Coordinate Change problem... I have
\begin{equation}
u(x,t)=\int_{0}^{t} v(x-ct+cs,s) ds 
\end{equation}
And,
\begin{equation}
v(x,t)=\int_{0}^{t} f(x+ct-cs,s)ds
\end{equation}
I'll show that
\begin{equation}
u(x,t)=\frac{1}{2c} \int_{0}^{t}\int_{x-ct+cs}^{x+ct-cs} f(y,s)dyds
\end{equation}
I don't know how to change coordinate...
Any help would appreciated.
 A: Performing the substitution, we have:
$$
u(x,t)=\int_0^t v(x-ct+cs,s) \, ds = \int_0^t \int_0^s f(x-ct+2cs-cr,r) \,dr\, ds 
$$
Now the region of integration is obviously the triangle given by the inequalities:
$$
0\le s \le t \qquad \qquad 0\le r \le s
$$
or equivalently:
$$
0\le r \le t \qquad \qquad r\le s \le t
$$
so swapping variables yields:
$$
\begin{aligned}
u(x,t)&= \int_0^t \int_r^t f(x-ct+2cs-cr,r) \,ds\, dr
\end{aligned}
$$
Now change $y=x-ct+2cs-cr$, and $dy=2c\, ds$ to obtain$^*$:
\begin{aligned}
u(x,t)&= \frac{1}{2c}\int_0^t \int_{x-ct+cr}^{x+ct-cr} f(y,r) \,dy\, dr
\end{aligned}
which is the required result under suitable variable renaming.
$^*$ Strictly speaking you have to change both variables and write the determinant of the jacobian of the transformation, so there is here some handwaving, that gives the same result though.
A: Substituting the definition of $v$ into the definition of $u$ we have
\begin{align}
u(x,t)&=\int_{0}^{t}\left(\int_0^sf\big((x-ct+cs)+cs-cr,r\big)\,dr \right)\,ds\\
&=\int_{0}^{t}\left(\int_0^sf\big(x+2cs-ct-cr,r\big)\,dr \right)\,ds
\end{align}
where we merely chose the letter $r$ to represent the variable of integration, rather than $s$. Now, let's change the order of integration:
\begin{align}
u(x,t)&=\int_{0}^{t}\left(\int_r^tf\big(x+2cs-ct-cr,r\big)\,ds \right)\,dr
\end{align}
Finally, for the inner integral, we use the substitution $y=x+2cs-ct-cr$, so $dy=2c\,ds$. Hence:
$$\int_r^tf\big(x+2cs-ct-cr,r\big)\,ds=\frac1{2c}\int_{x+cr-ct}^{x+ct-cr}f\big(y,r\big)\,dy,$$
so that
\begin{align}
u(x,t)&=\frac1{2c}\int_{0}^{t}\int_{x+cr-ct}^{x+ct-cr}f\big(y,r\big)\,dy \,dr.
\end{align}
To get the desired answer, it suffices to change the letter $r$ that represents the variable of integration of the outer integral to $s$.
