Cut an interpolating function

Question

I have some data

Tinterspike200 = {3.01026957638`, 5.314505776636686`, 
   10.494223363285943`, 16.585365853657912`};
Tinterspike400 = {2.5609756097561167`, 3.940949935815186`, 
   6.103979460847167`, 8.921694480102463`, 12.50962772785579`, 
   17.092426187419257`, 22.13093709884531`};
Tinterspike600 = {2.3748395378690628`, 3.177150192554557`, 
   4.358151476251605`, 6.059050064184852`, 8.401797175866495`, 
   11.206675224646983`, 14.80744544287548`, 18.58793324775353`, 
   22.310654685494224`, 26.78433889602054`};
Tinterspike800 = {2.2657252888318355`, 2.7856225930680356`, 
   3.4403080872913994`, 4.2105263157894735`, 5.7124518613607185`, 
   8.318356867779203`, 11.59178433889602`, 14.441591784338895`, 
   17.77920410783055`, 21.059050064184852`, 25.532734274711167`};
Tinterspike1000 = {2.1822849807445444`, 2.593068035943517`, 
   3.0680359435173297`, 3.5879332477535297`, 4.255455712451861`, 
   5.423620025673941`, 8.164313222079588`, 11.07188703465982`, 
   13.49165596919127`, 16.084724005134788`, 19.17201540436457`, 
   22.35558408215661`, 25.84724005134788`};

Nspikes200 = {1, 2, 3, 4};
Nspikes400 = {1, 2, 3, 4, 5, 6, 7};
Nspikes600 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
Nspikes800 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
Nspikes1000 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};
Istim = {200, 400, 600, 800, 1000};

which I arrange in the following form

(*DATA*)
data1 = 
  Table[{Istim[[1]], Tinterspike200[[i]], Nspikes200[[i]]}, {i, 1, 4}];
data2 = Table[{Istim[[2]], Tinterspike400[[i]], Nspikes400[[i]]}, {i, 
    1, 7}];
data3 = Table[{Istim[[3]], Tinterspike600[[i]], Nspikes600[[i]]}, {i, 
    1, 10}];
data4 = Table[{Istim[[4]], Tinterspike800[[i]], Nspikes800[[i]]}, {i, 
    1, 11}];
data5 = Table[{Istim[[5]], Tinterspike1000[[i]], 
    Nspikes1000[[i]]}, {i, 1, 13}];
data = Join[data1, data2, data3, data4, data5];
lpp = ListPointPlot3D[data, PlotStyle -> {PointSize[Large], Red}];

I define the following boundary

(*BOUNDARY*)
p1 = {Istim[[1]], Tinterspike200[[4]], Nspikes200[[4]]};
p2 = {Istim[[2]], Tinterspike400[[7]], Nspikes400[[7]]};
p3 = {Istim[[3]], Tinterspike600[[10]], Nspikes600[[10]]};
p4 = {Istim[[4]], Tinterspike800[[11]], Nspikes800[[11]]};
p5 = {Istim[[5]], Tinterspike1000[[13]], Nspikes1000[[13]]};
boundary = 
  Graphics3D[{Dashed, Thick, Red, Line[{p1, p2, p3, p4, p5}]}];

Then I interpolate the data obtaining

(*INTERPOLATION*)
{xmin, xmax} = MinMax[data[[All, 1]]];
{ymin, ymax} = MinMax[data[[All, 2]]];
dataInterp = {Most@#, Last@#} & /@ data;
Istim3D = Interpolation[dataInterp, InterpolationOrder -> 1]
plIstim3D = 
  Plot3D[Istim3D[x, y], {x, xmin, xmax}, {y, ymin, ymax}, 
   PlotStyle -> Opacity[0.8], 
   AxesLabel -> {"\!\(\*SubscriptBox[\(I\), \(stim\)]\)", 
     "Tempi interspikes", "Numero di spikes"}, PlotRange -> All, 
   ImageSize -> 800];
Show[lpp, boundary, plIstim3D, ImageSize -> 800]

The plot that I obtain is the following

Now my question is: I would like to cut the plot removing the part of surface, which goes beyond the red line, keeping the part of surface passing through my data. Is it possible to remove the part of the part of plot3d which goes beyond the dashed red line? If I have a curve defined by ParametricPlot3D, how can I achieve my objective?

