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- 1. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 Solution of Linear and Nonlinear Partial Differential Equations Using Mixture of Elzaki Transform and the Projected Differential Transform Method Tarig. M. Elzaki1* & Eman M. A. Hilal2 1. Mathematics Department, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University, Jeddah-Saudi Arabia. Mathematics Department, Faculty of Sciences, Sudan University of Sciences and Technology-Sudan. 2. Mathematics Department, Faculty of Sciences for Girles King Abdulaziz University Jeddah-Saudi Arabia * E-mail of the corresponding author: Tarig.alzaki@gmail.comAbstractThe aim of this study is to solve some linear and nonlinear partial differential equations using the newintegral transform "Elzaki transform" and projected differential transform method. The nonlinear terms canbe handled by using of projected differential transform method; this method is more efficient and easy tohandle such partial differential equations in comparison to other methods. The results show the efficiencyand validation of this method.Keywords: Elzaki transform, projected differential transform method, nonlinear partial differentialequations.1. IntroductionMany problems of physical interest are described by linear and nonlinear partial differential equations withinitial or boundary conditions, these problems are fundamental importance in science and technologyespecially in engineering. Some valuable contributions have already been made to showing differentialequations such as Laplace transform method [lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010),Lokenath Debnath and Bhatta (2006)], Sumudu transform [A.Kilicman and H.E.Gadain (2009), (2010)],differential transform method etc. Elzaki transform method is very effective tool for solve linear partialdifferential equations [Tarig Elzaki & Salih Elzaki (2011)]. In this study we use the projected differentialtransform method [Nuran Guzel and Muhammet Nurulay (2008), Shin- Hsiang Chang , Ling Chang (2008)] to decompose the nonlinear terms, this means that we can use both Elzaki transform and projecteddifferential transform methods to solve many nonlinear partial differential equations.1.1. Elzaki Transform:The basic definitions of modified of Sumudu transform or Elzaki transform is defined as follows, Elzakitransform of the function f (t ) is 50
- 2. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 ∞ t − E [ f (t ) ] = v ∫ f (t ) e v dt , t >0 0Tarig M. Elzaki and Sailh M. Elzaki in [2011], showed the modified of Sumudu transform [Kilicman &ELtayeb (2010)] or Elzaki transform was applied to partial differential equations, ordinary differentialequations, system of ordinary and partial differential equations and integral equations.In this paper, we combined Elzaki transform and projected differential transform methods to solve linearand nonlinear partial differential equations.To obtain Elzaki transform of partial derivative we use integration by parts, and then we have: ∂f (x , t ) 1 ∂ 2 f (x , t ) 1 ∂f (x , 0)E = T (x ,v ) −vf (x , 0) E = v 2 T (x ,v ) − f (x , 0) −v ∂t v ∂t ∂t 2 ∂f (x , t ) d ∂ 2f (x , t ) d 2 E = [T (x ,v )] E = 2 [T (x ,v )] ∂x dx ∂x 2 dxProof: To obtain ELzaki transform of partial derivatives we use integration by parts as follows: ∂f ∞ ∂f −t p −t ∂f −t p p −t Ε ( x ,t ) = ∫ v e dt = lim ∫ve v v dt = lim ve f (x , t ) − ∫ e v f (x , t )dt v ∂t 0 ∂t p →∞ 0 ∂t p →∞ 0 0 T ( x ,v ) = −vf ( x , 0 ) vWe assume that f is piecewise continuous and it is of exponential order. ∂f ∞ −t ∂f ( x , t ) ∂ ∞ −t = ∫ ve v ve v f ( x , t ) dt , using the Leibnitz rule to find: ∂x ∫Now Ε dt = ∂x 0 ∂x 0 ∂f d Ε = T ( x ,v ) ∂x dx ∂ 2f d 2 By the same method we find: Ε 2 = 2 T ( x ,v ) ∂x dx ∂2 f To find: Ε 2 ( x, t ) ∂t ∂fLet = g , then we have: ∂t ∂ 2f ∂g ( x , t ) Ε 2 ( x , t ) = Ε g ( x , t ) −vg x , 0 =Ε ( ) ∂t ∂t v 51
