Applied Mathematics E-Notes, 13(2013), 120-135 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On Explicit Representations Of Solutions Of Linear
Delay Systems
Felix Che Shuy
Received 24 May 2012
Abstract
Let A and B be real square matrices of dimension d
the system
X(t) = AX(t) + BX(t r); t
2 and r > 0. We consider
0;
where X is speci…ed on the interval [ r; 0] and give explicit solutions for the
system when the matrix A has a single eigenvalue, generalizing results of [7]. By
decoupling, we obtain explicit representations for solutions of a certain class of
these systems in which A has several distinct eigenvalues.
1
Introduction
Delay equations play an important role in mathematical modeling since the e¤ects of
delays are inherent in the dynamics of the evolution of many systems. A number of
applications of delay equations are discussed in Hauptmann et. al [3], Morelli et. al [6],
Decoa et. al [1] etc. and also in the monographs of Hale [2] and Smith [9].
A close look at the literature shows that explicit representations of solutions of delay
equations are known only in a few simple cases e.g. Küchler and Mensch [4], where a
formula appears for the one dimensional linear case with a single delay. Similar closed
form representations for the one dimensional equation with a single delay are also found
in a more recent paper by Khusainov et. al [5].
In [7] we gave a formula for the general solution of the two dimensional linear system
X(t) = AX(t) + BX(t r); t 0; where the matrix A has a single eigenvalue. The
results were used in [8] to give easily veri…able su¢ cient conditions for the stability of
the system.
Our aim in the present paper is to generalize the results of [7] to systems in Rd for
arbitrary d 2 and also to extend them to a certain class of these systems in which
A has several eigenvalues. The fundamental matrix solution which we present here
generalizes that known in the case of systems of ordinary di¤erential equations.
The rest of the paper is organized as follows: In Section 2 we introduce De…nitions.
We also prove a number of technical Lemmas which we use in Section 3. Our main
results appear in Section 3 and Section 4.
Mathematics Subject Classi…cations: 34K05, 34K12, 34K17.
of Mathematics, University of Bamenda, P.O. Box 39 Bambili, North West Region,
Cameroon (shu.felix@ubuea.cm)
y Department
120
F. C. Shu
2
121
Prerequisites
Let d 2; r > 0, A 2 M(d; d; R), B 2 M(d; d; R) where M(d; d; R) denotes the set of
d d real matrices, g : [ r; 0] ! Rd , f : [0; 1) ! Rd and consider the system
X(t) = AX(t) + BX(t
r) + f (t); t
0
(1)
X(t) = g(t); t 2 [ r; 0]:
(2)
DEFINITION 1.
(i) We say that (1)(2) is homogeneous if f
geneous.
0 otherwise we say that it is inhomo-
(ii) We say that the system is of Type I if A has a single eigenvalue otherwise we say
that it is of Type II.
REMARK 1. If the system is homogeneous then (1)(2) becomes
X(t) = AX(t) + BX(t
r); t
0
(3)
X(t) = g(t); t 2 [ r; 0]:
(4)
By a solution of (1)(2) we understand a function X which satis…es the following:
DEFINITION 2. A function X : [ r; 1) ! Rd is called a solution of (1) with
initial condition (2), if it is continuous, satis…es (1) Lebesgue almost everywhere on
[0; 1) and (2).
Let 1f0g denote the indicator function of the singleton set f0g de…ned for t 2 R by
1
0
1f0g (t) =
for t = 0;
for t =
6 0:
If f and g are integrable then the solution of (1) (2) is given by
X(t) := G(t)g(0) +
Z
0
G(t
r
s
r)Bg(s)ds +
Z
t
G(t
s)f (s)ds
(5)
0
(see Smith [9] for the derivation of (5)), where the M(d; d; R)-valued function G is
the fundamental matrix solution for the homogeneous system (3) in the sense of the
following De…nition:
DEFINITION 3. We call the function G : [ r; 1) ! M(d; d; R) the fundamental
matrix solution for the homogeneous system (3) if for any 2 Rd
X(t) := G(t) for t 2 [ r; 1)
122
Explicit Solutions of Linear Delay Systems
is a solution of (3) with the initial condition X(t) = 1f0g (t) for t 2 [ r; 0].
