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Fourier seroes and separation of variables

  1. Mar 22, 2008 #1
    1. The problem statement, all variables and given/known data
    Sorry for the mis-spelled title - it's "series".

    Please take a look at

    http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node21.html

    In step 2.60, when he wants to find the coefficient B_n, the argument in the sine-function does not contain a "2". In my book, the argument in the sine function is 2*pi*n*x/L, where L is the period. Why is this "2" not in the example?
     
    Last edited: Mar 22, 2008
  2. jcsd
  3. Mar 22, 2008 #2
    Is that because the period is 2L? I really don't get it.

    They do the same here (page 2, left side): http://walet.phy.umist.ac.uk/2C1/Handouts/Handout5.pdf

    They use the period L, instead of 2L? I hope you can explain this to me. The other reason might be that it's an even function, but it's not since they are finding B_n. So where am I going wrong?

    Thanks in advance.
     
    Last edited: Mar 22, 2008
  4. Mar 22, 2008 #3
    No hints :-)?
     
  5. Mar 22, 2008 #4
    It depends on the context. What is the physical context of the problem in your book? What are the boundary conditions there?
     
  6. Mar 22, 2008 #5
    The general formula for the Fourier-coefficient of an odd function, in our case B_n, is given by:

    B_n = [tex]\frac{2}{L}\int f(x)\cdot \sin (\frac{2\pi rx}{L})dx[/tex], where L is the period. The period in the first example (the one at eq. 2.60) has the period T = 2l. He integrates from 0 up to l, not 0 up to 2l. Why is that?

    And instead of 1/l, he has 2/l - again, I do not see why?
     
    Last edited: Mar 22, 2008
  7. Mar 22, 2008 #6
    Right, I'm saying it is determined when you set up the problem. For example, if I had a circle of length l, then I might want to change the total length to 2L for convenience. You would have to describe the problem in your book for me to really see what it is doing. The examples you listed are the standard approaches to a heat equation.
     
    Last edited: Mar 22, 2008
  8. Mar 22, 2008 #7
    The equaton I wrote for B_n was what I was referring to in my book, not an example.


    But I do not understand why he integrates from 0 to l, and not from 0 to 2l. He only integrates over half a period.

    I also do not understand why he multiplies with the factor 2/l, and not 1/l? This does not make sense, since the formula for B_n containts 2/L, and inserting L = 2l, we get 1/l.
     
  9. Mar 22, 2008 #8
    Alright, just back off from the equation for a moment and understand from where the equation comes from. The example from the first website has a heated rod, where the ends of the rods are fixed at x=0 and x=l. How would it make since to integrate beyond the boundary? It wouldn't.

    Now, as I was saying, the physics of the problem determines your solution, not plug and chug equations. These things need to be derived, not used.
     
    Last edited: Mar 22, 2008
  10. Mar 22, 2008 #9
    Thanks, I hadn't thought of that.

    Now I only have one concern. The "2" in the sine-argument - where does that go if the period is 0 to l?

    Thank you for your patience.
     
  11. Mar 22, 2008 #10
    Well, so when you get your separable ODEs you have (separation constant k^2)

    [tex]F(x) = c_1cos(kx) + c_2sin(kx)[/tex]

    So if you have the boundary conditions F(0)=0 and F(2L) = 0 then after imposing F(0) you would have

    [tex]F(x) = c_2sin(kx)[/tex]

    and with the second BC

    [tex]F(2L) = c_2sin(2Lk) = 0[/tex]

    So you might want to say that this means c_2 is zero, but you have to remember that if c_2 was zero then you wouldn't have a solution, the solution would just be zero. This means that the sine term has to be zero, which means that

    [tex] 2Lk = n\pi[/tex]

    or

    [tex] k = \frac{n \pi}{2L}[/tex]

    But if your boundary condition was at zero at x=L then you would have

    [tex] k_{other} = \frac{n \pi}{L}[/tex]
     
  12. Mar 22, 2008 #11
    It would be a whole different story if you were defined on -l to l also. My suggestion is to go over a few examples, and make sure you can set these up right and understand the ideas.
     
  13. Mar 23, 2008 #12
    I understand what you did in your second-last post.

    I am so confused now. We agree that the period T = 2L. But it does not make sense to use T = 2L, since the rod is only L long, so we have to rewrite.

