text:
\begin{document}

\IEEEoverridecommandlockouts
\title{The Degrees-of-Freedom of Multi-way Device-to-Device Communications is Limited by 2}
\author{
\IEEEauthorblockN{Anas Chaaban$^{*}$, Henning Maier$^{\dagger}$, Aydin Sezgin$^{*}$}
\IEEEauthorblockA{$^{*}$Institute of Digital Communication Systems, Ruhr-Universit\"at Bochum (RUB), Germany\\
Email: {anas.chaaban,aydin.sezgin@rub.de}\\
$^{\dagger}$Lehrstuhl f\"ur Theoretische Informationstechnik, RWTH Aachen, Germany,\\
Email: {maier@ti.rwth-aachen.de}
}
\thanks{
The work of A. Chaaban and A. Sezgin is supported by the German Research Foundation, Deutsche Forschungsgemeinschaft (DFG), Germany, under grant SE 1697/5.

The work of H. Maier is supported by the DFG  under grant PACIA-Ma 1184/15-2 and by the UMIC Research Centre, RWTH Aachen University.
}
}

\maketitle

\begin{abstract}
A 3-user device-to-device (D2D) communications scenario is studied where each user wants to send and receive a message from each other user. This scenario resembles a 3-way communication channel. The capacity of this channel is unknown in general. In this paper, a sum-capacity upper bound that characterizes the degrees-of-freedom of the channel is derived by using genie-aided arguments. It is further shown that the derived upper bound is achievable within a gap of 2 bits, thus leading to an approximate sum-capacity characterization for the 3-way channel. As a by-product, interesting analogies between multi-way communications and multi-way relay communications are concluded.
\end{abstract}

\section{Introduction}
The increase in the demand on data-rates in communication networks accompanied by the spectrum shortage has motivated researchers to seek new methods to combat these challenges. Several solutions have been proposed to overcome these challenges. Among the most promising solutions proposed to alleviate the limitations of the existing communication networks is a communication mode known as Device-to-device (D2D) communication.

D2D communication is defined as direct communication between mobile nodes in close proximity without incorporating a base station in the process (see \cite{AsadiWangMancuso} for a survey about D2D communications). It was first introduced in \cite{LinHsu}. Realizing that D2D communication helps offloading some data traffic from the cellular network, standardization bodies \cite{3GPPTR22803} are considering it as a potential component of future communication systems. Due to this fact, the research focus on D2D communication has increased recently. For instance, the potential of interference alignment \cite{Jafar} for D2D networks has been studied in \cite{ElkotbyElsayedIsmail}. Furthermore, a new spectrum sharing mechanism for D2D communication \alert{that outperforms state-of-the-art spectrum sharing mechanisms} has been proposed in \cite{NaderializadehAvestimehr}. \alert{At each time, this mechanism (which can also be implemented in a distributed way) schedules users that can achieve a near-optimal performance by communicating simultaneously while treating interference as noise.}

A D2D communication scenario between two nodes can be modelled by a two-way channel (TWC), where each node can communicate with the other node simultaneously. The TWC was introduced by Shannon in \cite{Shannon_TWC} where capacity upper and lower bounds were derived, and the capacity of some classes of TWC was characterized. If more than two nodes want to establish D2D communications, the scenario can be modelled by a multi-way channel. However, the extension of the TWC to more than two users has not been studied so far. Hence, information-theoretic results on the fundamental limits of communications over a multi-way channel are not available to date. The emergence of D2D communications calls upon establishing these limits. This direction is pursued in this paper.

We study a three-way channel consisting of three full-duplex nodes that want to communicate pair-wise simultaneously with each other. We call the resulting network the 3-D2D channel. We consider a Gaussian channel, i.e., where all nodes are disturbed by a Gaussian noise. Furthermore, we consider reciprocal channels where the channel gain between two nodes is the same in both directions. For this network, we derive novel upper bounds on the achievable rates which characterize the degrees-of-freedom (DoF) of the channel. Interestingly, while the cut-set bounds characterize the DoF of the TWC \cite{Han}, they do not characterize the DoF of the 3-D2D network, which is in turn characterized by our new bound. Furthermore, we prove the achievability of this sum-capacity upper bound within a gap of 2 bits. It turns out that the simple opportunistic strategy of letting the two users sharing the strongest channel communicate while leaving the third user silent suffices to achieve the sum-capacity of the 3-D2D channel within a constant gap.

As a by-product of this characterization, we obtain the following conclusion. Contrary to some networks where the DoF changes by increasing the number of users (such as the interference channel \cite{CadambeJafar_KUserIC}), the DoF of the multi-way channel does not change if we increase the number of users from two to three. Namely, the sum-capacity of the 3-D2D channel scales (as $\mathsf{SNR}$ increases) as twice the capacity of the strongest channel between two users (2 DoF). This behaviour is the same as that in the TWC \cite{Han}.

Furthermore, we discover the following analogies between multi-way communication (as in the 3-D2D channel) and multi-way relaying. It was observed in \cite{ChaabanSezginAvestimehr_YC_SC} that the DoF of the two-way relay channel does not change if we increase the number of users. As mentioned above, the same observation holds for multi-way communications when going from the 2-user to the 3-user case. Furthermore, it was observed in \cite{ChaabanSezginAvestimehr_YC_SC} that independent of the number of users, the optimal DoF can be achieved by letting the two strongest users communicate while leaving the remaining users silent. The same holds for the 3-D2D channel where the optimal DoF can be achieved by letting the two users sharing the strongest channel communicate. These analogies are interesting, and motivate the search for further analogies between the two types of networks.

The paper is organized as follows. The system model of the 3-D2D channel is described in Section \ref{Model}. The main result of the paper is given in Section \ref{MainResults}. Upper bounds on the capacity of the channel are given in Section \ref{UpperBounds} and transmission strategies are given in Section \ref{Achievability}. Finally, we conclude the paper with Section \ref{Conclusion}. Throughout the paper, we use $x^{n}$ to denote the length-$n$ sequence $(x(1),\cdots,x(n))$ and \alert{we use capital letters to denote random variables.} The function $C(x)$ is used to denote $\frac{1}{2}\log(1+x)$ for $x\geq0$, .

\section{System Model}
\label{Model}
The 3-D2D channel is a multi-way channel consisting of three users communicating simultaneously with each other. The channel is fully connected as shown in Fig. \ref{DChannelModel} and all nodes are full-duplex. Each user in the 3-D2D channel has two independent messages, one for each remaining user. That is, user 1 has messages $m_{12}$ and $m_{13}$ intended to users 2 and 3, respectively. Similarly user 2 has $m_{21}$ and $m_{23}$, and user~3 has $m_{31}$ and $m_{32}$. The message $m_{jk}$ is chosen uniformly from a message set $\mathcal{M}_{jk}=\{1,\cdots,2^{nR_{jk}}\}$, where $R_{jk}$ is the rate of the message and $n$ is the code length.

\begin{figure}[t]
\centering
\includegraphics[width=\columnwidth]{D_Channel}
\caption{Multi-way communication over the reciprocal 3-D2D channel. Each user sends a message to and receives a message from the two other users.}
\label{DChannelModel}
\end{figure}

To send his messages, user 1 sends a transmit signal $x_1^{n}$ of length $n$ symbols, whose $i$th symbol $x_1(i)\in\mathbb{R}$ is constructed from the messages $m_{12}$ and $m_{13}$ and the received symbols at user 1 up to time instant $i$, i.e., $y_1^{i-1}$, using an encoding function $\mathcal{E}_{1,i}$. Users 2 and 3 construct their signals similarly. Thus, we may write for each user $j$
\begin{align}
\label{Encoding}
x_j(i)=\mathcal{E}_{j,i}(m_{jk},m_{j\ell},y_j^{i-1}),
\end{align}
for distinct $j,k,\ell\in\{1,2,3\}$. The received signals are given by
\begin{align}
y_1(i)=h_3x_2(i)+h_2x_3(i)+z_1(i),\\
y_2(i)=h_3x_1(i)+h_1x_3(i)+z_2(i),\\
y_3(i)=h_2x_1(i)+h_1x_2(i)+z_3(i),
\end{align}
where $h_3,h_2,h_1\in\mathbb{R}$ are the (globally known) real-valued (static) channel coefficients, and $z_1,z_2,z_3\in\mathbb{R}$ represent the independent noises at users 1, 2, and 3, respectively, which are Gaussian with unit variance and i.i.d. over time. Note that the channels are assumed to be reciprocal, i.e., the channel gain between two users is the same in both directions. We assume without loss of generality that
\begin{align}
\label{Ordering}
h_3^2\geq h_2^2\geq h_1^2,
\end{align}
i.e., users 1 and 2 share the strongest channel. Each user has a power constraint $P$, i.e.,
\begin{align}
\sum_{i=1}^{n}\mathbb{E}[X_j(i)^{2}]\leq nP,\quad j\in\{1,2,3\}.
\end{align}

After receiving $y_j^{n}$, user $j$ decodes his desired messages $\hat{m}_{kj}$ and $\hat{m}_{\ell j}$ ($\{j,k,\ell\}=\{1,2,3\}$) using a decoding function $\mathcal{D}_j$ and his own messages $m_{jk}$ and $m_{j\ell}$, i.e.,
\begin{align}
\label{Decoder}
(\hat{m}_{kj},\hat{m}_{\ell j})=\mathcal{D}_j(y_j^{n},m_{jk},m_{j\ell}).
\end{align}
An error occurs if $m_{jk}\neq \hat{m}_{jk}$ for some $j\neq k$. The collection of message sets, encoders, and decoders defines a code for the 3-D2D channel denoted $(n,\vec{R})$ where $$\vec{R}=(R_{12},R_{13},R_{21},R_{23},R_{31},R_{32}),$$
and induces an error probability $P_{e,n}$.

A rate tuple $\vec{R}$ is said to be achievable if there exist a sequence of $(n,\vec{R})$ codes such that the $P_{e,n}$ can be made arbitrarily small by increasing $n$. In this paper, we are interested in the sum-capacity $C_\Sigma$ of the 3-D2D channel defined as the maximum achievable sum-rate $R_\Sigma=\sum_j\sum_{k\neq j}R_{jk}$. The main result of the paper is given in the following section.

\section{Main Result}
\label{MainResults}
The main result regarding the sum-capacity of the considered 3-D2D channel is presented in the following theorem.
\begin{theorem}
\label{Thm:Main}
The sum-capacity of the 3-D2D channel is bounded by
\begin{align}
\label{ApproxCap}
2C(h_3^{2}P)\leq C_\Sigma \leq 2C(h_3^{2}P)+2.
\end{align}
\end{theorem}
This theorem provides an approximate characterization of the sum-capacity of the given network within a gap of 2 bits. The proof of the theorem is given in Sections \ref{UpperBounds} where a novel genie-aided upper bound is derived leading to the right-hand side of \eqref{ApproxCap}, and in Section \ref{Achievability} where the achievability of the upper bound within a gap of 2 bits is shown leading to the left-hand side of \eqref{ApproxCap}.

The following interesting conclusions can be drawn from this theorem. First, the sum-capacity of the 3-D2D channel has the same scaling behaviour as that of the two-way channel~\cite{Han}. That is, both the 2-user case and the 3-user case have 2 DoF, where the DoF is defined as
\begin{align}
\text{DoF}=\lim_{\mathsf{SNR}\to\infty}\frac{C_\Sigma(\mathsf{SNR})}{\frac{1}{2}\log(\mathsf{SNR})}.
\end{align}
Thus, in a multi-way channel, the DoF stays constant at 2 if we increase the number of users from 2 to 3 (contrary to some channels such as the interference channel \cite{CadambeJafar_KUserIC} where the DoF depends on the number of users). Furthermore, the sum-capacity can be approached within a constant gap by letting the two users sharing the strongest channel (users 1 and 2) communicate while keeping the remaining user silent. That is, by letting users 1 and 2 communicate while keeping user 3 silent, we can achieve a sum-rate which is within 2 bits (at most) of the sum capacity. Interestingly, the same observations were concluded for multi-way relay communications in \cite{ChaabanSezginAvestimehr_YC_SC} where it was concluded that extending the two-way relay channel \cite{WilsonNarayananPfisterSprintson, AvestimehrSezginTse} to a three-way relay channel (Y-channel \cite{LeeLimChun,ChaabanSezgin_ISIT12_Y,ChaabanOchsSezgin}) preserves the same DoF, and that this scaling can be achieved by letting the two strongest users communicate while keeping the third user silent.

Next, we present new upper bounds for the 3-D2D channel which are necessary for proving Theorem \ref{Thm:Main}.

\section{Upper bounds}
\label{UpperBounds}
The cut-set bounds can be used to obtain an upper bound on the achievable rates in a multi-way channel. In fact, the cut-set bounds are tight for the 2-user case (the two-way channel) as shown in \cite{Han}. However, this is not the case for the 3-user case, i.e., the 3-D2D channel. For instance, in the 3-D2D channel, the rates of the messages from and to user 1 can be bounded by the cut-set bounds as
\begin{align}
R_{12}+R_{13}&\leq I(X_1;Y_2,Y_3|X_2,X_3)\\
R_{21}+R_{31}&\leq I(X_2,X_3;Y_1|X_1),
\end{align}
for some input distribution $p(x_1,x_2,x_3)$, where $X_j$ and $Y_j$ denote the input and output random variables. Similar bounds can be obtained for the other 2 users. By maximizing these bounds using the Gaussian input distribution, we can write
\begin{align}
R_{12}+R_{13}&\leq C(h_3^{2}P+h_2^{2}P)\\
R_{21}+R_{31}&\leq C(h_3^{2}P+h_2^{2}P).
\end{align}
Note that these bounds scale as $\frac{1}{2}\log(P)$ as $P$ grows. Similarly, we can bound the rates $R_{21}+R_{23}$,  $R_{12}+R_{32}$,  $R_{31}+R_{32}$, and $R_{13}+R_{23}$ with quantities that have the same scaling behaviour, i.e., they scale as $\frac{1}{2}\log(P)$. Note that this leads to a DoF of 3. It turns out however that the cut-set bounds are not tight for this network. In what follows, we provide bounds on the achievable rates that lead to a tighter sum-capacity upper bound.

\begin{lemma}
\label{Lemma:FirstBound}
An achievable rate for the 3-D2D channel must satisfy
\begin{align}
R_{21}+R_{31}+R_{32} &\leq C(h_3^2P+h_2^2P)+C\left(\frac{h_1^{2}}{h_2^{2}}\right).
\end{align}
\end{lemma}
\begin{proof}
The intuition for finding this bound is as follows. User 1 can decode $m_{21}$ and $m_{31}$ from $y_1^{n}$, $m_{12}$, and $m_{13}$ \eqref{Decoder}. We would like to enable user 1 to decode one more message (here $m_{32}$) by giving it the least possible side information. To guarantee this, we need to enable user 1 to construct $y_2^n$. Now note that if we give $m_{23}$ to user 1 as side information, then after decoding $m_{21}$, user 1 has all the information available to user 2 at time instant $i=1$, i.e., the message pair $(m_{21},m_{23})$. Thus, user 1 can generate the first symbol of $x_2^{n}$, i.e., $x_2(1)$ (cf. \eqref{Encoding}). Now by using $y_1(1)=h_3x_2(1)+h_2x_3(1)+z_1(1)$, user 1 can obtain $\tilde{y}_1(1)=h_2x_3(1)+z_1(1)$. After multiplying $\tilde{y}_1(1)$ by $\frac{h_1}{h_2}$, and  adding $h_3x_1(1)$, user 1 obtains
\begin{align}
\tilde{y}_2(1)=h_3x_1(1)+h_1x_3(1)+\frac{h_1}{h_2}z_1(1).
\end{align}
This is a less noisy version of $y_2(1)$. Note that in order to repeat this procedure for the following time instances ($i>1$), it is not enough to have $\tilde{y}_2(1)$. Rather, we need $y_2(1)$ exactly in order to produce $x_2(2)$ (which is generated from $m_{21}$, $m_{23}$, and $y_2(1)$ as in \eqref{Encoding}) which is essential for obtaining $y_2(2)$. In order to obtain $y_2(1)$ at user 1, we give him the signal $\tilde{z}_2^n=z_2^n-\frac{h_1}{h_2}z_1^n$. Now by adding $\tilde{y}_2(1)$ to $\tilde{z}_2(1)$, user 1 gets $y_2(1)$ and can generate $x_2(2)$. By repeating this procedure, user 1 can generate all instances of $y_2^n$. Then using the decoded $m_{21}$, the provided side information $m_{23}$, and the generated $y_2^{n}$, user 1 can decode $m_{32}$ just as user 2 can decode it. Therefore, we can write
\begin{align*}
n(R_{21}+R_{31}+R_{32}-\varepsilon_n)\leq I(\hat{\vec{M}}_{1},M_{32};Y_1^n,\tilde{Z}_2^n, \vec{M}_1,M_{23})
\end{align*}
by using Fano's inequality, where $\varepsilon_n\to0$ as $n\to\infty$, and where we used $\vec{M}_{1}$ and $\hat{\vec{M}}_{1}$ to denote the random vectors $(M_{12},M_{13})$ and $(M_{21},M_{31})$ indicating the message pairs sent and received at user 1, respectively. This bound can be written as
\begin{align}
& n(R_{21}+R_{31}+R_{32}-\varepsilon_n)\nonumber\\
& \stackrel{(a)}{\leq} I(\hat{\vec{M}}_{1},M_{32};Y_1^n,\tilde{Z}_2^n| \vec{M}_{1},M_{23})\\
& \stackrel{(b)}{=} \sum_{i=1}^{n} I(\hat{\vec{M}}_{1},M_{32};Y_1(i),\tilde{Z}_2(i)|\vec{M}_{1},M_{23},Y_1^{i-1},\tilde{Z}_2^{i-1})\nonumber\\
& \stackrel{(b)}{=} \sum_{i=1}^{n} I(\hat{\vec{M}}_{1},M_{32};Y_1(i)|\vec{M}_{1},M_{23},Y_1^{i-1},\tilde{Z}_2^{i-1})\nonumber\\
\label{TIE}
&\quad +\sum_{i=1}^{n} I(\hat{\vec{M}}_{1},M_{32};\tilde{Z}_2(i)| \vec{M}_{1},M_{23},Y_1^{i},\tilde{Z}_2^{i-1}),
\end{align}
where $(a)$ follows from the independence of the messages, and $(b)$ follows by using the chain rule. The first mutual information expression in \eqref{TIE} can be bounded as
\begin{align}
&I(\hat{\vec{M}}_{1},M_{32};Y_1(i)|\vec{M}_{1},M_{23},Y_1^{i-1},\tilde{Z}_2^{i-1})\nonumber\\
&\stackrel{(c)}{=}
h(Y_1(i)|\vec{M}_{1},M_{23},Y_1^{i-1},\tilde{Z}_2^{i-1})-h(Y_1(i)|\vec{M},Y_1^{i-1},\tilde{Z}_2^{i-1})\nonumber\\
&\stackrel{(d)}{\leq} h(Y_1(i))-h(Y_1(i)|\vec{M},Y_1^{i-1},\tilde{Z}_2^{i-1},Z_2^{i-1},Z_3^{i-1})\\
&\stackrel{(e)}{=} h(Y_1(i))-h(Z_1(i)|\vec{M},Z_1^{i-1},\tilde{Z}_2^{i-1},Z_2^{i-1},Z_3^{i-1})\\
&= h(Y_1(i))-h(Z_1(i)|\vec{M},Z_1^{i-1},Z_2^{i-1},Z_3^{i-1})\\
\label{IE1}
&\stackrel{(f)}{=} h(Y_1(i))-h(Z_1(i)),
\end{align}
where $(c)$ follows by using the definition of mutual information and by defining $\vec{M}=(M_{12},M_{13},M_{21},M_{23},M_{31},M_{32})$, $(d)$ follows
since conditioning does not increase entropy, $(e)$ follows since by knowing $\vec{M}$, $Z_2^{i-1}$, $Z_3^{i-1}$, and $Y_1^{i-1}$, all random variables required to determine $X_2^{i}$, and $X_3^{i}$ are given, and $(f)$ follows from the independence of the messages and the noise random variables. The second mutual information expression in \eqref{TIE} can be bounded as
\begin{align}
&I(\hat{\vec{M}}_{1},M_{32};\tilde{Z}_2(i)| \vec{M}_{1},M_{23},Y_1^{i},\tilde{Z}_2^{i-1})\nonumber\\
&= h(\tilde{Z}_2(i)|\vec{M}_{1},M_{23},Y_1^{i},\tilde{Z}_2^{i-1})-h(\tilde{Z}_2(i)| \vec{M},Y_1^{i},\tilde{Z}_2^{i-1})\nonumber\\
&\leq h(\tilde{Z}_2(i))-h(\tilde{Z}_2(i)| \vec{M},Y_1^{i},\tilde{Z}_2^{i-1},Z_2^{i-1},Z_3^{i-1})\\
&= h(\tilde{Z}_2(i))-h(\tilde{Z}_2(i)| \vec{M},Z_1^{i},\tilde{Z}_2^{i-1},Z_2^{i-1},Z_3^{i-1})\\
&= h(\tilde{Z}_2(i))-h(\tilde{Z}_2(i)| \vec{M},Z_1^{i},Z_2^{i-1},Z_3^{i-1})\\
&= h(\tilde{Z}_2(i))-h(Z_2(i)| \vec{M},Z_1^{i},Z_2^{i-1},Z_3^{i-1})\\
\label{IE2}
&= h(\tilde{Z}_2(i))-h(Z_2(i)),
\end{align}
which can be shown by using similar arguments as $(c)$-$(f)$ above. By substituting  \eqref{IE1} and \eqref{IE2} in \eqref{TIE} we obtain
\begin{align}
& n(R_{21}+R_{31}+R_{32}-\varepsilon_n)\nonumber\\
& \leq \sum_{i=1}^{n} h(Y_1(i))-h(Z_1(i))+h(\tilde{Z}_2(i))-h(Z_2(i))\\
& \leq \frac{n}{2}\log(1+h_3^{2}P+h_2^{2}P)+\frac{n}{2}\log\left(1+\frac{h_1^{2}}{h_2^{2}}\right),
\end{align}
which follows since the Gaussian distribution is a differential entropy maximizer. Now by dividing by $n$ and then letting $n\to\infty$, we obtain
\begin{align}
R_{21}+R_{31}+R_{32}&\leq C(h_3^{2}P+h_2^{2}P)+C\left(\frac{h_1^{2}}{h_2^{2}}\right),
\end{align}
which proves the statement of the lemma.
\end{proof}

Now, we have a bound on the sum $R_{21}+R_{31}+R_{32}$ which scales as $\frac{1}{2}\log(P)$ as $P$ grows. Next, we provide a bound on $R_{12}+R_{13}+R_{23}$ which complements the previous bound to a sum-capacity upper bound.

\begin{lemma}
\label{Lemma:SecondBound}
An achievable rate for the 3-D2D channel must satisfy
\begin{align}
R_{12}+R_{23}+R_{13} \leq C\left(h_3^2P+h_3^2\frac{h_1^2}{h_2^2}P\right)+\frac{1}{2}.
\end{align}
\end{lemma}
\begin{proof}
The derivation of the bound is similar to that in Lemma \ref{Lemma:FirstBound} with one difference, we start with enhancing user 3 by replacing the noise $z_3$ with $\frac{h_2}{h_3}z_3$ which is weaker than $z_3$ since $h_2^{2}\leq h_3^{2}$ (cf. \eqref{Ordering}). We denote the received signal of the enhanced receiver $y_3'$. Then, we give $m_{21}$ and $\tilde{z}_2^n=z_2^n-z_3^n$ to user 3 as side information. Now, after decoding $m_{23}$, user~3 has all the information available to user 2 at time instant $i=1$, i.e., the message pair $(m_{21},m_{23})$, and it can generate the first symbol of $x_2^{n}$, i.e., $x_2(1)$. By using $y_3'(1)=h_2x_1(1)+h_1x_2(1)+\frac{h_2}{h_3}z_3(1)$, user 3 can obtain $\tilde{y}_2(1)=h_2x_1(1)+\frac{h_2}{h_3}z_3(1)$. After multiplying $\tilde{y}_2(1)$ by $\frac{h_3}{h_2}$, and  adding $h_1x_3(1)$, user 3 obtains
\begin{align}
\tilde{y}_2(1)=h_3x_1(1)+h_1x_3(1)+z_3(1).
\end{align}
Now by adding $\tilde{y}_2(1)$ to $\tilde{z}_2(1)$, user 3 gets $y_2(1)$ and can generate $x_2(2)$. By repeating this procedure, user 3 can generate all instances of $y_2^n$, and then using $m_{21}$ and $m_{23}$, it can decode $m_{12}$ just as user 2 can decode it. Therefore, we can write
\begin{align*}
n(R_{12}+R_{23}+R_{13}-\varepsilon_n)\leq I(\vec{M}_{1},M_{23};Y_3'^n,\tilde{Z}_2^n, \hat{\vec{M}}_{1},M_{32})
\end{align*}
by using Fano's inequality, where $\varepsilon_n\to0$ as $n\to\infty$. Now by proceeding with similar steps as those in the proof of Lemma~\ref{Lemma:FirstBound} (i.e., $(c)$ to $(f)$), we can obtain
\begin{align}
R_{12}+R_{23}+R_{13} \leq C\left(h_3^2P+h_3^2\frac{h_1^2}{h_2^2}P\right)+\frac{1}{2},
\end{align}
which is the desired upper bound.
\end{proof}

Now we have the two components necessary for establishing our sum-capacity upper bound, given in the next theorem.
\begin{theorem}
\label{Thm:SCUB}
The sum-capacity of the 3-D2D channel satisfies
\begin{align}
\label{SCUB}
C_\Sigma\leq 2C(h_3^{2}P)+2.
\end{align}
\end{theorem}
\begin{proof}
To prove this theorem, we use the upper bound on the sum $R_{21}+R_{31}+R_{32}$ given in Lemma \ref{Lemma:FirstBound} which satisfies
\begin{align}
R_{21}+R_{31}+R_{32} &\leq C(h_3^2P+h_2^2P)+C\left(\frac{h_1^{2}}{h_2^{2}}\right)\\
&\quad\leq C(2h_3^2P)+C\left(1\right)\\
&\quad= C(2h_3^2P)+\frac{1}{2},
\end{align}
since $h_3^{2}\geq h_2^{2}\geq h_1^{2}$ \eqref{Ordering}, and the upper bound on the sum $R_{12}+R_{23}+R_{13}$ given in Lemma \ref{Lemma:SecondBound} which satisfies
\begin{align}
R_{12}+R_{23}+R_{13} &\leq C\left(h_3^2P+h_3^2\frac{h_1^2}{h_2^2}P\right)+\frac{1}{2}\\
&\leq C\left(2h_3^2P\right)+\frac{1}{2}.
\end{align}
By adding the two bounds, we get
\begin{align}
R_\Sigma &\leq 2C(2h_3^2P)+1<2C(h_3^2P)+2,
\end{align}
which proves that any achievable rate tuple must have a sum that satisfies $R_\Sigma \leq 2C(h_3^2P)+2$. Therefore, we obtain the desired sum-capacity upper bound.
\end{proof}
Clearly, this sum-capacity upper bound is tighter than that obtained from the cut-set bounds as $P$ increases. Namely, this bound behaves as $\log(P)$ as $P$ grows (2 DoF), in contrast to the cut-set bounds that behave as $\frac{3}{2}\log(P)$ (3 DoF). Next, we show that this upper bound is achievable within a constant gap.

\section{Transmission Strategies}
\label{Achievability}
The derived sum-capacity upper bound in Theorem \ref{Thm:SCUB} has the following desirable structure: it is equal to twice the capacity of the strongest channel ($h_3$) plus a constant. This directly indicates a near sum-rate optimal scheme for the 3-D2D channel. Namely, by allowing the two users sharing this strongest channel to communicate, we can achieve the upper bound within a constant gap.

Hence, let users 1 and 2 communicate via the channel $h_3$ while leaving user 3 silent. This reduces the 3-D2D channel to a two-way channel. As shown in \cite{Han}, the following rates are achievable in the resulting two-way channel
\begin{align}
R_{12}& \leq C(h_3^{2}P)\\
R_{21}& \leq C(h_3^{2}P).
\end{align}
By adding the two achievable rates, we conclude that the following sum-rate is achievable
\begin{align}
R_\Sigma& \leq 2C(h_3^{2}P).
\end{align}
Comparing this achievable sum-rate and the upper bound \eqref{SCUB} in Theorem \ref{Thm:SCUB}, we can see that this achievable sum-rate is within a gap of 2 bits of the sum-capacity upper bound. This leads to a sum-capacity characterization within a gap of 2 bits as follows
\begin{align}
2C(h_3^{2}P)\leq C_\Sigma \leq 2C(h_3^{2}P)+2.
\end{align}
This proves the main result of the paper given in Theorem \ref{Thm:Main}.

Although this transmission strategy suffices to show the achievability of the sum-capacity upper bound of the 3-D2D channel within a constant gap, we would like to additionally highlight the following interesting possibility. Consider a scenario where users 2 and 3 want to communicate with a rate that can not be supported by the channel $h_1$. Interestingly, if this rate can be supported by the channel $h_2$, then the two users can successfully communicate via user 1 as follows. Users 2 and 3 use nested-lattice codes \cite{NazerGastpar} to establish physical-layer network-coding \cite{WilsonNarayananPfisterSprintson} for bi-directional relaying via user 1. In other words, users 2 and 3 communicate via user 1 as in a two-way relay channel\footnote{Users 2 and 3 can also communicate via user 1 using quantize-forward as in \cite{AvestimehrSezginTse}}. This leads to the achievability of the rates
\begin{align}
R_{23},R_{32} \leq C\left(h_2^{2}P-\frac{1}{2}\right),
\end{align}
which is larger than the rates that can be achieved via the channel $h_1$ given by
\begin{align}
R_{23},R_{32} \leq C\left(h_1^{2}P\right),
\end{align}
as long as $h_2^{2}\geq h_1^{2}+\frac{1}{2P}$. This condition is guaranteed by \eqref{Ordering} at high $P$. While this strategy does not achieve the sum-capacity within a constant gap, it is useful for achieving high communication rates between users 2 and 3.

\section{Conclusion}
\label{Conclusion}
We studied the 3-user D2D channel (a three-way channel) and obtained its sum-capacity within a constant gap. While this required deriving a novel genie-aided upper bound, the achievability strategy is in fact simple; the sum-capacity is achievable within a constant gap by letting only two users communicate via the strongest channel. This insight is interesting since it shows that increasing the number of users in a multi-way communications channel \alert{from 2 to 3} does not increase the sum-capacity scaling behaviour of the channel. \alert{The authors believe that this conclusions extends to larger multi-way communications channels with more than 3 users.} Note that this is analogous to the multi-way relay channel where increasing the number of users also does not increase the capacity scaling behaviour. As an extension to this work, it would be interesting to find out if other analogies exist between multi-way channel and multi-way relay channels.

\bibliography{myBib}

\end{document}


TITLE: Entropy in ideal fluids
QUESTION [0 upvotes]: Can you derive a conservation law for entropy per unit mass? I know that there is a proof for this conservation.
My assumptions are:

$(\nabla p)/n = D \vec{v} + \vec{v} \cdot \nabla\vec{v}$ (Euler equation)
$\nabla \cdot (n\vec{v}) + D n = 0$ (continuity equation)
$p/n + u = Ts$ (chemical potential is zero)

Here I define:

$p$: pressure
$n$: mass density
$u$: internal energy per unit mass
$s$: entropy per unit mass
$D$: time derivative
$T$: temperature

REPLY [1 votes]: Yes, you can. You have given the equations in a typical set of primitive variables. If you transform the equations into another set of primitive variables, ie.
$$\vec{Q} = \lbrace p/(nc), u, v, w, s\rbrace$$
where $c$ is the speed of sound and $n$ is the density, you can end up with an conservation equation for entropy. If you take your equations and write them as a PDE of a vector $\vec{W} = \lbrace n, u, v, w, T\rbrace$ you can then transform from one set to another by computing the Jacobian matrix $\partial \vec{W}/\partial \vec{Q}$ and you will get your new conservation laws.


TITLE: Field momentum of Klein-Gordon Lagrangian
QUESTION [1 upvotes]: Given the Lagrangian $L$ of the field $\phi$ the field momentum $\Pi$ reads:
$$L_{KG}=-\frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2$$
$$\Pi=\frac{\partial L}{\partial(\partial_\mu\phi)}=\partial_\mu\phi$$
I dont see how the derivative above gives this result. How do we perform this derivative? The wrong way I thought is doing it like this:
$$\Pi=\frac{\partial L}{\partial(\partial_\mu\phi)}=-\frac{1}{2}\frac{\partial }{\partial\dot{\phi}}\dot{\phi}^2=-\dot{\phi}=-\partial_\mu\phi$$
PS: The Minkowski signature convention is $(-,+,+,+)$.

REPLY [1 votes]: $$L=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)-\frac{1}{2}\,m^2\,\phi^2$$
Where:
$\phi=\phi_i(t,x,y,z)$ and
$\partial_\mu\phi_i\equiv\frac{\partial\phi_i}{\partial x^\mu}$
$$\frac{\partial L}{\partial(\partial_\mu\phi)}=\partial^\mu\phi\quad\text{?}$$
you can "expand" the Lagrangian like this:
(signature $+,-,-,-$)
\begin{align*}
& {L}=\frac{1}{2}\left[\partial_0\phi\partial_0\phi-
\partial_1\phi\partial_1\phi-\partial_2\phi\partial_2\phi-
\partial_3\phi\partial_3\phi\right]-\frac{1}{2}\,m^2\phi^2\\\\
&\text{so}:\\\\
&\frac{\partial{L}}{\partial(\partial_0\phi)}=\partial_0\phi=\partial^0\phi\\
&\frac{\partial{L}}{\partial(\partial_1\phi)}=-\partial_1\phi=\partial^1\phi\\\\
&\text{and so on}\\\\
&\text{and}\\\\
&\frac{\partial{L}}{\partial\phi}=-m^2\,\phi\\\\
&\text{and hence the EL formula obtain:}\\\\
&\partial_\mu\partial^\mu\,\phi+m^2\phi=0\quad \surd
\end{align*}


TITLE: Show that the set of vectors are bases for $\mathbb{R}^3$
QUESTION [0 upvotes]: The set of vectors are:
S={ (1,6,4) , (2,4,1) , (-1,2,5) }

for it to be a base in $\mathbb{R}^3$ it must be linearly independent and span $\mathbb{R}^3$.
I first tried to solve it by taking its determinant and found that it's not zero, and therefore the set is linearly independent. Now how would I show that is spans? since none of them are scalar multiples of each other?

REPLY [0 votes]: Make a matrix $A$ by copying the given vectors as it columns.
For any given column vecor $b\in\mathbf{R}^3$, the linear $A(x\ y\ z)^T=b$ is solvable uniquely as you have verified the determinant is non-zero. (The solution vector is simply $A^{-1}b$.)
The solution values of $x,y,z$ provide the coefficients forming the linear combination of elements of $S$ giving (the arbitrary) $b$. So they span.


TITLE: Reflecting sunlight --- does this do any Work?
QUESTION [2 upvotes]: Photons have no mass, and yet via radiation pressure they do exert a force on whatever they reflect off of. Therefore the object must also exert an equal and opposite force on the photon.
So if Work is force times distance, this must do some kind of work. But I don't know how to calculate it because I don't know what distance to use. I think the work might be very high because going at the speed of light, then suddenly reversing direction, would be a delta v of 2c requiring extreme acceleration and thus extreme force (actually, extreme F/m, but photon mass is zero...).
(I have to throw this in): Also, why wouldn't this be a perpetual motion machine?

REPLY [1 votes]: Don't try and calculate it using forces, do it using energy and momentum instead. First conserve momentum:a number of photons strike a mirror on the earth and recoil. Technically they change their frequency by a tiny amount, but as far as conservation of momentum goes that is negligible. The mirror (plus the planet it is attached to) acquire an equal change in momentum in the opposite direction. From this you can calculate the change in velocity of the earth.
Then using that velocity you can calculate the change in KE of the earth (in the frame or was initially at rest), which by conservation of energy is going to give you the loss in energy of the photon. And obviously you can figure out the change in frequency from that.
It will be tiny. Really tiny. Which is why it is general ignored, because it almost never matters at all. But strictly speaking you are right that ignoring it means you are violating conservation of energy. Is just that doing it 'properly' makes no difference at all in most cases but is more work, do why not simplify the calculations and get the same result?


TITLE: The equation $x^3 + y^3 = z^3$ has no integer solutions - A short proof
QUESTION [37 upvotes]: Can someone provide the proof of the special case of Fermat's Last Theorem for $n=3$, i.e., that
$$
x^3 + y^3 = z^3,
$$
has no positive integer solutions, as briefly as possible?
I have seen some good proofs, but they are quite long (longer than a page) or use many variables. However, I would rather have an elementary long proof with many variables than a complex short proof.
Edit. Even if the bounty expires I will award one to someone if they have a satisfying answer.

REPLY [0 votes]: Eight years and one elementary number theory course later, I can give a brief but not as elementary sketch of an answer based on the ring of Eisenstein integers. They have a strong connection to cubes, considering the cubic root of unity $\omega$. Assume it is already established the Eisenstein integers $\mathbb Q(\omega)$ are a Euclidean domain, so UFD. This answer follows that in Dörrie's 100 great problems of elementary mathematics, but he spends some pages reproving results about Euclidean domains.
First off we can assume in $\alpha^3 + \beta^3 = \gamma^3$ that $\alpha, \beta, \gamma$ are coprime in the Eisenstein integers due to unique factorization. By trivially replacing $\gamma$ with $-\gamma$ the equation is $\alpha^3 + \beta^3 + \gamma^3 = 0$.
Now $i\sqrt 3 = \omega - \omega^2 = 2 \omega + 1$ is an Eisenstein prime by properties of norm. (Dörrie calls this $J-O$, where he defines $J = - \omega^2 = (1 + i \sqrt 3)/2, O = - \omega = (1-i \sqrt 3)/2$. He uses $\{J, O\}$ as a basis instead of $\{1,\omega\}$ or $\{\omega, \omega^2\}$ for some reason?)
By some basic computations and casework of $a,b$ mod 3 I'm too lazy to write up, if 3 divides $a + b$ then $i \sqrt 3$ divides $a + b \omega$, and more calculations show Lemma: if $i \sqrt 3$ doesn't divide $a + b\omega$, then $(a + b\omega)^3 \equiv \pm 1 \pmod 9$.
Now Lemma: $\alpha^3 + \beta^3 + \gamma^3 = 0$ implies $i \sqrt 3$ divides exactly one of $\alpha, \beta, \gamma$. This holds because suppose $i \sqrt 3$ didn't divide any of $\alpha, \beta, \gamma$. Then $\alpha^3 \equiv e, \beta^3 \equiv f, \gamma^3 \equiv g \pmod 9$ where $e^2 = f^2 = g^2 = 1$. But by casework mod 9, this is impossible. And we know $\alpha, \beta, \gamma$ are coprime so can't share a factor of $i \sqrt 3$.
Finally, $\alpha \beta \gamma \ne 0$ is impossible by some arguments of common divisor and infinite descent on the norm given in R. Andrew Ohana's more algebraic proof. Actually I should've just linked his proof instead of writing all the above, since the algebraic structure is more enlightening. But I'll leave the answer as a source of references and getting down ideas.