Explicit solutions of the system (1)(2) can therefore be obtained by determining G
explicitly. We shall …rst consider Type I systems which are homogeneous.
Consider the homogeneous system (3)(4) and assume that it is of Type I. Let 2 C
be the eigenvalue of A, then A admits a decomposition A = QJQ 1 , where J is the
Jordan canonical form of A i.e., if E denotes the d d identity matrix, then
J = E + M;
where M := (mij ),
i
(6)
2 f0; 1g and
mij :=
0;
i j or j
otherwise.
i;
i + 2;
Let Z := Q 1 X and H := Q 1 BQ, then in case of existence, solutions of (3)(4) can
be obtained by solving the system
Z(t)
Z(t)
= JZ(t) + HZ(t
= Q
1
r); t
0
(7)
g(t); t 2 [ r; 0]:
(8)
For the matrix M in (6), we de…ne M 0 = E. If z 2 Rd and m 2 f0; 1; 2; : : :g, then we
de…ne zm := M m z. Note that zm = 0 for m d.
If x; y 2 M(d; d; R), then we de…ne Tx (y) := xy and if A M(d; d; R), then we write
Tx A for fTx (y) : y 2 Ag. We de…ne I0 := I0j := fEg ; j = 0; : : : ; d 1. For k 1 and
j 2 f0; 1; : : : ; d 1g, we set Ikj := T(M j H) Ik 1 and Ik := [fIkj : j = 0; : : : ; d 1g. We
shall de…ne p(E) := 0; p(H) := p(M ) := 1 and if n 1, xi 2 fM; H; Eg; i = 1; : : : ; n,
n
P
then we de…ne p(x1
xn ) :=
p(xi ). If x := x1
xm = 0 for some m where
i=1
xi 2 fM; H; Eg; i = 1; : : : ; m then we call x a zero. If a set contains more than one
zero, then, where it is convenient, we shall jointly denote all the zeroes by a single
symbol 0. As an example fH; M d H; M d g shall be represented as fH; 0g where it is
convenient. Note however, that M d H and M d are distinct with p(M d H) = d + 1 and
m
z }| {
x.
p(M d ) = d. We shall write xm for the product x
The following observations will be useful:
LEMMA 1.
minfp(x) : x 2 Ik g = minfp(x) : x 2 Ik0 g
(9)
and
maxfp(x) : x 2 Ik g = maxfp(x) : x 2 Ikd
1
g:
(10)
PROOF.
Ik = [fT(M j H) Ik
1
: j = 0; : : : ; d
1g = [f(M j Hx) : x 2 Ik
1; j
= 0; : : : ; d
1g:
F. C. Shu
123
Therefore,
minfp(x) : x 2 Ik g = minfp(M 0 Hx) : x 2 Ik
1g
= minfp(x) : x 2 Ik0 g:
This proves (9). The argument proving (10) is similar.
LEMMA 2. Let j 2 f0; : : : ; d
1g, then for k
1
minfp(x) : x 2 Ikj g = j + k
(11)
and
maxfp(x) : x 2 Ikj g = j + 1 + (k
1)d:
(12)
PROOF. Let j 2 f0; : : : ; d 1g. If k = 1, then I1j = f(M j H)g. Therefore maxfp(x) :
x 2 I1j g = j + 1 = j + 1 + (k 1)d. Assume that the assertion is true for k = m: Then
for k = m + 1,
j
maxfp(x) : x 2 Im+1
g
=
=
maxfp(x) : x 2 T(M j H) Im g
j + 1 + maxfp(x) : x 2 Im g
d 1
= j + 1 + maxfp(x) : x 2 Im
g;
where the last equality follows from (10). By the assumption of the induction,
d 1
maxfp(x) : x 2 Im
g = (d
1) + 1 + (m
1)d = md = ((m + 1)
1)d:
This proves (12). (11) is proven similarly.
LEMMA 3. For k
0 and l 2 fk; : : : ; kdg, the set fx 2 Ik : p(x) = lg is non-empty.
PROOF. We shall also prove this statement by induction. If k = 0 then Ik = fEg,
dk = 0 and hence l 2 f0g. Also, fx 2 I0 : p(x) = 0g = fEg =
6 ;. Assume that the
statement is true for some k 0 then fx 2 Ik : p(x) = lg =
6 ; where l = k; : : : ; dk. We
now show that fx 2 Ik+1 : p(x) = lg =
6 ; where l = k + 1; : : : ; d(k + 1). Since
Ik+1 = [f(M j H)x : x 2 Ik ; j = 0; : : : ; d
1g;
fp(x) : x 2 Ik+1 g consists of the distinct elements of the following collection:
=
fj + 1 + p(x) : x 2 Ik and j 2 f0; : : : ; d
1gg
fj + 1 + m : m 2 fk; : : : ; dkg and j 2 f0; : : : ; d
1gg:
However, the distinct elements of this collection are the numbers in the set f(k +
1); : : : ; d(k + 1)g.
For typographical convenience, let us use the following notation:
k;l
d 1 j
:= fx 2 [j=1
Ik : p(x) = lg
124
Explicit Solutions of Linear Delay Systems
and
k;l
We note that
k;l
[
k;l
:= fx 2 Ik0 : p(x) = lg:
= fx 2 Ik : p(x) = lg. We have the following Remark:
REMARK 2.
(i) By Lemma 2, for j = 0; : : : ; d
1 and k
1,
fx 2 Ikj : p(x) < j + kg = ; and fx 2 Ikj : p(x) > j + 1 + (k
1)dg = ;:
So
(a)
k;k
= ; and
(b)
x 2 Ik0 : p(x) > (k
1)d + 1 = ;.
Also fx 2 Ik : p(x) < kg = ; and fx 2 Ik : p(x) > kdg = ; and hence
(ii) In view of Lemma 3, Ik is the disjoint union of the non-empty sets fx 2 Ik :
p(x) = lg where l = k; : : : ; kd.
(iii) For l 2 fk; : : : ; kdg,
d 2 j
fx 2 Ik : p(x) = lg = fx 2 [j=0
Ik : p(x) = lg [ fx 2 Ikd
1
: p(x) = lg:
Since TM Ikj = Ikj+1 ; minfp(x) : x 2 Ikd 1 g = k + d 1 and TM Ikd 1 = f0g for all
j = 0; : : : ; d 2, we observe that
8
< k;(l+1) [ F; l 2 fk; : : : ; kd 1g;
f0g;
l = kd;
TM fx 2 Ik : p(x) = lg =
:
;;
l k 1 or l kd + 1;
where
F :=
;;
f0g;
l <k+d 1
l 2 fk + d 1; : : : ; kd
1g:
0
(iv) From (ii) and the de…nition of Ik+1
,
TH fx 2 Ik : p(x) = lg =
3
(k+1);(l+1) ;
k l kd;
otherwise:
;;
Systems of Type I
In this Section, we give an explicit solution of the system (3)(4) when it is of Type I.
LEMMA 4. Let the matrices J and M be as in (6), d
arbitrary, then the system
Z(t)
Z(t)
= JZ(t) + HZ(t
r); t
2 and H 2 M(d; d; R) be
0
= z1f0g (t); t 2 [ r; 0]; z 2 R
(13)
d
(14)
F. C. Shu
125
admits a unique solution given by
8 t
[r]
dk
dP1
>
P
P
< P
(t kr)l+m
e (t kr)
x
0;
(l+m)! zm ; t
Z(t) :=
l=k fx2Ik :p(x)=lg m=0
k=0
>
:
z1f0g (t);
t 2 [ r; 0];
where zm = M m z.
PROOF. The uniqueness of the solution follows from the step method for solving the
equation. Also, it is easy to see from its de…nition, that Z is continuous. To complete
the proof, we will show that it satis…es (13) on the intervals (nr; (n+1)r); n = 0; 1; 2; : : :.
