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Int. J. Pure Appl. Sci. Technol., 14(1) (2013), pp. 16-26 International Journal of Pure and Applied Sciences and Technology ISSN 2229-6107 Available online at www.ijopaasat.in Research Paper First Order Linear Homogeneous Ordinary Differential Equation in Fuzzy Environment Sankar Prasad Mondal 1,*, Sanhita Banerjee 2 and Tapan Kumar Roy 3 1, 2, 3 Department of Mathematics, Bengal Engineering and Science University, Shibpur, Howrah- 711103, West Bengal, India * Corresponding author, e-mail: (sankar.mondal02@gmail.com) (Received: 21-7-12; Accepted: 18-12-12) Abstract: In this paper the solution procedure of first order linear homogeneous ordinary differential equation in fuzzy environment is described. It is discussed for three different cases. Here fuzzy numbers are taken as Generalized Triangular Fuzzy Numbers (GTFNs). Further two bio mathematical models are numerically illustrated. Keywords: Fuzzy Ordinary Differential Equation (FODE), Generalized Triangular fuzzy number (GTFN), strong solution, weak solution. 1. Introduction: In 1965 Zadeh [9] introduced fuzzy set theory. Fuzzy set is taken as a tool that makes things easy to illustrate some sort of vague as well as non stochastic uncertain model. Fuzzy differential equations (FDEs) are the natural way to model many systems under uncertainty. Fuzzy derivatives were first conceptualized by Chang and Zadeh [12]. The term was first described in 1978 by Kandel and Byatt [2]. There are many approaches to solve the FDE for fuzzy initial value problem. Buckley and Feuring [7,8] introduced two analytical methods for solving nth-order linear differential equations with fuzzy initial conditions. One is classical method and the other is extension principle method. Recently FDE has also used in many models such as HIV model [6], decay model [5], predator-prey model [10], etc. In this paper we have considered 1 st order linear homogeneous fuzzy ordinary differential equation and have described its solution procedure in section-3. In section-4 we have applied it for two biomathematical models (Growth and Decay model).

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 17 2. Preliminary Concepts: Definition 2.1: Generalized Triangular Fuzzy Number (GTFN): A Generalized Fuzzy number is called a Generalized Triangular Fuzzy Number if it is defined by =(,, ;)its membership function is given by ()= 0,,, =, 0, or, ()=,,,0 and the -cut of is = +, 0,,0< 1 where = and = Definition 2.2: Fuzzy Ordinary Differential Equation (FODE): Consider the 1 st Order Linear Homogeneous Ordinary Differential Equation (ODE) = with initial condition ( )=. The above ODE is called FODE if any one of the following three cases holds: (i) (ii) (iii) Only is a generalized fuzzy number (Type-I). Only k is a generalized fuzzy number (Type-II). Both k and are generalized fuzzy numbers (Type-III). Definition 2.3: Strong and Weak Solution of FODE: Consider the 1 st order linear homogeneous fuzzy ordinary differential equation = with ( )=. Here k or (and) be generalized fuzzy number(s). Let the solution of the above FODE be () and its -cut be (,)= (,), (,). If (,) (,) 0, where 0< 1 then () is called strong solution otherwise () is called weak solution and in that case the -cut of the solution is given by (,)=min (,), (,),max (,), (,). 3. Solution Procedure of 1 st Order Linear Homogeneous FODE The solution procedures of 1 st order linear homogeneous FODE of Type-I, Type-II and Type-III are described. Here fuzzy numbers are taken as GTFNs. 3.1 Solution Procedure of 1 st Order Linear Homogeneous FODE of Type-I Consider the initial value problem = with fuzzy Initial Condition (IC) ( )==(,, ;)..(3.1.1) Let () be a solution of FODE (3.1.1) with -cut (,)= (,), (,)

