# Heat Equation: Boundary Value Problem

1. Mar 8, 2012

### JonathanT

http://img821.imageshack.us/img821/7901/heatp.png [Broken]

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. Mar 8, 2012

### tjackson3

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?

3. Mar 8, 2012

### JonathanT

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.

4. Mar 8, 2012

### tjackson3

It would certainly be nice if v had no time dependence, but now it just becomes an inhomogeneous heat equation with homogeneous boundary conditions