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One Dimensional Diffusion Equation

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    Solve:

    [tex] \frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}, 0<x<\pi, t>0 [/tex]

    with initial condition

    [tex] u(x,0)=f(x)=\left\{\begin{array}{cc} 1,& 0\leq x< \pi/2 \\ 0, &\pi/2 \leq x < \pi \end{array}\right [/tex]

    and with non-homogeneous boundary conditions

    [tex] u(0,t) = 1, u(\pi,t)=0[/tex]

    3. The attempt at a solution

    I've tried this one a couple different ways, I tried separation of variables, and fourier series. I can get a solution (or infinitely many) to the equation, but I can't seem to make them fit the boundary/initial conditions both.

    My solution for the fourier series method is:

    [tex]u(x,0)=\frac{A_0}{2} + \sum^{\infty}_{n=1} e^{-kn^2t}(A_n \cos(nx) + B_n \sin(nx) [/tex]

    Any ideas how to make this (or any other solution) match both the boundary and initial conditions? Or is the problem incosistent?
     
  2. jcsd
  3. Mar 20, 2007 #2

    mjsd

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    Homework Helper

    it would be easier to see this if you express the function f(x), the initial condition, as a Fourier series.
     
  4. Mar 20, 2007 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There is a standard way of handling problems with non-homogeneous boundary conditions: "homogenize" them!

    The simple linear function [itex]1- \frac{1}{\pi}x[/itex] satisfies the boundary conditions. Now let [itex]v(x,t)= u(x,t)- 1+ \frac{1}{\pi}x[/itex]
    It is easy to see that v(x,t) also satisfies
    [tex] \frac{\partial v}{\partial t} = k\frac{\partial^2 v}{\partial x^2} [/tex]
    but the boundary conditions for v are v(0, t)= 0, [itex]v(\pi,t)= 0[/itex].
    Of course, you have to change the initial condition to
    [tex] v(x,0)=\left\{\begin{array}{cc} \frac{1}{\pi}x,& 0\leq x< \pi/2 \\ \frac{1}{\pi}x-1, &\pi/2 \leq x < \pi \end{array}\right [/tex]

    Now you can write v in a purely sine series:
    [tex]v(x,t)=\sum^{\infty}_{n=1} e^{-kn^2t}B_n \sin(nx) [/tex]

    Of course, as mjsd said, you will eventually have to write the initial condition as a Fourier series in order to find the Bn
     
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