1-D wave-equation and change of variables

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SUMMARY

The discussion focuses on the application of the change of variables in the 1-D wave equation, specifically using the transformations \( a = x + ct \) and \( b = x - ct \). The user, Niles, attempts to derive the condition \( \frac{{\partial^2 u}}{{\partial a \partial b}} = 0 \) using the chain rule for higher derivatives. Despite initial confusion, Niles recognizes the need to derive \( \frac{{du}}{{dx}} \) and \( \frac{{du}}{{dt}} \) with respect to \( dx \) and \( dt \), leading to the expression \( c^2 \frac{{\partial^2 u}}{{\partial x \partial a}} + c^2 \frac{{\partial^2 u}}{{\partial x \partial b}} = c \frac{{\partial^2 u}}{{\partial t \partial a}} - c \frac{{\partial^2 u}}{{\partial t \partial b}} \).

PREREQUISITES
  • Understanding of the 1-D wave equation
  • Familiarity with change of variables in differential equations
  • Knowledge of the chain rule for higher derivatives
  • Basic calculus, particularly partial derivatives
NEXT STEPS
  • Study the derivation of the wave equation using change of variables
  • Learn about the implications of the condition \( \frac{{\partial^2 u}}{{\partial a \partial b}} = 0 \)
  • Explore the application of the chain rule in multivariable calculus
  • Investigate the physical interpretations of the wave equation in various contexts
USEFUL FOR

Students and researchers in applied mathematics, physics, and engineering who are working with wave equations and transformations in differential equations.

Niles
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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}<br /> 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 }} \\ <br /> \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 }} \\ <br /> \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.
 
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I've changed the title. Thanks for letting us know that it was the preview that causes the problem.
 
Ok, my first attempt is quite bad. Instead of using the expression from Wikipedia, I can just derive du/dx and du/dt with respect to dx and dt, respectively. But still, this gives me:

[tex] c^2 \frac{{\partial ^2 u}}{{\partial x\partial a}} + c^2 \frac{{\partial ^2 u}}{{\partial x\partial b}} = c\frac{{\partial ^2 u}}{{\partial t\partial a}} - c\frac{{\partial ^2 u}}{{\partial t\partial b}}[/tex]

Is it possible to go from this to

[tex] <br /> \frac{{\partial ^2 u}}{{\partial a\partial b}} = 0<br /> [/tex]
?
 

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