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For $k\ne 0$, put $t=k^2>0$ and use the positive-definite Lyapunov function $V=x^2+t{y}^2$. Its derivative isFor $V>0$ set $u=x^2/V\in [0,1]$. The brace divided by $V^2$ is $A{u}^2+2Bu(1-u)+C(1-u{)}^2$, where $A=1$, $B=\beta (1+t)/(2t)$ and $C=1/t$. Since $\beta >2$, $B>A,C$. Completing the square gives its maximumHence $\dot{V}\le -2V+2M{V}^2$. If $V(0)<1/M$, this sublevel region is positively invariant: $V$ decreases, and $\dot{V}\le -2(1-MV(0))V$. The exponential bound makes $V\to 0$. The trajectory is bounded in a compact ellipse, so smoothness of the vector field also guarantees existence for every positive time. ThereforeFor $k=0$ the displayed strict inequality has no solutions; negative $k$ gives the same ellipse as $-k$. Taking the union of the nonempty ellipses is valid because each point needs only one Lyapunov certificate.

At $k=1$ the certified disk has squared radius $2/(\beta +1)$. Along $y=0$, the limiting threshold as $k^2\to \infty$ is $4(\beta -1)/{\beta }^2$, which is strictly larger, since their difference is $2({\beta }^2-2)/({\beta }^2(\beta +1))>0$ for $\beta >2$. Choose $x^2$ strictly between these thresholds and then a sufficiently large finite $k$. This point is certified by that ellipse but lies outside the $k=1$ disk. Thus the union gives a strictly larger certified region. It need not equal the whole basin of attraction.

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

For every $p<5$, the wave $Q_c$ has Orbital stability of a generalized KdV solitary wave in $H^1(\mathbb{R})$ modulo translation: for every $\epsilon >0$ there is $\delta >0$ such thatThe relevant stability slope has the correct sign because scaling giveswhich is strictly increasing in $c$ precisely for $p<5$. Together with the constrained variational characterization of $Q_c$, the conserved mass and energy provide a coercive Lyapunov function transverse to the translation direction.

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 $x_{\ast }$.

A Lyapunov function for an equilibrium is a continuously differentiable function that is positive definite there and whose derivative along nonconstant trajectories is negative definite in a neighbourhood.

For $V=x^2+k{y}^2$, positivity requires $k>0$, andThe quartic term is negligible sufficiently close to the origin, so $V$ is a local Lyapunov function exactly whenis positive definite. Its determinant condition isThe two roots of equality areTherefore

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.

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.

Let $A=c_{\infty }^{p-1}$ andUsing the transformed equation,Thus $J$ is a strict Lyapunov function away from equilibria.

On ${\mathbb{R}}_{+}$, $f(0)=0$, it decreases to its unique minimum at $v=c_{\infty }$, wherethen increases, crosses zero at $v=((p+1)/2{)}^{1/(p-1)}{c}_{\infty }$, and tends to $+\infty$.

Take the regular radial solution with $u(0)=1$. In the transformed variables it satisfiesso $J(s)\to 0$. Since $s_c>1$, the Lyapunov function immediately becomes negative. The trajectory cannot reach $v=0$, where $J={\dot{v}}^2\ge 0$, and $J<0$ confines it below the positive zero of $f$. Hence it remains positive and bounded for all $s$.

The identity $\dot{J}=-4(s_c-1){\dot{v}}^2$ and the LaSalle invariance principle force the omega-limit set to consist of equilibria. The negative limiting energy excludes $v=0$, leavingConsequentlyso $\alpha =2/(p-1)$. At infinity, $|u^{\prime }(r)|\sim a{c}_{\infty }{r}^{-a-1}$, andbecause $s_c>1$. Thus $\nabla u\notin L^2({\mathbb{R}}^d)$.

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 $V$.