Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 18, pp. 1–17.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
LINEAR AND LOGISTIC MODELS WITH TIME DEPENDENT
COEFFICIENTS
YOUNESS MIR, FRANÇOIS DUBEAU
Abstract. We sutdy the effects of some properties of the carrying capacity on
the solution of the linear and logistic differential equations. We present results
concerning the behaviour and the asymptotic behaviour of their solutions.
Special attention is paid when the carrying capacity is an increasing or a
decreasing positive function. For more general carrying capacity, we obtain
bounds for the corresponding solution by constructing appropriate subsolution
and supersolution. We also present a decomposition of the solution of the
linear, and logistic, differential equation as a product of the carrying capacity
and the solution to the corresponding differential equation with a constant
carrying capacity.
1. Introduction
In this article we shall study the solutions of the linear and logistic differential
equations defined respectively by
ẋ(t) = α(t) − β(t)x(t),
t ≥ t0 ,
(1.1)
and
ẋ(t) = β(t)x(t) − α(t)x2 (t), t ≥ t0 ,
(1.2)
where α(t) and β(t) are strictly positive and continuous functions on [t0 , +∞),
with t0 ∈ R. More precisely, we will consider the effect of α(t), β(t), and the ratio
k(t) = α(t)/β(t) on the behavior of the solutions of these models for any positive
initial value x(t0 ) > 0.
Considering that the mapping y : x 7→ y = x−1 transforms (1.2) into the linear
differential equation
ẏ(t) = α(t) − β(t)y(t), t ≥ t0 ,
we will focus our study on (1.1). Indeed, all results obtained for (1.1) could then
be directly applied to (1.2).
Considering the preceding mapping used to rely solutions of (1.2) to solutions
of (1.1), we will call the ratio k(t) = α(t)/β(t) the carrying capacity for the linear
model (1.1) despite the fact that this expression, for the logistic equation (1.2),
refer to β(t)/α(t) = 1/k(t).
2010 Mathematics Subject Classification. 91B62, 34G10, 34G20, 34K25, 00A71, 92D25.
Key words and phrases. Growth models; linear model; logistic model; carrying capacity;
product decomposition.
c
2016
Texas State University.
Submitted May 22, 2015. Published January 11, 2016.
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Y. MIR, F. DUBEAU
EJDE-2016/18
With a constant carrying capacity k(t) = k, these models, and many extensions
of them, have extensively been used to describe and improve the possible relationship between independent and dependent variables in terms of mathematical
equations. It happended in many fields of applied sciences, like ecology, sociology,
medicine, and other domains [7, 8, 10, 14, 16, 17, 23]. However, according to Coleman [2, 3, 4], and later to Meyer [11, 12], changes in the environment affect the
carrying capacity. Hence, modeling phenomena with unchanging carrying capacity
is often unrealistic. Several authors have reformulated these standards models with
constant carrying capacity to accommodate phenomena with varying [19, 20, 21],
logistically varying [11, 12], increasing [5, 6, 13] or sinusoidally varying [4, 3, 18]
carrying capacity. In [1], the author has given some examples on real situations
in which the carrying capacity k changes with time continuously. Many other
researchers have been interested in the problem of the existence and uniqueness
solution of the solution of (1.2) with bounded time dependent carrying capacity
[9, 15]. In [22], the author argued that it is difficult to make precise statement
about the asymptotic behaviour of the solution to a non-autonomous differential
equation when the coefficients α and β are time dependent functions. In [9] the
authors have proved that a monotone bounded carrying capacity is an attractor
forward in time of all positive solution of (1.2) in the sense that the limit at infinity
of the difference between any solution to the differential equation and the carrying
capacity vanishes.
Despite the intensive examples of real situations involving growth phenomena
with unbounded time dependent carrying capacity (see [1] for example), these situations has surprisingly received little attention in the literature compared to the
massive literature devoted to the problems with bounded coefficients.
The main purpose of this paper is to address this knowledge gap through a
qualitative study. We study in a thorough way the effect of an unbounded carrying capacity on the behaviour and the asymptotic behaviour of the solution of
the linear and logistic differential equations. We shall pay particular attention to
the cases when the carrying capacity k(t) is an increasing or decreasing positive
function. Moreover, the asymptotic behaviour of the solutions to these differential
equations is not well described when the carrying capacity k(t) is time dependent and unbounded. On this basis comes the second aim of this paper which
consists of reformulating the solutions x(t) of (1.1) and (1.2) as a product of a
simple function z̃(t) and a carrying capacity k̃(t) such that limt→+∞ z̃(t) = 1, and
limt→+∞ (x(t) − k̃(t)) = 0.
The present paper is organized as follows. In Section 2, we start by giving some
properties about the solution of the linear differential equation. Then, we present
some results on the behaviour and the asymptotic behaviour of its solution when
the time dependent coefficients are not necessary bounded. We also show that
when the carrying capacity is an increasing and unbounded function, the limit at
infinity of the difference between the solution of the linear differential equation
and the carrying capacity is not always equal to zero. In Section 3, we provide
monotonic bounds for the solution of (1.1) when k(t) is neither an increasing nor a
decreasing function. Section 4 addresses the problem of decomposing the solution
into a product of a carrying capacity and a simple analytic function. Finally, we
present a conclusion.
