Explanation to PDE $u_{tt}-c^2u_{xx}=0$ solution 
Exercise: Solve the Cauchy problem $u_{tt}-c^2u_{xx}=0$ with conditions $u(x,0)=g(x)$ and $u_t(x,0)=h(x)$, where $g(x)=0,\ h(x)=\begin{cases} 0,\ x<0  \\ 1,\ x\ge0
\end{cases}$.

Please ignore this 'solution' and see Felix Marin's solution
Solution: 
By D'Alembert formula I get 
$$u(x,t)=\dfrac{1}{2c}\int_{x-ct}^{x+ct}0 \,\mathrm{d}s=0,\quad x<0.$$
and
$$u(x,t)=\dfrac{1}{2c}\int_{x-ct}^{x+ct}1 \,\mathrm{d}s=t.\quad x\ge0.$$
Thus  the answer is $0,\ if\ x<0\ and\ t,\ if\ x\ge 0$. 

$\color{fuchsia}{Questions\ about\ the\ answer\ given\ by\ Marin.} $


*

*Why the integral of $F'(x)=-\dfrac{1}{2}h(x)$ is $F(x)-F(a)|_{a<0} $ ? Why $a<0$?

*Why $$\frac{-1}{2}\int_{a}^xh(ε) \,\mathrm{d}ε =\frac{-1}{2}H(x) \int_{0}^xh(ε) \,\mathrm{d}ε?$$
According to Wikipedia, $H(x)=\int_{-\infty}^x δ(ε) \,\mathrm{d}ε$.
Please help me to understand the solution of this Cauchy problem.

If there is an alternative solution, maybe an easier one, please show it :)
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets start from the very beginning. Lets
  $\ds{\mrm{u}\pars{x,t} = \mrm{F}\pars{x - t} + \mrm{G}\pars{x + t}}$ with the scaling $\ds{ct \mapsto t}$.

$$
0 = \mrm{u}\pars{x,0} = \mrm{F}\pars{x} + \mrm{G}\pars{x} \implies \mrm{G}\pars{x} = -\,\mrm{F}\pars{x}
$$
The solution becomes
$\ds{\mrm{u}\pars{x,t} = \mrm{F}\pars{x - t} - \mrm{F}\pars{x + t}}$ and
\begin{align}
&\mrm{h}\pars{x} = \mrm{u}_{t}\pars{x,0} = -\mrm{F}'\pars{x} - \mrm{F}'\pars{x} \implies \mrm{F}'\pars{x} = -\,{1 \over 2}\mrm{h}\pars{x}
\\[5mm] &\
\mrm{F}\pars{x} -
\left.\vphantom{\Large A}\mrm{F}\pars{a}\right\vert_{\ a\ <\ 0}= -\,{1 \over 2}\int_{a}^{x}\mrm{h}\pars{\xi}\,\dd\xi =
-\,{1 \over 2}\,\mrm{H}\pars{x}\int_{0}^{x}\mrm{h}\pars{\xi}\dd\xi =
-\,{1 \over 2}\,\mrm{H}\pars{x}x
\end{align}

where $\ds{\mrm{H}}$ is the Heaviside Step Function.


$$
\bbx{\mrm{u}\pars{x,t} =
{\mrm{H}\pars{x + t}\pars{x + t} - \mrm{H}\pars{x - t}\pars{x - t}\over 2}}
$$


A: $\def\d{\mathrm{d}}$By d'Alembert's formula,$$
u(x, t) = \frac{1}{2c} \int_{x - ct}^{x + ct} h(s) \,\d s.
$$
For $x \leqslant -ct$,$$
u(x, t) = \frac{1}{2c} \int_{x - ct}^{x + ct} h(s) \,\d s = \frac{1}{2c} \int_{x - ct}^{x + ct} 0 \,\d s = 0.
$$
For $-ct < x \leqslant ct$,$$
u(x, t) = \frac{1}{2c} \int_{x - ct}^{x + ct} h(s) \,\d s = \frac{1}{2c} \int_0^{x + ct} 1 \,\d s = \frac{1}{2c} (x + ct).
$$
For $x > ct$,$$
u(x, t) = \frac{1}{2c} \int_{x - ct}^{x + ct} h(s) \,\d s = \frac{1}{2c} \int_{x - ct}^{x + ct} 1 \,\d s = t.
$$
Thus$$
u(x, t) = \begin{cases}
0; & x \leqslant -ct\\
\dfrac{1}{2c} (x + ct); & -ct < x \leqslant ct\\
t; & x > ct
\end{cases}
$$
