TITLE: A "rearrangement" of a finite set
QUESTION [0 upvotes]: Let $T$ be a set of sets.
Let $\equiv$ be an equivalence relation on $\bigcup T$ defined by the formula $$a\equiv b \Leftrightarrow \forall X\in T:(a\in X\Leftrightarrow b\in X).$$
Let $S$ be a partition of $\bigcup T$ induced by the equivalence $\equiv$. In this question I've asked for a name of $S$. I was not told a name. So lets call $S$ the rearrangement of $T$.
Question: Can we prove that if $T$ is finite then $S$ is also finite?

REPLY [0 votes]: Having an equivalence class $C$, we can find the set $f \in 2^T$ of all $X \in
T$ such that $a \in X$ for all $a \in C$. $b \equiv a \Leftrightarrow \forall
X \in T : (a \in X \Leftrightarrow b \in X) \Leftrightarrow \forall X \in T :
(X \in f \Leftrightarrow b \in X)$. So $C = \left\{ b \in \bigcup T
\mid b \equiv a \right\}$ can be restored knowing $f$.
Consequently there are no more than $\operatorname{card} 2^T = 2^{\operatorname{card} T}$ classes.


TITLE: Simplifying Circuits with kirchoffs rules
QUESTION [0 upvotes]: enter image description here
I have spent such a long time looking at videos on simplifying circuits which I understand but even the complex ones that I am coming across seem more straight forward for finding equivalent resistances and simplifying the diagrams. Once I simplify this circuit I can calculate the currents at points A and B
If anyone could help with this it would be appreciated. Have attached an image as well.

REPLY [1 votes]: Because there are no places wehre two resistors are either purely in series or in parallel, "simplifying" this circuit is a dead end -- it is already as simple as any equivalent circuit can be.
Fortunately, calculating the currents and voltages is not too hard.  There are only three regions that could have different vlotages:  The one containing point A,
the one containing point B, and the one containing the lower left hand corner of your diagram, which I will call point C.  Since the zero-point of voltage is arbitrary, let's say that point C is at $0$ volts.
Then the 6 volt battery enforces
$$V_A = V_C + 6 = 6$$
and the 12 volt battery enforces
$$V_B = V_C + 12 = 12$$
So there will be a current, across the $20\Omega$ resistor, of $(12-6)/20 = .3$ amps going counterclockwise in your diagram.
And, since $V_B = V_C + 12$, there will be a current of $12/30 = .4$ amps going from B on the left across the  $20\Omega$ resistor and toward point C.
Finally, what is the current $I$ going between the positive end of the 12-volt battery and point B?  Well, it splits up into $.3$ amps going up to go through the  $20\Omega$ resistor, $.4$ amps going left to go through the  $30\Omega$ resistor, and some mount $I$ that will remain to go down toward the right side of the $60\Omega$ resistor. But $I$ is a current across a $60\Omega$ resistor that has a voltage drop of 12V; thus $I$ is  $.2$ amps.
Now we can determine all the other currents.  For example, the current in the short stretch going up from the juncture just to the left of the $60\Omega$ resistor, going up to the juncture just to the left of the 12 volt pattery, has to be $.2$ amps from the wirer coming from its right, plus the same $.3$ amps that went through the $60\Omega$ resistor and the 6 volt battery and comes in from the left, for a total of $.5$ amps.  This, plus $.4$ amps coming from above, tells you that the current going left to right through the 12 volt battery is .9 amps.
And just as a check, the current coming to $B$ from the left is $.9$ amps,. which ultimately gets divided into $.4$ through the $30\Omega$ resistor, $.3$ through the $20\Omega$ resistor, and the remaining $.2$ heading south to go through the through the $60\Omega$ resistor.


TITLE: What is the limit of this series?
QUESTION [2 upvotes]: assume  $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$
how compute $\lim_{n\to +\infty}x_n$?
Thanks for any hint

REPLY [4 votes]: Since the answer is straightforward if we exploit the Cesaro-Stolz theorem, here I want to present a solution that does not use this theorem.
Notice first that whenever $1<r<2$, we have
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ 0.$$
Indeed, the numerator is dominated by $n r^n$, thus we have
$$ 0 \leq \frac{\sum_{k＝1}^{n} \frac{r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{n(n+1)}{2} \left(\frac{r}{2}\right)^n \xrightarrow[]{n\to\infty} 0.$$
This shows that, the limit in question exists with value $\ell$ if and only if
$$ \lim_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} ＝ \ell.$$
But note that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} ＝ \sum_{k＝1}^{n} \int_r^2 x^{k-1} \; dx ＝ \int_r^2 \frac{x^{n} - 1}{x-1} \; dx. $$
This in particular shows that
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \leq \int_r^2 \frac{x^{n}-1}{r-1} \; dx \leq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{(r-1)(n+1)} $$
and
$$ \sum_{k＝1}^{n} \frac{2^k - r^k}{k} \geq \int_r^2 (x^{n}-1) \; dx \geq \frac{2^{n+1}-r^{n+1}-(n+1)(2-r)}{n+1} $$
These equalities show that
$$ 1\leq \liminf_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \limsup_{n\to\infty} \frac{\sum_{k＝1}^{n} \frac{2^k - r^k}{k}}{\frac{2^{n+1}}{n+1}} \leq \frac{1}{r-1}. $$
Now since $r$ was arbitrary, taking $r\to2^{-}$ yields the limit $\ell ＝ 1$.


TITLE: Solving $\frac{e^2M}{4\pi \epsilon_or^2}-\frac{mB}{r^{m+1}}=0$
QUESTION [0 upvotes]: From Principles of Electronic Materials and Devices by S.O. Kasap, Fourth Edition,  example 1.3,
$\frac{e^2M}{4\pi \epsilon_or^2}-\frac{mB}{r^{m+1}}=0$
is solved by
$r=\left[\frac{4\pi \epsilon_oBm}{e^2M}\right]^{1/(m-1)}$
I fail to see how this is true, even by substitution I am not convinced that this holds.
I need help approaching this; how to derive the formula for $r$.

REPLY [1 votes]: Supposing $r\neq 0$, just multiply your equation by $r^{m+1}$ and get
$$
\cfrac{r^{m+1}}{r^2}\cfrac{e^2M}{4\pi\epsilon_0}=mB
$$
Now $r^{m+1}/r^2=r^{m-1}$ and
$$
r^{m-1}=\cfrac{4\pi\epsilon_0mB}{e^2M}
$$
So it is sufficient to take the $(m-1)$-root which is equivalent to use the exponent $1/(m-1)$, i.e.:
$$
r=\left(\cfrac{4\pi\epsilon_0mB}{e^2M}\right)^{1/(m-1)}
$$


TITLE: connected topological manifold
QUESTION [0 upvotes]: There is a conclusion :Suppose $M$ is a connected topological manifold,then $M$ is path-connected.
We can define $S_p=\{q\in M:$there is a path from $p$ to $q\}$.Obviously,$S_p$ is nonempty since $p\in S_p$,If we prove that $S_p$ and $M\setminus S_p$ is open,then we can conclude that $M=S_p$ by connectness of $M$.
Suppose $q\in S_p$,then there is a coordinate chart $(U,\phi)$ around $q$ and $\phi(U)$ is open in $R^n$ for some $n$,how to prove $S_p$ is open by using the above fact?

