Question:
$$\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial t^2}=f(x,t) \qquad \qquad u(x,0)=\frac{\partial u}{\partial t}(x,0)=0$$
Solve this equation, writing the solution in the form
$$u(\phi,\psi)=\frac 12 \iint _\Delta f(x,t)dxdt$$
where $\phi = \phi(x,t)$, $\psi = \psi(x,t)$, and $\Delta$ is a triangle to be found.
Attempt:
So I chose the canonical change of variables
$$\phi = x+t \qquad \qquad \psi = x-t$$
and thus arrived at the canonical form
$$4\frac{\partial^2 u}{\partial \phi \partial \psi} = f(x,t) = f\Big(\frac{\phi + \psi}{2} \;,\; \frac{\phi - \psi}{2} \Big)$$
It follows that
$$u(\phi,\psi) = \frac 14 \int_.^\phi \int_.^\psi f\Big(\frac{u+v}{2} \;,\; \frac{u-v}{2} \Big) dudv + F(\phi)+G(\psi)$$
I can see why you should end up with a $\frac 12$ in front of the integral, since
$$d\phi d\psi = \bigg| \frac{\partial (\phi,\psi)}{\partial (x,t)} \bigg|dxdt = 2 \, dxdt$$
However, I am unsure of how to use the boundary conditions to find $F$ and $G$ and to fix the lower bounds of the integrals.
Any hints would be much appreciated. Thanks!