Int. J. Contemp. Mth. Sciences, Vol. 6, 2011, no. 11, 535-543 Solution to Fredholm Fuzzy Integrl Equtions with Degenerte Kernel M. M. Shmivnd, A. Shhsvrn nd S. M. Tri Fculty of Science, Islmic Azd University Borüjerd Brnch, Borüjerd, Irn m.shmivnd@iub.c.ir Abstrct The purpose of this pper is to present new method to solve fuzzy Fredholm integrl equtions with degenerte kernel. Our results re given to demonstrte the proposed method nd bsed on the concept concerning the crisp integrl equtions with degenerte kernel. Keywords: Degenerte Kernel; Fredholm Fuzzy Integrl Eqution 1 Introduction Fuzzy Fredholm integrl equtions hve been solved with different methods [1-7]. The most remrkble properties of fuzzy set re tht they re employed in mny different reserch fields rnging from rtificil intelligence nd robotics, imge processing, biologicl nd medicl science, pplied opertions reserch, economics nd geogerphy, quntum optics nd grvity, socilogy, psychology nd some more restricted topics. 2 Preliminries In this section the most used bsic nottions in fuzzy clcules re introduced. Definition 1. A fuzzy number is fuzzy set u : R [0, 1] which stisfies: 1. u is upper semi-continuous. 2. u(x) = 0 outside some intervl [c, d]. 3. there re rel numbers nd b: c b d for which, 3.1 u(x) is monotonic incresing on [c, ], 3.2 u(x) is monotonic decresing on [b, d],
536 M. M. Shmivnd, A. Shhsvrn nd S. M. Tri 3.3 u(x) =1, x b. An lterntive prmetric form of fuzzy number which is equivlent to definition 1 is s follows: fuzzy number u is completely determined by ny pir (u, u) of functions u(r) nd u(r), 0 r 1, which stisfying the following three conditions: i. u(r) is bounded monotonic incresing left-continuous function. ii. u(r) is bounded monotonic decresing left-continuous function. iii. u(r) u(r), 0 r 1. For rbitrry u =(u, u) nd k R we define ddition nd multipliction by k s: ddition: (u + v)(r) =u(r)+v(r), (u + v)(r) =u(r)+v(r), sclr multipliction: { ku(r), k 0 (ku)(r) = ku(r), k < 0 { ku(r), k 0 (ku)(r) = ku(r), k < 0 The collection of ll the fuzzy numbers with the ddition nd multipliction is denoted by E 1 nd is convex cone. Definition 2. For rbitrry fuzzy numbers u =(u, u) nd v =(v, v) the quntity D(u, v) = sup {mx( u(r) v(r), u(r) v(r) )} 0 r 1 is the distnce between u nd v. Definition 3. A function f : R E 1 is clled fuzzy function if for rbitrry fixed t 0 R nd ɛ>0 there exist δ>0 such tht t t 0 <δ= D(f(t),f(t 0 )) <ɛ, f is sid to be continuous. We now define the integrl of fuzzy function using the Riemnn integrl concept which is employed in the next section. Definition 4. Let f :[, b] E 1 for ech prtition P = {t 0,t 1,..., t n } of [, b] nd for rbitrry ξ i,t i 1 ξ i t i, 1 i n, let R p = f(ξ i )(t i t i 1 ), λ p = mx{ t i t i 1 :1 i n},
Solution to Fredholm fuzzy integrl equtions 537 tht 1 p n, then definite integrl of f(t) over [, b] is f(t)dt = lim λp 0 R p, provided tht this limit exists in the metric D. If the fuzzy function f(t) is continuous in the metric D, its definite integrl exists. Furthermore, ( f(t, r)dt) = f(t, r)dt, (1) ( f(t, r)dt) = Remrk 1. Let u(r) =(u(r), u(r)), S u (r) = u(r)+u(r), D u (r) = 2 f(t, r)dt. (2) 0 r 1, be fuzzy number, we tke u(r) u(r). 2 It is cler tht u(r) =S u (r) D u (r) nd u(r) =S u (r)+d u (r). Remrk 2. Let u(r) =(u(r), u(r)) nd v(r) =(v(r), v(r)), lso k nd s re rbitrry rel numbers. If w = ku + sv then 0 r 1, nd S w (r) =ks u (r)+ss v (r), D w (r) = k D u (r)+ s D v (r). Now by referring to remrk 1, we hve u(r) v(r) S u (r) S v (r) + D u (r) D v (r), u(r) v(r) S u (r) S v (r) + D u (r) D v (r). Hence for ll r [, b], mx{ u(r) v(r), u(r) v(r) } S u (r) S v (r) + D u (r) D v (r), so definition 2 yields D(u, v) sup { S u (r) S v (r) + D u (r) D v (r) }. 0 r 1 Subsequently, S u (r) S v (r) 0 nd D u (r) D v (r) 0 implies tht D(u, v) 0.
538 M. M. Shmivnd, A. Shhsvrn nd S. M. Tri 3 Illustrtion of the method In this section we present new method for solving the liner fuzzy Fredholm integrl eqution with degenerte kernel. The proposed pproch will be illustrted in terms of the following eqution: F (t) =f(t)+λ k(s, t)f (s)ds, (3) with λ>0. It is ssumed tht kernel k(s, t) is degenerte, tht is, k(s, t) = i (s)b i (t), where i (s) nd b i (t), i =1, 2,..., n, re linerly independent functions. In Eq. 3, if f is crisp function then the solution is crisp s well, nd in the cse tht f is fuzzy function, we hve Fredholm fuzzy integrl eqution of the second kind which my only process fuzzy solutions. Sufficient conditions for existence of unique solution to the eqution (1), where f is fuzzy function, re given in [5]. Now we introduce prmetric form of the fuzzy integrl Eqs. (3) with respect to definition 2. Let (f(t, r), f(t, r)) nd (F (t, r), F (t, r)) (0 r 1, t b) re prmetric form of f(t) nd F (t) respectively, then the prmetric form of the fuzzy integrl Eqs. (3) is s follows: k(s, t)f (s, r)ds, (4) k(s, t)f (s, r)ds, (5) where, { k(s, t)f (s, r), k(s, t) 0 k(s, t)f (s, r) = k(s, t)f (s, r), k(s, t) < 0 { k(s, t)f (s, r), k(s, t) 0 k(s, t)f (s, r) = k(s, t)f (s, r), k(s, t) < 0 By the bove ssumptions the Eqs. (4) nd (5) will become in the following forms respectively. i (s)b i (t)f (s, r)ds, (6)
Solution to Fredholm fuzzy integrl equtions 539 i (s)b i (t)f (s, r)ds. (7) If we denote by A i, i =1, 2,..., n the set of union on subintervls of [, b] tht i (s) is nonnegtive on these subintervls nd by B i, i =1, 2,..., n the set of union on subintervls of [, b] tht i (s) is negtive on these subintervls. It is cler tht A i B i =[, b], i =1, 2,..., n. Without loosing generlity, we suppose tht b i (t) is nonnegtive for fixed t nd 1 i n. By the bove ssumptions we cn rewrite Eqs. (6) nd (7) respectively s follows: ( I i A i ( I i A i I i i (s)b i (t)f (s, r)ds + J i B i I i i (s)b i (t)f (s, r)ds + J i B i By remrk 1, remrk 2 nd bove Eqs. we cn ffirm tht S F (t, r) =S f (t, r)+λ J i i (s)b i (t)f (s, r)ds), (8) J i i (s)b i (t)f (s, r)ds). (9) i (s)b i (t)s F (s, r)ds, (10) D F (t, r) =D f (t, r)+λ i (s) b i (t) D F (s, r)ds. (11) Note tht in the negtive cse of b i (t) for some i tht sme results s the bove equtions re obtined. It emerges tht the technique of solving Eqs. (8) nd (9) re dependent on the definitions of c i = d i = i (s)s F (s, r)ds, i (s) D F (s, r)ds. By substituting c i nd d i in (10) nd (11) respectively, we obtin S F (t, r) =S f (t, r)+λ c i b i (t), (12)
