Moreover, differentiating (15.112) with respect to t, we recover the original form
of (15.95). The confirmation is left for readers.
Remarks on Example 15.1: We become aware of several points in Example 15.1.
(i) The resolvent matrix given by (15.106) can be obtained by replacing t of
(15.104) with t À s. This is one of important characteristics of the resolvent
matrix that is derived from an exponential function of a constant matrix.
To see it, if A is a constant matrix, tA and ÀsA commute. By the definition (15.7)
of the exponential function we have
exp tA ¼ E þ tA þ
1
2!
t
2 A
2
þ Á Á Á þ
1
ν!
t
ν A
ν
þ Á Á Á,
ð15:113Þ
exp ÀsA
ð
Þ¼ E þ ÀsA
ð
Þþ
1
2!
Às
ð Þ
2 A
2
þ Á Á Á þ
1
ν!
Às
ð Þ
ν A
ν
þ Á Á Á:
ð15:114Þ
Both (15.113) and (15.114) are polynomials of a constant matrix A and,
hence, exp tA and exp(ÀsA) commute as well. Then, from Theorem 15.2 and
Property (1) of (15.28), we get
exp tA À sA
ð
Þ¼ exp tA
ð
Þ exp ÀsA
ð
Þ
½
¼ exp ÀsA
ð
Þ
½
exp tA
ð
Þ
¼ exp ÀsA þ tA
ð
Þ¼exp t À s
ð
ÞA
½
:
ð15:115Þ
That is, using (15.28), (15.72), and (15.83), we get
R s, t
ð Þ ¼ F s
ð Þ
À1 F t
ð Þ ¼ exp sA
ð Þ
À1 exp tA
ð Þ ¼ exp ÀsA
ð
Þexp tA
ð Þ
¼ exp t À s
ð
ÞA
½
¼F t À s
ð
Þ:
ð15:116Þ
This implies that once we get exp tA, we can safely obtain the resolvent matrix
by automatically replacing t of (15.72) with t À s. Also, exp tA and exp(ÀsA) are
commutative. In the present case, therefore, by exchanging the order of products
of F(s)
À1 and F(t) in (15.106), we have the same result as (15.106) such that
R s, t
ð Þ ¼ F t
ð ÞF s
ð Þ
À1 ¼
1
2
e
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
Àe
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
Àe
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
e
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
!
: ð15:117Þ
(ii) The resolvent matrix of the present example is real symmetric. This is because
if A is real symmetric, i.e., A
T
¼ A, we have
A
n
ð Þ
T ¼ A
T
À Á n ¼ A
n ,
ð15:118Þ
that is A
n is real symmetric as well. From (15.7), exp A is real symmetric
accordingly. In a similar manner, if A is Hermitian, exp A is also Hermitian.
15.3 System of Differential Equations
603
of (15.95). The confirmation is left for readers.
Remarks on Example 15.1: We become aware of several points in Example 15.1.
(i) The resolvent matrix given by (15.106) can be obtained by replacing t of
(15.104) with t À s. This is one of important characteristics of the resolvent
matrix that is derived from an exponential function of a constant matrix.
To see it, if A is a constant matrix, tA and ÀsA commute. By the definition (15.7)
of the exponential function we have
exp tA ¼ E þ tA þ
1
2!
t
2 A
2
þ Á Á Á þ
1
ν!
t
ν A
ν
þ Á Á Á,
ð15:113Þ
exp ÀsA
ð
Þ¼ E þ ÀsA
ð
Þþ
1
2!
Às
ð Þ
2 A
2
þ Á Á Á þ
1
ν!
Às
ð Þ
ν A
ν
þ Á Á Á:
ð15:114Þ
Both (15.113) and (15.114) are polynomials of a constant matrix A and,
hence, exp tA and exp(ÀsA) commute as well. Then, from Theorem 15.2 and
Property (1) of (15.28), we get
exp tA À sA
ð
Þ¼ exp tA
ð
Þ exp ÀsA
ð
Þ
½
¼ exp ÀsA
ð
Þ
½
exp tA
ð
Þ
¼ exp ÀsA þ tA
ð
Þ¼exp t À s
ð
ÞA
½
:
ð15:115Þ
That is, using (15.28), (15.72), and (15.83), we get
R s, t
ð Þ ¼ F s
ð Þ
À1 F t
ð Þ ¼ exp sA
ð Þ
À1 exp tA
ð Þ ¼ exp ÀsA
ð
Þexp tA
ð Þ
¼ exp t À s
ð
ÞA
½
¼F t À s
ð
Þ:
ð15:116Þ
This implies that once we get exp tA, we can safely obtain the resolvent matrix
by automatically replacing t of (15.72) with t À s. Also, exp tA and exp(ÀsA) are
commutative. In the present case, therefore, by exchanging the order of products
of F(s)
À1 and F(t) in (15.106), we have the same result as (15.106) such that
R s, t
ð Þ ¼ F t
ð ÞF s
ð Þ
À1 ¼
1
2
e
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
Àe
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
Àe
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
e
À tÀs
ð Þ
þ e
3 tÀs
ð Þ
!
: ð15:117Þ
(ii) The resolvent matrix of the present example is real symmetric. This is because
if A is real symmetric, i.e., A
T
¼ A, we have
A
n
ð Þ
T ¼ A
T
À Á n ¼ A
n ,
ð15:118Þ
that is A
n is real symmetric as well. From (15.7), exp A is real symmetric
accordingly. In a similar manner, if A is Hermitian, exp A is also Hermitian.
15.3 System of Differential Equations
603