Thread: Stability of a nonlinear system

1. Stability of a nonlinear system

Dear MHF members,

I have the following problem.
Problem. Find all the singular points of the system
$\dot{x}=xy+12$
$\dot{y}=x^{2}+y^{2}-25.$
Investigate the stability and determine the type of each singular point. $\rule{0.2cm}{0.2cm}$

I know that the critical points of the system are $(4,-3)$, $(3,-4)$, $(-4,3)$ and $(-3,4)$. My problem is determining the stability properties of these points. Please let me know if my idea below is correct. For instance for the point critical point $(4,-3)$, I linearize the equation at $(4,-3)$ and get
$\dot{x}=-3 (x - 4) + 4 (y + 3)$
$\dot{y}=8 (x - 4) - 6 (y + 3)$
This is possible since the remainders tend to zero, i.e.,
$\lim_{(x,y)\to(4,-3)}\frac{(x-4) (y+3)}{\sqrt{x^{2}+y^{2}}}=0$
and
$\lim_{(x,y)\to(4,-3)}\frac{25 + x(x-8) + y (y+6)}{\sqrt{x^{2}+y^{2}}}=0$.
Since the linearized equations eigenvalues $\lambda_{1,2}=(-9\pm\sqrt{137})/2$ are of opposite sign, we see that the critical point $(4,-3)$ is a saddle point and is therefore unstable. I will repeat the same steps above for each critical point.

Thanks a lot.
bkarpuz

2. Originally Posted by bkarpuz
Since the linearized equations eigenvalues $\lambda_{1,2}=(-9\pm\sqrt{137})/2$ are of opposite sign, we see that the critical point $(4,-3)$ is a saddle point and is therefore unstable. I will repeat the same steps above for each critical point.

Right, but its is sufficient to compute

$A=J_v(4,-3)=\begin{bmatrix}{\dfrac{{\partial v_1}}{{\partial x}}(4,-3)}&{\dfrac{{\partial v_1}}{{\partial y}}(4,-3)}\\{\dfrac{{\partial v_2}}{{\partial x}}(4,-3)}&{\dfrac{{\partial v_2}}{{\partial y}}(4,-3)}\end{bmatrix}=\begin{bmatrix}{-3}&{\;\;4}\\{\;\;8}&{-6}\end{bmatrix}$

whose eigenvalues are certainly

$\lambda=\dfrac{-9\pm \sqrt{137}}{2}$