Linearization.md

Linearization at a singular point

General theory

Given an ODE $\dot x=v(x)$, a point $a$ is called a singular point (a.k.a. zero, a.k.a. equilibrium state, a.k.a. critical point) of this ODE if $v(a)=0$. In order to study the behaviour of the solutions of $\dot x=v(x)$ near $x=a$, we can replace $v(x)$ with its linear part at $a$. More precisely, let $J$ be the jacobian matrix of $v$ at $a$: $$ J=\left[\frac{\partial v_i}{\partial x_j}(a)\right] $$

Then for $x\approx a$ we have $v(x)\approx J(x-a)$, so we can study $\dot y=Jy$ instead of $\dot x=v(x)$, where $y=x-a$. If the real parts of all eigenvalues of $J$ are nonzero, then the phase portraits of the linearized system near $y=0$ is “qualitatively the same” as the phase portrait of the original system near $x=a$.

Example 1: pendulum

Consider a pendulum (a massive bob on a weightless rod), see a diagram on Wikipedia. Its motion is given by $\ddot x=-\sin x$.

Reformulating as a first order ODE

Let us introduce $y=\dot x$, then we have \begin{align} \dot x&=y;\ \dot y&=-\sin x. \end{align}

Singular points

• We need to solve $y=0$, $-\sin x=0$.
• The solutions are $(x, y)=(\pi k, 0)$, $k\in\mathbb Z$.
• All the solutions $(2\pi k, 0)$ correspond to the bottom point of the circle, and all the solutions $((2k+1)\pi, 0)$ correspond to the top point of the circle.

Linearization at $(0, 0)$

The jacobian matrix is given by $$ J=\left.\begin{bmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\ \frac{\partial (-\sin x)}{\partial x} & \frac{\partial (-\sin x)}{\partial y} \end{bmatrix}\right|{x=0, y=0} = \left.\begin{bmatrix} 0 & 1\ -\cos x & 0 \end{bmatrix}\right|{x=0, y=0} = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}. $$

The eigenvalues are $\pm i$, their real parts are zeros, so we can't be sure yet that the original nonlinear system has qualitatively the same phase portrait (a center) at $(0, 0)$ as it linearization. Later we will discuss how to deal with this (spoiler: this nonlinear system has a center at $(0, 0)$).

Linearization at $(\pi, 0)$

The jacobian matrix is given by $$ J=\left.\begin{bmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\ \frac{\partial (-\sin x)}{\partial x} & \frac{\partial (-\sin x)}{\partial y} \end{bmatrix}\right|{x=\pi, y=0} = \left.\begin{bmatrix} 0 & 1\ -\cos x & 0 \end{bmatrix}\right|{x=\pi, y=0} = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}. $$

The eigenvalues are $\pm 1$, their real parts are nonzero, so and we know for sure that the phase portrait of the original nonlinear system near $(\pi, 0)$ is qualitatively the same as the phase portrait of the linearization, i.e., it has a saddle point with separatrices tangent to $x-\pi=\pm y$.

Linearizations at other points

Since $\sin$ is a $2\pi$-periodic function, the linearizations at $(2\pi k, 0)$ will be centers, and linearizations at $(2\pi k+\pi, 0)$ will be saddles.

import numpy as np
import matplotlib.pyplot as plt
Y, X = np.mgrid[-3:3:1000j, -4:5:1000j]
U, V = Y, -np.sin(X)
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 4))
ax1.streamplot(X, Y, U, V)
ax2.contour(X, Y, Y ** 2 / 2 - np.cos(X), 15)

Example 2: damped pendulum

Let's add some friction (damping) to our system. The new equation is $\ddot x=-\dot x-\sin x$.

Reformulating as a first order ODE

Let us introduce $y=\dot x$, then we have \begin{align} \dot x&=y;\ \dot y&=-\sin x-y. \end{align}

Singular points

• We need to solve $y=0$, $-\sin x-y=0$; adding equations we get $-\sin x=0$.
• The solutions are $(x, y)=(\pi k, 0)$, $k\in\mathbb Z$.
• All the solutions $(2\pi k, 0)$ correspond to the bottom point of the circle, and all the solutions $((2k+1)\pi, 0)$ correspond to the top point of the circle.

Linearization at $(0, 0)$

The jacobian matrix is given by $$ J=\left.\begin{bmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\ \frac{\partial (-\sin x-y)}{\partial x} & \frac{\partial (-\sin x-y)}{\partial y} \end{bmatrix}\right|{x=0, y=0} = \left.\begin{bmatrix} 0 & 1\ -\cos x & -1 \end{bmatrix}\right|{x=0, y=0} = \begin{bmatrix} 0 & 1 \ -1 & -1 \end{bmatrix}. $$

The eigenvalues are $-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$, their real parts are negative, so we know for sure that the original nonlinear system has qualitatively the same phase portrait (a stable focus) at $(0, 0)$ as its linearization.

Linearization at $(\pi, 0)$

The jacobian matrix is given by $$ J=\left.\begin{bmatrix} \frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\ \frac{\partial (-\sin x-y)}{\partial x} & \frac{\partial (-\sin x-y)}{\partial y} \end{bmatrix}\right|{x=\pi, y=0} = \left.\begin{bmatrix} 0 & 1\ -\cos x & -1 \end{bmatrix}\right|{x=\pi, y=0} = \begin{bmatrix} 0 & 1 \ 1 & -1 \end{bmatrix}. $$

The eigenvalues are $\frac{-1\pm\sqrt{5}}{2}$, their real parts are nonzero, so and we know for sure that the phase portrait of the original nonlinear system near $(\pi, 0)$ is qualitatively the same as the phase portrait of the linearization, i.e., it has a saddle point.

Linearizations at other points

Since $\sin$ is a $2\pi$-periodic function, the linearization at $(2\pi k, 0)$ will be a stable focus, and the linearization at $(2\pi k+\pi, 0)$ will be a saddle.

import numpy as np
import matplotlib.pyplot as plt
Y, X = np.mgrid[-3:3:1000j, -4:5:1000j]
U, V = Y, -np.sin(X)-Y
plt.streamplot(X, Y, U, V)