Solution of fractional ordinary differential equation by Kamal transform

Transcription

1 28; (2): ISSN: Maths 28; (2): Stats & Maths Receied: Acceted: Rachana Khandelwal Maharshi Arind Uniersity, Priyanka Choudhary Jaiur National Uniersity, Yogesh Khandelwal Jaiur National Uniersity, Solution of fractional ordinary differential equation by Kamal transform Rachana Khandelwal, Priyanka Choudhary and Yogesh Khandelwal Abstract The fractional calculus for the Kamal transform is introduced and some non- homogenous fractional ordinary differential equation soled by the Kamal transform. We hae to get multile shifting roerty and eriodic function of the Kamal transform. Keywords: Kamal transform, fractional differential equation, fractional deriaties. Introduction The Kamal transform has fundamental roerties which resented in this aer due to its simle formula and consequent and useful roerties. It is ery useful to sole intricate roblem in engineering mathematic and alied science. The Kamal transform can be effecting we sole fractional ordinary differential equation. The urose of this aer is to show the alicability of this interesting new transform its efficiency in soling the fractional ordinary differential equation. 2. Fundamental roerties of Fractional calculus and Kamal transform method In the field of ure and alied mathematics, the theory of fractional calculus lay a significant role different tyes of differential and integral equation are soled by fractional integrals and deriatie, in association with different integral transform. The descrition of deriatie of fractional order in the same of Abel-Riemann [2] (A-R) is gien by D α [f(t)] = { d t f(t) Γ(m α) dt m (t τ) α m+ dτ, m < α m d m f(t), α = m () dt m Where mε z + and αεr + and the integral oerator is defined by imlementing an integral of fractional order in Abel-Riemann D α [f(t)] = t Γ(α) (t τ) α f(τ)dτ, t >, α > (2) According to A-R, the integral oerator J α is J α [f(t)] = t Γ(α) (t τ) α f(τ)dτ, t >, α > () Corresondence Priyanka Choudhary Jaiur National Uniersity, ~279~

2 We hae J α t n = D α t n = Γ[+n] Γ[+n+α] tn+α (4) Γ[+n] Γ[+n+α] tn+α (5) The fractional calculus deriatie is gien by J α [D α f(t)] = f(t) k= f (k) () tk k! (6) 2. Kamal Transforam we can take set A the function is defined in the form t A = { f: ǀ f(t) ǀ < e j if t ɛ ( ) j x [, ), j =,2, ; j > } (7) Where, 2 may be finite or infinite and the constant must be finite. Then Kamal transform is K(f(t) ) = G() = f(t)e t dt, t, 2 (8) Deriatie of Kamal transform Let function f(t) then deriatie of f(t) with resect to t and the n th order deriatie of the same with resect to t are resectiely. Then Kamal transform of deriatie gien by [] K[f n (t)] = G() n n k= k n+ f k () (9) n =, 2,.. in equation (9) gie Kamal transform of first and second deriatie of f(t) with resect to t K[f (t)] = G() f() () K[ f (t)] = 2 G() f() f () () Conolution theorem Let f(t) and g(t) are two function then Kamal transform of conolution theorem of two function is gien by K (f g) = K( f) K(g) (2) Multile shift roerty Let the function f (t) in set A is multilied with shift function K[t f(t)] = 2 d d [u()] Proof: By definition of Kamal transform, K( f(t)) = f(t)e t dt = G() t () G() = f(t)e t dt (4) Differentiae both side with resect to, of equation (4), we hae G () = d d f(t)e t dt ~28~

3 G () = d d {f(t)e t }dt G () = f(t)e t ( t 2) dt G () = 2 (tf(t))e t dt G () = K[t f(t)] 2 K[t f(t)] = 2 d d [G()] (5) Theorem: The Kamal transform of a iece wise eriodic function f(t) with eriod is K [ f(t)] = e e t f(t) dt ; > (6) Proof: Let Function f(t) is said to be a eriodic function T> if f(t) = f( T + t ) = f(2t + t) =. = f( nt + t ) By definition K[ f(t)] = f(t)e t f(t) dt K[ f(t)] = e t f(t) dt + 2 e t f(t) dt + 2 e t f(t) dt + + n e t f(t) dt (7) Put t = u+ in second integral and u to t= u+(n-) in n th integral,in equation (7), Now new limit for each integral are to and equation (7) by eriodicity f(t + ) = f (t), f (t + 2) = f (t), and so on there for K[f(t)] = e u f (u) du + e (u+) f (u) du + e (u+2 f (u) du + K[f (t)] = e u [ + e + e 2 + e +.. ] f (u)du K[f (t)] = e e t f (t) dt ; > (8) The Kamal transform of eriodic function f(t) of eriod is obtained by integrating e t f(t) in the interal (, ) with resect to t and multily the resultant by the factor ( e ). Preosition : if f(t) is a function and G() is a Kamal transform then fractional integral for Kamal transform of order α is K[D α f (t)] = Γ(α) (α )! α G() Proof: The fractional integral foe the function (t), is D α [f (t)] = Γ(α) (t(α ) f (τ) (9) Taking the Kamal transform in the equation ~28~

