# 物理代写|电磁学代写electromagnetism代考|ELEC3104

## 物理代写|电磁学代写electromagnetism代考|Extraction of Vector Potentials—Vanishing Normal Trace

We consider now the case of divergence-free fields of $L^2(\Omega)$ with vanishing normal trace. As we already saw in Sect. $3.3$ for elements of $\boldsymbol{Z}_T(\Omega)$, if the domain $\Omega$ is not topologically trivial, one has to take cuts into account explicitly.

Theorem 3.5.1 Let $\Omega$ be a domain such that $(\text { Top }){I=0}$ or $(\text { Top }){I>0}$ is fulfilled. Then, given $v \in L^2(\Omega)$, it holds that
$$\left.\begin{array}{l} \operatorname{div} \boldsymbol{v}=0 \mathrm{in} \Omega, \ \boldsymbol{v} \cdot \boldsymbol{n}{\mid \Gamma}=0, \ \langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle{\Sigma_i}=0, \forall i \end{array}\right} \Longleftrightarrow\left{\begin{array}{l} \exists \boldsymbol{w} \in \boldsymbol{H}0(\operatorname{curl}, \Omega), \ \operatorname{div} \boldsymbol{w}=0, \ \langle\boldsymbol{w} \cdot \boldsymbol{n}, 1\rangle{H^{1 / 2}\left(\Gamma_k\right)}=0, \forall k \end{array} \quad \boldsymbol{v}=\operatorname{curl} w\right.$$
Moreover, $w$ is unique, and there exists $C>0$ independent of $v$ such that
$$|w|_{H(\operatorname{curl}, \Omega)} \leq C|v|_{L^2(\Omega)} .$$
Remark 3.5.2 Assuming that $v$ writes $v=\operatorname{curl} w$ with $w \in \boldsymbol{H}0(\mathbf{c u r l}, \Omega)$, one has $v \in \boldsymbol{H}_0$ (div, $\Omega$ ) according to Proposition 2.2.10. Now, using the functions $\left(\dot{r}_i\right){1 \leq i \leq I}$ as they are defined in Proposition 3.3.3, one obtains, by integrating by parts twice $((3.6)$, then $(2.20))$, for each $i$
\begin{aligned} \langle v \cdot \boldsymbol{n}, 1\rangle_{\Sigma_i} &=\sum_j\left\langle\boldsymbol{v} \cdot \boldsymbol{n},\left[\dot{r}i\right] \Sigma_j\right\rangle{\Sigma_j} \ &=\left(\operatorname{curl} w, \operatorname{grad} \dot{r}i\right){L^2(\dot{\Omega})}=\left(\operatorname{curl} w \mid \widehat{\operatorname{grad} \dot{r}_i}\right)=0 . \end{aligned}
In the case when (Top) $I=0$ is fulfilled, $Z_T(\Omega)={0}$, and there are no vanishing flux conditions for the field $v$ on the cuts.

Proof We note that the vector potential $w$, if it exists, is unique. Indeed, if $w_1$ and $w_2$ both fulfill all the conditions (3.10), then $\delta w:=w_1-w_2 \in X_N(\Omega), \operatorname{curl} \delta w=0$ and $\operatorname{div} \delta \boldsymbol{w}=0$ in $\Omega$, with $\langle\delta \boldsymbol{w} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0$, for all $k$. Hence, $\delta \boldsymbol{w}=0$, so uniqueness follows.
Next, introducing the (closed) subspace of $\boldsymbol{X}N(\Omega)$ : $$\boldsymbol{X}_N^{\Gamma}(\Omega):=\left{\boldsymbol{f} \in \boldsymbol{X}_N(\Omega):\langle\boldsymbol{f} \cdot \boldsymbol{n}, 1\rangle{H^{1 / 2}\left(\Gamma_k\right)}=0,1 \leq k \leq K\right}$$
one can solve the variational formulation ${ }^8$
$$\left{\begin{array}{l} \text { Find } w \in X_N^{\Gamma}(\Omega) \text { such that } \ \forall w^{\prime} \in X_N^{\Gamma}(\Omega),\left(\operatorname{curl} w \mid \operatorname{curl} w^{\prime}\right)+\left(\operatorname{div} w \mid \operatorname{div} w^{\prime}\right)=\left(v \mid \operatorname{curl} w^{\prime}\right) \end{array}\right.$$

## 物理代写|电磁学代写electromagnetism代考|Extraction of Vector Potentials—Complements

In the proofs of the results of Sects. 3.3-3.5, we remark that the fundamental results (extraction of scalar potentials at Theorem 3.3.1, respectively of vector potentials at Theorem 3.4.1) are obtained by continuation to $\mathbb{R}^3$, and direct estimates of the norms. On the other hand, all the other proofs rely on solving (well-posed) variational formulations, for which norm estimates are simply a consequence of the Lax-Milgram Theorem 4.2.8.
To obtain the compact imbedding results, the proofs-à la Weber [204]—that we proposed are obtained via the extraction of scalar and vector potentials. In Chap. 6 , we propose another, indirect proof, which relies on the continuous imbeddings of $\boldsymbol{X}_N(\Omega)$ (Sect. 6.1.6) and $\boldsymbol{X}_T(\Omega)$ (Sect. 6.2.6) into fractional-order Sobolev spaces $\boldsymbol{H}^s(\Omega)$, for some $s>0$ that depends only on the geometry of the domain $\Omega$.

