Author Topic: HA2 problem 3  (Read 1215 times)

Victor Ivrii

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HA2 problem 3
« 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. \begin{equation}
\left\{\begin{aligned}
&u_{tt}-9u_{xx}=0, \qquad &&x>0,\\
&u|_{t=0}=0, \qquad &&x>0,\\
&u_t|_{t=0}=\cos (x), \qquad &&x>0,\\
&u|_{x=0}=0, \qquad &&t>0.
\end{aligned}\right.
\label{eq-HA2.7}
\end{equation}


b. \begin{equation}
\left\{\begin{aligned}
&u_{tt}-9u_{xx}=0, \qquad &&x>0,\\
&u|_{t=0}=0, \qquad &&x>0,\\\\
&u_t|_{t=0}=\cos (x), \qquad &&x>0,\\
&u_x|_{x=0}=0, \qquad &&t>0.
\end{aligned}\right.
\label{eq-HA2.8}
\end{equation}


c. \begin{equation}
\left\{\begin{aligned}
&u_{tt}-9u_{xx}=0, \qquad &&x>0,\\
&u|_{t=0}=0, \qquad &&x>0,\\
&u_t|_{t=0}=\sin(x), \qquad &&x>0,\\
&u|_{x=0}=0, \qquad &&t>0. \end{aligned}\right.
\label{eq-HA2.9}
\end{equation}


d. \begin{equation}
\left\{\begin{aligned}
&u_{tt}-9u_{xx}=0, \qquad &&x>0,\\
&u|_{t=0}=0, \qquad &&x>0,\\
&u_t|_{t=0}=\sin(x), \qquad &&x>0,\\
&u_x|_{x=0}=0, \qquad &&t>0.
\end{aligned}\right.
\label{eq-HA2.10}
\end{equation}

Yiyun Liu

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Re: HA2 problem 3
« Reply #1 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) Let $0 < 3t < x$, 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.
« Last Edit: January 31, 2015, 05:27:02 AM by Victor Ivrii »