EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 3, 2019, 771-789 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Resolution of nonlinear convection - diffusion - reaction equations of Cauchy’s kind by the Laplace SBA method Youssouf Paré1,∗, Youssouf Minoungou1, Abdoul Wassiha Nébié1 1 Département de Mathématiques, UFR SEA, Université Joseph Ki-Zerbo, Ouagadougou, Burkina Faso Abstract. In this paper, we proposed the solution of few non linear partial differential equations modelling diffusion, convection and reaction problems Cauchy kind. The Laplace SBA method based on combination of Laplace’s transform, Adomian Decomposition Method(ADM), Picard principle and successive approximations is used for solving these equations. 2010 Mathematics Subject Classifications: 47H14, 34G20, 47J25, 65J15 Key Words and Phrases: Cauchy kind, diffusion-convection-reaction problems, Laplace-SBA method 1. Introduction Many problems are governed by partial differential equations, or by systems of partial differential equations. It is generally difficult to find exact solutions of these problems. In this article, we recall the definition, some results and determinate some models con- vection - diffusion - reaction non-linear equations with Cauchy condition by the Laplace - SBA method. The Laplace -SBA [6, 10] method is a numerical method based on the combination of the Laplace transform, the successive approximation method, the adomian decomposition method [1, 2, 7] and the Picard fixed point principle, the Laplace - SBA method avoids the difficulties associated with calculating Adomian polynomials unlike the Laplace - Adomian algorithm[8, 9], this method allows a rapid convergence towards the solution, if it exists. 2. Preliminary and definitions In this section, we recall some definitions and theorems ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3430 Email addresses: youssoufpare@gmail.com (Y. Paré), m.youl@yahoo.fr (Y. Minoungou), nebwass@yahoo.fr (A. W. Nébié) http://www.ejpam.com 771 c© 2019 EJPAM All rights reserved. Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 772 Definition 1. Let u : R+ 7→ C be a continuous function. The function L(u) defined by L (u) = ∫ ∞ 0 u(t)e−stdt. This theorem is a result that proves the existence and uniqueness of the solution of the functional equation Au = f Theorem 1. (Minty-Browder)[5, 6] Let E be a reflexive Banach space. Let A : E → E′ an application (non-linear) continues such that : 〈Au−Av;u− v〉 > 0 ∀u, v ∈ E, u 6= v lim ‖v‖→∞ 〈Av; v〉 ‖v‖ =∞ So for everything f ∈ E′, there is u ∈ E a unique solution to the equation Au = f 3. The numerical LAPLACE - SBA method In this section, we describe how to use the Laplace - SBA algorithm to solve the equations. Let us consider the following functional equation : Au = f (1) Where A : H → H, is an operator not necessarly linear