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LINEAR AND LOGISTIC MODELS
3
2. The linear differential equation
Let I = [t0 , +∞) be an interval such that t0 ∈ R and consider the linear differential equation
ẋ(t) = α(t) − β(t)x(t) = β(t)(k(t) − x(t)) t ≥ t0 ,
(2.1)
where α(t), β(t) : I → (0, +∞) are two continuous functions, and k(t) = α(t)/β(t).
Subject to the initial condition x(t0 ) = x0 , (2.1) has the unique solution given by
x(t) = e
−
Rt
t0
β(u)du
Z
t
x0 +
α(τ )e
Rτ
t0
β(u)du
dτ .
(2.2)
t0
Under the assumptions k(t0 ) = k0 , and
k∈
1
ACloc
(I)
Z t2
n
1
= k ∈ C(I) : k̇ ∈ Lloc (I), k(t2 ) = k(t1 ) +
k̇(τ )dτ,
t1
o
for all t1 , t2 ∈ I ,
(2.3)
Equation (2.2) takes the form
Z t
R
R
Ru
− t β(u)du
− t β(u)du
β(τ )dτ
x(t) = k(t) + (x0 − k0 )e t0
− e t0
k̇(u)e t0
du .
(2.4)
t0
In the case where α(t)/β(t) = k does not depend on time, (2.4) reduces to
x(t) = k + (x0 − k)e
−
Rt
t0
β(u)du
.
(2.5)
Regarding x0 and k we have the following situations:
• if x0 = k, from (2.5) we have x(t) = k for all t ∈ I;
• if x0 6= k, we have
lim x(t) = k + (x0 − k)e
t→+∞
−
R +∞
t0
β(u)du
= x∞ ,
and
* if x0 < k, then x(t) < x∞ for all t ∈ I, and the solution x(t) grows up
to x∞ ,
* if x0 > k, then x(t) > x∞ for all t ∈ I, and the solution x(t) decreases
to x∞ .
When α(t)/β(t) = k(t) depends on time, (2.1) has no constant solution and the
solution (2.2) may crosses k(t). It happens when ẋ(t) = 0. For t∗ ∈ I, let us
consider the closed interval
Jt∗ = {t ∈ I : t ≥ t∗ , and x(τ ) = x(t∗ ) for all t∗ ≤ τ ≤ t} = [t∗ , t∗∗ ],
where t∗∗ = supτ ∈I Jt∗ . Moreover, if t∗ < t∗∗ , then ẋ(t) = 0 for all t ∈ Jt∗ and
consequently, from (2.1) and (2.4) we have
x(t) = k(t),
and k̇(t) = 0 for all t ∈ Jt∗ .
(2.6)
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Y. MIR, F. DUBEAU
EJDE-2016/18
2.1. Increasing case. In this section, we suppose that k(t) is a non-decreasing
function
k(t1 ) ≤ k(t2 )
for all t1 , t2 ∈ I such that t1 < t2 .
(2.7)
We have the following result.
Lemma 2.1. Let k(t) = α(t)/β(t) : I → R be defined such that β(t) > 0 for all t.
Suppose that k satisfies the assumptions (2.3) and (2.7), and let x(t) be the solution
of (2.1) passing through the point (t0 , x0 ). If for some s ∈ I we have x(s) = k(s),
then x(t) ≤ k(t) for all t ≥ s. More precisely,
• x(t) = k(t), and ẋ(t) = 0, for all t ∈ Js , and
• x(t) < k(t), and ẋ(t) > 0, for all t > ts = sup Js .
Proof. Let x(t) be the solution of (2.1) passing through the point (t0 , x0 ), and let
s ∈ I such that x(s) = k(s). From (2.6), we have x(t) = k(t) for all t ∈ Js . Let
ts = sup Js ∈ Js , it follows that x(ts ) = k(ts ) = k(s) and k(t) > k(ts ) for all t > ts .
Thus, by considering the solution passing through the point (ts , x(ts )), we have
Z t
R
R
Ru
− tt β(u)du
− tt β(u)du
s
s
x(t) = k(t) + (x(ts ) − k(ts ))e
−e
k̇(u)e ts β(τ )dτ du
ts
Z
t
= k(t) −
k̇(u)e
Ru
t
β(τ )dτ
du.
ts
As k(t) > k(ts ) for all t > ts , it follows that k(ts ) < x(t) < k(t) for all t > ts .
The following result characterizes the behaviour of any solution of (2.1) in the
case where k(t) is non decreasing.
Theorem 2.2. Let k(t) and x(t) be defined as in Lemma 2.1. Also let k∞ =
limt→+∞ k(t). Then
(a) if x0 < k0 , then x(t) < k(t), and ẋ(t) > 0, for all t ∈ I;
(b) if x0 = k0 , then x(t) ≤ k(t), and ẋ(t) ≥ 0, for all t ∈ I;
(c) if x0 > k0 , we have two cases to consider
(i) if x0 ≤ k∞ , then it will exists some s > t0 where s = argmin x(t) such
that x(s) = k(s), and in this case x(t) decreases if t < s and increases
if t > s;
(ii) if x0 > k∞ , then either x(t) has the same behaviour as in (i), or
x(t) > k(t), and ẋ(t) < 0, for all t ∈ I.
Proof. The proofs of assertions (a) and (b) follow immediately from (2.4) and
Lemma 2.1. Let us prove (c).
