671
Internat. J. Math. & Math. Sci.
/oi.6 No.4 (1983) 671-703
ADVANCED DIFFERENTIAL EQUATIONS WITH
PIECEWISE CONSTANT ARGUMENT DEVIATIONS
S.M. SHAH
Department of Mathematics
University of Kentucky
Lexington, Kentucky 40502
and
JOSEPH WIENER
Department of Mathematics
Pan American University
Edinburg, Texas 78539
(Received November 15, 1983)
ABSTRACT.
Functional differential equations of advanced type with piecewise constant
argument deviations are studied.
They are closely related to impulse, loaded and,
especially, to difference equations, and have the structure of continuous dynamical
systems within intervals of unit length.
KEY WORDS AND PHRASES.
Difference Equation,
Functional
Differential Equation, Advanced Equation,
Piecewise Constant Deviation, Initial-Value Problem, Solution,
Existence, Uniqueness, Backward Continuation, Growth, Stability.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
i.
34K05, 34K10, 34K20, 39A10.
INTRODUCTION.
In [i] and [2] analytic solutions to differential equations with linear trans-
formations of the argument are studied.
point of the argument deviation.
The initial values are given at the fixed
Integral transformations establish close connec-
tions between entire and distributional solutions of such equations.
Profound links
exist also between functional and functional differential equations.
Thus, the study
of the first often enables one to predict properties of differential equations of
neutral type.
On the other hand, some methods for the latter in the special case
when the deviation of the argument vanishes at individual points has been used to
investigate functional equations [3].
Functional equations are directly related to
difference equations of a discrete (for example, integer-valued) argument, the theory
S.M. SHAH AND J. WIENER
672
of which has been very intensively developed in the book [4] and in numerous subse-
quent papers.
Bordering on difference equations are also impulse functional differ-
ential equations with impacts and switching, loaded equations (that is, those inclu-
ding values of the unknown solution for given constant values of the argument),
equations
x’(t)
f(t, x(t), x(h(t)))
[t], that is, having intervals of constancy, etc.
with arguments of the form
A sub-
stantial theory of such equations is virtually undeveloped [5]
In this article we study differential equations with arguments h(t)
h(t)
[t+n], where [t] denotes the greatest-integer function.
[t] and
Connections are
established between differential equations with piecewise constant deviations and
difference equations of an integer-valued argument.
may be included in our scheme too.
Indeed, consider the equation
ax(t) +
x’(t)
Impulse and loaded equations
a0x([t ])
+
alx([t+l])
(i.l)
and write it as
x’(t)
ax(t) +
Y
(a0x(i)
i----
where H(t)
i for t > 0 and H(t)
+
alx(i+l))(H(t-i)
0 for t < 0.
H(t-i-l)),
If we admit distributional deriva-
tives, then differentiating the latter relation gives
x"(t)
ax’(t) +
Z
+
(aoX(i)
where 6 is the delta functional.
alx(i+l))((t-i)
This impulse equation contains the values of the
unknown solution for the integral values of t.
considered.
for t > 0.
on
(_oo, 0]
(t-i-l)),
In the second section Eq. (I.I) is
The initial-value problem is posed at t
0, and the solution is sought
The existence and uniqueness of solution and of its backward continuation
is proved.
Furthermore, an important fact is established that the initial
condition may be posed at any point, not necessarily integral.
Necessary and suffi-
cient conditions of stability and asymptotic stability of the trivial solution are
determined explicitly via coefficients of the given equation, and oscillatory properties of solutions to
(i.I) are studied.
In the third part the foregoing results are generalized for equations with many
deviations and systems of equations.
We show that these equations are intrinsically
closer to difference rather than to differential equations.
In fact, the equations
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
673
considered in this paper have the structure of continuous dynamical systems within
intervals of unit length.
Continuity of a solution at a point joining any two con-
secutive intervals then implies recursion relations for the solution at such points.
The equations are thus similar in structure to those found in certain "sequential-
continuous" models of disease dynamics as treated in [6].
We also investigate a
class of systems that depending on their coefficients combine either equations of
retarded, neutral or advanced type.
In the last section linear equations with variable coefficients are studied.
First, the existence and uniqueness of solution on [0, ) is proved for systems with
continuous coefficients.
A simple algorithm of computing the solution by means of
continued fractions is indicated for a class of scalar equations.
+
estimate of the solutions growth as t
is found.
Then, a general
Special consideration is given
to the problem of stability, and for this purpose we employ a method developed
earlier in the theory of distributional and entire solutions to functional differen-
An existence criterion of periodic solutions to linear equations
tial equations.
Some nonlinear equations are also
with periodic coefficients is established.
tackled.
Consider the nth order differential equation with N argument delays
x
(m 0
(t)
x(mo-l)(t),
f(t, x(t)
x(mN) (t-rN(t))),
x(t-rN(t))
where all
Ti(t) _>
0 and n
x(ml)(t-Tl(t))
x(t-Tl(t))
max mi, 0
<_
i
<_ N.
Here x
tive of the function x(z) taken at the point z
(1.2)
(k)(t-Ti(t))
t-Ti(t).
is the kth deriva-
Often equations that can
be reduced to the form (1.2) by a change of the independent variable are also dis-
cussed.
A natural classification of functional differential equations has been
suggested in [7].
In Eq. (1.2) let
max mi, i
_<
i
_< N, and %
m0
D.
If % > 0,
then (1.2) is called an equation with retarded (lagging, delayed) argument.
case %
O, Eq. (1.2)
advanced type.
For % < O, (1.2) is an equation of
Retarded differential equations with piecewise constant delays (EPCD)
have been studied in
and [i0].
is of neutral type.
In the
[8, 9, i0].
Some results on neutral EPCD were announced in [9]
Together with the present paper, these works enable us to conclude that
all three types of EPCD share similar characteristics.
First of all, it is natural
to pose the initial-value problem for such equations not on an interval but at a
674
S.M. SHAH AND J. WIENER
number of individual points.
Secondly, in ordinary differential equations with a
continuous vector field the solution exists to the right and left of the initial
t-value.
For retarded functional differential equations, this is not necessarily the
case [ll].
Furthermore, it appears that advanced equations, in general, lose their
margin of smoothness, and the method of successive integration shows that after seve-
ral steps to the right from the initial interval the solution may even not exist.
However, two-sided solutions do
exist for all types of EPCD.
Finally, the striking
dissimilarities between the solutions growth of algebraic differential and difference
[12, 13].
equations are well-known
Since EPCD combine the features of both differen-
tial and difference equations, their asymptotic behavior as t
resembles in some
cases the solutions growth of differential equations, while in others it inherits the
In conclusion, we note that the separate study
properties of difference equations.
of Eq. (1.1) and more general equations with many argument deviations is motivated
not only by instructive purposes and the possibility of obtaining some deeper results
in this special case but, mainly, by the fact that
(I.I) possesses properties of both
advanced and neutral equations.
2.
EQUATIONS WITH CONSTANT COEFFICIENTS.
Consider the scalar initial-value problem
ax(t) +
x’(t)
a0x([t])
with constant coefficients.
+
alx([t+l])
x(0)
(2.1)
co
Here [t] designates the greatest-integer function.
We
introduce the following
A solution of Eq. (2.1) on [0, oo) is a function x(t) that
DEFINITION 2.1.
satisfies the conditions:
(i)
x(t) is continuous on
(ii)
The derivative x’(t) exists at each point t
exception of the points
(iii)
[0,oo).
[0, oo), with the possible
[t] e [0, o) where one-sided derivatives exist.
Eq. (2.1) is satisfied on each interval [n, n+l) C [0, oo) with integral
ends.
Denote
b0(t)
e
at
+
k
a- la0(eat
b
0(1)/(1
i),
b
bl(t)
1(1)).
a
-1
a
l(e
at
i),
(2.2)
675
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
Problem (2.1) has on
THEOREM 2.1.
{t}
a unique solution
,It]cO,
(bo({t}) + bl({t})))
x(t)
where
[0, oo)
(2.3)
is the fractional part of t, if
bl(1)
PROOF.
