Centrally Essential Factor Rings and
Subdirect Indecomposability

Oleg Lyubimtsev111Nizhny Novgorod State University, Nizhny Novgorod, Russia;
email<EMAIL_ADDRESS>., Askar Tuganbaev222National Research
University MPEI, Moscow, Russia; Lomonosov Moscow State University, Moscow,
Russia<EMAIL_ADDRESS>.

Abstract. Let $R$ be a ring and let $J(R)$, $C(R)$ be its Jacobson radical and
center, correspondingly. If $R$ is a centrally essential ring and the factor
ring $R/J(R)$ is commutative, then any minimal right ideal is contained in the
center $C(R)$. A right Artinian (or right Noetherian subdirectly
indecomposable) centrally essential ring is a right and left Artinian local
ring. We describe centrally essential Noetherian subdirectly indecomposable
rings and centrally essential rings with subdirectly indecomposable center. We
give examples of non-commutative subdirectly indecomposable, centrally
essential rings.

Key words. centrally essential ring, subdirectly indecomposable ring.

The work of Oleg Lyubimtsev is supported by Ministry of Education and Science
of the Russian Federation, project FSWR-2023-0034. The study of Askar
Tuganbaev is supported by grant of Russian Science Foundation (=RSF), project
22-11-00052, https://rscf.ru/en/project/22-11-00052.

MSC2020 database 16D25, 16R99

##  1 Introduction

We consider only associative unital non-zero rings. A ring $R$ is said to be
centrally essential if either $R$ is commutative or for every non-central
element $a$ of $R$, there exist non-zero central elements $x,y$ with $ax=y$.
It is clear that a ring $R$ with center $C(R)$ is centrally essential if and
only if the module $R_{C(R)}$ is an essential extension of the module
$C(R)_{C(R)}$. In addition, any commutative ring is centrally essential.

Remark 1.1. There exists a non-commutative ring $R$ such that all factor rings
of $R$ are centrally essential (see Example 3.3).

Remark 1.2. In [7, Proposition 2.8], it is proved that a right uniserial,
right Artinian, centrally essential ring is left Artinian.

Remark 1.3. There exist right and left subdirectly indecomposable rings $R$
which are not centrally essential. For example, let $V$ be a vector space over
a field $\mathbb{F}$ of characteristic $0$ or $p\neq 2$ with basis
$\\{e_{1},e_{2}\\}$, and let $\Lambda(V)$ be the exterior algebra of the space
$V$, i.e., $\Lambda(V)$ is an algebra with operation $\wedge$, generators
$e_{1},e_{2}$ and defining relations $e_{i}\wedge e_{j}+e_{j}\wedge
e_{i}=0\,\,\mbox{for all}\,\,i,j=1,2$. Then $R=\Lambda(V)$ is a non-
commutative $4$-dimensional $\mathbb{F}$-algebra with basis
$\\{1,e_{1},e_{2},e_{1}\wedge e_{2}\\}$ and the center $C(R)$ of $R$ is
$\\{q_{0}\cdot 1+q_{1}\cdot e_{1}\wedge e_{2}\,\mid\,q_{i}\in\mathbb{F}\\}$.
It is easy to see that $R$ is a right and left subdirectly indecomposable ring
with subdirectly indecomposable center. In addition, the ring $R$ is not
centrally essential; see [5, Theorem 1].

In connection to Remarks 1.1, 1.2 and 1.3, we prove Theorems 1.4, 1.5 and 1.6
which are the main results of this paper.

Theorem 1.4. Let $R$ be a centrally essential ring.

1. If the factor ring $R/J(R)$ is commutative, then any minimal right ideal $S$ of the ring $R$ is contained in the center $C(R)$. In particular, all minimal right ideals of the ring $R$ are ideals and its right socle is contained in the center of the ring $R$.

2. If $R$ is a right Artinian ring (or right Noetherian subdirectly indecomposable) ring, then $R$ is a right and left Artinian local ring.

3. If the center of the ring $R$ is subdirectly indecomposable, then the ring $R$ is right and left subdirectly indecomposable.

Theorem 1.5. If $R$ is a right Noetherian ring such that all factor rings of
$R$ are centrally essential, then $R$ is a subdirect product of local, right
and left Artinian rings.

Theorem 1.6. Let $R$ be a right Noetherian ring with center $C(R)=C$. We set
$\overline{C}=J(R)\backslash J(C)$.

1. If the ring $R$ is subdirectly indecomposable, then the ring $R$ is centrally essential if and only if $\text{Ann}_{\overline{C}}J(C)=\text{Ann}_{\overline{C}}J(R)=0$.

2. If the center $C$ of the ring $R$ is subdirectly indecomposable with core $H$, then the ring $R$ is centrally essential if and only if $rC\cap H\neq 0$ for any non-zero element $r\in R$.

We give some necessary definitions and notation.

For a ring $R$, the Jacobson radical of $R$ is denoted by $J(R)$. A module is
said to be subdirectly indecomposable if it has the least non-zero submodule.
A ring $R$ is said to be right subdirectly indecomposable (or $RSI$-ring [2])
if the intersection $H_{r}$ of all its non-zero right ideals is non-zero. The
left core $H_{\ell}$ of the ring $R$ is defined similarly. A ring $R$ is said
to be subdirectly indecomposable with core $H$ if the intersection $H$ of all
its non-zero ideals is non-zero. A ring $R$ is said to be centrally
subdirectly indecomposable if the center of $R$ is subdirectly indecomposable.

For a module $M$, the socle $Soc\,M$ is the sum of all simple submodules in
$M$; if $M$ does not contain simple submodules, then $Soc\,M=0$, by
definition. A module $M$ is said to be uniserial if any two submodules of $M$
are comparable with respect to inclusion.

For a ring $R$, the Lie ring $G(R)$ of $R$ consists of elements of the ring
$R$ with multiplication $[x,y]=xy-yx$ for any $x,y\in R$. An element $r\in R$
is said to be right regular (resp., left regular) if $rx\neq 0$ (resp.,
$xr\neq 0$) for every non-zero element $x\in R$. Right and left regular
elements are called regular. A non-zero element $r\in R$ is $C(R)$-torsion
free if $rc\neq 0$ for every non-zero element $c\in C(R)$. If $S$ and $T$ are
two subsets of the ring $R$, then $\text{Ann}_{S}T=\\{s\in S\,|\,sT=0\\}$. If
$\mathbb{F}$ is a field and $V$ is a vector $\mathbb{F}$-space with basis
$e_{1},\ldots e_{n}$, then the exterior algebra $\Lambda(V)$ is an algebra
with operation $\wedge$, generators $e_{1},\ldots e_{n}$, defining relations
$e_{i}\wedge e_{j}+e_{j}\wedge e_{i}=0\,\,\mbox{for all}\,\,i,j=1,\ldots n$,
and the basis

$\\{1,e_{i_{1}}\wedge e_{i_{2}}\wedge\ldots\wedge e_{i_{k}}\,\,\mid
k=1,2,\ldots n;\,\,1\leq i_{1}<i_{2}<\ldots i_{k}\leq n\\}.$

##  2 Proof of Main Results

Proposition 2.1. For a centrally essential ring $R$, the following conditions
are equivalent.

1) The ring $R$ is right subdirectly indecomposable.

2) The ring $R$ is left subdirectly indecomposable.

3) The ring $R$ is subdirectly indecomposable.

In addition, the cores $H_{r}$, $H_{\ell}$ and $H$ coincide in each of cases
1)–3).

Proof. Let the ring $R$ be right subdirectly indecomposable and let
$\\{J_{t}\\}_{t\in T}$ be the set of all non-zero left ideals of the ring $R$.
Since the ring $R$ is centrally essential, each of the left ideals $J_{t}$
contains a non-zero central element $c_{t}$. Then $0\neq\cap_{t\in
T}c_{t}R\subseteq\cap_{t\in T}J_{t}$. Therefore, 1) imples 2).

It is clear that 2) implies 3). Since each of the ideals $c_{t}R$ is two-
sided, it follows from 3) that 1) is true.

It is clear that $H_{r}=hR$ for any $0\neq h\in H_{r}$. Since $R$ is a
centrally essential ring, we have that $0\neq hc=d$ for some central elements
$c$ and $d$. Then $H_{r}=dR$ is a two-sided ideal. Therefore, $H_{r}=H$. The
relation $H_{\ell}=H$ can be proved similarly. $\square$

We remark that the analogue of Proposition 2.1 is not true for rings which are
not centrally essential. For example, the ring of all $n\times n$ ($n\geq 2$)
matrices over a division ring is subdirectly indecomposable but it is not
right or left subdirectly indecomposable. In addition, there exists a right
subdirectly indecomposable ring which is not left subdirectly indecomposable;
see [2, Example 4.5].

It follows from Proposition 2.1 that it is sufficient to restrict ourself to
centrally essential, subdirectly indecomposable rings in the study of right
(or left) subdirectly indecomposable, centrally essential rings.

Proposition 2.2. If $R$ is a centrally essential ring and the factor ring
$R/J(R)$ is commutative, then any minimal right ideal $S$ of the ring $R$ is
contained in $C(R)$. In particular, all minimal right ideals of the ring $R$
are ideals and $Soc\,R_{R}$ is contained in the center of the ring $R$.

Proof. The simple $R$-module $S_{R}$ is a cyclic module with non-zero
generator $s\in S$. We set $K=S\cap C(R)$. By assumption, there exist non-zero
central elements $x,y\in C(R)$ such that $0\neq sx=y\in S$. Then $S=yR$ and
$y\in K\neq 0$. Since $J(R)$ is the intersection of the annihilators of all
simple right $R$-modules, we have that $SJ(R)=0$. It follows from
commutativity of the ring $R/J(R)$ that $ab-ba\in J(R)$ for any elements
$a,b\in R$. In addition, $k(ab-ba)=0$ for any $k\in K$. Since $k\in C(R)$, we
have that $(ka)b=kba=b(ka)$. Therefore, $ka\in C(R)$. Further, it follows from
$ka\in S$ that $K$ is a right ideal of the ring $R$. Since $S$ is a minimal
right ideal, we have $S=K\subseteq C(R)$. $\square$

Proposition 2.3. Let $R$ be a centrally essential ring.

1. An element $r\in R$ is regular if and only if $r$ is $C(R)$-torsion free.

2, [7, Lemma 2.2 (2.2.1]. In the ring $R$, every one-sided zero-divisor is a
two-sided zero-divisor.

Proof. 1. It is clear that if $r$ is regular, then it is $C(R)$-torsion free.
Conversly, let $rs=0$ for some non-zero element $s\in R$. Then $s\notin C(R)$
and there exist $c,d\in C(R)$ such that $0\neq sc=d$. Since $r$ is
$C(R)$-torsion free, we have that $rd\neq 0$ and $rsc\neq 0$. This is a
contradiction.

2. If $rs=0$ for some $0\neq s\in R$, then 1 implies that $rd=dr=0$ for some $0\neq d\in C(R)$. $\square$

Proposition 2.4. Let $R$ be a right Noetherian, centrally essential,
subdirectly indecomposable ring with core $H$.

1. $R$ is a right Artinian local ring.

2. The factor ring $R/J(R)$ is commutative.

3. The core $H$ is contained in the center $C(R)$.

Proof. 1. It follows from Proposition 2.3(2) that any right regular element is
left regular. Then the assertion follows from [2, Theorem 2.1].

2. The assertion follows from 1 and [6, Proposition 2.4].

3. By 2, the ring $R/J(R)$ is commutative. Now we use Proposition 2.2. $\square$

Proposition 2.5. Let $R$ be a centrally essential ring.

