Systems and phase portraits

Chapter 9 - Several states move together

A system tracks a state vector, a list of state components, rather than one number:

$\mathbf{x}'=A\mathbf{x},\qquad \mathbf{x}=\begin{bmatrix}x\y\end{bmatrix}.$

At every point in the phase plane, whose axes are the two state components, the system assigns a velocity vector. A solution follows a trajectory, a path tangent to that field.

Lab: change the eigenvalue pattern

Linear phase-plane explorerClick to launch another trajectory.

System type

velocity fieldtrajectoriesequilibrium

Ask three questions for each preset:

Do trajectories approach or leave the origin?
Do they rotate?
Are any straight-line directions preserved?

Eigenvectors are invariant directions

If $A\mathbf{v}=\lambda\mathbf{v}$ , try a solution $\mathbf{x}=e^{\lambda t}\mathbf{v}$ . Then

$\mathbf{x}'=\lambda e^{\lambda t}\mathbf{v} =e^{\lambda t}A\mathbf{v} =A\mathbf{x}.$

Starting on a real eigenvector line keeps the trajectory on that line. Its real eigenvalue says whether motion grows, decays, or stays constant. A complex conjugate pair describes rotating motion in the real plane, not two real eigenvector lines.

Both real parts negativestable node or spiral sink

Nearby trajectories approach the equilibrium.

Both real parts positiveunstable node or spiral source

Nearby trajectories leave the equilibrium.

Opposite signssaddle

One direction contracts while another expands.

Pure imaginary paircenter in the ideal linear case

Trajectories orbit without decay or growth.

Worked example: an unstable node

Analyze $x'=3x+y,\ y'=x+3y$ eigen-directions

Matrix. $A=\begin{bmatrix}3&1\\1&3\end{bmatrix}$ .
Eigenvalues. $\det(A-\lambda I)=(3-\lambda)^2-1=0$ , so $\lambda=4,2$ .
Eigenvectors. For $\lambda=4$ , one eigenvector is $[1,1]^T$ . For $\lambda=2$ , one is $[1,-1]^T$ .
Build the family. $\mathbf{x}=C_1e^{4t}[1,1]^T+C_2e^{2t}[1,-1]^T$ .
Classify. Both eigenvalues are positive, so the origin is an unstable node, not a saddle. The $\lambda=4$ direction dominates for most initial states as $t\to\infty$ .
Sketch. Draw both eigenvector lines first. Other trajectories leave the origin and bend toward the faster $[1,1]^T$ direction.

Classification comes from eigenvalue signs, not from the visual appearance of the matrix. A saddle would require eigenvalues with opposite signs.

Backward-faded system

For $x'=x,\ y'=-2y$ :

matrix: $A=\underline{\hspace{7em}}$ ;
eigenpairs: $(\lambda_1,\mathbf v_1)=\underline{\hspace{7em}}$ , $(\lambda_2,\mathbf v_2)=\underline{\hspace{7em}}$ ;
general solution: $x=\underline{\hspace{4em}}$ , $y=\underline{\hspace{4em}}$ ;
classification: $\underline{\hspace{5em}}$ .

Check the portrait

$A=\operatorname{diag}(1,-2)$ . Eigenpairs are $(1,[1,0]^T)$ and $(-2,[0,1]^T)$ . Thus $x=C_1e^t$ , $y=C_2e^{-2t}$ . One direction grows and one decays, so the origin is a saddle.

Second-order equations are systems in disguise

Let $x_1=y$ and $x_2=y'$ . Then

$y''+ay'+by=0$

becomes

$\begin{bmatrix}x_1'\x_2'\end{bmatrix} = \begin{bmatrix}0&1\-b&-a\end{bmatrix} \begin{bmatrix}x_1\x_2\end{bmatrix}.$

The characteristic polynomial of this matrix is the same $\lambda^2+a\lambda+b$ from Chapter 7. The time trace and phase portrait are two views of one solution.

Classification. A 2×2 system has eigenvalues $-1$ and $3$ . What is the origin?

Error clinic

Error clinic: drawing component graphs in the phase plane

The phase plane plots $y(t)$ against $x(t)$ ; time is implicit along the trajectory. A time-series plot uses $t$ on the horizontal axis. Label axes before interpreting a curve.

Error clinic: classifying from trace alone

A negative trace does not guarantee stability when the determinant is negative. Opposite-sign eigenvalues always produce a saddle. Compute enough information to determine both eigenvalue signs or real parts.

Practice

Practice - diagonal system
9.1 Solve and sketch $x'=-x,\ y'=-3y$ . Which eigen-direction dominates late?
Detailed solution
Practice - coupled system
9.2 Find eigenvalues, eigenvectors, and the general solution for $\mathbf{x}'=\begin{bmatrix}2&1\\1&2\end{bmatrix}\mathbf{x}$ .
Detailed solution
Integrated - oscillator bridge
9.3 Convert $y''+2y'+5y=0$ into a first-order system. Use its eigenvalues to predict the phase portrait.
Detailed solution
Challenge - trace and determinant
9.4 For a real 2×2 matrix, eigenvalues satisfy $\lambda^2-\tau\lambda+\Delta=0$ , where $\tau$ is trace and $\Delta$ determinant. Build a qualitative classification map in the $(\tau,\Delta)$ -plane.
Detailed solution
Practice - extension
9.5 Solve $x'=-x+y,\ y'=-y$ , with $x(0)=0,\ y(0)=1$ , by finding $y$ first. Compute the matrix's eigenvalues and real eigenvector directions. Explain why one eigenvector mode alone cannot satisfy these initial conditions, and classify the origin.
Detailed solution
Challenge - transfer
9.6 For $x'=-4y,\ y'=x$ , with $x(0)=2,\ y(0)=0$ , find the solution and prove that $x^2+4y^2$ is conserved. Determine the orbit, rotation direction, and period. Is the origin attracting?
Detailed solution

Retrieve the representation bridge

Choose one oscillator setting from Chapter 7. Describe the same motion in four languages: roots, time trace, phase portrait, and physical behavior.

I clicked new initial states, marked eigen-directions first, and translated between time and phase views.

Next: Approximate a solution step by step and learn to see numerical error.

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Differential Equations as a Second Language