Group Abstract

Message Boards

WOLFRAM COMMUNITY

11.2K Views

3 Replies

3 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

Solving the Laplace equation with only Neumann boundary conditions

Jens Rix

Posted 11 years ago

Hey I want to solve the Laplace equation with Neumann boundary conditions on all boundaries, The solution seems random when I do not include any DirichletConditions though. Here is an example of the Laplace in cylindrical coordinates (with cylindrical symmetry). "mesh" and "region" are define on the region 0 <= r <= 20 and 0 <= z <= 30. T = NDSolveValue[ op == NeumannValue[r, z == 0] + NeumannValue[r, z == 30], u, {r, z} \[Element] mesh, Method -> {"FiniteElement", "MeshOptions" -> {"BoundaryMeshGenerator" -> "Continuation"}}]; Show[ ContourPlot[T[r, z], {r, z} \[Element] region , Mesh -> None, ColorFunction -> "TemperatureMap", Contours -> 40, AspectRatio -> Automatic, PlotRange -> All, Method -> {"GridLinesInFront" -> True}] ] "op" is: And I get the result: What do I do wrong?

Hey

I want to solve the Laplace equation with Neumann boundary conditions on all boundaries, The solution seems random when I do not include any DirichletConditions though.

Here is an example of the Laplace in cylindrical coordinates (with cylindrical symmetry). "mesh" and "region" are define on the region 0 <= r <= 20 and 0 <= z <= 30.

T = NDSolveValue[ 
   op == NeumannValue[r, z == 0] + NeumannValue[r, z == 30], 
   u, {r, z} \[Element] mesh, 
   Method -> {"FiniteElement", 
     "MeshOptions" -> {"BoundaryMeshGenerator" -> "Continuation"}}];
Show[
 ContourPlot[T[r, z], {r, z} \[Element] region , Mesh -> None, 
  ColorFunction -> "TemperatureMap", Contours -> 40, 
  AspectRatio -> Automatic, PlotRange -> All, 
  Method -> {"GridLinesInFront" -> True}]
 ]

"op" is: op function

And I get the result: plot

What do I do wrong?

POSTED BY: Jens Rix

3 Replies

Sort By:

Jens Rix

Posted 11 years ago

By "NeumannValue[r, z == 0]" I mean that the derivative perpendicular to the surface at z=0 is equal to the linear function f(r)=r. This works when I include a DirichletValue. Where are the conditions for the shank? When I don't specify anything on a boundary, Mathematica uses that NeumannValue[0,boundary]. Right? This is ill-posed problem as there is no unique solution in this case. Sorry, I didn't make myself clear. I am only going to use the gradient of the solution. When there is no unique solution in this case it means that adding a constant is a solution as well. If I could just specify a single point inside the region (and thereby not spoil the boundaries) I would get a unique solution. Would that be possible?