- 3. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 ∂ 2f 1 ∂f Ε 2 ( x , t ) = 2 T ( x , v ) − f ( x , 0 ) −v ( x , 0) ∂t v ∂tWe can easily extend this result to the nth partial derivative by using mathematical induction.1.2. Projected Differential Transform Methods:In this section we introduce the projected differential transform method which is modified of thedifferential transform method.Definition:The basic definition of projected differential transform method of function f (x 1 , x 2 ,LL , x n ) isdefined as 1 ∂ k f (x 1 , x 2 ,LL , x n ) f (x 1 , x 2 ,LL , x n −1 , k ) = (1) k ! ∂x n k x n =0 Such that f (x 1 , x 2 ,LL , x n ) is the original function and f (x 1 , x 2 ,LL , x n −1 , k ) is projectedtransform function.And the differential inverse transform of f (x 1 , x 2 ,LL , x n −1 , k ) is defined as ∞ f (x 1 , x 2 ,LL , x n ) = ∑ f (x 1 , x 2 ,LL , x n )(x − x 0 ) k (2) k =oThe fundamental theorems of the projected differential transform are 52
- 4. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012Theorems:(1) If z ( x1 , x2 ,......, xn ) = u ( x1 , x2 ,......, xn ) ± v ( x1 , x2 ,......, xn ) Then z ( x1 , x2 ,......, xn −1 , k ) = u ( x1 , x2 ,......, xn −1 , k ) ± v ( x1 , x2 ,......, xn −1 , k )( 2) Ifz ( x1 , x2 ,......, xn ) = c u ( x1 , x2 ,......, xn ) Then z ( x1 , x2 ,......, xn −1 , k ) = cu ( x1 , x2 ,......, xn −1 , k ) du ( x1 , x2 ,......, xn )( 3) Ifz ( x1 , x2 ,......, xn ) = dxn Then z ( x1 , x2 ,......, xn −1 , k ) = ( k + 1) u ( x1 , x2 ,....., xn −1 , k + 1) d n u ( x1 , x2 ,......, xn ) ( 4 ) Ifz ( x1 , x2 ,......, xn ) = n dxn Then z ( x1 , x2 ,......, xn−1 , k ) = ( k + n )! u ( x1 , x2 ,....., xn−1 , k + n ) k!( 5) If z ( x1 , x2 ,......, xn ) = u ( x1 , x2 ,......, xn ) v ( x1 , x2 ,......, xn ) k Then z ( x1 , x2 ,......, xn −1 , k ) = ∑ u ( x1 , x2 ,......, xn −1 , m ) v ( x1 , x2 ,......, xn −1 , k − m ) m =0( 6 ) If z ( x1 , x2 ,......, xn ) = u1 ( x1 , x2 ,......, xn ) u2 ( x1 , x2 ,......, xn ) ..... un ( x1 , x2 ,......, xn ) Then k k n−1 k3 k2z ( x1 , x2 ,......, xn −1 , k ) = ∑ ∑ ...... ∑ ∑ u ( x , x ,......, x 1 1 2 n −1 , k1 ) u2 ( x1 , x2 ,......, xn −1 , k2 − k1 ) kn −1 = 0 k n− 2 = 0 k 2 = 0 k1 = 0×.....un −1 ( x1 , x2 ,......, xn −1 , kn −1 − kn −2 ) un ( x1 , x2 ,......, xn −1 , k − kn −1 ) (7) If z ( x1 , x2 ,......, xn ) = x1q1 x2 2 ........xn n q q 1 k = qn Then z ( x1 , x2 ,......, xn −1 , k ) = δ ( x1 , x2 ,......, xn−1 , qn − k ) = 0 k ≠ qnNote that c is a constant and n is a nonnegative integer.2. Applications:Consider a general nonlinear non-homogenous partial differential equation with initial conditions of theform: Du (x , t ) + Ru (x , t ) + Nu (x , t ) = g (x , t ) (3) u (x , 0) = h (x ) , u t (x , 0) = f (x ) 53
- 5. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012Where D is linear differential operator of order two, R is linear differential operator of lessorder than D , N is the general nonlinear differential operator and g ( x , t ) is the source term.Taking Elzaki transform on both sides of equation (3), to get: E [Du (x , t )] + E [Ru (x , t )] + E [Nu (x , t )] = E [ g (x , t )] (4)Using the differentiation property of Elzaki transforms and above initial conditions, we have: E [u (x , t )] = v 2 E [ g (x , t )] + v 2 h (x ) + v 3f (x ) −v 2 E [Ru (x , t ) + Nu (x , t )] (5)Appling the inverse Elzaki transform on both sides of equation (5), to find: u (x , t ) = G (x , t ) − E −1 {v 2 E [ Ru (x , t ) + Nu (x , t )]} (6)Where G ( x , t ) represents the term arising from the source term and the prescribed initial conditions.Now, we apply the projected differential transform method. u (x , m + 1) = −E −1 {v 2 E [ A m + B m ]} , u (x , 0) = G (x , t ) (7)Where A m , B m are the projected differential transform of Ru (x , t ) , Nu (x , t ) .From equation (7), we have: u (x , 0) = G (x , t ) , u (x ,1) = − E −1 {v 2 E [ A 0 + B 0 ]} u (x , 2) = −E −1 {v 2 E [ A1 + B 1 ]} , u (x ,3) = − E −1 {v 2 E [ A 2 + B 2 ]} .....Then the solution of equation (3) is u (x , t ) = u (x , 0) + u (x ,1) + u (x , 2) + ........To illustrate the capability and simplicity of the method, some examples for different linear and nonlinearpartial differential equations will be discussed.Example 2.1:Consider the simple first order partial differential equation ∂y ∂y =2 +y , y (x , 0) = 6e −3x (8) ∂x ∂tTaking Elzaki transform of (8), leads to v E [ y (x , t )] = 6v 2e −3 x + E [Am − B m ] 2Take the inverse Elzaki transform to find, v y (x , m + 1) = E −1 { E [ Am − B m ] } , y (x , 0) = 6e −3x (9) 2 54
- 6. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 ∂y (x , m )Where Am = , B m = y (x , m ) are the projected differential transform of ∂x∂y (x , t ) , y (x , t ) . ∂xThe standard Elzaki transform defines the solution y (x , t ) by the series ∞ y (x , t ) = ∑ y (x , m ) m =0From equation (9) we find that: y (x , 0) = 6e −3x A 0 = −18e −3x , B 0 = 6e −3x , y (x ,1) = E =1 −12v 3e −3x = −12te −3x A1 = 36te −3x , B 1 = −12e −3x , y (x , 2) = E =1 24v 4e −3x = 12t 2e −3x .................................................................. y (x ,3) = −8t 3e −3xThe solution in a series form is given by y (x , t ) = 6e −3x − 12te −3x + 12t 2e −3x + ........ = 6e −3x e −2t = 6e − (3x + 2t )Example 2.2:Consider the following linear second order partial differential equation u xx + u tt = 0 , u (x , 0) = 0, u t (x , 0) = cos x (10)To find the solution we take Elzaki transform of equation (10) and making use of the conditions to find, ∂ 2u (x , m ) E [u (x , t ) ] = v cos x − v E [ A m ] , 3 2 Am = ∂x 2Take the inverse Elzaki transform we get: u (x , m + 1) = −E −1 {v 2 E [ A m ]} , u (x , 0) = 0 , u (x ,1) = t cos x (11)By using equation (11), we find that: t3 A1 = −t cos x ⇒ u (x , 2) = E −1 {v 2 E [t cos x ]} = cos x 3! t3 t 3 t 5 A 2 = − cos x ⇒ u (x ,3) = E −1 v 2 E cos x = cos x 3! 3! 5! . . . 55
- 7. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 t3 t5 Then the solution is u (x , t ) = cos x t + + + ..... = cos x sinh t 3! 5! Example 2.3:Consider the following second order nonlinear partial differential equation ∂u ∂u ∂ 2u 2 = +u , u (x .0) = x 2 (12) ∂t ∂x ∂x 2To find the solution take Elzaki transform of (12) and using the condition we get: E [u (x , t ) ] = v 2 x 2 + vE [ A m + B m ] h ∂u (x , m ) ∂u (x , h − m ) h ∂ 2u (h − m )Where A m =∑ , B m = ∑ u (x , m ) , are projected m =0 ∂x ∂x m =0 ∂x 2 ∂u ∂ 2u 2differential transform of , u ∂x ∂x 2Take the inverse Elzaki transform to get: u (x , m + 1) = E −1 {vE [ A m + B m ]} , u (x , 0) = x 2 (13)From equation (13) we find that: A 0 = 4x 2 , B 0 = 2x 2 ⇒ u (x ,1) = E −1 6x 2v 3 = 6x 2t A1 = 48x 2t , B 1 = 24x 2t ⇒ u (x , 2) = E −1 72x 2v 4 = 36x 2t 2 . . .Then the solution of equation (12) is x2 u (x , t ) = x 2 + 6x 2t + 36x 2t 2 + ..... = x 2 (1 − 6t ) −1 = 1 − 6tWhich is an exactly the same solution obtained by the Adomian decomposition method.Example 2.4:Let us consider the nonlinear partial differential equation ∂u ∂u ∂ 2u x +1 2 = 2u +u 2 2 , u (x , 0) = (14) ∂t ∂x ∂x 2Taking Elzaki transform of equation (14) and using the condition leads to: 56
