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Consider the following equation,

$x_{n+1}=f(x_n)$, with $f(x)= x\exp(\frac{2-x}{3})$

Calculate the two steady states $x^*_1$ and $x^*_2$ of the above map.

So based on what I've learnt to find the steady state you do the following:

Solve the equation $x^*=x^*\exp(\frac{2-x^*}{3})$ for $x^*$ and you will find the steady state(s).

However when attempting to do this myself I found only one steady state of $x^*=2$ which can't be correct as the question indicates there will be two solutions.. however I'm not sure where my knowledge gap is particularly.

Any help or hints would be appreciated.

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    You have assumed $x^*\neq 0$ to simplify!! Maybe $x^*=0$ is indeed a solution. – 2017-02-13

1 Answers 1

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2 is indeed one of the steady states, and to find it you probably did something akin to the following: Divide both sides by $x^*$: $$x^*=x^*e^{\frac{2-x}{3}}$$ $$1=e^\frac{2-x}{3}$$Then taking the logarithm $$0=\frac{2-x}{3}$$ and so $x=2$. However, in the first step I had to make an assumption about $x^*$. What was it?