Long time behavior for $u''(t) + u'(t) = \sin u(t) ,\ u(0)=1,\ u'(0)=0$

Asked Aug 16 '20 at 00:02

Active Aug 16 '20 at 02:24

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How can I prove it?

$\textbf{Problem.}$ $u(t)$ is solution of the following initial value problem. $$\begin{cases} u''(t) + u'(t) = \sin u(t) &\;\;(t>0),\\ u(0)=1,\;\; u'(0)=0 & \end{cases}$$ (1) Show that $u(t)$ and $u'(t)$ are bounded on $t>0$.

(2) Find $\lim\limits_{t\to\infty} u(t)$ with proof.

Resource: 2017 South Korea USCM P8.

My approach to solving the problem is to try to use the Laplace transform. $$ \mathcal{L}\{u''(t)\}+\mathcal\{u'(t)\}-\mathcal{L}\{\sin(u(t))\}=0$$ $$\left[s^{2}U(s)-su(0)-u'(0)\right]+\left[sU(s)-u(0)\right]-\mathcal{L}\{\sin(u(t))\}=0 $$ Now, using (IV) we can find that $$\left[s^{2}U(s)-s\right]+\left[sU(s)-1\right]-\mathcal{L}\{\sin(u(t))\}=0$$

but I don't know how to continue from there.

edited Aug 16 '20 at 02:24

Calvin Khor

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asked Aug 16 '20 at 00:02

We can get the bound $|u'(t)| \leq 1$, since $u'(b)>1$ would imply a interval $(a,b)$ where $u''(t)<0$ in the interval but $u'(a) = 1 < u'(b)$. – aschepler Aug 16 '20 at 01:02
uhm. I don't understand. – Aug 16 '20 at 01:27
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Does this answer your question? solving a particular form of Chini's equation – Calvin Khor Aug 16 '20 at 02:15

Long time behavior for $u''(t) + u'(t) = \sin u(t) ,\ u(0)=1,\ u'(0)=0$

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