research article analysis of a mathematical model of

Research ArticleAnalysis of a Mathematical Model of Emerging InfectiousDisease Leading to Amphibian Decline

Muhammad Dur-e-Ahmad Mudassar Imran and Adnan Khan

Department of Mathematics Lahore University of Management Sciences DHA Lahore Cantt 54792 Pakistan

Correspondence should be addressed to Muhammad Dur-e-Ahmad muhammaddureahmadlumsedupk

Received 9 November 2013 Revised 12 February 2014 Accepted 2 March 2014 Published 28 April 2014

Academic Editor Igor Leite Freire

Copyright copy 2014 Muhammad Dur-e-Ahmad et al This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We formulate a three-dimensional deterministic model of amphibian larvae population to investigate the cause of extinction dueto the infectious disease The larvae population of the model is subdivided into two classes exposed and unexposed depending ontheir vulnerability to disease Reproduction ratioR

0has been calculated and we have shown that ifR

0lt 1 the whole population

will be extinct For the case of R0gt 1 we discussed different scenarios under which an infected population can survive or be

eliminated using stability and persistence analysis Finally we also used Hopf bifurcation analysis to study the stability of periodicsolutions

1 Introduction

Worldwide catastrophic declines in the amphibian popula-tion are perhaps one of the most pressing and discussedproblems among the ecologists during the last two decadesAlthough many of these are attributable due to the habitatloss (see eg [1 2]) the majority have remained enigmatictill today Many hypotheses responsible for this decline havebeen documented in the literature such as adverse weatherpatterns [3 4] acid precipitation [5] environmental pollution[3] increased ultraviolet (UV-B) radiation [6] introductionof predators or competitors [7] infectious disease [3 8 9] ora combination of these

Recently infectious diseases have become one of theemergent factors behind rapid amphibian decline whichoften results in extinction of species for example the recentextinction of the golden toad in Costa Rica and somespecies of gastric-brooding frogs in Australia Investiga-tions reveal that the main cause behind the mass deathof these species is two infectious diseases chytridiomy-cosis in the rain forest of Australia and Central Americaand some parts of North America and iridoviral infec-tions in United Kingdom United States and Canada[3 10 11]

The larval stage of the amphibian is considered to bethe most vulnerable towards the spread of infectious diseaseMost of the larva population of tropical amphibian speciesremain alive for 12 to 18 months but some temperate speciesmay also survive as long as 3 years before metamorphosingRecently it has been observed both in Australia and CentralAmerica that larval amphibians infected with chytridiomy-cosis may exhibit disfigurement of their keratinized mouthparts and demonstrate a significant reduction in their growthand development which may eventually cause the completeextinction sometimes [3] Further in the case of reducedamphibian population this infection causes prolonging of theexistence of Batrachochytrium [12] and implicates the life-cycle stage as a reservoir host for the pathogen This kindof larval infection enhancing pathogen-mediated host pop-ulation extinction has also been reported for invertebrates[13] Therefore in this work we keep our focus on the larvalstage and investigate the dynamics of spread of disease amongthem

This work has been motivated by the recent work of [9]in which only susceptible and infected classes are consideredSince as discussed earlier larvae are the key players in thecase of density-dependent disease incidencewhich eventuallyleads to the host extinction therefore it is imperative to

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 145398 13 pageshttpdxdoiorg1011552014145398

2 Abstract and Applied Analysis

keep this component intact while studying the disease basedamphibian decline Since the main focus of the model isthe disease based larvae extinction therefore in order tolook at the complete scenario we assume that at the earlystage of their lives the larvae are surrounded by a safeenvironment which is free of disease like in a closed shell orsome protected holes For our model we denote this class by119871 At the later stage once these unexposed larvae enter intofree environment they become vulnerable to disease and wecall them susceptibles (denoted by 119878) These susceptibles canbe infected with the disease and enter into the infective classdenoted by 119868 Further these infected larvae can also recoverfrom the disease and return to the susceptible class Themodel is formulated in a way that the number of new infectedlarvae depends on the present number of susceptible andinfective larvae in the sense that the more larvae (susceptibleor infective) we have the more chance of spreading theinfections among the healthy or susceptible larvae will therebe that is the disease can be transferred from one to another

We organize this paper as follows First we describe themathematical model along with its underlying assumptionsThen we introduce many threshold quantities for examplethe reproduction ratios and the critical host density forthe disease establishments We also derive the number ofendemic equilibriums and discuss their stabilities Next wealso elaborate our findings using numerical simulationsFinally we give the conclusion and discuss the potentiallimitations and impact of our results

2 Formulation of Mathematical Model

In this mathematical model we assume that the disease onlyaffects the larval population So we divide the larvae 119871 intotwo categories the susceptibles larvae 119878 and the infectedlarvae 119868 Although this disease is transferable from one toanother the recovery is also possible and on recovering fromthe disease an infected larva will reenter into the class ofsusceptible larvae Further for ourmodel we also assume thatonly susceptibles can contribute to the reproduction processThus once the disease is spread the whole population will gotowards extinctions not only due to the illness but also due tothe lack of reproduction process

The model is as follows

1198711015840= 120573119878 minus 120587 (119871) 119871 minus 120583119871

1198781015840= 120587 (119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868

1198681015840= 120590119878119868 minus ]119868 minus 120572119868 minus 120588119868

(1)

Here119871 represents the unexposed class of the larval populationwhich is not yet entered into the susceptible class 119878 Here weassume that the transition from stage 119871 to stage 119878 requiresa certain minimum size This assumption is valid for manyspecies for example Daphnia magna a water flea where theindividuals typically have a length of 08mm at birth and alength of 25mm once they enter into the exposed class [14]Further in the case of some amphibians the body size atmetamorphosis is quite flexible (see eg [15]) therefore acertainminimum size still seems to be required in these cases

Table 1 Description of parameters in model (1)

Variables Description120573 Per capita birth rate of susceptibles120583 Per capita (natural) mortality rate of the larvae

] Per capita (natural) mortality rate of thesusceptible and infected larvae

120572Per capita mortality rate of the infected larvae dueto disease

120588 Recovery rate from disease

120587 (119871)Per capita transition rate This is the rate at whichthe susceptible larva matures

120590

Rate at which the disease is transferred frominfected to susceptible class when the contactbetween susceptible and infected individualsoccurs

Thus based on these evidences it is reasonable to assumethat the scarcity of resources will prolong the length of thelarval stage 119871 So if there are more larvae in stage 119871 and lessresources then it will take longer for a single larva in stage119871 to complete the transition into stage 119878 Therefore in ourmodel we assume that 120587(119871)119871 is nondecreasing in 119871 in orderto represent a strong negative feedback from the number oflarval population 119871 It is clear from the model that this classwill enter into the exposed susceptible class 119878 with the rate120583 which further can become infected with a rate ] Since inthe case of closed population the disease can transfer fromone individual to the other therefore by using the law of massaction we express this by120590119878119868 where120590 is disease transmissionrate Finally the last term in the third equation represents therecovery from disease with the rate 120588

The detailed description of parameters of our model isgiven in Table 1

Assumption 1 All the parameters are positive except 120572 whichcan also be zero 120587(119871) is a strictly decreasing nonnegativefunction of 119871 such that 120587(119871) rarr 0 as 119871 rarr infin

3 Existence Positivityand Boundedness of the Solution

Theorem 2 Assume that Assumption 1 is satisfied then fornonnegative initial data there are unique nonnegative solu-tions of the system which are defined for all nonnegative timesFurther all the solutions to the system are uniformly eventuallybounded

Proof Let 119865 = (1198911 1198912 1198913) and 119883 = (119909

1 1199092 1199093) = (119871 119878 119868)

where

1198911= 120573119878 minus 120587 (119871) 119871 minus 120583119871

1198912= 120587 (119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868


(2)

Abstract and Applied Analysis 3

Notice that all the partial derivatives 120597119891119894120597119909119895 119894 119895 = 1 2 3 are

continuous Further

119871 = 0 997904rArr 1198911= 120573119878 ge 0

119878 = 0 997904rArr 1198912= 120587 (119871) 119871 + 120588119868 ge 0

119868 = 0 997904rArr 1198913= 0

(3)

So 119883(119905) = (119871(119905) 119878(119905) 119868(119905)) isin 1198613 where 119861 = [0infin) for all

119905 ge 1199050ge 0 for which it is defined and whenever 119883(119905

0) isin 1198613

Thus all the solutions of the given system are nonnegative bypropositionA1 of [14]We conclude that there exists a uniquesolution to the above system with the values in R3

+and it is

defined in the interval [0 119887) 119887 isin (0infin] (Theorem A4 [14])and if 119887 lt infin then

sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) = infin (4)

To prove the first part of the theorem we need to show onlythat 119887 = infin Assume that 119887 lt infin then from model (1) we get

1198711015840le 120573119878

1198781015840le 120587 (119871) 119871 + 120588119868 minus 120590119878119868 le 120587 (0) 119871 + 120588119868 minus 120590119878119868

1198681015840le 120590119878119868

(5)

Let 120578 = max120587(0) 120573 120588 then we get

1198711015840+ 1198781015840+ 1198681015840le 120578 (119871 + 119878 + 119868)

997904rArr 119871 (119905) + 119878 (119905) + 119868 (119905) le [119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

997904rArr sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le sup0le119905lt119887

[119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

(6)

Since 119887 lt infin therefore

sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le 119862 lt infin (7)

This is a contradiction so 119887 = infinNow we show that the solutions are uniformly eventually

bounded Let us define 119881 = 119871 + 120585(119878 + 119868) where 120585 is to bedetermined later Then by using the system (1) we have

1198811015840= 120573119878 minus 120587 (119871) 119871 minus 120583119871 + 120585 (120587 (119871) 119871 minus ]119878 minus ]119868 minus 120572119868)

997904rArr 1198811015840= (120573 minus 120585]) 119878 + [(120585 minus 1) 120587 (119871) minus 120583] 119871 minus 120585 (] + 120572) 119868

(8)

By choosing 120585 = (120573]) + 1 we get

1198811015840= minus]119878 + (

120573

]120587 (119871) minus 120583)119871 minus 120585 (] + 120572) 119868 (9)

Since we know that 120587(119871) rarr 0 as 119871 rarr infin therefore foreach 120598 gt 0 there exist some 119871♯ gt 0 such that for all 119871 ge 119871

♯120587(119871) lt 120598 So we can divide the proof into two cases

Case 1 (119871 ge 119871♯) From (9) we get

1198811015840le minus ]119878 + (

120573

]120598 minus 120583)119871

le minus ]119878 minus

120583

2

119871 le minus

120583

2

119871♯

(10)

So if we take 119871♯ gt 2120583 then we get 1198811015840 lt minus1

Case 2 (119871 lt 119871♯)Using 120585 = (120573]) + 1 again from (9) we have

1198811015840le minus ]119878 +

120573

]120587 (119871) 119871 minus 120583119871 minus 120585 (] + 120572) 119868

le minus ]119878 +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus ](1

120585

(119881 minus 119871) minus 119868) +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus

]120585

119881 +

]120585

119871 + ]119868 +120573

]120587 (0) 119871

♯minus (

120573

]+ 1) (] + 120572) 119868

(11)

Simplification and neglecting the negative terms (involving119868) yields

997904rArr 1198811015840le minus

]120585

119881 + (

]120585

+

120573

]120587 (0)) 119871

♯ (12)

We find some 119881♯ such that 1198811015840 lt minus1 whenever 119881 ge 119881♯

Therefore in either case 1198811015840

lt minus1 This implies thatlim inf

119905rarrinfin119881(119905) le 119881

♯ Otherwise since 1198811015840(119905) le minus1 for all

sufficiently large time and 119881 would become negative in finitetime contradiction Further since 119881 ge 119871 + 119878 + 119868 and we haveshown that our system is weakly dissipative so it is dissipativeby Proposition 318 of [14]

4 Existence and Stability of Steady States

In this section we will discuss the existence and stability ofthe steady states of our model (1) We will have three steadystates the trivial steady states (0 0 0) the disease-free steadystate of the form ( 119878 0) and the interior steady state of theform (119871

lowast 119878lowast 119868lowast)The trivial steady state always exists To find

the disease-free steady state we proceed as follows From the1198681015840 equation of our main model (1) we have

(120590119878 minus ] minus 120572 minus 120588) 119868 = 0 (13)


From (13) we can say that either 119868 = 0 or 119878 = (]+120572+120588)120590 Letus suppose the disease-free case ( 119878 0) that is 119868 = 0 thenby first and second equations of model (1) we have

120573119878 minus 120587 () minus 120583 = 0

120587 () minus ]119878 = 0

(14)

Now by adding these two equations we get

119878 =

120583

(120573 minus ]) (15)

To find the value of we substitute the value of 119878 from (15)and 119868 = 0 in 119871

1015840 equation of model (1) We have

120573120583

120573 minus ] minus 120587 () minus 120583 = 0

[

120573120583

120573 minus ]minus 120587 () minus 120583] = 0

997904rArr = 0 or120573120583

120573 minus ]minus 120583 = 120587 ()

(16)

Notice that = 0 will give us trivial state (0 0 0) In order tofind the nontrivial state we will consider the case when = 0In this case we will have

120573120583

120573 minus ]minus 120583 = 120587 ()

120587 () =

120583]120573 minus ]

(17)

Equivalently

= 120587minus1

(

120583]120573 minus ]

) (18)

Thus we have two disease-free steady states that is (0 0 0)and ( (120583(120573minus ])) 0) where gt 0 is given by (17) Observethat there is a threshold condition

120573120583

120573 minus ]minus 120583 lt 120587 (0) (19)

We can rewrite the condition (120573 minus ])120587(0) lt ]120583 as(120573])(120587(0)(120587(0) + 120583)) lt 1 We refer this quantity asreproduction number and it is defined as

R0=

120573

]120587 (0)

120587 (0) + 120583

lt 1 (20)

Now we find the interior equilibrium (119871lowast 119878lowast 119868lowast) with

119871lowast 119878lowast 119868lowast

= 0 From the 1198681015840 equation of model (1) we get 119878lowast =(] + 120572 + 120588)120590

Adding the first two equations of our original model (1)gives

120573119878lowastminus 120583119871lowastminus ]119878lowast + 120588119868

lowastminus 120590119878lowast119868lowast= 0

997904rArr (120573 minus ]) 119878lowast + (120588 minus 120590119878lowast) 119868lowastminus 120583119871lowast= 0

(21)

Now by using the values of 119878lowast we get

(120573 minus ]) (] + 120572 + 120588)

120590

+ (120588 minus ] minus 120572 minus 120588) 119868 minus 120583119871 = 0

997904rArr (120573 minus ]) (] + 120572 + 120588) + 120590 (minus] minus 120572) 119868 minus 120590120583119871 = 0

(22)

Therefore we have

119868lowast=

120590120583119871lowastminus (120573 minus ]) (] + 120572 + 120588)

120590 (minus] minus 120572)

(23)

Nowweneed to find the value of119871lowast By substituting the valuesof 119878lowast and 119868

lowast in the first equation of model (1) we will have

120573

120590

(] + 120572 + 120588) minus (120587 (119871lowast) + 120583) 119871

lowast= 0

(120587 (119871lowast) + 120583) 119871

lowast=

120573

120590

(] + 120572 + 120588) = 120573119878lowast

(24)

Now we need to solve (24) numerically (depending onthe structure of the function 120587(119871)) to find the value of 119871lowastTherefore in this case the interior equilibrium is given by

(119871lowast

] + 120572 + 120588

120590


120590 (minus] minus 120572)

) (25)

The 119871lowast and 119878lowast are nonzero From the 1198781015840 equation of model (1)

we get

120587 (119871lowast) 119871lowastminus ]119878lowast + 119868

lowast(120588 minus 120590119878

lowast) = 0 (26)

If we solve this for 119868lowast we get

119868lowast=

120587 (119871lowast) 119871lowastminus ]119878lowast

] + 120572

(27)

This implies that 119868lowast gt 0 if

120587 (119871lowast) 119871lowastgt ]119878lowast (28)

This gives the condition for the existence of infected steadystate Observe that we can write (24) as

120590119878lowast

] + 120572 + 120588

= 1 (29)

Since 1(] + 120572 + 120588) is the average time a larva spends inthe infected stage and 120590119878

lowast is the rate at which one averageinfected larva infects a susceptible larva if they are at density119878lowast therefore we can express the left side of (29) as thereplacement ratio of the infectious disease from the infectedlarvae to the susceptible larvae when they have density 119878

lowastThat is

Rpar

(119878lowast) =

120590119878lowast

] + 120572 + 120588

(30)

The replacement ratio at the susceptible larval density is givenby

Rpar0

(119878) =

120590119878

] + 120572 + 120588

(31)

From the above results we have the following theorem forthe existence of the steady states


Theorem 3 (a) The trivial steady state (0 0 0) exists all thetime

(b) The disease-free state ( 119878 0) exists if and only ifR0gt

1(c)The infected state (119871lowast 119878lowast 119868lowast) exists if and only ifR

0gt 1

and 120587(119871lowast)119871lowastgt ]119878lowast

(d) LetRpar(L) gt 1 Then there is at least one steady state

with the infected larvae(e) Let Rpar

() lt 1 then there is no or two steady stateswith the infective larvae

Proof Proof of part (b) and part (c) follows directly from thediscussion above this theorem

(d) The conditionRpar() gt 1 can be written as

(120587 () + 120583) gt 120573

] + 120572 + 120588

120590

(32)

The intermediate value theorem implies that there is some1isin (0 ) which satisfies (24)(e) The conditionRpar

() lt 1 can be written as

(120587 () 119871+120583) lt 120573

] + 120572 + 120588

120590

(33)

If 120587(119871)119871 + 120583119871 is strictly increasing function for 119871 ge 0then there is no steady state in [0 ] Now assume thatthis function is not increasing and that there is at least onesolution

1gt 0 of (24) We can choose

1isin (0 119871

1] If

1= 1198711then we have two solutions counting multiplicities

They are steady state with the infectives if 1gt otherwise

that steady state does not exist with infectives If 1

lt 1198711

then we can choose some 119871 isin (1 1198711) This implies

120587 (1198711) 1198711+ 1205831198711gt 120587 (

1) 1+ 1205831= 120573(

] + 120572 + 120588

120590

)

120587 (1198711) 1198711+ 1205831198711gt 120573(

] + 120572 + 120588

120590

)

(34)

Therefore by the intermediate value theorem there exists asolution

2isin (1 119871) of (20) By the previous theorem

1

and 2are components of equilibria with positive infective

population

Theorem 4 Let 119871lowast gt 0 be a solution of (24) then 119871lowast satisfy

(29) if and only if 119871lowast lt with = 120587minus1(120583](120573 minus ]))

Proof Let 119871lowast lt Since 120587(119871) is strictly decreasing functionthis is equivalent to

120587 (119871lowast) gt 120587 () (35)

Since 120587() = ]120583(120573minus]) so we have the equivalent expression

120587 (119871lowast) gt

]120583120573 minus ]

(36)

Multiplying both sides with 119871lowast and rearranging the terms

imply

(120573 minus ]) 120587 (119871lowast) 119871lowastgt ]120583119871lowast (37)

At infected steady state (119871lowast 119878lowast 119868lowast) the 1198711015840 equation of model(1) can be written as

120583119871lowast= 120573119878lowastminus 120587 (119871

lowast) 119871lowast (38)

So we get

(120573 minus ]) 120587 (119871lowast) 119871lowastgt ] (120573119878lowast minus 120587 (119871

lowast) 119871lowast)

120587 (119871lowast) 119871lowastgt ]119878lowast

(39)

Similarly if 119871lowast ge then (31) will not be satisfied

It is also clear that (24) has one or three solutionsdepending on the choice of the parameters ] 120572 120588 120590 and 120573

Now we will discuss the stability of the trivial steady state(0 0 0) and the persistence of the populationThe persistenceof a population is defined as follows

Definition 5 The population 119871(119905) 119878(119905) 119868(119905) is called uni-formly strongly persistent if there exists 120598 gt 0 independent ofinitial data such that for all solutions of model (1) satisfying119871(0) gt 0 119878(0) gt 0 and 119868(0) gt 0 we have 119871(119905) 119878(119905) 119868(119905) gt 120598

for all sufficiently large 119905 It is robust uniform persistence ifmodel (1) depends continuously on a parameter 120578 isin R If 120578

0is

a parameter and uniform persistence holds with 120601(119905 119878 119871 119868)

replaced by 120601(119905 119878 119871 119868 120578) where 120601(119905 119909) is semiflow for all 120578in the neighborhood of 120578

0 then we say that the system (1) is

robust uniformly persistent

Theorem6 (a)The trivial state is locally asymptotically stableif and only ifR

0lt 1

(b) Assume thatR0gt 1 and A = 120587

1015840() + 120587() + 120583 gt 0

(or equivalently (120573])120587(0) + ] gt minus1205871015840()) Then the steady

state ( 119878 0) is locally asymptotically stable if and only if 119878 lt

(] + 120572 + 120588)120590 or 119878 lt 119878lowast

(c) Assume thatR0gt 1 andA = 120587

1015840(119871lowast)119871lowast+120587(119871lowast)+120583 gt 0

(or equivalently (120573])120587(0)+] gt minus1205871015840(119871lowast)119871lowast)Then the endemic

equilibrium is locally asymptotically stable(d) If R

0gt 1 then the disease is robustly uniformly

persistent that is for every parameter vector 1205780for the system

(1) there exists a neighborhoodN of 1205780and 120598 gt 0 such that

lim inf119905rarrinfin

min (119878120578(119905) 119868120578(119905)) gt 120598 120578 isin N (40)

for all solutions (119871120578(119905) 119878120578(119905) 119868120578(119905)) of model (1) correspondingto 120578 with (119878

120578(0) + 119868

120578(0)) gt 0

Proof The Jacobian matrix associated with our model (1) isgiven by

119869 =

[[[

[

minus1205871015840(119871) 119871 minus 120587 (119871) minus 120583 120573 0

1205871015840(119871) 119871 + 120587 (119871) minus] minus 120590119868 120588 minus 120590119878

0 120590119868 120590119878 minus ] minus 120572 minus 120588

]]]

]

(41)

(a) This Jacobian matrix at steady state (0 0 0) is

1198691= [

[

minus120587 (0) minus 120583 120573 0

120587 (0) minus] 120588

0 0 minus] minus 120572 minus 120588

]

]

(42)


The three eigenvalues are 1205821

= minus(] + 120572 + 120588) and theeigenvalues of the matrix

1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)

The trace and determinant are given by

trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0

det (1198601) = ]120587 (0) + ]120583 minus 120573120587 (0) = ]120583 minus (120573 minus ]) 120587 (0)

(44)

So the steady state (0 0 0) is locally asymptotically stable ifand only if (120573 minus ])120587(0) lt ]120583

(b) The Jacobian matrix at ( 119878 0) is

119869 =

[[[

[

minus1205871015840() minus 120587 () minus 120583 120573 0

1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)

One of the eigenvalues is 1205821= 120590119878minus]minus120572minus120588 which is negative

if 119878 lt (] + 120572 + 120588)120590 The other two eigenvalues come from

1198691= [

[

minus1205871015840() minus 120587 () minus 120583 120573

1205871015840() + 120587 () minus]

]

]

(46)

Note that

T = trace (1198691) = minus119860 minus ]

D = det (1198691) = (120590119878 minus ] minus 120572 minus 120588) [minus119860 (] + 120573) + 120573120583]

(47)

where 119860 = 1205871015840()119871 + 120587() + 120583

Notice thatD is positive if 119860 gt 0(c) The endemic equilibrium is given by (119871

lowast (] + 120572 +

120588)120590 (120590120583119871lowastminus (120573 minus ])(] + 120572 + 120588))120590(minus] minus 120572)) where 119871

lowast isthe solution of (120587(119871) + 120583)119871 = (120573120590)(] + 120572 + 120588)

The Jacobian matrix at this steady state is given by

Jlowast

=

[[[

[

minus1205871015840(119871lowast) 119871lowastminus 120587 (119871

lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871

lowast) minus] minus 120590119868

lowast120588 minus 120590119878

lowast

0 120590119868lowast

120590119878lowastminus ] minus 120572 minus 120588

]]]

]

(48)

Since 120590119878lowast minus ] minus 120572 minus 120588 = 0 and 120588 minus 120590119878lowast= minus(] + 120572) so we get

Jlowast= [

[

minusA 120573 0

A minus 120583 minus] minus 120590119868lowast

minus (] + 120572)

0 120590119868lowast

0

]

]

(49)

whereA = 1205871015840(119871lowast)119871lowast+ 120587(119871

lowast) + 120583

Now we have

T = trace (Jlowast) = minus (A + ] + 120590119868lowast) lt 0

D = det (Jlowast) = minus120590119868lowastA (] + 120572) lt 0

(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868

lowast) minus 120573 (A minus 120583)

(51)

whereJ119896119896is the determinant of the matrix of size 2 resulting

fromJlowast by removing the 119896th row and the kth column

997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868

lowast) minus 120573 (A minus 120583)]

= minusA120590119868lowast(] + 120572) minusA

2(] + 120590119868

lowast) +A120573 (A minus 120583)

minus (] + 120590119868lowast) 120590119868lowast(] + 120572)minus(] + 120590119868

lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868

lowast) minusA120573 (A minus 120583)

+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)

minus (] + 120590119868lowast) 120573 (A minus 120583)

(53)

Simplifying and rearranging the terms we get

= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2

minus 120573 (A minus 120583) [A + ] + 120590119868lowast]

ge minus 120573 (A minus 120583) [A + ] + 120590119868lowast]

(54)

Since 0 lt A = (120587(119871)119871)1015840

119871=119871lowast + 120583 so this implies 120583 minus A =

minus(120587(119871)119871)1015840

119871=119871lowast gt 0 by the assumption of the structure of 120587(119871)119871

(ie the casewhenwe have three interior equilibria and this isthe most right one) Thus the endemic equilibrium is locallyasymptotically stable

(d) Let 120578 be a fixed parameter and 119883 = 119909 isin R3119878 =

119868 = 0 Define 119872 = 119881 cap 119883 where 119881 is the same regiondefined inTheorem 2 in which all the solutions are uniformlyeventually bounded Notice that both119883 and119872 are positivelyinvariant and 119864

0= (0 0 0) isin 119872 that attracts all the

solutions in 119883 Considering 1198640as a periodic orbit (say eg

of period 119879 = 1) then our results will follow by applyingthe theorems discussed in [16] (see in particular Corollary(47) Proposition (41) and Theorem (32) of [16]) Now ifwe denote the last two equations of model (1) in the form


of a 2 times 2 matrix 119860(119909) then it is clear that 119860(1198640) has a

positive spectral radius provided R0gt 1 So condition (9)

of Corollary 47 in [17] is satisfied Also 119860(1198640) is irreducible

which using Theorem 11 in [18] implies that 119890119905119860(1198640) has allpositive entries for all 119905 gt 0 This implies that condition (1) ofthe abovementioned corollary is also satisfied

Theorem 7 119871(119905) 119878(119905) and 119868(119905) converge to zero as 119905 rarr infin ifand only ifR

0lt 1

Proof We will use ldquoLa Sallersquos Invariance Principlerdquo to provethe theorem

Case 1 ((120573minus]) lt 0)Consider the Lyapunov function119881 = 119871+

119878+119868 Since119871 119878 119868 ge 0 so119881 ge 0 is a continuously differentiablefunction with 119881(0) = 0 Now consider

119881lowast= + 119878 + 119868

= 120573119878 minus 120587 (119871) 119871 minus 120583119871 + 120587 (119871) 119871 minus ]119878

+ 120588119868 minus 120590119868119878 + 120590119878119868 minus ]119868 minus 120572119868 minus 120588119868

= (120573 minus ]) 119878 minus 120583119871 minus (] + 120572) 119868 le 0

(55)

Now define

119864 = (119871 119878 119868) | (120573 minus ]) 119878 minus 120583119871 minus (] + 120572) 119868 = 0 (56)

Further as we have

(120573 minus ]) 119878 minus 120583119871 minus (] + 120572) 119868 = 0

997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)

Since 119871 and 119868 are nonnegative and also 120573 minus ] lt 0 so we have119878 le 0 But from the condition that 119878 should be nonnegativewe get 119878 = 0 So we have

120583119871 + (] + 120572) 119868 = 0 (58)

As 120583 and ]+120572 are positive therefore we must have 119871 = 119868 = 0So (0 0 0) is the only subset of 119864 which is also one of the

steady states and so it is the largest invariant subset of119864 Sincewe have shown above that all the solutions of the above systemare bounded thus by the above result all the solutions to thegiven system converge to (0 0 0)

Case 2 ((120573 minus ]) = 0) Again consider 119881lowast = 119871 + 119878 + 119868 In thiscase we have

119881lowast= minus120583119871 minus (] + 120572) 119868 le 0 (59)

Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)

Thus from the nonnegativity of 119871 we conclude that 119871 = 0 andso we get 119868 = 0 and 119864 = (0 119878 0) 119878 ge 0since at (0 119878 0) weget from the first equation of our main model that 119878 = 0

Thus (0 0 0) is the only invariant subset of 119864 thereforeall the bounded solutions of the given system will convergeto (0 0 0)

Case 3 ((120573 minus ]) gt 0) Consider the Lyapunov function 119881 =

]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)

= ] (120573119878 minus 120587 (119871) 119871 minus 120583 (119871)) + 120573 (120587 (119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868)

+ 120573 (120590119878119868 minus ]119868 minus 120572119868 minus 120588119868)

= minus ]120587 (119871) 119871 minus ]120583119871 + 120573120587 (119871) 119871 minus 120573 (] + 120572) 119868

= (120573 minus ]) 120587 (119871) 119871 minus ]120583119871 minus 120573 (] + 120572) 119868

le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0

= (120573 minus ]) 120587 (119871) 119871 minus ]120583119871 minus 120573 (] + 120572) 119868 le 0

(62)

Define

119864 = (119871 119878 119868) (120573 minus ]) 120587 (119871) 119871 minus ]120583119871 minus 120573 (] + 120572) 119868 = 0 (63)

Since

(120573 minus ]) 120587 (119871) 119871 minus ]120583119871 minus 120573 (] + 120572) 119868 = 0

(120573 minus ]) 120587 (119871) minus ]120583 le (120573 minus ]) 120587 (0) minus ]120583 le 0

(64)

so we have 119868 le 0 Therefore from the nonnegativity of119868 we conclude that 119868 = 0 So we will have either 119871 = 0 or(120573 minus ])120587(119871) minus ]120583 = 0

Notice that if 119871 = 0 then 119871 gt 0 so from the strictlydecreasing property of 120587 120587(119871) lt 120587(0) and therefore

(120573 minus ]) 120587 (119871) minus ]120583 lt (120573 minus ]) 120587 (0) minus ]120583 le 0 (65)

Thus (120573 minus ])120587(119871) minus ]120583 lt 0 and the only possibility is 119871 = 0

and hence (0 119878 0) is the only subset of 119864 Again from the firstequation of our model (1) 1198711015840 = 120573119878 which implies 1198711015840 = 0 ifand only if 119878 = 0 Hence (0 0 0) is the only invariant subsetof 119864 therefore all the bounded solutions of the given systemwill converge to (0 0 0)

5 The Disease Transmission ModelSpecial Case

In this section we will assume that 120587(119871) = 120585(1 minus 120595119871) with120585 120595 gt 0 Also assume that 120587(119871) = (1 minus 119871) for 119871 le 1 120587(119871) isstrictly decreasing as long as it is strictly positive Let us define119871∙= 120595119871 119878

∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)


Now by using these substitutions our main model willbecome

1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)

On simplifying these equations we get

1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)

First we will make this system dimensionless Assume thatthe 120573 120585 120583 ] 120588 and 120572 are measured in 1time 119871 119878 and 119868

are measured in the units of population size say 119901 So 120595 ismeasured in 1119901 Also 120590 is measured in 1((119901)(time)) Nowdividing by 120585 both sides our system becomes

1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)

where 120573 = 120573120585 and vice versa Rewriting the above system

after renaming the parameters we will have

1198711015840= 120573119878 minus (1 minus 119871) 119871 minus 120583119871

1198781015840= (1 minus 119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868

1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)

Here we measured time in 120585minus1 and renamed the parame-

ters 120573 120583 ] 120588 and as 120573 120583 ] 120588 and 120572 120595 as 120590 and 119871

∙ 119878∙

and 119868∙as 119871 119878 and 119868 The above system (70) is dimensionless

Note that for the systemR0= (120573])(1(1 + 120583))

Again we will have three types of steady states that is(0 0 0) ( 119878 0) and (119871

lowast 119878lowast 119868lowast)

Theorem 8 (a) The trivial state (0 0 0) always exists(b) The disease-free steady state of the form ( 119878 0) exists

and is unique if and only ifR0gt 1

(c)The infected state (119871lowast 119878lowast 119868lowast) exists if and only ifR0gt 1

and (1 minus 119871lowast)119871lowastgt ]119878lowast

(d) If 120583 lt 1 then for the system (70) there exist one or morethan one interior steady state

If 120583 gt 1 then for the system (70) there exists only oneinterior steady state with the 119871 component given by 119897

1

Proof (b) Since we know that the ldquo119871rdquo component of disease-free steady state comes from

120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)

and the ldquo119878rdquo component is given by

119878 =

120583

120573 minus ]

=

120583

120573 minus ](1 minus

120583]120573 minus ]

)

(74)

This steady state will exist if and only if

0 lt 120583] lt 120573 minus ] (75)

(c) It directly follows from part (c) of Theorem 3(d) From the third equation of the system (70) we have

119878lowast=

] + 120572 + 120588

120590

(76)

As calculated earlier the ldquo119868rdquo component is given by

119868lowast=

(120573 minus ]) 119878lowast minus 120583119871lowast

] + 120572

=


] + 120572

(77)

It is clear that the feasibility condition for the existence of thatsteady state is


where 119871lowast lt as proved inTheorem 3 This is equivalent to

(1 minus 119871lowast) 119871lowastgt ]119878lowast (79)

Observe that this holds only if 119871lowast lt 1 since for 119871lowast ge 1 weget ]119878lowast lt 0 which is not true So we must have 119871

lowastlt 1 In

this case we get

119871lowast2

minus 119871lowast+ ]119878lowast lt 0 (80)

We can write the left-hand side of (80) as

(119871lowastminus 1198701) (119871lowastminus 1198702) lt 0 (81)

where

1198701=

1 minus radic1 minus 4]119878lowast

2

1198702=

1 + radic1 minus 4]119878lowast

2

(82)


Notice that 1198701lt 1198702and both of them will exist only if 119878lowast le

14] Further from (81) it is clear that 119871lowast minus 1198701and 119871

lowastminus 1198702

have the opposite signs So we have the following two cases

Case 1 (119871lowast minus 1198701gt 0and119871lowast minus 119870

2lt 0) This means 119871lowast gt 119870

1

and 119871lowastlt 1198702or equivalently119870

1lt 119871lowastlt 1198702

Case 2 (119871lowast minus 1198701lt 0 and 119871

lowastminus 1198702gt 0) This means 119871lowast lt 119870

1

and 119871lowastgt 1198702 Since119870

1lt 1198702 so this case is not possible Sowe

can have only one choice for 119871lowast that is 1198701lt 119871lowastlt 1198702(with

1198701and 119870

2given in (82)) for the existence of 119868lowast gt 0 The ldquo119871rdquo

component of the endemic steady state is the solution of theequation

120587 (119871lowast) 119871lowast+ 120583119871lowast= 120573119878lowast (83)

By using the value of 120587(119871) we get

(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)

This implies that for 119871 lt 1

1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)

which is a quadratic equation in 119871 and the roots of thisequation are given by

1198971=

(1 + 120583) minus radic(1 + 120583)2minus 4120573119878

lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)

It is clear that 0 lt 1198971lt 1198972 Also 119897

1will exist only if 119897

1lt =

1 minus (120583](120573 minus ])) lt 1 That is we have


lowastlt 2 (88)

Simplifying and rearranging the terms yield

(120583 minus 1) lt radic(1 + 120583)2minus 4120573119878

lowast (89)

We have the following two cases

Case i (120583 lt 1) For this case inequality (89) is automaticallysatisfied

Case ii (120583 ge 1) This is equivalent to 120583 minus 1 ge 0 Then onsquaring both sides of inequality (89) we get

1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)

Now consider 1198972 this will also exist only if 119897

2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)

which can be rewritten as

(120583 minus 1) lt minusradic(1 + 120583)2minus 4120573119878

lowast (93)

From the above equation notice that 1198972will exist only if 120583 lt 1

Taking square of both sides of this equation we get

1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)

which implies that 120583 lt 120573119878lowast

Theorem 9 (i)The steady states of the form ( 119878 0) are stableif and only ifR

0gt 1 and 120583 gt 2 minus 1

(ii) The endemic state 1198971is locally stable and 119897

2is unstable

whenever they exist

Proof (a)The proof directly follows fromTheorems 3 (b) and6 (b)

(b) The Jacobian matrix is given by

119869 =[[

[

minus (1 minus 2119871) minus 120583 120573 0

(1 minus 2119871) minus] minus 120590119868 120588 minus 120590119878

0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)

At the steady state (119871lowast 119878lowast 119868lowast) we get

119869lowast=[[

[

minus (1 minus 2119871lowast) minus 120583 120573 0

(1 minus 2119871lowast) minus] minus 120590119868

lowastminus (] + 120572)

0 120590119868lowast

0

]]

]

(96)

Assume that (119871lowast) = (1 minus 2119871lowast) + 120583 so we have

119869lowast=[[

[

minus120601 (119871lowast) 120573 0

120601 (119871lowast) minus 120583 minus] minus 120590119868


0 120590119868lowast

0

]]

]

(97)

the trace and determinant are given by

T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583

= 1 minus [(1 + 120583) minus radic(1 + 120583)2minus 4120573119878

lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)

Now we evaluate the determinant for 1198971and 1198972 At 119871lowast = 119897

1

T lt 0 and alsoD lt 0


At 119871lowast

= 1198972 we have D gt 0 so at least one of the

eigenvalues is positive and thus this equilibrium is unstableAs calculated earlier

D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871

lowast) minus 120583)

times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2

minus 120573 (120601 (119871lowast) minus 120583) [120601 (119871

lowast) + ] + 120590119868

lowast]

(103)

Since the only point of interest here is 119871lowast = 1198971(because 119897

2is

unstable) and 120601(1198971) gt 0 therefore we have

D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)

Now for the local asymptotic stability we want

120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)

simplification yields

1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt

4120573119878lowastminus 1

2

(106)

It can be easily shown that we can find such a 120583which satisfies(106) Further this is clear from (101) that

2120583 lt 4120573119878lowastminus 1 lt (1 + 120583)

2minus 1

997904rArr 1205832gt 0

(107)

Special Case Here we will discuss the special case when thereis no disease that is when 119868 = 0 Also assume that 120587(119871) =

(1 minus 119871) for 119871 le 1 Under this condition our system given by(70) will be reduced to a two-dimensional system and is givenby


1198781015840= (1 minus 119871) 119871 minus ]119878

(108)

Observe that (0 0) is a steady state for this system We canwrite the steady state form of this system into matrix form as

[minus (1 minus 119871) minus 120583 120573

(1 minus 119871) minus]] [119871

119878] = 0 (109)

for any nontrivial steady state 119871lowast 119878lowast gt 0 the determinant ofthis matrix should be zero that is

] (1 minus 119871) + 120583] minus 120573 (1 minus 119871) = 0 (110)

On rearranging the terms we get

(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)

This steady state (119871lowast 119878lowast) will exist if 120573 gt ] Now the

Jacobian matrix at (119871lowast 119878lowast) is given by

119869 = [minus (1 minus 2119871

lowast) minus 120583 120573

(1 minus 2119871lowast) minus]] (113)


T = minus (1 minus 2119871lowast) minus 120583 minus ] = minus [(1 minus 2119871

lowast) + 120583 + ]]

D = ] (1 minus 2119871lowast) + 120583] minus 120573 (1 minus 2119871

lowast)

= minus (120573 minus ]) (1 minus 2119871lowast) + 120583]

= minus (120573 minus ]) [(1 minus 119871lowast) minus 119871lowast] + 120583]

= minus (120573 minus ]) (1 minus 119871lowast) + 120583] + (120573 minus ]) 119871lowast

(114)

Notice that from (111) (120573 minus ])(1 minus 119871lowast) minus 120583] = 0 So we get

D = (120573 minus ]) 119871 gt 0 (115)

So this internal steady state is locally asymptotically stable if(1 minus 2119871

lowast) + 120583 + ] gt 0 that is when 119871

lowastlt (12)(1 + 120583 + ])

and unstable if (1 minus 2119871lowast) + 120583 + ] lt 0 that is when 119871

lowastgt

(12)(1 + 120583 + ]) Thus we have that 119871 = (12)(1 + 120583 + ]) isthe point of bifurcation

Theorem 10 Assume that the positive parameters 120583 and ]can be chosen such that 119871

lowastgt (12)(1 + 120583 + ]) then 119871

lowast

becomes the first coordinate of a unique nontrivial steady statethat is unstable spiral point Every solution that does not startat the nontrivial steady state or the origin converges towardsa periodic orbit In particular there exists an orbitally stableperiodic orbit

