞT 0
þ cc:
ð3:116Þ
Usually, when the excitation is stationary, one is primarily interested in stationary (post-transient) behavior. The homogeneous part of the solution describes
the transient response is therefore not needed.
The function A(T 1 ) is determined from the solvability condition (3.115) by first
expressing it in polar form:
A ¼
1
2
ae
iu
; aðT 1 Þ; uðT 1 Þ 2 R;
ð3:117Þ
Substituting into (3.115) we obtain, when separating real and imaginary parts:
a
0
¼ Àba;
u
0
¼
3
8
ca
2
:
ð3:118Þ
Solving for a and u one finds that
a ¼ a 0 e
ÀbT 1 ;
u ¼ u 0 À
3
8
a
2
0
c
2b
e
À2bT 1 ;
ð3:119Þ
where a 0 and u 0 are arbitrary constants. The first equation shows that a ! 0 as
T 1 ! ∞, since b > 0. Then by (3.117 also A ! 0, and by (3.113, (3.116) and
(3.108) the stationary response becomes h(s) = 0. Consequently the state h(s) = 0
is a stable equilibrium when x is away from 2, so that oscillations of the pendulum
damp out in time no matter how wildly the pendulum support is shaken up and
down.
Equations (3.118) are the modulation equations of this perturbation analysis.
They govern modulations in slow time T 1 of the amplitudes and phases of terms
oscillating harmonically in fast time T 0 . For determining stationary responses one
may attempt solving the modulation equations to see what happens with amplitudes
and phases as time approaches infinity. This approach was used above. Typically,
however, the modulation equations are not readily solved – and in fact do not need
to be solved for obtaining stationary responses. Instead one seeks the singular
points, i.e. the equilibriums, of the modulation equations – and examines their
stability with respect to slight disturbances. For the singular points it holds by
definition that a′ = u′ = 0, which describes a stationary response in which amplitudes and phases do not change in time. The following section will illustrate this
technique.
3.6 The Forced Response – Multiple Scales Analysis
133
1
8
cA
3 e
i3T 0 À
1
2
qx
1
x þ 2
Ae
iðx þ 1ÞT 0 þ
1
x À 2
Ae
iðxÀ1ÞT 0
þ cc:
ð3:116Þ
Usually, when the excitation is stationary, one is primarily interested in stationary (post-transient) behavior. The homogeneous part of the solution describes
the transient response is therefore not needed.
The function A(T 1 ) is determined from the solvability condition (3.115) by first
expressing it in polar form:
A ¼
1
2
ae
iu
; aðT 1 Þ; uðT 1 Þ 2 R;
ð3:117Þ
Substituting into (3.115) we obtain, when separating real and imaginary parts:
a
0
¼ Àba;
u
0
¼
3
8
ca
2
:
ð3:118Þ
Solving for a and u one finds that
a ¼ a 0 e
ÀbT 1 ;
u ¼ u 0 À
3
8
a
2
0
c
2b
e
À2bT 1 ;
ð3:119Þ
where a 0 and u 0 are arbitrary constants. The first equation shows that a ! 0 as
T 1 ! ∞, since b > 0. Then by (3.117 also A ! 0, and by (3.113, (3.116) and
(3.108) the stationary response becomes h(s) = 0. Consequently the state h(s) = 0
is a stable equilibrium when x is away from 2, so that oscillations of the pendulum
damp out in time no matter how wildly the pendulum support is shaken up and
down.
Equations (3.118) are the modulation equations of this perturbation analysis.
They govern modulations in slow time T 1 of the amplitudes and phases of terms
oscillating harmonically in fast time T 0 . For determining stationary responses one
may attempt solving the modulation equations to see what happens with amplitudes
and phases as time approaches infinity. This approach was used above. Typically,
however, the modulation equations are not readily solved – and in fact do not need
to be solved for obtaining stationary responses. Instead one seeks the singular
points, i.e. the equilibriums, of the modulation equations – and examines their
stability with respect to slight disturbances. For the singular points it holds by
definition that a′ = u′ = 0, which describes a stationary response in which amplitudes and phases do not change in time. The following section will illustrate this
technique.
3.6 The Forced Response – Multiple Scales Analysis
133