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\chapter{The $\phi^4$ kink on a wormhole spacetime} \label{chap:wormhole}

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\graphicspath{{Chapter2/Figs/Static/}{Chapter2/Figs/Dynamic/}}

The soliton resolution conjecture \cite{TT} states that solutions to solitonic equations with generic initial data should, after some non--linear behaviour, eventually resolve into a finite number of solitons plus a radiative term. This conjecture is intimately tied to soliton stability, which has been investigated for a number of solitonic equations, including that of $\phi^4$ theory on $\mathbb{R}^{1,1}$. In this chapter, we study a modification of this theory on a $3+1$ dimensional wormhole spacetime which has a spherical throat of radius $a$, with a focus on the stability properties of the modified kink. In particular, we prove that the modified kink is linearly stable, and compare its discrete spectrum to that of the $\phi^4$ kink on $\mathbb{R}^{1,1}$. We also study the resonant coupling between the discrete modes and the continuous spectrum for small but non--linear perturbations. Some numerical and analytical evidence for asymptotic stability is presented for the range of $a$ where the kink has exactly one discrete mode. This chapter is almost identical to the preprint \cite{me}.

\section{Background: the $\phi^4$ kink on $\mathbb{R}^{1,1}$}

One dimensional $\phi^4$ theory is well--documented in the literature (see for example \cite{Manton&Sutcliffe}). The aim of this section is to introduce some notation and some ideas about stability which will be useful when we come to consider the modified theory.

\subsection{Topological stability and the kink solution}
The action takes the form
\[
S=\int_{\R^2}\bigg( \frac{1}{2}\eta^{ab}\partial_a\phi\partial_b\phi + \frac{1}{2}(1-\phi^2)^2\bigg)dxdt,
\]
where $x^a=(t,x)$ are coordinates on $\mathbb{R}^{1,1}$ and $\eta^{ab}$ is the Minkowski metric with signature $(-,+)$. Note that it has two vacua, given by $\phi=\pm 1$. Finiteness of the associated conserved energy
\be \label{eq:E_integral_flat}
E = \int_\R \bigg( \frac{1}{2}(\phi_t)^2 + \frac{1}{2}(\phi_x)^2 + \frac{1}{2}(1-\phi^2)^2 \bigg)dx,
\ee
requires that the field lies in one of these two vacua in the limits $\phi_\pm=\mathrm{lim}_{x\rightarrow \pm \infty}[\phi(x)]$. We can thus classify finite energy solutions in terms of their topological charge $N = (\phi_+-\phi_-)/2$, which takes values in $\{-1,0,1\}$.

The equations of motion are
\begin{equation}
\label{eom:R11}
\phi_{tt}=\phi_{xx} + 2\phi(1-\phi^2)
\end{equation}
and we find a static solution $\phi=\mathrm{tanh}(x-c)$ which we call the flat kink. It interpolates between the two vacua and thus has topological charge $N=1$. The constant of integration $c$ can be thought of as the position of the kink. We will henceforth use $\Phi_0$ to denote the static kink at the origin, that is, $\Phi_0(x)=\mathrm{tanh}(x)$. Since any small departure from $\phi=\pm 1$ in the limits $x\rightarrow\pm\infty$ results in non--convergence of the integral (\ref{eq:E_integral_flat}), no finite energy deformation can affect $N$. For this reason, we say that the kink is \textit{topologically stable}.

\subsection{Linear stability}

A second notion of stability which will be important to our discussion is linear stability. On discarding non--linear terms, we find that small perturbations $\phi(t,x)=\Phi_0(x)+\mathrm{e}^{i\omega_0 t}v_0(x)$ satisfy the Schr\"odinger equation
\begin{equation}
\label{eq:zero_order}
L_0v_0:=-v_0^{\prime\prime} - 2(1-3\Phi_0^2)v_0 = \omega^2_0v_0.
\end{equation}
The potential $V_0(x)=-2[1-3\Phi_0(x)^2]$ exhibits a so--called ``mass gap'', meaning that it takes a finite positive value in the limits $x\rightarrow\pm\infty$. In this case, $V_0(\pm\infty)=4$. For $\omega_0^2>4$, (\ref{eq:zero_order}) admits a continuous spectrum of wave--like solutions.

%\begin{figure}
%\includegraphics[width=14cm]{flatQMpotential.pdf}
%\caption{\label{fig:flatQMpotential}The Schr\"odinger potential $V_0(r)$ appearing in the linear stability analysis of the $\phi^4$ kink on $\mathbb{R}^{1,1}$}.
%\end{figure}

In addition to its continuum states, the Schr\"odinger operator in (\ref{eq:zero_order}) has two discrete eigenvalues with normalisable solutions given by
\begin{equation}
\label{eq:flat_vib_modes}
\big(v_0(x),\omega_0\big) = \bigg(\frac{\sqrt{3}}{2}\mathrm{sech}^2(x),0\bigg) \quad \mathrm{and} \quad
\big(v_0(x),\omega_0\big) = \bigg(\frac{\sqrt{3}}{\sqrt{2}}\mathrm{sech}(x)\mathrm{tanh}(x),\sqrt{3}\bigg),
\end{equation}
where we have chosen the normalisation constant such that $\int_{-\infty}^{\infty}v_0^2(x)dx=1$.

The first of these is the zero mode of the kink. Its existence is guaranteed by the translation invariance of (\ref{eom:R11}), and up to a multiplicative constant it is equal to $\Phi_0^\prime(x)$. Excitation of this state corresponds to performing a Lorentz boost. In the non--relativistic limit, this amounts to replacing $\Phi_0(x)$ with $\Phi_0(x-vt)$ for some $v\ll 1$ \cite{Manton&Sutcliffe}.

The second normalisable solution, called an \textit{internal} or \textit{discrete} mode, has non--zero frequency $\omega_0$, and is thus time periodic. In the full non--linear theory, it decays through resonant coupling to the continuous spectrum \cite{Manton&Merabet}. This phenomenon is of considerable interest in non--linear PDEs, and was studied in a more general setting in \cite{SW98}. The corresponding process in the modified theory will be discussed in Section \ref{sec:dynamics}.

Linear stability of the kink is equivalent to the Schr\"odinger operator $L_0$ in (\ref{eq:zero_order}) having no negative eigenvalues, so that linearised perturbations cannot grow exponentially with time. One way to see that the kink is linearly stable is via the Sturm oscillation theorem:

\begin{theo}[Sturm]\label{th:sturm} Let $L$ be a differential operator of the form
\[
L=-\frac{d^2}{dx^2}+V(x)
\]
on the smooth square integrable functions $u$ on the interval $[0,\infty)$, with the boundary condition $u(0)=0$ (corresponding to even parity) or $u^\prime(0)=0$ (corresponding to odd parity). Let $\omega^2$ be an eigenvalue of $L$ with associated eigenfunction $u(x;\omega)$. Then the number of eigenvalues of $L$ (subject to the appropriate boundary conditions) which are strictly below $\omega^2$ is the number of zeros of $u(x;\omega)$ in $(0,\infty)$.
\end{theo}

Note that the symmetry of (\ref{eq:zero_order}) under $x\mapsto -x$ means that any solution on the interval $[0,\infty)$ has a corresponding solution on the interval $(-\infty,0]$, and these solutions can be pieced together to make a smooth solution on $(-\infty,\infty)$ as long as the boundary conditions at $x=0$ are chosen to ensure parity $\pm1$. Thus there is a one--to--one correspondence between solutions on $[0,\infty)$ and solutions on $(-\infty,\infty)$ which are smooth at $x=0$.
Since the eigenfunctions (\ref{eq:flat_vib_modes}) have no zeros on the interval $[0,\infty)$, it follows that there can be no eigenfunctions with $\omega_0^2<0$, and thus the kink is linearly stable.

\subsection{Asymptotic stability}
The final notion of stability that we will consider is that of asymptotic stability. Stated simply, asymptotic stability of the kink means that for sufficiently small initial pertubations, solutions of (\ref{eom:R11}) will converge locally to $\Phi_0(x)$ or its Lorentz boosted counterpart. This was proved in \cite{KowalczykMM} for odd perturbations, but has not been proved in the general case.

\subsection{Derrick's scaling argument}
Generalisation of the finite energy $\phi^4$ kink to higher dimensional Minkowski spacetimes is prohibited by a scaling argument due to Derrick \cite{Derrick}. Suppose $\Phi_n({\bf x})$ is a static, finite energy solution to the equation of motion of the $\phi^4$ theory on $\mathbb{R}^{1,n}$. Then it is a minimiser of the (static) energy
\[
E(\Phi_n)=\int\Big(\nabla\Phi_n({\bf x})\cdot\nabla\Phi_n({\bf x}) + U(\Phi_n)\Big)d^nx=:E_1+E_2,
\]
where we have split $E$ into the two components coming from the two different terms in the integrand. Now consider a spatial rescaling ${\bf x}\rightarrow\mu{\bf x}$, $\mu>0$ and define
\begin{align*}
e(\mu)= E(\Phi_n(\mu{\bf x}))=&\int\Big(\nabla(\Phi_n(\mu {\bf x}))\cdot\nabla(\Phi_n(\mu {\bf x})) + U(\Phi_n(\mu {\bf x}))\Big)d^dx \\
=& \int\Big(\mu^2\nabla\Phi_n(\mu {\bf x})\cdot\nabla\Phi_n(\mu {\bf x}) + U(\Phi_n(\mu {\bf x}))\Big)d^dx \\
=& \mu^{2-n}E_1 + \mu^{-n}E_2,
\end{align*}
where we have obtained the last line by a change of variables from ${\bf x}$ to $\mu{\bf x}$.

If $\Phi_n({\bf x})$ is a minimiser of $E$ then $\mu=1$ must also be a stationary point of $e(\mu)$. Evaluating the derivative yields
\[
e^\prime(\mu)= \begin{cases}
      -n\mu^{-n-1}E_2, & \text{if}\ n=2 \\
      (2-n)\mu^{1-n}E_1 -n\mu^{-n-1}E_2, & \text{otherwise}.
    \end{cases}
\]
Since $E_1$, $E_2$ and $\mu$ are all positive, the derivative can only have a zero only when $n$ and $2-n$ have the same sign, which only happens when $n=1$. We thus conclude that no static, finite energy solutions to the equations of motion exist for $n>1$.

In order to construct a higher dimensional $\phi^4$ kink, we must add curvature. In the next section we introduce a curved background, and show that a modified $\phi^4$ kink exists on this background. We will also examine a limit in which the modified kink reduces to the flat kink. In Section \ref{sec:linearised_pert} we consider linearised perturbations around the modified kink, proving that it is linearly stable and comparing its discrete spectrum to that of the flat kink. In Section \ref{sec:dynamics} we examine the mode of decay to the modified kink in the full non--linear theory, in particular the resonant coupling of its internal modes to the continuous spectrum.

\section{The static kink on a wormhole}

We now replace the flat $\mathbb{R}^{1,1}$ background with a wormhole spacetime $(M,g)$, where
\be
\nonumber
g=-dt^2+dr^2+(r^2+a^2)(d\vartheta^2 + \sin^2\vartheta d\varphi^2)
\ee
for some constant $a>0$, and $-\infty<r<\infty$. This spacetime was first studied by Ellis \cite{Ellis} and Bronnikov \cite{Bronnikov}, and has featured in a number of recent studies about kinks and their stability \cite{wavemaps,SG}. Note the presence of asymptotically flat ends as $r\rightarrow\pm\infty$, connected by a spherical throat of radius $a$ at $r=0$.

Our action is modified by the presence of a non-flat metric:
\be
\nonumber
S=\int\bigg( \frac{1}{2}g^{ab}\p_a\phi\p_b\phi + \frac{1}{2}(1-\phi^2)^2\bigg)\sqrt{-g}dx,
\ee
where $x^a$ are now local coordinates on $M$. Variation with respect to $\phi$ gives
%\be
%\nonumber
%\begin{split}
%\delta S & =  \int \big( g^{ab}\p_a\delta\phi\p_b\phi - 2\phi(1-\phi^2)\delta\phi\big)\sqrt{-g}dx \\
% & = \int \bigg(- \frac{1}{\sqrt{-g}}\p_a(g^{ab}\sqrt{-g}\p_b\phi)-2\phi(1-\phi^2)\bigg)\delta\phi\sqrt{-g}dx,
%\end{split}
%\ee
%yielding the equation of motion
\be
\label{eom.coordfree}
\square_g\phi+2\phi(1-\phi^2)=0
\ee
where $\square_g\phi=\frac{1}{\sqrt{-g}}\p_a(g^{ab}\sqrt{-g}\p_b\phi)$. We will always assume $\phi$ is independent of the angular coordinates $(\vartheta,\varphi)$, so (\ref{eom.coordfree}) can be written explicitly as
\be
\label{eom.coords}
\phi_{tt}=\phi_{rr}+\frac{2r}{r^2+a^2}\phi_r+2\phi(1-\phi^2).
\ee

The conserved energy in the theory is given by
\be
\nonumber
E=\int_{-\infty}^{+\infty} \bigg( \frac{1}{2} (\phi_t)^2 + \frac{1}{2} (\phi_r)^2 + \frac{1}{2}(1-\phi^2)^2\bigg)(r^2+a^2)dr,
\ee
which we require to be finite. This imposes the condition $\phi^2\rightarrow 1$ as $r\rightarrow\pm\infty$, so that the field lies at one of the two vacua at both asymptotically flat ends.

Static solutions $\phi(r)$ satisfy
\be
\label{eom.static}
\phi^{\prime\prime}+\frac{2r}{r^2+a^2}\phi^{\prime}=-\frac{d}{d\phi}\bigg(-\frac{1}{2}(1-\phi^2)^2\bigg),
\ee
which, imagining $r$ as a time coordinate, can be thought of as a Newtonian equation of motion for a particle at position $\phi$ moving in a potential $\mathcal{U}(\phi)=-(1-\phi^2)/2$, with a time dependent friction term.

In addition to the two vacuum solutions, we have a single soliton solution which interpolates between the saddle points at $(-1,0)$ and $(1,0)$ in the $(\phi,\phi^\prime)$ plane. Its existence and uniqueness among odd parity solutions follow from a shooting argument: suppose the particle lies at $\phi=0$ when $r=0$. If its velocity $\phi^\prime(0)$ is too small, it will never reach the local maximum of the potential at $\phi=1$, but if $\phi^\prime(0)$ is too large it will overshoot the maximum so that $\mathcal{U}(\phi)\rightarrow -\infty$ as $r\rightarrow\infty$, thus having infinite energy. Continuity ensures that there is some critical velocity $\phi^\prime(0)$ such that the particle reaches $\phi=1$ in infinite time and has zero velocity upon arrival. This corresponds to the non--trivial kink solution, which we call $\Phi(r)$. Time reversal implies that $\phi\rightarrow -1$ as $r\rightarrow -\infty$, and that the anti--kink $\phi(r)=-\Phi(r)$ is also a solution.

