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General Question: If I have an IVP with periodic and continuous initial condition, which rules the accuracy of the scheme - the manner in which we approximate spatial derivative or the acuuracy of the ODE solving scheme.

Specific Case: I'm solving

$u_t = u_{xx} + u$ ,

$x \in \mathbb{R} $, $t>0$, $u(t,x) = u(t, x + \pi )\forall x \in \mathbb{R} $

$u(t =0, x) =cos(x) + sin(2x)$

With the following approach - I discretisize $x$ to a grid $ x_0 ... x_N$, and then solve numerically a system of $N$ coupled ODE's.

It turns out that no matter how I approximate $u_{xx}$, the global error of the solution depends only on the scheme with which I solved the ODE - Forward Euler, Adam Beshfort, Runge Kutta etc.

I looked at this question in MO, and even when I took the spectral approach, it hasn't changed the global error.

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In general, when solving a PDE numerically, both the spatial and the temporal discretization will (of course) contribute to the local truncation error and hence to the global error. In many equations it is often the case that the spatial error is dominant. Of course, the choice of particular discretizations matters.

In the case of the reaction-diffusion system you're solving, it seems reasonable that the temporal error would dominate. The spatial operator is $\partial_{xx}$, which is dissipative, which means the errors committed in space will be rapidly damped out. On the other hand, the temporal discretization has to deal with the reaction term which leads to exponential growth of the solution, and errors committed there will be amplified. Another way to look at it is that the diffusive term will rapidly kill off all but the lowest-frequency Fourier components of the solution, leaving a slowly-varying (in space) solution that can be very accurately approximated.

I should add that you haven't given enough detail for me to be completely certain of what's going on (what are the boundary conditions; what spatial discretizations have you tried; how are you determining the error; what spatial and temporal mesh widths are you using, etc.). All of those factors could be relevant to the answer. But the above explanation seems the most likely one.

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  • $\begingroup$ I added boundary condition. Does it help? thnx $\endgroup$
    – Amir Sagiv
    Commented Oct 3, 2014 at 12:19
  • $\begingroup$ Yes, but it doesn't change the answer. $\endgroup$ Commented Oct 5, 2014 at 17:53

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