I would like to calculate the displacement function of a circular plane with a circular hole in the center, along which it is interlocked.

I thought about solving with the following code:

domain = ImplicitRegion[x^2 + y^2 <= 4^2 && x^2 + y^2 >= 2^2, {x, y}];
pde = D[w[x, y], {x, 4}] + 2 D[w[x, y], {x, 2}, {y, 2}] + D[w[x, y], {y, 4}] == 1;
bcs1 = DirichletCondition[w[x, y] == 0, x^2 + y^2 == 2^2];
bcs2 = DirichletCondition[D[w[x, y], {x, 2}] == 0, x^2 + y^2 == 2^2];
bcs3 = DirichletCondition[D[w[x, y], {y, 2}] == 0, x^2 + y^2 == 2^2];
NDSolveValue[{pde, bcs1, bcs2, bcs3}, w[x, y], {x, y} \[Element] domain]

but unfortunately I get the following error:

NDSolveValue::femcmsd: The spatial derivative order of the PDE may not exceed two.

Can you correct me, please? Thank you!


closed as off-topic by Henrik Schumacher, Coolwater, bbgodfrey, m_goldberg, LCarvalho Dec 3 '17 at 17:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – Henrik Schumacher, Coolwater, bbgodfrey
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ I'm voting to close this question as off-topic because the OP is asking for functionality that Mathematica does not currently support. $\endgroup$ – m_goldberg Dec 3 '17 at 1:36
  • $\begingroup$ Sometimes I want to delete some topics because I realize I'm wrong with the approach. Sorry. $\endgroup$ – TeM Dec 3 '17 at 12:23

It's partially explained in the similar question. Since you use regions, Mathematica has to go for FEM solver. Unfortunately, its FEM solver doesn't support derivatives greater than 2 (that's what the error message is about).


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