Stability Analysis of Plankton Ecosystem Model. Affected by Oxygen Deficit

Transcription

1 Applied Mathematical Sciences Vol no HIKARI Ltd wwwm-hikaricom Stability Analysis of Plankton Ecosystem Model Affected by Oxygen Deficit Yuriska Destania Department of Mathematics Bogor Agricultural University Jalan Meranti Kampus IPB Dramaga Bogor Indonesia Jaharuddin Department of Mathematics Bogor Agricultural University Jalan Meranti Kampus IPB Dramaga Bogor Indonesia Paian Sianturi Department of Mathematics Bogor Agricultural University Jalan Meranti Kampus IPB Dramaga Bogor Indonesia Copyright 2015 Yuriska Destania Jaharuddin and Paian Sianturi This article is distributed under the Creative Commons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited Abstract Plankton is one of the main components in the food chain system in water In this paper the plankton ecosystems studied by formulating a mathematical model in nonlinear ordinary differential equations system where the growth of plankton influenced by oxygen deficit This model includes four variables ie concentration of nutrients density of algae density of zooplankton and concentration of dissolved oxygen There are four equilibrium points obtained Based on stability analysis conducted one of the equilibrium points was stable under certain conditions Numerical simulations were conducted to convince that the ecosystem reaches a stable condition Mathematics Subject Classification: 93A30 93C15 93D05

2 4044 Yuriska Destania Jaharuddin and P Sianturi Keywords: Algae Dissolved oxygen Nutrients Oxygen deficit Zooplankton 1 Introduction Planktons are aquatic organisms that have particularly microscopic size usually float in the water motion capabilities are limited so easily swept away Based on the type plankton were divided into two groups namely plants (phytoplankton) and animals (zooplankton) Phytoplankton are capable of doing photosynthesis produce organic compound which is the source of energy for other aquatic organisms phytoplankton growth is influenced by the availability of nutrients (nitrate and phosphate) in the water In contrast a group of zooplankton is heterotrof its survival depends on phytoplankton Plankton need oxygen to survive Dissolved oxygen in the water can be derived from the atmosphere through the process of diffusion agitation of the water surface and the results of the photosynthesis of plants in the water Edward and Brindley [1] discusses mortality and behavior of the plankton population without involving the concentration of dissolved oxygen Specific research on the prediction of dissolved oxygen was done by Naik and Manjapp [4] with the involvement of oxygen deficit Also dissolved oxygen becomes an important discussion in the plankton ecosystem [378] Khare et al [2] presents the plankton ecosystem models involving nutrients algae zooplankton and dissolved oxygen Different from previous studies on the research of Khare et al [2] the growth of algae and zooplankton influenced by oxygen deficit The model is interesting to discuss because the role of algae as a supplier of oxygen through photosynthesis process have not been considered In this work the model used is a modification of model by Khare et al [2] Modifications conducted by considering the role of algae as supplier of oxygen from the photosynthesis process with reviewing the results of previous studies [3678] The paper is organized into six sections The first section is the background and purpose of this paper The second section describes the formulation of the model used The third section determine the equilibrium points and the stability properties of equilibrium points will be analyzed in sections four The fifth section is numerical simulation using certain values of the parameters The conclusions are provided in section six 2 Mathematical Formulation Formulation of the plankton ecosystem models such as a nonlinear ordinary differential equations system involving four variables namely the concentration of nutrients (n) the density of algae (a) the density of zooplankton (Z) and the concentration of dissolved oxygen (C) It is assumed that the nutrients from the outside get into the water at a rate of

3 Stability analysis of plankton ecosystem 4045 q and the depletion rate of nutrients by natural factors is αn The predation rate of nutrients by algae is β 1 Further assumed that the algae growth rate is proportional to the concentration of nutrients density of algae and oxygen deficit The natural death rate of algae is v 1 the predation rate of algae by zooplankton is α 2 The growth rate of zooplankton is proportional to the density of algae and zooplankton but inversely proportional with the oxygen deficit The natural death rate of zooplankton is v 3 The concentration of dissolved oxygen enters the system from a variety of sources assumed to p Algae give the supply of oxygen through photosynthesis process as big as λ 1 a The natural depletion rate of oxygen is v 2 C 0 is the saturation value of dissolved oxygen and C 0 C is the oxygen deficit Based on the above assumptions the plankton ecosystem models is given as follows: dn dt = q αn β 1na da dt = β 2 na (α 1 + C 0 C) v 1a α 2 az dz dt = dc dt = p v 2C + λ 1 a α 3 az (α 4 + C 0 C) v 3Z (1) with n(0) > 0 a(0) > 0 Z(0) > 0 C(0) > 0 β 2 and α 3 are positive constants α 1 and α 4 are saturation constants Furthermore we will show that the system (1) is bounded by following Lemma 1 Lemma 1 The set Ω = {(n a Z C)εR n + a + Z q C pδ m+λ 1 q } is δ m v 2 δ m a region of attraction for all solutions initiating in the positive octan where δ m = min {α v 1 v 2 } Proof Use the first three equations of system (1) we obtain d dn (n + a + Z) = dt dt + da dt + dz dt q αn v 1 a v 3 Z (2) Suppose that δ m = min {α v 1 v 2 } then the equation (2) can be written as d dt (n + a + Z) q δ m(n + a + Z) d dt (n + a + Z) + δ m(n + a + Z) q dw dt + δ mw q (3) where w = n + a + Z

