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Research ArticleDouble Laplace Transform Method for Solving Space andTime Fractional Telegraph Equations
Ranjit R Dhunde1 and G L Waghmare2
1Department of Mathematics Datta Meghe Institute of Engineering Technology and Research Wardha India2Department of Mathematics Government Science College Gadchiroli India
Correspondence should be addressed to Ranjit R Dhunde ranjitdhunderediffmailcom
Received 17 June 2016 Revised 23 September 2016 Accepted 5 October 2016
Academic Editor Irena Lasiecka
Copyright copy 2016 R R Dhunde and G L Waghmare This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
Double Laplace transform method is applied to find exact solutions of linearnonlinear space-time fractional telegraph equationsin terms of Mittag-Leffler functions subject to initial and boundary conditions Furthermore we give illustrative examples todemonstrate the efficiency of the method
1 Introduction
The telegraph equation developed byOliverHeaviside in 1880is widely used in Science and Engineering Its applicationsarise in signal analysis for transmission and propagation ofelectrical signals and also modelling reaction diffusion
In recent years great interest has been developed in frac-tional differential equation because of its frequent appearancein fluid mechanics mathematical biology electrochemistryand physics A space-time fractional telegraph equation isobtained from the classical telegraph equation by replacingthe time and space derivative terms by fractional derivatives
There are various methods developed to solve fractionaltelegraph equations Orsingher and Xuelei [1] and Orsingherand Beghin [2] considered the space and time fractional tele-graph equations and obtained the Fourier transform of theirfundamental solution Momani [3] and Garg and Sharma [4]used Adomian decomposition method developed by Ado-mian in [5] for solving homogeneous and nonhomogeneousspace-time fractional telegraph equation Chen et al [6]implemented separation of variables method for deriving theanalytical solutions of the nonhomogeneous time fractionaltelegraph equation under Dirichlet Neumann and Robinboundary conditions Huang [7] considered the combineFourier-Laplace transform to solve time fractional telegraphequation
Variational iteration method is proposed by He [8] andused by Sevimlican [9] for solving space and time fractionaltelegraph equations Yildirim in [10] used a homotopy per-turbation method and Das et al in [11] used a homotopyanalysis method to obtain approximate analytical solutionof fractional telegraph equation Garg et al [12] and Galue[13] used generalised differential transformmethod to derivethe solution of space-time fractional telegraph equations interms ofMittag-Leffler functions Srivastava et al [14] appliedreduced differential transform method to solve Caputo timefractional order hyperbolic telegraph equation
In recent years significant attention has been givenby many authors towards the study of fractional telegraphequations by using single Laplace transform combined withvariational iterationmethod homotopy analysismethod andhomotopy perturbation method Khan et al [15] Kumar etal [16] and Prakash [17] applied a combination of singleLaplace transform and homotopy perturbation method toobtain analytic and approximate solutions of the space-time fractional telegraph equations Alawad et al [18] useda combination of single Laplace transform and variationaliteration method for finding exact solutions of space-timefractional telegraph equations in terms of Mittag-Lefflerfunctions Kumar in [19] coupled single Laplace transformand homotopy analysis method for the solution of space-fractional telegraph equation
Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2016 Article ID 1414595 7 pageshttpdxdoiorg10115520161414595
2 International Journal of Mathematics and Mathematical Sciences
To our knowledge solving fractional partial differentialequations using the double Laplace transform is still seenin very little proportionate or no work is available in theliterature So the main objective of this paper is to findthe exact solutions of homogeneous and nonhomogeneousspace-time fractional telegraph equations in terms of Mittag-Leffler functions subject to initial and boundary conditionsby means of double Laplace transform
2 A Brief Introduction ofDouble Laplace Transform andCaputo Fractional Derivative
Let 119891(119909 119905) be a function of two variables 119909 and 119905 definedin the positive quadrant of the 119909119905-plane The double Laplacetransform of the function 119891(119909 119905) as given by Sneddon [20] isdefined by
119871119909119871 119905 119891 (119909 119905) = 119891 (119901 119904)= intinfin0119890minus119901119909 intinfin
0119890minus119904119905119891 (119909 119905) 119889119905 119889119909 (1)
whenever that integral exists Here 119901 and 119904 are complexnumbers
From this definition we deduce
119871119909119871 119905 [119891 (119909) 119892 (119905)] = 119891 (119901) 119892 (119904)= 119871119909 [119891 (119909)] 119871 119905 [119892 (119905)] (2)
The inverse double Laplace transform 119871119909minus1119871 119905minus1119891(119901 119904) =119891(119909 119905) is defined as in [21 22] by the complex double integralformula
119871minus1119909 119871minus1119905 119891 (119901 119904) = 119891 (119909 119905)= 12120587119894 int
119888+119894infin
119888minus119894infin119890119901119909119889119901 12120587119894 int
119889+119894infin
119889minus119894infin119890119904119905119891 (119901 119904) 119889119904 (3)
where 119891(119901 119904) must be an analytic function for all 119901 and 119904 inthe region defined by the inequalities Re119901 ge 119888 and Re 119904 ge 119889where 119888 and 119889 are real constants to be chosen suitably
The double Laplace transform formulas for the partialderivatives of an arbitrary integer order as in [23] are
119871119909119871 119905 120597119898119891 (119909 119905)120597119909119898 = 119901119898119891 (119901 119904)minus 119898minus1sum119895=0
119901119898minus1minus119895119871 119905 120597119895119891 (0 119905)120597119909119895
119871119909119871 119905 120597119899119891 (119909 119905)120597119905119899 = 119904119899119891 (119901 119904)minus 119899minus1sum119896=0
119904119899minus1minus119896119871119909 120597119896119891 (119909 0)120597119905119896
(4)
Definition 1 The Caputo fractional derivative of function119906(119909 119905) is defined in [18] as
120597120572119891 (119909 119905)120597119909120572 = 1Γ (119898 minus 120572) int119909
0(119909 minus 120585)119898minus120572minus1 120597119898119891 (120585 119905)120597120585119898 119889120585119898 minus 1 lt 120572 le 119898 119898 isin 119873
120597120573119891 (119909 119905)120597119905120573 = 1Γ (119899 minus 120573) int119905
0(119905 minus 120591)119899minus120573minus1 120597119899119891 (119909 120591)120597120591119899 119889120591
119899 minus 1 lt 120573 le 119899 119899 isin 119873
(5)
The double Laplace transform formulas for the partial frac-tional Caputo derivatives as in [23] are
119871119909119871 119905 120597120572119891 (119909 119905)120597119909120572 = 119901120572119891 (119901 119904)minus 119898minus1sum119895=0
119901120572minus1minus119895119871 119905 120597119895119891 (0 119905)120597119909119895
119871119909119871 119905 120597120573119891 (119909 119905)120597119905120573 = 119904120573119891 (119901 119904)minus 119899minus1sum119896=0
119904120573minus1minus119896119871119909 120597119896119891 (119909 0)120597119905119896
(6)
Definition 2 TheMittag-Leffler function is defined by
119864120572120573 (119905) = infinsum119896=0
119905119896Γ (120572119896 + 120573) 119905 120573 isin C R (120572) gt 0 (7)
The single Laplace transform of the function 119905120573minus1119864120572120573(120582119905120572)takes the form
119871 119905 119905120573minus1119864120572120573 (120582119905120572) = 119904120572minus120573119904120572 minus 120582 for |120582| lt 10038161003816100381610038161199041205721003816100381610038161003816 (8)
3 Double Laplace Transform Method
Consider the following general multiterms fractional tele-graph equation as in [18]
120597120572119906 (119909 119905)120597119909120572 = 119886120597120573119906 (119909 119905)120597119905120573 + 119887120597120574119906 (119909 119905)120597119905120574 + 119888119906 (119909 119905)+ ℎ (119909 119905)
1 lt 120572 120573 le 2 0 lt 120574 le 1 119909 119905 ge 0(9)
with initial conditions
119906 (119909 0) = 1198911 (119909) 119906119905 (119909 0) = 1198912 (119909) (10)
and boundary conditions
119906 (0 119905) = 1198921 (119905) 119906119909 (0 119905) = 1198922 (119905) (11)
Here 119886 119887 119888 are constants and ℎ(119909 119905) is given function
International Journal of Mathematics and Mathematical Sciences 3
Applying the double Laplace transform on both sides of(9) we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 119886 [119904120573119906 (119901 119904) minus 119904120573minus1119906 (119901 0) minus 119904120573minus2119906119905 (119901 0)]+ 119887 [119904120574119906 (119901 119904) minus 119904120574minus1119906 (119901 0)] + 119888119906 (119901 119904)+ ℎ (119901 119904)
(12)
where ℎ(119901 119904) = 119871119909119871 119905ℎ(119909 119905)Further applying single Laplace transform to initial (10)
and boundary conditions (11) we get
119906 (119901 0) = 1198911 (119901) 119906119905 (119901 0) = 1198912 (119901) 119906 (0 119904) = 1198921 (119904) 119906119909 (0 119904) = 1198922 (119904)
(13)
By substituting (13) in (12) and simplifying we obtain
119906 (119901 119904) = 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]
(14)
Applying inverse double Laplace transform to (14) we obtainthe solution of (9) in the form
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]]
(15)
Here we assume that the inverse double Laplace transform ofeach term in the right side of (15) exists
4 Illustrative Examples
In this section we demonstrate the applicability of theprevious method by giving examples
Example 1 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = 0 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) 1 lt 120572 le 2 119909 119905 ge 0
(16)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 119864120572 (119909120572) + 1199091198641205722 (119909120572) 119906119905 (119909 0) = 1198912 (119909) = minus [119864120572 (119909120572) + 1199091198641205722 (119909120572)] (17)
119906 (0 119905) = 1198921 (119905) = 119890minus119905119906119909 (0 119905) = 1198922 (119905) = 119890minus119905 (18)
a homogeneous space-fractional telegraph equationTaking single Laplace transform to initial (17) and bound-
ary conditions (18) we get
1198911 (119901) = ( 1119901 + 11199012) 119901120572119901120572 minus 1 1198912 (119901) = minus( 1119901 + 11199012) 119901120572119901120572 minus 1 1198921 (119904) = 1198922 (119904) = 1119904 + 1
(19)
Substituting above in (15) we get solution of (16)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 1119904 + 1+ 119901120572minus2 1119904 + 1 minus 119904 ( 1119901 + 11199012) 119901120572119901120572 minus 1+ ( 1119901 + 11199012) 119901120572119901120572 minus 1 minus ( 1119901 + 11199012) 119901120572119901120572 minus 1]]
(20)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119904 + 1) ( 1119901 + 11199012) 119901120572119901120572 minus 1] 119906 (119909 119905) = 119890minus119905 [119864120572 (119909120572) + 1199091198641205722 (119909120572)]
(21)
which agrees with the solution already obtained in [18]If we take 120572 = 2 then we get the exact solution of standard
telegraph equation
119906 (119909 119905) = 119890119909minus119905 (22)
Example 2 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = minus1199092 minus 119905 + 1 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) minus 1199092 minus 119905+ 1 1 lt 120572 le 2 119909 119905 ge 0
(23)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909)= 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572)
119906119905 (119909 0) = 1198912 (119909) = 1(24)
119906 (0 119905) = 1198921 (119905) = 119905119906119909 (0 119905) = 1198922 (119905) = 0 (25)
a space-fractional nonhomogeneous telegraph equation
4 International Journal of Mathematics and Mathematical Sciences
Taking single Laplace transform to initial (24) and bound-ary conditions (25) we get
1198911 (119901) = 21199013 minus 2119901 minus 2 119901120572minus3
119901120572 minus 1 + 2 119901120572minus1
119901120572 minus 1= 2 (1199012 minus 1)1199013 (119901120572 minus 1)
1198912 (119901) = 1119901 1198921 (119904) = 11199042 1198922 (119904) = 0
(26)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus21199013119904 minus 11199011199042 + 1119901119904 (27)
Substituting above in (15) we get solution of (23)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 11199042minus 119904 2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 1119901 minus
2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 21199013119904 minus 11199011199042
+ 1119901119904]]
(28)
Rearranging we have
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [[ 119901120572
1199011199042 minus 1119901minus 11199011199042 minus 1119901119904] + [
minus2119904 (1199012 minus 1)1199013 (119901120572 minus 1) minus
2 (1199012 minus 1)1199013 (119901120572 minus 1)
minus 21199013119904 + 2119901119904]]]
(29)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 11199011199042 +2 (1199012 minus 1)1199013119904 (119901120572 minus 1)]
119906 (119909 119905) = 119905 + 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572) (30)
which agrees with the solution already obtained in [18] for120574 = 1For 120572 = 2 then 119906(119909 119905) = 119905 + 1199092
Example 3 By substituting 119886 = 1 119887 = 1 119888 = 0 120572 = 2 120574 = 1and ℎ(119909 119905) = minus2119905(1199092 minus 119909)(1199051minus120573Γ(3 minus 120573) + 1) + 21199052 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597119906 (119909 119905)120597119905= 1205972119906 (119909 119905)1205971199092 + 2119905 (1199092 minus 119909)( 1199051minus120573Γ (3 minus 120573) + 1) minus 21199052
1 lt 120573 le 2 119909 119905 ge 0(31)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (32)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = minus1199052 (33)
a time fractional telegraph equation in [24]Taking single Laplace transform to initial (32) and bound-
ary conditions (33) we get
1198911 (119901) = 1198912 (119901) = 1198921 (119904) = 01198922 (119904) = minus21199043
(34)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 (35)
Substituting above in (15) we get solution of (31)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904) [minus21199043minus 2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 ]]
(36)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [( 21199013 minus 11199012) 21199043 ] 119906 (119909 119905) = (1199092 minus 119909) 1199052
(37)
Example 4 By substituting 119886 = 1 119887 = 1 119888 = 1 120572 = 2 andℎ(119909 119905) = 0 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120574119906 (119909 119905)120597119905120574 + 119906 (119909 119905) = 1205972119906 (119909 119905)1205971199092 1 lt 120573 le 2 12 lt 120574 le 1 119909 119905 ge 0
(38)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 119890119909 (39)
119906 (0 119905) = 1198921 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) 119906119909 (0 119905) = 1198922 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) (40)
a homogeneous time fractional telegraph equation in [25]
International Journal of Mathematics and Mathematical Sciences 5
Taking single Laplace transform to initial (39) and bound-ary conditions (40) we get
1198911 (119901) = 01198912 (119901) = 1119901 minus 1 1198921 (119904) = 1198922 (119904) = 119904120573minus120574minus2119904120573minus120574 + 1
(41)
