Author Topic: Q2  (Read 341 times)

Roro Sihui Yap

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Q2
« on: October 06, 2016, 08:39:26 PM »
$u_{tt}-4u_{xx}=0$           $x > 0, t > 0$


$u|_{t=0}=0 $
$u_t|_{t=0}=   \begin{cases}\cos(x) && |x| < \pi/2 \\0 &&|x| \ge \pi/2\end{cases}$

General solution : $$u(x,t)=\frac{1}{2}\bigl[g(x+ct)+g(x-ct)\bigr]+
\frac{1}{2c}\int_{x-ct}^{x+ct} h(y)\,dy = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy$$

when $(x-2t) \ge \frac{\pi}{2} $ and $(x+2t) \ge \frac{\pi}{2}   $
$u = \frac{1}{4}\int_{x-2t}^{x+2t} 0\,dy = 0$

when $|x-2t| < \frac{\pi}{2} $ and $(x+2t) \ge \frac{\pi}{2}   $
$u = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy = \frac{1}{4}\int_{x-2t}^{\frac{\pi}{2}} \cos(y)\,dy + \frac{1}{4}\int_{\frac{\pi}{2}}^{x+2t} 0 \,dy = \frac{1}{4}(1 - \sin(x-2t)) $

when $(x-2t) \le \frac{-\pi}{2} $ and $(x+2t) \ge \frac{\pi}{2}   $
$u = \frac{1}{4}\int_{x-2t}^{x+2t} h(y)\,dy = \frac{1}{4}\int_{\frac{\pi}{2}}^{x+2t} 0\,dy + \frac{1}{4}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(y)\,dy + \frac{1}{4}\int_{x-2t}^{\frac{-\pi}{2}} 0\,dy = \frac{1}{2} $

when $|x-2t| < \frac{\pi}{2} $ and $|x+2t| < \frac{\pi}{2}   $
$u = \frac{1}{4}\int_{x-2t}^{x+2t} \cos(y)\,dy = \frac{1}{4}(\sin(x+2t) - \sin(x-2t)) $

« Last Edit: October 06, 2016, 09:43:16 PM by Roro Sihui Yap »

Victor Ivrii

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Re: Q2
« Reply #1 on: October 06, 2016, 09:28:57 PM »
Great! Please use \cos, \sin etc
Also for long formulae better to use display math (in double dollars)