On the Existence of Positive Periodic Solution of an Amensalism Model
with Beddington-DeAngelis Functional Response
YANBO CHONG, SHANGMING CHEN, FENGDE CHEN
College of Mathematics and Statistics
Fuzhou University
No. 2, wulongjiang Avenue, Minhou County, Fuzhou
CHINA
Abstract: - A non-autonomous discrete amensalism model with Beddington-DeAngelis functional response is
proposed and studied in this paper. Sufficient conditions are obtained for the existence of positive periodic solution
of the system.
Key-Words: -Amensalism model; Positive periodic solution; Beddington-DeAngelis functional response
Received: September 29, 2021. Revised: May 27, 2022. Accepted: June 28, 2022. Published: July 18, 2022.
1 Introduction
Amensalism is an interaction in which an organism
inflicts harm to another organism without any costs
or benefits received by the other. In the past decade,
numerous works on the mutualism or commensalism
model has been published([1]-[25]). However, only
recently did scholars paid attention to the amensalism
model([26]-[36]). In 2019, Guan and Chen[26] pro-
posed the following two species amensalism model
with Beddington-DeAngelis functional response
dx1
dt =x1a1b1x1
cx2
mx1+nx1+ 1,
dx2
dt =x2a2b2x2.
(1)
The existence and stability of possible equilibria were
investigated. Under some additional assumptions, the
authors showed that there are two stable equilibria
which implies this system is not asymptotically stable.
Based on the stability analysis of equilibria, closed or-
bits and the saddle connection, they gave some com-
prehensive bifurcation and global dynamics of the
system.
It brings to our attention that the system (1) is an
autonomous ones. Model (1) is not well studied yet in
the sense that the model is with constant environmen-
t. The assumption that the environment is constant
is rarely the case in real life. Most natural environ-
ments are physically highly variable, and in response,
birth rates, death rates, and other vital rates of popu-
lations, vary greatly in time. Taking these factors into
consideration, then it is naturally to study the nonau-
tonomous case of system (1), i.e,
dx1
dt =x1a1(t)b1(t)x1
c(t)x2
m(t)x1+n(t)x2+ 1,
dx2
dt =x2a2(t)b2(t)x2.
(2)
It is well known that the discrete time models
governed by difference equations are more appropri-
ate than the continuous ones when the populations
have non-overlapping generations, corresponding to
system (2), we could propose the following discrete
nonautonomous amensalism model with Beddington-
DeAngelis functional response
x1(k+ 1) = x1(k)exp a1(k)b1(k)x1(k)
c(k)x2(k)
m(k)x1(k) + n(k)x2(k)+1},
x2(k+ 1) = x2(k)exp {a2(k)b2(k)x2(k),
(3)
where {bi(k)}, i = 1,2,{c(k)}{m(k)},{n(k)}are
all positive ω-periodic sequences, ωis a fixed posi-
tive integer, {ai(k)}are ω-periodic sequences, which
satisfies ai=1
ω
ω1
k=0
ai(k)>0, i = 1,2. Here we
assume that the coefficients of the system (3) are al-
l periodic sequences which having a common integer
period. Such an assumption seems reasonable in view
of seasonal factors, e.g., mating habits, availability
of food, weather conditions, harvesting, and hunting,
etc.
The aim of this paper is to obtain a set of sufficient
conditions which ensure the existence of positive pe-
riodic solution of system (3).
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2 Main Results
In the proof of our existence theorem below, we
will use the continuation theorem of Gaines and
Mawhin([37]).
Lemma 2.1 (Continuation Theorem) Let Lbe a
Fredholm mapping of index zero and let Nbe L-
compact on ¯
. Suppose
(a).For each λ(0,1), every solution xof Lx =
λNx is such that x∈ Ω;
(b).QNx = 0 for each xKerL and
deg{JQN, KerL, 0} = 0.
Then the equation Lx =Nx has at least one solution
lying in DomL ¯
.
Let Z, Z+, R and R+denote the sets of all inte-
gers, nonnegative integers, real numbers, and nonneg-
ative real numbers, respectively. For convenience, in
the following discussion, we will use the notation be-
low throughout this paper:
Iω={0,1, ..., ω 1},
g=1
ω
ω1
k=0
g(k), gu=max
kIω
g(k), gl=min
kIω
g(k),
where {g(k)}Ϊ is an ω-periodic sequence of real num-
bers defined for kZ.
Lemma 2.2[38] Let g:ZRbe ω-periodic, i. e.,
g(k+ω) = g(k).Then for any fixed k1, k2Iω,and
any kZ, one has
g(k)g(k1) +
ω1
s=0
|g(s+ 1) g(s)|,
g(k)g(k2)
ω1
s=0
|g(s+ 1) g(s)|.
Lemma 2.3 Assume that
¯a1>c
n(4)
holds, Then any solution (x
1, x
2)of the system of al-
gebraic equations
¯a1¯
b1exp{u1}
1
ω
ω1
k=0
c(k)exp{u2}
m(k)exp{u1}+n(k)exp{u2}+ 1 = 0,
¯a2¯
b2exp{u2}= 0.
(5)
satisfies
ln
¯a1c
n
b1u
1ln ¯a1
¯
b1
, u
2=ln ¯a2
¯
b2
,(6)
Proof. From the second equation of (5), it immedi-
ately follows that
u2=ln ¯a2
¯
b2
.(7)
From the first equation of system (5) we have
¯a1¯
b1exp{u1} 0,
thus
u1ln ¯a1
¯
b1
.(8)
From the first equation of system (5), we also have
0 = ¯a1¯
b1exp{u1}
1
ω
ω1
k=0
c(k)exp{u2}
m(k)exp{u1}+n(k)exp{u2}+ 1
¯a1¯
b1exp{u1} 1
ω
ω1
k=0
c(k)exp{u2}
n(k)exp{u2}
= ¯a1c
n¯
b1exp{u1}.
Thus
u1ln ¯a1c
n
¯
b1
.(9)
This ends the proof of Lemma 2.3.
We now reach the position to establish our main
result.
Theorem 2.1 Assume that (4) holds, then system (3)
admits at least one positive ω-periodic solution.
Proof. Let
xi(k) = exp{ui(k)}, i = 1,2,
so that system (3) becomes
u1(k+ 1) u1(k)
=a1(k)b1(k)exp{u1(k)}
H(u1(k), u2(k)),
u2(k+ 1) u2(k)
=a2(k)b2(k)exp{u2(k)}.
(10)
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where
H(u1(k), u2(k))
=c(k)exp{u2(k)}
m(k)exp{u1(k)}+n(k)exp{u2(k)}+ 1.
(11)
Define
l2=u={u(k)}, u(k) = (u1(k), u2(k))TR2.
For a= (a1, a2)TR2, define |a|=
max{|a1|,|a2|}. Let lωl2denote the subspace of
all ωsequences equipped with the usual normal form
u=max
kIω
|u(k)|. It is not difficult to show that lω
is a finite-dimensional Banach space. Let
lω
0={u={u(k)} lω:
ω1
k=0
u(k) = 0},
lω
c={u={u(k)} lω:u(k) = hR2, k Z},
then lω
0and lω
care both closed linear subspace of lω,
and
lω=lω
0lω
c, dimlω
c= 2.
Now let us define X=Y=lω,(Lu)(k) = u(k+
1) u(k). It is trivial to see that L is a bounded linear
operator and
KerL =lω
c, ImL =lω
0,
dimKerL = 2 = CodimImL.
Then it follows that Lis a Fredholm mapping of index
zero. Let
N(u1, u2)T= (N1, N2)T:= N(u, k),
where
N1=a1(k)b1(k)exp{u1(k)}
H(u1(k), u2(k)),
N2=a2(k)b2(k)exp{u2(k)}.
P x =1
ω
ω1
s=0
x(s), x X, Qy =1
ω
ω1
s=0
y(s), y Y.
It is not difficult to show that Pand Qare two con-
tinuous projectors such that
ImP =KerL and ImL =KerQ =Im(IQ).
Furthermore, the generalized inverse (to L)Kp:
ImLKerPDomLexists and is given by
Kp(z) =
k1
s=0
z(s)1
ω
ω1
s=0
(ωs)z(s).
Thus
QNx =1
ω
ω1
k=0
N(x, k),
Kp(IQ)Nx =
k1
s=0
N(x, s) + 1
ω
ω1
s=0
sN(x, s)
k
ω+ω1
2ωω1
s=0
N(x, s).
Obviously, QN and Kp(IQ)Nare continuous. S-
ince Xis a finite-dimensional Banach space, it is not
difficult to show that Kp(IQ)N(Ω) is compact for
any open bounded set X. Moreover, QN(Ω) is
bounded. Thus, Nis L-compact on any open bound-
ed set X. The isomorphism Jof ImQonto KerL
can be the identity mapping, since ImQ=KerL.
Now we are at the point to search for an appropri-
ate open, bounded subset in Xfor the application of
the continuation theorem. Corresponding to the oper-
ator equation Lx =λNx, λ (0,1), we have
u1(k+ 1) u1(k)
=λa1(k)b1(k)exp{u1(k)}
H(u1(k), u2(k)),
u2(k+ 1) u2(k)
=λ[a2(k)b2(k)exp{u2(k)}].
(12)
where H(u1(k), u2(k)) is defined by (11). Suppose
that u= (u1(k), u2(k))TXis an arbitrary solution
of system (12) for a certain λ(0,1). Summing on
both sides of (12) from 0 to ω1with respect to k,
we reach
ω1
k=0 a1(k)b1(k)exp{u1(k)}
H(u1(k), u2(k))= 0,
ω1
k=0
[a2(k)b2(k)exp{u2(k)}] = 0.
That is,
ω1
k=0 b1(k)exp{u1(k)}
+H(u1(k), u2(k))= ¯a1ω,
(13)
ω1
k=0
b2(k)exp{u2(k)}= ¯a2ω. (14)
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From (13) and (14), we have
ω1
k=0
|u1(k+ 1) u1(k)|
=λ
ω1
k=0 a1(k)b1(k)exp{u1(k)}
H(u1(k), u2(k))
ω1
k=0
|a1(k)|+ω1
k=0 b1(k)exp{u1(k)}
+H(u1(k), u2(k))
=ω1
k=0
|a1(k)|+ ¯a1ω
= ( ¯
A1+ ¯a1)ω,
(15)
ω1
k=0
|u2(k+ 1) u2(k)|
=λ
ω1
k=0 a2(k)b2(k)exp{u2(k)}
(¯
A2+ ¯a2)ω.
(16)
where ¯
A1=1
ω
ω1
k=0
|a1(k)|,¯
A2=1
ω
ω1
k=0
|a2(k)|.
Since {u(k)}={(u1(k), u2(k))T} X, there
exist ηi, δi, i = 1,2such that
ui(ηi) = min
kIω
ui(k), ui(δi) = max
kIω
ui(k).(17)
By (14), we have
exp{u2(η2)}
ω1
k=0
b2(k)¯a2ω.
So
u2(η2)ln ¯a2
¯
b2
.(18)
It follows from Lemma 2.2, (16) and (18) that
u2(k)u2(η2) + ω1
k=0
|u2(k+ 1) u2(k)|
ln ¯a2
¯
b2
+ ( ¯
A2+ ¯a2)ωdef
=K1,
(19)
From (14) we also have
exp{u2(δ2)}
ω1
k=0
b2(k)¯a2ω,
and so
u2(δ2)ln ¯a2
¯
b2
.(20)
It follows from Lemma 2.2, (16) and (20) that
u2(k)u2(δ2)ω1
k=0
|u2(k+ 1) u2(k)|
ln ¯a2
¯
b2
(¯
A2+ ¯a2)ωdef
=K2,
(21)
which together with (19) leads to
|u2(k)| max |K1|,|K2|def
=H2.(22)
It follows from (13) that
ω1
k=0
b1(k)exp{u1(η1)}
¯a1ωω1
k=0
H(u1(k), u2(k))
¯a1ω,
and so,
u1(η1)ln ¯a1
b1
.(23)
It follows from Lemma 2.2, (15) and (23) that
u1(k)u1(η1) + ω1
k=0
|u1(k+ 1) u1(k)|
ln ¯a1
b1
+ ( ¯
A1+ ¯a1)ωdef
=M1.
(24)
It follows from (13) that
ω1
k=0
b1(k)exp{u1(δ1)}
= ¯a1ωω1
k=0
H(u1(k), u2(k))
¯a1ωω1
k=0
c(k)
n(k)
¯a1ω(c
n)ω,
where (c
n) = 1
ω
ω1
k=0
c(k)
n(k).And so,
u1(δ1)ln ¯a1(c
n)
b1
,(25)
It follows from Lemma 2.2, (15) and (25) that
u1(k)u1(δ1)ω1
k=0
|u1(k+ 1) u1(k)|
ln ¯a1(c
n)
b1
(¯
A1+ ¯a1)ωdef
=M2.
(26)
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It follows from (24) and (26) that
|u1(k)| max |M1|,|M2|def
=H1.(27)
Clearly, H1and H2are independent on the choice of
λ.
It follows from (4) and Lemma 2.3 that any solu-
tion (x
1, x
2)of the system of algebraic equations
¯a1¯
b1exp{u1}
1
ω
ω1
k=0
c(k)exp{u2}
m(k)exp{u1}+n(k)exp{u2}+ 1 = 0,
¯a2¯
b2exp{u2}= 0
satisfies
ln
¯a1c
n
b1u
1ln ¯a1
¯
b1
, u
2=ln ¯a2
¯
b2
,(28)
Let H=H1+H2+H3,where H3>0is taken
sufficiently enough large such that
H3>
ln ¯a2
¯
b2
+
ln ¯a1
¯
b1
+
ln ¯a1c
n
b1
.
Let H=H1+H2+H3, and define
= u(k) = (u1(k), u2(k))TX:u< H.
