Critical point of autonomous system

A critical point of an Autonomous system ${\mathbf{x}}^{\prime }=\mathbf{F}(\mathbf{x})$ is a point ${\mathbf{x}}_0\in {\mathbb{R}}^n$ where $\mathbf{F}({\mathbf{x}}_0)=0$. The right-hand side vanishes, so a solution starting at ${\mathbf{x}}_0$ stays there forever.

Also called: equilibrium, fixed point, or steady state depending on context. In physics they’re “rest positions”; in linear-algebra contexts they’re “kernels” of $\mathbf{F}$.

For linear systems

For ${\mathbf{x}}^{\prime }=A\mathbf{x}$ (homogeneous linear autonomous), the critical points are exactly the solutions of $A\mathbf{x}=0$, i.e. the kernel of $A$.

Two cases:

• $\det A\ne 0$: kernel is just $\{0\}$. The origin is the unique critical point, isolated.
• $\det A=0$: kernel is a subspace of dimension $\ge 1$. There are infinitely many critical points (a line, a plane, etc.), none isolated.

For the affine case ${\mathbf{x}}^{\prime }=A\mathbf{x}+\mathbf{b}$, the critical points satisfy $A\mathbf{x}=-\mathbf{b}$. If $\det A\ne 0$, unique solution ${\mathbf{x}}^{\ast }=-A^{-1}\mathbf{b}$, isolated.

Isolated vs non-isolated

A critical point ${\mathbf{x}}_0$ is isolated if there’s a circle (ball, in higher dimensions) around it containing no other critical points.

Why this matters:

• Isolated critical points can be analyzed locally — eigenvalue methods, Lyapunov methods, all work.
• Non-isolated critical points (lines or higher-dimensional sets of equilibria) require special handling. Standard stability analysis doesn’t directly apply because perturbations don’t return to a single point; they slide along the equilibrium set.

Lemma: if an autonomous system has only finitely many critical points, each is isolated.

(If you have infinitely many critical points, by compactness they accumulate, so some are not isolated.)

Examples

Linear, $\det A\ne 0$: ${\mathbf{x}}^{\prime }=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\mathbf{x}$. Origin is the unique critical point, isolated. Saddle.

Linear, $\det A=0$: ${\mathbf{x}}^{\prime }=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\mathbf{x}$. Critical points satisfy $x+y=0$, the line $y=-x$. Every point on this line is a critical point; none isolated.

Nonlinear: $\{\begin{array}{l}{x}^{\prime }=x-x^2-xy\\ y^{\prime }=\frac{1}{2}y-\frac{3}{4}xy-\frac{1}{4}{y}^2\end{array}$

Setting $x^{\prime }=y^{\prime }=0$: from $x^{\prime }=0$, $x(1-x-y)=0$, so $x=0$ or $x+y=1$. Combined with $y^{\prime }=0$, you get specific isolated points — typically four for this kind of competing-species model.

Stability of critical points

For an isolated critical point ${\mathbf{x}}_0$, stability is the main question in qualitative ODE analysis: Stability of autonomous systems has the rigorous definitions, Phase plane behaviour the 2D classification.

For nonlinear systems, you typically:

1. Find all critical points by solving $\mathbf{F}(\mathbf{x})=0$.
2. Linearize around each critical point (compute the Jacobian).
3. Use the eigenvalues of the Jacobian to determine local stability (see Locally linear system).

Why “critical”

The term comes from the geometric interpretation: trajectories of the system flow according to the vector field $\mathbf{F}$. Where $\mathbf{F}=0$, the flow is “static”: solutions don’t move. These are the “critical” points in the sense of being singular or special points of the flow.

In the phase plane equation $dy/dx=g(x,y)/f(x,y)$, both numerator and denominator vanish at critical points, making the slope indeterminate. So critical points are also where the direction of the vector field is undefined — the singularities of the slope field.