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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>Properties of Admissible Set of an Optimal Non-destructive System Exploitation Problem in Some General Formalization</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Vladivir D. Mazurov</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <contrib contrib-type="author">
          <string-name>Alexander I. Smirnov</string-name>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>Krasovskii Institute of Mathematics and Mechanics UB RAS</institution>
          ,
          <addr-line>Ekaterinburg</addr-line>
          ,
          <country country="RU">Russia</country>
        </aff>
      </contrib-group>
      <fpage>359</fpage>
      <lpage>371</lpage>
      <abstract>
        <p>We study the properties of admissible strategies for the problem of using of a renewable resource system that does not lead to its destruction, in some general formulation. It is assumed that the system evolution is described by some iterative process. The conditions for the stable existence of this system, originally formulated in terms of the asymptotic properties of this iterative process, reduce to the existence of admissible controls for some mathematical programming problem. We analyzed the properties of its admissible set in terms of the dominant eigenvalues of some positively homogeneous maps. We obtained the conditions for the existence of optimal controls under the in uence of which the system can stably exist for an inde nitely long time. Introduced by author's generalization of the classical concept of the irreducibility of a non-linear map, the concept of local irreducibility, is essentially used. The problem under consideration has important applications in the tasks of rational exploitation of renewable natural resources, in particular, in the tasks of managing ecological populations.</p>
      </abstract>
      <kwd-group>
        <kwd>Concave programming</kwd>
        <kwd>Irreducible map</kwd>
        <kwd>Positive equilib- rium</kwd>
        <kwd>Rational exploitation of natural resources</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>-</title>
      <p>One of a vital problems that require the use of mathematical methods is a
problem of a rational use of renewable resources. Particularly acute in practice is
a problem of rational non-destructive ecosystem exploitation. This problem is
characterized by an abundance of speci c approaches for speci c natural
systems (see the review in [4]). In the matrix formulation, the approach to optimal
exploitation of a population in a stationary state has prevailed for a long time.
This approach uses as a structure of the population a stable distribution |
the structure of an eigenvector corresponding to the dominant eigenvalue of the
Copyright c by the paper's authors. Copying permitted for private and academic purposes.</p>
      <p>In: S. Belim et al. (eds.): OPTA-SCL 2018, Omsk, Russia, published at http://ceur-ws.org
projection matrix (that must be greater than unity). If at some point in time
the structure of the population coincides with a stable distribution, then it is
possible to remove the surplus without damage to the population, returning the
population to the previous state. This corresponds to the withdrawal of the same
proportion of all its constituent groups.</p>
      <p>One of the rst theoretical works in this direction was [2], where the concept
of sustainable yield was de ned, and the solvability of the problem of nding the
maximum level of permissible exploitation was proved. In work [1] some concepts
natural for the exploitation of ecosystems were formalized and the corresponding
statements were rigorously proved. The rst successful attempts to generalize
these results to the case of a nonlinear density dependence are related to papers
[11, 3] that were considered the dependence on density only for the rst age class.</p>
    </sec>
    <sec id="sec-2">
      <title>In [5] was considered the model with dependence on density for all age classes. We are developing here a general approach to this problem that considers the ecosystem as a discrete system, with its representation in the form of an iterative process</title>
      <p>
        xt+1 = F (xt);
t = 0; 1; 2; : : :
(
        <xref ref-type="bibr" rid="ref1">1</xref>
        )
where xt means the state of the system at time t = 0; 1; 2; : : :, the step operator
F transforms the system from the previous state to the following. With this
approach, one can naturally formalize the inadmissibility of excessive load on
the dynamic system condition.
      </p>
      <p>We will use the following notation: Rq+ denotes the non-negative orthant of
Rq; x y means y x 2 Rq+; x &lt; y means y x 2 int Rq+; int M | the
interior of the set M ; x y means x y and x 6= y; Z | the set of integers;
m; n = fi 2 Z j m i ng. Vector x = (x1; x2; : : : ; xq) can be written in
the short-hand form as x = (xi). The iteratives of the map F are denoted by
F t(x) (t = 1; 2; : : :), F 0(x) x.</p>
    </sec>
    <sec id="sec-3">
      <title>A controlled system is modelled by iterative process</title>
      <p>
        xt+1 = Fu(xt);
t = 0; 1; 2; : : : ;
(
        <xref ref-type="bibr" rid="ref2">2</xref>
        )
where x = (xi), u = (ui), Fu(x) = (F (x) u))+, a+ = maxfa; 0g, x+ = (xi+).
