example of an unstable fixed point for which the linearized dynamics are stable

Question

What would be an example of an unstable fixed point for which the linearized dynamics are stable?

Thanks in advance

Answer 1

It depends on what notion of stability you are using. Without loss of generality (by translating the coordinate system) we can assume our fixed point is the origin. Since you didn't indicate otherwise I'll also assume that the coefficients of your dynamical system is time independent.

Asymptotic stability

If we take the notion of stability to mean that for each initial data $f$, there exists a number $\epsilon > 0$ such that the dynamics with initial data $\epsilon f$ leads to solution $\lim_{t\to+\infty} x(t) = 0$. Then we have that

$$ \text{Linear stability } \implies \text{ nonlinear stability} $$

and the example you seek does not exist. This is due to the well-known fact that a linear system $\dot{x} = Ax$ is asymptotically stable if and only if all the eigenvalues of $A$ have strictly negative real parts. Thus for small data, the linear dynamics strictly dominates the nonlinear one, which leads to the Hartman-Grobman theorem (which can also be applied to more general hyperbolic fixed points).

Orbital stability

If, however, you are thinking in terms of orbital stability, then a simple example would be the dynamical system on $\mathbb{R}$ given by

$$ \dot{x} = x^3 $$

We have that $x(t) = 0$ is a fixed point. Its linearised dynamics is $\dot{x} = 0$, hence is trivially orbitally stable. (Recall that a fixed point $\bar{x}$ of a dynamical system is orbitally stable if for every $\epsilon > 0$ we can find a $\delta > 0$ such that if the initial data $|x_0 - \bar{x}| < \delta$ we have that the solution $|x(t) - \bar{x}| < \epsilon$ for all positive times $t$.) But solving the nonlinear ODE we get

$$ x(t) = \frac{x_0}{\sqrt{ 1 - 2 x_0^2 t}} $$

which blows-up in finite time for any non-zero initial data, and hence the nonlinear system is not orbitally stable.

Answer 2

A similar example to the one given by Willie Wong (but maybe less boring): perturbation of a center $$\frac{d\vec x}{dt} =\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}\vec x \tag1$$ which is stable but not asymptotically stable. To (1) we should add a nonlinear term that pushes solutions away from the origin. A linear term of this kind would be $f(\vec x)=\vec x$. To make it vanish to higher order at the origin, use something like $f(\vec x)=|\vec x|^2\vec x$ (squaring is optional, used only to avoid the square root). So, $$\frac{d\vec x}{dt} =\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}\vec x+ |\vec x|^2\vec x \tag2$$ is the desired example.

To prove that it is indeed unstable, observe that $$\frac{d}{dt}|\vec x|^2=2\left\langle \frac{d}{dt}\vec x,\vec x\right\rangle = 2|\vec x|^4$$