J. Math. Tokushima Univ.
Vol. 57 (2023), 85−92

85

Asymptotic Properties of Solutions forLanchester Type Models with Time Dependent
Coeﬃcients
By
Kosuke Ono
Department of Mathematical Sciences,
Tokushima University, Tokushima 770-8502, JAPANe-mail : k.ono@tokushima-u.ac.jp(Received September 30, 2023)
Abstract

We consider an ordinary diﬀerential system which is a so-called
Lanchester’s linear law model with time dependent coeﬃcients. We
study on asymptotic forms of solutions that decay to a point on thex-axis and y-axis.
2010 Mathematics Subject Classiﬁcation. 34A34, 34D05

1Introduction
In this paper, we consider the ordinary diﬀerential system of the form :{x′ (t) = −a(t)x(t)y(t)(1.1)
y ′ (t) = −b(t)x(t)y(t)where a(t) and b(t) are positive continuous functions on [0, ∞), and satisfy∫ t∫ ta(s) ds → ∞ and B(t) =b(s) ds → ∞(1.2)
A(t) =00

as t → ∞.
The initial value problem (1.1) with positive initial datax(0) = x0 > 0andy(0) = y0 > 0(1.3)

has non-negative solutions.
System (1.1) is known as one of Lanchester type models, which describes
many phenomena appearing in economics, logistics, biology, and so on.

Kosuke Ono86

In [5], F.W.Lanchester ﬁrst poposed system (1.1) to describe combat situations (see Taylor [9] for a review). System (1.1) is said a model of guerrillaengagements (see [2], [3], [10] and the references cited therein).
There are some mathematically treated reseach works for Lanchester typemodels (see [4] and [11] for Lanchester linear-law models, and see [6] and [8] for
Lanchester square-law models, and also see [1] and [7] for Lanchester models
with mixed forces).
First, we consider the special case where a(t) = α and b(t) = β for somepositive constants α > 0 and β > 0, that is,{x′ (t) = −αx(t)y(t)(1.4)
y ′ (t) = −βx(t)y(t)with initial data (1.3), then we have that A(t) = αt and B(t) = βt in (1.2).
Using the exchange ratio E = α/β of (1.4), we can easily see that (x(t) −Ey(t))′ = 0, and hence, x(t) − Ey(t) is a constant value which is denoted bysymbol M , that is,x(t) − Ey(t) = x0 − Ey0 = M.

(1.5)

Thus, we have from (1.4) and (1.5) thatx′ (t) = −αx(t)y(t) = −βx(t)(x(t) − M ) ,and moreover, by fundamental calculation we obtain the following representation fomula of solution (x(t), y(t)) of (1.4) :(1) When M = 0 (i.e. x0 = Ey0 ),−1x(t) = (x−10 + βt)y(t) = (y0−1 + αt)−1(1.6)(x0 /(x0 + EN ))EN e−EN βt
M=
1 − (1 − M/x0 )e−M βt(1 − (x0 /(x0 + EN ))e−EN βt(1.7)(y0 /(y0 + M/E))(M/E)e−(M/E)αt
N=−(M/E)αt
1 − (1 − (N/y0 ))e−N αt
1 − (y0 /(y0 + M/E))e(1.8)

andfor t ≥ 0.
(2) When M ̸= 0 (i.e. x0 ̸= Ey0 ),x(t) =andy(t) =where N = −M/E and hencex(t) − M =(1 − M/x0 )e−M βt
1 − (1 − M/x0 )e−M βt2(1.9)

Lanchester Type Models with Time Dependent Coefficients87

andy(t) − N =(1 − N/y0 )e−N αt
1 − (1 − N/y0 )e−N αt(1.10)

for t ≥ 0.
In what follows, ”f (t) ∼ g(t) as t → ∞” means that limt→∞ f (t)/g(t) = 1
for positive functions f (t) and g(t) deﬁned near +∞. Similary, for vector-valued
functions ”(f1 (t) , f2 (t)) ∼ (g1 (t) , g2 (t)) as t → ∞” means that fi (t) ∼ gi (t)as t → ∞, i = 1, 2.
Immediately, we can obtain from (1.6)–(1.10) the following decay properties
of solution (x(t), y(t)) of (1.4) :(i) When M = 0, (x(t), y(t)) → (0, 0) as t → ∞ and(x(t) , y(t)) ∼ ((βt)−1 , (αt)−1 )ast → ∞.

