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Global stability in n-dimensional discrete Lotka-Volterra predator-prey models

Abstract

There are few theoretical works on global stability of Euler difference schemes for two-dimensional Lotka-Volterra predator-prey models. Furthermore no attempt is made to show that the Euler schemes have positive solutions. In this paper, we consider Euler difference schemes for both the two-dimensional models and n-dimensional models that are a generalization of the two-dimensional models. It is first shown that the difference schemes have positive solutions and equilibrium points which are globally asymptotically stable in the two-dimensional cases. The approaches used in the two-dimensional models are extended to the n-dimensional models for obtaining the positivity and the global stability. Numerical examples are presented to verify the results.

MSC: 34A34, 39A10, 40A05.

1 Introduction

Consider the n-dimensional system

d x i d t = x i ( σ i r i + 1 j i 1 a i j x j i j n a i j x j ) ,
(1.1)

where r i >0, a i j >0 for 1i,jn, σ 1 =1, and σ i {1,1} for 2in.

The system equation (1.1) can be seen as a generalization of the two-dimensional Lotka-Volterra predator-prey model

d x d t =x( r 1 a 11 x a 12 y), d y d t =y( r 2 + a 21 x a 22 y),
(1.2)

where x and y denote the population sizes of prey and predator, respectively.

There are a number of works on investigating nonstandard finite difference schemes for the Lotka-Volterra competition models (see [1] and the references given there), but relatively few theoretical papers are published on discretized models of equation (1.2). In particular, to my knowledge, Euler difference schemes for equation (1.2) have not theoretically been studied for the global stability of the equilibrium points except a recent paper [2]. In Section 2, it is shown that the Euler difference scheme has positive solutions. In order to show the global asymptotic stability of the equilibrium point whose components are all positive, the paper [2] assumes that (0,0) is globally stable. Without using the assumption, we show the global stability of all of the equilibrium points in Section 3. In addition, we also analyze the Euler difference scheme for equation (1.2) with r 2 replaced by r 2 . We are interested in extending the method used in the two-dimensional discrete models to the n-dimensional discrete models for equation (1.1). In Section 4, we demonstrate the positivity and the global stability in the n-dimensional discrete cases. Numerical examples are given to verify the results of this paper.

2 Two-dimensional predator-prey model

In this section, we consider the Euler difference scheme for equation (1.2)

x 0 =x(0), y 0 =y(0), x k + 1 = F y k ( x k ), y k + 1 = G x k ( y k ),
(2.1)

where Δt is a time step size, r ˆ i = r i Δt, a ˆ i j = a i j Δt for 1i,j2, and

F y (x)=x(1+ r ˆ 1 a ˆ 11 x a ˆ 12 y), G x (y)=y(1 r ˆ 2 + a ˆ 21 x a ˆ 22 y).
(2.2)

Note that if τ 1 and τ 2 are positive constants such that

U 1 ( τ 2 )= 1 + r ˆ 1 a ˆ 12 τ 2 2 a ˆ 11 >0, U 2 ( τ 1 )= 1 r ˆ 2 + a ˆ 21 τ 1 2 a ˆ 22 >0,

then

F τ 2 (x) and  G τ 1 (y) are increasing on 0x< U 1 ( τ 2 ) and 0y< U 2 ( τ 1 ).
(2.3)

Let Δt satisfy max{ r ˆ 1 , r ˆ 2 }<1. Take positive constants χ 1 and χ 2 such that

r ˆ 2 + a ˆ 21 χ 1 1, r 1 a 11 χ 1 U 1 ( χ 2 ), r 2 + a 21 χ 1 a 22 χ 2 U 2 (0).
(2.4)

Theorem 2.1 Let Δt, χ 1 , χ 2 satisfy equation (2.4) with max{ r ˆ 1 , r ˆ 2 }<1. If ( x 0 , y 0 )(0, χ 1 )×(0, χ 2 ), then ( x k , y k )(0, χ 1 )×(0, χ 2 ) for all k with k1.

Proof Since

0< x 0 < χ 1 U 1 ( χ 2 )< U 1 ( y 0 ),0< y 0 < χ 2 U 2 (0)< U 2 ( x 0 ),

it follows from equation (2.3) that

x 1 = F y 0 ( x 0 )> F y 0 (0)=0, y 1 = G x 0 ( y 0 )> G x 0 (0)=0.

If r 1 a 11 x 0 a 12 y 0 0, then x 1 = F y 0 ( x 0 ) x 0 < χ 1 , and otherwise,

0< x 0 < r 1 a 12 y 0 a 11 < r ˆ 1 + ( r ˆ 1 a ˆ 12 y 0 ) 2 a ˆ 11 < 1 + r ˆ 1 a ˆ 12 y 0 2 a ˆ 11 = U 1 ( y 0 )

by the condition r ˆ 1 <1. Hence equation (2.3) implies that

x 1 = F y 0 ( x 0 )< F y 0 ( r 1 a 12 y 0 a 11 ) = r 1 a 12 y 0 a 11 < r 1 a 11 χ 1 .

Similarly if r 2 + a 21 x 0 a 22 y 0 0, then y 1 = G x 0 ( y 0 ) y 0 < χ 2 , and otherwise,

0< y 0 < r 2 + a 21 x 0 a 22 < ( r ˆ 2 + a ˆ 21 χ 1 ) + ( r ˆ 2 + a ˆ 21 x 0 ) 2 a ˆ 22 1 r ˆ 2 + a ˆ 21 x 0 2 a ˆ 22 = U 2 ( x 0 )

by the condition r ˆ 2 + a ˆ 21 χ 1 1. Thus equation (2.3) gives

y 1 = G x 0 ( y 0 )< G x 0 ( r 2 + a 21 x 0 a 22 ) = r 2 + a 21 x 0 a 22 < r 2 + a 21 χ 1 a 22 χ 2 .

