Understanding the qualitative behavior of 2D linear systems through geometric visualization
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
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:
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.
The origin is:
Key insight: For 2×2linear systems, the stability of the origin is completely determined by the eigenvalues of $A$.
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:
| 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 $\lambda < 0$ |
For a 2×2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, define:
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.
Step 1: Find eigenvalues
Factoring: $(\lambda + 2)(\lambda + 3) = 0$, so $\lambda_1 = -3, \lambda_2 = -2$.
Step 2: Check trace and determinant
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.
Step 1: Find eigenvalues
So $\lambda = \pm i$, purely imaginary with $\beta = 1$.
Step 2: Check trace and determinant
Point $(0, 1)$ on the boundary $\tau = 0$, $\Delta > 0$ → Center
Step 3: Solution structure
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.
Step 1: Find eigenvalues
So $\alpha = -1 < 0$ and $\beta = 2 \neq 0$.
Step 2: Check trace and determinant
$\tau < 0$, $\Delta > 0$, $\tau^2 - 4\Delta = 4 - 20 = -16 < 0$ (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.
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.
Classify the equilibrium of the system $\mathbf{x}' = \begin{pmatrix} 1 & -3 \\ 3 & 1 \end{pmatrix}\mathbf{x}$ and describe the phase portrait behavior.
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.
Statement: "A system with $\text{tr}(A) = 0$ and $\det(A) > 0$ must have a center at the origin."