This is the part of the question I do not understand. I'm given an nonlinear equation for a population, let's say $$ \dot{x}= x^{4}-3x^{3}+2 $$ So we are asked to find the equilibrium points, one is x=1 and considering small perturbation determine it's stability which is where I am lost. If someone can explain this I will really appreciate it.
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I'll write $\dot x = f(x) = x^4-3x^3+2$.
Consider the value of $f$ in a neighbourhood of 1.
If $f(a) > 0$ when $a > 1$ (for $a$ still sufficiently close to 1), then the values of $x$ will increase further since $\dot x > 0$. Hence the point can't be stable.
Similar reasoning applies when $f(b) < 0$ for $b < 1$ and sufficiently close to 1.
If, on the other hand, $f(a) < 0$ when $a > 1$, then after a small perturbation upwards, $x$ would tend to return to the value 1. Likewise for small perturbations downwards if $f(b) > 0$ for $b < 0$.
I would say that the previous paragraph is a reasonable definition of stability. Can you provide the definition you are using?
There's also the case where $f(a) = 0$ when $a > 1$. How to handle this depends on the definition of stability you have, but since $x$ would not tend to return to the initial value after perturbation, I would not call this stable.
Tommi
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F has a maximum at x0. If the system is disturbed by changing the energy by dE it remains 'close' to x0, so x0 is a position of stable equilibrium. That's all I have, thanks! – Samad Miah Aug 12 '15 at 11:44
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@SamadMiah Is my answer satisfactory? If not, can you be more specific with what you want? – Tommi Aug 12 '15 at 12:09
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Your answer is beyond satisfactory. Thanks! – Samad Miah Aug 12 '15 at 12:45