Substituting above in (15) we get solution of (38)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120574 minus 1) [119901 119904120573minus120574minus2(119904120573minus120574 + 1)+ 119904120573minus120574minus2(119904120573minus120574 + 1) minus 119904120573minus2 1(119901 minus 1)]]
(42)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901 minus 1) 119904120573minus120574minus2(119904120573minus120574 + 1)] 119906 (119909 119905) = 119890119909119905119864120573minus1205742 (minus119905120573minus120574)
(43)
which agrees with the solution already obtained in [25]
Example 5 By substituting 119886 = 119887 = 119888 = 1120572 = 2 and ℎ(119909 119905) =minus sinh119909(119905119899Γ(119899 + 1)) in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120573minus1119906 (119909 119905)120597119905120573minus1 + 119906 (119909 119905)= 1205972119906 (119909 119905)1205971199092 + sinh119909 119905119899Γ (119899 + 1)
1 lt 120573 le 2 119909 119905 ge 0(44)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (45)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = 119905119899+1205731198641119899+120573+1 (minus119905) (46)
a nonhomogeneous time fractional telegraph equation in[25]
Taking single Laplace transform to initial (45) and bound-ary conditions (46) we get
1198911 (119901) = 1198912 (119901) = 01198921 (119904) = 01198922 (119904) = 119904minus119899minus120573119904 + 1
(47)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus1(1199012 minus 1) 119904119899+1 (48)
Substituting above in (15) we get solution of (44)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120573minus1 minus 1) [ 119904minus119899minus120573
(119904 + 1)minus 1(1199012 minus 1) 119904119899+1]]
(49)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 1) 119904minus119899minus120573
(119904 + 1)] 119906 (119909 119905) = (sinh119909) 119905119899+1205731198641119899+120573+1 (minus119905)
(50)
which agrees with the solution already obtained in [25]
Example 6 Consider the following space-fractional-ordernonlinear telegraph equation
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 1199062 (119909 119905)minus 119890minus2119905 (119909 minus 1199092)2 minus 2119890minus119905 1199092minus120572Γ (3 minus 120572)
1 lt 120572 le 2 119909 119905 ge 0(51)
under the initial conditions
119906 (119909 0) = minus119906119905 (119909 0) = 119909 minus 1199092 (52)
and boundary conditions
119906 (0 119905) = 0119906119909 (0 119905) = 119890minus119905 (53)
Applying the double Laplace transform on both sides of (51)we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 1199042119906 (119901 119904) minus 119904119906 (119901 0) minus 119906119905 (119901 0) + 119904119906 (119901 119904)minus 119906 (119901 0) minus 2(119904 + 1) 1199013minus120572+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(54)
Further applying single Laplace transform to initial (52) andboundary conditions (53) we get
119906 (119901 0) = minus119906119905 (119901 0) = 11199012 minus 21199013 119906 (0 119904) = 0119906119909 (0 119904) = 1119904 + 1
(55)
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
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Stochastic AnalysisInternational Journal of
2 International Journal of Mathematics and Mathematical Sciences
To our knowledge solving fractional partial differentialequations using the double Laplace transform is still seenin very little proportionate or no work is available in theliterature So the main objective of this paper is to findthe exact solutions of homogeneous and nonhomogeneousspace-time fractional telegraph equations in terms of Mittag-Leffler functions subject to initial and boundary conditionsby means of double Laplace transform
2 A Brief Introduction ofDouble Laplace Transform andCaputo Fractional Derivative
Let 119891(119909 119905) be a function of two variables 119909 and 119905 definedin the positive quadrant of the 119909119905-plane The double Laplacetransform of the function 119891(119909 119905) as given by Sneddon [20] isdefined by
119871119909119871 119905 119891 (119909 119905) = 119891 (119901 119904)= intinfin0119890minus119901119909 intinfin
0119890minus119904119905119891 (119909 119905) 119889119905 119889119909 (1)
whenever that integral exists Here 119901 and 119904 are complexnumbers
From this definition we deduce
119871119909119871 119905 [119891 (119909) 119892 (119905)] = 119891 (119901) 119892 (119904)= 119871119909 [119891 (119909)] 119871 119905 [119892 (119905)] (2)
The inverse double Laplace transform 119871119909minus1119871 119905minus1119891(119901 119904) =119891(119909 119905) is defined as in [21 22] by the complex double integralformula
119871minus1119909 119871minus1119905 119891 (119901 119904) = 119891 (119909 119905)= 12120587119894 int
119888+119894infin
119888minus119894infin119890119901119909119889119901 12120587119894 int
119889+119894infin
119889minus119894infin119890119904119905119891 (119901 119904) 119889119904 (3)
where 119891(119901 119904) must be an analytic function for all 119901 and 119904 inthe region defined by the inequalities Re119901 ge 119888 and Re 119904 ge 119889where 119888 and 119889 are real constants to be chosen suitably
The double Laplace transform formulas for the partialderivatives of an arbitrary integer order as in [23] are
119871119909119871 119905 120597119898119891 (119909 119905)120597119909119898 = 119901119898119891 (119901 119904)minus 119898minus1sum119895=0
119901119898minus1minus119895119871 119905 120597119895119891 (0 119905)120597119909119895
119871119909119871 119905 120597119899119891 (119909 119905)120597119905119899 = 119904119899119891 (119901 119904)minus 119899minus1sum119896=0
119904119899minus1minus119896119871119909 120597119896119891 (119909 0)120597119905119896
(4)
Definition 1 The Caputo fractional derivative of function119906(119909 119905) is defined in [18] as
120597120572119891 (119909 119905)120597119909120572 = 1Γ (119898 minus 120572) int119909
0(119909 minus 120585)119898minus120572minus1 120597119898119891 (120585 119905)120597120585119898 119889120585119898 minus 1 lt 120572 le 119898 119898 isin 119873
120597120573119891 (119909 119905)120597119905120573 = 1Γ (119899 minus 120573) int119905
0(119905 minus 120591)119899minus120573minus1 120597119899119891 (119909 120591)120597120591119899 119889120591
119899 minus 1 lt 120573 le 119899 119899 isin 119873
(5)
The double Laplace transform formulas for the partial frac-tional Caputo derivatives as in [23] are
119871119909119871 119905 120597120572119891 (119909 119905)120597119909120572 = 119901120572119891 (119901 119904)minus 119898minus1sum119895=0
119901120572minus1minus119895119871 119905 120597119895119891 (0 119905)120597119909119895
119871119909119871 119905 120597120573119891 (119909 119905)120597119905120573 = 119904120573119891 (119901 119904)minus 119899minus1sum119896=0
119904120573minus1minus119896119871119909 120597119896119891 (119909 0)120597119905119896
(6)
Definition 2 TheMittag-Leffler function is defined by
119864120572120573 (119905) = infinsum119896=0
119905119896Γ (120572119896 + 120573) 119905 120573 isin C R (120572) gt 0 (7)
The single Laplace transform of the function 119905120573minus1119864120572120573(120582119905120572)takes the form
119871 119905 119905120573minus1119864120572120573 (120582119905120572) = 119904120572minus120573119904120572 minus 120582 for |120582| lt 10038161003816100381610038161199041205721003816100381610038161003816 (8)
3 Double Laplace Transform Method
Consider the following general multiterms fractional tele-graph equation as in [18]
120597120572119906 (119909 119905)120597119909120572 = 119886120597120573119906 (119909 119905)120597119905120573 + 119887120597120574119906 (119909 119905)120597119905120574 + 119888119906 (119909 119905)+ ℎ (119909 119905)
1 lt 120572 120573 le 2 0 lt 120574 le 1 119909 119905 ge 0(9)
with initial conditions
119906 (119909 0) = 1198911 (119909) 119906119905 (119909 0) = 1198912 (119909) (10)
and boundary conditions
119906 (0 119905) = 1198921 (119905) 119906119909 (0 119905) = 1198922 (119905) (11)