We will denote by fxt(x0; u)gt+=10 , xt(x0; u) = (xit(x0; u)) the realization of this
process starting from the initial vector x0. In [7] we formulated an optimization
problem for dynamical system (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ): determine
c~ = maxfc(u) j u 2 U g;
(
        <xref ref-type="bibr" rid="ref3">3</xref>
        )
where c(u) | the total e ect of using of system elements in the quantities
u1; u2; : : : ; uq, the set U is the closure of the set
      </p>
      <p>q</p>
      <p>U = fu 2 R+ j X0(u) 6= ?g
of controls. We will be more interested in the properties of the admissible set U ,
so for simplicity we shall assume that c(u) is linear nonnegative on Rq+ function.</p>
    </sec>
    <sec id="sec-4">
      <title>The control vector u is considered as admissible if there is at least one initial</title>
      <p>state x0, for which all units of the system stably exist for an inde nitely long
time:</p>
      <p>q</p>
      <p>X0(u) = fx0 2 R+ j intf xit(x0; u) &gt; 0 (8i 2 1; q)g</p>
    </sec>
    <sec id="sec-5">
      <title>Thus, the size of each unit of the system should not decrease with time to</title>
      <p>zero. As can be seen from this de nition, the set U formalizes the requirement
of the indestructability of the managed object.
the AmanpecFes,sai.rey.,ccoonnddititioionnfoNrF+U6 =6=??. Iins tthheisecxaissteenthcee soeftaUpocsointitvaeinsxaetdlepaositntthoef
trivial control: 0 2 U .</p>
    </sec>
    <sec id="sec-6">
      <title>Unfortunately, the set U is not always closed; thus, in problem (3), its closure is considered as the admissible set.</title>
      <sec id="sec-6-1">
        <title>Denote by u~ and Ue the optimal vector and optimal set of problem (3); and</title>
        <p>
          by Nu and Nu+ the sets of nonzero xed points and positive xed points of Fu,
respectively. It is assumed that the original model without control (
          <xref ref-type="bibr" rid="ref1">1</xref>
          ) has the
trivial equilibrium state: F (0) = 0.
        </p>
      </sec>
    </sec>
    <sec id="sec-7">
      <title>We are interested in the positive realizations of the iterative process (2) only,</title>
      <p>when xt(x0; u) &gt; 0 (8t = 1; 2; : : :). In this case Fu(x) = F (x) u &gt; 0 for all
x = xt(x0; u). Note that if x 2 Nu+, then Fu(x) &gt; 0 and F (x) = x + u also must
be ful lled.</p>
    </sec>
    <sec id="sec-8">
      <title>The main requirement imposed on the class of maps under consideration is</title>
      <p>a concavity or its weakening, a subhomogeneity [6]. The de nitions of
subhomogeneous and superhomogeneous maps, as well as other standard concepts used
below, are given, for example, in [8]. The subhomogeneity of a map make it
possible to depict adequately a so-called saturation e ect observed in many natural
systems with limited resources.</p>
    </sec>
    <sec id="sec-9">
      <title>In what follows, the concept of irreducibility (indecomposability) of a map</title>
      <p>is essentially used. Along with the classical concept of irreducibility [10], we will
also use its local analogues [12], such as a notion of irreducibility of a map at
zero. We give a classical de nition of the irreducibility of a map in a somewhat
di erent way, using the following sets:</p>
      <p>I+(x; y) = fj 2 1; q j xj &gt; yj g;</p>
      <p>I0(x; y) = fj 2 1; q j xj = yj g;
I+(x) = fi 2 1; q j xi &gt; 0g;</p>
      <p>I0(x) = fi 2 1; q j xi = 0g:
A map F is said to be reducible at a point y 2 Rq+, if
9x 2 Rq+ :
x
y;</p>
      <p>I0(x; y) 6= ?;</p>
      <p>I0(x; y)</p>
      <p>
        I0(F (x); F (y)):
(
        <xref ref-type="bibr" rid="ref4">4</xref>
        )
      </p>
    </sec>
    <sec id="sec-10">
      <title>A map F is said to be irreducible at point y, if it is not reducible at point y.</title>
    </sec>
    <sec id="sec-11">
      <title>A map reducible (or irreducible) at every point of set M is said to be reducible</title>
      <p>(or irreducible, respectively) on the set M .</p>
      <p>We consider separately the case of irreducibility at the point y = 0. A map
q q
F 2 fR+ 7! R+g is said to be reducible at point y = 0 (reducible at zero), if
9x 2 Rq+ :
x</p>
    </sec>
    <sec id="sec-12">
      <title>We note that the concept of reducibility at zero makes sense only for maps</title>
      <p>satisfying condition F (0) 6&gt; 0.
q</p>
      <sec id="sec-12-1">
        <title>The global irreducibility of a map means its irreducibility at all points R+</title>
        <p>and, in particular, irreducibility at zero.</p>
      </sec>
    </sec>
    <sec id="sec-13">
      <title>In a research of asymptotic properties of iterative processes a primitivity</title>
      <p>
        property of a map [10] is usually used. Although irreducibility at zero is a weaker
property in comparison with primitivity at zero, it also guarantees the
positivity of nonzero points of a map. Indeed, suppose, by way of contradiction, that
I0(x) 6= ?. Then from fi(x) = xi (8i 2 I0(x)) we get I0(x) I0(F (x)), which
conradicts, by (
        <xref ref-type="bibr" rid="ref5">5</xref>
        ), the irreducibility of F at zero. Therefore I0(x) = ? and
x &gt; 0.
