# Difference between a continuously differentiable function and a wave

• B
Summary:
What is the difference between an absolutely continuously differentiable function and a wave?
What is the difference between an absolutely continuously differentiable function and a wave? Are all absolutely continuously differentiable equations waves?

fresh_42
Mentor
A linear function is certainly smooth, in particular continuously differentiable, and it is not a wave function. These are distinct concepts. For approximations with wave functions, see https://en.wikipedia.org/wiki/Fourier_series

Is a wave function a subset of the group of absolutely continuously differentiable functions?

Every wave function can be considered as a superposition of plane waves, where each plane wave is absolutely continuously differentiable.

fresh_42
Mentor
Is a wave function a subset of the group of absolutely continuously differentiable functions?
The set of wave functions is. A single wave function is an element of the two sets. The group structure is only additive, which does not contain much information. So group is the wrong word here. We do not have multiplicative inverses. It is an algebra (I think. Depending on how wave function is defined. The continuously differentiable functions are.)

Right. My mistake on the terminology

Delta2
Homework Helper
Gold Member
To perplex things just a bit, the following little lemma might be of interest:

If ##f(x)## is a two times differentiable function of one real variable then $$f(\vec{k}\cdot\vec{x}-\omega t)$$ is a wave function of n+1 variables, ##\vec{k},\vec{x}\in \mathbb{R}^n##, ##\omega,t\in \mathbb{R}## for any ##n\in\mathbb{N}##

wrobel
I did not understand the question but answer:)
A function ##H(x-at)## is a solution of the wave equation as well but in the generalized sense
here ##H## is the Heaviside step function
##(x\in\mathbb{R})##

martinbn
To perplex things just a bit, the following little lemma might be of interest:

If ##f(x)## is a two times differentiable function of one real variable then $$f(\vec{k}\cdot\vec{x}-\omega t)$$ is a wave function of n+1 variables, ##\vec{k},\vec{x}\in \mathbb{R}^n##, ##\omega,t\in \mathbb{R}## for any ##n\in\mathbb{N}##
As pointed out, you can drop the differentiability and still have a weak solution to the wave equation.

Delta2
Homework Helper
Gold Member
As pointed out, you can drop the differentiability and still have a weak solution to the wave equation.
Sorry I don't understand, what do we mean by weak solution?

wrobel
Sorry I don't understand, what do we mean by weak solution?
For example, we shall say that a function ##u(t,x)\in L^1_{loc}(\mathbb{R}_+\times\mathbb{R})## is a weak solution to the equation ##u_{tt}=u_{xx}## if for any ##\psi\in\mathcal{D}(\mathbb{R}_+\times\mathbb{R}),\quad \mathbb{R}_+=\{\xi>0\mid\xi\in \mathbb{R}\}## it follows that
$$\int_{\mathbb{R}_+\times\mathbb{R}}u(t,x)(\psi_{tt}-\psi_{xx})dtdx=0$$

shock waves meet this definition

Delta2
Delta2
Homework Helper
Gold Member
Thanks @wrobel for making me remember that time is usually taken to be positive , as well as the concept of weak solution to a PDE.