Here 119886 119887 119888 are constants and ℎ(119909 119905) is given function
International Journal of Mathematics and Mathematical Sciences 3
Applying the double Laplace transform on both sides of(9) we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 119886 [119904120573119906 (119901 119904) minus 119904120573minus1119906 (119901 0) minus 119904120573minus2119906119905 (119901 0)]+ 119887 [119904120574119906 (119901 119904) minus 119904120574minus1119906 (119901 0)] + 119888119906 (119901 119904)+ ℎ (119901 119904)
(12)
where ℎ(119901 119904) = 119871119909119871 119905ℎ(119909 119905)Further applying single Laplace transform to initial (10)
and boundary conditions (11) we get
119906 (119901 0) = 1198911 (119901) 119906119905 (119901 0) = 1198912 (119901) 119906 (0 119904) = 1198921 (119904) 119906119909 (0 119904) = 1198922 (119904)
(13)
By substituting (13) in (12) and simplifying we obtain
119906 (119901 119904) = 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]
(14)
Applying inverse double Laplace transform to (14) we obtainthe solution of (9) in the form
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]]
(15)
Here we assume that the inverse double Laplace transform ofeach term in the right side of (15) exists
4 Illustrative Examples
In this section we demonstrate the applicability of theprevious method by giving examples
Example 1 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = 0 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) 1 lt 120572 le 2 119909 119905 ge 0
(16)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 119864120572 (119909120572) + 1199091198641205722 (119909120572) 119906119905 (119909 0) = 1198912 (119909) = minus [119864120572 (119909120572) + 1199091198641205722 (119909120572)] (17)
119906 (0 119905) = 1198921 (119905) = 119890minus119905119906119909 (0 119905) = 1198922 (119905) = 119890minus119905 (18)
a homogeneous space-fractional telegraph equationTaking single Laplace transform to initial (17) and bound-
ary conditions (18) we get
1198911 (119901) = ( 1119901 + 11199012) 119901120572119901120572 minus 1 1198912 (119901) = minus( 1119901 + 11199012) 119901120572119901120572 minus 1 1198921 (119904) = 1198922 (119904) = 1119904 + 1
(19)
Substituting above in (15) we get solution of (16)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 1119904 + 1+ 119901120572minus2 1119904 + 1 minus 119904 ( 1119901 + 11199012) 119901120572119901120572 minus 1+ ( 1119901 + 11199012) 119901120572119901120572 minus 1 minus ( 1119901 + 11199012) 119901120572119901120572 minus 1]]
(20)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119904 + 1) ( 1119901 + 11199012) 119901120572119901120572 minus 1] 119906 (119909 119905) = 119890minus119905 [119864120572 (119909120572) + 1199091198641205722 (119909120572)]
(21)
which agrees with the solution already obtained in [18]If we take 120572 = 2 then we get the exact solution of standard
telegraph equation
119906 (119909 119905) = 119890119909minus119905 (22)
Example 2 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = minus1199092 minus 119905 + 1 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) minus 1199092 minus 119905+ 1 1 lt 120572 le 2 119909 119905 ge 0
(23)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909)= 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572)
119906119905 (119909 0) = 1198912 (119909) = 1(24)
119906 (0 119905) = 1198921 (119905) = 119905119906119909 (0 119905) = 1198922 (119905) = 0 (25)
a space-fractional nonhomogeneous telegraph equation
4 International Journal of Mathematics and Mathematical Sciences
Taking single Laplace transform to initial (24) and bound-ary conditions (25) we get
1198911 (119901) = 21199013 minus 2119901 minus 2 119901120572minus3
119901120572 minus 1 + 2 119901120572minus1
119901120572 minus 1= 2 (1199012 minus 1)1199013 (119901120572 minus 1)
1198912 (119901) = 1119901 1198921 (119904) = 11199042 1198922 (119904) = 0
(26)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus21199013119904 minus 11199011199042 + 1119901119904 (27)
Substituting above in (15) we get solution of (23)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 11199042minus 119904 2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 1119901 minus
2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 21199013119904 minus 11199011199042
+ 1119901119904]]
(28)
Rearranging we have
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [[ 119901120572
1199011199042 minus 1119901minus 11199011199042 minus 1119901119904] + [
minus2119904 (1199012 minus 1)1199013 (119901120572 minus 1) minus
2 (1199012 minus 1)1199013 (119901120572 minus 1)
minus 21199013119904 + 2119901119904]]]
(29)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 11199011199042 +2 (1199012 minus 1)1199013119904 (119901120572 minus 1)]
119906 (119909 119905) = 119905 + 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572) (30)
which agrees with the solution already obtained in [18] for120574 = 1For 120572 = 2 then 119906(119909 119905) = 119905 + 1199092
Example 3 By substituting 119886 = 1 119887 = 1 119888 = 0 120572 = 2 120574 = 1and ℎ(119909 119905) = minus2119905(1199092 minus 119909)(1199051minus120573Γ(3 minus 120573) + 1) + 21199052 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597119906 (119909 119905)120597119905= 1205972119906 (119909 119905)1205971199092 + 2119905 (1199092 minus 119909)( 1199051minus120573Γ (3 minus 120573) + 1) minus 21199052
1 lt 120573 le 2 119909 119905 ge 0(31)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (32)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = minus1199052 (33)
a time fractional telegraph equation in [24]Taking single Laplace transform to initial (32) and bound-
ary conditions (33) we get
1198911 (119901) = 1198912 (119901) = 1198921 (119904) = 01198922 (119904) = minus21199043
(34)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 (35)
Substituting above in (15) we get solution of (31)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904) [minus21199043minus 2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 ]]
(36)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [( 21199013 minus 11199012) 21199043 ] 119906 (119909 119905) = (1199092 minus 119909) 1199052
(37)
Example 4 By substituting 119886 = 1 119887 = 1 119888 = 1 120572 = 2 andℎ(119909 119905) = 0 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120574119906 (119909 119905)120597119905120574 + 119906 (119909 119905) = 1205972119906 (119909 119905)1205971199092 1 lt 120573 le 2 12 lt 120574 le 1 119909 119905 ge 0
(38)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 119890119909 (39)
119906 (0 119905) = 1198921 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) 119906119909 (0 119905) = 1198922 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) (40)
a homogeneous time fractional telegraph equation in [25]
International Journal of Mathematics and Mathematical Sciences 5
Taking single Laplace transform to initial (39) and bound-ary conditions (40) we get
1198911 (119901) = 01198912 (119901) = 1119901 minus 1 1198921 (119904) = 1198922 (119904) = 119904120573minus120574minus2119904120573minus120574 + 1
(41)
Substituting above in (15) we get solution of (38)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120574 minus 1) [119901 119904120573minus120574minus2(119904120573minus120574 + 1)+ 119904120573minus120574minus2(119904120573minus120574 + 1) minus 119904120573minus2 1(119901 minus 1)]]
(42)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901 minus 1) 119904120573minus120574minus2(119904120573minus120574 + 1)] 119906 (119909 119905) = 119890119909119905119864120573minus1205742 (minus119905120573minus120574)
(43)
which agrees with the solution already obtained in [25]
Example 5 By substituting 119886 = 119887 = 119888 = 1120572 = 2 and ℎ(119909 119905) =minus sinh119909(119905119899Γ(119899 + 1)) in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120573minus1119906 (119909 119905)120597119905120573minus1 + 119906 (119909 119905)= 1205972119906 (119909 119905)1205971199092 + sinh119909 119905119899Γ (119899 + 1)