We can find $\Phi(r)$ numerically using a shooting method for the gradient at $r=0$. %For $a=1$ we find $\phi^\prime(0)=1.379602$. A plot of the corresponding soliton is shown in Figure \ref{fig:soliton}.
Figure \ref{fig:several_kinks} shows such numerically generated kinks for several values of $a$. Note that the absolute value of $\Phi(r)$ is always greater than or equal to that of the flat kink $\Phi_0(r)$, and that at fixed non--zero $r$, the absolute value of $\Phi$ decreases as $a$ increases. The reason for this will become clear in Section \ref{sec:linearised_pert}. In Section \ref{sec:large_a} we examine $\Phi(r)$ in the limit where $a$ is large, finding that it reduces to the flat kink $\Phi_0(r)$, and examining its departure from the flat kink at first order in $1/a^2$.

We again label the values at the boundary as
\[
\phi_\pm := \lim_{r\rightarrow\pm\infty} \Phi(r)\in \{\pm 1\}.
\]
Since no finite energy deformation can change the value of the topological charge $N=(\phi_+-\phi_-)/2 \in \{-1,0,1\}$, we again conclude that $\Phi(r)$ is topologically stable.

%\begin{figure}
%\includegraphics[width=14cm]{soliton.pdf}
%\caption{\label{fig:soliton}The kink solution $\Phi(r)$ for $a=1$.}
%\end{figure}

\begin{figure}
\includegraphics[width=14cm]{several_kinks.pdf}
\caption{\label{fig:several_kinks}The kink solution for several values of $a$, along with the flat kink $\Phi_0(r)$.}
\end{figure}

\subsection{Large $a$ limit} \label{sec:large_a}
As $a\rightarrow\infty$, equation (\ref{eom.coords}) becomes the standard equation (\ref{eom:R11}) for the flat kink. It is thus helpful to expand the modified kink in $\epsilon^2:=1/a^2$ for small $\epsilon^2$, since we can then solve both (\ref{eom.coords}) and (\ref{schrod}) analytically up to $\mathcal{O}(\epsilon^4)$. We shall denote the static kink by $\Phi_\epsilon(r)$ in this limit. It satisfies
\be
\label{eq.large_a}
\Phi_\epsilon^{\prime\prime}+\frac{2r\epsilon^2}{\epsilon^2 r^2+1}\Phi_\epsilon^{\prime}=-2\Phi_\epsilon(1-\Phi_\epsilon^2).
\ee

Setting $\Phi_\epsilon(r)=\Phi_0(r)+\epsilon^2\Phi_1(r)+\mathcal{O}(\epsilon^4)$ we obtain at order zero the equation (\ref{eom:R11}) of a static kink on $\mathbb{R}^{1,1}$. This has solution $\Phi_0(r)$, where we have chosen the kink at the origin to restrict to solutions with odd parity.

At order $\epsilon^2$ we find that $\Phi_1(r)$ must satisfy
\[
\Phi_1^{\prime\prime} + 2r\mathrm{sech}^2r=2\Phi_1(2-3\mathrm{sech}^2r).
\]
The unique solution which is odd and decays as $r\rightarrow\pm\infty$ is given by
\[
\Phi_1(r)=\frac{1}{24}\mathrm{sech}^2r(f_1(r)+f_2(r)+f_3(r)),
\]
where
\be
\nonumber
\begin{split}
f_1(r) &=r\big[3-8\mathrm{cosh}(2r)-\mathrm{cosh}(4r)\big], \\
f_2(r) &=\mathrm{sinh}(2r)\big[8\mathrm{log}(2\mathrm{cosh}(r))-1\big] + \mathrm{sinh}(4r)\mathrm{log}(2\mathrm{cosh}(r)), \\
f_3(r) &= \frac{\pi^2}{2} + 6r^2 +6\mathrm{Li}_2(-\mathrm{e}^{-2r}),
\end{split}
\ee
and $\mathrm{Li}_2(z)$ is the dilogarithm function, which is the special case $s=2$ of the polylogarithm
\[
\mathrm{Li}_s(z)=\sum_{k=1}^\infty \frac{z^k}{k^s}.
\]

To show that $\Phi_1(r)$ is odd, note that $\mathrm{sech}^2r$ is an even function, and that $f_1$ and $f_2$ are constructed from products of even and odd functions, and hence are odd. To see that $f_3$ is also odd, we use Landen identity for the dilogarithm:
\be
\nonumber
\begin{split}
\mathrm{Li}_2(-\mathrm{e}^{-2r})+\mathrm{Li}_2(-\mathrm{e}^{2r}) &= -\frac{\pi^2}{6} - \frac{1}{2}\big[\mathrm{log}(\mathrm{e}^{-2r})\big]^2 \\
&= -\frac{\pi^2}{6} - 2r^2,
\end{split}
\ee
thus verifying $f_3(r)+f_3(-r)=0$.

We now turn to the behaviour of $\Phi_1(r)$ as $r\rightarrow\infty$. Since $\mathrm{sech}^2r\sim 4\mathrm{e}^{-2r}$ for large $r$, we need only consider terms in the $\{f_i\}$ of order $\mathrm{e}^{2r}$ or higher. We first note that
\be
\nonumber
\begin{split}
\mathrm{log}(2\mathrm{cosh}r)&=\mathrm{log}(\mathrm{e}^r(1+\mathrm{e}^{-2r}))=r+\mathrm{log}(1+\mathrm{e}^{-2r}) \\
&=r+\mathrm{e}^{-2r} + \mathcal{O}(\mathrm{e}^{-4r}).
\end{split}
\ee
Then
\be
\nonumber
\begin{split}
f_1(r)&=-4r\mathrm{e}^{2r} -\frac{r}{2}\mathrm{e}^{4r}+\mathcal{O}(\mathrm{e}^r) \\
f_2(r)&=\frac{1}{2}\mathrm{e}^{2r}(8r+8\mathrm{e}^{-2r}-1)
+\frac{1}{2}\mathrm{e}^{4r}(r+\mathrm{e}^{-2r})+\mathcal{O}(\mathrm{e}^r) \\
&=4r\mathrm{e}^{2r} +\frac{r}{2}\mathrm{e}^{4r}+\mathcal{O}(\mathrm{e}^r),
\end{split}
\ee
so $f_1(r)+f_2(r)=\mathcal{O}(\mathrm{e}^r)$. Since $f_3(r)=\mathcal{O}(r^2)$ for large $r$, we see that $\Phi_1(r)$ vanishes as $r\rightarrow\infty$, as we expect. Note that its vanishing as $r\rightarrow-\infty$ then follows using parity. A plot of $\Phi_1(r)$ is shown in Figure \ref{fig:Phi_1}.

\begin{figure}
\includegraphics[width=14cm]{Phi_1.pdf}
\caption{\label{fig:Phi_1}The order $\epsilon^2$ perturbation to the static kink on $\mathbb{R}^{1,1}$.}
\end{figure}

\section{Linearised perturbations around the kink}
\label{sec:linearised_pert}
To study the linear stability of the kink, we first plug
\be
\label{eq:pert}
\phi(t,r)=\Phi(r)+w(t,r)
\ee
into equation (\ref{eom.coords}), discarding terms non--linear in $w$. Imposing the fact that $\Phi(r)$ satisfies (\ref{eom.static}), we find
\be
\nonumber
w_{tt}=w_{rr}+\frac{2r}{r^2+a^2}w_r+2w(1-3\Phi^2).
\ee
For $w(t,r)=\mathrm{e}^{i\omega t}(r^2+a^2)^{-1/2}v(r)$, this becomes a one--dimensional Schr\"odinger equation
\be
\label{schrod}
Lv:=(-\p_r\p_r+V(r))v=\omega^2 v,
\ee
where the potential is given by
\be
\label{QMpotential}
V(r)=\frac{a^2}{(r^2+a^2)^2}-2(1-3\Phi^2).
\ee

Figure \ref{fig:several_potentials} shows the potential $V(r)$ for several values of $a$. Note that for large $a$ it has a single well with a minimum at $r=0$, close to the potential $V_0$ corresponding to the flat kink. As $a$ decreases, the critical point at $r=0$ becomes a maximum with minima on either side, creating a double well. We find numerically that this happens at about $a=0.55$. Note that those potentials with $a<1/\sqrt{2}$ are everywhere positive.

\begin{figure}
\includegraphics[width=14cm]{several_potentials.pdf}
\caption{\label{fig:several_potentials}The potential of the 1--dimensional quantum mechanics problem arising from the study of stability of the soliton for values of $a$ between $a=10$ and $a=0.3$.}
\end{figure}

\begin{prop}
The kink solution $\Phi(r)$ is linearly stable with respect to spherically symmetric perturbations.
\end{prop}

{\bf Proof.} We first decompose the potential $V(r)$ in (\ref{schrod}) as $V=V_0+V_1+V_a$, where
\[
V_0 = -2[1-3\Phi_0(r)^2],\qquad
V_1 = 6[\Phi(r)^2-\Phi_0(r)^2],\qquad V_a = \frac{a^2}{(r^2+a^2)^2}.
\]
As discussed above, we know that the operator $L_0 = -\p_r\p_r + V_0$ has no negative eigenvalues. It then follows that $L$ itself has no negative eigenvalues as long as the functions $V_1(r)$ and $V_a(r)$ are everywhere non--negative.

The latter is obvious; to prove the former we recall that we can think of $\Phi(r)$ and $\Phi_0(r)$ as the trajectories of particles moving in a potential $\mathcal{U}(\phi)$, where $r$ is imagined as the time coordinate. The particle corresponding to $\Phi(r)$ suffers an increased frictional force compared to $\Phi_0(r)$, i.e.
\be
\label{eq:particles}
\Phi_0^{\prime\prime}=-\frac{\p\mathcal{U}}{\p\phi}\bigg|_{\phi=\Phi_0},\qquad
\Phi^{\prime\prime}+\frac{2r}{r^2+a^2}\Phi^\prime=-\frac{\p\mathcal{U}}{\p\phi}\bigg|_{\phi=\Phi}.
\ee
Both $\Phi$ and $\Phi_0$ interpolate between the maxima of $\mathcal{U}$ at $\phi=\pm 1$; reaching the minimum ($\phi=0$) when $r=0$.

Multiplying the equations (\ref{eq:particles}) by $\Phi_0^\prime$ and $\Phi^\prime$ respectively, then integrating from $r$ to $\infty$, we have that at every instant of time
\be
\label{eq:energy}
\frac{1}{2}(\Phi_0^\prime)^2 + \mathcal{U}(\Phi_0) = 0,\qquad
\frac{1}{2}(\Phi^\prime)^2 + \mathcal{U}(\Phi) = \int_r^\infty\frac{2r}{r^2+a^2}(\Phi^\prime)^2dr.
\ee
These equations are equivalent to conservation of energy for each of the particles. Note that the integral on the right hand side is non--negative for $r\geq 0$, and vanishes only at $r=\infty$. In particular, when $r=0$ we have $\mathcal{U}(\Phi)=\mathcal{U}(\Phi_0)=-1/2$, so $\Phi^\prime(0)>\Phi_0^\prime(0)$. This means $V_1(r)$ is initially increasing from zero.

For $V_1(r)$ to return to zero at some finite $r=r_0$, we would need that $\Phi(r_0)=\Phi_0(r_0)$ at a point where $\Phi^\prime(r_0)\leq\Phi_0^\prime(r_0)$. However, this is made impossible by equations (\ref{eq:energy}), since at such a point $\mathcal{U}(\Phi)=\mathcal{U}(\Phi_0)$ and the integral on the right hand side is positive. Hence $V_1(r)$ remains non--negative for all $r>0$, and thus for all $r$ since it is even in $r$.
\koniec

%{\bf Almost--proof relying on a bit of numerical evidence.} Suppose $v_g(r)$ is the ground state of (\ref{schrod}) and has $\mathcal{E}_g<0$. Our aim is to reach a contradiction. We know that $v_g(r)$ has even parity, and (by the Sturm oscillation theorem) no zeros, so we can make the choice $v_g(0)>0, v_g^\prime(0)=0$. Additionally, since it is a bound state, $v_g^\prime(r)\rightarrow 0$ from below as $r\rightarrow\infty$.

%First note that for $a<1/\sqrt{2}\approx 0.7$, $V(0)>0$. We find that in fact $V(r)$ is everywhere positive for $a<1/\sqrt{2}$ (see Figure \ref{fig:several_potentials} for the numerical evidence for this claim). By (\ref{schrod}) this implies that $v_g^{\prime\prime}(r)$ is everywhere negative. This is impossible, since $v_g^\prime(r)$ must vanish at $r=0$ and $r=\infty$. Thus there are no negative energy bound states for $a<1/\sqrt{2}$.

%For $a>1/\sqrt{2}$, we observe numerically that $V(r)$ only has one critical point: a minimum at $r=0$, and so is monotonically increasing for $0<r<\infty$. By (\ref{schrod}), this means that $v_g^{\prime\prime}(r)$ can have a maximum of one zero in this range, hence $v_g^\prime(r)$ cannot have any zeros other than those at $r=0$ and $r=\infty$. This implies that $v_g(r)$ is monotonically decreasing for $0<r<\infty$, which will be important in what follows.

%We now consider a Schr\"odinger problem with a potential $\widetilde{V}(r)$ differing from $V(r)$ by a function of $r$. Let $h(r)=\sqrt{a^2+r^2}\Phi^\prime(r)$. Note that since $\Phi(r)$ is monotonically increasing, $h(r)$ is everywhere positive. Taking the $r$ derivative of (\ref{eom.static}), we find that $h(r)$ satisfies
%\be
%\label{eq:schrod_tilde}
%\widetilde{L}h:=(-\p_r\p_r + \widetilde{V}(r))h = 0,
%\ee
%where
%\[
%\widetilde{V}(r) = V(r) - 2\frac{a^2-r^2}{(a^2+r^2)^2}.
%\]

% Since $v_g(r)$ is the ground state, it must be everywhere positive. Take $v_g(r)$ multiplied by (\ref{eq:schrod_tilde}) and subtract this from $h(r)$ multiplied by the Sch\"odinger equation for $v_g(r)$. We obtain
%\be
%\label{eq:-hv''+vh''}
%-hv_g^{\prime\prime} + v_gh^{\prime\prime} + 2\frac{a^2-r^2}{(a^2+r^2)^2}v_gh = \mathcal{E}_gv_gh.
%\ee

%One possible approach is to integrate (\ref{eq:-hv''+vh''}) from $r=-\infty$ to $r=\infty$. Note first that
%\[
%\int_{-\infty}^\infty \big(-hv_g^{\prime\prime} + v_gh^{\prime\prime}\big)dr = \big[v_gh^\prime-hv_g^\prime\big]^\infty_{-\infty}=0,
%\]
%since both $v_g(r)$, $h(r)$ and their derivatives go to zero as $r\rightarrow\pm\infty$. Secondly, we know that
%\[
%\int_{-\infty}^\infty hv_gdr > 0,
%\]
%because the integrand is a product of two functions which are everywhere positive, as noted above. Combining these observations with our assumption that $\mathcal{E}_g<0$, we deduce that
%\be
%\label{eq:integral_inequality}
%I:=\int_{-\infty}^\infty \frac{a^2-r^2}{(a^2+r^2)^2}hv_gdr < 0.
%\ee

%This integral conveniently splits into two parts. Define
%\[
%I_1 = \int_0^a \frac{a^2-r^2}{(a^2+r^2)^2}hv_gdr,\qquad
%I_2 = \int_a^\infty \frac{r^2-a^2}{(a^2+r^2)^2}hv_gdr,
%\]
%then $I = I_1 - I_2$. It is easy to see that both $I_1$ and $I_2$ are positive; we wish to show that $I_1>I_2$ for $a>1/\sqrt{2}$.

%We now use the fact that $v_g(r)$ is monotonically decreasing for $0<r<\infty$. Note also that using (\ref{eom.static}) we find
%\be
%\nonumber
%\begin{split}
%h^\prime(r) &= \frac{r}{\sqrt{a^2+r^2}}\Phi^\prime + \sqrt{a^2+r^2}\Phi^{\prime\prime} \\
%&= -\frac{r}{\sqrt{a^2+r^2}}\Phi^\prime -2\Phi(1-\Phi^2),
%\end{split}
%\ee
%which is negative in this range. Thus $h(r)$ is also monotonically decreasing. Returning to our integrals, this implies that
%\be
%\nonumber
%\begin{split}
%I_1 &> v_g(a)h(a)\int_0^a \frac{a^2-r^2}{(a^2+r^2)^2}dr = \frac{v_g(a)h(a)}{2a} \quad \mathrm{and} \\
%I_2 &< v_g(a)h(a)\int_a^\infty \frac{r^2-a^2}{(a^2+r^2)^2}dr = \frac{v_g(a)h(a)}{2a},
%\end{split}
%\ee
%so we can conclude that $I_1>I_2$ for $a>1/\sqrt{2}$, as required. We have thuse reached a contradiction: $L$ cannot have any bound states with negative energy. Thus $\Phi(r)$ is linearly stable.
%\koniec

\subsection{Finding internal modes numerically}
\label{sec:finding_bound_states_numerically}
Bound states of the potential (\ref{QMpotential}) correspond to internal modes of the kink like the odd solution of (\ref{eq:zero_order}) in (\ref{eq:flat_vib_modes}). In contrast, for frequencies greater than $\omega = 2$, solutions to (\ref{schrod}) are interpreted as radiation. It is possible to search for bound states of (\ref{QMpotential}) numerically. The method for this is as follows:
\begin{enumerate}
\item For the chosen value of $a$, generate the soliton $\Phi(r)$ as described above.
\item Calculate the potential $V(r)$.
\item For some initial guess of the eigenvalue $\omega^2$, integrate equation ({\ref{schrod}}) numerically, setting $v(0)=1$ and $v^\prime (0)=0$ to obtain even bound states and $v(0)=0,v^\prime(0)=1$ to obtain odd bound states.
\item Use a bisection method to find the value of $\omega^2$ for which a bound state exists.
\end{enumerate}
This procedure will only be effective within the range of $r$ for which $\Phi(r)$ is calculated.

For large $a$, the potential has both an even and an odd bound state which look qualitatively similar to the internal modes (\ref{eq:flat_vib_modes}) of the $\phi^4$ kink on $\mathbb{R}^{1,1}$. The bound states for several values of $a$ can be found in Figures \ref{fig:even_states} and \ref{fig:odd_states}. As $a$ decreases, the eigenvalues $\omega^2$ of the bound states increase, until they disappear into the continuous spectrum ($\omega^2>4$). This disappearance will be further discussed in Section \ref{sec:a_crit}. The internal mode frequencies are plotted against $a$ in Figure \ref{fig:omega_vs_a}. The choice of axis ticks will be motivated in Section \ref{sec:dynamics}.

%As $a$ decreases, the eigenvalues $\omega^2$ of the bound states increase, until they disappear into the continuous spectrum ($\omega^2>4$). The odd state disappears around $a=0.8$, and the even state disappears around $a=0.3$. A plot of both the even and odd bound state for $a=5$ shown in Figure \ref{fig:bound_states}. For $a=0.4$, a plot of the single even bound state is shown in Figure \ref{fig:bound_state}. Note that for small values of $a$, there is a  local maximum of the potential located at $r=0$, with local minima on either side. For $a=0.4$, this is reflected in the local maxima of the even bound state either side of $r=0$.

%\begin{figure}
%\includegraphics[width=14cm]{bound_states.pdf}
%\caption{\label{fig:bound_states}The even and odd bound states of the potential for $a=5$.}
%\end{figure}

%\begin{figure}
%\includegraphics[width=14cm]{bound_state.pdf}
%\caption{\label{fig:bound_state}The single even bound state of the potential $V(r)$ for $a=0.4$.}
%\end{figure}

\begin{figure}
\includegraphics[width=14cm]{several_even_modes.pdf}
\caption{\label{fig:even_states} Even bound states of the potential $V_0(r)$ and of the potential $V(r)$ for two different values of $a$.}
\end{figure}

\begin{figure}
\includegraphics[width=14cm]{several_odd_modes.pdf}
\caption{\label{fig:odd_states} Odd bound states of the potential $V_0(r)$ and of the potential $V(r)$ for two different values of $a$.}
\end{figure}

\begin{figure}
\includegraphics[width=14cm]{omega_vs_a.pdf}
\caption{\label{fig:omega_vs_a} The frequencies of the internal modes of the kink plotted against the wormhole radius $a$.}
\end{figure}

\subsection{Large $a$ limit}

We can also perturbatively expand the eigenvalues of the eigenvalue problem (\ref{schrod}). Consider solutions to (\ref{eom.coords}) of the form\footnote{Note that in Section \ref{sec:linearised_pert} we considered perturbations $v(r)$ which differ from $v_\epsilon(r)$ by a factor of $(r^2+a^2)^{-1/2}$, since such perturbations are described by a Schr\"odinger problem. Here it will be simpler to remove this factor; however there is a one--to--one correspondence between $v(r)$ and $v_\epsilon(r)$.} $\phi_\epsilon(r)=\Phi_\epsilon(r)+\mathrm{e}^{i\omega t}v_\epsilon(r)$, where $v_\epsilon$ is small.
These satisfy
\be
\label{eq:perturbed_large_a}
v_\epsilon^{\prime\prime} + \frac{2r\epsilon^2}{\epsilon^2r^2+1}v_\epsilon^\prime + 2(1-3\Phi_\epsilon^2)v_\epsilon = -\omega^2_\epsilon v_\epsilon.
\ee

Let $(v_\epsilon,\omega^2_\epsilon)$ be a solution to (\ref{eq:perturbed_large_a}) with
\[
\omega^2_\epsilon = \omega^2_0 + \epsilon^2\xi + \mathcal{O}(\epsilon^4) \quad \mathrm{and} \quad
v_\epsilon(r) = v_0(r) + \epsilon^2v_1(r) + \mathcal{O}(\epsilon^4).
\]
Our aim will be to find $\xi$. Substituting into (\ref{eq:perturbed_large_a}), at zero order we obtain the equation (\ref{eq:zero_order}) which controls the linear stability analysis of the $\phi^4$ kink on $\mathbb{R}^{1,1}$.

%Note that the first solution in (\ref{eq:flat_vib_modes}), the zero mode, is proportional to $\Phi_0^\prime(r)$. Its existence follows from translation invariance of (\ref{eq:zero_order}). It corresponds to an even bound state of the Schr\"odinger equation (\ref{schrod}) in the limit $a\rightarrow\infty$. The second solution in (\ref{eq:flat_vib_modes}) is the first non--trivial vibrational mode, and corresponds to an odd bound state of (\ref{schrod}) in the limit $a\rightarrow\infty$.

The terms of order $\epsilon^2$ in (\ref{eq:perturbed_large_a}) give us
\be
\label{eq:first_order}
v_1^{\prime\prime} + 2rv_0^\prime + 2(1-3\Phi_0^2)v_1 - 12\Phi_0\Phi_1v_0 = -\omega^2_0v_1 - \xi v_0.
\ee
We multiply equation (\ref{eq:first_order}) by $v_0$, and subtract from this $v_1$ multiplied by equation (\ref{eq:zero_order}). Integrating the result from $r=-\infty$ to $r=\infty$, we find
\[
\int_{-\infty}^{\infty}\big(v_1^{\prime\prime}v_0-v_0^{\prime\prime}v_1\big)dr
+ \int_{-\infty}^{\infty}2rv_0^\prime v_0dr
- 12\int_{-\infty}^{\infty}\Phi_0\Phi_1v_0^2dr
=-\xi.
\]
In the first term the integrand is a total derivative, and the second term is easily found to be $-1$ using integration by parts. We thus obtain
\be
\label{eq:xi}
\xi = 1 + 12\int_{-\infty}^{\infty}\Phi_0\Phi_1v_0^2dr,
\ee
which we can evaluate for each of the solutions (\ref{eq:flat_vib_modes}) using the symbolic computation facility in Mathematica. We find $\xi=2$ in the case of the zero mode and $\xi=\pi^2-7$ in the case of the first non--trivial vibrational mode. We can check these values by finding $(v,\omega)$ numerically as described in Section \ref{sec:finding_bound_states_numerically} for a range of small values of $\epsilon$ and comparing $\omega^2$ to the $\omega^2_0+\xi\epsilon^2$ predicted here. The corresponding plots are shown in Figures \ref{fig:xi_zero_mode} and \ref{fig:xi_other_mode}.

\begin{figure}
\includegraphics[width=14cm]{xi_zero_mode.pdf}
\caption{\label{fig:xi_zero_mode}A comparison of the predicted and numerical calculations for the energy of the zero mode as a function of $\epsilon^2$ for small $\epsilon$.}
\end{figure}

\begin{figure}
\includegraphics[width=14cm]{xi_other_mode.pdf}
\caption{\label{fig:xi_other_mode}A comparison of the predicted and numerical calculations for the energy of the odd vibrational mode as a function of $\epsilon^2$ for small $\epsilon$.}
\end{figure}

%The fact that both of these values of $\xi$ are positive supports the conjecture that the eigenvalue problem (\ref{schrod}) has no negative eigenvalues: we find no bound states for $a$ less than about $0.3$, and as $a$ approaches this critical value from above, the energy of the ground state is approaching $4$ from below, disappearing into the continuous spectrum. As $a\rightarrow\infty$ we find that the energy of the ground state approaches zero from above. In between these two extremes, all numerical calculations of the ground state reveal positive ground states energies between zero and $4$. The next section provides a further investigation into the conjecture that (\ref{schrod}) has no negative eigenvalues.

\subsection{Critical values of $a$} \label{sec:a_crit}

It is interesting to investigate the values of $a$ at which the internal modes disappear into the continuous spectrum. The larger of these, at which the odd internal mode disappears, we shall call $a_1$. The smaller one, at which the even internal mode disappears, we shall call $a_0$. %A useful way to gain some intuition about the value of $a_1$ is to approximate the potential $V(r)$ as a square well.

%\subsubsection{Square well approximation}

%\begin{figure}
%\includegraphics[width=14cm]{QMpotential.pdf}
%\caption{\label{fig:QMpotential}The potential (shown in blue) of the 1--dimensional quantum mechanics problem arising from the study of stability of the soliton in the case $a=1$, and a square well approximation to this potential with width $2r_0=2$ (shown in red).}
%\end{figure}

%Since $V(r)$ is not known explicitly, its spectrum is difficult to study. However, we can approximate the potential as a finite square potential well, as shown in figure \ref{fig:QMpotential}.

%A square well approximation to the potential $V(r)$ for $a=1$ is shown in figure \ref{fig:QMpotential}. The number of bound states of a square well depends on the width $2r_0$ and height $\Delta V$ of the well: they correspond to intersections in the region $x,y>0$ between the circle
%\be
%\label{squarewell_circle}
%x^2 + y^2 = \Delta Vr_0^2
%\ee
%and the lines $x\mathrm{tan}x=y$ (in the case of even parity) or $x\mathrm{cot}x =-y$ (in the case of odd parity). More detailed analysis can be found in Nick Dorey's quantum mechanics notes.

%For $a=1$ we have $\Delta V=5$, and I chose a width $r_0=1$, so the circle (\ref{squarewell_circle}) intersects the first two lines, resulting in  one even bound state and one odd bound state.
%For these choices, an intersection with the second line is nearly missed (since $\sqrt{2}$ is only just smaller than $\pi/2$), so it appears that for $a=1$ we are close to a critical value of $a$ where the odd bound state disappears into the continuous spectrum.
%Larger values of $a$ result in a potential which is both wider and deeper, so we expect more bound states.

%\subsubsection{Numerical estimates of the critical values of $a$}

The most convenient method of estimating $a_0$ and $a_1$ is based on the Sturm Oscillation Theorem \ref{th:sturm}. The points at which the even and odd internal modes disappear into the continuous spectrum are the points at which the zeros of the even and odd eigenfunctions of $L$ with $\omega^2=4$ disappear. We can thus examine the number of zeros of the odd eigenfunction with $\omega^2=4$ to determine the number of odd bound states with $\omega^2<4$. The critical value $a_1$ which we are searching for can then be found using a bisection method. An equivalent method using even bound states will yield an estimate of $a_0$.

