Boundedness of Solutions of a System of two Measure Di¤erential Equations with Constant Delay1 OYELAMI, Benjamin Oyediran and Sam O. ALE National Mathematical Centre, Abuja, Nigeria E-mail:
[email protected];
[email protected] Tel: +234-8073469337; +234-8036547613
1
Abstract
In this paper, boundedness of solutions is studied for a system of two measure di¤erential equations with constant delay that satisfy initial conditions of speci…c type .Necessary and su¢ciency conditions for uniform and ultimate boundedness of solutions of the two measure di¤erential equations with constant delay are obtained and are applied to a military model using marginalization and exclusion principles.
2
Introduction
Impulsive Di¤erential Equations are equations that are associated with small perturbations, such perturbations act for short period of time during the process of evolution. Impulses are noted to take the form of " pulses", "jumps", " shocks" etc. ([2],[6],[8],[11]& [12]). Recent investigations revealed that many real life phenomena can be explained using impulsive theory ([1],[6] &[12]). The study of impulsive systems has opened up another window to understand the behaviour of many real life processes. Examples are the economy of a country, the tempo in a battle…eld [6], pulses, neural network, and population under rapid changes and so on([6-15]). The boundedness of solution of impulsive di¤erential equations was studied in ([17]) and the stabilizability of impulsive systems have been studied and results obtained (see [1]&[12]).The Lyapunov-Razumkham method has been applied to obtain stabilizability results for impulsive delay di¤erential equations (see [12]). Measure di¤erential equations MDEs are equations developed to describe systems, in which the functions governing them are not continuous or integrable in the ordinary sense of the word, but are of bounded variation([8-11,14-15]). There are many delay impulsive models whose behaviour can be investigated through qualitative techniques, for example, the gonorrhea model under impulsive action (see [16]) and the impulsive population of …shes (see [11]). The use of qualitative techniques for example, boundedness of the solution may o¤er useful information about the behaviour and the real life implications of the most real life processes. 1 Keywords and phrases: Boundedness,delay,measure di¤erential equations, Lyapunov functions,marginalisation and exclusion principles. AMS subject classi…cation: 34K20, 34A37 and 34C60.
1
Application of impulsive systems using the MDEs approach has not received much needed attention of late if we consider the volume of publications made on the impulsive di¤erential equations . MDEs have many applications in real life which are not yet explored. For this reason we developed a military model to show one of the applications of MDEs. In this paper, boundedness of solutions of a system of two MDEs with constant delay that satisfy initial conditions of speci…c type is studied. We intend to obtain the necessary and su¢ciency conditions for the solutions of MDEs to be bounded with its various kinds and apply the results to a military model using the marginalization and exclusive principles(see [6-8]).
3
Preliminary Notes
Let R+ = [0; +1),J = [ h; +1); h > 0 and Rn be the n - dimensional Euclidean Space with elements of the form x = (x1 ; x2; :::; xn ) and the norm 1=2 n P x2k . Let fti gi=1;2;3:::n be the sequence of strictly increasing times in k=1
the interval [a; b] J and BV = BV (J; Rn ) be the set of functions with bounded variation de…ned on J taking values in Rn . i.e. '0 2 BV implies that the following condition is satis…ed by '0 : # " n X j'0 (ti ) '0 (ti 1 )j < +1: ('0 ; J) = sup a 0; h = constant.
(1)
Satisfying the initial conditions x(0) = '1 (0) , y(0) = '2 (0) and 'i (0) 2 BV (J) for i = 1; 2. Where fi ; gi : J BV (J) BV (J) ! Rn ; i = 1; 2; ui : J ! R+ ; i = 1; 2 are right continuous functions with bounded variation on the compact sub-interval 1 X 1 if t > 0 ak k , k = f J J and ui := t + ; for t 2 J. 0 if t 0 k=0
D(:) is the distribution derivative of (.) with respect to time t and assume
that fi ; i = 1; 2 are Lebesgue - Stieltjes integrable with respect to the measures dui ; i = 1; 2; ui are impulsive in J and gi are of bounded variation. Corresponding to the eq. (1) is the following equivalent perturbed system Dxt = f1 (t; xt (0); xt h (0))Du1 + g1 (t; xt (0); yt (0)); t > 0 Dyt = f2 (t; yt (0); yt h (0))Du2 + g2 (t; xt (0); yt (0)); t > 0 xt (0) = 1 (t); yt (0) = 2 (t); t 2 [ h; 0]
(2)
Where fi ; gi ; i = 1; 2 are Lebesque –Stieltjes integrable functions while gi are of bounded variation , fi and gi are lipschitz with respect to the second and the third variables respectively. De…nition 3 Assume that fi, , gi ,ui ,i = 1; 2 satis…ed the conditions stated in section 4,then any pair of functions xt and yt in BV is said to be the solutions to the measure di¤erential equations in eq.(1) if the following conditions are satis…ed: (i) xt and yt are di¤erentiable in the distribution sense as de…ned in the De…nition 1 and satis…ed the eq.(1) along with the prescribed initial conditions. Remark 1 The conditions imposed on f and gi ; i = 1; 2 above are made in order to guarantee the existence and the uniqueness of solutions of eq.(2) in the interval [ h; ] J,for > 0; h > 0 [see 4-6,11]. 3
Let (xt ; yt ) be the solution of eq. (2) such that xt = x(t; to + 0; 1 (0)) and yt = y(t; to + 0; 2 (0)) respectively. We introduce the Dini derivative of the Lyapunov function V (t; xt ; yt ) across the solution path of the eq. (2) as + D(2) V (t; xt ; yt ) = lim sup[V (t+h; xt+h +hf1 Du1 +hg1 ; yt+h +hf2 Du2 +hg2 ) h!0+
V (t; xt ; yt )] : + We write D(2 1 ) V (t; xt; ; yt ) whenever gi = 0 in eq.(2) and its corresponding Dini derivative as
+ D(2 V (t; xt ; yt )]. 1 ) V (t; xt ; yt ) = lim sup[V (t+h; xt+h +hf1 Du1 ; yt+h +hf2 Du2 ) + h!0
If there is no confusion in the notation of the Dini derivative and the distri+ bution derivative, we shall adopt the notation D(:) for the Dini derivative where (.) in the subscript we represent the solution path for which the derivative is taken. For the distribution derivative we will simply write D. We will make use of the following de…nitions: De…nition 4([11]) A. The trivial solution (xt (0); yt (0)) = 0 of eq. (1) is said to be equi-bounded if for any > 0 and t0 2 J there exists a continuous function = (t0 ; ) > 0 such that if j 1 (0)j + j 2 (0)j < then jxt ( 1 (0))j + jyt ( 2 (0))j < ; B. Uniformly bounded if from condition A above does not depend on to ; C. Ultimately bounded with the bound N if there exist numbers > 0 and T = T (t0 ; ) > 0 such that if j 1 (0)j+j 2 (0)j < then jxt ( 1 (0))j+jyt ( 2 (0))j < N for t > t0 + T ; D. Uniformly ultimately bounded with the bound N if T = T ( ) in the de…nition C above. The following auxiliary result relates the Dini derivative of a given Lyapunov function along the solution path of eq. (2) to the one whose Dini derivative is taken along the solution path of eq. (1).
