# Phase Potentials

Looking to get a explanation of the following solution.

$x'=sinx$ I understand that you must integrate $sinx$ and then take the negative of it making it $v(x)=cos(x) + c = cos(x)$ with $c=0$. However I am having problems understanding the equilibrium points.

The equilibrium points of this system are,

$$x^*=kz$$ $x*=kz$ k is even, this is unstable
$x^*=kz$ k is odd, are stable

Why is it unstable when even and stable when even?

• Have you looked at the direction field plot? For example, when $x^* = 0$, the direction field leaves that critical point (unstable). When it is odd, they come toward that point, like $x^* = \pi$, that is stable. When it is even, you will notice that they leave the critical point, that is unstable. Said differently, calculate the value of $x'$ on each side of a critical point and see that they are opposite signs. – Moo Feb 6 '17 at 21:01
• how would you know that $\pi$ is odd? – mp12345 Feb 6 '17 at 21:02
• could you submit a answer with this detail? – mp12345 Feb 6 '17 at 21:10

We are asked to determine the stability of the equilibrium points of

$$x' = \sin x$$

We solve

$$x' = \sin x = 0 \implies x = \pi~ n, n \in \mathbb{Z}$$

Now, we need to determine the stability of those critical points and will look at three cases, $n = 0, n = 1, n = 2$ to cover all bases.

We can look at the value of the slope on either side of the critical points to determine the direction field. For example, at $x^* = 0$, we have a table of $(x, x')$ as

$$(x, x') = (-1.01,-0.846832),(-0.76,-0.688921),(-0.51,-0.488177),(-0.26,-0.257081),(-0.01,-0.00999983),(0.24,0.237703),(0.49,0.470626),(0.74,0.674288),(0.99,0.836026)$$

Notice that points below $x^* = 0$, the slope is negative, hence the direction field points away from it. Notice that for points above $x^* = 0$, the slope is positive, hence the direction field points away from it. This means that this is an unstable critical point. Said another way, we are looking for the ranges for where sine is positive and negative to determine stability. Repeat this for the other two values of $n$.

A direction field plot shows stable for odd $n$ and unstable for even $n$

Update What would happen if we changed this to $x' = \cos(x)$? See the following (do the analysis so you know how)

• perfect that makes sense now, thank you very much – mp12345 Feb 6 '17 at 21:26
• so if we had cosx would it be the opposite? even would be stable and odd unstable? – mp12345 Feb 6 '17 at 21:32
• See update in answer. Slightly different critical points, but same idea. – Moo Feb 6 '17 at 21:41
• so would you still look at 0, $\pi$,$2\pi$? o being stable and pi being unstable – mp12345 Feb 6 '17 at 21:52
• The critical points are now $x^* = \pi n - \dfrac{\pi}{2}, n \in \mathbb{Z}$. The rest is the same approach. Notice the difference of those critical points in each diagram. The second seems shifted from the first and it is because the location of the critical points. – Moo Feb 6 '17 at 21:53