#  Some common fixed points results on metric spaces over topological modules

Ion Marian Olaru

###### Abstract

In this paper, we replace the real numbers by a topological R-module and
define R-metric spaces $(X,d)$. Also, we prove some common fixed point
theorems on R-module metric spaces. We obtain, as a particular case the Perov
theorem (see [3])

2010 Mathematical Subject Classification: 47H10

Keywords: R-metric spaces, fixed point theory, topological rings, topological
modules

##  1 R-metric spaces

In this section we shall define $R$-metric spaces and prove some properties.
All axioms for an ordinary metric space can be meaningfully formulated for an
abstract metric space, where the abstract metric takes values in a partially
ordered topological module of a certain type which will be defined below. Such
a space will be called $R$-metric space.

We begin this section by recalling a few facts concerning topological rings,
topological modules and partially ordered rings. Unless explicitly stated
otherwise all rings will be assumed to possess an identity element, denoted by
$1$.

######  Definition 1.1.

(see [5]) A topology $\tau$ on a ring $(R,+,\cdot)$ is a ring topology and
$R$, furnished with $\tau$, is a topological ring if the following conditions
hold:

  * (TR 1)

$(x,y)\rightarrow x+y$ is continuous from $R\times R$ to $R$;

  * (TR 2)

$x\rightarrow-x$ is continuous from $R$ to $R$;

  * (TR 3)

$(x,y)\rightarrow x\cdot y$ is continuous from $R\times R$ to $R$,

where $R$ is given topology $\tau$ and $R\times R$ the cartesian product
determined by topology $\tau$.

######  Definition 1.2.

(see [5]) Let $R$ be a topological ring, $E$ an R-module. A topology
$\mathcal{T}$ on $E$ is a R-module topology and E, furnished with
$\mathcal{T}$, is a topological R-module if the following conditions hold:

  * (TM 1)

$(x,y)\rightarrow x+y$ is continuous from $E\times E$ to $E$;

  * (TM 2)

$x\rightarrow-x$ is continuous from $E$ to $E$;

  * (TM 3)

$(a,x)\rightarrow a\cdot x$ is continuous from $R\times E$ to $E$,

where $E$ is given topology $\mathcal{T}$, $E\times E$ the cartesian product
determined by topology $\mathcal{T}$ and $A\times E$ the cartesian product
determined by topology of R and E.

By a partially ordered ring is meant a pair consisting of a ring and a
compatible partial order, denoted by $\preceq$(see [4]).

In the following we always suppose that $R$ is an ordered topological ring
such that $0\preceq 1$ and $E$ is a topological $R$-module.

######  Definition 1.3.

A subset $P$ of $E$ is called a cone if:

  * (i)

$P$ is closed, nonempty and $P\neq\\{0_{E}\\}$;

  * (ii)

$a,b\in R$, $0\preceq a$, $0\preceq b$ and $x,y\in P$ implies $a\cdot x+b\cdot
y\in P$;

  * (iii)

$P\cap-P=\\{0\\}.$

Given a cone $P\subset E$, we define on E the partial ordering $\leq_{P}$ with
respect to $P$ by

(1.1) $x\leq_{P}y\ if\ and\ only\ if\ y-x\in P.$

We shall write $x<_{P}y$ to indicate that $x\leq_{P}y$ but $x\neq y$, while
$x\ll y$ will stand for $y-x\in intP$(interior of $P$).

######  Example 1.1.

Let $R=\mathcal{M}_{n\times n}(\mathbb{R})$ be the ring of all matrices with n
rows and n columns with entries in $\mathbb{R}$ and $E=\mathbb{R}^{n}$. We
define the partial order $\preceq$ on $M_{n\times n}(\mathbb{R})$ as follows

$A\preceq B\ if\ and\ only\ if\ for\ each\ i,j=\overline{1,n}\ we\ have\
a_{ij}\leq b_{ij}.$

Then

  * (a)

the topology $\tau$, generated by matrix norm

$N:M_{n\times n}(\mathbb{R})\rightarrow\mathbb{R},$
$N(A)=\max\limits_{i=\overline{1,n}}\sum\limits_{j=1}^{n}|a_{ij}|,$

is a ring topology;

  * (b)

the standard topology $\mathcal{D}$ is a R-module topology on
$\mathbb{R}^{n}$;

  * (c)

$P=\\{(x_{1},x_{2},\cdots,x_{n})\in\mathbb{R}^{n}\mid x_{i}\geq
0,(\forall)i=\overline{1,n}\\}$ is a cone in E.

Indeed, Theorem 1.3 pp 2 of [5] leads us to $(a)$.

It is obvious that $(TM\ 1)$ and $(TM\ 2)$ are satisfied. Now we consider
$A_{n}\stackrel{{\scriptstyle\tau}}{{\rightarrow}}A$ and
$x_{n}\stackrel{{\scriptstyle\mathcal{D}}}{{\rightarrow}}x$ as
$n\rightarrow\infty$. Then

$\|A_{n}\cdot x_{n}-A\cdot x\|_{\mathbb{R}^{n}}=\|A_{n}\cdot x_{n}-A_{n}\cdot
x+A_{n}\cdot x-A\cdot x\|_{\mathbb{R}^{n}}\leq$
$\|A_{n}\cdot(x_{n}-x)\|_{\mathbb{R}^{n}}+\|(A_{n}-A)\cdot
x\|_{\mathbb{R}^{n}}\leq
N(A_{n})\|x_{n}-x\|_{\mathbb{R}^{n}}+N(A_{n}-A)\|x\|_{\mathbb{R}^{n}}\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\rightarrow}}0.$

Hence, $(TM\ 3)$ holds. Thus, we have obtained $(b)$. Finally, it easy to see
that $P$ is a cone in $E$.

In the following we always suppose that $E$ is a topological $R$-module, $P$
is a cone in $E$ with $intP\neq\emptyset$ and $\leq_{P}$ is a partial ordering
with respect to $P$.

######  Definition 1.4.

Let $X$ be a nonempty set. Suppose that a mapping

$d:X\times X\rightarrow E$

satisfies:

  * $(d_{1})$

$0_{E}\leq_{P}d(x,y)$ for all $x,y\in X$ and $d(x,y)=0_{E}$ if and only if
$x=y$;

  * $(d_{2})$

$d(x,y)=d(y,x)$, for all $x,y\in X$ ;

  * $(d_{3})$

$d(x,y)\leq_{P}d(x,z)+d(z,y)$, for all $x,y,z\in X$.

Then $d$ is called a R-metric on $X$ and $(X,d)$ is called a R-metric space.

######  Example 1.2.

Any cone metric space is a R-metric space.

######  Example 1.3.

Let $R=M_{n\times n}(\mathbb{R})$ be the ring of all matrices with n rows and
n columns with entries in $\mathbb{R}$, $E=\mathbb{R}^{n}$, $X=\mathbb{R}^{n}$
and

$P=\\{(x_{1},x_{2},\cdots,x_{n})\in\mathbb{R}^{n}\mid x_{i}\geq
0,(\forall)i=\overline{1,n}\\}$

a cone in E.

We define the partial order $\preceq$ on $M_{n\times n}(\mathbb{R})$ as
follows

$A\preceq B\ if\ and\ only\ if\ for\ each\ i,j=\overline{1,n}\ we\ have\
a_{ij}\leq b_{ij}.$

Then for all $A=(a_{ij})$, $a_{ij}>0$ we have that

$d:\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{n},$
$d(x,y)=(\sum\limits_{j=1}^{n}a_{1j}|x_{j}-y_{j}|,\cdots,\sum\limits_{j=1}^{n}a_{ij}|x_{j}-y_{j}|,\cdots,\sum\limits_{j=1}^{n}a_{nj}|x_{j}-y_{j}|)$

is a $R$-metric on $X$.

Indeed,

  * $(d_{1})$

Since $\sum\limits_{j=1}^{n}a_{ij}|x_{j}-y_{j}|\geq 0$ for all
$i=\overline{1,n}$, we have that $0\leq_{P}d(x,y)$ for all
$x,y\in\mathbb{R}^{n}$. Also, $d(x,y)=0$ involve that
$\sum\limits_{j=1}^{n}a_{ij}|x_{j}-y_{j}|=0$ which means that $x_{j}=y_{j}$
for all $j=\overline{1,n}$. It follows that $x=y$.

  * $(d_{2})$

It is obvious that $d(x,y)=d(y,x)$, for all $x,y\in\mathbb{R}^{n}$.

  * $(d_{3})$

Let be $x,y,z\in\mathbb{R}^{n}$. Since
$\sum\limits_{j=1}^{n}a_{ij}|x_{j}-y_{j}|\leq\sum\limits_{j=1}^{n}a_{ij}|x_{j}-z_{j}|+\sum\limits_{j=1}^{n}a_{ij}|z_{j}-y_{j}|$,
we have that $d(x,y)\leq_{P}d(x,z)+d(z,y)$, for all $x,y,z\in\mathbb{R}^{n}$.

In the following, we shall write $x\prec y$ to indicate that $x\preceq y$ but
$x\neq y$.

######  Remark 1.1.

We have that:

  * (i)

$intP+intP\subseteq intP$;

  * (ii)

$\lambda\cdot intP\subseteq intP$, where $\lambda$ is a invertible element of
the ring $R$ such that $0\prec\lambda$;

  * (iii)

if $x\leq_{P}y$ and $0\preceq\alpha$, then $\alpha\cdot x\preceq\alpha\cdot
y$.

Proof:

$(i)$ Let be $x\in intP+intP$. Then there exists $x_{1},x_{2}\in intP$ such
that $x=x_{1}+x_{2}$. It follows that there exists $V_{1}$ neighborhood of
$x_{1}$ and $V_{2}$ neighborhood of $x_{2}$ such that

$x_{1}\in V_{1}\subset P\ and\ x_{2}\in V_{2}\subset P.$

Since for each $x_{0}\in E$, the mapping $x\rightarrow x+x_{0}$ is a
homeomorphism of $E$ onto itself,we have that $V_{1}+V_{2}$ is a neighborhood
of $x$ with respect to topology $\mathcal{T}$. Thus, $x\in intP$.

$(ii)$ Let $0\prec\lambda$ be an invertible element of the ring $R$ and
$x=\lambda\cdot c$, $c\in intP$. It follows that there exists a neighborhood
$V$ of $c$ such that $c\in V\subset P$.

Since the mapping $x\rightarrow\lambda x$ is a homeomorphism of $E$ onto
itself, we have that $\lambda\cdot V$ is a neighborhood of $x$ with respect to
the topology $\mathcal{T}$. Thus, $x\in intP$.

$(iii)$ If $x\leq_{P}y$, then $y-x\in P$. It follows that for all
$0\preceq\alpha$ we have that $\alpha\cdot(y-x)\in P$ i.e. $\alpha\cdot
x\preceq\alpha\cdot y$.

In the following, unless otherwise specified, we always suppose that there
exists at least one a sequence $\\{\alpha_{n}\\}\subset R$ of invertible
elements such that $0\prec\alpha_{n}$ and $\alpha_{n}\rightarrow 0$ as
$n\rightarrow\infty$.

######  Remark 1.2.

Let $E$ be a $R$-topological module and $P\subset E$ a cone. We have that:

  * (i)

If $u\leq_{P}v$ and $v\ll w$, then $u\ll w$;

  * (ii)

If $u\ll v$ and $v\leq_{P}w$, then $u\ll w$;

  * (iii)

If $u\ll v$ and $v\ll w$, then $u\ll w$;

  * (iv)

If $0\leq_{P}u\ll c$ for each $c\in IntP$, then $u=0$;

  * (v)

If $a\leq_{P}b+c$ for each $c\in IntP$, then $a\leq_{P}b$;

  * (vi)

If $0\ll c$, $0\leq_{P}a_{n}$ and $a_{n}\rightarrow 0$ then there exists
$n_{0}\in\mathbb{N}$ such that $a_{n}\ll c$ for all $n\geq n_{0}$.

Proof:

  * (i)

We have to prove that $w-u\in intP$ if $v-u\in P$ and $w-v\in intP$. The
condition $(TM_{1})$ implies that there exists a neighborhood $V$ of $0$ such
that $w-v+V\subset P$. It follows that $w-u+V=(w-v)+V+(v-u)\subset P+P\subset
P$. Since for each $x_{0}\in E$ the mapping $x\rightarrow x+x_{0}$ is a
homeomorphism of $E$ onto itself we have that $w-u+V$ is a neighborhood of
$w-u$ with respect to the topology $\mathcal{T}$. Thus, $w-u\in int\ P$.

  * (ii)

Analogous with $(i)$.

  * (iii)

We have to prove that $w-u\in intP$ if $v-u\in intP$ and $w-v\in intP$. Since
we have $intP+intP\subset intP$, it easy to see that the above assertion is
satisfied.

  * (iv)

Let $\\{\alpha_{n}\\}_{n\in\mathbb{N}}\subset R$ be a sequence of invertible
elements such that for each $n\in\mathbb{N}$ we have $0\prec\alpha_{n}$ and
$\alpha_{n}\rightarrow 0$ as $n\rightarrow\infty$.

