text:
\begin{document}
\maketitle
\begin{abstract}
Two flavours Skyrmion will be analyzed analytically, in case of static and rotational Skyrme equations. Numerical solution of a nonlinear scalar field equation, i.e. the Skyrme equation, will be worked with finite difference method. This article is a more comprehensive version of \textit{SU(2) Skyrme Model for Hadron} which have been published at Journal of Theoretical and Computational Studies, Volume \textbf{3} (2004) 0407.\\\\
\textbf{Keywords}: \textit{SU(2) Skyrme model, hadron, topological soliton, finite difference method}.
\end{abstract}

\section{Introduction}
\textit{Nonlinear physics} is a general phenomenon in physics. It includes very long range from e.g. particle, nuclear, condensed matter, fluids, plasmas, biophysics (for example: nonlinear diffusion-reaction, DNA), to cosmology.

\textit{Soliton} is defined as classical solution of nonlinear wave equation which has properties: finite total energy, localized, nondispersive, stable with its profile of energy density distribution is like pulse centered in finite space \cite{mif04}.

\textit{Skyrmion} is a \textit{topological soliton} in three dimensions. Following, we will discuss soliton as \textit{baryon model}, especially \textit{Skyrme model} for hadron which is formulated in \textit{extended nonlinear Sigma model} \cite{mif04}, \cite{hans05}.

\section{Topological Solitons}
Suppose that $F_\phi=0$ is a differential equation or a system of differential equation involving a field or a set of fields denoted by $\phi$ which are functions of $D$ space coordinates $\vec{x}$ and a time coordinate $t$. We shall assume that $F_{\phi}=0$ is the equation of motion which results from some Lagrangian filed theory. We also assume that the system has a corresponding "energy" $\varepsilon_\phi$ and "energy density" $E_\phi(\vec{x},t)$ \cite{bal91}
\begin{equation}\label{1}
\varepsilon_\phi=\int d^Dx~E_\phi(\vec{x},t)
\end{equation}
where for all allowed field configurations $\phi$, $E_\phi$ is greater than or equal to zero. If $E_{\phi}$ is zero for all $\vec{x}$, we call $\phi$ \textit{a ground state} or \textit{a vacuum solution}, and denote it by $\phi_{vac}$. The vacuum solution need not be unique.

Now suppose that $\phi=\phi_{cl}$ is \textit{non-vacuum solution} to $F_\phi=0$. Following Coleman, we shall call $\phi_{cl}(\vec{x},t)$ a \textit{soliton solution} if the following properties are satisfied \cite{bal91}:
\begin{itemize}
\item[(1)]
$\varepsilon_{\phi=\phi_{cl}}$ is \textit{finite}.
\item[(2)]
$E_{\phi=\phi_{cl}}(\vec{x},t)$ is \textit{nonsingular (finite)} for all value of $\vec{x}$ and $t$, and \textit{localized} for all times $t$. A solution is localized at any time $t$ if there is a bounded region of space defined by $E_{\phi=\phi_{cl}}(\vec{x},t)\geq\delta$, $\delta$ being any arbitrary number, fulfilling
\begin{equation}\label{2}
0<\delta<\text{max}_{\vec{x}}~E_{\phi=\phi_{cl}}(\vec{x},t).
\end{equation}
We say that the solution is localized for all times $t$ if the bounded region can be chosen independent of $t$.
\item[(3)]
$\phi_{cl}(\vec{x},t)$ is \textit{nonsingular}.
\item[(4)]
$\phi_{cl}(\vec{x},t)$ is \textit{nondissipative}.\\
Concerning (4), a solution $\phi_{cl}(\vec{x},t)$ is considered dissipative if
\begin{equation}\label{3}
\text{lim}_{t\rightarrow\infty}\text{max}_{\vec{x}}~E_{\phi=\phi_{cl}}(\vec{x},t)=0.
\end{equation}
In our discussions, we will be concerned primarily with "static" solutions for which we denote $\phi_{cl}(\vec{x},t)$ by $\Phi_{cl}(\vec{x},t)$.

\textit{If we exclude the vacuum, then static, nonsingular, localized solutions are automatically nondissipative}. Thus for static solitons, we only require (1), (2) and (3).

We will, however, be interested in static solutions which satisfy yet another requirements; namely that they be
\item[(5)]
\textit{classically stable}.
We call a static solution "classically stable" if under all possible infinitesimal fluctuations $\delta\phi$ of the fields $\phi$ about the static solution,
\begin{equation}\label{4}
\varepsilon_{\phi_{cl}+\delta\phi}\geq\varepsilon_{\phi_{cl}}.
\end{equation}
\end{itemize}
Variational modes $\delta_0\phi$ which leave the energy unchanged are said to be "zero" frequency modes" or "zero modes". Thus, for zero modes, $\varepsilon_{\phi_{cl}+\delta_{0}\phi}=\varepsilon_{\phi_{cl}}$. \textit{Zero modes are generally associated with various symmetries of the soliton}. Excluding the zero modes, a classically stable solution is a local minimum of the energy.

Although the question of \textit{stability} is ultimately a dynamical one, \textit{topology} often plays an important role. We shall now elaborate on this point. Let $Q$ be \textit{the set} of all finite energy and nonsingular field configurations $\phi$ at some fixed time $t$. It is thus \textit{the configuration space} of the system. \textit{A subset} $Q_1$ of $Q$ is said to be \textit{(path-)connected} if any field $\phi_1$ of $Q_1$ can be \textit{continuously deformed} to any other field $\phi_1'$ in $Q_1$.

Two subsets $Q_1$ and $Q_2$ of $Q$ are said to be \textit{disconnected} if any field $\phi_1$ of $Q_1$ cannot be continuously deformed to any field $\phi_2$ in $Q_2$. We shall consider the case where $Q$ has $N>1$ disconnected components $Q_n$, each $Q_n$ being connected. $Q$ is the union of all these disconnected components,
\begin{equation}\label{5}
Q=\cup^N_{n=1}Q_n.
\end{equation}

Let $\phi(\vec{x})$ and $\phi'(\vec{x})$ be two \textit{fields} in $Q$. $\phi(\vec{x})$ and $\phi'(\vec{x})$ are \textit{homotopic} to each other, and write $\phi\sim \phi'$, if there exists a sequence of fields $\phi^{(\tau)}(\vec{x})$, $0\leq\tau\leq 1$, which is continuous in $\tau$ and $\vec{x}$, with $\phi^{(0)}(\vec{x})=\phi(\vec{x})$ and $\phi^{(1)}(\vec{x})=\phi'(\vec{x})$. With this definition, all fields within one component $Q_n$ are "homotopic" to each other, while a field $\phi^{(n)}$ belonging to $Q_n$ is not homotopic to a field $\phi^{(n')}$ belonging to $Q_{n'}$ when $n'\neq n$ \cite{bal91}.

We can treat all our examples assuming that $n$ is countable. We make this assumption: \textit{What is the physical significance of this classification?} Consider the initial conditions $(\phi^{(n)},d\phi^{(n)}/dt)$ at time $t=0$ for the equations of motion. (These equations for simplicity are assumed to be second order in time). Assume that $\phi^{(n)}\in Q_n$. After a lapse of time $T$, suppose that $\phi^{(n)}$ becomes $\phi'^{(n)}$ and $d\phi^{(n)}/dt$ becomes $d\phi'^{(n)}/dt$. Since time evolution is assumed to be a continuous operation, it follows that $\phi'^{(n)}$ is homotopic to $\phi^{(n)}$. Hence, $\phi'^{(n)}\in Q_n$. In other words, the value of $n$ associated with the field $\phi^{(n)}$ is a constant of the motion. Hence, if we define  $Q_0$ to be the sector which contains the vacuum solution $\phi_{vac}$, then the configurations $\phi^{(n)}\in Q_n,~n\neq 0$, cannot be time evolved to $\phi_{vac}$ or in fact, to any other $\phi^{(m)}\in Q_m$ with $m\neq n$. For such reasons, the sectors $Q_{n\neq0}$ of field configurations are said to be \textit{topological stable}. Soliton configurations which fulfill the five properties listed previously and which are in topologically stable sectors are called \textit{topological solitons} \cite{bal91}.

\section{Two Flavours Skyrmion as Hadron}
In particle physics, \textit{flavour} is a quantum number of elementary particles. In quantum chromodynamics (QCD), flavour is a global symmetry. In the electroweak theory, on the other hand, this symmetry is broken, and flavour changing processes exist \cite{wiki-flav}.

If there are two or more particles which have identical interactions, then they may be interchanged without affecting the physics. Any (complex) linear combination of these two particles give the same physics, as long as they are orthogonal or perpendicular to each other. In other words, the theory possesses symmetry transformations such as $M\left({u\atop d}\right)$, where $u$ and $d$ are the two fields, and $M$ is any $2\times 2$ unitary matrix with a unit determinant. Such matrices form a Lie group called $SU(2)$, \textit{special unitary group}. This is an example of flavour symmetry. The term "flavour" was first coined for use in the quark model of hadrons in 1968 \cite{wiki-flav}.

In order to understand the charge radius of nucleon which have size roughly 1 Fermi, Tony Hilton Royle Skyrme in 1962 proposed idea that strongly interacting particles (hadrons) were locally concentrated static solution of extended nonlinear Sigma (Chiral) model. In (3+1)-dimensions of space-time, we will observe Skyrme model which describes hadrons as solitons (Skyrmions) from nonlinear Sigma (chiral) model field theory in internal symmetrical group of SU(2): two flavours Skyrmion.

\textit{Skyrme's idea is unifying bosons and fermions in a common framework which provide a fundamental
fields model consisted of only pion. The nucleon was obtained, as a certain classical configuration of the pion fields. Skyrmion is a topological soliton object, i.e. solution to the classical field equation with localized energy density. Various atoms should correspond to vortices of different connectivities in some underlying liquid}.

Following, some ideas which are proposed by Skyrme in relations with baryon model \cite{mif04}, \cite{val92}, \cite{mift04}:
\begin{itemize}
\item[(1)] Meson field can take its value in $S_3$ manifold. As the result of this assumption, Skyrme discovers conserved quatity i.e. \textit{topological charge} or \textit{winding number} and he interpretates it as \textit{baryon number}.
\item[(2)] Solution of Skyrme field equation in spherical coordinate takes form:
\begin{equation}\label{6}
F_a (r,\theta,\phi) = g(r)n_a (\theta,\phi)
\end{equation}
where $g(r)$ is profile function which has spherical symmetric character and $n_a$ ($a=1,2,3$) is component of unit vector,
$\boldsymbol{\widehat n}$. $F_a (r,\theta,\phi)$ is called \textit{Skyrme ansatz} or \textit{ hedgehog}.
\item[(3)] Skyrme ansatz describes \textit{stable extended particle} with unit topological charge.
\item[(4)] Skyrme ansatz with unit topological charge \textit{can be quantized} as fermion, so it is possible to identify state of isotopic spin, $I$, and total spin, $J$, with nucleon doublet in case of  $I=J=\frac{1}{2}$, and with $\Delta$ resonance in case of $I=J=\frac{3}{2}$.
\end{itemize}

\section{SU(2) Skyrme Model}
SU(2) Skyrme model is very simple model, because it only consists of two flavours. This model is described by $U=U(x^\mu)$ function which has SU(2) group valued of (3+1)D space-time coordinates \cite{mif04,mift04,hans02}.

The dynamics is determined by \textit{action}
\begin{equation}\label{7}
S=\int d^4x~\mathcal{L},
\end{equation}
where
\begin{equation}\label{8}
\mathcal{L}=\text{Tr}\left[-\frac{F^2}{16}L_\mu L^\mu+\frac{1}{32a^2}\left[L_\mu,L_\nu\right]\left[L^\mu ,L^\nu\right]+\frac{F^2}{16}
M_\pi^2(U^{-1}+U-2I)\right]
\end{equation}
is the related \textit{Lagrangian density} and
\begin{equation}\label{9}
L_\mu=U^{-1}\partial_\mu U
\end{equation}
are \textit{left chiral currents}, where $\mathit{F}\cong{123}$ MeV is pion decay constant and $a$ is dimensionless constant. \textit{The first term in equation (\ref{8}) is the SU(2) chiral model Lagrangian density, the second term is Skyrme term for stabilizing solitonic solution. The last term descibes the mass term where $\mathit{M}_\pi$ is pion (meson) mass}.

Euler-Lagrange equation of SU(2) Skyrme model is derived from \textit{least action principle}
\begin{equation}\label{10}
\delta S = 0.
\end{equation}
If we take \textit{variation of action} to equation (\ref{7}), we obtain
\begin{equation}\label{11}
\delta S=\int d^4x~\delta\mathcal{L}.
\end{equation}
Substitute equation (\ref{8}) into (\ref{11}), and use least action principle (\ref{10}) \cite{mif04}
\begin{equation}\label{12}
\partial_\mu\left({L^\mu-\frac{1}{{a^2F^2}}\left[{L_\nu,\left[{L^\mu,L^\nu}\right]}\right]} \right)+\frac{1}{2}M_\pi^2(U-U^{-1})=0.
\end{equation}

In general case (nonstatic), energy of SU(2) Skyrme model is
\begin{equation}\label{13}
E = \int {d^3 } x~T^{00},
\end{equation}
where $T^{00}$ is \textit{energy-momentum tensor}. Explicitly,
\begin{eqnarray*}
E
&=&\int d^3x~\text{Tr}\left[-\frac{F^2}{16}L_a
L_a-\frac{1}{32a^2}[L_a,L_c][L_a,L_c]\right.\nonumber\\&&
\left.
+\frac{F^2}{16}L_o L_o+\frac{1}{16a^2}[L_o,L_a][L_o,L_a]-\frac{F^2}{16}
M_\pi^2(U^{-1}+U-2I)-\frac{F^2}{8}L_o L_o\right.\nonumber\\&&
\left.-\frac{1}{8a^2}[L_a,L_o][L_a,L_o]\right]\\
&=&E_{\text{static}}+E_{\text{rotation}}
\end{eqnarray*}\begin{equation}\end{equation}
where
\begin{equation}\label{15}
E_{\text{static}}=-\int{d^3x~\text{Tr}\left[{\frac{{F^2}}{{16}}L_a^2+\frac{1}{{32a^2}}\left[{L_a,L_c}\right]^2 +\frac{{F^2}}{{16}}M_\pi^2(U^{-1}+U-2I)}\right]}
\end{equation}
and
\begin{equation}\label{16}
E_{\text{rotation}}=-\int{d^3x~\text{Tr}\left[{\frac{{F^2 }}{{16}}L_0^2+\frac{1}{{16a^2}}\left[{L_0,L_a} \right]^2}\right]}.
\end{equation}

Solitonic properties of SU(2) Skyrme model for static energy in equation (\ref{15}) is studied by scaling the spatial coordinates
\begin{equation*}
x\rightarrow2\widetilde{x}/aF
\end{equation*}
and express energy in $F/4a$, i.e. by taking
\begin{equation*}
(F/4a)=(1/12\pi^2).
\end{equation*}
In this unit, equation (\ref{15}) becomes
\begin{equation}\label{17}
E_{\text{static}}=\frac{1}{{12\pi^2}}\int{d^3}x\left(-\frac{1}{2}\right)\text{Tr}\left[{L_a^2+\frac{1}{8}\left({\left[{L_a,L_c}\right]^2+m_\pi^2(U^{-1}+U-2I)}\right)}\right]
\end{equation}
where
\begin{equation}\label{18}
m_\pi=2M_\pi/aF.
\end{equation}
Euler-Lagrange equation (\ref{12}) in static case becomes
\begin{equation}\label{19}
\partial_a\left({L_a-\frac{1}{4}\left[{L_c,\left[{L_a,L_c}\right]}
\right]}\right)-\frac{{m_\pi^2}}{2}(U-U^{-1})=0.
\end{equation}

\section{Scale Stability}
Let us look at scale transformation below
\begin{equation}\label{20}
x\rightarrow \lambda x.
\end{equation}
We find that $\mathit L_a$ currents, by scale transformation (\ref{20}), transform into
\begin{equation}\label{21}
L_a(x)\to U^{-1}(\lambda x)\frac{\partial U(\lambda x)}{\partial x^a} =
\lambda L_a (\lambda x).
\end{equation}
The effect of scale transformation to static energy (\ref{17}), by ignoring pion mass term (because pion mass is small), is
\begin{equation}\label{22}
E\left[\lambda\right]_{\text{static}}=\frac{1}{\lambda}E_\sigma+\lambda E_{\text{Sky}}
\end{equation}
where $E_\sigma$ is \textit{static chiral energy term} and $E_{\text{Sky}}$ is \textit{Skyrme energy term}.

From equation (\ref{22}), we obtain
\begin{equation}\label{23}
\left.\frac{dE\left[\lambda\right]}{d\lambda}\right|_{\lambda=1}
=\left.\left(-\frac{1}{\lambda^2}E_\sigma+E_{\text{Sky}}\right)\right|_{\lambda=1}
=-E_\sigma+E_{\text{Sky}},
\end{equation}
and
\begin{equation}\label{24}
\left.\frac{d^2 E\left[\lambda\right]}{d\lambda^2}\right|_{\lambda=1}
=\frac{2}{\lambda^3}\left.{E_\sigma}\right|_{\lambda=1}=2E_\sigma.
\end{equation}
The requirement for \textit{extremum condition} is
\begin{equation}\label{25}
\frac{{dE\left[ \lambda  \right]}}{{d\lambda }} = 0.
\end{equation}
We apply extremum condition, (\ref{25}), to equation (\ref{23}), we obtain
\begin{equation}\label{26}
E_\sigma=E_{Sky},
\end{equation}
which it shows
\begin{equation}\label{27}
E_{\sigma}\geq 0.
\end{equation}
So that equation (\ref{24}) fulfills condition
\begin{equation}\label{28}
\frac{{d^2E\left[\lambda\right]}}{{d\lambda^2}}> 0.
\end{equation}
Equation (\ref{28}) is \textit{minimum stable condition} which implies that \textit{static energy (\ref{22}) is stable against scale perturbation}.

\section{Topological Charge}
Static energy of SU(2) Skyrme model can be expressed, by ignoring pion mass, as
\begin{equation}\label{29}
E_{\text{static}}=\frac{1}{{12\pi^2}}\int{d^3}x\left(-\frac{1}{2}\right)\text{Tr}\left[{\left({L_a\pm \frac{1}{4}\epsilon_{abc}\left[{L_b,L_c}\right]}\right)^2}\right]\mp\frac{1}{{24\pi^2}}\int {d^3}x~\epsilon_{abc}\text{Tr}~[L_aL_bL_c]
\end{equation}
where $\epsilon_{abc}$ is Levi-Civita symbol, $\epsilon_{abc}=\delta^{abc}_{123}$.
At energy lower bound
\begin{equation}\label{30}
E_{\text{static}}\geq B,
\end{equation}
where
\begin{equation}\label{31}
B=-\frac{1}{{24\pi^2}}\int{d^3 x\epsilon _{abc}}\text{Tr}~(L_aL_bL_c).
\end{equation}
$B$ \textit{integral, (\ref{31}), is independent of space-metric tensor}, i.e. topological quantity which is known as \textit{topological charge} of SU(2) Skyrme model.

\section{Static Skyrme Equation in Spherical Coordinate}
In spherical coordinate, $(r,\theta,\phi)$, static Skyrme equation has the following form \cite{mif04}
\begin{equation}\label{32}
\begin{split}
0
&=\partial_r\left(L_r-\frac{1}{4}\left\{\frac{1}{r^2}[L_\theta,[L_r,
L_\theta]]+\frac{1}{r^2\sin^2\theta}[L_\phi,[L_r,L_\phi]]\right\}\right)\\
&+\partial_\theta
\left(\frac{1}{r^2}L_\theta-\frac{1}{4}\left\{\frac{1}{r^2}[L_r,[L_\theta,L_r]]+\frac{1}{r^4\sin^2\theta}[L_\phi,[L_\theta,L_\phi]]\right\}\right)\\
&+\partial_\phi\left(\frac{1}{r^2\sin^2\theta}L_\phi-\frac{1}{4}\left\{\frac{1}{r^2\sin^2\theta}[L_r,[L_\phi,L_r]]+\frac{1}{r^4\sin^2\theta}[L_\theta,[L_\phi,L_\theta]]\right\}\right).
\end{split}
\end{equation}
Its solution takes form
\begin{equation}\label{33}
U(r)=\exp(iF_a(r,\theta,\phi)\sigma_a),
\end{equation}
where
\begin{equation}\label{34}
F_a(r,\theta,\phi)=g(r)n_a(\theta,\phi)
\end{equation}
is Skyrme ansatz, $g(r)$ is profile function, $n_a$, $a=1,2,3$ is component of unit vector, $\widehat{\boldsymbol{n}}$, in the internal space of SU(2), and $\sigma_a$ is Pauli matrix.

\section{Skyrmion Static Energy and Its Solution}
As a result of scaling spatial coordinates, by ignoring pion mass, static energy of SU(2) Skyrme model can be stated as \cite{mif04}
\begin{eqnarray}\label{35}
E_{\text{static}}=\frac{1}{12\pi^2}(4\pi)\int r^2 dr\left[\left(\frac{dg}{dr}\right)^2+\frac{2}{r^2
}\sin^2g\left(1+\left(\frac{dg}{dr}\right)^2\right)+\frac{1}{r^4}\sin^4g\right].
\end{eqnarray}
From equation (\ref{35}), by using least action principle
\begin{equation}\label{36}
\delta_g E_{static} = 0,
\end{equation}
we can derive Euler-Lagrange equation for profile function \cite{mif04}
\begin{equation}\label{37}
\frac{{d^2g}}{{dr^2}}\left[{1+\frac{2}{{r^2}}\sin^2g}\right]
+\left({\frac{{dg}}{{dr}}}\right)^2\left[{\frac{1}{{r^2}}\sin 2g}
\right]+ \left({\frac{{dg}}{{dr}}} \right)\left({\frac{2}{r}}\right)-\frac{1}{{r^2}}
\sin 2g-\frac{1}{{r^4}}\sin^2 g\sin 2g=0.
\end{equation}
Equation (\ref{37}) is second order of nonlinear differential equation.

The solution of equation (\ref{37}) will be worked numerically with finite difference method. In order that static energy (\ref{35}) has finite value at $r=0$ and $r=\infty$, then profile function $g(r)$ must fulfills \textit{boundary conditions}
\begin{equation}\label{38}
g(0)=\pi,~~~ g(\infty)=0.
\end{equation}

\section{Quantized Rotational Energy}
Skyrmion quantization is worked by involving time dependent of $U(\boldsymbol{r})$ as
\begin{equation}\label{41}
U(\boldsymbol{r})\to U(\boldsymbol{r},t)=A(t)U(\boldsymbol{r})A(t)^\dag,
\end{equation}
where
\[A(t)\in SU(2)_{\text{internal}},\]
\begin{equation}\label{42}
AA^\dag=A^\dag A=I
\end{equation}
$A$ is time dependent unitary matrix.

By using equation (\ref{41}), rotational energy can be stated as \cite{mif04}, \cite{mift04}
\begin{eqnarray}\label{43}
E_{\text{rotation}}
&=&-\left({\frac{{\pi F^2}}{6}\int\limits_0^\infty{dr~r^2}\sin^2g+\frac{{2\pi}}{{3a^2}}\int\limits_0^\infty{dr~r^2\sin^2g\left[{\left({\frac{{dg}}{{dr}}}\right)^2+\frac{1}{{r^2}}\sin^2g}\right]}}\right)\text{Tr}\left({R^{-1}\frac{{\partial R}}{{\partial t}}}\right)^2\nonumber\\
&=&\frac{1}{2}~I~\text{Tr}~\boldsymbol{\Omega}^2
\end{eqnarray}
where
\begin{equation}\label{44}
\text{Tr}~\boldsymbol{\Omega}^2=-\text{Tr}\left(R^{-1}\frac{\partial R}{\partial t}\right)^2
\end{equation}
with $\boldsymbol{\Omega}$ is angular velocity matrix of Skyrmion,
and
\begin{equation}\label{45}
I=2\left(\frac{\pi F^2}{6}\int^\infty_0dr~r^2\sin^2g+\frac{2\pi}{3a^2}\int^\infty_0 dr~r^2\sin^2g\left[\left(\frac{dg}{dr}\right)^2+\frac{1}{r^2}\sin^2g\right]\right)
\end{equation}
is Skyrmion moment of inertia.

\section{Finite Difference Method}
Assume that $g(x)$ is a continuous function of one variable. The value of this function is given only for discrete value, and equidistant $x$ \cite{ves}:
\begin{equation}\label{49}
g_k\equiv g(x_k),
\end{equation}
where $x_k\equiv x_0+k\Delta x$. Let us define quantity
\begin{equation}\label{50}
\Delta g_k\equiv g_{k+1}-g_k
\end{equation}
which is called \textit{forward difference} at point $x_k$. Apply (\ref{50}) for higher order of forward difference, we get
\begin{equation}\label{51}
\Delta^2g_k\equiv \Delta g_{k+1}-\Delta g_k=g_{k+2}-2g_{k+1}+g_k,
\end{equation}
\begin{equation}\label{52}
\Delta^3g_k\equiv \Delta^2 g_{k+1}-\Delta^2 g_k=g_{k+3}-3g_{k+2}+3g_{k+1}-g_k
\end{equation}
and so on. General form of forward difference is:
\begin{equation}\label{53}
\Delta^rg_k\equiv\Sigma_{i=0}^r(-1)^i\begin{pmatrix}
r\\i\end{pmatrix}g_{k+r-i}.
\end{equation}

\textit{Backward difference} is defined as
\begin{equation}\label{54}
\nabla g_k\equiv g_k-g_{k-1}
\end{equation}
and higher order of backward difference is
\begin{equation}\label{55}
\nabla^2g_k\equiv \nabla g_k-\nabla g_{k-1}=g_k-2g_{k-1}+g_{k-2}
\end{equation}
so on. General form of backward difference is
\begin{equation}\label{56}
\nabla^rg_k\equiv\Sigma_{i=0}^r(-1)^i\begin{pmatrix}
r\\i\end{pmatrix}g_{k-r+i}.
\end{equation}
Following, it is defined \textit{central difference} which is symmetric with $x_k$:
\begin{equation}\label{57}
\delta g_k\equiv g_{k+{1/2}}-g_{k-{1/2}}
\end{equation}
and higher order of central difference is
\begin{equation}\label{58}
\delta^2g_k=g_{k+1}-2g_k+g_{k-1}
\end{equation}
so on. General form of central difference is
\begin{equation}\label{59}
\delta^r g_k\equiv\Sigma_{i=0}^r(-1)^i\begin{pmatrix}
r\\i\end{pmatrix}g_{k-r/2+i}.
\end{equation}

If central difference is observed, $k+1/2$ index is out of the value which is served by discrete value. In order to obtain the integer value, mean value operator, $\mu$, is used to operate with $g(x)$ function as \textit{central mean}
\begin{equation}\label{60}
\mu g=\frac{1}{2}\left[g_{k+1/2}+g_{k-1/2}\right]
\end{equation}
and higher order of central mean is
\begin{equation}\label{61}
\mu^2g_k\equiv\frac{1}{2}\left[\mu g_{k+1/2}+\mu g_{k-1/2}\right]
\end{equation}
so on. \textit{Central mean of central difference} is defined as
\begin{eqnarray}\label{62}
\mu\delta g_k&\equiv& \frac{1}{2}\left[\delta g_{k+1/2}+\delta g_{k-1/2}\right]\nonumber\\
&=&\frac{1}{2}\left[g_{k+1}-g_{k-1}\right].
\end{eqnarray}

\section{Numerical Solution of Skyrme Equation}
Skyrme equation is stated in finite difference form, i.e. by substituting equation (\ref{54}), (\ref{58}), into Skyrme equation (\ref{37}) then we obtain
\begin{equation}\label{63}
\frac{\delta^2g_k}{\Delta r^2}\left[1+\frac{2}{r^2}\sin^2g_k\right]+\left(\frac{\nabla g_k}{\Delta r}\right)^2\left[\frac{1}{r^2}\sin~2g_k\right]+\left(\frac{\nabla g_k}{\Delta r}\right)\left(\frac{2}{r}\right)-\frac{1}{r^2}\sin~2g_k-\frac{1}{r^4}\sin^2g_k~\sin~2g_k=0.
\end{equation}
By reposisition of equation (\ref{63}), we get
\begin{eqnarray}
\delta^2g_k&=&\frac{-\left(\frac{\nabla g_k}{\Delta r}\right)^2\left[\frac{1}{r^2}\sin~2g_k\right]-\left(\frac{\nabla g_k}{\Delta r}\right)\left(\frac{2}{r}\right)+\frac{1}{r^2}\sin~2g_k+\frac{1}{r^4}\sin^2g_k\sin~2g_k}{\left[1+\frac{2}{r^2}\sin^2g_k\right]}\Delta r^2\nonumber\\
&=&\frac{-(\nabla g_k)^2\sin~2g_k-(\nabla g_k)2(\Delta r)r+\sin~2g_k\Delta r^2+\frac{1}{r^2}\sin^2g_k\sin~2g_k\Delta r^2}{r^2+2\sin^2g_k}
\end{eqnarray}
\begin{eqnarray}\label{65}
g_{k+1}&=&\frac{-(\nabla g_k)^2\sin~2g_k-(\nabla g_k)2(\Delta r)r+\sin~2g_k\Delta r^2+\frac{1}{r^2}\sin^2g_k\sin~2g_k\Delta r^2}{r^2+2\sin^2g_k}\nonumber\\
&-&g_{k-1}+2g_k.
\end{eqnarray}
The problems are initial values which are needed to be defined, i.e.
\begin{equation}\label{65.1}
g(r=0),~~~g(r=dr).
\end{equation}
Boundary conditions, which are given by Skyrme, are
\begin{equation}\label{65.2}
g(r=0)=\pi,~~~g(r\rightarrow\infty)=0.
\end{equation}

Apply the boundary conditions (\ref{65.2}) for initial values (\ref{65.1}). It is suitable for $g(r=0)$, and $g(r=dr)$ must be chosen, so that it is suitable for $g(r\rightarrow\infty)=0$. This method is known as \textit{shooting method} \cite{irwandi2006}. The ploting result, as shown in Figure 1: Profile function $\phi(r)$, can be compared with trial and error method, i.e.
\begin{equation}\label{66}
g(x)=4 \arctan~e^{-r}.
\end{equation}

\begin{figure}
\begin{centering}
\input{profil.tex}
\par\end{centering}
\caption{\label{cap:2-Fungsi-profile}Profile function $\phi(r)$}
\end{figure}
In order to calculate static energy (\ref{35}), the following integral is used
\begin{equation}\label{67}
E(g)=\frac{2\pi f_\pi}{e}\int_0^\infty dr\left[\left(r^2g'^2+2\sin^2g\right)+\sin^2g\left(2g'^2+\frac{\sin^2g}{r^2}\right)\right]
\end{equation}
where $g'=dg/dr$. Equation (\ref{67}), is solved numerically with \textit{trapezoid method} and it gives:
\begin{equation*}
23.2154~\pi f_\pi/e.
\end{equation*}
Let us calculate numerically $E_\text{static}$ (\ref{35}), using numerical value of $g(r)$
\begin{equation}\label{68}
E_{\text{static}}=\frac{1}{12\pi^2}(4\pi)\left(\frac{23.2}{2}\right).
\end{equation}
In the form of $F$ and $a$, using definition $(1/12\pi^2)=(F/4a)$, we obtain
\begin{equation}\label{69}
E_{\text{static}}=\frac{\pi F}{a}\left(\frac{23.2}{2}\right).
\end{equation}
The numerical value of $g(r)$ can be used to calculate
\begin{itemize}
\item[(1)] nucleon $(m_N)$ and delta static masses
$(m_\Delta)$;
\item[(2)] Skyrmion moment of inertia, $I$.
\end{itemize}

We get benefit from numerical value of profile function, $g(r)$, to calculate Skyrmion moment of inertia. It gives
\begin{equation}\label{46}
I=\frac{1}{4}\left(\frac{447}{Fa^3}\right).
\end{equation}
In quantum mechanics, angular momentum $\boldsymbol{J}$ is quantized as
\begin{equation}\label{47}
\boldsymbol{J}^2=j(j+1)\hbar^2
\end{equation}
where $j=0,\frac{1}{2},1,\frac{3}{2},2,\ldots$
\begin{equation*}
\hbar=\frac{h}{2\pi},
\end{equation*}
$h$ is Planck constant. Here, we use using natural unit, $\hbar=1=c$.

Rotational energy is
\begin{equation}\label{48}
E_{\text{rotation}} = \frac{\boldsymbol{J}^2}{2I}=\frac{j(j+1)}{2I}.
\end{equation}

\section{Hadron Mass}
Skyrmion is solution of Skyrme equation which has finite energy. Based on Einstein energy-mass formula, Skyrmion mass is \begin{equation}\label{70}
m=\frac{E_{static}+E_{rotation}}{c^2}=M_{static}+M_{rotation}.
\end{equation}
From equation (\ref{48}) and (\ref{70}), we obtain
\begin{equation}\label{71}
m= M_{static} + \frac{{j(j + 1)}}{2I}.
\end{equation}
Wess-Zumino quantization condition \cite{bal91}, requires:
\[j = \frac{1}{2},\frac{3}{2},\frac{5}{2},\ldots\]
\textit{It means that Skyrmion is fermion}.

In case of nucleon with spin $\frac{1}{2}$,
$j=\frac{1}{2}$, equation (\ref{71}) gives
\begin{equation}\label{72}
m_N = M_{static} + \frac{\frac{1}{2}(\frac{1}{2}+1)}{2I}
\end{equation}
By substituting the values of $F = 123$ MeV and $a = 4.95$ into (\ref{69}) and (\ref{46}), give
\begin{equation}\label{73}
M_{static}\cong 905.08~
\text{MeV}
\end{equation}
and
\begin{equation}\label{74}
I\cong 7.49\times 10^{-3}.
\end{equation}
Substitute (\ref{73}) and (\ref{74}) into (\ref{72}) gives the result
\begin{equation}\label{75}
m_N\cong 955.15~\text{MeV}.
\end{equation}

In case of delta particle with spin $\frac{3}{2}$, $j=\frac{3}{2}$,
equation (\ref{71}) gives
\begin{equation}\label{76}
m_\Delta=M_{static}+\frac{{\frac{3}{2}(\frac{3}{2}+1)}}{2I}.
\end{equation}
By substituting the values of $M_{static}$ and $I$ above, it gives
\begin{equation}\label{77}
m_\Delta\cong 1155.41~\text{MeV}.
\end{equation}

\section{Discussions}
The calculation of nucleon energy-mass gives the value about 955.15 MeV. The experimental value of nucleon energy-mass is about 939~MeV. The different value of nucleon energy-mass based on calculation and experiment is about 16.15 MeV.
The calculation of delta energy-mass gives the value about 1155.41 MeV. The experimental value of delta energy-mass is about 1232~MeV \cite{bal91}. The different value of delta energy-mass based on calculation and experiment is about 76.59 MeV.

The different values among calculation and experiment are becaused of SU(2) Skyrme model is very simple model, i.e. it consists of two flavours, $(u,d)$ or $(u,s)$ or $(d,s)$. It is necessary keeping in mind related with these different values \cite{hep54}:
\begin{itemize}
\item[(1)] Meson field Lagrangian only includes pseudoscalar field. Other low mass meson (vector) should be included.
\item[(2)] In nature, there are three flavour families than two "light" flavour families which must be included in more realistic formula.
\item[(3)] The effects of chiral and flavour symmetry breaking aren't calculated yet.
\item[(4)] $N_C$ (number of colour) correction of nucleon mass isn't included yet.
\end{itemize}

\section{Acknowledgment}
MH thank to Hans J. Wospakrik, Ph.D for great and eternal inspirations, Irwandi Nurdin and Denny Hermawanto for numerical solution. Toto Sudiro, Slamet for fruitful discussions. All kindly colleagues for their best supports. Physics Research Centre LIPI for research facilities, Andri, Ayu, Intan, Iim, Ike for their great hope.

In depth, MH thank to beloved Mother for continuous praying and motivation. Special thank to beloved ones, Ika Nurlaila for huge support and patience, Aliya Syauqina Hadi for her funny and curiosity.


TITLE: Polynomial with given group of symmetries
QUESTION [4 upvotes]: Let $f$ be a polynomial in $n$ variables and $G$ - its group of symmetries (group of permutations of variables wich left $f$ in place). I'm trying to such $f$ for given group $G$. I have troubles when $G\subset S_5$ is a cyclic group of fifth order. I have a fifth degree polynomial (for example, $x_1^2x_3x_4x_5+x_2^2x_1x_4x_5+x_3^2x_1x_2x_5+x_4^2x_1x_2x_3+x_5^2x_2x_3x_4$). Is it possible to find such $f$ with lower degree?

REPLY [2 votes]: Certainly $f$ cannot be linear: if $f = \displaystyle \sum_{i=1}^5 a_i x_i$ is invariant under cyclic permutation then the $a_i = a$ must all be equal to each other, and then $f$ is invariant under all permutations.
$f$ can be quadratic: let $f = \displaystyle \sum_{i=1}^5 (x_i x_{i+1} + 2 x_i x_{i+2})$ where the indices are considered $\bmod 5$.
The general question of whether such polynomials always exist can be addressed using representation theory. The problem reduces to showing that if $G$ is a finite group acting on a vector space $V$ and $H$ is a proper subgroup of $G$, then there exists some $d$ such that $\dim S^d(V)^G$ is strictly less than $\dim S^d(V)^H$, and this is a corollary of Molien's theorem. One then applies this theorem to some fixed group $H \subseteq S_n$, where $V$ is the permutation representation of $S_n$, and $G$ runs over all minimal subgroups of $S_n$ properly containing $H$.


text:
\begin{document}

\maketitle

\begin{abstract}

In this paper, we evaluate the efficacy, in a Hadoop setting,
of two coding schemes, both possessing an inherent
double replication of data. The two coding schemes
belong to the class of regenerating and
locally regenerating codes respectively, and these two classes are representative of recent
advances made in designing codes for the efficient storage of data in a distributed setting.
In comparison with triple replication, double replication permits a significant reduction in storage overhead,
while delivering good MapReduce performance under moderate work loads. The
two coding solutions under evaluation here, add only moderately to the storage
overhead of double replication, while simultaneously offering
reliability levels similar to that of triple replication.

One might expect from the property of inherent data duplication that the performance of these codes in executing a MapReduce job would be comparable to that of double replication.  However, a second feature of this class of code comes into play here, namely that under both coding schemes analyzed here, multiple blocks from the same coded stripe are required to be stored on the same node.   This concentration of data belonging to a single stripe negatively impacts MapReduce execution times.  However, much of this effect can be undone by simply adding a larger number of processors per node.  Further improvements are possible if one tailors the Map task scheduler to the codes under consideration.   We present both experimental and simulation results that validate these observations.

\end{abstract}

\vspace{-0.1in}

\section{Introduction}
\vspace{-0.1in}

Hadoop~\cite{hadoop} is an open-source platform dealing with distributed storage whose file system is known as the Hadoop distributed file system (HDFS).  The primary objective is to store a
collection of files in such a way that distributed computation on the stored data can be carried out efficiently under the MapReduce (MR)~\cite{mapreduce} paradigm. Any file that needs to be stored is divided into blocks, typically of size
$64-256$ MB and these blocks are replicated and are stored across the distributed storage network, such
that no two replicas of the same block are stored in the same storage node.

Replication of data, in addition to providing resiliency against irrecoverable data loss, supports efficient MR computation in two ways. Firstly, it ensures {\it availability} of data during transient node failures. Such failures, which do not cause data loss, are the norm~\cite{ford2010availability} in large-scale storage systems, and hence minimizing the number of repairs carried out to handle transient failures, can result in significant savings in network bandwidth~\cite{sathiamoorthy}. Secondly, replication helps to increase {\it data locality} for MR computation. A Hadoop node can typically perform $2$ to $8$ map tasks in parallel, depending on the number of processor cores available to it. As a result, in a system which is expected to handle multiple compute jobs simultaneously, the presence of replicas will increase the chance that any given map task can be assigned to  a node which contains the data block required by the task.  Such a task is called a {\it local} task. A non-local or {\it remote} task needs data to be fetched across the network to the node where the task is to be executed, leading to increased delay in job execution as well as increased network bandwidth usage~\cite{mapreduce}.

Triple replication is often used in a Hadoop system as it provides the desired level of resiliency as well as availability of data.  In comparison with triple replication, while double replication permits a significant reduction in storage overhead, and can potentially deliver good MapReduce performance under moderate work loads, it leaves the system vulnerable to irrecoverable data loss in the event of some patterns of two-node failure.  In an effort to avoid the large overheads associated with triple replication, storage-efficient erasure codes such as Reed-Solomon (RS) codes and some variants have recently been employed in Facebook's Hadoop clusters~\cite{hdfs_raid}, \cite{sathiamoorthy}. These codes store only a single copy of the data file and use parity blocks, as in RAID, to offer protection against failures. Application of these codes is thus limited to the storage of cold data, i.e., data on which MR jobs are rarely carried out.

In this paper, we evaluate the efficacy, in a Hadoop setting, of two coding schemes, both possessing an inherent
double replication of data. The two coding schemes are described in Section \ref{sec:pentagon}. Performance analysis of metrics like storage overhead and mean time to data loss (MTTDL), and also simulation results showing data locality for map tasks are presented in Section \ref{sec:analysis}. Details of the experimental set-up used for MR performance evaluation, and the associated findings can be found in Section \ref{sec:experiments}. Conclusions and future work are presented in Section \ref{sec:future} and an overview of the related literature is presented in  Section \ref{sec:related}.

\vspace{-0.1in}
\section{The Two Coding Schemes} \label{sec:pentagon}

\vspace{-0.1in}
\subsection{Pentagon Code}

The first code which we refer to here as the {\it pentagon code}, is a simple instance of a family of codes known as repair-by-transfer minimum-bandwidth regenerating codes~\cite{ShaRasKumRam_rbt}.   In the pentagon code, $9$ data blocks are encoded\footnote{In case, the file to be encoded is larger than $9$ blocks, the file is striped into sets of $9$ blocks each and each stripe is separately encoded.} into $20$ coded blocks and stored in $5$ nodes with $4$ blocks assigned to each node. Given the $9$ data blocks, a $10^{\text{th}}$ block which is an XOR parity of these $9$ blocks is first computed; the resulting $10$ blocks are replicated once more and stored in $5$ nodes as shown in Fig. \ref{fig:pentagon}. To explain the distribution of the blocks among the $5$ nodes, consider a fully connected graph with the $5$ target nodes as its vertices. Note that there are ${5 \choose 2} = 10$ edges in this graph.  Place each of the $10$ distinct blocks on the $10$ edges of this graph. The blocks stored in a node are then simply those associated to the edges that are incident on the particular node.  An important point to note here is that the pentagon code stores multiple blocks of a coded stripe in the same node. Such codes, in general, are known as array codes.

\begin{figure}[h]
  \centering
  \subfigure[Pentagon Code]{\label{fig:pentagon}\includegraphics[height=1.6in]{pentagon_code}}
  \hspace{0.05in}
  \subfigure[Heptagon-Local Code]{\label{fig:heptagon_local}\includegraphics[height=0.9in]{heptagon_local_code}}
  \vspace{-0.1in}
  \caption{Codes with inherent double replication of data.}
  \label{fig:regen_framework}
\end{figure}

It can be readily verified that the contents of any $3$ nodes suffice to recover all $9$ data blocks and
thus the code is resilient to $2$-node failure. Single-node repair is accomplished, simply by transferring to the replacement of the failed node, the blocks shared in common by the failed node with each of the remaining $4$ nodes. In the case of two node failures, nodes $N_1$ and $N_2$ say, $8$ blocks have to be recovered in all, out of which $6$ can be recovered just by copying the replicas of these blocks present in the remaining $3$ nodes. For the recovery of block $4$, nodes $N_3$, $N_4$ and $N_5$ internally compute the {\it partial parities} $P_3 = 3+6+P$, $P_4 = 2+8+9$ and $P_5 = 1+5+7$, respectively and transfer these partial parities to the first replacement node.  The replacement node computes $P_3 + P_4 + P_5$ and recovers block $4$, which is then copied to the second replacement node as well. Thus, the overall network data transfer incurred in repairing the two nodes (also known as repair bandwidth) is $10$ blocks. The repair bandwidth savings arising from the usage of partial parties is an intrinsic advantage possessed by array codes.

The construction of the pentagon code, suggests that a third coding scheme be included for comparison. Under this scheme, given $9$ data blocks, one computes a $10^{\text{th}}$ parity block as above and then duplicates each coded block to obtain a total of $20$ blocks.  These $20$ blocks are then stored across $20$ different nodes of the network.  This scheme is termed in the literature~\cite{xin2003reliability} as a RAID+mirroring scheme.  The code itself is termed as a $(10,9)$ RAID+m code.  The performance comparison we present in Section \ref{sec:analysis}, includes comparison with the RAID+m scheme.

\vspace{-0.1in}
\subsection{Heptagon-Local Code}

The heptagon-local code is built on top of a heptagon-code, which is the analogue of the pentagon code for $7$ nodes.  The heptagon code encodes $20$ data blocks into $42$ blocks and stores them in $7$ nodes, with each node hosting $6$ blocks following a placement rule similar to that applied in the case of the pentagon code. The storage overhead of the heptagon code is less than that of the pentagon code;  however it has a lower level of resiliency (see Table \ref{tab:bw_compare}).

The heptagon-local code, is an instance of a family of codes known as locally regenerating codes~\cite{KamPraLalKum}.  This family presents a simple way to increase the resiliency of the heptagon code, at the cost of a moderate increase in storage overhead. In this code, $40$ data blocks are encoded into $86$ blocks and stored in $15$ nodes, as follows: (1) The $40$ data blocks are first split into two sets of $20$ each, which are then individually encoded by two instances of the heptagon code taking care to ensure that the $7$ nodes chosen for the first heptagon are distinct from those chosen for the second heptagon. (2) Two \textit{global parity blocks} are computed as functions of all $40$ data blocks, and are placed in an additional {\em global parity node}. This computation involves Galois field arithmetic as in the case of RAID-6. In this code, the individual heptagon codes themselves are referred to as \textit{local codes}. In a rack-aware HDFS implementation, the two heptagons and the global parity node would be placed in three different racks.

The heptagon-local code can recover from any pattern of $3$ node erasures.  The failure of $1$ or $2$ nodes lying within a heptagon, can be handled \textit{locally}, i.e., by accessing other nodes within the same heptagon. In the case where $3$ nodes belonging to a single heptagon fail, the recovery will involve accessing the contents of the second heptagon as well as the contents of the global parity node. Here again, the repair bandwidth can be reduced by making use of partial parities.

\vspace{-0.1in}
\section{Performance Analysis for Hadoop} \label{sec:analysis}

\vspace{-0.1in}
\subsection{File System Performance}

A comparison of performance metrics is presented in Table \ref{tab:bw_compare}. The MTTDL values shown are computed assuming a $25$ node system, using standard node failure and repair models available in the literature~\cite{xin2003reliability}.

\begin{table}[h!]
\centering
\begin{tabular}{||c|c|c|c|c||}
\hline
Code & Storage & Code  &  MTTDL\\
&   Overhead & Length &  (in years)    \\
\hline
\hline
3-rep & 3x & 3&    1.20e+09 \\
\hline
pentagon & 2.22x &  5 &  1.05e+08 \\
\hline
heptagon & 2.1x & 7 &  2.68e+07 \\
\hline
heptagon-local & 2.15x & 15 &  8.34e+09 \\
\hline
$(10,9)$ RAID$+$m & 2.22x  &  20 & 2.03e+09 \\
\hline
$(12,11)$ RAID$+$m & 2.18x  &  24 & 6.50e+08 \\
\hline
\end{tabular}
\caption{Comparing storage overhead, code length and MTTDL of various coding schemes.}
\label{tab:bw_compare}
\end{table}

An important advantage of the proposed codes is the smaller value of their code length for a given value of storage overhead, in comparison with the corresponding RAID+m scheme. The code length specifies the number of data nodes over which a coded stripe is distributed.  For instance, both the pentagon and the $(10, 9)$ RAID+m code have a storage overhead of $2.22$; clearly between the two codes, only the pentagon code is feasible in a Hadoop system possessing just $20$ nodes. The length of the RAID+m solution can be decreased below $20$ only at the expense of increased storage overhead. For larger systems, the heptagon-local code turns out to be an attractive choice in terms of all three performance metrics.

The pentagon and heptagon-local codes also are efficient in terms of the repair-bandwidth needed during \text{on-the-fly} repair of a lost data block, during an MR computation. Imagine a situation where the two nodes which store the two replicas of a particular block are temporarily down, and a map task is initiated on this particular block. While the $(10, 9)$ RAID+m solution needs a repair bandwidth of $9$ blocks, a repair bandwidth of $3$ blocks suffices in the case of the pentagon code, arising from the pentagon code's ability to compute and make use of partial parities. Typically, in Hadoop, repair jobs themselves are performed as MR jobs and in such a case, computation of the partial parities can be incorporated into the MR-job through the use of ``combine" functions\footnote{Combine function is a mechanism, while performing MR jobs, to consolidate the outputs of the various map tasks which are performed on the same Hadoop node.}.

\vspace{-0.1in}
\subsection{Data Locality of Map Tasks}

We evaluate here the data locality under Map tasks of the pentagon and heptagon local codes in a moderately-loaded Hadoop system.  We quantify the notion of load as follows (with an example):  A $100$-node system that handles $250$ map tasks, with $4$ map slots per node, is said to be operating under a load of $\frac{250}{(4 \times 100)} \times 100 \ = \ 62.5\%$. By moderately loaded, we mean that the system is most of the time, operating on a $\leq100\%$ load.

The map-task-assignment problem can be modeled as a maximum-matching problem on a bipartite graph, with the tasks on one side and the nodes on the other. The edges on this graph indicate the nodes where the replicas of the blocks reside. We note that the choice of a particular coding solution essentially determines the manner in which the edges are incident on the vertices lying to the right, i.e., the nodes (see Fig.~\ref{fig:pta}).  From a practical point of view, maximum-matching algorithms are computationally intensive. Hadoop uses instead, a simple algorithm called delay scheduling for task assignment~\cite{zaharia2010delay}.

\begin{figure}[h!]
\begin{center}
\includegraphics[width=3in]{PTA.pdf}
\end{center}
\vspace{-0.25in}
\caption{The manner in which tasks are mapped onto nodes in the case of the pentagon code is shown here.}
\label{fig:pta}
\end{figure}

To compare the data locality of the various codes, we simulated the delay-scheduling algorithm in a $25$-node system, under various load conditions. The maximum-matching algorithm was also simulated as a benchmark. Simulation results are presented for the $2$-rep, pentagon and the heptagon codes, for the cases of $2$, $4$ and $8$ map slots per node (see Fig. \ref{fig:locality}). We note that the data locality of the heptagon-local code will be similar to that of the heptagon code, since the global parity node does not play a role in task assignment. Also, the locality of the $2$-rep systems is indicative of the locality of any of the RAID+m solutions.
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.0in]{scheduler.pdf}
\end{center}
\vspace{-0.25in}
\caption{Percentage data locality with $2$-rep, pentagon and heptagon-local codes, as a function of job load on a system, for the cases of $\mu = 2,4,8$ map slots per node. Also shown, is the improvement in locality obtained over the delay scheduler via the modified peeling algorithm, for the case of $\mu =4$.}
\label{fig:locality}
\end{figure}

\vspace{-0.15in}
We see from the plots that there is a significant loss in data locality with $2$ map slots per node, for the proposed coding schemes, with respect to double replication. This is an artifact of the fact that under both coding schemes analyzed here, multiple blocks from the same coded stripe are required to be stored on the same node (see Fig.~\ref{fig:pta}).   This concentration of data belonging to a single stripe negatively impacts MapReduce execution times.    Also, since the heptagon code has a greater concentration of data blocks in comparison with the pentagon code, it suffers more in this respect. However, as seen from the simulations, the loss in locality decreases with increasing number of map slots per node. For instance, both the pentagon and heptagon-local codes have locality greater than $90 \%$ at $100 \%$ load, with $8$ map slots. The plots also suggest that there is scope for improvement in locality by using task assignment algorithms, other than the delay scheduler.

A simple algorithm known as the {\em peeling algorithm} was proposed in \cite{xie2012degree}, for task-assignment problems. Appropriate modifications to this algorithm allows it to be used in pentagon or heptagon-coded Hadoop systems. A simulation using the modified peeling algorithm, for the case of $4$ map slots per node is also shown in Fig.~\ref{fig:locality}; the performance gains in locality are clearly evident.

\vspace{-0.1in}
\section{Experimental Set-up and Evaluation} \label{sec:experiments}
\vspace{-0.1in}
An implementation of the pentagon and the heptagon codes was carried out in HDFS, taking Facebook's open-source HDFS-RAID~\cite{hdfs_raid} module (hadoop-0.20) as the baseline software. The main challenge in the implementation was in handling the array-nature of these codes, as they necessitate the logical grouping of blocks within a node.  Preliminary experiments, to ascertain MR performance, were conducted on two different Hadoop systems, which were chosen to have different numbers of processor cores per Hadoop node.

\textit{Set-up $1$ (2 map slots)}: This set-up had $25$ data nodes and the hardware used for each of these nodes was a dual-core IBM laptop, having $3$ GB of RAM and $150$ GB of hard disk space. The Hadoop data block size was set to $128$ MB. Also, each node was configured with $2$ map and $1$ reduce slots. We tested the pentagon and the heptagon codes in this set-up.

\textit{Set-up $2$ (4 map slots)}: This had $9$ data nodes and each of these nodes was a server class machine having $4$ processor cores per node, $24$ GB of RAM and $2$ TB of hard disk space. The Hadoop data block size was set to $512$ MB. Also, each node was configured with $4$ map and $2$ reduce slots. We tested the pentagon code in this set-up.

In both the set-ups, an additional master node was used to host all the controllers namely NameNode, JobTracker and RaidNode. All machines ran Ubuntu $12.04$ for their operating system. Also, all nodes were configured to be part of a single rack and shared a private $10$ Gbps Ethernet LAN.

\vspace{-0.1in}
\subsection{MapReduce Performance}
\vspace{-0.1in}

The Terasort job was executed at various load points (from $25 \%$ to $100 \%$) and under various coding schemes. We calculate values of data locality, job execution time and network traffic during job execution, averaged over multiple runs. We use Hadoop's inbuilt delay-scheduling algorithm for map-task assignment, with the delay set such that every node has a chance to assign two (four) local map tasks in the first (second) set-up. Features such as cap-based load management and speculative execution were turned off.

The measurements are shown in Fig. \ref{fig:locality_setup1} and \ref{fig:locality_setup2}, respectively for the cases of the first and second set-up.  The following conclusions can be drawn from the plots: $(i)$ At moderate loads, the performance of $2$-rep is very close to that of $3$-rep. $(ii)$ The data locality curves, in both the set-ups, exhibit the same trend as seen in the simulations curves in Fig. \ref{fig:locality}. $(iii)$ It can be verified that the excess network traffic for either the pentagon or the heptagon code with respect to $2$-rep is almost entirely due to the corresponding loss in data locality. $(iv)$ Also, as expected, there is a substantial loss in performance in the case of $2$ processor cores; however with $4$ cores, we see that the pentagon code has performance very close to that of the $2$-rep code even at a load of $75 \%$.

\begin{figure}[h!]
\begin{center}
\includegraphics[width=3in]{experiments_2processors_per_node2.pdf}
\end{center}
\vspace{-0.25in}
\caption{Network traffic, data locality and job time for the Terasort job in set-up $1$, having $2$ map-slots per node.}
\label{fig:locality_setup1}
\end{figure}

\begin{figure}[h!]
\begin{center}
\includegraphics[width=3in]{experiments_4processors_per_node2.pdf}
\end{center}
\vspace{-0.25in}
\caption{Network traffic and data locality for the Terasort job in set-up $2$, having $4$ map-slots per node.}
\label{fig:locality_setup2}
\end{figure}

\section{Conclusions and Future Work} \label{sec:future}
\vspace{-0.1in}

In this work, we have implemented in Hadoop two coding schemes having inherent double replication of data, and have carried out a preliminary MR performance-evaluation under this coded system. Our measurements suggest that under moderate work loads, with an increased number of processors per node, the MR performance is comparable to that of double replication. In the next phase of our work, we plan to enhance the current implementation in two ways: $(i)$ Implement the heptagon-local code, and $(ii)$ implement the modified peeling algorithm, which was simulated (see Fig. \ref{fig:locality}), as an alternative to delay scheduler for map task assignment. We also plan to measure MR performance on a variety of work loads, for the enhanced system. Other important metrics, like encoding duration and MR performance in the presence of node failures (with the usage of partial parities) also need to be ascertained.

\vspace{-0.2in}
\section{Related Work}\label{sec:related}
\vspace{-0.15in}
Regenerating codes and locally repairable codes are respectively introduced in \cite{DimGodWuWaiRam} and \cite{GopHuaSimYek}. The authors of \cite{sathiamoorthy} describe an implementation of locally repairable codes in HDFS and evaluate the  savings in network traffic during node repairs, with these codes. A study of locally repairable codes in the context of Windows Azure storage can be found in \cite{HuaSimXu_etal_azure}, where it is shown that these codes are better choices than Reed-Solomon codes, in terms of their reliability vs storage overhead performance. Regenerating codes have been implemented in a multiple-cloud system in \cite{HuCheLeeTan}, and for HDFS in \cite{LiLinLee}. Both works focus on the problem of decreasing repair bandwidth. Statistics regarding node failures in Hadoop clusters with thousands of nodes have been reported in \cite{sathiamoorthy}, \cite{RasShaHotStorage}. A class of erasure codes for minimizing I/O during recovery and degraded reads with application to cloud-file systems has been proposed in \cite{khan2012rethinking}.
\vspace{-0.2in}

\section*{Acknowledgements}
\vspace{-0.1in}

The research of P. Vijay Kumar is supported in part by the National Science Foundation under Grant 0964507 and in part by the NetApp Faculty Fellowship. The works of V. Lalitha and Birenjith Sasidharan are supported by a TCS Research Scholarship.

\vspace{-0.1in}

{\scriptsize
\bibliographystyle{IEEEtran}
\bibliography{FAST_bib}
}

\end{document}

config: arxiv


TITLE: Discuss the convergence of the following integrals
QUESTION [0 upvotes]: I have to discuss the convergence of the following improper integrals:
$$ a) \int_{0}^{1} \frac{1}{(x-x^2)^\frac{1}{2} }\,dx $$
$$ b) \int_{0}^{1} \frac{x}{1-x^3}\,dx $$
$$ c) \int_{0}^1 \frac{\log(x)}{\sqrt x}\,dx $$
I am trying to use the comparison test (Note that I have to discuss the convergence without evaluating the integrals themselves), but I haven't found nice integrals to compare to yet. Any ideas, hints, or suggestions?

REPLY [2 votes]: hint
a)
Let $$f(x)=\frac{1}{\sqrt{x-x^2}}$$
Near $ 0$ ,
$$f(x)\sim \frac{1}{\sqrt{x}}$$
near $ 1$,
$$f(x)\sim \frac{1}{\sqrt{1-x}}$$
Both integrales $\int_0f $ and $ \int^1f $ converge, so $ \int_0^1f $ is convergent.
b)
Let $$ g(x)=\frac{x}{(1-x)(1+x+x^2)}$$
Near $ 1$,
$$g(x)\sim \frac{1}{3(1-x)}$$
thus
$$\int_0^1g \text{ diverges}$$
c)
With the substitution $ t=\sqrt{x} $,
the integrale becomes
$$4\int_0^1\ln(t)dt $$
which is known to be convergent.


TITLE: Comparison of G force in space with G force on Earth
QUESTION [1 upvotes]: I was reading another question on here where the general answer was that the G force would be the same on earth as it would be in space, this is the Q & A link:
Can a pilot in a small spaceship feel G force in space?
Now my question is, if the fastest speed of a fighter jet is around 1500 MPH, but the flight to the moon reached a point where Apollo 11 was travelling at 25,000 MPH, why was the G force bearable for the passengers of that ship?
Of course, the more we look at the possibilities of humans travelling through space, we have to think of the speed the orbiting spacecraft would need to leave Earths gravitational pull to get there, such as the Juno spacecraft that hit 165,000 MPH to generate enough speed to reach Jupiter. What would be the implications of human life being aboard a ship travelling at that speed?

REPLY [1 votes]: The main reason why high $g$ forces are generally not experienced in space as opposed to within the earth's atmosphere is modes of flight. Within the atmosphere, there is plenty of scope for high $g$ forces during fighter aircraft manoeuvres including aerobatics.
For fighter aircraft, which generally experience the highest $g$ forces for flight within the earth's atmosphere, the main causes of high $g$ forces are sharp turns, rapid acceleration/decceleration and upward acceleration. For a sharp turn at linear velocity $v$ and radius $r$, the $g$ force is acceleration $a$ in the equation:
$$a = {v^2\over r}$$
The larger $v$ is, the higher the $g$ force during the turn. Similarly, the sharper the turn, the smaller $r$ is, the greater the acceleration.
During the vertical ascent of a fighter aircraft, $g$ force is a sum of the earths gravitational force plus the acceleration of the fighter aircraft upwards:
$$g_{\small force} = g_{\small gravity} + a_{\small aircraft}$$
For space flight, up to about $3g$ are experienced in order to reach space. Once in space orbit, there are generally no manoeuvres that effect high $g$ forces. Low orbital vehicles like the ISS space station float around the earth at about 7.7 km/s. That is just enough to balance the earth's gravitational pull on the space station at its altitude ($408km$). The astronauts feel zero gravity inside this 'falling' vehicle. The thrusters on space vehicles are, unlike jet engine thrusters within the atmosphere, not used for aerobatics and steep manouvers, but to gently alter course at low $g$ forces.
If you intended to fly to a star outside our solar system very quickly (within months), then rapid acceleration causing high $g$ forces would be necessary.


TITLE: Number of real roots of the following polynomial
QUESTION [2 upvotes]: I have a cubic polynomial:
$f(x)=x^3+3x^2+3x+7$
I wanted to obtain number of real roots of the provided polynomial.
What i did: I took the derivative of the function which turned out to be $3x^2+6x+3$ which is  a perfect square after taking $3$ common. Hence, the function is strictly increasing. Also, it has two negative roots. Now, it's function can have at the most three roots using this monotonicity concept. Since, the polynomial is of order three, it can have at the most three roots. Using the rule of sign, we can also say that this polynomial will have either three negative roots or one negative roots. The three negative root situation can be ignored using the monotonicity. Only plausible situation is one negative roots. Hence, this polynomial will have one real root.
I think the number of steps that I took can be reduced. I am also using some redundant things please guide me. Are my steps correct to obtain the number of roots. If not, please add some point that I can think of while obtaining number of roots of a polynomial or of any other function whatsoever. Thanks in advance.

REPLY [6 votes]: $$  (x+1)^3 = -6  $$
..............................


TITLE: Integer partitions into restricted parts
QUESTION [1 upvotes]: Given a linear diophantine equation $$x_1+\dots+x_n=m\leq nn'$$ how many solutions does it have with each $x_i\in[0,n']\cap\mathbb Z$? Looking for asymptotics that parametrizes well with both $n$ and $n'$ over different ranges for both situations

$x_1\leq\dots\leq x_n$ and

unordered.

