solving a particular form of Chini's equation

Question

Solve the initial value problem $$u''(t)+u'(t)=\sin u(t)$$ with initial conditions $u(0)=1,u'(0)=0$, and hence show that $u(t),u'(t)$ is bounded for all $t>0$. Find $\displaystyle\lim_{t\to\infty}u(t)$.

I started with the substitution $u'(t)=p(t)$. Then by chain rule $\displaystyle{u''(t)=\frac{dp}{dt}=p\frac{dp}{du}}$. So our equation becomes $$p\frac{dp}{du}+p=\sin u \\ \implies \frac{dp}{du}=-1+\frac{\sin u}{p}$$ which (after a thorough search online) I identified as a particular form of Chini's equation. I calculated the Chini invariant as $C=\sec u$, which is not independent of $u$ and hence can not be solved easily. I wonder whether this equation is solvable by any form of algebraic transformation or not.

In case this equation is not solvable analytically, or at least in some closed form, how to check for boundedness of the solution? Any help is appreciated.

Answer 1

You can treat this as a mechanical problem with friction force $-u'$ and potential force $\sin u$ relating to a potential function $P(u)=1+\cos(u)=2\cos^2(u/2)$.

From this mechanical point-of-view it should be intuitive that the system will shed energy until it comes to rest at one of the minima $P(u)=0$, $u=\pi+2k\pi$, $k\in\Bbb Z$, of the potential function. This also implies the boundedness when starting with velocity zero and not at one of the maxima $2k\pi$ of the potential.

You can make this exact by using the sum of kinetic and potential energy as Lyapunov function $$V(u,v)=\frac12v^2+2\cos^2(u/2),$$ usually $v=u'$. Then its value along any solution evolve as $$ \dot V=\frac{d}{dt}V(u(t),u'(t))=u'(u''-\sin(u))=-u'^2. $$ Thus for any non-constant solution $V$ steadily decreases. As $V$ is bounded by $V(u(0),u'(0))=2\cos^2(1/2)$, $u$ is restricted to the interval $[1,2\pi-1]$ around the potential minimizer $\pi$.