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Suppose given a system of ODEs

$x' = sx^{r} - x\left(sx^{r}+ty^{r}\right) ;$

$y' = ty^{r} - y\left(sx^{r}+ty^{r}\right) ,$

such that $x+y\equiv 1$ and $s,t,r\in\mathbb{R}_{>0}$. The points $\left(x,y\right) =\left(1,0\right)$ and $\left(x,y\right) = \left(1,0\right)$ are equilibrium points for this system. If $r\neq 1$, then there is also another equilibrium point, but I'm going to forget about that for now. Intuitively, the points $\left(1,0\right)$ and $\left(0,1\right)$ should be unstable equilibria if $0<r<1$, but when I calculate the Jacobian of the map

$\left(x,y\right) \longmapsto \left( sx^{r} - x\left(sx^{r}+ty^{r}\right) , ty^{r} - y\left(sx^{r}+ty^{r}\right) \right)$,

I come up with

$\begin{pmatrix}srx^{r-1} - \left(r+1\right) sx^{r} - ty^{r} & -rtxy^{r-1} \\ -rsx^{r-1}y & rty^{r-1} - \left(r+1\right) ty^{r} - sx^{r}\end{pmatrix}.$

(I know that these two ODEs may be combined into one to avoid using the Jacobian matrix, but I'd like to try it this way.) Evaluating this matrix at $\left(0,1\right)$, for instance, gives

$\begin{pmatrix}-t & 0 \\ 0 & -t\end{pmatrix}$,

which has negative eigenvalues (since $t>0$). Doesn't this calculation imply that the point $\left(0,1\right)$ is stable? (Somehow it is very unexpected that this point should be stable.) Also, should the stability depend on $r$?

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