1. If the ring $R$ is right Artinian, then $R$ is left Artinian.

2. If $R$ is a right Noetherian, subdirectly indecomposable ring, then $R$ is a right and left Artinian ring.

3. If the ring $R$ is local and $rJ(C)=0$ for $r\in R$, then $r\in J(C)$.

4. If the center of the ring $R$ is subdirectly indecomposable, then $R$ is right and left subdirectly indecomposable.

Proof. 1. It follows from [6, Proposition 2.4] that the ring $R/J(R)$ is
commutative. Then $[x,y]\in J(R)$ for any $x,y\in G(R)$. In addition, the
radical $J(R)$ is nilpotent with some nilpotence index $n$. Then

$[x_{1},y_{1}]\cdot\ldots\cdot[x_{n},y_{n}]=0$

for all elements $[x_{i},y_{i}]\in G(R)$, $i=1,\ldots n$. In this case, the
Lie ring $G(R)$ is nilpotent. Then the result follows from [1, Theorem 2.2].

2. It follows from Proposition 2.4(1) that the ring $R$ is right Artinian. By 1, the ring $R$ is left Artinian.

3. We set $C=C(R)$. We assume that $rJ(C)=0$. Since $R$ is a centrally essential ring, we have $0\neq rc=d\in J(C)$. It follows from the relation $J(C)=C\cap J(R)$ that the element $c$ is invertible. Then $r=c^{-1}d\in J(C)$.

4. Let $M=\\{I_{k}\\}_{k\in K}$ be the set of all non-zero right ideals of the ring $R$. Since $R$ is a centrally essential ring, each of the ideals $I_{k}$ contains a non-zero central element $c_{k}$. We consider ideals $c_{k}C(R)$ of the center $C(R)$. Since the center is subdirectly indecomposable, we have $0\neq\cap_{k\in K}c_{k}C(R)\subseteq\cap_{k\in K}I_{k}$. $\square$

2.6. The proof of Theorem 1.4.

1. The assertion follows from Proposition 2.2.

2. The assertion follows from Proposition 2.4(1) and Proposition 2.5(1,2).

3. The assertion follows from Proposition 2.5(4). $\square$

2.7. The proof of Theorem 1.5.

Any ring is a subdirect product of subdirectly indecomposable rings.
Therefore, the right Noetherian ring $R$ is a subdirect product of right
Noetherian, subdirectly indecomposable, centrally essential rings $R_{i}$,
$i\in I$. It follows from Propositions 2.4(1) and 2.5(2) that all these rings
are right and left Artinian local rings. $\square$

2.8. The proof of Theorem 1.6. We recall that $R$ is a right Noetherian ring
with center $C$ and $\overline{C}=J(R)\backslash J(C)$.

1. If $R$ is a centrally essential ring, then it follows from Propositions 2.4(1) and 2.5(3) that $\text{Ann}_{\overline{C}}J(C)=0$. Then $\text{Ann}_{\overline{C}}J(R)=0$.

Conversely, let $0\neq r\notin C$ and $rJ(C)\neq 0$. By Proposition 2.4(1),
the ring $R$ is local; therefore, $r\in\overline{C}$ or $r$ is invertible.
Then $R$ is right Artinian and we can choose central elements
$c_{1},\ldots,c_{k-1}$ such that $rc_{1}\ldots c_{k-1}\neq 0$, but
$rc_{1}\ldots c_{k-1}c=0$ for all $c\in J(C)$. We set $c^{\prime}=c_{1}\ldots
c_{k-1}$. If $0\neq rc^{\prime}\notin J(C)$, then
$rc^{\prime}\in\text{Ann}_{\overline{C}}J(C)=0$. This is a contradiction.
Consequently, $rc^{\prime}$ is a non-zero central element and $R$ is a
centrally essential ring.

2. Let $R$ be a centrally essential ring and $r\notin H$. There exist central elements $c$ and $d$ such that $0\neq rc=d$. If $d\in H$, then it is nothing to prove. Let $d\notin H$. Since $R$ is a centrally subdirectly indecomposable ring, we have $dC\cap H\neq 0$, i.e., $0\neq dc^{\prime}=h$ for some $c^{\prime}\in C$ and $h\in H$. Then

$0\neq dc^{\prime}=(rc)c^{\prime}=r(cc^{\prime})=rc^{\prime\prime}\in rC,$

whence we obtain that $0\neq rc^{\prime\prime}\in rC\cap H$.

Conversely, for any $0\neq r\in R$, there exists an element $c\in C$ such that
$0\neq rc=h\in H\subseteq C$, i.e., $R$ is a centrally essential ring.
$\square$

##  3 Examples and Remarks

A ring is said to be invariant if all its one-sided ideals are two-sided
ideals.

Example 3.1. We give an example of an Artinian, centrally essential, right and
left subdirectly indecomposable, invariant ring $R$ such that $R$ is not a
centrally subdirectly indecomposable ring and all proper centrally essential
factor rings of $R$ are commutative. However,, there exists a proper factor
ring of the ring $R$ which is not subdirectly indecomposable or centrally
essential. A ring $R$ is a non-commutative $8$-dimensional algebra over an
arbitrary field $\mathbb{F}$ of characteristic $0$ or $p\neq 2$. If we take
$\mathbb{F}=\mathbb{Z}/p\mathbb{Z}$ for prime integer $p>2$, then the ring $R$
is finite.

Proof. Let $R=\Lambda(V)$ be the the exterior algebra with generators
$e_{1},e_{2},e_{3}$. Then $R$ is an $8$-dimensional $\mathbb{F}$-algebra with
basis

$\\{1,e_{1},e_{2},e_{3},e_{1}\wedge e_{2},e_{2}\wedge e_{3},e_{1}\wedge
e_{3},e_{1}\wedge e_{2}\wedge e_{3}\\}$

and $R$ is a non-commutative centrally essential ring (see details in [5,
Example 1]) with center

$C(R)=\\{q_{0}\cdot 1+q_{1}\cdot e_{1}\wedge e_{2}+q_{2}\cdot e_{2}\wedge
e_{3}+q_{3}\cdot e_{1}\wedge e_{3}+q_{4}\cdot e_{1}\wedge e_{2}\wedge
e_{3},\,q_{i}\in\mathbb{F}\\}.$

However, $I_{1}\cap I_{2}=\\{0\\}$ for central ideals $I_{1}=\langle
e_{1}\wedge e_{2}\rangle_{\mathbb{F}}$ and $I_{2}=\langle e_{1}\wedge
e_{2}\wedge e_{3}\rangle_{\mathbb{F}}$, i.e., the center of the ring $R$ is
not a subdirectly indecomposable ring. We remark that $R$ is an invariant ring
and the ideal $H=I_{2}$ is contained in the intersection of any ideals of $R$.
Consequently, $R$ is an Artinian, centrally essential, right and left
subdirectly indecomposable ring which is not centrally subdirectly
indecomposable.

We prove that all proper centrally essential factor rings of $R$ are
commutative. We consider the factor ring $\overline{R}=R/I$, where $I$ is a
non-zero ideal in $R$. If $e_{i}\notin I$ and $\overline{e_{i}}\notin
C(\overline{R})$ for some $i$, then
$\overline{e_{i}}J(C(\overline{R}))=\overline{0}$, since $I$ contains the core
$H$ and $e_{i}\wedge(e_{i}\wedge e_{j})=0$. Then the ring $\overline{R}$ is
not centrally essential, by Proposition 2.5(3). In addition, if, for example,
$I=\langle\overline{e_{1}}\wedge\overline{e_{3}},\overline{e_{1}}\wedge\overline{e_{2}}\wedge\overline{e_{3}}\rangle_{\mathbb{F}}$,
then the ring $\overline{R}$ is not subdirectly indecomposable. Indeed,
principal ideals generated by elements
$\overline{e_{1}}\wedge\overline{e_{2}}$ and
$\overline{e_{2}}\wedge\overline{e_{3}}$ have the zero intersection. $\square$

Example 3.2. Let $Q_{8}$ be the quaternion group, i.e., the group with two
generators $a$, $b$ and defining relations $a^{4}=1$, $a^{2}=b^{2}$ and
$aba^{-1}=b^{-1}$; e.g., see [3, Section 4.4]. We have

$Q_{8}=\\{e,a,a^{2},b,ab,a^{3},a^{2}b,a^{3}b\\},$

with center $C(Q_{8})=\\{e,a^{2}\\}$. We consider the group algebra
$R=\mathbb{Z}_{2}Q_{8}$ with center

$C(R)=\\{\lambda_{0}+\lambda_{1}a^{2}+\lambda_{2}(a+a^{3})+\lambda_{3}(b+a^{3}b)+\lambda_{4}(ab+a^{3}b),\quad\lambda_{i}\in\mathbb{Z}_{2}\\};$

e.g., see [9, Part 2]. Then $R$ is a ring with subdirectly indecomposable
center and core $H=\\{0,\widehat{Q}_{8}\\}$, where $\widehat{Q}_{8}=\sum_{g\in
Q_{8}}g$. In addition, $R$ is a centrally essential ring; see, [4]. It follows
from Theorem 1.6(2) that for every $0\neq x\in R$, there exists an element
$c\in C(R)$ such that $xc=\widehat{Q}_{8}$. Then the factor ring
$\overline{R}=R/H$ with basis
$\\{\overline{e},\overline{a},\overline{a}^{2},\overline{a}^{3},\overline{b},\overline{ab},\overline{a^{2}b}\\}$
and with center

$C(\overline{R})=\\{\alpha_{0}\overline{e}+\alpha_{1}\overline{a}^{2}+\alpha_{2}(\overline{a}+\overline{a}^{3})+\alpha_{3}(\overline{b}+\overline{a^{2}b})+\alpha_{4}(\overline{e}+\overline{a}+\overline{a}^{2}+\overline{a}^{3}+\overline{b}+\overline{a^{2}b}\\}$

is not a centrally essential ring. This property can be directly verified. For
example, let $x\in R$ and $xa\neq ax$ but
$\overline{x}\,\overline{a}=\overline{a}\,\overline{x}$ in $\overline{R}$.
Then $xa=ax+\widehat{Q}_{8}$, whence $\sum x_{i}a=\sum
ax_{i}+\widehat{Q}_{8}$. Since only basis elements $b$, $ab$, $a^{2}b$ and
$a^{3}b$ do not commute with the element $a$, we obtain the contradictory
relation

$\alpha_{1}b+\alpha_{2}ab+\alpha_{3}a^{2}b+\alpha_{4}a^{3}b=\widehat{Q}_{8},\quad\alpha_{i}\in\mathbb{Z}_{2}.$

The case $xb\neq bx$ is considered similarly, but
$\overline{x}\,\overline{b}=\overline{b}\,\overline{x}$. When we multiply the
element
$\overline{r}=(\overline{e}+\overline{a}+\overline{b}+\overline{ab})\notin
C(\overline{R})$ by central basis elements, we obtain either $\overline{r}$
(under multiplication by $\overline{a}^{2}$) or $\overline{0}$. Consequently,
$\overline{R}$ is not a centrally essential ring. Similarly, we can verify
that all centrally essential proper factor rings of the ring $R$ are
commutative.

Example 3.3. We give an example of a non-commutative ring $R$ with subdirectly
indecomposable center such that all factor rings of $R$ are centrally
essential. Let $F=\mathbb{Q}(x,y)$ be the field of rational functions. We
consider two partisl derivations $d_{1}=\dfrac{\partial}{\partial x}$ and
$d_{2}=\dfrac{\partial}{\partial x}$. Then the ring $R=T(F,F)$ of matrices

$\left\\{\left.\begin{pmatrix}f&d_{1}(f)&g\\\ 0&f&d_{2}(f)\\\
0&0&f\end{pmatrix}\;\;\right|\;\;f,g\in F\right\\}$

is a centrally essential ring; see [6]. In addition, $R$ is a ring with
subdirectly indecomposable center and with core

$H=\left\\{\left.\begin{pmatrix}0&0&g\\\ 0&0&0\\\
0&0&0\end{pmatrix}\;\;\right|\;\;g\in F\right\\}.$

Since $R/H\cong\mathbb{Q}(x,y)$ and $H$ is the the least ideal in $R$, the
ring $R$ does not have non-zero proper ideals not equal to $H$.

Remark 3.4. In [5, Proposition 3.3], it is proved that if $R$ is a centrally
essential local ring, then the ring $R/J(R)$ is commutative (in particular,
$R/J(R)$ is centrally essential). This is not true in general case; see [8,
Theorem 1.5].

