Re: Odd behaviour of solution of PDE

*To*: mathgroup at smc.vnet.net*Subject*: [mg116608] Re: Odd behaviour of solution of PDE*From*: Alan Ford <fabio.sattin at igi.cnr.it>*Date*: Mon, 21 Feb 2011 19:28:39 -0500 (EST)*References*: <201102211034.FAA22078@smc.vnet.net> <ijthmr$mn1$1@smc.vnet.net>

Hi Oliver, I had sent an answer to your question, providing a simple piece of code, but it seems the post was lost. So, rather than submitting the whole matter again, I would rather formulate my question as a more general issue. Suppose you want to integrate a PDE in time and in space, the range being a <= x <= b. You have to set boundary conditions. By mistake, you set one boundary condition not at x = b (the extremum of integration) but at x = b + eps (i.e, beyond it). Mathematica does not complain about it and returns a solution. My guess was that Mathematica did integrate from x = a to x = b + eps, in order to match boundary condition, and then returned solution only in the range (a,b). So, no harm, after all. However, it does not look like the case: in all cases I could benchmark mathematica result versus independent solutions, Mathematica simply computes a wrong solution. namely, at the first time step it jumps from the initial condition to a completely wrong value, and then relaxes with a trend quite similar to the correct one. At this time, I'm not asking for hints: it is now quite clear that, contrary to my belief in the first post, mathematica is definitely not providing a correct answer when I supply the "wrong" boundaries. But I am wondering why mathematica does not acknowledge that it is doing an illegal operation Fabio

**References**:**Odd behaviour of solution of PDE***From:*Alan Ford <fabio.sattin@igi.cnr.it>