4.1
Auxiliary Results
Lemma 1 Let the following conditions be satis…ed: H1: There exist non-negative constants k1 ; k2 ; k3 ; k4 ; l1 ; l2 ; m1 and m2 such that jf1 (t; x1t ; x1t
h)
f1 (t; x2t ; x2t
h )j
6 k1 jx1t
x2t j + k2 jx1t
yt2
h
hj
(3a)
And jf2 (t; yt1 ; yt1
h)
f2 (t; yt2 ; y 2 t
h )j
6 k3 jyt1
For x1t ; x2t ; yt1 ; yt2 2Sp . 4
yt2 j + k4 jyt1
h
yt2
hj
(3b)
H2: Let there exist constants m1 and m2 which are independent of xt ; xt and yt h such that i.e. jg1 (t; xt ; xt
h )j
6 m1 and jg2 (t; xt ; xt
h )j
h ; yt
6 m2
H3: There exist constants l1 and l2 such that jV (t; x1t ; yt1 ) Then
+ D(2) V (t; xt ; yt )
V (t; x2t ; yt2 )j 6 l1 jx1t
x2t j + l2 jyt1
+ (l1 m1 + l2 m2 ) + D(1) V (t; xt ; yt )
yt2 j
(4)
(5)
Proof 1 [V (t + h; xt+h + hf1 Du1 + hg1 ; yt+h + hf2 Du2 + hg2 ) h h!0+
+ D(2) V (t; xt ; yt ) = sup
V (t + h; xt+h + hf1 Du1 ; yt+h + hf2 D2 ) + V (t + h; xt+h + hf1 Du1 ; yt+h + hf2 Du2 ) 6 lim sup[l1 jxt+h + hf1 Du1 + hg1
yt+h
V (t; xt ; yt )]
hDu2 j
h!0+
+l2 jyt+h + hf2 Du2 + hg2
yt+h
hf2 Du2 j
+ y 6 l1 j g1 j + l2 j g2 j + D(2 1 ) V (t; xt ; t ):
Then
5
+ D(2) V (t; xt ; yt ) 6 (l1 m1 + l2 m2 ) + D(+21 ) V (t; xt ; yt ):
Main Results
We will derive the result that guarantees the necessary and su¢ciency conditions for uniform boundedness and ultimate boundedness of solutions of eq.(21 ): Theorem 1 Let the following conditions be satis…ed: H1: The condition H1 and H2 in the Lemma 1 H2: There exist functions a; b 2 K such that i. a(jxt j + jyt j) 6 V (t; xt ; yt ) 6 b(jxt j + jyt j) + V (t; xt ; yt ) 6 0 in J Sp Sp . ii.D(2) If the conditions H1 and H2 are satis…ed then trivial solution (xt ; yt ) = 0 of eq. (21 ) is uniformly bounded. Conversely, if trivial solution (xt ; yt ) = 0 of eq. (21 ) is uniformly bounded there exists at least one Lyapunov function V = V (t; xt ; yt )which is continuous in its arguments such that condition H2 is satis…ed. Proof Su¢ciency Given > 0; N > 0 and let there exist a; b 2 K such that a 1 (b( )) = N for j 1 (0)j + j 2 (0)j < . 5
Thus,integrating H2i we have V (t; xt ( 1 (0); yt ( 2 (0)) = V (0; 2 (0); 2 (0)). And from (H2i) & (H2ii) in the hypotheses we have < b( ); a(j xt j + j yt j) 6 V (t; xt ( 1 (0)); yt ( 2 (0))) b(j 1 (0)j + j 2 (0)j) since b 2 K thus jxt ( 1 (0))j + jyt ( 2 (0))j < a 1 (b( )) = N . Hence the trivial 0 solution of eq(2 )is uniformly bounded.This ends the proof of su¢ciency. Next, we prove the converse as follows: R0 (s) 2 (s)ds where (s) > 0 Take V (t; xt ( 1 (0)); yt ( 2 (0)) = jxt ( (0))j2 + h
for every s 2 [ h; 0). Let p =
(i)
sup t2[ h;1)
(i)
[jxi j; jyi j]; i = 1; 2 , K1 = 2P and K2 = 2hp
R0
(s)ds:We
h
will show that V (t; xt ( 1 (0)); yt ( 2 (0))) is locally Lipschitz with respect to xt and yt in Sp .Now let (t; x1t ; yt1 ); (t; x2t ; yt2 ) 2 J Sp Sp ,then V (t; x1t ; yt1 ) R0 2 V (t; x2t ; yt2 ) = jxt j2 jyt j2 + [ 22 (s) 1 (s)]ds h
Therefore jV
(t; x1t ; yt1 )
V
(t; yt2 ; yt 2 )j
6
2P [jx1t
yt2 j
+
Z0
j
2 (s)
1 (s)jds]
h
6 K1 jx1t
x2t j + K2 jyt1
yt2 j .