Proof Since we have shown earlier that all the solutions arebounded and we have only one interior steady state for thislimiting system which is also nondegenerate that is all theeigenvalues of the associated Jacobianmatrix evaluated at thissteady state point are nonzero so by Theorem A-15 of [19]there exists a locally asymptotically stable periodic orbit

So far we have shown that with proper choice of thevalues of 120583 and ] we can find 119871

lowastgt 0 which becomes the first

coordinate of an unstable spiral point or in particular there


exists a periodic solution Here we will use the approach ofthe theory of Hopf bifurcation and in particular the conceptsof supercritical and subcritical bifurcation to discuss thestability of these periodic orbits Actually we need to find aJacobian matrix of the form

[0 minus120574

120574 0] (116)

with 120574 gt 0Again consider the system equation (86) Define 119881 = 119878 +

(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573

[120573119878 minus (1 minus 119871) 119871 minus 120583119871]

= (1 minus 119871) 119871 minus ]119878 + ]119878 minus

]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573

[(120573 minus ]) (1 minus 119871) minus ]120583] (118)

Similarly

1198711015840= 120573(119881 minus

]120573

119871) minus (1 minus 119871) 119871 minus 120583119871

= 120573119881 minus ]119871 minus (1 minus 119871) 119871 minus 120583119871

(119)

This implies

1198711015840= 120573119881 minus (] + 120583) 119871 minus (1 minus 119871) 119871 (120)

Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573

[(120573 minus ]) (1 minus 119871) minus ]120583]) (121)

or

1199061015840=

119871

120574

[(120573 minus ]) (1 minus 119871) minus ]120583]

1198711015840= 120574119906 minus (] + 120583) 119871 minus (1 minus 119871) 119871

(122)

Observe that the 119871 component of the interior steady state 119871lowastof this system satisfies (120573 minus ])(1 minus 119871

lowast) minus ]120583 = 0 The Jacobian

matrix is given by

119869 =[[

[

0

1

120574

[(120573 minus ]) (1 minus 119871) minus ]120583] +119871

120574

[minus (120573 minus ])]

120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)

Evaluating the Jacobian matrix at the interior steady state(119906lowast 119871lowast) we get

119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])

120574 minus [(] + 120583) + (1 minus 2119871lowast)]

]]

]

(124)

Since we also know that at 119871lowast = 119871 (ie Hopf bifurcation

point) we have (] + 120583) + (1 minus 2119871) = 0 so at this point the

Jacobian matrix will become

119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)

Nowwe need to choose 120574 in such a way that 120574 = (119871120574)(120573minus])

which gives

120574 = radic119871(120573 minus ]) gt 0 (126)

So with this choice of 120574 we have Hopf bifurcation Nowassume that (1198651 1198652) is the vector field associated with (122)that is

1198651=

119871

120574



(127)

The stability of the bifurcating periodic orbit is determinedby the sign of the following number

119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)

Now we calculate these partial derivatives

1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574

(120573 minus ]) (1 minus 119871) minus ]120583 +

119871

120574


997904rArr 1198651

119871119871= minus

1

120574

(120573 minus ]) minus1

120574

(120573 minus ]) = minus

2

120574

(120573 minus ])

997904rArr 1198652

119871= minus (] + 120583) minus (1 minus 2119871)

997904rArr 1198652

119871119871= 2

(129)

So by substituting these values we get

119886 = minus

4

120574

(120573 minus ]) (130)

Since 119886 lt 0 so the bifurcating periodic orbits are asymp-totically stable that is we have the case of supercriticalbifurcation

6 Discussion

In this paper we have introduced and analyzed a model ofdisease based amphibian decline We found basic reproduc-tion number R

0 which guarantees the extinction of disease

population in the presence of disease as shown in Figure 1


L

S

I

20

18

16

14

12

10

8

6

4

2

0

0 50 100 150 200 250 300

Time

Popu

latio

n

Figure 1 This figure shows the time series of model (1) whenR0lt

1 It is clear that all the population will extinct in this case Theparameters used are 120573 = 0002 120583 = 005 ] = 002 120588 = 01 120590 =

0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

S

I

Figure 2The case ofR0gt 1 It is clear that under certain parame-

ter values only larvae and susceptible population will survive whilethe infected population will be extinct All parameter values are thesame except 120573 = 0208

We proved that only the disease-free steady state (0 0 0)

is stable and the stability of the other two steady statesis conditional and depends on some other environmentalfactors whenever R

0gt 1 We also have shown that not

only isR0the sufficient condition for the survival of infected

population but also for the case when R0gt 1 disease may

or may not persist depending on certain constraints (seeFigures 1 2 and 3) In our model we also have considereda small population size of 119871 The large population case will berelatively easy to handle in terms of the function 120587(119871) In thiscase since 120587(119871) is decreasing therefore for large populationsize 120587(119871lowast) asymp 0 in (24) and this will take the form

119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)

This is the case when we have large number of suscepti-bles

180

160

140

120

100

80

60

40

20

00 100 200 300 400 500

S

L

Figure 3 Phase portrait of 119871 versus 119878 The existence of a periodicsolution for model (1) is evident Parameters values are the same asof Figure 2

Further in our model we just consider the simple casewhen only the larvae in stage 119871 are subject to intrastagecompetition A future direction for this model is to considera rather more complicated case when both secure larvae 119871

and exposed larvae 119878 will compete for the resources In thiscase the term 120587(119871)119871 will be replaced by 120587(119871 119878)119871 which maycause some oscillatory effects due to the density dependentdevelopment rate for both 119871 and 119878

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] R M Lehtinen S M Galatowitsch and J R Tester ldquoConse-quences of habitat loss and fragmentation for wetland amphib-ian assemblagesrdquoWetlands vol 19 no 1 pp 1ndash12 1999

[2] S N Stuart J S Chanson N A Cox et al ldquoStatus and trendsof amphibian declines and extinctions worldwiderdquo Science vol306 no 5702 pp 1783ndash1786 2004

[3] P Daszak L Berger A A Cunningham A D Hyatt DEarl Green and R Speare ldquoEmerging infectious diseases andamphibian population declinesrdquo Emerging Infectious Diseasesvol 5 no 6 pp 735ndash748 1999

[4] S A Diamond G S Peterson J E Tietge and G T AnkleyldquoAssessment of the risk of solar ultraviolet radiation to amphib-ians III Prediction of impacts in selected northernmidwesternwetlandsrdquo Environmental Science and Technology vol 36 no 13pp 2866ndash2874 2002

[5] B A Pierce ldquoThe effect of acid precipitation on amphibiansrdquoEcotoxicology vol 2 no 1 pp 65ndash77 1993

[6] L E Licht and K P Grant ldquoThe effects of ultraviolet radiationon the biology of amphibiansrdquo American Zoologist vol 37 no2 pp 137ndash145 1997


[7] S J Hecnar and R T MrsquoCloskey ldquoThe effects of predatoryfish on amphibian species richness and distributionrdquo BiologicalConservation vol 79 no 2-3 pp 123ndash131 1997

[8] K R Lips F Brem R Brenes et al ldquoEmerging infectiousdisease and the loss of biodiversity in a Neotropical amphibiancommunityrdquo Proceedings of the National Academy of Sciences ofthe United States of America vol 103 no 9 pp 3165ndash3170 2006

[9] H R Thieme T Dhirasakdanon Z Han and R TrevinoldquoSpecies decline and extinction synergy of infectious diseaseand Allee effectrdquo Journal of Biological Dynamics vol 3 no 2-3pp 305ndash323 2009

[10] L Berger R Speare P Daszak et al ldquoChytridiomycosis causesamphibian mortality associated with population declines in therain forests of Australia andCentral AmericardquoProceedings of theNational Academy of Sciences of theUnited States of America vol95 no 15 pp 9031ndash9036 1998

[11] S E Drury R E Gough and A A Cunningham ldquoIsolationof an iridovirus-like agent from common frogs (Rana tempo-raria)rdquoThe Veterinary Record vol 137 no 3 pp 72ndash73 1995

[12] L J Rachowicz R A Knapp J A T Morgan et al ldquoEmerginginfectious disease as a proximate cause of amphibian massmortalityrdquo Ecology vol 87 no 7 pp 1671ndash1683 2006

[13] A Richards J CoryM Speight and TWilliams ldquoForaging in apathogen reservoir can lead to local host population extinctiona case study of a Lepidoptera-virus interactionrdquo Oecologia vol118 no 1 pp 29ndash38 1999

[14] H R Thieme Mathematics in Population Biology PrincetonSeries in Theoretical and Computational Biology PrincetonUniversity Press Princeton NJ USA 2003

[15] J P Collins ldquoIntrapopulation variation in the body size atmetamorphosis and timing of metamorphosis in the bullfrogRana castesbeianardquo Ecology vol 60 pp 738ndash749 1979

[16] P L Salceanu and H L Smith ldquoPersistence in a discrete-timestage-structured fungal disease modelrdquo Journal of BiologicalDynamics vol 3 no 2-3 pp 271ndash285 2009

[17] P L Salceanu ldquoRobust uniform persistence in discrete andcontinuous dynamical systems using Lyapunov exponentsrdquoMathematical Biosciences and Engineering vol 8 no 3 pp 807ndash825 2011

[18] H L Smith Monotone Dynamical Systems An Introduction tothe Theory of Competitive and Cooperative Systems vol 41 ofMathematical Surveys and Monographs American Mathemat-ical Society Providence RI USA 1995

[19] F Brauer and J A Nohel The Qualitative Theory of OrdinaryDifferential Equations Dover Books on Mathematics NewYork NY USA 1989

Submit your manuscripts athttpwwwhindawicom

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Algebra

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Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


keep this component intact while studying the disease basedamphibian decline Since the main focus of the model isthe disease based larvae extinction therefore in order tolook at the complete scenario we assume that at the earlystage of their lives the larvae are surrounded by a safeenvironment which is free of disease like in a closed shell orsome protected holes For our model we denote this class by119871 At the later stage once these unexposed larvae enter intofree environment they become vulnerable to disease and wecall them susceptibles (denoted by 119878) These susceptibles canbe infected with the disease and enter into the infective classdenoted by 119868 Further these infected larvae can also recoverfrom the disease and return to the susceptible class Themodel is formulated in a way that the number of new infectedlarvae depends on the present number of susceptible andinfective larvae in the sense that the more larvae (susceptibleor infective) we have the more chance of spreading theinfections among the healthy or susceptible larvae will therebe that is the disease can be transferred from one to another

