Download original notes
PAGE 1

6.1 Stability and Phase Plane

We focus on autonomous systems:

\[ \begin{cases} \frac{dx}{dt} = F(x,y) \\ \frac{dy}{dt} = G(x,y) \end{cases} \]
F, G do not contain \( t \) explicitly (even though \( x, y \) depend on \( t \))

These can be linear and homogeneous, for example:

\[ \begin{cases} \frac{dx}{dt} = -2x - y \\ \frac{dy}{dt} = -x - 2y \end{cases} \]
\[ \vec{x}' = A\vec{x} \]

or linear and nonhomogeneous, for example:

\[ \begin{cases} \frac{dx}{dt} = -2x - y + 5 \\ \frac{dy}{dt} = -x - 2y + 4 \end{cases} \]
PAGE 2

or nonlinear, for example:

\[ \begin{cases} \frac{dx}{dt} = y^2 \\ \frac{dy}{dt} = -x \end{cases} \]

The point \( (x,y) \) where \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \) is called a critical point or an equilibrium.

All linear homogeneous systems \( \vec{x}' = A\vec{x} \) have \( (0,0) \) and only \( (0,0) \) as the critical point.

For a nonhomogeneous linear system \( \vec{x}' = A\vec{x} + \vec{g} \) where \( \vec{g} \) is a constant vector will look like \( \vec{x}' = A\vec{x} \) by translated to a different "origin".

PAGE 3

Linear Systems and Phase Diagrams

For example,

\[ \left. \begin{aligned} x' &= -2x - y \\ y' &= -x - 2y \end{aligned} \right\} \vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x} \]

Phase diagram is an improper nodal sink.

\[ \left. \begin{aligned} x' &= -2x - y + 5 \\ y' &= -x - 2y + 4 \end{aligned} \right\} \vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} 5 \\ 4 \end{bmatrix} \]

Critical Point Calculation

Critical point:

\[ \begin{aligned} -2x - y + 5 &= 0 \\ -x - 2y + 4 &= 0 \\ &\vdots \\ x = 2, y &= 1 \end{aligned} \]

So, whatever happens in \( \vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x} \) stays the same but shifted to be centred at \( (2, 1) \).

PAGE 4

Comparison of Phase Portraits

Phase portrait for a homogeneous system centered at the origin (0,0) with trajectories flowing inward.
\[ \begin{aligned} x' &= -2x - y \\ y' &= -x - 2y \end{aligned} \]

Critical point (0,0)

Phase portrait for a nonhomogeneous system centered at (2,1) with identical inward trajectories.
\[ \begin{aligned} x' &= -2x - y + 5 \\ y' &= -x - 2y + 4 \end{aligned} \]

Critical point (2,1)

Notice the phase portraits are identical but the system with constant nonhomogeneous term has the "origin" shifted.

PAGE 5

Nonlinear Systems and Critical Points

A nonlinear system can have multiple critical points.

Example 1

For example,

\[ \begin{aligned} x' &= x(2-y) = 2x - xy \quad \text{(nonlinear)} \\ y' &= y(x-3) = -3y + xy \end{aligned} \]

Critical points occur where \( x' = 0 \) and \( y' = 0 \):

\[ (0,0), \, (3,2) \]

For many nonlinear systems, the solutions near each critical point resemble the solutions of a linear system but are generally unpredictable away from critical pts.

Another Example

\[ \begin{aligned} x' &= x^2 - y - 1 \rightarrow x' = 0 \rightarrow y = x^2 - 1 \\ y' &= x - y - 1 \rightarrow y' = 0 \rightarrow x - (x^2 - 1) - 1 = 0 \end{aligned} \]

Solving for \( x \):

\[ -x^2 + x = 0 \rightarrow x = 0, \, x = 1 \]

Critical pts: \( (0, -1), \, (1, 0) \)

PAGE 6
Vector field and trajectories for a nonlinear system with a saddle at the origin and a center at (3,2).
\[ \begin{aligned} x' &= x(2-y) \\ y' &= y(x-3) \end{aligned} \]

Critical points \( (0,0), (3,2) \)

Vector field and trajectories for a nonlinear system with a saddle at (0,-1) and a spiral sink at (1,0).
\[ \begin{aligned} x' &= x^2 - y - 1 \\ y' &= x - y - 1 \end{aligned} \]

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

Notice each critical point resembles a center, improper/proper nodal source/sink, saddle point, or spiral source/sink.

PAGE 7

Stability of Critical Points

Each critical pt is a solution that stays there but what about nearby solutions?

Stable Critical Points

If nearby solutions stay nearby, the crit. pt. is stable.

(center)

Asymptotically Stable Critical Points

If nearby solutions fall into critical pt, the crit. pt. is asymptotically stable.

(sink of any sort)

Unstable Critical Points

If nearby solutions run away, the critical pt. is unstable.

(source of any sort, saddle pt)

PAGE 8

Solving Nonlinear Systems

Nonlinear systems are usually not easy to solve.

Some can be solved:

\[ \begin{aligned} \frac{dx}{dt} &= -y^2 \\ \frac{dy}{dt} &= x \end{aligned} \]
\[ \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \]
\[ \frac{x}{-y^2} = \frac{dy}{dx} \]

separable

\[ \begin{aligned} x \, dx &= -y^2 \, dy \\ \int x \, dx &= \int -y^2 \, dy \\ \frac{1}{2}x^2 &= -\frac{1}{3}y^3 + C \end{aligned} \]
\[ \frac{1}{3}y^3 + \frac{1}{2}x^2 = C \]

each C gives one solution curve

but this is rarely possible w/ nonlinear sys in general