CHAPTER 6 MODEL FOR THE EFFECTS OF INDUSTRIALIZATION, POPULATION, PRIMARY - SECONDARY TOXICANTS ON DEPLETION OF FORESTRY RESOURCE INTRODUCTION The environmental problems in India are growing rapidly. Industrial pollution, soil erosion, deforestation, rapid industrialization, urbanization and land degradation are all worsening problems. Over- exploitation of the country resources be it land or water, the industrial process has resulted in environmental degradation of resources. Airborne pollutants can be classified broadly into two categories: primary and secondary. Primary pollutants are those that are emitted into the atmosphere by a source such as fossil fuels combustion from power plants, vehicle engines and industrial production, by combustion of biomass from agriculture and land clearing purposes, and by natural processes. Secondary pollutants are formed within the atmosphere when primary pollutants react with sunlight, oxygen, water and other chemicals present in the air. Atmospheric processes, including oxidation procedures, particle formation and equilibria, determine the fate of primary emission and, in most cases, the secondary product of these processes are the more important ones concerning their effects on human health and the quality of the environment,(Shukla and Dubey, [1996] ). So, the pollutants in both of their forms are a serious threat for the survival of the resource 145 biomass and exposed population and in order to regulate these pollutant wisely, we must assess the risk of the resource biomass and populations exposed to pollutants. Therefore, it is important to study the effects of pollutants on a resource-dependent biological population by making use of mathematical models. In view of the above considerations, in this chapter, a nonlinear mathematical model is proposed and analyzed for the survival of a resource-dependent biological population in the presence of two toxicants (primary and secondary). It is assumed that the density of the primary toxicant is increased by the population and industrialization in the environment and the secondary toxicant formed from primary toxicant, is more toxic. This situation is modeled by a system of five ordinary differential equations. Stability theory of nonlinear differential equations and a fourth order Runge-Kutta method are used to analyze and predict the behavior of the model. 6.1 THE MATHEMATICAL MODEL We consider an ecosystem where the resource biomass is being depleted due to the pressure of industrialization, population and primary-secondary toxicants in the environment. The system is assumed to be governed by the following differential equations: rB 0 B 2 dB rB N B IB, dt K B TP , TS r N2 dN rP B N P 0 1 IN , dt M TP , TS dTP QI , N 0TP 1 BT P 2 NTP gTP , dt 146 (6.1.1) dTS gTP 1TS 1 BTS 2 NTS , dt dI I r1 I 1 I B 2 I N . dt L B0 0, N 0 0, TP 0 0, TS 0 0, I 0 0. In model (6.1.1), B is the density of resource biomass, N is the density of the population, TP and TS are the densities of primary and secondary toxicants into the environment. I is the density of industrialization. It is assumed that the dynamics of the resource biomass, population and industrialization are governed by logistic type equations. It is also assumed that the growth rate of the resource biomass decreases with an increase in density of the population while its carrying capacity decreases with an increase in environmental concentration of primary-secondary toxicants. It is further assumed that growth rate of the population and industrialization increases as the density of resource biomass increases. is the depletion rate coefficient of the resource biomass due to the industrialization and is the corresponding growth rate coefficient of industrialization due to resource biomass. It is also considered that the emission of primary toxicant into the environment is industrialization and population dependent and the secondary toxicant which is transformed from the primary toxicant, is more toxic. The positive constant g is the transformation rate coefficient of primary toxicant into secondary toxicant in the environment. 1 and 2 are the growth rate coefficients of industrialization and population respectively due to their interaction. r1 is the intrinsic growth rate coefficient of industrialization. 