4 4046 Yuriska Destania Jaharuddin and P Sianturi Solve the equation (3) using integrating factor [5] thus w q + ke t δ m δ m As t we have 0 w q δ m 0 n + a + Z q (4) δ m Use the fourth equation of system (1) and the substitution the maximum values of a from equation (4) dc dt p v 2C + λ 1 q δ m dc dt + v 2C pδ m + λ 1 q (5) δ m Solve the equation (5) using integrating factor [5] thus 0 C pδ m + λ 1 q v 2 δ m + ke t v 2 As t we have 0 C pδ m+λ 1 q v 2 δ m 3 Equilibrium Analysis The equilibrium point of the system (1) is obtained from the condition dn dt = da 0 = 0 dz = 0 and dc = 0 The system generates four equilibria dt dt dt E i (n i a i Z i C i ) where i = i E 1 ( q 0 0 q 0 ) always exist α v 2 iii E 3 (n 3 a 3 0 C 3 ) ii E 2 (n 2 a 2 0 C 2 ) iv E 4 (n a Z C ) where a 2 and a 3 are positive root of the following quadratic equation β 1 v 1 λ 1 a 2 + (αv 1 λ 1 β 1 v 1 (v 2 α 1 + v 2 C 0 p))a + (β 2 v 2 q αv 1 (v 2 α 1 +v 2 C 0 p)) = 0 Given that the following conditions satisfied (i) D 0 (ii) β 1 v 1 (v 2 α 1 + v 2 C 0 p) αv 1 λ 1 > 0 (iii) β 2 v 2 q αv 1 (v 2 α 1 + v 2 C 0 p) > 0 (6)

5 Stability analysis of plankton ecosystem 4047 Using the value of a 2 and a 3 we get a positive value of C 2 n 2 and C 3 n 3 as follows C 2 = p + λ 1a 2 n v 2 = v 1(α 1 + C 0 C 2 ) 2 β 2 C 3 = p + λ 1a 3 ) n v 3 = v 1(α 1 + C 0 C 3 2 β 2 On condition of E 2 and E 3 zooplankton is not present For the equilibrium points E 4 n q = α + β 1 a a = v 3(v 2 α 4 + v 2 C 0 p) v 2 α 3 + v 3 λ 1 Z = β 2n 4 v 1 (α 1 + C 0 C ) α 2 (α 1 + C 0 C C = p + λ 1a ) v 2 The equilibrium points E 4 exists if β 2 n v 1 (α 1 + C 0 C ) > 0 At the equilibrium points E 4 all variables have a role in the system (1) As living organisms algae and zooplankton require oxygen to survive Furthermore the concentration of nutrients in the water also affects the growth of algae which is a source of food for zooplankton and the process of photosynthesis by algae is able to provide the supply of oxygen in the water The populations of algae is higher when the zooplankton is absent compared when the zooplankton is present [3] Thus a 2 > a and a 3 > a 4 Stability Analysis In this section we will discussed the stability behaviors of E 1 E 2 E 3 and E 4 in detail For that there will be a linearization process on the system (1) to produce a Jacobian matrix is given as follows: α β 1 a i β 1 n i J i = [ β 2 a i (α 1 +C 0 C i ) 0 β 2 n i (α 1 +C 0 C i ) v 1 α 2 Z i α 3 Z i (α 4 +C 0 C i ) 0 λ α 2 a i α 3 a i (α 4 +C 0 C i ) v 3 β 2 n i ai (α 1 +C 0 C i ) 2 α 3 a i Zi (α 4 +C 0 C i ) 2 0 v 2 ] For the equilibrium point E 1 The eigenvalues for J 1 are α v 2 v 3 dan β 2 v 2 q αv 1 (v 2 α 1 +v 2 C 0 p) αv 2 (α 1 +C 0 C 1 ) Since three eigenvalues of J 1 are negatives then stability of E 1 will depend on

