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PAGE 1

6.2 Linear and Almost Linear Systems (Continued)

last time:

\[\begin{aligned} x' &= F(x,y) \\ y' &= G(x,y) \end{aligned}\]

is almost/locally linear if near a critical pt \((x_0, y_0)\) the system acts like

\[\vec{x}' = A\vec{x} + \vec{g}(\vec{x})\]

such that

\[\lim_{(x,y) \to (x_0, y_0)} \frac{|\vec{g}|}{|\vec{x}|} = 0\]

for each \((x_0, y_0)\), there is a different \(A\) matrix near \((x_0, y_0)\).

example from last time:

\[\begin{aligned} x' &= -x + xy \\ y' &= -2y + 8xy \end{aligned}\]

cp: \((0,0)\), \((\frac{1}{4}, 1)\)

look at this one: \((\frac{1}{4}, 1)\)
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define \(u = x - \frac{1}{4}\), \(v = y - 1\), \(x = u + \frac{1}{4}\), \(y = v + 1\)

\(u' = x'\), \(v' = y'\)

system in terms of \(u, v\):

\[\begin{aligned} u' &= -(u + \frac{1}{4}) + (u + \frac{1}{4})(v + 1) = \frac{1}{4}v + uv \\ v' &= \dots = 8u + 8uv \end{aligned}\]
\[\begin{bmatrix} u' \\ v' \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{4} \\ 8 & 0 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} + \begin{bmatrix} uv \\ 8uv \end{bmatrix}\]

\(A\) near \((\frac{1}{4}, 1)\)

a more efficient way: linearization of the system near \((x_0, y_0)\)

Taylor series

\(x' = F(x,y)\)

\(y' = G(x,y)\)

cp: \((x_0, y_0)\)

\[F(x,y) = F(x_0, y_0) + \frac{\partial F}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial F}{\partial y}(x_0, y_0)(y - y_0) + \dots\]
\[G(x,y) = G(x_0, y_0) + \frac{\partial G}{\partial x}(x_0, y_0)(x - x_0) + \frac{\partial G}{\partial y}(x_0, y_0)(y - y_0) + \dots\]

Note on terms:

  • \(F(x_0, y_0) = 0\) and \(G(x_0, y_0) = 0\) by def. of crit. pt.
  • \((x - x_0) = u\)
  • \((y - y_0) = v\)
  • \(\dots\) represents higher order (not important)
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Linearized Systems and Jacobian Matrices

The system becomes:

\[ u' = F_x(x_0, y_0)u + F_y(x_0, y_0)v \]\[ v' = G_x(x_0, y_0)u + G_y(x_0, y_0)v \]
\[ \begin{bmatrix} u' \\ v' \end{bmatrix} = \begin{bmatrix} F_x & F_y \\ G_x & G_y \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} \]
linearized system near \( (x_0, y_0) \)
Jacobian matrix (function of \( x, y \))

Example

Given the system:

\[ x' = x^2 + y^2 - 6 = F \]\[ y' = x^2 - y = G \]

Finding critical points (cp):

\[ x^2 + y^2 - 6 = 0 \]\[ y + y^2 - 6 = 0 \]\[ y^2 + y - 6 = 0 \]\[ (y+3)(y-2) = 0 \]
\[ x^2 - y = 0 \rightarrow y = x^2 \]\[ y = -3, \quad y = 2 \]

Since \( x^2 = y \), we have \( x = \pm \sqrt{y} \). For \( y=2 \):

cp: \( (\sqrt{2}, 2) \), \( (-\sqrt{2}, 2) \)
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Linearize using the Jacobian matrix

\[ J(x, y) = \begin{bmatrix} F_x & F_y \\ G_x & G_y \end{bmatrix} = \begin{bmatrix} 2x & 2y \\ 2x & -1 \end{bmatrix} \]