      </p>
    </sec>
    <sec id="sec-14">
      <title>The local irreducibility properties for monotone increasing subhomogeneous maps are considered in [8, 9].</title>
    </sec>
    <sec id="sec-15">
      <title>We re ne our assumptions about properties of dynamical system (1). The</title>
      <p>
        step operator F is assumed to be nonnegative and concave on Rq+ (and hence it
is monotone increasing on Rq+). It is also natural to assume that the condition
for the absence of identically zero coordinates of F is satis ed:
(
        <xref ref-type="bibr" rid="ref6">6</xref>
        )
It is important that this property for monotone increasing subhomogeneous map
F implies that F (int Rq+) int Rq+. A su cient condition for (
        <xref ref-type="bibr" rid="ref6">6</xref>
        ) is the
irreducibility of F at zero. Since irreducibility at zero also guarantees the
positiveness of non-zero xed points, it is convenient to assume from now on that the
map F is irreducible at zero.
2
      </p>
      <p>Some Preliminary Results
For the subhomogeneous map F we can de ne the following positively
homogeneous maps:</p>
      <p>F0(x) = lim
!+0
1F ( x);</p>
      <p>F1(x) = lim
!+1
1F ( x);</p>
    </sec>
    <sec id="sec-16">
      <title>It is easy to see that</title>
    </sec>
    <sec id="sec-17">
      <title>De ne the following sets:</title>
      <p>F1(x)</p>
      <p>F (x)</p>
      <p>F0(x) (8x 2 Rq+):
PF+ = fx &gt; 0 j F (x) &gt; xg;</p>
      <p>
        QF+ = fx &gt; 0 j F (x) &lt; xg:
It is shown in [13] that for a subhomogeneous, monotone increasing and
irreq q
ducible at zero map F 2 fR+ 7! R+g the following properties are valid:
(F0) &gt; 1 , PF+ 6= ?;
(F1) &lt; 1 , QF+ 6= ?:
(
        <xref ref-type="bibr" rid="ref7">7</xref>
        )
(
        <xref ref-type="bibr" rid="ref8">8</xref>
        )
(
        <xref ref-type="bibr" rid="ref9">9</xref>
        )
Likewise, for the superhomogeneous, monotone increasing and irreducible at zero
map F the following properties are valid:
(F0) &lt; 1 , QF+ 6= ?;
(F1) &gt; 1 , PF+ 6= ?:
      </p>
    </sec>
    <sec id="sec-18">
      <title>We introduce the set and quantities</title>
      <p>MF (x) = f
&gt; 0 j F ( x =</p>
      <p>F (x)g;</p>
      <p>F (x) = inf MF (x); F (x) = sup MF (x);
where x is an arbitrary nonzero vector. Note that the set MF (x) always contains
unity as an element.</p>
      <p>
        Existence conditions of positive xed points for subhomogeneous maps on
q
cone R+ were studied in [13]. It follows from (
        <xref ref-type="bibr" rid="ref8">8</xref>
        ) due to [10, Theorem 10.3] that
the condition (F1) 1 (F0) is necessary for existence of a positive xed
point for monotone increasing, subhomogeneous and irreducible at zero map. A
similar condition with strict inequalities:
      </p>
      <p>(F1) &lt; 1 &lt; (F0)
is su cient for existence of a positive xed point. Su cient conditions with
(F1) = 1 or (F0) = 1 require the use of the global irreducibility property.</p>
    </sec>
    <sec id="sec-19">
      <title>With some work it can be shown that if F is monotone increasing, subho</title>
      <p>
        mogeneous, and condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed, then NF+ is bounded, convex and
contains a largest element xF . Moreover, a concave globally irreducible map
satisfying condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) has a unique positive xed point [13, Corollary 2.2.1].
      </p>
      <p>If Nu+ = ? for all u &gt; 0, then it follows from condition NF+ 6= ? that
U = f g</p>
      <p>
        0 . This case is uninteresting to us, therefore, in addition to condition
NF+ 6= ?, it is necessary to assume that Nu+ 6= ? for some u &gt; 0. In this case
F (x) = x + u &gt; x (8x 2 Nu+), and by (
        <xref ref-type="bibr" rid="ref10">10</xref>
        ), we have (F0) &gt; 1. For this reason,
the condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is assumed everywhere below. Then the set U is also convex,
contains some positive vector and with each positive vector u also contains the
segment [0; u]; those U is so-called downward set. Besides the following equalities
hold:
      </p>
      <p>q
U = fu 2 R+ j Nu+ 6= ?g;
q</p>
      <p>U = fu 2 R+ j Nu 6= ?g;</p>
      <p>
        Finally, it was shown in [7] that under above assumptions the problem of
mathematical programming
maxfc(u) j x = F (x)
u; x
0; u
0g
is solvable in the sense that c~ &lt; +1 and this value is attained on certain
admissible vectors u~, x~. Moreover, c~, u~ is a solution of problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) if and only if
c~, u~, x~ is a solution of problem (16).
