The question is about damped pendulum. There are two statements I don't understand or I'm not sure if my justification for them is correct. Could you say if I'm right?
The example is from a German book, Gewöhnliche Differentialgleichungen und dynamische Systeme, by Mathias Wilke & Jan Prüss. I've translated the question, but the original is here if you want to see it.
A damped pendulum has an equation $$V(u,v)=\frac12v^2+w^2(1-\cos u)$$
which is a Lyapunov function of $$\cases{u'=v & \cr v'=-\alpha v-w^2\sin u} \alpha, w\ge 0$$
It is written that $V(u,v)\to\infty$ for $|v|\to\infty$ which implies $|u'(t)|=|v(t)|\le M$. Is it because otherwise $V$ would be not decreasing?
$|u''(t)|\le M$ and multiplying it with the differential equation gives $\displaystyle\int_0^\infty u'(t)dt<\infty$. But why $u''+\alpha u'+w^2\sin u=0$ is true? Multiplying it with $u'$ gives $$u'\implies u'u''+u'\alpha u'+u'w^2\sin u=0\iff u''u'+u'w^2\sin u=-\alpha(u')^2$$
and integrating $$\int_0^\infty u''u'+u'w^2\sin u=\int_0^\infty-\alpha(u')^2.$$
The LHS now is $$\underbrace{\displaystyle\frac{(u')^2}{2}\Bigg|_{0}^{\infty}}_{<\infty\text{by (1)}}+\underbrace{\int u'w^2\sin u}_{\text{how to show $<\infty$}}\implies\int-\alpha(u')^2<\infty\implies\int_0^\infty u'(t)dt<\infty$$