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## 数学代写|变分法代写Calculus of Variations代考|The Stokes Theorem

Substituting Eq. (1.3.89) into Eq. (1.3.85), the Stokes formula expressed in the vector form can be obtained
$$\oint_L \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}=\iint_S \operatorname{rot} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{S}=\iint_S \operatorname{rot} \boldsymbol{a} \cdot \boldsymbol{n} \mathrm{d} S=\iint_S \nabla \times \boldsymbol{a} \mathrm{d} \boldsymbol{S}=\iint_S \boldsymbol{n} \cdot \nabla \times \boldsymbol{a} \mathrm{d} S$$
The Stokes formula is also called the Stokes theorem. The Stokes theorem revealed the transformation relations of the vector field line integral and surface integral. Thus it can be seen that the Green formula introduced before is just a special case of the Stokes theorem on a planar domain. Of course the Stokes theorem can also be directly proved, the proof method is similar to the one of the Gauss theorem. A proof is given below.

Proof Dividing the surface $S$ into $n$ elements of surface $\Delta S_1, \Delta S_2, \ldots, \Delta S_n$, the elements of perimeter which surround the elements of surface namely the closed curves are $\Delta L_1, \Delta L_2, \ldots, \Delta L_n$. Taking the $k$ th surface $\Delta S_k$ and the perimeter $\Delta L_k$, from the definition of rotation, there exists the following relationship $$\operatorname{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\frac{\oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}}{\Delta S_k}+\varepsilon_k$$
namely
$$\Delta S_k \mathrm{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}+\varepsilon_k \Delta S_k$$
where, the rotation is the value of the element of surface at a point $M$, the left-handed side expresses the flux on the element of surface $\Delta S_k ; \varepsilon_k$ is a small enough amount, and when $\Delta S_k \rightarrow 0, \varepsilon_k \rightarrow 0$. Summing the above expression with respect to $k$ from 1 to $n$, we obtain
$$\sum_{k=1}^n \Delta S_k \mathrm{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\sum_{k=1}^n \oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}+\sum_{k=1}^n \varepsilon_k \Delta S_k$$

## 数学代写|变分法代写Calculus of Variations代考|Fundamental Lemmas of the Calculus of Variations

For convenience of the later chapters to study variational problem, a few fundamental lemmas of the calculus of variations are introduced in the following.

Lemma 1.5.1 Let a function $f(x)$ be continuous in the interval $[a, b]$, an arbitrary function $\eta(x)$ has $n$th order continuous derivative in the interval $[a, b]$, and for a positive number $m(m=0,1, \ldots, n)$, when it satisfies the following conditions
$$\eta^{(k)}(a)=\eta^{(k)}(b)=0 \quad(k=0,1, \ldots, m)$$
if the integral
$$\int_a^b f(x) \eta(x) \mathrm{d} x=0$$
can always hold, there must be in the interval $[a, b]$
$$f(x) \equiv 0$$
Proof With the reduction to absurdity. If $f(x)$ does not identically vanish in the interval $[a, b]$, then it can be seen from the continuity of $f(x)$ that there is at least a point $\xi$ in the interval $(a, b)$, as shown in Fig. 1.4, such that $f(x) \neq 0$, might as well suppose $f(\xi)>0$, then there must exist a closed interval $\left[a_0, b_0\right]$ containing $\xi$, when $a0$. At this point, choosing the function $\eta(x)$ as follows
$$\eta(x)= \begin{cases}{\left[\left(x-a_0\right)\left(b_0-x\right)\right]^{2 n+2}} & x \in\left[a_0, b_0\right] \ 0 & x \notin\left[a_0, b_0\right]\end{cases}$$
Clearly Eq. (1.5.3) has the $n$th order continuous derivable function in the interval $[a, b]$, it satisfies the conditions $\eta^{(k)}(a)=\eta^{(k)}(b)=0(k=0,1, \ldots, m)$, hence, according to Eqs. (1.5.2) and (1.5.3), there is
$$\int_a^b f(x) \eta(x) \mathrm{d} x=\int_{a_0}^{b_0} f(x)\left(x-a_0\right)^{2 n+2}\left(b_0-x\right)^{2 n+2} \mathrm{~d} x>0$$

# 变分法代考

## 数学代写|变分法代写Calculus of Variations代考|The Stokes Theorem

$$\oint_L \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}=\iint_S \operatorname{rot} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{S}=\iint_S \operatorname{rot} \boldsymbol{a} \cdot \boldsymbol{n} \mathrm{d} S=\iint_S \nabla \times \boldsymbol{a} \mathrm{d} \boldsymbol{S}=\iint_S \boldsymbol{n} \cdot \nabla \times \boldsymbol{a} \mathrm{d} S$$

$$\operatorname{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\frac{\oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}}{\Delta S_k}+\varepsilon_k$$

$$\Delta S_k \operatorname{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}+\varepsilon_k \Delta S_k$$

$$\sum_{k=1}^n \Delta S_k \mathrm{rot} \boldsymbol{a} \cdot \boldsymbol{n}=\sum_{k=1}^n \oint_{\Delta L_k} \boldsymbol{a} \cdot \mathrm{d} \boldsymbol{L}+\sum_{k=1}^n \varepsilon_k \Delta S_k$$

## 数学代写|变分法代写Calculus of Variations代考|Fundamental Lemmas of the Calculus of Variations

$$\eta^{(k)}(a)=\eta^{(k)}(b)=0 \quad(k=0,1, \ldots, m)$$

$$\int_a^b f(x) \eta(x) \mathrm{d} x=0$$

$$f(x) \equiv 0$$

$$\eta(x)=\left{\left[\left(x-a_0\right)\left(b_0-x\right)\right]^{2 n+2} \quad x \in\left[a_0, b_0\right] 0 \quad x \notin\left[a_0, b_0\right]\right.$$

$$\int_a^b f(x) \eta(x) \mathrm{d} x=\int_{a_0}^{b_0} f(x)\left(x-a_0\right)^{2 n+2}\left(b_0-x\right)^{2 n+2} \mathrm{~d} x>0$$

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