Since for each invertible element $\lambda_{0}\in R$ the mapping
$x\rightarrow\lambda_{0}\cdot x$ is a homeomorphism of $E$ onto itself, we
have that if $V$ is a neighborhood of zero then $\lambda_{0}\cdot V$ is a
neighborhood of zero. Hence, $\alpha_{n}\cdot c\in intP$ for each
$n\in\mathbb{N}$.

Then $\alpha_{n}\cdot c-u\in intP$. It follows that
$\lim\limits_{n\rightarrow\infty}\alpha_{n}\cdot c-u=-u\in\overline{P}=P$.
Thus, $u\in P\cap-P=\\{0\\}$.

  * (v)

Analogous with $(iv)$.

  * (vi)

Let be $0\ll c$, $0\leq_{P}a_{n}$ and $a_{n}\rightarrow 0$. Since for each
invertible element $\lambda_{0}\in R$ the mapping
$x\rightarrow\lambda_{0}\cdot x$ is a homeomorphism of $E$ onto itself, it
follows that for all neighborhood $V$ of zero we have that $-V$ is a
neighborhood of zero. Now, $0\ll c$ implies that there exists a neighborhood
$U$ of zero such that $c+U\subset P$. Let $V=U\cap-U$ be a neighborhood of
zero. Since $a_{n}$ converges to zero, it follows that there exists
$n_{0}\in\mathbb{N}$ such that $a_{n}\in V$ for all $n\geq n_{0}$. Then we
have that $c-a_{n}\in c+V\subset c+U\subset P$ for all $n\geq n_{0}$. Thus,
$a_{n}\ll c$ for all $n\geq n_{0}$.

######  Definition 1.5.

Let $\\{x_{n}\\}$ be a sequence in a R-metric space $(X,d)$ and $x\in X$. We
say that:

  * (i)

the sequence $\\{x_{n}\\}$ converges to $x$ and is denoted by
$\lim\limits_{n\rightarrow\infty}x_{n}=x$ if for every $0\ll c$ there exists
$N\in\mathbb{N}$ such that $d(x_{n},x)\ll c$, for all $n>N$;

  * (ii)

the sequence $\\{x_{n}\\}$ is a Cauchy sequence if for every $c\in E$, $0\ll
c$ there exists $N\in\mathbb{N}$ such that $d(x_{m},x_{n})\ll c$ for all
$m,n>N$;

The R-metric space $(X,d)$ is called complete if every Cauchy sequence is
convergent.

From the above remark we obtain

######  Remark 1.3.

Let $(X,d)$ be a R-metric space and $\\{x_{n}\\}$ be a sequence in $X$. If
$\\{x_{n}\\}$ converges to $x$ and $\\{x_{n}\\}$ converges to $y$, then $x=y$.

Indeed, for all $0\ll c$

$d(x,y)\leq_{P}d(x,x_{n})+d(x_{n},y)\ll 2\cdot c.$

Hence, $d(x,y)=0$ i.e. $x=y$.

##  2 Common fixed points theorems

In this section we obtain several coincidence and common fixed point theorems
for mappings defined on a $R$-metric space.

######  Definition 2.1.

(see [1]) Let $f$ and $g$ be self maps of a set $X$. If $w=fx=gx$ for some
$x\in X$, then $x$ is called a coincidence point of $f$ and $g$, and $w$ is
called a point of coincidence of $f$ and $g$.

Jungck [2], defined a pair of self mappings to be weakly compatible if they
commute at their coincidence points.

######  Proposition 2.1.

(see [1]) Let $f$ and $g$ be weakly compatible self maps of a set $X$. If $f$
and $g$ have a unique point of coincidence $w=fx=gx$, then $w$ is the unique
common fixed point of $f$ and $g$.

Let $\mathcal{K}$ be the set of all $k\in R$, $0\preceq k$ which have the
property that there exists a unique $S\in R$ such that
$S=\lim\limits_{n\rightarrow\infty}(1+k+\cdots+k^{n})$.

######  Example 2.1.

Let be $A\in\mathcal{M}_{n\times n}(\mathbb{R}_{+})$ such that $\rho(A)<1$.
Then $A\in\mathcal{K}$.

######  Theorem 2.1.

Let $(X,d)$ be a R-metric space and suppose that the mappings
$f,g:X\rightarrow X$ satisfy:

  * (i)

the range of $g$ contains the range of $f$ and $g(X)$ is a complete subspace
of $X$;

  * (ii)

there exists $k\in\mathcal{K}$ such that $d(fx,fy)\leq_{P}k\cdot d(gx,gy)$ for
all $x,y\in X$.

Then $f$ and $g$ have a unique point of coincidence in $X$. Moreover, if $f$
and $g$ are weakly compatible then, $f$ and $g$ have a unique common fixed
point.

Proof: Let $x_{0}$ be an arbitrary point in $X$. We choose a point $x_{1}\in
X$ such that $f(x_{0})=g(x_{1}).$ Continuing this process, having chosen
$x_{n}\in X$, we obtain $x_{n+1}\in X$ such that $f(x_{n})=g(x_{n+1})$. Then

$d(gx_{n+1},gx_{n})=d(fx_{n},fx_{n-1})\leq_{P}k\cdot
d(gx_{n},gx_{n-1})\leq_{P}$ $\leq_{P}k^{2}\cdot
d(gx_{n-1},gx_{n-2})\leq_{P}\cdots\leq_{P}k^{n}\cdot d(gx_{1},gx_{0}).$

We denote $S_{n}=1+k+\cdots+k^{n}$and we get that

$d(gx_{n},gx_{n+p})\leq_{P}d(gx_{n},gx_{n+1})+d(gx_{n+1},gx_{n+2})+\cdots+d(gx_{n+p-1},gx_{n+p})\leq_{P}$
$\leq_{P}(k^{n}+k^{n+1}+\cdots+k^{n+p-1})\cdot
d(gx_{1},gx_{0})=(S_{n+p-1}-S_{n-1})\cdot d(gx_{1},gx_{0}),$

for all $p\geq 1$. Thus, via Remark 1.2 $(vi)$, we obtain that $gx_{n}$ is a
Cauchy sequence. Since $g(X)$ is complete, there exists $q\in g(X)$ such that
$gx_{n}\rightarrow q$ as $n\rightarrow\infty$. Consequently, we can find $p\in
X$ such that $g(p)=q$. Further, for each $0\ll c$ there exists
$n_{0}\in\mathbb{N}$ such that for all $n\geq n_{0}$

$d(gx_{n},fp)=d(fx_{n-1},f(p))\leq_{P}k\cdot d(gx_{n-1},gp)\ll c.$

It follows that $gx_{n}\rightarrow fp$ as $n\rightarrow\infty$. The uniqueness
of the limit implies that $fp=gp=q$. Next we show that $f$ and $g$ have a
unique point of coincidence. For this, we assume that there exists another
point $p_{1}\in X$ such that $fp_{1}=gp_{1}$. We have

$d(gp_{1},gp)=d(fp_{1},fp)\leq_{P}k\cdot d(gp_{1},gp)=k\cdot
d(fp_{1},fp)\leq_{P}k^{2}\cdot d(gp_{1},gp)\leq_{P}\cdots\leq_{P}k^{n}\cdot
d(gp_{1},gp).$

Let be $0\ll c$. Since $k^{n}\rightarrow 0$ as $n\rightarrow\infty$ it follows
that there exists $n_{0}\in\mathbb{N}$ such that $k^{n}\cdot d(gp_{1},gp)\ll
c$ for all $n\geq n_{0}$. Then $d(gp_{1},gp)\ll c$ for each $0\ll c$. Thus
$d(gp_{1},gp)=0$ i.e. $gp_{1}=gp$. From Proposition 2.1 $f$ and $g$ have a
unique common fixed point.

######  Remark 2.1.

The above theorem generalizes Theorem 2.1 of Abbas and Jungck [1], which
itself is a generalization of Banach fixed point theorem.

######  Corollary 2.1.

Let $(X,d)$ be a complete $R$-metric space and we suppose that the mapping
$f:X\rightarrow X$ satisfies:

  * (i)

there exists $k\in\mathcal{K}$ such that $d(fx,fy)\leq_{P}k\cdot d(x,y)$ for
all $x,y\in X$.

Then $f$ has in $X$ a unique fixed point point.

Proof: The proof uses Theorem 3.1 by replacing $g$ with the identity mapping.

From the above corollary using Example 1.1, we obtain the Perov fixed point
theorem (see [3])

######  Corollary 2.2.

Let $(X,d)$ be a complete $\mathcal{M}_{n\times n}(\mathbb{R}_{+})-$ metric
space and $E=\mathbb{R}^{n}$ and we suppose the mapping $f:X\rightarrow X$
satisfies:

  * (i)

there exists $A\in\mathcal{M}_{n\times n}(\mathbb{R}_{+})$ with $\rho(A)<1$
such that

$d(fx,fy)\leq_{P}A\cdot d(x,y),$

for all $x,y\in X$.

Then $f$ has in $X$ a unique fixed point point.

##  3 Comparison function

######  Definition 3.1.

Let P be a cone in a topological R-module E. A function $\varphi:P\rightarrow
P$ is called a comparison function if

  * (i)

$\varphi(0)=0$ and $\varphi(t)<_{P}t$ for all $t\in P-\\{0\\}$;

  * (ii)

$t_{1}\leq_{P}t_{2}$ implies $\varphi(t_{1})\leq_{P}\varphi(t_{2})$;

  * (iii)

$t\in intP$ implies $t-\varphi(t)\in intP$;

  * (iv)

if $t\in P-\\{0\\}$ and $0\ll c$, then there exists $n_{0}\in\mathbb{N}$ such
that $\varphi^{n}(t)\ll c$ for each $n\geq n_{0}$.

######  Example 3.1.

Let P be a cone in a topological R-module E and $\lambda\in\mathcal{K}$ such
that $0\prec 1-\lambda$. Then $\varphi:P\rightarrow P$, defined by
$\varphi(t)=\lambda\cdot t$ is a comparison function.

Indeed,

$(i)$ It is obvious that $\varphi(0)=0$ and $\varphi(t)<_{P}t$ for all $t\in
P-\\{0\\}$.

$(ii)$ if $t_{1}\leq_{P}t_{2}$ and $\lambda\in\mathcal{K}$ then
$\lambda\cdot(t_{2}-t_{1})\in P$. Thus $\varphi(t_{1})\leq_{P}\varphi(t_{2})$.

$(iii)$ We remark that if $\lambda\in\mathcal{K}$, then $1-\lambda$ is an
invertible element of the ring $R$. Now, let be $t\in intP$. Then
$(1-\lambda)\cdot t\in(1-\lambda)intP\subset intP$.

$(iv)$ Let be $t\in P-\\{0\\}$ and $0\ll c$. Then

$\varphi^{n}(t)=\lambda^{n}\cdot t\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\rightarrow}}0.$

We obtain, via Remark 1.2, that there exists $n_{0}\in\mathbb{N}$ such that
$\varphi^{n}(t)\ll c$ for each $n\geq n_{0}$.

Let $(X,d)$ be a $R$-metric space and let $\varphi:K\rightarrow K$ be a
comparison function. For a pair $(f,g)$ of self-mappings on $X$ consider the
following condition

  * (C)

for arbitrary $x,y\in X$ there exists $u\in\\{d(gx,gy),d(gx,fx),d(gy,fy)\\}$
such that $d(fx,fy)\leq_{P}\varphi(u)$.

######  Theorem 3.1.

Let $(X,d)$ be a R-metric space and let $f,g:X\rightarrow X$ such that

  * (i)

the pair $(f,g)$ satisfies the condition (C) for some comparison function
$\varphi$;

  * (ii)

$f(X)\subset g(X)$ and f(X) or g(X) is a complete subspace of X.

Then f and g have a unique point of coincidence in X. Moreover if $f$ and $g$
are weakly compatible, then $f$ and $g$ have a unique common fixed point.

Proof: Let $x_{0}$ be an arbitrary point in $X$. We choose a point $x_{1}\in
X$ such that $fx_{0}=gx_{1}.$ Continuing this process, having chosen $x_{n}\in
X$, we obtain $x_{n+1}\in X$ such that $fx_{n}=gx_{n+1}$.

$\ulcorner$ We shall prove that the sequence $\\{y_{n}\\}$, where
$y_{n}=fx_{n}=gx_{n+1}$( the so-called Jungck sequence ) is a Cauchy sequence
in $R$-metric space $(X,d)$.

