Recall that only the solutions of linear systems may be found explicitly. The problem is that in general real life problems may only be modeled by nonlinear systems. In this case, we only know how to describe the solutions globally (via nullclines). What happens around an equilibrium point remains a mystery so far. Here we propose the to discuss this problem. The main idea is to approximate a nonlinear system by a linear one (around the equilibrium point). Of course, we do hope that the behavior of the solutions of the linear system will be the same as the nonlinear one. This is the case most of the time (not all the time!).
Example. Consider the Van der Pol equation
This is a nonlinear equation. Let us translate this equation into a system. Set . Then we have
The equilibrium points reduce to the only point (0,0). Let us find the nullclines and the direction of the velocity vectors along them.
- The x-nullcline is given by
Hence the x-nullcline is the x-axis.
- The y-nullcline is given by
Hence the y-nullcline is the curve .
which happens to be a linear system. The eigenvalues of this system are . Hence the solutions of the linear system spiral away from the origin (since the real part is positive). So we suggest that the solutions of nonlinear system spiral away from the origin (look at the picture below)
Consider the autonomous system
And assume that is an equilibrium point. So we would like to find the closest linear system when (x,y) is close to . In order to do that we need to approximate the functions f(x,y) and g(x,y) when (x,y) is close to . This is a similar problem to approximating a real valued function by its tangent (around a point of course). From multivariable calculus, we get
when (x,y) is close to . Then the nonlinear system may be approximated by the system
But since is an equilibrium point, then we have . Hence we have
This is a linear system. Its coefficient matrix is
This matrix is called the Jacobian matrix of the system at the point .
Summary of the linearization technique.
Consider the autonomous system
and an equilibrium point.
- Find the partial derivatives
Write down the Jacobian matrix
- Find the eigenvalues of the Jacobian matrix.
- Deduce the fate of the solutions around the equilibrium point from the eigenvalues. For example,
- if the eigenvalues are negative or complex with negative real part, then the equilibrium point is a sink (that is all the solutions will dye at the equilibrium point). Note that if the eigenvalues are complex, then the solutions will spiral around the equilibrium point.
- If the eigenvalues are positive or complex with positive real part, then the equilibrium point is a source (that is all the solutions will move away from the equilibrium point). Note that if the eigenvalues are complex, then the solutions will spiral away from the equilibrium point.
- If the eigenvalues are real number with different sign (one positive and one negative), then the the equilibrium point is a saddle. In fact, there will be two solutions which approach the equilibrium point as , and two more solutions which approach the equilibrium point as . For the linear system theses solutions are lines, but for the nonlinear system they are not in general. These four solutions are called separatrix.
Example. Consider the equation of the pendulum
where is the damping coefficient. See the picture below.
The equilibrium points are , where . The angles , for , correspond to the pendulum at its lowest position, while , for , correspond to the pendulum at its highest position. The Jacobian matrix of the system
Let us concentrate on the equilibrium positions (0,0) and .
- For (0,0), the Jacobian matrix is
For the sake of illustration let us fix the parameters. For example,
- if we take (undamped pendulum), then the eigenvalues are which implies that the mass will oscillate around the lowest position in a periodic fashion.
- If (dumped pendulum), m=1, and l=1. Then the eigenvalues are
Since the real part is negative, the solutions will sink (dye) while oscillating around the equilibrium point. Here we have the same behavior for the linear and nonlinear system.
- For , the Jacobian matrix is
The eigenvalues are
Clearly we have two real eigenvalues with one positive and one negative. So the solutions will always get away from the equilibrium position except along one curve (the separatrix).