Let n = 0, then for t 2 (0; r), [ rt ] = 0; hence
d 1 m
X
t
zm :
m!
m=0
t
Z(t) = e
Therefore,
Z(t)
d 1 m
X
t
zm + e
m!
m=0
t
=
e
=
EZ(t) + e
Since
JZ(t) + HZ(t
and Z(t
t
t
d 1
X
d 1
X
tm 1
zm
(m 1)!
m=1
(15)
tm 1
zm :
(m 1)!
m=1
(16)
r) = M Z(t) + EZ(t) + HZ(t
r) = 0 for t 2 [0; r), we have to show that M Z(t) = e
then,
d 2 m
X
t
Zm+1 = e
M Z(t) = e
m!
m=0
t
t
d 1
X
r)
t
dP1
m=1
(17)
tm 1
(m 1)! zm .
But
tm 1
zm :
(m 1)!
m=1
The assertion is thus true for n = 0. Let now n 2 and assume that the formula holds
on ((n 1)r; nr). We will show that it holds on (nr; (n + 1)r). On (nr; (n + 1)r), we
have
Z(t) = e
t
d 1 j
X
t
j=0
j!
n
X
zj +
e
(t kr)
k=1
dk
X
X
l=k fx2Ik :p(x)=lg
x
d 1
X
(t
m=0
kr)l+m
zm :
(l + m)!
Therefore,
Z(t)
=
EZ(t)
+e
t
d 1
X
j=1
+
n
X
k=1
e
tj 1
zj
(j 1)!
(t kr)
dk
X
(18)
X
l=k fx2Ik :p(x)=lg
d 1
X
(t kr)(l 1)+m
x
zm :
((l 1) + m)!
m=0
(19)
126
Explicit Solutions of Linear Delay Systems
By (17), it su¢ ces to show that
M Z(t) + HZ(t
For t 2 (nr; (n + 1)r), t
r 2 ((n
M Z(t) + HZ(t
t
= e
d 1 j
X
j=0
+
n
X
e
(t kr)
(t r)
n
X1
r)
(20)
dk
X
d 1
X
(t
t
dP2
j=0
Mx
e
d 1
X
(t
m=0
(t (k+1)r)
dk
X
= e
t
X
dP1
j=1
(21)
(22)
Hx
tj 1
(j 1)! zj ,
d 1
X
(t
m=0
l=k fx2Ik :p(x)=lg
tj
j! zj+1
kr)l+m
zm
(l + m)!
r)j
Hzj
j!
k=1
Now (20) = e
X
l=k fx2Ik :p(x)=lg
j=0
+
1)r; nr), hence
t
M zj
j!
k=1
+e
r) = (18) + (19):
(k + 1)r)l+m
zm :
(l + m)!
(23)
which is (18). We will now show that
(21) + (22) + (23) = (19): We shall use the convention that if g is
Pa map taking values
in a set in which addition is de…ned and contains a zero, then
g(x) = 0.
fx2;g
(21)
=
n
X
e
(t kr)
k=1
=
=
n
X
dk
X1
l=k fx2
e
dk
X
(t kr)
X
k;(l+1) g
X
k=1
l=k+1 fx2
n
X
dk
X
e
(t kr)
k=1
x
k;l g
X
l=k fx2
x
x
k;l g
d 1
X
(t
m=0
kr)l+m
zm
(l + m)!
(24)
d 1
X
(t kr)(l 1)+m
zm
((l 1) + m)!
m=0
d 1
X
(t kr)(l 1)+m
zm ;
((l 1) + m)!
m=0
(25)
where (24) follows from Remark 2 (iii), while (25) follows from Remark 2 (i)(a). Since
I10 = fHg = fx 2 I10 : p(x) = 1g, it follows that
(22) = e
(t r)
d
X
X
l=1 fx241;l g
x
d 1
X
(t r)(l 1)+m
zm :
((l 1) + m)!
m=0
(26)
F. C. Shu
127
By Remark 2 (iv),
(23)
n
X1
=
e
(t (k+1)r)
k=1
n
X
=
l=k fx2
e
(t kr)
k=2
n
X
=
dk
Xd
l=k 1 fx2
(k 1)d+1
e
X
(t kr)
k=2
n
X
=
dk
X
l=k
e
(t kr)
k=2
dk
X
X
X
x
X
X
x
k;l g
x
k;l g
d 1
X
(t
m=0
k;(l+1) g
d 1
X
(t
m=0
(k+1);(l+1) g
fx2
l=k fx2
x
(k + 1)r)l+m
zm
(l + m)!
