Hurry said:
Consider
$\begin{align*}
& {{u}_{tt}}={{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0 \\
& u(x,0)=0,\text{ }x\in \mathbb{R} \\
& {{u}_{t}}(x,0)=\left\{ \begin{matrix}
\sin x,\text{ }\left| x \right|\le \pi \\
0,\text{ }x\notin [-\pi ,\pi ] \\
\end{matrix} \right.
\end{align*}
$
How can I solve this by using D'Lambert method?
Hi Hurry,
The d'Alembert's Solution for the given partial differential equation is, (Refer
this or
this.)
\[U(x,\,t)=\frac{1}{2}U(x-t,\,0)+\frac{1}{2}U(x+t,\,0)+\frac{1}{2}\int_{x-t}^{x+t}U_{t}(s,\,0)\,ds\]
Since \(U(x,\,0)=0~\forall~x\in\Re\) we get,
\begin{eqnarray}
U(x,\,t)&=&\frac{1}{2}\int_{x-t}^{x+t}U_{t}(s,\,0)\,ds\\
\end{eqnarray}
Case I: When, \(x-t\leq-\pi\mbox{ and }x+t\geq\pi\)
\begin{eqnarray}
U(x,\,t)&=&\frac{1}{2}\int_{-\pi}^{\pi}\sin(s)\,ds\\
&=&0
\end{eqnarray}
Case II: When, \(-\pi<x-t<\pi\mbox{ and }x+t\geq\pi\)
\begin{eqnarray}
U(x,\,t)&=&\frac{1}{2}\int_{x-t}^{\pi}\sin(s)\,ds\\
&=&\frac{1}{2}[1+\cos(x-t)]
\end{eqnarray}
Case III: When, \(-\pi<x-t<\pi\mbox{ and }-\pi<x+t<\pi\)
\begin{eqnarray}
U(x,\,t)&=&\frac{1}{2}\int_{x-t}^{x+t}\sin(s)\,ds\\
&=&\frac{1}{2}[-cos(x+t)+\cos(x-t)]
\end{eqnarray}
By the
Sum to product identity we get,
\[U(x,\,t)=\sin(x)\sin(t)\]
Case IV: When, \(x-t\leq-\pi\mbox{ and }-\pi<x+t<\pi\)
\begin{eqnarray}
U(x,\,t)&=&\frac{1}{2}\int_{-\pi}^{x+t}\sin(s)\,ds\\
&=&\frac{1}{2}[-1-\cos(x+t)]
\end{eqnarray}
Case V: Otherwise. (\([-\pi,\pi]\cap[x-t,x+t]=\varnothing\)
)
\[U(x,\,t)=0\]
Kind Regards,
Sudharaka.