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Re: Odd behaviour of solution of PDE
*To*: mathgroup at smc.vnet.net
*Subject*: [mg116646] Re: Odd behaviour of solution of PDE
*From*: Roland Franzius <roland.franzius at uos.de>
*Date*: Tue, 22 Feb 2011 06:24:51 -0500 (EST)
*References*: <201102211034.FAA22078@smc.vnet.net> <ijthmr$mn1$1@smc.vnet.net> <ijv26f$73a$1@smc.vnet.net>
Am 22.02.2011 02:06, schrieb Alan Ford:
> 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
The answer is given simply by the fact, that after constructing the
matrix of start values at t=0, the step at the boundary is too big to
give correct time step estimates there.
The difference scheme
f_(t+1) (x) = Mean_y h(x, f_t(y)) is used for inner points {x,y}.
At the outer boundary f simply stays fixed.
The difference quotient for the points next to the boundary are
completely wrong estimates then.
--
Roland Franzius
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