the scheme is as follows
$ u_j^{n+1} = u_j^{n-1} + 2r ( u_{j+1}^{n} - u_j^{n+1} - u_j^{n-1} + u_{j-1}^{n})$
using the anstaz $ u_{j}^{n} = A \xi^{n} e^{i \omega j}$ and using the identity
$ e^{iw} +e^{-iw} = 2 cosw $
rearranging i get
$ (1+2r) \xi^{2} = -4r \xi cosw-(1-2r) = 0 $ where r is r>0
for stability requires $ | \xi| \leq 1 $ and the roots distinct,
$ \xi \pm = \frac{1}{1+2r}( 2rcos\omega \pm \sqrt{4r^{2}cos^{2}\omega + 1-4r^{2}})$
simplify further using $ 4r^{2}cos^{2}w - 4r^{2} = -4r^{2}sin^{2}w$
$ \xi \pm = \frac{1}{1+2r}( 2rcos\omega \pm \sqrt{1-4r^{2}sin^{2}\omega })$
from this point I don't understand the solution in the book, if someone could explain it to me,
case 1 : $1-4r^{2}sin^{2}\omega$ <0 $ \xi _+ $ and $ \xi_ -$ are complex,
$| \xi _ -|^{2} = | \xi_+|^{2}$ = $\frac{4r^{2}cos^{2}w + 4r^{2}sin^{2}\omega + 1}{(1+2r)^{2}} $ = $ \frac{4r^{2} +1}{(1+2r)^{2}}$
$ 0 < \frac{4r^{2} +1}{(1+2r)^{2}} <1 $
therefore $| \xi _ -|\leq 1 $ and $| \xi_+| \leq 1 $
how have they concluded this and how do you get this inequality?
case 2: $1-4r^{2}sin^{2}\omega >0$
$\frac{1}{1+2r}( 2rcos\omega +\sqrt{4r^{2}cos^{2}\omega + 1-4r^{2}}) \leq \frac{2r cos\omega +1}{1+2r} \leq1$
therefore $\xi_+\leq1$ where we have used the fact that
$ 0< 1-4r^{2}sin^{2} \omega \leq 1$ how did they get this inequality ?
how does one conclude that $\xi +$ is less than or equal to one from the above?