# Need help with inequality to understand stability analysis

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?

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

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 $s=0$.

• Can you explain the first first line of your comment? What is AmM GM?? – italy Apr 26 '18 at 8:04
• 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 $y=1-p^2$. – Macavity Apr 26 '18 at 11:43

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 $p^{2} \leq 1$

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