(i) If k0 < x0 < k∞ , from (2.1), it follows that ẋ(t0 ) < 0. By continuity, we
also have ẋ(t) < 0 for all t provided that x(t) > k(t) which is satisfied at least
locally near t0 . We will prove that, there exists some s > t0 such that x(s) = k(s)
and x(t) increases for t > s with x(t) < k(t). Indeed, suppose that x(t) > k(t)
for all t > t0 . Thus, from (2.1), we have that ẋ(t) < 0 for all t ≥ t0 and hence,
k(t) < x(t) < x(t0 ) = x0 . By taking the limit at infinity we obtain k∞ ≤ x0 which
contradicts our assumption on x0 . If x0 = k∞ < ∞, by arguing as in the proof
above and by taking the limit at infinity we obtain x∞ = x0 which contradict the
fact that x(t) > k(t), i.e. ẋ(t) < 0, for all t > t0 .
EJDE-2016/18
LINEAR AND LOGISTIC MODELS
5
(ii) if x0 > k∞ > k0 , i.e. k∞ < ∞, we must have x∞ = limt→+∞ x(t) < ∞. In
Rt
addition, if we set limt→+∞ t0 β(u)du = l, from (2.4), we have
(x0 − k0 )e−l − (k∞ − k0 ) ≤ x∞ − k∞ ≤ (x0 − k∞ )e−l .
(2.8)
If l = +∞, from (2.8), it follows that x∞ ≤ k∞ and hence, x(t) has the same
asymptotic behaviour as in (i). If l < +∞, from (2.8), it follows that x∞ ≥ k∞
if x0 ≥ k0 + (k∞ − k0 )el , and the solution x(t) is always decreasing. If x0 ≤
k0 + (k∞ − k0 )el , the solution decreases or has the same asymptotic behaviour as
in (i).
The following result gives us some information about the asymptotic behaviour
of the solution to the linear problem (2.1) when the time dependent coefficients
α(t) and β(t) are not necessary bounded.
Theorem 2.3. Let k(t) and x(t) be defined as in Lemma 2.1 and set
k∞ = lim k(t),
x∞ = lim x(t).
t→+∞
t→+∞
Then
(a) If limt→+∞
(b) If limt→+∞
Rt
Rtt0
α(u)du < +∞, then x∞ < +∞.
α(u)du = +∞, then
Rt
(I) if limt→+∞ t0 β(u)du = l < +∞, then x∞ = +∞;
Rt
(II) if limt→+∞ t0 β(u)du = +∞, then x∞ = k∞ , and
(i) if k∞ < +∞, then limt→+∞ (x(t) − k(t)) = 0;
k̇(t)
if
(ii) if k∞ = +∞, then limt→+∞ (x(t) − k(t)) = − limt→+∞ β(t)
this limit exists.
Rt
Proof. (a) If limt→+∞ t0 α(u)du < +∞, then from the fact that α(t) = k(t)β(t),
Rt
and k(t) > k0 > 0 for all t ∈ I, it follows that limt→+∞ t0 β(u)du < +∞. Hence,
Z t
Z t
Ru
Rt
β(τ )dτ
β(u)du
t0
t0
lim
α(u)e
du ≤ lim e
α(u)du < +∞.
t→+∞
t0
t→+∞
t0
t0
Thus, from (2.2) it follows that x∞ < +∞.
Rt
(b) Let us suppose that limt→+∞ t0 α(u)du = +∞. As
Z t
Z t
Ru
β(τ )dτ
α(u)du ≤
α(u)e t0
du,
t0
t0
it follows that
Z
t
lim
t→+∞
α(u)e
Ru
t0
β(τ )dτ
du = +∞.
(2.9)
t0
Rt
(I) If limt→+∞ t0 β(u)du = l < +∞, then from (2.2), and (2.9), it follows that
limt→+∞ x(t) = ∞.
Rt
(II) If limt→+∞ t0 β(u)du = +∞, then from (2.2) and (2.9), and by using the
L’Hôpital’s rule we obtain
Rτ
Rt
β(u)du
α(τ )e t0
dτ
t0
Rt
x∞ = lim x(t) = lim
β(u)du
t→+∞
t→+∞
e t0
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Y. MIR, F. DUBEAU
EJDE-2016/18
Rt
α(t)e
= lim
t0
Rt
t→+∞
β(u)du
β(u)du
β(t)e t0
α(t)
= lim
t→+∞ β(t)
= lim k(t) = k∞ .
t→+∞
(i) If k∞ < ∞, we have that limt→+∞ (x(t) − k(t)) = x∞ − k∞ = 0.
Rt
(ii) If limt→+∞ k(t) = ∞, as limt→+∞ t0 β(u)du = +∞ it follows that
lim (x(t) − k(t))
t→+∞
= lim
t→+∞
(x0 −
Rt
= − lim
t0
R
α0 − Rtt β(u)du
− t β(u)du
− e t0
)e 0
β0
k̇(u)e
Ru
t0
Z
t
Ru
k̇(u)e
t0
β(τ )dτ
t0
β(τ )dτ
Rt
.
