Dynamical system of a mosquito population with distinct birth-death rates
aa r X i v : . [ m a t h . D S ] S e p DYNAMICAL SYSTEM OF A MOSQUITO POPULATION WITHDISTINCT BIRTH-DEATH RATES
Z.S. BOXONOV, U.A. ROZIKOV
Abstract.
We study the discrete-time dynamical systems of a model of wild mosquitopopulation with distinct birth (denoted by β ) and death (denoted by µ ) rates. The case β = µ was considered in our previous work. In this paper we prove that for β < µ themosquito population will die and for β > µ the population will survive, namely, the numberof the larvaes goes to infinite and the number of adults has finite limit αµ , where α > Introduction
In [1], [2], [3], [6], [9] and [10] (see also references therein) several kind of mathematicalmodels of mosquito population are studied. This paper is a continuation of our paper [11],where following [5] it was considered a model of the mosquito population. In the model awild mosquito population without the presence of sterile mosquitoes is considered. For thesimplified stage-structured mosquito population, due to the fact that the first three stages ina mosquitos life cycle are aquatic, it was grouped the three aquatic stages of mosquitoes intoone class and divide the mosquito population into only two classes, one of which consists ofthe first three stages that is called the larvae, denoted by x , and one of which consists of alladults, denoted by y .The birth rate is the oviposition rate of adults denoted by β ( · ); let the rate of emergencefrom larvae to adults be a function of the larvae with the form of α (1 − k ( x )), where α > ≤ k ( x ) ≤
1, with k (0) = 0 , k ′ ( x ) >
0, and lim x →∞ k ( x ) = 1,is the functional response due to the intraspecific competition [6]. Let the death rate oflarvae be d + d x , and the death rate of adults be constant µ . Then in the absence of sterilemosquitoes, and in case (as in [6]) k ( x ) = x x , β ( · ) = β the interactive dynamics for the wildmosquitoes are governed by the following system: dxdt = βy − αx x − ( d + d x ) x, dydt = αx x − µy. (1.1)Denote r = αβ ( α + d ) µ . (1.2)Theorem 3.1 in [5] states that: Mathematics Subject Classification.
Key words and phrases. mosquito; population; fixed point; periodic point; limit point. - if r ≤ r > x , y ) with x = p ( d + d ) − d ( α + d )(1 − r ) − d − d d , y = αx µ (1 + x ) , which is a globally asymptotically stable.Define the operator W : R → R by ( x ′ = βy − αx x − ( d + d x ) x + x,y ′ = αx x − µy + y (1.3)where α > , β > , µ > , d ≥ , d ≥ . In this paper we study the discrete time dynamical systems generated by (1.3). In [11] forthe evolution operator (1.3) the following results are obtained:- all fixed points of the evolution operator are found. Depending on the parametersthe operator may have unique, two and infinitely many fixed points;- under some conditions on parameters type of each fixed point is determined and thelimit points of the dynamical system are given.- for the case β = µ, d = d = 0 of parameters the full analysis of correspondingdynamical system is given.In this paper we consider the operator W (defined by (1.3)) for the case β = µ, d = d = 0and our aim is to study trajectories of any initial point from the invariant (under W ) set R .The case when d = 0 or d = 0 is not studied yet.2. Dynamics for β = µ, d = d = 0 . We assume β = µ, d = d = 0 (2.1)then (1.3) has the following form W : ( x ′ = βy − αx x + x,y ′ = αx x − µy + y. (2.2) Remark 1.
Note that the operator W well defined on R \ { x = − } . But to define adynamical system of continuous operator and interpret values of x and y as probabilities weassume x ≥ and y ≥ . Therefore, we choose parameters of the operator W to guaranteethat it maps R to itself. It is easy to see that if 0 < α ≤ , β > , < µ ≤ R to itself.A point z ∈ R is called a fixed point of W if W ( z ) = z .For fixed point of W the following holds. YNAMICAL SYSTEM OF MOSQUITO POPULATION 3
Proposition 1.
The operator W has a unique fixed point z = (0; 0) in R .Proof. We need to solve ( x = βy − αx x + x,y = αx x − µy + y. (2.4)It is easy to see that x = 0 , y = 0 . (cid:3) Now we shall examine the type of the fixed point.
Definition 1. (see [4] ) A fixed point s of an operator W is called hyperbolic if its Jacobian J at s has no eigenvalues on the unit circle. Definition 2. (see [4] ) A hyperbolic fixed point s called:1) attracting if all the eigenvalues of the Jacobi matrix J ( s ) are less than 1 in absolutevalue;2) repelling if all the eigenvalues of the Jacobi matrix J ( s ) are greater than 1 in absolutevalue;3) a saddle otherwise. To find the type of a fixed point of the operator (2.2) we write the Jacobi matrix: J ( z ) = J W = (cid:18) − α βα − µ (cid:19) . The eigenvalues of the Jacobi matrix are λ = 12 (cid:16) − α − µ + p ( α − µ ) + 4 αβ (cid:17) .λ = 12 (cid:16) − α − µ − p ( α − µ ) + 4 αβ (cid:17) . For type of z = (0 ,
0) the following lemma holds.
