Skip to main content

plotting - How to scale values for ColorFunction in SliceContourPlot3D?


Trying to help out in the question, Finite Elements 3D, I came across this behavior of SliceContourPlot3D:


op = -Laplacian[u[x, y, z], {x, y, z}] - 2;
uif = NDSolveValue[{op == 0, DirichletCondition[u[x, y, z] == 0, True}, 
   u, {x, y, z} ∈ Ball[{0, 0, 0}]];

SliceContourPlot3D[
 uif[x, y, z], "CenterPlanes", {x, y, z} ∈ MeshRegion@uif["ElementMesh"], 
 ColorFunction -> "Rainbow"]


Mathematica graphics


The range of colors does not go all the way down to violet/indigo. I expected the range of colors to be something like this:


With[{sel = Positive[uif["Grid"].{0.3, -0.4, -1}]},
 With[{pts = Pick[uif["Grid"], sel]},
  Graphics3D[
   GraphicsComplex[
    pts,
    {Point[Range@Length@pts, 
      VertexColors -> (ColorData["Rainbow"] /@ Rescale[Pick[uif["ValuesOnGrid"], sel]])

      ]}
    ]]
  ]]

Mathematica graphics


Is this a bug in SliceContourPlot3D? (Perhaps it's due to extrapolation from sampling outside the domain?) Is there a way to get it to rescale the colors the way I expected?


I'm using V10.3.0, Mac OSX.



Answer



It seems SliceContourPlot3D makes an initial sample, that is not quite right - i.e. outside of the region. Using


op = -Laplacian[u[x, y, z], {x, y, z}] - 2;

uif = NDSolveValue[{op == 0, 
    DirichletCondition[u[x, y, z] == 0, True]}, 
   u, {x, y, z} \[Element] Ball[{0, 0, 0}], 
   "ExtrapolationHandler" -> {(Indeterminate &),
              "WarningMessage" -> False}];

SliceContourPlot3D[
 uif[x, y, z], "CenterPlanes", {x, y, z} \[Element] 
  MeshRegion@uif["ElementMesh"], ColorFunction -> "Rainbow"]


enter image description here


On a related note, I am wondering if for FEM this ExtrapolationHandler should be the default - what do you think?


Comments

Post a Comment

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...
Post a Comment
Read more

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...
Post a Comment
Read more