Remark 3.5. If $R$ is not centrally essential, then the asserton of
Proposition 2.4(1) is not true: there exist right Noetherian, right
subdirectly indecomposable rings which are not right Artinian; see [2, Example
2.2].

Remark 3.6. If $R$ is a centrally essential, right and left subdirectly
indecomposable ring with core $H$, then either $H^{2}=0$ or $R$ is a field.
Indeed, if $H^{2}=H$, then $R$ is a division ring; see [2, Proposition 1.3].
Since $R$ is a centrally essential ring, it follows from [4, Proposition 3.3]
that $R$ is a field.

Open question 3.7. Is it true that there exists a centrally essential ring
which has a proper non-commutative centrally essential factor ring?

Open question 3.8. Is it true that a centrally essential subdirectly
indecomposable (or centrally subdirectly indecomposable) ring is invariant?

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# A Maximum Principle approach to deterministic Mean Field Games of Control
with Absorption ††thanks: This work was supported through the profile
partnership program between Humboldt-Universität zu Berlin and Princeton
University.

Paulwin Graewe111Deloitte Consulting GmbH, Kurfürstendamm 23, 10719 Berlin,
Germany; email<EMAIL_ADDRESS>Ulrich Horst222Department of Mathematics,
and School of Business and Economics, Humboldt-Universität zu Berlin, Unter
den Linden 6, 10099 Berlin, Germany; email<EMAIL_ADDRESS>Ronnie
Sircar333Department of Operations Research and Financial Engineering,
Princeton University, Sherrerd Hall, Princeton, NJ 08544; email:
<EMAIL_ADDRESS>
###### Abstract

We study a class of deterministic mean field games on finite and infinite time
horizons arising in models of optimal exploitation of exhaustible resources.
The main characteristic of our game is an absorption constraint on the
players’ state process. As a result of the state constraint the optimal time
of absorption becomes part of the equilibrium. This requires a novel approach
when applying Pontyagin’s maximum principle. We prove the existence and
uniqueness of equilibria and solve the infinite horizon models in closed form.
As players may drop out of the game over time, equilibrium production rates
need not be monotone nor smooth.

AMS Subject Classification: 93E20, 91B70, 60H30.

Keywords: stochastic control, mean field game, optimal exploitation, maximum
principle

##  1 Introduction

This paper establishes existence and uniqueness of equilibrium results for a
class deterministic mean field games (MFGs) arising in models of optimal
exploitation of exhaustible resources. MFGs provide a convenient tool for
analyzing complex strategic interactions among many players when each
individual player has only a small impact on the behavior of other players. In
a standard mean field game each player solves a control problem in which an
individual player’s payoff functional and the dynamics of the controlled state
process depend on the empirical distribution of the other players’ actions or
states. The existence of Nash equilibria in MFGs can be established by solving
either a coupled system of two partial differential equations (PDEs), a
backward Hamilton-Jacobi-Bellman equation determining the players’ utility,
and a forward Kolmogorov equation determining the evolution of the
distribution of states or actions, or by solving a system of McKean-Vlasov
forward-backward SDEs where the forward component describes the state dynamics
and the backward component describes the dynamics of the adjoint variable. We
refer to the monograph [1] for further background.

In the economics literature MFGs are often called anonymous games. First
introduced by Rosenthal [25] and Jovanovic and Rosenthal [22], anonymous games
have received renewed attention among economists in the last two decades; see
[2, 9, 19] and references therein. Huang, Malhamé and Caines [20] and Lasry
and Lions [23] independently introduced MFGs into the engineering and
mathematical literature. Ever since, MFGs have become an important driver of
mathematical innovation, especially in the areas of PDEs and backward
stochastic equations. Complementing the theoretical work on MFGs, there is a
by now substantial literature where anonymous and mean field games have been
successfully applied to an array economic and engineering problems, ranging
from network security and traffic networks [14, 26] and systemic risk
management [5], to portfolio liquidation [10, 11, 13, 21] and oil and energy
production in competitive markets [6, 7, 15, 16, 17].

Oligopoly models of markets with a small number of competitive players that
compete on the amount of output they produce go back to the classical work of
Cournot [8] in 1838. These have typically been static (or one-period) models,
where the existence and construction of a Nash equilibrium have been
extensively studied. Dynamic Cournot models for energy production in a
competitive market have been proposed and analyzed by many authors in recent
years; we refer to [24] for a survey on energy production models. In these
models market participants are often endowed with limited amounts of
exhaustible resources such as oil, coal or natural gas that they choose to
extract for sale. The models may lead to continuous time single player control
problems [3], finite player nonzero-sum differential games [18], or continuum
mean field games [6, 7]. In the context of nonzero-sum dynamic games between
finitely many players the computation of a solution is a challenging problem,
typically involving coupled systems of nonlinear PDEs, with one value function
per player, and existence theory is sparse. MFGs allow one to handle certain
types of competition in the continuum limit of an infinity of small players by
solving either systems of PDEs or forward-backward SDEs.

One of the key characteristics of games with exhaustible resources is that
resources may run out and change the structure of the game. In a stochastic
setting, the strict absorption constraint is often ignored, assuming that
resource levels fluctuate randomly and recover as soon as they reach positive
levels again. The situation is very similar in models of optimal portfolio
liquidation. When asset portfolios are liquidated on behalf of third parties,
short positions are not allowed legally. Nonetheless, these constraints are
usually ignored for reasons of mathematical tractability. The non-negativity
constraint may be binding even in deterministic games as shown in [11].

Incorporating non-negativity constraints on the state processes leads to MFGs
with absorbing boundaries. The literature on MFGs with absorbing state
dynamics is sparse. Models of mean field type without strategic interactions
have been applied to model interactive default times by various authors.
Graber and Bensoussan [15] consider a modification of the model of Chan and
Sircar [6] on a bounded domain. While the restriction to bounded domains
simplifies the mathematical analysis it seems undesirable from an economic
perspective. We impose only non-negativity constraints but restrict ourselves
to deterministic settings. MFGs with absorbing state constraints in stochastic
settings have recently been considered by Campi and Fischer in [4]. Their
model heavily relies on diffusive state dynamics, though.

We consider a deterministic MFG model of optimal exploitation of exhaustible
resources in which players with zero resources drop out of the game before the
terminal time in equilibrium. Both finite and the infinite horizon models are
studied. As we will see there can be very different equilibrium strategies
depending on whether a terminal time is imposed or not. The fact players may
exit the game before the terminal time requires a novel approach to the
maximum principle associated with an individual player’s optimization problem.
Due to the non-negativity constraint on the state process, the Hamiltonian
depends on the initial condition, while the terminal value of the adjoint
process is unknown. The latter is very similar to portfolio liquidation models
where the terminal condition of the adjoint process is unknown, due to the
liquidation constraint; see [11, 12] for details. In order to overcome this
problem we consider the Hamiltonian only up a candidate optimal exploitation
time. Obtaining such candidate requires a candidate terminal condition for the
adjoint process. In our deterministic setting the candidate terminal condition
and hence the candidate optimal exploitation time can be given in closed form;
the stochastic case is much more involved and left for future work.

The key observation in solving the MFG is that a certain best response
function is independent of the mean field equilibrium. More precisely, we
prove that the dynamics of the amount of resource extracted by an individual
player up to any given time by using a strategy that would be optimal if the
optimal exhaustion time was equal to that given time is independent of the
equilibrium mean production rate. Moreover, we show that the dynamics follows
an ODE that depends only the initial distribution of the resource levels and
the risk free interest rate. The ODE can be solved in closed form. The
solution is strictly monotone and the optimal exhaustion time is given by the
inverse function. In the infinite horizon case where the equilibrium mean
production rate converges to zero as time increases to infinity, this allows
us to fully describe the unique MFG equilibrium in terms of the said ODE. In
the finite horizon case the equilibrium mean production rate at the terminal
time is unknown. This results in an additional fixed point problem that can
easily be solved. Thus we provide among the very few explicit solutions to
MFGs outside the linear-quadratic framework.

To illustrate our main ideas we first consider in Section 2 the benchmark case
of a single monopolist oil producer. We fully characterize optimal
exploitation strategies. In particular, we prove that full exploitation may
not be optimal in finite horizon problems. The MFG of optimal extraction is
analyzed in Section 3. We first determine the equilibrium time of exploitation
as a function of the competitors’ strategies and own initial resources.
Subsequently, we determine the equilibrium exploitation times and strategies.
We illustrate by two simple examples that equilibrium exploitation rates do
need not be monotone, nor smooth.

##  2 The Monopoly Case

As a motivation for our general analysis we illustrate in this section our
main ideas in the framework of monopolist oil producer. We fix a time horizon
$T\in(0,\infty]$ and denote the monopolist’s resource at time $0$ by $x_{0}$.
The monopolist extracts the resource according to a measurable rate function
(control) $q:[0,T]\to[0,\infty)$. A control is called admissible if the
corresponding state process

$X_{t}^{q}:=x_{0}-\int_{0}^{t}q_{u}\,du$

is non-negative on $[0,T]$. Following [6, 7], we assume that the price
function $p:[0,T]\to[0,1]$ when the production rate $q$ is employed is given
by $p=1-q$. For a given constant discount rate $r>0$, the monopolist’s value
function is then defined by

$u(x_{0}):=\sup_{q\geq 0}J(q),$ (2.1)

where

$J(q):=\int_{0}^{\tau_{q}\wedge T}e^{-rt}q_{t}(1-q_{t})\,dt$

is the monopolist’s discounted revenue, and

$\tau_{q}=\tau_{q}(x_{0}):=\inf\left\\{t\in[0,T]:X^{q}_{t}=0\right\\}$

denotes the time of depletion of the resource under the control $q$. We put

$\tau_{q}=+\infty\quad\mbox{if}\quad\tau_{q}>T.$

Due to the dependence of $\tau_{q}$ on the control, the Hamiltonian depends on
the initial state. This calls for a non-standard approach when applying the
maximum principle. We can rewrite the value function as

$u(x_{0})=\sup_{q\geq
0}\int_{0}^{T}e^{-rt}q_{t}(1-q_{t})\mathbbm{1}_{\\{X^{q}_{t}>0\\}}\,dt,$ (2.2)

and so the Hamiltonian is given by

$H(x,q,y)=q(1-q)\mathbbm{1}_{\\{x>0\\}}-qy.$

The maximum principle formally states that if $q^{*}$ is an optimal control
and $X^{*}$ the induced state process, then there exists an adjoint process
$Y$, to be interpreted as $\dot{u}(X^{*}_{t})$, that follows the dynamics

$\begin{split}-\dot{Y}_{t}&=-rY_{t}+\text{``}\partial_{x}H(X_{t}^{*},q_{t}^{*},Y_{t})\text{''},\quad
0\leq t<T\\\ Y_{T}&=\text{``}\dot{u}(X_{T}^{*})\text{''},\end{split}$

and that satisfies the maximum condition

$q^{*}_{t}=\operatorname*{arg\,max}_{\bar{q}\in\mathbb{R}_{+}}H(X_{t}^{*},\bar{q},Y_{t})=\frac{1}{2}\left(1-Y_{t}\right)^{+}.$
(2.3)

However, $\partial_{x}H(x_{0},q,y)=0$ for $x_{0}\neq 0$, while
$\partial_{x}H(0,q,y)$ does not exist. Moreover, the terminal condition of $Y$
is not well defined; it will turn out that the derivative $\dot{u}(X_{T}^{*})$
does not exist. The way to overcome this problem is to apply the maximum
principle only up to the time $\tau^{*}:=\tau_{q^{*}}$ where $q^{*}$ is the
candidate optimal strategy. In order to obtain the candidate optimal strategy
we will determine the value $Y_{\tau^{*}}$ of the adjoint process at the time
of depletion and then define the adjoint process on the entire time interval
as

$Y_{t}=Y_{\tau^{*}}e^{-r(\tau^{*}-t)}\quad\mbox{for}\quad t\in[0,T].$ (2.4)

We need to distinguish three different cases, depending on whether full
exploitation occurs strictly before time $T$, exactly at the terminal time or
whether full exploitation is not optimal. In what follows, we derive a
characterization of the candidates for the optimal exploitation time and
terminal condition of the adjoint process in terms of the initial resource
level.

