Skip to main content

finite element method - Non-linear Poisson equation over non-rectangular domain



I need to solve non-linear Poisson equation


Laplacian[u[x, y], {x, y}] == u[x, y]^2


Over a non-rectangular domain


The problem in short: non-linear Poisson equation over rectangular domain runs OK, and linear Poisson equation over non-rectangular domain runs OK, but not the non-linear over non-rectangular.


The domain is


boundaries = {-y, .25^2 - (x)^2 - y^2, -x, y - 1, x - 1};

\[CapitalOmega]in = 
ImplicitRegion[And @@ (# <= 0 & /@ boundaries), {x, y}];


Show[RegionPlot[\[CapitalOmega]in], 
ContourPlot[
Evaluate[Thread[boundaries == 0]], {x, 0., 1}, {y, 0, 1.}, 
ContourStyle -> {Purple, Green, Red, Blue, Purple}], 
PlotRange -> {{0.0, 1}, {0., 1.}}, AspectRatio -> Automatic]

with simple boundary conditions


  Conditions = {DirichletCondition[u[t, x, y] == 1, 
  boundaries[[1]] == 0.],
  DirichletCondition[u[t, x, y] == 1, boundaries[[2]] == 0],

  DirichletCondition[u[t, x, y] == 1, boundaries[[3]] == 0.],
  DirichletCondition[u[t, x, y] == 1, boundaries[[4]] == 0.],
  DirichletCondition[u[t, x, y] == 1, boundaries[[5]] == 0.],
  u[0, x, y] == 1};

I try to run a relaxation scheme


 Eq = Laplacian[u[t, x, y], {x, y}] - u[t, x, y]^2

sol = NDSolveValue[{Eq == Derivative[1, 0, 0][u][t, x, y], 
Conditions}, u, {t, 0, 1}, {x, y} \[Element] \[CapitalOmega]in, 

Method -> {"MethodOfLines", Method -> "Automatic", 
"DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}}]

The problem is "Nonlinear coefficients are not supported in this version of NDSolve".


The linear Poisson equation runs OK


Eq = Laplacian[u[t, x, y], {x, y}] - u[t, x, y]

sol = NDSolveValue[{Eq == Derivative[1, 0, 0][u][t, x, y], 
 Conditions}, u, {t, 0, 1}, {x, y} \[Element] \[CapitalOmega]in, 
Method -> {"MethodOfLines", Method -> "Automatic", 

"DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}}]

ContourPlot[sol[1, x, y], {x, y} \[Element] \[CapitalOmega]in, 
 ColorFunction -> "TemperatureMap", Contours -> 50, 
 AspectRatio -> Automatic]

Also, non-linear over a rectangular domain runs OK:


boundaries = {-y, -x, y - 1, x - 1};
\[CapitalOmega]in = 
ImplicitRegion[And @@ (# <= 0 & /@ boundaries), {x, y}];


 Show[RegionPlot[\[CapitalOmega]in], 
 ContourPlot[
 Evaluate[Thread[boundaries == 0]], {x, 0., 1}, {y, 0, 1.}, 
 ContourStyle -> {Purple, Green, Red, Blue, Purple}], 
 PlotRange -> {{0.0, 1}, {0., 1.}}, AspectRatio -> Automatic] 

Conditions = {DirichletCondition[u[t, x, y] == 1, 
 boundaries[[1]] == 0.],
 DirichletCondition[u[t, x, y] == 1, boundaries[[2]] == 0],

 DirichletCondition[u[t, x, y] == 1, boundaries[[3]] == 0.],
 DirichletCondition[u[t, x, y] == 1, boundaries[[4]] == 0.],
 u[0, x, y] == 1};

 sol = NDSolveValue[{Eq == Derivative[1, 0, 0][u][t, x, y], 
  Conditions}, u, {t, 0, 1}, {x, 0, 1}, {y, 0, 1}, 
 Method -> {"MethodOfLines", Method -> "Automatic", 
 "DifferentiateBoundaryConditions" -> {True, "ScaleFactor" -> 1}}]

 ContourPlot[sol[1, x, y], {x, y} \[Element] \[CapitalOmega]in, 

 ColorFunction -> "TemperatureMap", Contours -> 50, 
 AspectRatio -> Automatic]


Comments

Post a Comment

Popular posts from this blog

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

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....
Post a Comment
Read more