Problem 1.3 #6

Heat flow in a metal rod with a unit internal heat source is governed by the problem
$$u_t = ku_{xx} + 1, \: \: u(0,t)=0, \: u(1,t) = 1.$$
What will be the steady state temperature in the bar after a long time?
Does it matter that no initial condition is given?

A good answer should include a graph of the steady state function together with a qualitative explanation of why it's reasonable.

The truly Awesome Answer might include an animation of the evolution from some particular initial condition!

I put together a plot of the solution using an initial condition of $u(x,0)=0$. That said, I'm at a loss as to how to explain its behavior. It would be greatly appreciated if someone would explain it to me!

I think that plot is good. We can find a solution based off the form:

$\text{u(x,t) = }
e^{-t} \sin \left(\sqrt{k} x\right)$

From the way this situation is governed the solution for the temperature might look like the following:

$\text{u(x,t) = }
ke^{-t} \sin \left(\sqrt{k} x\right)-\frac{x^2}{2k}$

More generally for $u_t\text{=k}u_{\text{xx}}+q$, for arbitrary $q$ our solution becomes:

$\text{u(x,t) = }
ke^{-t} \sin \left(\sqrt{k} x\right)-\frac{x^2}{2k}{q}$

If we had $q=10$ we get the following steady state: ***

Since our solution is some temperature plus some constant temperature $q$ the entire function $\text{u(x,t)}$ shifted by $q$ up (or down). The boundary conditions which are holding $\text{u(x,t)}$ are still present so the steady state should bulge more the closer to $x=1/2$. It should also be more pronounced for higher values of $q$.

Initial conditions for a similar effect:
$u_t
\text{=k}
u_{\text{xx}}
\text{ - 1, u(0,t) = 1, u(1,t) =1, u(x,0)=1}$

*** New user, can't upload gifs/images. Will edit

Great explanation! I figured it had to be the case that the constant was shifting the function up while keeping the boundaries locked, but I wasn't sure how to justify that.