  * •

$\tau^{*}<T\leq\infty$. In this case the maximum condition (2.3) suggests that
$Y_{\tau^{*}}=1$ to ensure that $q^{*}\equiv 0$ in $[\tau^{*},T]$ and hence
that $Y_{t}=e^{-r(\tau^{*}-t)}\mathbbm{1}_{\\{t\leq\tau^{*}\\}}$ and

$q^{*}_{t}(x_{0})=\frac{1}{2}\left(1-e^{-r(\tau^{*}-t)}\mathbbm{1}_{\\{t\leq\tau^{*}\\}}\right).$

Then $\tau^{*}$ is determined by the identity

$x_{0}=\int_{0}^{\tau^{*}}q_{t}^{*}\,dt=\frac{1}{2}\int_{0}^{\tau^{*}}(1-e^{-r(\tau^{*}-t)})\,dt=\frac{\tau^{*}}{2}-\frac{1}{2r}(1-e^{-r\tau^{*}}).$

This gives

$\tau^{*}(x_{0})=2x_{0}+\frac{1+\mathbb{W}(-e^{-1-2rx_{0}})}{r},$ (2.5)

where $\mathbb{W}$ denotes the principal branch of the Lambert-W function,
defined as the inverse function of $xe^{x}$, restricted to the range
$[-1,\infty)$ and the domain $[-e^{-1},\infty)$. This recovers the formula
found by dynamic programming methods for the $T=\infty$ case in [6,
Proposition 3].

  * •

$\tau^{*}=T<\infty$. In this case we expect the terminal value $Y_{T}(x_{0})$
to be obtained by the identity

$x_{0}=\int_{0}^{T}q_{t}^{*}\,dt=\frac{1}{2}\int_{0}^{T}(1-Y_{T}(x_{0})e^{-r(T-t)})\,dt=\frac{T}{2}-\frac{Y_{T}(x_{0})}{2r}(1-e^{-rT}),$

which yields

$Y_{T}(x_{0})=\frac{r(T-2x_{0})}{1-e^{-rT}},\quad\mbox{ and }\quad
q_{t}^{*}(x_{0})=\frac{1}{2}\left(1-\frac{r(T-2x_{0})}{1-e^{-rT}}e^{-r(T-t)}\right).$
(2.6)

Note that $q^{*}_{T}>0$ in general, and that $Y\geq 0$ only if $x_{0}\leq
T/2$. This suggests that full exploitation is not optimal if $x_{0}>T/2$.

  * •

$\tau^{*}=+\infty$. In this case, we expect the problem to be locally
independent of $x_{0}$, and hence that $Y\equiv 0$. This yields
$q^{*}\equiv\frac{1}{2}$ and therefore necessarily $x_{0}\geq T/2$ and
$T<\infty$.

Using the above obtained terminal values $Y_{T}(x_{0})$ we now define the
adjoint process $Y_{t}$ by (2.4) for every $t\in[0,T]$. By construction
$Y_{t}\geq 0$ for all $t\in[0,T]$ and $Y_{t}\geq 1$ for $t\geq\tau^{*}$.
Moreover, $Y_{t}\equiv 0$ if $\tau^{*}>T$ or $x_{0}=T/2$. In particular, for
any admissible control $q$, the Hamiltonian satisfies

$H(X^{q}_{t},q_{t},Y_{t})\leq 0\quad\mbox{on}\quad[\tau^{*},T].$ (2.7)

We can now prove that the above constructed candidate optimal strategy is
indeed optimal.

######  Theorem 2.1 (Verification Theorem).

The optimal control at time $t\in[0,T]$ to the control problem (2.2) with
initial state $x_{0}>0$ is given by

$q^{*}_{t}(x_{0})=\frac{1}{2}\begin{cases}(1-e^{-r(\tau^{*}(x_{0})-t)})\mathbbm{1}_{\\{t\leq\tau^{*}(x_{0})\\}},&0\leq
x_{0}\leq\frac{T}{2}-\frac{1}{2r}(1-e^{-rT});\\\
1-Y_{T}(x_{0})e^{-r(T-t)},&\frac{T}{2}-\frac{1}{2r}(1-e^{-rT})\leq x_{0}\leq
T/2;\\\ 1,&x_{0}\geq T/2;\end{cases}$

where $\tau^{*}(x_{0})$ and $Y_{T}(x_{0})$ are given by (2.5) and (2.6),
respectively.

###### Proof.

For notational convenience we drop the dependence of $\tau^{*},Y_{T}$ and
$q^{*}$ on $x_{0}$. A direct computation verifies that $q^{*}$ is admissible.
Let $q$ be another admissible control and $\tau_{q}$ be the corresponding
exhaustion time. We put

$\bar{\tau}_{q}:=\tau_{q}\wedge
T\quad\mbox{and}\quad\bar{\tau}^{*}:=\tau^{*}\wedge T.$

By definition of the adjoint process $e^{-rt}Y_{t}\equiv Y_{0}$ on $[0,T]$
where $Y_{0}\geq 0$ and $Y_{0}=0$ if $\tau^{*}>T$. Moreover, for any
admissible control $q$

$q^{*}_{t}=\operatorname*{arg\,max}_{\bar{q}\in\mathbb{R}_{+}}H(X^{q}_{t},\bar{q},Y_{t})\quad\mbox{on}\quad[0,\tau^{*}\wedge\tau^{q}].$
(2.8)

We now distinguish two cases.

  * •

The case $\tau^{*}\geq\tau_{q}$. In this case it follows from (2.8) that

$\displaystyle J(q^{*})-J(q)$
$\displaystyle=\int_{0}^{\bar{\tau}^{*}}e^{-rt}q^{*}_{t}(1-q^{*}_{t})\,dt-\int_{0}^{\bar{\tau}_{q}}e^{-rt}q_{t}(1-q_{t})\,dt$
$\displaystyle=\int_{0}^{\bar{\tau}_{q}}e^{-rt}(H(X^{*}_{t},q^{*}_{t},Y_{t})-H(X^{q}_{t},q_{t},Y_{t})+(q^{*}_{t}-q_{t})Y_{t})\,dt$
$\displaystyle\quad+\int_{\bar{\tau}_{q}}^{\bar{\tau}^{*}}e^{-rt}(H(X^{*}_{t},q^{*}_{t},Y_{t})+q^{*}_{t}Y_{t})\,dt$
$\displaystyle\geq\int_{0}^{\bar{\tau}_{q}}(q^{*}_{t}-q_{t})Y_{t}e^{-rt}\,dt+\int_{\bar{\tau}_{q}}^{\bar{\tau}^{*}}q^{*}_{t}Y_{t}e^{-rt}\,dt$
$\displaystyle=Y_{0}\left(\int_{0}^{\bar{\tau}^{*}}q^{*}_{t}\,dt-\int_{0}^{\bar{\tau}_{q}}q_{t}\,dt\right)\color[rgb]{0,0,0}$
$\displaystyle=0$

where the last equality follows from the fact that $Y_{0}=0$ if $\tau^{*}>T$
and that the term in parenthesis vanishes if $\bar{\tau}^{*}\leq T$.

  * •

The case $\tau^{*}<\tau_{q}$. In this case, it follows again from (2.8) that

$\displaystyle J(q^{*})-J(q)$
$\displaystyle=\int_{0}^{\bar{\tau}^{*}}e^{-rt}(H(X^{*}_{t},q^{*}_{t},Y_{t})-H(X^{q}_{t},q_{t},Y_{t})+(q^{*}_{t}-q_{t})Y_{t})\,dt$
$\displaystyle\quad-\int_{\bar{\tau}_{*}}^{\bar{\tau}^{q}}e^{-rt}(H(X_{t},q_{t},Y_{t})+q_{t}Y_{t})\,dt$
$\displaystyle\geq\int_{0}^{\bar{\tau}^{*}}(q^{*}_{t}-q_{t})Y_{t}e^{-rt}\,dt-\int_{\bar{\tau}^{*}}^{\bar{\tau}^{q}}e^{-rt}(H(X^{q}_{t},q_{t},Y_{t})+q_{t}Y_{t})\,dt.$

In view of Equation (2.7), the Hamiltonian is non-positive on $[\tau^{*},T]$
and hence

$J(q^{*})-J(q)\geq
Y_{0}\left(\int_{0}^{\bar{\tau}^{*}}q^{*}_{t}\,dt-\int_{0}^{\bar{\tau}^{q}}q_{t}\,dt\right).$

If $\tau^{*}>T$, then $Y_{0}=0$. Else, $Y_{0}\geq 0$,
$\int_{0}^{\bar{\tau}^{*}}q^{*}_{t}dt=x_{0}$ and
$\int_{0}^{\bar{\tau}^{q}}q_{t}\,dt=x_{0}-X^{q}_{\bar{\tau}^{q}}$. As a
result,

$J(q^{*})-J(q)\geq 0.$

∎

Having determined the optimal control allows us to compute the value function.

######  Corollary 2.2.

The value function to the control problem (2.1) is given by the smooth
function

$u(x_{0})=\frac{1}{4r}\begin{cases}\left(1+\mathbb{W}(-e^{-1-2rx_{0}})\right)^{2},&0\leq
x_{0}\leq\frac{T}{2}-\frac{1-e^{-rT}}{2r};\\\
1-e^{-rT}-\frac{r^{2}(T-2x_{0})^{2}}{e^{rT}-1},&\frac{T}{2}-\frac{1-e^{-rT}}{2r}\leq
x_{0}\leq\frac{T}{2};\\\ 1-e^{-rT},&x_{0}\geq\frac{T}{2}.\end{cases}$

This corresponds with the solution to the dynamic programming equation in the
$T=\infty$ case found in [18, Section 5].

##  3 The Mean Field Game

To motivate the form of demand functions that we are going to use in the
continuum MFG, we first introduce a finite market with $N$ oil producers that
compete for market share in a one-period game. Associated to each firm
$i\in\\{1,\dots,N\\}$ are variables $p_{i}\in\mathbb{R}$ and
$q_{i}\in\mathbb{R}_{+}$ representing the price and quantity, respectively. In
the Cournot model, players choose quantities as a strategic variable in non-
cooperative competition with the other firms, and the market determines the
price of each good. The market model is specified by linear inverse demand
functions, which give prices as a function of quantity produced. The firms are
suppliers, and so quantities are nonnegative. For $q\in\mathbb{R}^{N}_{+}$,
the price received by player $i$ is $p_{i}=P_{i}(q)$ where

$P_{i}(q)=1-\left(q_{i}+\epsilon\bar{q}_{i}\right),\quad\mbox{where}\quad\bar{q}_{i}=\frac{1}{N-1}\sum_{j\neq
i}q_{j},\,\,i=1,\cdots,N,\quad\mbox{and}\quad 0\leq\epsilon<N-1.$ (3.1)

The inverse demand functions are decreasing in all of the quantities, and
$\epsilon$ measures the strength of interaction between players. In the linear
model (3.1), some of the prices $p_{i}=P_{i}(q)$ may be negative, meaning
player $i$ produces so much that he has to pay to have his goods taken away,
but negative prices do not arise in competitive equilibrium. Moreover, the
goods are similar but differentiated, meaning each player potentially receives
a different price as there will be some residual loyalty to obtaining the
product from individual suppliers. Most crucially, the interaction is of mean
field type: $p_{i}$ is affected by the mean production of the other players,
and players $j$ and $k$ ($j,k\neq i$) are exchangeable as far as player $i$ is
concerned.

