Difference between a continuously differentiable function and a wave

  • #1
redtree
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TL;DR 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?
 

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  • #3
redtree
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Is a wave function a subset of the group of absolutely continuously differentiable functions?
 
  • #4
redtree
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Every wave function can be considered as a superposition of plane waves, where each plane wave is absolutely continuously differentiable.
 
  • #5
fresh_42
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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
redtree
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Right. My mistake on the terminology
 
  • #7
Delta2
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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
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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
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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
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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
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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
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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.
 

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