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In a very simple example I try to define a rectangular region

u = 1/Sqrt[2]; 
reg = ImplicitRegion[(-1 <= x <= 1 + u) && (-1 <= y <= 0 ) , {x, y}] 
RegionPlot[ reg  ]

enter image description here

Obviously the plot isn't complete.

What's wrong? Thanks!

Ulrich Neumann
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  • I'm surprised that it returned anything. The documentation for RegionPlot shows its argument as a predicate rather than a region. For example, RegionPlot[reg[[1]], {x, -1, 1 + u}, {y, -1, 0}]. For the region use Region[reg, Frame -> True, AspectRatio -> 1] – Bob Hanlon Dec 22 '19 at 20:42
  • @BobHanlon Thanks, but documentation of ImplicitRegion show several examples like \[ScriptCapitalR] =ImplicitRegion[ x^6 - 5 x^4 y + 3 x^4 y^2 + 10 x^2 y^3 + 3 x^2 y^4 - y^5 + y^6 ==0, {x, y}];RegionPlot[\[ScriptCapitalR]] – Ulrich Neumann Dec 22 '19 at 20:49
  • Repeatable in version 12.0 on Win 10 Ent. Post to WRI support to see if it is a bug. – Edmund Dec 22 '19 at 21:16
  • I can confirm the issue on V12.0.0 (macOS). DiscretizeRegion[reg, Frame -> True] and RegionMeasure[reg] work correctly though. – Roman Dec 22 '19 at 21:20
  • I discovered these flaws of RegionPlot[] back in version 11.3 - see https://mathematica.stackexchange.com/questions/180959/solver-for-unsteady-flow-with-the-use-of-the-navier-stokes-and-mathematica-fem – Alex Trounev Dec 22 '19 at 23:24
  • @AlexTrounev Thanks for the very interesting link. I've to elaborate... – Ulrich Neumann Dec 23 '19 at 11:49
  • @UlrichNeumann You are welcome! – Alex Trounev Dec 23 '19 at 13:08

1 Answers1

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  • Method -> {"DiscretizationMethod" -> "Symbolic"} work for this case.
u = 1/Sqrt[2];
reg = ImplicitRegion[(-1 <= x <= 1 + u) && (-1 <= y <= 0), {x, y}]
RegionPlot[reg, Method -> {"DiscretizationMethod" -> "Symbolic"}]

enter image description here

cvgmt
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