(2.4)
i.
(t) are solutions of Eq. (2 l) on the interAssuming that x (t) and x
n
n+l
[n, n+l) and [n+l, n+2), respectively, satisfying the
vals
Xn+ l(n+l)
since
Cn+ l,
conditions x (n)
n
c
n
and
we have
X’n(t) aXn(t) + a0c n + alCn+I,
Xn(n+l) Xn+ l(n+l). The general solution
(2.5)
of this equation on the given inter-
val is
x (t)
n
e
a(t-n)
c
(i + a
-lao
-I
(a0Cn
+
alcn+l)
Putting here t
with an arbitrary constant c.
c
a
)c
n
n gives
-1
+
a
+
bl(t-n )cn+ I
alcn+l
and
Xn(t)
For t
b0(t-n)Cn
(2.6)
n+l we have
Cn+l
b0(1)Cn + bl(1)Cn+ I,
n _> 0
and inequality (2.4) implies
Cn+l
1
b0(1)
bl(1) Cn
With the notations (2.2), this is written as
Xc
Cn+ 1
Hence,
c
n > 0.
n
%n c 0
n
This result together with (2.6) yields (2.3).
implicit assumption a
# 0.
(I +
0
i
which is the limiting case of
[0
0, then
If a
a
x(t)
Formula (2.3) was obtained with the
+ aI
a
I
I + a
{t})(l
(2.3) as a
al)[t] Co’
0.
0
The uniqueness of solution (2.3) on
) follows from its continuity and from the uniqueness of the problem xn (n)
for (2.5) on each interval
equivalent to
[n, n+l].
c
n
It remains to observe that hypothesis (2.4) is
676
S.M. SHAH AND J. WIENER
a
In the particular case
00=
# a/(e
1
b0(1)
a
I).
0, and formula (2.3) holds true assuming
0 we have %
i.
REMARK.
x(t)
bl(1)
for b0(1)
0, and
COROLLARY.
+
tially as t
PROOF.
ix(t) <_
i, two possibilities may occur:
If
bo(1)
# 0 implies
0 problem (2.1) has infinitely many solutions.
The solution of (2.1) cannot grow to infinity faster than exponen-
.
b0({t})
The values
mlll
the case
[t]
bl({t})
and
(2.3) are bounded.
in
Therefore,
with some constant m
The solution of (2.1) has a unique backward continuation on
THEOREM 2.2.
(_oo, O] given by formula (2.3) if
b0(1)
PROOF.
vals
If x
-n
(t) and x
-n+l
(2.7)
# 0.
(t) denote the solutions of Eq. (2 I) on the
I-n,-n+l) and [-n+l,-n+2), respectively, satisfying x -n (-n)
X_n+l(-n+l)
X_n(-n+l)
then by virtue of the condition
C_n+l,
c
-n
inter-
and
X_n+l(-n+l),
it
follows from the equation
x’-n (t)
ax
-n
(t) +
a0c -n
+
alC-n+l
that
x
(t)
-n
e
where
(l+a
c
a(t+n)
-la0
a
c
c
-i
(a0c -n
+
a
Ic -n+l
+a -i alC-n+l
-n
Therefore,
x
-n
(t)
(e
a(t+n)
With the notations
+ (e a(t+n)
a-la0
)c
-n
+ a -I a i (e a (t+n)
l) C_n+l.
(2.2), we have
X_n(t)
Putting t
I)
+
bo(t+n)C_n
bl(t+n)C-n+l
-n+l gives
b
Cun+l
O(1)c_ n
+
b
l(1)c_n+ I
and
1
C
-n
b
I(I)
b (i)
0
C-n+l
Finally, we write
c
-n
=I -i c -n+l
n> I
(2.8)
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
6V7
and
c
%-nc 0
-n
which together with (2.8) proves (2.3) for t < 0.
The initial-value problem for Eq. (2.1) may be posed at any point, not necessa-
rily integral.
THEOREM 2.3.
If conditions (2.4) and (2.7) hold, and
bo({t0})
0)
then the problem x(t
PROOF.
x
0
%bl({t0})
+
# 0,
for Eq. (2.1) has a unique solution on (_oo, ).
By virtue of (2.4) and (2.7), the problem x(0)
unique solution on (-co, oo) given by (2.3).
(b0({t0})
x(t0)
c
for (2.1) has a
o
Hence,
%bl({to}))%[tO]c 0
+
and
c
(b0({t0})
o
+
%bl({t0}))-l%-[t0]Xo
It remains to substitute this result in (2.3), to obtain
x(t)
THEOREM 2.4.
(b0({t})
+
+ oo,
[t]-[t0]
x
O.
0 of Eq. (2.1) is stable (respectively, asympto-
The solution x
tically stable) as t
-I
bl({t}))(b0{t 0} + bl({t0}))
if and only if
I%I
_<
i (respectively,
I%I
< i).
Proof follows directly from (2.3).
THEOREM 2.5.
0 of Eq. (2.1) is stable (respectively, asympto-
The solution x
+ oo,
tically stable) as t
(a + a
0
if and only if
+
al)(a I
a
a(e
0
e
a
+
a
i))_> 0
(2.9)
I
(respectively, > 0).
PROOF.
The inequality
I%1
<_ i can be
I
If I
bl(1)
written as
bo(1)
b I(I)
< I.
> O, then
b
I(I)
i <_ b
0(I)
<_ I
b
I(I)
and
Since (2.10) implies
gives
bl(1) + b0(1) < i, bl(1) b0(1) < i.
bl(1) <_ i, we analyze only (2.10). Taking
(2.10)
into account
(2.2)
S .M. SHAH AND J. WIENER
6 78
a
a
-i
-i
al(ea
a
al(e
a
i) + e
i)
+
a
e
-i
a
-i
a
a
i) <_ I,
a
I) < i.
a0(e
a0( e
From here, we have
a
-i
(a0 +
al)( e
a
a
i) < i- e
that is,
a
+
a
+ a I _< 0,
0
(2.11)
and
-i
a
(a
a0)(ea
I
i)
<_
e
a
+
i
which is equivalent to
a
ao- a(eea + ii)
alIf I
b
I(I)
< 0
(2.12)
< 0, then
bl(1)
These inequalities imply
a
a
b0(1) >
b1(1) > 1.
-la l(e a
-I
a
l(e
+
i) + e
a
i)
e
a
bl(1)
i,
b0(1)
_> 1.
Therefore, we consider only
a-la0 ea
+
a
a
-i
a
0( e
i) _> i,
a
i) _> I.
These relations yield inequalities opposite to (2.11) and (2.12) and prove the theorem
COROLLARY. The solution x
0 of the equation
ax(t) +
x’(t)
a0x([t])
(2.13)
is stable (asymptotically stable) iff the inequalities
-a(e
+ l)/(e a
a
<_-
i) <_ a
0
a
(strict inequalities) take place.
THEOREM 2.6.
In each interval (n, n+l) with integral ends the solution of Eq.
(2.1) with the condition x(O)
t
n
c
=n+l/n
a
0 has precisely one zero
0 #
a
0
+
alea
a+a0+a I
if
(a0
+
e
aae
a
i
If (2.14) is not satisfied and a
zeros in [0,
)(al
0
#
e
aa
-aea/(e a
> 0.
(2.14)
i
i), c o # O, then solution (2.3) has no
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
PROOF.
# 0,
For
c
the equation x(t)
O # O,
bo({t}) + %bl({t})
679
0 is equivalent to
O.
Hence,
bo({t})/bl({t}) bo(1)/(bl(1)
I)
and
e
a{t)
a
a-lao(ea(t)
+
-I a
I
(ea(t}
i)
e
i)
a
a- la0
a
+
-i
al(e a-
e
a
l)
i)
i
It follows from here that
a{t}
e
(a
0
alea)/(a + a0 + al).
+
If a > 0, then
i < (a
0
+
alea /(a + a0 +
a
I)
< ea
and, by virtue of (2.4), the equality sign on the left must be omitted.
