The space of solutions of the classical wave equation

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SUMMARY

The classical wave equation in one dimension, represented as \(\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}\), has an infinite-dimensional solution space formed by the basis functions \(\sin{\omega t}\) and \(\cos{\omega t}\). Every solution can be expressed in the form \(f(x+vt)+g(x-vt)\), indicating a change of basis rather than a contradiction. The construction of functions \(f\) and \(g\) relies on Fourier series, which are derived from Sturm-Liouville theory and real analysis, confirming the completeness of the basis in the Hilbert Space \(L^2[0,1]\).

PREREQUISITES
  • Understanding of the classical wave equation and its mathematical formulation.
  • Familiarity with Fourier series and their application in function approximation.
  • Knowledge of Sturm-Liouville theory and its relevance to eigenvalue problems.
  • Basic concepts of Hilbert Spaces, particularly \(L^2[0,1]\).
NEXT STEPS
  • Study Sturm-Liouville theory to understand eigenvalue problems and their solutions.
  • Learn about Fourier analysis and its application in constructing functions from basis sets.
  • Explore the properties of Hilbert Spaces, focusing on completeness and dimensionality.
  • Investigate the relationship between Taylor series and Fourier series in function approximation.
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ShayanJ
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Consider the classical wave equation in one dimension:
<br /> \frac{\partial^2 \psi}{\partial x^2}=\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}<br />
It is a linear equation and so the set of its solutions forms a vector space and because this space is a function space,its dimensionality is infinite.
Also,because \sin{\omega t} and \cos{\omega t} are solutions to the aforementioned equation,every other solution can be formed by a Fourier series,which means {\sin{n\omega t}}_1^{\infty} and { \cos{\omega t}}_1^{\infty} form a basis for the vector space of the solutions of the classical wave equation.
We know that the number of base elements of a vector space shouldn't vary between different bases,but about the classical wave equation,we can tell that every function of the form f(x+vt)+g(x-vt) is a solution.
If it is also right that every solution of the classical wave equation can be written in the formf(x+vt)+g(x-vt),then it seems that we have a basis with only two elements,in contrast to the sines and cosines which make a infinite set of base elements!and this seems to be a contradiction.
Can anyone help?
Thanks
 
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Shyan said:
If it is also right that every solution of the classical wave equation can be written in the formf(x+vt)+g(x-vt),then it seems that we have a basis with only two elements

But f,g - you have to use the basis to construct them!
 
UltrafastPED said:
But f,g - you have to use the basis to construct them!

You're describing a change of basis!
Every element of a base set can be constructed from a linear combination of the elements of another base set!

The sentence you quoted means that I can choose e.g. ln(x-vt) and e^{x+vt} as a basis!

May be there are solutions that are not of the form f(x+vt)+g(x-vt)!
This solves the problem!
 
No, he's right. Every solution is of that functional form. Every solution to the wave equation has a forward traveling wave and a backward traveling wave.

However, the context of your conclusion solution is what is setting you off. First look at sturm-liouville theory, then learn some real and Fourier analysis. The basics are that you construct this f and g from the Fourier series, just as you construct any other vector from a basis, which is determined by solving the separable eigenvalue equations to obtain the eigenvectors and applying the boundary conditions to obtain the eigenvalues. The sin(npix/L) and cos(npix/L) sequences form a basis in L^2[[0,1]:1] (if I remember correctly?) which is a Hilbert Space and is infinite dimensional. Pretty much, it is complete in the since that the sum of Fourier terms can converge to any periodic function in the interval [0,1] in x. They are a lot like taylor series in that sense.
 

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