Review / Part I — Explicit & Qualitative Methods / Ch 9
2D Systems: Node / Spiral / Saddle / Center
Ch 9 — Part I — Explicit & Qualitative Methods
For the 2D linear system \(\mathbf{x}' = A \mathbf{x}\) with \(A \in \mathbb{R}^{2\times 2}\), the geometry of trajectories near the origin is classified by the eigenvalues of \(A\) (equivalently, by \(T = \operatorname{tr}(A)\) and \(D = \det(A)\)).
Trace-determinant summary
The characteristic polynomial is \(\lambda^2 - T\lambda + D\). Let \(\Delta = T^2 - 4D\).
| Condition | Eigenvalues | Portrait |
|---|---|---|
| \(D < 0\) | real, opposite sign | saddle |
| \(D > 0, \Delta > 0, T < 0\) | real, both \(<0\) | stable node |
| \(D > 0, \Delta > 0, T > 0\) | real, both \(>0\) | unstable node |
| \(D > 0, \Delta < 0, T < 0\) | complex, \(\Re<0\) | stable spiral |
| \(D > 0, \Delta < 0, T > 0\) | complex, \(\Re>0\) | unstable spiral |
| \(D > 0, T = 0\) | purely imaginary | center (closed orbits) |
| \(\Delta = 0\) | repeated real | degenerate/improper node |
Rule of thumb
Sign of real part ⇒ stability; complex vs real ⇒ spiral vs node; sign flip on eigenvalues ⇒ saddle.
Practice Problems
Classify the phase portrait of \(A = \begin{pmatrix}-1 & 2 \\ 0 & -3\end{pmatrix}.\)
Hint
\(T = -4, D = 3\). Triangular ⇒ eigenvalues on diagonal.
Solution
Eigenvalues \(-1, -3\), both real negative ⇒ stable node.
Answer: Stable node.
Classify \(A = \begin{pmatrix}0 & -2 \\ 2 & 0\end{pmatrix}.\)
Hint
\(T = 0, D = 4\).
Solution
Purely imaginary eigenvalues \(\pm 2i\): center.
Answer: Center.
Classify \(A = \begin{pmatrix}1 & 4 \\ 1 & 1\end{pmatrix}.\)
Hint
\(T = 2, D = 1 - 4 = -3\).
Solution
\(D < 0\) ⇒ saddle.
Answer: Saddle.
For what values of \(a\) does \(\mathbf{x}' = \begin{pmatrix}a & 1 \\ -1 & a\end{pmatrix}\mathbf{x}\) have a stable spiral?
Hint
Eigenvalues \(a \pm i\).
Solution
Need complex with negative real part: \(a < 0\).
Answer: \(a < 0.\)