Phase Portraits and Stability Analysis

What a phase portrait shows

A phase portrait draws solution trajectories of a 2D system $\mathbf{x}' = \mathbf{F}(\mathbf{x})$ in the $(x_1, x_2)$-plane. Unlike a plot of $x_1$ or $x_2$ versus $t$, it plots $x_2$ versus $x_1$ directly, with the time direction shown by arrows.

The beauty: you can read off the long-term behaviour of the system — stability, type of equilibrium, oscillation, blow-up — from the geometry of the trajectories. For linear systems $\mathbf{x}' = A \mathbf{x}$, this geometry is entirely determined by the eigenvalues of $A$.

The classification chart (memorise this)

For the 2D linear system $\mathbf{x}' = A \mathbf{x}$ with eigenvalues $\lambda_1, \lambda_2$ of $A$:

Real eigenvalues:

• Both $> 0$ — unstable node (all trajectories exit)
• Both $< 0$ — stable node (all trajectories enter)
• Opposite signs — saddle (two 1-D invariant manifolds; all other trajectories diverge)
• Both equal — degenerate or star node (depends on whether $A$ has a basis of eigenvectors)

Complex conjugate eigenvalues $\lambda = \alpha \pm i \beta$:

• $\alpha > 0$ — unstable spiral (outward)
• $\alpha < 0$ — stable spiral (inward)
• $\alpha = 0$ — centre (closed orbits; neither attracting nor repelling)

Equivalent to reading off the trace ($\tau = \lambda_1 + \lambda_2$) and determinant ($\Delta = \lambda_1 \lambda_2$) of $A$: $\lambda = (\tau \pm \sqrt{\tau^2 - 4\Delta})/2$ gives the same classification with $\Delta > 0$, $\tau < 0$ for stable node / spiral; $\Delta < 0$ for saddle; etc. This is the "$\tau$-$\Delta$ diagram" classification.

The given system

$A = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}.$

This arises from the 2nd-order ODE $y'' + 3 y' + 2 y = 0$ rewritten as the system $x_1 = y$, $x_2 = y'$: $x_1' = x_2 = y', \qquad x_2' = y'' = -3 y' - 2 y = -2 x_1 - 3 x_2.$

So studying the phase portrait of this system is the same as studying the phase portrait of the damped oscillator $y'' + 3 y' + 2 y = 0$.

Step-by-step analysis

Step 1 — Eigenvalues.

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

Factor: $(\lambda + 1)(\lambda + 2) = 0 \implies \lambda_1 = -1, \; \lambda_2 = -2$.

Step 2 — Classify.

Both real, both negative ⇒ stable node (asymptotically stable; all trajectories enter the origin). The origin is an attractor.

Trace $\tau = -3 < 0$, determinant $\Delta = 2 > 0$, and discriminant $\tau^{2} - 4 \Delta = 9 - 8 = 1 > 0$ — two real distinct roots, both negative; classification matches.

Step 3 — Eigenvectors (to draw the portrait).

For $\lambda_1 = -1$: $(A + I) \mathbf{v} = \mathbf{0}$: $\begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix} \mathbf{v} = \mathbf{0} \implies v_1 + v_2 = 0 \implies \mathbf{v}_1 = (1, -1)^{T}.$

For $\lambda_2 = -2$: $(A + 2 I) \mathbf{v} = \mathbf{0}$: $\begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix} \mathbf{v} = \mathbf{0} \implies 2 v_1 + v_2 = 0 \implies \mathbf{v}_2 = (1, -2)^{T}.$

Step 4 — General solution.

$\mathbf{x}(t) = C_1 e^{-t} \begin{pmatrix} 1 \\ -1 \end{pmatrix} + C_2 e^{-2 t} \begin{pmatrix} 1 \\ -2 \end{pmatrix}.$

Reading the portrait

Both exponentials decay, so every trajectory flows toward the origin. The slower mode ($\lambda_1 = -1$) dominates asymptotically — as $t \to \infty$ the $e^{-2t}$ term shrinks faster and trajectories approach the origin tangent to $\mathbf{v}_1 = (1, -1)^{T}$.