It is clear that verifies requirement (a) in Lemma
2.1. When uKerL =R2,uis constant
vector in R2with ||u|| =B. Then
QNu
=
¯a1¯
b1exp{u1} 1
¯a2¯
b2exp{u2}
= 0.
where
1=1
ω
ω1
k=0
c(k)exp{u2}
m(k)exp{u1}+n(k)exp{u2}+ 1.
In order to compute the Brouwer degree, let us con-
sider the homotopy
Hµu=µQNu + (1 µ)Gu, (29)
where
Gu =¯a1¯
b1exp{u1}
¯a2¯
b2exp{u2}.
From the definition of H, it follows that 0 /Hµ(
KerL)for µ[0,1].In addition, one can easily show
that the algebraic equation Gu = 0 has a unique so-
lution in R2. Note that J=Isince ImQ =KerL,
by the invariance property of homotopy, direct calcu-
lation produces
deg(JQN, KerL, 0)
=deg(QN, KerL, 0)
=deg(G, KerL, 0) = sgnΓ= 1 = 0,
where
Γ = ¯
b1¯
b2exp{u
1}exp{u
2}
and deg(., ., .)is the Brouwer degree. By now
we have proved that verifies all requirements in
Lemma 2.1. Hence (4) has at least one solution
(u
1(k), u
2(k))Tin DomL ¯
.And so, system (3)
admits a positive periodic solution (x
1(k), x
2(k))T,
where x
i(k) = exp{u
i(k)}, i = 1,2, This completes
the proof of the claim.
3 Numeric simulations
Now let us consider the following two examples.
Example 3.1.
x1(k+ 1) = x1(k)exp 1.5x1(k)
(2 + sin(πk))x2(k)
1 + x2(k) + 0.1x1(k),
x2(k+ 1) = x2(k)exp 1.5
(3 + cos(πk +π
3))x2(k).
(30)
Corresponding to system (3), here we choose a1(k) =
1.5, b1(k) = 1, c(k) = 2 + sin(πk), m(k) =
0.1, n(k) = 1, a2(k) = 1.5, b2(k) = 3+cos(πk+π
3).
One could easily check that the condition of Theo-
rem 2.1 holds, and consequently, system (30) admits
at least one positive 2-period solution. Numeric sim-
ulations (Fig.1, Fig. 2 ) also support this assertion.
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time n
0 2 4 6 8 10 12 14 16 18 20
solution x1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Figure 1: Dynamic behaviors of the first compo-
nent x1in system (30) with the initial condition
(x(0), y(0)) = (0.5,0.5),(1,1),(1.5,1.5) and
(2,2), respectively.
time n
0 2 4 6 8 10 12 14 16 18 20
solution x2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Figure 2: Dynamic behaviors of the second com-
ponent x2in system (30) with the initial condi-
tion (x(0), y(0)) = (0.5,0.5),(1,1),(1.5,1.5)
and (2,2), respectively.
Example 3.2.
x1(k+ 1) = x1(k)exp 3x1(k)
(2 + sin(πk))x2(k)
1 + x2(k) + 0.1x1(k),
x2(k+ 1) = x2(k)exp 3
(3 + cos(πk +π
3))x2(k),
(31)
Corresponding to system (3), here we change
a1(k), a2(k)to 3, other coefficients are the same as
system (30). Numeric simulations (Fig.3, Fig. 4 )
show that system (31) admits one positive periodic so-
lution. However, the other solutions need more time
to approach to the periodic solution.
time n
0 5 10 15 20 25 30 35 40 45 50
solution x1
0
1
2
3
4
5
6
7
8
9
10
Figure 3: Dynamic behaviors of the first compo-
nent x1in system (31) with the initial condition
(x(0), y(0)) = (0.5,0.5),(1,1),(1.5,1.5) and
(2,2), respectively.
time n
0 5 10 15 20 25 30 35 40 45 50
solution x2
0
1
2
3
4
5
6
7
8
9
10
Figure 4: Dynamic behaviors of the second com-
ponent x2in system (31) with the initial condi-
tion (x(0), y(0)) = (0.5,0.5),(1,1),(1.5,1.5)
and (2,2), respectively.
4 Discussion
In this paper, we proposed a discrete amensilism mod-
el with with Beddington-DeAngelis functional re-
sponse, by using the coincidence degree theory, suf-
ficient conditions which ensure the existence of posi-
tive periodic sequences solution are established. Nu-
meric simulations are carried out to show the feasibil-
ity of the main result.
We mention here that we did not investigate the
stability property of the system, however, numeric
simulations (Fig.1, 2, 3 and 4) showed that the pe-
riodic solution is unique and globally asymptotically
stable in system (30) and (31). We leave this for future
investigation.
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2022.21.64
Yanbo Chong, Shangming Chen, Fengde Chen
E-ISSN: 2224-2880
579