The case
a+ao+al>0
(2.15)
leads to
a
aea/(e a
>
0
i),
a
> a/(e a
I
i).
(2.16)
By adding inequalities (2.16) it is easy to see that (2.15) is a consequence of
(2.16). If
a > 0, a
+ a 0 + a I < 0,
then
(a + a0 + a I) e
a
< a
+
0
alea <
a
+
a
0
+
and
a
0
Again, the inequality a
with
+
a
i), a I < a/(e
aea/(e a
<a
0
+ aI <
(2.17)
i).
0 follows from (2.17).
The case a < O, together
(2.4), gives
e
a
< (a
0
+
a
ale
)/(a + a 0 + a I) < I,
and assumption (2.15) yields (2.16).
And if a
Hypothesis (2.14) can also be written as
k<O.
COROLLARY.
x(t)
+
a
0
+ a I < 0,
then we obtain (2.17).
i) > 0 which is equivalent to
b0(1)(bl(1)
- -
In each interval (n, n+l) the solution
(e
a0
a{t}(l +-
a0 )(ea(l
of (2.13) satisfying the condition x(0)
c
o
+-a0
a0
[t]
c
O
0 has precisely one zero
(2.18)
S.M. SHAH AND J. WIENER
680
t
n
a
=n+lln
a
0
a+a 0
if
a
0
aea/(e a
<
(2.19)
i).
If (2.19) is not satisfied, solution (2.18) has no zeros in
[0, ).
From the last two theorems we obtain the following decomposition of the space
(a,
ao, al).
If
i.
(a
ae
+
0
e
solution (2.3) with c
lira x(t)
0 as t
a
a
a
(al
-I
a
a(e + i)
0
e
a
> 0,
(2.20)
-i
0 has precisely one zero in each interval (n, n+l) and
O #
+ o.
/
In fact, the solution possesses the required properties iff -I < % < O, that is
%(% + i) < O.
(2.2), we have
With the notations
bo(1)(bo(1)
bl(1)
+ i) < 0 which
yields (2.20).
2.
If
(a + a
solution (2.3) with c
O
#
0
+
al)(a 0
aea
+
< 0,
e
1
0 has no zeros in (0, ) and lim x(t)
The properties take place iff 0 < % < i.
%(%
Hence,
0 as t
+
.
(2.21)
we consider the inequality
i) < 0 leading to (2.21).
3.
If
(a
a
I
e
a
I
solution (2.3) with c
(al
a(e
a
0
e
a
+ i)
a
< 0,
(2.22)
i
0 has precisely one zero in each interval (n, n+l) and is
O #
unbounded on [0, ).
The proof follows from the inequality % <
0
+
al)(a I
solution (2.3) with c
o #
a
e
a
< 0,
i) < 0.
(2 23)
i
0 has no zeros in (0, oo) and is unbounded.
Relation (2.23) results from % > i which can be written as
I(I)
i)
If
(a + a
b
(bl(1)
I) < 0 and gives (2.22).
(bl(l -b0(1)
4.
i which is equivalent to
(bl(1)
l)(b0(1)
+
681
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
5.
If a
a
I
a(e
0
e
a
+ i)
a0 +
O,
I
e
a ea
# 0, solution (2.3) with
c
i
o #
0
has precisely one zero in each interval (n, n+l) and is bounded on [0, oo) (but does
-i.
m) since, in this case, %
a
a
ae
a(e + i)
a
0, a 0 + a
0
a
not vanish as t
6.
If a
I
i
e
i
e
a/(e
0, then a I
a
i), and
problem (2.1) has infinitely many solutions.
7.
%
+ a0 + a I
If a
0, a I # a/(e
a
I), then x(t)
c
Indeed, here we have
o
i and
(b0({t})
x(t)
8.
bl({t}))c 0
+
Finally, if a
+
0
e
ae
a
a
ea{t} (a
-i
a
i
+ a0 +
a
=0, a 1
e
a
a
I)
a
al)c 0.
0
# 0, then x(t)
I
b0(t)c 0,
0, for t >_ i.
0 <_ t < I, and x(t)
For Eq. (2.13) we have the following decomposition of the plane (a,
If -a(e
i.
a
+ l)/(e a
i) < a
-aea/(e a
<
0
0 as t
If
-aea/(e a
0
0 has
0
0 as t
< -a, then x(t) has no zeros in (0, oo) and lim x(t)
/o.
3.
interval
4.
For a
<-a(ea +
0
(n, n+l) and
For a
l)/(e
a
i) the solution has precisely one zero in each
[0, oo).
is unbounded on
> -a the solution has no zeros in (0, oo) and is unbounded.
0
The case a
on
i) < a
a0).
i), solution (2.18) with c
precisely one zero in each interval (n, n+l) and lim x(t)
2.
for
-a yields
0
[0, I), and x(t)
x(t)
0 on [i, ).
c
If a
o
For a
0
a
a
-ae /(e
-a(ea +
0
l)/(e
I), then x(t)
a
bo(t)c 0
i) the solution x(t)
if bounded and oscillatory.
3.
SOME GENERALIZATIONS.
For the problem
x’(t)
in which
A,
A0,
A
I
are r x r
B0(t)
THEOREM 3.1.
Ax(t) +
e
At
A0x([t])
+
AlX([t+l]),
x(0)
c
o
(3.1)
matrices and x is an r-vector, let
+ (eAt
I)A-IA0
If the matrices A and I
(3.1) has on [0, ) a unique solution
Bl(t)
BI(1)
(e
At
I)A-IAI.
are nonsingular, then problem
S.M. SHAH AND J. WIENER
682
BI(1))-IBo(1))(I Bl(1))-[t]B0[t](1)c0
(B0({t}) + Bl({t})(I
x(t)
and this solution cannot grow to infinity faster than exponentially.
Consider the equation
ax(t) +
x’(t)
+
a0x([t])
+
alx([t+l])
a2x([t+2]),
a
2
# 0
(3.2)
with the initial conditions
x(O)
Let h
where
2
and
I
b0(t)
c
x(1)
o
c
(3.3)
1.
be the roots of the equation
and
2
b2(1)%
bl(t) are
b2(t)
+
b0(1)
0,
(3.4)
given in (2.2) and
a
-i
a
2
(eat
x(t)
50({t})c[t
c[t]
(hi
i).
(3.3) has on [0, oo) a unique solution
Problem (3.2)
THEOREM 3.2.
i) +
(bl(1)
bl({t})c[t+l
+
+
b2({t})c[t+2 ],
(3.5)
where
It]
(2c0 Cl)
+
[t]
%1c0)2 )/(2 hi)’
(Cl
(3.6)
and this solution cannot grow to infinity faster than exponentially.
PROOF.
For n < t < n+l, Eq. (3.2) takes the form
ax(t) +
x’(t)
+
a0x(n)
+
alx(n+l
a2x(n+2
with the general solution
x(t)
a(t-n) c
e
-I
a
(a0x(n)
+
alx(n+l)
+
a2x(n+2))
Hence, a solution xn (t) of Eq. (3.2) on the given
where c is an arbitrary constant.
interval satisfying the conditions
x(n)
x(n+l)
Cn,
en+l, x(n+2)
Cn+ 2
is
x (t)
n
e
a (t-n)
a
c
-I
(a0c n + alCn+I
To determine the value of c, put t
c
-la0
(i + a
c
n
+
a
+
b
-i
+
a2Cn+2).
n; then
alcn+l
+
a
-i
a
2Cn+2
and
Xn(t)
b
0(t-n)c n
l(t-n)cn+ I
+ b 2(t-n)cn+ 2.
By virtue of the relation
x (n+l)
n
we have
Xn+l (n+l)
Cn+l
(3.7)
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
b0(1)Cn
Cn+l
+
683
bl(1)Cn+ I + b2(1)Cn+2
whence
+
b2(1)Cn+ 2
(bl(1)
l)Cn+I + b0(1)c n
0
n > 0
(3 8)
We look for a nontrivial solution of this difference equation in the form c
n
xn.
