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A NUMERICAL METHOD FOR APPROXIMATING THE SOLUTION OF A LOTKA-VOLTERRA SYSTEM WITH TWO DELAYS

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A NUMERICAL METHOD FOR APPROXIMATING THE SOLUTION OF A LOTKA-VOLTERRA SYSTEM WITH TWO DELAYS
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  STUDIA UNIV. “BABES¸–BOLYAI”, MATHEMATICA, Volume  L , Number 1, March 2005 A NUMERICAL METHOD FOR APPROXIMATING THE SOLUTIONOF A LOTKA-VOLTERRA SYSTEM WITH TWO DELAYS DIANA OTROCOL Abstract . In this paper, using the step method, we established the exis-tence and uniqueness of solution for the system (1.2) with initial condition(1.3). The aim of this paper is to present a numerical method for thissystem. 1.  The statement of the problem Consider the following Lotka-Volterra type delay differential system:  x  i ( t ) =  x i ( t ) r i ( t )  c i − a i x i ( t ) − n  j =1 m  k =0 a kij x j ( τ  kij ( t ))  , t ≥ t 0 ,  1 ≤ i ≤ nx i ( t ) =  φ i ( t ) ≥ 0 , t ≤ t 0  and  φ i ( t 0 )  >  0 ,  1 ≤ i ≤ n (1)There have been many studies on this subject (see [2], [5], [7]). In particular, for n  = 2 , r i ( t )  ≡  1 , a i  = 0 and  τ  kij ( t ) =  t − τ  kij ,  1  ≤  i,j  ≤  2 ,  0  ≤  k  ≤  m , the factthat time delays are harmless for the uniform persistence of solutions, is establishedby Wang and Ma for a predator-prey system, by Lu and Takeuchi and Takeuchi forcompetitive systems.Recently, Saito, Hara and Ma [7] have derived necessary and sufficient condi-tions for the permanence (uniform persistence) and global stability of a symmetricalLotka-Volterra-type predator-prey system with  a i  >  0 , i  = 1 , 2 and two delays.For a nonautonomous competitive Lotka-Volterra system with no delays, re-cently Ahmad and Lazer have established the average conditions for the persistence, Received by the editors: 19.07.2004.2000  Mathematics Subject Classification.  34L05, 47H10. Key words and phrases.  Differential equation, delay, the step method. 99  DIANA OTROCOL which are weaker than those of Gopalsamy and Tineo and Alvarez for periodic oralmost-periodic cases.In this paper, using the step method [6], we established the existence anduniqueness of solution for the following system  x  ( t ) =  f  1 ( t,x ( t ) ,y ( t ) ,x ( t − τ  1 ) ,y ( t − τ  2 )) y  ( t ) =  f  2 ( t,x ( t ) ,y ( t ) ,x ( t − τ  1 ) ,y ( t − τ  2 )) , t ∈ [ t 0 ,b ] , t 0  < b  (2)with initial condition  x ( t ) =  ϕ ( t ) , t ∈ [ t 0 − τ  1 ,t 0 ] y ( t ) =  ψ ( t ) , t ∈ [ t 0 − τ  2 ,t 0 ](3)Here  τ  1  and  τ  2  are constants with  τ  1  ≥  0 , τ  2  ≥  0 , τ  1  ≤  τ  2  and  ϕ,ψ  are continuousfunctions.On the basis of these results, the aim of this paper is to present a numericalmethod for obtaining the solutions of system (2) with initial condition (3).2.  