Fixed Points of 2D Dynamical System

Question

I am trying to find the fixed points (i.e. where $\dot{x}=\dot{y}=0$) of the following dynamical system:

$\dot{x} = \sin(x-y)-\sin(x)$

$\dot{y} = \sin(y-x)-\sin(y)$

Using the fact that $\sin(-x)=-\sin(x)$, I reduce the system to:

$\sin(x)=-\sin(y)$.

This tells me that $y=-x$ is our set of fixed points. But when I plug in, $x = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$ into the system of equations, I do not get that $\dot{x}=0$ or $\dot{y}=0$. Can someone explain what I am doing wrong?

Answer 1

First of all, in order to have an idea of the behavior of this differential system, I have obtained, using a Matlab script, the graphics below. One can see in particular on this figure the $2 \pi$ periodicity in $x$ and $y$ of the solutions.

Point $x=\pi/2,y=-\pi/2$ is not a solution (the arrows are not zero at this point), but there is a rather close solution as we will see it.

Your error comes from the fact that, having a system of 2 equations with two unknowns

$$\begin{cases}\sin(x-y)&=&\sin(x)\\ \sin(y-x)&=&\sin(y)\end{cases} \ \ \iff$$

$$\tag{1}\begin{cases}\sin(x-y)&=&\sin(x)& \ \ \ (a)\\ \sin(x-y)&=&\sin(-y)& \ \ \ (b)\end{cases}$$

you have reduced it to a single equation $\sin(x)=-\sin(y)$, and thus obtained an erroneous conclusion.

Here is a way to solve system (1). As

$$\sin(u)=\sin(v) \ \ \ \iff \ \ \ u=v+2k\pi \ \text{or} \ u=\pi-v+2k\pi $$ system (1) is equivalent to four "coupled" solutions:

$$\begin{cases} x-y&\equiv&x \ &\text{and}& \ x-y&\equiv&-y \ &\implies& x&=&2k\pi,& y&=&2\ell\pi,\\ x-y&\equiv&\pi-x \ &\text{and}& \ x-y&\equiv&-y \ &\implies& x&=&2k\pi,& y&=&\pi+2\ell\pi,\\ x-y&\equiv&x \ &\text{and}& \ x-y&\equiv&\pi+y \ &\implies& x&=&\pi+2k\pi,& y&=&2\ell\pi,\\ x-y&\equiv&\pi-x \ &\text{and}& \ x-y&\equiv&\pi+y \ &\implies& &&&&&(\text{see below.}) \end{cases}$$

for $k \in \mathbb{Z}$

Detailed solution of the fourth case:

If $x-y=\pi-x+2K\pi$ and $x-y=\pi+y+2L\pi$, we can infer that $y=-x +2M\pi$. Then $2x=\pi-x+2K\pi$. Thus

$$x=\frac{\pi}{3}+k\frac{2\pi}{3}, y=-\frac{\pi}{3}-k\frac{2\pi}{3} \ \ \text{with the same} \ k.$$

for $k \in \mathbb{Z}$, as can be verified on the graphics.

Remark: on this figure, some integral solutions have been displayed.

Answer 2

I like the other answer, but just to give a simple answer to your question, it's a system of 2 equations and 2 variables. So once you've solved for one variable, $y=-x$ in your case, just plug it back in to the original equations to solve for the other. Taking the first equation, with $y=-x$, we get $\sin(2x)-\sin(x)=0$ which is equivalent to $2\sin(x)\cos(x) - \sin(x) = 2\sin(x)(\cos(x)-\frac{1}{2}) = 0$, which is true when $\sin(x) = 0$, so $x = n\pi$, or when $\cos(x) = \frac{1}{2}$, so $x = 2\pi n \pm \frac{\pi}{3}$. Plug it back into that equation for $y$ you got and that will give you your set of fixed points:

$(n\pi,-n\pi), \ (2\pi n-\frac{\pi}{3},\frac{\pi}{3} - 2\pi n), \ (2\pi n + \frac{\pi}{3}, -2\pi n - \frac{\pi}{3})$