Finally we obtain, if ( x 0 , y 0 )(0, χ 1 )×(0, χ 2 ), then

( x 1 , y 1 )(0, χ 1 )×(0, χ 2 ).

By the principle of mathematical induction, the proof is completed. □

From now on, we assume that ( x 0 , y 0 )(0, χ 1 )×(0, χ 2 ) and ( x k , y k ) for k1 denote the solutions of equation (2.1). For simplicity of notation, we write for all k instead of for all k with k1 when there is no confusion.

Remark 2.2 Theorem 2.1 gives for all k

0< x k < χ 1 < U 1 ( χ 2 )< U 1 ( y k ),0< y k < χ 2 < U 2 (0)< U 2 ( x k ).

Hence it follows from equation (2.3) that for every fixed ( x k , y k )

F y k (x) and  G x k (y) are increasing on 0x< χ 1  and 0y< χ 2 .
(2.5)

Let f(x)= r 1 a 11 x a 12 and g(x)= r 2 + a 21 x a 22 . Since f and g are decreasing and increasing, respectively, it follows from equation (2.4) that for all k

f 1 ( y k )< f 1 (0) χ 1 ,max { g ( f 1 ( 0 ) ) , g ( x k ) } g( χ 1 ) χ 2 .
(2.6)

Set D=(0, χ 1 )×(0, χ 2 ), and let S i for 1i4 denote the four areas

S 1 = { ( x , y ) D | g ( x ) y < f ( x ) } , S 2 = { ( x , y ) D | y f ( x ) , y < g ( x ) } , S 3 = { ( x , y ) D | f ( x ) < y g ( x ) } , S 4 = { ( x , y ) D | y f ( x ) , y > g ( x ) } .

Remark 2.3 Let ( x k , y k ) 1 i 4 S i for some k. The following can be obtained by using equation (2.5), equation (2.6), and the definitions of F y k ( x k ) and G x k ( y k ).

  1. (a)

    Suppose ( x k , y k ) S 1 . Since g( x k ) y k <f( x k ), we have

    x k + 1 = F y k ( x k ) > F f ( x k ) ( x k ) = x k , x k + 1 = F y k ( x k ) < F y k ( f 1 ( y k ) ) = f 1 ( y k ) , y k + 1 = G x k ( y k ) G g 1 ( y k ) ( y k ) = y k < f ( x k + 1 ) .
  2. (b)

    Suppose ( x k , y k ) S 2 . This gives y k f( x k ) and y k <g( x k ), and then

    x k + 1 = F y k ( x k ) F f ( x k ) ( x k ) = x k , y k + 1 = G x k ( y k ) > G g 1 ( y k ) ( y k ) = y k , y k + 1 < G x k ( g ( x k ) ) = g ( x k ) < g ( x k + 1 ) .
  3. (c)

    Suppose ( x k , y k ) S 3 . Since f( x k )< y k g( x k ), we have

    x k + 1 = F y k ( x k ) < F f ( x k ) ( x k ) = x k , x k + 1 > F y k ( f 1 ( y k ) ) = f 1 ( y k ) , y k + 1 = G x k ( y k ) G g 1 ( y k ) ( y k ) = y k > f ( x k + 1 ) .
  4. (d)

    Suppose ( x k , y k ) S 4 , which means that y k f( x k ) and y k >g( x k ). Then

    x k + 1 = F y k ( x k ) F f ( x k ) ( x k ) = x k , y k + 1 = G x k ( y k ) < G g 1 ( y k ) ( y k ) = y k , y k + 1 > G x k ( g ( x k ) ) = g ( x k ) > g ( x k + 1 ) .

Therefore ( x k , y k ) in S 1 , S 2 , S 3 , and S 4 moves to

the lower right, upper right, upper left, and lower left parts
(2.7)

of S 1 S 2 , S 2 S 3 , S 3 S 4 , and S 4 S 1 , respectively (see Figures 1 and 2).

Figure 1
figure 1

Two-dimensional models with ( σ 1 , σ 2 )=(1,1). The dark circles denote initial conditions (see Examples 3.2, 3.6, and 3.7 for details). (a) r 1 a 21 r 2 a 11 <0 and the limits of the four trajectories are ( r 1 a 11 ,0)=(10,0). (b) r 1 a 21 r 2 a 11 >0. The solutions of equation (2.1) rotate finitely many times around the limit θ=( θ 1 , θ 2 )=(1.25,0.25). (c) r 1 a 21 r 2 a 11 >0. The solutions of equation (2.1) rotate infinitely many times around the limit θ=(1.287,28.713).

Figure 2
figure 2

Two-dimensional models with ( σ 1 , σ 2 )=(1,1). The dark circles denote initial conditions (see Example 3.10). (a) r 1 a 22 r 2 a 12 <0 and the limits of the four trajectories are (0, r 2 a 22 )=(0,2). (b)  r 1 a 22 r 2 a 12 >0. The solutions of equation (3.10) rotate finitely many times around the limit ϑ=(0.455,10.455). (c) r 1 a 22 r 2 a 12 >0. The solutions of equation (3.10) rotate infinitely many times around the limit ϑ=(0.683,29.317).

Set S 5 = S 1 , and use the notation x N , y N S i + 1 for 1i4 and a positive integer N to denote both ( x k , y k ) S i for all k with 0k<N and ( x N , y N ) S i + 1 . Then equation (2.7) implies the following theorem.

Theorem 2.4 Let the assumptions of Theorem  2.1 hold. Suppose 1i4.

If ( x 0 , y 0 ) S i ,then ( x k , y k ) S i  for all k or  x N i , y N i S i + 1  for some  N i .

3 Dynamics of the two-dimensional predator-prey models

In this section, we first consider dynamics of the Euler difference scheme for equation (1.2) and next for equation (1.1) with n=2 and ( σ 1 , σ 2 )=(1,1). Let θ 1 = r 1 a 22 + r 2 a 12 a 12 a 21 + a 11 a 22 and θ 2 = r 1 a 21 r 2 a 11 a 12 a 21 + a 11 a 22 . For calculating the limits of x k and y k , we use the inequalities

lim k x k >0, lim k y k >0,
(3.1)

which is equivalent to lim k ( x k , y k )=( θ 1 , θ 2 ).

