4.3 Phase Portraits and Stability Analysis

Understanding the qualitative behavior of 2D linear systems through geometric visualization

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Learning Objectives

Understand phase portraits as geometric representations of system trajectories

Classify equilibrium points based on eigenvalue analysis

Determine stability of the origin as an equilibrium point

Use the trace-determinant plane to classify linear systems

Interpret and sketch phase portraits for 2×2 systems

What is a Phase Portrait?

A phase portrait is a plot of trajectories in the $x_1$-$x_2$ phase plane for a dynamical system. For a 2×2 linear system

\mathbf{x}' = A\mathbf{x}

each solution $\mathbf{x}(t) = [x_1(t), x_2(t)]^T$ traces a curve (trajectory) in the phase plane as $t$ varies. The arrows on these curves indicate the direction of motion as time increases.

A phase portrait captures the qualitative behavior of the system:

Do trajectories spiral inward or outward?
Are there stable or unstable directions?
What is the long-term behavior as $t \to \infty$?

What is Stability?

An equilibrium point is a point where $\mathbf{x}' = \mathbf{0}$. For the linear system $\mathbf{x}' = A\mathbf{x}$, the origin $\mathbf{x} = \mathbf{0}$ is always an equilibrium.

Stability Definitions

The origin is:

Stable if for any $\epsilon > 0$, there exists $\delta > 0$ such that $\|\mathbf{x}_0\| < \delta$ implies $\|\mathbf{x}(t)\| < \epsilon$ for all $t \geq 0$.
Asymptotically stable if it is stable AND $\lim_{t \to \infty} \mathbf{x}(t) = \mathbf{0}$ for all nearby initial conditions.
Unstable if it is not stable.

Key insight: For 2×2linear systems, the stability of the origin is completely determined by the eigenvalues of $A$.

Classification of Equilibrium Points

The behavior of a 2×2 linear system is determined entirely by the eigenvalues $\lambda_1, \lambda_2$ of the coefficient matrix $A$. Below is a comprehensive classification:

Stable Node

$\lambda_1, \lambda_2 \in \mathbb{R}$, $\lambda_1 < \lambda_2 < 0$

All trajectories approach the origin along straight line directions (eigenvectors). Motion is exponential decay with rate $\max(|\lambda_1|, |\lambda_2|)$.

Asymptotically Stable

Unstable Node

$\lambda_1, \lambda_2 \in \mathbb{R}$, $0 < \lambda_1 < \lambda_2$

All trajectories diverge from the origin along eigenvector directions. Exponential growth away from origin.

Unstable

Saddle Point

$\lambda_1, \lambda_2 \in \mathbb{R}$, $\lambda_1 < 0 < \lambda_2$

The stable manifold (along $\mathbf{v}_1$) attracts trajectories; the unstable manifold (along $\mathbf{v}_2$) repels them. Most trajectories diverge.

Unstable

Stable Spiral (Sink)

$\lambda = \alpha \pm \beta i$, $\alpha < 0, \beta \neq 0$

Trajectories spiral inward toward the origin. The rotation rate is $\beta$ (rad/s), the decay rate is $e^{\alpha t}$.

Asymptotically Stable

Unstable Spiral (Source)

$\lambda = \alpha \pm \beta i$, $\alpha > 0, \beta \neq 0$

Trajectories spiral outward from the origin. Exponential growth with oscillatory behavior.

Unstable

Center (Elliptic)

$\lambda = \pm \beta i$, $\beta > 0$ (purely imaginary)

All trajectories form closed ellipses around the origin. The system oscillates with constant period $T = 2\pi/\beta$.

Stable (but NOT asymptotically)

Star Node (Degenerate)

Repeated eigenvalue: $\lambda$ with 2 independent eigenvectors ($A = \lambda I$)

All trajectories are straight lines through the origin. Either all converge (if $\lambda < 0$) or diverge (if $\lambda > 0$).

Stable if $\lambda < 0$

Improper Node

Repeated eigenvalue: $\lambda$ with only 1 independent eigenvector

Trajectories approach the origin along the eigenvector direction, but not straight lines. One "fast" and one "slow" mode.

Asymptotically Stable if $\lambda < 0$

Summary Table

Type	Eigenvalues	Condition	Stability
Stable Node	$\lambda_1, \lambda_2 \in \mathbb{R}$	$\lambda_1 < \lambda_2 < 0$	Asymptotically Stable
Unstable Node	$\lambda_1, \lambda_2 \in \mathbb{R}$	$0 < \lambda_1 < \lambda_2$	Unstable
Saddle Point	$\lambda_1, \lambda_2 \in \mathbb{R}$	$\lambda_1 < 0 < \lambda_2$	Unstable
Stable Spiral	$\lambda = \alpha \pm \beta i$	$\alpha < 0, \beta \neq 0$	Asymptotically Stable
Unstable Spiral	$\lambda = \alpha \pm \beta i$	$\alpha > 0, \beta \neq 0$	Unstable
Center	$\lambda = \pm \beta i$	$\beta > 0$	Stable
Star Node	Repeated $\lambda$	2 independent eigenvectors	Stable if $\lambda < 0$
Improper Node	Repeated $\lambda$	1 independent eigenvector	Stable if $\lambda < 0$

The Trace-Determinant Plane

For a 2×2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, define:

\tau = \text{tr}(A) = a + d \quad \text{and} \quad \Delta = \det(A) = ad - bc

The characteristic polynomial is $\lambda^2 - \tau\lambda + \Delta = 0$, with discriminant $D = \tau^2 - 4\Delta$.