1 lt 120573 le 2 119909 119905 ge 0(44)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (45)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = 119905119899+1205731198641119899+120573+1 (minus119905) (46)
a nonhomogeneous time fractional telegraph equation in[25]
Taking single Laplace transform to initial (45) and bound-ary conditions (46) we get
1198911 (119901) = 1198912 (119901) = 01198921 (119904) = 01198922 (119904) = 119904minus119899minus120573119904 + 1
(47)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus1(1199012 minus 1) 119904119899+1 (48)
Substituting above in (15) we get solution of (44)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120573minus1 minus 1) [ 119904minus119899minus120573
(119904 + 1)minus 1(1199012 minus 1) 119904119899+1]]
(49)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 1) 119904minus119899minus120573
(119904 + 1)] 119906 (119909 119905) = (sinh119909) 119905119899+1205731198641119899+120573+1 (minus119905)
(50)
which agrees with the solution already obtained in [25]
Example 6 Consider the following space-fractional-ordernonlinear telegraph equation
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 1199062 (119909 119905)minus 119890minus2119905 (119909 minus 1199092)2 minus 2119890minus119905 1199092minus120572Γ (3 minus 120572)
1 lt 120572 le 2 119909 119905 ge 0(51)
under the initial conditions
119906 (119909 0) = minus119906119905 (119909 0) = 119909 minus 1199092 (52)
and boundary conditions
119906 (0 119905) = 0119906119909 (0 119905) = 119890minus119905 (53)
Applying the double Laplace transform on both sides of (51)we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 1199042119906 (119901 119904) minus 119904119906 (119901 0) minus 119906119905 (119901 0) + 119904119906 (119901 119904)minus 119906 (119901 0) minus 2(119904 + 1) 1199013minus120572+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(54)
Further applying single Laplace transform to initial (52) andboundary conditions (53) we get
119906 (119901 0) = minus119906119905 (119901 0) = 11199012 minus 21199013 119906 (0 119904) = 0119906119909 (0 119904) = 1119904 + 1
(55)
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Mathematics and Mathematical Sciences 3
Applying the double Laplace transform on both sides of(9) we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 119886 [119904120573119906 (119901 119904) minus 119904120573minus1119906 (119901 0) minus 119904120573minus2119906119905 (119901 0)]+ 119887 [119904120574119906 (119901 119904) minus 119904120574minus1119906 (119901 0)] + 119888119906 (119901 119904)+ ℎ (119901 119904)
(12)
where ℎ(119901 119904) = 119871119909119871 119905ℎ(119909 119905)Further applying single Laplace transform to initial (10)
and boundary conditions (11) we get
119906 (119901 0) = 1198911 (119901) 119906119905 (119901 0) = 1198912 (119901) 119906 (0 119904) = 1198921 (119904) 119906119909 (0 119904) = 1198922 (119904)
(13)
By substituting (13) in (12) and simplifying we obtain
119906 (119901 119904) = 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]
(14)
Applying inverse double Laplace transform to (14) we obtainthe solution of (9) in the form
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 119886119904120573 minus 119887119904120574 minus 119888) [119901120572minus11198921 (119904)+ 119901120572minus21198922 (119904) minus 119886119904120573minus11198911 (119901) minus 119886119904120573minus21198912 (119901)minus 119887119904120574minus11198911 (119901) + ℎ (119901 119904)]]
(15)
Here we assume that the inverse double Laplace transform ofeach term in the right side of (15) exists
4 Illustrative Examples
In this section we demonstrate the applicability of theprevious method by giving examples
Example 1 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = 0 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) 1 lt 120572 le 2 119909 119905 ge 0
(16)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 119864120572 (119909120572) + 1199091198641205722 (119909120572) 119906119905 (119909 0) = 1198912 (119909) = minus [119864120572 (119909120572) + 1199091198641205722 (119909120572)] (17)
119906 (0 119905) = 1198921 (119905) = 119890minus119905119906119909 (0 119905) = 1198922 (119905) = 119890minus119905 (18)
a homogeneous space-fractional telegraph equationTaking single Laplace transform to initial (17) and bound-
ary conditions (18) we get
1198911 (119901) = ( 1119901 + 11199012) 119901120572119901120572 minus 1 1198912 (119901) = minus( 1119901 + 11199012) 119901120572119901120572 minus 1 1198921 (119904) = 1198922 (119904) = 1119904 + 1
(19)
Substituting above in (15) we get solution of (16)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 1119904 + 1+ 119901120572minus2 1119904 + 1 minus 119904 ( 1119901 + 11199012) 119901120572119901120572 minus 1+ ( 1119901 + 11199012) 119901120572119901120572 minus 1 minus ( 1119901 + 11199012) 119901120572119901120572 minus 1]]
(20)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119904 + 1) ( 1119901 + 11199012) 119901120572119901120572 minus 1] 119906 (119909 119905) = 119890minus119905 [119864120572 (119909120572) + 1199091198641205722 (119909120572)]
(21)
which agrees with the solution already obtained in [18]If we take 120572 = 2 then we get the exact solution of standard
telegraph equation
119906 (119909 119905) = 119890119909minus119905 (22)
Example 2 By substituting 119886 = 1 119887 = 1 119888 = 1 120573 = 2 120574 = 1and ℎ(119909 119905) = minus1199092 minus 119905 + 1 in (9)
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 119906 (119909 119905) minus 1199092 minus 119905+ 1 1 lt 120572 le 2 119909 119905 ge 0
(23)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909)= 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572)
119906119905 (119909 0) = 1198912 (119909) = 1(24)
119906 (0 119905) = 1198921 (119905) = 119905119906119909 (0 119905) = 1198922 (119905) = 0 (25)
a space-fractional nonhomogeneous telegraph equation
4 International Journal of Mathematics and Mathematical Sciences
Taking single Laplace transform to initial (24) and bound-ary conditions (25) we get
1198911 (119901) = 21199013 minus 2119901 minus 2 119901120572minus3
119901120572 minus 1 + 2 119901120572minus1
119901120572 minus 1= 2 (1199012 minus 1)1199013 (119901120572 minus 1)
1198912 (119901) = 1119901 1198921 (119904) = 11199042 1198922 (119904) = 0
(26)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus21199013119904 minus 11199011199042 + 1119901119904 (27)
Substituting above in (15) we get solution of (23)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 11199042minus 119904 2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 1119901 minus
2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 21199013119904 minus 11199011199042
+ 1119901119904]]
(28)
Rearranging we have
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [[ 119901120572
1199011199042 minus 1119901minus 11199011199042 minus 1119901119904] + [
minus2119904 (1199012 minus 1)1199013 (119901120572 minus 1) minus
2 (1199012 minus 1)1199013 (119901120572 minus 1)
minus 21199013119904 + 2119901119904]]]
(29)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 11199011199042 +2 (1199012 minus 1)1199013119904 (119901120572 minus 1)]
119906 (119909 119905) = 119905 + 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572) (30)
which agrees with the solution already obtained in [18] for120574 = 1For 120572 = 2 then 119906(119909 119905) = 119905 + 1199092
Example 3 By substituting 119886 = 1 119887 = 1 119888 = 0 120572 = 2 120574 = 1and ℎ(119909 119905) = minus2119905(1199092 minus 119909)(1199051minus120573Γ(3 minus 120573) + 1) + 21199052 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597119906 (119909 119905)120597119905= 1205972119906 (119909 119905)1205971199092 + 2119905 (1199092 minus 119909)( 1199051minus120573Γ (3 minus 120573) + 1) minus 21199052