kr)l+m
zm
(l + m)!
d 1
X
(t kr)(l 1)+m
zm
((l 1) + m)!
m=0
d 1
X
(t kr)(l 1)+m
zm ;
((l 1) + m)!
m=0
(27)
where (27) follows from Remark 2 (i)(b). Therefore,
(21) + (22) + (23)
=
=
(25) + (26) + (27)
n
X
e
(t kr)
k=1
+
n
X
e
n
X
e
(t kr)
x
k;l g
dk
X
X
l=k fx2
k=1
=
X
l=k fx2
(t kr)
k=1
=
dk
X
dk
X
d 1
X
(t kr)(l 1)+m
zm
((l 1) + m)!
m=0
x
k;l g
X
d 1
X
(t kr)(l 1)+m
zm
((l 1) + m)!
m=0
x
l=k fx2Ik :p(x)=lg
(19):
d 1
X
(t kr)(l 1)+m
zm
((l 1) + m)!
m=0
In the notation of Section 2 and Lemma 4 we have the following
THEOREM 1. For t
0, de…ne
t
K(t) :=
[r]
X
k=0
e
(t kr)
dk
X
X
l=k fx2Ik :p(x)=lg
QxQ
1
d 1
X
(t
m=0
kr)l+m
QM m Q
(l + m)!
1
;
then the fundamental matrix solution for the homogeneous system (3) is given by
G(t) =
K(t)
for t 0
E1f0g (t) for t 2 [ r; 0]:
PROOF. Let 2 Rd and de…ne X(t) := G(t) ; t 2 [ r 1). We have to show
that X(t) satis…es (3) Lebesgue a.e. on [0; 1) with X(t) = 1f0g (t); t 2 [ r; 0]. For
128
Explicit Solutions of Linear Delay Systems
t 2 [ r; 0], X(t) = G(t) = E1f0g (t) = 1f0g (t). By Lemma 4,
8 t
[r]
dk
dP1
>
P
P
< P
(t kr)l+m
e (t kr)
x
0;
(l+m)! zm ; for t
Z(t) :=
l=k fx2Ik :p(x)=lg m=0
k=0
>
:
z1f0g (t);
for t 2 [ r; 0]
satis…es (7) Lebesgue almost everywhere on [0; 1) with the initial condition Z(t) =
z1f0g (t); t 2 [ r; 0]. In particular, if z = Q 1 ; then X(t) = QZ(t) satis…es (3)
Lebesgue a.e. on [0; 1) with X(t) = 1f0g (t); t 2 [ r; 0]. Now for t 0,
t
QZ(t)
=
[r]
X
e
(t kr)
k=0
=
dk
X
X
QxQ
d 1
X
(t
m=0
l=k fx2Ik :p(x)=lg
K(t) = G(t) :
1
kr)l+m
QM m Q
(l + m)!
1
(28)
Therefore X(t) = G(t) for t 2 [ r 1) satis…es (3) Lebesgue a.e. on [0; 1) with
X(t) = 1f0g (t) for t 2 [ r; 0].
From Theorem 1 we obtain the following Corollary which generalizes formula 2.3
in [4]:
COROLLARY 1. If the matrix A is diagonal then the fundamental matrix solution
for the homogeneous system (3) is given by
8 [t]
r
< P
kr)k k
e (t kr) (t k!
B ; for t 0;
G(t) =
: k=0
E1f0g (t);
for t 2 [ r; 0]:
PROOF. If A is diagonal i.e. i = 0 for all i, then M m = 0 for all m 1 and hence
(28) implies
8 t
[r]
dk
>
l
P
P
< P
e (t kr)
QxQ 1 (t l!kr) E ; for t 0;
X(t) =
k=0
l=k fx2Ik :p(x)=lg
>
:
1f0g (t);
for t 2 [ r; 0]:
Since
fx 2 Ik : p(x) = lg = f0g
for l = k + 1; : : : ; dk; it follows that
t
G(t) =
[r]
X
k=0
Now note that B = QHQ
1
e
(t kr)
QH k Q
1 (t
kr)k
; t
k!
