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

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...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...