Thus V (t; xt ; yt ) is locally Lipschitz with respect to xt and yt . Next, we show that V (t; xt ; yt ) is continuous in its arguments. We observe that
V (t + h; x1 (t + h + s); y1 (t + s + h)) V (t; x(t + s); y(t + s)) = V (t + h; x1 (t + h + s); y1 (t + s + h)) V (t + h; x1 (t + s); y1 (t + s)) + V (t + h; x1 (t + s); y1 (t + s)) V (t; x(t + s); y(t + s))
K1 j(x1 (t + s + h) x1 (t + s)j + K2 jy1 (t + s + h) y1 (t + s)j + jV (t + h; x1 (t + s); y1 (t + s)) V (t; x(t + s); y(t + s))j: (8) But jx1 (t + s + h)
x1 (t + s)j 6
t+s+h Z
K1 jx1 (2s + h)
x1 (t + 2s)jds
t+s
+
t+s+h Z
K1 jx1 (t + 2s)
x1 (t + 2s
h)jds
t+s
+
Z
t+s+h
K3 jy1 (2s + h)
t+s
6
y1 (t + 2s)jds
+
t+s+h Z
K2 jy1 (t + 2s)
y1 (t + 2s
h)jds
(9)
t+s
+
Thus as h ! 0 , the right hand side of the inequality in (9) tends to zero thus, jV (t + h; x1 (t + s + h); y1 (t + s + h)) V (t; x(t + s); y(t + s))j 6 jV (t + h; x1 (t + s + h); y1 (t + s + h)) V (t + h; x1 (t + s); y1 (t + s))j ! 0 as h ! 0. Therefore, V (t; xt ; yt ) is continuous in its arguments. Let Z 0 2 2 M (t) = V (t; xt ; yt ) = jxt j + 2 (s)ds h
Thus
M (t + h) = V (t + h; x(t + s + h); y(t + s + h)) = jx(t + s + h)j2 +
Z0
2 2 (s + h)ds
h
It follows that M (t) = jx(t + s + h)j2
M (t + h)
jx(t + s)j2 +
R0
[
2 2 (s
+ h)
2 1 (s)]ds
h
Thus 1 D M (t) = sup [jx(t + s + h)j2 h +
2
jx(t + s)j +
1 6 lim sup [2pjx(t + s + h) h h!0+
+
[
2 2 (s
[
2 2 (s
+ h)
2 1 (s)]]ds
h
h!0+
Z0
Z0
x(t + s)j
2 2 (s)]ds]
+ h)
h
= 2plim sup[jx(t + s + h)
x(t + s)j +
h!0+
! Thus
Z0
j
2 (s
+ h)
2 (s)jds]
h
+
0 as h ! 0 : + D(2 1 ) V (t; xt ; yt ) 6 0
(10)
To end the proof, we assume that (xt ; yt ) = 0 is uniformly ultimate bounded, then there exist > 0 and N > 0 such that jxt j+jyt j < N for j 1 (0)j+j 2 (0)j < . Now de…ne a(jxt j + jyt j) = jxt j2 + jyt j2 ; b (j 1 (0)j2 + j 2 (0)j2 ) = h (jxt0 j2 + jyt0 j2 ),h := max[1; h]; h > 0. 7
From the inequality in (10) 2
V (t; xt ; yt ) 6 V (t0 + 0; xt0 ; yt0 ) = jxt0 j +
Z0
2 2 (s)ds
h
6 h (jxto j2 + jyto j2 ) 6 h (jxto j + jyto j)2 = b (j
2 1 (0)j
+j
2 2 (0)j )
.
