Exercise · ZW-STABILITY-OF-AN-EQUILIBRIUM-009 · ★★★ Advanced
The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}0&1\\-1&-1\end{pmatrix}\) is:
Choose one option.
Show answer
- C. Asymptotically stable
Explanation
\(\lambda^2+\lambda+1=0\) gives \(\lambda=\tfrac{-1\pm i\sqrt3}{2}\), with real part \(-\tfrac12<0\). A zero on the diagonal does not matter: what counts is the real part of the eigenvalues.
Review the concept: Stability of an equilibrium
An equilibrium \(\mathbf x^*\) is stable if, for every tolerance \(\varepsilon\), there is a radius \(\delta\) such that solutions starting within \(\delta\) of it never move farther than \(\varepsilon\).
\[\|\mathbf x(0)-\mathbf x^*\|<\delta\;\Longrightarrow\;\|\mathbf x(t)-\mathbf x^*\|<\varepsilon\quad\text{for all }t\ge0\]
More practice problems: stability of an equilibrium
- ZW-STABILITY-OF-AN-EQUILIBRIUM-001 The origin of mathbf x'=Amathbf x with A=[-1 0; 0 -3] is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-002 The origin of mathbf x'=Amathbf x with A=[2 1; 0 -1] is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-003 The origin of mathbf x'=Amathbf x with A=[0 1; -1 0] is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-004 The origin of mathbf x'=Amathbf x with A=[-1 2; -2 -1] is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-005 The origin of mathbf x'=Amathbf x with A=[1 -2; 2 1] is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-006 A 2×2 matrix A has trace -3 and determinant 2. The origin of mathbf x'=Amathbf x is:
- ZW-STABILITY-OF-AN-EQUILIBRIUM-007 In y'=y(1-y), which equilibrium is asymptotically stable?
- ZW-STABILITY-OF-AN-EQUILIBRIUM-008 The system x'=ax,\ y'=-y has an asymptotically stable origin exactly when:
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