REPLY [2 votes]: If $q \in S_p$ then, as you said, there is a coordinate chart $(U,\phi)$ such that $U \ni q$ and $\phi(U)$ is -without loss of generality- path connected (indeed you can consider an open ball $B$  -which is path connected- of $R^n$ containing $\phi(q)$ and contained in $\phi(U)$ and then restrict your chart to $\phi^{-1}(B)\subset U$).
Hence, if $q' \in U$, you can consider a path in $\phi(U)$ connecting $\phi(q)$ and $\phi(q')$. Then you consider the inverse image of this path through $\phi$ in order to get a path in $U$ connecting $q$ and $q'$ ($\phi$ is bijective). Hence $q'\in S_p$ because you already know that there is a path connecting $p$ and $q$. And this proves that $U \subset S_p$. As $U$ is open and $q \in U$, we are done.
More briefly, as $\phi(U)$ can be choosen to be path connected and as $\phi$ is an homeomorphism, $U$ can be choosen to be path connected.


TITLE: Prove or disprove that for any $a, b, n\in \mathbb{N}^0$ satisfying $a ≤b≤n$,$ C(n,b)C(b, a) = C(n, a)C(n-a, b-a)$
QUESTION [0 upvotes]: I am having trouble figuring out how to prove this that claim is true or false. To be very honest, combination problems are not my strongest forte at all. So, I am having difficulty trying to start this.
Any tips or suggestions on how to start this proof would be greatly appreciated!

REPLY [1 votes]: HINT: Imagine that you have $n$ white balls numbered $1$ through $n$. You want to paint $a$ of the red and $b-a$ of them blue. You can do this in (at least) two different ways.

You can choose $a$ balls and paint them red. Then you can choose $b-a$ of the remaining white balls and paint them blue.
You can choose $b$ balls to be colored. Then from those you can choose $a$ to paint red, after which you will paint the remaining $b-a$ chosen balls blue.

Each of these corresponds to one side of the identity that you want to prove (which is true). Can you see which is which, and explain why? That would prove the identity, since its two sides would just be two different legitimate ways of calculating the same thing.
Note: I like combinatorial proofs of combinatorial identities and think that they are very often more informative than more computational proofs, so I tend to emphasize them, but Integrand’s suggestion in the comments also works perfectly well.


text:
\begin{document}

\centerline{}

\selectlanguage{english}

\title{Electromagnetic shielding by thin periodic structures
and the Faraday cage effect
}

\selectlanguage{english}
\begin{center}
\Large{Electromagnetic shielding by thin periodic structures
and the Faraday cage effect.}
\end{center}
\begin{center}{\large{B\'erang\`ere Delourme$^{(1)}$, David P.\ Hewett$^{(2)}$}} \end{center}
\footnotesize{(1)  Laboratoire Analyse G\'eom\'etrie et Applications (LAGA), Universit\'e Paris 13,Villetaneuse, France, delourme@math.univ-paris13.fr }\\
\footnotesize {(2) Department of Mathematics, University College London, London, United Kingdom, d.hewett@ucl.ac.uk}
\normalsize

\medskip

\begin{abstract}

\selectlanguage{english}

\noindent In this note we consider the scattering of electromagnetic waves (governed by the time-harmonic Maxwell equations) by a thin periodic layer of perfectly conducting obstacles. The size of the obstacles and the distance between neighbouring obstacles are of the same small order of magnitude $\delta$.
By deriving homogenized interface conditions for three model configurations, namely
(i) discrete obstacles,
(ii) parallel wires,
(iii) a wire mesh,
we show that the limiting behaviour as $\delta\to0$ depends strongly on the topology of the periodic layer, with full shielding (the so-called ``Faraday cage effect'') occurring only in the case of a wire mesh.
\end{abstract}
\vskip 0.5\baselineskip

\selectlanguage{english}

\section{Introduction}\label{}
\noindent
The ability of wire meshes to block electromagnetic waves (the celebrated ``Faraday cage'' effect) is well known to physicists and engineers. Experimental investigations into the phenomenon date back over 180 years to the pioneering work of Faraday \cite{Faraday}, and the effect is routinely used to block or contain electromagnetic fields in countless practical applications. (An everyday example is the wire mesh in the door of a domestic microwave oven, which stops microwaves escaping, while letting shorter wavelength visible light pass through it.)

But, somewhat remarkably, a rigorous mathematical analysis of the effect does not appear to be available in the literature.

The mathematical richness of the Faraday cage effect was highlighted in an recent article by one of the authors \cite{ChapmanHewettTrefethen}, where
a number of different mathematical approaches were applied to the

2D electrostatic version of the problem.
In particular it was shown in \cite{ChapmanHewettTrefethen} how modern techniques of homogenization and matched asymptotic expansions could be used to derive effective interface conditions that accurately capture the shielding effect.
These results were generalised to the
2D electromagnetic case (TE- and TM polarizations) in \cite{HewettHewitt}, and related approximations for similar problems have also been studied recently by other authors, e.g.\ \cite{Holloway,MarigoMaurel}.
However, as far as we are aware, an analysis of the full 3D electromagnetic version of the problem with perfectly conducting scatterers (the focus of the current note) has not previously been performed. (We note that related approximations have been presented for thin layers of dielectric obstacles in \cite{DelourmeWellPosedMax,DelourmeHighOrderMax}.)

In this note we consider full 3D electromagnetic scattering by a thin periodic layer of small, perfectly conducting obstacles. We derive leading-order homogenized interface conditions for three model configurations, namely where the periodic layer comprises (i) discrete obstacles,
(ii) parallel wires, and (iii) a wire mesh.
Our results verify that

the effective behaviour depends strongly on the topology of the periodic layer, with shielding of arbitrarily polarized waves occurring only in the case of a wire mesh. We note that analogous observations have been made in the related setting of volume homogenization in \cite{SchweizerUrban}.

Our analysis assumes that the obstacles/wires making up the thin periodic layer are of approximately the same size/thickness as the separation between them. The case of very small obstacles/thin wires is expected to produce different interface conditions, analogous to those derived in \cite{ChapmanHewettTrefethen,HewettHewitt} in the 2D case. But we leave this case for future work.

\section{Statement of the problem}\label{}
\noindent
Our objective is to derive effective interface conditions for electromagnetic scattering by a thin periodic layer of equispaced perfectly-conducting obstacles on the interface $\Gamma = \{\mathbf{x} = (x_1, x_2, x_3) \in \R^3 : x_3 =0\}$.

Let  $\HatOmegaHole \in \R^3$  be the canonical  obstacle described by one of the following three cases (see Fig.~\ref{HatOmegaHolen}):
\begin{enumerate}
\item $\HatOmegaHole$ is a simply connected Lipschitz domain whose closure is contained in

$(0,1)^2 \times (-\frac{1}{2}, \frac{1}{2})$.
\item

$\HatOmegaHole = [0,1] \times (\frac{3}{8}, \frac{5}{8}) \times (-\frac{1}{8}, \frac{1}{8}) $, \ie~a wire (of square section) parallel to the direction $\be_1$.
\item

$\HatOmegaHole = \{ [0,1] \times (\frac{3}{8}, \frac{5}{8}) \times (-\frac{1}{8}, \frac{1}{8}) \} \cup \{ (\frac{3}{8}, \frac{5}{8}) \times [0,1]\times (-\frac{1}{8}, \frac{1}{8})\}$, \ie~a cross-shape domain made of the union of two perpendicular wires (one parallel to $\be_1$ and the other parallel to $\be_2$).
\end{enumerate}
\begin{figure}[htbp]
        \centering
        \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=0.8\textwidth, trim = 8cm 20cm 8cm 4.5cm, clip]{BPer3D_Casi.pdf}
                \caption{Case (i)}
                \end{center}
        \end{subfigure}
        \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=0.8\textwidth,trim = 8cm 20cm 8cm 4.5cm, clip]{BPer3D_Casii.pdf}
                \caption{Case (ii)}
                \end{center}
        \end{subfigure}
         \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=0.8\textwidth, trim = 8cm 20cm 8cm 4.5cm, clip]{BPer3D_Casiii.pdf}
                \caption{Case (iii)}
                \end{center}
        \end{subfigure}
              \caption{The canonical obstacle $\HatOmegaHole$ in the three cases under consideration.}\label{HatOmegaHolen}
 \end{figure}
\noindent We construct the thin layer as a union of scaled and shifted versions of the canonical obstacle $\HatOmegaHole$.
For $\delta>0$ we define $\mathscr{L}^{\delta}\subset \R^2\times[-\delta/2,\delta/2]$ by
\[
\mathscr{L}^{\delta} =
{\rm int}\left( \bigcup_{(i,j) \in \Z^2}

\delta\left\lbrace \overline{\HatOmegaHole} + i \mathbf{e}_1 + j \mathbf{e}_2  \right\rbrace\right).

\]

Our domain of interest is then $\Omega^\delta =
\R^3
\setminus  \overline{\mathscr{L}^{\delta}}$ (cf. Fig~\ref{OmegaDelta}), and we define $\Gamma^\delta = \partial \Omega^\delta$.

\begin{figure}[htbp]
        \centering
        \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=1\textwidth, trim = 3.5cm 10.5cm 2.5cm 5cm, clip]{BInfini_Casi.pdf}
                \caption{Case (i) - discrete obstacles}
                \end{center}
        \end{subfigure}
        \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=1\textwidth,trim = 3.5cm 10.5cm 2.5cm 5cm, clip]{BInfini_Casii.pdf}
                \caption{Case (ii) - parallel wires}
                \end{center}
        \end{subfigure}
         \begin{subfigure}[b]{0.3\textwidth}
        \begin{center}
                \includegraphics[width=1\textwidth, trim = 3.5cm 10.5cm 2.5cm 5cm, clip]{BInfini_Casiii.pdf}
                \caption{Case (iii) - wire mesh}
                \end{center}
        \end{subfigure}
              \caption{The domain $\Omega^\delta$ in the three cases under consideration.}\label{OmegaDelta}
 \end{figure}

\noindent On the domain $\Omega^\delta$ we consider the solution $\bu^\delta$ of the Maxwell equations
\begin{equation}\label{MaxwellEquation}
\rot \rot \bu^\delta - \omega^2 \varepsilon \bu^\delta = \mathbf{f} \quad  \mbox{in }  \Omega^\delta,
\end{equation}
where $\omega>0$ and $\varepsilon\in \mathbb{C}$,
subject to the perfectly conducting boundary condition
\begin{equation}\label{BoundaryCondition}
\bu^\delta \times \bn = 0 \quad \mbox{on} \quad \Gamma^\delta.
\end{equation}

For analytical convenience we avoid any complications arising from far-field behaviour by assuming that

${\rm Re}[\varepsilon] > 0$ and ${\rm Im}[\varepsilon] > 0$.
The assumption that ${\rm Im}[\varepsilon] > 0$

could be relaxed to ${\rm Im}[\varepsilon] \geq 0$ at the expense of some technical modifications, including the imposition of an appropriate radiation condition.
We also assume that the support of $\mathbf{f}$ does not intersect the interface $\Gamma$.
Then, given $\mathbf{f} \in \left(L^2(\Omega^\delta)\right)^3$, the Lax-Milgram Lemma ensures that Problem~(\ref{MaxwellEquation})-(\ref{BoundaryCondition}) has a unique solution $\bu^\delta$ in the standard function space

\begin{equation}\label{DefinitionHrot}
H(\rot ;  \Omega^\delta) = \left\{\bv \in (L^2(\Omega^\delta))^3: \rot \bv \in (L^2(\Omega^\delta))^3 \right\},
\end{equation}
equipped with the usual graph norm
$
\| \bv  \|_{H(\rot ;  \Omega^\delta)} = \big( \| \bv  \|_{\left(L^2(\Omega^\delta)\right)^3}^2 + \| \rot \bv \|_{\left(L^2(\Omega^\delta)\right)^3}^2 \big)^{1/2}$.
Moreover, one can prove that there exists $C>0$, independent of $\delta$, such that
\begin{equation}\label{Stability}
\| \bu^\delta \|_{H(\rot ;  \Omega^\delta)} \leq C \| \mathbf{f}  \|_{\left(L^2(\Omega^\delta)\right)^3}, \qquad \text{for all }0<\delta<1.
\end{equation}
The objective of this work is to identify formally the limit $\bu_0$ of $\bu^\delta$ as $\delta$ tends to $0$.  This limit solution is defined in the

union of two distinct domains $\Omega^\pm = \{\bx\in\R^3 :\pm x_3>0\}$, whose common interface is $\Gamma$. Our main result is the following:

\begin{theorem}\label{Prop1} The limit solution  $\bu^0$ satisfies the Maxwell equations
\begin{equation}\label{EquationVolumeu0}
\rot \rot \bu^0 - \omega^2 \varepsilon \bu^0 = \mathbf{f} \quad \mbox{in }  \Omega^+ \cup \Omega^-,
\end{equation}
together with the following interface conditions on $\Gamma$:
\begin{enumerate}
\item[] Case (i): $[ \bu_0 \times \be_3 ]_{\Gamma} = \boldsymbol{0}$ and $[\rot \bu_0  \times \be_3]_{\Gamma}=0$.

\item[] Case (ii): $\bu_0 \cdot \be_1 = 0$ on $\Gamma$, $[ \bu_0 \cdot \be_2]_{\Gamma}=0$, and $[ (\rot \bu_0 \times \be_3)\cdot \be_2 ]_{\Gamma} = 0$.
\item[] Case (iii): $\bu_0 \times \be_3=\boldsymbol{0}$ on $\Gamma$.