In the MFG version, there is a continuum of players, say oil producers, who
are labeled by their reserves at time. Initial reserves are distributed
according to the probability measure $\mu$ on $[0,\infty)$. Each producer
extracts oil in continuous time at a rate $q_{t}\geq 0$, and the price
received by this producer is $P(q_{t},Q_{t})=1-q_{t}-\varepsilon Q_{t}$, where
$Q_{t}$ is the mean production rate of all the players, and $\varepsilon\geq
0$ quantifies the degree of interaction between the players. In the following,
the time horizon $T\leq\infty$.

###  3.1 The best response function

In a first step, we consider the representative player’s best response given
the aggregate production of all the other players. Aggregate production is
described by a nonnegative and absolutely continuous function
$Q:[0,T]\color[rgb]{0,0,0}\rightarrow[0,\frac{1}{2+\varepsilon}]$. The
derivative of $Q$ exists a.e. and is denoted by $\dot{Q}$.

######  Remark 3.1.

It is in fact sufficient here to simply assume $Q\leq 1/\varepsilon$ which is
immediately evident to guarantee nonnegative prices. The a priori bounds
$0\leq Q\leq 1/(2+\varepsilon)$ are motivated by the following observation:
for any candidate optimal control we have

$0\leq q\leq\operatorname*{arg\,max}_{q\geq 0}\\{q(1-\varepsilon
Q-q)\\}=\frac{1}{2}(1-\varepsilon Q)^{+},$

where the right-hand side is the optimal control given infinite reserves. Any
aggregate production function $Q$ leading to a solution to the MFG therefore
necessarily satisfies

$0\leq Q\leq\frac{1}{2}(1-\varepsilon Q)\quad\mbox{and so}\quad
Q\leq\frac{1}{2+\varepsilon}.$

We assume throughout that the aggregate production function satisfies the
following compatibility condition. We will see that this condition guarantees
that each player fully exploits her initial resources if $T$ is large enough.
The assumption will be satisfied in equilibrium.

######  Assumption 1.

There exists $\delta>0$ such that the aggregate exploitation rate
$Q:[0,T]\color[rgb]{0,0,0}\rightarrow[0,\frac{1}{2+\varepsilon}]$ satisfies
the compatibility condition

$1-\varepsilon Q+\frac{\varepsilon}{r}\dot{Q}\geq\delta>0.$ (3.2)

######  Remark 3.2.

Our compatibility condition is equivalent to
$\frac{d}{dt}(e^{-rt}Q_{t})>\frac{1}{\varepsilon}\frac{d}{dt}(e^{-rt})$. Since
we expect $Q$ to be decreasing in equilibrium (production slows as resources
run out), the condition (3.2) puts a lower bound on how quickly that may occur
over time.

Let us now consider a representative producer with any initial state
$x_{0}\in[0,\infty)$ at time $0$. We call a control $q:[0,T]\to[0,1]$
admissible if the state process

$X_{t}^{q}:=x_{0}-\int_{0}^{t}q_{u}\,du$

is always non-negative. For $T<\infty$ or $T=\infty$, depending on whether the
finite or infinite horizon case is considered, the value function for the
representative producer with respect to a given function $Q$ is defined by

$u^{Q}(x_{0})=\sup_{q\geq 0}J^{Q}(q),\quad\mbox{where }\quad
J^{Q}(q):=\int_{0}^{\tau_{q}^{Q}\wedge T}e^{-rt}q_{t}(1-\varepsilon
Q_{t}-q_{t})\,dt,$ (3.3)

and the exhaustion time $\tau^{Q}_{q}=\tau^{Q}_{q}(x_{0})$ is

$\tau_{q}^{Q}:=\inf\left\\{t>0\mid X_{t}^{q}=0\right\\}.$

By analogy to the single player case, the corresponding Hamiltonian is given
by

$H(x,q,y)=q(1-\varepsilon Q-q)\mathbbm{1}_{\\{x>0\\}}-qy.$

####  3.1.1 Heuristics by the maximum principle

As in the single player case Pontryagin’s maximum principle formally states
that if $q^{Q}$ is an optimal control, with associated optimal state process
$X^{q^{Q}}$ and vanishing time $\tau^{Q}$, then there exists an adjoint
process $Y$ given by

$Y_{t}=Y_{\tau^{Q}}e^{-r(\tau^{Q}-t)},\quad t\geq 0$ (3.4)

and $q^{Q}$ satisfies the maximum condition

$q^{Q}_{t}=\operatorname*{arg\,max}_{\bar{q}\in\mathbb{R}_{+}}H(X^{q^{Q}}_{t},\bar{q},Y_{t})=\frac{1}{2}\left(1-\varepsilon
Q_{t}-Y_{t}\right)^{+}.$ (3.5)

The challenge is again that the terminal condition of the adjoint process is
unknown. As in the single player case we distinguish again three different
cases depending on whether full exploitation is optimal or not.

  * •

The case $\tau^{Q}<T\leq\infty$. In this case we expect $q^{Q}_{\tau^{Q}}=0$,
and so, from (3.5), the terminal condition of the adjoint process is
$Y_{\tau^{Q}}=1-\varepsilon Q_{\tau^{Q}}$. Hence, from (3.4), we have

$Y_{t}=(1-\varepsilon Q_{\tau^{Q}})e^{-r(\tau^{Q}-t)},\quad t<\tau^{Q},$

and the optimal strategy is

$q^{Q}_{t}(x_{0})=\frac{1}{2}\left(1-\varepsilon Q_{t}-(1-\varepsilon
Q_{\tau^{Q}(x_{0})})e^{-r(\tau^{Q}(x_{0})-t)}\right),\quad t<\tau^{Q}.$ (3.6)

  * •

The case $\tau^{Q}=T<\infty$. In this case the terminal value $Y_{T}(x_{0})$
of the adjoint process is obtained by the identity

$x_{0}=\int_{0}^{T}q_{t}^{Q}(x_{0})\,dt=\frac{1}{2}\int_{0}^{T}\left(1-\varepsilon
Q_{t}-Y_{T}(x_{0})e^{-r(T-t)}\right)\,dt,$

which yields

$Y_{T}(x_{0})=\frac{1}{\beta_{T}}\left(\frac{1}{2}\int_{0}^{T}(1-\varepsilon
Q_{t})\,dt-
x_{0}\right),\quad\mbox{where}\quad\beta_{T}:=\frac{1-e^{-rT}}{2r}.$ (3.7)

In particular, the terminal condition depends on the initial resource level
and $Y\geq 0$ if and only if

$x_{0}\leq\eta^{Q}(T),\quad\mbox{where}\quad\eta^{Q}(T):=\frac{1}{2}\int_{0}^{T}(1-\varepsilon
Q_{t})\,dt.$ (3.8)

In this case, the candidate optimal strategy is

$q^{Q}_{t}(x_{0})=\frac{1}{2}\left(1-\varepsilon
Q_{t}-Y_{T}(x_{0})e^{-r(T-t)}\right),\quad t<T.$

  * •

The case $T<\infty$ and $\tau^{Q}>T$. In this case we expect the optimization
problem to be independent of $x_{0}$ and hence that $Y\equiv 0$. From (3.5),
this leads to the candidate optimal strategy

$q^{Q}_{t}(x_{0})=\frac{1}{2}\left(1-\varepsilon Q_{t}\right).$

Thus, this case occurs when $x_{0}\geq\eta^{Q}(T)$.

####  3.1.2 Verification Result

In order to carry out the verification argument it will be convenient to
define the function

$\xi^{Q}(t):=\frac{1}{2}\int_{0}^{t}\\{1-\varepsilon Q_{s}-(1-\varepsilon
Q_{t})e^{-r(t-s)}\\}\,ds,$ (3.9)

which is the amount of resource extracted up to time $t$ by using the strategy
(3.6) that would be optimal if the exhaustion time $\tau^{Q}$ was equal to
$t$. The function $\xi^{Q}$ allows us to determine the optimal exploitation
time as a function of the initial resource level. In fact the derivative of
$\xi^{Q}$ is given by

$\dot{\xi}^{Q}(t)=\frac{1}{2}\left\\{1-\varepsilon
Q_{t}+\frac{\varepsilon}{r}\dot{Q}_{t}\right\\}(1-e^{-rt}).$ (3.10)

The compatibility condition (3.2) guarantees that
$\dot{\xi}^{Q}(t)\geq\frac{\delta}{2}(1-e^{-rt})>0$. As a result, the inverse
$(\xi^{Q})^{-1}$ is well-defined on $[0,\xi^{Q}(T)]$. By Assumption 1,
$\xi^{Q}(\infty)=\infty$. Moreover, we see that

$\xi^{Q}(\tau^{Q}(x_{0}))=\int_{0}^{\tau^{Q}(x_{0})}q^{Q}_{s}(x_{0})\,ds,$
(3.11)

and hence that $\tau^{Q}(x_{0})$ is determined implicitly by the identity

$\xi^{Q}(\tau^{Q}(x_{0}))=x_{0},\quad\mbox{i.e.}\quad\tau^{Q}(x_{0})=(\xi^{Q})^{-1}(x_{0})\quad\mbox{for}\quad
x_{0}\leq\xi^{Q}(T).$

The preceding heuristics suggest that an optimal exploitation strategy for a
given initial resource level $x_{0}$ and a given aggregate production function
$Q$ that satisfies our compatibility condition can be obtained by first
computing the optimal exploitation time by inverting the function $\xi^{Q}$
introduced in (3.9) on its range $[0,\xi^{Q}(T)]$, and then applying the
maximum principle on $[0,\tau^{Q}(x_{0})]$. The following result shows that
the above heuristics do indeed give us an optimal control.

######  Theorem 3.3 (Verification Theorem).

Let $Q:[0,T]\to\mathbb{[}0,\frac{1}{2+\varepsilon}]$ be an absolutely
continuous function that satisfies the compatibility condition (3.2). Define
the function $\xi^{Q}:[0,T]\to[0,\xi^{Q}(T)]$ by (3.9), and let
$\tau^{Q}=(\xi^{Q})^{-1}$.

  * •

$T=\infty$. For initial state $x_{0}\in[0,\infty)$, $\tau^{Q}(x_{0})<\infty$,
and the optimal control is given by

$q^{Q}_{t}(x_{0})=\frac{1}{2}\\{1-\varepsilon Q_{t}-(1-\varepsilon
Q_{\tau^{Q}(x_{0})})e^{-r(\tau^{Q}(x_{0})-t)}\\}\mathbbm{1}_{\\{t\leq\tau^{Q}(x_{0})\\}}.$
(3.12)

  * •

$T<\infty$. The optimal control for the initial state $x_{0}\in[0,\infty)$ is
given by

$q^{Q}_{t}(x_{0})=\begin{cases}\frac{1}{2}\\{1-\varepsilon
Q_{t}-(1-\varepsilon
Q_{\tau^{Q}(x_{0})})e^{-r(\tau^{Q}(x_{0})-t)}\\}\mathbbm{1}_{\\{t\leq\tau^{Q}(x_{0})\\}},&x_{0}\in[0,\xi^{Q}(T)]\\\
\frac{1}{2}\left(1-\varepsilon
Q_{t}-Y_{T}(x_{0})e^{-r(T-t)}\right)&x_{0}\in[\xi^{Q}(T),\eta^{Q}(T)],\\\
\frac{1}{2}(1-\varepsilon Q_{t}),&\mbox{else},\end{cases}$ (3.13)

where $Y_{T}(x_{0})$ and $\eta^{Q}(T)$ are defined in (3.7) and (3.8),
respectively.

