# Solution on strong partition of $2$-balanced regular multipartite
tournaments

Jiangdong Ai, Fankang He, Yihang Liu Corresponding author. School of
Mathematical Sciences and LPMC, Nankai University, Tianjin 300071, P.R. China.
Email<EMAIL_ADDRESS>School of Mathematical Sciences and LPMC, Nankai
University, Tianjin 300071, P.R. China. Email:
<EMAIL_ADDRESS>of Mathematical Sciences and LPMC, Nankai
University, Tianjin 300071, P.R. China. Email<EMAIL_ADDRESS>
###### Abstract

We call a partition of a $c$-partite tournament into tournaments of order $c$
is strong if each tournament is strongly connected. The strong partition
number denoted as $ST(r)$, represents the minimum integer $c^{\prime}$ such
that every regular $r$-balanced $c$-partite tournament has a strong partition
with $c\geq c^{\prime}$.

Figueroa, Montellano-Ballesteros and Olsen showed the existence of $ST(r)$ for
all $r\geq 2$ and proved that $5\leq ST(2)\leq 7$. In this note, we establish
that $ST(2)=6$ and we also show the unique $2$-balanced $5$-partite tournament
which has no strong partition.

Keywords: Multi-partite tournament; Strong partition; Control
AMS classification: 05C70, 05C20

##  1 Introduction

Notation follows [2]. Here, we provide only the definitions relevant to this
note. A digraph $D$ is _strong_ if there is a dipath between every ordered
vertex pair of $V(D)$. A _strong- $k$-partition_ of a digraph $D=(V,A)$ is a
partition $(V_{1},...,V_{k})$ of $V$ into $k$ non-empty disjoint sets such
that the induced digraphs of these vertex subsets $D[V_{1}],...,D[V_{k}]$ are
strong. For $c\geq 3$, a _$c$ -partite tournament_ is a digraph obtained from
a complete $c$-partite graph by orienting each edge. Let $G$ be a $c$-partite
tournament with partite sets $\\{V_{i}\\}_{i=1}^{c}$. We say $G$ is _$r$
-balanced_ if all the partite sets are of the same size $r$. We denote a
$r$-balanced $c$-partite tournament $G$ by $G_{r,c}$. A partition of $G_{r,c}$
into maximal tournaments is a spanning subgraph of $G_{r,c}$ formed by $r$
pairwise vertex-disjoint tournaments of order $c$. In the rest of this paper,
a partition of $G_{r,c}$ always means a partition of $G_{r,c}$ into maximal
tournaments.

Being strong is a basic property in digraphs. Partitioning a digraph into
exactly $k$ strong subdigraphs has been extensively studied. Bang-Jensen,
Cohen and Havet [1] proved that determining whether a digraph $D$ can be
partitioned into $k$ strong subgraphs is $\mathcal{NP}$-hard. Reid [4] proved
that any $2$-strong $n$-vertex tournament with $n\geq 8$ can be partitioned
into two strong subtournaments. Now we pay attention to partitioning a regular
balanced multipartite tournament, instead of tournaments, into strong maximal
tournaments. This problem also comes from Volkmann [5]: How close to regular
must a c-partite tournament be to guarantee a strong $c$-vertex tournament?
And in a subsequent paper, Volkmann and Winzen [6] conjectured that each
vertex of a regular $c$-partite tournament with $c\geq 5$ partite sets is
contained in a strong tournament of order $c$. This conjecture is proved by G.
Xu, S. Li, H. Li and Q. Guo [7] for the case when $c\geq 16$. Partitioning a
regular balanced $c$-partite tournament into strong maximal tournaments is a
stronger question, compared with the former conjecture, of whether this
$c$-partite tournament has many disjoint strong $c$-vertex tournaments.

A partition of $G_{r,c}$ is called a _strong partition_ (st-partition for
short) if every maximal tournament of this partition is strong, otherwise it
is called a _non-strong partition_. The _st-partition number_ , denoted by
$ST(r)$, is the minimum integer $c^{\prime}$ such that every $G_{r,c}$ has a
st-partition if $c\geq c^{\prime}$. This function was first defined by
Figueroa, Montellano-Ballesteros and Olsen [3], and they showed this function
is well defined, i.e. such $c^{\prime}$ exists for any integer $r\geq 2$.
Furthermore, they find a rough range for $ST(2)$.

######  Theorem 1.1 ([3]).

$5\leq ST(2)\leq 7$.

In this note, we determine the exact value of $ST(2)$.

######  Theorem 1.2.

$ST(2)=6$.

##  2 Preliminaries

A $2$-balanced $5$-partite tournament $G_{2,5}$ is called _nearly regular_ if
for any $v\in V(G_{2,5})$ we have $|d^{+}(v)-d^{-}(v)|\leq 2$. In this paper,
all the $2$-balanced $5$-partite tournaments we considered are nearly regular,
without special statements. In this section, we explore the structure of such
$2$-balanced $5$-partite tournaments. Furthermore, we give an intuitive idea
to construct a regular $2$-balanced $5$-partite tournament $G_{2,5}$ with no
st-partition, which implies that $ST(2)\geq 6$.

######  Definition 2.1 (control①).

Let $G_{2,c}$ be a regular $2$-balanced $c$-partite tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[c]$. For two distinct partite sets
$V_{i}$ and $V_{j}$, We say that $V_{i}$ _control①_ $V_{j}$, denoted by
$V_{i}\stackrel{{\scriptstyle 1}}{{\to}}V_{j}$, if there are exactly two arcs
from $V_{i}$ to $V_{j}$ and $|N^{+}(v_{i,1})\cap V_{j}|=|N^{+}(v_{i,2})\cap
V_{j}|=|N^{+}(v_{j,1})\cap V_{i}|=|N^{+}(v_{j,2})\cap V_{i}|=1$.

######  Definition 2.2 (control②).

Let $G_{2,c}$ be a regular $2$-balanced $c$-partite tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[c]$. For two distinct partite sets
$V_{i}$ and $V_{j}$, we say that $V_{i}$ _control②_ $V_{j}$, denoted by
$V_{i}\stackrel{{\scriptstyle 2}}{{\to}}V_{j}$, if there are exactly two arcs
from $V_{i}$ to $V_{j}$ and $|N^{+}(v_{j,1})\cap V_{i}|\neq
1,|N^{+}(v_{j,2})\cap V_{i}|\neq 1$.

######  Definition 2.3 (control③, control④).

Let $G_{2,c}$ be a regular 2-balanced $c$-partite tournament with partite sets
$V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[c]$. For two distinct partite sets $V_{i}$
and $V_{j}$, we say that $V_{i}$ _control③_ $V_{j}$, denoted by
$V_{i}\stackrel{{\scriptstyle 3}}{{\to}}V_{j}$, if there are exactly three
arcs from $V_{i}$ to $V_{j}$. We say that $V_{i}$ _control④_ $V_{j}$, denoted
by $V_{i}\stackrel{{\scriptstyle 4}}{{\to}}V_{j}$, if there are exactly four
arcs from $V_{i}$ to $V_{j}$.

We want to emphasize that throughout the entirety of the text, we may
repeatedly substitute ’control’ for ’controls’, disregarding grammatical
correctness.

The above four relationships between two partite sets of $G_{2,c}$ are shown
in Figure 1. We say a vertex $u\in V$ is _good to $V_{i}$_ if $u\to v_{i,1}$
and $v_{i,2}\to u$ or $u\to v_{i,2}$ and $v_{i,1}\to u$. And $V_{i}$ is _good
to $V_{j}$_, if both $v_{i,1}$ and $v_{i,2}$ are good to $V_{j}$. Note that if
$V_{i}$ is good to $V_{j}$ then $V_{i}\stackrel{{\scriptstyle 1}}{{\to}}V_{j}$
or $V_{i}\stackrel{{\scriptstyle 2}}{{\to}}V_{j}$. Let $v$ be a vertex in $D$,
the _minimum degree_ of $v$ is $\delta(v):=\min\\{d^{+}(v),d^{-}(v)\\}$ and
the _minimum degree_ of $D$ is $\delta(D):=\min_{v\in V(D)}\\{\delta(v)\\}$. A
vertex $v$ is _regular_ if $d^{+}(v)=d^{-}(v)$, otherwise we say $v$ is
_irregular_. Note that if $V_{i}$ is good to $V_{j}$, then
$V_{i}\stackrel{{\scriptstyle 1}}{{\to}}V_{j}$ or
$V_{i}\stackrel{{\scriptstyle 2}}{{\to}}V_{j}$.

$v_{i,1}$$v_{j,2}$$v_{i,2}$$v_{j,1}$ (a) $V_{i}\stackrel{{\scriptstyle
1}}{{\to}}V_{j}$

$v_{i,1}$$v_{j,2}$$v_{i,2}$$v_{j,1}$ (b) $V_{i}\stackrel{{\scriptstyle
2}}{{\to}}V_{j}$

$v_{i,1}$$v_{j,2}$$v_{i,2}$$v_{j,1}$ (c) $V_{i}\stackrel{{\scriptstyle
3}}{{\to}}V_{j}$

$v_{i,1}$$v_{j,2}$$v_{i,2}$$v_{j,1}$ (d) $V_{i}\stackrel{{\scriptstyle
4}}{{\to}}V_{j}$

Figure 1: Four relationships between two partite sets

######  Lemma 2.4 (Folklore).

Let T be a tournament of order c. If T is non-strong then
$\delta(T)\leq\left\lfloor\frac{c-2}{4}\right\rfloor$.

Note that for a tournament $T$ on $5$ vertices, $T$ is non-strong if and only
if $\delta(T)=0$.

######  Lemma 2.5 ([3]).

The number of partitions of any $G_{r,c}$ into maximal tournaments is
$(r!)^{c-1}$

Given a nearly regular $2$-balanced $5$-partite tournament $G_{2,5}$ with
partite sets $V_{i}=\\{v_{i,1},v_{i,2}\\}$ for $i=[5]$. Let $N_{1}$ be the
number of st-partitions of $G_{2,5}$. Let $N_{2}$ be the number of non-strong
partitions of $G_{2,5}$. By Lemma 2.5, the number of partitions of $G_{2,5}$
into maximal tournaments is $16$, i.e. $N_{1}+N_{2}=16$. Choose a vertex $u\in
V(G_{2,5})$ and a partition $\tau$, let $T_{u,\tau}$ denote the maximal
tournament containing $u$ under partition $\tau$. Let $n(u)$ be the number of
partitions $\tau$ such that $\delta_{T_{u,\tau}}(u)=0$. Then $n(u)\leq 2$. Let
$n(V_{i})$ be the number of non-strong partitions $\tau$ such that there
exists a vertex $u\in V_{i}$ with $\delta_{T_{u,\tau}}(u)=0$. Then
$n(V_{i})\leq n(v_{i,1})+n(v_{i,2})\leq 4$. Note that
$N_{2}\leq\sum_{i=1}^{5}n(V_{i})$ since for each non-strong partition $\tau$
of $G_{2,5}$, there exists a maximal tournament $T$ under partition $\tau$
such that $T$ is non-strong, i.e. there is a vertex $u\in V(T)$ with
$\delta_{T}(u)=0$. If $\sum_{i=1}^{5}n(V_{i})<16$, then $G_{2,5}$ has a st-
partition. The idea to construct a regular $5$-partite tournament $G_{2,5}$
with no st-partition is to make $\sum_{i=1}^{5}n(V_{i})$ as large as possible.

We say $V_{i}$ is _regular_ if both $v_{i,1}$ and $v_{i,2}$ are regular in
$G_{2,5}$; $V_{i}$ is _semi-regular_ if there is exactly one regular vertex
$u\in V_{i}$ in $G_{2,5}$; $V_{i}$ is _irregular_ if both $v_{i,1}$ and
$v_{i,2}$ are irregular in $G_{2,5}$. We have the following useful
observations.

######  Observation 2.6.

Let $G_{2,5}$ be a nearly regular $2$-balanced $5$-tournament. For a vertex
$u\in V_{i}$, $n(u)\in\\{0,2\\}$. In particular, if $u$ is regular, $n(u)=2$
if and only if $u$ is good to any other partite sets. Moreover, if $u$ is
irregular, $n(u)=2$ if and only if $u$ is good to the other three partite
sets.

By studying a nearly regular $2$-balanced $5$-partite tournament $G_{2,5}$, we
attain the following two lemmas.

######  Lemma 2.7.

Let $G_{2,5}$ be a nearly regular $2$-balanced $5$-tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[5]$. For any (semi-)regular partite set
$V_{i}$ in $V(G_{2,5})$, if $n(V_{i})=4$, then $V_{i}$ control② some partite
set $V_{j}$ where $j\neq i$.

###### Proof.

Assume the contrary that no partite set $V_{j}$ is controlled② by $V_{i}$.

When $V_{i}$ is regular, since $n(V_{i})=4$, we have
$n(v_{i,1})=n(v_{i,2})=2$. Then $v_{i,1}$ and $v_{i,2}$ are good to other
partite sets. So $V_{i}$ is good to other partite sets. Since there is no
partite set being controlled② by $V_{i}$, then $V_{i}$ control① all other
partite sets. For any partition $\tau$ of $G_{2,5}$, if
$\delta_{T_{v_{i,1},\tau}}(v_{i,1})=0$ , then
$\delta_{T_{v_{i,2},\tau}}(v_{1,2})=0$. So $n(V_{i})=2$. It’s a contradiction.

When $V_{i}$ is semi-regular, there is a vertex in $V_{i}$, say $v_{i,1}$,
such that $|d^{+}(v_{i,1})-d^{-}(v_{i,1})|=2$. W.l.o.g., let
$d^{+}(v_{i,1})-d^{-}(v_{i,1})=2$. Since $n(V_{i})=4$, we have
$n(v_{i,1})=n(v_{i,2})=2$. So $v_{i,2}$ is good to all other partite sets, and
$v_{i,1}$ is good to all other partite sets except for one. Since there is no
partite set being controlled② by $V_{i}$, then $V_{i}$ control③ $V_{j}$ such
that $N^{+}(v_{i,1})\supseteq V_{j}$ and control① $V_{k}$, $k\neq i,j$. But in
this case, $n(V_{i})=3$, it’s a contradiction.

######  Lemma 2.8.

Let $G_{2,5}$ be a nearly regular $2$-balanced $5$-tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[5]$. If a partite set $V_{i}$ in
$V(G_{2,5})$ is controlled② by two other partite sets, then $n(V_{i})=0$. In
particular, if a regular part $V_{i}$ in $V(G_{2,5})$ is controlled② by
another partite set, then $n(V_{i})=0$.

###### Proof.

Let $V_{j}\stackrel{{\scriptstyle 2}}{{\to}}V_{i}$ and
$V_{\ell}\stackrel{{\scriptstyle 2}}{{\to}}V_{i}$. We consider the following
two cases.

Case 1: $N^{+}(V_{j})\cap V_{i}=N^{+}(V_{\ell})\cap V_{i}$.
W.l.o.g., let $N^{+}(V_{j})\cap V_{i}=v_{i,1}$. Note that in this case,
$N^{-}(V_{j})\cap V_{i}=N^{-}(V_{\ell})\cap V_{i}=v_{i,2}$. As $G_{2,5}$ is
nearly regular, there is another partite set $V_{q}(q\neq i,j,\ell)$ such that
$V_{q}\subseteq N^{+}(v_{i,1})$. Then there is no partition $\tau$ such that
$T_{v_{i,1},\tau}$ with $\delta(T_{v_{i,1},\tau})=0$, i.e $n(v_{i,1})=0$.
Similar argument holds for $v_{i,2}$, we have $n(v_{i,2})=0$. Thus, we
conclude that $n(V_{i})=0$.

