# Non linear phase portrait

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

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 can someone give me 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) this doesn't tell me much

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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 –  Amzoti Feb 1 '13 at 3:52
Alright i removes the second question, thanks for the advice. –  Faust7 Feb 1 '13 at 3:54
Have you tried calculating the vector field at some simple points and plotting it and seeing what you get –  user45150 Feb 1 '13 at 7:31
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. –  Faust7 Feb 1 '13 at 21:00

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 $$\begin{array}{cccc} x>0&y>0&x'>0&y'<0\\ x>0&y<0&x'>0&y'<0\\ x<0&y>0&x'>0&y'>0\\ x<0&y<0&x'>0&y'>0 \end{array}$$