0:
.
We also have the following generalization of formula 2.4 in [4]
F. C. Shu
129
COROLLARY 2. If the matrix A is diagonalizable then the solution of (1) (2) with
integrable g and f is given for t 0 by
X(t) := G(t)g(0) + B
Z
0
G(t
s
r)g(s)ds +
r
Z
t
G(t
s)f (s)ds:
0
PROOF. The assertion follows from (5) and the fact that if A is diagonalizable then
G and B commute.
We will now give an example illustrating the considerations above: Consider the
Type I system
x(t)
= Ax(t) + Bx(t
r); t
x(t)
= g(t); t 2 [ r; 0];
0;
where r > 0;
0
5
A=@ 8
4
3
5
3
The only eigenvalue of A is
0
1
J =@ 0
0
0
1
0
1
0
1
2
4 A;B = @ 0
0
3
= 1, Q
1
0
1
0
1 A;Q = @ 0
2
1
For t 2 [0; r), K(t) = et
1
P
m=0
1
0
0
1
1
0 1
0
t
0 A and g(t) = @ 0 A .
0
0
AQ = J where
2
4
2
tm
m
1
m! QM Q
1
0
0 A and Q
1
1
= et [E + tQM Q
0
1
=@ 0
2
1
1
2
1
4
3
2
] since M 2 = 0.
Therefore the fundamental matrix solution is given on [ r; r) by
et E + tQM Q
E1f0g (t);
G(t) :=
1
; for t 2 [0; r);
for t 2 [ r; 0]:
Since
QM Q
1
0
4
=@ 8
4
3
6
3
1
0
0 A:
1
1
0 1
2
1
4 A and g(t) = t @ 0 A ;
2
0
130
Explicit Solutions of Linear Delay Systems
we see that for t 2 [0; r),
X(t)
= G(t)g(0) +
=
=
=
Z
Z
Z
t r
K(t
r
t r
et
s r
r
t r
set
s
r
0
1
Z
1
@
A
0
=
0
=
[et (1
t
r
r)
and for t 2 [r; 2r);
K(t)
=
Z
1
X
e(t
k=0
kr)
0
G(t
s
r)Bg(s)ds
r
s r)Bg(s)ds
2
4E + (t
s
20
0
4
r) @ 8
4
13 0
2
1
4 A5 @ 0
2
0
13
3
6
3
1
0
1
4
r 4@
0 A + (t s r) @ 8 A5 ds
0
4
0
1
Z
4
t r
r
set s r (t s
set s r ds + @ 8 A
r
4
1
0
1
(t r + 1)] @ 0 A + [(2 + t r) f2 + tr
0
3k
X
X
QxQ
1
m=0
l=k fx2Ik :p(x)=lg
= et [E + tQM Q
1
] + et
r
1
X
(t
3
X
X
0
0
1
r)ds
t
kr)(l+m)
QM m Q
(l + m)!
QxQ
l=1 fx2I1 :p(x)=lg
1
1 0 1
0
1
0 A s @ 0 A ds
0
0
1
4
rget ] @ 8 A
4
0
1
1
X
(t r)m+l
QM m Q
(m
+
l)!
m=0
1
since I0 = fEg. Now I10 = fHg; I11 = f(M H)g; I12 = fM 2 Hg and so I1 = fH; (M H);
(M 2 H)g, p(E) = 0, p(H) = 1; p(M H) = 2 and p(M 2 H) = 3. However M 2 H = 0.
