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

[yungman]

For wave equation:


\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)

D'Alembert Mothod:

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 \;\;

Why the book call f(x\;-\;ct)\; ,\; f(x\;+\;ct) odd extention of f(x)?

 

[yungman]

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.

 

[LCKurtz]

In general they are not odd extensions of f; they are translates. So there must be more about the context that is missing.

 

[yungman]

In general they are not odd extensions of f; they are translates. So there must be more about the context that is missing.

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.

Thanks

Alan