If $y_{N}=y_{N+1}$ for some $N\in\mathbb{N}$ then $y_{m}=y_{N}$ for each $m>N$
and the conclusion follows. Indeed, we prove by induction arguments that

(3.1) $y_{N+k}=y_{N+k+1},(\forall)k\in\mathbb{N}.$

For $k=0$ we have $y_{N}=y_{N+1}$. Let (3.1) hold for all $k=\overline{0,i}$.
Then

$d(y_{N+i+1},y_{N+i+2})=d(fx_{N+i},fx_{N+i+1})\leq_{P}\varphi(u),$

where

$u\in\\{d(gx_{N+i},gx_{N+i+1}),d(gx_{N+i},fx_{N+i}),d(gx_{N+i+1},fx_{N+i+1})\\}=$
$\\{d(y_{N+i-1},y_{N+i}),d(y_{N+i-1},y_{N+i}),d(y_{N+i},y_{N+i+1})\\}=\\{0\\}.$

Hence, $d(y_{N+i+1},y_{N+i+2})\leq_{P}\varphi(u)=0$ i.e.
$y_{N+i+1}=y_{N+i+2}$. Q.E.D

Suppose that $y_{n}\neq y_{n+1}$ for each $n\in\mathbb{N}$. The condition (C)
implies that

$d(y_{n},y_{n+1})=d(fx_{n},fx_{n+1})\leq_{P}\varphi(u),$

where

$u\in\\{d(gx_{n},gx_{n+1}),d(fx_{n},gx_{n}),d(fx_{n+1},gx_{n+1})\\}=\\{d(y_{n-1},y_{n}),d(y_{n},y_{n-1}),d(y_{n+1},y_{n})\\}.$

The case $u=d(y_{n+1},y_{n})$ is impossible, since this would imply

$d(y_{n+1},y_{n})\leq_{P}\varphi(d(y_{n+1},y_{n}))<_{P}d(y_{n+1},y_{n}).$

Thus, $u=d(y_{n},y_{n-1})$ and

$d(y_{n+1},y_{n})\leq_{P}\varphi(d(y_{n},y_{n-1}))\leq_{P}\cdots\leq_{P}\varphi^{n}(d(y_{1},y_{0})).$

Using Remark 1.2 (i) and property $(iv)$ of the comparison function we obtain
that for all $0\ll\varepsilon$ there exists $n_{0}\in\mathbb{N}$ such that

(3.2) $d(y_{n},y_{n+1})\ll\varepsilon,\ (\forall)n\geq n_{0}.$

Now, let be $0\ll c$. Then, using property $(iii)$ of the comparison function,
we get that

(3.3) $d(y_{n},y_{n+1})\ll c-\varphi(c),\ (\forall)n\geq n_{0}.$

Let us fix now $n\geq n_{0}$ and let us prove that

(3.4) $d(y_{n},y_{k+1})\ll c,\ (\forall)k\geq n.$

Indeed, for $k=n$ we have

$d(y_{n},y_{n+1})\ll c-\varphi(c)\leq_{P}c.$

Hence,

$d(y_{n},y_{n+1})\ll c.$

Let (3.4 ) hold for some $k\geq n$. Then we have

$d(y_{n},y_{k+2})\leq_{P}d(y_{n},y_{n+1})+d(y_{n+1},y_{k+2})\ll$
$c-\varphi(c)+d(fx_{n+1},fx_{k+2})\leq_{P}c-\varphi(c)+\varphi(u),$

where

$u\in\\{d(gx_{n+1},gx_{k+2}),d(gx_{n+1},fx_{n+1}),d(gx_{k+2},fx_{k+2})\\}.$

Consider now the following three possible cases:

  * Case 1:

$u=d(gx_{n+1},gx_{k+2})$. Then

$\varphi(u)=\varphi(d(gx_{n+1},gx_{k+2}))=\varphi(d(y_{n},y_{k+1}))\leq_{P}\varphi(c).$

From the above relation it follows that,

$d(y_{n},y_{k+2})\ll
c-\varphi(c)+\varphi(u)\leq_{P}c-\varphi(c)+\varphi(c)=c.$

Hence, $d(y_{n},y_{k+2})\ll c.$

  * Case 2:

$u=d(gx_{n+1},fx_{n+1})=d(y_{n},y_{n+1})$. Then

$\varphi(u)\leq_{P}\varphi(d(y_{n},y_{n+1}))\leq_{P}\varphi(c-\varphi(c))\leq_{P}\varphi(c).$

From the above relation we get that,

$d(y_{n},y_{k+2})\ll
c-\varphi(c)+\varphi(u)\leq_{P}c-\varphi(c)+\varphi(c)=c.$

Hence, $d(y_{n},y_{k+2})\ll c.$

  * Case 3:

$u=d(gx_{k+2},fx_{k+2})$. Then

$\varphi(u)=\varphi(d(gx_{k+2},fx_{k+2}))=\varphi(d(y_{k+1},y_{k+2}))\leq_{P}\varphi(c-\varphi(c))\leq_{P}\varphi(c).$

From the above relation we get that,

$d(y_{n},y_{k+2})\ll
c-\varphi(c)+\varphi(u)\leq_{P}c-\varphi(c)+\varphi(c)=c.$

Hence, $d(y_{n},y_{k+2})\ll c.$

So, it has been proved by induction that $\\{y_{n}\\}$ is a Cauchy
sequence.$\lrcorner$

Since, by assumption, $f(X)$ or $g(X)$ is a complete subspace of $X$, we
conclude that there exists $q\in g(X)$ such that
$y_{n}=fx_{n}=gx_{n+1}\rightarrow q$ as $n\rightarrow\infty$ and there exists
$p\in X$ such that $q=gp$.

$\ulcorner$ We claim that $q=fp$. Indeed, if we suppose that $d(fp,q)\neq 0$,
then we have

$d(fp,q)\leq_{P}d(fp,fx_{n})+d(fx_{n},q)\leq_{P}\varphi(u)+d(y_{n},q),$

where

$u\in\\{d(gp,gx_{n}),d(gp,fp),d(gx_{n},fx_{n})\\}$

Let $0\ll c$. At least one of the following three cases holds for infinitely
many $n\in\mathbb{N}$:

  * Case 1:

$u=d(gp,gx_{n})$. Then, there exists $n_{0}(c)\in\mathbb{N}$ such that for all
$n\geq n_{0}(c)$

$d(fp,q)\leq_{P}\varphi(d(gp,gx_{n}))+d(y_{n},q)<_{P}d(q,y_{n-1})+d(y_{n},q)\ll
2\cdot c.$

It follows that $d(fp,q)=0$, which is a contradiction.

  * Case 2:

$u=d(gp,fp)=d(fp,q)$ . Then we have

$d(fp,q)\leq_{P}\varphi(d(q,fp))+d(y_{n},q)\ll\varphi(d(q,fp))+c.$

Thus, $d(fp,q)\leq_{P}\varphi(d(q,fp))$. So, by using of the properties $(ii)$
and $(iv)$ of the comparison function, we obtain that there exists
$n_{0}\in\mathbb{N}$ such that $d(fp,q)\leq_{P}\varphi^{n}(d(fp,q))\ll c$ i.e.
$d(fp,q)=0$, which is a contradiction.

  * Case 3:

$u=d(gx_{n},fx_{n})=d(y_{n-1},y_{n})$. Then, there exists
$n_{0}(c)\in\mathbb{N}$ such that for all $n\geq n_{0}(c)$ we have

$d(fp,q)\leq_{P}\varphi(d(y_{n-1},y_{n}))+d(y_{n},q)<_{P}d(y_{n-1},y_{n})+d(y_{n},q)\ll
2\cdot c,$

i.e. $d(fp,q)=0$, which is a contradiction.

It follows that $fp=gp=q$ i.e. $p$ is a coincidence point of the pair $(f,g)$
and $q$ is a point of coincidence.$\lrcorner$

$\ulcorner$ Next we show that $f$ and $g$ have a unique point of coincidence.
For this we assume that there exists another point $p_{1}\in X$ such that
$fp_{1}=gp_{1}$. If we suppose that $d(fp_{1},fp)\neq 0$ we get that
$d(fp_{1},fp)\leq_{P}\varphi(u)$, where

$u\in\\{d(gp_{1},gp),d(gp_{1},fp_{1}),d(gp,fp)\\}=\\{d(gp_{1},gp),0\\}.$

In both possible cases a contradiction easily follows :
$d(fp_{1},fp)\leq_{P}\varphi(d(fp_{1},fp))<_{P}d(fp_{1},fp)$ or
$d(fp_{1},fp)\leq_{P}\varphi(0)=0$. We conclude that the mappings $f$ and $g$
have a unique point of coincidence. From Proposition 2.1 $f$ and $g$ have a
unique common fixed point.$\lrcorner$

## References

  * [1] M. Abbas, G. Jungck,Common fixed point results for noncommuting mappings without continuity in cone metric spaces, J. Math.Anal.Appl. 341 (2008)
  * [2] G. Jungck, Common fixed points for noncontinuous nonself maps on non-metric spaces, Far East J. Math. Sci.(FJMS) 4(1996) 199-215.
  * [3] A. I. Perov, On the Cauchy problem for a system of ordinary differential equations, Pviblizhen. Met. Reshen. Differ. Uvavn., vol. 2, pp. 115 134, 1964.
  * [4] Stuart A. Steinberg, Lattice-ordered rings and modules, Springer Science + Business Media, LLC 2010, DOI 10.1007/978-1-4419-1721-8-1.
  * [5] Seth Warner, Topological rings, North-Holland Math. Studies 178, 1993.

Department of Mathematics,
Faculty of Science,
University ”Lucian Blaga” of Sibiu,
Dr. Ion Ratiu 5-7, Sibiu, 550012, Romania
E-mail<EMAIL_ADDRESS>


# A mean field problem approach for the double curvature prescription problem

Luca Battaglia  Luca Battaglia
Università degli Studi Roma Tre
Dipartimento di Matematica e Fisica
Via della Vasca Navale 84
00146 Roma, Italy<EMAIL_ADDRESS>and  Rafael López–Soriano
Rafael López–Soriano
Universidad de Granada
Departamento de Análisis Matemático
Campus Fuentenueva
18071 Granada, Spain<EMAIL_ADDRESS>
###### Abstract.

In this paper we establish a new mean field-type formulation to study the
problem of prescribing Gaussian and geodesic curvatures on compact surfaces
with boundary, which is equivalent to the following Liouville-type PDE with
nonlinear Neumann conditions:

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta u+2K_{g}=2Ke^{u}&\text{in
}\Sigma\\\ \partial_{\nu}u+2h_{g}=2he^{\frac{u}{2}}&\text{on
}\partial\Sigma.\end{array}\right.$

We provide three different existence results in the cases of positive, zero
and negative Euler characteristics by means of variational techniques.

###### Key words and phrases:

Prescribed curvature problem, conformal metric, variational methods.
The authors certify they have NO conflict of interest.

###### 2010 Mathematics Subject Classification:

35J20, 58J32.

##  1\. Introduction

Let $\displaystyle\Sigma$ be a compact surface with boundary equipped with a
metric $\displaystyle g$. We consider the following Neumann boundary value
problem

(1.1) $\left\\{\begin{array}[]{ll}-\Delta u+2K_{g}=2Ke^{u}&\text{in }\Sigma\\\
\partial_{\nu}u+2h_{g}=2he^{\frac{u}{2}}&\text{on
}\partial\Sigma,\end{array}\right.$

where $\displaystyle\Delta=\Delta_{g}$ is the Laplace–Beltrami operator in
$\displaystyle(\Sigma,g)$, $\displaystyle\partial_{\nu}$ is the normal
derivative with $\displaystyle\nu$ being the outward normal vector to
$\displaystyle\partial\Sigma$ and $\displaystyle K_{g},K:\Sigma\to\mathbb{R}$
and $\displaystyle h_{g},h:\partial\Sigma\to\mathbb{R}$ are smooth.
This equation has a special interest due to its geometric meaning. In fact, it
is equivalent to prescribing at the same time the Gaussian curvature in
$\displaystyle\Sigma$ and the geodesic curvature on
$\displaystyle\partial\Sigma$. More precisely, given a metric
$\displaystyle\tilde{g}=ge^{u}$ conformal to $\displaystyle g$, if
$\displaystyle K_{g},K$ are the Gaussian curvatures and $\displaystyle
h_{g},h$ are the geodesic curvatures of $\displaystyle\partial\Sigma$ with
respect to the metrics $\displaystyle g,\tilde{g}$, then $\displaystyle u$
satisfies (1.1).

Problem (1.1) can be seen as a natural generalization of the following
Liouville-type equation on closed surfaces

(1.2) $-\Delta u+2K_{g}=2Ke^{u}\qquad\text{in }\Sigma,$

which is equivalent to prescribing the Gaussian curvature $\displaystyle K$ on
$\displaystyle\Sigma$.
Problem (1.2) has been very widely studies for several decades. For a survey
on this topic we refer to the Chapter 6 in [1] and the references therein.
On the other hand, literature concerning problem (1.1) is not as wide, unless
in some special cases. For instance, the case of zero prescribed geodesic
curvature $\displaystyle h\equiv 0$ has been treated in [5, 12, 14], whereas
the case of zero prescribed Gaussian curvature $\displaystyle K\equiv 0$ in
[6]. The special case of constant curvatures $\displaystyle K\equiv
K_{0},h\equiv h_{0}$ has been studied in [4]. In that case, solutions have
been explicitely classified when $\displaystyle\Sigma$ is a disk (see [11, 18,
9]) or an annulus ([10]).
In the case of non-constant curvatures, problem (1.1) has been studied in [15]
for a wide range of cases, under the assumption of negative Gaussian curvature
$\displaystyle K(x)<0$ and non-positive Euler characteristic
$\displaystyle\chi(\Sigma)\leqslant 0$. The authors provided existence,
uniqueness and non-existence results as well as a blow-up analysis.

