# 2.9 Introduction to Finite Elements

## 2.9.5 Calculation of the Finite Element Weighted Residual

The implementation of the finite element method requires finding the weak form of the residual for each weight function $$\phi _ j(x)$$. For the diffusion problem the $$j^{th}$$ weighted residual is

 $r(\tilde{T}, \phi _ j) \equiv \left[\phi _ j\, k \tilde{T}_ x\right]^{L/2}_{-L/2} - \int _{-L/2}^{L/2} {\phi _ j}_ x\, k \tilde{T}_ x\, dx + \int _{-L/2}^{L/2} \phi _ j f\, dx. \label{equ:Rj_ dif1d}$ (2.211)

We will consider the evaluation of the term $$\int _{-L/2}^{L/2} \phi _{j_ x}k\tilde{T}_{x} dx$$ below. The approximation of $$\int _{-L/2}^{L/2} \phi _{j_ x} f dx$$ will be discussed in Section 2.10.1; for now, we assume that this integral can be calculated analytically. For now, consider the term $$\int _{-L/2}^{L/2} {\phi _ j}_ x\, k \tilde{T}_ x\, dx$$. While it is a global integral (i.e., an integral over the entire domain), in reality $$\phi _ j(x)$$ is non-zero only over the two elements that include node $$j$$:

 $\Rightarrow \int _{-L/2}^{L/2} {\phi _ j}_ x\, k \tilde{T}_ x\, dx = \int _{x_{j-1}}^{x_{j+1}} {\phi _ j}_ x\, k \tilde{T}_ x\, dx.$ (2.212)

The derivative of basis function $$j$$ is

 ${\phi _ j}_ x(x) = \left\{ \begin{array}{cl} 0, & \mbox{for } x < x_{j-1}, \\[0.1in] \frac{1}{\Delta x_{j-1}}, & \mbox{for } x_{j-1} < x < x_{j},\\[0.1in] \frac{-1}{\Delta x_{j}}, & \mbox{for } x_{j} < x < x_{j+1},\\[0.1in] 0, & \mbox{for } x > x_{j+1}. \end{array}\right. \label{equ:phi_ x_ linear}$ (2.213)

Thus,

 $\int _{x_{j-1}}^{x_{j+1}} {\phi _ j}_ x\, k \tilde{T}_ x\, dx = \frac{1}{\Delta x_{j-1}} \int _{x_{j-1}}^{x_{j }} k \tilde{T}_ x\, dx - \frac{1}{\Delta x_{j }} \int _{x_{j }}^{x_{j+1}} k \tilde{T}_ x\, dx.$ (2.214)

Next, take the derivative of $$\tilde{T}$$:

 $\tilde{T}_ x(x) = \sum _{i=1}^{N+1} a_ i {\phi _ i}_ x(x).$ (2.215)

For the integral in element $$j-1$$, the only non-zero contributions to $$\tilde{T}_ x$$ are from $$\phi _{j-1}$$ and $$\phi _{j}$$, specifically,

 $\mbox{In element }j-1: \qquad \tilde{T}_ x = a_{j-1}{\phi _{j-1}}_ x + a_{j}{\phi _{i}}_ x = \frac{a_ j - a_{j-1}}{\Delta x_{j-1}}.$ (2.216)

Similarly,

 $\mbox{In element }j: \qquad \tilde{T}_ x = a_{j}{\phi _{j}}_ x + a_{j+1}{\phi _{j+1}}_ x = \frac{a_{j+1}-a_ j}{\Delta x_{j}}.$ (2.217)

Substituting these expressions for the derivatives into the integral gives

 $\int _{x_{j-1}}^{x_{j+1}} {\phi _ j}_ x\, k \tilde{T}_ x\, dx = \frac{a_{j}-a_{j-1}}{(\Delta x_{j-1})^2} \int _{x_{j-1}}^{x_{j }} k \, dx - \frac{a_{j+1}-a_{j}}{(\Delta x_{j })^2} \int _{x_{j }}^{x_{j+1}} k \, dx.$ (2.218)

At this point, we must integrate $$k(x)$$ in each element. Efficient numerical methods to approximate this integral are discussed in Section 2.10.1. For the situation in which $$k$$ is constant throughout the domain the integral reduces to the following expression.

 $\mbox{For }k = \mbox{constant}: \qquad \int _{x_{j-1}}^{x_{j+1}} {\phi _ j}_ x\, k \tilde{T}_ x\, dx = k\frac{a_{j}-a_{j-1}}{\Delta x_{j-1}}- k\frac{a_{j+1}-a_{j}}{\Delta x_{j }}.$ (2.219)

## Exercise

For constant spacing $$\Delta x$$, what does the integral $$\int _{x_{j-1}}^{x_{j+1}} {\phi _ j}_ x\, \tilde{T}_ x\, dx$$ equal?

Exercise 1

Answer: Setting $$\Delta x_{j-1} = \Delta x_{j+1} = \Delta x$$, we get the same approximation for the second derivative that we saw in the section of finite differences. Is that a coincidence?