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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

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  • $\begingroup$ 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 Jan 28 '15 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 Jan 28 '15 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 Jan 28 '15 at 22:07
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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.

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  • $\begingroup$ 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 Jan 29 '15 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 Jan 29 '15 at 23:45

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