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Solution to Volterra Singular Integral Equations and Non Homogenous Time Fractional Pdes

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Solution to Volterra Singular Integral Equations and Non Homogenous Time Fractional Pdes Gen. Math. Notes, Vol. 14, No. 1, January 2013, pp. 6-20 ISSN 2219-7184; Copyright © ICSRS Publication, 2013 www.i-csrs.org Available free online at http://www.geman.in Solution to Volterra Singular Integral Equations and Non Homogenous Time Fractional PDEs A. Aghili 1 and H. Zeinali 2 1,2 Department of Applied Mathematics Faculty of Mathematical Sciences, University of Guilan, P.O. Box- 1841, Rasht – Iran 1E-mail: [email protected] 2E-mail: [email protected] (Received: 8-10-12 / Accepted: 19-11-12) Abstract In this work, the authors implemented Laplace transform method for solving certain partial fractional differential equations and Volterra singular integral equations. Constructive examples are also provided to illustrate the ideas. The result reveals that the transform method is very convenient and effective. Keywords : Non-homogeneous time fractional heat equations; Laplace transform; Volterra singular integral equations. 1 Introduction In this work, the authors used Laplace transform for solving Volterra singular integral equations and PFDEs. Solution to Volterra Singular Integral… 7 The Laplace transform is an alternative method for solving different types of PDEs. Also it is commonly used to solve electrical circuit and systems problems. In this work, the authors implemented transform method for solving the partial fractional heat equation which arise in applications. Several methods have been introduced to solve fractional differential equations, the popular Laplace transform method, [ 1 ] , [ 2 ] , [ 3 ], [ 4 ] , and operational method [ 10]. However, most of these methods are suitable for special types of fractional differential equations, mainly the linear with constant coefficients. More detailed information about some of these results can be found in a survey paper by Kilbas and Trujillo [10]. Atanackovic and Stankovic [5],[6]and Stankovic [20] used the Laplace transform in a certain space of distributions to solve a system of partial differential equations with fractional derivatives, and indicated that such a system may serve as a certain model for a visco elastic rod. Oldham and Spanier I, [13] and [14] , respectively, by reducing a boundary value problem involving Fick’s second low in electro analytic chemistry to a formulation based on the partial Riemann – Liouville fractional with half derivative. Oldham and Spanier [14] gave other application of such equations for diffusion problems. K.Sharma et al, in [18],derive a solution of a generalized fractional Volterra − integral equation involving K4 function with the help of the Sumudo transform. Wyss [22] considered the time fractional diffusion and wave equations and obtained the solution in terms of Fox functions. 1.1 Definitions and Notations Laplace transform of function f( t ) is as follows ∞ Lft{()}= e− s t ftdtFs () : = (). ∫0 If Lft{ ()}= Fs () , then L−1{ F ( s )} is given by + ∞ 1 c i ft()= eFsdsst (), π ∫ 2 i c− i ∞ Where F(s) is analytic in the half- plane Re(s ) > c . For n−1 <α ≤ n ,one gets [15],[16] n −1 C α= α − α −k − 1 ( k ) LDft{0 t ()} sFs ()∑ s f (0). k =0 8 A. Aghili et al. Theorem 1.1 (Effros’s Theorem [ 9 ]) Let Lft{ ()}= Fs () and Lut{( ,)}τ= Us ()exp( − τ qs ()) and assuming φ (s ), q ( s ) are analytic, then one has ∞ L( f()(,)τ ut τ d τ ) = UsFqs ()(()). ∫0 1 α Example 1.1 Let us assume that, U( s ) = and q( s ) = s , then one has s α exp(−τ s α ) L{(,)} u t τ = , s α which leads to ut(,)τ= tWα−1 (,; − αατ − t − α ). Then, we α obtain ∞ 1τ− αατ −−α τ = F ( s ) L1−α ∫ fW()(,; td ) α . t 0 s Provided that the integral in bracket converges absolutely. Like the Fourier transform, the Laplace transform is used in a variety of applications. Perhaps the most common usage of the Laplace transform is in the solution of initial value problems. However, there are other situations for which the properties of the Laplace transform are also very useful, such as in the evaluation of certain integrals and in the solution of fractional singular integral equations of Volterra type. In the following, we may show some applications of integral transform in evaluating certain integrals. Lemma 1.1 The following relations hold true 2π 2 π− α ϕϕ= ϕϕπ = 1 - ∫exp(xd cos ) ∫ exp( xd cos ) 2 Ix0 ( ), 0 −α 2π− α ϕ+ ϕϕπ =2 + 2 2 - ∫ exp(x cos y sin ) d 2 Ixy0 ( ), −α Solution to Volterra Singular Integral… 9 2π ∞ ϕ ϕ = 1 3 - ∫ I0 ( cos ) d ∑ 2 3 k . = ( 2k )!