Consider the (nonlinear) system, $ \left\{ \begin{array}{c} \dot x=|y|,\\ \dot y=-x. \end{array} \right.$

Sketch the phase portrait of the system.

I have tried to sketch for cases $y > 0$ and $y < 0$.

This a decoupled system and I want to think it looks like an absolute value graph with everything moving away there, and I have no idea what this looks like. Could someone give me a hint?

I have tried to solve each system separately by splitting the variables and integrating. I get $x(t)=|y|(t+c)$ and $y(t)=-x(t+c)$ but this doesn't tell me much.

  • $\begingroup$ Welcome to MSE! To keep questions and answers together, the etiquitte is to have one question at a time unless related. Look at MathJax in FAQ for formatting. Regards $\endgroup$
    – Amzoti
    Feb 1 '13 at 3:52
  • $\begingroup$ Alright i removes the second question, thanks for the advice. $\endgroup$
    – Faust
    Feb 1 '13 at 3:54
  • $\begingroup$ Have you tried calculating the vector field at some simple points and plotting it and seeing what you get $\endgroup$
    – user45150
    Feb 1 '13 at 7:31
  • $\begingroup$ I am not sure how to do that I have never seen a non linear system graphed before. And there isn't one in my linear differential equations textbook. $\endgroup$
    – Faust
    Feb 1 '13 at 21:00
  • $\begingroup$ "this a decoupled system" Quite the contrary, the system is very much coupled since $x'$ involves $y$ and $y'$ involves $x$. $\endgroup$
    – Did
    Nov 10 '15 at 8:07

The easiest method I have found for graphing phase planes is as follows.

  1. Find the nullclines of $x$ and $y$. These are the lines where $x'$ and $y'$ equal zero. They are important because in each region separated by the nullclines, the sign of $x'$ and $y$ doesn't change. Furthermore, on the nullclines, you know that vector field is parallel to one of the coordinate axes on the nullcline. In your case, $|y|=0\Rightarrow y=0$ is the $x$ nullcline, this is where $x'=0$ For the $y$ nullcline, $-x=0\Rightarrow x=0$ is where $y'=0$.

  2. In each region, evaluate $x'$ and $y'$ to determine the direction of the vectors in that region. Your nullclines split the graph into 4 regions: $x>0,y>0$; $x>0,y<0$; $x<0,y>0$; and $x<0,y<0$. Checking each of the regions $$ \begin{array}{cccc} x>0&y>0&x'>0&y'<0\\ x>0&y<0&x'>0&y'<0\\ x<0&y>0&x'>0&y'>0\\ x<0&y<0&x'>0&y'>0 \end{array} $$

Here is an image for the final result (on a mobile device, so I can't edit the unnecessary parts out): enter image description here

  • 1
    $\begingroup$ That is really really weird looking... my prof must be crazy to think I'd know how to draw that! I have a dynamical systems andnonlinear/chaos textbook for ant or class I am going to see if it mentions these nullclines and give it a try. Ty for explaining this $\endgroup$
    – Faust
    Feb 1 '13 at 22:48

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