and H is a Hilbert space adequatly chosen given the operator A. Let : A = L+R+N (2) Where L is an inversible operator in the Adomian sense, R the linear remainder and N a nonlinear operator. The equation (1) therefore becomes : Lu+Ru+Nu = f (3) Applying the Laplace transform Lt, on both sides of equations (3) , we have : L [Lu] + L [Ru] + L [Nu] = L (f) (4) We suppose : L(.) = ∂m ∂tm (.) By using the differentiation property of Laplace transform, we then obtain Lt [ ∂mu ∂tm ] = smLt (u)− m−1∑ k=0 skum−k−1(0, x) (5) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 773 The equation (4) become smLt (u) = −Lt (Ru)− Lt (Nu) + m−1∑ j=0 sjum−j−1(0, x) + Lt (f) (6) Using the successive approximations, we have ⇒ smLt ( uk ) = −Lt ( Ruk ) − Lt ( Nuk−1 ) + m−1∑ j=0 sjum−j−1(0, x) + Lt (f) (7) Substituting uk(x, t) par ∑ n≥0 ukn(x, t) et gk−1 = Nuk−1, we obtain sm ∑ n≥0 Lt ( ukn ) = − ∑ n≥0 Lt ( Rukn ) − Lt (gk−1) + m−1∑ j=0 sjum−j−1(0, x) + Lt (f) (8) We deduce the following Laplace - SBA algorithm[9]:  Lt(uk0) = − 1 sm Lt(gk−1) + 1 sm m−1∑ j=0 sjum−j−1(0, x) + 1 sm Lt (f) , k ≥ 1 Lt(ukn+1) = − 1 sm Lt ( Rukn ) , n ≥ 0, k ≥ 1 (9) Applying the inverse transform L−1t Laplace, we have: uk0 = L−1t  1 sm m−1∑ j=0 pjum−j−1(0, x) + L−1t [ 1 sm Lt (f) ] − L−1t ( 1 sm Lt(gk−1) ) , k ≥ 1 ukn+1 = −L−1t [ 1 sm Lt ( Rukn )] , n ≥ 0, k ≥ 1 (10) The Picard principle is then applied to the equation (6) : let u0 be such that N ( u0 ) = 0, for k = 1 we get : u10 = L−1t  1 sm m−1∑ j=0 pjum−j−1(0, x) + L−1t [ 1 sm Lt (f) ] − L−1t ( 1 sm Lt(g0) ) u1n+1 = −L−1t [ 1 sm Lt ( Ru1n )] (11) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 774 If the series ( +∞∑ n=0 u1n ) converges, then u1 = +∞∑ n=0 u1n. For k = 2, we get :  u20 = L−1t  1 sm m−1∑ j=0 sjum−j−1(0, x) + L−1t [ 1 sm Lt (f) ] − L−1t ( 1 sm Lt(g1) ) u2n+1 = −L−1t [ 1 sm Lt ( Ru1n )] (12) If the series ( +∞∑ n=0 u2n ) converges, then u2 = +∞∑ n=0 u2n. This process is repeated to k. If the series ( +∞∑ n=0 ukn ) converges, then u2 = +∞∑ n=0 ukn, therefore u = lim k→+∞ uk is the solution of the equation (1) . At each stage k ≥ 1 , we check that : N ( uk ) = 0. 4. Application of Laplace SBA In this section, we propose the solution of some equations of convection - reaction - diffusion non-linear[4][3] by the Laplace - SBA method, we show that the choice of operator L is very decisive for the convergence speed of the algorithm. Example 1. : Let us consider the following nonlinear diffusion equation (E) :  ∂u ∂t = ∂2u ∂x2 − u2∂u ∂x + u3 u (0, x) = ex with u = u(t, x), (t, x) ∈ [0,+∞[× R. Applying the Laplace SBA method, we have uk0(x, t) = L−1t [ 1 s u (0, x) + 1 s Lt (gk−1) ] ukn+1(x, t) = L−1t [ 1 s Lt ( R(ukn) )] , n ≥ 0 Where Lu = ∂u ∂t , Ru = ∂2u ∂x2 and gk−1 = Nuk−1 = − ( uk−1 )2 ∂uk−1 ∂x + ( uk−1 )3 Determinate ukn(t, x), for n ≥ 0 We take u0 = 0 u10(t, x) = L−1t ( 1 s ex ) + L−1t ( 1 s Lt (g0) ) ⇒ u10(t, x) = ex u11(t, x) = L−1t [ 1 s Lt ( R(u10) )] Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 775 ⇒ u11(t, x) = L−1t ( 1 s2 ex ) = tex u12(t, x) = L−1t [ 1 s Lt ( R(u11) )] ⇒ u12(t, x) = L−1t ( 1 s3 ex ) = t2 2! ex u13(t, x) = L−1t [ 1 s Lt ( R(u12) )] ⇒ u13(t, x) = L−1t ( 1 s4 ex ) = t3 3! ex u14(t, x) = L−1t [ 1 s Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( 1 s5 ex ) = t4 4! ex u15(t, x) = L−1t [ 1 s Lt ( R(u14) )] ⇒ u15(t, x) = L−1t ( 1 s6 ex ) = t5 5! ex In recursive way, we deduce u1n(t, x) = 1 n! tnex Then u1(t, x) = ∑ n≥0 u1n(t, x) = ∑ n≥0 1 n! tnex ⇒ u1(t, x) = ex+t So Nu1 = − ( u1 )2 ∂ (u1) ∂x + ( u1 )3 Nu1 = − ( ex+t )3 + ( ex+t )3 = 0 Step by step, we obtain u1(t, x) = u2(t, x) = ... = uk(t, x) = ex+t The exact solution of model is u(t, x) = lim k→+∞ uk(t, x) = ex+t Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 776 Conclusion : The exact solution of model is u(t, x) = ex+t Example 2. Let us consider the following nonlinear diffusion equation (E) :  ∂u ∂t = ε ∂2u ∂x2 + λ ∂u ∂x + u3 + u2 ∂2u ∂x2 u(0, x) = sinx , (t, x) ∈ Ω = [0,+∞[× R (13) with u = u(t, x), (t, x) ∈ [0,+∞[× R, ε > 0 et λ > 0. Applying the Laplace - SBA method, we get uk0(t, x) = L−1t [ 1 s u(0, x) + 1 s Lt (gk−1) ] ukn+1(t, x) = L−1t [ 1 s Lt ( R(ukn) )] , n ≥ 0 (14) where Lu = ∂u ∂t , Ru = ε ∂2u ∂x2 + λ ∂u ∂x2 and gk−1 = Nuk−1 = ( uk−1 )3 + ( uk−1 )2 ∂2uk−1 ∂x2 . Let us calculate ukn(t, x), for n ≥ 0 : we take u0 = 0 u10(t, x) = L−1t ( 1 s sinx ) + L−1t ( 1 s Lt (g0) ) ⇒ u10(t, x) = sinx u11(t, x) = L−1t [ 1 s Lt ( R(u10) )] ⇒ u11(t, x) = L−1t ( λ cosx− ε sinx s2 ) ⇒ u11(t, x) = λt cosx− εt sinx u12(t, x) = L−1t [ 1 s Lt ( R(u11) )] ⇒ u12(t, x) = L−1t ( −λ (tε cosx+ tλ sinx)− ε (tλ cosx− tε sinx) s3 ) ⇒ u12(t, x) = −1 2 (sinx) t2λ2 − (cosx) t2λε+ 1 2 (sinx) t2ε2 u13(t, x) = L−1t [ 1 s Lt ( R(u12) )] Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 777 ⇒ u13(t, x) = L−1t ( − (cosx)λ3 + 3 (sinx) t2λ2ε+ 3 (cosx) t2λε2 − (sinx) ε3 s4 ) ⇒ u13(t, x) = − 1 3! (cosx) t3λ3 + 1 2! (sinx) t3λ2ε+ 1 2! (cosx) t3λε2 − 1 3! (sinx) t3ε3 u14(t, x) = L−1t [ 1 s Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( (sinx)λ4 + 4 (cosx)λ3ε− 6 (sinx)λ2ε2 − 3 (cosx)λε3 − (cosx)λε3 + (sinx) ε4 s5 ) ⇒ { u14(t, x) = 1 24 (sinx) t4λ4 + 1 6 (cosx) t4λ3ε− (sinx) t4λ2ε2 −1 6 (cosx) t4λε3 + 1 24 (sinx) t4ε4 u15(t, x) = L−1t [ 1 s Lt ( R(u14) )] ⇒  u15(t, x) = L−1t ( 1 24 (cosx)λ5 − 5 24 (sinx)λ4ε− 7 6 (cosx)λ3ε2 s6 ) +L−1t ( 7 6 (sinx)λ2ε3 + 5 24 (cosx)λε4 − 1 24 (sinx) ε5 s6 ) ⇒ { u15(t, x) = 1 24 (cosx) t5λ5 − 5 24 (sinx) t5λ4ε− 7 6 (cosx) t5λ3ε2 + 7 6 (sinx) t5λ2ε3 + 5 24 (cosx) t5λε4 − 1 24 (sinx) t5ε5 Step by step, we then deduce u1(t, x) =  sinx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 + ... ) +λt cosx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 + ... ) −1 2 (λt)2 