Lyapunov function ($V$)

A Lyapunov function near an equilibrium $x^{\ast }$ is continuously differentiable, satisfies $V(x^{\ast })=0$ and $V(x)>0$ away from $x^{\ast }$, and has orbital derivative

The orbital derivative of a differentiable function $V$ along the autonomous system $\dot{x}=f(x)$ isIt is the ordinary derivative of $V(x(t))$ along each trajectory.

A Lyapunov function is positive definite at an equilibrium $x_{\ast }$ when $V(x_{\ast })=0$ and $V(x)>0$ for nearby $x\ne x_{\ast }$. Thus $x_{\ast }$ is a strict local minimum of $V$.

An equilibrium is Lyapunov stable if every neighbourhood contains a smaller neighbourhood whose forward trajectories remain in the original neighbourhood for all time.

Orbital stability asks that a disturbance remain close to an invariant orbit or symmetry family, rather than to one fixed representative. In a complex Landau amplitude equation with $\beta =0$, $\mu >0$, $\lambda >0$, the circle $|A|=\sqrt{\mu /\lambda }$ attracts radial disturbances while every phase on it remains stationary. The family is attracting as a set, but its individual points do not have individual asymptotic stability because their phase is neutral.

A positive-definite Lyapunov function with nonpositive orbital derivative proves Lyapunov stability of the equilibrium. Indeed, fix a sufficiently small ball $B_{\epsilon }$ in the function's domain and putContinuity and $V(0)=0$ give a ball $B_{\delta }$ on which $V<\alpha$. Since $V$ cannot increase along a trajectory, one starting in $B_{\delta }$ cannot cross the sphere $\Vert x\Vert =\epsilon$, on which $V\ge \alpha$. It therefore remains in $B_{\epsilon }$ for all forward time.

A positive-definite Lyapunov function with strictly negative orbital derivative away from the equilibrium proves asymptotic stability.

An equilibrium is asymptotically stable when it is Lyapunov stable and every trajectory starting sufficiently nearby converges to it.

A positive-definite quadratic form defines ellipsoidal sublevel sets and often turns a polynomial vector field into an exact factored orbital derivative.

If a Lyapunov function is nonincreasing on a sublevel set, trajectories cannot cross its boundary outward.

A smooth gradient flow confined to a compact invariant sublevel set has finite total integral of $\Vert \nabla V{\Vert }^2$ when $V$ is bounded below. Compactness supplies uniform continuity of that squared norm along the trajectory, so it tends to zero. Every accumulation point is a critical point. If the trapped component contains exactly one critical point, the trajectory converges to it. This supplies a direct convergence proof in place of an inference solely from an energy-contour picture.

Suppose a compact connected component $C$ of $\{x:V(x)\le c\}$ contains an equilibrium $x_{\ast }$ and $\dot{V}<0$ throughout $C\setminus \{x_{\ast }\}$. Then $C$ is an invariant sublevel set, and every trajectory in $C$ converges to $x_{\ast }$ by the LaSalle invariance principle. Thus the interior of $C$ lies in the basin of attraction of $x_{\ast }$. The largest such component supplies the strongest basin estimate available from that Lyapunov function alone.

For $\dot{z}=Az+\Vert z{\Vert }^{2}z$, suppose a positive-definite matrix $P$ solves $A^{T}P+PA=-2I$. The quadratic function $V=z^{T}Pz$ then obeys $\dot{V}=2\Vert z{\Vert }^2(V-1)$. Its unit ellipsoid separates decay toward the origin from escape. In a planar system with nonzero angular speed it is an invariant periodic orbit.

A boundary point with $\dot{V}=0$ can still enter the sublevel set; LaSalle analysis decides whether it belongs to an invariant zero-derivative trajectory.

On a compact positively invariant set where $\dot{V}\le 0$, every trajectory approaches the largest invariant subset of $\{\dot{V}=0\}$.

For $\ddot{x}=-U^{\prime }(x)-\mu \dot{x}$ with $\mu >0$,If $U$ is radially unbounded, every trajectory is bounded, and LaSalle's principle reduces its omega-limit set to invariant points with zero velocity.

Forthe mechanical energy isThe origin is a center for $k=0$ and an asymptotically stable equilibrium for $k>0$, while $(\pm 1,0)$ are saddles. For positive damping, their stable manifolds form the boundary between the basin of the origin and the two escape regions.

For $k>0$, the trajectory beginning at $(x,y)=(1,y_0)$ with $y_0>0$ remains in $x>1$, $y>0$ and obeysIt enters every strip $0<y<\epsilon$: otherwise $\dot{E}=-k{y}^2\le -k{\epsilon }^2$ would eventually make the nonnegative energy negative.