[ET Trac] [Einstein Toolkit] #2030: Multi-block boundaries leave uninitialized boundary points

Einstein Toolkit trac-noreply at einsteintoolkit.org
Wed Apr 19 11:53:51 CDT 2017

#2030: Multi-block boundaries leave uninitialized boundary points
  Reporter:  eschnett               |       Owner:                     
      Type:  defect                 |      Status:  confirmed          
  Priority:  major                  |   Milestone:                     
 Component:  EinsteinToolkit thorn  |     Version:  development version
Resolution:                         |    Keywords:                     

Comment (by eschnett):

 Synchronization does not play a role here. (Let's be exact about
 terminology to avoid confusion.) Yes, there is part of the grid -- an edge
 -- that is both an inter-patch boundary and a symmetry boundary.

 Let's also assume that we look at the lower left corner of a 2D grid for
 simplicity. Let's assume that the x<0 boundary is a multi-block
 interpolation boundary, and that the y<0 boundary is a symmetry boundary.
 Initially, the interior of the grid (x>0, y>0) is defined.

 The multi-block interpolation requires an interpolation stencil. It can
 never fill the y<0 part of the grid, since the stencil doesn't fit. It can
 only fill the (x<0, y>0) boundary, and for this, it requires all (x>0)
 points to be defined so that the stencil can be evaluated. That means it
 will look at (y<0) points, so it needs to be applied after the symmetry

 The symmetry boundary can define all (y<0) points, and requires the
 respective (y>0) points to be defined for the same x coordinate. So
 initially, since all boundaries are undefined, we can only set the (x>0,
 y<0) part of the boundary via the symmetry condition. Since we don't have
 any values for (x<0, y>0), we cannot define the lower left corner (x<0,
 y<0) with a symmetry condition (yet).

 However, after applying the symmetry condition and defining the (x>0, y<0)
 boundary, now all points with (x>0) are defined. We can now apply the
 multi-block interpolation, which fills part of the (y<0) boundary. Since
 the stencil has a finite size, this defines the (x<0, y>0) points, leaving
 the lower left corner still undefined.

 And now we can, in the final step, apply the symmetry condition again,
 this time to define the missing (x<0, y<0) corner.

Ticket URL: <https://trac.einsteintoolkit.org/ticket/2030#comment:11>
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