6 4048 Yuriska Destania Jaharuddin and P Sianturi the sign of β 2 v 2 q αv 1 (v 2 α 1 +v 2 C 0 p) αv 2 (α 1 +C 0 C 1 ) As we know that α v 2 α 1 > 0 and C 0 > C 1 so αv 2 (α 1 + C 0 C 1 ) > 0 Furthermore we know too that ( β 2 v 2 q αv 1 (v 2 α 1 + v 2 C 0 p)) > 0 whenever E 2 exists (6) Thus it can be concluded that E 1 unstable For the equilibrium point E 2 From matrix J 2 we may easly note that one of the eigenvalues of J 2 is α 3 a 2 v 3 (α 4 +C 0 C 2 ) (α 4 +C 0 C 2 ) As we know that α 4 > 0 and C 0 > C 2 so (α 4 + C 0 C 2 ) Stability of E 2 will depend on the sign of (α 3 a 2 v 3 (α 4 + C 0 C 2 )) α 3 a 2 v 3 (α 4 + C 0 C 2 ) = (v 2α 3 + v 3 λ 1 )a 2 v 3 (v 2 α 4 + v 2 C 0 p) v 2 We have already noted that a 2 > a this implies that α 3 a 2 v 3 (α 4 + C 0 C 2 ) > (v 2α 3 + v 3 λ 1 )a v 3 (v 2 α 4 + v 2 C 0 p) v 2 By substituting a will we get that α 3 a 2 v 3 (α 4 + C 0 C 2 ) > 0 Thus it can be concluded that E 2 unstable For the equilibrium point E 3 From matrix J 3 we may easly note that one of the eigenvalues of J 3 is α 3 a 3 v 3 (α 4 +C 0 C 3 ) (α 4 +C 0 C 3 ) As we know that α 4 > 0 and C 0 > C 3 so (α 4 + C 0 C 3 ) Stability of E 3 will depend on the sign of (α 3 a 3 v 3 (α 4 + C 0 C 3 )) In a similar way as E 2 we will get that α 3 a 3 v 3 (α 4 + C 0 C 3 ) > 0 and E 3 is unstable too For the equilibrium point E 4 As we cannot say much about the stability behavior of E 4 from matrix J 4 we study the stability behavioor of this equilibrium by using Lyapunov s method Theorem 1 The equilibria E 4 is nonlinear stable in Ω if the following conditions are statisfied m 1 β 2 [ α 1 + C 0 pδ β m+λ 1 q 1 n ] v 2 δ m m 2 α 3 [ (α 4 + C 0 pδ m+λ 1 q ) ] v 2 δ m 2 2 < 2m 1αα 2 3 < m α 2 6

7 Stability analysis of plankton ecosystem 4049 [ m 2 α 3 a [ (α 4 + C 0 pδ m+λ 1 q ) (α v 2 δ 4 + C 0 C ) ] m m 1 β 2 n (α 1 + C 0 pδ m+λ 1 q v 2 δ m ) (α 1 + C 0 C ) + m 3λ 1 ] 2 2 < m 1m 3 v 2 α 2 2 < m 1m 3 v 2 α 2 3 Proof: To prove this theorem we consider the following positive definite function V = 1 2 (n n ) 2 + m 1 (a a a In a a ) + m 2 (Z Z Z In Z Z ) m 3(C C ) 2 (7) where m 1 m 2 and m 3 are positive constant that is chosen appropriately Differentiating equation (7) to obtain dv dt = (n n ) dn dt + m 1 (a a ) da a dt + m 2 (Z Z ) dp Z dt + m 3(C C ) dc dt (8) Substitution the equation of system (1) into equation (8) and then do some algebraic manipulation and use the properties of inequality a 2 + b 2 2ab Thus obtained the following inequality: where dv dt β 1a(n n ) 2 α(n n ) 2 m 3 v 2 (C C ) 2 m 1α 2 2 (a a ) 2 m 1α 2 2 (Z Z ) 2 m 1 β 2 + ( α 1 + C 0 C β 1n ) (n n )(a a ) m 1 β 2 n + ( (α 1 + C 0 C)(α 1 + C 0 C ) + m 3λ 1 ) (a a )(C C ) + m 2α 3 a (Z Z )(C C ) (α 4 + C 0 C)(α 4 + C 0 C ) + m 2α 3 (a a )(Z Z ) α 4 + C 0 C dv dt β 1a(n n ) r 11(n n ) 2 + r 12 (n n )(a a ) 1 2 r 22(a a ) r 22(a a ) 2 + r 23 (a a )(Z Z ) 1 2 r 33(Z Z ) r 33(Z Z ) 2 + r 34 (Z Z )(C C ) 1 2 r 44(C C ) r 22(a a ) 2 + r 24 (a a )(C C ) 1 2 r 44(C C ) 2

