text:
\section{Stability and convergence analysis}\label{sec:stab_conv}
{In this section, we perform a stability and convergence analysis of the multirate RKC method introduced in \cref{sec:mrkca}. We will show stability of the scheme on the multirate test equation \eqref{eq:mtesteq} and on a $2\times 2$ model problem. Then we prove its first-order accuracy.}

\subsection{Stability analysis}\label{sec:stabanalysis}
{First, we prove that the mRKC method is stable when it is applied to the multirate test equation \eqref{eq:mtesteq}, which is sufficient when the Jacobians of $\ff$ and $\fs$ are simultaneously triangularizable. Then, we also show stability for a $2\times 2$ model problem where the Jacobians of $\ff$ and $\fs$ are not simultaneously triangularizable, and hence the stability analysis cannot be reduced to \eqref{eq:mtesteq}.}

\subsubsection*{Stability analysis for the multirate test equation}
Since \eqref{eq:defu} is approximated numerically, the stability analysis performed in \cref{sec:stab_modeq} is no longer valid; indeed, $\varphi(z)$ is now replaced by a numerical approximation with different stability properties. Hence, we now compute a closed expression for $u_\eta$ given $y$, as in \eqref{eq:solutesteq} for $u(\eta)$.
We denote by
\begin{equation}\label{eq:defPm}
P_m(z)=a_m T_m(\vz+\vu z)
\end{equation}
the stability polynomial of the $m$-stage RKC scheme, with $a_m,\vz,\vu$ from \cref{eq:defv01}.
The next lemma computes the solution $u_\eta$ of \cref{eq:defbue} in the case of the multirate test equation \cref{eq:mtesteq}.

\begin{lemma}\label{lemma:closedbue}
	Let $\lambda,\zeta\leq 0$, $\fs(y)=\zeta y$, $\ff(y)=\lambda y$, $\eta>0$, $m\in\Nb$ and $y\in\Rb$. Then, the solution $u_\eta$ of \cref{eq:defbue}, is given by
	\begin{equation}\label{eq:numutesteq}
	u_\eta = (P_m(\eta \lambda)+ \Ps_m(\eta\lambda)\eta\zeta)y,
	\end{equation}
	where $P_m(z)$ is given in \cref{eq:defPm},
	\begin{equation}\label{eq:PhimTfrac}
	\Ps_m(z) = \frac{P_m(z)-1}{z} \quad \text{for } z\neq 0
	\end{equation}
	and $\Ps_m(0)=1$ is defined by continuous extension.
\end{lemma}
\begin{proof}
A generalization of \cite[Proposition 3.1]{HaW02} to the equation $u'=\lambda u+\zeta y$ (instead of $u'=\lambda u$) yields
\begin{equation}
	u_\eta = P_m(z)y+\eta b^\top (I-zA)^{-1}\mathds{1}\zeta y,
\end{equation}
where $z=\eta\lambda$, $I\in \Rb^{m\times m}$ is the identity matrix, $A,b$ are the coefficients of the Butcher tableau of the $m$-stage RKC scheme and $\mathds{1}\in \Rb^m$ is a vector of ones. Since $P_m(z)=1+z b^\top (I-zA)^{-1} \mathds{1}$ \cite[Proposition 3.1]{HaW02} the result follows.
\end{proof}

Note the similarity between \cref{eq:solutesteq} and \cref{eq:numutesteq}, with $e^z,\varphi(z)$ replaced by $P_m(z),\Ps_m(z)$, respectively.
In \cref{fig:Phim}, we also observe that $\Ps_m(z)$ and $\varphi(z)$ share similar stability properties. Indeed, $\Ps_m(z)$ is the numerical counterpart of $\varphi(z)$,

yet with the exponential replaced by the stability polynomial -- compare \cref{eq:defphi,eq:PhimTfrac}.
\begin{figure}
		\begin{subfigure}[t]{\subfigsize\textwidth}
			\centering
			\begin{tikzpicture}[scale=\plotimscale]
			\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth, ymin=-0.2, ymax=1,xmax=0,xmin=-136,legend columns=1,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0,1)},anchor=north west},legend cell align={left},
			xlabel={$z$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
			\addplot[color=\colorone,line width=\plotlinewidth pt,mark=\markone,mark repeat=32,mark phase=20,mark size=\plotmarksizeu pt] table [x=z,y=phi,col sep=comma]
			{data/text/Phi_8.csv};\addlegendentry{$\varphi(z)=(e^z-1)/z$}
			\addplot[color=\colortwo,line width=\plotlinewidth pt,mark=\marktwo,mark repeat=32,mark phase=4,mark size=\plotmarksizeu pt] table [x=z,y=Phim,col sep=comma]
			{data/text/Phi_8.csv};\addlegendentry{$\Ps_8(z)=(P_8(z)-1)/z$}
			\addplot[black,dashed,line width=\plotdashedlinewidth pt,domain=-0.2:1] (-123.91,x);
			\end{axis}
			\end{tikzpicture}

		\end{subfigure}\hfill
		\begin{subfigure}[t]{\subfigsize\textwidth}
			\centering
			\begin{tikzpicture}[scale=\plotimscale]
			\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth, ymin=-0.2, ymax=1,xmax=0,xmin=-172.26,legend columns=1,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0,1)},anchor=north west},legend cell align={left},
			xlabel={$z$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
			\addplot[color=\colorone,line width=\plotlinewidth pt,mark=\markone,mark repeat=32,mark phase=20,mark size=\plotmarksizeu pt] table [x=z,y=phi,col sep=comma]
			{data/text/Phi_9.csv};\addlegendentry{$\varphi(z)=(e^z-1)/z$}
			\addplot[color=\colortwo,line width=\plotlinewidth pt,mark=\marktwo,mark repeat=32,mark phase=4,mark size=\plotmarksizeu pt] table [x=z,y=Phim,col sep=comma]
			{data/text/Phi_9.csv};\addlegendentry{$\Ps_{9}(z)=(P_{9}(z)-1)/z$}
			\addplot[black,dashed,line width=\plotdashedlinewidth pt,domain=-0.2:1] (-156.82,x);
			\end{axis}
			\end{tikzpicture}
		\end{subfigure}
	\caption{Illustration of $\varphi(z)$ and $\Ps_m(z)$ for $m=8$ (left) and $m=9$ (right). The dashed line indicates the end of the stability domain.}
	\label{fig:Phim}
\end{figure}

We can now compute the stability polynomial of the mRKC scheme. From \cref{eq:defbfe,eq:numutesteq,eq:PhimTfrac}, we get
\begin{equation}\label{eq:bfetesteq}
\bfe(y)= \frac{1}{\eta}(\Pi_m(\eta\lambda)+\Ps_m(\eta\lambda)\eta\zeta-1)y  =\Ps_m(\eta\lambda)(\lambda+\zeta)y,
\end{equation}
which is the numerical counterpart of $\fe$ in \cref{eq:fetesteq}. Now, we insert \cref{eq:bfetesteq} into \cref{eq:defmrkc}, which leads to
\begin{equation}
y_{n+1} = \Pe_s(\tau \Ps_m(\eta\lambda)(\lambda+\zeta))y_n,
\end{equation}
with $R_s(z)$ the stability polynomial of the $s$-stage RKC scheme defined in \cref{sec:rkc}, and hence motivates the following definition.
\begin{definition}
	Let $s,m\in\Nb$, $\tau>0$ be a step size, $\eta>0$ and $\lambda,\zeta\leq 0$. The stability polynomial of the $(s,m)$-stage mRKC scheme \eqref{eq:defmrkc}--\eqref{eq:defbue} is defined as
	\begin{equation}\label{eq:defstabpol}
	R_{s,m}(\lambda,\zeta,\tau,\eta)=R_s(\tau \Ps_m(\eta\lambda)(\lambda+\zeta)),
	\end{equation}
with $R_s(z)$ as in \cref{eq:stabpolrkc} and $\Phi_m(z)$ as in \cref{eq:PhimTfrac}.
\end{definition}
The following lemma is the discrete version of \cref{lemma:cont_stab} and is needed to prove stability of the mRKC scheme in \cref{thm:stab_mrkc} below. Its proof is purely technical and postponed to \cref{app:proofs}.

\begin{lemma}\label{lemma:disc_stab}
	Let $m\in\Nb$ and $w\leq 0$. There exists $\overline\varepsilon_m>0$ such that for $\varepsilon\leq\overline\varepsilon_m$, $\Ps_m(z)(z+w)\in [w,0]$ for all $z\in [-\ell_m^\varepsilon,0]$ if, and only if, $\Ps_m'(0)|w|\geq 1$, i.e.  $|w|\geq 2/P_m''(0)$ since $\Ps_m'(0)=P_m''(0)/2$.
\end{lemma}
For $\varepsilon=0$, it holds $2/\Pi_m''(0)=6 m^2/(m^2-1)>6$. In the continuous setting, the condition on $w$ in \cref{lemma:cont_stab} was $|w|\geq 2$. For the discrete mRKC scheme, however, $|w|>6$ is necessary because of the milder slope of $\Ps_m(z)$ at the origin, see \cref{fig:Phim}.
\begin{remark}\label{rem:scalesep}
	In the case of scale separation, $z$ is bounded away from the origin and the value of $\Phi_m(z)$ thus considerably smaller than $1$, see \cref{fig:Phim}. Hence, the condition $\Ps_m(z)(z+w)\in [w,0]$ is already satisfied for $|w|\geq 2$ and a slightly larger damping $\overline\varepsilon=0.1$, so that the larger value $|w|\geq 2/P_m''(0)\approx 6$, required by \cref{lemma:disc_stab}, is no longer necessary. By allowing $|w|\geq 2$ instead of $|w|\geq 2/P_m''(0)$ in \cref{thm:stab_mrkc} below, we find that the weaker stability conditions \cref{eq:defsmetaweak} already guarantee stability in the case of scale separation -- see \cite[Section 3.4.5]{Ros20} for further details.
\end{remark}
\begin{theorem}\label{thm:stab_mrkc}
Let $\overline\varepsilon_m$ be as in \cref{lemma:disc_stab} and, for $\varepsilon\geq 0$, let $\varepsilon_m=\min\{\varepsilon,\overline{\varepsilon}_m\}$. Let $\lambda\leq 0$ and $\zeta<0$. Then, for all $\tau>0,s,m$ and $\eta$ such that
\begin{align}\label{eq:defsmetascalar}
\tau |\zeta| &\leq\ell_s^\varepsilon, & \eta|\lambda| &\leq \ell^{\varepsilon_m}_m &\text{with}  && \eta \geq \frac{6\tau}{\ell_s^\varepsilon}\frac{m^2}{m^2-1},
\end{align}
$|\Pe_{s,m}(\lambda,\zeta,\tau,\eta)|\leq 1$, i.e. the mRKC scheme is stable.
\end{theorem}
\begin{proof}
If $\tau \Ps_m(\eta\lambda)(\lambda+\zeta)\in [-\ell_s^\varepsilon,0]$ then $|\Pe_{s,m}(\lambda,\zeta,\tau,\eta)|=|\Pe_s(\tau \Ps_m(\eta\lambda)(\lambda+\zeta))|\leq 1$. Hence, it is sufficient to prove the equivalent condition:
\begin{align}
\Phi_m(\eta\lambda)(\eta\lambda+\eta\zeta)&\in [w(\eta),0], & \mbox{with} && w(\eta)&=-\frac{\eta}{\tau}\ell_s^\varepsilon.
\end{align}
Since $\eta\lambda\in [-\ell_m^{\varepsilon_m},0]$, it holds $|\Pi_m(\eta\lambda)|\leq 1$ and from \eqref{eq:PhimTfrac} we thus deduce that $\Ps_m(\eta\lambda)\geq 0$. Furthermore, \eqref{eq:defsmetascalar} yields $\eta\zeta\geq w(\eta)$ which implies
\begin{equation}
0\geq \Phi_m(\eta\lambda)(\eta\lambda+\eta\zeta)\geq  \Phi_m(z(\eta))(z(\eta)+w(\eta)),
\end{equation}
with $z(\eta)=\eta\lambda$. Hence, it is sufficient to show that $\Phi_m(z(\eta))(z(\eta)+w(\eta))\in [w(\eta),0]$ for all $z(\eta)\in [-\ell^{\varepsilon_m}_m,0]$. From \cref{lemma:disc_stab}, we know that
\begin{equation}
|w(\eta)| \geq  \frac{2}{\Pi_m''(0)}
\end{equation}
is necessary and sufficient. Since $T_m'(v_0)^2/(T_m(v_0)T_m''(v_0))$ is decreasing for $v_0\geq 1$ (see \cref{lemma:ddRincr}), we infer from the definition of $\eta$ in \eqref{eq:defsmetascalar} that
\begin{equation}
|w(\eta)|\geq 6\frac{m^2}{m^2-1}=\frac{2\, T_m'(1)^2}{T_m(1)T_m''(1)}\geq \frac{2\, T_m'(v_0)^2}{T_m(v_0)T_m''(v_0)}=\frac{2}{\Pi_m''(0)}. \tag*{\qedhere}
\end{equation}
\end{proof}

In the continuous setting in \cref{sec:stab_modeq}, $\eta$ directly depends on $\fs$; indeed, the
condition $|\varphi(\eta\lambda)(\lambda+\zeta)|\leq |\zeta|$ implies $\eta\geq 2/|\zeta|$ (see \cref{thm:stifflinscamod}). Therefore, $\eta$ could rapidly grow as $\zeta\to 0$. In contrast, for the mRKC method, $\eta$ depends only indirectly on $\fs$: $\eta$ depends on the $s$-stage RKC method, which in turn depends on $\fs$. This indirect dependence of $\eta$ on $\fs$ creates a ``protective buffer'', which prevents the explosion of $\eta$ as $\zeta\to 0$; indeed, $\ell_s^\varepsilon\geq 2$ for all $s\in\Nb$.

The restriction $\varepsilon\leq \overline\varepsilon_m$ is necessary for proving \cref{lemma:disc_stab}, but probably not needed in practice. Indeed, we have verified numerically that for any $\varepsilon \geq 0$, $\Ps_m(z)(z+w)\in [w,0]$ for all $z\in [-\ell_m^\varepsilon,0]$ if, and only if, $|w|\geq 2/\Pi_m''(0)$. Hence, we can suppose $\varepsilon_m=\varepsilon$ in \eqref{eq:defsmetascalar} and replace $\ell_s^\varepsilon,\ell_m^{\varepsilon_m}$ by $\beta s^2,\beta m^2$, respectively, which yields \eqref{eq:defsmeta}.
In \cref{fig:Rsm}, we display the stability polynomial $R_{s,m}(\lambda,\zeta,\tau,\eta)$ for $s=5$ and $m=3$ as a function of $\lambda$ for $\varepsilon=0.05$ or $\varepsilon=1$. Here, we set $\tau=1$, $\eta$ to its lower bound in \eqref{eq:defsmetascalar}, and $\zeta=-\ell_s^\varepsilon,-\ell_s^\varepsilon/2$ or $0$. Since $|R_{s,m}(\lambda,\zeta,\tau,\eta)|\leq 1$, the mRKC method is always stable.
\begin{figure}
		\begin{subfigure}[t]{\subfigsize\textwidth}
			\centering
			\begin{tikzpicture}[scale=\plotimscale]
			\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth, ymin=-1.0, ymax=1.0,xmax=0,xmin=-125.08,legend columns=3,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0.5,1.05)},anchor=south},legend cell align={left},
			xlabel={$\lambda$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
			\addplot[color=\colorone,line width=\plotlinewidth pt,mark=\markone,mark repeat=24,mark phase=0,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm1,col sep=comma]
			{data/text/R_5_3_damp_0p05.csv};\addlegendentry{$\zeta=-\ell^\varepsilon_s$}
			\addplot[color=\colortwo,line width=\plotlinewidth pt,mark=\marktwo,mark repeat=24,mark phase=8,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm2,col sep=comma]
			{data/text/R_5_3_damp_0p05.csv};\addlegendentry{$\zeta=-\ell^\varepsilon_s/2$}
			\addplot[color=\colorthree,line width=\plotlinewidth pt,mark=\markthree,mark repeat=24,mark phase=16,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm3,col sep=comma]
			{data/text/R_5_3_damp_0p05.csv};\addlegendentry{$\zeta=0$}
			\end{axis}
			\end{tikzpicture}
			\caption{Small damping $\varepsilon=0.05$.}
			\label{fig:Rsma}
		\end{subfigure}\hfill
		\begin{subfigure}[t]{\subfigsize\textwidth}
			\centering
			\begin{tikzpicture}[scale=\plotimscale]
			\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth, ymin=-1.0, ymax=1.0,xmax=0,xmin=-54.36,legend columns=3,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0.5,1.05)},anchor=south},legend cell align={left},
			xlabel={$\lambda$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
			\addplot[color=\colorone,line width=\plotlinewidth pt,mark=\markone,mark repeat=24,mark phase=0,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm1,col sep=comma]
			{data/text/R_5_3_damp_1.csv};\addlegendentry{$\zeta=-\ell^\varepsilon_s$}
			\addplot[color=\colortwo,line width=\plotlinewidth pt,mark=\marktwo,mark repeat=24,mark phase=8,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm2,col sep=comma]
			{data/text/R_5_3_damp_1.csv};\addlegendentry{$\zeta=-\ell^\varepsilon_s/2$}
			\addplot[color=\colorthree,line width=\plotlinewidth pt,mark=\markthree,mark repeat=24,mark phase=16,mark size=\plotmarksizeu pt] table [x=lambdas,y=Rsm3,col sep=comma]
			{data/text/R_5_3_damp_1.csv};\addlegendentry{$\zeta=0$}
			\end{axis}
			\end{tikzpicture}
			\caption{High damping $\varepsilon=1$.}
			\label{fig:Rsmb}
		\end{subfigure}
	\caption{Stability polynomial $R_{s,m}(\lambda,\zeta,\tau,\eta)$ of the mRKC method vs. $\lambda$ for $\zeta=-\ell_s^\varepsilon,-\ell_s^\varepsilon/2$ or $0$ and $s=5$, $m=3$, $\tau=1$, $\eta$ as in \cref{eq:defsmetascalar} and damping $\varepsilon=0.05$ (left) or $\varepsilon=1$ (right).}
	\label{fig:Rsm}
\end{figure}

\subsubsection*{Stability analysis for a $2\times 2$ model problem}
Here, we consider a $2\times 2$ linear model problem where the Jacobians of $\ff$ and $\fs$ are not simultaneously triangularizable. Then, the stability analysis cannot be reduced to the scalar multirate test equation \eqref{eq:mtesteq}, yet we shall show that the same stability conditions still hold. Moreover, we introduce a coupling term between the fast and slow variables and show that the same stability conditions are necessary even when the coupling is weak.

Thus, we consider the system of differential equations
\begin{align}\label{eq:twodim}
y'&=Ay, &\mbox{with} && A&=\begin{pmatrix}
\zeta & \sigma\\ \sigma & \lambda
\end{pmatrix}
\end{align}
and $y(0)=y_0\in\Rb^2$. We let $\lambda,\zeta<0$, $\sigma\in\Rb$ the coupling term, and assume that $\sigma^2\leq\lambda\zeta$ to ensure that both eigenvalues of $A$ are negative or zero. We note $D\in\Rb^{2\times 2}$ the diagonal matrix satisfying $D_{11}=0$ and $D_{22}=1$ and consider the splitting defined by $\ff(y)=A_F\, y$ and $\fs(y)=A_S\, y$, where
\begin{align} \label{eq:tauAfAc1}
A_F &:=DA = \begin{pmatrix}
0 & 0 \\ \sigma & \lambda
\end{pmatrix},
&
A_S &:= (I-D)A =\begin{pmatrix}
\zeta & \sigma \\ 0 & 0
\end{pmatrix}.
\end{align}
Observe that $\rhof=|\lambda|$ and $\rhos=|\zeta|$. The matrices $A_F,A_S$ are simultaneously triangularizable if, and only if, they have a common eigenvector, which occurs only for $\sigma=0$ or $\sigma^2=\lambda\zeta$. We set $\sigma=0.1\sqrt{\lambda\zeta}$, so that the present stability analysis cannot be reduced to the scalar multirate test equation \cref{eq:mtesteq}. Furthermore, as the eigenvalues of $A$ are negative or zero for all $|\sigma|\leq \sqrt{\lambda\zeta}$, the current coupling $\sigma=0.1\sqrt{\lambda\zeta}$ can be considered to be weak when compared to the maximal coupling $\sqrt{\lambda\zeta}$.

Given $y\in \Rb^2$, we obtain $\bfe(y)$ by replacing $\lambda,\zeta$ in \eqref{eq:bfetesteq} by $A_F,A_S$, respectively. This yields
\begin{align}\label{eq:defAeta}
\bfe(y) &= A_\eta y, &\mbox{with}&& A_\eta&=\Ps_m(\eta A_F)A y,
\end{align}
and since $\Ps_m$ is a polynomial, $A_\eta$ is well-defined. From \eqref{eq:defmrkc} it follows $y_{n+1} = \Pe_s(\tau A_\eta)y_n$. If the eigenvalues of $\tau A_\eta$ are in the interval $[-\ell_s^\varepsilon,0]$, the mRKC method is stable. For convenience, we set $\tau=1$, $|\zeta|=\ell_s^\varepsilon$ with $s=10$, and also fix $m=8$ and $\eta=\frac{6\tau}{\ell_s^\varepsilon}\frac{m^2}{m^2-1}$ (as in \eqref{eq:defsmetascalar}). Then, the mRKC method is stable if the spectral radius $\rho_\eta$ of $A_\eta$ satisfies $\rho_\eta\leq |\zeta|$ for all $\eta\lambda\in [-\ell_m^\varepsilon,0]$, or equivalently $\eta\rhoe\leq \eta|\zeta|= |w|$.

In \cref{fig:stabweakcoupa}, we display $\eta\rhoe$ as a function of $z=\eta\lambda\in [-\ell_m^\varepsilon,0]$ and observe that $\eta\rhoe\leq |w|$; thus, the mRKC scheme is stable. Hence, the stability conditions \eqref{eq:defsmeta} guarantee stability of the scheme even though the Jacobians of $\ff,\fs$ are not simultaneously triangularizable.

Next, in \cref{fig:stabweakcoupb}, we consider a value of $\eta$ smaller than that dictated by \eqref{eq:defsmetascalar}. For $\overline\eta=0.9\eta$, we again display $\overline\eta\rho_{\overline\eta}$ as a function of $\overline z=\overline\eta\lambda\in [-\ell_m^\varepsilon,0]$. Then, a small region of instability appears for $\overline z$ close to zero, where $\overline\eta\rho_{\overline\eta}>|\overline w|$. Hence, the stability conditions \eqref{eq:defsmetascalar} are necessary even for systems of equations with a weak coupling $\sigma=0.1\sqrt{\lambda\zeta}$, where $\sqrt{\lambda\zeta}$ corresponds to the maximal coupling strength.
Similar instabilities as in \cref{fig:stabweakcoupb} occur for even weaker couplings
$\sigma=0.01\sqrt{\lambda\zeta}$, $\sigma=0.001\sqrt{\lambda\zeta}$ and for larger $\overline\eta=0.95\eta$.

\begin{figure}
		\begin{subfigure}[t]{\subfigsize\textwidth}
			\centering
			\begin{tikzpicture}[scale=\plotimscale,every pin/.style={fill=white}]
			\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth,xmax=0,xmin=-124,legend columns=1,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0,1)},anchor=north west},legend cell align={left},
			xlabel={$\phantom{\overline z}z\phantom{\overline z}$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
			\addplot[color=\colorone,line width=\plotlinewidth pt,mark=none] table [x=etalambda,y=rhoB,col sep=comma]
			{data/text/stab_weak_coupling_a.csv};\addlegendentry{$\eta\rhoe$}
			\addplot[color=black,dashed,line width=\plotdashedlinewidth pt] table [x=etalambda,y=w,col sep=comma]
			{data/text/stab_weak_coupling_a.csv};\addlegendentry{$|w|$}

			\coordinate (pta) at (axis cs:-7.3,6.0952);
			\coordinate (ptb) at (axis cs:-30,6.078);
			\end{axis}
			\draw[] (pta)node[circle,anchor=west,draw]{}--(ptb)node[anchor=east,draw,circle,fill=white,inner sep=0,outer sep=0]{
				\begin{tikzpicture}[scale=\zoomsize,trim axis left,trim axis right]
				\begin{axis}[tiny,hide axis,xlabel={},ylabel={},ticks=none,xmin=-1,xmax=0]
				\addplot[color=\colorone,line width=\zoomlinewidth pt,mark=none] table [x=etalambda,y=rhoB,col sep=comma]
				{data/text/stab_weak_coupling_a_zoom.csv};
				\addplot[color=black,dashed,line width=\zoomdashedlinewidth pt] table [x=etalambda,y=w,col sep=comma]
				{data/text/stab_weak_coupling_a_zoom.csv};
				\end{axis}
				\end{tikzpicture}
			};
		\end{tikzpicture}
		\caption{$\eta\rhoe$ vs. $z$, for $\eta$ as in \cref{eq:defsmetascalar}.}
		\label{fig:stabweakcoupa}
	\end{subfigure}\hfill
	\begin{subfigure}[t]{\subfigsize\textwidth}
		\centering
		\begin{tikzpicture}[scale=\plotimscale,every pin/.style={fill=white}]
		\begin{axis}[height=\aspectratio*\plotimsized\textwidth,width=\plotimsized\textwidth,xmax=0,xmin=-124,legend columns=1,legend style={draw=\legendboxdraw,fill=\legendboxfill,at={(0,1)},anchor=north west},legend cell align={left},
		xlabel={$\overline z$}, ylabel={},label style={font=\normalsize},tick label style={font=\normalsize},legend image post style={scale=\legendmarkscale},legend style={nodes={scale=\legendfontscale, transform shape}},grid=none]
		\addplot[color=\colorone,line width=\plotlinewidth pt,mark=none] table [x=etalambda,y=rhoB,col sep=comma]
		{data/text/stab_weak_coupling_b.csv};\addlegendentry{$\overline\eta\rho_{\overline\eta}$}
		\addplot[color=black,dashed,line width=\plotdashedlinewidth pt] table [x=etalambda,y=w,col sep=comma]
		{data/text/stab_weak_coupling_b.csv};\addlegendentry{$|\overline w|$}

		\coordinate (ptc) at (axis cs:-7.3,5.4857);
		\coordinate (ptd) at (axis cs:-30,5.47);
		\end{axis}

		\draw[] (ptc)node[circle,anchor=west,draw]{}--(ptd)node[anchor=east,draw,circle,fill=white,inner sep=0,outer sep=0]{
			\begin{tikzpicture}[scale=\zoomsize,trim axis left,trim axis right]
			\begin{axis}[tiny,hide axis,xlabel={},ylabel={},ticks=none,xmin=-1,xmax=0]
			\addplot[color=\colorone,line width=\zoomlinewidth pt,mark=none] table [x=etalambda,y=rhoB,col sep=comma]
			{data/text/stab_weak_coupling_b_zoom.csv};
			\addplot[color=black,dashed,line width=\zoomdashedlinewidth pt] table [x=etalambda,y=w,col sep=comma]
			{data/text/stab_weak_coupling_b_zoom.csv};
			\end{axis}
			\end{tikzpicture}
		};
	\end{tikzpicture}
	\caption{$\overline\eta\rho_{\overline\eta}$ vs. $z$, for $\overline\eta=0.9 \eta$.}
	\label{fig:stabweakcoupb}
\end{subfigure}
\caption{Verification that conditions \cref{eq:defsmetascalar} are sufficient and necessary for the stability of mRKC applied to the $2\times 2$ test problem \cref{eq:twodim}.}
\label{fig:stabweakcoup}
\end{figure}

\subsection{Convergence analysis}
We end this section by proving that the mRKC scheme is first-order accurate.
\begin{theorem}\label{thm:firstorder}
The mRKC scheme is first-order accurate.
\end{theorem}

\begin{proof}
We estimate the local error after one step. From \cref{def:fe} with $y$ replaced by $y_0$ in \cref{eq:defu} follows $u(\eta) = y_0 + \eta f(y_0)+\bigo{\eta^2}$ and thus $\fe(y_0) = f(y_0)+\bigo{\eta}$.
Let $y(\tau)$ and $\ye(\tau)$ be the solutions of \cref{eq:ode,eq:odemod} at time $\tau$, respectively, then
\begin{equation}
	y(\tau)-\ye(\tau) = \tau(f(y_0)-\fe(y_0))+\bigo{\tau^2} = \bigo{\eta\tau+\tau^2}.
\end{equation}
Let $\bar y_\eta(\tau)$ be the solution of $\bar y_\eta' = \bfe(\bar y_\eta)$ with $\bar y_\eta(0)=y_0$ and $\bfe$ as in \cref{eq:defbfe,eq:defbue}. Since the RKC scheme \cref{eq:defbue} is first-order accurate then $u_\eta=u(\eta)+\bigo{\eta^2}$ and $\bfe(y_0)=\fe(y_0)+\bigo{\eta}$, which yields
\begin{equation}
	\ye(\tau) -\bar y_\eta(\tau) = \tau(\fe(y_0)-\bfe(y_0))+\bigo{\tau^2} = \bigo{\eta\tau+\tau^2}.
\end{equation}
Finally, let $y_1$ be the solution after one step of the mRKC scheme \cref{eq:defmrkc,eq:defbfe,eq:defbue}, which can also be seen as the solution after one step of the RKC scheme applied to $\bar y_\eta' = \bfe(\bar y_\eta)$. Using the fact that the RKC scheme \cref{eq:defmrkc} has first-order accuracy then
\begin{equation}
	\bar y_\eta(\tau) -y_1 =  \bigo{\tau^2}.
\end{equation}
By triangular inequality we obtain $|y(\tau)-y_1|= \bigo{\eta\tau+\tau^2}$ and from \cref{eq:defsmeta} follows $\eta\leq 8\tau$, thus $|y(\tau)-y_1|= \bigo{\tau^2}$ and the scheme is first-order accurate.
\end{proof}
Typically $s\gg 1$, i.e. $\eta\ll\tau$, and the error made when approximating $f$ by the averaged force $\fe$ is negligible. In fact, we observe that the difference between the RKC and the mRKC solutions in our numerical experiments in \cref{sec:numexp} is always very small.


TITLE: Algebra. Permutations with composition proof.
QUESTION [2 upvotes]: Set $X := \lbrace 1,2,3 \rbrace$. Denote the idenity $\text{id}_X$ of $X$ by 1. Define $1\phi :=1$, $2\phi :=3$, $3\phi = 2$, $1\psi =2$, $2\psi = 1$, $3\psi = 3$. Prove that $\phi \psi \phi= \psi \phi \psi$, abbreviate $\delta := \phi \psi \phi$, and finish the following table by computing all composites.
$$
\begin{array}{l||c|c|c|c|c|c}
 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline \hline
1 & 1 & \phi & \psi &  & & \\
\hline
\phi & \phi & 1 & \phi \psi &  & & \\
\hline
\psi & \psi & \psi \phi & & & & \\
\hline
\phi \psi  & & & & & & \\
\hline
\psi \phi & & & & & & \\
\hline
\delta & & & & & &
\end{array}
$$
Quick question how do I show they are equal? Do I just do it the old fashion way by choosing an element in $X$ and just computing 6 times or is there a more formal way to go about this?
For the table I can not get it to show up, but how would I evaluate things like $\phi^2$ or $\phi \psi$ ?

REPLY [3 votes]: The objects of interest here are the functions acting on the set $X$, which can be combined by composition
$$
a \alpha \beta = \beta(\alpha(a)) \quad a \in X \quad  \alpha, \beta \in X^X = \{ \varphi \,\left|\, \varphi : X \to X \right. \}
$$
the combined composition function is what is meant by the short hand product
$$
\alpha \beta = \beta \circ \alpha
$$
To get the first foot on the ground one looks how a function acts on the elements of the set. One finds that certain compositions of functions result in the same mapping of set elements like some others and thust one identifies them as the same composition. Like $\phi^2=1$ (the identity function).
Using tuples to act on several elements at once to avoid writing down three individual equations:
$$
(1,2,3) \varphi =: \left( \varphi(1), \varphi(2), \varphi(3) \right) = (x, y, z) \iff \\
\varphi(1) = x \wedge \varphi(2) = y \wedge \varphi(3) = z
$$
one gets:
$$
(1,2,3) \phi = (1,3,2) \\
(1,2,3) \psi = (2,1,3) \\
(1,2,3) \phi \psi = (1,3,2) \psi = (2,3,1) \\
(1,2,3) \psi \phi = (2,1,3) \phi = (3,1,2) \\
(1,2,3) \phi \psi \phi = (2,3,1) \phi = (3,2,1) \\
(1,2,3) \psi \phi \psi = (3,1,2) \psi = (3,2,1) \\
\implies \phi \psi \phi = \psi \phi \psi =: \delta
$$
The functions act as permutations on $X$, there are $3!=6$ different possible.
From here on one should mostly need the associative law of function composition
$$
(\alpha \beta) \gamma =
\gamma \circ (\beta \circ \alpha) =
(\gamma \circ \beta) \circ \alpha =
\alpha (\beta \gamma)
$$
together with the already discovered identities:
$$
\delta^2 = \delta \delta =
(\phi \psi \phi)(\phi \psi \phi) =
\phi (\psi (\phi \phi) \psi) \phi =
\phi(\psi \psi) \phi =
\phi \phi =
1 \\
\delta \phi = (\phi \psi \phi) \phi = \phi \psi \\
\phi \delta = \phi (\phi \psi \phi) = \psi \phi \\
\delta \psi = (\phi \psi \phi) \psi = \phi (\psi \phi \psi) =
\phi \delta  = \psi \phi \\
\psi \delta = \psi (\psi \phi \psi) = \phi \psi
$$
Putting all this in the table gives:
$$
\begin{array}{l||c|c|c|c|c|c}
 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline \hline
1 & 1 & \phi & \psi & \phi \psi & \psi \phi & \delta \\
\hline
\phi & \phi & 1 & \phi \psi & \psi & \delta & \psi \phi \\
\hline
\psi & \psi & \psi \phi & 1 & \delta & \phi & \phi \psi \\
\hline
\phi \psi  & \phi \psi & \delta & \phi & \psi \phi & 1 & \psi \\
\hline
\psi \phi & \psi \phi & \psi & \delta & 1 & \phi \psi & \phi \\
\hline
\delta & \delta & \phi \psi & \psi \phi & \phi & \psi & 1
\end{array}
$$

REPLY [2 votes]: $\phi^2$ can be calculated by taking the compositum of the function $\phi$ with itself. this means that $$1(\phi^2) = (1\phi)\phi = 1\phi = 1\\
2(\phi^2) = (2\phi)\phi = 3\phi = 2\\
3(\phi^2) = (3\phi)\phi = 2\phi = 3,$$
which proves that $\phi^2=1$  (which is also written in your table).


TITLE: Evaluating a summation
QUESTION [1 upvotes]: I am trying to solve a homework question in which a part involves the evaluation of a summation.
The summation is:
$$
\sum_{i=0}^n2^{2i+1}
$$
and this is my attempt which i am stuck at. Any lead into this would be help full.
$$
\sum_{i=0}^n2^{2i+1} = \sum_{i=0}^n (2^{2i})(2) = 2\sum_{i=0}^n2^{2i}
$$
when i expand this out like so
$$
2\sum_{i=0}^n2^{2i} = (2)(1+2^2+2^4+2^6 + ... + 2^{2n})
$$
I don't see or know any existing summation that can help me solve this or any way to use pertubation to solve this. Again any lead into helping me out will go a long way
Thanks in advance!

REPLY [1 votes]: You could have a direct calculation by using the index of the summation. Let
$$
S=\sum_{i=0}^n2^{2i+1}=2\sum_{i=0}^n2^{2i}.
$$
We have
$$
2S=2\sum_{i=0}^n2^{2i+1}=\sum_{i=0}^n2^{2i+2}=\sum_{i=1}^{n+1}2^{2i}=\sum_{i=0}^n2^{2i}-1+{2}^{2n+2}=\frac{S}{2}-1+2^{2n+2}.
$$
Solve the equation with variable $S$ we obtain
$$
S=\frac{2}{3}(2^{2n+2}-1).
$$


TITLE: Which function is appropriate to analyse, using parseval's identity, to sum $\frac{1}{n^2-a^2}$?
QUESTION [1 upvotes]: If I wanted to find: $$\sum_{n=0}^\infty\frac{1}{n^2+a^2}$$
For some non-zero real number $a$, then my task is powerfully done with Fourier series:
My derivation:

The complex Fourier coefficients for the function $\exp(ax)$ over $[-\pi,\pi]$ are given by:
$$c_n=\frac{1}{2\pi}\int_{-\pi}^\pi\exp(x(a-in))\,\mathrm{d}x=\frac{(-1)^n}{\pi}\cdot\frac{a+in}{a^2+n^2}\cdot\sinh(\pi a)$$
And we have:
$$\sum_{n\in\Bbb Z}\frac{\sinh^2(\pi a)}{\pi^2}\cdot\frac{1}{a^2+n^2}=\sum_{n\in\Bbb Z}|c_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi\exp(2ax)\,\mathrm{d}x=\frac{1}{2\pi a}\sinh(2\pi a)\\\sum_{n\in\Bbb Z}\frac{1}{n^2+a^2}=\frac{\pi}{2a}\frac{2\sinh(\pi a)\cosh(\pi a)}{\sinh^2(\pi a)}=\frac{\pi}{a}\cdot\coth(\pi a)\\\cdots\\\sum_{n=0}^\infty\frac{1}{n^2+a^2}=\frac{1}{2a^2}\cdot(1+\pi a\coth\pi a)$$

A formula I was proud to find (and corroborate on Desmos). Eagerly, I then went to derive a closed formula for the sum of $\frac{1}{n^2-a^2}$, where $a$ is a real but not an integer, and hit a brick wall. Analysing the function $\exp(iax)$ seems promising, but in the expression for $|c_n|^2$, one gets a term something like: $$\frac{1}{(n-a)^2}$$ Which is not quite what I am trying to sum! I am not well educated in Fourier series, so I was wondering if anyone knew an appropriate function to analyse, to yield: $$\sum_{n=0}^\infty\frac{1}{n^2-a^2},\,a\in\Bbb R\setminus\Bbb Z$$
Indeed, if one comes across a sum in the wild, (supposing it's Fourier-summable), how does one think of the right function to analyse? Are there patterns to spot, or is it down to experience?

REPLY [2 votes]: $\sum_{n=0}^\infty 1/(n^2 + z^2)$ is an analytic function on $\mathbb C \backslash  \mathbb Z i$.  So is $(1+\pi z \coth(\pi z))/(2 z^2)$.  If they agree for real nonzero $z$, then they must agree on the whole domain, in particular for $z=ia$.  Thus we should have
$$ \sum_{n=0}^\infty \frac{1}{n^2-a^2} = \frac{1+\pi i a \coth(\pi i a)}{2 (ia)^2} =- \frac{1+\pi a \cot(\pi a)}{2a^2} $$


TITLE: Expectation of Wiener Measure
QUESTION [2 upvotes]: Fix $T>0$. Write $C$ for the space of all continuous functions $f:[0,T] \to \mathbb{R}$. For a standard Brownian motion $B = \{B_t\}_{t \in [0,T]}$ defined on probability space $(\Omega,F,P)$, let $W$ be the classical Wiener measure. That is, we treat $B$ as a random variable $B: \Omega \to C$ and $B(\omega)\in C$ is the continuous sample path, so $W$ is the distribution law of $B$: if $E \subseteq C$ is a measurable set, then $W(E) = P(\{\omega \in \Omega | B(\omega) \in E\})$.
Now fix $t \in [0, T]$. I am asked to compute this integral:
\begin{equation*}
\int_{C}\phi(t)dW(\phi)
\end{equation*}
How do I even start? I have no idea how to approach such a problem. In a special case which I integrate on a finite set $E=\{\phi_1,\cdots,\phi_n\} \subseteq C$, I can write:
\begin{equation*}
\int_{E}\phi(t)dW(\phi) = \sum_{i=1}^n\phi_i(t)W(\{\phi_i\}) = \sum_{i=1}^n\phi_i(t) P\{B(\omega) = \phi_i\}
\end{equation*}
Even in such a special case I don't know how to calculate $P\{B(\omega) = \phi_i\}$ .

REPLY [2 votes]: This all may become simpler to grasp if you add some formalism. You have a measure $W$ on $C$ and you are asked to compute an integral
$$
\int_C e_t(\phi)W(\mathrm d\phi)
$$
where $e_t:C \to \Bbb R$ is the evalution map  that assigns to each function $\phi\in C$ its value $\phi(t) \in \Bbb R$. So you can treat $(C,\mathcal C, W)$ as a probability space and hence $e_t$ would just be a random variable, whose expectation you are asked to find. You very well know the distribution of this random variable:
$$
W(\phi:e_t(\phi) \in A) = P(\omega: B(t, \omega) \in A) = \mathcal N(0, t)(A).
$$
The rest should be easy.


TITLE: Relationship between Radius of curvature and Instantaneous Axis of Rotation
QUESTION [5 upvotes]: Consider a Ring (radius $r$) rolling without slipping with angular speed $\omega$ on the surface. We want to find the radius of curvature of the top most point.
1.The radius of curvature(R): Regardless of the actual path that the particle travels, a particle at every instant can be thought of as tracing a circle, of radius=$R$. The top most point has, at the current instant, Velocity = $2\omega r \hat{i}$, and acceleration =-$\omega^2 R\hat{j}$. Using the radius of curvature formula: $R= v^2/a_{normal}$, we get $R=4r$.Which matches my textbook, and also fact that the radius of curvature of a cycloid, which is the path that any point on the rim of the ring traces, is $4r.$

The Instantaneous axis of rotation(IAR): Each point on a rigid body can be thought of as tracing a circle at an instant, about a point known as the instantaneous center of rotation(c). The axis passing through C is the IAR.

It seems to me from these two statements that the distance of a point from the IAR must be equivalent to its radius of curvature. This is clearly not the case, as in a ring which is rolling without slipping , the IAR is at a distnace $2r$ from the top-most point.
What is the flaw in my interpretation?
Edit: Radius of curvature and Instantaneous Axis of Rotation seems quite similar. However, the top-voted answer has simply tried to explain $R=2r$, using the formula $v^2/a_{normal}$.This answer doesnt explain that why we get two different answers for the seemingly same thing.

REPLY [4 votes]: The key difference between Instantaneous Axis Of Rotation(IOAR) and Radius Of Curvature(ROC) is
IOAR: The point about which all points of a rigid body have perpendicular velocities at a
particular instant.
ROC: $R= v^2/a_\mathrm{normal}$ where the velocity is perpendicular to the center of curvature which is at a distance $R$ from the particle whose ROC we are calculating.
The confusion (I assume) came from the fact that the particle seems to circle both IOAR and ROC. However for the IOAR we will not be able to satisfy the equation  $R= v^2/a_\mathrm{normal}$ where $R$ is the distance from the IOAR. But this equation will give the one and only ROC. However its velocity is perpendicular to both the line joining ROC and IOAR (by definition which also happens to be why you may have confused the two). Therefore the instantaneous axis of rotation cannot be used to calculate radius of curvature.
I would also like to note that while instantaneous axis of rotation depends on the motion at an instant, radius of curvature depends on the trajectory of the individual particle of the rigid body which we are observing.


TITLE: When are we required to use the Wess-Zumino term?
QUESTION [3 upvotes]: I was recently reading about non-Abelian bosonization, and I had a question concerning the Wess-Zumino term. In particular, I have been reading this short introduction by Ivan Karmazin, which states that

A non-abelian bosonisation introduced by Witten in 1983 allows  to  translate any fermi  theory  into local bose theory while having all of the original symmetries conserved.

To ensure that the field theory is scale invariant, we add the Wess-Zumino action
$\Gamma[g]=\frac{1}{24\pi}\int_B d^3 y \epsilon^{ijk}{\bf Tr}\left(
g^{-1} \partial_i g g^{-1}\partial_j g g^{-1}\partial_k g
\right)$
where $g\in SU(N)$. My question is when, precisely, do we need to include the Wess-Zumino term when we bosonize a theory of 1+1 D massless fermions? Does the Wess-Zumino term mainly needed to describe a nonlinear sigma model, or is it more general?

REPLY [2 votes]: Alas, the book report you relied on glosses over the crucial geometrical peculiarity/pathology of the plain chiral model, which makes it unsuitable for bozonizing free nonabelian fermions.
I'll emphasize the point @Qmechanic and the original paper make, without using light-cone language and by paring down all superfluous aspects to focus on a primitive bare minimum.
So just look at right-invariant, so left currents, but the argument will go through for the right currents, mutatis mutandis.
In 2d bosonization it is important that the conserved symmetry currents identified have the same geometric properties, over and above mere conservation. (Ultimately they have to satisfy the same current algebras, but these trail the geometry.)
A free fermion theory has a left-chiral current,
$$
L^\mu= \bar \psi \gamma^\mu (1-\gamma^5)\psi,
$$
crucially self-dual,
$$
\tilde L^\mu \equiv \epsilon^{\mu \nu } L_\nu=  \bar \psi \epsilon^{\mu \nu }\gamma_\nu (1-\gamma^5)\psi \\
= \bar \psi \gamma^5 \gamma^\mu (1-\gamma^5)\psi = \bar \psi \gamma^\mu (1-\gamma^5)\psi  =L^\mu,
$$
being, of course, conserved, $ \partial\cdot L= \partial\cdot \tilde L=0$.
It had been know for a long time that the conserved currents of the plain (Gürsey, 1960) chiral model fail this requirement miserably.  That is, for example, the density
$$
{\cal L} \propto \operatorname {Tr} \partial_\mu  g^{-1}  \partial^\mu  g
$$
has EOM
$$
\partial_\mu (g\partial^\mu g^{-1}) =0  \Longrightarrow \\
\Box g+ g \partial_\mu g^{-1} \partial^{\mu} g= 0=\Box g + \partial_\mu g \partial^\mu g^{-1} g,
$$
where $\partial g ~g^{-1}+ g \partial g^{-1}=0$.
The evident Noether current of a left transformation $g\mapsto U g$ is the (right-invariant)
$$
L'_\mu= g \partial_\mu g^{-1},
$$
conserved by the above EOM,
$$
\partial ^\mu  L'_\mu =0,
$$
(evident from the parity transform $g\leftrightarrow g^{-1}$ of the above EOM form.)
But it is zero-curvature, so not self-dual,
$$
\partial \cdot \tilde L'= \epsilon^{\mu\nu} \partial_\mu g   \partial_\nu g^{-1}= -L'\cdot \tilde L' .
$$
(N.B. This is a central property in 2d nonabelian duality, and inapplicable/irrelevant to 4d, where these models have been used since the 60s.)
Witten chose to cut this Gordian knot by noting these two relations combine to one,
$$
\partial^\mu L_\mu'' =0
$$
for a self-dual current,
$$
L_\mu ''\equiv L_\mu ' + \tilde L'_\mu = \tilde L_\mu '' ~~.
$$

But wait!This is not conserved, as seen above, for the chiral model. Could one, instead, augment the action so it is?

That is, how does one obtain an extended action with EOM
$$
\partial g \cdot \partial g^{-1} + g \Box g^{-1} + \epsilon^{\mu \nu} \partial_\mu g \partial_\nu g^{-1}=0 ~~?
$$
Witten noted this would actually be identical to the parity image right current conservation below!
The short bare-bones-simplified answer is augmenting the chiral action by his eponymous topological term for a special fixed coupling, extending the $g(x)$ s to $\bar g(y)$ s in a solid ball
B with the $S^2$ sphere as a boundary. This is probably not the place to review this WZW term, whose analog summarizes the flavor-chiral anomalies of low-energy QCD. The crucial property is that the variation of the WZW term is a surface term in the ball, so it is local  on the sphere (!). That is the full action for a special coupling is proportional to
$$
\operatorname{Tr} \Biggl (  \int_{S^2} d^2x  ~~  \partial^\mu g \partial_\mu g^{-1} -\frac{2}{3} \int_B d^3y ~~\epsilon^{ijk}    \bar g ^{-1} \partial_i \bar g   ~\bar g ^{-1} \partial_j \bar g ~\bar g ^{-1} \partial_k \bar g       \Biggr ),
$$
has variation
$$
\bbox[yellow]{
2\operatorname{Tr}  \int_{S^2} d^2x ~ g\delta g^{-1}  \Biggl ( \partial_\mu (g \partial^\mu g^{-1} )+\epsilon^{\mu \nu } \partial_\mu g   \partial_\nu  g^{-1} \Biggr )},
$$
yielding the above equations of motion. You haven't lived if you skipped this most instructive of exercises. This is how conserved self-dual currents are achieved.
The analog for the parity-image right currents holds for the opposite relative sign of the WZW term, since these are anti-self-dual, $R_\mu ''= g^{-1} \partial^\mu g -\epsilon^{\mu \nu } g^{-1}   \partial_\nu  g =-\tilde R''_\mu$.
Indeed, pursuing the current algebra confirms the wisdom of the choice of $L''$ and $R''$ and whence of the isospin (vector) and axial currents.
Once the geometry is thus fixed, the RG has no choice but stay put at a fixed point (geometrostasis), so the theory does not renormalize, just like the free fermion equivalent theory.


TITLE: Optimal tiling for a collection of partitions
QUESTION [2 upvotes]: I'm interested in a possible generalization of Tiling relation on the set of partitions (the question has only been partially answered).
Let $x$ be an infinite set and let $\text{Part}(x)$ be the collection of
all partitions of $x$. Moreover, let $P \in \text{Part}(x)$
and $t\subseteq x$. We set
$$P_{[t]} = \{p\in P:p\cap t \neq \emptyset\}.$$
We define the tiling relation
on $\text{Part}(x)$ by
$$ P \triangleleft Q \textrm{ if and only if for all } S\subseteq
P\textrm{ we have } \mathsf{card}(S) \leq \mathsf{card}(Q_{[\bigcup S]}).$$
In other words, the relation $P\triangleleft Q$ holds if no subset $S$ of $P$
is covered by a subset of $Q$ having a smaller cardinality than $S$.
Let $\mathcal{A}\subseteq \text{Part}(x)$ be a non-tempty collection of partitions on the set $x$ such that there is a set $M$ of subsets of $x$ so that every member of $\mathcal{A}$ refines $M$ (that is, for every $A\in\mathcal{A}$ and $a\in A$ there is $m\in M$ with $a\subseteq m$).
Question. Is there a partition $Z\in\text{Part}(x)$ such that

$Z$ refines $M$ and
$Z \triangleleft A$ for all $A\in\mathcal{A}$?

REPLY [1 votes]: 1) This question (as posed) has a simple negative answer. Just take any countable set $x$ and put $M$ be the family of finite subsets of $x$. Then for the family $\mathcal A$ of all possible partitions of $x$ into finite subsets there is no partition $Z$ of $x$ into finite sets such that $Z\triangleleft A$ for every $A\in\mathcal A$.
Indeed, any partition $Z$ of $x$ into finite sets is infinite. So, we can take any 2-element subfamily $S=\{z,z'\}\subset Z$ and take any partition $A\in\mathcal A$ such that $\{z,z'\}\in A$. Then $|S|=2>1=|A_{[\cup S]}|$, witnessing that $Z\not\triangleleft A$.
2) A bit more complicated counterexample can be constructed as follows. Consider the family $\mathcal A$ of all possible partitions of the set $\mathbb P:=\mathbb R\setminus\mathbb Q$ of irrational numbers into compact susbets. Let $M$ be the family of all compact subsets of $\mathbb P$, endowed with the Vietoris topology. It is well-known that the space $M$ is separable and metrizable.
Then there is no partition $Z$ of $\mathbb P$ into compact subsets such that $Z\triangleleft A$ for all $A\in\mathcal A$.
Indeed, since the space $\mathcal P$ is not $\sigma$-compact, each partition $Z$ of $\mathbb P$ into non-empty compact subsets of $\mathbb P$ is uncountable and considered as a subspace of the hyperspace $M$ is not discrete. Hence $Z$ contains a sequence $\{z_n\}_{n\in\omega}\subset Z$ of pairwise distinct elements that converges to a compact set $z_\omega\in Z$. Then the set $S:=\{z_n\}_{n\le\omega}$ has compact union in $\mathbb P$ and hence $\bigcup S\in A$ for some partition $A$ of $\mathbb P$ into compact sets. Taking into account that $|S|=\omega>|A_{[\cup S]}|=1$, we conclude that $Z\not\triangleleft A$.
3) In fact, the gap between $|S|$ and $|A_{[\cup S]}|$ can be arbitrarily high.
Indeed, for any infinite regular cardinal $\kappa$ consider the set $x=\kappa$ and the family $M$ of subsets of $x$ of cardinality $<\kappa$. Let $\mathcal A$ be the family of all possible partitions of $x$ into subsets of cardinality $<\kappa$. Then for any partition $Z\in\mathcal A$ and any subfamily $S\subset Z$ of cardinality $|S|<\kappa$ there union $\bigcup S$ has cardinality $<\kappa$ and hence is an eleemnt of some partition $A\in\mathcal A$, for which $|A_{[\cup S]}|=1$.


TITLE: groups, generators, and orders
QUESTION [0 upvotes]: I think I have a problem connecting the dots here:
Suppose $G$ is any group, and $g,h \in G$ and $h \in \langle g\rangle$. If $ord(g) = s$ and $ord(h) = t$, need to prove $t$ divides $s$.
I understand that $g$ is a generator of $G$ and for every $x \in G, x = g^n$ for some integer $n$. Does it have to be the case $G$ is cyclic? Since $ord(g) = s$, then $\langle g\rangle$ = $\{e, g, g^2,…g^{s-1}\}$ and $|\langle g\rangle | = s$. Since $h \in \langle g\rangle$, then $h$ is some power of $g$, or $h = g^m$, where $0 \le m \le s-1$. Since $ord(h) = t$, we can substitute $h^t =(g^m)^t = g^{mt} = e$. We also have $s$ being the least positive integer s.t. $g^s = e$ and $m < s$. So does it mean from the above that $t$ divides $s$?

REPLY [2 votes]: Clearly $(g^m)^s=(g^s)^m=e^m=e$, so $t\le s$. Suppose $t$ does not divide $s$, so $\gcd(t,s)<t$.It is a theorem that there exist integers $a,b$ such that $at+bs=\gcd(t,s)$. Thus
$$g^{m\gcd(t,s)}=g^{amt+bms}=((g^m)^t)^a(g^s)^{bm}=e^ae^{bm}=e$$
so $t$ cannot be the smallest positive integer such that $(g^m)^t=e$.


TITLE: Problem in understanding the proof of a theorem
QUESTION [0 upvotes]: I can not understand the proof of Theorem 4 on page 2 of this document.

Theorem 4: Let $(x_n)$ be a bounded sequence with lower and upper limits $\underline{x}$ and $\overline{x}$,
  respectively. There are subsequences of $(x_n)$ converging to $\underline{x}$ and to $\overline{x}$.
Proof: We will prove the assertion for the upper limit. The assertion for the lower limit is similar.
We construct a subsequence $(x_{n_k})$ inductively so that
  $$
\left|x_{n_k} − \overline{x}\right| <\frac{1}{k}.
$$
By Lemma 3, there is an $n_1\in \Bbb N$ such that $$\left|x_{n_1} − \overline{x}\right| < 1.$$ Now suppose $n_1,\dots, n_k$ have been chosen. By Lemma 3, there is an $n_{k+1} > n_k$
  such that $$|x_{n_{k+1}} − \overline{x}| <\frac{1}{k+1},$$
  which completes the inductive step. Finally, it follows from (3) and the Squeeze Theorem that
  $$x_{n_k}\to\overline{x},$$
  which completes the proof.

I do not understand how it follows from Lemma 3 that there is $n_1\in\mathbb{N}$ such that $|x_{n_1}-\overline{x}|<1$.
I also do not understand how it follows from Lemma 3 that there is an $n_{k+1}>n_k$ such that $|x_{n_{k+1}}-\overline{x}|<\frac{1}{k+1}$.
Can someone kindly explain it to me?

REPLY [0 votes]: Lemma 3 says the following: "Let $(x_n)$ be a bounded sequence with lower and upper limits $\underline{x}$ and $\overline{x}$, respectively. For every $\epsilon > 0$ and every $K\in\Bbb N$ there are $m, n > K$ such that
$|x_m − \underline{x}| < \epsilon$ and $|x_n − \overline{x}| < \epsilon$."
To find $n_1$, let $\epsilon = 1 > 0$ and let $K = 1$. Then the lemma says that there exists $n > 1 = K$ such that $\left|x_n - \overline{x}\right| < 1$. Let this $n$ be your $n_1$. You can do this because you know lemma 3 holds for any $\epsilon > 0$, and $1$ is certainly greater than $0$. You can also find $n$ for any $K\in\Bbb N$, and $1\in\Bbb N$.
Next, to find $n_{k+1}$ given $n_1,\dots, n_k$, let $\epsilon = \frac{1}{k+1} > 0$ and let $K = n_k$. Lemma 3 then tells you that there is an $n > n_k = K$ such that $\left|x_n - \overline{x}\right| < \frac{1}{k+1}$. You can take this $n$ to be your $n_{k+1}$. Again, the lemma allows this, because $\epsilon = \frac{1}{k+1} > 0$ and $K = n_k\in\Bbb N$ and the lemma applies to any $\epsilon > 0$ and any $K\in\Bbb N$.


text:
\begin{document}

\raggedbottom

\maketitle
\thispagestyle{empty}
\pagestyle{empty}

\begin{abstract}
In this paper, we address pursuit-evasion problems involving multiple pursuers and multiple evaders.
The pursuer and the evader teams are assumed to be heterogeneous, in the sense that each team has agents with different speed capabilities.
The pursuers are all assumed to be following a constant bearing strategy.
A \emph{dynamic divide and conquer} approach, where at every time instant each evader is assigned to a set of pursuers based on the instantaneous positions of all the players, is introduced to solve the multi-agent pursuit problem.
In this regard, the corresponding multi-pursuer single-evader problem is analyzed first.
Assuming that the evader can follow any strategy, a dynamic task allocation algorithm is proposed for the pursuers.
The algorithm is based on the well-known Apollonius circle and allows the pursuers to allocate their resources in an intelligent manner while guaranteeing the capture of the evader in minimum time.
The proposed algorithm is then extended to assign pursuers in multi-evader settings that is proven to capture all the evaders in finite time.
\end{abstract}

\section{Introduction} \label{sec:intro}

Coordination strategies for unmanned aerial vehicles (UAVs)
has been an active area of research especially in the realm of multi-agent systems over the past decade \cite{Huang2013, Nex2014, Yuan2015}, having
numerous applications, including agriculture, aerial surveying, fire detection, disaster management, weather monitoring, and commercial product delivery.
Recent analyses of the commercial UAV market show that their use is expected to grow manyfold over the coming years,
as aerial drones are becoming a household product, used for recreational and industrial purposes alike\footnote{\url{http://www.businessinsider.com/commercial-uav-market-analysis-2017-8}}.
These advancements suggest an urgent need to explore designs for airspace safety systems that can regulate the traffic and usage of UAVs in a large scale.
Similarly, UAVs already play a major role in military engagement scenarios, and their use as part of swarm tactics (encirclement, coordinated attack, search and rescue,
perimeter defense) promises to change future battlefield operations.
It should therefore come as no surprise that a great amount of work has been devoted over the past decade to study coordination strategies of multi-agent UAV problems \cite{Rizk2018}.
To this end, UAV coordination strategies that formulate the problem as
a multi-player pursuit-evasion (PE) game offer solutions that address many of the challenges involving multi-agent systems such as of collision avoidance, surveillance and target acquisition~\cite{Mylvaganam2014, LasFargeas2015}.

The literature for multi-pursuer multi-evader (MPME) problems is actually limited.
In most cases, some form of heuristic is introduced in order to make the problem tractable.
Ge et al.~\cite{Ge2006} proposed three approaches, which include hierarchical decomposition, moving horizon hierarchical decomposition, and cooperative control.
Li et al.~\cite{Li2005} also explored a hierarchical approach, while Jin and Qu \cite{Jin2011} proposed a heuristic task allocation algorithm.
Extensions to the MPME problem includes problems with incomplete information \cite{Antoniades2003}, nonlinear dynamics \cite{Stipanovic2010}, and a mix of continuous and discrete observations \cite{Stipanovic2009}.
However, finding scalable algorithms which can be implemented in real-life MPME scenarios is still an open problem~\cite{Vieira2009a, Pierson2017}.

This work aims at extending current solution techniques for MPME games involving large teams of UAVs by
developing implementable, scalable solutions
based on a decomposition of the original MPME problem to a sequence of simpler multi-pursuer single-evader (MPSE) problems.
A major enabler for this decomposition is a new result that allows us to characterize each pursuer as \emph{active} or \emph{redundant} for each evader.
Only the relevant pursuers participate in the MPSE pursuit of each evader.
The identification and classification of each pursuer as active or redundant makes use of the classical tool of the Apollonius circle \cite{isaacs1999differential}.
Previously, the proposed MPME formulation was analyzed with the pursuit and the evading teams being homogeneous \cite{makkapati2019optimal}.
In this paper, we generalize the results in Ref. \cite{makkapati2019optimal} to include heterogeneous teams of agents with guarantees on finite-time capture.
The videos for the simulation results discussed in this paper can be found on the web\footnote{\url{https://youtu.be/KTH9lmdUdRs}}.

\section{Motivation and Problem Setup}

Consider a group of $n$ agents (pursuers) guarding a given area of interest.
The objective of the agents is to pursue and intercept  $m$ (where typically $m \le n$) intruders (or evaders) that may be detected in this area.
Some of the relevant questions that arise while solving this problem include: a)Which pursuer(s) should go after which evader(s)?
b) How many pursuers should chase each intruder (evader) in order to capture it in the shortest time possible?

Obtaining the answers to the previous questions in their most general form is elusive at this point.
Addressing them involves solving a multi-player dynamic game, eventually demanding the solution to a high-dimensional partial differential equation, with the dimensionality increasing as the number of players ($n+m$).
In order to proceed and mitigate this problem, the following assumptions are made in this work.
\begin{enumerate}
  \item[A1:] The pursuers are all faster compared to the evaders.

  \item[A2:] The pursuers know the instantaneous positions and the velocities of all the evaders, and each pursuer follows a constant bearing strategy with respect to the evader to which it is assigned.
\end{enumerate}
The rationale behind these assumptions is as follows.
Under assumption A1, a constant bearing strategy guarantees capture.
Furthermore, constant bearing strategy has been implemented successfully in various aerial defense systems \cite{Shneydor}.

Formally, we address a pursuit-evasion problem in the Euclidean plane that involves $n$ pursuers and $m$ evaders.
The pursuers' objective is to capture all the evaders.
Capture occurs when one or more pursuers enter the capture zone of an evader (assumed here to be a disk of radius $\epsilon > 0$ centered at the instantaneous position of the evader).
Let
$P_i$ denote the $i^{th}$ pursuer and let $E_j$ denote the $j^{th}$ evader.
Let also $\Pset = \{P_1,P_2,\ldots,P_n\}$ denote the set of pursuers and, similarly, let $\Eset = \{E_1,E_2,\ldots,E_m\}$ denote the set of evaders.
The equations of motion of all the agents are given below
\begin{align}
\dot{x}_i &= u_i\cos \theta_i, \quad \dot{y}_i = u_i\sin \theta_i, \quad i \in  \Pset, \label{eqn:p1} \\
\dot{x}_j &= v_j\cos \theta_j, \quad \dot{y}_j = v_j\sin \theta_j, \quad j \in \Eset, \label{eqn:e}
\end{align}
where $p_i = (x_i,y_i) \in \mathbb{R}^2$, and $e_j = (x_j, y_j) \in \mathbb{R}^2$ denote the positions of pursuer $P_i$, and
evader $E_j$, respectively, and $\theta_i$ and $\theta_j$ denote the heading angles (control inputs) for the pursuers and the evaders, respectively.
In (\ref{eqn:p1}) and (\ref{eqn:e}), $u_i$ and $v_j$ are the speeds of $P_i$ and $E_j$, which are assumed to be constant with $\underset{i \in \mathcal{P}}{\min}~u_i>\underset{j \in \mathcal{E}}{\max}~v_j$ (A1).

A potential approach  to solve complicated MPME problems is a \emph{dynamic} ``divide and conquer'' approach,
where the pursuers are divided into several groups corresponding to the evader they pursue at each instant of time.
In essence, such  divide and conquer strategies formulate the original MPME problem as
a sequence of several (simpler) MPSE problems~\cite{Ge2006}.
This approach leads to decentralized (although likely suboptimal) solutions.
By analyzing the associated MPSE problems, one may arrive at an efficient dynamic task-allocation algorithm of pursuers to evaders.
To this end, we first present an approach for task allocation in MPSE problems using Apollonius circles in Section \ref{sec:strategies}.

\begin{figure}[htb!]
  \centering
  \includegraphics[width=0.4\textwidth]{game_scenario_1}
  \caption{Schematic of the proposed multi-pursuer single-evader problem.}
  \label{fig:game_scene}
\end{figure}

A schematic of the proposed PE problem with one evader and multiple pursuers, that follow a constant bearing strategy--henceforth referred to as the MPSE problem--is shown in
Fig.~\ref{fig:game_scene}.
Since the pursuers are assumed to be following a constant bearing strategy, the problem can be analyzed by tracking the relative distances between the pursuers and the evader.
In this regard, the dynamics can be written in the form,
\begin{align}
\dot{r}_i &= v \cos(\thetae - \varphi_i) - u_i \cos(\theta_i - \varphi_{i}), \qquad i \in \Pset,
\end{align}
where $r_i$ is the relative distance between pursuer $P_i$ and the evader, and $\varphi_{i} = \text{atan2}(\ye-y_i,\xe-x_i)$ is the corresponding line of sight (LoS) angle.
For the MPSE problem, we drop the subscripts for the evader's speed $v$, and will use $E$ instead of $j$ (when required) to denote the single evader in this setting.
Furthermore, in the MPSE problem, we indicate the pursuers using the subscripts directly and the set $\mathcal{P} = \{1,2,\dots,n\}$.
Note that in the case of a constant bearing strategy the bearing angle between a pursuer and the evader remains constant until the time of capture.
Using this fact, the instantaneous heading of pursuer $P_i$ ($\theta_i$, $i \in \Pset$) can be obtained from the relation,
\begin{align}
u_i \sin(\theta_i - \varphi_{i}) &= v \sin(\thetae - \varphi_{i}), \label{eq:constri}
\end{align}
which is a function of the instantaneous heading of the evader $\thetae$. The above relation has two possible solutions for each $\theta_i$, given $\thetae$, and the solution for which $\dot{r}_i < 0$ is chosen.

\section{Task Allocation in Multi-Pursuer Single-Evader Problems}\label{sec:strategies}

In this section, we address task allocation in the MPSE problem by dynamically categorizing the pursuers into active and redundant.
The proofs for the lemmas presented in this section can be found in Ref. \cite{makkapati2019optimal}.
First, the formal definitions for time-of-capture and the corresponding capturing pursuer set, that are used to define active and redundant pursuers, in the MPSE problem are provided below.

\vspace*{1ex}

\begin{definition} \label{def:tc}
  For a given evading strategy, \emph{the time-to-capture $t_c~(\geq 0)$} is the minimum time so that there is at least one pursuer in the capture zone of the evader, and \emph{the capturing pursuer set $\mathcal{P}_c \subset \Pset$} is the set of pursuers that are in the capture zone of the evader at $t_c$.
\end{definition}

\vspace*{1ex}

The time-to-capture $t_c$ is always finite since the pursuers follow a constant bearing strategy.
Note that at the time of capture, one or more pursuers can be in the capture zone of the evader.
Therefore, $1 \leq \text{card}[\mathcal{P}] \leq n$, where card$[\cdot]$ represents the cardinality of the set.
Now, the following definitions establish the notions of active and redundant pursuers.

\vspace*{1ex}

\begin{definition}
  If there exists an evading strategy for which $P_i \in \mathcal{P}_c$, then $P_i$ is an \emph{active pursuer}.
  Otherwise, $P_i$ is a \emph{redundant pursuer}.
\end{definition}

\vspace*{1ex}

Given the instantaneous positions of the pursuers and the evader, it is of interest to find a condition to verify whether a pursuer is active or redundant.
To this end, Apollonius circles are employed in this work.
The Apollonius circle for a pursuer-evader pair is the locus of points where capture occurs, for all possible initial headings of a non-maneuvering evader, given the initial positions of the
pursuer/evader pair and assuming that the pursuer follows a constant bearing strategy, see Fig.~\ref{fig:curves}.
For the MPSE problem, the Apollonius circle of the pair $P_i$-$E_j$ is denoted as $\mathcal{A}_{ij}$.
It has  its center at $O_i\bigg(\dfrac{x_j - \rho_{ij} x_i}{1 - \rho_{ij}^2},\dfrac{y_j - \rho_{ij} y_i}{1 - \rho_{ij}^2} \bigg)$ and radius $d_{ij}=\dfrac{\rho_{ij}\|p_i-p_j\|}{1-\rho_{ij}^2}$, where $\rho_{ij} = v_j/u_i$ (speed ratio) \cite{Ramana2017a}.
The Apollonius circles evolve in time as the players move, but the time dependencies will be dropped for the sake of brevity.
Let $T_{ij}$ be the closest point to the evader on the Apollonius circle where collision occurs when the evader goes head-on with the pursuer, as shown in Fig.~\ref{fig:curves}.
Therefore the distance of $T_{ij}$ from the evader is ${v_j\|p_i - p_j\|}/{(u_i+v_j)}$.
Next, we define \emph{Apollonius boundary} and analyze its properties to identify the active and the redundant pursuers.
\begin{figure}[htb!]
  \centering
  \includegraphics[width=0.4\textwidth]{apollonius_circle}
  \caption{Apollonius circle. Simulation parameters: $u_i = 1$, $v=0.6$, $p_i(0) = (0,0)$, $\Xe = (1,0)$.}
  \label{fig:curves}
\end{figure}

\begin{definition} \label{def:AB}
  Given the positions all the pursuers and the evader $E_j$ at time $t \geq 0$,
  the \emph{Apollonius boundary} around the evader $E_j$ at time $t$ is the set of points
  \begin{align*}
      \mathcal{B}^t_j = \{ X \in {\textstyle \cup_{i=1}^{n}} \mathcal{A}_{ij} ~|~ \mathcal{M}(e_j, X) \cap \big( {\textstyle \cup_{i=1}^{n}} \mathcal{A}_{ij} \big) = \{X\} \},
  \end{align*}
  where $\mathcal{M}(e_j,X)$ denotes the set of points on the line segment with endpoints $e_j$ (position of $E_j$) and $X$ at time $t$.
\end{definition}

\vspace*{1ex}

In other words, the Apollonius boundary is the set of points that belong to the union of all the instantaneous Apollonius circles corresponding to the evaders and, in addition,
each such point is the closest to the evader along its respective line-of-sight originating from the evader.
Note that the Apollonius boundary evolves with time as the Apollonius circles also evolve with time.
The subscript $j$ will be dropped for the rest of this section, since we are dealing with the MPSE problem.
\begin{figure}[htb!]
  \centering
  \includegraphics[width=0.4\textwidth]{hetero_bound_CB}
  \caption{Apollonius boundary. Simulation parameters: $u_1 = 0.8$, $u_2 = 1$, $u_3 = 1.2$, $v=0.7$}
  \label{fig:boundaries}
\end{figure}

\subsection{Apollonius circle based Active Pursuer Check}

The algorithm that is developed to identify active/redundant pursuers in this work is based on the following conjecture.