1 lt 120573 le 2 119909 119905 ge 0(31)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (32)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = minus1199052 (33)
a time fractional telegraph equation in [24]Taking single Laplace transform to initial (32) and bound-
ary conditions (33) we get
1198911 (119901) = 1198912 (119901) = 1198921 (119904) = 01198922 (119904) = minus21199043
(34)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 (35)
Substituting above in (15) we get solution of (31)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904) [minus21199043minus 2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 ]]
(36)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [( 21199013 minus 11199012) 21199043 ] 119906 (119909 119905) = (1199092 minus 119909) 1199052
(37)
Example 4 By substituting 119886 = 1 119887 = 1 119888 = 1 120572 = 2 andℎ(119909 119905) = 0 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120574119906 (119909 119905)120597119905120574 + 119906 (119909 119905) = 1205972119906 (119909 119905)1205971199092 1 lt 120573 le 2 12 lt 120574 le 1 119909 119905 ge 0
(38)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 119890119909 (39)
119906 (0 119905) = 1198921 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) 119906119909 (0 119905) = 1198922 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) (40)
a homogeneous time fractional telegraph equation in [25]
International Journal of Mathematics and Mathematical Sciences 5
Taking single Laplace transform to initial (39) and bound-ary conditions (40) we get
1198911 (119901) = 01198912 (119901) = 1119901 minus 1 1198921 (119904) = 1198922 (119904) = 119904120573minus120574minus2119904120573minus120574 + 1
(41)
Substituting above in (15) we get solution of (38)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120574 minus 1) [119901 119904120573minus120574minus2(119904120573minus120574 + 1)+ 119904120573minus120574minus2(119904120573minus120574 + 1) minus 119904120573minus2 1(119901 minus 1)]]
(42)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901 minus 1) 119904120573minus120574minus2(119904120573minus120574 + 1)] 119906 (119909 119905) = 119890119909119905119864120573minus1205742 (minus119905120573minus120574)
(43)
which agrees with the solution already obtained in [25]
Example 5 By substituting 119886 = 119887 = 119888 = 1120572 = 2 and ℎ(119909 119905) =minus sinh119909(119905119899Γ(119899 + 1)) in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120573minus1119906 (119909 119905)120597119905120573minus1 + 119906 (119909 119905)= 1205972119906 (119909 119905)1205971199092 + sinh119909 119905119899Γ (119899 + 1)
1 lt 120573 le 2 119909 119905 ge 0(44)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (45)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = 119905119899+1205731198641119899+120573+1 (minus119905) (46)
a nonhomogeneous time fractional telegraph equation in[25]
Taking single Laplace transform to initial (45) and bound-ary conditions (46) we get
1198911 (119901) = 1198912 (119901) = 01198921 (119904) = 01198922 (119904) = 119904minus119899minus120573119904 + 1
(47)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus1(1199012 minus 1) 119904119899+1 (48)
Substituting above in (15) we get solution of (44)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120573minus1 minus 1) [ 119904minus119899minus120573
(119904 + 1)minus 1(1199012 minus 1) 119904119899+1]]
(49)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 1) 119904minus119899minus120573
(119904 + 1)] 119906 (119909 119905) = (sinh119909) 119905119899+1205731198641119899+120573+1 (minus119905)
(50)
which agrees with the solution already obtained in [25]
Example 6 Consider the following space-fractional-ordernonlinear telegraph equation
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 1199062 (119909 119905)minus 119890minus2119905 (119909 minus 1199092)2 minus 2119890minus119905 1199092minus120572Γ (3 minus 120572)
1 lt 120572 le 2 119909 119905 ge 0(51)
under the initial conditions
119906 (119909 0) = minus119906119905 (119909 0) = 119909 minus 1199092 (52)
and boundary conditions
119906 (0 119905) = 0119906119909 (0 119905) = 119890minus119905 (53)
Applying the double Laplace transform on both sides of (51)we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 1199042119906 (119901 119904) minus 119904119906 (119901 0) minus 119906119905 (119901 0) + 119904119906 (119901 119904)minus 119906 (119901 0) minus 2(119904 + 1) 1199013minus120572+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(54)
Further applying single Laplace transform to initial (52) andboundary conditions (53) we get
119906 (119901 0) = minus119906119905 (119901 0) = 11199012 minus 21199013 119906 (0 119904) = 0119906119909 (0 119904) = 1119904 + 1
(55)
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Mathematics and Mathematical Sciences
Taking single Laplace transform to initial (24) and bound-ary conditions (25) we get
1198911 (119901) = 21199013 minus 2119901 minus 2 119901120572minus3
119901120572 minus 1 + 2 119901120572minus1
119901120572 minus 1= 2 (1199012 minus 1)1199013 (119901120572 minus 1)
1198912 (119901) = 1119901 1198921 (119904) = 11199042 1198922 (119904) = 0
(26)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus21199013119904 minus 11199011199042 + 1119901119904 (27)
Substituting above in (15) we get solution of (23)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [119901120572minus1 11199042minus 119904 2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 1119901 minus
2 (1199012 minus 1)1199013 (119901120572 minus 1) minus 21199013119904 minus 11199011199042
+ 1119901119904]]
(28)
Rearranging we have
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904 minus 1) [[ 119901120572
1199011199042 minus 1119901minus 11199011199042 minus 1119901119904] + [
minus2119904 (1199012 minus 1)1199013 (119901120572 minus 1) minus
2 (1199012 minus 1)1199013 (119901120572 minus 1)
minus 21199013119904 + 2119901119904]]]
(29)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 11199011199042 +2 (1199012 minus 1)1199013119904 (119901120572 minus 1)]
119906 (119909 119905) = 119905 + 1199092 minus 2 minus 211990921198641205723 (119909120572) + 21198641205721 (119909120572) (30)
which agrees with the solution already obtained in [18] for120574 = 1For 120572 = 2 then 119906(119909 119905) = 119905 + 1199092
Example 3 By substituting 119886 = 1 119887 = 1 119888 = 0 120572 = 2 120574 = 1and ℎ(119909 119905) = minus2119905(1199092 minus 119909)(1199051minus120573Γ(3 minus 120573) + 1) + 21199052 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597119906 (119909 119905)120597119905= 1205972119906 (119909 119905)1205971199092 + 2119905 (1199092 minus 119909)( 1199051minus120573Γ (3 minus 120573) + 1) minus 21199052
1 lt 120573 le 2 119909 119905 ge 0(31)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (32)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = minus1199052 (33)
a time fractional telegraph equation in [24]Taking single Laplace transform to initial (32) and bound-
ary conditions (33) we get
1198911 (119901) = 1198912 (119901) = 1198921 (119904) = 01198922 (119904) = minus21199043
(34)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 (35)
Substituting above in (15) we get solution of (31)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904) [minus21199043minus 2( 21199013 minus 11199012)( 11199043minus120573 + 11199042 ) + 2 21199011199043 ]]
(36)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [( 21199013 minus 11199012) 21199043 ] 119906 (119909 119905) = (1199092 minus 119909) 1199052
(37)
Example 4 By substituting 119886 = 1 119887 = 1 119888 = 1 120572 = 2 andℎ(119909 119905) = 0 in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120574119906 (119909 119905)120597119905120574 + 119906 (119909 119905) = 1205972119906 (119909 119905)1205971199092 1 lt 120573 le 2 12 lt 120574 le 1 119909 119905 ge 0
(38)
