text:
\section{Introduction}
\subsection{Background}
A Euclidean (or spherical or hyperbolic) polyhedral surface is a triangulated surface $(S,T)$ with a metric, called a polyhedral metric, so that each triangle in the triangulation is isometric to a Euclidean (or spherical or hyperbolic) triangle. Let $E$, $V$ be the set of edges and vertices of $(S,T)$, the polyhedral metric can be identified as a function $d:E\to\mathbb{R}_{+}$.
The rigidity problem asks which geometric data on $(S,T)$ can determine $d$ up to isometry. Over the past decades, there are many results related to this topic. See e.g. \cite{Luo3,Luo1,Luo2,Guo,Xu,Glick,GT,Gu1,Gu2,Xu2}.
Let $\mathbb{E}^{2}$, $\mathbb{H}^{2}$, $\mathbb{S}^{2}$ be the Euclidean, the hyperbolic, and the spherical $2$-dimensional geometries, respectively. Luo \cite{Luo3,Luo1} introduced the following two families of discrete curvatures.
\begin{definition}\label{def1}
Let $(S,T)$ be a closed triangulated surface. Given a $\mathrm{K}^{2}$ polyhedral metric $d$ on $(S,T)$ where $\mathrm{K}^{2}=\mathbb{E}^{2}$, or $\mathbb{H}^{2}$, or $\mathbb{S}^{2}$, the $\phi_{h}$-curvature of $d$ is defined by a function $\phi_{h}: E\rightarrow\mathbb{R}$ sending an edge $e$ to
\[\phi_{h}(e)=\int_{a}^{\pi/2}\sin^{h}t\textit{d}t+\int_{a'}^{\pi/2}\sin^{h}t\textit{d}t.\]
The $\psi_{h}$-curvature of $d$ is defined by a function $\psi_{h}: E\rightarrow\mathbb{R}$ sending an edge $e$ to
\[\psi_{h}(e)=\int_{0}^{(b+c-a)/2}\cos^{h}t\textit{d}t+\int_{0}^{(b'+c'-a')/2}\cos^{h}t\textit{d}t,\]
where $a,a'$ are the angles facing the edge $e$ and $b,c,b',c'$ are inner angles adjacent to the edge $e$.
\end{definition}
\begin{comment}
\begin{figure}[htbp]
\centering
\hypertarget{fig.1}{}
\begin{tikzpicture}[scale=0.9]
\coordinate  (C) at (0.3,1.5);
\coordinate  (B) at (1.9,0.1);
\coordinate  (D) at (4,2.3);
\coordinate  (A) at (2.2,2.5);
\draw
 (A) -- node[left] [font=\small]{$e$} (B);
 \draw
 (B) -- (C);
 \draw
 (C) --  (A);
\draw
 (A) -- (D);
 \draw
 (D) -- (B);
\tkzLabelAngle[pos = 0.4][font=\small](B,C,A){$a$}
\tkzLabelAngle[pos =-0.4][font=\small](A,D,B){$a'$}
\tkzLabelAngle[pos = 0.4][font=\small](C,A,B){$b$}
\tkzLabelAngle[pos=0.4][font=\small](A,B,C){$c$}
\tkzLabelAngle[pos = 0.4][font=\small](B,A,D){$b'$}
\tkzLabelAngle[pos = 0.5][font=\small](D,B,A){$c'$}
\end{tikzpicture}
\caption{\small Edge-angle relation in Definition \ref{def1}.}
\end{figure}
\end{comment}

\begin{figure}[htbp]
\centering

\includegraphics[scale=0.3]{figure1}
\caption{\small Edge-angle relation in Definition \ref{def1}.}
\label{fig1}

\end{figure}

Remark that some choices for the parameter $h$ are particularly interesting. For example, when $h=0$, one reads
\[\phi_{0}(e)=\pi-a-a'\quad\text{and}\quad\psi_{0}(e)=\frac{1}{2}(b+b'+c+c'-a-a').\]
The geometric meaning of them is related to the dihedral angle along edges of a hyperbolic polyhedron associated to the polyhedral surface. In addition, the curvature
\[\phi_{-2}(e)=\cot a+\cot a'\]
is the discrete (cotangent) Laplacian operator on a polyhedral surface. See \cite{Pin,Bo2,Zeng} for more information.
Moreover, a Euclidean or hyperbolic polyhedral surface is Delaunay if and only if $\psi_{h}(e)\geq0$ for all $e\in E$.
\begin{theorem}[Luo]\label{th1.1}
Let $(S,T)$ be a closed triangulated surface. For any $h\in\mathbb{R}$, the following statements hold:
\begin{itemize}
\item[$(a)$] A Euclidean polyhedral metric on $(S,T)$ is determined up to scaling by its $\phi_{h}$-curvature or $\psi_{h}$-curvature.
\item[$(b)$] A hyperbolic polyhedral metric on $(S,T)$ is determined up to isometry by its $\psi_{h}$-curvature.
\item[$(c)$] A spherical polyhedral metric on $(S,T)$ is determined up to isometry by its $\phi_{h}$-curvature.
\end{itemize}
\end{theorem}
\begin{remark}
Rivin \cite{Ri1} and Leibon \cite{Le} proved Theorem \ref{th1.1}$(a)$, Theorem \ref{th1.1}$(b)$ for $h=0$, respectively. In Euclidean geometry, it is easy to see $a+b+c=\pi$, $a'+b'+c'=\pi$, which implies $\phi_{h}(e)=\psi_{h}(e)$.
\end{remark}
A circle packing metric on $(S,T)$ is a polyhedral metric $d:E\to\mathbb{R}_{+}$
so that there is a map $r:V\to\mathbb{R}_{+}$ satisfying $d(uv)=r(u)+r(v)$ whenever the edge $uv$ has end vertices $u$ and $v$. The $k_{h}$-curvature of the circle packing metric was introduced by Luo \cite{Luoarxiv,Luo1}.
\begin{definition}
Let $(S,T)$ be a closed triangulated surface. Given a circle packing metric $r$ on $(S,T)$, the $k_{h}$-curvature of $r$ is defined by the function $k_{h}:V\rightarrow\mathbb{R}$ sending a vertex $v$ to
\[k_{h}(v)=(2-\frac{m}{2})\pi-\sum_{i=1}^{m}\int_{\pi/2}^{\theta_{i}}\tan^{h}(t/2)\textit{d}t,\]
where $\theta_{1},\cdots,\theta_{m}$ are all inner angles at $v$.
\end{definition}
Notice that $k_{0}(v)=2\pi-\sum_{i=1}^{m}\theta_{i}$ is the classical discrete curvature. If $e$ is an edge having $v$ as a vertex, we denote it by $v\prec e$. It is easy to show $\sum\nolimits_{v\prec e}\psi_{0}(e)=2\pi-k_{0}(v)$.

\begin{theorem}[Luo]\label{th1.2}
Let $(S,T)$ be a closed triangulated surface. For any $h\in\mathbb{R}$, the following statements hold:
\begin{itemize}
\item[$(a)$] A Euclidean circle packing metric on $(S,T)$ is determined up to scaling by its $k_{h}$-curvature.
\item[$(b)$] A hyperbolic circle packing metric on $(S,T)$ is determined up to isometry by its $k_{h}$-curvature.
\end{itemize}
\end{theorem}
\begin{remark}
Theorem \ref{th1.2}$(a)$ for $h=0$ was proved by Andreev \cite{An} and Thurston \cite{Thu}. Theorem \ref{th1.2}$(b)$ for $h=0$ was proved by Thurston \cite{Thu}.
\end{remark}

\subsection{Main results}
The polyhedral surfaces in above works are required to be closed. One may ask whether analogous results still hold for bordered polyhedral surfaces. Actually, there are two questions.
The first one is parallel to this for closed case: Whether a polyhedral metric is determined by its discrete curvature. This was answered by the works of Ba-Zhou \cite{Ba}, which shows a similar result to Theorem \ref{th1.1}.
The second question asks whether a polyhedral metric is determined by its discrete curvature on interior edges together with its boundary value. We mention that this problem plays a significant role in exploring rigidity of cyclic polygons. See Corollary \ref{co} for details.

For an edge $e$ of a bordered polyhedral surface $(S,T)$, we call $e$ a boundary edge if $e$ belongs to the boundary of $S$. Otherwise, we call $e$ an interior edge. Similarly, we can define boundary vertices and interior vertices. For a polyhedral metric $d:E\to\mathbb{R}_{+}$ on $(S,T)$, the boundary value of $d$ is the restriction of $d$ to $E_{\partial}$, where $E_{\partial}$ is the set of boundary edges. In this paper we prove the following result.

\begin{theorem}\label{mt1}
Let $(S,T)$ be a bordered triangulated surface. For any $h\in\mathbb{R}$, the following statements hold:
\begin{itemize}
\item[$(a)$] A Euclidean polyhedral metric on $(S,T)$ is determined up to isometry by its boundary value and $\phi_{h}$-curvature (or $\psi_{h}$-curvature) on interior edges.
\item[$(b)$] A hyperbolic polyhedral metric on $(S,T)$ is determined up to isometry by its boundary value and $\psi_{h}$-curvature on interior edges.
\item[$(c)$] A spherical polyhedral metric on $(S,T)$ is determined up to isometry by its boundary value and $\phi_{h}$-curvature on interior edges.
\end{itemize}
\end{theorem}
A Euclidean (or hyperbolic) polygon is said to be cyclic if it can be inscribed into a Euclidean (or hyperbolic) circle. For a cyclic polygon, by adding proper interior edges, we can make it into a bordered polyhedral surface (see Figure \ref{fig2}). Notice that the boundary value of this bordered polyhedral surface is exactly the lengths of sides of the cyclic polygon and its $\psi_{0}$-curvature on each interior edge equals to zero. As a corollary of Theorem \ref{mt1}, we immediately obtain the following result. See e.g. \cite{Pe,Pi,Sc,Ko} for other proofs.
\begin{comment}
\begin{figure}[htbp]
\centering

\includegraphics[scale=0.3]{figure2}
\caption{\small Triangulated cyclic hexagon}
\label{fig2}

\end{figure}
\end{comment}
\begin{figure}[htbp]
\centering
\begin{minipage}{0.35\textwidth}
\centering
\begin{tikzpicture}[scale=0.25]
\coordinate  (A) at (5,0);
\coordinate  (B) at (2.5,4.3301270);
\coordinate  (C) at (-2.5,4.3301270);
\coordinate  (D) at (-5,0);
\coordinate  (E) at (-2.5,-4.3301270);
\coordinate  (F) at (2.5,-4.3301270);
\draw (0,0) circle (5);
\draw(A)--(B)--(C)--(D)--(E)--(F)--(A);
\end{tikzpicture}
\end{minipage}
\begin{tikzpicture}
\draw[->] (0,0) -- (1,0);
\end{tikzpicture}
\begin{minipage}{0.35\textwidth}
\centering
\begin{tikzpicture}[scale=0.25]
\coordinate  (A) at (5,0);
\coordinate  (B) at (2.5,4.3301270);
\coordinate  (C) at (-2.5,4.3301270);
\coordinate  (D) at (-5,0);
\coordinate  (E) at (-2.5,-4.3301270);
\coordinate  (F) at (2.5,-4.3301270);
\draw (0,0) circle (5);
\draw(A)--(B)--(C)--(D)--(E)--(F)--(A);
\draw[dashed](B)--(D)--(F)--(B);
\end{tikzpicture}
\end{minipage}
\caption{\small Triangulated cyclic hexagon}
\label{fig2}
\end{figure}
\begin{corollary}\label{co}
A Euclidean (or hyperbolic) cyclic polygon is determined up to isometry by its lengths of sides.
\end{corollary}
In \cite{Luo3,Luo1}, Luo asked whether $\phi_{h}$-curvature determines a hyperbolic polyhedral metric. Below we give a part answer to this question. We say a triangulated surface $(S,T)$ is stripped if every triangle of $T$ has at least one boundary edge. For example, the left (resp. right) in Figure \ref{fig3} is a stripped (resp. non-stripped) triangulated surface.

\begin{comment}
\begin{figure}[htbp]
\centering

\includegraphics[scale=0.3]{figure3}
\caption{\small Stripped and non-stripped triangulated surfaces}
\label{fig3}

\end{figure}
\end{comment}
\begin{figure}[htbp]
\centering
\begin{minipage}{0.420\textwidth}
\centering
\begin{tikzpicture}[scale=0.6]
\coordinate   (C) at (0.8,2);
\coordinate   (B) at (2.2,0.1);
\coordinate   (D) at (4.4,2);
\coordinate   (A) at (2.7,2.1);
\coordinate   (E) at (4,0);
\coordinate   (F) at (3.5,-1.5);
\coordinate   (G) at (5.5,-1.5);
\coordinate   (H) at (5.7,0);
\coordinate   (I) at (1.9,3.6);
\draw(A)--(B)--(C)--(A);
\draw
 (A) -- (D);
\draw
(D) -- (B)--(E)--(D);
\draw
(B)--(F)--(E);
\draw
(E)--(G)--(F);
\draw
(E)--(H)--(G);
\draw
(C)--(I)--(A);
\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.420\textwidth}
\centering
\begin{tikzpicture}[scale=0.6]
\coordinate   (C) at (0.8,2);
\coordinate   (B) at (2.2,0.1);
\coordinate   (D) at (4.4,2);
\coordinate   (A) at (2.7,2.1);
\coordinate   (E) at (4,0);
\coordinate   (F) at (3.5,-1.5);
\coordinate   (G) at (5.5,-1.5);
\coordinate   (H) at (5.7,0);
\coordinate   (I) at (1.9,3.6);
\coordinate   (J) at (0,0.5);
\draw[line width=1.2]
(A)--(B)--(C);
\draw[line width=1.2]
(C)--(A);
\draw
(A)--(D);
\draw[line width=1.2]
(D)--(B);
\draw
(E)--(D);
\draw[line width=1.2]
(E)--(B);
\draw[line width=1.2]
(E)--(D);
\draw
(B)--(F);
\draw
(F)--(E);
\draw
(E)--(G)--(F);
\draw
(E)--(H)--(G);
\draw
(C)--(I)--(A);
\draw
(D)--(H);
\draw
(C)--(B)--(J)--(C);
\end{tikzpicture}
\end{minipage}
\caption{\small Stripped and non-stripped triangulated surfaces}
\label{fig3}
\end{figure}
\begin{theorem}\label{1.7}
Let $(S,T)$ be a stripped triangulated surface. For any $h\in\mathbb{R}$, a hyperbolic polyhedral metric on $(S,T)$ is determined up to isometry by its boundary value and $\phi_{h}$-curvature of interior edges.
\end{theorem}
For a circle packing metric $r:V\to\mathbb{R}_{+}$, the boundary value is the restriction of $r$ to $V_{\partial}$, where $V_{\partial}$ is the set of boundary vertices.
\begin{theorem}\label{mt2}
Let $(S,T)$ be a bordered triangulated surface. For any $h\in\mathbb{R}$, the following statements hold:
\begin{itemize}
\item[$(a)$] A Euclidean circle packing metric on $(S,T)$ is determined by its boundary values and $k_{h}$-curvature of interior vertices.
\item[$(b)$] A hyperbolic circle packing metric on $(S,T)$ is determined by its boundary values and $k_{h}$-curvature of interior vertices.
\end{itemize}
\end{theorem}

Theorem \ref{mt2} can be applied to study the rigidity of planar circle packing. A planar circle packing is a connected collection of circles on Euclidean plane whose interiors are disjoint. The contact graph of a circle packing is the graph having a vertex for each circle, and an edge for every pair of circles that are tangent. If there exists a bordered triangulated surface whose nerve is isomorphic to the contact graph of the circle packing, the circle packing is called a triangle-type circle packing. Suppose $P$ is a triangle-type circle packing. If the vertex of a circle in its contact graph is corresponding to a boundary point of the bordered triangulated surface under the isomorphism, the circle is called a boundary circle in $P$. As a corollary Theorem \ref{mt2}, we obtain the following well-known result regarding the rigidity of planar circle packing.
\begin{comment}
\begin{figure}[htbp]
\centering

\includegraphics[scale=0.3]{figure4}
\caption{\small A circle packing and its contact graph}
\label{fig4}

\end{figure}
\end{comment}
\begin{figure}[htbp]
\centering
\begin{minipage}{0.420\textwidth}
\centering
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\hspace{4cm}
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\end{tikzpicture}
\end{minipage}
\caption{\small A circle packing and its contact graph}
\label{fig4}
\end{figure}
\begin{corollary}\label{cocp}
A triangle-type planar circle packing on the Euclidean plane is determined up to isometry by the size of boundary circles in the sense of keeping the contact graph unchanged.
\end{corollary}
The variational principle is applied to proofs of the above theorems. The use of variational principle in the study of triangulated surfaces first appeared in the work of Colin de Verdi\`{e}re \cite{Co} to give a new proof Koebe-Andreev-Thurston's circle packing theorem. Similar methods have been used by Rivin \cite{Ri1,Ri2}, Leibon \cite{Le}, Bobenko-Springborn \cite{Bo1} in other situations. We are also highly influenced by the works of Luo \cite{Luoarxiv,Luo3,Luo1}, Luo-Dai-Gu~\cite{L.D.G}, Guo-Luo \cite{Luo2}, Guo \cite{Guo}, Zhou \cite{Zhou1,Zhou2}, Xu \cite{Xu,Xu2}, Ge-Hua-Zhou \cite{Ge1,Ge2} and others.

