[Developers] Robin boundary condition in WaveToys is broken

Erik Schnetter schnetter at uni-tuebingen.de
Thu Jan 8 17:51:57 CST 2004


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The WaveToy examples set a bad example by not passing an options table 
to the boundary conditions that they apply.  They state in a comment 
that this is because the use all default values.  This is wrong in the 
case of a Robin boundary condition, which assumes by default a value at 
infinity of 1.0, whereas Wavetoy needs 0 (for the initial data that 
exist).

One solution would be to add parameters to WaveToy so that the user can 
select the value at infinity depending on the physical system.  I would 
prefer changing the default in thorn Boundary.  This default was 
introduced about a year ago when the new boundary mechanism was 
introduced; previously, the user always had to specify a value.

I don't know why the Robin boundary condition test case does not catch 
this.  I assume it is because it uses a very coarse resolution and puts 
the boundary so close in that a Robin boundary condition is not 
justified.

Is it okay to change the default for the Robin boundary condition?  A 
value of zero at infinity seems to be more generic than a value of one.

- -erik

- -- 
Erik Schnetter <schnetter at aei.mpg.de>   http://www.aei.mpg.de/~eschnett/

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