text:
\begin{document}

\title{Illusory Shapes via Phase Transition}

\author{Yoon Mo Jung  \and Jianhong Jackie Shen}

\institute{
	Yoon Mo Jung \at
		Department of Computational Science and Engineering, Yonsei University, Korea \\
			\email{ ymjung@yonsei.ac.kr }
		\and
	Jackie Shen (corresponding author) \at
		Department of Industrial and Systems Engineering, University of Illinois, IL 61801, USA \\
		Tel.:  (646) 725-0065 \\
		Fax:  (217) 244-5705 \\
	\email{ jhshen@illinois.edu}
}

\date{\em \small Dedicated to Gil Strang for His 80th Birthday}

\maketitle

\begin{abstract}
We propose a new variational illusory shape (VIS) model via phase fields and phase transitions. It is inspired by the first-order variational illusory contour (VIC) model proposed by Jung and Shen [{\em J. Visual Comm. Image Repres.}, {\bf 19}:42-55, 2008].
Under the new VIS model, illusory shapes are represented by phase values close to 1 while the rest by values close to 0. The 0-1 transition is achieved by an elliptic energy with a double-well potential, as in the theory of $\Gamma$-convergence. The VIS model is non-convex, with the zero field as its trivial global optimum. To seek visually meaningful local optima that can induce  illusory shapes, an iterative algorithm is designed and its convergence behavior is closely studied. Several generic numerical examples confirm the versatility of the model and the algorithm.
\keywords{ Illusory Shapes \and Phase Transition \and Null Hypothesis \and Convergence}
\end{abstract}

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

The intriguing phenomenon of {\em illusory contours and shapes} (ICS) has been actively pursued in contemporary brain science, neurophysiology, neuroimaging, neural networks, computational neuroscience, and psychophysics~\cite{vision:fMRI2IC-LOC-stanley03,vision:IC-neurons-Lee01,vision:nature-v1forIC-grosof93,vision:neuralnet-interp-fukushima10,vision:nonlinearfeedback-leveille10,vision:salient2IC-yoshino06,vision_Lee_Mumford03}. We first quote  from the recent review paper by Murray and Herrmann~\cite{vision:review-IC-murray13}(2013) to highlight its importance:
\begin{quote} \em
  ``The progressive development of these models ({\em for ICS}) and the data supporting them in many regards highlight what may be considered a general tendency in neuroscience {\em over the past 30 or so years}, namely, progression from ... individual neurons ... to the discrete localization of brain functions ...''
\end{quote}
That is, the persistent scientific interest in ICS has been representative in modern neuroscience, and has profoundly enhanced the {\em systemic} approach to studying complex vision tasks. ICS is fundamentally a system-level vision phenomenon, as contrast to simpler
tasks (e.g., detection of edgelets) by individual simple or complex neurons.

Kanizsa's Triangle illustrated in Fig.~\ref{Fig:Kaniza} is a classic example for ICS. The white triangle in the middle pops out despite that there exist no {\em modal} borders against the white background. Neuronal measurements in the visual cortices of cats or monkeys revealed direct firing responses along such modal edges~\cite{illu_heydt84,illu_heydt89}. This is the {\em very}  nature of illusion, namely, the remarkable ability in reconstruction and re-organization from given visual information with internal structures.

In terms of systems and control,  such ICS phenomena have provided the ideal class of input signals that help identify the complex structures and functionalities of the primate vision system, including, e.g., the bottom-up and top-down schema~\cite{vision_Lee_Mumford03,vision_Wu_Zhu11,vision_Han_Zhu09}. Treating the multi-tier visual cortices and pathways as systems is fundamental. In particular, it implies that models have to be based on {\em systemic} estimations and decisions, instead of isolated neuronal firing patterns. Mathematically, this naturally invites Bayesian decision making over networks, or system-wide optimization~\cite{book_vision_bayes_knill}.

Deterministic variational optimization could be considered as the low temperature limit of the Bayesian decision framework. As in statistical mechanics~\cite{book_mechanics_chandler}, under the low-temperature limit, geometric regularities emerge from statistical gaseous patterns (corresponding to the textural patterns in imaging and vision)~\cite{bImage_ChanShen}. The variational PDE methodology is particularly powerful in handling geometric regularities such as the distance or the curvature~\cite{bImage_AubertKorn,bImage_ChanShen,aubves,luminita_BV}. For ICS, there have been several  variational PDE models in the literature~\cite{illu_sarmalset02,illu_chazhu_levelset03,illu_zhuchan_shape05,illu_jungshen}, as partially reviewed in our earlier work~\cite{illu_jungshen}.

In~\cite{illu_jungshen}, based on the decomposition into {\em real} and {\em imaginary} components, we have proposed a first-order variational illusory contour (VIC) model, which is implemented by the supervised level-set method of Osher and Sethian~\cite{oshset}. Among all the variational-PDE models for ICS that the authors are aware of, this VIC model has the lowest complexity and thus allows detailed analysis of the illusory shapes (as local optima). This will be briefly recapped in Section~\ref{sec:contourmodel}.

The current work has been closely inspired by~\cite{illu_jungshen} and seeks a variational illusory shape (VIS) model based on phase transitions. An illusory shape is ideally represented by the phase value 1.0 and the rest by 0.0. The machinery of $\Gamma$-convergence and phase transitions~\cite{gamma_AmbrosioTor90,gamma_AmbrosioTor92} allows us to construct a phase-field energy that is closely related to the VIC model introduced in~\cite{illu_jungshen}. The new VIS model is more self-contained from modeling to computing. More details are explained in Sections~\ref{sec:represent} and~\ref{sec:new_model}.

Like most phase-field models~\cite{gamma_March92,gamma_MarchDozio97,gamma_ShenPCMS05,gamma_ShenSoftMS06}, the proposed nonlinear VIS model is not convex. The zero field is the global optima but uninteresting. Visually meaningful local optima are hence sought after via an iterative algorithm. The design and analysis of the algorithm will be elaborated in Section~\ref{sec:algorithm}.

In Section~\ref{sec:num_examples}, we summarize the entire algorithm via a pseudo code, and present several generic numerical examples to illustrate the versatility of the model and the algorithm. The work is finally concluded in Section~\ref{sec:conclusion}, where the limitations of the model are also addressed.

\section{Phase Field Representation of Illusory Shapes}
\label{sec:represent}
Given a configuration $Q$ in a visual field $\Omega$,
there may emerge an illusory shape $S$ outside $Q$: $S\subseteq \Omega \setminus Q$.
Fig.~\ref{Fig:Kaniza} shows the classic example of \textit{Kanizsa's Triangle}, of which $Q$ consists of three pac-man disks and $S$ refers to the white triangle encompassed and  induced by them.
Mathematically, we assume that $\Omega$ is a bounded Lipschitz convex open domain in $\mathbb{R}^2$, e.g., a Cartesian rectangle or a disk, and that $Q \subset \Omega$ is a {\em compact} Lipschitz sub-domain.

Define the ideal binary phase field $z(\bx)$, with $\bx =(x_1, x_2) \in \Omega$:
\begin{equation*}
z(\bx) =
\left\{
\begin{array}{cl}
  1, & \bx \in S, \\
  0, &  \bx \in \Omega \setminus S.
\end{array}
\right.
\end{equation*}
If $S$ is given, $z(\bx)$ is exactly its indicator function,
and a pixel $\bx \in \Omega$ belongs to $S$ if and only if its phase value is $1$.

\begin{figure}[!ht]
 \centering{
 \includegraphics[width=3cm]{figure_1.eps}
 \caption{\textit{Kanizsa's Triangle}. The white illusory triangle $S$ in the middle is induced by the black configuration $Q$ consisting of three pac-man disks.}
 \label{Fig:Kaniza} }
\end{figure}

In reality, neither $S$ nor $z(\bx)$ is known \textit{a priori}. The phase field approach
attempts to  reconstruct a nontrivial phase field $z(\bx)$'s that can induce illusory shapes. Since binary fields are hard to work with
both theoretically and computationally, one resorts to {\em continuous} fields that are the mollified versions of the binary ones.  They complete the $0$-$1$ phase transition along a tubular neighborhood of the desirable sharp boundaries. Such continuous fields are thus called the {\em diffusive} phase fields in the literature.

For the current task, we impose the following two natural conditions:
\begin{enumerate}[(a)]
 \item $z|_Q \equiv 0$, i.e., illusory shapes can only emerge outside $Q$.

 \item $z|_{\partial \Omega} \equiv 0$. This is because, for variational illusion models like~\cite{illu_jungshen}, one can prove that illusory  shapes can only emerge {\em within} the convex hull of $Q$, and thus also within the interior of $\Omega$.
\end{enumerate}
The main challenge then becomes how to encourage the phase $1.0$ (or close to 1.0) to emerge for configurations that can induce illusory shapes.

\section{Variational Contour Model of Jung and Shen}
\label{sec:contourmodel}

In this section, we briefly review the first-order variational illusory contour (VIC) model by Jung and Shen~\cite{illu_jungshen}. In the next section, the new phase-field model is to be built upon this VIC model.

For simplicity, below we assume all curves to be  piecewise smooth.
Let $\Gamma$ be any simple closed (Jordan) curve on $\Omega \setminus Q^\circ$.
In~\cite{illu_jungshen}, it is decomposed to the {\em real} and {\em imaginary} parts by:
\begin{equation*}
\Gamma_{\re} = \Gamma \cap \partial Q, \quad \Gamma_{\im} = \Gamma \setminus \partial Q.
\end{equation*}
If $\Gamma$ is a genuine illusory contour such as the three sides of Kanizsa's Triangle,
the imaginary part $\Gamma_{\im}$ corresponds to the illusory interpolant.
In Figure \ref{Fig:Kaniza}, for instance, $\Gamma_{\re}$ corresponds to the three corner turns along the disk inducers, and $\Gamma_{\im}$ to the three illusory {\em modal} segments.

For any such a curve $\Gamma$, Jung and Shen proposed the following variational contour model based on the decomposed ``energy''~\cite{illu_jungshen}:
\begin{equation}\label{Eold1}
 E[\Gamma] = E[\Gamma|a,\,b] = a\int_{\Gamma_{\re}}ds + b\int_{\Gamma_{\im}}ds,
\end{equation}
for any pair of properly defined weights $a$ and $b$ with $0<a<b$.
Illusory contours are then defined as the \emph{local} minima.

Compared with several other models~\cite{illu_sarmalset02,illu_chazhu_levelset03,illu_zhuchan_shape05}, the {VIC} model~\eqref{Eold1} is simple and yet powerful enough, as demonstrated by the generic  examples in~\cite{illu_jungshen}. Analytically, the authors were able to establish detailed geometric properties, among which the most important is Theorem 2.11 in~\cite{illu_jungshen}.

\begin{theorem}[Characterization of a Local Minimum] \label{T1:characterization}
 Let $Q$ be a generic compact configuration on an open Lipschitz domain $\Omega$. Assume that $\partial Q$ is piece-wise smooth with finitely many corners (but no kinks).  Let  $\theta_{\min{}},\,
 \theta^\ast_{\min{}} >0$ denote its minimum outer and inner spans. Suppose a given simple closed Jordan curve
 $\Gamma \in \mathcal{C}$ satisfies the following structural conditions:
 \begin{enumerate}[(i)]
 \item (Imaginary Behavior) each connected component $\gamma_\im$ of $\Gamma_\im$ is a
        straight line segment, and no two distinct components share a common hinge;
 \item (Junction Behavior) at any junction point $z \in J[\Gamma]$, the turn $\phi_z <
         \pi/2$, and the idle angle $\phi^\idle_z \ge \pi/2$.
 \end{enumerate}
 Let $\phi_{\max{}}$ denote the maximum turn on $J[\Gamma]$. Then there exists a critical ratio
 $r_c=r_c(\theta_{\min{}}, \phi_{\max{}})< 1$, such that for any $\alpha$ and $\beta$ with
 $r=\alpha/\beta < r_c$, $\Gamma$ is a local minimum to the energy $E[\cdot \mid \alpha,\, \beta]$.
\end{theorem}

We refer to~\cite{illu_jungshen} for the definitions of spans, hinges, turns, and idle angles. Such detailed characterizations are much harder to establish for more complex illusion models~\cite{illu_sarmalset02,illu_chazhu_levelset03,illu_zhuchan_shape05} .

\begin{figure}[!ht]
 \centering{
 \includegraphics[width=7cm]{figure_2}
 \caption{An illusory contour emerges from a complex configuration $Q$ of objects, as captured by the supervised level-set method of Osher and Sethian~\cite{oshset} for the VIC model in  Eqn.~\eqref{Eold1} proposed by Jung and Shen~\cite{illu_jungshen}. }
 \label{Fig1}
 }
\end{figure}

In~\cite{illu_jungshen},  illusory shapes (as the local minima to $E[\Gamma|a,\,b]$)
have been computed via a supervised level-set scheme of Osher and Sethian~\cite{oshset},
as applied to the approximate energy (Eqn.~(7) in~\cite{illu_jungshen}):
\begin{equation}
E_\sigma[\Gamma|\alpha,\, \beta] = \alpha\int_{\Gamma}ds + \beta\int_{\Gamma} g(\lvert \nabla \chi_{Q, \sigma} (s)\rvert)\, ds,
\end{equation}
with $\alpha = a + b$, $\beta = b$ for the targeted energy $E[\Gamma|a,\, b]$ in Eqn.~\eqref{Eold1}. Here $\sigma \ll 1$ denotes a small diffusion or mollification scale, so that the binary indicator $\chi_Q$ is mollified to a smooth approximation $\chi_{Q, \sigma}$.
The function $g$ could be any positive function satisfying: $g(0^+) = 1$, $g(+\infty) = 0$, for example, $g(p) =\exp{(-p^2)}$ or
$g(p) = 1/(1+p^2)$. It was shown in~\cite{illu_jungshen} that for any admissible contour $\Gamma$,
$E_\sigma[\Gamma|\alpha,\, \beta] \rightarrow E[\Gamma|a,\, b]$ as $\sigma \to 0^+$. The critical ratio $r_c$ for $a/b$ in Theorem \ref{T1:characterization} now transfers to  $R_c = r_c/(1-r_c)$ for $\alpha/\beta$, accordingly.

\section{A Phase Transition Model for Illusory Shapes}
\label{sec:new_model}
In this section, we develop a new phase transition model for illusory shapes, based on the discussion in the preceding two sections.

We first define the inducing scalar field via $G(\bx) = \alpha +\beta g(\lvert \nabla \chi_{Q, \sigma} (\bx)\rvert)$, for any $\bx\in \Omega$.
Alternatively, one could employ any optimal phase field $z^{\text{MS}}$ from the phase transition approximation to the Mumford-Shah model~\cite{gamma_AmbrosioTor90,gamma_AmbrosioTor92}:
\begin{equation*}
G(\bx) = \alpha + \beta z^{\text{MS}}_\sigma(\bx), \quad \bx \in \Omega.
\end{equation*}
The basic requirements for $z^{\text{MS}}_\sigma(\bx)$ are:
(i) the phase transition bandwidth is small: $\sigma \ll 1$, (ii) $u_0(\bx) = \chi_{Q}(\bx)$  is used as the input to the Mumford-Shah model, and (iii) $z^{\text{MS}}_\sigma(\bx)\in [0, \, 1]$ with $z^{\text{MS}}_\sigma \simeq 0$ along $\partial Q$ and $\simeq 1$  away from $\partial Q$.

