Lyapunov Stability, Invariant Sets & Regions of Attraction

For $\dot x=f(x)$ with equilibrium shifted to $x=0$, a Lyapunov function is an energy-like scalar whose value constrains motion. Instead of integrating every initial condition, show that trajectories cannot climb its level sets and, under stronger conditions, must descend toward the equilibrium.

A decreasing $V$ makes a sublevel set invariant when the derivative condition holds on its boundary. The largest certified set need not equal the true basin.

Stability notions #

An equilibrium is:

• stable in the sense of Lyapunov if every sufficiently small initial displacement remains small for all future time;
• asymptotically stable if it is stable and nearby trajectories converge to it;
• exponentially stable if convergence is bounded by an exponential rate;
• globally asymptotically stable if all initial states in the state space converge, subject to existence of solutions.

Attraction without stability is possible in pathological systems, so “trajectories seem to return” is not the full definition.

Direct method #

Let $V(x)$ be continuously differentiable in a neighborhood $D$ of the origin. If

$$V(0)=0,\qquad V(x)>0\quad(x\ne0),$$

and

$$\dot V(x)=\nabla V(x)^Tf(x)\le0,$$

then the origin is stable. If $\dot V(x)<0$ for every nonzero $x$ in $D$, it is locally asymptotically stable. Bounds of the form

$$c_1\|x\|^2\le V(x)\le c_2\|x\|^2,qquad \dot V\le-c_3\|x\|^2$$

imply local exponential stability.

For a global claim, the inequalities must hold globally and $V$ normally must be radially unbounded—$V(x)\to\infty$ as $\|x\|\to\infty$—or another argument must establish bounded trajectories.

Semidefinite derivative and LaSalle #

If $\dot V\le0$ but equals zero away from the equilibrium, strict descent is unavailable. LaSalle's invariance principle says bounded trajectories approach the largest invariant subset of

$$\{x:\dot V(x)=0\}.$$

For the damped pendulum, mechanical energy has

$$\dot E=-b\dot\theta^2\le0.$$

The set $\dot E=0$ contains every state with instantaneous zero velocity, but most immediately accelerate and leave. The invariant subset determines the actual limit behavior. Replacing this reasoning with “$\dot V\le0$, therefore asymptotically stable” is a common error.

Worked one-dimensional basin #

Consider

$$\dot x=-x+x^3.$$

Equilibria are $x=0,\pm1$. Choose

$$V(x)=\frac12x^2.$$

Then

$$\dot V=x(-x+x^3)=-x^2+x^4=-x^2(1-x^2).$$

Thus $\dot V<0$ for $0<|x|<1$ and $\dot V>0$ for $|x|>1$. The sublevel set $V<1/2$, equivalent to $|x|<1$, lies inside the region of attraction of the origin; here it is exact. The same quadratic candidate does not support a global claim.

This example shows why the proof region belongs in the result.

Linear Lyapunov equation #

For $\dot x=Ax$, if $A$ is Hurwitz, then for any $Q=Q^T>0$ there is a unique $P=P^T>0$ solving

$$A^TP+PA=-Q.$$

Then $V=x^TPx$ has

$$\dot V=-x^TQx<0.$$

Conversely, existence of such $P$ characterizes Hurwitz stability. This converts a pole condition into an energy certificate and generalizes naturally to uncertainty, switching, and nonlinear local analysis.

Numerically, solve the Lyapunov equation with a stable algorithm; do not form Kronecker inverses casually for ill-conditioned large systems. Check symmetry, positive definiteness, residual $A^TP+PA+Q$, and scaling.

Certified invariant sets and regions of attraction #

A sublevel set

$$\Omega_\rho=\{x:V(x)\le\rho\}$$

is positively invariant if trajectories on its boundary point inward or tangent, for example if $\dot V\le0$ throughout it. If the origin is the only invariant set with $\dot V=0$, trajectories in $\Omega_\rho$ converge to it.

To estimate a region of attraction:

1. choose a candidate $V$ from physical energy, a linearized Lyapunov equation, or structured search;
2. find the largest $\rho$ for which the set respects state constraints and $\dot V<0$ (or a valid LaSalle condition);
3. prove the inequality over the continuous set, not just sampled grid points;
4. compare with simulation from boundary-adjacent initial conditions as a falsification check.

The certified set is an inner approximation. Failure to certify outside it is not proof of instability.

Control Lyapunov functions #

For $\dot x=f(x)+g(x)u$, a control Lyapunov function requires that for each nonzero state there exists an admissible $u$ making $\dot V<0$. Constructing a feedback law also requires continuity, actuator bounds, and robustness. An unconstrained $u$ that makes the derivative negative may be physically impossible.

Similarly, barrier functions certify avoidance rather than convergence by preventing trajectories from crossing a boundary. Stability and safety certificates can conflict when the stabilizing action violates a barrier or input limit; controller synthesis must satisfy both.

Common Lyapunov functions and uncertainty #

If one $V$ decreases for every model in an uncertainty family, it certifies robust stability for that family. The condition is sufficient and can be conservative: stable members may lack a common quadratic Lyapunov function. For switched systems, a common function can certify arbitrary switching; mode-dependent functions require explicit switching/dwell conditions.

Proof discipline #

State every Lyapunov claim with:

• the exact closed-loop dynamics and equilibrium;
• the domain and differentiability assumptions;
• positive-definiteness and radial-unboundedness conditions actually proved;
• the expression for $\dot V$ and where its sign holds;
• any LaSalle invariant-set argument;
• state/input constraints and resulting certified sublevel set;
• whether the result is local, regional, global, asymptotic, or exponential.

Lyapunov analysis is powerful because it replaces trajectory enumeration with a scalar certificate. Omitting the certificate's domain turns that power into overclaiming.