0387986995Basic TheoryC

78 IV. GENERAL THEORY OF LINEAR SYSTEMS

In this case, SN = NS = N and N2 = O. Let Vj = Image (PI(A)) (j = 1,2).

Then, by virtue of (1) of Lemma IV-1-10, Pj(A)p" = p" for all IT E Vj (j=1,2).

		14	1	and	0
Furthermore, V1 is spanned by 19			and V2 is spanned by 0		1	Set
		12	0		-2
14	1	0

Pa = 19	0	1	. Then,
12	0	-2
			11	0	0	Pp 1 NPo = 2 [00
Po 1 SPo =			0	1	0		1
			0	0	1	0	1

-1 .

-1

It is noteworthy that there is only one linearly independent eigenvector x =

for the eigenvalue A2 = 1.

It is not difficult to make a program for calculation of S and N with a computer. For more examples of calculation of S and N, see [HKSI.

IV-2. Homogeneous systems of linear differential equations

In this section, we explain the basic results concerning the structure of solutions of a homogeneous system of linear differential equations given by

(IV.2.1)

dt = A(t)y',

where the entries of the n x n matrix A(t) are continuous on an interval I = It : a < t < b). Let us prove the following basic theorem.

Theorem 1V-2-1. The solutions of (1V2. 1) forms an n-dimensional vector space over C.

We break the entire proof into three observations.

Observation IV-2-2. Any linear combination of a finite number of solutions of (IV.2.1) is also a solution of (IV.2.1). We can prove the existence of n linearly independent solutions of (IV.2.1) on the interval Z by using Theorem I-3-5 with n linearly independent initial conditions at t = to. Notice that each column vector of a solution Y of the differential equation

(IV.2.2)

dt = A(t)Y

on an n x n unknown matrix Y is a solution of system (IV.2. 1). Therefore, construct- ing an invertible solution Y of (IV.2.2), we can construct n linearly independent

solutions of (IV.2.1) all at once. If an n x n matrix Y(t) is a solution of equation

(IV.2.2)on an interval I={t:a<t<b}and Y(t)EGL(n,C)for all tE1,then

Y(t) is called a fundamental matrix solution of system (1V.2.1) on.T. Furthermore, n columns of a fundamental matrix solution Y(t) of (IV.2.2) are said to form a

fundamental set of n linearly independent solutions of (IV.2.1) on the interval T.

2. HOMOGENEOUS SYSTEMS OF LINEAR DIFF. EQUATIONS

79

Observation IV-2-3. Let 4(t) be a solution of (IV.2.2) on Z. Also, let *(t) be a solution of the adjoint equation of (IV.2.2):

(IV.2.3)	dZ	= -ZA(t)
	dt

on the interval Z, where Z is an n x n unknown matrix. Then,

-W(t)A(t)4s(t) + $(t)A(t)4;(t) = 0.

dt

This implies that the matrix %P(t)4i(t) is independent of t. Therefore, W(t)4i(t) =

%P(r)t(r) for any fixed point r E Z and for all t E Z. Note that the initial values

44(r) and %P(r) at t = r can be prescribed arbitrarily. In particular, in the case

when 4?(r) E GL(n,C), by choosing %P(r) =

we obtain WY(t)4i(t) = In

for all t E Z. Thus, we proved the following lemma.

Lemma IV-2-4. Let an n x n matrix 4i(t) be a solution of (IV.2.2) on the interval

Z. Then, 45(t) is invertible for all t E I (i.e., a fundamental matrix solution of

(IV.2.1)) if 4i(r) is invertible for some r E Z. Furthermore, 4 (t)-1 is the unique solution of (IV.2.8) on Z satisfying the initial condition Z(r) = 4i(r)-1.

Observation IV-2-5. Denote by 4?(t; r) the unique solution of the initial-value problem

(IV.2.4)	-	= A(t)Y,	Y(r) = In,
		dt

where r E Z. Then, 1(t; r) E GL(n, C) for all t E Z. The general structure of solutions of (IV.2.1) and (IV.2.2) are given by the following theorem, which can be easily verified.