\end{enumerate}
\end{theorem}
Let us make a few comments on this result. First, we emphasize that the nature of the limit problem depends strongly on the topology of the thin layer of obstacles $\mathscr{L}^{\delta}$. In case (iii), where $\mathscr{L}^{\delta}$ comprises a wire mesh, we observe the ``Faraday cage effect'', where the effective interface $\Gamma$ is a solid perfectly conducting sheet. Hence if the support of $\mathbf{f}$ lies in $ \Omega^+$ (above the layer $\mathscr{L}^{\delta}$), then $\bu_0=\mathbf{0}$ in $\Omega_-$.

In other words, despite the holes in its structure, the layer $\mathscr{L}^{\delta}$ shields the domain $\Omega^-$ from electromagnetic waves of all polarizations.
At the opposite extreme, in case (i), where $\mathscr{L}^{\delta}$ comprises discrete obstacles, the interface is transparent and there is no shielding effect. In the intermediate case (ii), where $\mathscr{L}^{\delta}$ comprises an array of parallel wires, one observes polarization-dependent shielding. Fields polarized parallel to the wire axis are shielded, whereas those polarized perpendicular to the wire axis are not.

Note that this case (ii) includes as a subcase the simpler two-dimensional situation studied in \cite{HewettHewitt,Holloway,MarigoMaurel} where the fields are invariant in the direction of the wire axis.

\noindent The remainder of this note is dedicated to the formal proof of Theorem~\ref{Prop1}. The proof is based on the construction of an asymptotic expansion of $\bu^\delta$ using the method of matched asymptotic expansions (cf.~\cite{MazyaNazarovPlam}). To simplify the computation, we work with the first order formulation of~\eqref{MaxwellEquation}, introducing the magnetic field $\bh^\delta =  \frac{1}{\ri \omega} \rot \bu^\delta$ (see e.g.\ \cite{Monk}) and obtaining
\begin{equation}\label{MaxwellOrdre1}
\begin{cases}
- \ri \omega \bh^\delta  + \rot \bu^\delta = 0  & \mbox{in }  \Omega^\delta,\\
 - \ri \omega \bu^\delta  - \rot \bh^\delta = -\frac{1}{\ri \omega} \mathbf{f}   & \mbox{in }  \Omega^\delta,
 \end{cases} \quad \bu^\delta  \times \bn = 0  \mbox{ and }  \bh^\delta\cdot \bn = 0\mbox{ on } \Gamma^\delta.
\end{equation}
Far from the periodic layer $\mathscr{L}^{\delta}$, we construct an expansion of $\bh^\delta$ and $\bu^\delta$ of the form
 \begin{equation}\label{FFExpansion}
 \bh^\delta =  \bh_0(\bx) + \delta \bh_1(\bx) + \cdots,  \quad   \bu^\delta =  \bu_0(\bx) + \delta \bu_1(\bx) + \cdots, \quad \bx = (x_1, x_2, x_3),
 \end{equation}
 and, in the vicinity of $\mathscr{L}^{\delta}$,
 \begin{equation}\label{NFExpansion}
  \bh^\delta =  \bH_0(x_1, x_2, \frac{\bx}{\delta}) + \delta \bH_1(x_1, x_2, \frac{\bx}{\delta}) + \cdots,  \quad   \bu^\delta =  \bU_0(x_1, x_2, \frac{\bx}{\delta})+ \delta \bU_1(x_1, x_2, \frac{\bx}{\delta}) + \cdots, \quad
 \end{equation}
 where, for $i \in \{0,1\}$, $\bH_i(x_1, x_2,y_1, y_2, y_3)$ and $\bU_i(x_1, x_2,y_1, y_2, y_3)$ are assumed to be $1$-periodic in both $y_1$ and $y_2$.  Near and far field expansions communicate through so-called matching conditions, which ensure that the far and near field expansions coincide in some intermediate areas. Since we are only interested in the leading order terms, it is sufficient to consider only the $O(1)$ matching conditions, namely
 \begin{equation}\label{MatchingConditionOrdre0}
 \lim_{x_3 \rightarrow 0^\pm} \bh_0 = \lim_{y_3 \rightarrow \pm \infty}\bH_0 \quad \mbox{and } \quad \lim_{x_3 \rightarrow 0^\pm} \bu_0 = \lim_{y_3 \rightarrow \pm \infty}\bU_0.
 \end{equation}
 Inserting~\eqref{FFExpansion} into~\eqref{MaxwellEquation} and separating the different powers of $\delta$ directly gives~\eqref{EquationVolumeu0}. To obtain the interface conditions, we have to study the problems satisfied by $\bU_0$ and $\bH_0$:
 \begin{equation}\label{NearFieldProblemOrder0}
\begin{cases}
\rot_y \bU_0  = 0  & \mbox{in } \mathscr{B}_\infty, \\
 \dive_y \bU_0 =  0 & \mbox{in } \mathscr{B}_\infty, \\
 \bU_0 \times \bn =  0 & \mbox{on } \partial \mathscr{B}_\infty, \\
 \end{cases} \quad  \begin{cases}
\rot_y \bH_0  = 0  & \mbox{in } \mathscr{B}_\infty, \\
 \dive_y \bH_0 =  0 & \mbox{in } \mathscr{B}_\infty, \\
 \bH_0 \cdot \bn =  0 & \mbox{on } \partial \mathscr{B}_\infty,\\
 \end{cases}
 \mathscr{B}_\infty =  \Omega^1 = \R^3\setminus \overline{\mathscr{L}^1} .

 \end{equation}
 \section{The spaces $K_N(\mathscr{B}_\infty)$ and $K_T(\mathscr{B}_\infty)$}
\noindent

Denoting by $\mathscr{B}$ the restriction of $\mathscr{B}_\infty$ to the strip $(0, 1)^2 \times (-\infty,\infty)$,

we introduce the

spaces
\begin{multline}
\mathscr{H}_{N}(\mathscr{B}_\infty) = \left\{
\bu \in H_{\rm loc}({\rot};\mathscr{B}_\infty) \cap H_{\rm loc}(\dive;\mathscr{B}_\infty):

\; \bu \mbox{ is }  1\mbox{-periodic in } y_1  \mbox{ and }  y_2,  \right.
\\  \frac{\bu_{|\mathscr{B}}}{\sqrt{1+(y_3)^2}} \in (L^2(\mathscr{B}))^3, \quad   \rot \bu_{|\mathscr{B}}  \in (L^2(\mathscr{B}))^3,  \quad \xdiv \bu_{|\mathscr{B}} \in L^2(\mathscr{B}),    \quad \bu \times \bn = 0 \; \mbox{on} \; \partial \mathscr{B}_\infty  \big\} ,

\end{multline}
\begin{multline}
\mathscr{H}_{T}(\mathscr{B}_\infty) = \left\{
\bh \in H_{\rm loc}({\rot};\mathscr{B}_\infty) \cap H_{\rm loc}(\dive;\mathscr{B}_\infty):

