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Rearranging wave equation

  1. Jun 20, 2015 #1

    The wave equation given: [itex]{1\over{c^2}} {\partial^2 \phi\over{\partial t^2}} = \Delta \phi [/itex] with [itex] r = \sqrt{x^2+y^2+z^2}[/itex] needs to be rearranged, so that [itex]{1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}} [/itex].

    Are there any tricks to obtain this result?
  2. jcsd
  3. Jun 20, 2015 #2


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    This is only possible if you have a wave which does not depend on the angular coordinates in spherical coordinates, i.e., a spherical wave. Then you need express the Laplace operator in spherical coordinates.
  4. Jun 20, 2015 #3
    Yes, this works!

    Is it possible to find solutions using the second formula?

  5. Jun 20, 2015 #4
    it sure is but you're gonna need to specify some initial conditions otherwise your solution can be any periodic function whatsoever that solves the wave equation...
  6. Jun 21, 2015 #5
    I would like to show, that if [itex] \phi [/itex] is a solution to the equation than each partial derivative of [itex] \phi [/itex] is also a solution.
    I am failing to show that just by plugging the derivative in. How can i do that?
  7. Jun 23, 2015 #6
    The product rule should give you this result
  8. Jun 23, 2015 #7
    Do you mean product rule in this formula>?
    [itex]{1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}} [/itex].
  9. Jun 23, 2015 #8
    could you link the exact wording of the question?
    without actually doing it my intuition suggests that to check that a partial derivative of phi solves the equation, make the substitution $$\phi \rightarrow \frac{\partial\phi}{\partial r}$$
    or $$\phi \rightarrow \frac{\partial\phi}{\partial t}$$
    and then apply the product rule.
    Last edited: Jun 23, 2015
  10. Jun 23, 2015 #9
    The question is to show, that if [itex]\phi[/itex] solves the equation [itex]{1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}} [/itex]
    than so should every partial derivative $$\frac{\partial\phi}{\partial x }, \frac{\partial\phi}{\partial y }, \frac{\partial\phi}{\partial z }, \frac{\partial\phi}{\partial t }.$$
    And it seems that the product rule has to be applied here for more than 20 times, but i still dont get the result.
  11. Jun 24, 2015 #10
    maybe write the general solution for the product ##r\phi## then take derivatives?
  12. Jun 25, 2015 #11
    Yes, but the general solution is not known in this case
  13. Jun 26, 2015 #12
    The general solution is ##\phi = \frac{Au(r-ct)}{r} + \frac{Bv(r+ct)}{r} ##
  14. Jun 26, 2015 #13
    Perhaps I have misunderstood the problem but anyway. Since r is independent the partial derivative of ##\phi## with time is obviously a solution.
    ## \partial^{2}_{t}(r\partial_{t}\phi) = \partial_{t}(\partial_{t}^{2} (r\phi)) = \partial^{2}_{r}(r\partial_{t}\phi) = \partial_{t}(\partial_{r}^{2} (r\phi)) ##. Now do the same with the other partial derivatives. I think we get something along the lines of ## \partial_{t}^{2}(\cos(\theta)\phi) = \partial_{r}^{2}(\cos(\theta)\phi) ## in 2 dimensions.
  15. Jun 27, 2015 #14
    Well , yes the part with the time derivative is obvious. The other part is not.
    Why would there be a cosine(theta) ?
  16. Jun 28, 2015 #15
    It would be from ## \partial_{x}r = \cos(\phi) ##. Isn't that correct?
  17. Jun 28, 2015 #16
    oh you mean $${\partial r\over {\partial x}} = {x \over r} = {r \cos(\theta) \over {r}} $$
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