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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 $\theta$.

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  • $\begingroup$ The expression $3-4e^{-i\theta}+e^{-2i\theta}$ is equal to zero if $\theta=0$, which is within your allowed range, so there is no upper bound on the value of $r$. $\endgroup$ Sep 22 at 0:18
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    $\begingroup$ note that $\theta =\pi$ gives $G=1-2r$ so one needs $r \le 1$; are you sure that there is such $r$ as some computations suggest that for every $r \in (0, 1]$ there is $\theta_r$ (given by $\cos \theta_r=\frac{4-4r}{4-3r}$) st $|G(\theta_r)| >1$ $\endgroup$
    – Conrad
    Sep 22 at 2:30
  • $\begingroup$ @Conrad, thank you a lot. I only ask you to write your comment as a solution for me to close the question :) Another thing, can you give me some hints on how you arrived to that expression? $\endgroup$
    – Fernando
    Sep 22 at 20:16
  • $\begingroup$ Done as requested - computed the maximum value of $|G|^2$ for $0 \le r \le 1$; the same analysis works for $r >1$ where we now we still have a $\theta_r$ for $1 \le r \le 8/7$ and the maximum of $|G|^2$ is the bigger of $\frac{r^3}{4(4-3r)}+1, (2r-1)^2$ (which coincide at $r=8/7$ not surprisingly as $\theta_{8/7}=\pi$), so most likely the maximum is still $\frac{r^3}{4(4-3r)}+1$) while for $r >8/7$ there is no more $\theta_r$ as $|\frac{4-4r}{4-3r}| >1$ so the maximum of $|G|$ is just $2r-1$ $\endgroup$
    – Conrad
    Sep 23 at 0:54

1 Answer 1

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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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