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Let $x_t = f (x_{t-1})$ a difference equation of order $1$, with $f(x) = ux(1-x)$; $u \in (0; 4)$.

  • Show that $x^* = 0$ is always a steady state for all $u$ .

  • Compute its positive steady state (meaning x? > 0) depending on .

  • Give two different values of $u$ such that in one case the steady state is stable and another for which it is unstable.

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  • $\begingroup$ ask yourself: what does steady state mean? what doest stability mean? this is a simple question. $\endgroup$ – Ethan Oct 28 '13 at 16:37
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First question: $x^* = 0$ is a steady state

You can check this just by inserting $x=0$ in $f(x)$, it is always zero regardless of $u$.

Second question: positive steady states

We look for $x > 0$ such that $x = f(x) = ux(1-x)$... it turns out to be $x=1 - \frac{1}{u}$, which is positive only for $u > 1$.

Third question: stability

In order to have stability, we need the derivative of $f(x)$ in $x^*$ to be lower than one. Given the formula for steady state in point 2, $f'(x^*)$ is equal to $2 - u$. So, any example with $u < 1$ will have an unstable steady state (and the other way around, $u > 1$ implies stable steady state).

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