Chapter 4 · Systems of DEs

Phase Portraits & Stability Analysis

Classifying System Behavior Without Solving

Dr. Mohamed Mabrok · Qatar University

What is a Phase Portrait?

Definition

A phase portrait plots trajectories of $\mathbf{x}(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix}$ in the $x_1$-$x_2$ plane as $t$ increases.

Key Properties

      • For $\mathbf{x}' = A\mathbf{x}$, the origin is always an equilibrium point ($\mathbf{x}' = \mathbf{0}$)

      • Arrows show direction of motion as $t$ increases

      • Questions we answer: Do trajectories spiral? Converge? Diverge?

Visual Interpretation

      Each curve represents one solution trajectory. The pattern of these curves (nodes, spirals, saddles) tells us the complete long-term behavior of the system without solving it!

What is Stability?

Three fundamental definitions:

Stable

Nearby solutions stay near origin

Small perturbations don't escape

Asymptotically Stable

Stable AND solutions → 0 as $t \to \infty$

All solutions approach origin

Unstable

Not stable

Solutions can escape to infinity

Key Insight for Linear Systems

      $$\text{For } \mathbf{x}' = A\mathbf{x}, \text{ stability is } \mathbf{\mathbf{COMPLETELY}} \text{ determined by eigenvalues of } A$$
    
      If ALL eigenvalues have negative real parts → asymptotically stable!

Classification Overview: 8 Equilibrium Types

The eigenvalues of $A$ determine which of 8 types the equilibrium belongs to:

Stable Node

$\lambda_1 < \lambda_2 < 0$ (both negative)

Unstable Node

$\lambda_1 > \lambda_2 > 0$ (both positive)

Saddle Point

$\lambda_1 < 0 < \lambda_2$ (opposite signs)

Stable Spiral

$\alpha \pm \beta i$ with $\alpha < 0$

Unstable Spiral

$\alpha \pm \beta i$ with $\alpha > 0$

Center

$\pm \beta i$ (purely imaginary)

Star Node

Repeated $\lambda$ with 2 indep. eigvecs

Improper Node

Repeated $\lambda$ with 1 eigvec

Stable Node

Eigenvalue Condition

      $$\lambda_1 < \lambda_2 < 0 \quad \text{(both real, both negative)}$$

Behavior

      • All trajectories approach origin along eigenvector directions

      • The "faster" eigenvector (larger $|\lambda|$) dominates initially

      • The "slower" eigenvector dominates near the origin

      • Asymptotically stable

Visual Pattern

      Node-like shape with all arrows pointing inward to the origin. Trajectories follow eigenvector directions, faster ones along the "steep" eigenvector, slower ones along the "gentle" eigenvector.

Unstable Node

Eigenvalue Condition

      $$0 < \lambda_1 < \lambda_2 \quad \text{(both real, both positive)}$$

Behavior

      • All trajectories diverge from origin

      • Reverse of the stable node case

      • Solutions grow exponentially along both eigenvector directions

      • Unstable

Visual Pattern

      Node-like shape with all arrows pointing outward from the origin. This is the opposite of the stable node.

Saddle Point

Eigenvalue Condition

      $$\lambda_1 < 0 < \lambda_2 \quad \text{(one negative, one positive)}$$

Behavior

      • Stable manifold along $\mathbf{v}_1$ (attracts solutions)

      • Unstable manifold along $\mathbf{v}_2$ (repels solutions)

      • Most trajectories diverge to infinity

      • ALWAYS unstable

Key Warning

      $$\det(A) < 0 \quad \Rightarrow \quad \text{Saddle point (always!)}$$
    
      If the determinant is negative, you have a saddle — eigenvalues must have opposite signs.

Stable Spiral (Sink)

Eigenvalue Condition

      $$\lambda = \alpha \pm \beta i \quad \text{with } \alpha < 0, \beta \neq 0$$

Behavior

      • Trajectories spiral INWARD toward origin

      • Decay rate: $e^{\alpha t}$ (exponential contraction)

      • Rotation rate: $\beta$ rad/s

      • Asymptotically stable

Physical Analogy

      Like an underdamped oscillator: the system oscillates while losing energy at an exponential rate.

Unstable Spiral (Source)

Eigenvalue Condition

      $$\lambda = \alpha \pm \beta i \quad \text{with } \alpha > 0, \beta \neq 0$$

Behavior

      • Trajectories spiral OUTWARD from origin

      • Growth rate: $e^{\alpha t}$ (exponential expansion)

      • Rotation rate: $\beta$ rad/s

      • Unstable

Visual Pattern

      Growing oscillation — spirals get larger and larger as they move away from the origin.

Center

Eigenvalue Condition

      $$\lambda = \pm \beta i \quad \text{(purely imaginary)}$$

Behavior

      • Closed elliptical orbits around origin

      • Period: $T = \frac{2\pi}{\beta}$

      • Stable but NOT asymptotically stable (solutions don't approach origin)

      • Solutions maintain constant amplitude forever

Physical Analogy

      Like an undamped oscillator or harmonic motion: the system perpetually orbits the equilibrium with no energy dissipation.

Degenerate Cases: Star Node & Improper Node

When $\lambda$ is a repeated eigenvalue, we have two special cases:

Star Node

Repeated $\lambda$ with 2 independent eigenvectors

Example: $A = \lambda I$

All trajectories are straight lines through origin

If $\lambda < 0$: stable. If $\lambda > 0$: unstable.

