Systems and stability · Definition

Stability of an equilibrium

differential equationssystemsstability

Before you start you need to know

It answers the practical question an equilibrium leaves open: if something disturbs it slightly, does the system return, stay close, or move away?

Definition

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 \(\varepsilon\).

\[\|\mathbf x(0)-\mathbf x^*\|<\delta\;\Longrightarrow\;\|\mathbf x(t)-\mathbf x^*\|<\varepsilon\quad\text{for all }t\ge0\]

There are three behaviors. An equilibrium is unstable if some nearby solution ends up far away. It is stable if nearby ones stay close. And it is asymptotically stable if, in addition, they tend to it as \(t\to\infty\); this last one is what you expect from a pendulum with friction at its lowest position.

For \(\mathbf x'=A\mathbf x\) the origin is decided by the eigenvalues \(\lambda\) of \(A\). If all have negative real part, it is asymptotically stable; if any has positive real part, it is unstable. If some real parts are zero and none is positive, it is stable but not asymptotically (eigenvalues \(\pm i\beta\) give centers, with closed trajectories). In dimension two, trace and determinant suffice: both eigenvalues have negative real part if and only if the trace is negative and the determinant positive.

Guiding example: in \(y'=y(1-y)\), the derivative \(f'(y^*)\) decides. At \(y^*=0\) it equals \(1>0\) and solutions move away; at \(y^*=1\) it equals \(-1<0\) and solutions approach. In a nonlinear system the same idea applies to the Jacobian matrix at the equilibrium, and the result holds when no eigenvalue has zero real part.

Typical mistake: thinking stable means "tends to the equilibrium". A center is stable without solutions approaching. This concept closes the course's path: it relates eigenvalues, phase portrait and the long-term behavior of a model.

Think about it

Can the origin of \(\mathbf x'=A\mathbf x\) be asymptotically stable if \(A\) is \(2\times2\) and \(\det A<0\)?

No. If \(\det A<0\), the two eigenvalues are real with opposite signs, and the positive one produces solutions that move away: it is a saddle, unstable.

Exercises 10

  1. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}-1&0\\0&-3\end{pmatrix}\) is:
  2. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}2&1\\0&-1\end{pmatrix}\) is:
  3. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) is:
  4. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\) is:
  5. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}1&-2\\2&1\end{pmatrix}\) is:
  6. A \(2\times2\) matrix \(A\) has trace \(-3\) and determinant \(2\). The origin of \(\mathbf x'=A\mathbf x\) is:
  7. In \(y'=y(1-y)\), which equilibrium is asymptotically stable?
  8. The system \(x'=ax,\ y'=-y\) has an asymptotically stable origin exactly when:
  9. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}0&1\\-1&-1\end{pmatrix}\) is:
  10. The origin of \(\mathbf x'=A\mathbf x\) with \(A=\begin{pmatrix}0&0\\0&-2\end{pmatrix}\) is:

References

  • William E. Boyce, Richard C. DiPrima, Douglas B. Meade, Elementary Differential Equations and Boundary Value Problems, 12th edition, Wiley, 2021. §9.2
  • Dennis G. Zill, A First Course in Differential Equations with Modeling Applications, 11th edition, Cengage, 2018. §10.2
  • Steven H. Strogatz, Nonlinear Dynamics and Chaos, 2nd edition, Westview Press, 2015. §5.2