Clearly,V (t; xt ; yt ) > a (j 1 (0)j2 + j 2 (0)j2 ): If x" < pxt for t 2 [ h; 0] ; y" < pyt , p is a positive real number and de…ne a (jxt j + jyt j) =min[1; p2 h]:(jxt j2 + jyt j2 ) Then 2
V (t; xt ; yt ) = jxt j +
Z0
2 2 (s)ds
h
2
> jxt j + j jxt j2 +
Z0
pyt2 (s)dsj
h 2
jyt j2 h > a (jxt j2 + jyt j2 )
Thus a (jxt j + jyt j) 6 V (t; xt ; yt ) 6 b ((j
2 1 (0)j
+j
2 2 (0))j ))
This ends the proof of the theorem. Theorem 2 Let the following conditions be satis…ed: H1: The conditions in Theorem 1 with exception of condition H2ii replaced by + D(2 1 ) V (t; xt ; yt ) 6
C(j
1 (0)j
+j
2 (0)j)on
L
Where C(:) 2 K; H2: V (t + ; xt ; yt )
L(V (t; xt ; yt )); 2 [ h; 0] on
L
Then the trivial solution (xt ; yt ) = 0 of eq. (21 ) is ultimately bounded in J but uniformly bounded in [ h; t0 ]; t0 > 0:Conversely ,if trivial solution (xt ; yt ) = 0 of eq. (21 ) is ultimately bounded in J or uniformly bounded in [ h; t0 ]; t0 > 0 then there exists at least a Lyapunov V (t; xt ; yt ) for which all the conditions H1 and H2 in the theorem are satis…ed. Proof Suppose on the contrary that eq. (21 ) is nowhere ultimately bounded in J. Then given > 0 such that j 1 (0)j + j 2 (0)j < , take = max [0; 2to h L(b( ))C 1 ( )]; t0 > 0: Let T( ) =
8 < :
h + L(b( ))=C( ); t 2 [ h; ]; t0 > 0 (12) h + a1 L(b( ))=C( ); t 2 [ h; ]; t0 6 0 8
And ( )=
8 < :
C( ) t0 ; t0 > 0 (13) ( a a 1 )L(b( ))=C( ); 8a > 1; t0 6 0.
It is obvious that ( ) > 0 for every V (t; xt ; yt ) 6 V ( h;
1(
h);
> 0 and for (t; xt ; yt ) 2
2(
t0 +T Z ( )
h))
C(j
1 (0)j
L
+j
then 2 (0)j)ds
h
6 L(V (t + ;
1 (0);
< L(b( ))
2 (0))
C( )(t0 +T ( ) + h)
C( )(t0 +T ( ) + h) ( ) < 0.
6
A contradiction of non-negativity of V (t; xt ; yt ) therefore the trivial solution of eq.(21 ) must be ultimately bounded . Conversely In order to establish this we must exhibit a Lyapunov function V (t; xt ; yt ) that satis…es all the properties (i.e. H1-H2 in the theorem). Take V (t; xt ; yt ) = sup[j xt+ j; jyt+ j]e (14) >0
and let Thus
(t; x1t ; yt2 ); (t; x2t ; yt2 )
V (t; x1t ; yt1 )
2J
Sp
Sp .
1 V (t; x2t ; x2t ) = sup[jx1t+ + yt+ j
2 j]e jx2t+ + yt+
>0
6 sup[jx1t+
1 x2t+h + yt+h
6 sup[jx1t+h
1 x2t+h j + jyt+h
>0
>0
2 yt+h j]e 2 yt+h j]e
.
It is not di¢cult to show that sup[jx1t+h
2 yt+h j]e
1 x2t+h j + jyt+h
>0
6 K1 jx1t+h
1 x2t+h j + K2 jyt+h
2 yt+h j for 0
< 1.
Therefore, V (t; xt ; yt ) is Lipschitz with respect to xt and yt Similarly, D+ M (t) = lim
1
h!0+ h
fsup[V (t + h; xt+h ; yt+h )
V (t; xt ; yt )]g
>0
1 = lim+ f sup[jx(t + + s + h) h >0 h!0
y(t + + s + h)j
jx(t + s) + y(t + s)]e
g:
But jx(t + + s)
y(t + + s)j 6 j
1 (0)j
+j
2 (0)j
+
t+ Z +s
[jf1 jDu1 + jf2 jDu2 ]ds
h
=j
1 (0)j
+j 9
2 (0)j
+ ( )
(16)
j
Where ( ) is the integral in the eq. (16), also jx(t + s) + y(t + s)j 6 j 1 (0)j + 2 (0)j + (0). Thus the left hand side of of the inequality in (16) is less than or equal to lim+
h!0
1 f sup [j h > h
sup(j
1 (0)j
1 (0)j
+j
+j
2 (0)j
2 (0)j
+ ( )]e
+ (0))e
(
h)
g:
>0
Thus 1 lim+ supf sup(j !0
=
1 (0)j
+j
2 (0)j
+ ( )
(0))e
(e
1)g
>0
sup(
1 (0)j
+j
2 (0)j)e
>0
= Therefore
M (0): + D(2 M (0) 1 ) M (t) 6 + D(21 ) V (t; xt ; yt ) 6 V (t;
1 (0);
1 (0)).
Corollary 1 Suppose V 2 C(R+ S S ; R+ ),V (t; xt ; yt ) is locally Lipschitz with respect to time t for every xt and yt in Sp such that: H1: There exist a; b1 ; b2 2 K such that a(j xt j + j yt j) 6 V (t; xt ; yt ) b1 (j 1 (0)j + j H2:
+ D(2 1 ) V (t; xt ; yt ) 6 0 in
2 (0)j)) + b2 (j 1 (0)j + j 2 (0)j)
.
Then the trivial solution of eq. (21 ) is ultimately bounded in . Proof Given > 0 take such that = max(b1 ( )+b2 ( )) and j 1 (0)j+j 2 (0)j < and N = a 1 (b1 ( ) + b2 ( )).Then we can show, just like theorem 1, that jxt j + jyt j < N for j
1 (0)j
+j
2 (0)j
< .
(17)
Remark 2 Corollary 1 is very useful for practical purpose whenever it is not easy to apply condition H2ii in Theorem 1 directly. However,it is not an easy task to establish the converse of this corollary . The following theorem gives the su¢ciency condition for boundedness of solution of eq. (1). Theorem 3 Let the following conditions be satis…ed: H1: There exist a; b 2 K such that
10
a(jxt j + jyt j)
6 V (t; xt ; yt )
6 b(j
1 (0)j
+j
2 (0)j)
For every (t; xt ; yt ) 2 J BV (J) BV (J); g0 2 BV (J) such that is the maximal solution of the equation
(18) (t; to ; ut )
u= = go (t; u; ut ); ut_ 0 = u0 > 0
(19)
existing on t0 < t < t ; t 2 [tk ; tk+1 ]; k = 1; 2; 3; ::: H2: D+ V (t;
1 (0);
2 (0))
6 g(t;
1 (0);
2 (0))
on
Then the boundedness of the solution of eq. (19) implies the boundedness of solution of eq. (2). Proof Let (t; t0 + 0; 0 ) = (t) be the maximal solution of eq.(19) such that (t0 ; t0 + 0; u0 ) = > u0 = u(t; t0 + 0; ut0 ).To establish the proof take N = b( ), it is enough to show that (t; n0 ; s) 6 a(N ) for any given bound N for the solution of the comparison equation in eq.(19). From eq.(19) we have u(t) 6 u0 +
Zt
go (s; u(s); us )ds 6 u0 +
t0
Zt
go (s; (s; to +0; us ); (s+t; t0 +0; us ))ds:
t0
That is u(t;
1 (0);
2 (0))
6 u(t0 +0;
1 (0);
2 (0))
+
Zt
go (s;
1 (0);
1 (0);
2 (0))
2 (0))ds
t0
6 (t; t0 + 0;
0 ):
: From H1 in the corollary 1 we have a(jxt j + jyt j) 6 V (t; xt ; yt )
6
0
+
Zt
go (s; (s);
6 V (t;
s )ds
6 (t;
0 ; s):
t0
Thus jxt j + jyt j 6 a
1
( ) 6 b(j
1 (0)j
+j
2 (0)j)
6 b( ) = N
