EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 1750-1759 ISSN 1307-5543 – ejpam.com Published by New York Business Global Analytical solution of the Ginzburg-Landau equation Yanick Alain Servais Wellot1,∗, Gires Dimitri Nkaya2 1 Mathematics, Ecole Normale Supérieure, Université Marien Ngouabi, Brazzaville, Congo 2 Mathematics, Faculté des Sciences et Technique, Université Marien Ngouabi, Brazzaville, Congo Abstract. In this paper, we will construct the solution of the Landau-Ginzburg equation by the Adomian decomposition method. This method avoids linearization of space and discretization of time, it often gives a good approximation of the exact solution. 2020 Mathematics Subject Classifications: 47H14, 34G20, 47J25, 65J15 Key Words and Phrases: Adomian method, Ginzburg-Landau equation. 1. Introduction The Landau equation, also called the Fokker-Planck-Landau equation, is a nonlinear partial differential equation that describes the motion of particles in a plasma. This equation was obtained by Lev Davidovitch Landau in 1936 from the Bolzmann equation [3, 4]. In 1950, Ginzburg and Landau proposed an extension of the Landau energy to describe the superconductor in the presence of a magnetic field. In this article, we are interested in the difference in these two models try to describe the field applied to systems of partial differential equations with initial conditions. Then we apply the Adomian Decomposition Method [1, 2], an analytical method that allows us to obtain an exact solution when it exists without space linearization and time discretization. However, the existence and uniqueness of solution of the Ginzburg-Landau equation has been proved in [5–7]. In the mathematical and physical context, explanations are being made available to the research community. During the last two decades, the mathematics of superconductivity has been the subject of intense activity [8, 10, 13, 14]. The Ginzburg-Landau function is a commonly used model to describe the behavior of a superconductor involving, a wave function (called order parameter) and a vector field (called magnetic potential), both defined on an open set [9]. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4551 Email addresses: yanick.wellot@umng.cg (Y.A.S. Wellot), mail:giresnkaya@gmail.com (G. D. Nkaya) https://www.ejpam.com 1750 © 2022 EJPAM All rights reserved. Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1751 2. Proposition On the basis of a suitable choice of physical data the model can be written in a mathematical way in the form [11]: ∂u(x, t) ∂t = (1 + iα) ∂2u(x, t) ∂x2 + γu(x, t)− (1 + ib)|u(x, t)|2nu(x, t)− (1− 4i)|u(x, t)|4nu(x, t) n ⩾ 1, i2 = −1, α, γ, b are real u(x, 0) = f(x). (1) 3. About Adomian Decomposition Method The Adomian Decomposition Method allows to solve functional problems of different types: algebraic equations, differential, integral, integro-differential and partial differential equations. It was introduced in the 1980s by Professor , GEORGES ADOMIAN ( 1922 - 1996). It is a decomposition method which consists in seeking the solution in the form of a con- vergent series . It has the advantage of not linearizing and discretizing the equations. This method allows to determine the exact solution of the problem if it exists or to give an approximation of the solution of the problem by preserving the physical properties of the studied phe- nomenon. Let us solve the functional equation : Au = h, (2) where A is an operator of a real Hilbert space H, u ∈ H an unknown function and h ∈ H a known function. The principle of the Adomian Decomposition Method consists in decomposing the operator A into two parts : one linear and the other non-linear: A = L+R︸ ︷︷ ︸ linear + N︸︷︷︸ nonlinear , (3) where L is assumed to be invertible in the Adomian sense. A in (2) leads to Lu+Ru+Nu = h ; (4) By applying L−1 to (4), we obtain the canonical form of Adomian: u = θ + L−1h − L−1Ru− L−1Nu, (5) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1752 where θ verifies the relationship Lθ = 0 ( θ is the integration constant if L is a differential operator). As for the Adomian algorithm, it is obtained by assuming that the solution of (2) has the form of a series u = ∞∑ n=0 un. (6) The nonlinear operator also has the form of a series Nu = ∞∑ n=0 An. (7) An are Adomian polynomials. That are defined by the formula [5]: An = 1 n! [ dn dλn N ( +∞∑ i=0 λiui )] λ=0 , n = 0, 1, 2, ... (8) λ is a real parameter introduced by convenience then we obtain ∞∑ n=0 un = θ + L−1h − L−1 [ R ( ∞∑ n=0 un )] − L−1 [ ∞∑ n=0 An ] . (9) Assuming that the ∞∑ n=0 un series and ∞∑ n=0 An{ u0 = θ + L−1(h ) un+1 = −L−1 [R (un)]− L−1 (An) ,∀n ≥ 0 (10) Equation (10) allows to compute all un recursively, then the analytical solution of (2) is defined by the sum of un: u = ∞∑ n=0 un (11) If φn = n−1∑ n=0 un is the truncated series, and if ∞∑ n=0 un is convergent, we have : u = lim n→+∞ φn (12) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1753 4. Application 4.1. Problem 1 We consider the following initial value problem [11]:  ∂u(x, t) ∂t − (1 + i) ∂2u(x, t) ∂x2 + (1 + 2i)|u(x, t)|2u(x, t)− 3u(x, t) + (1− 4i)|u(x, t)|4u(x, t) = 0 u(x, 0) = eix, (13) with u(x, t) a complex function. Notation: |u|2 = uu; ⇒ |u|2u = u2u |u|4 = u2u2 ⇒ |u|4u = u3u2 The Adomian decomposition method consists to look the solution of u(x, t) in the form: u(x, t) = u(x, 0) + (1 + i) ∫ t 0 ∂2u(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 |u(x, s)|2u(x, s)ds+ 3 ∫ t 0 u(x, s)ds− (1− 4i) ∫ t 0 |u(x, s)|4u(x, s)ds (14) Let’s look for the solution in the following form: u(x, t) = +∞∑ n=0 un(x, t) (15) with N = N1 +N2, and let us note N1u = |u(x, t)|2u(x, t) N2u = |u(x, t)|4u(x, t) } (16) Suppose that N1u = ∞∑ n=0 An(x, t) and N2u = ∞∑ n=0 Bn(x, t)  (17) with An = 1 n!  dn dλn N1 +∞∑ j=0 λjuj  λ=0 , n = 0, 1, 2, ... (18) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1754 and Bn = 1 n!  dn dλn N2 +∞∑ j=0 λjuj  λ=0 , n = 0, 1, 2, ... (19) λ is a real parameter introduced