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

Hence we reduced the problem to checking signs. Obviously Mu and Ms
have the vector field (x2, sx2− f (x1), 0) as tangent vector. Consider another tan-
gent vector given by (a ± , b ± , c=1). We want to take the third coordinate to be
1 to assure that the vector is linearly independent from the vector field and this
is justified since (ds) ′ = 0 (Check!). We can make the two vectors orthonormal
using the Gram-Schmidt algorithm. We conclude that the forms P ± ... are equal -
up to a positive normalization factor N> 0 - to the corresponding 2×2 subde-
terminant of the 2×3 matrix
⎛ ⎞
x2 sx2− f (x1) 0
⎜⎝ ⎟⎠
a ± b ± 1
In particular for P ± x1s this means:
P ±
x2 0
x1s=N
a ±
=
Nx2
1
From the fast subsystem it follows that x2>0 on Mu and Ms. Therefore P ± x1s=
Nx2> 0 and equation (2.46) can be simplified to:
P ±′
x1x2 = sPx1x2+ N(x2) 2
(2.47)
The next step is to look at the signs of P ± x1x2 . Observe that Mu and Ms both have
a line of equilibrium points with a tangent vector in the s-direction; this follows
directly from the construction as we have appended the equation s ′ = 0. For
any plane containing such a line the form Px1x2 must vanish. Suppose the time
variable in the differential equations we have written down so far is t. Then if
t→∞ we must have P − x1x2
→ 0 due to the location of the line of equilibrium
points. This implies that Px1x2< 0 since equation (2.47) has a positive right-hand
side for t sufficiently large. Similarly it follows that if t→−∞ then P + x1x2→ 0 and
so P + x1x2 > 0. This shows the sign condition we want.
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