1. Not finding help here? Sign up for a free 30min 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!

Heat equation

  1. Apr 18, 2007 #1
    1. The problem statement, all variables and given/known data

    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


    User Avatar
    Staff Emeritus
    Science Advisor

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

Have something to add?

Similar Discussions: Heat equation
  1. Heat Equation (Replies: 1)

  2. Heat equation (Replies: 2)

  3. The heat equation (Replies: 17)