Theorem 3.1 Let the assumptions of Theorem  2.1 hold. Suppose r 1 a 21 r 2 a 11 0. Then ( x k , y k ) satisfies the following dynamics with the limit ( r 1 a 11 ,0).

  1. (a)

    If ( x 0 , y 0 ) S 1 , then ( x k , y k ) S 1 for all k.

  2. (b)

    If ( x 0 , y 0 ) S 4 , then ( x k , y k ) S 4 for all k or x N 1 , y N 1 S 1 for some N 1 .

  3. (c)

    If ( x 0 , y 0 ) S 3 , then x N 2 , y N 2 S 4 for some N 2 .

Proof (a) Since f 1 (0)= r 1 a 11 r 2 a 21 = g 1 (0), the set S 2 is empty, and then Theorem 2.4 gives ( x k , y k ) S 1 for all k. Thus equation (2.7) shows that x k is bounded and increasing, and y k is bounded and decreasing. Hence lim k x k >0 and lim k y k 0. Finally lim k y k =0 since otherwise 0< lim k y k = θ 2 0, which is a contradiction. Therefore lim k x k = r 1 a 11 by using equation (2.1) with both lim k x k >0 and lim k y k =0.

(b) Theorem 2.4 and (a) in this theorem show that it suffices to show lim k ( x k , y k )=( r 1 a 11 ,0) in the case ( x k , y k ) S 4 for all k, where lim k x k 0 and lim k y k 0 by equation (2.7). Then lim k y k =0 since otherwise the last equation in equation (2.1) gives lim k x k >0, and hence equation (3.1) implies lim k y k = θ 2 0, producing a contradiction to lim k y k >0.

Note that y k >f( x k ) for all k since ( x k , y k ) S 4 for all k. Thus if lim k x k =0, then lim k y k lim k f( x k )= r 1 a 12 >0, which contradicts to lim k y k =0. Hence lim k x k >0. Consequently, lim k x k = r 1 a 11 by using equation (2.1) with lim k y k =0.

(c) Assume, to the contrary, that ( x k , y k ) S 3 for all k. Then equation (2.7) implies that x k and y k have positive limits, and hence equation (3.1) gives 0< lim k y k = θ 2 0, which is a contradiction. Therefore x N 2 , y N 2 S 4 for some N 2 , which gives lim k ( x k , y k )=( r 1 a 11 ,0) by (b) in this theorem. □

Example 3.2 Consider the discrete system

x k + 1 = x k + x k (505 x k 2 y k )Δt, y k + 1 = y k + y k (100+5 x k 5 y k )Δt

with Δt=0.001 and ( χ 1 , χ 2 )=(80,90). Then equation (2.4) and the conditions r 1 a 21 r 2 a 11 0 and max{ r ˆ 1 , r ˆ 2 }<1 in Theorem 3.1 are satisfied. For the four initial conditions

(2,15) S 1 ,(28,5) S 3 ,(10,20) S 4 ,(15,0.01) S 4 ,

Figure 1(a) shows that the solution ( x k , y k ) satisfies Theorem 2.1 and Theorem 3.1 with the limit ( r 1 a 11 ,0)=(10,0). In the case ( x 0 , y 0 )=(15,0.01) S 4 , Figure 1(a) shows that for 0i25,000 and 2,000k25,000

10<10+9 10 39 = x k x i + 1 x i

which implies that ( x k , y k ) S 4 for all k.

In order to show the global asymptotic stability of the equilibrium point θ=( θ 1 , θ 2 ), the linearized system of equation (2.1) at θ is used: Consider the Jacobian matrix of T(x,y)=( F y (x), G x (y)) at θ

D T θ = ( 1 θ 1 a 11 Δ t θ 1 a 12 Δ t θ 2 a 21 Δ t 1 θ 2 a 22 Δ t ) .
(3.2)

Letting J= 1 Δ t (D T θ I) with the 2×2 identity matrix I, we have tr(J)= θ 1 a 11 θ 2 a 22 , det(J)= θ 1 θ 2 ( a 11 a 22 + a 12 a 21 ), and the eigenvalues of D T θ

1+ 1 2 { tr ( J ) ± tr 2 ( J ) 4 det ( J ) } Δt.
(3.3)

The following lemma is used for showing that the equilibrium point θ of the nonlinear system x k + 1 =T( x k ) is locally asymptotically stable.

Lemma 3.3 Suppose r 1 a 21 r 2 a 11 >0. Let J= 1 Δ t (D T θ I). Then all of the eigenvalues of D T θ have magnitude less than 1 if one of the following is true.

  1. (a)

    tr 2 (J)4det(J).

  2. (b)

    tr 2 (J)<4det(J) and det(J)Δt<tr(J).

In order to find conditions for when ( x n , y n ) rotates finitely or infinitely many times around θ, we need the following theorem about a hyperbolic point: A point p of x k + 1 =T( x k ) is called hyperbolic if all of the eigenvalues of D T p have nonzero real parts.

Theorem 3.4 (Hartman-Grobman theorem for maps)

Let p be a hyperbolic fixed point of x k + 1 =H( x k ), where H is a continuously differentiable function defined on a neighborhood of 0 R n for n1. Then there exist neighborhoods U of p, V of the hyperbolic fixed point 0 of x k + 1 =D H p ( x k ), and a homeomorphism h:VU such that H(h(x))=h(D H p (x)) for all xV.

Theorem 3.4 (see [3] for the proof) states that the nonlinear system x k + 1 =H( x k ) is topologically equivalent to the linearized system x k + 1 =D H p ( x k ) of the nonlinear system at p.

Theorem 3.5 Let the assumptions of Theorem  2.1 hold and let J= 1 Δ t (D T θ I). Suppose r 1 a 21 r 2 a 11 >0 and ( x 0 , y 0 ) 1 i 4 S i .