      </p>
      <p>
        We now turn to properties of the admissible set for this problem. Proceeding
from its basic interpretation (
        <xref ref-type="bibr" rid="ref10">10</xref>
        ), we will be interested rst of all in properties
and conditions for existence of the positive equilibria x~ for the control u~.
(
        <xref ref-type="bibr" rid="ref11">11</xref>
        )
(
        <xref ref-type="bibr" rid="ref12">12</xref>
        )
(
        <xref ref-type="bibr" rid="ref13">13</xref>
        )
(
        <xref ref-type="bibr" rid="ref14">14</xref>
        )
(15)
(16)
      </p>
      <p>
        Some Properties of the Admissible Set of Problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        )
Note that the set Nu for u 2 U contains the largest element xu and the map
x(u) : u ! xu is monotone decreasing on U . It can also be shown that x(u)
inherits the concavity of the map F .
      </p>
    </sec>
    <sec id="sec-20">
      <title>We introduce the set</title>
      <p>
        q
D = fu 2 R+ j Nu 6= ?; Nv = ? (8v &gt; u)g
(17)
that forms part of a boundary of U . The set D is of interest in connection
with the fact that it contains all potentially optimal vectors of the problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ):
Ue D. We divide this set into two disjoint parts by the criterion of the presence
or absence of common points with U :
      </p>
    </sec>
    <sec id="sec-21">
      <title>From equalities (15) we obtain the following representation of D00:</title>
      <p>D0 = D n D00;</p>
      <p>D00 = U n U:
q</p>
      <p>D00 = fu 2 R+ j Nu 6= ?; Nu+ = ?g:
We note that, by de nition of D00, for elements of Nu with u 2 D00 the following
property is true:</p>
      <p>xu 2= int Rq+ (8u 2 D00):</p>
    </sec>
    <sec id="sec-22">
      <title>Directly from (17){(19) the following representation follows:</title>
      <p>D0 = fu j Nu+ 6= ?; Nv = ? (8v &gt; u)g:
The representations (19), (21) can be clari ed. Indeed, it follows from (21), in
view of (19), that D0 U . Since D0 D, we get D0 D \ U . On the other
hand, D \ U D0, because u 2 U together with the second equality in (18)
means u 2= D00. Hence, taking into account u 2 D and the rst equality in (18),
we obtain u 2 D0. Therefore, D0 = D \ U .</p>
      <sec id="sec-22-1">
        <title>Further, from the second equality in (18) it follows that D n U D00. On the</title>
        <p>other hand, D00 D n U , since D00 D and D00 \ U = ?. Therefore D00 = D n U .</p>
      </sec>
    </sec>
    <sec id="sec-23">
      <title>Thus, we have the following equalities:</title>
      <p>D0 = D \ U;</p>
      <p>
        D00 = D n U:
These equalities show that the positive optimal vector x~ for the optimal control
u~ of the problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) exist if and only if u~ 2 D0. Using (22), we can re ne that
U n D = U n D0.
      </p>
      <sec id="sec-23-1">
        <title>The set U n D does not a fortiori contain the optimal vectors of the prob</title>
        <p>
          lem (
          <xref ref-type="bibr" rid="ref3">3</xref>
          ). We give without proof the following assertion characterizing this set.
        </p>
        <p>
          q
Lemma 1. If F is concave on R+, irreducible at zero and the condition (
          <xref ref-type="bibr" rid="ref14">14</xref>
          )
is satis ed, then u 2 U n D if and only if there exists v &gt; u belonging to the
same set: v 2 U n D; or equivalently, if and only if there exists x &gt; 0 such that
Fu(x) &gt; x.
(18)
(19)
(20)
(21)
(22)
Let us de ne the maps y 2 fR+ 7! Rq+g, y 2 f[0; y] 7! Rq+g, where y 2 Nu,
q
y(x) = ( iy(x)), y(x) = ( yi(x)), with the following equalities:
y(x) = F (x + y)
        </p>
        <p>F (x);
y(x) = F (y)</p>
        <p>F (y
x):
(23)
Obviously, xu xF ; therefore we can assume that y, which plays the role of a
parameter here, varies in [0; xF ].</p>
        <p>
          It is easy to see that the maps y; y inherit the monotonicity of the map F ,
iass awlseoll caosntchaveeporenseRnq+ce, aonfda tthrievimalapxeyd ipsocinotnvxex=o0n. [F0u; ryt]h.eTrmheorreef,orteh,efomratphesye
maps there exist positively homogeneous (of the rst degree) maps ( y)0; ( y)0
de ned for F in (
          <xref ref-type="bibr" rid="ref7">7</xref>
          ). For their components the following equalities hold:
( iy)0(x) = fi0(y; x);
( yi)0(y; x) =
fi0(y; x) (8i 2 1; q);
where fi0(y; x) is one-sided directional derivative of the function fi(x) at the
point y in the direction x.
