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I have this

$$ | \xi |^{2} = 1 - 4p^{2}(1-p^{2})s^{4}$$

where $s = \sin\left(\frac\omega 2 \right)$. The method is said to be stable if $ | \xi|\leq1$.

From here I am supposed to deduce that this scheme is stable for $ -1 \leq p \leq 1$, but I do not know where to go from here... Can anyone suggest how?

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  • $\begingroup$ What is the premise and what is the conclusion precisely? $\endgroup$ – DeepSea Apr 25 '18 at 18:23
  • $\begingroup$ VON Neumann stability analysis, what do you mean by given~? $\endgroup$ – italy Apr 25 '18 at 18:24
  • $\begingroup$ math.stackexchange.com/questions/2743887/… this is the full question and I am stuck at this point $\endgroup$ – italy Apr 25 '18 at 18:33
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Hints:

  1. If $p^2\in[0,1]$ then $4p^2(1-p^2)\in [0, 1]$, say by AM-GM.

  2. If $p^2>1$, then $(1-p^2)$ is negative, so your LHS is $>1$ unless $s=0$.

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  • $\begingroup$ Can you explain the first first line of your comment? What is AmM GM?? $\endgroup$ – italy Apr 26 '18 at 8:04
  • $\begingroup$ AM-GM refers to the inequality between Arithmetic Mean and Geometric Mean. If that’s not familiar, note $(x+y)^2=(x-y)^2+4xy\geqslant 4xy$. Now set $x=p^2$ and $y=1-p^2$. $\endgroup$ – Macavity Apr 26 '18 at 11:43
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Is this a correct analysis?

$ 1-4p^{2}(1-p^{2}) s^{4} \leq 1 $

$ -4p^{2}(1-p^{2}) s^{4} \leq 0 $

$ p^{2}(1-p^{2}) \geq 0$ Since sine is bounded between 0 and 1

$p^{2} \geq 0 $ is always true. So $ 1-p^{2} \geq 0 $ gives $p^{2} \leq 1 $

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    $\begingroup$ Yes, it is fine. To be completely exhaustive, one should distinguish the case $s=0$ from $s\neq 0$ when dividing by $s^4$. $\endgroup$ – Harry49 Apr 27 '18 at 10:06
  • $\begingroup$ Thank you, also since this equals $ \xi^{2} $ do i have to also look at $ \xi_{\pm}$? $\endgroup$ – italy Apr 27 '18 at 10:32
  • $\begingroup$ Proving $|\xi|^2 \leq 1$ is fine since equivalent to $|\xi| \leq 1$. $\endgroup$ – Harry49 Apr 27 '18 at 12:18

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