Then
b2 (1)In+2
and
X
satisfies
+
1)X n+l +
(bl(1)
(3.4).
If the roots
1
b0(1)X
and
n
X2
0,
of (3.4) are different, the general
solution of (3.8) is
c
n
with arbitrary constants k
In particular, for n
n
kil I + k212
n
and k
I
0 and n
k
1
+ k2
In fact, it satisfies (3.8) for all integral n.
2.
1 this formula gives
c
Ilk I + X2k2
O
c
1,
and
(%2c0- Cl)/(% 2
k
I
c
It]
%1 ),
k
(c
%ic0)/(%2I
These results, together with (3.7), establish (3.5) and (3.6).
If
2
%1 ).
l, then
%1-- %2
It-l]
I
(c l[t]
which is the limiting case of (3.6) as
implicit assumption a
# 0.
If a
c[t +
x(t)
%Co[t-l])
/
0, then
(a0c[t
+ alc[t+l +
which is the limiting case of (3.5) as a
[0, m) follows
Formula (3.5) was obtained with the
%2 %1"
a2c[t+2]){t},
The uniqueness of solution (3.5) on
0.
from its continuity and from the uniqueness of the coefficients c
for each n >_ 0.
n
The conclusion about the solution growth is an implication of the
estimates for expressions (3.6).
THEOREM 3.3.
continuation on
If
b0(l
# 0, the solution of (3.2)
(-oo, O] given by (3.5)
(3.3) has a unique backward
(3.6).
Theorem 2.3 establishes an important fact that the initial-value problem for Eq.
(2.1) may be posed
at any point, not necessarily integral.
A similar proposition is
true also for Eq. (3.2).
THEOREM 3.4.
2
If
80
+ (1-
+
b0(1)81
2
b0 (I)
# 0 and
bl(1))8081
+
b0(1)(l
+ ((1
bl(1))
bi(i))8182
2
2b0(i))8082
2
2
+ b0(1)B
# 0,
2
(3.9)
684
S.M. SHAH AND J. WIENER
where
B0 b0({t0})’ B1 bl({t0})’ B2 b2({t0})’
then the initial-value problem
x0, x(t 0 +
x(t0)
i)
x
I
for (3.2) has a unique
solution on (-oo,
PROOF.
With the notation
[ t]
(t)
(h
)l(h2
i
we obtain from (3.6) the equations
c[t0+i] =-b0(1)(t0-1+i)c0 + (t0+i)cl,
Substituting these values in (3.5) for t
{t
o + i}
{t
t
and t
O
o +
t
+ (t 0 +
i)2)c 0
((t0)B 0
+ #(t + i) i + #(t 0 +
0
2)B2)c I
(-b0(1)((t0) 0
+ #(t 0 + i) + (t 0 +
i
2)2)c 0
I)B
+
0
(to)B I
+ ((t 0 + I) 0 + (t 0 + 2)B + (t 0 +
I
for the unknowns c
and c
(t 0 + m)
I. Computations
(t 0 + n)
(t 0 + p)
(t 0 + n + p
o
m-n
h-n
m-p)
hi
for any integers m, n, p.
(3.6),
o
1 and taking into account
+
Xo,
+
(3.i0)
3)B2)c I
x
I
show that
0(I)) [t0]+n+p-m(h-P
-(b
m)
)/(h 2
h i)
2
This result leads to the conclusion that the determinant
of (3.10) coincides with the left-hand side of (3.9).
coefficients c
0, i, 2, 3.
gives the system
o
(-b0(1)((t 0
+
i
and c
I
via the initial values x
to find the solution
THEOREM 3.5.
+
,
and x
I
we can express the
and substitute in (3.5) and
x(t).
0 of Eq. (3.2) is stable (respectively, asympto-
The solution x
tically stable) as t
0
Hence,
if and only if
lhl
< i (respectively,
lhl
< i), i
1,2.
Proof follows from (3.6) and from the boundedness of the values b.({t}).
THEOREM 3.6.
0 of Eq. (3.2) is asymptotically stable if
The solution x
(a0
+
e
ae
a
a
)a
1
(3.11)
< 0
2
and either of the following hypotheses is satisfied:
(i)
(a + a 0 + a + a 2)(a
I
I
a
e
a
)<0,
1
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
(a
(ii)
a
I
e
ao
)(_
a
I
a(e
+
a2
al
e
a
+ i)
a
685
< 0,
-i
where the first factors in (i) and (ii) retain the sign of a 2.
D
2
it follows from (3.11) that
b.(1),
With the notations b.
PROOF.
bl )2
(1
> O,
4b0b 2
hence the roots
hi,
of (3.4) are real.
(I- b
2
If a
b
I
b
D)/2b
2
I%iI
> 0, the inequalities
2
i
+
I
< 1 are equivalent to
D < 2b
I +
2
(3.12)
and
D >
I
2b
(3.13)
2.
From (3.12) we obtain
b
+ 2b 2 > I,
I
b
+ bI +
0
b
> I,
2
and (3.13) yields
b
The inequalities b
b
0
< 0 and b
+
I
2b
I
2b
< i,
2
< b
2
0
b
+ bI +
0
b
+ bI
b
and
b
2
2
0
< I.
+ bI
> 0 which are consequences of (3.11) and a
2
b
2
2
< b
> O.
I
2b
2
contradict to
Therefore, it remains
to consider only
b
0
+ bI +
b
2
> i,
b
+ bI
0
b
2
< i,
that is,
b
If i
b
b
0
2
> I
>
I
b
I
b
0
0
+
>
bl, b 2
bl, then
b
b
i
< i.
I
+ b I.
0
In this case, a sufficient condition of
asymptotic stability is
b
0
+ b I + b 2 > i,
which, in terms of the coefficients
b
I
<
b
1
0
+
bl,
then b
b
The case b
0
I
aj,
b
< i
I
coincides with hypothesis (i).
If i
b
0
> I, and (ii) follows from the inequalities
I
> i,
b
0
+ bI
b
2
< I.
> 0, b < 0 is treated similarly.
2
Let x (t) be a solution of the equation
N
x’(t)
ax(t) +
Z
i=0
a.x([t+i])
N > 2
N # 0,
a
(3.14)
S.M. SHAH AND J. WIENER
686
[n,
with constant coefficients on the interval
n
+ i).
If the initial conditions for
(3.14) are
+ i)
x(n
_<
0 < i
Cn+i,
N
then we have the equation
N
ax (t)
n
x’(t)
n
+
7,
a-Cn+’zz
i=0
the general solution of which is
x (t)
n
e
N
l
a(t-n) c-
a
-i
i--O
a#0
aiCn+ i
For t=n this gives
N
c
7,
c
n
a
-i
a-Cn+-zz
and
eat_n,t, c n +
x (t)
n
Taking into account that
a
e c
Cn+
a
(e (t-n)
Xn+ l(n+l)
N
+ l
e
a
(3.15)
a.c
z n+i
a
i=0
Xn(n+l)
n
N
I
I)
we obtain
Cn+l
l)a-i a i c n+i
i=0
With the notations
b
0
e
a
+
a-la0(ea
I),
b.I
a
-I a.(e a
i
i)
i > i
this equation takes the form
bNCn+N + bN-ICn+N-I
+
b2Cn+2+ (bl-l)Cn+l + b0Cn
+
Its particular solution is sought as c
bN%N + bN_I%N-I +
If all roots
+
%1’ 2’
AN
Cn
kl% + k2% +
b{%
(3.16)
then
n
2
0.
+
(bl-l)%
+
b
0
0.
(3.17)
of (3.17) are simple, the general solution of (3.16) is
given by
+
with arbitrary constant coefficients.
posed at any N consecutive points.
the solution to (3.14) for t
x(i)
c
i,
i
%’
(3.18)
The initial-value problem for (3.14) may be
Thus we consider the existence and uniqueness of
0 satisfying the conditions
0, i
N- i.
