Determine the classification of a fixpoint of a vector field

Question

Hi I have following problem. I have to show for the vector field:

$$ \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix}y^3 - 4x \\ y^3 - y - 3x \end{pmatrix} $$

that there are fixpoints at $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix} -2 \\ -2 \end{pmatrix}$, and $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$, which I have but I cannot figure out how to determine which type of fixpoints they are mathematically, I could only determine it graphically. Has someone an idea on how to show that point 0,0 is a stable fixpoint and the other two are unstable?

Greetings Max

Answer 1

Let

$X(x, y) = \begin{pmatrix} y^3 - 4x \\ y^3 - y- 3x \end{pmatrix} \tag 1$

denote our vector field. Then the Jacobian matrix of $X(x, y)$ is

$J_X(x, y) = \begin{bmatrix} -4 & 3y^2 \\ -3 & 3y^2 -1 \end{bmatrix}; \tag 2$

at the critical point $(0, 0)$ of $X(x, y)$ this becomes

$J_X(0, 0) = \begin{bmatrix} -4 & 0 \\ -3 & -1 \end{bmatrix}; \tag 3$

since $J_X(0, 0)$ is lower tringular, we may directly read off the eigenvalues from the diagonal as being $-4$ and $-1$, which are both negative; thus in accord with the usual theorems on the subject $(0, 0)$ is a stable fixed point of the flow of $X$. At both $(-2, -2)$ and $(2, 2)$, we find

$J_X = \begin{bmatrix} -4 & 12 \\ -3 & 11 \end{bmatrix}; \tag 4$

it is well-known, and easy to derive, that the eigenvalues of a $2 \times 2$ matrix such as $J_X$ satisfy

$\mu^2 - \text{Tr}(J_X) \mu + \det J_X = 0; \tag 5$

we have

$\text{Tr}(J_X) = 7; \; \det J_X = -8; \tag 6$

it then follows from the quadratic formula that

$\mu = \dfrac{1}{2}(7 \pm \sqrt{81}) = -1, 8; \tag 7$

the critical points $(-2, -2)$ and $(2, 2)$ are thus saddles, hence unstable, having each as they do a positive eigenvalue.

The above shows how it's done: we linearize the system at the critical points and then find the eigenvalues, which, provided they have non-zero real part as they do here, then determine the type and stability of the zeroes of $X(x, y)$. Though the technique is elementary, the rigorous mathematics behind it can take one quite deep; see for example the stable manifold theorem.

Answer 2

Try to linearize the vector field. Then the Jacobian (it's eigenvalues) at each point will give you informations about the stability of the point. For example, if eigenvalues have negative real part then the point is stable.