(1.11)(ii) When M > 0, (x(t), y(t)) → (M, 0) as t → ∞ and(log(x(t) − M ) , log y(t)) ∼ (−M βt , −(M/E)αt) ast→∞(1.12)and x(t) − M = O(e−M βt ) and y(t) = O(e−(M/E)αt ).
(iii) When N = −M/E > 0, (x(t), y(t)) → (0, N ) as t → ∞ and(log x(t) , log(y(t) − N )) ∼ (−EN βt , −N αt) ast→∞(1.13)and x(t) = O(e−EN βt ) and y(t) − N = O(e−N αt ).
Remark. When the time dependent coeﬃcients a(t) and b(t) in (1.1) satisfya(t)/b(t) = const > 0 for t ≥ 0, we can obtain the similar representation
formula of solution (x(t), y(t)) of (1.1) replaced αt and βt in (1.6)–(1.10) byA(t) and B(t), respectively.
In [4], Ito, Ogiwara and Usami have derived the following asymptotic forms(1.14) and (1.15) of solution (x(t), y(t)) of (1.1) decaying to the origin (0, 0),like (1.11) for (1.1) with constant coeﬃcients :(i) If a(t) and b(t) satisfy (1.2) and limt→∞ a(t)/b(t) = const > 0, then(x(t) , y(t)) ∼ (B(t)−1 , A(t)−1 )ast → ∞.

(ii) If a(t) and b(t) are of class C 1 and satisfy()′a(t)≤ 0 for large tb(t)anda(t)B(t)= k > 0 andt→∞ A(t)b(t)

lim

lim

t→∞

3(a(t)B(t)A(t)b(t))′B(t)= 0,b(t)(1.14)

Kosuke Ono88

then(x(t) , y(t)) ∼ (kB(t)−1 , k −1 A(t)−1 )ast → ∞.

(1.15)

However, there is no known research work related to asymptotic forms ofsolutions of (1.1) decaying to a point other than the origin (0, 0), like (1.12)
and (1.13) for (1.4) with constant coeﬃcients.
The notations we use in this paper are standard. Positive constants will bedenoted by C and will change from line to line.

2Results

We will give asymptotic forms of solutions of (1.1) decaying to a point on
the x-axis and y-axis.
Theorem 2.1 Let E, M and N be constants. Assume that a(t) and b(t) satisfy(1.2) anda(t)= E > 0.
t→∞ b(t)
lim

(2.1)

Then, we have the following :(i) For M > 0, every solution (x(t), y(t)) of (1.1) decaying to (M, 0) has
the asymptotic form(log(x(t) − M ) , log y(t)) ∼ (−M B(t) , −(M/E)A(t))

ast → ∞.

(2.2)(ii) For N > 0, every solution (x(t), y(t)) of (1.1) decaying to (0, N ) has
the asymptotic form
(log x(t) , log(y(t) − N )) ∼ (−EN B(t) , −N A(t))

ast → ∞.

(2.3)

Proof. (i) Let M > 0 and (x(t), y(t)) → (M, 0) as t → ∞. By L’Hospital’srule, we have from (2.1) thatx(t) − M
x′ (t)−a(t)x(t)y(t)= lim ′
= lim
t→∞
t→∞
t→∞
y(t)
y (t)
−b(t)x(t)y(t)a(t)=E,= limt→∞ b(t)lim

and hence, we obtain from (2.1) and (2.4) thatlog(x(t) − M )(x(t) − M )−1 x′ (t)
= lim
t→∞
t→∞
−M B(t)−M b(t)(x(t) − M )−1 (−a(t)x(t)y(t))
= lim
t→∞
−M b(t)a(t) y(t) x(t)
= 1,
= lim
t→∞ b(t) x(t) − M Mlim4(2.4)

Lanchester Type Models with Time Dependent Coefficients89

which implies log(x(t) − M ) ∼ −M B(t) as t → ∞.
On the other hand, by L’Hospital’s rule again, we have from (2.1) thatlog y(t)y(t)−1 y ′ (t)y(t)−1 (−b(t)x(t)y(t))
= lim
= lim
t→∞ −(M/E)A(t)t→∞ −(M/E)a(t)t→∞−(M/E)a(t)b(t) x(t)= E lim= 1,t→∞ a(t) Mlim

which implies log y(t) ∼ −(M/E)A(t) as t → ∞.
(ii) Next, let N > 0 and (x(t), y(t)) → (0, N ) as t → ∞. By L’Hospital’srule, we have from (2.1) thatlog x(t)x(t)−1 x′ (t)x(t)−1 (−a(t)x(t)y(t))
= lim
= lim
t→∞ −EN B(t)t→∞ −EN b(t)t→∞−EN b(t)1a(t) y(t)=lim
= 1,E t→∞ b(t) Nlim

which implies log x(t) ∼ −EN B(t) as t → ∞.
On the other hand, by L’Hospital’s rule again, we have from (2.1) thatx(t)x′ (t)a(t)= lim ′= lim
=E,t→∞ y(t) − N
t→∞ y (t)t→∞ b(t)
lim

(2.5)

and hence, we obtain from (2.1) and (2.5) that(y(t) − N )−1 y ′ (t)log(y(t) − N )
= lim
t→∞
t→∞
−N A(t)−N a(t)(y(t) − N )−1 (−b(t)x(t)y(t))
= lim
t→∞
−N a(t)b(t) x(t) y(t)= 1,= limt→∞ a(t) y(t) − N Nlim

which implies log(y(t) − N ) ∼ −N A(t) as t → ∞. □Theorem 2.2 Let M be a constant. Assume that a(t) and b(t) are of class C 1and satisfy (1.2) and(a(t)b(t))′≤0for large t .