\;\bh \mbox{ is 1-periodic in } y_1  \mbox{ and } y_2 , \right.
\\  \frac{\bh_{|\mathscr{B}}}{\sqrt{1+(y_3)^2}} \in (L^2(\mathscr{B}))^3, \quad   \rot \bh_{|\mathscr{B}}  \in (L^2(\mathscr{B}))^3,  \quad \xdiv \bh_{|\mathscr{B}} \in L^2(\mathscr{B}),    \quad \bh \cdot \bn = 0 \; \mbox{on} \;
\partial \mathscr{B}_\infty

\big\},
\end{multline}

both of which include periodic vector fields in $H_{\rm loc}({\rot};\mathscr{B}_\infty) \cap H_{\rm loc}(\dive;\mathscr{B}_\infty) $ that tend to a constant vector as $|y_3|\to\infty$. Investigation of \eqref{NearFieldProblemOrder0} requires the characterization of the so-called normal and tangential cohomology spaces $K_N(\mathscr{B}_\infty)$ and $K_T(\mathscr{B}_\infty)$ defined by (see \cite{DaugeBernardiAmrouche})
\begin{equation}
 K_N(\mathscr{B}_\infty) = \left\{ \bu \in \mathscr{H}_{N}(\mathscr{B}_\infty) ,  \rot \bu = 0, \dive \bu =0 \right\},\; K_T(\mathscr{B}_\infty) = \left\{ \bh \in \mathscr{H}_{T}(\mathscr{B}_\infty) ,  \rot \bh = 0, \dive \bh =0 \right\}.
\end{equation}
This characterization involves the representation of elements of $K_N(\mathscr{B}_\infty)$ and $K_T(\mathscr{B}_\infty)$ as gradients of harmonic scalar potentials, constructed by solving certain variational problems in the space
\begin{equation}\label{W1}
 \mathcal{W}_{1}(\mathscr{B_\infty}) = \big\{ p \in H^1_{\rm loc}(\mathscr{B}_{\infty}): \;p \mbox{ is 1-periodic in } y_1 \mbox{ and } y_2, \frac{p_{|\mathscr{B}}} {\sqrt{1+(y_3)^2}} \in L^2(\mathscr{B}),   \nabla p_{|\mathscr{B}}  \in L^2(\mathscr{B})  \big\},
\end{equation}
and variants of it.
In each case the existence and uniqueness of the potential follows

from the Lax-Milgram Lemma.

While we do not reproduce the proofs here, we remark that the unbounded nature of the domain $\mathscr{B}$ requires us, when verifying coercivity of the requisite bilinear forms, to appeal to the inequality

\begin{align}
\label{}
\left\|\frac{p}{\sqrt{1+(y_3)^2}}\right\|_{L^2(\mathscr{B_+})} \leq 2 \| \nabla p\|_{L^2(\mathscr{B_+})}, \qquad \mathscr{B_+} = \mathscr{B}\cap\{y_3>0\},
\end{align}
valid if $p\in C^\infty(\overline{\mathscr{B}_+})$, $p/\sqrt{1+(y_3)^2} \in L^2(\mathscr{B_+})$, $\nabla p\in L^2(\mathscr{B_+})$ and $p=0$ in a neighbourhood of $\{y_3=0\}$,
which is an elementary consequence of the Hardy inequality~\cite[Lemma 2.5.7]{Nedelec}
\begin{align}
\label{}
\int_0^\infty t^{-2}|\varphi(t)|^2\,{\rm d} t \leq 4 \int_0^\infty |\varphi'(t)|^2\,{\rm d} t, \qquad \varphi\in C_0^\infty ((0,\infty)).
\end{align}

\subsection{Characterization of $K_N(\mathscr{B}_\infty)$}
\noindent To characterize $K_N(\mathscr{B}_\infty)$ we first define two functions $p^\pm_{3} \in  H^1_{\rm loc}(\mathscr{B}_\infty)$, $1$-periodic in $y_1$ and $y_2$,
such that
$$
\begin{cases}
- \Delta p^\pm_3 = 0 &   \mbox{in }\mathscr{B}_\infty,\\
p^\pm_3 = 0 & \mbox{on } \partial \mathscr{B}_\infty, \\
\end{cases}
\quad \lim_{y_3 \rightarrow \pm \infty} \nabla p^\pm_3  =  \be_3,   \quad
\lim_{y_3 \rightarrow \mp \infty} \nabla p^\pm_3  = 0.
$$
Then, in case (i) we introduce the functions $\tilde{p}_1\in \mathcal{W}_{1}(\mathscr{B_\infty})$ and $p_1 \in  H^1_{\rm loc}(\mathscr{B}_\infty)$,  such that

