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I am given a problem to define the functions $\phi_1(x, y)$, $\phi_2(x, y)$ and $\phi_3(x, y)$ for a single triangular element as a reference to do element assembly. However I'm not sure why the answer is:

φ1(x, y) := 1 − (1/2 x) − (1/2 y)
φ2(x, y) := (1/2 x) − (1/2 y)
φ3(x, y) := y

I understand that the φ1(x, y) is a representation of the vertex at point 1, which can be represented by φ1(x, y) := ax + by + c . However how do you obtain the coefficients of the functions φ1(x, y) := 1 − (1/2 x) − (1/2 y) ? How can you create a function out of a vertex?

The problem is as below:

A two-dimensional triangle, cf. Figure 1 on the left, is given as reference element. Piecewise linear functions are defined and shall be used as test functions; each function takes the value 1 at exactly one corner of the triangle and 0 at all other corners. Define the three functions $\phi_1(x, y)$, $\phi_2(x, y)$ and $\phi_3(x, y)$ for a single triangular element and compute their gradients.

Figure 1: Left: reference triangle. Right: domain which is discretised by triangles of same shape (and size) as on the left.

Figure 1: Left: reference triangle. Right: domain which is discretised by triangles of same shape (and size) as on the left.

Korf
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essy.lin
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1 Answers1

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The three basis functions for linear finite elements are given by functions \begin{equation} \phi_i = a_i x + b_i y + c_i \quad i \in \{ 1, 2, 3 \} \end{equation} and by the conditions of each $\phi_i$ being 1 at one of the triangle vertices and 0 on the other two vertices.

So for $\phi_1$ you plug in these conditions and get a system of linear equations for coefficients $a_1, b_1, c_1$ that reads \begin{align} 1 =& \phi_1(0, 0) = c_1, \\ 0 =& \phi_1(1, 1) = a_1 + b_1 + c_1, \\ 0 =& \phi_1(2, 0) = 2a_1 + c_1. \end{align} You solve this system and get the coefficients and then do the same for $\phi_2$ and $\phi_3$.

This is the main idea although there are some tricks to skip the computation and directly guess the solution.

Korf
  • 697