1 , 2 and 1 , 2 are the depletion rate coefficients of primary and secondary toxicants due to resource biomass and population, respectively. 0 and 1 are the natural washout rate 147 coefficients of the primary and secondary toxicants from the environment, respectively. The constant, 0 1, is a fraction, which represent the magnitude of transformation of primary toxicant into secondary toxicant. Model (6.1.1) is derived from following assumptions: (H1): The function rB N denotes the specific growth rate of resource biomass which decreases as N increases. Hence we take rB 0 rB 0 0, rB N 0 for N 0. (H2): The function K B TP , TS represents the maximum density of resource biomass which the environment can support in the presence of primary and secondary toxicants, and it also decreases as TP and TS increases. Hence we take K B 0,0 K B 0 0, K B TP , TS 0, TP K B TP , TS 0 TS for TP 0, TS 0. (H3): The function rP B denotes the growth rate coefficient of the population and it increases as the resource biomass density increases. Hence we take rP 0 rP 0 0, rP B 0 for B 0. (H4):The function M TP , TS represents the maximum density of population which the environment can support in the presence of primary and secondary toxicants, and it also decreases as TP and TS increases. Hence we take M 0,0 M 0 0, M TP , TS 0, TP M TP , TS 0 for TP 0, TS 0. TS (H5): The function QI , N is the rate of introduction of toxicant into the environment which increases as I and N increase. Hence we take 148 QI , N QI , N 0, 0 for I 0, N 0. I N Q0,0 Q0 0, To analyze the model (6.1.1), we need the bounds of dependent vatiables involved. For this we find the region of attraction in the following lemma. 6.2 BOUNDEDNESS OF SOLUTIONS Lemma (6.2.1): Suppose that assumptions (H1) - (H5) hold. Then all solutions of system (6.1.1) are bounded within the region , B, N , TP , TS , I : 0 B K B 0 ,0 N N a ,0 TP TS Qm , 0 I I a where Qm QI a , N a , min 0 g g , 1 . Proof: Proof is analogous to the proof of lemma (3.1.1) of chapter 3. 6.3 EQUILIBRIUM ANALYSIS The system (6.1.1) may have eight nonnegative equilibria in the B N TP TS I Q0 Q0 g Q0 Q0 g space, namely E1 0,0, , ,0 , E2 0,0, , , L , 0 g 1 ( 0 g ) 0 g 1 ( 0 g ) ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ˆ ˆ ˆ ˆ E3 0, N , TP , TS ,0 , E4 0, N , TP , TS , I , E5 Bˆ ,0, TˆP , TˆS ,0 , E6 Bˆ , Nˆ , TˆP , TˆS ,0 , E 7 B,0, TP , TS , I and E * B * , N * , TP* , TS* , I * . The existence of E1 and E2 is obvious. We prove the existence of the other equilibria. ~ ~ ~ Existence of E3 0, N , TP , TS ,0 : ~ ~ ~ In this case, N , TP and TS are the positive solutions of the following equations. ~ ~ ~ N M TP , TS , (6.3.1) 149 ~ TP ~ Q 0, N ~ f 1 ( N ), say, ~ 0 2N g (6.3.2) ~ ~ gf1 ( N ) ~ TS ~ f 2 ( N )., say, 1 2 N (6.3.3) ~ ~ ~ It is noted that from equation (6.3.2) and (6.3.3) that TP and TS are functions of N ~ only. To show the existence of E 3 , we define a function F1 ( N ) as follows ~ ~ ~ ~ F1 ( N ) N M ( f1 ( N ), f 2 ( N )). (6.3.4) From equation (6.3.4), we note that Q0 gQ0 0. F1 (0) M , 0 g 1 ( 0 g ) Q(0, N a ) gQ(0, N a ) 0. F1 N a N a M , 0 g 2 N a ( 0 g 2 N a )( 1 2 N a ) ~ ~ ~ Thus there exists a root N in the interval 0 N N a given by F1 N 0. ~ Now, the sufficient condition for E 3 to be unique is F1 ( N ) 0, where ~ ~ ~ F1 ( N ) 1 M 1 f 1 ( N ) M 2 f 2 ( N ) 0. ( g )Q2 Q0 2 ~ where f1 ( N ) 0 , ~ ( 1 1 N g ) 2 (6.3.5) ~ ~ ~ ~ g ( 1 2 N ) f1 ( N ) 2 f1 ( N ) f 2 (N ) . ~ ( 1 2 N ) 2 ~ ~ ~ With this value of N , the value of TP and TS can be found from equations (6.3.2) and (6.3.3), respectively. This completes the existence of E3 . ~ ~ ~ ~ ~ ~ ~ ~ Existence of E 4 0, N , TP , TS , I : ~ ~ ~ ~ ~ ~ ~ ~ In this case N , TP , TS and I are the solutions of the following equations: 150 ~ ~ ~ rP 0 N ~ rP 0 I 0, ~ 1 ~ ~ ~ M (TP , TS ) (6.3.6) ~ ~ ~ ~ ~ ~ ~ ~ ~~ ~ ~ Q( I , N ) 0TP 2 NTP gTP 0, (6.3.7) ~ ~ ~ ~ ~ ~~ ~ gTP 1TS 2 NTS 0, (6.3.8) ~ ~ ~ L(r 2 N ) ~ I g1 ( N ), say, r Using the value of (6.3.9) ~ ~ I from equation (6.3.9) in equations (6.3.7) and (6.3.8), we obtain ~ ~ ~ ~ ~ ~ Q( g1 ( N ), N ) ~ ~ TP g ( N ), say, ~ 2 ~ 0 2N g (6.3.10) ~ ~ ~ ~ ~ gQ2 ( N ) ~ TS g ( N ), say, ~ 3 ~ 1 2 N (6.3.11) ~ ~ ~ ~ ~ ~ It is noted from equations (6.3.9), (6.3.10) and (6.3.11) that I , TP and TS are functions ~ ~ ~ ~ of N only. To show the existence of E 4 , we define a function F2 ( N ) as follows ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ F2 ( N ) rP0 N (rP 0 1 g1 ( N )) M ( g 2 ( N ), g 3 ( N )). (6.3.12) From equation (6.3.12), we note that QL,0 gQL,0 0. F2 0 rP 0 1 L M , g g 1 0 0 F2 N a rP 0 N a rP 0 1 g1 N a M g 2 N a , g 3 N a 0 where g1 ( N a ) L(r 2 N a ) Q( g1 ( N a ), N a ) gg 2 ( N a ) , g2 (Na ) , g3 (N a ) . r 0 2 Na g 1 2 N a ~ ~ ~ ~ ~ ~ Thus there exists a root N in the interval 0 N N m given by F2 N 0. 151 ~ ~ Now, a sufficient condition for E4 to be unique is F2 ( N ) 0, where ~ ~ ~ ~ ~ ~ ~ ~ F2 ( N ) rP 0 L 1 2 M ( g 2 ( N ), g 3 ( N )) (rP 0 1 g1 ( N ) r ~ ~ ~ ~ (M 1 g 2 ( N ) M 2 g 3 ( N )) 0. (6.3.13) ( g )(Q1 L 2 Q2 r ) Q( L,0)r 2 ~ ~ where g 2 ( N ) 0 , ~ ~ ( 0 2 N g ) 2 ~ ~ g3 (N ) ~ ~ ~ ~ ~ ~ g ( 1 2 N ) g 2 ( N ) 2 g 2 ( N ) ~ ~ ( 1 2 N ) 2 . ~ ~ ~ ~ ~ ~ ~ With this value of N , values of I , TP and TS can be found from equations (6.3.9), (6.3.10) and (6.3.11), respectively. This completes the existence of E4 . Existence of E5 Bˆ ,0, TˆP , TˆS ,0 : In this case Bˆ , TˆP , TˆS are the solutions of the following equations Bˆ K B TˆP , TˆS , TˆP Q0 0 1 Bˆ g (6.3.14) h1 ( Bˆ ), say, (6.3.15) gh1 ( Bˆ ) TˆS h2 ( Bˆ ), say, ˆ 1 1 B (6.3.16) It is noted from equations (6.3.15) and (6.3.16) that TˆP and TˆS , are functions of B̂ only. To show the existence of E 5 , we define a function F3 ( Bˆ ) as follows F3 ( Bˆ ) Bˆ K B .(h1 ( Bˆ ), h2 ( Bˆ )). (6.3.17) From equation (6.3.17), we note that 152 Q0 gQ0 0. F3 0 K B , 0 g 1 0 g F3 K B 0 K B1 Q0 gQ0 K B2 0, 0 g 1 K B 0 ( 1 1 K B 0 )( 0 g 1 K B 0 ) Thus there exists a root B̂ , in the interval 0 Bˆ K B 0 , given by F3 ( Bˆ ) 0. Now, the sufficient condition for E 5 to be unique is F3 ( Bˆ ) 0, where F3 ( Bˆ ) 1 K B1 Q0 1 gQ0 ( g1 1 0 1 1 ) K B2 0. 2 ˆ ( 0 1 B g ) ( 0 1 Bˆ g ) 4 (6.3.18) With this value of B̂ , value of TˆP and TˆS , can be found from equations (6.3.15) and (6.3.16), respectively. ˆ ˆ ˆ ˆ Existence of E6 ( Bˆ , Nˆ , TˆP , TˆS ) : ˆ ˆ ˆ ˆ In this case, Bˆ , Nˆ , TˆP , TˆS are the solutions of the following equations ˆ rB ( Nˆ ) ˆ rB 0 Bˆ 0, ˆ ˆ K B (TˆP , TˆS ) (6.3.19) ˆ rP ( Bˆ ) ˆ rP 0 Nˆ 0, ˆˆ ˆˆ M (TP , TS ) (6.3.20) ˆ ˆ ˆˆ ˆˆ ˆ Q(0, Nˆ ) 0TˆP 1 Bˆ TˆP 2 Nˆ TˆP gTˆP 0, ˆ ˆ ˆˆ ˆˆ gTˆP 1TˆS 1 Bˆ TˆS 2 Nˆ TˆS 0. (6.3.21) (6.3.22) From the equation (6.3.21) and (6.3.22), we have ˆ TˆP ˆ Q(0, Nˆ ) ˆ ˆ d1 ( Bˆ , Nˆ ), say, ˆ ˆ 0 1 Bˆ 2 Nˆ g 153 (6.3.23) ˆ ˆ gd1 ( Bˆ , Nˆ ) ˆˆ ˆ ˆ TS d 2 ( Bˆ , Nˆ ), say. ˆ ˆ 1 1 Bˆ 2 Nˆ (6.3.24) Using values of TP and TS from (6.3.23) and (6.3.24) in equations (6.3.19) and (6.3.20) respectively, we get ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ F4 ( Bˆ , Nˆ ) (rB 0 rB1 Nˆ ) ( K B 0 K B1d1 ( Bˆ , Nˆ ) K B 2 d 2 ( Bˆ , Nˆ )) rB 0 Bˆ 0, (6.3.25) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ F5 ( Bˆ , Nˆ ) (rP 0 rP1 Bˆ ) ( M 0 M 1d1 ( Bˆ , Nˆ ) M 2 d 2 ( Bˆ , Nˆ )) rP 0 Nˆ 0, (6.3.26) From (6.2.25), we note the following ˆ ˆ ˆ when Nˆ 0, then Bˆ Bˆ a , where ˆ d11 Bˆ a2 d12 Ba d13 0, where d11 1 1 K B 0 , d12 K B 0 (1 0 1 1 g1 Q0 1 K B1 ) 1. d13 K B 0 ( 1 0 g 1 ) K B1Q0 1 Q0 gK B 2 . ˆ ˆ Let D1 ( Bˆ a ) d11 Bˆ a2 d12 Ba d13 , D1 (0) d13 0, D1 ( K B 0 ) d11 K B 0 d12 K B 0 d13 0. ˆ ˆ ˆ Thus there exists a root Bˆ a in the interval 0 Bˆ a K 0 given by D1 ( Bˆ a ) 0. ˆ ˆ Now, the sufficient condition for Bˆ a to be unique is D1 ( Bˆ a ) 0. where ˆ ˆ D1 ( Bˆ a ) 2d11 Bˆ a d12 0. F4 ˆ ˆ ˆ ˆ ˆ [( rB 0 rB1 Nˆ ) ( K B1 d1 ( Bˆ , Nˆ ) K B 2 d 2 ( Bˆ , Nˆ )) rB 0 ], ˆˆ B 154 (6.3.27) F4 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ rB1 K (d1 ( Bˆ , Nˆ ), d 2 ( Bˆ , Nˆ )) rB ( Nˆ )( K B1 d1 ( Bˆ , Nˆ ) K B 2 d 2 ( Bˆ , Nˆ )) ˆˆ N (6.3.28) Now, from (6.3.27) and (6.3.28) we get F4 ˆˆ ˆˆ B N . ˆ F4 Nˆ ˆ Bˆ It is clear that ˆ Bˆ 0, if either ˆˆ N (i) F4 F4 0 and 0, or ˆ ˆ Bˆ Nˆ (ii) F4 F4 0 and 0, ˆˆ ˆ B Nˆ (6.3.29) From (6.3.26), we note the following ˆ ˆ ˆ when Bˆ 0, then Nˆ Nˆ b , where ˆ ˆ v11 Nˆ b2 v12 Nˆ b v13 0, where v11 M 0 2 2 M 1Q2 2 , v12 M 0 ( 2 0 2 1 g 2 ) M 1 (Q0 2 Q2 1 ) M 2 gQ2 1. v13 M 0 ( 0 1 g 1 ) M 1Q0 1 M 2 gQ0 . ˆ ˆ ˆ Let D2 ( Nˆ b ) v11 Nˆ b2 v12 Nˆ b v13 , D2 (0) v13 0, D2 ( N a ) v11 N a2 v12 N a v13 0, ˆ ˆ ˆ Thus there exists a root Nˆ b in the interval 0 Nˆ b N a given by D2 ( Nˆ b ) 0. 155 ˆ ˆ Now, the sufficient condition for Nˆ b , to be unique is D2 ( Nˆ b ) 0. where ˆ ˆ D2 ( Nˆ b ) 2v11 Nˆ b v12 0. F5 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ rP1 M (d1 ( Bˆ , Nˆ ), d 2 ( Bˆ , Nˆ )) rP ( Bˆ )( M 1 d1 ( Bˆ , Nˆ ) M 2 d 2 ( Bˆ , Nˆ )), ˆˆ B (6.3.30) F5 ˆ ˆ ˆ ˆ ˆ [( rP 0 rP1 Bˆ ) ( M 1 d1 ( Bˆ , Nˆ ) M 2 d 2 ( Bˆ , Nˆ )) rP 0 ] 0. ˆ Nˆ (6.3.31) Now, from (6.3.30) and (6.3.31) we get ˆ Bˆ ˆ Nˆ F5 ˆˆ N . F5 ˆ Bˆ ˆ Bˆ It is clear that 0, if either ˆˆ N (i) F5 F5 0 and 0, or ˆ ˆ Bˆ Nˆ (ii) F5 F54 0 and 0, ˆˆ ˆˆ B N (6.3.32) ˆ ˆ Thus the two isoclines (6.3.25) and (6.3.26) intersects at a unique ( Bˆ , Nˆ ) if in ˆ ˆ addition to conditions (6.3.29) and (6.3.32), the inequality Bˆ b Nˆ b holds. Knowing ˆ ˆ ˆ ˆ the value of Bˆ and Nˆ , we get TˆP and TˆS , can be calculated from equations (6.3.23) and (6.3.24). This completes the existence of E6 . Existence of E 7 B,0, TP , TS , I : In this case B, TP , TS , I are the solutions of the following equations. 156 rB 0 B rB 0 I 0, K B TP , TS (6.3.33) Q( I ,0) 0TP 1 BTP gTP 0, (6.3.34) gTP 1TS 1 BTS 0, (6.3.35) I r1 1 - B 0. L (6.3.36) From equation (6.3.34), (6.3.35) and (6.3.36) we get, B e1 ( B), say. I L1 r1 Q e1 ( B),0 TP e2 ( B), say, 0 1 B g (6.3.37) (6.3.38) ge2 ( B) TS e3 ( B), say. 1 1 B (6.3.39) From equations (6.3.37), (6.3.38) and (6.3.39) we note that TP , TS and I are functions of B only. To show the existence of E 7 , we define a function F6 ( B ) as follows F6 ( B) rB 0 B (rB 0 e1 ( B)) K B (e2 ( B), e3 ( B)). (6.3.40) From equation (6.3.40), we note that QL,0 gQL,0 0. F6 (0) rB 0 L K B , 0 g 1 0 g F6 K B 0 rB 0 ( K B1e2 ( K B 0 ) K B 2 e3 ( K B 0 )) e1 ( K B 0 ) K B e2 K B 0 , e3 K B 0 0. where K B 0 e1 ( K B 0 ) L1 r1 , e2 ( K B 0 ) ge2 ( K B 0 ) Q0 Q1e1 ( K B 0 ) . , e3 ( K B 0 ) 1 1 K B 0 0 1 K B 0 g Thus there exists a root B, in the interval 0 B K B 0 , given by F6 ( B) 0. 157 Now, the sufficient condition for E7 to be unique is F6 ( B) 0, where L K B (e2 ( B), e3 ( B)) (rB 0 e1 ( B)) ( K B1e2 ( B) K B 2 e3 ( B)). where F6 ( B) rB 0 r Q L ( 0 g ) 1r1Q( L,0) where e2 ( B) 1 , r1 ( 0 1 B g ) 2 e3 ( B ) g 2 (e2 ( B)( 1 B) e2 ( B) 1 ). ( 1 1 B) With this value of B, value of I , TP , and TS can be found from equations (6.3.37), (6.3.38) and (6.3.39) respectively. This completes the existence of E7 . Existence of E * B * , N * , TP* , TS* , I * . In this case, B* , N * , TP* , TS* , I * are the solutions of following equations. rB ( N * ) rB 0 B* I * 0, * * K B (TP , TS ) (6.3.41) rP ( B* ) rP 0 N * 1 I * 0, * * M (T , T ) (6.3.42) P S Q( I * , N * ) 0TP* 1 B*TP* 2 N *TP* gTP* 0, (6.3.43) gTP* 1TS* 1 B*TS* 2 N *TS* 0, (6.3.44) I* B* N * 0. r1 1 2 L (6.3.45) From the equation (6.3.45), we have I* L r1 B* 2 N * s1 ( B* , N * ), say. r1 With this value of I *, and from the equation (6.3.43) and (6.3.44), we have 158 (6.3.46) TP* TS* Q( s1 ( B* , N * ), N * ) s 2 ( B* , N * ), * * ( 0 g 1 B 2 N ) gs 2 ( B* , N * ) ( 1 1 B* 2 N * ) s3 ( B* , N * ), say, say. (6.3.47) (6.3.48) Using values of I * , TP* and TS* in equations (6.3.41) and (6.3.42) respectively, we get F7 ( B* , N * ) (rB ( N * ) s1 ( B* , N * )) K ( s 2 ( B* , N * ), s3 ( B* , N * )) rB 0 B* , (6.3.49) F8 ( B* , N * ) (rP ( B* ) 1 s1 ( B* , N * )) M ( s 2 ( B* , N * ), s3 ( B* , N * )) rP 0 N *. (6.3.50) From (6.3.49), we note the following when N * 0, then B * Be* , where 3 2 m11 Be* m12 Be* m13 Be* m14 0, where Q L m12 rB 0 L K B 0 1 1 rB 0 1 ( K B1 1 K B 2g 1 ) rB 0 ( 0 1 1 1 g1 ) r1 QL L m11 1 1 rB 0 K B 0 1 1 1 ( K B1 1 K B 2g 1 ), r1 r1 L QL K B 0 ( 0 1 1 1 g1 ) K B1 Q0 1 Q1 L1 1 1 r1 r1 QL K B 2g 1Q0 Q1 L 1 1 ( 0 g ), r1 159 Q1 L 1 K B 0 ( 0 1 1 1 g1 ) K B1 Q( L,0) 1 r1 m13 rB 0 L rB 0 1 ( 0 g ) Q L 1 K B 2g 1 (Q( L,0) 0 r 1 L K B 0 1 ( 0 g ) K B1 1Q( L) K B 2g (Q( L)( 0 g ), r1 m14 rB 0 L K B 0 1 ( 0 g ) K B1 Q( L) 1 K B 2g Q( L)( 0 g . 3 2 Let M 1 ( Be* ) m11 Be* m12 Be* m13 Be* m14 , M1 (0) m14 0, M 1 ( K B 0 ) m11 K B3 0 m12 K B2 0 m13 K B 0 m14 0. Thus there exists a root B *E in the interval 0 Be* K B 0 given by M 1 ( Be* ) 0. Now, the sufficient condition for B e* to be unique is M 1 ( Be* ) 0. where 2 M 1 ( Be* ) 3m11 Be* 2m12 Be* m13 0, s ( B* , N * )) K ( s ( B* , N * ), s ( B* , N * )) r 1 2 3 B0 F7 * B * * * * * * * (rB ( N ) s1 ( B , N ))( K B1 s 2 ( B , N ) K B 2 s3 ( B , N ) (6.3.51) (r s ( B* , N * )) K ( s ( B* , N * ), s ( B* , N * )) B1 1 2 3 F7 * N * * * * * * * (rB ( N ) s1 ( B , N ))( K B1 s 2 ( B , N ) K B 2 s3 ( B , N ) (6.3.52) Now, from (6.3.51) and (6.3.52) we get F7 * * B N . F7 N * B * 160 It is clear that B* 0, if either * N (i) F7 F7 0 and 0, or * B N * (ii) F7 F7 0 and 0. B* N * (6.3.53) From (6.3.50), we note the following when B* 0, then N * N e* , where 3 2 n11 N e* n12 N e* n13 N e* n14 0, where n11 (rP 0 2 2 M 2gQ2 2 ), Q L Q L n12 rP 0 1 L M 0 2 2 M 1 1 2 2 Q2 2 ) M 2g 1 2 2 Q2 2 r1 r1 rP 0 ( 0 21 2 1 g 21 ) M 0 ( 0 2 2 2 g 2 ) M 1 2 Q0 Q1 L 2 1 Q2 r1 M 2g 2 Q1 L 2 1 , r1 n13 rP 0 Q1 L 2 1 Q ( L, 1 ) 2 ) M 0 (( 0 g ) 2 2 1 ) M 1 r1 1 L M 2g ( 0 g ) Q1 L 2 Q2 Q ( L,0) 2 r 1 2 1 Q2 M 0 1 ( 0 2 g ) M 1 1 Q0 Q1 L L r1 r ( g ), 1 2 P0 1 0 r1 Q1 L M 2g ( 0 g ) Q0 Q1 r 1 n14 rP 0 1 L M 0 ( 0 g ) 1 M 1 1Q( L,0) M 2g ( 0 g )Q( L,0). 161 3 2 Let M 2 ( N e* ) n11 N e* n12 N e* n13 N e* n14 , M 2 (0) n14 0, M 2 ( N a ) n11 N a3 n12 N a2 n13 N a n14 0. Thus there exists a root N e* in the interval 0 N e* N a given by M 2 ( N e* ) 0. Now, the sufficient condition for N e* , to be unique is M 2 ( N e* ) 0, where 2 M 2 ( N e* ) 3n11 N e* 2n12 N e* n13 0. F8 (rP1 ( B* ) 1 s1 ( B* , N * )) M ( s 2 ( B* , N * ), s3 ( B* , N * )) * B (rP ( B* ) 1 s1 ( B* , N * ))( M 1 s2 ( B* , N * ) M 2 s3 ( B* , N * )), (6.3.54) F8 1 s1 ( B* , N * )) M (s 2 ( B* , N * ), s3 ( B* , N * )) rP 0 N * (rP ( B* ) 1 s1 ( B* , N * ))( M 1 s2 ( B* , N * ) M 2 s3 ( B* , N * )). (6.3.55) Now, from (6.3.54) and (6.3.55) we get F8 N * B * . F8 N * B * It is clear that (i) B* 0, if either N * F8 F8 0 and 0, or B* N * (6.3.56) 162 (ii) F8 F8 0 and 0. * * B N Thus the two isoclines (6.3.49) and (6.3.50) intersects at a unique ( B* , N * ) if in addition to conditions (6.3.53) and (6.3.56), the inequality Be* N e* , holds. Knowing the value of B* and N * , we get I *, TP* and TS* , can be calculated from equations (6.3.46), (6.3.47) and (6.3.48). This completes the existence of E * . 6.4. STABILITY ANALYSIS 6.4.1 Local Stability To discuss the local stability of system (6.1.1), we compute the variational matrix of system (6.1.1).The entries of general variational matrix are given by differentiating the right hand side of system (6.1.1) with respect to B, N , TP , TS , and I i.e. v11 r N M ( E ) P1 1TP T 1 S I Q 2 2 TP rB 0 B 2 K B1 K B2 TP , TS rP 0 N 2 M 1 M 2 TP , TS 0 1 B 2 N g 2TS g 2I 0 rB1 B v 22 where, v11 rB N B 1N , Q1 ( 1 1 B 2 N ) 0 0 v33 rB 0 B 2 K B 2 K B2 TP , TS rP 0 N 2 M 2 M 2 TP , TS 0 2rB 0 B 2rP 0 N I , v22 rP B 1I , K B TP , TS M TP , TS 2I v33 r1 1 B 2 N . L The variational matrix M ( E1 ) at equilibrium point E1 is given by 163 rB 0 0 0 0 0 rP 0 0 0 Q 1Q0 Q2 2 0 0 g 0 M ( E1 ) g 0 g 0Q g Q g 1 0 2 0 g 1 1 0 g 1 0 g 0 0 0 0 0 0 Q1 . 0 r1 From M ( E1 ) , we note that characteristic roots namely, rB 0 , rP 0 and r1 are positive, giving a saddle point which is stable in the TP TS plane and unstable in the B N I , space. Therefore, equilibrium point E1 is unstable. The variational matrix M ( E2 ) at equilibrium point E2 is given by 0 0 0 rB 0 L rP1 N rP 0 1 L 0 0 Q 1Q0 Q2 2 0 0 g 0 M E 2 g 0 g 0Q g Q g 1 0 2 0 g 1 1 0 g 1 0 g L 2L 0 0 0 0 Q1 , 0 r1 From M ( E1 ) , we note that characteristic roots namely, rB 0 L, and rP 0 1 L are positive, giving a saddle point which is stable in the TP TS I space and unstable in the B N , plane. Therefore, equilibrium point E 2 , is unstable. The variational matrix M ( E3 ) at equilibrium point E 3 is given by ~ rB N ~ rP1 N M ( E3 ) ~ T 1 P ~ 1TS 0 0 ~ rP 0 N ~ ~ M TP , TS ~ Q 2 2 TP ~ 2TS 0 0 ~ rP 0 N 2 M 1 ~ ~ M 2 TP , TS ~ 0 2N g g 0 ~ rP 0 N 2 M 2 ~ ~ M 2 TP , TS 0 ~ ( 1 2 N ) 0 0 164 0 ~ 1N , Q1 0 ~ r1 2 N ~ ~ From M ( E3 ) , we note that characteristic roots namely, rB N and r1 2 N are positive, giving a saddle point which is stable in the N TP TS space and unstable in the B I , plane. Therefore, equilibrium point E 3 is unstable. The variational matrix M ( E4 ) at equilibrium point E4 is given by ~ ~ ~ ~ r ( N ) I 0 B ~ ~ ~ rP 0 N 2 ~ r N P1 ~ ~ ~ ~ M TP , TS ~ ~ ~ ~ M (E4 ) 1TP Q 2 2 TP ~ ~ ~ ~ T T 1 S 2 S ~ ~ ~ ~ I 2I 0 0 ~ ~ ~2 ~ rP 0 N M 1 rP 0 N 2 M 2 ~ ~ ~ ~ ~ ~ ~ ~ M 2 TP , TS M 2 TP , TS ~ ~ 0 2 N g 0 ~ ~ g ( 1 2 N ) 0 0 , Q1 0 ~ ~ r1 I L 0 ~ ~ 1N ~ ~ ~ ~ From M ( E4 ) , we note that characteristic root namely, rB ( N ) I is positive, giving a saddle point which is stable in the N TP TS I space and unstable in the B , direction. Therefore, equilibrium point E4 is unstable. The variational matrix M ( E5 ) at equilibrium point E 5 is given by rB 0 Bˆ 2 ˆ ˆ K B T P , TS 0 M ( E5 ) ˆ 1TP Tˆ 1 S 0 rB1 Bˆ rP Bˆ rB 0 Bˆ 2 K B1 K 2 Tˆ , Tˆ B P S 0 Q2 2TˆP Tˆ 0 1 Bˆ g 0 0 g 2 S rB 0 Bˆ 2 K B 2 K 2 Tˆ , Tˆ Bˆ B P S 0 0 , 0 Q1 ( 1 1 Bˆ ) 0 0 r1 Bˆ From M ( E5 ) , we note that characteristic roots namely, rP Bˆ and r1 Bˆ are positive, giving a saddle point which is stable in the B TP TS space and unstable in the N I , plane. Therefore, equilibrium point E 5 is unstable. 165 The variational matrix M ( E6 ) at equilibrium point E6 is given by ˆ rB 0 Bˆ K Tˆˆ , Tˆˆ B P S ˆ rP1 Nˆ M ( E6 ) 1TˆˆP ˆ 1TˆS 0 ˆ rB1 Bˆ ˆ rP 0 Nˆ ˆ ˆ M TˆP , TˆS ˆˆ Q 2 2 TP ˆ 2TˆS ˆ rB 0 Bˆ 2 K B1 ˆ ˆ K B2 TˆP , TˆS ˆˆ 2 r N M ˆ rB 0 Bˆ 2 K B 2 ˆ ˆ K B2 TˆP , TˆS ˆˆ 2 r N M ˆ ˆ M 2 TˆP , TˆS ˆ ˆ M 2 TˆP , TˆS d11 0 g d 22 0 0 P0 0 ˆ ˆ where d11 0 1 Bˆ 2 Nˆ g , 1 P0 2 ˆˆ 1N , Q1 0 ˆ ˆ r1 Bˆ 2 Nˆ ˆ Bˆ ˆ ˆ d 22 1 1 Bˆ 2 Nˆ . From M ( E6 ) , we note that characteristic root namely, ˆ ˆ r1 Bˆ 2 Nˆ is positive, giving a saddle point which is stable in the B N TP TS space and unstable in the I direction. Therefore, E6 is unstable. The variational matrix M ( E7 ) at equilibrium point E7 is given by rB 0 B K T B P , TS 0 M ( E7 ) T 1 P 1TS I rB 0 B 2 K B1 rB1 B K B2 TP , TS rP B 1 I 0 Q 2 2 TP 0 1 B g 2TS g 2I 0 rB 0 B 2 K B 2 K B2 TP , TS 0 B 0 0 Q1 , ( 1 1 B) 0 r1 I 0 L From M ( E7 ) , we note that characteristic root namely, rP B 1 I is positive, giving a saddle point which is stable in the B TP TS I space and unstable in the N direction. Therefore, E7 is unstable. 166 The variational matrix M (E*) at equilibrium point E * is given by * M (E ) rB 0 B* K B TP* , TS* 2 rB 0 B * K B1 K 2 T *,T * rB1 B* B rP 0 N * M T *,T * P S 2 1TP* Q2 2TP* 1TS* 2TS* g I * 2I* 0 P S rP 0 N * M 1 M 2 T *,T * Q I * , N * rP1 N * P S 2 rB 0 B* K B 2 K 2 T *,T * B P S 2 rP 0 N * M 2 M 2 T *,T * P 0 TP* gTP* TS* 0 S * B 1N * Q1 , 0 * rI 1 L In the following theorem we show that E * is locally asymptotically stable. Theorem 6.4.1: In addition to assumptions (H1) – (H5), let the following inequalities hold rP1 N * 1TP* 1TS* I * rB 0 B* , K B (TP* , TS* ) rB1 B* Q2 - 2TP* 2TS* 2 I * rP 0 N * , M (T * , T * ) P (6.4.1) (6.4.2) S 2 2 K B1 M1 Q( I * , N * ) rB 0 B* rP 0 N * g , K B2 (TP* , TS* ) M 2 (TP* , TS* ) TP* (6.4.3) 2 2 K B2 M2 gTP* * * rB 0 B rP 0 N , K B2 (TP* , TS* ) M 2 (TP* , TS* ) TS* (6.4.4) r1 I * * * B 1 N Q1 . L (6.4.5) then E * is locally asymptotically stable. 167 Proof: If inequalities (6.4.1) – (6.4.5) hold, then by Gerschgorin’s theorem (Lancaster and Tismenetsky, 1985), all eigenvalues of M (E * ) have negative real parts and interior equilibrium E * is locally asymptotically stable. 6.4.2. Global Stability Theorem (6.4.2): In addition to the assumption (H1) – (H5), let rB ( N ), rP ( B), K B TP , TS , M TP , TS and Q ( I , N ) satisfy the conditions 0 rB N 1 , 0 -rP B 2 , M n M TP , TS M 0 , K m K B (TP , TS ) K B 0 ,0 0 M m2 , TS K K Q Q M 3 ,0 4 , 0 B k1 , 0 B k 2 , 0 m1, , in for I N TP TS TP some positive constants 1 , 2 , 3 , 4 , k1 , k 2 , K 0 , K m , M 0 , M n , m1 , m2 . (6.4.6) Then if the following inequalities hold rB 0 rP 0 , 4 K (T * , T * ) M (T * , T * ) B P S P S 1 2 2 1 (6.4.7) 2 k 1Qm rB 0 K B 0 1 2 Km rB 0 1 0 g 1 B* 2 N * , * * 4 K B (TP , TS ) k 1Qm rB 0 K B 0 2 2 Km 1 rB 0 1 1 B* 2 N * , * * 3 K B (TP , TS ) 2 rB 0 r1 , 3 K (T * , T * ) L B P S 2 1 (6.4.8) (6.4.9) (6.4.10) 2 rP 0 m 1 4 2 Qm rP 0 N m 12 ( 0 g 1 B* 2 N * ), * * 4 M (T , T ) Mn P S 168 (6.4.11) 2 rP 0 m 1 2 Qm rP 0 N m 22 ( 1 1 B* 2 N * ), M n 3 M (T * , T * ) P S (6.4.12) rP 0 r1 , * * 3 M (T , T ) L 1 2 2 1 P (6.4.13) S g 2 1 ( 1 1 B* 2 N * )( 0 g 1 B* 2 N * ), (6.4.14) 3 32 1 r1 ( 0 g 1 B* 2 N * ), 3L (6.4.15) E * is globally asymptotically stable with respect to all solutions initiating in the positive orthant . Proof: Consider the following positive definite function about E * B N 1 TP TP* N N * N * ln V B, N , TP , TS , I B B* B* ln B* N* 2 2 1 I . (TS TS* ) 2 I I * I * ln * 2 I Differentiating V with respect to time t, we get dTS I I * dI dTP dV B B* dB N N * dN TP TP* TS TS* . dt B dt N dt dt dt I dt Substituting values of dI dB dN dP1 dP2 and from the system of equation (6.1.1) , , , dt dt dt