2
$\begingroup$

I'm going through question 5 the past exam here. For starters, I'm trying to find and classify the critical points of the system $$\begin{align} x'&=y, \\ y'&=x^2-y-\epsilon. \end{align}$$ I've found that:

  • The system has no critical points if $\epsilon<0$.
  • The system has a critical point at $(0,0)$ if $\epsilon=0$.
  • The system has two critical points at $(\pm \sqrt{\epsilon},0)$ if $\epsilon>0$.

I've managed to classify the critical points, except for the following boundary cases:

  1. $(0,0)$ when $\epsilon=0$, as the linearized system has zero determinant.
  2. $(-\sqrt{\epsilon},0) $ when $\epsilon=1/64$, which falls on the parabola in the trace-det critical point picture.

All I have is computer generated trajectories, which make me suspect that $(0,0)$ is a saddle. I'd like to know whether my list of boundary cases is corret, and within the list, what is the type of the critical points.

Thank you!

$\endgroup$
1
$\begingroup$

There is a theorem which says so long as the critical (or equilibrium) point is not a center then the nature of this critical point in the system $x',y'$, is the same as that point in the linearization i.e,

$$\tilde{x}' = p_x(x_0,y_0) \cdot x + p_y(x_0,y_0) \cdot y$$

$$ \tilde{y}' = q_x(x_0,y_0) \cdot x + q_y(x_0,y_0) \cdot y$$

where $x' = p(x,y)$ and $y' = q(x,y)$. Now you have a linear system. Find the associated matrix to this system, compute it's eigenvalues and use the Painleve Analysis.

$\textbf{Update}$: My professor refereed to it as Painless analysis, but it seems to also be reference as linearization technique for equilibrium point analysis. In any case, here are two links 1,2.

$\endgroup$
1
$\begingroup$
  1. In the first case the Jacobian matrix is $$ \left(\begin{array}{rr} 0&1\\0&-1 \end{array}\right). $$ As one of its eigenvalues is $0$, the Hartman-Grobman theorem is not applicable to this case, thus, we can say nothing about the structure of the critical point of the nonlinear system. The phase portrait of this system is enter image description here (the blue trajectories are stable, the red trajectories are unstable). We can see here that this is some kind of Saddle-sink.

  2. In the second case the Jacobian matrix is equal to $$ \left(\begin{array}{rr} 0&1\\-\frac14&-1 \end{array}\right), $$ its eigenvalues are $(-\frac12,-\frac12)$. As the eigenvalues have nonzero real part, the Hartman-Grobman theorem is applicable; hence, according to the diagram, this critical point is (homeomorphic to) a degenerate sink.

$\endgroup$
1
$\begingroup$

Solving $x' = 0 = y'$, we get \begin{aligned} x^2 &= \epsilon \\ y &= 0 \, . \end{aligned} Therefore, no critical point (or equilibrium point) is obtained with $\epsilon < 0$.

The origin is the only critical point when $\epsilon = 0$. One eigenvalue of the flow's Jacobian matrix at $(0,0)$ is $-1$, the other is zero. Therefore, this critical point is non hyperbolic. Below is the phase plane, which may confirm that the origin is a saddle-point:

$\epsilon = 0$

When $\epsilon > 0$, two critical points $(\pm\sqrt{\epsilon},0)$ arise. Let us compute the eigenvalues of the flow's Jacobian matrix at the critical points. Depending on the sign of the characteristic polynomial's determinant $\Delta^\pm = 1\pm 8\sqrt{\epsilon}$, the eigenvalues of the flow's Jacobian matrix at the critical points may be complex conjugate or real.

  • if $0 < \epsilon < 1/64$, the determinant $\Delta^\pm$ at both critical points is positive. Therefore, all four eigenvalues $-\frac{1}{2} \pm \sqrt{\frac{1}{4}\pm 2\sqrt{\epsilon}}$ are real. At the critical point $(-\sqrt{\epsilon},0)$, both eigenvalues are negative, whereas at the critical point $(\sqrt{\epsilon},0)$, the eigenvalues have opposite signs.
  • if $\epsilon = 1/64$, the determinant $\Delta^\pm$ is zero at the critical point $(-\sqrt{\epsilon},0)$. The double eigenvalue is negative. At the critical point $(\sqrt{\epsilon},0)$, the determinant $\Delta^\pm$ is positive, and the eigenvalues still have opposite signs.
  • if $\epsilon > 1/64$, the determinant $\Delta^\pm$ is negative at the critical point $(-\sqrt{\epsilon},0)$. The eigenvalues are complex conjugates with negative real part. At the critical point $(\sqrt{\epsilon},0)$, the determinant $\Delta^\pm$ is positive, and the eigenvalues still have opposite signs.

In all these cases, all eigenvalues have nonzero real part, so that according to the Hartman-Grobman theorem, the system will behave similarly to the linearized system in the vicinity of critical points. Below is the phase plane in the case $\epsilon = 1$:

$\epsilon = 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.