\vspace*{1ex}

\begin{conjec} \label{conjecture:active}
  Given the positions of all the players in an MPSE problem at time $0 \leq t < t_c$, and assuming that the pursuers follow a constant bearing strategy, pursuer $P_i$ is active at time $t$ if $\mathcal{B}^t \cap \mathcal{A}_i \neq \varnothing$, and is redundant otherwise.
\end{conjec}

\vspace*{1ex}

The conjecture implies that a pursuer is active at time $0 \leq t < t_c$ if and only if its corresponding Apollonius circle is part of the Apollonius boundary at that instant.
The conjecture is inspired from the fact that the region in which the capture point lies in is bounded by the instantaneous Apollonius circle for any strategy of the evader.
Note that if a pursuer is active at time $t'$, it need not remain active for all $t > t'$.
But if a pursuer is redundant at time $t'$, it will remain redundant for all $t > t'$.
The following lemmas based on Conjecture~\ref{conjecture:active} provide simple checks to determine whether a pursuer is active or redundant.

\vspace*{1ex}

\begin{lemma} \cite{makkapati2019optimal} \label{lemma:one}
  Given the positions of the players in the MPSE problem at time $0 \leq t < t_c$, pursuer $P_i$ is the only active pursuer if and only if the conditions
  \begin{align}
  \mathcal{A}_i \cap \big( {\textstyle \cup_{j=1,\,j \neq i}^{n}} \mathcal{A}_j \big) &= \varnothing, \label{eq:lemma1_cond1}\\
  \mathcal{M}(e,T_i) \cap \big( {\textstyle \cup_{j=1,\,j \neq i}^{n}} \mathcal{A}_j \big) &= \varnothing, \label{eq:lemma1_cond2}
  \end{align}
  are satisfied, where $T_i$ is the closest point to the evader on the Apollonius circle $\mathcal{A}_i$.
  Furthermore, if conditions (\ref{eq:lemma1_cond1}) and (\ref{eq:lemma1_cond2}) are not satisfied, then $P_i$ is a redundant pursuer.
\end{lemma}

\vspace*{1ex}

\begin{lemma} \cite{makkapati2019optimal} \label{lemma:two}
  Given the positions of the players in the MPSE problem at time $0 \leq t < t_c$, and that the Apollonius circle $\mathcal{A}_i$ intersects at least one  of the other Apollonius circles.
  Then, pursuer $P_i$ is an active pursuer if and only if there exists $X \in \mathcal{I}_i$ such that $\mathcal{M}(e,X) \cap \big( {\textstyle \cup_{j=1}^{n}} \mathcal{A}_j \big) = \{X\}$,
  where $\mathcal{I}_i$ is the set of intersection points between $\mathcal{A}_i$ and the rest of the Apollonius circles.
\end{lemma}

\vspace*{1ex}

The set of intersection points $\mathcal{I}_i$ can be obtained analytically given the instantaneous positions of all the players \cite{Weisstein}.
The above two lemmas can be used to verify if a pursuer is active or redundant.
In this regard, Algorithm~\ref{algo:active} below, named Apollonius circle based Active Pursuer Check (AAPC), can be employed to check the status of each pursuer.
The time complexity of the algorithm is of order $O(n^2)$, since the maximum number of intersections between any two circles is  two.
Note that by dynamically allocating the task of capturing the evader using AAPC (where at every instant the active pursuers keep pursuing the evader while the redundant pursuers do not react), the pursuers as a group will be able to capture the evader in minimum time.
Furthermore, if a pursuer becomes redundant at any point of time $0 \leq t < t_c$, it remains redundant after that (i.e., till capture occurs).

\begin{algorithm}
  \caption{Apollonius circle based Active Pursuer Check (AAPC)}\label{algo:active}
  \begin{algorithmic}[1]
    \Require Positions of all the players ($p_1$,\dots,$p_n$,$e$,$i$)
    \Ensure Status of pursuer $P_i$
    \Procedure{obtain\_status}{$p_1$,\dots,$p_n$,$e$,$i$}
    \State intersection = 0
    \State status = redundant
    \For  {$j=1$ to $n$ and $j \neq i$}
    \State Obtain the set of intersection points $\mathcal{I}_{ij}$
    \If {$\mathcal{I}_{ij} \neq \varnothing$}
    \State intersection = 1
    \For  {$\ell=1$ to $\text{card}[\mathcal{I}_{ij}]$}
    \State boundary = 1
    \For  {$k=1$ to $n$ and $k \neq i,\,j$}
    \If {$\mathcal{M}(e,X_\ell)$ intersects $\mathcal{A}_k$}
    \State boundary = 0
    \EndIf
    \EndFor
    \If {boundary = 1}
    \State status = active
    \State break from outermost loop.
    \EndIf
    \EndFor
    \EndIf
    \EndFor
    \If {intersection = 0}
    \State status = active
    \For {$j=1$ to $n$ and $j \neq i$}
    \If {$\mathcal{M}(e,T_i)$ intersects $\mathcal{A}_j$}
    \State status = redundant
    \State \textbf{break}
    \EndIf
    \EndFor
    \EndIf
    \State \Return status
    \EndProcedure
  \end{algorithmic}
\end{algorithm}

\subsection{Numerical Simulations} \label{subsec:numsim}

In this section simulations of pursuer allocation using AAPC involving five pursuers and one evader are presented.
There are three different pursuers with speeds $u_i \in \{0.8,1,1.2\}$, and the speed of the evader is set to
$v = 0.6$.
The radius of capture is chosen as $\epsilon = 0.1$.
The evader follows a form of blind evasion strategy with switching times that are predefined~\cite{Morgan2016}.
At each switching time the evader randomly chooses a heading from a set of allowable headings.
The allowable headings set that is specific to the example showcased in this work is $\{-\pi/4,~\pi/2,~3\pi/4\}$.

\begin{figure}[htb!]
  \centering
  \subfigure[Initial Apollonius circles]{\includegraphics[width=0.39\textwidth]{hetero_sim_apollo}}  \quad
  \subfigure[Trajectories]{\includegraphics[width=0.39\textwidth]{hetero_sim_traj}}
  \caption{Results obtained using AAPC for task allocation in the case of CB.}
  \label{fig:sim_cb}
\end{figure}

Fig.~\ref{fig:sim_cb}(a) shows the initial positions of all the players along with the corresponding Apollonius circles.
The triangle denotes the initial position of the evader and the square markers denote the initial positions of the pursuers.
It can be observed that at the initial time, the pursuers identified with the colors red, magenta, green, and blue are the active pursuers, as their corresponding Apollonius circles are part of the Apollonius boundary.
Fig.~\ref{fig:sim_cb}(b) shows the trajectories of all the players.
It can be seen that the green pursuer finally captures the evader, and the rest of the three pursuers, which are  initially active, become redundant as time progresses.
The cyan pursuer, which is redundant at the initial time, does not move at all.

\section{Apollonius Allocation Algorithm}\label{sec:mpme}

In this section, the AAPC is extended to solve MPME problems.
Given the positions of all the players at some instant of time, the set of evaders for which a pursuer is active can be obtained using AAPC.
Note that at a given time instant, a pursuer can be classified as active by more than one evader or no evader whatsoever.
In the case where a pursuer is classified as active for more than one evader, one can break the tie by assigning the pursuer to the evader that can be captured in minimum possible time i.e., $\mathrm{argmin}_j \|p_i - e_j\|/(u_i + v_j)$.

\vspace*{1ex}

\begin{remark}
    The aforementioned criterion for breaking a tie is equivalent to choosing the nearest evader when the teams are assumed to be homogeneous \cite{makkapati2019optimal}.
\end{remark}

\vspace*{1ex}

Using this criterion, the following algorithm can be used for pursuer allocation in MPME problems.

\vspace{1em}

\emph{Apollonius Allocation (A2) Algorithm:}

At a given time instant $0 \leq t \leq t_c$, let $\Eset_f$ be the set of evaders that are yet to be captured, and let $\Eset_c$ be the set of evaders that have already been  captured.
Note that $\mathcal{E} = \Eset_f \cup \Eset_c$.
Given the current positions of all the players,
let $\mathscr{I}: \Eset_f \rightarrow 2^{\Pset}$ be the initial allocation function that maps each evader $E_j$ (in $\Eset_f$) to its set of active pursuers obtained by considering the positions of all the pursuers.
That is, for a given $j \in \Eset_f$, $\mathscr{I}(j)$ is a subset of $\Pset$.
Furthermore, $\Pset_a = \cup_{j \in \Eset_f}\mathscr{I}(j)$ denotes the set of all the active (or assigned) pursuers according to the
initial allocation function $\mathscr{I}$.
Given the initial allocation function $\mathscr{I}$, let now $\mathscr{J}: \Pset \rightarrow 2^{\Eset_f}$ be the dual function defined by
$\mathscr{J}(i) = \{ j \in \Eset_f: \mathscr{I}(j) = i \}$.
In other words, $\mathscr{J}$ maps each pursuer to the set of the evaders to which it is allotted as per $\mathscr{I}$.
Next, we define the \emph{final allocation function} $\mathscr{F}$ and the \emph{intermediate allocation function} $\mathscr{G}$
as follows.

\begin{enumerate}[(a)]

  \item
  If card$[\mathscr{J}(i)] \leq 1$, for all $i  \in \Pset$, then let $\mathscr{F} = \mathscr{I}$.
  Otherwise, let $\mathscr{G}: \Eset \rightarrow 2^{\Pset}$ be defined as $\mathscr{G}(j) = \Big\{ i \in \mathscr{I}(j) : j = \underset{k \in \mathscr{J}(i)}{\mathrm{argmin}}~\dfrac{\|p_i - e_k\|}{u_i + v_k} \Big\}$.
  The function $\mathscr{G}$ maps each evader to a set of pursuers in accordance to the mapping $\mathscr{I}$,
  such that each active pursuer is assigned to the nearest evader among its assigned ones.
  Note that $\mathscr{G}(j)$ can be an empty set for some $j$, i.e., an evader can end up be unassigned as per $\mathscr{G}$.

  \item
  Let $\Pset_u  = \Pset \backslash \Pset_a$ be the set of unassigned pursuers.
  Now for each evader $E_j$, find the active pursuers considering the positions of the pursuers that are only in the set $\mathscr{G}(j)~\cup~\Pset_u$,
  and obtain an \emph{updated allocation function} $\mathscr{I}'$ and its corresponding dual $\mathscr{J}'$.

  \item Repeat steps (a) and (b), by replacing $\mathscr{I}$ and $\mathscr{J}$ with $\mathscr{I}'$ and $\mathscr{J'}$, respectively, until $\mathscr{F}$ is obtained.

\end{enumerate}

Note that in step (a) of the algorithm, ties with multiple assignments of the same pursuer are broken using distance as the metric.
Furthermore, if each pursuer is assigned to only one evader or if it remains unassigned, then the initial allocation function $\mathscr{I}$ is also the final one.
In any other case, once the intermediate allocation function $\mathscr{G}$ is obtained in step (b) of the algorithm,
the set of unassigned pursuers according to $\mathscr{I}$ is obtained.
In step (b), an updated allocation function $\mathscr{I}'$ is obtained by checking for active pursuers among the set of unassigned pursuers coupled with the pursuers assigned as per $\mathscr{G}$, for each evader.
Because one of the unassigned pursuers (in the set $\Pset_u$) can become active to the evaders that have lost one or more pursuers during the tie break in step (a).
With $\mathscr{I}'$ and its corresponding dual $\mathscr{J}'$, steps (a) and (b) are repeated until each pursuer has only one (or none) assignment.
Once an evader is captured, it is removed from the set $\Eset_f$ and added to the set $\Eset_c$.

\begin{figure*}[htb!]
  \centering
  \subfigure[$t$ = 0 (At initial time)]{\includegraphics[width=0.32\textwidth]{sim_mpme_t0}}
  \subfigure[$t$ = 1.8 (Three evaders are captured)]{\includegraphics[width=0.32\textwidth]{sim_mpme_t18}}
  \subfigure[$t$ = 2.75 (All evaders are captured)]{\includegraphics[width=0.32\textwidth]{sim_mpme_t275}}
  \caption{Plots showing the positions and trajectories of the players in a multi-pursuer (squares) multi-evader (triangles) problem at different time instants.}
  \label{fig:sim_mpme}
\end{figure*}

The above algorithm is run at every time instant to obtain the allocation function $\mathscr{F}$, given the players' current positions, until all the evaders are captured,
i.e., until $\mathcal{E}_f$ is empty.
The algorithm provides a potentially sub-optimal solution, but it is scalable for any number of pursuers and evaders.
The algorithm guarantees capture of all $m$ evaders as is shown in Theorem~\ref{Theorem:main} below.
In order to prove this theorem, several preparatory results are needed.

\vspace*{1ex}

\begin{definition}\label{def:CST}
  Given the instantaneous positions of the players in the MPME problem at time $t \geq 0$, the \emph{current shortest time} (CST) is defined by $t_s = \underset{(i,j) \in \Pset \times \Eset_f}{\min}~\dfrac{\|p_i - e_j\|}{u_i + v_j}$.
\end{definition}

\vspace*{1ex}

\begin{lemma}
  At a given time instant $t \geq 0$, $i^* \subseteq \mathscr{F}(j^*)$, where $(i^*,j^*) = \underset{(i,j) \in \Pset \times \Eset_f}{\mathrm{argmin}}~\dfrac{\|p_i - e_j\|}{u_i + v_j}$, and $\mathscr{F}$ is the final allocation function of A2 algorithm. \label{lemma:MPME_2}
\end{lemma}

\vspace*{1ex}

\begin{proof}
    From Definition~\ref{def:CST} it is understood that $t_s$ represents the minimum possible time taken to capture an agent in the evading team by an agent in the pursuing team, when the corresponding evader goes head-on with the pursuer.
    Therefore, if it can be shown that the Apollonius boundary around the evader $j^*$ contains part of the Apollonius circle $\mathcal{A}_{i^*j^*}$ (i.e., $\mathcal{B}^t_{j^*} \cap \mathcal{A}_{i^*j^*} \neq \varnothing$, $\forall
    ~t \geq 0$), then the pursuer $i^*$ will be assigned to the evader ${j^*}$, even when there is a tie.
    Since the Apollonius circle denotes the capture points for a non-maneuvering evader, the point $T_{i^*j^*}$, with the length of the line segment $E_{j^*}T_{i^*j^*} = v_{j^*}t_s$, is the closest capture point to evader $j^*$ along the corresponding line of sight.
    Therefore $T_{i^*j^*} \in \mathcal{B}^t_{j^*}$, and hence, $\mathcal{B}^t_{j^*} \cap \mathcal{A}_{i^*j^*} \neq \varnothing$, $\forall~t\geq 0$.
\end{proof}

\vspace*{1ex}

\begin{lemma}  \label{lemma:MPME_3}
  Assuming the pursuers are assigned to the evaders using A2, at any given time $t \geq 0$, CST will converge to zero in finite time, and hence at least one evader will be captured in finite time.
\end{lemma}

\vspace*{1ex}

\begin{proof}
  From Lemma~\ref{lemma:MPME_2}, pursuer $i^*$ (corresponding to the CST) is always assigned to evader $j^*$. Since all the pursuers are faster compared to the evaders, and since they follow a constant bearing strategy, d$t_s/$dt $\leq (v_{j^*} - u_{i^*})/(v_{j^*} + u_{i^*}) < 0$, for all $t \geq 0$~\cite{Shneydor}.
  Furthermore, as the initial CST is finite, the CST converges to zero in finite time.
  Hence, capture of one evader is guaranteed in finite time.
\end{proof}

\vspace*{1ex}

\begin{theorem} \label{Theorem:main}
  The A2 algorithm guarantees capture of all the evaders in finite time.
\end{theorem}

\vspace*{1ex}

\begin{proof}
  The result immediately follows from Lemmas \ref{lemma:MPME_2} and \ref{lemma:MPME_3}. Note that the CST is updated (from zero) every time a capture occurs, and the captured evader is removed from the list of participating players. Also, the number of evaders are finite.
\end{proof}

\vspace*{1ex}

Fig.~\ref{fig:sim_mpme} demonstrates the performance of A2 algorithm for 10 pursuers and 5 evaders.
For the sake of simulation, the evading is assumed to be homogeneous.
The simulation parameters remain the same as in Section~\ref{subsec:numsim}.
In Fig.~\ref{fig:sim_mpme}, the red triangles indicate the current positions of the evaders that are not captured and the magenta ones are the evaders that are captured.
The blue squares indicate the current positions of the active pursuers and the cyan ones indicate the redundant pursuers.
In all three plots, the Voronoi partition of the domain with the evaders as generators is also included for reference.