Therefore
K(t)
= et [E + tQM Q
+et
] + et
QM HQ
1
= et [E + tQM Q
1
+
r
1
1
QHQ
1
1
X
(t r)m+1
QM m Q
(m
+
1)!
m=0
1
X
(t r)m+2
QM m Q
(m
+
2)!
m=0
]+
1
X
(t r)m+2 t
e
(m + 2)!
m=0
= et [E + tQM Q
r
1
X
(t r)m+1 t
e
(m + 1)!
m=0
r
] + (t
QM HM m Q
r)et
r
r
1
1
QHM m Q
1
1
QHQ
1
+
r)2
(t
2
et
r
Q(HM + M H)Q
1
F. C. Shu
131
where the last equality is explained by the fact that QM HM Q
evaluated above, QHQ 1 = B and one can check that
1
Q(HM + M H)Q
0
8
=@ 8
4
1
= 0. QM Q
1
was
1
2
0 A:
4
5
4
4
Therefore,
K(t) = et [E + tQM Q
1
r)et
] + (t
r
QHQ
1
+
r)2
(t
2
et
r
Q(HM + M H)Q
1
and
X(t)
=
=
Z
Z
+
=
Z
0
G(t
r)Bg(s)ds =
r
0
s et
s r
r
(t
fE + (t
2r)2
s
2
0
t s r
se
r
et
s
s 2r
0
K(t
s
1
r)QM Q
g + (t
Q(HM + M H)Q
1
Z
1
A
@
0
+
ds
0
0
r)Bg(s)ds
r
0
et
s r
s(t
r
s
0
2r)et
1
s 2r
0
+ et
+ et
et
e
ret + et
r
2r
t r
r
(t
r
r)2
2
(t 2r)2
2
f r(t
f r(t
+
(t
r) + t
0
r) + t
r)2
(t
2
3g
(t
QHQ
1
@ 0 A ds
0
1
0
4
s r)ds @ 8 A
4
1
1
Z 0
1
(t s 2r)2 t
+
s(t s 2r)et s 2r ds @ 0 A +
s
e
2
r
r
0
1
0
4
et ft tr + r 2g et r ft r 2g @ 8 A
4
Z
=
s
Z
et
2r
ft
2r
r) + 1
s
0
2g
e
t 2r
ft
2r
2g
1
8
2r
ds @ 8 A
4
1
1
2g @ 0 A
0
2r) + 1
0
1
8
@ 8 A:
4
The process can then be continued on the interval [kr; (k + 1)r); k
2.
1
0
132
4
Explicit Solutions of Linear Delay Systems
Systems of Type II
In this Section, we will use the results of Section 3 to give an explicit representation for
the fundamental matrix solution for the homogeneous system (3) and hence a solution
of (1)(2) for a certain class of Type II systems.
In what follows all vectors will be understood to be column vectors and we will
write x for the transpose of the vector x. Let d 2 and consider the system (3)(4).
There exists an invertible matrix Q 2 M(d; d; R) such that A = QJQ 1 , where for
some n 1,
J = Diag(J11 ; : : : ; Jnn );
a block-diagonal matrix such that for each i = 1; : : : ; n, Jii is a square matrix having
the same structure as the matrix J in (6). For i = 1; : : : ; n, let di be the dimension of
Jii and de…ne
i
:=
1;
d1 +
+ di
1
+ 1;
i=1
i = 2; : : : ; n
and
i
:= d1 +
+ di :
Let H = (hij )ij=1;:::;d . The matrix J induces a partition of H into sub-matrices- H =
(Hij )ij=1;:::;n where Hij := (hkl ) k= i ;:::; i . With this notation we have the following:
l=
j ;:::; j
LEMMA 5. Assume that Hij = 0 for all i < j and for i = 1; : : : ; n let Gii denote
the fundamental matrix solution for the system
Wi (t) = Jii Wi (t) + Hii Wi (t
r); t
0:
For i = 1; 2; : : : ; n successively, t 0 and each j 2 f1; : : : ; ng de…ne the sub-matrices
Fij (t) by
8
0
for j > i;
>
>
< G (t)
for i = j;
ii
Fij (t) :=
iP1 R
t
>
>
F (t s)Hil Flj (s r)ds for i > j;
:
0 ii
l=j
and let F (t) := (Fij (t))ij=1;:::;n . For z 2 Rd ; let
Z(t) :=
F (t)z
for t 0;
z1f0g (t) for t 2 [ r; 0]:
Then Z solves (13)(14).
