Sumudu Transform of Hilfer-Prabhakar Fractional Derivatives and Applications
aa r X i v : . [ m a t h . C A ] S e p Sumudu Transform of Hilfer-Prabhakar FractionalDerivatives and Applications
S. K. Panchal, Amol D. Khandagale,Pravinkumar V. Dole.
Department of mathematics,Dr. Babasaheb Ambedkar Marathwada University,Aurangabad-431004 (M.S.) India.E-mail ID - [email protected]@[email protected]
July 2016.
Abstract
In this paper the Sumudu transforms of Hilfer-Prabhakar fractionalderivative and regularized version of Hilfer-Prabhakar fractional derivative are ob-tained. These results are used to obtain relation between them involving Mittag-Leffler function. Also these results are applied to solve some problems in physics.Here the solutions of problems involving Hilfer-Prabhakar fractional derivative andregularized version of Hilfer-Prabhakar fractional derivative are obtained by usingFourier and Sumudu transform techniques.
Keywords:
Fractional calculus, Hilfer-Prabhakar fractional derivative, Mittag-Leffler function, Integral transforms. A
33, 42 A
38, 42 B The concept of integral transforms is originated from the Fourier integral formula.The importance of integral transforms is that they provide powerful operationalmethod for solving initial and boundary value problems. The operational calculusof integral transform is used to solve the differential and integral equations arisingin applied mathematics, mathematical physics and Engineering science. K. S. Miller,B. Ross, Podlubny Igor, A. A. Kilbas, J. J. trujillo used Laplace transform approach1to solve Cauchy type fractional differential equations and F. Mainardi solved the vis-coelastic problems by Laplace techniques. Recently, Watugala G. K. [5] introduced anew integral transform in 1990 known as Sumudu transform. Watugala G. K. solvedfractional differential equations by using Sumudu transform techniques.The Prabhakar integral [3] is defined by modifying Riemann-Liouville integral op-erator by extending its kernel with a three-parameter Mittag-Leffler function. TheHilfer-Prabhakar fractional derivative and its Caputo like regularized counterpartwere first introduced in [4]. In this paper the Sumudu transforms of Hilfer-Prabhakarfractional derivative and regularized version of Hilfer-Prabhakar fractional deriva-tive are obtained. These results are used to obtain relation between them involvingMittag-Leffler function and it is also used to obtain the solutions of non-homogeneousCauchy type fractional differential equations [4] in which Hilfer-Prabhakar fractionalderivative and regularized version of Hilfer-Prabhakar fractional derivative are in-volved.
In this section we gives some definitions, theorems and lemma, which are used inthe paper.
Definition 2.1 [5] Consider a set A defined as, A = (cid:26) f ( t ) / ∃ M, τ , τ > , | f ( t ) | ≤ M e | t | τj if t ∈ ( − j × [0 , ∞ ) (cid:27) (2.1) For all real t ≥ the Sumudu transform of function f ( t ) ∈ A is defined as, S [ f ( t )]( u ) = Z ∞ u e − tu f ( t ) dt, u ∈ ( − τ , τ ) (2.2) and is denoted by F ( u ) = S [ f ( t )]( u ) . Definition 2.2 [5] The function f ( t ) in (2.1) is called inverse Sumudu transform of F ( u ) and is denoted by, f ( t ) = S − [ F ( u )]( t ) (2.3) and the inversion formula for Sumudu transform is given by [5] , f ( t ) = S − [ F ( u )]( t ) = 12 πi Z γ − i ∞ γ − i ∞ u e tu f ( u ) du (2.4) For Re ( u ) > γ and γ ∈ C . Definition 2.3 [3] The three parameter Mittage-Leffter function introduced by Prab-hakar is of the form E γα,β ( z ) = ∞ X k =0 Γ( r + k )Γ( r )Γ( αk + β ) z k k ! (2.5) for α, β, γ ∈ C and Re ( α ) > . Definition 2.4 [2] Let f ( x ) be a function defined on ( −∞ , ∞ ) and be piecewise con-tinuous in each finite partial interval and absolutely integrable in ( −∞ , ∞ ) then F [ f ( x )]( p ) = Z ∞−∞ e − ipx f ( x ) dx (2.6) is called Fourier transform of f ( x ) and is denoted by F [ f ( x )]( p ) = f ( p ) .The function f ( x ) called the inverse Fourier transform of f ( p ) , is defined as f ( x ) = F − [ f ( p )]( x ) = 12 π Z ∞−∞ f ( p ) e ipx dp (2.7) and is denoted by f ( x ) = F − [ f ( p )]( x ) . Definition 2.5 [3] (Prabhakar Integral)
Let f ǫL [0 , b ]; 0 < t < b < ∞ . The prabhakar integral is defined as, E γρ,µ,ω, + f ( t ) = Z t ( t − y ) µ − E γρ,µ [ ω ( t − y ) ρ ] f ( y ) dy = ( f ∗ e γρ,µ,ω )( t ) (2.8) where * denote the convolution operation; ρ, µ, ω, γ ∈ C ; Re ( ρ ) , Re ( µ ) > and e γρ,µ,ω ( t ) = t µ − E γρ,µ [ ωt ρ ] . (2.9)For n ∈ N , we denote by AC n [ a, b ] the space of the real valued function f ( t )which have continuous derivative up to order ( n −
1) on [ a, b ] such that f ( n − ( t )belongs to the space of absolutely continuous functions AC [ a, b ], AC n [ a, b ] = (cid:26) f : [ a, b ] → R ; d n − dx n − f ( x ) ∈ AC [ a, b ] (cid:27) . Definition 2.6 [4] ( Hilfer-Prabhakar Fractional Derivative)
Let µ ∈ (0 , ν ∈ [0 , and let f ∈ L [0 , b ]; 0 < t < b < ∞ ; ( f ∗ e − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω )( t ) ∈ AC [0 , b ] . The Hilfer-Prabhakar fractional derivative of f ( t ) of order µ denoted by D γ,µ,νρ,ω, + f ( t ) is defined as, D γ,µ,νρ,ω, + f ( t ) = (cid:18) E − γνρ,ν (1 − µ ) ,ω, + ddt (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f (cid:1)(cid:19) ( t ) (2.10) where γ, ω ∈ R , ρ > and E ρ, ,ω, + f = f . In order to consider the Cauchy problems in which initial condition dependingonly on the function and its integer-order derivative, we use the regularized versionof Hilfer-Prabhakar fractional derivative defined as below.
Definition 2.7 [4] (Regularized Version of Hilfer-Prabhakar FractionalDerivative)
For f ∈ AC [0 , b ] , < t < b < ∞ ; µ ∈ (0 , ν ∈ [0 , ; γ, ω ∈ R , ρ > . The regular-ized version of Hilfer-Prabhakar fractional derivative of f ( t ) denoted by C D γ,µρ,ω, + f ( t ) is defined as, C D γ,µρ,ω, + = (cid:18) E − γνρ,ν (1 − µ ) ,ω, + E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + ddt f (cid:19) ( t ) (2.11)= (cid:18) E − γρ,ν (1 − µ ) ,ω, + ddt f (cid:19) ( t ) . (2.12) Theorem 2.1 [5] Let F ( u ) and G ( u ) be Sumudu transforms of f ( t ) and g ( t ) respec-tively. The Sumudu transforms of convolution of f and g is S (cid:2) ( f ∗ g )( t ) (cid:3) ( u ) = uF ( u ) G ( u ) , (2.13) where ( f ∗ g )( t ) = R t f ( t ) g ( t − τ ) dτ . Lemma 2.1 [1] Let α, β, λ ∈ R and α > , β > , n ∈ N . The Sumudu transform offunction e γρ,µ,ω ( t ) defined in (2 . is, S (cid:2) e γρ,µ,ω ( t ) (cid:3) ( u ) = u ( β − (1 − λu α ) δ . (2.14) Lemma 3.1