sinx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 + ... ) − 1 3! (λt)3 cosx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 ... ) + 1 4! (λt)4 sinx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 ... ) + 1 5! (λt)5 cosx ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + 1 4! (εt)4 − 1 5! (εt)5 + 1 6! (εt)6 + ... ) +... We obtain u1 (t, x) =  sinx ( 1− 1 2! (λt)2 + 1 4! (λt)4 + ... ) × ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + ... ) + cosx ( λt− 1 3! (λt)3 + 1 5! (λt)5 + ... ) × ( 1− εt+ 1 2! (εt)2 − 1 3! (εt)3 + ... ) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 778 u1 (t, x) = sinx cos (λt) e−εt + cosx sin (λt) e−εt = e−εt sin (x+ λt) The solution to step k = 1 is u1 (t, x) = e−εt sin (x+ λt) So, we have Nu1 = ( u1 )3 + ( u1 )2 ∂2u1 ∂x2 Nu1 = ( e−εt sin (x+ λt) )3 − (e−εt sin (x+ λt) )3 = 0 We deduce by recursive way : u1 (t, x) = u2 (t, x) = ... = uk (t, x) = e−εt sin (x+ λt) and we obtain the exact solution u (t, x) = lim k→+∞ uk (t, x) = e−εt sin (x+ λt) Conclusion : The exact solution of model is u (t, x) = e−εt sin (x+ λt) Example 3. : Let us consider the following nonlinear diffusion equation  ∂u ∂t = λ ∂u ∂x + γu+ un + un−1 ∂2u ∂x2 ; n ≥ 1 u(0, x) = cosx , (t, x) ∈ Ω = [0,+∞[× R (15) Applying the Laplace SBA method, we deduce uk0(t, x) = L−1t [ 1 s u(x, 0) + 1 s Lt (gk−1) ] ukn+1(t, x) = L−1t [ 1 s Lt ( R(ukn) )] , n ≥ 0 (16) Where Lu = ∂u ∂t , Ru = λ ∂u ∂x + γu and gk−1 = Nuk−1 = ( uk−1 )n + ( uk−1 )n−1 ∂2uk−1 ∂x2 Let us calculate the terms ukn(t, x) for n ≥ 0. We take u0 = 0 u10(t, x) = L−1t ( 1 s cosx ) + L−1t ( 1 s Lt (g0) ) ⇒ u10(t, x) = cosx Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 779 u11(t, x) = L−1t [ 1 s Lt ( R(u10) )] ⇒ u11(t, x) = L−1t ( γ cosx− λ sinx s2 ) ⇒ u11(x, t) = γt cosx− λt sinx u12(t, x) = L−1t [ 1 s Lt ( R(u11) )] ⇒ u12(t, x) = L−1t ( − (cosx)λ2 − 2 (sinx)λγ + (cosx) γ2 s3 ) ⇒ u12(t, x) = −(λt)2 2! (cosx)− λγt2 (sinx) + (γt)2 2! (cosx) u13(t, x) = L−1t [ 1 s Lt ( R(u12) )] ⇒ u13(t, x) = L−1t ( (sinx)λ3 − 3 (cosx)λ2γ − 3 (sinx)λγ2 + (cosx) γ3 s3 ) ⇒ u13(t, x) = 1 3! (sinx) t3λ3 − 1 2! (cosx) t3λ2γ − 1 2! (sinx) t3λγ2 + 1 3! (cosx) t3γ3 u14(t, x) = L−1t [ 1 s Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( (cosx)λ4 + 4 (sinx) t3λ3γ − 6 (cosx)λ2γ2 − 4 (sinx) t3λγ3 + (cosx) γ4 s3 ) ⇒ u14(t, x) = 1 4! (cosx) t4λ4+ 1 3! (sinx) t4λ3γ−1 4 (cosx) t4λ2γ2− 1 3! (sinx) t4λγ3+ 1 4! (cosx) t4γ4 u15(t, x) = L−1t [ 1 s Lt ( R(u14) )] ⇒  u15(t, x) = L−1t ( − (sinx)λ5 + 5 (cosx)λ4γ + 10 (sinx)λ3γ2 − 10 (cosx)λ2γ3 s3 ) +L−1t ( −5 (sinx) t4λγ4 + (cosx) γ5 s3 ) ⇒  u15(t, x) = − 1 5! (sinx) t5λ5 + 1 4! (cosx) t4λ4γ + 2 4! (sinx) t5λ3γ2 − 2 4! (cosx) t5λ2γ3 − 1 4! (sinx) t5λγ4 + 1 5! (cosx) t5γ5 we deduce Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 780 u10 (t, x) = cosx u11 (t, x) = γt cosx−λt sinx u12 (t, x) = −(λt)2 2! (cosx)−λγt2 (sinx) + (γt)2 