PROOF. For i = 1; : : : ; n let Si be the system
Wi (t) = Jii Wi (t) + Hii Wi (t
r) +
i 1
X
Hij Wj (t
r); t
0;
(29)
j=1
Wi (t) =
i 1f0g (t);
t 2 [ r; 0];
(30)
where i := (z i ; : : : ; z i ) . Each system Si is a system of Type I. Under the assumptions above, solving (13)(14) reduces to solving the n Type I systems, Si ; i = 1; : : : ; n.
F. C. Shu
133
In particular ((W1 (t)) ; : : : ; (Wn (t)) ) = Z(t) where Z is the solution of (13)(14).
If we solve the systems fSi g successively in the order S1 ; S2 ; : : : ; Sn then S1 is a homogeneous system of Type I while the systems S2 ; : : : ; Sn are inhomogeneous Type I
systems.
Let 1 ; : : : ; n be the distinct eigenvalues of A such that Jii corresponds to i .
Let Ei be the identity matrix of dimension di and Jii = i Ei + Mi , i = 1; : : : ; n.
j
j
Further let Ii0
= fEi g for all j = 0; : : : ; di 1, Iik
= T(M j Hii ) Ii(k 1) and Iik =
j
[fIik
: j = 0; : : : ; di
By Lemma 4,
i
1g:
Ki (t);
Ei 1f0g (t);
Gii (t) =
t 0;
t 2 [ r; 0];
where
t
Ki (t) :=
[r]
X
e
i (t
kr)
k=0
di k
X
X
x
l=k fx2Iik :p(x)=lg
dX
i 1
(t
m=0
kr)l+m m
Mi :
(l + m)!
By (5) the solution of (29), (30) is given by
Wi (t) = Gii (t)
i+
i 1Z
X
j=1
t
Gii (t
s)Hij Wj (s
r)ds
(31)
0
and hence W (t) = ((W1 (t)) ; : : : ; (Wn (t)) ) solves
W (t)
= JW (t) + HW (t
W (t)
= z1f0g (t); t 2 [ r; 0]:
We will now show that for t
r); t
0;
r,
W (t) = Z(t):
(32)
It is obvious that W (t) = Z(t) for all t 2 [ r; 0]. We will now show that W (t) = Z(t)
for t > 0.
To do this, we show by induction that the i-th vector Wi (t) on the left hand side of
(32) equals the i-th vector on the right hand side, i = 1; : : : ; n. For i = 1, using (31),
W1 (t) = G11 (t) 1 = F11 (t) 1 . Assume that the assertion is true for some i; 1 i
n 1. We now show that it is true for i + 1. Again by (31) and de…ning '(i; t; s; l) :=
Fii (t s)Hil ,
Wi+1 (t)
=
G(i+1)(i+1) (t)
i+1 +
i Z
X
F(i+1)(i+1) (t)
i+1 +
i Z
X
l=1
G(i+1)(i+1) (t
s)H(i+1)l Wl (s
0
l=1
=
t
0
t
'(i + 1; t; s; l)Wl (s
r)ds:
r)ds
134
Explicit Solutions of Linear Delay Systems
By the assumption of the induction,
i Z
X
l=1
t
'(i + 1; t; s; l)Wl (s
r)ds
=
0
i Z
X
l=1
=
t
'(i + 1; t; s; l)
i X
l Z
X
'(i + 1; t; s; l)Flj (s
r) j ds
'(i + 1; t; s; l)Flj (s
r)ds
t
0
i X
i Z
X
i
X
r) j ds
j=1
j=1 l=j
=
Flj (s
0
l=1 j=1
=
l
X
t
j
0
F(i+1)j (t) j :
j=1
Hence Wi+1 (t) =
i+1
P
j=1
F(i+1)j (t) j :
THEOREM 2. Under the assumptions of Lemma 5 the fundamental matrix solution
for the homogeneous system (3) is given by
G(t) :=
QF (t)Q
E1f0g (t)
1
t 0
t 2 [ r; 0]:
PROOF. The proof is similar to the proof of Theorem 1 if we use Lemma 5.
REMARK 3. Similar to Theorem 2 an explicit representation of the fundamental
matrix solution for the homogeneous system (3) can be obtained if Hij = 0 for all i > j.
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