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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}). $

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    yes you are right.2012-11-26

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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 $\frac{\partial^2 f}{\partial r\partial s}=\frac{\partial^2 f}{\partial s\partial r}$

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    @TheBaj: Since you are new to the site, I wanted to let you know that you should accept the answer that answers your question (don't get me wrong - I'm not telling you to accept my answer, but in general, once you got a satisfying answer to a question asked on this site you should accept it, so that other users will know that you got your answer). More on accepting answers can be found here: http://meta.math.stackexchange.com/questions/3286/how-do-i-accept-an-answer2012-11-26