Proposition 2.
The type of the fixed point z for (2.2) are as follows: i) if β < µ then z is attracting; ii) if β > µ then z is saddle;Proof. For attractiveness we should have | λ , | = (cid:12)(cid:12)(cid:12)(cid:12) (cid:16) − α − µ ± p ( α − µ ) + 4 αβ (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) < . (2.5)The inequality (2.5) is equivalent to the following α + µ + p ( α − µ ) + 4 αβ < , < α + µ − p ( α − µ ) + 4 αβ < . (2.6)From the condition (2.3) of the parameters α, µ it follows that α + µ ≤
2. If β < µ then α + µ > p ( α − µ ) + 4 αβ and the system of inequalities (2.6) holds. In the case β > µ wehave λ > | λ | < (cid:3) The following theorem describes the trajectory of any point ( x (0) , y (0) ) in R . Z.S. BOXONOV, U.A. ROZIKOV
Theorem 1.
For the operator W given by (2.2) (i.e. under condition (2.1)) and for anyinitial point ( x (0) , y (0) ) ∈ R the following hold lim n →∞ x ( n ) = ( , if β < µ, + ∞ , if β > µ lim n →∞ y ( n ) = ( , if β < µ, αµ , if β > µ where ( x ( n ) , y ( n ) ) = W n ( x (0) , y (0) ) , with W n is n -th iteration of W .Proof. Let β < µ.
Then there exists k > β · k = µ . Denote c ( n ) = x ( n ) + y ( n ) and c ( n )0 = k · x ( n ) + y ( n ) , where x ( n ) , y ( n ) defined by the following x ( n ) = βy ( n − − αx ( n − x ( n − + x ( n − ,y ( n ) = αx ( n − x ( n − − µy ( n − + y ( n − . (2.7)Both sequences { c ( n ) } and { c ( n )0 } are monotone and bounded, i.e.,0 ≤ ... ≤ c ( n ) ≤ c ( n − ≤ ... ≤ c (0) , ≤ ... ≤ c ( n )0 ≤ c ( n − ≤ ... ≤ c (0)0 . Thus { c ( n ) } and { c ( n )0 } have limit point, denote the limits by c ∗ and c ∗ respectively. Conse-quently, the following limits exist x ∗ = lim n →∞ x ( n ) = 11 − k lim n →∞ ( c ( n ) − c ( n )0 ) = 11 − k ( c ∗ − c ∗ ) ,y ∗ = lim n →∞ y ( n ) = c ∗ − x ∗ . and by (2.7) we have x ∗ = βy ∗ − αx ∗ x ∗ + x ∗ , y ∗ = αx ∗ x ∗ − µy ∗ + y ∗ , i.e., x ∗ = 0 , y ∗ = 0 . Let β > µ.
The proof is based on the following three lemmas.
Lemma 1.
For any parameters satisfying (2.3) and for arbitrary initial point ( x (0) , y (0) ) in R the sequence y ( n ) (defined in (2.7)) is bounded: ≤ y ( n ) ≤ ( y (0) , if y (0) > αµαµ , if y (0) < αµ . Proof.
Note that on the set R + the function y ( x ) = x x is increasing and bounded by 1.Therefore y ( n ) = αx ( n − x ( n − + (1 − µ ) y ( n − ≤ α + (1 − µ ) y ( n − ≤ α + (1 − µ )( α + (1 − µ ) y ( n − ) ≤ α + α (1 − µ ) + (1 − µ ) ( α + (1 − µ ) y ( n − ) ≤ ... ≤ α + α (1 − µ ) + α (1 − µ ) + ... + α (1 − µ ) n − +(1 − µ ) n y (0) = αµ + (1 − µ ) n ( y (0) − αµ ) . Thus if the initial point y (0) > αµ then for any m ∈ N we have 0 ≤ y ( m ) ≤ y (0) . Moreover,if y (0) < αµ then for any m ∈ N we have 0 ≤ y ( m ) ≤ αµ . (cid:3) Lemma 2.
