# 2D formulation for Electrostatic Problems

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 Revision as of 14:52, 30 October 2009 (view source)JMora (Talk | contribs)← Older edit Revision as of 14:57, 30 October 2009 (view source)JMora (Talk | contribs) Newer edit → Line 28: Line 28: [/itex] [/itex] − − − − − We will apply the [[Poisson's_Equation#Weighted_Residual_Method_.28WRM.29 | residual formulation]] based on the Weighted Residual Method (WRM). − − − ::$− { − \int_{\Omega} W(x,y,z) r_{\Omega} \partial \Omega − + \oint_{\Gamma} \overline{W}(x,y,z) r_{\Gamma} \partial \Gamma=0 − } −$ Line 46: Line 33: with: with: − ::$W(x,y,z) \,$ and $\overline{W}(x,y,z)$ the weighting functions. − ::$\frac{}{} r_{\Omega} = A(\hat V) \ne 0 \quad in \quad \Omega$ + ::$V (x,y,z) \cong \hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i$ − ::$\frac{}{} r_{\Gamma} = B(\hat V) \ne 0 \quad in \quad \Gamma$ + ::$\mathbf{N} = + \begin{bmatrix} + N_1 \\ + \, \\ + N_2 \\ + \, \\ + \vdots \\ + \, \\ + N_n + \end{bmatrix} +$ − Where $\hat V \,$ is the numerical approach of the unknown $V \,$: + ::$\mathbf{B_i}= + \begin{bmatrix} + \frac{\partial N_i}{\partial x} \\ + \, \\ + \frac{\partial N_i}{\partial y} \\ + \, \\ + \frac{\partial N_i}{\partial z} + \end{bmatrix} +$ − − ::$V (x,y,z) \cong \hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i$

## Revision as of 14:57, 30 October 2009

The 2D Electrostatic Poisson's equation given by the governing PDE and its boundary conditions:

$A(V) = \left[ \frac{\partial}{\partial x}\cdot \left( \varepsilon_{x} \cdot \frac{\partial}{\partial x}\right) + \frac{\partial}{\partial y}\cdot \left(\varepsilon_{y} \cdot \frac{\partial}{\partial y} \right) \right]V(x,y) + \rho_S = 0 ~~ in ~ \Omega$

$B(V) = \begin{cases} \left . V - \bar V = 0 \right |_{\Gamma_{V}} & in ~ \Gamma_{\varphi} \\ \left . \hat n \vec{D} - \bar D_n = 0 \right |_{\Gamma_{q}} & in ~ \Gamma_{q} \\ \left . \frac{\partial V}{\partial r} \right |_{\Gamma_{\infty}} \approx - \frac{V}{r} & in ~ \Gamma_{\infty} \end{cases}$

can be written as (see the General formulation for Electrostatic Problems):

$\mathbf{K}^{(e)}= \int_{\Omega^{(e)}} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \partial \Omega^{(e)} + \oint_{\Gamma_{\infty}^{(e)}} \mathbf{N^T} \alpha \mathbf{N} \partial \Gamma_{\infty}^{(e)}$
$\mathbf{f}^{(e)}= \int_{\Omega^{(e)}} \mathbf{N^T} \rho_v \partial \Omega^{(e)} - \oint_{\Gamma_q^{(e)}} \mathbf{N^T} \bar D_n \partial \Gamma_q^{(e)} - \oint_{\Gamma_V^{(e)}} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V^{(e)}$

with:

$V (x,y,z) \cong \hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i$

$\mathbf{N} = \begin{bmatrix} N_1 \\ \, \\ N_2 \\ \, \\ \vdots \\ \, \\ N_n \end{bmatrix}$

$\mathbf{B_i}= \begin{bmatrix} \frac{\partial N_i}{\partial x} \\ \, \\ \frac{\partial N_i}{\partial y} \\ \, \\ \frac{\partial N_i}{\partial z} \end{bmatrix}$

This is:

${ \int_{\Omega} W \left [ \nabla^T \mathbf{\varepsilon} \nabla \hat V + \rho_v \right ] \partial\Omega + \oint_{\Gamma} \overline{W} \left [\mathbf{n}^T \mathbf{\varepsilon} \nabla \hat V + \bar \mathbf{q} + \alpha V \right ] \partial \Gamma=0 }$

with $\alpha = \frac{1}{r}$ the infinit condition factor and $\bar \mathbf{q}$ the field produced whe $V \,$ is fixed by $\bar V \,$.

The weak form of this expression can be obtained using the integration by parts. In addition, if     $\bar W = - W \,$:

${ \int_{\Omega} \nabla^T W^T \mathbf{\varepsilon} \nabla \hat V \partial \Omega + \oint_{\Gamma_{\infty}} W^T \alpha V \partial \Gamma_{\infty} = \int_{\Omega} W^T \rho_v \partial \Omega - \oint_{\Gamma_q} W^T \bar D_n \partial \Gamma_q - \oint_{\Gamma_V} W^T \mathbf{q_n} \partial \Gamma_V }$

Remembering that:

$\hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i = \mathbf{N} \mathbf{a}^{(e)}$

$\nabla \hat V = \nabla \mathbf{N} \mathbf{a}^{(e)} = \mathbf{B} \mathbf{a}^{(e)}$

is the gradient potential with:

$\mathbf{B}= \left [ \mathbf{B_1, B_2 ... B_n} \right ]$

and:

$\mathbf{B_i}= \begin{bmatrix} \frac{\partial N_i}{\partial x} \\ \, \\ \frac{\partial N_i}{\partial y} \\ \, \\ \frac{\partial N_i}{\partial z} \end{bmatrix}$

The electric field and electric displacement field can be written as follows:

$\mathbf{q} = - \mathbf{B} \mathbf{a}^{(e)} \qquad \mathbf{q'} = - \mathbf{\varepsilon} \mathbf{B} \mathbf{a}^{(e)}$

We will now use the Galerkin Method $W_i(x) \equiv N_i(x) \,$. So, finally, the integral expression ready to create the matricial system of equations is:

${ \int_{\Omega} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \mathbf{a} \partial \Omega + \oint_{\Gamma_{\infty}} \mathbf{N^T} \alpha \mathbf{N} \mathbf{a} \partial \Gamma_{\infty} = \int_{\Omega} \mathbf{N^T} \rho_v \partial \Omega - \oint_{\Gamma_q} \mathbf{N^T} \bar D_n \partial \Gamma_q - \oint_{\Gamma_V} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V }$

$\mathbf{K} \mathbf{a} \,= \mathbf{f}$

Note that K is a coefficients matrix that depends on the geometrical and physical properties of the problem, a is the vector with the n unknowns to be obtained and f is a vector that depends on the source values and boundary conditions.

$\mathbf{K}^{(e)}= \int_{\Omega^{(e)}} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \partial \Omega^{(e)} + \oint_{\Gamma_{\infty}^{(e)}} \mathbf{N^T} \alpha \mathbf{N} \partial \Gamma_{\infty}^{(e)}$
$\mathbf{f}^{(e)}= \int_{\Omega^{(e)}} \mathbf{N^T} \rho_v \partial \Omega^{(e)} - \oint_{\Gamma_q^{(e)}} \mathbf{N^T} \bar D_n \partial \Gamma_q^{(e)} - \oint_{\Gamma_V^{(e)}} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V^{(e)}$