Case 2: $N^{+}(V_{j})\cap V_{i}\neq N^{+}(V_{\ell})\cap V_{i}$.
W.l.o.g., let $N^{+}(V_{j})\cap V_{i}=v_{i,1}$. we have $V_{j}\subseteq
N^{-}(v_{i,1}),V_{\ell}\subseteq N^{+}(v_{i,1})$ and $V_{\ell}\subseteq
N^{-}(v_{i,2}),V_{j}\subseteq N^{+}(v_{i,2})$. Then $n(v_{i,1})=0$ and
$n(v_{i,2})=0$, which yields that $n(V_{i})=0$.

Now let $V_{i}$ be regular, and $V_{j}\stackrel{{\scriptstyle
2}}{{\to}}V_{i}$. W.l.o.g., let $N^{+}(v_{i,1})\supseteq V_{j}$ and
$N^{-}(v_{i,2})\supseteq V_{j}$. In order to make $v_{i,1}$ regular, there is
another partite set $V_{q}(q\neq i,j)$ such that $V_{j}\subseteq
N^{+}(v_{i,1})$. Then there is no partition $\tau$ such that
$T_{v_{i,1},\tau}$ with $\delta_{T_{v_{i,1},\tau}}(v_{i,1})=0$, i.e
$n(v_{i,1})=0$. Similar argument holds for $v_{i,2}$, we have $n(v_{i,2})=0$.
Thus, $n(V_{i})=0$.

We say a partition $\tau$ of $G_{2,5}$ is a _good partition_ if one of the two
maximal tournaments is strong and the other is non-strong with exactly one
vertex whose minimum degree is $0$ under $\tau$.

######  Lemma 2.9.

Let $G_{2,5}$ be a nearly regular $2$-balanced $5$-tournament. If $G_{2,5}$
has no st-partition, then $G_{2,5}$ has at least 12 good partitions.

###### Proof.

By Lemma 2.5, the number of partitions of $G_{2,5}$ into maximal tournaments
is $16$. Under these partitions, there are $32$ maximal tournaments. And the
number of non-strong maximal tournaments is at most $\sum_{u\in
V(G_{2,5})}n(u)\leq 10\times 2=20$. Each partition contains at least one non-
strong maximal tournament since $G_{2,5}$ has no st-partition. Therefore,
$G_{2,5}$ has at least $12$ good partitions.

##  3 The unique regular $2$-balanced $5$-partite tournament with no st-
partition

We first explore the structure of regular $2$-balanced $5$-partite tournaments
with no st-partition.

######  Theorem 3.1.

Let $G_{2,5}$ be a regular $2$-balanced $5$-partite tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[5]$. If $G_{2,5}$ has no st-partition,
then there is a partite set $V_{i}$ such that $V_{j}\stackrel{{\scriptstyle
2}}{{\to}}V_{i}$ for all $j\in[5]\setminus\\{i\\}$. Furthermore, the other
four partite sets control① each other.

###### Proof.

Since $G_{2,5}$ has no st-partition, $\sum_{i=1}^{5}n(V_{i})\geq 16$. So,
there is a partite set $V_{j}$ such that $n(V_{j})=4$. By Lemma 2.7, there
exists a partite set $V_{i}$ such that $V_{j}\stackrel{{\scriptstyle
2}}{{\to}}V_{i}$. By the fact that $V_{i}$ is regular and Lemma 2.8,
$n(V_{i})=0$. In order to make $\sum_{i=1}^{5}n(V_{i})\geq 16$, we have
$n(V_{j})=4$ for all $j\in[5]$ and $j\neq i$. So, for any vertex
$u\in{G_{2,5}}/{\\{v_{i,1},v_{i,2}\\}}$, $u$ is good to all other partite
sets. Then $V_{j}$ is good to partite sets for all $j\in[5],j\neq i$. So
$V_{i}$ is controlled② by all other four partite sets. And the other four
partite sets controll① each other.

Let $G_{1}$ and $G_{2}$ be the digraphs as shown in Figure 2.

$v_{1,1}$$v_{1,2}$$v_{2,1}$$v_{2,2}$$v_{3,1}$$v_{3,2}$$v_{4,1}$$v_{4,2}$ (a)
$G_{1}$

$v_{1,1}$$v_{1,2}$$v_{2,1}$$v_{2,2}$$v_{3,1}$$v_{3,2}$$v_{4,1}$$v_{4,2}$ (b)
$G_{2}$

Figure 2: Suitable $G_{2,4}$

######  Lemma 3.2.

Let $G_{2,4}$ be a regular $2$-balanced $4$-partite tournament with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[4]$. If all partite sets control① each
other, then $G_{2,4}$ is isomorphic to $G_{1}$ or $G_{2}$.

###### Proof.

In this proof, we say that $V_{i}$ control① $V_{j}$ if $v_{i,1}$ dominates
$v_{j,1}$ and $v_{i,2}$ dominates $v_{j,2}$, otherwise we say $V_{j}$ control①
$V_{i}$. Now we construct an auxiliary tournament $T(G_{2,4})$ of $G_{2,4}$ to
record the relationships between partite sets of $G_{2,4}$. Let
$V(T(G_{2,4}))=\\{V_{1},V_{2},V_{3},V_{4}\\}$, and $V_{i}V_{j}$ is an arc of
$T(G_{2,4})$ if and only if $V_{i}$ control① $V_{j}$. Note that the
construction can be reversed: given a $4$-vertex tournament $T$ and a
numbering of its vertex set $\\{v_{1},v_{2},v_{3},v_{4}\\}$, there is a
regular $2$-balanced $4$-partite tournament $G_{2,4}$ with partite sets
$V_{i},i\in[4]$ such that $T(G_{2,4})=T$. Let $G^{\prime}_{2,4}$ be the
$2$-balanced $4$-partite tournaments obtained from $G_{2,4}$ by replacing
partite set $V_{4}=\\{v_{4,1},v_{4,2}\\}$ by
$V^{\prime}_{4}=\\{v^{\prime}_{4,1},v^{\prime}_{4,2}\\}$ in which
$v^{\prime}_{4,1}=v_{4,2}$ and $v^{\prime}_{4,2}=v_{4,1}$. Then
$G^{\prime}_{2,4}$ is isomorphic to $G_{2,4}$. And the auxiliary graph
$T(G^{\prime}_{2,4})$ can be obtained from $T(G_{2,4})$ by reversing arcs
incident to $V_{4}$.

There are exactly four kinds of $4$-vertex tournament up to isomorphism as
shown in Figure 3. Note that $T_{1}$ can be obtained from $T_{2}$ by reversing
all arcs incident to $v_{4}$ and $T_{3}$ is isomorphic to the tournament
obtained from $T_{4}$ by reversing all arcs incident to $v_{3}$. Thus, the
corresponding $G_{2,4}$ of $T_{1}$ is isomorphic to the corresponding
$G_{2,4}$ of $T_{2}$ and the corresponding $G_{2,4}$ of $T_{3}$ is isomorphic
to the corresponding $G_{2,4}$ of $T_{4}$.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$ (a) $T_{1}$

$v_{1}$$v_{2}$$v_{3}$$v_{4}$ (b) $T_{2}$

$v_{1}$$v_{2}$$v_{3}$$v_{4}$ (c) $T_{3}$

$v_{1}$$v_{2}$$v_{3}$$v_{4}$ (d) $T_{4}$

Figure 3: Four kinds of 4-vertex tournaments

The corresponding $G_{2,4}$ of $T_{1}$ is $G_{1}$ and the corresponding
$G_{2,4}$ of $T_{3}$ is $G_{2}$. This means that such $G_{2,4}$ is isomorphic
to $G_{1}$ or $G_{2}$.

######  Theorem 3.3.

There is a unique regular 2-balanced 5-partite tournament $H_{0}$ with no st-
partition.

###### Proof.

Suppose the partite sets of $G_{2,5}$ are
$V_{i}=\\{v_{i,1},v_{i,2}\\},i\in[5]$. By Lemma 3.1, we know that there is a
partite set of $G_{2,5}$, say $V_{5}$, being controlled② by some other partite
sets. As $V_{5}$ is regular, the arcs incident to $V_{5}$ would be as shown in
Figure 4.

$v_{5,1}$$v_{5,2}$$v_{i,1}$$v_{i,2}$$v_{j,1}$$v_{j,2}$$v_{k,1}$$v_{k,2}$$v_{\ell,2}$$v_{\ell,2}$
Figure 4: Arcs incident to $V_{5}$

Let $G:=G_{2,5}\setminus V_{5}$. Then $G$ is a regular $2$-balanced
$4$-partite tournament and the partite sets of $G$ control① each other. By
Lemma 3.2, we have that $G$ is isomorphic to $G_{1}$ or $G_{2}$.

But for $G_{2}$, there is a partition
$V_{a}=\\{v_{1,1},v_{2,2},v_{3,1},v_{4,2}\\},V_{b}=\\{v_{1,2},v_{2,1},v_{3,2},v_{4,1}\\}$
such that both $D[V_{a}]$ and $D[V_{b}]$ are strong. Then
$V_{a}=\\{v_{1,1},v_{2,2},v_{3,1},v_{4,2},v_{5,1}\\},V_{b}=\\{v_{1,2},v_{2,1},v_{3,2},v_{4,1},v_{5,2}\\}$
would be a st-partition of $G_{2,5}$. Now let $H_{0}$ be the joint of $G_{1}$
and partite sets $V_{5}=\\{v_{5,1},v_{5,2}\\}$ such that $V_{5}$ is
controlled② by all other partite sets. Note such $H_{0}$ is unique up to
isomorphism. Any partition of $H_{0}$ contains exactly one non-strong maximal
tournament. Therefore, $H_{0}$ is the unique regular $2$-balanced $5$-partite
tournament with no st-partition, up to isomorphism.

Thus, we have the following lemma.

######  Lemma 3.4.

$ST(2)\geq 6$.

##  4 Every regular $2$-balanced $6$-partite tournament has a st-partition

Given a regular $2$-balanced $6$-partite tournament $G_{2,6}$ with partite
sets $V_{i}=\\{v_{i,1},v_{i,2}\\}$ for $i=[6]$, observe that after deleting
$V_{6}$ from $G_{2,6}$, we get a nearly regular $2$-balanced $5$-partite
tournament $G_{2,5}$ with partite sets $V_{i}=\\{v_{i,1},v_{i,2}\\},i=[5]$.
The following lemmas are useful in exploring the structure of $G_{2,6}$.

######  Lemma 4.1.

For any irregular part $V_{i}$ in $V(G_{2,5})$ where $i\in[5]$, if
$n(V_{i})=4$, then $V_{i}$ control② another partite set $V_{j}$ or $V_{i}$ is
controlled② by $V_{6}$.

###### Proof.

W.l.o.g., let $V_{1}$ be an irregular partite set in $G_{2,5}$. If
$d^{+}_{G_{2,5}}(v_{1,1})-d^{-}_{G_{2,5}}(v_{1,1})=2$ and
$d^{+}_{G_{2,5}}(v_{1,2})-d^{-}_{G_{2,5}}(v_{1,2})=-2$, we have that $V_{1}$
is controlled② by $V_{6}$. So, we only need to consider the case that the in-
degree( or out-degree) of $v_{1,1}$ and $v_{1,2}$ are $5$. For convenience, we
assume that $d^{+}_{G_{2,5}}(v_{1,1})=d^{+}_{G_{2,5}}(v_{1,2})=5$. Observe
that $n(v_{1,i})=2$ if and only if $v_{1,i}$ is good to exactly three partite
sets. If $V_{1}$ is good to $V_{j}$, $V_{1}$ will control① or control②
$V_{j},j\in[5]$ and $i\neq j$. Then we consider the following two cases.

Case 1: $v_{1,1}$ and $v_{1,2}$ are good to the same part.
If $V_{1}$ doesn’t control② any partite set, $V_{1}$ control④ a partite set
and control① other three partite sets. Observe that $n(V_{1})$ is $2$ in this
case, it’s a contradiction.

Case 2: $v_{1,1}$ and $v_{1,2}$ are good to different partite sets.
So, $V_{1}$ control③ two different partite sets. If $V_{1}$ doesn’t control②
any partite set, then $V_{1}$ will control① other two partite sets. Observe
that $n(V_{1})=2$ in this case, it’s also a contradiction.

Therefore, $v_{i}$ control② another part $V_{j}$ or $V_{i}$ is controlled② by
$V_{6}$.

######  Lemma 4.2.

Let $G_{2,6}$ be a regular 2-balanced $6$-partite tournament. If $G_{2,6}$ has
no st-partition, then it contains a $2$-balanced $5$-partite tournament
$G_{2,5}$ with no st-partition as a subgraph.

###### Proof.

Assume the contrary that for any $2$-balanced 5-partite tournament $G_{2,5}$
in $G_{2,6}$, there exists a st-partition of $G_{2,5}$. We claim that the
relations between any two partite sets are control① or control③ in $G_{2,6}$.
Let $G_{i}$ be the $2$-balanced 5-partite tournament we get after deleting
$V_{i}$ from $G_{2,6}$. By assumption, $G_{i}$ has a st-partition into $T_{1}$
and $T_{2}$. Then there exists a vertex $u\in V_{i}$ such that
$N^{+}_{G_{2,6}}(u)=V(T_{1}),N^{-}_{G_{2,6}}(u)=V(T_{2})$ or
$N^{+}_{G_{2,6}}(u)=V(T_{2}),N^{-}_{G_{2,6}}(u)=V(T_{1})$. Otherwise we can
add $v_{i,1}$ to $T_{1}$ and $v_{i,2}$ to $T_{2}$, then we get a st-partition
of $G_{2,6}$, a contradiction. Note that such vertex $u\in V_{i}$ is a vertex
that is good to each other partite set in $G_{2,6}$, we call such vertex a
good vertex; and if both vertices of $V_{i}$ are good vertices, we call
$V_{i}$ a good partite set. Then for every partite set, there is a good
vertex. Thus, a partite set neither control② nor control④ other partite sets,
otherwise we can not find a good vertex in each partite set. Moreover, a
partite set can control③ at most two partite sets. Now, we need the following
claim.

######  Claim 4.3.

Let $G_{2,5}:=G_{2,6}\setminus V_{\ell}$, then
$\sum_{i\neq\ell}n_{G_{2,5}}(V_{i})\geq 14$. Moreover, if
$\sum_{i\neq\ell}n_{G_{2,5}}(V_{i})=14$ and $G_{2,6}$ has no st-partition,
then $V_{\ell}$ is good to each other partite set in $G_{2,6}$.

###### Proof of Claim 4.3.

Suppose that $\tau$ is a st-partition of $G_{2,5}$ and $\tau$ partitions
$G_{2,5}$ into $T^{\tau}_{1}$ and $T^{\tau}_{2}$. Since $G_{2,6}$ has no st-
partition, then there is vertex $u\in V_{\ell}$ such that
$N^{+}(u)=V(T^{\tau}_{1}),N^{-}(u)=V(T^{\tau}_{2})$ or
$N^{-}(u)=V(T^{\tau}_{1}),N^{+}(u)=V(T^{\tau}_{2})$. This means that $u$ is a
good vertex. Now assume that $\sum_{i\neq\ell}n_{G_{2,5}}(V_{i})\leq 13$, then
there are at least three st-partitions of $G_{2,5}$. This implies that there
are at least three different vertices in $V_{\ell}$ since for different st-
partitions $\tau$ and $\tau^{\prime}$, we have $V(T^{\tau}_{1})\neq
V(T^{\tau^{\prime}}_{1}),V(T^{\tau}_{2})\neq V(T^{\tau^{\prime}}_{2})$ and
$V(T^{\tau}_{1})\neq V(T^{\tau^{\prime}}_{2})$. While $|V_{\ell}|=2$, and so
it is a contradiction. Thus, $\sum_{i\neq\ell}n_{G_{2,5}}(V_{i})\geq 14$. And
if $\sum_{i\neq\ell}n_{G_{2,5}}(V_{i})=14$, then there are two st-partitions
of $G_{2,5}$. So both two vertices of $V_{\ell}$ are good vertices.

