Mathematical Modelling of the Transmission Dynamics of ...downloads.hindawi.com/journals/jam/2019/2490313.pdf · JournalofAppliedMathematics S V E I R Q V S Λ E kQ Q qI ( t + r)I

Research ArticleMathematical Modelling of the Transmission Dynamics ofContagious Bovine Pleuropneumonia with Vaccination andAntibiotic Treatment

Achamyelesh Amare Aligaz and Justin ManangoW Munganga

Department of Mathematical Sciences University of South Africa South Africa

Correspondence should be addressed to Justin Manango W Munganga jmwmungangagmailcom

Received 18 September 2018 Revised 16 December 2018 Accepted 3 January 2019 Published 3 February 2019

Academic Editor Urmila Diwekar

Copyright copy 2019 Achamyelesh Amare Aligaz and Justin Manango W Munganga This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution and reproduction in any mediumprovided the original work is properly cited

In this paper we present a mathematical model for the transmission dynamics of Contagious Bovine Pleuropneumonia (CBPP)by considering antibiotic treatment and vaccination The model is comprised of susceptible vaccinated exposed infectiouspersistently infected and recovered compartments We analyse the model by deriving a formula for the control reproductionnumberR119888 and prove that forR119888 lt 1 the disease free equilibrium is globally asymptotically stable thus CBPP dies out whereasfor R119888 gt 1 the unique endemic equilibrium is globally asymptotically stable and hence the disease persists Thus R119888 = 1 actsas a sharp threshold between the disease dying out or causing an epidemic As a result the threshold of antibiotic treatment is120572lowast119905 = 01049 Thus without using vaccination more than 8545 of the infectious cattle should receive antibiotic treatment orthe period of infection should be reduced to less than 815 days to control the disease Similarly the threshold of vaccination is120588lowast = 00084 Therefore we have to vaccinate at least 80 of susceptible cattle in less than 495 days to control the disease Usingboth vaccination and antibiotic treatment the threshold value of vaccination depends on the rate of antibiotic treatment 120572119905 andis denoted by 120588120572119905 Hence if 50 of infectious cattle receive antibiotic treatment then at least 50 of susceptible cattle should getvaccination in less than 738 days in order to control the disease

1 Introduction

Contagious Bovine Pleuropneumonia (CBPP) is a majorconstraint to cattle production in the key pastoral regionsof Africa (see [1ndash3] for more details) It is caused byMycoplasma mycoides subspecies mycoides (Mmm) thatattacks the lungs and the membranes of cattle and waterbuffalo It is transmitted by direct contact between an infectedand a susceptible animal which becomes infected by inhalingdroplets disseminated by coughing It causes high morbidityand mortality losses to cattle which leads to economic crisis(see [4ndash7] for more details) Cost of control of CBPP is alsoa major concern in African countries [6 8] Since someanimals can carry the diseasewithout showing signs of illnesscontrolling the spread is more difficult In many countriesin sub-Saharan Africa CBPP control is based on vaccinationalone but this strategy does not eradicate the disease [9]

In [10] we presented and analysed a five-compartmentalmathematical model of the transmission dynamics of CBPPwithout any intervention having the objective of identifyingparameters that have significant role in changing the dynam-ics of the diseaseAs a result fromelasticity analysis we foundthat the effective contact rate 120573 and the rate of recovery 120572119903are the top two parameters that control the dynamics of thedisease in such a way that as the value of 120573 decreases andthe value of 120572119903 increasesR0 decreases and can be made lessthan one as a result the disease can be controlled Howeverwe know that vaccination is one of the ways of reducing theeffective contact rate (120573) and antibiotic treatment is one wayof reducing infection by increasing the recovery rate

Thus in this paper we consider vaccination and antibi-otic treatment as a controlling tool of CBPP and presenta compartmental model for the transmission dynamics ofCBPP containing six compartments susceptible vaccinated

HindawiJournal of Applied MathematicsVolume 2019 Article ID 2490313 10 pageshttpsdoiorg10115520192490313

2 Journal of Applied Mathematics

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Figure 1 A compartmental model for the transmission dynamics of CBPP with antibiotic treatment and vaccination

exposed infectious persistently infected and recovered com-partments Antibiotic treatment is considered in the modelby incorporating rate of recovery of treated cattle suchthat treated cattle move from infectious compartment torecovered compartment at a rate of 120572119905

The objective of this paper is to determine the bettercontrol method out of vaccination antibiotic treatment anda combination of both We derive the formula for the controlreproduction number R119888 and determine the number ofcattle to be vaccinated and treated independently and incombination which will enable us to choose the feasibleand effective controlling method in our context Numericalsimulations are performed using MATLAB

This paper is structured as follows In Section 2 wepresent a mathematical model of the dynamics of CBPPwith vaccination and antibiotic interventions In Section 3we prove the well-posedness of the model We calculateequilibria of the system and rigorously derive a formula of thecontrol reproduction numberR119888 in Section 4 Stability anal-ysis of the DFE and EE is presented in Section 5 we presentparameter values andnumerical simulations in Section 6 andlastly we draw the conclusions and remarks in Section 7

2 Mathematical Model

We model the transmission dynamics of Contagious BovinePleuropneumonia (CBPP) In this model we assume inter-vention by vaccination and antibiotic treatment Thus thecompartmental model is consisting of susceptible vacci-nated exposed infectious persistently infected and recov-ered classes as shown in Figure 1 We assume an openpopulation with a total number 119873 at time 119905 where allnewborn animals are born into susceptible class (119878) at rate119887 Susceptible cattle move to vaccinal immune class (119881) ata rate 120588 Cattle in vaccinal immune class can lose vaccinalimmunity and return back to susceptible class at a rate120596 Susceptible animals move to the exposed compartment(119864) at a rate 120573(119868119873) Cattle in the exposed compartmentmove to the infectious compartment (119868) at a rate 120574 Naturalmortality occurs at a rate 120583 and results in losses from all sixcompartments However we assume that death due to thedisease does not occur The infectious cattle either naturallyheal or receive antibiotic treatment and enter directly into therecovered (119877) compartment at a rate 120572119903 and 120572119905 respectively

or they pass through a process of sequestration and enterinto persistently infected (119876) compartment at a rate 120572119902Cattle in persistently infected compartment are encapsulatedand infected but not infectious As sequestra resolve andorbecome noninfected then the animals in persistently infectedcompartment move to the recovered (119877) compartment ata rate 120595 Recovered cattle remain recovered for life timeInfected sequestra can occasionally be reactivated and inthis instance the animal will transition from the persistentlyinfected (119876) compartment back to the infectious (119868) compart-ment at a rate 119896 We assume randommixing of all individualsin the population The total number of population at time 119905119873 is given by119873 = 119878(119905) + 119881(119905) + 119864(119905) + 119868(119905) + 119876(119905) + 119877(119905) Theflow diagram of the model is shown in Figure 1

The differential equationmodel is given by system (1)ndash(7)

119889119878119889119905 = 120583119873 + 120596119881 minus 120573119878119868119873 minus 120588119878 minus 120583119878 (1)

119889119881119889119905 = 120588119878 minus 120596119881 minus 120583119881 (2)

119889119864119889119905 = 120573119878119868119873 minus 120574119864 minus 120583119864 (3)

119889119868119889119905 = 120574119864 + 119896119876 minus (120572119903 + 120572119905) 119868 minus 120572119902119868 minus 120583119868 (4)

119889119876119889119905 = 120572119902119868 minus 119896119876 minus 120595119876 minus 120583119876 (5)

119889119877119889119905 = (120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 (6)

with initial condition

(119878 (0) 119881 (0) 119864 (0) 119868 (0) 119876 (0) 119877 (0))= (1198780 1198810 1198640 1198680 1198760 1198770) (7)

3 Well-Posedness of the System

Let119883(119905) = (119878(119905) 119881(119905) 119864(119905) 119868(119905) 119876(119905) 119877(119905)) and119891 Ω 997888rarr Y

119883 997891997888rarr 1198831015840 (8)

Journal of Applied Mathematics 3

provided that Y sube 1198776 Ω is a compact subset of 1198776 suchthat Ω = 119909 = (119878(119905) 119881(119905) 119864(119905) 119868(119905) 119876(119905) 119877(119905)) isin 1198776 119878(119905) + 119881(119905) + 119864(119905) + 119868(119905) + 119876(119905) + 119877(119905) le 1198730 = 119873(0) and119891 = (1198911 1198912 1198913 1198914 1198915 1198916) where

1198911 (119883) = 119889119878119889119905 = 120583119873 + 120596119881 minus 120573119878119868119873 minus 120588119878 minus 120583119878 (9)

1198912 (119883) = 119889119881119889119905 = 120588119878 minus 120596119881 minus 120583119881 (10)

1198913 (119883) = 119889119864119889119905 = 120573119878119868119873 minus 120574119864 minus 120583119864 (11)

1198914 (119883) = 119889119868119889119905 = 120574119864 + 119896119876 minus (120572119903 + 120572119905) 119868 minus 120572119902119868 minus 120583119868 (12)

1198915 (119883) = 119889119876119889119905 = 120572119902119868 minus 119896119876 minus 120595119876 minus 120583119876 (13)

1198916 (119883) = 119889119877119889119905 = (120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 (14)

Then (9)ndash(14) can be written of the form

1198831015840 (119905) = 119891 (119883 (119905)) 119883 (0) = (1198780 1198810 1198640 1198680 1198760 1198770) isin Ω (15)

Theorem 1 System (1)ndash(6) has a unique solution X(t) whichis positive and bounded if 119891 is given by (15) and the initialconditions119883(0) = (1198780 1198810 1198640 1198680 1198760 1198770) is nonnegativeProof See [10]

4 Equilibria and ControlReproduction Number

41 Equilibria of the System

Proposition 2 Model (1)-(6) has at least two equilibriumpoints the disease free equilibrium and at least one endemicequilibrium

Proof The equilibria of the system are obtained by solvingequations

119889119878119889119905 = 119889119881119889119905 = 119889119864119889119905 = 119889119868119889119905 = 119889119876119889119905 = 119889119877119889119905 = 0 (16)

From (9)ndash(14) we have

120583119873 + 120596119881 minus 119878(120573119868119873 + 120588 + 120583) = 0 (17)

120588119878 minus 119881 (120596 + 120583) = 0 (18)

120573119878119868119873 minus 119864 (120574 + 120583) = 0 (19)

120574119864 + 119896119876 minus 119868 (120572119903 + 120572119905 + 120572119902 + 120583) = 0 (20)

120572119902119868 minus 119876 (119896 + 120595 + 120583) = 0 (21)

(120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 = 0 (22)

From (21)

119876 = 120572119902119868119896 where 119896 = 119896 + 120595 + 120583 (23)

Putting (23) into (20) yields

119864 = (119896120572 minus 119896120572119902) 119868120574119896 where 120572 = 120572119903 + 120572119905 + 120572119902 + 120583 (24)

And putting (23) into (22) we find that

119877 = ((120572119903 + 120572119905) 119896 + 120595120572119902) 119868120583119896 (25)

Similarly putting (24) into (19) gives

120573119878119868119873 minus 120574 (119896120572 minus 119896120572119902) 119868120574119896 = 0 lArrrArr (26)

120573119868( 119878119873 minus 120574 (119896120572 minus 119896120572119902)120573120574119896 ) = 0 (27)

Then we have the following two cases for solution of (27)

Case 1 If 120574(119896120572minus119896120572119902)120573120574119896 ge 1 then 119868 = 0 is the only solutionof (27) and

Case 2 If 120574(119896120572 minus 119896120572119902)120573120574119896 lt 1 then 119868 = 0 or 119878 = 120574(119896120572 minus119896120572119902)119873120573120574119896 are the solutions of (27)For Case 1 when 119868 = 0 = 1198680 let 119878 = 1198780 119881 = 1198810 119864 = 1198640 119876 =1198760 and 119877 = 1198770 for (17)ndash(22) Then from (23)ndash(25) we have

1198760 = 1198640 = 1198770 = 0 (28)

And from (17) and (18)

1198810 = 120588119873120588 + 120596 (29)

and

1198780 = 120596119873120588 + 120596 (30)

Therefore from (28)ndash(30) 1198830 = (1198780 1198810 1198640 1198680 1198760 1198770) =(120596119873(120588 + 120596) 120588119873(120588 + 120596) 0 0 0 0) is the disease free equi-librium (DFE)

ForCase 2 we are donewhen 119868 = 0Andwhen 119878 = 120574(119896120572minus119896120572119902)119873120573120574119896 = 119878lowast let 119881 = 119881lowast 119864 = 119864lowast 119868 = 119868lowast 119876 = 119876lowast and119877 = 119877lowast for (17)ndash(22) Then (18) gives that

119881lowast = 120588120574 (119896120572 minus 119896120572119902)119873120596120573120574119896 (31)


Finally putting 119878lowast and 119881lowast into (17) we get119868lowast = (120573120574119896120596 + 120574 (120588 + 120596) (119896120572 minus 119896120572119902)) 120583119873

120573120596120574 (119896120572 minus 119896120572119902) (32)

Hence 119883lowast = (119878lowast 119881lowast 119864lowast 119868lowast 119876lowast 119877lowast) is an endemic equilib-rium (EE) where 119864lowast 119868lowast 119876lowast 119877lowast are as in (24) (32) (23) and(25) respectively

42 The Control Reproduction Number (R119888) Due to thepresence of control measures we will use the term controlreproduction number (R119888) instead of the commonly usedbasic reproduction number (R0) As explained in [12] weuse the next generation matrix to calculate the control repro-duction number Compartments 119864 119868 and 119876 are consideredto be the disease compartments and 119878 119881 and 119877 are thenondisease compartments We set F = (F1F2F3)119879 andV = (V1V2V3)119879 where F119894 represents the rate of newinfections in the 119894119905ℎ disease compartment V+119894 being thetransfer rate of individuals into compartment 119894 by all othermeans whileVminus119894 represents the transfer rate of individual outof compartment 119894 Assuming1198830 to be the DFE we have

F = [[[[

12057311987811986811987300]]]]

V =Vminus minusV

+ = [[[[

120574119864120572119868 minus 120574119864 minus 119896119876119896119876 minus 120572119902119868

]]]]

119865 = [120597F119894120597119909119895 (1198830)] =[[[[[

0 1205731198780119873 00 0 00 0 0

]]]]]

119881 = [120597V119894120597119909119895 (1198830)] =[[[[

120574 0 0minus120574 120572 minus1198960 minus120572119902 119896

]]]]

and

119881minus1 =[[[[[[[[[

1120574 0 0120574119896

120574 (119896120572 minus 119896120572119902)119896

119896120572 minus 119896120572119902119896

119896120572 minus 119896120572119902120572119902120574120574 (119896120572 minus 119896120572119902)

120572119902119896120572 minus 119896120572119902

120572119896120572 minus 119896120572119902

]]]]]]]]]

(33)

Therefore

119865119881minus1

= 1205731198780119873[[[[[[

120574119896120574 (119896120572 minus 119896120572119902)

119896119896120572 minus 119896120572119902

119896119896120572 minus 1198961205721199020 0 0

0 0 0

]]]]]] (34)

ThereforeR119888 = 120588(119865119881minus1) = 1198792+radic(1198792)2 minus 119863 where119879 and119863 are trace and determinant of thematrix 119865119881minus1 Since119863 = 0R119888 = 119879 = 120573120596120588 + 120596 ( 120574119896

120574 (119896120572 minus 119896120572119902)) (35)

Equivalently

R119888 = ( 120596120588 + 120596)(119896 (120572 minus 120572119905) minus 119896120572119902

119896120572 minus 119896120572119902 )R0 (36)

whereR0 = 120573120574119896120574(119896(120572minus120572119905) minus 119896120572119902) is the basic reproductionnumber as derived in [10] and 120596(120588 + 120596) is the proportion ofcattle that survive the vaccination class and the control repro-duction number R119888 is the average number of secondarycases caused by an infected individual over the course ofinfectious period in the presence of vaccination and antibiotictreatment We observe thatR119888 ltR0

5 Stability Analysis

51 Stability Analysis of the Disease Free Equilibrium (DFE)

511 Local Stability Analysis of the DFE

Theorem 3 (see [12]) If 1198830 is a DFE of the model given by(1)ndash(7) then 1198830 is locally asymptotically stable ifR119888 lt 1 andunstable ifR119888 gt 1 whereR119888 is defined by (36)Proof See [12]

512 Global Stability Analysis of the DFE

Theorem4 IfR119888 lt 1 then disease free equilibrium (120596119873(120596+120588) 120588119873(120596+120588) 0 0 0 0) is globally asymptotically stable inΩIf R119888 gt 1 then the DFE is unstable the system is uniformlypersistent and there is at least one equilibrium in interior ofΩ where Ω = (119878 119881 119864 119868 119876 119877) isin 1198776+ 119878 ge 0 119881 ge 0 119864 ge0 119868 ge 0 119876 ge 0 119877 ge 0 119878 + 119881 + 119864 + 119868 + 119876 + 119877 le 1198730 =119873(0)Proof We use matrix-theoretic method as explained in [13]We assume 119909 = (119864 119868 119876)119879 and 119910 = (119878 119881 119877)119879 Andconsidering 119865 119881 and 119881minus1 as in Section 42 we set

119891 (119909 119910) = (119865 minus 119881) 119909 minusF (119909 119910) +V (119909 119910)

= [[[[[

1205731198780119868119873 minus 12057311987811986811987300

]]]]]= 120573119868119873

[[[[[

( 120596119873120596 + 120588 minus 119878)00

]]]]] (37)

where 1198780 = 120596119873(120596 + 120588) is as in (30)


And

119881minus1119865 = 1205731198780119873[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(38)

= 120573120596120596 + 120588[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(39)

= 120573120574120596120574 (120596 + 120588)[[[[[[[[[

0 1120574 00 119896(119896120572 minus 119896120572119902) 0

0 120572119902(119896120572 minus 119896120572119902) 0

]]]]]]]]]

(40)

We observe that 119865 ge 0 119881minus1 ge 0 and 119891(119909 119910) ge 0 when 120588 = 0and119891(119909 (120596119873(120596+120588) 120588119873(120596+120588) 0)119879) = 0 inΩ Sincematrix119881minus1119865 is reducible we use Theorem 21 of [13] to constructa Lyapunov function Let 120596119879 = (V1 V2 V3) ge 0 be the lefteigenvector of nonnegativematrix119881minus1119865 corresponding to theeigenvalueR119888 Then

(V1 V2 V3) 119881minus1119865 = 119877119888 (V1 V2 V3) (41)

such that

(V1 V2 V3) 119881minus1119865 = ( 120573120596120574(120596 + 120588) 120574) (0 119909 0) where (42)

119909 = V1120574 + 119896V2(119896120572 minus 119896120572119902) +120572119902V3

(119896120572 minus 119896120572119902) (43)

and

119877119888 (V1 V2 V3) = 120573120596120574(120596 + 120588) 120574 ( 119896120572119896 minus 119896120572119902)(V1 V2 V3) (44)

Thus from (41)-(44) we find that V1 = V3 = 0 and V2 isin 119877+Hence 120596119879 = (0 V2 0) By Theorem 21 of [13]

119871 = 120596119879119881minus1119909= V2( 120574119896119864

120574 (119896120572 minus 119896120572119902) +119896119868

119896120572 minus 119896120572119902 +119896119876

119896120572 minus 119896120572119902)(45)

is the Lyapunov function for the system when R119888 lt 1Since 1198711015840 = (R119888 minus 1)120596119879119909 minus 120596119879119881minus1119891(119909 119910) = 0 implies that119909 = 0 and 119910 = (120596119873(120596 + 120588) 120588119873(120596 + 120588) 0)119879 it follows that

(120596119873(120596 + 120588) 120588119873(120596 + 120588) 0 0 0 0) is the only invariant setin Ω when 119909 = 0 and 119910 = (120596119873(120596 + 120588) 120588119873(120596 + 120588) 0)119879Thus by LaSallersquos invariance principle the DFE (120596119873(120596 +120588) 120588119873(120596 + 120588) 0 0 0 0) is globally asymptotically stable inΩ when R119888 lt 1 If R119888 gt 1 then 1198711015840 gt 0 for 119909 = 0 and119910 = (120596119873(120596 + 120588) 120588119873(120596 + 120588) 0)119879 Hence by continuity1198711015840 gt 0 in the neighbourhood of the DFE implies that theDFE is unstable whenR119888 gt 1 Instability of the DFE impliesuniform persistence of (1)ndash(7) Uniform persistence and thepositive invariance of the compact set Ω imply the existenceof a unique EE of (1)ndash(7)

52 Global Stability Analysis of the Endemic Equilibrium (EE)

Theorem 5 If R119888 gt 1 then the endemic equilibrium 119883lowastof (1)ndash(7) is unique and globally asymptotically stable in theinterior of ΩProof We use a graph-theoretic method as explained in [13]Thus for construction of a Lyapunov function set 1198631 =119878 minus 119878lowast minus 119878lowast ln(119878119878lowast) 1198632 = 119881 minus 119881lowast minus 119881lowast ln(119881119881lowast) 1198633 =119864 minus 119864lowast minus 119864lowast ln(119864119864lowast) 1198634 = 119868 minus 119868lowast minus 119868lowastln(119868119868lowast) 1198635 =119876 minus 119876lowast minus 119876lowastln(119876119876lowast) and 1198636 = 119877 minus 119877lowast minus 119877lowastln(119877119877lowast)And putting (119878lowast 119881lowast 119864lowast 119868lowast 119876lowast 119877lowast) into (17)ndash(22) we findthat 120583119873 = 120573119878lowast119868lowast119873 + 119878lowast(120588 + 120583) minus 120596119881lowast 120596 + 120583 = 120588119878lowast119881lowast120574 + 120583 = 120573119878lowast119868lowast119873119864lowast 120572119903 + 120572119902 + 120583 = (120574119864lowast + 119896119876lowast)119868lowast119896+120595+120583 = 120572119902119868lowast119876lowast and 120583 = (120572119903119868lowast +120595119876lowast)119877lowast We use theseequalities and the inequality 1minus119909+ ln 119909 le 0 in differentiationof1198631 1198632 1198633 1198634 1198635 and1198636 with respect to 119905 as follows

11986310158401 = (1 minus 119878lowast119878 )(120583119873 + 120596119881 minus 120573119878119868119873 minus 119878 (120588 + 120583))= (1 minus 119878lowast119878 )(120573119878

lowast119868lowast119873 + 119878lowast (120588 + 120583) minus 120596119881lowast + 120596119881minus 120573119878119868119873 minus 119878 (120588 + 120583)) = 120596119881lowast (1 minus 119878lowast119878 ) ( 119881119881lowast minus 1)+ 119878lowast (120588 + 120583) (1 minus 119878lowast119878 ) (1 minus 119878119878lowast ) + 120573119878lowast119868lowast119873 (1minus 119878lowast119878 ) (1 minus 119878119868119878lowast119868lowast ) = 120596119881lowast ( 119881119881lowast minus 119878lowast119881119878119881lowast + 119878lowast119878minus 1) + (120588 + 120583) (120596 + 120583)119881lowast

120588 (1 minus S119878lowast minus 119878lowast119878 + 1)+ 120573119878lowast119868lowast119873 (1 minus 119878119868119878lowast119868lowast minus 119878lowast119878 + 119868119868lowast)le (120588 + 120583) (120596 + 120583)119881lowast120588 ( 119881119881lowast minus 119878lowast119881119878119881lowast minus 119878119878lowast + 1)+ 120573119878lowast119868lowast119873 (1 minus 119878119868119878lowast119868lowast minus 119878lowast119878 + 119868119868lowast)le (120588 + 120583) (120596 + 120583)119881lowast120588 ( 119881119881lowast minus 119878119878lowast minus ln(119878lowast119878 )


1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31

Figure 2 The weighted digraph (119866 119860) constructed for model (1)ndash(6)

minus ln( 119881119881lowast )) + 120573119878lowast119868lowast119873 (1 minus 119878119868119878lowast119868lowast minus 119878lowast119878 + 119868119868lowast)š 1198861211986612 + 1198861411986614

11986310158402 = (1 minus 119881lowast119881 ) (120588119878 minus (120596 + 120583)119881) = (1 minus 119881lowast119881 )(120588119878minus 120588119878lowast119881119881lowast ) = 120588119878lowast (1 minus 119881lowast119881 )( 119878119878lowast minus 119881119881lowast )= 120588119878lowast ( 119878119878lowast minus 119881119881lowast minus 119881lowast119878119881119878lowast + 1) le 120588119878lowast ( 119878119878lowast minus 119881119881lowast+ ln 119881119881lowast + ln 119878lowast119878 ) š 1198862111986621

11986310158403 = (1 minus 119864lowast119864 )(120573119878119868119873 minus (120574 + 120583) 119864) = (1 minus 119864lowast119864 )sdot (120573119878119868119873 minus 120573119878lowast119868lowast119864119873119864lowast ) = (1 minus 119864lowast119864 )(120573119878lowast119868lowast119873 )( 119878119868119878lowast119868lowastminus 119864119864lowast ) = 120573119878lowast119868lowast119873 ( 119878119868119878lowast119868lowast minus 119864119864lowast minus 119864lowast119878119868119864119878lowast119868lowast + 1)le 120573119878lowast119868lowast119873 ( 119878119868119878lowast119868lowast minus 119864119864lowast + ln 119864119864lowast minus ln 119878119868119878lowast119868lowast )= 120573119878lowast119868lowast119873 ( 119878119868119878lowast119868lowast minus ln 119878119868119878lowast119868lowast minus 119868119868lowast + ln 119868119868lowast)+ 120573119878lowast119868lowast119873 (minus 119864119864lowast + ln 119864119864lowast + 119868119868lowast minus ln 119868119868lowast) š 1198863111986631+ 1198863411986634

11986310158404 = (1 minus 119868lowast119868 ) (120574119864 + 119896119876 minus 119868 (120572119903 + 120572119902 + 120583)) = (1minus 119868lowast119868 )(120574119864 + 119896119876 minus 119868 (120574119864lowast + 119896119876lowast)

119868lowast ) = 120574119864lowast ( 119864119864lowastminus 119868119868lowast minus 119868lowast119864119868119864lowast + 1) + 119896119876lowast ( 119876119876lowast minus 119868119868lowast minus 119868lowast119876119868119876lowast + 1)le 120574119864lowast ( 119864119864lowast minus 119868119868lowast + ln 119868119868lowast minus ln 119864119864lowast ) + 119896119876lowast ( 119876119876lowastminus 119868119868lowast + ln 119868119868lowast minus ln 119876119876lowast) š 1198864311986643 + 1198864511986645

11986310158405 = (1 minus 119876lowast119876 ) (120572119902119868 minus 119876 (119896 + 120595 + 120583)) = (1 minus 119876lowast119876 )

sdot (120572119902119868 minus 119876(120572119902119868lowast

119876lowast )) = 120572119902119868lowast ( 119868119868lowast minus 119876119876lowast minus 119868119876lowast119868lowast119876+ 1) le 120572119902119868lowast ( 119868119868lowast minus 119876119876lowast + ln 119876119876lowast minus ln 119868119868lowast)š 1198865411986654

11986310158406 = (1 minus 119877lowast119877 ) (120572119903119868 + 120595119876 minus 120583119877) = (1 minus 119877lowast119877 )(120572119903119868

+ 120595119876 minus 119877 (120572119903119868lowast + 120595119876lowast)119877lowast ) = 120572119903119868lowast ( 119868119868lowast minus 119877119877lowastminus 119877lowast119868119877119868lowast + 1) + 120595119876lowast ( 119876119876lowast minus 119877119877lowast minus 119877lowast119876119877119876lowast + 1)le 120572119903119868lowast ( 119868119868lowast minus 119877119877lowast + ln 119877119877lowast minus ln 119868119868lowast ) + 120595119876lowast ( 119876119876lowastminus 119877119877lowast + ln 119877119877lowast minus ln 119876119876lowast) š 1198866411986664 + 1198866511986665

(46)

where 11988612 = ((120588 + 120583)(120596 + 120583)120588)119881lowast = (120588 + 120583)119878lowast 11988614 = 11988631 =11988634 = 120573119878lowast119868lowast119873 11988621 = 120588119878lowast 11988643 = 120574119864lowast 11988645 = 119896119876lowast 11988654 = 120572119902119868lowast11988664 = 120572119903119868lowast 11988665 = 120595119876lowast and all other 119886119894119895 = 0 such that theweight matrix is 119860 = [119886119894119895]6times6 where 119886119894119895 gt 0 is the weight ofarc(119895 119894) Thus the associated weighted digraph (119866 119860) for themodel given by system of (1)ndash(6) is presented in Figure 2

Along each directed cycle 11986612 + 11986621 = 0 11986634 + 11986643 = 0and 11986645 + 11986654 = 0 Therefore by Theorem 35 of [13] thereexists 119888119894 1 le 119894 le 6 such that 119881 = sum6119894=1 119888119894119863119894 is a Lyapunovfunction for (1)ndash(6) where the relations between 119888119894rsquos can bederived from Theorems 33 and 34 of [13] that 119889+(2) = 1implies 119888111988612 = 119888211988621 119889+(3) = 1 implies 119888411988643 = 1198883(11988631 + 11988634)and 119889minus(5) = 1 implies 119888511988654 = 119888411988645 + 119888611988665 Hence 1198882 =(1198861211988621)1198881 1198884 = ((11988631 + 11988634)11988643)1198883 and 1198886 = ((11988654119886431198885 minus11988645(11988631 + 11988634)1198883)1198866511988643) And 1198811015840 = sum6119894=1 1198881198941198631015840119894 = 0 implies119883 = 119883lowast Hence the largest invariance set for (1)ndash(7) where1198811015840 = 0 is the singleton set 119883lowast

Thus proving uniqueness and global asymptotic stabilityof Xlowast in interior ofΩ provided thatR119888 gt 1


Table 1 Description of model parameters and their values indicating baselines ranges and references Units are daysminus1 unless otherwisedefined lowastProportions

Parameter Description Baseline value Value range andreferences

120573 Effective contact rate 0126 007 to 013 [11]119901119890 Vaccination efficacylowast 065 05 to 08 [11]119901V Proportion vaccinatedlowast 05 [11]120598 Vaccination efficiencylowast 08 [11]119901 Proportion immunizedlowast 120598 times 119901V times 119901119890 [11]

120588 Rate of vaccination11990173days assumed

120596 Rate of loss of vaccinal immunity 13 times 365 000078 to 00011 [11]120574 Transition rate from exposed to infectious compartment 16 times 7 18 times 7 to 14 times 7 [11]

120572119903 Natural recovery rate of infectious cattle 14 times 56 14 times 142 to 14 times 170 [11]

120572119902 Rate of sequestrum formation of infectious cattle 3120572119903 [11]

120572119905 Rate of recovery of treated cattle 128 minus 156128 minus 142 to 128 minus 170

assumed

119896 Rate of sequestrum reactivation 000009 000007 minus 000011[11]120595 Rate of sequestrum resolution 00075 00068 to 00079 [11]120583 Mortality rate 15 times 365

