The equation that I am dealing with is $\frac{x_{n+1}-x_{n-1}}{2 \,\Delta t}=i \omega x_n$ (i.e $F(x_n)=\frac{dx}{dt}=i \omega x_n$).
From this I have calculated the amplification factor to be $A=i \omega \,\Delta t \pm \sqrt {1- \omega^2 \,\Delta t^2}$.
In turn, I get the result $|A|^2=1$. Is this correct?
I am trying to consider two different cases for the timestep: Firstly $\omega \,\Delta t<1$ and secondly $\omega \,\Delta t>1$. However, I am confused how to do this as I have $|A|^2=1$. Does this mean that the numerical solution is absolutely stable for all $\Delta t$, or is the stability still conditional on the size of $\omega \,\Delta t$? Because then the $\sqrt{1- \omega^2 \, \Delta t^2}$ part of $A$ would change in sign for the two cases.
Any help on how to analyse the numerical stability for these two cases would be greatly appreciated!
$|A|=1$ is only true as long as $\sqrt{1−ω^2Δt^2}$ stays real, i.e., as long as $ωΔt<1$. In the other case, $$|A|=ωΔt+\sqrt{ω^2Δt^2-1}>1$$ for the larger of the solutions.