In either way, one has
\begin{equation*}
G(\bx) \simeq
\left\{
\begin{array}{ll}
\alpha, &  \bx \,\, \text{near or along}\,\, \partial Q,\\
\alpha + \beta, & \text{otherwise}.
\end{array}
\right.
\end{equation*}
We call $G$ a   \emph{canyon} function associated with the given $Q$, since near $\partial Q$  the values of $G$ quickly drop from $\alpha + \beta$ to $\alpha$. For the canyon effect to be more salient, it is natural to require the dropping size $\beta \gg \alpha$, which is also in accordance with the critical ratio requirement $\alpha/\beta <R_c$ in the preceding section.

Let $\Sov$ denote the Sobolev space of functions on $\Omega$ with zero boundary traces.
It hosts all diffusive phase fields discussed in Section~\ref{sec:represent}.
For any phase field $z\in \Sov$, we introduce the following energy:
\begin{equation}\label{model:phase}
E_\epsilon[z] = \int_\Omega \left[\frac{\epsilon}{2} \lvert\nabla z \rvert^2+ \frac{(1-z)^2z^2}{2\epsilon} \right]\cdot G\,d\bx
 + \lambda \int_\Omega \chi_Q \frac{z^2}{2\epsilon}\,d\bx,
\end{equation}
where the positive weight $\lambda$ is in the same order as  of $G$.
When $\alpha < \beta = O(1)$ so that $G=O(1)$, we simply set $\lambda =1$ in all our computational examples later on. The small ``diffusivity'' parameter $\epsilon$ defines the intended transition bandwidth between 0.0 and 1.0.

The two terms have been motivated  as follows. The second term is:
\begin{equation*}
\lambda \int_\Omega \chi_Q \frac{z^2}{2\epsilon}\,d\bx = \lambda \int_Q \frac{z^2}{2\epsilon}\,d\bx.
\end{equation*}
Thus for $\epsilon \ll 1$, it acts as a soft way to enforce $z  = 0$ on $Q$, which is the condition (a) imposed in Section~\ref{sec:represent}.  {Also notice that the condition (b) is met automatically since $z \in \Sov$.}

For the first term in $E_\epsilon[z]$, define the Borel measure $\mu_\epsilon$ for any given $z\in  \Sov$ by:
\begin{equation*}
d\mu_\epsilon = \left[\frac{\epsilon}{2} \lvert\nabla z \rvert^2+ \frac{(1-z)^2z^2}{2\epsilon}\right] \, d\bx.
\end{equation*}
For any smooth Jordan curve $\Gamma$ with arclength element $ds$,
we show semi-heuristically that $d\mu_\epsilon \rightarrow \frac{1}{6}ds$ as $\epsilon \rightarrow 0^+$, or more generally via the 1-dimensional Hausdorff measure $d\mathcal{H}^1_\Gamma$ that $d\mu_\epsilon \rightarrow \frac{1}{6}d\mathcal{H}^1_\Gamma$  (with $\mathcal{H}^1_\Gamma(A) = \mathcal{H}^1(\Gamma \cap A)$) for a properly
designed sequence of phase fields {$\{z_\epsilon\}_\epsilon$}.
Assume that $\Gamma$ is the $0$-level set of a smooth function $f(\bx)$: $\Gamma = f^{-1}(0)$, and that $f$ is \emph{regular} along $\Gamma$ in the sense that $\nabla f(\omega) \neq 0$ for all $\omega \in \Gamma$. With at most a sign flip, one could further assume that  $\nabla f(\omega)$ points towards the inside of $\Gamma$. Then by the Tubular Neighborhood Theorem~\cite{booth86}, there exists some $\delta_0 >0$ such that the map:
$$\varphi: (\omega, n)\in \Gamma \times (-2\delta_0, 2\delta_0)
\quad \rightarrow \quad \bx = \omega + n \cdot \frac{\nabla f}{\lvert\nabla f \rvert} \in \Omega, $$
is a diffeomorphism between $\Gamma \times (-2\delta_0, 2\delta_0)$ and an open tubular neighborhood $B_{2\delta_0}$ of $\Gamma$ in $\Omega$.
Let $S$ denote the logistic function:
$$S(t) = \frac{1}{1+e^{-t}}, \quad t\in \mR^1.$$
For any $\epsilon \ll \delta_0$, we then construct a special phase field $z_\epsilon \in \Sov$:
\[
z_\epsilon(\bx) =
\left\{
\begin{array}{ll}
S\left(\frac{n}{\epsilon}\right), & \bx = \varphi(\omega, n)\in B_{\delta_0},\\
S\left(\frac{\delta_0}{\epsilon}\right), &  \bx \notin  B_{\delta_0} \,\, \text{and} \,\, \bx \in \inte(\Gamma),\\
S\left(-\frac{\delta_0}{\epsilon}\right), &   \bx\notin B_{\delta_0} \,\, \text{and} \,\, \bx \in \exte(\Gamma).
\end{array}
\right.
\]
Here $\inte(\Gamma)$ and $\exte(\Gamma)$ denote the interior and exterior domains of $\Gamma$, which are well defined according to the renowned Jordan Curve Theorem in topology~\cite{topo_thmJordan_hales2007}.

One can show that for any \emph{continuous} function $\phi$ on $\Omega$:
\begin{equation}
\int_\Omega \phi\,d\mu_\epsilon \rightarrow \frac{1}{6}\int_\Omega \phi \,d\mathcal{H}^1_\Gamma
= \frac{1}{6}\int_\Gamma \phi \,ds, \quad \text{as}\quad \epsilon \rightarrow 0^+,
\end{equation}
where the multiplier $\frac{1}{6}$ is the total variation of $\displaystyle F(z) = \frac{z^2}{2} -  \frac{z^3}{3}$ from $z = 0$ to $z=1$, and $F(z)$ is the primitive of $z(1-z)$ so that $F'(z)=z(1-z)$. As a result, for example, if $\Gamma$ represents the illusory Kanizsa Triangle in Fig.~\ref{Fig:Kaniza} and $\phi=G$, as $\varepsilon \to 0$,  one has
\[
6\int_\Omega G\,d\mu_\epsilon \simeq a\int_{\Gamma_{\re}}ds + b\int_{\Gamma_{\im}}ds,
\quad a=\alpha, \; b= \alpha + \beta.
\]
This gives a phase-field interpretation of the model by Jung and Shen~\cite{illu_jungshen}.
A similar argument could also be found in~\cite{gamma_ShenPCMS05}, for example.
But rigorous treatment is only offered by the theory of $\Gamma$-convergence
approximation~\cite{bGamma_Braides02,bGamma_DalMaso92}.

In summary, the phase transition model $E_\epsilon[z]$ for $z\in \Sov$ proposed in Eqn.~\eqref{model:phase} generalizes the contour model $E[\Gamma]$ of Jung and Shen~\cite{illu_jungshen} in Eqn.~\eqref{Eold1} to continuous phase fields. It also implements the two conditions (a) and (b) imposed in Section~\ref{sec:represent}.

One notices that $z_* \equiv 0$ is the global minimum of $E_\epsilon[z]$ but uninteresting. We thus define any local minimum $z_\epsilon \neq z_*$ to be an \emph{illusory phase field}, and
\begin{equation}\label{IlluShape}
S_\epsilon = \{ \bx\in \Omega: z_\epsilon > 1/2\}
\end{equation}
to be the associated \emph{illusory shape}.

\begin{theorem}\label{Thm2}
Suppose $z_\epsilon$ is a local minimum of $E_\epsilon[z]$ and is not always $0$ on $\Omega$.
Then the associated illusory shape $S_\epsilon \neq \emptyset$.
\end{theorem}
\begin{proof}
Otherwise assume $S_\epsilon = \emptyset$. Then $z_\epsilon \le 1/2$ for any $\bx \in \Omega$.
For any $t\in(0, \,1)$, define $z^{(t)}(\bx) = tz_\epsilon$. Then $z^{(t)} \in \Sov$,
and for $t$ close to $1$,  $z^{(t)}$ is a perturbation to $z^{(1)} = z_\epsilon$. One has:
\begin{equation}\label{eqn:perturb}
E_\epsilon[z^{(t)}] = t^2\left[\int_\Omega \frac{\epsilon}{2} \lvert\nabla z_\epsilon \rvert^2 \cdot G\,d\bx
 +\lambda \int_\Omega \chi_Q \frac{z_\epsilon^2}{2\epsilon}\,d\bx   \right]
 +\int_\Omega \frac{\Phi(tz_\epsilon)}{2\epsilon} \cdot   G\,d\bx,
\end{equation}
where $\Phi(z) = (1-z)^2z^2$ is the double-well potential.

Since $\Phi(z)$ is decreasing on $[-\infty,\, 0]$ and increasing on $(0, \, 1/2]$, and $z_\epsilon(\bx) \le 1/2$
for any $\bx \in \Omega$, one has:
$$  \Phi( tz_\epsilon(\bx)) \le  \Phi(z_\epsilon(\bx)), \quad \text{for any } \bx \in \Omega \text{ and } t\in(0,\, 1).$$
Thus \eqref{eqn:perturb} implies that for any $t\in(0,\, 1)$, $E_\epsilon[z^{(t)}] \le E_\epsilon[z_\epsilon]$. In fact, one must have $E_\epsilon[z^{(t)}] < E_\epsilon[z_\epsilon]$. This is because
$$ \int_\Omega \frac{\epsilon}{2} |\nabla z_\epsilon |^2 \cdot G\,d\bx
\ge \frac{\epsilon\alpha}{2} \int_\Omega  \lvert\nabla z_\epsilon \rvert^2 \,d\bx
= \frac{\epsilon\alpha}{2} |z_\epsilon|^2_1,$$
with $\alpha =\min G > 0$. Notice that in $\Sov$, the first-order homogeneous semi-norm $|\cdot|_1$ is actually a norm due to the zero trace~\cite{adafou}. Since it has been assumed that $z_\epsilon \neq 0 $ in $\Sov$, one must have $|z_\epsilon|^2_1> 0$.
Thus the first term alone in \eqref{eqn:perturb} shows $E_\epsilon[z^{(t)}] < E_\epsilon[z_\epsilon]$ for $t\in(0,\, 1)$. This contradicts to the assumption that $z_\epsilon$ is a local minimum since $z^{(t)}$'s are its small perturbations for $t \simeq 1$.
\closeProof

\section{Null Hypotheses and An Iterative Algorithm}
\label{sec:algorithm}

We now design a specific iterative algorithm to seek visually meaningful local minima of the proposed energy:
\begin{equation}\label{model:phase2}
E_\epsilon[z] = \int_\Omega \left[\frac{\epsilon}{2} \lvert\nabla z \rvert^2+ \frac{(1-z)^2z^2}{2\epsilon} \right]\cdot G\,d\bx
 + \lambda \int_\Omega \chi_Q \frac{z^2}{2\epsilon}\,d\bx,
\end{equation}
for $z\in\Sov$, given $G \in \mathcal{C}(\Omega)$ and $G \ge 0$.
For the current work, $G$ is specifically constructed as in the opening of the preceding section. In particular, we assume that
\begin{equation}\label{condG}
\alpha = \min G > 0\quad \text{and} \quad \alpha + \beta = \max G,
 \text{ with } \beta \ge \alpha.
\end{equation}
For optimization, $\lambda$ could be ``absorbed'' into $G$ via $G/\lambda$.
Thus we assume $\beta = O(1)$, and $\lambda = 1$.

In the literature of non-convex optimization, there has been much discussion on finding the global optima (e.g., stochastic or deterministic annealing~\cite{comput_annealing_kirkpatrick1983}). But no universal methodologies exist for locating the local optima that are of practical interest. Generally it has to be problem specific, and the two key components are:
\begin{enumerate}[(i)]
\item the choice of an initial guess or state, and
\item the design of an iterative searching algorithm.
\end{enumerate}

For the initial guess, as motivated by hypothesis testing in statistics~\cite{book_stat_elem_freedman}, we work with the following \emph{null hypothesis}:
\begin{equation}\label{hypothesis}
\text{``There indeed exists an illusory shape somewhere outside $Q$."}
\end{equation}
As the information of the illusory shape is unknown {\em a priori},
we start with following initial guess:
\begin{equation}\label{eqn:iniguess}
z_0(\bx) = 1\cdot(1-\chi_Q) + 0\cdot \chi_Q = 1-\chi_Q.
\end{equation}
It conservatively assigns phase $1$ to all pixels outside the given configuration $Q$. Strictly speaking, $z_0\notin \Sov$. This is not an issue since $z_0$ is to be used in an iterative algorithm: $z_n \rightarrow z_{n+1}$, and $z_n$'s generated afterwords all belong to $\Sov$ for $n = 1, 2, \cdots$.
Alternatively, one may apply diffusion to $z_0:\,\, u_t =\Delta u, \,\, u|_{\partial \Omega} = 0, \,\, u|_{t=0} = z_0$,
and use $u(\cdot, \delta)\in \Sov$ instead as the initial guess for some $\delta \ll 1$.

Next, to design the iterative algorithm, we first compute the Euler-Lagrange equation of $E_\epsilon$ in Eqn.~\eqref{model:phase2}:
$$-\nabla\cdot(\epsilon G \nabla)z +\frac{G}{\epsilon}(z-3z^2+2z^3) + \frac{\lambda \chi_Q}{\epsilon}\cdot z = 0,$$
which can be rearranged as:
\begin{equation}\label{eqn:pde}
-\nabla\cdot(\epsilon^2 G \nabla)z + (G(1+2z^2)+\lambda \chi_Q)z = 3Gz^2.
\end{equation}
This is a nonlinear elliptic equation on $\Omega$ with boundary condition $z|_{\partial \Omega} = 0$.
As discussed earlier, the global minimum  $z_\ast\equiv 0$ is a solution.
We are interested in the non-zero solutions that are associated with the local minima of $E_\epsilon[z]$.

Our proposed iterative algorithm is to solve the following linear elliptic equation for $z_{n+1}$, given a current guess $z_n$:
\begin{equation}\label{eqn:linpde}
\begin{aligned}
&-\nabla\cdot(\epsilon^2 G \nabla)z + g_nz = f_n, \quad z|_{\partial \Omega} = 0, \quad \text{with}\\
& g_n = G(1+2z^2_n)+\lambda \chi_Q,\quad \mbox{and}\; f_n=3Gz^2_n.
\end{aligned}
\end{equation}

\begin{theorem}
Suppose that $G$ satisfies Eqn.~\eqref{condG} and $z_n \in \mathbf{L}^\infty(\Omega)$.
Then a weak solution $z = z_{n+1}\in \Sov$ to Eqn.~\eqref{eqn:linpde} always exists and is unique in the sense that, for any $u \in \Sov$,
\begin{equation}\label{weakform}
(\epsilon^2 G\nabla z_{n+1}, \, \nabla u) + (g_nz_{n+1}, \, u) = (f_n,\, u),
\end{equation}
where $(\cdot,\, \cdot)$ denotes the canonical inner product in $\mathbf{L}^2(\Omega,\, \mR^k)$.
\end{theorem}
\begin{proof}
In $\Sov$, define the symmetric bilinear function $\langle \cdot,\, \cdot \rangle_1$ via:
$$\langle u,\, v \rangle_1  = (\epsilon^2 G\nabla u, \, \nabla v) + (g_n u, \, v),\quad u,v\in \Sov,$$
and denote the canonical inner product in $\Sov$ by $\langle \cdot,\, \cdot \rangle_0 $, which is defined by:
$$\langle u,\, v \rangle_0  = (\nabla u, \, \nabla v) + (u, \, v).$$
By Eqn.~\eqref{condG},
$$\epsilon^2\alpha (\nabla u, \, \nabla u) \le (\epsilon^2 G\nabla u, \, \nabla u) \le \epsilon^2(\alpha+\beta)(\nabla u, \, \nabla u),$$
and by the definition of $g_n$ in~\eqref{eqn:linpde},
$$\alpha (u, \, u) \le  (G u, \, u)  \le  (g_nu, \, u) \le \big((\alpha+\beta)(1+2\|z_n\|^2_\infty)+\lambda\big)  (u, \, u). $$
Therefore, $\langle \cdot,\, \cdot \rangle_1$ is an inner product equivalent to $\langle u,\, v \rangle_0$. Noticing that
$$|(f_n,\, u)| \le 3(\alpha+\beta)\|z_n\|^2_\infty(1,\, |u|) \le 3(\alpha+\beta)\|z_n\|^2_\infty\sqrt{|\Omega|}\sqrt{\langle u, \,u \rangle_0}, $$
$(f_n,\, \cdot)$ must be a continuous linear function for $\langle \cdot ,\, \cdot \rangle_0$ and thus also $\langle \cdot ,\, \cdot \rangle_1$.
Applying Riesz representation theorem~\cite{fol} to $\langle \cdot , \cdot \rangle_1$ and $(f_n,\, \cdot)$,
one concludes that there exists a unique $z_{n+1} \in \Sov$, such that
$$\langle z_{n+1}, \, u \rangle_1 \equiv (f,\, u),\quad \text{for any } u\in \Sov,$$
which is precisely Eqn.~\eqref{weakform}.
\closeProof

We also have an energy-form description for $z_{n+1}$ under given $z_n$.

\begin{proposition}\label{Prop1}
$z = z_{n+1}\in\Sov$ is the unique weak solution to Eqn.~\eqref{eqn:linpde}
if and only if $z_{n+1} = \arg\min_z E_\epsilon[z|z_n]$, where
\begin{equation}\label{model:phase3}
E_\epsilon[z|z_n] = \int_\Omega \left[\frac{\epsilon}{2} \lvert\nabla z \rvert^2+ (1+2z_n^2)\frac{(z-\gamma_n)^2}{2\epsilon} \right]\cdot G\,d\bx
 + \lambda \int_\Omega \chi_Q \frac{z^2}{2\epsilon}\,d\bx,
\end{equation}
with $\gamma_n = 3z^2_n/(1+2z^2_n)$. Notice that $E_\epsilon[\cdot |z_n]$ is strictly convex.
\end{proposition}
\begin{proof} By definition, for any $z, u \in \Sov$,
\begin{equation} \label{E:expansion}
E_\epsilon[z+u|z_n] = E_\epsilon[z|z_n] + E^*_\epsilon[u|z_n] + J_\epsilon[z, u|z_n],
\end{equation}
with
$$E^*_\epsilon[u|z_n] = \int_\Omega \left[\frac{\epsilon}{2} \lvert\nabla u \rvert^2+ (1+2z_n^2)\frac{u^2}{2\epsilon} \right]\cdot G\,d\bx
 + \lambda \int_\Omega \chi_Q \frac{u^2}{2\epsilon}\,d\bx, $$
and
$$J_\epsilon[z, u|z_n] = (\epsilon G \cdot \nabla z, \, \nabla u) +  \frac{1}{\epsilon}(G(1+2z_n^2)(z-\gamma_n), \, u)
+\frac{\lambda}{\epsilon}( \chi_Q \cdot z, \, u).$$
Since $E^*_\epsilon[u|z_n]\ge 0$, $E^*_\epsilon[tu|z_n]=t^2E^*_\epsilon[u|z_n]$, and
$J_\epsilon[z, tu|z_n] = tJ_\epsilon[z, u|z_n]$, we conclude from Eqn.~\eqref{E:expansion} by using $t \ll 1$ that
$$z_{n+1} = \arg\min_z E_\epsilon[z|z_n]\quad \text{iff}\quad J_\epsilon[z_{n+1}, u|z_n] = 0,
\,\,\text{for any } u\in \Sov.$$
On the other hand,
$$\epsilon \cdot J_\epsilon[z, u|z_n] = (\epsilon^2 G\nabla z, \, \nabla u) +  (g_nz,\, u)-(f_n,\, u),$$
which is precisely Eqn.~\eqref{weakform} with $z=z_{n+1}$.  Thus $z_{n+1}=\mathrm{argmin} E_\epsilon[z | z_n]$ if and only if it satisfies Eqn.~\eqref{weakform}. {Moreover,  $E^*_\epsilon[u|z_n] > 0$ for $u \neq 0$ guarantees strict convexity.} This completes the proof.
\closeProof

This leads to the following desirable property for $z_n$'s to faithfully approximate $0$-$1$ binary phases.
\begin{proposition}\label{Prop2}
Let $z_{n+1}\in \Sov$ be the unique weak solution to the iterative algorithm in Eqn.~\eqref{eqn:linpde}
given $z_n$. If $z_n(\bx) \in [0, 1]$ for any $\bx\in\Omega$, so must be $z_{n+1}$.
\end{proposition}
\begin{proof} Given $z_{n+1} \in \Sov$, define truncation $z^{[0, 1]}_{n+1}$ by
$$
z^{[0, 1]}_{n+1}(\bx) =
\left\{
\begin{array}{ll}
z_{n+1}(\bx), & \text{if} \quad 0 < z_{n+1}< 1,\\
1, & \text{if}\quad z_{n+1}\ge 1,\\
0, &  \text{if}\quad z_{n+1} \le 0.
\end{array}
\right.
$$
Then $z^{[0, 1]}_{n+1}\in \Sov$, and it is well known~\cite{adafou} that truncation does not increase
the norm of a gradient in Sobolev spaces. Then one must have:
$$\int_\Omega \frac{\epsilon}{2} \left\lvert\nabla z^{[0, 1]}_{n+1} \right\rvert^2 G\,d\bx
\le \int_\Omega \frac{\epsilon}{2} \lvert\nabla z_{n+1} \rvert^2G\,d\bx. $$
Since $z_n\in[0,1]$, one has $z_n^2 \le 1$, and for the scalar field $\gamma_{n}$ in Eqn.~\eqref{model:phase3},
$$\gamma_n = \frac{3z_n^2}{1+2z_n^2}\le 1.$$
The one must have for any $\bx\in \Omega$,
$$(z^{[0, 1]}_{n+1}(\bx) - \gamma_n )^2 \le (z_{n+1}(\bx) - \gamma_n )^2\quad \text{and}\quad
\left(z^{[0, 1]}_{n+1}(\bx)\right)^2 \le (z_{n+1}(\bx))^2. $$
Examining the expression in Eqn.~\eqref{model:phase3} thus shows:
$$E_\epsilon\left[z^{[0, 1]}_{n+1}\big|z_n\right]\le E_\epsilon[z_{n+1}|z_n]. $$
Due to the uniqueness result from the preceding proposition,
$$z_{n+1}(\bx)\equiv z^{[0, 1]}_{n+1}(\bx)\in [0, 1], \quad \bx\in\Omega.$$
This completes the proof.
\closeProof

The next theorem reveals that the phase field sequence $(z_n)$ defined by the iterative algorithm in~\eqref{eqn:linpde} is indeed an energy decreasing sequence.

\begin{theorem}\label{thm4}
Let $\{z_n\}^\infty_{n=1}$ be the sequence of phase fields generated by the algorithm in Eqn.~\eqref{eqn:linpde}, started from the null hypothesis $z_0$ in Eqn.~\eqref{eqn:iniguess}. Then
$$E_\epsilon[z_1]\ge E_\epsilon[z_2]\ge \cdots \ge E_\epsilon[z_n] \ge \cdots $$
Furthermore, for $n \ge 1$,
\begin{equation}\label{eqn:difference}
E_\epsilon[z_n]- E_\epsilon[z_{n+1}] \ge \int_\Omega \frac{G}{2\epsilon}
(z_{n+1} - z_n)^2\left(2z_{n}+4z_{n+\frac{1}{2}}\left(1-z_{n+\frac{1}{2}}\right)\right)d\bx,
\end{equation}
where $z_{n+\frac{1}{2}} =  (z_n + z_{n+1})/2$.
\end{theorem}
\begin{proof}
By the null hypothesis $z_0$ and Proposition~\ref{Prop2}, we have
$$0 \le z_n(\bx) \le 1, \quad \text{for all } \bx\in\Omega \quad\text{and}\quad n = 1, 2, \cdots.$$
Then $z_{n+\frac{1}{2}}(\bx) =  (z_n(\bx) + z_{n+1}(\bx))/2\in [0, 1]$ for any $\bx \in \Omega$.
Thus Eqn.~\eqref{eqn:difference} implies $E_\epsilon[z_n]\ge E_\epsilon[z_{n+1}]$ for $n\ge 1$ since the
integrand is nonnegative. It suffices to establish~\eqref{eqn:difference} for $n\ge 1$.
By Proposition~\ref{Prop1}, since $z_{n+1} = \arg\min_z E_\epsilon[z|z_n]$, one has
$$ E_\epsilon[z_{n+1}|z_n] \le E_\epsilon[z_n|z_n]. $$
This spells out to be:
$$\begin{aligned}
 &\int_\Omega \frac{\epsilon G}{2} \lvert\nabla z_{n+1} \rvert^2 \,d\bx
 + \lambda \int_\Omega \chi_Q \frac{z_{n+1}^2}{2\epsilon}\,d\bx
 + \int_\Omega \frac{G}{2\epsilon} \left((1+2z_n^2)z_{n+1}^2 - 6z_n^2 z_{n+1}\right) d\bx\\
  \le & \int_\Omega \frac{\epsilon G}{2} \lvert\nabla z_n \rvert^2\, d\bx
 + \lambda \int_\Omega \chi_Q \frac{z_{n}^2}{2\epsilon}\,d\bx
 + \int_\Omega \frac{G}{2\epsilon}  \left((1+2z_n^2)z_{n}^2 - 6z_n^3\right) d\bx.
\end{aligned} $$
Then by the definition of $E_\epsilon[z]$ in Eqn.~\eqref{model:phase2},
$$ \begin{aligned}
& E_\epsilon[z_n]- E_\epsilon[z_{n+1}] \\
 \ge & \int_\Omega \frac{G}{2\epsilon} \left((1+2z_n^2)z_{n+1}^2 - 6z_n^2 z_{n+1}\right) d\bx
- \int_\Omega \frac{G}{2\epsilon} \left((1+2z_n^2)z_{n}^2 - 6z_n^3\right) d\bx \\
&+ \int_\Omega \frac{G}{2\epsilon} z_n^2(1-z_n)^2 \,d\bx
-\int_\Omega \frac{G}{2\epsilon} z_{n+1}^2(1-z_{n+1})^2 \,d\bx.
\end{aligned}$$
Expanding and re-organizing the integrands on the right, one arrives at the inequality~\eqref{eqn:difference}.
\closeProof

\begin{proposition}\label{Prop3}
Following the same assumptions of Theorem \ref{thm4},
we conclude that there must exist some $E_*\ge 0$,
such that $\lim_{n \rightarrow \infty} E_\epsilon[z_n] = E_*$.
\end{proposition}
This is because any bounded monotonic sequence must converge.

\begin{proposition}\label{Prop4}
If  $E_* = \lim_{n \rightarrow \infty} E_\epsilon[z_n] = 0$,  $\{z_n\}^\infty_{n=1}$ must converge to the
global minimum $z_* \equiv 0$ in $\Sov$.
\end{proposition}
\begin{proof}
By definition of $E_\epsilon[\cdot]$ in Eqn.~\eqref{model:phase2}, and the assumptions on $G$ in Eqn.~\eqref{condG},
$$E_\epsilon[z] \ge \frac{\epsilon \alpha}{2} \int_\Omega  |\nabla z|^2 \,d\bx
\ge \frac{\epsilon \alpha}{2}C_\Omega \|z\|^2_{\Sov},$$
where $C_\Omega$ is a positive constant only depending on the domain. The second inequality holds since the trace along $\partial \Omega$ vanishes~\cite{adafou}. Thus $E_\epsilon[z_n]\rightarrow 0$
implies that  $z_n \rightarrow 0$ in $\Sov$.
\closeProof

Let $\rho_n = E_\epsilon[z_n]- E_\epsilon[z_{n+1}]$
denote the successive energy improvement. Then
$$\sum^\infty_{n =1} \rho_n = E_\epsilon[z_1] - E_* < \infty.$$
The series $(\rho_n)$ are said to converge in a generalized quadratic power-law (GQPL) if
$$\sum^\infty_{n =1}\sqrt{\rho_n} < \infty.$$
For instance, $\displaystyle \rho_n = 1/(1+n)^{2+\delta}$ or $1/ ((1+n)^2 \log^{2+\delta}(1+n))$ for any $\delta > 0$.

\begin{proposition}
Suppose $\{E_\epsilon[z_n]\}^\infty_{n=1}$ converges to some $E_* > 0$ in GQPL.
Suppose there exists
a measurable set $S \subseteq\Omega$ with a positive Lebesgue measure,
some constant $C>0$ and integer $N$ such that for any $n>N$,
$$z_n(\bx) \ge C, \quad \text{for }  \bx \in S. $$
Then $\{z_n|_S\}^\infty_{n=1}$ converges in $\mathbf{L}^2(S)$.
\end{proposition}
\begin{proof}
For $n > N$,
$$2z_n + 4z_{n+\frac{1}{2}}(1-z_{n+\frac{1}{2}}) \ge 2 z_n \ge 2C.$$
Then Eqn.~\eqref{eqn:difference} leads to:
$$\rho_n \ge \int_S \frac{G}{2\epsilon} (z_{n+1}-z_n)^2 (2C)\,d\bx
\ge \frac{\alpha C}{\epsilon}\|z_{n+1}-z_n\|^2_{\mathbf{L}^2(S)}.$$
Thus,
$$\|z_{n+1}-z_n\|_{\mathbf{L}^2(S)} \le \sqrt{\frac{\epsilon}{\alpha C}} \cdot \sqrt{\rho_n}.$$
Since $\sum^\infty_{n =1}\sqrt{\rho_n} < \infty$, it implies that {$\{z_n\}_{n=1}^\infty$} is a Cauchy sequence in $\mathbf{L}^2(S)$.
\closeProof

The result can be intuitively interpreted as follows for the half-way threshold $C = 0.5$.
One starts with the null hypothesis (via $z_0$) that the illusory shape is $S_\epsilon = \Omega\setminus Q$.
As the iteration progresses, some pixels are rejected if $z_n(\bx) < 0.5$.
But suppose there exists a positive set $S$, such that all $z_n$'s after some $N$
consistently vote for it in the sense of $z_n(\bx) > 0.5$. Then the voting must be ``directional'' or converging, and $S$ has to be part of the final illusory shape.

\section{Numerical Implementation and Examples}
\label{sec:num_examples}
In this section, we briefly describe the numerical scheme of the proposed model and
present several generic computational examples.

Algorithm~\ref{algo} below offers a pseudocode block describing the major computational steps for the proposed model. For the initial guess $z_0$, we have adopted the null hypothesis~\eqref{hypothesis}. The core iteration formula for updating $z_{n+1}$ from $z_n$ follows  Eqn.~\eqref{eqn:linpde}.
Convergence analysis of the algorithm has been partially given in the preceding section.

\begin{algorithm}
\caption{Illusory Shapes via Phase Transition} \label{algo}
\begin{algorithmic}
\STATE {\bf input}: the configuration $Q$ represented via its indicator $\chi_Q$;
\STATE {\bf initialize:} $z_0(\bx) = 1\cdot(1-\chi_Q) + 0\cdot \chi_Q = 1-\chi_Q$ as the null hypothesis;
\STATE {\bf pre-process:} the canyon funtion $G(\bx) = \alpha +\beta g(\lvert \nabla \chi_{Q, \sigma} (\bx)\rvert)$ or $\alpha +\beta z^{\mathrm{MS}}_\sigma(\bx)$;
\WHILE{$\Vert z_{n+1} - z_{n} \Vert > \delta $}
\STATE Solve for $z_{n+1}$ from:
\begin{equation*}\label{eqn:algo}
\begin{aligned}
& -\nabla\cdot(\epsilon^2 G \nabla)z + g_nz = f_n, \quad\text{with}\\
& g_n = \,G(1+2z^2_n)+\lambda \chi_Q,\, f_n=3Gz^2_n,\text{ and } z|_{\partial \Omega} = 0.
\end{aligned}
\end{equation*}
\ENDWHILE
\end{algorithmic}
\end{algorithm}

Therefore, the core of the algorithm is an elliptic solver, which can be found in the standard literature of computational PDE's~\cite{book_numerical_Sauer}. Some typical parameters generating the examples herein are given as follows:
\[
	\alpha =0.1;\; \beta =1.0;\; \lambda =1.0;\;\; \mbox{and}\;\; \epsilon = (2\sim 4) \cdot h.
\]
Here $h$ denotes the grid/pixel size and is defined in such a way that the longest size of the image domain $\Omega$ is always normalized to the unit length.
The convergence tolerance is set to be $\delta =10^{-6}$.

\begin{figure}[h!]
  \centering
  \subfigure[Kaninza Triangle]{
   \includegraphics[width=3.5cm]{figure_3a}
  }
  \subfigure[Canyon Function $G$]{
   \includegraphics[width=3.5cm]{figure_3b}
  }
  \subfigure[Null Hypothesis $z_0(\bx)$]{
   \includegraphics[width=3.5cm]{figure_3c}
  }\\
  \subfigure[$20$ Iterations]{
   \includegraphics[width=3.5cm]{figure_3d}
  }
  \subfigure[$100$ Iterations]{
   \includegraphics[width=3.5cm]{figure_3e}
  }
  \subfigure[Final Illusory Shape]{
   \includegraphics[width=3.5cm]{figure_3f}
  }
  \caption{The model and algorithm on the Kanizsa Triangle.}
  \label{fig:triangle}
\end{figure}
Fig.~\ref{fig:triangle} shows the numerical simulation on the classic example of Kanizsa's Triangle. In all the panels (b-f), the white colors represent phase values close to $1.0$ while the black ones to $0.0$. Panel (b) shows the canyon function that is fed into the model and algorithm. Panel (c) shows the null hypothesis $z_0(\bx)$ that conservatively assigns phase $1.0$ to all pixels outside $Q$. Panels (d) and (e) show two intermediate iterations before final convergence. In Panel (f) the final phase field is plotted after numerical convergence. It successfully captures the illusory triangle (up to the numerical precision).  Fig.~\ref{fig:circle} and Fig.~\ref{fig:butterfly_lady} demonstrate another two examples with more complex layouts.

\begin{figure}[h!]
  \centering
  \subfigure[Illusory Disk]{
   \includegraphics[width=3.7cm]{figure_4a}
  }
  \subfigure[Intermediate Snapshot]{
   \includegraphics[width=3.7cm]{figure_4b}
  }
  \subfigure[Final Result]{
   \includegraphics[width=3.7cm]{figure_4c}
  }
  \caption{The model and algorithm on an example with an illusory disk.}
  \label{fig:circle}
\end{figure}

\begin{figure}[h!]
  \centering
  \subfigure[Illusory Lady]{
   \includegraphics[height=4cm]{figure_5a}
  } \quad \quad \quad \quad
  \subfigure[Converged Result]{
   \includegraphics[height=4cm]{figure_5b}
  }
  \caption{The Butterfly Lady, with a more complex configuration $Q$. }
  \label{fig:butterfly_lady}
\end{figure}

Fig.~\ref{fig:twoobjs} shows an example consisting of two disjoint illusory shapes: an ellipse and a triangle. As a result, the phase field sequence $(z_n)$ is expected to experience a topological splitting operation during the iteration. Like the renowned level-set methodology of Osher and Sethian~\cite{oshset}, the phase-field approach is also very versatile in handling region merging or splitting.

\begin{figure}[h!]
  \centering
  \subfigure[Illusory Ellipse and Triangle]{
   \includegraphics[width=5cm]{figure_6a}
  }
  \subfigure[Converged Result from the Model]{
   \includegraphics[width=5cm]{figure_6b}
  }
  \caption{Like the level-set methodology of Osher and Sethian~\cite{oshset}, the phase-field approach is versatile in handling topological changes like region splitting. }
  \label{fig:twoobjs}
\end{figure}

\section{Conclusion}
\label{sec:conclusion}
Inspired by the first-order variational illusory contour (VIC) model proposed in~\cite{illu_jungshen}, we have proposed a variational illusory shape (VIS) model based on the tool of phase transitions. The VIS model represents an illusory shape via phase values close to 1.0, and the rest by values close to 0.0. The phase transition is achieved by a variational energy formulated in the current work.

As for most non-quadratic phase transition models~\cite{gamma_March92,gamma_MarchDozio97,gamma_ShenPCMS05,gamma_ShenSoftMS06}, the proposed VIS model is non-convex. The zero field is the global optimum but uninteresting. To seek visually meaningful local optima, we have designed an iterative algorithm with a suitable initial guess, which could be considered as the {\em null hypothesis} in statistical testing. The null hypothesis assumes that there exists an illusory shape outside the given configuration. The iterative algorithm lets the pixels  ``vote'' collectively, until reaching the final consistent and stationary decision. Some key behaviors of the algorithm have been revealed through our analysis. And several generic numerical examples show the versatility of the proposed model and algorithm.

As in~\cite{illu_jungshen}, such lower-order models allow one to develop detailed analysis, but are necessarily limited in terms of applicability or performance. For example, illusory interpolation is often done via straight lines. Nevertheless, they help point towards more complex high-order models involving the curvature feature or Euler's elasticas~\cite{bImage_ChanShen,chakanshe,mum_elastica,illu2014_kangzhushen}, for example.

\begin{acknowledgements}
Jung has been supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) of Korea (2012R1A1A1015492). Shen has been supported by the National Science Foundation (NSF) of USA.
\end{acknowledgements}

config: arxiv


TITLE: Find degree function with max at $(2,44)$ and min at $(4,36)$
QUESTION [3 upvotes]: A function has a local maximum at $(2,44)$ and a local minimum value at $(4,36)$. Find a degree function that has these qualities.

I know that the two important points for the $f (x)$ are $2$ and $4$ so $f'(x)=(x-2)(x-4)=x^2-6x+8$ so $f(x)$ should be equal to $f(x)=\dfrac{x^3}3 -3x^2 +8x+d$
but now when I try to find the value of d and it doesn't satisfy the points the answer in the textbook answer key says the answer may vary and that's all they said no example equation so I want to know is this function possible or not and if it is possible I want an example of it with a little bit of explanation of where I was wrong?
I got the answer for anyone who is wondering the answer is that as  $f(x)=a[\dfrac{x^3}3 -3x^2 +8x]+d$ you can find values of a and d and get your equation now!

REPLY [0 votes]: The question is a bit confusing. It was not clear if it is a single variable function or dual-variable function. As the points $(2,44)$ and $(4,36)$ are in the form of $(x,y)$, I consider $y=f(x)$ as the form. Thus, it is a function in one variable.
Now that $(2,44)$ and $(4,36)$ are two points which the function satisfies:

$f'(x=2)=0$ because $(2,44)$ is a local maxima. And, $f'(x=4)=0$ because $(4,36)$ is a local minima.
$f(x=2)=44$ and $f(x=4)=36$.

From condition 1,
$$f'(x) = a (x-2)(x-4) \text{ where } a \ne 0$$
$$ \Rightarrow f(x) = a \bigg( \frac{x^3}{3}-3x^2+8x   \bigg)+d$$
From condition 2, $$f(x=2)=44 \Rightarrow \frac{20}{3}a+b=44$$ and $$f(x=4)=36 \Rightarrow \frac{16}{3}a+b=36$$
Together they yield $$a=6 \text{ and } b=4$$
Thus, the function is $2x^3-18x^2+48x+4$


TITLE: Determine whether the following series converges or diverges
QUESTION [3 upvotes]: I'm stuck on this problem:
$$1 + {1\cdot2\over1\cdot3} + {1\cdot2\cdot3\over1\cdot3\cdot5}+ {1\cdot2\cdot3\cdot4\over 1\cdot3\cdot5\cdot7} +\cdots$$
I've simplified the numerators $n!$ but can't figure out how to represent the denominators.
How do I go about solving this?

REPLY [2 votes]: Let us try to write down in a slightly simpler than above form the general term:
$$a_n:=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{1\cdot 3\cdot 5\cdot\ldots\cdot (2n-1)}=\frac{n!2\cdot 4\cdot 6\cdot\ldots\cdot 2n}{(2n)!}=\frac{2^n(n!)^2}{(2n)!}$$
and now just mimic what experiment did (D'Alembert's test):
$$\frac{a_{n+1}}{a_n}=\frac{2^{n+1}[(n+1)!]^2}{(2n+2)!}\,\frac{(2n)!}{2^n(n!)^2}=\frac{2(n+1)^2}{(2n+1)(2n+2)}\xrightarrow [n\to\infty]{}\frac{2}{2\cdot 2}=\frac{1}{2}<1$$and the series converges.


TITLE: Is any dual metrizable locally convex space a Frechet space?
QUESTION [7 upvotes]: [I have posted this question on MSE some time ago, but received no answer.]
The title basically says all of it.
If a normed space $F$ is a dual of a normed space $E$, then $F$ is a Banach space. I wonder if the same holds for Frechet spaces.
The strong dual $F$ of a locally convex space $E$ is complete, once $E$ is bornological, but I am not sure if this is the case here. Perhaps the completion of $E$ is though.

REPLY [9 votes]: I am not 100% clear what you are asking, but I will answer according to two interpretations:
a) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a locally convex TVS. Need $F$ be complete?
The answer to this question is no, by the following counterexample, where we obtain an incomplete normed space as the strong dual of a locally convex space.
Take $E$ to be $B(\mathcal{H})$, the bounded operators on an an infinite dimensional Hilbert space $\mathcal{H}$. The topology on $B(\mathcal{H})$ we will use is the strong operator topology (SOT, i.e. pointwise convergence of Hilbert space norm). It is not hard to show that, as a vector space, $F = E'$ is $\mathrm{FinRank}(\mathcal{H})$, the finite-rank operators (see, for example, Takesaki's Theory of Operator Algebras I, Theorem II.2.6). To work out what the strong dual topology on $F$ is, we need to know what the bounded sets are for the SOT on $B(\mathcal{H})$.
To this end, we can use the completeness of $\mathcal{H}$ and apply the uniform boundedness principle (see Schaefer's Topological Vector Spaces, III.4.2), showing SOT-bounded sets are exactly equicontinuous sets, a.k.a. norm-bounded sets in $B(\mathcal{H})$. Therefore the strong dual topology on $F$ is that defined by the dual norm, which is the trace-class norm on $\mathrm{FinRank}(\mathcal{H})$. As $\mathcal{H}$ is infinite-dimensional, $F$ is not complete in this norm, as there are trace-class operators that are not of finite rank.
(Added in edit: Here, the definition of strong dual topology on $F$ that I used was the unique locally convex topology having the polars of the bounded sets of $E$ as basic 0-neighbourhoods. It can also mean (as it does in Schaefer) the topology obtained by taking the polars of weakly (i.e. $\sigma(E,F)$) bounded sets instead. In this case, as $\sigma(E,F) = \sigma(B(\mathcal{H}),\mathrm{FinRank}(\mathcal{H}))$ is the weak operator topology, these are the WOT-bounded sets. Fortunately, WOT-bounded sets can be shown to be SOT-bounded by applying the uniform boundedness principle pointwise, so the two definitions of strong topology on $F$ coincide.)
b) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a metrizable TVS. Need $F$ be complete?
The answer to this question is yes, and also that $F$ is a Banach space (or "Banachable" space, without choosing any norm in particular). This is a consequence of this answer, plus the fact that the dual of a normed space is a Banach space.


TITLE: Simple vector algebra question for perpendicular lines
QUESTION [0 upvotes]: I'm actually a little ashamed to ask this,
Suppose A and B are not 0. Consider a line "l" whose cartesian equation is $Ax+ By + D = 0$. Suppose that $P_0 = (x_0,y_0)$ does not lie on "l". Show that $n = (A,B)$ is a vector that is perpendicular to "l"
I am trying to do this without the dot-product which makes it even more embarassing. So I know that in order for two lines to be consider perpendicular that either their slopes have to be negative reciprocals of each other or in dot-product form that they equal 0. How can I use that point and the normal "n" to show the vector is perpendicular?

REPLY [1 votes]: If I'm understanding your question correctly, you want to show that the line Ax + By + D = 0 is perpendicular to the vector (A,B). I don't know what you intend $P_0$ to be used for.
At any rate, since you assume A and B are not zero, you can rewrite this as $y = mx + b$ to see that the slope of the line is -A/B. But the vector (A,B) has slope B/A. So the product of the slopes is -1. (Note that D is irrelevant - changing it translates the line, but doesn't change its slope.)


TITLE: Reference needed for lower bound on number of guards in three-dimensional art gallery guarding
QUESTION [4 upvotes]: During my research (writing my master's thesis) I've stumbled across the book Art Gallery Theorems and Algorithms by O’Rourke.
In chapter 10 (pdf available on the above site), section 10.2.2. he shows a lower bound of $\Omega\left(n^{3/2}\right)$ on the number of guards needed in three-dimensional art gallery guarding. He says (bold marking is by myself):

In fact, we describe in this section a polyhedron constructed by Seidel that has these two properties.

I believe that the construction, as presented in the book, is not fully correct and thus I tried to find the original work by Seidel.
Sadly there is no reference in this section and the only reference in the book with Seidel is Constructing Arrangements of Lines and Hyperplanes with Applications by H. Edelsbrunner, J. O’Rourke, and R. Seidel which does not include the construction.
I have looked through Seidels Google Scholar page but there seems to be nothing relevant.

Do you have a source or know anything more about this construction?
What could be a good and accepted way for me to obtain information? Contacting either of the two?

REPLY [2 votes]: To my knowledge, Seidel's construction has only been published in O'Rourke's book and nowhere else.
In one of his papers, Seidel even refers to O'Rourke's book for a description of his own construction. He writes on top of page 253 of his joint paper with Jim Ruppert:
"In his book [9, p. 255] O'Rourke describes n-vertex three-dimensional polyhedra that require $\Omega(n^{3/2})$ guards."

Jim Ruppert, Raimund Seidel: On the Difficulty of Triangulating Three-Dimensional Nonconvex Polyhedra. Discrete & Computational Geometry 7: 227-253 (1992)


text:
\begin{document}

\title[A criterion for the logarithmic differential operators]{A criterion for the logarithmic differential operators to be generated by vector fields}
\author{Mathias Schulze}
\address{Mathias Schulze\\
Department of Mathematics\\
Oklahoma State University\\
401 MSCS\\
Stillwater, OK 74078\\
USA}
\email{mschulze@math.okstate.edu}

\begin{abstract}
We study divisors in a complex manifold in view of the property that the algebra of logarithmic differential operators along the divisor is generated by logarithmic vector fields.
We give
\begin{itemize}
\item a sufficient criterion for the property,
\item a simple proof of F.J. Calder{\'o}n--Moreno's theorem that free divisors have the property,
\item a proof that divisors in dimension $3$ with only isolated quasi--homogeneous singularities have the property,
\item an example of a non--free divisor with non--isolated singularity having the property,
\item an example of a divisor not having the property, and
\item an algorithm to compute the V--filtration along a divisor up to a given order.
\end{itemize}
\end{abstract}

\keywords{free divisor, hyperplane arrangement, logarithmic differential operator, symmetric algebra, V--filtration}

\subjclass{32C38, 13A30}

\thanks{The author is grateful to M.~Granger for many valuable discussions and comments and to F.J.~Castro--Jim\'enez, L.~Narv\'aez--Macarro, and J.M.~Ucha--Enr\'iquez for explaining their results and ideas.}

\maketitle
\tableofcontents

\section{Logarithmic comparison theorem for free divisors}

Let $X$ be a complex manifold of dimension $n\ge2$, $\O$ the ring holomorphic functions on $X$, and $\Omega^\bullet$ the complex of holomorphic differential forms.
Grothendieck's Comparison Theorem states that the De~Rham system $\O$ is regular \cite[Thm.\,2.3.4]{Meb89}.
This is equivalent to the fact that, for any divisor $D\subset X$, the natural morphism
\[
\xymat{
\Omega^\bullet(*D)=\DR(\O(*D))\ar[r]&\RR i_*i^{-1}\DR(\O)=\RR i_*\C_U,
}
\]
where $i$ is the inclusion $U=X\backslash D\subset X$, is a quasi--isomorphism.
Let  $\Omega^\bullet(\log D)\subset\Omega^\bullet(*D)$ be the subcomplex of logarithmic differential forms along $D$ \cite[Def.\,1.2]{Sai80}.
The above statement raises the question whether the inclusion $\Omega^\bullet(\log D)\subset\Omega^\bullet(*D)$ is also a quasi--isomorphism.
That is: Can one compute the cohomology of the complement of $D$ by logarithmic differential forms along $D$?
This turns out to be a property of $D$ called the logarithmic comparison theorem or simply LCT.
It is an open problem to characterize the divisors for which LCT holds.

Let $\Theta=\Der_\C(\O)$ be the $\O$--module of holomorphic vector fields on $X$ and $\Der(\log D)\subset\Theta$ be the $\O$--submodule of logarithmic differential operators along $D$ \cite[Def.\,1.4]{Sai80}.
A divisor $D$ is called free if $\Der(\log D)$ is a locally free $\O$--module.
Let $\D$ be the $\O$--algebra of differential operators on $X$ with holomorphic coefficients and let $F$ be the increasing filtration on $\D$ by the order of differential operators.
Let $\V^D$ be the V--filtration along $D$ on $\D$ as defined in Section \ref{20} such that  $\V^D_0=\D(\log D)$ is the $\O$--algebra of logarithmic differential operators along $D$.
F.J.~Calder{\'o}n--Moreno \cite[Thm.\,1]{Cal99} proves that, for a free divisor $D$,  $\V^D_0$ is generated by vector fields, that is $\V^D_0=\O[\Der(\log D)]$.
Let $S_D$ be the decreasing filtration on $\D$ which is locally defined by $S_D^k=f^k\cdot\D$ where $f\in\O$ such that $D=(f)$.
By Corollary \ref{24}, the induced filtration $S_D$ on $\V^D_0$ defined by $S_D^k\V^D_0=\V^D_0\cap( f^k\cdot\D)$ reflects the embeddings $\V^D_k\subset\D$.
If $\V^D_0=\O[\Der(\log D)]$ then $(\V^D_0,S_D)$ is a filtered $(\V^D_0,\V^D)$--module.

F.J.~Calder{\'o}n--Moreno and L.~Narv{\'a}ez--Macarro \cite[Cor.\,4.2]{CN05} prove that LCT holds for a free divisor $D$ if and only if the complex
\[
\D\overset\LL\otimes_{\D(\log D)}\O(D)=\D\otimes_{\D(\log D)}\SC^\bullet_{\D(\log D)}(\O(D))
\]
is concentrated in degree $0$ and the natural multiplication morphism
\[
\xymat{
\D\otimes_{\D(\log D)}\O(D)\ar[r]^-{\epsilon_D}&\O(*D)
}
\]
is an isomorphism.
Injectivity of $\epsilon_D$ is locally equivalent to $\Ann_\D(\frac{1}{f})$ being generated by operators of order $1$ where $f\in\O$ such that $D=(f)$.
For any divisor $D$, T.~Torrelli proves that the latter condition already implies surjectivity of $\epsilon_D$ \cite[Prop.\,1.3]{Tor04} and conjectures that it is even equivalent to LCT \cite[Conj.\,1.11]{Tor04}.
A problem to verify this conjecture for a free divisor $D$ consists in $\D\otimes_{\D(\log D)}\SC^\bullet_{\D(\log D)}(\O(D))$ not being $F$--strict  in general \cite[Rem.\,4.2.4]{Cal99}.
So grading by $F$ does not reduce the problem to a commutative one.
But both properties of $D$ in question can be characterized in terms of $S_D$--strictness:
On the one hand, exactness of $\D\otimes_{\D(\log D)}\SC^\bullet_{\D(\log D)}(\O(D))$ in degree $k$ is equivalent to $S_D$--strictness of the differential of $\SC^\bullet_{\D(\log D)}(\O(D))$ from degree $k-1$ to degree $k$.
On the other hand, injectivity of $\epsilon_D$ is equivalent to $S_D$--strictness of the last differential of $\D\otimes_{\D(\log D)}\SC^\bullet_{\D(\log D)}(\O(D))$.

A solution of the LCT problem seems to require a deeper understanding of the V--filtration in general.
There are many questions:
\begin{itemize}
\item What are properties of the $\V^D_k$?
\item When is $\V^D_0$ generated by vector fields?
\item When is $\V^D_0$ locally finitely generated?
\item What are properties of the embeddings $\V^D_k\subset\D$?
\end{itemize}
We shall approach the first two questions in this article.

\section{V--filtration along subvarieties and divisors}\label{20}

Let $Y\subset X$ be a subvariety in $X$ and let $\I\subset\O$ be its ideal.
The V--filtration $\V^Y$ along $Y$ is the increasing filtration on $\D$ defined by
\[
\V^Y_k=\{P\in\D\mid\forall l\in\Z:P(\I^l)\subset\I^{l-k}\}
\]
for all $k\in\Z$.
We shall omit the index $Y$ if it is clear from the context.
Clearly $\V_k\cdot\V_l\subset\V_{k+l}$ for all $k,l\in\Z$.
Hence $\V_0$ is an $\O$--algebra and $\V_k$ is an $\V_0$--module for all $k\in\Z$.

\begin{exa}\label{2}
Let $x_1,\dots,x_m,y_1,\dots,y_n$ be coordinates on $X=\C^{m+n}$.
\begin{enumerate}
\item For the submanifold $Y=\{y=0\}$,
\[
\V^Y_k=\bigl\{P=\sum_{j_1-i_1+\cdots+j_n-i_n\le k}P_{i,j}(x,\partial_x)y_1^{i_1}\partial_{y_1}^{j_1}\cdots y_n^{i_n}\partial_{y_n}^{j_n}\in\D\bigr\}.
\]
\item For the normal crossing divisor $D=(y_1\cdots y_n)$,
\[
\V^D_k=\bigl\{P=\sum_{j_1-i_1,\dots,j_n-i_n\le k}P_{i,j}(x,\partial_x)y_1^{i_1}\partial_{y_1}^{j_1}\cdots y_n^{i_n}\partial_{y_n}^{j_n}\in\D\bigr\}.
\]
\end{enumerate}
\end{exa}

Denote the complement of the singularities of $Y$ by
\[
\xymat{
U_Y=X\backslash\Sing(Y)\ar[r]^-{i_Y}&X.
}
\]
We shall omit the index $Y$ if it is clear from the context.
The V--filtration along a divisor has a special property.

\begin{prp}\label{1}
Let $D\subset X$ be a divisor.
Then $\V^D=(i_D)_*i_D^{-1}\V^D$.
\end{prp}

\begin{proof}
We may assume that $D=(f)$ for some $f\in\O$ by the local nature of the statement.
Since $\V_k\subset\D$ and $\D$ is a locally free $\O$--module,
\[
i_*i^{-1}\V_k\subset i_*i^{-1}\D=\D.
\]
Since $\O\cdot f^{l-k}$ is a free $\O$--module, $P\in i_*i^{-1}\V_k$ implies
\[
P(g\cdot f^l)\in i_*i^{-1}(\O\cdot f^{l-k})=\O\cdot f^{l-k}
\]
for all $g\in\O$ and $l\in\Z$ and hence $P\in\V_k$.
\end{proof}

\begin{cor}\label{24}
Let $D=(f)\subset X$ with $f\in\O$ be a divisor.
Then
\[
\V_k=
\begin{cases}
f^{-k}\V_0, & k\le0,\\
f^{-k}(\V_0\cap f^k\D), & k\ge1.
\end{cases}
\]
\end{cor}

\begin{proof}
The equalities in question hold on $U_D$ by Example \ref{2} (2) and hence on $X$ by Proposition \ref{1}.
\end{proof}

Denote the symbol map for $F$ by
\[
\xymat{
\D\ar@{->>}[r]^-\sigma&\gr^F\D
}.
\]
The decomposition $F_1\D=\O\oplus\Theta$ defines the $\O$--module $\Der(\log Y)\subset\Theta$ of logarithmic vector fields along $Y$ by
\[
F_1\V_0=\O\oplus\Der(\log Y).
\]
This definition simplifies to
\[
\Der(\log Y)=\{\theta\in\Theta\mid \theta(\I)\subset\I\}
\]
by the Leibniz rule and implies involutivity of $\Der(\log Y)$, that is
\[
[\Der(\log Y),\Der(\log Y)]\subset\Der(\log Y).
\]

\begin{exa}\label{3}
Let $x_1,\dots,x_m,y_1,\dots,y_n$ be coordinates on $X=\C^{m+n}$.
\begin{enumerate}
\item For the submanifold $Y=\{y=0\}$,
\[
\Der(\log D)=\O\langle\partial_{x_1},\dots,\partial_{x_m}\rangle+\O\langle y_i\partial_{y_j}\mid1\le i,j\le n\rangle.
\]
\item For the normal crossing divisor $D=(y_1\cdots y_n)$,
\[
\Der(\log D)=\O\langle\partial_{x_1},\dots,\partial_{x_m},y_1\partial_{y_1},\dots,y_n\partial_{y_n}\rangle.
\]
\end{enumerate}
\end{exa}

Let $\O[\Der(\log Y)]\subset\D$ be the image of the tensor algebra
\[
\xymat{
\T_\C\Der(\log Y)\ar[r]^-{\gamma_Y}&\D.
}
\]
Then at least $\O[\Der(\log Y)]\subset\V_0^Y$.

\begin{cor}\label{7}
Let $D\subset X$ be a divisor.
Then $\V_0^D=\O[\Der(\log D)]$ if and only if $\O[\Der(\log D)]=(i_D)_*i_D^{-1}\O[\Der(\log D)]$.
\end{cor}

\begin{proof}
By Examples \ref{2} and \ref{3}, $\V_0^D=\O[\Der(\log D)]$ on $U_D$.
Hence the claim follows from Proposition \ref{1}.
\end{proof}

A divisor $D\subset X$ is called free if $\Der(\log D)$ is a locally free $\O$--module.
By K.~Saito \cite[Cor.\,1.7]{Sai80}, $\Der(\log D)$ is reflexive and hence all divisors in dimension $n=2$ are free.
By Example \ref{3} (2), normal crossing divisors are free.
In particular, any divisor $D$ is free on $U_D$.

