Finding the Critical points of a system

Question

Find all critical points of the system

$y_1'= y_1(10-y_1-y_2)$

$y_2'= y_2(30-2y_1-y_2)$

then classify them as stable, asymptotically stable, or unstable.

I need help with this particular question, as you may see, the only problem is that I have NO idea how to determine the critical points of the system , nonetheless, I have a thought, if the critical points are the eigenvalues of the system, then can't I turn this into a matrix and determine the eigenvalues from the matrix, or does this not work? How would I go about determining these critical points as eigenvalues. Any suggestion of literature that may help or explanations would be immensely appreciated.

Answer 1

The stationary or critical points are the points where the solutions through them are constant. Which means that the derivatives are zero.

In $y_1'=0=y_2'$ you get trivially $(y_1,y_2)=(0,0)$ as solution, then setting only one component to zero $(y_1,y_2)=(0,30)$ and $(y_1,y_2)=(10,0)$ and then as fourth point the solution of the linear system of the second factors.

Answer 2

Critical points are points where $(y'_1, y_2') = (0, 0)$, so in your case these are the possible cases

$y_1 = 0$ and $y_2 = 0$

Let's call that point ${\bf x}_1 = (0, 0)$

$y_1 = 0$ and $30−2y_1−y_2 = 0$

This leads to $y_2 = 30$, let's call that point ${\bf x}_2 = (0, 30)$

$10−y_1−y_2 = 0$ and $y_2 = 0$

This leads to $y_1 = 10$, let's call that point ${\bf x}_3 = (10, 0)$

$10−y_1−y_2 = 0$ and $30−2y_1−y_2 = 0$

In this case the solution is ${\bf x}_4 = (20, -10)$

To test the stability, you just need to calculate the Jacobian matrix and evaluate its eigenvalues at each of the four locations above.

$$ J = \pmatrix{\partial y_1'/\partial y_1 & \partial y_1'/\partial y_2 \\ \partial y_2'/\partial y_1 & \partial y_2'/\partial y_2} $$

The plot below will help you confirm that the numbers you calculate for the eigenvalues make sense