1-D wave-equation and change of variables

Click For Summary
The discussion centers on the application of a change of variables in the 1-D wave equation, specifically using a = x + ct and b = x - ct. The user, Niles, is attempting to derive the result that ∂²u/(∂a∂b) = 0 but is struggling to connect his current equations to this outcome. He initially referenced the chain rule for higher derivatives but later realized that a more straightforward approach might be needed. Despite his efforts, he finds himself with a complex expression that does not simplify to the desired result. The conversation highlights the challenges of applying variable transformations in wave equations.
Niles
Messages
1,834
Reaction score
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:

<br /> \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}<br />

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:

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

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.
 
Physics news on Phys.org
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:

<br /> 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}}<br />

Is it possible to go from this to

<br /> <br /> \frac{{\partial ^2 u}}{{\partial a\partial b}} = 0<br /> <br />
?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
5
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
6K
  • · Replies 11 ·
Replies
11
Views
1K