Solving and plotting the 2-D Lorenz Equation

Question

We are asked to first solve and then plot the phase plane of \begin{align} \begin{cases} \dot{x}=\sigma x - \sigma y\\ \dot{y} = \rho x-y \end{cases}, \ \sigma, \ \rho >0. \end{align}

Now the textbook way of going at this is to derive the first line, replace in second line and as such remove one variable.

Doing this we get

\begin{align} \ddot{x} = \sigma \dot{x}-\sigma \dot{y} = \sigma \dot{x}-\sigma \rho x+\sigma y = \sigma \dot{x} - \sigma \rho x + \sigma x-\dot{x}\\ \implies \ddot{x}+(1-\sigma)\dot{x}+\sigma (\rho-1)x=0\\ \implies P(\lambda)=\lambda^2+(1-\sigma)\lambda+\sigma(\rho-1)=0\\ \implies \Delta=\sigma^2+\sigma(2-4\rho)+1. \end{align} At this point we have two variables $\sigma, \rho$. How exactly are we supposed to continue? Take a painstakingly $16$ (!!!) numbers of cases?

I am sure there is a faster way for this, please someone enlighten me.

Thank you.

Answer 1

Since you are able to use Mathematica, here is an approach.

You can manually run or use the two play buttons to see the changes based on the two parameters. You could also choose static settings, copy all those and make a movie. Just paste the following code into your notebook.

  Manipulate[
  StreamPlot[{\[Sigma] x - \[Sigma] y, \[Rho] x - y}, {x, -5, 
  5}, {y, -5, 5}, 
  PlotLabel -> 
  Row[{"\[Sigma] = ", \[Sigma], 
  " ,  \[Rho] = ", \[Rho]}]], {\[Sigma], 0, 10, 0.25}, {\[Rho], 0, 
 10, 0.25}]

You will get a window like this

You can also solve the system and do plots from that. For example, https://mathematica.stackexchange.com/questions/15858/bifurcation-diagrams-for-multiple-equation-systems or https://mathematica.stackexchange.com/questions/60633/help-in-bifurcation-diagram.

Lastly, you might be able to draw a two-parameter bifurcation diagram.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\left\{% \begin{array}{rcrcr} \ds{\dot{\on{x}}\pars{t}} & \ds{=} & \ds{s\on{x}\pars{t}} & \ds{-} & \ds{s\on{y}\pars{t}} \\ \ds{\dot{\on{y}}\pars{t}} & \ds{=} & \ds{\rho\on{x}\pars{t}} & \ds{-} & \ds{\on{y}\pars{t}} \end{array}\right.} \end{align} Note that I replaced $\ds{\sigma\ \mbox{by}\ s}$ to avoid any confusion with $\ds{Pauli\ Matrices\ \sigma_{x},\ \sigma_{y},\ \sigma_{z}\ \mbox{o/y}\ \vec{\sigma}}$. The above expression can be rewritten as $$ \totald{}{t}{\on{x}\pars{t} \choose \on{y}\pars{t}} = \pars{\begin{array}{rr}\ds{s} & \ds{-s} \\ \ds{\rho} & \ds{-1}\end{array}}{\on{x}\pars{t} \choose \on{y}\pars{t}} = \pars{-\,{1 \over \tau}\,{\bf 1} + \vec{a}\cdot\vec{\sigma}} {\on{x}\pars{t} \choose \on{y}\pars{t}} $$ $\ds{\bf 1}$ is the $\ds{2 \times 2\ identity\ matrix}$ and $\left\{\begin{array}{rcl} \ds{\tau} & \ds{=} & \ds{2 \over \ds{1 - s}} \\ \ds{\vec{a}} & \ds{=} & \ds{{r - s \over 2}\,\hat{x} - {r + s \over 2}\,\ic\,\hat{y} + {s + 1 \over 2}\,\hat{z}} \end{array}\right.$

Therefore, \begin{align} {\on{x}\pars{t} \choose \on{y}\pars{t}} = \expo{-t/\tau}\exp\pars{\vec{a}\cdot\vec{\sigma}\,t}{\on{x}\pars{0} \choose \on{y}\pars{0}} \end{align} $\ds{\exp\pars{\vec{a}\cdot\vec{\sigma}\,t}}$ satisfies $\ds{\pars{\partiald[2]{}{t} - \kappa^{2}}\exp\pars{\vec{a}\cdot\vec{\sigma}\,t} = 0{\bf 1}}$ \begin{align} & \mbox{where}\ k = \root{\vec{a}\cdot\vec{a}} = \root{{1 \over 4}\, s^{2} + \pars{{1 \over 2} - \rho}s + {1 \over 4}} \\[5mm] & \mbox{Moreover,}\ \left.\exp\pars{\vec{a}\cdot\vec{\sigma}\,t} \right\vert_{\, t\ =\ 0}\ = {\bf 1}\quad\mbox{and}\quad \left.\totald{\exp\pars{\vec{a}\cdot\vec{\sigma}\,t}}{t} \right\vert_{\, t\ =\ 0}\ = \vec{a}\cdot\vec{\sigma} \end{align} $$ {\on{x}\pars{t} \choose \on{y}\pars{t}} = \expo{-t/\tau}\ \overbrace{\bracks{\cosh\pars{\kappa t}{\bf 1} + {\sinh\pars{\kappa t} \over \kappa}\vec{a}\cdot\vec{\sigma}}} ^{\ds{\equiv {\bf A}}} {\on{x}\pars{0} \choose \on{y}\pars{0}} $$ $$ \rule{0pt}{3mm} $$ $$ {\bf A} = \pars{% \begin{array}{cc} \ds{\cosh\pars{\kappa t} + {s + 1 \over 2\kappa}\sinh\pars{\kappa t}} & \ds{{\rho - s \over 2\kappa} - {\rho + s \over 2\kappa}\sinh\pars{\kappa t}} \\[2mm] \ds{{\rho - s \over 2\kappa} + {\rho + s \over 2\kappa}\sinh\pars{\kappa t}} & \ds{\cosh\pars{\kappa t} - {s + 1 \over 2\kappa}\sinh\pars{\kappa t}} \end{array}}\phantom{AAAAAAAA} $$