540 M. M. Shmivnd, A. Shhsvrn nd S. M. Tri D F (t, r) =D f (t, r)+λ By substituting (12) into c i we obtin c i b i (t) = = b i (t) b i (t) therefore we get b i (t){c i d i b i (t). (13) i (s)s F (s, r)ds i (s)(s f (s, r)+λ i (s)(s f (s, r)+λ in similr mnner we get b i (t) {d i i (s) (D f (s, r)+λ c k b k (s))ds, c k b k (s))ds} =0, d k b k (s) )ds} =0. Since the functions b i (t) nd consequently b i (t), independent, therefore, c i d i i (s)(s f (s, r)+λ i (s) (D f (s, r)+λ i =1, 2,..., n re linerly c k b k (s))ds =0, (14) d k b k (s) )ds =0. (15) For these computtions in the sense of the unknowns, we simplify the problems by using the following nottions f (1) i = f (2) i = i (s)s f (s, r)ds, (1) ik i (s) D f (s, r)ds, (2) ik where the constnt f (1) i,f (2) i, (1) ik (14) nd (15) become respectively c i λ nd (2) ik = i (s)b k (s)ds, = i (s) b k (s) ds, (1 i, k n) re known then Eqs. (1) ik c k = f (1) i, i =1, 2,..., n, (16)
Solution to Fredholm fuzzy integrl equtions 541 d i λ (2) ik d k = f (2) i, i =1, 2,..., n, (17) which re two systems of n lgebric Eqs. for the unknowns c i nd d i. Therefore the problem reduces to finding the quntities c i nd d i, i =1, 2,..., n. The determinnts of these systems re two polynomils in term of λ of degree t most n. For ll vlues of λ for which the determinnts re not equl to zero the lgebric systems (16) nd (17) hve solution nd thereby the fuzzy integrl Eqs. (3) hve unique solutions. We cn use Eqs. (12) nd (13) nd remrk 1 to obtin the fuzzy solution of the problem. 4 Experimentl results This section illustrtes how to implement the proposed method in order to obtin the solutions of the fuzzy Fredholm integrl equtions with degenerte kernel. Exmple 1 [7]: Consider the fuzzy Fredholm integrl eqution with f(t, r) = sin( t 2 )(13 15 (r2 r)+ 2 15 (4 r3 r)), f(t, r) = sin( t 2 )( 2 15 (r2 + r)+ 13 15 (4 r3 r)), k(s, t) = 1 10 sin(s) sin( t ), 0 s, t 2π, 2 lso =0, b = 2π nd λ=1. Using the ssumptions described in the previous sections, we hve S f (t, r) = 1 2 sin( t 2 )(4 + r2 r 3 ) D f (t, r) = 11 30 sin( t 2 )(4 2r r2 r 3 ). By considering 1 (s) = 1 sin(s),b 10 1(t) = sin( t ) nd using Eqs. (16) nd (17), 2 we hve c 1 =0,d 1 = 4 30 (4 2r r2 r 3 ). From Eqs. (9) nd (10), S f (t, r) =S g (t, r) = 1 2 sin( t 2 )(4 + r2 r 3 ), D f (t, r) =D g (t, r)+d 1 b 1 (t) = 1 2 sin( t 2 )(4 2r r2 r 3 )
542 M. M. Shmivnd, A. Shhsvrn nd S. M. Tri hence by using remrk 1, the solution is obtined f(t, r) =(4 r r 3 ) sin( t 2 ), f(t, r) =(r + r 2 ) sin( t 2 ), which is the exct solution of the problem. Exmple 2 [constructed]: Consider the fuzzy Fredholm integrl eqution with f(t, r) =rt r r 2, f(t, r) =r rt t 2, k(s, t) =s + t, 0 s, t 1, nd =0, b=1 nd λ=1. Using remrk 1, we hve S f (t, r) = t 2, D f (t, r) =r rt. By considering 1 (s) =s, 2 (s) =1,b 1 (t) =1,b 2 (t) =t nd using Eqs. (16) nd (17) we obtin c 1 = 17 6, c 2 =5, d 1 = 3r, d 2 =5r, from Eqs.(9) nd (10) S f (t, r) nd D f (t, r) re obtined S f (t, r) = t 2 +5t + 17 6, D f (t, r) = 2r 6rt. Hence by using remrk 1, the solution is obtined f(t, r) = t 2 +5t +6rt +2r + 17 6, f(t, r) = t 2 +5t 6rt 2r + 17 6, which is the exct solution of the problem.
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