Table of contents

Linear Systems - Part B

1. Analyze a general two dimensional system according to the parameter

Let us summarize all the cases we've discussed:
  1. Two real eigenvalues \lambda_1, \lambda_2:
    • If \lambda_1 \ne \lambda_2, there are always two independent eigenvectors. The system can be a stable or unstable node, saddle, or degnerate if one of the eigenvalue is 0.
    • If \lambda_1 = \lambda_2 and there are two eigenvectors, the system is a stable or unstable star.
    • If \lambda_1 = \lambda_2 but there is only one eigenvector, the system is a degenerate node.
  2. Two conjugate complex eigenvalues \alpha \pm \beta i.
    • If \alpha = 0, the system is an elliptic centre.
    • If \alpha \ne 0, the system is a stable or unstable focus.

1.1. An example

Example 1.1.
Analyze the two-dimensional system
\dot{R} = - a R + b J, \quad \dot{J} = b R - a J,
describe the possible phase portraits according to the parameter values a, b > 0.
Example 1.2.
Analyze the two-dimensional system
\dot{R} = - a R + b J, \quad \dot{J} = - b R + a J,
describe the possible phase portraits according to the parameter values a, b > 0.

1.2. A simple two parameter bifurcation diagram

A bifurcation diagram is a picture that list all the different type of phase portrait that can appear with respect to values of the parameter. For example, for the system
\dot{x}_1 = a x, \quad \dot{x}_2 = dx,
the bifurcation diagram looks like:
simple-bd.svg

Figure 1. Bifurcation diagram for separable systems

We can learn a lot from this diagram.
  • The "nondegenerate" types correspond to regions in the diagram, indicating that they are open conditions. If the parameter value change slightly, the phase portrait type does not change.
  • The "degenerate" types correspond to lines or curves in the diagram, often as borders of two regions. A small change will move it into one of the nondegenerate types.
  • The diagram also list the relation between different types. As the parameter value changes, the type of the phase portrait may change from one to another, often going through a degenerate type in the middle. This is the essence of bifurcation analysis.

1.3. A bifurcation diagram for the general two-dimensional system

For the general two dimensional linear system
\bmat{\dot{x}_1 \\ \dot{x}_2} = \bmat{a & b \\ c & d} \bmat{x_1 \\ x_2},
there are four parameters. Given that it's very hard to understant 4-dimensional graphs, we need to lower the dimension. Note that the characteristic polynomial
\det (A - \lambda I) = \lambda^2 - (a + d) \lambda + (ad - bc),
at least the eigenvalues depends only on two values:
\tau = a + d \quad \text{ (the trace)},
and
\Delta = ad - bc \quad \text{ (the determinant)}.
The discriminant of the characteristic polynomial is
\tau^2 - 4 \Delta.
Moreover, according to Vieta's formula,
\lambda_1 + \lambda_2 = \tau, \quad \lambda_1 \lambda_2 = \Delta.
We can use information on \tau and \Delta to determine the type of the phase portrait.
  • \tau^2 - 4\Delta > 0. There are two distince real eigenvalues \lambda_1, \lambda_2.
    • If \Delta = \lambda_1 \lambda_2 < 0 then \lambda_1, \lambda_2 have different signs. The phase portrait is a saddle.
    • If \Delta = \lambda_1 \lambda_2 > 0 and \tau = \lambda_1 + \lambda_2 > 0, then \lambda_1, \lambda_2 have the same sign and are positive. This is an unstable node.
    • If \Delta = \lambda_1 \lambda_2 > 0 and \tau = \lambda_1 + \lambda_2 < 0, then \lambda_1, \lambda_2 have the same sign and are negative. This is a stable node.
      Note that \tau cannot be 0 if \Delta = \lambda_1 \lambda_2 > 0.
    • If \Delta = \lambda_1 \lambda_2 = 0, then one of the eigenvalue is zero.
  • If \tau^2 - 4\Delta < 0, then \lambda_1, \lambda_2 = \alpha \pm i \beta. In this case we note that \tau = \lambda_1 + \lambda_2 = 2\alpha.
    • If \tau > 0, this is a unstable focus.
    • If \tau < 0, this is a stable focus.
    • If \tau = 0, this is an elliptic centre.
  • If \tau^2 - 4\Delta = 0, then \lambda_1 = \lambda_2. There are two possibilities that cannot be distinguished from \tau and \Delta alone.
    • If there are two independent eigenvectors, we have a stable or unstable star.
    • If there is only one independent eigenvector, we have a stable or unstable node.
In the following diagram, we list all the regions in a bifurcation diagram. The degenerate types appear at the border between different regions.
all-linear-bd.svg

