3.2 The Weighted Residual Method

The weighted residual method is demonstrated for the scalar function $u$ and can be applied analogously to the vector one $\vec{u}$ . Usually the unknown function $u$ of (3.1) cannot be found analytically, therefore, it is approximated by

$$\tilde{u}(\vec{r}) = \sum_{j=1}^{n}c_jN_j(\vec{r}) + v(\vec{r}) \simeq u(\vec{r}).$$ (3.5)

In (3.5) $v(\vec{r})$ is a known function which fulfills exactly the Dirichlet boundary condition on $\mathcal{A}_D$

$$v(\vec{r}) = u(\vec{r}) \mathrm{on} \mathcal{A}_D.$$ (3.6)

The basis (also called form or shape) functions $N_j(\vec{r}), j\in[1;n]$ build a set of linear independent known functions which vanish on the Dirichlet boundary $\mathcal{A}_D$ . Thus (3.6) is satisfied for each arbitrary set of coefficients $c_j, j\in[1;n]$ . The coefficients $c_j$ must be determined in such a way that the function $\tilde{u}$ approximates the solution of (3.1) as exactly as possible. The basis functions should be formulated in such way that each solution can be approximated with arbitrary accuracy, if a sufficiently large number of basis functions is used. After substitution of $\tilde{u}$ for $u$ in (3.1) a nonzero residual is obtained in general

$$R(\vec{r}) = \mathcal{L}\left[\tilde{u}(\vec{r})\right] - f(\vec{r}) \neq 0.$$ (3.7)

To find a good approximation $\tilde{u}$ for $u$ it is required to minimize the residual (3.7). The weighted residual method finds the unknown coefficients $c_j$ by weighting the residual (3.7). This is performed by choosing a set of linear independent weighting (called test or trial) functions $W_i(\vec{r})$ , $i\in[1;n]$ and by enforcing the condition

$$\int_{\mathcal{V}}W_i(\vec{r})R(\vec{r}) \mathrm{d}V = 0.$$ (3.8)

Insertion of (3.7) in (3.8) gives

$$\int_{\mathcal{V}}W_i(\vec{r})\left\{\mathcal{L}\left[\tilde{u}(\vec{r})\right] - f(\vec{r})\right\} \mathrm{d}V = 0, i\in[1;n],$$ (3.9)

which leads to the following expression to obtain the coefficients $c_j$

$$\int_{\mathcal{V}}W_i(\vec{r})\left\{\mathcal{L}\left[\sum_{j=1}^... ...c{r}) + v(\vec{r})\right] - f(\vec{r})\right\} \mathrm{d}V = 0, i\in[1;n].$$ (3.10)

Since $\mathcal{L}$ is a linear differential operator (3.10) becomes

$$c_j\sum_{j=1}^{n}\int_{\mathcal{V}}W_i(\vec{r})\mathcal{L}\left[N... ...c{r}) - \mathcal{L}\left[v(\vec{r})\right]\right\} \mathrm{d}V, i\in[1;n],$$ (3.11)

which corresponds to a linear equation system

$$\left[K\right]\left\{c\right\} = \left\{d\right\}.$$ (3.12)

The Matrix $[K]$ and the right hand side vector $\{d\}$ are given by the expressions

$$\begin{split}K_{ij} & = \int_{\mathcal{V}}W_i(\vec{r})\mathca... ...t]\right\} \mathrm{d}V, i\in[1;n], j\in[1;n]. \end{split}$$ (3.13)