# 2.11 The Finite Element Method for Two-Dimensional Diffusion

## 2.11.3 Differentiation using the Reference Element

Measurable Outcome 2.17

To find the derivative of $$\tilde{T}$$ with respect to $$x$$ (or similarly $$y$$) within an element, we differentiate the three nodal basis functions within the element:

 $$\displaystyle \tilde{T}_ x$$ $$\displaystyle =$$ $$\displaystyle \frac{\partial }{\partial x}\left(\sum _{j=1}^{3} a_ j\phi _ j\right),$$ (2.271) $$\displaystyle =$$ $$\displaystyle \sum _{j=1}^{3} a_ j\frac{\partial \phi _ j}{\partial x}.$$ (2.272)

To find the $$x$$-derivatives of each of the $$\phi _ j$$'s, the chain rule is applied:

 $\frac{\partial \phi _ j}{\partial x} = \frac{\partial \phi _ j}{\partial \xi _1}\frac{\partial \xi _1}{\partial x} + \frac{\partial \phi _ j}{\partial \xi _2}\frac{\partial \xi _2}{\partial x}.$ (2.273)

Similarly, to find the derivatives with respect to $$y$$:

 $\frac{\partial \phi _ j}{\partial y} = \frac{\partial \phi _ j}{\partial \xi _1}\frac{\partial \xi _1}{\partial y} + \frac{\partial \phi _ j}{\partial \xi _2}\frac{\partial \xi _2}{\partial y}.$ (2.274)

The calculation of the derivatives of $$\phi _ j$$ with respect to the $$\xi$$ variables gives:

 $$\displaystyle \frac{\partial \phi _1}{\partial \xi _1}$$ $$\displaystyle =$$ $$\displaystyle -1, \qquad \frac{\partial \phi _1}{\partial \xi _2} = -1,$$ (2.275) $$\displaystyle \frac{\partial \phi _2}{\partial \xi _1}$$ $$\displaystyle =$$ $$\displaystyle 1, \qquad \frac{\partial \phi _2}{\partial \xi _2} = 0,$$ (2.276) $$\displaystyle \frac{\partial \phi _3}{\partial \xi _1}$$ $$\displaystyle =$$ $$\displaystyle 0, \qquad \frac{\partial \phi _3}{\partial \xi _2} = 1.$$ (2.277)

The only remaining terms are the calculation of $$\frac{\partial \xi _1}{\partial x}$$, $$\frac{\partial \xi _2}{\partial x}$$, etc. which can be found by differentiating Equation (2.270),

 $$\displaystyle \frac{\partial \vec{\xi }}{\partial \vec{x}}$$ $$\displaystyle =$$ $$\displaystyle \left(\begin{array}{cc} x_2-x_1 & x_3-x_1 \\ y_2-y_1 & y_3-y_1\end{array}\right)^{-1},$$ (2.278) $$\displaystyle =$$ $$\displaystyle \frac{1}{J} \left(\begin{array}{cc} y_3-y_1 & -(x_3-x_1) \\ -(y_2-y_1) & x_2-x_1\end{array}\right),$$ (2.279)

where

 $J = (x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1).$ (2.280)

Note that the Jacobian, $$J$$, is equal to twice the area of the triangular element.