by convenience, we obtain the following Adomian poly- nomials: An = 1 n! dn dλn  ( +∞∑ i=0 λiui )2 ( +∞∑ j=0 λjuj )  λ=0 (20) and Bn = 1 n! dn dλn  ( +∞∑ i=0 λiui )3 ( +∞∑ j=0 λjuj )2  λ=0 (21) This gives A0 = u20u0. (22) A1 = u20u1 + 2u0u1u0. (23) A2 = u20u2 + 2u0u0u2 + 2u1u1u0 + u21u0. (24) A3 = u20u3 + 2u0u0u3 + 2u0u1u2 + 2u0u1u2 + 2u0u1u2 + u21u1. (25) and B0 = u30u0 2. (26) B1 = 3u1u 2 0u 2 0 + 2u30u1u0. (27) B2 = 3u2u 2 0u 2 0 + 3u21u0u 2 0 + 6u1u 2 0u1u0 + 2u30u2u0 + u30u 2 1. (28) The equation of problem (1), becomes: +∞∑ n=0 un(x, t) = u(x, 0) + (1 + i) +∞∑ n=0 ∫ t 0 ∂2un(x, s) ∂x2 ds− (1 + 2i) +∞∑ n=0 ∫ t 0 An(x, s)ds+ 3 +∞∑ n=0 ∫ t 0 un(x, s)ds− (1− 4i) +∞∑ n=0 ∫ t 0 Bn(x, s)ds. (29) From the above equation (4.1) we get the Adomian canonical form. This leads us to obtain the following Adomian algorithm (30) below: u0(x, t) = u(x, 0) un+1(x, t) = (1 + i) ∫ t 0 ∂2un(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 An (x, s) ds+ 3 ∫ t 0 un(x, s)ds− (1− 4i) ∫ t 0 Bn (x, s) ds (30) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1755 Either:  u0(x, t) = eix un+1(x, t) = (1 + i) ∫ t 0 ∂2un(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 An (x, s) ds+ 3 ∫ t 0 un(x, s)ds− (1− 4i) ∫ t 0 Bn (x, s) ds (31) u1(x, t) = (1 + i) ∫ t 0 ∂2u0(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 A0ds+ 3 ∫ t 0 u0(x, s)ds− (1− 4i) ∫ t 0 B0ds (32) With A0 = B0 = eix, we can easily calculate u1(x, t) u1(x, t) = −(1 + i)teix + 3teix − (1 + 2i)teix − (1− 4i)teix = iteix (33) The same calculation procedure leads us to obtain A1 as follows A1 = u20u1 + 2u0u0u0 = iteix. (34) Then B1 = 3u1u 2 0u 2 0 + 2u30u1u0 = iteix. (35) Thus, the simple formula to calculate the expression u2(x, t) follows from this. Let it be: u2(x, t) = (1 + i) ∫ t 0 ∂2u1(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 A1ds+ 3 ∫ t 0 u1(x, s)ds− (1− 4i) ∫ t 0 B1ds (36) u2(x, t) = [−i(1 + i) + 3i− (1 + 2i)(i)− (1− 4i)(i)] t2 2 eix (37) Thus: u2(x, t) = − t2 2 eix (38) We proceed in the same way for the expression of u3(x, t). u3(x, t) = (1 + i) ∫ t 0 ∂2u2(x, s) ∂x2 ds− (1 + 2i) ∫ t 0 A2ds+ 3 ∫ t 0 u2(x, s)ds− (1− 4i) ∫ t 0 B2ds (39) With A2 = u20u2 + 2u0u0u2 + 2u1u1u0 + u21u0 = − t2 2 eix, (40) and B2 = 3u2u 2 0u 2 0 + 3u21u0u 2 0 + 6u1u 2 0u1u0 + 2u30u2u0 + u30u 2 1 = − t2 2 eix. (41) Therefore: u3(x, t) = [1 + i+ 1 + 2i− 3 + 1− 4i] t3 3! eix = −i t3 3! eix. (42) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1756 Thus we have:  u0(x, t) = eix u1(x, t) = iteix u2(x, t) = (it)2 2! eix u3(x, t) = (it)3 3! eix ... un(x, t) = (it)n n! eix (43) Hence, the exact solution of the Ginzburg-Landau equation is: u(x, t) = +∞∑ n=0 un (x, t) = +∞∑ n=0 (it)n n! eix = ei(x+t). (44) 4.2. Problem 2 We consider the following initial value problem [12, 15]:  ∂u(x, y, t) ∂t − (1 + 2i) [ ∂2u(x, y, t) ∂x2 + ∂2u(x, y, t) ∂y2 ] + (1 + 2i)|u(x, y, t)|2u(x, y, t)− γu(x, y, t) = 0 u(x, y, 0) = e i π 3 (x+ y) , (45) With u(x, y, t) a complex function, γ ∈ R. According to the Adomian decomposition