The Sumudu transform of Hilfer-Prabhakar fractional derivative (2.10) is, S (cid:18) D γ,µ,νρ,ω, + f ( t ) (cid:19) ( u ) = S (cid:18) E − γνρ,ν (1 − µ ) ,ω, + ddt (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f (cid:1)(cid:19) ( u )= u − µ (1 − ωu ρ ) γ F ( u ) − u ν (1 − µ ) − (1 − ωu ρ ) γν (cid:20) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:21) t =0 + . (3.1) Proof:
Taking Sumudu transform of Hilfer-Prabhakar fractional derivative (2.10)and using (2.8), (2.9), (2.13) and (2.14), we have, S (cid:18) D γ,µ,νρ,ω, + f ( t ) (cid:19) ( u ) = S (cid:18)(cid:18) E − γνρ,ν (1 − µ ) ,ω, + ddt (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f (cid:1)(cid:19) ( t ) (cid:19) ( u )= S (cid:18)(cid:18) e − γνρ,ν (1 − µ ) ,ω ∗ ddt (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f (cid:1)(cid:19) ( t ) (cid:19) ( u )= uS (cid:18) t ν (1 − µ ) − E − γνρ,ν (1 − µ ) ( ωt ρ ) (cid:19) ( u ) S (cid:18) ddt (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f (cid:1) ( t ) (cid:19) ( u )= uu ν (1 − µ ) − (1 − ωu ρ ) γν (cid:26) S (cid:2) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:3) ( u ) − (cid:2) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:3) t =0 + u (cid:27) = u ν (1 − µ ) − (1 − ωu ρ ) γν S (cid:18)(cid:0) e − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω ∗ f (cid:1) ( t ) (cid:19) ( u ) − u ν (1 − µ ) − (1 − ωu ρ ) γν (cid:20) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:21) t =0 + = u ν (1 − µ ) − (1 − ωu ρ ) γν uS (cid:18) t (1 − ν )(1 − µ ) − E ρ, (1 − ν )(1 − µ ) ( ωt ρ ) (cid:19) ( u ) S [ f ( t )]( u ) − u ν (1 − µ ) − (1 − ωu ρ ) γν (cid:20) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:21) t =0 + = u γ (1 − µ ) (1 − ωu ρ ) γν u (1 − ν )(1 − µ ) − (1 − ωu ρ ) γ (1 − ν ) S [ f ( t )]( u ) − u ν (1 − µ ) − (1 − ωu ρ ) γν (cid:20) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:21) t =0 + . Thus we get the required result (3.1).
Lemma 3.2
The Sumudu transforms of the regularized version of Hilfer-Prabhakarfractional derivative (2.11) of order µ is, S (cid:18) C D γ,µρ,ω, + f ( t ) (cid:19) ( u ) = u − µ (1 − ωu ρ ) γ (cid:18) F ( u ) − f (0 + ) (cid:19) = u − µ (1 − ωu ρ ) γ F ( u ) − u − µ (1 − ωu ρ ) γ f (0 + ) . (3.2) Proof:
Taking Sumudu transforms of regularized version of Hilfer-Prabhakar frac-tional derivative (2.11) of order µ and using (2.8), (2.9), (2.13) and (2.14). We have, S (cid:18) C D γ,µρ,ω, + f ( t ) (cid:19) ( u ) = S (cid:18)(cid:18) E − γνρ,ν (1 − µ ) ,ω, + (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + ddt f (cid:1)(cid:19) ( t ) (cid:19) ( u )= S (cid:18)(cid:18) e − γνρ,ν (1 − µ ) ,ω ∗ (cid:0) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + ddt f (cid:1)(cid:19) ( t ) (cid:19) ( u )= uS (cid:18) t ν (1 − µ ) − E − γνρ,ν (1 − µ ) ( ωt ρ ) (cid:19) ( u ) S (cid:18) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + ddt f ( t ) (cid:19) ( u )= uu ν (1 − µ ) − (1 − ωu ρ ) γν S (cid:18)(cid:18) e − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω ∗ ddt f (cid:19) ( t ) (cid:19) ( u )= u ν (1 − µ ) (1 − ωu ρ ) γν uS (cid:18) t (1 − ν )(1 − µ ) − E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ( ωt ρ ) (cid:19) ( u ) S (cid:18) ddt f ( t ) (cid:19) ( u )= u ν (1 − µ ) (1 − ωu ρ ) γν uu (1 − ν )(1 − µ ) − (1 − ωu ρ ) γ (1 − ν ) (cid:26) S (cid:2) f ( t ) (cid:3) ( u ) − f (0 + ) u (cid:27) Thus we get the required result (3.2).