  1. (a)

    If tr 2 (J)4det(J), then ( x k , y k ) rotates finitely many times around θ in the counterclockwise direction and finally stays in one of S i for 1i4 with lim k ( x k , y k )=θ.

  2. (b)

    If the four inequalities tr 2 (J)<4det(J), det(J)Δt<tr(J),

    Δt max 1 i 2 { a i 1 ( θ 1 + χ 1 ) + a i 2 ( θ 2 + χ 2 ) } <0.5,
    (3.4)
4Δt { a 21 ( a 11 + a 12 ) + a 12 ( a 21 + a 22 ) } max 1 i , j 2 a i j <min{ a 21 a 11 , a 12 a 22 },
(3.5)

are satisfied, then ( x k , y k ) rotates infinitely many times around θ in the counterclockwise direction with lim k ( x k , y k )=θ.

Proof (a) Since all of the eigenvalues of D T θ in equation (3.3) are positive numbers less than 1, the fixed point θ of ( x k + 1 , y k + 1 )=T( x k + 1 , y k + 1 ) is hyperbolic and the solutions of the linearized system of equation (2.1) at θ rotate finitely many times around (0,0), converging to (0,0). Hence Theorem 3.4 with Theorem 2.4 gives the proof of (a) without showing lim k ( x k , y k )=θ.

For obtaining the limit of ( x k , y k ) consider the case ( x k , y k ) S i for i=1 and all sufficiently large k. Then equation (2.7) yields the result that x k is increasing with the upper bound θ 1 less than r 1 a 11 , and hence 0< lim k x k < r 1 a 11 . Consequently lim k y k >0, since otherwise lim k x k = r 1 a 11 , which is a contradiction. Therefore we obtain equation (3.1).

In the case ( x k , y k ) S i for i=2 and all sufficiently large k, it follows from equation (2.7) that x k and y k are both increasing and bounded, which gives equation (3.1).

Similarly, the other two cases for i=3 and 4 can be proved by using equation (3.1), equation (2.7), and the method of proof by contradiction.

(b) Since all of the eigenvalues of D T θ in equation (3.3) have positive real parts and magnitude less than 1 by det(J)Δt<tr(J), the fixed point θ is hyperbolic and the solutions of the linearized system of equation (2.1) at θ rotate infinitely many times around (0,0), converging to (0,0). Hence (b) is proved by using Theorem 3.4 and Theorem 2.4. It remains to show that lim k ( x k , y k )=θ. Consider a function V k defined by

V k = a 21 ( x k θ 1 ln x k )+ a 12 ( y k θ 2 ln y k )
(3.6)

for all the solutions ( x k , y k ). Letting θ 1 k = θ 1 x k , θ 2 k = θ 2 y k , and Δ x k = x k + 1 x k , we have Δ x k =( a ˆ 11 θ 1 k + a ˆ 12 θ 2 k ) x k and Δ y k =( a ˆ 21 θ 1 k + a ˆ 22 θ 2 k ) y k by equation (2.1). Then the Mean Value Theorem gives for some α, β with 0<α,β<1

V k + 1 V k = a 21 Δ x k ( 1 θ 1 Δ ln x k Δ x k ) + a 12 Δ y k ( 1 θ 2 Δ ln y k Δ y k ) = a 21 ( a ˆ 11 θ 1 k + a ˆ 12 θ 2 k ) ( x k θ 1 x k α Δ x k + x k ) + a 12 ( a ˆ 21 θ 1 k + a ˆ 22 θ 2 k ) ( y k θ 2 y k β Δ y k + y k ) .
(3.7)

Note that

x k α Δ x k + x k = 1 α ( a 11 θ 1 k + a 12 θ 2 k ) Δ t α ( a 11 θ 1 k + a 12 θ 2 k ) Δ t + 1 1 ( C 1 θ 1 k + C 2 θ 2 k ) Δ t , y k β Δ y k + y k = 1 β ( a 21 θ 1 k + a 22 θ 2 k ) Δ t β ( a 21 θ 1 k + a 22 θ 2 k ) Δ t + 1 1 ( C 3 θ 1 k + C 4 θ 2 k ) Δ t ,

where equation (3.4) gives

max 1 i 4 | C i |<2 max 1 i , j 4 a i j .
(3.8)

Then equation (3.7) becomes

V k + 1 V k ( a 21 a ˆ 11 C 5 Δt) θ 1 k 2 ( a 12 a ˆ 22 C 6 Δt) θ 2 k 2 ,
(3.9)

where C 5 = a 21 { a ˆ 11 (| C 1 |+| C 2 |)+ a ˆ 12 | C 1 |}+ a 12 { a ˆ 21 (| C 3 |+| C 4 |)+ a ˆ 22 | C 3 |}, C 6 = a 21 { a ˆ 11 | C 2 |+ a ˆ 12 (| C 1 |+| C 2 |)}+ a 12 { a ˆ 21 | C 4 |+ a ˆ 22 (| C 3 |+| C 4 |)}, and

max{ C 5 , C 6 }<4Δt { a 21 ( a 11 + a 12 ) + a 12 ( a 21 + a 22 ) } max 1 i , j 2 a i j

by equation (3.8). Hence equation (3.9) together with equation (3.5) becomes

V k + 1 V k C 7 ( θ 1 k 2 + θ 2 k 2 ) Δt

for a positive constant C 7 .

Now assume, to the contrary, that ( x k , y k ) does not converge to θ. Since θ is locally asymptotically stable by the linearization method (see [4]), the assumption implies that θ 1 k 2 + θ 2 k 2 has a positive lower bound. Then there exists a positive constant C such that for all k with k0

V k + 1 V k Cand hence V k V 0 kC.

