Substitution in general double integral

Question

Is possible to go from $$\int_0^t\int_0^s f(x-ct+2cs-cr,r) \, \Bbb{d}r \, \Bbb{d}s$$ to $$\frac{1}{2c}\int_0^t\int_{x-c(t-z)}^{x+c(t-z)} f(y,z)\,\Bbb{d}y\,\Bbb{d}z,$$ where $c\in \Bbb{R}$, $x\in \Bbb{R}$, $t\gt 0$, with some smart substitution or am I completely wrong? What I am trying to do is to get the general solution of the nonhomogenous 1D wave equation with zero initial conditions $$u_{tt}-c^2u_{xx}=f(x,t), $$ $$u(x,0)=0,u_t(x,0)=0, $$ by solving an equivalent problem, which is system of two nonhomogenous transport equations $$u_t+cu_x=v(x,t), $$ $$v_t-cv_x=f(x,t) $$ and now I am stuck with this integral.

Answer 1

The trick here is to use Fubini's first, finding that your integral is equivalent to $$ \int_0^t\int_r^tf(x-ct+2cs-cr,r)\mathrm ds\mathrm dr $$ then perform your substitution $$ y=x-ct+2cs-cr\implies \mathrm dy=2c\mathrm ds $$ and your upper limit is $$ y=x+ct-cr=x+c(t-r) $$ and your lower limit is $$ y=x-ct+cr=x-c(t-r) $$ yielding your final integral $$ \frac{1}{2c}\int_0^t\int_{x-c(t-r)}^{x+c(t-r)}f(y,r)\mathrm dy \mathrm dr $$