I have the simple, but not so simple system:
\begin{equation} x'=-3x+y^2+2 \\ y'=x^2-y^2 \end{equation}
and its critical points, found by solving the system by multiplying the lower by 3, and summing up, are $(x,y)=(-1,1), (1,1), (1,-1), (-1,-1)$.
I would solve this ideally by setting up the matrix
\begin{vmatrix} -3-\lambda & 1 \\ 1 & -1-\lambda \end{vmatrix}
and solving for the determinant. But that +2 in the first equation messes things up. How do I proceed here?
Thanks!
For $P(x,y)$ and $Q(x,y)$.
$P(x,y)=−3x+y^2+2$
$P(x,y)=x^2−y^2$
We have the solution by finding the Jacobian,
\begin{vmatrix} \frac{\partial P}{\partial x}|_{(x_1,y_1)} & \frac{\partial P}{\partial y}|_{(x_1,y_1)} \\ \frac{\partial Q}{\partial x}|_{(x_1,y_1)} & \frac{\partial Q}{\partial y}|_{(x_1,y_1)} \end{vmatrix}
where we obtain:
\begin{vmatrix} -3 & 2y \\ 2x &-2y \end{vmatrix}
Insert for each critical points:
CP : (1,1):
$DetA_1$=\begin{vmatrix} -3 & 2\cdot 1 \\ 2\cdot 1 &-2\cdot 1 \end{vmatrix} =2.
Since 2 > 0, the system is expansive, from the critical point, and thus unstable. For a CP lesser than 0, the system is contractive, and it is therefore stable.
For CP (-1,1): Det A= 10 > 0, Unstable.
For CP (1,-1): Det A= -2 < 0, Stable.
For CP (-1,-1): Det A= -10 < 0, Stable.
Then, for the second set of Critical points:
$CP_2=(x,y)\cdot 2=(1,1)\cdot 2=(2,2)$
where 2 is the constant from the inhomogenous term.
We now have:
CP : (2,2):
$Det A_2$=\begin{vmatrix} -3 & 2\cdot 2 \\ 2\cdot 2 &-2\cdot 2 \end{vmatrix} =4.
Since 4 > 0, the system is expansive, from the critical point, and thus unstable. For a CP lesser than 0, the system is contractive, and it is therefore stable.
For CP (-2,2): Det A_2= 38 > 0, Unstable.
For CP (2,-2): Det A_2= 4 > 0, Unstable.
For CP (-2,-2): Det A_2= -32 < 0, Stable.