Improper Node

Repeated $\lambda$ with only 1 eigenvector

Uses generalized eigenvector

Solution has form $te^{\lambda t}$ terms

Trajectories are tangent to eigenvector near origin

Stability Rule

      For both types: if $\lambda < 0$ → stable, if $\lambda > 0$ → unstable.

The Trace-Determinant (τ-Δ) Plane

For a $2 \times 2$ system $\mathbf{x}' = A\mathbf{x}$ where $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

Define Trace and Determinant

      $$\tau = \text{tr}(A) = a + d \quad \text{and} \quad \Delta = \det(A) = ad - bc$$
    
      These are related to eigenvalues by the characteristic equation: $\lambda^2 - \tau\lambda + \Delta = 0$

The Discriminant

      $$D = \tau^2 - 4\Delta$$
    
      This determines the nature of eigenvalues: $D > 0$ (real), $D = 0$ (repeated real), $D < 0$ (complex).

Regions in (τ, Δ) plane

      • $\Delta < 0$: Saddle

      • $\Delta > 0, D > 0, \tau < 0$: Stable Node

      • $\Delta > 0, D > 0, \tau > 0$: Unstable Node

      • $\Delta > 0, D < 0, \tau < 0$: Stable Spiral

      • $\Delta > 0, D < 0, \tau > 0$: Unstable Spiral

      • $\tau = 0, \Delta > 0$: Center

Summary: All 8 Cases at a Glance

Type	Eigenvalues	Stability	Visual
Stable Node	$\lambda_1 < \lambda_2 < 0$	Asymp. Stable	Inward node
Unstable Node	$0 < \lambda_1 < \lambda_2$	Unstable	Outward node
Saddle	$\lambda_1 < 0 < \lambda_2$	Unstable	Hyperbolic cross
Stable Spiral	$\alpha \pm \beta i, \alpha < 0$	Asymp. Stable	Inward spiral
Unstable Spiral	$\alpha \pm \beta i, \alpha > 0$	Unstable	Outward spiral
Center	$\pm \beta i$	Stable	Closed orbits
Star Node	Repeated $\lambda$, 2 eigvecs	Depends on sign	Radial lines
Improper Node	Repeated $\lambda$, 1 eigvec	Depends on sign	Tangent curves

Worked Example

Example 1: Classify the System

Classify and sketch the phase portrait for: $A = \begin{pmatrix} -3 & 1 \\ 0 & -2 \end{pmatrix}$

STEP 1 Find eigenvalues

\text{Upper triangular} \Rightarrow \lambda_1 = -3, \quad \lambda_2 = -2

STEP 2 Compute trace and determinant

\tau = -3 + (-2) = -5, \quad \Delta = (-3)(-2) - 0 = 6

STEP 3 Classify

\text{Both } \lambda < 0 \Rightarrow \text{ Stable Node}

STEP 4 Find eigenvectors

\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

Conclusion

      Trajectories approach origin along both eigenvector directions. Asymptotically stable.

Worked Example

Example 2: Center Equilibrium

Classify: $A = \begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$

STEP 1 Find eigenvalues via characteristic equation

\det(A - \lambda I) = \lambda^2 + 4 = 0 \quad \Rightarrow \quad \lambda = \pm 2i

STEP 2 Identify $\alpha$ and $\beta$

\text{Purely imaginary} \Rightarrow \alpha = 0, \quad \beta = 2

STEP 3 Classify

\text{Purely imaginary} \Rightarrow \text{ Center}

STEP 4 Period

T = \frac{2\pi}{\beta} = \frac{2\pi}{2} = \pi

Conclusion

      Closed elliptical orbits with period $\pi$. Stable but NOT asymptotically stable.

Worked Example

Example 3: Saddle Point

Classify: $A = \begin{pmatrix} 1 & 3 \\ 1 & -1 \end{pmatrix}$

STEP 1 Find eigenvalues

\det(A - \lambda I) = \lambda^2 - 4 = 0 \quad \Rightarrow \quad \lambda = \pm 2

STEP 2 Classify

\text{One positive, one negative} \Rightarrow \text{ Saddle Point}

STEP 3 Find eigenvectors

\lambda_1 = -2: \mathbf{v}_1 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}, \quad \lambda_2 = 2: \mathbf{v}_2 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}

Conclusion

      Stable manifold along $\mathbf{v}_1$ (attracts), unstable manifold along $\mathbf{v}_2$ (repels). Most trajectories escape to infinity. Always unstable!

Summary: Quick Decision Flowchart

Step-by-Step Procedure

Compute eigenvalues of $A$

Are both eigenvalues real and negative? → Stable Node

Is one positive, one negative? → Saddle Point (always unstable!)

Are both eigenvalues real and positive? → Unstable Node

Are eigenvalues complex with $\alpha < 0$? → Stable Spiral

Are eigenvalues complex with $\alpha > 0$? → Unstable Spiral

Are eigenvalues purely imaginary? → Center (stable, not asymptotically)

Is $\lambda$ repeated? Check multiplicity of eigenvectors → Star or Improper Node

Key Takeaway

      $$\text{Eigenvalues tell you } \mathbf{\mathbf{EVERYTHING}} \text{ about long-term system behavior}$$
    
      No need to solve the differential equations! The eigenvalue structure immediately reveals the phase portrait structure.

Next Steps in the Course

      Applications to real systems: Two-tank mixing problems, spring-mass systems, RLC circuits, and biological models. All use the same classification framework!