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I have two systems in $\mathbb R^2$

(1) $\dot x=x^2-1$, $\dot y=y^2-1$

(2) $\dot x=y^2-1$, $\dot y=x^2-1$

How is it possible to sketch the phase portrait for each of the systems in $\mathbb R^2$ without linearizing about the fixed points?

EDIT:The phase portrait for (2) has the following form:

enter image description here

To be honest, I do not see why it has exactly this kind of form (without usng the methods of linearization). Using the same approach as Amzoti one can write down the slope as $\frac{dy}{dx}=\frac{(y+1)(y-1)}{(x+1)(x-1)}$ and get the same critical points, but why is the phase portrait completely different from the first one?

Ulrich Otto
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1 Answers1

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I will talk you through the first one because this is about as hard that one can make it.

We are given:

$$\tag 1 x'=x^2-1, y'=y^2-1$$

From $(1)$, we can determine the critical points by setting $x' =0, y'=0$, yielding the four points:

$$(x,y) = (-1,-1),(-1,1),(1,-1),(1,1)$$

We can write $(1)$ as the slope as:

$$\dfrac{y'}{x'} = \dfrac{dy}{dx} = \dfrac{y^2 - 1}{x^2 - 1} = \dfrac{(y+1)(y-1)}{(x+1)(x-1)}$$

Now we can see what happens to this slope at the critical points:

  • At $y=1$, the slope is $0$ (that is, horizontal).
  • At $y=-1$, the slope is $0$ (that is, horizontal).
  • At $x=1$, the slope is infinite (that is, vertical).
  • At $x=-1$, the slope is infinite (that is, vertical).

From the above, we are going to have five regions of interest.

We can now consider the direction and magnitude of the slopes and plot those in the $x-y$ plane for a bunch of $(x,y)$ points (a reasonable number for each region). From each of the critical points and the slopes, we can now plot points by following the slopes.

For example, look at the point $x=2$ and then $y = \{2, \ldots 10\}$. We get the following slopes:

$$\dfrac{dy}{dx} = (2,1),(2, 8/3),(2,5),(2, 8),(2, 35/3),(2, 16),(2, 21),(2, 80/3),(2, 33)$$

What do you notice about the slope for this fixed $x$ and incrementing $y$, it is a positive slope that is increasing. If you locate the $x$ and then look at the magnitude (green arrow). If you draw a solution (the blue) line in that vicinity of those points, what do you notice, it goes away from the fixed point we found at $(1,1)$. Now, repeat this for the other regions and taking all of this information, you get a sense of the system. You might also have a look at the Hamiltonian (every bit of qualitative information helps).

You could also try isoclines, but this is just a lot of work.

For the first system, try the above and see if you can reason out this phase portrait and then try your hand at the second one.

enter image description here

Amzoti
  • 56,093
  • Now that's a pretty phase portrait! +1 – amWhy Oct 28 '13 at 00:50
  • Thanks, I am not sure I understand the portrait completely. You are saying "We can now consider the direction and magnitude of the slopes and plot those in the x−y plane for a bunch of (x,y) points". How do you do that for a specific point? – Ulrich Otto Oct 28 '13 at 00:51
  • @amWhy: It is very interesting looking and has five regions to it. – Amzoti Oct 28 '13 at 00:51
  • @UlrichOtto: I added details to my answer. Was none of this covered in your class? Maybe your instructor is using a different approach, so use whatever you are comfortable with. Regards – Amzoti Oct 28 '13 at 01:05
  • Thanks, I will try the second one tomorrow and will show it to you – Ulrich Otto Oct 28 '13 at 01:10
  • Ok, I edited my starting post, may you can take a look at it, that would be great. – Ulrich Otto Oct 28 '13 at 15:55