Thus, as a solution of the homogeneous boundary conditions [i.e., u 0
ð Þ ¼ _
u 0
ð Þ ¼ 0]
we get
u t
ð Þ ¼
1
mω
Z t
0
e
À
r
2m tÀτ
ð
Þ e
iΩt sin ω t À τ
ð
Þdτ,
ð10:212Þ
where t is an arbitrary positive or negative time. Equation (10.212) shows that with
the real part we have an external field cosΩt and that with the imaginary part we have
an external field sinΩt. To calculate (10.212) we use
sin ω t À τ
ð
Þ¼
1
2i
e
iω tÀτ
ð
Þ
À e
Àiω tÀτ
ð
Þ
h
i
:
ð10:213Þ
Then, the equation can readily be solved by integration of exponential functions,
even though we have to do somewhat lengthy (but straightforward) calculations.
Thus for the real part (i.e., the external field is cosΩt), we get a solution
Cu t
ð Þ ¼
1
m
r
m
ΩsinΩt þ
1
m
r
2m
2
cos Ωt À
1
m
Ω
2
À ω
2
À
Á
cos Ωt
þ
1
m
e
À
r
2m t
Ω
2
À ω
2
À
Á
cos ωt À
r
2m
1
ω
Ω
2
þ ω
2
À
Á
sin ωt
h
À
r
2m
2
cos ωt À
r
2m
3 1
ω
sin ωt,
ð10:214Þ
where C is a constant [i.e., a constant denominator of u(t)] expressed as
C ¼ Ω
2
À ω
2
À
Á 2 þ
r
2
2m 2 Ω
2
þ ω
2
À
Á þ
r
2m
4 :
ð10:215Þ
For the imaginary part (i.e., the external field is sinΩt) we get
Cu t
ð Þ ¼ À
1
m
Ω
2
À ω
2
À
Á
sin Ωt À
1
m
r
m
ΩcosΩt þ
1
m
r
2m
2
sin Ωt
þ
1
m
e
À
r
2m t Ω
ω
Ω
2
À ω
2
À
Á
sin ωt þ
r
m
Ωcosωt þ
r
2m
2 Ω
ω
sin ωt
!
:
ð10:216Þ
In Fig. 10.5 we show an example that depicts the positions of a damped oscillator as
a function of time. In Fig. 10.5a, an amplitude of an envelope gradually diminishes
with time. An enlarged diagram near the origin (Fig. 10.5b) clearly reflects the initial
conditions u 0
ð Þ ¼ _
u 0
ð Þ ¼ 0 . In Fig. 10.5, we put m ¼ 1 [kg], Ω ¼ 1
1
s
 Ã
, ω ¼
0:94
1
s
 Ã
, and r ¼ 0:006
kg
s
 Ã
. In the above calculations, if
r
m is small enough (i.e.,
damping is small enough), the third order and fourth order of
r
m may be ignored and
the approximation is precise enough.
10.6 Initial Value Problems (IVPs)
423
ð Þ ¼ _
u 0
ð Þ ¼ 0]
we get
u t
ð Þ ¼
1
mω
Z t
0
e
À
r
2m tÀτ
ð
Þ e
iΩt sin ω t À τ
ð
Þdτ,
ð10:212Þ
where t is an arbitrary positive or negative time. Equation (10.212) shows that with
the real part we have an external field cosΩt and that with the imaginary part we have
an external field sinΩt. To calculate (10.212) we use
sin ω t À τ
ð
Þ¼
1
2i
e
iω tÀτ
ð
Þ
À e
Àiω tÀτ
ð
Þ
h
i
:
ð10:213Þ
Then, the equation can readily be solved by integration of exponential functions,
even though we have to do somewhat lengthy (but straightforward) calculations.
Thus for the real part (i.e., the external field is cosΩt), we get a solution
Cu t
ð Þ ¼
1
m
r
m
ΩsinΩt þ
1
m
r
2m
2
cos Ωt À
1
m
Ω
2
À ω
2
À
Á
cos Ωt
þ
1
m
e
À
r
2m t
Ω
2
À ω
2
À
Á
cos ωt À
r
2m
1
ω
Ω
2
þ ω
2
À
Á
sin ωt
h
À
r
2m
2
cos ωt À
r
2m
3 1
ω
sin ωt,
ð10:214Þ
where C is a constant [i.e., a constant denominator of u(t)] expressed as
C ¼ Ω
2
À ω
2
À
Á 2 þ
r
2
2m 2 Ω
2
þ ω
2
À
Á þ
r
2m
4 :
ð10:215Þ
For the imaginary part (i.e., the external field is sinΩt) we get
Cu t
ð Þ ¼ À
1
m
Ω
2
À ω
2
À
Á
sin Ωt À
1
m
r
m
ΩcosΩt þ
1
m
r
2m
2
sin Ωt
þ
1
m
e
À
r
2m t Ω
ω
Ω
2
À ω
2
À
Á
sin ωt þ
r
m
Ωcosωt þ
r
2m
2 Ω
ω
sin ωt
!
:
ð10:216Þ
In Fig. 10.5 we show an example that depicts the positions of a damped oscillator as
a function of time. In Fig. 10.5a, an amplitude of an envelope gradually diminishes
with time. An enlarged diagram near the origin (Fig. 10.5b) clearly reflects the initial
conditions u 0
ð Þ ¼ _
u 0
ð Þ ¼ 0 . In Fig. 10.5, we put m ¼ 1 [kg], Ω ¼ 1
1
s
 Ã
, ω ¼
0:94
1
s
 Ã
, and r ¼ 0:006
kg
s
 Ã
. In the above calculations, if
r
m is small enough (i.e.,
damping is small enough), the third order and fourth order of
r
m may be ignored and
the approximation is precise enough.
10.6 Initial Value Problems (IVPs)
423