# Solving Initial Value problem ignoring the time-derivative

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

• 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. – Brian Borchers Jan 28 '15 at 20:29
• @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 $u(x,y,t) = \sin(2\pi x t)\cos(2\pi y)$. – Eff Jan 28 '15 at 20:33
• @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. – Brian Borchers Jan 28 '15 at 22:07

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.
• 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. – Eff Jan 29 '15 at 5:34