# Sketching phase portrait

$\dot{x}=-2x-2y$

$\dot{y}=-x-3y$

Equilibrium point is $(0,0)$. Eigenvalues are $\lambda_+=-1$ and $\lambda_-=-4$ which have corresponding eigenvectors $2\choose -1$ and $1 \choose 1$ respectively. The point is a stable node.

So when sketching the diagram, there will be a line through the origin with gradient $1$ and a line through the origin with gradient $-1/2$. The arrows on these lines are going to be inwards towards the equilibrium point. How do I sketch the phase paths and isoclines?

• find the nullclines
Commented Apr 17, 2015 at 0:57
• I have never heard of that before Commented Apr 17, 2015 at 0:58
• The fixed point is of this type. Since $\lambda_-<\lambda_+$, the trajectories are tangent to the eigenvector $2\choose -1$ hence the linked image rotated by $-90$° is your phase plane.
– Did
Commented Apr 17, 2015 at 8:18
• why did you rotate it? in our solutions, it is not rotated. Commented Apr 17, 2015 at 13:40
• Surely you noted that this is an image randomly picked on the internet. You should rotate it so that the two eigenvectors on the image roughly correspond to the two eigenvectors in your case. (Unrelated: Please use @.)
– Did
Commented Apr 18, 2015 at 8:20

To begin with your analysis, you should first find the regions of the plane that have a zero partial derivative. Namely $\dot x, \dot y = 0$.

Setting $\dot x = -2x - 2y = 0 \implies y = -x$ is a set of points where $\dot x = 0$.
Setting $\dot y =-x - 3y = 0\ \ \implies y = -\frac{x}3$ is a set of points where $\dot y = 0$.

While a point on the line $y=-x \implies \dot x = 0$, we can evaluate $\dot y$ along this point
to get $\dot y = -x-3(-x) = 2x$, so along this line, $x>0\implies\dot y >0, x<0\implies \dot y<0.$

A similar argument can be made about evaluating $\dot x$ along the curve which has $\dot y = 0$.
Drawing these lines, and a vector field along these 'nullclines' can give a lot of information about the behavior of systems of differential equations.

• I really don't understand. what would $\dot{y}=-4x$ even look like on a xy plane? since it is not simply $y=-4x$/ Commented Apr 17, 2015 at 13:42
• You don't need to worry about what it looks like, but now you have a function where you can evaluate what the slope is at every point. So if we consider $(x,y) = (1,-1)$ we know that $\dot x = 0$ and $\dot y < 0$. Commented Apr 17, 2015 at 13:51
• so $\dot{y}<0$ and $\dot{x}=0$ but what does that mean? how can you tell the slope by these two? Commented Apr 17, 2015 at 14:15
• You normally would want to think about $\frac{\dot y}{\dot x}$ for the slope but since $\dot x = 0$ here you would just have a slope pointing straight downwards along $y = -x$ Commented Apr 17, 2015 at 14:20
• Do at point (1,-1) it goes straight downwards and then curves left then along the line of $2\choose -1$ towards the equilibrium point? Commented Apr 17, 2015 at 14:24