One problem with this method is that we need the number of zeros in the interval $(0,\infty)$, and the shooting method we use to generate $\Phi(r)$ and $V(r)$ is only accurate up to a finite value of $r$. Since zeros of the eigenfunction with $\omega^2=4$ disappear at $r=\infty$, this limits the accuracy with which we can determine $a_0$ and $a_1$.

For the finite integration range which is accessible based on the shooting method, the odd state disappears at $a_1 \approx 0.8$ and the even state disappears at $a_0 \approx 0.3$.

For a potential $\mathcal{V}(r)$ which decays sufficiently quickly as $r\rightarrow\pm\infty$, it is well known that the condition
\be
\label{eq:integral_condition}
I:=\int_{-\infty}^{\infty}\mathcal{V}(r)dr<0
\ee
is sufficient to ensure that the operator $-\p_r\p_r + \mathcal{V}(r)$ has at least one bound state. In fact, the condition $I\leq 0$ is sufficient \cite{BarrySimon}. However, (\ref{eq:integral_condition}) is not a necessary condition: there are potentials which have at least one bound state where (\ref{eq:integral_condition}) is not satisfied. It is interesting to investigate the disappearance of our ground state in this context.

For us the relevant choice is $\mathcal{V}(r)=V(r)-4$ so that $\mathcal{V}(r)$ vanishes at the boundaries. We then examine the value of this integral for the critical value $a=a_0$ when the ground state disappears. We find that $I\approx0$ at the critical value of $a_0 \approx 0.3$ given above. We can also search numerically for the value of $a$ at which $I=0$; this also occurs at around $a_0 \approx 0.3$. Thus our results would be consistent with the conjecture that (\ref{schrod}) has no bound states for $I>0$.

\section{Resonant coupling of the internal modes to the continuous spectrum} \label{sec:dynamics}
We now move on to consider time dependent perturbations of the form
\[
\phi(t,r) = \Phi(r) + (r^2+a^2)^{-1/2}w(t,r),
\]
where we consider non--linear terms in $w(t,r)$. Substituting into (\ref{eom.coords}) we find
\be
\label{eq:nonsmall_pert}
w_{tt} = - Lw + f(w),
\ee
where we have defined
\be \label{eq:f}
f(w) = - \frac{6w^2\Phi}{\sqrt{a^2+r^2}} - \frac{2w^3}{a^2+r^2},
\ee
suppressing the dependence of $f$ on $r$ to simplify the notation. We will not have much need for the expression for $f$ other than to note that it contains terms which are quadratic and cubic in $w$.

%If the value of $a$ is such that no internal modes are present in the spectrum of the Schr\"odinger operator $L$, then $|w(t)|\sim t^{-3/2}$ as $t\rightarrow\infty$.
If $a$ is large enough to allow internal modes, then these can only decay through resonant coupling to the continuous spectrum of $L$. The analogous process of decay to the $\phi^4$ kink on $\mathbb{R}^{1,1}$ was discussed in \cite{Manton&Merabet}, and the general theory was developed in \cite{SW98}. In the following sections we investigate this decay in the case of a single internal mode, before comparing our result with numerical data.

\subsection{Conjectured decay rate in the presence of a single internal mode} \label{sec:conjecture}
In this section we follow the analysis in \cite{SG}. Looking at Figure \ref{fig:omega_vs_a}, we note that for $a\in(0.3,0.8)$ we have
\be \label{eq:spec_L}
\mathrm{spec}\ L=\{\omega^2\}\cap [m^2,\infty), \qquad \omega^2<m^2<4\omega^2
\ee
where $m^2=4$. As above, we denote the unique normalised eigenfunction of $L$ by $v$, so that $Lv=\omega^2v$. We will use $\langle\cdot,\cdot\rangle$ to denote the usual inner product on $\R$.

We decompose the perturbation as
\be
w(t,r) = \alpha(t)v(r) + \eta(t,r),
\ee
where $v(r)$ refers to the single even internal mode of the kink and $\eta$ is a superposition of states from the continuous spectrum of $L$. Where there is only one internal mode present, its frequency $\omega$ always lies in the upper half of the mass gap: $1<\omega<2$. This is important because it means that $2\omega$ lies within the continuous spectrum.

We substitute this into (\ref{eq:nonsmall_pert}) and project onto and away from the internal mode direction, obtaining the following equations for $\alpha$ and $\eta$:
\begin{align}
\ddot{\alpha} + \omega^2\alpha &= \langle v, f(\alpha v + \eta) \rangle \label{eq:alpha}\\
\ddot{\eta} + L\eta &= P^\perp f(\alpha v + \eta),\label{eq:eta}
\end{align}
where $P^\perp$ is the projection onto the space of eigenstates of $L$ which are orthogonal to $v$, given by
\be \label{eq:Pperp}
P^\perp\psi = \psi - \langle v,\psi\rangle v.
\ee

These equations have initial conditions $\alpha(0)$ and $\eta(0,r)$ such that
\begin{align*}
\phi(0,r) &= \Phi(r) + (r^2+a^2)^{-1/2}(\alpha(0)v(r) + \eta(0,r)), \quad \mathrm{and} \\
\dot{\phi}(0,r) &= (r^2+a^2)^{-1/2}(\dot{\alpha}(0)v(r) + \dot{\eta}(0,r)).
\end{align*}

In the following analysis we investigate decay of $\alpha(t)$. Equation (\ref{eq:alpha}) has a homogeneous solution consisting of oscillations with frequency $\omega$. Since $2\omega$ lies within the continuous spectrum of $L$, there will be a resonant interaction between these oscillations and the radiation modes in $\eta$ with frequencies $\pm 2 \omega$, arising from the term of order $\alpha^2$ in the right hand side of (\ref{eq:eta}). Thus, to leading order, (\ref{eq:eta}) is a driven wave equation with driving frequency $2\omega$. This resonant part of $\eta$ will have a back--reaction on $\alpha$ through (\ref{eq:alpha}), which will result in decay of the internal mode oscillations.

We now define $\alpha_1=\alpha,\ \omega\alpha_2=\dot{\alpha}_1$  so that (\ref{eq:alpha}) becomes
\[
\begin{cases} \nonumber
\dot{\alpha}_1 = \omega\alpha_2, \\
\dot{\alpha}_2 = - \omega\alpha_1 + \frac{1}{\omega}\langle v, f(\alpha_1 v+\eta)\rangle,
\end{cases}
\]
or equivalently
\be \label{eq:A}
\dot{A} = -i\omega A + \frac{i}{\omega}\bigg< v,f\bigg(\frac{1}{2}(A + \bar{A})v+\eta\bigg)\bigg>,
\ee
where $A=\alpha_1 + i\alpha_2$. Next we write $\eta_1=\eta,\ \eta_2=\dot{\eta}_1$, converting (\ref{eq:eta}) to
\[ \label{eq:eta2}
\begin{cases}
\dot{\eta}_1 =\eta_2, \\
\dot{\eta}_2 = -L\eta_1 + P^\perp f\big( \frac{1}{2} (A+\bar{A}) v + \eta_1\big).
\end{cases}
\]

We will regard the right hand sides of (\ref{eq:A}) and (\ref{eq:eta2}) as power series in $A$ and $\eta$. Terms which we expect to be higher order will not be treated rigorously; for this reason, our analysis will produce only a conjecture about the decay rate. Numerical evidence concerning the conjecture will be discussed in Section \ref{sec:numerics}.

It will be helpful to introduce the notation $\cO_p(A,\eta)$ to mean terms of at least order $p$ in $A,\bar{A},\eta_1,\eta_2$, so that $A^2$, $\eta_1^2 $ and $\bar{A}\eta_1$ are all examples of terms which are $\cO_2(A,\eta)$. Currently, the coupling between (\ref{eq:A}) and (\ref{eq:eta2}) is $\cO_2(A,\eta)$. We will write
\[
f\bigg( \frac{1}{2} (A+\bar{A}) v + \eta_1\bigg) = \sum_{k+l\geq 2} f_{kl}A^k\bar{A}^l + \sum_{\substack{k+l\geq 1 \\ n\geq 1}} f_{kln}\eta_1A^k\bar{A}^l
\]
where $k,l,n$ are non--negative, to elucidate the lowest order terms in (\ref{eq:eta2}). Note that $f_{kl}$ and $f_{kln}$ are decaying functions of $r$ defined by (\ref{eq:f}). We can then write
\[
P^\perp\bigg[ f\bigg( \frac{1}{2} (A+\bar{A}) v + \eta_1\bigg) \bigg] = \sum_{k+l=2} P^\perp[f_{kl}]A^k\bar{A}^l + \sum_{k+l=1} P^\perp [f_{kl1}\eta_1]A^k\bar{A}^l + \cO_3(A,\eta).
\]
%In particular, note that
%\[
%G_{02}=G_{20}=-P^\perp\bigg[\frac{3v(r)^2\Phi(r)}{2\sqrt{a^2+r^2}}\bigg] \quad \mbox{and} \quad
%F_{kl1}(\eta_1) = -\bigg< v(r),\frac{6v(r)\Phi(r)}{\sqrt{a^2+r^2}}\eta_1\bigg>.
%\]

Terms in (\ref{eq:A}) with imaginary coefficients correspond to rotation in the complex plane, and thus to oscillatory behaviour in $\alpha$. At first order, $A$ oscillates with frequency $\omega$. This is exactly the behaviour expected in the linearised theory discussed in Section \ref{sec:linearised_pert}. In fact, a priori, all the terms in the power series for $\dot{A}$ have coefficients which are purely imaginary.

The next step in our analysis will be to attempt a change of variable $\eta_i\mapsto\tilde{\eta}_i$ in (\ref{eq:eta2}) so that its right hand side is $\cO_3(A,\tilde{\eta})$, meaning $\tilde{\eta}$ is $\cO(A^3)$. It will turn out that the required change of variables is complex. The result will be a term in (\ref{eq:A}) which is $\cO(A^3)$ and has a real coefficient. This will be the lowest order term with a real coefficient, and thus the key resonant damping term. %The remaining terms in (\ref{eq:A}) which are $\cO_2(A,\tilde{\eta})$ will be proportional to $A\tilde{\eta}_1$ and $\tilde{\eta}_1^2$. These will be ignored, because $\tilde{\eta}$ is at least third order in $A$.

We write the change of variables as
\begin{equation} \label{eq:eta_change}
\eta_1 = \tilde{\eta}_1 + \sum_{k+l=2}b_{kl}A^k\bar{A}^l, \qquad \eta_2 = \tilde{\eta}_2 + \sum_{k+l=2}c_{kl}A^k\bar{A}^l,
\end{equation}
where $b_{kl}$ and $c_{kl}$ are functions of $r$ which are so far undetermined. Differentiating with respect to time and using (\ref{eq:A}), we find
\begin{align*}
\dot{\eta}_1 &= \dot{\tilde{\eta}}_1 - i\omega\sum_{k+l=2}b_{kl}(k-l)A^k\bar{A}^l+\cO_3(A,\tilde{\eta}), \\[5pt]
 \dot{\eta}_2 &= \dot{\tilde{\eta}}_2 - i\omega\sum_{k+l=2}c_{kl}(k-l)A^k\bar{A}^l + \cO_3(A,\tilde{\eta}).
\end{align*}
We equate these to the right hand sides of (\ref{eq:eta2}), substituting from (\ref{eq:eta_change}) and requiring that
\be \label{eq:eta_tilde}
\dot{\tilde{\eta}}_1 = \tilde{\eta}_2 + \cO_3(A,\tilde{\eta}), \qquad \dot{\tilde{\eta}}_2 = -L\tilde{\eta}_1 + \cO_3(A,\tilde{\eta}).
\ee
This yields
\[
-i\omega b_{kl}(k-l)=c_{kl} \qquad\mbox{and}\qquad -i\omega c_{kl}(k-l)=-Lb_{kl} + P^\perp[f_{kl}]
\]
for $k+l=2$, where we have discarded
\begin{align*}
\sum_{k+l=1} P^\perp[f_{kl1}\eta_1]A^k\bar{A}^l &= \sum_{k+l=1} P^\perp [f_{kl1}\tilde{\eta}_1]A^k\bar{A}^l + \sum_{\substack{k+l=1 \\ p+q=2}} P^\perp [f_{kl1}b_{pq}]A^{k+p}\bar{A}^{l+q} \\
&= \cO_3(A)
\end{align*}
because $\tilde{\eta}$ is at least third order in $A$.

The change of variables (\ref{eq:eta_change}) is now given by the solution to
\be \label{eq:solve_for_b}
\big(L-\omega^2(k-l)^2\big)b_{kl}=P^\perp[f_{kl}].
\ee
Because of the spectrum of $L$ given in (\ref{eq:spec_L}), for $(k,l)\in(2,0)\cup (0,2)$ the solution $b_{kl}$ is in general a complex function of $r$, whilst for $k=l=1$ the solution is real and decaying. The reason for this can be understood using the variation of parameters method for inhomogeneous ordinary differential equations.

Let $g(r)$ be such that $\langle g,g \rangle$ is finite, and $\lambda\geq 0$ a constant. The general solution of
\[
(L-\lambda^2)b(r)=g(r)
\]
is given by
\[
b(r)=Z_2(r)\int_{-\infty}^r \frac{1}{W(r^\prime)}Z_1(r^\prime)g(r^\prime)dr^\prime + Z_1(r)\int_r^\infty \frac{1}{W(r^\prime)} Z_2(r^\prime)g(r^\prime) dr^\prime,
\]
where $\{Z_1,Z_2\}$ is a basis for solutions to the homogeneous equation with Wronskian $W(r)=Z_1Z_2^\prime -Z_2Z_1^\prime$. The basis must be chosen so that the above integrals converge. %The solution $b(r)$ will behave like $Z_1$ in the limit $r\rightarrow\infty$, and will behave like $Z_2$ in the limit $r\rightarrow -\infty$.
%limits of integration must be chosen so as to satisfy the boundary conditions on $b$, which in our case are inherited from the boundary conditions on $w$. For large $|r|$, solutions to (\ref{eq:nonsmall_pert}) are of the wave--like form $\mathrm{e}^{-i(\Omega t \pm \xi r)}$, where $\xi^2=m^2+\Omega^2$. We will require that there are no ingoing waves at the boundaries.