We organize this paper as follows First we describe themathematical model along with its underlying assumptionsThen we introduce many threshold quantities for examplethe reproduction ratios and the critical host density forthe disease establishments We also derive the number ofendemic equilibriums and discuss their stabilities Next wealso elaborate our findings using numerical simulationsFinally we give the conclusion and discuss the potentiallimitations and impact of our results

2 Formulation of Mathematical Model

In this mathematical model we assume that the disease onlyaffects the larval population So we divide the larvae 119871 intotwo categories the susceptibles larvae 119878 and the infectedlarvae 119868 Although this disease is transferable from one toanother the recovery is also possible and on recovering fromthe disease an infected larva will reenter into the class ofsusceptible larvae Further for ourmodel we also assume thatonly susceptibles can contribute to the reproduction processThus once the disease is spread the whole population will gotowards extinctions not only due to the illness but also due tothe lack of reproduction process

The model is as follows

1198711015840= 120573119878 minus 120587 (119871) 119871 minus 120583119871

1198781015840= 120587 (119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868


(1)

Here119871 represents the unexposed class of the larval populationwhich is not yet entered into the susceptible class 119878 Here weassume that the transition from stage 119871 to stage 119878 requiresa certain minimum size This assumption is valid for manyspecies for example Daphnia magna a water flea where theindividuals typically have a length of 08mm at birth and alength of 25mm once they enter into the exposed class [14]Further in the case of some amphibians the body size atmetamorphosis is quite flexible (see eg [15]) therefore acertainminimum size still seems to be required in these cases

Table 1 Description of parameters in model (1)

Variables Description120573 Per capita birth rate of susceptibles120583 Per capita (natural) mortality rate of the larvae

] Per capita (natural) mortality rate of thesusceptible and infected larvae

120572Per capita mortality rate of the infected larvae dueto disease

120588 Recovery rate from disease

120587 (119871)Per capita transition rate This is the rate at whichthe susceptible larva matures

120590

Rate at which the disease is transferred frominfected to susceptible class when the contactbetween susceptible and infected individualsoccurs

Thus based on these evidences it is reasonable to assumethat the scarcity of resources will prolong the length of thelarval stage 119871 So if there are more larvae in stage 119871 and lessresources then it will take longer for a single larva in stage119871 to complete the transition into stage 119878 Therefore in ourmodel we assume that 120587(119871)119871 is nondecreasing in 119871 in orderto represent a strong negative feedback from the number oflarval population 119871 It is clear from the model that this classwill enter into the exposed susceptible class 119878 with the rate120583 which further can become infected with a rate ] Since inthe case of closed population the disease can transfer fromone individual to the other therefore by using the law of massaction we express this by120590119878119868 where120590 is disease transmissionrate Finally the last term in the third equation represents therecovery from disease with the rate 120588

The detailed description of parameters of our model isgiven in Table 1

Assumption 1 All the parameters are positive except 120572 whichcan also be zero 120587(119871) is a strictly decreasing nonnegativefunction of 119871 such that 120587(119871) rarr 0 as 119871 rarr infin

3 Existence Positivityand Boundedness of the Solution

Theorem 2 Assume that Assumption 1 is satisfied then fornonnegative initial data there are unique nonnegative solu-tions of the system which are defined for all nonnegative timesFurther all the solutions to the system are uniformly eventuallybounded

Proof Let 119865 = (1198911 1198912 1198913) and 119883 = (119909

1 1199092 1199093) = (119871 119878 119868)

where

1198911= 120573119878 minus 120587 (119871) 119871 minus 120583119871

1198912= 120587 (119871) 119871 minus ]119878 + 120588119868 minus 120590119878119868


(2)



continuous Further

119871 = 0 997904rArr 1198911= 120573119878 ge 0

119878 = 0 997904rArr 1198912= 120587 (119871) 119871 + 120588119868 ge 0

119868 = 0 997904rArr 1198913= 0

(3)



0) isin 1198613


+and it is


sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) = infin (4)


1198711015840le 120573119878

1198781015840le 120587 (119871) 119871 + 120588119868 minus 120590119878119868 le 120587 (0) 119871 + 120588119868 minus 120590119878119868

1198681015840le 120590119878119868

(5)


1198711015840+ 1198781015840+ 1198681015840le 120578 (119871 + 119878 + 119868)

997904rArr 119871 (119905) + 119878 (119905) + 119868 (119905) le [119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

997904rArr sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le sup0le119905lt119887

[119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

(6)


sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le 119862 lt infin (7)





(8)


1198811015840= minus]119878 + (

120573

]120587 (119871) minus 120583)119871 minus 120585 (] + 120572) 119868 (9)




1198811015840le minus ]119878 + (

120573

]120598 minus 120583)119871


120583

2

119871 le minus

120583

2

119871♯

(10)



1198811015840le minus ]119878 +

120573

]120587 (119871) 119871 minus 120583119871 minus 120585 (] + 120572) 119868

le minus ]119878 +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus ](1

120585

(119881 minus 119871) minus 119868) +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus

]120585

119881 +

]120585

119871 + ]119868 +120573

]120587 (0) 119871

♯minus (

120573

]+ 1) (] + 120572) 119868

(11)


997904rArr 1198811015840le minus

]120585

119881 + (

]120585

+

120573

]120587 (0)) 119871

♯ (12)




119905rarrinfin119881(119905) le 119881










120573119878 minus 120587 () minus 120583 = 0

120587 () minus ]119878 = 0

(14)


119878 =

120583

(120573 minus ]) (15)



120573120583


[

120573120583


997904rArr = 0 or120573120583

120573 minus ]minus 120583 = 120587 ()

(16)


120573120583

120573 minus ]minus 120583 = 120587 ()

120587 () =

120583]120573 minus ]

(17)

Equivalently

= 120587minus1

(

120583]120573 minus ]

) (18)


120573120583

120573 minus ]minus 120583 lt 120587 (0) (19)


R0=

120573

]120587 (0)

120587 (0) + 120583

lt 1 (20)








(21)


(120573 minus ]) (] + 120572 + 120588)

120590



(22)

Therefore we have

119868lowast=


120590 (minus] minus 120572)

(23)



120573

120590

(] + 120572 + 120588) minus (120587 (119871lowast) + 120583) 119871

lowast= 0

(120587 (119871lowast) + 120583) 119871

lowast=

120573

120590

(] + 120572 + 120588) = 120573119878lowast

(24)


(119871lowast

] + 120572 + 120588

120590


120590 (minus] minus 120572)

) (25)


we get


lowast(120588 minus 120590119878

lowast) = 0 (26)


119868lowast=


] + 120572

(27)




120590119878lowast

] + 120572 + 120588

= 1 (29)



lowastThat is

Rpar

(119878lowast) =

120590119878lowast

] + 120572 + 120588

(30)


Rpar0

(119878) =

120590119878

] + 120572 + 120588

(31)






0gt 1







(120587 () + 120583) gt 120573

] + 120572 + 120588

120590

(32)



(120587 () 119871+120583) lt 120573

] + 120572 + 120588

120590

(33)



1isin (0 119871

1] If




lt 1198711


120587 (1198711) 1198711+ 1205831198711gt 120587 (

1) 1+ 1205831= 120573(

] + 120572 + 120588

120590

)

120587 (1198711) 1198711+ 1205831198711gt 120573(

] + 120572 + 120588

120590

)

(34)



1


population




120587 (119871lowast) gt 120587 () (35)


120587 (119871lowast) gt

]120583120573 minus ]

(36)


imply





So we get




(39)






0is






0lt 1


1015840() + 120587() + 120583 gt 0



(] + 120572 + 120588)120590 or 119878 lt 119878lowast









min (119878120578(119905) 119868120578(119905)) gt 120598 120578 isin N (40)


120578(0) + 119868

120578(0)) gt 0


119869 =

[[[

[




]]]

]

(41)


1198691= [

[

minus120587 (0) minus 120583 120573 0

120587 (0) minus] 120588


]

]

(42)




1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)


trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0


(44)



119869 =

[[[

[


1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)



1198691= [

[


1205871015840() + 120587 () minus]

]

]

(46)

Note that



(47)

where 119860 = 1205871015840()119871 + 120587() + 120583


lowast (] + 120572 +




Jlowast

=

[[[

[


lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871


lowast120588 minus 120590119878

lowast

0 120590119868lowast


]]]

]

(48)


Jlowast= [

[

minusA 120573 0


minus (] + 120572)

0 120590119868lowast

0

]

]

(49)


lowast) + 120583

Now we have



(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868


(51)



997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868



2(] + 120590119868

lowast) +A120573 (A minus 120583)


lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868


+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)


(53)


= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2



(54)

Since 0 lt A = (120587(119871)119871)1015840


minus(120587(119871)119871)1015840














0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





L

S

I

20

18

16

14

12

10

8

6

4

2

0

0 50 100 150 200 250 300

Time

Popu

latio

n



0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

S

I









119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


180

160

140

120

100

80

60

40

20

00 100 200 300 400 500

S

L






References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











continuous Further

119871 = 0 997904rArr 1198911= 120573119878 ge 0

119878 = 0 997904rArr 1198912= 120587 (119871) 119871 + 120588119868 ge 0

119868 = 0 997904rArr 1198913= 0

(3)



0) isin 1198613


+and it is


sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) = infin (4)


1198711015840le 120573119878

1198781015840le 120587 (119871) 119871 + 120588119868 minus 120590119878119868 le 120587 (0) 119871 + 120588119868 minus 120590119878119868

1198681015840le 120590119878119868

(5)


1198711015840+ 1198781015840+ 1198681015840le 120578 (119871 + 119878 + 119868)

997904rArr 119871 (119905) + 119878 (119905) + 119868 (119905) le [119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

997904rArr sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le sup0le119905lt119887

[119871 (0) + 119878 (0) + 119868 (0)] 119890120578119905

(6)


sup0le119905lt119887

119871 (119905) + 119878 (119905) + 119868 (119905) le 119862 lt infin (7)





(8)


1198811015840= minus]119878 + (

120573

]120587 (119871) minus 120583)119871 minus 120585 (] + 120572) 119868 (9)




1198811015840le minus ]119878 + (

120573

]120598 minus 120583)119871


120583

2

119871 le minus

120583

2

119871♯

(10)