Direction of approach. Trajectories align with the slow eigenvector as they approach the fixed point. The fast eigenvector $\mathbf{v}_2$ is the direction along which trajectories leave at $t \to -\infty$.

For our system, trajectories "stream in" toward the origin tangent to the line through $(1, -1)$ — the slow stable manifold. The fast stable manifold $\mathbf{v}_2$ direction $(1, -2)$ matters more at early times, further from the origin.

Nullclines — where one component is zero

Drawing the curves $F_1 = 0$ (where $x_1' = 0$ — horizontal tangents in the phase plane) and $F_2 = 0$ (where $x_2' = 0$ — vertical tangents) gives a coarse skeleton of the flow. For our linear system:

• $x_1' = x_2 = 0$ on the $x_1$-axis (trajectories cross horizontally there).
• $x_2' = -2 x_1 - 3 x_2 = 0$ on $x_2 = -\frac{2}{3} x_1$ (trajectories cross vertically there).

These lines partition the plane into regions where the signs of $x_1'$ and $x_2'$ are fixed, so you can sketch the direction of flow in each region without solving.

Non-linear systems — linearisation at fixed points

For a non-linear system $\mathbf{x}' = \mathbf{F}(\mathbf{x})$ with a fixed point $\mathbf{x}^{*}$ where $\mathbf{F}(\mathbf{x}^{*}) = 0$, near $\mathbf{x}^{*}$ the behaviour is approximated by the linear system $\mathbf{x}' \approx J(\mathbf{x}^{*}) (\mathbf{x} - \mathbf{x}^{*}),$ where $J = \partial \mathbf{F} / \partial \mathbf{x}$ is the Jacobian matrix. So you classify each fixed point by computing the Jacobian there and reading its eigenvalues — "Hartman–Grobman theorem" makes this rigorous when eigenvalues have nonzero real parts.

This is how you draw phase portraits of non-linear systems piece by piece: identify fixed points, linearise each, classify, then sketch global trajectories connecting them.

Physical meaning — damped oscillator

Our system is the 2D phase space of the damped linear oscillator $y'' + 3 y' + 2 y = 0$. The damping coefficient $c = 3$ versus natural frequency $\omega_0 = \sqrt{2}$: we have $c^{2} = 9 > 4 \omega_0^{2} = 8$, so the oscillator is over-damped. The two decay rates $1$ and $2$ correspond to two different exponential decay modes — no oscillation because both eigenvalues are real.

Change the damping: $y'' + y = 0$ is a centre (pure oscillation, eigenvalues $\pm i$). $y'' + 2 y' + 2 y = 0$ is a stable spiral (eigenvalues $-1 \pm i$).

Common mistakes

• Classifying by trace/determinant wrong. Stable requires $\tau < 0$ and $\Delta > 0$. Just $\tau < 0$ isn't enough (could be a saddle).
• Reading direction of spiral wrong. The sign of the imaginary part of the eigenvalue gives the rotational direction through a specific reference, but the overall direction depends on how you plot axes — always sanity-check with a specific trajectory point.
• Using eigenvalues of the wrong matrix for non-linear systems. You need the Jacobian at the fixed point, not the matrix at an arbitrary point.
• Drawing trajectories that cross. For smooth vector fields (linear or non-linear), uniqueness of solutions means trajectories do not intersect (except possibly at fixed points, where they arrive asymptotically).

Try it in the visualization

Place a cursor at any starting point in the phase plane and watch the trajectory unfurl toward the origin. Overlay the two eigenvector directions as dashed lines — see trajectories approach tangent to the slow eigenvector. Slide the coefficients of $A$ to morph between portrait types (node ↔ spiral ↔ saddle).