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Proving an exponential function obeys the wave equation

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]y(x,t)=De^{-(Bx-Ct)^{2}}[/itex] obeys the wave equation


    2. Relevant equations
    The wave equation:
    [itex]\frac{d^{2}y(x,t)}{dx^{2}}=\frac{1}{v^{2}}\frac{d^{2}y(x,t)}{dt^{2}}[/itex]



    3. The attempt at a solution
    1: [itex]y(x,t)=De^{-u^{2}}; \frac{du}{dx}=B; \frac{du}{dt}=-C[/itex]
    2: [itex]\frac{dy(x,t)}{dx}=-2uBDe^{-u^{2}};

    \frac{d^{2}y(x,t)}{dx^{2}}=4u^{2}B^{2}De^{-u^{2}}[/itex]
    3: [itex]\frac{dy(x,t)}{dt}=2uCDe^{-u^{2}};

    \frac{d^{2}y(x,t)}{dt^{2}}=4u^{2}C^{2}De^{-u^{2}}[/itex]
    Then I'm stuck, I think I might have done something wrong but I can't see what.
    I think v=C (from (x-vt)) and that would cancel the [itex]C^{2}[/itex] when I substitute into the wave equation, but then I would be left with a [itex]B^{2}[/itex]
     
  2. jcsd
  3. Jan 9, 2014 #2
    The C in the exponent is not the speed of the wave. It has dimensions of s^(-1).
    The speed of the wave will be a function of C and D. Which you can find from the equation.
     
  4. Jan 9, 2014 #3

    BruceW

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    also, the equation
    [tex]\frac{dy(x,t)}{dx}=-2uBDe^{-u^{2}}[/tex]
    is correct, but the next equation
    [tex]\frac{d^{2}y(x,t)}{dx^{2}}=4u^{2}B^{2}De^{-u^{2}}[/tex]
    is not correct. It looks like you've done the derivative of the exponential, but what about the ##u## (in the previous equation), doesn't it depend on x also?
     
  5. Jan 11, 2014 #4
    Sorry it has taken so long. I will come back to this question I'm just in the middle of moving right now :S
    Thanks for replies :)
     
  6. Jan 11, 2014 #5

    BruceW

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    no worries :) hope the move goes well
     
  7. Jan 11, 2014 #6

    HallsofIvy

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    You did not differentiate the first "u". It should be
    [itex]\frac{\partial y(x,t)}{\partial x}= -2B^2De^{-u^2}- u^2B^2De^{-u^2}[/itex]

     
  8. Jan 11, 2014 #7

    dextercioby

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    I know that you;re supposed to differentiate like crazy to get to the expected solution, but you can surprise your professor if you make the following change of variables in your PDE: x-vt = u; x+vt = v. Then you can fully solve the original PDE and make certain simplifying assumptions to find the function you're given in the statement.
     
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