The present paper is built up as follows: In next section, we establish some preliminary results. In Sections \ref{3}, \ref{4}, we construct several convex energy functionals on polyhedral surfaces. In Section \ref{5}, applying variational principle, we derive Theorem \ref{mt1}, Corollary \ref{co}, Theorem \ref{1.7} and Theorem \ref{mt2}. The last two sections contain appendixes regarding to some results about trigonometry and differential forms. Throughout this paper, we always assume the polyhedral surface is connected unless otherwise stated.
\section{Preliminaries}
\subsection{Basic results about cosine law}\label{section2.1}
We first introduce some basic results regarding to the cosine law function. Let $l_{1}$, $l_{2}$, $l_{3}$ and $\theta_{1}$, $\theta_{2}$, $\theta_{3}$ denote three edges and corresponding inner angles of a triangle. The cosine law expresses the $\theta_{i}$ in terms of $l_{1}$, $l_{2}$, $l_{3}$. See Appendix \ref{AppA} for details. So we regard $\theta_{i}$ as the function of $l_{1}$, $l_{2}$, $l_{3}$, called cosine law function. The following three lemmas establish some basic properties of cosine law function. please refer to \cite{chow,L.D.G,Zhou1,Ge1,Ge2} for more information.
\begin{lemma}\label{2.1}
Let cosine law function be as above. Then
\[
\frac{\partial\theta_{i}}{\partial l_{i}}=\frac{m_{i}}{\sin\theta_{i}m_{j}m_{k}},\qquad\frac{\partial\theta_{i}}{\partial l_{j}}=-\cos\theta_{k}\frac{\partial\theta_{i}}{\partial l_{i}},
\]
where $m_{i}=l_{i}$ (resp. $m_{i}=\sinh l_{i}$, $m_{i}=\sin l_{i}$) in Euclidean (resp. hyperbolic, spherical) triangles.
\end{lemma}
\begin{lemma}\label{silemma}
Let cosine law function be as above. Set $r_{i}=\frac{1}{2}(l_{j}+l_{k}-l_{i})$. Then
\[
\frac{1}{s_{i}}\frac{\partial\theta_{i}}{\partial r_{j}}=\frac{1}{s_{j}}\frac{\partial\theta_{j}}{\partial r_{i}},
\]
where $s_{j}=r_{j}$ (resp. $s_{j}=\sinh r_{j}$) in Euclidean (resp. hyperbolic) triangles.
\end{lemma}
\begin{lemma}\label{2.3}
Let cosine law function be as above. In hyperbolic triangles,
\[\begin{aligned}
\frac{\partial(\theta_{i}-\theta_{j}-\theta_{k})}{\partial l_{i}}&=c\tanh l_{i}/2(\cosh l_{1}+\cosh l_{2}+\cosh l_{3}+1),\\
\frac{\partial(\theta_{i}-\theta_{j}-\theta_{k})}{\partial l_{j}}&=c\coth l_{j}/2(-\cosh l_{i}-\cosh l_{j}+\cosh l_{k}+1),
\end{aligned}\]
where $c$ is a positive constant.
\end{lemma}
Readers can deduce the results by some computation. We omit the proof here.
The following two lemmas play a crucial role in the proof of the convexity of energy functionals constructed in Section \ref{3}.
\begin{lemma}\label{hypermatrix}
Let $l_{1}$, $l_{2}$, $l_{3}$ denote the three edges of a hyperbolic triangle. Set $\mathrm{P}=[p_{st}]$, where \[
p_{ii}=\tanh^{2}\frac{l_{i}}{2}(\cosh l_{1}+\cosh l_{2}+\cosh l_{3}+1),\quad
p_{ij}=-\cosh l_{i}-\cosh l_{j}+\cosh l_{k}+1.
\]
Then $\mathrm{P}$ is positive definite.
\end{lemma}
\begin{proof}
Denote $\theta_{i}$ be the corresponding angle of $l_{i}$. Write $\zeta=-\frac{1}{2}(\theta_{i}+\theta_{j}+\theta_{k})$ and $\alpha_{i}=\frac{1}{2}(\theta_{i}-\theta_{j}-\theta_{k})$. Some tedious computation gives
\[\begin{aligned}
\tanh^{2}{\frac{l_{i}}{2}}(\cosh l_{1}+\cosh l_{2}+\cosh l_{3}+1)&=\frac{-4\cos\zeta\sin\zeta\cos^{2}\alpha_{i}}{\sin\theta_{1}\sin\theta_{2}\sin\theta_{3}},\\
-\cosh l_{i}-\cosh l_{j}+\cosh l_{k}+1&=\frac{4\cos\zeta\cos \alpha_{i}\cos \alpha_{j}\sin \alpha_{k}}{\sin\theta_{1}\sin\theta_{2}\sin\theta_{3}}.
\end{aligned}\]
Set $\mathrm{D}=[d_{st}]$, where $d_{ii}=\cos\alpha_{i}$, $d_{ij}=0$ and $\mathrm{Q}=[q_{st}]$, where $q_{ii}=-\sin\zeta$, $q_{ij}=\sin\alpha_{k}$.
Then
\[
\mathrm{P}=\frac{4\cos\zeta}{\sin\theta_{1}\sin\theta_{2}\sin\theta_{3}}\mathrm{D}\mathrm{Q}\mathrm{D}.
\]
Now it remains to prove the positive-definiteness of $\mathrm{Q}$. We will prove it by showing that determinant of sequential principal minor of $\mathrm{Q}$ is positive. Obviously, we have $-\sin\zeta>0$.
The determinant of $2$-principal minor of $\mathrm{Q}$ equals to
\[\begin{aligned}
&(\sin\zeta+\sin\alpha_{3})(\sin\zeta-\sin \alpha_{3})\\
=&\ 4\sin\frac{\zeta+\alpha_{3}}{2}\cos\frac{\alpha_{1}+\alpha_{2}}{2}\cos\frac{\zeta+\alpha_{3}}{2}\sin\frac{\alpha_{1}+\alpha_{2}}{2}\\
=&\sin(\zeta+\alpha_{3})\sin(\alpha_{1}+\alpha_{2})\\
=&\sin(\theta_{1}+\theta_{2})\sin\theta_{3}>0.
\end{aligned}\]
Some tedious computation yields
\[
\begin{aligned}
\det\mathrm{Q}&=\sin\zeta\cos^{2}\zeta-\frac{1}{2}\cos\zeta\sum\nolimits_{i=1}^{3}\sin2\alpha_{i}\\
&=\frac{1}{2}\cos\zeta\left(\sin2\zeta-\sum\nolimits_{i=1}^{3}\sin2\alpha_{i}\right)\\
&=\cos\zeta\big(\cos(\theta_{1}-\theta_{2})-\cos(\theta_{1}+\theta_{2})\big)\sin\theta_{3}\\
&=4\cos\left(\frac{\theta_{1}+\theta_{2}+\theta_{3}}{2}\right)\sin\theta_{1}\sin\theta_{2}\sin\theta_{3}>0.
\end{aligned}
\]
We thus finish the proof.
\end{proof}
\begin{lemma}\label{A}
Let $\theta_{1}$, $\theta_{2}$, $\theta_{3}$ be inner angles of a triangle in Euclidean or spherical geometry. Set $\mathrm{M}=[m_{st}]$, where $p_{ii}=1$, $p_{ij}=-\cos\theta_{k}$. Then $\mathrm{M}$ is semi-positive definite (resp. positive definite) in Euclidean (resp. spherical) geometry.
\end{lemma}
\begin{remark}
Lemma \ref{hypermatrix} was also proved in \cite{Luoarxiv,L.D.G,Luo1}. Comparing with the proof of in \cite{Luoarxiv,L.D.G,Luo1}, our proof is purely computational.
\end{remark}
\subsection{Differential forms on the moduli spaces of triangles}
In this subsection, we introduce several closed differential forms defined on the moduli space of geometric triangles. One refers to Appendix \ref{AppB} for some basic results about differential forms. The moduli space of a geometric triangle under polyhedral metric can be written as
\[\begin{aligned}
\Omega^{E}_{3}=\Omega^{H}_{3}&=\left\{(l_{1},l_{2},l_{3})\in\mathbb{R}^{3}_{+}\ \big|\ l_{i}+l_{j}>l_{k},\{i,j,k\}=\{1,2,3\}\right\},\\
\Omega^{S}_{3}&=\left\{(l_{1},l_{2},l_{3})\in(0,\pi)^{3}\ \big|\ l_{i}+l_{j}>l_{k},\ l_{1}+l_{2}+l_{3}<2\pi\right\},
\end{aligned}\]
in Euclidean, hyperbolic and spherical geometry, respectively. If we restrict the length of one edge, such as $l_{3}$ is a positive constant, then the moduli space of the triangle is the subset of $\mathbb{R}^2_{+}$, denote as $\Omega^{E}_{2}$, $\Omega^{H}_{2}$ and $\Omega^{S}_{2}$. Similarly, if $l_{2}, l_{3}$ are constants, we write the moduli space as $\Omega^{E}_{1}$, $\Omega^{H}_{1}$ and $\Omega^{S}_{1}$.
Specifically, if the polyhedral metric is a circle packing metric, the moduli space of triangles can be regard as the admissible space of the radius assignment, which is $\mathbb{R}^3_{+}$. The moduli space of triangles which fix the radius assignments of one, (resp. two) points is $\mathbb{R}^2_{+}$ (resp. $\mathbb{R}_{+}$). Set
\[\begin{aligned}
             &\omega^{E}_{\phi,s}:=\sum_{i=1}^{s}\frac{\int_{\pi/2}^{\theta_{i}}\sin^{h}{t}\mathrm{d}t}{l_{i}^{h+1}}\mathrm{d}l_{i},\quad
             \omega^{H}_{\psi,s}:=\sum_{i=1}^{s}\frac{\int_{0}^{\alpha_{i}}\cos^{h}{t}\mathrm{d}t}{\tanh^{h+1}{l_{i}/2}}\mathrm{d}l_{i},\quad
             \omega^{S}_{\phi,s}:=\sum_{i=1}^{s}\frac{\int_{\pi/2}^{\theta_{i}}\sin^{h}{t}\mathrm{d}t}{\sin^{h+1}{l_{i}}}\mathrm{d}l_{i},\\
             &\omega^{H}_{\phi,s}:=\sum_{i=1}^{s}\frac{\int_{\pi/2}^{\theta_{i}}\sinh^{h}{t}\mathrm{d}t}{\sinh^{h+1}l_{i}}\mathrm{d}l_{i},\quad
             \eta^{E}_{s}:=\sum_{i=1}^{s}\frac{\int_{\pi/2}^{\theta_{i}}\tan^{h}{\frac{t}{2}}\mathrm{d}t}{r_{i}^{-h+1}}\mathrm{d}r_{i},\quad\
             \eta^{H}_{s}:=\sum_{i=1}^{s}\frac{\int_{0}^{\theta_{i}}\tan^{h}\frac{t}{2}\mathrm{d}t}{\sinh^{-h+1}r_{i}}\mathrm{d}r_{i},
\end{aligned}\]
where $\alpha_{i}=\frac{1}{2}(\theta_{i}-\theta_{j}-\theta_{k})$ and $r_{i}=\frac{1}{2}(l_{j}+l_{k}-l_{i})$.
The first four differential forms are defined on the moduli space of triangles for polyhedral metric. The last two differential forms are defined on the moduli space of triangles for circle packing metric. Using Lemma \ref{2.1}, Lemma \ref{silemma}, Lemma \ref{2.3} and proposition \ref{tan1}, the following result can be verified easily.
\begin{lemma}\label{path}
The differential forms defined above are closed.
\end{lemma}
\begin{remark}
When $s=3$, the differential forms $\omega^{E}_{\phi,s}$, $\omega^{H}_{\psi,s}$, $\omega^{S}_{\phi,s}$, $\eta^{E}_{s}$ and $\eta^{H}_{s}$ first appeared in the work of Luo \cite{Luoarxiv}, which also proved the closeness of these differential forms. For the need to prove our main results, we consider above differential forms when $s=1,2$ and define $\omega^{H}_{\phi,s}$.
\end{remark}

\section{Energy functionals on the moduli spaces of polyhedral metrics}\label{3}
In this section, we construct energy functionals defined on the moduli space of the polyhedral surface such that the gradient of the energy functionals realizes a map from polyhedral metrics to discrete curvatures. The main technical tool is the following well-known fact from analysis. We omit the proof here.
\begin{lemma}\label{importantlemma}
If $\Omega$ is open convex in $\mathbb{R}^{n}$ and $W:\Omega\rightarrow\mathbb{R}$ is a smooth strictly convex function with positive definite Hessian matrix, then $\nabla W:\Omega\rightarrow\mathbb{R}^{n}$ is a smooth embedding.
\end{lemma}
\subsection{Energy functionals on the moduli spaces of triangles}\label{Section3.1}
We start from the energy functional defined on the moduli spaces of triangles. According to the number of interior edges of the triangle of polyhedral surface, we divide the triangles into four types. If the polyhedral surface contains a triangle with no interior edges, Theorem \ref{mt1}, Theorem \ref{1.7} and Theorem \ref{mt2} hold obviously. So we construct energy functionals on the other three types of triangles. We begin by introducing a continuous map defined on the moduli space of triangles. Let $\tau$ be a Euclidean (or hyperbolic or spherical) triangle on a bordered triangulated surface and let $l=(l_{1},l_{2},l_{3})\in\mathbb{R}^{3}_{+}$ represent the edge length of $\tau$. Then the continuous map $\xi$ is defined as
\[\xi(l)=\big(\xi_{h}(l_{1}),\xi_{h}(l_{2}),\xi_{h}(l_{3})\big).\]
where
\[\xi_{h}(t)=\left\{
             \begin{aligned}
             &\frac{-t^{-h}}{h},\ h\neq0,&\tau\ \text{is a Euclidean triangle}, \\
             &\ln{t},\quad h=0,&\tau\ \text{is a Euclidean triangle},\\
             &\int^{t}_{1}\tanh^{-h-1}\frac{x}{2}\mathrm{d}x,&\tau\ \text{is a hyperbolic triangle},\\
             &\int^{t}_{1}\sin^{-h-1}{x}\mathrm{d}x.&\tau\ \text{is a spherical triangle}.
             \end{aligned}
\right.\]
When $\tau$ is a hyperbolic triangle, another continuous map $\gamma$ is defined as
\[\gamma(l)=\big(\gamma_{h}(l_{1}),\gamma_{h}(l_{2}),\gamma_{h}(l_{3})\big)\]
where
\[\gamma_{h}(t)=\int^{t}_{1}\sinh^{-h-1}{x}\mathrm{d}x.\]
If we consider the moduli space of triangles with one or two interior edges, similar map can be defined on $\Omega^{E}_{s}$, $\Omega^{H}_{s}$, $\Omega^{S}_{s}$ when $s=1,2$. We still denote them by $\xi$ and $\gamma$. Then the following result is straightforward because of the monotonicity of $\xi_{h}(t)$, $\gamma_{h}(t)$.
\begin{lemma}\label{homeomorphism}
The continuous maps $\xi$, $\gamma$ defined in above are homeomorphic to its image.
\end{lemma}
Now we introduce several energy functionals on the moduli space of triangles. For $s=1,2,3$, we define line integral function
\[
             \begin{aligned}
             F^{E}_{s}(u):=\int^{u}_{-h}\omega^{E}_{\phi,s}, &\ u\in \xi(\Omega^{E}_{s}),\quad
             F^{H}_{s}(u):=\int^{u}_{0}\omega^{H}_{\psi,s}, \  u\in\xi(\Omega^{H}_{s}),\\
             F^{S}_{s}(u):=\int^{u}_{0}\omega^{S}_{\phi,s}, &\  u\in\xi(\Omega^{S}_{s}),\quad
             G^{H}_{s}(u):=\int^{u}_{0}\omega^{H}_{\phi,s}, \  u\in\gamma(\Omega^{H}_{s}).
             \end{aligned}
\]
By Poincar\'{e}'s Lemma and Lemma \ref{path}, above functions are well-defined. The following two lemmas investigate the convexity of these energy functionals.
\begin{remark}
Function $F^{E}_{s}(u)$, $F^{H}_{s}(u)$ and $F^{S}_{s}(u)$ were first introduced in \cite{Luoarxiv} when $s=3$.
\end{remark}
\begin{lemma}\label{form}
The Hessian matrix of the four energy functionals defined above can be represented as $c\mathrm{D}\mathrm{A}\mathrm{D}$, where $\mathrm{D}$ is a diagonal matrix in which each element of the main diagonal are positive, $c$ is a positive number, and $s=2,3$.
\end{lemma}
\begin{proof}
Let us first consider the Hessian matrix of $F^{E}_{3}(u)$. A routine substitution gives
\[\begin{aligned}
\frac{\partial^{2}F^{E}_{3}(u)}{\partial u_{i}\partial u_{j}}&=\sin^{h}\theta_{i}\frac{\partial\theta_{i}}{\partial u_{j}}\\
&=\sin^{h}\theta_{i}\frac{\partial\theta_{i}/\partial l_{j}}{\xi'_{h}(l_{j})}.
\end{aligned}\]
By Lemma \ref{2.1}, we deduce
\begin{equation}\label{2-derivative}
\frac{\partial^{2}F^{E}_{3}(u)}{\partial u_{i}^{2}}=m_{1}l^{2h+2}_{i},\quad\quad
\frac{\partial^{2}F^{E}_{3}(u)}{\partial u_{i}\partial u_{j}}=m_{1}l^{h+1}_{i}l^{h+1}_{j}\cos\theta_{k},
\end{equation}
where
\[m_{1}=\frac{\sin^{h-1}\theta_{i}}{l^{h}_{i}l_{j}l_{k}}\]
is independent of the index. Set $\mathrm{D_{1}}=[a_{st}]$, where $a_{ij}=0$, $a_{ii}=l_{i}^{h+1}$. Recall that $\mathrm{M}=[m_{st}]$, where $p_{ii}=1$, $p_{ij}=-\cos\theta_{k}$. Then the Hessian matrix of $F^{E}_{3}(u)$ can be represented as $c\mathrm{D_{1}}\mathrm{M}\mathrm{D_{1}}$. By Lemma \ref{2.1}, the second partial derivative of $F^{S}_{3}(u)$ and $G^{H}_{3}(u)$ have the same form as (\ref{2-derivative}). Therefore, taking a different positive number and diagonal matrix, the Hessian matrix of $F^{S}_{3}(u)$ and $G^{H}_{3}(u)$ can be expressed by $c\mathrm{D}\mathrm{M}\mathrm{D}$.

Next we compute the Hessian matrix of $F^{H}_{3}(u)$. Remember that $\alpha_{i}=\frac{1}{2}(\theta_{i}-\theta_{j}-\theta_{k})$. It is easy to see that
\[\begin{aligned}
\frac{\partial^{2}F^{H}_{3}(u)}{\partial u_{i}\partial u_{j}}&=\cos^{h}\alpha_{i}\frac{\partial \alpha_{i}}{\partial u_{j}}\\
&=\frac{\cos^{h}\alpha_{i}}{\gamma'_{h}(l_{j})}\frac{\partial\alpha_{i}}{\partial l_{j}}.
\end{aligned}\]
By Lemma \ref{2.3} we obtain
\[\begin{aligned}
\frac{\partial^{2}F^{H}_{3}(u)}{\partial u_{i}^{2}}&=m_{2}\tanh^{2h+2}\frac{l_{i}}{2}(\cosh l_{1}+\cosh l_{2}+\cosh l_{3}+1),\\
\frac{\partial^{2}F^{\mathrm{H}}_{3}(u)}{\partial u_{i}\partial u_{j}}&=m_{2}\tanh^{h}\frac{l_{i}}{2}\tanh^{h}\frac{l_{j}}{2}(-\cosh l_{i}-\cosh l_{j}+\cosh l_{k}+1),
\end{aligned}\]
where \[m_{2}=\frac{\cos^{h}\alpha_{i}}{\tanh^{h}\frac{l_{i}}{2}\sin\theta_{i}\sinh l_{j}\sinh l_{k}}.\] Lemma \ref{tan1} yields that $m_{2}$ is independent of the index. Set $\mathrm{D_{2}}=[b_{st}]$, where $b_{ij}=0$, $b_{ii}=\tan^{h}\frac{l_{i}}{2}$. Now the Hessian matrix of $F^{H}_{3}(u)$ can be represented as $c\mathrm{D_{2}}\mathrm{P}\mathrm{D_{2}}$, where $\mathrm{P}$ is the matrix defined in Lemma \ref{hypermatrix} and $c$ is a positive number.

Now we use $\mathrm{A'}$ to represent the submatrix of $\mathrm{A}$, obtained by removing the last row and the last column of matrix $\mathrm{A}$. When $s=2$, similar reasoning yields that the Hessian matrix of $F^{E}_{2}(u)$, $F^{S}_{2}(u)$ and $G^{H}_{2}(v)$ can be represented as $c\mathrm{D_{1}'}\mathrm{M'}\mathrm{D'_{1}}$, and the Hessian matrix of $F^{H}_{2}(u)$ can be represented as $c\mathrm{D_{2}'}\mathrm{P'}\mathrm{D'_{2}}$.
where $c$ is a positive number.
\end{proof}
We call a smooth function defined in an open set $U\subset\mathbb{R}^{n}$ locally convex (or locally strictly convex) if its Hessian matrix is positive semi-definite (or positive definite) at each point. If $U$ is convex, the function is called a convex or strictly convex function. Hence the local convexity of $F^{E}_{s}(u)$, $F^{H}_{s}(u)$, $F^{S}_{s}(u)$ and $G^{H}_{s}(u)$ is equivalent to the positive-definiteness of $\mathrm{M}$ and $\mathrm{P}$. By Lemma \ref{hypermatrix} and Lemma \ref{A}, we have the following result.
\begin{lemma} \label{locconv}
Let $F^{E}_{s}(u)$, $F^{H}_{s}(u)$, $F^{S}_{s}(u)$, $G^{H}_{s}(u)$ be as above.
\begin{itemize}
\item[$(a)$]The integral $F^{E}_{3}$ is locally convex, and $F^{H}_{3}$ , $F^{S}_{3}$ is locally strictly convex.
\item[$(b)$]The integral $F^{E}_{s}$, $F^{H}_{s}$, $F^{S}_{s}$, $G^{H}_{s}$ is locally strictly convex when $s=1,2$.
\end{itemize}
\end{lemma}
\subsection{Extension of locally convex functionals}
We establish some simple facts on extending locally convex energy functionals to convex functionals in this subsection. Suppose $A$ is a subspace of a topological space $X$ and $f:A\rightarrow Y$ is continuous. If there exists a continuous function $F:X\rightarrow Y$ such that $F|_{A}=f$ and $F$ is a constant function on each connected component of $X-A$, then we say $f$ can be extended continuously by constant functions to $X$. In this subsection, we define $J=\mathbb{R}_{+}$ in Euclidean and hyperbolic geometry and $J=(0,\pi)$ in spherical geometry.
\begin{lemma}\cite[Lemma 2.2]{Luo3}\label{ex.cont.3}
Let $\theta_{i}$, $l_{i}$ ($i=1,2,3$) denote the three angles and corresponding edges of a triangle. Then the cosine law function $\theta_{i}(l)$ ($i=1,2,3$) can be extended continuously by constant functions to a function $\widetilde{\theta}_{i}(l)$ ($ i=1,2,3$) on $J^{3}$.
\end{lemma}
\begin{proof}
Set
\[l_{jk}=\left\{(l_{1},l_{2},l_{3})\in J^{3}\ \big|\ l_{j}+l_{k}=l_{i}\right\}.\]
As $l\to l_{jk}$, we obatin
$\cos\theta_{i}\to-1$ and $\cos\theta_{j}$, $\cos\theta_{k}\to1.$
Hence we have
$\theta_{i}\to\pi$, $\theta_{j}$, $\theta_{k}\to0$
as $(l_{1},l_{2},l_{3})\to l_{jk}$.
Meanwhile, set
\[l_{123}=\left\{(l_{1},l_{2},l_{3})\in J^{3}\ \big|\ l_{1}+l_{2}+l_{3}= 2\pi\right\}.\]
In spherical geometry, as $l\to l_{123}$, we obtain
$\cos\theta_{i}\to1$, $\theta_{i}\to2\pi$,
where $i\in\{1,2,3\}$.
Thus $\theta_{i}(l)$ can be extended to
\[
\widetilde{\theta}_{i}(l)=\left\{
\begin{aligned}
&\theta_{i},&l\in\Omega^{E}_{3},
\\
&\pi,&l\in\Omega_{jk},
\\
&0,&l\in\Omega_{ij}\cup \Omega_{ik}.
\\
\end{aligned}
\right.
\]
on $J^{3}$ in Euclidean, hyperbolic geometry, and extended to
\[
\widetilde{\theta}_{i}(l)=\left\{
\begin{aligned}
&\theta_{i},&l\in\Omega^{S}_{3},
\\
&\pi,&l\in(\Omega_{jk}\cup\Omega_{123}),
\\
&0,&l\in(\Omega_{ij}\cup\Omega_{ik}),
\end{aligned}
\right.
\]
on $J^{3}$ in spherical geometry, where
\[
\Omega_{jk}=\left\{(l_{1},l_{2},l_{3})\in J^{3}\ \big|\ l_{j}+l_{k}\leq l_{i}\right\},\quad\Omega_{123}=\left\{(l_{1},l_{2},l_{3})\in J^{3}\ \big|\ l_{1}+l_{2}+l_{3}\geq 2\pi\right\}.
\]\end{proof}
Inspired by Lemma \ref{ex.cont.3}, we obtain the following observation. The proof is analogous to the proof of Lemma \ref{ex.cont.3}, we omit it here.
\begin{lemma}\label{ex.cont.2}
Let $\theta_{i}$, $l_{i}$ ($i=1,2,3$) denote three angles and corresponding edges of a triangle. For $i=1,2,3$, let $\theta_{i}(l)$ be cosine law function.
\begin{itemize}
\item[$(a)$] Suppose $l_{3}$ is a positive constant. Then $\theta_{i}(l_{1},l_{2})$ can be extended continuously by constant functions to $\widetilde{\theta}_{i}(l_{1},l_{2})$ on $J^{2}$ for $i=1,2,3$.
\item[$(b)$] Suppose $l_{2}$, $l_{3}$ are positive constants. Then $\theta_{i}(l_{1})$ can be extended continuously by constant functions to $\widetilde{\theta}_{i}(l_{1})$ on $J^{1}$ for $i=1,2,3$.
\end{itemize}
\end{lemma}
Substituting $\theta_{i}$ in $\omega^{E}_{\phi,s}$, $\omega^{H}_{\phi,s}$, $\omega^{S}_{\phi,s}$, $\omega^{H}_{\psi,s}$ to $\widetilde{\theta}_{i}$, we obtain four new differential forms $\widetilde{\omega}^{E}_{\phi,s}$, $\widetilde{\omega}^{H}_{\phi,s}$, $\widetilde{\omega}^{S}_{\phi,s}$ and $\widetilde{\omega}^{H}_{\psi,s}$.
\begin{lemma}\label{closeness}
Differential forms $\widetilde{\omega}^{E}_{\phi,s}$, $\widetilde{\omega}^{H}_{\phi,s}$, $\widetilde{\omega}^{S}_{\phi,s}$ and $\widetilde{\omega}^{H}_{\psi,s}$ are closed on $J^{s}$.
\end{lemma}
\begin{proof}
First, we want to show that $\widetilde{\omega}^{E}_{\phi,s}$ is closed. By Proposition \ref{B1} , we need to verify the following three propositions.
\begin{itemize}
\item[$(i)$] Differential form $\widetilde{\omega}^{E}_{\phi,s}$ is continuous on $J^{s}$ and closed on $\Omega_{s}^{E}$.
\item[$(ii)$] Differential form $\widetilde{\omega}^{E}_{\phi,s}$ is closed on each connected component of $J^{s}\setminus\Omega_{s}^{E}$.
\item[$(iii)$] The boundary of $\Omega_{s}^{E}$ is a $(s-1)$-dimensional smooth submanifold in $J^{s}$.
\end{itemize}
By Lemma \ref{ex.cont.3}, Lemma \ref{ex.cont.2}, $(i)$ is proved. Proposition $(iii)$ follows from the construction of $\Omega_{s}^{E}$. Proposition $(ii)$ is trivial because $\widetilde{\theta}_{i}$ is constant on each connected components of $J^{s}\setminus\Omega_{s}^{E}$. The proof of closeness of $\widetilde{\omega}^{H}_{\phi,s}$, $\widetilde{\omega}^{S}_{\phi,s}$ and $\widetilde{\omega}^{H}_{\psi,s}$ is quite similar to the proof of the closeness of $\widetilde{\omega}^{E}_{\phi,s}$.
\end{proof}
Similarly, substituting $\omega^{E}_{\phi,s}$, $\omega^{H}_{\psi,s}$, $\omega^{S}_{\phi,s}$ and $\omega^{H}_{\phi,s}$ in $F^{E}_{s}$, $F^{H}_{s}$, $F^{S}_{s}$ and $G^{H}_{s}$ to $\widetilde{\omega}^{E}_{\phi,s}$, $\widetilde{\omega}^{H}_{\psi,s}$, $\widetilde{\omega}^{S}_{\phi,s}$ and $\widetilde{\omega}^{H}_{\phi,s}$, we obtain four new energy functionals $\widetilde{F}^{E}_{s}$, $\widetilde{F}^{H}_{s}$, $\widetilde{F}^{S}_{s}$ and $\widetilde{G}^{H}_{s}$. By Lemma \ref{closeness}, these four energy functionals are well-defined on $\xi(J^{s})$ and $\gamma(J^{s})$. Now we want to prove the convexity of these four energy functionals.
\begin{lemma}\label{convex}
The energy functionals $\widetilde{F}^{E}_{s}$, $\widetilde{F}^{H}_{s}$, $\widetilde{F}^{S}_{s}$ are convex on $\xi(J^{s})$ when $s=1,2,3$ and the energy functional $\widetilde{G}^{H}_{s}$ are convex on $\gamma(J^{s})$ when $s=1,2$.
\end{lemma}
\begin{proof}
First, we want to show that $\widetilde{F}^{E}_{s}$ is convex on $\xi(J^{s})$. Owing to Proposition \ref{B2}, we need to check the following three propositions.
\begin{itemize}
\item[$(i)$] Differential form $\widetilde{\omega}^{E}_{h,s}$ is continuous on $J^{s}$.
\item[$(ii)$] The boundary of $\xi(\Omega^{E}_{s})$ is a real analytic codimension-$1$ submanifold in $\xi(J^{s})$.
\item[$(iii)$] Function $\widetilde{F}^{E}_{s}$ is locally convex on $\xi(\Omega^{E}_{s})$ and each connected component of $\xi(J^{s})\setminus \overline{\xi(\Omega_{s}^{E})}$.
\end{itemize}
By Lemma, \ref{ex.cont.3}, Lemma \ref{ex.cont.2}, $(i)$ is proved. Combining the definition of $\Omega^{E}_{s}$ and Lemma \ref{homeomorphism}, $(ii)$ is established. Lemma \ref{locconv} yields the locally convexity of $\widetilde{F}^{E}_{s}$ on $\xi(\Omega^{E}_{s})$ directly. It is easily can be seen that
\[\xi(J^{s})\setminus \overline{\xi(\Omega^{E}_{s})}=\interior\xi(\Omega_{12})\cup\interior \xi(\Omega_{23})\cup\interior\xi(\Omega_{13}).\]
From
\[\frac{\partial\widetilde{F}^{E}_{s}}{\partial u_{i}}=\int^{\widetilde{\theta}_{i}}_{\pi/2}\sin^{h}t\mathrm{d}t\]
we know that $\widetilde{F}^{E}_{s}$ is a linear function on $\interior\xi(\Omega_{ij})$. Thus $(iii)$ is proved. Similar deduction gives the convexity of $\widetilde{F}^{H}_{s}$, $\widetilde{F}^{S}_{s}$ and $\widetilde{G}^{H}_{s}$.
\end{proof}

\section{Energy functionals on the moduli spaces of circle packing metrics}\label{4}
Following the idea of the previous section, we will construct the convex functionals on the moduli space of triangles for circle packing metric. We still begin by introducing a continuous map defined on the moduli space of triangles. Let $\tau$  be a triangle on a bordered triangulated surface. If all points of $\tau$ are interior points, denote $r=(r_{1},r_{2},r_{3})\in\mathbb{R}^{3}_{+}$ as the circle packing metric on $\tau$. Then the continuous map $\gamma$ is defined as
\[g(r)=\left(g_{h}(r_{1}),g_{h}(r_{2}),g_{h}(r_{3})\right).\]
where
\[g_{h}(t)=\left\{
             \begin{aligned}
             &\frac{t^{h}}{h},  h\neq0, &\tau\ \text{is a Euclidean triangle}, \\
             &\ln{t},  h=0, &\tau\ \text{is a Euclidean triangle},\\
             &\int^{t}_{1}\sinh^{h-1}\frac{x}{2}\mathrm{d}x, &\tau\ \text{is a hyperbolic triangle}.
             \end{aligned}
\right.\]
Similarly, the continuous map $g$ can be defined on the moduli space of the triangles which have one or two interior points. Because $g_{h}(t)$ is monotonic, the continuous map $g$ is homeomorphic to its image.

Now it is ready to introduce convex functionals defined on the moduli space of triangles. Define line integral function
\[
             C^{E}_{s}(u):=-\int^{u}\eta^{E}_{s}, u\in g(\mathbb{R}^{3}_{+}),\quad
             C^{H}_{s}(u):=-\int^{u}\eta^{H}_{s}, u\in g(\mathbb{R}^{3}_{+}).
\]
By Poincar\'{e} Lemma and Lemma \ref{path}, above functions are well-defined.
\begin{lemma}\label{cpcon}
Let $C^{E}_{s}(u)$, $C^{H}_{s}(u)$ be as above.
\begin{itemize}
\item[$(a)$]The integral $C^{E}_{3}(u)$ is convex, $C^{E}_{s}(u)$ is strictly convex when $s=1,2$.
\item[$(b)$]The integral $C^{H}_{s}(u)$ is strictly convex for $s=1,2,3$.
\end{itemize}
\end{lemma}
\begin{proof}
First we prove $(a)$. It is easy to see that
\[\begin{aligned}
\frac{\partial^{2}C^{E}_{s}}{\partial u_{i}\partial u_{j}}&=-\frac{1}{2}\tan^{h}\frac{\theta_{i}}{2}\frac{\partial\theta_{i}}{\partial u_{j}}\\
&=-\frac{1}{2}\tan^{h}\frac{\theta_{i}}{2}\frac{\partial\theta_{i}/\partial r_{j}}{g'_{h}(r_{j})}\\
&=-\frac{1}{2}(\tan\frac{\theta_{i}}{2}r_{i})^{h}(r_{i}r_{j})^{-h}r_{j}\frac{\partial\theta_{i}}{\partial r_{j}}.
\end{aligned}\]
By Proposition \ref{tan1},  $\tan\frac{\theta_{i}}{2}r_{i}$ is a positive number independent of the indices. Hence the Hessian matrix of $C^{E}_{3}(u)$ is congruent to $[a_{st}]$ where
\[a_{ij}=-r_{j}\frac{\partial\theta_{i}}{\partial r_{j}}.\]
Note that
\begin{equation}\label{aii}\begin{aligned}
a_{ii}&=-r_{i}\left(\frac{\partial\theta_{i}}{\partial l_{i}}+\frac{\partial\theta_{i}}{\partial l_{k}}\right)\\
&=r_{i}\frac{\partial\theta_{i}}{\partial l_{i}}(\cos\theta_{k}+\cos\theta_{j})>0.
\end{aligned}\end{equation}
From Lemma \ref{silemma} we obtain that $a_{ij}<0$. Hence,
\begin{equation}\label{diag}\begin{aligned}
|a_{ii}|-|a_{ji}|-|a_{ki}|&=a_{ii}+a_{ji}+a_{ki}\\
&=r_{i}\frac{\partial(\theta_{i}+\theta_{j}+\theta_{k})}{\partial r_{i}}=0.
\end{aligned}\end{equation}
Thus the matrix $[a_{st}]$ is positive semi-definite due to the diagonal dominate property. Thus the Hessian matrix of $C^{E}_{3}$ is positive semi-definite, which yields that $C^{E}_{3}(u)$ is convex. From (\ref{diag}) we also derive that
\[
|a_{ii}|-|a_{ji}|>0,
\]
which yields that the Hessian matrix of $C^{E}_{2}$ is positive definite. Thus $C^{E}_{2}(u)$ is strictly convex. From (\ref{aii}) we know
\[\frac{\mathrm{d}^{2}C^{E}_{1}(u)}{\mathrm{d}u^{2}}>0,\]
which yields that $C^{E}_{1}(u)$ is strictly convex.

Next we prove $(b)$. An analogous analysis derives that the Hessian matrix of $C^{H}_{3}(u)$ are congruent to $[b_{st}]$ where
\[b_{ij}=-\sinh r_{j}\frac{\partial\theta_{i}}{\partial r_{j}}.\]
From $b_{ii}>0, b_{ij}<0$ and the Gauss-Bonnet formula we know that
\[\begin{aligned}
|b_{ii}|-|b_{ji}|-|b_{ki}|&=b_{ii}+b_{ji}+b_{ki}\\
&=-\sinh r_{i}\frac{\partial(\theta_{i}+\theta_{j}+\theta_{k})}{\partial r_{i}}\\
&=\sinh r_{i}\frac{\partial}{\partial r_{i}}\text{Area}(\triangle)>0,
\end{aligned}\]
which yields the strict convexity of $C^{H}_{3}(u)$. A similar deduction to the proof of $(a)$ gives the strict convexity of $C^{H}_{1}(u)$, $C^{H}_{2}(u)$.
\end{proof}

\section{Proof of main results}\label{5}
Let us prove our main results via variational principle in this section. We begin by introducing some notations that will be used in the proofs. For any polyhedral metric $d$ on $(S,T)$, denote
\[l_{d}=\big(d(e_{1}),\cdots,d(e_{n})\big)\] as the coordinate of $d$, where $\{e_{i}\}^{n}_{i=1}$ are all interior edges of $(S,T)$. Denote $u_{d}=\xi(l_{d})$
as the $u$-coordinate of $d$, where $f$ is the homeomorphic map defined in Section \ref{Section3.1}. Suppose that $\tau$ is a triangle in $T$. We write the coordinate of interior edges of $\tau$ in $d$ as $l^{\tau}_{d}$, and set $u^{\tau}_{d}=\xi(l^{\tau}_{d})$.
For $s=1,2,3$, denote
\[T_{s}=\left\{\tau\in T\ |\ \#s\ \text{edges in}\ \tau\ \text{are interior edges}\right\}.\]

Next, we construct some functions to be used in the proofs. They are constructed under the assumption $\cup_{s=1}^{3}T_{s}\neq\emptyset$. Define a function $W^{E}:\xi_{h}(\mathbb{R}_{+})^{n}\rightarrow\mathbb{R}$ by
\[W^{E}(u_{d})=\sum\nolimits_{s=1}^{3}\sum\nolimits_{\tau\in T_{s}}\widetilde{F}^{E}_{s}(u^{\tau}_{d}).\]
For each interior edge $e_{i}$, there exist two triangles
$\tau_{1}^{i},\ \tau_{2}^{i}\in\cup_{s=1}^{3}T_{s}$
such that $\tau_{1}^{i}\cap\tau_{2}^{i}=e_{i}$. Hence we have
\[\begin{aligned}
\left.\frac{\partial W^{E}}{\partial u_{i}}\right|_{u_{d}}&=\frac{\partial\widetilde{F}^{E}_{s}(u^{\tau^{i}_{1}}_{d})}{\partial u_{i}}+\frac{\partial\widetilde{F}^{E}_{s'}(u^{\tau^{i}_{2}}_{d})}{\partial u_{i}}\\
&=\int^{\widetilde{\theta}_{1}^{i}}_{\pi/2}\sin^{h}t\mathrm{d}t+\int^{\widetilde{\theta}_{2}^{i}}_{\pi/2}\sin^{h}t\mathrm{d}t.
\end{aligned}\]
where $\theta_{1}^{i}$, $\theta_{2}^{i}$ are angles facing $e_{i}$ in $\tau_{1}^{i}$, $\tau_{2}^{i}$, separately. Note that
\[\left.\frac{\partial W^{E}}{\partial u_{i}}\right|_{u_{d}}=\phi_{h}(e_{i}).\]
It follows that
\[\nabla W^{E}(u_{d})=\big(\phi_{h}(e_{1}),\cdots,\phi_{h}(e_{n})\big).\]
Suppose that $u_{d_{1}}$ and $u_{d_{2}}$ are $u$-coordinates of two polyhedral metrics. Set
\[f^{E}(t)=W^{E}\left(tu_{d_{1}}+(1-t)u_{d_{2}}\right).\]
\begin{lemma}\label{near}
Suppose $d_{1}$, $d_{2}$ are two different polyhedral metrics on $(S,T)$. Then $f^{E}(t)$ is strictly convex for $t$ near $0$,$1$.
\end{lemma}
\begin{proof}
For each $\tau\in T_{s}$, we would like to consider
\[f^{\tau}_{s}(t)=\widetilde{F}^{E}_{s}\left(tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}\right)\]
If $u_{d_{1}}^{\tau}=u_{d_{2}}^{\tau}$, $f^{\tau}_{s}(t)$ is a constant. Hence
\[f^{E}(t)=\sum\nolimits_{s=1}^{3}\sum\nolimits_{\tau\in T'_{s}}f^{\tau}_{s}(t)+c,\]
where $T'_{s}=\left\{\tau\in T_{i}\ \big|\ u_{d_{1}}^{\tau}\neq u_{d_{2}}^{\tau}\right\}.$ Then we have
\[{f^{E}}''(t)=\sum\nolimits_{s=1}^{3}\sum\nolimits_{\tau\in T'_{s}}{f^{\tau}_{s}}''(t).\]
It is easy to compute that if $tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}\in\xi\left(\Omega^{E}_{s}\right)$, we obtain
\begin{equation}
\label{''}
{f^{\tau}_{s}}''(t)=\left\{
\begin{aligned}
&\left(u^{\tau}_{d_{1}}-u^{\tau}_{d_{2}}\right)^{\mathrm{T}}\Hessian\widetilde{F}^{E}_{s}\left(tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}\right)\left(u^{\tau}_{d_{1}}-u^{\tau}_{d_{2}}\right),&s=2,3,\\
&\left(u^{\tau}_{d_{1}}-u^{\tau}_{d_{2}}\right)^2\widetilde{F}^{E}_{s}{''}\left(tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}\right),&s=1.
\end{aligned}
\right.
\end{equation}
If $tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}\notin\xi\left(\Omega^{E}_{s}\right)$, then ${f^{\tau}_{s}}''(t)=0$.
Recall that $\xi$ is a homomorphic map from the moduli space of triangles to its image. So $u_{d_{1}}^{\tau}$, $u_{d_{2}}^{\tau}$ are interior points. Hence, for each $\tau\in \cup_{s=1}^{3} T'_{s}$, there exists $\varepsilon_{\tau}>0$ such that $tu^{\tau}_{d_{1}}+(1-t)u^{\tau}_{d_{2}}$ is $u$-coordinate of a Euclidean metric for any $t\in(0,\varepsilon_{\tau})\cup(1-\varepsilon_{\tau},1)$. Set $\varepsilon_{0}=\min\left\{\varepsilon_{\tau}\ |\ \tau\in T'\right\}.$
According to (\ref{''}) and the Hessian matrix of $F^{E}_{s}$, we derive the following two claims.
\begin{itemize}
\item[$(i)$]  ${f^{\tau}_{3}}''(t)=0$ near $0,1$ if and only if $u_{d_{1}}^{\tau}=cu_{d_{2}}^{\tau}(c\neq0)$.
\item[$(ii)$] ${f^{\tau}_{s}}''(t)=0$ near $0,1$ if and only if $u_{d_{1}}^{\tau}=u_{d_{2}}^{\tau}$ for $s=1,2$.
\end{itemize}
Note that polyhedral surface is connected. That means the polyhedral metric $d_{1}$, $d_{2}$ cannot satisfy both $u_{d_{1}}^{\tau}=cu_{d_{2}}^{\tau}$ for $\tau\in T'_{3}$ and $u_{d_{1}}^{\tau}=u_{d_{2}}^{\tau}$ for $\tau\in T'_{1}\cup T'_{2}$, which yields ${f^{E}}''(t)>0$ for any $t\in(0,\varepsilon_{0})\cup(1-\varepsilon_{0},1)$.
Then $f^{E}(t)$ is strictly convex on $(0,\varepsilon_{0})\cup(1-\varepsilon_{0},1)$.
\end{proof}
\begin{proof}[\textbf{Proof of Theorem \ref{mt1}}]
We only prove Theorem \ref{mt1}$(a)$. Theorem \ref{mt1}$(b)$,$(c)$ can be obtained by a similar analysis. If $\cup_{i=s}^{3}T_{s}=\emptyset$, Theorem \ref{mt1}$(a)$ holds obviously. So we suppose $\cup_{i=s}^{3}T_{s}\neq\emptyset$. First we prove $(a)$. Let $d_{1}$, $d_{2}$ be two Euclidean polyhedral metrics that share equal boundary value and $\phi_{h}$-curvature on the interior edges. It suffices to show that $d_{1}=d_{2}$. Let $a\in\mathbb{R}^{n}$ be the $\phi_{h}$-curvature of $d_{1}$, $d_{2}$ on interior edges, where each vector component of $a$ represents the $\phi_{h}$-curvature of an interior edge. Then
\[\widetilde{W}^{E}(u)=W^{E}(u)-a\cdot u\]
is convex and $u_{d_{1}}$, $u_{d_{2}}$ are critical points of $\widetilde{W}^{E}(u)$. Hence
\[\widetilde{W}^{E}\big(tu_{d_{1}}+(1-t)u_{d_{2}}\big)=W^{E}\big(tu_{d_{1}}+(1-t)u_{d_{2}}\big)-a\cdot\big(tu_{d_{1}}+(1-t)u_{d_{2}}\big)\]
is a constant. It follows that
\[f^{E}(t)=W^{E}\big(tu_{d_{1}}+(1-t)u_{d_{2}}\big)\]
is a linear function if $u_{d_{1}}\neq u_{d_{2}}$. Combining with Lemma \ref{near}, we know that $u_{d_{1}}=u_{d_{2}}$. Recall that $f_{1}$ is a homeomorphic map to its image. Thus $l_{d_{1}}=l_{d_{2}}$. That means $d_{1}$ and $d_{2}$ are identical on the interior edges of $(S,T)$. This proves $(a)$.
\end{proof}
\begin{lemma}\label{cyc}
Let $Q$ be a cyclic quadrilateral and $e$ be diagonal of $Q$. Then $\psi_{0}(e)=0$.
\end{lemma}
\begin{proof}
Denote four vertexes as $A$, $B$, $C$, $D$ and denote $M$ be the center of the circumcircle. Then $\triangle MAB$, $\triangle MBC$, $\triangle MCD$ and $\triangle MDA$ are isosceles. As a result, the sum of the opposite angles of the quadrilateral is equal. Thus $\psi_{0}(e)=0$.
\end{proof}
\begin{proof}[\textbf{Proof of Corollary \ref{co}}]
Let $P$ be a Euclidean or hyperbolic cyclic polygon and $T$ be a triangulation of $P$. Lemma \ref{cyc} yields that $\psi_{0}(e)=0$ for each inner edge $e$. Recall that $\phi_{0}(e)=\psi_{0}(e)$ in Euclidean geometry, Thus for each inner edge $e$ we have $\phi_{0}(e)=0$ (resp. $\psi_{0}(e)=0$) if $P$ is a Euclidean (resp. hyperbolic) cyclic polygon. By Theorem \ref{mt1} $(a)$, $(b)$, the result holds immediately.
\end{proof}
\begin{proof}[\textbf{Proof of Theorem \ref{1.7}}]
If $\cup_{s=1}^{3}T_{s}=\emptyset$, Theorem \ref{1.7} holds obviously. So we suppose that $\cup_{s=1}^{3}T_{s}\neq\emptyset$. Let $d_{1}$ and $d_{2}$ be two hyperbolic polyhedral metrics that have the equal boundary value and $\phi_{h}$-curvature on the interior edges.  It suffices to show that $d_{1}=d_{2}$. Here we set $u_{d}=\gamma(l_{d})$, where $\gamma$ is defined in Section \ref{Section3.1}. Denote the number of interior edges by $n$, then we define a function $V^{E}:\gamma_{h}(\mathbb{R})^{n}\rightarrow\mathbb{R}$ by
\[V^{E}(u_{d})=\sum\nolimits_{s=1}^{2}\sum\nolimits_{\tau\in T_{s}}\widetilde{G}^{E}_{s}(u^{\tau}_{d}).\]
Polyhedral surface $(S,T)$ does not exist interior triangles, which means $T_{3}=\emptyset$. Hence we obtain
\[\nabla V^{E}(u_{d})=\big(\phi_{h}(e_{1}),\cdots,\phi_{h}(e_{n})\big).\]
We denote $b\in\mathbb{R}^{n}$ be the $\phi_{h}$-curvature of $d_{1}$, $d_{2}$ on interior edges. Then
\[\widetilde{V}^{E}(u)=V^{E}(u)-b\cdot u\]
is convex and $u_{d_{1}}$, $u_{d_{2}}$ are critical points of $V^{E}(u)$. By similar analysis as the proof of Theorem \ref{mt1} we conclude that $d_{1}=d_{2}$.
\end{proof}

Let us introduce some similar notions before we prove Theorem \ref{mt2}. For each circle packing metric $r$ on $(S,T)$, denote
$p_{r}=\big(r(v_{1}),\cdots,r(v_{n})\big)$
as the coordinate of interior points, where $\{v_{j}\}^{n}_{j=1}$ are all interior points of $(S,T)$. Suppose that $\tau$ is a triangle in $T$. We write the coordinate of radius assignment of interior points of $\tau$ as $p^{\tau}_{r}$, and set $q^{\tau}_{r}=g(p^{\tau}_{r})$. One can recall Section \ref{4} for the definition of homeomorphism $g$. Here we denote
\[T_{s}=\left\{\tau\in T\ \big|\ \#s\ \text{points in}\ \tau\ \text{are interior points}\right\}.\]
Define a function $U^{E}:g_{h}(\mathbb{R}_{+})^{|n|}\rightarrow\mathbb{R}$ by
\[U^{E}(q_{r})=\sum\nolimits_{s=1}^{3}\sum\nolimits_{\tau\in T_{s}}C^{E}_{s}(q^{\tau}_{r}).\]
Similarly, we can define $U^{H}(q_{r})$. Note that the triangulated surface is bordered. Hence $T_{1}\cup T_{2}\neq\emptyset$. By Lemma \ref{cpcon}, the following result is straightforward.
\begin{lemma}\label{cpconv}
Function $U^{E}(q_{r})$, $U^{H}(q_{r})$ are strictly convex on $g_{h}(\mathbb{R}_{+})^{|n|}$.
\end{lemma}
\begin{proof}[\textbf{Proof of Theorem \ref{mt2}}]
If all points are on the boundary, then Theorem \ref{mt2} holds obviously. So we assume that $\cup_{s=1}^{3}T_{s}\neq\emptyset$. By the construction of $U^{E}$ we obtain
\[\nabla U^{E}(q_{r})=\left(-\sum\nolimits_{i=1}^{m_{1}}\int_{\pi/2}^{\theta_{i}^{1}}\tan^{h}(t/2)\textit{d}t,\cdots,-\sum\nolimits_{i=1}^{m_{n}}\int_{\pi/2}^{\theta_{i}^{n}}\tan^{h}(t/2)\textit{d}t\right),\]
where $\theta_{1}^{j},\cdots,\theta_{m_{j}}^{j}$ are all inner angles at vertex $v_{j}$. Combining Lemma \ref{importantlemma} and Lemma \ref{cpconv}, we obtain a homeomorphic map from the coordinate of $r$ to its $k_{h}$-curvature. This map is injective, obviously. Therefore, if two Euclidean circle packing metrics have the equal $k_{h}$-curvature on the interior points, then there are equal on the interior points. This proves $(a)$. The proof of $(b)$ can be completed by using the same analysis to $U^{H}(q_{r})$.
\end{proof}
\begin{proof}[\textbf{Proof of Corollary \ref{cocp}}]
Let $P$ be a triangle-type circle packing. There is a natural way to build the bordered triangulated surface $(S,T)_{P}$ that isomorphic to the contact graph of $P$. The vertex set of $(S,T)_{P}$ is the set of centers of circles in $P$, denote as $V_{P}$. The edge set of $(S,T)_{P}$ is the set of edges joining the centers of each pair of mutually tangent circles in $P$. For any radius assignment $R:V\to\mathbb{R}_{+}$ of $P$, one can induce a circle packing metric on $(S,T)_{P}$ by $r_{R}(v)=R(v)$ for $v\in V_{P}$. Suppose $R_{1}$, $R_{2}$ are two radius assignments of $P$ (keeping the contact graph of $P$ unchanged) that equal on each boundary circle. Then $r_{R_{1}}$, $r_{R_{2}}$ are two circle packing metrics on $(S,T)_{P}$ having equal boundary value. Because $P$ is a planar circle packing, $(S,T)_{P}$ is a bordered triangulated surface on Euclidean plane. Recall that $k_{0}(v)=2\pi-\sum_{i=1}^{m}\theta_{i}$, where $\theta_{1},\cdots,\theta_{1}$ are inner angles at $v$. Hence $k_{0}$-curvature of $r_{R_{1}}$, $r_{R_{2}}$ for each interior point of $(S,T)_{P}$ equals to zero. By Theorem\ref{mt2}, we obtain $r_{R_{1}}=r_{R_{2}}$. Thus $R_{1}=R_{2}$. This completes the proof.
\end{proof}

\begin{appendices}
\section{Some formulas in Trigonometry}\label{AppA}
This section is devoted to some results from trigonometry.
\begin{proposition}[Cosine law]
Let $l_{1}$, $l_{2}$, $l_{3}$ be three edges of a triangle and $\theta_{1}$, $\theta_{2}$, $\theta_{3}$ are the corresponding inner angles.
\begin{itemize}
\item[$(a)$]In Euclidean geometry, we have
\[\cos\theta_{i}=\frac{-l^2_{i}+l^2_{j}+l^2_{k}}{2l_{j}l_{k}},\quad\cos l_{i}=\frac{\cos\theta_{i}+\cos\theta_{j}\cos\theta_{k}}{\cos\theta_{j}\cos\theta_{k}}.\]
\item[$(b)$]In hyperbolic geometry, we have
\[\cos\theta_{i}=\frac{-\cosh l_{i}+\cosh l_{j}\cosh l_{k}}{\sinh l_{j}\sinh l_{k}},\quad\cosh l_{i}=\frac{\cos\theta_{i}+\cos\theta_{j}\cos\theta_{k}}{\sin\theta_{j}\sin\theta_{k}}.\]
\item[$(c)$]In spherical geometry, we have
\[\cos\theta_{i}=\frac{\cos l_{i}-\cos l_{j}\cos l_{k}}{\sin l_{j}\sin l_{k}},\quad\cosh l_{i}=\frac{\cos\theta_{i}+\cos\theta_{j}\cos\theta_{k}}{\sin\theta_{j}\sin\theta_{k}}.\]
\end{itemize}
\end{proposition}
\begin{proposition}[Tangent law]\label{tan1}
Let $l_{1}$, $l_{2}$, $l_{3}$ be three edges of a triangle and $\theta_{1}$, $\theta_{2}$, $\theta_{3}$ are the corresponding inner angles. Set
\[h_{i}=\cos \alpha_{i}\coth\frac{l_{i}}{2},\ h'_{i}=r_{i}\tan\frac{\theta_{i}}{2}\ s_{i}=\cos \alpha_{i}\cot\frac{l_{i}}{2},\ s'_{i}=\sinh r_{i}\tan\frac{\theta_{i}}{2},\]
where $i\in\{1,2,3\}$ and $\alpha_{i}=\frac{1}{2}(\theta_{i}-\theta_{j}-\theta_{k})$, $r_{i}=\frac{1}{2}(l_{j}+l_{k}-l_{i})$. Then $h_{i}$, $h'_{i}$ (resp. $s_{i}$, $s'_{i}$) are positive numbers independent of indices in hyperbolic (resp. spherical) geometry.
\end{proposition}
\section{Basic results about closed forms}\label{AppB}
In this section, we give a simple introduction to some results on differential forms and convex functions constructed by differential forms. One refers to \cite{Luo3,Bo3} for more background.

A differential 1-form $\omega=\sum^{n}_{i=1}f_{i}(x)\mathrm{d}x_{i}$ is said to be continuous in an open set $U\subset\mathbb{R}^{n}$ if each $f_{i}(x)$ is continuous on $U$. A continuous 1-form is called closed if $\int_{\partial\gamma}\omega=0$ for each piecewise $C^{1}$-smooth null homologous loop $\gamma$ in $U$. By path independence for line integrals, the closeness of continuous forms is equivalent to
\[\frac{\partial f_{i}(x)}{\partial x_{j}}=\frac{\partial f_{j}(x)}{\partial x_{i}}\]
for $i,j\in\{1,\cdots,n\}$.

\begin{proposition}\label{B1}
Suppose $X$ is an open set in $\mathbb{R}^{n}$ and $A\subset X$ is an open subset bounded by a smooth $(n-1)$-dimensional submanifold in $X$. If $\omega=\sum^{n}_{i=1}f_{i}(x)\mathrm{d}x_{i}$ is a continuous 1-form on $X$ such that $\omega|_{A}$ and $\omega|_{X-\overline{A}}$ are closed where $\overline{A}$ is the closure of $A$ in $X$, then $\omega$ is closed in $X$.
\end{proposition}
\begin{proposition}\label{B2}
Suppose $X\subset\mathbb{R}^{n}$ is an open convex set and $A\subset X$ is an open subset of $X$ bounded by a codimension-1 real analytic submanifold in $X$. If $\omega=\sum^{n}_{i=1}f_{i}(x)\mathrm{d}x_{i}$ is a continuous closed 1-form on $X$ such that $F(x)=\int^{x}_{a}\omega$ is locally convex in $A$ and in $X-\overline{A}$, then $F(x)$ is convex in $X$.
\end{proposition}
\end{appendices}
\section{Acknowledgement}
The authors would like to thank their advisor, Ze Zhou, for his encouragement and helpful discussions. They also would like to thank NSF of China (No.11631010) and Postgraduate Scientific Research Innovation Project of Hunan Province  (CX20210397) for financial support.

redpajama_set_name: RedPajamaArXiv


text:
The Gaming Table:
its Votaries and Victims
(Vol I & II)
Andrew Steinmetz.
This title, running to 880 pages over two volumes, was originally published in 1870 and is one of around 50,000 titles that has been "digitised" by those nice volunteers at Project Gutenberg - and at the time of writing the digital download is available completely free of charge either directly from the PG site or via Amazon (in Kindle format).
A good read? I thought so. Lots of anecdotes, yarns and gambling related stories - in the main from around a hundred and fifty years up to the date of publication from across Europe as well as the States, India and the Far East. Prepare yourself for lots of tales: of degenerate gambling amongst the gentry, lost fortunes, ruin, ladies of title settling their gambling debts in the bed chamber, debtors prisons, rakes, "sharpers", cheats, ne'r-do-wells, naive young gentlemen with more money than sense, rigged decks, degenerate gambling amongst the lower orders, lotteries, some infamous names of the time, horses, dice, early wheels of doom, private clubs, temptation, degenerate gambling . . . well, you get the picture? It touches on a lot. If you want fairly solid proof that an awful lot of the gambling scene hasn't really changed in the last two hundred years (apart from the fact that most of it's now licenced and the Chancellor gets a cut of at least some of the action - oh, and the communications technology angle), then this is an entertaining trip into the past.
The only disappointed, I found, was the lack of a conclusion or summary chapter to round the whole thing off. The last 30 pages of volume two is dedicated to a range of card tricks (of the type some do to amuse and entertain others - no TV, radio or interweb thingy in those days), and after that nothing - which is a great shame as the content frequently alludes to the vulnerabilities and foibles of the many who fall prey to the spectre of addictive gambling. A concluding chapter would have rounded this extensive piece of work off nicely.
Andrew Steinmetz (1816-1877) was a barrister by profession and practiced at the bar in London up until his retirement. He authored several other titles, including Japan and her People and The Romance of Duelling across two volumes, with his signature work being History of the Jesuits which comprises three volumes.
Here are the links to the downloads of The Gaming Table:
http://www.gutenberg.org/ebooks/466 - Vol I
http://www.gutenberg.org/ebooks/531 - Vol II
Amazon Kindle - Vol I
Amazon Kindle - Vol II

redpajama_set_name: RedPajamaCommonCrawl


text:
The 2021 Emmy Nominations Are Revealed On Tuesday
The 2021 Emmy nominations are revealed this morning and will reflect a strange year for television. The pandemic, of course, forced many shows to delay or suspend production, even as people staying home gave them more time to watch. Glen Weldon, co-host of NPR's Pop Culture Happy Hour, is with us. Glen, good morning.
GLEN WELDON, BYLINE: Hey. Good morning, Steve.
INSKEEP: I'm one of those people who had more time to watch series. So what are some that might get honored here? We'll start with dramas.
WELDON: Well, you're going to look for the usual suspects, of course - "The Crown," "The Handmaid's Tale," "This Is Us." I personally would love to see "Pose" get some long-overdue love now that it's finished up its run and also "Lovecraft Country," which was just cancelled by HBO, but it was a really ambitious show. I think "Bridgerton" has a shot, though Emmy voters might consider it a bit light and frothy for serious drama consideration. But as a nerd, you know, this is the first year that Disney+'s Marvel shows might be up for consideration. It's going to be interesting to see if the Emmy voters embrace them or if they just treat them like they treat every other sci-fi or genre show and ignore it.
INSKEEP: (Laughter).
WELDON: If they do take them into consideration, "The Falcon And The Winter Soldier" would be nominated in this category, drama. But "WandaVision" was submitted as a limited series for some reason. And if a Marvel show breaks through, then maybe a "Star Wars" show could, too. Don't count out "The Mandalorian." I would love to see it get something besides just the technical special effects recognition stuff like this gets.
INSKEEP: I appreciate you identifying as a nerd. I think that's brave. I like you coming out in that way. That's really good.
WELDON: (Laughter).
INSKEEP: What about the best comedy series?
WELDON: Well, this isn't a case of the usual suspects because we don't have a season of "Marvelous Mrs. Maisel" this year, and that thing is an Emmy sponge, so that leaves room for new nominations. For comedy, look for "Ted Lasso" to be a stone-cold lock. Look for it to pick up a boatload of nominations today. I'm personally hoping that "Hacks" gets some of the love it deserves. And you can probably count on a "Black-ish" nomination. And Emmys really seem to love "The Kominsky Method" on Netflix. And if Peacock's "Girls5eva" doesn't get nominated here - I'm hoping it does - look for it in the writing category because, man, the joke density on that show is hugely impressive.
INSKEEP: Glen, you said something earlier about limited series category. I guess that's what it sounds like - something short?
WELDON: Yeah. I mean, in a lot of cases, the shows that broke through, the ones that we used to call water-cooler shows, are - all belong to this category, chief among them "Mare Of "Easttown," which is going to have a very strong showing not only because it's good, but also because it had its cultural moment right at the time when Emmy voters - in that window of time when Emmy voters were voting for these nominations. I suspect to see "The Queen's Gambit" and certainly "The Underground Railroad." They'd be nuts if they didn't.
As I mentioned, it'll be interesting to see what Emmy voters make of "WandaVision." That's a television series that, for most of its run, was really about the format and structure of the American sitcom - television about television; Emmy voters love that. But what I'm personally pulling for is "I May Destroy You" on HBO. That is a very smart, very funny, harrowing, really well-observed portrayal of sexual assault and its aftermath. It stars Michaela Coel. More people need to see that show, and a nomination would help.
INSKEEP: And is that why the Emmys remain relevant - they just call attention to things that maybe have been even slightly overlooked?
WELDON: Even more so today than ever because there is more television today than there ever has been, and winning an Emmy can help a show really cut through the noise and find an audience. Now, that's going to be less important for an institution like "The Crown." But for a new show like "Ted Lasso" - look; if universal critical praise and word of mouth hasn't sold you on that show yet, then maybe an Emmy nomination will.
INSKEEP: Glen, it's always a pleasure hearing from you. Thanks so much.
WELDON: Thank you. My pleasure.
INSKEEP: Glen Weldon is a host of NPR's Pop Culture Happy Hour.
(SOUNDBITE OF AMBIENT JAZZ ENSEMBLE'S "RISE") Transcript provided by NPR, Copyright NPR.


text:
TB Alliance Announces Two New Members to its Board of Directors
TB Alliance Announces Two New Members to its Board of Directors:
Dr. Mario Raviglione of WHO and Shalini Sharp of Ultragenyx
TB Alliance, an international not-for-profit organization with the mission of developing improved tuberculosis (TB) treatments, announced the appointment of Mario Raviglione, MD, director of the Global TB Programme at the World Health Organization (WHO), and Shalini Sharp, the chief financial officer of Ultragenyx Pharmaceutical Inc., to its Board of Directors.
"Tuberculosis is a leading killer worldwide. The war against TB will only be won with new and better tools and treatments," said Dr. Raviglione. "I'm deeply honored to join the Board of TB Alliance to contribute my knowledge and expertise, particularly as the organization prepares for late-stage testing and the potential introduction of new, innovative and game-changing TB treatments."
"The TB Alliance's product development partnership model is critical to the delivery of superior treatment options to patients in need," said Ms. Sharp. "It is exciting to be working with an organization that has the potential to impact the course of this devastating disease and save many lives."
Dr. Raviglione began working for the World Health Organization in 1991 and became director of the Global TB Programme in 2003. In his role, he is responsible for setting the policies and standards on global TB control, coordinating technical support, monitoring the global TB pandemic, and translating new evidence into policies and practice. Dr. Raviglione serves on a variety of scientific committees, has published extensively on the TB epidemic, and has won numerous awards for his work on TB control. Dr. Raviglione graduated from the University of Turin in Italy in 1980, and trained in internal medicine and infectious diseases in New York and Boston, where he was appointed an AIDS clinical research fellow at Beth Israel Hospital, Harvard Medical School.
Ms. Sharp has been the chief financial officer and senior vice president, finance, of Ultragenyx, a clinical-stage biotechnology company developing medicines to treat rare diseases, since 2012. She is a member of the Board of Directors of Agenus Inc. (formerly Antigenics Inc.), a publicly traded biotechnology firm, where she served as chief financial officer from 2006 to 2012. Prior to Agenus, Ms. Sharp held strategic planning and corporate finance roles at Elan Pharmaceuticals. She has also worked with McKinsey & Company and Goldman Sachs, specializing in pharmaceuticals and medical devices. Ms. Sharp holds both a BA and an MBA from Harvard University.
"We are pleased to welcome Dr. Raviglione and Ms. Sharp to the Board of the TB Alliance," said Carlos Morel, MD, chairman of the board for TB Alliance and director of the Center for Technological Development in Health (CDTS) of FIOCRUZ, Brazil. "Their expertise will be invaluable as the organization launches an unprecedented number of trials and initiatives."
Tuberculosis is a global pandemic, killing someone every 21 seconds—nearly 1.5 million in 2013 alone, according to the WHO. More than 95 percent of TB deaths occur in low- and middle-income countries, and TB is among the top five causes of death for women aged 15 to 44. Today's TB drug regimens take too long to cure the disease and, especially in the case of multidrug-resistant TB, are complicated to administer and can be toxic. New drug regimens are urgently needed to control TB and stop the spread of drug-resistant TB.
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Lint your code - The easiest way to apply the conventions is to install overcommit.
Add tests! - Your patch won't be accepted if it doesn't have tests.
Document any change in behaviour - Make sure the README and any other relevant documentation are kept up-to-date.
Consider our release cycle - We try to follow semver. Randomly breaking public APIs is not an option.
Create topic branches - Don't ask us to pull from your master branch.
Ensure, that you have Chef DK installed.


This is our inspiration board. Medals on the left are mine, on the right Mr CTs. The empty pins near the top are there to give me a push to take part in more events and collect more bling (I'm a full on magpie for medals!). Honestly I'll do almost anything for a bit of ribbon and chunk of metal!
If anyone has a spare @theparishwalk medal from 2014 I've got a gap on my board because I got half way up the sloc and had to retire from heat stroke, so didn't get medal at Rushen 😢😭😢 That gap haunts me to this day, the 2013 & 2015 medals mock that gap!


text:
La Prva savezna liga SFRJ 1983-1984, conosciuta semplicemente come Prva liga 1983-1984, fu la 55ª edizione del campionato jugoslavo di calcio, la 38ª dopo la fine della seconda guerra mondiale. La vittoria finale fu appannaggio della , al suo quindicesimo titolo. È stata l'unica edizione in cui sono state rappresentate tutte le 6 repubbliche (Slovenia, Croazia, Bosnia Erzegovina, Serbia, Montenegro e Macedonia) e le 2 province autonome (Voivodina e Kosovo).

Nel ranking UEFA 1983-84 la Prva liga si piazzò al 9º posto (8º nel quinquennale 1979-1984).

Squadre partecipanti

Profili

Provenienza

Classifica finale

Legenda:
      Campione di Jugoslavia, qualificata alla Coppa dei Campioni 1984-1985
      Vincitrice della coppa nazionale, qualificata alla Coppa delle Coppe 1984-1985
      Qualificate alla Coppa UEFA 1984-1985
      Retrocesse in Druga Liga 1984-1985

Note:
Due punti a vittoria, uno a pareggio, zero a sconfitta.
In caso di arrivo di due o più squadre a pari punti, la graduatoria viene stilata secondo la differenza reti delle squadre interessate.
Coppe minori:Qualificata alla Coppa Mitropa 1984-1985 : .Qualificata alla Coppa dei Balcani per club 1984-1985 : nessuna.Qualificate alla Coppa Piano Karl Rappan 1984 : nessuna.

Classifica marcatori

|- style="background-color: #ffffff;"
! colspan="5" style="text-align: right;" | Fonte: sportsport.ba

Risultati

Tabellone

Calendario

 Fonti: exyufudbal e rsssf

Squadra campione

Tomislav Ivković (34 presenze)
Marko Elsner (32)
Miloš Šestić (29)
Boško Ǵurovski (23)
Miroslav Šugar (22)
Milan Jovin (21)
Đorđe Milovanović (21)
Mitar Mrkela (20)
Milko Ǵurovski (27)
Jovica Nikolić (22)
Zoran Banković (19)
Allenatore: Gojko Zec

Riserve
Ivan Jurišić (19), Dragan Miletović (17), Milan Janković (16), Ljubiša Stojanović (15), Ranko Đorđić (14), Rajko Janjanin (14), Goran Milojević (14), Žarko Đurović (13), Nedeljko Milosavljević (12), Dragić Komadina (9), Miodrag Krivokapić (4), Cvijetin Blagojević (3), Zlatko Krmpotić (2), Radoslav Žugić (1), Radomir Savić (1), Slavko Radovanović (1)
Fonti: sportsport.ba, worldfootball.net e crvenazvezdafk.com

Affluenza

Note

Voci correlate
Druga Liga 1983-1984
Kup Maršala Tita 1983-1984

Collegamenti esterni
  Statistiche stagione 1983-84 su historical-lineups.com
  Stagione 1983-1984 su eu-football.info
  Stagione 1983-1984 su HRnogomet
  Stagione 1983-1984 su sportsport.ba
  Le stagioni 1983-1988 su FSG Zrenjanin
  Le stagioni 1981-1990 su claudionicoletti
  Risultati su exYUfudbal
  Stagione 1983-1984 su RSSSF

Calcio nel 1983
Calcio nel 1984
1984
Campionato jugoslavo di calcio

redpajama_set_name: RedPajamaWikipedia


The CIC believes that research and innovation are of great importance to the sustainable development of the Hong Kong construction industry. Hence, CIC is committed to working closely with industry stakeholders to drive innovation and initiate practical research projects. The CIC Research and Technology Development Fund (R&D Fund) is set up to provide financial support for research projects which can benefit the Hong Kong construction industry through practical application of the research outcomes, covering the areas of (i) Construction Safety and Health; (ii) Construction Productivity; (iii) Environment and Sustainability; (iv) Construction Procurement; and (v) Building Information Modelling (BIM).
The Research Policy herein attached provides details of the CIC research framework and CIC funding and/or technical support to research institutes.


Today I added webmention.io to my website's header. This will allow me to now receiving webmentions from other sites that support it. I haven't gone through the task of actually collecting and displaying them, but the first step is making sure that someone can send me a webmention (A comment, like, rsvp or reaction).
I have two goals for webmentions. The first is to build my website to start feteching webmentions from webmention.io and to start displaying them at the bottom of posts.
The second is to build a native iOS app that can receiving webmentions from a service like webmention.io (or a person's personal website) and display it as an iOS alert. This would be integrated into the indigenous app that I'm working on, but probably won't be integrated until Version 2 in Summer 2017. That would help do for the IndieWeb what the native Twitter and Facebook clients do in allowing you to be notified about what people are talking to you.