1198811015840le minus ]119878 +

120573

]120587 (119871) 119871 minus 120583119871 minus 120585 (] + 120572) 119868

le minus ]119878 +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus ](1

120585

(119881 minus 119871) minus 119868) +

120573

]120587 (0) 119871

♯minus 120585 (] + 120572) 119868

le minus

]120585

119881 +

]120585

119871 + ]119868 +120573

]120587 (0) 119871

♯minus (

120573

]+ 1) (] + 120572) 119868

(11)


997904rArr 1198811015840le minus

]120585

119881 + (

]120585

+

120573

]120587 (0)) 119871

♯ (12)




119905rarrinfin119881(119905) le 119881










120573119878 minus 120587 () minus 120583 = 0

120587 () minus ]119878 = 0

(14)


119878 =

120583

(120573 minus ]) (15)



120573120583


[

120573120583


997904rArr = 0 or120573120583

120573 minus ]minus 120583 = 120587 ()

(16)


120573120583

120573 minus ]minus 120583 = 120587 ()

120587 () =

120583]120573 minus ]

(17)

Equivalently

= 120587minus1

(

120583]120573 minus ]

) (18)


120573120583

120573 minus ]minus 120583 lt 120587 (0) (19)


R0=

120573

]120587 (0)

120587 (0) + 120583

lt 1 (20)








(21)


(120573 minus ]) (] + 120572 + 120588)

120590



(22)

Therefore we have

119868lowast=


120590 (minus] minus 120572)

(23)



120573

120590

(] + 120572 + 120588) minus (120587 (119871lowast) + 120583) 119871

lowast= 0

(120587 (119871lowast) + 120583) 119871

lowast=

120573

120590

(] + 120572 + 120588) = 120573119878lowast

(24)


(119871lowast

] + 120572 + 120588

120590


120590 (minus] minus 120572)

) (25)


we get


lowast(120588 minus 120590119878

lowast) = 0 (26)


119868lowast=


] + 120572

(27)




120590119878lowast

] + 120572 + 120588

= 1 (29)



lowastThat is

Rpar

(119878lowast) =

120590119878lowast

] + 120572 + 120588

(30)


Rpar0

(119878) =

120590119878

] + 120572 + 120588

(31)






0gt 1







(120587 () + 120583) gt 120573

] + 120572 + 120588

120590

(32)



(120587 () 119871+120583) lt 120573

] + 120572 + 120588

120590

(33)



1isin (0 119871

1] If




lt 1198711


120587 (1198711) 1198711+ 1205831198711gt 120587 (

1) 1+ 1205831= 120573(

] + 120572 + 120588

120590

)

120587 (1198711) 1198711+ 1205831198711gt 120573(

] + 120572 + 120588

120590

)

(34)



1


population




120587 (119871lowast) gt 120587 () (35)


120587 (119871lowast) gt

]120583120573 minus ]

(36)


imply





So we get




(39)






0is






0lt 1


1015840() + 120587() + 120583 gt 0



(] + 120572 + 120588)120590 or 119878 lt 119878lowast









min (119878120578(119905) 119868120578(119905)) gt 120598 120578 isin N (40)


120578(0) + 119868

120578(0)) gt 0


119869 =

[[[

[




]]]

]

(41)


1198691= [

[

minus120587 (0) minus 120583 120573 0

120587 (0) minus] 120588


]

]

(42)




1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)


trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0


(44)



119869 =

[[[

[


1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)



1198691= [

[


1205871015840() + 120587 () minus]

]

]

(46)

Note that



(47)

where 119860 = 1205871015840()119871 + 120587() + 120583


lowast (] + 120572 +




Jlowast

=

[[[

[


lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871


lowast120588 minus 120590119878

lowast

0 120590119868lowast


]]]

]

(48)


Jlowast= [

[

minusA 120573 0


minus (] + 120572)

0 120590119868lowast

0

]

]

(49)


lowast) + 120583

Now we have



(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868


(51)



997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868



2(] + 120590119868

lowast) +A120573 (A minus 120583)


lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868


+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)


(53)


= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2



(54)

Since 0 lt A = (120587(119871)119871)1015840


minus(120587(119871)119871)1015840














0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0 50 100 150 200 250 300

Time

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0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

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L

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











120573119878 minus 120587 () minus 120583 = 0

120587 () minus ]119878 = 0

(14)


119878 =

120583

(120573 minus ]) (15)



120573120583


[

120573120583


997904rArr = 0 or120573120583

120573 minus ]minus 120583 = 120587 ()

(16)


120573120583

120573 minus ]minus 120583 = 120587 ()

120587 () =

120583]120573 minus ]

(17)

Equivalently

= 120587minus1

(

120583]120573 minus ]

) (18)


120573120583

120573 minus ]minus 120583 lt 120587 (0) (19)


R0=

120573

]120587 (0)

120587 (0) + 120583

lt 1 (20)








(21)


(120573 minus ]) (] + 120572 + 120588)

120590



(22)

Therefore we have

119868lowast=


120590 (minus] minus 120572)

(23)



120573

120590

(] + 120572 + 120588) minus (120587 (119871lowast) + 120583) 119871

lowast= 0

(120587 (119871lowast) + 120583) 119871

lowast=

120573

120590

(] + 120572 + 120588) = 120573119878lowast

(24)


(119871lowast

] + 120572 + 120588

120590


120590 (minus] minus 120572)

) (25)


we get


lowast(120588 minus 120590119878

lowast) = 0 (26)


119868lowast=


] + 120572

(27)




120590119878lowast

] + 120572 + 120588

= 1 (29)



lowastThat is

Rpar

(119878lowast) =

120590119878lowast

] + 120572 + 120588

(30)


Rpar0

(119878) =

120590119878

] + 120572 + 120588

(31)






0gt 1







(120587 () + 120583) gt 120573

] + 120572 + 120588

120590

(32)



(120587 () 119871+120583) lt 120573

] + 120572 + 120588

120590

(33)



1isin (0 119871

1] If




lt 1198711


120587 (1198711) 1198711+ 1205831198711gt 120587 (

1) 1+ 1205831= 120573(

] + 120572 + 120588

120590

)

120587 (1198711) 1198711+ 1205831198711gt 120573(

] + 120572 + 120588

120590

)

(34)



1


population




120587 (119871lowast) gt 120587 () (35)


120587 (119871lowast) gt

]120583120573 minus ]

(36)


imply





So we get




(39)






0is






0lt 1


1015840() + 120587() + 120583 gt 0



(] + 120572 + 120588)120590 or 119878 lt 119878lowast









min (119878120578(119905) 119868120578(119905)) gt 120598 120578 isin N (40)


120578(0) + 119868

120578(0)) gt 0


119869 =

[[[

[




]]]

]

(41)


1198691= [

[

minus120587 (0) minus 120583 120573 0

120587 (0) minus] 120588


]

]

(42)




1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)


trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0


(44)



119869 =

[[[

[


1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)



1198691= [

[


1205871015840() + 120587 () minus]

]

]

(46)

Note that



(47)

where 119860 = 1205871015840()119871 + 120587() + 120583


lowast (] + 120572 +




Jlowast

=

[[[

[


lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871


lowast120588 minus 120590119878

lowast

0 120590119868lowast


]]]

]

(48)


Jlowast= [

[

minusA 120573 0


minus (] + 120572)

0 120590119868lowast

0

]

]

(49)


lowast) + 120583

Now we have



(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868


(51)



997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868



2(] + 120590119868

lowast) +A120573 (A minus 120583)


lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868


+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)


(53)


= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2



(54)

Since 0 lt A = (120587(119871)119871)1015840


minus(120587(119871)119871)1015840














0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0

0 50 100 150 200 250 300

Time

Popu

latio

n



0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


180

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References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra













0gt 1







(120587 () + 120583) gt 120573

] + 120572 + 120588

120590

(32)



(120587 () 119871+120583) lt 120573

] + 120572 + 120588

120590

(33)



1isin (0 119871

1] If




lt 1198711


120587 (1198711) 1198711+ 1205831198711gt 120587 (

1) 1+ 1205831= 120573(

] + 120572 + 120588

120590

)

120587 (1198711) 1198711+ 1205831198711gt 120573(

] + 120572 + 120588

120590

)

(34)



1


population




120587 (119871lowast) gt 120587 () (35)


120587 (119871lowast) gt

]120583120573 minus ]

(36)


imply





So we get




(39)






0is






0lt 1


1015840() + 120587() + 120583 gt 0



(] + 120572 + 120588)120590 or 119878 lt 119878lowast









min (119878120578(119905) 119868120578(119905)) gt 120598 120578 isin N (40)


120578(0) + 119868

120578(0)) gt 0


119869 =

[[[

[




]]]

]

(41)


1198691= [

[

minus120587 (0) minus 120583 120573 0

120587 (0) minus] 120588


]

]

(42)




1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)


trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0


(44)



119869 =

[[[

[


1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)



1198691= [

[


1205871015840() + 120587 () minus]

]

]

(46)

Note that



(47)

where 119860 = 1205871015840()119871 + 120587() + 120583


lowast (] + 120572 +




Jlowast

=

[[[

[


lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871


lowast120588 minus 120590119878

lowast

0 120590119868lowast


]]]

]

(48)


Jlowast= [

[

minusA 120573 0


minus (] + 120572)

0 120590119868lowast

0

]

]

(49)


lowast) + 120583

Now we have



(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868


(51)



997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868



2(] + 120590119868

lowast) +A120573 (A minus 120583)


lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868


+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)


(53)


= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2



(54)

Since 0 lt A = (120587(119871)119871)1015840


minus(120587(119871)119871)1015840














0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0

0 50 100 150 200 250 300

Time

Popu

latio

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0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












1198601= [

minus120587 (0) minus 120583 120573

120587 (0) minus]] (43)


trace (1198601) = minus (] + 120583 + 120587 (0)) lt 0


(44)



119869 =

[[[

[


1205871015840() + 120587 () minus] 120588 minus 120590119878


]]]

]

(45)



1198691= [

[


1205871015840() + 120587 () minus]

]

]

(46)

Note that



(47)

where 119860 = 1205871015840()119871 + 120587() + 120583


lowast (] + 120572 +




Jlowast

=

[[[

[


lowast) minus 120583 120573 0

1205871015840(119871lowast) 119871lowast+ 120587 (119871


lowast120588 minus 120590119878

lowast

0 120590119868lowast


]]]

]

(48)


Jlowast= [

[

minusA 120573 0


minus (] + 120572)

0 120590119868lowast

0

]

]

(49)


lowast) + 120583

Now we have



(50)

Further we have

Jlowast

11= 120590119868lowast(] + 120572)

Jlowast

22= 0

Jlowast

33= A (] + 120590119868


(51)



