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How can we sketch by hand the phase portrait of a system of nonlinear ODEs like the following?

$$\begin{align} \dot{x} &= 2 - 8x^2-2y^2\\ \dot{y} &= 6xy\end{align}$$

I can easily find the equilibria, which are

$$\left\{ (0, \pm 1), \left(\pm \frac{1}{2}, 0\right) \right\}$$

The corresponding stable subspace for $\left(\pm \frac{1}{2}, 0\right)$ is

$$\mbox{span} \left\{ \left(\frac{2i}{\sqrt{6}}, 1 \right), \left(-\frac{2i}{\sqrt{6}}, 1 \right) \right\}$$

and the unstable subspace for $(0, \pm 1)$ is

$$\mbox{span} \left\{ (0, 1), (1, 0) \right\}$$

respectively. But I can't see how to use these pieces of information to sketch the phase portrait. Any help would really be appreciated!

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  • $\begingroup$ It would be useful (and a time-saver) if you listed the equilibrium points and the stabilities that you found. Generally, that's where all the work comes in to solving these kinds of problems $\endgroup$ – Brenton Dec 18 '15 at 6:00
  • $\begingroup$ There are two separate phase portrait plots for each motion.. $ (x,\dot x),(y,\dot y).$ $\endgroup$ – Narasimham Dec 18 '15 at 6:09
  • $\begingroup$ @Narasimham I suspect that you have slightly different understanding of what phase portrait is... $\endgroup$ – Evgeny Dec 18 '15 at 6:10
  • $\begingroup$ @Brenton: I added the information you need. Please help with details if you can:) $\endgroup$ – user177196 Dec 18 '15 at 6:17
  • $\begingroup$ @Evgeny Could be. I took a phase portrait to mean a relation between position and velocity for each of $x,y$ component motions. $\endgroup$ – Narasimham Dec 18 '15 at 6:37
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The basic process is to find the critical points, evaluate each critical point by finding eigenvalues/eigenvectors using the Jacobian, determine and plot $x$ and $y$ nullclines, plot some direction fields and use all of this type of information to draw the phase portrait.

You can see two different views of this process at this website and notes.

For your particular problem

$$x' = 2 - 8x^2-2y^2 \\ y' = 6xy$$

We find the critical points where we simultaneously get $x' = 0, y' = 0$ so

$$(x, y) = (0, -1), (0, 1), \left(-\dfrac{1}{2}, 0\right), \left(\dfrac{1}{2}, 0\right)$$

The Jacobian is

$$J(x, y) = \begin{bmatrix}\dfrac{\partial x'}{\partial x} & \dfrac{\partial x'}{\partial y}\\\dfrac{\partial y'}{\partial x} & \dfrac{\partial y'}{\partial y}\end{bmatrix} = \begin{bmatrix}-16 x & -4y\\6y & 6x\end{bmatrix}$$

Evaluate eigenvalue/eigenvector for each critical point

$J(0, -1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\mp i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral

$J(0, 1) \implies \lambda_{1,2} = \pm 2 i \sqrt{6}, v_{1,2} = \left(\pm i \sqrt{\frac{2}{3}}, 1\right) \implies$ spiral

$J(-\frac{1}{2}, 0) \implies \lambda_{1,2} = (8, -3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle

$J(\frac{1}{2}, 0) \implies \lambda_{1,2} = (-8, 3), v_{1} = (1,0), v_2 = (0, 1) \implies$ saddle

Using all the above (critical points, eigenvalues/eigenvectors, x-nullcline (red and black curves), y-nullcline (green curve), direction fields, etc.), you can now sketch the phase portrait. Exercise - make sure to add direction fields from the two sets of notes linked above so you understand how to do that. The phase portrait will look like:

enter image description here

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  • $\begingroup$ Awesome!! Thanks so much for your help! Might you please help me with this problem? math.stackexchange.com/questions/1580742/… $\endgroup$ – user177196 Dec 18 '15 at 9:23
  • $\begingroup$ Btw, do you know where i can find how to sketch solutions to a linear system $\dot{x}=Ax$, A=3*3 matrix? $\endgroup$ – user177196 Dec 18 '15 at 9:50
  • $\begingroup$ @user177196: Cannot help with other problem. For a description of the 3D process no 9as it is the same as 2D basically), but for tools, you can look at users.dimi.uniud.it/~gianluca.gorni/Mma/Mma.html for MMA and stacking plots like mathematica.stackexchange.com/questions/687/…. Maple also has a similar command mapleprimes.com/questions/35774-Phase-Portrait-In-3D $\endgroup$ – Moo Dec 18 '15 at 12:51
  • $\begingroup$ What do you mean by no9? $\endgroup$ – user177196 Dec 18 '15 at 14:31
  • $\begingroup$ Sorry, just woke up :-), that was I do not know of any notes, but it is basically the same as 2D. These 3D variants are more done through analysis because it is typically hard to get anything 3D that is meaningful. If you have a 3D example in mind, post it as a new question and I'd give 3D a shot. $\endgroup$ – Moo Dec 18 '15 at 14:40

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