$$
\begin{cases}
- \Delta \tilde{p}_1 = 0 &   \mbox{in }\mathscr{B}_\infty,\\
\tilde{p}_1 =  - \mathscr{P} \mathscr{R} y_1 & \mbox{on }  \partial \mathscr{B}_\infty, \\
\end{cases} \quad \mbox{and} \quad p_1 = \tilde{p}_1 + y_1.
$$
 Here, for any function $u \in L^2_{\rm loc}(\mathscr{B}_\infty)$, $\mathscr{R} u$ denotes its restriction to $\mathscr{B}$, while for any function $u \in  L^2_{\rm loc}(\mathscr{B})$, $\mathscr{P}u$ denotes its periodic extension to  $\mathscr{B}_\infty$.
Similarly, in cases (i) and (ii) we introduce the functions $\tilde{p}_2\in \mathcal{W}_{1}(\mathscr{B_\infty})$ and $p_2 \in  H^1_{\rm loc}(\mathscr{B}_\infty)$,  such that
$$
\begin{cases}
- \Delta \tilde{p}_2 = 0 &   \mbox{in }\mathscr{B}_\infty,\\
\tilde{p}_2 =  - \mathscr{P} \mathscr{R} y_2 & \mbox{on }  \partial \mathscr{B}_\infty, \\
\end{cases} \quad \mbox{and} \quad p_2= \tilde{p}_2 + y_2.
$$
We emphasize that it is not possible to construct $\tilde{p}_1$ in cases (ii) and (iii), and it is not possible to construct $\tilde{p}_2$ in case (iii).  An adaptation of the proof of \cite[Proposition 3.18]{DaugeBernardiAmrouche} leads to the following result:
\begin{proposition}\label{PropositionKN}~\\[-3ex]
 \begin{enumerate}
 \item[]{Case (i)}:  $K_N$ is the space of dimension $4$  given by $K_N(\mathscr{B}_\infty)  = span \left\{   \nabla  {p}_1,  \nabla  {p}_2 ,\nabla {p}^-_3,  \nabla p^+_3  \right\}.$
 \item[]{Case (ii)}: $K_N$ is the space of dimension $3$ given by
$
 K_N(\mathscr{B}_\infty)=  span \left\{   \nabla{p}_2, \nabla {p}^-_3,  \nabla p^+_3 \right\}.
 $
 \item[] {Case (iii)}: $K_N$ is the space of dimension $2$ given by
$
  K_N(\mathscr{B}_\infty)=  span \left\{   \nabla {p}^-_3,  \nabla p^+_3 \right\}.$
\end{enumerate}
\end{proposition}
\begin{proof}[Sketch of the proof in case (ii)]   First, one can verify directly that the family $\left\{   \nabla{p}_2, \nabla {p}^-_3,  \nabla p^+_3\right\}$ is linearly independent (using the limit of $\nabla{p}_2$ and $\nabla {p}^\pm_3$  as $y_3$ tends to $\pm\infty$). Moreover, it is clear that $\nabla{p}_2$ and $\nabla {p}^\pm_3$ belong to $K_N(\mathscr{B}_\infty)$. Now, let $\bu \in  K_N(\mathscr{B}_\infty)$. Since $\mathscr{B}_\infty$ is connected, there exists $p \in H^1_{\rm loc}(\mathscr{B}_\infty)$, unique up to the addition of a constant, such that $\bu = \nabla p$. (This follows e.g.\ from applying \cite[Theorem 3.37]{Monk} on an increasing sequence of nested subsets of $\mathscr{B}_\infty$ after extension of $\mathbf{u}$ by zero inside $\R^3\setminus\overline{\mathscr{B}_\infty}$.) Moreover, $\nabla p$ is periodic and there exists a real sequence $(c_j)_{j\in \Z}$ such that
$$
-\Delta p =0 \;\mbox{in } \mathscr{B}_\infty, \quad p = c_j \; \mbox{on } \partial \mathscr{B}_{\infty,j}  =  \partial \mathscr{B}_\infty \cap \{ j < y_2 < (j+1) \}.
$$
Because $\nabla p$ is periodic and $\frac{\bu_{|\mathscr{B}}}{\sqrt{1+(y_3)^2}} \in (L^2(\mathscr{B}))^3$, there exists four constants $\alpha_1$, $\alpha_2$, $\alpha_3^\pm$ such that $$\tilde{p} = p - \alpha_1 y_1 - \alpha_2 y_2 - \sum_{\pm} \alpha_3^\pm p^\pm_3 \in  \mathcal{W}_1( \mathscr{B}_\infty).$$
Since
$\tilde{p}  = c_j  - \alpha_1 y_1 - \alpha_2 y_2$ on $\partial \mathscr{B}_{\infty,j}$, the periodicity of $\tilde{p}$ in $y_1$ implies that $\alpha_1 = 0$, while its periodicity in $y_2$ leads to $c_j = c_0 + \alpha_2 j$.  As a result,
$$
\tilde{p}  = c_0 - \alpha_2 (y_2 -j) \quad \; \mbox{on } \partial \mathscr{B}_{\infty,j}.
$$
Since $\tilde{p}$ is harmonic, we deduce that $\tilde{p} = c_0 + \alpha_2 \tilde{p}_2$, and hence that $p =c_0+ \alpha_2 p_2 + \sum_{\pm} \alpha_3^\pm p^\pm_3$, which completes the proof.
Cases (i) and (iii) follow similarly.

\end{proof}
\subsection{Characterization of $K_T(\mathscr{B}_\infty)$}
\noindent First, let us define $q_3 \in H^1_{\rm loc}(\mathscr{B}_{\infty})$ as the unique function such that
$$
\begin{cases}
- \Delta q_3 = 0 &   \mbox{in }\mathscr{B}_\infty,\\
\partial_{\bn} q_3= 0 & \mbox{on } \partial\mathscr{B}_\infty, \\
\end{cases}
\quad \lim_{y_3 \rightarrow \pm \infty} \nabla q_3  =  \be_3, \quad \lim_{y_3 \rightarrow + \infty} q_3 - y_3  = 0.
$$
Then for $i \in \{1, 2\}$ we introduce the functions $\tilde{q}_i \in \mathcal{W}_{1}(\mathscr{B_\infty})$ and $q_i \in  H^1_{\rm loc}(\mathscr{B}_\infty)$ such that
$$
\begin{cases}
- \Delta \tilde{q_i} = 0 &   \mbox{in }\mathscr{B}_\infty,\\
\partial_{\bn} \tilde{q_i}=  - \be_i \cdot \bn & \mbox{on } \partial\mathscr{B}_\infty,
\end{cases}

\quad \lim_{y_3 \rightarrow + \infty} \tilde{q_i}  = 0,  \quad \mbox{and} \quad
q_i = \tilde{q}_i + y_i.
$$
In case (ii) we introduce a set of `cuts' $\Sigma$ defined by
$$
\Sigma = \bigcup_{j \in \Z} \Sigma_j,  \quad \mbox{where} \quad \Sigma_j = \Sigma_0 + j \be_2, \quad
\Sigma_0  = (-\infty,\infty)\times(-\tfrac{3}{8},\tfrac{3}{8})\times\{0\}.

$$
Similarly, in case (iii) we introduce the cuts
$$
\Sigma = \bigcup_{(i,j) \in \Z^2} \Sigma_{ij},  \quad \mbox{where} \quad \Sigma_{ij} =   \Sigma_{0 0} + i \be_1 +  j \be_2,   \quad
\Sigma_{0 0}= (-\tfrac{3}{8},\tfrac{3}{8})^2\times\{0\}.

$$
In both cases, $\mathscr{B}_\infty \setminus \Sigma$ is then the union of the two simply connected domains $
\mathscr{B}_\infty^\pm = \left( \mathscr{B}_\infty \setminus \Sigma \right) \cap \{ \pm y_3 > 0\} $.
We denote by $\mathcal{W}_1(\mathscr{B}_\infty^\pm)$ the space defined by formula~\eqref{W1} replacing $\mathscr{B}_\infty$ with $\mathscr{B}_\infty^\pm$.
In case (ii) we let  $ \qtwoperpm= \left( ( \qtwoperpm)_+ ,( \qtwoperpm)_-\right) \in \mathcal{W}_1(\mathscr{B}_\infty^+) \times  \mathcal{W}_1(\mathscr{B}_\infty^-)$ be the unique solutions to \begin{equation}\label{p2plus}
\begin{cases}
- \Delta \qtwoperpm = 0  & \mbox{in } \mathscr{B}_\infty\setminus \Sigma  , \\
\partial_\bn  \qtwoperpm = -\be_2 \cdot \bn  & \mbox{on } \partial \mathscr{B}_\infty^\pm \cap \partial \mathscr{B}_\infty ,\\
\partial_\bn  \qtwoperpm = 0  & \mbox{on } \partial \mathscr{B}_\infty^\mp\cap \partial \mathscr{B}_\infty,
\end{cases}\quad
\begin{cases}
 [ \qtwoperpm ]_{\Sigma_j} =  \pm (j-y_2), \\
[ \partial_{y_3}   \qtwoperpm]_{\Sigma_j} = 0 ,
\end{cases}
\lim_{y_3 \rightarrow +\infty}   \qtwoperpm =0 ,
 \quad \quad \end{equation}
and we define $q^\pm_2 = \qtwoperpm + y_2 1_{ \mathscr{B}_\infty^\pm}$,
 $1_{ \mathscr{B}_\infty^\pm}$ being the indicator function of $\mathscr{B}_\infty^\pm$. In case (iii) the functions $q^\pm_2$ are defined similarly, except that we replace $\Sigma_{j}$ by $\Sigma_{ij}$ in the jump conditions.

