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Heat equation

  1. Apr 18, 2007 #1

    I am solving the heat equation [tex] 1/a^2\theta_t = \theta_x

    boundary conditions are [tex] \theta(0,t) = \theta(L,t) = 0 t > 0 [/tex]

    initial conditions are [tex] \theta(x,0) = T_0sin(x\pi/L) [/tex]

    now I have derived the steady solution to be 0 and I have derived that the general solution will be of the form

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

    I am next required to determine the sequence [tex] {B_n} \\

    \mbox{now usually I would proceed in the following manner, using the initial conditions}\\
    \theta(x,0) = \sum_{n=1}^\infty B_n sinn\pix/L = T_0sin(\frac{\pix}{L})[/tex]

    I would then multiply both sides by [tex] \int_{0}^{L}sin\frac{mx\pi}{L} [/tex]

    but instead i decided first to separted term [tex] B_1 [/tex] and cancel term sin(mx\pi/L)

    so I proceeded using the cancellation by the orthogonality conditions and derived the following

    [tex] B_mL/2 = (T_0 -B_1)\int_{0}^{L}sin\frac{mx\pi}{L}\\

    = \frac{(B_1 -T_0)}{m\pix}\left[cosmx\pi/L\right]_{0}^{L}\\
    = \frac{B_1-T_0}{m\pi}\left[(-1)^m -1] [/tex]

    you see the problem is I end up with [tex] B_1 [/tex] undetermined
    [tex] B_1 = \frac{-4T_0}{m\pi -4} [/tex]

    where did I go wrong/ what should I have done differently??:frown:

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Apr 18, 2007
  2. jcsd
  3. Apr 18, 2007 #2


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    You didn't do anything wrong except that you guarenteed you would not be able to determine B1 when you "decided first to separted term [itex]B_1[/itex] and cancel term sin(mx\pi/L)"!

    There are many different ways to do this but I would probably expand [itex]T_0 sin(\pi x/L)[/itex] in a Fourier sine series in [itex]\pi x/L[/itex] and equate coefficients.
  4. Apr 18, 2007 #3
    mmm...I am not familiar with this method, you see the reason I took out term B1 was because I knew it would allow me to proceed by multplying by

    [tex] sin\frac{m\pix}{L} [/tex]and then integrating from 0 to L. Without taking out term B1 this would eventually lead me to do the following integral
    [tex] L/2B_m = T_0\int_{0}^{L} sin(\frac{m\pix}{L})sin(\frac{pix}{L})dx [/tex]

    is this also a valid method of solution and if so How do I do the integral? Do I have to use some sort of trig identity
    Last edited: Apr 18, 2007

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