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# 数学代写|有限元方法作业代写finite differences method代考|Convergence of the finite element method

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## 数学代写|有限元方法作业代写finite differences method代考|A one-dimensional example

Consider the following positive definite, self-adjoint, two-point boundary-value problem:
$$-\frac{d}{d x}\left(p(x) \frac{d u}{d x}\right)+q(x) u=f(x), \quad a0 \quad \text { and } \quad q(x) \geq 0 \quad \text { for } \quad a<x<b .$$
It follows from the one-dimensional form of the functional (2.44) of Section $2.6$ that the solution $u_{0}$ of eqn (6.1) subject to eqn (6.2) minimizes
$$I[u]=\int_{a}^{b}\left{p\left(u^{\prime}\right)^{2}+q u^{2}-2 u f\right} d x .$$
It is convenient at this stage to introduce the following inner product notations:
\begin{aligned} (u, v) &=\int_{a}^{b} u v d x \ A(u, v) &=\int_{a}^{b}\left(p u^{\prime} v^{\prime}+q u v\right) d x . \end{aligned}

## 数学代写|有限元方法作业代写finite differences method代考|Two-dimensional problems involving Poisson’s equation

For two-dimensional elements, error bounds are found in terms of the element diameters; for example, the diameter of a triangle is the length of the longest side and the diameter of a quadrilateral is the length of the longer diagonal.
Consider a discretization of some two-dimensional region $D$ by means of triangles and suppose that $h$ is the maximum diameter for these triangles. Error analysis in this case is far more complicated than for the two-point boundaryvalue problem of Section $6.1$, and only a statement of the error bounds is given here; the interested reader is referred to the books by Strang and Fix (1973) and Wait and Mitchell (1985) for more details and further references.
The norm used here is given by
$$|u|_{p}=\left[\iint_{D}\left{u^{2}+\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}+\left(\frac{\partial^{2} u}{\partial x^{2}}\right)^{2}+\ldots+\left(\frac{\partial^{p} u}{\partial y^{p}}\right)^{2}\right} d x d y\right]^{1 / 2} .$$
The error $e=u_{0}-\tilde{u}$ may be shown to satisfy an inequality of the form
$$|e|_{1} \leq C h^{2} \max \left(\left|u_{x x}\right|,\left|u_{x y}\right|,\left|u_{y y}\right|\right),$$
i.e. just as in the one-dimensional case, the norm of the error behaves like $h^{2}$ as $h \rightarrow 0$.

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|Isoparametric elements: numerical integration

The isoparametric concept was introduced in Section 4.6. There, it was noted that the forms of the integrands obtained are usually too complicated to be evaluated analytically and are invariably obtained numerically using Gauss quadrature. Consequently, another source of error is introduced. The effect of numerical integration is considered in this section. Typically, for Poisson’s equation, the integrals involved in computing $k_{i j}^{e}$ are of the form of eqn (3.40),
$$\iint_{[e]} k(x, y)\left(\frac{\partial N_{i}^{e}}{\partial x} \frac{\partial N_{j}^{e}}{\partial x}+\frac{\partial N_{i}^{e}}{\partial y} \frac{\partial N_{j}^{e}}{\partial y}\right) d x d y$$
This expression was obtained using Galerkin’s method and can be considered to come from an expression of the form
Convergence of the finite element method
227
$$\iint_{[e]} k(x, y) \operatorname{grad} \tilde{u}^{e} \cdot \operatorname{grad} W d x d y$$
where $\tilde{u}^{e}$ is a trial function and $W$ a weighting function. In the case where $W$ is a linear polynomial, the integral is of the form
$$\iint_{[e]} k(x, y)\left(c_{1} \frac{\partial \tilde{u}^{e}}{\partial x}+c_{2} \frac{\partial \tilde{u}^{e}}{\partial y}\right) d x d y$$

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|A ONE-DIMENSIONAL EXAMPLE

−ddX(p(X)d在dX)+q(X)在=F(X),一种0 和 q(X)≥0 为了 一种<X<b.

I[u]=\int_{a}^{b}\left{p\left(u^{\prime}\right)^{2}+q u^{2}-2 u f\right} d x 。I[u]=\int_{a}^{b}\left{p\left(u^{\prime}\right)^{2}+q u^{2}-2 u f\right} d x 。

(在,在)=∫一种b在在dX 一种(在,在)=∫一种b(p在′在′+q在在)dX.

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|TWO-DIMENSIONAL PROBLEMS INVOLVING POISSON’S EQUATION

|u|_{p}=\left[\iint_{D}\left{u^{2}+\left(\frac{\partial u}{\partial x}\right)^{2}+\left (\frac{\partial u}{\partial y}\right)^{2}+\left(\frac{\partial^{2} u}{\partial x^{2}}\right)^{2 }+\ldots+\left(\frac{\partial^{p} u}{\partial y^{p}}\right)^{2}\right} d x d y\right]^{1 / 2} 。|u|_{p}=\left[\iint_{D}\left{u^{2}+\left(\frac{\partial u}{\partial x}\right)^{2}+\left (\frac{\partial u}{\partial y}\right)^{2}+\left(\frac{\partial^{2} u}{\partial x^{2}}\right)^{2 }+\ldots+\left(\frac{\partial^{p} u}{\partial y^{p}}\right)^{2}\right} d x d y\right]^{1 / 2} 。

|和|1≤CH2最大限度(|在XX|,|在X是|,|在是是|),

## 数学代写|有限元方法作业代写FINITE DIFFERENCES METHOD代考|ISOPARAMETRIC ELEMENTS: NUMERICAL INTEGRATION

4.6 节介绍了等参概念。在那里，人们注意到获得的被积函数的形式通常太复杂而无法进行分析评估，并且总是使用高斯求积以数值方式获得。因此，引入了另一个误差源。本节考虑数值积分的影响。通常，对于泊松方程，计算中涉及的积分ķ一世j和是 eqn 的形式3.40,
∬[和]ķ(X,是)(∂ñ一世和∂X∂ñj和∂X+∂ñ一世和∂是∂ñj和∂是)dXd是

227
∬[和]ķ(X,是)毕业⁡在~和⋅毕业⁡在dXd是

∬[和]ķ(X,是)(C1∂在~和∂X+C2∂在~和∂是)dXd是