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

 

[blue_raver22]

The book refers to the functions f(x-ct) and f(x+ct) as odd extensions of the function f(x) because they exhibit odd symmetry about the origin. This can be seen by substituting -x for x in the equation for f(x-ct) and f(x+ct), which results in f(-x+ct) and f(-x-ct) respectively. This means that these functions are equal to their negative counterparts, f(-x+ct) = -f(x-ct) and f(-x-ct) = -f(x+ct). This odd symmetry is a key characteristic of odd functions, which are defined as functions that satisfy f(-x) = -f(x).

In the context of the D'Alembert method for solving the wave equation, the odd extensions of f(x) are used to create a solution that satisfies the initial conditions of the equation. This is because the odd symmetry of these functions ensures that they have a zero value at the origin, which is necessary for the solution to satisfy the initial condition u(x,0) = f(x). Additionally, the odd extensions are used in the integration term of the solution to ensure that the solution satisfies the initial condition for the derivative, \frac{\partial u}{\partial t}(x,0) = g(x).

Overall, the odd extensions of f(x) are an important component of the D'Alembert method for solving the wave equation because they allow for a solution that satisfies the initial conditions of the equation. They are called odd extensions because they exhibit odd symmetry, which is a key characteristic of odd functions.