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 $xy$ plane (which in this context is called the phase plane), with the independent variable $t$ as time. 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 system. In the animated version of this page, you can see the moving points as well as the trajectories. But on paper, the best we can 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. One reason that this is important is because, as we will see shortly, it will be very useful in the study of nonlinear systems. The classification will not be quite complete, because we'll leave out the cases where 0 is an eigenvalue of $A$.

The first step in the classification is to find 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 real numbers (assuming as usual that our matrix $A$ has real entries). The classification will depend mainly on $p$ and $q$, and we make a chart of the possibilities in the $pq$ plane.

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:

• If $p^2 - 4 q > 0$, there are two distinct real eigenvalues. This occurs below the parabola $p^2 - 4 q = 0$ in the $pq$ plane.
• If $p^2 - 4 q = 0$, there is one real eigenvalue (a double eigenvalue). This occurs on the parabola.
• If $p^2 - 4 q < 0$, there are two complex eigenvalues (complex conjugates of each other). This occurs in the region above the parabola.

Each of these cases has subcases, depending on the signs (or in the complex case, the sign of the real part) of the eigenvalues. Note 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 opposite sign. We will ignore the possibility of $q=0$, as that would mean 0 is an eigenvalue.

The sum of the eigenvalues is $-p$, so if they have 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 $${\bf x}= {\rm e}^{r t} {\bf u}$$ that is always on the ray from the origin in the direction of the eigenvector ${\bf u}$. The solution $${\bf x}= - {\rm e}^{rt} {\bf u}$$ is on the ray in the opposite direction. 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 hand, the eigenvector is not so useful.

In addition to a classification on the basis of what the curves look like, we will want to discuss the stability of the origin as an equilibrium point.

• An equilibrium point of a system is a point $(x,y)$ where the system says $x'$ and $y'$ are both 0. Thus if a solution starts out at an equilibrium point, it stays there for all time.
• An equilibrium point is stable if, whenever a solution starts close enough to that equilibrium point, it stays close forever (i.e. for $t > 0$).
• An equilibrium point is unstable if it is not stable. This means that it is possible for a solution to start arbitrarily close to that equilibrium point and eventually leave the neighbourhood of that point.
• An equilibrium point is an attractor if every solution that starts close enough to that equilibrium point approaches it in the limit $t \to \infty$. Every attractor is stable, but as we will see there are stable equilibrium points that are not attractors.

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 have 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 go off to $\infty$ approach the direction of the ``fast'' eigenvector.

  This case is called a node. It is an attractor.

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

  

• If $p^2 - 4 q > 0$, $q > 0$ and $p < 0$, we have two positive eigenvalues. The picture 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 $\infty$ approach the direction of the ``fast'' eigenvector. This is also a node, but it is unstable. Here is the picture for the matrix $$\pmatrix{1 & 1\cr 0 & 2\cr}$$, which has characteristic polynomial $r^2 - 3 r + 2$. The eigenvalues are $1$ (slow) and $2$ (fast), corresponding to eigenvectors $$\pmatrix{1\cr 0\cr}$$ and $$\pmatrix{1\cr 1\cr}$$ respectively. Note 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).

  

• If $p^2 - 4 q > 0$ and $q < 0$, we have one positive and one negative eigenvalue. Again there are straight-line trajectories corresponding to the eigenvectors, with the motion outwards for the positive eigenvalue and inwards 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 straight-line trajectory for the positive eigenvalue.

  This is called a saddle. It is unstable. Note that if you start on the straight line in the direction of the negative eigenvalue you do approach the equilibrium point as $t \to \infty$, but if you start off this line (even very slightly) you end up going off to $\infty$.

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

  

• If $p^2 - 4 q = 0$, we have only one eigenvalue $-p/2$ (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 two linearly independent eigenvectors, every nonzero vector is an eigenvector. Therefore we have straight-line trajectories in all directions. The motion is always inwards 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 $p < 0$.

     Here is the picture for the matrix $$I = \pmatrix{1 & 0\cr 0 & 1\cr}$$, 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 same picture except with time reversed.

     

  2. If there is only one linearly independent eigenvector, there is only one straight line. The other trajectories are curves, which come in to the origin tangent to the straight line trajectory and curve around to the opposite direction. Trajectories on opposite sides of the straight line form an ``S'' shape. The way to tell whether it is a forwards S or backwards S is to look 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 $p < 0$.

     Here is the picture for the matrix $$A = \pmatrix{2 & 1\cr -1 & 0\cr}$$, which has characteristic polynomial $r^2 - 2 r + 1$, eigenvalue 1 and eigenvector $$\pmatrix{1 \cr -1\cr}$$. It is unstable. Note that the trajectories above 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 example, $$\pmatrix{0\cr 1\cr}$$, which is $$A \pmatrix{0\cr 1\cr} = \pmatrix{ 1\cr 0\cr}$$. Since that points to the right, it's easy to see the S must be forwards.

     

• 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 $${\rm e}^{-pt/2}$$ times some combinations of $\sin \mu t$ and $\cos \mu t$. The picture is a spiral, also known as a focus. It is an attractor if $p > 0$, as the factor $${\rm e}^{-pt/2}$$ makes all solutions approach the origin as $t \to \infty$, and unstable if $p < 0$, as in that case the factor $${\rm e}^{-pt/2}$$ makes all solutions (except the one starting at the equilibrium point itself) go off to $\infty$ as $t \to \infty$. We can calculate a velocity vector to check if the motion is clockwise or counterclockwise.

  Here is the picture for the matrix $$\pmatrix{3 & 5\cr -8 & -1\cr}$$, 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 $$\pmatrix{0\cr 1\cr}$$ is $$\pmatrix{5 \cr -1\cr}$$, which is to the right.

  

• Finally, if $p^2 - 4 q < 0$ 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 near 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 not an attractor. Again we can calculate a velocity vector to see whether the motion is clockwise or counterclockwise.

  Here is the picture for the matrix $$\pmatrix{2 & 5\cr -8 & -2\cr}$$, which has characteristic polynomial $r^2 + 36$ and eigenvalues $\pm 6 i$. Again you can check that the motion is clockwise by noting that the velocity vector at $$\pmatrix{0\cr 1\cr}$$ is $$\pmatrix{5\cr -2\cr}$$, which is to the right.