β(u)du
e t0
In addition, as k(t) is an non decreasing function, we have for all t > t0 ,
t→+∞
k̇(u) ≤ k̇(u)e
thus
Z
k(t) − k(t0 ) ≤
and limt→+∞
Rt
k̇(u)e
Ru
Ru
t
k̇(u)e
t0
β(τ )dτ
Ru
t0
du ≤ k̇(u)e
β(τ )dτ
Rt
t0
β(τ )dτ
,
du ≤ (k(t) − k(t0 ))e
Rt
t0
β(τ )dτ
,
t0
β(τ )dτ
du = +∞. Thus
Ru
Rt
β(τ )dτ
t0
k̇(u)e
k̇(t)
= − lim
,
lim x(t) − k(t) = − lim t0 R t
β(u)du
t→+∞ β(t)
t→+∞
t→+∞
t0
e
if this limit exists.
t0
t0
Example 2.4 ([1], Linear asymptote). Let β(t) = a, and k(t) = pt + q, where
a > 0, p > 0, and q ≥ 0. From (2.4) we have
Z t
−a(t−t0 )
x(t) = e
x0 +
aea(x−t0 ) (px + q)dx ,
t0
= k(t) − p/a + (x0 − (k0 − p/a))e−a(t−t0 ) ,
where x0 = x(t0 ) and k0 = k(t0 ). From Lemma 2.2 we have
(1) If x0 < k0 , then x(t) is increasing and x(t) < k(t) for all t. In this case x(t)
is convex if x0 > k0 − p/a and concave if not.
(2) If x0 ≥ k0 , then x(t) is convex and intersects k(t) at t∗ = t0 + ln(1 + ap (x0 −
k0 ))/a, with t∗ = argmin x(t). In this case, x(t) increases if t > t∗ and
decreases if not.
Moreover we have
lim (x(t) − k(t)) = − lim
t→+∞
t→+∞
k̇(t)
b
=− .
β(t)
a
In the next example, we give some hypothesis on β(t) which ensure that the limit
at infinity between the solution to the differential equation (2.1) and a curvilinear
carrying capacity k(t) vanishes.
EJDE-2016/18
LINEAR AND LOGISTIC MODELS
7
Example 2.5 (Curvilinear asymptote). Let α(t), and β(t) be two continuous functions such that
α(t)
k(t) =
= ptλ + q,
β(t)
with t > 0, p > 0, q ≥ 0, and suppose that
β(t)
≥ c > 0, for all t > 0,
tγ
with 0 < λ < γ + 1. The unique solution to the initial value problem
ẋ(t) = α(t) − β(t)x(t),
t ≥ t0 ,
x(t0 ) = x0 ,
has the following asymptotic property,
lim (x(t) − k(t)) = 0.
t→+∞
Indeed, as β(t) ≥ ctγ , and k(t)β(t) = α(t) it follows that
Z +∞
Z +∞
β(τ )dτ =
α(τ )dτ = +∞.
t0
t0
From Theorem 2.3, and limt→+∞ k(t) = +∞, it follows that limt→+∞ (x(t)−k(t)) =
k̇(t)
− limt→+∞ β(t)
. In addition, we have
0≤
k̇(t)
k̇(t)
λp λ−(γ+1)
≤ γ =
t
.
β(t)
ct
c
As 0 < λ < γ + 1, it follows that limt→+∞
k̇(t)
β(t)
= 0. Thus,
lim (x(t) − k(t)) = − lim
t→+∞
t→+∞
k̇(t)
= 0.
β(t)
2.2. Decreasing case. In this section, we suppose that k(t) is a non increasing
function
k(t1 ) ≥ k(t2 ) for all t1 , t2 ∈ I such that t1 < t2 .
(2.10)
We have the following result.
Lemma 2.6. Let k(t) = α(t)/β(t) : I → R+ be defined such that α(t) > 0, and
β(t) > 0 for all t. Suppose that k(t) satisfies the assumptions (2.3) and (2.10), and
let x(t) be the solution of (2.1) passing through the point (t0 , x0 ). If for some s ∈ I
we have x(s) = k(s), then x(t) ≥ k(t) for all t ≥ s. More precisely,
• x(t) = k(t), and ẋ(t) = 0, for all t ∈ Js , and
• x(t) > k(t), and ẋ(t) < 0, for all t > ts = sup Js .
Proof. Let x(t) be the solution of (2.1) passing through the point (t0 , x0 ), and let
s ∈ I such that x(s) = k(s). From (2.6), we have x(t) = k(t) for all t ∈ Js . Let
ts = sup Js ∈ Js , it follows that x(ts ) = k(ts ) = k(s) and k(t) < k(ts ) for all t > ts .
Thus, by considering the solution passing through the point (ts , x(ts )), we have
Z t
Rt
Rt
Ru
x(t) = k(t) + (x(ts ) − k(ts ))e− ts β(u)du − e− ts β(u)du
k̇(u)e ts β(τ )dτ du ,
ts
Z
t
= k(t) −
k̇(u)e
ts
Ru
t
β(τ )dτ
du > k(t).
8
Y. MIR, F. DUBEAU
EJDE-2016/18
The following result characterizes the behaviour of any solution of (2.1) in the
case where k(t) is decreasing.
