Determine the stability for the point critical point $(0,0)$ to a solution $u(t)$ to the equation $$u''+u^5=0$$

Apparently, this cannot be determined using linearization. What other options do I have?


Define $v=u'$ and look at the system

$$\begin{cases} u' =v \\ v'=-u^5 \end{cases}$$

Consider the Lyapunov function $L(u,v)=\tfrac12 v^2+\tfrac16 u^6$. Then $L(0,0)=0$ and $L(u,v)>0$ in a neighborhood around the equilibrium. Further,

$$\frac{d}{dt} L=\langle \nabla L,\binom{u'}{v'}\rangle=\langle \binom{u^5}{v},\binom{v}{-u^5} \rangle=u^5v-u^5v=0$$

hence $(0,0)$ is a stable equilibrium (but not asymptotically stable).

  • 3
    $\begingroup$ This is more for anyone who reads your solution and wants to know how to find a Lyapunov functional for this problem. You can essentially get the Lyapunov functional above by multiplying the ODE by $u'$ and integrating. This yields the conserved quantity $$\frac{u'^{2}}{2} + \frac{u^{6}}{6} = C$$ Also, (+1). $\endgroup$ – Mattos Dec 17 '17 at 15:25

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