This is a contradiction, since lim k ( V 0 kC)= and V k is bounded by Theorem 2.1. □

Example 3.6 Consider the discrete system

x k + 1 = x k + x k (2.752 x k y k )Δt, y k + 1 = y k + y k (1+ x k y k )Δt

and the four initial conditions

(0.9,0.15) S 1 ,(1.25,0.12) S 2 ,(1.4,0.2) S 3 ,(1.3,0.35) S 4

with Δt=0.001 and ( χ 1 , χ 2 )=(100,100). Then r 1 a 21 r 2 a 11 >0, max{ r ˆ 1 , r ˆ 2 }<1, and θ=(1.25,0.25). Since tr(J)=2.75 and det(J)=0.938, the condition tr 2 (J)4det(J) in Theorem 3.5(a) is satisfied. Figure 1(b) shows the dynamics in Theorem 3.5(a) with the limit θ.

Example 3.7 Consider the discrete system

x k + 1 = x k + x k (30 x k y k )Δt, y k + 1 = y k + y k (1+ x k 0.01 y k )Δt,

and the four initial conditions ( x 0 , y 0 )

(0.5,15) S 1 ,(5,15) S 2 ,(7,30) S 3 ,(0.5,0.45) S 4

with Δt=0.0001 and ( χ 1 , χ 2 )=(100,100). Then r 1 a 21 r 2 a 11 >0, max{ r ˆ 1 , r ˆ 2 }<1, and θ=(1.287,28.713). Since tr(J)=1.574 and det(J)=37.327, the conditions in Theorem 3.5(b) are satisfied. Figure 1(c) shows the dynamics in Theorem 3.5(b) with the limit θ.

In the remainder of this section we consider the Euler difference scheme for equation (1.1) with n=2 and ( σ 1 , σ 2 )=(1,1)

x k + 1 = x k (1+ r ˆ 1 a ˆ 11 x k a ˆ 12 y k ), y k + 1 = y k (1+ r ˆ 2 + a ˆ 21 x k a ˆ 22 y k ),
(3.10)

which has the three nonzero equilibrium points

( r 1 a 11 , 0 ) , ( 0 , r 2 a 22 ) ,ϑ= ( r 1 a 22 r 2 a 12 a 12 a 21 + a 11 a 22 , r 1 a 21 + r 2 a 11 a 12 a 21 + a 11 a 22 ) .

Replace r 2 in Section 2 with r 2 . For example, g(x)= r 2 + a 21 x a 22 is replaced with g(x)= r 2 + a 21 x a 22 . Then Theorem 2.1, equation (2.7), and Theorem 2.4 remain true. In the case r 1 a 22 r 2 a 12 0, the set S 1 is empty, and hence we can prove the following theorem, which corresponds to Theorem 3.1.

Theorem 3.8 Let the assumptions of Theorem  2.1 hold with r 2 replaced by r 2 and let g(x)= r 2 + a 21 x a 22 . Suppose r 1 a 22 r 2 a 12 0. Then the solution ( x k , y k ) of equation (3.10) satisfies the following dynamics with the limit (0, r 2 a 22 ).

  1. (a)

    If ( x 0 , y 0 ) S 4 , then ( x k , y k ) S 4 for all k.

  2. (b)

    If ( x 0 , y 0 ) S 3 , then ( x k , y k ) S 3 for all k or x N 1 , y N 1 S 4 for some N 1 .

  3. (c)

    If ( x 0 , y 0 ) S 2 , then x N 2 , y N 2 S 3 for some N 2 .

Remark 3.9 The Jacobian matrix of equation (3.10) at ϑ equals D T ϑ as defined in equation (3.2), and then Theorem 3.5 remains true if r 1 a 21 r 2 a 11 >0, r 2 , and θ in Theorem 3.5 are replaced with r 1 a 22 r 2 a 12 >0, r 2 , and ϑ, respectively. Therefore only the two equilibrium points (0, r 2 a 22 ) and ϑ of equation (3.10) are globally asymptotically stable.

Example 3.10 Let E=( r 1 , a 11 , a 12 , r 2 , a 21 , a 22 ). Consider the Euler difference scheme for equation (3.10) with Δt=0.001.

  1. (a)

    E=(1,1,1,2,1,1) with the three initial conditions (0.2,0.5) S 2 , (2,1) S 3 , and (0.2,3) S 4 . Then the conditions in Theorem 3.8 are satisfied. Figure 2(a) shows that the three trajectories converge to (0, r 2 a 22 )=(0,2) as in Theorem 3.8. Replacing the values r 1 =1 in E with r 1 =0.0000001 and letting ( x 0 , y 0 )=(0.0001,1.5), we have y i y i + 1 y k =20.75 10 36 <2 for 0i125,000 and 50,000k125,000, which imply ( x k , y k ) S 3 for all k.

  2. (b)

    E=(1.5,1,0.1,1,0.1,0.1) with the four initial conditions (0.41,10.8) S 1 , (0.41,10.1) S 2 , (0.58,10.2) S 3 , and (0.58,10.9) S 4 . Then ϑ=(0.455,10.455) and r 1 a 22 r 2 a 12 =0.05>0. Since tr(J)=1.5 and det(J)=0.523, the condition tr 2 (J)4det(J) in Theorem 3.5(a) is satisfied. Figure 2(b) shows that solutions rotate finitely many times around the limit ϑ.

  3. (c)

    E=(30,1,1,0.1,2,0.05) with the four initial conditions (0.1,15) S 1 , (5,15) S 2 , (5,30) S 3 , and (0.5,43) S 4 . Then ϑ=(0.683,29.317) and r 1 a 22 r 2 a 12 =1.4>0. Since tr(J)=2.149 and det(J)=41.044, the conditions in Theorem 3.5(b) are satisfied. Figure 2(c) shows that the spiral trajectories rotate infinitely many times around the limit ϑ.

4 n-Dimensional predator-prey models

In this section, we consider the Euler difference scheme for equation (1.1)

x k + 1 i = F x k i ( x k i ) for 1k,1in,
(4.1)

where x k i denotes x k 1 ,, x k i 1 , x k i + 1 ,, x k n and

F x k i ( x k i ) = x k i ( 1 + σ i r ˆ i + 1 j i 1 a ˆ i j x k j i j n a ˆ i j x k j )
(4.2)

with σ 1 =1 and σ i {1,1} for 2in.

Let ζ i denote ζ 1 ,, ζ i 1 , ζ i + 1 ,, ζ n . Note that if ζ 1 ,, ζ n are positive constants such that for 1in

U i ( ζ i ) = 1 + σ i r ˆ i + 1 j i 1 a ˆ i j ζ j i + 1 j n a ˆ i j ζ j 2 a ˆ i i >0,

then

F ζ i ( ζ i ) is increasing on 0 ζ i < U i ( ζ i ) .
(4.3)

Assume that there exist positive constants χ i such that for 1in

χ i < 1 + σ i r ˆ i i + 1 j n a ˆ i j χ j 2 a ˆ i i ,
(4.4)
σ i r i + 1 j i 1 a i j χ j a i i < χ i ,
(4.5)
σ i r ˆ i + 1 j i 1 a ˆ i j χ j <1,
(4.6)

which is the generalization of equation (2.4).

Theorem 4.1 Let Δt and χ i satisfy equations (4.4)-(4.6). If ( x 0 1 ,, x 0 n ) 1 i n (0, χ i ), then ( x k 1 ,, x k n ) 1 i n (0, χ i ) for all k.

Proof It follows from equation (4.4) and the definition of U i that for 1in

0< x 0 i < χ i < 1 + σ i r ˆ i i + 1 j n a ˆ i j χ j 2 a ˆ i i < U i ( x 0 i ) .

Then equation (4.3) with ζ i = x 0 i gives

x 1 i = F x 0 i ( x 0 i ) > F x 0 i (0)=0.

If σ i r i + 1 j i 1 a i j x 0 j i j n a i j x 0 j 0 for 1in, then

x 1 i x 0 i < χ i ,

and otherwise, we have 0< x 0 i < f i ( x 0 i ) with f i ( x 0 i )= σ i r i + 1 j i 1 a i j x 0 j i + 1 j n a i j x 0 j a i i .

Since 0< f i ( x 0 i )< U i ( x 0 i ) by equation (4.6), it follows from equation (4.3) with ζ i = x 0 i that

x 1 i = F x 0 i ( x 0 i ) < F x 0 i ( f i ( x 0 i ) ) = f i ( x 0 i ) < σ i r i + 1 j i 1 a i j χ j a i i < χ i ,

where the last inequality is obtained by equation (4.5). Therefore we find that if ( x 0 1 ,, x 0 n ) 1 i n (0, χ i ), then

( x 1 1 , , x 1 n ) 1 i n (0, χ i ),

and hence mathematical induction completes the proof. □

From now on, we consider the global asymptotic stability of the equilibrium point of equation (4.1) whose components are all positive. Let σ i j =1 if 1i<jn, and otherwise, σ i j =1. Assume that

the inverse matrix of the  n × n  matrix  ( σ i j a i j )  exists and ( σ i j a i j ) 1 ( σ 1 r 1 σ 2 r 2 σ n r n ) T  is a positive matrix ,
(4.7)

where ( σ 1 r 1 σ 2 r 2 σ n r n ) T is the transpose of 1×n matrix ( σ 1 r 1 σ 2 r 2 σ n r n ).

Let

θ T = ( σ i j a i j ) 1 ( σ 1 r 1 σ 2 r 2 σ n r n ) T ,

which gives for 1in

σ i r i + 1 j i 1 a i j θ j i j n a i j θ j =0.
(4.8)

Then equation (4.7) is the generalization of the condition r 1 a 21 r 2 a 11 >0 in both Lemma 3.3 and Theorem 3.5.

In order to use the linearized system of equation (4.1) at θ, consider the Jacobian matrix of T( x k 1 ,, x k n )=( F x k 1 ( x k 1 ),, F x k n ( x k n )) at θ

D T θ = ( 1 θ 1 a 11 Δ t θ 1 a 12 Δ t θ 1 a 1 n Δ t θ 2 a 21 Δ t 1 θ 2 a 22 Δ t θ 2 a 2 n Δ t θ n a n 1 Δ t θ n a n 2 Δ t 1 θ n a n n Δ t ) .

Since the global stability of θ implies that θ is locally stable, we need to assume the condition for the local stability of θ:

All of the eigenvalues of D T θ  have magnitude less than 1.
(4.9)

Let J= 1 Δ t (D T θ I) for the n×n identity matrix I and let P(λ) be the characteristic polynomial of the matrix J. Note that 1+ λ J Δt are the eigenvalues of D T θ for the roots λ J of P(λ). Thus equation (4.9) is equivalent to the condition

all of the roots of P(λ) are negative or have negative real parts,
(4.10)
and |Re( λ J )|> Δ t 2 λ J 2  if Re( λ J )<0 and Im( λ J )0,
(4.11)

where λ J 2 = Re 2 ( λ J )+ Im 2 ( λ J ), Re( λ J ) and Im( λ J ) are the real and imaginary parts of λ J , respectively. Hence equation (4.10) together with equation (4.11) is the generalization of Lemma 3.3.

Remark 4.2 The condition equation (4.9) with 1+ λ J Δt0 implies that θ of equation (4.1) is asymptotically stable by using Theorem 3.4 with the following two facts: first, θ is a hyperbolic point since all of the eigenvalues of D T θ are of the form 1+ λ J Δt, which is nonzero. Second, equation (4.9) is equivalent to the fact that the linearized system of equation (4.1) is asymptotically stable (see Theorem 3.3.20 in [5]).

