Analytic Solution of Fuzzy Second Order Differential Equations under H-Derivation

Transcription

1 Teory of Approximation and Applications Vol. 11, No. 1, (016), Analytic Soltion of Fzzy Second Order Differential Eqations nder H-Derivation Lale Hoosangian a, a Department of Matematics, Dezfl Branc, Islamic Azad University, Dezfl, Iran. Received 14 Marc 016; accepted April 017 Abstract In tis paper, te soltion of linear second order eqations wit fzzy initial vales are investigated. Te analytic general soltions of tem sing a first soltion is fonded. Te parametric form of fzzy nmbers to solve te second order eqations is applied. Te soltions are searced in for cases. Finally te example is got to illstrate more and te soltions are sown in figres for for cases. Key words: Second order eqation, Fzzy initial vales, H-derivations, General soltion. Corresponding ators mail: l-oosangian@yaoo.com(l. Hoosangian)

2 1 Introdction Te differential eqations of second order are one of te eqations tat cannot be solved simply, ts it is necessary to classify tem in special eqations, for example linear and non-linear eqations. Differential eqation wit constant mltipliers is one of te linear eqations and te Caci differential eqation is te nonlinear one. Tese eqations are received in te formlation of applied matematics problems, bt in natre te fzzy second eqations are cagt for example in pysics problems, mecanical problems and etc. Te H-derivative of fzzy nmber-valed fnction was introdced for solving a fzzy first order eqation in [9]. Under tis setting, te existence and niqeness of a soltion of fzzy differential eqations were stdied in [3,4]. Under H-derivative. Te nmerical metod for solving differential eqations is stdied by Pallingkinis trog Rnge-Kta metod, [1]. Te strong general differentiable was introdced in [4]. Tis concept allows s to solve te problem of H-derivative. Te existence of fzzy differential eqations of second order are stdied by Allaviranloo et al., in [1] and ten by Zang [7], Under te general H-derivative, [16]. Kastan et al. stdied solving second order eqations nder bondary vale problem by te general H-derivative, [14]. Allaviranloo and Hoosangian searced fzzy second order derivations more, investigated te relationsips between fzzy second order derivations and fond te soltion of fzzy constant mltipliers and Caci second order eqations wit fzzy initial vales []. Some nmerical metods for solving fzzy second order derivations are stdied in [10, 17, 19, 5] In tis paper, general derivatives are sed to find new soltions for initial vale problem of fzzy linear second order eqations wit fzzy initial vales. Indeed, wit generalized differentiability, te soltion for a larger class of tem tan sing H-derivative. In Section, some needed concepts are reviewed. In Section 3, an analytic metod in order to find fzzy second order nder H-differential is introdced and or formla for finding soltions of fzzy Hakara differential is obtained and for cases reac te general soltions is searced. Nmerical examples are given to sow more of or metod. In section 4, nmerical examples of all cases are given to sow more. 100

3 Basic concepts Te basic definitions of a fzzy nmber are given in [9,11 13] as follows: Definition.1 is named a fzzy nmber in parametric form tat is sown wit a pair (, ) of fnctions (r), (r), 0 r 1, wic satisfy te following reqirements: 1. (r) is a bonded non-decreasing left continos fnction in (0, 1], and rigt continos at 0,. (r) is a bonded non-increasing left continos fnction in (0, 1], and rigt continos at 0, 3. (r) (r), 0 r 1. Definition. For arbitrary ũ = ((r), (r)) and ṽ = (v(r), v(r)), 0 r 1, and scalar k, it is defined addition, sbtraction, scalar prodct by k and mltiplication are respectively as follows: addition : + v(r) = (r) + v(r), + v(r) = (r) + v(r) sbtraction : v(r) = (r) v(r), v(r) = (r) v(r) scalar prodct: (k(r), k(r)), k 0 kũ = (k(r), k(r)), k < 0 mltiplication : v(r) = max{(r)v(r), (r)v(r), (r)v(r), (r)v(r)} v(r) = min{(r)v(r), (r)v(r), (r)v(r), (r)v(r)} Definition.3 Let (r) = [(r), (r)], 0 r 1 be a fzzy nmber, let c = (r) + (r) d (r) (r) = It is clear tat d (r) 0 and (r) = c (r) d (r) and (r) = c (r)+ d (r) Definition.4 Let (r) = [(r), (r)], v(r) = [v(r), v(r)], 0 r 1 are two fzzy nmbers and also k, s are two arbitrary real nmbers. If 101