Then, we consider the following two cases:

Case 1: There are no partite sets $V_{i}$ and $V_{j}$ such that
$V_{i}\stackrel{{\scriptstyle 3}}{{\to}}V_{j}$ in $G_{2,6}$.
Let $G:=D[\bigcup_{1\leq i\leq 5}V_{i}]$, then $G$ is a regular $2$-balanced
$5$-partite tournament. Note that $n(V_{i})=2$ for all $i\in[5]$ since for any
two partite sets $V_{1}$ and $V_{2}$, we have $V_{i}\stackrel{{\scriptstyle
1}}{{\to}}V_{j}$. Then $\sum_{i=1}^{5}n_{G}(V_{i})=10$, a contradiction to
Claim 4.3.

Case 2: There are partite sets $V_{1}$ and $V_{2}$ such that
$V_{1}\stackrel{{\scriptstyle 3}}{{\to}}V_{2}$ in $G_{2,6}$.
Let $D$ be a digraph. Now we define the _co-out-neighborhood_ of a vertex
subset $V_{0}$ is $N_{co}^{+}(V_{0})=\cap_{v\in V_{0}}N^{+}(v)$, and the _co-
in-neighborhood_ of a vertex subset $V_{0}$ is $N_{co}^{-}(V_{0})=\cap_{v\in
V_{0}}N^{-}(v)$.

As $V_{1}$ is regular, $V_{1}$ is controlled③ by another partite set, say
$V_{3}$, such that $N^{-}_{co}(V_{2})\cap V_{1}=N^{+}_{co}(V_{3})\cap V_{3}$.
And as $V_{3}$ is regular, $V_{3}$ is controlled③ by another partite set
$V_{j},j\neq 1,3$, such that $N^{-}_{co}(V_{1})\cap
V_{3}=N^{+}_{co}(V_{j})\cap V_{3}$.

Subcase 2.1: $j=2$.
If $N^{-}_{co}(V_{3})\cap V_{2}\neq N^{+}_{co}(V_{1})\cap V_{2}$, then as
$V_{2}$ is regular in $G_{2,6}$, there must be another two partite sets
$V_{\ell}$ and $V_{\ell^{\prime}}$ such that $V_{2}$ is controlled③ by
$V_{\ell}$ and $V_{2}$ control③ $V_{\ell^{\prime}}$. Besides,
$N^{-}_{co}(V_{3})\cap V_{2}=N^{+}_{co}(V_{\ell})\cap V_{2}$ and
$N^{-}_{co}(V_{\ell^{\prime}})\cap V_{2}=N^{+}_{co}(V_{1})\cap V_{2}$. Then
there remains a partite set $V_{k}$ such that
$k\not\in\\{1,2,3,\ell,\ell^{\prime}\\}$. Let $G:=G_{2,6}\setminus V_{k}$,
then $n_{G}(V_{1})=0,n_{G}(V_{2})\leq 2$ and $n_{G}(V_{3})\leq 2$. As a
result, $\sum_{i\neq k}n_{G}(V_{i})\leq 0+2\times 2+2\times 4=12$, a
contradiction to Claim 4.3. Now we consider the case when
$N^{-}_{co}(V_{3})\cap V_{2}=N^{+}_{co}(V_{1})\cap V_{2}$. If for any
$V_{j},j\in\\{4,5,6\\}$, $V_{j}$ does not control③ any other partite sets, let
$G:=D[\bigcup_{1\leq i\leq 5}V_{i}]$. Then $n_{G}(V_{i})\leq 2$ for all
$i\in[5]$. As a result, $\sum_{i}^{5}n_{G}(V_{i})\leq 10$, a contradiction to
Claim 4.3. If there is a partite set $V_{k},k\in\\{4,5,6\\}$ such that $V_{k}$
control③ some partite set $V_{\ell},\ell\neq k$, let $G:=G_{2,6}\setminus
V_{k}$, then $n_{G}(V_{1})\leq 2,n_{G}(V_{2})\leq 2$ and $n_{G}(V_{3})\leq 2$.
So $\sum_{i\neq k}n_{G}(V_{i})\leq 14$. Observe that $V_{k}$ is not a good
partite set in $G_{2,6}$ since $V_{k}$ control③ $V_{\ell}$. This contradicts
to Claim 4.3.

Subcase 2.2. $j\geq 4$.
W.l.o.g., let $j=4$. Again, because $V_{4}$ is regular, $V_{4}$ is controlled③
by another partite set $V_{\ell},\ell\neq 3$, such that $N^{-}_{co}(V_{1})\cap
V_{3}=N^{+}_{co}(V_{j})\cap V_{3}$. If $\ell\neq 2$, let $G:=G_{2,6}\setminus
V_{2}$. Then $n_{G}(V_{1}),n_{G}(V_{3}),n_{G}(V_{4})\leq 2$, and so
$\sum_{i\neq 2}n_{G}(V_{i})\leq 3\times 2+2\times 4=14$. Observe that $V_{2}$
is not good in $G_{2,6}$, since $V_{2}$ is controlled③ by $V_{1}$. It is a
contradiction to Claim 4.3. Now it remains to consider the case when $\ell=2$.
If $N^{-}_{co}(V_{4})\cap V_{2}=N^{+}_{co}(V_{1})\cap V_{2}$, let
$G:=D[\bigcup_{1\leq i\leq 5}V_{i}]$. Then $n_{G}(V_{i})\leq 2$ for all
$i\in[4]$ and $n_{G}(V_{5})\leq 4$. Therefore $\sum_{i=1}^{5}n_{G}(V_{i})\leq
12$, a contradiction to Claim 4.3.

######  Lemma 4.4.

Let $G_{2,6}$ be a regular $2$-balanced $6$-partite tournament. If $G_{2,6}$
has no st-partition, there are two partite sets $V_{i},V_{j}$ in $G_{2,6}$
such that $V_{i}$ control② $V_{j}$.

###### Proof.

By lemma 4.2, there is a $2$-balanced $5$-partite graph $G_{2,5}\subseteq
G_{2,6}$ and this $G_{2,5}$ has no st-partition. Thus, the sum of $n(V_{i})$
over all partite sets $V_{i}$ of $G_{2,5}$ is at least $16$. By the Pigeon
Hole Principle, there is a partite set $V_{i}$ of $G_{2,5}$ with $n(V_{i})=4$.
Then by Lemma 2.7 and Lemma 4.1, we get the desired result.

######  Theorem 4.5.

There is no regular $2$-balanced $6$-partite tournament with no st-partition.

###### Proof.

Assume the contrary that there is a regular $2$-balanced $6$-tournament
$G_{2,6}$ with no st-partition. By Lemma 4.4, there is a partite set, say
$V_{5}$, control② another partite set, say $V_{6}$. Then, there exists a
vertex $u\in V_{6}$ such that $v_{5,1}\to u$ and $v_{5,2}\to u$. W.l.o.g., let
$u$ be $v_{6,1}$. Since $v_{6,1}$ is regular in $G_{2,6}$, there is another
partite set $V_{j},j\neq 5,6$ such that $v_{6,1}\to v_{j,1}$ and $v_{6,1}\to
v_{j,2}$.

For the other vertex $v_{6,2}\in V_{6}$, we have $v_{6,2}\to v_{5,1}$ and
$v_{6,2}\to v_{5,2}$. As $v_{6,2}$ is regular, there is another partite set
$V_{j^{\prime}},j^{\prime}\neq 5,6$ such that $v_{j^{\prime},1}\to v_{6,2}$
and $v_{j^{\prime},2}\to v_{6,2}$. Deleting $V_{6}$ from $G_{2,6}$, we get a
nearly regular $2$-balanced $5$-tournament, denoted by $G_{2,5}$. Suppose
$\tau$ is a st-partition of $G_{2,5}$, then $\tau$ partitions $G_{2,5}$ into
two maximal tournaments, denoted by $T^{\tau}_{1}$ and $T^{\tau}_{2}$
respectively. Note that $G_{2,5}$ has no st-partiton, otherwise
$V(G_{2,6})=W_{1}\cup W_{2}$, in which $W_{1}=\\{v_{6,1}\\}\cup
V(T^{\tau}_{1})$ and $W_{2}=\\{v_{6,2}\\}\cup V(T^{\tau}_{2})$, would be a st-
partiton for $G_{2,6}$.

By Lemma 2.9, there are at least $12$ good partitions for $G_{2,5}$. For every
good partition $\tau$ for $G_{2,5}$, $\tau$ partitions $G_{2,5}$ into one
strong maximal tournament, denoted by $T^{\tau}_{1}$, and one non-strong
maximal tournament, denoted by $T^{\tau}_{2}$, with exactly one vertex
$v_{1}\in V(T^{\tau}_{2})$ having minimum degree $0$ in $V(T^{\tau}_{2})$.

######  Claim 4.6.

Such vertex $v_{1}$ is regular in $G_{2,5}$.

###### Proof of Claim 4.6.

Assume the contrary that $v_{1}$ is irregular in $G_{2,5}$, w.l.o.g., let
$d^{+}_{T^{\tau}_{2}}(v_{1})=4$ and name the other four vertices in
$T^{\tau}_{2}$ as $v_{2},v_{3},v_{4}$ and $v_{5}$ respectively. As $v_{1}$ is
regular in $G_{2,6}$, we have that $v_{6,1}\to v_{1}$ and $v_{6,2}\to v_{1}$.
Since $V_{6}$ is controlled② by $V_{5}$, neither $v_{6,1}$ nor $v_{6,2}$ is
good in $G_{2,6}$. Since $v_{5,1}\to v_{6,1}$ and $v_{5,2}\to v_{6,1}$, there
is a vertex $V(T_{2}^{\tau})$, say $v_{2}$, such that $v_{2}\to v_{6,1}$. If
there is a vertex in $V(T^{\tau}_{2}\setminus\\{v_{1},v_{2}\\})$, say $v_{3}$,
such that $v_{3}\to v_{2}$. Then there is a cycle $v_{1}v_{2}v_{6,1}$. Suppose
$D[v_{3}v_{4}v_{5}]$ forms a cycle. Note that $v_{1}\to v_{5}$ and $v_{3}\to
v_{2}$. Then $D[v_{6,1}\cup V(T^{\tau}_{2})]$ is a strong tournament. Recall
that $T_{1}^{\tau}$ is strong, $D[v_{6,2}\cup V(T^{\tau}_{1})]$ is also
strong. This yields a st-partition of $G_{2,6}$, a contradiction. Suppose
$D[v_{3}v_{4}v_{5}]$ forms a transitive triangle and let $u$ and $u^{\prime}$
be the sink and source of $D[v_{3}v_{4}v_{5}]$ respectively. Since $\tau$ is a
good partition, then $\delta_{T^{\tau}_{2}}(v_{i})\geq 1$ for $2\leq i\leq 5$.
Then there is an arc from $u$ to $\\{v_{1},v_{2},v_{6,1}\\}$. Together with
$v_{1}\to u^{\prime}$, we have $D[v_{6,1}\cup V(T^{\tau}_{2})]$ is a strong
tournament which will lead to that $G_{2,6}$ has a st-partition as before. As
a result, there is no vertex $v\in\\{v_{3},v_{4},v_{5}\\}$ such that $v\to
v_{2}$. Since $\delta_{T^{\tau}_{2}}(v_{i})\geq 1$ for all $2\leq i\leq 5$,
then $v_{3}v_{4}v_{5}$ would be a cycle in this case. And note that
$v_{6,1}v_{1}v_{2}$ is also a cycle. Then all arcs between these two cycles
should be from $\\{v_{3},v_{4},v_{5}\\}$ to $\\{v_{6,1},v_{1},v_{2}\\}$,
otherwise $D[v_{6,1}\cup V(T^{\tau}_{2})]$ is a strong tournament which will
lead to that $G_{2,6}$ has a st-partition as before. Then consider the
neighborhood of $v_{6,1}$, we have $v_{6,1}\to v_{1},v_{2}\to
v_{6,1},v_{6,1}\to v_{3},v_{6,1}\to v_{4}$ and $v_{6,1}\to v_{5}$, as shown in
Figure 5. By similar arguments for $v_{6,2}$, we have $v_{6,2}\to
v_{1},v_{2}\to v_{6,2},v_{6,2}\to v_{3},v_{6,2}\to v_{4}$ and $v_{6,2}\to
v_{5}$, as shown in Figure 5. This means that for any partite set
$V_{i},i\in[5]$, there is a vertex $u\in V_{i}$ such that $N^{+}(u)\supseteq
V_{6}$ and $N^{-}(u)\supseteq V_{6}$. Then no partite set $V_{i},i\in[5]$
control② $V_{6}$, otherwise both $v_{i,1}$ and $v_{i,2}$ are good to $V_{6}$.
This contradicts the fact that $V_{6}$ is controlled② by another partite set.

$v_{6,1}$$v_{6,2}$$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$ Figure 5: Arcs between
$V_{6}$ and $V(T^{\tau}_{2})$

Now, we modify the function $n(\cdot)$ to $n^{\prime}(\cdot)$ to obtain a
better approximation of the number of non-strong partitions $N_{2}$ for
$G_{2,5}$. First, we classify the non-strong partitions of $G_{2,5}$ into two
categories. Let $\tau$ be a non-strong partition of $G_{2,5}$. For a fixed
$V_{i}$, we say $\tau$ is a _useful partition_ if either there is a regular
vertex $u\in V_{i}$ and $\delta_{T_{u,\tau}}(u)=0$ or if $V_{i}$ is irregular
and $\delta_{T_{v_{i,1},\tau}}(v_{i,1})=\delta_{T_{v_{i,2},\tau}}(v_{i,2})=0$;
we say $\tau$ is a _useless partition_ if there is exactly one vertex $v\in
V_{i}$ such that $\delta_{T_{v,\tau}}(v)=0$ and $v$ is irregular. Let
$n_{1}(V_{i})$ be the number of useful partitions for $V_{i}$, and
$n_{2}(V_{i})$ be the number of useless partitions for $V_{i}$. Note that
$n(V_{i})=n_{1}(V_{i})+n_{2}(V_{i})$. Define
$n^{\prime}(V_{i})=n_{1}(V_{i})+\frac{1}{2}n_{2}(V_{i})$.

######  Claim 4.7.

$\sum_{i=1}^{5}n^{\prime}(V_{i})\geq 16$.

###### Proof of Claim 4.7.

We prove this claim by contradiction. Suppose that
$\sum_{i=1}^{5}n^{\prime}(V_{i})<16$. Then, there is a non-strong partition
$\tau$ such that $\tau$ is not a useful partition for any $V_{i}$, $i\in[5]$.
In addition, $\tau$ is a useless partition for exactly one partite set, say
$V_{j}$. Now, we have that $\tau$ is a good partition. However, this
contradicts Claim 4.6.

Note that $n^{\prime}(V_{i})\leq n(V_{i})$. Then by Lemma 2.7 and Lemma 2.8,
we can derive the following four facts about $n^{\prime}(V_{i})$.

######  Fact 4.8.

For any regular partite set $V_{i}$ in $G_{2,5}$, $n^{\prime}(V_{i})\leq 4$.
And if $n^{\prime}(V_{1})=4$, then $V_{i}$ control② another partite set
$V_{j}$.

######  Fact 4.9.

For any semi-regular partite set $V_{i}$ in $G_{2,5}$, $n^{\prime}(V_{i})\leq
3$. And if $n^{\prime}(V_{1})=3$, then $V_{i}$ control② another partite set
$V_{j}$.

######  Fact 4.10.

For any irregular partite set $V_{i}$ in $G_{2,5}$, $n^{\prime}(V_{i})\leq 2$.

######  Fact 4.11.

If a partite set $V_{i}$ in $G_{2,5}$ is controlled② by two other partite
sets, then $n^{\prime}(V_{i})=0$. In particular, if a regular partite set
$V_{i}$ in $G_{2,5}$ is controlled② by another partite set, then
$n^{\prime}(V_{i})=0$. And if a semi-regular partite set $V_{i}$ in $G_{2,5}$
is controlled② by another partite set, then $n^{\prime}(V_{i})\leq 1$.

Next, we give an upper bound on $\sum_{i=1}^{5}n^{\prime}(V_{i})$.

######  Claim 4.12.

$\sum^{5}_{i=1}n^{\prime}(V_{i})\leq 16$

###### Proof of Claim 4.12.

By above facts, we have that if $n^{\prime}(V_{i})=4$, then $V_{i}$ is regular
and $V_{i}$ control② $V_{j}$ for some $j$.

Case 1: there is only one partite set $V_{1}$ such that $n^{\prime}(V_{1})=4$.
So, $V_{1}$ is regular and $V_{1}$ control② $V_{j}$. By Fact 4.10 and Fact
4.11, we have $n^{\prime}(V_{j})\leq 2$. So
$\sum^{5}_{i=1}n^{\prime}(V_{i})\leq 4+3+3+3+n^{\prime}(V_{j})<16$.

Case 2: there are two partite sets $V_{1},V_{2}$ such that
$n^{\prime}(V_{1})=n^{\prime}(V_{2})=4$.
So, $V_{1}$ and $V_{2}$ control① each other. If $V_{1}$ and $V_{2}$ control②
the same partite set $V_{j}$, then by Fact 4.10, $n^{\prime}(V_{j})=0$. So
$\sum^{5}_{i=1}n^{\prime}(V_{i})\leq 4+4+3+3+0<16$. If $V_{1}$ and $V_{2}$
control② different partite sets $V_{k},V_{l}$, then $n^{\prime}(V_{k})\leq 2$
and $n^{\prime}(V_{l})\leq 2$. So, $\sum^{5}_{i=1}n^{\prime}(V_{i})\leq
4+4+2+2+3<16$.

Case 3: there are three partite sets $V_{1},V_{2},V_{3}$ such that
$n^{\prime}(V_{1})=n^{\prime}(V_{2})=n^{\prime}(V_{3})=4$.
By Fact 4.8, $V_{1},V_{2},V_{3}$ control① each other and all of them will
control② a partite set. So, there is a partite set $V_{j}$ being controlled②
by two different partite sets. By Fact 4.10, we have $n^{\prime}(V_{j})=0$. So
$\sum^{5}_{i=1}n^{\prime}(V_{i})\leq 4+4+4+3+0<16$.

Case 4: there are four partite sets $V_{1},V_{2},V_{3},V_{4}$ such that
$n^{\prime}(V_{1})=n^{\prime}(V_{2})=n^{\prime}(V_{3})=n^{\prime}(V_{4})=4$.
In this case, $V_{1},V_{2},V_{3},V_{4}$ control① each other and all of them
control② $V_{5}$. Then $\sum_{i=1}^{5}n^{\prime}(V_{i})=16$.

Combine the lower bound and upper bound on $\sum_{i=1}^{5}n^{\prime}(V_{i})$,
we have $\sum_{i=1}^{5}n^{\prime}(V_{i})=16$. In this case,
$V_{1},V_{2},V_{3},V_{4}$ control① each other and all of them control②
$V_{5}$. Note that this structure is the same as the one shown in Theorem 3.1.
By similar arguments in Section 3, we have $G_{2,5}$ is isomorphic to $H_{0}$
which is constructed in Theorem 3.3.

Then $G_{2,5}$ is a regular $2$-balanced $5$-partite tournament. While
$G_{2,6}$ is also regular, then for any vertex $u\in V(G_{2,5})$, $u$ is good
to $V_{6}$. Recall that for any vertex $v\in V_{6}$, we can find $i,j\in[5]$
such that $V_{i}\subseteq N^{+}(v)$ and $V_{j}\subseteq N^{-}(v)$.

However, we could provide a st-partition of such $G_{2,6}$ as follows. Now, we
change the labels of the vertices of $G_{2,5}$ so that they match those of
$H_{0}$. Let $N^{+}(v_{4,1})\cap V_{6}=\\{v_{6,1}\\}$. Then both
$D[\\{v_{i,1}:i\in[6]\\}]$ and $D[\\{v_{i,2}:i\in[6]\\}]$ are strong maximal
tournaments and so it is a st-partition.

$v_{6,1}$$v_{5,1}$$v_{1,1}$$v_{2,1}$$v_{3,1}$$v_{4,1}$ Figure 6: An example to
explain that there is a st-partition of $G_{2,6}$

###### Proof of Theorem 1.2.

By Theorem 1.1, Lemma 3.4 and Theorem 4.5, we have $ST(2)=6$.

## Acknowledgement

This work was partially supported by the National Natural Science Foundation
of China (No. 12161141006), the Natural Science Foundation of Tianjin (No.
20JCJQJC00090), and the Fundamental Research Funds for the Central
Universities, Nankai University (No. 63231193).

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# On blow-up conditions for solutions of systems of quasilinear second-order
elliptic inequalities

A.A. Kon’kov  Department of Differential Equations, Faculty of Mechanics and
Mathematics, Moscow Lomonosov State University, Vorobyovy Gory, Moscow, 119992
Russia<EMAIL_ADDRESS>and  A.E. Shishkov  Center of Nonlinear
Problems of Mathematical Physics, RUDN University, Miklukho-Maklaya str. 6,
Moscow, 117198 Russia<EMAIL_ADDRESS>
###### Abstract.

We study systems of the differential inequalities

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
F_{1}(x,u_{2})&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq F_{2}(x,u_{1})&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$

where $n\geq 2$ and $A_{i}$ are Caratheodory functions such that

$C_{1}|\xi|^{p_{i}}\leq\xi A_{i}(x,\xi),\quad|A_{i}(x,\xi)|\leq
C_{2}|\xi|^{p_{i}-1},\quad i=1,2,$

with some constants $C_{1},C_{2}>0$ and $p_{1},p_{2}>1$ for almost all
$x\in{\mathbb{R}}^{n}$ and for all $\xi\in{\mathbb{R}}^{n}$, $n\geq 2$. For
non-negative solutions of these systems we obtain exact blow-up conditions.

###### Key words and phrases:

Global solutions; Nonlinearity; Blow-up

###### 1991 Mathematics Subject Classification:

35B44, 35B08, 35J30, 35J70

##  1\. Introduction

We consider systems of the differential inequalities

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
F_{1}(x,u_{2})&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq F_{2}(x,u_{1})&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$ (1.1)

where $n\geq 2$ and $A_{k}$ are Caratheodory functions such that

$C_{1}|\xi|^{p_{k}}\leq\xi A_{k}(x,\xi),\quad|A_{k}(x,\xi)|\leq
C_{2}|\xi|^{p_{k}-1},\quad k=1,2,$ (1.2)

with some constants $C_{1},C_{2}>0$ and $p_{1},p_{2}>1$ for almost all
$x\in{\mathbb{R}}^{n}$ and for all $\xi\in{\mathbb{R}}^{n}$. In so doing, the
functions $F_{1}$ and $F_{2}$ are assumed to be non-negative on
${\mathbb{R}}^{n}\times[0,\infty)$, positive on
${\mathbb{R}}^{n}\times(0,\infty)$, and non-decreasing with respect to the
last argument on the interval $[0,\varepsilon]$ for some real number
$0<\varepsilon<1$.

An ordered pair $(u_{1},u_{2})\in W_{p_{1},loc}^{1}({\mathbb{R}}^{n})\times
W_{p_{2},loc}^{1}({\mathbb{R}}^{n})$ is called a solution of (1.1) if
$F_{1}(x,u_{2}),F_{2}(x,u_{1})\in L_{1,loc}({\mathbb{R}}^{n})$ and, moreover,

$\int_{{\mathbb{R}}^{n}}A_{1}(x,\nabla
u_{1})\nabla\varphi\,dx\geq\int_{{\mathbb{R}}^{n}}F_{1}(x,u_{2})\varphi\,dx$

and

$\int_{{\mathbb{R}}^{n}}A_{2}(x,\nabla
u_{2})\nabla\varphi\,dx\geq\int_{{\mathbb{R}}^{n}}F_{2}(x,u_{1})\varphi\,dx$

for any non-negative function $\varphi\in C_{0}^{\infty}({\mathbb{R}}^{n})$.

Without loss of generality, we can assume that solutions of (1.1) satisfy the
relations

$\operatorname*{ess\,inf}\limits_{{\mathbb{R}}^{n}}u_{k}=0,\quad k=1,2.$ (1.3)

If this is not the case, we replace $u_{k}$ by $u_{k}-\alpha_{k}$, where

$\alpha_{k}=\operatorname*{ess\,inf}\limits_{{\mathbb{R}}^{n}}u_{k},\quad
k=1,2.$ (1.4)

After this replacement, the left-hand sides of (1.1) obviously do not change
and the right-hand sides transform to $F_{1}(x,u_{2}+\alpha_{2})$ and
$F_{2}(x,u_{1}+\alpha_{1})$.

The absence of solutions of differential equations and inequalities, which is
known as the blow-up phenomenon, has been studied by many authors [1–14]. In
most cases, these studies were limited to power-law nonlinearity. In our
paper, we consider the case of the general nonlinearity. In so doing, we
manage to strengthen results of [1, 3].

Note that the only relevant case is $n>\operatorname{max}\\{p_{1},p_{2}\\}$.
Indeed, if $n\leq p_{k}$ for some $k\in\\{1,2\\}$, then any non-negative
solution of the inequality

$-A_{k}(x,\nabla u_{i})\geq 0\quad\mbox{in }{\mathbb{R}}^{n}$

is a constant [12].

Below it is assumed that $\theta>1$ is some given real number and

$f_{k}(r,\zeta)=\operatorname*{ess\,inf}\limits_{x\in B_{\theta r}\setminus
B_{r/\theta}}F_{k}(x,\zeta),\quad r,\zeta>0,\;k=1,2.$ (1.5)

Let us denote by $B_{r}$ the open ball of radius $r>0$ centered at zero. We
also assume that for any real numbers $\varepsilon_{*},\zeta_{*}>0$ there
exist a real number $r_{*}>0$ and functions
$q_{k}:[r_{*},\infty)\to[0,\infty)$ and
$g_{k}:[\zeta_{*},\infty)\to(0,\infty)$, $k=1,2$, satisfying the following
conditions:

  1. $(a)$

for all $r\in[r_{*},\infty)$ and $\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\leq\varepsilon\quad\mbox{and}\quad\varepsilon_{*}r^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(r,\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\leq\varepsilon$

we have

$f_{2}\left(r,\varepsilon_{*}r^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(r,\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\right)\geq
q_{1}(r)g_{1}(\zeta);$ (1.6)

  2. $(b)$

for all $r\in[r_{*},\infty)$ and $\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\leq\varepsilon\quad\mbox{and}\quad\varepsilon_{*}r^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(r,\left(\frac{\zeta}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\leq\varepsilon$

we have

$f_{1}\left(r,\varepsilon_{*}r^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(r,\left(\frac{\zeta}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\right)\geq
q_{2}(r)g_{2}(\zeta).$

##  2\. Main results

######  Theorem 2.1.

Suppose that $n>\operatorname{max}\\{p_{1},p_{2}\\}$. Also let for any real
numbers $\varepsilon_{*},\zeta_{*}>0$ there be a real number $r_{*}>0$, non-
decreasing functions $g_{1},g_{2}:[\zeta_{*},\infty)\to(0,\infty)$, and
locally bounded measurable functions $q_{1},q_{2}:[r_{*},\infty)\to[0,\infty)$
such that $(a)$ and $(b)$ are valid and, moreover,

$\int_{\zeta_{*}}^{\infty}\frac{d\zeta}{g_{k}(\zeta)}<\infty$ (2.1)

and

$\int_{r_{*}}^{\infty}r^{n-1}q_{k}(r)\,dr=\infty$ (2.2)

for some $k\in\\{1,2\\}$. Then any non-negative solution of (1.1), (1.3) is
identically zero.

The proof of Theorem 2.1 is given in Section 3. A partial case of (1.1) is the
systems

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
a_{1}(x)u_{2}^{\lambda_{1}}&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq
a_{2}(x)u_{1}^{\lambda_{2}}&\mbox{in }{\mathbb{R}}^{n},\end{aligned}\right.$
(2.3)

where $a_{1}$ and $a_{2}$ are non-negative measurable functions and
$\lambda_{1},\lambda_{2}\geq 0$ are real numbers.

######  Corollary 2.1.

Suppose that $n>\operatorname{max}\\{p_{1},p_{2}\\}$ and

$\frac{\lambda_{1}\lambda_{2}}{(p_{1}-1)(p_{2}-1)}>1.$ (2.4)

Also let there be $k\in\\{1,2\\}$ such that

$\int_{1}^{\infty}r^{\mu_{k}}\alpha_{k}(r)\,dr=\infty,$ (2.5)

where

$\mu_{1}=n-1+\frac{\lambda_{2}p_{1}}{p_{1}-1}-\frac{\lambda_{1}\lambda_{2}(n-p_{2})}{(p_{1}-1)(p_{2}-1)},$
$\mu_{2}=n-1+\frac{\lambda_{1}p_{2}}{p_{2}-1}-\frac{\lambda_{2}\lambda_{1}(n-p_{1})}{(p_{2}-1)(p_{1}-1)},$
$\alpha_{1}(r)=\left(\operatorname*{ess\,inf}\limits_{B_{\theta r}\setminus
B_{r/\theta}}a_{1}\right)^{\lambda_{2}/(p_{1}-1)}\operatorname*{ess\,inf}\limits_{B_{\theta
r}\setminus B_{r/\theta}}a_{2},,$ (2.6)

and

$\alpha_{2}(r)=\left(\operatorname*{ess\,inf}\limits_{B_{\theta r}\setminus
B_{r/\theta}}a_{2}\right)^{\lambda_{1}/(p_{2}-1)}\operatorname*{ess\,inf}\limits_{B_{\theta
r}\setminus B_{r/\theta}}a_{1}$ (2.7)

for some real number $\theta>1$. Then any non-negative solution of (2.3),
(1.3) is identically zero.

###### Proof.

We assume, without loss of generality, that (2.5) holds for $k=1$. If $k=2$,
then all reasoning is absolutely similar. For systems (2.3), inequality (1.6)
takes the form

$\varepsilon_{*}^{\lambda_{2}}\alpha_{1}(r)r^{\lambda_{2}p_{1}/(p_{1}-1)}\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{\lambda_{1}\lambda_{2}/((p_{1}-1)(p_{2}-1))}\geq
q_{1}(r)g_{1}(\zeta).$

Thus, using Theorem 2.1 with

$g_{1}(\zeta)=\zeta^{\lambda_{1}\lambda_{2}/((p_{1}-1)(p_{2}-1))}\quad\mbox{and}\quad
q_{1}(r)=\varepsilon_{*}^{\lambda_{2}}r^{\lambda_{2}p_{1}/(p_{1}-1)-\lambda_{1}\lambda_{2}(n-p_{2})/((p_{1}-1)(p_{2}-1))}\alpha_{1}(r),$

we complete the proof. ∎

######  Corollary 2.2.

Suppose that $n>\operatorname{max}\\{p_{1},p_{2}\\}$ and

$\frac{\lambda_{1}\lambda_{2}}{(p_{1}-1)(p_{2}-1)}\leq 1.$ (2.8)

If at least one of the following two relations is valid:

$\operatorname*{lim\,sup}\limits_{r\to\infty}r^{n+\lambda_{2}p_{1}/(p_{1}-1)-\lambda(n-p_{2})}\alpha_{1}(r)>0,$
(2.9)
$\operatorname*{lim\,sup}\limits_{r\to\infty}r^{n+\lambda_{1}p_{2}/(p_{2}-1)-\lambda(n-p_{1})}\alpha_{2}(r)>0,$
(2.10)

where $\lambda>1$ is a real number and the functions $\alpha_{1}$ and
$\alpha_{2}$ are defined by (2.6) and (2.7) for some real number $\theta>1$,
then any non-negative solution of (2.3), (1.3) is identically zero.

###### Proof.

We assume that (2.9) is fulfilled. In the case where (2.10) holds, our
reasoning is absolutely similar. Let $r_{i}$, $i=1,2,\ldots$, be a sequence of
real numbers such that $r_{i}\to\infty$ as $i\to\infty$ and, moreover,

$r_{i}^{n+\lambda_{2}p_{1}/(p_{1}-1)-\lambda(n-p_{2})}\alpha_{1}(r_{i})>\gamma$

with some constant $\gamma>0$ for all $i=1,2,\ldots$. Putting

$\tilde{\alpha}_{1}(r)=\left(\operatorname*{ess\,inf}\limits_{B_{\tilde{\theta}r}\setminus
B_{r/\tilde{\theta}}}a_{1}\right)^{\lambda_{2}/(p_{1}-1)}\operatorname*{ess\,inf}\limits_{B_{\tilde{\theta}r}\setminus
B_{r/\tilde{\theta}}}a_{2},$

where $1<\tilde{\theta}<\theta$ is some real number, we obviously obtain

$r_{i}^{n+\lambda_{2}p_{1}/(p_{1}-1)-\lambda(n-p_{2})}\inf_{(r_{i}\theta/\tilde{\theta},r_{i}\tilde{\theta}/\theta)}\tilde{\alpha}_{1}>\gamma$

for all $i=1,2,\ldots$. Hence,

$\int_{r_{0}}^{\infty}r^{n-1}q_{1}(r)\,dr=\infty,$

where

$q_{1}(r)=r^{\lambda_{2}p_{1}/(p_{1}-1)-\lambda(n-p_{2})}\tilde{\alpha}_{1}(r).$

Since,

$\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{\lambda_{1}\lambda_{2}/((p_{1}-1)(p_{2}-1))}\geq\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{\lambda}$

for all real numbers $r>0$ and $\zeta>0$ satisfying the condition

$\frac{\zeta}{r^{n-p_{2}}}\leq 1,$ (2.11)

we also have

$\tilde{\alpha}_{1}(r)r^{\lambda_{2}p_{1}/(p_{1}-1)}\left(\frac{\zeta}{r^{n-p_{2}}}\right)^{\lambda_{1}\lambda_{2}/((p_{1}-1)(p_{2}-1))}\geq
q_{1}(r)\zeta^{\lambda}$ (2.12)

for all real numbers $r>0$ and $\zeta>0$ satisfying (2.11). It can bee seen
that for systems of the form (2.3) inequality (2.12) is equivalent to (1.6)
with $g_{1}(\zeta)=\varepsilon_{*}^{\lambda_{2}}\zeta^{\lambda}$ and $\theta$
replaced by $\tilde{\theta}$. Thus, to complete the proof, it remains to apply
Theorem 2.1. ∎

######  Example 2.1.

Consider the system

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
u_{2}^{\lambda_{1}}&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq u_{1}^{\lambda_{2}}&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$ (2.13)

where $n>\operatorname{max}\\{p_{1},p_{2}\\}$ and $\lambda_{1},\lambda_{2}\geq
0$ are real numbers. By Corollaries 2.1 and 2.2, if

$\frac{n(p_{1}-1)(p_{2}-1)}{\lambda_{1}\lambda_{2}}\geq\operatorname{min}\left\\{n-p_{1}-(p_{1}-1)\frac{p_{2}}{\lambda_{2}},n-p_{2}-(p_{2}-1)\frac{p_{1}}{\lambda_{1}}\right\\},$
(2.14)

then any non-negative solution of (2.13), (1.3) is identically zero. Indeed,
in the case where (2.4) is valid, this follows immediately from Corollary 2.1.
Let (2.8) be fulfilled. We obviously have

$n+\lambda_{2}p_{1}/(p_{1}-1)>n-p_{2};$

therefore, there exists a real numbers $\lambda>1$ such that

$n+\lambda_{2}p_{1}/(p_{1}-1)-\lambda(n-p_{2})>0.$

Thus, to show the triviality of any non-negative solution of (2.13), (1.3), it
is sufficient to use Corollary 2.2.

Note that (2.14) coincides with the analogous condition obtained in [1,
Theorem 4.6].

######  Example 2.2.

Let us examine the case of critical exponents $\lambda_{1}$ and $\lambda_{2}$
in (2.14). Namely, consider the system

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
u_{2}^{\lambda_{1}}\log^{\sigma_{1}}\left(e+\frac{1}{u_{2}}\right)&\mbox{in
}{\mathbb{R}}^{n},\\\ &-\operatorname{div}A_{2}(x,\nabla u_{2})\geq
u_{1}^{\lambda_{2}}\log^{\sigma_{2}}\left(e+\frac{1}{u_{1}}\right)&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$ (2.15)

where $n>\operatorname{max}\\{p_{1},p_{2}\\}$, the real numbers
$\lambda_{1},\lambda_{2}>0$ satisfy (2.4),
$\sigma_{1},\sigma_{2}\in{\mathbb{R}}$ and, moreover,

$\frac{n(p_{1}-1)(p_{2}-1)}{\lambda_{1}\lambda_{2}}=n-p_{2}-(p_{2}-1)\frac{p_{1}}{\lambda_{1}}$

and

$\frac{n(p_{1}-1)(p_{2}-1)}{\lambda_{1}\lambda_{2}}<n-p_{1}-(p_{1}-1)\frac{p_{2}}{\lambda_{2}}.$

If $u_{1}=0$ or $u_{2}=0$, then it is assumed that the right-hand side of the
corresponding inequality in (2.15) is equal to zero. It can be verified that
for any real numbers $\varepsilon_{*},\zeta_{*}>0$ there is a real number
$r_{*}>0$ such that condition (a) is satisfied with some positive continuous
functions

$g_{1}(\zeta)\asymp\frac{\zeta^{\lambda_{1}\lambda_{2}/((p_{1}-1)(p_{2}-1))}}{\log^{|\sigma_{2}|+\lambda_{2}|\sigma_{1}|/(p_{1}-1)}\zeta}\quad\mbox{and}\quad
q_{1}(r)\asymp\frac{1}{r^{n}}\log^{\sigma_{2}+\lambda_{2}\sigma_{1}/(p_{1}-1)}r.$

Thus, in accordance with Theorem 2.1 if

$\sigma_{2}+\frac{\lambda_{2}\sigma_{1}}{p_{1}-1}\geq-1,$

then any non-negative solution of (2.15), (1.3) is identically zero.

##  3\. Proof of Theorem 2.1

Let us agree to denote by $C$, $\sigma$, and $\varkappa$ various positive
constants that can depend only on $n$, $p_{1}$, $p_{2}$, $C_{1}$, $C_{2}$,
$\varepsilon$, $\theta$, and $\lambda$. In so doing, by $\chi_{\omega}$ we
mean the characteristic function of a set $\omega\subset{\mathbb{R}}^{n}$,
i.e.

$\chi_{\omega}(x)=\left\\{\begin{array}[]{ll}1,&x\in\omega,\\\
0,&x\not\in\omega.\end{array}\right.$

Assume that $A$ is a Caratheodory function satisfying the uniform ellipticity
condition

$C_{1}|\xi|^{p}\leq\zeta A(x,\xi),\quad|A(x,\xi)|\leq C_{2}|\xi|^{p},\quad
p>1,$ (3.1)

for almost all $x\in{\mathbb{R}}^{n}$ and for all $\xi\in{\mathbb{R}}^{n}$. We
say that $v\in W_{p,loc}^{1}(\omega)$ is a solution of the inequality

$\operatorname{div}A(x,\nabla v)\geq a(x)\quad\mbox{in }\omega,$ (3.2)

where $\omega\subset{\mathbb{R}}^{n}$ is a non-empty open set and $a\in
L_{1,loc}(\omega)$, if

$-\int_{\omega}A(x,\nabla v)\nabla\varphi\,dx\geq\int_{\omega}a(x)\varphi\,dx$

for any non-negative function $\varphi\in C_{0}^{\infty}(\omega)$. A solution
of the inequality

$-\operatorname{div}A(x,\nabla u)\geq a(x)\quad\mbox{in }\omega,$ (3.3)

where $\omega\subset{\mathbb{R}}^{n}$ is a non-empty open set and $a\in
L_{1,loc}(\omega)$, is defined in a similar way.

######  Lemma 3.1 (Generalized Kato’s inequality).

Let $v\in W_{p,loc}^{1}(\omega)$ be a solution of (3.2). Then
$v_{+}=\chi_{\omega_{+}}v$ satisfies the inequality

$\operatorname{div}A(x,\nabla v_{+})\geq\chi_{\omega_{+}}(x)a(x)\quad\mbox{in
}\omega,$

where $\omega_{+}=\\{x\in\omega:v(x)>0\\}$.

The proof of Lemma 3.1 is given in [9, Lemma 4.2]. Putting $v=\varepsilon-u$
in Lemma 3.1, we obtain the following statement.

######  Corollary 3.1.

Let $u$ be a solution of (3.3). Then
$u_{\varepsilon}=\chi_{\omega_{\varepsilon}}u+(1-\chi_{\omega_{\varepsilon}})\varepsilon$
satisfies the inequality

$-\operatorname{div}A(x,\nabla
u_{\varepsilon})\geq\chi_{\omega_{\varepsilon}}(x)a(x)\quad\mbox{in
}{\mathbb{R}}^{n},$

where $\omega_{\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u(x)<\varepsilon\\}$.

We also need the three lemmas below.

######  Lemma 3.2 (Weak Harnack inequality).

Let $u\geq 0$ be a solution of the inequality

$-\operatorname{div}A(x,\nabla u)\geq 0\quad\mbox{in }{\mathbb{R}}^{n},$ (3.4)

where $n>p$. Then

$\left(\frac{1}{\mathop{\rm mes}\nolimits
B_{2r}}\int_{B_{2r}}u^{\lambda}\,dx\right)^{1/\lambda}\leq
C\operatorname*{ess\,inf}\limits_{B_{r}}u$

for all $\lambda\in(0,n(p-1)/(n-p))$ and $r\in(0,\infty)$.

######  Lemma 3.3.

Let $u\geq 0$ be a solution of the inequality

$-\operatorname{div}A(x,\nabla u)\geq a(x)\quad\mbox{in }{\mathbb{R}}^{n},$
(3.5)

where $a\in L_{1,loc}({\mathbb{R}}^{n})$ is a non-negative function. If
$u^{\lambda}\in L_{1,loc}({\mathbb{R}}^{n})$ for some
$\lambda\in(p-1,\infty)$, then

$\frac{1}{\mathop{\rm mes}\nolimits B_{r}}\int_{B_{r}}a(x)\,dx\leq
Cr^{-p}\left(\frac{1}{\mathop{\rm mes}\nolimits B_{2r}\setminus
B_{r}}\int_{B_{2r}\setminus B_{r}}u^{\lambda}\,dx\right)^{(p-1)/\lambda}$

for all $r\in(0,\infty)$.

######  Lemma 3.4.

Let $u\geq 0$ be a solution of (3.4) such that

$\operatorname*{ess\,inf}\limits_{{\mathbb{R}}^{n}}u=0,$ (3.6)

where $n>p$. Then

$\lim_{r\to\infty}\frac{\mathop{\rm mes}\nolimits
B_{r}\setminus\omega_{\varepsilon}}{\mathop{\rm mes}\nolimits B_{r}}=0,$

where $\omega_{\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u(x)<\varepsilon\\}$.

Lemmas 3.2 and 3.3 are proved in [15, 16, 17] and [1], respectively. Lemma 3.4
or its equivalent statement can be found in [2, Lemma 3.1]. This lemma is a
direct consequence of Lemma 3.2.

Corollary 3.1 and Lemmas 3.2 and 3.3 with
$\lambda\in(p-1,p)\cap(0,n(p-1)/(n-p))$ imply the following assertion.

######  Corollary 3.2.

Let $u\geq 0$ be a solution of (3.5), where $n>p$ and $a\in
L_{1,loc}({\mathbb{R}}^{n})$ is a non-negative function. Then

$\frac{1}{\mathop{\rm mes}\nolimits B_{r}}\int_{\omega_{\varepsilon}\cap
B_{r}}a(x)\,dx\leq
Cr^{-p}\left(\operatorname*{ess\,inf}\limits_{B_{r}}u_{\varepsilon}\right)^{p-1}$

for all $r\in(0,\infty)$, where
$\omega_{\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u(x)<\varepsilon\\}$ and
$u_{\varepsilon}=\chi_{\omega_{\varepsilon}}u+(1-\chi_{\omega_{\varepsilon}})\varepsilon.$

###### Proof of Theorem 2.1.

Assume the converse. Let $(u_{1},u_{2})$ be a non-negative solution of (1.1),
(1.3) that is not equal to zero. According to Lemma 3.2, both the functions
$u_{1}$ and $u_{2}$ are positive almost everywhere in ${\mathbb{R}}^{n}$.

