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

1. Feb 25, 2010

### 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)?

2. Feb 27, 2010

### 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.

3. Feb 28, 2010

### LCKurtz

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

4. Mar 2, 2010

### yungman

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

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