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We consider the differential equation $$\dot{x}=-\epsilon x\sin^{2}t$$ subject to initial condition $x(0)=x_{0}$ and condition $0<\epsilon\ll1$. Separating variables and integrating shows $$x(t)=x_{0}\exp\left\{\frac{\epsilon}{4}\left(\sin2t-2t\right)\right\}.$$ Define a new quantity $$\bar{x}(t)=\frac{1}{2\pi}\int_{t-\pi}^{t+\pi}x(\tau)d\tau.$$ We want to show that $$x(t)=\bar{x}(t)+O(\epsilon).$$ The integral defining $\bar{x}$ is clearly impossible to carry out analytically. On the other hand, it is easy to see $x(0)=x_{0}$ and $x'(0)=0$ so that integrating the first few terms of the Taylor expansion about $0$ seems promising. This gives \begin{eqnarray*} \bar{x}(t) &=& \frac{1}{2\pi}\int_{t-\pi}^{t+\pi}\left(x_{0}+\frac{x''(\xi)}{2}t^{2}\right)dx\\ &=& \frac{x_{0}}{2\pi}\left(2\pi\right) + \frac{x''(\xi)}{12\pi}\left(6\pi t^{2}+2\pi^{3}\right)\\ &=& \frac{x''(\xi)}{2}t^{2}+\frac{x''(\xi)}{6}\pi^{2}+x_{0}. \end{eqnarray*} where $\xi\in(0,t)$.

At this point, however, I am stuck, and perhaps this was not the best way to go out proving the statement in the title of the post. Any suggestions? I'm clearly overlooking something, because this shouldn't be that difficult (it's only part (b) in a multi-part exercise).

(EDIT):

Perhaps we should expand about $\epsilon=0$ to obtain:

$$ x(t,\epsilon)=x_{0}+\epsilon\frac{x_{0}}{4}\left(\sin2t-2t\right)+O(\epsilon^{2}). $$ If we integrate this expression, we obtain

\begin{eqnarray*} \bar{x(t)} &=& \frac{1}{2\pi}\int_{t-\pi}^{t+\pi}x(\tau)d\tau\\ &=& \frac{1}{2\pi}\int_{t-\pi}^{t+\pi}\left(x_{0}+\epsilon\frac{x_{0}}{4}\left(\sin2\tau-2\tau\right)+O(\epsilon^{2})\right)d\tau\\ &=& x_{0}\left[\frac{\tau}{2\pi}\right]_{\tau=t-\pi}^{\tau=t+\pi}+\frac{\epsilon x_{0}}{8\pi}\left[-\frac{1}{2}\cos2\tau-\tau^{2}\right]_{\tau=t-\pi}^{\tau=t+\pi}+O(\epsilon^{2})\\ &=& x_{0}-\epsilon\frac{x_{0}}{2}t+O(\epsilon^{2}) \end{eqnarray*}.

But I still don't see how this gives the desired result.

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What do you mean by $x(t)=\bar{x}(t)+O(\epsilon)$? How can this hold at $t=0$? How do you define $\bar{x}$ for $t \in [0,\pi)$? –  copper.hat Aug 11 '12 at 20:34
    
To be frank, I'm not quite sure. This was what I was given, and that was what I was asked to prove. Perhaps treating $x$ as a function of $\epsilon$ and $t$, then expanding about $\epsilon=0$ is the way to go? At least then you get an expression involving $O(\epsilon^{k})$ instead of $O(t^{k})$.... –  Taylor Aug 11 '12 at 21:24
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up vote 0 down vote accepted

Looking to my edit, the answer was right before me....

\begin{eqnarray*} |x(t)-\bar{x}(t)| &=& \left|\left(x_{0}+\epsilon\frac{x_{0}}{4}\left(\sin2t-2t\right)+O(\epsilon^{2})\right)-\left(x_{0}-\epsilon\frac{x_{0}}{2}t+O(\epsilon^{2})\right)\right|\\ &=&\left|\epsilon\frac{x_{0}}{4}\left(\sin2t-2t+2t\right)\right|\\ &=&\epsilon\left|\frac{x_{0}}{4}\sin2t\right|\\ &=& O(\epsilon) \end{eqnarray*}

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