How to Draw Phase Plane Trajectory

Phase Portraits of Linear Systems

Consider a $2 \times 2$ linear homogeneous organisation ${\bf x}' = A {\bf x}$ . We think of this every bit describing the motion of a bespeak in the $xy$ aeroplane (which in this context is called the phase plane), with the independent variable $t$ as fourth dimension. The path travelled by the point in a solution is called a trajectory of the arrangement. A picture of the trajectories is called a phase portrait of the system. In the animated version of this page, you can see the moving points too as the trajectories. But on paper, the all-time we can exercise is to use arrows to indicate the direction of motion.

In this section we study the qualitative features of the phase portraits, obtaining a classification of the different possibilities that can ascend. One reason that this is important is considering, equally we will see shortly, information technology will be very useful in the written report of nonlinear systems. The classification will non be quite consummate, because we'll leave out the cases where 0 is an eigenvalue of $A$ .

The first step in the nomenclature is to find the characteristic polynomial, $\mbox{det}\,(A - r I)$ , which will exist a quadratic: nosotros write it as $r^2 + p r + q$ where $p$ and $q$ are real numbers (assuming every bit usual that our matrix $A$ has existent entries). The classification will depend mainly on $p$ and $q$ , and nosotros make a chart of the possibilities in the $pq$ plane.

Now nosotros look at the discriminant of this quadratic, $p^2 - 4 q$ . The sign of this determines what blazon of eigenvalues our matrix has:

$Image stabil.gif$

Each of these cases has subcases, depending on the signs (or in the complex case, the sign of the real role) of the eigenvalues. Notation that $q$ is the production of the eigenvalues (since $r^2 + p r + q = (r - r_1)(r - r_2)$ ), so for $p^2 - 4 q > 0$ the sign of $q$ determines whether the eigenvalues take the aforementioned sign or opposite sign. We will ignore the possibility of $q=0$ , as that would mean 0 is an eigenvalue.

The sum of the eigenvalues is $-p$ , then if they take the same sign this is reverse to the sign of $p$ . If the eigenvalues are circuitous, their real role is $-p/2$ .

Some other important tool for sketching the phase portrait is the post-obit: an eigenvector ${\bf u}$ for a real eigenvalue $r$ corresponds to a solution $\displaystyle {\bf x}= {\rm e}^{r t} {\bf u}$ that is e'er on the ray from the origin in the management of the eigenvector ${\bf u}$ . The solution $\displaystyle {\bf x}= - {\rm e}^{rt} {\bf u}$ is on the ray in the reverse management. If $r > 0$ the motion is outward, while if $r < 0$ it is inward. As $t \to -\infty$ (if $r > 0$ ) or $+\infty$ (if $r < 0$ ), these trajectories approach the origin, while as $t \to +\infty$ (if $r > 0$ ) or $-\infty$ (if $r < 0$ ) they go off to $\infty$ . For complex eigenvalues, on the other paw, the eigenvector is not then useful.

In improver to a classification on the basis of what the curves wait like, we will want to hash out the stability of the origin as an equilibrium point.

Hither, and so, is the nomenclature of the phase portraits of $2 \times 2$ linear systems.

If $p^2 - 4 q > 0$ , $q > 0$ and $p > 0$ , we take two negative eigenvalues. There are straight-line trajectories corresponding to the eigenvectors. The other trajectories are curves, which come up in to the origin tangent to the ``slow'' eigenvector (respective to the eigenvalue that is closer to 0), and as they go off to $\infty$ arroyo the direction of the ``fast'' eigenvector.
This example is called a node. Information technology is an attractor.

Here is the picture for the matrix $\displaystyle \pmatrix{-1 & -1\cr 0 & -2\cr}$ , which has feature polynomial $r^2 + 3 r + 2$ . The eigenvalues are $-1$ (deadening) and $-2$ (fast), corresponding to eigenvectors $\displaystyle \pmatrix{1\cr 0\cr}$ and $\displaystyle \pmatrix{1\cr 1\cr}$ respectively.

$Image anode.gif$
If $p^2 - 4 q > 0$ , $q > 0$ and $p < 0$ , we have ii positive eigenvalues. The picture is the aforementioned as in the previous case, except with the arrows reversed (going outward instead of inward). Again the curved trajectories come in to the origin tangent to the ``slow'' eigenvector (corresponding to the eigenvalue that is closer to 0), and as they become off to $\infty$ approach the management of the ``fast'' eigenvector. This is also a node, but information technology is unstable. Here is the picture for the matrix $\displaystyle \pmatrix{1 & 1\cr 0 & 2\cr}$ , which has characteristic polynomial $r^2 - 3 r + 2$ . The eigenvalues are $1$ (wearisome) and $2$ (fast), corresponding to eigenvectors $\displaystyle \pmatrix{1\cr 0\cr}$ and $\displaystyle \pmatrix{1\cr 1\cr}$ respectively. Note that the picture is exactly the aforementioned every bit what we had for the attractor node, except that the direction of time is reversed (the animation is run backwards).
$Image node.gif$
If $p^2 - 4 q > 0$ and $q < 0$ , we have i positive and one negative eigenvalue. Again there are straight-line trajectories respective to the eigenvectors, with the motion outwards for the positive eigenvalue and inwards for the negative eigenvalue. These are the but trajectories that arroyo the origin (in the limit equally $t \to -\infty$ for the positive and $t \to +\infty$ for the negative eigenvalue). The other trajectories are curves that come in from $\infty$ asymptotic to a straight-line trajectory for the negative eigenvalue, and go dorsum out to $\infty$ asymptotic to a straight-line trajectory for the positive eigenvalue.
This is called a saddle. It is unstable. Notation that if yous first on the straight line in the direction of the negative eigenvalue you do approach the equilibrium point as $t \to \infty$ , but if you offset off this line (fifty-fifty very slightly) you end up going off to $\infty$ .