TITLE: Can I solve an integral (or other tough problem) by playing with knots?
QUESTION [39 upvotes]: I've seen that in calculating things in knot theory that involves a lot of hard looking integrals and matrices, even though the knots themselves appear fairly simple.
So is there some way in which this works backwards, where I might have a really large determinant or complicated integral that can be represented as a knot, and by tying, untying, or rearranging the knot I can solve the problem on that "knot domain" and then transfer it back into the lingo of integrals and matrices, kind of like you might imagine using a laplace transform?

REPLY [13 votes]: I like your question.  Unfortunately, I do not have a general answer, but I do have a specific example where it is possible.  You can look more into this if you wish in Rolfsen, Knots and Links under linking numbers, which is my reference.  A link is just a few disjoint knots, which may be "linked" together.  There is something called a Gauss Integral $\ell k (J,K)$ defined for a link which has components $J$ and $K$ which is
$$ \frac{1}{4\pi} \int_J \int_K \frac{ (x'-x)(dxdz'-dzdy')+(y'-y)(dzdx'-dxdz')+(z'-z)(dxdy'-dydx') }{ [(x'-x)^2+(y'-y)^2+(z'-z)^2]^\frac{3}{2} } $$
where $(x,y,z)\in J$ and $(x',y',z')\in K.$ This is obviously an integral you do not want to have to work out.  Luckily, Rolfsen gives seven equivalent ways of computing the linking number.  The most common of which is the following: For an oriented link diagram of $J\cup K$, only look at the crossing where $K$ is the overarc (on top).  Then rotate the picture until $K$ is "pointing" up at one of the crossing. If $J$ goes from right to left, assign a $+1$ to this crossing. If $J$ goes from left to right, a $-1$. The sum of these is your linking number, up to a sign convention. So, you never have to do the integral!
This is, as I said, very specific.  But maybe it gives some intuition.


TITLE: Evaluating integral $\int\frac{2(\cos x + \sin x)}{\sin 2x} dx$
QUESTION [0 upvotes]: Evaluate
$$\int\frac{2(\cos x + \sin x)}{\sin 2x} dx$$
I have tried a few times and came up with an answer
$$-\ln | \csc x + \cot x| +\frac 12\ln |\tan x + \sec x| + C$$
which is incorrect. I just need someone to explain what is wrong with my answer.

REPLY [1 votes]: Use $\sin2x =2\sin x\cos x$,
$$\int\frac{2(\cos x + \sin x)}{\sin 2x} dx
=\int\frac1{\sin x} dx+ \int \sec x dx $$
$$=\ln \tan \frac x2 + \ln (\tan x + \sec x) + C$$
where the two integrals are obtained as follows
$$\int\frac{dx}{\sin x}=\int\frac{dx}{2\sin \frac x2\cos\frac x2}
=\int\frac{\sec^2\frac x2dx}{2\tan\frac x2}=\ln \tan \frac x2 $$
$$\int \sec x dx = \int \frac{\sec x\tan x + \sec^2 x}{\tan x + \sec x}dx
=\int \frac{d(\tan x+ \sec x)}{\tan x + \sec x}=  \ln (\tan x + \sec x)$$


TITLE: localization preserves free finite resolution?
QUESTION [0 upvotes]: If any finite generated module over ring $R$ has a finite free resolution. Is it true that any finite generated module over $S^{-1}R$ has a finite free resolution ?
If it's not please give me a counterexample.
Thanks.

REPLY [2 votes]: Let $M$ be a finitely generated module over the localization. It contains an $R$-submodule $N$ which is finitely generated and whose localization is precisely $M$ itself. The hypothesis implies that $N$ has a finite free resolution. Localizing that resolution gives you a finite free resolution of $M$.


TITLE: tensor product of algebras over a field
QUESTION [1 upvotes]: Let $k$ be a commutative field and $E,F$ two commutative, associative,
unital $k$-algebras whose respective unit elements $e,e'$ are
non-zero. Let $\eta_E:K\rightarrow E$ and $\eta_F:K\rightarrow F$ be
the canonical ring homomorphisms such that $\eta_E(1)=e$ and
$\eta_F(1)=e'$. These mappings are injective and allow us to identify
$K$ with a subfield of $E$ (resp. $F$).
Let $u:E\rightarrow E\otimes_KF$ and $v:F\rightarrow E\otimes_KF$ be
the canonical homomorphisms such that $u(x)=x\otimes e'$ and
$v(y)=e\otimes y$, for all $x\in E$ and $y\in F$. These mappings are
injective and allow us to identify $E$ and $F$ with subalgebras of
$E\otimes_KF$, both having as unit element the unit element $e\otimes
 e'$.
In $E\otimes_K F$, we have $E\cap F=K$.

Can someone please translate $E\cap F=K$ into expressions without identifications? I think $u(E)$ replaces $E$ and $v(F)$ replaces $F$.
What should replace $K$? Maybe $u(\eta_E(K))?$ But then let $z\in u(E)\cap v(F)$. Then $e\otimes y=z=x\otimes e'$ for some $x\in E$ and $y\in F$. How to show that this is in $u(\eta_E(K))$?

REPLY [1 votes]: For $K$, use
$$u(\eta_E(K))=v(\eta_F(K))=\{c\cdot e\otimes e':c\in K\}\,.$$


TITLE: Radius and domain of convergence for $\sum _{n=1}^{\infty}2^n x^{n^2}$
QUESTION [3 upvotes]: Another Question on Radius of convergence :
Calculate Radius and domain of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$
I used the formula $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|$ to calculate radius of convergence.
I have $\lim_{n\rightarrow \infty} |\frac{a_n}{a_{n+1}}|=\lim_{n\rightarrow \infty} |\frac{2^n}{2^{n+1}}|=\frac{1}{2}$
So,radius of convergence is $\frac{1}{2}$ for $\sum _{n=1}^{\infty}2^n x^n$ but how would i use this to calculate radius of convergence for $$\sum _{n=1}^{\infty}2^n x^{n^2}$$
If i have series something like $\sum _{n=1}^{\infty}2^n (x^n)^2$
I would have just write radius of convergence is $\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$
But in this case i do not know what to do.. please help me to see this.
Another point is I some how tried to use : For $\sum_{n=1}^{\infty}a_nx^n$ radius would be $R$ where
$$\frac{1}{R}=\lim \sup \sqrt[n]{|a_n|}$$
My book says that if I have some thing like  $$\sum _{n=1}^{\infty}a_n x^{n^2}$$ I could use same formula with $$\frac{1}{R}=\lim \sup \sqrt[n^2]{|a_n|}$$
I am not very sure if i can really use it but if i use that, I would get :
$$\frac{1}{R}=\lim \sup \sqrt[n^2]{2^n}=\sqrt{2}$$
SO, radius of convergence would be $\frac{1}{\sqrt{2}}$
which is not much different from what i have got with previous case of $\sum _{n=1}^{\infty}2^n (x^n)^2$
I would like to know more about something similar to this..
Even if i have $\sum _{n=1}^{\infty}a_n x^{n^2}$ or $\sum _{n=1}^{\infty}a_n x^n$, the radius of convergenc is not changing.
Please help me to look into this in more detail.
Thank you

REPLY [1 votes]: Define the sequence
$$
a_n=\left\{\begin{matrix} 2^{\sqrt{n}} &\text{if $n=m^2$ for some $m\in\mathbb{N}$}\\
0 &\text{otherwise}
\end{matrix}
\right.
$$
The series in the OP can be expressed as
$$\sum_na_nx^n$$
The radius of convergence is $\frac{1}{L}$ where $L=\limsup_n\sqrt[n]{|a_n|}$ which in this case is
$$L=\lim_{m\rightarrow\infty}\big(2^m)^{1/m^2}=\lim_{m\rightarrow\infty}2^{1/m}=1$$
Thus the series converges for all $x$ such that $|x|<1$.


text:
\begin{document}

\title[Eigenvalues of elliptic functional differential systems]{Eigenvalues of elliptic functional differential systems via a Birkhoff--Kellogg type theorem}

\date{}

\author[G. Infante]{Gennaro Infante}
\address{Gennaro Infante, Dipartimento di Matematica e Informatica, Universit\`{a} della
Calabria, 87036 Arcavacata di Rende, Cosenza, Italy}
\email{gennaro.infante@unical.it}

\begin{abstract}
Motivated by recent interest on Kirchhoff-type equations, in this short note we utilize a classical, yet very powerful, tool of nonlinear functional analysis in order to investigate the existence of positive eigenvalues of systems of elliptic functional differential equations. An example is presented to illustrate the theory.
\end{abstract}

\subjclass[2010]{Primary 35J47, secondary 35B09, 35J57, 35J60, 47H10}

\keywords{Positive solution, nonlocal elliptic system, functional boundary condition, cone, Birkhoff-Kellogg type theorem}

\maketitle

\section{Introduction}
A well known result in nonlinear analysis is the Birkhoff-Kellogg invariant-direction Theorem~\cite{B-K-1922}. In the case of an infinite-dimensional normed linear space $V$ this theorem reads as follows.
\begin{thm}~\cite[Theorem~6.1]{GD}
Let $U$ be a bounded open neighborhood of $0$ in an infinite-dimensional normed linear space $(V, \| \, \|)$, and let $T:\partial U \to V$ be a compact map satisfying $\|T(x)\| \geq \alpha$ for some $\alpha > 0$ for every $x$ in $\partial U$. Then there exist $x_0 \in \partial U$ and $ \lambda_{0} \in (0,+\infty)$ such that $x_0 = \lambda_{0} F(x_0).$
\end{thm}
The invariant direction Theorem has been object of deep studies in the past, with applications and extensions in several directions, we refer the reader to \cite{applbook, BK, Casey, Fitz-Petr, Fitz-Petr2, Fu-Pe-Vi, Fu-Vi, guosun, Krasno, Kryszewski, Ivar-Charlie, Oregan} and references therein. In particular, we highlight that~\cite{Fitz-Petr, Krasno, Ivar-Charlie} provide interesting applications to the existence of eigenvalues and eigenfunctions of elliptic boundary value problems.