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 18 and () = +, 0,,0< 1 Here we solve the given problem for >0 and <0 respecively. Case 3.1.1:- When >0 The FODE (3.1.1) becomes (,) = (,) ( =1,2) (3.1.2) The solution is (,)= + ( ), (,)= ( ). (3.1.3) Here (,)= ( ) >0, (,)= ( ) <0 and (,)= ( ) = (,). So the solution is a generalized fuzzy number. The -cut of the strong solution is (,) = +, ( ). So the solution of (3.1.1) is ()= ( ). Case 3.1.2:- when <0, let = where m is a positive real number. Then the FODE (3.1.1) becomes (,) = (,), (,) = (,) The general solution is (,)= + + ( ) + 1 + ( ) (3.1.4) (,)= 1 2 + + ( ) Now 1 + ( ) (3.1.5) (,) = ( ) + + ( ) (,)= ( ) + ( ) and (,)= ( ) = (,) Here three cases arise. Case 3.1.2.1:- When = Then (,)>0, (,)<0 and (,)= (,)

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 19 Hence [ + ( ) + 1 + ( ), + ( ) 1 + ( ) ] is the -cut of the strong solution of the FODE (3.1.1). So, ()= ( ) + 0 ( ) ( ) where 0 =( 1,0,1;) be a symmetric GTFN is the solution of (3.1.1) Case 3.1.2.2:- When < then (,)<0. So in this case we get the strong solution if (,)>0 i.e. > + log Hence [ + + ( ) + 1 + ( ), + + ( ) 1 + ( ) ] is the -cut of the strong solution of the FODE (3.1.1) if > + log. Case 3.1.2.3:- When > then (,)>0 So in this case we get the strong solution if (,)<0 i.e. > + log Hence [ + + ( ) + 1 + ( ), + + ( ) 1 + ( ) ] is the -cut of the strong solution of the FODE (3.1.1) if > + log. For case Case 3.1.2.2, Case 3.1.2.3 the solution is ()= Γ ( ) + two symmetric GTFNs. 0 ( ) where Γ=( +,2, + ;), 0 =( 1,0,1;) are 3.2 Solution Procedure of 1 st Order Linear Homogeneous FODE of Type-II Consider the initial value problem = with IC ( )= where =(,, ;).(3.2.1) Let () be the solution of FODE (3.2.1) Let (,)= (,), (,) be the -cut of the solution and the -cut of be = +, 0,,0< 1

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 20 Here we solve the given problem for >0 and <0 respecively. Case 3.2.1: when >0 The FODE (3.2.1) becomes (,) = () (,) for =1,2 (3.2.2) From (3.2.2) we get the -cut of the solution as (,)= ( ), (,)= ( )... (3.2.3) Here and and (,) = ( ) ( ) >0 (,) = ( ) ( ) <0 (,)= ( ) = (,) Hence the -cut of the strong solution of FODE (3.2.1) is (,)=[ ( ), ( ) ].(3.2.4) Case 3.2.2:- When <0, let =, where = (,, ;) is a positive GTFN. So () = (), ()= +, 0,,0< 1 The FODE (3.2.1) becomes (,) = () (,) and (,) = () (,) The general solution is a generalized fuzzy number with -cut (,)= 1 () () () ()( ) +1+ () () () ()( ) = 1 ( )( ) +1+ ( )( )..(3.2.5) (,)= () () 1 () () () ()( ) +1+ () () () ()( ) = 1 ( )( ) +1+ ( )( )..(3.2.6) Now if (,)>0, (,)<0 and (,) (,) then we get the strong solution.

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 21 3.3 Solution Procedure of 1 st Order Linear Homogeneous FODE of Type- III Consider the initial value problem = with fuzzy IC ( )==(,, ;), where =(,, ;) Let () be the solution of FODE (3.3.1). Let (,)= (,), (,) be the -cut of the solution. Also = +, 0,,0< 1 and () = +, 0,,0< 1 Let =min(,) Here we solve the given problem for >0 and <0 respecively. Case 3.3.1:- when >0.(3.3.1) The FODE (3.3.1) becomes (,) = +, (,) for =1,2.. (3.3.2) The solution is a generalized fuzzy number with -cut (,)= + ( ) and (,)= ( )..(3.3.3)....(3.3.4) Now if (,)>0, (,)<0 and (,) (,) then we get the strong solution. Case 3.2.2:- when <0 then = where =(,, ;) is a positive GTFN. Then () = +, 0,,0< 1 Let =min(,) The FODE (3.3.1) becomes (,) and (,) = (,) = + (,) (3.3.5) (3.3.6) The general solution is a generalized fuzzy number with -cut

