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

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Chapter 7<br />

Delay <strong>Differential</strong> <strong>Equation</strong>s<br />

7.1 Basic <strong>The</strong>ory<br />

Definition 7.1 (Delay <strong>Differential</strong> <strong>Equation</strong>). Let<br />

0 ≤ τ 1 < τ 2 < · · · < τ m (7.1)<br />

be a set of m positive numbers representing the delays, the largest delay being denoted<br />

by τ m . If y appears explicitly without any delay then τ 1 = 0. <strong>The</strong>n we may define<br />

the scalar delay differential equation (DDE) as<br />

y ′ (t) = f(t, y(t − τ 1 ), y(t − τ 2 ), . . . , y(t − τ m )) (7.2)<br />

y(t) = g(t), t 0 − τ m ≤ t ≤ t 0 (7.3)<br />

<strong>The</strong>orem 7.1. Let D ∈ [t 0 , t 0 + β) × R m be convex; suppose that f : D ↦→ R is<br />

continuous on D; and that g(t) : [t 0 − τ m ] ↦→ D. <strong>The</strong>n if r 1 ≠ 0 the DDE (equations<br />

7.2, 7.3) has a unique solution on [t 0 , t 0 + α), where 0 < α ≤ β. Furthermore, if<br />

α < β then y(t) approaches the boundary of D as t → α. If τ 1 = 0 then the DDE<br />

has at most one solution.<br />

Proof. <strong>The</strong> proof is constructive. We consider two cases:<br />

Case 1: τ 1 ≥ 0: For t 0 ≤ t ≤ τ 1 + t 0 , each y(t − τ i ) = g(t − τ i ), hence we have an<br />

explicit expression for f(t), viz.,<br />

y ′ (t) = f(t, g(t − τ 1 ), g(t − τ 2 ), . . . , g(t − τ m )) (7.4)<br />

y(0) = g(0) (7.5)<br />

Hence we can integrate on (0, t) where t < τ 1 ,<br />

y(t) =<br />

∫ t<br />

t 0<br />

y ′ (t)dt = g(0) +<br />

∫ t<br />

t 0<br />

f(x, g(x − τ 1 ), . . . , g(x − τ m ))dx (7.6)<br />

This uniquely determines y(t), t 0 ≤ t ≤ τ 1 . So we can repeat the process on<br />

[t 0 + τ 1 , t 0 + min(β, 2τ 1 )]. We can repeat this process until we reach β. This<br />

process is called the method of steps.Thus by construction, a solution exists.<br />

135

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