    I understand everything up till 2.60. The things I do not understand are:

    1) The period is 2L, so we get [tex]\frac{2}{2L}\int f(x)\cdot \sin (\frac{2\pi rx}{2L})dx [/tex] = [tex]\frac{1}{L}\int f(x)\cdot \sin (\frac{\pi rx}{L})dx[/tex].

    We have the boundaries from 0 to 2L, but me must rewrite that since the rod has length L. How can we possibly rewrite, since we are dealing with an odd function? So here, the integral times 2 from 0 to L is NOT the same as the integral from 0 to 2L.

    So, where does the "2" infront of the integral come from?

    That is all I am missing now. Am I wrong about what I wrote in 1?

    Again, thanks for being patient. I have looked at other examples (e.g. http://walet.phy.umist.ac.uk/2C1/Handouts/Handout5.pdf), and they do exactly the same, and I do not see why.
    I am getting quite mad at myself for not understanding this - I am just following the "rules" in my book.
     
  14. Mar 23, 2008 #13
    All of the examples are right, if you can understand them then maybe you can understand your book example. Also, if you want to post the way your book does its example, then I can figure out the discrepancy.

    Your way isn't quite right. If we have a "period," I would prefer to look at it as a physical length because it really isn't a period, of 2L then we would get a general solution

    [tex]u(x,t) = \sum_{n=1}^\infty B_n sin(\frac{n \pi}{2L})exp[-(\frac{n \pi}{2L})^2 t][/tex]

    and with an initial condition defined as f(x) such that

    [tex]u(x,0)=\sum_{n=1}^\infty B_n sin(\frac{n \pi}{2L}) = f(x)[/tex]

    then to exploit orthogonality we will want to multiply both sides by an the integral of sin with m indices

    [tex]\int_0^{2L} f(x) sin(\frac{m \pi}{2L}) dx = \int_0^{2L}\sum_{n=1}^\infty B_n sin(\frac{n \pi}{2L}) sin(\frac{m \pi}{2L}) dx[/tex]

    So the right side, from orthogonality, is going to be a kronecker delta times L (the same is if you had integral 0 to L multiplied by two, since you could easily just redefine the length of 2L to be L), which gives:

    [tex]\frac{1}{L} \int_0^{2L} f(x) sin(\frac{n \pi}{2L}) dx = B_n[/tex]
     
  15. Mar 23, 2008 #14
    I've heard of Kronecker delta function once in linear algebra, but it was just a quick remark given by the lecturer, so I would not know how to use it here.

    Is another way of looking at it that [tex]\frac{1}{L}\int f(x)\cdot \sin (\frac{\pi rx}{L})[/tex] (limits from 0 to 2L and period 2L) is a product between two odd functions (since f(x) has only sine in its Fourier series it must be odd and sin(...) is an odd function), which results in an even function, and therefore we can use the limits 0 to L (which is what we want because of the rod's length) and then we multiply with two, because even functions have that special ability?

    Is that consistent?

    EDIT: The example in my book is actually an exercise - http://books.google.dk/books?id=9p6...OTs7wZK&sig=zi1tcn6GTishIvYo30egcs3lY_o&hl=da - and the way I see it, they do as I wrote just above (in this post). So it would be great if I am right :-)
     
    Last edited: Mar 23, 2008
  16. Mar 23, 2008 #15
    We don't necessarily know that f(x) is an even or odd function, it could even be neither even nor odd.

    Maybe let's start with the exact problem (I don't know which of the exercises it is), and just work through it.
     
  17. Mar 23, 2008 #16
    Well, don't we see that f(x) is odd because it can be written as a Fourier-series with only a sine-term in it - hence it must be odd?

    If this is not true, I will write what I have found out for problem 19.15.
     
  18. Mar 23, 2008 #17
    That would be true if we had a full fourier series defined from -L to L, because odd functions will integrate to zero over the bounds and an odd times an even is an odd; hence, the function f(x) would have to be odd. But that is not true in this case simply because of where we have defined the boundaries. The sine term comes out of the dirichlet boundary conditions, the actual physics of the problem.
     
  19. Mar 24, 2008 #18
    Well, I can just choose to look at the area from -L to L instead of 0 to 2L?

    I am right when I say that it does not matter whether we look at 0 and T or -T/2 and T/2 (T being the period)?

    I hope you do not think of me as being difficult.
     
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