dt dt in the above equation and after doing some algebraic manipulations and considering functions, r N r N * B B , B N N N * * rB N , N N *, N N* 169 (6.4.16) r B r B * P P , P B B B * * rP B , B B*, (6.4.17) B B* Q I , N Q I * , N , I I *, * I I Q1 I , N Q I * , N , I I *, I (6.4.18) 1 1 K T , T K B TP* , TS B P S , * T T B1 TP , TS P P K B TP* , TS 1 , K 2 T *,T TP B P S (6.4.19) TP TP* , TP TP* , 1 1 * * * K B (TP , TS ) K B (TP , TS ) , TS TS* B 2 (TP* , TS ) K B TP* , TS* 1 , K 2 T *,T * TS B P S (6.4.20) 1 1 M (T , T ) P S M (TP* , TS ) , * T T P1 (TP , TS ) P P M TP* , TS* 1 , M 2 T *,T TP P S TS TS* , TS TS* TP TP*, (6.4.21) TP TP* 1 1 * * * M (TP , TS ) M (TP , TS ) , TS TS* P 2 (TP* , TS ) M TP* , TS* 1 , M 2 T *,T * TS P S 170 TS TS* , (6.4.22) TS TS* Q 2 Q I *, N Q I *, N * , * * N N I ,N Q I *, N * , N N N *, (6.4.23) N N *, we get dV 1 1 a11 ( B B* ) 2 a12 ( B B* )( N N * ) a 22 ( N N * ) 2 dt 4 4 1 1 a11 ( B B* ) 2 a13 ( B B* )(TP TP* ) a33 (TP TP* ) 2 , 4 4 1 1 a11 ( B B* ) 2 a14 ( B B* )(TS TS* ) a 44 (TS TS* ) 2 , 4 3 1 1 a11 ( B B* ) 2 a15 ( B B* )( I I * ) a55 ( I I * ) 2 , 4 3 1 1 a 22 ( N N * ) 2 a 23 ( N N * )(TP TP* ) a33 (TP TP* ) 2 , 4 4 1 1 a 22 ( N N * ) 2 a 24 ( N N * )(TS TS* ) a 44 (TS TS* ) 2 , 4 3 1 1 a 22 ( N N * ) 2 a 25 ( N N * )( I I * ) a55 ( I I * ) 2 , 4 3 1 1 a33 (TP TP* ) 2 a34 (TP TP* )(TS TS* ) a 44 (TS TS* ) 2 , 4 3 1 1 a33 (TP TP* ) 2 a35 (TP TP* )( I I * ) a55 ( I I * ) 2 , 4 3 where a11 rB 0 rP 0 , a12 B N P ( B), a22 , a23 - rP 0 N P1 (TP , TS ), K B (TP* , TS* ) M (TP* , TS* ) a44 1 1 B* 2 N * , a55 r1 , a15 , a24 rP 0 N P 2 (TP* , TS ) 2TS , L 171 a35 Q1 I , N , a13 1TP rB 0 B P1 (TP , TS ), a14 1TS rB 0 B 2 (TP* , TS ) a 25 1 2 , a33 0 g 1 B* 2 N * , a34 g. Then sufficient conditions for dV to be negative definite are that the following dt inequalities hold a122 1 1 1 1 1 2 2 a11a 22 , a132 a11a33 , a142 a11a 44 , a15 a11a55 , a 23 a 22 a33 , 4 4 3 3 4 1 1 1 1 2 2 2 2 a 24 a 22 a 44 , a 25 a 22 a55 . a34 a33 a 44 , a 35 a 33 a 55 . 3 3 3 3 (6.4.24) Now, from (6.4.6) and mean value theorem, we note that B N 1 , P B 2 , Q1 I , N 3 , Q 2 I *, N 4 , P1 TP , TS P 2 (TP* , TS ) m2 Mn 2 , B1 TP , TS k1 Km 2 , B 2 (TP* , TS ) k2 Km 2 . m1 Mn 2 , (6.4.25) Further, we note that the stability conditions (6.4.7) - (6.4.15) as stated in theorem (6.4.2), can be obtained by maximizing the left-hand side of inequalities (6.4.24). This completes the proof of theorem (6.4.2). 6.5 NUMERICAL SIMULATIONS AND DISCUSSION To facilitate the interpretation of our mathematical findings by numerical simulation, we integrated system (6.1.1) using fourth order Runge - Kutta method. We take the following particular form of the functions involved in the model (6.1.1): rB N rB 0 rB1 N , rP B rP 0 rP1 B, K B TP , TS K B 0 K B1TP K B 2TS , M TP , TS M 0 M 1TP M 2TS , QI , N Q0 Q1 I Q2 N . 172 (6.5.1) Now we choose the following set of values of parameters in model (6.1.1) and equation (6.5.1). rB 0 11, rB1 0.2, K B 0 12.2, K B1 0.1, K B 2 0.3, 0.01, rP 0 20, rP1 0.1, M 0 10, M 1 0.1, M 2 0.2, 1 0.02, Q0 20, Q1 0.3, Q2 0.2, 0 14, 1 0.001, 2 0.08, g 5, 0.5, 1 17, 1 0.6, 2 0.1, r1 9, l 5, 0.1, 2 0.2, K m 0.001, k1 0.2, k 2 0.01, m1 0.02, m2 0.01, M n 1.3, 1 0.2, 2 0.1, 3 1, 4 0.1. (6.5.2) With the above values of parameters, we note that condition for the existence of E * are satisfied, and E * is given by B* 9.6912, N * 10.3966, TP* 1.2140, TS* 0.1272, I * 6.6936. (6.5.3) It is further noted that all conditions of local stability (6.4.1) – (6.4.5), global stability (6.4.7) – (6.4.15) are satisfied for the set of values of parameters given in (6.5.2). In Figures (1) – (3), the primary and secondary toxicants for different rate of emission of toxicants by natural sources or by industrialization and population Q0 , Q1 and Q2 , against time are plotted. From these plots, we can infer that as the rate of emission of toxicants either directly or due to industrialization or by population increases, equilibrium densities of both primary and secondary toxicants increases as expected. These figures, show that the qualitative behaviors of both primary and secondary toxicants are same whereas the quantitative behaviors of both toxicants are different as the concentration of toxicants in the environment increases. The density of primary toxicant is more in the environment than density of secondary toxicant by increase in concentration of toxicants due to population and industrialization. Figure (4), shows 173 the dynamics of resource-biomass for different depletion rate coefficient of resource biomass due to industrialization , w.r.t time ‘t’. This shows that density of resourcebiomass decreases as , increases. It is also noted that the resource-biomass density initially increases w.r.t time t and after certain time it settle down to its steady state. 