Time-Scale Decomposition: The slow variable only has a single unstable steady state?

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I'm completing a time-scale decomposition of a set of equations. The original set of two equations is

$$ S'= 3k_1S^3E_T + (3k_{-1}-3k_1S^3)E_1 $$ $$ E_1'=-k_1S^3E_T+(k_{-1}+k_2+k_1S^3)E_1 $$

I was able to nondimensionalize the equations to the following:

$$ \frac{ds}{d\tau}=3s^3+3(\lambda-s^3)e $$ $$ \epsilon\frac{de}{d\tau}=-s^3+(K+s^3)e $$

Now I'm trying to do a time-scale decomposition, with $\epsilon$ in front of the "slow variable", then $e$ is the slow variable. So the first step in a time-scale decomposition would be to look at the second equation $\frac{de}{d\tau} $ and assume that $s$ is a constant.

So I have a linear equation that crosses the $e'$ axis (y-axis) at $-s^3$ and the $e$ axis (x-axis) at $ \frac{s^3}{K+s^3} $. But this implies that the only steady state is at $e = \frac{s^3}{K+s^3} $, which is an unstable steady state.

Does this mean that I cannot do the time-scale decomposition for the other equation? Normally we would assume instantaneous convergence to the stable steady state, but in this case I can't.

What is the next step (or have I gone as far as I can)?

THANK YOU!

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I was able to fix the system!

I changed one of my parameters to include a negative that way my system became

$$ \frac{ds}{d\tau}=-3s^3-3(\lambda-s^3)e $$ $$ \epsilon\frac{de}{d\tau}=s^3-(K+s^3)e $$

Then my time-scale decomposition gave me a stable steady state at $ e = \frac{s^3}{K+s^3} $ and I was able to finish the decomposition, which showed that s decays to 0 with time.