3 The Elzaki The Elzaki variational iteration method In this work, we assume that L is an operator of the first order ǝ/ǝt in the equation (2). Let us take Elzaki transform of both sides and apply the differ- entiation property of Elzaki transform. Then we have and E[Lu(x,t)]+E[Nu(x,t)] = E[g(x,t)] Eu(x,t)]=vE[g(x,t)] + vh(x) - vE[Nu(x,t)]. Applying the inverse Elzaki transform of both sides of the equation, we have u(x,t) =G(x,t) E¹{vE[Nu(x,t)]}, Tarig Joon Kim where G(x, t) represents the terms arising from the source term and the pre- scribed initial condition. Taking the first partial derivative with respect to t we have u(x,t)-G(x,t) + E-{vE[Nu(x, t)]}= 0. E By the correction function of the irrational method, we have a Un+1(x,t) = u(x,t) — [ { (Un),(x, 8) ———G(x, s)+7 - {(un),(x, 8) — — — G(x, 8) + E¹{vE[Nu,(x,s)}}}ds, Os Or alternately Un+1(x,t) = G(x,t) - E{vE[Nun(x, t)]}. Thus, we can obtain the solution u by u(x,t) = lim u,(x,t). 11-00 (3)