        </p>
        <p>The maps y, y (and, hence, their dominant eigenvalues) are related by the
followings inequalities:
(24)
(25)
y(x)</p>
        <p>y(x) (8x 2 [0; y]);
(( y)0)</p>
        <p>(( y)0):
( y)0(x) = ( y)0(x) = F 0(y)x;
(( y)0) = (( y)0) = (F 0(y)):
If F is di erentiable at point y and F 0 is its derivative, then the following
equations are valid:</p>
        <sec id="sec-23-1-1">
          <title>The following relations between the sets of xed points for the maps Fu, y; y and the sets (9) for these maps follow directly from their de nitions.</title>
          <p>
            q
Lemma 2. Suppose that F is concave on R+, irreducible at zero and the
condition (
            <xref ref-type="bibr" rid="ref14">14</xref>
            ) is satis ed. Then the following properties are valid:
(
            <xref ref-type="bibr" rid="ref1">1</xref>
            ) x 2 P +y ; y 2 Nu ) x + y 2 PF+u ;
(
            <xref ref-type="bibr" rid="ref2">2</xref>
            ) x 2 Q+y ; y 2 Nu; x &lt; y ) y x 2 PF+u ;
(
            <xref ref-type="bibr" rid="ref3">3</xref>
            ) x 2 N y ; y 2 Nu ) x + y 2 Nu;
(
            <xref ref-type="bibr" rid="ref4">4</xref>
            ) x 2 N y ; y 2 Nu; x y ) y x 2 Nu;
(
            <xref ref-type="bibr" rid="ref5">5</xref>
            ) x 2 Nu; y 2 Nu; y x ) x y 2 N y :
(
            <xref ref-type="bibr" rid="ref6">6</xref>
            ) x 2 Nu; y 2 Nu; y x ) x y 2 N +x :
          </p>
        </sec>
        <sec id="sec-23-1-2">
          <title>We show that y and y inherit also the global irreducibility of the map F .</title>
          <p>
            Lemma 3. Suppose that F is concave on Rq+ and the condition (
            <xref ref-type="bibr" rid="ref14">14</xref>
            ) is
satised. If the map F is globally irreducible, then the map y (y 2 Rq+) is globally
irreducible, and the map y (y 2 int Rq+) is irreducible on [0; y).
Proof. Let F be globally irreducible and suppose for contradiction that y is
q
not globally irreducible for some y 2 R+. By (
            <xref ref-type="bibr" rid="ref4">4</xref>
            ) this means that there exist
x; x0 such that x0 x, I0(x; x0) 6= ? and i(x; y) = i(x0; y) for i 2 I0(x; x0).
Then for z = x + y, z0 = x0 + y we have: z0 z, I0(z; z0) = I0(x; x0) 6= ? and
fi(z) = fi(z0) for all i 2 I0(z; z0). Hence F is not globally irreducible, which
contradicts our assumptions. Thus, y is globally irreducible. The proof of the
irreducibility for y is carried out analogously. The proof is complete.
          </p>
        </sec>
      </sec>
    </sec>
    <sec id="sec-24">
      <title>We proceed to describe the set D by means of the maps (23).</title>
      <p>
        q
Theorem 1. Suppose that F is concave on R+, irreducible at zero and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. If u 2 U , then u 2= D if and only if (( xu )0) &lt; 1.
Proof. Necessity. Using (15) and Lemma 1 for u 2 U nD, we obtain that Nv+ 6= ?
for some v &gt; u. Then, denoting xu xv &gt; 0 by x, we get: xu (x) = F (xu)
F (xu x) = F (xu) F (xv) = (xu+u) (xv +v) = x+(u v) &lt; x, i. e. xu (x) &lt; x.
It then follows, by (
        <xref ref-type="bibr" rid="ref11">11</xref>
        ) due to convexity (and, hence, superhomogenity) of the
map xu , that (( xu )0) &lt; 1.
      </p>
      <sec id="sec-24-1">
        <title>Su ciency. If (( y)0) &lt; 1, then, by (11), there exists x, such that 0 &lt; x &lt; y</title>
        <p>
          and x 2 Q+y . Hence, due to the assertion (
          <xref ref-type="bibr" rid="ref2">2</xref>
          ) of Lemma 2, we obtain y x 2 PF+u ,
that is, Fu(y x) &gt; y x. But this means, by Lemma 1, that u 2 U n D, as
required.
        </p>
      </sec>
    </sec>
    <sec id="sec-25">
      <title>Now we can obtain a characterization of the set D in terms of the maps (23).</title>
      <p>
        q
Corollary 1. Suppose that F is concave on R+, irreducible at zero and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. If u 2 U , then u 2 D if and only if (( xu )0) 1.
      </p>
    </sec>
    <sec id="sec-26">
      <title>The following statement gives a su cient condition for an admissible vector not to belong to the set D.</title>
      <p>
        q
Theorem 2. Suppose that F is concave on R+, globally irreducible and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. If u 2 U and (( x)0) &gt; 1 for some x 2 Nu, then
u 2 U n D.