Here is the picture for the matrix $\displaystyle \pmatrix{1 & -2\cr 0 & -1\cr}$ , which has characteristic polynomial $r^2 - 1$ . The eigenvalues are $1$ and $-1$ , respective to eigenvectors $\displaystyle \pmatrix{1\cr 0\cr}$ and $\displaystyle \pmatrix{1\cr 1\cr}$ respectively.

$Image saddle.gif$
If , we have simply one eigenvalue (a double eigenvalue). There are two cases here, depending on whether or non at that place are ii linearly independent eigenvectors for this eigenvalue.
1. If there are two linearly independent eigenvectors, every nonzero vector is an eigenvector. Therefore we take directly-line trajectories in all directions. The motion is e'er inward if the eigenvalue is negative (which means $p > 0$ ), or outwards if the eigenvalue is positive ( $p < 0$ ). This is called a singular node. It is an attractor if $p > 0$ and unstable if $p < 0$ .
  Hither is the picture for the matrix $\displaystyle I = \pmatrix{1 & 0\cr 0 & 1\cr}$ , which has feature polynomial $r^2 - 2 r + 1$ and eigenvalue $1$ . Information technology is unstable. For the matrix $-I$ nosotros would have an attractor: the same movie except with time reversed.
  
  $Image snode.gif$
2. If at that place is just one linearly independent eigenvector, there is only one straight line. The other trajectories are curves, which come in to the origin tangent to the directly line trajectory and curve around to the opposite direction. Trajectories on reverse sides of the direct line form an ``Southward'' shape. The mode to tell whether it is a forwards South or backwards S is to look at the direction of the velocity vector ${\bf x}'$ at some bespeak off the straight line.
  This is chosen a degenerate node. Again, it is an attractor if $p > 0$ and unstable if $p < 0$ .
  
  Here is the picture for the matrix $\displaystyle A = \pmatrix{2 & 1\cr -1 & 0\cr}$ , which has feature polynomial $r^2 - 2 r + 1$ , eigenvalue 1 and eigenvector $\displaystyle \pmatrix{1 \cr -1\cr}$ . It is unstable. Notation that the trajectories in a higher place the straight line $y=-x$ are come up out of the origin heading to the left along that line, and those below the line come up out heading to the right. Thus the South is forwards. To check this, you could calculate the velocity vector at, for example, $\displaystyle \pmatrix{0\cr 1\cr}$ , which is $\displaystyle A \pmatrix{0\cr 1\cr} = \pmatrix{ 1\cr 0\cr}$ . Since that points to the right, it's easy to see the Southward must exist forwards.
  
  $Image dnode.gif$
If $p^2 - 4 q < 0$ and $p \ne 0$ , nosotros have complex eigenvalues $-p/2 \pm i \mu$ . The solutions are of the class $\displaystyle {\rm e}^{-pt/2}$ times some combinations of $\sin \mu t$ and $\cos \mu t$ . The picture is a spiral, too known as a focus. It is an attractor if $p > 0$ , as the factor $\displaystyle {\rm e}^{-pt/2}$ makes all solutions arroyo the origin every bit $t \to \infty$ , and unstable if $p < 0$ , as in that case the factor $\displaystyle {\rm e}^{-pt/2}$ makes all solutions (except the one starting at the equilibrium point itself) go off to $\infty$ as $t \to \infty$ . We tin can summate a velocity vector to check if the motion is clockwise or counterclockwise.
Here is the picture for the matrix $\displaystyle \pmatrix{3 & 5\cr -8 & -1\cr}$ , which has characteristic polynomial $r^2 -2 r + 37$ and eigenvalues $1 \pm 6 i$ . It is unstable. To check that the motion is clockwise, you could note that the velocity vector at $\displaystyle \pmatrix{0\cr 1\cr}$ is $\displaystyle \pmatrix{5 \cr -1\cr}$ , which is to the right.
Finally, if $p^ii - 4 q < 0$ and $p = 0$ , nosotros have pure imaginary eigenvalues $\pm \sqrt{q} i$ . The solutions involve combinations of $\sin(\sqrt{q} t)$ and $\cos (\sqrt{q} t)$ . These are all periodic, with menstruation $2\pi/\sqrt{q}$ . The trajectories plow out to exist ellipses centred at the origin. The picture is known equally a centre. Since a solution that starts about the origin simply goes around and around the same ellipse, never getting any closer to or farther from the equilibrium than the closest and farthest points on the ellipse, this equilibrium is stable but not an attractor. Again we can summate a velocity vector to run into whether the motion is clockwise or counterclockwise.
Here is the moving picture for the matrix $\displaystyle \pmatrix{2 & 5\cr -8 & -2\cr}$ , which has feature polynomial $r^2 + 36$ and eigenvalues $\pm 6 i$ . Once again you tin check that the motion is clockwise by noting that the velocity vector at $\displaystyle \pmatrix{0\cr 1\cr}$ is $\displaystyle \pmatrix{5\cr -2\cr}$ , which is to the right.

$Image centre.gif$