1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     

    Attached Files:

    • G_1.bmp
      G_1.bmp
      File size:
      252.7 KB
      Views:
      101
  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

    User Avatar
    Science Advisor
    Homework Helper

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

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook