text:
\section{Introduction}
We consider a Brownian particle bound to a surface which contains a small trap denoted $\Gamma_\varepsilon$. The narrow capture problem deals with the time required for such a particle to first encounter the trap. This time is called the first-passage time and is denoted $\tau_{\Gamma_{\varepsilon}}$. Starting from an initial location $x$ on the surface, the \emph{expected time} that a particle will wander before being captured by the trap is called the mean first-passage time and is denoted $u_\varepsilon(x)$.

The narrow capture problem along with the closely related narrow escape problem (where the traps are small windows on the otherwise reflective boundary of the search domain) have been used as simple, prototypical models for various processes involving diffusive search such as a diffusing ion inside a cell escaping through an ion channel on the cell membrane (see \cite{pillay2010asymptotic, holcman2004escape, schuss2007narrow, benichou2008narrow, bressloff2013stochastic} and references therein). While we highlight some results below, we refer the reader to \cite{holcman2015stochastic, bressloff2014stochastic, HS2014} for a more comprehensive description of results along with their applications to cellular and molecular biology.

On a flat, bounded domain $\Omega$, it was shown in \cite{holcman2004escape} and \cite{singer2006narrow} that the mean escape time had the leading order expansion $u_\varepsilon \sim -|\Omega|\log\varepsilon + O(1)$ as $\varepsilon \to 0$.
Within \cite{Singer2006}, similar expansions were derived for the sphere and annulus with absorbing windows near singular points. More precisely, it was found that in geometric configurations where the absorbing window was near a corner, the mean escape time had an expansion of the form $u_\varepsilon \sim \frac{|\Omega|}{\alpha}\log\varepsilon + O(1)$ where $\alpha$ denotes the angle of the corner. Furthermore, it was found that when the absorbing window was near a cusp point, the mean escape time had an expansion given by $u_\varepsilon \sim \frac{|\Omega|}{\varepsilon}+O(1)$. It is emphasized that in the above works, the leading order term was determined and the extent to which the remainder terms were understood was $O(1)$.

In \cite{pillay2010asymptotic}, a matched asymptotic method was employed to determine the $O(1)$ term in the expansion in terms of a certain Green's function that encoded information on the geometry of $\Omega$, the locations of the traps, and the initial position $x$. This method, developed in \cite{ward1993strong}, effectively summed all logarithmic correction terms in the expansion of $u_\varepsilon$, with the resulting error term being transcendentally small in $\varepsilon$. In \cite{bressloff2015escape}, a more detailed model was considered in which the windows were allowed to open and close stochastically, more closely mimicking the behavior of cell ion channels. Other quantities of interest aside from the mean first passage time include the variance of the first passage time \cite{KTCW2015} as well as the so-called extreme first passage time, the minimum search time achieved by a large group of searchers \cite{madrid2020competition}.

In three dimensions, the narrow escape time from a Euclidean, bounded domain $\Omega$ with one circular trap of radius $\varepsilon$ on its boundary was shown \cite{singer2006narrow, singer2008narrow} to have the leading order expansion $u_\varepsilon \sim |\Omega|(4\varepsilon)^{-1}(1-\varepsilon \pi^{-1} H\log\varepsilon ) + O(1)$, where $H$ is the mean curvature of $\partial \Omega$ at the center of the trap. In \cite{cheviakov2010asymptotic}, a matched asymptotic analysis similar to that employed in \cite{pillay2010asymptotic} was used to compute the $O(1)$ correction term in the expansion for the special case in which $\Omega$ was the unit ball. In \cite{NTT2021}, the calculation of this correction term was generalized using geometric microlocal methods to Riemannian 3-manifolds with smooth boundary containing either a circular or elliptical trap.

The narrow capture problem also has wide applications in cellular biology \cite{CW2011,CSW2009}. For example, a diffusing molecule must arrive at a localized signaling region within a cell or on its surface before a signaling cascade can be initiated. In another example, a T cell may diffuse in search of an antigen-presenting cell to trigger an immune response.  In this latter example, determining the duration of this search  is relevant to understanding immune response time \cite{CSW2009, delgado2015conditional}.

The matched asymptotic methods of \cite{ward1993strong} that were used for the aforementioned narrow escape problems have also been successfully applied to narrow capture problems in Euclidean metrics. In earlier works, the closely related problem of computing the fundamental Neumann eigenvalue $\lambda_0$ for the Laplacian in Euclidean two- and three-dimensional domains with small traps was considered in \cite{cheviakov2011optimizing, kolokolnikov2005optimizing}. The spatial average of the mean first passage time was shown to be proportional to $1/\lambda_0$ (see \cite{kolokolnikov2005optimizing, ward1993strong}), and in \cite{RC2019}, a numerical algorithm was employed to optimize this quantity with respect to configurations of traps located in the domain.

Extensions of these works include computing the full probability distribution (i.e., all moments as opposed to just the first)  of the first passage time \cite{lindsay2016hybrid, B20212d, B20213d, bressloff2022narrow}, stochastic resetting \cite{B20212d, B20213d, bressloff2022narrow}, moving traps \cite{tzou2015mean, lindsay2017optimization, iyaniwura2021simulation}, partially absorbing traps \cite{LBW2017, bressloff2022narrow}, traps grouped in clusters \cite{KTCW2015, iyaniwura2021asymptotic}, and the effect of advection \cite{KTCW2015, NTT2021force}.

In a non-Euclidean  geometry, \cite{CSW2009} considered the mean first passage time of a Brownian particle on a sphere containing small absorbing traps. Explicit results were obtained through employment of the aforementioned matched asymptotic method along with the known analytic formula for the Neumann Green's function for the Laplacian on sphere. The spherical geometry considered was meant to approximate the geometry of a cell with receptor clusters on its surface awaiting the arrival of surface-bound signaling molecules. A more detailed model of a cell, however, would be non-spherical. In fact, a cell's geometry  can be crucial to the manner in which it serves its function \cite{fernandes2019cell}. It is with this motivation that we develop here a rigorous mathematical framework for narrow capture problems posed on non-Euclidean and non-spherical geometries.

We now mathematically formulate the narrow capture problem. Let $(M,g)$ be a compact, connected, orientable, Riemannian surface with smooth boundary, $\partial M$. First, we assume the boundary to be empty and calculate the associated mean sojourn time as well as its spatial average. When the manifold is of non-empty boundary, we assume, without loss of generality that $M$ is a connected open subset of a compact orientable, Riemannian manifold $(\tilde{M}, g)$ without boundary. Let $(X_t, \mathbb{P}_t)$ be the Brownian motion on $M$ generated by the (negative) Laplace-Beltrami operator $\Delta_g=-d^*d$. We use $\Gamma_\varepsilon\subset M$ to denote a trap with radius $\varepsilon>0$ and we denote by $\tau_{\Gamma_{\varepsilon}}$ the first time the Brownian motion $X_t$ hits $\Gamma_{\varepsilon}$, that is
$$\tau_{\Gamma_{\varepsilon}} := \inf \{ t\geq 0: X_t \in \Gamma_{\varepsilon}\}.$$
Within the narrow capture problem we wish to derive an asymptotic as $\varepsilon \to 0$ for the mean first-passage (sojourn) time which is defined as the expected value: $$u_\varepsilon(x) = \mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x].$$
An associated quantity of interest is the spatial average of the mean first-passage time:
$$|M_\varepsilon|^{-1} \int_{M_\varepsilon} \mathbb{E}[\tau_{\Gamma_{\varepsilon, a}} | X_0 = x] dvol_g(x),$$
where $M_\varepsilon:= M\setminus \Gamma_\varepsilon$ and $|M_\varepsilon|$ is the Riemannian volume of $M_\varepsilon$ with respect to the metric $g$.

Many works have been devoted to this topic, especially in applied mathematics. In \cite{LL1986}, the mean first-passage time for diffusing particles on a surface of the sphere with one absorbing trap was considered. They obtained the asymptotic, up to the bounded term, for mean first passage time and its average. These results were generalised, in \cite{CSW2009}, for the case of several traps. In \cite{CW2011,LBW2017}, the three-dimensional version of this problem was studied. For domains in $\mathbb{R}^3$, they obtain the asymptotic formulas in terms of capacitance, by using the method of matched asymptotic expansions. We also mention works \cite{BEW2008,RC2019,TW2000,IWMW2021,B20212d,B20213d,SWF2007}, where the authors investigate related problems.

Despite the large number of works on this topic, there are still many questions regarding more general geometries. In this direction, the goal of this paper is to investigate the narrow capture problem for the Riemannian surface. Similar to \cite{AKK2012}, we use a layer potential method, however by adjoining this method with techniques originating from geometric microlocal analysis, we can extend the results, as well as the method to more general geometries, similar to the extension to broader classes of geometries for the narrow escape problem in \cite{NTT2021,NTT2021force}. As mentioned previously, we will consider empty and non-empty boundary cases. For the sake of conciseness, we will present the results and required Green's function for the $\partial M = \emptyset$ case here. For the associated \textit{Neumann Green's function} and results for the $\partial M\neq \emptyset$, see Section 5.
\begin{equation}\label{G function}
\Delta_g E(x,y) = - \delta_y(x) + \frac{1}{|M|}, \quad E(x,y) = E(y,x), \quad \int_M E(x,y) dvol_g(y) = 0.
\end{equation}
It was already known, see for example \cite{NTT2021} and \cite{taylor2}, that near the diagonal the Green function satisfies
\begin{equation}\label{dec}
	E(x,y) = -\frac{1}{2\pi} \log d_g(x,y) + P_{-4}(x,y),
\end{equation}
for some $P_{-4}(x,y) \in C^1(M\times M)$ which is infinitely smooth away from the set $\{(x,y)\in M\times M\mid x = y\}$. (In fact, in the language of pseudodifferential operators, we will see that $P_{-4}(x,y)$ is the Schwartz kernel of a pseudodifferential operator of degree $-4$.)

This expansion allows us to obtain the following asymptotic for the narrow capture of Brownian particles in a small trap:
\begin{theorem}\label{main}
	Let $(M,g)$ be a closed orientable Riemannian surface. Fix $x_0\in M$ and let $\Gamma_{\varepsilon}:=B_\varepsilon(x_0)$ be a geodesic ball centred at $x_0$ of geodesic radius $\varepsilon >0$.\\
	\\
	i) For each $x\notin B_\varepsilon(x_0)$, the first-passage time satisfies the  following asymptotic formula, as $\varepsilon\rightarrow 0$,
	\begin{equation*}
	\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x] = -\frac{|M|}{2\pi}\log\varepsilon + |M|P_{-4}(x_0,x_0) - |M| E(x,x_0) + r_{\varepsilon}(x) + O(\varepsilon\log\varepsilon).
	\end{equation*}
	for some function $r_\varepsilon$ such that $\|r_\varepsilon\|_{C(K)} \leq C_{K} \varepsilon$ for any compact $K\subset M$ for which $K\cap \Gamma_\varepsilon = \emptyset$. The Green function $E(x,y)$ is given by \eqref{G function} and $P_{-4}(x_0,x_0)$ is the evaluation at $(x,y)=(x_0,x_0)$ of the $C^1(M\times M)$ function $P_{-4}(x,y)$ in \eqref{dec}.\\
	\\
	ii) Let $M_\varepsilon=M\setminus \Gamma_{\varepsilon}$, then the spatial average of the mean first-passage time satisfies the asymptotic formula, as $\varepsilon\rightarrow 0$,
	\begin{equation*}
	\frac{1}{|M_\epsilon|}\int_{M_{\varepsilon}}\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = y]dvol_g(y) = -\frac{|M|}{2\pi}\log\varepsilon +|M|P_{-4}(x_0,x_0) + O(\varepsilon\log\varepsilon).
	\end{equation*}
\end{theorem}

In Section $5$ Theorem \ref{main2} we will prove a similar result for $\partial M\neq \emptyset$ with reflection boundary conditions for the Brownian motion. In this setting, we will use instead the Neumann Green's function $E(x,y)$. See Section \ref{proof of thm with bry} for details.

This paper is structured in the following manner. In Section \ref{Prelim}, we introduce some notation and the geometric framework with which we will be operating. Section \ref{Green function} deals with investigating the singular structure of the Green's function on a Riemannian surface without boundary. In Section \ref{proof of thm without bry}, We make use of the derived Greens function to prove Theorem \ref{main}. In Section \ref{proof of thm with bry}, we consider the analogous problem in the setting of a manifold with boundary and impose a reflecting boundary condition for our Brownian motion. The result will be stated in Theorem \ref{main2}.

\section{Preliminaries}\label{Prelim}
Throughout this paper, $(M,g)$ be a compact connected orientable Riemannian surface with smooth boundary, $\partial M$ which could be empty.The corresponding volume form and geodesic distance are denoted by $dvol_g$ and $d_g(\cdot,\cdot)$, respectively.  By $|M|$ we denote the volume of $M$.

For fixed $x\in M$, we will denote by $B_\rho(x)$ the geodesic ball of radius $\rho>0$ centred at $x$. In what follows $\rho$ will always be smaller than the injectivity radius of $( M, g)$ and the distance from $x$ to $\partial M$. We let $\mathbb{D}_\rho$ be the Euclidean ball in $\mathbb{R}^2$ of radius $\rho$ centred at the origin.

In this work, we will often use the geodesic coordinates constructed as follows. For fixed $x_0\in M$ and orthonormal tangent vectors $E_1, E_2 \in T_{x_0}M$, write $t=(t_1,t_2) \in \mathbb{D}_\rho$ and define
\begin{equation}\label{geod coord}
x(t;x_0):= \exp_{x_0}(t_1E_1 + t_2E_2)
\end{equation}
where $\exp_{x_0} (V)$ denotes the time $1$ map of $g$-geodesics with initial point $x_0$ and initial velocity $V\in T_{x_0} M$. The coordinate $t\in \mathbb{D}_\rho \mapsto x(t; x_0)$ is then an $g$-geodesic coordinate system for a neighbourhood of $x_0$ on $M$.

We will also use the re-scaled version of this coordinate system. For $\varepsilon >0$ sufficiently small we define the (re-scaled) $g$-geodesic coordinate by the following map
\begin{equation}\label{res coord}
x^{\varepsilon}(\cdot ; x_0) : t = (t_1, t_2) \in \mathbb{D} \mapsto x(\varepsilon t; x_0) \in B_\varepsilon(x_0),
\end{equation}
where $ \mathbb{D}$ is the unit disk in $\mathbb{R}^2$.

In the subsequent sections, we denote the centre of the "trap" by $x_0\in M$, which will be considered as fixed. We will use the following notations. We set $\Gamma_\varepsilon := B_{\varepsilon}(x_0)$
\begin{equation*}
M_{\varepsilon}= M_\varepsilon(x_0) := M \setminus \Gamma_\varepsilon
\end{equation*}
and denote by $h=h(\varepsilon,x_0)$ the metric on $\partial M_\varepsilon$, induced by the trivial embedding of $\partial M_\varepsilon$ into $M_\varepsilon$. The corresponding volume form is denoted by $dvol_h$. Further, we set $z\in \partial M_\varepsilon \mapsto \nu_z$ to be an outward pointing normal for $M_\varepsilon$. Finally, we let $|M_\varepsilon|$, $|\partial \Gamma_\varepsilon|$ be the volumes of $M_\varepsilon$ and $\partial \Gamma_\varepsilon$ with respect to $g$ and $h$.

\section{Green's function}\label{Green function}
 Within this section we assume that $\partial M$ is empty and we consider the Green function on $M$, which is the fundamental solution to the Laplace equation:
\begin{equation}
\label{no boundary green}
\Delta_g E(x,y) = - \delta_y(x) + \frac{1}{|M|}, \quad E(x,y) = E(y,x), \quad \int_M E(x,y) dvol_g(y) = 0.
\end{equation}

For a fixed $x_0\in M$ and set $\Gamma_\varepsilon = B_{\varepsilon}(x_0)$ we consider the following function
\begin{equation*}
	I_{\varepsilon}(x_0,x) := \int_{\Gamma_\varepsilon} E(x,y) dvol_g(y).
\end{equation*}
for $x\in M_\epsilon$. We will need to know about the singular behaviour of $\partial_{\nu_x} E(\cdot,\cdot)$, $I_\varepsilon(x_0,\cdot)$, and $\partial_{\nu_x}I_{\varepsilon}(x_0,\cdot)$ on $\partial \Gamma_\varepsilon$ as we approach neighbourhoods of the diagonal. To investigate these, we recall the singularity structure of $E(\cdot,\cdot)$:

\begin{proposition}\label{sing structure E}
	The Green function $E(x,y)$ and has the following singularity structure near the diagonal
	\begin{equation*}
	E(x,y) = -\frac{1}{2\pi} \log d_g(x,y) + P_{-4}(x,y),
	\end{equation*}
	where $P_{-4}(x,y)\in C^1(M\times M)$ is infinitely differentiable off the diagonal $\{x=y\}$.
\end{proposition}
We will prove proposition 3.1 in Section \ref{proof of lemma 3.1} as it involves the use of pseudodifferential operators.

As the distance function plays a crucial role in the Green's function $E(x,y)$, it is useful to derive asymptotics for them in the appropriate coordinate systems:

\begin{lemma}
Let
$$d_g^*(s,t): = d_g(x(s,x_0),x(t,x_0)).$$
where $t  = (t_1, t_2) \in \mathbb{D}_\rho \mapsto x(t,x_0)$ is the coordinate system defined in \eqref{geod coord}. Then we have that
\begin{equation*}
	d_g^*(s,t) = |s - t| + |s - t| F\left(t,\frac{s - t}{|s - t|}, |s - t|\right)
	\end{equation*}
for some smooth function $F(t,\omega, r)\in C^\infty(\mathbb{D}_\rho \times S^1\times[0,2\rho])$  which is $O(t) + O(r)$.
\end{lemma}

\begin{proof}
By Lemma 4.8 of \cite{lefeuvretensor}, if $t\mapsto x(t,x_0)$ is any coordinate system, there exists a matrix $H_{j,k}(s,t)$ smooth in $(s,t)$ such that
\begin{eqnarray}
\label{dstar}
d_{g}^*(s,t)^2 = \sum_{j=1}^2H_{j,k}(s,t)(s_j - t_j) (s_k - t_k),
\end{eqnarray}
where $H_{j,k}(t,t)  = g_{j,k}(t)$ is the coordinate expression for the metric tensor $g$. Since the coordinate system \eqref{geod coord} is the geodesic coordinate system, we have that  $g_{j,k}(t) =\delta_{j,k} + O(|t|^2)$. So we get $H_{j,k}(t, t) = \delta_{j,k} + O(|t|^2)$. Taylor expand $H_{j,k}(s,t)$ around $s=t$ and insert the resulting expression into \eqref{dstar} we get
$$d_{g}^*(s,t) = |s-t| + |s-t| F\left(t, \frac{s-t}{ |s-t|},  |s-t|\right)$$
for some smooth function $F\in C^\infty(\mathbb{D}_\rho \times S^1\times[0,r_0])$ which is $O(t) + O(r)$.
\end{proof}

The following distance expression in the rescaled normal coordinates given by \eqref{res coord} was stated in Corollary 2.6 of \cite{NTT2021}:
\begin{lemma}
\label{d inverse}
For the coordinates given by \eqref{res coord},
\begin{equation*}
		d^{-1}_g(x^\varepsilon(s,x_0), x^\varepsilon(t,x_0)) = \varepsilon^{-1}|t - s|^{-1} + \varepsilon |t - s|^{-1}A(\varepsilon, s,r,\omega)
	\end{equation*}
for some smooth function $A$ in the variables $(\varepsilon,s,r,\omega)\in [0,\varepsilon_0]\times\mathbb{D}\times\mathbb{R}\times S^1$, where $r=|t-s|$ and $\omega = \frac{t-s}{|t-s|}$.
\end{lemma}

In the next two lemmas, we investigate the properties of $I_{\varepsilon}(x_0,\cdot)$
\begin{lemma}\label{I}
	The following estimate holds
	\begin{equation}\label{sing of I}
		\sup_{x\in \partial \Gamma_\varepsilon} I_{\varepsilon}(x_0, x) = O(\varepsilon^2 \log\varepsilon), \qquad
		\text{as } \varepsilon \rightarrow 0.
	\end{equation}
\end{lemma}

\begin{proof}
	Due to Proposition \ref{sing structure E} it is sufficient to prove that
	\begin{equation*}
		\sup_{x\in \partial \Gamma_\varepsilon} \int_{\Gamma_\varepsilon} \log d_g(x,y) dvol_g(y) = O(\varepsilon^2 \log\varepsilon).
	\end{equation*}
	We consider $\varepsilon>0$ sufficiently small, so that $\log (10\varepsilon) < 0$. Then, for $x\in \partial \Gamma_\varepsilon$,
	\begin{equation*}
	\left|\int_{\Gamma_\varepsilon} \log d_g(x,y) dvol_g(y)\right| =
	\left|\int_{B_{\varepsilon}(x_0)} \log d_g(x,y) dvol_g(y)\right| \leq \left|\int_{B_{2\varepsilon}(x)} \log d_g(x,y) dvol_g(y)\right|.
	\end{equation*}
	For $\varepsilon>0$ sufficiently small we can find $\rho>3\varepsilon$ which is smaller than the injectivity radius. We will use the coordinate system given by
	\begin{equation*}
		\mathbb{D}_{\rho} \ni (s_1,s_2) \mapsto x(s_1,s_2;x_0),
	\end{equation*}
	defined in Section \ref{Prelim}. We recall that $s=(s_1,s_2)$ and $t=(t_1,t_2)$ and let
$$d_g^*(s,t): = d_g(x(s,x_0),x(t,x_0)).$$
Lemma 3.2 tells us that
	\begin{equation*}
	d_g^*(s,t) = |s - t| + |s - t| F\left(t,\frac{s - t}{|s - t|}, |s - t|\right)
	\end{equation*}
	for some smooth function $F$ which is $O(|t|) + O(|s - t|)$. Therefore, for sufficiently small $\varepsilon >0$, we can choose $\rho>0$ small enough so that for all $s,t \in \mathbb{D}_\rho$,
	\begin{equation*}
	\frac{1}{2}|s - t| \leq d_g^*(t,s) \leq 2|s - t|.
	\end{equation*}
	Furthermore, we choose $C > 0$ such that $\sqrt{\textrm{det}( g_{j,k}(s))} \leq C$ for $s\in \mathbb{D}_{\rho}$. Therefore, for $x = x(t,x_0)\in \partial \Gamma_\varepsilon$, we estimate
	\begin{align*}
		\left|\int_{B_{2\varepsilon}(x)} \log d_g(x,y) dvol_g(y)\right|
		&\leq \left| C\int_{d_g^*(t,s)\leq 2\varepsilon} \log\left(2|s - t|\right) ds \right|\\
		&\leq \left| C\int_{|s - t|\leq 4\varepsilon} \log\left(2|s - t|\right) ds \right| = O(\varepsilon^2 \log\varepsilon).
	\end{align*}
\end{proof}

\begin{lemma}\label{partial I}
	The following estimate holds
	\begin{equation}\label{sing of partialI}
	\sup_{x\in \partial \Gamma_\varepsilon} \partial_{\nu_x} I(x_0,x) = O(\varepsilon),
	\qquad \text{as } \varepsilon \rightarrow 0.
	\end{equation}
\end{lemma}

\begin{proof}
	Let us use the coordinate system $x(t,x_0)$. Note that in these coordinates the volume form for $M$ is given by
	\begin{equation}
		dvol_g(y) = (1 + \varepsilon V_{\varepsilon}(s))ds_1\wedge ds_2
	\end{equation}
	for some smooth function $V_\varepsilon (s)$ whose derivatives of all orders are bounded uniformly in $\varepsilon$. We also note that in these coordinates, we have
	\begin{equation}\label{dist}
	d_g(x(t;x_0), x(s;x_0))^2 = \sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)
	\end{equation}
	where $t=(t_1,t_2)$, $s=(s_1,s_2)$, and $G_{\alpha,\beta}(s,t)$ is a smooth function on $\mathbb{D}\times\mathbb{D}$ such that $G_{\alpha,\beta}(s,s) = \delta_{\alpha}^{\beta} + O(|s|^2)$ for $s$ near $0$. Then, by Proposition 2.8 in \cite{taylor2}, we know
	\begin{align*}
	E(x(t;x_0), x(s;x_0)) = &-\frac{1}{4\pi} \log \left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right)\\
	&+ q_2(s,s-t) + p_2(s,s-t) \log |s - t|  +R(s,t).
	\end{align*}
	Here $p_2(x,z)$ is a polynomial homogeneous of degree $2$ in $z$, with the coefficients that are bounded, together with their $x$-derivatives. A function $q_2(x,z)$ is smooth on $\mathbb{R}^2\setminus\{0\}$ and homogeneous of degree $2$ in $z$. Finally, $R\in C^2(\mathbb{R}^2_s \times \mathbb{R}^2_t)$.

	Let us use the polar coordinates
	\begin{equation*}
		t = (r\cos\theta, r\sin \theta) \qquad s = (r'\cos\theta', r'\sin \theta').
	\end{equation*}
	We note that $x(\{|t|= \varepsilon;x_0\}) = \partial \Gamma_\varepsilon$ and $r \mapsto ( r\cos\theta, r \sin\theta)$ is the parametrization of unit speed geodesic issued from the origin. Therefore, since $\partial_{\nu_x}$ is the inward normal of $\partial \Gamma_\varepsilon$ for $x = ( r\cos\theta, r \sin\theta)\in \partial \Gamma_\varepsilon$, it follows from Gauss Lemma that $\Phi_* \partial_{\nu_x} = -\partial_r \in T_{(r\cos\theta, r \sin\theta)}\mathbb{R}^2$. Therefore
	\begin{align*}
	\partial_{\nu_x}E(x, y) = \partial_r \Bigg[-\frac{1}{4\pi} \log &\left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right)\\
	&+ q_2(s,s-t) + p_2(s,s-t)\log |s - t|  +R(s,t)\Bigg].
	\end{align*}
	Therefore
	\begin{align*}
	\partial_x I(x_0,x) = &\int_{\mathbb{D}_\varepsilon}\partial_r \Bigg[-\frac{1}{4\pi} \log \left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right)\\
	&+ q_2(s,s-t) + p_2(s,s-t)\log |s - t|  +R(s,t)\Bigg](1 + \varepsilon V_\varepsilon(s))ds.
	\end{align*}
	From the properties of functions $q_2$, $p_2$, and $R$, mentioned above, it follows that
	\begin{equation*}
	\int_{\mathbb{D}_\varepsilon} \partial_r \left( q_2(s,s-t) + p_2(s,s-t)\log |s - t|  +R(s,t) \right)(1 + \varepsilon V_\varepsilon(s))ds = O(\varepsilon^2)
	\end{equation*}
	uniformly on $t$. Hence, we have
	\begin{align}\label{I in eucl}
		\nonumber\partial_x I(x_0,x) = &-\frac{1}{4\pi} \int_{\mathbb{D}_\varepsilon}\partial_r  \log \left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right)ds\\
		& -\frac{\varepsilon}{4\pi} \int_{\mathbb{D}_\varepsilon}\partial_r  \log \left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right) V_\varepsilon(s)ds + O(\varepsilon^2)
	\end{align}
	The first integral of the right-hand side is equal to
	\begin{align}
	\nonumber \int_{0}^{2\pi}&\int_{0}^{\varepsilon} \frac{2\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}r'dr'd\theta'\\
	&+ \int_{0}^{2\pi}\int_{0}^{\varepsilon} \frac{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		\partial_rG
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}r'dr'd\theta',
	\end{align}
	where $G=G(r,\theta,r',\theta')$ is two by two matrix with entries $\{G_{\alpha,\beta}(s,t)\}$ with $t = t(r,\theta)$ and $s = s(r',\theta')$. Since $x\in \partial B_\varepsilon(x_0)$, we take $r=\varepsilon$. Then, if we change the variable $r' \mapsto \varepsilon r'$, the last expression becomes
	\begin{align*}
	\varepsilon\int_{0}^{2\pi}&\int_{0}^{1} \frac{2\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - r'\cos\theta'\\
		\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		\cos\theta - r'\cos\theta' & \sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - r'\cos\theta'\\
		\sin\theta - r'\sin\theta'
		\end{bmatrix}}r'dr'd\theta'\\
	&+ \varepsilon^2\int_{0}^{2\pi}\int_{0}^{1} \frac{\begin{bmatrix}
		\cos\theta - r'\cos\theta' & \sin\theta - r'\sin\theta'
		\end{bmatrix}
		\partial_rG
		\begin{bmatrix}
		\cos\theta - r'\cos\theta'\\
		\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		\cos\theta - r'\cos\theta' & \sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - r'\cos\theta'\\
		\sin\theta - r'\sin\theta'
		\end{bmatrix}}r'dr'd\theta'.
	\end{align*}
	Note that we have integrable singularity at point $(r',\theta') = (1,\theta)$ and integrals are bounded uniformly on $\theta$. Therefore, the last expression is $O(\varepsilon)$ as $\varepsilon\rightarrow 0$ uniformly on $\theta$. Since $V_\varepsilon(s)$ is bounded uniformly on $\varepsilon$, the second term of \eqref{I in eucl} is $O(\varepsilon^3)$.
\end{proof}

Next, we obtain the singularity structure of $\partial_{\nu_x} E(\cdot,\cdot)$ in a neighbourhood of $x_0$:

\begin{lemma}\label{sing str of E}
	Let $B_{\varepsilon}(x_0)$ be the geodesic ball with radius $\varepsilon$ centred at $x_0$. Then
	\begin{equation*}
	\partial_{\nu_x} E(x,y)\left.\right|_{x,y\in \partial \Gamma_\varepsilon} = \frac{1}{4\pi\varepsilon} + Q_{\varepsilon}(x,y),
	\end{equation*}
	for some function $Q_\varepsilon$ such that
	\begin{equation*}
	\sup_{x\in \partial \Gamma_\varepsilon} \int_{\partial \Gamma_\varepsilon}  Q_{\varepsilon}(x,y) dvol_h(y) = O(\varepsilon).
	\end{equation*}
\end{lemma}

\begin{proof}
	We begin as in the prove of Lemma \ref{partial I}. We repeat all steps until we derive
	\begin{align}\label{partial x E 2}
	\nonumber\partial_{\nu_x}E(x, y) = \partial_r \Bigg[-\frac{1}{4\pi} \log &\left(\sum_{\alpha,\beta=1}^{2}G_{\alpha,\beta}(s,t)(s_\alpha - t_\alpha)(s_\beta - t_\beta)\right)\\
	&+ q_2(s,s-t) + p_2(s,s-t) \log |s - t| +R(s,t)\Bigg].
	\end{align}
	We recall that $p_2(x,z)$ is a polynomial homogeneous of degree $2$ in $z$, with the coefficients that are bounded, together with their $x-$derivatives. A function $q_2(x,z)$ is smooth on $\mathbb{R}^2\setminus\{0\}$ and homogeneous of degree $2$ in $z$. Finally, $R\in C^2(\mathbb{R}^2_s \times \mathbb{R}^2_t)$. These conditions imply that
	\begin{equation*}
		 \int_0^{2\pi}\partial_r \left[q_2(s,s-t) + p_2(s,s-t) \log |s - t| +R(s,t)\right] \left.\right|_{r=r'=\varepsilon}d\theta' = O(1)
	\end{equation*}
	as $\varepsilon\rightarrow 0$ uniformly on $\theta$.

	Next, we investigate the first term of the right-hand side of \eqref{partial x E 2}, which can be written as follows
	\begin{multline*}
	\frac{1}{4\pi} \frac{2\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}\\
	+\frac{1}{4\pi} \frac{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		\partial_rG
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		r\cos\theta - r'\cos\theta' & r\sin\theta - r'\sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		r\cos\theta - r'\cos\theta'\\
		r\sin\theta - r'\sin\theta'
		\end{bmatrix}}.
	\end{multline*}
	Since $x$, $y\in \partial \Gamma_\varepsilon$, we take $r=r'=\varepsilon$, so that the last expression becomes
	\begin{multline*}
	\frac{1}{2\pi\varepsilon} \frac{\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix}}\\
	+\frac{1}{4\pi} \frac{\begin{bmatrix}
		\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
		\end{bmatrix}
		\partial_rG
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix}}{\begin{bmatrix}
		\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix}}.
	\end{multline*}
	Note that the last term belongs to $L^{\infty}(S^1_{\theta} \times S^1_{\theta'})$ uniformly in $\varepsilon$. Therefore, it remains to show that
	\begin{equation}\label{it remains to show}
		\frac{1}{2\pi\varepsilon} \frac{\begin{bmatrix}
			\cos\theta & \sin\theta
			\end{bmatrix}
			G
			\begin{bmatrix}
			\cos\theta - \cos\theta'\\
			\sin\theta - \sin\theta'
			\end{bmatrix}}{\begin{bmatrix}
			\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
			\end{bmatrix}
			G
			\begin{bmatrix}
			\cos\theta - \cos\theta'\\
			\sin\theta - \sin\theta'
			\end{bmatrix}}
		= \frac{1}{4\pi\varepsilon} + L_\varepsilon(\theta, \theta')
	\end{equation}
	for some function $L_{\varepsilon}$ such that
	\begin{equation*}
		\int_{0}^{2\pi}L_\varepsilon(\theta, \theta')d\theta' = O(1), \qquad \text{as } \varepsilon\rightarrow 0,
	\end{equation*}
	uniformly on $\theta$. Let $J(\theta,\theta')$ be the left-hand side of \eqref{it remains to show}. We denote $J_{1} : = J \chi_{|\theta - \theta'|<\varepsilon}$ and $J_{2} : = J \chi_{|\theta - \theta'|>\varepsilon}$, where $\chi$ is an indicator function of the corresponding set.

	To investigate $J_1$, we will use Taylor expansion for its numerator and denominator at $\theta'=\theta$. We recall that $G_{j,k}(x,x) = g_{j,k}(x)$, for $r=r'=\varepsilon$, we get
	\begin{equation}\label{G}
	G = g +\varepsilon R_{\varepsilon}(\theta,\theta') (\theta - \theta'),
	\end{equation}
	where $R_\varepsilon$ is two by two matrix with $C^{\infty}(S^1_{\theta} \times S^1_{\theta'})$ entries and $g = \{g_{j,k}(\varepsilon\cos \theta, \varepsilon\sin \theta)\}_{k,j}^2$. Furthermore, we express $g$ in the following way
	\begin{equation}\label{g}
	g = I + \Gamma(\varepsilon,\theta),
	\end{equation}
	where $I$ is two by two identity matrix and $\Gamma$ is two by two matrix with interiors $O(\varepsilon^2)$. Therefore, by applying Taylor expansion at $\theta = \theta'$, we obtain
	\begin{multline*}
	\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	G
	\begin{bmatrix}
	\cos\theta - \cos\theta'\\
	\sin\theta - \sin\theta'
	\end{bmatrix}
	=
	\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	g
	\begin{bmatrix}
	- \sin\theta\\
	\cos\theta
	\end{bmatrix}\\
	+\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	g
	\begin{bmatrix}
	K_1(\theta,\theta')\\
	K_2(\theta,\theta')
	\end{bmatrix}(\theta-\theta')^2+ O(\varepsilon) O( |\theta-\theta'|^2),
	\end{multline*}
	for some $K=(K_1,K_2) \in L^{\infty}\left(S^1\times S^1\right)^2$. Note that the normal vector on $\{|t| = \varepsilon\}$ is given by $\cos\theta\partial_{t_1} + \sin\theta\partial_{t_2}$ at the point $(\varepsilon\cos\theta, \varepsilon\sin\theta)$, while the tangent is given by $-\sin\theta\partial_{t_1} + \cos\theta\partial_{t_2}$. Therefore the first term of the right-hand side of the last equation is zero, so that
	\begin{align*}
	\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	G
	\begin{bmatrix}
	\cos\theta - \cos\theta'\\
	\sin\theta - \sin\theta'
	\end{bmatrix}
	=\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	g
	\begin{bmatrix}
	K_1\\
	K_2
	\end{bmatrix}(\theta-\theta')^2 + O(\varepsilon)O( |\theta-\theta'|^2).
	\end{align*}
	Similarly, by using \eqref{G} and \eqref{g}, we show that
	\begin{align*}
	\begin{bmatrix}
	\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
	\end{bmatrix}
	G
	\begin{bmatrix}
	\cos\theta - \cos\theta'\\
	\sin\theta - \sin\theta'
	\end{bmatrix}
	=2(\theta - \theta')^2 + O(\varepsilon)O( |\theta-\theta'|^2).
	\end{align*}
	The last two estimates imply that
	\begin{equation*}
	\int_{0}^{2\pi} J_1d\theta' = \frac{1}{2\pi\varepsilon} \int_{0}^{2\pi}
	\frac{\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		K_1\\
		K_2
		\end{bmatrix} + O(\varepsilon)}
	{2 + O(\varepsilon)}\chi_{|\theta - \theta'|<\varepsilon}(\theta') d\theta' = O(1)
	\end{equation*}
	as $\varepsilon\rightarrow0$ uniformly in $\theta$.

	Next, we will investigate $J_2$. From \eqref{G} and \eqref{g}, it follows that
	\begin{equation*}
		\begin{bmatrix}
		\cos\theta & \sin\theta
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix} = 1 - \cos(\theta - \theta') + O(\varepsilon^2)O(|\theta - \theta'|) + O(\varepsilon)O(|\theta - \theta'|^2).
	\end{equation*}
	In the region $\{|\theta - \theta'|>\varepsilon\}$, we can rewrite this
	\begin{equation*}
	\begin{bmatrix}
	\cos\theta & \sin\theta
	\end{bmatrix}
	G
	\begin{bmatrix}
	\cos\theta - \cos\theta'\\
	\sin\theta - \sin\theta'
	\end{bmatrix} = 1 - \cos(\theta - \theta') + O(\varepsilon)O(|\theta - \theta'|^2).
	\end{equation*}
	Similarly,
	\begin{equation*}
		\begin{bmatrix}
		\cos\theta - \cos\theta' & \sin\theta - \sin\theta'
		\end{bmatrix}
		G
		\begin{bmatrix}
		\cos\theta - \cos\theta'\\
		\sin\theta - \sin\theta'
		\end{bmatrix}
		= 2 - 2\cos(\theta - \theta') + O(\varepsilon)O(|\theta - \theta'|^3).
	\end{equation*}
	Therefore, we have
	\begin{align*}
		J_2 &= \frac{1}{2\pi\varepsilon}\left(\frac{1}{2} + \frac{O(\varepsilon)O(|\theta - \theta'|^2)}{2 - 2\cos(\theta - \theta') + O(\varepsilon)O(|\theta - \theta'|^3)}\right)\\
		&= \frac{1}{4\pi\varepsilon} + \frac{O(1)}{\frac{2 - 2\cos(\theta - \theta')}{(\theta - \theta')^2} + O(\varepsilon)O(|\theta - \theta'|)}
	\end{align*}
	Since $(2 - 2\cos(\theta - \theta')) (\theta - \theta')^{-2}$ is a positive and continuous function of $\theta'\in [0,2\pi]$, we conclude that
	\begin{equation*}
		\int_{0}^{2\pi}\left(J_2 - \frac{1}{4\pi\varepsilon}\right)d\theta' = O(1) \qquad \text{as } \varepsilon\rightarrow 0,
	\end{equation*}
	uniformly in $\theta$. The Lemma is proved.
\end{proof}

\section{Narrow capture problem on the surface without boundary}\label{proof of thm without bry}
In this section, we prove Theorem \ref{main}. We start by recalling the formulation of the problem. Let $(X_t, \mathbb{P}_x)$ be the Brownian motion on a boundaryless manifold $M$ starting at $x$, generated by $\Delta_g$. For $x_0\in M$ and  $\varepsilon>0$, let $\Gamma_{\varepsilon} =  B_\varepsilon(x_0)$ be a small geodesic ball centred at fixed point $x_0\in M$.
Denote by $\tau_{\Gamma_{\varepsilon}}$ the first time the Brownian motion $X_t$ hits $\Gamma_{\varepsilon}$, that is
\begin{equation*}
\tau_{\Gamma_{\varepsilon}} := \inf \{ t\geq 0: X_t \in \Gamma_{\varepsilon}\}.
\end{equation*}
We aim to investigate the mean first-passage time and its average:
\begin{equation*}
\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x],
\qquad |M_\varepsilon|^{-1} \int_{M_\varepsilon} \mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x] dvol_g(x).
\end{equation*}
where $M_{\varepsilon}:=M\setminus \Gamma_\varepsilon$. Namely, we want to derive asymptotic expansion for these quantities as $\varepsilon\to 0$. It is known that $\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x]$ satisfies the following boundary value problem, see for instance Appendix A in \cite{NTT2021},
\begin{equation}\label{bvp}
\begin{cases}
\Delta_g u_\varepsilon = - 1 & \text{on } M_{\varepsilon};\\
u_\varepsilon = 0 & \text{on } \partial M_{\varepsilon}=\partial \Gamma_\varepsilon,
\end{cases}
\end{equation}
which gives the compatibility condition
\begin{equation}\label{comp cond}
\int_{\partial \Gamma_\varepsilon}\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = - |M_{\varepsilon}|.
\end{equation}

To prove Theorem \ref{main}, we will need the following auxiliary result.
\begin{proposition}\label{partial_u}
	Let $u_{\varepsilon}$ be the solution of \eqref{bvp}, then
	\begin{equation*}
	\partial_{\nu}\left.u_{\varepsilon}\right|_{\partial \Gamma_\varepsilon} = -\frac{|M_{\varepsilon}|}{2\pi\varepsilon} + W_\varepsilon.
	\end{equation*}
	for some $W_\varepsilon\in O_{L^{\infty}(\partial \Gamma_\varepsilon)}(1)$ as $\varepsilon\rightarrow 0$.
\end{proposition}

\begin{proof}
	By using Green's identity, we obtain
	\begin{equation}\label{u_partialu_eq}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) - u_{\varepsilon}(x) + \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = I_{\varepsilon}(x_0,x).
	\end{equation}
	We take $\partial_{\nu_x}$ and restrict to $\partial \Gamma_\varepsilon$
	\begin{equation*}
	- \partial_{\nu_x}u_{\varepsilon}(x) + \partial_{\nu_x} \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = \partial_{\nu_x}I_{\varepsilon}(x_0,x),
	\end{equation*}
	and hence, by Lemma \ref{partial I}, we derive
	\begin{equation*}
	- \partial_{\nu_x}u_{\varepsilon}(x) + \partial_{\nu_x} \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	By Proposition 11.3 of \cite{taylor2},
	\begin{equation*}
	- \frac{1}{2} \partial_{\nu}u_{\varepsilon}(x) +  \int_{\partial \Gamma_\varepsilon}\partial_{\nu_x}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	Therefore, from Lemma \ref{sing str of E}, it follows
	\begin{equation*}
	\frac{1}{2} \partial_{\nu}u_{\varepsilon}(x) =  \frac{1}{4\pi\varepsilon}\int_{\partial \Gamma_\varepsilon}\partial_{\nu}u_{\varepsilon}(y)dvol_h(y)  + \int_{\partial \Gamma_\varepsilon}Q_{\varepsilon}(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y)+ O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	Hence, the compatibility condition \eqref{comp cond} gives
	\begin{equation}
\label{epsilonpartialnu}
	\varepsilon \partial_{\nu}u_{\varepsilon}(x) =  - \frac{|M_{\varepsilon}|}{2\pi}   + 2\varepsilon\int_{\partial \Gamma_\varepsilon}Q_{\varepsilon}(x,y)\partial_{\nu}u_{\varepsilon}(y) dvol_h(y) + O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon^2).
	\end{equation}
	Next, we estimate
	\begin{eqnarray*}
	\sup_{x\in \partial \Gamma_\varepsilon}\left|\varepsilon\int_{\partial \Gamma_\varepsilon}Q_{\varepsilon}(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y)\right|
	&\leq& \sup_{x\in \partial \Gamma_\varepsilon}\left|\varepsilon \partial_{\nu}u_{\varepsilon}\right|
	\sup_{x\in \partial \Gamma_\varepsilon} \int_{\partial \Gamma_\varepsilon}\left|Q_{\varepsilon}(x,y)\right|dvol_h(y)\\
&\leq& C \varepsilon^2\sup_{x\in \partial \Gamma_\varepsilon}\left| \partial_{\nu}u_{\varepsilon}\right|
	\end{eqnarray*}
The last estimate comes from Lemma \ref{sing str of E}. Combine this estimate with \eqref{epsilonpartialnu} we obtain that
	\begin{equation*}
	\varepsilon \partial_{\nu_x}u_{\varepsilon}(x) = - \frac{|M_{\varepsilon}|}{2\pi} + O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{main}]
	We first prove ii) then proceed with i). By Proposition \ref{partial_u}, we can express
	\begin{equation*}
	\partial_{\nu}\left.u_{\varepsilon}\right|_{\partial \Gamma_\varepsilon} = -\frac{|M_{\varepsilon}|}{2\pi\varepsilon} + W_{\varepsilon}
	\end{equation*}
\begin{eqnarray}
\label{W bound}
\|W_\varepsilon\|_{L^\infty(\partial\Gamma_\epsilon)} \leq C.
\end{eqnarray}
uniformly in $\epsilon>0$. Then, for $x\in M_{\varepsilon}\setminus\partial \Gamma_\varepsilon$, \eqref{u_partialu_eq} gives
	\begin{multline*}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) - u_{\varepsilon}(x) - \frac{|M_{\varepsilon}|}{2\pi\varepsilon}\int_{\partial \Gamma_\varepsilon}E(x,y)dvol_h(y) + \int_{\partial \Gamma_\varepsilon}E(x,y)W_{\varepsilon}(y)dvol_h(y)\\
	= I_{\varepsilon}(x_0,x),
	\end{multline*}
	or equivalently
	\begin{multline}\label{before restriction}
		\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) = u_{\varepsilon}(x) + \frac{|M_{\varepsilon}|}{2\pi\varepsilon}\int_{\partial \Gamma_\varepsilon}E(x,x_0)dvol_h(y)\\
		+ \frac{|M_{\varepsilon}|}{2\pi\varepsilon}\int_{\partial \Gamma_\varepsilon}(E(x,y) - E(x,x_0))dvol_h(y)
		-\int_{\partial \Gamma_\varepsilon}E(x,y)W_{\varepsilon}(y)dvol_h(y) + I_{\varepsilon}(x,x_0).
	\end{multline}
	To compute the left-hand side, we restrict this to $\partial \Gamma_\varepsilon$ where $u_{\varepsilon} = 0$. We note that Proposition \ref{sing structure E} combined with Lemma \ref{dstar} shows that in the coordinates \eqref{geod coord} the leading  singuarlity of the Green's function $E(x(s,x_0,), x(t, x_0))$ is of the form
$$ E(x(s,x_0), x(t,x_0)) =C\log|t-s| + L^\infty(\mathbb{D}_\rho\times \mathbb{D}_\rho)$$
Combine this with Lemma \ref{I} and \eqref{W bound} gives
	\begin{equation*}
	\sup_{x\in\partial \Gamma_\varepsilon}\left|\int_{\partial \Gamma_\varepsilon}E(x,y)W_{\varepsilon}(y)dvol_h(y)\right| = O(\varepsilon\log\varepsilon), \qquad \sup_{x\in\partial \Gamma_\varepsilon}\left|I_{\varepsilon}(x_0,x)\right| = O(\varepsilon^2\log\varepsilon)
	\end{equation*}
	as $\varepsilon\rightarrow 0$. Therefore, restricting \eqref{before restriction} to $\partial \Gamma_\varepsilon$ and using Proposition \ref{sing structure E}, we obtain
	\begin{multline}\label{ave u}
		\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) = -\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{4\pi^2\varepsilon}\log\varepsilon +\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{2\pi\varepsilon}P_{-4}(x_0,x_0)\\
		+ \frac{|M_{\varepsilon}|}{2\pi\varepsilon}\int_{\partial \Gamma_\varepsilon}(E(x,y) - E(x,x_0))dvol_h(y)\left.\right|_{x\in \partial\Gamma_\varepsilon} + O_{L^\infty(\partial \Gamma_{\varepsilon})}(\varepsilon\log\varepsilon).
	\end{multline}
	Let us examine the third term of the right-hand side
	\begin{multline}\label{difference of E}
		\int_{\partial \Gamma_\varepsilon}(E(x,y) - E(x,x_0))dvol_h(y)\\
		 = \int_{\partial \Gamma_\varepsilon}(\log d_g(x,y) - \log d_g(x,x_0))dvol_h(y) + \int_{\partial \Gamma_\varepsilon}(P_{-4}(x,y) - P_{-4}(x,x_0))dvol_h(y),
	\end{multline}
	where $x\in \partial\Gamma_\varepsilon$. Joint differentiability of $P_{-4}$ gives
	\begin{equation}
\label{estimate the smooth part}
	\sup_{x\in \partial\Gamma_\varepsilon}\left|	\int_{\partial \Gamma_\varepsilon}(P_{-4}(x,y) - P_{-4}(x,x_0))dvol_h(y)\right| \leq C\varepsilon^2.
	\end{equation}
	To investigate the first term of the right-hand side of \eqref{difference of E}, we use the coordinate system $x^{\varepsilon}(\cdot,x_0)$ with $x = x^{\varepsilon}(t,x_0)$ and $y = x^{\varepsilon}(s,x_0)$. Let $d\sigma(s)$ be the pull back of the volume form $dvol_h(y)$ under $s\mapsto x^{\varepsilon}(s,x_0)$, then
	\begin{equation*}
	dvol_h(y) = \varepsilon(1+v_\varepsilon(s))d\sigma(s)
	\end{equation*}
	for some smooth function $\|v_{\varepsilon}\|_{L^\infty(\partial\Gamma_\varepsilon)} \leq C\varepsilon$. By Lemma \ref{d inverse} we have
	\begin{equation*}
		d^{-1}_g(x^\varepsilon(t,x_0),x^\varepsilon(s,x_0)) = \varepsilon^{-1}|t - s|^{-1} + \varepsilon |t - s|^{-1}A(\varepsilon, s,r,\omega)
	\end{equation*}
	for some smooth function $A(\epsilon, s, r,\omega)$ in the variables $(\varepsilon,s,r,\omega)\in [0,\varepsilon_0]\times\mathbb{D}\times\mathbb{R}\times S^1$, where $r=|t-s|$ and $\omega = \frac{t-s}{|t-s|}$. Therefore, the first term of \eqref{difference of E} becomes
	\begin{align}\label{difference of ln}
		\nonumber\int_{\partial \Gamma_\varepsilon}(\log d_g(x,y) - \log d_g(x,x_0))dvol_h(y)
		&=\varepsilon \int_{\partial\mathbb{D}} \log\left(\frac{1}{\varepsilon^{-1}|t - s|^{-1} + \varepsilon |t - s|^{-1}A}\right)(1+v_\varepsilon(s))d\sigma(s)\\
		&=\varepsilon \int_{\partial\mathbb{D}} \log\left(\frac{\varepsilon |t - s|}{1 + \varepsilon^2A}\right)(1+v_\varepsilon(s))d\sigma(s).
	\end{align}
	We have trivially that
	\begin{equation}
\label{epslogeps term}
		\varepsilon\left| \int_{\partial\mathbb{D}}\log\left(\frac{\varepsilon}{1 + \varepsilon^2 A(\varepsilon, s, r,\omega)}\right) (1+v_\varepsilon(s))d\sigma(s)\right| \leq C\varepsilon\log\varepsilon
	\end{equation}
	and
	\begin{align}
\label{epsilonsquared estimate}
		\left|\varepsilon \int_{\partial\mathbb{D}}\log|t - s| v_\varepsilon(s)d\sigma(s)\right|& \leq \varepsilon\int_{\partial\mathbb{D}} \left|\log|t - s|v_\varepsilon(s)\right|d\sigma(s) \leq C \varepsilon^2
	\end{align}
for all $t\in \partial\mathbb{D}$. Therefore, inserting the estimates \eqref{epslogeps term} and \eqref{epsilonsquared estimate} into \eqref{difference of ln} gives that for $t\in \partial\mathbb{D}$,
	\begin{align*}
		\int_{\partial \Gamma_\varepsilon}(\log d_g(x,y) - \log d_g(x,x_0))dvol_h(y) &=\varepsilon  \int_{\partial\mathbb{D}}\log|t - s|d\sigma(s) +  O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon\log\varepsilon)\\
		&=\frac{\varepsilon}{2}\int_{0}^{2\pi} \log(2-2\cos \theta')d\theta'
		+O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon\log\varepsilon)\\
		& = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon\log\varepsilon)
	\end{align*}

This combined with \eqref{estimate the smooth part} and \eqref{difference of E} gives
$$\left|\int_{\partial \Gamma_\varepsilon}(E(x,y) - E(x,x_0))dvol_h(y)\right| \leq C\varepsilon \log\varepsilon.$$
Inserting this estimate into \eqref{ave u} implies
	\begin{multline*}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) = -\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{4\pi^2\varepsilon}\log\varepsilon +\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{2\pi\varepsilon}P_{-4}(x_0,x_0)\\
	+ O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon\log\varepsilon).
	\end{multline*}
	We take the supremum norm over $\partial \Gamma_\varepsilon$ to obtain
	\begin{multline}\label{av}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) = -\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{4\pi^2\varepsilon}\log\varepsilon +\frac{|M_{\varepsilon}||\partial \Gamma_\varepsilon|}{2\pi\varepsilon}P_{-4}(x_0,x_0)\\
	+  O(\varepsilon\log\varepsilon).
	\end{multline}
	Next, we note that $|\Gamma_\varepsilon| = |B_\varepsilon(x_0)| = O(\varepsilon^2)$ and
	\begin{equation*}
		|\partial \Gamma_\varepsilon|=|\partial B_\varepsilon(x_0)| = \int_{\partial B_\varepsilon(x_0)}dvol_h(y) = \int_{\partial \mathbb{D}}\varepsilon (1 + v_{\varepsilon}(s))d\sigma(s) = 2\pi \varepsilon +O (\varepsilon^2).
	\end{equation*}
	This gives us part ii) of Theorem \ref{main}.

	Let us put \eqref{av} into \eqref{u_partialu_eq} to we obtain
	\begin{multline*}
	u_{\varepsilon}(x) = -\frac{|M|}{2\pi}\log\varepsilon + |M_{\varepsilon}|P_{-4}(x_0,x_0) \\
	+\int_{\partial M_{\varepsilon}}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) - I_{\varepsilon}(x_0,x) + O(\varepsilon\log\varepsilon).
	\end{multline*}
	as $\varepsilon \rightarrow 0$. It remains to show that for a fixed compact set $K\subset\subset M_\varepsilon$,
	\begin{equation}
\label{volume term}
	\int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) - I_{\varepsilon}(x_0,x) = - |M| E(x,x_0) + r_{\varepsilon}(x).
	\end{equation}
for some $r_\varepsilon(x)$ whose $L^\infty(K)$ norm is of order $\varepsilon$.

To do this, let us fix any compact set $K\subset M$ which does not contain $x_0$, so that $K\cap \Gamma_\varepsilon$ is empty for sufficiently small $\varepsilon>0$. Then $E(\cdot,\cdot)$ is smooth in $K\times \Gamma_\varepsilon$, and hence we estimate
	\begin{equation*}
	\sup_{x\in K} I(x_0,x) = \sup_{x\in K} \int_{\Gamma_\varepsilon} E(x,y) dvol_g(y) = O(\varepsilon^2) \qquad \text{as } \varepsilon \rightarrow 0.
	\end{equation*}
	Next, we write
	\begin{multline*}
	\int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = \int_{\partial \Gamma_\varepsilon}(E(x,y) - E(x,x_0))\partial_{\nu}u_{\varepsilon}(y)dvol_h(y)\\
	+E(x,x_0) \int_{\partial \Gamma_\varepsilon}\partial_{\nu}u_{\varepsilon}(y)dvol_h(y).
	\end{multline*}
The first integral can be estimated using Proposition \ref{partial_u} and the smoothness of $E(\cdot,\cdot)$ in $K\times \Gamma_\varepsilon$to give

$$ \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^\infty(K)}(\varepsilon) +  E(x,x_0) \int_{\partial \Gamma_\varepsilon}\partial_{\nu}u_{\varepsilon}(y)dvol_h(y). $$
Now use the compatibility condition \eqref{comp cond} to get for $x\in K$,
	\begin{equation*}
	 \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = - |M_\varepsilon| E(x,x_0) + O_{L^\infty(K)}(\varepsilon) = - |M| E(x,x_0) + O_{L^\infty(K)}(\varepsilon).
	\end{equation*}
	This gives Part i) of Theorem \ref{main}.

\end{proof}

\section{Narrow capture problem on the surface with boundary}\label{proof of thm with bry}
Here, we consider the same problem for the case when the surface has a smooth boundary , $\partial M$, which reflects the particle. Without loss of generality, we assume that $M$ is an connected open subset of a compact orientable Riemannian manifold $(\tilde{M}, g)$ without boundary. Let $\tilde{E}(x,y)$ be the Green's function on $\tilde{M}$, given by \eqref{no boundary green}. The Neumann Green's function $E(x,y)$ is given by, for $x\in M^0$,
\begin{equation}\label{Neumann Green function}
	\begin{cases}
		\Delta_{g,y}E(x,y) = -\delta_x(y) + \frac{1}{|M|}, & \text{for } y\in M,\\
		\partial_{\nu_y}E(x,y)=0, & \text{for } y\in \partial M,\\
		\int_M E(x,y)dvol_g(y) = 0.
	\end{cases}
\end{equation}
We can obtain this function by setting $E = \tilde{E} - C$, where the correction term $C(x,y)$ is the solution to the boundary value problem, for $x\in M^0$,
\begin{equation*}
	\begin{cases}
		\Delta_{g,y}C(x,y) = \frac{1}{|\tilde{M}|} - \frac{1}{|M|}, & \text{for } y\in M,\\
		\partial_{\nu_y}C(x,y) = \partial_{\nu_y}\tilde{E}(x,y), & \text{for } y\in \partial M,\\
		\int_M C(x,y)d_g(y) = \int_M \tilde{E}(x,y)dvol_g(y).
	\end{cases}
\end{equation*}
Therefore, for $U\subset \subset M$ away from the boundary, that is $\mathrm{dist}_g(U, \partial M) > 0$, it follows
\begin{equation*}
	C = \tilde{E} - E \in C^{\infty}(\bar{U} \times \bar{U}).
\end{equation*}
Hence, we can decompose
\begin{equation}\label{dec2}
E(x,y) = -\frac{1}{2\pi} \log d_g(x,y) + P_{-4}(x,y),
\end{equation}
where $P_{-4}(x,y) \in C^1(\bar{U}\times \bar{U})$. Moreover, since we are considering the centre of the trap, $x_0$, to be fixed in the interior of $M$ with a sufficiently small $\varepsilon >0$, we have that
\begin{equation*}
	\sup_{x\in \partial \Gamma_\varepsilon} \int_{\Gamma_\varepsilon} C(x, y) dvol_g(y) = O(\varepsilon^2),
	\qquad
	\sup_{x\in \partial \Gamma_\varepsilon} \partial_{\nu_x} \int_{\Gamma_\varepsilon} C(x, y) dvol_g(y) = O(\varepsilon^2),
\end{equation*}
\begin{equation*}
	\sup_{x\in \partial \Gamma_\varepsilon} \partial_{\nu_x} \int_{\partial \Gamma_\varepsilon} C(x, y) dvol_h(y) = O(\varepsilon),
\end{equation*}
as $\varepsilon\rightarrow 0$. As a result, we obtain the following analogues of Lemmas \ref{I}-\ref{sing str of E} for the function
\begin{equation*}
	I_\varepsilon(x_0,x) := \int_{\Gamma_\varepsilon}E(x, y) dvol_g(y),
\end{equation*}
\begin{lemma}\label{analogue}
	As $\varepsilon\rightarrow 0$, we have that
	\begin{equation*}
		\sup_{x\in \partial \Gamma_\varepsilon} I_{\varepsilon}(x_0, x) = O(\varepsilon^2 \log\varepsilon),
		\qquad
		\sup_{x\in \partial \Gamma_\varepsilon} \partial_{\nu_x} I(x_0,x) = O(\varepsilon),
	\end{equation*}
	\begin{equation*}
		\partial_{\nu_x} E(x,y)\left.\right|_{x,y\in \partial \Gamma_\varepsilon} = \frac{1}{4\pi\varepsilon} + Q_{\varepsilon}(x,y),
	\end{equation*}
	for some function $Q_\varepsilon$ such that
	\begin{equation*}
	\sup_{x\in \partial \Gamma_\varepsilon} \int_{\partial \Gamma_\varepsilon}  Q_{\varepsilon}(x,y) dvol_h(y) = O(\varepsilon).
	\end{equation*}
\end{lemma}

Note that, in this case, $\partial M \neq \emptyset$, the mean first-passage time $\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x]$ satisfies the following mixed boundary value problem, see Appendix in \cite{NTT2021},
\begin{equation}\label{bvp2}
	\begin{cases}
		\Delta_g u_\varepsilon = 1 & \text{on } M_\varepsilon,\\
		u_\varepsilon = 0, & \text{on } \partial \Gamma_\varepsilon,\\
		\partial_{\nu} u = 0, & \text{on } \partial M.
	\end{cases}
\end{equation}
which gives the compatibility condition
\begin{equation}\label{comp cond 2}
\int_{\partial \Gamma_\varepsilon}\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = - |M_{\varepsilon}|.
\end{equation}
As a result, we have the following analogue of Proposition \ref{partial_u}
\begin{proposition}\label{partial_u_2}
	Let $u_{\varepsilon}$ be the solution of \eqref{bvp2}, then
	\begin{equation*}
	\partial_{\nu}\left.u_{\varepsilon}\right|_{\partial \Gamma_\varepsilon} = -\frac{|M_{\varepsilon}|}{2\pi\varepsilon} + W_\varepsilon.
	\end{equation*}
	for some $W_\varepsilon\in O_{L^{\infty}(\partial \Gamma_\varepsilon)}(1)$ as $\varepsilon\rightarrow 0$.
\end{proposition}
\begin{proof}
	By the Green's identity,
	\begin{multline*}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) - u_{\varepsilon}(x) - I_{\varepsilon}(x_0,x)\\
	= \int_{\partial M_{\varepsilon}}\partial_{\nu_y}E(x,y)u_{\varepsilon}(y)dvol_h(y) - \int_{\partial M_{\varepsilon}}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y).
	\end{multline*}
	Using the boundary condition for both $E$ and $u_\varepsilon$, we obtain
	\begin{equation*}
	\frac{1}{|M|}\int_{M_{\varepsilon}}u_{\varepsilon}(y)dvol_g(y) - u_{\varepsilon}(x) + \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = I_{\varepsilon}(x_0,x).
	\end{equation*}
	Next, as in Proposition \ref{partial_u}, we take $\partial_{\nu_x}$, restrict to $\partial M_\varepsilon$, and use Lemma \ref{analogue} to obtain
	\begin{equation*}
	- \partial_{\nu}u_{\varepsilon}(x) + \partial_{\nu} \int_{\partial \Gamma_\varepsilon}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	By Proposition 11.3 of \cite{taylor2},
	\begin{equation*}
	- \frac{1}{2} \partial_{\nu}u_{\varepsilon}(x) + \int_{\partial \Gamma_\varepsilon}\partial_{\nu_x}\tilde{E}(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) - \int_{\partial \Gamma_\varepsilon}\partial_{\nu_x}C(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon)
	\end{equation*}
	so that
	\begin{equation*}
	- \frac{1}{2} \partial_{\nu}u_{\varepsilon}(x) +  \int_{\partial \Gamma_\varepsilon}\partial_{\nu_x}E(x,y)\partial_{\nu}u_{\varepsilon}(y)dvol_h(y) = O_{L^{\infty}(\partial \Gamma_\varepsilon)}(\varepsilon).
	\end{equation*}
	Further, we repeat the steps of the proof of Proposition \ref{partial_u} with replacing Lemma \ref{sing str of E} to Lemma \ref{analogue}.
\end{proof}
Repeating the proof of Theorem \ref{main} by replacing Proposition \ref{partial_u} with \ref{partial_u_2} and Lemmas \ref{I}-\ref{sing str of E} with \ref{analogue} yields the following theorem
\begin{theorem}\label{main2}
	Let $(M,g,\partial M)$ be a compact, connected and orientable Riemannian surface with smooth boundary. Fix $x_0\in M^0$ and let $\Gamma_{\varepsilon}:=B_\varepsilon(x_0)$ be a geodesic ball centered at $x_0$ of geodesic radius $\varepsilon >0$ such that $\partial \Gamma_\varepsilon \cap \partial M = \emptyset$ .\\
	\\
	i) For each $x\notin \Gamma_\varepsilon$, the first-passage time satisfies the  following asymptotic formula, as $\varepsilon\rightarrow 0$,
	\begin{multline*}
	\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = x] = -\frac{|M|}{2\pi}\log\varepsilon + |M|P_{-4}(x_0,x_0)
	- |M| E(x,x_0) + r_{\varepsilon}(x) + O(\varepsilon\log\varepsilon).
	\end{multline*}
	for some function $r_\varepsilon$ such that $\|r_\varepsilon\|_{C(K)} \leq C_{K} \varepsilon$ for any compact $K\subset M$ for which $K\cap \Gamma_\varepsilon = \emptyset$. The Neumann Green's function $E(x,y)$ is given by \eqref{Neumann Green function} and $P_{-4}(x_0,x_0)$ is the evaluation at $(x,y)=(x_0,x_0)$ of the kernel $P_{-4}(x,y)$ in \eqref{dec2}.\\
	\\
	ii) Let $M_\varepsilon=M\setminus \Gamma_{\varepsilon}$, then the spatial average of the mean first-passage time satisfies the asymptotic formula, as $\varepsilon\rightarrow 0$,
	\begin{equation*}
	\frac{1}{|M|}\int_{M_{\varepsilon}}\mathbb{E}[\tau_{\Gamma_{\varepsilon}} | X_0 = y]dvol_g(y) = -\frac{|M|}{2\pi}\log\varepsilon +|M|P_{-4}(x_0,x_0)+  O(\varepsilon\log\varepsilon).
	\end{equation*}
\end{theorem}

\begin{appendix}
\section{Proof of Proposition \ref{sing structure E}}
\label{proof of lemma 3.1}
In this section we provide a brief outline for the necessary aspects of the theory of pseudo-differential operators. For a greater in-depth description of $\mathit{\Psi DO}$, we refer to the reader to \cite{Hor3}, \cite{taylor2} or \cite{doi:10.1142/4047}. After the basic elements of $\mathit{\Psi DO}$s have been described, we offer a proof for Proposition \ref{sing structure E}.
\subsection{Overview of Pseudo-differential operators ($\mathit{\Psi DO}$) on Manifolds}
Let $p(x,\xi) \in C^\infty (T^*\mathbb{R}^n)$. We call $p(x,\xi)$ a standard symbol of order $m$ if for all $m\in \mathbb{R}$, the following estimate holds uniformly
\begin{align*}
    |D^\alpha_x D^\beta_\xi p(x,\xi)| \lesssim \langle\xi\rangle^{m-|\beta|}
\end{align*}
for every multi-index $\alpha,\beta\in \mathbb{N}^n$. Should $p(x,\xi)$ be an order $m$ standard symbol, we say that $p(x,\xi) \in S^m_{1,0}(T^*\mathbb{R}^n)$.
\begin{Rem}
    We use $D_\xi := -i \partial_\xi, D_x := -i\partial_x$ and $\langle\xi\rangle:= (1+|\xi|^2)^{1/2}$.
\end{Rem}
Of particular interest is a subspace of $S^m_{1,0}(T^*\mathbb{R}^n)$ known as the \emph{classical symbols of order $m$} denoted by $S^m_{cl}(T^*\mathbb{R}^n)$. Such symbols are defined via a homogeneity requirement on the asymptotic expansion of $p(x,\xi)$
\begin{equation}
    p(x,\xi) \sim \sum_{j=0}^\infty p_{m-j}(x,\xi). \label{homogeneous expansion}
\end{equation}
where $p_{m-j}(x,\xi)$ are homogeneous of order $m-j$ in the fiber for all $x\in \mathbb{R}^n$. i.e. $p_{m-j}(x,\lambda \xi) = \lambda^{m-j}p_{m-j}(x,\xi)$ for $\lambda, |\xi|\geq 1$. The above expansion \eqref{homogeneous expansion} is an asymptotic expansion in the sense that
\begin{align*}
    p(x,\xi) - \sum_{j=0}^N p_{m-j}(x,\xi) \in S^{m-N-1}_{1,0}(T^*\mathbb{R}^n).
\end{align*}
If $p(x,\xi)\in S^m_{1,0}(T^*\mathbb{R}^n)$, we can define an operator $p(x,D):C_c^\infty(\mathbb{R}^n) \rightarrow \mathcal{D}'(\mathbb{R}^n)$ which is given locally by the following expression
\begin{equation}
    p(x,D)v:=\int_{\mathbb{R}^n} e^{i\xi\cdot x} p(x,\xi) \widehat{v}(\xi) d\xi. \label{local psido rep}
\end{equation}
Such an operator is called an $m$-th order pseudo-differential operator and we say that $p(x,D) \in \Psi^m_{1,0}(\mathbb{R}^n)$. We can also define $\Psi^m_{cl}(\mathbb{R}^n)$ by requiring $p(x,\xi)\in S^m_{cl}(T^*\mathbb{R}^n)$ in \eqref{local psido rep}. Furthermore, we can uniquely extend $p(x,D)$ to a bounded linear operator $p(x,D):H^k(\mathbb{R}^n) \rightarrow H^{k-m}(\mathbb{R}^n)$ for $k\in \mathbb{R}$. We also define the space of \emph{smoothing operators}, pseudo-differential operators with smooth kernels \emph{along the diagonal} as
\begin{align*}
    \Psi^{-\infty}(\mathbb{R}^n) := \bigcup_{m\in \mathbb{R}} \Psi^m(\mathbb{R}^n).
\end{align*}
Smoothing operators arise as pseudo-differential operators of symbols belonging to the space defined by
\begin{align*}
    S^{-\infty}(T^*\mathbb{R}^n) = \bigcup_{m\in \mathbb{R}} S^m(T^*\mathbb{R}^n).
\end{align*}
We also have that if $p(x,D)\in \Psi^m(\mathbb{R}^n)$ and $q(x,D)\in \Psi^l(\mathbb{R}^n)$, then $p(x,D)q(x,D) \in \Psi^{m+l}(\mathbb{R}^n)$. The way such composition is defined is via a \emph{symbol calculus}. The symbol for $p(x,D)q(x,D)$, denoted by $(p\#q)(x,\xi)$ is given by
\begin{equation}
    (p\#q)(x,\xi) \sim \sum_\mu \frac{i^{|\mu|}}{\mu!} D_\xi^\mu p(x,\xi)D_x^\mu q(x,\xi). \label{product formula}
\end{equation}
where $\mu\in \mathbb{N}^n$ denotes a multi-index. The derivation for this formula can be found in \cite{taylor2}, Chapter 7, Section 3. Another important aspect of pseudo-differential operators, which will largely be used in the proof of proposition 3.1 is the notion of elliptic parametrices. First, if $p(x,D)\in \Psi^m(\mathbb{R}^n)$, we say that $p(x,D)$ is elliptic if the following lower bound estimate holds for constants $C,R>0$
\begin{align*}
    |p(x,\xi)| \geq C(1+|\xi|)^m, \;\ \text{for} |\xi| \geq R.
\end{align*}
If $p(x,D)$ is elliptic, then the following theorem holds
\begin{theorem}
    If $p(x,D)\in \Psi^m(\mathbb{R}^n)$ is elliptic, then there exists a $q(x,D),\tilde{q}(x,D)\in \Psi^{-m}(\mathbb{R}^n)$ such that
    \begin{align*}
        p(x,D)q(x,D) &= I + \Psi^{-\infty}(\mathbb{R}^n) \\
        \tilde{q}(x,D)p(x,D) &= I + \Psi^{-\infty}(\mathbb{R}^n)
    \end{align*}
\end{theorem}
The proof for the above theorem can be found in \cite{taylor2}, Chapter 7, Section 4. The operators $q(x,D),\tilde{q}(x,D)$ are known as right and left parametrices of $p(x,D)$ respectively. Furthermore, it is a straightforward corollary that $q(x,D) = \tilde{q}(x,D)+\Psi^{-\infty}(\mathbb{R}^n)$.
\\~\\
In lieu of the pseudo-differential theory on $\mathbb{R}^n$, there is a natural extension to $C^\infty$-manifolds. Let $M$ be a closed manifold. An operator $A:C^\infty(M)\rightarrow \mathcal{D}'(M)$ is said to belong to $\Psi^m_{1,0}(M)$ if there is an atlas $(U_j,\varphi_j)$ covering $M$, with $\varphi_j :U_j \rightarrow V_j\subset \mathbb{R}^n$ and a partition of unity $\{\chi_j\}$ subordinate to the atlas covering such that the following operator
\begin{align*}
    u\mapsto (\chi_kA\chi_j\varphi_j^*u)\circ \varphi_k^{-1}.
\end{align*}
belongs to $\Psi^m_{1,0}(\mathbb{R}^n)$. Similarly, if $a\in C^\infty(T^*M)$, we say that $a\in S^m_{1,0}(T^*M)$ if
\begin{align*}
    \chi_j\circ \varphi_j^{-1}a(\varphi_j^{-1}(\cdot),\varphi_j^*\cdot) \in S^m_{1,0}(T^*\mathbb{R}^n).
\end{align*}
The classical pseudo-differential operators and symbols on $M$ are defined in the same way.
\subsection{Proof of Proposition 3.1}
Since $\Delta_g\in \Psi^2_{cl}(M)$ elliptic, we have that as a result of Theorem A.2, there is a parametrix $P\in \Psi^{-2}_{cl}(M)$ satisfying the following equation
\begin{align*}
    \Delta_gP = I +\Psi^{-\infty}(M)
\end{align*}
Furthermore, as a corollary of Borel's lemma ,\cite{doi:10.1142/4047}, we can express the Schwartz kernel of $P$ as
\begin{align*}
    P(x,y) = \sum_{j=0}^\infty P_{-2-j}(x,y)
\end{align*}
where $P_{-2-j}\in \Psi^{-2-j}_{cl}(M)$. A standard first order parametrix construction indicates that we can choose for $x$ near $y$
\begin{align*}
    P_{-2}(x,y) = -\frac{1}{2\pi}\log d_g(x,y)
\end{align*}
So, our claim is that $P_{-3} = 0$. This problem, reduces to showing that
\begin{equation}
    P-P_{-2} \in \Psi^{-4}_{cl}(M) \label{Reduction 1}
\end{equation}
Left composition of $\Delta_g$ with \eqref{Reduction 1} results in the following equivalent formulation
\begin{equation}
    \Delta_g P_{-2} -I \in \Psi^{-2}_{cl}(M) \label{Reduction 2}
\end{equation}
Self-adjointness of $P_{-2}$ and $\Delta_g$imply that \eqref{Reduction 2} is equivalent to
\begin{equation}
    P_{-2}\Delta_g - I \in \Psi^{-2}_{cl}(M) \label{Adjoint}
\end{equation}
Should \eqref{Adjoint} be true, then we would infer that the expansion for $P_{-2}\Delta_g$ consists of no $-1$ order pseudo-differential operator. This is equivalent to requiring the principle symbol, which is homogeneous of degree $-1$ satisfy the following
\begin{align*}
    \sigma_{-1}(P_{-2}\Delta_g-I)(y,\eta) = 0, \,\ \text{for all} \,\ (y,\eta)\in T^*M
\end{align*}
In order to attain the above requirement, we show that $\sigma_{-1}(P_{-2}\Delta_g-I)$ can be bounded from above in the following manner
\begin{equation}
    |\sigma_{-1}(P_{-2}\Delta_g-I)(y_0,\tau \eta_0)| \lesssim \tau^{-2} \label{Estimate for principle symbol}
\end{equation}
for $\tau \rightarrow \infty$ and fixed $(y_0,\eta_0)\in S^*M$. Since the decay is radially symmetric and is independent of the choice of $y_0$, \eqref{Estimate for principle symbol} implies that $\sigma_{-1}(P_{-2}\Delta_g-I)$ vanishes on $T^*M$. Now, we let $\Phi:V \rightarrow U$ be a Riemann normal co-ordinate chart, centered at $y_0$ for which $\Phi(0) = y_0\in U\subset M$. Let $A:C_c^\infty(\mathbb{R}^2)\rightarrow \mathcal{D}'(\mathbb{R}^2)$ and $B:C_c^\infty(\mathbb{R}^2)\rightarrow \mathcal{D}'(\mathbb{R}^2)$ denote the pull-back operators for $P_{-2}$ and $\Delta_g$ by $\Phi$ respectively. Then, by the invariance of principle symbols under symplectomorphism, we have that
\begin{align*}
    \sigma_{-1}(P_{-2}\Delta_g - I)(y_0,\eta_0) = \sigma_{-1}(AB-I)(0,\xi)
\end{align*}
If $a(t,\xi)$ and $b(t,\xi)$ denote the symbols of $A$ and $B$, then by \eqref{product formula}, we have that
\begin{align*}
    (a\#b)(t,\xi) = a(t,\xi)b(t,\xi) -i\sum_{|\mu|=1} D^\mu_\xi a(t,\xi) D^\mu_t b(t,\xi) + S^{-2}_{cl}(T^*\mathbb{R}^2)
\end{align*}
Furthermore, if we restrict $t=0$, since we are working in Riemannian normal co-ordinates, we have that $\left.D^\mu_t b(t,\xi)\right\vert_{t=0}=0$, which is shown in \cite{lee}, and thus
\begin{align*}
    (a\#b)(0,\xi) = a(0,\xi)b(0,\xi) + S^{-2}_{cl}(T^*\mathbb{R}^2)
\end{align*}
Since the symbol $a(0,\xi)$ is given by the Schwartz kernel of $A$, where
\begin{align*}
    a(0,\xi) = -\frac{1}{2\pi} \int_{\mathbb{R}^2} e^{-i\xi\cdot t} \log|t| dt = |\xi|^{-2}
\end{align*}
we have that
\begin{align*}
    (a\#b)(0,\xi) = 1 + S^{-2}_{cl}(T^*\mathbb{R}^2)
\end{align*}
This implies that
\begin{align*}
    |(a\#b)(0,\xi)-1| \lesssim \langle\xi\rangle^{-2}\implies \sigma_{-1}(AB-I)(0,\xi) = 0
\end{align*}
The last equality thus implies that $\sigma_{-1}(P_{-2}\Delta_g-I)(y,\eta) = 0$ for all $(y,\eta)\in T^*M$.
\end{appendix}
\begin{comment}
\section{to do}
2. Add REF for BVP.\\
4. Interchange diff and int.\\
5. Preliminaries add lemmas.\\

\begin{itemize}
\item $I+N$ or $-I+N$
\item with boundary
\item up to the bounded term
\item how we use Green identity: the function is given on whole manifold, while we use Green identity to part of it
\end{itemize}

\end{comment}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


I can stop lurking on this forum now and openly participate again.
hopefully a fair bit of reliable driving enjoyment straight away.
Shame you missed Syncropendence last weekend. But its Camper Jam this weekend and the Syncronauts will be there doing demo drives. A good time to put the Tristar through its paces?
Its a cool truck.... Have you been to see it yet?
He owned it years ago, then a couple of local guys to me owned it. The previous owner used it as his daily workhorse, he was a lumberjack of sorts.
I dont think he really knew what he had. Sounds like Mike has done a fair amount of work on it since.
When i looked at it, there was oil in the water but i believe that Mike fixed it.
I wish i bought it when it was local to me! ha ha!
Good luck with the purchase.
Good luck with it. He has probably raped all the best bits off it and given you the less good parts for top money. Its his style.
Winchweight wrote: Shame you missed Syncropendence last weekend. But its Camper Jam this weekend and the Syncronauts will be there doing demo drives. A good time to put the Tristar through its paces?
Cool, thanks for the info and opportunity.
Think I will start with a few trips to the supermarket..
Sorry thats a sort of joke!
my playgrounds are pretty fun to drive around and full of stuff and trailers.
probably somewhere wild rather than a crowded VW event.
bene wrote: Its a cool truck.... Have you been to see it yet?
I won't really know what I have until I have been driving it a couple of months.
silverbullet wrote: Good luck with it. He has probably raped all the best bits off it and given you the less good parts for top money. Its his style.
The one part that he paid for in full because there was nothing to haggle. He still tried.
ha ha! cool.... a quite rare truck is what you have!
Baz, wasn't it you that nearly bought it when it was for sale round the corner from me?
bene wrote: Baz, wasn't it you that nearly bought it when it was for sale round the corner from me?
That's right Ben, I didn't buy it then thanks to your feedback ,didn't buy it now either which means I have 9k + to spend !! Lots of booze or maybe a T4 Syncro ?
Syncrobaz wrote: or maybe a T4 Syncro ?
I tried that one, they are a bit better than a T4 but not much.
Its certainly not got the balance and traction you have been used to.
I could get mine in and out of the places I needed to, eventually.
which is dodgy on narrow twisty, muddy, stone walled lanes.
I very nearly bought one 10 years ago, syncro's were a lot cheaper then..


Over the sound of the music we hear Brad asking Diana if she wants to look at his paintings.
But Brad doesn't look very happy with this idea. 'Oh, would you, Mrs… er, Willow?' he says.
And off Mum goes, looking a little annoyed that Brad doesn't want to show his pictures to her himself.
A loud song is playing on the stereo, and I can't hear what Brad and Di are saying any longer. I'm just about to get up and go for a walk, or do anything that takes me away from Brad, when Barry comes in.
'Has the party got going yet?' he asks, sitting on the arm of my chair.
'It has for some people,' I say, looking at Di and Brad.
I must tell you I've never felt less like dancing in all my life. Yet somehow, there I am, with the rest of them, moving my body to the music. Brad doesn't look as if he wants to dance either. Actually, he isn't very good at it. It's quite surprising actually, the way he looks so… well, unnatural is as good a way of describing it as any. He doesn't seem to have a sense of rhythm at all. Whereas Barry's always been a really good dancer, much better than I am.
'Do you mind if I borrow Barry for a salsa?' Diana asks me when a South American song starts playing, and I shake my head.
Brad and I stand and watch. If you've ever seen a couple salsa dancing, you'll know how close the man and woman's bodies get to each other. They touch almost all the time. In fact, sometimes salsa dancing can look a lot like… making love.
'Of course she isn't!' I say, but suddenly, watching Di and Barry together, I feel really… well, strange. And somehow I can't help remembering how they were laughing together in the kitchen earlier on.
'Perhaps we should make them jealous,' Brad suggests, and next thing I know, he's taken me into his arms and he's bending his head and… he's kissing me!
There is a long silence, broken only by the ringing of the doorbell. Then suddenly I hear someone laughing. I don't realise at first that it's me.
'm sorry. Oh, I'm so sorry, Brad,' I say, but it's no good, I can't help it. You see, the white picture looks so much better with a little bit of colour on it. Bright Christmassy colour.
And suddenly everyone's laughing. Everyone: me, Barry, Di, Mum and even Christopher. In fact the only person not laughing is Brad.
The doorbell's still ringing. When we open the door, we find three people waiting outside: a man, a woman, and… Tania.
'Mummy!' shouts Christopher. 'Daddy!' And the man pushes past Tania to pick up his son and lift him high into the air.
When Christopher comes back down to earth we all crowd round him in a big happy group. Somehow Tania manages to push her way past us.
We all leave soon after that. Christopher goes home with his parents, Diana gives my mother a lift back to Rottingdean, and I… well, I go with Barry.
At first we're rather quiet with each other. I think we both feel a little shy. Either that, or we both know that whatever we say to each other in the next hour or so is really important.
Finally, Barry looks at me. 'Do you feel like going for a walk?' he asks, and I nod.
We both look down at my unsuitable shoes. 'These are fine,' I say.
'There might be mud,' Barry says doubtfully.
'It's OK,' I say. The fact is, I just want to be with Barry. In rain, in mud. Anywhere. I know you might find that difficult to believe after all I've said about him, but it's true.
You see, when I saw Barry and Di dancing like that, it did feel really strange. Oh, I know it was only a dance to them, but it did make me think. For the first time I thought about what it would be like if I didn't see Barry any more. What it would be like if he got a new girlfriend. It wasn't a pleasant thought at all. And suddenly I knew why I'd never managed to tell Barry it was over between us. I hadn't really wanted it to be over.
'For a moment there, I thought you were attracted to Brad,' Barry says as we walk along a path through the mud.
'No!' I say, taking his hand, and, strange as it sounds, it isn't a lie. Honestly!
I know I've wanted Brad to kiss me for weeks. I know I said I wanted to marry him and have his babies. But the fact is, when he kissed me… nothing happened. I didn't feel anything. Not one thing.
Diana was right. I had a crush on Brad. I was attracted to him because of his good looks, and I thought that meant I was in love with him. But the truth is, I didn't know him at all. These last few days I've had the chance to get to know him much better, and… well, I now realise I've never seen him properly. In my mind, I made Brad into the type of person I wanted him to be. But that Brad never really existed.
Brad's never been interested in me. He's not even very interested in his nephew. Brad is only interested in Brad. In fact, he's really rather shallow.
I look down at myself. I'm covered in mud. It isn't only on my legs, but all over my clothes too. 'It doesn't matter,' I say.


text:
Габриэлла Джамманко (, родилась 13 июня 1977 года в Палермо) — итальянский политик, депутат Палаты депутатов Италии от партии «Вперёд, Италия».

Биография
Получила образование в сфере коммуникаций в 2003 году, став профессиональной журналисткой. Работала на телевидении Палермо ведущей передачи «Opinion leader» и редактором телеканала TG4, в том числе и в передаче «Sipario».

Габриэлла Джамманко была избрана в Палату депутатов 29 апреля 2008 года по итогам парламентских выборов от округа Сицилия 1 по списку партии «Народ свободы». Она заседала в VII комиссии (по культуре, науке и образованию) и в XI комиссии (по государственному и частному труду). В мае 2008 года она вместе с Нунцией Де Джироламо оказалась в центре внимания журналистов, поскольку обе женщины обменивались на заседании Палаты депутатов записками с Сильвио Берлускони. С 2008 года Джамманко — вице-президент .

В июне 2011 года кандидат на пост мэра Багерии Бартоло ди Сальво, поддерживаемый партиями «Народ свободы» и «Народ Италии завтра», объявил о том, что Джамманко станет его помощницей в Совете в случае победы, но проиграл на выборах кандидату от партии «Будущее и свобода для Италии» и «Союза центра» Винченцо Ло Мео. В 2012 году Джамманко на муниципальных выборах в Палермо отказалась войти в штаб Массимо Косты, баллотировавшегося от «Народа свободы».

5 марта 2013 года Джамманко была переизбрана на очередных выборах от округа Сицилия 1 от той же партии «Народ свободы», войдя в правоцентристскую коалицию Сильвио Берлускони 19 марта. С 16 ноября 2013 года состоит в партии «Вперёд, Италия», образовавшейся после распада «Народа свободы». С 21 июля 2015 года работает секретарём X комиссии (по производственной деятельности, торговле и туризму).

Одной из близких подруг Габриэллы является Мариярозария Росси. Пресса также приписывала Джамманко роман с Сильвио Берлускони.

Примечания

Литература

Ссылки
 Официальный сайт
 Профиль на сайте Палаты депутатов Италии

Члены Палаты депутатов Италии
Члены партии «Народ свободы»
Члены партии «Вперёд, Италия»
Женщины-политики Италии
Телеведущие Италии

redpajama_set_name: RedPajamaWikipedia


In this article, we will explain how you can automatically push any feedback that comes in from your Usabilla for Apps feedback forms to a Slack Channel using the Slack Mail integration.
A Usabilla account that can edit the Usabilla for Apps feedback forms.
A Slack account with admin rights for enabling a Slack integration.
A Premium Slack Team Account to have the ability to enable integrations.
A Slack channels that you can use to push Usabilla for Apps Feedback too.
First, we're going to set up an integration in Slack that will allow us to push any Usabilla for Apps feedback to Slack using the Slack Mail integration. You can enable the Slack integration by going to the URL below.
Tip: Make sure that you're logged in and that you cleared the adding of the integration with your colleagues. Letting your colleagues know also makes sure that they know what to do with the incoming feedback.
On the web page, you'll see a search box. In this search box, you'll need to search for "mail". Doing so will give you a couple of search results including an integration called "Email". Click on the integration "Email" to go to the integration page.
On the integration page, you'll need to click on the "add configuration". Doing so will open a new page that will allow you to add the integration to your Slack and to configure it. First, you'll be asked to select a Slack channel that will be used for the integration. You're completely free to pick whatever channel you want. After this step, you'll be redirected to the "Integration Settings" page. On this page, you see a couple of key elements.
The first thing that you'll see is that an email address will be displayed. This email address will be used for pushing the Usabilla for Apps feedback, so keep the email address nearby for when we configure the Usabilla for App forms. The email address will also be emailed to you.
The second thing that you'll see is the "Post to channel" you can use this setting to change or channel that will be used for the integration.
The third thing that you'll see is the "Descriptive Label". This is an optional field but we do recommend that you describe here for which Apps the integration is used.
In addition to the 3 aforementioned settings, you'll have the option to customize the integration. In order to make integration to be clear for everyone in your organization, we highly recommend that you customize it. Adding a descriptive name allows to instantly see from which Usabilla for App feedback form the feedback was pushed. Having different icons makes the overall experience a little more polished and allows you to instantly visually see which integration pushed the feedback.
Tip: For the sake of maintainability and organization we recommend that you set up separate "Mail" integrations for each Usabilla for Apps feedback form that you have.
When you're logged into Usabilla, go to the Apps Setup section. In this section click on the Usabilla for Apps feedback form for which you want to setup up the automatic push. Clicking one of the forms will fold-open a details pane. In the details pane, click on the "edit" link to edit your Usabilla for Apps feedback form.
In the edit page of your Usabilla for Apps feedback form, you'll need to make a change in the advanced settings. You can access the advanced settings by clicking on the "Advanced" button in the top-right.
Within the "Advanced Settings", you'll see a checkbox that says "I want to push all feedback items to an email address.". Tick this box and in the text-field below the checkbox fill in the email address from the Slack Mail integration. After adding the email address hit the "ok" button at the bottom of the modal and save your Usabilla for Apps feedback form.
That's it, any feedback coming in from you Usabilla for Apps feedback form will automatically be pushed to your Slack channel of choice using the mail integration. Keep in mind that you can only enable one email address for pushing feedback from a Usabilla for Apps feedback form.
If you want to send different feedback to different Slack channels we highly recommend using multiple Usabilla for Apps feedback forms to get more control of where the feedback is pushed through. You can use multiple Usabilla for Apps feedback forms with a single Slack Mail integration.