In order to study problem (1.1), some information can be easily deduced from
the Gauss-Bonnet theorem. In fact, integrating both sides of (1.1) and then by
parts, one obtains:

(1.3)
$\int_{\Sigma}Ke^{u}+\int_{\partial\Sigma}he^{\frac{u}{2}}=\int_{\Sigma}K_{g}+\int_{\partial\Sigma}h_{g}=2\pi\chi(\Sigma),$

which imposes necessary conditions on the functions $\displaystyle K,h$.
In particular, if $\displaystyle\chi(\Sigma)>0$, that is $\displaystyle\Sigma$
is topologically equivalent to the unit disk, then $\displaystyle K$ or
$\displaystyle h$ must be positive somewhere in $\displaystyle\Sigma$ or
$\displaystyle\partial\Sigma$, respectively. Similarly, if
$\displaystyle\chi(\Sigma)<0$, then $\displaystyle K$ or $\displaystyle h$
must be negative somewhere. On the other hand, if
$\displaystyle\chi(\Sigma)=0$ then either $\displaystyle K$ or $\displaystyle
h$ must change sign, or one of the two is everywhere non-negative and the
other is everywhere non-positive.

It can be seen, using the Uniformization Theorem, that one can always
prescribe zero geodesic curvature and constant Gaussian curvature equal to
$\displaystyle\mathrm{sign}(\chi(\Sigma))$ (more details can be found in [15],
Proposition 3.1). Geometrically, this is equivalent to working on a half-
sphere in the case $\displaystyle\chi(\Sigma)>0$, on a cylinder in the case
$\displaystyle\chi(\Sigma)=0$ and on a domain of the hyperbolic plane in the
case $\displaystyle\chi(\Sigma)<0$.
Therefore, one can assume without losing generality that $\displaystyle
K_{g}\equiv\frac{2\pi\chi(\Sigma)}{|\Sigma|},h_{g}\equiv 0$ and consider the
following problem:

(1.4) $\left\\{\begin{array}[]{ll}-\Delta
u+\frac{4\pi\chi(\Sigma)}{|\Sigma|}=2Ke^{u}&\text{in }\Sigma\\\
\partial_{\nu}u=2he^{\frac{u}{2}}&\text{on }\partial\Sigma,\end{array}\right.$

From now on, we will focus on problem (1.4) instead of (1.1) without any
further comment.

In [15] solutions to (1.4) are found as critical points of the energy
functional

$\displaystyle\mathcal{I}(u):=\int_{\Sigma}\left(\frac{|\nabla
u|^{2}}{2}+\frac{4\pi\chi(\Sigma)}{|\Sigma|}u-2Ke^{u}\right)-4\int_{\partial\Sigma}he^{\frac{u}{2}}.$

Such a formulation turns out to be very convenient when both $\displaystyle K$
and $\displaystyle\chi(\Sigma)$ are non-positive, but it is not clear how it
can be extended to other cases.
In this paper we propose an alternative and equivalent formulation to attack
problem (1.4), which seems to be more useful in several cases.
In particular, we will write problem (1.4) in an equivalent mean field form,
which is somehow more suitable to a variational analysis. This approach is a
very widely-used tool to attack problem (1.2) on closed surfaces and have also
been used in [6, 5, 12, 14] for problem (1.4) when either $\displaystyle K$ or
$\displaystyle h$ identically vanishes.
Another mean field formulation for problem (1.4) was given in [7], although it
seems to be working only if both curvature $\displaystyle K,h$ and
$\displaystyle\chi(\Sigma)$ are all positive. A double mean field problem
related to (1.4) was also studied by the authors in [3]. However, such results
of existence do not apply for the geometrical problem.
On the other hand, the mean field formulation here presented is totally
equivalent in most of the cases and in all cases produces solutions for
problem (1.4). This will be discussed in further detail in Section 2.

The new mean field energy functional is defined on an open subset of
$\displaystyle H^{1}(\Sigma)$, which in general does not coincide with the
whole space, and it has different forms depending whether
$\displaystyle\chi(\Sigma)$ is positive, zero or negative. The key role played
by the sign of the Euler characteristic is consistent with the existence
results from [15] as well as for many results concerning problem (1.2).
In the case of a disk the new energy functional shares many similarities with
Liouville-type problem where only one curvature is prescribed. In particular,
using Trudinger-Moser type inequalities, one can show that the energy is
bounded from below though not coercive. Section 3 elaborates further on this
aspect.
Quite surprisingly, the picture is rather different for surfaces of non-
positive genus. In fact, the main nonlinear term is now related to the
interaction between the two nonlinear terms, which can have different signs.
In particular, in the case of negative $\displaystyle K(x)$, an important role
will be played by the quotient function
$\displaystyle\mathfrak{D}:\partial\Sigma\to\mathbb{R}$, defined by:

(1.5) $\mathfrak{D}(x):=\frac{h(x)}{\sqrt{|K(x)|}}.$

The importance of $\displaystyle\mathfrak{D}(x)$ has already emphasized in
[15, 2].

We are now in position to state the main theorems of this work.
We start with the case $\displaystyle\chi(\Sigma)>0$. Here, we need to
prescribe a positive curvature $\displaystyle K$ in order to have coercivity
of the energy functional on its domain of definition, which may have a
nonempty boundary where the energy does not diverge; this is a new difficulty
arising from this formulation. However, this does not seem to be a mere
technicality, since when $\displaystyle K$ is not positive the problem may
have no solutions, even in the case of constant prescribed curvatures on the
disk (see [18]).
On the other hand, to get coercivity we also need to assume some symmetry on
$\displaystyle K,h$. In fact, this allows to look for solutions in the space
of functions with the same symmetry of the prescribed curvatures, where one
gets an improved constant in the Trudinger-Moser inequality. The role of
symmetry was first highlighted in [17] for problem (1.2) and then in [7] for
problem (1.4).
To be more precise, we will assume $\displaystyle\Sigma$ to be the upper half-
sphere

(1.6)
$\Sigma:=\left\\{(x_{1},x_{2},x_{3})\in\mathbb{R}^{3}:\,x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1,x_{3}>0\right\\},$

which we just showed to be non-restrictive, and we assume $\displaystyle K,h$
to be symmetric with respect to a subgroup of the orthonormal group of the
plane $\displaystyle SO(2)$ having no fixed points on the unit circle
$\displaystyle\mathbb{S}^{1}=\partial\Sigma$. This is equivalent to assume
$\displaystyle K,h$ to be $\displaystyle k$-symmetric for some positive
integer $\displaystyle k\in\mathbb{N}$ with $\displaystyle k\geqslant 2$
(which one may also assume to be prime, without losing generality), namely:

$\displaystyle K(\rho_{k}x)=K(x)\quad\forall\,x\in\Sigma,\qquad
h(\rho_{k}y)=h(y)\quad\forall\,y\in\partial\Sigma,$

where

(1.7)
$\rho_{k}(x_{1},x_{2},x_{3})=\left(x_{1}\cos\frac{2\pi}{k}-x_{2}\sin\frac{2\pi}{k},x_{1}\sin\frac{2\pi}{k}+x_{2}\cos\frac{2\pi}{k},x_{3}\right).$

######  Theorem 1.1.

Assume $\displaystyle\Sigma$ is given by (1.6) and:

  * •

$\displaystyle K(x)\geqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$ and $\displaystyle K,h$ are $\displaystyle
k$-symmetric.

Then problem (1.4) admits a solution.

Let us point out that with respect to the results given in [7], the previous
theorem does not include any obstruction regarding the sign of the curvature
$\displaystyle h$.

In the case $\displaystyle\chi(\Sigma)=0$ we have two possible scenarios. If
$\displaystyle K$ is somewhere positive, then the nonlinear term is everywhere
negative, hence the energy functional is coercive and the problem admits a
minimizing solution.
On the other hand, if $\displaystyle K$ is always negative, then we require a
condition on the quantity $\displaystyle\mathfrak{D}$ to deal with the
nonlinear term and get minimizers. This is actually a particular case of
Theorem 1.2 in [15]. In both scenarios we shall make some assumptions on the
sign of $\displaystyle h$ in order to avoid loss of coercivity in the boundary
of the domain of the energy functional, as in Theorem 1.1.

######  Theorem 1.2.

Assume $\displaystyle\chi(\Sigma)=0$ and one of the following occurs:

  1. (1)

$\displaystyle K(x)>0$ for some $\displaystyle x\in\Sigma$ and $\displaystyle
h(y)\leqslant 0$ for all $\displaystyle y\in\partial\Sigma$, $\displaystyle
h\not\equiv 0$;

  2. (2)

$\displaystyle K(x)\geqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$ and $\displaystyle\int_{\partial\Sigma}h<0$;

  3. (3)

$\displaystyle K(x)\leqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$ and $\displaystyle h\not\equiv
0,h(y)>0,\mathfrak{D}(y)<1$ for all $\displaystyle y\in\partial\Sigma$.

Then problem (1.4) admits a solution.

Finally, in the case $\displaystyle\chi(\Sigma)<0$ we get minimizing solutions
under an assumption on $\displaystyle\mathfrak{D}$, similarly as Theorem 1.2.
This is the same result as Theorem 1.1 in [15], although obtained with
different arguments.

######  Theorem 1.3.

Assume $\displaystyle\chi(\Sigma)<0$ and:

  * •

$\displaystyle K(x)\leqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$ and $\displaystyle\mathfrak{D}(y)<1$ for all
$\displaystyle y\in\partial\Sigma$.

Then problem (1.4) admits a solution.

We point out that Theorem 1.3 includes the case of a non-positive
$\displaystyle h(y)\leqslant 0$, which can be equivalently written as
$\displaystyle\mathfrak{D}(y)\leqslant 0$ for all $\displaystyle
y\in\partial\Sigma$.

The paper is structured as follows. In Section 2 we introduce the new
equivalent formulation for problem, (1.4), the underlying energy functional
and an appropriate functional setting. In Section 3 we assume
$\displaystyle\chi(\Sigma)>0$ and we prove Theorem 1.1. In Section 4 we assume
$\displaystyle\chi(\Sigma)\leqslant 0$, which requires different arguments
with respect to the previous case, and we provide the proofs for Theorems 1.2
and 1.3.

##  2\. Mean field formulation and setting

We start by introducing a new mean field problem and showing its equivalence
with problem (1.4).
Actually, the mean field problem will be a stronger formulation, in the sense
that its solvability will imply the solvability of problem (1.4), but in most
of the cases we will consider the two problems are in fact equivalent.

######  Proposition 2.1.

Problem (1.4) has a solution if the following problem has a solution:

(2.1) $\left\\{\begin{array}[]{ll}-\Delta
u+\frac{4\pi\chi(\Sigma)}{|\Sigma|}=2C(u)^{2}Ke^{u}&\text{in }\Sigma\\\
\partial_{\nu}u=2C(u)he^{\frac{u}{2}}&\text{on }\partial\Sigma,\\\
C(u)>0\end{array}\right.$

where $\displaystyle C(u)$ is a constant defined by:

(2.2)
$C(u):=\left\\{\begin{array}[]{ll}\frac{4\pi\chi(\Sigma)}{\sqrt{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}+8\pi\chi(\Sigma)\int_{\Sigma}Ke^{u}}+\int_{\partial\Sigma}he^{\frac{u}{2}}}&\text{if
}\chi(\Sigma)>0\\\
-\frac{\int_{\partial\Sigma}he^{\frac{u}{2}}}{\int_{\Sigma}Ke^{u}}&\text{if
}\chi(\Sigma)=0\\\
\frac{4\pi|\chi(\Sigma)|}{\sqrt{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}-8\pi|\chi(\Sigma)|\int_{\Sigma}Ke^{u}}-\int_{\partial\Sigma}he^{\frac{u}{2}}}&\text{if
}\chi(\Sigma)<0.\end{array}\right.$

Moreover, if either $\displaystyle\chi(\Sigma)=0$ or
$\displaystyle\chi(\Sigma)K(x)>0$ for all $\displaystyle x\in\Sigma$, then the
two problems are equivalent.

###### Proof.

Let $\displaystyle u$ be a solution of (2.1). Then for any $\displaystyle
c\in\mathbb{R}$, the function $\displaystyle v=u+c$ satisfies:

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
v+\frac{4\pi\chi(\Sigma)}{|\Sigma|}=2C(u)^{2}Ke^{v-c}&\text{in }\Sigma\\\
\partial_{\nu}u=2C(u)he^{\frac{v-c}{2}}&\text{on
}\partial\Sigma,\end{array}\right.$

which solves (1.4) with the choice $\displaystyle c=2\log C(u)$.

Now, suppose that $\displaystyle u$ is a solution of (1.4). Then, by (1.3) one
gets
$\displaystyle\int_{\partial\Sigma}he^{\frac{u}{2}}=2\pi\chi(\Sigma)-\int_{\Sigma}Ke^{u}$
which, in the definition (2.2) of $\displaystyle C(u)$, gets $\displaystyle
C(u)=1$ in the cases $\displaystyle\chi(\Sigma)=0$ or
$\displaystyle\chi(\Sigma)K>0$. Therefore $\displaystyle u$ solves (2.1). ∎

######  Remark 2.2.

Observe that $\displaystyle C(u)$ is one of the roots of the second order
equation

$\displaystyle
C^{2}(u)\int_{\Sigma}Ke^{u}+C(u)\int_{\partial\Sigma}he^{\frac{u}{2}}=2\pi\chi(\Sigma).$

The choice of one root rather than the other is the reason why problems (1.4)
and (2.1) are not totally equivalent.
One may verify that in (2.2) we chose the largest of the two roots, which is a
convenient choice since, in virtue of the equivalence of both problems, we
want it to be positive.

######  Remark 2.3.

Let us point out that the condition $\displaystyle C(u)>0$ is necessary in
order to guarantee the equivalence between the original problem and the mean
field problem, otherwise a solution of (2.1) may be trivial. For instance, if
$\displaystyle\chi(\Sigma)=0$ and $\displaystyle\int_{\partial\Sigma}h=0$,
then $\displaystyle u=0$ would be a solution of (2.1) whereas the constant
solution does not verify (1.4) unless $\displaystyle K\equiv 0,h\equiv 0$.

Such a mean field formulation admits an energy functional which, in many
cases, can be treated rather easily with respect to the direct formulation.
This is one of the main reasons why this new formulation has been introduced.

######  Proposition 2.4.

Solutions to (2.1) are critical points on the space $\displaystyle
H_{\chi(\Sigma)}$ of the following energy functional:

(2.3) $\mathcal{J}(u):=\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-F_{\chi(\Sigma)}\left(\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}\right);$

here,

$\displaystyle
H_{\chi(\Sigma)}:=\left\\{u\in\overline{H}^{1}(\Sigma):\,\left\\{\begin{array}[]{ll}\int_{\Sigma}Ke^{u}>-\frac{1}{8\pi\chi(\Sigma)}\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)_{+}^{2}&\mbox{if
}\chi(\Sigma)>0\\\
\int_{\Sigma}Ke^{u}\int_{\partial\Sigma}he^{\frac{u}{2}}<0&\mbox{if
}\chi(\Sigma)=0\\\
\int_{\Sigma}Ke^{u}<-\frac{1}{8\pi|\chi(\Sigma)|}\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)_{-}^{2}&\mbox{if
}\chi(\Sigma)<0\end{array}\right.\right\\};$

$\displaystyle t_{+}:=\max\\{t,0\\},t_{-}:=\max\\{-t,0\\}$ denote respectively
the positive and negative part of a real number $\displaystyle t$,

$\displaystyle\overline{H}^{1}(\Sigma):=\left\\{u\in
H^{1}(\Sigma):\,\int_{\Sigma}u=0\right\\},$

and

$\displaystyle
F_{\chi(\Sigma)}(\alpha,\beta):=\left\\{\begin{array}[]{ll}8\pi\chi(\Sigma)\left(\log\left(\sqrt{\beta^{2}+8\pi\chi(\Sigma)\alpha}+\beta\right)+\frac{\beta}{\sqrt{\beta^{2}+8\pi\chi(\Sigma)\alpha}+\beta}\right)&\text{if
}\chi(\Sigma)>0\\\ -2\frac{\beta^{2}}{\alpha}&\mbox{if }\chi(\Sigma)=0\\\
8\pi|\chi(\Sigma)|\left(-\log\left(\sqrt{\beta^{2}-8\pi|\chi(\Sigma)|\alpha}-\beta\right)+\frac{\beta}{\sqrt{\beta^{2}-8\pi|\chi(\Sigma)|\alpha}-\beta}\right)&\mbox{if
}\chi(\Sigma)<0\end{array}\right..$

###### Proof.

One easily verifies that $\displaystyle F_{\chi(\Sigma)}(\alpha,\beta)$ is
well defined and smooth provided $\displaystyle\alpha,\beta$ are such that
$\displaystyle u\in H_{\chi(\Sigma)}$. Therefore, the critical points
$\displaystyle u$ of $\displaystyle\mathcal{J}$ verify

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u+\mu=\partial_{\alpha}F_{\chi(\Sigma)}\left(\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}\right)Ke^{u}&\text{in
}\Sigma\\\
\partial_{\nu}u=\frac{1}{2}\partial_{\beta}F_{\chi(\Sigma)}\left(\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}\right)he^{\frac{u}{2}}&\text{on
}\partial\Sigma,\end{array}\right.$

for some $\displaystyle\mu\in\mathbb{R}$. By the definition of $\displaystyle
F_{\chi(\Sigma)}(\alpha,\beta)$ and $\displaystyle C(u)$ in (2.2) one sees
that, in all three cases of $\displaystyle\chi(\Sigma)$, one has

$\displaystyle\partial_{\alpha}F_{\chi(\Sigma)}\left(\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}\right)=2C(u)^{2},\qquad\partial_{\beta}F_{\chi(\Sigma)}\left(\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}\right)=4C(u).$

Finally, integrating by parts one gets
$\displaystyle\mu=\frac{4\pi\chi(\Sigma)}{|\Sigma|}$, therefore a critical
point $\displaystyle u$ solves (2.1). ∎

We now need to verify that the energy functional is defined in a non-empty
space. This will be true under no assumption on $\displaystyle K,h$ other than
the necessary ones to get solutions.

######  Lemma 2.5.

The space $\displaystyle H_{\chi(\Sigma)}$ is non-empty if and only if:

  * $\displaystyle(1)$

if $\displaystyle\chi(\Sigma)>0$, $\displaystyle K(x)>0$ for some
$\displaystyle x\in\Sigma$ or $\displaystyle h(y)>0$ for some $\displaystyle
x\in\partial\Sigma$;

  * $\displaystyle(2)$

if $\displaystyle\chi(\Sigma)=0$, $\displaystyle K(x)h(y)<0$ for some
$\displaystyle x\in\Sigma,y\in\partial\Sigma$;

  * $\displaystyle(3)$

if $\displaystyle\chi(\Sigma)<0$, $\displaystyle K(x)<0$ for some
$\displaystyle x\in\Sigma$ or $\displaystyle h(y)<0$ for some $\displaystyle
x\in\partial\Sigma$.

Moreover, if such conditions are not satisfied, then problem (1.4) has no
solution.

###### Proof.

It is clear in all cases that the space is empty if the assumptions on
$\displaystyle K,h$ are not satisfied. If this holds true when
$\displaystyle\chi(\Sigma)>0$, then for any solution of (1.4) one has
$\displaystyle\int_{\Sigma}Ke^{u},\int_{\partial\Sigma}he^{\frac{u}{2}}$ both
negative, in contradiction with the Gauss-Bonnet formula (1.3). Similarly, the
assumptions on $\displaystyle K,h$ are necessary to get solutions in the case
$\displaystyle\chi(\Sigma)\leqslant 0$.
Let us now show the converse. In case $\displaystyle(1)$, if $\displaystyle K$
is positive somewhere, then we can choose a positive $\displaystyle\varphi$
supported in the region where $\displaystyle K>0$ and set
$\displaystyle\delta:=\inf_{\\{\varphi\geqslant 1\\}}K>0$ so that

$\displaystyle\mathrm{spt}(\varphi)\subset\\{K>0\\},\qquad\\{\varphi\geqslant
1\\}\subset\\{K\geqslant\delta\\},$

so that for $\displaystyle C\gg 1$ large enough one has

$\displaystyle\int_{\Sigma}Ke^{C\varphi}\geqslant\int_{\\{K\geqslant\delta\\}}Ke^{C\varphi}+\int_{\\{K<0\\}}K\geqslant\delta|\\{K\geqslant\delta\\}|e^{C}-\|K\|_{\infty}|\Sigma|>.$

Taking into account its definition, that implies the non-emptiness of
$\displaystyle H_{\chi(\Sigma)}$. Similarly, if $\displaystyle K<0$ everywhere
and $\displaystyle h>0$ in a portion of $\displaystyle\partial\Sigma$, we
choose a positive $\displaystyle\varphi$ supported in a small neighborhood of
such portion and set $\displaystyle\delta^{\prime}:=\inf_{\\{\varphi\geqslant
1\\}\cap\partial\Sigma}h>0$ in such a way that

$\displaystyle|\mathrm{spt}(\varphi)|\leqslant\varepsilon,\qquad\mathrm{spt}(\varphi)\cap\partial\Sigma\subset\\{h>0\\},\qquad\\{\varphi\geqslant
1\\}\cap\partial\Sigma\subset\\{h\geqslant\delta^{\prime}\\},$

for some $\displaystyle\varepsilon>0$. Therefore, taking $\displaystyle C\gg
1$ large enough and then $\displaystyle\varepsilon\ll 1$ in dependence on
$\displaystyle C$ one has

$\displaystyle\displaystyle\int_{\partial\Sigma}he^{\frac{C\varphi}{2}}-\sqrt{-8\pi\chi(\Sigma)\int_{\Sigma}Ke^{C\varphi}}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\int_{\\{h\geqslant\delta\\}}he^{\frac{C\varphi}{2}}-\int_{\\{h<0\\}}h-\sqrt{8\pi|\chi(\Sigma)|\left(\int_{\mathrm{spt}(\varphi)}|K|e^{\frac{C\varphi}{2}}+\int_{\Sigma\setminus\mathrm{spt}(\varphi)}|K|\right)}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\delta|\\{h\geqslant\delta\\}|e^{\frac{C}{2}}-\|h\|_{\infty}|\partial\Sigma|-\sqrt{8\pi|\chi(\Sigma)|\|K\|_{\infty}\varepsilon}e^{\frac{C\|\varphi\|_{\infty}}{2}}$
$\displaystyle\displaystyle>$ $\displaystyle\displaystyle 0.$

In case $\displaystyle(2)$, if $\displaystyle h(x)<0<K(x)$ for the same
$\displaystyle x\in\partial\Sigma$, then one can take $\displaystyle\varphi$
satisfying

$\displaystyle\displaystyle\mathrm{spt}(\varphi)\subset\\{K>0\\},$
$\displaystyle\displaystyle\\{\varphi\geqslant
1\\}\subset\\{K\geqslant\delta\\},$
$\displaystyle\displaystyle\mathrm{spt}(\varphi)\cap\partial\Sigma\subset\\{h<0\\},$
$\displaystyle\displaystyle\\{\varphi\geqslant
1\\}\cap\partial\Sigma\subset\\{h\leqslant-\delta\\};$

hence one easily gets, for large $\displaystyle C$,

$\displaystyle\displaystyle\int_{\Sigma}Ke^{C\varphi}\geqslant\delta|\\{K\geqslant\delta\\}|e^{C}-\|K\|_{\infty}|\Sigma|>0,$
$\displaystyle\displaystyle\int_{\partial\Sigma}he^{\frac{C\varphi}{2}}\leqslant-\delta|\\{h\leqslant-\delta\\}|e^{\frac{C}{2}}-\|h\|_{\infty}|\partial\Sigma|<0;$

the same argument works if $\displaystyle K(x)<0<h(x)$ at one boundary point.
On the other hand, if $\displaystyle K,h$ have the same sign on the whole
boundary, say positive, we take $\displaystyle\varphi_{1}$ supported where
$\displaystyle K<0$ and $\displaystyle\varphi_{2}$ supported close to the
boundary, as before. Since $\displaystyle\varphi_{1}|_{\partial\Sigma}\equiv
0$, then
$\displaystyle\int_{\partial\Sigma}he^{\frac{C_{1}\varphi_{1}+C_{2}\varphi_{2}}{2}}=\int_{\partial\Sigma}he^{\frac{C_{2}\varphi_{2}}{2}}>0$
for large $\displaystyle C_{2}$; on the other hand, if $\displaystyle C_{1}$
is also large enough in dependence of $\displaystyle C_{2}$, then

$\displaystyle\displaystyle\int_{\Sigma}Ke^{C_{1}\varphi_{1}+C_{2}\varphi_{2}}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\int_{\\{K\leqslant-\delta\\}}Ke^{C_{1}\varphi_{1}}+\int_{\mathrm{spt}(\varphi_{2})}Ke^{C_{2}\varphi_{2}}+\int_{\Sigma\setminus(\mathrm{spt}(\varphi_{1})\cup\mathrm{spt}(\varphi_{2}))}K$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle-\delta|\\{K\leqslant-\delta\\}|e^{C_{1}}+\|K\|_{\infty}\left(e^{C_{2}\|\varphi_{2}\|_{\infty}}+|\Sigma|\right)$
$\displaystyle\displaystyle<$ $\displaystyle\displaystyle 0.$

The case $\displaystyle\chi(\Sigma)<0$ can be treated in the same way as
$\displaystyle\chi(\Sigma)>0$ by just changing sign to both $\displaystyle K$
and $\displaystyle h$. ∎

##  3\. The case $\displaystyle\chi(\Sigma)>0$

In this section we will focus on the case $\displaystyle\chi(\Sigma)>0$ to
prove Theorem 1.1.
In this case, problem (2.1) can be attacked variationally by means of
Trudinger-Moser type inequalities. In fact, in the energy functional (2.3) the
main term will be given by logarithm, since the other is uniformly bounded
from above, as follows by the following elementary lemma.