F.J.~Calder\'on--Moreno \cite[Thm.\,1]{Cal99} proves that $\V^D_0=\O[\Der(\log D)]$ for a free divisor.
We give a simple proof of this result.

\begin{cor}
Let $D\subset X$ be a free divisor.
Then $\V^D_0=\O[\Der(\log D)]$.
\end{cor}

\begin{proof}
By Lemma \ref{23} and grading by $F$, $\O[\Der(\log D)]$ is a locally free $\O$--module and hence Corollary \ref{7} applies.
\end{proof}

\begin{lem}\label{23}
Let $R$ be a domain and let $P_1,\dots,P_n\in R\cdot T_1\oplus\dots\oplus R\cdot T_n=R^n$ be $R$--linearly independent.
Then $R[P_1,\dots,P_n]\subset R[T_1,\dots,T_n]$ is a polynomial ring.
\end{lem}

\begin{proof}
Write $P_i=\sum_j p_{i,j}T_j$ with $p_{i,j}\in R$.
Then by assumption $p=\det(p_{i,j})\ne0$ and hence $R_p[P_1,\dots,P_n]$ is a polynomial ring.
Since $R$ is a domain, $\xymat{R\ar[r]&R_p}$ is injective and hence $R[P_1,\dots,P_n]$ is a polynomial ring.
\end{proof}

In general it is not clear if, or under which conditions, $\V_0^Y$ is a locally finite $\O$--algebra.
Even to compute $F_k\V_0^Y$ is a problem since the definition involves infinitely many conditions.
The following result allows one to compute $F_k\V_0^D$ algorithmically.

\begin{prp}\label{5}
Let $x_1,\dots,x_n$ be coordinates on $X=\C^n$.
Let $D=(f)\subset X$ with $f\in\O$ be a divisor.
Then, for $P\in F_d\D$, $P\in\V^D_k$ if and only if
\begin{equation}\label{4}
\forall\alpha\in\N^n,l\in\N:|\alpha|+l\le d\Rightarrow P(x^\alpha f^l)\in\O\cdot f^{l-k}.
\end{equation}
\end{prp}

\begin{proof}
Let $0\ne P\in F_d\D$ and assume that condition (\ref{4}) holds.
For $l\in\N$, the vector space $\C[x_1,\dots,x_n]_{\le d-l}$ is invariant under $x\mapsto Ax+a$ for $a\in\C^n$ and $A\in\GL_n(\C)$.
Hence, at a smooth point $y$ of $D$, condition (\ref{4}) holds for coordinates $x_1,\dots,x_n$ at $y$ such that $\partial_{x_n}(f)(y)\ne0$.
Then $y_1,\dots,y_{n-1},t=x_1,\dots,x_{n-1},f$ are coordinates at $y$ such that
\[
\forall\beta\in\N^{n-1},l\in\N:|\beta|+l\le d\Rightarrow P_y(y^\beta t^l)\in\O_y\cdot t^{l-k}.
\]
Write $P_y=\sum_{|\beta|+l\le d}p_{\beta,l}\partial_y^\beta\partial_t^l$ with $p_{\beta,l}\in\O_y$ and choose $\gamma\in\N^{n-1}$ and $m\in\N$ such that $|\gamma|+m$ is minimal with $p_{\gamma,m}\ne0$.
Then
\[
\gamma!m!p_{\gamma,m}=P(y^\gamma t^m)\in\O_y\cdot t^{m-k}
\]
and hence $p_{\gamma,m}\partial_y^\gamma\partial_t^m\in\V_{k,y}$ by Example \ref{2} (2).
By increasing induction on $|\gamma|+m$, this implies $P_y\in\V_{k,y}$ for all $y\in U_D$ and hence $P\in\V_k$ by Proposition \ref{1}.
\end{proof}

\begin{exa}\label{9}
Let $x,y,z$ be coordinates on $\C^3$ and
\[
f=xyz(x+y+z)(x+2y+3z).
\]
Then $D=(f)\subset\C^3$ is a central generic hyperplane arrangement.
Let
\[
Q=(x+y+z)(x+2y+3z)(3zy^2\partial_y^2+(x+4y-3z)yz\partial_y\partial_z-4yz^2\partial_z^2).
\]
Then $Q\in F_2\V^D_0$ by a {\sc Singular} \cite{GPS05} computation using Proposition \ref{5}.
We shall see in Example \ref{16} that $Q\notin F_2\O[\Der(\log D)]$.
\end{exa}

There is another special property of the V--filtration along a divisor.

\begin{prp}
Let $D\subset X$ be a divisor.
Then $\depth_x(\V^D_k)\ge2$ for all $x\in X$ and $k\in\Z$.
\end{prp}

\begin{proof}
Let $x\in X$ and $D_x=(f)$ with $f\in\O_x$.
Since $\O_x$ is torsion free and $\depth(\O_x)\ge2$, there is an $\O_x$--sequence $a_1,a_2\in\m_x$ such that $a_1$ is different from all irreducible factors of $f$.
Let $P\in\V_{k,x}$ with $a_2\cdot P\in a_1\cdot\V_{k,x}\subset a_1\cdot\D_x$.
Then $P\in a_1\cdot\D_x$ since $\D_x$ is a free $\O_x$--module.
But $P(g\cdot f^l)\in\O_x\cdot f^{l-k}$ implies $(a_1^{-1}\cdot P)(g\cdot f^l)\in\O_x\cdot f^{l-k}$ by the choice of $a_1$ for all $g\in\O$ and $l\in\Z$ and hence $P\in a_1\cdot\V_{0,x}$.
Then $a_1,a_2\in\m_x$ is a $\V_{k,x}$--sequence and hence $\depth_x(\V_k)\ge2$.
\end{proof}

\section{Symmetric algebra of logarithmic vector fields}

The condition in Corollary \ref{7} is difficult to verify in general.
Therefore we focus on a case in which it still holds after grading by $F$.
There is a commutative diagram of graded algebras
\[
\xymat@C=-20pt{
&\T_\C\Der(\log Y)\ar@{->>}[ld]_-{\Sigma}\ar[rd]^-{\gr\gamma_Y}\\
\Sym_\O\Der(\log Y)\ar[rr]^-{\alpha_Y}\ar@{->>}[d]_-{\pi_Y}&&\gr^F\O[\Der(\log Y)]\\
\Rees_\O\Der(\log Y)\ar[rr]^\cong&&\O[\sigma(\Der(\log Y)]\ar@{^(->}[u]
}.
\]

\begin{lem}\label{12}
If $\alpha_Y$ is an isomorphism then
\[
\Sym_\O\Der(\log Y)=(i_Y)_*i_Y^{-1}\Sym_\O\Der(\log Y)
\]
implies $\O[\Der(\log Y)]=(i_Y)_*i_Y^{-1}\O[\Der(\log Y)]$.
\end{lem}

\begin{proof}
There is a commutative diagram
\[
\xymat@C=-40pt{
\Sym_\O\Der(\log Y)\ar[rr]\ar[d]^-{\alpha_Y}&&i_*i^{-1}\Sym_\O\Der(\log Y)\ar[d]^-{i_*i^{-1}\alpha_Y}\\
\gr^F\O[\Der(\log Y)]\ar@{^(->}[rr]\ar@{^(->}[rd]&&i_*i^{-1}\gr^F\O[\Der(\log Y)]\\
&\gr^Fi_*i^{-1}\O[\Der(\log Y)]\ar@{^(->}[ru]
}.
\]
Then the claim follows by induction on $\deg(P)$ for $P\in i_*i^{-1}\O[\Der(\log Y)]$.
\end{proof}

\begin{lem}\label{10}
$\alpha_Y$ is an isomorphism if and only if $\pi_Y$ is injective.
\end{lem}

\begin{proof}
Assume that $\pi_Y$ is injective.
An element of $\gr^F\O[\Der(\log Y)]$ is of the form $\sigma(\gamma_Y(P))$ where $P\in\T_\C\Der(\log Y)$.
Write $P=P_0\oplus\cdots\oplus P_d$ where $d=\deg(P)$.
If $\sigma(\gamma_Y(P))\notin\img\alpha_Y$ then $(\gr\gamma_Y)(P_d)=(\gr\gamma_Y)(P)=0$ and hence $P_d\in\ker\Sigma$ by injectivity of $\pi_Y$.
By definition of $\Sym_\O$, this implies that $P_d$ is in the two--sided ideal generated by the relations $\xi\otimes\eta-\eta\otimes\xi$ and $\xi\otimes(a\eta)-(a\xi)\otimes\eta$ where $\xi,\eta\in\Der(\log Y)$ and $a\in\O$.
But
\[
\gamma_Y(\xi\otimes\eta-\eta\otimes\xi)=\xi\eta-\eta\xi=[\xi,\eta]\in\Der(\log Y)
\]
by involutivity of $\Der(\log Y)$ and
\[
\gamma_Y(\xi\otimes(a\eta)-(a\xi)\otimes\eta)=\xi a\eta-a\xi\eta=[\xi,a]\eta=\xi(a)\eta\in\Der(\log Y).
\]
This means that
\begin{align*}
\xi\otimes\eta-\eta\otimes\xi&\equiv[\xi,\eta]\mod\ker\gamma_Y,\\
\deg(\xi\otimes\eta-\eta\otimes\xi)&>\deg([\xi,\eta]),\\
\xi\otimes(a\eta)-(a\xi)\otimes\eta&\equiv\xi(a)\eta\mod\ker\gamma_Y,\\
\deg(\xi\otimes(a\eta)-(a\xi)\otimes\eta)&>\deg(\xi(a)\eta).
\end{align*}
Hence $\gamma_Y(P)=\gamma_Y(P')$ and $\deg(P)<\deg(P')$ for some $P'\in\T_\C\Der(\log Y)$.
Then the claim follows by induction on $d=\deg(P)$.
\end{proof}

\begin{exa}\label{16}
Let $D$ and $Q$ be as in Example \ref{9}.
Then a {\sc Singular} \cite{GPS05} computation shows that $\pi_D$ is injective and that
\[
\sigma(Q)\notin\alpha_D\bigl(\Sym_\O^2\Der(\log D)\bigr).
\]
By Lemma \ref{10}, this implies $Q\notin F_2\O[\Der(\log D)]$ and hence, by Example \ref{9}, $\O[\Der(\log D)]\subsetneq\V^D_0$.
\end{exa}

By the following general statement, injectivity of $\pi_Y$ is equivalent to $\O$--torsion freeness of $\Sym_\O\Der(\log Y)$.

\begin{lem}\label{11}
Let $R$ be a domain and $M$ a finitely presented torsion free $R$--module.
Then the following are equivalent:
\begin{enumerate}
\item $\Sym_RM$ is $R$--torsion free.
\item $\Sym_RM$ is a domain.
\item $\xymat{\Sym_RM\ar@{->>}[r]^-{\pi_M}&\Rees_RM}$ is injective.
\end{enumerate}
\end{lem}

\begin{proof}
Assume that $\Sym_RM$ is $R$--torsion free.
Let $K=Q(R)$ be the fraction field of $R$.
Then $M\otimes_RK\cong K^d$ where $d=\rk(M)$.
By choosing a basis of $K^d$ and clearing denominators, one can embed $M\subset R^d$.
Then
\[
\Sym_R(M)\otimes_RK\cong\Sym_{R\otimes_RK}(M\otimes_RK)\cong\Sym_K(K^d)
\]
is a domain and hence $\Sym_RM$ is a domain since $R$ is a domain.
Applying $\Sym_R$ to the inclusion $M\subset R^d$ yields
\[
\xymat@C=-10pt{
\Sym_RM\ar[rr]^-\phi\ar@{->>}[rd]_-{\pi_M}&&\Sym_R(R^d)\\
&\Rees_RM\ar@{^(->}[ru]
}
\]
and $\ker(\phi)\otimes_RK=0$ since
\[
\Sym_R(M)\otimes_RK\cong\Sym_K(K^d)\cong\Sym_R(R^d)\otimes_RK.
\]
A presentation
\[
\xymat{
R^m\ar[r]^-{(a_{i,j})}&R^n\ar[r]&M\ar[r]&0
}
\]
of $M$ defines an isomorphism
\[
\Sym_RM\cong R[T_1,\dots,T_n]/J
\]
where $J=\langle\sum_ja_{i,j}T_j\rangle$ is a prime ideal since $\Sym_RM$ is a domain.
Since $\Sym_R(R^d)$ is a domain, $\ker\phi$ lifts to a prime ideal $Q\subset R[T_1,\dots,T_n]$.
Then $J\subset Q$, $Q\cap R=0$, and $J\otimes_RK=Q\otimes_RK$ implies $J=Q$ and hence $\ker\pi_M=\ker\phi=0$.
\end{proof}

\begin{exa}
Let $D_4\subset\C^4$ be the central generic hyperplane arrangement defined in Section \ref{18}.
Then one can compute that the coordinates are zero divisors on $\Sym_\O^2A_4$ at $0$.
By Lemmata \ref{10}, \ref{11}, and \ref{17}, this implies that $\alpha_{D_4}$ is not an isomorphism.
\end{exa}

A divisor $D\subset X$ is called Euler homogeneous if locally $\chi(f)=f$ for some $\chi\in\Der(\log D)$ and $f\in\O$ such that $D=(f)$.
In this case, $\chi$ is called an Euler vector field and
\[
\Der(\log D)\cong\O\cdot\chi\oplus\Ann_\Theta(f).
\]
If $\Der(\log D)\cong\O\cdot\chi\oplus A$ then $\Sym_\O\Der(\log D)\cong\Sym_\O(A)[\chi]$.
For an Euler homogeneous divisor $D$, this implies
\[
\Sym_\O\Der(\log D)\cong\Sym_\O(\Ann_\Theta(f))[\chi].
\]

\begin{prp}\label{13}
Let $D\subset X$ be a divisor such that $\Sym_\O\Der(\log D)$ is $\O$--torsion free.
Then $\V^D_0=\O[\Der(\log D)]$ follows from
\[
\Sym_\O\Der(\log D)=(i_D)_*i_D^{-1}\Sym_\O\Der(\log D).
\]
If $D$ is Euler homogeneous and $A=\Ann_\Theta(f)$ or $A\oplus\O\cdot\chi\cong\Der(\log D)$ then the latter is equivalent to $\Sym_\O A=(i_D)_*i_D^{-1}\Sym_\O A$.
\end{prp}

\begin{proof}
This follows from Corollary \ref{7}, Lemmata \ref{12}, \ref{10}, and \ref{11}, and the preceding remarks.
\end{proof}

\section{Depth and torsion of symmetric algebras}

Using a theorem of G.~Scheja \cite{Sch61} on extension of coherent analytic sheaves, we shall give sufficient conditions for $\V^D_0=\O[\Der(\log D)]$ in terms of the depth and torsion of the symmetric algebras in Proposition \ref{13}.

\begin{thm}\label{14}
Let $D\subset X$ be a divisor such that $\Sym_\O\Der(\log D)$ is $\O$--torsion free.
Let $Z\subset\Sing(D)$ be a closed subset such that $\V^D_0=\O[\Der(\log D)]$ on $X\backslash Z$.
Then $\V^D_0=\O[\Der(\log D)]$ on $X$ if
\[
\depth_z\bigl(\Sym^k_\O\Der(\log D)\bigr)\ge\dim_z(Z)+2
\]
for all $z\in Z$ and $k\in\N$.
In particular, this holds if $D$ is Euler homogeneous, $A=\Ann_\Theta(f)$ or $A\oplus\O\cdot\chi\cong\Der(\log D)$, and
\[
\depth_z\bigl(\Sym^k_\O A\bigr)\ge\dim_z(Z)+2
\]
for all $z\in Z$ and $k\in\N$.
\end{thm}

\begin{proof}
This follows from \cite[Satz\,I-III]{Sch61} and Proposition \ref{13}.
\end{proof}

We shall apply a criterion by C.~Huneke \cite{Hun81} for the torsion freeness of symmetric algebras.

