How to Draw Phase Plane Using Eigenvectors

Phase Portraits of Linear Systems

Consider a $2 \times 2$ linear homogeneous system ${\bf x}' = A {\bf x}$ . Nosotros recall of this equally describing the movement of a point in the $xy$ aeroplane (which in this context is called the phase plane), with the contained variable $t$ as time. The path travelled past the point in a solution is called a trajectory of the organization. A picture of the trajectories is called a phase portrait of the system. In the blithe version of this page, you can see the moving points as well equally the trajectories. But on paper, the best nosotros tin do 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 arise. 1 reason that this is important is considering, as nosotros volition see presently, it will be very useful in the study of nonlinear systems. The classification will not exist quite consummate, considering nosotros'll leave out the cases where 0 is an eigenvalue of $A$ .

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

At present nosotros look at the discriminant of this quadratic, $p^2 - 4 q$ . The sign of this determines what type 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 part) of the eigenvalues. Notation that $q$ is the product of the eigenvalues (since $r^2 + p r + q = (r - r_1)(r - r_2)$ ), and so for $p^2 - 4 q > 0$ the sign of $q$ determines whether the eigenvalues have the same sign or opposite sign. We will ignore the possibility of $q=0$ , every bit that would hateful 0 is an eigenvalue.

The sum of the eigenvalues is $-p$ , and then if they accept the same sign this is opposite to the sign of $p$ . If the eigenvalues are complex, their real part is $-p/2$ .

Another important tool for sketching the phase portrait is the following: an eigenvector ${\bf u}$ for a real eigenvalue $r$ corresponds to a solution $\displaystyle {\bf x}= {\rm e}^{r t} {\bf u}$ that is always on the ray from the origin in the direction of the eigenvector ${\bf u}$ . The solution $\displaystyle {\bf x}= - {\rm e}^{rt} {\bf u}$ is on the ray in the reverse direction. If $r > 0$ the move is outward, while if $r < 0$ it is inward. Equally $t \to -\infty$ (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 circuitous eigenvalues, on the other hand, the eigenvector is not so useful.

In addition to a nomenclature on the basis of what the curves expect like, we will want to discuss the stability of the origin every bit an equilibrium point.

Here, then, is the classification of the phase portraits of $2 \times 2$ linear systems.

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

Here is the motion-picture show for the matrix $\displaystyle \pmatrix{-1 & -1\cr 0 & -2\cr}$ , which has characteristic polynomial $r^2 + 3 r + 2$ . The eigenvalues are $-1$ (tedious) and $-2$ (fast), respective 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 two positive eigenvalues. The motion-picture show is the same 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 go off to $\infty$ arroyo the direction of the ``fast'' eigenvector. This is also a node, but information technology is unstable. Hither 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 $1$ (irksome) and $2$ (fast), corresponding to eigenvectors $\displaystyle \pmatrix{1\cr 0\cr}$ and $\displaystyle \pmatrix{1\cr 1\cr}$ respectively. Notation that the picture is exactly the same as 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 one positive and one negative eigenvalue. Again there are straight-line trajectories respective to the eigenvectors, with the motion outwards for the positive eigenvalue and inward for the negative eigenvalue. These are the but trajectories that approach the origin (in the limit every bit $t \to -\infty$ 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 back out to $\infty$ asymptotic to a directly-line trajectory for the positive eigenvalue.
This is called a saddle. It is unstable. Annotation that if you outset on the straight line in the direction of the negative eigenvalue you do approach the equilibrium bespeak as $t \to \infty$ , but if you lot starting time off this line (fifty-fifty very slightly) you finish 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 $-1$ , corresponding to eigenvectors $\displaystyle \pmatrix{1\cr 0\cr}$ and $\displaystyle \pmatrix{1\cr 1\cr}$ respectively.

$Image saddle.gif$
If , we accept only 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 ii linearly independent eigenvectors, every nonzero vector is an eigenvector. Therefore we have direct-line trajectories in all directions. The move is e'er inward if the eigenvalue is negative (which ways $p > 0$ ), or outwards if the eigenvalue is positive ( $p < 0$ ). This is called a atypical node. Information technology is an attractor if $p > 0$ and unstable if $p < 0$ .
  Here is the moving picture for the matrix $\displaystyle I = \pmatrix{1 & 0\cr 0 & 1\cr}$ , which has feature polynomial $r^2 - 2 r + 1$ and eigenvalue $1$ . It is unstable. For the matrix $-I$ we would have an attractor: the aforementioned picture except with time reversed.
  
  $Image snode.gif$
2. If there is only 1 linearly contained eigenvector, in that location is merely i directly line. The other trajectories are curves, which come up in to the origin tangent to the directly line trajectory and curve around to the opposite direction. Trajectories on opposite sides of the straight line class an ``S'' shape. The way to tell whether it is a forwards S or backwards South is to wait at the direction of the velocity vector ${\bf x}'$ at some point off the straight line.
  This is called a degenerate node. Again, it is an attractor if $p > 0$ and unstable if $p < 0$ .
  
  Hither is the picture for the matrix $\displaystyle A = \pmatrix{2 & 1\cr -1 & 0\cr}$ , which has characteristic polynomial $r^2 - 2 r + 1$ , eigenvalue 1 and eigenvector $\displaystyle \pmatrix{1 \cr -1\cr}$ . It is unstable. Note that the trajectories in a higher place the straight line $y=-x$ are come out of the origin heading to the left along that line, and those below the line come out heading to the right. Thus the S is forwards. To check this, you could calculate the velocity vector at, for instance, $\displaystyle \pmatrix{0\cr 1\cr}$ , which is $\displaystyle A \pmatrix{0\cr 1\cr} = \pmatrix{ 1\cr 0\cr}$ . Since that points to the correct, it's piece of cake to run across the S must be forwards.
  
  $Image dnode.gif$
If $p^2 - 4 q < 0$ and $p \ne 0$ , we have complex eigenvalues $-p/2 \pm i \mu$ . The solutions are of the form $\displaystyle {\rm e}^{-pt/2}$ times some combinations of $\sin \mu t$ and $\cos \mu t$ . The picture is a screw, also known every bit a focus. Information technology is an attractor if $p > 0$ , every bit the factor $\displaystyle {\rm e}^{-pt/2}$ makes all solutions approach the origin as $t \to \infty$ , and unstable if $p < 0$ , as in that instance the gene $\displaystyle {\rm e}^{-pt/2}$ makes all solutions (except the one starting at the equilibrium signal itself) go off to $\infty$ as $t \to \infty$ . Nosotros can calculate a velocity vector to check if the motion is clockwise or counterclockwise.
Here is the picture show for the matrix $\displaystyle \pmatrix{3 & 5\cr -8 & -1\cr}$ , which has feature polynomial $r^2 -2 r + 37$ and eigenvalues $1 \pm 6 i$ . It is unstable. To check that the motion is clockwise, yous could notation 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 $p^two - 4 q < 0$ and $p = 0$ , we 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 menstruum $2\pi/\sqrt{q}$ . The trajectories plow out to be ellipses centred at the origin. The pic is known every bit a eye. Since a solution that starts nigh the origin just 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 non an attractor. Once again nosotros can calculate a velocity vector to see whether the movement 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 $\pm 6 i$ . Again you tin can bank 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.

$Image centre.gif$