# 数学代写|有限元方法代写Finite Element Method代考|EG55M1

#### Doug I. Jones

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

While the dependent variable changes from $u$ to $u+\delta u$ the functional changes from $I(u)$ to $I(u)+\delta I(u)$. The variation represents a change from a stationary function to a function close to it. Therefore, in order to find the stationary function the variation of the functional would have to be zero. Let’s consider the variation of a functional that is a function of $u$ and $u_{, x}$, such as Eq. (3.21). Let’s also assume that this functional does not vary on its boundaries. The stationary value of such a function, can be found by setting the variation of the functional to zero,
$$\delta I(u)=\delta \int_{x_0}^{x_L} F\left(x, u, u_{, x}\right) d x=\int_{x_0}^{x_L} \delta F\left(x, u, u_{, x}\right) d x=0$$
The variation of the function $F$ inside the functional can be expressed as follows $[2]$
$$\delta F=\frac{\partial F}{\partial x} \delta x+\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_{, x}} \delta u_{, x}$$
However, the variational operator $\delta$ measures the change of the value of the functional as the function $u$ is varied. In fact it is not concerned with the effects of point-to-point changes in $u$. Thus the first term in Eq. (3.32) is zero, and the variational derivative of $F\left(x, u, u_{, x}\right)$ becomes,
$$\delta F=\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_{, x}} \delta u_{, x}$$
It is instructive to note the similarities and the differences between the variational derivative operator and the total differential operator. Recall the total differential $d f$ of a function $f=f(x, y, z)$ is given as follows:
$$d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z$$
and, the total differential $d F$ of the function $F\left(x, u, u_{, x}\right)$ is given as follows:
$$d F=\frac{\partial F}{\partial x} d x+\frac{\partial F}{\partial u} d u+\frac{\partial F}{\partial u_x} d u_{, x}$$
As stated above the variation sought by the variational operator is on the effects of the value of the function but not on the effects of the point-to-point variations of the function. This is sometimes stated as the variational operator $\delta$ seeks a variation of $F$ at a fixed $x$ location.

## 数学代写|有限元方法代写Finite Element Method代考|Properties of functionals

A generic functional $I(u)$ is linear if it satisfies the following condition:
$$I(\alpha u+\beta v)=\alpha I(u)+\beta I(v)$$
where $\alpha$ and $\beta$ two arbitrary scalars and $u$ and $v$ are two arbitrary functions. A generic functional $B(u, v)$ of two dependent variables is bilinear if both dependent variables can be shown to be linear separately. The linearity of the first argument can be shown if the following statement can be satisfied,
$$B\left(\alpha u_1+\beta u_2, v\right)=\alpha B\left(u_1, v\right)+\beta B\left(u_2, v\right)$$

where $u_1$ and $u_2$ are two arbitrary functions. Similarly the linearity of the second argument can be shown if,
$$B\left(u, \alpha v_1+\beta v_2\right)=\alpha B\left(u, v_1\right)+\beta B\left(u, v_2\right)$$
can be satisfied for any dependent variables $v_1$ and $v_2$. A bilinear functional is symmetric if the following statement can be demonstrated:
$$B(u, v)=B(v, u)$$

# 有限元方法代考

## 数学代写|有限元方法代写有限元法代考|函数导数

$$\delta I(u)=\delta \int_{x_0}^{x_L} F\left(x, u, u_{, x}\right) d x=\int_{x_0}^{x_L} \delta F\left(x, u, u_{, x}\right) d x=0$$

$$\delta F=\frac{\partial F}{\partial x} \delta x+\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_{, x}} \delta u_{, x}$$

$$\delta F=\frac{\partial F}{\partial u} \delta u+\frac{\partial F}{\partial u_{, x}} \delta u_{, x}$$
。注意变分导数算子与全微分算子之间的异同是有指导意义的。函数$f=f(x, y, z)$的总微分$d f$如下:
$$d f=\frac{\partial f}{\partial x} d x+\frac{\partial f}{\partial y} d y+\frac{\partial f}{\partial z} d z$$
，函数$F\left(x, u, u_{, x}\right)$的总微分$d F$如下:
$$d F=\frac{\partial F}{\partial x} d x+\frac{\partial F}{\partial u} d u+\frac{\partial F}{\partial u_x} d u_{, x}$$

## 数学代写|有限元方法代写有限元法代考|泛函的性质

$$I(\alpha u+\beta v)=\alpha I(u)+\beta I(v)$$

$$B\left(\alpha u_1+\beta u_2, v\right)=\alpha B\left(u_1, v\right)+\beta B\left(u_2, v\right)$$

，则可以证明第一个参数的线性

$$B\left(u, \alpha v_1+\beta v_2\right)=\alpha B\left(u, v_1\right)+\beta B\left(u, v_2\right)$$

$$B(u, v)=B(v, u)$$

## 有限元方法代写

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