(3.19)
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
tions with
N- I and c
0, i,
Then letting n
the unknowns
x(n) in (3.18) we get a system of equa-
n
Vandermond’s determinant det
()
which is different from zero.
Hence,
If some roots of (3.17) are
are uniquely determined by (3.19).
kj
687
multiple, the general solution of (3.16) contains products of exponential functions
The limiting case of (3.15) as a
by polynomials of certain degree.
solution of (3.14) when a
THEOREM 3.7.
0 gives the
We proved
0.
Problem (3.14)
(3.19) has a unique solution on [0, ).
This
solution cannot grow to infinity faster than exponentially.
If (3.17) has a zero root of multiplicity j, then (3.16) contains only
REMARK.
the terms with
Cn+j,
The substitution of n for n
Cn+N.
the difference equation (3.16) to N
+
j reduces the order of
In this case, the solution x(t) for t >_ j
j.
x(N-l).
depends only on the initial values x(j), x(j+l)
Consider the initial-value problem
N
Ax(t) + l
x’(t)
i=0
matrices A and A
with constant r x r
then (3.20) is a retarded equation.
And if A
type.
i
i
A.x
I
ahd
"([t]),
i)"
r-vector x.
For A # 0 and A
i
c
If A
0(i
i
(3.20)
0
I,
0, for all i
2), it is of neutral
2, (3.20) is an advanced equation.
0, for some i
N
THEOREM 3.8.
x(0)
If the matrices A and B
I
Z
i=l
AiAi-I
are nonsingular, then
problem (3.2on [0, ) has a unique solution
x(t)
E({t})E [t]
E(t)
I + (e
(3.21)
(1)c0,
where
At
I)S
and
S
PROOF.
n
+ I
A-IB-I(A + A0).
Assuming that x (t) is a solution of Eq. (3.20) on the interval n < t <
n
with the initial conditions
x(i)(n+)
Cni
0
i
N
we have
X’n(t)
Axn(t) +
N
Z
i=O
AiCni.
(3.22)
S.M. SHAH AND J. WIENER
688
eA’t-nc(
Since the general solution
of the homogeneous equation corresponding to
(3.22) is analytic, we may differentiate (3.22) and put t
n.
This gives
N
c
=Ac
nl
From the latter relation, c
g
ACn0 +
Cnl
i= 2
n, i-i
Ai
nl
AiCni,
i-0
lCnl
N
and from the first it follows that
N
Cnl
(A +
cnl
Ai-IB -I (A + A0)Cn0,
A0)Cn0
+
A.A
i-1
nl
i-1
Hence,
i
_>
i.
The general solution of (3.22) is
x (t)
n
e
A (t-n)
N
l
c
A-IA.c
i=0
For t
with an arbitrary vector c.
Xn(t)
I nl
n we get c
and
SCn0
(3.23)
n)CnO.
E(t
Thus, on [n, n+l) there exists a unique solution of Eq. (3.20) with the initial condition x(n)
x(n-l)
Cn_l,
we have
0
Xn_l(t)
The requirement
For the solution of (3.20) on [n-l, n) satisfying
given by (3.23).
Cn0
Xn_l(n)
Cn0
E(t
Xn(n)
+
n
implies
1)Cn_l, 0"
Cn0 E(1)Cn_l,
0
from which
n
E (1)c
0"
Together with (3.23) this result proves the theorem.
The uniqueness of solution
(3.21) on [0, ) follows from its continuity and from the uniqueness of the problem
x(n)
CnO
for (3.20) on each interval
triction det A
# 0
in deriving formula
THEOREM 3.9.
[n, n+l).
It is easy to see that the res-
is required not in the proof of existence and uniqueness
(3.21).
If the matrix B is nonsingular, problem (3.20) on [0, oo) has a
unique solution
x(t)
V0(t)c 0,
x(t)
V
0 <_ t < I
i
where
bu.t only
[t]
(t)
Vk_ l(k)c0,
k=[t]
t > i
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
k(t)
V
e
A(t-k)
eA(t_s
+
689
Cds
k
and
B
C
-I
(A + A0)
A.
The solution of (3.20) cannot grow to infinity faster than exponen-
COROLLARY.
tially.
PROOF.
Since
IVk_l(k)
lV[t
<_ m and
(t)
ll
<_ m, where m
(I +
lCl l)e IAI I,
we have
Ix( )ll
mt+l
<_
I=oII"
It is well known that if the spectrum of the matrix A lies in the open
REMARK.
left halfplane, then for any function f(t) bounded on [0, ) all solutions of the
nonhomogeneous equation
Ax(t) + f(t)
x’(t)
[0, oo).
are bounded on
In general, this is not true for (3.20) which is a nonhomo-
geneous equation with a constant free term on each interval [n, n+l).
For instance,
the solution
x(t)
e-{t})6)(l +
(i + (i
e-l)6)[t]
(i
of the scalar problem
x"(t) =-x(t) + (i + 6)x([t]),
is unbounded on
[0, =)
x(0)
i,
6 > 0
since
Ix(t) _>
+ (i
(i
e
-i) 6) [tl
The reason for this is the change of the free term as t passes through integral
values.
THEOREM 3.10.
If, in addition to the hypotheses of Theorem 3.8, the matrix E(1)
is nonsingular, then the solution of
(3.20) has a unique backward continuation on
(-oo, 0] given by formula (3.21).
PROOF.
condition x
If x
-n
(-n)
-n
(t) denotes the solution of (3.20) on [-n, -n+l) with the initial
c
-nO’ then
X_n(t)
E(t +
n)C_n0,
and changing n to n-i gives
X_n+l(t)
E(t + n
l)c
-n+l, 0"
Since
x
-n
(- n + I)
x
-n+l
(- n + i)
c
-n+l, 0
690
S.M. SHAH AND J. WIENER
we get the relation
E-l(1)C-n+l,
C-n0
0
which proves the theorem.
If the matrices A, B, E(1) and
THEOREM 3.11.
Eq. (3.20) with the condition
x(t)
PROOF.
From (3.21) we
x(t)
x(t0)
x
0
E({t0})
are nonslngular, then
has on (-, 0o) a unique solution
E({t})g[t]-[t0](1)E-l({t0})x 0.
have
o E-[t0](1)x([t0] and
E({t})g[t]-[t0](1)x([t0]).
c
Furthermore, from (3.23) it follows that
E-l({t0})x0.
x([t0]
It remains to substitute this result in the previous relation.
EQUATIONS WITH VARIABLE COEFFICIENTS
4.
Along with the equation
N
A(t)x(t) +
x’(t)
x(i)
c i,
7.
i=0
(4.1)
matrices and x is an r-vector, we also consider
A(t)x(t).
(4.2)
If A(t) is continuous, the problem x(0)
U(t)c0,
N >_ 2, (0 <_ t < )
N- i
0
i
the coefficients of which are r x r
x’(t)
Ai(t)x([t+i]),
c
O
for (4.2) has a unique solution x(t)
where U(t) is the solution of the matrix equation
U’(t)
A(t)U(t), U(0)
The solution of the problem x(s)
c
O
s E
(4.3)
I.
[0, ) for (4.2)
is represented in the
form
x(t)
u(t)u
-i
(s)c 0.
Let
B0n(t)
U(t)(u-l(n)
Bin(t)
U(t)
THEOREM 4.1.
1210, ),
+
U
-I
(s)A0(s)ds)
u-l(s)Ai(s)ds,
i
i
Problem (4.1) has a unique solution on 0
N.
(4.4)
_<
t <
if A(t) and A.(t)
1
and the matrices
BNn(n+l)
are nonsingular for n
O, 1,
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
<_
On the interval n
PROOF.
