0
$\begingroup$

I am looking at a heat initial value problem \begin{align} \frac{\partial u}{\partial t}-\nabla^2u = f\quad&\text{in}\quad \Omega\times(0,T)\\ u = g \quad&\text{on}\quad \partial\Omega\times(0,T)\\ u = u_0\quad&\text{in}\quad \Omega\times\{0\} \end{align} which I have succesfully solved with the finite element method along with a time-stepping scheme, so that is not my question. My question is however, it seems to work to just solve the boundary value problem \begin{align} -\nabla^2u = f\quad&\text{in}\quad \Omega\\ u = g \quad&\text{on}\quad \partial\Omega\\ \end{align} for various fixed times, e.g. $f(x,y,0),f(x,y,0.1),$ etc. This seems to give a good approximation of the solution to the initial value problem, and I don't understand why. This simply ignores the time-derivative it seems to me (of course, the functions $f$ are the same for both problems so the "information" of the time-derivative is not lost, but it still confuses me). Is this simply the method of lines?

Thanks

Improve this question
$\endgroup$
3
  • $\begingroup$ If you remove the time dependence in the boundary conditions, how quickly does $u(x,y,t)$ approach a stead state solution? It could be that this time scale is much shorter than the longer time scale of changes in the boundary conditions. $\endgroup$
    – Brian Borchers
    Commented Jan 28, 2015 at 20:29
  • $\begingroup$ @BrianBorchers I don't quite understand what you mean. It could be that $u(x,y,t)$ does not converge to anything steady as $t\to\infty$ for example for $u(x,y,t) = \sin(2\pi x t)\cos(2\pi y)$. $\endgroup$
    – Eff
    Commented Jan 28, 2015 at 20:33
  • $\begingroup$ @Elf. Yes of course. What I'm asking is if for your particular problem the solution does reach a steady state very rapidly compared to changes in the boundary conditions. e.g. if for any reasonable initial condition, the time dependent equation effectively reachies a steady state in 1.0e-15 seconds, while the boundary conditions are changing slowly on a time scale of 0.1 seconds, then you'd see the behavior you've described. $\endgroup$
    – Brian Borchers
    Commented Jan 28, 2015 at 22:07

1 Answer 1

Reset to default
1
$\begingroup$

If $f$ and $g$ are functions that depend on the spatial variable, but not on time, then $u(x,t)$ converges to a function $\bar u(x)$ that satisfies the Poisson equation $$ -\Delta u=f, $$ together with $u|_{\partial\Omega}=g$. It does so regardless of the initial conditions. In other words, the solution of the heat equation is, in a sense, not very interesting if you consider the "long-term" behavior.

What constitutes "long-term" depends on the eigenvalues of the Laplace operator on the domain you're on, as well as the decomposition of the initial condition and the right hand side term with regard to the eigenvectors of the Laplace operator. It may simply be that in the situation you describe, you are always in the long-term case and so you do not observe behavior that is fundamentally different from just the Laplace equation.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you for the answer and for giving me something new to study (the last paragraph). I have generally (for testing) chosen a solution a priori, e.g. $u(x,y,t) = \text{exp}(-100(x-t)^2-100(y-0.5)^2)$, and then computed $f$ from the known solution. So it will depend on time. $\endgroup$
    – Eff
    Commented Jan 29, 2015 at 5:34
  • $\begingroup$ Take any book on PDEs that discusses the solution of the heat equation as a sum over eigenfunctions and you will understand how my answer came about. $\endgroup$
    – Wolfgang Bangerth
    Commented Jan 29, 2015 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.