Jacobian linearization of trigonometric functions

Question

If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$: $$\dot{x_1} = x_2$$ $$\dot{x_2} = \cos(x_1)$$

I can apply the small angle approximation to find the matrix A: $$ A = \pmatrix{0&1\\1&0} $$

However, if I apply Jacobian linearization and take the partial derivatives of $\dot{x}$, I get: $$ A = \pmatrix{0&1\\-\sin(x_1)&0}$$

Evaluated at the origin, I get: $$ A = \pmatrix{0&1\\0&0} $$

Seeing as how the two different linearization techniques yield different results, is one more valid than the other?

Answer 1

Your small angle approximation is wrong and your Jacobian linearization is incomplete.

If $x_1 \approx 0$ then $\cos(x_1) \approx 1$ and so the system

$$ \begin{align} \dot{x_1} &= x_2 \\ \dot{x_2} &= \cos(x_1) \\ \end{align} $$

behaves close to $(0,0)$ approximately like

$$ \begin{align} \dot{x_1} &= x_2 \\ \dot{x_2} &= 1 \\ \end{align} $$

If you put this in matrix form:

$$ \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

This is the "small angle approximation" and as you can see this is essentially the same as the Jacobian linearization. Remember that the Jacobian linearization of a nonlinear system $\dot{x} = f(x)$ at $x_0$ is

$$ \dot{x} = f(x_0) + A(x - x_0) $$

where $A$ is the Jacobian matrix of $f$ at $x_0$. If you insert your system and your point you get:

$$ \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -\sin(0) & 0 \end{pmatrix} \begin{pmatrix} x_1 - 0 \\ x_2 - 0 \end{pmatrix} + \begin{pmatrix} 0 \\ \cos(0) \end{pmatrix} $$

which is just:

$$ \begin{pmatrix} \dot{x}_1 \\ \dot{x}_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

so the same as the small angle approximation.

Your mistake was that you neglected the fact that you linearized at a non-equilibrium point. Usually we linearize at an equilibrium and then $f(x_0) = 0$, so the constant term can be neglected.

But you linearize at $(0, 0)$ which is not an equilibrium point of your system because $\dot{x_2} \neq 0$ at $x_1 = 0$, so the system is not at rest at the origin.

Answer 2

"Approximation" in our case means Taylor's theorem, i.e. $$ f(x) = f(0) + Df(0)x $$ where $f(x) := (x_2, \cos(x_1))$. This is where $Df(0)$, i.e. your second matrix appears. So this is a linearization in the normal sense. This type of linearization is actually useful when we want to determine stability of an equilibrium point.

The first approach I would call incoherent, especially since you leave $x_2$ untouched. This is not how linearization in analysis is understood. Linear maps have the property that they vanish in $0$. So what we do, is to always approximate $$ x \mapsto f(x) - f(0) $$ by a linear map since the former vanishes in $0$, too. This is the idea behind Taylor's theorem. We just accept, that this is actually an affine approximation of $f(x)$ since it simply has no major ramifications in practice.

But what you did, is to find a linear map that approximates $f(x)$ only. But since $f(0) \neq 0$ and $A0 = 0$, this might yield huge errors. To my knowledge (which is of course limited), the first one has no application whatsoever, but I would love to be corrected if I were wrong.