Unstable fixed point

Question

Consider the system

$\dot{x} = x(1-4x^2-y^2)-\frac{1}{2}y(1+x) $

$\dot{y} = y(1-4x^2-y^2)-2x(1+x) $

Show that origin is an unstable fixed point

I made $\dot{x} = 0$ and $\dot{y}=0$ and $\dot{x} = \dot{y}$ . Then found an equation,$x=0,y=\frac {\sqrt3}{2}$ but when I did y=0, i found imaginary x. Then I did $\frac{\dot{y}}{\dot{x}}$. I cant continue now on.

I am editing, actually adding one more question.I confused because of that

Here is it,

By considering, $\dot{V}$ , wherer $V=(1-4x^2-y^2)^2$, show all trajectories approach the ellipse $4x^2+y^2=1$ as t goes to infinity

Answer 1

Hints:

• Find the Jacobian, $J(x,y)$
• Evaluate the eigenvalues of the Jacobian at the point $(x,y) = (0,0)$.

Note: It looks like you were trying to find all of the fixed points. I am not sure if you were supposed to find all of the fixed (critical) points, but there are four of them, which includes the point $(x,y) = (0,0)$.

The fixed points are:

• $x = -\dfrac{1}{8}, y = -\dfrac{1}{4}$
• $x = 0, y = 0$
• $x = \dfrac{1}{16} (1-\sqrt{65}), y = \dfrac{1}{8} (\sqrt{65}-1)$
• $x = \dfrac{1}{16} (1+\sqrt{65}), y = \dfrac{1}{8} (-1-\sqrt{65})$

Here is a phase portrait showing these.

Updates to Question

• You are given $V=(1-4x^2-y^2)^2$.
• Evaluate $V' = 2(1-4x^2-y^2)(-8x x' - 2yy')$
• Substitute in $x'$ and $y'$ and simplify.
• Do you notice anything that looks like the ellipse $4x^2+y^2=1$ in the phase portrait?
• This is what this part of the exercise is after (notice how all of the trajectories approach this ellipse).

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Linearize the equations: $$ \dot{x} \sim \half\pars{x - y}\,,\quad\dot{y} \sim y - 2x\qquad\imp\qquad {\dot{x} \choose \dot{y}} \sim \pars{% \begin{array}{rr} \half & -\,\half \\[2mm] -2 & 1 \end{array}}{x \choose y} $$ The matrix eigenvalues $\braces{\lambda}$ are given by: $\ds{\pars{\half - \lambda}\pars{1 - \lambda} - 1 = 0\quad\imp\quad\lambda^{2} - {3 \over 2}\,\lambda + \half = 0}$: $$ \lambda_{\pm} = {3/2 \pm \root{\vphantom{\large A}9/4 - 2} \over 2} = {3/2 \pm 1/2 \over 2} = \left\lbrace% \begin{array}{rcl} \lambda_{+} & = & 1 \\[1mm] \lambda_{-} & = & \half \end{array}\right. $$ You'll have some behaviors $\propto \expo{t}$ and $\propto \expo{t/2}$ which 'blows up' $x$ and $y$ when $x, y \sim 0$.