The $(\tau, \Delta)$ plane is divided into regions corresponding to different equilibrium types:

How to use the diagram: Compute $\tau$ and $\Delta$ for your system, plot the point $(\tau, \Delta)$. The region containing that point determines the equilibrium type.

Worked Examples

Solution

Step 1: Find eigenvalues

\det(A - \lambda I) = \det\begin{pmatrix} -3-\lambda & 1 \\ 0 & -2-\lambda \end{pmatrix} = (-3-\lambda)(-2-\lambda) = \lambda^2 + 5\lambda + 6 = 0

Factoring: $(\lambda + 2)(\lambda + 3) = 0$, so $\lambda_1 = -3, \lambda_2 = -2$.

Step 2: Check trace and determinant

\tau = -3 + (-2) = -5, \quad \Delta = (-3)(-2) - (1)(0) = 6

Point $(-5, 6)$ in the $(\tau, \Delta)$ plane: $\tau < 0$ and $\Delta > 0$ with $\tau^2 - 4\Delta = 25 - 24 = 1 > 0$ (real eigenvalues) → Stable Node

Step 3: Find eigenvectors

For $\lambda_1 = -3$: $(A + 3I)\mathbf{v} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

For $\lambda_2 = -2$: $(A + 2I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}\mathbf{v} = \mathbf{0}$ gives $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

Step 4: Sketch phase portrait

Interpretation: Both eigenvalues are negative, so all trajectories decay exponentially to the origin. The decay is faster along the $v_1$ direction (rate $e^{-3t}$) than along the $v_2$ direction (rate $e^{-2t}$). The origin is asymptotically stable.

Solution

Step 1: Find eigenvalues

\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & 1 \\ -1 & -\lambda \end{pmatrix} = \lambda^2 + 1 = 0

So $\lambda = \pm i$, purely imaginary with $\beta = 1$.

Step 2: Check trace and determinant

\tau = 0 + 0 = 0, \quad \Delta = (0)(0) - (1)(-1) = 1

Point $(0, 1)$ on the boundary $\tau = 0$, $\Delta > 0$ → Center

Step 3: Solution structure

\mathbf{x}(t) = c_1 \begin{pmatrix} \cos t \\ -\sin t \end{pmatrix} + c_2 \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}

Step 4: Phase portrait

Interpretation: All trajectories are circles centered at the origin. The period is $T = 2\pi/1 = 2\pi$. The origin is stable but not asymptotically stable — nearby solutions stay near the origin but don't converge to it.

Solution

Step 1: Find eigenvalues

\det(A - \lambda I) = \det\begin{pmatrix} -1-\lambda & -4 \\ 1 & -1-\lambda \end{pmatrix} = (-1-\lambda)^2 + 4 = \lambda^2 + 2\lambda + 5 = 0

\lambda = \frac{-2 \pm \sqrt{4-20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i

So $\alpha = -1 < 0$ and $\beta = 2 \neq 0$.

Step 2: Check trace and determinant

\tau = -1 + (-1) = -2, \quad \Delta = (-1)(-1) - (-4)(1) = 1 + 4 = 5

$\tau < 0$, $\Delta > 0$, $\tau^2 - 4\Delta = 4 - 20 = -16 < 0$ (complex eigenvalues with negative real part) → Stable Spiral

Step 3: Phase portrait

Interpretation: Trajectories spiral inward toward the origin. The decay rate is $e^{-t}$ and the rotation rate is $2$ rad/s. The origin is asymptotically stable. This is the underdamped case in oscillator problems.

Interactive Phase Portrait Explorer

Enter the elements of a 2×2 matrix $A$ to visualize its phase portrait. The system will automatically compute eigenvalues and classify the equilibrium type.

Matrix $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$

$a_{11}$

$a_{12}$

$a_{21}$

$a_{22}$

Preset Examples

Trace (τ)

—

Determinant (Δ)

—

Eigenvalue 1

—

Eigenvalue 2

—

Classification

—

📝 Exam-Style Practice Problems

These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the hint and the step-by-step solution.

Problem 1: Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} 1 & -3 \\ 3 & 1 \end{pmatrix}\mathbf{x}$ ▼

Classify the equilibrium of the system $\mathbf{x}' = \begin{pmatrix} 1 & -3 \\ 3 & 1 \end{pmatrix}\mathbf{x}$ and describe the phase portrait behavior.

Hint

Compute the trace and determinant. $\tau = 1 + 1 = 2 > 0$ and $\Delta = 1 + 9 = 10 > 0$. What does $\tau > 0$ and $\Delta > 0$ tell you about the nature of the eigenvalues?

Solution

Compute $\tau = 1 + 1 = 2$ and $\Delta = (1)(1) - (-3)(3) = 1 + 9 = 10$.

