# 2.11 The Finite Element Method for Two-Dimensional Diffusion

## 2.11.2 Reference Element and Linear Elements

Measurable Outcome 2.15, Measurable Outcome 2.16

In multiple dimensions, a common practice in defining the polynomial functions within an element is to transform each element into a canonical, or so-called “reference" element. Figure 2.47 shows the mapping commonly used for triangular elements which maps a generic triangle in \((x,y)\) into a right triangle in \((\xi _1, \xi _2)\).

Figure 2.47: Transformation of a generic triangular element in \((x,y)\) into the reference element in \((\xi _1,\xi _2)\).

In the reference element space, the nodal basis for linear polynomials will be one at one of the nodes, and reduce linearly to zero at the other nodes. These functions are

| \(\displaystyle \phi _1(\xi _1, \xi _2)\) | \(\displaystyle =\) | \(\displaystyle 1 - \xi _1 - \xi _2, \label{equ:phi1_2d}\) |  | (2.262)
| \(\displaystyle \phi _2(\xi _1, \xi _2)\) | \(\displaystyle =\) | \(\displaystyle \xi _1, \label{equ:phi2_2d}\) |  | (2.263)
| \(\displaystyle \phi _3(\xi _1, \xi _2)\) | \(\displaystyle =\) | \(\displaystyle \xi _2. \label{equ:phi3_2d}\) |  | (2.264)

Then, within the element, the solution \(\tilde{T}\) in the \((\xi _1,\xi _2)\) space is the combination of these three basis functions multiplied by the corresponding nodal coefficients,

\[\tilde{T}(\xi _1,\xi _2) = \sum _{j=1}^{3} a_ j \phi _ j(\xi _1,\xi _2). \label{equ:Txi}\] | (2.265)

Using Equation (2.265), the value of \(\tilde{T}\) can be found at any \((\xi _1,\xi _2)\). To find the \((x,y)\) locations in terms of the \((\xi _1, \xi _2)\), we can expand them using the nodal locations and the nodal basis functions, i.e.,

\[\vec{x}(\xi _1,\xi _2) = \sum _{j=1}^{3} \vec{x}_ j \phi _ j(\xi _1,\xi _2).\] | (2.266)

Since the \(\phi _ i(\xi _1,\xi _2)\) are linear functions of \(\xi _1\) and \(\xi _2\), this amounts to a linear transformation between \((\xi _1,\xi _2)\) and \((x,y)\). Specifically, substituting the nodal basis functions gives

| \(\displaystyle \vec{x}(\xi _1,\xi _2)\) | \(\displaystyle =\) | \(\displaystyle \vec{x}_1 \left(1 - \xi _1 - \xi _2\right) + \vec{x}_2\xi _1 + \vec{x}_3\xi _2,\) |  | (2.267)
| \(\displaystyle \Rightarrow \vec{x}(\xi _1,\xi _2)\) | \(\displaystyle =\) | \(\displaystyle \vec{x}_1 + \left(\vec{x}_2-\vec{x}_1\right)\xi _1 + \left(\vec{x}_3-\vec{x}_1\right)\xi _2.\) |  | (2.268)

This can be written in a matrix notation as,

\[\left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{c} x_1 \\ y_1 \end{array}\right) + \left(\begin{array}{cc} x_2-x_1 & x_3-x_1 \\ y_2-y_1 & y_3-y_1\end{array}\right) \left(\begin{array}{c} \xi _1 \\ \xi _2 \end{array}\right) \label{equ:xi_ to_ x}\] | (2.269)

This equation can be inverted to also determine \((\xi _1,\xi _2)\) as a function of \((x,y)\),

\[\left(\begin{array}{c} \xi _1 \\ \xi _2 \end{array}\right) = \left(\begin{array}{cc} x_2-x_1 & x_3-x_1 \\ y_2-y_1 & y_3-y_1\end{array}\right)^{-1} \left(\begin{array}{c} x-x_1 \\ y-y_1 \end{array}\right) \label{equ:x_ to_ xi}\] | (2.270)