\section{Conclusion} \label{sec:conclude}

Under the assumption that the pursuers are faster than the evader(s), and that they follow a constant bearing strategy, workable solutions for multi-pursuer single-evader (MPSE) and multi-pursuer multi-evader (MPME) problems are provided.
A dynamic allocation algorithm for the pursuers that is independent of the evader's strategy has been proposed to solve the MPSE problem.
The proposed algorithm is based on the notion of active/redundant pursuers, and employs the concept of Apollonius circles.
The algorithm is further extended to solve MPME problems for any number of pursuers and evaders.
These algorithms ensure capture of all the evaders either in an MPSE or an MPME setting in finite time.
Several extensions of this work are possible.
For example,
the computational requirements can be reduced by having an estimate of when the assignment can change to avoid unnecessary calculations at every time instant.
It would also be very interesting to extend
the notion of Apollonius circles to account for turn-radius constraints for all the players, but this would probably end up being a major challenge.

\vspace*{1ex}

config: arxiv


text:
\begin{document}
\title{Wave Support Theorem \\and Inverse Resonant Uniqueness on the Line}
\author{Lung-Hui Chen$^1$}\maketitle\footnotetext[1]{General Education Center, Ming Chi University of Technology, New Taipei City, 24301, Taiwan. Email:
mr.lunghuichen@gmail.com.}\maketitle
\begin{abstract}
In the paper, we experimentally study the inverse problem with the resonant scattering determinant. We analyze the structure of characteristics of perturbed linear waves. Assuming there is the common part of potential perturbation propagating along the same strips, we estimate the common part of the perturbed wave, and its Fourier transform.
 We deduce the partial inverse uniqueness from the Nevanlinna type of representation theorem.
\\MSC: 34B24/35P25/35R30.
\\Keywords:  scattering; resonance;  Schr\"{o}dinger equation; inverse problem; Nevanlinna theorem.
\end{abstract}
\section{Introduction}
Let us consider the Schr\"{o}dinger equation
\begin{equation}\label{1.1}
-\frac{d^{2}}{dx^{2}}+V(x),\,x\in\mathbb{R},\,V(x)\in L_{comp}^{1}([a,b]),\,a<0<b,\,|a|\ll|b|,
\end{equation}
where we assume the potential is effectively support on $[a,b]$, that is, $[a,b]$ is the minimal convex hull that contains the support of $V^{j}$.
The scattering matrix is of the form
\begin{equation}\label{S}
S(k)=\left(\begin{array}{cc}\frac{ik}{\hat{X}(k)} & \frac{\hat{Y}(k)}{\hat{X}(k)} \vspace{7pt}\\\frac{\hat{Y}(-k)}{\hat{X}(k)} & \frac{ik}{\hat{X}(k)}\end{array}\right).
\end{equation}
The scattering matrix $S(k)$ is meromorphic in $\mathbb{C}$, and its poles in $\{\Im k>0\}$ are the square roots of $L^{2}$-eigenvalues of~(\ref{1.1}). In this paper, we understand $\hat{X}(k)$ through the one-dimensional wave equation
\begin{eqnarray} \label{AA}
\left\{
\begin{array}{ll}
\big(D_{x}^{2}-D_{y}^{2}+V(x)\big)A_{\pm}(x,y)=0;\vspace{5pt}\\
A_{\pm}(x,y)=\delta(x-y),\,\pm x\gg0,
\end{array}
\right.
\end{eqnarray}
where $D_{y}A_{-}(x,y)=X(y-x)+Y(y+x)$.
In particular \cite[p.\,727]{Mellin},
\begin{eqnarray}\label{1122}
&&X(x)-\delta'(x)+\frac{\int V(t)dt}{2}\delta(x)\in L^{1}(\mathbb{R})\cap L^{\infty}(\mathbb{R});\\
&&Y(y)-\frac{V(y/2)}{4}\in L^{1}(\mathbb{R})\cap L^{\infty}(\mathbb{R}).\label{511}
\end{eqnarray}
Thus, $A_{\pm}(x,y)$ satisfies the wave equation with $x$ taking the place of time (this choice is dictated by the forcing condition imposed) in~(\ref{AA}). The uniqueness part follows from the energy estimates of the wave equation \cite{Dya,Mellin,Tang,Zworski2}.

\par
In this paper, we consider the complex analysis of entire function $\hat{X}(k)$ and $\hat{Y}(k)$  which are represented in form of
\begin{eqnarray}\label{1.6}
&&\hat{X}(k)=\int_{-2(b-a)}^{0}X(x)e^{-ikx}dx;\\\label{1.7}
&&\hat{Y}(k)=\int_{2a}^{2b}Y(y)e^{-iky}dy,\,k\in\mathbb{C},
\end{eqnarray}
that the unitary identity holds in $\mathbb{C}$:
\begin{equation}\label{U}
\hat{X}(k)\hat{X}(-k) = k^{2} + \hat{Y}(k)\hat{Y}(-k) ,
\end{equation}
\par
More importantly, we consider experimentally
\begin{equation}\label{SS}
\det S(k):=\frac{-\hat{X}(-k)}{\hat{X}(k)}.
\end{equation}
as scattering data in this paper inspired its simpler analytic structure.
\par
We define for potential $V^{j}$, $j=1,2$,\,$0<r\ll1$,
\begin{eqnarray}\label{1.10}
&&\hat{X}^{j}_{1}(k):=\mathcal{F}\{X^{j}\chi_{[2a,0]}(y-b)\}(k);\\\label{1.11}
&&\hat{X}^{j}_{2}(k):=\mathcal{F}\{X^{j}\chi_{[2a-2r,2a]}(y-b)\}(k);\\\label{1.12}
&&\hat{X}^{j}_{3}(k):=\mathcal{F}\{X^{j}\chi_{[2a-2b,2a-2r]}(y-b)\}(k);\\
&&\hat{Y}^{j}_{1}(k):=\mathcal{F}\{X^{j}\chi_{[2a,0]}(y+b)\}(k);\\
&&\hat{Y}^{j}_{2}(k):=\mathcal{F}\{X^{j}\chi_{[0,2r]}(y+b)\}(k);\\
&&\hat{Y}^{j}_{3}(k):=\mathcal{F}\{X^{j}\chi_{[2r,2b]}(y+b)\}(k),
\end{eqnarray}
in which $\chi_{[2a,0]}(x)$ is the characteristic function defined on $[2a,0]$, and so on. The support of linear waves $D_{y}A^{j}_{-}(x,y)$ is illustrated in the shaded areas in Figure \ref{F}. Most importantly, $X^{j}\chi_{[2a-2r,2a]}(y-b)$ in not in the domain of influence of $V^{j}\chi[a,0]$ when time variable $x= b$. The function $D_{y}A^{j}_{-}(x,y)=X(y-x)+Y(y+x)$ satisfies the wave equation for $x\geq b$. For $x\leq a$, we have $D_{y}A^{j}_{-}(x,y)=-\delta'(x-y)$. If $b=0$, we firstly see the support of $X^{j}\chi_{[2a,0]}(y)$. For $b\gg0$, we find the red triangle in the center of diagram that shows the support of the wave solution that is not affected by the $V^{j}\chi[a,0]$, and simply depends on $V^{j}\chi[0,b]$. We refer more detailed analysis construction and characteristics analysis to Figure \ref{F}.
\begin{figure}\begin{center}\includegraphics[scale=0.23]{support.pdf}\caption{characteristics of linear waves}\label{F}\end{center}\end{figure}

\par
Let
$$\det S^{j}(k):=\frac{-\hat{X}^{j}(-k)}{\hat{X}^{j}(k)},\,j=1,2,
$$
be the corresponding scattering determinant of of potential $V^{j}$.
\begin{theorem}\label{11}
If $V^{1}(x)\equiv V^{2}(x)$ on non-empty $[0,b]\subset[a,b]$, and $\det S^{1}(k)=\det S^{2}(k)$, then $V^{1}(x)\equiv V^{2}(x)$ on $[a,b]$.
\end{theorem}
In literature, when there are no bound states, the potential is determined by the reflection coefficients by Faddeev's theory \cite{DT79,Fa64}.
In inverse resonance problem, we consider to determine the potential $V$ from the resonances of~(\ref{1.1}) which includes the square root of $L^{2}$-eigenvalues. The inverse resonance problem of Schr\"{o}dinger operator on the half line has been studied in \cite{Korotyaev1,Korotyaev3,Zworski2}. In the half line case, the unique recovery of the potential from the eigenvalues and resonances is justified in \cite{Korotyaev1}. However, in the full line case, the inverse resonance problems mainly remained open for a long time. It is known that the potential cannot be solely determined by the eigenvalues and resonances. Specifically, Zworski \cite{Zworski2} proved the uniqueness theorem for the symmetric potentials along with certain isopolar results. Furthermore, Korotyaev \cite{Korotyaev1,Korotyaev3} applied the value distribution theory in complex analysis to prove that all eigenvalues and resonances, and a signed sequence can uniquely determine the potential $V$.

\section{Lemmata}
\begin{lemma}\label{21}
If $F(x)$ is of bounded variation on $(-\infty,\infty)$, then $F(x)$ is constant except on some finite interval if and only if $f(z)=\int_{-\infty}^{\infty}e^{izt}dF(t)$ is an entire functional of exponential type; and if $(a,b)$ is the smallest interval outside which $F(x)$ is a constant, then $a=h_{f}(-\frac{\pi}{2})$ and $b=h_{f}(\frac{\pi}{2})$.
\end{lemma}
\begin{proof}
We refer the proof to Boas \cite[p.\,109]{Boas}.
\end{proof}
Here, we note that the indicator diagram of $\hat{X}^{j}(k)$ is the line set $i[-2(b-a),0]$ on the imaginary axis, and its length is $|a|+|b|$.
\begin{lemma}
The length of indicator diagram of $\hat{X}^{j}(k)$ is $2(b-a)$.
\end{lemma}
\begin{proof}
We use Lemma \ref{21} and~(\ref{1.6}) to conclude the result.
\end{proof}
\begin{lemma}
$\hat{X}^{j}_{1}(k)$ has indicator function $|a||\sin\theta|$;\,$\hat{X}^{j}_{2}(k)$ has indicator function $|r||\sin\theta|$;\,$\hat{X}^{j}_{3}(k)$ has indicator function $|b-r||\sin\theta|$.
\end{lemma}
\begin{proof}
It is straightforward by Lemma \ref{21},~(\ref{1.10}),~(\ref{1.11}), and~(\ref{1.12}).
\end{proof}
\begin{lemma}\label{24}
$\hat{X}^{j}_{j'}(k)$ has only finite zeros in $\mathbb{C}^{+}$, and infinitely many in $\mathbb{C}^{-}$.
\end{lemma}
\begin{proof}
This is well-known in the literature, say, \cite{Dya}.
\end{proof}
\begin{lemma}\label{id}
If $V^{1}\equiv V^{2}$ on $[0,b]$, then $\hat{X}^{1}_{2}(k)\equiv\hat{X}^{1}_{2}(k)$, and $\hat{Y}^{1}_{2}(k)\equiv\hat{Y}^{1}_{2}(k)$.
\end{lemma}
\begin{proof}
Let us discuss the Figure \ref{F}. Let us first take $b=0$, so the wave starting inside the triangle: $(0,0)$, $(a,a)$, $(0,2a)$ propagate in parallel strips between the shaded area between the lines pointing at $(a,a)$ and $(0,2a)$ and the lines pointing at $(a,a)$ and $(0,0)$. The wave function $X^{j}_{1}$ and $Y^{j}_{1}$, $j=1,2$, are defined on the those two strips correspondingly. Now we consider $x\in[0,b]$, if $V^{1}_{2}\equiv V^{2}_{2}$ on $[0,b]$,  then the wave function $Y^{1}_{2}\equiv Y^{2}_{2}$ are defined on the strips between the lines contain $(0,0)$ and $(b,b)$. Similarly, the wave function $X^{1}_{2}\equiv X^{2}_{2}$ are defined on the strips between the lines contain $(0,2a)$ and $(b,2a-b)$. Now we deduce from~(\ref{1.10}) that $\hat{X}^{1}_{2}(k)\equiv\hat{X}^{1}_{2}(k)$. Similar arguments works for $\hat{Y}^{1}_{2}(k)\equiv\hat{Y}^{1}_{2}(k)$.

