MHB Solving the D'Lembert Method with Multiple Conditions

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Consider

$\begin{aligned} & {{u}_{tt}}=9{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\
& u(x,0)=\left\{ \begin{align}
& 1,\text{ }x\in [1,2] \\
& 0,\text{ }x\notin [1,2] \\
\end{align} \right. \\
& {{u}_{t}}(x,0)=0,
\end{aligned}
$

then determine the points of the semiplane $t>0$ where $u(x,t)=0.$

Okay I know the D'Lembert's formula, but I don't know how to apply it since having $u(x,0)$ defined by two conditions.
Thanks!
 
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Markov said:
Consider

$\begin{aligned} & {{u}_{tt}}=9{{u}_{xx}},\text{ }x\in \mathbb{R},\text{ }t>0, \\
& u(x,0)=\left\{ \begin{align}
& 1,\text{ }x\in [1,2] \\
& 0,\text{ }x\notin [1,2] \\
\end{align} \right. \\
& {{u}_{t}}(x,0)=0,
\end{aligned}
$

then determine the points of the semiplane $t>0$ where $u(x,t)=0.$

Okay I know the D'Lembert's formula, but I don't know how to apply it since having $u(x,0)$ defined by two conditions.
Thanks!
RE-write your IC as

$u(x,0) = H(x-1)-H(x-2)$

then apply the D'Almbert Formula.
 
Thank you Jester! I'm sorry but I'm a bit lost on how applying D'lembert's formula, do I need to apply it for $H(x-1)$ and $H(x-2)$ ? How to do so?

Much appreciated!
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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