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Piecewise initial condition heat equation

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    I have the solution to the heat equation, with the BC's and everything but the IC applied. So I am just trying to solve for the coefficients, the solution without the coefficients is
    $$u(x,t) = \sum_{n=1}^{\infty} A_n\sin(nx)e^{-n^2t}$$
    If the initial condition is ##u(x,0) = f(x)## such that $$f(x) = \begin{cases}
    0 & 0 < x < \frac{\pi}{3} \\
    100 & \frac{\pi}{3} < x < \frac{2\pi}{3} \\
    0 & \frac{2\pi}{3} < x < \pi
    \end{cases}
    $$
    I used the formula $$A_m = \frac{2}{\pi}\int_0^\pi f(x)\sin(mx)dx=\frac{200}{m\pi}\bigg[\cos(\frac{\pi}{3}m) - cos(\frac{2\pi}{3}m)\bigg]$$

    I couldn't find a pattern in the coefficients other than all the even indices go to ##0##. Also is this even correct? When I try to graph this at ##t=0## it isn't giving me the piecewise function ##f(x)##. Is it just that I didn't use enough terms to make it noticeable? Also how would I show that at any time ##t>0## the temperature distribution in the rod achieves a local maximum at ##x=\pi/2##?
     
    Last edited: Feb 2, 2016
  2. jcsd
  3. Feb 4, 2016 #2

    RUber

    User Avatar
    Homework Helper

    That's right. Use more terms. I used 100, and it was still pretty rough. Fourier sums like your solution work well for smooth functions, abrupt changes like that require a lot of terms to reproduce.
     
  4. Feb 4, 2016 #3
    $$\cos(\frac{\pi}{3}m) - cos(\frac{2\pi}{3}m)=\cos\left(\frac{\pi}{3}m\right)-2\cos^2\left(\frac{\pi}{3}m\right)+1$$
    m=1: ##\frac{1}{2}-\frac{1}{2}+1=1##
    m = 2: ##-\frac{1}{2}-\frac{1}{2}+1=0##
    m=3: ##-1-2+1=-2##
    m=4: ##-\frac{1}{2}-\frac{1}{2}+1=0##
    m=5: ##1##
    m=6: ##0##
    m=7: ##1##
    m=8: ##0##
    m=9: ##-2##

    There are 3 separate summations.

    For m=6k+1 (k=0,1,2,...), the coeff is 1
    For m = 6k+3 (k=0,1,2,..), the coeff is -2
    For m = 6k+5 (k=0,1,2,...), the coeff is 1
     
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