The Klein-Gordon equation in anti-de Sitter spacetime - Seminario ...

Klein-Gordon equation in anti-de Sitter spacetime 281(27)u(x,t) ==∫ e t −10∫ 10[ ]ϕ(x+z)+ϕ(x−z) K 1 (z,t)dz[]ϕ(x+φ(t)s)+ϕ(x−φ(t)s) K 1 (φ(t)s,t)φ(t)ds,where φ(t)=e t − 1. The proof of the theorem is divided into several steps.PROPOSITION 2. The solution u=u(x,t) of the Cauchy problem (9) for whichϕ 0 (x)=0 and ϕ 1 (x)=ϕ(x) can be represented as followsu(x,t) =(28)∫ t0]db[14e −t/2 e b/2 (2+b)+16 1 be−3t/2 e b/2 (e b − e t )(1+4M 2 )[]× ϕ(x+e t − e b )+ϕ(x−e t + e b )∫ t ∫ x+e t −e b [ ( ∂ 2 ]+ db dyϕ(y)b e 2b E(x−y,t;0,b)−M0 x−(e t −e b )∂y) 2 E(x−y,t;0,b) .Proof. We look for the solution u=u(x,t) of the form u(x,t)=w(x,t)+ tϕ(x). Then(9) impliesw tt − e 2t w xx + M 2 w=te 2t ϕ (2) (x)−M 2 tϕ(x),w(x,0)=0, w t (x,0)=0.We set f(x,t)= te 2t ϕ (2) (x)−M 2 tϕ(x) and due to Theorem 3 obtainwherew(x,t)= ˜w(x,t)−M 2 ∫ t0∫ x+e t −e bbdb dyϕ(y)E(x−y,t;0,b),x−(e t −e b )˜w(x,t) :=∫ t0∫ x+e tbe 2b −e bdb dyϕ (2) (y)E(x−y,t;0,b).x−(e t −e b )Then we integrate by parts:˜w(x,t) =∫ t0[be 2b db ϕ (1) (x+e t − e b )E(e b − e t ,t;0,b)]−ϕ (1) (x−e t + e b )E(e t − e b ,t;0,b)∫ t ∫ x+e t− be 2b −e bdb dyϕ (1) (y) ∂0x−(e t −e b ) ∂y E(x−y,t;0,b).But ϕ (1) (x+e t − e b )=−e −b ∂ ∂b ϕ(x+et − e b ), ϕ (1) (x−e t + e b )=e −b ∂ ∂b ϕ(x−et + e b ).Then, E(e b − e t ,t;0,b)=E(−e b + e t ,t;0,b) due to (19), and we obtain