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I have the following nonlinear dynamical system:

\begin{align} \frac{dX}{dt} &= Y(\gamma - 2X) \\\ \frac{dY}{dt} &= - \gamma X + X^2- Y^2 + 1 \end{align}

And I'm trying to understand what type of bifurcation is happening at the point $\gamma = 2$. I'm fairly new to dynamical analysis and not sure what would be the best way to approach this specific problem.

So far what I tried is to look at the jacobian: $$ \mathbf{J} = \begin{bmatrix} -2Y & \gamma \\[1ex] % <-- 1ex more space between rows of matrix -\gamma + 2X & -2Y \end{bmatrix} $$ And when $\gamma =2$ the equilibrium point of the system is $(X = 1, Y = 0)$, which gives the following: $$ \mathbf{J}= \begin{bmatrix} 0 & 2 \\[1ex] % <-- 1ex more space between rows of matrix 0 & 0 \end{bmatrix} $$ The problem is that I'm not sure how to continue from here, since the eigenvalue is zero and I don't see a way to characterize the bifurcation.

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  • $\begingroup$ Find a textbook, which explains the details? $\endgroup$
    – Artem
    Jun 28, 2021 at 20:44
  • $\begingroup$ Do you mean, you want to know about all the bifurcations in the $XY$-plane, or is there a particular point you want to know about? If the latter, I'd start by finding out what point you're interested in, expanding $X,Y$ about there and comparing it to examples from Wikipedia's "Bifurcation Theory". I'd also use a computer plotter, like Mathematica's StreamPlot[]. $\endgroup$
    – David
    Jun 28, 2021 at 21:04
  • $\begingroup$ The type of bifurcation happening here is not the stuff you find in textbooks, it is more unique. I've tried linear stability analysis of the Jacobian matrix near the equilibrium point but it didn't lead me to anything - that's why I posted the question $\endgroup$ Jun 28, 2021 at 21:28
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    $\begingroup$ Really? Then I would suggest to add everything that you tried so far. $\endgroup$
    – Artem
    Jun 28, 2021 at 22:06

1 Answer 1

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We're looking for stationary points: $$ \left\{\begin{aligned} & y (\gamma - 2x) = 0\\ & -\gamma x + x^2 - y^2 + 1 = (x - 0.5 \gamma)^2 - y^2 + (1-0.25 \gamma^2) = 0 \end{aligned}\right. . $$

There are two possibilities: $y = 0$ or $x = 0.5\gamma$.

In the first case we have: $$ \left\{\begin{aligned} & \varnothing, \quad \gamma < 2\\ & (1,0), \quad \gamma = 2\\ & (0.5\gamma \pm \sqrt{0.25\gamma - 1}, 0), \quad \gamma > 2 \end{aligned}\right. . $$

In the second case we have: $$ \left\{\begin{aligned} & (0.5\gamma, \pm \sqrt{1 - 0.25\gamma}), \quad \gamma < 2\\ & (1,0), \quad \gamma = 2\\ & \varnothing, \quad \gamma > 2 \end{aligned}\right. . $$

Jacobian: $$ J(x,y) = \begin{pmatrix} -2y & \gamma - 2x \\ 2x - \gamma & -2y \end{pmatrix}. $$

Hence:

  1. point $(0.5\gamma, \sqrt{1 - 0.25\gamma})$, $\gamma < 2$, is a stable star (the Jacobian has two negative equal eigenvalues);
  2. point $(0.5\gamma, -\sqrt{1 - 0.25\gamma})$, $\gamma < 2$, is an unstable star (the Jacobian has two positive equal eigenvalues);
  3. points $(0.5\gamma \pm \sqrt{0.25\gamma - 1}, 0)$, $\gamma > 2$, are unstable focuses (the Jacobian has complex eigenvalues with positive real part);
  4. point $(1,0)$, $\gamma = 2$, is a degenerate equilibrium point (the Jacobian has zero eigenvalue, $J(1,0) = 0$, $\gamma = 2$).

When $\gamma = 2$, we have the following equation: $$ \left\{\begin{aligned} & \dot{\xi} = - 2 \eta \xi\\ & \dot{\eta} = \xi^2 - \eta^2 \end{aligned}\right. , $$ where $\xi = x - 1$ and $\eta = y$. This point doesn't have a special name, I think. Degenerate points are topologically classified by their sectors. This point has two elliptic sectors (they contain homoclinic loops). You can find more in the literature listed here.

The scenario of the bifurcation is simple: nodes merge with each other into one degenerate point $(1,0)$, and then this point splits into two focuses.

The bifurcation is already atypical, it is not Morse--Smale, because the point $(1,0)$, $\gamma = 2$, has infinitely many homoclinic orbits.

two nodes

the degenerate equilibrium point

two focuses

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  • $\begingroup$ Thank you! Can you please explain how you classified the points using the jacobian? Also how the $\eta$ and $\zeta$ equation helped to understand the bifurcation $\endgroup$ Jun 29, 2021 at 3:06
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    $\begingroup$ @ValientProcess I made an edit about the jacobian $\endgroup$ Jun 29, 2021 at 8:58
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    $\begingroup$ @ValientProcess actually, $\eta$ and $\xi$ didn't help me much, but this equation is some kind of a standard form for such degenerate point. The picture in this case were much more useful. $\endgroup$ Jun 29, 2021 at 9:01

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