For $k=l=1$, so that $\lambda^2=0$ and hence $\lambda^2<m^2$, we can choose a basis such that $W=1$ and $Z_1,Z_2$ are both real, and they decay to zero in the limits $r\rightarrow -\infty$ and $r\rightarrow\infty$ respectively. Then
\be \label{eq:decaying_soln}
b_{11}(r) = Z_2(r)\int_{-\infty}^r Z_1(r^\prime)P^\perp[f_{11}](r^\prime)dr^\prime + Z_1(r)\int_r^{\infty} Z_2(r^\prime)P^\perp [f_{11}](r^\prime) dr^\prime.
\ee
For $\lambda^2\geq m^2$, we cannot choose a real solution in general. In the case $(k,l)\in(2,0)\cup (0,2)$, we take as a basis the Jost functions $\{j_\pm\}$, defined by
\begin{equation*}
j_\pm(r) \sim e^{\pm i\xi r} \ \mbox{as} \ r\rightarrow\infty,
\end{equation*}
where $\xi=\sqrt{4\omega^2-m^2}$. Their Wronskian is then $W(j_+,j_-)=-2i\xi$, and we write the solution as
\be \label{eq:oscillating_soln}
b_{02}(r) = b_{20}(r) = \frac{ij_-(r)}{2\xi}\int_{-\infty}^r j_+(r^\prime)P^\perp[f_{20}](r^\prime)dr^\prime + \frac{ij_+(r)}{2\xi}\int_r^{\infty} j_-(r^\prime)P^\perp[f_{20}](r^\prime) dr^\prime.
\ee

Finally, we use (\ref{eq:decaying_soln}) and (\ref{eq:oscillating_soln}) to change variable $\eta_i\mapsto\tilde{\eta}_i$ in (\ref{eq:A}), obtaining
\be \label{eq:A2}
\dot{A} = -i\omega A + \frac{i}{\omega}\Bigg(\sum_{2\leq k+l\leq 3}\langle v, f_{kl}\rangle A^k\bar{A}^l + \sum_{\substack{k+l=1 \\ p+q=2}} \langle v, f_{kl1}b_{pq}\rangle A^{k+p}\bar{A}^{l+q} + \cO(A^4)\Bigg),
\ee
where we have ignored terms containing $\tilde{\eta}_1$ since these are at least fourth order in $A$. We can now see that, of the terms which we have written explicitly, the only ones that can give a real contribution to $\dot{A}$ are those containing $b_{02}$ and $b_{20}$. We thus find
\begin{align*}
\frac{d}{dt}|A|^2 &= \dot{A}\bar{A} + A\dot{\bar{A}} = 2\mathbb{R}\mathrm{e}[\dot{A}\bar{A}] \\
&= \frac{2}{\omega} \mathbb{R}\mathrm{e}\Bigg[i\sum_{k+l=1}\bigg( \langle v, f_{kl1}b_{20}\rangle A^{k+2}\bar{A}^{l+1} + \langle v, f_{kl1}b_{02}\rangle A^{k}\bar{A}^{l+3} \bigg) \Bigg] + \cO(A^5) \\
&= -\frac{2}{\omega} \mathbb{I}\mathrm{m}\bigg[\langle v, f_{101}b_{20}\rangle(A^3\bar{A} + A^2\bar{A}^2 + A\bar{A}^3 + \bar{A}^4) \bigg] + \cO(A^5).
\end{align*}
In particular, the term $A^2\bar{A}^2=|A|^4$ is real and non--oscillating, giving a contribution
\[
\frac{d}{dt}|A|^2 \sim -\frac{2}{\omega} \mathbb{I}\mathrm{m}\big[\langle v, f_{101}b_{20}\rangle\big]|A|^4.
\]
The terms $A^3\bar{A}$, $A\bar{A}^3$ and $\bar{A}^4$, on the other hand, would be expected to oscillate at frequencies $2\omega$ and $4\omega$ at first order, and thus time average to zero.

Hence we conclude
\be \label{eq:decay_behaviou}
|A|\sim \bigg(\Gamma t + \frac{1}{|A(0)|^2}\bigg)^{-1/2}, \qquad \Gamma:=\frac{2}{\omega} \mathbb{I}\mathrm{m}\big[\langle v, f_{101}b_{20}\rangle\big].
\ee

The constant $\Gamma$ is a function of $a$ which can be calculated explicitly. Using (\ref{eq:f}) and (\ref{eq:oscillating_soln}) gives
\begin{align*}
\langle v, f_{101}b_{20}\rangle = \int_{-\infty}^{\infty}dr\ v(r)f_{101}(r)\bigg(&\frac{ij_-(r)}{2\xi}\int_{-\infty}^r j_+(r^\prime)P^\perp[f_{20}](r^\prime)dr^\prime \\
&+ \frac{ij_+(r)}{2\xi}\int_r^{\infty} j_-(r^\prime)P^\perp[f_{20}](r^\prime) dr^\prime\bigg).
\end{align*}
We now use the facts that $f_{20}=vf_{101}/4$, and $P^\perp[f_{20}]=f_{20}-\langle v,f_{20}\rangle v$. Note that $f_{20}$ is an odd function of $r$, so in fact $\langle v,f_{20}\rangle=0$ and so $P^\perp[f_{20}]=f_{20}$. We thus obtain
\begin{align*}
\langle v, f_{101}b_{20}\rangle = \frac{2i}{\xi}\bigg(&\int_{-\infty}^{\infty}dr
\int_{-\infty}^r dr^\prime f_{20}(r)j_-(r)f_{20}(r^\prime)j_+(r^\prime) \\
&+ \int_{-\infty}^{\infty}dr\int_r^{\infty}dr^\prime f_{20}(r)j_+(r)f_{20}(r^\prime)j_-(r^\prime)\bigg).
\end{align*}
The two double integrals are integrals over complementary halves of the $(r,r^\prime)$ plane, and thus sum to a single integral over the full plane. Hence
\[
\langle v, f_{101}b_{20}\rangle = \frac{2i}{\xi}\int_{-\infty}^{\infty}f_{20}(r)j_+(r)dr
\int_{-\infty}^\infty  f_{20}(r^\prime)j_-(r^\prime) dr^\prime=\frac{2i}{\xi}|\langle f_{20}, j_+ \rangle|^2,
\]
since $j_\pm$ are complex conjugates.

Combining this with (\ref{eq:decay_behaviou}) gives
\[
\Gamma = \frac{4}{\omega\xi}|\langle f_{20}, j_+\rangle|^2.
\]
The so--called Fermi Golden Rule then reads
\[
|\langle f_{20}, j_+\rangle|\neq 0.
\]

\subsection{Numerical investigation of the conjectured decay rate}
\label{sec:numerics}

\begin{figure}
\includegraphics[width=14cm]{ic_comparison_0002.pdf}
\caption{\label{fig:decay} The decay of internal mode oscillations for various initial conditions when $a=0.5$.}
\end{figure}

In order to integrate the PDE (\ref{eom.coords}) to large times $t$, we employ the method of hyperboloidal foliations and scri--fixing \cite{hyp_foliations}. This is necessary because surfaces of constant $t$ can never intersect with future null infinity, as they always reach the boundary of the spacetime at spatial infinity. In order to integrate out to large times at large $r$, i.e. to reach future null infinity, we define
\[
s = \frac{t}{a} - \sqrt{\frac{r^2}{a^2}+1}, \qquad y=\mathrm{arctan}\Big(\frac{r}{a}\Big),
\]
resulting in the hyperbolic equation
\be
\label{eq:s_and_y}
\p_s\p_sF + 2\mathrm{sin}(y)\p_y\p_sF + \frac{1 + \mathrm{sin}^2(y)}{\mathrm{cos}(y)}\p_sF = \mathrm{cos}^2(y)\p_y\p_yF + 2a^2\frac{F(1-F^2)}{\mathrm{cos}^2(y)}
\ee
for $F(s,y)=\phi(t,r)$. Then surfaces of constant $s$ approach right future null infinity $\mathcal{J}^+_R$ along outgoing null cones of constant retarded time $t-r$, and left future null infinity $\mathcal{J}^+_L$ along outgoing null cones of constant advanced time $t+r$. For more details see \cite{SG,hyp_foliations}.

We solve the corresponding initial value problem at space--like hypersurfaces of constant $s$, specifying $\phi(s=0,y)$ and $\p_s\phi(s=0,y)$. No boundary conditions are required, since the principal symbol of (\ref{eq:s_and_y}) degenerates to $\p_s(\p_s\pm 2\p_y)$ as $y\rightarrow\pm \pi/2$, so there are no ingoing characteristics. This reflects the fact that no information comes in from future null infinity.

Following \cite{wavemaps, sym_hyp} we define the auxiliary variables
\[
\Psi = \p_yF, \qquad \Pi = \p_sF + \mathrm{sin}y\p_yF
\]
to obtain the first order symmetric hyperbolic system
\begin{align}
\p_sF &= \Pi - \Psi\mathrm{sin}y \\
\label{eq:Psi} \p_s\Psi &= \p_y(\Pi-\Psi\mathrm{sin}y) \\
\p_s\Pi &= \p_y(\Psi-\Pi\mathrm{sin}y) + 2\mathrm{tan}y(\Psi-\Pi\mathrm{sin}y) + 2a^2\frac{F(1-F^2)}{\mathrm{cos}^2y},
\end{align}
which we solve numerically using the method of lines. Kreiss--Oliger dissipation is required to reduce unphysical high--frequency noise. We also add the term $-0.1(\Psi-\p_yF)$ to the right hand side of equation (\ref{eq:Psi}) to suppress violation of the constraint $\Psi=\p_yF$.

We are interested in the range of values $a_0<a<a_1$ for which the kink has exactly one internal mode. We find that, for fixed but arbitrary $y$, $F(s,y)$ oscillates in $s$ with a frequency close to the internal mode frequency, and that these oscillations tend towards a decay rate of $s^{-1/2}$, as we expect from Section \ref{sec:conjecture}. Plots demonstrating this decay at $y=0$ for $a=0.5$ are shown in Figure \ref{fig:decay}. Note that $\phi(0,s)$ is used as a proxy for the internal mode amplitude, and we use a log--log scale to elucidate the dependence on $s^{-1/2}$ in the large $s$ limit. The lines are labelled in the legend by the initial conditions which produced them, with the exception of the gradient line $4.2s^{-1/2}$. The constant $4.2$ is related to $\Gamma$ as defined in (\ref{eq:decay_behaviou}).

\subsection{Expected decay rates in other regimes}
From Figure \ref{fig:omega_vs_a}, we see that the second internal mode appears before the frequency of the first internal mode moves out of the range $(m/2,m)$. In the presence of more than one internal mode, we expect complicated coupling between their amplitudes, making the behaviour at large times very difficult to predict. However, if we restrict to odd initial data, the solution to (\ref{eq:nonsmall_pert}) remains odd. This means that the even internal mode can never be excited, so the system effectively has only one internal mode. In this case, the analysis in Section \ref{sec:conjecture} still applies, since the frequency of the odd internal mode always lies in the range $(m/2,m)$, so we expect its amplitude to decay like $s^{-1/2}$.

For general initial data and wormhole radii $a>a_1$, we cannot produce a concrete conjecture about the decay rate. However, we expect the behaviour of the system to depend on the locations of the two internal mode frequencies within the mass gap. The analysis in Section \ref{sec:conjecture} for a single internal mode suggests that for frequencies less than $m/2$, a real contribution to $\dot{A}$ does not appear until at least $\mathcal{O}(A^4)$. In this case, we require another change of variable in the radiation equation to rule out the contribution of the radiative term at higher order. We then expect to solve an equivalent of (\ref{eq:solve_for_b}) where $k+l=3$ to find the required change of variable. If $9\omega^2<m^2$ so that the solution is still real, we can proceed by induction, changing the variable until the solution is complex. A real contribution to (\ref{eq:A2}) will only be obtained for $k+l=N$ such that $N^2\omega^2>m^2$. This would mean a real contribution to $\dot{A}$ at $\mathcal{O}(A^{N+1})$, and hence result in a decay rate of $s^{-1/N}$. Further detail can be found in \cite{SG}.

Although the presence of a second internal mode complicates the dynamics, we still expect the smallest $N$ such that the even internal mode frequency $\omega$ satisfies $N^2\omega^2>m^2$ to be an important factor in the behaviour at large $s$. The axis ticks in Figure \ref{fig:omega_vs_a} show the value of $N$ for a range of $a$.


\documentclass{article}
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\begin{document}
\subsection{Schur complement preconditioner}

We start our analysis with an ideal version of~\eqref{eq:mhd_pc_ls}, $\mathcal{M}_{\rm schurMH}$, which corresponds
to combining ${\mathcal M}_{\rm NS}$ in \eqref{eq:NS_precond} with $F$ replacing $\widehat{F}$, together with ${\mathcal M}_{\rm idealMX}$ in~\eqref{eq:maxwell_pc_ideal}, along with a Schur complement based on the coupling terms: % to precondition the Maxwell sub-system:
\begin{equation}
\label{eq:mhd_pc_schur}
\mathcal{M}_{\rm schurMH} =
\left(
\begin{array}{cccc}
  F + C^T ( M + D^T L^{-1} D)^{-1} C  & B^T & C^T & 0\\
0 & -\widehat{S} & 0 & 0 \\
0 & 0 & M + D^T L^{-1} D & 0\\
0 & 0 & 0 & L
\end{array}
\right).
\end{equation}

\begin{theorem}
\label{thm:mhd_outer_schur}
The matrix $\mathcal{M}_{\rm schurMH}^{-1} \mathcal{K}_{\rm MH} $ has an eigenvalue $\lambda = 1$ with algebraic multiplicity of at least $n_b+n_c$ where $n_c$ is the dimension of the nullspace of $C$ and an eigenvalue $\lambda = -1$ with  algebraic multiplicity of at least $m_b$. The eigenvector corresponding to $\lambda=1$ is: $(u_c, -\widehat{S}^{-1}B u_c, b, L^{-1} D b)$ with $u_c$ in the nullspace of $C$ and $b$ anything.
\end{theorem}

\begin{proof}
The corresponding eigenvalue problem is
 \begin{equation*}
 \hspace{-3mm}   % Chen: a small manipulation to avoid black thick vertical line; remove later
\left(
\begin{array}{cccc}
F & B^T & C^T & 0\\
B & 0 & 0& 0\\
-C & 0& M & D^T\\
0 & 0 & D & 0
\end{array}
\right)
\left(
\begin{array}{c}
u\\
p\\
b\\
r
\end{array}
\right)
=
\lambda
\left(
\begin{array}{cccc}
  F + C^T ( M + D^T L^{-1} D)^{-1} C  & B^T & C^T & 0\\
0 & -\widehat{S} & 0 & 0 \\
0 & 0 & M + D^T L^{-1} D & 0\\
0 & 0 & 0 & L
\end{array}
\right)
\left(
\begin{array}{c}
u\\
p\\
b\\
r
\end{array}
\right).
\end{equation*}
The four block rows can be written as
\begin{eqnarray}
    \label{eq:outer_row1_schur} (1-\lambda) (Fu + B^T p+ C^T b) - \lambda C^T(M+D^TL^{-1} D)^{-1}Cu   &=& 0,\\
    \label{eq:outer_row2_schur} B u &=& -\lambda \widehat{S}\, p,\\
    \label{eq:outer_row3_schur} (\lambda -1) C u + (1 - \lambda) M b - \lambda D^T L^{-1} D b + D^T r &=& 0,\\
    \label{eq:outer_row4_schur} D b &=& \lambda L r.
\end{eqnarray}

If $\lambda = 1$, \eqref{eq:outer_row1_schur} is satisfied if
$$C^T(M+D^TL^{-1} D)^{-1}Cu = 0.$$ This only happens when $u = 0$ or $u \in
\mbox{Null}(C)$. Using $u_c$ to denote the nullspace vector of $C$ then \eqref{eq:outer_row2_schur} simplifies to:
$$p = -\widehat{S}^{-1}B u_c.$$
Equation~\eqref{eq:outer_row4_schur} leads to $r = L^{-1} D b.$
If this holds, \eqref{eq:outer_row3_schur} is readily satisfied.
Therefore, $(u_c, -\widehat{S}^{-1} B u_c, b, L^{-1} D b)$ is an eigenvector corresponding to $\lambda=1$. There exist $n_c$ linearly independent such $u$'s and $n_b$ linearly independent such $b$'s. There are at least $n_c+n_b$ linearly independent nonzero vectors satisfying the eigenvalue problem when $\lambda = 1$. It follows that $\lambda = 1$ is an eigenvalue with algebraic multiplicity of at least $n_c + n_b$.

