plotting - why this call to StreamPlot returns unevaluated? (no error message, no beep)

I am really baffled by this. Call to StreamPlot does not return empty plot, nor an error message, nor a beep. It just return unevaluated.

I do not think I've seen something like this before. What would the cause for this?

Is this expected behavior?

Normally, when a plot can not be generated, either an error is returned or empty plot.

Mathematica graphics

ClearAll[x, y, f]; 
f = (x*y - Sqrt[-1 + x^2 + y^2])/(-1 + x^2); 
StreamPlot[{1, f}, {x, -2, 2}, {y, -2, 2}]

btw, it will plot OK when changing the -1 to 1 in the above, under the square root:

Mathematica graphics

So it seems due to value of the function becomes complex over some region. But normally when this happens, an empty plot is returned, right?

V 12 on windows 10.

Answer

I think it should give an error message, perhaps. I agree with the OP's guess that problem stems from the complex values of the vector field. It may be that it's because the field is real around the boundaries and then becomes complex. Here is a workaround:

StreamPlot[
 Evaluate[ConditionalExpression[#, # ∈ Reals] & /@ {1, f}],
 {x, -2, 2}, {y, -2, 2}]

Update: Here is another that seems to be more robust. It zeros out the complex-valued parts, and zero vectors are not drawn nor lead to stream lines:

StreamPlot[Boole[f ∈ Reals] {1, f}, {x, -2, 2}, {y, -2, 2}]

Comments

plotting - Filling between two spheres in SphericalPlot3D

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plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

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