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1-D wave equation

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Express the solution
    P(t, x1) = cos(!t − kx1)
    as the superposition of two complex exponentials. Show that each complex
    exponential is also a solution of the 1-D wave equation.

    2. Relevant equations
    just that THETA=P

    whoops, made a type

    3. The attempt at a solution
    im not really sure how to tackle this, i do know that a solution to the wave equation added to another solution is also a solution.
    Last edited: Jan 28, 2007
  2. jcsd
  3. Jan 27, 2007 #2
    well, for the first part, what is the complex exponential equivalent of cos?

    Hint: Remeber that Euler's Formula igives [tex]e^{i\theta}=cos\theta + isin\theta[/tex] and [tex]e^{-{i\theta}}=cos\theta - isin\theta[/tex]. How can you derive a formula for [tex]cos\theta[/tex] from this (ie how do you get cos in terms of _only_ complex exponentials.

    You know that from the first part you should be able to express your wave function as a sum of two complex exponentials. Take each one and sub it into the wave equation and see if the LS equates to the RS for each one. It may also be useful to remember that [tex]\frac{\omega}{k}=v[/tex] where v is the velocity of the wave.
  4. Jan 27, 2007 #3
    The solutions to the wave equation are harmonic functions. Their sums, differences and scalar multiples are also solutions to the wave equation.

    You should use Euler's formula to write the cosine as a sum of two complex exponentials.

    EDIT: crossposted with Warr
    Last edited: Jan 27, 2007
  5. Jan 28, 2007 #4

    Gib Z

    User Avatar
    Homework Helper

    Ahh dont you guys love it when you see a formula for the first time, and it looks obscure and you don't really understand it, eg Euler formula. And then you take a second to think, and from Eulers Formula you can get the good old cos^2 x + sin^2 x=1? I love it :p
  6. Jan 28, 2007 #5
    guys, thanks for the help but im so lost in all this stuff. Is there somewhere where i can learn about harmonic functions to get up to speed. i did crappy in the math course before the geophys course im taking so im pretty lost as it is. i do not want the answer (obviously) but an easier explanation leading me down the right path please....
  7. Jan 29, 2007 #6
    You can't get easier than this:

    Try adding the two equations.
  8. Jan 29, 2007 #7
    ok, i get (in words i dont know how to type it fancy...)

    cos (theta) = [ e^j(theta) + e^-j(theta) ] /2

    then just sub back in for (theta)

    because adding euler together was adding 2 complex equations right?
    Last edited: Jan 29, 2007
  9. Jan 30, 2007 #8
    The problem asks to express the solution [tex]P(t,x_{1})=cos({\omega}t-kx_{1})[/tex] as the superposition of two complex exponentials and to show that each complex exponential is also a solution of the 1-D wave equation. Now using Euler's formula you found that

    [tex]cos({\omega}t-kx_{1}) = \frac{e^{j({\omega}t-kx_{1})}+e^{-j({\omega}t-kx_{1})}}{2}[/tex]


    Superposition , if you didn't know, means algebraic sum.

    So what would be the two complex exponentials which summed give you [tex]P(t,x_{1})=cos({\omega}t-kx_{1})[/tex] ?

    I'm sure you know the answer. Next you need to verify that the two complex exponentials are also solutions to the wave equation, which is very easy since you know the expression of the 1D wave equation.
    Last edited: Jan 30, 2007
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