8 4050 Yuriska Destania Jaharuddin and P Sianturi r 11 = 2α r 12 = r 23 = m 1 β 2 α 1 + C 0 C β 1n r 22 = m 1α 2 3 m 2α 3 a α 4 + C 0 C r 33 = m 1α 2 2 r 34 = r 44 = m 3 v 2 r 24 = m 1 β 2 n (α 1 + C 0 C)(α 1 + C 0 C ) + m 3λ 1 m 2 α 3 a (α 4 + C 0 C)(α 4 + C 0 C ) Based on Lyapunov s stability method the equilibria E 4 is stable if dv 0 dt Sufficient conditions so dv 0 is to satisfy the following inequality: dt r 12 2 < r 11 r 22 r 23 2 < r 22 r 33 r 34 2 < r 33 r 44 r 24 2 < r 22 r 44 (9) The equilibria E 4 is stable if conditions (9) are met 5 Numerical Simulation In this work we conducted some numerical computation by choosing the following value for model system in (1) These parameter values were originated of Khare et al [2] and Misra [3] q = 3 mg l -1 day -1 α = 01 day -1 β 1 = 05 l mg -1 day -1 β 2 = 035 day -1 α 1 = 051 mg l -1 C 0 = 30 mg l -1 v 1 = 0009 day -1 α 2 = 041 l mg -1 day -1 α 3 = 033 day -1 α 4 = 03 mg l -1 v 3 = 001 day -1 p = 24 mg l -1 day -1 v 2 = 3 day -1 λ 1 = 02 day -1 It is found that the under above set of parameter conditions for existence of interior equilibrium E 4 are satisfied and E 4 is given by n = 6862 n = 0674 Z = 0239 C = 8045 The eigenvalue of the matrix J 4 corresponding to this equilibrium E 4 are obtained as which are all negative With the above values of parameter we note that the conditions for nonlinear stability in Theorem 1 are statisfied In Figure 1 we can see that the system (1) approaching the point E4 with asymptotically The simulation was conducted with some initial value for sufficient long period At the beginning of the simulation algae populations declined drastically in a short time This is due to oxygen deficit and predation by zooplankton The decline of the population of algae and oxygen deficit result in a decrease zooplankton population In contrast the concentration of nutrients have increased sharply due to predation by algae decreased After a certain time the system will experience a turning point and reach a stable condition at E 4

9 Stability analysis of plankton ecosystem Conclusions Figure 1 The field of solution for E 4 In this paper a nonlinear mathematical model for plankton ecosystems has been modified and analyzed The model exhibits four equilibrium points Stability analysis shows that one of the equilibrium points will be stable under certain conditions Numerical simulations give the same result that this ecosystem will reach the stable condition Numerical simulations also show the dynamics that occur until the ecosystem reaches a stable condition References [1] A M Edwards J Brindley Zooplankton mortality and the dynamical behavior of plankton population models Bulletin of Mathematical Biology 61 (1999)

10 4052 Yuriska Destania Jaharuddin and P Sianturi [2] S Khare S Kumar and C Singh Modelling effect of the depleting dissolved oxygen on the existence of interacting planktonic population Elixir Appl Math 55 (2013) [3] A K Misra Mathematical modeling and analysis of eutrophication of water bodies caused by nutrients Nonlinear Analysis: Modelling and Control 12 (2007) [4] V K Naik S Manjapp Prediction of dissolved oxygen through mathematical modeling Int J Environ Res 4 (2010) [5] J C Robinson An Introduction to Ordinary Differential Equations Cambridge University Press [6] G Sakalauskienė Dissolved oxygen balance model for Neris Nonlinear Analysis: Modelling and Control 6 (2001) [7] J B Shukla A K Misra and P Chandra Mathematical modelling and analysis of the depletion of dissolved oxygen in eutrophied water bodies affected by organic pollutants Nonlinear Analysis: Real world Applications 9 (2008) [8] J B Shukla A K Misra and P Chandra Modeling and analysis of the algal bloom in a lake caused by discharge of nutrients App Math and Comp 196 (2008) Received: March ; Published: May