16 times 365 to 145 times 365[11] and estimated

guess

119887 Birth rate 15 times 36516 times 365 to 145 times 365

[11] and estimated

6 Parameter Values andNumerical Simulations

61 Parameter Values Most of the parameter values used inthis paper are explained in Table 1 Sections 22 and 23 of [14]and Table 1 of [1] We assume that the life expectancy of cattleis in average 5 years then the value of 120583 and 119887 is taken to be1(5 times 365) 120573 = 0126 the incubation period between 4 and8 weeks with mean value of 6 weeks yields 120574 = 1(6 times 7)without applying antibiotic treatment the infection periodis between 6 and 10 weeks with mean value of 8 weeks and120572119902 = 3120572119903 then120572119903 = 1(4times56) the persistently infected periodgiven in a range of 18ndash21 weeks with an average period of19 weeks with 4months times 2 reactivations per month for 582cases gives 119896 = 00009 and120595 = 00075 the rate of vaccination120588 = 120598119901V119901119890119905 where 120598 is the efficiency of vaccine 119901V is theproportion vaccinated 119901119890 is efficacy of the vaccine and 119905 isthe period of vaccination and vaccinal immunity lasts for 3

years which implies that 120596 = 1(3times365) When we introduceantibiotic treatment at a rate of 120572119905 the period of infection(56 days) will be reduced to some new period 119875 such that120572119903 + 120572119905 + 120572119902 = 1119875 implies 120572119905 = 1119875 minus 156 Since R119888 = 1acts as a sharp threshold between the disease dying out orcausing an epidemic we find that the threshold of antibiotictreatment is given by 120572lowast119905 = 120573120574(R0 minus 1)(120574 + 120583)R0 = 01049where R0 is the basic reproduction number as in [10] Thisimplies that without using vaccination more than 8545 ofthe infectious cattle should receive antibiotic treatment or theperiod of infection should be reduced to less than 815 daysto control the disease And threshold of vaccination is alsogiven by 120588lowast = (120598 times 119901V times 119901119890)119905 = (120596 + 120583)(R0 minus 1) = 00084where 119905 is period of vaccination which can be interpretedthat at least 80 of susceptible cattle should get vaccinationin less than 495 days in order to control the disease howeversince the proportion to be vaccinate 119901V depends on 119905 a singlevalue of 120588 can have many practical interpretation For the last


SusceptiblesExposedInfectiousPersistently-infectedRecovered

1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

Figure 3 Number of cattle in each compartment for parametervalues in Table 1 with 120572119903 = 156 (as in [10]) and 1198680 = 1 1198780 =499 1198640 = 1198760 = 1198770 = 0 giving approximate equilibrium values119878lowast = 127 119864lowast = 12 119868lowast = 9 119876lowast = 15 119877lowast = 337 andR0 = 39399

option in applying both vaccination and antibiotic treatmentthe threshold value of vaccination depends on the rate ofantibiotic treatment 120572119905 and is given by 120588120572

119905

= (120596 + 120583)(((119896(120572 minus120572119905) minus 119896120572119902)(119896120572 minus 119896120572119902))R0 minus 1) Thus if we introduce bothantibiotic treatment and vaccination in the population suchthat 50 of infectious cattle receive antibiotic treatment orthe period of infection is reduced to 28 days then at least50 of susceptible cattle should get vaccination in less than738 days in order to control the disease Mathematically itmeans that for 120572119905 = 128 minus 156 = 156 we should take 120588 =(065 times 05 times 08)73 to makeR119888 lt 1 to control the diseaseParameter values considering both antibiotic treatment andvaccination are summarized in Table 1

(i) All parameter values used in this paper and in [10] arethe same except the value of 120572119903 which is taken in [10]as 156 instead of 1(4 times 56)

62 Numerical Simulations Initially we consider a herd sizeof 500 cattle population which is consisting of an infectiouscattle and 499 susceptible cattle with individual animals as theepidemiological units of interest For the same assumptionand parametric values the result obtained in this papercoincides with the result obtained in [10] in this case R0 =39399 as shown in Figure 3 Using parameter values inTable 1model (1)-(6) is numerically solved If no interventionis considered the population goes extinct withR0 = 67462as shown in Figure 4 Figures 5 and 6 show the number ofcattle in each compartment when we consider interventionby antibiotic treatment without vaccination and vice versarespectively in both cases R119888 lt 1 Lastly considering bothantibiotic treatment and vaccination the number of cattle in

SusceptiblesVaccinatedExposedInfectiousPersistently-infectedRecovered

1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

Figure 4 Number of cattle in each compartment with baselineparameter values in Table 1 with 120588 = 120572119905 = 0 1198680 = 1 1198780 = 499 and1198810 = 1198640 = 1198790 = 1198760 = 1198770 = 0 giving approximate equilibrium values119878lowast = 74 119864lowast = 10 119868lowast = 12 119876lowast = 21 119877lowast = 383 andR0 = 67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218

Figure 5 Number of cattle in each compartment with baselineparameter values in Table 1 with the assumption that 857 ofinfectious cattle receive antibiotic treatment within 8 days (120572119905 =18 minus 156) without vaccinating healthy cattle (120588 = 0) 1198680 = 11198780 = 499 and 1198810 = 1198640 = 1198760 = 1198770 = 0 giving approximateequilibrium values 119878lowast = 498 119864lowast = 119868lowast = 119876lowast = 0 119877lowast = 2 andR119888 = 09822


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906


Figure 6 Number of cattle in each compartment with baselineparameter values in Table 1 with the assumption that 80 ofsusceptible cattle are vaccinated within 49 days (120588 = (065 lowast 08 lowast08)49) without treating infectious cattle (120572119905 = 0) 1198680 = 1 1198780 = 499and 1198810 = 1198640 = 1198760=1198770 = 0 giving approximate equilibrium values119878lowast = 73 119881lowast = 421 119864lowast = 119868lowast = 119876lowast = 0 119877lowast = 6 andR119888 = 09906

each compartment at time 119905 is plotted in Figure 7 in this caseR119888 = 099213 for parametric values in Table 1

7 Conclusion and Remarks

In this paper we presented compartmental model and differ-ential equations for the transmission dynamics of CBPP withinterventionWe calculated the equilibriumof the system andto study the behaviour of the disease we derived a formulafor the control reproduction numberR119888 We proved that forR119888 lt 1 the DFE is globally asymptotically stable thus CBPPdies out whereas forR119888 gt 1 the EE is globally asymptoticallystable and hence the disease persists in all the populationsHence R119888 = 1 acts as a sharp threshold between the diseasedying out or causing an epidemic Without considering anyintervention the model in this paper coincides with themodel studied in [10] in both papers R0 = 39399 when120572119903 = 156 see Figure 3 Similarly when 120572119903 = 1(4 times56) which is the right value and without any interventionR0 = 67462 see Figure 4 Hence without any interventionthe disease persists in all the population However we cancontrol the disease by giving antibiotic treatment to 857of infectious cattle without vaccinating any of healthy cattlesee Figure 5 As a second option we can also control thedisease by vaccinating 80 of susceptible cattle within aperiod of 49 days without treating any of infectious cattlesee Figure 6 Finally for parametric values in Table 1 withthe assumption that 50 of susceptible are vaccinated within


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213

Figure 7 Number of cattle in each compartment with baselineparameter values in Table 1 with the assumption that 50 of infec-tious cattle receive antibiotic treatment or the period of infectionis reduced to 28 days (120572119905 = 128 minus 156) 50 of susceptible getvaccinationwithin 73 days (120588 = (05times08times065)73) 1198680 = 1 1198780 = 499and 1198810 = 1198640 = 1198790 = 1198760 = 1198770 = 0 giving approximate equilibriumvalues 119878lowast = 144 119881lowast = 350 119864lowast = 119868lowast = 119876lowast = 0 119877lowast = 6 andR119888 = 09921

a period of 73 days and 50 of infectious cattle are treatedR119888 can be made less than one and hence we can controlthe disease see Figure 7 In all the above three interventionmethodsR119888 lt 1 and hence the disease can be controlled byproperly applying the methods as explained above howeverdue to lack of awareness time and financial and logisticconstraint the first two methods do not look feasible in thecontext of developing countries Therefore we recommendthat vaccination with antibiotic treatment is the best way tocontrol the disease which is in agreement with the resultof [1] Since the proportion to be vaccinated 119901V and 119905 areindependent variables of 120588 a given value of 120588 can have manypractical interpretation Therefore practical implementationof the value of 120588 can be adjusted based on availability cost ofcontrol and time value

Data Availability

Data are available in the literature

Disclosure

An earlier version of this paper was presented at the Inter-national Conference in Mathematical Methods and Modelsin Biosciences and Young Scientists Biomath 2017 SkukuzaCamp Kruger Park South Africa


Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] J C Mariner J McDermott J A P Heesterbeek G ThomsonP L Roeder and S W Martin ldquoA heterogeneous populationmodel for contagious bovine pleuropneumonia transmissionand control in pastoral communities of East Africardquo PreventiveVeterinary Medicine vol 73 no 1 pp 75ndash91 2006

[2] J O Onono B Wieland and J Rushton ldquoEstimation ofimpact of contagious bovine pleuropneumonia on pastoralistsin Kenyardquo Preventive Veterinary Medicine vol 115 no 3-4 pp122ndash129 2014

[3] E M Vilei and J Frey ldquoDetection of Mycoplasma mycoidessubsp mycoides SC in bronchoalveolar lavage fluids of cowsbased on a TaqMan real-time PCR discriminating wild typestrains from an lppQ- mutant vaccine strain used for DIVA-strategiesrdquo Journal of Microbiological Methods vol 81 no 3 pp211ndash218 2010

[4] N Abdela and N Yune ldquoSeroprevalence and distribution ofcontagious bovine pleuropneumonia in ethiopia update andcritical analysis of 20 years (1996-2016) reportsrdquo Frontiers inVeterinary Science vol 4 article 100 2017

[5] M J Otte R Nugent and A Mcleod ldquoTransboundary animaldiseases assessment of socio-economic impacts and institu-tional responses food and agriculture organizationrdquo Technicalreport Transboundary animal diseases Assessment of socio-economic impacts and institutional responses FAO RomeItaly 2004

[6] A Ssematimba J Jores and J C Mariner ldquoMathematicalmodelling of the transmission dynamics of contagious bovinepleuropneumonia reveals minimal target profiles for improvedvaccines and diagnostic assaysrdquo PLoS ONE vol 10 articlee0116730 no 2 2015

[7] N E Tambi W O Maina and C Ndi ldquoAn estimation of theeconomic impact of contagious bovine pleuropneumonia inAfricardquo Revue Scientifique Et Technique vol 25 no 3 pp 999ndash1011 2006

[8] M Lesnoff G Laval P Bonnet and A Workalemahu ldquoAmathematical model of contagious bovine pleuropneumonia(CBPP) within-herd outbreaks for economic evaluation of localcontrol strategies An illustration from a mixed crop-livestocksystem in Ethiopian highlandsrdquo Animal Research vol 53 no 5pp 429ndash438 2004

[9] V Dupuy L Manso-Silvan V Barbe et al ldquoEvolutionary His-tory of Contagious Bovine Pleuropneumonia Using Next Gen-eration Sequencing of Mycoplasma mycoides Subsp mycoidesldquoSmall Colonyrdquordquo PLoS ONE vol 7 no 10 article e46821 2012

[10] A A Aligaz and J M W Munganga ldquoAnalysis of a mathemati-calmodel of the dynamics of contagious bovine pleuropneumo-niardquo inMathematical Methods and Models in Biosciences vol 1pp 64ndash80 Biomath Forum Sofia Bulgaria 2017

[11] J C Mariner A Araba and S Makungu ldquoConsultancy onthe dynamics of CBPP endemism and the development ofeffective control eradication strategies for pastoral comunitiesfinal data collection reportrdquo in The Community Animal Healthand Participatory Epidemiology Unit of the African Union interAfrica Bureau for Animal Resources Nairobi Kenya 2003

[12] P van denDriessche and JWatmough ldquoReproduction numbersand sub-threshold endemic equilibria for compartmental mod-els of disease transmissionrdquoMathematical Biosciences vol 180pp 29ndash48 2002

[13] Z Shuai and P van denDriessche ldquoGlobal stability of infectiousdisease models using Lyapunov functionsrdquo SIAM Journal onApplied Mathematics vol 73 no 4 pp 1513ndash1532 2013

[14] J C Mariner J McDermott J A P Heesterbeek G Thomsonand S W Martin ldquoA model of contagious bovine pleurop-neumonia transmission dynamics in East Africardquo PreventiveVeterinary Medicine vol 73 no 1 pp 55ndash74 2006

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S

V

E I R

Q

SV

ΛE

kQ Q

qI

(t + r)ISI

N

Figure 1 A compartmental model for the transmission dynamics of CBPP with antibiotic treatment and vaccination

exposed infectious persistently infected and recovered com-partments Antibiotic treatment is considered in the modelby incorporating rate of recovery of treated cattle suchthat treated cattle move from infectious compartment torecovered compartment at a rate of 120572119905

The objective of this paper is to determine the bettercontrol method out of vaccination antibiotic treatment anda combination of both We derive the formula for the controlreproduction number R119888 and determine the number ofcattle to be vaccinated and treated independently and incombination which will enable us to choose the feasibleand effective controlling method in our context Numericalsimulations are performed using MATLAB

This paper is structured as follows In Section 2 wepresent a mathematical model of the dynamics of CBPPwith vaccination and antibiotic interventions In Section 3we prove the well-posedness of the model We calculateequilibria of the system and rigorously derive a formula of thecontrol reproduction numberR119888 in Section 4 Stability anal-ysis of the DFE and EE is presented in Section 5 we presentparameter values andnumerical simulations in Section 6 andlastly we draw the conclusions and remarks in Section 7

2 Mathematical Model

We model the transmission dynamics of Contagious BovinePleuropneumonia (CBPP) In this model we assume inter-vention by vaccination and antibiotic treatment Thus thecompartmental model is consisting of susceptible vacci-nated exposed infectious persistently infected and recov-ered classes as shown in Figure 1 We assume an openpopulation with a total number 119873 at time 119905 where allnewborn animals are born into susceptible class (119878) at rate119887 Susceptible cattle move to vaccinal immune class (119881) ata rate 120588 Cattle in vaccinal immune class can lose vaccinalimmunity and return back to susceptible class at a rate120596 Susceptible animals move to the exposed compartment(119864) at a rate 120573(119868119873) Cattle in the exposed compartmentmove to the infectious compartment (119868) at a rate 120574 Naturalmortality occurs at a rate 120583 and results in losses from all sixcompartments However we assume that death due to thedisease does not occur The infectious cattle either naturallyheal or receive antibiotic treatment and enter directly into therecovered (119877) compartment at a rate 120572119903 and 120572119905 respectively

or they pass through a process of sequestration and enterinto persistently infected (119876) compartment at a rate 120572119902Cattle in persistently infected compartment are encapsulatedand infected but not infectious As sequestra resolve andorbecome noninfected then the animals in persistently infectedcompartment move to the recovered (119877) compartment ata rate 120595 Recovered cattle remain recovered for life timeInfected sequestra can occasionally be reactivated and inthis instance the animal will transition from the persistentlyinfected (119876) compartment back to the infectious (119868) compart-ment at a rate 119896 We assume randommixing of all individualsin the population The total number of population at time 119905119873 is given by119873 = 119878(119905) + 119881(119905) + 119864(119905) + 119868(119905) + 119876(119905) + 119877(119905) Theflow diagram of the model is shown in Figure 1

The differential equationmodel is given by system (1)ndash(7)

119889119878119889119905 = 120583119873 + 120596119881 minus 120573119878119868119873 minus 120588119878 minus 120583119878 (1)

119889119881119889119905 = 120588119878 minus 120596119881 minus 120583119881 (2)

119889119864119889119905 = 120573119878119868119873 minus 120574119864 minus 120583119864 (3)

119889119868119889119905 = 120574119864 + 119896119876 minus (120572119903 + 120572119905) 119868 minus 120572119902119868 minus 120583119868 (4)

119889119876119889119905 = 120572119902119868 minus 119896119876 minus 120595119876 minus 120583119876 (5)

119889119877119889119905 = (120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 (6)

with initial condition

(119878 (0) 119881 (0) 119864 (0) 119868 (0) 119876 (0) 119877 (0))= (1198780 1198810 1198640 1198680 1198760 1198770) (7)

3 Well-Posedness of the System

Let119883(119905) = (119878(119905) 119881(119905) 119864(119905) 119868(119905) 119876(119905) 119877(119905)) and119891 Ω 997888rarr Y

119883 997891997888rarr 1198831015840 (8)




1198912 (119883) = 119889119881119889119905 = 120588119878 minus 120596119881 minus 120583119881 (10)

1198913 (119883) = 119889119864119889119905 = 120573119878119868119873 minus 120574119864 minus 120583119864 (11)



1198916 (119883) = 119889119877119889119905 = (120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 (14)


1198831015840 (119905) = 119891 (119883 (119905)) 119883 (0) = (1198780 1198810 1198640 1198680 1198760 1198770) isin Ω (15)






119889119878119889119905 = 119889119881119889119905 = 119889119864119889119905 = 119889119868119889119905 = 119889119876119889119905 = 119889119877119889119905 = 0 (16)


120583119873 + 120596119881 minus 119878(120573119868119873 + 120588 + 120583) = 0 (17)

120588119878 minus 119881 (120596 + 120583) = 0 (18)

120573119878119868119873 minus 119864 (120574 + 120583) = 0 (19)

120574119864 + 119896119876 minus 119868 (120572119903 + 120572119905 + 120572119902 + 120583) = 0 (20)

120572119902119868 minus 119876 (119896 + 120595 + 120583) = 0 (21)

(120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 = 0 (22)

From (21)

119876 = 120572119902119868119896 where 119896 = 119896 + 120595 + 120583 (23)


119864 = (119896120572 minus 119896120572119902) 119868120574119896 where 120572 = 120572119903 + 120572119905 + 120572119902 + 120583 (24)


119877 = ((120572119903 + 120572119905) 119896 + 120595120572119902) 119868120583119896 (25)



120573119868( 119878119873 minus 120574 (119896120572 minus 119896120572119902)120573120574119896 ) = 0 (27)




1198760 = 1198640 = 1198770 = 0 (28)


1198810 = 120588119873120588 + 120596 (29)

and

1198780 = 120596119873120588 + 120596 (30)



119881lowast = 120588120574 (119896120572 minus 119896120572119902)119873120596120573120574119896 (31)



120573120596120574 (119896120572 minus 119896120572119902) (32)



F = [[[[

12057311987811986811987300]]]]

V =Vminus minusV

+ = [[[[


]]]]

119865 = [120597F119894120597119909119895 (1198830)] =[[[[[

0 1205731198780119873 00 0 00 0 0

]]]]]

119881 = [120597V119894120597119909119895 (1198830)] =[[[[


]]]]

and

119881minus1 =[[[[[[[[[

1120574 0 0120574119896

120574 (119896120572 minus 119896120572119902)119896

119896120572 minus 119896120572119902119896

119896120572 minus 119896120572119902120572119902120574120574 (119896120572 minus 119896120572119902)

120572119902119896120572 minus 119896120572119902

120572119896120572 minus 119896120572119902

]]]]]]]]]

(33)

Therefore

119865119881minus1

= 1205731198780119873[[[[[[

120574119896120574 (119896120572 minus 119896120572119902)

119896119896120572 minus 119896120572119902

119896119896120572 minus 1198961205721199020 0 0

0 0 0

]]]]]] (34)


120574 (119896120572 minus 119896120572119902)) (35)

Equivalently

R119888 = ( 120596120588 + 120596)(119896 (120572 minus 120572119905) minus 119896120572119902

119896120572 minus 119896120572119902 )R0 (36)








119891 (119909 119910) = (119865 minus 119881) 119909 minusF (119909 119910) +V (119909 119910)

= [[[[[

1205731198780119868119873 minus 12057311987811986811987300

]]]]]= 120573119868119873

[[[[[

( 120596119873120596 + 120588 minus 119878)00

]]]]] (37)

where 1198780 = 120596119873(120596 + 120588) is as in (30)


And

119881minus1119865 = 1205731198780119873[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(38)

= 120573120596120596 + 120588[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(39)

= 120573120574120596120574 (120596 + 120588)[[[[[[[[[

0 1120574 00 119896(119896120572 minus 119896120572119902) 0

0 120572119902(119896120572 minus 119896120572119902) 0

]]]]]]]]]

(40)


(V1 V2 V3) 119881minus1119865 = 119877119888 (V1 V2 V3) (41)

such that

(V1 V2 V3) 119881minus1119865 = ( 120573120596120574(120596 + 120588) 120574) (0 119909 0) where (42)

119909 = V1120574 + 119896V2(119896120572 minus 119896120572119902) +120572119902V3

(119896120572 minus 119896120572119902) (43)

and

119877119888 (V1 V2 V3) = 120573120596120574(120596 + 120588) 120574 ( 119896120572119896 minus 119896120572119902)(V1 V2 V3) (44)


119871 = 120596119879119881minus1119909= V2( 120574119896119864

120574 (119896120572 minus 119896120572119902) +119896119868

119896120572 minus 119896120572119902 +119896119876

119896120572 minus 119896120572119902)(45)









1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31












(46)













assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018











1198912 (119883) = 119889119881119889119905 = 120588119878 minus 120596119881 minus 120583119881 (10)

1198913 (119883) = 119889119864119889119905 = 120573119878119868119873 minus 120574119864 minus 120583119864 (11)



1198916 (119883) = 119889119877119889119905 = (120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 (14)


1198831015840 (119905) = 119891 (119883 (119905)) 119883 (0) = (1198780 1198810 1198640 1198680 1198760 1198770) isin Ω (15)






119889119878119889119905 = 119889119881119889119905 = 119889119864119889119905 = 119889119868119889119905 = 119889119876119889119905 = 119889119877119889119905 = 0 (16)


120583119873 + 120596119881 minus 119878(120573119868119873 + 120588 + 120583) = 0 (17)

120588119878 minus 119881 (120596 + 120583) = 0 (18)

120573119878119868119873 minus 119864 (120574 + 120583) = 0 (19)

120574119864 + 119896119876 minus 119868 (120572119903 + 120572119905 + 120572119902 + 120583) = 0 (20)

120572119902119868 minus 119876 (119896 + 120595 + 120583) = 0 (21)

(120572119903 + 120572119905) 119868 + 120595119876 minus 120583119877 = 0 (22)

From (21)

119876 = 120572119902119868119896 where 119896 = 119896 + 120595 + 120583 (23)


119864 = (119896120572 minus 119896120572119902) 119868120574119896 where 120572 = 120572119903 + 120572119905 + 120572119902 + 120583 (24)


119877 = ((120572119903 + 120572119905) 119896 + 120595120572119902) 119868120583119896 (25)



120573119868( 119878119873 minus 120574 (119896120572 minus 119896120572119902)120573120574119896 ) = 0 (27)




1198760 = 1198640 = 1198770 = 0 (28)


1198810 = 120588119873120588 + 120596 (29)

and

1198780 = 120596119873120588 + 120596 (30)



119881lowast = 120588120574 (119896120572 minus 119896120572119902)119873120596120573120574119896 (31)



120573120596120574 (119896120572 minus 119896120572119902) (32)



F = [[[[

12057311987811986811987300]]]]

V =Vminus minusV

+ = [[[[


]]]]

119865 = [120597F119894120597119909119895 (1198830)] =[[[[[

0 1205731198780119873 00 0 00 0 0

]]]]]

119881 = [120597V119894120597119909119895 (1198830)] =[[[[


]]]]

and

119881minus1 =[[[[[[[[[

1120574 0 0120574119896

120574 (119896120572 minus 119896120572119902)119896

119896120572 minus 119896120572119902119896

119896120572 minus 119896120572119902120572119902120574120574 (119896120572 minus 119896120572119902)

120572119902119896120572 minus 119896120572119902

120572119896120572 minus 119896120572119902

]]]]]]]]]

(33)

Therefore

119865119881minus1

= 1205731198780119873[[[[[[

120574119896120574 (119896120572 minus 119896120572119902)

119896119896120572 minus 119896120572119902

119896119896120572 minus 1198961205721199020 0 0

0 0 0

]]]]]] (34)


120574 (119896120572 minus 119896120572119902)) (35)

Equivalently

R119888 = ( 120596120588 + 120596)(119896 (120572 minus 120572119905) minus 119896120572119902

119896120572 minus 119896120572119902 )R0 (36)








119891 (119909 119910) = (119865 minus 119881) 119909 minusF (119909 119910) +V (119909 119910)

= [[[[[

1205731198780119868119873 minus 12057311987811986811987300

]]]]]= 120573119868119873

[[[[[

( 120596119873120596 + 120588 minus 119878)00

]]]]] (37)

where 1198780 = 120596119873(120596 + 120588) is as in (30)


And

119881minus1119865 = 1205731198780119873[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(38)

= 120573120596120596 + 120588[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(39)

= 120573120574120596120574 (120596 + 120588)[[[[[[[[[

0 1120574 00 119896(119896120572 minus 119896120572119902) 0

0 120572119902(119896120572 minus 119896120572119902) 0

]]]]]]]]]

(40)


(V1 V2 V3) 119881minus1119865 = 119877119888 (V1 V2 V3) (41)

such that

(V1 V2 V3) 119881minus1119865 = ( 120573120596120574(120596 + 120588) 120574) (0 119909 0) where (42)

119909 = V1120574 + 119896V2(119896120572 minus 119896120572119902) +120572119902V3

(119896120572 minus 119896120572119902) (43)

and

119877119888 (V1 V2 V3) = 120573120596120574(120596 + 120588) 120574 ( 119896120572119896 minus 119896120572119902)(V1 V2 V3) (44)


119871 = 120596119879119881minus1119909= V2( 120574119896119864

120574 (119896120572 minus 119896120572119902) +119896119868

119896120572 minus 119896120572119902 +119896119876

119896120572 minus 119896120572119902)(45)









1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31












(46)













assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018










120573120596120574 (119896120572 minus 119896120572119902) (32)



F = [[[[

12057311987811986811987300]]]]

V =Vminus minusV

+ = [[[[


]]]]

119865 = [120597F119894120597119909119895 (1198830)] =[[[[[

0 1205731198780119873 00 0 00 0 0

]]]]]

119881 = [120597V119894120597119909119895 (1198830)] =[[[[


]]]]

and

119881minus1 =[[[[[[[[[

1120574 0 0120574119896

120574 (119896120572 minus 119896120572119902)119896

119896120572 minus 119896120572119902119896

119896120572 minus 119896120572119902120572119902120574120574 (119896120572 minus 119896120572119902)

120572119902119896120572 minus 119896120572119902

120572119896120572 minus 119896120572119902

]]]]]]]]]

(33)

Therefore

119865119881minus1

= 1205731198780119873[[[[[[

120574119896120574 (119896120572 minus 119896120572119902)

119896119896120572 minus 119896120572119902

119896119896120572 minus 1198961205721199020 0 0

0 0 0

]]]]]] (34)


120574 (119896120572 minus 119896120572119902)) (35)

Equivalently

R119888 = ( 120596120588 + 120596)(119896 (120572 minus 120572119905) minus 119896120572119902

119896120572 minus 119896120572119902 )R0 (36)








119891 (119909 119910) = (119865 minus 119881) 119909 minusF (119909 119910) +V (119909 119910)

= [[[[[

1205731198780119868119873 minus 12057311987811986811987300

]]]]]= 120573119868119873

[[[[[

( 120596119873120596 + 120588 minus 119878)00

]]]]] (37)

where 1198780 = 120596119873(120596 + 120588) is as in (30)


And

119881minus1119865 = 1205731198780119873[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(38)

= 120573120596120596 + 120588[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(39)

= 120573120574120596120574 (120596 + 120588)[[[[[[[[[

0 1120574 00 119896(119896120572 minus 119896120572119902) 0

0 120572119902(119896120572 minus 119896120572119902) 0

]]]]]]]]]

(40)


(V1 V2 V3) 119881minus1119865 = 119877119888 (V1 V2 V3) (41)

such that

(V1 V2 V3) 119881minus1119865 = ( 120573120596120574(120596 + 120588) 120574) (0 119909 0) where (42)

119909 = V1120574 + 119896V2(119896120572 minus 119896120572119902) +120572119902V3

(119896120572 minus 119896120572119902) (43)

and

119877119888 (V1 V2 V3) = 120573120596120574(120596 + 120588) 120574 ( 119896120572119896 minus 119896120572119902)(V1 V2 V3) (44)


119871 = 120596119879119881minus1119909= V2( 120574119896119864

120574 (119896120572 minus 119896120572119902) +119896119868

119896120572 minus 119896120572119902 +119896119876

119896120572 minus 119896120572119902)(45)









1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31












(46)













assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018









And

119881minus1119865 = 1205731198780119873[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(38)

= 120573120596120596 + 120588[[[[[[[[[

0 1120574 00 120574119896120574 (119896120572 minus 119896120572119902) 0

0 120572119902120574120574 (119896120572 minus 119896120572119902) 0

]]]]]]]]]

(39)

= 120573120574120596120574 (120596 + 120588)[[[[[[[[[

0 1120574 00 119896(119896120572 minus 119896120572119902) 0

0 120572119902(119896120572 minus 119896120572119902) 0

]]]]]]]]]

(40)


(V1 V2 V3) 119881minus1119865 = 119877119888 (V1 V2 V3) (41)

such that

(V1 V2 V3) 119881minus1119865 = ( 120573120596120574(120596 + 120588) 120574) (0 119909 0) where (42)

119909 = V1120574 + 119896V2(119896120572 minus 119896120572119902) +120572119902V3

(119896120572 minus 119896120572119902) (43)

and

119877119888 (V1 V2 V3) = 120573120596120574(120596 + 120588) 120574 ( 119896120572119896 minus 119896120572119902)(V1 V2 V3) (44)


119871 = 120596119879119881minus1119909= V2( 120574119896119864

120574 (119896120572 minus 119896120572119902) +119896119868

119896120572 minus 119896120572119902 +119896119876

119896120572 minus 119896120572119902)(45)









1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31












(46)













assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018









1 2 3 4 5 6a21

a12

a14

a34 a54

a64

a43 a45

a65

a31












(46)













assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018

















assumed



guess


[11] and estimated






1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018










1000 2000 3000 4000 5000 60000Time (in days)２0=39399

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle



119905





1000 2000 3000 4000 5000 60000Time (in days)２0=67462

0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle


1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=098218



0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018









0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=09906







0

50

100

150

200

250

300

350

400

450

500

Num

ber o

f cat

tle

1000 2000 3000 4000 5000 6000 70000Time (in days)２＝=099213



Data Availability


Disclosure





References






















Journal of












Journal of







Volume 2018






Volume 2018











References






















Journal of












Journal of







Volume 2018






Volume 2018















Journal of












Journal of







Volume 2018






Volume 2018








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Mathematical Modelling of the Transmission Dynamics of ...downloads.hindawi.com/journals/jam/2019/2490313.pdf · JournalofAppliedMathematics S V E I R Q V S Λ E kQ Q qI ( t + r)I