+
t < n
691
i we have the equation
N
A(t)x(t) +
x’(t)
Its solution x (t) satisfying the condition x(n)
c is given by the expression
n
n
U(t)(U-I (n)Cn +
x (t)
n
Xn(n+l)
Hence, the relation
Cn+I
(4.4), this
With the notations
BNn(n+l)Cn+N
+
U- I (s) Z Ai(S)Cn+ids).
i=0
n
+-n
implies
Cn+I
n+l
I (s)
U
(4.5)
N
i=07" Ai(S)Cn+ids).
difference equation takes the form
+
BNn(n+l)
Since the matrices
N
t
Xn+l(n+l)
U(n+l) (U-I (n)Cn
x(n+i)
Ai(t)Cn+ i, Cn+
l
i=0
B2n(n+l)Cn+ 2
+
(BIn(n+l)-l)Cn+l+ B0n(n+l)Cn
are nonsingular for all n
solution c (n >N) provided that the values
n
Co,
_> O, there
exists a unique
are prescribed.
CN_ I
0.
Substitu-
ting the vectors cn in (4.5) yields the solution of (4.1).
For the scalar problem
a(t)x(t) +
x’(t)
x(O)
+
ao(t)x([t])
Co, x(1)
with coefficients continuous on
c
[0, m)
+
al(t)x([t+l])
a2(t)x([t+2])
I
we can indicate a simple algorithm of compu-
According to (4.4) and (4.5), we have
ting the solution.
Bon (t)cn + Bin (t) Cn+l + B2n(t) Cn+2
x (t)
n
(4.6)
with
t
U(t)
exp(
a(s)ds).
0
The coefficients c
n
satisfy the equation
+
B2n(n+l)Cn+2
(Bln(n+l)
l)Cn+ I
+
Bon(n+l)cn
0
n > 0
Denote
d
0(n+l)
-B0n(n+l )/B2n(n+l), d l(n+l)
n
Then from the relation
Cn+l
n
(I
Bln(n+l ))/B2n(n+l),
S.M. SHAH AND J. WIENER
692
cN+2
d
l(n+l) cn+ I
rn+I
d
l(n+l)
+
d
0(n+l) cn
it follows that
+
do (n+l)
r
n
which yields
r
r
+
dl(2)
2
+
I(I)
d
I
d
0(1)/r0,
d0(2)/r I
d
+
I(2)
dI
d0(2)
do(l)
(i) +
r
0
and continuing this procedure leads to the continued-fraction expansion
r
dl(n)
n
+
d
do(n) do(n-l)
l(n-1)+ d l(n-2)+
do(2) do(l)
1(1)+ Cl/C 0
d
and to the formula
Cn
rlr2
rn-lC0
for the coefficients of solution (4.6).
Assume that the
THEOREM 4.2.
0
<_
t <
,
A(t),
matrices
A0(t), +/-A’(t)
are continuous on
and let
(t)
max
]A(s> I]
max]
i(t)
IAi(s) II
0
_<
s _< t,
0, I.
i
[0, ),
If for all t
l(t)e
(t)
<_
q < I,
(4.7)
then the solution of the problem
A(t)x(t) +
x’(t)
A0(t)x([t])
+ AI (t)x([t+l]), x(0)
c
O
(4.8)
satisfies the estimate
lx(t)
II <-- (l-q)-t-l(o(t+l) + l)t+l e (t+l)(t+l) llc011,
PROOF.
According to
with the condition x(n)
x (t)
n
Since x (n+l)
n
c
n
solution x (t) of Eq. (4.8) on n < t < n
n
(4.9)
+
i
is given by
B0n (t)cn + Bin (t) Cn+l"
(n+l)
x
n+l
Cn+ I
(4.5), the
0 <_t < o.
Cn+l,
(4.10)
we have
Bon(n+l)cn + Bln(n+l)Cn+ I,
and
(I
Bln(n+l))Cn+ I Bon(n+l)Cn"
(4.11)
693
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
U(t, s)
The matrix
U(t)U
-I
(s) is the solution of Eq. (4.3) satisfying U(s, s)
I.
Therefore, it is the sum of the absolutely convergent series
I +
U(t, s)
+
A(Sl)dSl
S
i
A(Sl)dSl
+
A(s2)ds 2
S
S
whence
flU(t,
s)
_< e (t-s) (t)
II
and turning to (4.4) we find that for n <_ t _< n
lB0n(t) II
_<
+ l)e
(0(t)
+ i,
_< el(t)e(t)
(t), llBln(t) ll
(4 12)
Bln(n+l)
By virtue of (4.7) and the latter inequality, the matrices I
have
inverses for all n > 0 and
Bln(n+l))-lll _< I + lBln(n+l) II + llBln(n+l) II 2 +
II(I
q)-l.
_< (i
From the relation
Bln(n+l))-IB On (n+l)c
(I
Cn+ I
n
it follows that
lCn+ll
q)-l(000(n+l) + l)e(n+l) llc n I]
< (i
and
lcn If
+ i) n
q)-n(0(n)
<_ (1
end(n)
ICol I.
The application of this result together with (4.12) to (4.10) yields
llXn(t) ll
0(t)
--< (e
(o(t) +
q)-n(o(n)
i)(I
q)-n-le(t) (0 (n+l)
+ q(l
+ i) nen(n)
+ I) n+le (n+l) (n+l)
This inequality implies (4 9) since the functions (t)
0(t)
q)-t
and (I
are
increasing.
THEOREM 4.3.
The solution of
ax(t) +
x’(t)
+
tends to zero as t
a0(t) al(t)
stant,
some t
o
a0(t)x([t])
+
al(t)x([t+l])
x(0)
c
if the following hypotheses are satisfied:
E
C[O, ); (ii)
a
+
suplal(t)
ql <
i on
(4.13)
O
(i) a is con-
[0, oo) and,
starting with
0,
a(e
e
a
i
-i
ql
<
m0<
M
0
<
aql
a
e
a
-i
(4 14)
S.M. SHAH AND J. WIENER
694
where m
ao(t),
inf
0
M
>_ t o
t
(4.10), with
The solution of Eq. (4.13) is given by
Proof.
Bon(t)
and n _< t < n
+
e
t
a(t-n)
+
e
a(t-s)
n
Since the functions
i.
remains to show that lira c
(4.11)
ao(t),
sup
0
0 as n
n
.
and
Bin(t)
a (t-s)
al(S)ds
are bounded on [0, ), it
The condition x (n+l)
implies
c
n+l
n
and
r__.c,
u
n
r.r...,
c
Bon(t)
e
B0, n_l(n)/(l BI, n-l(n))
Cn/Cn_ I
from which
t
a0(s)ds, Bin(t)
where
r
i
(e
a
+
e
a(i-s)
i-i
i
ao (s) ds)/(i
ea(i-s) aI (s)ds).
i-i
By virtue of (i) and the first part of (il),
Ii
e
a(i-s)
I (s)ds.l
a
i-I
Therefore, I
j _>
to,
> 1
i-i
_> 1
ql"
Let j be a natural number such that
0 for all i >_ I.
BI, i_l(i)
la l(s) Ids
then due to (4.14) we have
I
BO i-l(1) ea +
B0,
i-i
i
i-I
ea(i-S)ao(s)ds < ea + MO(ea
a
(i) _> e +
mo(ea
a
(1) < e
a(e a
a
(i) > e
a( ea + I-
l)/a,
i
l)/a,
_> j
and
B0,
BO,
i-i
l)/a
aql(e
ql)( e
a
a
l)/a(e a
l)/a( ea
i)
i)
I
ql
(I
ql ).
Hence,
{B0,
+/-_(i)
< q(
II BI, f_l(f)
ql
>_
i
ql’ 0 <_ q <
i
and
]r+/-]
<_q,
-
i_>j.
On the other hand, for i < j we use the inequalities
ISo,
i_Ci) l-<
+
e
and
r l< Pl/(
1
p,
)/
ql
P-
suplao()I,
[o, )
695
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
[a0(t)
The boundedness of the function
of (4.14).
[0, =o)
on
is a result of its continuity and
To finish the proof, we write
Icnl Iol
I+/-l ]--[ ]ril < pj-1 qn-j {Col
i=l
and here the second factor vanishes as n
m.