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 22 (,)= + ( ) + + + ( )..(3.3.7) and (,)= + + ( ) + ( )......(3.3.8) Here also if (,)>0, (,)<0 and (,) (,) then we get the strong solution. 4. Applications: 4.1: Growth Model: Problem: A culture initially has about numbers of bacteria. The rate of growth is proportional to the number of bacteria present. The constant of proportionality is. Determine the solution when only or only k or both and k is a GTFN. Solution:- The problem is governed by the FODE = subject to (=0)=.(4.1.1) Let () be the solution of FODE (4.1.1) with -cut (,), (,). Case1: If =(3 10,5 10,7 10 ;0.8) and =0.4055 then (,)=(3 10 +2.5 10 α)., (,)=(7 10 2.5 10 α). Now for =2 we plot (,) and (,) in the graph shown below: Fig 4.1: Rough sketch of (,) and (,)

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 23 From the above graph we see that for this particular value of t, (,) is an increasing function, (,) is a decreasing function and (,0.8)= (,0.8). Hence we get that this is a strong solution. Case2: If =5 10 and =(0.3055,0.4055,0.5055;0.7) then (,)=5 10 (..α), (,)=5 10 (..α) Now for =2 we plot (,) and (,) in the graph shown below: Fig 4.2: Rough sketch of (,) and (,) Hence this is strong solution. Case3: If =(3 10,5 10,7 10 ;0.8) and =(0.3055,0.4055,0.5055;0.7) then (,)=(3 10 +2.857 10 α) (..α) (,)=(7 10 2.857 10 α) (..α) Now for =2 we plot (,) and (,) in the graph shown below: Hence we see that this solution is a strong solution. Fig 4.3: Rough sketch of (,) and (,) 4.2: Decay Model: Problem:- Suppose biochemical oxygen demand (BOD) in water is of amount. The rate of decay is proportional to the amount of dissolved oxygen in water at present. The constant of proportionality is. Determine the solution when only or only k or both and k is a GTFN. Solution:- The problem is governed by the FODE = subject to (=0)=.(4.2.1)

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 24 Let () be the solution of FODE (4.2.1) with -cut (,), (,). Case1: If =(90,100,120;0.7) and = 0.038 = where =0.038 then (,)= (210 14.2857). +15(1.4286 1). (,)= (210 14.2857). 15(1.4286 1). Now for =16 we plot (,) and (,) in the graph shown below: Fig 4.4: Rough sketch of (,) and (,) From the above graph we see that for this particular value of t, (,) is an increasing function, (,) is a decreasing function and (,0.7)= (,0.7). Hence this is strong solution. Case2: If =100 and = =(0.028,0.038,0.048;0.7) then (,)= 501..α..α (..α)(..α) + 1+..α..α (..α)(..α) (,)= 50..α..α 1..α..α (..α)(..α) + 1+..α..α (..α)(..α) Now for =16 we plot (,) and (,) in the graph shown below: Fig 4.5: Rough sketch of (,) and (,)

Int. J. Pure Appl. Sci. Technol., 14(1) (2013), 16-26 25 Hence we get that this is a strong solution. Case3: If =(90,100,120;0.8) and = =(0.028,0.038,0.048;0.7) then (,)= (90+14.2857α) (120 28.5714α)..α..α (..α)(..α) + (90+14.2857α)+(120 28.5714α)..α..α (..α)(..α) (,)=..α..α (90+14.2857α) (120 28.5714α)..α..α (..α)(..α) +(90+14.2857α)+ (120 28.5714α)..α..α (..α)(..α) Now for =16 we plot (,) and (,) in the graph shown below: Hence this solution is a strong solution. 5. Conclusion and Future Work: Fig 4.6: Rough sketch of (,) and (,) In this paper we have solved first order linear homogeneous ordinary differential equation in fuzzy environment. Also one can repeat the same method for first order linear non-homogeneous ordinary differential equation in fuzzy environment. This process can be applied for any economical or biomathematical model and problems in engineering and physical sciences. Acknowledgement The authors are thankful to the editor and referees for their most valuable comments which have substantially improved the presentation of this paper.

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