We put
$\Omega_{\varepsilon}=\Omega_{1,\varepsilon}\cap\Omega_{2,\varepsilon}$, where
$\Omega_{k,\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u_{k}(x)<\varepsilon\\}$,
$k=1,2$. By Lemma 3.4,

$\lim_{r\to\infty}\frac{\operatorname{mes}B_{r}\setminus\Omega_{k,\varepsilon}}{\operatorname{mes}B_{r}}=0,\quad
k=1,2.$

Hence, there is a real number $r_{0}>0$ such that

$\mathop{\rm mes}\nolimits\Omega_{\varepsilon}\cap B_{\theta r}\setminus
B_{r}\geq Cr^{n},$ (3.7)

for all $r\geq r_{0}$. Denote

$E_{1}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}F_{2}(x,u_{1})\,dx,\quad r>0,$

and

$E_{2}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}F_{1}(x,u_{2})\,dx,\quad r>0.$

Also let $r_{i}=\theta^{i}r_{0}$, $i=1,2,\ldots$. Corollary 3.2 implies the
estimates

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}}F_{1}(x,u_{2})\,dx\leq
Cr_{i}^{-p_{1}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1,\varepsilon}\right)^{p_{1}-1},\quad
i=1,2,\ldots,$

and

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}}F_{2}(x,u_{1})\,dx\leq
Cr_{i}^{-p_{2}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2,\varepsilon}\right)^{p_{2}-1},\quad
i=1,2,\ldots,$

where

$u_{k,\varepsilon}=\chi_{\Omega_{k,\varepsilon}}u_{k}+(1-\chi_{\Omega_{k,\varepsilon}})\varepsilon,\quad
k=1,2,$ (3.8)

whence it follows that

$\operatorname*{ess\,inf}\limits_{\Omega_{1,\varepsilon}\cap
B_{r_{i}}}u_{1}=\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1,\varepsilon}\geq\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\quad
i=1,2,\ldots,$ (3.9)

and

$\operatorname*{ess\,inf}\limits_{\Omega_{2,\varepsilon}\cap
B_{r_{i}}}u_{2}=\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2,\varepsilon}\geq\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)},\quad
i=1,2,\ldots.$ (3.10)

Taking into account (3.9) and the fact that ${F_{1}(x,\cdot)}$ is a non-
decreasing function on the interval $[0,\varepsilon]$ for almost
$x\in{\mathbb{R}}^{n}$, we obtain

$F_{2}(x,u_{1}(x))\geq
F_{2}\left(x,\operatorname*{ess\,inf}\limits_{\Omega_{1,\varepsilon}\cap
B_{r_{i}}}u_{1}\right)\geq
F_{2}\left(x,\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)$

for almost all $x\in\Omega_{1,\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$.
This immediately implies that

$\displaystyle\frac{1}{\operatorname{mes}\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}F_{2}(x,u_{1})\,dx$
$\displaystyle\qquad{}\geq\operatorname*{ess\,inf}\limits_{x\in\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}F_{2}\left(x,\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right),\quad
i=1,2,\ldots,$

whence in accordance with (3.7) and the definition of $f_{2}$ we have

$E_{1}(r_{i})-E_{1}(r_{i-1})\geq
Cr^{n}f_{2}\left(r,\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)$
(3.11)

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$.

Analogously, taking into account (3.7) and (3.10) and the definition of
$f_{1}$, we obtain

$E_{2}(r_{i})-E_{2}(r_{i-1})\geq
Cr^{n}f_{1}\left(r,\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$
(3.12)

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. The last inequality, in
particular, implies that

$E_{2}(r_{i})\geq
Cr^{n}f_{1}\left(r,\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. Combining this with (3.11), we
arrive at the estimate

$E_{1}(r_{i})-E_{1}(r_{i-1})\geq Cr^{n}f_{2}\left(r,\sigma
r^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(r,\left(\frac{\varkappa
E_{1}(r_{i})}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\right)$

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. In the same way, it can be
shown that

$E_{2}(r_{i})-E_{2}(r_{i-1})\geq Cr^{n}f_{1}\left(r,\sigma
r^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(r,\left(\frac{\varkappa
E_{2}(r_{i})}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\right)$

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$.

Thus, there are an integer $i_{0}>0$, non-decreasing functions
$g_{1},g_{2}:[\zeta_{*},\infty)\to(0,\infty)$, and locally bounded measurable
functions $q_{1},q_{2}:[r_{*},\infty)\to[0,\infty)$ such that

$E_{s}(r_{i})-E_{s}(r_{i-1})\geq Cr^{n}q_{s}(r)g_{s}(\varkappa
E_{s}(r_{i})),\quad s=1,2,$

for all $i>i_{0}$ and $r\in(r_{i-1},r_{i})$ and, moreover, conditions (2.1)
and (2.2) are valid with some $k\in\\{1,2\\}$, where $r_{*}=r_{i_{0}}$ and
$\zeta_{*}=\varkappa\operatorname{min}\\{E_{1}(r_{i_{0}}),E_{2}(r_{i_{0}})\\}$.
We obviously have

$\frac{E_{k}(r_{i})-E_{k}(r_{i-1})}{g_{k}(\varkappa E_{k}(r_{i})),}\geq
Cr^{n}q_{k}(r)$

for all $i>i_{0}$ and $r\in(r_{i-1},r_{i})$, whence in accordance with the
inequalities

$\int_{E_{k}(r_{i-1})}^{E_{k}(r_{i})}\frac{d\zeta}{g_{k}(\varkappa\zeta))}\geq\frac{E_{k}(r_{i})-E_{k}(r_{i-1})}{g_{k}(\varkappa
E_{k}(r_{i})),}$

and

$\sup_{r\in(r_{i-1},r_{i})}r^{n}q_{k}(r)\geq
C\int_{r_{i-1}}^{r_{i}}r^{n-1}q_{k}(r)\,dr$

it follows that

$\int_{E_{k}(r_{i-1})}^{E_{k}(r_{i})}\frac{d\zeta}{g_{k}(\varkappa\zeta))}\geq
C\int_{r_{i-1}}^{r_{i}}r^{n-1}q_{k}(r)\,dr,\quad i=i_{0}+1,i_{0}+2,\ldots.$

Summing the last expression over all $i>i_{0}$, we obtain

$\int_{E_{k}(r_{i_{0}})}^{\infty}\frac{d\zeta}{g_{k}(\varkappa\zeta))}\geq
C\int_{r_{i_{0}}}^{\infty}r^{n-1}q_{k}(r)\,dr.$

This contradicts (2.1) and (2.2). ∎

##  4\. Some remarks and generalizations

As in the previous section, we denote by $C$, $\sigma$, and $\varkappa$
various positive constants that can depend only on $n$, $p_{1}$, $p_{2}$,
$C_{1}$, $C_{2}$, $\varepsilon$, $\theta$, and $\lambda$.

### 4.1

Consider systems of the form

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
a_{1}(x)h_{1}(u_{2})&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq a_{2}(x)h_{2}(u_{1})&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$

where $a_{1},a_{2}:{\mathbb{R}}^{n}\to(0,\infty)$ and
$h_{1},h_{2}:[0,\infty)\to[0,\infty)$ are measurable functions such that
$h_{1}$ and $h_{2}$ are positive on the interval $(0,\infty)$ and non-
decreasing on $[0,\varepsilon]$ for some real number $0<\varepsilon<1$. It
does not present any particular problem to verify that for these systems
Theorem 2.1 remains valid with (1.5) replaced by

$f_{k}(r,\zeta)=\frac{h_{k}(\zeta)}{\left(\frac{1}{\operatorname{mes}B_{\theta
r}\setminus B_{r/\theta}}\int_{B_{\theta r}\setminus
B_{r/\theta}}a_{k}^{-\lambda}(x)\,dx\right)^{1/\lambda}},\quad
r,\zeta>0,\;k=1,2,$ (4.1)

where $\lambda>0$ is a real number. In so doing, if

$\int_{B_{\theta r}\setminus B_{r/\theta}}a_{k}^{-\lambda}(x)\,dx=\infty$

for some $r\in(0,\infty)$ and $k\in\\{1,2\\}$, then we assume that
$f_{k}(r,\zeta)=0$ for all $\zeta\in(0,\infty)$.

Indeed, inequality (3.9) yields

$h_{2}(u_{1}(x))\geq
h_{2}\left(\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)$

for almost all $x\in\Omega_{1,\varepsilon}\cap B_{r_{i}}$, whence it follows
that

$\displaystyle\frac{1}{\operatorname{mes}\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}h_{2}^{\lambda/(1+\lambda)}(u_{1})\,dx$
$\displaystyle\qquad{}\geq
h_{2}^{\lambda/(1+\lambda)}\left(\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right),\quad
i=1,2,\ldots.$

Combining this with (3.7) and the estimate

$\displaystyle\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}h_{2}^{\lambda/(1+\lambda)}(u_{1})\,dx=\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}a_{2}^{-\lambda/(1+\lambda)}(x)a_{2}^{\lambda/(1+\lambda)}(x)h_{2}^{\lambda/(1+\lambda)}(u_{1})\,dx$
$\displaystyle\quad{}\leq\left(\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}a_{2}^{-\lambda}(x)\,dx\right)^{1/(1+\lambda)}\left(\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}a_{2}(x)h_{2}(u_{1})\,dx\right)^{\lambda/(1+\lambda)}$

which follows from Hölder’s inequality, we obtain

$\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}a_{2}(x)h_{2}(u_{1})\,dx\geq\frac{Cr_{i}^{n}h_{2}\left(\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)}{\left(\frac{1}{\operatorname{mes}B_{r_{i}}\setminus
B_{r_{i-1}}}\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}a_{2}^{-\lambda}(x)\,dx\right)^{1/\lambda}},\quad i=1,2,\ldots.$

The last formula obviously implies (3.11) with

$E_{1}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}a_{2}(x)h_{2}(u_{1})\,dx,\quad
r>0,$

and $f_{2}$ defined by (4.1).

Repeating the previous reasoning with (3.9) replaced by (3.10), we also have

$\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}a_{1}(x)h_{1}(u_{2})\,dx\geq\frac{Cr_{i}^{n}h_{1}\left(\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)}{\left(\frac{1}{\operatorname{mes}B_{r_{i}}\setminus
B_{r_{i-1}}}\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}a_{1}^{-\lambda}(x)\,dx\right)^{1/\lambda}},\quad i=1,2,\ldots.$

This in turn implies (3.12) with

$E_{2}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}a_{1}(x)h_{1}(u_{2})\,dx,\quad
r>0,$

and $f_{1}$ defined by (4.1).

### 4.2

An interesting case is when $F_{1}$ and $F_{2}$ on the right in (1.1) are non-
decreasing functions with respect to the last argument on the whole interval
$[0,\infty)$. Fix some real number $0<\varepsilon<1$. Also let for any real
numbers $\varepsilon_{*},\zeta_{*}>0$ there be a real number $r_{*}>0$ and
functions $q_{k}:{\mathbb{R}}^{n}\setminus B_{r_{*}}\to[0,\infty)$ and
$g_{k}:[\zeta_{*},\infty)\to(0,\infty)$, $k=1,2$, satisfying the following
conditions:

  1. $(a^{\prime})$

for almost all $x\in{\mathbb{R}}^{n}\setminus B_{r_{*}}$ and for all
$\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\leq\varepsilon\quad\mbox{and}\quad\varepsilon_{*}|x|^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(|x|,\left(\frac{\zeta}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\leq\varepsilon$

we have

$F_{2}\left(x,\varepsilon_{*}|x|^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(|x|,\left(\frac{\zeta}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\right)\geq
q_{1}(x)g_{1}(\zeta);$ (4.2)

  2. $(b^{\prime})$

for almost all $x\in{\mathbb{R}}^{n}\setminus B_{r_{*}}$ and for all
$\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)}\leq\varepsilon\quad\mbox{and}\quad\varepsilon_{*}|x|^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(|x|,\left(\frac{\zeta}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\leq\varepsilon$

we have

$F_{1}\left(x,\varepsilon_{*}|x|^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(|x|,\left(\frac{\zeta}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\right)\geq
q_{2}(x)g_{2}(\zeta),$ (4.3)

where

$f_{k}(r,\zeta)=\frac{1}{\operatorname{mes}B_{r}}\int_{B_{r}}F_{k}(x,\zeta)\,dx,\quad
r,\zeta>0,\;k=1,2.$

######  Theorem 4.1.

Suppose that $n>\operatorname{max}\\{p_{1},p_{2}\\}$ and, moreover, the
functions $F_{1}$ and $F_{2}$ are non-negative and non-decreasing with respect
to the last arguments on the set ${\mathbb{R}}^{n}\times[0,\infty)$ and
positive on ${\mathbb{R}}^{n}\times(0,\infty)$. Also let for any real numbers
$\varepsilon_{*},\zeta_{*}>0$ there be a real number $r_{*}>0$, non-decreasing
functions $g_{1},g_{2}:[\zeta_{*},\infty)\to(0,\infty)$, and locally bounded
measurable functions $q_{1},q_{2}:{\mathbb{R}}^{n}\setminus
B_{r_{*}}\to[0,\infty)$ such that $(a^{\prime})$ and $(b^{\prime})$ are valid.
If there exist $k\in\\{1,2\\}$ such that (2.1) holds and

$\int_{{\mathbb{R}}^{n}\setminus B_{r_{*}}}q_{k}(x)\,dx=\infty,$ (4.4)

then any non-negative solution of (1.1), (1.3) is identically zero.

###### Proof.

Arguing by contradiction, we assume that $(u_{1},u_{2})$ is a non-negative
solution of (1.1), (1.3) that is not equal to zero. By Lemma 3.2, both the
functions $u_{1}$ and $u_{2}$ are positive almost everywhere in
${\mathbb{R}}^{n}$. We take a real number $r_{0}>0$ such that

$\operatorname*{ess\,inf}\limits_{B_{r_{0}}}u_{k}<\varepsilon,\quad k=1,2.$
(4.5)

In view of (1.3), such a real number $r_{0}$ obviously exists. Let us put
$r_{i}=2^{i}$, $i=1,2,\ldots$.

Applying Lemmas 3.2 and 3.3 with $\lambda\in(p-1,p)\cap(0,n(p-1)/(n-p))$, we
obtain

$\frac{1}{\mathop{\rm mes}\nolimits
B_{r_{i}}}\int_{B_{r_{i}}}F_{1}(x,u_{2})\,dx\leq
Cr_{i}^{-p_{1}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1}\right)^{p_{1}-1},\quad
i=1,2,\ldots,$

and

$\frac{1}{\mathop{\rm mes}\nolimits
B_{r_{i}}}\int_{B_{r_{i}}}F_{2}(x,u_{1})\,dx\leq
Cr_{i}^{-p_{2}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\right)^{p_{2}-1},\quad
i=1,2,\ldots,$

whence it follows that

$\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1}\geq\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\quad
i=1,2,\ldots,$ (4.6)

and

$\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\geq\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)},\quad
i=1,2,\ldots,$ (4.7)

where

$E_{1}(r)=\int_{B_{r}}F_{2}(x,u_{1})\,dx,\quad r>0,$

and

$E_{2}(r)=\int_{B_{r}}F_{1}(x,u_{2})\,dx,\quad r>0.$

Inequalities (4.6) and (4.7) imply the estimates

$F_{2}(x,u_{1}(x))\geq
F_{2}\left(x,\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1}\right)\geq
F_{2}\left(x,\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)$

and

$F_{1}(x,u_{2}(x))\geq
F_{1}\left(x,\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\right)\geq
F_{1}\left(x,\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$

for almost all $x\in B_{r_{i}}$, $i=1,2,\ldots$. Integrating them over
$B_{r_{i}}$, we have

$E_{1}(r_{i})\geq\operatorname{mes}B_{r}f_{2}\left(r_{i},\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right),\quad
i=1,2,\ldots,$

and

$E_{2}(r_{i})\geq\operatorname{mes}B_{r}f_{1}\left(r_{i},\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right),\quad
i=1,2,\ldots.$

Consequently, one can assert that

$F_{2}(x,u_{1}(x))\geq F_{2}\left(x,\sigma
r_{i}^{p_{1}/(p_{1}-1)}f_{1}^{1/(p_{1}-1)}\left(r_{i},\left(\frac{\varkappa
E_{1}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\right)$

and

$F_{1}(x,u_{2}(x))\geq F_{1}\left(x,\sigma
r_{i}^{p_{2}/(p_{2}-1)}f_{2}^{1/(p_{2}-1)}\left(r_{i},\left(\frac{\varkappa
E_{2}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)}\right)\right)$

for almost all $x\in B_{r_{i}}$, $i=1,2,\ldots$. In view of conditions
$(a^{\prime})$ and $(b^{\prime})$, this yields

