<s>[INST] Inequality with trigonometric relations performing a Von Neumann analysis, I arrived to the relation
\begin{equation}
E_m(t+\Delta t)=\underbrace{\left[1-\frac{r}{4}\left(3-4e^{-i\theta}+e^{-2i\theta}\right)\right]}_{G}E_m(t),
\end{equation}
where $\theta\in[-\pi,\pi]$ and $r>0$. The question is to find the largest $r$ such that the factor $G$ has absolute value less or equal than 1, i.e., $|G|\leq 1$.
I know that I can estimate $r$ numerically, but the idea is to find some analytical relation. I have spent quite some time looking to that expression without success, so, any help will be appreciated.
Update: The inequality $|G|\leq 1$ has to be valid for all values of $\theta$.
 [/INST] Note that for $z=e^{-i\theta}$ one has $G(z)=1-\frac{r}{4}(z^2-4z+3)$ so by the polar C-R the maximum of $|G(z)|$ when $|z|=1$ is attained at points where $\frac{zG'(z)}{G(z)} \ge 0$ (which of course means $\frac{zG'(z)}{G(z)}$ real)
Now $G'(z)=\frac{r}{2}(2-z)$ is non zero on $|z|=1$ so the above is equivalent to $\frac{G(z)}{zG'(z)} > 0$ and simplifying $r/2 >0$ one gets $\frac{G(z)}{2z-z^2}>0$ hence one needs $\Im (G(z) (2\bar z-\bar z^2)) =0$ so using $z\bar z=1$ one needs
$$\Im (2\bar z-\bar z^2-\frac{r}{4}(2z+6\bar z+4\bar z-3\bar z^2))=0$$
But $z=e^{-i\theta}$ so $\Im (2z+10\bar z)=8\sin \theta$ hence we get:
$2\sin \theta -\sin 2\theta-2r\sin \theta +\frac{3r}{4} \sin 2\theta=0$ which means $\sin \theta =0$ or $2-2\cos \theta-2r+\frac{3r}{2}\cos \theta=0$
In particular for $\theta =\pi$ one gets that $z=-1$ and $G(-1)=1-2r$ hence we need $r \le 1$.
For $r=1$ one gets $\cos \theta =0$ as the other roots of the maximum modulus equation so $z =\pm i$ and $G(\pm i)=1-\frac{1}{4}(2 \pm 4i)$ has modulus greater than $1$ so $r<1$ and we need to study the points where $\cos \theta =\frac{4-4r}{4-3r}=f(r)$ so $\sin^2 \theta =1-f(r)^2, \cos 2\theta =2f(r)^2-1, \sin 2\theta=2f(r)\sin \theta$
Now (separating in real and imaginary parts) $$|G(r,\theta)|^2=(1-\frac{r}{2}(1-\cos \theta)^2)^2+\frac{r^2(2-\cos \theta)^2\sin^2 \theta}{4}$$ so substituting as above, one gets
$$|G_{\max}(r)|^2 \ge (1-\frac{r^3}{2(4-3r)^2})^2+\frac{r^2}{4}(2-\frac{4-4r}{4-3r})^2(1-\frac{(4-4r)^2}{(4-3r)^2})$$ or by further simplifications
$$|G_{\max}(r)|^2 \ge  \frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}$$
But (with the help of Wolfram Alpha) we note that:
$$\frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}=\frac{(2-r)^2(r+4)}{4(4-3r)}=\frac{r^3}{4(4-3r)}+1$$
which shows that $$|G(r, \theta)|^2 \ge \frac{r^3}{4(4-3r)}+1$$ for $0 \le r \le 1$ and actually the analysis above shows that is the maximum for $0 \le r \le 1$ and it is bigger than $1$
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