In case (iii) we additionally introduce the functions $ \qoneperpm= \left( (  \qoneperpm)_+ ,(  \qoneperpm)_-\right) \in \mathcal{W}_1(\mathscr{B}_\infty^+) \times  \mathcal{W}_1(\mathscr{B}_\infty^-)$ as the unique solutions to
\begin{equation}\label{p2plus}
\begin{cases}
- \Delta \qoneperpm= 0  & \mbox{in } \mathscr{B}_\infty\setminus \Sigma  , \\
\partial_\bn  \qoneperpm = -\be_1 \cdot \bn  & \mbox{on } \partial \mathscr{B}_\infty^\pm\cap \partial \mathscr{B}_\infty,\\
\partial_\bn  \qoneperpm= 0  & \mbox{on } \partial \mathscr{B}_\infty^\mp\cap \partial \mathscr{B}_\infty,
\end{cases} \;\;
\begin{cases}
[ \qoneperpm ]_{\Sigma_{ij}} =  \pm (i-y_1), \\
[ \partial_{y_3}   \qoneperpm]_{\Sigma_{ij}} = 0, \\
\end{cases} \;
\lim_{y_3 \rightarrow +\infty}   \qoneperpm =0,
\end{equation}
and we define $q^\pm_1 =  \qoneperpm + y_1 1_{ \mathscr{B}_\infty^\pm}$. Then, adapting the proof of \cite[Proposition 3.14]{DaugeBernardiAmrouche} one obtains the following result:
 \begin{proposition}\label{PropositionKT}~\\[-3ex]
  \begin{enumerate}
 \item[]{Case (i)}:  $K_T$ is the space of dimension $3$ given by
 $
 K_T(\mathscr{B}_\infty)  = span \left\{\nabla  {q}_1,  \nabla  {q}_2 ,  \nabla {q}_{3}  \right\}.
 $
 \item[]{Case (ii)}: $K_T$ is the space of dimension $4$ given by
$
 K_T(\mathscr{B}_\infty)=  span \left\{ \nabla  {q}_1,  \nabla  {q}^+_2 ,  \nabla  {q}^-_2, \nabla {q}_{3}\right\}.
 $
 \item[] {Case (iii)}: $K_T$ is the space of dimension $5$ given by
$
  K_T(\mathscr{B}_\infty)=  span \left\{  \nabla  {q}^+_1 ,  \nabla  {q}^-_1, \nabla  {q}^+_2 ,  \nabla  {q}^-_2, \nabla {q}_{3}\right\}.
 $
\end{enumerate}
\end{proposition}
\begin{proof}[Sketch of the proof in case (ii)]
As in the proof of Proposition~\ref{PropositionKN}, it is not difficult to prove that the family $\{ \nabla q_1, \nabla q_2^+, \nabla q_2^-, \nabla q_3\}$ is linearly independent and that its elements belong to  $K_T(\mathscr{B}_\infty)$. Then,
let $\bh = (\bh_+, \bh_-) \in K_T(\mathscr{B}_\infty)$. Since $\mathscr{B}_\infty^\pm$ are simply connected, there exists $q = (q_+, q_-) \in H^1_{\rm loc}(\mathscr{B}_\infty^+)\times H^1_{\rm loc}(\mathscr{B}_\infty^-)$ and a real sequence $(c_j)_{j\in \Z}$ such that
$$
\bh_\pm = \nabla q_\pm,
\quad \Delta q = 0   \mbox{ in } \mathscr{B}_\infty\setminus \Sigma, \quad  \partial_\bn q = 0 \mbox{ on } \partial  \mathscr{B}_\infty,
\quad [ q ]_{\Sigma_j} = c_j, \quad [\partial_{y_3}q ]_{\Sigma_j} =0.
$$
Since $\nabla q$ is periodic and $\frac{\bh_{|\mathscr{B}}}{\sqrt{1+(y_3)^2}} \in (L^2(\mathscr{B}))^3$, there exist five constants  $\alpha_1^\pm$, $\alpha_2^\pm$ and $\alpha_3$ such that
$$
\tilde{q}_\pm= q_\pm- \alpha_1^\pm y_1 -  \alpha_2^\pm y_2  - \alpha_3 q_3 \in  \mathcal{W}_1( \mathscr{B}_\infty^\pm).
$$
Because $\tilde{q}=(\tilde{q}_+,\tilde{q}_-)$ satisfies $[ \tilde{q}]_{\Sigma_j} = c_j - [\alpha_1] y_1 - [\alpha_2] y_2$ and $\tilde{q}$ is periodic, we find that $[\alpha_1]=0$, and $
c_j = c_0 +  j [\alpha_2]$  for each $j \in \Z$. To conclude, it suffices to note that $\hat{q} = q - \alpha_1 q_1  - \sum_{\pm} \alpha_2^\pm q_{2, \pm} - \alpha_3 q_3$ is periodic and satisfies
$$
\begin{cases}
- \Delta \hat{q}= 0 & \mbox{in }  \mathscr{B}_\infty \setminus \Sigma, \\
\partial_\bn \hat{q} = 0 & \mbox{in }  \partial \mathscr{B}_\infty, \\
\end{cases} \quad
\begin{cases}
[ \hat{q}]_{\Sigma_j} = c_0,    \\
[ \partial_\bn  \hat{q}]_{\Sigma_j} =0, \\
\end{cases} \quad
\int_{\Sigma_j} \partial_{y_3}  \hat{q}  = 0,
$$
which proves that $\hat{q}$ is constant in each of $\mathscr{B}_\infty^\pm$.
\end{proof}
 \section{Formal proof of Theorem~\ref{Prop1}}
 \noindent We treat the three cases separately. In case (i), using  Propositions~\ref{PropositionKN}-\ref{PropositionKT}, we have
 $$
 \bU_0 = \sum_{i=1}^2 a_i(x_1, x_2) \nabla p_i + \sum_{\pm}  a_3^\pm \nabla p^\pm_3  \quad \mbox{and} \quad \bH_0 = \sum_{i=1}^3 b_i(x_1, x_2) \nabla q_i.
 $$
 The behaviour at infinity of the functions $p_i$ and $q_i$ and the matching conditions~\eqref{MatchingConditionOrdre0} then imply
 $$
 a_i = (\bu_0)_i^\pm(x_1, x_2,0) \quad b_i  = (\bh_0)_i^\pm(x_1, x_2,0)  \quad \forall  i \in \{ 1, 2 \},
 $$
and, consequently (by~\eqref{MaxwellOrdre1}), that

$[\bu_0\times \be_3]_\Gamma = 0$
and $[ \rot \bu_0 \times \be_3]_\Gamma=0$. In case (ii) we have
$$
\bU_0 = a_2(x_1, x_2) \nabla p_2 + \sum_{\pm}  a_3^\pm \nabla p^\pm_3  \quad \mbox{and} \quad \bH_0 = b_1(x_1, x_2) \nabla q_1  + \sum_{\pm} b_2^\pm(x_1, x_2) \nabla q_2^\pm +b_3(x_1, x_2) \nabla q_3,
$$
which, together with the matching conditions~\eqref{MatchingConditionOrdre0}, leads to
$
(\bu_0)_1^\pm(x_1,x_2) = 0$,  $[(\bu_0)_2]_\Gamma = 0$, $[(\bh_0)_1]_\Gamma =0$.
Finally, in case (iii) we have
$
\bU_0  = \sum_{\pm}  a_3^\pm \nabla p^\pm_3$,
which implies that $(\bu_0)_i^\pm(x_1, x_2) =0$ for $i =1 $ or $2$.