$F_{2}(x,u_{1}(x))\geq q_{1}(x)g_{1}(\varkappa E_{1}(r_{i}))$

and

$F_{1}(x,u_{2}(x))\geq q_{2}(x)g_{2}(\varkappa E_{2}(r_{i}))$

for almost all $x\in B_{r_{i}}\setminus B_{r_{i-1}}$,
$i=i_{0}+1,i_{0}+2,\ldots$, where $i_{0}\geq 0$ is some integer. Integrating
further the last two inequalities over $B_{r_{i}}\setminus B_{r_{i-1}}$, we
obtain

$E_{1}(r_{i})-E_{1}(r_{i-1})\geq g_{1}(\varkappa
E_{1}(r_{i}))\int_{B_{r_{i}}\setminus B_{r_{i-1}}}q_{1}(x)\,dx,\quad
i=i_{0}+1,i_{0}+2,\ldots,$ (4.8)

and

$E_{2}(r_{i})-E_{2}(r_{i-1})\geq g_{2}(\varkappa
E_{2}(r_{i}))\int_{B_{r_{i}}\setminus B_{r_{i-1}}}q_{2}(x)\,dx,\quad
i=i_{0}+1,i_{0}+2,\ldots.$ (4.9)

In can be assumed without loss of generality that (2.1) and (4.4) are valid
for ${k=1}$. It follows from (4.8) that

$\sum_{i=i_{0}+1}^{\infty}\frac{E_{1}(r_{i})-E_{1}(r_{i-1})}{g_{1}(\varkappa
E_{1}(r_{i}))}\geq\int_{{\mathbb{R}}^{n}\setminus B_{r_{i_{0}}}}q_{1}(x)\,dx,$

whence in accordance with the evident estimates

$\int_{E_{1}(r_{i-1})}^{E_{1}(r_{i})}\frac{d\zeta}{g_{1}(\varkappa\zeta)}\geq\frac{E_{1}(r_{i})-E_{1}(r_{i-1})}{g_{1}(\varkappa
E_{1}(r_{i}))},\quad i=i_{0}+1,i_{0}+2,\ldots,$

we have

$\int_{E_{1}(r_{i_{0}})}^{\infty}\frac{d\zeta}{g_{1}(\varkappa\zeta)}\geq\int_{{\mathbb{R}}^{n}\setminus
B_{r_{i_{0}}}}q_{1}(x)\,dx.$

This contradicts (2.1) and (4.4).

To complete the proof, it remains to note that, in the case where (2.1) and
(4.4) are valid for $k=2$, we repeat our reasoning with (4.8) replaced by
(4.9). ∎

### 4.3

The above method is suitable for non-negative solutions of the inequality

$-\operatorname{div}A(x,\nabla u)\geq F(x,u)\quad\mbox{in }{\mathbb{R}}^{n}$
(4.10)

for which (3.6) is valid, where $A$ is a Caratheodory function satisfying the
uniform ellipticity condition (3.1) and the function $F$ is non-negative on
${\mathbb{R}}^{n}\times[0,\varepsilon]$, positive on
${\mathbb{R}}^{n}\times(0,\varepsilon)$, and non-decreasing with respect to
the last argument on the interval $[0,\varepsilon]$ for some real number
$0<\varepsilon<1$.

Denote

$f(r,\zeta)=\operatorname*{ess\,inf}\limits_{x\in B_{\theta r}\setminus
B_{r/\theta}}F(x,\zeta),\quad r,\zeta>0,$ (4.11)

where $\theta>1$ is some given real number. We shall assume that for any real
number $\zeta_{*}>0$ there exist a real number $r_{*}>0$ and functions
$q:[r_{*},\infty)\to[0,\infty)$ and $g:[\zeta_{*},\infty)\to(0,\infty)$
satisfying the following condition:

  1. $(c)$

for all $r\in[r_{*},\infty)$ and $\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{r^{n-p}}\right)^{1/(p-1)}\leq\varepsilon$

we have

$f\left(r,\left(\frac{\zeta}{r^{n-p}}\right)^{1/(p-1)}\right)\geq
q(r)g(\zeta).$

######  Theorem 4.2.

Suppose that $n>p$. Also let for any real number $\zeta_{*}>0$ there exist a
real number $r_{*}>0$, a locally bounded measurable function
$q:[r_{*},\infty)\to[0,\infty)$, and a non-decreasing function
$g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying condition $(c)$ and, moreover,

$\int_{\zeta_{*}}^{\infty}\frac{d\zeta}{g(\zeta)}<\infty$ (4.12)

and

$\int_{r_{*}}^{\infty}r^{n-1}q(r)\,dr=\infty.$ (4.13)

Then any non-negative solution of (4.10), (3.6) is identically zero.

###### Proof.

By contradiction, we assume that (4.10), (3.6) has a solution $u\geq 0$ that
is not identically zero. According to Lemma 3.3, this solution is positive
almost everywhere in ${\mathbb{R}}^{n}$. We denote
$u_{\varepsilon}=\chi_{\Omega_{\varepsilon}}u+(1-\chi_{\Omega_{\varepsilon}})\varepsilon,$
where $\Omega_{\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u(x)<\varepsilon\\}$ and
$\chi_{\Omega_{\varepsilon}}$ is the characteristic function of the set
$\Omega_{\varepsilon}$. Also let

$E(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}F(x,u)\,dx,\quad r>0.$

By Lemma 3.4 there is a real number $r_{0}>0$ such that (3.7) is valid for all
$r\geq r_{0}$. As in the proof of Theorem 2.1, we put $r_{i}=\theta^{i}r_{0}$,
$i=1,2,\ldots$. Corollary 3.2 yields

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}}F(x,u)\,dx\leq
Cr_{i}^{-p}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{\varepsilon}\right)^{p-1},\quad
i=1,2,\ldots,$

whence it follows that

$\operatorname*{ess\,inf}\limits_{\Omega_{\varepsilon}\cap
B_{r_{i}}}u=\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{\varepsilon}\geq\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)},\quad
i=1,2,\ldots.$ (4.14)

This in turn implies the estimate

$F(x,u(x))\geq
F\left(x,\operatorname*{ess\,inf}\limits_{\Omega_{\varepsilon}\cap
B_{r_{i}}}u\right)\geq
F\left(x,\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)$

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$.
Consequently,

$\displaystyle\frac{1}{\operatorname{mes}\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}F(x,u)\,dx$
$\displaystyle\qquad{}\geq\operatorname*{ess\,inf}\limits_{x\in\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}F\left(x,\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right),\quad
i=1,2,\ldots,$

whence in accordance with (3.7) and (4.11) we have

$E(r_{i})-E(r_{i-1})\geq Cr^{n}f\left(r,\left(\frac{\varkappa
E(r_{i})}{r^{n-p}}\right)^{1/(p-1)}\right)$

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. By condition $(c)$, this
implies that

$\frac{E(r_{i})-E(r_{i-1})}{g(\varkappa E(r_{i}))}\geq Cr^{n}q(r)$ (4.15)

for all $r\in(r_{i-1},r_{i})$, $i=i_{0}+1,i_{0}+2,\ldots$, where $i_{0}\geq 0$
is some integer. Combining the last estimate with the inequalities

$\int_{E(r_{i-1})}^{E(r_{i})}\frac{d\zeta}{g(\varkappa\zeta))}\geq\frac{E(r_{i})-E(r_{i-1})}{g(\varkappa
E(r_{i}))}$ (4.16)

and

$\sup_{r\in(r_{i-1},r_{i})}r^{n}q(r)\geq
C\int_{r_{i-1}}^{r_{i}}r^{n-1}q(r)\,dr,$

we obtain

$\int_{E(r_{i-1})}^{E(r_{i})}\frac{d\zeta}{g(\varkappa\zeta))}\geq
C\int_{r_{i-1}}^{r_{i}}r^{n-1}q(r)\,dr,\quad i=i_{0}+1,i_{0}+2,\ldots.$

Summing the last expression over all integers $i>i_{0}$, one can conclude that

$\int_{E(r_{i_{0}})}^{\infty}\frac{d\zeta}{g(\varkappa\zeta))}\geq
C\int_{r_{i_{0}}}^{\infty}r^{n-1}q(r)\,dr.$

Thus, we arrive at a contradiction with (4.12) and (4.13). ∎

For $F(x,\zeta)=a(x)h(\zeta)$, Theorem 4.2 remains valid with (4.11) replaced
by

$f(r,\zeta)=\frac{h(\zeta)}{\left(\frac{1}{\operatorname{mes}B_{\theta
r}\setminus B_{r/\theta}}\int_{B_{\theta r}\setminus
B_{r/\theta}}a^{-\lambda}(x)\,dx\right)^{1/\lambda}},\quad r,\zeta>0,$ (4.17)

where $\lambda>0$ is some real number. To see this, it suffices to repeat the
reasoning given in paragraph 4.1. Really, from (4.14), it follows that

$h(u(x))\geq
h\left(\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)$

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$.
Consequently, we have

$\displaystyle\frac{1}{\operatorname{mes}\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}h^{\lambda/(1+\lambda)}(u)\,dx$
$\displaystyle\qquad{}\geq
h^{\lambda/(1+\lambda)}\left(\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right),\quad
i=1,2,\ldots.$

Combining this with (3.7) and the estimate

$\displaystyle\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}h^{\lambda/(1+\lambda)}(u)\,dx=\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}a^{-\lambda/(1+\lambda)}(x)a^{\lambda/(1+\lambda)}(x)h^{\lambda/(1+\lambda)}(u)\,dx$
$\displaystyle\quad{}\leq\left(\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus
B_{r_{i-1}}}a^{-\lambda}(x)\,dx\right)^{1/(1+\lambda)}\left(\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}}a(x)h(u)\,dx\right)^{\lambda/(1+\lambda)}$

which follows from Hölder’s inequality, we obtain

$\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}a(x)h(u)\,dx\geq\frac{Cr_{i}^{n}h\left(\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)}{\left(\frac{1}{\operatorname{mes}B_{r_{i}}\setminus
B_{r_{i-1}}}\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}a^{-\lambda}(x)\,dx\right)^{1/\lambda}},\quad i=1,2,\ldots.$

The last relation implies (4.15) with

$E(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}a(x)h(u_{1})\,dx,\quad r>0,$

and $f$ defined by (4.17).

Theorem 4.2 can be strengthen if we assume that the function $F$ is non-
negative and non-decreasing with respect to the last argument on the whole set
${\mathbb{R}}^{n}\times[0,\infty)$ and positive on
${\mathbb{R}}^{n}\times(0,\infty)$. Namely, let for any real number
$\zeta_{*}>0$ there be a real number $r_{*}>0$ and functions
$q:{\mathbb{R}}^{n}\setminus B_{r_{*}}\to[0,\infty)$ and
$g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying the following condition:

  1. $(c^{\prime})$

for almost all $x\in{\mathbb{R}}^{n}\setminus B_{r_{*}}$ and for all
$\zeta\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta}{|x|^{n-p}}\right)^{1/(p-1)}\leq\varepsilon$

we have

$F\left(x,\left(\frac{\zeta}{|x|^{n-p}}\right)^{1/(p-1)}\right)\geq
q(x)g(\zeta).$

Here $\varepsilon\in(0,1)$ can be an arbitrary real number. We should only
assume that $\varepsilon$ does not depend on $\zeta_{*}$.

######  Theorem 4.3.

Suppose that $n>p$ and the function $F$ is non-negative and non-decreasing
with respect to the last argument on the set
${\mathbb{R}}^{n}\times[0,\infty)$ and positive on
${\mathbb{R}}^{n}\times(0,\infty)$. Also let for any real number $\zeta_{*}>0$
there be a real number $r_{*}>0$, a locally bounded measurable function
$q:{\mathbb{R}}^{n}\setminus B_{r_{*}}\to[0,\infty)$, and a non-decreasing
function $g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying condition
$(c^{\prime})$. If (4.12) is valid and, moreover,

$\int_{{\mathbb{R}}^{n}\setminus B_{r_{*}}}q(x)\,dx=\infty,$ (4.18)

then any non-negative solution of (4.10), (3.6) is identically zero.

###### Proof.

By contradiction, let there be a non-negative solution of (4.10), (3.6) that
is not identically zero. According to Lemma 3.3, this solution is positive
almost everywhere in ${\mathbb{R}}^{n}$. We put

$E(r)=\int_{B_{r}}F(x,u)\,dx,\quad r>0.$

By (3.6), there is a real number $r_{0}>0$ such that

$\operatorname*{ess\,inf}\limits_{B_{r_{0}}}u<\varepsilon.$

We also denote $r_{i}=2^{i}$, $i=1,2,\ldots$. Lemmas 3.2 and 3.3 lead to the
estimate

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{B_{r_{i}}}F(x,u)\,dx\leq
Cr_{i}^{-p}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u\right)^{p-1},\quad
i=1,2,\ldots,$

whence it follows that

$\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u\geq\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)},\quad
i=1,2,\ldots.$

This in turn yields

$F(x,u(x))\geq
F\left(x,\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u\right)\geq
F\left(x,\sigma\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)$

for almost all $x\in B_{r_{i}}$, $i=1,2,\ldots$. Hence, taking into account
condition $(c^{\prime})$, we have

$F(x,u(x))\geq q(x)g(\varkappa E(r_{i}))$

for almost all $x\in B_{r_{i}}\setminus B_{r_{i-1}}$,
$i=i_{0}+1,i_{0}+2,\ldots$, where $i_{0}\geq 0$ is some integer. Integrating
the last inequalities over $B_{r_{i}}\setminus B_{r_{i-1}}$, one can conclude
that

$E(r_{i})-E(r_{i-1})\geq g(\varkappa E(r_{i}))\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}q(x)\,dx$

or, in other words,

$\frac{E(r_{i})-E(r_{i-1})}{g(\varkappa E(r_{i}))}\geq\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}q(x)\,dx\quad i=i_{0}+1,i_{0}+2,\ldots.$

Combining this with (4.16), we obtain

$\int_{E(r_{i-1})}^{E(r_{i})}\frac{d\zeta}{g(\varkappa\zeta))}\geq\int_{B_{r_{i}}\setminus
B_{r_{i-1}}}q(x)\,dx\quad i=i_{0}+1,i_{0}+2,\ldots.$

Thus, summing the last expression over all $i>i_{0}$, we arrive at the
inequality

$\int_{E(r_{i_{0}})}^{\infty}\frac{d\zeta}{g(\varkappa\zeta))}\geq
C\int_{{\mathbb{R}}^{n}\setminus B_{r_{i_{0}}}}q(x)\,dx$

which contradicts (4.12) and (4.18). ∎

### 4.4

Consider the systems

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
F_{1}(x,u_{1},u_{2})&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq F_{2}(x,u_{1},u_{2})&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$ (4.19)

where $A_{1}$ and $A_{2}$ are Caratheodory functions satisfying the uniform
ellipticity condition (1.2) and, moreover, the functions $F_{1}$ and $F_{2}$
are non-negative and non-decreasing with respect to the last two arguments on
the set ${\mathbb{R}}^{n}\times[0,\varepsilon]\times[0,\varepsilon]$ and
positive on ${\mathbb{R}}^{n}\times(0,\varepsilon)\times(0,\varepsilon)$ for
some real number $0<\varepsilon<1$.