      </p>
      <p>
        Proof. By (
        <xref ref-type="bibr" rid="ref10">10</xref>
        ) and Lemma 3, there exists x &gt; 0 such that x(x) &gt; x. It then
follows, since assertion (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) of Lemma 2, that Fu(x0) &gt; x0 for x0 = y + x. But this
means, by Lemma 1, that u 2 U n D. The proof is complete.
q
Corollary 2. Suppose that F is concave on R+, globally irreducible and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. If u 2 D and y 2 Nu then (( y)0) 1.
      </p>
    </sec>
    <sec id="sec-27">
      <title>Combining Theorem 1, Corollary 1 and Corollary 2, we obtain the following characteristic of U for the di erentiable map F .</title>
      <p>
        q
Corollary 3. Suppose that F is concave, di erentiable on R+, and the condition
(
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. If the map F is also irreducible at zero, then
If, in addition, the map F is globally irreducible, then
(F 0(xu)) &lt; 1 (8u 2 U n D):
(F 0(xu)) = 1 (8u 2 U \ D):
(26)
Proof. If u 2 U n D, then the inequality (26) follows from Theorem 1 due to (25).
If u 2 U \ D, then, by Corollary 1, we obtain (F 0(xu)) 1. On the other hand,
according to Corollary 2, we have (F 0(xu)) 1, so (F 0(xu)) = 1. The proof
is complete.
      </p>
    </sec>
    <sec id="sec-28">
      <title>Using the property 5 of Lemma 2 and Lemma 3, we obtain the following im</title>
      <p>
        portant property about the impossibility of a partial coincidence for coordinates
of xed points of Fu in the case of the globally irreducible map F :
8x; y 2 Nu : x
y ) x &lt; y:
(28)
We will give now one more characteristic feature of the elements of Nu for
u 2 U n D. The following assertion supplements the conclusion of Theorem 2.
q
Theorem 3. Suppose that F is concave on R+, global irreducible and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. Then the following properties are valid:
(( xu )0) &lt; 1;
      </p>
      <p>(( x)0) &gt; 1 (8 u 2 U n D; x 2 Nu n fxug):
Proof. The rst inequality follows from Theorem 1 and inequalities (24). Further,
if x 2 Nu n fxug, then, by assertion 5 of Lemma 2, x0 = xu x is the xed point
of x that is positive by (28). Hence we obtain (( x)0) 1 [10, Theorem 10.3].</p>
      <p>
        Suppose, by way of contradiction, that (( x)0) = 1. Then due to [13,
Theorem 2.2.11], we get (0; 1) M x (x0) (see (
        <xref ref-type="bibr" rid="ref12">12</xref>
        ){(
        <xref ref-type="bibr" rid="ref13">13</xref>
        )). We show that the condition
      </p>
      <sec id="sec-28-1">
        <title>2 M x (x0) is satis ed if and only if the following equality is true:</title>
        <p>
          F ((
          <xref ref-type="bibr" rid="ref1">1</xref>
          )x + (x + x0)) = (
          <xref ref-type="bibr" rid="ref1">1</xref>
          )F (x) +
        </p>
        <p>
          F (x + x0) (8
2 (0; 1)):
(29)
Indeed, we have: 2 M x (x0) , x( x0) = x(x0) , F ( x0 + x) F (x) =
(F (x0 + x) F (x)) , F ( x0 + x) = (
          <xref ref-type="bibr" rid="ref1">1</xref>
          )F (x) + F (x + x0). Since x0 + x =
(
          <xref ref-type="bibr" rid="ref1">1</xref>
          )x + (x + x0), we obtain the equality (29). Taking into account that
x + x0 = xu, we obtain from here the equality
        </p>
        <p>
          F ((
          <xref ref-type="bibr" rid="ref1">1</xref>
          )x +
xu) = (
          <xref ref-type="bibr" rid="ref1">1</xref>
          )F (x) +
        </p>
        <p>F (xu) (8
2 (0; 1)):
(30)
By assertion 6 of Lemma 2, x0 is a xed point of xu too. We show now that,
moreover, if (( x)0) = 1, then all points x0 ( 2 (0; 1]) are also xed points
of the map xu .</p>
        <p>
          Indeed, since xu x0 = (
          <xref ref-type="bibr" rid="ref1">1</xref>
          )xu + x, we have xu ( x0) = F (xu) F (xu
x0) = F (xu) F ((
          <xref ref-type="bibr" rid="ref1">1</xref>
          )xu + x): The equality (30) holds for all 2 (0; 1), so
we get: xu ( x0) = F (xu) (
          <xref ref-type="bibr" rid="ref1">1</xref>
          )F (xu) F (x) = (xu + u) (x + u) =
x0; so that really xu ( x0) = x0 (8 2 (0; 1]): These equalities mean that
( xu )0(x0) = x0, thus (( xu )0) 1 [10, Theorem 10.3]. But then it follows, by
Corollary 1, that u 2 D. This contradiction shows that (( x)0) &gt; 1. The proof
is complete.
