# 数学代写|有限元方法代写Finite Element Method代考|SEM712

## 数学代写|有限元方法代写Finite Element Method代考|Weighted residual integral

Let us first define the residual of the governing equation. For example, the residual of Eq. (3.1) is defined as follows:
$$R(u)=\mathcal{L}(u)-f=0 \text { in } \Omega$$
Note that the residual, $R$ is zero, if the exact solution $u$ of the boundary value problem is known. Otherwise, if we have an approximate solution, the residual will have an error $\varepsilon$ which can be represented as follows:
$$R(\tilde{u})=\varepsilon \neq 0$$
The weighted-residual statement of the original differential equation or simply the weighted-residual integral (WRI) is a bilinear functional defined as follows:
$$I(u, w)=\int_{\Omega} w R d \Omega=0$$
where $w$ is a weight function. This can be any nonzero integrable function that helps to find an approximate solution to the problem.
For Eq. (3.3a), the residual becomes,
$$R=\left[\frac{d}{d x}\left(a \frac{d u}{d x}\right)+c u+q\right]=0 \text { in } x_0<x<x_L$$
and the corresponding weighted residual integral is,
$$I(u, w)=\int_{x_0}^{x_L} w R d x=\int_{x_0}^{x_L} w\left[\frac{d}{d x}\left(a \frac{d u}{d x}\right)+c u+q\right] d x=0$$

## 数学代写|有限元方法代写Finite Element Method代考|Boundary conditions

The order of differentiation of the dependent variable $u$ in the weighted residual statement can be reduced by using integration by parts until the differentiation is distributed evenly between the dependent variable $u$ and the weight function $w$. The boundary terms that results from integration by parts represent the admissible boundary conditions of the problem.

Let’s demonstrate this on the example boundary value problem whose weighted residual integral form is given by Eq. (3.52). In order to distribute the differentiation operator evenly between the weight function $w$ and the independent variable $u$ let’s use integration by parts on the first term,
\begin{aligned} &\int_{x_0}^{x_L}\left[w \frac{d}{d x}\left(a \frac{d u}{d x}\right)+c w u+w q\right] d x=0 \ &w\left(x_L\right)\left(a \frac{d u}{d x}\right){x_L}-w\left(x_0\right)\left(a \frac{d u}{d x}\right){x_0}-\int_{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x \equiv 0 \end{aligned}
Note that in order for Eq. (3.53a) to be satisfied, each one of its terms must vanish separately. Thus it can be easily shown that on the boundaries the following conditions must be satisfied,
at $x=x_0$ : either $w\left(x_0\right)=0$
or
$$\left.a\left(x_0\right) \frac{d u}{d x}\right|_{x_0}=0$$ and
at $x=x_L$ : either $w\left(x_L\right)=0$
or
$$\left.a\left(x_L\right) \frac{d u}{d x}\right|_{x_L}=0$$

## 数学代写|有限元方法代写有限元法代考|加权残差积分

$$R(u)=\mathcal{L}(u)-f=0 \text { in } \Omega$$

$$R(\tilde{u})=\varepsilon \neq 0$$

$$I(u, w)=\int_{\Omega} w R d \Omega=0$$

$$R=\left[\frac{d}{d x}\left(a \frac{d u}{d x}\right)+c u+q\right]=0 \text { in } x_0<x<x_L$$
，对应的加权残差积分为，
$$I(u, w)=\int_{x_0}^{x_L} w R d x=\int_{x_0}^{x_L} w\left[\frac{d}{d x}\left(a \frac{d u}{d x}\right)+c u+q\right] d x=0$$

## 数学代写|有限元方法代写有限元法代考|边界条件

\begin{aligned} &\int_{x_0}^{x_L}\left[w \frac{d}{d x}\left(a \frac{d u}{d x}\right)+c w u+w q\right] d x=0 \ &w\left(x_L\right)\left(a \frac{d u}{d x}\right){x_L}-w\left(x_0\right)\left(a \frac{d u}{d x}\right){x_0}-\int_{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x \equiv 0 \end{aligned}

$$\left.a\left(x_0\right) \frac{d u}{d x}\right|_{x_0}=0$$和

$$\left.a\left(x_L\right) \frac{d u}{d x}\right|_{x_L}=0$$

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