The existence and uniqueness of solution We consider the system (2) with initial condition (3) and we established theexistence and uniqueness of the solution for the problem (2) + (3).We have x  ∈  C  [ t 0 − τ  1 ,b ] ∩ C  1 [ t 0 ,b ] y  ∈  C  [ t 0 − τ  2 ,b ] ∩ C  1 [ t 0 ,b ]If we suppose that(i)  f  i  ∈ C  ([ t 0 ,b ] × R 4 ) , i  = 1 , 2(ii) | f  i ( t,u 1 ,v 1 ,u,v ) − f  i ( t,u 2 ,v 2 ,u,v ) |≤ L 1 | u 1 − u 2 | +  L 2 | v 1 − v 2 | , ∀ u 1 ,u 2 ,v 1 ,v 2 ,u,v  ∈ R , ∀ t ∈ [ t 0 ,b ](ii’) | f  i ( t,u 1 ,u 2 ,u 3 ,u 4 ) − f  i ( t,v 1 ,v 2 ,v 3 ,v 4 ) | ≤  L ( | u 1 − v 1 |  +  | u 2 − v 2 |  + | u 3 − v 3 | + | u 4 − v 4 | ) ∀ u i ,v i  ∈ R ,i  = 1 , 4 , ∀ t ∈ [ t 0 ,b ]then the following result is given. 100  A NUMERICAL METHOD Theorem 1.  We consider the system (2) with initial condition (3). If the conditions (i) and (ii) are satisfied, then the problem (2)+(3) has a unique solution.Proof.  We use the step method. t ∈ [ t 0 ,t 0  +  τ  1 ]  x  ( t ) =  f  1 ( t,x ( t ) ,y ( t ) ,ϕ ( t − τ  1 ) ,ψ ( t − τ  2 )) y  ( t ) =  f  2 ( t,x ( t ) ,y ( t ) ,ϕ ( t − τ  1 ) ,ψ ( t − τ  2 )) x ( t 0 ) =  ϕ ( t 0 ) y ( t 0 ) =  ψ ( t 0 )So we have the Cauchy problem with  f  i  continuous functions,  i  = 1 , 2. But f  i ( t, · , · ,u,v ) : R → R  are Lipschitz. Then it results from the Cauchy theorem that: ∃ !  x 1  ∈  C  1 [ t 0 ,t 0  +  τ  1 ] ∃ !  y 1  ∈  C  1 [ t 0 ,t 0  +  τ  1 ]solution of the problem (2) + (3). t ∈ [ t 0  +  τ  1 ,t 0  + 2 τ  1 ]  x  ( t ) =  f  1 ( t,x ( t ) ,y ( t ) ,x 1 ( t − τ  1 ) ,y 1 ( t − τ  2 )) y  ( t ) =  f  2 ( t,x ( t ) ,y ( t ) ,x 1 ( t − τ  1 ) ,y 1 ( t − τ  2 )) x ( t 0  +  τ  1 ) =  x 1 ( t 0  +  τ  1 ) y ( t 0  +  τ  1 ) =  y 1 ( t 0  +  τ  1 ) ⇒ ∃ ! x 2  ∈ C  [ t 0  +  τ  1 ,t 0  + 2 τ  1 ] ⇒ ∃ ! y 2  ∈ C  [ t 0  +  τ  1 ,t 0  + 2 τ  1 ]solution of the problem (2) + (3). t ∈ [ t 0  +  nτ  1 ,t 0  +  τ  2 ]  x  ( t ) =  f  1 ( t,x ( t ) ,y ( t ) ,x n ( t − τ  1 ) ,y n ( t − τ  2 )) y  ( t ) =  f  2 ( t,x ( t ) ,y ( t ) ,x n ( t − τ  1 ) ,y n ( t − τ  2 )) x ( t 0  +  nτ  1 ) =  x n ( t 0  +  nτ  1 ) y ( t 0  +  nτ  1 ) =  y n ( t 0  +  nτ  1 ) 101  DIANA OTROCOL ⇒ ∃ ! x n +1  ∈ C  [ t 0  +  nτ  1 ,t 0  +  τ  2 ] ⇒ ∃ ! y n +1  ∈ C  [ t 0  +  nτ  1 ,t 0  +  τ  2 ]So we obtained:( x ( t ) ,y ( t )) =  ( x 1 ( t ) ,y 1 ( t )) , t ∈ [ t 0 ,t 0  +  τ  1 ]( x 2 ( t ) ,y 2 ( t )) , t ∈ [ t 0  +  τ  1 ,t 0  + 2 τ  1 ] ... ( x n +1 ( t ) ,y n +1 ( t )) , t ∈ [ t 0  +  nτ  1 ,t 0  +  τ  2 ]solution of the problem (2) + (3). Remark 1.  