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Fourier series representation

  1. Jul 8, 2010 #1
    Hi,

    When I solve the diffusion equation for a spherically symmetric geometry in spherical coordinates I obtain the following general solution (after application of the boundary conditions).

    [tex] T(r,t) = \sum_{n=1}^{\infty}\, \frac{A_n}{r}\sin(\lambda_nr)\exp(-\alpha\lambda_n^2t)[/tex]

    So to determine [tex]A_n[/tex] I apply the Initial condition [tex]T(r,0) = f(r)[/tex] which gives me

    [tex] T(r,0) = f(r) = \sum_{n=1}^{\infty} \frac{A_n}{r} \sin(\lambda_nr)[/tex]

    My problem is the [tex]1/r[/tex] term in the above equation. If this term wasn't there this is exactly the same as the half-range Fourier series.

    However I have seen a document (please find it at the bottom) online which still writes A_n as

    [tex] A_n = \frac{2}{b}\int_0^b\, f(r')\sin\left(\frac{n\phi{r'}}{b}\right)r'dr'[/tex]

    this looks very much like the Fourier coefficient of a half-range Fourier series, however, notice the extra [tex]r[/tex] term in there, which is not in the standard Fourier coefficient of a half-range Fourier series. I don't understand how this particular form of Fourier series is obtained?

    http://www.docstoc.com/docs/22184820/Heat-Conduction-in-Cylindrical-and-Spherical-Coordinates-I
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jul 8, 2010 #2
    Hi,
    Can you post the original PDE with the initial conditions.
     
  4. Jul 8, 2010 #3

    nicksauce

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    Just multiply both sides by r, and then you have

    [tex]rf(r) = \sum A_n\sin{\lambda r}[/tex]

    Then you can use the usual method.
     
  5. Jul 8, 2010 #4
    Right.

    Geometrically the idea is that [tex] rf(r)[/tex] is to be viewed as a function, which one presumes to be square integrable over some interval of interest, and then one seeks the representation in that Hilbert space with respect to a Hilbert space basis consisting of sines and cosines. The coefficients are then the inner produce of [tex] rf(r)[/tex] with the sine or cosine function resulting in the integral noted earlier.
     
  6. Jul 9, 2010 #5
    Thanks for the replies.

    So does this mean that now the Fourier coefficient has to be written as

    [tex] A_n = \frac{2}{b}\int_0^b\, rf_0(r)\sin(\lambda{r}) \, dr[/tex]

    And when I substitute it back in to the original equation I have to divide by [tex]r[/tex]

    i.e.

    [tex] f(r) = \sum_{n=1}^{\infty}\left[ \frac{1}{r} \left(\frac{2}{b}\int_0^b\, rf_0(r)\sin(\lambda{r})\right) \, dr \right] \sin(\lambda{r})e^{-\alpha\lambda^2t}[/tex]
     
  7. Jul 9, 2010 #6
    No. You need to adjust the sin functions so that they are orthonormal over the interval. This is all about Hilbert spaces, so you need a Hilbert space basis (aka complete orthonormal set, although the completeness only comes when you throw in the cosines as well).
     
  8. Jul 9, 2010 #7
    Sorry to sound daft here but can't I take

    [tex] g(r) = rf(r)[/tex]

    and then

    [tex] A_n = \frac{2}{b}\int_{0}^{b}\, g(r)\sin(\lambda{r})\, dr [/tex]
     
  9. Jul 9, 2010 #8
    Like I said, not in general. You need to have selected b so that
    [tex]0=\int_{0}^{b}\, sin(\lambda_n{r})\, dr [/tex] and [tex]1=\int_{0}^{b}\, sin^2(\lambda_n{r})\, dr [/tex]

    That is the condition for orthogonormality, of the sine functions that you are using in your Hilbert space basis.

    Think about it in terms of taking the inner product of the series representation with each of the sine functions indifvidually. You need that inner product to pick out the desired coefficient. (This is not completely rigorous, but can be made so with a small bit of work and involving a wee bit of topology, and it gets across the main idea.)
     
  10. Jul 9, 2010 #9
    Let's back up a moment and talk about what the theory of Fourier series really involves.

    There are different ways to get at the theory but the most illuminating way is to look at it in terms of Hilbert spaces.

    So, what is a Hilbert space ? It is just a vector space with a inner product (aka dot product) that is complete (if you don't know what complete is don't worry about it for the moment, it just means that when you go to take a limit the limit is likely to exist).

    Now, in any ordinary vector space if you want to find the component of one vector in the direction of some unit vector you just take the dot product. That is exactly the case for Fourier series. You start with a "complete orthornormal set" of functions defined on some interval [a,b] That complete orthonormal set is the set of functions of the form

    [tex]\frac {1}{b-a}cos(\frac{2 \pi n x}{b-a}) ,\frac{1}{b-a}sin(\frac{2 \pi n x}{b-a}) [/tex].

    The inner product of two functions [tex] f[/tex] and [tex]g[/tex] is

    [tex]<f,g>= \int_a^b f(x)g(x)dx [/tex]

    For ease of notation I will assume in what follows that [a,b] =[0,1]

    So if [tex] f(x) = \sum A_n cos(2 \pi n x) + B_n sin ( 2 \pi n x)[/tex]

    Then [tex]<f(x), cos(2 \pi n x)> = A_n[/tex] and [tex]<f(x), sin(2 \pi n x)> = B_n[/tex]

    The key here is that

    [tex]<cos(2 \pi n x),cos(2 \pi m x> = 0[/tex] if [tex] m \ne n[/tex]

    [tex]<cos(2 \pi n x),cos(2 \pi m x> = 1 [/tex] if [tex] m= n[/tex],

    [tex]<cos(2 \pi n x),sin(2 \pi m x> = 0 [/tex] for all [tex]m,n [/tex],

    [tex]<sin(2 \pi n x),sin(2 \pi m x> = 0[/tex] if [tex] m \ne n[/tex],

    and [tex]<sin(2 \pi n x),sin(2 \pi m x> = 1 [/tex] if [tex] m =n[/tex],
     
  11. Jul 9, 2010 #10
    Thanks very much for that, I was trying to find out more about Hilbert spaces and how it's used for PDEs.
    I've yet to figure out what you've explained but I do like mathematical rigour.
     
  12. Jul 9, 2010 #11
    Halmos wrote a small book, An Introductinon to Hilbert Space that is pretty good and very readable. Beyond that you would need a good book on functional analysis, like Rudin's book, which also give the rudiments of distribution theory which is need for PDEs. It is a pretty big subject.
     
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