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

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when a=0<strong>and</strong>ǫ= 0. We also have<br />

X ′ ⎛ ⎞⎛<br />

⎞ ⎛ ⎞<br />

Λ+Γ 0 0 0 X1 aR5+ bR6<br />

0 Γ 0 0 X2 bR7+ǫR8<br />

= +<br />

0 0 Γ 0 X3 bR9+ǫR10<br />

⎜⎝ ⎟⎠⎜⎝<br />

⎟⎠ ⎜⎝ ⎟⎠<br />

0 0 0 0 X4 ǫR11+ǫR12<br />

⎛<br />

a(ΛaX1−Λy1X2−Λy2X3)+b(ΓbX1+Γy1Z1+Γy2Z2)<br />

= BX+<br />

⎜⎝<br />

b(ΓaZ1+ΓbX2−Γy2X4)+ǫ(−g1aX1+ g1y1X2+ g1y2X3)<br />

b(ΛaZ2+ΛbX3+Λy1X4)+ǫ(−g2aX1+ g2y1X2+ g2y2X3)<br />

ǫ(−g1aZ1− g1bX2+ g1y2X4+ g1aZ2+ g2bX3+ g2y1X4)<br />

⎞<br />

⎟⎠<br />

(2.27)<br />

Hence for a=0, b=0 <strong>and</strong>ǫ= 0 we must haveη2= 0 which completes the proof<br />

of (i). For (ii) note that<br />

⎛<br />

Γ 0 0 0<br />

0 Γ−Λ 0 0<br />

B−Λ(Id)=<br />

(2.28)<br />

0 0 Γ−Λ 0<br />

⎜⎝ ⎟⎠<br />

0 0 0 −Λ<br />

Recall thatΛ≥Λ0>0 <strong>and</strong>Γ≤Γ00; using this observation (ii) follows. We can<br />

group the terms on the right-h<strong>and</strong> side in (2.26) as follows:<br />

F1(Z1, X, t) := a(ΛaZ1+ΛbX2−Λy2X4)<br />

G1(Z, X, t) := ǫ(g1bX1+ g1y1Z1+ g1y2Z2)<br />

F2(Z2, X, t) := a(ΛaZ2+ΛbX3+Λy1X4)<br />

G2(Z, X, t) := ǫ(g2bX1+ g2y1Z1+ g2y2Z2)<br />

Using the triangle inequality we arrive at the estimates required in (iii). Now<br />

set E(X, t) equal to all the terms of X ′ − BX that do not vanish at b=0 <strong>and</strong>ǫ= 0.<br />

43<br />

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