Systems and phase portraits
Chapter 9 - Several states move together
A system tracks a state vector rather than one number:
At every point in the phase plane, the system assigns a velocity vector. A solution is a trajectory tangent to that field.
Lab: change the eigenvalue pattern
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
Starting on an eigenvector line keeps the trajectory on that line. The eigenvalue says how motion along that direction grows, decays, or oscillates.
stable node or spiral sinkNearby trajectories approach the equilibrium.
unstable node or spiral sourceNearby trajectories leave the equilibrium.
saddleOne direction contracts while another expands.
center in the ideal linear caseTrajectories orbit without decay or growth.
Worked example: a saddle
- 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.
The title deliberately invites a mistake. Classification comes from eigenvalue signs, not from the visual appearance of the matrix.
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
becomes
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
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.
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?
- Practice - coupled system
9.2 Find eigenvalues, eigenvectors, and the general solution for \(\mathbf{x}'=\begin{bmatrix}2&1\\1&2\end{bmatrix}\mathbf{x}\).
- Integrated - oscillator bridge
9.3 Convert \(y''+2y'+5y=0\) into a first-order system. Use its eigenvalues to predict the phase portrait.
- 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.
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.