/
n > j
+ 1
The method of proof suggests also the
following
The solution of (4.13) tends to zero as t
THEOREM 4.4.
a0(t) al(t)
a is constant, the functions
_>
and, starting with some t O
al(t)
a
x(i)(0)
tends to zero as t
N
F.
i=0
ax(t) +
x’(t)
i) < m0 <_ M0 <
/
+
ai(t)x([t+i])
[0, )
i--
ci,
N- i
0,
if the following hypotheses are satisfied.
(ii)
The functions a.(t)(i
1
The function
as t
/
(4.15)
N _> 2
The coefficient a is constant, the functions
Proof.
for all t e
a.
(i)
aN(t)
if the coefficient
The solution of the scalar problem
THEOREM 4.5.
(iii)
<_ 0
+
0,
a
-a(e + l)/(e
lira
e C[O, ),
/
.
aN(t)
0, i,
N
ai(t)
E
C[0,
I) are bounded on [0,
is positive, monotonically increasing, and
The solution of Eq. (4.15) on the interval n
t< n+l is given by the
formula
N
7.
x (t)
n
i=0
Bin(t)Cn+
i,_ Xn(n)_
c
n
where
Bon(t)
ca(t-n)+
J
t
n
ea(t-S)a0(s)ds
(i
Since the functions
c
n
0 as n
oo.
Bi, [t](t)
i,
B
in
(t)
I
t
e
n
a(t-s)
ai(s) ds,
N).
are bounded on
[0, =o),
it remains to prove that
The continuity of the solution to (4.15) implies x (n+l)
n
c
n+l
and leads to the equation
N-I
Cn+N
Z
i=O
d
i(n+l) Cn+i,
(4.16)
S.M. SHAH AND J. WIENER
696
with
d
l(n+l)
-Bin(n+l)/BNn(n+l)
Bln(n+l))/BNn(n+l), di(n+l)
(i
1).
N
0, 2,
(i
We have
BNn
(n+l)
n+l
]n
e
a (n+l-s)
hence, by virtue of (iii),
n
oo.
a
aN(s) ds
aN(n)
a
n <
n
< n
> 0 for all n >_ 0 and lira
BNn(n+l)
+ I,
BNn(n+l)
as
On the other hand, due to (ii), the values
n+l
Bin (n+l)
e
a(n+l-s)
u,
a.ts)es,
1
-n
are uniformly bounded for all n >_ 0 since
IBin(n+l)
<_
elal suplai(t)I,
t
[0, o).
Therefore,
Idi (n+l) --< M/aN(n ),
with some constant M > 0.
6n
The positive variable
M/aN(n)
N-I
ICn+N! <
To evaluate the coefficients c
From (4.16) it follows that
oo.
is monotonically decreasing to zero as n
r.
n
i=0
(4.17)
Cn+i
we employ the method developed in
n
[14] for the
study of distributional solutions to functional differential equations.
Mn
max
cjl,
0 _< j
_<
Denote
n.
Then
Cn+Nl
<--
n
NMn+N_ I.
Starting with some natural k,
N<I,
n
Cn+N
<
Mn+N-1
and
Mn+N Mn+N_I,
n > k.
Hence,
Mn+N Mk+N,
n
>_ k.
(4.18)
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
697
The application of (4.18) to (4.17) successively yields
Ck+N+l I-<
6k+iNMk+N, ;Ck+N+21 --< k+2NMk+N
Ck+2Nl --< 6k+NN+N{6
Since the sequence
is
n
decreasing it follows that
ICk+N+i+ I I<--6k+IN+N,
Now we put n
...,
k+N+l, k+N+2,
O,
i
N
i.
k+2N in (4.17) and use the latter inequalities to
obtain
ICk+2N+l
--<
(6k+iN)2 +N’ ICk+2N+21 -< (6k+iN)2 +N’
2
Ck+BNl <-- (6k+iN) Mk+N,
that is,
Ck+2N+l+il < (6k+IN)2 Mk+N
N
0,
i
I.
Continuation of the iteration process shows that
j
ICk.+jN+l+i
(k+IN) Mk+N,
Ck+j N+l+i
+0 asj
N
0,
for all natural j and i
I, and the inequality
6k+IN
< i implies
/oo.
THEOREM 4.6.
matrices of period
Assume that the coefficients in (4.1) are continuous periodic
If there exists a periodic solution x(t) of period i to Eq.
I.
I is an eigenvalue of the matrix
(4.1), then %
N
Y
S
i=0
where
Bi0(t)
are given in
matrices
BN0(1)
and
B00(1)
x(i)
has a unique solution on
PROOF.
o #
(4.4), and x(O)
is a corresponding eigenvector.
Conversel
1 is an eigenvalue of S and c o is a corresponding eigenvector and if the
if
c
Bio(1)
0.
are nonsingular, then Eq. (4.1) with the conditions
Co,
i
O,
(4.19)
N- 1
< t < oo, and it is 1-periodic.
Suppose that x(t) is a solution of period i to Eq. (4.1) and let x(0)
Since
x(n)
x(O), for all integral n, we have
(4.5), the assumption c o
0 would mean x(t)
0.
Cn
Putting t
Co
and, according to
n+l in (4.5) gives
S.M. SHAH AND J. WIENER
698
c
From the relation A(t+n)
U(t)Cn,
Hence, U(t+n)
U(0)C
yields U(n)
[n+l u-l(s)Ai(s)
7.
iffi0 -n
and to determine the matrix C
n
U(n).
n
Ai(t)
(s)At(s)ds
ds) c
0
A(t) it follows that U(t+n) is a solution of Eq. (4.3).
we put here t
0 which
Now we can write
U(t)U(n), U(n+I)U-I (n)
Ai(t+n)
U-
+
that is C
n
U(t+n)
and since
N
U(n+l)(u-l(n)
O
U(1),
u-l(t+n)
u-l(n)u-l(t)
we observe that
0
u-l(n)
U-l(s-t)Ai(s)ds
0
U-l(s)i(s)ds
The refo re,
N
(U(1) +
1
7. U(1)
i=0
I u-l(s)Ai(s)ds
I)co
0
0
and, taking into account (4.4), we conclude that
N
Z
i=O
Bio(1)
I)c
O.
0
(4.20)
This proves the first part of the theorem.
According to (4.5), the solution of problem (4.1) for t > 0 is
U(t)(u-l([t])c[t]
x(t)
+
i=O
It
U-I(s)Ai(s) ds) c it+i]),
(4.21)
and if the coefficients of (4.1) are 1-perlodlc, it has a unique backward continuation on
(_oo, 0] given by the same formula (4.21).
Indeed, on the interval [-n, -n+l)
Eq. (4.1) becomes
N
X’_n(t)
where
C_n+i
x
-n
x
-n
(t)
A(t)X_n(t)
(-n+i), and
+
7. A
i‘
i=0
_(t)c_n+i
its solution
U(t)(u-l(-n)c -n
can be written in the form (4.21).
+
7.
i=O
-n
u-l(s)Ai(s)ds)c_n+i
It remains to show that the coefficients
(4.22)
c[t
satisfy the same difference equation for both positive and negative values of t.
From the derivation of (4.20) we note that
DIFFERENTIAL EQUATIONS WITH PIECEISE CONSTANT ARGUHENT DEVIATIONS
699
N
Z
Cn+l
i=O
(4.23)
n > O.
Bio(1)Cn+ i,
For t =-n+l formula (4.22) gives
C
Since
U(-n/l) (U
-n+l
A(t-n)
-1
N
I
"
(-n)c -n +
i=O
-n+l
--n
U- 1 (s)Ai(s)ds)c_n+f)
Ai(t-n) Ai(t), we have
U(1), u-l(t-n)
U(t)U(-n), U(-h+l)U-l(-n)
A(t) and
U(t-n)
u-l(-n)u-l(t)
and
I
I
f-n+l U-I (s)A. (s) ds
J-n
U-1
0
(s-n)Ai(s)ds)
U
-i
I
I u-l(s)Ai(s)ds
(-n)
0
The equat ion
C-n+ 1
U(1)(c
for the coefficients c
-n
-n
l(s)Af(s)ds)c_n/f
l
i=0
/
(4.24)
coincides with Eq. (4.23) for c
The representation of
n
(4.24) as
N
B00(1)C_n
C_n+l
shows that since the matrix
mine the values c
tion to
B00(1)
is nonsingular, conditions
for all n > I.