4 w = k + sv ten w c (r) = k c (r) + sv c (r) w d (r) = k d (r) + s v d (r) Definition.5 Let x, y R F. If tere exists z R F sc tat x = y +z ten z is called te H-differential of x, y and it is denoted x y. Definition.6 [3] A fnction F : I R F, I = (a, b), is called H differentiable on t I if for > 0 sfficiently small tere exist te H differences F (t 0 + ) F (t 0 ), F (t 0 ) F (t 0 ) and an element F (t) R F sc tat: 0 = lim 0 D( F (t 0 + ) F (t 0 ), F (t)) = lim 0 D( F (t 0) F (t 0 ), F (t)) Definition.7 [3] Let F : I R F and t 0 I. F is differentiable at t 0 if tere is F (t 0 ) R F sc tat eiter (i)for > 0 sfficiently close to 0, te H-differences F (t 0 + ) F (t 0 ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 + ) F (t 0 ) = lim 0 F (t 0 ) F (t 0 ) = F (t) or (ii)for > 0 sfficiently close to 0, te H-differences F (t 0 ) F (t 0 + ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 ) F (t 0 + ) = lim 0 F (t 0 ) F (t 0 ) = F (t) or (iii)for > 0 sfficiently close to 0, te H-differences F (t 0 + ) F (t 0 ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 + ) F (t 0 ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) or (iv)for > 0 sfficiently close to 0, te H-differences F (t 0 ) F (t 0 + ) 10

5 and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 ) F (t 0 + ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) Definition.8 Let F : I R F. For fix t 0 I we say F is differentiable of second-order at t 0, if tere is F (t 0 ) R F sc tat eiter (i)for > 0 sfficiently close to 0, te H-differences F (t 0 + ) F (t 0 ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 + ) F (t 0 ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) or (ii)for > 0 sfficiently close to 0, te H-differences F (t 0 ) F (t 0 + ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 ) F (t 0 + ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) or (iii)for > 0 sfficiently close to 0, te H-differences F (t 0 + ) F (t 0 ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 + ) F (t 0 ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) or (iv)for > 0 sfficiently close to 0, te H-differences F (t 0 ) F (t 0 + ) and F (t 0 ) F (t 0 ) exist and te following limits lim 0 F (t 0 ) F (t 0 + ) = lim 0 F (t 0 ) F (t 0 ) = F (t 0 ) Teorem.1 [1] If f : [a, b] R F R F is continos and let t 0 [a, b]. A mapping x : [a, b] R F is a soltion to te initial vale problem x = f(t, x(t, r), x (t, r)), x(t 0 ) = k 1, x (t 0 ) = k if and only if x and x are continos and satisfy one of te following conditions: (a) x(t) = k (t t 0 ) + t t 0 ( t t 0 f(s, x(s), x (s))ds)ds + k 1 103

6 were x and x are (i)-differentials, or (b) x(t) = ( 1)(k (t t 0 ) ( 1) t t 0 ( t t 0 f(s, x(s), x (s))ds)ds) + k 1 were x and x are (ii)-differentials, or (c) x(t) = ( 1)(k (t t 0 ) + t t 0 ( t t 0 f(s, x(s), x (s))ds)ds) + k 1 were x is te (i)-differential and x is te (ii)-differential, or (b) x(t) = k (t t 0 ) ( 1) t t 0 ( t t 0 f(s, x(s), x (s))ds)ds + k 1 were x is te (ii)-differential and x is te (i)-differential. proof: See teorem (3.1) in Ref. [1]. Teorem. [3] Let [t 0, T ] E E E be continos and sppose tat tere exist M 1, M > 0 sc tat d(f(t, x 1, x ), f(t, y 1, y )) M 1 d(x 1, y 1 ) + M d(x, y ) for all t [t 0, T ] and x 1, x, y 1, y R F. Ten initial vale problem mentioned in teorem (.9) as a niqe soltion on [t 0, T ] for eac case (i) or (ii). 3 Analytic Soltion of Fzzy Linear Second Order Eqation In tis section an analytic metod for solving fzzy linear second order eqation wit fzzy initial vale is stdied. Fzzy linear second order eqation wit fzzy initial vales is considered by following: (t) + p(t) (t) = q(t)(t), (a) = 0, (a) = 0 tat p(t) and q(t) be two crisp fnctions. If initial vales are fzzy nmbers and tere is one fzzy soltion 1 (t), it can to be considered for cases. 104

7 3.1 Case(1) If p(t) and q(t) are two crisp fnctions and positive and and are considered (i)-differentiable ten: (t) + p(t) (t) = q(t)(t), (t) + p(t) (t) = q(t)(t), (a) = 0, (a) = 0, (a) = 0, (a) = 0 (3.1) Now te following system can be denoted: (t) + p(t) 0 = 0 q(t) (t) 0 p(t) q(t) 0 were U(t) = (t), P (t) = p(t) 0, Q(t) = q(t) 0 (t) 0 p(t) 0 q(t) ten it can be written te system (1) in te following eqation: U (t) + P (t)u (t) = Q(t)U(t) (3.) Now if U 1 (t) = 1(t) Ten it is considered tat te soltion of () is 1 (t) given: U = U 1 ( 1 e P (t)dt )dt U1 105

8 3. Case() If p(t) and q(t) be two crisp fnction and are positive and and are considered (ii)-differentiable ten it can to denote two systems: and (t) + p(t) (t) = q(t)(t), (a) = 0, (a) = 0 (t) + p(t) (t) = q(t)(t), (a) = 0, (a) = 0, Now te following system can be denoted: (t) + p(t) 0 = 0 q(t) (t) 0 p(t) q(t) 0 (3.3) (3.4) were U(t) = (t), P (t) = p(t) 0, R(t) = 0 q(t) (t) 0 p(t) q(t) 0 Ten it given: U (t) + P (t)u (t) = R(t)U(t) Ten by sing (3) and (4) togeter it denoted: c (t) + p(t) c (t) = q(t) c (t), c (a) = c 0, c (a) = c 0, (3.5) 106

9 and d (t) + p(t) d (t) = q(t) d (t), d (a) = d 0, d (a) = d 0, Now by solving (5) and (6) togeter and sing case (1): c = c 1 1 ( ( c 1) e p(t)dt )dt, d = d 1 1 ( ( d 1) e p(t)dt )dt (3.6) were c = +, c = + and c = + also d =, d = and d =. If we consider c 1 = 1 + 1, d 1 = 1 1. Now by definition (.3) we can find and. Now te general soltion is (t) = c 1 1 (t) + c (t) tat c 1 and c are two fzzy nmbers. 3.3 Case(3) If p(t) and q(t) be two crisp positive fnction and tose are considered tat is (i)-differentiable and is (ii)-differentiable ten it can be denoted two systems: Now te following system can be denoted: (t) = p(t) (t) + q(t)(t), (0) = 0, (3.7) (0) = 0 and Te it given te following: (t) = p(t) (t) + q(t)(t), (0) = 0, (0) = 0, (t) + p(t) 0 = 0 q(t) (t) 0 p(t) q(t) 0 (3.8) 107

10 were U(t) = (t), S(t) = 0 p(t), R(t) = 0 q(t) (t) p(t) 0 q(t) 0 Ten te following system is gotten: U (t) + P (t)u (t) = R(t)U(t) Now by solving (7) and (8) togeter we ave c (t) + p(t) c (t) = q(t) c (t), c (a) = c 0, c (a) = c 0, (3.9) and Ten we ave c = c 1 d (t) p(t) d (t) = q(t) d (t), d (a) = d 0, d (a) = d 0, 1 ( ( c 1) e p(t)dt )dt, d = d 1 1 ( p(t)dt ( d 1) e )dt (3.10) were c = +, c = +, c = + and d = also d =, d = and c 1 = 1 + 1, d 1 = 1 1. Now by definition (.3) it can be to find and. Ts (t) = c 1 1 (t)+c (t) is te general soltion. 3.4 Case(4) If p(t) and q(t) be two crisp positive fnction and tese are considered tat is (ii)-differentiable and is (i)-differentiable ten two following 108

11 systems are denoted: (t) = p(t) (t) + q(t)(t), (0) = 0, and (0) = 0 (t) = p(t) (t) + q(t)(t), (0) = 0, (0) = 0, Ts te following system is fonded: (t) + p(t) 0 = 0 q(t) (t) 0 p(t) q(t) 0 (3.11) (3.1) were U(t) = (t), S(t) = 0 p(t), Q(t) = q(t) 0 (t) p(t) 0 0 q(t) Briefly: U (t) + S(t)U (t) = Q(t)U(t) Now by solving (11) and (1) togeter te following eqations are given: c (t) + p(t) c (t) = q(t) c (t), c (a) = c 0, c (a) = c 0, (3.13) and d (t) + p(t) d (t) = q(t) d (t), d (a) = d 0, d (a) = d 0, (3.14) 109

12 Ten we ave c = c 1 ( 1 c 1 e p(t)dt )dt, d = d 1 ( 1 d 1 e p(t)dt )dt were c = +, c = + and c = + also d =, d =, d = and c 1 = 1 + 1, d 1 = 1 1. Now by definition (.3) we can find and and (t) = c 1 1 (t)+c (t) is te general soltion. 4 Examples Example 4.1 Consider te following second order differential eqation wit initial vale: (t) + 1 t (t) = 1 (t) t (1) = [α 1, 3 α] (4.1) (1) = [6α 5, 5 4α] ten by analytic metod it can given: In case (1): Te first soltion of tis eqation is 1 (t) = t, ten (t) = 1. Te t soltion of (15) is: (t) = [(α 3)t + (4α )( 1 t ), (4 3α)t + (1 α)( 1 t )] In case (): Te first soltion of tis eqation are c 1(t) = t and d 1(t) = 1 t, d (t) = t3 4, c (1) = 1, d (1) = 1 α, c (1) = α, d (1) = 5 5α. Te soltion of (15) is given by: (t) = [( α + 1 )t + (α 1)( 1 t ) (6 6α)t3 4, (α + 1 )t + (6 6α) t3 4 ] 110

13 In case (3): Te first soltion of tis eqation are c 1(t) = t and d 1(t) = t, d (t) = tlnt, c (1) = 1, d (1) = 1 α, c (1) = α, d (1) = 5 5α. Te soltion of (15) is denoted by following: (t) = [( 3α 1 )t+(α 1)( 1 t α )+(4α 4)tlnt, (3 )t+(α 1)( 1 t )+(4 4α)tlnt] In case (4): Te first soltion of tis eqation are c 1(t) = t and d 1(t) = t 1+, d (t) = t 1, c (1) = 1, d (1) = 1 α, c (1) = α, d (1) = 5 5α. Te soltion of (15) is in te following: (t) = [( α+1 )t+(α 1)( 1 t )+( 4 )(α 1)t 1 ( 6 )(α 1)t +1, ( α+1 )t+ (α 1)( 1 t ) ( 4 )(α 1)t 1 + ( 6 )(α 1)t +1 ] 111

14 Fig(1): Case (1) in example (4.1) Fig(): Case () in example (4.1) Fig(3): Case (3) in example (4.1) Fig(4): Case (4) in example (4.1) 11

15 5 Conclsion In tis work, fzzy second order differential eqations wit fzzy initial vales considered. Parametric fzzy nmber sed to find te general soltion wit first given soltion nder Hakara derivations in te formla in for cases. Ten by sing fzzy initial vales for soltions are fond. References [1] T.Allaviranloo, N.Kiani, N.Barkordari,Toward te existence and niqeness of soltion of second order fzzy differential eqations, Information sciences 179, 009. [] Allaviranloo. T, Hoosangian. L, Fzzy Generalized H-Differential and Applications to Fzzy Differential Eqations of Second-Order, Intelligent and Fzzy Systems, 6 (014) [3] B.Bede, S.G.Gal, Generalizations of differentiability of fzzy nmber vale fnction wit applications to fzzy differential eqations, Fzzy sets and systems151, 005. [4] B. Bede, I.J. Rdas, A.L. BencsikFirst order linear fzzy differential eqations nder genaralizes differentiability, Information Sciences 177, 007. [5] J. Bckley, T. Fering, M.D. Jimenez-Gamero, Fzzy differential eqation, Fzzy Sets and Systems 110, 000. [6] Y. Calco-Cano, H. Roman-Flores,Fzzy differential eqations wit generalized derivative, 7t ANFIPS International Conference IEEE, 008. [7] Y. Calco-Cano, H. Roman-Flores, On new soltions of fzzy differntial eqations, Caos solitons and fractals 38, 008. [8] Y. Calco-Cano, H. Roman-Flores, M.D. Jimenez-Gamero, Generalized derivative and π derivative for set-valed fnctions, Information Science,

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