Theorem IV-2-6. The C"-valued functwn y(t) = 44(t; r)it is the unique solution of the initial-value problem

A(t)y",	y(r) =1,
dt

where it E C. Also, the n x n matrix Y = 4i(t; r)r is the unique solution of the initial-value problem

dY = A(t)Y,	Y(r) = I
dt

where r E

Theorem IV-2-1 is a corollary of Theorem IV-2-6. 0

Remark IV-2-7.

(1)The general form of a fundamental matrix solution of (IV.2.1) is given by Y(t) = 4i(t; r)r, where r E GL(n, C).

(2)If a fundamental matrix solution is given by Y(t) = 4'(t; r)r, then Y(r) = F.

Hence,

(IV.2.5)

0(t; ,r) =

Y(t)Y(r)-1

(t, r E Z)

80 W. GENERAL THEORY OF LINEAR SYSTEMS

for any fundamental matrix solution Y(t). In particular,

(IV.2.6)	1(t;r) _		for t, r, r1 E Z.
(3) In the case when A(t) is a scalar (i.e., n = 1), we obtain easily
(IV.2.7)		1(t; r) = exp I JA(s)ds)]t
		t	1
In the general case, we define exp I J A(s)ds] by
		V,r
exp[B(t)] = In		+M		B(t) = Jt A(s)ds.
		B(t)"',	where
		m=1 T11.

However, generally speaking, (IV.2.7) holds only in the case when B(t) and

B'(t) = A(t) commute. In particular, 4(t; r) = exp[(t - r )AJ if A = A(t) is independent of t. In §IV-3, we shall explain how to calculate exp((t - r)AJ, using the S-N decomposition of A. Also, (IV.2.7) holds in the case when

A(t) is diagonal on the interval I. A less trivial case is given in Exer- cise IV-9. It is easy to see that B(t) and A(t) commute if A(t) is a 2 x 2

upper-triangular matrix with an eigenvalue of multiplicity 2. For exam- ple, the matrix A(t) = [coStt I satisfies the requirement. In this

case, 4b(t; r) = exp I J t A(s)ds = exp (jsin t o sin r r, ) sint - sin

1 t 1 r

exp(sin t -sin r) 10

1 J1

t

(4)det Y(t) = det Y(r) exp if trA(s)ds) if Y(t) satisfies (IV.2.2), where det A and trA are the determinant and trace of the matrix A. This formula is

known as Abel's formula (cf. (CL, p. 28]).

Proof.
Regarding detY(t) as a function of n column vectors {y'1(t),...					of
Y(t), set det Y(t) = 7(g, (t), ... , y (t)). Then,
	d det Y(t)	- [:	P(...
(IV.2.8)	dt	L	, A(t)ym(t), ... ).
		M=1
Denote the right-hand side of (IV.2.8) by G(y'1(t),...				Then, 9 is

multilinear and alternating in y1(t),... ,j,,(t). Furthermore, 9 = trA(t) if

Y(t) = I,,. Therefore, 9 = tr A(t) det Y(t). Solving the differential equation

ddet Y(t) = tr A(t) det Y(t), we obtain Abel's formula. 0

dt

3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS 81

IV-3. Homogeneous systems with constant coefficients

For an n x n matrix A, we define exp[A] by

(IV.3.1)	exp[A] = In + E h Ah.
	h=1

It is easy to show that the matrix exp[A] satisfies the condition exp[A + B] = exp[A] exp[B] if A and B commute. This implies that exp[A] is invertible and

(exp[A])-1 = exp[-A]. Thus, we obtain a fundamental matrix solution Y = exp[tA]

of the system d = Ay with a constant matrix A by solving the initial-value

problem

(IV.3.2)

= AY,

Y(0) = In.

This, in turn, implies that the unique solution of the initial-value problem

(IV.3.3)

d = Ay,

y(T) = P

is given by y = exp[(t - r)A]p, where p E C' is a constant vector.

In this section, we explain how to calculate exp[tA] for a given constant matrix A, using the S-N decomposition of A. Assume that an n x n matrix A has k distinct eigenvalues Al, A2, ... , Ak. Let A = S + N be the S-N decomposition of A. Also, let Pj (A) (,j = 1, 2,... , k) be the projections defined in §IV-1 (cf. (IV.1.4)). Then,

	k	k
(IV.3.4)	In = > PJ(A), S =	AjPJ(A), N = A - S,
	J=1	)=1

and
(IV.3.5) PJ(A)Ph(A)	0 (A)		h
		if	h	36
				=

j

j	(3, h = 1, 2, .... k).