2! (cosx) u13 (t, x) = 1 3! (sinx) t3λ3− 1 2! (cosx) t3λ2γ − 1 2! (sinx) t3λγ2 + 1 3! (cosx) t3γ3 u14 (t, x) = 1 4! (cosx) t4λ4+ 1 3! (sinx) t4λ3γ − 1 4 (cosx) t4λ2γ2− 1 3! (sinx) t4λγ3 + 1 4! (cosx) t4γ4 u15 (t, x) = − 1 5! (sinx) t5λ5 + 1 4! (cosx) t4λ4γ + 2 4! (sinx) t5λ3γ2 − 2 4! (cosx) t5λ2γ3− 1 4! (sinx) t5λγ4 + 1 5! (cosx) t5γ5 By recursive way, we have  u1 (t, x) = cosx ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) − (λt)2 2! (cosx) ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) + (λt)4 4! (cosx) ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) + ... −λt sinx ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) + (λt)3 3! sinx ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) −(λt)5 5! sinx ( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) + ...  u1 (t, x) ' cosx ( 1− (λt)2 2! + (λt)4 4! + ... )( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) − sinx ( λt− (λt)3 3! + (λt)5 5! + ... )( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) We obtain u1 (t, x) = eγt cosx cos (λt)− eγt sinx sin (λt) = eγt cos (x+ λt) So, we have Nu1 = ( u1 )n + ( u1 )n−1 ∂2u1 ∂x2 Nu1 = ( eγt cos (x+ λt) )n − (eγt cos (x+ λt) )n eγt cos (x+ λt) = 0 By recursive way, we have u1 (t, x) = u2 (t, x) = ... = uk (t, x) = eγt cos (x+ λt) Conclusion : The exact solution of the model is u (t, x) = lim k→+∞ uk (t, x) = eγt cos (x+ λt) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 781 Example 4. Let us consider the following nonlinear diffusion equation (E) :  ∂u ∂t = λ ∂u ∂x + γu+ u2 − ( ∂2u ∂x2 )2 ; n ≥ 1 u(0, x) = sinx , (t, x) ∈ Ω = [0,+∞[× R (17) Applying the Laplace SBA method, we obtain uk0(t, x) = L−1t [ 1 s u(x, 0) + 1 p Lt (gk−1) ] ukn+1(t, x) = L−1t [ 1 s Lt ( R(ukn) )] , n ≥ 0 (18) Where Lu = ∂u ∂t , Ru = λ ∂u ∂x + γu and gk−1 = Nuk−1 = ( uk−1 )2 − (∂2uk−1 ∂x2 )2 Let us determinate ukn(t, x) for n ≥ 0 We take u10 = 0 u10(t, x) = L−1t ( 1 s sinx ) + L−1t ( 1 s Lt (g0) ) ⇒ u10(t, x) = sinx u11(t, x) = L−1t [ 1 s Lt ( R(u10) )] ⇒ u11(t, x) = L−1t ( λ cosx+ γ sinx s2 ) ⇒ u11(t, x) = λt cosx+ γt sinx u12(t, x) = L−1t [ 1 s Lt ( R(u11) )] ⇒ u12(t, x) = − (sinx) (λt)2 2! + (cosx)λγt2 + (sinx) (γt)2 2! u13(t, x) = L−1t [ 1 s Lt ( R(u12) )] ⇒ u13(t, x) = L−1t ( − (cosx)λ3 − 3 (sinx)λ2γ + 3 (cosx)λγ2 + (sinx) γ3 s4 ) ⇒ u13(t, x) = − 1 3! (cosx) t3λ3 − 1 2 (sinx) t3λ2γ + 1 2 (cosx) t3λγ2 + 1 3! (sinx) t3γ3 u14(t, x) = L−1t [ 1 s Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( (sinx)λ4 − 4 (cosx)λ3γ − 6 (sinx)λ2γ2 + 4 (cosx)λγ3 + (sinx) γ4 s5 ) ⇒ u14(t, x) = 1 4! (sinx) t4λ4− 1 3! (cosx) t4λ3γ−1 4 (sinx) t4λ2γ2+ 1 3! (cosx) t4λγ3+ 1 4! (sinx) t4γ4 u15(t, x) = L−1t [ 1 s Lt ( R(u14) )] Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 782 ⇒ u15(t, x) = L−1t ( (cosx)λ5 + 5 (sinx)λ4γ − 10 (cosx)λ3γ2 − 10 (sinx)λ2γ3 + 5 (cosx)λγ4 + (sinx) γ5 s6 ) ⇒ u15(t, x) = 1 5! (cosx) t5λ5+ 1 