method, the equation (45) ∂u(x, y, t) ∂t − (1 + 2i)∆u(x, y, t) + (1 + 2i)|u(x, y, t)|2u(x, y, t)− γu(x, y, t) = 0 (46) u(x, y, t) = u(x, y, 0)+ ∫ t 0 (1+2i)∆u(x, y, s)ds−(1+2i) ∫ t 0 |u(x, y, s)|2u(x, y, s)ds+γ ∫ t 0 u(x, y, s)ds) = 0 (47) Let’s put Nu = |u(x, y, t)|2u(x, y, t), we have u(x, y, t) = u(x, y, 0)+ ∫ t 0 (1+2i)∆u(x, y, s)ds−(1+2i) ∫ t 0 Nu(x, y, s)ds+γ ∫ t 0 u(x, y, s)ds) = 0 (48) Y. A. S. Wellot, G. D. Nkaya / Eur. J. Pure Appl. Math, 15 (4) (2022), 1750-1759 1757 Let us look for the solution of (45) in the form of a series and the nonlinear part u(x, y, t) = ∞∑ n=0 un(x, y, t) and Nu(x, y, t) = ∞∑ n=0 An(x, y, t) (49) with A0 = u20u0. (50) A1 = u20u1 + 2u0u1u0. (51) A2 = u20u2 + 2u0u0u2 + 2u1u1u0 + u21u0. (52) We obtain the following canonical form: ∞∑ n=0 un(x, y, t) = u(x, y, 0)+(1+2i) ∫ t 0 ∆un(x, y, s)ds−(1+2i) ∫ t 0 Ands+γ ∫ t 0 un(x, y, s)ds (53) From (53), for γ = 1 + 2π2 9 (because γ is a constant), we obtain the following Adomian algorithm: u0(x, y, t) = e i π 3 (x+ y) un+1(x, y, t) = ∫ t 0 (1 + 2i)∆un(x, y, s)ds− (1 + 2i) ∫ t 0 Ands+ (1 + 2π2 9 ) ∫ t 0 un(x, y, s)ds (54) For n = 0, u1(x, y, t) = ∫ t 0 (1+2i)∆u0(x, y, s)ds− (1+2i) ∫ t 0 A0ds+(1+ 2π2 9 ) ∫ t 0 u0(x, y, s)ds (55) With ∆u0(x, y, t) = −2π2 9 e i π 3 (x+ y) and A0 = e i π 3 (x+ y) (55), gives us u1(x, y, t) = −2i ( 1 + 2π2 9 ) te i π 3 (x+ y) For n = 2, u2(x, y, t) = ∫ t 0 (1+2i)∆u1(x, y, s)ds− (1+2i) ∫ t 0 A1ds+(1+ 2π2 9 ) ∫ t 0 u1(x, y, s)ds (56) With ∆u1(x, y, t) = 4iπ 2 9 (1+ 2π2 9 )te i π 3 (x+ y) and A1 = −2i(1+ 2π2 9 )te i π 3 (x+ y) So u2(x, y, t) = −4 ( 1 + 2π2 9 )2 t2 2 e i π 3 (x+ y) (57) REFERENCES 1758 For n = 3, u3(x, y, t) = ∫ t 0 (1+2i)∆u2(x, y, s)ds− (1+2i) ∫ t 0 A2ds+(1+ 2π2 9 ) ∫ t 0 u2(x, y, s)ds (58) With ∆u2(x, y, t) = 8π2 9 ( 1 + 2π2 9 )2 t2 2 e i π 3 (x+ y) and A2 = −4 ( 1 + 2π2 9 )2 t2 2 e i π 3 (x+ y) So u3(x, y, t) = 8i ( 1 + 2π2 9 )3 t3 3! e i π 3 (x+ y) (59) Then u(x, y, t) = ∞∑ n=0 un(x, y, t) = u0(x, y, t) + u1(x, y, t) + u2(x, y, t) + u3(x, y, t) + ... (60) u(x, y, t) = [ 1− 2i ( 1 + 2π2 9 ) t− 4 ( 1 + 2π2 9 )2 t2 2 + 8i ( 1 + 2π2 9 )3 t3 3 + ... ] e i π 3 (x+ y) (61) This implies u(x, y, t) = ∞∑ n=0 [ −2i ( 1 + 2π2 9 )]n n! e i π 3 (x+ y) . (62) Then the exact solution of the Landau-Ginzburg equation in dimension two is: u(x, y, t) = e i π 3 (x+ y)− 2 ( 1 + 2π2 9 ) t  . (63) 5. Conclusion In this paper, the Adomian decomposition method has been used to solve the complex model of the Landau Ginzburg equation. In order to show the importance and applicability of the proposed method, the Adomian polynomial manipulated in this article allowed us to find the exact solution of the studied problem. References [1] George Adomian. A review of the decomposition method in applied mathematics. Journal of mathematical analysis and applications. [2] George Adomian. Solving frontier problems of physics: the decomposition method. Springer Science & Business Media. REFERENCES 1759 [3] Kamel Attar. 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