Thank you very much for your help.

Answer 1

(*DATA*)data1 = 
  Table[{Istim[[1]], Tinterspike200[[i]], Nspikes200[[i]]}, {i, 1, 4}];
data2 = Table[{Istim[[2]], Tinterspike400[[i]], Nspikes400[[i]]}, {i, 
    1, 7}];
data3 = Table[{Istim[[3]], Tinterspike600[[i]], Nspikes600[[i]]}, {i, 
    1, 10}];
data4 = Table[{Istim[[4]], Tinterspike800[[i]], Nspikes800[[i]]}, {i, 
    1, 11}];
data5 = Table[{Istim[[5]], Tinterspike1000[[i]], 
    Nspikes1000[[i]]}, {i, 1, 13}];
data = Join[data1, data2, data3, data4, data5];
lpp = ListPointPlot3D[data, PlotStyle -> {PointSize[Large], Red}];

(*BOUNDARY*)p1 = {Istim[[1]], Tinterspike200[[4]], Nspikes200[[4]]};
p2 = {Istim[[2]], Tinterspike400[[7]], Nspikes400[[7]]};
p3 = {Istim[[3]], Tinterspike600[[10]], Nspikes600[[10]]};
p4 = {Istim[[4]], Tinterspike800[[11]], Nspikes800[[11]]};
p5 = {Istim[[5]], Tinterspike1000[[13]], Nspikes1000[[13]]};
ifun = Interpolation[{p1[[1 ;; 2]], p2[[1 ;; 2]], p3[[1 ;; 2]], 
   p4[[1 ;; 2]], p5[[1 ;; 2]]}, InterpolationOrder -> 1]
boundary = 
  Graphics3D[{Dashed, Thick, Red, Line[{p1, p2, p3, p4, p5}]}];

...

plIstim3D = 
  Plot3D[Istim3D[x, y], {x, xmin, xmax}, {y, ymin, ymax}, 
   PlotStyle -> Opacity[0.8], 
   AxesLabel -> {"\!\(\*SubscriptBox[\(I\), \(stim\)]\)", 
     "Tempi interspikes", "Numero di spikes"}, PlotRange -> All, 
   ImageSize -> 800, 
   RegionFunction -> 
    Function[{x, y, z}, ifun[x] > y > 0 && 200 < x && x < 1000]];
Show[lpp, boundary, plIstim3D, ImageSize -> 600]

Answer 2

Perhaps the easiest way is to plot over {x, y} ∈ boundarypolygon, but due to the different scales in the x and y directions, the polygon is not meshed very well.

(*BOUNDARY*)
datasets = {data1, data2, data3, data4, data5};
pp = datasets[[All, -1]];
qq = datasets[[All, 1]];
bdycoord = Join[data1, pp[[2 ;; 4]], Reverse@data5, qq[[4 ;; 1 ;; -1]]];
boundary = Graphics3D[{Dashed, Thick, Red, Line[bdycoord]}];

Plot3D[Istim3D[x, y],
 {x, y} ∈ Polygon[Most /@ bdycoord],
 PlotStyle -> Opacity[0.8]]

It can be improved by setting PlotPoints -> {75, 1500}, but that seems excessive. A better result is obtained by rescale the polygon for meshing and then scaling back.

xc = Transpose[Most /@ bdycoord];
bdy = Polygon@Transpose[Rescale /@ xc];
bdymesh = ToElementMesh[bdy, MaxCellMeasure -> 1/1000];
bdymesh = ToElementMesh[
   "Coordinates" -> Transpose[
     MapThread[
      Rescale[#1, {0, 1}, #2] &,
       {Transpose@bdymesh@"Coordinates", MinMax /@ xc}]],
   "MeshElements" -> bdymesh@"MeshElements"
   ];
plIstim3D = Plot3D[Istim3D[x, y],
   {x, y} ∈ bdymesh,
   PlotStyle -> Opacity[0.8]];
Show[lpp, boundary, plIstim3D, PlotRange -> All]

Note the lower edge was trimmed, even though the OP did not ask for it. Adjust the definition of the polygon to set the lower edge as desired. (I thought trimming the domain so that there is no extrapolation would be more desirable, but I might be wrong.)