For sequences x ( n ) and y ( n ) the following statements hold: For any n ∈ N and β > µ the inequalities x ( n ) > x ( n +1) and y ( n ) > y ( n +1) can not besatisfied at the same time. If x ( m − < x ( m ) , y ( m − < y ( m ) for some m ∈ N then x ( m ) < x ( m +1) , y ( m ) < y ( m +1) . If x ( m − > x ( m ) , y ( m − < y ( m ) for any m ∈ N then for β > µ the inequalities x ( m ) > x ( m +1) , y ( m ) < y ( m +1) can not be satisfied at the same time. If x ( m − < x ( m ) , y ( m − > y ( m ) for any m ∈ N then for β > µ the inequalities x ( m ) < x ( m +1) , y ( m ) > y ( m +1) can not be satisfied at the same time. For any m ∈ N the inequalities x ( m − < x ( m ) , x ( m ) > x ( m +1) , y ( m − > y ( m ) , y ( m )
3) Assume in case when x ( m − > x ( m ) , y ( m − < y ( m ) hold for any m ∈ N then for β > µ the inequalities x ( m ) > x ( m +1) , y ( m ) < y ( m +1) are satisfied at the same time. Thensince x ( n ) is decreasing and bounded; y ( n ) is increasing and bounded (see Lemma 1)there exist their limits x ∗ , y ∗ = 0 respectively. By (2.7) we obtain ( βy ∗ = αx ∗ x ∗ αx ∗ x ∗ = µy ∗ i.e. x ∗ = 0 , y ∗ = 0 . This contradiction shows that if for any m ∈ N one has x ( m − >x ( m ) , y ( m − < y ( m ) then for β > µ the inequalities x ( m ) > x ( m +1) , y ( m ) < y ( m +1) cannot be satisfied at the same time (see Fig 1). Z.S. BOXONOV, U.A. ROZIKOV
4) Assume if for any m ∈ N the inequalities x ( m − < x ( m ) , y ( m − > y ( m ) hold then for β > µ the inequalities x ( m ) < x ( m +1) , y ( m ) > y ( m +1) are satisfied at the same time,i.e. x ( m ) is increasing and y ( m ) is decreasing. Let x ( m +1) − x ( m ) = ∆ ( m ) , y ( m +1) − y ( m ) = − δ ( n ) , ( x ( m +1) − x ( m ) ) + ( y ( m +1) − y ( m ) ) = ∆ ( m ) − δ ( m ) < ( β − µ ) y (0) . Since { ∆ ( m ) } is decreasing, { δ ( m ) } is increasing and ∆ ( m ) − δ ( m ) > β > µ we conclude that the sequence { ∆ ( m ) − δ ( n ) } is decreasing and bounded from below.Thus { ∆ ( m ) − δ ( m ) } has a limit and since y ( m ) has limit we conclude that x ( m ) has afinite limit. By (2.7) we havelim m →∞ x ( m ) = 0 , lim m →∞ y ( m ) = 0 . But this is a contradiction to lim m →∞ x ( m ) = 0. This completes proof of part 4 (seeFig 2).5) Assume for any m ∈ N one has x ( m − < x ( m ) , x ( m ) > x ( m +1) , y ( m − > y ( m ) , y ( m )
There exists n such that the sequences x ( n ) and y ( n ) are increasing for n ≥ n and x ( n ) is unbounded from above. YNAMICAL SYSTEM OF MOSQUITO POPULATION 7
Figure 2. α = 0 . , β = 0 . , µ = 0 . , x (0) = 0 . , y (0) = 2 Figure 3. α = 0 . , β = 0 . , µ = 0 . , x (0) = 0 . , y (0) = 0 . Proof.
Monotonicity of x ( n ) and y ( n ) follow from Lemma 2. Consider ( x ( n +1) − x ( n ) ) + ( y ( n +1) − y ( n ) ) = ( β − µ ) y ( n ) ( x ( n +2) − x ( n +1) ) + ( y ( n +2) − y ( n +1) ) = ( β − µ ) y ( n +1) ... ( x ( n − − x ( n − ) + ( y ( n − − y ( n − ) = ( β − µ ) y ( n − ( x ( n ) − x ( n − ) + ( y ( n ) − y ( n − ) = ( β − µ ) y ( n − (2.9)Adding equations of (2.9) we get( x ( n ) − x ( n ) ) + ( y ( n ) − y ( n ) ) = ( β − µ )( y ( n ) + y ( n +1) + ... + y ( n − + y ( n − ) (2.10) Z.S. BOXONOV, U.A. ROZIKOV
Let y ( n ) (see Lemma 1) is bounded by θ . By (2.10) we have x ( n ) = x ( n ) + y ( n ) − y ( n ) + ( β − µ )( y ( n ) + y ( n +1) + ... + y ( n − + y ( n − ) > x ( n ) + y ( n ) − θ + ( β − µ )( n − n ) y ( n ) . For β > µ from lim n →∞ ( x ( n ) + y ( n ) − θ + ( β − µ )( n − n ) y ( n ) ) = + ∞ it follows that x ( n ) is not bounded from above. (cid:3) Thus for β > µ the sequence y ( n ) has limit y ∗ (see Lemma 1). Consequently, by (2.7) andlim n →∞ x ( n ) = + ∞ we get y ∗ = αµ . Theorem is proved. (cid:3) Biological interpretation of our result is clear: for β < µ the mosquito population will dieand for β > µ the population will survive. Comparing our results of Theorem 1 with resultsof the continuous time dynamics (mentioned in the Introduction) one can see that they arethe same.A point z in W is called periodic point of W if there exists p so that W p ( z ) = z . Thesmallest positive integer p satisfying W p ( z ) = z is called the prime period or least period ofthe point z. Theorem 2.
For p ≥ the operator (2.2) does not have any p -periodic point in the set R . Proof.
This is a corollary of Theorem 1. Here we give an alternative proof. Let us firstdescribe periodic points with p = 2 on R , in this case the equation W ( W ( z )) = z. That is x = β ( αx x − µy + y ) + ( βy − αx x + x )(1 − α βy − αx x + x ) ,y = α ( βy − αx x + x )1+ βy − αx x + x + (1 − µ )( αx x − µy + y ) . (2.11)Simple calculations show that the last equation is equivalent to the following αx x = ( µ − y, αx x ≥ , ( µ − y ≤ . (2.12)The only solution to (2.12) is x = 0 , y = 0. Thus the operator (2.2) does not have any twoperiodic point in the set R . Now we show that W does not have any periodic point (except fixed). Rewrite operator(2.2) in normalized form: U : x ′ = (1+ x )( x + βy ) − αx (1+ x )( x +( β − µ +1) y ) ,y ′ = αx +(1+ x )(1 − µ ) y (1+ x )( x +( β − µ +1) y ) . (2.13)Denote S = { ( x, y ) ∈ R : x + y = 1 } . From conditions (2.3) one gets x ′ ≥ y ′ ≥ x ′ + y ′ = 1 . Hence U : S → S. Remark 2.
The operators which map S to itself have been extensively studied (see for example [7] , [8] , [10] and the references therein). YNAMICAL SYSTEM OF MOSQUITO POPULATION 9
Using x + y = 1, from (2.13) we get T : x ′ = (1 − β ) x + (1 − α ) x + β ( µ − β ) x + x + β − µ + 1 . (2.14)Denote S ∗ = [0 , . Proposition 3.
If conditions (2.3) are satisfied then the function T (defined by (2.14)) maps S ∗ to itself.Proof. We want to show that if x ∈ S ∗ then x ′ = T ( x ) ∈ S ∗ . Let a = (1 − β ) x + (1 − α ) x + β, b = ( µ − β ) x + x + β − µ + 1Then h ( x ) = b − a = ( µ − x + αx + 1 − µ. Since h (0) = 1 − µ > h (1) = α > x ∈ [0 ,
1] there is h ( x ) ≥ , i.e., x ′ ≤ . By a = (1 − β ) x + (1 − α ) x + β = (1 + x )( x + β (1 − x )) − αx ≥ ,b = ( µ − β ) x + x + β − µ + 1 = (1 + x )( x + ( β − µ + 1)(1 − x )) > x ′ ≥ , y ′ ≥ . Therefore under conditions (2.3) we get T : S ∗ S ∗ . (cid:3) Let us first describe periodic points of U with p = 2. In this case the equation U ( U ( z )) = z can be reduced to description of 2-periodic points of the function T defined in (2.14), i.e.,tosolution of the equation T ( T ( x )) = x. (2.15)Note that the fixed points of T are solutions to (2.15), to find other solution we consider theequation T ( T ( x )) − xT ( x ) − x = 0 , simple calculations show that the last equation is equivalent to the following Ax + Bx + C = 0 . (2.16)where A = (1 − β )( β −
2) + ( β − µ + 1)( β − µ ) ,B = ( β − β − µ − α + 2) − β ( β − µ ) ,C = ( β − µ + 1)( α + µ − β −
2) + β ( β − . By (2.3) we have A + B + C < B < C <
0. Therefore since x ≥ < S ∗ . Thus function (2.14) doesnot have any 2-periodic point in S ∗ . Since T is continuous on S ∗ by Sharkovskii’s theorem([4], [12]) we have that T p ( x ) = x does not have solution (except fixed) for all p ≥
2. Henceit follows that the operator (2.2) has no periodic points (except fixed) in the set R . (cid:3) Acknowledgements
We thank both referees for their useful comments.
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Z. S. Boxonov, Institute of mathematics, 81, Mirzo Ulug‘bek str., 100170, Tashkent, Uzbek-istan.
E-mail address : [email protected] U. A. Rozikov, Institute of mathematics, 81, Mirzo Ulug‘bek str., 100170, Tashkent, Uzbek-istan.
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