Hence the proof. Corollary 2 H1: Suppose V 2 C(R+ Sp Sp ; R+ ); V (t; xt ; yt ) is locally Lipschitz with respect to xt , and yt in Sp such that: H2: There exist a; b1 ; b2 2 K such that 11
DV (t; xt ; yt ) 6 0 in Then the solution of eq:(21 ) is ultimately bounded in . Proof Given 1 , 2 , > 0 such that = max (b1 ( 1 ), b2 ( 2 )) and j 1 (0)j + j 2 (0)j < . Take N = a 1 (2 ), then we can show just like theorem 1, that jxt j + jyt j < N for j 1 (0)j + j 2 (0)j < thus the proof follows immediately.
6
Application
In [6] & [8] military models were considered and analyzed using quantitative technique. Here we will consider a more general kind of MDEs military model to the one in [6] and it is of delay type . Meanwhile,we will introduce some terminologies that are needed for the study as follows: Consider two groups (Units) involve in a battle. The commander of each Unit must apply some strategies to the battle. In order to model the scenario in the battle …elds e¤ectively, in [6]& [8] we introduced the reinforcement function, Exclusion and Marginalization principles into the military model and impulsive HIV-1 model([7]). The reinforcement function is a kind of strategic function each Unit must use in order to win the battle. This reinforcement function must depend of the number soldiers to be featured in the battle…eld and the predicted population of the enemy together with the type of armour the opponent will be using. Exclusion principle is all about how to reduce the in‡uence of the opponent in the battle to zero. This can be done by cordoning the opponent away from getting military support to reinforce the Unit with troops, food, armour and so on .Marginalization is all about how to reduce the reinforcement of the opponent’s Unit with food and military supply to a negligible e¤ect([8]). Now consider the model Dx(t) =
1
x(t)(x(t)
x(t
h ))Du1 +
1
y(t) g1 (t; x(t); y(t)); t > 0
Dy(t) =
2
y(t)(y(t)
y(t
h))Du2 +
2
x(t) g2 (t; x(t); y(t)); t > 0
(E1)
x(t) = 1 (t); y(t) = 2 (t); t 2 [ h; 0]; x(t0 ) = x0 and y(t0 ) = y0 : Where D(:) is the distribution derivative of (:) . xt (t) and yt (t) are the populations of the soldiers in the Unit one and the Unit two respectively. ui = t +
1 X
1 if t > 0 aik Hk (t); i = 1; 2; Hk (t) = f 0 if t 6 0
k=1 i;
i i = 1; 2 are some relevant rate constants. gi = gi (t; xt ; yt ) are reinforcement functions[see [6]] for x(t); y(t) 2 BV (J); gi : J BV (J) BV (J) ! R+ . In relation to eq. (E1), we say that:
12
1. The Unit one excludes the Unit two if g2 = 0 .The Unit one will be said to marginalize the Unit two if there exists a small positive real number 21 such that jg1 j 0 always,therefore 2 1 x(t)x(t h) < 0. + Thus D(E1) V (t; xt ; yt ) < 0. It is easy to show that following estimate is valid a(r1 ) 6 V (t; xt ; yt ) 6 b(r2 ) where r1 =min(1; ah); r = jxt j2 +jyt j2 ; r2 = r max(1; ah).By application of theorem 1, the proof follows. Theorem 5 Let the following conditions be satis…ed: H1: The Unit one excludes the Unit two such that Sx