4 K[D α f (t)] = K [ Γ(α) (t(α ) f (τ)] K[D α f (t)] = Γ(α) (α )! α G() (2) Proosition 2: If function f(t) and G() is Kamal transform then fractional deriatie for Kamal transform of order n is K[f n (t)] = G() n n k n+ f k ( ) k= (22) If f(t) is a function and G() is Kamal transform then Riemann-Liouille fractional deriatie is K [ n y(t) ] = K[D n f(t)] = [ G() t n n (2) n k= n (k+) [D n k (f(t)) ] t= ]. Alication of Kamal transform In this section, we discuss the solution of fractional ordinary differential equation using general roerties with initial condition.. Sole the non - homogeneous fractional ordinary differential equation as, n y(t) t n = 2 y(t) t 2 + y(t) t + y(t) + a (24) And initial condition, () = f(), (25) We aling the Kamal transform of equation (24), K [ n y(t) t n ] = K [ 2 y(t) t 2 + y(t) t + y(t) + a ] [ G() n n (k+) n k= [D n k (f(t)) ] t= ] = G() f() 2 f () + G() f() + a [ G() n k= n (k+) [D n k (f(t)) ] t= ] = n n G() f() 2 n f () + n G() n f() + n a G() n k= n (k+) [D n k (f(t)) ] t= = n 2 G() n f() n f () + n G() n f() + n a G() [ n 2 n ] = n k= n (k+) [D n k (f(t)) ] t= n f() n f () n f() + n a (26) Sole the equation (26) and find out the alue of G(), with initial condition..2 Sole the non - homogeneous fractional ordinary differential equation as, D 2y(t) + D y(t) = + t With initial condition, y() = y() =, and [D 2 (f(t)) ] t= ] = Solution; Gien equation; ~282~

5 D 2y(t) + D y(t) = + t (27) And, y() = y () =, and [D 2 (f(t)) ] t= ] = (28) We alying the Kamal transform both side equation (27) K[D 2y(t)] + K[D y(t)] = K [ ] + K[ t ] 2 2 [ G() 2 (k+) [D 2 K (f(t)) ] t= ] + G() f() = + 2 Now taking the k=, then 2 [ G() [D 2 (f(t)) t= ] + G() f() = [ G() [ G() 2 We alying the initial condition, G() = [D 2 (f(t)) t= ]] + G() f() = + 2 We alying the iners Kamal transform both side for the alue of y(t) K [G() = K [[ + 2 ]] We get exact solution by the Kamal transform method as follows: y(t) = t + 2! t2. Sole the non - homogeneous fractional ordinary differential equation as, D 2y(t) + y(t) = 2 t + t 2, and [D 2 (f(t)) ] t= = ] Solution: Gien equation; D 2y(t) + y(t) = 2 t + t 2 (29) With initial condition; [D 2 (f(t)) ] t= = ] () We alying the Kamal transform both side equation (29), K [ D 2y(t) ] + K [y(t)] = K [ 2 t + t 2] 2 2 [ G() 2 (k+) k= [D 2 k (f(t)) ] t= ] + G() = ! Now taking the k=, ~28~

6 2 [ G() 2 [D 2 [f(t) ] t= ] + G() = ! 2 Aly the initial condition, 2 G() + G() = G() [ + ] = 2 2 [ + 2 ] 2 G() = 2 2 Aly the inerse Kamal transform for the alue of y(t) k [G()] = k [ 2 2 ] We get exact solution by the Kamal transform method as follows: y(t) = 2 t 4. Discussion and Conclusion We obsered that Kamal transform soles fractional ordinary differential equation with a few comutations as well as time unlike the Lalace transform and other we obsered that the Kamal transform is defind on the interal [, ]. We hae alied Kamal transform for fractional ordinary differential equation as well as eriodic function. It is found that the Kamal transform has an extensie affinity with the solutions of differential and integral equations, and more secifically with the Fractional differential equations which has been the centre forum of this aer. We found that the solution of fraction ordinary differential equation can be obtained in the form of distribution fractional ordinary differential equations when distributed Kamal transform are inoked. 5. Acknowledgements The authors would like to thank the J.N.U. and the anonymous referees for their reading of the original manuscrit and for their comments and suggestions that greatly imroe the resentation of this work. 6. References. Abdelilah K, Hassan Sedeeg. The New Integral, Kamal Transform, Adances in Theoretical and Alied Mathematics, 26; (4): Herbert Kreyszig, Edward J Norminton. Adanced Engineering Mathematics Ohio State Uniersity Columbus, Ohio, Tenth edition, 2.. Joel L Sciff. The Lalace Transform theory and Alication, sringer, Kim Hwa Joon. The time shifting for Elzaki transform. International Journal of Pure and Alied Mathematics. 2; 87(2): Mohnad Mahgoub AM. The new integral transform Mahgoub transform, Adances in theoretical and Alied Mathematics. 26; (4): Ali SS, Chaudhary MS. On a new integral transform and solution of some integral equations. International Journal of Pure and Alied Mathematics. 2; 7(): William F Trench. Elementary Differential Equations, Trinity Uniersity, 2. ~284~