        </p>
        <sec id="sec-28-1-1">
          <title>In conclusion, we give several assertions about a cardinality of set Nu with</title>
          <p>u 2 D. If xu is not unique in Nu, then it follows from (28) that xu &gt; xu
(8xu 2 Nu n fxug). In this case, we can de ne the set</p>
          <p>
            Lu = cofxu; xug;
where co M is the convex hull of a set M . We note that, as follows from the
assertion 5 of Lemma 2 and from Theorem 3, for u 2 U n D the set Nu cannot
contain the entire segment Lu, along with points xu; xu. But for the D0 the
situation is di erent. The following assertion shows that Nu with u 2 D is either
a singleton or an in nite set.
q
Theorem 4. Suppose that F is concave on R+, global irreducible and the
condition (
            <xref ref-type="bibr" rid="ref14">14</xref>
            ) is satis ed. Then the following property holds:
u 2 D; xu 2 Nu n fxug ) Nu
          </p>
          <p>
            Lu:
Proof. Due to the assertion 5 of Lemma 2, the map xu has a non-zero xed
point x = xu xu. The map xu is global irreducible (see Lemma 3), therefore,
this xed point is positive. Due to (
            <xref ref-type="bibr" rid="ref8">8</xref>
            ), we get: ( xu )0(x) xu (x) = x, hence
(( xu )0) 1 [10, Theorem 10.3]. But, the Corollary 2 gives the opposite
inequality, so that (( xu )0) = 1. In this case, the equation xu = 0 must be
ful lled, where xu is de ned by (
            <xref ref-type="bibr" rid="ref13">13</xref>
            ) (see [13, Theorem 2.2.11]). This means
that xu ( x) = xu (x) (8 2 [0; 1]). Therefore, we obtain for the set of positive
xed points of xu that N +xu f x j 2 [0; 1]g. By assertion 3 of Lemma 2 this
means that Nu fxu + x j 2 [0; 1]g = fxu + (xu xu) j 2 [0; 1]g = Lu.
          </p>
        </sec>
      </sec>
    </sec>
    <sec id="sec-29">
      <title>The proof is complete.</title>
    </sec>
    <sec id="sec-30">
      <title>Taking into account properties (20), (28), we obtain from the Theorem 4 the following assertion.</title>
      <p>
        q
Corollary 4. Suppose that F is concave on R+, global irreducible and the
condition (
        <xref ref-type="bibr" rid="ref14">14</xref>
        ) is satis ed. Then the following properties hold:
jNuj 2 f1; +1g (8u 2 D0);
jNuj = 1 (8u 2 D00):
(31)
If F is also strictly concave on [0; xF ], then jNuj = 1 (8u 2 D).
      </p>
      <p>In conclusion, let us give an example showing the essentiality of the
requirement of global irreducibility in the above statements, which used this
assumption. This example uses a generalization of so-called Leslie's model [14] of the
following form:</p>
      <p>(t+1) = fi(at);
xi;1
xi(;tj++11) =</p>
      <p>(t)
i;j xi;j
(i 2 1; m; j 2 1; n
1);
(32)
where at = Pim=1 Pjn=1 i;j xi(;tj), i;j &gt; 0, i;j 0, xi(;tj) 0 (8i 2 1; m; j 2 1; n).
The functions fi(a) (i 2 1; m) will be assumed to be nonnegative and concave
on R+. The step operator F (x) = (fi;j (x)) of this model has the following
components:
fi;1(x) = fi(a(x));
fi;j+1(x) =
i;j xi;j (i 2 1; m; j 2 1; n
1);
(33)
where a(x) = Pim=1 Pjn=1 i;j xi;j . Thus, the map F inherits a concavity and
monotonicity of functions fi(a).</p>
    </sec>
    <sec id="sec-31">
      <title>The admissible set U in this case, in addition to the requirement of non</title>
      <p>negativity of variables, is given by the following constraints:
xi;1 = fi(a(x))
ui;1; xi;j+1 = i;jxi;j
ui;j+1 (i 2 1; m; j 2 1; n
1); (34)
Let us introduce the following notation:</p>
      <p>m
(a) = X (i)fi(a); q(u) = (q; u);
i=1
(a) = (a)
a;</p>
      <p>(35)
n
where ( ; ) denotes the scalar product, qj(i) = P
k=j</p>
      <p>
        k 1
i;k Q
`=j
s
Q a` = 1 for i &gt; k), (i) = q1(i), q = (q(
        <xref ref-type="bibr" rid="ref1">1</xref>
        ); q(
        <xref ref-type="bibr" rid="ref2">2</xref>
        ); : : : ; q(m)) , q(i) = (q1(i); q2(i); : : : ; qn(i))
`=r
(i 2 1; m; j 2 1; n) (k; `; r; s 2 Z).
