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Heat Equation: Boundary Value Problem

  1. Mar 8, 2012 #1
    http://img821.imageshack.us/img821/7901/heatp.png [Broken]

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    I'm having difficulty with the boundary conditions on this problem. I don't need a solution or a step by step. I've just never solved a boundary condition like this.

    Its the u(pi,t) = cos(t) that is giving me difficulty

    I tried getting a steady state solution for this. However, I end up with

    v(x) = (x/pi)*Cos(t)

    which makes no sense because v(x) should not be dependent on 't.'

    I can't make it homogeneous in order to solve it by separation of variables. Any advice would be greatly appreciated.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 8, 2012 #2
    Your "steady state" is correct. You shouldn't expect a time-independent steady state for a problem like this since you have time-dependent forcing from the boundary. To see that, imagine the wave equation equivalent. If you have a string with some initial condition, but you keep shaking one end of the string indefinitely, you don't expect it to have a true steady state, right?

    Anyway, the trick to solving a heat equation with inhomogeneous boundary conditions is to introduce a change of variables that allows you to make it homogeneous; namely,

    v(x,t) = u(x,t) - u0(x,t)

    where u0(x,t) is any function that satisfies the boundary conditions, i.e. u0(0,t) = 0, u0(pi,t) = cos(t). Can you think of any function that would accomplish this?
     
  4. Mar 8, 2012 #3
    Hmmm, I guess I got confused by v(x) having cos(t) in it. My book literally says "find a function v(x), independent of 't.'

    I'll keep working it. I'll return if I need help. Or if I figure it out.
     
  5. Mar 8, 2012 #4
    It would certainly be nice if v had no time dependence, but now it just becomes an inhomogeneous heat equation with homogeneous boundary conditions
     
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