The two matrices S and N commute.

Denote by V, the image of the mapping PJ(A) : C"			Cn (cf. Lemma IV-I-10).
It is known that Sp = \,)5 for pin VJ . Hence, Sl p" = \1)5 and
		+00	earh=1
exp[tS]p' =	1 + E (A2t)n }ic
		n-i h
On the other hand, exp(tN] = In +		A Nh since N is nilpotent. Therefore,
		h=1
exp[tA)# = exp[t(S + N)]p' = expitN] exp[tS]p" = e'\'` exp[tN]P
(IV.3.6)	[In	n- i th
= eA'`	+	p"	for fl E Vj.

n=1

82 IV. GENERAL THEORY OF LINEAR SYSTEMS

Applying (IV.3.4) and (IV.3.6) to a general p E C", we derive

n-10

(IV.3.7)	exp[tA]p =	C.Njt In + E hl Nh		P, (A)p'	for 9E C".
		j=1	h=I
Thus, we proved the following theorem.
Theorem IV-3-1. The matrix exp[tA] is calculated by formula
		k	n-1 th	Nh Pj (A).
(IV.3.8)	exp[tA] _ > ea+`		In + F,
		=l	h=1 T!

Since the general solution of the differential equation

(IV.3.9)

is given by (IV.3.7), the following important result is obtained.

Theorem IV-3-2.

(i)If R(,\,) < 0 for j = 1, 2, ... , k, then every solution of (IV. 3.9) tends to 06 as t -' +00,

(ii)if R(Aj) > 0 for some j, some solutions of (IV.3.9) tend to 00 as t - +00,

(iii)every solution of (IV.3.9) is bounded for t > 0 if and only if R(AE) < 0 for j = 1, 2, ... , k and NP, (A) = 0 if R(A)) = 0.

Now, we illustrate calculation of exp[tA] in two examples. Note that in the case when A has nonreal eigenvalues, we must use complex numbers in our calculation.

Nevertheless, if A is a real matrix, then exp[tA] is also real. Hence, at the end of our calculation, we obtain real-valued solutions of (IV.3.9) if A is real.

-2	1	0
Example IV-3-3. Consider the matrix A = 0	-2 0 . The characteristic
3	2	1

polynomial of A is pA(A) _ (A -1)(A+2)2. By using the partial fraction decomposi-

tion of	1		(A + 2)2 - (A + 5)(A - 1)					(A + 2)2
	PA(A), we derive 1 =			9	. Setting P1(A) =			9
and P2(A)		+ 5)(A - 1), we obtain
		9
		0	0	0	1	0	0
		P1(A) = 0	0 0, P2(A) =		0	1	0
		1	1	1	-1	-1	0

Set
-2	0	0	0	1	0
S = PI(A) - 2P2(A) = 0	-2 0 , N = A - S =		0	0	0.

3

3 1

0 -1 0

3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS 83

Note that N2 = 0. Hence,

exp[tA] = e'[ 13 + tN ] PI(A) + e-21 [ 13 + tN ] P2(A)

e-2t	to-2t		0
0	e-2c		0
et - e-2t	et - (1 +	t)e-2t	et

The solution of the initial-value problem dt = Ay, y(0) = y is y(t) = exp[tA]i).

To find a solution satisfying the condition	Jim y(0) = 0, we must choose it so that
	t +m

P1(A)ij = 0. Such an iJ is given by it = P2(A)c", where c is an arbitrary constant vector in C3.

0	-1	1
Example IV-3-4. Next, consider the matrix A = 1	0	-1 . The charac-
-1	1	0

teristic polynomial of A is PA(A) = A(A2 + 3) = A(A - if)(A + i\/3-). Using the

partial fraction decomposition of				1	, we obtain
				PA (A)
1 =	(A- if)(A+ if) - A(A+ if)- sA(A- if}.
Setting
P1(A) = 3(A2 + 3),			P2(A)		IA(A + if), P3(A)		A(A - if),
we obtain						1-if 1+if
1	1	1	1		1 -2
							1-if
Pi(A) = -	1	1	1 , P2(A) = -- 1+if			-2
3	1	1	1		6 1-if 1+iv		-2

and P3(A) is the complex conjugate of P2(A). If we set

S = (if)P2(A) - (if)P3(A),

then S = A. This implies that N = 0. Thus, we obtain

exp(tA] = P1(A) + e,.''P2(A) + e-;J1tP3(A)

= P1(A) + 23t (e'3tP2(A)).