\smallskip
\begin{rem}
We point out that our formal proof can be made rigorous by justifying the asymptotic expansions~\eqref{FFExpansion}-\eqref{NFExpansion}. This can be done a posteriori by constructing an approximation of $\bu^\delta$ on $\Omega^\delta$  (based on the truncated series~\eqref{FFExpansion}-\eqref{NFExpansion}) and using the stability estimate~\eqref{Stability} (see \cite{MazyaNazarovPlam}).  However, this would require us to identify the terms of order $1$ in the expansions, which is beyond the scope of this note.
\end{rem}

config: arxiv


TITLE: Show that the Frattini subgroup is the set of nongenerators of $G$.
QUESTION [1 upvotes]: The Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$. (If there are none, then $\Phi(G)=G$.) We say that an element $g\in G$ is a nongenerator if whenever $\langle X\cup\{g\}\rangle=G$, we have $\langle X\rangle=G$ for subsets $X\subseteq G$. If $G$ is finite, show that $\Phi(G)$ is the set of nongenerators of $G$.
Proof: Let $N$ be the set of nongeneratos of $G$. We need to show that $\Phi(G)=N$.
(1) $\Phi(G)\subseteq N$
$\Phi(G)=\cap \mathcal{H}$ where $\mathcal{H}$ is maximal subgroup. We need to show that all $g\in \Phi(G)$ is a nongenerator of $G$. We will do this by contradiction.
i.e. We will show that there exists $g\in \Phi(G)$ such that $g$ is not generator.
i.e. We will show that there exists $X$, $\langle X\cup\{g\}\rangle=G$ and $\langle X\rangle\neq G$.
At this point all the definitions have me confused can some clear things up for me and help me finish the proof?

REPLY [0 votes]: Hint: Assume $X$ does not generate $G$, but $X \cup \{g\}$ does. Let $H$ be the subgroup generated by $X$, so that $g \not\in H$. In general, you can use Zorn's lemma to find a maximal subgroup $K \supseteq H$ such that $g \not\in K$. If $G$ is finite (as in your question), Zorn's lemma isn't needed: you can take $K \supseteq H$ to be subgroup of maximal order such that $g \not\in K$. Either way: is $K$ a maximal subgroup of $G$?


TITLE: Uniform convergence of the series $\sum\limits_n \frac{\log(1+n^2x^2)}{n^2}$ on $[0,1]$
QUESTION [1 upvotes]: Is the series $$\sum_1^\infty \frac{\log(1+n^2x^2)}{n^2}$$ uniformly convergent on $[0,1]$? I tried to use comparison test but unable to bound this by any convergent series. Any suggestion will be very helpful.

REPLY [5 votes]: Note that
$$
\log(1+n^2x^2)\leq \log(1+n^2)
$$
for $x\in [0,1]$ by monotonicity of the log. Use Weirstrass M-test to conclude.
edit: To see that the series
$$
\sum\frac{\log(1+n^2)}{n^2}
$$
converges note that
$$
\frac{\log(1+n^2)}{n^2}\sim \frac{\log n}{n^2}<\frac{n^\epsilon}{n^2}=\frac{1}{n^{2-\epsilon}}
$$
for any epsilon greater than zero and correspondingly large $n$ (a very useful fact about the logarithm). Probably going to want to select an $\epsilon<1$.


TITLE: Constrained optimization: problem to establish an inequality
QUESTION [0 upvotes]: Consider the vector z. The problem is:
$\max {z^T x}$, subject to $\ x^T P x \leq 1$, where $P$ is a positive definite symmetric matrix.
Show that the optimal value is $\sqrt{z^TP^{-1}z}$. Once this is proof use this to establish the inequaity:
$\ (x^Tz)^2 \leq (x^TPx)(z^TP^{-1}z)$.
I am having problems to proof the inequality.

REPLY [1 votes]: Define $|||x|||=\sqrt{x^TPx}$. The optimization constraint is then $|||x|||\le 1$. The vector
$$
\hat x=\frac{x}{|||x|||}
$$
satisfies the constraint ($|||\hat x|||=1$), hence,
$$
z^T\hat x\le\max z^Tx=\sqrt{z^TP^{-1}z}.
$$
We are done after some rearranging the inequality.


TITLE: Show spectrum for operator $S$ and $T$ coincide given that there is some $R$ s.t $T=R^{-1} S R$
QUESTION [0 upvotes]: Im really stuck on the following excersice in functional analysis.
Show spectrum for operator $S$ on some Banach space $Y$ and $T$ on $X$ coincide given that there is some inverable $R$ s.t $T=R^{-1} S R$.
Im thinking bounded inverse theorem in some way might help but without any luck.
Could I get some hits please?

REPLY [2 votes]: Hint: $\lambda I - T=R^{-1}(\lambda I-S)R$


TITLE: How to build an explicit polyhedral representation of $P_n$
QUESTION [0 upvotes]: I am having difficulty with the following question.
Let $P_n = \{x\in\mathbb{R}^n \; : \; |x_i| \leq 1, i\leq n, \sum_i |x_i| \leq 2 \}$.
Build an explicit polyhedral representation of $P_n$ that is linear. That is, a representation of form
$P_n = \{x\in\mathbb{R}^n : \exists u\in\mathbb{R}^m \text{ s.t. } Ax + Bu \leq c \}$, where $dim(u)$ and $dim(c)$ are linear in $n$ (also $A, B, c, m$ can depend on $n$).
I have a lot of difficulty with questions asking to building explicit polyhedral representations in general. I think I would see how to do these types of questions in general if I saw how specifically to build a polyhedral representation in this case.

REPLY [1 votes]: In other words, you want to linearize the absolute values.  Introduce variable $u_i$ to represent $|x_i|$.  The desired linear constraints in $x$ and $u$ are
\begin{align}
 x_i - u_i &\le 0 \tag1 \\
-x_i - u_i &\le 0 \tag2 \\
u_i &\le 1 \tag3 \\
\sum_i u_i &\le 2 \tag4
\end{align}
Constraints $(1)$ and $(2)$ enforce $|x_i| \le u_i$.  Constraint $(3)$ enforces $|x_i| \le 1$.  Constraint $(4)$ enforces $\sum_i |x_i| \le 2$.