If $\lambda = -1$, \eqref{eq:outer_row4_schur} leads to $r = -L^{-1} D b.$
Substituting it into~\eqref{eq:outer_row3_schur}, we obtain $Cu = Mb.$ If $b=Gs$ is a discrete gradient, with the gradient matrix~$G$ defined in~\eqref{eq:gradient}, then $Mb = 0$ and $C^T b = 0$. If we take $u=0$, then $Cu = 0$ and the requirement $Cu = Mb$ is satisfied.
If $u=0$ and $b=Gs$ is a discrete gradient, equation~\eqref{eq:outer_row1_schur} becomes $B^Tp = 0$. Since $B$ has full row rank, this implies $p=0$.

Therefore, if $b=Gs$ is a discrete gradient, then $(0, 0, Gs, -L^{-1}D Gs)$ is an eigenvector corresponding to $\lambda = -1$. According to the discrete Helmholtz decomposition~\eqref{eq:helmholtz}, there are $m_b$ discrete gradients. Therefore $\lambda = -1$ is an eigenvalue with algebraic multiplicity at least  $m_b$.
\end{proof}

\begin{remark}
In our experiments, we have observed the eigenvalue $\lambda = 1$ has algebraic multiplicity of exactly $n_u+n_b$ and the eigenvalue $\lambda = -1$ has algebraic multiplicity of exactly $m_b$. Proving this may be difficult, though, due to complicated generalized eigenvalue problems that arise in the calculation.
\end{remark}

Theorem~\ref{thm:mhd_outer_schur} shows that $\mathcal{M}_{\rm idealMH}^{-1} \mathcal{K}_{\rm MH}$ has tightly clustered eigenvalues $\lambda=\pm 1$.  Since we have provided explicit expressions for the eigenvectors associated with $\lambda= \pm 1$, we know that the geometric multiplicities of these two eigenvalues are equal to their algebraic multiplicity.
We thus expect a good convergence behavior for  $\mathcal{M}_{\rm schurMH}^{-1} \mathcal{K}_{\rm MH}$.

\end{document}


\documentclass[a4paper, 11 pt, article, accentcolor=tud7b]{tudreport}

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{longtable}

\title{CNuVS Exercise 1}
\author{Nils Rollshausen, Daniel Drodt}
\subtitle{Nils Rollshausen, Daniel Drodt}

\begin{document}
	\maketitle
	\section{Dijkstra's Algorithm}

	\begin{longtable}[c]{l|ll|ll|ll|ll|ll|ll}
		\caption{}
		\label{Dijkstra}\\
		Visited vertices    & $P_2$ & $C_2$ & $P_3$ & $C_3$ & $P_4$ & $C_4$ & $P_5$ & $C_5$ & $P_6$ & $C_6$ & $P_7$ & $C_7$ \\ \hline
		\endfirsthead
		%
		\endhead
		%
		-                   & -    & $\infty$     & -    & $\infty$     & -    & $\infty$     & -    & $\infty$     & -    & $\infty$     & -    &  $\infty$    \\
		1                   & -    & $\infty$     & -    & $\infty$     & -    & $\infty$     & 1    & 1    & -    & $\infty$     &  1    & 4    \\
		1, 5                & -    & $\infty$     & 5    & 5    & -    & $\infty$     & 1    & 1    & -    & $\infty$     & 1    & 4    \\
		1, 5, 7             & 7    & 9    & 5    & 5    & 7    & 11   & 1    & 1    & 7    & 8    & 1    & 4    \\
		1, 5, 7, 3          & 7    & 9    & 5    & 5    & 7    & 11   & 1    & 1    & 3    & 7    & 1    & 4    \\
		1, 5, 7, 3, 6       & 7    & 9    & 5    & 5    & 7    & 11   & 1    & 1    & 3    & 7    & 1    & 4    \\
		1, 5, 7, 3, 6, 2    & 7    & 9    & 5    & 5    & 2    & 10   & 1    & 1    & 3    & 7    & 1    & 4    \\
		1, 5, 7, 3, 6, 2, 4 & 7    & 9    & 5    & 5    & 2    & 10   & 1    & 1    & 3    & 7    & 1    & 4
	\end{longtable}

	\section{Multiplexing}
	\subsection*{a) Internet without Multiplexing}
	Without multiplexing, every single connection to a popular site would require a dedicated wire to that destination. If, for example, Google had a total of 500 incoming wires at a datacenter, a maximum of 500 users could connect to Google at any given time.
	\subsection*{b) TDM vs FDM}
	Time-division Multiplexing (TDM) works by compressing data into a smaller timeframe, buffering those packets and interleaving packets of multiple incoming connections on a shared high-bandwidth link before reconstructing them into multiple low-bandwidth connections again. In contrast to that, Frequency-division Multiplexing (FDM) actually transmits data from both connections at the same time, though on different, non-interfering frequencies over the same medium.
	\subsection*{c) CDM and SDM}
	Code Disivion Multiplexing is roughly analogous to FDM in a digital signal. It uses non-interfering modulation codes (orthogonal vectors) for different connections to transmit multiple data streams over a single connection.
  Space Division Multiplexing refers to modifying the spatial direction of a wireless transmission (e.g. of a cellular base station) to match the physical position of the receiving device, thereby limiting the amount of noise / interference that other devices are exposed to.
\end{document}


\subsection{Analysis of variance (ANOVA) table}


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\hypersetup{hypertex=true, colorlinks=true, linkcolor=black, anchorcolor=black, citecolor=black}
\title{\Large \textbf{VG441 Problem Set 3}\\
\author{\textbf{Pan, Chongdan ID:516370910121}\\
}
}
\begin{document}
\maketitle
\section{Problem 1}
\quad
The problem can be consider as a MILP in the following ways:
$$\min \sum_{i=1}^{m}s_i$$
\\s.t.
$$s_i=\{0, 1\},\forall i=0,1\cdots m$$
\begin{center}
    \[x_{ij}=\begin{cases}
        1 & e_j\in S_i\\
        0 & e_j\notin S_i
    \end{cases}\]
\end{center}
$$y_j=\max\{s_1x_{1j},s_2x_{2j}\cdots s_ix_{ij} \}=1,\forall j=0,1\cdots n$$
$i$ is the number of sets, $s_i$ represents whether we select it, $s_i=1$ if we select the $i_{th}$ set, otherwise 0
$j$ is the number of elements, and $x_{ij}$ represents whether $e_j\in S_i$, if it's covered by $S_i$, then 1, otherwise 1. $y_j$ represents whether the $j_{th}$ element $e_j$ is covered by our selection, it's 1 if covered, otherwise 0. Hence all $y_j$ should be 1 because we want to cover all elements.
\par Our objective is to minimize the sum of $x_{s_i}$, which represents the number of set selected.
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.5]{P1.png}
    \caption{Result from Gurobi}
\end{figure}
\par According to solution from Gurobi, we get that we should choose $S_1,S_3,S_4,S_5$ so that all elements are covered, the minimize number of set is 4.
\section{Problem 2}
For a Fractional Knapsack problem, the object is:
$$\max\sum_{i=1}^{n}v_i x_i$$
\\s.t.
$$\sum_{i=1}^{n}s_i x_i\leq B$$
$$0\leq x_i\leq 1,\forall i$$
\\According to greedy algorithm, we need to create a new list  $[\frac{v_1}{s_1}\geq \frac{v_2}{s_2}\cdots\geq \frac{v_i}{x_i}]$, and the solution is $x=[1, 1\cdots 1,x_k,0\cdots,0]$ such that $\sum_{i=1}^{k-1}s_i + x_ks_k=B$, then the value of object function is $\sum_{i=1}^{k-1}v_i + x_kv_k$.
\\Assume we can have a better solution by taking items with total $\bigtriangleup s$ from the greedy algorithm solutions and add items with total $\bigtriangleup s'$ to it. $\bigtriangleup s\geq\bigtriangleup s'\geq 0$ because $\sum_{i=1}^k s_i\leq B$
\\The value we take with $\bigtriangleup s$ is bounded by $\frac{v_k}{s_k}\bigtriangleup s\leq\bigtriangleup v\leq \frac{v_1}{s_1}\bigtriangleup s$
\\The value we can add with $\bigtriangleup s'$ is also bounded by $0\leq\bigtriangleup v'\leq \frac{v_k}{s_k}\bigtriangleup s'\leq \frac{v_k}{s_k}\bigtriangleup s$
\\Hence the net value we can add $\bigtriangleup v'-\bigtriangleup v\leq\frac{v_k}{s_k}\bigtriangleup s-\frac{v_k}{s_k}\bigtriangleup s=0$
\\Therefore, we can't add any value to the solution from greedy algorithm anymore, and it gives the optimal solution.
\newpage
\section*{Python Code}
\begin{verbatim}
    import numpy as np
    import pandas as pd
    from gurobipy import *
    import matplotlib.pyplot as plt

    # 读取数据
    data = pd.DataFrame(pd.read_csv('D:\PANDA\Study\VG441\Homework\Problem Set 3\Data.csv'))
    data = data.iloc[:, 1:]
    m = data.shape[0]
    n = data.shape[1]
    # 初始化数据
    X = []
    t = []
    for i in range(m): X.append(list(data.loc[i]))
    WW = Model()
    s = WW.addVars(m, vtype=GRB.BINARY, name="set_if_selected") # m sets
    y = WW.addVars(n, name="element_if_covered") # n elements covered
    WW.setObjective(quicksum(s), GRB.MINIMIZE)
    for i in range(0, m):
        t.append(WW.addVars(n))
        WW.addConstrs(t[i][j] == s[i]*X[i][j] for j in range(n))
    WW.addConstrs(y[j] == max_(t[i][j] for i in range(m)) for j in range(n))
    WW.addConstrs(y[j] == 1 for j in range(n))
    WW.optimize()
    print('s=',WW.getAttr('X',s).values())
\end{verbatim}
\end{document}


%
% originally set up on Feb. 13th, 2013
%
%
\chapter{Numerical Analysis}
%
%
In this chapter we will talk bout the numerical analysis methods used in building
the computational chemistry software. we will discuss the ideas,
theorems and algorithms in the numerical analysis in general; reader could refer
to the following material for more details. For understanding the mathematical
idea and theorems, you can refer to the book given by David R. Kincaid and
E. Ward Cheney\cite{Kincaid_2002_numerical_analysis}; for the book discussing the
algorithms and code in details, you can refer to the book of ``Numerical Recipes''
\cite{numerical_recipes}.

Additionally, we note that in this chapter we use small letter to indicate the vectors,
and capitalized letter to indicate the matrix.

\section{Floating Point}

Floating point describes a method of representing an approximation
of a real number in a way that can support a wide range of values.
Today, the most important standard to represent a floating point
is IEEE 754 Standard.

A real number could be expressed according to scientific notation as:
\begin{equation}
\label{Floating_Point_eq:1}
 123.45678 = +0.12345678\times 10^{+3}
\end{equation}
Such expression is divided into two parts: one is the mantissa,
$+0.12345678$; the other is exponent, $10^{+3}$. Here we note that
for mantissa we have it in the range of $\frac{1}{10}$ and $1$.

In computer, since all of data is stored in binary way; therefore
the real number in computer is actually expressed as:
\begin{equation}
\label{Floating_Point_eq:2}
 x = \pm q\times 2^{m}
\end{equation}
where $q$ is between $\frac{1}{2}$ and $1$ if x is not 0.

Now how we express a real number in computer system? Take the single
precision as an example, for each real number the total storage is
32 bits; and the structure for usage is like:
\begin{description}
 \centering
 \item[Sign of mantessa] 1 bit;
 \item[Sign of exponent] 1 bit;
 \item[exponent($|m|$)]  7 bits;
 \item[mantissa($|q|$)]  23 bits
\end{description}
Therefore, from such structure we can derive the range the single precision.
Since $2^{7} = 128$,  so the range for a single precision is between $2^{-127}$
and $2^{127}$ in binary, which is approximated as $10^{-38}$ and $10^{38}$.

Compared with the range of single precision, the accuracy for a single
precision number is much lower. The total bits used to express the
mantissa is 23, where $2^{23}$(in binary) $\approx 10^{7}$ therefore it means that
for a single precision there's only maximum of 7 significant digits
could be expressed. Since in \ref{Floating_Point_eq:1} the mantissa is between
0.1 and 1; therefore for relatively larger number(for example, 12345.456),
we may lose important accuracy if we store it in single precision. This is
very important when we consider the basic data types.