Using

(e'' t(1 + if)) = cos(ft) - \/3-sin(ft),

2 (e'-1t(1 - if)) wa(ft) + f sin(ft),

84	IV. GENERAL THEORY OF LINEAR SYSTEMS

we find

a(t) b(t) c(t) exp[tA] = 1 c(t) a(t) b(t)

3 b(t) c(t) a(t)

where

a(t) = I + 2cos(ft),

b(t) = 1 - {cos(ft) + f sin(ft)} ,

c(t) = 1 - {wa(ft) - f sin(ft))

Remark IV-3-5. Fhnctions of a matnx In this remark, we explain how to define functions of a matrix A.

I. A particular case: Let A0, I,,, and N be a number, the n x n identity matrix, and an n x n nilpotent matrix, respectively. Also, consider a function f (X) in a neighborhood of A0. Assume that f (A) has the Taylor series expansion (i.e., f is analytic at A0)

(h)

f(A) = f(A0) + 1 f h?Ao)(A - Ao)h.

h=1

In this case, define f (,\o1 + N) by

=

f(h)PLO)		.A	= f(Ao)II +	n-1 f(h)(AO)	NA.
f(AoI. + N) = f(Ao)I. +	t	A		h.
h=1	h.
				h=1
n-1	(h)
Since N is nilpotent, the matrix >2f			is also nilpotent. Therefore, the

h=1

characteristic polynomial pf(A0,+N)(A) of f(AoI + N) is

Pf(aol-N)(A) = (A - 1(A0))".

II. The general case: Assume that the characteristic polynomial PA(A) of an n x n matrix A is

PA(A) = (A - .l1)m1(A - A 2 ) ' - 2 ... (A - Ak)mr.

where A1,... , Ak are distinct eigenvalues of A. Construct P, (A) (j = 1, ... , k), S, and N as above. Then,

A = (A1In + N)P1(A) + (A21n + N)P2(A) + ... + 0k1n + N)Pk(A).

Therefore,

A' = (A11n + N)1P1(A) + (A2In + N)'P2(A) 4- ... + (Ak1n + N)'Pk(A)

for every integer P.

3. HOMOGENEOUS SYSTEMS WITH CONSTANT COEFFICIENTS 85

Assuming that a function f (A) has the Taylor series expansion

f(A) = f(A3)	0f(h)(A,)(A-A,)''
	00

h=1	h!

at A = A., for every j = 1, ... , k, we define f (A) by

f(A) = f (A11, + N)Pk(A) + f (A21n + N)P2(A)

(IV.3.10)

+ ... + f (Akin + N)Pk(A).

Since P2(S) = P,(A) (cf. Observation IV-1-15), this definition applied to S yields

f(S) = f(AIIn)P1(A) + f(A21.)P2(A) + -' + f(AkII)Pk(A)

and f (A) - f (S) has a form N x (a polynomial in S and N). Therefore, f (A) - f (S) is nilpotent. Furthermore, f (S) and f (A) commute. This implies that

f(A) = f(S) + (f(A) - f(S))

is the S-N decomposition of f (A). Thus,

Pf(A)(A) = pf(s)(A) = (A - f(A1))m' ... (A -

f(Ak))m

Example IV-3-6. In the case when f (A) = log(A), define log(A) by

log(A) = log(A1In+N)Pk(A) + log(A21n+N)Pk(A) + ... + log(Akln+N)Pk(A),

where we must assume that A is invertible so that A, # 0 for all eigenvalues of A.

