Example 15.6 Let us consider the following system of differential equations:
ð _
x t
ð Þ _
y t
ð ÞÞ ¼ x y
ð Þ
0 Àω
ω 0
þ a 0
ð
Þ:
ð15:155Þ
The matrix
0 Àω
ω 0
is characterized by an anti-Hermitian operator. This type
of operator frequently appears in Lie groups and Lie algebras that we will deal with
in Chap. 20. Here we define an operator D as
D ¼
0 À1
1 0
or ωD ¼
0 Àω
ω 0
:
ð15:156Þ
The calculation procedures will be seen in Chap. 20, and so we only show the
result here. That is, we have
exp tωD ¼
cos ωt À sin ωt
sin ωt
cos ωt
:
ð15:157Þ
We seek the resolvent matrix R s, t
ð Þ such that
R s, t
ð Þ ¼
cos ω t À s
ð
Þ Àsin ω t À s
ð
Þ
sin ω t À s
ð
Þ
cos ω t À s
ð
Þ
:
ð15:158Þ
The implication of (15.158) is simple; synthesis of a rotation of ωt and the inverse
rotation Àωs is described by ω(t À s). Physically, (15.155) represents the motion of a
point mass under an external field.
We routinely obtained the above solution using (15.90). If, for example, we solve
(15.155) under the BCs for which the point mass is placed at rest at the origin of the
xy-plane at t ¼ 0, we get a solution described by
x ¼
a
ω
sin ωt, y ¼
a
ω
cos ωt À 1
ð
Þ :
ð15:159Þ
Deleting t from (15.159), we get
x
2
þ y þ
a
ω
2 ¼
a
ω
2 :
ð15:160Þ
In other words, the point mass performs a circular motion as shown in Fig. 15.2.
In Example 15.1 we mentioned that if the matrix A that defines the system of
differential equations varies as a function of t, then the solution method based upon
exp tA does not work. Yet, we have an important case where A is not a constant
matrix. Let us consider a next illustration.
610
15 Exponential Functions of Matrices