Question: Find under what conditions of $n$ is $y=n\left(\frac{\pi }{4}\right)$ stable or unstable given $y'=\sin(4y)$.
I actually have the solutions given by my lecturer, but there are still few things that I am unclear about.
His solutions:
Take $n$ to be even ($2m$) . Then $y\:=\:\frac{2m\pi }{4}=\frac{m\pi }{2}$
If $\frac{m\pi}{4}-\frac{\pi}{4}<y<\frac{m\pi}{2}$
$2m\pi -\pi <4y<2m\pi $
$y'=\sin(4y)<0$
My question is how did he arrive at this inequality? and why did he minus $\frac{\pi }{4}$?
And he also did:
$\frac{m\pi }{2}<y<\frac{m\pi}{2}+\frac{\pi}{4}$
$2m\pi < 4y < 2m\pi + \pi $
$y' = \sin(4y) > 0 $
Also how did he arrive at this inequality and why did he add $\frac{\pi }{4}$?
What about the odd numbers? He has done nothing about the odd numbers.
Step 1, (that you already solved) was to find all the equilibrium points, that is where $y' = 0$. This is at $y = n \frac{\pi}{4}$ for all integers $n$.
At all other $y \in \mathbb{R}$, either $y' > 0$ or $y' < 0$.
An equilibrium point $y = n \frac{\pi}{4}$ is stable if the region "above" it is $y' < 0$ and the region "below" it is $y' > 0$.
Your prof has shown for any even $n = 2m$, $y = m \frac{\pi}{2}$ is unstable because the region "above", namely $(m \frac{\pi}{2}, m \frac{\pi}{2}+ \frac{\pi}{4})$, satisfies $y' > 0$, and the region "below", namely $(m \frac{\pi}{2}- \frac{\pi}{4},m \frac{\pi}{2})$, satisfies $y' < 0$.
The add/subtract $\frac{\pi}{4}$ come from that's how far to the next equilibrium point.
The odds are the stable ones. See if you can work out why.