How To Draw Phase Diagrams Linear Algebra

Phase Portraits of Linear Systems

Consider a $2 \times 2$ linear homogeneous system ${\bf x}' = A {\bf x}$ . We think of this as describing the motion of a point 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 system. A picture of the trajectories is called a phase portrait of the organisation. In the animated version of this folio, you lot can see the moving points too as the trajectories. But on newspaper, the best we can do is to utilize arrows to indicate the management of motion.

In this section nosotros study the qualitative features of the phase portraits, obtaining a classification of the unlike possibilities that can arise. One reason that this is of import is considering, every bit we will see soon, it will exist very useful in the study of nonlinear systems. The classification will not be quite complete, because nosotros'll leave out the cases where 0 is an eigenvalue of $A$ .

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

Now we 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 $r^2 + p r + q = (r - r_1)(r - r_2)$ ), so for $p^2 - 4 q > 0$ the sign of $q$ determines whether the eigenvalues have the same sign or reverse sign. We will ignore the possibility of $q=0$ , as that would hateful 0 is an eigenvalue.

The sum of the eigenvalues is $-p$ , so if they accept the same sign this is reverse to the sign of $p$ . If the eigenvalues are circuitous, their existent function is $-p/2$ .

Another important tool for sketching the phase portrait is the post-obit: an eigenvector ${\bf u}$ for a existent 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 opposite management. If $r > 0$ the movement is outward, while if $r < 0$ it is inward. As $t \to -\infty$ (if $r > 0$ ) or $+\infty$ (if $r < 0$ ), these trajectories arroyo 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 hand, the eigenvector is not so useful.

In addition to a classification on the basis of what the curves look like, we will desire to discuss the stability of the origin as 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 take two negative eigenvalues. There are straight-line trajectories corresponding to the eigenvectors. The other trajectories are curves, which 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 direction of the ``fast'' eigenvector.
This case is called a node. It is an attractor.

Here is the moving-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 $-1$ (ho-hum) 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 two positive eigenvalues. The picture is the same every bit 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 every bit they go off to $\infty$ arroyo the direction of the ``fast'' eigenvector. This is also a node, only it is unstable. 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 $1$ (tiresome) 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 every bit what we had for the attractor node, except that the direction of time is reversed (the blitheness 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 corresponding to the eigenvectors, with the motion outwards for the positive eigenvalue and in for the negative eigenvalue. These are the only trajectories that approach the origin (in the limit as $t \to -\infty$ 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. Information technology is unstable. Note that if you lot start on the straight line in the management of the negative eigenvalue you practise arroyo the equilibrium point as $t \to \infty$ , but if you lot start off this line (even very slightly) you end upwardly 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 , nosotros accept simply one eigenvalue (a double eigenvalue). There are two cases here, depending on whether or not there are two linearly independent eigenvectors for this eigenvalue.
1. If there are 2 linearly independent eigenvectors, every nonzero vector is an eigenvector. Therefore we have directly-line trajectories in all directions. The motility is always inwards if the eigenvalue is negative (which means $p > 0$ ), or outwards if the eigenvalue is positive ( $p < 0$ ). This is chosen a singular node. It is an attractor if $p > 0$ and unstable if $p < 0$ .
  Here is the picture for the matrix $\displaystyle I = \pmatrix{1 & 0\cr 0 & 1\cr}$ , which has characteristic polynomial $r^2 - 2 r + 1$ and eigenvalue $1$ . It is unstable. For the matrix $-I$ we would have an attractor: the aforementioned movie except with time reversed.
  
  $Image snode.gif$
2. If there is but i linearly independent eigenvector, there is simply i directly line. The other trajectories are curves, which come up in to the origin tangent to the straight line trajectory and curve around to the reverse management. Trajectories on reverse sides of the direct line grade an ``S'' shape. The fashion to tell whether it is a forwards S or backwards South is to look at the management of the velocity vector ${\bf x}'$ at some point off the straight line.
  This is called a degenerate node. Over again, information technology is an attractor if $p > 0$ and unstable if $p < 0$ .
  
  Hither is the picture show 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. Annotation that the trajectories above the directly line $y=-x$ are come out of the origin heading to the left along that line, and those beneath the line come up out heading to the right. Thus the S is forwards. To check this, you could summate 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 correct, it'southward like shooting fish in a barrel to run across the S must exist forrad.
  
  $Image dnode.gif$
If $p^two - 4 q < 0$ 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 moving picture is a spiral, also known as a focus. Information technology is an attractor if $p > 0$ , as the gene $\displaystyle {\rm e}^{-pt/2}$ makes all solutions approach the origin as $t \to \infty$ , and unstable if $p < 0$ , equally in that instance the factor $\displaystyle {\rm e}^{-pt/2}$ makes all solutions (except the 1 starting at the equilibrium point itself) go off to $\infty$ equally $t \to \infty$ . We can calculate 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 $1 \pm 6 i$ . Information technology is unstable. To bank 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 $p = 0$ , 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 period $2\pi/\sqrt{q}$ . The trajectories turn out to be ellipses centred at the origin. The picture is known as a centre. Since a solution that starts nearly the origin simply goes around and around the same ellipse, never getting whatsoever closer to or further from the equilibrium than the closest and farthest points on the ellipse, this equilibrium is stable but non an attractor. Once more we tin summate a velocity vector to see whether the motion is clockwise or counterclockwise.
Here is the picture for the matrix $\displaystyle \pmatrix{2 & 5\cr -8 & -2\cr}$ , which has characteristic polynomial $r^2 + 36$ and eigenvalues $\pm 6 i$ . Again yous tin can check that the movement 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 correct.

$Image centre.gif$