Difference between a continuously differentiable function and a wave

  • #1
265
7
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?
 

Answers and Replies

  • #3
265
7
Is a wave function a subset of the group of absolutely continuously differentiable functions?
 
  • #4
265
7
Every wave function can be considered as a superposition of plane waves, where each plane wave is absolutely continuously differentiable.
 
  • #5
15,565
13,677
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.)
 
  • #6
265
7
Right. My mistake on the terminology
 
  • #7
Delta2
Homework Helper
Insights Author
Gold Member
4,578
1,857
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}##
 
  • #8
wrobel
Science Advisor
Insights Author
877
611
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})##
 
  • #9
martinbn
Science Advisor
2,495
902
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.
 
  • #10
Delta2
Homework Helper
Insights Author
Gold Member
4,578
1,857
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?
 
  • #11
wrobel
Science Advisor
Insights Author
877
611
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
 
  • #12
Delta2
Homework Helper
Insights Author
Gold Member
4,578
1,857
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.
 

Related Threads on Difference between a continuously differentiable function and a wave

  • Last Post
Replies
1
Views
807
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
627
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
7
Views
7K
Top