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Consider the Laplace equation on a single triangular domain with a Dirichlet condition on two of the sides and a Neumann condition on the remaining side. I am using a linear element ... $\mathbb{P}_1$ which means (I think) that the solution will be a single linear function on the triangle. The Dirichlet conditions on two sides determine the value of the solution on the three corners of the triangle and hence, by linearity, on the entire triangular domain. But that suggests that the solution is entirely independent of the Neumann condition. It would seem that such a boundary value problem is overdetermined. That doesn't sit well with my intuition.

It this situation (a single triangular element) simply degenerate? That is, can I be sure that domains with more than one triangle (still using linear solution on each triangle) will yield uniquely determined solutions?

dan

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  • $\begingroup$ For linear finite elements, Dirichlet conditions are enforced on vertices, not edges, while the Neumann condition only determines the normal derivative on this edge. That's still three degrees of freedom for the three unknown coefficients of $a_0 + a_1 x + a_2 y$. (The solution is affine, not linear.) $\endgroup$ Jul 10, 2016 at 22:20
  • $\begingroup$ @ChristianClason, I don't understand. The (affine) Dirichlet conditions on the two sides of the triangle before discretizing yields a Dirichlet condition on each the three vertices after discretizing. Adding to that the Neumann condition on the third side gives four conditions in total for the three unknowns in $a_0 + a_1x + a_2y$. What am I missing? $\endgroup$
    – sitiposit
    Jul 11, 2016 at 2:13
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    $\begingroup$ If you impose Dirichlet conditions on all three vertices, you indeed cannot impose Neumann conditions on that element. In that sense, a single triangle is degenerate; FEM approximations are only reasonable in the limit as the mesh size goes to zero (i.e., if there are enough triangles to resolve the different boundary parts). $\endgroup$ Jul 11, 2016 at 7:21
  • $\begingroup$ I think that your thoughts simply imply that the solution is in general not affine in the case of mixed boundary conditions. $\endgroup$
    – user35593
    Jul 11, 2016 at 9:55
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    $\begingroup$ ...in fact, solving a PDE is not meaningful on a single element unless you use a very high polynomial degree (which basically amounts to using a global spectral method). $\endgroup$ Jul 11, 2016 at 10:19

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I'll summarize the comments above that settled my confusion;

An affine function on a triangle cannot, in general, satisfy a Dirichlet condition on two sides and a Neumann condition on the third side.

Thanks to Christian and user35593!

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  • $\begingroup$ Just to pick a nit: An affine function very well can satisfy a set of such boundary conditions, just not any set -- i.e., you cannot prescribe an arbitrary set of conditions and hope to find an affine function satisfying them, but there exist specific compatible conditions which are satisfied by an affine function (namely, those obtained by taking an arbitrary affine function and then computing the respective Dirichlet and Neumann traces). $\endgroup$ Sep 9, 2016 at 15:11

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