# 物理代写|电磁学代写electromagnetism代考|PHYS3040

## 物理代写|电磁学代写electromagnetism代考|Problems Without Constraints

First, we introduce the eigenvalue problem ${ }^{13}$ :
$$\left{\begin{array}{l} \text { Find }(\mathrm{e}, \lambda) \in(\mathcal{V} \backslash{0}) \times \mathbb{R} \text { such that } \ \forall \mathrm{v} \in \mathcal{V}, a(\mathrm{e}, \mathrm{v})=\lambda(\mathrm{e}, \mathrm{v}) \mathcal{H} . \end{array}\right.$$
According to Corollary $4.5 .12$, there exist a non-decreasing sequence of strictly positive eigenvalues $\left(\lambda_i\right){i \in \mathbb{N}}$ and a sequence of eigenfunctions $\left(\mathrm{e}_i\right){i \in \mathbb{N}}$ that are a Hilbert basis of $\mathcal{H}$ and such that $\left(\lambda_i^{-1 / 2} e_i\right){i \in \mathbb{N}}$ is a Hilbert basis for $\mathcal{V}$. This leads to the definition of a scale $\left(\mathcal{V}^s\right){s \in \mathbb{R}}$ of Hilbert spaces, the A-Sobolev spaces.
Definition 4.4.1 Let $s \in \mathbb{R}$. The space $\mathcal{V}^s$ is:

• if $s \geq 0$, the subspace of $\mathcal{H}$ characterised by the condition
$$\sum_{i \in \mathbb{N}} u_i \mathrm{e}i=\mathrm{u} \in \mathcal{V}^s \Longleftrightarrow|u|{\mathcal{V}^s}^2:=\sum_{i \in \mathbb{N}} \lambda_i^s\left|u_i\right|^2<+\infty,$$
which defines its canonical norm;
• if $s<0$, the dual of $\mathcal{V}^{-s}$ with respect to the pivot space $\mathcal{H}$.
Then, we summarise some properties of this scale. The proofs are left to the reader.
Proposition 4.4.2 The following statements hold true:
1. $\mathcal{V}^0=\mathcal{H}, \mathcal{V}^1=\mathcal{V}, \mathcal{V}^2=D(\mathrm{~A}), \mathcal{V}^{-1}=\mathcal{V}^{\prime}$, algebraically and topologically.
2. For all $i \in \mathbb{N}$ and $s \in \mathbb{R}, \mathrm{e}i \in \mathcal{V}^s$. Furthermore, the sequence $\left(\mathrm{e}_i^s\right){i \in \mathbb{N}}:=$ $\left(\lambda_i^{-s / 2} \mathrm{e}i\right){i \in \mathbb{N}}$ is a Hilbert basis for $\mathcal{V}^s$.
3. For all $t<s \in \mathbb{R}, \mathcal{V}^s$ is densely and compactly imbedded in $\mathcal{V}^t$.
4. Let $s \in \mathbb{R}$ and $u \in \mathcal{V}^s$. The scalar $u_i$ equivalently defined as
$$u_i=\left\langle\mathrm{u}, \mathrm{e}i^{-t}\right\rangle{\mathcal{V}^{-t}}=\lambda_i^{-t / 2}\left(\mathrm{u}, \mathrm{e}i^t\right){\mathcal{V}^t}$$
does not depend on $t \leq s$. Of course, if $\mathrm{u} \in \mathcal{H}, u_i$ coincides with the coordinate of $\mathrm{u}$ on the basis $\left(\mathrm{e}i\right){i \in \mathbb{N}}$.
5. As a consequence of items 2 and 4, an element of an A-Sobolev space admits $a$ renormalised expansion $\mathrm{u}=\sum_{i \in \mathbb{N}} u_i \mathrm{e}_i$, which converges in $\mathcal{V}^s$ under the condition (4.29).

## 物理代写|电磁学代写electromagnetism代考|Problems with Constraints

Now, we proceed to the framework of constrained problems. We thus consider a sesquilinear form $b$ on $\mathcal{V} \times \mathcal{Q}$, satisfying the inf-sup condition (4.10), its kernel $\mathcal{K}$ and $\mathcal{L}$ is the closure of $\mathcal{K}$ within $\mathcal{H}$. Furthermore, we assume the double orthogonality property of Definition 4.3.17. We begin by deducing two fundamental consequences of this property.

Lemma 4.4.5 Assume that the sesquilinear, continuous and Hermitian form a fulfills property (4.15) with $v=0$, and that the double orthogonality property holds between $\mathcal{V}$ and $\mathcal{H}$. Then, for any $\mathrm{v} \in \mathcal{V}^s$ with $s \geq 0$, its $\mathcal{H}$-orthogonal projections $\mathrm{v}{|} \in \mathcal{L}$ and $\mathrm{v}{\perp} \in \mathcal{L}^{\perp}$ belong to $\mathcal{V}^s$, with $\left|\mathrm{v}{|}\right|{\mathcal{V}^s}^2+\left|\mathrm{v}{\perp}\right|{\mathcal{V}^s}^2=| \mathrm{v}_{\mathcal{V}^s}^2$.

