finding $h$ that make absolute instability for Euler method

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: $$\phi(x) = x + h \beta x (1- \frac{x}{N})$$ I use $h=6$ in this method but if I use a $h$ which is a little bit bigger (for example $h=20$), I have an absolute instability error. I want to find the value of $h$ from which this absolute error is shown?
Thanks in advanced

Answer 1

Divide the equation by $N$ to get a new equation in $x=I/N$ that does not contain $N$. Replace $k=h\beta$, so that what remains is the more simple equation $$ x_{n+1}=x_n+kx_n(1-x_n)=x_n(1+k-kx_n) $$ Now finally divide by $(1+k)$ and set $y=kx/(1+k)$ to get the discrete logistic map $$ y_{n+1}=(1+k)y_n(1-y_n) $$ You can read off the stability of that iteration from any plot of the Feigenbaum diagram.

In summary, for $k\in[0,2]$ you get convergent behavior, after that periodic solutions, looking increasingly chaotic after $k=2.82..$

Answer 2

If $\phi(x)$ is a contraction in

$$ x_{k+1}=\phi(x_k) $$

then following with

$$ x_{k+1}-x_k = \phi(x_k) - \phi(x_{k-1}) = \phi'(\xi)(x_k-x_{k-1}) $$

to have a fixed point is sufficient that $|\phi'(\xi)|<1$ or

$$ -1\lt 1-h\frac{\beta x_k}{N}+h \beta\left(1-\frac{x_k}{N}\right)< 1 $$