######  Lemma 3.1.

Let $\displaystyle\chi>0$ be a fixed positive number. Then, for any
$\displaystyle\alpha,\beta\in\mathbb{R}$ such that

$\displaystyle\alpha>-\frac{1}{8\pi\chi}\beta_{+}^{2}$

one has

$\displaystyle\frac{\beta}{\sqrt{\beta^{2}+8\pi\chi\alpha}+\beta}\leqslant 1.$

Trudinger-Moser inequalities are a widely-used tool for the problem of
prescribing one curvature (see [6, 12] and [17] for closed surfaces) and the
formulation introduced in the previous section allows us to extend it to the
prescription of both curvatures. Such results will quickly provide an ad hoc
Trudinger-Moser inequality for problem (2.1).

######  Proposition 3.2.

There exists $\displaystyle C>0$ such that for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$ there holds:

(3.1)
$\log\left(\sqrt{\left(\int_{\partial\Sigma}e^{\frac{u}{2}}\right)^{2}+8\pi\int_{\Sigma}e^{u}}+\int_{\partial\Sigma}e^{\frac{u}{2}}\right)\leqslant\frac{1}{16\pi}\int_{\Sigma}|\nabla
u|^{2}+C.$

###### Proof.

From Corollary 2.5 in [6] there exists $\displaystyle C>0$ such that, for any
$\displaystyle u\in\overline{H}^{1}(\Sigma)$:

(3.2) $\log\int_{\Sigma}e^{u}\leqslant\frac{1}{8\pi}\int_{\Sigma}|\nabla
u|^{2}+C;$

on the other hand, from Corollary 2.6 in [7] we get (see also [12]):

(3.3)
$\log\int_{\partial\Sigma}e^{\frac{u}{2}}\leqslant\frac{1}{16\pi}\int_{\Sigma}|\nabla
u|^{2}+C.$

Therefore, from the elementary inequality

(3.4)
$\sqrt{\beta^{2}+8\pi\alpha}+\beta\leqslant\left(\sqrt{1+8\pi}+1\right)\max\\{\sqrt{\alpha},\beta\\},\qquad\forall\alpha,\beta>0$

we deduce

$\displaystyle\displaystyle\log\left(\sqrt{\left(\int_{\partial\Sigma}e^{\frac{u}{2}}\right)^{2}+8\pi\int_{\Sigma}e^{u}}+\int_{\partial\Sigma}e^{\frac{u}{2}}\right)$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\max\left\\{\frac{1}{2}\log\int_{\Sigma}e^{u},\log\int_{\partial\Sigma}e^{\frac{u}{2}}\right\\}+\log\left(\sqrt{1+8\pi}+1\right)$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\frac{1}{16\pi}\int_{\Sigma}|\nabla
u|^{2}+C+\log\left(\sqrt{1+8\pi}+1\right),$

which concludes the proof. ∎

######  Corollary 3.3.

If $\displaystyle\chi(\Sigma)=1$ then the energy functional
$\displaystyle\mathcal{J}(u)$ is uniformly bounded from below for
$\displaystyle u\in H_{1}$.

###### Proof.

We define

(3.5)
$\alpha:=\int_{\Sigma}Ke^{u},\qquad\beta:=\int_{\partial\Sigma}he^{\frac{u}{2}}.$

Using Lemma 3.1, Proposition 3.2 and the uniform boundedness of $\displaystyle
h,K$ we get:

$\displaystyle\displaystyle\mathcal{J}(u)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-8\pi\log\left(\sqrt{\beta^{2}+8\pi\alpha}+\beta\right)-8\pi\frac{\beta}{\sqrt{\beta^{2}+8\pi\chi\alpha}+\beta}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-8\pi\log\left(\sqrt{\left(\int_{\partial\Sigma}e^{\frac{u}{2}}\right)^{2}+8\pi\int_{\Sigma}e^{u}}+\int_{\partial\Sigma}e^{\frac{u}{2}}\right)$
$\displaystyle\displaystyle-$ $\displaystyle\displaystyle
4\pi\log\max\left\\{\|h\|_{\infty}^{2},\|K\|_{\infty}\right\\}-8\pi$
$\displaystyle\displaystyle\geqslant$ $\displaystyle\displaystyle-C.$

∎

Boundedness from below, however, does not ensure the existence of minimizing
solutions. In fact, using standard test functions, one can easily see that the
energy functional is not coercive, see Proposition 3.2 in [3].
This is the reason why we assumed $\displaystyle K,h$ to be symmetric in
Theorem 1.1: symmetry allows one to take a better constant in the inequality
(3.1) (a so-called improved Trudinger-Moser inequality) which yields
coercivity, hence minimizers. The argument is similar to [17] and [7],
Proposition 2.12.
To this purpose, we introduce a subset of
$\displaystyle\overline{H}^{1}(\Sigma)$ sharing the same $\displaystyle
k$-symmetry as $\displaystyle K,h$:

$\displaystyle\widetilde{H}_{k}:=\left\\{u\in\overline{H}^{1}(\Sigma):\,u(\rho_{k}x)=u(x)\,\text{for
a.e. }x\in\Sigma\right\\},$

where now $\displaystyle\Sigma$ is the half-sphere (1.6) and
$\displaystyle\rho_{k}$ is the rotation of $\displaystyle\frac{2\pi}{k}$ given
by (1.7).

######  Proposition 3.4.

For any $\displaystyle\varepsilon>0$, there exists $\displaystyle
C_{\varepsilon}>0$ such that for any $\displaystyle u\in\widetilde{H}_{k}$
there holds:

(3.6)
$\log\left(\sqrt{\left(\int_{\partial\Sigma}e^{\frac{u}{2}}\right)^{2}+8\pi\int_{\Sigma}e^{u}}+\int_{\partial\Sigma}e^{\frac{u}{2}}\right)\leqslant\frac{1+\varepsilon}{32\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}$

###### Proof.

We start by estimating the boundary integral.
We cover $\displaystyle\partial\Sigma$ with open balls of radius
$\displaystyle\delta\leqslant\frac{\sqrt{2\left(1-\cos\frac{2\pi}{k}\right)}}{2}$,
so that for any $\displaystyle x\in\partial\Sigma$ the ball of radius
$\displaystyle 2\delta$ does not intersect its image under the rotation of
$\displaystyle\frac{2\pi}{k}$, namely

(3.7) $B_{2\delta}(x)\cap B_{2\delta}(\rho_{k}x)=\emptyset.$

Due to compactness, one has
$\displaystyle\partial\Sigma\subset\cup_{i=1}^{N}B_{\delta}(x_{i})$ for some
$\displaystyle x_{1},\dots,x_{N}\in\partial\Sigma$ and, up to relabeling,
$\displaystyle\frac{\int_{B_{\delta}(x_{1})\cap\partial\Sigma}e^{\frac{u}{2}}}{\int_{\partial\Sigma}e^{\frac{u}{2}}}\geqslant\frac{1}{N}$.
Because of symmetry, namely $\displaystyle u\in\widetilde{H}_{k}$, one has

$\displaystyle\frac{\int_{B_{\delta}\left(\rho_{k}^{j}x_{1}\right)\cap\partial\Sigma}e^{\frac{u}{2}}}{\int_{\partial\Sigma}e^{\frac{u}{2}}}=\frac{\int_{B_{\delta}(x_{1})\cap\partial\Sigma}e^{\frac{u}{2}}}{\int_{\partial\Sigma}e^{\frac{u}{2}}}\geqslant\frac{1}{N},\qquad
j=1,\dots,k,$

where
$\displaystyle\rho_{k}^{j}(x_{1})=\underbrace{\rho_{k}\circ\dots\circ\rho_{k}}_{j}(x_{1})$.
In view of (3.7), the balls $\displaystyle
B_{\delta}\left(\rho_{k}^{j}x_{1}\right)$ are at a distance greater or equal
than $\displaystyle 2\delta$ with respect to each other, therefore we are in
position to apply Corollary 2.11 from [7] to get:

(3.8)
$\log\int_{\partial\Sigma}e^{\frac{u}{2}}\leqslant\frac{1+\varepsilon}{16k\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}\leqslant\frac{1+\varepsilon}{32\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}.$

As for the interior integral, one of the following occurs (or both):

(3.9) Either
$\displaystyle\displaystyle\frac{\int_{B_{\frac{1}{2}}(0)}e^{u}}{\int_{\Sigma}e^{u}}\geqslant\frac{1}{2},$
(3.10) or $\displaystyle\displaystyle\frac{\int_{\Sigma\setminus
B_{\frac{1}{2}}(0)}e^{u}}{\int_{\Sigma}e^{u}}\geqslant\frac{1}{2}.$

In case inequality (3.9) occurs, then as in Proposition 2.2 from [16] we get

(3.11) $\log\int_{\Sigma}e^{u}\leqslant\log\int_{B_{\frac{1}{2}}(0)}e^{u}+\log
2\leqslant\frac{1+\varepsilon}{16\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}.$

Putting together (3.8) and (3.11) and using (3.4) as in the proof of
Proposition 3.2 we obtain (3.6).

On the other hand, if (3.10) occurs, then we may argue as for the boundary
integral to get $\displaystyle\delta>0,N\in\mathbb{N}$ and $\displaystyle k$
symmetric points $\displaystyle x_{1},\rho_{k}x_{1},\dots,\rho_{k}^{k-1}x_{1}$
such that

$\displaystyle\displaystyle B_{2\delta}\left(\rho_{k}^{i}x_{1}\right)\cap
B_{2\delta}\left(\rho_{k}^{j}x_{1}\right)=\emptyset$
$\displaystyle\displaystyle i,j=1,\dots,k,\,i\neq j;$
$\displaystyle\displaystyle\frac{\int_{B_{\delta}\left(\rho_{k}^{j}x_{1}\right)\cap\left(\Sigma\setminus
B_{\frac{1}{2}}(0)\right)}e^{u}}{\int_{\Sigma}e^{u}}\geqslant\frac{1}{2}\frac{\int_{B_{\delta}\left(\rho_{k}^{j}x_{1}\right)\cap\left(\Sigma\setminus
B_{\frac{1}{2}}(0)\right)}e^{u}}{\int_{\Sigma\setminus
B_{\frac{1}{2}}(0)}e^{u}}\geqslant\frac{1}{2N}$ $\displaystyle\displaystyle
i=1,\dots,k.$

From Corollary 2.9 in [7] one gets

$\displaystyle\log\int_{\Sigma}e^{u}\leqslant\frac{1+\varepsilon}{8k\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}\leqslant\frac{1+\varepsilon}{16\pi}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon},$

and the conclusion follows as in the previous case. ∎

Proposition 3.6 gives coercivity at infinity of the energy functional.
However, one needs also to take care of the fact that the energy functional
will not be defined on the whole space $\displaystyle\overline{H}^{1}(\Sigma)$
(or on $\displaystyle\widetilde{H}_{k}$) but rather on its subspace
$\displaystyle H_{1}$, introduced in Proposition 2.4. Therefore, in general
coercivity does not suffice to get global minimizers since
$\displaystyle\mathcal{J}(u)$ may not grow to $\displaystyle+\infty$ as
$\displaystyle u$ approaches the boundary of the domain

$\displaystyle\partial
H_{1}=\left\\{u\in\overline{H}^{1}(\Sigma):\,\int_{\Sigma}Ke^{u}=-\frac{1}{8\pi}\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)_{+}^{2}\right\\}.$

It is not clear, in the general case, which assumptions should be made on
$\displaystyle K,h$ in order to prevent minimizing sequences to approach
$\displaystyle\partial H_{1}$. However, the assumption $\displaystyle K>0$
fixes this issue since it implies $\displaystyle\partial H_{1}=\emptyset$.

###### Proof of Theorem 1.1.

Since $\displaystyle K\geqslant 0,K\not\equiv 0$, then
$\displaystyle\int_{\Sigma}Ke^{u}>0$ for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$, therefore $\displaystyle
H_{1}=\overline{H}^{1}(\Sigma)$.
If $\displaystyle u\in\widetilde{H}_{k}$, then by Proposition 3.4 we get, for
any $\displaystyle\varepsilon>0$:

$\displaystyle\displaystyle\mathcal{J}(u)$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-8\pi\log\left(\sqrt{\left(\int_{\partial\Sigma}e^{\frac{u}{2}}\right)^{2}+8\pi\int_{\Sigma}e^{u}}+\int_{\partial\Sigma}e^{\frac{u}{2}}\right)-C$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1-\varepsilon}{4}\int_{\Sigma}|\nabla
u|^{2}-C.$

By choosing $\displaystyle\varepsilon<1$ we get
$\displaystyle\mathcal{J}(u)\to+\infty$ as $\displaystyle\int_{\Sigma}|\nabla
u|^{2}\to+\infty$, that is $\displaystyle\mathcal{J}$ is coercive on
$\displaystyle\widetilde{H}_{k}$. Since $\displaystyle\mathcal{J}$ is also
continuous on $\displaystyle\widetilde{H}_{k}$, direct methods from calculus
of variations give the existence of some minimizer.
Finally, since $\displaystyle\widetilde{H}_{k}$ is a natural constraint for
$\displaystyle\mathcal{J}$, such a minimizer $\displaystyle u$ will solve
(2.1). ∎

Similarly as [7], Theorem 1.1 is stable under perturbation and still holds
true if $\displaystyle K,h$ are not $\displaystyle k$-symmetric but still
uniformly close to a $\displaystyle k$-symmetric function.