- 8. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012 x +1 E [u (x , t ) ] = v 2 + vE [ 2A m + B m ] 2 Taking the inverse Elzaki transform to find: x +1 u (x , m + 1) = E −1 {vE [ 2A m + B m ]} , u (x , 0) = (15) 2 h k ∂u (x , h − m ) ∂u (x , h − k ) A m = ∑ ∑ u (x , m ) k =0 m = 0 ∂x ∂xWhere h k ∂ 2u (x , h − k ) B m = ∑ ∑ u (x , m ) u (x , h − m ) k =0 m =0 ∂x 2From equation (15) we have: x +1 x +1 3(x + 1) 2 u (x , 0) = , u (x ,1) = t , u (x , 2) = t ,........... 2 2 8Then the solution of equation (14) is x +1 1 x +1 (1 − t ) 2 = − u (x , t ) = 2 2 1−t3. ConclusionThe solution of linear and nonlinear partial differential equations can be obtained using Elzaki transformand projected differential transform method without any discretization of the variables. The results for allexamples can be obtained in series form, and all calculations in the method are very easy. This technique isuseful to solve linear and nonlinear partial differential equations.References[1] S. lslam, Yasir Khan, Naeem Faraz and Francis Austin (2010), Numerical Solution of LogisticDifferential Equations by using the Laplace Decomposition Method, World Applied Sciences Journal 8(9):1100-1105.[2] Nuran Guzel and Muhammet Nurulay (2008), Solution of Shiff Systems By using DifferentialTransform Method, Dunlupinar universities Fen Bilimleri Enstitusu Dergisi, ISSN 1302-3055, PP. 49-59.[3] Shin- Hsiang Chang , I. Ling Chang (2008), A new algorithm for calculating one – dimensionaldifferential transform of nonlinear functions, Applied Mathematics and Computation 195, 799-808.[4] Tarig M. Elzaki (2011), The New Integral Transform “Elzaki Transform” Global Journal of Pure andApplied Mathematics, ISSN 0973-1768, Number 1, pp. 57-64.[5] Tarig M. Elzaki & Salih M. Elzaki (2011), Application of New Transform “Elzaki Transform” to PartialDifferential Equations, Global Journal of Pure and Applied Mathematics, ISSN 0973-1768,Number 1, pp.65-70.[6] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Connections Between Laplace and Elzaki transforms,Advances in Theoretical and Applied Mathematics, ISSN 0973-4554 Volume 6, Number 1, pp. 1-11.[7] Tarig M. Elzaki & Salih M. Elzaki (2011), On the Elzaki Transform and Ordinary Differential EquationWith Variable Coefficients, Advances in Theoretical and Applied Mathematics. ISSN 0973-4554 Volume 6, 57
- 9. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012Number 1, pp. 13-18.[8] Lokenath Debnath and D. Bhatta (2006). Integral transform and their Application second Edition,Chapman & Hall /CRC.[9] A.Kilicman and H.E.Gadain (2009). An application of double Laplace transform and Sumudu transform,Lobachevskii J. Math.30 (3), pp.214-223.[10] J. Zhang (2007), A Sumudu based algorithm m for solving differential equations, Comp. Sci. J.Moldova 15(3), pp – 303-313.[11] Hassan Eltayeb and Adem kilicman (2010), A Note on the Sumudu Transforms anddifferential Equations, Applied Mathematical Sciences, VOL, 4, no.22,1089-1098[12] Kilicman A. & H. ELtayeb (2010). A note on Integral transform and Partial Differential Equation,Applied Mathematical Sciences, 4(3), PP.109-118.[13] Hassan ELtayeh and Adem kilicman (2010), on Some Applications of a new Integral Transform, Int.Journal of Math. Analysis, Vol, 4, no.3, 123-132.[14] N.H. Sweilam, M.M. Khader (2009). Exact Solutions of some capled nonlinear partial differentialequations using the homotopy perturbation method. Computers and Mathematics with Applications 582134-2141.[15] P.R. Sharma and Giriraj Methi (2011). Applications of Homotopy Perturbation method to Partialdifferential equations. Asian Journal of Mathematics and Statistics 4 (3): 140-150.[16] M.A. Jafari, A. Aminataei (2010). Improved Homotopy Perturbation Method. InternationalMathematical Forum, 5, no, 32, 1567-1579.[17] Jagdev Singh, Devendra, Sushila (2011). Homotopy Perturbation Sumudu Transform Method forNonlinear Equations. Adv. Theor. Appl. Mech., Vol. 4, no. 4, 165-175. 58
- 10. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012Table 1. Elzaki transform of some Functions f (t ) E [ f (t )] = T (u ) 1 v2 t v3 tn n ! v n +2 e at v2 1 − av sinat av 3 1 + a 2v 2 cosat v2 1 + a 2v 2 59

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