\begin{prp}\label{22}
Let $R$ be a Noetherian domain and let
\[
\xymat@C=20pt{
0\ar[r]&R\ar[rr]^-{(a_1,\dots,a_m)^t}&&R^m\ar[r]&M\ar[r]&0
}
\]
be a resolution of $M$.
If $\grade(I)\ge k+1$ for $I=\langle a_1,\dots,a_m\rangle$ then
\[
\depth(I,\Sym_R(M))\ge k.
\]
\end{prp}

\begin{proof}
We proceed by induction on $k$.
By \cite[Prop.\,2.1]{Hun81}, $\grade(I)\ge 2$ implies that $\Sym_R(M)$ is $R$--torsion free.
If $k\ge2$ then $\grade(I/a)\ge k$ for some $a\in I$ and hence
\[
\xymat{
0\ar[r]&R/a\ar[rr]^-{([a_1],\dots,[a_m])^t}&&(R/a)^m\ar[r]&M/a\ar[r]&0
}
\]
is a resolution of $M$.
Since $\Sym_{R/a}(M/a)\cong\Sym_R(M)/a$, the induction hypothesis applies.
\end{proof}

\begin{thm}\label{21}
Let $D\subset X$ be an Euler homogeneous divisor and $A=\Ann_\Theta(f)$ or $A\oplus\O\cdot\chi\cong\Der(\log D)$.
Let $Z\subset\Sing(D)$ be a closed subset such that $\V^D_0=\O[\Der(\log D)]$ on $X\backslash Z$.
For $z\in Z$, let
\[
\xymat{
0\ar[r]&\O_z\ar[rr]^-{(a_{z,1},\dots,a_{z,m})^t}&&\O_z^m\ar[r]&A_z\ar[r]&0
}
\]
be a resolution of $A_z$ such that
\[
\grade(\langle a_{z,1},\dots,a_{z,m}\rangle)\ge\dim_z(Z)+3.
\]
Then $\V^D_0=\O[\Der(\log D)]$ on $X$.
\end{thm}

\begin{proof}
This follows from Theorem \ref{14}, Proposition \ref{22}, and \cite[Prop.\,2.1]{Hun81}.
\end{proof}

\begin{cor}
Let $X$ be a complex manifold of dimension $3$ and let $D\subset X$ be a divisor with only isolated quasi--homogeneous singularities.
Then $\V^D_0=\O[\Der(\log D)]$.
\end{cor}

\begin{proof}
We may assume that $X\subset\C^3$ is an open neighbourhood of $0$, $D=(f)$ with $f\in\O$, and $\Sing(D)=\{0\}$.
Let $x_1,x_2,x_3$ be coordinates on $X$.
Then $\partial(f)=\partial_1(f),\partial_2(f),\partial_3(f)\in\m_0$ is an $\O_0$--sequence.
Hence the Koszul--complex
\[
\xymat@C=10pt@R=10pt{
0\ar[r]&\O_0\ar[rr]^-{\partial(f)^t}&&\O_0^3\ar[rr]\ar[rd]&&\O_0^3\ar[rr]&&\O_0\ar[rr]&&\O_0/\langle\partial(f)\rangle\ar[r]&0\\
&&&&\Ann_{\Theta_0}(f)\ar[ru]\ar[rd]\\
&&&0\ar[ru]&&0
}
\]
is exact and induces a resolution of $\Ann_{\Theta_0}(f)$.
Then the claim follows from Theorem \ref{21} with $Z=\Sing(D)$.
\end{proof}

Our criterion also applies to some cases of non--isolated singularities.

\begin{exa}
Let $D_3\subset\C^3$ be the central generic hyperplane arrangement defined in Section \ref{18}.
Then $D_3$ is not a free divisor and has a non--isolated singularity at $0$.
By Lemma \ref{19} and Proposition \ref{17}, $A_3\cong\O^3/\O\cdot(x_1,x_2,x_3)$ and $A_3\oplus\O\cdot\chi\cong\Der(\log D_3)$.
Then, by Examples \ref{2} (2) and \ref{3} (2) on $\Sing(D_3)\backslash\{0\}$ and Theorem \ref{21} for $A=A_3$ and $Z=\{0\}$, $\V^{D_3}_0=\O[\Der(\log D_3)]$.
\end{exa}

Our approach may fail in dimension $n>3$ even for quasi--homogeneous isolated singularities.

\begin{exa}
Let $x_1,x_2,x_3,x_4$ be coordinates on $\C^4$ and
\[
f=x_1^2+x_2^2+x_3^2+x_4^2.
\]
Then $D=(f)\subset\C^4$ has a quasi--homogeneous isolated singularity at $0$.
One can compute that the coordinates are zero divisors on $\Sym_\O^2\Ann_\O(f)$ at $0$.
By Lemmata \ref{10} and \ref{11} this implies that $\alpha_{D}$ is not an isomorphism.
\end{exa}

\section{Example of generic hyperplane arrangements}\label{18}

We shall provide some background for the examples in the previous sections.
Let $x_1,\cdots,x_n$ be coordinates on $\C^n$ and
\[
f_n=x_1\cdots x_n(x_1+\cdots+x_n).
\]
Then $D_n=(f_n)\subset\C^n$ is a central generic hyperplane arrangement.
Let $\chi=\sum_ix_i\partial_i$ be the Euler vector field,
\[
\eta_{i,j}=x_ix_j(\partial_i-\partial_j)\in\Der(\log D_n)
\]
for $i<j$, and $A_n=\O\langle\eta_{i,j}\rangle$.
By J.~Wiens \cite[Thm.\,3.4]{Wie01},
\[
\Der(\log D_n)=\O\cdot\chi+A_n
\]
with a minimal number of generators.
Let
\[
\sigma_{i,j,k}=x_i\eta_{j,k}-x_j\eta_{i,k}+x_k\eta_{i,j}\in\syz(\eta_{i,j})
\]
for $i<j<k$ and choose a monomial ordering refining $\partial_1<\cdots<\partial_n$.

\begin{lem}\label{19}
$(\eta_{i,j})$ is a standard basis of $A_n$ and $\syz(\eta_{i,j})=\langle\sigma_{i,j,k}\rangle$.
\end{lem}

\begin{proof}
This follows from Buchberger's criterion \cite[Thm.\,1.7.3]{GP02}.
\end{proof}

\begin{prp}\label{17}
$\Der(\log D_n)=\O\cdot\chi\oplus A_n$.
\end{prp}

\begin{proof}
It suffices to verify that no syzygy of $\chi$ and the $\eta_{i,j}$ involves $\chi$.
One can obtain the syzygies from a standard basis computation \cite[Alg.\,2.5.4]{GP02}.
The first s--polynomials $x_k\chi-\eta_{k,n}$ and $x_j\eta_{i,k}-x_i\eta_{j,k}$ have a zero $\partial_n$ component.
Hence only a sequence of s--polynomials starting with $x_k\chi-\eta_{k,n}$ can contribute to syzygies involving $\chi$ and the coefficient of $\chi$ remains a monomial.
Each element in such a sequence has exactly one monomial involving $x_n$.
Since the $\partial_2,\dots,\partial_{n-1}$ are leading components of the $\eta_{i,j}$, the sequence terminates with a non--zero element $a_k\partial_1\equiv x^{\alpha_k}\chi\mod A_n$.
By the same reason, $\O\cdot\partial_1\oplus A_n$ is a direct sum and hence $x^{\alpha_j}a_k=x^{\alpha_k}a_j$.
This implies that the coefficient of $\chi$ in any syzygy is zero.
\end{proof}

\bibliographystyle{amsalpha}
\bibliography{cldo}

\end{document}


TITLE: Solving the wave equation, with boundary conditions, in the sense of distributions (Generalized functions)
QUESTION [7 upvotes]: After learning some distribution theory, I find that in my book, all PDEs given as examples are in free space (without any boundary conditions). I wonder if distribution theory can be used to tackle PDEs with boundary conditions.
To be more specific, let's consider this problem. Let there be a string of length $\pi$ with both ends fixed. Transverse waves can be produced on the string, satisfying the classical wave equation
$$
\partial^2_t u(x,t)=c^2\partial^2_x u(x,t).
$$
The boundary conditions are $u(0)=u(\pi)=0$.
Now, let impose a wired initial condition: let's pluck the string in the middle, so initially, the string is at rest, in the position
$$
u(x,0)=A(\pi/2-|x-\pi/2|), A\in \mathbb R.
$$
As one can see, the initial condition is not everywhere differentiable. However, $u$ can be seen as an element of $\mathcal D'(\mathbb R)$ or $\mathcal S'(\mathbb R)$, the space of (tempered) distributions. The differential equation therefore make sense in the sense of distributions.
Using Fourier transform and convolution, we can manage to get a solution, IF there are no boundary conditions. However, in this situation, I do not know how to state the boundary condition in term of distributions.
So, my question now is: can we make sense out of this problem, possibly in the sense of distributions, and solve the equation?
Edit: we can use Fourier series expansion to solve this, but then I don't feel it really a way of "understanding" how it really works - after all, the original equation ceases to make sense when it is not differentiable. I want to somehow have some formalism in making sense of the derivative of a function which is not differentiable. Possibly weak derivative?
Edit: Fourier transform over a bounded interval doesn't seem to be obvious to define; it appears that Fourier series are really easier.

REPLY [1 votes]: It is possible to use Fourier series here, by making use of the periodicity and the continuity of the initial condition. Indeed, the problem can be extended to $x$ in $\Bbb R$ by periodization (successive identical strings with fixed ends of length $\pi$). Consider the spatially $\pi$-periodic function $$u(x,t) = \sum_n c_n(t)\, \text{e}^{2\text{i}nx} \, ,$$ which is written as a spatial Fourier series. Its restriction to $x\in [0,\pi]$ may solve the problem if the boundary condition $\sum_n c_n(t) = 0$ is satisfied. Also, at time $t=0$, the Fourier coefficients $c_n(0)$ must be the Fourier coefficients of the initial condition $u(\cdot, 0)$. Similarly, their derivatives $c'_n(0)$ are the Fourier coefficients of the null function, i.e. $c'_n(0) = 0$. Injecting this function in the PDE gives
$$
c_n''(t) + 4n^2c^2 c_n(t) = 0
$$
for all $n$,
by uniqueness of Fourier series. Therefore, we have $c_n(t) = c_n(0)\cos(2nct)$. The same approach can be followed for other triangular signals, Gaussian signals, rectangular signals, etc.
Instead of the initial Fourier series, we could have written the Fourier transform representation $$u(x,t) = \frac1{2\pi} \int_{\Bbb R} \hat u(k,t)\, \text{e}^{\text{i}kx}\,\text d k \, ,$$ where $\hat u$ is the spatial Fourier transform of $u$.


TITLE: Cauchy but not convergent sequence in $L^1$
QUESTION [0 upvotes]: Show that the $\arctan(nx)$ is Cauchy but not convergent in the $1$-norm of the space $C^1 [-1,1]$, where the standard $1$-norm is $||f||_1 = \int \limits _0^1 |f(x)| dx$

REPLY [0 votes]: Hint: Write $f_n(x) = \arctan (nx)$. Try to show that $\{f_n\}$ converge in $L^1$ norm to the function
$$f(x) = \begin{cases} \pi/2 &\text{if } x\in [0,1], \\-\pi/2 &\text{if } x\in [-1,0). \end{cases}$$
That is, try to show
$$ \int_{-1}^1 |f_n(x) - f(x)| dx \to 0$$
as $n\to \infty$. You might consider splitting the integral into $[-1,0]$ and $[0,1]$ part.


TITLE: Clarification of solution set of a system of linear equations in three variables.
QUESTION [1 upvotes]: I am reviewing Jordan Canonical form:
I have a situation where I have two linearly independent eigenvectors for a 3x3 matrix and I am trying to find the third.
In the example it is saying that by solving the following:
$$ \begin{bmatrix}
    0 & 1 & 1  \\
    0 & 0 & 0 \\
    0 &0 & 0
  \end{bmatrix}     \cdot  \begin{bmatrix} x  \\y \\ z\end{bmatrix} = \begin{bmatrix} 1  \\0 \\ 0\end{bmatrix} $$
we will get the solution set $\,\,x \begin{bmatrix} 1  \\0 \\ 0\end{bmatrix} + y \begin{bmatrix} 0  \\1 \\ -1\end{bmatrix} + \begin{bmatrix} 0  \\0 \\ 1\end{bmatrix}$
But for whatever reason I am not understanding why that is the solution set. Looking for some explanation here.

REPLY [1 votes]: We get that the solutions are $\{(x,y,-y)+(0,0,1)|x,y\in\Bbb R\}$.  Just back substitute to get the solution to the associated homogeneous system; and note that $(0,0,1)$ is a particular solution.
That is, it's correct.  The general form for the solution of a linear system is $\text{general solution}=\{\text{particular solution} + \text{elements of the kernel}\}$.


TITLE: Evaluting $\lim_{(x,y) \to(0,0)} x \cdot \ln{(x^2+2y^2)}$
QUESTION [1 upvotes]: How can I find such limit:
$$\lim_{(x,y)\to(0,0)} x \cdot \ln{(x^2+2y^2)}$$?

REPLY [2 votes]: Using polar coordinates $x=\rho \cos \phi$ and $y=\rho \sin \phi$ we have
$$|\log (x^2+2y^2)|=|\log \rho^2(1+\sin^2 \phi)|\le 2|\log \rho|+|\log 2|$$
and
$|x|\le \rho$
Therefore,
$$|x\log(x^2+2y^2)|\le \rho \left(2|\log \rho|+|\log 2|\right)\to 0$$
as $\rho \to 0$.


TITLE: Showing two Joint Random Variables are Independent
QUESTION [0 upvotes]: I have two random variables, $X,Y\sim$ Geo$(p)$.  Define two more random variables, $U=\min{(X,Y)}$ and $V=X-Y$.  I am tasked with showing that $U$ and $V$ are independent.
This means I need the joint distribution $f_{U,V}(u,v)$.  So
$$f_{U,V}(u,v)=P(U=u,V=v)=P(\min{(X,Y)}=u, X-Y=v)$$
This is confusing me.  So i thougth I look at the support of $U$ and $V$.  Since $X,Y$ is geometric, then the minimum of $X$ and $Y$ is also over the same support of $X$ and $Y$ which would just be $\mathbb{N}$.  But the support of $V$ would then be $\mathbb{Z}$ since $V$ can take on negative values ($x=3, y=4$, for example).  So do I need to consider cases of $V$ then and compare the distributions of all three?
For example, if $V>0$, then $X>Y$ and $\min{(X,Y)}=Y$.  If $V<0$, then $Y>X$ and $\min{(X,Y)}=X$.  For the first case,
$$V>0\Rightarrow  P(\min{(X,Y)}=u, X-Y=v)=P(Y=u, X=v+y)=P(X=v+u,Y=u)$$
So since $X,Y$ are independent,
$$f_{U,V}(u,v)=P(X=v+u,Y=u+v)=P(X=u+v)P(Y=u)=p(1-p)^{u+v}p(1-p)^u$$
So $f_{U,V}(u,v)=p^2(1-p)^{2u+v}$.  Is this correct? And is this the correct approach to now consider $V<0$ and then $V=0$?