Theorem 2.7. Let k(t) and x(t) be defined as in Lemma 2.6. Also let k∞ =
limt→+∞ k(t) and x∞ = limt→+∞ x(t). Then
(a) if x0 > k0 , then x(t) > k(t), and ẋ(t) < 0, for all t ∈ I;
(b) if x0 = k0 , then x(t) ≥ k(t), and ẋ(t) ≤ 0, for all t ∈ I;
(c) if 0 ≤ x0 < k0 , we have two cases to consider
(i) if x0 ≥ k∞ , then it will exists some s > t0 where s = argmax x(t) such
that x(s) = k(s), and in this case x(t) increases if t < s and decreases
if t > s;
(ii) if x0 < k∞ , then either x(t) has the same behaviour as in (i), or
x(t) < k(t), and ẋ(t) > 0, for all t ∈ I.
Proof. The proofs of the assertions (a) and (b) follow immediately from (2.4) and
Lemma 2.6. Let us prove (c).
(i) If k0 > x0 > k∞ , from (2.1), it follows that ẋ(t0 ) > 0. By continuity, we also
have ẋ(t) > 0 for all t provided that x(t) < k(t). We will prove that, there exists
some s > t0 such that x(s) = k(s) and x(t) decreases for t > s with x(t) > k(t).
Indeed, suppose that x(t) < k(t) for all t > t0 . Thus, from (2.1), we have that
ẋ(t) > 0 for all t ≥ t0 and hence, k(t) > x(t) > x(t0 ) = x0 . By taking the limit at
infinity we obtain k∞ ≥ x0 which contradicts our assumption on x0 .
If x0 = k∞ , by arguing as in the proof above and by taking the limit at infinity
we obtain x∞ = x0 which contradict the fact that x(t) < k(t), i.e. ẋ(t) > 0, for all
t > t0 .
Rt
(ii) Suppose that 0 ≤ x0 < k∞ < k0 . If we set limt→+∞ t0 β(u)du = l, from
(2.4), we have
(x0 − k∞ )e−l ≤ x∞ − k∞ ≤ (x0 − k0 )e−l − (k∞ − k0 ).
(2.11)
If l = +∞, from (2.11), it follows that x∞ ≥ k∞ and hence, x(t) has the same
asymptotic behaviour as in (i). If l < +∞, from (2.11), it follows that x∞ ≤ k∞
if x0 ≤ k0 + (k∞ − k0 )el , and the solution x(t) is always increasing. If x0 ≥
k0 + (k∞ − k0 )el , the solution increases or has the same asymptotic behaviour as
in (i).
The following result gives us some information about the asymptotic behaviour
of the solution to the linear problem (2.1) when the time dependent coefficients
α(t) and β(t) are not necessary bounded.
Theorem 2.8. Let k(t) and x(t) be defined as in Lemma 2.6. Also let k∞ =
limt→+∞ k(t), and x∞ = limt→+∞ x(t). Then
Rt
(a) If limt→+∞ t0 α(u)du = +∞, then x∞ = k∞ ≥ 0.
Rt
(b) If limt→+∞ t0 α(u)du < +∞, then
Rt
(I) if limt→+∞ t0 β(u)du = +∞, then x∞ = k∞ = 0;
Rt
(II) if limt→+∞ t0 β(u)du = l < +∞, then
k∞ + (x0 − k∞ )e−l ≤ x∞ ≤ k0 + (x0 − k0 )e−l .
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Rt
α(u)du = +∞, as α(t) = k(t)β(t), and k(t) ≤ k0 for
Rt
Rt
all t ∈ I, it follows that limt→+∞ t0 β(u)du = +∞. In addition, t0 α(u)du ≤
R
u
Rt
β(τ )dτ
α(u)e t0
du, hence
t0
Z t
Ru
β(τ )dτ
α(u)e t0
du = +∞.
lim
Proof. (a) If limt→+∞
t0
t→+∞
t0
Thus, by using L’Hôpital’s rule we have
Rt
x∞ = lim x(t) = lim
t→+∞
t→+∞
= lim
t→+∞
α(τ )e
t0
eR
α(t)e
Rt
t0
t
t0
Rt
Rτ
t0
β(u)du
dτ
β(u)du
β(u)du
β(u)du
β(t)e t0
α(t)
= lim
t→+∞ β(t)
= lim k(t) = k∞ ≥ 0.
t→+∞
Rt
(b) Let us suppose that limt→+∞ t0 α(u)du < +∞.
Rt
(I) If limt→+∞ t0 β(u)du = +∞, then from (2.2) and the dominated convergence
Theorem, it follows that
Z t
Rt
lim x(t) = lim
α(τ )e− τ β(u)du dτ = 0.
t→+∞
t→+∞
t0
As x∞ ≥ k∞ ≥ 0 itR follows that k∞ = 0.
t
(II) If limt→+∞ t0 β(u)du = l < +∞, then from (2.4) the result follows.
Example 2.9 (Hyperbolic asymptote, first case). Let k(t) = 1/(pt+q), and α(t) =
1/(t + 1)γ where p > 0, q > 0, and x(t) be the solution of (2.1) passing through
the point (t0 , x0 ) where t0 = 0. From Theorem 2.8 we have
Rt
If γ ≤ 1 then limt→+∞ t0 α(u)du = +∞ and hence x∞ = k∞ = 0.
Rt
1
.
If γ > 1, then α(t) is decreasing with α(0) = 1 and limt→+∞ t0 α(u)du = γ−1
γ
Moreover, β(t) = (pt + q)/(t + 1) , β(0) = q, and we have
(
Z t
p
q−p
+ γ−1
if γ > 2,
l = γ−2
lim
β(u)du =
t→+∞ t
+∞
if 1 < γ ≤ 2.