###### Proof.

For notational convenience we drop the dependence of $q^{Q}$ and $\tau^{Q}$ on
the initial state $x_{0}$. The proof is similar to the monopoly case; we give
it for completeness. First, we recall from the argument following (3.10), that
$\tau^{Q}$ is well-defined. Moreover, it is straightforward to check that i)
the candidate $q^{Q}$ in (3.12) and (3.13) are of the form

$q_{t}^{Q}=\frac{1}{2}(1-\varepsilon Q_{t}-Y_{t})^{+}$ (3.14)

where the adjoint process $Y$ is defined in (3.4) , and ii) are admissible. In
particular $q^{Q}$, and the corresponding state process $X^{q^{Q}}$ are non-
negative. Denoting $\bar{\tau}_{q^{Q}}:=\tau_{q^{Q}}\wedge T$, we see from the
formulas (3.12) and (3.13) that $\bar{\tau}_{q^{Q}}=\tau^{Q}\wedge T$. In
order to verify the optimality of the strategy $q^{Q}$ for the first case in
(3.13), we fix an admissible control $q$, whose exhaustion time is $\tau_{q}$,
with the convention $\tau_{q}=\infty$ if the strategy does not exhaust all of
the resource. We denote by $\bar{\tau}_{q}:=\tau_{q}\wedge T$. Admissibility
requires that $\int_{0}^{\bar{\tau}_{q}}q_{t}\,dt\leq x_{0}$. Moreover, since
$q^{Q}$ satisfies the maximum condition (3.5),

$H(x,q^{Q}_{t},Y_{t})-H(x,q_{t},Y_{t})\geq 0\quad\mbox{for all}\quad x>0.$
(3.15)

  * •

The case $\tau_{q}\leq\tau_{q^{Q}}$. In this case it follows from (3.15) and
$Y_{t}e^{-rt}=Y_{0}$ that

$\displaystyle J^{Q}(q^{Q})-J^{Q}(q)$
$\displaystyle=\int_{0}^{\bar{\tau}_{q}}e^{-rt}(H(X^{q^{Q}}_{t},q^{Q}_{t},Y_{t})-H(X^{q}_{t},q_{t},Y_{t})+(q^{Q}_{t}-q_{t})Y_{t})\,dt$
$\displaystyle~{}~{}+\int_{\bar{\tau}_{q}}^{\bar{\tau}_{q^{Q}}}e^{-rt}(H(X^{q^{Q}}_{t},q^{Q}_{t},Y_{t})+q^{Q}_{t}Y_{t})\,dt$
$\displaystyle\geq\int_{0}^{\bar{\tau}_{q}}(q^{Q}_{t}-q_{t})Y_{t}e^{-rt}\,dt+\int_{\bar{\tau}_{q}}^{\bar{\tau}_{q^{Q}}}q^{Q}_{t}Y_{t}e^{-rt}\,dt$
$\displaystyle=Y_{0}\left(\int_{0}^{\bar{\tau}_{q^{Q}}}q^{Q}_{t}\,dt-\int_{0}^{\bar{\tau}_{q}}q_{t}\,dt\right)\color[rgb]{0,0,0}.$

As in the monopoly case, $Y_{0}=0$ if $\tau^{Q}>T$. If
$\tau_{q}\leq\tau^{Q}\leq T$, then the term in parenthesis vanishes. In both
cases

$J^{Q}(q^{Q})-J^{Q}(q)\geq 0.$

  * •

The case $\tau_{q}>\tau^{Q}$. In this case it follows again from (3.15) that

$\displaystyle J^{Q}(q^{Q})-J^{Q}(q)$
$\displaystyle=\int_{0}^{\bar{\tau}_{q^{Q}}}e^{-rt}(H(X^{q^{Q}}_{t},q^{Q}_{t},Y_{t})-H(X^{q}_{t},q_{t},Y_{t})+(q^{Q}_{t}-q_{t})Y_{t})\,dt$
$\displaystyle~{}~{}-\int_{\bar{\tau}_{q^{Q}}}^{\bar{\tau}_{q}}e^{-rt}(H(X^{q}_{t},q_{t},Y_{t})+q_{t}Y_{t})\,dt$
$\displaystyle\geq\int_{0}^{\bar{\tau}_{q^{Q}}}e^{-rt}(q_{t}^{Q}-q_{t})Y_{t}(x_{0})\,dt-\int_{\bar{\tau}_{q^{Q}}}^{\bar{\tau}_{q}}e^{-rt}(H(X^{q}_{t},q_{t},Y_{t})+q_{t}Y_{t})\,dt.$

Using that $Y\geq 1$ on $[\tau^{q^{Q}},T]$, and hence that $H\leq 0$ for all
$(q,t)\in\mathbb{[}0,\infty)\times[\tau^{q^{Q}},T]$, we arrive at

$J^{Q}(q^{Q})-J^{Q}(q)\geq
Y_{0}\left(\int_{0}^{\bar{\tau}_{q^{Q}}}q_{t}^{Q}\,dt-\int_{0}^{\bar{\tau}_{q}}q_{t}\,dt\right).$

If $\tau^{q^{Q}}>T$, then $Y_{0}=0$. Else, $Y_{0}\geq 0$,
$\int_{0}^{\bar{\tau}_{q^{Q}}}q_{t}^{Q}\,dt=x_{0}$ while
$\int_{0}^{\bar{\tau}_{q}}q_{t}\,dt\leq x_{0}$ because $\tau_{q}>\tau^{Q}$. As
a result,

$J^{Q}(q^{Q})-J^{Q}(q)\geq 0.$

∎

It follows from the verification theorem that all the optimal production rates
$q_{t}^{Q}$ satisfy the same ODE, albeit on possibly different time horizons
and with possibly different terminal conditions. Moreover, from (3.10), the
mapping $x_{0}\mapsto\tau^{Q}(x_{0})$ is strictly increasing and so

$\\{x_{0}\geq 0:\tau^{Q}(x_{0})<t\\}=\\{x_{0}\geq 0:x_{0}<\xi^{Q}(t)\\}.$

In particular, if we denote by $S_{\mu}(x_{0})=\int_{x_{0}}^{\infty}\mu(dx)$
the survival function associated with the initial distribution $\mu$, then the
proportion of players with remaining resources at time $t\in[0,T]$ is given by
$S_{\mu}(\xi^{Q}(t))$. With this, we have the following corollary.

######  Corollary 3.4.

For all $x_{0}>0$ and all functions $Q$ that satisfy the compatibility
condition (3.2), the optimal rate $q^{Q}_{t}(x_{0})$ and the aggregate
production rate

$Q_{t}^{Q}:=\int_{\xi^{Q}(t)}^{\infty}q^{Q}_{t}(x_{0})\,d\mu(x_{0}),\quad
0\leq t\leq T$ (3.16)

satisfy the ODEs

$\displaystyle q^{Q}_{t}(x_{0})-\frac{1}{r}\dot{q}^{Q}_{t}(x_{0})$
$\displaystyle=\frac{1}{2}\left\\{1-\varepsilon
Q_{t}+\frac{\varepsilon}{r}\dot{Q}_{t}\right\\}$ (3.17) $\displaystyle
Q_{t}^{Q}-\frac{1}{r}\dot{Q}_{t}^{Q}$
$\displaystyle=\frac{1}{2}\left\\{1-\varepsilon
Q_{t}+\frac{\varepsilon}{r}\dot{Q}_{t}\right\\}S_{\mu}(\xi^{Q}(t)),$ (3.18)

on the time interval $[0,\tau^{Q}(x_{0})\wedge T]$. In the case $T=+\infty$,
the terminal condition for (3.17) is $q^{Q}_{\tau^{Q}(x_{0})}=0$, while for
(3.18), $\lim_{t\to\infty}Q_{t}=0$. If $T<\infty$, then the optimal production
rate has terminal condition

$q^{Q}_{\tau^{Q}(x_{0})\wedge T}(x_{0})=\begin{cases}0&\mbox{ if
}x_{0}\in[0,\xi^{Q}(T)]\\\ \frac{1}{2}\left(1-\varepsilon
Q_{T}-Y_{T}(x_{0})\right)&\mbox{ if }x_{0}\in[\xi^{Q}(T),\eta^{Q}(T))]\\\
\frac{1}{2}(1-\varepsilon Q_{T})&\mbox{ else,}\end{cases}$ (3.19)

where $Y_{T}(x_{0})$ was given in (3.7). In this case, the terminal condition
for the aggregate production rate is

$Q^{Q}_{T}=\frac{1}{2}(1-\varepsilon
Q_{T})S_{\mu}(\xi^{Q}(T))-\frac{\eta^{Q}(T)}{\beta_{T}}\left(S_{\mu}(\xi^{Q}(T))-S_{\mu}(\eta^{Q}(T))\right)+\frac{1}{\beta_{T}}\int_{\xi^{Q}(T)}^{\eta^{Q}(T)}x_{0}\mu(dx_{0}).$
(3.20)

###### Proof.

The ODE (3.17) follows from differentiating the expression (3.14) for $q^{Q}$
(the positive part being unnecessary on $t\in[0,\tau^{Q}(x_{0})\wedge T]$),
using the ODE for $Y$, and substituting back for $Y$ in terms of $Q$ and
$q^{Q}$. The ODE (3.18) for $Q^{Q}$ follows from integrating (3.17) with
respect to $\mu(x_{0})$ over $(\xi^{Q}(t),\infty)$. The terminal conditions
for $q^{Q}$ follow from (3.12) and (3.13). The terminal condition for $Q^{Q}$
comes from integrating (3.19) with respect to $\mu$ over $(0,\infty)$. ∎

######  Remark 3.5.

Under the compatibility condition, we have $\xi^{Q}(T)\uparrow\infty$ as
$T\uparrow\infty$. As a result, $\lim_{T\to\infty}Q^{Q}_{T}=0$. This shows
that the infinite horizon case can indeed be viewed as a limiting case when
$T\uparrow\infty$.

###  3.2 Solution to the MFG

In this section we prove the existence of a solution to the mean field game.
In terms of the optimal controls $q^{Q}$ we are looking for a fixed point of
the mapping (3.16). We consider the infinite and the finite horizon case
separately. The main difference is that full exploitation occurs in
equilibrium if $T=+\infty$, but may not occur if $T<\infty$. It turns out that
the infinite horizon case can be solved in closed form. In view of Corollary
3.4 , we have the following characterization.

######  Corollary 3.6.

Any fixed point $Q^{*}=Q^{Q^{*}}$ satisfies the dynamics

$Q^{*}-\frac{1}{r}\dot{Q}^{*}=\frac{S_{\mu}\circ\xi^{Q^{*}}}{2+\varepsilon
S_{\mu}\circ\xi^{Q^{*}}},$

and satisfies the compatibility condition (3.2).

###### Proof.

Setting $Q^{Q}$ and $Q$ in (3.18) to $Q^{*}$ and re-arranging leads to (3.6),
which is equivalent to

$1-\varepsilon Q^{*}+\frac{\varepsilon}{r}\dot{Q}^{*}=\frac{2}{2+\varepsilon
S_{\mu}\circ\xi^{Q^{*}}}\,,$ (3.21)

from which compatibility of $Q^{*}$ easily follows. ∎

For any fixed point $Q^{*}$ the equations (3.10) and (3.21) yield the
following ODE for $\xi^{Q^{*}}$:

$\displaystyle\dot{\xi}^{Q^{*}}(t)=\frac{1-e^{-rt}}{2+\varepsilon
S_{\mu}\circ\xi^{Q^{*}}},\quad\xi^{Q^{*}}_{t}(0)=0.$ (3.22)

The key observation is that (3.22) does not depend on $Q^{*}$. We prove below
that the (unique) solution to the MFG can be defined in terms of the unique
solution $\xi^{*}$ to the above ODE.