997904rArr

3

sum

119896=1

J119896119896

= 120590119868lowast(] + 120572) +A (] + 120590119868


T3

sum

119896=1

Jlowast

119896119896= minus (A + 120590119868

lowast)

times [120590119868lowast(] + 120572) +A (] + 120590119868



2(] + 120590119868

lowast) +A120573 (A minus 120583)


lowast)A (] + 120590119868

lowast)

+ (] + 120590119868lowast) 120573 (A minus 120583)

(52)

Finally

D minusT3

sum

119896=1

Jlowast

119896119896= A2(] + 120590119868


+ (] + 120590119868lowast) 120590119868lowast(] + 120572)

+ (] + 120590119868lowast)A (] + 120590119868

lowast)


(53)


= A2(] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572) +A(] + 120590119868

lowast)2



(54)

Since 0 lt A = (120587(119871)119871)1015840


minus(120587(119871)119871)1015840














0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0

0 50 100 150 200 250 300

Time

Popu

latio

n



0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

S

I









119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


180

160

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120

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References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra















0lt 1




119881lowast= + 119878 + 119868




(55)

Now define


Further as we have


997904rArr 119878 =

120583119871 + (] + 120572) 119868

120573 minus ]

(57)


120583119871 + (] + 120572) 119868 = 0 (58)





Define

119864 = (119871 119878 119868) | minus120583119871 minus (] + 120572) 119868 = 0 (60)

Since we have

minus120583119871 minus (] + 120572) 119868 = 0 997904rArr 119871 =

minus (] + 120572) 119868

120583

997904rArr 119871 le 0

(61)




]119871 + 120573(119878 + 119868) Then

119881lowast= ] + 120573 ( 119878 + 119868)





le (120573 minus ]) 120587 (119871) 119871 minus ]120583119871

le ((120573 minus ]) 120587 (0) minus ]120583) 119871 le 0


(62)

Define


Since



(64)








∙= 120595119878 and 119868

∙= 120595119868 This implies

119871 =

119871∙

120595

119878 =

119878∙

120595

119868 =

119868∙

120595

(66)



1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





L

S

I

20

18

16

14

12

10

8

6

4

2

0

0 50 100 150 200 250 300

Time

Popu

latio

n



0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

S

I









119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


180

160

140

120

100

80

60

40

20

00 100 200 300 400 500

S

L






References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1198711015840

∙

120595

= 120573

119878∙

120595

minus 120585 (1 minus 119871∙)

119871∙

120595

minus

120583

120595

119871∙

1198781015840

∙

120595

= 120585 (1 minus 119871∙)

119871∙

120595

minus

]120595

119878∙+

120588

120595

119868∙minus

120590

1205952119878∙119868∙

1198681015840

∙

120595

=

120590

1205952119878∙119868∙minus (] + 120572 + 120588)

119868∙

120595

(67)


1198711015840

∙= 120573119878∙minus 120585 (1 minus 119871

∙) 119871∙minus 120583119871∙

1198781015840

∙= 120585 (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120590

120595

119878∙119868∙

1198681015840

∙=

120590

120595

119878∙119868∙minus (] + 120572 + 120588) 119868

∙

(68)



1

120585

1198711015840

∙=

120573119878∙minus (1 minus 119871

∙) 119871∙minus 120583119871∙

1

120585

1198781015840

∙= (1 minus 119871

∙) 119871∙minus ]119878∙+ 120588119868∙minus

120595

119878∙119868∙

1

120585

1198681015840

∙=

120595

119878∙119868∙minus (] + + 120588) 119868

∙

(69)





1198681015840= 120590119878119868 minus (] + 120572 + 120588) 119868

(70)



∙ 119878∙











1


120587 () =

120583]120573 minus ]

(71)

this implies

1 minus =

120583]120573 minus ]

(72)

which gives

= 1 minus

120583]120573 minus ]

(73)


119878 =

120583

120573 minus ]

=

120583


120583]120573 minus ]

)

(74)


0 lt 120583] lt 120573 minus ] (75)


119878lowast=

] + 120572 + 120588

120590

(76)


119868lowast=


] + 120572

=


] + 120572

(77)






lowastlt 1 In

this case we get

119871lowast2




where

1198701=


2

1198702=


2

(82)




lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





L

S

I

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18

16

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12

10

8

6

4

2

0

0 50 100 150 200 250 300

Time

Popu

latio

n



0001 120572 = 001

700

600

500

400

300

200

100

0

0 100 200 300 400 500 600 700 800

Time

Popu

latio

n

L

S

I









119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


180

160

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120

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References




























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












lowastminus 1198702




1


1lt 119871lowastlt 1198702



1




1198701and 119870





(1 minus 119871) 119871 + 120583119871 = 120573119878lowast (84)


1198712minus 119871 (120583 + 1) + 120573119878

lowast= 0 (85)


1198971=


lowast

2

(86)

1198972=

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowast

2

(87)



1lt =



lowastlt 2 (88)



lowast (89)




1 + 1205832minus 2120583 lt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (90)

This implies that

120573119878lowastlt 120583 (91)


2lt 1 from (89)

we get

(1 + 120583) + radic(1 + 120583)2minus 4120573119878

lowastlt 2 (92)



lowast (93)



1 + 1205832minus 2120583 gt 1 + 120583

2+ 2120583 minus 4120573119878

lowast (94)





2is unstable

whenever they exist



119869 =[[

[



0 120590119868 120590119878 minus (] + 120572 + 120588)

]]

]

(95)


119869lowast=[[

[




0 120590119868lowast

0

]]

]

(96)


119869lowast=[[

[

minus120601 (119871lowast) 120573 0



0 120590119868lowast

0

]]

]

(97)


T = minus (120601 (119871lowast) + ] + 120590119868

lowast)

D = minus120590119868lowast(] + 120572) 120601 (119871

lowast)

(98)

Observe that

120601 (1198971) = (1 minus 2119897

1) + 120583


lowast] + 120583

(99)

This implies

120601 (1198971) = radic(1 + 120583)

2minus 4120573119878

lowastgt 0 (100)

Similarly

120601 (1198972) = minusradic(1 + 120583)

2minus 4120573119878

lowastlt 0 (101)


1



At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0001 120572 = 001

700

600

500

400

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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Algebra










At 119871lowast



D minusT3

sum

119896=1

119869119896119896

= 1206012(119871lowast) (] + 120590119868

lowast) + (] + 120590119868

lowast) 120590119868lowast(] + 120572)

+ 120601 (119871lowast) (] + 120590119868

lowast)2minus 120573 (120601 (119871


times [120601 (119871lowast) + ] + 120590119868

lowast]

(102)

This implies

D minusT3

sum

119896=1

119869119896119896

gt 120601 (119871lowast) (] + 120590119868

lowast)2


lowast) + ] + 120590119868

lowast]

(103)


2is


D minusT3

sum

119896=1

119869119896119896

gt minus120573 (120601 (1198971) minus 120583) [120601 (119897

1) + ] + 120590119868

lowast] (104)


120601 (1198971) minus 120583 lt 0

997904rArr radic(1 + 120583)2minus 4120573119878

lowastlt 120583

997904rArr 1 + 1205832+ 2120583 minus 4120573119878

lowastlt 1205832

(105)


1 + 2120583 lt 4120573119878lowast

997904rArr 120583 lt


2

(106)



2minus 1

997904rArr 1205832gt 0

(107)




1198781015840= (1 minus 119871) 119871 minus ]119878

(108)



(1 minus 119871) minus]] [119871

119878] = 0 (109)




(] minus 120573) (1 minus 119871) + 120583] = 0 (111)

This implies

120583] = (120573 minus ]) (1 minus 119871)

120583

(1 minus 119871)

=

120573

]minus 1

(112)



119869 = [minus (1 minus 2119871





lowast) + 120583 + ]]


lowast)




(114)


D = (120573 minus ]) 119871 gt 0 (115)



lowastlt (12)(1 + 120583 + ])


lowastgt



lowastgt (12)(1 + 120583 + ]) then 119871

lowast








[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0001 120572 = 001

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600

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400

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0 100 200 300 400 500 600 700 800

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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Volume 2014




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Journal of


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Algebra











[0 minus120574

120574 0] (116)


(]120573)119871 This implies

1198811015840= (1 minus 119871) 119871 minus ]119878 +

]120573



]120573

(1 minus 119871) 119871 minus

120583]120573

119871

(117)

This implies

1198811015840=

119871

120573


Similarly

1198711015840= 120573(119881 minus

]120573



(119)

This implies


Let 119906 = (120573120574)119881 So we have

1199061015840=

120573

120574

(

119871

120573


or

1199061015840=

119871

120574



(122)



matrix is given by

119869 =[[

[

0

1

120574


120574


120574 minus [(] + 120583) + (1 minus 2119871)]

]]

]

(123)


119869 =[[

[

0 minus

119871lowast

120574

(120573 minus ])


]]

]

(124)




119869 =[[

[

0 minus

119871

120574

(120573 minus ])

120574 0

]]

]

(125)


which gives

120574 = radic119871(120573 minus ]) gt 0 (126)


1198651=

119871

120574



(127)


119886 = 120574 [1198651

119906119906119906+ 1198651

119906119871119871+ 1198652

119906119906119871+ 1198652

119871119871119871] + 1198651

119906119871(1198651

119906119906+ 1198651

119871119871)

minus 1198652

119906119871(1198652

119906119906+ 1198652

119871119871) minus 1198651

1199061199061198652

119906119906+ 1198651

1198711198711198652

119871119871

(128)


1198651

119906119906119906= 0 119865

1

119906119871119871= 0 119865

2

119906119906119871= 0 119865

2

119871119871119871= 0

1198651

119906119871= 0 119865

2

119906119871= 0 119865

1

119906119906= 0

1198651

119871=

1

120574


119871

120574


997904rArr 1198651

119871119871= minus

1

120574


120574


2

120574

(120573 minus ])

997904rArr 1198652


997904rArr 1198652

119871119871= 2

(129)


119886 = minus

4

120574

(120573 minus ]) (130)


6 Discussion





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0001 120572 = 001

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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Volume 2014




Journal of











Journal of


Function Spaces






Algebra










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0001 120572 = 001

700

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0 100 200 300 400 500 600 700 800

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119871lowast=

120573

120590120583

(] + 120572 + 120588) =

120573

120583

119878lowast (131)


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Volume 2014




Journal of











Journal of


Function Spaces






Algebra






























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









research article analysis of a mathematical model of

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