The space of solutions of the classical wave equation

In summary, the classical wave equation in one dimension is a linear equation with an infinite-dimensional vector space of solutions. The solutions can be expressed as a Fourier series with sine and cosine functions as a basis. However, it is also possible to construct a basis with only two elements, f(x+vt) and g(x-vt), which may seem contradictory. However, this is resolved by understanding that every solution can be written in the form f(x+vt)+g(x-vt) and that these functions can be constructed from the Fourier series basis.
  • #1
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Consider the classical wave equation in one dimension:
[itex]
\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{v^2} \frac{\partial^2 \psi}{\partial t^2}
[/itex]
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 [itex] \sin{\omega t} [/itex] and [itex] \cos{\omega t} [/itex] are solutions to the aforementioned equation,every other solution can be formed by a Fourier series,which means [itex] {\sin{n\omega t}}_1^{\infty}[/itex] and [itex]{ \cos{\omega t}}_1^{\infty} [/itex] 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 [itex] f(x+vt)+g(x-vt) [/itex] is a solution.
If it is also right that every solution of the classical wave equation can be written in the form[itex] f(x+vt)+g(x-vt) [/itex],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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  • #2
Shyan said:
If it is also right that every solution of the classical wave equation can be written in the form[itex] f(x+vt)+g(x-vt) [/itex],then it seems that we have a basis with only two elements

But f,g - you have to use the basis to construct them!
 
  • #3
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. [itex] ln(x-vt) [/itex] and [itex] e^{x+vt} [/itex] as a basis!

May be there are solutions that are not of the form [itex] f(x+vt)+g(x-vt) [/itex]!
This solves the problem!
 
  • #4
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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