📂Dynamics

Linearization of Nonlinear Systems

Buildup

Let’s say a vector field is given as a differential equation with respect to space $\left( X, \left\| \cdot \right\| \right)$ and function $f : X \to X$. $$\dot{x} = f(x)$$ When the fixed point of such an autonomous system $\overline{x}$ is given, it is essential to use a method called linearization to understand the stability around it. The idea is to analyze the system linearly around the fixed point, when viewing the system as a whole. This aligns with the idea of handling multi-dimensional nonlinear maps.

Let’s assume that a point very close to the fixed point $\overline{x}$, denoted by $x$, is represented as follows. $$x:= \overline{x} + \xi$$ Then, its Taylor expansion would be $$\dot{x} = \overline{x} ' + \xi ' = f \left( \overline{x} \right) + D f \left( \overline{x} \right) \xi + O \left( \left\| \xi \right\|^{2} \right)$$ Of course, since $\overline{x} ' = f \left( \overline{x} \right)$, by eliminating the second and third terms and rearranging for $\xi '$ $$\xi ' = D f \left( \overline{x} \right) \xi + O \left( \left\| \xi \right\|^{2} \right)$$ $D$ is the Jacobian. Here, if $\left\| \xi \right\|^{2}$ is sufficiently small, i.e., if $x$ is very close to $\overline{x}$, the last term can be ignored.

Definition

This process is called linearization, and the new differential equation obtained in this manner is referred to as the associated linear system. $$\xi ' = D f \left( \overline{x} \right) \xi$$

Explanation

Limitations of Linearization

It is important to note that the Lyapunov stability of a fixed point in the linearized system does not guarantee stability in the original system. For illustration, let’s examine the linearization of the following nonlinear system: $$\begin{align*} \dot{x} =& f_{1} (x , y) = -y + x \left( x^{2} + y^{2} \right) \\ \dot{y} =& f_{2} (x , y ) = x + y \left( x^{2} + y^{2} \right) \end{align*}$$ Obviously, the origin $\overline{(x,y)} = (0,0)$ becomes a fixed point of the above system.

---

Part 1.

For the linear system, the associated linear system $\xi ' = D f \left( \overline{x} \right) \xi$ is derived as follows. $$\begin{align*} \begin{bmatrix} \xi_{1} \\ \xi_{2} \end{bmatrix} ' =& \begin{bmatrix} {{ \partial f_{1} } \over { \partial x }} & {{ \partial f_{1} } \over { \partial y }} \\ {{ \partial f_{2} } \over { \partial x }} & {{ \partial f_{2} } \over { \partial y }} \end{bmatrix}_{\overline{(x,y)} = (0,0)} \begin{bmatrix} \xi_{1} \\ \xi_{2} \end{bmatrix} \\ =& \begin{bmatrix} 3 x^{2} + y^{2} & -1 + 2 x y \\ 1 + 2x y & x^{2} + 3 y^{2} \end{bmatrix}_{ \overline{(x,y)} = (0,0)} \begin{bmatrix} \xi_{1} \\ \xi_{2} \end{bmatrix} \\ =& \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \xi_{1} \\ \xi_{2} \end{bmatrix} \end{align*}$$ If we tidy up $\xi ' = D f \left( \overline{x} \right) \xi$, it would look like this. $$\begin{align*} \xi_{1} ' =& - \xi_{2} \\ \xi_{2} ' =& \xi_{1} \end{align*}$$

---

Part 2. Stability in the Linear System

Let’s now check the Lyapunov stability of $\overline{x}$ in the linearized system. It’s apparent that all solutions of this system are quite simple; considering $D f \left( \overline{x} \right)$ as a linear transformation, because of the matrix’s rotation transformation nature, all points rotate in a counterclockwise direction around the origin, maintaining a constant distance $r>0$.

Definition of Lyapunov Stability: Let’s denote $t_{0} \in \mathbb{R}$. If for every $$\left\| \overline{x} \left( t_{0} \right) - y \left( t_{0} \right) \right\| < \delta \implies \left\| \overline{x}(t) - y(t) \right\| < \varepsilon \qquad , t > t_{0}$$ other solution $y(t)$ satisfies $\delta ( \varepsilon ) > 0$, then $\overline{x}(t)$ is said to be stable.

According to the definition, for the fixed point $\overline{(x,y)} = (0,0)$ to be stable, $$\left\| \xi (t_{0}) \right\| < \delta \implies \left\| \xi (t) \right\| < \varepsilon \qquad , t > t_{0}$$ must exist whenever $\varepsilon > 0$ is given. Since all other solutions of this system rotate around the origin, maintaining a constant distance irrespective of time, $\delta (\varepsilon) = \varepsilon$ exists trivially. Therefore, we can say that $\overline{(x,y)} = (0,0)$ is stable in this linear system.

---

Part 3. Stability in Nonlinear Systems

Although not a typical method for verifying stability in nonlinear systems, in this case, it’s relatively easy to check stability via polar coordinate transformation. Consider representing $x, y$ in polar coordinates as follows.

$$x = r \cos \theta \\ y = r \sin \theta$$ Then, since $r^{2} = x^{2} + y^{2}$, differentiating both sides with respect to time $t$ gives $$\begin{align*} & 2r \dot{r} = 2 x \dot{x} + 2y \dot{y} \\ \implies& r\dot{r} = x \left[ -y + x \left( x^{2} + y^{2} \right) \right] + y \left[ x + y \left( x^{2} + y^{2} \right) \right] \\ \implies& r \dot{r} = r \cos \theta \left[ - r \sin \theta + r \cos \theta \cdot r^{2} \right] + r \sin \theta \left[r \cos \theta + r \sin \theta \cdot r^{2} \right] \\ \implies& r \dot{r} = r^{4} \cos^{2} \theta + r^{4} \sin^{2} \theta \\ \implies& \dot{r} = r^{3} \end{align*}$$ Similarly, differentiating the polar coordinate transformation $x=r \cos \theta$ with respect to time $t$ yields $$\begin{align*} & \dot{x} = \dot{r} \cos \theta - r \theta^{\prime}\sin \theta \\ \implies& - r \sin \theta + r^{3} \cos \theta = r^{3} \cos \theta - \left( r \sin \theta \right) \dot{\theta} \\ \implies& 1 = \dot{\theta} \end{align*}$$ Organizing this, $$\begin{align*} \dot{r} =& r^{3} \\ \dot{\theta} =& 1 \end{align*}$$ According to the polar coordinate representation, except for the fixed point $\overline{(x,y)} = (0,0)$, $r>0$ is maintained, so $\dot{r}>0$ as well. This means all solutions except the fixed point rotate around the origin by a constant $\theta$, drawing increasingly larger spirals over time. Thus, even if very close to the fixed point, they eventually move away from the origin, making this nonlinear system unstable.

Asymptotic Stability

The example above shows that stability in a linear system does not guarantee stability in a nonlinear system. [NOTE: This is also referred to as linear stability in some cases.] However, for hyperbolic fixed points, this is equivalent, making linearization still a powerful tool.

Associated Linearization System and Asymptotic Stability¹: $\dot{x} = f(x)$ 의 고정점 $\overline{x}$ 에 대해 $D f (\overline{x})$ 의 모든 고유값들의 실수부가 음수면 $\overline{x}$ 는 점근적 안정성을 가진다.

This theorem is widely known and used in most analyses of dynamical systems. Its proof requires the existence of a Lyapunov function² and algebraic tricks³, which are not necessarily straightforward to demonstrate directly.