Stability of two equilibrium points

Question

Having the system: $$x'=-x+xy \ \ \ \ \ y'=-2y+3y^2 $$ Find the equilibria and study the stability. I've found the equilibria as being $(0,0)$ and $(1,\frac{2}{3})$ but I don't know how to study their stability.

Answer 1

For the critical points you just have to solve $$\left\{\begin{array}{cc}-x+xy&=0\\-2y+3y^2&=0\end{array}\right.$$ and you get $(0,0)$ and $\left(0,\frac{2}{3}\right)$. For their stability you should always check first if Hartman's Theorem applies.

Let $F$ be $F(x,y)=(-x+xy,-2y+3y)$ and now consider the Jacobian matix of $F$.

$$DF(x,y)=\left(\begin{array}{cc}-1+y&x\\0&-2+6y\end{array}\right)$$

For each point you should now evaluate the matrix:

• $(0,0)$

$$DF(0,0)=\left(\begin{array}{cc}-1&0\\0&-2\end{array}\right)$$

Thus the origin is an attractor.

• $\left(0,\frac{2}{3}\right)$

$$DF\left(0,\frac{2}{3}\right)=\left(\begin{array}{cc}- \frac{1}{3}&0\\0&2\end{array}\right)$$

So $\left(0,\frac{2}{3}\right)$ is a saddle.

You should check Hartman's Theorem and the Trace Determinant Diagram.

Answer 2

Setting

$X(x, y) = -x + xy \tag 1$ and

$Y(x, y) = -2y + 3y^2, \tag 2$

the given system is written

$\dot x = X(x, y), \tag 3$

$\dot y = Y(x, y). \tag 4$

The equilibria occur where

$X(x, y) = Y(x, y) = 0 \tag 5$

that is, as correctly observed by Moo in his comment to the question itself, at

$(x, y) = (0, 0), \; \left (0, \dfrac{2}{3} \right ). \tag 6$

To investigate the nature of these critical points, we invoke the Jacobian matrix of the system,

$J(x, y) = \begin{bmatrix} \dfrac{\partial X(x, y)}{\partial x} & \dfrac{\partial X(x, y)}{\partial y} \\ \dfrac{\partial Y(x, y)}{\partial x} & \dfrac{\partial Y(x, y)}{\partial y} \end{bmatrix} = \begin{bmatrix} y - 1 & x \\ 0 & 6y - 2 \end{bmatrix}; \tag 7$

thus we have

$J(0, 0) = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}; \tag 8$

since both eigenvalues of $J(0, 0)$ are negative real, $(0, 0)$ is an attractive critical point of (3)-(4); we also have

$J\left (0, \dfrac{2}{3} \right ) = \begin{bmatrix} -\dfrac{1}{3} & 0 \\ 0 & 2 \end{bmatrix}, \tag 9$

and since now $J\left (0, \dfrac{2}{3} \right )$ has one positive and one negative eigenvalue, we conclude that $\left (0, \dfrac{2}{3} \right )$ is a saddle point of the system.