( k !) 2 0 k 0 Proof. 1 – In order to show the above relation, let us introduce the function 2π gx( )= ∫ exp( x cosϕ ) d ϕ , 0 we first calculate the Laplace transform of g( x ) as following, +∞ 2π Lgx{()}=∫ exp( − sxdx ) ∫ exp(cos xϕ ) d ϕ , 0 0 changing the order of integration and simplifying to obtain π +∞ π 2 2 dϕ Lgx{ ( )}= dϕ exp( −+ sxx cos ϕ ) dx = ∫ ∫ ∫ ϕ − 0 0 0 cos s iϕ = At this point we introduce the change of variable e z and simplifying to get π 2 dϕ dz L{ g ( x )} = = ∫cosϕ −s ∫ iz2 − 2 szi + 0z = 1 The value of the complex integral after using Cauchy integral formula is dz 2π Lgx{ ( )}= = = 2π LIx { ( )} ∫ 2 − + 2 0 z =1 iz2 sz i s −1 2- We can rewrite the left side of the equation in the following form x y 2π− α x2+ y 2 ( cosθ + sin θ ) 22 22 I= ∫ exy+ xy + d θ , −α x y we introduce a new variable α such that =sin,α = cos α then xy22+ xy 22 + π− α 2 2 2 I= ∫ ex+ y sin(α + θ ) d θ, −α and again introducing the new variable α+ θ = ϕ leads to 2π 2 2 I= ex+ y sin ϕ dϕ =2 π Ixy (2 + 2 ). ∫0 0 10 A. Aghili et al. It is obvious that if we set y = 0, then we get the relationships 1 and 2. 3- Let us define a new function by the integral 2π = θ θ Ix( )∫ I0 ( 2 x cos ) d , 0 taking Laplace transform of the above relation ,we get cos θ π π θ 2e s 1 2 cos 1 1 = =s θ = LIx{ ( )}∫ ( ) ∫ edI 0 ( ). 0ss 0 ss On the other hand, one has the following expansion for modified Bessel’s function of order zero ∞ 2k = y I0 ( y )∑ 2 2 k , k =0 (k !) 2 so we have π ∞ 2k 2 θ θ = x I0( 2 x cos ) d . ∫0 ∑ 2 2 k k =0 ( 2k )!( k !) 2 If we set x = 1in the above relation, we get the desired equation. π 4 Lemma 1.2 Let us assume that LftFs( ( ))= ( ) =∫ exp{ − s (sec2α + csc 2 α )} d α . 0 Show that the following relations hold true 1 1 - f( t )= , 2t t − 4 π 2 - F( s )= erf ( 2 s ), 2 π 2 e 3 - ∫ tanα exp(− tan2 α )d α =− Ei ( − 1), 0 2 x e t where Ei( x ) = ∫ dt is exponential integral and the relation Eix(− ) =−Γ (0, x ) −∞ t for x > 0 in which Γ(z , x ) is incomplete Gamma function. Proof: 1- We have π 4s 4 −( ) = sin2 2 θ θ Fs( )∫ e d , 0 Solution to Volterra Singular Integral… 11 and consequently applying Bromwich's integral, we get π c+ i ∞ 1 4 −4s ( ) 1 2 ft()= ( esin 2 θ dedsθ ),st π ∫ ∫ 2 i c− i ∞ 0 changing the order of integration, leads to π 4 4 ft()=δ ( t − ). d θ ∫ 2 θ 0 sin 2 4 Now, we introduce the new variable t− = w , then we get sin2 2 θ 1 f( t )= . − 2t t 4 2- We take Laplace transform of the above equation to obtain +∞ e−st Lft{()}= dt . ∫ − 4 2t t 4 Now we introduce the new variable t−4 = u 2 to get +∞ −su 2 − e L{ f ( t )}= e4s du . ∫ 2 + 0 u 4 +∞ 2 e−su π At this point, let us assume that I( s )= du , I (0) = then ∫ 2 + 0 u 4 4 +∞− 2 +∞ +∞ − 2 2 su su u e−su 2 e I′( s )=− du =− e du + 4 ∫u2 +4 ∫ ∫ u 2 + 4 0 0 0 1 π = − + 4I . 2 s Solving the above ODE we obtain π I( s )= e4s erf (2 s ), 4 and consequently π F( s )= erf (2 s ). 4 12 A. Aghili et al. Lemma 1.3 If k( x ) and ϕ(x ) are Laplace transformable functions then we have the following relationship +∞ L∫ kxt(− )()ϕ tdt =−Φ Ks ( )(), s x where K( s ),Φ ( s ) are Laplace transforms of functions k(− x ),ϕ ( x ) respectively. Proof: See [ 9 ][ 17 ]. 2 Solution to Volterra Singular Integral Equations Laplace transform can be used to solve certain types of Volterra singular integral equations. Problem 2.1 Let us consider fractional Volterra singular integral equation of the form, +∞ c Dfxα ()=+− gx ()λ ∫ kxtftdt ( )(), f (0)0, = (2.1) x in which kxt(,)= kx ( − t ) is the kernel and g( x ) is assumed to be a Laplace transformable function. + ∞ 1c i G ( s ) Then (2.1) has the formal solution f() x= { } est ds π∫ λ − − α 2ic− i ∞ Kss ( ) Solution: Let Lfx(())= FsLgx (),(()) = GsLk (),(( −= x )) Ks () be the Laplace transforms of fx( ), gxk ( ), (− x ) , respectively, then by using Lemma 1.3 one gets the following relationship, sFsα ()= Gs () +λ K ( − sFs )(). (2.2) So one can write, −G( s ) F( s )= , λK(− s ) − s α (2.3) and consequently by Bromwich's integral we get the following relation, c+ i ∞ − = 1G ( s ) st f() x {α }, e ds (2.4) 2πi∫ λ Kss (− ) − c− i ∞ which can be solved by the use of Residue theorem.
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