######  Proposition 3.7.

If the initial distribution has a bounded density or is concentrated on
finitely many points, then the preceding ODE has a unique global solution. The
solution can be expressed in terms of the functions
$\phi,\psi:[0,\infty)\to[0,\infty)$ defined by

$\phi(t):=\frac{rt-1+e^{-rt}}{r}\quad\mbox{and}\quad\psi(x):=2x+\varepsilon\int_{0}^{x}S_{\mu}(\xi)\,d\xi,$

respectively as

$\xi^{*}:=\psi^{-1}\circ\phi$ (3.23)

###### Proof.

If $\mu$ has a bounded density, then the mapping
$x\mapsto\frac{1-e^{-rt}}{2+\varepsilon S_{\mu}(x)}$ is Lipschitz continuous.
If $\mu$ is concentrated on finitely many points, then this mapping is locally
independent of $x$ and hence can be solved iteratively. The functions $\phi$
and $\psi$ are well-defined, strictly increasing, surjective, and satisfy

$\dot{\phi}(t)=1-e^{-rt}\quad\mbox{and}\quad\psi^{\prime}(x)=2+\varepsilon
S_{\mu}(x).$

Now, a direct computation verifies that $\xi^{*}$ satisfies the desired ODE. ∎

Note that the inverse of $\phi$ is, in terms of the principle branch of the
Lambert-$\mathbb{W}$ function,

$\phi^{-1}(x)=x+\frac{1+\mathbb{W}(-e^{-1-rx})}{r}.$

####  3.2.1 Infinite horizon

We are now ready to solve the MFG in the infinite horizon case. In terms of
the functions $\phi$ and $\psi$, and their inverses we obtain

  * •

$\xi^{*}:=\psi^{-1}\circ\phi$

  * •

$\tau^{*}:={\xi^{*}}^{-1}=\phi^{-1}\circ\psi$

In view of (3.21) we can express the unique solution to the ODE (3.17) with
terminal condition (3.18) fully in terms of these functions as

$\displaystyle
q_{t}^{*}(x_{0}):=\int_{t\wedge\tau^{*}(x_{0})}^{\tau^{*}(x_{0})}\frac{re^{-r(s-t)}}{2+\varepsilon
S_{\mu}(\xi^{*}(s))}\,ds.$

The following theorem verifies that this production rate does indeed solve the
MFG in the infinite horizon case.

######  Theorem 3.8.

The aggregate production rate

$\displaystyle Q_{t}^{*}:=\int_{0}^{\infty}q_{t}^{*}(x_{0})\,d\mu(x_{0})$

is absolutely continuous, takes values in $[0,\frac{1}{2+\varepsilon}]$ and
satisfies the compatibility condition (3.2) as well as the fixed point
property $Q^{Q^{*}}=Q^{*}$. Hence, $Q^{*}$ is the unique solution to the MFG
(that is absolutely continuous and satisfies the compatibility condition).

###### Proof.

To verify $Q^{*}\in[0,\frac{1}{2+\varepsilon}]$ we use the following
representation by Fubini:

$\displaystyle Q_{t}^{*}$
$\displaystyle=\int_{t}^{\infty}\frac{rS_{\mu}(\xi^{*}(s))e^{-r(s-t)}}{2+\varepsilon
S_{\mu}(\xi^{*}(s))}\,ds\leq\frac{S_{\mu}(\xi^{*}(t))}{2+\varepsilon
S_{\mu}(\xi^{*}(t))}\leq\frac{1}{2+\varepsilon}.$ (3.24)

Next, by construction $\xi^{*}$ satisfies the ODE

$\dot{\xi}^{*}(t)=\frac{1-e^{-rt}}{2+\varepsilon
S_{\mu}(\xi^{*}(t))},\quad\xi^{*}(0)=0$ (3.25)

and $q^{*}$ is absolutely continuous in the time variable with

$q^{*}_{t}(x_{0})-\frac{1}{r}\dot{q}_{t}^{*}(x_{0})=\frac{1}{2+\varepsilon
S_{\mu}(\xi^{*}(t))}\mathbbm{1}_{\\{t<\tau^{*}(x_{0})\\}}.$ (3.26)

The latter yields that $Q^{*}$ is absolutely continuous with

$Q_{t}^{*}-\frac{1}{r}\dot{Q}_{t}^{*}=\frac{S_{\mu}(\xi^{*}(t))}{2+\varepsilon
S_{\mu}(\xi^{*}(t))},$ (3.27)

which is equivalent to

$\frac{1}{2}\left\\{1-\varepsilon
Q_{t}^{*}+\frac{\varepsilon}{r}\dot{Q}_{t}^{*}\right\\}=\frac{1}{2+\varepsilon
S_{\mu}(\xi^{*}(t))}.$

As a result, $Q^{*}$ satisfies the compatibility condition. Moreover, in view
of (3.10) and (3.25),

$\xi^{*}=\xi^{Q^{*}}\quad\mbox{and hence}\quad\tau^{*}=\tau^{Q^{*}}.$

>From this we conclude that $q^{Q^{*}}=q^{*}$ and hence the fixed point
property $Q^{Q^{*}}=Q^{*}$. Applying the verification result Theorem 3.3
finally verifies that $Q^{*}$ is the unique solution the MFG. ∎

While the individual production rate $q^{*}$ may not be monotone in general as
illustrated below, it is a direct consequence of (3.24) and (3.27) that the
aggregate production rate is nonincreasing.

######  Corollary 3.9.

The aggregate production rate $Q^{*}$ is nonincreasing.

####  3.2.2 Finite horizon

In the finite horizon case an additional challenge emerges. While we can still
solve the ODE (3.17) using (3.21), due to the dependence of the second and
third terminal conditions in (3.19) on $Q_{T}$ we obtain the equilibrium
production rate only up to its terminal value.444In the infinite horizon game
both the dynamics and the terminal conditions were defined in terms of
$\xi^{*}$ and its inverse $\tau^{*}$. This is no longer the case in the finite
horizon game. An additional fixed point argument on the terminal value
$Q^{*}_{T}$ is required to solve the game. In terms of the yet to be
determined terminal condition $Q^{*}_{T}$ we do know that

$Q^{*}_{t}=Q_{T}^{*}e^{-r(T-t)}+\int_{t}^{T}\frac{rS_{\mu}(\xi^{*}(s))e^{-r(s-t)}}{2+\varepsilon
S_{\mu}(\xi^{*}(s))}\,ds.$

In terms of

$\psi^{Q}(x_{0}):=\frac{1}{\beta_{T}}\left(x_{0}-\xi^{Q}(T)\right)$ (3.28)

we obtain for any admissible process $Q$,

$\displaystyle Q_{T}^{Q}$
$\displaystyle=\frac{1}{2}\int_{\xi^{Q}(T)}^{\xi^{Q}(T)+\beta_{T}(1-\varepsilon
Q_{T})}\psi^{Q}(x_{0})\,d\mu(x_{0})+\frac{1}{2}\int^{\infty}_{\xi^{Q}(T)+\beta_{T}(1-\varepsilon
Q_{T})}(1-\varepsilon Q_{T}^{Q})\,d\mu(x_{0})$
$\displaystyle=\frac{1}{2}\int_{\xi^{Q}(T)}^{\infty}\\{\psi^{Q}(x_{0})\wedge(1-\varepsilon
Q_{T}^{Q})\\}\,d\mu(x_{0}).$

Putting

$\psi^{*}(x_{0}):=\frac{1}{\beta_{T}}\left(x_{0}-\xi^{*}(T)\right)$

we obtain for any aggregate equilibrium production rate function $Q^{*}$ that

$\displaystyle
Q_{T}^{*}=\frac{1}{2}\int_{\xi^{*}(T)}^{\infty}\\{\psi^{*}(x_{0})\wedge(1-\varepsilon
Q_{T}^{*})\\}\,d\mu(x_{0}).$

This means that $Q_{T}^{*}$ is a fixed point of the map
$\Gamma:[0,\frac{1}{2}]\to[0,\frac{1}{2}]$ definied by

$\Gamma(\mathcal{Q})=\frac{1}{2}\int_{\xi^{*}(T)}^{\infty}\\{(\psi^{*}(x_{0})\wedge\\{1-\varepsilon\mathcal{Q}\\})\vee
0\\}\,d\mu(x_{0}),$

where we added the nonnegative cut-off (being redundant for any fixed point)
to obtain a map on $[0,\frac{1}{2}]$. Moreover, is $\Gamma$ nonincreasing and
continuous. Hence, it admits a unique fixed point
$\mathcal{Q}^{*}=\Gamma(\mathcal{Q}^{*})$. Since

$\Gamma(\mathcal{Q}^{*})\leq\frac{1}{2}S_{\mu}(\xi^{*}(T))(1-\varepsilon\mathcal{Q}^{*}),$

it follows that

$\mathcal{Q}^{*}\leq\frac{S_{\mu}(\xi^{*}(T))}{2+\varepsilon
S_{\mu}(\xi^{*}(T))}.$

This is important to verify that the induced $Q^{*}$ takes indeed values in
$[0,\frac{1}{2+\varepsilon}]$. For $t\in[0,T]$,

$\displaystyle Q^{*}_{t}$
$\displaystyle=\mathcal{Q}^{*}e^{-r(T-t)}+\int_{t}^{T}\frac{rS_{\mu}(\xi^{*}(s))e^{-r(s-t)}}{2+\varepsilon
S_{\mu}(\xi^{*}(s))}\,ds$
$\displaystyle\leq\frac{S_{\mu}(\xi^{*}(t))}{2+\varepsilon
S_{\mu}(\xi^{*}(t))}e^{-r(T-t)}+\frac{S_{\mu}(\xi^{*}(t))}{2+\varepsilon
S_{\mu}(\xi^{*}(t))}\int_{t}^{T}re^{-r(s-t)}\,ds$
$\displaystyle=\frac{S_{\mu}(\xi^{*}(t))}{2+\varepsilon S_{\mu}(\xi^{*}(t))}.$

The fact that $Q^{*}$ satisfies the compatibility condition has already been
established in Corollary 3.6. Let us summarize.

######  Theorem 3.10.

In terms of the above definitions, the function $Q^{*}$ is absolutely
continuous, takes values in $[0,\frac{1}{2+\varepsilon}]$, and satisfies the
compatibility condition (3.2) and the fixed point property

$Q^{Q^{*}}=Q^{*}.$

Hence, $Q^{*}$ is the unique solution to the finite-time MFG (that is
absolutely continuous and satisfies the compatibility condition).

###  3.3 Examples

We close this section with two examples with infinite time horizon that
illustrate that equilibrium production rates need not be monotone, nor smooth.

####  3.3.1 Discrete initial distribution

Let us assume that the population of producers splits into two groups within
which producers are identical. Producers in Group 1 have an initial resource
$x_{1}$; producers in Group 2 have an initial resource $x_{2}$. That is, the
initial distribution is given by

$\mu=(1-p_{2})\delta_{x_{1}}+p_{2}\delta_{x_{2}}$

for $0\leq x_{1}\leq x_{2}<\infty$ and $0\leq p_{2}\leq 1$ where $p_{2}$
denotes the proportion of producers that belong to Group 2. We refer to a
representative producer in Group $i$ as Player $i$ where $i=1,2$. The
equilibrium production rates can be computed in closed form. The equilibrium
production rate of Player 1 is given by

$q_{t}^{*}(x_{1})=\frac{1-e^{-r(\tau_{1}-t)}}{2+\varepsilon}\mathbbm{1}_{\\{t\leq\tau_{1}\\}},$
(3.29)

where $\tau_{1}:=\phi^{-1}((2+\varepsilon)x_{1})$, while the equilibrium
production rate of Player 2 is given by

$q_{t}^{*}(x_{2})=\begin{cases}\dfrac{1-e^{-r(\tau_{1}-t)}}{2+\varepsilon}+\dfrac{e^{-r(\tau_{1}-t)}-e^{-r(\tau_{2}-t)}}{2+\varepsilon
p_{2}},&t\leq\tau_{1},\\\ \dfrac{1-e^{-r(\tau_{2}-t)}}{2+\varepsilon
p_{2}}\mathbbm{1}_{\\{t\leq\tau_{2}\\}},&t\geq\tau_{1},\end{cases}$

where $\tau_{2}:=\phi^{-1}(\varepsilon(1-p_{2})x_{1}+(2+\varepsilon
p_{2})x_{2})$. We notice that the equilibrium rate of Player 2 is continuous,
but not differentiable at time $\tau_{1}$. Figure 1 illustrates the
equilibrium production rates $q^{*}_{t}(x_{i})$ and resources
$X^{*}_{t}(x_{i})$ of both players. While both Players initially produce at
the same rate, Player 1 produces at a decreasing rate while Player 2 initially
produces at an increasing rate, and then at a decreasing rate once Player 1
has run out of resources. Player 2 anticipates the fact Player 1 will
eventually drop out of the market; her production rate reaches it peak at the
time Player 1’s resources have been depleted.

