The Computable Differential Equation Lecture ... - Bruce E. Shapiro

138 CHAPTER 7. DELAY DIFFERENTIAL EQUATIONS
Since f is strongly Lipshitz,
and therefore
|f(t, y 1 , . . . , y m ) − f(, ỹ 1 , . . . , ỹ m )| ≤ K max
1≤j≤m |y j − ỹ j | (7.36)
|y(t) − ỹ(t)| ≤ |θ(t 0 ) − ˜θ(t 0 )| + K
≤ v(t 0 ) + K
∫ t
∫ t
Applying the definition of v(t) one more time,
By Gronwall’s inequality,
t 0
max |θ(x − τ j) − ˜θ(x − τ j )|dx (7.37)
1≤j≤m
t 0
v(x)dx (7.38)
v(t) ≤ v(t 0 ) + K
from which 7.30 follows immediately.
∫ t
t 0
v(x)dx (7.39)
∫ t
v(t) ≤ v(t 0 ) exp v(x)dx
t 0
(7.40)
7.2 Method Of Steps
Consider the DDE with a single delay,
y ′ (t) = f(t − τ, y(t − τ)) (7.41)
y(t) = g(t), t 0 − τ ≤ t ≤ t 0 (7.42)
where we will assume that g(t 0 − τ) = g(t 0 ). Then on the interval [t 0 , t 0 + τ] we
have
∫ t
y(t) = g(t 0 ) + f(s − τ, y(s − τ))ds (7.43)
t 0
= g(t 0 ) +
∫ t
t 0
f(s − τ, g(s − τ))ds (7.44)
The right hand side is fully explicit and can be evaluated. This gives the solution
on the first “step,” [t 0 , t 0 + τ]. Next, assume we have already solved the DDE on
[t 0 , t 0 + kτ] for some integer k > 0. (This statement is immediately true for k = 1
once we have solved the first interval). Then on subsequent steps [t 0 + kτ, t 0 + (k +
1)τ], for k = 1, 2, . . . ,
y(t) = y(t 0 + kτ) +
∫ t
t 0 +kτ
f(s − τ, y(s − τ))ds (7.45)
Since the integrals are evaluated in order, i.e., the integral is solved on the interval
[t 0 + (k − 1)τ, t 0 + kτ] first, everything on the right-hand side of the equation is
already known, so the integral can be evaluated.
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007