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

For the Liénard system $\dot{x}=z-F(x)$, $\dot{z}=-g(x)$ with $F=x^3/3-2x$, $g=x^3-x$, and $V^{\prime }=g$, the displayed function is coercive up to a constant. Differentiation gives $\dot{W}=-gF-z^2+Fz+xg\le -z^2/2-5x^6/18+8x^4/3-x^2$. Its derivative is negative outside a compact set, providing bounded forward dynamics even though the original oscillator has negative damping near the origin. The coordinate $z=y+F(x)$ exposes this restoring and damping balance.

A Lyapunov functional assigns a scalar to the entire state of an evolution equation and is monotone along its trajectories. For a gradient flow $A_T=\delta V/\delta \overline{A}$, the identity $dV/dT=2\int |A_T{|}^2$ follows from the variational chain rule, with boundary terms removed by periodicity or decay. A spatial density is not itself a Lyapunov functional unless all remaining coordinates are integrated or averaged.

For the conserved-mean convection amplitude equations before amplitude reduction, the functional $V=⟨\mu |\nabla T{|}^2/2-(\Delta T{)}^2/2-|\nabla T{|}^4/4⟩$ generates a gradient flow with increasing $V$. Periodic integration by parts identifies its variational gradient with the evolution equation. Completing the square gives $V={\mu }^2/4-⟨(|\nabla T{|}^2-\mu {)}^2/4+(\Delta T{)}^2/2⟩\le {\mu }^2/4$. Hence ${\int }_0^{\infty }⟨T_t^2⟩dt<\infty$ along a globally smooth solution. Under compactness of the orbit, its limiting states are stationary; a nonstationary periodic orbit is impossible.

A nonconstant smooth periodic roll obeys $\mu ⟨T_0^{\prime 2}⟩-⟨T_0^{\prime \prime 2}⟩-⟨T_0^{\prime 4}⟩=0$. The orthogonal-roll perturbation raises the increasing-gradient functional for poorly conducting convection by the displayed amount. The variance is strictly positive: a nonconstant periodic function has a derivative that vanishes somewhere and is nonzero somewhere. The Hessian of $V$ therefore has a positive direction. Its self-adjoint linearized evolution operator has a positive Rayleigh quotient and hence a positive eigenvalue, proving transverse linear instability wherever a nonconstant roll exists.

For $\dot{x}=x(1-y)$ and $\dot{y}=x^2-y$, on either half-plane $x\ne 0$ the Lyapunov function $L$ is nonnegative, has compact sublevel sets contained in that half-plane, and satisfies $\dot{L}=-2(y-1{)}^2$. The only invariant points of $\{y=1\}$ are $(1,1)$ and $(-1,1)$. The invariance argument therefore makes each half-plane converge to its corresponding stable focus. The stable set of the saddle at $(0,0)$ is exactly the line $x=0$.

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 $V$.

Near an equilibrium, a positive-definite scalar function has value zero at that point and is strictly positive at every other nearby point. This scalar notion differs from positive definiteness of a kernel.

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 periodic orbit has orbital asymptotic stability if nearby trajectories remain close and their distance to the orbit tends to zero. A persistent phase shift along the orbit is allowed.

For $\dot{r}=r(a-g(r))$, $\dot{\theta }=1$, a positive root $a=g(r_0)$ gives a circular periodic orbit of period $2\pi$. The radial exponent is $-r_{0}g\prime (r_0)$, so $g\prime (r_0)>0$ gives orbital asymptotic stability and $g\prime (r_0)<0$ gives instability. The phase direction is neutral; a periodic orbit is not an asymptotically stable individual point of the flow.

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.

An equilibrium has exponential stability when sufficiently small displacements satisfy an estimate $\Vert x(t)-x_{\ast }\Vert \le C{e}^{-\alpha t}\Vert x(0)-x_{\ast }\Vert$ for some $C,\alpha >0$.

An equilibrium point is quasi-asymptotically stable when some neighborhood of initial conditions has forward solutions converging to it. This attraction condition alone does not include Lyapunov stability. Asymptotic stability requires both attraction and Lyapunov stability. A positive Lyapunov function with nonpositive derivative and no nonstationary invariant trajectory in its zero-derivative set proves both when its sufficiently small sublevel sets are compact and forward invariant.

For $\dot{e}=Ae$, differentiation gives $d\Vert e{\Vert }^2/dt=e^T(A+A^T)e$. A uniform negative-definiteness bound with $\kappa >0$ yields exponential contraction by direct integration of this differential inequality. This is a sufficient criterion for asymptotic stability, stronger than requiring all eigenvalues of $A$ to have negative real parts. A merely semidefinite symmetric part needs additional analysis.

A Hurwitz stable matrix has all eigenvalues strictly in the open left half-plane. For $\dot{x}=Ax$, this is equivalent to $e^{At}x\to 0$ for every initial vector. Jordan blocks contribute polynomial factors times $e^{\lambda t}$, which decay exactly when their eigenvalues have negative real part. Translating an affine equilibrium reduces its attractivity to this same condition.

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.