(2.6)

Then, for M > 0, every solution (x(t), y(t)) of (1.1) decaying to (M, 0) hasx(t) − M = O(e−M B(t) ) .

5(2.7)

Kosuke Ono90

In addition, if there exists a positive constant K such thata(t)2 eM B(t)= const > 0 ,t→∞ b(t)2 eKA(t)(2.8)y(t) = O(e−KA(t) ) .

(2.9)

limthenProof.
Since x(t) → M as t → ∞ and (−x(t))′ = (a(t)/b(t))(−y(t))′ , itfollows that∫ ∞∫ ∞a(s)′(−y(s))′ dsx(t) − M =(−x(s)) ds =b(s)tt)′∫ ∞(a(s)a(t)y(t) +y(s) ds=b(t)b(s)
tfor large t, and from (2.6) that there exists t1 > 0 such thaty(t) ≥b(t)(x(t) − M )a(t)for t ≥ t1 .

Then we havex′ (t) = −a(t)x(t)y(t) ≤ b(t)x(t)(x(t) − M )for t ≥ t1 . Solving this diﬀerential inequality of a separeble type on [t1 , t], weobtainx(t) x1 − M
1log≥ B(t) − B1M
x1 x(t) − M
andx(t) ≤M≤ Ce−M B(t) ,1 − (1 − M/x1 )e−M (B(t)−B1 )where we use symbols x1 = x(t1 ) and B1 = B(t1 ), and hence,x(t) − M ≤(1 − M/x1 )e−M (B(t)−B1 )≤ Ce−M B(t)1 − (1 − M/x1 )e−M (B(t)−B1 )(2.10)

for t ≥ t1 , which implies (2.7).
On the other hand, since y(t) → 0 as t → ∞ and (−y(t))′ = (b(t)/a(t))·(−(x(t) − M ))′ , it follows that∫ ∞∫ ∞b(s)′′y(t) =(−(x(s) − M )) ds(−y(s)) ds =a(s)tt)′∫ ∞(b(t)b(s)(x(t) − M ) +(x(s) − M ) ds
(2.11)=a(t)a(s)t6Lanchester Type Models with Time Dependent Coefficients91

for large t. Here, since it follows from (2.8) that0≤b(t) −M B(t)a(t) b(t)2 −M B(t)e=e
≤ Ce−KA(t)a(t)b(t) a(t)2

for larte t ,(2.12)

and from (b(t)/a(t))′ = −(b(t)/a(t))2 (a(t)/b(t))′ ≥ 0 for large t and (2.8) that)′∫ ∞(b(s)e−M B(s) ds0≤a(s)t
∫ ∞b(t) −M B(t)b(s)=−eM b(s)e−M B(s) ds+a(t)a(s)t∫ ∞b(t) −M B(t)e+CKa(s)e−KA(s) ds≤−a(t)t∫ ∞b(t) −M B(t)=−e+C(−e−KA(s) )′ dsa(t)t≤ Ce−KA(t)for large t ,(2.13)

we obtain from (2.10)–(2.13) thaty(t) ≤ Ce−KA(t) + Ce−KA(t)for large t ,which implies (2.9). □Remark. (i) When a(t) = α > 0 and b(t) = β > 0, we see that (a(t)/b(t))′ = 0
and the limit value of (2.8) is α2 /β 2 > 0 by taking K = M β/α.
(ii) When a(t) = (1 + t)−1 and b(t) = (e + t)−1 , we see that (a(t)/b(t))′ < 0and the limit value of (2.8) is e−M > 0 by taking K = M .
By the similar argument of Theorem 2.2 we have the following theorem.
Theorem 2.3 Let N be a constant. Assume that a(t) and b(t) are of class C 1and satisfy (1.2) and()′b(t)≤ 0 for large t .
a(t)Then, for N > 0, every solution (x(t), y(t)) of (1.1) decaying to (0, N ) hasy(t) − N = O(e−N A(t) ) .
In addition, if there exists a positive constant K such thatb(t)2 eN A(t)= const > 0 ,t→∞ a(t)2 eKB(t)limthenx(t) = O(e−KB(t) ) .

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