Alternating Proof of Lemma (3.2) : Taking Sumudu transforms of regularizedversion of Hilfer-Prabhakar fractional derivative (2.12) of order µ and using (2.8),(2.9), (2.13), (2.14). We have, S (cid:18) C D γ,µρ,ω, + f ( t ) (cid:19) ( u ) = S (cid:18) E − γρ,ν (1 − µ ) ,ω, + ddt f (cid:19) ( u )= S (cid:18)(cid:18) e − γρ, (1 − µ ) ,ω ∗ ddt f (cid:19) ( t ) (cid:19) ( u )= uS (cid:18) t − µ E − γρ, (1 − µ ) ( ωt ρ ) (cid:19) ( u ) S (cid:18) ddt f ( t ) (cid:19) ( u )= uu − µ (1 − ωu ρ ) γ (cid:26) S (cid:2) f ( t ) (cid:3) ( u ) − f (0 + ) u (cid:27) = u − µ (1 − ωu ρ ) γ (cid:18) F ( u ) − f (0 + ) (cid:19) . Observing that for absolutely continuous function f ∈ AC [0 , b ], (cid:2) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:3) t =0 + = 0 (3.3)then the result (3.1) becomes, S (cid:18) D γ,µ,νρ,ω, + f ( t ) (cid:19) ( u ) = u − µ (1 − ωu ρ ) γ F ( u ) (3.4)Substituting this value (3.4) in (3.2), we get, S (cid:18) C D γ,µρ,ω, + f ( t ) (cid:19) ( u ) = S (cid:18) D γ,µ,νρ,ω, + f ( t ) (cid:19) ( u ) − u − µ (1 − ωu ρ ) γ f (0 + ) (3.5)taking inverse Sumudu transform of (3.5), we get the relation between Hilfer-Prabhakarfractional derivative and regularized version of Hilfer-Prabhakar fractional derivativein terms of Mittag-leffter function as below, C D γ,µρ,ω, + f ( t ) = D γ,µ,νρ,ω, + f ( t ) − t − µ E − γρ, (1 − µ ) ( ωt ρ ) f (0 + ) , (3.6)for f ∈ AC [0 , b ]. Theorem 4.1
The solution of Cauchy problem D γ,µ,νρ,ω, + y ( x ) = λ E δρ,µ,ω, + y ( x ) + f ( x ) , (4.1) (cid:2) E − γ (1 − ν ) ρ, (1 − ν )(1 − µ ) ,ω, + f ( t ) (cid:3) t =0 + = K, (4.2) where f ( x ) ∈ L [0 , ∞ ) ; µ ∈ (0 , , ν ∈ [0 , ; ω, λ ∈ C ; x, ρ > , K, γ, δ ≥ , is y ( x ) = K ∞ X n =0 λ n x ν (1 − µ )+ µ (2 n +1) − E γ − γν + n ( δ + γ ) ρ,ν (1 − µ )+ µ (2 n +1) ( ωx ρ ) + ∞ X n =0 E γ + n ( δ + γ ) ρ, (2 n +1) µ,ω, + f ( x ) . (4.3) Proof:
Let Y ( u ) and F ( u ) denote the Sumudu transform of y ( x ) and f ( x ) respec-tively, Now taking Sumudu transform of (4.1) and using (2.8), (2.9), (2.13), we get S (cid:18) D γ,µ,νρ,ω, + y ( x ) (cid:19) ( u ) = λS (cid:18) E δρ,µ,ω, + y ( x ) (cid:19) ( u ) + S (cid:18) f ( x ) (cid:19) ( u )= λS (cid:18)(cid:0) e δρ,µ,ω ∗ y (cid:1) ( x ) (cid:19) ( u ) + F ( u )= λuS (cid:18) x µ − E δρ,µ ( ωx ρ ) (cid:19) ( u ) S (cid:0) y ( x ) (cid:1) ( u ) + F ( u )From (2.14),(3.1) and (4.2), we get, u − µ (1 − ωu ρ ) γ Y ( u ) − u ν (1 − µ ) − (1 − ωu ρ ) γν K = λuu µ − (1 − ωu ρ ) − δ Y ( u ) + F ( u ) (cid:18) u − µ (1 − ωu ρ ) γ − λu µ (1 − ωu ρ ) − δ (cid:19) Y ( u ) = u ν (1 − µ ) − (1 − ωu ρ ) γν K + F ( u ) Y ( u ) = (cid:18) Ku ν (1 − µ ) − (1 − ωu ρ ) γν u − µ (1 − ωu ρ ) γ + F ( u ) u − µ (1 − ωu ρ ) γ (cid:19) (cid:18) − λu µ (1 − ωu ρ ) − δ u − µ (1 − ωu ρ ) γ (cid:19) = (cid:18) Ku ν (1 − µ )+ µ − (1 − ωu ρ ) γν − γ + F ( u ) u µ (1 − ωu ρ ) − γ (cid:19) ∞ X n =0 λ n u nµ (1 − ωu ρ ) − n ( δ + γ ) = K ∞ X n =0 u ν (1 − µ )+ µ (2 n +1) − (1 − ωu ρ ) − n ( δ + γ )+ γν − γ + F ( u ) ∞ X n =0 u µ (2 n +1) (1 − ωu ρ ) − n ( δ + γ ) − γ . Taking inverse Sumudu transform on both side of above equation, we get the requiredsolution (4.3).
Theorem 4.2
The solution of Cauchy problem C D γ,µρ, − ω, + G ( v, t ) = − λ (1 − v ) G ( v, t ) , f or | v |≤ G ( v,
0) = 1 , (4.5) with t > , φ, λ > , γ ≥ , < ρ ≤ , < µ ≤ , is G ( v, t ) = ∞ X n = o ( − λ ) n (1 − v ) n t nµ E nγρ,nµ +1 ( − ωt ρ ) . (4.6) Proof:
Taking Sumudu transform of (4.4) with respect to t and using (3.2) and (4.5). S (cid:18) C D γ,µρ, − ω, + G ( v, t ) (cid:19) ( v, q ) = − λ (1 − v ) S (cid:18) G ( v, t ) (cid:19) ( v, q ) q − µ (1 + ωq ρ ) γ S (cid:18) G ( v, t ) (cid:19) ( v, q ) − q − µ (1 + ωq ρ ) γ G ( v,
0) = − λ (1 − v ) S (cid:18) G ( v, t ) (cid:19) ( v, q ) (cid:20) q − µ (1 + ωq ρ ) γ + λ (1 − v ) (cid:21) S (cid:18) G ( v, t ) (cid:19) ( v, q ) = q − µ (1 + ωq ρ ) γ S (cid:18) G ( v, t ) (cid:19) ( v, q ) = q − µ (1 + ωq ρ ) γ q − µ (1 + ωq ρ ) γ (cid:18) λ (1 − v ) q − µ (1+ ωq ρ ) γ (cid:19) S (cid:18) G ( v, t ) (cid:19) ( v, q ) = ∞ X n =0 ( − λ ) n (1 − v ) n q nµ (1 + ωq ρ ) − nγ Taking inverse Sumudu transform on both side of above equation, we get the requiredsolution (4.6).
