2.11 The Finite Element Method for Two-Dimensional Diffusion

2.11.2 Reference Element and Linear Elements

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)