We consider the system (2) with initial condition (3). If the conditions (i) si (ii’) are satisfied, then the problem (2)+(3) has a unique solution which can be obtained by the method of successive approximations. 3.  The approximation of the solution We consider the system (2) with initial condition (3)This problem is equivalent with the delayed integral Volterra equations: x ( t ) =  ϕ ( t ) , t ∈ [ t 0 − τ  1 ,t 0 ] ϕ ( t 0 ) +   tt 0 f  1 ( s,x ( s ) ,y ( s ) ,x ( s − τ  1 ) ,y ( s − τ  2 ))d s, t ∈ [ t 0 ,b ] y ( t ) =  ψ ( t ) , t ∈ [ t 0 − τ  2 ,t 0 ] ψ ( t 0 ) +   tt 0 f  2 ( s,x ( s ) ,y ( s ) ,x ( s − τ  1 ) ,y ( s − τ  2 ))d s, t ∈ [ t 0 ,b ]where  f  i  ∈ C  ([ t 0 ,b ] × R 4 ) , i  = 1 , 2.We suppose that the hypotheses of Remark 1 are satisfied. Then the problem(2) + (3) has a unique solution x  ∈  C  [ t 0 − τ  1 ,t 0 ] ∩ C  1 [ t 0 ,b ] y  ∈  C  [ t 0 − τ  2 ,t 0 ] ∩ C  1 [ t 0 ,b ] . 102  A NUMERICAL METHOD Let ( α,β  ) be the solution, which, by virtue of Remark 1, can be obtained by successiveapproximation method. So, we have α ( t ) =  ϕ ( t ) , t ∈ [ t 0 − τ  1 ,t 0 ] β  ( t ) =  ψ ( t ) , t ∈ [ t 0 − τ  2 ,t 0 ]For  t ∈ [ t 0 ,b ] we have:  α 0 ( t ) =  ϕ ( t ) β  0 ( t ) =  ψ ( t )(4)  α 1 ( t ) =  ϕ ( t 0 ) +   tt 0 f  1 ( s,α 0 ( s ) ,β  0 ( s ) ,α 0 ( s − τ  1 ) ,β  0 ( s − τ  2 )) dsβ  1 ( t ) =  ψ ( t 0 ) +   tt 0 f  2 ( s,α 0 ( s ) ,β  0 ( s ) ,α 0 ( s − τ  1 ) ,β  0 ( s − τ  2 )) ds...  α m ( t ) =  ϕ ( t 0 ) +   tt 0 f  1 ( s,α m − 1 ( s ) ,β  m − 1 ( s ) ,α m − 1 ( s − τ  1 ) ,β  m − 1 ( s − τ  2 )) dsβ  m ( t ) =  ψ ( t 0 ) +   tt 0 f  2 ( s,α m − 1 ( s ) ,β  m − 1 ( s ) ,α m − 1 ( s − τ  1 ) ,β  m − 1 ( s − τ  2 )) ds To obtain the sequence of successive approximations (4), it is necessary tocalculate the integrals which appear in the right-hand side. In general, this problemis difficult. We shall use the trapezoidal rule.Let an interval [ a,b ] ⊆ R  be given, and the function  f   ∈ C  2 [ a,b ].Divide the interval [ a,b ] by points a  =  x 0  < x 1  < x 2  < ... < x n  =  b into  n  equal parts of length ∆ x  =  b − an  .Then we have the trapezoidal formula:    ba f  ( x ) dx  =  b − a 2 n  f  ( a ) +  f  ( b ) + 2 n − 1  i =1 f  ( x i )  +  r n ( f  ) (5)where  r n ( f  ) is the remainder of the formula.To evaluate the approximation error of the trapezoidal formula there existsthe following result. Theorem 2.  For every function   f   ∈  C  2 [ a,b ] ,  the remainder   r n ( f  )  from the trape-zoidal formula (5), satisfies the ineguality: | r n ( f  ) |≤  ( b − a ) 3 12 n 2  = max x ∈ [ a,b ] | f   ( x ) |  (6) 103
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