Theorem 4.3
The solution of Cauchy problem D γ,µ,νρ,ω, + u ( x, t ) = K ∂ ∂x u ( x, t ) , (4.7) (cid:20) E − γ (1 − ν ) ρ, (1 − ν ) , (1 − µ ) ,ω, + u ( x, t ) (cid:21) t =0 + = g ( x ) , (4.8)lim x →±∞ u ( x, t ) = 0 , (4.9) with µ ∈ (0 , , ν ∈ [0 , ; x, ω ∈ R ; t, ρ > ; K, γ ≥ , is u ( x, t ) = 12 π Z ∞−∞ dp e ipx b g ( p ) ∞ X n =0 ( − k ) n p n t µ ( n +1) − ν ( µ − − E γ ( n +1 − ν ) ρ,µ ( n +1) − ν ( µ − ( ωt ρ ) . (4.10) Proof:
Let u ( x, q ) denote the Sumudu transform of u ( x, t ) and b u ( p, t ) denote theFourier transform of u ( x, t ). Taking Fourier-Sumudu transform of (4.7) and using(3.1), (4.8), (4.9) we get, q − µ (1 − ωq ρ ) γ b u ( p, q ) − q ν (1 − µ ) − (1 − ωq ρ ) γν b g ( p ) = − Kp b u ( p, q ) (cid:18) q − µ (1 − ωq ρ ) γ + Kp (cid:19) b u ( p, q ) = q ν (1 − µ ) − (1 − ωq ρ ) γν b g ( p ) b u ( p, q ) = q ν (1 − µ ) − (1 − ωq ρ ) γν q − µ (1 − ωq ρ ) γ (cid:18) Kp q − µ (1 − ωq ρ ) γ (cid:19) b g ( p ) b u ( p, q ) = b g ( p ) q µ + ν (1 − µ ) − (1 − ωq ρ ) γν − γ ∞ X n =0 ( − K ) n p n q nµ (1 − ωq ρ ) − nγ b u ( p, q ) = b g ( p ) ∞ X n =0 ( − K ) n p n q ν (1 − µ )+ µ ( n +1) − (1 − ωq ρ ) γν − γ − nγ (4.11)Taking inverse Fourier transform of above equation (4.11) we get, u ( x, q ) = 12 π Z ∞−∞ dp e ipx b g ( p ) ∞ X n =0 ( − K ) n p n q ν (1 − µ )+ µ ( n +1) − (1 − ωq ρ ) γν − γ − nγ (4.12)Taking inverse Sumudu transform of above equation (4.12), we get the required solu-tion (4.10). Theorem 4.4
The solution of Cauchy problem C D γ,µρ,ω, + u ( x, t ) = K ∂ ∂x u ( x, t ) , (4.13) u ( x, + ) = g ( x ) , (4.14)lim x →±∞ u ( x, t ) = 0 , (4.15) with µ ∈ (0 , , x, ω ∈ R , t, K, ρ > , γ ≥ , is u ( x, t ) = 12 π Z ∞−∞ dp e ipx b g ( p ) ∞ X n =0 ( − k ) n p n t nµ E nγρ,nµ +1 ( ωt ρ ) . (4.16) Proof:
Let u ( x, q ) denote the Sumudu transform of u ( x, t ) and b u ( p, t ) denote theFourier transform of u ( x, t ). Now taking Fourier-Sumudu transform of (4.13) andusing (3.2), (4.14), (4.15) we get, q − µ (1 − ωq ρ ) γ b u ( p, q ) − q − µ (1 − ωq ρ ) γ b g ( p ) = − K p b u ( p, q ) (cid:18) q − µ (1 − ωq ρ ) γ + K p (cid:19)b u ( p, q ) = q − µ (1 − ωq ρ ) γ b g ( p ) b u ( p, q ) = q − µ (1 − ωq ρ ) γ q − µ (1 − ωq ρ ) γ (cid:18) K p q − µ (1 − ωq ρ ) γ (cid:19) b g ( p ) b u ( p, q ) = b g ( p ) ∞ X n =0 ( − K ) n p n q nµ (1 − ωq ρ ) − nγ (4.17)Taking inverse Fourier transform of (4.17) , we get, u ( x, q ) = 12 π Z ∞−∞ dp e ipx b g ( p ) ∞ X n =0 ( − K ) n p n q nµ (1 − ωq ρ ) − nγ (4.18)Taking inverse Sumudu transform of (4.18), we get required solution (4.16).0 References [1] Chaurasia V. B. L., Dubey R. S., Belgacem F. B. M. ;
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