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Consider the autonomous ODE $$x' = x(e^{-x}-x^2).$$

  1. Determine the stability at the trivial equilibrium $x_1 = 0$.
  2. Show that $f$ has exactly one equilibrium $x_2$ in $(0, 1)$. Determine the stability of $x_2$.
  3. Use the Newton's method to find an approximation of $x_2$ with $x_0 = 1$.

My question is in the second part. For the first part, we let $v(x) = x(e^{-x}-x^2)$. Direct computation shows that $v'(x)=-3x^2 + (1-x)e^{-x}$. Hence, $v'(0)=1>0$. We conclude that $x_1$ is an unstable equilibrium.

Next, I try to show that there is at least one equilibrium in $(0, 1)$ by showing that $v(0)v(1)<0$ while $v(0)$ and $v(1)$ are not the roots. However, how should I proceed to show the uniqueness of such an equilibrium?

Note: I also have no problem in part 3. I just include it for the sake of completeness of the problem, or in case someone suggests a search for $x_2$ by numerical methods, which is not the purpose of part 2.

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2 Answers 2

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The uniqueness is immediate, since $e^{-x}$ decreases and $x^2$ increases (on $(0,1)$).

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You can show that the second derivative is strictly negative, so by concavity it must be constant in the interval or those are the only two steady states. Pick one other point and calculate it to show it's not constant in the interval.

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