Quiz prep (Tr/Det 1)

markmark
edited October 2020 in Problems

Consider the system:

$$
\begin{align}
x' &= x - x y^2 \\
y' &=x+y.
\end{align}
$$

First, find the equilibria of the system.

Then, for each equilibrium:

  • Find the linearization (expressed as the Jacobian matrix) of the system at that equilibrium, and
  • Use the determinant and trace of the Jacobian to classify as an
    • Unstable node,
    • Unstable spiral,
    • Stable spiral,
    • Stable node, or
    • Saddle point

Note that the Det/Tr analysis of a linear system is described in detail in section 4.5 of our text, as well as on this webpage. Linearization is described in detail in section 5.1 of our text.

Comments

  • ChristianDonaldsonChristianDonaldson

  • markmark

    As we discussed in class, this looks pretty good though I'd definitely recommend that you use a Tr/Det analysis to classify the equilibria.

    Taking a closer look, though, I noticed a couple of other issues.

    It seems to me that $x+y=0$ implies that $y=-x$; so, I'm not sure why you've got $(1,1)$ as an equilibrium. (Note that the term is equilibrium, not critical point.)
    I think the equilibria are $(1,-1)$, $(0,0)$, and $(-1,1)$.

    Also, I guess you're missing a minus sign in the upper right term of your Jacobian matrix.

    Again and most importantly, a Tr/Det analysis is much simpler than an eigenvalue analysis, if you want to answer these specific quesiotns

    For example, $\det J(\pm 1,\mp 1)=-2<0$; therefore, we have saddle points.

    The classification at the origin is just a little more involved; since $\det J(0,0)=1$ and $\text{Tr}(0,0)=2$, we're in the upper right quadrant of the Tr/Det plane meaning we have an unstable node or spiral. Since $4\times\text{Det}=\text{Tr}^2$, we've it's an unstable node, though it's right on the cusp of being a spiral.

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