# Toronto Math Forum

## APM346-2015S => APM346--Home Assignments => HA2 => Topic started by: Victor Ivrii on January 27, 2015, 09:35:24 PM

Title: HA2 problem 3
Post by: Victor Ivrii on January 27, 2015, 09:35:24 PM
By method of continuation combined with D'Alembert formula solve each of
the following four problems (a)--(d).

a.
\left\{\begin{aligned}
\end{aligned}\right.
\label{eq-HA2.7}

b.
\left\{\begin{aligned}
\end{aligned}\right.
\label{eq-HA2.8}

c.
\left\{\begin{aligned}
\label{eq-HA2.9}

d.
\left\{\begin{aligned}
\end{aligned}\right.
\label{eq-HA2.10}

Title: Re: HA2 problem 3
Post by: Yiyun Liu on January 29, 2015, 09:00:33 PM
(a) Let $0 < 3t < x$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int_{x - 3t}^{x + 3t} {\cos x'dx'}
= \frac{1}{6}\left[ {\sin \left( {x + 3t} \right) - \sin (x - 3t)} \right]
= \frac{1}{3}\cos x\sin 3t.
\end{equation*}
Let $0 < x < 3t$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int_{3t - x}^{x + 3t} {\cos x'dx'}  = \frac{1}{6}\left[ {\sin (x + 3t) - \sin (3t - x)} \right]
= \frac{1}{3}\sin x\cos 3t.
\end{equation*}

So,
\begin{equation*}
u(x,t)=\left\{ \begin{aligned}
&\frac{1}{6}\cos x\sin 3t && 0<3t<x,\\ & \frac{1}{3}\sin x\cos 3t. &&0<x<3t. \end{aligned}\right. \end{equation*} (b) Let0 < 3t < x$, then$u(x,t) = \frac{1}{3}\sin 3t\cos x$. Let$0 < x < 3t$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int_0^{3t + x} {\cos x'dx' + \frac{1}{6}} \int\limits_0^{3t - x} {\cos x'dx'} = \frac{1}{6}\sin (x + 3t) + \frac{1}{6}\sin (3t - x) = \frac{1}{3}\sin 3t\cos x \end{equation*} (c)$0 < 3t < x$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int\limits_{x - 3t}^{x + 3t} {\sin x'dx'} = - \frac{1}{6}\cos (x + 3t) + \frac{1}{6}\cos (x - 3t) = \frac{1}{3}\cos x\cos 3t. \end{equation*} Let$0 < x < 3t$. Then \begin{equation*} u(x,t) = \frac{1}{6}\int_{3t - x}^{x + 3t} \sin x'dx' = \frac{1}{6}\left[ {\cos (3t - x) - \cos (x + 3t)} \right] = \frac{1}{3}\cos x\cos 3t \end{equation*} (d)$0 < 3t < x$, then$u(x,t) = \frac{1}{3}\cos x\cos 3t$. Let$0 < x < 3t\$. Then
\begin{equation*}
u(x,t) = \frac{1}{6}\int\limits_0^{x + 3t} {\sin x'dx' + \frac{1}{6}} \int\limits_0^{3t - x} {\sin x'dx'}
= \frac{1}{3} - \frac{1}{6}\left[ {\cos (x + 3t) + \cos (3t - x)} \right]
= \frac{1}{3}(1 - \cos x\cos 3t)
\end{equation*}

I fixed LaTeX: too short lines and too small vertical spacing made what you wrote difficult to read. --V.I.