REPLY [3 votes]: Since $m$ is a dummy variable (i.e. a bound variable) and $n,n'$ are "real" variables (i.e. they are free) perhaps we should rewrite the problem accordingly as
$``$compute the following
$$ f(y,z) = \#\left\lbrace  (x_1,... , x_y )\mid  x_1 + ... + x_y = m,\  x_i \in \mathbb{N},\  m \leq yz ,\  i < j \implies x_i \leq x_j \right\rbrace."$$
We can let $f'$ be the number if we omit the condition $i < j \implies x_i \leq x_j$ (i.e. $f$ corresponds to partitions and $f'$ corresponds to compositions). Intuition tells us that $f$ will likely have a closed form solution and $f'$ will probably only have solution expressible by generating functions (which are perfect for asymptotics). Let us confirm our intuitions:

Compositions (i.e. $f'$)

It is straightforward to see that if we let
$$g(y,m) = \#\left\lbrace (x_1,..., x_y ) \mid  x_1 + ... + x_y = m, \  x_i \in \mathbb{N}\right\rbrace,$$
then we have the identity
$$f'(y,z) = \sum_{m=0}^{yz}g(y,m).$$
It is well known, see the famous stars and bars method in feller's introduction to probability theory, that
$$g(y,m)= \binom{y+m-1}{m},$$

so that  $$f'(y,z) = \sum_{m=0}^{yz}\binom{y+m-1}{m}=
\binom{yz+y}{yz},$$
(or, using your notation, $f'(n,n')=\binom{nn'+n}{nn'}$)

where the last identity can be derived as a consequence of Chu Shih-Chieh's identity, see example 2.5.1 of Chuan-Chong & Khee-Meng's text on combinatorics. I also strongly recommend taking a peek at Flajolet & Sedgewick's text on analytic combinatorics for asymptotics and the more abstract symbolic method/species style which is necessary for the more difficult analysis of $f$.

Partitions (i.e. $f$)

It is straightforward to see that if we let
$$g(y,m) = \#\left\lbrace  (x_1,... , x_y ) \mid x_1 + ... + x_y = m, \  x_i \in \mathbb{N}, \  i < j \implies x_i \leq x_j\right\rbrace,$$
then we have the identity
$$f(y,z) = \sum_{m=0}^{yz}g(y,m).$$
It is well known that if we define the generating function $\mathcal{G}_y$ as
$$\mathcal{G}_y(x) = \sum_{m \in \mathbb{N}} g(y,m) x^m,$$
then we have that
$$\mathcal{G}_y(x) = \prod_{k=1}^{y}\frac{1}{1-x^k},$$
see Flajolet & Sedgewick's text on analytic combinatorics or Andrews elementary text on partitions. One way to see this is to notice the following famous theorem attributed to Euler

The number of partitions of a number $n$ into at most $l$ parts is equal to the number of partitions of a number $n$ into parts all bounded by $l$

and the result follows by elementary generating function magic. Finally, theorem 5.1.1 of Chuan-Chong & Khee-Meng's text on combinatorics states that
$$\mathcal{A}(x) = \sum_{n \in \mathbb{N}} a_n x^n \implies \frac{1}{1-x}\mathcal{A}(x) = \sum_{n \in \mathbb{N}} \left(\sum_{k \leq n} a_k \right) x^n;
$$
therefore, if we define the generating function $\mathcal{F}_y$ as
$$\mathcal{F}_{y}(x) = \sum_{n \in \mathbb{N}} f(y,n) x^n,$$

then we have that $$\mathcal{F}_{y}(x) = \frac{1}{1-x}\mathcal{G}_y(x)
 =  \frac{1}{1-x}\prod_{k=1}^{y}\frac{1}{1-x^k}. $$

More explicitly we have that
$$f(y,z) = [x^{yz}] \mathcal{F}_{y}(x) =[x^{yz}] \left(\frac{1}{1-x}\prod_{k=1}^{y}\frac{1}{1-x^k}\right)$$
where the operator $[x^{k}] $ is defined as follows:
$$\mathcal{A}(x) = a_0+a_1x+ ... +a_nx^n+ ... \implies [x^{k}]\mathcal{A}(x) = a_k.$$ For the asymptotics please consult Flajolet & Sedgewick's text on analytic combinatorics where you will find a wealth of information and techniques for extracting the asymptoics of the coefficients of $\mathcal{F}_{y}(x) $.


TITLE: Derivative of Kronecker product of vector with itself
QUESTION [2 upvotes]: I'm struggling with the following problem. Suppose $\pmb{x}$ and $\pmb{y}$ are vectors of the same length and $\pmb{y}$ is not a function of $\pmb{x}$. What is the following derivative?
$$
\frac{\partial}{\partial \pmb{x}} (\pmb{y} - \pmb{x}) \otimes (\pmb{y} - \pmb{x})
$$
My thought was to write use $\pmb{z} = \pmb{y} - \pmb{x}$ and $\pmb{f} = \pmb{z} \otimes \pmb{z}$ and derive first:
\begin{align}
d\pmb{f} &= ((d\pmb{z}) \otimes \pmb{z}) + (\pmb{z} \otimes (d\pmb{z})) \\
&= (\pmb{I} \otimes \pmb{z})d\pmb{z} + (\pmb{z} \otimes \pmb{I})d\pmb{z} \\
&= ((\pmb{I} \otimes \pmb{z}) + (\pmb{z} \otimes \pmb{I}))d\pmb{z} \\
\frac{\partial \pmb{f}}{\partial \pmb{z}} &= (\pmb{I} \otimes \pmb{z}) + (\pmb{z} \otimes \pmb{I})
\end{align}
and then obtain by chain rule:
$$
\frac{\partial}{\partial \pmb{x}} (\pmb{y} - \pmb{x}) \otimes (\pmb{y} - \pmb{x}) = -\left( (\pmb{I} \otimes (\pmb{y} - \pmb{x})) + ((\pmb{y} - \pmb{x}) \otimes \pmb{I}) \right)
$$
Which seems sensble. However, this is part of a Hessian I am deriving, and it's corresponding transpose element I derived to be:
$$
-2\left(\pmb{I} \otimes (\pmb{y} - \pmb{x})\right)
$$
Which is very similar but not the same. Am I missing something obvious?

REPLY [2 votes]: Let's clear out some definitions first.
If $f:z \to f(x)$ is a matrix valued function and there is a function $D_f$ such that
$$
f(z+h) = f(z) + D_f(z,h) + o(\|h\|)
$$
Then $D_f$ is the differential of $f$. If there exists a matrix valued function $A(z)$ such that
$$ D_f(z,h) = A(z) h$$
Then $A(z)$ is the derivative of $f$.
(This is sometimes called the first identification theorem; see for instance Magnus and Neudecker, 1999).
In the case at hand, we have
$$f(z+h) = (z+h)\otimes (z+h) = \underbrace{z\otimes z}_{f(z)} + \underbrace{(h\otimes z) + (z\otimes h)}_{D_f(z,h)} + \underbrace{h\otimes h}_{o(\|h\|)}$$
So, by the definition $D_f(z,h) = (h\otimes z) + (z\otimes h)$ is the differential of $z\otimes z$. Now we can use the identification theorem to say that, since
$$ D_f(z,h) = \big[(I\otimes z) + (z \otimes I)\big]h = A(z) h$$
the matrix
$$ A(z) =  (I\otimes z) + (z \otimes I)$$
is the derivative of $z \otimes z$.
So in the case at hand, the same reasoning brings you to the correct derivative
$$ \frac{\partial}{\partial x}(y-x)\otimes (y-x) = \big(I\otimes(x-y)\big) + \big((x-y) \otimes I\big)$$
which is what you found.
Check the other half of the Hessian: there has to be something wrong there!


TITLE: A calculus problem regarding continuity and uniform continuity
QUESTION [1 upvotes]: A function $f : \mathbb{R} \to \mathbb{R} $ is continuous at the origin $x=0$ and we have
$$ f(0) = 0 $$
$$ \forall x_1 \in \mathbb{R} ,\forall x_2 \in \mathbb{R}: f(x_1 + x_2 ) \leq f(x_1) + f(x_2) $$
does it follow that $f$ is uniformly continuous?

REPLY [2 votes]: We have $f(x+h)\leq f(x) + f(h)$, so$$f(x+h) - f(x) \leq f(h)$$
Similarly we have  $f(x) =f(x+h -h)\leq f(x+ h) + f(-h)$, so$$f(x+h) - f(x) \geq -f(-h)$$
Now we have $$-f(-h)\leq f(x+h) - f(x) \leq f(h)$$
so $$|f(x+h) - f(x)| \leq \max\{|f(h)|, |f(-h)|\}$$
Now use the continuity at $x=0$ to conclude:
$f(0) =0$ and $f$ is continuous at $x=0$ means for any $\epsilon >0$, $\exists \delta >0$ such that for all $|h| < \delta$, $|f(h)|= |f(h) - f(0)|< \epsilon$
So once $|h| < \delta$, we have $|f(h)|< \epsilon$ and $|f(-h)|< \epsilon$, so
$$|f(x+h) - f(x)| \leq \max\{|f(h)|, |f(-h)|\} < \epsilon, \forall x\in \mathbb{R}$$
So $f$ is uniformly continuous.