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We are given the equations:

\begin{align} \dot{x}& =\mu \, x +y+y^3 \\ \dot{y}& =2x-2y+xy^2+\gamma \, x^2y \end{align}

The question at hand is to determine whether there is some sort of reflection symmetry in this system of equations, and thus to predict what kind of bifurcation occurs as $\mu$ is altered. For some background info, I have already determined that when $\mu < -1$, the fixed point $(0,0)$ is a stable node, when $\mu = -1$ it is still a stable node, and when $\mu > -1$, the fixed point is a saddle point.

Here's what I did, and I'm wondering if it's legitimate/correct:

If there is a reflection symmetry situation occurring, then both $x(t)$, $y(t)$ and $-x(t)$, $-y(t)$ will be solutions/will yield fixed points, so we get, by hypothesis, that:

\begin{align} \dot{x} & =\mu x +y+y^3 \\ \dot{y}& =2x-2y+xy^2+\gamma x^2y \end{align}

and

\begin{align} \dot{x}&= -\mu \, x -y-y^3 \\ \dot{y}&= -2x+2y-xy^2-\gamma \, x^2y \end{align}

thus

\begin{align} \dot{x}&=-\mu \, x -y-y^3=-\dot{x} \\ \dot{y}&=-2x+2y-xy^2-\gamma \, x^2y=-\dot{y} \end{align}

which yields: $\dot{x}=-\dot{x}\Longrightarrow \dot{x}=0,$ and the same thing for $\dot{y}$

Does this show that there is a reflection symmetry? I think it should, since this shows that if $(x,y)$ yields fixed points, so does $(-x,-y)$, and this would imply a pitchfork bifurcation, due to the change in stability of $(0,0)$ as $\mu$ is changed, and the fact that the only fixed point is $(0,0)$ when $\mu=-1$, but I am unsure if my reasoning is correct.

Thoughts?

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