Case (iii):
Suppose that the damping is relatively weak such that the characteristic equation
has two complex roots. Let us examine further details of this case following the
prescriptions of IVPs. We divide (10.198) by m for the sake of easy handling of the
differential equation such that
d
2 u
dt 2 þ
r
m
du
dx
þ
k
m
u ¼
1
m
d x
ð Þ:
Putting
ω
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
À
r 2
4m 2 þ
k
m
r
,
ð10:202Þ
we get a fundamental set of solutions described as
u t
ð Þ ¼ exp À
rt
2m
exp Çiωt
ð
Þ:
ð10:203Þ
Given BCs, following (10.172) we get as a Green’s function
G t, τ
ð Þ ¼
u 2 t
ð Þu 1 τ
ð Þ À u 1 t
ð Þu 2 τ
ð Þ
W u 1 τ
ð Þ, u 2 τ
ð Þ
ð
Þ
Θ t, τ
ð Þ
¼
1
ω
e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
ÞΘ t, τ
ð Þ,
ð10:204Þ
where u 1 t
ð Þ ¼ exp À
rt
2m
À
Á
exp iωt
ð Þ and u 2 t
ð Þ ¼ exp À
rt
2m
À
Á
exp Àiωt
ð
Þ.
We examine whether G(t, τ) is eligible for the Green’s function as follows:
dG
dt
¼
1
ω
À
r
2m
e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
Þþωe
À
r
2m tÀτ
ð
Þ cos ω t À τ
ð
Þ
h
i
Θ t, τ
ð Þ
þ
1
ω
e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
Þδ t À τ
ð
Þθ τ
ð Þ þ δ τ À t
ð
Þθ Àτ
ð Þ
½
Š :
ð10:205Þ
The second term of (10.205) vanishes because sinω(t À τ)δ(t À τ) ¼ 0 Á δ(t À τ) ¼ 0.
Thus,
d
2 G
dt 2 ¼
1
ω
À
r
2m
2
e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
ÞÀ
r
m
ωe
À
r
2m tÀτ
ð
Þ cos ω t À τ
ð
Þ
Àω
2 e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
ÞŠΘ t, τ
ð Þ
þ
1
ω
À
r
2m
e
À
r
2m tÀτ
ð
Þ sin ω t À τ
ð
Þþωe
À
r
2m tÀτ
ð
Þ cos ω t À τ
ð
Þ
h
i
 δ t À τ
ð
Þθ τ
ð Þ þ δ τ À t
ð
Þθ Àτ
ð Þ
½
Š :
ð10:206Þ
10.6 Initial Value Problems (IVPs)
421