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June 24th, 2014, 06:31 PM   #1
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Does the stability of a trivial fixed point change?

Course: Self-Study
Textbook: A Course in Mathematical Biology

Question: At what value of a does the stability of the trivial fixed point change?

I have this equation $f(x) = axe^{-x}$, where $a = e^r$.

To find the fixed points, $f(b) = b$, then we have
$$\begin{align}
abe^{-b} &= b \\
abe^{-b} - b &= 0 \\
b\left[ae^{-b} - 1 \right] &= 0
\end{align}$$
Thus, our fixed points are $b_1=0$ (trivial) or $b_2=\ln(a)$ (nontrivial).

To determine the stability of the fixed points, we have $$f'(x) = ae^{-x}[1-x]$$ so that
$$\begin{align}
f'(b_1) &= f'(0) = a = e^r\\
f'(b_2) &= f'(\ln(a)) = 1 - \ln(a) = 1 -r
\end{align}$$
Since $f'(b_1) = e^r$, $f'(b_1)>1$, thus the trivial fixed point, $b_1$, is always unstable. So the absolute value of $f'(b_1)$ is never less than one. So the stability of the trivial fixed point never changes, right?
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June 25th, 2014, 03:08 AM   #2
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I'm not 100% sure if you're right, but if you are, it's worth pointing out that your conclusions are based on the assumption that r is positive. Is this the case?
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June 25th, 2014, 04:48 AM   #3
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Yes, I think so.

\begin{align*}
f^\prime(x) \ge 1 \implies \frac{f(x+h) - f(x)}{h} &\ge 1 &\text{for small $h$} \\
f(x+h) - f(x) &\ge h & h \ge 0 \\
F(x+h) - f(x) &\le h & h \le 0 \\
\end{align*}
so a small perturbation in $x$ grows with iterations and $f(x)$ is unstable.

I think your result holds for any $r$, but I could be wrong.
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Last edited by v8archie; June 25th, 2014 at 04:55 AM.
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June 25th, 2014, 10:26 AM   #4
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Quote:
Originally Posted by Benit13 View Post
I'm not 100% sure if you're right, but if you are, it's worth pointing out that your conclusions are based on the assumption that r is positive. Is this the case?
Yes, that should've been part of my info. My apologies.
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