DearMHFmembers,

I have the following problem.

Problem.Find all the singular points of the system

$\displaystyle \dot{x}=xy+12$

$\displaystyle \dot{y}=x^{2}+y^{2}-25.$

Investigate the stability and determine the type of each singular point. $\displaystyle \rule{0.2cm}{0.2cm}$

I know that the critical points of the system are $\displaystyle (4,-3)$, $\displaystyle (3,-4)$, $\displaystyle (-4,3)$ and $\displaystyle (-3,4)$. 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 $\displaystyle (4,-3)$, I linearize the equation at $\displaystyle (4,-3)$ and get

$\displaystyle \dot{x}=-3 (x - 4) + 4 (y + 3)$

$\displaystyle \dot{y}=8 (x - 4) - 6 (y + 3)$

This is possible since the remainders tend to zero, i.e.,

$\displaystyle \lim_{(x,y)\to(4,-3)}\frac{(x-4) (y+3)}{\sqrt{x^{2}+y^{2}}}=0$

and

$\displaystyle \lim_{(x,y)\to(4,-3)}\frac{25 + x(x-8) + y (y+6)}{\sqrt{x^{2}+y^{2}}}=0$.

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

Thanks a lot.

bkarpuz