######  Theorem 3.5.

Assume $\displaystyle\Sigma$ is given by (1.6) and:

  * •

$\displaystyle K(x)\geqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$ and there exists $\displaystyle
K_{0}(x),h_{0}(x)$ $\displaystyle k$-symmetric and
$\displaystyle\varepsilon>0$ small enough such that
$\displaystyle\|K-K_{0}\|_{\infty}+\|h-h_{0}\|_{\infty}<\varepsilon$.

Then problem (1.4) admits a solution.

The proof will be skipped, since it is similar to [7], Section 4.

We conclude this section by pointing out that problem (2.1) can be generalized
to a general mean field problem in the same spirit as the ones studied in [13,
8] on closed surfaces. Consider, for $\displaystyle\lambda>0$,

(3.12) $\left\\{\begin{array}[]{ll}-\Delta u=\lambda\left(\frac{\lambda
Ke^{u}}{2\left(\sqrt{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}+\lambda\int_{\Sigma}Ke^{u}}+\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}}-\frac{1}{|\Sigma|}\right)&\text{in
}\Sigma\\\
\partial_{\nu}u=\lambda\frac{he^{\frac{u}{2}}}{\sqrt{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}+\lambda\int_{\Sigma}Ke^{u}}+\int_{\partial\Sigma}he^{\frac{u}{2}}}&\text{on
}\partial\Sigma,\\\ \end{array}\right.$

As in Proposition 2.4, solutions of (3.12) are critical point of the energy
functional

(3.13) $\mathcal{J}_{\lambda}(u):=\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-\lambda\log\left(\sqrt{\beta^{2}+\lambda\alpha}+\beta\right)-\lambda\frac{\beta}{\sqrt{\beta^{2}+\lambda\alpha}+\beta},$

defined in the subdomain $\displaystyle H_{1}$ with
$\displaystyle\alpha,\beta$ as in (3.5).
In the case $\displaystyle\chi(\Sigma)=1$ we just considered, problems (3.12)
coincides with (2.1) when $\displaystyle\lambda=8\pi$, that is the limiting
case when the energy functional is bounded from below but not coercive.
However, in the presence of conical singularities, $\displaystyle\chi(\Sigma)$
may be lower or higher than $\displaystyle 1$, corresponding to
$\displaystyle\lambda<8\pi$ or $\displaystyle\lambda>8\pi$, namely the
functional may be coercive or unbounded and one may find global minimizer or
min-max critical points. The authors will address this problem in a subsequent
paper.

##  4\. Trace inequality and the case $\displaystyle\chi(\Sigma)\leqslant 0$

In the cases $\displaystyle\chi(\Sigma)=0$ and $\displaystyle\chi(\Sigma)<0$
the variational approach is rather similar to each other, but totally
different from the previously studied case $\displaystyle\chi(\Sigma)>0$.
In fact, here the logarithmic term is not present at all when
$\displaystyle\chi(\Sigma)=0$ and it is bounded from above when
$\displaystyle\chi(\Sigma)<0$. On the other hand, the extra term may be
unbounded from above and one needs some inequalities to control such a term.
To this purpose, we have the following Proposition, in the same spirit as
Lemma 3.2 in [15].

######  Proposition 4.1.

Assume $\displaystyle K(x)\leqslant 0$ for any $\displaystyle x\in\Sigma$ and
let $\displaystyle\mathfrak{D}$ be defined by $\displaystyle\eqref{d}$. Then,
for any $\displaystyle\varepsilon>0$ there exists $\displaystyle
C_{\varepsilon}$ such that for any $\displaystyle u\in H^{1}(\Sigma)$ one has:

(4.1)
$\frac{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}}{\int_{\Sigma}|K|e^{u}}\leqslant\frac{\left(\mathfrak{D}_{M}+\varepsilon\right)^{2}}{4}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon},\qquad\text{where
}\mathfrak{D}_{M}:=\max_{x\in\partial\Sigma}|\mathfrak{D}(x)|.$

###### Proof.

Fix $\displaystyle\varepsilon>0$ and take a partition of the unit
$\displaystyle\\{\phi_{j}\\}_{j}$ of $\displaystyle\partial\Sigma$ with
$\displaystyle\mathrm{spt}(\phi_{j})$ so small that
$\displaystyle\frac{\max_{\mathrm{spt}(\phi_{j})}|h|}{\min_{\mathrm{spt}(\phi_{j})}\sqrt{|K|}}\leqslant\mathfrak{D}_{M}+\frac{\varepsilon}{2}$,
which follows from the uniform continuity. Then, take a smooth vector field
$\displaystyle\Psi:\Sigma\to\mathbb{R}^{2}$ such that
$\displaystyle\Psi|_{\partial\Sigma}=\nu$ and apply the divergence theorem to
$\displaystyle\Psi\phi_{j}e^{u}$; one has

$\displaystyle\int_{\partial\Sigma}|h|e^{\frac{u}{2}}\leqslant
C\int_{\Sigma}e^{\frac{u}{2}}+\frac{\mathfrak{D}_{M}+\frac{\varepsilon}{2}}{2}\sum_{j}\int_{\Sigma}\sqrt{|K|}\phi_{j}e^{\frac{u}{2}}|\nabla
u|.$

Then, applying the Cauchy-Schwartz inequality and the elementary inequality

$\displaystyle
A+B\leqslant\sqrt{(1+\varepsilon)A^{2}+\left(1+\frac{1}{\varepsilon}\right)B^{2}},$

one gets:

$\displaystyle\displaystyle\left|\int_{\partial\Sigma}he^{\frac{u}{2}}\right|$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\frac{\mathfrak{D}_{M}+\frac{\varepsilon}{2}}{2}\int_{\Sigma}\sqrt{|K|}e^{\frac{u}{2}}|\nabla
u|+C\int_{\Sigma}\sqrt{|K|}e^{\frac{u}{2}}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\sqrt{\int_{\Sigma}|K|e^{u}}\left(\frac{\mathfrak{D}_{M}+\frac{\varepsilon}{2}}{2}\sqrt{\int_{\Sigma}|\nabla
u|^{2}}+C\right)$ $\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\sqrt{\left(\int_{\Sigma}|K|e^{u}\right)\left(\frac{\left(\mathfrak{D}_{M}+\varepsilon\right)^{2}}{4}\int_{\Sigma}|\nabla
u|^{2}+C_{\varepsilon}\right)}$

∎

######  Remark 4.2.

Proposition 4.1 is sharp in the sense that the constant
$\displaystyle\frac{\mathfrak{D}_{M}^{2}}{4}$ in (4.1) cannot be improved in
any case.
To see this, fix a small region $\displaystyle\Gamma\subset\partial\Sigma$ and
its tubular neighborhood $\displaystyle\Omega\subset\Sigma$ in such a way
that, up to changing sign to $\displaystyle h$, which does not change
inequality (4.1),
$\displaystyle\frac{h(y)}{\sqrt{|K(x)|}}\geqslant\mathfrak{D}_{M}-\varepsilon$
for any $\displaystyle x\in\Omega,y\in\Gamma$. We may also assume that, for
some $\displaystyle\delta>0$, the diffeomorphism
$\displaystyle\Phi:\Omega\leftrightarrow[-\delta,\delta]\times[0,\delta]$ is
$\displaystyle\varepsilon$-close to the identity in $\displaystyle
C^{1}$-norm.
Then, we take

$\displaystyle u_{n}:=v_{n}\circ\Phi,\qquad\mbox{with}\qquad
v_{n}(s,t):=\eta(s,t)\xi_{n}(t),\qquad\xi_{n}(t):=-2\log\left(1+nt\right)$

and $\displaystyle\eta\in C^{\infty}_{0}([-\delta,\delta]\times[0,\delta])$
such that

$\displaystyle 0\leqslant\eta\leqslant
1,\qquad\eta_{[-\delta+\varepsilon,\delta-\varepsilon]\times[0,\delta-\varepsilon]}\equiv
1,\qquad\|\nabla\eta\|_{\infty}\leqslant\frac{C}{\varepsilon};$

therefore,

$\displaystyle\displaystyle\int_{\partial\Sigma}he^{\frac{u_{n}}{2}}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle(1-C\varepsilon)\int_{-\delta}^{\delta}\left(h\circ\Phi^{-1}\right)(s,0)e^{\frac{v_{n}(s,0)}{2}}\mathrm{d}s$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle(1-C\varepsilon)\left(\min_{\Gamma}h\right)\int_{-\delta+\varepsilon}^{\delta+\varepsilon}e^{\frac{\xi_{n}(0)}{2}}\mathrm{d}s$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle(1-C\varepsilon)\left(\min_{\Gamma}h\right)2\delta;$
$\displaystyle\displaystyle\int_{\Sigma}|K|e^{u_{n}}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)\int_{-\delta}^{\delta}\mathrm{d}s\int_{0}^{\delta}\left(|K|\circ\Phi^{-1}\right)(s,t)e^{v_{n}(s,t)}\mathrm{d}t$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)\max_{\Omega}|K|\int_{-\delta}^{\delta}\mathrm{d}s\int_{0}^{\delta}e^{\xi_{n}(t)}\mathrm{d}t$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)\max_{\Omega}|K|\frac{2\delta}{n};$
$\displaystyle\displaystyle\int_{\Sigma}|\nabla u_{n}|^{2}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)\int_{-\delta}^{\delta}\mathrm{d}s\int_{0}^{\delta}|\nabla
v_{n}(s,t)|^{2}\mathrm{d}t$ $\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)\int_{-\delta}^{\delta}\mathrm{d}s\int_{0}^{\delta}\left((1+\varepsilon)\eta_{n}(s,t)^{2}\xi_{n}^{\prime}(t)^{2}+\left(1+\frac{1}{\varepsilon}\right)\xi_{n}(t)^{2}|\nabla\eta_{n}(s,t)|^{2}\right)\mathrm{d}t$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon)2\delta\int_{0}^{\delta}\left(\xi_{n}(t)^{2}+\frac{C}{\varepsilon^{3}}\xi_{n}(t)^{2}\right)\mathrm{d}t$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle(1+C\varepsilon){8\delta
n}+C_{\varepsilon}\log^{2}n.$

Therefore, if one takes $\displaystyle\mathfrak{D}_{0}<\mathfrak{D}_{M}$ and
$\displaystyle\varepsilon$ small enough,

$\displaystyle\frac{\left(\int_{\Sigma}he^{\frac{u_{n}}{2}}\right)^{2}}{\int_{\Sigma}|K|e^{u_{n}}}-\frac{\mathfrak{D}_{0}^{2}}{4}\int_{\Sigma}|\nabla
u_{n}|^{2}\geqslant\left((1-C\varepsilon)\left(\mathfrak{D}_{M}-\varepsilon\right)^{2}-\mathfrak{D}_{0}^{2}\right)2\delta
n-C_{\varepsilon}\log^{2}n\underset{n\to+\infty}{\to}+\infty.$

More surprisingly, in some cases one cannot even take
$\displaystyle\varepsilon=0$ in inequality (4.1).
In fact, consider the case $\displaystyle h\equiv 1,K\equiv-1$ and assume
$\displaystyle\partial\Sigma$ has a connected component $\displaystyle\Gamma$
where the geodesic curvature $\displaystyle h_{g}$ has positive average and
take, similarly as before, $\displaystyle
u_{n}:=v_{n}\circ\Phi,v_{n}(s,t):=\zeta(t)\xi_{n}(t)$ with

$\displaystyle\zeta\in C^{\infty}([0,\delta]),\qquad 0\leqslant\zeta\leqslant
1,\qquad\zeta_{\left[0,\frac{\delta}{2}\right]}\equiv 1.$

Since $\displaystyle\det D\Phi(s,t)=1-h_{g}(s)t+O\left(t^{2}\right)$, we have

$\displaystyle\displaystyle\int_{\partial\Sigma}e^{\frac{u_{n}}{2}}$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle|\partial\Sigma|;$
$\displaystyle\displaystyle\int_{\Sigma}e^{u_{n}}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\int_{\partial\Sigma}\mathrm{d}s\int_{0}^{\delta}e^{\xi_{n}(t)}\left(1-h_{g}(s)t+O\left(t^{2}\right)\right)\mathrm{d}t$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\frac{|\partial\Sigma|}{n}-\left(\int_{\partial\Sigma}h\right)\frac{\log
n}{n^{2}}+O\left(\frac{1}{n^{2}}\right);$
$\displaystyle\displaystyle\int_{\Sigma}|\nabla u_{n}|^{2}$
$\displaystyle\displaystyle\leqslant$
$\displaystyle\displaystyle\int_{\partial\Sigma}\mathrm{d}s\int_{0}^{\delta}\left(\zeta^{\prime}(t)\xi_{n}(t)+\zeta(t)\xi_{n}^{\prime}(t)\right)^{2}\left(1-h_{g}(s)t+O\left(t^{2}\right)\right)\mathrm{d}t$
$\displaystyle\displaystyle\int_{\partial\Sigma}\mathrm{d}s\int_{0}^{\delta}\xi_{n}^{\prime}(t)^{2}\left(1-h_{g}(s)t+O\left(t^{2}\right)\right)\mathrm{d}t$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle
4|\partial\Sigma|n-4\left(\int_{\partial\Sigma}h\right)\log n+O(1).$

Therefore,

$\displaystyle\frac{\left(\int_{\Sigma}e^{\frac{u_{n}}{2}}\right)^{2}}{\int_{\Sigma}e^{u_{n}}}-\frac{1}{4}\int_{\Sigma}|\nabla
u_{n}|^{2}=2|\partial\Sigma|\left(\int_{\partial\Sigma}h\right)\log
n+O(1)\underset{n\to+\infty}{\to}+\infty.$

It would be interesting to find out under which condition on the surface
$\displaystyle\Sigma$ inequality (4.1) holds true with
$\displaystyle\varepsilon=0$ and, in the cases when it does not hold true,
what type of correction terms should be added.

