Bsrgvty

This is an offspring of this question.

How to get a nice, smooth, uniform plot of the following? I.e. with no horizontal lines, and no ragged boundary at the top. I went with PlotPoints up to 400 and I'm dissapointed. What I'm actually after is a nicely Exported .pdf.

ParametricPlot[1 + a1/(-1 + a2), 
 1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/π, 
 a2 >= -1 && 1 + a1 >= a2 && a1 + a2 <= 1, a1, -2, 2, a2, -1, 
 1, Frame -> True, PlotRange -> 0, 2, 0, 1, 
 AspectRatio -> 1/GoldenRatio, PlotPoints -> 150] // Quiet

To avoid the artifacts from singularities and jumps, we can take a somewhat manual approach.

Notice that the bottom boundary is formed from a2 == -1, the top boundary is a horizontal line formed as a2 -> 1 from the left, and the left boundary is a vertical line formed as a2 sweeps from -1 to 1.

So we can get a clean graphic by plotting the bottom boundary by fixing a2 == -1, extracting the points, and adding the upper left corner to form a polygon.

bdplot = With[a2 = -1,
 ParametricPlot[1 + a1/(-1 + a2), 1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/π, a1, -2, 2, 
 Frame -> True, PlotRange -> 0, 2, 0, 1, AspectRatio -> 1/GoldenRatio] // Quiet
]

pts = Append[MeshCoordinates[DiscretizeGraphics[bdplot]], 0, 1];

poly = Polygon[FindShortestTour[pts][[2, 1 ;; -2]]];

Graphics[GraphicsComplex[pts, EdgeForm[], Hue[0.6, 0.3, 0.95], poly], Frame -> True, AspectRatio -> 1/GoldenRatio]

Now notice that your constraint is not needed:

Reduce[a2 >= -1 && 1 + a1 >= a2 && a1 + a2 <= 1, a1]

(-1 <= a2 < 1 && -1 + a2 <= a1 <= 1 - a2) || (a2 == 1 && a1 == 0)

We see the constraint says a1 should range from -1 + a2 to 1 - a2 instead of -2 to 2. If we plot for many fixed values of a2, we see we'd have the same plot if all a2 were sampled:

Show@Table[
 ParametricPlot[1 + a1/(-1 + a2), 1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/π, a1, -1 + a2, 1 - a2, 
 Frame -> True, PlotRange -> 0, 2, 0, 1, AspectRatio -> 1/GoldenRatio, PlotPoints -> 100] // Quiet,
 a2, Range[-1, 1, .01] /. -1. -> -0.999, 1. -> 0.999
]

As discussed by Chip Hurst, the lower boundary of the region can be obtained by setting a2=-1. Therefore, this boundary is parametrized by a1 only (let it be called $(A,T)$):

reg = With[a2 = -1, 1 + a1/(-1 + a2), 1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/[Pi]]

1 - a1/2, 1 - (2 ArcCsc[2/Sqrt[2 + a1]])/[Pi]

This can be solved to get a1 as a function of A:

sol = Solve[A == reg[[1]], a1][[1]]

a1 -> -2 (-1 + A)

and inserted into T to obtain a function $T(A)$, Then the plotting is done with Filling:

Plot[reg[[2]] /. sol, A, 0, 2, Frame -> True, Filling -> Top, PlotStyle -> None]

The region can then be describe with e.g. ImplicitRegion.

Try option RegionFunction inside ParametricPlot together with the Option MaxRecursions.

The second plot argument 1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/[Pi] is only defined for 1 + a1 >= a2, that's why I only consider this restriction!

ParametricPlot[ 1 + a1/(-1 + a2) ,1 - 2 ArcCsc[2/Sqrt[1 + a1 - a2]]/[Pi] , a1, -2, 2, a2, -1, 1,Frame -> True, PlotRange -> 0, 2 , 0, 1 ,AspectRatio -> 1/GoldenRatio, Evaluated -> True, MaxRecursion -> 4,PlotPoints->50, FrameLabel -> a1, a2,RegionFunction -> Function[x,y,a1, a2, -a1 + a2 <= 1 ]]