REPLY [1 votes]: $U$ and $V$ are independent if and only if $f_{U,V}(u,v)$ can be written as
$f(u)f(v)$.
Now our variables are
$$
X = \left\{ \begin{array}{cl}  U&V\geq 0\\
U+V& V<0\end{array}\right.\\
Y = \left\{ \begin{array}{cl}  U-V&V\geq 0\\
U& V<0\end{array}\right.
$$
Then using our knowledge of $f_X(x)$ and $f_Y(y)$ we can fill in  $f_{U,V}(u,v)$ as
$$
 f_{U,V}(u,v) = \left\{ \begin{array}{cl}  (1-p)^{u-1}p(1-p)^{u-v-1}p &v\geq 0\\
(1-p)^{u+v-1}p(1-p)^{u-1}p & v<0\end{array}\right.
$$
and these expressions can be written as
$$
 f_{U,V}(u,v) =   (1-p)^{2u+|v|-2}p^2
= \left[(1-p)^{2u-1}p \right] \left[(1-p)^{|v|-1}p \right]=f(u)f(v)
$$


TITLE: Connection on a Principal bundle and transition functions, as in Hitchin's notes
QUESTION [5 upvotes]: This is along the lines of this question

Gerbes are not just topological objects: we can do differential geometry with them too. We shall next describe what a connection on a gerbe is.
To begin with, let’s look at a connection on a line bundle which is given
by transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow S^1\subseteq \mathbb{C}^*$. A connection on consists of $1$ forms $A_\alpha$ defined on $U_\alpha$ such that on a twofold intersection $U_\alpha\cap U_\beta$ we have $iA_\alpha-iA_\beta=g_{\alpha\beta}^{-1}dg_{\alpha\beta}$

I know what is a connection $1$ form (on principal bundle) but not as in above version. I am trying to relate what I know with what is given here.
Let $(P,M,\pi)$ be a principal  $G$ bundle with $\mathfrak{g}=T_eG$ and transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G$.

Definition : A connection form on  $P$ is a $\mathfrak{g}$ valued $1$
form $\omega$ on $P$ such that

$\omega(p)(A^*(p))=A$ for all $A\in \mathfrak{g}$ and $p\in P$
$(\delta_g^*\omega)(p)(v)=Ad_{g^{-1}}(\omega(p)(v))$ for all $p\in P,g\in G$ and $v\in T_pP$.

Given a connection $1$ form $\omega$ on $P$ I am trying to associate a collection  of $1$ forms $\{A_\alpha\}$ with some compatability conditions (here $A_\alpha$ is a $1$ form on $U_\alpha$).
Given local trivialization $\psi_\alpha$, we have a section of $\pi$ namely $\sigma_\alpha:U_\alpha\rightarrow P$ defined as $\sigma_\alpha(x)=\psi_\alpha^{-1}(x,e)$. Given local trivialization $\psi_\beta$, we have a section of $\pi$ namely $\sigma_\beta:U_\beta\rightarrow P$ defined as $\sigma_\beta(x)=\psi_\beta^{-1}(x,e)$.
Suppose $x\in U_\alpha\cap U_\beta$ then, we have $\sigma_\alpha(x)\in \pi^{-1}(x)$ and $\sigma_\beta(x)\in \pi^{-1}(x)$. Thus, there exists $g\in G$ (depending on $x$) such  that $\sigma_\beta(x)=\sigma_\alpha(x)g$. Given $x\in U_\alpha\cap U_\beta$ there is an obvious choice for an element of $G$ namely $g_{\alpha\beta}(x)$. I  see that we have $\sigma_\beta(x)=\sigma_\alpha(x)g_{\alpha\beta}(x)$ for all $x\in U_{\alpha}\cap U_\beta$ i.e., $\sigma_\beta=\sigma_\alpha g_{\alpha\beta}$.
Given a $1$ form $\omega$ on $P$ and we can pull back $\omega$ to $U_\alpha$ under $\sigma_\alpha$ to get $1$ form $\omega_\alpha=\sigma_\alpha^*\omega$ on $U_\alpha$ similarly we can pull back to $U_\beta$ to get $1$ form $\omega_\beta=\sigma_\beta^*\omega$  on $U_\beta$.
As $\sigma_\alpha$ and $\sigma_\beta$ are related by $\sigma_\alpha=\sigma_\beta g_{\alpha\beta}$, one can expect that $\omega_\alpha$ and $\omega_\beta$ are related some how. Given $g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G$ we can produce a $1$ form on $U_\alpha\beta$ as pull back of $\theta$ on $G$ i.e., the canonical $1$ form on $G$ which is a left invariant $1$ form determined by $\theta(e)(A)=A$ for all $A\in \mathfrak{g}$. Let us denote pull back of $\theta$ to $U_\alpha\cap U_\beta$ by $\theta_{\alpha\beta}$. Then, I am expecting some compatibility  relation between $1$ forms $\omega_\alpha,\omega_\beta$ and $\theta_{\alpha\beta}$ that should come from
$\sigma_\alpha=\sigma_\beta g_{\alpha\beta}$.
EDIT : I could see in Kobayshi and Nomizu that these local connection forms $\sigma_\alpha$ are related by
$$ \sigma_\beta(p)(X_p)=ad(g_{\alpha\beta}(p)^{-1})(\sigma_\alpha(p)(X_p))+\theta_{\alpha\beta}(p)(X_p) \text{ on } U_\alpha\cap U_\beta.$$
In short, we have
$$\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+\theta_{\alpha\beta} \text{ on } U_\alpha\cap U_\beta$$
I see that people use $g_{\alpha\beta}^{-1}dg_{\alpha\beta}$ (notation of Maurer Cartan differential of a map $g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G$) to denote what we have called as $\theta_{\alpha\beta}$. So, in this notation, we have
$$\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+g_{\alpha\beta}^{-1}d g_{\alpha\beta}\text{ on } U_\alpha\cap U_\beta.$$
Question : I am trying to understand how $\sigma_\beta=ad(g_{\alpha\beta}^{-1})\sigma_\alpha+g_{\alpha\beta}^{-1}d g_{\alpha\beta}\text{ on } U_\alpha\cap U_\beta$ on principal bundles has its similar part as $iA_\alpha-iA_\beta=g_{\alpha\beta}^{-1}dg_{\alpha\beta}$ in case of line bundles.
Any suggestion/reference on how to see this is welcome.

REPLY [4 votes]: A  principal $G$-bundle $P\to M$ can be described by an open cover $(U_\alpha)$ of $M$  and a cocycle  $g_{\beta\alpha}: U_\alpha\cap U_\beta\to G$.  The total space  is  the quotient of the disjoint union of the spaces $U_\alpha\times G$ via the equivalence relation
$$ U_\alpha\times g\ni (x, g')\sim (y, g'')\in U_\beta\times G $$
iff
$$ x=y,\;\;g''=g_{\beta\alpha}(x)\cdot g'. $$
A section of a $G$ bundle is then a collection of smooth maps
$$\sigma_\alpha: U_\alpha\to G $$
satisfying the gluing conditions
$$ \sigma_\beta(x)=g_{\beta\alpha}(x)\cdot \sigma_\alpha(x),\;\;\forall x\in U_\alpha\cap U_\beta. $$


TITLE: Problem with combination of discrete and continuous random variables
QUESTION [2 upvotes]: I'm pretty new to probability and a question is giving me some troubles.

A binary information source produces $0$ and $1$ with equal probability. The output of the source, denoted as $X$, is transmitted via an additive white Gaussian noise (AWGN) channel. The output of the channel, denoted as $Y$, satisfies $Y = X + N$, where the random noise $N$ has the distribution $N(0,0.1)$.
a) If the channel outputs $0.2$, what is the probability that $0$ is the source output and what is the probability that $1$ is the source output
b) If the channel output is $0.2$ and you are required to make a guess on the source output, what is your guess and why?

I've tried solving this directly and looking at it generally, and from both perspectives it isn't making a lot of sense to me.
Directly, it's asking to find $P(X=0 \mid Y = 0.2)$ which I assume can be solved for like any other conditional problem, using
$$f_{X\mid Y}(x\mid y) = \frac{f(x,y)}{f_Y(y)}$$
Since X seems like it's best expressed using a Bernoulli distribution, I don't know how to actually evaluate this at all. To find the joint PDF of X and Y (which I think I need to find the conditional probability of X given Y = y) I would need to combine a Bernoulli distribution and a Normal distribution. I can't find any examples of this and I'm not entirely sure it's possible at all seeing as one is discrete and one is continuous. This is where my solution falls apart.
Looking at it more generally though, I have a feeling that there might not be a solution at all. The source is perfectly binary, so it will only produce EXACTLY 0 or 1. And the output from the channel has to be EXACTLY 0.2. For that to be true, doesn't the noise have to be EXACTLY 0.2 or 0.8? And since the noise is a continuous distribution, isn't the probability of either case 0? Obviously it's more likely for N to be around 0.2 than around 0.8, but I can't think of how to actually evaluate the difference since they're both exact numbers and there is no region/error given or anything. And the fact that part b) implies that part a) have an ambiguous answer makes me think this might be true. The problem is I'm not really comfortable with probability, so I don't know if my logic is sound here or if I'm missing something.
Any help is greatly appreciated

REPLY [1 votes]: The density of $Y$, given that $X=0$ is $\text{constant}\cdot e^{-y^2/(2(0.1))}$ (assuming you meant $0.1$ is the the variance, not the standard deviation.  Then density of $Y$, given that $X=1$, is $\text{constant}\cdot e^{-(y-1)^2/(2\cdot0.1)}$.  So the likelihood function is given by
\begin{align}
L(x) & = \text{constant}\cdot \left.\begin{cases} e^{-y^2/(2\cdot0.1)} & \text{if }x=0, \\  e^{-(y-1)^2/(2\cdot0.1)} & \text{if }x=1,  \end{cases} \right\} \\[10pt]
& = \text{constant}\cdot e^{-y^2/(0.2)} \left.\begin{cases} 1 & \text{if }x=0, \\ e^{(2y-1)/0.2} & \text{if } x=1.\end{cases}\right\}
\end{align}
The expression in front doesn't depend on whether $x=0$ or $x=1$, so it's a "constant" as a function of $x$; it doesn't change as $x$ changes.
To get the posterior distribution, multiply the prior by the likelihood and then normalize (and remember that the prior is uniform in this case):
$$
\begin{cases} \dfrac{1}{1+e^{(2y-1)/0.2}} & \text{if }x=0 \\[12pt]
\dfrac{e^{(2y-1)/0.2}}{1+e^{(2y-1)/0.2}} & \text{if }x=1  \end{cases}
$$
Now plut in $y=0.2$ and you've got it.
One can express this as
$$
\operatorname{logit}P(X=1\mid Y=y) = \frac{2y-1}{0.2} = 10y-5,
$$
where
$$
\operatorname{logit} p = \log_e \frac{p}{1-p}.
$$


TITLE: Algorithms for finding a Tutte-Berge maximizer
QUESTION [3 upvotes]: Let $G=(V,E)$ be a graph. The Tutte-Berge formula states that a maximum matching on $G$ has size $$\nu(G)=\frac12\left(|V|-\max_{U\subset V}(o(V-U)-|U|)\right)$$ with $o(V-U)$ the number of connected components of $V-U$ of odd size.
Question: How much is known about algorithms for finding $U\subset V$ maximizing $o(V-U)-|U|$? I am having trouble finding any resources about this question.
I did come up with a polynomial time algorithm by myself. I determined that you can construct such $U$ by iteratively finding a vertex $v\in V$ such that $\nu(V-\{v\})=\nu(V)-1$. This requires solving $O(|V|^2)$ maximum matching problems, and thus runs in polynomial time. However, I would like to know if there is any faster algorithm.

REPLY [1 votes]: It is not necessary to do $O(|V|^2)$ passes of a maximum matching algorithm; a single pass is enough if we're careful about it, use the right algorithm, and look at the additional data that algorithm generates.
Suppose you go looking for a maximum matching in $G$ using the blossom algorithm. This takes us through some number of steps of finding augmenting paths and contracting odd cycles (blossoms). At the end of the algorithm, we find a maximum matching $M'$ in some smaller graph $G'$, and by expanding the blossoms again we can obtain a maximum matching $M$ in $G$. At this stage:

Right before expanding the blossoms, we can find a Tutte–Berge maximizer in $G'$;
After we expand the blossoms, it remains a Tutte–Berge maximizer in $G$.

Finding a Tutte–Berge maximizer in $G'$
Here, the blossom algorithm has built a forest $F$ in $G'$ (called an $M'$-alternating forest) in which each tree component has a root in $S$: the set of vertices not covered by the matching. We say that a vertex is an inner vertex if its distance in $F$ from a root is odd, and an outer vertex if the distance is even. By the algorithm, all inner vertices have degree $2$ in $F$: they have one parent and one child, and the edge from each inner vertex to its child is an edge of $M'$.
This much should hold at every step of the blossom algorithm. When we've reached the end, we also know the following:

No outer vertex has neighbors outside $F$; otherwise, we could use those neighbors to expand the forest.
Two outer vertices cannot be adjacent. If there were an edge between two outer vertices in different components of $F$, we would find an $M'$-augmenting path; if there were an edge between two outer vertices in the same component of $F$, we would find a blossom.

In this case, let $U$ be the set of inner vertices: this set $U$ is a Tutte–Berge maximizer in $G'$! Why? Well, by the observations above, the outer vertices can only have neighbors in $U$, so in particular, they are odd components of $G' - U$. The number of outer vertices is exactly $|U| + |S|$, because each outer vertex in the forest $F$ is either a root or the unique child of an inner vertex. Therefore $o(G' - U) - |U|$ is exactly $|S|$, the number of vertices uncovered by a maximum matching in $G'$.
Returning to $G$
We wanted to find a Tutte–Berge maximizer in $G$, not $G'$. To do this, we expand the blossoms, one at a time, and see what happens to $U$.
Each blossom is an odd cycle containing $k$ inner vertices and $k+1$ outer vertices of an alternating forest; after the blossom is contracted, it becomes a single outer vertex of an alternating forest in the smaller graph. Therefore, as we expand the blossoms, the set $U$ is left untouched: it started out consisting of inner vertices only, and as we expand, it continues to consist of inner vertices only (though it will no longer contain all the inner vertices).
Meanwhile, the odd components in the complement of $U$ will remain odd components. As we expand the blossoms, they might get bigger. However, they'll remain isolated from each other, and replacing a vertex by an odd cycle will not change their parity.
Therefore $o(G-U) - |U|$ in the original graph $G$ will have the same value as $o(G'-U) - |U|$ did in $G'$. The number of vertices uncovered by the maximum matching also remains the same: expanding a blossom of length $2k+1$ adds $2k$ vertices, but also adds $k$ edges to the matching. We conclude that $U$ is still a Tutte–Berge maximizer in $G$.
A worse but lazier solution
If we stick to treating the blossom algorithm (or any alternative) as a black box, we can still do better and identify a Tutte–Berge maximizer after only solving $O(|V|)$ maximum matching problems.
Let $D$ be the set of all inessential vertices of $G$: the vertices that are left uncovered by some maximum matching in $G$. A vertex $v \in D$ can be identified by the property that $G-v$ and $G$ have maximum matchings of the same size. Next, split $V(G)-D$ into sets $A$ and $C$, defined to be the essential vertices adjacent to some vertex in $D$, and the other essential vertices.
The partition $A \cup C \cup D$ is the Gallai–Edmonds decomposition of $G$. In particular, the set $A$ is a Tutte–Berge maximizer.