###### Proof of Theorem 1.2.

  1. (1)

Since $\displaystyle h\leqslant 0,h\not\equiv 0$, then
$\displaystyle\int_{\partial\Sigma}he^{\frac{u}{2}}<0$ for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$, therefore

$\displaystyle
H_{0}=H_{0}^{\prime}:=\left\\{u\in\overline{H}^{1}(\Sigma):\,\int_{\Sigma}Ke^{u}>0\right\\}.$

By definition, for any $\displaystyle u\in H_{0}^{\prime}$ one has

$\displaystyle\mathcal{J}(u)=\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}+2\frac{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}}{\int_{\Sigma}Ke^{u}}\geqslant\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}\underset{\|u\|\to+\infty}{\to}+\infty.$

We will now show that $\displaystyle\mathcal{J}$ diverges to
$\displaystyle+\infty$ also when $\displaystyle u$ approaches the boundary of
the domain

$\displaystyle\partial
H_{0}^{\prime}=\left\\{u\in\overline{H}^{1}(\Sigma):\,\int_{\Sigma}Ke^{u}=0\right\\};$

this will imply $\displaystyle\mathcal{J}$ is coercive on $\displaystyle
H_{0}^{\prime}$, hence it has a global minimizer solving (2.1). Assume
$\displaystyle H_{0}^{\prime}\ni
u_{n}\underset{n\to+\infty}{\to}u_{0}\in\partial H_{0}^{\prime}$; by weak
continuity, one has

$\displaystyle\int_{\partial\Sigma}he^{\frac{u_{n}}{2}}\underset{n\to+\infty}{\to}\int_{\partial\Sigma}he^{\frac{u_{0}}{2}}<0,\qquad\int_{\Sigma}Ke^{u_{n}}\underset{n\to+\infty}{\to}\int_{\Sigma}Ke^{u_{0}}=0,$

therefore

$\displaystyle\mathcal{J}(u_{n})\geqslant
2\frac{\left(\int_{\partial\Sigma}he^{\frac{u_{n}}{2}}\right)^{2}}{\int_{\Sigma}Ke^{u_{n}}}\underset{n\to+\infty}{\to}+\infty,$

which concludes the proof.

  2. (2)

We have $\displaystyle\int_{\Sigma}Ke^{u}>0$ for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$, therefore $\displaystyle\mathcal{J}(u)$ is
defined for any $\displaystyle u\in\overline{H}^{1}(\Sigma)$. Arguing as in
case $\displaystyle(1)$, $\displaystyle\mathcal{J}$ is coercive on
$\displaystyle\overline{H}^{1}(\Sigma)$, hence it has a minimizer which solves

(4.2) $\left\\{\begin{array}[]{ll}-\Delta
u=2\frac{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}}{\left(\int_{\Sigma}Ke^{u}\right)^{2}}Ke^{u}&\text{in
}\Sigma\\\
\partial_{\nu}u=-2\frac{\int_{\partial\Sigma}he^{\frac{u}{2}}}{\int_{\Sigma}Ke^{u}}he^{\frac{u}{2}}&\text{on
}\partial\Sigma.\\\ \end{array}\right..$

However, solutions to (4.2) may not solve (2.1), because $\displaystyle
C(u)=-\frac{\int_{\partial\Sigma}he^{\frac{u}{2}}}{\int_{\Sigma}Ke^{u}}$ may
be zero or negative, both of which we want to rule out.
If $\displaystyle C(u)=0$, then the only solution to (4.2) is $\displaystyle
u\equiv 0$, which would give

$\displaystyle\int_{\partial\Sigma}h=-C(u)\int_{\Sigma}K=0,$

in contradiction with the assumptions (see also Remark 2.3). Suppose now
$\displaystyle C(u)<0$; since (4.2) is invariant under changing sign to
$\displaystyle h$ and $\displaystyle C(u)$ is odd with respect to
$\displaystyle h$, then $\displaystyle u$ solves

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta u=2C(u)^{2}Ke^{u}&\text{in
}\Sigma\\\ \partial_{\nu}u=2C(u)(-h)e^{\frac{u}{2}}&\text{on
}\partial\Sigma,\\\
C(u)=-\frac{\int_{\partial\Sigma}(-h)e^{\frac{u}{2}}}{\int_{\Sigma}Ke^{u}}>0\end{array}\right..$

Therefore, due to Proposition 2.1, problem (1.4) with $\displaystyle-h$
instead of $\displaystyle h$ has a solution:

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta u=2Ke^{u}&\text{in }\Sigma\\\
\partial_{\nu}u=-2he^{\frac{u}{2}}&\text{on
}\partial\Sigma,\end{array}\right..$

However, multiplying the equation by $\displaystyle e^{-\frac{u}{2}}$ and
integrating by parts gets

$\displaystyle 0>-\int_{\Sigma}|\nabla
u|^{2}e^{-\frac{u}{2}}=\int_{\Sigma}Ke^{\frac{u}{2}}-\int_{\partial\Sigma}h>-\int_{\partial\Sigma}h,$

again contradicting the assumptions. This shows that the minimizer on
$\displaystyle\overline{H}^{1}(\Sigma)$ actually solves (2.1).

  3. (3)

We have $\displaystyle\int_{\Sigma}Ke^{u}<0$ and
$\displaystyle\int_{\partial\Sigma}he^{\frac{u}{2}}>0$ for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$, therefore $\displaystyle
H_{0}=\overline{H}^{1}(\Sigma)$, hence in order to get minimizing solutions,
we suffice to show coercivity at infinity. To this purpose, we take a small
$\displaystyle\varepsilon$ such that
$\displaystyle\mathfrak{D}(y)+\varepsilon<1$ for any $\displaystyle
y\in\partial\Sigma$ and apply Proposition 4.1:

$\displaystyle\mathcal{J}(u)=\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-2\frac{\left(\int_{\partial\Sigma}he^{\frac{u}{2}}\right)^{2}}{\int_{\Sigma}|K|e^{u}}\geqslant\underbrace{\frac{1-\left(\mathfrak{D}_{M}+\varepsilon\right)^{2}}{2}}_{>0}\int_{\Sigma}|\nabla
u|^{2}-C_{\varepsilon}\underset{\|u\|\to+\infty}{\to}+\infty.$

Then we find a minimizing solution by applying the direct method.

∎

Theorem 1.2 can be extended by dropping the assumptions on the sign of
$\displaystyle h$, but if $\displaystyle h$ is allowed to change sign we do
not know if problem (1.4) can be solved for $\displaystyle h$ or
$\displaystyle-h$.

######  Theorem 4.3.

Assume $\displaystyle\chi(\Sigma)=0$ and:

  * •

$\displaystyle K(x)\leqslant 0$ for all $\displaystyle x\in\Sigma$,
$\displaystyle K\not\equiv 0$, $\displaystyle\int_{\partial\Sigma}h\neq 0$ and
$\displaystyle|\mathfrak{D}(y)|<1$ for all $\displaystyle y\in\partial\Sigma$.

Then, problem (1.4) admits a solution for $\displaystyle h$ or
$\displaystyle-h$.

As in Theorem 1.2, this is a sub-case of Theorem 1.2 in [15], which gives
existence of solutions under assuming $\displaystyle\int_{\partial\Sigma}h>0$.

###### Proof.

As in case $\displaystyle(3)$ in Theorem 1.2, $\displaystyle\mathcal{J}$ is
well-defined and coercive on $\displaystyle\overline{H}^{1}(\Sigma)$ and, as
in case $\displaystyle(2)$, it has a minimizers solving (4.2). $\displaystyle
C(u)=0$ is excluded since it would give $\displaystyle\int_{\Sigma}h=0$, so
one may have either $\displaystyle C(u)>0$ or $\displaystyle C(u)<0$: in the
former case, (1.4) has a solution for $\displaystyle h$; in the latter case,
(1.4) has a solution for or $\displaystyle-h$. ∎

In the case $\displaystyle\chi(\Sigma)<0$ the arguments are similar as the
case $\displaystyle\chi(\Sigma)=0$.
In fact, even though the nonlinear terms $\displaystyle
F_{\chi(\Sigma)}(\alpha,\beta)$ is different than $\displaystyle
F_{0}(\alpha,\beta)$, it has essentially the same asymptotic behavior for
positive $\displaystyle\beta$, whereas for negative $\displaystyle\beta$ it is
bounded from below. To be more precise, we have the following elementary
lemma.

######  Lemma 4.4.

For any $\displaystyle\varepsilon>0$ there exists $\displaystyle
C_{\varepsilon}>0$ such that the function

(4.3)
$f(t):=F_{\chi(\Sigma)}(-1,t)=8\pi|\chi(\Sigma)|\left(-\log\left(\sqrt{t^{2}+8\pi|\chi(\Sigma)|}-t\right)+\frac{t}{\sqrt{t^{2}+8\pi|\chi(\Sigma)|}-t}\right)$

verifies

$\displaystyle f(t)\leqslant(2+\varepsilon)t_{+}^{2}+C_{\varepsilon}.$

In view of lemma 4.4, the case $\displaystyle\chi(\Sigma)<0$ can still be
treated using Proposition 4.1.

###### Proof of Theorem 1.3.

From assuming $\displaystyle K\leqslant 0,K\not\equiv 0$ we get
$\displaystyle\int_{\Sigma}Ke^{u}<0$ for any $\displaystyle
u\in\overline{H}^{1}(\Sigma)$, hence $\displaystyle
H_{\chi(\Sigma)}=\overline{H}^{1}(\Sigma)$.
We may write the energy functional as

$\displaystyle\mathcal{J}(u)=\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}+4\pi|\chi(\Sigma)|\log(-\alpha)+4\pi|\chi(\Sigma)|\log(8\pi)-f\left(\frac{\beta}{\sqrt{-\alpha}}\right),$

where $\displaystyle\alpha,\beta$ are as in (3.5) and $\displaystyle f$ is as
in (4.3).
The first nonlinear term is uniformly bounded from below on
$\displaystyle\overline{H}^{1}(\Sigma)$ because of the Jensen’s inequality:

$\displaystyle\int_{\Sigma}(-K)e^{u}=\int_{\Sigma}e^{u+\log|K|}\geqslant|\Sigma|e^{\frac{1}{|\Sigma|}\int_{\Sigma}(u+\log|K|)}=|\Sigma|e^{\frac{1}{|\Sigma|}\int_{\Sigma}\log|K|}.$

By assumption, we have

$\displaystyle\mathfrak{D}_{M}^{+}:=\max_{x\in\partial\Sigma}\frac{h(x)}{\sqrt{|K(x)|}}=\max_{x\in\partial\Sigma}\frac{h_{+}(x)}{\sqrt{|K(x)|}}<1;$

therefore, using Lemma 4.4 and Proposition 4.1 with $\displaystyle h_{+}$
instead of $\displaystyle h$ we get, for $\displaystyle\varepsilon$ small
enough:

$\displaystyle\displaystyle\mathcal{J}(u)$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}+f\left(\frac{\beta}{\sqrt{-\alpha}}\right)-C$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-(2+\varepsilon)\frac{\beta_{+}^{2}}{-\alpha}-C_{\varepsilon}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\frac{1}{2}\int_{\Sigma}|\nabla
u|^{2}-(2+\varepsilon)\frac{\left(\int_{\partial\Sigma}h_{+}e^{\frac{u}{2}}\right)^{2}}{\int_{\Sigma}|K|e^{u}}-C_{\varepsilon}$
$\displaystyle\displaystyle\geqslant$
$\displaystyle\displaystyle\underbrace{\frac{2-(2+\varepsilon)\left(\mathfrak{D}_{M}^{+}+\varepsilon\right)^{2}}{4}}_{>0}\int_{\Sigma}|\nabla
u|^{2}-C_{\varepsilon}$
$\displaystyle\displaystyle\underset{\|u\|\to+\infty}{\to}$
$\displaystyle\displaystyle+\infty,$

hence $\displaystyle\mathcal{J}$ has minimizers which solve (2.1). ∎

Acknowledgements: The first author is partially supported by the group INdAM-
GNAMPA project “Proprietà qualitative delle soluzioni di equazioni ellittiche”
CUP_E53C22001930001.
The second author is currently supported by the grant Juan de la Cierva
Incorporación fellowship (JC2020-046123-I), funded by
MCIN/AEI/10.13039/501100011033, and by the European Union Next Generation
EU/PRTR. He is also partially supported by Grant PID2021-122122NB-I00 funded
by MCIN/AEI/ 10.13039/501100011033 and by “ERDF A way of making Europe”.
The authors wish to thank David Ruiz for many fruitful discussions and
suggestions.

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