Suppose I have a pair of 2 non-linear differential equations of the form: $$\begin{matrix} \frac{dy}{dt}=f(x,y)\\ \frac{dx}{dt}=g(x,y) \end{matrix}$$ Equilibrium points are where the trajectory ends up on, when plotted on the $x-y$ plane.

What are the qualitative differences between a 'stable node' and a 'stable spiral'?

Are they both stable?


2 Answers 2


Yes both stable nodes and stable spirals are stable equilibria (as indicated by their names). And they are qualitatively very similar to one another.

The only real difference between the two is that solution trajectories in the $x-y$ phase plane for a stable spiral tend to spiral around the equilibrium before they are "sucked" into it. This is analogous to water spiralling around a sink hole as it flows into it.

On the other hand a stable focus just "sucks" solution trajectories directly into it - you could think of this as water flowing into a sink but without any (or relatively little) rotation.

To illustrate the point I've added phase plane portraits below showing examples of both equilibrium types. These plots have come from solving two simple linear ODE systems. Each system of ODEs has an equilibrium at $(x,y)=(0,0)$ and as you can see the solution trajectories on the LHS plot spiral into their equilibrium point. Whilst on the RHS plot the trajectories flow directly into their equilibrium point.

enter image description here


If we have two coupled nonlinear ODEs

$$\dot{x}_1 = f_1 (x_1, x_2) \qquad \qquad \dot{x}_2 = f_2 (x_1, x_2)$$

and the origin is an equilibrium point, i.e., $f_1 (0,0) = f_2 (0,0) = 0$, then let the linearization of vector field $f$ near the origin be

$$\dot{\mathrm{x}} = \mathrm{A} \mathrm{x}$$

Let $\lambda_1, \lambda_2$ be the eigenvalues of $\mathrm{A}$. If the eigenvalues of $\mathrm{A}$ have negative real parts, $\mbox{Re} (\lambda_i) < 0$, then the origin is a stable equilibrium point. If the eigenvalues have negative real parts and

  • zero imaginary parts, $\mbox{Im} (\lambda_i) = 0$, then the origin is a stable node.
  • nonzero imaginary parts, $\mbox{Im} (\lambda_i) \neq 0$, then the origin is a stable spiral.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .