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The direct First Lyapunov theorem says that a continuously differentiable positive-definite function $V$ near an equilibrium, with $\dot{V}\le 0$, proves stability. LaSalle invariance principle says that in a compact positively invariant set, trajectories approach the largest invariant subset of $\{\dot{V}=0\}$. Under the terminology in which “first theorem” means the linearization theorem, eigenvalues with negative real part imply asymptotic stability, while zero-real-part eigenvalues leave the test inconclusive. Here the linearization has eigenvalues $0,-k$, so the direct criterion is needed.

Write $y=\dot{x}$ and chooseNear zero, $V=y^2/2+x^4/4+O(x^6)$ is positive definite. Along solutions,The component containing the origin of a sufficiently small closed sublevel set is compact, lies in $|x|<\pi$ and is positively invariant. The set $\dot{V}=0$ has $y=0$; to remain there also requires ${\sin }^{3}x=0$, so its largest invariant subset is just the origin. The two theorems establish the origin is asymptotically stable.

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

SetThe stated identity givesHence, along the system,Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne 0$. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel setOn $\{\dot{E}=0\}$ we have $y=0$. A trajectory can remain there only ifwhich occurs at $x=0,\pm 1$. The choice $c<1/4=U(\pm 1)$ excludes the two latter points, so the largest invariant subset of $\{\dot{E}=0\}\cap {\Omega }_c$ is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every $k>0$,This is an instance of damped mechanical energy as a Lyapunov function.

For $\dot{x}=f(x)$, a fixed point $x^{\ast }$ satisfies $f(x^{\ast })=0$. It is Lyapunov stable when, for every $\epsilon >0$, there is $\delta >0$ such that

A Lyapunov function on a neighbourhood of $x^{\ast }$ is a continuously differentiable function $V$ such thatThe First Lyapunov theorem states that such a function makes $x^{\ast }$ Lyapunov stable. If $\dot{V}<0$ away from $x^{\ast }$, then $x^{\ast }$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^{\ast }|=\epsilon$, compactness and positive definiteness giveContinuity at $x^{\ast }$ supplies $\delta <\epsilon$ such that $|x-x^{\ast }|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot{V}<0$ away from $x^{\ast }$, take a sufficiently small compact sublevel set of $V$. The LaSalle invariance principle says that every trajectory in it approaches the largest invariant subset of $\{\dot{V}=0\}$, which is just $\{x^{\ast }\}$. Hence the equilibrium is also asymptotically stable.

Put $y=\dot{x}$. The first-order system isAt an equilibrium point of a dynamical system, $y=0$ and $x(1-x^2{)}^2=0$, so the three fixed points are

DefineThen $U^{\prime }(x)=2x(1-x^2{)}^2=-F(x)$ andMoreover,so $E$ is positive definite at the origin. It is therefore a damped mechanical energy as a Lyapunov function, and the First Lyapunov theorem proves that the origin is Lyapunov stable.

Since $U(\pm 1)=1/3$, choose a compact energy sublevel $\{E\le c\}$ with $0<c<1/3$. In this set, $\dot{E}=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm 1$. Thus the largest invariant subset of $\{\dot{E}=0\}$ is the origin. The LaSalle invariance principle proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its omega-limit set insideAn omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.

A $C^1$ function $V$ near an equilibrium at the origin is a Lyapunov function if $V(0)=0$, $V(x)>0$ for $x\ne 0$, andThe First Lyapunov theorem says these conditions imply stability. The second says that if the last inequality is strict away from the origin, then the origin is asymptotically stable.

To prove the first, fix a sufficiently small ball of radius $\epsilon$. Positive definiteness and compactness giveContinuity at zero gives $\delta >0$ such that $|x_0|<\delta$ implies $V(x_0)<m$. Since $V$ cannot increase along a trajectory, that trajectory cannot meet the sphere $|x|=\epsilon$, where $V\ge m$. It therefore remains in the $\epsilon$-ball for all forward time, which is stability.