id	author	title	date	pages	extension	mime	words	sentence	flesch	summary	cache	txt
ejpam-4551	Wellot, Yanick Alain Servais; Nkaya, Gires Dimitri 	Analytical Solution of the Ginzburg-Landau Equation	2022	10	.pdf	application/pdf	2904	186	78	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. (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)	cache/ejpam-4551.pdf	txt/ejpam-4551.txt