On the other hand, the storage for double precision is like this:
\begin{description}
 \centering
 \item[Sign of mantessa] 1 bit;
 \item[Sign of exponent] 1 bit;
 \item[exponent($|m|$)]  10 bits;
 \item[mantissa($|q|$)]  52 bits
\end{description}
The range for double precision is between $10^{-308}$ and $10^{308}$, and the
accuracy for double precision has 15-16 figures.

\subsection{Chopping, Rounding up, Underflow and Overflow}
%
%
Sometimes it's useful to find a nearby number for a given real number stored on
computer. For example, the number is:
\begin{equation}
 n = (0.a_{1}a_{2}\cdots a_{n})_{2}\times 2^{m}
\end{equation}
If we drop the last digit of $a_{n}$ then such approximation is ``chopping''. If we
increase the $a_{n}$ it's called rounding up. It's easy to prove that the relative
error for both chopping and rounding up is smaller than $\dfrac{1}{2}2^{-n}$.

Occasionally, sometime the a provide real number may be outside the range of the
exponent. That means, its exponent is larger than the range. Such case is called
``overflow''. If the number smaller than the lower bound of the range, it's called
``underflow''. In this case, the number could be safely set to 0 then the program
could still continue; however, if the number's exponent is larger than the higher
bound of the range, this case is ``overflow'' and the state of the number is ``NAN''
which means ``not a number''.

\subsection{Error Propagation in Arithmetic}
%
%
\label{error_propagation_numerical}
In statistics, the error(either absolute error or relative error)\footnote{
See \url{https://en.wikipedia.org/wiki/Error_propagation}} are propagating in
the way of ``accumulation''. That means, the error is usually accumulating in
the arithmetic process when we handle the floating point numbers. However, such
drawback could be conquered by introducing higher precision way to store the
real number (I think that is also why we need FPU(Floating-point unit) to
handle the floating point operations). Therefore, usually for a programmer we
do not need to generally care about the accuracy lose in arithmetic operation.

However, in some cases the lose of accuracy is introduced because of the ``
inappropriate'' programming way. Here we will introduce some examples.

The first example is to subtract two nearby real numbers. For example, if we
have two number of $a_{1}$ and $a_{2}$:
\begin{equation}
 a_{1} = 1.234556   a_{2} = 1.234557
\end{equation}
both of the two numbers are stored in single precision way, that means; we have
6 significant figures. However, the difference between $a_{1}-a_{2}$, is
$1.0\times 10^{6}$ therefore the difference is already lying in the error range.
The thing is that if we want this difference to be very accurate, since it's
in the error range therefore we may not be able to retain the accurate value.
Therefore, such manipulation should be desiring special attention.

The second example, compared with first one is more troublesome. It's called
``numerical instability'' operations. It means that the numerical accuracy
is significantly increase because of the way we express the formula. For example,
if we have a recursive expression:
\begin{equation}
 x_{n+1} = \frac{13}{3}x_{n} - \frac{4}{3}x_{n-1} (n\geq 1)
\end{equation}
if we set that $x_{0}$ is 1, $x_{1}$ is $\dfrac{1}{3}$; then the series
should be more and more approach to 0. However, if we use single precision
to process the operation then it turns out that the error will surpass the
result in some point then the result will gradually get larger until to infinity!
Therefore, such recursive expression is ``numerical sensitive'' and could be
suffer from the error propagating.

The reason for this case to be numerical instable is that as $n$ gets larger,
the difference between $\dfrac{13}{3}x_{n}$ and $\dfrac{4}{3}x_{n-1}$ is getting
smaller. Therefore at some point, the numerical error will surpass the
real calculation result because two nearby number subtraction. Therefore, to use
the recursive relation in numerical calculation it's necessary to keep en eye
on the error propagation.

\section{Solution of Non-Linear Equations}

In practical application, we usually meet the problem that to solve the non-linear
equation like below:
\begin{equation}
 x - a\sin x = b
\end{equation}
It's root could be got iteratively. Therefore, in this section we will discuss about
the method that how to iteratively solve the non-linear equations.

\subsection{Newton-Raphson Method and Quasi-Newton Method}
%
%
%
Newton-Raphson method is the simplest case to derive a solution for linear equations.
Firstly, let's look at this method in one dimension.

Suggest that we have a function of $f(x)$, and we want to get its solution of $x$
when $f(x) = 0$. However, $x$ maybe difficult to be solved due to the complicated form
of $f(x)$. Therefore, we use iterative way to get $x$.

Let's assume that $x_{0}$ is a good approximation to the $x$, then we have:
\begin{equation}\label{Newton-Raphson_eq:1}
 f(x) = f(x_{0} + h) \approx f(x_{0}) + hf^{'}(x_{0})
\end{equation}
Since $f(x) = 0$, then we have:
\begin{equation}
 \label{Newton-Raphson_eq:2}
 f(x_{0}) + hf^{'}(x_{0}) = 0 \Rightarrow h = -\frac{f(x_{0})}{f^{'}(x_{0})}
\end{equation}
Because $h$ is the difference between the new estimated solution $x_{1}$ and
the original approximated solution $x_{0}$, therefore we have:
\begin{equation}
 \label{Newton-Raphson_eq:3}
 h = -\frac{f(x_{0})}{f^{'}(x_{0})} \Rightarrow x_{1} = x_{0} - \frac{f(x_{0})}{f^{'}(x_{0})}
\end{equation}
If we solve the equation in iterative way, then it's easy to see that:
\begin{equation}
 \label{Newton-Raphson_eq:4}
 x_{n} = x_{n-1} - \frac{f(x_{n-1})}{f^{'}(x_{n-1})}
\end{equation}
until the difference between $x_{n}$ and $x_{n-1}$ is small enough.

This method is quadratic convergence method, it's easy to see that the error is estimated
as:
\begin{equation}
 \label{newton_raphson_error}
 e_{n+1} = \frac{1}{2}\frac{f^{''}(\xi)}{f^{'}(\xi)}e_{n}^{2}
\end{equation}
That means, the successive error would be the square of its previous error, so the error
lowers down in quadratic way.

The above procedure only concerns the solution for the given function $f(x)$, however; if
we want to calculate its first derivatives, for example $f^{'}(x) = 0$ then how can we do?
To answer this question we may use quasi-Newton method, which is a step further beyond the
Newton-Raphson method.

Let's further consider the second derivatives of $f(x)$, by using the Taylor expansion,
we can have:
\begin{equation}
 \label{Newton-Raphson_eq:5}
 f(x) = f(x_{0}) + f^{'}(x_{0})h + \frac{1}{2}f^{''}(x_{0})h^{2}
\end{equation}
Now let's take the directive for $x$, since $h = x - x_{0}$ then we can see that
\begin{equation}
 \label{Newton-Raphson_eq:6}
 f^{'}(x) = f^{'}(x_{0}) + f^{''}(x_{0})h = 0
\end{equation}
It leads to the result that:
\begin{equation}
\label{Newton-Raphson_eq:7}
 h = -\frac{f^{'}(x_{0})}{f^{''}(x_{0})} \Rightarrow x = x_{0} - \frac{f^{'}(x_{0})}{f^{''}(x_{0})}
\end{equation}
All in all, we have:
\begin{equation}
\label{Newton-Raphson_eq:8}
 x_{n} = x_{n-1} - \frac{f^{'}(x_{n-1})}{f^{''}(x_{n-1})}
\end{equation}

It's easy to extend the above equation into multi-dimensions variable:
\begin{equation}
\label{Newton-Raphson_eq:9}
 \mathbf{x_{n}} = \mathbf{x_{n-1}} - H^{-1}G
\end{equation}
Here $H$ is the Hessian matrix, and $G$ is the derivative vectors for the $\mathbf{x_{n-1}}$
vector.

Generally, the convergence for the Newton-Raphson method and quasi-Newton method is fast,
it converges in quadratic way: as the method converges on the root, the difference between
the root and the approximation is squared (the number of accurate digits roughly doubles)
at each step.

However, this method relies on the point that the initial guess of $x_{0}$ should be a good
approximation to the root. If a bad guess is given then the Newton-Raphson method may not
converge.

\subsection{Secant Method}
%
%
%
In Newton-Raphson method, the iterative root is evaluated as:
\begin{equation}
 x_{n} = x_{n-1} - \frac{f(x_{n-1})}{f^{'}(x_{n-1})}
\end{equation}
The problem here is that we need the first derivatives for the $f(x)$. If the derivatives
is hard to get, then the Newton-Raphson method could be suffer from this deficiency.

One way to solve this problem is to replace the derivative term of $f^{'}(x)$ as:
\begin{equation}
 f^{'}(x_{n-1}) = \frac{f(x_{n-1}) - f(x_{n-2})}{x_{n-1} - x_{n-2}}
\end{equation}
Then the above equation is turning into:
\begin{equation}
\label{secant_eq:1}
 x_{n} = x_{n-1} - \frac{f(x_{n-1})(x_{n-1} - x_{n-2})}{f(x_{n-1}) - f(x_{n-2})}
\end{equation}
This new equation is called ``Secant'' method.

One of important question related to the new method is that how fast it's converging to
the real root. Now let's try to evaluate it. Suggest that the error is $e_{n} = x_{n} -r$,
where $r$ is the real root; then we have:
\begin{align}
\label{secant_eq:2}
 e_{n} &= e_{n-1} - \frac{f(x_{n-1})(x_{n-1} - x_{n-2})}{f(x_{n-1}) - f(x_{n-2})}  \nonumber \\
       &= e_{n-1} - \frac{(x_{n-1} - x_{n-2})}{1 - \frac{f(x_{n-2})}{f(x_{n-1})}} \nonumber \\
       &= e_{n-1} - \frac{(e_{n-1} - e_{n-2})}{1 - \frac{f(x_{n-2})}{f(x_{n-1})}} \nonumber \\
       &= \frac{f(x_{n-2})}{f(x_{n-2})-f(x_{n-1})}e_{n-1} +
\frac{f(x_{n-1})}{f(x_{n-1})-f(x_{n-2})}e_{n-2} \nonumber \\
       &= \frac{f(x_{n-1})e_{n-2} - f(x_{n-2})e_{n-1}}{f(x_{n-1})-f(x_{n-2})}
\end{align}

Now let's use Taylor expansion:
\begin{equation}
 f(x_{n}) = f(r) + e_{n}f^{'}(r) + \frac{1}{2}e^{2}_{n}f^{''}(r)
\end{equation}
Then the above expression turns into (remember that $f(r) = 0$):
\begin{align}
\label{secant_eq:3}
 e_{n} &= \frac{\frac{1}{2}f^{''}(r)(e_{n-1}-e_{n-2})e_{n-1}e_{n-2}}{(e_{n-1}-e_{n-2})
\Big ( f^{'}(r) +\frac{1}{2}f^{''}(r)(e_{n-1}+e_{n-2}) \Big ) } \nonumber \\
&= \frac{\frac{1}{2}f^{''}(r)e_{n-1}e_{n-2}}{f^{'}(r) +\frac{1}{2}f^{''}(r)(e_{n-1}+e_{n-2})}
\end{align}

Here we meet a dilemma, since the size of $\frac{1}{2}f^{''}(r)(e_{n-1}+e_{n-2})$ is
approximated to be $-f^{'}(r)$ according to the Taylor expansion.
From this point, we can see that the difficulty arises because that we are trying to
evaluate the $f(x_{n-1})-f(x_{n-2})$, which should be avoided.

Let's try to multiply $\dfrac{x_{n-1} - x_{n-2}}{x_{n-1} - x_{n-2}}$ on \ref{secant_eq:2}:
\begin{align}
\label{secant_eq:4}
 e_{n}  &= \frac{x_{n-1} - x_{n-2}}{f(x_{n-1})-f(x_{n-2})}
\frac{f(x_{n-1})e_{n-2} - f(x_{n-2})e_{n-1}}{x_{n-1} - x_{n-2}} \nonumber \\
&= \frac{x_{n-1} - x_{n-2}}{f(x_{n-1})-f(x_{n-2})}
\frac{f(x_{n-1})/e_{n-1} - f(x_{n-2})/e_{n-2}}{x_{n-1} - x_{n-2}}e_{n-1}e_{n-2}
\end{align}
such transformation is suggested by the dead end of \ref{secant_eq:3}, that we know there should
be the $e_{n-1}e_{n-2}$ in the final expression. On the other hand, such transformation would
allow us to use the Taylor expansion:
\begin{equation}
 f(x_{n}) = f(r) + e_{n}f^{'}(r) + \frac{1}{2}e^{2}_{n}f^{''}(r)
\end{equation}
Now we try to get $f(x_{n})/e_{n}$:
\begin{equation}
 f(x_{n})/e_{n} = f^{'}(r) + \frac{1}{2}e_{n}f^{''}(r)
\end{equation}
Therefore \ref{secant_eq:4} is finally becoming:
\begin{equation}
 e_{n}  = \frac{1}{2}\frac{f^{''}(r)}{f^{'}(r)}e_{n-1}e_{n-2} = Ce_{n-1}e_{n-2}
\end{equation}
More detailed analysis reveals that the convergence rate is about 1.6, which is
lower than the quadratic method of 2.

\section{Solution of Linear Equations}
%
%
%
In this section, we will briefly talk about the solution of linear equations.
Generally, to solve a linear equation is divided into two types of methods:
\begin{enumerate}
 \item directly solving the equations
 \item iteratively solving the equations(indirect way)
\end{enumerate}

For the direct solving process, the key idea for the solution of linear equations,
is actually the decomposition of the left hand side matrix(this is also called
matrix factorization). Although we could apply the basic Cramer's rule to solve
the linear equations, however; the matrix factorization way is much faster.

Why such roundabout way could be faster than the direct Cramer's rule?
Usually in solving the linear equations, we decompose the matrix into orthogonal
matrix and/or triangular matrix. For orthogonal matrix, it's inverse is just its
transpose form, therefore the expensive matrix inverse is avoided. On the other
hand, from the following content(Gaussian elimination, see the back-substitution
etc.) we can see that triangular matrix is very easy to combine and separate with
the vectors.

The matrix factorization methods commonly used here for solving linear equations
are referred as:
\begin{enumerate}
 \item LU decomposition (L and U are two triangular matrices)
 \item SVD(single value decomposition) decomposition
       (two orthogonal matrices plus the diagonal matrix)
 \item QR decomposition (Q is orthogonal matrix and R is triangular matrix)
\end{enumerate}

For an arbitrary matrix it's decomposition methods could be divided into two types
according to the matrix is singular or not. When the given matrix is non-singular,
then the LU decomposition could be the best choice; on the other hand, when the given
matrix is singular, the SVD or QR is necessary for solving the linear equation. Furthermore,
we will discuss the symmetrical matrix decomposition, since this is the most cases we
encounter in the computational chemistry.