0
Hence,
• If 1 < γ ≤ 2, then x∞ = k∞ = 0.
• If γ > 2, then
1
x0 e−l ≤ x∞ ≤ x0 e−l + (1 − e−l ).
q
Example 2.10 (Hyperbolic asymptote, second case). Let k(t) = 1/(ptλ + q), and
α(t) = 1/tγ where λ > 0, p > 0, q > 0, and x(t) be the solution of (2.1) passing
through the point (t0 , x0 ) where t0 = 1. From Theorem 2.8 we have
Rt
• If γ ≤ 1 then limt→+∞ t0 α(u)du = +∞ and hence x∞ = k∞ = 0.
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Y. MIR, F. DUBEAU
EJDE-2016/18
Rt
• If γ > 1, then α(t) is decreasing with α(1) = 1 and limt→+∞ t0 α(u)du =
1
λ
γ
γ−1 < +∞. Moreover, β(t) = (pt + q)/t , and we have
(
Z t
p
q
l = γ−λ−1
+ γ−1
if γ > λ + 1,
β(u)du =
lim
t→+∞ t
+∞
if 1 < γ ≤ λ + 1.
0
Hence,
• If 1 < γ ≤ λ + 1, then x∞ = k∞ = 0.
• If γ > λ + 1, then
1
x0 e−l ≤ x∞ ≤ x0 e−l + (1 − e−l ).
q
3. Subsolution and supersolution
The following theorem gives us some information on the boundedness of the
solution x(t) of (2.1) passing through a point (t0 , x0 ) when the carrying capacity
k(t) = α(t)/β(t) is bounded above and below by given positive functions.
Theorem 3.1. Let k : I → (0, +∞) be defined such that α(t) > 0, and β(t) > 0
for all t, and let x(t) be the solution of (2.1) passing through the point (t0 , x0 ).
Suppose that there exist two functions k, k : I → (0, +∞) such that
k(t) ≤ k(t) ≤ k(t),
for all t ≥ t0 .
(3.1)
In addition, let x(t) and x(t) be respectively the solutions of the auxiliary problems
ẋ(t) = β(t)(k(t) − x(t)),
(3.2)
ẋ(t) = β(t)(k(t) − x(t)),
with x(t0 ) = x(t0 ) = x(t0 ). Then, we have
x(t) ≤ x(t) ≤ x(t),
for all t ≥ t0 .
(3.3)
Proof. The solutions (t) = x(t) − x(t), and (t) = x(t) − x(t) of the auxiliary
problems
˙ (t) = β(t)((k(t) − k(t)) − (x(t) − x(t)))),
˙ (t) = β(t)((k(t) − k(t)) − (x(t) − x(t))),
with (t0 ) = (t0 ) = 0 are respectively
Z t
R
Rτ
− t β(u)du
β(u)dudτ
(t) = e t0
β(τ )(k(τ ) − k(τ ))e t0
,
−
(t) = e
Rt
t0
β(u)du
t0
t
Z
β(τ )(k(τ ) − k(τ ))e
Rτ
t0
β(u)dudτ
.
t0
From (3.1) it follows that (t) ≥ 0, (t) ≥ 0, and hence the inequalities (3.3)
follow.
The following corollary is an immediate consequence of Theorem 3.1.
Corollary 3.2 (Increasing case). Let k, k : I → (0, +∞) be defined by
k(t) = max{k(τ ) : t0 ≤ τ ≤ t},
and
k(t) = min{k(τ ) : τ ≥ t},
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and let x(t) and x(t) be respectively the solutions of the auxiliary problems (3.2),
with x(t0 ) = x(t0 ) = x(t0 ). Then,
x(t) ≤ x(t) ≤ x(t),
for all t ≥ t0 .
(3.4)
Proof. It will be noted that k(t) and k(t) are respectively the least upper and the
greatest lower bounds of the sets of all increasing function f (t), such that k(t) ≤
f (t), respectively k(t) ≥ f (t). Hence, k(t) and k(t) are positive and increasing
functions with k(t) ≤ k(t) ≤ k(t) for all t ≥ t0 . From Theorem 3.1 the inequalities
(3.4) follow immediately.
Example 3.3. Let k(t) = pt + q(1 + r sin(ωt)) and β(t) = a, where p ≥ 0, q ≥ 0,
0 < r ≤ 1, ω > 0, and a > 0. From (2.4), and Example 2.4 we have
rqω
[φ(t) − φ(t0 )e−a(t−t0 ) ],
x(t) = k(t) − p/a + (x0 − (k0 − p/a))e−a(t−t0 ) − 2
a + ω2
where x0 = x(t0 ), k0 = k(t0 ), and φ(t) = (ω sin(ωt) + a cos(ωt)). It will be noted
that, for a large values of t, we have
rqω
x(t) − k(t) ≈ −p/a − 2
φ(t).
a + ω2
On the other hand, let k (t) and e
k(t) be defined by
e
k (t) = pt + q(1 − r), and e
k(t) = pt + q(1 + r).
e
Obviously, k and e
k are non decreasing functions and satisfy the inequalities
e
k (t) ≤ k(t) ≤ k(t) ≤ k(t) ≤ e
k(t), for all t ≥ t0 .
e
It follows from Corollary 3.2, that
where
x(t) ≤ x(t) ≤ x
e(t),
e
for allt ≥ t0 ,
x(t) = k (t) − p/a + (x − (k − p/a))e−a(t−t0 ) ,
e
e
e0
e0
e
x
e(t) = k(t) − p/a + (e
x0 − (e
k0 − p/a))e−a(t−t0 ) ,
are respectively the solutions of (2.1) with carrying capacities k (t) and e
k(t) respece
tively. Figures 1 and 2 illustrates respectively the cases p > 0 and p = 0.