1.8 1.6 1.4 TP(Q0=30) 1 TP(Q0=20) P T ,T S 1.2 0.8 0.6 TS(Q0=30) TS(Q0=20) 0.4 0.2 0 0 10 20 30 40 50 time (t) 60 70 80 90 100 Figure 1, Graph of TP and TS versus t for different Q0 and other values of parameters are same as in equation (6.5.2). 174 1.6 1.4 1.2 1 TP(Q1=0.8) P T ,T S TP(Q1=0.3) 0.8 0.6 0.4 TS(Q1=0.8) TS(Q1=0.3) 0.2 0 0 20 40 60 80 100 time (t) Figure 2, Graph of TP and TS versus t for different Q1 and other values of parameters are same as in equation (6.5.2) 1.5 TP(Q2=0.6) 1 P T ,T S TP(Q2=0.2) 0.5 TS(Q2=0.6) TS(Q2=0.2) 0 0 10 20 30 40 50 time (t) 60 70 80 90 100 Figure 3, Graph of TP and TS versus t for different Q2 and other values of parameters are same as in equation (6.5.2) 175 Figure 4, Graph of B versus t for different and other values of parameters are same as in equation (6.5.2). 10 9.8 Resource Biomass ( B ) 9.6 9.4 9.2 Presence of Secondary toxicant, TS Absence of Secondary toxicant, TS 9 8.8 8.6 8.4 8.2 8 0 1 2 3 4 5 time ( t ) 6 7 8 9 10 Figure 5, Graph of B versus t for presence and absence of secondary toxicant and other values of parameters are same as in equation (6.5.2). 176 Figure (5), shows the dynamics of resource biomass w.r.t time ‘t’ for presence and absence of secondary toxicant. It is found that equilibrium level of resource biomass is greater in the absence of secondary toxicant than in the presence of it. Figures (6) and (7), shows the effects of K B1 and K B 2 , on the dynamics of resource-biomass w.r.t time ‘t’. In both cases the carrying capacity of resource-biomass decreases with increase in density of primary and secondary toxicant. These figures also show that primary pollutant has more adverse effect on the resource-biomass carrying capacity for a larger period than secondary toxicant. Similar behavior can be seen in Figures (8) and (9), which is plotted between population and time ‘t’ for different values of M 1 and M 2 , respectively. Figure 6, Graph of B versus t for different k B1 and other values of parameters are same as in equation (6.5.2). 177 Figure 7, Graph of B versus t for different k B 2 and other values of parameters are same as in equation (6.5.2). Figure 8, Graph of N versus t for different M1 and other values of parameters are same as in equation (6.5.2). 178 Figure 9, Graph of N versus t for different M 2 and other values of parameters are same as in equation (6.5.2). 6.6 CONCLUSION In this chapter, a nonlinear mathematical model to study the effects of industrialization, population, primary–secondary toxicants on depletion of forestry resource is proposed and analyzed. It is assumed that primary toxicant is emitted into the environment with a constant prescribed rate as well as its growth is increased by increase in density of population and industrialization. Further, a part of primary toxicant is transformed into secondary toxicant, which is more toxic, both affecting the resource biomass and population simultaneously. Criteria for local stability, instability and global stability are obtained by using stability theory of differential equation. It is found that if the densities of industrialization and population increases, then the density of primary toxicant become very large into the environment due to 179 which the densities of resource biomass and population decreases & it settle down at its equilibrium level whose magnitude is lower than its original carrying capacity. From the analysis it is also observed that in the absence of secondary toxicant, the equilibrium level of resource biomass is more, than in the presence of it. It is also found that due to high level of primary toxicant in the environment, it led in large transformation of secondary toxicant, which decreases the densities of resource biomass and population more than the case of single toxicant. 180
© Copyright 2026 Paperzz