Figure 1: $r=.02$, $\varepsilon=2$, $x_{1}=50$, $x_{2}=100$, $p_{2}=.5$

####  3.3.2 Exponential initial distribution

Figure 2 illustrates the equilibrium production rates $q^{*}_{t}(i)$ for
players with initial resources $x_{i}=1,2,3$ for the benchmark case of an
exponential initial distribution. For absolutely continuous initial
distributions players “gradually” drop out of the market so equilibrium
production rates do not display kinks. However, we still observe non-
monotonicity of production rates. Players with larger resources do again
anticipate that players with lower reserves will eventually drop out of the
game. For them it is optimal to produce at higher rates once the number of and
the price pressure from their competitors decreases.

Figure 2: $\mu\sim\text{Exp}(\lambda)$, $r=\varepsilon=\lambda=1$

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# Probing new physics at the LUXE experiment

Arka Santra
Weizmann Institute of Science
234, Herzl Street, Rehovot, Israel-7610001
for the LUXE Collaboration

###### Abstract

The proposed Laser Und XFEL (LUXE) Experiment at DESY, Hamburg, aims to probe
QED in the strong field non-perturbative regime. This regime will be created
in collisions between high-intensity laser pulses and high-energy electron or
photon beams coming from the Eu.XFEL setup. This experiment comes with a
unique opportunity to probe physics beyond the standard model. In this
proceeding, it is described that with the help of the large photon flux
generated at LUXE, axion-like-particles can be probed up to a mass of 350 MeV
and with a photon coupling of $3\times 10^{-6}$ GeV-1. This reach in parameter
space is comparable to the projected reach of future experiments like FASER2
in the HL-LHC and NA62 in the dump mode.

##  1 Introduction

The standard model (SM) of particle physics is very successful in describing
most of the interactions in nature. Yet it does not describe the complete
picture of nature, for example dark matter, hierarchy problem and cosmological
baryon asymmetry etc. This motivates the search for beyond the SM (BSM)
particles/interactions at different energy scales and different kind of
experiments. In spite of all the efforts, we are yet to find any “smoking gun”
of a direct observation of BSM. This suggests that we need to use new
experimental techniques which may open alternative ways to explore the
uncharted regions.

The non-perturbative and non-linear quantum electrodynamics (QED), such as
Schwinger pair production in the presence of a strong electric field [1, 2, 3]
is still largely unexplored. The proposed LUXE (Laser Und XFEL Experiment)
experiment [4] wants to inspect this strong field QED (SFQED) with the help of
the interaction between the high energetic electron beam from the European
XFEL (Eu.XFEL) facility and very intense lasers. An intense laser can be used
as an effective optical dump for electrons where electrons emit a large flux
of hard photons. Since the photons negligibly interact with the
electromagnetic field of the laser, the hard photons can free stream in the
laser after production. With an additional forward detector, the LUXE
experiment can probe a well-motivated uninvestigated parameter space of a new
spin-0 particle (scalar or pseudo-scalar) which couples to photons. This is
indicated as LUXE-NPOD: New Physics at Optical Dump [5]. In this proceeding,
it will be shown that this setup can probe scalar and pseudo-scalar particle
masses up to $\mathcal{O}(350)$ MeV and decay constants of
$\mathcal{O}(10^{5}-10^{6})$ GeV-1.

##  2 Probing the New Physics

The focus of this proceeding is the production of new spin-0 particles
motivated by many extensions of SM. The pseudo-scalar particles, denoted
generically as axion-like-particle (ALP), can naturally address the strong CP
problem [6, 7], address the Heirarchy problem [8, 9], can be a valid dark
matter candidate [10, 11, 12]. The effective ALP ($a$) interaction with
photons and electrons are given by the following Lagrangian:

$\mathcal{L}_{a}=\frac{a}{4\Lambda_{a}}F_{\mu\nu}\tilde{F}^{\mu\nu}+ig_{ae}a\bar{e}\gamma^{5}e,$
(1)

where
$\tilde{F}_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\alpha\beta}F^{\alpha\beta}$,
$g_{ae}$ is the coupling between ALP and electrons and $\Lambda_{a}$ is the
scale of interaction between ALP and the SM particles. For this discussion, we
take $g_{ae}$ to be 0, so ALPs can decay only to photons. The decay rate of
the ALP into two photons is shown in the following equation:

$\Gamma_{a\rightarrow 2\gamma}=\frac{m_{a}^{3}}{64\pi\Lambda_{a}^{2}}.$ (2)

The interaction of the scalar particle ($\phi$) with the photons and electrons
can be written in a similar fashion [5].

###  2.1 The Concept of Optical Dump

In the LUXE experiment, high flux of free streaming hard photons emitted from
the electron-laser interaction region will be used to search for the weakly
interacting spin-0 particles; this is the so-called optical dump. This simply
refers to the use of the laser as an effective dump for the incoming
electrons. In order to successfully meet the criteria of the optical dump
where electrons will emit free streaming hard photons, the time scales of the
problem should have the following relation:

$\omega_{L}^{-1}\ll\tau_{\gamma}\lesssim t_{L}\ll\tau_{ee},$ (3)

where $\omega_{L}$ is the laser angular frequency, $\tau_{\gamma}$ is the
typical timescale of the high intensity Compton scattering, $t_{L}$ is the
laser pulse length and $\tau_{ee}$ is the typical pair production time scale.
These numbers relevant to the LUXE experiment are as follows:

  * •

the laser wavelength will be 800 nm; that makes $\omega_{L}^{-1}\sim 0.4$ fs
[4],

  * •

the typical timescale of non-linear Compton scattering is $\mathcal{O}(10)$
fs,

  * •

the laser pulse length for the LUXE experiment is expected to be
$\sim\mathcal{O}(10-200)$ fs,

  * •

the pair production timescale relevant for the LUXE experiment is
$\mathcal{O}(10^{4}-10^{6})$ fs [4].

Hence the relation (3) will hold for the LUXE setup.

The $\omega_{L}^{-1}$ is the shortest timescale in the problem - this allows
the laser to be treated as a background field to a leading order. Since
$\tau_{ee}$ is the highest timescale here, the emitted photons can be
considered as free streaming. Finally as $\tau_{\gamma}$ is shorter than
$t_{L}$, the laser behaves as a thick target for the electrons. The
combination of the last two points effectively establishes the LUXE experiment
as the source of free hard photons from the electrons, just like a dump but
made of lasers, hence optical dump.

###  2.2 The LUXE-NPOD proposal

After the production of free photons in the optical dump, the photons will
propagate to a physical dump and interact with the nucleus of the dump to
produce the $X$ particle ($X$ can be a scalar or pseudo-scalar as discussed in
Sec. 2). Here we focus on the Primakoff production of X’s:

$\gamma+\ N\rightarrow\ N\ +\ X,$ (4)

where N is nucleus of the dump and $\gamma$ is the photon coming from the
optical dump. In the proposed setup, the dump length is $L_{D}$ and the
position of it is $\sim$13 m away from the electron-laser interaction point in
LUXE. The long-lived X particles will travel some distance before decaying
into $\gamma\gamma$. Therefore there will be an empty volume of length $L_{V}$
left after the physical dump where $X$ can decay into $\gamma\gamma$. This two
photon signature is the signal for the BSM detector system. The expected
number of $X$ particles produced at the proposed setup shown in Fig. 1 is
given by [13, 14, 5]:

$N_{X}\thickapprox\mathcal{L}_{eff}\int
dE_{\gamma}\frac{dN_{\gamma}}{dE_{\gamma}}\sigma_{X}(E_{\gamma})\left(e^{-\frac{L_{D}}{L_{X}}}-e^{-\frac{L_{V}+L_{D}}{L_{X}}}\right)\mathcal{A}.$
(5)

In the above equation, $\sigma_{X}({E_{\gamma}})$ is the Primakoff cross-
section which varies as the square of the nuclear charge $Z^{2}$ of the dump,
$\mathcal{L}_{eff}$ is the effective luminosity, $E_{\gamma}$ is the incoming
photon energy spectrum, $\mathcal{A}$ is the angular acceptance and efficiency
of the detector. The differential photon flux per electron,
$\frac{dN_{\gamma}}{dE_{\gamma}}$ as a function of incoming photon energy
spectrum for two different phases of the LUXE experiment is shown in Fig. 2.
The laser intensity will be 40 TW for phase-0 of the run, and this will be
increased to 350 TW for the phase-1 of the run. The $L_{X}\equiv
c\tau_{X}p_{X}/m_{X}$ is the propagation length of the X particle with
$\tau_{X}$, $p_{X}$ and $m_{X}$ being its proper lifetime, momentum and mass,
respectively.

Figure 1: A schematic diagram showing the LUXE-NPOD proposal. The free photons
streaming from the laser and electron interaction point hit the dump producing
the ALPs. The two photons coming from the decay of ALP in the decay volume of
$L_{V}$ are captured by the electromagentic calorimeter kept at the end of the
setup. The muon detector is used to detect the charged muons produced during
the SM processes during the shower in the dump. The magnet in the $L_{V}$ is
kept to bend away the background charged particles and reduce their effect.
Figure 2: The emitted photon spectrum for phase-0 in blue compared with that
of the phase-1 in black. The perturbative Bremsstrahlung spectrum with
electron energy 16.5 GeV and target length of 0.01 of electron radiation
length is shown in red. In this plot, the primary photons are those which come
from the electron-laser interaction point, and the secondary photons come from
the shower in the dump.

##  3 Results

The background for this search consists of (i) charged particles - electrons,
muons and hadrons, (ii) fake photons where neutrons are misidentified as
photons and (iii) real photons, from the EM/hadronic interaction close to the
end of the dump. The rates of different background particles are estimated
using Geant4 simulation setup. The background charged particles can be bent
away by a magnetic field of $B\approx 1$ T over an active length of $\sim 1$
m. The muon detector will be used to identify muons which may still reach the
BSM detector setup. Neutrons can be rejected by looking at their arrival time
at the detector because the signal arrives much faster than neutron. The
rejection power of the full selection criteria can be kept $\lesssim 10^{-3}$.
It is possible to keep neutron to photon fake rate below $10^{-3}$ by
utilizing modern detector technology. For a one year of data taking period
($\sim 10^{7}$ s) and the choice of detector technology, the number of
background event is expected to be restricted to less than 1 [5]. Hence this
search is considered effectively background free. The projection of the
sensitivity of LUXE-NPOD (effectively the number of expected signal events
$N_{X}=3$, which is the 95% CL equivalent for background free search) is shown
in Fig. 3 in the effective coupling vs the scalar/pseudoscalar mass plane. In
the phase-0 (phase-1) setup, LUXE can probe an unexplored region in the mass
range from 50 MeV to 250 MeV (40 MeV to 350 MeV) and $1/\Lambda_{X}>4\times
10^{-6}$ GeV-1 ($1/\Lambda_{X}>2\times 10^{-6}$ GeV-1). The reach of the LUXE-
NPOD is comparable to the projection of other future experiments such as NA62
(in dump mode) [15] and FASER2 in the HL-LHC run [16].

Figure 3: The projected reach of the LUXE-NPOD proposal in the effective
coupling ($1/\Lambda_{X}$) vs the mass ($m_{X}$) parameter space. The phase 0
(1) result is shown in blue (black). The currently existing bounds are shown
in grey region.

##  4 Conclusion

In this proceeding, a novel way of searching for feebly interacting spin-0
scalar or pseudoscalar particle has been described. This search uses free
streaming hard photons coming from the collision between high energetic
electrons and intense laser pulse - laser pulse is used as an effective dump.
This hard photon flux can be directed to a physical dump to produce weakly
interacting spin-0 scalar or pseudoscalar particles. The signature of the BSM
particles (two hard photons) can be detected by an EM calorimeter in an
effectively background free way. The BSM detector setup can be used in a
parasitic mode in the proposed LUXE experiment at DESY without interfering
with the main physics goal of this experiment.

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