Stability and Phase Plane Analysis

9.1 Autonomous Systems

For $\mathbf{x}" = \mathbf{f}(\mathbf{x})$ A critical point $\mathbf{x}^*$ satisfies $\mathbf{f}(\mathbf{x}^*) = \mathbf{0}$ .

9.2 Linearization and Stability

Let $A = J\mathbf{f}(\mathbf{x}^*)$ be the Jacobian at the critical point. The eigenvalues of $A$ Determine the local stability:

Eigenvalues of $A$	Type	Stability
Both real, negative	Stable node	Asymptotically stable
Both real, positive	Unstable node	Unstable
Real, opposite signs	Saddle point	Unstable
Complex, $\mathrm{Re}(\lambda) \lt 0$	Stable spiral	Asymptotically stable
Complex, $\mathrm{Re}(\lambda) > 0$	Unstable spiral	Unstable
Purely imaginary	Center	(Marginally) stable

9.3 Lyapunov Stability

Definition. A critical point $\mathbf{x}^*$ is stable if for every $\varepsilon > 0$ There Exists $\delta > 0$ such that $\|\mathbf{x}(0) - \mathbf{x}^*\| \lt \delta$ implies $\|\mathbf{x}(t) - \mathbf{x}^*\| \lt \varepsilon$ for all $t > 0$ .

It is asymptotically stable if it is stable and $\mathbf{x}(t) \to \mathbf{x}^*$ as $t \to \infty$ .

Theorem 9.1 (Lyapunov). If there exists a continuously differentiable function $V$ (a Lyapunov Function) such that $V(\mathbf{x}^*) = 0$ , $V(\mathbf{x}) > 0$ for $\mathbf{x} \neq \mathbf{x}^*$ And $\dot{V} \leq 0$ in a neighbourhood of $\mathbf{x}^*$ Then $\mathbf{x}^*$ is stable. If $\dot{V} \lt 0$ for $\mathbf{x} \neq \mathbf{x}^*$ Then $\mathbf{x}^*$ is asymptotically stable.

9.4 Worked Example: Linearization

Problem. Find and classify the critical points of $x' = x - y$ , $y' = x^2 + y^2 - 1$ .

Solution

Solution. Set $x' = 0$ and $y' = 0$ :

$x - y = 0 \implies y = x$

$x^2 + x^2 - 1 = 0 \implies 2x^2 = 1 \implies x = \pm 1/\sqrt{2}$

Critical points: $(1/\sqrt{2}, 1/\sqrt{2})$ and $(-1/\sqrt{2}, -1/\sqrt{2})$ .

The Jacobian is $J = \begin{pmatrix} 1 & -1 \\ 2x & 2y \end{pmatrix}$ .

At $(1/\sqrt{2}, 1/\sqrt{2})$ : $J = \begin{pmatrix} 1 & -1 \\ \sqrt{2} & \sqrt{2} \end{pmatrix}$ .

$\mathrm{tr}(J) = 1 + \sqrt{2} > 0$ , $\det(J) = \sqrt{2} + \sqrt{2} = 2\sqrt{2} > 0$ .

$\tau^2 - 4\Delta = (1 + \sqrt{2})^2 - 8\sqrt{2} = 3 + 2\sqrt{2} - 8\sqrt{2} = 3 - 6\sqrt{2} \lt 0$ .

Complex eigenvalues with positive real part: unstable spiral.

At $(-1/\sqrt{2}, -1/\sqrt{2})$ : $J = \begin{pmatrix} 1 & -1 \\ -\sqrt{2} & -\sqrt{2} \end{pmatrix}$ .

$\mathrm{tr}(J) = 1 - \sqrt{2} \lt 0$ , $\det(J) = -\sqrt{2} + \sqrt{2} = 0$ .

Wait, $\det(J) = (1)(-\sqrt{2}) - (-1)(-\sqrt{2}) = -\sqrt{2} - \sqrt{2} = -2\sqrt{2} \lt 0$ .

Negative determinant: saddle point (unstable). $\blacksquare$

9.5 Phase Portraits for 2D Nonlinear Systems

For the nonlinear system $\mathbf{x}' = \mathbf{f}(\mathbf{x})$ The Hartman-Grobman theorem States that near a hyperbolic critical point (one where the Jacobian has no eigenvalues on the Imaginary axis), the nonlinear phase portrait is topologically equivalent to the linearized one.

Procedure for sketching phase portraits:

Find all critical points by solving $\mathbf{f}(\mathbf{x}) = \mathbf{0}$ .
Compute the Jacobian $J$ at each critical point.
Classify each critical point using the eigenvalue analysis from Section 4.9.
Sketch the local behaviour near each critical point.
Connect the local pictures using nullclines ( $x' = 0$ and $y' = 0$ curves).

9.6 Limit Cycles and Poincaré-Bendixson

A limit cycle is an isolated closed periodic orbit. Limit cycles are inherently nonlinear Phenomena --- linear systems cannot have isolated closed orbits.

Theorem 9.2 (Poincaré-Bendixson). If a trajectory of a $C^1$ planar system is confined to a Closed bounded region $R$ that contains no critical points, then the trajectory approaches a closed Periodic orbit as $t \to \infty$ .

Remark. The Poincaré-Bendixson theorem is specific to two dimensions. In three or more Dimensions, much more complex behaviour (chaos) is possible.

Example: Van der Pol oscillator. The equation

$x'' + \mu(x^2 - 1)x' + x = 0$

With $\mu > 0$ has a unique stable limit cycle. This system models electrical circuits with Nonlinear resistance and arises in biology (cardiac rhythms, neuron firing).

9.7 Worked Example: Lotka-Volterra Analysis

Problem. Analyze the stability of the Lotka-Volterra system $x' = x(2 - y)$ , $y' = y(x - 1)$ .

Solution

Solution. Critical points: $(0, 0)$ and $(1, 2)$ .

Jacobian: $J = \begin{pmatrix} 2 - y & -x \\ y & x - 1 \end{pmatrix}$ .

At $(0, 0)$ : $J = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$ . Eigenvalues $2$ and $-1$ : saddle point (unstable).

At $(1, 2)$ : $J = \begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}$ . $\det(J) = 2 > 0$ , $\mathrm{tr}(J) = 0$ . Eigenvalues $\pm i\sqrt{2}$ : center.

Remark. For the linearized system, the center is (marginally) stable. However, for the Nonlinear Lotka-Volterra system, the trajectories are actually closed orbits surrounding $(1, 2)$ . This can be verified using the first integral $H = x - \ln x + 2\ln y - y$ Which is constant Along trajectories. $\blacksquare$

9.8 Competing Species

The competing species model is:

$x' = x(r_1 - a_{11}x - a_{12}y), \quad y' = y(r_2 - a_{21}x - a_{22}y)$

Where $r_i > 0$ are growth rates and $a_{ij} > 0$ are competition coefficients. The four critical Points are $(0, 0)$ , $(r_1/a_{11}, 0)$ , $(0, r_2/a_{22})$ And the coexistence point $(x^*, y^*)$ where both $x'$ and $y'$ vanish.

The stability of the coexistence point determines whether both species survive. If $a_{11}a_{22} > a_{12}a_{21}$ Coexistence is stable; otherwise, one species drives the other To extinction (competitive exclusion).

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