# 物理代写|空气动力学代写Aerodynamics代考|MAE589

## 物理代写|空气动力学代写Aerodynamics代考|Variational Foundations of the Finite Element Method

In order to clarify the variational foundations of the finite element method, we examine in this section the one-dimensional elliptic equation
$$L u=f$$
on a domain from $a$ to $b$, where
$$L u=-\frac{d}{d x} p \frac{d u}{d x}+q u$$
with $p(x) \geq p_{\min }>0, q(x) \geq 0$, and Dirichlet or Neumann boundary conditions
\begin{aligned} &u(a)=u_L \text { or } u^{\prime}(a)=0 \ &u(b)=u_R \text { or } u^{\prime}(b)=0 \end{aligned}
either at both ends, or one at each end. This is the most general second order selfadjoint linear operator. We seek a solution $u(x)$ in a space $\mathcal{S}$ of twice-differentiable functions that satisfy any Dirichlet boundary conditions that may be specified.
The solution of this problem minimizes the functional
$$I(u)=\frac{1}{2} \int_a^b\left(p u^{\prime 2}+q u^2-2 f u\right) d x .$$
Suppose that $u(x)$ is modified by the addition of $\delta u(x)=\epsilon v(x)$, where $\epsilon$ is small. The corresponding variation of $I$ is
\begin{aligned} \delta I &=I(u+\delta u)-I(u) \ &=\epsilon V_1+\epsilon^2 V_2, \end{aligned}

where
$$V_1=\int_a^b\left(p u^{\prime} v^{\prime}+q u v-f v\right) d x$$
and
$$V_2=\frac{1}{2} \int_a^b\left(p v^2+q v^2\right) d x \geq 0$$

## 物理代写|空气动力学代写Aerodynamics代考|Error estimates based on the variational form

An error estimate for the solution error $u-u_h$ can be derived by the following argument. First, since the trial space $\mathcal{S}_h$ is a subspace of the solution space $\mathcal{S}, u$ satisfies
$$a\left(u, v_h\right)=\left(f, v_h\right)$$
for all $v_h$ in $\mathcal{S}_h$, and hence it follows from (3.41) and (3.42) that
$$a\left(u-u_h, v_h\right)=0 .$$
In other words, the error is orthogonal to all test functions $v_h$ in $\mathcal{S}_h$. But this implies that $u_h$ also minimizes $a\left(u-u_h, u-u_h\right)$ over $\mathcal{S}_h$ since
$$a\left(u-u_h-\epsilon v_h, u-u_h+\epsilon v_h\right)=a\left(u-u_h, u-u_h\right)-2 \epsilon a\left(u-u_h, v_h\right)+\epsilon^2 a\left(v_h, v_h\right) \text {, }$$
where the middle term vanishes according to the orthogonality condition (3.44). Thus $u_h$ minimizes the energy $\frac{1}{2} a\left(u-u_h, u-u_h\right)$ of the error, and provided that $a(u, u)$ meets the requirements for a legitimate norm, we can regard this as a norm of the error.
Also, we can substitute $u_h$ for $v_h$ in (3.44) to obtain
$$a\left(u-u_h, u_h\right)=0 .$$
Hence
$$a\left(u, u_h\right)=a\left(u_h, u_h\right)$$
and
\begin{aligned} a\left(u-u_h, u-u_h\right) &=a(u, u)-2 a\left(u, u_h\right)+a\left(u_h, u_h\right) \ &=a(u, u)-a\left(u_h, u_h\right) . \end{aligned}
Thus the energy of the error is equal to the error in the energy, and it can also be seen that if the error is not zero, the discrete energy $a\left(u_h, u_h\right)$ always underestimates the true energy $a(u, u)$.

## 物理代写|空气动力学代写空气动力学代考|有限元方法的变分基础

$$L u=f$$

$$L u=-\frac{d}{d x} p \frac{d u}{d x}+q u$$
with $p(x) \geq p_{\min }>0, q(x) \geq 0$，以及Dirichlet或Neumann边界条件
\begin{aligned} &u(a)=u_L \text { or } u^{\prime}(a)=0 \ &u(b)=u_R \text { or } u^{\prime}(b)=0 \end{aligned}

$$I(u)=\frac{1}{2} \int_a^b\left(p u^{\prime 2}+q u^2-2 f u\right) d x .$$

\begin{aligned} \delta I &=I(u+\delta u)-I(u) \ &=\epsilon V_1+\epsilon^2 V_2, \end{aligned}

where
$$V_1=\int_a^b\left(p u^{\prime} v^{\prime}+q u v-f v\right) d x$$
and
$$V_2=\frac{1}{2} \int_a^b\left(p v^2+q v^2\right) d x \geq 0$$

## 物理代写|空气动力学代写空气动力学代考|基于变分形式的误差估计

$$a\left(u, v_h\right)=\left(f, v_h\right)$$
，因此从(3.41)和(3.42)得出
$$a\left(u-u_h, v_h\right)=0 .$$
。换句话说，误差与$\mathcal{S}_h$中的所有测试函数$v_h$正交。但这意味着$u_h$也使$a\left(u-u_h, u-u_h\right)$比$\mathcal{S}_h$最小，因为
$$a\left(u-u_h-\epsilon v_h, u-u_h+\epsilon v_h\right)=a\left(u-u_h, u-u_h\right)-2 \epsilon a\left(u-u_h, v_h\right)+\epsilon^2 a\left(v_h, v_h\right) \text {, }$$
，其中根据正交条件(3.44)，中项消失。因此$u_h$使误差的能量$\frac{1}{2} a\left(u-u_h, u-u_h\right)$最小化，并且假设$a(u, u)$满足合法范数的要求，我们可以将其视为误差的范数。

$$a\left(u-u_h, u_h\right)=0 .$$

$$a\left(u, u_h\right)=a\left(u_h, u_h\right)$$

\begin{aligned} a\left(u-u_h, u-u_h\right) &=a(u, u)-2 a\left(u, u_h\right)+a\left(u_h, u_h\right) \ &=a(u, u)-a\left(u_h, u_h\right) . \end{aligned}

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