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multiple time scale dynamics with two fast variables and one slow ...

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since|a| is exp<strong>one</strong>ntially small,|b|≤δ <strong>and</strong> ˆZ≤1 (see Lemma 2.5.4). Then we<br />

can use (2.32) <strong>and</strong> get<br />

<br />

ˆXi ≤ ¯M ˆX0e<br />

<br />

≤ ¯M<br />

1<br />

<br />

≤ ¯M<br />

1<br />

¯Kǫ e−µ 2 t + 4K<br />

µ (ǫ+δ)<br />

<br />

t<br />

e s β1(r)dr<br />

<br />

β2(s)ds<br />

t<br />

0 β1(s)ds<br />

t<br />

+<br />

0<br />

¯Kǫ e−µ 2 t t<br />

+ e<br />

0<br />

−µ<br />

2 (t−s) 2K(ǫ+δ)ds<br />

<br />

(2.38)<br />

Looking at (2.38) we see that the first term is exp<strong>one</strong>ntially small for T1= O(1/ǫ)<br />

<strong>and</strong> so we haveˆX≤O(ǫ+δ). Since|a| is exp<strong>one</strong>ntially small we also have<br />

α(t)+2C|a|≤K2e −c2/ǫ =:κ (2.39)<br />

for some K2, c2> 0. Using (2.31) <strong>and</strong> the usual Gronwall Lemma we compute<br />

| ˆZi(t)| ≤ | ˆZi|e κt +<br />

t<br />

0<br />

e κ(t−s) κds<br />

= | ˆZi(0)|e κt + e κt − 1 (2.40)<br />

But sinceκis exp<strong>one</strong>ntially small <strong>and</strong> t≤O(1/ǫ) we see thatκt must be expo-<br />

nentially small. But e κt − 1 ≈ κt which means that the second term of (2.40)<br />

is exp<strong>one</strong>ntially small. In Lemma 2.5.2 we showed that| ˆZi(0)| is exp<strong>one</strong>nen-<br />

tially small <strong>and</strong> this implies that| ˆZi| is exp<strong>one</strong>ntially small. This establishes the<br />

<strong>two</strong> estimates we want for Step 1 as long as we can show that ˆZ ≤ 1 <strong>and</strong><br />

ˆX≤1/( ¯Kǫ). Indeed, we know that| ˆZi(0)| is exp<strong>one</strong>nentially small <strong>and</strong> so there<br />

is a <strong>time</strong> 0

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