Inverted cells for 3d limited regions.

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Inverted cells for 3d limited regions.

Hello,

I did a quick search and found some information on inverted cells (see "Reciprocal Problems" topic for that) but it didn't seem to address the issues I have been having. For some reason the code below results in an inverted cell. If one takes out the limited region then no inverted cell results. You will notice I defined both regions as "limited". This is because in the actual code we are working with they need to be limited regions. However, even if you change the first limited region to a region, the problem persists (one can easily comment/uncomment lines in the code to see this). This seems to be a very simple geometry so I'm not sure why I'm getting this error and what might be done to fix this. Hope to hear from someone on this soon.

Jared

TITLE 'Inverted_cell_issues' { the problem identification }
COORDINATES cartesian3 { coordinate system, 1D,2D,3D, etc }
VARIABLES { system variables }
u { choose your own names }
SELECT { method controls }
ngrid = 1
aspect = 100
curvegrid = off
order = 3
DEFINITIONS { parameter definitions }
fadein = 100
delta = 0.1
k
z0 = 0
z1 = 1/3
z2 = 2/3
sm_slope = 0.01
z3 = if y > -2 then (y-(-2))/(3/2)*z2+((-1/2)-y)/(3/2)*1 else 1+sm_slope*(-2-y)
EXTRUSION z = z0,z1,z2,z3
EQUATIONS { PDE's, one for each variable }
div(k*grad(u))=0 { one possibility }
BOUNDARIES { The domain definition }

LIMITED REGION 1 { For each material region }
LAYER 1 k = 1
LAYER 2 k = 1
LAYER 3 k = 1
! REGION 1 k = 1
START (0,-2) arc (center = 0,0) to (1,-sqrt(4-1)) line to (1,-1/2) line to (0,-1/2) line to close
LIMITED REGION Layer 2 k = 2 START (0,-2) arc (center = 0,0) to (1/3,-sqrt(4-1/9)) line to (1/3,-1/2) line to (0,-1/2) line to close
MONITORS { show progress }
PLOTS { save result displays }
CONTOUR(u) on z = 1/2
elevation(z2) from (-1/2,1) to (1,1)
END
Jared Barber

Posts: 47
Joined: Thu Jan 24, 2013 10:37 am

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