The following corollary is an immediate consequence of Theorem 3.1.
Corollary 3.4 (Decreasing case). Let k, k : I → (0, +∞) be defined by
k(t) = max{k(τ ) : τ ≥ t}
and
k(t) = min{k(τ ) : t0 ≤ τ ≤ t},
and let x(t) and x(t) be respectively the solutions of the auxiliary problems (3.2),
with x(t0 ) = x(t0 ) = x(t0 ). Then
x(t) ≤ x(t) ≤ x(t),
for all t ≥ t0 .
The proof can be done in a similar way as done in the proof of Corollary 3.2.
Hence it is omitted.
12
Y. MIR, F. DUBEAU
(a) x0 < k0
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(b) x0 > k0
Figure 1. Representation of the solution x(t) (solid line) with its
lower and upper bound solutions x(t) and x
e(t) (long dashed line),
e
and the carrying capacity k(t) (black line) when p > 0.
(a) x0 < k0
(b) x0 > k0
Figure 2. Representation of the solution x(t) (solid line) with its
lower and upper bound solutions x(t) and x
e(t) (long dashed line),
e
and the carrying capacity k(t) (black line) when p = 0.
Example 3.5. Let k(t) = pe−σ(t−t0 ) (1 + r sin(ωt)) and β(t) = a, where p > 0,
σ ≥ 0, 0 < r ≤ 1, ω > 0, and a > 0. Two cases would be considered. If a 6= σ, the
solution of the problem (2.1) is
x(t) = e−a(t−t0 )
ap
rap
(e(a−σ)(t−t0 ) − 1) +
(e(a−σ)(t−t0 ) ϕ(t) − ϕ(t0 ))
a−σ
(a − σ)2 + ω 2
+ x0 e−a(t−t0 ) ,
where ϕ(t) = (a − σ) sin(ωt) − ω cos(ωt). Otherwise, the solution is
r
r
x(t) = e−a(t−t0 ) ap(t − cos(ωt)) − ap(t0 − cos(ωt0 )) + x0 .
ω
ω
In both cases, let k (t) and e
k(t) be defined by
e
k (t) = pe−σ(t−t0 ) (1 − r),
e
e
k(t) = pe−σ(t−t0 ) (1 + r).
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13
Obviously, k and e
k are non increasing functions and satisfy the inequalities
e
k (t) ≤ k(t) ≤ k(t) ≤ k(t) ≤ e
k(t), for all t ≥ t0 .
e
It follows from Corollary 3.4, that
x(t) ≤ x(t) ≤ x
e(t),
e
for all t ≥ t0 ,
where
x(t) =
e
x
e(t) =
(1 − r)ap −σ(t−t0 )
e
− e−a(t−t0 ) + x0 e−a(t−t0 ) ,
a−σ
(1 + r)ap −σ(t−t0 )
e
− e−a(t−t0 ) + x0 e−a(t−t0 ) ,
a−σ
are respectively the solutions of (2.1) with carrying capacities k (t) and e
k(t) respece
tively for the case a 6= σ. In the case a = σ, x(t) and x
e(t) are given by
e
x(t) = e−a(t−t0 ) (ap(1 − r)(t − t0 ) + x0 ),
e
An illustration is given in Figure 3.
(a) x0 < k0
x
e(t) = e−a(t−t0 ) (ap(1 + r)(t − t0 ) + x0 ).
(b) x0 > k0
Figure 3. Representation of the solution x(t) (solid line) with its
lower and upper bound solutions x(t) and x
e(t) (long dashed line),
e
and the carrying capacity k(t) (black line) when a 6= σ.
4. A product decomposition of the solution
In this section, we present a decomposition for the solution of (1.1) and (1.2) as
the product of the carrying capacity and the solution to a corresponding differential
equation with a constant carrying capacity. The next two theorems present these
results.
Theorem 4.1 (Linear equation). Let α, β ∈ C 1 (I; (0, +∞)), k(t) = α(t)/β(t) > 0
1
for all t ≥ t0 , and k(t) ∈ ACloc
(I). We have the following product decomposition
of the solution.
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Y. MIR, F. DUBEAU
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(i) The unique solution to
k̇(t) ˙
x̃(t)
= α(t) − β(t) −
x̃(t),
k(t)
x̃(t0 ) = x̃0 ,
t ≥ t0 ,
(4.1)
is
x̃(t) = k(t)ω(t),
(4.2)
where ω(t) is the solution of
ω̇(t) = β(t)(1 − ω(t)),
t ≥ t0 ,
(4.3)
ω(t0 ) = x̃0 /k0 .
(ii) Let β̃(t) be the solution of
α̇(t)
˙
+ β(t) β̃(t),
β̃(t) = 1 −
α(t)
t ≥ t0 ,
(4.4)
β̃(t0 ) > 0.