-n
(4.25)
iZIBi0(1)C_n+i=
(4.19) uniquely deter-
This proves the existence and uniqueness of solu-
Eq. (4.1) satisfying (4.19).
To establish its periodicity, we put n
0 in
(4.23) and invoking (4.19) obtain
N-I
7.
i=0
l)c
Bi0(1)
0
+
BNO(1)cN
Subtracting this relation from (4.20) gives
n
I,
0.
BN0(1)(c 0
c
N)
0, and cN
c
o
For
it follows from (4.23) that
N-I
r.
i=O
Bio(1)
l)c
0
+
BNO(1)CN_1
and combining this result with (4.20) yields
confirms that
Cn
c
o
for all n
(4.25) together with (4.20).
_> O.
CN_ I
To prove that
O,
c
0.
C_n
Continuing this procedure
c
o
for n _>
-I,
we consider
Finally, on each interval [n, n+l) problem (4.1)
(4.19) is reduced to the same equation
S.M. SHAH AND J. WIENER
7OO
N
7
A(t)x(t) +
x’(t)
x(n)
Ai(t))c0,
i=0
cn
which concludes the proof.
The initial-value problem
N
A(t)x(t) +
x’(t)
matrices A(t) and
with r x r
7 A.(t)x (i)
i=O
cN-I[o,
Ai(t)
(It]), x(O)
c
(4.26)
o
oo) and r-vector x(t), can be treated
Assuming that x (t) is a solution of Eq. (4.26) for n < t < n +
similarly to (3.20).
n
i satisfying the conditions
x
(i)
(n)
N
0,
i
Cni
we have
N
x’(t)
n
A(t)x (t) +
n
7.
i= 0
(4 27)
Ai(t)cni
Since the general solution of the homogeneous equation corresponding to
times differentiable, we may differentiate
(4.27) i
(4.27)
i times and put t
n.
is N
This
gives
i 1
c
ni
j=0
(i-.J l)A(i-l-j) (0)Cnj
N
+ Z
k=O
i-l) (O)Cnk
I,
i
(4.28)
N.
The formula for x (t) is
n
U(t)(u-l(n)Cn0 +
Xn(t)
where U(t) is the solution of
Cn+l
U(n+l)
0
Nit
Y.
i=0
U
n
-l(s)A i(s)cn ids)
(4.3), and
(u-l(n)Cn0 +
Xn(n+l)
[n+Iu-i (s)Ai
Co0
c
o
(4.29),
e
A(t) and A.(t)
1
the vectors c
ni"
(i
nl
n
0
implies
(4.29)
ids).
N) and substitute
i,
to obtain a recursion relation between
is given we can find
THEOREM 4.6.
(s)c
i 0 -n
Now we solve the system (4.28) for the unknowns c
these values in
Cn+l,
Cn0
(and
Cni
for all n
_> I.
Cn+l,
0
and
Cn0.
Since
Thus, we proved
Problem (4.26) has a unique solution on [0, o) if the matrices
cN-1 [0,
oo) and system (4.28) has a unique solution with respect to
701
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
Suppose the equation
THEOREM 4.7.
A(t)x(t) + B(t)x([t+N]) + f(t, x([t]), x([t+l]),
x’(t)
x([t+N-1])),(4.30)
(N -->2)
satisfies the following hypotheses.
(i)
A(t), B(t) e C[0, oo), and f
The matrices
is a continuous function in the
space of its variables.
(ii)
The
matrices|L+Iu-l(t)B(t)dt
are nonsingular for all n
In
>_ 0,
where U(t) is
the solution of (4.3).
Then (4.30) with the initial conditions (4.1) has a unique solution on [0, oo).
<_
t < n
+ I Eq. (4.30) takes the form
PROOF.
On n
x’(t)
n
A(t)x (t) + B(t)c
f(t
n
n+N +
x(j),
c.
j
c
n
Cn+N_l)
Cn+l’
n, n+l,
n+N.
Hence,
x (t)
n
U(t)(U -I (n)Cn +
+
[tu-l(s)f(s,
t
n
U- I (s)B(S)Cn+N ds +
Cn+I, ..., Cn+N_l)dS).
cn,
"n
For t
n
+
i we have
Cn+ I
U(n+l) (U -I (n)Cn +
+
I+Iu-l(s)f(s,
n+l
n
cn,
U- I (s)B(s)ds)cn+N +
Cn+I
Hypothesis (ii) allows to solve (4.31) for
Cn+N
(4.31)
Cn+N_l) ds).
and to find all coefficients c
n
successively by employing conditions (4.1).
In conclusion, we consider the scalar equation
x’(t)
f(x(t), x([t]), x([t+l])), x(0)
Co,
(0 <_ t < oo).
(4.32)
If in the ordinary differential equation with parameters
x’
f(x, %, ),
f(x, %, ) e C(R 3)
f(x, %, ) # 0 everywhere, then there
F(x, %, )
t
exists a general integral
+ g(%, ),
(4.33)
702
S.M. SHAH AND J. WIENER
with an arbitrary function
g(%, ).
If x(t) can be extended over all t >_ 0, then for
the solution x (t) of (4.32) on n < t < n
n
X’n
f(Xn’ Cn’ Cn+I),
Hence, if %
Cn,
n’ Cn,
For t
we have the equation
(4.34)
Cn.
then
Cn+I,
F(x
n(n)
x
+ I
Cn+l
t
+ g(cn’
Cn+l
Cn+l
n
+ g(cn
Cn+l)
c
t
+
n this gives
F(c
c
n
n
and
F(Xn, Cn,
n+l
F(Cn, Cn,
c
(4.35)
n.
n+l
Now we take the solution of (4.32) on [n+l, n+2) that satisfies x(n+l)
Xn(n+l)
Xn+l(n+l)
F(Cn+I,
Cn+ I)
Since
n+l in (4.35) and get
we put t
cn,
Cn+I.
I.
F(Cn, Cn’ Cn+l
This equation can be written as
cn+l
Cn
dx
f(x,
cn,
(4.36)
1,
Cn+I)
which also follows directly by integrating (4.34) from t
we relax the condition that
always suppose f(x,
0 for some c e
f(x,
Cn, Cn+I) #
_(Cn, Cn+ I)_
,
)
0, x e
0 everywhere.
(Cn, Cn+l
0, then xn (t)
virtue of the condition x (n+l)
n
c
n
Cn, Cn+ I)
0.
f(c,
Cn, Cn+l
leads to the conclusion that the constant function
tion does not satisfy the condition
n
After this,
It is easy to see that we may
since the assumption
c is a solution of Eq. (4.34) on the interval n
f(c
n+l.
n to t
Xn(n)
Cn
<_
+
t < n
because c #
i.
Cn.
Cn+l,
we have
However, this soluOn the other hand, if
(4.34) on [n, n+l).
c is a solution of problem
n
(n+l)
x
n+l
Xn(t)
Cn
Cn+l
and
By
f(Cn, Cn’
Thus, we proved
THEOREM 4.8.
,
Assume the following hypotheses:
) e C(R 3) and f(x, x, x)
(i)
f(x,
(ii)
The solutions of Eq. (4.33) can be extended over
[0, ).
(iii)
Eq. (4.36) has a unique solution with respect to
Cn+ I.
0 for all real x.
Then problem (4.32) has a unique solution on
[0, )
given by formula (4.35) for
DIFFERENTIAL EQUATIONS WITH PIECEWISE CONSTANT ARGUMENT DEVIATIONS
t e
703
[n, n+l], where
F(x,
If
f(Cn, Cn, cn)
the condition x(n)
,
)
dx
f(x,
0, then x(t)
c
.,,^
c
n
)
is a solution on
[n, ) of Eq. (4 32)
with
n
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