## 数学代写|有限元方法代写Finite Element Method代考|The weak form

The weak form of a differential equation is obtained by applying the prescribed boundary conditions to the weighted residual integral, following the integration by parts. For example, recall the boundary conditions of the example boundary value problem in Eqs. (3.3b), (3.3c),
at $x=x_0: \quad u\left(x_0\right)=u_0$
at $x=x_L:\left.a\left(x_L\right) \frac{d u}{d x}\right|{x_L}=Q_L$ and the weighted residual integral for this example is given in Eq. (3.53b). Note that an essential boundary condition is specified at $x=x_0$. This implies that $\delta u\left(x_0\right)=0$ and therefore we should use $w\left(x_0\right)=0$. A natural boundary condition is specified on the other boundary, $x=x_L$. This implies that the quantity $a\left(x_L\right) u_r\left(x_L\right)$ must take the value $Q_L$. These observations are implemented in Hin. (3.53h) which hecomes. $$w\left(x_L\right)(\underbrace{\left(\left.a\left(x_L\right) \frac{d u}{d x}\right|{x_L}\right)}{=Q_L}-\underbrace{w\left(x_0\right)}{=0}\left(\left.a\left(x_0\right) \frac{d u}{d x}\right|{x_0}\right)-\int{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x=0$$
The weak form of the boundary value problem described in Example $3.1$ then becomes.
$$w\left(x_L\right) Q_L-\int_{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x=0$$
It should be noted that the natural boundary condition is now incorporated in the integral formulation. This will be helpful later when we develop solution methods.

## 数学代写|有限元方法代写Finite Element Method代考|Relationship between the weak form and functionals

We showed in the previous section that the weight function $w$ and the variation of the dependent variable $\delta u$ behave in the same way. Therefore, as $w$ is an arbitrary function, we can use $\delta u$ in its place. The weak form of the boundary value problem given in Example $3.1$ becomes,
$$-\int_{x_0}^{s_L}\left[a \frac{d u}{d x} \frac{d \delta u}{d x}-c u \delta u\right] d x+\int_{x_0}^{s_L} q \delta u d x+Q_L \delta u\left(x_L\right)=0$$
Noting that $\frac{1}{2} \delta\left(u^2\right)=u \delta u$ and $\frac{1}{2} \delta\left(u_{, x}^2\right)=u_{, x} \delta u_{, x}$ the first term is modified as follows:
$$-\int_{x_0}^{x_L} \frac{1}{2} \delta\left[a\left(\frac{d u}{d x}\right)^2-c u^2\right] d x+\int_{x_0}^{x_L} q \delta u d x+\delta u\left(x_L\right) Q_L=0$$

Using the commutation property of the variational derivative the following relationship is obtained,
$$\delta\left[-\frac{1}{2} \int_{x_0}^{x_L}\left(a\left(\frac{d u}{d x}\right)^2-c u^2\right) d x+\int_{x_0}^{x_L} q u d x+u\left(x_L\right) Q_L\right]=0$$
Note that the term in the square brackets is a functional $I\left(u_1, u_{, x}\right)$ and Eq. (3.58) represents the variation of the functional $\delta I$. This result indicates that the weak-integral-form of the boundary value problem, e.g. Eq. (3.57), is equal to mimimum of the variation of the functional $I$ associated with the problem.
For solid mechanics problems $I\left(u, u_{, x}\right)$ is the total potential energy of the system. The statement
$$\delta I\left(u, u_{, x}\right)=0$$
represents the minimum total potential energy principle: of all admissible functions $u$, that which makes the total potential energy I(u) a minimum, also satisfies the differential equation of equilibrium and natural boundary conditions.

## 数学代写|有限元方法代写有限元法代考|弱形式

at $x=x_0: \quad u\left(x_0\right)=u_0$
at $x=x_L:\left.a\left(x_L\right) \frac{d u}{d x}\right|{x_L}=Q_L$，本例的加权残差积分如式(3.53b)所示。注意，在$x=x_0$上指定了一个必要的边界条件。这意味着$\delta u\left(x_0\right)=0$，因此我们应该使用$w\left(x_0\right)=0$。另一个边界$x=x_L$上指定了一个自然边界条件。这意味着数量$a\left(x_L\right) u_r\left(x_L\right)$必须取$Q_L$的值。这些观察结果在Hin中实现。(3.53小时)。$$w\left(x_L\right)(\underbrace{\left(\left.a\left(x_L\right) \frac{d u}{d x}\right|{x_L}\right)}{=Q_L}-\underbrace{w\left(x_0\right)}{=0}\left(\left.a\left(x_0\right) \frac{d u}{d x}\right|{x_0}\right)-\int{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x=0$$

$$w\left(x_L\right) Q_L-\int_{x_0}^{x_L}\left[a \frac{d u}{d x} \frac{d w}{d x}-c w u-w q\right] d x=0$$

## 数学代写|有限元方法代写Finite Element Method代考|Relationship between the weak form and functionals

$$-\int_{x_0}^{s_L}\left[a \frac{d u}{d x} \frac{d \delta u}{d x}-c u \delta u\right] d x+\int_{x_0}^{s_L} q \delta u d x+Q_L \delta u\left(x_L\right)=0$$

$$-\int_{x_0}^{x_L} \frac{1}{2} \delta\left[a\left(\frac{d u}{d x}\right)^2-c u^2\right] d x+\int_{x_0}^{x_L} q \delta u d x+\delta u\left(x_L\right) Q_L=0$$

$$\delta\left[-\frac{1}{2} \int_{x_0}^{x_L}\left(a\left(\frac{d u}{d x}\right)^2-c u^2\right) d x+\int_{x_0}^{x_L} q u d x+u\left(x_L\right) Q_L\right]=0$$

$$\delta I\left(u, u_{, x}\right)=0$$

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