Here we make use of the following Birkhoff-Kellogg type result, which is set in cones. This is a special case of a result due to Krasnosel'ski\u{i} and Lady\v{z}enski\u{\i}~\cite{Kra-Lady}, see also~\cite[Theorem~5.5]{Krasno}.
Before stating the result, we recall that a cone $C$ of a real Banach space $(X,\| \, \|)$ is a closed set with $C+C\subset C$, $\mu C\subset C$ for all $\mu\ge 0$ and $C\cap(-C)=\{0\}$, and we introduce the following notation:
$$C_{r}:=\{x\in C: \|x\|<r\},\,
\overline{C}_{r}:=\{x\in C: \|x\|\leq r\},\ \text{and}\ \partial
C_{r}:=\{x\in C: \|x\|=r\}.$$

\begin{thm}\label{B-K}
Let $(X,\| \, \|)$ be a real Banach space, let $T:\overline{C}_{r}\to C$ be compact and suppose that
$$\inf_{x\in \partial C_{r}}\|Tx\|>0.$$
Then there exist $\lambda_{0}\in (0,+\infty)$ and $x_{0}\in \partial C_{r}$
such that $x_{0}=\lambda_{0} Tx_{0}$.
\end{thm}

By means of Theorem~\ref{B-K} we discuss the solvability, with respect to the parameter $\lambda$, of the following system of second order elliptic functional differential equations subject to functional boundary conditions (BCs)
\begin{equation}
 \label{nellbvp-intro}
 \left\{
\begin{array}{ccc}
L_i u_i=\lambda f_i(x, u, Du, w_i [u]), & \text{in}\ \Omega, & i=1,2,\ldots,n, \\
 B_i u_i=\lambda \zeta_i (x)h_{i}[u], & \text{on}\ \partial \Omega, & i=1,2,\ldots,n,
\end{array}
\right.
\end{equation}
where $\Omega \subset \mathbb{R}^{n}$ is a bounded domain with a sufficiently smooth boundary, $L_i$ is a strongly uniformly elliptic operator, $B_i$ is a first order boundary operator,
 $u=(u_1,\dots, u_n)$,
 $Du=(\nabla u_1,\dots, \nabla u_n) $, $f_i$ are continuous functions, $\zeta_i$ are sufficiently regular functions, $w_{i}$ and $h_{i}$ are suitable compact functionals.

The class of systems occurring in~\eqref{nellbvp-intro} is fairly general and allows us to deal with nonlocal problems of Kirchhoff-type.
This is a very active area of research, a typical example of a Kirchhoff-type problem is
\begin{equation} \label{nonlocal-ToFu}
-M\left(\int_{\Omega} |\nabla u|^2\,dx\right)\Delta u= f(x,u), \ x\in \Omega,\quad u=0\ \text{on}\ \partial \Omega,
\end{equation}
which has been investigated by Ma in his survey~\cite{ToFu}. An extension to systems of the BVP~\eqref{nonlocal-ToFu} has been considered by Figueiredo and Su\'{a}rez~\cite{Giovany}, namely
\begin{equation} \label{SysGio}
\left\{
\begin{array}{ll}
-M_1\left(\int_{\Omega} |\nabla u_1|^2\,dx\right)\Delta u_1= f_1(x,u_1,u_2),& x \text{ in }\Omega , \\
-M_2\left(\int_{\Omega} |\nabla u_2|^2\,dx\right)\Delta u_2= f_2(x,u_1,u_2), & x \text{ in }\Omega , \\
u_1=u_2=0 & \text{on }\partial \Omega .
\end{array}
\right.
\end{equation}
The approach employed in~\cite{Giovany} is the sub-supersolution method. The system~\eqref{SysGio} has been studied also
by the sub-supersolution method in~\cite{BoGuChAl, MeBoGuAl, BoBoDj, BoGu}, while variational methods were employed in~\cite{FurOliSi, LouQin, Nguyen, ZhangSun}.

Note that there has been also interest in Kirchhoff-type problems with gradient terms appearing within the nonlinearities, we mention the recent papers by
Alves and Boudjeriou~\cite{AlBo}, Yan and co-authors~\cite{YaOrAg}, Chen~\cite{Chen} and references therein.

The framework of~\eqref{nellbvp-intro} allows us to deal with non-homogenous BCs of functional type. In the case of nonlocal elliptic equations, non-homogeneous BCs have been investigated by Wang and An~\cite{WangAn}, Morbach and Corr\~{e}a~\cite{MorCor} and by the author~\cite{gi-jepeq}.
The formulation of the functionals occurring in~\eqref{nellbvp-intro} allows us to consider \emph{multi-point} or \emph{integral} BCs.
There exists a wide literature on this topic,
 we refer the reader to the reviews~\cite{Cabada1, Conti, rma, sotiris, Stik, Whyburn} and the papers~\cite{Goodrich3, Goodrich4, kttmna, ktejde, Pao-Wang, Picone, jw-gi-jlms}.

Here we discuss, under fairly general conditions, the existence of positive eigenvalues with corresponding non-negative eigenfunctions for the system~\eqref{nellbvp-intro} and illustrate how these results can be applied in the case of nonlocal elliptic systems. Our results are new and complement previous results of the author~\cite{gi-jepeq}, by allowing the presence of gradient terms within the nonlinearities and the functionals. The results also complement the ones in~\cite{sBaCgI}, by considering more general nonlocal elliptic systems.

\section{Eigenvalues and eigenfunctions}
In what follows, for every $\hat{\mu}\in (0,1)$ we denote by $C^{\hat{\mu}}(\overline{\Omega})$ the space of all $\hat{\mu}$-H\"{o}lder continuous functions $g:\overline{\Omega}\to \mathbb{R}$ and, for every $k\in \mathbb{N}$, we denote by $C^{k+\hat{\mu}}(\overline{\Omega})$ the space of all functions $g\in C^{k}(\overline{\Omega})$ such that all the partial derivatives of $g$ of order $k$ are $\hat{\mu}$-H\"{o}lder continuous in $\overline{\Omega}$ (for more details see \cite[Examples~1.13 and~1.14]{Amann-rev}).

We make the following assumptions on the domain $\Omega$ and the operators $L_i$ and $B_i$ and the functions $\zeta_i$ that occur in~\eqref{nellbvp-intro}
 (see \cite[Section~4 of Chapter~1]{Amann-rev} and \cite{Lan1, Lan2})):
\begin{enumerate}
\item $\Omega\subset \R^m$, $m\ge 2$, is a bounded domain such that its boundary $\partial \Omega$ is an $(m-1)$-dimensional $C^{2+\hat{\mu}}-$manifold for some $\hat{\mu}\in (0,1)$, such that $\Omega$ lies locally on one side of $\partial \Omega$ (see \cite[Section 6.2]{zeidler} for more details).
\item $L_i$ is a the second order elliptic operator given by
\begin{equation*}
L_i u(x)=-\sum_{j,l=1}^m a_{ijl}(x)\frac{\partial^2 u}{\partial x_j \partial x_l}(x)+\sum_{j=1}^m a_{ij}(x) \frac{\partial u}{\partial x_j} (x)+a_i (x)u(x), \quad \mbox{for $x\in \Omega$,}
\end{equation*}
where $a_{ijl},a_{ij},a_i\in C^{\hat{\mu}}(\overline{\Omega})$ for $j,l=1,2,\ldots,m$, $a_i(x)\ge 0$ on $\bar{\Omega}$, $a_{ijl}(x)=a_{ilj}(x)$ on $\bar{\Omega}$ for $j,l=1,2,\ldots,m$. Moreover $L_i$ is strongly uniformly elliptic; that is, there exists $\bar{\mu}_{i0}>0$ such that
$$\sum_{j,l=1}^m a_{ijl}(x)\xi_j \xi_l\ge \bar{\mu}_{i0} \|\xi\|^2 \quad \mbox{for $x\in \Omega$ and $\xi=(\xi_1,\xi_2,\ldots,\xi_m)\in\R^m$.}$$

\item $B_i$ is a boundary operator given by
$$B_i u(x)=b_i(x)u(x)+\delta_i \frac{\partial u}{\partial \nu}(x) \quad \mbox{for $x\in\partial \Omega$},$$
where $\nu$ is an outward pointing and nowhere tangent vector field on $\partial \Omega$ of class $C^{1+\hat{\mu}}$ (not necessarily a unit vector field), $\frac{\partial u}{\partial \nu}$ is the directional derivative of $u$ with respect to $\nu$, $b_i:\partial \Omega \to \R$ is of class $C^{1+\hat{\mu}}$ and moreover one of the following conditions holds:
\begin{enumerate}
\item $\delta_i =0$ and $b_i(x)\equiv 1$ (Dirichlet boundary operator).
\item $\delta_i =1$, $b_i(x)\equiv 0$ and $a_i(x)\not\equiv 0$ (Neumann boundary operator).
\item $\delta_i =1$, $b_i(x)\ge 0$ and $b_i(x)\not\equiv 0$ (Regular oblique derivative boundary operator).
\end{enumerate}

