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Solving the Laplace equation with only Neumann boundary conditions

Jens Rix

Posted 10 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

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Jens Rix

Posted 10 years ago

POSTED BY: Jens Rix

Nasser M. Abbasi

Nasser M. Abbasi, student

Posted 10 years ago

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. You can't have Neumann boundary conditions on all boundaries. This is ill-posed problem as there is no unique solution in this case. One of the boundaries must be Dirichlet.

POSTED BY: Nasser M. Abbasi

Udo Krause

Udo Krause, NOrganization

Posted 10 years ago

First, what does it mean to say NeumannValue[r, z == 0] taken into account that r is an independent variable? Then there are not enough conditions to avoid randomness in a way. To see that take manual's example NDSolveValue[{\!\( \SubsuperscriptBox[\(\[Del]\), \({x, y}\), \(2\)]\(u[x, y]\)\) == NeumannValue[1., x >= 0.35], DirichletCondition[u[x, y] == 0., x <= -0.3]}, u, {x, y} \[Element] Disk[]] Plot3D[%[x, y], {x, y} \[Element] Disk[]] and kick the Dirichlet condition out. The result is (in 2D) In[9]:= NDSolveValue[{\!\( \SubsuperscriptBox[\(\[Del]\), \({x, y}\), \(2\)]\(u[x, y]\)\) == NeumannValue[1., x >= -1.]}, u, {x, y} \[Element] Disk[]] During evaluation of In[9]:= NDSolveValue::femibcnd: No DirichletCondition or Robin-type NeumannValue was specified; the result may be off by a constant value. >> Out[9]= InterpolatingFunction[{{-1., 1.}, {-1., 1.}}, <>] and this NDSolveValue::femibcnd is pretty reasonable. In 3D the question is: Where are the conditions for the shank?

First, what does it mean to say

NeumannValue[r, z == 0]

taken into account that r is an independent variable?

Then there are not enough conditions to avoid randomness in a way. To see that take manual's example

NDSolveValue[{\!\(
\*SubsuperscriptBox[\(\[Del]\), \({x, y}\), \(2\)]\(u[x, y]\)\) == 
   NeumannValue[1., x >= 0.35], 
  DirichletCondition[u[x, y] == 0., x <= -0.3]}, u, {x, y} \[Element] 
  Disk[]]
Plot3D[%[x, y], {x, y} \[Element] Disk[]]

and kick the Dirichlet condition out. The result is (in 2D)

In[9]:= NDSolveValue[{\!\(
\*SubsuperscriptBox[\(\[Del]\), \({x, y}\), \(2\)]\(u[x, y]\)\) == 
   NeumannValue[1., x >= -1.]}, u, {x, y} \[Element] Disk[]]

During evaluation of In[9]:= NDSolveValue::femibcnd: No DirichletCondition or Robin-type NeumannValue was specified; the result may be off by a constant value. >>

Out[9]= InterpolatingFunction[{{-1., 1.}, {-1., 1.}}, <>]

and this NDSolveValue::femibcnd is pretty reasonable. In 3D the question is: Where are the conditions for the shank?

POSTED BY: Udo Krause

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