1.2 Linear System v.s Nonlinear System

Linear System

Consider the dynamical system $\dot{x} (t) = F (x (t))$ in $Ω \subset R^{n}$ .
If $F$ consists of only linear functions, then the system is called linear, otherwise the system is nonlinear.

Let $A \in R^{n \times n}$ be the coefficient matrix of a linear system, the system can be denoted by

\dot{x} (t) = A x (t) .

Harmonic Oscillation

\ddot{x} (t) = - x

\begin{aligned} ⟹ \dot{x} & = y \\ \dot{y} & = - x \end{aligned}

\begin{array}{r} [\begin{array}{c} \dot{x} \\ \dot{y} \end{array}] = \underset{A}{\underset{⏟}{[\begin{array}{c} 0 & 1 \\ - 1 & 0 \end{array}]}} [\begin{array}{c} x \\ y \end{array}] \end{array}

Solution:

det (A - λ I) = λ^{2} + 1, λ = \pm i

For $λ_{1} = - i$ :

[\begin{array}{ccc} i & 1 & 0 \\ - 1 & i & 0 \end{array}] \overset{R_{1} \leftarrow (- i) R_{1}}{\to} [\begin{array}{ccc} 1 & - i & 0 \\ - 1 & i & 0 \end{array}] \overset{R_{2} \leftarrow R_{2} + R_{1}}{\to} [\begin{array}{ccc} 1 & - i & 0 \\ 0 & 0 & 0 \end{array}]

$x_{1} - i x_{2} = 0 \Rightarrow x_{1} = i x_{2}$
So

(x_{1}, x_{2}) = x_{2} (i, 1), Nullspace = span {(i, 1)} .

For $λ_{2} = i$

[\begin{array}{ccc} - i & 1 & 0 \\ - 1 & - i & 0 \end{array}] \overset{R_{1} \leftarrow i R_{1}}{\to} [\begin{array}{ccc} 1 & i & 0 \\ - 1 & - i & 0 \end{array}] \overset{R_{2} \leftarrow R_{2} + R_{1}}{\to} [\begin{array}{ccc} 1 & i & 0 \\ 0 & 0 & 0 \end{array}]

$x_{1} + i x_{2} = 0 \Rightarrow x_{1} = - i x_{2}$

(x_{1}, x_{2}) = x_{2} (- i, 1), Nullspace = span {(- i, 1)} .

From the eigen-decomposition:

x (t) = c_{1} e^{i t} [\begin{matrix} 1 \\ i \end{matrix}] + c_{2} e^{- i t} [\begin{matrix} 1 \\ - i \end{matrix}] .

\begin{aligned} e^{i t} [\begin{array}{c} 1 \\ i \end{array}] & = [\begin{array}{c} \cos t + i \sin t \\ i \cos t - \sin t \end{array}], \\ e^{- i t} [\begin{array}{c} 1 \\ i \end{array}] & = [\begin{array}{c} \cos t - i \sin t \\ i \cos t + \sin t \end{array}] . \end{aligned}

Add them (real part):

\begin{aligned} [\begin{array}{c} \cos t + i \sin t \\ i \cos t - \sin t \end{array}] + [\begin{array}{c} \cos t - i \sin t \\ i \cos t + \sin t \end{array}] & = [\begin{array}{c} (\cos t + i \sin t) + (\cos t - i \sin t) \\ (i \cos t - \sin t) + (i \cos t + \sin t) \end{array}] \\ = [\begin{array}{c} 2 \cos t \\ 2 i \cos t \end{array}] . \end{aligned}

Divide by 2 to normalize:

\Rightarrow x_{1} (t) = [\begin{matrix} \cos t \\ i \cos t \end{matrix}] .

Subtract them (imag part), multiply by $i$ :

\begin{aligned} i ([\begin{array}{c} \cos t + i \sin t \\ i \cos t - \sin t \end{array}] - [\begin{array}{c} \cos t - i \sin t \\ i \cos t + \sin t \end{array}]) & = i [\begin{array}{c} (\cos t + i \sin t) - (\cos t - i \sin t) \\ (i \cos t - \sin t) - (i \cos t + \sin t) \end{array}] \\ = i [\begin{array}{c} 2 i \sin t \\ - 2 \sin t \end{array}] \\ = [\begin{array}{c} - 2 \sin t \\ - 2 i \sin t \end{array}] . \end{aligned}

Divide by (-2) for a cleaner solution:

\Rightarrow x_{2} (t) = [\begin{matrix} \sin t \\ i \sin t \end{matrix}] .

Combine:

Now you have two independent solutions:

x (t) = C_{1} [\begin{matrix} \cos t \\ i \cos t \end{matrix}] + C_{2} [\begin{matrix} \sin t \\ i \sin t \end{matrix}] .

$⟹ x (t) = C_{1} \cos t + C_{2} \sin t$

Pendulum

\dot{θ} = p, \dot{p} = - \sin θ

image-2.png|267x259

Double Pendulum

image-3.png|238x270
Is harder to understand

A nonlinear system is generally more challenging than a linear system (e.g., oscillator vs. pendulum).
A system with more degrees of freedom is typically more difficult to analyze than a system with fewer degrees of freedom (e.g., pendulum vs. double pendulum).

We can try to find

Fixed points: Points from which the system remains at rest.

Periodic orbits: Orbits that return to the initial point after a certain (finite) time.

Homoclinic/heteroclinic orbits: Orbits that connect a fixed point to itself or to another fixed point over infinite time.