Using nullclines to plot phase portraits

[Aero6]

I'm plotting phase portraits and have a question about determining the direction of nullclines.

The systematic method that I'm using to plot phase portraits is:

1) find the nullclines
2) determine the direction of the arrows on the nullclines
3) find the eigenvalues
4)find the eigen vectors
5)plot the eigen vectors on phase portrait
6) determine the direction of the arrows on the eigen vectors according
7) draw the solutions to the equations by following the arrows on your nullclines and your eigenvectors

The two equations I have are:

Xsub1 = x subscript 1
Xsub2 = x subscript 2

Equation A) dXsub1\dt = 3Xsub1 - 2(Xsub2)

Equation B) dXsub2\dt = 2Xsub1 - Xsub2


when finding the nullclines, I set each derivative equation equal to 0.

for equation:

Substitution A) (2)(Xsub1) = Xsub2

Substitution B) (3\2)(Xsub1) = Xsub2


After plotting these two nullclines in my Xsub1 , Xsub2 plane, I tried to determine the direction of the arrows that you put on each nullcline (these arrows are what I follow when trying to plot my solution)


NOTE: Xsub2 corresponds to the vertical axis on my graph and Xsub1 corresponds to the horizontal axis on my graph.

Since the origin on my graph is my equilibrium solution, I looked to the right of the nullcline for dXsub1\dt. Since we are looking at dXsub1\dt =0 (this is the nullcline I plotted), Xsub1 is not changing so I can only move vertically along the nullcline for dXsub1\dt.


To determine the direction, I first substituted 'Substitution A' into 'Equation B' and solved for Xsub1 and got :

dXsub2\dt = -Xsub1 negative values of Xsub1 would mean my arrows along this
nullcline point to the right, correct?

positive values of Xsub1 would mean my arrows along this
nullcline would point to the left because my overall
dXsub2\dt would be negative, correct?

Then I substituted 'Substitution B' into 'Equation A' and solved for Xsub1 and got:

dXsub1\dt = (.5)(Xsub1) negative values for Xsub1 mean my arrows point to down
along this nullcline, because a negative value for Xsub1 would
make dXsub1dt negative, correct?

positive values for Xsub1 means that my arrows would point up
along this nullcline because this would make dXsub1\dt positive?


When I looked at this example and another from class, the arrows go the direction opposite of how I've written them here and that's why I'm confused about how you determine the direction of the arrows on your nullclines.


Thank you

 

[jvicens]

for sharing your systematic method for plotting phase portraits. It seems like you have a good understanding of the process. However, I noticed some confusion in determining the direction of the arrows on the nullclines. Let me explain it in a bit more detail.

Firstly, it is important to understand that the nullclines are the lines on which the derivatives are equal to zero. Therefore, when you plot the nullclines, you are essentially plotting the points where the system is at equilibrium. This means that the direction of the arrows on the nullclines is determined by the direction in which the system will move away from the equilibrium point.

Now, let's look at your example equations. For equation A, you correctly found that the nullcline is given by 2Xsub1 = Xsub2. This means that when Xsub2 is equal to 2Xsub1, the system is at equilibrium. To determine the direction of the arrows on this nullcline, you need to look at the derivative dXsub2\dt. You correctly substituted Xsub2 = 2Xsub1 and found that dXsub2\dt = -Xsub1. This means that when Xsub1 is positive, the system will move downwards (towards the equilibrium point) and when Xsub1 is negative, the system will move upwards (away from the equilibrium point). Therefore, the arrows on this nullcline will point downwards when Xsub1 is positive and upwards when Xsub1 is negative.

Similarly, for equation B, the nullcline is given by 3\2Xsub1 = Xsub2. Again, when Xsub2 is equal to 3\2Xsub1, the system is at equilibrium. To determine the direction of the arrows on this nullcline, you need to look at the derivative dXsub1\dt. You correctly substituted Xsub2 = 3\2Xsub1 and found that dXsub1\dt = 0.5Xsub1. This means that when Xsub1 is positive, the system will move to the right (away from the equilibrium point) and when Xsub1 is negative, the system will move to the left (towards the equilibrium point). Therefore, the arrows on this nullcline will point to the right when Xsub1 is positive and to the left when Xsub1 is negative.

I hope this clarifies the direction of the arrows on the null