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6.1 Stability and the Phase Plane

We will look at autonomous systems:

\[ \begin{cases} \frac{dx}{dt} = F(x, y) \\ \frac{dy}{dt} = G(x, y) \end{cases} \]
F, G don't depend on t explicitly (even though x, y are implicit functions of t)

They can be linear and homogeneous, for example:

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

or nonhomogeneous and linear, for example:

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

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or nonlinear (homogeneous or non), for example:

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

A critical point \( (x, y) \) is where \[ \frac{dx}{dt} = \frac{dy}{dt} = 0 \]

\( \rightarrow \) it is an equilibrium solution

All linear homogeneous systems \( \vec{x}' = A\vec{x} \) have only one critical point \( \rightarrow (0, 0) \)

For \( \vec{x}' = A\vec{x} + \vec{g} \) where \( \vec{g} \) is a constant vector,

there is only one critical point but moved to a different location, but the phase diagram remains the same as \( \vec{x}' = A\vec{x} \)

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Analysis of Linear Systems and Critical Points

Let's look at the following system of differential equations:

\[\begin{cases} x' = -2x - y \\ y' = -x - 2y \end{cases} \rightarrow \vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x}\]

This system represents an improper nodal sink with a critical point at \((0,0)\).

Nonhomogeneous System Shift

Now, consider the system with added constant terms:

\[\begin{cases} x' = -2x - y + 5 \\ y' = -x - 2y + 4 \end{cases} \rightarrow \vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x} + \begin{bmatrix} 5 \\ 4 \end{bmatrix}\]

To find the critical point (crit. pt.), we set the derivatives to zero:

\[\begin{aligned} x' = 0 & \implies -2x - y + 5 = 0 \\ y' = 0 & \implies -x - 2y + 4 = 0 \end{aligned}\]

Solving this system of equations:

\(x = 2, \quad y = 1\)

The new critical point is at \((2, 1)\).

The phase diagram is the same as \(\vec{x}' = \begin{bmatrix} -2 & -1 \\ -1 & -2 \end{bmatrix} \vec{x}\), but shifted to be centered at \((2, 1)\) instead of \((0, 0)\).

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Comparison of Phase Portraits

Phase portrait of a nodal sink centered at the origin (0,0) for the homogeneous system.
\[\begin{cases} x' = -2x - y \\ y' = -x - 2y \end{cases}\]

Critical point \((0,0)\)

Phase portrait of an identical nodal sink shifted to be centered at the critical point (2,1).
\[\begin{cases} x' = -2x - y + 5 \\ y' = -x - 2y + 4 \end{cases}\]

Critical point \((2,1)\)

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

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Nonlinear Systems and Critical Points

A nonlinear system can have multiple critical pts.

For example,

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

Crit. pts: \((0, 0), (3, 2)\)

Another example,

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

To find critical points, set the derivatives to zero:

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

Crit. pts: \((0, -1), (1, 0)\)

Phase diagram of a nonlinear is generally complicated but near each crit. pt the phase diagram will resemble that of a linear sys.

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Phase Diagrams of Nonlinear Systems

Phase portrait for x' = x(2-y), y' = y(x-3) showing a saddle point at (0,0) and a center at (3,2).
\[\begin{aligned} x' &= x(2 - y) \\ y' &= y(x - 3) \end{aligned}\]

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

Phase portrait for x' = x^2-y-1, y' = x-y-1 showing a spiral sink at (0,-1) and a saddle point 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.

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Stability of Critical pts

if solutions near a crit. pt stay near a crit. pt, the crit. pt is said to be stable (center)

if solutions fall into a crit. pt. → asymptotically stable
(any sort of sink)

if solutions run away → unstable
(any sort of source or saddle pt)

General idea for nonlinear sys:

look at what happens near each critical pt

nonlinear sys are often hard to solve:

\[ \begin{aligned} x' &= \tan(xy) + e^{y^2} \\ y' &= \sin^2(x+y) \end{aligned} \]
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there are cases where we can turn the system into a differential eq. we can solve

\[ \left. \begin{aligned} \frac{dx}{dt} &= -y^2 \\ \frac{dy}{dt} &= x \end{aligned} \right\} \quad \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx} = \frac{x}{-y^2} \quad \text{Separable} \]
\[ -y^2 \, dy = x \, dx \]
\[ -\frac{1}{3}y^3 = \frac{1}{2}x^2 + C \]
\[ \frac{1}{2}x^2 + \frac{1}{3}y^3 + C = 0 \]

implicit solution

(can be used to graph each solution curve on the phase plane)