Figure 2. A bifurcation diagram using \tau and \Delta, regions only.

2. Some notes about higher dimensional linear systems

Let A be an n\times n matrix, we briefly discuss the linear system
\dot{\bx} = A \bx.
We know that the characteristic equation
\det(A - \lambda I) = 0
is a nth order polynomial equation, and therefore always has n solutions in \C counting multiplicities. However, when there are repeated eigenvalues, there may not exists n independent eigenvectors. Here we will only discuss the case where all eigenvalues are distinct. In this case, we can always find n independent eigenvectors
A \bv_1 = \lambda_1 \bv_1, \cdots, A \bv_n = \lambda_n \bv_n.
When an eigenvalue / eigenvector pair (\lambda, \bv) is real, then
\bx(t) = C e^{\lambda t} \bv
is a solution. In this case, the solution either grow or shrink exponentially along the direction \bv.
If we have a complex eigenvalue \lambda = \alpha + i\beta and eigenvector \bv + i \bw, then we have two parameter family of solutions
\bx(t) = e^{\alpha t} \left( (C_1 \cos(\beta t) - C_2 \sin (\alpha t)) \bv + (C_1 \sin(\beta t) + C_2 \cos(\beta t)) \bw \right).
We have a spiral solution in the plane formed by \bv and \bw: if \alpha > 0, it is an unstable spiral; if \alpha < 0, a stable spiral; if \alpha = 0, an elliptic center.

2.1. 3 dimensional systems with distinct real or complex eigenvalues, without zero eigenvalues

We discuss all the possibility that can happen in a 3 dimensional linear system.
3 real eigenvalues:
  • \lambda_1 < \lambda_2 < \lambda_3 < 0.
    \bx(t) = C_1 e^{\lambda_1 t} \bv_1 + C_2 e^{\lambda_2 t} \bv_2 + C_3 e^{\lambda_3 t} \bv_3.
    All solutions converges to 0 in positive time. Furthermore,
    \bx(t) = e^{\lambda_3 t} \left( C_1 e^{(\lambda_1 - \lambda_3) t} \bv_1 + C_2 e^{(\lambda_2 - \lambda_3) t} \bv_2 + C_3 \bv_3 \right).
    We see that all solutions will turn toward \bv_3 as t \to \infty.
    This is the 3 dimension version of the stable node.
  • 3 negative eigenvalues. This is the unstable node, the picture is the time-reversal of the stable node.
  • \lambda_1 < 0 < \lambda_2 < \lambda_3. This is the 3 dimensional version of the saddle. There is a two-dimensional expanding, and 1 dimensional contracting direction.
  • \lambda_1 < \lambda_2 < 0 < \lambda_3. Similar to above, but with 1d expanding and 2d contracting.
1 real and a conjugate pair of complex eigenvalues. We have \lambda_1 \in \R, and \lambda_2, \lambda_3 = \alpha \pm i \beta.
  • \lambda_1 > 0, \alpha > 0. Unstable focus.
  • \lambda_1 < 0, \alpha < 0. Stable focus.
  • \lambda_1 < 0, \alpha > 0. Spiraling saddle.
Example 2.1.
Convert the 3rd order equation
x'''(t) + x''(t) + x'(t) + x(t) = 0
to a system, then sketch its phase portrait.