Proof Let e be a solution to (4.28). Taking a test function $\mathrm{v}{|} \in \mathcal{K}$ and using the double orthogonality, one obtains $a\left(\mathrm{e}{|}, \mathrm{v}{|}\right)=\lambda\left(\mathrm{e}{|}, \mathrm{v}{|}\right) \mathcal{H}$. Again invoking the double orthogonality, one arrives at: $a\left(\mathrm{e}{|}, \mathrm{v}\right)=\lambda(\mathrm{e} |, \mathrm{v}){\mathcal{H}}, \quad \forall \mathrm{v} \in \mathcal{V}, \quad$ and similarly, $\quad a\left(\mathrm{e}{\perp}, \mathrm{v}\right)=\lambda\left(\mathrm{e}{\perp}, \mathrm{v}\right){\mathcal{H}}$
In other words, the projections onto $\mathcal{K}$ and $\mathcal{K}^{\perp}$ of any eigenfunction are either an eigenfunction, or zero. Thus, the Hilbert basis $\left(\mathrm{e}i\right){i \in \mathbb{N}}$ can be chosen such that all its elements belong either to $\mathcal{K}$ or to $\mathcal{K}^{\perp}$. Let $I_{|}$(respectively $I_{\perp}$ ) be the set of indices $i$ such that $\mathrm{e}i \in \mathcal{K}$ (respectively $\mathrm{e}_i \in \mathcal{K}^{\perp}$ ). Then, we have: $$\forall \mathrm{v}=\sum{i \in \mathbb{N}} v_i \mathbf{e}i \in \mathcal{H}, \quad \mathrm{v}{|}=\sum_{i \in I_I} v_i \mathbf{e}i \quad \text { and } \quad \mathrm{v}{\perp}=\sum_{i \in I_{\perp}} v_i \mathbf{e}_i$$
The conclusion follows using the property (4.29).

## 物理代写|电磁学代写electromagnetism代考|Problems Without Constraints

$\$ \$$Find (\mathrm{e}, \lambda) \in(\mathcal{V} \backslash 0) \times \mathbb{R} such that \forall \mathrm{v} \in \mathcal{V}, a(\mathrm{e}, \mathrm{v})=\lambda(\mathrm{e}, \mathrm{v}) \mathcal{H} 址确的。 \ \$$

• 如果 $s \geq 0$, 的子空间 $\mathcal{H}$ 以条件为特征
$$\sum_{i \in \mathbb{N}} u_i \mathrm{e} i=\mathrm{u} \in \mathcal{V}^s \Longleftrightarrow|u| \mathcal{V}^{s 2}:=\sum_{i \in \mathbb{N}} \lambda_i^s\left|u_i\right|^2<+\infty,$$
它定义了它的规范规范;
• 如果 $s<0$ ，对偶 $\mathcal{V}^{-s}$ 关于枢轴空间 $\mathcal{H}$.
然后，我们总结了这个量表的一些性质。证明留给读者。
命题 4.4.2 以下陈述成立:
1. $\mathcal{V}^0=\mathcal{H}, \mathcal{V}^1=\mathcal{V}, \mathcal{V}^2=D(\mathrm{~A}), \mathcal{V}^{-1}=\mathcal{V}^{\prime}$ ，代数和拓扑。
2. 对所有人 $i \in \mathbb{N}$ 和 $s \in \mathbb{R}, \mathrm{e} i \in \mathcal{V}^s$. 此外，序列 $\left(\mathrm{e}_i^s\right) i \in \mathbb{N}:=\left(\lambda_i^{-s / 2} \mathrm{e} i\right) i \in \mathbb{N}$ 是苃尔伯特基 $\mathcal{V}^s$.
3. 对所有人 $t<s \in \mathbb{R}, \mathcal{V}^s$ 密密麻麻地嵌入 $\mathcal{V}^t$.
4. 让 $s \in \mathbb{R}$ 和 $u \in \mathcal{V}^s$. 标量 $u_i$ 等效定义为
$$u_i=\left\langle\mathrm{u}, \mathrm{e} i^{-t}\right\rangle \mathcal{V}^{-t}=\lambda_i^{-t / 2}\left(\mathrm{u}, \mathrm{e} i^t\right) \mathcal{V}^t$$
不依赖于 $t \leq s$. 当然，如果 $\mathrm{u} \in \mathcal{H}, u_i$ 与坐标相吻合 $\mathrm{u}$ 基本上来说 $(\mathrm{e} i) i \in \mathbb{N}$.
5. 作为第 2 项和第 4 项的结果， $\mathrm{A}-S \mathrm{Sobolev}$ 空间的一个元责承认 $a$ 重整化扩展 $\mathrm{u}=\sum_{i \in \mathbb{N}} u_i \mathrm{e}_i$ ，收玫于 $\mathcal{V}^s$

## 物理代写|电磁学代写electromagnetism代考|Problems with Constraints

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