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In a simple pde-problem, which I solve in a predefined mesh using NDSolveValue I would like to plot the result in the mesh NDSolve uses:

<< NDSolve`FEM`
netz = ToElementMesh[Rectangle[], "MeshElementType" -> TriangleElement, MaxCellMeasure -> {"Length" -> .1}]

u = NDSolveValue[{Laplacian[\[Psi][x, y], {x, y}] ==1 + \[Psi][x, y] + NeumannValue[0, x == 1] ,\[Psi][0, y] == 0, \[Psi][x, 0] == 0, \[Psi][x, 1] == 0}, \[Psi] ,Element[{x, y}, netz], Method -> "FiniteElement"]

If I plot the result u[x,y] 

Plot3D[u[x, y], {x, 0, 1}, {y, 0, 1}, Mesh -> All]

Vierecknetz I get a foursided mesh even though the solution mesh 

u["ElementMesh"]["Wireframe"]

is triangular. How can I force Plot3D to show the right mesh?

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    $\begingroup$ A good point. This is easy to make with the ContourPlot such as Show[{ContourPlot[u[x, y], {x, 0, 1}, {y, 0, 1}], netz["Wireframe"] }] but this is not what you are asking about. $\endgroup$
    – Alexei Boulbitch
    Dec 15, 2017 at 14:53
  • $\begingroup$ @Alexei Boulbitch: Thanks, something like Plot3D[u[x, y] , Element[{x, y}, netz]] would help $\endgroup$
    – Ulrich Neumann
    Dec 15, 2017 at 16:20

2 Answers 2

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You can use:

Plot3D[u[x, y], {x, y} \[Element] u["ElementMesh"], Mesh -> All, 
 MaxRecursion -> 0]

enter image description here

In version 11.0 there was a bug that you can work around with something like:

Plot3D[u[x, y], {x, y} \[Element] 
  MeshRegion[u["ElementMesh"]]["MakeLinear"], Mesh -> All, 
 MaxRecursion -> 0]
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  • $\begingroup$ Thank you, this is the answer I'm looking for. In Mathematica v11.0.1.0 I get the error Plot3D::idomdim: {x,y}\[Element]u[ElementMesh] does not have a valid dimension as a plotting domain. Which is the version you use? $\endgroup$
    – Ulrich Neumann
    Dec 28, 2017 at 12:43
  • $\begingroup$ @UlrichNeumann, I am using 11.2 and I vaguely remember that there was such a bug. See if the edit helps you. $\endgroup$
    – user21
    Dec 28, 2017 at 13:00
  • $\begingroup$ That's it, thank you! $\endgroup$
    – Ulrich Neumann
    Dec 28, 2017 at 13:18
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Plot3D will resample points. If you want to maintain the mesh structure of netz, you could specify z-coords using u function. I specify "MeshOrder" -> 1 to get triangles:

<< NDSolve`FEM`
netz = ToElementMesh[Rectangle[], 
  "MeshElementType" -> TriangleElement, 
  MaxCellMeasure -> {"Length" -> .1}, "MeshOrder" -> 1]

u = NDSolveValue[{Laplacian[\[Psi][x, y], {x, y}] == 
    1 + \[Psi][x, y] + NeumannValue[0, x == 1], \[Psi][0, y] == 
    0, \[Psi][x, 0] == 0, \[Psi][x, 1] == 0}, \[Psi], 
  Element[{x, y}, netz], Method -> "FiniteElement"]

triangles = u["ElementMesh"]["MeshElements"][[1, 1]];
coords = u["ElementMesh"]["Coordinates"];

Graphics3D[
 GraphicsComplex[{##, u[##]} & @@@ 
   coords, {EdgeForm[{GrayLevel[0.2]}], 
   Directive[Specularity[GrayLevel[1], 3], RGBColor[
    0.880722, 0.611041, 0.142051], 
    Lighting -> {{"Ambient", RGBColor[
       0.30100577, 0.22414668499999998`, 
        0.090484535]}, {"Directional", RGBColor[
       0.2642166, 0.18331229999999998`, 0.04261530000000001], 
       ImageScaled[{0, 2, 2}]}, {"Directional", RGBColor[
       0.2642166, 0.18331229999999998`, 0.04261530000000001], 
       ImageScaled[{2, 2, 2}]}, {"Directional", RGBColor[
       0.2642166, 0.18331229999999998`, 0.04261530000000001], 
       ImageScaled[{2, 0, 2}]}}], Polygon[triangles]}], 
 BoxRatios -> {1, 1, .4}]
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  • $\begingroup$ Thank you for your answer. Trying to understand the look at the 3D-Plot shows a regular mesh structure with holes inside??? In triangles I would have expected some triples of knots , but I found sublists of length 6. $\endgroup$
    – Ulrich Neumann
    Dec 15, 2017 at 15:46
  • $\begingroup$ Now I got it! The holes disappear with Polygon[triangles[[All,{1,2,3}]] ] . But the solution obtained in this way is unfortunately only a workaround. What I'm looking for is the projection of the triangle mesh onto the surface u[x,y] $\endgroup$
    – Ulrich Neumann
    Dec 15, 2017 at 16:11
  • $\begingroup$ did you specify "MeshOrder" -> 1 in netz? $\endgroup$
    – halmir
    Dec 15, 2017 at 18:43
  • $\begingroup$ Not in my first attempt... $\endgroup$
    – Ulrich Neumann
    Dec 16, 2017 at 12:45

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