\item $\zeta_i \in C^{2-\delta_i+\hat{\mu}}(\partial\Omega)$.
\end{enumerate}

It is known that, under the previous conditions (see \cite{Amann-rev}, Section 4 of Chapter 1), a strong maximum principle holds,
given $g\in C^{\hat{\mu}}(\bar{\Omega})$, the~BVP
 \begin{equation}
 \label{eqelliptic}
 \left\{
\begin{array}{ll}
  L_{i}u(x)=g(x), & x\in \Omega, \\
 B_{i}u(x)=0, & x\in \partial \Omega,
\end{array}
\right.
\end{equation}
admits a unique classical solution $u\in C^{2+\hat{\mu}}(\bar{\Omega})$ and, moreover, given $\zeta_i \in C^{2-\delta_i+\hat{\mu}}(\partial\Omega)$ the~BVP
 \begin{equation}
 \label{eqellipticbc}
 \left\{
\begin{array}{ll}
  L_{i}u(x)=0, & x\in \Omega, \\
 B_{i}u(x)=\zeta_i(x), & x\in \partial \Omega,
\end{array}
\right.
\end{equation}
also admits a unique solution $\gamma_i \in C^{2+\hat{\mu}}(\bar{\Omega})$.

In order to investigate the solvability of the system~\eqref{nellbvp-intro}, we make use of the cone of non-negative functions $\hat{P}=C(\bar{\Omega},\R_+)$. The solution operator associated to the BVP~\eqref{eqelliptic}, $K_i:C^{\hat{\mu}}(\bar{\Omega})\to C^{2+\hat{\mu}}(\bar{\Omega})$, defined as $K_{i}g=u$ is linear and continuous. It is also known (see \cite{Amann-rev}, Section 4 of Chapter 1) that $K_{i}$ can be extended uniquely to a continuous, linear and compact operator (that we denote again by the same name) $K_{i}:C(\bar{\Omega})\to C^{1}(\bar{\Omega})$ that leaves the cone $\hat{P}$ invariant, that is $K_{i}(\hat{P})\subset \hat{P}$.

We utilize the space $C^{1}(\bar{\Omega},\R^n)$, endowed with the norm
$$\|u\|_{1}:=\displaystyle\max \{\|u_i\|_{\infty}, \|\partial_{x_j} u_i \|_{\infty}: i=1,2,\ldots,n,\ j=1,2,\ldots,m\},$$ where $\|z\|_{\infty}=\displaystyle\max_{x\in \bar{\Omega}}|z(x)|,$ and consider the cone $P=C^{1}(\bar{\Omega},\R^n_+)$. We rewrite the elliptic system~\eqref{nellbvp-intro} as a fixed point problem, by considering the operators $T,\Gamma:C^1(\bar{\Omega}, \R^n) \to C^1(\bar{\Omega},\R^n)$ given by
\begin{align}
T(u):=(K_i F_i(u))_{i=1..n},\quad
\Gamma (u):=(\gamma_i h_{i}[u] )_{i=1..n},
\end{align}
where $K_i$ is the above mentioned extension of the solution operator associated to~\eqref{eqelliptic},
 $\gamma_i \in C^{2+\hat{\mu}}(\overline{\Omega})$ is the unique solution of the BVP~\eqref{eqellipticbc} and
$$F_i(u)(x):=f_i(x,u(x),Du(x), w_i[u]),\ \text{for}\ u\in C^{1}(\bar{\Omega}, I)\ \text{and}\ x\in \bar{\Omega}.$$

\begin{defn} We say that $\lambda$ is an~\emph{eigenvalue} of the system~\eqref{nellbvp-intro} if there exists
 $u\in C^1(\bar{\Omega})$ with $\|u\|_1>0 $ such that the pair $(u,\lambda)$ satisfies the operator equation
\begin{equation}\label{chareq}
u=\lambda(Tu+\Gamma u)=\lambda( K_i F_i(u)+ \gamma_i h_{i}[u] )_{i=1..n}.
\end{equation}
If the pair $(u,\lambda)$ satisfies~\eqref{chareq} we say that $u$ is an~\emph{eigenfunction} of the system~\eqref{nellbvp-intro} corresponding to the eigenvalue $\lambda$. If, furthermore, the components of $u$ are non-negative, we say that $u$ is a \emph{non-negative eigenfunction} of the system~\eqref{nellbvp-intro}.
\end{defn}

\begin{thm}\label{eigen}Let $\rho \in (0,+\infty)$ and assume the following conditions hold.
\begin{itemize}

\item[(a)] For every $i=1,2,\ldots,n$, $w_{i}: \overline{P}_{\rho} \to \mathbb{R}$ is continuous and there exist
$\underline{w}_{i,\rho}, \overline{w}_{i,\rho}\in \R$ such that
\begin{equation}\label{w-cond}
\underline{w}_{i,\rho}\leq w_{i}[u]\leq \overline{w}_{i,\rho},\ \text{for every}\ u\in \overline{P}_{\rho}.
\end{equation}

\item[(b)]
For every $i=1,2,\ldots,n$, $f_i\in C(\Pi_{\rho}, \R)$ and there exist $\delta_i \in C(\bar{\Omega},\R_+)$ such that
\begin{equation}\label{f-cond}
f_{i}(x,u,v,w)\ge \delta_{i, \rho}(x),\ \text{for every}\ (x,u,v,w)\in \Pi_{\rho},
\end{equation}
where $$\Pi_{\rho}:=\bar{\Omega}\times [0,\rho]^n \times [-\rho,\rho]^{m\times n}\times [\underline{w}_{i,\rho}, \overline{w}_{i,\rho}].$$

\item[(c)] For every $i=1,2,\ldots,n$, $\zeta_i \in C^{2-\delta_i+\hat{\mu}}(\partial\Omega)$, $\zeta_i\geq 0$,
$h_{i}: \overline{P}_{\rho} \to \R$ is continuous and bounded. Let $\eta_{i, \rho}\in [0,+\infty)$ be such that
\begin{equation}\label{h-cond}
h_{i}[u]\geq \eta_{i, \rho},\ \text{for every}\ u\in \overline{P}_{\rho} .
\end{equation}

\item[(d)] There exist $i_0\in \{1,\ldots,n\}$ and $ \phi_{i_0, \rho} \in (0,+\infty)$ such that
\begin{equation}\label{c-cond}
\|K_{i_0}(\delta_{i_0, \rho})+ \eta_{i_0, \rho} \gamma_{i_0}\|_{\infty}\geq \phi_{i_0, \rho}.
\end{equation}
\end{itemize}

Then the system~\eqref{nellbvp-intro} has a positive eigenvalue with an associated eigenfunction $u\in \partial P_{\rho}$.
\end{thm}
\begin{proof}
Due to the assumptions above, the operator $T+\Gamma$ maps $\overline{P}_{\rho}$ into $P$ and is compact (by construction, the map $F$ is continuous and bounded and $\Gamma$ is a finite rank operator). Take $u\in \partial P_{\rho}$, then for every $x\in \bar{\Omega}$ we have
\begin{equation}\label{lwest}
K_{i_0} F_{i_0} u(x)+ \gamma_{i_0}(x) h_{i_0}[u]
\ge  K_{i_0}(\delta_{i_0, \rho})(x) + \eta_{i_0, \rho} \gamma_{i_0}(x) .
\end{equation}
Taking the supremum for $x\in \bar{\Omega}$ in~\eqref{lwest} we obtain
\begin{equation}\label{lwest2}
\| Tu+\Gamma u\|_{1}\geq \|T_{i_0}u+\Gamma_{i_0} u\|_{\infty}\geq \|K_{i_0}(\delta_{i_0, \rho})+ \eta_{i_0, \rho} \gamma_{i_0}\|_{\infty}\geq \phi_{i_0, \rho}.
\end{equation}
Note that the RHS of \eqref{lwest2} does not depend on the particular $u$ chosen. Therefore we have
$$
\inf_{u\in \partial P_{\rho}}\| Tu+\Gamma u\|\geq \phi_{i_0, \rho}>0,
$$
and the result follows by Theorem~\ref{B-K}.
\end{proof}
\begin{rem}
Note that we have chosen to use \emph{inequalities} in~\eqref{w-cond}-\eqref{c-cond}; this is due that, in applications, it is often easier and somewhat more efficient to use~\emph{estimates} on the nonlienarieties involved. Furthermore note that, in our reasoning, what really matters is that some positivity occurs in one component of the system, either in the nonlinearity $f_i$ or in the functional $h_i$.
\end{rem}
The following Corollary provides a sufficient condition for the existence of an unbounded set of eigenfunctions for the system~\eqref{nellbvp-intro}.
\begin{cor}\label{CorEig}
In addition to the hypotheses of Theorem~\ref{eigen}, assume that $\rho$ can be chosen arbitrarily in $(0,+\infty)$. Then for every $\rho$ there exists a non-negative eigenfunction $u_{\rho}\in \partial P_{\rho}$ of the system~\eqref{nellbvp-intro} to which corresponds a $\lambda_{\rho} \in (0,+\infty)$.
\end{cor}