subject to the initial and boundary conditions119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 119890119909 (39)
119906 (0 119905) = 1198921 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) 119906119909 (0 119905) = 1198922 (119905) = 119905119864120573minus1205742 (minus119905120573minus120574) (40)
a homogeneous time fractional telegraph equation in [25]
International Journal of Mathematics and Mathematical Sciences 5
Taking single Laplace transform to initial (39) and bound-ary conditions (40) we get
1198911 (119901) = 01198912 (119901) = 1119901 minus 1 1198921 (119904) = 1198922 (119904) = 119904120573minus120574minus2119904120573minus120574 + 1
(41)
Substituting above in (15) we get solution of (38)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120574 minus 1) [119901 119904120573minus120574minus2(119904120573minus120574 + 1)+ 119904120573minus120574minus2(119904120573minus120574 + 1) minus 119904120573minus2 1(119901 minus 1)]]
(42)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901 minus 1) 119904120573minus120574minus2(119904120573minus120574 + 1)] 119906 (119909 119905) = 119890119909119905119864120573minus1205742 (minus119905120573minus120574)
(43)
which agrees with the solution already obtained in [25]
Example 5 By substituting 119886 = 119887 = 119888 = 1120572 = 2 and ℎ(119909 119905) =minus sinh119909(119905119899Γ(119899 + 1)) in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120573minus1119906 (119909 119905)120597119905120573minus1 + 119906 (119909 119905)= 1205972119906 (119909 119905)1205971199092 + sinh119909 119905119899Γ (119899 + 1)
1 lt 120573 le 2 119909 119905 ge 0(44)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (45)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = 119905119899+1205731198641119899+120573+1 (minus119905) (46)
a nonhomogeneous time fractional telegraph equation in[25]
Taking single Laplace transform to initial (45) and bound-ary conditions (46) we get
1198911 (119901) = 1198912 (119901) = 01198921 (119904) = 01198922 (119904) = 119904minus119899minus120573119904 + 1
(47)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus1(1199012 minus 1) 119904119899+1 (48)
Substituting above in (15) we get solution of (44)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120573minus1 minus 1) [ 119904minus119899minus120573
(119904 + 1)minus 1(1199012 minus 1) 119904119899+1]]
(49)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 1) 119904minus119899minus120573
(119904 + 1)] 119906 (119909 119905) = (sinh119909) 119905119899+1205731198641119899+120573+1 (minus119905)
(50)
which agrees with the solution already obtained in [25]
Example 6 Consider the following space-fractional-ordernonlinear telegraph equation
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 1199062 (119909 119905)minus 119890minus2119905 (119909 minus 1199092)2 minus 2119890minus119905 1199092minus120572Γ (3 minus 120572)
1 lt 120572 le 2 119909 119905 ge 0(51)
under the initial conditions
119906 (119909 0) = minus119906119905 (119909 0) = 119909 minus 1199092 (52)
and boundary conditions
119906 (0 119905) = 0119906119909 (0 119905) = 119890minus119905 (53)
Applying the double Laplace transform on both sides of (51)we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 1199042119906 (119901 119904) minus 119904119906 (119901 0) minus 119906119905 (119901 0) + 119904119906 (119901 119904)minus 119906 (119901 0) minus 2(119904 + 1) 1199013minus120572+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(54)
Further applying single Laplace transform to initial (52) andboundary conditions (53) we get
119906 (119901 0) = minus119906119905 (119901 0) = 11199012 minus 21199013 119906 (0 119904) = 0119906119909 (0 119904) = 1119904 + 1
(55)
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Mathematics and Mathematical Sciences 5
Taking single Laplace transform to initial (39) and bound-ary conditions (40) we get
1198911 (119901) = 01198912 (119901) = 1119901 minus 1 1198921 (119904) = 1198922 (119904) = 119904120573minus120574minus2119904120573minus120574 + 1
(41)
Substituting above in (15) we get solution of (38)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120574 minus 1) [119901 119904120573minus120574minus2(119904120573minus120574 + 1)+ 119904120573minus120574minus2(119904120573minus120574 + 1) minus 119904120573minus2 1(119901 minus 1)]]
(42)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901 minus 1) 119904120573minus120574minus2(119904120573minus120574 + 1)] 119906 (119909 119905) = 119890119909119905119864120573minus1205742 (minus119905120573minus120574)
(43)
which agrees with the solution already obtained in [25]
Example 5 By substituting 119886 = 119887 = 119888 = 1120572 = 2 and ℎ(119909 119905) =minus sinh119909(119905119899Γ(119899 + 1)) in (9)
120597120573119906 (119909 119905)120597119905120573 + 120597120573minus1119906 (119909 119905)120597119905120573minus1 + 119906 (119909 119905)= 1205972119906 (119909 119905)1205971199092 + sinh119909 119905119899Γ (119899 + 1)
1 lt 120573 le 2 119909 119905 ge 0(44)
subject to the initial and boundary conditions
119906 (119909 0) = 1198911 (119909) = 0119906119905 (119909 0) = 1198912 (119909) = 0 (45)
119906 (0 119905) = 1198921 (119905) = 0119906119909 (0 119905) = 1198922 (119905) = 119905119899+1205731198641119899+120573+1 (minus119905) (46)
a nonhomogeneous time fractional telegraph equation in[25]
Taking single Laplace transform to initial (45) and bound-ary conditions (46) we get
1198911 (119901) = 1198912 (119901) = 01198921 (119904) = 01198922 (119904) = 119904minus119899minus120573119904 + 1
(47)
Taking double Laplace transform of ℎ(119909 119905) we haveℎ (119901 119904) = minus1(1199012 minus 1) 119904119899+1 (48)
Substituting above in (15) we get solution of (44)
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 119904120573 minus 119904120573minus1 minus 1) [ 119904minus119899minus120573
(119904 + 1)minus 1(1199012 minus 1) 119904119899+1]]
(49)
Simplifying we obtain
119906 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(1199012 minus 1) 119904minus119899minus120573
(119904 + 1)] 119906 (119909 119905) = (sinh119909) 119905119899+1205731198641119899+120573+1 (minus119905)
(50)
which agrees with the solution already obtained in [25]
Example 6 Consider the following space-fractional-ordernonlinear telegraph equation
120597120572119906 (119909 119905)120597119909120572 = 1205972119906 (119909 119905)1205971199052 + 120597119906 (119909 119905)120597119905 + 1199062 (119909 119905)minus 119890minus2119905 (119909 minus 1199092)2 minus 2119890minus119905 1199092minus120572Γ (3 minus 120572)
1 lt 120572 le 2 119909 119905 ge 0(51)
under the initial conditions
119906 (119909 0) = minus119906119905 (119909 0) = 119909 minus 1199092 (52)
and boundary conditions
119906 (0 119905) = 0119906119909 (0 119905) = 119890minus119905 (53)
Applying the double Laplace transform on both sides of (51)we get
119901120572119906 (119901 119904) minus 119901120572minus1119906 (0 119904) minus 119901120572minus2119906119909 (0 119904)= 1199042119906 (119901 119904) minus 119904119906 (119901 0) minus 119906119905 (119901 0) + 119904119906 (119901 119904)minus 119906 (119901 0) minus 2(119904 + 1) 1199013minus120572+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(54)
Further applying single Laplace transform to initial (52) andboundary conditions (53) we get
119906 (119901 0) = minus119906119905 (119901 0) = 11199012 minus 21199013 119906 (0 119904) = 0119906119909 (0 119904) = 1119904 + 1
(55)
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 International Journal of Mathematics and Mathematical Sciences
By substituting (55) in (54) and simplifying we obtain
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= 1199011205721199012 (119904 + 1) minus 119904 [ 11199012 minus 21199013 ] + [ 11199012 minus 21199013 ]minus [ 11199012 minus 21199013 ] minus 2119901120572(119904 + 1) 1199013+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
(56)
