Htydjtk

While looking at How do I Plot a Divergence?, I thought to suggest the following as a solution to the OP's problem (where potdistr is an InterpolatingFunction solution to a PDE returned by NDSolve):

VectorDensityPlot[

 Evaluate[Grad[potdistr[x, y], {x, y}]],

 {x, -0.01, 0.11}, {y, -0.005, 0.053}]

I surprised that it worked once and then failed on subsequent calls. It seems to be connected to InterpolatingFunction and values being set for x and y. Here is a minimal example:

field = {Interpolation[Range[4]^2/4][y], x};

VectorDensityPlot[field, {x, 1, 4}, {y, 1, 4}]

Subsequent calls generate a InterpolatingFunction::dmval extrapolation warning message and the vector field is constant.

The problem does not occur if field = {y^2/4, x} is used. It does not occur for VectorPlot, DensityPlot, ContourPlot, or Plot3D.

What's going on? Is it a bug? Is there a way to get it to work?

Additional info:

$Version

(*  "11.3.0 for Mac OS X x86 (64-bit) (January 22, 2018)"  *)

Filed as [CASE:4228039]

What is happening is that x and y are being set equal to numeric values. (One might notice that the color for x and y changes from blue to black, but my eyes have trouble seeing that for single-letter variables.) For some reason, these values are outside the domain specified in the plot. I think this must be a bug and have reported it to WRI.

Clear[x, y]

{x, y}

field = {Interpolation[Range[4]^2/4][y], x};

VectorDensityPlot[field, {x, 1, 4}, {y, 1, 4}]

{x, y}

One possible workaround is to clear the variables after plotting with Clear[x, y]. Another is to use Block:

Block[{x, y}, VectorDensityPlot[field, {x, 1, 4}, {y, 1, 4}]]

Further, it does not seem restricted only to InterpolatingFunction. The following has the same issue, and, further, VectorStyle is ignored:

Clear[x, y, ff]

{x, y}

ff[xx_, yy_] := {yy^2/4, xx};

VectorDensityPlot[ff[x, y], {x, 1, 4}, {y, 1, 4}, VectorStyle -> Red]

{x, y}

(*  output is the same as above, same coloring  *)

Pre-evaluating ff[x, y] gives the desired plot:

VectorDensityPlot[Evaluate@ff[x, y], {x, 1, 4}, {y, 1, 4}, VectorStyle -> Red]

This works...once you have potdistr, run this to get a Function

pdg = Evaluate[Grad[potdistr[#1, #2], {#1, #2}]] &

You can then run the plot function repeatedly, no problem, and go back and run previous statements (you couldn't when x, y were getting set).

VectorDensityPlot[pdg[x,y], {x, 0, 0.1}, {y, 0, 0.05}]

VectorDensityPlot[pdg[x,y], {x, 0, 0.1}, {y, 0, 0.05}]

VectorDensityPlot[pdg[x,y], {x, 0, 0.1}, {y, 0, 0.05}]

You can also run this kludgy version, but it is slow.

VectorDensityPlot[Evaluate[Grad[potdistr[#1, #2], {#1, #2}]] &[x, y], 

                  {x, 0, 0.1}, {y, 0, 0.05}]

I was interested in the not very useful looking StreamPlot I was getting.

StreamPlot[pdg[x, y], {x, 0, 0.1}, {y, 0, 0.05}]

One thing I found odd is that if you look at the domain for potdistr versus the plot ranges people are using, there should be some extrapolation going on. Shouldn't we get warnings for that?