Stability from the linearisation · Differential Equations

Intuition

The linearisation decides the local picture whenever it can. If every eigenvalue of the Jacobian has negative real part the equilibrium is asymptotically stable, and if one has positive real part it is unstable, whatever the neglected terms do; saddles stay saddles, nodes stay nodes and spirals stay spirals. The one case it cannot decide is the borderline: purely imaginary eigenvalues, a linear centre, which the smallest nonlinear term can turn into a spiral winding either way.

A barometer forecast. When the pressure is falling fast, the forecast is reliable whatever the details; when it holds exactly on the boundary, the details decide, and the barometer cannot.

plane diagram: a-linear-centre-that-winds-in

$x'=-y+ax(x^{2}+y^{2})$ , $y'=x+ay(x^{2}+y^{2})$ with $a=-0.1$ . Its Jacobian at the origin is a centre, yet every solution winds slowly in: $r'=ar^{3}<0$ .

plane diagram: the-same-centre-winding-out

The same system with $a=0.1$ : the same Jacobian, the same linear centre, and every solution winds out. The linearisation cannot tell these two apart.

When the linearisation decides

Let $J$ be the Jacobian at an equilibrium of a system with continuous second derivatives. If every eigenvalue of $J$ has negative real part, the equilibrium is asymptotically stable; if some eigenvalue has positive real part, it is unstable. A saddle of $J$ is a saddle of the system, and nodes and spirals keep their type. When the eigenvalues are purely imaginary the linearisation decides nothing: the equilibrium may be a centre, a stable spiral or an unstable spiral. These are the theorems of Lyapunov and of Hartman and Grobman, used here without proof.

\begin{aligned}&\operatorname{Re}\lambda<0\ \text{for every eigenvalue}\\&\Longrightarrow\ \text{asymptotically stable}\end{aligned}

Deciding stability

Pendulum with friction, $\theta'=\omega$ , $\omega'=-\sin\theta-c\omega$ with : at the Jacobian has trace and determinant , so the rest position is asymptotically stable.

A linear centre decides nothing

\begin{aligned}&x'=-y+ax(x^{2}+y^{2}),\ y'=x+ay(x^{2}+y^{2})\\&\Longrightarrow\ J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\ \text{and}\ r'=ar^{3}\end{aligned}

The added terms are cubic, so their first derivatives vanish at the origin and every system of the family has the same Jacobian, a centre. Differentiate the squared distance from the origin: the rotation cancels, and the cubic terms leave a times r to the fourth. So the distance shrinks when a is negative and grows when a is positive — a stable and an unstable spiral with one and the same linearisation.

Proof steps

J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\ \text{for every}\ a

The cubic terms have zero first derivatives at the origin.

rr'=xx'+yy'

Differentiate r² = x² + y² and halve it.

rr'=a\bigl(x^{2}+y^{2}\bigr)^{2}=ar^{4}

The rotation terms cancel; the cubic ones remain.

r'=ar^{3}

So the distance falls for negative a and grows for positive a, while the linearisation is the same centre.

Applications

Practice

Real Parts Decide

All eigenvalues of the Jacobian with negative real part: asymptotically stable. One with positive real part: unstable. Purely imaginary: undecided.

\lambda=\pm i\beta\ \Longrightarrow\ \text{no conclusion}

Try it

The Jacobian at an equilibrium has eigenvalues $-1\pm2i$ . What is the equilibrium of the nonlinear system?