Eigenvalues from $\lambda^2 - 2\lambda + 10 = 0$:

\lambda = \frac{2 \pm \sqrt{4-40}}{2} = \frac{2 \pm \sqrt{-36}}{2} = 1 \pm 3i

Classification: Complex eigenvalues with $\alpha = 1 > 0$ → Unstable Spiral

Behavior: Trajectories spiral outward from the origin with angular frequency $3$ rad/s and exponential growth rate $e^t$.

Problem 2: Convert and classify $y'' + 4y' + 5y = 0$ ▼

A mass-spring-damper system is described by $y'' + 4y' + 5y = 0$. Convert this to a first-order system, find its eigenvalues, and classify the equilibrium.

Hint

Let $x_1 = y$ and $x_2 = y'$. Then $x_1' = x_2$ and $x_2' = -5x_1 - 4x_2$. Build the matrix and compute eigenvalues.

Solution

Step 1: Convert to system

\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -5 & -4 \end{pmatrix}\mathbf{x}

Step 2: Compute trace and determinant

$\tau = 0 + (-4) = -4$, $\Delta = (0)(-4) - (1)(-5) = 5$

Step 3: Find eigenvalues

\lambda^2 + 4\lambda + 5 = 0 \Rightarrow \lambda = \frac{-4 \pm \sqrt{16-20}}{2} = -2 \pm i

Classification: Complex with $\alpha = -2 < 0$ → Stable Spiral (Underdamped)

Physical meaning: The mass oscillates with decaying amplitude, returning to equilibrium.

Problem 3 (True/False): $\operatorname{tr}(A) = 0$ and $\det(A) > 0$ forces a center ▼

Statement: "A system with $\text{tr}(A) = 0$ and $\det(A) > 0$ must have a center at the origin."

Hint

With $\tau = 0$, the eigenvalues satisfy $\lambda^2 + \Delta = 0$. If $\Delta > 0$, what are the eigenvalues?

Solution

Answer: TRUE

Explanation: If $\tau = 0$, the characteristic equation becomes $\lambda^2 + \Delta = 0$. If $\Delta > 0$, then $\lambda^2 = -\Delta < 0$, giving $\lambda = \pm i\sqrt{\Delta}$ (purely imaginary with $\beta = \sqrt{\Delta} > 0$). This is exactly the definition of a center.

The trajectories are closed ellipses, and the equilibrium is stable but not asymptotically stable.

Practice Problems

Test your understanding with these multiple-choice questions. Click a choice to see if you're correct.

Q1: Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}\mathbf{x}$.

Correct answer: (B) Stable node. $\tau = -4$, $\Delta = 4 - 1 = 3$, and $\tau^2 - 4\Delta = 16 - 12 = 4 > 0$, so the eigenvalues are real and distinct: $\lambda = -1, -3$. Both negative and real → stable (attracting) node, not a spiral.

Q2: Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\mathbf{x}$.

Correct answer: (D) Saddle point. $\Delta = (1)(1) - (2)(2) = -3 < 0$. A negative determinant always means one positive and one negative eigenvalue (here $\lambda = 3$ and $\lambda = -1$) — that is a saddle, and it is always unstable. When $\Delta < 0$ you never need the trace.

Q3: A $2\times 2$ system has $\tau = -2$ and $\Delta = 5$. What type of equilibrium is the origin?

Correct answer: (A) Stable spiral. The discriminant is $\tau^2 - 4\Delta = 4 - 20 = -16 < 0$, so the eigenvalues are complex: $\lambda = -1 \pm 2i$. Since $\operatorname{Re}\lambda = \tau/2 = -1 < 0$, trajectories spiral inward.

Q4: The equation $y'' + 9y = 0$ becomes $\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -9 & 0 \end{pmatrix}\mathbf{x}$. What is the equilibrium at the origin?

Correct answer: (C) Center. $\tau = 0$ and $\Delta = 9 > 0$, so $\lambda^2 + 9 = 0$ and $\lambda = \pm 3i$ — purely imaginary. Trajectories are closed ellipses: the undamped oscillator neither gains nor loses energy. The equilibrium is stable but not asymptotically stable.

Q5: In the trace–determinant plane, what does the parabola $\tau^2 = 4\Delta$ separate?

Correct answer: (B). The quantity $\tau^2 - 4\Delta$ is the discriminant of $\lambda^2 - \tau\lambda + \Delta = 0$. Above the parabola ($\tau^2 < 4\Delta$) the eigenvalues are complex → spirals or centers; below it ($\tau^2 > 4\Delta$) they are real → nodes or saddles. Stability is decided by the sign of $\tau$, and saddles by $\Delta < 0$ — different boundaries.

Q6: Classify the equilibrium of $\mathbf{x}' = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\mathbf{x}$.

Correct answer: (D). $\tau = 4$, $\Delta = 4$, so $\tau^2 - 4\Delta = 0$: the system sits exactly on the parabola with the repeated eigenvalue $\lambda = 2$. The matrix is defective (only one independent eigenvector, $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$), so the origin is a degenerate node, and $\lambda = 2 > 0$ makes it unstable.