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Homework Help: PDE: heat equation

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    u[itex]_{t}[/itex]=3u[itex]_{xx}[/itex] x=[0,pi]
    u(0,t)=u(pi,t)=0
    u(x,0)=sinx*cos4x

    2. Relevant equations



    3. The attempt at a solution

    with separation of variables and boundry conditions I get:

    u(x,t)= [itex]\sum[/itex]B[itex]_{n}[/itex]e[itex]^-3n^{2)}}*sinnx[/itex]

    u(x,0)=sinx*cos4x

    f(x)=sinx*cos4x=[itex]\sum[/itex]B[itex]_{n}*sinnx[/itex]

    And here is where I am stuck! I tried computing B[itex]_{n} by computing it like a Fourier coeff. of f(x) but all I got was zero... I don't really know where to go from here.

    I'm having a hard time with Fourier analysis, that's why I have bombarded this forums with question these last couple of days. I really appreciate the help I get.
     
  2. jcsd
  3. Sep 16, 2011 #2
    Hey,
    It might help to consult the literature on this; Try http://en.wikipedia.org/wiki/Heat_equation#equation_6".

    And as for your case, in particular, can you please show us how did you integrate the Coefficient? It should be something along the lines of:
    [tex]
    B_n \propto \int_{0}^{\pi}f(x)\sin(nx\pi)
    [/tex]
    as per your statement, certainly not zero!

    Daniel
     
    Last edited by a moderator: Apr 26, 2017
  4. Sep 16, 2011 #3
    Bn= 2/L *integral sinx*cos4x*sinnpix/L

    L should be pi since my intervall is 0 to pi, or have I misunderstood something? Then integral becomes:

    http://www.wolframalpha.com/input/?i=integrate+sinx*cos4x*sin(n*x)

    sin(n+5)pi should be zero for all n?
     
    Last edited by a moderator: Apr 26, 2017
  5. Sep 17, 2011 #4
    Last edited by a moderator: Apr 26, 2017
  6. Sep 17, 2011 #5
    Thanks dude, this problem is driving me insane.
     
  7. Sep 17, 2011 #6
    Okay,
    Firstly I solved your problem numerically, and received the following graph, tested it using two schemes, and it's mighty accurate!(Pat on the back here :)).
    Since we're getting zero here, one naturally inclines to use a Taylor expansion of the series, around Pi.
    In the case of Sin(n*x), we'll get, up to the second order:
    That the expansion of the integral, at large, gives:
    [itex]B_n = \frac{(15 + n^2) \sin(n \pi)}{((-5 + n) (-3 + n) (3 + n) (5 + n))} [/itex]
    This resolves, for most N, as:
    [itex]B_n = \frac{-n(x-\pi)^3}{3} [/itex]
    I would advise you to plot the resulting function using this expansion, and see whether it matches the attached diagram.
    Lets hope it works!
    Daniel
     

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  8. Sep 18, 2011 #7
    Thanks! Pat on back is well deserved =). Will try this. Do you think there is a way to solve this analytical?
     
    Last edited: Sep 18, 2011
  9. Sep 18, 2011 #8
    The Fourier solution method is very discreet and is only effective for particular cases. I would advise you to use, in such PDEs and those beyond these(i.e Hyperbolic ones etc.) Numerical methods..
    Daniel
     
  10. Sep 20, 2011 #9
    I tried this problem again this morning and solved after 5 minutes. I am sooooo stupid.

    sinx*cos4x = 1/2*(sin3x+sin5x)

    So: ∑Bn∗sinnx = 1/2(sin3x+sin5x)

    ---> Bn1sinn1x+Bn2+sinn2=1/2(sin3x+sin5x)

    Bn1=Bn2=1/2

    n1=3
    n2=5

    And its solved. That is all.
     
  11. Sep 20, 2011 #10
    Well done!
    I had a feeling this had to sit with reducing your compound expression with the cosines and sines...
    Congratulations, and kudos!
    Daniel
     
  12. Sep 20, 2011 #11

    Ray Vickson

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    Science Advisor
    Homework Helper

    Why does your summation formula for u(x,t) not have t in it anywhere?

    RGV
     
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