How to sketch the phase portrait near the critical point at the origin.

Question

A linear system and its general solution.

$dx/dt$ = $6x - 2y$

$dy/dt$ = $4x + 2y$

It has a general solution of this:

$$\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = A\begin{bmatrix} cos(2t) \\ cos(2t)+sin(2t) \end{bmatrix} e^{4t} + B\begin{bmatrix} sin(2t) \\ sin(2t) - cos(2t) \end{bmatrix}e^{4t}$$

Sketch the phase portrait near the critical point at the origin. Discuss the type and stability of the critical point.

I don't know how to approach this since I'm used to drawing in the 2 eigenvectors and figuring out if they face in or out against the origin. Then I would draw the orbits. How do I do this one?

Answer 1

Here is a phase portrait for this system.

• What do you notice about the direction fields?
• What are your eigenvalues telling you about stability?
• What is happening to the solution curves as $t$ is increasing?

Update

Lets find the critical point, that is, where $x'$ and $y'$ are simultaneouly equal to zero.

We have:

• $6x - 2y = 0$
• $4x + 2y = 0$

This leads to a single critical point at $(0,0)$.

Next, the eigenvalues for this system are:

$$\lambda_{1,2} = 4 \pm 2i$$

Since we have complex conjugate, with positive real part, these are unstable spirals and that is what the phase portrait is telling us. Here is a handy stabilty reference classifying eigenvalues.

So, we need to learn to do three things, find eigenvalues, find critical points and draw phase portraits.

Next, look at your $x(t)$ and $y(t)$ result. What do you notice as $t$ increases? Now, what if you plot $x(t)$ versus $y(t)$? You get the phase portrait showing that as $t$ increases, you get directions leaving the critical point off to infinity and this is unstable.