Let us look at log(AoI,, + N) more closely, assuming that A0 0 0. Since

/	=		+O°	(-I)mm+1
log(Ao + u) = 1000) + log t 1 + -o)		logl o) +		I o
\\			m=1
we obtain
	n-1 (-1)m+1
log(.1oIn + N) = log(Ao)ln +			m
	m=1
It is not difficult to show that exp[log(A)J	= A. In fact, since
(log(A))m = (log(A11n + N))mPk(A)		+ (log(A21n + N))mp2(A)
+ ... + (log(Akln + N))mPk(A),
it is sufficient to show that exp[1og(Aoln + N)J		= A01,, + N. This can be proved by
using exp[log(Ao +,u)] =A0 +,u.

86 IV. GENERAL THEORY OF LINEAR SYSTEMS

Observation IV-3-7. In the definition of log(A) in Example N-3-6, we used log(A,). The function log(A) is not single-valued. Therefore, the definition of log(A) is not unique.

Observation IV-3-8. Let A = S + N be the S-N decomposition of A. If A is invertible, S is also invertible. Therefore, we can write A as A = S(II + M), where

M = S' N = NS-1. Since S and N commute, two matrices S and M commute.

Furthermore, M is nilpotent. Using this form, we can define log(A) by

log(A) = log(S) + log(Ih + M),

where
log(S) = Iog(A1)P1(A) + log(A2)P2(A) +	. + log(Ak)Pk(A)
and
(1)m+1
log(IR + Al) = E - m	Mm.
M=1

This definition and the previous definition give the same function log(A) if the same definition of log(A,) is used.

3	4	3
Example N-3-9. Let us calculate sin(A) for A = 2	7	4 (cf. Example
-4 8		3

IV-1-19). The matrix A has two eigenvalues 11 and 1. The corresponding projections are

0	14	7		-14	-7
	25	25		25	25
P1(A) = 0	L9	L9	P2(A) =	0	50
				2s
0	2512	506		-12	19
	25	25		0	25
				5

Define S = 11P1(A) + P2(A) and N = A - S. Then N2 = 0. Also,

( sin(11 + x) = -0.99999 + 0.0044257x + 0.499995x2 + 0(x3 ), t sin(1 + x) = 0.841471+0.540302x-0.420735x 2 + 0(x3).

Therefore,

sin(A) _ (-0.9999913 + 0.0044257N)P1(A) + (0.84147113 + 0.540302N)P2(A)

1.92207 -1.8957 -0.407549

1.0806 -1.42252 -0.591695

-2.1612 0.845065 0.1834

It is known that sin x has the series expansion

+00

sin x = 1 (2h +)1)I T2h+1

4. SYSTEMS WITH PERIODIC COEFFICIENTS

87

Therefore, we can also define sin(A) by

sin(A) -- -+ (2h(-1)h+ 1)! A2h+1

However, this approximation is not quite satisfactory if we notice that

		_1h	= -117.147.
sin(11) = -0.99999	and	112h+1
		(2h + 1)!

h=O

IV-4. Systems with periodic coefficients

In this section, we explain how to construct a fundamental matrix solution of a system

(IV.4.1)	dy	= A(t)y
	dt

in the case when the n x n matrix A(t) satisfies the following conditions:

(1)entries of A(t) are continuous on the entire real line R,

(2)entries of A(t) are periodic in t of a (positive) period w, i.e.,

(IV.4.2)

A(t + w) = A(t)

for t E R.

Look at the unique n x n fundamental matrix solution 4'(t) defined by the initialvalue problem

dY	= A(t)Y,	Y(0) = In
(IV.4.3)
.it

Since 4i '(t + w) = A(t + w)4S(t + w) = A(t)4'(t + w) and +(0 + w) = 4t(w), the matrix 41(t +w) is also a fundamental matrix solution of (IV.4.3). As mentioned in

(1) of Remark IV-2-7, there exists a constant matrix r such that 44(t +w) = 4i(t)r and, consequently, r = 4?(w). Thus,

(IV.4.4)

41(t + w) = 4(t)t(w)

for t E R.

Setting B = w-' log(4?(w)J (cf. Example IV-3-6), define an n x n matrix P(t) by

(IV.4.5)

P(t) = d'(t) exp(-tBJ.

Then, P(t + w) = ((t + w) exp(-(t + w)BI = 4i(t)0(w) exp(-wB) exp(-tBJ =

4'(t) exp(-tBJ = P(t). This shows that P(t) is periodic in t of period w. Thus, we

proved the following theorem.