At critical point \( (\sqrt{2}, 2) \):

\[ J(\sqrt{2}, 2) = \begin{bmatrix} 2\sqrt{2} & 4 \\ 2\sqrt{2} & -1 \end{bmatrix} \rightarrow \vec{x}' = \begin{bmatrix} 2\sqrt{2} & 4 \\ 2\sqrt{2} & -1 \end{bmatrix} \vec{x} \]
  • \( \lambda \approx 4.8, -3 \)
  • Saddle pt, unstable
  • not sensitive to perturbation
  • linearization does not "lie" about this being a saddle pt

At critical point \( (-\sqrt{2}, 2) \):

\[ J(-\sqrt{2}, 2) = \begin{bmatrix} -2\sqrt{2} & 4 \\ -2\sqrt{2} & -1 \end{bmatrix} \]
  • \( \lambda \approx -2 \pm 3.2i \)
  • Spiral sink
  • asympt. stable
  • not sensitive to perturbation
A 2D Cartesian coordinate system with x and y axes. A spiral sink is sketched in the second quadrant.
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Phase Portrait of a Nonlinear System

The following figure displays the phase portrait for a system of nonlinear differential equations. The vector field is represented by small black arrows, and several trajectories are shown as solid red curves. The axes range from -4 to 4 in both the horizontal and vertical directions.

Phase portrait showing a vector field with red trajectories spiraling around critical points on a grid from -4 to 4.

The trajectories exhibit complex behavior, including spiraling around a central point near the origin and diverging in other regions of the phase plane, characteristic of nonlinear dynamics.

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Example: Nonlinear System Analysis

System Definition

\[ \begin{aligned} x' &= 2xy = F \\ y' &= 1 - x^2 + y^2 = G \end{aligned} \]

Critical points (cp): \( (1, 0), (-1, 0) \)

Complex plane plot showing eigenvalues as dots with arrows indicating movement relative to the imaginary axis.

Jacobian Matrix

\[ J(x, y) = \begin{bmatrix} 2y & 2x \\ -2x & 2y \end{bmatrix} \]

Analysis at \( (1, 0) \)

\[ J(1, 0) = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \quad \lambda = \pm 2i \]

Linearization says center, stable but true behavior of the nonlinear system may or may not be a center.

Analysis at \( (-1, 0) \)

\[ J(-1, 0) = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \quad \lambda = \pm 2i \]

Same story.

To determine true behavior, we can graph the nonlinear phase portrait or solve the system (this one can be solved using sub \( u = y^2/x \)).

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Linearization and Phase Portraits

Linearization doesn't "lie" but not always the case in general.

A phase portrait on a Cartesian grid from -4 to 4 on both axes, showing a vector field with two focal points and curved trajectories.
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Example

\[ \begin{aligned} x' &= -3y + ay(x^2 + y^2) \\ y' &= 3x + ay(x^2 + y^2) \end{aligned} \]

\( a \) is some constant

cp: \( (0,0) \)

\[ J(x,y) = \begin{bmatrix} 2axy & -3 + ax^2 + 3ay^2 \\ 3 + 2axy & ax^2 + 3ay^2 \end{bmatrix} \]
\[ J(0,0) = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} \]

Linearization lost \( a \) completely

  • Linearization cannot capture effect of \( a \)
  • Linearization: center, stable

cannot trust: may or may not still be a center

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Phase Portraits of Linearized and Nonlinear Systems

Linearized System

The phase portrait for the linearized system shows a stable center. The trajectories form closed concentric loops around the equilibrium point, indicating periodic behavior without decay or growth.

Phase portrait of a linearized system showing concentric circular trajectories around a stable center point.

Linearized (stable center)

Nonlinear System: Case \( a = 1 \)

For the nonlinear system with \( a = 1 \), the equilibrium point becomes an unstable spiral point. Trajectories spiral outward away from the center.

Phase portrait of a nonlinear system with a=1 showing trajectories spiraling outward from an unstable spiral point.

Nonlinear \( a = 1 \) (unstable spiral point)

Nonlinear System: Case \( a = -1 \)

For the nonlinear system with \( a = -1 \), the equilibrium point is an asymptotically stable spiral point. Trajectories spiral inward toward the center.

Phase portrait of a nonlinear system with a=-1 showing trajectories spiraling inward toward a stable spiral point.

Nonlinear \( a = -1 \) (asymptotically stable spiral point)