= Solution
Intersect the two <nullcline> families. The axes yield $(0,0)$ and $(1,0)$, and $y=1$ with $xy=2$ yields $(2,1)$. The possible remaining intersection would have $x+y=1$ and $xy=2$, requiring $x^2-x+2=0$, whose <discriminant> is $-7$. Hence there are no further real <equilibria>.
The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}
(1-y)(1-2x-y)&x(x+2y-2)\\
-y^2&2-2xy
\end{pmatrix}.
$$
Evaluate it at each <equilibrium>:
* At $(0,0)$ it is $\operatorname{diag}(1,2)$. Both <eigenvalues> are positive, so the point is an \b[<unstable node>].
* At $(1,0)$ it is $\begin{pmatrix}-1&-1\\0&2\end{pmatrix}$. The <eigenvalues> $-1,2$ have opposite signs, so this is a \b[<saddle point>]. Its stable direction is $(1,0)$ and its unstable direction can be taken as $(-1,3)$.
* At $(2,1)$ it is $\begin{pmatrix}0&4\\-1&-2\end{pmatrix}$. Its <characteristic polynomial> is $\lambda^2+2\lambda+4$, so $\lambda=-1\pm i\sqrt3$. This is an \b[<stable focus>], or spiral sink.
All three are <hyperbolic equilibria>, so their classifications persist for the nonlinear system. Immediately to the right of $(2,1)$ on $y=1$, the velocity is vertically downward, showing that the attracting spiral turns \b[clockwise].
Back to article page