      </p>
      <p>Multiplying equalities (34) by i;j and summing, we obtain for a = a(x) the
equation
i;` (by convention
q(u) = (a):
(36)
Thus, if xu 2 Nu, then (36) holds, which for a given u can be regarded as an
equation for a. This equation, due to the concavity of (a), can have no more
than two solutions for (a) 6= = maxa (a).</p>
      <p>Example 1. Consider the system of constraints
xi;1 = fi(a)</p>
      <p>1
ui;1; xi;2 = 2 xi;1
ui;2;
a = x1;1 + x1;2 + x2;1 + x2;2; xi;j
0; ui;j
0 (i; j = 1; 2);
where f1(a) = 6a(1 + a) 1, f2(a) = minfa; 1g are non-negative monotone
increasing concave functions of a nonnegative argument. We nd from (35):
3 3 9a
(a) = 2 f1(a) + 2 f2(a) = 1 + a
+
3</p>
      <p>minfa; 1g;
2
3 3 9a 3
q(u) = 2 u1;1 + u1;2 + 2 u2;1 + u2;2; (a) = 1 + a + 2 minfa; 1g a:</p>
    </sec>
    <sec id="sec-32">
      <title>As shown in [14], condition (14) for the model (32) is equivalent to condition</title>
      <p>0(+1) &lt; 1 &lt; 0(0):
It is easy to see that in our particular case this condition is satis ed. Next, here
F 2 fR+ 7! R+g has the following form: F (x) = (f1(a(x); 21 x1;1; f2(a(x)); 12 x2;1).</p>
      <p>4 4
This map clearly satis es all assumptions of previous propositions, with the
exception of the global irreducibility. The function (a) reaches its maximum value
= 11=2 for a = 2.</p>
      <p>We take for illustration u = (16=5; 0; 1=10; 0) and show that 2 D00.
Calculate q(u) = 27=5 for this vector. Solving the equation (a) = 27=5, we obtain
solutions a1 = 3=2, a2 = 13=5. Further, we nd x(a1; u) = (2=5; 1=5; 9=10; 0),
x(a2; u) = (17=15; 17=30; 9=10; 0); so that Nu = fx(a1; u); x(a2; u)g. Therefore,
xu = x(a2; u), and from (19) it follows that u 2 D00. Thus, we get jNuj = 2
for u 2 D00, so that the second part of the conclusion (31) of Corollary 4 is not
satis ed. In addition, we see that the property (28) is also violated for these
vectors.</p>
    </sec>
    <sec id="sec-33">
      <title>The matrix F 0(x) has the form of a generalized Leslie's matrix:</title>
      <p>2f10 (a) f10 (a) f10 (a) f10 (a)3
L(a) = 664f120=(a2) f20 (a) f20 (a) f20 (a)775 ;</p>
      <p>0 0 0
0
0
1=2
0
where a = a(x) (see (33)). To compare the dominant eigenvalue of this matrix
with unity, it is not necessary to nd this value, because the followig equation
is valid [14]:
sgn( (L(a))
1) = sgn( 0(a)
1);
where sgn(x) = jxj=x for x 6= 0, and sgn(0) = 0.</p>
      <p>For x = x(a1; u) we obtain (F 0(x)) = (L(a1)) &gt; 1 as (a1) &gt; 1. We see
that the conclusions of Theorem 2 and of Corollary 2 for the chosen vector u 2 D
also do not hold.</p>
    </sec>
    <sec id="sec-34">
      <title>Thus, we have demonstrated that the requirement of the global irreducibility in the above statements is essential.</title>
      <p>4</p>
      <p>
        Conclusion
In this paper we investigated the properties of the admissible controls of the
problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) that can be optimal with appropriate presetting of the objective
function. In the context of the interpretation of this problem, from which we
proceeded, the question of presence of positive equilibrium x for the iterative
process (
        <xref ref-type="bibr" rid="ref2">2</xref>
        ) with given admissible control u is essential. We showed that such
controls must be contained in the set D0 de ned by (18). In the case of the
di erentiability of the map F , these controls are distinguished from the others,
by Corollary 3, with simple characteristic property (27). It turns out that the
dominant eigenvalue at the equilibrium point xu should equal unity. In addition,
as Corollary 4 shows, only for u 2 D0 the set Nu can contain an in nite number
of elements.
      </p>
    </sec>
    <sec id="sec-35">
      <title>The question whether in our assumptions this set is non-empty remains gen</title>
      <p>
        erally open. But in some cases, for some known (and su ciently general)
mathematical models used in the modeling of biological communities, it is sometimes
possible to establish that D0 6= ?. Preliminary studies show that this takes place
for the above generalization of Leslie's model, as was the case in Example 1. It
can be shown that in the assumption of concavity of all functions fi(x), the set
D0 in this model is always nonempty. Particularly important and, to a degree,
unexpected is the fact that the set D0 here is part of some hyperplane. Other
key properties of the admissible set of problem (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) for this and other models call
for further investigations.
      </p>
    </sec>
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