Author Topic: TT1-P2  (Read 1545 times)

Victor Ivrii

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TT1-P2
« on: October 21, 2015, 08:48:29 PM »
(a) Find solution $u(x,t)$ to
\begin{align}
&u_{tt}-u_{xx}= \frac{x}{(x^2+1)^2},\\
&u|_{t=0}=u_t|_{t=0}=0.
\end{align}

(b) Find $\lim _{t\to +\infty} u(x,t)$.

Xi Yue Wang

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Re: TT1-P2
« Reply #1 on: October 22, 2015, 12:39:17 AM »
For part a) we have $$ c = 1,\ g(x) = 0 = h(x),\ f(x,t) = \frac{x}{(x^2 + 1)^2}\\Then\ u(x,t) = \frac{1}{2}\int_{0}^{t} \int_{x-(t-t')}^{x+(t-t')} \frac{x'}{(x'^2+1)^2} dx'dt'\\ = \frac{1}{-4}\int_{0}^{t} \frac{1}{x'^2+1}\Big|_{x-(t-t')}^{x+(t-t')} dt'\\= \frac{1}{-4}\int_{0}^{t} \frac{1}{(x+(t-t'))^2+1} - \frac{1}{(x-(t-t'))^2+1)} dt'\\ for\ \frac{1}{4}\int_{0}^{t} \frac{1}{(x-(t-t'))^2+1)} dt'\\ take\ z = x - t + t'\ and\ dz = dt'\\ = \frac{1}{4}\int_{x-t}^{x} \frac{1}{z^2+1} dz\\= \frac{1}{4}[\arctan(x) - \arctan(x-t)]\\similar\ way\ to\ get\ \frac{1}{-4}\int_{0}^{t} \frac{1}{(x+(t-t'))^2+1}\\Finally\ u(x,t) = \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]$$
For par b) $$\lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]\\as\ t\to +\infty, x+t\to +\infty, x-t\to -\infty\\ hence, \arctan(x+t)\to \frac{\pi}{2} + 2k\pi, \arctan(x-t)\to -\frac{\pi}{2} + 2k\pi\\ \lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t) - \arctan(x-t)]\\= \frac{1}{4}[\frac{\pi}{2} + 2k\pi + \frac{\pi}{2} - 2k\pi] = \frac{\pi}{4}$$

I am not sure this is correct. :-[

Rong Wei

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Re: TT1-P2
« Reply #2 on: October 22, 2015, 05:43:13 PM »
Xi Yue, We have different answers. I'm not sure which one is correct.

Zaihao Zhou

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Re: TT1-P2
« Reply #3 on: October 22, 2015, 06:08:05 PM »
I have the same answer as Rong Wei.

Victor Ivrii

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Re: TT1-P2
« Reply #4 on: November 01, 2015, 10:08:32 AM »
Xi Yue Wang, somewhere you lost $\arctan (x)$