\end{proof}

\section{Proof of Theorem \ref{11}}

We start with the assumption of Theorem \ref{1.1}
$$\det S^{1}(k)\equiv\det S^{2}(k),$$
that is
\begin{equation}\label{3.1}
\frac{\hat{X}^{1}(-k)}{\hat{X}^{1}(k)}\equiv\frac{\hat{X}^{2}(-k)}{\hat{X}^{2}(k)}.
\end{equation}
Using Lemma \ref{24} and comparing the poles on both sides of~(\ref{3.1}), $\hat{X}^{1}(k)$ and $\hat{X}^{2}(k)$ have common zeros in $\mathbb{C}^{-}$, except finite ones in $\mathbb{C}^{+}$.
If $\sigma$ is a common zero of $\hat{X}^{1}(k)$ and $\hat{X}^{2}(k)$, then
\begin{equation}
\hat{X}^{1}_{1}(\sigma)
+\hat{X}^{1}_{3}(\sigma)=\hat{X}^{2}_{1}(\sigma)+\hat{X}^{2}_{3}(\sigma)=0,
\end{equation}
by Lemma \ref{id}. Now we want to show that
\begin{equation}\label{3.2}
\hat{X}^{1}(k):=\hat{X}^{1}_{1}(k)+\hat{X}^{1}_{2}(k)
+\hat{X}^{1}_{3}(k)\equiv\hat{X}^{2}_{1}(k)+\hat{X}^{2}_{2}(k)+\hat{X}^{2}_{3}(k)=:\hat{X}^{2}(k).
\end{equation}
That is,
\begin{equation}\label{3.22}
\hat{X}^{1}_{1}(k)
+\hat{X}^{1}_{3}(k)\equiv\hat{X}^{2}_{1}(k)+\hat{X}^{2}_{3}(k).
\end{equation}
By Lemma \ref{id},~(\ref{3.2}) and~(\ref{3.22}) should have same density of common zeros.
\par
Let us count the zero set of
\begin{equation}
F(k):=\hat{X}^{j}(k),\,j=1,2,
\end{equation}
and the zero set of
\begin{equation}\label{3.3}
G(k):=\hat{X}^{1}_{1}(k)
+\hat{X}^{1}_{3}(k)-\hat{X}^{2}_{1}(k)-\hat{X}^{2}_{3}(k).
\end{equation}
Using ~(\ref{119}) and~(\ref{120}) in Appendix, we deduce that
\begin{eqnarray}
&&h_{F}(\theta)=(b-a)|\sin\theta|;\\
&&h_{G}(\theta)=\max\{|a|,|b-r|\}|\sin\theta|=|b-r||\sin\theta|.
\end{eqnarray}
Using Theorem \ref{C}, $F(z)$  has zero density $\frac{(b-a)}{\pi}$, and $G(k)$  has identical zero density $\frac{b-r}{\pi}$. That contradicts to Lemma \ref{id}. Then, we deduce that
$$
\hat{X}^{1}_{1}(k)
+\hat{X}^{1}_{3}(k)\equiv\hat{X}^{2}_{1}(k)+\hat{X}^{2}_{3}(k)
.$$
Due to Lemma \ref{id}, we have
$$\hat{X}^{1}(k)\equiv\hat{X}^{2}(k).$$
\par
Using~(\ref{U}), we have
\begin{equation}\label{312}
\hat{X}^{j}(k)\hat{X}^{j}(-k)=k^{2}+\hat{Y}^{j}(k)\hat{Y}^{j}(-k).
\end{equation}
Thus, we obtain
\begin{equation}\label{313}
\hat{Y}^{1}(k)\hat{Y}^{1}(-k)\equiv\hat{Y}^{2}(k)\hat{Y}^{2}(-k),\,k\in\mathbb{C}.
\end{equation}
Equivalently,
$$|\hat{Y}^{1}(k)|^{2}=|\hat{Y}^{2}(k)|^{2},\,k\in\mathbb{R},$$
and we apply Theorem \ref{NL} to deduce
\begin{equation}\label{3100}
\hat{Y}^{1}(k)\prod_{n=1}^{\infty}\frac{1-\frac{k}{\overline{a}^{1}_{n}}}{1-\frac{k}{a^{1}_{n}}}=e^{i\gamma}\hat{Y}^{2}(k)\prod_{n=1}^{\infty}\frac{1-\frac{k}{\overline{a}^{2}_{n}}}{1-\frac{z}{a^{2}_{n}}},
\end{equation}
where $\{a_{n}^{j}\}$ are the zeros of $\hat{Y}^{j}(k)$ in $\mathbb{C}^{+}$. The Blaschke product
$$\prod_{n=1}^{\infty}\frac{1-\frac{k}{\overline{a}^{j}_{n}}}{1-\frac{k}{a^{j}_{n}}}$$ is a function of zero type and of zero density of zeros. We refer the detail to \cite{Boas}. Again, the zero density of~(\ref{3100}) is $\frac{b-a}{\pi}$. Thus, $\hat{Y}^{1}(k)$ and $\hat{Y}^{2}(k)$ have common zero of density $\frac{b-a}{\pi}$. That is,
\begin{equation}
\hat{Y}^{1}(\sigma)=\hat{Y}^{2}(\sigma),\,\forall\sigma\in\Sigma,
\end{equation}
in which $\Sigma$ is a set of density $\frac{b-a}{\pi}$. That is,
\begin{equation}
\hat{Y}^{1}_{1}(\sigma)+\hat{Y}^{1}_{2}(\sigma)+\hat{Y}^{1}_{3}(\sigma)=\hat{Y}^{2}_{1}(\sigma)+\hat{Y}^{2}_{2}(\sigma)+\hat{Y}^{2}_{3}(\sigma),\,\forall\sigma\in\Sigma.
\end{equation}
That is, by Lemma \ref{id},
\begin{equation}
\hat{Y}^{1}_{1}(\sigma)+\hat{Y}^{1}_{3}(\sigma)=\hat{Y}^{2}_{1}(\sigma)+\hat{Y}^{2}_{3}(\sigma),\,\forall\sigma\in\Sigma,
\end{equation}
which however could have a zero set of density $\frac{b-a-r}{\pi}$.
Hence, $\hat{Y}^{1}_{1}(k)+\hat{Y}^{1}_{3}(k)\equiv\hat{Y}^{2}_{1}(k)+\hat{Y}^{2}_{3}(k)$, and then
$$\hat{Y}^{1}(k)\equiv\hat{Y}^{2}(k).$$
Therefore, we deduce that $V^{1}$ and $V^{2}$ have the same scattering matrix;
\begin{equation}
S^{1}(k)\equiv S^{2}(k).
\end{equation}
Using Zworski \cite[Proposition\,8]{Zworski}, which says that the potential function with compact support is determined by the scattering matrix, we deduce that
\begin{eqnarray}
V^{1}(x)\equiv V^{2}(x).
\end{eqnarray}
This proves the theorem.

\section{Appendix}
We review some results from complex analysis  \cite{Boas,Levin2}.
\begin{definition}
Let $f(z)$ be an entire function. Let
\begin{equation}\nonumber
M_f(r):=\max_{|z|=r}|f(z)|.
\end{equation}
An entire function of $f(z)$ is said
to be a function of finite order if there exists a positive
constant $k$ such that the inequality
\begin{equation}\nonumber
M_f(r)<e^{r^k}
\end{equation}
is valid for all sufficiently large values of $r$. The greatest
lower bound of such numbers $k$ is called the order of the entire
function $f(z)$. By the type $\sigma$ of an entire function $f(z)$
of order $\rho$, we mean the greatest lower bound of positive
number $A$ for which asymptotically we have
\begin{equation}\nonumber
M_f(r)<e^{Ar^\rho}.
\end{equation}
That is,
\begin{equation}\nonumber
\sigma_{f}:=\limsup_{r\rightarrow\infty}\frac{\ln M_f(r)}{r^\rho}.
\end{equation}  If $0<\sigma_{f}<\infty$, then we say
$f(z)$ is of normal type or mean type. For $\sigma_{f}=0$, we say $f(z)$ is of minimal type.
\end{definition}
We refer the details to \cite{Levin2}.
\begin{definition}\label{33}
Let $f(z)$ be an integral function of finite order $\rho$ in the
angle $[\theta_1,\theta_2]$. We call the following quantity as the
indicator function of function $f(z)$.
\begin{equation}\nonumber
h_f(\theta):=\lim_{r\rightarrow\infty}\frac{\ln|f(re^{i\theta})|}{r^{\rho}},
\,\theta_1\leq\theta\leq\theta_2.
\end{equation}
\end{definition}
The type of a function is connected to the maximal value of indicator function.
\begin{definition}\label{d}
The following quantity is called the width of the indicator
diagram of entire function $f$:
\begin{equation}\label{22222}
d=h_f(\frac{\pi}{2})+h_f(-\frac{\pi}{2}).
\end{equation}
\end{definition}

\begin{definition}\label{255}
Let $f(z)$ be an integral function of order $1$, and let
$n(f,\alpha,\beta,r)$ denote the number of the zeros of $f(z)$
inside the angle $[\alpha,\beta]$ and $|z|\leq r$. We define the
density function as
\begin{equation}\nonumber
\Delta_f(\alpha,\beta):=\lim_{r\rightarrow\infty}\frac{n(f,\alpha,\beta,r)}{r},
\end{equation}
and
\begin{equation}\nonumber
\Delta_f(\beta):=\Delta_f(\alpha_0,\beta),
\end{equation}
with some fixed $\alpha_0\notin E$ such that $E$ is at most a
countable set \cite{Boas,Levin,Levin2}. In particular, we denote the density function of $f$ on the open right/left half complex plane as $\Delta^{+}_{f}$/$\Delta^{-}_{f}$ respectively. Similarly, we can define the set density of a zero set $S$. Let $n(
S,r)$ be the number of the discrete elements of $S$ in $\{|z|<r\}$. We define
\begin{equation}\nonumber
\Delta_S:=\lim_{r\rightarrow\infty}\frac{n(S,r)}{r},
\end{equation}

\end{definition}
\begin{theorem}[Cartwright]\label{C}
Let $f$ be an entire function of exponential type with zero set $\{a_{k}\}$. We assume $f$ satisfies one of the
following conditions:
\begin{equation}\nonumber
\mbox{ the integral
}\int_{-\infty}^\infty\frac{\ln^+|f(x)|}{1+x^2}dx\mbox{ exists}.
\end{equation}
\begin{equation}\nonumber
|f(x)|\mbox{ is bounded on the real axis}.
\end{equation}
Then
\begin{enumerate}
\item all of the zeros of the function $f(z)$, except possibly
those of a set of zero density, lie inside arbitrarily small
angles $|\arg z|<\epsilon$ and $|\arg z-\pi|<\epsilon$, where the
density
\begin{equation}
\Delta_f(-\epsilon,\epsilon)=\Delta_f(\pi-\epsilon,\pi+\epsilon)=\lim_{r\rightarrow\infty}
\frac{n(f,-\epsilon,\epsilon,r)}{r}
=\lim_{r\rightarrow\infty}\frac{n(f,\pi-\epsilon,\pi+\epsilon,r)}{r},
\end{equation}
is equal to $\frac{d}{2\pi}$, where $d$ is the width of the
indicator diagram in~(\ref{22222}). Furthermore, the limit
$\delta=\lim_{r\rightarrow\infty}\delta(r)$ exists, where
$$
\delta(r):=\sum_{\{|a_k|<r\}}\frac{1}{a_k};
$$
\item moreover,
\begin{equation}\nonumber
\Delta_f(\epsilon,\pi-\epsilon)=\Delta_f(\pi+\epsilon,-\epsilon)=0;
\end{equation}
\item the function $f(z)$ can be represented in the form
\begin{equation}\nonumber
f(z)=cz^me^{i\kappa
z}\lim_{r\rightarrow\infty}\prod_{\{|a_k|<r\}}(1-\frac{z}{a_k}),
\end{equation}
where $c,m,\kappa$ are constants and $\kappa$ is real;
\item the indicator
function of $f$ is of the form
\begin{equation}
h_f(\theta)=\sigma|\sin\theta|.
\end{equation}
\end{enumerate}
\end{theorem}
We refer the Cartwright's theory to Levin \cite[p,251]{Levin}.
\begin{lemma}\label{36}
Let $f$, $g$ be two entire functions. Then the following two
inequalities hold.
\begin{eqnarray}
&&h_{fg}(\theta)\leq h_{f}(\theta)+h_g(\theta),\mbox{ if one limit exists};\label{119}\\\label{120}
&&h_{f+g}(\theta)\leq\max_\theta\{h_f(\theta),h_g(\theta)\},
\end{eqnarray}
where the equality in~(\ref{119}) holds if one of the functions is of completely regular growth, and secondly the equality~(\ref{120}) holds if the indicator of the two summands are not equal at some $\theta_0$.
\end{lemma}
\begin{proof}
 We can find
the details in \cite{Levin}.
\end{proof}

\begin{lemma}\label{38}
The Fourier transform $\hat{X}(z)$ as in~(\ref{1.6}) is of Cartwright class, and the function can be represented in the form
$$\hat{X}(z)=cz^{m}e^{i\delta z}\lim_{R\rightarrow\infty}\prod_{|\sigma_{n}|<R}(1-\frac{z}{\sigma_{n}}),\,z=x+iy,$$
where $\delta\in\mathbb{R}$, and the following integral converges:
\begin{equation}\label{2100}
\int_{-\infty}^{\infty}\frac{\ln^{+}|\hat{X}(x)|}{1+x^{2}}dx<\infty.
\end{equation}
Similar results hold for $\hat{Y}(k)$.
\end{lemma}
\begin{proof}
We refer the definition of Cartwright class to \cite{Levin,Levin2}.
\end{proof}
\begin{theorem}[Nevanlinna-Levin]\label{NL}
If the function $F(z)$ is holomorphic and of exponential type in the half-plane $\Im z\geq0$, and if~(\ref{2100}) holds, then
\begin{enumerate}
\item \begin{equation}\nonumber
F(z)\prod_{k=1}^{\infty}\frac{1-\frac{z}{\overline{a}_{k}}}{1-\frac{z}{a_{k}}}=e^{i\gamma}e^{u(z)+iv(z)},
\end{equation}
where $$u(z)=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\ln|F(t)|}{(t-x)^{2}+y^{2}}dt+\sigma^{+}_{F} y,$$  $\sigma^{+}_{F}=h_{F}(\frac{\pi}{2})$, $v(z)$ is the harmonic conjugate of $u(z)$, and $\{a_{k}\}$ are the zeros of the function $F(z)$ in the half-plane $\Im z>0$;
\item
 $$
\ln|F(z)|=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\ln|F(t)|}{(t-x)^{2}+y^{2}}dt+\sigma^{+}_{F} y+\ln|\chi(z)|,\,z=x+iy,
$$
where
$$\chi(z)=\prod_{k=1}^{\infty}\frac{1-\frac{z}{a_{k}}}{1-\frac{z}{\overline{a}_{k}}}.$$
\end{enumerate}
\end{theorem}
\begin{proof}
We refer the proof to \cite[p.\,240]{Levin}.
\end{proof}