In this section we want to expand one of the cases from the
previous section a little bit. In the
previous section we look at the following heat problem.
Now, there is nothing inherently wrong with this problem,
but the fact that we’re fixing the temperature on both ends at zero is a little
unrealistic. The other two problems we
looked at, insulated boundaries and the thin ring, are a little more realistic
problems, but this one just isn’t all that realistic so we’d like to extend it
What we’d like to do in this section is instead look at the
In this case we’ll allow the boundaries to be any fixed
temperature, or . The problem here is that separation of
variables will no longer work on this problem because the boundary conditions
are no longer homogeneous. Recall that
separation of variables will only work if both the partial differential
equation and the boundary conditions are linear and homogeneous. So, we’re going to need to deal with the boundary
conditions in some way before we actually try and solve this.
Luckily for us there is an easy way to deal with them. Let’s consider this problem a little
bit. There are no sources to
add/subtract heat energy anywhere in the bar.
Also our boundary conditions are fixed temperatures and so can’t change
with time and we aren’t prescribing a heat flux on the boundaries to
continually add/subtract heat energy.
So, what this all means is that there will not ever be any forcing of
heat energy into or out of the bar and so while some heat energy may well
naturally flow into our out of the bar at the end points as the temperature
changes eventually the temperature distribution in the bar should stabilize out
and no longer depend on time.
Or, in other words it makes some sense that we should expect
that as our temperature distribution, should behave as,
where is called the equilibrium temperature.
Note as well that is should still satisfy the heat equation and boundary
conditions. It won’t satisfy the initial
condition however because it is the temperature distribution as whereas the initial condition is at . So, the equilibrium temperature distribution
This is a really easy 2nd order ordinary differential
equation to solve. If we integrate twice
and applying the boundary conditions (we’ll leave this to
you to verify) gives us,
Okay, just what does this have to do with solving the
problem given by (1)
above? We’ll let’s define the function,
where is the solution to (1) and
is the equilibrium temperature for (1).
Now let’s rewrite this as,
and let’s take some derivatives.
In both of these derivatives we used the fact that is the equilibrium temperature and so is
independent of time t and must
satisfy the differential equation in (2).
What this tells us is that both and must satisfy the same partial differential
equation. Let’s see what the initial
conditions and boundary conditions would need to be for .
So, the initial condition just gets potentially messier, but
the boundary conditions are now homogeneous!
The partial differential equation that must satisfy is,
how to solve this in the previous section and so we the solution is,
where the coefficients are given by,
The solution to (1) is
and the coefficients are given above.