Remark 4.3 Using Routh-Hurwitz criteria, we can find conditions for both r i and a i j to satisfy equation (4.10). Let P(λ)= λ n + a 1 λ n 1 ++ a n 1 λ+ a n with real constants a i for 1in. Define the n Hurwitz matrices H i for 1in

H 1 =( a 1 ), H 2 = ( a 1 1 a 3 a 2 ) ,, H n = ( a 1 1 0 0 0 a 3 a 2 a 1 0 0 a 5 a 4 a 3 a 2 0 0 0 0 0 a n ) .

Then equation (4.10) is equivalent to det( H i )>0 for 1in. Thus Routh-Hurwitz criteria for n=2 and 3 are

n = 2 : a 1 > 0  and  a 2 > 0 , n = 3 : a 1 > 0 , a 3 > 0 , and  a 1 a 2 > a 3 .
(4.12)

If we assume equation (4.9) with 1+ λ J Δt0, then θ is asymptotically stable (see Remark 4.2). Hence the following lemma can be proved.

Lemma 4.4 Let r i , a i j , Δt satisfy equations (4.4)-(4.7), (4.10), and (4.11) with 1+ λ J Δt0. If lim k ( x k 1 ,, x k n )θ, then there exists a positive constant C such that

max 1 i n ( θ i x k i ) 2 >C>0for all k.

We need to find constants α i that play the same role as a 12 and a 21 in equations (3.6), (3.7), and (3.9): If x= a 21 and y= a 12 , then

x ( a 11 θ 1 n 2 + a 12 θ 1 n θ 2 n ) y ( a 21 θ 1 n θ 2 n + a 22 θ 2 n 2 ) = ( θ 1 n θ 2 n ) ( x a 11 1 2 ( x a 12 y a 21 ) 1 2 ( x a 12 y a 21 ) y a 22 ) ( θ 1 n θ 2 n ) T 0 .

Hence we choose positive constants α i such that the matrices

A i j = ( α i a i i n 1 2 ( α i a i j α j a j i ) n 1 2 ( α i a i j α j a j i ) α j a j j ) for 1i<jn
(4.13)

are positive definite. Note that all of the eigenvalues of A i j are positive if and only if

α i a i i α j a j j > ( n 1 2 ) 2 ( α i a i j α j a j i ) 2 ,

which is equivalent to

( n 1 ) 2 a j i 2 ( α j α i ) 2 2 { ( n 1 ) 2 a i j a j i + 2 a i i a j j } α j α i + ( n 1 ) 2 a i j 2 <0.

Hence consider the quadratic equation in the variable t

( n 1 ) 2 a j i 2 t 2 2 { ( n 1 ) 2 a i j a j i + 2 a i i a j j } t+ ( n 1 ) 2 a i j 2 =0,
(4.14)

which has two different solutions denoted by m i j and M i j with m i j < M i j for 1i<jn. Therefore in order that the eigenvalues of A i j are positive, assume that there exist positive constants α i such that for 1i<jn

m i j < α j α i < M i j .
(4.15)

Lemma 4.5 Suppose that there exist positive constants α i (1in) satisfying equation (4.15). Let λ i j be the minimum of the two positive eigenvalues of A i j for 1i<jn in equation (4.13). Then for all (x,y) R 2

( x y ) A i j ( x y ) T λ i j ( x 2 + y 2 ) .
(4.16)

Assume that

Δt max 1 i n 1 j n a i j ( θ j + χ j )< 1 2 ,
(4.17)
2 Δ t 1 s n α s θ s { ( max 1 i n a s i ) n + ( 1 j n a s j ) } max 1 i , j n a i j < min 1 i n i + 1 j n λ i j n 1 ,
(4.18)

which are the generalization of equations (3.4) and (3.5), respectively.

Theorem 4.6 Let the assumptions in Lemma  4.4 hold and let equations (4.15), (4.17), and (4.18) be satisfied. If ( x 0 1 ,, x 0 n ) 1 i n (0, χ i ), then lim k ( x k 1 ,, x k n )=θ.

Proof Consider a function

V k = 1 i n α i ( x k i θ i ln x k i )

for all the solutions x k 1 ,, x k n of equation (4.1). The Mean Value Theorem with θ i k = θ i x k i and equation (4.8) shows that there exist constants c i (0,1) such that

V k + 1 V k = 1 i n α i ( x k + 1 i x k i ) ( 1 θ i ln x k + 1 i ln x k i x k + 1 i x k i ) = 1 i n α i Θ i k ( x k i θ i x k i c i ( x k + 1 i x k i ) + x k i )
(4.19)

with

Θ i k = 1 j i 1 a ˆ i j θ j k + a ˆ i i θ i k + i + 1 j n a ˆ i j θ j k .
(4.20)

Note that

x k i c i ( x k + 1 i x k i ) + x k i = 1 c i Θ i k Δ t + 1 1 1 n c i θ k Δt,
(4.21)

where c i = c i a i c i Θ i k Δ t + 1 for 1i1, c i = c i a i c i Θ i k Δ t + 1 for in, and

| c i |2max{ a i j |1i,jn}
(4.22)

by equation (4.17). Then equation (4.19) with (4.21) becomes

V k + 1 V k = 1 i n α i Θ i k θ i k +Δt ( 1 i n α i Θ i k 1 n θ i c i θ k ) .
(4.23)

It follows from equations (4.20) and (4.13) that

1 i n α i Θ i k θ i k = 1 i n 1 j < i α i a ˆ i j θ i k θ j k + 1 i n α i a ˆ i i θ i k 2 + 1 i n i < j n α i a ˆ i j θ i k θ j k = 1 i < j n ( α i a ˆ i j α j a ˆ j i ) θ i k θ j k + 1 i < j n 1 n 1 ( α i a ˆ i i θ i k 2 + α j a ˆ j j θ j k 2 ) = Δ t n 1 1 i < j n ( θ i k θ j k ) A i j ( θ i k θ j k ) T
(4.24)

and

1 i n α i Θ i k 1 n θ i c i θ k Δt 1 i n C i θ i k 2 ,
(4.25)

where

C i = 1 s n α s θ s { a s i ( 1 n | c s | ) + ( 1 j n a s j ) | c s i | } .
(4.26)

Hence substituting equations (4.24), (4.25), and Lemma 4.5 into equation (4.23) yields

V k + 1 V k Δt 1 i n { ( 1 n 1 i < j n λ i j ) C i Δ t } θ i k 2 .
(4.27)

Using equations (4.26), (4.22), and (4.18), we obtain 1 n 1 i + 1 j n λ i j C i Δt>0, and then equation (4.27) becomes

V k + 1 V k C 1 Δt 1 i n θ i k 2
(4.28)

for some positive constant C 1 . Now assume, to the contrary, that ( x k 1 ,, x k n ) does not converge to θ as k goes to infinity. Then combining equation (4.28) with Lemma 4.4, we see that there exists a positive constant C 2 such that

V k + 1 V k C 2 for all k0,

and hence

V k V 0 k C 2 for all k0.

This is a contradiction, since lim k ( V 0 k C 2 )= and V k is lower bounded for all k by using the boundedness of the solutions x k i for 1in. □

Example 4.7 Consider the three-dimensional scheme

x k + 1 1 = x k 1 + x k 1 ( 2 x k 1 2 x k 2 2 x k 3 ) Δ t , x k + 1 2 = x k 2 + x k 2 ( 2 + 3 x k 1 2 x k 2 2 x k 3 ) Δ t , x k + 1 3 = x k 3 + x k 3 ( 2 + 2 x k 1 + x k 2 x k 3 ) Δ t ,
(4.29)

which is equation (4.1) with n=3, σ 1 =1, σ 2 = σ 3 =1, and Δt=0.0001. The initial condition ( x 0 1 , x 0 2 , x 0 3 ) is (1.5,1.5,1.0).

  1. (a)

    Let χ 1 = r 1 a 11 +1=3, χ 2 = a 21 χ 1 a 22 =4.5, and χ 3 = a 31 χ 1 + a 32 χ 2 a 33 =10.5. Then equations (4.4), (4.5), and (4.6) are satisfied.

  2. (b)

    The condition (4.7) is satisfied and θ=(1,0.25,0.25).

  3. (c)

    The conditions (4.10), (4.11), and (4.12) are satisfied, since P(λ)= λ 3 + a 1 λ 2 + a 2 λ+ a 3 = λ 3 +1.75 λ 2 +3.5λ+1 with the three roots 0.710+1.590i, −0.330, and 0.7101.509i.

  4. (d)

    The values α 1 = α 2 = α 3 =1 satisfy equation (4.15), since the three equations a j i 2 t 2 (2 a i j a j i + a i i a j j )t+ a i j 2 =0 (1i<j3) have solutions {0.378,1.178}, {0.610,1.640}, and {0.764,5.236}, respectively.

  5. (e)

    The inequalities (4.17) and (4.18) are satisfied, since λ 12 =0.382, λ 13 =1, and λ 23 =0.382.

Hence the conditions in Theorem 4.6 are satisfied. Therefore the solutions of equation (4.29) are positive and the equilibrium point θ is globally asymptotically stable, which are demonstrated in Figure 3(a).

Figure 3
figure 3

Three-dimensional models with σ 1 =1 and σ 2 , σ 3 {1,1}. The four thicker curves denote the trajectories of the solutions of equation (4.1) with n=3 and the other curves are the projections of the trajectories in two planes among x 1 x 2 , x 1 x 3 , and x 2 x 3 planes. The four larger stars denote the limits θ (see Examples 4.7 and 4.8). (a) ( σ 2 , σ 3 )=(1,1) and θ=(1,0.25,0.25). (b) ( σ 2 , σ 3 )=(1,1) and θ=(1,2,3). (c) ( σ 2 , σ 3 )=(1,1) and θ=(1,2,3). (d) ( σ 2 , σ 3 )=(1,1) and θ=(1,2,3).

Example 4.8 Let E=( r 1 , a 11 , a 12 , a 13 , r 2 , a 21 , a 22 , a 23 , r 3 , a 31 , a 32 , a 33 ). Consider the following three difference schemes for equation (1.1) with n=3, Δt=0.0001, and σ 1 =1.

  1. (a)

    σ 2 =1, σ 3 =1, and E=(3.5,1,0.5,0.5,0.5,3,0.5,0.5,1,1,0.5,1) with the initial condition (4,3,2) (see Figure 3(b)).

  2. (b)

    σ 2 =1, σ 3 =1, and E=(3.5,1,0.5,0.5,4,1,1,1,2,1,2,1) with the initial condition (3,1,1) (see Figure 3(c)).

  3. (c)

    σ 2 =1, σ 3 =1, and E=(3.5,1,0.5,0.5,4,11,1,1,1,0.5,1) with the initial condition (1.5,1,1) (see Figure 3(d)).

Set χ 1 =15, χ 2 =25, χ 3 =65, θ=(1,2,3), and ( α 1 , α 2 , α 3 )=(1,0.99,0.792) for the three difference schemes. The points ( λ 12 , λ 13 , λ 23 ) in (a), (b), and (c) are (0.403,0.467,0.058), (0.505,0.586,0.289), and (0.505,0.586,0.289), respectively. Hence the conditions in Theorem 4.6 are satisfied. Consequently, the solutions of the three difference schemes are positive and the equilibrium point θ is globally asymptotically stable, which are demonstrated in Figure 3(b), (c), and (d).

Remark 4.9 In the n-dimensional cases, we only consider the equilibrium point which components are all positive. Thus a future study is to investigate dynamics on the other equilibrium points with some zero components.

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Acknowledgements

This work was supported by the 2013 Research Fund of University of Ulsan.

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Choo, S. Global stability in n-dimensional discrete Lotka-Volterra predator-prey models. Adv Differ Equ 2014, 11 (2014). https://doi.org/10.1186/1687-1847-2014-11

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Keywords

  • Euler discrete schemes
  • global stability
  • predator-prey models