4! (sinx) t5λ4γ− 2 4! (cosx) t5λ3γ2− 2 4! (sinx) t5λ2γ3+ 1 4! (cosx) t5λγ4+ 1 5! (sinx) t5γ5 We deduce u10 (t, x) = sinx u11 (t, x) = λt cosx+ γt sinx u12 (t, x) = − (sinx) (λt)2 2! + (cosx)λγt2 + (sinx) (γt)2 2! u13 (t, x) = − 1 3! (cosx) t3λ3−1 2 (sinx) t3λ2γ + 1 2 (cosx) t3λγ2 + 1 3! (sinx) t3γ3 u14 (t, x) = 1 4! (sinx) t4λ4 − 1 3! (cosx) t4λ3γ − 1 4 (sinx) t4λ2γ2 + 1 3! (cosx) t4λγ3 + 1 4! (sinx) t4γ4 u15 (t, x) = 1 5! (cosx) t5λ5 + 1 4! (sinx) t5λ4γ − 2 4! (cosx) t5λ3γ2 − 2 4! (sinx) t5λ2γ3 + 1 4! (cosx) t5λγ4 + 1 5! (sinx) t5γ5 In recursive way, we obtain  u1 (t, x) = sinx ( 1 + γt+ 1 2 t2γ2 + ... ) − (λt)2 2! (sinx) ( 1 + γt+ 1 2 t2γ2 + ... ) + (λt)4 4! (sinx) ( 1 + γt+ 1 2 t2γ2 + ... ) + ... +λt cosx ( 1 + γt+ 1 2 t2γ2 + ... ) − (λt)3 3! (cosx) ( 1 + γt+ 1 2 t2γ2 + ... ) + (λt)5 5! (cosx) ( 1 + γt+ 1 2 t2γ2 + ... ) + ... ⇒  u1 (t, x) = sinx ( 1− (λt)2 2! + (λt)4 4! + ... )( 1 + γt+ (γt)2 2 + (γt)3 3! + ... ) + cosx ( λt− (λt)3 3! + (λt)5 5! + ... )( 1 + γt+ 1 2 t2γ2 + 1 3! t3γ3 + ... ) we deduce u1 (t, x) = −eγt sinx cos (λt) + eγt cosx sin (λt) = eγt sin (x+ λt) So, we have Nu1 = u2 − ( ∂2u ∂x2 )2 = ( eγt sin (x+ λt) )n − (eγt sin (x+ λt) )n−1 eγt sin (x+ λt) = 0 Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 783 Step by step, we deduce u1 (t, x) = u2 (t, x) = ... = uk (t, x) = eγt sin (x+ λt) Conclusion : The exact solution of the model is u (t, x) = lim k→+∞ uk (t, x) = eγt sin (x+ λt) Example 5. Let us consider the following non linear convection - diffusion reaction equa- tion (E) :  ∂u ∂t = ε ∂2u ∂x2 + λ ∂u ∂x + γu+ u3 + ( ∂2u ∂x2 )3 u(0, x) = sinx , (t, x) ∈ Ω = [0,+∞[× R (19) Applying the Laplace SBA method, we have uk0(t, x) = L−1t [ 1 (s− γ) u(x, 0) + 1 (s− γ) Lt (gk−1) ] ukn+1(t, x) = L−1t [ 1 (s− γ) Lt ( R(ukn) )] , n ≥ 0 (20) Where Lu = ∂u ∂t − γu, Ru = ε ∂2u ∂x2 + λ ∂u ∂x and Nu = u3 + ( ∂2u ∂x2 )3 In this example, we take the linear operator L in this form to accelerate convergence Let us determinate the terms ukn(t, x) for n ≥ 0 We take u0(t, x) = 0 u10(t, x) = L−1t ( 1 (s− γ) sinx ) + L−1t ( 1 (s− γ) Lt (g0) ) ⇒ u10(t, x) = eγt sinx u11(t, x) = L−1t [ 1 (s− γ) Lt ( R(u10) )] ⇒ Lt ( R(u10) ) = λ cosx− ε sinx (s− γ) ⇒ u11(t, x) = L−1t ( λ cosx− ε sinx (s− γ)2 ) ⇒ u11(t, x) = (λt cosx− εt sinx) eγt u12(t, x) = L−1t [ 1 p Lt ( R(u11) )] Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 784 ⇒ Lt ( R(u11) ) = (sinx)λ2 − 2 (cosx)λε+ (sinx) ε2 (s− γ)2 ⇒ u12(t, x) = L−1t ( − (sinx)λ2 − 2 (cosx)λε+ (sinx) ε2 (s− γ)3 ) ⇒ u12(t, x) = −(λt)2 2! (sinx)− t2 (cosx)λε+ (εt)2 2! (sinx) u13(t, x) = L−1t [ 1 p Lt ( R(u12) )] ⇒ u13(t, x) = L−1t ( 3λε2 cosx− ε3 sinx− λ3 cosx+ 3λ2ε sinx (s− γ)4 ) ⇒ u13(t, x) = ( − 1 3! t3λ3 cosx+ 1 2! t3λε2 cosx+ 1 2! t3λ2ε sinx− 1 3! t3ε3 sinx ) etγ u14(t, x) = L−1t [ 1 p Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( λ4 sinx+ ε4 sinx− 6λ2ε2 sinx− 4λε3 cosx+ 4λ3ε cosx (s− γ)5 ) ⇒ u14(t, x) = 1 4! t4λ4etγ sinx+ 1 4! t4ε4etγ sinx− t 4 4 λ2ε2etγ sinx− 1 3! t4λε3etγ cosx+ 1 3! t4λ3εetγ cosx u15(t, x) = L−1t [ 1 p Lt ( R(u14) )] Lt ( R(u14) ) = λ5 cosx− ε5etγ sinx− 10λ3ε2 cosx+ 10λ2ε3 sinx+ 5λε4 cosx− 5λ4ε sinx (s− γ)5 ⇒ u15(t, x) = L−1t ( λ5 cosx− ε5 sinx− 10λ3ε2 cosx+ 10λ2ε3 sinx+ 5λε4 cosx− 5λ4ε sinx (s− γ)6 ) ⇒  u15(t, x) = 1 5! t5λ5etγ cosx− 1 5! t5ε5etγ sinx− 10 5! t5λ3ε2etγ cosx+ 10 5! t5λ2ε3etγ sinx + 1 4! t5λε4etγ cosx− 1 4! t5λ4εetγ sinx we deduce Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 785 u10 (t, x) = eγt sin x u11 (t, x) = (λt cosx− εt sinx) eγt u12 (t, x) = −(λt)2 2! eγt (sinx)− t2λεeγt (cosx) + (εt)2 2! eγt (sinx) u13 (t, x) = − 1 3! t3λ3eγt cosx+ 1 2! t3λε2eγt cosx+ 1 2! t3λ2εeγt sinx− 1 3! t3ε3eγt sinx u14 (t, x) = 1 4! t4λ4etγ sinx+ 1 4! t4ε4etγ sinx− t4 4 λ2ε2etγ sinx− 1 3! t4λε3etγ cosx+ 1 3! t4λ3εetγ cosx u15 (t, x) = 1 5! t5λ5etγ cosx− 1 5! t5ε5etγ sinx− 10 5! t5λ3ε2etγ cosx+ 10 5! t5λ2ε3etγ sinx+ 1 4! t5λε4etγ cosx − 1 4! t5λ4εetγ sinx Step by step, we deduce u1 (t, x) = eγt sinx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) − (λt)2 2! eγt sinx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + (λt)4 4! eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + ... λteγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) − (λt)3 3! eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + (λt)5 5! eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + ... We obtain  u1 (t, x) = eγt sinx ( 1− (λt)2 2! + (λt)4 4! + ... )( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) eγt cosx ( λt− (λt)3 3! + (λt)5 5! + ... )( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) u1 (t, x) = eγt sinxcos (λt) e−εt + eγt cosxsin (λt) e−εt = e(γ−ε)t sin (x+ λt) So, we have Nu1 = u3 + ( ∂2u ∂x2 )3 Nu1 = ( e(γ−ε)t sin (x+ λt) )3 + ( −e(γ−ε)t sin (x+ λt) )3 = 0 In recursive way, we have u1 (t, x) = u2 (t, x) = ... = uk (t, x) = e(γ−ε)t sin (x+ λt) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 786 The exact solution of the model is u (t, x) = lim k→+∞ uk (t, x) = e(γ−ε)t sin (x+ λt) Example 6. Let us consider the following non linear convection diffusion reaction model (E) :  ∂u ∂t = ε ∂2u ∂x2 + λ ∂u ∂x + γu+ un + un−1 ∂2u ∂x2 ; n ≥ 1 u(0, x) = cosx , (t, x) ∈ Ω = [0,+∞[× R (21) Where u = u(t, x) with x ∈ R and t ≥ 0 Applying the Laplace SBA method, we obtain uk0(t, x) = L−1t [ 1 (s− γ) u(0, x) + 1 (s− γ) Lt (gk−1) ] ukn+1(t, x) = L−1t [ 1 (s− γ) Lt ( R(ukn) )] , n ≥ 0 (22) With Lu = ∂u ∂t − γu, Ru = ε ∂2u ∂x2 +λ ∂u ∂x and Nu = un−un−1∂ 2u ∂x2 In this example, we put the linear operator L in the form ∂u ∂t − γu to accelerate convergence. Let us determinate the terms ukn (x, t) for n ≥ 0 We take u0(t, x) = 0 u10(t, x) = L−1t ( 1 (s− γ) cosx ) + L−1t ( 1 (s− γ) Lt (g0) ) ⇒ u10(t, x) = eγt cosx u11(t, x) = L−1t [ 1 (s− γ) Lt ( R(u10) )] ⇒ u11(t, x) = L−1t ( −ε cosx− λ sinx (s− γ)2 ) ⇒ u11(t, x) = (−εt cosx− λt sinx) eγt u12(t, x) = L−1t [ 1 p Lt ( R(u11) )] ⇒ Lt ( R(u11) ) = − (cosx)λ2 + 2 (sinx)λε+ (cosx) ε2 (s− γ)2 ⇒ u12(t, x) = L−1t ( − (cosx)λ2 + 2 (sinx)λε+ (cosx) ε2 (s− γ)3 ) Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 787 ⇒ u12(t, x) = −(λt)2 2! eγt (cosx) + t2eγt (sinx)λε+ (εt)2 2! eγt (cosx) u13(t, x) = L−1t [ 1 p Lt ( R(u12) )] ⇒ Lt ( R(u12) ) = λ3 sinx− ε3 cosx+ 3λ2ε cosx− 3λε2 sinx (s− γ)3 ⇒ u13(t, x) = L−1t ( λ3 sinx− ε3 cosx+ 3λ2ε cosx− 3λε2 sinx (s− γ)4 ) ⇒ u13(t, x) = 1 3! t3λ3etγ sinx− 1 3! t3ε3etγ cosx+ 1 2! t3λ2εetγ cosx− 1 2! t3λε2etγ sinx u14(t, x) = L−1t [ 1 s Lt ( R(u13) )] ⇒ u14(t, x) = L−1t ( λ4 cosx+ ε4 cosx− 6λ2ε2 cosx+ 4λε3 sinx− 4λ3ε sinx (s− γ)5 ) ⇒ u14(t, x) = 1 4! t4λ4etγ cosx+ 1 4! t4ε4etγ cosx− t 4 4 λ2ε2etγ cosx+ 1 3! t4λε3etγ sinx− 1 3! t4λ3εetγ sinx u15(t, x) = L−1t [ 1 p Lt ( R(u14) )] R(u14) = ε ∂2u14 ∂x2 + λ ∂u14 ∂x ⇒ u15(t, x) = L−1t ( 10λ2ε3 cosx− λ5 sinx− ε5 cosx+ 10λ3ε2 sinx− 5λ4ε cosx− 5λε4 sinx (s− γ)6 ) ⇒  u15(t, x) = − 1 5! t5λ5etγ sinx− 1 5! t5ε5etγ cosx+ 10 5! t4λ3ε2etγ sinx− 1 4! t4λ4εetγ cosx − 1 4! t4λε4etγ sinx+ 10 5! t4λ2ε3etγ cosx We deduce u10 (t, x) = eγt cosx u11 (t, x) = (−εt cosx− λt sinx) eγt u12 (t, x) = −(λt)2 2! eγt (cosx) + t2eγt (sinx)λε+ (εt)2 2! eγt (cosx) u13 (t, x) = 1 3! t3λ3etγ sinx− 1 3! t3ε3etγ cosx+ 1 2! t3λ2εetγ cosx− 1 2! t3λε2etγ sinx u14 (t, x) = 1 4! t4λ4etγ cosx+ 1 4! t4ε4etγ cosx− t4 4 λ2ε2etγ cosx+ 1 3! t4λε3etγ sinx− 1 3! t4λ3εetγ sinx u15 (t, x) = − 1 5! t5λ5etγ sinx− 1 5! t5ε5etγ cosx+ 10 5! t4λ3ε2etγ sinx− 1 4! t4λ4εetγ cosx− 1 4! t4λε4etγ sinx + 10 5! t4λ2ε3etγ cosx Step by step, we obtain Y. Paré, Y. Minoungou, A. W. Nébié / Eur. J. Pure Appl. Math, 12 (3) (2019), 771-789 788 u1 (t, x) = eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) − (λt)2 2! eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + (λt)4 4! eγt cosx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + ... −λteγt sinx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + (λt)3 3! eγt sinx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) −(λt)5 5! eγt sinx ( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) + ... we obtain u1 (t, x) = eγt cosx ( 1− (λt)2 2! + (λt)4 4! + ... )( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) −eγt sinx ( λt− (λt)3 3! + (λt)5 5! + ... )( 1− εt+ (εt)2 2! − (εt)3 3! + .... ) We deduce u1 (t, x) = eγt cosxcos (λt) e−εt − eγt sinxsin (λt) e−εt = e(γ−ε)t cos (x+ λt) So, we have Nu1 = ( u1 )n + ( u1 )n−1 ∂2u1 ∂x2 Nu1 = ( e(γ−ε)t cos (x+ λt) )n + ( e(γ−ε)t cos (x+ λt) )n−1 ( −e(γ−ε)t cos (x+ λt) ) = 0 In recursive way, we deduce u1 (t, x) = u2 (t, x) = ... = uk (t, x) = e(γ−ε)t sin (x+ λt) Conclusion : The exact solution of the model is u (t, x) = lim k→+∞ uk (t, x) = e(γ−ε)t sin (x+ λt) 5. 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