Delay Dierential Equations (DDE)
There will be a few homework assignments, a computer project and a presentation of
a paper (on tghe last week of classes). For the computer project, you need to know how to
run Matlab (use the command runmatlab o
(vii) Replacing x(t r) by its Taylor polynomial in r may lead to wrong conclusion.
x (t) = 2x(t) + x(t r).
Since a = 2 < 0 and 0 < b = 1 < |a|, we know that solutions converge to 0
exponentially as t . Now, consider the ODE
x (t) = 2x(t
(iii) For a < 0 and 0 < |b| < |a|, solution x(t) 0, exponentially, as t .
This is the case of so called diagonally dominant, meaning that the instantaneous/
nodelay term is somehow stronger than the term with delay. Also, it says the trivial
For a nonautonoumous ODE, an initial value problem takes the form
for x(t) Rn
x (t) = f (t, x(t)
x(t0 ) = x0
where t0 R and x0 Rn . For DDE, its initial value problem looks like
x (t) = f (t, xt )
where R plays the role of t
x (t) = f (t, xt )
where r > 0, R, C and D R C are given and f : D Rn is continuous.
Picture 1 is t x(t) in R Rn . Picture 2 is t xt in R C .
Given the pair (, ) R C , we dene C ([ r, ), Rn ) by
( + t) = (0),
for t 0
BACKWARDS CONTINUATION OF SOLUTIONS.
Consider the example
x (t) = a(t)x(t r).
Let C . In order to have a solution x(t) dened on (r , 0], where > 0, such that
x(t) = (t) for t [r, 0], we need x (0) = a(0)x(r), i.e (0) = a(0)(r). Thus, the
IC () s
CONTINUATION OF SOLUTIONS.
Let x : [ r, + ] Rn ( > 0) and x : [ r, + ] Rn ( > 0) be two solutions
of RFDE(f ) through (, ). We say that x is a continuation of x if and x(t) = x(t)
for all t [ r, + ]. By using Zorns lemma, we see that there is a
Proof of Lemma 4. T is well dened by Lemma 3. Next, since |f | < M , we have
|T (, , f, y )(t1 ) T (, , f, y )(t2 )|
|f ( + s, ys + +s )| ds M |t1 t2 |
and |T (, , f, y )(t)| M , for t, t1 , t2 I . In fact, the inequalities hold for all t