(119901120572 minus 1199042 minus 119904) 119906 (119901 119904)= [ 1199011205721199012 (119904 + 1) minus 1199041199012 ] minus [ 2119901120572(119904 + 1) 1199013 minus 21199041199013 ]+ 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]
119906 (119901 119904)= [ 11199012 minus 21199013 ] 1(119904 + 1)+ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)
2]
(57)
Applying inverse double Laplace transform of (57) we get
119906 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [1199062 (119909 119905) minus 119890minus2119905 (119909 minus 1199092)2]]
(58)
Now we apply the Iterative method as in [26]
119906 (119909 119905) = infinsum119894=0
119906119894 (119909 119905) (59)
Substituting (59) in (58) we get
infinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[[infinsum119894=0
119906119894 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]
]]]
(60)
The nonlinear term119873 is decomposed as
[infinsum119894=0
119906119894 (119909 119905)]2
= [1199060 (119909 119905)]2
+ infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]
(61)
Substituting (61) in (60) we getinfinsum119894=0
119906119894 (119909 119905) = (119909 minus 1199092) 119890minus119905 + 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905)]2 minus 119890minus2119905 (119909 minus 1199092)2]]
+ 119871minus1119909 119871minus1119905 [[1(119901120572 minus 1199042 minus 119904)
sdot 119871119909119871 119905 [[infinsum119894=1
[[[ 119894sum119896=0
119906119896 (119909 119905)]2
minus [ 119894minus1sum119896=0
119906119896 (119909 119905)]2]]]]]]
(62)
Then we define the recurrence relations as
1199060 (119909 119905) = (119909 minus 1199092) 119890minus1199051199061 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)119871119909119871 119905 [[1199060 (119909 119905)]2
minus 119890minus2119905 (119909 minus 1199092)2]] = 01199062 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905)]2 minus [1199060 (119909 119905)]2]] = 0
1199063 (119909 119905) = 119871minus1119909 119871minus1119905 [ 1(119901120572 minus 1199042 minus 119904)sdot 119871119909119871 119905 [[1199060 (119909 119905) + 1199061 (119909 119905) + 1199062 (119909 119905)]2minus [1199060 (119909 119905) + 1199061 (119909 119905)]2]] = 0
(63)
and so onTherefore we obtain the solution of (51) as follows
119906 (119909 119905) = 119890119909minus2119905 (64)
This is the required exact solution of (51)
5 Conclusion
We have applied double Laplace transform to obtain exactsolutions of linearnonlinear space-time fractional telegraphequations All of the examples considered show that doubleLaplace transform method is capable of reducing the volumeof computational work as compared to other methods Itmay be concluded that DLT technique solves the problemswithout using Adomian polynomials Lagrange multipliervalue Hersquos polynomials and small parameters
Competing Interests
The authors declare no competing interests regarding thepublication of this paper
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Mathematics and Mathematical Sciences 7
References
[1] E Orsingher and Z Xuelei ldquoThe space-fractional telegraphequation and the related fractional telegraph processrdquo ChineseAnnals Mathematics B vol 24 no 1 pp 45ndash56 2003
[2] E Orsingher and L Beghin ldquoTime-fractional telegraph equa-tions and telegraph processes with Brownian timerdquo ProbabilityTheory and Related Fields vol 128 no 1 pp 141ndash160 2004
[3] S Momani ldquoAnalytic and approximate solutions of the space-and time-fractional telegraph equationsrdquo Applied Mathematicsand Computation vol 170 no 2 pp 1126ndash1134 2005
[4] M Garg and A Sharma ldquoSolution of space-time fractional tele-graph equation by Adomian decompositionmethodrdquo Journal ofInequalities and Special Functions vol 2 no 1 pp 1ndash7 2011
[5] G Adomian Solving Frontier Problems of PhysicsTheDecompo-sitionMethod vol 60 Kluwer Academic Publishers DordrechtThe Netherlands 1994
[6] J Chen F Liu and V Anh ldquoAnalytical solution for the time-fractional telegraph equation by the method of separatingvariablesrdquo Journal of Mathematical Analysis and Applicationsvol 338 no 2 pp 1364ndash1377 2008
[7] F Huang ldquoAnalytical solution for the time-fractional telegraphequationrdquo Journal of Applied Mathematics vol 2009 Article ID890158 9 pages 2009
[8] J-H He ldquoVariational iteration methodmdasha kind of non-linearanalytical technique some examplesrdquo International Journal ofNon-Linear Mechanics vol 34 no 4 pp 699ndash708 1999
[9] A Sevimlican ldquoAn approximation to solution of space andtime fractional telegraph equations by Hersquos variational iterationmethodrdquo Mathematical Problems in Engineering vol 2010Article ID 290631 10 pages 2010
[10] A Yildirim ldquoHersquos homotopy perturbation method for solvingthe space- and time-fractional telegraph equationsrdquo Interna-tional Journal of Computer Mathematics vol 87 no 13 pp2998ndash3006 2010
[11] S Das K Vishal P K Gupta and A Yildirim ldquoAn approxi-mate analytical solution of time-fractional telegraph equationrdquoApplied Mathematics and Computation vol 217 no 18 pp7405ndash7411 2011
[12] M Garg P Manohar and S L Kalla ldquoGeneralized differentialtransformmethod to space-time fractional telegraph equationrdquoInternational Journal of Differential Equations vol 2011 ArticleID 548982 9 pages 2011
[13] L Galue ldquoSolution of some fractional order telegraph equa-tionsrdquo Revista Colombiana de Mathematicas vol 48 no 2 pp247ndash267 2014
[14] VK SrivastavaMKAwasthi andMTamsir ldquoRDTMsolutionof Caputo time fractional-order hyperbolic telegraph equationrdquoAIP Advances vol 3 no 3 2013
[15] Y Khan J Diblik N Faraz and Z Smarda ldquoAn efficientnew perturbative Laplace method for space-time fractionaltelegraph equationsrdquo Advances in Difference Equations vol2012 article 204 2012
[16] D Kumar J Singh and S Kumar ldquoAnalytic and approxi-mate solutions of space-time fractional telegraph equations vialaplace transformrdquoWalailak Journal of Science and Technologyvol 11 no 8 pp 711ndash728 2014
[17] A Prakash ldquoAnalytical method for space-fractional telegraphequation by homotopy perturbation transform methodrdquo Non-linear Engineering vol 5 no 2 pp 123ndash128 2016
[18] F A Alawad E A Yousif and A I Arbab ldquoA new techniqueof Laplace variational iteration method for solving space-time fractional telegraph equationsrdquo International Journal ofDifferential Equations vol 2013 Article ID 256593 10 pages2013
[19] S Kumar ldquoA new analytical modelling for fractional telegraphequation via Laplace transformrdquo Applied Mathematical Mod-elling vol 38 no 13 pp 3154ndash3163 2014
[20] I N SneddonTheUse of Integral Transforms TataMcgrawHill1974
[21] L Debnath and D Bhatta Integral Transforms and TheirApplications CRC Press Taylor amp Francis Group Boca RatonFla USA 3rd edition 2015
[22] L Debnath ldquoThe double Laplace transforms and their prop-erties with applications to functional integral and partialdifferential equationsrdquo International Journal of Applied andComputational Mathematics vol 2 no 2 pp 223ndash241 2016
[23] A M Anwar F Jarad D Baleanu and F Ayaz ldquoFractionalCaputo heat equation within the double Laplace transformrdquoRomanian Journal of Physics vol 58 no 1-2 pp 15ndash22 2013
[24] S Sarwar and M M Rashidi ldquoApproximate solution of two-term fractional-order diffusion wave-diffusion and telegraphmodels arising in mathematical physics using optimal homo-topy asymptotic methodrdquo Waves in Random and ComplexMedia vol 26 no 3 pp 365ndash382 2016
[25] R Joice Nirmala and K Balachandran ldquoAnalysis of solutionsof time fractional telegraph equationrdquo Journal of the KoreanSociety for Industrial and AppliedMathematics vol 18 no 3 pp209ndash224 2014
[26] R R Dhunde and G L Waghmare ldquoDouble Laplace trans-form combined with Iterative method for solving non-lineartelegraph equationrdquo Journal of the Indian Mathematical Societyvol 83 no 3-4 pp 221ndash230 2016
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of