Differential equation D'Alembert Approach with coordinate transformation.

How does one get from $$\frac{d^2f}{dz^2} - c^2 \frac{d^2f}{dt^2} = 0$$

with $f$ being $f(z,t)$, by performing a coordinate transformation to get $f(r,s)$ with $r=z-ct$ and $s=z+ct$, to $$\frac{d^2f(r,s)}{dz^2}=\frac{d^2f}{dr^2} +2\frac{d^2f}{drds} + \frac{d^2f}{ds^2}.$$ and $$\frac{d^2f(r,s)}{dt^2}=c^2(\frac{d^2f}{dr^2} -2\frac{d^2f}{drds} + \frac{d^2f}{ds^2}).$$

-
There should be $drds$ in both denumerators, not $dr^2ds^2$ –  Dennis Gulko Nov 26 '12 at 23:20
yes you are right. –  TheBaj Nov 26 '12 at 23:25

Using the chain rule we have: $$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}$$ From here: \begin{align*}\frac{\partial^2 f}{\partial t^2}&=\frac{\partial}{\partial t}\left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}\right)\\&=\left(\frac{\partial^2 f}{\partial r^2}\frac{\partial r}{\partial t}+\frac{\partial^2 f}{\partial s\partial r}\frac{\partial s}{\partial t}\right)\frac{\partial r}{\partial t}+\frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial t^2}+\left(\frac{\partial^2 f}{\partial r\partial s}\frac{\partial r}{\partial t}+\frac{\partial^2 f}{\partial s^2}\frac{\partial s}{\partial t}\right)\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial t^2}\end{align*} Similarly for $\frac{\partial^2 f}{\partial z^2}$.
Now, just plug in the information you have: $$\frac{\partial r}{\partial t}=-c, \hspace{10pt} \frac{\partial s}{\partial t}=c, \hspace{10pt} \frac{\partial^2 r}{\partial t^2}=\frac{\partial^2 s}{\partial t^2}=0, \hspace{10pt}$$ And remember that if $\frac{\partial^2 f}{\partial r\partial s},\frac{\partial^2 f}{\partial s\partial r}$ are continuous, then $\frac{\partial^2 f}{\partial r\partial s}=\frac{\partial^2 f}{\partial s\partial r}$