Chef Kevin Gillespie of Atlanta's Gunshow wants root vegetables like turnips and carrots to have some time in the spotlight, out of the shadow of a holiday roast.
To help make Gillespie's dream a reality, here are six out-of-the-box recipes for root vegetables that require zero roasting or mashing.
David Chang tosses root vegetables like Jerusalem artichokes, parsnips, lotus root, carrots and potatoes with maple syrup before stir-frying them.
These earthy vegetable chips are great as a salty snack or for adding crunch to stewy braised dishes.
Sweet potatoes make an excellent addition to tomato sauce—they are sweet and creamy, and balance acidic tomatoes perfectly.
These crisp, mild turnips are ready after just two days of room temperature pickling.
Michael Symon slices vegetables like radishes, golden beets and carrots super-thin for a crisp, multicolored salad.
This incredible pizza is topped with buttery braised cabbage, pureed sunchokes, crisp pan-fried sunchokes and gooey Emmental cheese.


Women of these nationalities have far more differences than just distinctions in Chinese and Japanese girl names or using of Chinese vs Japanese script.
Japanese women usually try to keep a low profile. That's why they are very quiet and polite in social situations. If you meet them one day, you will probably be very surprised by the fact that Japanese women have very high-pitched voices regardless of their age. So that's common not only in anime.
Chinese vs Japanese women, on the contrary, are much more self-confident and assertive ladies. They are impulsive, undisciplined and are able to perform a reckless act in public.
In some situations they may even act like men. If a Chinese woman doesn't want to talk to you or have a relationship with you, in most cases she tells you about it without feeling any embarrassment and provide her opinion on you.
Chinese women don't even wish to act more ladylike. They almost never wear feminine outfits, high-heels or put on a makeup. And the worst part about this is that they don't have a feminine walk. It's very difficult to distinguish Chinese women from their male compatriots when they walk down the street.
The same is true for their character. It's a very mannish one and it's common among all the levels of the Chinese population.
Japanese women, in contrast, believe that going outside without makeup is considered to be bad form. Even if they are at home and dressed in housecoats, makeup is a sacred thing.
With reference to clothes, Japanese women have a deep understanding of style. They wear only famous brands, even if it's just a tracksuit. Chinese vs Japanese females lag behind them in this regard. Their clothes can be called peculiar and improper. Chinese women can easily go grocery shopping in the evening wearing pajamas. But it's unacceptable for a Japanese woman.
In spite of the fact that Confucianism has had a huge impact on the mentality of both Chinese and Japanese women for a long time, today's women of these countries are very different from each other.
The point is that women's position in Chinese society changed enormously when communists came to power in China in 1921 and founded the Peoples Republic of China. Since then the status of a woman was raised.
Nowadays Chinese vs Japanese women are treated as equal to men. It means that they can enjoy the same rights and get the same jobs as men. By the way, China is one of few East Asian countries where the question of gender equality isn't an extremely acute one.
With regard to Japan, local women still adhere to strict Confucian traditions according to which women aren't permitted to compete with men in any spheres and have to obey their husbands.
If a Japanese woman marries and gives birth to a child, it's not appropriate for her to go back to work. She has to take responsibility for the home and children.
Chinese vs Japanese women are characterized by yellow skin color. That's why they don't like to sunbathe at the beach. The dermis of Japanese women is lighter than Chinese females' one. Moreover, they actively use cosmetics to get rid of inborn yellowish skin.
It's considered that Japanese women's faces have an elongated shape, whereas Chinese females have faces which are circular in shape.
Chinese females have wide cheekbones, and Japanese ones have big mouths. As concerns the form of their noses, Chinese women have flat ones, whereas Japanese females have sharp noses.
The features of these women's eyes can also help us to identify their nationality. Although both Chinese and Japanese women have Asian eyes and pronounced epicanthus (the folds of upper eyelids), it's quite possible to tell them apart – Japanese women have punched eyes like fishes and they are wide.
Chinese female eyes, for their part, are narrower than Chinese females' ones and their outer corners are raised.
With regard to eyebrows, Japanese women have straight ones and they are therefore forced to give them a desired shape. Chinese females' eyebrows have a natural bending. That's why they can do without putting on a makeup.
As concerns proportions of the body, it's worth mentioning here that Chinese women aren't foreign to international women's entertainment called "I'm getting thinner", but they aren't embarrassed to admit that they like to eat. In their view it's not that bad to be a puffy girl. It's even cool. If you call a Chinese girl plumpy or chubby, it sounds like a compliment.
Nevertheless, you should never call them strong. In this case you risk making enemies, because such a claim is perceived as the accusation of masculinity.
There are a lot of full-figured women in China, but they mainly don't suffer from excess weight. Chinese girls eat a lot of vegetables and different greenery, and they are foreign to chocolate and sweets which contributes to their good figures.
They know which products are useful for women's beauty and fertility. For example, these are ginger, jojoba, black sesame and black raw sugar.
In terms of Japanese women's body shapes, it can be said that their figures are slender. It's not just about genetics. They often do sports and eat healthy. A balanced meal is a key to good health. It's rare to find a chubby Japanese woman.
They have very small breasts which match an "A" bra size or even less. Many of Japanese females therefore use a push-up bra. If you see a woman with big breast in Japan, she is most likely a mother of many children.
As a rule, Japanese women have a beautiful waistline. They also have slim legs which are sometimes slightly uneven and almost total absence of buttocks.
Good-looking and educated Chinese girls with a good job don't rush to get married not because of feminist view, but because of the absence of deserving candidates. Brides in China are in short supply. That's why beautiful Chinese women know their value and are willing to date only rich and handsome men.
Chinese vs Japanese women often choose their partners themselves and offer them to start dating. They don't fear to manage the situation and waste their time waiting for a man's first step. However a man should be careful in this case, since Chinese wives have a firm hand over their husbands. They like giving orders, making scenes, being jealous and controlling their men.
Japanese women, for their part, don't strive to get married, since there is no social security guarantees for women during pregnancy and in the post-natal period in Japan. That's why they strive to build a career, live alone and avoid marriage.
Sometimes Japanese women treat foreigners like movie stars. They can be surprised by blue eyes, fair hair and white skin. Sometimes Japanese women ask foreigners' permission to touch their hair and then admire its softness.
The narrowness of European faces and the size of our eyes usually amaze them. Sometimes they find it unnatural that the color of eyes and hair can be different.
Foreign men seem giant to them. Tall men thrill them.
Chinese vs Japanese women, on the contrary, don't face a cultural shock at the sight of a foreigner. Their attitude to them is pragmatic and consuming. The appearance of any foreigner causes a desire to receive something from him, whether a good deal, a contract, money or an invitation.
Which girls do you mostly like: Chinese or Japanese? Or, maybe, Korean? Write in comments, please, or send us your own story!
Next Post:TOP-10 most greedy women of all time!


Duplicate Attribute Resiliency is a feature in Azure Active Directory designed to eliminate friction caused by UserPrincipalName and ProxyAddress conflicts when running one of Microsoft's synchronization tools. The UserPrincipalName and ProxyAddress attributes are generally required to be unique across all User, Group, or Contact objects in a given Azure Active Directory directory.
If the attribute is not required, like a ProxyAddress, Azure Active Directory simply quarantines the conflict attribute and proceeds with the object creation or update.


Our grandparents are the first persons to spoil us since birth. They have taken care of our parents when they were young. Yet, they love their grandchildren more than their child.
We feel remorse whenever we hear news about grandparents being abused and taken advantaged by bad people.
In a recent viral post from Nathaniel Casupang, an old couple from Lucban, Quezon was robbed by unidentified burglars from their home while sleeping. The robbers are able to steal four thousand pesos from Lola Tessa and Lolo Roberto. Moreover, the burglars also hit them with an adverse object.
Casupang is a concerned neighbor of the old couple. He posted the incident on social media and so many have expressed sympathy and offered help for Lolo and Lola.
Nathaniel posted an update regarding the present condition of the two. He also mentioned that alongside private netizens, some church priests, GMA news, and Ruru Madrid send out financial help. Casupang is thankful and grateful because blessings are outpouring for the old couple.
Despite receiving help from various persons and organizations, there were people who have expressed doubts about Casupang's intention to help Lola Tessa and Lolo Roberto.
However, he proved his good intentions, even invited the people to drop by in their place to see the couple. A netizen confirmed that Nathaniel is a neighbor of the two and personally knows the victims. The confirmation brushes off the doubts of some netizens claiming that Casupang is just using the old couple to shell out money.
The old couple is currently living in a private home all thanks to those who have extended help. As of today, the robbers are still unidentified. Lolo Roberto and Lola Tessa are now recovering from wounds inflicted by the robbers.


With over two decades of experience in the Life Science industry across all functions along the value chain plus strong IT capabilities, we have the experience, creativity and know-how to help you achieve your strategic and tactical goals.
Since our objective is to give our clients a keen competitive edge, our solutions do not stop at state-of-the-art strategy, processes and systems.
Together with our customers, we define and implement the services and technology to bridge the divide between merely competing and winning.


Create your own Prezi, PowToon, or Google Presentation and post it on your blog.
Use Diigo to bookmark at least three presentations and/or presentation sites of interest.
Take a look at the guides for Graphic Organizers and Mind Maps.
Try to create a simple mindmap or graphic organizer and post it on your blog.


Twitter (NYSE:TWTR) is ringing in its eighth birthday and as part of the anniversary, the company has reintroduced a feature that allows users to look back at their very first tweet.
Here's the issue: Most users stick with Twitter long enough to post a first tweet, but a new report from Twopcharts, a website that monitors Twitter account activity, notes that just 11 percent of users who were new to the site in 2012 are still tweeting today. The website, while very popular, doesn't seem to be an indispensable tool for its users in quite the same way that rival sites such as Facebook (NASDAQ:FB) continue to be.
According to The Wall Street Journal, which first reported the story on Friday, about 40 percent of the 20 million accounts that are created on Twitter each month post at least one tweet during their first month on the site. Amsterdam-based Twopcharts estimates that just a quarter of those new accounts will still be tweeting next year, on Twitter's ninth birthday.
Twitter's user retention problems aren't new, the Journal notes, reporting the issue has plagued the social media site, which went public with a much-hyped IPO earlier last year, for years now.
Part of the reasoning behind Twitter's retention problems could be the fact that the site can be more confusing for new users compared with other social media platforms. The fact that Twitter users are limited to just 140 characters means that longtime Twitter users often communicate in a kind of shorthand that can throw new users for a loop.
Additionally, features like hashtags and at-replies aren't always intuitive for new users, making it difficult for them to connect with other users in a discussion. For this reason, Twitter has apparently been considering ditching its trademark hashtags and at-replies altogether.
Twopcharts reports that there are currently more than 1.5 billion Twitter accounts, with 955 million of those still in existence today, but a mere 13 percent of those 955 million have posted at least one tweet in the last 30 days.
Worse news is that Twopcharts findings also indicate that accounts made in recent years are actually less likely to continue to be active than those made by early adopters in previous years (anywhere between 2006-2011). The Wall Street Journal says that 25 percent of the accounts created in 2008 are still active and tweeting today, while just 10.7 percent of accounts created in 2012 are likely still tweeting just more than a year later.
Will Twitter Chuck Hashtags and At-Replies?


"The myriad ambiguous signs and letters in the street and fragmented specular images on shop-window displays are, for me, crucial elements in creating peculiar relationships with the world of things, implying the emergence of the imagination.
Through these elements, the infinite boundary of the imagination unleashes the city as unaccustomed, disoriented, and reencountered.
My art practice focuses on the rediscovery of psychological heteromorphic identification, the presence of invisible substance, and fantastic visual experiences emanating from serendipitous moments of glitch in the city. The work implies the opening of the urban landscape to a distinct poetics of the city in which mythology, sign, symbol, voice, text, and trace occurs."
Kyung Hwa Shon's work explores the reciprocal relation between a city and the imagination, where the city is a field of excavation where everything is buried, hidden, and remaining undiscovered.
Kyung Hwa Shon Shon is recipient of numerous prizes, most recently the SOLO Award 2015 and the Alternative Space Loop Art Prize.
Paper | Publication | Performance, Lychee One Gallery, London, UK, 2016; WW Solo Award: The City of Fragments, WW Contemporary Art, London Art Fair 2016, London, UK; 'UK/RAINE', Saatchi Gallery, London, UK, 2015.
Kyung Hwa Shon (born 1983) lives and works in London and Seoul. She is currently studying for a PhD in Painting at the Royal College of Art.


How do you want your business to be perceived by potential and existing customers? A powerful brand identity can be the difference between low or high conversion rate. The very best brands combine clear visual engagement with an intelligent message. We help our clients to deliver both of these essential aspects when developing a brand identity.
Brand identity is an extremely important factor the the image of your company and what you represent. Whether you want to reinvent, update or create a successful brand identity, our team of creatives will work alongside you to produce a powerful brand that people will come to recognise and connect with.
Your business name, logo design, typeface design and tagline all contribute to the overall finish of your brand identity, at Media Street, we have created fantastic brand identity's for a range of companies since 2009. These factors must represent your key business goals and identify the customers as well as understanding how you would like your customers to perceive your company.
We work with you to understand the kind of people you want to attract, by doing this, we create a brand identity that specifically connects with a key target audience.
Media Street will help you to separate your business from the competition with a brand that people won't forget in a hurry. Enhance your customer perceptions and get in touch with us today.