# 物理代写|电磁学代写electromagnetism代考|PHYC20014

## 物理代写|电磁学代写electromagnetism代考|Extraction of Scalar Potentials and Consequences

We consider first the case of curl-free fields of $L^2(\Omega)$. Let us begin with the fundamental result proven ${ }^4$ in [117, Chapter I] and in [164, Chapter 3].

Theorem 3.3.1 Let $\Omega$ be either a domain, or a bounded, open and connected set with a pseudo-Lipschitz boundary. Assume that $\Omega$ is topologically trivial. Then, given $v \in \boldsymbol{L}^2(\Omega)$, it holds that
$$\operatorname{curl} v=0 \text { in } \Omega \Longleftrightarrow \exists p \in H^1(\Omega), v=\operatorname{grad} p .$$
The scalar potential $p$ is unique up to a constant, and $|p|{H^1(\Omega)}=|v|{L^2(\Omega)^{\circ}}$.
Next, we have the more general result below, proven in [9] for smooth cuts. We provide the main steps of the proof, which is slightly different than the one proposed in [9], according to the assumptions we made on the regularity of the cuts.

Theorem 3.3.2 Let $\Omega$ be a domain such that $(\text { Top }){I>0}$ is fulfilled. Then, given $v \in$ $L^2(\Omega)$, it holds that $$\text { curl } v=0 \text { in } \Omega \Longleftrightarrow \exists \dot{p} \in P(\dot{\Omega}), v=\widehat{\operatorname{grad}} \dot{p}$$ The scalar potential $\dot{p}$ is unique up to a constant, and $|\dot{p}|{H^1(\dot{\alpha})}=|v|_{L^2(\Omega)^{\circ}}$
Proof We remark that, if there exists $\dot{p} \in P(\dot{\Omega})$ such that $v=\widehat{\operatorname{grad} \dot{p}}$, then obviously curl $v=0$ in $\dot{\Omega}$. One can then prove that curl $v=0$ in $\Omega$ by using the tangential gradient $\operatorname{grad}{\Gamma}$ of Proposition 3.1.20, which leads easily to $[\pi \uparrow v]{\Sigma_i}=\operatorname{grad}{\Gamma}\left([\dot{p}]{\Sigma_i}\right)=0$, for $1 \leq i \leq I$, plus Proposition $2.2 .32$ to conclude. Conversely, one first uses Theorem $3.3 .1$ in $\dot{\Omega}$ to find $\dot{p} \in H^1(\dot{\Omega})$ such that $v=$ grad $\dot{p}$ in $\dot{\Omega}$. Then, as $v$ belongs to $\boldsymbol{H}$ (curl, $\Omega$ ), it follows that $[\pi \top v]{\Sigma_i}=0$, for all $i$. Using again the tangential gradient as defined in Proposition 3.1.20, we find that $\operatorname{grad}{\Gamma}\left([\dot{p}]{\Sigma_i}\right)$ is zero, for all $i$. In other words, one has $[\dot{p}]{\Sigma_i}=c s t^i$, for $1 \leq i \leq I$, i.e., $\dot{p} \in P(\dot{\Omega})$.
Finally, we note that the uniqueness of $\dot{p}$ (up to a constant) follows from the fact that $\dot{\Omega}$ is connected.

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

We consider now the case of divergence-free fields of $L^2(\Omega)$, for which one can prove the fundamental result below.
Theorem 3.4.1 Let $\Omega$ be a domain. Then, given $v \in L^2(\Omega)$, it holds that
Furthermore, there exists $C>0$ such that for all $v$, one may choose a vector potential $w$ that fulfills
$$|w|_{H^1(\Omega)} \leq C|v|_{L^2(\Omega)^*}$$
Remark 3.4.2 Assuming that $v$ writes $v=\operatorname{curl} w$ with $w \in H(\operatorname{curl}, \Omega)$, let us briefly comment on the conditions $\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0$, for $0 \leq k \leq K$. For $k>0$, define $q_k \in H^1(\Omega)$ such that $q_k$ is a basis function of $Q_N(\Omega)$. Then, one obtains by integrating by parts twice:
$$\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=\left\langle\boldsymbol{v} \cdot \boldsymbol{n}, q_k\right\rangle_{H^{1 / 2}(\Gamma)}=\left(\boldsymbol{v} \mid \operatorname{grad} q_k\right)=\left(\operatorname{curl} w \mid \operatorname{grad} q_k\right)=0,$$
because $\operatorname{grad} q_k \in \boldsymbol{H}0($ curl, $\Omega$ ). On the other hand, for $k=0$, one has simply $$\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle{H^{1 / 2}\left(\Gamma_0\right)}=-\sum_{1 \leq k \leq K}\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0 .$$
Proof We use the notations of Sect. 3.2. The result is obtained in four steps:

1. Define ${ }^6\left(q_{\ell}\right){0 \leq \ell \leq K}$ by: $-q_0 \in H{z m v}^1\left(\Omega_0\right)$ s.t. $\Delta q_0=0$ in $\Omega_0, \partial_n q_0=v \cdot n$ on $\Gamma_0, \partial_n q_0=0$ on $\partial \mathcal{O}$

## 物理代写|电磁学代写电磁学代考|标量势和结果的提取

$$\operatorname{curl} v=0 \text { in } \Omega \Longleftrightarrow \exists p \in H^1(\Omega), v=\operatorname{grad} p .$$

## 物理代写|电磁学代写电磁学代考|向量势的提取

$C>0$ 对于所有人来说 $v$，我们可以选择一个向量势 $w$ 它满足
$$|w|_{H^1(\Omega)} \leq C|v|_{L^2(\Omega)^*}$$3.4.2假设 $v$ 写 $v=\operatorname{curl} w$ 用 $w \in H(\operatorname{curl}, \Omega)$，让我们简要地评论一下条件 $\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=0$，为 $0 \leq k \leq K$。对于 $k>0$，定义 $q_k \in H^1(\Omega)$ 如此这般 $q_k$ 基函数是 $Q_N(\Omega)$。然后通过两次分部积分得到:
$$\langle\boldsymbol{v} \cdot \boldsymbol{n}, 1\rangle_{H^{1 / 2}\left(\Gamma_k\right)}=\left\langle\boldsymbol{v} \cdot \boldsymbol{n}, q_k\right\rangle_{H^{1 / 2}(\Gamma)}=\left(\boldsymbol{v} \mid \operatorname{grad} q_k\right)=\left(\operatorname{curl} w \mid \operatorname{grad} q_k\right)=0,$$

1. 定义${ }^6\left(q_{\ell}\right){0 \leq \ell \leq K}$通过:$-q_0 \in H{z m v}^1\left(\Omega_0\right)$ s.t. $\Delta q_0=0$在$\Omega_0, \partial_n q_0=v \cdot n$上$\Gamma_0, \partial_n q_0=0$上$\partial \mathcal{O}$

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