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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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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
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up vote 1 down vote accepted

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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thanks . that's what i was thinking, i just didn't remember the rule to back it up. –  TheBaj Nov 26 '12 at 23:31
    
@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: meta.math.stackexchange.com/questions/3286/… –  Dennis Gulko Nov 26 '12 at 23:50
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