# In control theory, why do we linearize around the equilibrium for a nonlinear system?

For example, in these notes:

In the first example with the pendulum, they define the equilibrium as where the pendulum is at the vertical position (x=0), with a angular velocity of 0 (x'=0) and the input torque is 0 (u=0). Why do we want to study the behaviour of a system at rest? Why not linearize somewhere else. Couldn't linearization be performed at a state where the system is not at rest in case you wanted to calculate for information in such a configuration?

In general, you can linearize around any known solution. The idea is that once a solution $\theta_0(t)$ is known, nearby solutions $\theta$ approximately follow a linear equation: namely, writing $\theta (t) = \theta_0(t) + h(t)$ we get $$I\theta''+Mgl\sin\theta \approx (I\theta_0''+Mgl\sin\theta_0) + Ih''+(Mgl\cos\theta_0 )h$$ which leads to approximate linear equation $Ih''+(Mgl\cos\theta_0 )h =0$ because $(I\theta_0''+Mgl\sin\theta_0)=u$.

The catch is: do you know $\theta_0$ to begin with? An equilibrium solution is easy to find. Finding a generic solution... well, that's just the original problem.

But you will occasionally see linearization along a non-constant periodic orbit called a limit cycle or even an arbitrary trajectory. This is generally referred to as tracking-control.

• The main point being that linear systems are usually much nicer to investigate and easier to find solutions. Jun 21 '15 at 23:30
• That would be the main point if the question was "why linearize?"... but it was "why linearize around an equilibrium, rather than somewhere else?"
– user147263
Jun 21 '15 at 23:32

I believe the answer to your question is simply Hartman-Grobman Theorem. Wikipedia states the theorem wonderfully so I just quote:

Basically the theorem states that the behaviour of a dynamical system near a hyperbolic equilibrium point is qualitatively the same as the behaviour of its linearization near this equilibrium point provided that no eigenvalue of the linearization has its real part equal to 0. Therefore when dealing with such fixed points one can use the simpler linearization of the system to analyze its behaviour.

We don't have any clue if linearization around some random point will provide insights about the local behavior of the system.

Linearization around an equilibrium point (where the derivative of the full state vector is zero) tells you how the system behaves for small deviations around the point. It is easier than looking at the nonlinear system, because the 0-order term of the Taylor series is null, and the terms of order 2 and higher are dominated by the 1st-order term. Similar reasoning applies to linearization around a trajectory.

At other points, the 0-order term of the Taylor series is not zero, so the expansion doesn't give useful qualitative information, and linearization doesn't make analysis easier. Geometrically, wherever the system is not at an equilibrium the trajectories can be rectified - the Flow Box Theorem.