I've got a rather long question, and i've done a lot of it, but i've hit a small hurdle and i'm not really sure what it wants me to work out.

I have:

$\displaystyle \dot{x_1}=x_1(x_2-x_3)$

$\displaystyle \dot{x_2}=x_2(x_3-x_1)$

$\displaystyle \dot{x_3}=x_3(x_1-x_2)$

and i'm considering it's behaviour on the plane $\displaystyle x_1+x_2+x_3=1$.

The question is this:

I decided to set $\displaystyle x_3=0$ so I would just be working in the $\displaystyle x_1, x_2$ plane.On each line $\displaystyle x_i=0$, $\displaystyle i=1,2,3$ the dynamics can first be reduced to two equations in two variables (since $\displaystyle x_i$ is identically equal to 0), and then by using $\displaystyle x_1+x_2+x_3=1$, to one equation in one variable.

Write down the equation and solve it.

Find the Poincare map for a time step of t=1.

Describe the dynamics on such a line.

Substitutiing $\displaystyle x_3=0$ into the equations gives:

$\displaystyle \dot{x_1}=x_1 x_2$

$\displaystyle \dot{x_2}=-x_1 x_2$

$\displaystyle \dot{x_3}=0$

Then, using $\displaystyle x_1+x_2+x_3=1$ we get:

$\displaystyle \dot{x_1}=x_1(1-x_1)$

$\displaystyle \dot{x_2}=-x_2(1-x_2)$

$\displaystyle \dot{x_3}=0$

This shows that $\displaystyle x_3=0$ for all values of $\displaystyle x_1$ and $\displaystyle x_2$ (since it start at 0 and stays where it is).

We can solve $\displaystyle x_1$ and $\displaystyle x_2$ to get:

$\displaystyle x_1(t)=\frac{Ae^t}{1-Ae^t}$ and $\displaystyle x_2(t)=\frac{-Be^t}{1-Be^t}$ where A and B are constants.

So my solution is:

$\displaystyle \begin{pmatrix} {x_1} \\ {x_2} \end{pmatrix}=\begin{pmatrix} {\frac{Ae^t}{1-Ae^t}} \\ {\frac{-Be^t}{1-Be^t}} \end{pmatrix}$ <----- Is this right?

So now I move on to try and find the Poincare map. I know that:

$\displaystyle \begin{pmatrix} {\dot{x_1}} \\ {\dot{x_2}} \\ \dot{s} \end{pmatrix}=\begin{pmatrix} {x_1(1-x_1)} \\ {-x_2(1-x_2)} \\ 1 \end{pmatrix}$

Now I have to make it $\displaystyle 2 \pi$ periodic so I get:

$\displaystyle \begin{pmatrix} {\dot{x_1}} \\ {\dot{x_2}} \\ \dot{s} \end{pmatrix}=\begin{pmatrix} {x_1(1-x_1)} \\ {-x_2(1-x_2)+f(cos(2 \pi s))} \\ 1 \end{pmatrix}$

From here I would just integrate the LHS, but I run into problems with $\displaystyle \dot{x_2}$. I tried choosing $\displaystyle f(cos( 2 \pi s))=cos(2 \pi s)$ but I still couldn't integrate that.

Am I approaching the Poincare map in the correct way?