We shall assume that for any real number $\zeta_{*}>0$ there exist a real
number $r_{*}>0$ and functions $q:[r_{*},\infty)\to[0,\infty)$ and
$g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying the following condition:

  1. $(d)$

for all $r\in[r_{*},\infty)$ and $\zeta_{1},\zeta_{2}\in[\zeta_{*},\infty)$
such that

$\left(\frac{\zeta_{1}}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\leq\varepsilon\quad\mbox{and}\quad\left(\frac{\zeta_{2}}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\leq\varepsilon$

we have

$f\left(r,\left(\frac{\zeta_{1}}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\zeta_{2}}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\geq
q(r)g(\zeta_{1}+\zeta_{2}),$

where

$f(r,\zeta_{1},\zeta_{2})=\operatorname*{ess\,inf}\limits_{x\in B_{\theta
r}\setminus
B_{r/\theta}}(F_{1}(x,\zeta_{1},\zeta_{2})+F_{2}(x,\zeta_{1},\zeta_{2})),\quad
r,\zeta_{1},\zeta_{2}>0,$

with some given real number $\theta>1$.

######  Theorem 4.4.

Suppose that $n>p$. Also let for any real number $\zeta_{*}>0$ there exist a
real number $r_{*}>0$, a locally bounded measurable function
$q:[r_{*},\infty)\to[0,\infty)$, and a non-decreasing function
$g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying conditions $(d)$, (4.12), and
(4.13). Then any non-negative solution of (4.19), (1.3) is identically zero.

###### Proof.

Assume by contradiction that $(u_{1},u_{2})$ is a non-negative solution of
(4.19), (1.3) not equal to zero. In view of Lemma 3.2, the functions $u_{1}$
and $u_{2}$ are positive almost everywhere in ${\mathbb{R}}^{n}$. As in the
proof of Theorem 2.1, we denote
$\Omega_{\varepsilon}=\Omega_{1,\varepsilon}\cap\Omega_{2,\varepsilon}$, where
$\Omega_{k,\varepsilon}=\\{x\in{\mathbb{R}}^{n}:u_{k}(x)<\varepsilon\\}$,
$k=1,2$. By Lemma 3.4, there is a real number $r_{0}>0$ such that (3.7) holds
for all $r\geq r_{0}$. We put $r_{i}=\theta^{i}r_{0}$, $i=1,2,\ldots$. Also
let

$E_{1}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}F_{1}(x,u_{1},u_{2})\,dx,\quad
r>0,$

and

$E_{2}(r)=\int_{\Omega_{\varepsilon}\cap B_{r}}F_{2}(x,u_{1},u_{2})\,dx,\quad
r>0.$

Corollary 3.2 allows us to assert that

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}}F_{1}(x,u_{1},u_{2})\,dx\leq
Cr_{i}^{-p_{1}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1,\varepsilon}\right)^{p_{1}-1},\quad
i=1,2,\ldots,$

and

$\frac{1}{\mathop{\rm mes}\nolimits B_{r_{i}}}\int_{\Omega_{\varepsilon}\cap
B_{r_{i}}}F_{2}(x,u_{1},u_{2})\,dx\leq
Cr_{i}^{-p_{2}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2,\varepsilon}\right)^{p_{2}-1},\quad
i=1,2,\ldots,$

where the functions $u_{k,\varepsilon}$, $k=1,2$, are defined by (3.8). This
immediately yields

$\operatorname*{ess\,inf}\limits_{\Omega_{1,\varepsilon}\cap
B_{r_{i}}}u_{1}=\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1,\varepsilon}\geq\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\quad
i=1,2,\ldots,$ (4.20)

and

$\operatorname*{ess\,inf}\limits_{\Omega_{2,\varepsilon}\cap
B_{r_{i}}}u_{2}=\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2,\varepsilon}\geq\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)},\quad
i=1,2,\ldots.$ (4.21)

Hence, taking into account the fact that $F_{1}$ and $F_{2}$ are non-
decreasing functions with respect to the last two arguments on the set
${\mathbb{R}}^{n}\times[0,\varepsilon]\times[0,\varepsilon]$, we obtain

$\displaystyle F_{k}(x,u_{1}(x),u_{2}(x))\geq
F_{k}\left(x,\operatorname*{ess\,inf}\limits_{\Omega_{1,\varepsilon}\cap
B_{r_{i}}}u_{1},\operatorname*{ess\,inf}\limits_{\Omega_{2,\varepsilon}\cap
B_{r_{i}}}u_{2}\right)$ $\displaystyle\qquad{}\geq
F_{k}\left(x,\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right),\quad
k=1,2,$ (4.22)

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$. This
implies that

$\displaystyle F_{1}(x,u_{1}(x),u_{2}(x))+F_{2}(x,u_{1}(x),u_{2}(x))$
$\displaystyle\qquad{}\geq f\left(r,\left(\frac{\varkappa
E_{1}(r_{i})}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\varkappa
E_{2}(r_{i})}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}\setminus B_{r_{i-1}}$
and for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. Integrating the last
expression, we have

$\displaystyle\int_{\Omega_{\varepsilon}\cap B_{r_{i}}\setminus
B_{r_{i-1}}}(F_{1}(x,u_{1},u_{2})+F_{2}(x,u_{1},u_{2}))\,dx$
$\displaystyle\qquad{}\geq Cr_{i}^{n}f\left(r,\left(\frac{\varkappa
E_{1}(r_{i})}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\varkappa
E_{2}(r_{i})}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$

for all $r\in(r_{i-1},r_{i})$, $i=1,2,\ldots$. Taking further condition $(d)$
into account, we arrive at estimate (4.15) with

$E(r)=E_{1}(r)+E_{2}(r),\quad r>0.$ (4.23)

Thus, to complete the proof, it remains to repeat the argument given in the
proof of Theorem 4.2. ∎

Theorem 4.4 can be strengthen if the $F_{1}$ and $F_{2}$ are non-decreasing
functions with respect to the last two arguments on the whole set
${\mathbb{R}}^{n}\times[0,\infty)\times[0,\infty)$.

In this case, it should be assumed that for any real number $\zeta_{*}>0$
there exist a real number $r_{*}>0$ and functions $q:{\mathbb{R}}^{n}\setminus
B_{r_{*}}\to[0,\infty)$ and $g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying the
following condition:

  1. $(d^{\prime})$

for almost all $x\in{\mathbb{R}}^{n}\setminus B_{r_{*}}$ and for all
$\zeta_{1},\zeta_{2}\in[\zeta_{*},\infty)$ such that

$\left(\frac{\zeta_{1}}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)}\leq\varepsilon\quad\mbox{and}\quad\left(\frac{\zeta_{2}}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\leq\varepsilon$

we have

$\displaystyle
F_{1}\left(x,\left(\frac{\zeta_{1}}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\zeta_{2}}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$
$\displaystyle\qquad{}+F_{2}\left(x,\left(\frac{\zeta_{1}}{|x|^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\zeta_{2}}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\geq
q(x)g(\zeta_{1}+\zeta_{2}).$

In so doing, we do not make any assumptions regarding the real number
$\varepsilon\in(0,1)$, except that it does not depend on $\zeta_{*}$.

######  Theorem 4.5.

Suppose that $n>p$ and the functions $F_{1}$ and $F_{2}$ are non-negative and
non-decreasing with respect to the last two arguments on the set
${\mathbb{R}}^{n}\times[0,\infty)\times[0,\infty)$ and positive on
${\mathbb{R}}^{n}\times(0,\infty)\times(0,\infty)$. Also let for any real
number $\zeta_{*}>0$ there be a real number $r_{*}>0$, a locally bounded
measurable function $q:{\mathbb{R}}^{n}\setminus B_{r_{*}}\to[0,\infty)$, and
a non-decreasing function $g:[\zeta_{*},\infty)\to(0,\infty)$ satisfying
conditions $(d^{\prime})$, (4.12), and (4.18). Then any non-negative solution
of (4.19), (1.3) is identically zero.

###### Proof.

By contradiction, let $(u_{1},u_{2})$ be a non-negative solution of (4.19),
(1.3) that is not equal to zero. According to Lemma 3.2, both the functions
$u_{1}$ and $u_{2}$ are positive almost everywhere in ${\mathbb{R}}^{n}$. In
vie of (1.3), there exists a real number $r_{0}>0$ such that (4.5) is valid
for all $r\geq r_{0}$. Also let $r_{i}=\theta^{i}r_{0}$, $i=1,2,\ldots$.

We denote

$E_{1}(r)=\int_{B_{r}}F_{1}(x,u_{1},u_{2})\,dx,\quad r>0,$

and

$E_{2}(r)=\int_{B_{r}}F_{2}(x,u_{1},u_{2})\,dx,\quad r>0.$

Corollary 3.2 implies the estimates

$\frac{1}{\mathop{\rm mes}\nolimits
B_{r_{i}}}\int_{B_{r_{i}}}F_{1}(x,u_{1},u_{2})\,dx\leq
Cr_{i}^{-p_{1}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1}\right)^{p_{1}-1},\quad
i=1,2,\ldots,$

and

$\frac{1}{\mathop{\rm mes}\nolimits
B_{r_{i}}}\int_{B_{r_{i}}}F_{2}(x,u_{1},u_{2})\,dx\leq
Cr_{i}^{-p_{2}}\left(\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\right)^{p_{2}-1},\quad
i=1,2,\ldots,$

whence it follows that

$\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1}\geq\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\quad
i=1,2,\ldots,$

and

$\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\geq\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)},\quad
i=1,2,\ldots.$

Taking into account the fact that $F_{1}$ and $F_{2}$ are non-decreasing
functions with respect to the last two arguments on the set
${\mathbb{R}}^{n}\times[0,\infty)\times[0,\infty)$, we have

$\displaystyle F_{k}(x,u_{1}(x),u_{2}(x))$ $\displaystyle{}\geq
F_{k}\left(x,\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{1},\operatorname*{ess\,inf}\limits_{B_{r_{i}}}u_{2}\right)$
$\displaystyle{}\geq
F_{k}\left(x,\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right),\quad
k=1,2,$

for almost all $x\in B_{r_{i}}\setminus B_{r_{i-1}}$, $i=1,2,\ldots$. By
condition $(d^{\prime})$, this implies that

$F_{1}(x,u_{1}(x),u_{2}(x))+F_{2}(x,u_{1}(x),u_{2}(x))\geq
q(x)g(\varkappa(E_{1}(r_{i})+E_{2}(r_{i})))$

for almost all $x\in B_{r_{i}}\setminus B_{r_{i-1}}$, $i=1,2,\ldots$.
Integrating the last inequality over $B_{r_{i}}\setminus B_{r_{i-1}}$, we
obtain (4.15) with $E$ defined by (4.23). Thus, it remains to repeat the
argument given in the proof of Theorem 4.2. ∎

### 4.5

In (4.19), let the functions $F_{1}$ and $F_{2}$ do not depend on $x$. In
other words, we deal with the system

$\left\\{\begin{aligned} &-\operatorname{div}A_{1}(x,\nabla u_{1})\geq
F_{1}(u_{1},u_{2})&\mbox{in }{\mathbb{R}}^{n},\\\
&-\operatorname{div}A_{2}(x,\nabla u_{2})\geq F_{2}(u_{1},u_{2})&\mbox{in
}{\mathbb{R}}^{n},\end{aligned}\right.$ (4.24)

where $A_{1}$ and $A_{2}$ are Caratheodory functions satisfying the uniform
ellipticity condition (1.2). In so doing, the functions $F_{1}$ and $F_{2}$
are non-negative and non-decreasing with respect to every of their arguments
on the set $[0,\varepsilon]\times[0,\varepsilon]$ and positive on
$(0,\varepsilon)\times(0,\varepsilon)$, where $0<\varepsilon<1$ is some real
number.

We shall assume that for any real number $\zeta_{*}>0$ there exist real
numbers $r_{*}>0$ and $p>1$ and a function $f:(0,\infty)\to(0,\infty)$
satisfying the following condition:

  1. $(e)$

for all $r\in[r_{*},\infty)$ and $\zeta_{1},\zeta_{2}\in[\zeta_{*},\infty)$
such that

$\left(\frac{\zeta_{1}}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)}\leq\varepsilon\quad\mbox{and}\quad\left(\frac{\zeta_{2}}{|x|^{n-p_{2}}}\right)^{1/(p_{2}-1)}\leq\varepsilon$

we have

$\displaystyle
F_{1}\left(\left(\frac{\zeta_{1}}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\zeta_{2}}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$
$\displaystyle\qquad{}+F_{2}\left(\left(\frac{\zeta_{1}}{r^{n-p_{1}}}\right)^{1/(p_{1}-1)},\left(\frac{\zeta_{2}}{r^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)\geq
f\left(\left(\frac{\zeta_{1}+\zeta_{2}}{r^{n-p}}\right)^{1/(p-1)}\right).$

######  Theorem 4.6.

Suppose that $n>\operatorname{max}\\{p_{1},p_{2}\\}$. Also let for any real
number $\zeta_{*}>0$ there exist real numbers $r_{*}>0$ and $p>1$ and a non-
decreasing function $f:(0,\infty)\to(0,\infty)$ satisfying condition $(e)$
and, moreover,

$\int_{0}^{\varepsilon}\frac{f(\zeta)\,d\zeta}{\zeta^{1+n(p-1)/(n-p)}}=\infty.$

Then any non-negative solution of (4.24), (1.3) is identically zero.

###### Proof.

Let us follow the same argument and notation as in the proof of Theorem 4.4.
The only difference is that $F_{1}$ and $F_{2}$ are now independent of the
spatial variable. Taking into account (4.22), we have

$\displaystyle F_{1}(u_{1}(x),u_{2}(x))+F_{1}(u_{1}(x),u_{2}(x))$
$\displaystyle\quad{}\geq
F_{1}\left(\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$
$\displaystyle\qquad{}+F_{2}\left(\sigma\left(\frac{E_{1}(r_{i})}{r_{i}^{n-p_{1}}}\right)^{1/(p_{1}-1)},\sigma\left(\frac{E_{2}(r_{i})}{r_{i}^{n-p_{2}}}\right)^{1/(p_{2}-1)}\right)$

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$. In
view of condition $(e)$, this implies that

$F_{1}(u_{1}(x),u_{2}(x))+F_{1}(u_{1}(x),u_{2}(x))\geq
f\left(\varkappa\left(\frac{E_{1}(r_{i})+E_{2}(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)$

for almost all $x\in\Omega_{\varepsilon}\cap B_{r_{i}}$, $i=1,2,\ldots$.
Integrating the last relation over $\Omega_{\varepsilon}\cap
B_{r_{i}}\setminus B_{r_{i-1}}$, we arrive in accordance with (3.7) at the
following estimate:

$E(r_{i})-E(r_{i-1})\geq
Cr_{i}^{n}f\left(\varkappa\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right),\quad
i=1,2,\ldots,$

where the function $E$ is defined by (4.23). In particular, one can assert
that $n>p$; otherwise we obtain

$\operatorname*{lim\,inf}\limits_{i\to\infty}\frac{E(r_{i})}{r_{i}^{n}}\geq
C\operatorname*{lim\,inf}\limits_{i\to\infty}f\left(\varkappa\left(\frac{E(r_{i})}{r_{i}^{n-p}}\right)^{1/(p-1)}\right)>0,$

which contradicts (4.20) and (4.21). Thus, to complete the proof, it suffices
to repeat the argument given in the proof of [11, Theorem 2.1]. ∎

### 4.6

Let us note in conclusion that the method outlined in our paper is obviously
suitable for systems containing more then two inequalities. It is also not
difficult to transfer all the above results to Carnot groups.

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