# D'Lambert problem

#### Hurry

##### New member
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?

#### Sudharaka

##### Well-known member
MHB Math Helper
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.