3.3.2 Galerkin's Method for the Vector Differential Operator

If the vector differential operator $\vec{\mathcal{L}}$ from (3.3) is used the corresponding excitation function in (3.1) must be a vector function as well which is written as $\vec{f}\left(\vec{r}\right)$ .

$$\vec{\mathcal{L}}\left[\vec{u}\left(\vec{r}\right)\right] = \vec{f}\left(\vec{r}\right).$$ (3.20)

The approximation (3.5) corresponds to

$$\tilde{\vec{u}}(\vec{r}) = \sum_{j=1}^{n}c_j\vec{N}_j(\vec{r}) + \vec{v}(\vec{r}) \simeq \vec{u}(\vec{r}),$$ (3.21)

where $\vec{v}(\vec{r})$ is a known function which exactly fulfills the Dirichlet boundary condition on $\mathcal{A}_D$ . The basis functions $\vec{N}_j(\vec{r}), j\in[1;n]$ build a set of linear independent known functions. Their tangential components vanish on the Dirichlet boundary $\mathcal{A}_D$ . The residual is given analogously to (3.7) by

$$\vec{R}(\vec{r}) = \vec{\mathcal{L}}\left[\tilde{\vec{u}}\left(\vec{r}\right)\right] - \vec{f}(\vec{r}) \neq 0$$ (3.22)

and does not vanish in general. In this case the weighting functions in Section 3.2 must also be merged into vector functions $\vec{W}_i$ and the dot product must be used instead of scalar multiplication.

$$\int_{\mathcal{V}}\vec{W}_i(\vec{r})\cdot\vec{R}(\vec{r}) = \int_... ...{u}}(\vec{r})\right] - \vec{f}(\vec{r})\right\} \mathrm{d}V = 0, i\in[1;n]$$ (3.23)

or

$$\int_{\mathcal{V}}\vec{W}_i(\vec{r})\cdot\vec{\mathcal{L}}\left[\... ...cal{V}}\vec{W}_i(\vec{r})\cdot\vec{f}(\vec{r}) \mathrm{d}V = 0, i\in[1;n].$$ (3.24)

With these considerations the weighted residual method described in section 3.2 will also lead to the linear equation system (3.12) where the matrix $[K]$ and the right hand side $\{d\}$ are given by

$$\begin{split}K_{ij} & = \int_{\mathcal{V}}\vec{W}_i(\vec{r})\... ...t]\right\} \mathrm{d}V, i\in[1;n], j\in[1;n]. \end{split}$$ (3.25)

After using the Galerkin approach $\vec{W}_i(\vec{r}) = \vec{N}_i(\vec{r})$ , substituting the vector differential operator $[\vec{\mathcal{L}}]$ from (3.3), and applying the first vector theorem of Green

$$\begin{split}& \quad \int_{\mathcal{V}}\vec{W}\cdot\left\{\ve... ...c{\nabla}\times\vec{u})\right]\right\} \mathrm{d}A \end{split}$$ (3.26)

to the first integral from the left hand side of (3.24), the following is obtained:

$$\begin{split}\int_{\mathcal{V}}\vec{W}_i\cdot\vec{\mathcal{L}... ..._i\cdot\utilde{b}\cdot\tilde{\vec{u}} \mathrm{d}V. \end{split}$$ (3.27)

Since the tangential component of $\vec{N}_i$ on the Dirichlet boundary $\mathcal{A}_D$ is zero the boundary integral of (3.27) can be written as

$$\begin{split}& \quad - \int_{\partial\mathcal{V}}\vec{n}\cdot... ...\times\tilde{\vec{u}})\right]\right\} \mathrm{d}A. \end{split}$$ (3.28)

Similarly to (3.18) for the scalar function $u$ and its approximation $\tilde{u}$ , the expression $\vec{n}\times\left[\utilde{a}\cdot(\vec{\nabla}\times\tilde{\vec{u}})\right]$ corresponds with the boundary condition $\vec{u}_n$ on $\mathcal{A}_N$  [35,39,40]

$$\vec{n}\times\left[\utilde{a}\cdot(\vec{\nabla}\times\tilde{\vec{... ...\utilde{a}\cdot(\vec{\nabla}\times\vec{u})\right] \mathrm{on} \mathcal{A}_N.$$ (3.29)

Consequently for $[K]$ and $\{d\}$ it can be written

$$\begin{split}K_{ij} & = \int_{\mathcal{V}}(\vec{\nabla}\times... ...t\vec{u}_n \mathrm{d}A, i\in[1;n], j\in[1;n], \end{split}$$ (3.30)

where $\vec{\mathcal{L}}$ is given by (3.3). For both, the scalar case (3.19) and the vector one (3.30), the matrix $[K]$ is symmetric.

The previous considerations are based on a three-dimensional domain. For the two-dimensional case similar formulas can be written. This will be explained by an example using Gauß's law:

$$\int_{\mathcal{V}}\vec{\nabla}\cdot\vec{u} \mathrm{d}V = \int_{\partial\mathcal{V}}\vec{n}\cdot\vec{u} \mathrm{d}A.$$ (3.31)

Let the three-dimensional domain $\mathcal{V}$ be a cylinder with an arbitrary basal plane $\mathcal{A}$ (with boundary $\partial\mathcal{A}$ ) and the height $h$ .

The function $\vec{u}$ is only defined in the two-dimensional domain of the cylinder basal plane. Thus it has no normal component to the cylinder basal plane and does not depend on the height. Taking these considerations into account the left and the right hand side of (3.31) can be written as

$$\begin{split}\int_{\mathcal{V}}\vec{\nabla}\cdot\vec{u} \mat... ...artial\mathcal{A}}\vec{n}\cdot\vec{u} \mathrm{d}s, \end{split}$$

which leads to Gauß's law for the two-dimensional case

$$\int_{\mathcal{A}}\vec{\nabla}\cdot\vec{u} \mathrm{d}A = \int_{\partial\mathcal{A}}\vec{n}\cdot\vec{u} \mathrm{d}s.$$

In a similar way the formulas for the three-dimensional case used in this work can be easily rewritten for two-dimensional regions. Thereby the three-dimensional domains $\mathcal{V}$ are replaced by two-dimensional ones $\mathcal{A}$ . The boundaries $\partial\mathcal{V}$ of the three-dimensional regions are replaced by the boundaries $\partial\mathcal{A}$ of the corresponding two-dimensional ones. The integration variable must also be changed accordingly.