Let us set
k̃(t) = α(t)β̃(t),
(4.5)
and let ω̃(t) be the solution of
˙
ω̃(t)
=
1
(1 − ω̃(t)),
β̃(t)
t ≥ t0 ,
(4.6)
ω̃(t0 ) = ω̃0 = x0 /k̃0 .
Then the solution of (2.1) passing through the point (t0 , x0 ) is
x(t) = k̃(t)ω̃(t),
(4.7)
where k̃(t) is a solution of (2.1) passing through the point (t0 , k̃0 ). Moreover,
(
Rt
0
if limt→+∞ t0 β(u)du = +∞,
Rt
lim (x(t) − k̃(t)) =
(4.8)
t→+∞
(x0 − k̃0 )e−l if limt→+∞ t0 β(u)du = l.
Proof. (i) From (2.2), the unique solution of (4.1) is
Z t
R
Rτ
k̇(u)
k̇(u)
(β(u)− k(u) )du
− t (β(u)− k(u) )du
x̃0 +
dτ
x̃(t) = e t0
α(τ )e t0
=e
−
Rt
t0
β(u)du ln
e
k(t)
k0
t0
t
Z
x̃0 +
α(τ )e
Rτ
t0
β(u)du − ln
e
k(τ )
k0
dτ
t0
Z t
k(t) − Rtt β(u)du α(τ ) Rtτ β(u)du =
e 0
x̃0 + k0
e 0
dτ
k0
t0 k(τ )
Rt
k(t) − Rtt β(u)du β(u)du
=
e 0
x̃0 + k0 (e t0
− 1)
k0
− R t β(u)du x̃0
= k(t) 1 +
− 1 e t0
k0
= k(t)ω(t),
where ω(t) is the solution of the initial value problem (4.3).
(4.9)
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(ii) From (4.5) and (4.4), we show that k̃(t) is the solution of (2.1) because
˙
˙
k̃(t) = α̇(t)β̃(t) + α(t)β̃(t) = α(t) − β(t)k̃(t).
(4.10)
Then using (4.7), (4.10) and (4.6), we have
˙
˙
ẋ(t) = k̃(t)ω̃(t) + k̃(t)ω̃(t)
k̃(t)
= α(t) − β(t)k̃(t) ω̃(t) +
(1 − ω̃(t))
β̃(t)
= α(t)ω̃(t) − β(t)k̃(t)ω̃(t) + α(t) − α(t)ω̃(t)
= α(t) − β(t)k̃(t)ω̃(t)
= α(t) − β(t)x(t).
Thus, x(t) is the solution of (2.1) passing through the point (t0 , x0 ).
In addition, since both x(t) and k̃(t) are solution of (2.1), from (2.2) we have
x(t) − k̃(t) = (x0 − k̃0 )e
−
Rt
t0
β(u)du
and (4.8) follows directly.
The results presented in Theorem 4.1 can then be applied to the logistic equation
(1.2) and we have the following theorem.
Theorem 4.2 (Logistic equation). Let α, β ∈ C 1 (I; (0, +∞)), k(t) = α(t)/β(t) >
1
0 for all t ≥ t0 , and suppose k(t) ∈ ACloc
(I). We have the following product
decomposition of the solution.
(i) The unique solution to
k̇(t) ˙
x̃(t)
= β(t) −
x̃(t) − α(t)x̃2 (t),
k(t)
x̃(t0 ) = x̃0 ,
t ≥ t0 ,
(4.11)
is
x̃(t) =
w(t)
,
k(t)
(4.12)
where w(t) is the solution of
ẇ(t) = β(t)w(t)(1 − w(t)),
t ≥ t0 ,
w(t0 ) = k0 x̃0 .
(4.13)
(ii) Let β̃(t) be the solution of
α̇(t)
˙
β̃(t) = (
+ β(t))β̃(t) − β̃ 2 (t),
α(t)
t ≥ t0 ,
(4.14)
β̃(t0 ) > 0.
Let us set
β̃(t)
,
α(t)
and let w̃(t) be the solution of the initial value problem
k̃(t) =
˙
w̃(t)
= β̃(t)w̃(t)(1 − w̃(t)),
w̃(t0 ) = k̃0 x̃0 .
t ≥ t0 ,
(4.15)
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Y. MIR, F. DUBEAU
EJDE-2016/18
Then the solution of (1.2) passing through the point (t0 , x0 ) is
x(t) = k̃(t)w̃(t),
(4.16)
where k̃(t) is the solution of (1.2) passing through the point (t0 , k̃0 ).
Proof. (i) and (ii) are directly obtained from Theorem 4.1 using the application
x 7→ y = x−1 .
Conclusion. The goal of this study has been to make a qualitative study of the
solutions of the linear and logistic differential equations. We have obtained new
results on the behaviour and the asymptotic behaviour of any solution to these
differential equations in the case where the coefficients are time dependent. We
have studied the monotone case and also the non monotone case when it is possible
to construct subsolution and supersolution. Finally we obtain a product decomposition of the solution for some special form of these models.
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Youness Mir
Département de mathématiques, Université de Sherbrooke,
2500 Boulevard de l’Université, Sherbrooke (Qc), J1K 2R1, Canada
E-mail address: youness.mir@usherbrooke.ca
François Dubeau
Département de mathématiques, Université de Sherbrooke,
2500 Boulevard de l’Université, Sherbrooke (Qc), J1K 2R1, Canada
E-mail address: francois.dubeau@usherbrooke.ca