\subsection{Gaussian Elimination}
%
%
A typical linear equation could be generally expressed as:
\begin{equation}
 \label{Solution_Linear_Equations_eq:1}
 \begin{pmatrix}
a_{11}  &  a_{12}  & \cdots  & a_{1n} \\
a_{21}  &  a_{22}  & \cdots  & a_{2n} \\
\cdots  &  \cdots  & \cdots  & \cdots \\
a_{m1}  &  a_{m2}  & \cdots  & a_{mn} \\
\end{pmatrix}
\begin{pmatrix}
 x_{1} \\
 x_{2} \\
 \cdots \\
 x_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
 b_{1} \\
 b_{2} \\
 \cdots \\
 b_{m} \\
\end{pmatrix}
\end{equation}
Which is in short, we can write it as:
\begin{equation}
 \label{Solution_Linear_Equations_eq:2}
 Ax = b
\end{equation}

It's easy to see that the following elemental operations does not change the
solution of $x$ in the above equation:
\begin{enumerate}
 \item interchanging any two rows and columns of matrix $A$ (if we interchange
 the rows, we note that $x$ and $b$ should change accordingly);
 \item multiply any row of matrix $A$ with a factor of $\lambda$;
 \item adding/subtracting any two rows of matrix $A$.
\end{enumerate}

Therefore, we can use the above operations to solve the equation in general,
this is called ``Gaussian Elimination''. Such algorithm is to convert the above
equation in the following form:
\begin{equation}
 \label{Solution_Linear_Equations_eq:3}
 \begin{pmatrix}
a_{11}^{'}  &  a_{12}^{'}  & \cdots  & a_{1n}^{'} \\
0           &  a_{22}^{'}  & \cdots  & a_{2n}^{'} \\
\cdots      &  \cdots      & \cdots  & \cdots \\
0           &  0           & \cdots  & a_{mn}^{'} \\
\end{pmatrix}
\begin{pmatrix}
 x_{1} \\
 x_{2} \\
 \cdots \\
 x_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
 b_{1}^{'} \\
 b_{2}^{'} \\
 \cdots \\
 b_{m}^{'} \\
\end{pmatrix}
\end{equation}
that is to say, the matrix $A$ is transformed into a upper triangular form. Then
we could use back-substitution to get the vector of $x$:
\begin{equation}
 \label{Solution_Linear_Equations_eq:4}
 x_{i} = \frac{1}{a_{ii}^{'}}\left( b_{i}^{'} - \sum_{j=i+1}^{n}a_{ij}^{'}x_{j}\right)
\end{equation}
This is called ``back-substitution'' is because that the solution of $x$ is solved
recursively from $x_{n}, x_{n-1}$ to $x_{1}$. Similarly, we can also transform the
matrix into lower triangular form and use ``forward-substitution'' to get the solution
of $x$.

In applying the Gaussian elimination method, we note that we should make sure that
all of $A_{ii}$ should be at least larger than ZERO(the element of $A_{ii}$ is
referred as pivot or pivot element). this could be achieved by interchanging the
columns of matrix $A$. In practice, the requirement of larger than 0 is not enough
for picking up the pivot element in terms of the rounding error. usually it should be
the largest value among the given row, however; there's no general algorithm to assign
the pivot element(see book of numerical recipes for more details, section 2.1.2).

We note, that the Gaussian elimination method is a very stable method for solving the
linear equations. This is also applied to the matrix without full rank(in this case, after
transformation some row of matrix $A$ would be all zero). It's complexity is approximated
as $O(n^{3})$. On the other hand, the disadvantage for this algorithm is that the solving
process requires the existence of the right hand side of the equation.

\subsection{LU Decomposition}
%
%
Now let's consider a special extension of Gaussian elimination method. In this method,
the matrix $A$ shown above in general performs the operation to the rows step by step; all of
these operations actually could be represented by a lower triangular form of matrix. Therefore,
it implies a new matrix factorization method where the matrix $A$ could be decomposed into
a product of lower triangular matrix and a upper triangular matrix:
\begin{equation}
 \label{Solution_Linear_Equations_eq:5}
  \begin{pmatrix}
a_{11}  &  a_{12}  & \cdots  & a_{1n} \\
a_{21}  &  a_{22}  & \cdots  & a_{2n} \\
\cdots  &  \cdots  & \cdots  & \cdots \\
a_{m1}  &  a_{m2}  & \cdots  & a_{mn} \\
\end{pmatrix} =
\begin{pmatrix}
l_{11}  &  0       & \cdots  &  0     \\
l_{21}  &  l_{22}  & \cdots  &  0     \\
\vdots  &  \vdots  & \ddots  & \vdots \\
l_{m1}  &  l_{m2}  & \cdots  & l_{mn} \\
\end{pmatrix}
\begin{pmatrix}
u_{11}  &  u_{22}  & \cdots  & u_{1n} \\
0       &  u_{22}  & \cdots  & u_{2n} \\
\vdots  &  \vdots  & \ddots  & \vdots \\
0       &  0       & \cdots  & u_{mn} \\
\end{pmatrix}
\end{equation}
We note that this is only applied when $m=n$
in the matrix $A$.

Furthermore, it could be proved that the LU decomposition
should be used when $A$ is non-singular(full rank). To be
precise, this is not strictly true. The exact requirement
to perform LU decomposition is that the first $k$th leading
principal minor is not zero for matrix $A$ if it's rank is $k$.

To carry out the LU decomposition in a given matrix is very easy.
The practical algorithm to perform such decomposition is called
Doolittle factorization or Crout factorization. For the Doolittle
factorization, it's general algorithm could be got as:
\begin{verbatim}
for k = 1, 2, ... n;
   l_{kk} = 1;
   for j = k, k+1, .. n;
      u_{kj} = a_{kj} - sum_{s=1}^{k-1}l_{ks}u_{sj}
   end
   for i = k+1, k+2, .. n;
      l_{ik} = (a_{ik} - sum_{s=1}^{k-1}l_{is}u_{sk})/u_{kk}
   end
end
\end{verbatim}

Since the LU decomposition is essentially equal to the Gaussian
elimination, therefore its complexity is similar with the Gaussian
elimination process.

\subsection{Cholesky Decomposition}
%
%
%
In the above LU decomposition process, if the matrix $A$ is symmetrical;
that means $A^{T} = A$; then it's easy to know that $U = L^{T}$ therefore
$A = LL^{T}$. This is called Cholesky decomposition.

Usually, the matrix for applying Cholesky decomposition is positively defined;
that means all of eigenvalues for the matrix $A$ are positive. The matrix element
in the $L$ could be evaluated as:
\begin{align}
 L_{ii} &= \left( A_{ii} - \sum_{k=1}^{i-1}L_{ik}^{2}\right) \nonumber \\
 L_{ij} &= \frac{1}{L_{jj}}\left( A_{ij} - \sum_{k=1}^{j-1}L_{ik}L_{jk} \right)
\end{align}

We note that the complexity for Cholesky decomposition is faster than LU decomposition.
Approximately it's two times fast(see numerical recipes, section 2.9).

\subsection{Pivoting in LU Decomposition and Cholesky Decomposition}
%
%
%
In the above paragraph, we have slightly discussed the pivoting used in Gaussian elimination.
The improper pivoting process may result in wrong solution in applying the algorithms. In
the book given by Kincaid, in section 4.3 of pivoting there's a very good example to illustrate
that why the improper pivoting would cause huge errors.

Here we just note that in practice the LU decomposition is heavily relying on the proper pivoting,
but the Cholesky decomposition is very stable. It just failed when the given matrix is not
positively defined.

\subsection{Single Value Decomposition}
%
%
%
the above algorithms shares a common feature, that the matrix could be applied should be
non-singular. However, the singular matrix is very common in computational chemistry.
For example, the matrix formed based on the auxiliary basis set functions could be a
singular matrix. Therefore, how to deal with the matrix which is in singular type? From
this section we will briefly discuss it.

Single value decomposition could be expressed as:
\begin{align}
 \label{Solution_Linear_Equations_eq:6}
 \begin{pmatrix}
a_{11}  &  a_{12}  & \cdots  & a_{1n} \\
a_{21}  &  a_{22}  & \cdots  & a_{2n} \\
\vdots  &  \vdots  & \ddots  & \vdots \\
a_{m1}  &  a_{m2}  & \cdots  & a_{mn} \\
\end{pmatrix}
&=
\begin{pmatrix}
U_{11}  &  U_{12}  & \cdots  & U_{1m} \\
U_{21}  &  U_{22}  & \cdots  & U_{2m} \\
\vdots  &  \vdots  & \ddots  & \vdots \\
U_{m1}  &  U_{m2}  & \cdots  & U_{mm} \\
\end{pmatrix} \times \\ \nonumber
&
\begin{pmatrix}
w_{11}  &  0       & \cdots  & 0      \\
0       &  w_{22}  & \cdots  & 0      \\
\vdots  &  \vdots  & \ddots  & \vdots \\
0       &  0       & \cdots  & \ddots \\
\end{pmatrix}
\begin{pmatrix}
V_{11}  &  V_{12}  & \cdots  & V_{1n} \\
V_{21}  &  V_{22}  & \cdots  & V_{2n} \\
\vdots  &  \vdots  & \ddots  & \vdots \\
V_{m1}  &  V_{m2}  & \cdots  & V_{nn} \\
\end{pmatrix}
\end{align}
Where the $U$ and $V$ are orthogonal matrix:
\begin{align}
 \label{Solution_Linear_Equations_eq:7}
 UU^{T} &= I \nonumber \\
 VV^{T} &= I
\end{align}
Mathematically, the $U$ is the eigenvectors of $AA^{T}$,
and $V$ is eigenvectors of $A^{T}A$.

The middle matrix of $W$ is some rectangular diagonal matrix,
it's diagonal elements are in a range of $1,min(M,N)$.
We note, the diagonal elements
are the square of eigenvalues of matrix $AA^{T}$ and $A^{T}A$:
\begin{align}
 \label{Solution_Linear_Equations_eq:8}
 AA^{T} &= UWV^{T}VW^{T}U^{T} \nonumber \\
        &= UWW^{T}U^{T}
\end{align}
similar facts hold for the $A^{T}A$ matrix.

The SVD method is numerically very stable, but the time consumption
for SVD algorithm perhaps is the most among all of matrix factorization
methods. The detailed time consumption evaluation could be referred to
the math library.

The SVD method could be used to factorize a singular matrix. For example,
if the matrix $A$ is square matrix, then we can write its SVD decomposition
as $A = UWV^{T}$, where both $U$ and $V$ are square matrix. The $W$ will appear
as:
\begin{equation}
 \label{Solution_Linear_Equations_eq:9}
 W =
 \begin{pmatrix}
w_{11}  &  0       & \cdots  & 0      \\
0       &  w_{22}  & \cdots  & 0      \\
\vdots  &  \vdots  & \ddots  & \vdots \\
0       &  0       & \cdots  & W_{nn} \\
\end{pmatrix}
\end{equation}

Suggest that $w_{ii}, w_{i+1,i+1} \cdots w_{nn}$ are all zero, then here
how can we evaluate $A^{-1}$, and more generally, the $A^{p}$ where $p$
could be any integer?

To answer this question, we need to review the mathematical meanings of SVD
method. The matrix of $U$ and $V$, are actually composed into a set of
orthogonal basis sets for the $K^{M}$ space and $K^{N}$ space. The SVD
means that for a given vector(here we consider $A$'s column),
we pick up an orthogonal basis set in $K^{M}$ space(the $U$'s columns)
and it's ith vector(the ith column of $U$) is mapping into another vector
in $K^{N}$ space(the column of $V$) by a positive value. This value is the element
of $w_{i}$.

Furthermore, If we consider a general space, that means an arbitrary group of
vectors in space of $K^{N}$, then if there's linear dependency in the vectors
the mapping factor between $K^{M}$ space and $K^{N}$ space would be zero. Therefore,
we can see that if the given matrix has linear dependency, we can simply drop the
vectors in $U$ and $V$(that means to find a linear independent set in the two spaces)
and perform the following operations\footnote{In the book of numerical recipes in
chapter 2 section 2.6.2 there's another interpretation for the above words, more precisely
and strictly. Author could refer to the material for more information}.

Therefore, to evaluate the inverse of matrix $A$, if $A$ is singular and there's no
strict inverse for $A$; we could approximate the $A^{-1}$ as:
\begin{equation}
 \label{Solution_Linear_Equations_eq:10}
 A = UWV^{T} \Rightarrow A^{-1} \approx V_{m,k}\times W_{k,k}U^{T}_{k,m}
\end{equation}
Where $k$ is the rank for matrix $A$. The matrix generated by such method is also
called Moore–Penrose pseudoinverse\footnote{see the web page of
\url{http://en.wikipedia.org/wiki/Moore\%E2\%80\%93Penrose_pseudoinverse}}.

\subsection{Symmetrical Matrix Decomposition}
%
%
%
For a symmetrical matrix, if it's non-singular the best way to factorize the matrix
is the Cholesky decomposition. However, usually in computational chemistry it's very hard
to determine a given matrix is singular or not. For example, it's hard to judge that
whether there's linear dependency in the given basis set functions. Therefore, we usually
take another way.

From the spectrum's theorem, it's easy to know that for any given symmetrical matrix,
it can be decomposed into:
\begin{equation}
  \label{Solution_Linear_Equations_eq:11}
  A = UDU^{T}
\end{equation}
$U$ is some orthogonal matrix, and $D$ is diagonal matrix. The columns of $U$ represent
the orthogonal set of eigenvectors for matrix $A$, $D$ contains its corresponding $n$
eigenvalues. From mathematical point of view, such decomposition is to transform the
vectors of $A$ ($A$'s columns) into another orthogonal set in the same space.

Therefore, by referring the point in the above section of SVD decomposition; it's easy
to know that even though $A$ is singular, we could apply the similar tricks to get its
Moore–Penrose pseudoinverse:
\begin{equation}
  \label{Solution_Linear_Equations_eq:11}
  A^{-1} = U_{m,k}D_{k}U^{T}_{k,m}
\end{equation}
just by simply dropping it's zero eigenvectors.