Skip to main content

mesh - How to create subregions for the NDSolve FEM Solver



I am trying to create a 2d region consisting of two subregions. The inner region has several holes, where boundary conditions are applied. The figure shows the idea.


Geometry with two regions


I have tried to create this region using various region functions, but without success. The only approach that has worked so far is to replace the circles with many-sided polygons and use ToBoundaryMesh, specifying all of the points and lines that make up the polygons. This approach allows an ElementMesh to be generated, but it seems overly complex for such a simple geometry. In addition, the solution I get to Laplace's equation for this mesh looks unphysical.


Here is the mesh generated this way:


Element Mesh of region


The code to generate the mesh:


Needs["NDSolve`FEM`"]
nSides = 48;  (*number of sides for each circle*)
nCircles = 12; (*number of circles in the model *)
(*Function to create points for a single circle:*)

circlePts[{x_, y_}, r_] :=
Map[{x + r*Cos[#], y + r*Sin[#]} &, Range[0, 2 π - (2 π)/nSides, (2 π)/nSides]]

(*Generate list of coordinates and connectivity*)
cPts = Flatten[Map[circlePts[#[[1]], #[[2]]] &, circList], 1]; 
connect = Partition[Riffle[Range[nSides], RotateLeft[Range[nSides]]], 2];
nn = (Range[nCircles] - 1)*nSides;
bigConnect = Map[LineElement[connect + #] &, nn];

(*Create boundary mesh*)

bmesh = ToBoundaryMesh[
"Coordinates" -> cPts, "BoundaryElements" -> bigConnect];
bmesh["Wireframe"]

(*Create 2d mesh*)
mesh = ToElementMesh[bmesh, "RegionHoles" -> circList[[3 ;; -1, 1]]];
mesh["Wireframe"]

(*Set up boundary conditions*)
bcs = Join[{DirichletCondition[u[x, y] == 0, x^2 + y^2 >= 0.150^2]}, 

Map[DirichletCondition[
 u[x, y] == 
  20, (x - #[[1, 1]])^2 + (y - #[[1, 2]])^2 == #[[2]]^2] &, 
circList]
]

(*Solve the model*)
op=-Laplacian[u[x,y],{x,y}];
Subscript[Γ, D]=bcs;
uif=NDSolveValue[{op==0,Subscript[Γ, D]},u,{x,y}∈mesh]


ContourPlot[uif[x, y], {x, y} ∈ mesh, 
ColorFunction -> "Temperature", AspectRatio -> Automatic, 
PlotRange -> All, Contours -> 10]

Contour plot of solution


Is there a more elegant way to create subregions?



Answer



Here is a clean way to do it. The idea is to specify all circular boundary regions and then to explicitly set those that are region holes, such that those are not meshed.


Needs["NDSolve`FEM`"]

\[CapitalOmega] = ImplicitRegion[(9/10)^2 <= x^2 + y^2 <= 1^2, {x, y}];
ToElementMesh[\[CapitalOmega]]["Wireframe"]

enter image description here


ToElementMesh[\[CapitalOmega], "RegionHoles" -> None]["Wireframe"]

enter image description here


disk[{x0_, y0_}, r_] := ((x + x0)^2 + (y + y0)^2 <= (r)^2)
crds = {{-1/2, 0}, {1/2, 1/2}};
sd = Or @@ (disk[#, 1/8] & /@ crds);

\[CapitalOmega]2 = 
  ImplicitRegion[Or[(9/10)^2 <= x^2 + y^2 <= 1^2, sd], {x, y}];
ToElementMesh[\[CapitalOmega]2, 
  "BoundaryMeshGenerator" -> {"Continuation"}]["Wireframe"]

enter image description here


ToElementMesh[\[CapitalOmega]2, 
  "BoundaryMeshGenerator" -> {"Continuation"}, 
  "RegionHoles" -> -crds]["Wireframe"]


enter image description here


You could also refine one of the sub regions:


(mesh = ToElementMesh[\[CapitalOmega]2, 
    "BoundaryMeshGenerator" -> {"Continuation"}, 
    "RegionHoles" -> -crds, 
    "RegionMarker" -> {{{0, 0}, 1, 0.01}, {{19/20, 0}, 2, 
       0.001}}])["Wireframe"]

enter image description here


And visualize with different colors



mesh["Wireframe"[
  "MeshElementStyle" -> {FaceForm[Green], FaceForm[Red]}]]

enter image description here


But strictly speaking the sub regions as not necessary. as the PDE does not have a discontinuity.


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

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}]
Post a Comment
Read more