- #1
Niles
- 1,866
- 0
Homework Statement
Hi all.
I have the 1-D wave-equation, and I wish to make a change of variables, where a = x+ct and b = x-ct. I get:
[tex]
\begin{array}{l}
c^2 \frac{{\partial ^2 u}}{{\partial x^2 }} = c^2 \left[ {\frac{{d^2 u}}{{da^2 }}\left( {\frac{{da}}{{dx}}} \right)^2 + \frac{{du}}{{da}}\frac{{d^2 a}}{{dx^2 }}} \right] + c^2 \left[ {\frac{{d^2 u}}{{db^2 }}\left( {\frac{{db}}{{dx}}} \right)^2 + \frac{{du}}{{db}}\frac{{d^2 b}}{{dx^2 }}} \right] = c^2 \frac{{d^2 u}}{{da^2 }} + c^2 \frac{{d^2 u}}{{db^2 }} \\
\frac{{\partial ^2 u}}{{\partial t^2 }} = \left[ {\frac{{d^2 u}}{{da^2 }}\left( {\frac{{da}}{{dt}}} \right)^2 + \frac{{du}}{{da}}\frac{{d^2 a}}{{dt^2 }}} \right] + \left[ {\frac{{d^2 u}}{{db^2 }}\left( {\frac{{db}}{{dt}}} \right)^2 + \frac{{du}}{{db}}\frac{{d^2 b}}{{dt^2 }}} \right] = c^2 \frac{{d^2 u}}{{da^2 }} + c^2 \frac{{d^2 u}}{{db^2 }} \\
\end{array}
[/tex]
For this I have used the chain rule for higher derivates (for second derivates, from Wikipedia: http://en.wikipedia.org/wiki/Chain_rule). The result I wish to get is:
[tex]
\frac{{\partial ^2 u}}{{\partial a\partial b}} = 0
[/tex]
I can't quite see how I would get this. Am I on the right track here?
Cheers,
Niles.EDIT: Ok, now this is the second thread in a row I am doing this: I preview the thread, and the title resets itself. The original title was: "1-D wave-equation and change of variables". If a moderator can insert the proper title, I would be grateful.