Thus, defining F(t) as
F t
ð Þ
cos e b sin ωt
sin e b sin ωt
À sin e b sin ωt
cos e b sin ωt
0
B
@
1
C
A,
ð15:178Þ
and taking account of (15.174), we recast the homogeneous equation (15.173) as
dF t
ð Þ
dt
¼ F t
ð ÞA,
A
0
b cos ωt
Àb cos ωt
0
:
ð15:179Þ
Equation (15.179) is formally the same as (15.47), even though F(t) is not given
in the form of exp tA. This enables us to apply the general scheme mentioned in Sect.
15.3.2 to the present case. That is, we are going to address the given problem using
the method of variation of constants such that
x t
ð Þ ¼ k t
ð ÞF t
ð Þ,
ð15:78Þ
where we define x(t) (ξ(t) ζ(t)) and k(t) (u(t) v(t)) as a variable constant.
Hence, using (15.80)–(15.90) we should be able to find the solution of (15.172)
described by
x t
ð Þ ¼ x t 0
ð Þℜ t 0 , t
ð Þþ
Z t
t 0
ds b s
ð Þℜ s, t
ð Þ
½
Š ,
ð15:90Þ
where b(s) (acosωt 0), i.e., the inhomogeneous term in (15.172). The resolvent
matrix R s, t
ð Þ is expressed as
R s, t
ð Þ ¼ F s
ð Þ
À1 F t
ð Þ ¼
cos f t
ð Þ À f s
ð Þ
½
Š
sin f t
ð Þ À f s
ð Þ
½
Š
À sin f t
ð Þ À f s
ð Þ
½
Š cos f t
ð Þ À f s
ð Þ
½
Š
,
ð15:180Þ
where f(t) is given by (15.177). With F(s)
À1 , we simply take a transpose matrix of
(15.178), because it is an orthogonal matrix. Thus, we obtain
F s
ð Þ
À1 ¼
cos e b sin ωs
À sin e b sin ωs
sin e b sin ωs
cos e b sin ωs
0
B
@
1
C
A:
The resolvent matrix (15.180) possesses the properties described as (15.85)–
(15.87). These can readily be checked. Also, we become aware that F(t) of
614
15 Exponential Functions of Matrices