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

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52 CHAPTER 3. APPROXIMATE SOLUTIONS<br />

<strong>The</strong>orem 3.4. Fundamental Inequality. Suppose that f is continuous and Lipshitz<br />

(with constant K) on a set R that includes the point (t 0 , y 0 ). Let h > 0, and<br />

suppose that y (1) , y (2) are admissible ɛ 1 - and ɛ 2 -approximations on |t−t 0 | < h. <strong>The</strong>n<br />

|y (2) (t) − y (1) (t)| ≤ e K|t−t 0| |y (2) (t 0 ) − y (1) (t 0 )| + ɛ 1 + ɛ 2<br />

K<br />

( )<br />

e K|t−t0| − 1<br />

(3.75)<br />

Proof. Since y (1) and y (2) are ɛ-approximations with ɛ 1 and ɛ 2 respectively, we have<br />

Let<br />

∣<br />

∣<br />

dy (1)<br />

dt<br />

dy (2)<br />

dt<br />

− f(t, y (1) )<br />

∣ ≤ ɛ 1 (3.76)<br />

− f(t, y (2) )<br />

∣ ≤ ɛ 2 (3.77)<br />

(3.78)<br />

ɛ = ɛ 1 + ɛ 2 (3.79)<br />

p(t) = y (2) (t) − y (1) (t) (3.80)<br />

<strong>The</strong>n<br />

dp<br />

∣ dt ∣ ≤ ∣ ∣f(t, y (1) ) − f(t, y (2) ) ∣ + ɛ (3.81)<br />

≤ K ∣ ∣<br />

∣y (1) − y (2) + ɛ (3.82)<br />

≤ K|p| + ɛ (3.83)<br />

except for a possible finite number of locations where dp/dt is undefined.<br />

Suppose that (t) ≠ 0, t 0 < t ≤ t 0 + h. <strong>The</strong>n since its sign does not change, p(t) is<br />

either always positive or always negative. Suppose it is positive. <strong>The</strong>n p(t) > 0, and<br />

Multiply the equation by e −Kt and rearrange to give<br />

<strong>The</strong> left hand side is exactly<br />

p ′ (t) ≤ Kp(t) + ɛ (3.84)<br />

e −Kt [ p ′ (t) − Kp(t) ] ≤ ɛe −Kt (3.85)<br />

d (<br />

p(t)e<br />

−Kt ) = e −Kt [ p ′ (t) − Kp(t) ] (3.86)<br />

dt<br />

Hence<br />

d (<br />

p(t)e<br />

−Kt ) ≤ ɛe −Kt (3.87)<br />

dt<br />

Math 582B, Spring 2007<br />

California State University Northridge<br />

c○2007, B.E.<strong>Shapiro</strong><br />

Last revised: May 23, 2007

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