We now show the applicability of the above results in the context of systems of nonlocal elliptic equations with functional BCs.
\begin{ex}
Take $\Omega=\{x\in \mathbb{R}^2 : \|x\|_2<1\}$ and consider the system
\begin{equation} \label{example}
\left\{
\begin{array}{ll}
-\bigl(e^{u_2(0)}+\int_{\Omega}| \nabla u_1|^2\,dx\bigr)\Delta u_1=\lambda e^{u_1}(1+| \nabla u_2|^2),& \text{in }\Omega , \\
-e^{(\int_{\Omega}| \nabla u_1|^2+| \nabla u_2|^2\,dx)}\Delta u_2=\lambda u_2^2| \nabla u_1|^2, & \text{in }\Omega , \\
u_1=\lambda h_{1}[(u_1,u_2)],\ u_2=\lambda h_{2}[(u_1,u_2)], & \text{on }\partial \Omega ,
\end{array}
\right.
\end{equation}
where
$$
h_{1}[(u_1,u_2)]= u_1(0)+\bigl( \frac{\partial u_2}{\partial x_1}(0)\bigr)^{2}\ \mbox{and} \ h_{2}[(u_1,u_2)]=(u_1(0))^2+\int_{\Omega}| \nabla u_2|^2\,dx.
$$
Denote by $\hat{1}$ the function equal to $1$ on $\bar{\Omega}$.
Note that for $i=1,2$, $K_{i}(\hat{1})=\frac{1}{4}(1-x_1^2-x_2^2)$, where $x=(x_1,x_2)$, and $\|K_{i}(\hat{1})\|_{\infty}=\frac{1}{4}$. Furthermore note that we may take $\gamma_{1}=\gamma_{2}\equiv 1$.

 We fix $\rho\in (0,+\infty)$ and consider
\begin{align*}
f_1(u_1,u_2, \nabla u_1, \nabla u_2, w_1[(u_1,u_2)]):&=e^{u_1}(1+| \nabla u_2|^2)w_1[(u_1,u_2)],\\
f_2(u_1,u_2, \nabla u_1, \nabla u_2, w_2[(u_1,u_2)]) :&= u_2^2| \nabla u_1|^2w_2[(u_1,u_2)],
\end{align*}
where
\begin{align*}
w_1[(u_1,u_2)]:&=\Bigl(e^{u_2(0)}+\int_{\Omega}| \nabla u_1|^2\,dx\Bigr)^{-1}, \\
w_2[(u_1,u_2)]:&=e^{-(\int_{\Omega}| \nabla u_1|^2+| \nabla u_2|^2\,dx)}.
\end{align*}

In this case we may take
$$[\underline{w}_{1,\rho}, \overline{w}_{1,\rho} ]=[(2\pi \rho^2+e^{\rho})^{-1},1],\ [\underline{w}_{2,\rho}, \overline{w}_{2,\rho} ]=[e^{-4\pi \rho^2},1],$$
$$\delta_{1, \rho}(x)\equiv (2\pi \rho^2+e^{\rho})^{-1},\ \delta_{2, \rho}(x)\equiv 0,\ \eta_{1, \rho}=\eta_{2, \rho}=0,$$
and therefore we get
$$
\|K_{1}(\delta_{1, \rho})+ \eta_{1, \rho} \gamma_{1}\|_{\infty}=(8\pi \rho^2+4e^{\rho})^{-1}= \phi_{1, \rho}>0.
$$
Thus we can apply Corollary~\ref{CorEig}, obtaining a uncountably many pairs  $(u_{\rho}, \lambda_{\rho})$ of  non-negative eigenfunctions and positive eigenvalues for the system~\eqref{example}.
\end{ex}

\section*{Acknowledgement}
G. Infante was partially supported by G.N.A.M.P.A. - INdAM (Italy).

config: arxiv


TITLE: Why is the magnetic flux proportional to the current (Inductance)?
QUESTION [1 upvotes]: I'm trying to understand the reasoning behing this formula, or how to get there, if possible, using Maxwell's equations:
$$\Phi_i = L_i I_i + M_{ij} I_j$$
From the wikipedia page of inductance I got this formula which is basically faraday's law:
$$v(t) = - \frac{d\Phi(t)}{dt}$$
But then they somehow get to this one which is what I don't understand:
$$v(t) = L \frac{d i(t)}{dt}$$
I've tried doing some things with the Maxwell equations but I couldn't really get to anything interesting.

REPLY [2 votes]: Your first formula is in fact one possible definition of $L_i$ and $M_{ij}$, so it's not surprising that you're having a hard time deriving it.
The reasoning is simple: The magnetic field generated by a circuit is proportional to the current in the circuit. The magnetic field at circuit $i$ is just the sum of the magnetic fields generated by each circuit, including the same circuit $i$, and each of these is proportional to the current of the corresponding circuit. We can write the contribution of circuit $j$ as
$$\mathbf{B}_j(\mathbf{r}) = I_j \tilde{\mathbf{B}}_j(\mathbf{r})$$
where $\tilde{\mathbf{B}}_j$ is some function that depends on the shape of circuit $j$ but not on its current. The flux across circuit $i$ is then
$$\Phi_i = I_i \iint_\text{circuit $i$} \tilde{\mathbf{B}}_i \cdot \mathbf{dS} + \sum_{j\neq i} I_j \iint_\text{circuit $i$} \tilde{\mathbf{B}}_j \cdot \mathbf{dS}.$$
Each of these integrals depends only on the geometry of the circuits but not on their currents, and they're what we call the inductances:
$$\Phi_i = L_i I_i + \sum_{j \neq i} M_{ij} I_j.$$
Your last formula follows if we take just one circuit; in that case we have $\Phi = LI$ and $d\Phi/dt = L\, dI/dt$. Note that the voltage in your formula seems to be missing a sign.


TITLE: How to calculate $\int_{0}^4\int_{y/2}^{2} e^{x^2} dxdy$
QUESTION [0 upvotes]: http://www4a.wolframalpha.com/Calculate/MSP/MSP36251d183i6d402226cb000031ag27ead8g7fa5i?MSPStoreType=image/gif&s=44&w=244.&h=42.
I can't make any strides in this problem so I typed it in on wolfram alpha to see if there is even a viable solution (apparently there is). The link above will give you a better picture of the problem. If anyone could help with this at all I'd greatly appreciate it.

REPLY [0 votes]: Swap the integral terms:
$$\int_{0}^2 \int_{0}^{2x} e^{x^2}\,dy\,dx = \int_{0}^2 2xe^{x^2}\,dx$$
That latter integral you should be able to compute using a normal substitution.
The swap above is due to the fact that:
$$\left\{(x,y)\mid y\in[0,4], x\in[y/2,2]\right\}=\left\{(x,y)\mid x\in[0,2], y\in[0,2x]\right\}$$


TITLE: Compactification by ends
QUESTION [5 upvotes]: I am trying to solve Exercise 1.19 in Spivak's Volume 1 of "A comprehensive introduction to Differential Geometry".
After the definition of an 'end' of $X$, a topology on the new space $X\cup \epsilon(X)$ is defined "by choosing as neighborhoods of an end $\epsilon_0$
the sets
$N_C(\epsilon_0)= \epsilon_0(C)\cup \{\text{ends }\epsilon : \epsilon(C)=\epsilon_0(C)\}$,
for all compact C."
Now, with this, I assume that he defines $V$ to be a neighborhood of $x$ in $X$ if there exists an open subset $U$ of $X$ with $x\in U\subset V$ and now with this system of neighborhoods at any point in $X$ or end in $\epsilon(X)$, there is only one topology on $X\cup \epsilon(X)$ generated by it. Namely, $V$ is open in $X\cup \epsilon(X)$ iff
$V$ is a neighborhood of each one of its points.
I assume this is the topology we are talking about. Am I right?
I am trying to prove that $X\cup \epsilon(X)$ is a compact space for $X$ a connected, locally connected, locally compact Hausdorff space, but for one case this seems to be untrue.
Take the subspace of the plane defined by $X=\{(x,0) \mid x\in \mathbb{R}\}\cup \bigcup_{n\in\mathbb{Z}}\{(n,y)\mid y\geq 0\}$, this space satisfies all the properties listed before. Moreover, it has and end $\epsilon_n$ for each "ramification" $\{(n,y)\mid y\geq 0,n\in\mathbb{Z}\}$. However, we could cover $X$ with any collection of open subsets of $X$ (don't include any of the newly introduced neighborhoods of ends), and join it to the cover $\epsilon(X)$ with the cover ${N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)}_{n\in \mathbb{Z}}$ of $\epsilon(X)$, and each $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$ contains exactly one end, $\epsilon_n$.
Furthermore, $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$ is obviously a neighborhood of each of its points, so it is open in $X\cup \epsilon(X)$. Hence, from the resulting open cover of $X\cup \epsilon(X)$ cannot be extracted a finite subcover, because it would have just a finite amount of $N_{\overline{B(n,1/2)\cap X}}(\epsilon_n)$, and hence it would include just a finite amount of ends, while $\epsilon(X)$ is infinite.
So, $X\cup \epsilon(X)$ is not compact.
This leads me to think I'm not working with the right topology. Any light on this?
Thanks in advance.

REPLY [3 votes]: Your definition is correct.  What you are missing is that $X$ has two more ends besides the ends $\epsilon_n$: it has an end $\epsilon_{\infty}$ on which $x\to\infty$ and an end $\epsilon_{-\infty}$ on which $x\to-\infty$.  That is, if $C\subset X$ is compact, $\epsilon_\infty(C)$ is you the (unique) component of $X\setminus C$ in which the $x$ coordinate is unbounded above and $\epsilon_{-\infty}(C)$ is the (unique) component of $X\setminus C$ in which the $x$ coordinate is unbounded below.  A neighborhood of $\epsilon_{\infty}$ must contain all but finitely many of the $\epsilon_n$.  So if you were to extend your open cover by adding a set containing $\epsilon_\infty$, you would only have finitely many $\epsilon_n$ left uncovered, breaking your counterexample.