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Why the book call f(x+ct) and f(x-ct) odd extension of D'Alembert Method?

  1. Feb 25, 2010 #1
    For wave equation:

    [tex]\frac{\partial^2 u}{\partial t^2} \;=\; c^2\frac{\partial^2 u}{\partial x^2} \;\;,\;\; u(x,0)\; =\; f(x) \;\;,\;\; \frac{\partial u}{\partial t}(x,0) \;=\; g(x)[/tex]

    D'Alembert Mothod:

    [tex] u(x,t)\; = \;\frac{1}{2} f(x\;-\;ct)\; +\; \frac{1}{2} f(x\;+\;ct)\; +\; \frac{1}{2c} \int_{x-ct}^{x+ct} \; g(s) ds \;\;[/tex]

    Why the book call [tex]f(x\;-\;ct)\; ,\; f(x\;+\;ct)[/tex] odd extention of f(x)?
  2. jcsd
  3. Feb 27, 2010 #2
    Anyone please?

    I want to clarify, I am not asking what is an odd extension of a function. I want to know why the book claimed f(x+ct) and f(x-ct) are odd extension of f(x) in D'Alembert Method.
  4. Feb 28, 2010 #3


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    In general they are not odd extensions of f; they are translates. So there must be more about the context that is missing.
  5. Mar 2, 2010 #4
    Thanks for your answer. This is from Partial Differential Equations and Boundary Value Problem by Nakhle Asmar. It is very specificly said it is odd extension!! I don't understand this either.


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