The additional knowledge on the regularity of elements of $\boldsymbol{X}_N(\Omega)$ and $\boldsymbol{X}_T(\Omega)$ will be used there. The compact imbedding results are then consequences of Proposition 2.1.43.
If one is looking for a vector potential that does not necessarily belong to $\boldsymbol{H}^1(\Omega$ ) for divergence-free fields, one has the result below, which “symmetrizes” the roles of $\boldsymbol{X}_T(\Omega)$ and $\boldsymbol{X}_N(\Omega)$.

Theorem 3.6.1 Let $\Omega$ be a domain such that $(\text { Top }){I=0}$ or $(\text { Top }){I>0}$ is fulfilled. Then, given $v \in L^2(\Omega)$, it holds that
$$\left.\begin{array}{l} \operatorname{div} \boldsymbol{v}=0 \mathrm{in} \Omega, \ \langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0, \forall k \end{array}\right} \Longleftrightarrow\left{\begin{array}{l} \exists \boldsymbol{w} \in \boldsymbol{H}0(\operatorname{div}, \Omega), \ \operatorname{div} \boldsymbol{w}=0, \ \langle\boldsymbol{w} \cdot \boldsymbol{n}, 1\rangle{\Sigma_i}=0, \forall i \end{array} \quad \boldsymbol{v}=\mathbf{c u r l} w\right.$$
Moreover, $w$ is unique, and there exists $C>0$ independent of $v$ such that
$$|w|_{H(\operatorname{curl}, \Omega)} \leq C|v|_{L^2(\Omega)^{\circ}}$$
Remark 3.6.2 In the case when (Top) $I=0$ is fulfilled, the result holds without the vanishing flux conditions on the cuts for the vector potential! In this case, we recall that $Z_T(\Omega)$ is reduced to ${0}$ (Proposition 3.3.11).

Proof The uniqueness of the vector potential $w$, if it exists, follows from the second Weber inequality. Indeed, if $w_1$ and $w_2$ both fulfill all the conditions (3.11), then $\delta w:=w_1-w_2 \in X_T(\Omega), \operatorname{curl} \delta w=0$ and $\operatorname{div} \delta w=0$ in $\Omega$, with $\langle\delta w \cdot n, 1\rangle_{\Sigma_i}=0$, for all $i$. Hence, $\delta w=0$, so uniqueness follows.

## 物理代写|电磁学代写电磁代考|矢量电位的提取-消失的法向迹

$$\left.\begin{array}{l} \operatorname{div} \boldsymbol{v}=0 \mathrm{in} \Omega, \ \boldsymbol{v} \cdot \boldsymbol{n}{\mid \Gamma}=0, \ \langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle{\Sigma_i}=0, \forall i \end{array}\right} \Longleftrightarrow\left{\begin{array}{l} \exists \boldsymbol{w} \in \boldsymbol{H}0(\operatorname{curl}, \Omega), \ \operatorname{div} \boldsymbol{w}=0, \ \langle\boldsymbol{w} \cdot \boldsymbol{n}, 1\rangle{H^{1 / 2}\left(\Gamma_k\right)}=0, \forall k \end{array} \quad \boldsymbol{v}=\operatorname{curl} w\right.$$
$w$ 是唯一的，又是存在的 $C>0$ 独立于 $v$ 这样
$$|w|_{H(\operatorname{curl}, \Omega)} \leq C|v|_{L^2(\Omega)} .$$

\begin{aligned} \langle v \cdot \boldsymbol{n}, 1\rangle_{\Sigma_i} &=\sum_j\left\langle\boldsymbol{v} \cdot \boldsymbol{n},\left[\dot{r}i\right] \Sigma_j\right\rangle{\Sigma_j} \ &=\left(\operatorname{curl} w, \operatorname{grad} \dot{r}i\right){L^2(\dot{\Omega})}=\left(\operatorname{curl} w \mid \widehat{\operatorname{grad} \dot{r}_i}\right)=0 . \end{aligned}

one可以求解变分公式${ }^8$
$$\left{\begin{array}{l} \text { Find } w \in X_N^{\Gamma}(\Omega) \text { such that } \ \forall w^{\prime} \in X_N^{\Gamma}(\Omega),\left(\operatorname{curl} w \mid \operatorname{curl} w^{\prime}\right)+\left(\operatorname{div} w \mid \operatorname{div} w^{\prime}\right)=\left(v \mid \operatorname{curl} w^{\prime}\right) \end{array}\right.$$

## 物理代写|电磁学代写电磁学代考|矢量电位的提取-补体

$$\left.\begin{array}{l} \operatorname{div} \boldsymbol{v}=0 \mathrm{in} \Omega, \ \langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0, \forall k \end{array}\right} \Longleftrightarrow\left{\begin{array}{l} \exists \boldsymbol{w} \in \boldsymbol{H}0(\operatorname{div}, \Omega), \ \operatorname{div} \boldsymbol{w}=0, \ \langle\boldsymbol{w} \cdot \boldsymbol{n}, 1\rangle{\Sigma_i}=0, \forall i \end{array} \quad \boldsymbol{v}=\mathbf{c u r l} w\right.$$

$$|w|_{H(\operatorname{curl}, \Omega)} \leq C|v|_{L^2(\Omega)^{\circ}}$$

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