find the value of instablity error from which this value shows instability

Question

I have an Euler method that has this form:
$$\hat{I}(t_{n+1}) = \hat{I}(t_{n})+h\beta \hat{I}(t_{n})(1-\frac{\hat {I}(t_{n})}{N})$$ which can also be written like $$\hat{I}(t_{n+1})=\phi (\hat{I}(t_{n}))$$ where $\phi (x)$ is the iteration function down below: $$\phi(x) = x + h \beta x (1- \frac{x}{N})$$ We know that iteration $$x_{n+1}=\phi(x_{n})$$ converges to $x^*$ if $\phi(x^*)=x^*$ and $|\phi'(x)|<1$ for $x$ around $x^*$.
Here we have $\phi(N) = N$ and $\phi'(x) = 1 +h\beta (1- \frac{2x}{N})$, so $\phi'(N)=1-h\beta$.
So if $h\beta < 2$, we have $$\lim_{n\to\infty} \hat{I}(t_{n})= N = \lim_{t\to\infty}I(t)$$ Here $\beta=0.125$ and $N=10000$.
So my question is: what is the condition for which the iteration converges to $N$? Or in another word, what is the value of $h$ from which this problem become instable?

Answer 1

Assuming everything up until $\phi'(N)=1-h\beta$ is fine, we should have $|\phi'(N)|<1$, not $\phi'(N)<1$.

This leaves us with $-1<1-h\beta<1$ or $0<h\beta<2$.

Solving for $h$ when $\beta>0$ then gives us the requirement that $0<h<2\beta^{-1}$.

For $\beta=0.125$, we get $0<h<16$.