# Toronto Math Forum

## APM346-2016F => APM346--Tests => Q2 => Topic started by: Roro Sihui Yap on October 06, 2016, 08:39:26 PM

Title: Q2
Post by: Roro Sihui Yap 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 $|x+2t| < \frac{\pi}{2}$
$u = \frac{1}{4}\int_{x-2t}^{x+2t} \cos(y)\,dy = \frac{1}{4}(\sin(x+2t) - \sin(x-2t))$

Title: Re: Q2
Post by: Victor Ivrii 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)