# What is known about C0 triangular finite elements with nonstandard mesh point placement?

I'm curious about the general case, but for ease of explaining lets just take the case of a $$P^2(\Omega)$$ approximation. For simplicity, let's also just consider the reference element $$(0,0), (0,1), (1,0)$$. To satisfy the global continuity condition, each edge must have three points on it. Thus to have a conforming FE we must use the vertices as three of the six mesh points.

The main question then: Where can the other three mesh points be placed without ruining anything? Clearly, the midpoint of each edge is the standard choice. Similarly, it is clear that each edge must have one point. We can write the last three points as $$(0,\alpha_1), (\alpha_2,0),(1-\alpha_3,\alpha_3)$$. For what values, if any, of $$\alpha_1,\alpha_2,\alpha_3\in(0,1)$$ does the finite element posses the following:

1. The finite element is conforming (For a second order elliptic BVP).
2. The finite element has optimal interpolation properties.
3. Nothing else in the FEM I haven't thought of breaks.

Note: I'm already aware that doing as I described makes it impossible to use one reference element for the whole mesh. This does not strike me as a large downside--because for reasonably low order elements it would still be possible to hard code several parent elements with hardly any additional computational cost.

A reference and brief comments would be an ample reply. I've been unable to find any information on this. Every FE text I have encountered simply chooses the standard mesh points and moves along.

Almost all of the statements about the convergence of the finite element are only about the finite element space, not about what specific basis you choose for it. As a consequence, you are for example free to use an equidistant set of nodes to generate basis functions for the usual $$P_k$$ of $$Q_k$$ elements, or a set of nodes that correspond to Chebyshev or Legendre points: The space is the same, and consequently so are the convergence properties, it's just that the basis (and consequently the entries in the matrix and the right hand side vector) is different.

The only place where the choice of node points enter the consideration is whether (i) you have chosen them so that the set of basis functions spans the space, and (ii) what the condition number of the matrix is. The former is often packaged as the question of "unisolvency". The latter matters from a practical perspective because a matrix with bad condition number is difficult to solve with; as a consequence, for example, we don't use equidistant node points for higher order finite elements.

None of this is specific to your question about $$C^1$$ elements, of course, but it applies there just like for any other element.

• Absolutely great answer, which is marvelous in its simplicity. Of course the error only depends on the space! So the idea then would be as the $\alpha_i \rightarrow 0$ (or 1) the basis functions become nearly identical and the system becomes nearly singular. This answer is already amazing, but do you know of any work on these types of elements, or do you have a general feeling of what $\alpha_i$ might be admissible, practically? Thinking about what you said it seems probable that much of $(0,1)$ would be admissible--as long as all points have some "breathing room". Sep 14 at 12:28
• As $\alpha \to 0$ or $1$, your condition number deteriorates, and so "what values are admissible" comes down to how tolerant you are of matrices with bad condition numbers. Sep 14 at 16:43
• On the question of what is used: I see no particularly good reason to use anything other than $\alpha=1/2$, and so don't expect anyone to use anything other than that. Just because you can doesn't mean that you should. Sep 14 at 16:44

Complementing Wolfgang answer, if you enumerate your nodes as $$(0, 0), (1, 0), (0, 1), (\alpha_1, 0), (1 - \alpha_3, \alpha_3), (0, \alpha_2)$$ you get the following basis functions:

\begin{align} &\operatorname{N_{0}}{\left(x,y \right)} = 1 + \frac{y^{2}}{\alpha_{2}} + \frac{y \left(- \alpha_{2} - 1\right)}{\alpha_{2}} + \frac{x^{2}}{\alpha_{1}} + \frac{x \left(- \alpha_{1} - 1\right)}{\alpha_{1}} + \frac{x y \left(\alpha_{1} + \alpha_{2}\right)}{\alpha_{1} \alpha_{2}}\, ,\\ &\operatorname{N_{1}}{\left(x,y \right)} = \frac{\alpha_{1} x}{\alpha_{1} - 1} - \frac{x^{2}}{\alpha_{1} - 1} + \frac{x y \left(- \alpha_{1} - \alpha_{3} + 1\right)}{\alpha_{1} \alpha_{3} - \alpha_{3}}\, ,\\ &\operatorname{N_{2}}{\left(x,y \right)} = \frac{\alpha_{2} y}{\alpha_{2} - 1} + \frac{x y \left(\alpha_{2} - \alpha_{3}\right)}{\alpha_{2} \alpha_{3} - \alpha_{2} - \alpha_{3} + 1} - \frac{y^{2}}{\alpha_{2} - 1}\, ,\\ &\operatorname{N_{3}}{\left(x,y \right)} = \frac{x^{2}}{\alpha_{1}^{2} - \alpha_{1}} + \frac{x y}{\alpha_{1}^{2} - \alpha_{1}} - \frac{x}{\alpha_{1}^{2} - \alpha_{1}}\, ,\\ &\operatorname{N_{4}}{\left(x,y \right)} = - \frac{x y}{\alpha_{3}^{2} - \alpha_{3}}\, ,\\ &\operatorname{N_{5}}{\left(x,y \right)} = \frac{x y}{\alpha_{2}^{2} - \alpha_{2}} + \frac{y^{2}}{\alpha_{2}^{2} - \alpha_{2}} - \frac{y}{\alpha_{2}^{2} - \alpha_{2}}\, . \end{align}

So, as long as your $$\alpha_i$$ are different from 0 or 1 you are good to go.

• "So, as long as your $\alpha_i$ are different from 0 or 1" -- which is exactly the unisolvency condition. Sep 14 at 3:06
• Yes, I should have explicitly stated that. Sep 14 at 4:16