Toronto Math Forum
APM3462015F => APM346Tests => Test 1 => Topic started by: Victor Ivrii 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 $\lim _{t\to +\infty} u(x,t)$.

For part a) we have $$ c = 1,\ g(x) = 0 = h(x),\ f(x,t) = \frac{x}{(x^2 + 1)^2}\\Then\ u(x,t) = \frac{1}{2}\int_{0}^{t} \int_{x(tt')}^{x+(tt')} \frac{x'}{(x'^2+1)^2} dx'dt'\\ = \frac{1}{4}\int_{0}^{t} \frac{1}{x'^2+1}\Big_{x(tt')}^{x+(tt')} dt'\\= \frac{1}{4}\int_{0}^{t} \frac{1}{(x+(tt'))^2+1}  \frac{1}{(x(tt'))^2+1)} dt'\\ for\ \frac{1}{4}\int_{0}^{t} \frac{1}{(x(tt'))^2+1)} dt'\\ take\ z = x  t + t'\ and\ dz = dt'\\ = \frac{1}{4}\int_{xt}^{x} \frac{1}{z^2+1} dz\\= \frac{1}{4}[\arctan(x)  \arctan(xt)]\\similar\ way\ to\ get\ \frac{1}{4}\int_{0}^{t} \frac{1}{(x+(tt'))^2+1}\\Finally\ u(x,t) = \frac{1}{4}[\arctan(x+t)  \arctan(xt)]$$
For par b) $$\lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t)  \arctan(xt)]\\as\ t\to +\infty, x+t\to +\infty, xt\to \infty\\ hence, \arctan(x+t)\to \frac{\pi}{2} + 2k\pi, \arctan(xt)\to \frac{\pi}{2} + 2k\pi\\ \lim_{t\to +\infty} u(x,t) = \lim_{t\to +\infty} \frac{1}{4}[\arctan(x+t)  \arctan(xt)]\\= \frac{1}{4}[\frac{\pi}{2} + 2k\pi + \frac{\pi}{2}  2k\pi] = \frac{\pi}{4}$$
I am not sure this is correct. :[

Xi Yue, We have different answers. I'm not sure which one is correct.

I have the same answer as Rong Wei.

Xi Yue Wang, somewhere you lost $\arctan (x)$