1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Heat Equation: Boundary Value Problem

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

    Uploaded with ImageShack.us

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook